Computer Algebra Recipes

Richard H. Enns 

George C. McGuire 

Computer Algebra Recipes 

An Advanced Guide to 

Scientific Modeling

Richard H. Enns 

Simon Fraser University 

Department of Physics 

Burnaby, B.C. V5A 1S6 

Canada 

renns@sfu.ca 

Cover design by Mary Burgess. 

Library of Congress Control Number: 2006936017 

ISBN-10: 0-387-25768-3 e-ISBN-10: 0-387-49333-6 

ISBN-13: 978-387-25768-6 e-ISBN-13: 978-0-387-49333-6 

Printed on acid-free paper. 

George C. McGuire 

University College of Fraser Valley 

Department of Physics 

Abbotsford, BC V2S 7M9 

Canada 

george.mcguire@ucfv.ca 

c○2007 Springer Science+Business Media, LLC 

All rights reserved. This work may not be translated or copied in whole or in part without the written 

permission of the publisher (Springer Science+Business Media LLC, 233 Spring Street, New York, 

NY 10013, USA) and the author, except for brief excerpts in connection with reviews or scholarly 

analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, 

computer software, or by similar or dissimilar methodology now known or hereafter developed is 

forbidden. 

The use in this publication of trade names, trademarks, service marks and similar terms, even if they 

are not identified as such, is not to be taken as an expression of opinion as to whether or not they are 

subject to proprietary rights. 

987654321 

springer.com (EB)

PREFACE 

A computer algebra system (CAS) not only has the number \crunching" and 

plotting capability of traditional computing languages such as Fortran and C, 

but also allows one to perform the symbolic manipulations and derivations 

required in most mathematically based science and engineering courses. To 

introduce students in these disciplines to CAS-based mathematical modeling 

and computation, the authors have previously developed and classroom tested 

the text Computer Algebra Recipes: A Gourmet's Guide to the Mathematical 

Models of Science [EM01] based on the Maple CAS. Judging by course evaluations 

and reader feedback, the response to this book and the computer algebra 

approach to modeling has been very favorable. With the release of several new 

versions of Maple since this text was published and the authors' accumulation 

of many insightful comments and helpful suggestions, a second up-dated edition 

seemed expedient. However, incorporating all the changes would make 

an already lengthy book even longer. So the topics of the Gourmet's Guide 

have been reorganized into two new stand-alone volumes, an already-published 

Introductory Guide [EM06] and this Advanced Guide. 

In this book, we explore mathematical models involving linear and nonlinear 

ordinary and partial di®erential equations (ODEs and PDEs). This volume, 

which may be used either as a course text or for self-study, features an eclectic 

collection of Maple computer algebra worksheets, or \recipes," that are systematically 

organized to illustrate graphical, analytical, and numerical techniques 

applied to ODE/PDE-based scienti¯c modeling. No prior knowledge of Maple 

is assumed, the early recipes introducing the reader to the basic Maple syntax, 

the subsequent recipes introducing further Maple commands and structure on 

a need-to-know basis. 

The recipes are fully annotated in the text and in most cases presented as 

\stories" or in a historical context. Each recipe typically takes the reader from 

the analytic formulation of an interesting mathematical model to its analytic 

or numerical solution and ¯nally to either a static or animated graphical visualization 

of the answer. Every recipe is followed by a set of problems that can 

be used to check one's understanding or develop the topic further. For your 

convenience, the recipes are included on a CD located in the inside back cover. 

v

Contents 

PREFACE v 

INTRODUCTION 1 

A. Computer Algebra Systems . . . . . . . . . . . . . . . . . . . . . . 1 

B. Computer Algebra Recipes . . . . . . . . . . . . . . . . . . . . . . . 2 

C. Introductory Recipe: Boys Will Be Boys . . . . . . . . . . . . . . . 3 

D. Maple Help . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 

E. How to Use This Text . . . . . . . . . . . . . . . . . . . . . . . . . 9 

I THE APPETIZERS 11 

1 Phase-Plane Portraits 13 

1.1 Phase-Plane Portraits . . . . . . . . . . . . . . . . . . . . . . . . 13 

1.1.1 Romeo and Juliet . . . . . . . . . . . . . . . . . . . . . . . 18 

1.1.2 There's No Damping Vectoria's Romantic Heart . . . . . 23 

1.1.3 Van der Pol's Limit Cycle . . . . . . . . . . . . . . . . . . 28 

1.2 Three-Dimensional Autonomous Systems . . . . . . . . . . . . . 32 

1.2.1 The Period-Doubling Route to Chaos . . . . . . . . . . . 33 

1.2.2 The Oregonator . . . . . . . . . . . . . . . . . . . . . . . 40 

1.2.3 RÄossler's Strange Attractor . . . . . . . . . . . . . . . . . 44 

2 Phase-Plane Analysis 47 

2.1 Phase-Plane Analysis . . . . . . . . . . . . . . . . . . . . . . . . . 47 

2.1.1 Foxes Munch Rabbits . . . . . . . . . . . . . . . . . . . . 51 

2.1.2 The Mona Lisa of Nonlinear Science . . . . . . . . . . . . 58 

2.1.3 Mike Creates a Higher-Order Fixed Point . . . . . . . . . 67 

2.1.4 The Gnus and Sung of Erehwon . . . . . . . . . . . . . . 73 

2.1.5 A Plethora of Points . . . . . . . . . . . . . . . . . . . . . 78 

2.2 Three-Dimensional Autonomous Systems . . . . . . . . . . . . . 82 

2.2.1 Lorenz's Butter°y . . . . . . . . . . . . . . . . . . . . . . 82 

2.3 Numerical Solution of ODEs . . . . . . . . . . . . . . . . . . . . . 88 

vii

viii CONTENTS 

2.3.1 Finite Di®erence Approximations . . . . . . . . . . . . . . 89 

2.3.2 Rabbits and Foxes: The Sequel . . . . . . . . . . . . . . . 91 

2.3.3 Glycolytic Oscillator . . . . . . . . . . . . . . . . . . . . . 96 

2.3.4 Fox Rabies Epidemic . . . . . . . . . . . . . . . . . . . . . 101 

II THE ENTREES 107 

3 Linear ODE Models 109 

3.1 First-Order Models . . . . . . . . . . . . . . . . . . . . . . . . . . 110 

3.1.1 How's Your Blood Pressure? . . . . . . . . . . . . . . . . 110 

3.1.2 Greg Arious Nerd's Problem . . . . . . . . . . . . . . . . 115 

3.2 Second-Order Models . . . . . . . . . . . . . . . . . . . . . . . . . 118 

3.2.1 Daniel Encounters Resistance . . . . . . . . . . . . . . . . 118 

3.2.2 Meet Mr. Laplace . . . . . . . . . . . . . . . . . . . . . . 121 

3.2.3 Jennifer's Formidable Series . . . . . . . . . . . . . . . . . 126 

3.3 Special Function Models . . . . . . . . . . . . . . . . . . . . . . . 130 

3.3.1 Jennifer Introduces a Special Family . . . . . . . . . . . . 131 

3.3.2 The Vibrating Bungee Cord . . . . . . . . . . . . . . . . . 137 

3.3.3 Mathieu's Spring . . . . . . . . . . . . . . . . . . . . . . . 142 

3.3.4 Quantum-Mechanical Tunneling . . . . . . . . . . . . . . 144 

4 Nonlinear ODE Models 149 

4.1 First-Order Models . . . . . . . . . . . . . . . . . . . . . . . . . . 150 

4.1.1 An Irreversible Reaction . . . . . . . . . . . . . . . . . . . 150 

4.1.2 The Struggle for Existence . . . . . . . . . . . . . . . . . 152 

4.1.3 The Bad Bird Equation . . . . . . . . . . . . . . . . . . . 161 

4.2 Second-Order Models . . . . . . . . . . . . . . . . . . . . . . . . . 164 

4.2.1 Patches Gives Chase . . . . . . . . . . . . . . . . . . . . . 164 

4.2.2 Oh What Sounds We Hear! . . . . . . . . . . . . . . . . . 168 

4.2.3 Vectoria Feels the Force and Hits the Bottle . . . . . . . . 175 

4.2.4 Golf Is Such an \Uplifting" Experience . . . . . . . . . . 179 

4.3 Variational Calculus Models . . . . . . . . . . . . . . . . . . . . . 185 

4.3.1 Dress Design, the Erehwonese Way . . . . . . . . . . . . . 185 

4.3.2 Queen Dido Wasn't a Dodo . . . . . . . . . . . . . . . . . 191 

4.3.3 The Human Fly Plans His Escape Route . . . . . . . . . . 195 

4.3.4 This Would Be a Great Amusement Park Ride . . . . . . 201 

5 Linear PDE Models. Part 1 207 

5.1 Checking Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 207 

5.1.1 The Palace of the Governors . . . . . . . . . . . . . . . . 207 

5.1.2 Play It, Sam . . . . . . . . . . . . . . . . . . . . . . . . . 211 

5.1.3 Three Easy Pieces . . . . . . . . . . . . . . . . . . . . . . 215 

5.1.4 Complex, Yet Simple . . . . . . . . . . . . . . . . . . . . . 220 

5.2 Di®usion and Laplace's Equation Models . . . . . . . . . . . . . . 223

CONTENTS ix 

5.2.1 Freeing Excalibur . . . . . . . . . . . . . . . . . . . . . . . 223 

5.2.2 Aussie Barbecue . . . . . . . . . . . . . . . . . . . . . . . 227 

5.2.3 Benny's Solution . . . . . . . . . . . . . . . . . . . . . . . 231 

5.2.4 Hugo and the Atomic Bomb . . . . . . . . . . . . . . . . . 236 

5.2.5 Hugo Prepares for His Job Interview . . . . . . . . . . . . 241 

6 Linear PDE Models. Part 2 247 

6.1 Wave Equation Models . . . . . . . . . . . . . . . . . . . . . . . . 247 

6.1.1 Vectoria Encounters Simon Legree . . . . . . . . . . . . . 247 

6.1.2 Homer's Jiggle Test . . . . . . . . . . . . . . . . . . . . . 251 

6.1.3 Vectoria's Second Problem . . . . . . . . . . . . . . . . . 254 

6.1.4 Sound of Music? . . . . . . . . . . . . . . . . . . . . . . . 257 

6.2 Semi-in¯nite and In¯nite Domains . . . . . . . . . . . . . . . . . 261 

6.2.1 Vectoria's Fourth Problem . . . . . . . . . . . . . . . . . . 261 

6.2.2 Assignment Complete! . . . . . . . . . . . . . . . . . . . . 263 

6.2.3 Radioactive Contamination . . . . . . . . . . . . . . . . . 266 

6.2.4 \Play It, Sam" Revisited . . . . . . . . . . . . . . . . . . 270 

6.3 Numerical Simulation of PDEs . . . . . . . . . . . . . . . . . . . 274 

6.3.1 Freeing Excalibur the Numerical Way . . . . . . . . . . . 275 

6.3.2 Enjoy the Klein{Gordon Vibes . . . . . . . . . . . . . . . 278 

6.3.3 Vectoria's Secret . . . . . . . . . . . . . . . . . . . . . . . 281 

III THE DESSERTS 285 

7 The Hunt for Solitons 287 

7.1 The Graphical Hunt for Solitons . . . . . . . . . . . . . . . . . . 290 

7.1.1 Of Kinks and Antikinks . . . . . . . . . . . . . . . . . . . 290 

7.1.2 In Search of Bright Solitons . . . . . . . . . . . . . . . . . 293 

7.1.3 Can Three Solitons Live Together? . . . . . . . . . . . . . 296 

7.2 Analytic Soliton Solutions . . . . . . . . . . . . . . . . . . . . . . 299 

7.2.1 Follow That Wave! . . . . . . . . . . . . . . . . . . . . . . 300 

7.2.2 Looking for a Kinky Solution . . . . . . . . . . . . . . . . 304 

7.2.3 We Have Solitons! . . . . . . . . . . . . . . . . . . . . . . 306 

7.3 Simulating Soliton Collisions . . . . . . . . . . . . . . . . . . . . 308 

7.3.1 To Be or Not to Be a Soliton . . . . . . . . . . . . . . . . 308 

7.3.2 Are Diamonds a Kink's Best Friend? . . . . . . . . . . . . 312 

8 Nonlinear Diagnostic Tools 319 

8.1 The Poincare Section . . . . . . . . . . . . . . . . . . . . . . . . . 319 

8.1.1 A Rattler Signals Chaos . . . . . . . . . . . . . . . . . . . 320 

8.1.2 Hamiltonian Chaos . . . . . . . . . . . . . . . . . . . . . . 323 

8.2 The Power Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . 329 

8.2.1 Frank N. Stein's Heartbeat . . . . . . . . . . . . . . . . . 332 

8.2.2 The Rattler Returns . . . . . . . . . . . . . . . . . . . . . 334

x CONTENTS 

8.3 The Bifurcation Diagram . . . . . . . . . . . . . . . . . . . . . . 337 

8.3.1 Pitchforks and Other Bifurcations . . . . . . . . . . . . . 338 

8.4 The Lyapunov Exponent . . . . . . . . . . . . . . . . . . . . . . . 342 

8.4.1 Mr. Lyapunov Agrees . . . . . . . . . . . . . . . . . . . . 343 

8.5 Reconstructing an Attractor . . . . . . . . . . . . . . . . . . . . . 345 

8.5.1 Putting Humpty Dumpty Together Again . . . . . . . . . 346 

8.5.2 Random Is Random . . . . . . . . . . . . . . . . . . . . . 349 

8.5.3 Butter°y Reconstruction . . . . . . . . . . . . . . . . . . . 351 

Epilogue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 

Bibliography 355 

Index 361

INTRODUCTION 

A. Computer Algebra Systems 

Man is still the most extraordinary computer of all. 

John F. Kennedy, former American president. Speech, 21 May 1963. 

Unlike traditional programming languages such as Fortran and C, a computer 

algebra language such as Maple allows one to compute not only with numbers, 

but also with symbols, formulas, equations, and so on. Using a computer 

algebra system (CAS), symbolic computation can be done on the computer, replacing 

the traditional pen-and-paper approach with the keyboard/mouse and 

computer display. By entering short, simple, transparent commands on the 

computer keyboard (which will be referred to as the \classic" approach), or by 

selecting mathematical symbols from a palette with the mouse, the CAS user 

can quickly and accurately generate symbolic input and output on the computer 

screen. Mathematical operations such as di®erentiation, integration, and 

series expansion of functions can be done analytically on the computer. 

Because it also has numerical capability, a CAS allows the student or the 

researcher to tackle all aspects of mathematical modeling, from analytic derivation 

and manipulation of the model equations to the analytic or numerical 

solution of those equations, to the plotting or animation of the results. One of 

the most powerful computer algebra systems currently available is Maple 10, 

which will be used in this text. Useful reference books to this CAS are the 

Maple user manual [Map05] and the introductory and advanced programming 

guides [MGH + 05]. 

In the two volumes of Computer Algebra Recipes, we present classic Maple 

worksheets, or \recipes," that demonstrate how a CAS can serve as a valuable 

adjunct tool in easily deriving, solving, plotting, and exploring interesting, 

modern scienti¯c models chosen from a wide variety of disciplines ranging from 

the physical and biological sciences to the social sciences and engineering. The 

present book is the second volume and concentrates on mathematically more 

advanced models involving linear and nonlinear ordinary and partial di®erential 

equations (ODEs and PDEs). The classic Maple worksheet interface, which 

requires less computer memory than the standard interface, is used to generate 

all the mathematical and graphical output shown in this text. 

1

2 INTRODUCTION 

B. Computer Algebra Recipes 

The mathematics is not there till we put it there. 

Sir Arthur Eddington, The Philosophy of Physical Science, 1939 

The ODE and PDE recipes in our computer algebra \menu" have been organized 

into three sections of increasing mathematical sophistication. The Appetizers 

illustrate phase-plane portraits and analysis, the Entrees deal with 

linear and nonlinear ODEs and linear PDEs, and the Desserts feature the 

hunt for solitons and some nonlinear diagnostic tools. The recipes are intended 

to ful¯ll not only a useful and serious pedagogical purpose but also to titillate 

and stimulate the reader's intellect and imagination. Associated with each 

recipe is an important scienti¯c model or method and usually some historical 

background or an interesting story featuring an engineering or science student 

who will guide you through the steps of the recipe. These storybook characters 

are ¯ctitious composites of some of the more likeable, industrious, and brighter 

students that the authors have had the privilege of teaching over the years. 

Every topic or story in the text contains the Maple code or recipe to explore 

that particular topic. To make life easier for you, all recipes have been placed 

on the CD-ROM enclosed within the back cover of this text. The recipes are 

ordered according to the chapter, section, and subsection (story) number. For 

example, the recipe 01-2-3, entitled RÄossler's Strange Attractor, is associated 

with Chapter 1, Section 2, Subsection 3. Although the recipes can be directly 

accessed on the CD by clicking on the appropriate worksheet number, it is 

recommended that you access them through the menu index ¯le, 00menu. 

All recipes may be conveniently accessed from this menu using the hyperlinks. 

Complete instructions on how to do this may be found in the menu ¯le. 

The computer code on the CD is unannotated, so you will have to read the 

text in order to understand what the code is trying to accomplish. The code has 

been imported into the text and here is accompanied by detailed explanations 

of the underlying modeling concepts and computational methods. 

The recommended procedure for using this text is ¯rst to read a given 

topic/story for overall comprehension and enjoyment. If you are having any 

di±culty in understanding a piece of the text code, then you should execute 

the corresponding Maple worksheet and try variations on the code. Keep in 

mind that the same objective may often be achieved by a di®erent combination 

of Maple commands from those used by the authors. After reading the topic, 

you should execute the worksheet (if you have not already done so) to make 

sure the code works as expected. At this point feel free to explore the topic. 

Try rotating any three-dimensional graphs or running any animations in the 

¯le. See what happens when changes in the model or Maple code are made and 

then try to interpret any new results. This book is intended to be open-ended 

and merely serve as a guide to what is possible in mathematical modeling using 

a CAS, the possibilities being limited only by your own background and desires. 

Each topic or story is self-contained and generally done completely, from

INTRODUCTORY RECIPE 3 

the derivation to the solution to the plot and accompanied by a thorough discussion 

of the steps and results. Since arriving at the answer is more important 

in our opinion than the method used, one will encounter recipes in which analytic 

derivation of the model equations occurs, followed by a numerical solution 

because an analytic solution doesn't exist. Although brief introductions, which 

generally include some de¯nitions of terminology and short explanations of underlying 

concepts, are given for each main topic area, this text is not intended 

to teach you everything that you want to know, for example, about methods of 

solving ODEs or PDEs. Neither is it intended to teach you about the myriad 

subject areas of science or engineering. Instead, it is meant to serve as a guide 

to how these topics and areas can be handled using a CAS. However, this book 

is not just any ordinary guide. It is a gourmet's guide! It presupposes that the 

reader has learned or is about to learn about various scienti¯c models and/or 

methods, and we are providing the computer algebra tools to enable you to 

solve complex scienti¯c problems more easily, to attain greater understanding, 

and to explore the frontiers of science that interest you. 

At the end of most recipe subsections, there are related problems where you, 

the reader, can check your mastery of the scienti¯c computation and computer 

algebra techniques presented in the recipes. The problems also allow you to 

explore new frontiers and challenge you to invent and solve \What happens if 

:::?" problems. The purpose of this text is not only to teach computer-assisted 

computational techniques useful to engineering and science students, but to 

whet the student's curiosity and put some fun back into the pursuit of a science 

education. For maximum satisfaction and learning, it demands an interactive 

approach by the reader. Although the stories were designed to be interesting 

or amusing to read, the Maple recipes must be run, the models explored, and 

the problems solved. Some things never change in the learning process! 

C. Introductory Recipe: Boys Will Be Boys 

Giving money and power to government is like giving 

whiskey and car keys to teenage boys. 

P. J. O'Rourke, American journalist (1947{) 

To give you some idea of what a typical computer algebra recipe looks like 

and to introduce some basic Maple syntax, consider the problem that follows. 

The recipe that solves this problem is not on the CD, so after reading this section 

you should open up classic Maple 10 on your computer, type in the recipe, 

and execute it. 

Richard's grandson Daniel throws a small ball with an initial speed of 15 

m/s towards a 3:5-meter-high fence located 20 meters from the ball's initial 

position. The ball leaves Daniel's hand at a height of 2 m above the level 

ground and just clears the top of the fence. The gravitational acceleration is 

g =9:8 m/s 2 . The ball may be regarded as a point particle and air resistance 

neglected.


(a) At what angle Á with the horizontal is the ball thrown? 

(b) How long does it take the ball to reach the fence? 

(c) Plot the entire trajectory, then animate the motion of the ball, including 

the fence in the animation. 

Let's choose the origin to be on the ground below the initial position of the 

ball and take the x-coordinate to be horizontal and the y-coordinate vertical. 

To begin the recipe, we ¯rst clear Maple's internal memory of any previously 

assigned values (other worksheets may be open with numerical values given to 

some of the same symbols being used in the present recipe). This is done by 

typing in the restart command after the opening prompt ( >) symbol, ending 

the command with a colon (:), and pressing Enter (which generates a new 

prompt symbol) on the computer keyboard. 

> restart: 

All Maple command lines must be ended with either a colon, which suppresses 

any output, or a semicolon (;), which allows the output to be viewed. 

Next, the given parameter values are speci¯ed. For example, the initial 

x-coordinate of the ball is entered, the symbolic name xb being placed to the 

left of the Maple assignment operator (:=). The numerical value (0) of the 

coordinate is placed on the right-hand side of the operator and the output 

suppressed here with a command-ending colon. Assigned quantities can be 

mathematically manipulated. In a similar manner, the numerical values of the 

ball's initial y-coordinate (yb), the horizontal location (xf) of the fence, the 

fence's height (yf), the initial speed (V) of the ball, and the magnitude of the 

gravitational acceleration (g) are entered. Because the command entries are 

short, we have chosen to place them all on the same prompt line, separating 

the entries by a space for reading clarity. 

> xb:=0: yb:=2: xf:=20: yf:=3.5: V:=15: g:=9.8: 

Using the symbol * for multiplication, we express the horizontal (vx) and vertical 

(vy) components of the ball's initial velocity in terms of the unknown angle 

Á. The Maple input syntax phi is used to generate the Greek letter Á in the 

output. Note that the assigned value (15) of V is automatically substituted. 

> vx:=V*cos(phi); vy:=V*sin(phi); 

vx := 15 cos(Á) vy := 15 sin(Á) 

Using the standard kinematic relations [Oha85], we calculate the ball's x and y 

coordinates at arbitrary time t. The symbols +, -, /, and^ are used for addition, 

subtraction, division, and exponentiation. Note that the decimal coe±cient of 

t 2 in the output is given to 10 digits, Maple's usual default accuracy. 

> x:=xb+vx*t; y:=yb+vy*t-(1/2)*g*t^2; 

x := 15 cos(Á) t y := 2 + 15 sin(Á) t ¡ 4:900000000 t 2 

Setting x=xf in the solve command, the time t = tf for the ball to reach the 

fence is determined in terms of Á. 

> tf:=solve(x=xf,t); #time to reach fence


tf := 4 1 

3 cos(Á) 

A comment, pre¯xed by the pound sign #, has been added to the command 

line. Short comments are useful for later reference or for others to read and 

understand the purpose of a Maple command. 1 

A transcendental equation eq for Á results on evaluating y at t = tf and 

equating the result to yf: 

> eq:=eval(y,t=tf)=yf; 

20 sin(Á) 8:711111111 

eq := 2 + ¡ 

cos(Á) cos(Á) 2 =3:5 

The transcendental equation is solved for Á, the result being labeled ©. 

> Phi:=solve(eq,phi); #angles in radians 

©:=0:9917653601; 0:6538908144; ¡2:149827293; ¡2:487701839 

Four angles, expressed in radians, are generated in the above output ©, the 

default accuracy being 10 digits. Since the initial angle must be above the horizontal, 

only the positive answers are acceptable as solutions for this problem. 

Since there are two positive results, this means that Daniel could throw the ball 

at two di®erent angles to just clear the fence. To proceed, we shall select one of 

the positive answers, say the second one in ©. This is done by entering Phi[2]. 

You can look at the ¯rst positive angle by changing this entry to Phi[1]. Ifdesired, 

the angle Á ¼ 0:65 radians can be converted to degrees using the convert 

command with units as the second argument. 

> phi:=Phi[2]; theta:=convert(phi,units,radian,degree); 

Á := 0:6538908144 μ := 37:46518392 

In this case, Daniel throws the ball at an angle μ ¼ 37 1 degrees to the horizontal. 

2 

It should be mentioned that the convert command is very useful for converting 

an expression from one form to another, the form of conversion being dictated 

by the choice of second argument. To see the types of conversions possible with 

Maple, click on the convert command in the worksheet, then on Help in the 

tool bar, and ¯nally on Helponconvert. 

Using the °oating-point evaluation command, evalf, we numerically evaluate 

the time to reach the fence, which is found to be about 1.68 seconds. 

> tf:=evalf(tf); #time to reach fence 

tf := 1:679846933 

To plot the entire trajectory, the time T for the ball to hit the ground must 

be determined. This is accomplished by setting y = 0 and solving for t, which 

produces two answers. 

> T:=solve(y=0,t); 

T := ¡0:1981184874; 2:060197767 

1 Longer or more detailed comments may be inserted into a worksheet by clicking on 

Insert in the tool bar at the top of the computer screen, then on Execution Group, on 

either Before Cursor or After Cursor, onText, and ¯nally typing in the comments.


The negative answer is the time at which the ball would have had to been 

thrown if it had started from ground level. The second answer is the relevant 

one here and is now selected. The ball hits the ground after 2.06 seconds. 

> T2:=T[2]; #time to hit ground 

T2 := 2:060197767 

The entire trajectory of the ball is now plotted using the plot command. The 

ball's horizontal and vertical coordinates must be entered as part of a Maple 

\list," the list entries being separated by commas and the entire list enclosed 

in square brackets. Maple preserves the order of items in a list. In the present 

case, the list entries are the x-coordinate, the y-coordinate, and the time range. 2 

If no optional arguments are speci¯ed in the plot command, a ¯gure will be 

generated in which Maple chooses its own horizontal and vertical scaling. Constrained 

scaling can be achieved by including the option scaling=constrained 

as an additional argument. 3 

On executing the plot command, we generate a computer plot similar to 

that reproduced in Figure 1, with y plotted vertically and x horizontally. 

> plot([x,y,t=0..T2],scaling=constrained); 

6 

4 

2 

0 

5 10 15 20 25 

Figure 1: Trajectory of the ball. 

Note that the horizontal and vertical axes are not labeled in the ¯gure. Axis 

labels can be added by including the option labels=["x(in m)","y(in m)"] 

in the plot command. Try it and see. 4 

If you wish to learn more about the plot command and its optional arguments, 

click the left mouse button on the plot command and then on Help 

at the top of the computer screen. Clicking on the entry Help on plot opens 

up a help page with information about this command structure. The various 

2 Thisisaparametric plot, because x and y depend parametrically on the time t. 

3 If this optional argument is omitted, constrained scaling can be achieved in another way. 

Click your left mouse button on the computer plot. This opens a tool bar at the top of the 

computer screen with various options. Clicking on Projection and then on Constrained 

produces a plot with constrained scaling. 

4 Thedoublequotesinthelabels option indicate a Maple \string," a sequence of characters 

that has no value other than itself. A string cannot be assigned to, and will always evaluate to, 

itself. Omit the double quotes and note what happens when the plot command is executed.


plotting options available can be found by clicking on the underlined hyperlink 

plot/details that appears on the help page, and then on plot[options]. Try 

including some other options in the plot command. For example, change the 

color of the ball's trajectory from the default red to, say, blue. 

The ¯nal part of the problem is to animate the trajectory of the ball with 

the vertical fence included. The fence is now plotted, but not displayed, being 

assigned the name fence. The coordinates of the bottom and the top of the 

fence are entered as Maple lists and then formed into another list, i.e., one has 

a \list of lists." The fence will be plotted as a vertical (default red) line. 

> fence:=plot([[xf,0],[xf,yf]]): 

To produce specialized plots, such as an animated one, we must access the plots 

library package. This is done by entering the command with(plots). The 

preface with always indicates a Maple library package is being \loaded" into 

the worksheet. Library packages are extremely important, since they contain 

approximately 90% of Maple's mathematical knowledge. Normally, we would 

end the command with a colon, but here a semicolon is used to see what plot 

commands are contained in the plots library. 5 

> with(plots); 

Warning, the name changecoords has been redefined 

[Interactive; animate; animate3d; animatecurve; arrow; changecoords;:::] 

The animate command, appearing in the above list, will be used to animate the 

motion of the ball. The syntax is animate(plot command,[plot arguments], 

time range,options). Thepointplot command will plot the ball as a point. 

The plot arguments symbol=circle and symbolsize=14 instruct Maple to plot 

the point as a size-14 circle. 6 Included in the plot arguments list are the ball's 

x- andy-coordinates given as a list of lists. The time range is t =0toT2 .In 

the options, the number7 of frames is taken to be 200, the frames being equally 

spaced in time. The option background=fence causes the fence to appear as 

background in each time frame. Finally, the scaling is constrained. 

> animate(pointplot,[[[x,y]],symbol=circle,symbolsize=14], 

t=0..T2,frames=200,background=fence,scaling=constrained); 

When the animate command line is executed on the computer, the initial frame 

of the animation will appear on the screen. Clicking on the picture with the left 

mouse button places the picture in a viewing box and opens up an animation bar 

at the top of the screen. The animation is started by clicking on the arrowhead 

( ¤ ) and stopped by clicking on the square ( 2 ). 

5 Only a partial list of plot commands is shown here in the text, as indicated by the dots. 

Note that a warning message is also produced that informs us that the name changecoords 

has been rede¯ned in the current release of Maple. If desired, warnings can be removed by 

inserting the command interface(warnlevel=0) prior to loading the library package. From 

now on, all such warnings will generally be arti¯cially removed in the text. 

6 The default symbol is a diamond and the default size is 10. 

7 The default number is 25 frames. 200 frames produces a smoother animation.


D. Maple Help 

We teachers can only help the work going on, 

as servants wait upon a master. 

Maria Montessori, Italian educator, The Absorbent Mind, 1949. 

We have already seen in the introductory recipe how Maple's Help can be used 

to learn more about the Maple commands that appear in the text recipes. If 

you wish to ¯nd out what other help is available, click on Help at the top of 

the computer screen and then on the entry Using Help. A help page opens 

with a number of hyperlinks that you should explore. 

Two of the more important hyperlinks are entitled Perform a Topic 

Search and Perform a Full Text Search. Here we shall give two simple examples 

of using these searches, leaving the full descriptions of the search types 

for you to read. It should be noted that neither type of search is case-sensitive. 

Our ¯rst example illustrates a topic search, which locates help based on 

a keyword that you specify. For example, suppose that you wanted the correct 

form of the command for taking a square root. Click on Help, thenon 

Topic Search, making sure that the Auto-search box is selected. Depending 

on the programming language used, the square root command could be sqr, 

sqrt, :::, ::: On typing the ¯rst couple of letters, sq, in the Topic box, Maple 

will display all the commands starting with sq. Double click on sqrt or, alternatively, 

single click on sqrt and then on OK. A description of the square root 

command will appear on the screen. You can then close the Help window and 

proceed with programming your recipe. 

The second example illustrates a full-text search. Suppose, for example, 

that you wish to ¯nd the command for analytically or numerically solving an 

ODE. In the Help window, click on Full Text Search. Then type ode in 

the Word(s) boxandclickonSearch. Double clicking on dsolve produces 

a description of the dsolve command for solving ODEs along with several 

examples. If, for example, you want to know how to ¯nd a numerical solution, 

click on the hyperlink dsolve,numeric. 

The approach employed in the introductory recipe may also be used to ¯nd 

information about unfamiliar mathematical functions that appear in the Maple 

output. If, for example, the output contained the word \EllipticF," you may 

¯nd out what this function is by clicking on the word to highlight it, then on 

Help, and ¯nally on Help on EllipticF. You will ¯nd that EllipticF refers to 

the incomplete elliptic integral of the ¯rst kind, which is de¯ned in the Help 

page. The same Help window may also be opened by typing in a question mark 

followed by the word and a semicolon, e.g., ?EllipticF;. 

Maple's Help is certainly not perfect, and on occasion you might feel frustrated, 

but generally it is helpful and should be consulted whenever you get 

stuck with Maple syntax or are seeking just the right command to accomplish 

a certain mathematical task.

HOW TO USE THIS TEXT 9 

E. How to Use This Text 

Begin at the beginning ... and go on till you come to the end. 

Lewis Carroll, Alice's Adventures in Wonderland (1865) 

The recommended procedure for most readers, particularly for someone who 

is new to CASs in general and to Maple in particular, is to follow the advice 

given by the king to the white rabbit in Alice's Adventures in Wonderland. 

Start with the Appetizers, thengoontotheEntrees, and ¯nish o® with 

the Desserts. In the early recipes of the Appetizers you will be introduced 

to more of the basic features of the Maple system and see further examples of 

Maple's Help. Keep in mind that although we have made every e®ort to keep 

this book self-contained, it is impossible for us to teach you everything you 

would wish to know about programming with the Maple CAS. After all, that 

it is not the primary purpose of this text. Further, the choice of Maple library 

packages and associated command structures is to some extent dictated by the 

choice of subject matter. Since the topics of Computer Algebra Recipes have 

now been split between two volumes, you might wish to consult the Introductory 

Guide for use of library packages not emphasized in this Advanced Guide. And, 

at the risk of sounding immodest, the former volume also contains a wonderful 

selection of scienti¯c models. 

Of course, if you are already a Maple expert, feel free to pick and choose 

those topics and recipes in this volume that interest you or are relevant to your 

own scienti¯c tastes or goals. 

No matter what approach to using this text is taken, we hope that you 

will enjoy the wide range of interdisciplinary topics and stories that we have 

presented. Before beginning your intellectual journey through this book, let us 

paraphrase a well-known saying from the world of sports with these words of 

advice: 

You can't learn the great game of scienti¯c modeling 

by being a spectator. You must play the game! 

We trust that as you sample and explore the various recipes on which our 

menu is based, you will enjoy the intellectual \feast" that we have prepared 

andpresentedinthisAdvanced Guide to the Mathematical Models of Science. 

Bon Appetit! 

Richard and George, 

Your CAS chefs

Part I 

THE APPETIZERS 

A man ceases to be a beginner in any given science 

and becomes a master in that science when he has 

learned that ... he is going to be a beginner all his life. 

R. G. Collingwood, British philosopher, The New Leviathan, 1942 

In a time of drastic change it is the learners who inherit 

the future. The learned usually ¯nd themselves equipped 

to live in a world that no longer exists. 

Eric Ho®er, American philosopher, Re°ections on the Human Condition, 1973 

To know yet to think that one does not know is best; 

Not to know yet to think that one knows 

will lead to di±culty. 

Lao-Tzu, Chinese philosopher, 6th century BC 

11

Chapter 1 

Phase-Plane Portraits 

Every portrait that is painted with feeling 

is a portrait of the artist, not of the sitter. 

Oscar Wilde, Anglo-Irish playwright, novelist, and poet (1854{1900) 

Consider a system of two ¯rst-order coupled ODEs of the general structure 

_X ´ dX 

= P (X;Y ); _ 

dY 

Y ´ = Q(X;Y ); (1.1) 

dt dt 

where P and Q are known functions of the dependent variables X and Y ,and 

the independent variable has been taken to be the time t. In other model 

equations, the independent variable could be a spatial coordinate, e.g., the 

Cartesian coordinate x. For compactness, the dot notation of (1.1) will often 

be used in our text discussion for time derivatives, one dot denoting d=dt, two 

dots standing for d2 =dt2 , and so on. Superscripted primes on the dependent 

variable indicate a spatial derivative, e.g., Y 0 ´ dY=dx, Y 00 ´ d2Y=dx2 ,etc. 

The 2-dimensional ODE system (1.1) is said to be autonomous, meaningthat 

P and Q do not depend explicitly on t. If there is an explicit dependence on the 

independent variable, the equations are said to be nonautonomous. Our goal in 

the following section is to illustrate a simple graphical procedure for exploring 

all possible solutions of equations (1.1) for speci¯c forms of P and Q. In the 

second section, we shall show that a 2-dimensional nonautonomous system of 

ODEs can be cast into a 3-dimensional autonomous system and present some 

interesting examples of the latter. 

1.1 Phase-Plane Portraits 

Some biological models of competing species are naturally of the standard form 

(1.1). For example, a simple ODE model of the temporal evolution of interacting 

rabbit and fox populations might be given by the system 

_r =2r ¡ 0:04 rf; f _ = ¡f +0:01 rf; (1.2) 

where r(t) andf(t) are the numbers of rabbits and foxes per unit area at time 

t. If the \interaction" terms involving rf are omitted in (1.2), the remaining 

13

14 CHAPTER 1. PHASE-PLANE PORTRAITS 

ODEs are linear (¯rst order) in r and f, respectively, and are said to be linear 

ODEs. Inclusion of the higher-order interaction terms changes the equations 

into nonlinear ODEs. That is to say, P (r; f ) ´ 2 r ¡0:04 rf and Q(r; f) ´¡f + 

0:01 rf are nonlinear functions of the dependent variables r and f. Nonlinear 

models, such as this one, are usually di±cult or impossible to solve analytically, 

so that one must resort to graphical or numerical means to obtain a solution. 

Other models, involving second-order ODEs, can often be recast into the 

standard form, e.g., models arising from Newton's second law of the structure 

Äx = F (x; _x); (1.3) 

where x is the displacement and F is the force per unit mass. Setting _x = v, 

which is the velocity, (1.3) can be expressed as the coupled ¯rst-order system 

_x = v; _v = F (x; v): (1.4) 

For example, consider the damped, simple harmonic oscillator (SHO) equation 

Äx +2° _x + ! 2 0 

x =0; (1.5) 

describing the oscillations of a mass m attached to a light spring (spring constant 

k) that obeys Hooke's law and experiences a frictional force (damping coe±cient 

°) linear in the velocity. The characteristic frequency !0 is equal to p k=m. 

On setting _x = v, this second-order linear ODE may be rewritten in standard 

form with P (x; v) ´ v and Q(x; v) ´ F (x; v) =¡2 °v¡ ! 2 0 x. 

Whether linear or nonlinear, a graphical approach can be used to view all 

possible solutions of those ODE systems that can be put into the standard form. 

This graphical procedure has proved especially important in the investigation 

of nonlinear systems, less so for linear systems, since the latter can usually be 

solved analytically. Since equations (1.1) do not depend explicitly on t, the 

independent variable may be eliminated by dividing one equation by the other 

to form the ratio 

dY 

dX 

Q(X;Y ) 

= : (1.6) 

P (X;Y ) 

If the Y -versus-X plane is considered, this ratio represents the slope of the 

trajectory of the ODE system at the point (X,Y ) in the plane. The X-Y plane 

is referred to as the phase plane and the trajectory as a phase-plane trajectory. 

For the rabbits{foxes system the phase plane shows the fox number (f) versus 

the rabbit number (r), while for the SHO the phase plane is a plot of velocity 

(v) against displacement (x). 

The time evolution of any possible motion of the ODE system may be pictured 

by systematically ¯lling the phase plane with a grid of uniformly spaced 

arrows indicating the direction of increasing time and the slope at each grid 

point. For a given set of initial conditions, the subsequent temporal evolution 

of the system may be traced out by moving from one arrow to the next and 

drawing an appropriate line for the trajectory. Pictures created in this manner 

are called phase-plane portraits. Since an arrow at any point in the phase plane 

is tangent to the trajectory at that point, the grid of arrows is referred to as 

the tangent ¯eld.

1.1. PHASE-PLANE PORTRAITS 15 

The phase-plane analysis of ODE systems, particularly those that are nonlinear, 

can be aided by ¯rst locating the stationary or ¯xed points (X0,Y0) of 

the system and then identifying their topological nature, i.e., the characteristic 

shape of the trajectories in the vicinity of the ¯xed points. At these points all 

the derivatives are zero. Thus, from equation (1.1) they are found by solving 

P (X0;Y0) =0; Q(X0;Y0) =0: (1.7) 

At a stationary point, it follows from equation (1.6) that the slope of the trajectory 

is of the form 0=0 and thus indeterminate. At any other point (referred 

to as an ordinary point) of the phase plane, a trajectory has a de¯nite slope. 

For linear systems that can be expressed in standard form there will be only 

one stationary point. For example, setting P = Q =0fortheSHOyieldsa 

stationary point at the origin of the v versus x phase plane. 

For nonlinear systems, there can be more than one ¯xed point. As an 

example, let's use Maple to determine these points for the rabbits{foxes system. 

> restart: 

The forms of P and Q for equations (1.2) are entered, 

> P:=2*r-0.04*r*f; 

> Q:=-f+0.01*r*f; 

P := 2 r ¡ 0:04 rf 

Q := ¡f +0:01 rf 

and the solve command applied to the set of equations 1 P =0,Q =0. A 

Maple \set" is enclosed in \curly" (f g) brackets. Unlike a Maple list, the 

order of the items is not preserved in a Maple set. The unknowns r and f are 

also entered as a set. 

> solve(fP=0,Q=0g,fr,fg); 

fr =0:; f =0:g; fr =100:; f =50:g 

From the output, we see that there are two stationary points, one at the origin 

of the f-r phase plane and the other at r =100rabbits,f =50foxes. 

Every stationary point has a certain topology in its neighborhood that dictates 

the nature of the phase-plane trajectories near that point. Thus, identifying 

the nature of each stationary point was an important task historically, since 

it allowed investigators in the precomputer era to sketch all the possible phase{ 

plane trajectories from a knowledge of the location of the stationary points 

and their types. In the modern computer era we can let software packages like 

Maple do the graphing and analysis for us. 

For ODE systems that can be put into standard form, what types of stationary 

points are possible? It turns out that there are only four types of so-called 

simple 2 stationary points, which are schematically illustrated in Figure 1.1. For 

the ¯rst three types (the vortex, focal, and nodal points) we shall 

1 Note that it isn't necessary to explicitly set P and Q equal to zero, since Maple will 

automatically assume that this is so unless you specify otherwise. 

2 The precise de¯nition of the phrase \simple" will be given in Chapter 2.


V 

N 

Figure 1.1: Curves near a vortex (V ), focal (F ), nodal (N), saddle (S) point. 

use the SHO equation expressed in standard form as an explanatory tool, assuming 

that the reader already has some idea of the qualitative nature of the 

solutions as the damping coe±cient ° is varied. 

For ° = 0, a SHO released from rest outside the origin will oscillate inde¯nitely 

about the origin with no change in amplitude. If the velocity is plotted 

versus the displacement, the phase-plane trajectories near the origin will qualitatively 

look like those shown in the top left of Figure 1.1, the arrows indicating 

the direction of increasing time. As the SHO passes through the origin, the velocity 

will be a maximum, going to zero as the system reaches its turning points. 

For each choice of initial displacement, a di®erent closed loop will be traced out 

in the phase plane. In this case, the origin (labeled V )issaidtobeavortex 

point or center. Since a continuum of initial conditions is possible, a vortex 

point is surrounded by a continuum of closed loops. 

For small ° below some critical threshold, the SHO will oscillate about the 

origin with an ever-decreasing amplitude, asymptotically approaching the origin 

as t !1. The origin is said to be a stable equilibrium point. Two representative 

trajectories are shown in the top right of Figure 1.1. As time progresses, each 

trajectory approaches the stationary point F at the origin along a (di®erent) 

spiral path. The point F is an example of a stable spiral point or focal point. 

An unstable focal point would be one for which the trajectory winds o® the 

A 

1 

A 

3 

F 

S 

A 

4 

A 

2


stationary point as t !1. 

As ° is increased, a critical damping is reached, beyond which the behavior 

of the SHO system changes. The SHO will no longer oscillate inde¯nitely about 

the origin but eventually approach it from a de¯nite direction as t !1.Thisis 

schematically illustrated in the bottom left of Figure 1.1 for six di®erent initial 

conditions. The origin, labeled N, isinthiscaseanexampleofastablenodal 

point. Foranunstablenodalpoint,thetimearrowswouldbereversed. 

The fourth type of possible ¯xed point, which will be illustrated in the 

following Romeo and Juliet recipe, is called a saddle point. 

Figure 1.2: Saddle-point topography. 

If you have done any alpine hiking, you might know that a \saddle" is a ridge 

between two mountain peaks or summits as illustrated in Figure 1.2, acquiring 

its name because the local topography resembles that of a horse saddle. The 

saddle point is the low point in the saddle, where in two opposite directions you 

would go downhill toward the two valleys and in the two transverse directions 

uphill toward the two peaks. If one characterized these four directions by arrows 

pointing in the direction of decreasing elevation, the arrows would point away 

fromthesaddlepointinthe¯rstcaseandtowarditinthesecond.Ifgivenan 

in¯nitesimal nudge, a particle would tend to move away from the saddle point 

in the ¯rst case and toward it in the second. 

A saddle point (labeled S) in the phase plane is a two-dimensional analogue 

of this topology, as schematically indicated in the bottom right of Figure 1.1, the 

arrows pointing in the direction of increasing time. The four trajectories labeled 

A1S, A2S, A3S, andA4S are examples of so-called separatrixes, dividing the 

area about the saddle point into four distinct regions with the trajectories in 

each region evolving with time in the directions indicated. A representative 

point approaches S along A1S and A2S as t ! +1, while it departs from S 

along A3S and A4S at t = ¡1.


1.1.1 Romeo and Juliet 

I am convinced we do not only love ourselves in others 

but hate ourselves in others too. 

G. C. Lichtenberg, German physicist, philosopher (1742{1799) 

The mathematician Steven Strogatz [Str88] [Str94] has suggested a simple dynamic 

model to create di®erent scenarios for the love a®air between Romeo and 

Juliet. In his model, R(t) andJ(t) represents Romeo's love/hate for Juliet and 

Juliet's love/hate for Romeo, respectively, at time t. Positive values of R and 

J indicate love, while negative values indicate hate. The love a®air equations 

take the form 

_R(t) =aR+ bJ; J(t) _ =cR+ dJ; 

where a, b, c, anddarereal coe±cients that may have either sign. For the sake 

of de¯niteness, let's take a =2,b =1,c = ¡1, and d = ¡2 in the following 

problem, leaving other coe±cient values for you to explore. 

(a) Is the ODE system linear or nonlinear? Locate the ¯xed point(s). 

(b) Create a tangent ¯eld plot and identify the nature of the ¯xed point(s). 

(c) Create a phase-plane portrait that contains the four trajectories corresponding 

to the following initial conditions: (i) R(0) = ¡0:25, J(0) = 1, 

(ii) R(0) = ¡0:27, J(0) = 1, (iii) R(0) = 0:27, J(0) = ¡1, and ¯nally 

(iv) R(0) = 0:25, J(0) = ¡1. Consider t =0to4timeunits. 

(d) Plot R versus t over the interval t = 0 to 2 for initial condition (i). 

(e) Derive analytic solutions for R(t) andJ(t) for initial condition (i). 

To plot the tangent ¯eld and create the phase-plane portrait, the dfieldplot 

and phaseportrait commands, respectively, will be used. These specialized 

di®erential equation plotting tools are contained in the DEtools library package, 

which is now loaded. The colon may be replaced with a semicolon to 

display the complete list of available commands in this package. 

> restart: with(DEtools): 

The general love a®air di®erential equations are entered in de1 and de2, each 

¯rst-order derivative with respect to t being entered with the diff command. 

> de1:=diff(R(t),t)=a*R(t)+b*J(t); 

de1 := d 

R(t) =a R(t)+b J (t) 

dt 

> de2:=diff(J(t),t)=c*R(t)+d*J(t); 

de2 := d 

J (t) =c R(t)+d J (t) 

dt 

Before proceeding further with solving our problem, a few words should 

be said about Maple's derivative command. Using exp, ln, andsin to enter 

the exponential, natural logarithm, and sine functions, suppose that Romeo's 

love/hate for Juliet depended on time in the following way.


> Romeo:=exp(-t)*ln(1+t)*sin(t)/sqrt(1+t^2); 

Romeo := e(¡t) ln(1 + t)sin(t) 

p 1+t 2 

The ¯rst derivative of Romeo with respect to t is now taken, the \inert" form, 

Diff, being used on the left-hand side to display the derivative, the \active" 

form, diff, employed on the right to explicitly perform the di®erentiation. 

> derivative:=Diff(Romeo,t)=diff(Romeo,t); 

derivative := d 

dt 

μ 

(¡t) e ln(1 + t)sin(t) 

p = ¡ 

1+t2 e(¡t) ln(1 + t)sin(t) 

p 

1+t2 + e(¡t) sin(t) 

(1 + t) p 1+t2 + e(¡t) ln(1 + t)cos(t) 

p ¡ 

1+t2 e(¡t) ln(1 + t)sin(t) t 

(1 + t2 ) (3=2) 

The extension of Maple's syntax to higher-order derivatives is straight-forward. 

The active form of the second derivative of Romeo is diff(Romeo,t,t) or, 

alternatively, diff(Romeo,t$2). Asanexercise,youshouldcalculate,say,the 

seventh time derivative of Romeo and see how easy it is to generate the new 

answer. A calculation by hand would be very tedious and prone to error. 

Returning to our problem, since the right-hand side (rhs) of both de1 and 

de2 depend linearly on R and J, the ODEs are linear. By visual inspection, 

there is one stationary point at R = J = 0. This is con¯rmed by applying the 

solve command to the rhs of the two ODEs. 

> statpoint:=solve(frhs(de1),rhs(de2)g,fR(t),J(t)g); 

statpoint := fJ (t) =0; R(t) =0g 

The given coe±cient values are now speci¯ed, 

> a:=2: b:=1: c:=-1: d:=-2: 

which are automatically substituted into the two ODEs. 

> de1; de2; 

d 

R(t) =2R(t)+J (t) 

dt 

d 

J (t) =¡R(t) ¡ 2 J (t) 

dt 

The dfieldplot command is used to plot the tangent ¯eld. 

> dfieldplot([de1,de2],[R(t),J(t)],t=0..1,R=-1..1,J=-1..1, 

dirgrid=[30,30],arrows=MEDIUM); 

The ¯rst and second arguments are Maple lists of the ODEs and dependent 

variables. The time range has been taken to be from t =0to1andthe 

plotting range for both R and J is from ¡1 to+1. Thedirgrid option speci¯es 

the number of horizontal and vertical mesh points (30 £ 30 here) to use for 

the tangent arrows. The mimimum is 2 £ 2 and the default is 20 £ 20. The 

option arrows=MEDIUM produces \full" arrowheads, the default being \half" 

arrowheads. To learn more about dfieldplot, highlight this command with 

your mouse, then click on Help, andonHelp on d¯eldplot.


J 

1 

0.5 

–1 –0.5 0.5 1 

R 

–0.5 

–1 

Figure 1.3: Tangent ¯eld for Romeo and Juliet's love a®air. 

The tangent ¯eld for Romeo and Juliet's love a®air is shown in Figure 1.3. Inspecting 

the graph and comparing with Figure 1.1, it is seen that the stationary 

point at the origin is a saddle point. By changing the coe±cient values, the 

nature of the stationary point and therefore the nature of the love a®air can be 

altered. This is left as a problem at the end of the section for you to explore. 

The four initial conditions are now entered, 

> ic1:=(R(0)=-0.25,J(0)=1): ic2:=(R(0)=-0.27,J(0)=1): 

ic3:=(R(0)=0.27,J(0)=-1): ic4:=(R(0)=0.25,J(0)=-1): 

and the phaseportrait command is used to create the phase-plane portrait for 

the four initial conditions. These conditions are entered as a list of lists. The 

time step size is taken to be 0.05. The default is to divide the time range into 20 

equal steps. So the default step size here would be 4=20 = 0:2. Again MEDIUM 

arrows are chosen, which are colored red. The trajectories are colored blue using 

the linecolor option. A complete list of options may be found under DEplot, 

whose Help page may be accessed though the topic search. 

> phaseportrait([de1,de2],[R(t),J(t)],t=0..4,[[ic1],[ic2], 

[ic3],[ic4]],stepsize=0.05,dirgrid=[30,30],R=-1..1,J=-1..1, 

arrows=MEDIUM,color=red,linecolor=blue); 

The phase-plane portrait for Romeo and Juliet's love a®air is reproduced in 

Figure 1.4. Asymptotically, the trajectories approach the separatrixes of the 

saddle point at the origin, the separatrixes dividing the phase plane into four 

di®erent \°ow" regions for the tangent arrows. Thus, for example, one can see 

that for any initial condition in the lower right region, R will remain positive 

and J negative. Romeo's love for Juliet is unrequited!


J 

1 

0.5 

–1 –0.5 0 0.5 

R 

1 

–0.5 

–1 

Figure 1.4: Phase-plane portrait for the love a®air. 

By introducing the option scene=[t,R] into phaseportrait, a plot of R 

vs. t can be produced. The resulting picture for the ¯rst initial condition is 

shown in Figure 1.5, Romeo's initial hate turning to love. Changing R to J in 

the scene option will generate J versus t. Try it and see how Juliet responds. 

> phaseportrait([de1,de2],[R(t),J(t)],t=0..2,[[ic1]], 

stepsize=0.2,scene=[t,R],color=red,linecolor=blue); 

R 

0.6 

0.4 

0.2 

0 

–0.2 

0.5 1 1.5 2 

t 

Figure 1.5: R versus t for the love a®air.


Now, the ODEs in Strogatz's model are linear with constant coe±cients, so 

they can be solved analytically. If proceeding by hand, one can solve de1 for 

J(t) and substitute the result into de2 , yielding a second-order ODE in R(t) 

alone. Assuming a solution for R(t) oftheforme ®t yields a quadratic equation 

in ®, withtworoots®1and ®2. Then R would involve a linear combination 

of e ®1 t ®2 t and e , the two arbitrary coe±cients being evaluated with the initial 

condition. With R(t) completely determined, J(t) is then easily evaluated. 

The Maple di®erential equation solve command, dsolve, hassuchstandard 

analytic methods of attack built into its solution algorithm. Applying this 

command to the ODE set, subject to the ¯rst initial condition, generates the 

following analytic solutions for R(t) andJ(t). 

> solution:=dsolve(fde1,de2,ic1g,fR(t),J(t)g); 

( Ã 

1 

solution := R(t) =¡ 

2 ¡ 7 p ! 

3 p3 

( e 

24 

p Ã 

3 t) 1 

+ 

2 + 7 p ! 

3 p3 

(¡ e 

24 

p 3 t) 

Ã 

1 

¡ 2 

2 ¡ 7 p ! 

3 

e 

24 

(p Ã 

3 t) 1 

¡ 2 

2 + 7 p ! 

3 

e 

24 

(¡p3 t) ; 

Ã 

1 

J (t) = 

2 ¡ 7 p ! 

3 

e 

24 

(p Ã 

3 t) 1 

+ 

2 + 7 p ! 

3 

e 

24 

(¡p ) 

3 t) 

In this case, one can see by inspecting the output that ® = § p 3. If the analytic 

forms of either R(t) andJ(t) are to be further manipulated, the solution must 

be assigned. Otherwise, entering R(t) or J(t) will not produce the analytic 

solutions, but just the symbols R(t) andJ(t) in the output. 

> assign(solution): 

Having assigned the solution, it can now be checked by, for example, subtracting 

the right-hand side (rhs) ofde1 from the left-hand side (lhs) ofde1 and 

simplifying with the simplify command. The result is 0 as expected. 

> check:=simplify(lhs(de1)-rhs(de1)); 

check := 0 

You can check that a zero result occurs if de1 is replaced with de2. 

PROBLEMS: 

Problem 1-1: Recasting into ¯rst-order ODEs 

Recast each of the following second-order ODEs into a ¯rst-order ODE system 

and identify P and Q. Which systems are nonlinear? For each autonomous 

system, locate all the stationary points. All parameters are real and positive. 

(a) Airy equation: y 00 ¡ xy =0; 

(b) \soft" spring equation: Äx +(1¡ x2 ) x =0; 

(c) Hermite equation: y 00 ¡ xy0 + ny =0; 

(d) Rayleigh equation: Äx ¡ ² (1 ¡ _x 2 )_x + x =0;


(e) con°uent hypergeometric equation: xy00 +(° ¡ x) y 0 ¡ ®y =0; 

(f) plane pendulum equation: Ä μ +sinμ =0. 

Problem 1-2: Symbolic di®erentiation 

Display and evaluate the following derivatives: 

(a) d5 

dx 5 

³ 

x6 ln(x)cos(x) e ¡x2´ , (b) d9 

dx 9 

μ 

11 

x 

p 

tanh(2 x) 

1+x4 Problem 1-3: Alternative love a®airs 

For each of the following possible love a®airs involving Romeo and Juliet: 

² produce a tangent ¯eld plot using the dfieldplot command and identify 

each stationary point and its stability; 

² produce a phase-plane portrait with the speci¯ed initial condition; 

² use the phaseportrait command and scene options to plot J(t) andR(t); 

² use the dsolve command to derive the analytic solution; 

² discuss how the love a®air evolves with time. 

(a) _ R = ¡2 R + J, J _ = ¡R ¡ 2 J, R(0) = ¡1, J(0) = 1. 

(b) _ R = J, J _ = ¡R, R(0) = 5, J(0) = ¡2. 

(c) _ R = J, J _ = ¡R + J, R(0) = 0:1, J(0) = 0. 

(d) _ R =2R + J, J _ = R +2J, R(0) = 0:2, J(0) = ¡0:1. 

1.1.2 There's No Damping Vectoria's Romantic Heart 

Don't waste time trying to break a man's heart; be satis¯ed if you 

can just manage to chip it in a brand new place. 

Helen Rowland, American journalist, A Guide to Men, \Syncopations," 1922 

Vectoria, a physics major at the Metropolis Institute of Technology (MIT), 

is working on her computer algebra assignment while waiting for a phone call 

from her boyfriend Mike, who is returning from a summer job with an archaeological 

dig in a remote area of Asia. In particular, she is asked to use the 

DEplot command to illustrate the change in the phase-plane trajectory of the 

damped simple harmonic oscillator (SHO) as the damping coe±cient is varied. 

Entering DEplot in Maple's Topic Search and clicking OK, Vectoria ¯nds 

that the DEplot command is contained in the DEtools library package, which 

she now loads. 


The SHO equation results on applying Newton's second law of motion to a 

unit mass acted on by a Hooke's-law restoring force, Fhooke = ¡! 2 x(t), and 

a velocity-dependent Stokes's drag force, Fdrag = ¡¯ (dx=dt). Here x(t) isthe 

displacement of the mass from equilibrium, ! is the frequency, and ¯ is the


damping or drag coe±cient. These two forces are now entered, square brackets 

being used for each assigned force to produce the relevant subscript. 

> F[hooke]:=-omega^2*x(t); F[drag]:=-beta*diff(x(t),t); 

Fhooke := ¡! 2 μ 

d 

x(t) Fdrag := ¡¯ x (t) 

dt 

Equating the second time derivative of x(t) to the sum of the forces generates 

the SHO equation, de. 

> de:=diff(x(t),t,t)=F[hooke]+F[drag]; 

de := d2 

dt2 x(t) =¡!2 μ 

d 

x(t) ¡ ¯ x (t) 

dt 

Vectoria now relates the velocity v(t) to the displacement in de2 , 

> de2:=diff(x(t),t)=v(t); 

de2 := d 

x (t) =v(t) 

dt 

and substitutes this result into de to create a coupled system (de2 and de3 )of 

two ¯rst-order ODEs, linear in x and v. 

> de3:=subs(de2,de); 

de3 := d 

dt v(t) =¡!2 x (t) ¡ ¯ v(t) 

To plot the phase-plane trajectory, the parameter values must be speci¯ed. 

Vectoria takes ! = 1 and three di®erent ¯ values, namely, b1 =0, b2 =0:2, and 

b3 =3. 

> omega:=1: b[1]:=0; b[2]:=0.2; b[3]:=3; 

b1 := 0 b2 := 0:2 b3 := 3 

A \do loop" is used to create a phase-plane portrait using the DEplot command 

for each of the three ¯ values.Thegeneralsyntaxforadoloopis 

for from by to 

while do end do 

where the is the main body (the DEplot command here) 

of the do loop. In the following do loop, is the index i, the ¯rst 

is 1, the second is missing so i automatically increments 

by 1, the third is 3, and there is no conditional while 

present. The syntax for the DEplot command is the same as for the 

phaseportrait command used in the Romeo and Juliet recipe, the time 

rangeherebeingfromt =0to50. 

> for i from 1 to 3 do 

> DEplot([de2,subs(beta=b[i],de3)],[x(t),v(t)],t=0..50, 

[[x(0)=1,v(0)=0]],stepsize=0.05,x=-1.1..1.1,v=-1.1..1.1, 

dirgrid=[30,30],arrows=MEDIUM,color=red,linecolor=blue): 

> end do;


On execution of the do loop, the phase-plane portraits shown in Figures 1.6 to 

1.8, are produced. For zero damping (¯ =b1 =0), the phase-plane trajectory 

v 

1 

0.5 

–1 –0.5 0 0.5 1 

x 

–0.5 

–1 

Figure 1.6: Phase-plane portrait for ¯ =0. 

displayed in Figure 1.6 is a closed loop circling a vortex ¯xed point at the 

origin. The tangent ¯eld is also shown, the arrows pointing in the direction of 

increasing time. 

For ¯ =0:2, the phase-plane trajectory shown in Figure 1.7 is a spiral that 

v 

1 

0.5 

–1 –0.5 0.5 1 

x 

–0.5 

–1 

Figure 1.7: Phase-plane portrait for ¯ =0:2.


asymptotically approaches a stable focal point at the origin. This behavior is 

characteristic of underdamping. 

v 

1 

0.5 

–1 –0.5 0 0.5 x 1 

–0.5 

–1 

Figure 1.8: Phase-plane portrait for ¯ =3. 

Finally, for ¯ = 3, the phase-plane trajectory shown in Figure 1.8 approaches a 

stable nodal point at the origin, a behavior characteristic of overdamping. Itis 

left as an exercise for you to determine the critical damping threshold between 

under- and overdamping. 

Unlike the phaseportrait command, DEplot mayalsobeusedtoproduce 

solution curves, but not tangent ¯elds, for single higher-order ODEs. 

Taking, say, ¯ = b2 =0:2, Vectoria now uses DEplot to generate the x(t) 

solution curve shown in Figure 1.9 for the second-order ODE de, giventhe 

initial condition 3 x(0) = 1, _x(0)=0. Thetimerangeisfromt =0to50,and 

the time step size in the underlying numerical scheme is taken to be 0:05. 

> beta:=b[2]: 

> DEplot(de,x(t),t=0..50,[[x(0)=1,D(x)(0)=0]],stepsize=0.05); 

The numerically derived solution curve in Figure 1.9 decreases in amplitude in 

an oscillatory manner, again characteristic of the underdamped SHO. 

Finally, since the SHO equation de is linear with constant coe±cients, an 

analytic solution can be easily obtained for x(t) usingthedsolve command. 

> dsolve(fde,x(0)=1,D(x)(0)=0g,x(t)); 

x (t) = 1 p 

11 e 

(¡ 

33 

t 

10 ) sin 

Ã 

3 p 11 t 

10 

! 

t 

+ e 

(¡ 10 ) cos 

Ã 

3 p ! 

11 t 

10 

3 Note that the derivative condition _x(0) = 0 is entered as D(x)(0)=0, where D is the 

di®erential operator. The di®erential operator D is more general than diff. It can represent 

derivatives evaluated at a point and can di®erentiate procedures.


x(t) 

1 

0.8 

0.6 

0.4 

0.2 

0 

–0.2 

–0.4 

–0.6 

–0.8 

10 20 30 40 50 

t 

Figure 1.9: x versus t for ¯ =0:2. 

Vectoria leaves it as a problem for you to obtain the analytic solution for 

the critically damped case. 

Unfortunately, Vectoria must leave us for now, since her cell phone has just 

rung. Mike's plane has just landed and she's o® to the Metropolis International 

Airport to pick him up after he clears immigration and customs. 

PROBLEMS: 

Problem 1-4: Critical damping 

Given ! =1,what¯value corresponds to critical damping of the SHO? Make 

a phase-plane portrait for this case using the DEplot command and the initial 

condition x(0) = 1, _x(0) = 0. Use this command to plot x(t). Then obtain the 

analytic solution. 

Problem 1-5: Competition for the same food supply 

Two biological species competing for the same food supply are described by the 

following nonlinear population number equations: 

_ 

N1 =(4¡ 0:0002 N1 ¡ 0:0004 N2) N1; 

_ 

N2 =(2¡ 0:00015 N1 ¡ 0:00005 N2) N2: 

(a) Locate all the stationary points. 

(b) Create a tangent ¯eld plot that includes all the stationary points. Identify 

thenatureofthesepoints. 

(c) Create a phase-plane portrait that includes several representative trajectories 

that support your identi¯cation of the stationary points. 

(d) Use the scene option to plot N1(t) andN2(t). 

(e) Attempt to obtain an analytic solution of the ODE system.


1.1.3 Van der Pol's Limit Cycle 

In order to be able to set a limit to thought, we should have to ¯nd 

both sides of the limit thinkable (i.e., we should have to be able to 

think what cannot be thought). 

Ludwig Wittgenstein, Austrian philosopher (1889{1951) 

In the ¯rst two recipes, the ODEs were linear and, because they had constant 

coe±cients, were easily solved analytically. In this recipe, we look at the historically 

important nonlinear Van der Pol ODE, for which no analytic solution 

exists. Balthasar Van der Pol was a Dutch electrical engineer who pioneered 

the development of experimental nonlinear dynamics in the 1920s and 1930s 

using electrical circuits and discovered several important nonlinear phenomena. 

For example, he found that certain nonlinear circuits containing vacuum 

tubes could begin to spontaneously oscillate even though the energy source 

was constant, the oscillations evolving into a stable cycle, now called a limit 

cycle. When these circuits were driven with a signal whose frequency was near 

that of the limit cycle, the resulting periodic response shifted its frequency to 

that of the driving signal. The circuit became entrained to the driving signal. 

Entrainment is the basis of the modern pacemaker, which is used to stabilize 

irregular heart beats, or arrhythmias. 

In the September 1927 issue of the journal Nature, Van der Pol and van der 

Mark reported that an \irregular noise" was heard at certain driving frequencies, 

probably one of the ¯rst experimental reports of deterministic chaos. 4 

Here, we shall look at a modern electrical circuit [Cho64] that is governed by 

the Van der Pol equation and can produce his limit cycle. The circuit, involving 

a battery (voltage VB), inductor L, resistorR, capacitor C, and a tunnel diode 

D, is shown on the left of Figure 1.10. The tunnel diode has a nonlinear current 

(iD)-voltage (VD) curve similar to that shown on the right. 

L 

V R C D 

B 

i D 

i 

i 

S 

point of 

inflection 

Figure 1.10: Left: tunnel diode circuit. Right: Current-voltage curve for diode. 

4 This historical information is taken from the IEEE History Center (www.ieee.org). 

D 

V S 

V 

D


The battery voltage VB is adjusted to coincide with the in°ection point VS of 

the iD vs. VD curve, i.e., VB = VS. Near this operating point, one may write 

i = ¡av+ bv3 ,wherei = iD ¡ iS and v = VD ¡ VS, andaand b are positive. 

The governing Van der Pol (VdP) ODE, which will presently be derived, is 

Äx ¡ ² (1 ¡ x 2 )_x + x =0; ² > 0; (1.8) 

with x proportional to v. Equation (1.8) is just the simple harmonic oscillator 

equation for unit frequency and mass with an amplitude-dependent damping 

term. For x1 the damping is positive, tending to reduce the 

oscillations. The negative damping is responsible for the growth of any small 

spontaneous circuit \noise" into stable oscillations, i.e., into a stable limit cycle. 

Let's now derive the VdP equation and demonstrate the growth of a small 

input signal into a limit cycle for a typical tunnel diode, 1N3719, for which 

a =0:05 and b =1:0 in SI units. 

The DEtools and PDEtools packages are loaded. The former contains the 

DEplot3d command, which is a three-dimensional generalization of the DEplot 

command. The PDEtools package contains the dchange command, which will 

allow us to easily make a somewhat complicated variable transformation. 

> restart: with(DEtools): with(PDEtools): 

The time-dependent tunnel diode current and voltage expressions are entered. 

> i[D]:=i[S]-a*v(t)+b*v(t)^3; V[D]:=V[S]+v(t); 

iD := iS ¡ a v(t)+b v(t) 3 VD := VS + v(t) 

The voltage drop across both the resistor R and the capacitor C is the same as 

across the diode D, i.e., VR = VD and VC = VD. The voltage drop across the 

inductor L is VL = VB ¡ VD = VS ¡ VD, the latter form being entered. 

> V[R]:=V[D]: V[C]:=V[D]: V[L]:=V[S]-V[D]; 

VL := ¡v(t) 

By Ohm's law, the current through the resistor is iR = VR=R. The current 

through the capacitor is iC = C (dVC=dt). 

> i[R]:=V[R]/R; i[C]:=C*diff(V[C],t); 

iR := VS 

μ 

+ v(t) 

d 

iC := C 

R 

dt v(t) 

 

Using Kirchho®'s current rule, eq1 states that the current leaving L must be 

equal to the sum of the currents entering R, C, andD. 

> eq1:=-i[L](t)+i[R]+i[C]+i[D]=0; #Kirchhoff's current rule 

eq1 := ¡iL(t)+ VS 

μ 

+ v(t) d 

+ C 

R dt v(t) 

 

+ iS ¡ a v(t)+b v(t) 3 =0 

Di®erentiating eq1 with respect to t eliminates the in°ection point current iS. 

> eq2:=expand(diff(eq1,t)/C); 

d 

eq2 := dt v(t) 

CR ¡ 

d 

dt iL(t) 

μ 

d 

μ 2 a 

d dt 

+ v(t) ¡ 

C dt2 v(t) 

 

3 b v(t) 

+ 

C 

2 

μ 

d 

dt v(t) 

 

=0 

C


From the de¯nition of inductance, one has diL=dt = VL=L, which is substituted 

into eq2 . This yields a second-order ODE entirely in terms of the potential v(t). 

> de:=subs(diff(i[L](t),t)=V[L]/L,eq2); 

de := v(t) 

CL + 

d 

dt v(t) 

CR + 

μ 

d 

μ 2 a 

d dt 

v(t) ¡ 

dt2 v(t) 

 

3 b v(t) 

+ 

C 

2 

μ 

d 

dt v(t) 

 

=0 

C 

Then de is put into more compact form by collecting the ¯rst derivatives. The 

resulting ODE is the unnormalized form of the VdP equation. 

> de1:=collect(de,diff(v(t),t)); #unnormalized VdP equation 

μ μ 

1 a 3 b v(t)2 d 

de1 := ¡ + 

CR C C dt v(t) 

 

+ v(t) 

CL + 

μ 2 d 

v(t) =0 

dt2 To obtain the normalized (dimensionless) VdP equation, a transformation will 

be made to new variables. First a characteristic frequency ! = 1= p LC is 

introduced by making the following substitution into de1 . 

> de2:=subs(v(t)/(C*L)=omega^2*v(t),de1); 

μ μ 

1 a 3 b v(t)2 d 

de2 := ¡ + 

CR C C dt v(t) 

 

+ ! 2 μ 2 d 

v(t)+ v(t) =0 

dt2 Inspecting the structure of de2 , we are led to introduce a dimensionless time 

¿ = !t,andvoltagex(¿) = p (3 b) v(t)= p (a ¡ 1=R). The transformation from 

the \old" (t; v(t)) to the \new" (¿; x(¿)) variables is entered. 

> tr:=ft=tau/omega,v(t)=x(tau)*sqrt(a-1/R)/sqrt(3*b)g: 

The dchange command allows us to apply the transformation to de2. The 

result is then multiplied by the factor p (3 b)=(! 2 p (a ¡ 1=R). 

> sqrt(3*b)*dchange(tr,de2,[x(tau),tau])/(omega^2*sqrt(a-1/R)): 

Using the ditto operator, %, to refer 5 to the last computed result, we collect 

dx(¿)=d¿ terms and factor the result. 

> de3:=collect(%,diff(x(tau),tau),factor); 

μ 

d 

(x (¿) ¡ 1) (x (¿)+1)(aR¡ 1) x (¿) 

μ 

2 

d¿ d 

de3 : 

+ x (¿)+ x(¿) =0 

!CR 

d¿ 2 

Introducing the dimensionless parameter ² =(aR¡ 1)=(!CR), the normalized 

Van der Pol equation results. 

> vdp:=subs((a*R-1)=epsilon*(omega*C*R),de3); #VdP equation 

μ 

d 

vdp := (x (¿) ¡ 1) (x (¿)+1)² 

d¿ x(¿) 

μ 

2 d 

+ x (¿)+ x (¿) =0 

d¿ 2 

To make a phase-plane picture, the second-order VdP equation is now rewritten 

as two ¯rst-order ODEs in de4 and de5 ,bysettingy(¿) ´ dx(¿)=d¿. 

> de4:=diff(x(tau),tau)=y(tau); de5:=subs(de4,vdp); 

5 To refer to the second-to-last expression, use %%, and so on.


de4 := d 

x(¿) =y(¿) 

d¿ 

μ 

d 

de5 := (x (¿) ¡ 1) (x (¿)+1)² y(¿)+x (¿)+ 

d¿ y(¿) 

 

=0 

With a =0:05 entered for the tunnel diode 1N3719, a necessary condition for 

a limit cycle to occur is that ²>0orR>1=a =1=0:05 = 20 ohms. We take 

R =55ohms,L =25:0 £ 10 ¡3 henries, and C =10 ¡6 farads, and calculate ! 

and ². 

> a:=0.05: R:=55: L:=25.0*10^(-3): C:=10^(-6): 

> omega:=1/sqrt(L*C); epsilon:=(a*R-1)/(omega*C*R); 

! := 6324:555320 ² := 5:030896278 

The VdP equation has a ¯xed point at the origin of the y vs. x phase plane. 

Let's choose an initial condition close to this point, viz., x(0) = 0:1, y(0) = 0. 

> ic:=x(0)=0.1,y(0)=0: 

Instead of plotting the trajectory in two dimensions using either the DEplot 

or phaseportrait commands, the solution curve corresponding to the initial 

condition can be drawn in the three-dimensional ¿ vs. x vs. y space using 

DEplot3d with the option scene=[tau,x,y]. The line color of the trajectory 

is allowed to vary with ¿. The resulting trajectory appears in a 3-dimensional 

viewing box similar to that shown in Figure 1.11. The viewing box can be 

rotatedonthecomputerscreen,byclickingontheboxanddraggingwiththe 

> DEplot3d([de4,de5],[x(tau),y(tau)],tau=0..60, 

scene=[tau,x,y],[[ic]],stepsize=0.01,linecolor=tau); 

y 

4 

0 

–4 

–2 

–1 

x 

0 

1 

2 

0 

10 

20 

30 

40 t 

50 

60 

Figure 1.11: Evolution of the VdP trajectory onto a limit cycle.


mouse. The angular coordinates, μ and Á, of the viewing box appear in a 

small window near the top left of the computer screen, the default angles being 

45 ± ,45 ± . The option orientation=[angle,angle], with the values of the two 

angles speci¯ed, can be inserted into DEplot3d if some other default orientation 

is desired. For example, choose the angles to be μ =0,Á =90toseethephase 

plane. 

The trajectory evolves away from the vicinity of the origin onto a closed 

loop, the limit cycle. You can check that the limit cycle will be obtained no 

matter what the choice of initial condition. 6 Can you identify the stationary 

point at the origin? 

The nonlinear Van der Pol equation does not have an analytic solution. 

Applying the dsolve command to the ODE system, de4 and de5 , subject to 

the initial condition, 

> dsolve(fde4,de5,icg,fx(tau),y(tau)g); 

produces no output for x(t) andy(t). Only a few nonlinear ODEs of physical 

interest have analytic solutions. We shall see a few examples in Chapter 4. 

PROBLEMS: 

Problem 1-6: Di®erent initial conditions 

With all other parameters as in the text recipe, show for a number of di®erent 

initial conditions that all trajectories wind onto the limit cycle. Take the 

orientation that shows the y versus x phase plane. 

Problem 1-7: Varying the resistance 

With all other parameters as in the text recipe, investigate the behavior of the 

VanderPolequationastheresistanceRis varied. Choose an orientation that 

shows x versus ¿. Discuss the results. 

Problem 1-8: Tangent ¯eld 

In the text recipe, use the phaseportrait command instead of DEplot3d to 

make a phase-plane portrait with the tangent ¯eld included. 

1.2 Three-Dimensional Autonomous Systems 

Although a three-dimensional plot was produced in the last example, we were 

still dealing with a two-dimensional autonomous system. Let's now consider a 

general three-dimensional autonomous ODE system of the structure 

_X = P (X;Y; Z); Y _ = Q(X;Y; Z); Z _ = R(X;Y; Z); (1.9) 

with P , Q, andRknown functions of the three dependent variables X, Y ,and 

Z. Some systems of physical interest are naturally of this structure, while the 

two-dimensional nonautonomous ODE system 

_X = P (X;Y; t); Y _ = Q(X;Y; t); (1.10) 

can be recast into the form (1.9) by setting R = 1 and imposing Z(0) = 0. 

6 Ifyoustarttoofaro®thelimitcycle,thetimerangemayhavetobeincreased.

1.2. THREE-DIMENSIONAL AUTONOMOUS SYSTEMS 33 

1.2.1 The Period-Doubling Route to Chaos 

Chaos often breeds life, when order breeds habit. 

Henry Adams, American historian (1838{1918) 

Nonautonomous nonlinear ODEs, such as Du±ng's equation, 

Äx +2° _x + ®x+ ¯x 3 = F cos(!t); (1.11) 

have played a very important role in the development of nonlinear dynamics. 

Du±ng's equation is a model for the motion of a viscously damped (damping 

coe±cient °) spring that is subject to a nonlinear restoring force f = ¡®x¡ 

¯x3 and is being driven by a periodic force of amplitude F and frequency !. 

Depending on the signs and magnitudes of ® and ¯, various descriptive names 

are usually applied to Du±ng's equation: 

² hard-spring Du±ng equation: ®>0, ¯>0; 

² soft-spring Du±ng equation: ®>0, ¯0. 

Setting _x = y, Du±ng's equation can be written in the 2-dimensional nonautonomous 

form (1.10) with P ´ y and Q ´¡2 °y¡ ®x¡ ¯x3 + F cos(!t). 

It can be made autonomous by introducing a third dependent variable, z, and 

expressing Du±ng's equation as the three-dimensional system 

_x = y; _y = ¡2 °y¡ ®x¡ ¯x 3 + F cos(z); _z = !; with z(0) = 0: (1.12) 

After a transient time interval, the Du±ng system can, not unexpectedly, 

display a periodic oscillation in response to the periodic driving term. A more 

surprising result is that it can exhibit highly irregular, or chaotic, oscillatory 

motion that is essentially unpredictable, even though the Du±ng equation is 

deterministic. In contrast to the periodic regime, there is an extreme sensitivity 

to initial conditions in the chaotic domain. 

The Du±ng ODE is not the only dynamical system to exhibit chaotic behavior. 

In general, for chaos to occur in a dynamical system, two ingredients are 

necessary, namely that some nonlinearity be present and that the system have at 

least three dynamical dependent variables (i.e., be at least three-dimensional). 

The study of chaotic behavior is a nonlinearly growing ¯eld, and it is not our 

intention to explore it in any depth in this text, although some useful diagnostic 

tools are brie°y presented in the Desserts. 

In the following recipe, Jennifer, a mathematician at MIT, will illustrate 

the so-called period-doubling route to chaos for the Du±ng system. This refers 

to a sequence of period doublings (halving of the frequency response) that are 

observed when a \control" parameter is increased, ultimately ending in a chaotic 

regime. This period-doubling scenario is not the only route to chaos, but it is a 

very common one in the study of driven nonlinear ODE systems as well as other 

nonlinear systems that are naturally three- (or more) dimensional in nature.


For a given Du±ng spring system with speci¯ed initial conditions, there are 

two control parameters that could be varied in the driving force, namely the 

frequency ! and the amplitude F . Jennifer decides to hold ! ¯xed, and study 

how the Du±ng system responds as F is increased. 

After loading the plots and DEtools packages, Jennifer unprotects °, 

> restart: with(plots): with(DEtools): unprotect(gamma): 

which she would like to use for the damping coe±cient. Otherwise, Maple 

treats the entry of gamma as a request for Euler's constant. Ifthisconstant 

is unfamiliar to you, consult Maple's Help. 

For the sake of de¯niteness, Jennifer considers the inverted spring system 

with the parameter values ® = ¡1, ¯ =1,° =0:25, and ! =1, 

> alpha:=-1: beta:=1: gamma:=0.25: omega:=1: 

and the four force amplitudes F1 =0:325, F2 =0:35, F3 =0:356, and F4 =0:42. 

> F[1]:=0.325: F[2]:=0.35: F[3]:=0.356: F[4]:=0.42: 

The F values were selected to illustrate distinctly di®erent responses of the 

spring system to the driving force. To gain a preliminary understanding of 

what motions are possible, Jennifer decides to derive the potential energy function 

V (x) and plot it. This may be accomplished by entering the anharmonic 

(nonlinear) restoring force f = ¡®x¡ ¯x3 , 

> f:=-alpha*x-beta*x^3; #anharmonic restoring force 

f := x ¡ x3 and performing the inde¯nite integral V = ¡ R fdxusing the int command. 

> V:=-int(f,x); 

V := ¡ 1 

2 x2 + 1 

4 x4 

The potential energy V is plotted over the range x = ¡1:5 to1:5, 

> plot(V,x=-1.5..1.5,tickmarks=[3,2],labels=["x","V"]); 

–1 

x 

1 

V 

–0.2 

Figure 1.12: Double-well potential for an inverted-spring Du±ng equation.


the resulting picture being shown in Figure 1.12. The potential is commonly 

referred to as the double-well potential. There are two minima, at x = ¡1 and 

x = +1, at which points V = ¡ 1 

4 , separated by a maximum at x =0,where 

V = 0. In the absence of any driving force (set F =0)ordamping(° =0),the 

two minima correspond to vortex points, and the maximum is a saddle point. 

In this case, the spring system will oscillate in one of the two potential wells 

provided that the total energy is less than zero. For a total energy greater than 

zero, the oscillations will be back and forth between the two potential wells. 

These possible motions can be con¯rmed by making a phase-plane portrait for 

° =0andF = 0 in the Du±ng equation. 

To make this portrait, Jennifer inserts the time-dependence of the displacement 

by changing the variable x to x(t) in the restoring force, 

> f:=subs(x=x(t),f): 

and introduces the velocity dx=dt = y(t) ineq0 . 

> eq0:=diff(x(t),t)=y(t): 

The following three initial conditions are considered. 

> ic1:=x(0)=0.09,y(0)=0: ic2:=x(0)=-0.09,y(0)=0: 

ic3:=x(0)=-1.5,y(0)=0: 

These conditions should produce undamped oscillatory motion in the right, 

left, and both potential wells, respectively. To con¯rm this, Jennifer applies 

the phaseportrait command to the coupled system eq0 and dy=dt = f. 

> phaseportrait([eq0,diff(y(t),t)=f],[x(t),y(t)],t=0..100, 

[[ic1],[ic2],[ic3]],stepsize=0.1,,x=-1.5..1.5,color=red, 

linecolor=blue,arrows=MEDIUM): 

1 

y 

–1 x 1 

–1 

Figure 1.13: Phase portrait for inverted Du±ng equation for ° =0andF =0.


The resulting Figure 1.13 exhibits closed-loop trajectories in the phase plane 

characteristic of the predicted possible oscillatory motions. 

The inclusion of damping and a nonzero force amplitude, on the other hand, 

will make the physical behavior much harder to predict. For (° 6= 0), the 

two vortices will change to stable focal or nodal points, and in the absence 

ofanyenergysource(F = 0) the spring system would asymptotically (t !1) 

approach one of the two minima. With the inclusion of the driving term (F 6=0), 

which periodically pumps energy into the nonlinear system, periodic solutions 

are again possible but their nature much more di±cult to forecast, since the 

periodicity of the response depends on the amplitude F chosen (for ¯xed !) 

and the initial conditions. 

To determine what happens, Jennifer now uses a do loop to construct a 

phase-plane graph for each of the four di®erent F values. 


Du±ng's equation is entered for the ith force value, the result being labeled as 

the ith equation. 

> eq[i]:=diff(y(t),t)+2*gamma*y(t)-f=F[i]*cos(omega*t); 

In the phaseportrait command, Jennifer starts the time range at t =100 

to eliminate any transient response of the system. The ¯rst initial condition is 

chosen, which corresponds to starting the system from rest in the right potential 

well of Figure 1.12, slightly displaced to the right of the central peak that 

separates the two wells. To create a phase-plane portrait for the nonautonomous 

case, the scene=[x,y] option is selected. 

> gr[i]:=phaseportrait([eq0,eq[i]],[x(t),y(t)],t=100..250, 

[[ic1]],scene=[x,y],stepsize=0.1,color=red,linecolor=blue); 

> end do: 

The four graphs produced by this do loop are grouped together in a 2 £ 2array 

> Graphs:=array(1..2,1..2,[[gr[1],gr[2]],[gr[3],gr[4]]]): 

and then displayed, the resulting picture being shown in Figure 1.14. 

> display(Graphs,tickmarks=[2,2]): 

For the ¯rst three phase-plane portraits, corresponding to F1 = 0:325, 

F2 =0:35, and F3 =0:356, the Du±ng system executes qualitatively di®erent 

periodic motions in the right potential well of Figure 1.12. Because of the 

mathematical uniqueness of the solutions for given initial conditions, \true" 

phase-plane trajectories do not cross at ordinary points of the phase plane. But 

here there are apparent \crossings" of the trajectories, the number increasing 

with increasing F . The crossings are an artifact of using a phase plane to 

represent the motion of the driven system. In fact, as mentioned earlier, the 

inverted spring system is actually a three-dimensional autonomous system. The 

trajectories do not cross when plotted in the three-dimensional x-y-z space. 

For F4 =0:42, the driving-force amplitude is su±ciently large that the system 

clearly oscillates between both of the potential wells, its motion in the phase 

plane appearing to be quite chaotic.


y y 

0 x 1 

0 x 1 

y 

0 x 1 

Figure 1.14: Phase-plane portraits for F1 =0:325 (top left), F2 =0:35 (top 

right), F3 =0:356 (bottom left), and F4 =0:42 (bottom right). 

To produce a deeper understanding, Jennifer reruns the ¯le with the scene 

option in the ith graph, gr[i], replaced with scene=[t,x] and the time range 

shortened to t = 100 to 160. The four displacement (x(t)) curves in Figure 1.15 

result, each corresponding to the matching phase-plane portrait in Figure 1.14. 

For F1 =0:325, the inverted spring responds periodically at exactly the 

driving frequency, the period being T =2¼=! =6:28. If the period is written 

as T = n (2 ¼=!), then n = 1 for this case, and the motion is referred to as a 

period-one response. 

For F2 =0:35, the spring has a repeat period that is twice that of the driving 

term, i.e., one has n =2andthereforeaperiod-two response. Notice that now 

the system alternates each half-cycle between di®erent maximum values of the 

displacement. 

For F3 =0:356, the repeat period is four times as large, corresponding to 

period four. AsF was increased, the period doubled from period one to period 

twotoperiodfour. AsF is further increased, this period doubling will continue 

–1 

y 

x 

1


until the repeat period is so large that the motion appears to be chaotic. This 

is what has happened for F4 =0:42. 

1 

x x 

100 

1 

x 

100 

t 150 100 t 

t 

150 

x 

1 

1 

–1 

100 

t 

150 

150 

Figure 1.15: Displacement x versus time t for F1 =0:325 (top left), F2 =0:35 

(top right), F3 =0:356 (bottom left), and F4 =0:42 (bottom right). 

The scenario that Jennifer has outlined for the inverted Du±ng equation is 

a commonly observed phenomenon for forced oscillator systems and is referred 

to in the literature as the period-doubling route to chaos. 

PROBLEMS: 

Problem 1-9: Unforced damped motion 

In the text example, keep all equation coe±cients the same (leaving ° =0:25) 

but take the forcing amplitude F to be zero. 

(a) If the inverted spring system is initially at rest in the right potential well 

with total energy E =+ 1 

4 , to what stationary point does it asymptotically 

evolve? Make a phase-plane portrait and a plot of x versus t. 

(b) Identify the stationary point.


(c) How long does it take for the system to be within 1% of this point? 

(d) If the spring system is initially at rest in the right potential well, what is 

the minimum value that the total energy must have so that it asymptotically 

approaches the stationary point in the left potential well? 

Problem 1-10: Period 8 

In the text example, determine an approximate F for which period 8 occurs. 

Problem 1-11: A di®erent response 

In the text example, what e®ect does changing the driving frequency to ! =2 

haveonthefourgraphs?IdentifytheperiodresponseforeachFvalue. Problem 1-12: Varying frequency 

With all other parameters the same as in the text, but with F =0:42, study the 

response of the inverted Du±ng system as ! is varied over the range between 

zero and one. Interpret the results in each case. 

Problem 1-13: Varying the damping coe±cient 

With all other parameters the same as in the text, investigate the e®ect on the 

four graphs when the damping coe±cient is reduced to ° =0:125. Identify the 

period response for each F value. Repeat with ° =0:0625. 

Problem 1-14: Varying the force law 

Execute the text recipe with the x3 term in the force law replaced with x5 and 

discuss how this change a®ects the results. Then try some larger F values (all 

other parameters remaining the same) and determine the period response of 

each solution. 

Problem 1-15: Nonharmonic Du±ng oscillator 

For the nonharmonic Du±ng oscillator with ® =0,¯ =1,° =0:04, ! =1, 

F =0:2, x(0) = 0:25, and y(0) ´ _x(0) = 0, determine the period response 

of the solution. Use both scene=[x,y] and scene=[t,x] before making your 

conclusion. What is the period response if x(0) = 0:2, all other parameter 

values remaining the same? 

Problem 1-16: Another forced oscillator 

Determine the period response of the forced-oscillator equation 

Äx +0:7 _x + x 3 =0:75 cos t; 

subject to the initial condition x(0) = _x(0) = 0. Explore the change in period 

response of the solution as the force amplitude is varied. What type of Du±ng 

equation is the above equation? 

Problem 1-17: Three-dimensional plots 

Instead of the planar plots presented in the text recipe, make use of the DEplot3d 

command to make three-dimensional plots in the t versus x versus y space. Plot 

the trajectory for each F value separately, choosing an orientation in each case 

that gives the best view.


1.2.2 The Oregonator 

Everybody's youth is a dream, a form of chemical madness. 

F. Scott Fitzgerald, American writer (1896{1940) 

The Belousov{Zhabotinski (BZ) chemical reaction is now probably one of the 

best known of the chemical oscillators. However, because it was contrary to 

the then current belief that all solutions of reacting chemicals must go monotonically 

to equilibrium, Belousov could not initially get his chemical oscillator 

discovery published in any Soviet journal. Only years later, when his results 

were con¯rmed by Zhabotinski, was he given due recognition for his discovery. 

For his pioneering research work he was awarded, along with Zhabotinski, the 

Soviet Union's highest medal, but unfortunately for him, he had passed away 

10 years earlier. 

The BZ reaction may be achieved [Tys76] by dissolving 4.292 g of 0.28 M 

malonic acid 7 and 0.175 g of 0.002 M cerium ammonium nitrate in 150 ml of 1 

M sulfuric acid and stirring well. The solution will initially be yellow, then turn 

clear after a few minutes. Then, on adding 1.415 g of 0.063 M sodium bromate, 

the solution will oscillate between yellow and clear with a period of about one 

minute. A more dramatic color change between red and blue can be achieved 

by adding a few ml of 0.025 M ferroin. 

Field, K}orÄos, and Noyes [FKN72][FN74] were able to measure the periodic 

oscillations in the Br ¡ concentration and the Ce 4+ /Ce 3+ ratio in the BZ reaction. 

They then isolated the ¯ve most important reactions in the complicated 

chemistry that was taking place and created a kinetic model called the Oregonator, 

the name re°ecting the location of where the research was carried out. 

Labeling the concentration of BrO ¡ 

3 as A, thatofHBrO2 as X, thatofBr ¡ 

as Y ,thatofCe 4+ as Z, and all other less-important chemical species as ¤, the 

relevant chemical reactions in the Oregonator model are as follows: 

A + Y + ¤ 

X + Y + ¤ 

A + X + ¤ 

k1 

! X + ¤ 

k2 

! ¤ 

k3 

! 2 X +2Z + ¤ (1.13) 

2 X k4 

! A + ¤ 

Z + ¤ 

k5 

! hY + ¤ 

Here the ki denote the rates of reaction and h is a numerical \fudge factor" 

introduced because of the severe truncation of the full set of equations describing 

the complicated chemistry. How sensitive the results are to the value chosen 

for h will be left as a problem. 

Ignoring the less-important chemical species and noting that the depletion 

of A can be neglected, i.e., A is constant, the rate equations for the production 

7 Belousov [Bel58] used citric acid, but malonic acid is now commonly substituted.


of X, Y , Z are 

_X = k1 AY ¡ k2 XY + k3 AX ¡ k4 X 2 ; 

_Y = ¡k1 AY ¡ k2 XY + hk5 Z; 

_Z =2k3 AX ¡ k5 Z: 

(1.14) 

In writing down equations (1.14), use has been made of the following empirical 

rule: When two substances react to produce a third, the reaction rate is proportional 

to the product of the concentrations of the two substances. Thus, for 

example, the structure of the _ X equation can be easily understood. In the ¯rst 

chemical reaction, the rate of producing X is +k1 AY. In the second reaction, 

there is a decrease in X, the rate contribution being ¡k2 XY. The third reaction 

provides a positive contribution +k3 AX. Finally, noting that 2 X in the 

fourth reaction is treated as X + X, this reaction provides a rate decrease given 

by ¡k4 X 2 . The other rate equations may be similarly understood. The factor 

of 2 in the _ Z equation appears because in the third reaction, two of Z appear 

for each net (2 X ¡ X) oneofX. 

The nonlinear rate equations (1.14) can be converted into a normalized 

form that reduces the number of parameters. Introducing a normalized time 

¿ =(k1A) t and concentrations 

x =(k2X)=(k1 A); y =(k2Y )=(k3 A); z =(k2k5 Z)=(2 k1 k3 A 2 ); 

and positive parameters, 

² = k1=k3; p =(k1A)=k5; q =(k1k4)=(k2 k3); 

the Oregonator equations reduce to 

² _x(¿) =x + y ¡ qx 2 ¡ xy; _y(¿) =¡y +2hz¡ xy; p _z(¿) =x ¡ z: 

As an illustrative example, we will look at the onset of oscillations in the 

Oregonator model for ² =0:03, p =2,q =0:006, h =0:75 and initial (normalized) 

concentrations x(0) = 1, y(0) = 1, and z(0) = 20. 

After loading the plots package, 

> restart: with(plots): 

the three governing di®erential equations are entered, 

> de1:=epsilon*diff(x(t),t)=x(t)+y(t)-q*x(t)^2-x(t)*y(t); 

μ 

d 

de1 := ² x (t) = x(t)+y(t) ¡ q x (t) 

dt 2 ¡ x(t) y(t) 

> de2:=diff(y(t),t)=-y(t)+2*h*z(t)-x(t)*y(t); 

de2 := d 

y(t) =¡y(t)+2h z(t) ¡ x (t) y(t) 

dt 

> de3:=p*diff(z(t),t)=x(t)-z(t); 

μ 

d 

de3 := p 

dt z(t) 

 

= x(t) ¡ z(t)


along with the parameter values 

> epsilon:=0.03: p:=2: q:=0.006: h:=0.75: 

and the initial condition. 

> ic:=x(0)=1,y(0)=1,z(0)=20: 

The set of nonlinear ODEs cannot be solved analytically, so we seek a numerical 

solution for the set of three unknowns using the dsolve command with the 

numeric option. Unless otherwise speci¯ed, the default numerical scheme is the 

Runge{Kutta{Fehlberg 45 (rkf45) algorithm. See Burden and Faires. [BF89] 

> sol:=dsolve(fde1,de2,de3,icg,fx(t),y(t),z(t)g,numeric): 

The 3-dimensional text plot command is used to place the blue-colored word 

\start" near the starting point of the trajectory. The three numbers in the list 

are the x-, y-, and z-coordinates where the word is to be placed. 

> tp:=textplot3d([1,1,21,"start"],color=blue): 

The odeplot command enables us to plot the numerical solution over the time 

interval0to20timeunits. Theminimumnumberofplottingpointsistaken 

to be 3000 in order to obtain a smooth curve. The default number is 50. 

> gr:=odeplot(sol,[x(t),y(t),z(t)],0..20,numpoints=3000, 

thickness=2,axes=frame,labels=["x","y","z"], 

tickmarks=[3,3,3]): 

The above two plots are then displayed together, producing Figure 1.16. 

> display(ftp,grg); 

30 

z 

20 

10 

0 

5 

y 

10 

15 

Figure 1.16: Evolution of Oregonator system onto limit cycle in phase space. 

The trajectory in the 3-dimensional phase space evolvesontoaclosedloop, 

characteristic of an oscillatory solution. No matter what the starting point 

start 

100 

50 

x 

0


(initial condition), the trajectory will wind onto the same loop, indicating that 

the loop is a stable 3-dimensional limit cycle. 

The odeplot command is used to plot the concentrations x(t), y(t), and z(t) 

as red, blue, and green curves respectively, with a title indicating this included. 

> odeplot(sol,[[t,x(t),color=red],[t,y(t),color=blue], 

[t,z(t),color=green]],0..20,numpoints=3000,tickmarks=[3,3], 

thickness=2,title="red=x, blue=y, green=z"); 

x, y, z 

100 

50 

0 

red=x, blue=y, green=z 

10 20 

t 

Figure 1.17: Oscillatory behavior of HBrO2 (x), Br ¡ (y), and Ce 4+ (z). 

The black-and-white version is shown in Figure 1.17, the tallest curve being x(t), 

the intermediate curve z(t), and the shortest curve y(t). Each curve reaches its 

maximum amplitude at a di®erent time. 

PROBLEMS: Problem 1-18: Oregonator limit cycle 

Con¯rm that a limit cycle results in the Oregonator model, regardless of the 

initial (nonzero) concentrations. 

Problem 1-19: Another chemical oscillator 

The rate equations for a certain chemical oscillator are 

A k1 

! X 

B + X k2 

! Y + ¤ 

2 X + Y 

k3 

! 3 X 

X k4 

! ¤ 

where the concentrations A and B of species A and B are held constant. 

(a) Using the empirical rule for chemical reactions, write down the rate equations 

for X and Y .


(b) Convert the rate equations into a normalized form by setting ¿ = k4 t, 

x= p (k3=k4) X, y = p (k3=k4) Y , a= p (k3=k4)(k1=k4) A, b=(k2=k4) B. 

(c) Taking a =1,b =2:5, x(0) = y(0) = 0:1, produce a 3-dimensional plot 

showing x(t) vs.y(t) vs.t. Choose an orientation that clearly shows a 

periodic orbit. 

(d) Con¯rm the limit-cycle nature by trying a few di®erent initial normalized 

concentrations. 

Problem 1-20: Oregonator fudge factor 

In the text recipe for the Oregonator model, the fudge factor was taken to be 

h =0:75. Exploring the range h =0:1 toh = 1, with all other conditions the 

same, determine whether a limit cycle occurs. Comment on the sensitivity of 

the model on h. 

1.2.3 RÄossler's Strange Attractor 

Strange as it may seem, no amount of learning can cure stupidity, 

and formal education positively forti¯es it. 

Stephen Vizinczey, Hungarian novelist (1933{) 

The following 3-dimensional nonlinear ODE system, due to RÄossler [RÄ76], 

_x = ¡(y + z); _y = x + ay; _z = b + z (x ¡ c); (1.15) 

can display a variety of di®erent trajectories in the x-y-z phase space, depending 

on the values assigned to the parameters a, b, andcand the initial condition. 

The following recipe produces one of the more interesting trajectories. 

After loading the DEtools library package, 


the right-hand sides of the _x, _y, and_zequations are entered and assigned the 

names P , Q, andR, respectively. 

> P:=-(y+z); Q:=x+a*y; R:=b+z*(x-c); 

P := ¡y ¡ z Q := x + ay R:= b + z (x ¡ c) 

We take the parameter values to be a =0:2, b =0:2, and c =5:7. 

> a:=0.2: b:=0.2: c:=5.7: 

The number and locations of the ¯xed points are determined by solving the 

three equations P =0,Q =0,andR =0forx, y, andz. Listsareusedsothat 

the order x; y; z of the ¯xed-point coordinates is maintained in the output. 

> points:=solve([P=0,Q=0,R=0],[x,y,z]); 

points := [ [x =0:007026204834; y= ¡0:03513102417; z=0:03513102417]; 

[x =5:692973795; y= ¡28:46486898; z=28:46486898] ] 

There are two ¯xed points, one of which is near the origin. We take our initial 

condition to be x(0) = 0:1, y(0) = 0:1, z =0:1, i.e., near this ¯xed point. The


trajectory that results if the initial condition is near the other ¯xed point is left 

as a problem. 

> ic:=[x(0)=0.1,y(0)=0.1,z(0)=0.1]: 

To enter the relevant ODEs, the dependent variables x, y, andz must be made 

time-dependent. This is done in the following command line. 

> vars:=fx=x(t),y=y(t),z=z(t)g: 

Substituting the variables into P , Q, andR, and equating to dx=dt, dy=dt, and 

dz=dt, yields the RÄossler system of ODEs. 

> sys:=diff(x(t),t)=subs(vars,P),diff(y(t),t)=subs(vars,Q), 

diff(z(t),t)=subs(vars,R); 

sys := d 

d 

x (t) =¡y(t) ¡ z(t); y(t) =x(t)+0:2y(t); dt dt 

d 

z (t) =0:2+z (t)(x (t) ¡ 5:7) 

dt 

Choosing the option scene=[x,y,z] in the DEplot3d command, we plot the 

trajectory in x-y-z space over the time interval t = 0 to 150, subject to the 

given initial condition. The step size is taken to be 0.01 in order to obtain a 

smooth curve. The trajectory is colored with the zhue shading option, and a 

particular orientation of the viewing box is chosen. 

> DEplot3d([sys],[x(t),y(t),z(t)],t=0..150,[ic],scene=[x,y,z], 

stepsize=0.01,shading=zhue,orientation=[-120,60], 

tickmarks=[3,3,3],thickness=1); 

20 

z 

10 

5 

y 

0 

–5 

–10 

Figure 1.18: RÄossler's strange attractor. 

0 

x 

10


The resulting picture is shown in Figure 1.18. The trajectory unwinds in a 

spiral fashion from its starting point near the origin, indicating that the ¯xed 

point at the origin is an unstable focal point. As time progresses, the trajectory 

is attracted to a localized region of the phase space where it traces out a 

never-repeating (chaotic) path. This is an example of a strange attractor, the 

word strange being introduced historically because it was not like a \normal" 

attractor (e.g., a focal point). Strange attractors also have the property that 

they have noninteger, or fractal, dimensions. If you wish to learn more about 

strange attractors and fractal patterns, this topic is discussed at length in the 

Introductory Guide. 

PROBLEMS: Problem 1-21: Second ¯xed point 

Run the text recipe with an initial condition near the second ¯xed point. What 

is the probable nature of this ¯xed point? What is the nature of the resulting 

trajectory as time progresses? 

Problem 1-22: Varying c 

Holding all other parameters as in the text recipe, explore the behavior of the 

RÄossler system as the coe±cient c is varied. Interpret the results.

Chapter 2 

Phase-Plane Analysis 

The more important the subject and the closer it cuts to the bone of 

our hopes and needs, the more we are likely to err in establishing a 

framework for analysis. 

Stephen Jay Gould, American paleontologist and science historian (1941{2002) 

In the ¯rst chapter, the reader has seen examples of phase-plane portraits for 

two-dimensional autonomous ODE systems of the structure 

_x = P (x; y); _y = Q(x; y); (2.1) 

where P and Q were speci¯ed real functions. Given the mathematical forms 

of P and Q, the number and locations of the ¯xed points is easily established, 

either analytically or numerically. Quite generally, the topological nature of a 

¯xed point can then be determined by examining the °ow of tangent arrows in 

its vicinity and/or the temporal evolution of a nearby trajectory. For nonlinear 

systems, this was done with numerically based graphing commands. In this 

chapter, we will complement this approach by introducing phase-plane analysis, 

which involves analytically examining the nature of the trajectories at ordinary 

points lying near each ¯xed point. The method can be generalized [Hay64] to 

three-dimensional systems, but becomes considerably more complicated. 

2.1 Phase-Plane Analysis 

Consider an ordinary point (x; y) lying near a ¯xed point (x0;y0). From equation 

(2.1), the general expression for the slope of a trajectory at (x; y) is 

dy=dx = Q(x; y)=P(x; y): (2.2) 

At a stationary point, Q(x0;y0) =P (x0;y0) = 0, while at ordinary points, 

although either Q or P may be zero (corresponding to zero slope or in¯nite 

slope), they are not zero simultaneously. For ordinary points close to a given 

¯xed point, we can write x = x0 + u, y = y0 + v, whereuand v are small, so 

that equation (2.2) becomes 

dy 

dx = Q(x0 + u; y0 + v) 

: (2.3) 

P (x0 + u; y0 + v) 

47

48 CHAPTER 2. PHASE-PLANE ANALYSIS 

Taylor expanding the right-hand side of (2.3) in powers of u and v yields 

dy dv 

= 

dx du = cu+ dv+ c 0 u2 + d 0 v2 + f 0 uv+ ¢¢¢ 

au+ bv+ a 0 u2 + b 0 v2 + e 0 ; (2.4) 

uv+ ¢¢¢ 

where the coe±cients 

μ μ μ 

μ 

@P 

@P 

@Q 

@Q 

a ´ 

; b ´ 

; c ´ 

; d ´ 

; 

@x 

@y 

@x 

@y 

x0;y0 

x0;y0 

x0;y0 

x0;y0 

etc., are real since Q, P were assumed to be real. 

For a linear model, only linear terms in u and v will be present in equation 

(2.4), the coe±cients a 0 ´ (@2P=@x2 )x0;y0, b 0 ´ (@2P=@y2 )x0;y0, etc., being 

identically zero. On the other hand, for a nonlinear model, higher-order terms 

corresponding to some of these coe±cients being nonzero must be present. 

A simple stationary point for a nonlinear model is one in the neighborhood 

of which the qualitative behavior of the trajectories is correctly described by 

retaining only the linear terms in u and v in equation (2.4), so that 

dv cu+ dv 

= : (2.5) 

du au+ bv 

Clearly, if a, b, c, anddarenonzero and u and v are su±ciently small, then 

equation (2.5) should be a good approximation to equation (2.4), the higherorder 

terms in u and v making only small corrections that, except for the 

vortex, 1 do not qualitatively change the nature of the trajectories. 

If, on the other hand, c and d (or a and b) both vanish, then higher-order 

terms should be kept in the numerator (or denominator). Even for a, b, c, and 

d all nonzero, one can have au + bv =0andcu+ dv =0foruand v 6= 0 

(u, v = 0 corresponds to the ¯xed point (x0;y0) of interest), in which case 

higher-order terms should be kept in both the numerator and denominator. A 

nontrivial solution of cu+ dv =0,au+ bv = 0, can occur only if the system 

has zero determinant: 

¯ ¯ 

¯ c d ¯ 

¯ a b ¯ = bc¡ ad=0: (2.6) 

If this occurs, the stationary point is no longer simple (i.e., it is not determined 

by linear terms in u and v alone). Since setting either c and d or a and b equal 

to zero also makes bc¡ ad vanish, it follows that a simple ¯xed point can occur 

if bc¡ ad 6= 0. In the neighborhood of such a stationary point, the trajectories 

are described by equation (2.5), so their nature is completely determined by 

the four coe±cients a, b, c, andd. 

Next we shall establish that there are only four types of simple ¯xed points 

for the two-dimensional phase plane, namely the vortex, focal, nodal, and saddle 

points introduced in Chapter 1. The expression (2.5) for dv=du can be thought 

of as resulting from a pair of coupled ¯rst-order linear ODEs, viz., 

_u = au+ bv; _v = cu+ dv: (2.7) 

1 Even the smallest corrections can change a vortex into a focal point.

2.1. PHASE-PLANE ANALYSIS 49 

Solving for v in the ¯rst equation and substituting into the second yields the 

second-order linear ODE 

Äu + p _u + qu=0; with p ´¡(a + d); q ´ ad¡ bc: (2.8) 

Since this ODE has constant coe±cients, a solution of the form u = e¸t is 

sought. Substituting u into (2.8) yields the quadratic auxiliary equation 

¸ 2 + p¸+ q =0; with two roots; ¸1;2 = ¡ p 1 p 

§ p2 ¡ 4 q: (2.9) 

2 2 

These roots may be either real or complex. Since the general solution u is a 

linear combination of e ¸1 t and e ¸2 t ,itisclearthatu ! 0ast !1if the 

real parts of both ¸1 and ¸2 arenegativeandu !1if either one (or both) of 

the roots has a positive real part. For the former, the stationary point will be 

stable, while for the latter it will be unstable. 

For simple ¯xed points, ad¡ bc ´ q 6= 0, so a zero ¸ root is not possible. 

The case q = 0 corresponds to a higher-order stationary point, which will be 

illustrated later. The possible roots ¸1, ¸2, which dictate the topological nature 

of the ¯xed point, depend on the relative size and signs of p and q. First, let's 

consider q>0andp 6= 0. Three cases have to be examined: 

² For p 2 > 4q, both the roots ¸1 and ¸2 arerealandofthesamesign, 

negative for p>0andpositiveforp0, the general solution 

for u is of the structure u = Ae ¡j¸1j t + Be ¡j¸2j t (A, B are arbitrary 

constants), while for p0 

and unstable nodal points for p0isthe 

critically damped SHO. 

² For p2 ¡ 4 q0, and unstable focal points for p


For q>0, now consider the case p =0. Thetworoots,¸1 and ¸2, arenow 

purely imaginary, viz., ¸1;2 = § iq, and the general solution is of the undamped 

oscillatory form u = A cos(qt)+B sin(qt). The solution is characteristic of 

trajectories in the vicinity of a vortex point. 

Finally, we examine the situation in which q0 has been labeled as 

vortices and focal points, rather than vortices alone. This is because the analysis 

for the vortices is not de¯nitive for nonlinear models, 2 sincewehavekept 

only ¯rst-order terms in u and v in the Taylor expansion (2.4). Higher-order 

terms in the expansion may turn vortices into focal points. With the Taylor expansion 

option available in Maple, we could, of course, keep higher-order terms 

in an attempt to distinguish between the two types of ¯xed points. This can be 

done for individual cases, but it is di±cult to make \global" statements that 

2 For linear ODE models, recall that higher-order terms are not present and one can have 

vortices only for p =0andq>0. 

q


apply to all nonlinear systems. A simple global theorem, which is left for the 

reader to prove, is due to Poincare: 

Suppose that for the system of equations _x = P (x; y), _y = Q(x; y), thefunctions 

P (x; y), Q(x; y) satisfy, in the neighborhood of the stationary point O, the 

conditions for O to be a vortex or a focus. If P (x; y) and Q(x; y) satisfy the 

conditions P (x; ¡y) =¡P (x; y), Q(x; ¡y) =Q(x; y), thenO is a vortex. 

In some situations, P and Q may not satisfy the above conditions, yet O is a 

vortex. Poincare's theorem represents a su±cient condition for the existence of 

a vortex, but is not a necessary condition. 

2.1.1 Foxes Munch Rabbits 

As for life, it is a battle ... 

Marcus Aurelius Antonius, Roman emperor and philosopher (AD 121{180) 

In mathematical biology there has been a great deal of interest in predator{prey 

systems in which certain animal species (the predator) survive by munching or 

crunching on one or more others (the prey). As a simple example, suppose that 

a species of fox survives by eating jackrabbits in the rolling hills of Rainbow 

County. The rabbits in turn subsist on the available vegetation, of which we 

shall assume there is an adequate supply. A model of this predator{prey interaction 

can be built up phenomenologically. Let's call f(t) andr(t) the fox and 

jackrabbit numbers per unit area (acre, hectare, or whatever) at time t. 

If no foxes were present, the rabbit population would increase, the rate 

of increase assumed to be proportional to the number of rabbits present, i.e., 

_r(t) = A1 r(t), with the rate constant A1 positive. On the other hand, if 

no rabbits were present, the foxes would starve to death and their numbers 

decrease, the rate equation being _ f(t) =¡A2 f(t), with A2 > 0. 

With both species present, the probability of an interaction will be proportional 

to the product r(t) f(t) of the population numbers. For the foxes the 

interaction will be positive in nature, but negative for the rabbits. Thus, the 

simple phenomenological model takes the following form: 

_r = A1 r ¡ B1 rf; f _ = ¡A2 f + B2 rf; (2.10) 

with the interaction coe±cients B1 and B2 positive. In practice, mathematical 

biologists create more realistic models with other factors taken into consideration 

and the coe±cient values determined from observational data. Despite its 

simple appearance, this set of nonlinear ODEs cannot be solved analytically. 

To begin the recipe, the DEtools library package is loaded because a phaseplane 

portrait will be constructed. 


We enter the following numerical values: A1 =2, A2 =1, B1 =3=100, and 

B2 =1=100, for the coe±cients. The coe±cient numbers are strictly arti¯cial,


and you should feel free to experiment with di®erent positive values. 

> A[1]:=2: A[2]:=1: B[1]:=3/100: B[2]:=1/100: 

The nonlinear functions P =A1 r ¡ B1 rf and Q=¡A2 f + B2 rf, corresponding 

to the right-hand sides of the ODEs in (2.10), are entered as \functional 

operators" using the \arrow notation." 

> P:=(r,f)-> A[1]*r-B[1]*r*f; 

P := (r; f ) ! A1 r ¡ B1 rf 

> Q:=(r,f)-> -A[2]*f+B[2]*r*f; 

Q := (r; f) !¡A2f + B2 rf 

The \arrow" (->) in the above inputs is formed on the keyboard by entering a 

\hyphen" followed by a \greater than" sign. The operation or \procedure" on 

the right-hand side of each arrow will be applied when the two3 variables r and 

f on the left-hand side are supplied as arguments to P and Q. For example, 

let's take r =100andf =10andcalculateP (100; 10). 

> P(100,10); #example 

170 

The procedure on the rhs of P has been applied with the coe±cient values 

automatically substituted, namely, 2 £ 100 ¡ (3=100) £ 100 £ 10 = 170. 

To classify the stationary points, we need to evaluate a, b, c, andd, which 

respectively involve the partial derivatives @P=@r, @P=@f, @Q=@r, and@Q=@f 

evaluated at the stationary points. A functional operator F is introduced to 

di®erentiate an arbitrary quantity X(r; f ) with respect to a variable v. 

> F:=(X,v)->diff(X(r,f),v): 

Then F is used to calculate the four relevant partial derivatives of P and Q. 

> a:=F(P,r): b:=F(P,f): c:=F(Q,r): d:=F(Q,f): 

The results have been labeled a, b, c, andd, but remember that the derivatives 

must still be evaluated at each stationary point. So let's locate the stationary 

points by solving P (r; f) =0andQ(r; f)=0forr and f. 

> sol:=solve(fP(r,f)=0,Q(r,f)=0g,fr,fg); 

½ 

sol := fr =0;f=0g; f = 200 

3 ;r=100 

¾ 

There are two ¯xed points, one at the origin (r =0, f = 0) and the second at 

r =100, f =200=3=662. By choosing one of the ¯xed points and assigning the 

3 

solution, the coordinates will automatically be substituted into the expressions 

that follow. Let's select the nonzero ¯xed point, which is the second solution 

here. As a check, the coordinates r0 and f0 of the ¯xed point are displayed. 

> assign(sol[2]); r0:=r; f0:=f; 

r0 := 100 f0 := 200 

3 

Then the values of a, b, c, anddare determined for the assigned ¯xed point. 

3 Any (¯nite) number of variables may be used.


> a:=a; b:=b; c:=c; d:=d; 

a := 0 b := ¡3 c := 2 

d := 0 

3 

The quantities p = ¡(a + d) andq = ad¡ bc are calculated. 

> p:=-(a+d); q:=a*d-b*c; 

p := 0 q := 2 

Referring to Figure 2.1, these values of p and q indicate that the stationary 

point must be either a vortex or a focal point. Instead of always referring 

back to the p-q picture in the text when tackling other examples, it is more 

convenient to create the picture directly in the code and place the (q, p) point 

on it. The range of q and p in the ¯gure will be set to be from R = ¡3 to 

+3. For other stationary-point problems, this range may have to be adjusted. 

Further, the display command will be used to superimpose a number of graphs 

in the same ¯gure. This command is found in the plots library package, which 

is now loaded. 

> R:=3: with(plots): 

The ¯rst graph, assigned the name gr1, usesthepointplot command to plot 

the point (q, p) (entered as a list) as a size-20 blue box. The default size is 10. 

> gr1:=pointplot([q,p],symbol=box,symbolsize=20,color=blue): 

The second graph, gr2, plotsthetwobranchesp = § p 4 q of the parabola that 

divides the focal point and nodal point regions in the p-q diagram. Note that 

q is entered as qq (which ranges from ¡R to R), because q has already been 

assigned a speci¯c value. The resulting parabola is represented by a thick (the 

default thickness is 0) red curve on the computer screen. The minimum number 

of plotting points is taken to be 250, the default being 50. 

> gr2:=plot([sqrt(4*qq),-sqrt(4*qq)],qq=-R..R, 

numpoints=250,thickness=2,color=red): 

The third graph, gr3, generates a thick green line along the p=0 axis between 

q =0 andq =R and along the q =0 axis between p=¡R and p=+R. 

> gr3:=plot([[[0,0],[R,0]],[[0,-R],[0,R]]],style=line, 

color=green,thickness=3): 

Using the textplot command, the fourth graph labels the various regions of 

the p-q diagram. Here sFP stands for stable Focal Point, uFP for unstable Focal 

Point, sNP for stable Nodal Point, uNP for unstable Nodal Point, SP for Saddle 

Point, and V and FP for Vortices and Focal Points. The two numbers in each 

list are determined by trial and error and indicate the horizontal and vertical 

locations of each string of letters. 

> gr4:=textplot([[1,0.9,"sFP"],[1,-0.9,"uFP"], 

[0.7,2.5,"sNP"],[0.7,-2.5,"uNP"],[-1,0.8,"SP"], 

[-1,-0.8,"SP"],[1,0.2,"V and FP "]]): 

All four graphs are superimposed with the display command, the axes being 

labeled, and the minimum number of tickmarks along each axis speci¯ed. 

> display(fgr1,gr2,gr3,gr4g,labels=["q","p"],tickmarks=[3,3]);


SP 

SP 

p 

2 

–2 

sNP 

sFP 

V and FP 

–2 2 q 

uFP 

uNP 

Figure 2.2: p-q diagram for the second ¯xed point. 

The resulting p-q diagram is shown in Figure 2.2, the location of the box at 

p =0,q = 2 indicating that the second stationary point is either a vortex or a 

focal point. As you may verify, assigning the ¯rst ¯xed point would move the 

box to the saddle-point region of the p-q diagram. 

So, is the second ¯xed point a vortex or a focal point? Let's apply Poincare's 

theorem. Forming P (r0 + u; f0 ¡ v)+P (r0 + u; f0 + v) andQ(r0 + u; f0 ¡ v) ¡ 

Q(r0 + u; f0 + v) and expanding, 

> expand(P(r0+u,f0-v)+P(r0+u,f0+v)); 

0 

> expand(Q(r0+u,f0-v)-Q(r0+u,f0+v)); 

¡ uv 

50 

we see that the ¯rst relation yields zero, but the second doesn't. So Poincare's 

theorem is inconclusive here. A phase-plane portrait that includes the two ¯xed 

points should settle this issue. 

To proceed, the quantities r and f are unassigned from their previous values, 

> unassign('r','f'): 

and the rabbits{foxes equations generated using the operators P and Q. 

> req:=diff(r(t),t)=P(r(t),f(t)); 

req := d 

3 

r(t) =2r(t) ¡ r(t) f (t) 

dt 100


> feq:=diff(f(t),t)=Q(r(t),f(t)); 

feq := d 

1 

f (t) =¡f (t)+ r(t) f (t) 

dt 100 

A phase-plane portrait operator PP is formed to generate a phase-plane portrait 

for the rabbits{foxes equations over the time interval t = 0 to 10 time units, 

given an initial population of 100 rabbits and 5 foxes (per unit area). The scene 

variables x and y and the linecolor c must be speci¯ed. Instead of the default 

color red, the tangent ¯eld arrows are colored blue. 

> PP:=(x,y,c)->phaseportrait([req,feq],[r(t),f(t)], 

t=0..10,[[r(0)=100,f(0)=5]],stepsize=0.01,scene=[x,y], 

color=blue,linecolor=c,arrows=MEDIUM,dirgrid=[25,25]): 

The operator PP is used to plot a red-colored phase-plane trajectory in the f 

versus r plane, the resulting picture being shown in Figure 2.3. 

> PP(r,f,red); 

250 

200 

f 

100 

50 

0 

100 200 r 400 500 600 

Figure 2.3: The phase-plane portrait of the rabbits{foxes interaction. 

The phase-plane trajectory is a closed loop around the stationary point at 

r0 =100,f0 =66 2 

3 , indicating that this stationary point is probably4 avortex, 

not a focal, point. The behavior of the trajectory and the appearance of the 

tangent ¯eld near the origin is consistent with the origin being a saddle point. 

The temporal evolution of the rabbit and fox population numbers can be 

observed by entering PP(t,r,blue) and PP(t,f,red) and superimposing the 

blue and red curves using the display command. This produces Figure 2.4, 

the taller curve in this black-and-white rendition being for the rabbits. 

4 One can feel more con¯dent about this conclusion by considering other initial conditions 

and taking longer time intervals.


> display([PP(t,r,blue),PP(t,f,red)],labels=["t","r,f"]); 

600 

500 

400 

r,f 

200 

100 

0 

2 4 6 8 10 

t 

Figure 2.4: Periodic variation in the rabbits{foxes population numbers. 

The periodic variation in population numbers is clearly seen, the two curves 

being slightly out of phase with each other as one would intuitively expect. 

You might argue that this cyclic result depends on the particular values chosen 

for the coe±cients in the model. As you may con¯rm, choosing other positive 

values shifts the vortex-point location but doesn't destroy the periodicity. 

quantity 

160 

120 

80 

40 

snowshoe hare 

lynx 

1845 1855 1865 1875 1885 1895 1905 1915 1925 1935 

year 

Figure 2.5: Trading records of fur catches for the Hudson's Bay Company. 

Although the above predator{prey model is a phenomenological and oversimpli¯ed 

model of reality, the cyclic variations in population numbers that it 

predicts is a feature that has been observed in nature for di®erent predator{ 

prey interactions. For example, this can be seen in Figure 2.5, which shows the 

trading records for the period 1845 to 1935 of fur catches by trappers working in 

the Canadian north for the Hudson's Bay Company. In this case the lynx were 

the predators and the snowshoe hares the prey. Of course, the periodic vari-


ations observed in the lynx and snowshoe hare data curves are not as smooth 

and regular as in our idealized mathematical model. 

PROBLEMS: 

Problem 2-1: Iron core inductor 

Consider the simple circuit shown in Figure 2.6, consisting of a charged capacitor 

C connected to a coil of N turns wrapped around an iron core. The current i 

C 

i 

iron core 

inductor 

Figure 2.6: Iron core inductor circuit. 

versus °ux © relation for the iron core inductor has the form i = N ©=L0+A © 3 , 

where L0 is the self-inductance of the coil, © is the °ux threading through one 

turn of the coil, and A>0. 

(a) Using Kirchho®'s voltage rule, show that the governing ODE is given by 

Ä©+® ©+¯ © 3 =0; 

where ® and ¯ are left for you to identify. 

(b) Reexpress the ODE in a dimensionless form with ® and ¯ scaled out. 

(c) Analytically show that the origin of the phase plane is a vortex. Con¯rm 

with a phase-plane ( _ © versus ©) portrait containing a representative orbit. 

(d) Usethesceneoptiontoplot©(t). 

Problem 2-2: Competing armies 

The armies of two warring countries are modeled by the following equations: 

_ 

C1 = ®C1 ¡ ¯C1 C2; 

_ 

C2 =(® +1)C2 ¡ °¯C1 C2; 

with ® and ¯ both positive and °>1. Here C1 and C2 are the numbers of 

individuals in the armies of countries 1 and 2. 

(a) Discuss the model equations and how the model could be improved. 

(b) Analytically locate and identify all the stationary points. 

(c) Taking ® = 5, ° = 1:15, and ¯ = 1=2500, make a tangent ¯eld plot 

that includes all stationary points and some representative trajectories. 

Discuss possible outcomes on the basis of this plot. 

(d) Using appropriate scene options, create plots of C1(t) andC2(t).


Problem 2-3: Vortex or focal point? 

You are told that the following system has either a vortex or a focal point at 

the origin. Analytically determine which it is and support your conclusion by 

creating a suitable phase-plane portrait. 

_x = y + x (x 2 + y 2 ); _y = ¡x + y (x 2 + y 2 ) 

2.1.2 The Mona Lisa of Nonlinear Science 

Opinion is like a pendulum .... If it goes past the center of gravity on 

one side, it must go a like distance on the other. 

Arthur Schopenhauer, German philosopher (1788{1860) 

In the world of art, one of the most famous portraits ever painted is that of 

a woman with an enigmatic smile, referred to as the Mona Lisa. The artist 

was the Italian Leonardo da Vinci (1452{1519), who was not only a painter, 

but also a sculptor, architect, musician, and scientist. If Leonardo were alive 

today, he would probably appreciate on esthetic, as well as scienti¯c, grounds 

one of the most important phase-plane portraits of nonlinear science, that of 

the simple plane pendulum. If we may make a puny pun, the plane pendulum 

is not so plain as its name implies. 

θ 

L 

mgsinθ 

m 

θ 

mg 

Figure 2.7: A simple plane pendulum. 

What is a simple plane pendulum? It can be modeled as a small mass m 

attached to the end of a very light rigid rod of length L that is allowed to 

swing freely in a circular arc in the vertical plane as shown in Figure 2.7. Our


goal is to derive the equation of motion, locate and classify all of the possible 

stationary points, and determine all possible solutions of the simple pendulum. 

The necessary packages to carry out this program are now loaded. 

> restart: with(plots): with(DEtools): with(PDEtools): 

The plots library package contains the display command for superimposing 

plots and the textplot command for placing text on a ¯gure. The DEtools 

package is needed in order to use the DEplot command to produce a phaseplane 

portrait. Finally, the PDEtools package contains the dchange command, 

which will enable us to make variable changes in the equation of motion. 

Since the coe±cient ° will be introduced as a normalized damping coe±cient, 

it must be unprotected from its Maple assignment as Euler's constant. 

> unprotect(gamma); 

If μ(t) is the angle in radians that the pendulum rod makes with the vertical 

at time t, g is the acceleration due to gravity, and the air resistance is assumed 

to be proportional to the angular velocity dμ=dt with damping coe±cient ¡, 

Newton's second law, applied in the direction tangent to the circular arc, yields 

> eq1:=m*L*diff(theta(t),t,t)=-m*g*sin(theta(t)) 

-Gamma*diff(theta(t),t); 

μ 

μ 2 d d 

eq1 := mL μ(t) = ¡mgsin(μ(t)) ¡ ¡ 

dt2 dt μ(t) 

 

On dividing eq1 by mg and expanding, eq2 results. 

> eq2:=expand(eq1/(m*g)); 

μ 

μ 2 d d 

L μ(t) 

¡ 

dt2 dt 

eq2 := 

= ¡sin(μ(t)) ¡ 

g 

μ(t) 

 

mg 

Noting that the radian is actually a dimensionless unit, it is clear from the ¯rst 

term in eq2 that p L=g has the units of time. The reciprocal of this quantity 

is the characteristic frequency ! of the plane pendulum, which is entered. 

> omega:=sqrt(g/L); 

r 

g 

! := 

L 

A new dimensionless time variable ¿ = !tis introduced in the following transformation 

tr. Since the angle μ is already dimensionless, the transformation 

of eq2 into a completely dimensionless form is quite trivial. If Maple is used, 

however, the dependent variable μ(t) must be replaced by a new symbol, say, 

£(¿), in the variable transformation. The \old" variables are placed on the left 

of the transformation set, the \new" variables on the right. 

> tr:=ft=tau/omega,theta(t)=Theta(tau)g; 

8 

>< 

tr := μ(t) =£(¿); t= 

>: 

¿ 

9 

>= 

r 

g >; 

L


Thevariabletransformationofeq2 is implemented in eq3 with the dchange 

command. The ¯rst argument is the variable transformation tr, the second 

argument the equation to be transformed, the third argument the new variables, 

and the last optional argument is to simplify the result. 

> eq3:=dchange(tr,eq2,[Theta(tau),tau],simplify); 

eq3 := d2 

μ 

d 

sin(£(¿)) mg+¡ 

d¿ 

£(¿) =¡ 

d¿ 2 £(¿) 

r 

g 

L 

mg 

The equation is further simpli¯ed by de¯ning a dimensionless damping coef- 

¯cient ° through the relation ¡ = 2 °mg=!. Inclusion of the factor 2 is a 

matter of taste, a choice often made for damped harmonic oscillator systems. 

Substituting this expression for ¡ into eq3 and expanding, 

> eq4:=expand(subs(Gamma=2*gamma*m*g/omega,eq3)); 

eq4 := d2 

μ 

d 

£(¿) =¡sin(£(¿)) ¡ 2 ° 

d¿ 2 d¿ £(¿) 

 

yields eq4 , the dimensionless equation of motion for the plane pendulum. 

To make a phase-plane portrait, this second-order nonlinear ODE is reexpressed 

as two coupled ¯rst-order equations by setting the angular velocity 

d£=d¿ = V (¿) ineq5 , and substituting eq5 into eq4 . 

> eq5:=diff(Theta(tau),tau)=V(tau); 

> eq6:=subs(eq5,eq4); 

eq5 := d 

£(¿) =V (¿) 

d¿ 

eq6 := d 

V (¿) =¡sin(£(¿)) ¡ 2 °V (¿) 

d¿ 

From the right-hand sides of eq5 and eq6 ,theformsofPand Q needed to 

analyze the possible stationary points are easily identi¯ed and now entered. 

> P:=V; Q:=-sin(Theta)-2*gamma*V; 

P := V Q := ¡sin(£) ¡ 2 °V 

The derivatives @P=@£, @P=@V , @Q=@£, @Q=@V are calculated in a, b, c, d. 

> a:=diff(P,Theta): b:=diff(P,V): c:=diff(Q,Theta): d:=diff(Q,V): 

The quantities a, b, c, andd still have to be evaluated at the stationary points, 

which are found by solving P =0,Q = 0 for the unknowns £ and V . 

> sol:=solve(fP,Qg,fTheta,Vg); 

sol := f£ =0;V =0g 

There is a ¯xed point at the origin of the phase plane, corresponding to the 

pendulum at rest with the supporting rod oriented vertically downward (μ =0 

in Figure 2.7). This is obvious on physical grounds, since the ¯xed point arises 

because there is a zero net force on the mass m in this position. The downward 

pull of gravity on m is balanced by the upward tension in the supporting rod.


But the net force on m would also be zero if the pendulum were oriented 

vertically upward (μ = ¼). If the mass also has zero initial angular velocity, it 

will remain at rest, so the point (μ = ¼, V =0)mustbeanotherstationary 

point in the phase plane. Clearly, this second ¯xed point is one of unstable 

equilibrium, since the slightest nudge would cause the pendulum to move away 

from μ = ¼. So, why didn't the solve command produce a second solution 

to the equations P = 0, Q = 0? The reason is that the sine function is 

a transcendental function, so that in order to obtain the stationary points, 

one must solve an inverse transcendental function. To return the entire set of 

stationary points, we must make use of the following command line. 

> _EnvAllSolutions:=true: 

On applying the solve command a second time to the equations P =0,Q =0, 

> Sol:=solve(fP,Qg,fTheta,Vg); assign(Sol); 

Sol := f£ =¼ Z1 ;V =0g 

and noting that the symbol Z1 in the output stands for integer value, the 

complete family of stationary points is obtained for the simple pendulum. This 

new solution, Sol, is assigned. 

For Z1 =0 and Z1 = 1, the ¯xed points (μ =0,V =0) and (μ = ¼, V =0) 

result. The other stationary points corresponding to Z1 =2; 3; 4; ::: and 

Z1 =¡1; ¡2; ¡3; ::: re°ect the mathematical periodicity of the sine function, 

which has a period 2 ¼. Physically, for example, the angular position μ =2¼ is 

the same as μ = 0. In our analysis of the types of stationary points, we shall 

therefore concentrate on the points (μ =0,V =0) and (μ =¼, V =0). 

Continuing, the quantities p = ¡(a + d) andq = ad¡ bc are evaluated, 

> p:=-(a+d); q:=a*d-b*c; 

p := 2 ° q := (¡1) Z1 

and found to be p =2°, while q =+1forZ1 = 0 (or any even integer) and 

q = ¡1 forZ1 = 1 (or any odd integer). Let's label the former q value as q1 

and the latter as q2 . 

> q1:=1: q2:=-1: 

Three di®erent ° values, °1 =0,°2 =0:1, and °3 =1:25, are entered, the 

numbers selected to reveal di®erent physical behaviors of the pendulum. 

> gamma[1]:=0: gamma[2]:=0.1: gamma[3]:=1.25: 

In the following do loop, 


two plotting points, corresponding to the stationary points at (μ =0,V =0) and 

at (μ =¼, V =0), are calculated for each ° value, 

> pt1:=[q1,subs(gamma=gamma[i],p)]; 

> pt2:=[q2,subs(gamma=gamma[i],p)]; 

and three plots formed, the ¯xed points represented by size-20 blue circles. 

> pl[i]:=plot(fpt1,pt2g,style=point,symbol=circle, 

symbolsize=20,color=blue):


> end do: 

Now the rest of the p-q diagram is created. The parabola p 2 =4q is plotted as 

a thick red line over the range q = ¡2 to+2. Notethatthe\dummyvariable" 

r is used below instead of q as the latter has already been assigned. 

> pl[4]:=plot([sqrt(4*r),-sqrt(4*r)],r=-2..2, 

numpoints=250,thickness=2,color=red): 

The segments q =0 to +2 along the q-axis, and p = ¡3 to+3alongthep-axis, 

are plotted as thick green lines. 

> pl[5]:=plot([[[0,0],[2,0]],[[0,-3],[0,3]]],style=line, 

color=green,thickness=3): 

The various regions of the p-q plane are labeled in the same manner as the 

previous recipe. 

> pl[6]:=textplot([[1,0.8,"sFP"],[1,-0.8,"uFP"], 

[0.7,2.5,"sNP"],[0.7,-2.5,"uNP"],[-1,0.8,"SP"], 

[-1,-0.8,"SP"],[1.7,0.2,"V/FP"]]): 

The six plots are superimposed to produce the entire p-q diagram in Figure 2.8. 

> display(fseq(pl[i],i=1..6g,labels=["q","p"],tickmarks=[2,3]); 

SP 

–2 

q 

V/FP 

2 

SP 

2 

p 

–2 

sNP 

uNP 

sFP 

uFP 

Figure 2.8: p-q diagram for a simple pendulum. 

From bottom to top, the pairs of points in Figure 2.8 correspond to ° =0, 

° = 0:1, and ° = 1:25. In all three cases, the point on the left, which is 

associated with the ¯xed point (μ = ¼, V = 0), is an unstable saddle point. 

Because of the mathematical periodicity of the sine function, saddle points


will also occur at (μ = ¡¼; §3 ¼; :::, V =0). The¯xedpointsontheright 

correspond to (μ =0,V =0)aswellas(μ = §2 ¼; :::, V =0). For° =0, 

the ¯xed point is a vortex or a focal point. Applying Poincare's theorem, the 

reader will be able to con¯rm that it is a vortex. For ° =0:1, the damping 

coe±cient is su±ciently small that (μ =0,V = 0) is a stable focal point, while 

for ° =1:25 the damping is su±ciently large that a stable nodal point results. 

With the stationary point analysis completed, a phase-plane portrait is now 

constructed for each of the ° values. First, the values of £ and V are unassigned. 

> unassign('Theta','V'): 

Two initial conditions are chosen, both corresponding to pulling the pendulum 

to the left of the vertical so that the initial angle is ¡2 radians (about ¡115 ± ). 

For ic1 the pendulum is given an initial angular velocity of 2:75 radians per 

second, while for ic2 it is released from rest. 

> ic1:=Theta(0)=-2,V(0)=2.75; 

ic1 := £(0) = ¡2; V (0) = 2:75 

> ic2:=Theta(0)=-2,V(0)=0; 

ic2 := £(0) = ¡2; V (0) = 0 

A functional operator de is formed to evaluate the di®erential equation eq6 for 

the ith ° value. 

> de:=i->eval(eq6,gamma=gamma[i]): 

Another arrow operator, G, is introduced which uses the DEplot command 

to produce a phase-plane portrait for the two initial conditions. Using the 

linecolor option, the trajectories are colored red. A tangent ¯eld of fullheaded 

blue-colored arrows is also included. The argument i =1,2,or3must 

be provided for the operator G. 

> G:=i->DEplot([eq5,de(i)],[Theta(tau),V(tau)],tau=0..25, 

[[ic1],[ic2]],Theta=-2..8.5,V=-2..3.5,stepsize=0.05, 

color=blue,linecolor=red,arrows=MEDIUM, 

dirgrid=[30,30]): 

Entering G(1) and G(2), 

> G(1); G(2); 

produces Figures 2.9 and 2.10, corresponding to ° =0and° =0:1, respectively. 

In Figure 2.9, the vortex ¯xed points at V =0and£=0and2¼can be 

clearly seen, as well as the saddle point at V =0,£=¼. 

The closed curve encircling the vortex point at the origin is for the second 

initial condition (zero initial velocity). In this case, the mass has insu±cient 

initial energy to swing over the top, i.e., past the saddle point. The direction 

of increasing time is indicated by the tangent-¯eld arrows. 

For the ¯rst initial condition (nonzero initial velocity), on the other hand, 

the initial energy is su±ciently large that the pendulum can swing over the top. 

Since no damping is present, the pendulum continues to move in the positive 

μ-direction, alternately speeding up and slowing down, but never approaching 

a stationary point.


V 

3 

–2 8 

Theta 

–2 

Figure 2.9: Phase-plane diagram for ° =0. 

V 

3 

–2 8 

Theta 

–2 

Figure 2.10: Phase-plane diagram for ° =0:1. 

Figure 2.10, corresponding to ° =0:1, displays a completely di®erent behavior. 

Due to the presence of small damping, the vortices have turned into stable 

focal points. For ic2 , the pendulum initially overshoots μ = 0 but eventually 

winds onto the origin. For ic1 , the pendulum makes it over the top once, before 

winding onto the stationary point (μ =2¼, V = 0). Physically, of course, this


point corresponds to the vertically downward position of the pendulum rod. 

The third graph, for ° =1:25, is shown in Figure 2.11. Because the trajectories 

turn out to be con¯ned to a limited region of the phase plane, we use the 

display command to better control the viewing range of G(3) and e®ectively 

magnify the region of interest. Care must be taken in doing this, since the 

number of tangent-¯eld arrows is correspondingly reduced and their directions 

may not accurately re°ect the directions of the trajectories. 

> display(G(3),view=[-2..1,-0.5..3],tickmarks=[3,3]); 

–2 –1 1 

Theta 

Figure 2.11: Phase-plane diagram for ° =1:25. 

The damping is now su±ciently large that both trajectories approach the stable 

nodal point at the origin. One would have to increase the initial angular velocity 

to see any over-the-top behavior before things settle down to a stable nodal 

point. Here, the two trajectories merge along a common path for large times. 

From an artistic, as well as a scienti¯c, viewpoint Leonardo da Vinci would 

undoubtedly have appreciated the many di®erent faces revealed by the, perhaps 

not so plain, plane pendulum. 

PROBLEMS: 

Preamble: For each of the nonlinear ODE systems in the following set of 

problems, carry out the following steps: 

(1) Use the p-q diagram approach to locate and identify the stationary points 

of the system, using Poincare's theorem where necessary. 

3 

V 

2 

1


(2) In each case make a phase-plane portrait with all the ¯xed points included 

as well as the tangent ¯eld and some representative trajectories. 

(3) Plot the temporal evolution of the dependent variable(s) for the above 

trajectories, using the appropriate scene option. 

(4) Answer any additional questions posed for the system. 

Problem 2-4: Hard and soft springs 

For a hard (soft) spring, the displacement x from equilibrium is described by 

Äx + ! 2 (1 § ® 2 x 2 ) x =0; with !>0; ®>0: 

The plus sign is for the hard spring, the minus sign for the soft spring. Carry 

out the steps listed in the preamble for each spring type. Discuss the origin and 

nature of the restoring force leading to each equation. 

Problem 2-5: Eardrum equation 

The displacement x of an eardrum is described by the model equation 

Äx + x ¡ x 2 =2=0: 

Carry out the steps listed in the preamble. 

Problem 2-6: Some nonlinear systems 

Carry out the steps listed in the preamble for each of the following systems: 

(a) _x = x2 ¡ y3 , _y =2x (x2 ¡ y2 ); 

(b) _x = ¡x, _y =1¡ x2 ¡ y2 ; 

(c) _x = x (1 ¡ x2 ¡ 6 y2 ), _y = y (1 ¡ 3 x2 ¡ 3 y2 ). 

Problem 2-7: SIR model of infectious diseases 

The study of disease occurrence is called epidemiology. There are basically three 

types of deterministic models for infectious diseases that are spread by direct 

person-to-person contact. One of these models is referred to as the SIR model, 

the name being an acronym for the three population categories in the model. 

The S refers to the number of susceptibles who have not yet caught the disease, 

the I to the number of infectibles who have become infected with the disease, 

and R to the number of removables who have had the disease and are immune 

to catching that disease again. The disease being considered is such that very 

few people die from it. The SIR model equations are given by 

_S = b ¡ aIS; I _ = aIS¡ cI; R _ = cI; 

where b = 1 is the constant birth rate, a =0:001 is the interaction coe±cient 

between susceptibles and infectibles, and c = 0:1is the rate of increase of 

the removables. More generally, the model will have natural deaths in each 

category. The initial conditions are S(0) = 100, I(0) = 1, and R(0) = 0. 

Carry out the steps listed in the preamble. Note: To use the symbol I, ¯rst 

enter the command interface(imaginaryunit=j), which assigns j to stand 

for p ¡1, rather than I. Discuss problems with the SIR model.


2.1.3 Mike Creates a Higher-Order Fixed Point 

I never learn anything talking. 

I only learn things when I ask questions. 

Lou Holtz, American football coach 

Echoing Lou Holtz, Vectoria asks her mathematician boyfriend Mike, \Can 

you give me a simple physical example of a higher-order stationary point?" 

\Sure," Mike replies, \I can build up a phenomenological force law model 

for a nonlinear spring system that will display a higher-order ¯xed point. To 

do this, let's consider a light (weightless) spring suspended from one end with 

a unit mass attached to the other. Since you are into using computer algebra, 

I will do so as well. Let me load the following Maple library packages, that I 

am sure we will need, onto your computer. 

> restart: with(DEtools): with(plots): 

If the spring is stretched by only a small amount x from the equilibrium position, 

the force F required to de°ect the unit mass is given by Hooke's law F = kx 

with the spring constant k being positive. The restoring force is, of course, of 

the opposite sign. As you already know, the equation of motion is then just the 

SHO equation, which has periodic solutions. There is only a single stationary 

point, namely a vortex, located at the origin of the velocity{displacement phase 

plane. Physically, the stationary point corresponds to the situation that both 

the velocity and the force F (and therefore the acceleration) are equal to zero. 

By integrating F with respect to x, the potential energy V is given by the 

parabolic curve V = 1 

2 kx2 . The stationary point is at x =0,thebottomof 

the potential well. 

If we stretch the spring even more, the displacement can be su±ciently large 

that higher-order terms in a Taylor expansion of F (x) should be included, thus 

leading to a nonlinear equation of motion. If x is not too large, we need keep 

terms only up to third order in the Taylor expansion of F (x). I will write the 

polynomial force law in the form F = kx¡ gx 2 + hx 3 with k, g, andh all 

positive. 

> F:=k*x-g*x^2+h*x^3; #deflecting force 

F := kx¡ gx 2 + hx 3 

The associated potential energy V is again easily obtained by integrating F ." 

> V:=int(F,x); 

V := 1 

2 kx2 ¡ 1 

3 gx3 + 1 

4 hx4 

\Holdonamoment,Mike. Idon'tseewhyyouwenttothirdorderinx in 

the force law and why you took the quadratic term to be negative?" 

\Let me answer your last question ¯rst. If all three terms in F were positive, 

then F (x) would have only one real root at x = 0, which clearly still corresponds 

to a vortex. A higher-order stationary point is not possible in this case, no 

matter what the size of the coe±cients. Now let's make the quadratic term


negative. If the coe±cient g is su±ciently large, then as x is increased from 

zero, F will grow linearly at small x due to the kxterm, then begin to decrease 

at intermediate x due to the ¡gx 2 term, and again increase at larger x due 

to the hx 3 term. Since F is a cubic polynomial, there exists the possibility of 

F = 0 having three real x roots and therefore three stationary values. Since 

we can still look at very small vibrations around the origin where the Hooke's 

law contribution predominates, there will still be a vortex at the origin of the 

phase plane. So this still leaves two other possible nonzero stationary points 

corresponding to the nonzero x roots of F = 0. If you substitute numbers and 

play around with the cubic polynomial, the other two stationary points will be 

simple stationary points unless the coe±cients are such that the two nonzero 

roots coalesce into a single degenerate root. When coalescence takes place, the 

F curve will just be tangent to the x-axis at the location of the degenerate root. 

So we must impose the condition that dF(x0)=dx =0,aswellasF (x0) =0,in 

order to adjust the coe±cients to give a degenerate root x0. It is this degenerate 

situation that gives rise to a higher-order ¯xed point, as I will now show you. 

We would not have obtained a degenerate root and therefore a higher-order 

stationary point if I had not kept the cubic term in F . 

Let's calculate the derivative of the force F with respect to x, 

> derF:=diff(F,x); 

derF := k ¡ 2 gx+3hx 2 

and substitute x = x0 into both F =0andderF =0. 

> eq1:=subs(x=x[0],fF=0,derF=0g); 

eq1 := fk ¡ 2 gx0 +3hx 2 0 =0;kx0 ¡ gx 2 0 + hx3 0 =0g 

Solving the set of equations contained in eq1 for the coe±cients g and h, 

> sol:=solve(eq1,fg,hg); 

sol := 

yields g =2k=x0 and h = k=x 2 0 

½ 

g =2 k 

x0 

;h= k 

¾ 

. Assigning the solution and collecting the 

coe±cients of k in F yields the force law necessary for coalescence of the nonzero 

roots to occur. 

> assign(sol): F:=collect(F,k); 

μ 

2 x2 

F := x ¡ + 

x0 

x3 

x2 

k 

0 

In order to plot both the force and potential energy curves as a function of x, 

I will choose some speci¯c values for x0 and k. What do you suggest?" 

\Oh, I think that we will get a good idea of the physical behavior if we take 

x0 =1andk =1." 

\OK," Mike replies, \I will evaluate F and V with your suggested numbers, 

> F:=eval(F,fx[0]=1,k=1g); 

F := x ¡ 2 x 2 + x 3 

x 2 0


> V:=eval(V,fx[0]=1,k=1g); 

V := 1 

2 x2 ¡ 2 

3 x3 + 1 

4 x4 

and now plot the force and potential energy in the same ¯gure. To distinguish 

the two curves on the computer screen and in a black-and-white text rendition, 

I will use di®erent colors and line styles for the two curves. Using a list format 

to preserve order, the following plot command will generate a dashed red curve 

for F and a solid blue curve for V ." 

> plot([F,V],x=-0.25..1.5,color=[red,blue], 

linestyle=[DASH,SOLID],labels=["x","F,V"]); 

For the reader's bene¯t, the resulting picture observed by Mike and Vectoria 

on executing the plot command is reproduced in Figure 2.12. 

F,V 

0.2 

–0.2 

–0.4 

Figure 2.12: De°ecting force F , dashed line; potential energy V , solid line. 

Mike continues, \At the degenerate root location, x0 = 1, the (solid) potential 

energy curve is horizontal, corresponding to the (dashed) force curve 

touching zero. The equation of motion takes the form 

Äx = ¡F = ¡x +2x 2 ¡ x 3 ; (2.11) 

so that on setting the velocity _x equal to y, wecanidentifythefunctionsPand Q needed for stationary point analysis. 

> P:=-F; Q:=y; 

P := ¡x +2x2 ¡ x3 Q := y 

To evaluate a, b, c, andd, the following partial derivatives are calculated. 

> a:=diff(P,x): b:=diff(P,y): c:=diff(Q,x): d:=diff(Q,y): 

Solving the simultaneous equations P =0,Q =0, 

x 

1


> sol:=solve(fP,Qg,fx,yg); 

sol := fx =0;y=0g; fy =0;x=1g; fy =0;x=1g 

yields the expected root at the origin and the twofold degenerate root (x =1, 

y = 0). Since we already know that the former is a vortex, let's select the 

second (or third) solution and assign it for later use. 

> assign(sol[2]); 

Then, calculating p and q using the standard formulas, 

> p:=-(a+d); q:=a*d-b*c; 

p := ¡1 q := 0 

we obtain p = ¡1 andq =0. Sinceqis zero, the degenerate root is a higherorder 

stationary point as I predicted." 

\OK, I understand your construction of a higher-order ¯xed point. But what 

do the trajectories look like in the phase plane, and how does the displacement 

x(t) behave in this case?" 

\To answer your questions, I ¯rst have to unassign x and y from their 

stationary-point values, 

> unassign('x','y'); 

and then enter the associated ¯rst-order ODEs. 

> ODEs:=fdiff(x(t),t)=y(t),diff(y(t),t)=-x(t)+2*x(t)^2-x(t)^3g; 

ODEs := 

½ 

d 

d 

x(t) =y(t); 

dt dt y(t) =¡x(t)+2x(t)2 ¡ x (t) 3 

¾ 

In order to put the stationary points in the phase-plane diagram, let's create a 

plot for the two points, using size-20 red circles to represent their locations. 

> gr1:=plot([[0,0],[1,0]],style=point,symbol=circle, 

symbolsize=20,color=red): 

Employing the DEplot command, we can produce a phase-plane portrait for 

di®erent initial conditions." 

After some experimentation, Mike comes up with four di®erent initial conditions, 

which are included in the following DEplot command line. He chooses 

blue arrows with two barbs on the head (using arrows=MEDIUM) for the tangent 

¯eld and colors the trajectories red with the linecolor option. The time range 

is taken from t = 0 to 20 and the time step size equal to 0:05. 

> gr2:=DEplot(ODEs,[x(t),y(t)],t=0..20,x(t)=-1..2.5, 

y(t)=-1.5..1.5,[[x(0)=-0.4,y(0)=1],[x(0)=-0.35,y(0)=0], 

[x(0)=-0.3,y(0)=0],[x(0)=-0.2,y(0)=0]],stepsize=0.05, 

color=blue,linecolor=red,arrows=MEDIUM,dirgrid=[20,20]): 

Putting the two graphs gr1 and gr2 together with the display command 

> display(fgr1,gr2g,tickmarks=[2,3]); 

and controlling the tick marks on the coordinate axes results in the phase portrait 

shown in Figure 2.13. The four trajectories, corresponding to the four 

initial conditions, are clearly seen.


1 

–1 

Figure 2.13: Phase-plane diagram for the nonlinear spring system. 

\That's a nice plot, Mike. I can see the two stationary points and the 

change in shape of the trajectories from the inner one to the outer one. Close 

to the origin, the tangent ¯eld is characteristic of a vortex. The inner closed 

trajectory for the initial condition x(0) = ¡0:2 andy = 0 is a distorted circle 

cycling around this vortex point. As the loops grow larger, corresponding to 

the other initial conditions, the higher-order terms in the polynomial force law 

become more important and there is even more distortion of the closed loops. 

For x(0) = ¡0:35 and y(0) = 0, the third-largest loop encloses both stationary 

points. For the degenerate stationary point at x =1,y = 0, the tangent 

¯eld to the left of the point looks like that for a saddle point, while on the right 

it looks like that for a vortex. Does this type of higher-order ¯xed point have 

aname?" 

\This hybrid stationary point is, not surprisingly, referred to as a saddlevortex 

¯xed point. For other examples involving higher-order stationary points, 

di®erent hybrid combinations are possible, and are best studied on a case-bycase 

basis. 

As the trajectory moves further away from both stationary points, as in 

the case of the outer loop in Figure 2.13, generated by the initial condition 

x(0) = ¡0:4 andy(0) = 1, the positive cubic term in the force law predominates 

and periodic motion about both stationary points still occurs. 

You also asked about what x(t) lookslike. Thex(0) = ¡0:35, y = 0 

trajectory looks as if it could be interesting. To see x(t), we can use the DEplot 

2


command again, but this time with the scene=[t,x] option." 

> DEplot(ODEs,[x(t),y(t)],t=0..50,[[x(0)=-0.35,y(0)=0]],scene 

=[t,x],stepsize=.05,color=blue,linecolor=red,axes=normal); 

1.2 

1 

0.8 

x 

0.6 

0.4 

0.2 

–0.2 

–0.4 

10 20 30 40 50 

t 

Figure 2.14: Displacement of the nonlinear spring as a function of time. 

Again, we have reproduced what Mike and Vectoria see on the computer screen 

in Figure 2.14. 

Mike continues, \Two features displayed by the x(t) curveleapoutatme, 

Vectoria. The shape is not sinusoidal and it is highly asymmetric about the 

origin. The deviation away from sinusoidal behavior is a signal that nonlinear 

terms are present in the force law. The asymmetry arises speci¯cally from the 

quadratic term, which doesn't reverse sign as the spring system passes through 

the origin." 

\That was great, Mike. I have learned a lot from this example. It's Friday 

and I feel like wrapping the week up by going to that little Greek restaurant on 

West 4th for supper. How about you?" 

\With a glass or two of Tsantali to drink, that sounds good. Save the ¯le, 

shut down your computer, and let's be o®." 

PROBLEMS: 

Problem 2-8: Verhulst predator{prey equations 

The population densities of prey (variable x) and predators (variable y) are 

governed by the following nonlinear ODEs: 

_x = x ¡ Ax 2 ¡ Bxy; _y = y ¡ Cy 2 + Dxy; 

with the coe±cients A, B, C, andDall positive.


(a) What do the terms involving A and C represent physically? 

(b) Show that the nature of the stationary points depends on whether C>B, 

C = B, orCBand C


g(0) = 0:5 ands(0) = 1? Can the gnus and sung get along su±ciently well that 

both groups will survive, or is the interaction such that only one will survive? 

Although this competition problem is similar to that for the rabbits and 

foxes, it is complicated by the appearance of higher-order polynomial and transcendental 

terms on the right-hand sides of the evolution equations. This makes 

locating the stationary points more of a challenge than in the earlier examples. 

To answer the question, the society has called on the services of the preeminent 

scientist and conservationist Dr. Eiram Eiruc. Dr. Eiruc begins her 

computer analysis by loading the plots and DEtools packages, 

> restart: with(plots): with(DEtools): 

as well as the given coe±cient values. 

> a[g]:=2: a[s]:=1: b[g]:=0.03: b[s]:=0.01: c[g]:=0.1: c[s]:=0.1: 

From equations (2.12), the functions P and Q are identi¯ed and entered. 

> P:=a[g]*g-b[g]*g*s-c[g]*g^6*s^7; 

P := 2 g ¡ 0:03 gs¡ 0:1 g 6 s 7 

> Q:=-a[s]*s+b[s]*g*s+c[s]*g^5*exp(-s); 

Q := ¡s +0:01 gs+0:1 g 5 e (¡s) 

The stationary points can be obtained by setting P = Q =0. However,the 

simultaneous solution of these equations is nontrivial because of their complexity. 

Before carrying out a numerical search for the roots, Eiram uses the 

implicitplot command with boxed axes to graph the functions P =0,Q =0. 

The intersection points will then correspond to the stationary points. The 

ranges of g and s are determined by trial and error. 

> implicitplot(fQ=0,P=0g,g=-6..6,s=-10..10,grid=[60,60], 

numpoints=5000,tickmarks=[4,2],axes=box); 

s 

0 

–10 

–6 –4 –2 0 g 2 4 6 

Figure 2.15: Graphically solving P =0,Q = 0 to locate stationary points.


The result is shown in Figure 2.15. Since the population densities cannot be 

negative, the only stationary points of interest to Eiram are the one that appears 

to be at the origin and the one in the g>0, s>0quadrant. Eiram 

could, of course, obtain approximate values by clicking the mouse arrow on the 

intersection points in the computer screen plot. More accurate numbers are 

found by using the °oating point solve command. However, since the equations 

P =0andQ = 0 are quite nonlinear, the fsolve command will yield only one 

real root if no range is speci¯ed for the variables. When more than one root is 

present, a range must be given that includes the root of interest. 

For example, by taking g = ¡1 to1inthefsolve command, 

> sol[1]:=fsolve(fP,Qg,fg,sg,g=-1..1); 

sol 1 := fg =0:; s =0:g 

Eiram con¯rms that there is indeed a stationary point at the origin. Referring 

to Figure 2.15, she then chooses the range g =1to3, 

> sol[2]:=fsolve(fP,Qg,fg,sg,g=1..3); 

sol 2 := fg =1:902803669; s=0:9669140476g 

and ¯nds that the second stationary point is at g ¼ 1:90 gnus and s ¼ 0:97 

sung (per unit area). 

The relevant partial derivatives of P and Q for identifying the stationary 

points are calculated. 

> a:=diff(P,g): b:=diff(P,s): c:=diff(Q,g): d:=diff(Q,s): 

A do loop is formed that will enable Eiram to carry out the identi¯cation process 

for each of the two relevant stationary points. 


The \coordinates" of the ith stationary point are assigned. 

> assign(sol[i]): 

The quantities p = ¡(a + d), q = ad¡ bc,andr = p 2 ¡ 4 q are calculated for 

the ith stationary point. 

> p[i]:=-(a+d); q[i]:=a*d-b*c; r[i]:=p[i]^2-4*q[i]; 

Then g and s are unassigned, 

> unassign('g','s'); 

and the do loop ended, 

> end do; 

yielding the following results: 

p1 := ¡1: q1 := ¡2: r1 := 9: 

p2 := 11:78445045 q2 := 87:09248243 r2 := ¡209:4966573 

Having the Springer Intergalactic edition of this text, Eiram refers to the p-q 

diagram given in Figure 2.1. Noting that q1 < 0, she concludes that the stationary 

point at the origin is a saddle point. For the second nonzero stationary 

point, p2 > 0andq2 > 0, so this ¯xed point is either a stable focal point or a 

stable nodal point. But r2 < 0, so it is actually a stable focal point.


Eiram will con¯rm this analysis by making a phase-plane portrait. She forms 

the gnus and sung ODEs by inserting the g and s time-dependence in the P , Q 

functions and entering dg(t)=dt = P (g(t);s(t)) and ds(t)=dt = Q(g(t);s(t)). 

> gnus:=diff(g(t),t)=subs(fg=g(t),s=s(t)g,P); 

gnus := d 

dt g(t) =2g(t) ¡ 0:03 g(t) s(t) ¡ 0:1 g(t)6 s(t) 7 

> sung:=diff(s(t),t)=subs(fg=g(t),s=s(t)g,Q); 

sung := d 

dt s(t) =¡s(t)+0:01 g(t) s(t)+0:1g(t)5 e (¡s(t)) 

A functional operator F is formed to produce a phase-plane portrait for the 

gnus and sung, given that initially there were 0:5 gnus and 1 sung (per unit 

area), and looking at a time span of 10 years. The scene variables x and y, and 

the line color c, must be supplied as arguments. 

> F:=(x,y,c)->phaseportrait([gnus,sung],[g(t),s(t)],t=0..10, 

[[g(0)=0.5,s(0)=1]],scene=[x,y],g=0..3,s=0..2,stepsize 

=0.01,color=blue,linecolor=c,arrows=MEDIUM, 

dirgrid=[20,20]): 

Then, entering F(g,s,red) produces the phase-plane portrait shown in Figure 

2.16, the tangent arrows being colored blue and the trajectory red on the 

computer screen. 

> F(g,s,red); 

2 

s 

1 2 3 

g 

Figure 2.16: Phase-plane portrait for the gnus{sung interaction. 

As expected, the trajectory winds onto the stable focal point. The tangent 

arrows near the origin are indicative of a saddle point.


Next, Eiram will show the time evolution of the two animal populations. 

First she uses the textplot commandtocreatenamelabelstobeaddedtothe 

¯nal ¯gure. 

> tp:=textplot([[1.2,1.1,"sung"],[1.2,2,"gnus"]]): 

She then uses the display command and the functional operator F to produce 

Figure 2.17, showing the temporal evolution of the gnus and sung populations 

for a period of two years. On the computer screen the gnus' population density 

is colored red, the sung's population density blue. 

> display(fF(t,g,red),F(t,s,blue),tpg,labels=["t",""], 

tickmarks=[3,3],view=[0..2,0..3]); 

3 

2 

1 

0 

gnus 

sung 

1 t 2 

Figure 2.17: Temporal evolution of gnus and sung. 

So Eiram concludes that if all conditions remain the same, then the two groups 

will ultimately live in relative harmony with each other, since both gnus and 

sung survive. However, she notes that the gnus will gain the upper hand in 

terms of population density over their more backward relatives, even though 

initially the density of sung was twice that of the gnus. 

PROBLEMS: 

Problem 2-10: Create your own model 

Create your own complicated model of the gnus{sung interaction by modifying 

the last term in each of the equations. Follow the procedure in the text recipe 

and determine the fate of the two populations in your model. Feel free to 

experiment with parameter values and initial conditions.


2.1.5 A Plethora of Points 

The sure conviction that we could if we wanted to 

is the reason so many good minds are idle. 

G. C. Lichtenberg, German physicist, philosopher (1742{1799) 

We have asked Jennifer, the MIT mathematician, to provide us with a recipe 

that locates and identi¯es all the ¯xed points of the nonlinear ODE system 

_x = x (1 ¡ x 2 ¡ 6 y 2 ); _y = y (1 ¡ 3 x 2 ¡ 3 y 2 ); 

making use of conditional logic statements, and to support the identi¯cation 

with a tangent-¯eld plot containing all the stationary points. Here is her recipe. 

After loading the required library packages, 


Jennifer identi¯es the rhs of the _x and _y equations as the functions P and Q 

needed for the ¯xed-point analysis. She then enters P and Q. 

> P:=x*(1-x^2-6*y^2); Q:=y*(1-3*x^2-3*y^2); 

P := x (1 ¡ x2 ¡ 6 y2 ) Q := y (1 ¡ 3 x2 ¡ 3 y2 ) 

Jennifer introduces a functional operator f for di®erentiating a given function 

X with respect to an arbitrary variable v. 

> f:=(X,v)->diff(X,v): 

Making use of the functional operator f, the partial derivatives a = @P=@x, 

b = @P=@y, c = @Q=@x, andd = @Q=@y are calculated. 

> a:=f(P,x): b:=f(P,y): c:=f(Q,x): d:=f(Q,y): 

The partial derivatives must be evaluated at each ¯xed point. To locate these 

points, the equations P =0andQ = 0 are solved for x and y. Lists are used 

here, because the order of x and y in sol must be preserved for later use. 

> sol:=solve([P=0,Q=0],[x,y]); 

sol := [ [x =0;y=0]; [x =0;y=RootOf(¡1+3 Z 2 ; label = L1 )]; 

[x =1;y=0]; [x = ¡1; y=0]; 

[x =RootOf(5Z 2 ¡ 1); y=RootOf(¡2+15 Z 2 )] ] 

In the solution, RootOf( ) is a placeholder for the roots of the polynomial 

functiongivenineachincludedargument. Jenniferwantstoextractallthe 

roots explicitly. To this end, she ¯rst determines the number of operands (here 

the number of lists) in sol using the nops command. 

> N:=nops(sol); 

N := 5 

The number of operands is 5, which is easily con¯rmed by visual inspection of 

sol. The roots can be explicitly obtained by applying the allvalues command 

to the ith entry in sol, and using the seq command (running from i =1toN) 

to generate the sequence of roots. The results are again put into a list, so the 

number of operands in the new list of lists can be extracted.


> sol2:=[seq(allvalues(sol[i]),i=1..N)]; 

" 

" 

p # " 

p # 

3 

3 

sol2 := [x =0;y=0]; x =0;y= ; x =0;y= ¡ ; 

3 

3 

" p 

5 

[x =1;y=0]; [x = ¡1; y=0]; x = 

5 ;y= 

p # 

30 

; 

15 

" p p # " p 

5 30 

5 

x = ;y= ¡ ; x = ¡ 

5 15 

5 ;y= 

p # " p p ## 

30 

5 30 

; x = ¡ ;y= ¡ 

15 

5 15 

In the sol2 list of lists, there are clearly nine lists of x and y values, corresponding 

to the coordinates of nine stationary points. There is indeed a plethora 

of ¯xed points in this example. The number of operands (lists here) will be 

needed, so this number is extracted with the nops command. 

> N2:=nops(sol2); 

N2 := 9 

Functional operators are formed to numerically evaluate p = ¡(a + d) and 

q = ad¡ bc for the ith entry in sol2 . 

> p:=i->evalf(eval(-(a+d),sol2[i])): 

> q:=i->evalf(eval(a*d-b*c,sol2[i])): 

Next, Jennifer creates a do loop, running from i =1toN2 = 9, to identify the 

nature of each ¯xed point and to plot each point. 

> for i from 1 to N2 do 

The ith entry (list of x and y coordinates) of sol2 is entered, and then the x and 

y coordinates are extracted separately. For example, taking i =1,sol2[1,1] 

extracts the ¯rst entry (x = 0) in the ¯rst list of sol2 , while sol2[1,2] produces 

the second entry (y = 0). The right-hand sides are extracted to give the x and 

y coordinates of the ¯xed point for plotting purposes. 

> sol2[i]; X[i]:=rhs(sol2[i,1]); Y[i]:=rhs(sol2[i,2]); 

The values of p, q, andr = p2 ¡ 4 q are determined for the ith ¯xed point. Jennifer 

uses the concatenation operator || to attach numbers to these quantities. 

For example, the outputs of p||1, p||2, etc., would be of the form p1 , p2 ,etc. 

Concatenation is a useful alternative way of numbering quantities. 

> p||i:=p(i); q||i:=q(i); r||i:=simplify(p||i^2-4*q||i); 

A conditional statement is now entered that will classify the ith ¯xed point 

according to the region of the p-q diagram in which it is located. The general 

syntax for a conditional statement is 

if then 

elif then 

else 

end if 

where the elif (else if) and else phrases are optional.


> if q||i0 and p||i>0 and r||i0 and p||i>0 and r||i>=0 then s||i:=stablenodal; 

elif q||i>0 and p||i0 and p||i=0 then s||i:=vortexorfocal; 

else s||i:=higherorder; 

end if: 

The ith stationary point is plotted as a size-20 blue circle, 

> gr[i]:=pointplot([[X[i],Y[i]]],symbol=circle, 

symbolsize=20,color=blue); 

and the conditional statement ended, a colon being used to suppress the output. 

> end do: 

Using the sequence command, the x and y ¯xed-point coordinates and ¯xedpoint 

type are now displayed for all nine ¯xed points. For example, the ¯xed 

point at x =0,y = 0, is an unstable nodal point. 

> seq([X[i],Y[i],s||i],i=1..N2); 

" p # " p # 

3 

3 

[0; 0; unstablenodal]; 0; ; stablenodal ; 0; ¡ ; stablenodal ; 

3 3 

" p 

5 

[1; 0; stablenodal] ; [¡1; 0; stablenodal]; 

5 ; 

p # 

30 

; saddle ; 

15 

" p p # " p 

5 30 

5 

; ¡ ; saddle ; ¡ 

5 15 5 ; 

p # " p p # 

30 

5 30 

; saddle ; ¡ ; ¡ ; saddle 

15 5 15 

The two relevant ODEs are entered, with the time dependence of x and y being 

substituted into P and Q. 

> xeq:=diff(x(t),t)=subs(fx=x(t),y=y(t)g,P); 

xeq := d 

dt x (t) =x(t)(1¡ x (t)2 ¡ 6 y(t) 2 ) 

> yeq:=diff(y(t),t)=subs(fx=x(t),y=y(t)g,Q); 

yeq := d 

dt y(t) =y(t)(1¡ 3 x(t)2 ¡ 3 y(t) 2 ) 

The dfieldplot command is used to plot the tangent ¯eld for this ODE system, 

the resulting graph being assigned the name gr[N2+1]. 

> gr[N2+1]:=dfieldplot([xeq,yeq],[x(t),y(t)],t=0..10, 

x=-1.2..1.2,y=-1.2..1.2,dirgrid=[26,26],arrows=MEDIUM): 

The above graph assignment allows Jennifer to use the sequence command to 

plot all the stationary points and the tangent ¯eld in the same ¯gure. 

> display(fseq(gr[i],i=1..N2+1)g,tickmarks=[3,3]);


The resulting picture is shown in Figure 2.18. The plot con¯rms the identi¯cation 

of the ¯xed points, e.g., the unstable nodal point at the origin. 

1 

–1 1 

–1 

Figure 2.18: A plethora of ¯xed points. 

PROBLEMS: 

Problem 2-11: Lots of ¯xed points 

Consider the coupled ODE system 

_x = y (1 + x ¡ y 2 ); _y = x (1 + y ¡ x 2 ): 

Using conditional logic statements, locate and identify all the ¯xed points. Make 

a tangent-¯eld plot containing all these stationary points. 

Problem 2-12: Not so many ¯xed points 

Locate and identify all the ¯xed points of the ODE system 

_x =16x 2 +9y 2 ¡ 25; _y =16x 2 ¡ 16 y 2 ; 

making use of conditional logic statements. Support the identi¯cation with a 

tangent-¯eld plot containing all the stationary points. 

Problem 2-13: Squid and herring 

The major food source for squid is herring. If S and H are the numbers of 

squid and herring, respectively, per acre of seabed, the interaction between the 

two species can be modeled [Sco87] by the system (with time in years) 

_H = k1 H ¡ k2 H 2 ¡ k3 HS; S _ = ¡k4 S ¡ k5 S 2 + bk3 HS; 

with k1 =1:1, k2 =10 ¡5 , k3 =10 ¡3 , k4 =0:9, k5 =10 ¡4 ,andb =0:02. 

(a) Using conditional logic statements, locate and identify all the stationary 

points of the squid{herring system.


(b) Make a phase-plane portrait that shows all the stationary points and 

includes some representative trajectories. Discuss possible outcomes for 

di®erent ranges of initial populations. 

(c) Suppose that every last squid were removed from the area occupied by 

the herring and from all surrounding areas. Would the herring population 

increase inde¯nitely without bound or would there be an upper limit on 

the number of herring per unit area? If you believe the latter would occur, 

what is that number? 

(d) If the squid-free situation just described had persisted for many years, 

how many squid would there be two years later if a pair of fertile squid 

were introduced into the area? 

2.2 Three-Dimensional Autonomous Systems 

The analysis of the general autonomous 3-dimensionsal system, 

_X = P (X;Y; Z); Y _ = Q(X;Y; Z); Z _ = R(X;Y; Z); (2.13) 

where P , Q, andRmay be nonlinear functions of the dependent variables X, Y , 

and Z, can be tackled in a manner similar to the 2-dimensional case. However, 

the analytic identi¯cation of the ¯xed points is nontrivial, and we refer the 

interested reader to the texts by Jackson [Jac90] and Hayashi [Hay64]. 

We shall be content to look at one speci¯c example, locating the ¯xed points, 

establishing their stability, and using a graphical approach to identify their 

nature. What better example is there than Lorenz's famous chaotic \butter°y"? 

2.2.1 Lorenz's Butter°y 

The butter°y's attractiveness derives not only from colors 

and symmetry: deeper motives contribute to it. 

Primo Levi, Italian chemist, author, Other Peoples Trades,\Butter°ies," 1985 

In a famous paper, Edward Lorenz [Lor63] discussed the possibility of very-longrange 

weather forecasting. Starting with the nonlinear Navier{Stokes PDEs of 

°uid mechanics, Lorenz attempted to model thermally driven convection in the 

earth's atmosphere. The atmosphere was treated as a °at °uid layer that is 

heated from below by the surface of the earth and cooled from above by heat 

radiation into outer space. Lorenz managed to reduce the original set of PDEs 

to the following set of three nonlinear ODEs: 

_x = ¾ (y ¡ x); _y = rx¡ y ¡ xz; _z = xy¡ bz; (2.14) 

where x is proportional to the convective velocity, y to the temperature di®erence 

between ascending and descending °ows, and z to the mean convective heat


°ow. The positive coe±cients ¾ and r are the Prandtl and reduced Rayleigh 

numbers, respectively, and b>0 is related to the wave number. 

The Lorenz equations (2.14) must be solved numerically. On doing so, 

Lorenz discovered that a very small change in the initial condition could lead 

to dramatically di®erent long-term behavior of the numerical solutions. This 

could have implications for very-long-range weather forecasting, because present 

conditions are never known exactly due to the inevitable inaccuracy and incompleteness 

of weather observations. One might argue that this conclusion was 

model dependent, particularly because the original PDEs were so severely simpli¯ed, 

but in fact Lorenz found that this sensitivity to initial conditions was a 

general feature of nonlinear systems. 

In addition to the sensitivity feature, Lorenz's ODE system is well known 

because its trajectory in x-y-z phase space is attracted to a localized region 

where it traces out a never-repeating (chaotic) path whose shape somewhat 

resembles the two wings of a butter°y. This is another example of a strange 

attractor, the attractor di®ering in appearance from the RÄossler strange attractor 

seen in Chapter 1. Lorenz's butter°y attractor is illustrated in the following 

recipe, along with a stability analysis of the ¯xed points. 

The DEtools and LinearAlgebra packages are loaded. The former is needed 

because we shall be using the DEplot and DEplot3d commands to plot x(t) 

and the butter°y attractor, respectively. The LinearAlgebra package contains 

commands to construct and manipulate matrices and vectors and solve linear algebra 

problems. Various commands (e.g., GenerateMatrix and Eigenvalues) 

in this package will be used for the stability analysis. 

> restart: with(DEtools): with(LinearAlgebra): 

The right-hand sides of the three Lorenz equations are entered and assigned the 

names P , Q, andR, to be consistent with equations (2.13). 

> P:=sigma*(y-x); Q:=r*x-y-x*z; R:=x*y-b*z; 

P := ¾ (y ¡ x) Q := rx¡ y ¡ xz R:= xy¡ bz 

To produce Lorenz's butter°y, we take ¾ =10,b =8=3, and r =28. 

> sigma:=10: b:=8/3: r:=28: 

To locate the ¯xed points, the equations P =0,Q =0,andR =0aresolved 

for x, y, andz. 

> sol:=solve([P,Q,R],[x,y,z]); 

sol := [ [x =0;y=0;z=0]; [ x =6RootOf(¡2+ Z 2 ; label = L3 ); 

y =6RootOf(¡2+ Z 2 ; label = L3 );z =27]] 

The number N of operands in sol is determined. 

> N:=nops(sol); 

N := 2 

The allvalues command is used to remove the RootOf that appeared in sol. 

> sol2:=[seq(allvalues(sol[i]),i=1..N)];


sol2 := [ [x =0;y=0;z=0]; [ x =6 p 2;y=6 p 2;z =27]; 

[ x = ¡6 p 2;y= ¡6 p 2;z =27]] 

Thenumberof¯xedpointsisequaltothenumberN2 of operands in sol2 . 

> N2:=nops(sol2); 

N2 := 3 

So the Lorenz system has three ¯xed points. The stability of these ¯xed points 

will now be determined by extending the procedure used in the phase-plane 

analysis. A functional operator f is introduced to substitute the coordinates 

x + u, y + v, z + w of an ordinary point, near a ¯xed point (x; y; z), into a 

dependent variable V ,whereV will be taken to be P , Q, andR. 

> f:=V->simplify(subs(fx=x+u,y=y+v,z=z+wg,V)): 

Assuming that u, v, andw are su±ciently small, only linear terms in these 

variables will be retained in P , Q, andR. Since P is already linear, entering 

f(P) produces the linear algebraic form shown in eq1 . However, on entering 

f(Q) the nonlinear term uw will be present. This term can be removed with 

the remove command, which is done in eq2 . Similarly, the nonlinear term uv 

is removed from f(R) in eq3 . 

> eq1:=f(P); eq2:=remove(has,f(Q),u*w); 

eq3:=remove(has,f(R),u*v): 

eq1 := 10 y +10v ¡ 10 x ¡ 10 u 

eq2 := 28 x +28u ¡ y ¡ v ¡ xz¡ xw¡ uz 

8 z 8 w 

eq3 := xy+ xv+ uy¡ ¡ 

3 3 

Let's take i = 1, so that assigning the ith operand in sol2 corresponds to taking 

the ¯xed point coordinates to be x =0,y =0,z = 0. The stability of the other 

two ¯xed points may be examined by choosing i =2ori =3. 

> i:=1: assign(sol2[i]): #choose i 

For i =1,eq1 , eq2 ,andeq3 reduce to the following linear forms: 

> eq1:=eq1; eq2:=eq2; eq3:=eq3; 

8 w 

eq1 := 10 v ¡ 10 u eq2 := 28 u ¡ v eq3 := ¡ 

3 

So, the linearized ODEs at an ordinary point near the ¯xed point at the origin 

are 

du 

dv 

dw 8 w 

=10v ¡ 10 u; =28u ¡ v; = ¡ 

dt dt dt 3 : 

The stability is determined by assuming that u; v; w ' e¸t . For the ODEs 

to have a nontrivial solution, a determinantal equation must be satis¯ed that 

when expanded produces a cubic equation in ¸ with three roots. To have a 

stable ¯xed point, the real part of ¸ must be negative for all three roots. If any 

root has a real part that is positive, the solution will grow with time, leading 

to an unstable situation.


Let's now carry out this analysis using a matrix approach. Equating eq1 to 

¸u, eq2 to ¸v,andeq3 to ¸w,theGenerateMatrix command 

> (A,B):=GenerateMatrix([eq1=lambda*u,eq2=lambda*v, 

eq3=lambda*w],[u,v,w]); 

2 

3 

¡10 ¡ ¸ 10 0 2 3 

0 

6 

A; B := 

28 ¡1 ¡ ¸ 0 7 

4 

¡8 

5 ; 4 0 5 

0 0 ¡ ¸ 0 

3 

yields the relevant 3 £ 3matrixAandzero column vector B such that 

2 

u 

3 2 

0 

3 

A 4 v 5 = B ´ 4 0 5 : 

w 

0 

The determinant of A is computed, yielding a cubic polynomial in ¸. 

> det:=Determinant(A); 

det := 720 + 722 41 

¸ ¡ 

3 3 ¸2 ¡ ¸3 The determinant is set equal to zero, solved for ¸, and the answer put in °oatingpoint 

form. 

> lambdas:=evalf(solve(det=0,lambda)); 

lambdas := ¡2:666666667; 11:82772345; ¡22:82772345 

Three real roots are obtained, with two of them being negative, but the third 

one positive. Thus, the ¯xed point at the origin is unstable. It should be noted 

that the same result can be obtained by setting ¸ =0inA and obtaining the 

eigenvalues by using the Eigenvalues command. In this case, the ¸ values are 

expressed as a column matrix or vector. 

> eiv:=evalf(Eigenvalues(eval(A,lambda=0))); 

2 

¡2:666666667 

3 

eiv := 4 11:82772345 

¡22:82772345 

5 

To see how the trajectory evolves from the vicinity of the ¯xed point, the 

variables x, y, z are unassigned from their previous values, 

> unassign('x','y','z'): 

and an initial condition near the ¯xed point (the origin for i = 1) chosen. 

> ic:=[x(0)=rhs(sol2[i,1])+0.1,y(0)=rhs(sol2[i,2])+0.1, 

z(0)=rhs(sol2[i,3])+0.1]: 

ic := [x(0) = 0:1;y(0) = 0:1;z(0) = 0:1] 

The system of Lorenz ODEs is entered by introducing time dependence into 

the x, y, z variables, 

> var:=(x=x(t),y=y(t),z=z(t)):


and substituting these variables into P , Q, andR, whichareequatedtodx=dt, 

dy=dt, anddz=dt, respectively. 

> sys:=diff(x(t),t)=subs(var,P),diff(y(t),t)=subs(var,Q), 

diff(z(t),t)=subs(var,R); 

sys := d 

d 

x (t) =10y(t) ¡ 10 x(t); y(t) =28x (t) ¡ y(t) ¡ x (t) z (t); 

dt dt 

d 

8 

z(t) =x (t) y(t) ¡ 

dt 3 z(t) 

The DEplot command is used with the scene option scene=[t,x] to plot x 

versus t for the system of ODEs (subject to the initial condition), the line color 

being allowed to change with time as the time increases. 

> DEplot([sys],[x(t),y(t),z(t)],t=0..60,[ic],scene=[t,x], 

stepsize=0.01,linecolor=t,thickness=1); 

The black-and-white version of the plot is shown in Figure 2.19. 

20 

x 

10 

0 

–10 

10 20 30 40 50 60 

t 

Figure 2.19: x versus t for an initial condition near the origin. 

There is no apparent repeat pattern as x increases with t, themotionbeing 

chaotic. Longer times can be taken to con¯rm this. Sensitivity to the initial 

condition can be checked by changing the initial condition slightly and observing 

that the x versus t curve changes dramatically. Although the ODEs are 

deterministic, one's intuition is inadequate to predict what the value of x will 

be at large times if the initial condition is changed slightly. This is in contrast 

to a linear ODE system, where a slight change in initial condition will produce 

only a small change in the solution. 

The DEplot3d command is now used to plot the trajectory in the x-y-z 

space, the line color again being allowed to change with t.


> DEplot3d([sys],[x(t),y(t),z(t)],t=0..50,[ic],scene=[x,y,z], 

stepsize=0.01,linecolor=t,orientation=[-23,68]): 

40 

30 

z 

20 

10 

–10 

x 

0 10 

–20 –10 0 10 20 

y 

Figure 2.20: Lorenz's butter°y. 

Qualitatively, the trajectory leaves the origin in a ¯xed direction, implying that 

the ¯xed point is an unstable nodal point. The trajectory then evolves into a 

localized region of phase space, producing a nonrepeating path that ¯lls in the 

region so that the pattern resembles (to an active imagination, especially when 

seen in color on the computer screen) the wings of a butter°y. 

By choosing i = 2 and 3, you can use the recipe to determine the nature of 

the other two ¯xed points and observe the strange attractor that results. 

PROBLEMS: 

Problem 2-14: Other ¯xed points 

Use the text recipe to investigate the other two ¯xed points for the Lorenz 

system. Identify their probable nature and generate x versus t as well as the 

phase-space trajectory for an initial condition near each ¯xed point. 

Problem 2-15: Sensitivity 

Use the text recipe to explore the sensitivity to initial conditions for the Lorenz 

system. 

Problem 2-16: The Oregonator 

Consider the Oregonator system 

² _x = x + y ¡ qx 2 ¡ xy; _y = ¡y +2hz¡ xy; p _z = x ¡ z; 

with ² =0:03, p =2,q =0:006, h =0:75. Show analytically that the origin is 

an unstable ¯xed point.


Problem 2-17: The RÄossler system 

Consider the RÄossler system 

_x = ¡(y + z); _y = x + ay; _z = b + z (x ¡ c); 

with a, b, c>0. Analytically show that there are no stationary points for 

c p 4 ab. Find analytic expressions for 

the latter points. Linearizing the system in the vicinity of these ¯xed points, 

¯nd the cubic equation for the roots ¸. Taking a = b =0:2 and allowing c to 

take on the values c =2:4, 3:5, 4:0, 5:0, 8:0, solve the cubic equation for the 

roots and determine the stability in each case. 

Taking x(0) = 0:1, y(0) = 0:1, z(0) = 0, show by plotting the trajectories in 

phase space that each c value leads to a qualitatively di®erent behavior of the 

solution. Identify the behavior by plotting y(t). 

Problem 2-18: Multiple stationary points 

For the following 3-dimensional system, 

_x =(1¡ z)[(4¡ z 2 )(x 2 + y 2 ¡ 2 x + y)+4(2x ¡ y) ¡ 4]; 

_y =(1¡ z)[(4¡ z 2 )(xy¡ x ¡ zy)+4(x + zy) ¡ 2 z ]; 

_z = z 2 (4 ¡ z 2 )(x 2 + y 2 ); 

locate all of the stationary points and determine the stability of each point. 

Explore the nature of the ¯xed points and the behavior of this system by making 

suitable 3-dimensional x-y-z plots. 

2.3 Numerical Solution of ODEs 

In the opening chapter as well as this one, we have encountered nonlinear ODE 

models for which analytic solutions simply do not exist, in which case use must 

be made of Maple command structures that solve the given ODE(s) numerically. 

In this section, we would like to discuss brie°y the basis for the Runge{Kutta{ 

Fehlberg 45 (RKF45) method that Maple employs as its default scheme in the 

dsolve(ODE,numeric) command for numerically solving ODEs. 

All such numerical schemes make use of the idea of replacing the ODE with 

a di®erence equation in which each derivative is replaced with a ¯nite di®erence 

approximation involving a ¯nite step size. The step size may either be ¯xed or, 

as in the RKF45 method, variable, taking bigger steps in °atter regions and 

smaller steps in steeper regions. The di®erence equation can be iterated using 

adoloopprocedure. 

The common ¯nite di®erence approximations for ¯rst and second derivatives 

are introduced in the ¯rst recipe. Our intention in this and subsequent recipes 

is only to give you the \°avor" of the subject of numerical analysis, referring 

you to texts such as Burden and Faires [BF89] or Numerical Recipes [PFTV89] 

for more complete treatments.

2.3. NUMERICAL SOLUTION OF ODES 89 

2.3.1 Finite Di®erence Approximations 

A tool knows exactly how it is meant to be handled, while the user of 

the tool can only have an approximate idea. 

Milan Kundera, Czech author, critic (1929{) 

Consider a general function y(x) as schematically depicted by the solid curved 

line in Figure 2.21. The independent variable has been taken to be x, butit 

could just as well be the time t. It is desired to ¯nd, say, the ¯rst and second 

Q 

x–h 

h 

y(x–h) 

P 

y(x) 

h 

R 

y(x+h) 

x x+h 

Figure 2.21: Obtaining ¯nite di®erence approximations to derivatives. 

derivatives of y at an arbitrary point P on the curve located at the horizontal 

position x. Assuming that h is small, two neighboring points R and Q located 

at x + h and x ¡ h, respectively, are considered. At point R, the vertical height 

is y(x + h) andatQ it is y(x ¡ h). 

Since h is assumed to be small, y(x + h) can be Taylor expanded in powers 

of h about h = 0, neglecting terms of, say, order h 4 (i.e., O(h 4 )) and higher. 

> restart: 

> eq1:=y(x+h)=taylor(y(x+h),h=0,4); 

eq1 := y(x + h) = 

y(x)+D(y)(x) h + 1 

2 (D(2) )(y)(x) h2 + 1 

6 (D(3) )(y)(x) h3 +O(h4 ) 

The O(h4 )termineq1 can be removed with the convert(polynom) command. 

> eq1b:=convert(eq1,polynom); 

eq1b := y(x + h) =y(x)+D(y)(x) h + 1 

2 (D(2) )(y)(x) h2 + 1 

6 (D(3) )(y)(x) h3 Similarly y(x ¡ h) is Taylor expanded and the fourth-order term removed.


> eq2:=y(x-h)=convert(taylor(y(x-h),h=0,4),polynom); 

eq2 := y(x ¡ h) =y(x) ¡ D(y)(x) h + 1 

2 (D(2) )(y)(x) h2 ¡ 1 

6 (D(3) )(y)(x) h3 On subtracting the second expansion, eq2 , from the ¯rst, eq1b, 

> eq3:=eq1b-eq2; 

eq3 := y(x + h) ¡ y(x ¡ h) =2D(y)(x) h + 1 

3 (D(3) )(y)(x) h 3 

the h 2 terms cancel, leaving terms involving h and h 3 on the right-hand side of 

eq3 . If the latter term is now dropped by substituting h 3 =0intoeq3 , 

> eq3b:=subs(h^3=0,eq3); 

eq3b := y(x + h) ¡ y(x ¡ h) =2D(y)(x) h 

the relative error in eq3b will be O(h3 )=O(h) =O(h2 ). Solving eq3b for D(y)(x) 

yields the central di®erence approximation (CDA) y 0 CDA to the ¯rst derivative. 

To use the symbol y 0 in the assignment, y 0 must be enclosed in left quotes. 

> `y 0 `[CDA]:=solve(eq3b,D(y)(x)); 

y 0 1 y(x + h) ¡ y(x ¡ h) 

CDA := 

2 h 

The quantity y 0 CDA is just the slope of the chord QR that approaches the slope 

of the tangent to the y(x) curveatPas h ! 0. The reader will recognize 

this limiting result as the usual de¯nition for the ¯rst derivative. The ¯nite 

di®erence approximation consists in taking h to be small, but ¯nite. 

Now, the above CDA is not the only way of representing y 0 .Theforward- 

to the ¯rst derivative results on dropping the 

di®erence approximation y 0 FDA 

h2 and h3 terms in eq1b and solving for D(y)(x). 

> eq4:=subs(fh^2=0,h^3=0g,eq1b); 

eq4 := y(x + h) =y(x)+D(y)(x) h 

> `y 0 `[FDA]:=solve(eq4,D(y)(x)); 

y 0 FDA 

The backward-di®erence approximation y 0 BDA 

> eq5:=subs(fh^2=0,h^3=0g,eq2); 

y(x + h) ¡ y(x) 

:= 

h 

is similarly obtained from eq2 . 

eq5 := y(x ¡ h) =y(x) ¡ D(y)(x) h 

> `y 0 `[BDA]:=solve(eq5,D(y)(x)); 

y 0 y(x ¡ h) ¡ y(x) 

BDA := ¡ 

h 

The forward- and backward-di®erence approximations, both of which have errors 

of O(h), correspond to approximating y 0 at P by the slopes of the chords 

PRand QP, respectively. For a given (small) value of h, the FDA and BDA are 

not as accurate as the CDA. Despite this, the FDA is more commonly used to 

represent ¯rst time derivatives in solving initial value problems than the CDA, 

because, as will be seen, it allows one to explicitly advance forward in time from 

t = 0. Numerical ODE-solving procedures based on the FDA are called explicit


schemes. If the step size is too large, explicit ¯xed-step numerical schemes tend 

to become numerically unstable, so variable-step schemes or implicit schemes 

are also commonly used. Implicit schemes are based on the BDA. 

How does one represent the second derivative? As with the ¯rst derivative 

thereismorethanonewaytorepresenty 00 by a ¯nite di®erence approximation. 

However, the most commonly used form is obtained by adding eq1b and eq2 

and solving for the second derivative. 

> `y 00 `:=solve(eq1b+eq2,(D@D)(y)(x)); 

y 00 := 

y(x + h)+y(x ¡ h) ¡ 2 y(x) 

h 2 

This central di®erence approximation, which has an error O(h 2 ), is often used in 

numerically solving di®usion and wave equation models, as shall be illustrated 

in Chapter 6. A more accurate approximation to y 00 is left as a problem. 

PROBLEMS: 

Problem 2-19: Alternative second-derivative approximation 

By also Taylor expanding y(x +2h) andy(x ¡ 2 h) show that 

y 00 [¡y(x+2h)+16y(x+h) ¡ 30 y(x)+16y(x ¡ h) ¡ y(x ¡ 2 h)] 

(x)= 

(12 h2 +O(h 

) 

4 ) 

is an alternative ¯nite di®erence approximation to the second derivative. 

Problem 2-20: Fourth-derivative approximation 

Show that 

y 0000 [y(x +2h) ¡ 4 y(x + h)+6y(x) ¡ 4 y(x ¡ h)+y(x ¡ 2 h)] 

(x) = 

h4 +O(h 2 ) 

is a ¯nite di®erence approximation to the fourth derivative. 

2.3.2 Rabbits and Foxes: The Sequel 

A journey of a thousand miles must begin with a single step. 

Lao-tzu, Chinese philosopher (sixth century BC) 

Now that we have some idea of how to approximate ¯rst and second derivatives 

by ¯nite di®erences, numerical ODE-solving algorithms can be created of 

varying accuracy and CPU (central processing unit) time. The ultimate goal, 

of course, is to produce a highly accurate numerical scheme for a given ODE 

that takes as little CPU time as possible. Generally, more accurate schemes 

allow larger step sizes to be used but involve more function evaluations on each 

step. Larger step sizes reduce the CPU time, while more function evaluations 

increase the time. Unfortunately, the step size can be increased only so far before 

the solution becomes quite inaccurate or even displays wild oscillations, an 

indication of numerical instability. How the best compromise between accuracy 

and time may be reached is the subject matter of numerical analysis courses


and beyond the scope of this text. Here, we shall be content to present a few 

examples that illustrate some of the more important ideas. 

The ¯rst recipe involves revisiting those natural foes, the sly foxes and pesky 

jackrabbits of Rainbow County. In this sequel, we will not employ the phaseplane 

portrait approach but instead replace each time derivative with a forwarddi®erence 

approximation and convert the ODEs to ¯nite di®erence equations. 

Recall that the predator{prey equations were of the general structure 

_r = ar¡ brf ´ R(r; f ); f _ = ¡cf + drf ´ F (r; f); (2.15) 

where r(t) andf(t) aretherabbitandfoxnumbers(perunitarea)attimet. 

For the coe±cient values we shall take a =2,b =0:01, c =1,andd =0:01, 

and r(0) = 300 rabbits and f(0) = 150 foxes as initial conditions. 

To solve this set of ¯rst-order nonlinear ODEs, we can proceed as follows. 

If h is the time step size, then the time on step (n + 1) is related to the time 

on step n by tn+1 = tn + h. Introducing the notation rn ´ r(tn), fn ´ f(tn), 

rn+1 ´ r(tn+1), and fn+1 ´ f(tn+1), the FDA to the two time derivatives is 

_r = (rn+1 ¡ rn) 

and 

h 

_ f = (fn+1 ¡ fn) 

: 

h 

The overall accuracy of the numerical scheme then depends on how R(r; f )and 

F (r; f ) are approximated. Historically, the oldest and simplest approximation 

is to evaluate these functions on the nth time step, reducing the ODEs to 

rn+1 = rn + hR(rn;fn); fn+1 = fn + hF(rn;fn): (2.16) 

This ¯nite di®erence approximation, which allows one to advance forward a 

single step at a time, is referred to as the forward Euler algorithm. 

The step size h must be taken to be small to obtain an accurate approximation 

to the solution of the original ODEs. How small? A common approach 

to answering this question is to cut the step size in two and see what e®ect it 

has on the answer. This procedure can be repeated until a su±ciently accurate 

answer for your purposes has been attained. Some words of caution should be 

o®ered, however. One cannot keep cutting the step size in two inde¯nitely for 

a ¯xed number of digits or one will encounter round-o® error, theanswerin 

fact becoming less accurate. Further, to execute the worksheet out to the same 

total time will take more and more CPU time as the step size is decreased. 

Let's now solve equations (2.16), ¯rst loading the plots package and entering 

the coe±cient values. 


> a:=2: b:=0.01: c:=1: d:=0.01: 

The initial time, initial rabbit number, and initial fox number are speci¯ed and 

thetimestepsizetakentobeh =0:02. We take N = 1000 steps, so the total 

elapsed time for the numerical run will be 0:02 £ 1000 = 20 time units. 

> t[0]:=0: r[0]:=300: f[0]:=150: h:=0.02: N:=1000: 

Although the Maple system normally uses 10 digits as its default, the number of 

digits is speci¯ed in case it is desired to increase this number, e.g., to lessen the


round-o® error mentioned earlier. To compare the speed of di®erent numerical 

schemes or algorithms, the computer CPU time at which the execution of the 

algorithm begins is recorded using the time command. The CPU time at which 

the execution is ¯nished will also be recorded, the CPU time for the algorithm 

then being the di®erence. The CPU time will depend on the computer being 

used, the times quoted here being for a 3-GHz PC. 

> Digits:=10: begin:=time(): 

Functional operators are formed to calculate R and F for speci¯ed x and y. 

> R:=(x,y)->a*x-b*x*y: F:=(x,y)->-c*y+d*x*y: 

A do loop is used to iterate the di®erence equations (2.16) and the time. 

> for n from 0 to N do 

> r[n+1]:=r[n]+h*R(r[n],f[n]); 

> f[n+1]:=f[n]+h*F(r[n],f[n]); 

> t[n+1]:=t[n]+h; 

A 3-dimensional plotting point is formed for the nth time step. 

> pnt[n]:=[t[n],r[n],f[n]]; 

> end do: 

On ending the loop, the temporal evolution of the solution will be observed by 

creating a three-dimensional plot using a sequence of the plotting points from 

n =0ton = N and the pointplot3d command. 

> pointplot3d([seq(pnt[n],n=0..N)],symbol=cross,color=red, 

axes=normal,labels=["t","r","f"],tickmarks=[2,3,3]); 

20 

t 

600 

f 

400 

200 

0 0 

200 

r 

400 

Figure 2.22: Forward Euler solution of the rabbits{foxes equations.


The resulting spiral trajectory, corresponding to a cyclic variation in population 

numbers, is shown in Figure 2.22. The elapsed CPU time for the run, 

> CPUtime:=(time()-begin)*seconds; 

CPUtime := 0:130 seconds 

was only a fraction of a second. 

Instead of this \¯rst principles approach" to numerically solving the rabbits{ 

foxes equations, one can alternatively \dial up" the forward Euler approximation 

as follows. The dependent variables r and f are unassigned, 

> unassign('r','f'): begin2:=time(): 

and the ODE system is entered. 

> sys:=diff(r(t),t)=R(r(t),f(t)),diff(f(t),t)=F(r(t),f(t)): 

The initial population numbers are entered as the initial condition. 

> ic:=r(0)=300,f(0)=150: 

The numerical ODE solve command is used with the \classical" forward Euler 

method included as an option. The step size is 0:02, the same as before. 

> sol:=dsolve(fsys,icg,fr(t),f(t)g,type=numeric, 

method=classical[foreuler],stepsize=0.02): 

By using the odeplot command and specifying that 1000 time steps are to be 

taken, exactly the same plot (not displayed here) as in Figure 2.22 is produced. 

> odeplot(sol,[t,r(t),f(t)],0..20,numpoints=1000,style=point, 

symbol=cross,axes=normal,labels=["t","r","f"], 

tickmarks=[2,3,3]); 

If the three-dimensional picture produced by either code is rotated to show the 

r{f plane, the phase-plane trajectory is as shown in Figure 2.23. 

600 

400 

f 

200 

0 200 400 

r 

Figure 2.23: Phase-plane trajectory for the rabbits{foxes system. 

Although it appears that the trajectory is an outward-growing spiral, this conclusion 

would not be correct. The trajectory really should be a vortex, i.e., a


closed loop. The incorrect behavior is due to cumulative error arising from use 

of the Euler approximation (which has low accuracy) and the ¯nite step size. 

Keeping in mind the danger of round-o® error for a ¯xed number of digits, the 

growth of the spiral can in principle be reduced by reducing h. But in practice, 

the CPU time goes up so rapidly that the Euler method is not used for serious 

numerical calculations. In the next recipe, the classical fourth-order Runge{ 

Kutta method is introduced, a much more accurate numerical scheme than the 

Euler method for a given step size. 

Finally, the CPU time for the forward Euler dial-up method is determined, 

> CPUtime2:=(time()-begin2)*seconds; 

CPUtime2 := 0:330 seconds 

and is actually slightly longer than for the ¯rst-principles calculation. 

PROBLEMS: 

Problem 2-21: Van der Pol equation 

By setting _x = y, the second-order Van der Pol (VdP) equation 

Äx ¡ ² (1 ¡ x 2 )_x + x =0 

may be written as a coupled system of two ¯rst-order ODEs. Choosing ² =5 

and x(0) = _x(0) = 0:1, solve the VdP system for h =0:01 and t =0to15using 

the ¯rst principles Euler method. Rotate your plot to produce a phase-plane 

portrait solution and compare the result with what would be obtained using 

the dial-up Euler option of the dsolve command. 

Problem 2-22: White dwarf equation 

In his theory of white dwarf stars, Chandrasekhar [Cha39] introduced the nonlinear 

equation 

x (d 2 y=dx 2 )+2(dy=dx)+x (y 2 ¡ C) 3=2 =0; 

with the boundary conditions y(0) = 1 and y 0 (0) = 0. Write the second-order 

equation as two ¯rst-order equations and solve the system using the ¯rst principle's 

Euler method. Take h =0:01 and 10-digit accuracy, and numerically 

compute y(x) over the range 0 · x · 4withC =0:1 and plot the result. (Hint: 

Start at x =0:01 to avoid any problem at the origin.) 

Problem 2-23: Baleen whales 

May [May80] has discussed the solution of the following normalized equation 

describing the population of sexually mature adult baleen whales: 

_x(t) =¡ax(t)+bx(t ¡ T )(1 ¡ (x(t ¡ T )) N ): 

Here x(t) is the normalized population number at time t, a and b are the 

mortality and reproduction coe±cients, T is the time lag necessary to achieve 

sexual maturity, and N is a positive parameter. If the term 1 ¡ (x(t ¡ T )) N is 

negative, then this term is to be set equal to zero. Taking a =1,b =2,T =2, 

step size h =0:01, and 4000 time steps, use the ¯rst principles Euler method to 

solve numerically for x(t ¡ T )versusx(t) andforx(t) for(a)N =3:0, and (b) 

N =3:5. Plot your results. For (a) you should observe a period-one solution,


and for (b) a period-two solution. Discuss how this interpretation can be made 

from your plots. 

Problem 2-24: Arti¯cial example 

Consider the nonlinear equation 

dy=dx = xy(y ¡ 2); 

with y(0) = 1. Taking h =0:02 and 10-digit accuracy, solve for y(x) outto 

x = 3 using the dial-up Euler's method and plot the result. 

Problem 2-25: Modi¯ed Euler algorithm 

If the rabbits{foxes system is written for brevity as _r = R(r; f )and_ f = F (r; f ), 

the modi¯ed Euler algorithm for solving the equations is 

rn+1 = rn + h 

2 (R1[n]+R2[n]); fn+1 = fn + h 

2 (F1[n]+F2[n]); 

where 

tn+1 = tn + h; R1[n] ´ R(rn;fn); F1[n] ´ F (rn;fn); 

R2[n] ´ R(rn + hR1[n];fn + hF1[n]); F2[n] ´ F (rn + hR1[n];fn + hF1[n]): 

Taking the same parameter values as in the text recipe, but N twice as large, 

solve the rabbits{foxes equations using the modi¯ed Euler algorithm given 

above. Rotate your plot to show the r{f phase plane and compare the result 

with that for the Euler method. 

Problem 2-26: Onset of numerical instability 

Investigate the onset of numerical instability as h is increased in the modi¯ed 

Euler algorithm of the previous problem. 

2.3.3 Glycolytic Oscillator 

Give me another horse! 

William Shakespeare, King Richard III (1564{1616) 

The Euler algorithm, which is the simplest example of an explicit ¯xed-step 

method, is not a very accurate numerical procedure, being of order h accuracy 

(error O(h2 )). A systematic approach to developing more accurate explicit numerical 

schemes are the ¯xed-step Runge{Kutta (RK) methods [BF89], which 

still use the FDA approximation for the ¯rst derivatives but create better approximations 

to the functions on the right-hand side of the ODEs. 

Consider an ODE system of the general form 

_x = X(t; x; y); _y = Y (t; x; y); (2.17) 

where X and Y are known functions of the arguments. Note that any secondorder 

ODE of the general structure Äx = Y (t; x; _x) can be put into this standard 

form by setting _x = y ´ X. 

In the Euler method, the functions X and Y are evaluated only once on 

each step. The general RK approach is to increase the accuracy by using more


evaluations of the functions X and Y , the extra evaluations being carried out at 

intermediate points between tn and tn + h in such a way that an nth-order RK 

scheme is equivalent to a Taylor series expansion of the functions truncated at 

terms of order n. The interested reader is referred to numerical analysis texts 

such as Burden and Faires [BF89] for the details. 

Here it will su±ce to illustrate the application of the fourth-order Runge{ 

Kutta (RK4) method, which has been the historical \workhorse" of ¯xed-step 

methods. This method, which involves four function evaluations of X and Y on 

each step, is of order h 4 in accuracy (error O(h 5 )). Explictly, the RK4 method 

applied to equation (2.17) takes the following form: 

xn+1 = xn + 1 

6 (K1 +2K2 +2K3 + K4); yn+1 = yn + 1 

6 (L1 +2L2 +2L3 + L4); 

with K1 = hX(tn; xn; yn); 

K2 = hX(tn + h=2; xn + K1=2; yn + L1=2) ; 

K3 = hX(tn + h=2; xn + K2=2; yn + L2=2) ; 

K4 = hX(tn + h; xn + K3; yn + L3); 

and L1 to L4 obtained by replacing X with Y in K1 to K4. 

Since this RK4 scheme is O(h4 ) accurate, one can either use the same step 

size as in the Euler method, producing an answer of greater accuracy, or, alternatively, 

use a larger step size to speed up the calculation. However, h can be 

increased only so much before the Taylor expansion on which the RK4 method 

is based begins to break down and wild numerical oscillations in the output 

occur. All explicit schemes are prone to this numerical instability if the step 

size is made too large. 

Runge{Kutta methods of even greater accuracy than RK4 can be generated, 

but these involve even more evaluations of X and Y , which increases the CPU 

time. The RK4 method is usually deemed to be the optimum ¯xed-step explicit 

method that best combines good accuracy with reasonable CPU time. 

As an illustration of the RK4 method, let's consider solving the nonlinear 

ODEs describing the glycolytic oscillator. Living cells obtain energy by breaking 

down sugar, a process known as glycolysis. In yeast cells, this process proceeds 

in an oscillatory way with a period of a few minutes. A simple system of model 

equations proposed by Sel'kov [Sel68] to describe the oscillations is as follows: 

_x = ¡x + ay+ x 2 y; 

_y = b ¡ ay¡ x 2 (2.18) 

y: 

Here x and y refer to the concentrations of adenosine diphosphate (ADP) and 

fructose-6-phosphate (F6P), respectively, and a and b are positive constants. 

We shall use the nominal values a=0:05 and b= 0:5, and take x0 =3, y0 =3.


After loading the plots package, specifying the number of digits, and recording 

the beginning time, 

> restart: with(plots): Digits:=10: begin:=time(): 

the parameter values are entered and N = 1000 steps of size h =0:05 are 

considered. 

> t[0]:=0: x[0]:=3: y[0]:=3: a:=0.05: b:=0.5: N:=1000: h:=0.05: 

The functional, or arrow, operator is used to produce the right-hand side of 

each ODE. Whatever forms are inserted for x and y in the entries X(x,y) and 

Y(x,y) will be operated on as indicated on the right-hand side of the arrow. 

> X:=(x,y)-> -x+a*y+x^2*y; 

X := (x; y) !¡x + ay+ x2 y 

> Y:=(x,y)->b-a*y-x^2*y; 

Y := (x; y) ! b ¡ ay¡ x2 y 

Similarly, operators are introduced to perform the four function evaluations, 

where f will be taken to be X and Y . Note that X and Y do not depend 

explicitly on time here. 

> k1:=f->h*f(x[n],y[n]): 

> k2:=f->h*f(x[n]+K1[n]/2,y[n]+L1[n]/2): 

> k3:=f->h*f(x[n]+K2[n]/2,y[n]+L2[n]/2): 

> k4:=f->h*f(x[n]+K3[n],y[n]+L3[n]): 

Using the above operators, the fourth-order Runge{Kutta algorithm is iterated 

from n =0toN = 1000 and a plotting point produced at each step. 


> K1[n]:=k1(X); L1[n]:=k1(Y); 

> K2[n]:=k2(X); L2[n]:=k2(Y); 

> K3[n]:=k3(X); L3[n]:=k3(Y); 

> K4[n]:=k4(X); L4[n]:=k4(Y); 

> x[n+1]:=x[n]+(K1[n]+2*K2[n]+2*K3[n]+K4[n])/6; 

> y[n+1]:=y[n]+(L1[n]+2*L2[n]+2*L3[n]+L4[n])/6; 

> t[n+1]:=t[n]+h; 

> pnt[n]:=[t[n],x[n],y[n]]; 

> end do: 

Using the spacecurve command, the sequence of points is plotted as a solid 

blue line with normal axes. The tickmarks are controlled, axis labels added, 

and a particular orientation of the 3-dimensional viewing box chosen. 

> spacecurve([seq(pnt[n],n=0..N)],color=blue,axes=normal, 

tickmarks=[2,2,2],labels=["t","x","y"], 

orientation=[23,71]); 

The resulting picture is shown in Figure 2.24. It may be rotated on the computer 

screen by dragging with the mouse.


50 

t 

2 

y 

0 

Figure 2.24: Temporal evolution of a glycolytic oscillator. 

The spiral behavior characteristic of temporal oscillations is clearly seen. If 

the three-dimensional ¯gure is rotated on the computer screen to show the x-y 

phase plane, a closed loop will be observed. One can also view x versus t and 

y versus t. The CPU time for the ¯rst-principles RK4 calculation is 



about one-half a second. 

After unassigning the variables x and y, 

> unassign('x','y'): begin2:=time(): 

exactly the same 3-dimensional plot is produced (not displayed here) by using 

the dsolve command with the option method=classical[rk4]. 

> sys:=diff(x(t),t)=X(x(t),y(t)),diff(y(t),t)=Y(x(t),y(t)): 

> sol:=dsolve(fsys,x(0)=3,y(0)=3g,fx(t),y(t)g,type=numeric, 

method=classical[rk4],stepsize=0.05): 

> odeplot(sol,[t,x(t),y(t)],0..50,axes=normal,numpoints=1000, 

color=blue,style=line,tickmarks=[2,2,2], 

labels=["t","x","y"],orientation=[23,71]); 

> CPUtime2:=(time()-begin2)*seconds; 

CPUtime2 := 0:300 seconds 

The CPU time for the dial-up routine is slightly shorter than the ¯rst-principles 

calculation here. 

x 

5


PROBLEMS: 

Problem 2-27: Harvesting of ¯sh 

In suitably normalized units, the e®ect of ¯shing on the normalized population 

number x of a single species of ¯sh with a limited food supply can be described 

by the following nonlinear ODE, 

_x = x (1 ¡ x) ¡ Hx=(a + x); 

where H is the harvesting coe±cient and a is a parameter. For H =0,the 

remaining ODE is known as the logistic equation. Show that this equation has 

an analytic solution and plot the result for a =0:2 andx(0) = 0:1. Discuss the 

behavior of the solution. 

Then, using the ¯rst-principles RK4 method with step size h =0:1, numerically 

investigate this equation as the harvesting coe±cient H is increased from 

zero. Plot your results and discuss the change in behavior as H increases. 

Problem 2-28: Bucky the beaver 

Bucky the beaver attempts to swim across a river by steadily aiming at a target 

point directly across the river. The river is 1 km wide and has a speed of 1 

km/hr, while Bucky's speed is 2 km/hr. In Cartesian coordinates, Bucky is 

initially at (x =1;y = 0), while the target point is at (0; 0). Bucky is initially 

swept an in¯nitesimal distance downstream but recovers almost instantly to 

continue his swimming motion. His equations of motion are 

2 x 

2 y 

_x = ¡ p ; _y =1¡ p 

x2 + y2 x2 + y2 : 

(a) Justify the structure of the equations. 

(b) Using the ¯rst-principles RK4 method with h =0:01, determine how long 

it takes Bucky to reach the target point. 

(c) Determine the analytic solution y(x) for Bucky's path across the river. 

(d) Plot the analytic and numerical solutions together for Bucky's path. 

Problem 2-29: Chemical reaction 

Consider the irreversible chemical reaction 

2K2Cr2O7 +2H2O+3S! 4KOH + 2Cr2O3 +3SO2, with initially N1 = 2000 molecules of potassium dichromate (K 2Cr 2O 7), N2 = 

2000 molecules of water (H 2O), and N3 = 3000 atoms of sulphur (S). The 

number X of potassium hydroxide (KOH) molecules at time t sisgivenbythe 

rate equation 

_X = k (2 N1 ¡ X) 2 (2 N2 ¡ X) 2 (4 N3=3 ¡ X) 3 ; 

with k =1:64 £ 10 ¡20 s ¡1 and X(0) = 0. Determine X(t) byusingthe¯rstprinciples 

RK4 method with h =0:001. How many KOH molecules are present 

at t =0:1 s?att =0:2 s?


2.3.4 Fox Rabies Epidemic 

Thought is an infection. In the case of certain thoughts, it becomes 

an epidemic. 

Wallace Stevens, American poet (1879{1955) 

As already mentioned, if the step size h is made too large, ¯xed-step explicit 

methods not only become increasingly inaccurate, but will become numerically 

unstable with the solution eventually \blowing up." One approach to avoiding 

this instability is to make use of the backward-di®erence approximation to the 

derivative and create an implicit or semi-implicit numerical scheme. 

In this example, we will illustrate this approach by solving the nonlinear 

ODEs for the fox rabies epidemic 7 model of Anderson and coworkers [AJMS81] 

using the backward Euler algorithm. The ideas presented here can be extended 

to higher-order RK methods. 

The rabies epidemic model was applied to the fox population in central Europe, 

the epidemic believed to have originated in Poland in 1939. In this model, 

the fox population is divided into three categories: susceptibles (population 

density x foxes/km 2 ) are foxes that are currently healthy but are susceptible 

to catching the virus; infected (density y) arefoxesthathavecaughtthevirus 

but are not yet capable of passing on the virus; infectious (density z) arefoxes 

that are capable of infecting the susceptibles. The model has no category of 

recovered immune foxes because very few, if any, survive after acquiring the 

rabies virus. For other diseases with a lower mortality rate, one would add a 

recovered category and therefore an additional modeling equation. The relevant 

equations for the fox rabies epidemic are 

_x = ax¡ (b + gN) x ¡ ¯xz´ X; 

_y = ¯xz¡ (¾ + b + gN) y ´ Y; 

_z = ¾y¡ (® + b + gN) z ´ Z; 

(2.19) 

with N = x + y + z being the total fox density. The meaning of the various 

coe±cients in (2.19) and their estimated values is as follows: 

Symbol Meaning Value 

a average per capita birth rate 1/year 

b average per capita intrinsic death rate 0:5/year 

¯ rabies transmission coe±cient 79:67 km 2 /year 

¾ 1=¾ = average latent period (¼ 28 to 30 days) 13/year 

® death rate of rabid foxes 73/year 

g gN represents increased death rate at large 0.1-5 km 2 /year 

N due to depletion of food supply 

7 Modeling the spread of diseases is extensively discussed in Murray [Mur89].


Using the BDA for the time derivative to connect time step n to the previous 

step n ¡ 1 and approximating the rhs with its value at step n, thebackward 

Euler approximation to equations (2.19) is 

xn ¡ xn¡1 

h 

= Xn; yn ¡ yn¡1 

h 

= Yn; 

zn ¡ zn¡1 

h 

= Zn; 

or letting n ! n + 1 and substituting the forms of X, Y ,andZ, 

xn+1 = xn + h (a ¡ b) xn+1 ¡ hgNn+1 xn+1 ¡ h¯xn+1 zn+1 

yn+1 = yn + h¯xn+1 zn+1 ¡ h (¾ + b) yn+1 ¡ hgNn+1 yn+1 

zn+1 = zn + h¾yn+1 ¡ h (® + b) zn+1 ¡ hgNn+1 zn+1 

(2.20) 

with Nn+1 = xn+1 + yn+1 + zn+1. 

This algorithm is nonlinear in the unknowns xn+1, yn+1, andzn+1, andis 

an example of an implicit algorithm because one cannot explicitly express the 

unknowns in terms of the known values xn, yn, andzn. The standard approach 

is to make the scheme semi-implicit by linearizing it as follows. For a nonlinear 

function f(xn+1; zn+1), we Taylor expand thus: 

μ 

μ 

@f 

@f 

f(xn+1;zn+1) =f(xn;zn)+(xn+1¡xn) 

+(zn+1¡zn) 

+¢¢¢ 

@x 

@z 

xn;zn 

xn;zn 

For our fox-rabies example, this gives us, for example, 

f ´ (xz) n+1 = xn zn + zn(xn+1 ¡ xn)+xn(zn+1 ¡ zn): (2.21) 

Applying this procedure to all of the quadratic terms on the rhs of (2.20) and 

gathering all the \new" values xn+1, etc.,onthelhs,weobtain 

with 

A1n xn+1 + B1n yn+1 + C1n zn+1 = xn + D1n; 

A2n xn+1 + B2n yn+1 + C2n zn+1 = yn + D2n; 

A3n xn+1 + B3n yn+1 + C3n zn+1 = zn + D3n; 

(2.22) 

A1n =1+h (b ¡ a)+hg(Nn + xn)+h¯zn; B1n =hgxn; C1n =h (¯ + g) xn; 

A2n =hgyn ¡ h¯zn; B2n =1+h (¾ + b)+hg(Nn + yn); C2n =hgyn ¡ h¯xn; 

A3n =hgzn; B3n =hgzn ¡ h¾; C3n =1+h (® + b)+hg(Nn + zn); 

D1n =hgNn xn + h¯xn zn; D2n =hgNn yn ¡ h¯xn zn; D3n =hgNn zn: 

We now have three linear equations (2.22), with known numerical coe±cients 

determined from the previous step, for the three unknowns xn+1, yn+1, zn+1.


The semi-implicit algorithm is now implemented in the following recipe, 

taking a total of 5000 time steps of size h =0:004. 

> restart: Digits:=10: total:=5000: h:=0.004: 

The parameter values are entered, with g at the lower end of its range. 

> a:=1: b:=0.5: beta:=79.67: g:=0.1: alpha:=73: sigma:=13: 

Initially, we take x0 =4,y0 =0:4, and z0 =0:1. 

> t[0]:=0: x[0]:=4: y[0]:=0.4: z[0]:=0.1: 

The initial plotting point is formed, and the starting time of the do loop that 

iterates the algorithm is recorded. 

> pnt[0]:=[t[0],x[0],y[0]]: begin:=time(): 

> for n from 0 to total do 

The various coe±cients are evaluated on the nth step. 

> N[n]:=x[n]+y[n]+z[n]; 

> A1[n]:=1+h*(b-a)+h*g*(N[n]+x[n])+h*beta*z[n]; 

> B1[n]:=h*g*x[n]; 

> C1[n]:=h*(beta+g)*x[n]; 

> D1[n]:=h*g*N[n]*x[n]+h*beta*x[n]*z[n]; 

> A2[n]:=h*g*y[n]-h*beta*z[n]; 

> B2[n]:=1+h*(sigma+b)+h*g*(N[n]+y[n]); 

> C2[n]:=h*g*y[n]-h*beta*x[n]; 

> D2[n]:=h*g*N[n]*y[n]-h*beta*x[n]*z[n]; 

> A3[n]:=h*g*z[n]; 

> B3[n]:=h*g*z[n]-h*sigma; 

> C3[n]:=1+h*(alpha+b)+h*g*(N[n]+z[n]); 

> D3[n]:=h*g*N[n]*z[n]; 

The linear equations for xn+1, yn+1, zn+1 are entered, 

> E1:=A1[n]*x[n+1]+B1[n]*y[n+1]+C1[n]*z[n+1]=x[n]+D1[n]; 

> E2:=A2[n]*x[n+1]+B2[n]*y[n+1]+C2[n]*z[n+1]=y[n]+D2[n]; 

> E3:=A3[n]*x[n+1]+B3[n]*y[n+1]+C3[n]*z[n+1]=z[n]+D3[n]; 

and numerically solved. 

> sol[n+1]:=(fsolve(fE1,E2,E3g,fx[n+1],y[n+1],z[n+1]g)); 

The solution is assigned and the values of xn+1, yn+1, andzn+1 are recorded 

at time tn+1 = tn + h. 

> assign(sol[n+1]); 

> x[n+1],y[n+1],z[n+1]; 

> t[n+1]:=t[n]+h; 

Finally, a plotting point is formed and the loop ended. 

> pnt[n+1]:=[t[n+1],x[n+1],y[n+1]]; 

> end do:


Ona3-GHzPC,theCPUtimetoexecutetheloop 

> cpu:=time()-begin; 

cpu := 10:565 

is about 11 seconds. 

The points are now plotted, the resulting curve being shown in Figure 2.25. 

The three-dimensional curve may be rotated on the computer screen by dragging 

the viewing box with the mouse. What does the model calculation predict? 

> plots[spacecurve]([seq(pnt[n],n=0..total)],color=black, 

axes=normal,tickmarks=[3,2,3],labels=["t","x","y"]); 

20 

t 

10 

1.5 

1 

0.5 

Figure 2.25: Semi-implicit solution of fox rabies epidemic model. 

You can con¯rm that the algorithm remains stable for much larger h, but 

becomes increasingly inaccurate. In general, semi-implicit schemes are not guaranteed 

to be stable, but usually are. However, such schemes are costly in terms 

of computing time if high accuracy is required. For systems of nonlinear ODEs, 

this is usually the case, so variable- or adaptive-step explicit schemes are preferred 

by researchers. Such schemes adjust the step size so that larger steps are 

used when the \terrain" is °atter and smaller steps when it becomes steeper. 

Maple's default ODE algorithm is based on the Runge{Kutta{Fehlberg 45 

(RKF45) scheme, one of the most widely used adaptive step methods. Without 

getting into the details, which can be found in standard numerical analysis 

texts [BF89], the code uses both the fourth-order and ¯fth-order Runge{Kutta 

schemes and compares the results of both algorithms on each step, using the 

di®erence in answers as a measure of the error. The step size is then adjusted 

to maintain some maximum tolerance on the error. 

0 

y 

2 

x 

4


PROBLEMS: 

Problem 2-30: The arms race 

Rapoport [Rap60] has proposed the following system of model equations to 

describe the arms race between two nations (or two groups of nations), 

_x = ¡m1 x + a1 y + b1 y 2 ; 

_y = ¡m2 y + a2 x + b2 x 2 ; 

where x and y are the defense budgets (in suitable units of currency) of the two 

nations and all constants are positive. Explain the terms in the model. 

Apply the semi-implicit backward Euler method to Rapoport's model, taking 

the nominal values m1 =0:5, a1 =1,b1 =0:02, m2 =0:4, a2 =0:1, 

b2 =0:05, x(0) = y(0) = 0:5. Try di®erent step sizes and plot the results in t 

vs. x vs. y space. Discuss the results. 

Problem 2-31: Van der Pol oscillator 

The Van der Pol equation, 

Äx ¡ ² (1 ¡ _x 2 )+x =0; 

with ² =5:03, x(0) = 0:1, and _x(0) = 0 was the subject of recipe 01-1-3. 

Choosing a suitable step size, numerically solve this ODE using the semi-implicit 

backward Euler method and compare with the results obtained in 01-1-3. 

Problem 2-32: Second-order-accurate scheme 

Given a system of ¯rst-order nonlinear ODEs with a representative equation 

of the structure _x = X(x; : : :), one can create a second-order-accurate semiimplicit 

numerical scheme by using the backward-di®erence approximation for 

the derivative and replacing the Euler approximation Xn on the rhs with the 

average (Xn + Xn+1)=2. 

Derive a second-order-accurate semi-implicit algorithm for the RÄossler system 

of Section 1.2.3. Taking a = b =0:2, c =5:0, x(0) = 4:0, y(0) = z(0) = 0, 

and h =0:05, determine the behavior of the system up to t = 100 and plot the 

trajectory in x-y-z space. How do your results compare with those obtained 

with recipe 01-2-3 for the same parameter values? 

Problem 2-33: Sti® ODE system 

A sti® ODE system is one for which there are two or more very di®erent time or 

spatial scales of the independent variable. Numerical instability can occur for 

sti® systems solved with ¯xed-step explicit schemes unless the step size is taken 

to be shorter than (about) the shortest time scale for the system. Consider the 

following set of coupled linear ODEs, 

_x =998x +1998y; _y = ¡999 x ¡ 1999 y; 

subject to the initial conditions x(0) = 1 and y(0) = 0. 

(a) Analytically solve the system for x(t) andy(t). 

(b) Determine the two characteristic times in the solutions and con¯rm that 

they are very di®erent from each other.


(c) Using the dial-up RK4 algorithm, con¯rm the statement on numerical 

instability and the characteristic time scales. 

(d) Using the backward-di®erence approximation for the time derivatives and 

the Euler approximation on the right-hand side, show that the numerical 

instability can be \cured." Choose a step size such that when plotted 

using a point style, the numerical points lie on the analytic curves for 

x(t) andy(t). 

Problem 2-34: Forced sti® system 

Repeat the steps of the previous problem for the forced system 

_x =9x +24y +5cost ¡ 1 

1 

sin t; _y = ¡24 x ¡ 51 y ¡ 9cost + sin t; 

3 3 

subject to the initial conditions x(0) = 4=3 andy(0) = 2=3.

Part II 

THE ENTREES 

That is the essence of science: 

ask an impertinent question, 

and you are on the way to a pertinent answer. 

Jacob Bronowski, British scientist, author, The Ascent of Man, 1973 

The whole of science is nothing more 

than a re¯nement of everyday thinking. 

Albert Einstein, physics Nobel laureate, Out of My Later Years, 1950 

The e®ort to understand the universe is one 

of the very few things that lifts human life a little 

above the level of farce, and gives it 

some of the grace of tragedy. 

Steven Weinberg, American physicist, The First Three Minutes, 1977 

107

Chapter 3 

Linear ODE Models 

Among all the mathematical disciplines the theory of di®erential 

equations is the most important. It furnishes the explanation of all 

those elementary manifestations of nature which involve time. 

Sophus Lie, Norwegian mathematician (1842{1899) 

In the two chapters of the Appetizers, we presented only a few examples of 

linear ODEs, because the graphical and numerical techniques were more suitable 

for nonlinear systems where generally exact analytic solutions simply do 

not exist. Also, modern research often involves nonlinear ODE (and PDE) 

systems, a subject that is almost completely neglected in undergraduate mathematics 

training. So, one of the goals of this text is to partially ¯ll in this 

\hole," showing how a computer algebra system may be used to explore nonlinear 

models without getting \buried" in messy and complicated mathematical 

details. However, we would be remiss if we did not provide some coverage of 

linear di®erential equation systems, both ordinary and partial. The former are 

covered in this chapter, while linear PDEs are dealt with in Chapters 5 and 6. 

In between, we shall explore nonlinear ODEs further in Chapter 4. 

For linear ODE systems with constant coe±cients, it is always possible to 

¯nd exact analytic solutions in terms of \elementary" mathematical functions 

such as sines and cosines, logs, and exponentials. Analytic solutions are also 

possible for linear ODEs with variable coe±cients that are of the so-called 

Sturm{Liouville [MW71] type. Included in this classi¯cation are such \famous" 

ODEs as Bessel's equation and the Legendre and Hermite equations, to mention 

just a few. The solutions are expressed in terms of \special" functions, de¯ned 

in terms of either in¯nite series or ¯nite polynomials. 

In this chapter, the emphasis will be on illustrating how the dsolve command 

can be used to easily generate analytic solutions for some physically 

interesting linear ODE systems, with both constant and variable coe±cients. 

We will not go into mathematical methods here, leaving this to your mathematics 

education. However, if you would like to see computer algebra recipes that 

apply mathematical methods to solving ODEs, you are referred to Computer 

Algebra Recipes for Mathematical Physics [Enn05]. 

109

110 CHAPTER 3. LINEAR ODE MODELS 

3.1 First-Order Models 

The most general nth-order linear ODE can be written in the form 

a0(t) dn x(t) 

dt n 

+ a1(t) dn¡1 x(t) 

dt n¡1 

dx(t) 

+ ¢¢¢+ an¡1(t) 

dt + an(t) x(t) =h(t); (3.1) 

the equation being labeled as linear because each term on the left-hand side is 

linear, or of ¯rst order, in the dependent variable x. For the sake of de¯niteness, 

the independent variable has been taken to be the time t here, but could be a 

spatial variable. If h(t) = 0, the di®erential equation is said to be homogeneous, 

otherwise it is inhomogeneous. 

In the following two recipes, we look at two examples of inhomogeneous 

¯rst-order ODEs, the ¯rst involving constant coe±cients, the second containing 

variable coe±cients. 

3.1.1 How's Your Blood Pressure? 

Amid the pressure of great events, a general principle gives no help. 

Georg Hegel, German philosopher (1770{1831) 

On measuring your blood pressure, your doctor will give you two numbers. 

A normal blood pressure for humans is 120/80, the ¯rst number specifying 

the maximum (systolic) pressure (in units of mm of Hg) on the arterial walls, 

the second number the minimum (diastolic) pressure. The blood pressure is 

generated by the beating of the heart, its variation controlled by the aorta. 

The aorta is the large blood vessel into which the arterial blood °ows on 

leaving the heart. During the systolic phase of the heartbeat cycle, blood is 

pumped under pressure from the heart into one end of the aorta, whose walls 

then stretch in order to accommodate the blood. The diastolic phase then 

follows during which there is no °ow of blood into the aorta, its walls elastically 

contracting. The blood is then squeezed out of the aorta and around the body's 

circulatory system. The following recipe presents a simple model of the variation 

of blood pressure due to the beating heart and the aorta. 

The entry infolevel[dsolve] will provide some information about the 

dsolve command, the amount of information generally increasing as the integer 

(which can vary from 1 to 5) speci¯ed on the right of the colon is increased. 

> restart: infolevel[dsolve]:=2: 

Let V (t) be the volume of the aorta and p(t) the pressure within it at time t. 

Assuming that the aorta expands linearly with increasing p, thenV = V0 + Cp, 

where V0 and C are constants. The parameter C, called the compliance, isa 

measure of the stretchability of the aorta. Mentally di®erentiating this relation 

produces ode1 . 

> ode1:=diff(V(t),t)=C*diff(p(t),t); 

ode1 := d 

V (t) =C 

dt 

μ d 

dt p(t)

3.1. FIRST-ORDER MODELS 111 

Therateofchangeofvolume(dV=dt) of the aorta is equal to the di®erence 

between the rate f(t) at which blood is pumped into the aorta by the heart and 

the rate p(t)=R at which blood is pumped out of the aorta into the circulatory 

system. The constant R is referred to as the systemic resistance. Thus, the 

second ODE is is now given by ode2 . 

> ode2:=diff(V(t),t)=f(t)-p(t)/R; 

ode2 := d 

p(t) 

V (t) =f (t) ¡ 

dt R 

An ODE (ode3 ) for pressure alone results on substituting ode1 into ode2 . 

> ode3:=subs(ode1,ode2); 

μ 

d 

ode3 := C 

dt p(t) 

 

= f (t) ¡ p(t) 

R 

As a simple model of the forcing function f(t), let's assume that during the 

systolic (pumping) phase f(t) =A sin(¼t=¿), with A the amplitude and ¿ the 

duration of this phase, and f(t) = 0 during the diastolic (nonpumping) phase. 

So f(t) is a piecewise forcing function. If T is the time for one complete cycle 

(time between heart beats), the piecewise function for two heartbeats can be 

entered using the following piecewise command. 

> f(t):=piecewise(t0, R>0, A>0, and P>0. 

> sol:=dsolve(fode3,p(0)=Pg,p(t)) 

assuming (tau>0,T>tau,C>0,R>0,A>0,P>0) ; 

The assuming command applies the assumption only to the command line in 

which it appears. If you wish to apply assumptions to the entire work sheet, 

you can use the assume command at the beginning of the work sheet. In either


case, the assumptions are determined by trial and error, i.e., looking at the 

form of the answer, which is now given. 

Methods for ¯rst order ODEs: 

| Trying classi¯cation methods | 

trying a quadrature 

trying 1st order linear 

< ¡t 

e 

( RC 

>: 

)μ 

P + AC¼¿R2 

¿ 2 + ¼ 2 R 2 C 2 

AC¼¿R 

¡ 

2 μ 

¼t 

cos 

¿ 

¿ 2 + ¼2R2C 2 + 

AR¿ 2 μ 

¼t 

sin 

¿ 

¿ 2 + ¼2R2 ; t · ¿ 

C2 8 

< 

AC¼¿R 

: 

2 ¿¡t 

e 

( RC ) 

¿ 2 + ¼2R2 ¡t 

+ e( RC 

C2 )μ 

P + AC¼¿R2 

¿ 2 + ¼2R2C 2 

 

; t · T 

¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢ 

Intheabovesolution,wehaveshownonlytheanalyticform(thusthedots) 

for the ¯rst cycle to save on text space. You can observe the complete solution 

by executing the recipe on your computer. Because of the inclusion of the 

infolevel[dsolve] command, we are told in the output that Maple recognizes 

ode3 as a ¯rst-order linear ODE and presumably uses a standard method for 

solving such an ODE. You may object to this \black box" nature of the dsolve 

command, but obtaining the answer easily in our opinion is more important 

than the mechanical details leading up to it. However, if the methodology is 

important to you, we shall mimic a hand calculation in the next recipe. 

To make a plot, we shall take the parameter values1 to be T = 1 second (60 

heartbeats per minute), ¿ =0:15 s, C =0:002 liters/mm Hg, P =80mmHg, 

and R = 1056 mm Hg/liter/second. 

> tau:=0.15: T:=1: C:=0.002: P:=80: R:=1056: 

The amplitude parameter A must be such that the right-hand side (rhs) of the 

solution must be the same at t = T as at t = 0, i.e., the heartbeat is periodic. 

> eq:=eval(rhs(sol),t=0)=eval(rhs(sol),t=T); 

432:0593387 A¼ 

eq := 80: = 

+49:82624166 

0:0225 + 4:46054400 ¼2 The above equation is numerically solved for A. 

> A:=fsolve(eq,A); 

A := 0:9791418590 

The pressure is then given by the rhs of sol, the °oating-point evaluation commandbeingusedtoexpresstheanswertofourdigits. 

> Pressure:=evalf(rhs(sol),4); 

1These approximate values are taken from the Internet.


Pressure:= © 103:4 e (¡0:4735 t) ¡ 23:36 cos(20:95 t)+0:5280 sin(20:95 t); t· 0:15 

f 103:4 e (¡0:4735 t) +23:36 e (0:07102¡0:4735 t) ;t· 1: 

f 103:4 e (¡0:4735 t) +23:36 e (0:07102¡0:4735 t) (0:4735¡0:4735 t) 

+23:36 e 

¡ 23:36 cos(¡20:95 + 20:95 t)+0:5280 sin(¡20:95 + 20:95 t); t· 1:15 

f 103:4 e (¡0:4735 t) +23:36 e (0:07102¡0:4735 t) (0:4735¡0:4735 t) 

+23:36 e 

+23:36 e (0:5445¡0:4735 t) ; 1:15 plot(f(t),t=0..2*T,labels=["t","f(t)"],tickmarks=[3,4]); 

0.8 

0.6 

f(t) 

0.4 

0.2 

0 

1 t 2 

Figure 3.1: Piecewise forcing function f(t). 

The blood pressure variation is plotted over the same interval. 

> plot(Pressure,t=0..2*T,labels=["t","Pressure"]); 

120 

110 

Pressure 

100 

90 

80 

0 0.2 0.4 0.6 0.8 1 t 1.4 1.6 1.8 2 

Figure 3.2: Time variation of the blood pressure.


The resulting picture is shown in Figure 3.2, the plot capturing the gross behavior 

of the variation in blood pressure with time. More detailed behavior can 

be obtained by using more sophisticated models. You might try an Internet 

search to learn more about these models. 

PROBLEMS: 

Problem 3-1: More heartbeats 

Modify the forcing function in the text recipe to include four heartbeats. Do an 

Internet search on blood pressure and discuss how the model could be improved. 

Problem 3-2: An RL circuit 

A simple electrical circuit consists of a resistor of R ohms (−) in series with an 

inductor of L henries (H) and a battery providing a constant voltage v (in volts 

(V)). If i(t) is the current in the circuit at time t, the voltage drop across the 

resistor is Ri (Ohm's law) and across the inductor is L (di=dt). 

(a) Making use of Kirchho®'s voltage rule, which states that the sum of the 

voltage drops is equal to the supplied voltage, derive the ¯rst-order ODE 

for i(t) and solve it, given that the initial current is zero. 

(b) Taking L =4H,R =12−,andv = 60 V, plot the solution over a time 

interval that brings the current to within 1% of its asymptotic value. 

(c) The battery is replaced with a generator that produces a variable voltage 

of v(t) = 60 sin(30 t) volts, the resistor and inductor retaining the same 

values as in part (b). If the initial current is zero, determine the current 

i(t) fort¸0. Plot the current over a time interval that brings the current 

to within 1% of steady state. 

Problem 3-3: Learning curves for widget production 

Learning curves are used by scientists interested in learning theory. A learning 

curve is a plot of the performance P (t) of someone learning a skill as a function 

of the training time t. A simple model of learning curve is to assume that the 

rate dP=dt at which performance improves is proportional to M ¡ P (t), where 

M is the maximum level of performance. Determine the analytic form of P (t) 

and discuss what this model implies. 

Two new workers (Jimbo and Jumbo) are hired for an assembly line producing 

widgets. Jimbo (Jumbo) produces 25 (35) widgets during the ¯rst hour and 

45 (50) widgets the second hour. Estimate the maximum number of widgets 

per hour that each is capable of producing and plot their learning curves. 

Problem 3-4: Population growth 

A deer population initially numbers 1000 and has a growth rate of 0.5 when t 

is measured in months. The population is \harvested" throughout the year at 

therateofh (2 + cos(¼t=6)) per month, where h is a constant. Determine the 

deer population number as a function of t. What value of h would lead to a 

zero population growth over the 12-month period from t =0tot =12? Plot 

the deer-population curve for this 12-month period.


3.1.2 Greg Arious Nerd's Problem 

It will be found, in fact, that the ingenious are always fanciful, 

and the truly imaginative never otherwise than analytic. 

Edgar Allan Poe, American writer, TheMurdersintheRueMorgue,1841 

Greg Arious Nerd, an eminent mathematician at the Erehwon Institute of Technology 

(EIT), has assigned the following problem for his undergraduate class to 

solve using computer algebra. They are not to use dsolve to analytically solve 

the relevant ODE, but instead use command structures found in the DEtools 

library package that mimic the steps in a hand calculation. 

Starting from rest, Evil Knievel weevil experiences a time-dependent drive 

force Fdrive = t2 e ¡t andadragforceFdrag = ¡(t2 =(1 + t)) v(t), per unit mass. 

Mimicking a hand calculation, determine Evil's velocity v(t) attimetand plot 

it over the interval t = 0 to 10 Erehwonian time units. What distance does Evil 

travel in this interval? 

Here is Professor Nerd's answer key. Loading the DEtools package, 


the drag and drive forces are entered. 

> F[drag]:=-(t^2/(1+t))*v(t); F[drive]:=t^2*exp(-t); 

Fdrag := ¡ t2 v(t) 

Fdrive := t 

1+t 

2 e (¡t) 

Applying Newton's second law, the ODE governing Evil Knievel's velocity is 

> ode:=diff(v(t),t)=F[drag]+F[drive]; 

ode := d 

dt v(t) =¡t2 v(t) 

1+t + t2 e (¡t) 

Although not requested, Nerd applies the odeadvisor command to ode. This 

command will classify the ODE and inclusion of the help option causes a relevant 

Help page with useful hyperlinks to be opened. The Help page should be 

closed when one is ¯nished reading it. 

> odeadvisor(ode,help); #close Help page 

[ linear] 

In this case, ode is classi¯ed as linear, which is obvious by inspection. 

Proceeding by hand, one would look for the integrating factor2 of the ¯rstorder 

inhomogeneous ODE. With Maple, the integrating factor IF for ode is 

obtained by entering intfactor(ode), which is then simpli¯ed. 

> IF:=intfactor(ode); 

t2 

IF := e 

( 2 ¡ t +ln(1+t)) 

> IF:=simplify(IF); 

R 

2 f(t) dt. 

For a general linear ¯rst-order ODE, _x + f(t) x = g(t), the integrating factor is e


( t (t¡2) 

IF := (1 + t) e 2 ) 

With the integrating factor known, the standard mathematical procedure is to 

multiply ode by IF and integrate. This is easily accomplished with the ¯rst 

integral (firint) command. 

> sol1:=firint(ode*IF); 

μ 

 

t (t¡2) t (t¡2) 

sol1 := e 

( ) 2 + e 

( ) 2 t v(t) ¡ 3 te (1=2 t2 ¡ 2 t) ¡ 4 e (1=2 t2 ¡ 2 t) 

+ 5 

2 I p ¼e (¡2) p μ 

1 

2erf 

2 I p 2 t ¡ p 

2 I ¡ t2 e (1=2 t2 ¡ 2 t) + C1 =0 

In the solution output, I stands for p ¡1, erf for the error function, 3 and C1 

is the integration constant to be determined. Then, sol1 is solved for v(t), the 

result (assigned the name V ) not being shown here in the text. 

> V:=solve(sol1,v(t)); 

Evil Knievel's initial velocity is zero. So the constant C1 is determined by 

evaluating the velocity V at t = 0, equating the result to 0, and solving for 

C1 . The constant will be automatically substituted into V . 

> _C1:=solve(eval(V,t=0)=0,_C1); 

C1 := 4 + 5 

2 I p ¼e (¡2) p 2erf( p 2 I) 

On simplifying V , Evil Knievel's velocity is now determined at arbitrary time 

t ¸ 0. This is a nontrivial result to derive by hand. 

> Vel:=simplify(V); 

Vel := 1 

μ 

6 te 

2 

+2t 2 e 

t (t¡4) 

( 2 

t (t¡4) 

( 2 

) +8e ( 

t (t¡4) 

2 ) p (¡2) ¡ 5 I ¼e p μ 

1 

2erf 

) p 

¡ 8 ¡ 5 I ¼e (¡2) p 2erf( p 

2 I) e 

2 I p 2 t ¡ p 

2 I 

t (t¡2) 

(¡ 2 

) ± (1 + t) 

Now, since the original ODE was completely real, the velocity must also be 

completely real. It can be converted to a real form using the Re (real part of) 

command, assuming that t>0. The new function er¯ is the imaginary error 

function, 4 which is a real function here. 

> Vel:=Re(Vel) assuming t>0; 

Vel := 1 

Ã 

6 te 

2 

+2t 2 e 

t (t¡4) 

( 2 

t (t¡4) 

( 2 

R 

3 2 u 

De¯ned as erf(u) = p¼ 

) +8e ( 

t (t¡4) 

2 ) p (¡2) +5 ¼e p Ã p 

2 t 

2er¯ 

) p (¡2) ¡ 8+5 ¼e p 2er¯( p 

2) e 

0 e¡t2 

dt: 

4 De¯ned as er¯(u) =¡I erf(Iu)= 2 

p¼ 

R u 

0 et2 

dt. 

t (t¡2) 

(¡ 2 

2 ¡ p 2 

! 

) ± (1 + t)


Evil Knievel's velocity is plotted over the time interval t = 0 to 10 time units, 

labels being added to the graph. 

> plot(Vel,t=0..10,labels=["t","Vel"],tickmarks=[3,3]); 

0.3 

Vel 

0.2 

0.1 

0 

2 4 6 8 10 

t 

Figure 3.3: Evil Knievel's velocity pro¯le. 

The resulting picture is shown in Figure 3.3, the velocity curve increasing from 

zero to a maximum and then decreasing to zero again as t !1. 

The distance traveled in the time interval t = 0 to 10 is equal to the area 

under the velocity curve between these limits, this area obtained by performing 

the integration R 10 

Vel dt. However, if the integer 10 is used, the integral will 

0 

not be evaluated. A °oating-point answer could be obtained by apply the evalf 

command. An alternative way used here in Professor Nerd's answer key is to 

enter the upper limit as 10.0, which forces a °oating-point evaluation of the 

integral. It does not imply any increase in accuracy. 

> distance:=int(Vel,t=0..10.0); #note floating point number 

distance := 0:9943356328 

So, Evil Knievel travels just under 1 Erehwonian spatial unit in the time interval. 

PROBLEMS: 

Problem 3-5: Direct solution 

Solve Professor Nerd's problem directly with the dsolve command and check 

that the answer is identical to that derived in the text recipe. 

Problem 3-6: Di®erent drive force 

The drive force in the text recipe is replaced with Fdrive =sin(t) t2 e ¡t . Determine 

the velocity as a function of time and plot the result over the interval 

t = 0 to 10. What is Evil Knievel's maximum velocity and at what time does 

this occur? What is his maximum displacement from the origin in this interval 

and at what time does this occur?


3.2 Second-Order Models 

3.2.1 Daniel Encounters Resistance 

Belief like any other moving body follows the path of least resistance. 

Samuel Butler, English writer (1835{1902) 

In the introductory recipe, recall that Richard's grandson Daniel threw a small 

ball with an initial speed of 15 m/s toward a 3:5-meter-high fence located 20 

meters from the ball's initial position. The ball left his hand at a height of 2 

m above the level ground and just cleared the top of the fence. The gravitational 

acceleration has the value g =9:8 m/s2 . The ball was regarded as a point 

particle and air resistance was neglected. 

In the recipe, we determined the angle with the horizontal that the ball was 

thrown, the time for the ball to reach the fence, and animated the motion of 

the ball with the fence included. 

In the following recipe, we shall make the above model calculation more 

realistic by including the e®ect of air resistance on the ball. Provided that the 

ball is smooth and its velocity ~v is not too high, the air resistance is governed by 

Stokes's resistance law, the drag force taking the form ~ Fdrag = ¡mk~v, where 

m is the mass of the ball and the drag coe±cient k is positive. 

As previously, the origin is chosen to be on the ground below the initial 

position of the ball and the x-coordinate is taken to be horizontal and the 

y-coordinate vertical. The plots package, needed for the animation, is loaded. 


With air resistance included, Newton's second law of motion yields the following 

ODE in the x direction. 

> xeq:=diff(x(t),t,t)=-k*diff(x(t),t); 

xeq := d2 

μ 

d 

x (t) =¡k x (t) 

dt2 dt 

Although xeq is a second-order ODE, it clearly can be reduced to a ¯rst-order 

ODE by integrating both sides once, and then solved by the same \hand" 

procedure used in the last recipe. However, it's quicker to solve for x(t) using 

the dsolve command, subject to the initial condition x(0) = xb and _x(0) = 

V cos(Á). Here xb is the initial x-coordinate of the ball, V the initial speed, 

and Á the initial angle of the velocity with the horizontal. For the time being, 

everything is kept symbolic, the parameter values being substituted later. After 

applying the dsolve command, we take the rhs of the resulting answer, yielding 

the x-coordinate of the ball at time t ¸ 0. 

> x:=rhs(dsolve(fxeq,x(0)=xb,D(x)(0)=V*cos(phi)g,x(t))); 

xb k + V cos(Á) 

x := ¡ 

k 

V cos(Á) e(¡kt) 

k 

Taking xf to be the x-coordinate of the fence, we ¯nd the time tf for the ball 

to reach the fence by equating x to xf and solving for t.

3.2. SECOND-ORDER MODELS 119 

> tf:=solve(x=xf,t); 

μ 

xb k + V cos(Á) ¡ xf k 

ln 

V cos(Á) 

tf := ¡ 

k 

In the y direction, the gravitational force on the ball must be included as well 

as air resistance. Newton's second law yields the ODE given in yeq. 

> yeq:=diff(y(t),t,t)=-k*diff(y(t),t)-g; 

yeq := d2 

μ 

d 

y(t) =¡k 

dt2 dt y(t) 

 

¡ g 

Solving yeq with the dsolve command, subject to the initial condition y(0) = 

yb, _y(0) = V sin(Á), where yb is the ball's initial y-coordinate, and taking the 

rhs, yields y. 

> y:=rhs(dsolve(fyeq,y(0)=yb,D(y)(0)=V*sin(phi)g,y(t))); 

y := ¡ e(¡kt) (V sin(Á) k + g) 

¡ gt 

k 2 

k + yb k2 + V sin(Á) k + g 

k2 Evaluating y at t=tf , and equating to the fence height yf , yields the following 

formidable looking transcendental equation eq for the unknown angle Á. 

> eq:=eval(y,t=tf)=yf; 

(xb k + V cos(Á) ¡ xf k)(V sin(Á) k + g) 

eq := ¡ 

k2 V cos(Á) 

μ 

xb k + V cos(Á) ¡ xf k 

g ln 

V cos(Á) 

+ 

+ yb k2 + V sin(Á) k + g 

k 2 

k 2 

= yf 

To solve the transcendental equation for Á, thegivenvaluesxb =0m,yb =2m, 

xf =20m,yf =3:5m, V =15m/s,andg =9:8m/s 2 , are entered. The drag 

coe±cient is taken to be k =0:01 s ¡1 . 

> xb:=0: yb:=2: xf:=20: yf:=3.5: V:=15: g:=9.8: k:=0.01: 

The transcendental equation must be solved numerically. This is done in the 

following two command lines, two di®erent search ranges (in radians) being 

speci¯ed for Á to determine the two possible angles, ©1 and ©2, atwhichthe 

ball can be thrown to just clear the fence. Making use of the ditto operator, we 

convert the ¯rst angle to degrees. 

> Phi[1]:=fsolve(eq,phi,0..0.8); evalf(convert(%,degrees)); 

©1 := 0:6696475750 38:36797980 degrees 

> Phi[2]:=fsolve(eq,phi,0.8..Pi/2); 

©2 := 0:9693759172 

Without air resistance, it was previously found that the smaller of the two 

possible angles was 37.47 degrees, about 1 degree less than the value found 

above. It is left as an exercise for you to compare the values for the upper 

angle, with and without air resistance. 

To animate the ball, the ¯rst angle ©1 will be selected.


> phi:=Phi[1]: #select angle 

In this case, the time tf to reach the fence is calculated 

> tf:=evalf(tf); 

tf := 1:715218642 

to be about 1.72 s, slightly more than the 1.68 s without air resistance. 

Setting y = 0, we determine the time T for the ball to hit the ground, 

> T:=solve(y=0,t); 

T := 2:090175405; ¡0:1946627301 

the positive angle (¯rst solution here) being selected. 

> T2:=T[1]; #choose positive time 

T2 := 2:090175405 

The ball hits the ground after 2.09 seconds, compared to 2.06 seconds without 

air resistance. 

The fence is plotted as a quite thick (default color red) line. 

> fence:=plot([[xf,0],[xf,yf]],style=line,thickness=3): 

Using exactly the same syntax as in the introductory recipe, the motion of the 

ball is animated with the fence as background. 

> animate(pointplot,[[[x,y]],symbol=circle,symbolsize=14], 

t=0..T2,frames=200,background=fence,scaling=constrained); 

To see this animation, execute the recipe on your computer, click on the opening 

frame,andthenonthestartarrowintheMapletoolbar. 

PROBLEMS: 

Problem 3-7: Maximum drag coe±cient 

With all other parameters the same as in the text recipe, what is the maximum 

value of the drag coe±cient k such that the ball just clears the fence? What 

are the two corresponding initial angles? 

Problem 3-8: How high? 

If the drag coe±cient k is equal to 0:1s ¡1 , how high can the fence be for the 

ball to just clear it, all other parameters the same as in the text recipe? What 

are the two initial angles in this case? 

Problem 3-9: A falling raindrop 

For a sphere of diameter d meters moving in air, the approximate value of the 

constant k in Stokes's linear resistance law, ~ Fdrag = ¡k~v newtons, is given by 

k =1:55 £ 10 ¡4 d. For a small spherical raindrop (density ½ =103kg/m3 , 

d =10 ¡4 m) falling from rest, determine the distance through which it falls in 

t seconds and its velocity then. Plot the distance and velocity separately over 

atimeintervalt = 0 to the time at which the velocity reaches 99 percent of its 

terminal (asymptotic) value.


3.2.2 Meet Mr. Laplace 

The weight of evidence for an extraordinary claim must be 

proportioned to its strangeness (known as the principle of Laplace). 

Pierre-Simon Laplace, French mathematician (1749{1827) 

Pierre-Simon Laplace was one of the most in°uential scientists in history. Indeed, 

he has been referred to as the Newton of France. Not only did he make 

outstanding contributions to astronomy, for example, mathematically investigating 

the stability of the solar system, he developed probability theory as a 

coherent body of knowledge from a set of miscellaneous problems, and played a 

leading role in forming the modern discipline of mathematical physics. One of 

his most important contributions to the latter is the Laplace transform, which 

canbeusedtosolvelinearODEs. 

The Laplace transform of a function f(t) isde¯nedas 

L(f(t)) ´ F (s) = 

Z 1 

0 

f(t)e ¡st dt: 

By integrating by parts and assuming that e ¡stf(t) ! 0ast !1,itiseasy 

to show that 

³ ´ 

³ ´ 

L f(t) _ = sF(s) ¡ f(0) and L Äf(t) = s 2 F (s) ¡ sf(0) ¡ _ f(0): 

The Laplace transform method of solving a linear ODE (system) with constant 

coe±cients is to Laplace transform the ODE, solve the resulting algebraic 

equation for F (s), and then perform the inverse Laplace transform to obtain 

the solution f(t). In the era before computer algebra existed, tables of Laplace 

transforms and their inverses were almost as common as integral tables, and 

one of their main uses was in solving linear ODEs. With the Maple computer 

algebra system, these tables are obsolete, since Maple has an integral transform 

library package that includes the Laplace transform and its inverse. 

The use of this package is now demonstrated in the ¯rst of the two recipes 

that follow. It is desired to solve the following forced oscillator ODE for x(t): 

Äx +4_x +13x =sin(t); with x(0) = 1 and _x(0) = ¡5: 

Entering the integral transform package and using a semicolon to display 

its contents, we see that it contains the Laplace transform (laplace) command 

and its inverse (invlaplace). 

> restart: with(inttrans); 

[addtable; fourier; fouriercos; fouriersin; hankel; hilbert; invfourier; 

invhilbert; invlaplace; invmellin; laplace; mellin; savetable] 

The forced oscillator ode is entered, 

> ode:=diff(x(t),t,t)+4*diff(x(t),t)+13*x(t)=sin(t); 

μ μ 

2 d d 

ode := x (t) +4 x (t) +13x(t) =sin(t) 

dt2 dt


along with the initial conditions. 

> x(0):=1: D(x)(0):=-5: 

The alias commandisusedtoreplacelaplace(x(t); t; s), which would otherwise 

appear in the output, with the symbol F .ThatistosayFwill be the Laplace 

transform of x(t), the transformed independent variable being s. 

> alias(F=laplace(x(t),t,s)): 

The Laplace transform of ode is performed, 

> eq:=laplace(ode,t,s); 

eq := s 2 F +1¡ s +4sF +13F = 1 

s 2 +1 

and the algebraic equation eq solved for F . 

> F:=solve(eq,F); 

F := 

s (¡s + s 2 +1) 

s 4 +14s 2 +4s 3 +4s +13 

The inverse Laplace transform of F is calculated, yielding the analytic solution 

(labeled X) of the forced oscillator ODE. 

> X:=invlaplace(F,s,t); 

X := ¡ 1 3 1 

cos(t)+ sin(t)+ (123 cos(3 t) ¡ 121 sin(3 t)) e(¡2 t) 

40 40 120 

The steps carried out above mimic a hand calculation. An easier way to derive 

exactly the same form of x(t) istousethedsolve commandwiththeLaplace 

transform option, i.e., include method=laplace. 

> dsolve(fode,x(0)=1,D(x)(0)=-5g,x(t),method=laplace); 

x (t) =¡ 1 3 1 

cos(t)+ sin(t)+ (123 cos(3 t) ¡ 121 sin(3 t)) e(¡2 t) 

40 40 120 

The implementation of the Laplace transform method with Maple can be extended 

to systems of linear ODEs, as is demonstrated in the following problem, 

a problem that is quite challenging to do by hand. 

Masses m1 and m2, with equilibrium positions at x=2 andx=5, arefreeto move horizontally on a smooth surface (the x-axis). The mass m1 is connected 

to a ¯xed wall on its left by a linear spring (spring constant k) andonitsright 

to m2 with an identical spring. A driving force F =f sin(!t)actstotheright 

on m2. A °uid resistance is present, given by Stokes's drag law, Fdrag =¡av. 

(a) Derive the governing ODEs for the displacements x1(t) andx2(t) ofm1 

and m2 from equilibrium. Taking m1 =2, m2 =1, k =1, a =1, ! =2, 

f =2, x1(0)=1, x2(0)=0, _x1(0)=0, and _x2(0) =0, solve the ODEs using 

the Laplace transform option in the dsolve command. 

(b) Extract the steady-state and transient parts of x1 and x2 separately and 

plot them. Discuss the results. 

(c) Animate the motion of m1 and m2 about their equilibrium positions over 

a time interval for which the transients become small.


Loading the plots package, we use the alias command. Entering x1, x2, m1, 

m2 will produce the subscripted quantities x1, x2, m1, andm2in the output. 


> alias(x[1]=x1,x[2]=x2,m[1]=m1,m[2]=m2): 

When m2 is displaced from equilibrium by amount x2 at time t in, say, the 

positive x-direction, it exerts a force on m1 through the connecting spring. If 

m1 is displaced from equilibrium by amount x1(t), the force exerted on m1 by 

m2 is k (x2(t) ¡ x1(t)). The spring connected to the wall will exert a force 

¡kx1(t) in the opposite direction. Including damping, Newton's second law 

yields the ODE ode1 governing the displacement x1(t) ofm1. 

> ode1:=m1*diff(x1(t),t,t) 

=k*(x2(t)-x1(t))-k*x1(t)-a*diff(x1(t),t); 

μ 

μ 2 d d 

ode1 := m1 x1(t) = k (x2(t) ¡ x1(t)) ¡ k x1(t) ¡ a 

dt2 dt x1(t) 

 

The mass m1 will exert an equal and opposite force on m2 through the connecting 

spring, i.e., ¡k (x2(t) ¡ x1(t)). Including the damping and driving forces, 

the displacement x2(t) ofm2 is given by the ODE ode2 . 

> ode2:=m2*diff(x2(t),t,t) 

=-k*(x2(t)-x1(t))-a*diff(x2(t),t)+f*sin(omega*t); 

μ 

μ 2 d d 

ode2 := m2 x2(t) = ¡k (x2(t) ¡ x1(t)) ¡ a 

dt2 dt x2(t) 

 

+ f sin(!t) 

One has a system of two coupled, linear, second-order, inhomogeneous ODEs. 

To solve them, let's ¯rst enter the given parameter values, 

> m1:=2: m2:=1: k:=1: a:=1: omega:=2: f:=2: 

and the initial conditions. 

> ic:=x1(0)=1,x2(0)=0,D(x1)(0)=0,D(x2)(0)=0: 

The set of ODEs is analytically solved for x1(t) andx2(t), subject to the initial 

conditions, using the dsolve command with the Laplace transform option. 

The same answer could be obtained by mimicking the hand calculation in the 

previous example. This is left as a problem. 

> sol:=dsolve(fode1,ode2,icg,fx1(t),x2(t)g,method=laplace); 

sol := fx1(t) = 36 

26 

cos(2 t)+ sin(2 t) 

493 493 

¡ 1 X 

( (56380 + 339955 ® + 193087 ® 

194242 

®=%1 

2 + 188250 ® 3 ) e ( ®t) )g; 

fx2(t) =¡ 164 228 

cos(2 t) ¡ sin(2 t) 

493 493 

¡ 1 X 

( (107923 + 533529 ® + 249954 ® 

194242 

®=%1 

2 + 293736 ® 3 ) e ( ®t) )g 

%1 := RootOf(2 Z 4 +3 Z 3 +5 Z 2 +3 Z +1)


The sine and cosine terms in the solution survive as t ! 1 and represent 

the steady-state parts, while the exponential terms decay to zero and thus 

correspond to the transient parts of the solution. The summation is over the four 

roots of the quartic polynomial given in the subexpression %1. To manipulate 

the solution further, it is now assigned. 

> assign(sol): 

Using the remove command, the exponential terms are removed from x1(t) and 

x2(t) to yield the steady-state parts, x1ss and x2ss, separately. 

> x1ss:=remove(has,x1(t),exp); x2ss:=remove(has,x2(t),exp); 

x1ss := 36 

26 

228 

cos(2 t)+ sin(2 t) x2ss := ¡164 cos(2 t) ¡ sin(2 t) 

493 493 493 493 

The select command is used to extract the exponential terms from x1(t) and 

x2(t), thus producing the transient parts separately. 

> x1tr:=select(has,x1(t),exp): x2tr:=select(has,x2(t),exp): 

Because the roots of a quartic polynomial must be found, the numerical °oatingpoint 

evaluation command is applied to each of the transient parts. The complex 

evaluation command is then used to simplify the outputs, which are given 

by x1tr and x2tr. 

> x1tr:=evalc(evalf(x1tr)); x2tr:=evalc(evalf(x2tr)); 

x1tr := 0:5591296681 e (¡0:3923340189 t) cos(0:3903466128 t) 

+0:9914981558 e (¡0:3923340189 t) sin(0:3903466128 t) 

+0:3678480196 e (¡0:3576659811 t) cos(1:226572647 t) 

¡ 0:1154210539 e (¡0:3576659811 t) sin(1:226572647 t) 

x2tr := 0:6802842310 e (¡0:3923340189 t) cos(0:3903466128 t) 

+1:721342502 e (¡0:3923340189 t) sin(0:3903466128 t) 

¡ 0:3476270302 e (¡0:3576659811 t) cos(1:226572647 t) 

+0:3225193055 e (¡0:3576659811 t) sin(1:226572647 t) 

You might have thought our earlier identi¯cation of the exponential terms as 

the transient contribution a bit premature, but now one can clearly see that 

they decay to zero as t goes to in¯nity. 

The steady-state and transient contributions are now plotted separately. 

Lists are used here so that the steady-state and transient parts of x1(t) are 

colored red on the computer screen, while the corresponding parts for x2(t) are 

colored blue. 

> plot([x1ss,x2ss],t=0..20,color=[red,blue],tickmarks=[3,4]); 

> plot([x1tr,x2tr],t=0..20,color=[red,blue],tickmarks=[3,4]); 

The black-and-white versions of the two pictures are shown in Figure 3.4, the 

steady-state contributions on the left, the transient contributions on the right.


0.4 

0.2 

0 

–0.2 

–0.4 

10 t 20 

1 

0.8 

0.6 

0.4 

0.2 

0 

10 t 20 

Figure 3.4: Steady-state (left) and transient (right) contributions. 

In the steady-state picture, the shorter of the two oscillatory curves corresponds 

to x1(t), the taller one to x2(t). The motion of mass m1 is approximately 180 

degrees out of phase with m2. The period of oscillation of both masses is 

identical with that of the driving force. 

In the transient picture, the curve that rises above 1 is x2(t). Both curves 

decay to essentially zero in less than 20 time units. 

To animate the solution, the complete (steady-state plus transient) timedependent 

displacements are added to the given equilibrium coordinates. 

> X1:=2+x1ss+x1tr: X2:=5+x2ss+x2tr: 

The motion of the two masses is now animated, each mass being represented 

by a size-10 (default color red) circle. To produce motion along the horizontal 

axis, the horizontal coordinate of each mass is entered in a list with the vertical 

coordinate set equal to zero. The time range is such that the transient parts 

vanish, revealing the steady-state motion. You can take a larger time interval 

if you desire. To obtain a smooth animation, 200 frames are used. 

> animate(f[X1,0],[X2,0]g,t=0..20,frames=200,style=point, 

symbol=circle,symbolsize=20,tickmarks=[3,2]); 

Execute the command on your computer to see the animated motion. 

PROBLEMS: 

Problem 3-10: Hand calculation 

Making use of the integral transform package, mimic a hand calculation to 

derive the solution for the second example. 

Problem 3-11: Transform package approach 

Use the integral transform package and the Laplace transform method to solve 

the following ODEs. Con¯rm the solutions using the method= laplace option 

in the dsolve command. Plot each solution over a suitable time range that 

includes the steady-state regime. Extract the transient part of the solution and


estimate the time interval over which it lasts. 

(a) _y +5y =cos(t)+e ¡t ; y(0) = 1; 

(b) Äy ¡ 5_y +6y =0; y(0) = 2; _y(0) = 5; 

(c) 9Äy +6_y + y =5; y(0) = 6; _y(0) = 1; 

(d) 5Äy +2_y + y =sin 2 (t); y(0) = ¡3; _y(0) = 1; 

(e) _x = ¡4 x + y +3; _y = ¡4 x ¡ 4 y +5; x(0) = 1; y(0) = ¡1; 

(f) Äy +_y =2cos4 (t) e ¡t ; y(0) = 0; _y(0) = ¡2; 

(g) Äy +2_y = t cos(t) e ¡t + t2 cos(2 t) e ¡2 t ; y(0) = 0; _y(0) = 0. 

3.2.3 Jennifer's Formidable Series 

Isn't life a series of images that change as they repeat themselves? 

Andy Warhol, American pop artist (1928{1987) 

Jennifer is currently teaching a course on ordinary di®erential equations at 

MIT. As part of an assignment, she has assigned the following problem to be 

done with the Maple computer algebra system. 

Consider the following second-order linear ODE with variable coe±cients, 

2 xy 00 (x)+(1¡ 2 x) y 0 (x) ¡ x 2 y(x) =0; y(0) = 1;y 0 (0) = 0: 

(a) Show that Maple is unable to produce a closed-form analytic solution. 

(b) Derive a procedure for numerically determining y(x) andy0 (x) atanarbitrary 

x value. Evaluate y and y 0 at X =4. 

(c) Derive a series solution for y(x), keeping su±cient terms to achieve 1% 

agreement with the numerical answer at X. 

(d) Plot the series derived in part (c) and the numerical solution together in 

thesame¯gureovertherangex =0toX. 

Vectoria, who is in Jennifer's class, has submitted the following recipe to solve 

the problem. In order to have the ODE appear on the computer screen in the 

prime notation used in the problem wording, she loads the PDEtools package, 

which contains the declare command for achieving this notation. 

> restart: with(plots): with(PDEtools): 

As indicated by the output, the following declare command will cause y(x) to 

be displayed as y and introduce primes for the derivatives with respect to x. 

> declare(y(x),prime=x): 

y(x) will now be displayed as y 

derivatives with respect to xoffunctionsofonevariable 

will now be displayed with 0


On entering the given di®erential equation, Vectoria observes that the output 

in de is expressed in terms of the prime notation. 

> de:=2*x*diff(y(x),x,x)+(1-2*x)*diff(y(x),x)-x^2*y(x)=0; 

de := 2 x y 00 +(1¡ 2 x) y 0 ¡ x2 y =0 

The initial condition is speci¯ed, as well as the value of X and a parameter d 

that will control the spacing of the numerical points in the graph. 

> ic:=y(0)=1,D(y)(0)=0: X:=4: d:=8: 

Using dsolve, an attempt is made to analytically solve de for y(x). 

> dsolve(fde,icg,y(x)); #no analytic solution 

No output is generated, Maple being unable to provide an analytic answer. Although 

she has now omitted it, Vectoria had included infolevel[dsolve]:=5; 

prior to the dsolve command. On doing so she was able to view a very lengthy 

list of unsuccessful ODE solving methods tried by Maple. If you wish to see 

these attempts, include the above infolevel command. 

To generate a procedure for numerically evaluating y and y 0 at arbitrary x, 

Vectoria includes the numeric option in dsolve and requests that the output 

be given as a \list procedure." 

> numsol:=dsolve(fde,icg,y(x),numeric,output=listprocedure); 

numsol := [x =(proc(x) ::: end proc); y=(proc(x) ::: end proc); 

y 0 =(proc(x) ::: end proc)] 

Evaluating y(x) withthenumsol procedure will generate a numerical answer 

for y at x when the value of x is supplied as the argument of Ynum de¯ned 

below. Then, entering Ynum(X) gives the numerical value of y at X (4, here), 

the answer being given to 18 digits, more than the \normal" 10-digit accuracy. 

> Ynum:=eval(y(x),numsol): Ynum(X); 

41:1065491999371986 

Thus, y ¼ 41 at x = X = 4. A similar command structure is used to numerically 

evaluate the derivative of y, yielding y 0 ¼ 76 at X. 

> Ynumder:=eval(diff(y(x),x),numsol): Ynumder(X); 

75:6306643796735756 

Vectoria will use a do loop to systematically calculate series solutions of de as 

a function of the order n, terms of order xn and larger being neglected. The 

do loop will include a conditional statement that will terminate the loop when 

the absolute percentage di®erence j100 (ynum(X)¡yseries(X))=ynum(X)j drops 

below one percent. As a \seed number" to implement the conditional statement, 

she calculates the absolute percentage di®erence between the starting value of 

y(0) = 1 and the numerical value of y at x = X =4. 

> percent[0]:=abs(evalf(100*(Ynum(X)-1)/Ynum(X))); #seed 

percent 0 := 97:56729762 

The do loop runs from 1 to 50, the number 50 being chosen to be large enough 

to achieve a percentage error below one percent. The loop will calculate the


series solution to order n as long as the percentage error in the previous order 

is above 1 percent. 

> for n from 1 to 50 while percent[n-1]>1 do 

The order of the series is taken to be n. The error in the series is of order x n , 

terms of this order and larger being neglected. If the order is not speci¯ed, the 

default order is 6. 

> Order:=n: 

The di®erential equation is solved for y(x), subject to the initial condition, 

with the series option included. The dsolve command uses several methods 

when trying to ¯nd a series solution to an ODE or ODE system. When initial 

conditions are given, such as in this problem, the series is calculated at the 

given point. Otherwise, the series is calculated at the origin. The ¯rst method 

used is a Newton iteration, the second involves a direct substitution to generate 

a system of equations which must be solved, while the third is the method of 

Frobenius discussed in standard ODE texts. Useful references may be found by 

entering the topic dsolve,series in Maple's Topic Search. 

> sol[n]:=dsolve(fde,icg,y(x),'series'); 

To remove the \order of" term that appears in the output of the last command 

line, the convert( ,polynom) command is applied to the right-hand side of the 

nth order solution. The largest term in the series for Yn is x n¡1 . 

> Y[n]:=convert(rhs(sol[n]),polynom); 

The absolute percentage error at x = X is calculated for each order, 

> percent[n]:=abs(evalf(100*(Ynum(X)-eval(Y[n],x=X))/Ynum(X))); 

and the do loop ended with a colon to suppress the lengthy output. 

> end do: 

The loop stops at N = n ¡ 1, in this case order 17, for which the percentage 

error (to 3 digits) is 0.696 percent. As you can check by either replacing the 

colonwithasemicoloninthedolooportakingN = n¡ 2, the percentage error 

in order 16 is 1.31 percent. Order 17 is the ¯rst order for which the percentage 

error drops below one percent. The corresponding polynomial representation 

of the solution is given by Y17. 

> N:=n-1; evalf(percent[N],3)*percent; Y[N]:=Y[N]; 

N := 17 0:696 percent 

Y17 := 1 + 1 

15 x3 + 1 

70 x4 + 4 

1575 x5 + 29 

20790 x6 + 43 

126126 x7 61 

+ 

1001000 x8 

16013 

+ 

1033782750 x9 498307 

+ 

152770117500 x10 7710323 

+ 

14115958857000 x11 

2117887 

+ 

21250934424720 x12 96270661 

+ 

5534097506437500 x13 121568353967 

+ 

46021554863534250000 x14 

1141494706493 

+ 

2859910909376771250000 x15 2799853112879 

+ 

47283860368362618000000 x16


Using the sequence command, we graph the numerical solution from x =0 to 

X in steps of size 1=d. Each numerical point is plotted as a size-16 blue circle. 

> gr1:=plot([seq([i/d,Ynum(i/d)],i=0..d*X)],style=point, 

symbol=circle,symbolsize=16,color=blue): 

The series solution is plotted as a solid thick curve over the same range of x, 

> gr2:=plot(Y[N],x=0..X,thickness=2): 

and the two graphs superimposed to produce Figure 3.5. 

> display(fgr1,gr2g,labels=["x","y"]); 

40 

30 

y 

20 

10 

0 1 2 x 3 4 

Figure 3.5: Comparison of numerical (circles) and series (curve) solutions. 

Of course, because of the imposition of the conditional statement, the ¯t of the 

series solution to the numerical points is very good. 

Vectoria hopes that Jennifer will like her solution to this problem. By 

modifying Vectoria's recipe, you should be able to solve the following problems, 

which also appeared on Jennifer's assignment. 

PROBLEMS: 

Problem 3-12: What's special about this ODE? 

Consider the following linear second-order ODE: 

xy 00 (x)+y 0 (x)+xy(x) =0; y(0) = 1;y 0 (0) = 0: 

(a) Show that Maple is able to produce a closed-form analytic solution. Identify 

the ODE and the \special" function that appears in the solution. 

(b) Derive a series solution for y(x), keeping su±cient terms to achieve 1% 

agreement with the analytic answer at x = X =10. 

(c) Plot the series derived in part (b) and the analytic solution together in the 

same ¯gure over the range x =0toX, representing the analytic answer 

by appropriately sized and spaced circles.


Problem 3-13: Another ODE 

Consider the following linear second-order ODE: 

xy 00 (x)+2y 0 (x)+xy(x) =0; y(0) = 1;y 0 (0) = 0: 

(a) Show that Maple is able to produce a closed-form analytic solution. 

(b) Derive a series solution for y(x), keeping su±cient terms to achieve 1% 

agreement with the analytic answer at x = X =20. 

(c) Plot the series derived in part (b) and the analytic solution together in the 

same ¯gure over the range x =0toX, representing the analytic answer 

by appropriately sized and spaced circles. 

Problem3-14: Variationonthetextproblem 

Carry out the same steps as in Jennifer's text problem for the ODE 

y 00 (x) ¡ xy 0 (x) ¡ 1=0; y(0) = 1;y 0 (0) = 0: 

3.3 Special Function Models 

Many boundary-value problems of mathematical physics lead to variable-coe±cient 

ODEs of the Sturm{Liouville (S{L) form, 

· 

d 

p(x) 

dx 

dy(x) 

¸ 

¡ q(x) y(x) =¡¸w(x) y(x); (3.2) 

dx 

where p(x), q(x), and w(x) are real functions of the spatial variable x, ¸ is a 

real constant, and certain conditions on y and its ¯rst derivative are speci¯ed 

at the boundaries. Often the only solution to such a problem is the trivial 

solution y(x) = 0, but for special values (called the eigenvalues) ¸1, ¸2, ::: of 

¸, nontrivial solutions (called the eigenfunctions) y1, y2, ::: can occur. 

For certain choices of the functions p(x), q(x), w(x), and ¸, the well-known 

Bessel, Legendre, Hermite, Chebyshev, Mathieu, etc., ODEs result. The solutions 

of these ODEs, which take the form of in¯nite series or ¯nite polynomials 

in x, have been historically labeled as special functions, to distinguish them 

from such ordinary functions as the sine, cosine, log, and exponential functions. 

Special-function solutions can also occur for initial-value problems with x 

replaced with the time t. The systematic study of the S{L ODE and special 

functions and their properties is beyond the scope of this text. The interested 

reader is referred to standard mathematical texts (e.g., [MW71], [Boa83]) for 

the theoretical aspects and to [Enn05] for recipes on applications. Here, we shall 

be content to introduce two of the more famous members of the S{L family and 

then look at two special-function models.

3.3. SPECIAL FUNCTION MODELS 131 

3.3.1 Jennifer Introduces a Special Family 

\That's a great deal to make one word mean," Alice said in a 

thoughtful tone. \When I make a word do a lot of work like that," 

said Humpty Dumpty, \I always pay it extra." 

Lewis Carroll, Through the Looking Glass, 1872 

Although we shall not emulate Humpty Dumpty and pay the word \special" 

extra for making it work so hard in this section, we would tell Alice, if she were 

here, that the phrase \special" function does encompass a great many functions 

with their associated mathematical properties. The Handbook of Mathematical 

Functions by Abramowitz and Stegun [AS72], for example, contains over 

1000 pages dealing with the properties of special functions. This handbook, 

produced by the U.S. National Bureau of Standards, is one of the standard 

reference books for these functions and has been reprinted many times since it 

was ¯rst issued. 

We have asked our MIT mathematician friend Jennifer to introduce two of 

the more prominent members of this special family of functions. 

Letting the letter L stand for ¸, Jennifer forms a functional operator SL for 

generating speci¯c cases of the Sturm{Liouville ODE when the forms of p, q, 

w, andLare supplied as arguments. 


> SL:=(p,q,w,L)->diff(p*diff(y(x),x),x)-q*y(x)=-L*w*y(x); 

μ μ μ 

d d 

SL := (p; q; w; L) ! p 

dx dx y(x) 

 

¡ q y(x) =¡Lwy(x) 

A second functional operator sol is introduced to provide the general analytic 

solution y(x) of a speci¯ed ODE. 

> sol:=ode->dsolve(ode,y(x)): 

Now Jennifer will make two choices for p, q, w, andLthat lead to probably 

the best-known two members of the family of special functions. Taking p = x, 

q = ¡x, w =1=x, andL = ¡m2 as arguments in the Sturm{Liouville operator 

produces ode1 . 

> ode1:=SL(x,-x,1/x,-m^2); 

μ 

d 

ode1 := 

dx y(x) 

μ 

2 d 

+ x y(x) + x y(x) = 

dx2 m2 y(x) 

x 

To illustrate that Maple is able to identify this ODE, Jennifer loads the DEtools 

package and applies the odeadvisor command to ode1 . 

> with(DEtools): odeadvisor(ode1); 

[ Bessel] 

So, ode1 is Bessel's di®erential equation, whose general solution y(x) 

> sol(ode1); 

y(x) = C1 BesselJ(m; x)+ C2 BesselY(m; x)


is a linear combination of a Bessel function of the ¯rst kind (BesselJ(m; x)) and 

of the second kind (BesselY(m; x)), with C1 and C2 arbitrary constants. In 

traditional mathematical notation, the Bessel functions are written as Jm(x) 

and Ym(x), m being referred to as the order of the Bessel function. Highlighting 

BesselJ(m; x) orBesselY(m; x) on the computer screen will open a Help page 

with some (but not much) information about these special functions. 

If you wish to know what other special functions are known to Maple, execute 

the following inifcns (initially known mathematical functions) command 

line and use the hyperlinks provided in the lengthy list of functions. Remember 

to close the Help page when you're done. 

> ?inifcns; 

The Bessel functions are actually in¯nite Frobenius power series solutions of 

the Bessel ODE, i.e., a series expansion about x = 0 is sought of the form 

y(x) = P1 m=0 cm xm+s . The allowed values of the index s and the forms of the 

coe±cients cm are determined5 by substituting y(x) intotheBesselODE. 

The series command can be used to obtain the form of the Frobenius series. 

For example, since the most commonly occurring order in physical problems 

involves positive-integer values of m, Jennifer calculates the Frobenius series of 

Jm(x) form = 0 to 2, terms of order x8 and higher being omitted here. 

> seq(J[m]=series(BesselJ(m,x),x=0,8),m=0..2); 

J0 =1¡ 1 

4 x2 + 1 

64 x4 ¡ 1 

2304 x6 +O(x8 ); 

J1 = 1 1 

x ¡ 

2 16 x3 + 1 

384 x5 ¡ 1 

18432 x7 +O(x8 ); 

J2 = 1 

8 x2 ¡ 1 

96 x4 + 1 

3072 x6 +O(x8 ) 

Note that at x =0,wehaveJ0 =1andJ1 = J2 =0. Asiseasilycon¯rmedby 

increasing the range of m in the seq command, Jm(0) = 0 for m =3; 4 ::: 

Looking at the above truncated series does not convey much of an idea of 

the shapes of J0(x), J1(x), and J2(x). Jennifer will now plot these functions 

and attach appropriate identifying labels to each curve. First, in the graph gr1 

she plots Jm(x) overtherangex =0to20form =0to2. 

> gr1:=plot(fseq(BesselJ(m,x),m=0..2)g,x=0..20,thickness=2): 

She uses the textplot command in gr2 to place appropriate labels on the 

Bessel function curves. The horizontal and vertical coordinates of the names 

(entered as strings) were determined by observing the picture generated by gr1. 

> gr2:=textplot([[1.5,0.9,"J0"],[3,0.6,"J1"],[5,0.4,"J2"]]): 

The two graphs are superimposed to yield Figure 3.6. 


The Jm(x) are oscillatory with amplitudes that decrease with increasing x. 

Unlike sin(x) orcos(x), the Jm(x) do not cross the horizontal axis at equal 

5 A recipe is given in Computer Algebra Recipes for Mathematical Physics [Enn05].


1 

0 

J0 

J1 

J2 

10 20 

x 

Figure 3.6: Bessel functions of the ¯rst kind of order 0, 1, and 2. 

intervals of x. The zeros of a particular Bessel function, for example J2, canbe 

approximately determined by clicking the mouse on the computer screen plot 

with the cursor placed at the location of a zero. The coordinates of the cursor 

are displayed in a small window at the top left of the computer screen. Moreaccurate 

values of the Bessel J2 zeros can be determined with the BesselJZeros 

command. The ¯rst six zeros are now determined to 4-digit accuracy. 

> Zeros:=evalf(BesselJZeros(2,0..5),4); 

Zeros := 0; 5:136; 8:417; 11:62; 14:80; 17:96 

What do the integer-order Bessel functions of the second look like? Jennifer ¯rst 

examines the analytic forms of the Frobenius series for Y0(x), Y1(x), and Y2(x), 

terms of order x8 and higher again being omitted. Because the expressions are 

lengthy, only Y0 isshownhereinthetext. 

> seq(Y[m]=series(BesselY(m,x),x=0,8),m=0..2); 

2(¡ln(2) + ln(x)) 

Y0 =( + 

¼ 

2 ° 

¼ )+ 

0 

B 

@¡ 1 ¡ln(2) + ln(x) ¡ 

¡ 

2 ¼ 

1 

1 

° 

+ 

2 2C 

A x 

¼ 

2 

0 

3 

B 

+ @¡ 64 

¡ ° 

32 

¼ 

+ 1 

32 

0 

¡ 

B 

+ @¡ 

11 ° 

+ 

6912 1152 

¼ 

1 

¡ln(2) + ln(x) C 

A x 

¼ 

4 

¡ 1 

1152 

1 

¡ln(2) + ln(x) C 

A x 

¼ 

6 +O(x8 ) 

The constant ° in the above output is the Euler{Mascheroni constant. It is


de¯ned as 

° = lim 

n!1 

Ã nX 

i=1 

! 

1 

¡ ln(n) ¼ 0:5772157: 

i 

Because of the ln(x) terms,Y0(x) divergesto¡1 at x = 0. A similar behavior 

occurs for Y1(x), Y2(x), etc. Because of this divergence, positive integer-order 

Bessel functions of the second kind are rejected in physical problems (where 

they often occur) in regions that include the origin. 

Again the truncated series do not reveal the shapes of Y0(x), Y1(x), etc. 

In the next three command lines, Jennifer plots Ym(x) over the range x =0 

to 20 for m =0; 1; 2 and adds identifying labels to the curves. Note that in 

the display command, she limits the vertical range of the ¯nal ¯gure to be 

between ¡1 and +1 so the oscillations are clearly seen. 

> gr3:=plot(fseq(BesselY(m,x),m=0..2)g,x=0..20,thickness=2): 

> gr4:=textplot([[2,.65,"Y0"],[3.7,.55,"Y1"],[6,.4,"Y2"]]): 

> display(fgr3,gr4g,view=[0..20,-1..1],tickmarks=[2,2]); 

1 

0 

–1 

Y0 

Y1 

Y2 

10 x 20 

Figure 3.7: Bessel functions of the second kind of order 0, 1, and 2. 

The Bessel functions Y0(x), Y1(x), and Y2(x) areshowninFigure3.7.Aswith 

the Bessel functions of the ¯rst kind, the zeros are not evenly spaced along the 

x-axis. Their locations can be determined with the fsolve command. 

As a second example of a special-function solution to a S{L ODE, Jennifer 

takes p = 1 ¡ x 2 , q = 0, w = 1, and L = n (n + 1) as arguments in the 

Sturm{Liouville functional operator and then derives the general solution. 

> ode2:=SL(1-x^2,0,1,n*(n+1)); sol(ode2); 

μ 

d 

ode2 := ¡2 x 

dx y(x) 

 

+(1¡ x2 μ 

2 d 

) y(x) = ¡n (n +1)y(x) 

dx2 y(x) = C1 LegendreP(n; x)+ C2 LegendreQ(n; x)


The general solution y(x) is a linear combination of a Legendre function of 

the ¯rst kind (LegendreP(n; x)) and of the second kind (LegendreQ(n; x)), n 

indicating the order. In problems of physical interest, n often takes on positiveinteger 

values, the Legendre functions of the ¯rst kind then being ¯nite polynomials, 

denoted by the symbol Pn(x), with x ranging6 from x = ¡1 to+1. 

Jennifer now derives the ¯rst few Legendre polynomials, 

> polynomials:=seq(P[n]=simplify(LegendreP(n,x)),n=0..4); 

3 x2 1 

polynomials := P0 =1;P1 = x; P2 = ¡ 

2 2 ;P3 = 5 

2 x3 ¡ 3 

2 x; 

P4 = 35 

8 x4 ¡ 15 

4 x2 + 3 

8 

and plots them with identifying labels added to the curves. 

> gr5:=plot(fseq(LegendreP(n,x),n=0..4)g,x=-1..1,thickness=2): 

> gr6:=textplot([[.3,.9,"P0"],[.5,.65,"P1"],[.2,-.55,"P2"], 

[-.5,.55,"P3"],[.15,.45,"P4"]]): 


P3 

1 

–1 

P4 

P2 

P0 

P1 

–1 1 

x 

Figure 3.8: The Legendre polynomials Pm(x) form =0to4. 

The Legendre polynomials are shown in Figure 3.8. They take on the values 

¡1 and +1 at the ends of the range. 

What do the Legendre functions of the second kind look like? For physical 

problems where x varies from ¡1 to +1, the following environment command 

line must be entered to pick out the correct solution branch. 

> _EnvLegendreCut:=1..infinity: 

TheanalyticformsofQm(x) are then calculated for m =0,1,and2. 

> seq(Q[m]=simplify(LegendreQ(m,x)),m=0..2); 

6 The variable x can be the cosine of a polar angle, whose range is from 0 to ¼ radians. So 

x =cosμ varies from ¡1 to+1.


Q0 = 1 

1 

ln(x +1)¡ 

2 2 ln(1 ¡ x); Q1 = 1 

1 

x ln(x +1)¡ x ln(1 ¡ x) ¡ 1; 

2 2 

Q2 = ¡ 1 

ln(x +1)+1 ln(1 ¡ x)+3 

4 4 4 x2 ln(x +1)¡ 3 

4 x2 3 x 

ln(1 ¡ x) ¡ 

2 

The Qm(x) formequal to zero or a positive integer diverge to in¯nity at x = ¡1 

and +1 and must therefore be rejected as being unphysical. 

PROBLEMS: 

Problem 3-15: A Sturm{Liouville equation 

Show that the following ODE is of the Sturm{Liouville form: 

y 00 (x) ¡ 2 xy 0 (x)+2ny(x) =0: 

Determine the general solution of this ODE and plot the included special functions 

over a suitable range of the independent variable x. 

Problem 3-16: Recurrence Formula 

Bessel functions of di®erent orders can be related through recurrence relations. 

Use Maple to prove the following Bessel function recurrence relations. 

² Jm¡1(x)+Jm+1(x) =(2m=x) Jm(x); 

² 4 d2Jm(x) dx 2 = Jm+2(x) ¡ 2 Jm(x)+Jm¡2(x). 

Problem 3-17: Bessel function solutions 

Find the general solution of each of the following ODEs in terms of Bessel 

functions and identify the order. Identify any other new functions that occur. 

(a) y00 (x)+y(x)= p x =0; 

(b) y00 (x)+xy(x) =0, Hint: Useconvert( ,Bessel); 

 

¡ x 8 

5 y(x) =0; 

(c) d2 

dx 2 

μ 

x 16 

5 d2 y 

dx 2 

Problem 3-18: Orthogonality 

An important general property that all solutions yn(x) of the Sturm{Liouville 

equation corresponding to a given ¸n possess is orthogonality. Provided that 

y(x) ory 0 (x) orp(x) vanishes at the endpoints a and b of the range (referred 

to as Sturm{Liouville boundary conditions), then 

Z b 

a 

w(x) ym(x) yn(x) dx =0; for m 6= n; (3.3) 

where w(x) isreferredtoastheweight function. Con¯rm the orthogonality 

property for Legendre functions of the second kind of orders 2 and 3 over the 

range x = ¡1 to +1. Which of the possible Sturm{Liouville boundary conditions 

is satis¯ed?


3.3.2 The Vibrating Bungee Cord 

Wisdom consists in being able to distinguish among dangers 

and make a choice of the least harmful. 

Niccolμo Machiavelli, from The Prince (1469{1527) 

While on vacation in Rainbow County, Jennifer is nervously watching her sister 

Heather getting ready to bungee jump from an abandoned railway bridge 

spanning a deep gorge. At present, the uniform elastic bungee cord has no one 

attached to its lower end, but is simply hanging vertically downward and displaying 

small vibrations transverse to its length. This reminds Jennifer that the 

famous mathematician Daniel Bernoulli ¯rst studied this vibrational problem 

nearly 300 years ago, and was able to solve it. To put a modern spin on an old 

problem, Jennifer recently showed her mathematical physics class a computer 

algebra derivation of the solution. 

With Jennifer's permission, we shall now reproduce her treatment. To aid 

in understanding the physics of the problem, a free-body diagram is shown in 

Figure 3.9, which shows the relevant forces on the bungee cord and introduces 

the notation that will be used. 

y 

y+dy 

T () y θ 

ψ ( y,t) ψ( 

y+dy,t ) 

dy 

ds 

dψ 

T( 

y+dy) 

Figure 3.9: Free-body diagram for a segment of vibrating bungee cord. 

Jennifer begins her recipe by loading the plots and plottools library packages, 

which will be needed for the animation of the transverse vibrations of the 

vertical cord.


> restart: with(plots): with(plottools): 

Measuring y vertically downward, the origin y = 0 is chosen to be at the top 

end of the bungee cord (length L) where it is attached to the bridge. If the cord 

has a mass density ² per unit length, and g is the acceleration due to gravity, 

the tension T in the cord is given by T = ²g(L ¡ y), which is entered. 

> T:=epsilon*g*(L-y); 

T := ²g(L ¡ y) 

At the top end (y = 0), the cord has a tension T = ²gL, because the cord must 

support its entire weight. At the bottom end, y = L andthetensioniszero. 

The cord is now allowed to undergo a small transverse displacement dÃ(y; t) 

at time t from the equilibrium position. The above expression for T will still 

be valid. To understand this, consider the cord segment of arc length 

ds = p (dy) 2 +(dÃ) 2 = p 1+(dÃ=dy) 2 dy 

showninFigure3.9. Lettingμ be the angle with the vertical, and noting that 

the forces still balance in the y-direction, then 

(T cos μ) y ¡ (T cos μ) y+dy =(²ds) g; (3.4) 

with cos μ = dy=ds. Assuming that dÃ=dy ¿ 1, one has ds ¼ dy and cos μ ¼ 1, 

so the vertical force equation (3.4) reduces to 

T (y) ¡ T (y + dy) =(²dy) g: (3.5) 

Taylor expanding the left-hand side of (3.5) to the ¯rst power in dy, and dividing 

both sides by dy, yields @T=@y = ¡²g. The expression T = ²g(L ¡ y) follows 

on integrating and evaluating the constant at y =0. 

In the Ã-direction, Newton's second law yields 

(T sin μ) y+dy ¡ (T sin μ) y =(²ds) @2Ã @t2 (3.6) 

with sin μ = dÃ=ds. For dÃ=dy ¿ 1, sin μ ¼ @Ã=@y, partial derivatives being 

used because one also has time as an independent variable. So (3.6) becomes 

μ 

T (y) @Ã 

μ 

¡ T (y) 

@y y+dy 

@Ã 

 

=(²dy) 

@y y 

@2Ã ; (3.7) 

@t2 or, on Taylor expanding the left-hand side for small dy, dividing by dy, and 

taking the limit dy ! 0, 

μ 

@ 

T (y) 

@y 

@Ã 

 

= ² 

@y 

@2Ã : (3.8) 

@t2 This equation of motion is now entered, 

> eq:=diff(T*diff(psi(y,t),y),y)=epsilon*diff(psi(y,t),t,t); 

μ 

μ μ 

2 

2 

@ 

@ @ 

eq := ¡²g Ã(y; t) + ²g(L ¡ y) Ã(y; t) = ² Ã(y; t) 

@y @y2 @t2


the expression for the tension being automatically substituted. 

The equation of motion is a linear partial di®erential equation. Solving linear 

PDEs is the subject matter of Chapters 5 and 6. The PDE may be converted 

into an ODE by assuming a solution of the form Ã(y;t) =X(y)cos(!t), with 

! taken to be a positive angular frequency. 

> psi(y,t):=X(y)*cos(omega*t); 

Ã(y; t) :=X (y)cos(!t) 

With the assumed solution automatically substituted, the resulting output of 

eq is divided by cos(!t) and expanded to produce the ODE eq2 . 

> eq2:=expand(eq/cos(omega*t)); 

μ μ μ 

2 

2 

d 

d d 

eq2 := ¡²g X (y) + ²g X (y) L ¡ ²g X (y) y = ¡² X (y) ! 

dy dy2 dy2 2 

The second derivative terms are collected in eq2 , 

> eq3:=collect(eq2,diff(X(y),y,y)); 

μ μ 2 d d 

eq3 := (²gL¡ ²gy) X (y) ¡ ²g 

dy2 dy 

and a general analytic solution to eq3 obtained. 

 

X (y) = ¡² X (y) ! 2 

> sol:=dsolve(eq3,X(y)); 

μ r 

μ r 

L ¡ y 

L ¡ y 

sol := X (y) = C1 BesselJ 0; 2 ! + C2 BesselY 0; 2 ! 

g 

g 

The general solution is a linear combination of zeroth-order Bessel functions of 

the ¯rst and second kinds with arbitrary constants C1 and C2 . The Bessel 

Y0 function diverges to ¡1 at y = L so must be removed on physical grounds. 

> X:=remove(has,rhs(sol),BesselY); 

μ r 

L ¡ y 

X := C1 BesselJ 0; 2 ! 

g 

To remove the arbitrary constant C1 ,theop command is used to select the 

second operand in X. 

> X:=op(2,X); 

μ r 

L ¡ y 

X := BesselJ 0; 2 

Taking the bungee cord length to be L = 30 meters, its density ² = 1 

2 kilogram/meter, 

and g =9:8 meter/second 2 ,theformofX is as follows: 

> L:=30: g:=9.8: epsilon:=1/2: X:=X; 

X := BesselJ(0; 2 p 3:061224489 ¡ 0:1020408163 y!) 

The transverse displacement X of the cord at y = 0 is zero, which is entered as 

a boundary condition, bc. 

g 

 

! 

> bc:=eval(X,y=0)=0; 

bc := BesselJ(0; 3:499271060 !) =0


The boundary condition implies that there are certain allowed frequencies (eigenfrequencies) 

! of vibration, each frequency corresponding to a possible normal 

mode solution. A general transverse motion of the cord will involve a linear combination 

of normal modes, the combination depending on the initial conditions 

imposedonthecord. 

The eigenfrequencies will now be determined. First, the op command is 

used to extract the ¯rst element of the second argument on the lhs of bc. 

> c:=op([2,1],lhs(bc)); 

c := 3:499271060 

Using the BesselJZeros command, the ¯rst 10 eigenfrequencies are numerically 

evaluated, and assigned. 

> freq:=seq(omega[n]=evalf(BesselJZeros(0,n)/c),n=1..10); 

freq := !1 =0:6872361462; !2 =1:577493717; !3 =2:473008739; 

!4 =3:369711645; !5 =4:266865142; !6 =5:164236682; 

!7 =6:061730076; !8 =6:959298411; !9 =7:856916100; 

!10 =8:754568007 

> assign(freq): 

A functional operator for creating the nth normal mode, with the time dependence 

included, is formed. 

> mode:=n->eval(X,omega=omega[n])*cos(omega[n]*t): 

As an explicit example, choosing n = N = 3 produces the normal mode X3 

corresponding to the third eigenfrequency. 

> N:=3: X[N]:=mode(N); 

X3 := BesselJ(0; 4:946017478 p 3:061224489 ¡ :1020408163 y)cos(2:473008739 t) 

The normal mode is now animated, the plot being rotated by ¡ ¼ 

2 radians using 

the rotate command, so that the animated string is hanging vertically in equilibrium, 

rather than being pictured as horizontal. The scaling is constrained, 

so that the transverse displacement is small compared to the length of the cord. 

The axes are also removed so that the small vibrations are better viewed. 

> rotate(animate(plot,[X[N],y=0..L],t=0..10,color=red, 

frames=100,scaling=constrained,axes=none),-Pi/2); 

The animation can be observed on your computer by executing the above command 

line, clicking on the computer plot, and on the start arrow in the tool 

bar. You should also look at the other transverse vibrational modes as well, by 

changing the value of N from 3 to some other integer between 1 and 10. Or, if 

you want, you could generate even higher-integer normal modes. 

While we have been looking at Jennifer's computer algebra treatment of the 

transverse vibrations of the cord, Heather has completed her bungee jump, the 

largest of the vertical oscillations bringing her head to within a meter of the 

river far below the bridge. Looking rather pale, Heather attempts to persuade 

Jennifer to also do a bungee jump, but Jennifer sensibly declines.


PROBLEMS: 

Problem 3-19: Vibrations of a weighted bungee cord 

Determine the transverse normal modes of vibration of the bungee cord if a 

mass M = 60 kg hangs from the lower end. Animate one of the normal modes. 

Problem 3-20: A sti®ening spring 

A vibrating spring is governed by the ODE 

Äx(t)+a _x(t)+kx(t) =0; 

where the spring coe±cient is k = b + cedt , and initially x(0) = A, _x(0) = 0. 

Determine the analytic form of x(t). Taking a = p 5, b = 1 

4 , c =1,d =1, 

and A =1,plotx(t) over a time interval for which the oscillations e®ectively 

vanish. Determine the threshold on ® for critical damping. 

Problem 3-21: The growing pendulum 

Consider a pendulum that consists of a point mass m at the bottom end of 

a light supporting rod of length L that is allowed to move in a vertical plane 

about a pivot point at its top end. Suppose that L increases at a steady rate, 

i.e., L = L0 + vt,wherev>0 is a constant speed and t the time. Letting 

the rod make an angle μ(t) with the vertical and neglecting drag, use Newton's 

second law to derive the ODE for small μ. Solve the ODE for L0 =1 meter, 

g =9:8 m/s2 , μ(0)=¼=6 radians, _ μ(0) = 0 radians/s, and v =0:5 m/s.Animate 

the motion of the pendulum arm (representing it as a thick line) over the time 

interval t=0 to 100 seconds, taking 100 frames and using constrained scaling. 

Problem 3-22: Onset of bending 

A thin, vertical steel wire of length L and circular cross section of radius a is 

clamped at its bottom and is free at its top. Let μ be the angular de°ection of 

the wire from the vertical at a distance y from the top. If L is small, the wire 

is stable in the vertical position, i.e., μ = 0 for all values of y. As L increases, 

there is a critical value Lcr beyond which the wire is unstable and will bend 

from the vertical. 

The relevant ODE for small angular displacements μ is 

d2μ dy2 = ¡c2 yμ; where c = 2 

r 

½g 

: (3.9) 

a Y 

Here ½ is the mass density, g is the acceleration due to gravity, and Y is Young's 

modulus. 

(a) Determine the solution of the ODE, subject to the boundary conditions 

μ =0 aty = L (wire clamped at bottom) and dμ=dy =0 at y =0 (wireis free at top). 

(b) Show that the onset of bending occurs at Lcr ¼ (2:8=c) 2=3 . 

(c) Determine Lcr for a steel (Y =2:1 £ 1011 N/m2 and ½=7800 kg/m3 )wire 

of radius 1 mm. Take g =9:8 m/s2 .


3.3.3 Mathieu's Spring 

In every tyrant's heart there springs in the end 

This poison, that he cannot trust a friend. 

Aeschylus, Greek dramatist, Prometheus, in Prometheus Bound (525{456 BC) 

Emile Mathieu (1835{1890), a French mathematician, introduced the special 

functions that bear his name as solutions to the problem of determining the 

transverse vibrations of an elliptically shaped elastic membrane. As illustrated 

in the following recipe, Mathieu functions also occur for the vibrations of a unit 

mass attached to the end of a linear spring having a time-dependent spring 

coe±cient k = a + b cos(ct), where a, b, andcare real constants. The mass is 

allowed to slide on a smooth horizontal surface, but experiences a viscous drag 

force given by Stokes's law, Fdrag = ¡¡(dx=dt), where x(t) is the displacement 

of the mass from equilibrium at time t and ¡ is the damping coe±cient. 

To see what mathematical approach Maple uses in solving the relevant ODE, 

the infolevel[dsolve] command is set to 2. 


The mathematical form of the spring coe±cient is entered. 

> k:=a+b*cos(c*t); 

k := a + b cos(ct) 

A functional operator ode is introduced to generate the ODE governing the 

motion of the unit mass for a given value of the damping coe±cient ¡. 

> ode:=Gamma->diff(x(t),t,t)+Gamma*diff(x(t),t)+k*x(t)=0; 

μ μ 

2 d d 

ode := ¡ ! x(t) +¡ x (t) + k x(t) =0 

dt2 dt 

An operator X is formed to analytically solve ode for x(t) for a speci¯ed ¡ value 

and the initial condition x(0) = A, _x(0) = 0. 

> X:=Gamma->rhs(dsolve(fode(Gamma),x(0)=A,D(x)(0)=0g,x(t))): 

As speci¯c parameter values, let's take a =10,b =2,c =2,andA =1. 

> a:=10: b:=2: c:=2: A:=1: 

The analytic solution for, say, ¡ = 2 is generated. 

> sol:=simplify(X(2)); #example 

Methods for second order ODEs: 


¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢ 

¡> Trying a Liouvillian solution using Kovacic's algorithm 

Trying a solution in terms of special functions: 

¡> Bessel 

¡> elliptic 

¡> Legendre 

¡> Whittaker


¡> hypergeometric 

¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢ 

¡> Mathieu 


3.3.4 Quantum-Mechanical Tunneling 

If we see light at the end of the tunnel, 

It'sthelightoftheoncomingtrain. 

Robert Lowell, American poet commenting on pessimism (1917{1977) 

The SchrÄodinger equation [Gri95] describing the one-dimensional motion of a 

particle of mass m moving in a potential V (x) attimet is 

¡ ¹h2 @ 

2 m 

2 ª(x; t) 

@x2 + V (x)ª(x; t) =I ¹h 

@ª(x; t) 

; (3.10) 

@t 

where ª(x; t)isthewave function, I = p ¡1, and ¹h=h=(2 ¼), where h is Planck's 

constant. Assuming a stationary-state solution, ª(x; t)=Ã(x) e ¡IEt=¹h , reduces 

the PDE (3.10) to the time-independent SchrÄodinger ODE, 

d2Ã(x) dx2 2 m 

+ 2 (E ¡ V (x)) Ã(x) =0 (3.11) 

¹h 

The probability of ¯nding the particle between x and x + dx at time t is 

jª(x; t)j2 dx = jÃ(x)j2 dx ´ P (x) dx. The total probability of ¯nding the particle 

somewhere in the range x = ¡1 to +1 is R +1 

P (x) dx =1. 

¡1 

Although classically impossible, a particle with energy E incident on a ¯nite 

potential barrier of maximum height Vmax >Ehas a nonzero probability of 

quantum-mechanically tunneling through the barrier to the opposite side. This 

is the basis of ®-emission, inwhich® particles are able to escape from the nucleus 

through the nuclear potential barrier. For a particle incident on a barrier 

with incident amplitude Ãinc, re°ected amplitude Ãrefl, andtransmittedamplitude 

Ãtrans, one can de¯ne the re°ection coe±cient R = jÃreflj2 =jÃincj2 and 

the transmission coe±cent T = p (E ¡ Vtrans)=(E ¡ Vinc) jÃtransj2 =jÃincj2 . 

The factor 2 m=¹h 2 can be removed from equation (3.11) by rescaling the 

spatial variable, letting x now stand for ( p 2 m=¹h) x. Usingthe\scaled"ODE, 

our goal in this recipe is to determine the re°ection and transmission coe±cients 

for a particle of energy E > 0 incident on a potential barrier V (x) =ax2 , 

with a = 1, located between x = 0 and x = L = 1. Outside the barrier, 

V (x) = 0, i.e., Vtrans = Vinc =0soT = jÃtransj2 =jÃincj2 . Then we will plot 

both coe±cients as a function of the energy E. 

The DEtools library package is loaded, because it contains the expsols command, 

which will enable us to generate exponential solutions to the SchrÄodinger 

ODE rather than the default sine and cosine solutions that would otherwise occur 

in the regions xL,whereV (x) =0. 


The parameter values a =1andL = 1 are entered, along with the assumed 

condition E>0. 

> a:=1: L:=1: assume(E>0): 

The scaled SchrÄodinger ODE is entered for V (x) =0, 

> de1:=diff(psi(x),x,x)+E*psi(x);


μ 

2 d 

de1 := Ã(x) + EÃ(x) 

dx2 and two independent exponential solutions obtained. 

> sol1:=expsols(de1,psi(x)); 

h 

sol1 := e (pExI) (¡I ;e p i 

Ex) 

Remembering that the spatial forms are to be mentally multiplied by the time 

factor e ¡IEt=¹h , the ¯rst term in sol1 corresponds to a plane wave traveling to 

the right, the second term to a plane wave traveling to the left. In region 1 

(x psi1:=sol1[1]+A*sol1[2]; 

Ã1 :=e (p ExI) + Ae (¡I p Ex) 

In region 3 (x >L), there is only a transmitted wave traveling to the right, 

which is now entered. The transmission coe±cient will be T = jBj 2 . 

> psi3:=B*sol1[1]; 

Ã3 :=Be (p ExI) 

In the barrier (0


There are four unknown coe±cients, A, B, C1 ,andC2, sofourequations 

are needed to determine them. Continuity of the amplitudes Ã1 andÃ2 at 

x = 0 is imposed in the ¯rst boundary (or matching) condition bc1 .Thelatter 

waveform must be carefully handled. One cannot simply evaluate Ã2 atx =0, 

because the error message \division by zero" will appear. One must take the 

limit as x approaches zero from the right, i.e., from x>0. The quantity ¡ 

appearing in the output is the gamma function. 

> bc1:=eval(psi1,x=0)=limit(psi2,x=0,right); 

C2 

bc1 := 1 + A = 

p ¼ 

μ 

3 E 

¡ ¡ 

4 4 

In the second boundary condition, Ã2 andÃ3 areequatedatx = L. 

> bc2:=eval(psi2,x=L)=eval(psi3,x=L); 

μ 

C2 HermiteH ¡ 1 

 

E 

+ ; 1 

2 2 

= Be (pEI) μ 

E 1 

bc2 := C1 WhittakerM ; ; 1 + 

4 4 

E 

2 

(¡1=2+ 2 ) e (1=2) 

Because the second derivative is ¯nite, the ¯rst derivative (slope) of the wave 

formsmustbecontinuousatx = 0. This condition is imposed in bc3 ,the 

limiting process again being used for the derivative dÃ2=dx. 

> bc3:=eval(diff(psi1,x),x=0)=limit(diff(psi2,x),x=0,right); 

bc3 := p EI¡ A p EI = 

μ 

5 E 

E 

C1 ¡ ¡ 2 

(¡1=2+ 2 

4 4 

) E 

¡ C2 2 

(¡3=2+ 2 ) p E 

¼ + C2 2 

(¡3=2+ 2 ) p ¼E 

μ 

5 E 

E 

¡ ¡ 2 

(¡1=2+ 2 

4 4 

) 

The slope continuity condition is also imposed at x = L (lengthy output not 

shown here). 

> bc4:=eval(diff(psi2,x),x=L)=eval(diff(psi3,x),x=L); 

The four boundary conditions are solved for the four unknown coe±cients, and 

the solution is assigned. 

> sol3:=solve(fbc1,bc2,bc3,bc4g,fA,B,_C1,_C2g): assign(sol3): 

The re°ection and transmission coe±cients are calculated, the very lengthy 

expressions being suppressed with line-ending colons. Noting that the amplitudes 

are complex, these coe±cients are given by R = jAj2 = A £ A ¤ and 

T = jBj2 = B £ B ¤ , where the asterisk denotes the complex conjugate. The 

conjugate command is used to enter the complex conjugates. 

> R:=A*conjugate(A): T:=B*conjugate(B): 

The re°ection and transmission coe±cients are now plotted over the energy 

range E = 0 to 1, the two curves being colored blue and red on the computer 

screen. The black-and-white version is shown in Figure 3.10.


> plot([R,T],E=0..1,color=[blue,red],tickmarks=[3,3], 

view=[0..1,0..1]); 

1 

0.8 

0.6 

0.4 

0.2 

0 

0.2 0.4 E 0.8 1 

Figure 3.10: Re°ection and transmission coe±cients versus energy E. 

The transmission coe±cient rapidly increases from 0 and approaches 1 as E 

is increased. Conversely, the re°ection coe±cient rapidly decreases with increasing 

E. By energy conservation, the sum of the re°ection and transmission 

coe±cients should sum to 1 for all values of the energy E. From the ¯gure, this 

appears to be the case, the con¯rmation being left for you as a problem. Note 

that the re°ection coe±cient is not equal to zero at E =1,eventhoughthe 

energy of the incoming particle is equal to that at the top of the barrier. 

PROBLEMS: 

Problem 3-25: R+T=1 

Con¯rm that the sum of the re°ection and transmission coe±cients is equal to 

1forallE. You may do this either graphically or analytically. 

Problem 3-26: Equality 

Determine the energy at which the re°ection and transmission coe±cients are 

equal by (i) using the mouse, (ii) using the fsolve command. 

Problem 3-27: Exponential barrier 

Determine the transmission and re°ection coe±cients as a function of energy 

E if the barrier has the form V (x) =Ve ¡x=L between x =0andL =1, 

with V = 1. Outside the barrier region, V (x) = 0. Plot the re°ection and 

transmission coe±cients in the same ¯gure and discuss how the results compare 

with those obtained in the text recipe.

Chapter 4 

Nonlinear ODE Models 

The elegant body of mathematical theory pertaining to linear 

systems ... tends to dominate even moderately advanced university 

courses. The mathematical intuition so developed ill equips the 

student to confront the bizarre behavior exhibited by the simplest of 

...nonlinear systems. Yet such nonlinear systems are surely the rule, 

not the exception .... Not only in research, but also in the everyday 

world of politics and economics, we would all be better o® if more 

people realized that simple nonlinear systems do not necessarily 

possess simple dynamic properties. 

Robert M. May, mathematical biologist, Nature, Vol. 261, 459 (1976) 

In Chapter 1, phase-plane portraits were used to explore some simple nonlinear 

ODE models whose temporal evolution could not have been predicted, even 

qualitatively, before the portraits were numerically constructed. An example 

was the period-doubling route to chaos exhibited by the Du±ng equation 

Äx +2° _x + ®x+ ¯x 3 = F cos(!t) (4.1) 

when the amplitude F of the driving force was increased, the other parameters 

being held ¯xed. If the nonlinear term, ¯x 3 , were not present, this \bizarre" 

period-doubling behavior would not even be possible. If we were to change the 

various coe±cient values in (4.1), the response of the nonlinear system would 

in general be entirely di®erent and not easily predicted on the basis of mathematical 

or physical intuition alone. To aid in the qualitative understanding of 

the behavior of nonlinear ODE systems such as this one, the concepts of ¯xed 

points and phase-plane analysis were discussed in Chapter 2. 

One might well ask whether there are mathematical techniques for obtaining 

the exact analytic solutions to nonlinear ODEs, and if so, whether Maple 

can be used to ¯nd these solutions. The answer is that the vast majority of 

nonlinear ODEs of interest to physicists and engineers do not possess exact 

analytic solutions. Only a handful of ODEs of physical interest can be solved 

exactly, and for those equations Maple can be used to ¯nd the solutions. Some 

examples of ¯rst-order nonlinear ODEs for which this is the case are illustrated 

in the following section. 

149

150 CHAPTER 4. NONLINEAR ODE MODELS 

4.1 First-Order Models 

4.1.1 An Irreversible Reaction 

I shall use the phrase \time's arrow" to express this one-way property 

of time which has no analogue in space. 

Arthur Eddington, British astrophysicist (1882{1944) 

As our ¯rst example, we consider the following problem, which the reader was 

previously asked to solve numerically (see Problem 2-29). 

Consider the irreversible chemical reaction 

2K 2Cr 2O 7 +2H 2O+3S ! 4KOH+2Cr 2O 3 +3SO 2 

with initially N1 molecules of potassium dichromate (K 2Cr 2O 7), N2 molecules of 

water (H2O), N3 atoms of sulphur (S), and 0 molecules of potassium hydroxide 

(KOH). The number x of KOH molecules at time t seconds is given by the 

nonlinear rate equation 

_x = k (2 N1 ¡ x) 2 (2 N2 ¡ x) 2 (4 N3=3 ¡ x) 3 

with k =1:64 £ 10 ¡20 s ¡1 . Taking N1 = 2000, N2 = 2000, and N3 = 3000, 

analytically determine the number of KOH molecules at arbitrary time t>0 

and plot the solution for the ¯rst second after the chemical reaction begins. 

How many KOH molecules are present at 0:2 seconds? How many are present 

in the limit t !1? 

To observe Maple's method of solution, the infolevel[dsolve] command 

is set equal to 2, and the rate equation entered. 


> de:=diff(x(t),t)=k*(2*N1-x(t))^2*(2*N2-x(t))^2 

*(4*N3/3-x(t))^3; 

de := d 

dt x (t) =k (2 N1 ¡ x(t))2 (2 N2 ¡ x(t)) 2 

μ 

4 N3 

3 

3 

¡ x(t) 

Theratecoe±cientk and the initial molecule numbers are speci¯ed. 

> k:=1.64*10^(-20): N1:=2000: N2:=2000: N3:=3000: 

The di®erential equation de is analytically solved for x(t), subject to x(0) = 0. 

> sol:=dsolve(fde,x(0)=0g,x(t)); 





trying Bernoulli 

trying separable 


equation is separable. Thatistosay,itcanbewrittenintheformdx=dt = 

f(t)=g(x), where f(t) andg(x) are given functions. So, on rearranging and 

integrating, one has R g(x) dx = R f(t) dt, the answer following if the integrals 

can be performed (which is the case here). 

The right-hand side of the solution sol can be converted into a functional 

operator x using the unapply command. Then entering x(t) will evaluate x at 

the speci¯ed value of t. 

> x:=unapply(rhs(sol),t); 

4000 5 

x := t !¡ 

(2=3) 

+4000 

(251904 t +625) (1=6) 

The number x(t) of KOH molecules at time t is plotted over the time interval 

t = 0 to 1 second. 

> plot(x(t),t=0..1,labels=["t","x"]); 

2500 

2000 

1500 

x 

1000 

500 

0 

0.2 0.4 0.6 0.8 1 

t 

Figure 4.1: Number of KOH molecules as a function of time. 

The number of KOH molecules present after 0.2 seconds is determined, 

> number:=evalf(x(0.2)); 

number := 2079:400844 

and rounded o® to the nearest integer. 

> KOH:=round(number); 

KOH := 2079 

So, 2079 KOH molecules are present 0.2 seconds after the reaction begins. 

The number of KOH molecules appears to be leveling o® in Figure 4.1 as 

the time increases. From the analytic form of the solution, clearly the limiting 

number N is 4000 KOH molecules as t !1. This can be con¯rmed by applying 

the following limit command to x(t): 

> N:=limit(x(t),t=infinity); 

N := 4000


PROBLEMS: 

Problem 4-1: Falling basketball 

A spherical object (diameter d meters and mass m kg) falling from rest experiences 

[FC99] a drag force Fdrag = ¡Av¡ Bv2 newtons, where v is the velocity 

in m/s and A =1:55 £ 10 ¡4 d, B =0:22 d2 . Derive the nonlinear ODE governing 

the velocity of the falling sphere. If the sphere is a basketball of diameter 

25 cm and mass 0.60 kg, analytically determine v(t). Does Maple recognize the 

ODE as being separable? Plot v(t) and show that the ball will reach a terminal 

velocity. Determine the terminal velocity. How long does it take the basketball 

to come within 1 percent of the terminal velocity? 

Problem 4-2: It's separable 

Consider the ODE dy 

dx = 2 x3 y ¡ y4 x4 ; y(1) = 5: 

¡ 2 xy3 Does this ODE appear to be separable? Show that assuming y(x) =xz(x)leads 

to a separable equation for z(x). Making use of this transformation, determine 

y(x) and plot the solution, starting at x = 1, over the range for which it remains 

real. What is the x value at the upper end of this real range? 

4.1.2 The Struggle for Existence 

The mathematics of uncontrolled growth are frightening. A single 

cell of the bacterium E. coli would, under ideal circumstances, divide 

[in two] every twenty minutes .... it can be shown that in a single 

day, one cell of E. Coli could produce a super-colony equal in size 

and weight to the entire planet Earth. 

Michael Crichton, The Andromeda Strain (1969) 

A classic experiment in microbiology is to grow yeast, or other microorganisms, 

in a nutrient broth inside a test tube or °ask at a suitable ¯xed temperature. 

As an assignment associated with her microbiology course in the premed 

program at MIT, Heather has been asked to create a Maple worksheet that illustrates 

the solution of simple model equations describing the growth of yeast 

in a test tube. She is to search the literature and ¯nd realistic numbers for 

the parameter values and use these to create suitable plots of the solutions. 

Consulting her older sister Jennifer who, recall, is a mathematics faculty member 

at MIT, Heather is guided to the text Mathematical Models in Biology by 

Leah Edelstein-Keshet [EK88]. This interesting and easy-to-read book describes 

some yeast-growing experiments and associated model equations. 

After loading the library plots package, 


Heather considers the simplest model of yeast growth, which would apply if 

there were an unlimited supply of nutrient. She lets N(t) be the yeast popula-


tion density at time t>0andk>0 the rate of reproduction. If k is taken to 

be a constant, the following linear yeast equation, YE, results: 

> YE:=diff(N(t),t)=k*N(t); 

YE := d 

N (t) =k N (t) 

dt 

This growth equation is historically known as Malthus's law, named in honor 

of Thomas Malthus (1766{1834), who published a pamphlet (entitled Essay on 

Population) on population growth in 1798. Although Heather can solve this 

simple linear ODE in her head, she lets Maple generate the solution, subject to 

the initial condition N (0) = No. 

> ic:=N(0)=No; 

ic := N (0) = No 

> sol:=dsolve(fYE,icg,N(t)); 

sol := N (t) =No e (kt) 

Malthus's law leads to exponential growth of the yeast population density. 

Having read The Andromeda Strain, Heather is curious as to how many 

Escherichia coli there would be at the end of 24 hours if the exponential solution 

prevailed. According to the Crichton quotation, the doubling time is 

20 minutes, or 1=3 of an hour. In the following command line, Heather takes 

N(t =1=3)=No = 2 in the exponential solution, 

> eq:=2=exp(k/3); #time in hours 

k 

( 

eq := 2 = e 3 ) 

and numerically solves for the reproductive rate constant, labeled k1 . 

> k1:=fsolve(eq,k); 

k1 := 2:079441542 

Using the Malthus solution, at the end of 24 hours the number of E. coli would 

have grown from a single bacterium (No =1), 

> Number:=1*exp(k1*24); #in 24 hours 

Number := 0:4722366529 1022 to about 1022 bacteria. Talk about explosive growth! Heather wonders how 

accurate the Malthus solution is. After all, she need not have formulated the E. 

coli growth as a continuous-time ODE, but instead could calculate the growth 

directly. In a 24-hour period there would be 24 £ 3 = 72 doublings. At the end 

of 24 hours, the number of E. coli is given by the output of the following line: 

> Number2:=2^(24.0*3); 

Number2 := 0:4722366483 1022 The two numbers di®er only in the eighth decimal place. 

Although 1022 E. coli is a large number, Heather notes that since an E. coli 

bacterium weighs about 10 ¡12 gm, their total weight is only 10 10 gm, much less 

than the 6 £ 1027 gm weight of the earth. Crichton's conclusion seems wrong.


This is interesting, but Heather realizes that she should get back to the yeast 

problem. Such unlimited growth as seen above clearly cannot occur in a test 

tube experiment with only a ¯nite amount of nutrient available. It seems more 

reasonable that the growth would depend on the amount of nutrient left in the 

test tube. To this end, Heather modi¯es her original model by assuming that 

the reproductive rate k is not a constant but is proportional to the concentration 

C(t) of nutrient, the proportionality constant being labeled K. 

> k:=K*C(t); 

k := K C (t) 

The new yeast equation then is given by NYE. 

> NYE:=YE; 

NYE := d 

N (t) =K C (t) N (t) 

dt 

As the yeast density increases, the nutrient concentration must decrease. Heather 

assumes that the rate of decrease of C(t) is proportional to the rate of increase of 

N(t), the positive proportionality constant being labeled a. The concentration 

equation (CE) then is as follows. 

> CE:=diff(C(t),t)=-a*diff(N(t),t); 

CE := d 

μ 

d 

C (t) =¡a N (t) 

dt dt 

The problem now involves two coupled ODEs, NYE and CE, but the general 

solution of the latter is easily obtained, C1 being the integration constant. 

> dsolve(CE,C(t)); 

C (t) =¡a N (t)+ C1 

The last output is then substituted into the new yeast equation. 

> NYE:=subs(%,NYE); 

NYE := d 

N (t) =K (¡a N (t)+ C1) N (t) 

dt 

It is traditional in mathematical biology to make the form of the equation 

notationally simpler, so Heather substitutes C1 = ab and K = r=(ab)into 

NYE ,whereband r are new constants. 

> subs(f_C1=a*b,K=r/(a*b)g,NYE); 

d r (¡a N (t)+ab) N (t) 

N (t) = 

dt ab 

To cancel the constant a, thesimplify command is applied. 

> NYE:=simplify(%); 

NYE := d r (¡N (t)+b) N (t) 

N (t) = 

dt b 

The resulting nonlinear di®erential equation is known as the logistic equation 

and was ¯rst studied by the mathematician Verhulst in 1838. 

According to Edelstein-Keshet's text, the logistic equation can be solved in 

a straightforward way, and the solution is quoted in her book. If an analytic


solution exists, Heather reasons that she should be able to use the dsolve 

command to ¯nd it. Setting the infolevel[dsolve] command to be 2, NYE 

is solved for N(t), subject to the same initial condition as for Malthus's law. 

> infolevel[dsolve]:=2: 

> yeastdensity:=dsolve(fNYE,icg,N(t)); 






yeastdensity:=collect(yeastdensity,exp); 

No b 

yeastdensity := N (t) = 

(¡No + b) e (¡rt) + No 

This mathematical form is referred to as the logistic curve. Forpositiver, N(t) 

approaches the value b as t !1. Unlike Malthus's law, the logistic equation 

predicts \saturation" of the yeast number to a constant value at large times, 

rather than uncontrolled growth. 

Heather notes that by loading the DEtools package, the general Bernoulli 

solution to NYE can also be obtained using the bernoullisol command. 

> with(DEtools): bernoullisol(NYE,N(t)); 



second. With the initial population density N0 =0:5329 known, the data from 

the ¯rst experiment were used to ¯nd the parameter values b and r to use in 

the logistic-model solution for each yeast variety. Using a least-squares ¯tting 

approach, Gause found that for the kephir, r =0:0607 and b =5:80, while for 

the cerevisiae r =0:2183 and b =13:0. The ¯rst experiment was carried out 

over a time period of 160 hours. Heather decides to plot the logistic curve for 

the kephir, so enters the parameter values, 

> b:=5.80: No:=0.5329: r:=0.0607: yeastdensity; 

N (t) = 

3:090820 

5:2671 e (¡0:0607 t) +0:5329 

and applies the plot command, adding a title to the picture. 

> plot(rhs(yeastdensity),t=0..160,labels=["t","A"], 

color=blue,thickness=2,tickmarks=[2,4], 

title="Amount of yeast (A) vs. hours (t)"); 

Amount of yeast (A) vs. hours (t) 

5 

4 

A 

3 

2 

1 

0 100 

t 

Figure 4.2: Growth curve for the yeast kephir. 

Thelogisticgrowthcurveofthekephir population is shown in Figure 4.2. The 

plot for the cerevisiae would be similar in appearance. 

Gause's second experiment studied the competition between the two yeasts 

for the available nutrient when they were grown together in the same test tube. 

To explore this case Heather labels the kephir population density and parameters 

with the subscript k, while the subscript c is used for the cerevisiae. To 

account for the interaction between the two species of yeast, Gause assumed 

that the probability of an interaction is proportional to the product of the 

population densities, i.e., to NkNc, and that the competition for nutrient was 

detrimental to both populations. Accordingly, Gause described the growth of


the two interacting species by the following pair of coupled nonlinear equations: 

Nk _ = rk (bk ¡ Nk ¡ ¯kc Nc) Nk=bk; Nc _ = rc (bc ¡ Nc ¡ ¯ck Nk) Nc=bc; (4.3) 

with the interaction parameters experimentally determined to be ¯kc =0:439 

and ¯ck =3:15. This set of nonlinear ODEs cannot be solved analytically, but 

the behavior of the two competing yeast populations can easily be determined 

by making a phase-plane portrait of Nk versus Nc. 

Heather unassigns k so that it can be used as a subscript to label the kephir 

population and sets the infolevel[dsolve] command to zero. 

> unassign('k'): infolevel[dsolve]:=0: 

The values determined by Gause for the coe±cients in (4.3) are entered, 

> r[k]:=0.0607: b[k]:=5.8: beta[kc]:=0.439: r[c]:=0.2183: 

b[c]:=13.0: beta[ck]:=3.15: 

as well as the system (sys) ofequations. 

> sys:=diff(N[k](t),t)=r[k]*(b[k]-N[k]-beta[kc]*N[c])*N[k]/b[k], 

diff(N[c](t),t)=r[c]*(b[c]-N[c]-beta[ck]*N[k])*N[c]/b[c]; 

sys := d 

dt Nk(t) =0:01046551724 (5:8 ¡ Nk(t) ¡ 0:439 Nc(t)) Nk(t); 

d 

dt Nc(t) =0:01679230769 (13:0 ¡ Nc(t) ¡ 3:15 Nk(t)) Nc(t) 

Heather takes Nk(0)=Nc(0)=0:5 inthephaseportrait command, 

> phaseportrait([sys],[N[k](t),N[c](t)],t=0..250, 

[[N[k](0)=0.5,N[c](0)=0.5]],stepsize=0.1,linecolor=blue); 

N[c] 

8 

6 

4 

2 

0 

1 2 4 5 

N[k] 

Figure 4.3: Phase portrait showing competition between kephir and cerevisiae.


producing the phase-plane picture shown in Figure 4.3. In the actual experiment, 

the time interval was only about 50 hours. On reducing the time range 

to be from t =0to50inthephaseportrait command line, Heather ¯nds that 

the curve terminates in the vicinity of the maximum value for Nc. Inthiscase, 

both species of yeast survived over the 50 hours of the experiment, with the 

cerevisiae dominating over the kephir. However, assuming that enough nutrient 

was available, she can see that ultimately only one species would survive over 

250 hours and it isn't the cerevisiae, but rather the kephir. This is because the 

interaction coe±cient ¯kc is much smaller than ¯ck. In this struggle for existence, 

there would be only the victorious kephir and the vanquished cerevisiae. 

To see the explicit time evolution of each yeast population, Heather uses the 

phase-plane portrait command again, but now creates two plots, one with the 

scene option scene=[t,N[k]], the other with scene=[t,N[c]]. 

> plot1:=phaseportrait([sys],[N[k],N[c]],t=0..300, 

[[N[k](0)=0.5,N[c](0)=0.5]],stepsize=0.1, 

scene=[t,N[k]],color=red,linecolor=blue): 

> plot2:=phaseportrait([sys],[N[k],N[c]],t=0..300, 

[[N[k](0)=0.5,N[c](0)=0.5]],stepsize=0.1, 

scene=[t,N[c]],color=red,linecolor=red): 

On superimposing the two plots with the display command, Figure 4.4 results. 

> display(fplot1,plot2g,tickmarks=[3,3]); 

8 

6 

N 

4 

2 

0 

100 200 300 

t 

Figure 4.4: Time evolution of the kephir (saturable curve) and the cerevisiae. 

By clicking with the mouse on the plot, the crossover point for the two population 

densities occurs at about 164 hours. With her assignment completed, 

Heather is grateful that her sister was able to recommend a useful text. She 

will have to treat her to a Starbucks latte the next time they get together.


PROBLEMS: 

Problem 4-3: Nonlinear diode circuit 

A linear capacitor with capacitance C is connected in series with a nonlinear 

diode that has a current (i){voltage (v) relationoftheformi = av+ bv2 ,with 

the coe±cients a and b both positive. The voltage across the capacitor at time 

t =0isv = V . Derive the nonlinear ODE governing this circuit. Introducing 

the dimensionless variables y = v(t)=V , ¿ = at=C,and¯ = bV=a,rewritethe 

ODE in dimensionless form. Analytically solve this ODE, demonstrating that 

Maple recognizes the ODE as a Bernoulli equation. For ¯ = 2, plot the solution 

overthetimeinterval¿ =0to2. 

Problem 4-4: Nonlinear diode revisited 

Suppose that the current{voltage relation for the nonlinear diode in the previous 

problem is given by the more general relation i = av+bvn ,wheren =2,3,4,5, 

::: Derive the corresponding general dimensionless ODE and analytically solve 

it for arbitrary n. For the same parameters and time range as in the previous 

problem, produce a single plot that shows the solutions for n =2to5. Discuss 

the e®ect of increasing n. 

Problem 4-5: A potpourri of Bernoulli equations 

For each of the following nonlinear ODEs (with prime indicating an x-derivative): 

² con¯rm that it is of the Bernoulli type; 

² analytically solve the ODE for y(0) = 1; 

² plot the solution y(x) overtherangex =0to5. 

(a) y 3 y 0 + x ¡1 y 4 = x; (b) y 0 + y = xy 3 ; (c) y ¡ y 0 =3y 3 e ¡2 x ; 

(d) y 3 +3y 2 y 0 =4. 

Problem 4-6: General solution 

Determine the general solution of the following Bernoulli equation: 

xy 0 + y = x 3 y 6 : 

Problem 4-7: Laser beam competition 

In the theory of stimulated thermal scattering [BEP71], the intensities IL and IS 

of two interacting collinear laser beams traveling in the z-direction are governed 

by the following pair of coupled ¯rst-order nonlinear ODEs: 

dIL 

dz = ¡gILIS ¡ ®IL; dIS 

dz = gILIS ¡ ®IS; 

where the gain coe±cient g>0andtheabsorption coe±cient ® ¸ 0. 

(a) By adding the two equations and eliminating IL, show that a single 

Bernoulli equation can be obtained for the intensity IS. 

(b) Analytically solve this Bernoulli equation for IS(z). 

(c) Plot ln(IS(z)=IS(0)) for z =0to10cmgiventhatIS(0)=IL(0) = 0:01, 

gIL(0) = 1 cm ¡1 ,and(a)® =0,(b)® =0:5 cm ¡1 . Discuss the results.


Problem 4-8: Variable transformation 

Explicitly show that the general Bernoulli equation (4.2) can be transformed 

into a linear ODE by making use of the transformation p =1=yn¡1 .Herenis an arbitrary positive integer greater than or equal to two. 

Problem 4-9: Flu epidemic in Spuzzum 

In the small town of Spuzzum (population N0 = 2000 people), the number 

N(t) of people infected with the °u after t weeks is governed by the logistic 

equation, N _ =0:002 N (N0 ¡ N): If two people are initially infected, derive 

the analytic formula for N(t): How many weeks does it take before 75% of 

Spuzzum's population is infected with the °u? Plot N(t) forthetimethatit 

takes 90% of the population to be infected. 

Problem 4-10: Hare today, gone tomorrow 

The growth of a snowshoe hare population is governed by the logistic equation, 

_N = rN(N0 ¡ N); with the rate constant r =0:25=N0, time being measured 

in months. A viral epidemic of deadly myxamatosis strikes the population, 

suddenly reducing the hare number to 1% of the equilibrium number N0. How 

many months does it take for the snowshoe hare population to climb back up to 

a level of 50% of its equilibrium number? Plot N(t)=N0 over this time interval. 

Problem 4-11: Harvesting of blue whales 

In the absence of harvesting, the blue whale population number N(t)isgoverned 

by the logistic equation. If a constant rate h of harvesting is included, the 

equation becomes 

N(t) _ =a (M ¡ N(t)) N(t) ¡ h; 

where M is the maximum number of whales in the absence of harvesting and a is 

a positive constant. With t in years, it has been estimated that a =6£10 ¡7 and 

M=200,000. The steady-state (equilibrium) population occurs when _ N =0. 

(a) Plot h versus N for the steady state. Find the maximum value hmax of 

h by: (i) clicking the mouse on the maximum point of the curve, and 

(ii) working with the algebraic form and then substituting numbers. The 

quantity hmax is called the maximum sustained yield. What happens to 

the whale population for h À hmax? 

(b) Using the dsolve command, solve the logistic equation for the given parameters 

and N(0)=100,000, h = 7000. How many years would it take for 

the blue whale population to go extinct? It is because of overharvesting 

that the blue whale population has been nearly wiped out. 

Problem 4-12: Gompertz's law 

Gompertz's law is an alternative nonlinear growth model to the logistic equation 

and has been used to model the growth of cancerous tumors. The Gompertz 

ODE is of the form 

_y = ky ln(ym=y); 

where k and ym are positive constants. Derive the general analytic solution 

to the Gompertz ODE. If k =0:1, ym =0:5, and y(0) = 0:1, determine the 

analytic form for y(t) and plot the solution for t = 0 to 75. What is the 

interpretation of the constant ym?


Problem 4-13: The von Bertalan®y model of growth 

The limited-growth von Bertalan®y model is governed by the nonlinear ODE 

_y =3ky 2=3 (ym ¡ y 1=3 ); 

with k and ym positive constants. Derive the general analytic solution to the 

von Bertalan®y ODE. If k =0:1, ym =0:5, and y(0) = 0:1, determine the 

analytic form for y(t) and plot the solution for t =0to75. 

Problem 4-14: The Michaelis{Menton equation 

The nonlinear Michaelis{Menton model equation 

_y = ¡ay=(b + y); 

with a, b positive constants, describes the rate at which an enzyme reaction 

takes place. Here y(t) is the amount of substrate that is being transformed by 

the enzyme at time t. Takinga = b = y(0) = 1, determine the analytic form of 

y(t). Plot the analytic solution for t =0to5. 

4.1.3 The Bad Bird Equation 

Everyone pushes a falling fence. 

Chinese proverb about failure. 

In an article by Peastrel and coworkers in the The Physics of Sports [PLA92], 

the experimental data for the distance y (in meters) that a badminton bird 

(\bad" bird, for short) falls in t seconds from rest is well described by 

y = 

V 2 

g ln 

μ 

cosh 

μ gt 

V 

 

; (4.4) 

where V is the terminal velocity and g is the acceleration due to gravity. Vectoria 

will now show you how equation (4.4) can be easily derived, using Newton's 

second law and assuming that the drag force is given by Newton's law of air 

resistance, viz.,Fres = ¡kmv2 ,withkthe drag coe±cient, m the mass of 

the bird, and v the speed. The quadratic dependence on v is appropriate for 

turbulent air °ow. For the falling bird, the turbulence arises because of the 

bird's \feathers." Stokes's law of resistance (Fres / v) prevails for laminar (i.e., 

smooth) °ow. 2 

Vectoria begins by loading the DEtools package, 


and entering Newton's law of air resistance. 

> F[res]:=-k*m*v(t)^2; 

Fres := ¡kmv(t) 2 

2 The more general resistance law is a combination of linear and quadratic terms in the 

velocity. Which term dominates depends on the shape of the moving object and the magnitude 

of the velocity.


The equation of motion follows on applying Newton's second law to the motion 

of the bad bird. Equating the net force on the bird, due to the pull of gravity 

and the drag force, to the bird's mass times its acceleration produces eq. 

> eq:=m*g+F[res]=m*diff(v(t),t); 

eq := mg¡ kmv(t) 2 μ 

d 

= m 

dt v(t) 

 

The force equation is simpli¯ed by dividing eq through by the mass m. 

> eq2:=simplify(eq/m); 

eq2 := g ¡ k v(t) 2 = d 

dt v(t) 

Since the terminal velocity V is reached when the upward and downward forces 

balance, so that the acceleration is zero, then g = kV2 .Thus,thedragcoe±cient 

k may be eliminated by substituting k = g=V 2 into eq2 . 

> ode:=subs(k=g/V^2,eq2); #Riccati equation 

g v(t)2 

ode := g ¡ 

V 2 

d 

= 

dt v(t) 

Vectoria recognizes the resulting ¯rst-order nonlinear ode as a speci¯c example 

of Riccati's equation, which has the general structure 

_x + ax 2 + f1(t) x + f2(t) =0; (4.5) 

with a a constant. In the present case, Vectoria identi¯es x ´ v, a ´ k = g=V 2 , 

f1 =0,andf2 = ¡g. She recalls that Riccati's equation can be reduced to a 

linear ODE involving a new dependent variable z(t) by introducing the transformation 

z(t) =exp(a R t 

0 x(t) dt): However, it is easier to derive the general 

form of v(t) by applying the riccatisol command to ode. The square brackets 

are included at the end of the command line to remove the brackets that would 

otherwise enclose the output. 

> sol:=riccatisol(ode,v(t))[]; 

sol := v(t) =¡ 1 

0 r 

Bt 

¡ 

tan B 

@ 

2 

4 g2 

V 2 

r 

C1 ¡ 

¡ 

2 

4 g2 

V 2 

1 

r 

C 

2 A ¡ 4 g2 

2 V 

V 2 

g 

The integration constant C1 is determined by evaluating the right-hand side 

of the solution at t = 0, setting the result to zero (since the bird starts from 

rest), and solving for the constant. 

> _C1:=solve(eval(rhs(sol),t=0)=0,_C1); 

C1 := 0 

The integration constant is zero. The velocity expression follows on simplifying 

the rhs of sol, assuming that g>0, V > 0, and t>0. 

> v:=simplify(rhs(sol)) assuming g>0,V>0,t>0;


μ 

tg 

v := tanh V 

V 

To determine y(t), a second ODE is formed by equating dy=dt to v. To make 

the ¯nal expression appear in the desired form, the convert command is used 

to reexpress the hyperbolic tangent in terms of the hyperbolic sine and cosine. 

> ode2:=diff(y(t),t)=convert(v,sincos); 

ode2 := d 

μ 

tg 

sinh V 

V 

y(t) = μ 

dt tg 

cosh 

V 

Analytically solving ode2 for y(t), subject to the initial condition y(0) = 0, 

> dsolve(fode2,y(0)=0g,y(t)); 

V 

y(t) = 

2 μ μ 

tg 

ln cosh 

V 

g 

produces an output that is identical in structure to equation (4.4), thus completing 

Vectoria's task. 

PROBLEMS: 

Problem 4-15: Return velocity 

A ball of unit mass is thrown vertically upward near the earth's surface with an 

initial speed v(0) = U. Assuming that Newton's law of air resistance prevails, 

show that after the ball rises to its maximum height and begins to fall it passes 

its initial position with a velocity v = (UV)=( p U 2 + V 2 ); where V is the 

terminal velocity. Hint: Reexpress the acceleration as 

dv 

dt = 

μ μ μ 

dv dx 

dv(x) 

= v(x) 

dx dt 

dx 

and note that at the maximum height the speed is zero. 

Problem 4-16: A Riccati equation 

Consider the following Riccati equation: 

y 0 (x)+y(x)=x + ay(x) 2 + b =0; 

with the initial condition y(0) = A, wherea, b, andAare real constants. 

(a) Analytically solve the ODE. Does the answer depend on the value of A? 

(b) What does the general solution look like if no initial condition is speci¯ed? 

Explain what happens mathematically when y(0) = A is imposed. 

(c) Taking a =5andb =2,plotthesolutionfory(0) = A over the range 

x = 0 to 2, using the plot option view=[0..2,-5..5]. 

(d) Explain the origin of the singular points in the graph in terms of the 

behavior of Bessel functions.


Problem 4-17: Nonlinear drag on Lake Ogopogo 

A boat is launched on Lake Ogopogo with initial speed v0. Thewaterexertsa 

drag force F (v) =¡aebv ,witha>0, b>0, thus slowing the boat down. 

(a) Find an analytic expression for the speed v(t). 

(b) Determine the time it takes for the boat to come to rest. 

(c) How far does the boat travel along Lake Ogopogo before coming to rest? 

4.2 Second-Order Models 

The vast majority of second-order nonlinear models that physicists and engineers 

are interested in must be solved numerically. However, we begin this 

section with a few examples of models that lead to nonlinear ODEs having exact 

analytic solutions. Historically, these models have appeared in many equivalent 

guises, so keep this in mind when you read the stories. 

4.2.1 Patches Gives Chase 

Man ... cannot learn to forget, but hangs on the past: 

however far or fast he runs, that chain runs with him. 

Friedrich Nietzsche, German philosopher (1844{1900) 

A loveable beagle, named Patches, is patrolling a °at farm ¯eld, sni±ng contentedly 

at gopher holes, when she spots her mistress, Heather, walking at 

constant speed along a straight road at the edge of the ¯eld. Patches then 

runs at constant speed toward Heather in such a way as to aim always at her 

with her sensitive beagle nose. With distances in km, Patches is initially at 

(x =1;y= 0) and Heather at (0; 0). The road is described by the equation 

x = 0, and the ratio of Heather's speed to Patches' speed is r. 

(a) Derive the nonlinear ODE describing Patches' path y(x). 

(b) Analytically solve the ODE for y(x). If Heather walks at 3 km/h and 

Patches runs at 8 km/h, plot y(x) using constrained scaling. 

(c) How many minutes does it take the dog to reach her mistress? How far 

has Heather walked in this time? How far has Patches run? 

(d) Animate the motion of Patches and Heather, including a tangent line to 

Patches' instantaneous position, which points towards Heather. Again, 

use constrained scaling. 

At some instant in time, Patches' coordinates are (x; y(x)), while Heather's 

are (0; h). Patches runs towards Heather in such a way as to aim always at her. 

Therefore the slope of the tangent to Patches' path is dy=dx =(h¡y(x)=(0¡x), 

whichisnowenteredineq.


> restart: 

> eq:=diff(y(x),x)=(h-y(x))/(0-x); 

eq := d 

¡ y(x) 

y(x) =¡h 

dx x 

We solve eq for Heather's vertical coordinate h. 

> h:=solve(eq,h); 

μ 

d 

h := ¡ 

dx y(x) 

 

x + y(x) 

Heather's speed is r times that of Patches, so dh=dt = r (ds=dt), where t is time 

and ds = p 1+(dy=dx) 2 dx is an element of arclength along Patches' path. 

But then dh = rds,ordh=dx = r (ds=dx) =r p 1+(dy=dx) 2 . Entering this 

last relation yields the relevant second-order nonlinear ODE for y(x), 

> ode:=diff(h,x)=r*sqrt(1+diff(y(x),x)^2); 

μ 

s 

μ 2 d d 

ode := ¡ y(x) x = r 1+ 

dx2 dx y(x) 

2 

the expression for h having been automatically substituted. Although ode is 

nonlinear, it can be analytically solved. Patches is initially at x =1;y=0and 

the tangent line there has zero slope. These starting conditions are used in the 

dsolve command. 

> sol:=dsolve(fode,y(1)=0,D(y)(1)=0g,y(x)); 

x xxr 

sol := y(x) =¡ 

¡ 

2(¡1+r) xr 2(1+r) + 

r 

; 

¡1+r2 x xxr 

y(x) = 

+ 

2(¡1+r) xr 2(1+r) ¡ 

r 

¡1+r2 Two forms of the solution are generated, the ¯rst corresponding to Heather 

moving downward (in the negative y direction) along the x = 0 line, the second 

corresponding to moving upward. The upward (second one here) solution is 

selected and simpli¯ed with respect to powers of x. 

> y:=simplify(rhs(sol[2]),power); 

y := x(1¡r) x(1+r) 

+ 

2(¡1+r) 2(1+r) ¡ 

r 

¡1+r2 Evaluating y at x = 0 yield's Heather's vertical coordinate when Patches reaches 

her. Notice that as r approaches 1 from below, Y goes to in¯nity, i.e., Patches 

cannot catch Heather in a ¯nite distance. Obviously, Patches must run faster 

than Heather can walk to catch her. 

> Y:=eval(y,x=0); 

r 

Y := ¡ 

The given ratio r = 3 

8 

> r:=3/8: y:=y; Y:=evalf(Y); 

¡1+r 2 

is entered, and y and Y are automatically evaluated.


4 x(5=8) 

y := ¡ + 

5 

4 x(11=8) 

+ 

11 

24 

55 

Y := 0:4363636364 

Heather has walked about 0.44 km before Patches catches her. The times T 

in hours and T2 in minutes for Patches to reach Heather are calculated, along 

with the distance Sdog that Patches runs. The absolute value is taken for T , 

because Y will be negative if the downward solution branch is chosen. 

> T:=abs(Y)/3; T2:=T*60*minutes; Sdog:=8*T; 

T := 0:1454545455 T2 := 8:727272730 minutes Sdog := 1:163636364 

The time for Patches to catch Heather is about 0:15 hours, or 8:73 minutes. 

(Patches is rather slow!) Patches has run about 1:16 km. 

A labeled plot of y is now created with constrained scaling, 

> pl:=plot(y,x=0..1,thickness=2,tickmarks=[3,3], 

labels=["x","y"],scaling=constrained): 

and, loading the plots package, displayed in Figure 4.5. 

> with(plots): display(pl); 

0.4 

y 

0.2 

0 

0.2 0.4 0.6 0.8 1 

x 

Figure 4.5: Patches' path is the curved line. 

Finally, an animation of Heather's and Patches' motion is produced. The animation 

will contain N = 40 frames, the frames labeled from i =0toN ¡ 1. 

The Student[Calculus1] library package is loaded so that the tangent line to 

Patches' position on the path can be drawn for each frame. 

> N:=40: with(Student[Calculus1]): 

The following functional operator X will calculate Patches' horizontal coordinate 

for the ith frame. For i =0; 1; 2; etc., X(0) = 1; X(1) = 1 ¡ 1=40 = 

0:975; X(2) = 0:95, etc. 

> X:=i->1-i/N:


The operator hh calculates Heather's y-coordinate for the ith frame. The sign 

command is included to pick up the correct sign for the coordinate, whether 

Heather's motion is upward or downward. 

> hh:=i->sign(-Y)*evalf(int(r*sqrt(1+diff(y,x)^2),x=1..X(i))): 

The pointplot command is used to graph Heather's and Patches' positions for 

the ith frame as size-16 blue circles. 

> gr:=i->pointplot([[X(i),eval(y,x=X(i))],[0,hh(i)]], 

symbol=circle,symbolsize=16,color=blue): 

The command Tangent(y,x=X(i),0..1) generates the equation of the line 

tangent to y at the point x = X(i), the third argument giving the range of x. 

A plot of the tangent line is produced by including the arguments showtangent 

and output=plot. 

> gr2:=i->Tangent(y,x=X(i),0..1,showtangent,output=plot, 

thickness=2): 

All the graphs are superimposed for the ith frame, using the display command. 

> Gr:=i->display([pl,gr(i),gr2(i)],view=[0..1,0..Y]): 

Themotionisanimatedusingthedisplay command again, with the option 

insequence=true. 

> display(seq(Gr(i),i=0..N-1),insequence=true, 

scaling=constrained); 

On execution of the above command line, the opening frame of the animation 

appears on the computer screen. Click on the plot and on the start arrow to 

see the animation. The equation of the tangent curve is displayed at the top of 

each frame, along with the coordinates at which the tangent line is calculated. 

PROBLEMS: 

Problem 4-18: Heather slows down 

On seeing Patches begin to run toward her, Heather slows down, her speed 

being 3 x km/h, where x is Patches' horizontal coordinate. Thus, when Patches 

is at x = 1, Heather's speed is 3 km/h, but when Patches reaches her (x =0) 

Heather is standing still. If Patches still runs at a constant speed of 8 km/h, 

analytically determine the equation y(x) describing Patches' path and plot it. 

How far has Heather walked when Patches catches her? How far has Patches 

run? How long does it take Patches to reach Heather? 

Problem 4-19: Curves of pursuit 

Do an Internet search on curves of pursuit and discuss the history and types of 

pursuit curves that have been studied in the mathematical literature. Also see 

Davis [Dav62]. Create a recipe that generates the pursuit curve of your choice 

and motion along it, including the instantaneous tangent line. Depending on 

your choice, you may have to solve the relevant ODE numerically.


4.2.2 Oh What Sounds We Hear! 

A man falls in love through his eyes, a woman through her ears. 

Woodrow Wyatt, British journalist, To the Point,\The Ears Have It," 1981 

Wyattarguesthatwhatissaidtoawomanbyaman,andwhatshebelieves 

about his status, is usually more important than the super¯ciality of good looks. 

At the end of the nineteenth century, the famous German scientist Heinrich 

Helmholtz knew that the ear perceives frequencies that are not present in the 

incident acoustic waves. He attributed this to an asymmetric nonlinear response 

of the eardrum's tympanic membrane and developed a simple mechanical model 

for the eardrum's vibrations. 

In the present century, a nervous young mathematics major by the name of 

Mike has an hour or so to fret over his upcoming date with Vectoria. Why is 

Mike nervous? This is the night that he plans to reveal his serious intentions for 

her hand, and he is worried about possible rejection. To calm him down, let's 

try to take his mind o® Vectoria by asking him to reproduce Helmholtz's derivation. 

Helmholtz's model involves applying Newton's second law to generate a 

phenomenological nonlinear ODE describing the one-dimensional displacement 

x(t) of the freely vibrating eardrum about the equilibrium position x =0. 

Reluctantly acquiescing to our request, Mike begins by loading the plots 

library package, and then formally Taylor expanding the restoring force per 

unit mass, F (x), about x = 0 out to third order in x. 


> Force:=taylor(F(x),x=0,3); 

Force := F (0) + D(F )(0) x + 1 

2 (D(2) )(F )(0) x2 +O(x3 ) 

The coe±cients D(F )(0) and D (2) (F )(0) in the above output stand for the ¯rst 

and second derivatives of F with respect to x, evaluated at x = 0, respectively. 

Mike then assumes that the displacement x is su±ciently small that only terms 

to second order in x have to be retained. So he removes the O(x3 )term. 

> Force:=convert(Force,polynom); 

Force := F (0) + D(F )(0) x + 1 

2 (D(2) )(F )(0) x2 Why keep the second-order term? The term that is linear in x corresponds to 

Hooke's law, while the quadratic term is the ¯rst nonlinear correction to Hooke's 

law and is responsible for the asymmetric response noted by Helmholtz. As the 

eardrum vibrates about the equilibrium position x = 0, the linear (x) termin 

the force changes sign with x, but the quadratic (x2 ) term does not change sign. 

In equilibrium the restoring force must vanish, so Mike sets F (0) = 0. He 

also makes the symbolic substitutions D(F )(0) = ¡! 2 0 and D(2) (F )(0) = ¡2 ¯, 

> Force:=subs(fF(0)=0,D(F)(0)=-omega[0]^2,(D@@2)(F)(0)= 

-2*betag,%); 

Force := ¡!0 2 x ¡ ¯x 2


with !0 and ¯ both positive. Using Newton's second law and substituting 

the explicit time dependence into the force yields the relevant second-order 

nonlinear ODE. 

> ode:=diff(x(t),t,t)-subs(x=x(t),Force)=0; 

μ 2 d 

ode := 

x (t) 

dt2 + !0 2 x (t)+¯ x(t) 2 =0 

If ¯ is set equal to zero, then the familiar linear simple harmonic oscillator 

equation results, the eardrum vibrating at the characteristic frequency !0. In 

this case, the period (time for one complete oscillation) is T =2¼=!0. With 

¯ 6= 0, the above ODE is nonlinear, but can be solved analytically, as Mike 

will demonstrate. Since he is already furtively looking at his watch, we will not 

insist that Mike look for realistic parameter values for the eardrum, but instead 

allow him to set !0 =1and¯ =1,sotheode has the following form. 

> omega[0]:=1: beta:=1: ode:=ode; 

μ 2 d 

ode := x(t) + x(t)+x (t) 

dt2 2 =0 

Applying the dsolve command to ode, 

> solution:=dsolve(ode,x(t)); 

Z x(t) 

3 

solution := p d a ¡ t ¡ C2 =0; 

¡9 a2 ¡ 6 a3 +9 C1 

Z x (t) 

¡ 

3 

p ¡9 a 2 ¡ 6 a 3 +9 C1 d a ¡ t ¡ C2 =0 

produces two general implicit solutions for the time, with positive and negative 

square roots. To obtain a positive period, Mike chooses to work with the positive 

square root, which is the ¯rst solution for this particular run. 

> answer:=solution[1]; #choose positive square root 

answer := 

Z x (t) 

3 

p d a ¡ t ¡ C2 =0 

¡9 a2 ¡ 6 a3 +9 C1 

The constant C2 in the answer merely shifts the origin of time and so can be 

disregarded or set equal to zero. The other arbitrary constant C1 canbeevaluated 

by specifying an initial amplitude x(0) = A with zero initial velocity there. 

The amplitude sets the total energy E of the system. For a one-dimensional 

mechanical system having mass m and speed v, thetotalenergyisthesumof 

the kinetic energy and potential energy U(x), i.e., 

E = 1 

2 mv2 + U(x): (4.6) 

But this can be rewritten as 

v = dx 

dt 

r 

2 

= § (E ¡ U(x)); (4.7) 

m


or, on integrating, Z x(t) 

dx 

t = § p : (4.8) 

0 (2=m)(E ¡ U(x)) 

Comparing the result in answer with the positive square root in (4.8), the 

potential energy U(x) for the present problem can be taken (to within an overall 

proportionality constant) as 

U(x) =9x 2 +6x 3 : (4.9) 

Now Mike di®erentiates the answer with respect to time, 

> eq:=diff(answer,t); 

μ 

d 

3 

dt 

eq := 

x(t) 

 

p ¡ 1=0 

¡9 x (t) 2 ¡ 6 x (t) 3 +9 C1 

thus removing the unwanted coe±cient C2 from the analysis. Next, he solves 

eq for dx=dt, i.e., for the velocity vel. 

> vel:=solve(eq,diff(x(t),t)); 

vel := 1 p 

¡9 x(t) 2 ¡ 6 x(t) 3 +9 C1 

3 

and removes the time dependence for later integration purposes, replacing x(t) 

with the dummy variable y. 

> vel:=subs(x(t)=y,vel); 

p 

¡9 y2 ¡ 6 y3 +9 C1 

vel := 

3 

The unknown constant C1 can be determined by imposing the condition that 

the velocity must be zero at the turning point y = A, 

> condition:=eval(vel,y=A)=0; 

p 

¡9 A2 ¡ 6 A3 +9 C1 

condition := 

=0 

3 

and solving for the constant. The velocity expression is then determined. 

> _C1:=solve(condition,_C1); vel:=vel; 

C1 := A2 + 2 

3 A3 

p 

¡9 y2 ¡ 6 y3 +9A2 +6A3 vel := 

3 

To proceed any further, the amplitude A must be given a speci¯c value. Mike 

, which is automatically substituted into vel. 

enters the nominal value A = 1 

3 

> A:=1/3: vel:=vel; 

vel := 

r 

¡9 y 2 ¡ 6 y 3 + 11 

9 

3


In vel, the total energy is identi¯ed to be E =11=9. To get a qualitative feeling 

for the eardrum motion, Mike plots the potential energy U as a thick red line 

over the range x = ¡1:6 to0:6, 

> U:=plot(9*x^2+6*x^3,x=-1.6..0.6,color=red,thickness=2): 

and the total energy E as a thick green horizontal line between the points 

(x = ¡2; E=11=9) and (x =1;E=11=9). 

> E:=plot([[-2,11/9],[1,11/9]],style=line,color=green, 

thickness=2): 

The two energy curves are superimposed in Figure 4.6. 

> display(fU,Eg,tickmarks=[3,3],labels=["x","Energy"]); 

-1 x 

4 

Energy 

Figure 4.6: Asymmetric potential well for the eardrum model equation. 

The horizontal total energy line in Figure 4.6 intersects the potential energy 

curve at three x values. The two intersection points inside the potential well 

centered at x =0aretheturning points, the eardrum system oscillating between 

these two points if it is placed at one of these two points with zero velocity. 

The turning point on the right is the input value x(0) = A = 1 

3 . Because the 

potential well is slightly asymmetric, the other turning point on the left is not 

at x = ¡ 1. 

The other turning point is easily determined by ¯rst ¯nding all the 

3 

intersection points using the solve command. 

> points:=solve(vel=0,y); 

points := 1 11 

; ¡ 

3 12 + 

p 

33 11 

; ¡ 

12 12 ¡ 

p 

33 

12 

Three solutions are produced, corresponding to the three intersection points 

of the total energy line with the potential energy curve. The ¯rst answer corresponds 

to the turning point on the far right, and the third answer to the 

intersection on the far left. So the second answer is selected as corresponding 

to the second turning point for motion inside the potential well. This turning 

point is labeled B. 

2 

0


> B:=points[2]; 

B := ¡ 11 

12 + 

p 

33 

12 

If ¯ = 0, the potential well would have been symmetric about x =0,andthe 

second turning point would have been at x = ¡ 1 

3 ¼¡0:333. For ¯ =1,the 

ppotential well is asymmetric and the second turning point is at B =(¡11 + 

33)=12 ¼¡0:438. Clearly, the period of oscillation will be somewhat longer 

than the period 2 ¼=!0 ¼ 6:28forthelinearcase. 

Mike calculates the period T for the nonlinear situation by multiplying the 

time it takes the system to go from B to A by 2, i.e, T =2 R A 

(1=vel) dy. 

B 

> T:=2*int(1/vel,y=B..A); 

Ãs 

15 ¡ 

24 EllipticK 

T := 

p 33 

15 + p ! 

33 

p p 

30 + 2 33 

The period is expressed in terms of the complete elliptic integral K(k) ofthe 

¯rst kind, which is de¯ned as 

Z ¼=2 

d® 

K(k) ´ p 

0 1 ¡ k2 sin 2 : (4.10) 

® 

Maple expresses 

q 

the complete elliptic integral in the form EllipticK(k), so in 

this case k = (15 ¡ p 33)=(15 + p 33): The elliptic integral can be evaluated 

numerically, 

> T:=evalf(T); 

T := 6:747679332 

so the period is T ¼ 6:75, which is about 7 1% 

longer than for the linear case. 

2 

To perform the necessary integration to obtain the analytic solution, it is 

necessary to assume that the displacement satis¯es x>Band x assume(x>B,x sol:=t=int(1/vel,y=B..x); #implicit solution 

Ã s 

45 x +33¡ 

12 EllipticF 2 

sol := t = 

p 33 + 3 p 33 x 

(15 ¡ p 33) (11 + p 33 + 12 x) ; 

s 

15 ¡ p 33 

15 + p ! 

33 

p p 

30 + 2 33 

The implicit solution is expressed in terms of the incomplete elliptic integral of 

the ¯rst kind, u ´ F (Á; k), which is de¯ned as follows: 

Z Á 

Z sin Á 

d® 

dy 

u ´ F (Á; k) = p = p 

0 1 ¡ k2 2 

sin ® 0 1 ¡ y2 p 1 ¡ k2 : (4.11) 

y2 Maple expresses the incomplete elliptic integral in the form EllipticF(z; k) with 

z ´ sin Á. The complete elliptic integral K(k) corresponds to taking Á = ¼=2.


An explicit solution is obtained by solving sol for x. 

> x:=solve(sol,x); 

x := ¡ 1 88 %1 + 8 %1 

3 

p 11 p 3 ¡ 77 ¡ 3 p 33 

32 %1 ¡ 43 ¡ 5 p 33 

%1 := JacobiSN 

Ã 

t 

p 

30 + 2 p 33 

; 

12 

s 

15 ¡ p 33 

15 + p 33 

The formidable-looking answer, which is challenging to derive with pen and 

paper,isgivenintermsoftheJacobian elliptic sine function, whichMaple 

expresses in the form JacobiSN(u; k). When written by hand, the elliptic sine 

function is often written as sn(u), the argument k not being explicitly expressed. 

How is sn(u) de¯ned? The upper limit Á in the ¯rst integral of equation 

(4.11) is referred to as the \amplitude" of u and is written as Á =am(u). 

Then, for a given k value, the elliptic sine function, sn(u), is de¯ned by 

sn(u) =sinÁ = sin(am(u)): (4.12) 

For k =0,itiseasytoverifythatsn(u) reduces to sin(u). As k is increased, 

the elliptic sine function deviates away from the \ordinary" sine function shape. 

To learn more about the Jacobian elliptic sine function, Mike suggests that you 

try some of the relevant problems that follow this recipe. 

The solution is now plotted over the time interval t = 0 to 30, displaying 

asymmetric oscillation about the equilibrium point. 

> plot(x,t=0..30,thickness=2); 

0.2 

0 

–0.2 

–0.4 

!2 

5 10 15 t 20 25 30 

Figure 4.7: Asymmetric oscillations of the eardrum. 

Unfortunately, Mike has to leave to pick Vectoria up, so let's wish him luck 

in his quest for this fair damsel's hand and hope that he will return soon to 

explore some more interesting recipes with us.


PROBLEMS: 

Problem 4-20: Di®erent amplitude 

Taking the initial conditions x(0) = ¡ 4 

5 ,_x(0) = 0 in the text recipe, determine 

the analytic form of the period and its numerical value. Determine the analytic 

solution and plot it over a suitable time range. 

Problem 4-21: Maximum energy for bounded motion 

What is the maximum total energy E for which oscillatory motion of the 

eardrum can occur in the text recipe? What are the numerical values of the 

turning points? 

Problem 4-22: Higher-order correction to the period 

For the freely vibrating eardrum, also keep the cubic term in the Taylor expansion 

of the force law and express the restoring force per unit mass as 

F = ¡! 2 0 x ¡ ¯x2 ¡ °x3 . Taking ! =1,¯ = 3 1 

1 

4 , ° = 2 ,andA = 3 determine 

the period of vibrations of the eardrum. By how much is the period 

changed with respect to that predicted by the linear Hooke's law? 

Problem 4-23: Plotting the elliptic sine function 

Plot the Jacobian elliptic sine function sn(u) over the range u =0to20inthe 

same graph for k =0:1, k =0:5, k =0:9, and k =0:995. Discuss the behavior 

of sn(u) ask is increased. 

Problem 4-24: Plotting the elliptic cosine function 

In analogy to the elliptic sine function, the Jacobian elliptic cosine function of u 

is de¯ned as cn(u) =cosÁ =cos(am(u)): Noting that the Maple command for 

cn(u) isJacobiCN(u,k), plot the elliptic cosine function over the range u =0 

to 20 in the same graph for k =0:1, k =0:5, k =0:9, and k =0:995. Discuss 

the behavior of cn(u) ask is increased. 

Problem 4-25: Properties of elliptic functions 

Using the de¯nition of cn(u) in the previous problem and de¯ning another 

elliptic function dn(u) = p 1 ¡ k2 sin 2 Á = p 1 ¡ k2 sn2 (u); prove the following: 

(a) d 

(cn(u)) = ¡ sn(u) dn(u); 

du 

(b) d2 

du 2 (cn(u)) = (2 k2 ¡ 1) cn(u) ¡ 2 k 2 cn 3 (u); 

Z 

(c) cn(u) du = 1 

k arccos(dn(u)). 

Problem 4-26: Vibrating hard spring 

The equation of motion for a \hard" spring is 

Äx(t)+(1+a 2 x(t) 2 ) x(t) =0: 

Analytically determine the period and the solution x(t) for the hard spring, 

given the initial conditions x(0) = A, _x(0) = 0. Plot the solution over several 

cycles for a = A =1.


4.2.3 Vectoria Feels the Force and Hits the Bottle 

Man your ships, and may the force be with you. 

Film line from Star Wars, writtenbyGeorgeLucas,1977 

Our fair damsel, Vectoria, has felt the force of Mike's love and can't wait to tell 

her girlfriends tonight about becoming engaged. Perhaps a bottle of sparkling 

Asti Spumante wine would be in order then, but in the meantime a homework 

assignment in her electromagnetics course beckons and an entirely di®erent sort 

of bottle awaits her immediate consideration. 

She has been asked to create a recipe that illustrates the movement of a 

charged particle in a magnetic bottle. The phrase \magnetic bottle" refers to 

the carefully designed arrangements of magnetic ¯elds that are used for the 

con¯nement of extremely hot plasmas studied in experiments on thermonuclear 

fusion. Basically, the magnetic ¯eld is approximately uniform in the center of 

the \bottle" but gets stronger and pinches inward at the ends of the bottle. 

A charged particle starting in the central region, and given an initial velocity 

toward one end of the bottle, spirals around a ¯eld line traveling toward that 

end. As it approaches the end, the orbits become smaller and smaller, and 

¯nally at some point the charge re°ects o® the \end" of the bottle and reverses 

direction. It then travels to the other end of the bottle and repeats the scenario. 

In her recipe, Vectoria will make use of the Lorentz force law and a magnetic 

¯eld con¯guration of her own mathematical design. A particle of charge q 

moving with velocity ~v in an external electric ¯eld ~ E and magnetic ¯eld ~ B will 

experience the Lorentz force ~ F = q ~ E + q (~v £ ~ B). If present, the electric ¯eld 

produces a contribution to the Lorentz force in the direction of ~ E, while the 

magnetic force contribution is perpendicular to the velocity vector. So, the 

electric force will accelerate the charge, while the magnetic force de°ects it. 

Vectoria begins her recipe by loading the VectorCalculus and plots packages, 

the former being needed for creating and manipulating vectors. 

> restart: with(VectorCalculus): with(plots): 

Using Cartesian coordinates, with êx,êy, andêzpointing along the x-, y-, and zaxes 

respectively, the position vector ~r = x(t)êx + y(t)êy + z(t)êz of the charge 

at time t is entered, and its velocity ~v = d~r=dt and acceleration ~a = d~v=dt 

calculated. The position vector is entered here with the \shorthand" syntax, 

the longer form being Vector([x(t),y(t),z(t)]). The default output is in 

terms of Cartesian unit vectors, although Maple does not place \hats" on them. 

> r:=; v:= diff(r,t); a:=diff(v,t); 

r := x (t)ex + y(t)ey + z(t)ez 

μ 

d 

v := 

dt x(t) 

μ 

d 

ex + 

dt y(t) 

μ 

d 

ey + 

dt z(t) 

 

ez 

μ μ μ 

2 

2 

2 

d d d 

a := x(t) ex + y(t) ey + z(t) ez 

dt2 dt2 dt2


The radial distance R = p x(t) 2 + y(t) 2 at time t is entered. The coordinates 

and initial conditions will be chosen so that spirals are traced out in the radial 

direction with translation of the spirals along the z-axis. 

> R:=sqrt(x(t)^2+y(t)^2); 

R := p x(t) 2 + y(t) 2 

For the magnetic bottle ¯elds, Vectoria takes the electric ¯eld to be zero, and 

starts to build up the magnetic ¯eld structure, inputting the x-, y-, and zcomponents 

but with the mathematical forms of the radial function Br and 

longitudinal function Bz not yet speci¯ed. 

> E:=; B:=; 

E := 0 ex 

Br x (t) 

B := ¡ p 

x(t) 2 + y(t) 2 ex ¡ 

Br y(t) 

p x(t) 2 + y(t) 2 ey + Bz ez 

TheLorentzforceonthechargeq is calculated, the short-hand form v&xBbeing 

used to enter the cross product ~v £ ~ B. The long form is CrossProduct(v,B). 

> F:=q*E+q*(v &x B); 

0 

μ 

B 

F := q B 

d 

@ dt y(t) 

μ 

d 

 

dt 

Bz + 

z(t) 

1 

Br y(t) 

C 

p C 

x (t) 2 + y(t) 2 

0 

B 

+q B 

@ ¡ 

μ 

d 

dt z(t) 

 

Br x(t) 

p 

x(t) 2 + y(t) 2 ¡ 

A ex 

1 

μ 

d C 

x (t) BzC 

dt 

A ey 

0 

B 

+q B 

@ ¡ 

μ 

d 

dt x(t) 

 

Br y(t) 

p 

x(t) 2 + y(t) 2 + 

μ 

d 

dt y(t) 

1 

Br x (t) 

C 

p C 

x(t) 2 + y(t) 2 

A ez 

Taking the mass of the charge to be m, Newton's second law gives the system 

of three component equations. Only the ¯rst ODE is shown here in the text. 

> sys:=seq(m*a[i]=F[i],i=1..3); 

0 

μ μ 2 d B 

sys := m x(t) = q B 

d 

dt2 @ dt y(t) 

μ 

d 

 

dt 

Bz + 

z(t) 

1 

Br y(t) 

C 

p C 

x(t) 2 + y(t) 2 A ;::: 

Vectoria uses the piecewise command to build up the functional forms of Br 

and Bz . The structure of Br is such that there is no radial component of the 

magnetic ¯eld between z = ¡1 andz = +1. In this region, the magnetic ¯eld 

is completely in the z-direction. Outside this central region, the radial ¯eld


component is allowed to vary with z according to a hyperbolic cosine function. 

The function Bz is taken to be constant inside the region z = ¡0:5 to0:5 and 

allowed to grow linearly stronger outside this region. 

> Br:=piecewise(z(t)>1,cosh(z(t)),z(t) Bz:=piecewise(abs(z(t)) B1:=subs(fx(t)=x,z(t)=z,y(t)=0g,B[1]): 

B3:=subs(fx(t)=x,z(t)=zg,B[3]): 

She uses the fieldplot command to represent the direction and magnitude 

of the ¯eld with thick red arrows. A cross-sectional plot of the symmetrical 

magnetic ¯eld is shown in Figure 4.8. Note that the size of the arrows is an 

indication of ¯eld strength, bigger arrows for stronger ¯elds. The ¯eld con¯guration 

shown in the ¯gure should allow magnetic bottle behavior. 

> fieldplot([B3,B1],z=-1.5..1.5,x=-1..1,arrows=THICK,grid= 

[15,15],color=red,scaling=constrained,tickmarks=[3,3]); 

x 

1 

–1 0 

1 

z 

–1 

Figure 4.8: Magnetic ¯eld con¯guration to produce a magnetic bottle. 

To animate the motion of a charge in her magnetic bottle, Vectoria chooses the 

nominal mass and charge values m =1andq = 1 and the initial conditions


x(0) = 0:5, y(0) = 0, z(0) = 0, _x(0) = 0, _y(0) = ¡2, and _z(0) = ¡0:2. 

> m:=1: q:=1: 

> ic:=(x(0)=0.5,y(0)=0,z(0)=0,D(x)(0)=0,D(y)(0)=-2,D(z)(0)=0.2): 

Maple is not able to produce an explicit closed-form analytic solution for the 

nonlinear ODE system, sys, so a numerical solution is generated, the output 

being given as a \listprocedure." 

> sol:=dsolve(fsys,icg,fx(t),y(t),z(t)g,numeric, 

output=listprocedure): 

Evaluating x(t), y(t), and z(t) withsol will allow Vectoria to calculate the 

charge's position at an arbitrary time t. 

> X:=eval(x(t),sol): Y:=eval(y(t),sol): Z:=eval(z(t),sol): 

For example, the charge's x-coordinate at t = 2 time units is now evaluated. 

> X(2); 

¡0:0727500262205896948 

Vectoria will now animate the motion of the charge in the magnetic bottle. She 

takes the total time to be T =24:0time units and will have N =200time 

steps in her animation. The time step size T=N is then calculated. 

> T:=24.0: N:=200: step:=T/N; 

step := 0:1200000000 

The spacecurve command is used to plot (but not display) the entire trajectory 

overthetimeintervalt =0toT , the trajectory being colored with the zhue 

option. To obtain a smooth curve, the minimum number of plotting points is 

taken to be 1000. 

> sc:=spacecurve([X(t),Y(t),Z(t)],t=0..T,shading=zhue, 

numpoints=1000): 

Using a do loop running from n =0toN, the charge is plotted on each time step 

in 3 dimensions as a size-18 red circle superimposed on the entire trajectory. 


> t:=step*n; 

> pp[n]:=pointplot3d([X(t),Y(t),Z(t)],symbol=circle, 

symbolsize=18,color=red); 

> gr[n]:=display(fsc,pp[n]g); 

> end do: 

The motion of the charge in the magnetic bottle is animated with the display 

command, using the insequence=true option. To see the animation execute 

the worksheet, click on the plot, and on the start arrow in the tool bar. 

> display(seq(gr[n],n=0..N),insequence=true,axes=frame, 

labels=["x","y","z"]); 

Vectoria is quite pleased that her piecewise model mimics magnetic bottle behavior. 

But now she has to forsake this bottle for another one, as she rushes 

o® to join her girlfriends and share a bottle of Asti Spumante.


PROBLEMS: 

Problem 4-27: Crossed electric and magnetic ¯elds 

Consider a charge q and mass m moving in the crossed electric and magnetic 

¯elds described by ~ E = E0 sin(!t)êz, ~ B = B0 êx, whereE0, B0, and! are real 

constants. Using the Lorentz force law, derive the system of ODEs governing 

the motion of the charge. Is the system linear or nonlinear? Analytically derive 

the solution of the system, given the initial conditions x(0) = 0, y(0) = 0, 

z(0) = 0, _x(0) = 0, _y(0) = 0, and _z(0) = 0. Taking m =1,q =1,! =5:1, 

E0 =2,andB0 =2:3, animate the motion of the charge (represented as a 

circle), superimposing the motion on the entire trajectory. 

Problem 4-28: Altering the magnetic bottle ¯eld 

In the text recipe, experiment with di®erent functional forms of Br and Bz and 

discuss the e®ect of your alterations on magnetic bottle behavior. Optional: 

Create a continuous ¯eld con¯guration that produces magnetic bottle behavior 

and check that div ~ B=0. 

4.2.4 Golf Is Such an \Uplifting" Experience 

If you watch a game, it's fun. If you play it, it's recreation. 

If you work at it, it's golf. 

Bob Hope, American comedian, Readers Digest, October 1958. 

Heather's girlfriends have invited her to go gol¯ng, but they still need another 

woman to make up a foursome for the Thursday afternoon ladies' day at the 

Metropolis Country Club. Heather assures them that she can probably twist 

her older sister Jennifer's arm to go with them. Jennifer, who has been busy 

preparing lectures for her applied mathematics course, has had little time to 

play golf recently, but is persuaded to go on the basis that a gourmet bu®et 

supper will be available in the clubhouse after the 18 holes are completed. 

Although it won't help her to play golf any better, Jennifer's inborn curiosity 

about how she would go about realistically modeling the °ight of a golf ball 

is piqued, so she trots over to the university science library to look up some 

articles on the physics of golf [Erl83], [MA88], [MH91]. She recalls from her 

undergraduate physics courses that the trajectory of a golf ball was calculated 

assuming that there was no atmospheric drag on the ball, and also not taking 

any possible uplift due to backspin of the ball into account. Jennifer wonders 

what e®ect the inclusion of these two contributions might have on the ball's 

°ight path? For example, can the uplift compensate su±ciently for the aerodynamic 

drag to enable the ball to °y farther? After photocopying the cited 

references, she makes a beeline back to her o±ce, where she commences to simulate 

the °ight of the golf ball with both drag and uplift included. Fortunately, 

the references also provide actual parameter values that can be used in her 

computer modeling.


As she was taught way back in ¯rst-year physics, Jennifer begins by making 

a free-body diagram showing all the forces acting on the golf ball. This diagram 

is reproduced in Figure 4.9. 

φ 

golf ball 

F 

F F 

drag 

lift 

θ 

V 

gravity 

Figure 4.9: Free-body diagram for a golf ball, including drag and lift. 

Three forces are included, the pull of gravity ~ Fgravity downward, the drag force 

~Fdrag in the direction opposite to the instantaneous velocity vector ~ V , and 

the lift force ~ Flift in a direction perpendicular to ~ V . For simplicity, Jennifer 

assumes that the ball's spin axis is in the z-direction and that the translational 

motion is in the x-y plane with x horizontal and y vertical. She makes a call to 

the plots and VectorCalculus packages, 

> restart: with(plots): with(VectorCalculus): 

and enters the velocity vector ~ V = Vx êx + Vy êy. 

> Velocity:=; 

Velocity := Vx ex + Vy ey 

Since the speed V will be needed, it is calculated by taking the dot product of 

the velocity vector with itself and then taking the square root. The dot product 

is entered with the shorthand3 dot syntax. The unit vector (assigned the name 

uv) in the direction of ~ V is then equal to the velocity divided by the speed. 

> Speed:=sqrt(Velocity . Velocity); u[v]:=Velocity/Speed; 

uv := q 

q 

Speed := 

Vx 

Vx 2 + Vy 2 

ex + 

V 

x 

Vx 2 + Vy 2 

q 

3 The long form is DotProduct(Velocity,Velocity). 

V 

y 

Vy 

Vx 2 + Vy 2 

ey


Referring to the freebody diagram, sin(μ) =Vy=V and cos(μ) =Vx=V . 

> sin(theta):=V[y]/Speed; cos(theta):=V[x]/Speed; 

sin(μ) := q 

Vy 

Vx 2 + Vy 2 

cos(μ) := q 

Vx 

Vx 2 + Vy 2 

The unit vector ^ Á pointing in the lift direction is related to the Cartesian unit 

vectors êx and êy pointing along the x- andy-directions, respectively, by the 

relation ^ Á = ¡ sin μ êx +cosμ êy. This relation is entered and labeled uÁ. 

> u[phi]:=; 

uÁ := ¡ q 

Vy 

Vx 2 + Vy 2 

ex + 

q 

Vx 

Vx 2 + Vy 2 

ey 

The gravitational force ~ Fgravity = ¡mgêy, wherem is the mass of the ball and 

g is the acceleration due to gravity. 

> F[gravity]:=; 

Fgravity := ¡mgey 

According to the photocopied references, both the drag force ~ Fdrag and the 

lift force ~ Flift are proportional to the square of the speed. Jennifer labels the 

proportionality constants (per unit mass) as Kdrag and Klift, respectively, and 

enters the two forces. 

> F[drag]:=-K[drag]*m*Speed^2*u[v]; 

q 

Fdrag := ¡Kdrag m 

Vx 2 + Vy 2 Vx ex ¡ Kdrag m 

q 

Vx 2 + Vy 2 Vy ey 

> F[lift]:=K[lift]*m*Speed^2*u[phi]; 

q 

Flift := ¡Klift m Vx 2 + Vy 2 q 

Vy ex + Klift m Vx 2 + Vy 2 Vx ey 

The net force ~ F acting on the golf ball is the vector sum of the three forces. 

> F:=F[gravity]+F[drag]+F[lift]: 

Now, Vx = dx(t)=dt and Vy = dy(t)=dt, wherex(t) andy(t) are the horizontal 

and vertical coordinates of the golf ball at time t. 

> V[x]:=diff(x(t),t): V[y]:=diff(y(t),t): 

Newton's second law of motion is applied in the x- andy-directions. 

> xeq:=diff(V[x],t)=simplify(F[1]/m); 

xeq := d2 

s 

μ 2 μ 

d d 

x(t) =¡ x (t) + 

dt2 dt dt y(t) 

2μ μ μ 

d 

d 

Kdrag x (t) + Klift 

dt dt y(t) 

 

> yeq:=diff(V[y],t)=simplify(F[2]/m);


yeq := d2 

y(t) =¡g ¡ Kdrag 

dt2 +Klift 

s μ d 

s μ d 

x (t) 

dt 

x (t) 

dt 

2 

2 

μ 

d 

+ 

dt y(t) 

2 μ 

d 

dt y(t) 

 

μ 

d 

+ 

dt y(t) 

2 μ 

d 

dt x(t) 

 

Jennifer's task is now to solve the coupled nonlinear ODEs, xeq and yeq, numerically, 

since they do not have analytic solutions. To accomplish this, the 

parameter values must be entered. From the references, she ¯nds that the drag 

and lift parameters for a British golf ball of mass m =0:046 kg and radius 

r =0:0207 m launched at an initial speed V0 =61m/sandwithabackspinof 

about 60 rev/s are approximately equal to each other with a value 0:28. The 

density ½ of dry air at 1 atm pressure and 20 ± Cis½ =1:21 kg/m 3 . Jennifer 

takes g =9:81 m/s and the initial launch angle of the ball to be 16 ± . 

> rho:=1.21: m:=0.046: r:=0.0207: g:=9.81: V0:=61: Angle:=16: 

Drag:=0.28: Lift:=0.28: 

According to the references, the drag and lift coe±cient are given by Kdrag = 

½¼r 2 Drag=(2 m) andKlift = ½¼r 2 Lift=(2 m). These two coe±cients are numerically 

evaluated and temporarily labeled KDrag and KLift. 

> KDrag:=evalf(rho*Pi*r^2*Drag/(2*m)); 

KLift:=evalf(rho*Pi*r^2*Lift/(2*m)); 

KDrag := 0:004957310686 KLift := 0:004957310686 

The initial angle is converted from degrees into radians, 

> Theta:=convert(Angle*degrees,radians); 

4 ¼ 

£:= 

45 

and the initial velocity of the golf ball calculated. 

> Vo:=evalf(V0*); 

Vo := 58:63696345 ex +16:81387871 ey 

The ball starts with a horizontal velocity component of 58:6 m/s and a vertical 

component of 16:8 m/s. Taking the golf ball to be initially positioned at the 

origin of the coordinate system, the initial conditions are then entered. 

> ic:=x(0)=0,D(x)(0)=Vo[1],y(0)=0,D(y)(0)=Vo[2]: 

Jennifer wishes to compare the path followed by the ball when both drag and 

lift are included with the trajectory that results when both contributions are 

absent. So in the following do loop, n = 2 corresponds to the former, while 

n = 1 produces the latter situation. 

> for n from 1 to 2 do 

For n = 1 in the following command line, Kdrag =0andKlift = 0, while n =2 

produces Kdrag = KDrag and Klift = KLift. 

> K[drag]:=(n-1)*KDrag: K[lift]:=(n-1)*KLift: 

The nonlinear ODEs xeq and yeq are solved numerically for x(t) andy(t),


> sol:=dsolve(fxeq,yeq,icg,fx(t),y(t)g,type=numeric); 

and the odeplot command used to plot the trajectory of the ball. The time 

interval for the golf ball to strike the ground again is di®erent for the two 

trajectories. Without drag and lift, the time interval is about 3:42 s, while 

the time interval lengthens to about 7 s when these contributions are both 

included. Jennifer allows for this in the following command line. The linestyle 

option, linestyle=3-n, isusedtoproduceasolidcurveforthesituationthat 

both lift and drag are included (n = 2), and a dashed line when they are not. 

> pl[n]:=odeplot(sol,[x(t),y(t)],0..3.42+3.6*(n-1), 

labels=["x","y"],linestyle=3-n,thickness=2): 

> end do: 

On ending the do loop, Jennifer muses about what usually happens when she 

goes gol¯ng. Invariably, she ends up in the rough, in a sand trap, or has to 

shoot over a tree. So in her simulation, she places a thick brown \tree" 30 

meters high at a distance of 120 meters from where the ball is struck. 

> tree:=plot([[120,0],[120,30]],color=brown,thickness=3): 

Thetwopossibletrajectoriesaresuperimposedwiththetreeplottoproduce 

> display([seq(pl[i],i=1..2),tree],scaling=constrained); 

y 

30 

20 

10 

0 

100 x 200 

Figure 4.10: Solid curve: drag and lift included. Dashed curve: no drag or lift. 

Figure 4.10. Jennifer notes a number of important di®erences between the two 

trajectories. Despite the e®ect of drag, her simulation indicates that a golf ball 

in the real world rises higher (thus, clearing the \tree" here) and consequently 

travels farther because of the lift on the golf ball due to backspin. This is not 

too surprising, since the ball stays in the air about twice as long as when both 

drag and lift are neglected. She further notices that the ball's path actually 

bends upward before reaching the zenith of its trajectory, a phenomenon that 

she has seen on the golf course, especially when the club pro is playing. In 

her model calculation the ball has traveled about 203 m (222 yards), hardly a 

long shot by the standards of someone like Tiger Woods, who averages about 

300 yards a drive. However, she is con¯dent that she could modify the initial 

conditions to mimic one of Tiger's shots. Undoubtedly Heather, who is a big 

Tiger Woods fan, will ask her to do it when she learns about the simulation.


As much fun as she has had, Jennifer suddenly realizes that she had better 

spend a little time hitting balls down at the local driving range before she goes 

out with Heather and her friends. Otherwise, she might end up proving to be 

an embarrassment to her sister. Now where did she put those golf clubs? 

PROBLEMS: 

Problem 4-29: The Bo®o brothers go gol¯ng 

Syd and Benny Bo®o have taken an afternoon o® from their realty business 

and gone gol¯ng at the Metropolis Country Club. After a good approach shot 

on a level fairway, Benny Bo®o has placed his golf ball 100 m from the hole. 

He selects a 9-iron and hits the ball perfectly, landing practically at the hole. 

If both lift and drag are included and all other parameters of the text recipe 

prevail, at what angle to the horizontal (in degrees) was the ball hit? How high 

did the ball rise? 

Problem 4-30: Syd Bo®o makes a hole in one 

Syd and Benny Bo®o's golf match continues and they move onto the next hole, 

a par 3. The tee-o® point is elevated 20 m above the hole, the ball having to 

travel over an intervening lily pond. The horizontal distance from the tee-o® 

position to the hole is 160 m. Syd miraculously scores a hole in one, which 

means that drinks are on him at the clubhouse after the match. Assuming 

that both lift and drag are included and all other parameters of the text recipe 

prevail, at what angle did Syd's ball ascend in order for him to make the hole 

in one? How high did the ball rise relative to the hole? If the green is circular 

(radius 5 m) with the hole at its center, and the pond stretches up to the edge of 

the green, would the ball have ended up in the pond if lift had been neglected? 

Or would the ball have overshot the green? 

Problem 4-31: Rocky Mountain high 

At a Rocky Mountain golf resort, the elevation is such that the density of air is 

½ =1:0 kg/m3and g =9:8 m/s2 .Howmuchfartherwouldtheballtravelthan 

for the sea-level data given in the text recipe? 

Problem 4-32: Horizontal drive 

A golf ball that has been teed up 3 cm above the ground is hit badly and leaves 

the tee horizontally. Including lift and drag and taking all other parameters as 

in the text recipe, plot the trajectory of the ball and determine where it strikes 

the ground, assuming that the fairway is level. How does the distance compare 

to the situation in which there is no lift? 

Problem 4-33: Vertical shot 

As part of its testing routine, a golf ball manufacturing company tests the 

performance of one of its golf balls by shooting it vertically into the air with 

the same spin and other parameters as in the text recipe. If lift and drag are 

included, where would the golf ball land?

4.3. VARIATIONAL CALCULUS MODELS 185 

4.3 Variational Calculus Models 

The models in this section are based on the following mathematical framework 

developed by the mathematicians Euler and Lagrange. Consider a function 

y(x) with¯xedvaluesy(x0) =y0 and y(x1) =y1 at two distinct points A and 

B, respectively. Form a speci¯c function F of x, y(x) andy 0 ´ dy(x)=dx, i.e., 

F = F (x; y; y 0 ). In our examples, F will be determined by the speci¯cation of 

the problem. Among all functions y(x) connectingA and B, ¯ndthosethat 

give an extremum (minimum or maximum, usually) to the integral 

I[ y ]= 

Z x1 

x0 

F (x; y(x);y 0 (x)) dx: (4.13) 

The value I of the integral depends on the functional form chosen for y(x). 

Rather than trying di®erent y(x) ad in¯nitum, it turns out that the y(x) that 

yields an extremum value for I must satisfy the Euler{Lagrange equation [FC99] 

μ 

@F d @F 

¡ 

@y dx @y 0 

 

=0: (4.14) 

The resulting ODE is often nonlinear in nature, but may be susceptible to 

analytic solution. If not, it can be solved numerically. 

4.3.1 Dress Design, the Erehwonese Way 

Haute Couture should be fun, foolish and almost unwearable. 

Christian Lacroix, French couturier (1951{) 

On the planet Erehwon the leading dress designer, Gino Spoo¯a, has created a 

new design, called the lampshade look, 4 as shown in Figure 4.11, to accommo- 

Figure 4.11: The lampshade look in dress design. 

4 The supporting straps are not displayed. 

x 

y


date the shapes of its most beautiful (by Erehwon standards) female citizens. 

Being an amateur scientist on the side and also to keep down material cost, he 

wantsthesurfaceareaofclothintheone-piecedresstobeaminimum. The 

smaller upper ring at x = 0 is of radius y = 1 sretem (the local unit of length), 

and the lower, larger, ring at x = 1 sretem is of radius y = 2 sretem. To answer 

the question of what the shape of the surface should be to minimize the area, 

Gino consults one of Erehwon's most prominent mathematicians, Greg Arious 

Nerd, who says that this problem was solved many millennia ago. Greg then 

proceeds to outline the solution to the problem for his dress-designing friend. 

For the bene¯t of readers who may not understand the peculiar Northern ErehwonesedialectspokenbyGreg,weshalltranslatewhathehastosay. 

The 

authors are not totally °uent in this language, so bear with us if we occasionally 

are in doubt about the exact words and thus resort to paraphrasing what Greg 

says instead of reporting his remarks verbatim. 

\It's all very simple," Greg starts out. \The surface area of the dress can be 

generated by revolving the curve y(x) aboutthex-axis. So what 5 is the shape 

of y(x)? 

ds 

y 

y(x) 

Figure 4.12: Determining the surface area of the dress. 

Referring to Figure 4.12, an element dA of surface area is given by 

dA =(2¼y) ds =2¼y p (dx) 2 +(dy) 2 =2¼y p 1+(y 0 ) 2 dx; (4.15) 

where the prime indicates a derivative with respect to x. So the surface area 

between x =0andx =1is 

A =2¼ 

Z 1 

0 

y p 1+(y 0 ) 2 dx: (4.16) 

5Author's note: This appears to be a rhetorical question on Greg's part, since Gino hasn't 

opened his mouth yet. 

x


The goal is to ¯nd the shape y(x) that minimizes A." 

\I know that, Greg," Gino interjects, \but get to the point. As my business 

associates from Earth say, time is money. What is the bottom line? How many 

square sretem of cloth will I need?" 

A little perturbed by this sign of impatience, Greg's dialect has become 

stronger and his response is di±cult for us to precisely translate, but the gist of 

it goes something like this. Long ago, some ancestors on his mother's side had 

adopted a trial-and-error approach to ¯nding y(x). This consisted in choosing 

some form for y(x) and performing the integration to calculate A. Then the 

venerable ancestors would choose another form and see whether it generated a 

smaller value for A. With a touch of hyperbole, we think, Greg said that some 

kept repeating this process until either they became residents in the Erehwon 

asylum or inductees into the Erehwon Academy of Mathematics. According 

to his unbiased (?) version of history, a better approach was discovered long 

ago by one of Greg's illustrious ancestors on his father's side, Hil Arious Nerd. 

Evidently, Hil independently discovered the same method as developed by Euler 

and Lagrange on far-distant Earth. The method is as follows. 

Label the integrand of the area integral as F , i.e., F =2¼y p 1+(y0 ) 2 . 

The curve y(x) that minimizes the area is the solution of the Euler{Lagrange 

equation(knownastheNerdequationonErehwon) 

@F d @F 

¡ =0: (4.17) 

@y dx @y 0 

Solving this type of problem by hand has been a standard assignment given in 

the past by perverse professors on Erehwon6 to their students. 

It appears that Greg has calmed down so that we can understand him better, 

so let's pick up Greg's detailed narrative once again. 

\It'snowfrownedupontoin°ictmentalpainonstudents(anddressdesigners) 

by performing hand calculations when computers can do all the algebra for 

you with no errors. At the Erehwon Institute of Technology we use the Elpam 

computer algebra system, so let's use it to crack your problem. 

I will begin by loading the VariationalCalculus library package which contains 

the EulerLagrange command for generating the left-hand side of the 

Euler{Lagrange (Nerd) equation (4.17). 

> restart: with(VariationalCalculus): 

Next let's enter the integrand, F = y(x) p 1+(dy(x)=dx) 2 ,omittingthefactor 

of 2 ¼, which would ultimately cancel out of the equation, 

> F:=y(x)*sqrt(1+diff(y(x),x)^2); 

s 

μ 

d 

F := y(x) 1+ 

dx y(x) 

2 

The EulerLagrange command is applied to F , the second argument (x) being 

the independent variable, the third argument the dependent variable. The 

result is then simpli¯ed. 

6 Author's note: This quaint, but archaic, custom is still widely practiced on Earth.


> eq:=simplify(EulerLagrange(F,x,y(x))); 

8 

>< 

y(x) 

eq := s 

μ 

d 

>: 1+ 

dx y(x) 

μ 

d 

1+ 

dx 

 

= K1; 

2 y(x) 

2 μ 

2 d 

¡ y(x) y(x) 

dx2 Ã μ 

d 

1+ 

dx y(x) 

9 

>= 

! 

2 

(3=2) 

>; 

Two results are generated in eq. The expression not involving the constant K1 

is just the lhs of the Euler{Lagrange equation. It can be extracted using the 

remove command to remove the expression containing K1 from eq.Inclusionof 

the brackets at the end of the command line removes the brackets that would 

otherwise enclose the following answer. 

> eq2:=remove(has,eq,K[1])[]; 

μ 

d 

1+ 

dx 

eq2 := 

y(x) 

2 μ 

2 d 

¡ y(x) y(x) 

dx2 Ã μ 

d 

1+ 

dx y(x) 

! 

2 

(3=2) 

The relevant governing nonlinear ODE that must be solved for y(x) follows on 

setting the numerator of eq2 equal to zero. This ODE would have been tedious 

to derive by hand. 

> ode:=numer(eq2)=0; #relevant ODE 

μ 

d 

ode := 1 + 

dx y(x) 

2 μ 

2 d 

¡ y(x) y(x) =0 

dx2 If proceeding by hand, one could reduce the second-order ode to a ¯rst-order 

ODE by setting p = dy=dx, sothatd2y=dx2 = dp=dx =(dy=dx)(dp=dy) = 

p (dp=dy). The variables can then be separated and the integrations performed, 

yielding p ´ dy=dx with one arbitrary constant. But this ¯rst integral is already 

contained in eq, being the term involving the constant K1. This ¯rst integral 

can be extracted from eq using the following select command: 

> eq3:=select(has,eq,K[1])[]; #first integral 

y(x) 

eq3 := s 

μ 

d 

1+ 

dx y(x) 

 

= K1 

2 

The ¯rst-order ODE eq3 is solved for y(x), a second constant C1 appearing in 

the solution, which contains two equivalent answers (not displayed). 

> sol:=dsolve(eq3,y(x)); 

I will select the ¯rst answer in sol and simplify the rhs of the result. 

> y:=simplify(rhs(sol[1])); 

y := 1 

2 (K1 2 2(x¡ C1) x¡ C1 

( ) (¡ ) 

+ e K1 ) e K1


We now have the general curve y that will minimize the area, but we have to 

determine the two constants K1 and C1 . Evaluating y at x = 0 and setting 

the result to 1 yields the ¯rst boundary condition, while evaluating y at x =1 

and setting the result to 2 yields the second one. The two boundary conditions 

are then solved numerically for K1 and C1 and the solution is assigned. 

> bc1:=eval(y,x=0)=1: bc2:=eval(y,x=1)=2: 

> sol2:=fsolve(fbc1,bc2g,fK[1],_C1g); assign(sol2): 

sol2 := fK1 =0:9499888273; C1 = ¡0:2581775581g 

On expanding and simplifying, y is now completely determined, 

> y:=simplify(expand(y)); 

y := 0:3438580549 e (¡1:052643959 x) (1:052643959 x) 

+0:6561419450 e 

and can be plotted, the vertical view being controlled." 

> plot(y,x=0..1,view=[0..1,0..2.5],tickmarks=[4,4]); 

2.5 

2 

1.5 

1 

0.5 

0 

0.2 0.4 x 0.8 1 

Figure 4.13: The curve y(x) that minimizes the surface area of the dress. 

\That's O.K., Greg," Gino interrupts, \but a 3-dimensional picture of the dress 

would be nicer, and more importantly, what is the minimum surface area?" 

\Since you're getting impatient, let's load the Student[Calculus1] package, 

> with(Student[Calculus1]): 

and use the SurfaceOfRevolution command to obtain the minimum area A. 

> A:=evalf(SurfaceOfRevolution(y,x=0..1)); 

A := 13:06167468 

The minimum surface area of your lampshade dress is about 13 square sretem. 

Using the same command, but with the option output=plot included, 

> SurfaceOfRevolution(y,x=0..1,output=plot); 

will yield the desired surface of revolution, i.e., your \dress," with the form of 

y(x) included. The dress will be oriented horizontally on the computer screen 

but can be `dragged' into a vertical position."


PROBLEMS: 

Problem 4-34: Other curves 

Calculate the surface areas for dresses generated with some other curves y(x) 

joining the endpoints (x = 0, y = 1) and (x = 1, y = 2). How do these 

areas compare with the area obtained in the text recipe? Plot the surfaces of 

revolution generated by the curves. 

Problem 4-35: Brachistochrone 

Consider the smooth curve y(x) joining the two points A and B in Figure 4.14. 

A 

y 

bead 

x 

B 

(x ,y) 

0 0 

Figure 4.14: Brachistochrone problem. 

(a) If a small bead slides without friction between A and B, whaty(x) corresponds 

to the shortest time? This famous brachistochrone (\shortest 

time" in Greek) problem was proposed and solved by Jakob and Johann 

Bernoulli. Hint: First show that the time is given by 

t = 

Z x0 

0 

p (1+(y 0 ) 2 )=(2 gy) dx; 

where g is the acceleration due to gravity. 

(b) Let the coordinates of B be x0 =0:5 m,y0 =0:5 mfortheremainderof 

this problem. What is the shortest time for the bead to slide from point 

A at the origin to point B? 

(c) Plot the path of the bead from A to B. 

(d) What is the speed of the bead when half the shortest time has elapsed? 

(e) What is the position of the bead at this latter time? 

Problem 4-36: Fermat's principle and mirages 

Fermat's principle states that a ray of light in a medium with a variable refractive 

index will follow the path that requires the shortest traveling time. For 

a two-dimensional situation, write down the integral that must be minimized 

to obtain such a path. Note that the speed of light in a medium of refractive 

index n is c=n, wherec is the vacuum speed of light. If the refractive index 

varies vertically from ground level according to the formula n = n0 (1 + ®y), 

with n0 > 0and®>0, determine the equation for the path taken by a light


ray. By plotting a light ray path for physically reasonable parameter values, 

discuss how your answer may be related to the phenomenon of mirages. 

Problem 4-37: A di®erent refractive index 

Making use of Fermat's principle, prove that a light ray will follow a semicircular 

path in a medium whose refractive index n(x; y) equals1=y. Plot a typical path. 

Problem 4-38: Geodesic 

The geodesic between two points is the curve that gives the shortest distance. 

Show that the geodesics on the surface of a sphere are the arcs of great circles. 

(A great circle is a curve resulting from the intersection of a sphere with a plane 

passing through the center of that sphere.) Create a three-dimensional plot of 

the geodesic between New York City and Sydney, Australia, assuming that the 

Earth is spherical. You will have to look up the necessary parameter values. 

4.3.2 Queen Dido Wasn't a Dodo 

Mathematics is the queen of the sciences. 

Carl Friedrich Gauss, German mathematician (1777{1855) 

According to the Aeneid, written by the Roman poet Virgil in the ¯rst century 

BC, the Phoenician Queen Dido was able to convince the North African 

ruler King Jambas to give her as much land as she could enclose with an oxhide. 

Being rather clever, she had the hide cut into very thin strips, the ends 

stitched together, and was able to stake out a sizable area along the Mediterranean 

coast on which she built the city of Carthage (now Tunis). Queen Dido's 

problem was to lay out the joined oxhide strips that had a total ¯xed length in 

such a way as to maximize the area enclosed. Can you suggest what shape the 

perimeter traced out by the joined strips might take? Problems of this type 

that involve maximizing an area enclosed by a perimeter of ¯xed length are 

called isoperimetric (constant perimeter) problems. 

A recipe is now given to solve Queen Dido's problem. A strip of oxhide of 

¯xed length L>ais connected at its ends to the points (x =0,y =0)and 

(x = a, y = 0). The area enclosed by a strip of shape y(x) andthex-axis 

between x =0andx = a is A = R a 

y(x) dx and the length of the strip is given 

by L = R a 

p 0 

1+(y0 ) 2 7 dx: If, say, a = 1 milion and L =1:5 milion, what is 

0 

the shape y(x) that maximizes the area subject to the length constraint? Plot 

y(x) and determine the value of the maximum area. 

If the integrands of A and L are labeled as F and G, respectively, the shape 

y(x) may be found by solving the Euler{Lagrange equation of the previous 

subsection with F replaced by FF = F + ¸G,where¸ is an undetermined 

parameter. See, e.g., references [MW71] and [Boa83]. 

The VariationalCalculus package is loaded, 

> restart: with(VariationalCalculus): 

7 A milion is a Roman mile, so this recipe involve one very big oxhide!


and the integrands F and G are entered. 

> F:=y(x); G:=sqrt(1+diff(y(x),x)^2); 

s 

μ 

d 

F := y(x) G := 1+ 

dx y(x) 

2 

Then FF = F + ¸G is formed. 

> FF:=F+lambda*G; 

s 

μ 

d 

FF := y(x)+¸ 1+ 

dx y(x) 

2 

The EulerLagrange commandisappliedtoFF . 

> eq:=EulerLagrange(FF,x,y(x)); 

8 

μ 

d 

>< ¸ 

dx 

eq := 1+ 

>: 

y(x) 

2 μ 

2 d 

y(x) 

dx2 Ã μ 

d 

1+ 

dx y(x) 

μ 

2 d 

¸ y(x) 

dx2 ! ¡ s 

2 

(3=2) μ 

d 

1+ 

dx y(x) 

 

; 

2 

s 

μ 

d 

y(x)+¸ 1+ 

dx y(x) 

μ 

d 

2 

dx 

¡ 

y(x) 

2 

¸ 

s 

μ 

d 

1+ 

dx y(x) 

9 

>= 

 

= K1 

2 

>; 

The ¯rst integral is extracted from eq by selecting the term that has K1 in it, 

and removing the brackets that would otherwise appear in ode. 

> ode:=select(has,eq,K[1])[]; 

s 

μ 

d 

ode := y(x)+¸ 1+ 

dx y(x) 

μ 

d 

2 

dx 

¡ 

y(x) 

2 

¸ 

s 

μ 

d 

1+ 

dx y(x) 

 

= K1 

2 

A general analytic solution to the ¯rst-order nonlinear ode is sought, yielding 

two answers that di®er by the sign in front of the square root. 

> sol:=dsolve(ode,y(x)); 

p 

sol := y(x) =K1 + ¸2 ¡ C1 2 ¡ x2 +2x C1 ; 

p 

y(x) =K1 ¡ ¸2 ¡ C1 2 ¡ x2 +2x C1 

The rhs of, say, the negative square root result (second one) is chosen. 

> Y:=rhs(sol[2]); 

p 

Y := K1 ¡ ¸2 ¡ C1 2 ¡ x2 +2x C1


The boundary conditions that Y =0atx =0andatx = a are imposed in bc1 

and bc2 , which are then solved for K1 and C1 insol2 . 

> bc1:=eval(Y,x=0)=0; bc2:=eval(Y,x=a)=0; 

p 

bc1 := K1 ¡ ¸2 ¡ C1 2 =0 

p 

bc2 := K1 ¡ ¸2 ¡ C1 2 ¡ a2 +2a C1 =0 

> sol2:=solve(fbc1,bc2g,fK[1],_C1g); 

sol2 := f C1 = a 

2 ;K1 

p 

4 ¸2 ¡ a2 = 

g 

2 

Assigning sol2 ,weobtainYin the following form. 

> assign(sol2): Y:=Y; 

p r 

4 ¸2 ¡ a2 Y := 

¡ ¸ 

2 

2 ¡ 1 

4 a2 ¡ x2 + xa 

To determine ¸, the constraint condition R a 

G(y(x) =Y ) dx = L is imposed. 

0 

> eq2:=simplify(int(eval(G,y(x)=Y),x=0..a))=L; 

r 

¸ 

eq2 := ¡ 

2 

¡4 ¸2 + a2 p p 

¡4 ¸2 + a2 (¡ln(¡a + ¡4 ¸2 + a2 ) 

+ln(a + p ¡4 ¸2 + a2 )) = L 

The values a =1andL =1:5 are entered, and eq2 numerically solved for ¸, 

the search range ¸ =0to0:6being speci¯ed. 

> a:=1: L:=1.5: lambda:=fsolve(eq2,lambda=0..0.6); 

¸ := 0:5014101096 

The ¯nal form of Y is then displayed, and jY j plotted in Figure 4.15. 

> Y:=Y; plot(abs(Y),x=0..a,scaling=constrained); 

Y := 0:03757789244 ¡ p 0:0014120980 ¡ x2 + x 

0.4 

0.3 

0.2 

0.1 

0 

0.2 0.4 0.6 0.8 1 

x 

Figure 4.15: Shape of the oxhide strip that maximizes the enclosed area.


TheabsolutevalueofY was plotted, because in some executions of the work 

sheet Y will be negative because the ordering of the answers in sol is reversed. 

The shape of the strip that maximizes the area is a circular arc. 

The area R a 

Ydxis now calculated, the absolute value being taken to avoid 

0 

a possible negative area if the \wrong" solution is selected in sol. 

> A:=abs(int(Y,x=0..a)); 

A := 0:3572686358 

The maximum area is about 0:36 square milions. Without doing any mathematical 

calculation, can you suggest why this is a maximum and not a minimum? 

PROBLEMS: 

Problem 4-39: Maximum volume of a solid 

Acurvey(x) of length 2 is drawn between the points (0, 0) and (1, 0) in such 

a way that the solid obtained by rotating the curve about the x-axis has the 

largest possible volume. 

(a) Determine y(x). 

(b) Plot y(x) over the range x =0to1. 

(c) What is the value of y at x =0:5? 

(d) Make a three-dimensional plot of the solid. 

(e) What is the volume of the solid? 

Problem 4-40: The catenary curve 

Consider a uniform cable of length L =1:5 km and mass per unit length ² =1 

kg/m suspended between the two endpoints (¡a=2, b) and(a=2, b), where 

a = b =1km. 

(a) Determine the equilibrium shape (referred to as a catenary curve) of the 

cable. Hint: The potential energy of the cable will be at a minimum when 

the cable has its equilibrium shape. Take g =10m/s2 . 

(b) Plot the equilibrium shape of the cable. 

(c) If the cable crosses a very deep Himalayan gorge with the river located a 

distance b below the endpoints of the cable, what is the distance between 

the minimum in the cable and the river? 

(d) What is the distance down to the river from a point one-third of the way 

along the cable? 

(e) What force is exerted on the supports at the endpoints of the cable? 

(f) What length of cable should be used if the sag in the middle of the cable 

is not to exceed 50 m?


4.3.3 The Human Fly Plans His Escape Route 

A product of the untalented, sold by the unprincipled 

to the utterly bewildered. 

Al Capp, American cartoonist, comment on abstract art (1907{1979) 

Mr. X, whom the popular press have dubbed the \human °y," plans to scale the 

outside vertical wall of one of Metropolis's tallest skyscrapers, the Metropolis 

Stock Exchange, using a combination of his rock-climbing skills and suction 

cups. Knowing that once the climb has begun, the police will quickly learn of 

his exploit and try to capture him, Mr. X plans to execute a daring escape. 

He intends to evade the law enforcement o±cers by sliding down a thin, but 

strong, wire to a yet-undetermined window at a lower elevation in the slanted 

roof of the Museum of Modern Art directly across the street (see Figure 4.16). 

The window will be left open so that Mr. X can zoom through to the room in- 

y 

Figure 4.16: Mr. X sliding down wire with a police helicopter hovering nearby. 

side, brake rapidly, and make a (hopefully) relatively soft landing on the °oor. 

Catching the police by surprise and taking advantage of the museum's many 

exits, Mr. X hopes once again to avoid capture and expand his ever-growing 

legend. However, being somewhat cautious, despite an otherwise harebrained 

scheme, Mr. X decides to check the details of the escape route by conferring 

with his mathematician friend Mike. 

\Mike, the roof of the stock exchange is slightly over 200 meters above 

street level and the police will surely be waiting for me on the rooftop and in 

the street below. They will probably even have a police helicopter hovering 

nearby. If I were to connect a wire to the stock exchange outer wall at the 200 

meter point, what shape should the wire have, and what point on the slanting 

x


museum surface should it be connected to, in order to minimize the time of 

descent? What is the time of descent and what speed would I have acquired 

on reaching the slanted museum roof? Fortunately, the museum roof has many 

windows to let in light and I could arrange for the appropriate window to be 

left open. A few relevant facts are as follows. The museum roof slants at 45 ± 

to the horizontal. The street is 50 meters wide, and the vertical section of the 

museum wall adjoining the slanted roof is 50 meters tall." 

\You're crazy," Mike replies, \but you do pose an interesting mathematical 

problem. I can tackle the solution using the Euler{Lagrange equation. Referring 

to Figure 4.16, let's choose to measure x to the right and y downward from the 

point at which the wire is attached to the stock exchange building. We need 

to ¯nd a general expression for the time of descent along the wire. Let the 

equation of the wire be y(x). Neglecting friction and equating the increase 

in kinetic energy of a falling mass to its decrease in potential energy yields a 

speed v = p 2 gy,wheregistheacceleration due to gravity. But v = ds=dt, 

where ds = p 1+(dy=dx) 2 dx is an element of arc length along the wire. If the 

(unknown) coordinates of the contact point on the museum roof are (x1 , y1 ), 

the time of descent will be given by 

Z p 

x1 

1+(dy=dx) 2 

T = p dx: 

0 2 gy 

Since the factor p 2 g will cancel out, the integrand to use in the Euler{Lagrange 

equation can be taken to be 

F =( p 1+(dy=dx) 2 )= p y: 

I will now use Maple to solve the problem. Loading the necessary library 

packages, I enter the integrand F . 

> restart: with(VariationalCalculus): with(plots): 

> F:=sqrt((1+diff(y(x),x)^2)/y(x)); 

v 

u 

t 

F := 

1+ 

μ 

d 

dx y(x) 

2 

y(x) 

The EulerLagrange command is applied to F , and the ¯rst integral result, involving 

the constant K1, is selected from the output (not shown) and simpli¯ed 

in ode. 

> EL:=EulerLagrange(F,x,y(x)); 

> ode:=simplify(select(has,EL,K[1])[]); 

1 

ode := v 

u 

t 1+ 

μ 

d 

dx y(x) 

= K1 

2 

y(x) 

y(x) 

A general solution to the ¯rst-order nonlinear ode is sought.


> sol:=dsolve(ode,y(x)); #implicit solution 

q 

y(x) ¡ K1 

sol := x + 

2 y(x) 2 

¡ 

K1 

1 

0p 

K1 B 

arctanB 

@ 

2 

2 

μ 

y(x) ¡ 1 1 

2 K1 2 

 

q 

y(x) ¡ K1 2 y(x) 2 

1 

C 

A 

p 

K1 K1 2 

¡ C1 =0; 

q 

y(x) ¡ K1 

x ¡ 

2 y(x) 2 0p 

K1 B 

arctan B 

@ 

+ 1 

K1 2 

2 

μ 

y(x) ¡ 1 1 

2 K1 2 

 

q 

y(x) ¡ K1 2 y(x) 2 

1 

C 

A 

p 

K1 K1 2 

¡ C1 =0 

A formidable-looking implicit solution has been produced, with two forms 

present. It turns out that we must select the form having the positive square 

root (the ¯rst solution for this particular run). For later convenience, I will also 

replace the awkward constant symbols K1 and C1 with1= p A and B. 

> ans:=subs(fK[1]=1/sqrt(A),_C1=Bg,sol[1]); #select + root 

ans := x + p r 

A y(x) ¡ y(x)2 

0r 

1 A 

p B (y(x) ¡ 

A arctan B A 2 

@ 

1 

¡ 

A 2 

) 

r 

y(x) ¡ y(x)2 

1 

C 

A 

r 

A 

¡ B =0 

1 

A 

To proceed any further with the implicit answer, its square-root structure suggests 

a trigonometric substitution of the form y = A sin 2 μ,whereμisan angular 

parameter not connected to any geometrical feature in our picture. Here A and 

B are assumed to be positive, as are sinμ and cos μ as well. 

> assume(A>0,B>0,sin(theta)>0,cos(theta)>0): 

> Y:=y(x)=A*sin(theta)^2; 

Y := y(x) =A sin(μ) 2 

Then Y is substituted into the answer, and simpli¯ed with the trig option. 

> sol1:=simplify(subs(Y,ans),trig); 

sol1 := x + A sin(μ)cos(μ)+ 1 

μ 

1 ¡1+2cos(μ) 

A arctan 

2 2 

2 

¡ B =0 

sin(μ)cos(μ) 

Then we solve sol1 for x and combine with respect to trig terms. 

> x:=combine(solve(sol1,x),trig); 

x := ¡ 1 

μ 

1 

cos(2 μ) 

A sin(2 μ) ¡ A arctan 

+ B 

2 2 sin(2 μ) 

So, now we have expressions for the coordinates x and y of our sought-after 

curve in terms of A, B, andμ. If A and B can be found and the range of μ


determined, the curve that minimizes the time of descent will be known. 

At the starting point x = 0, we shall choose to set the parameter μ equal 

to zero. Some care must be taken in evaluating the arctan term at μ =0. The 

limit must be taken from the positive side (i.e., from the \right") of μ =0. 

> eq:=limit(x,theta=0,right)=0; 

eq := B ¡ A¼ 

4 =0 

The resulting eq is easily solved for the constant B. 

> sol2:=B=solve(eq,B); 

sol2 := B = A¼ 

4 

The solution sol2 is assigned and x displayed. 

> assign(sol2); x:=x; 

x := ¡ 1 

μ 

1 

cos(2 μ) 

A sin(2 μ) ¡ A arctan 

+ 

2 2 sin(2 μ) 

A¼ 

4 

We still have to determine the constant A and the range of μ. Now it gets a bit 

tricky. Wehavenoideayetwhatthecoordinates(x1 ; y1 ) of the endpoint on 

the museum roof should be. 

Hand me that copy of Mathematical Methods in Physics [MW71]. Ah, here 

we go. The case in which one endpoint of the sought-after curve is ¯xed and 

the other endpoint is allowed to vary along a line g(x; y) = 0 is considered. 

It is shown that if the Euler{Lagrange function F is of the structure8 F = 

f(x; y) p 1+(y0 ) 2 ; which it is in our case, then the condition for determining 

the unknown endpoint is that the slope of y(x) must satisfy the condition y 0 = 

(@g=@y)=(@g=@x) atthatpoint. Butthisisjustamathematicalstatement 

that the curve y(x) of quickest descent must intersect the destination curve 

g(x; y) = 0 at right angles. Here the equation for the slanting museum roof is 

the straight line g(x; y) =y + x ¡ 200 = 0: But both @g=@y and @g=@x are 

equal to 1, so we have y 0 = 1 at the museum roof. Since the parameter μ has 

been introduced, the slope of y(x) mustbecalculatedintermsofμ. 

> slope:=simplify(diff(rhs(Y),theta)/diff(x,theta)); 

slope := cos(μ) 

sin(μ) 

Setting the slope to 1, the value £ that the parameter μ must have at the 

museum roof is determined, assuming that μ is positive. 

> Theta:=solve(slope=1,theta) assuming theta>0; 

£:= ¼ 

4 

8If F is not of this structure, the endpoint condition is more complicated, taking the form 

³ 

0 @F 

F ¡ y 

@y 0 

´ 

@g @F 

¡ 

@y @y 0 

@g 

@x =0:


Evaluating x at μ =£mustyieldx1 , 

> eq1:=x1=eval(x,theta=Theta); 

eq1 := x1 = ¡ 1 1 

A + 

2 4 A¼ 

while evaluating the rhs of Y at μ =£mustyieldy1 . 

> eq2:=y1=eval(rhs(Y),theta=Theta); 

eq2 := y1 = A 

2 

Finally, the straight-line equation for the museum roof at (x1; y1 )isentered, 

> eq3:=x1+y1=200; 

eq3 := x1 + y1 =200 

and the three equations solved for A, x1, andy1 . 

> sol3:=fsolve(feq1,eq2,eq3g,fA,x1,y1g); assign(sol3): 

sol3 := fA =254:6479090; y1 =127:3239545; x1 =72:67604557g 

Thus, the endpoint of the wire on the museum roof has coordinates x1 ¼ 72:7, 

y1 ¼ 127:3. The endpoint is about 127 meters below the starting point on the 

stock exchange wall. I hope you realize, X, that this means that you will be 

dropping more than 40 stories as you slide along the wire! 

I will plot the curve (colored blue in graph gr1) of minimum descent to the 

museum roof (colored red in gr2).Thewireandroofarelabeledingr3. 

> gr1:=plot([x,-rhs(Y)+200,theta=0..Theta],color=blue): 

> gr2:=plot([[50,0],[50,50],[125,125]],color=red,thickness=2): 

> gr3:=textplot(f[40,140,"wire"],[100,85,"roof"]g): 

> display(fgr1,gr2,gr3g,scaling=constrained,tickmarks=[3,3]); 

200 

100 

0 

wire 

roof 

50 100 

Figure 4.17: Shape of wire that minimizes the descent time to the roof.


OK, X, there's your curve shown in Figure 4.17. You can arrange for a window 

to be left open at the point where the curve intersects the museum roof. 

Now that we know the distance through which you will be dropping, we 

can easily calculate your velocity from the expression v = p 2 g y1. Takingthe 

gravitational acceleration to be g =9:81 m/s 2 , 

> g:=9.81: vel:=sqrt(2*g*y1); 

vel := 49:98095625 

you will be traveling at almost 50 m/s as you pass through the window on 

the museum roof, if you haven't been braking. This is an upper bound on 

your speed, of course, because we have completely neglected friction and air 

resistance. I can see by the look on your face that you don't quite appreciate 

how fast this is. So, let me convert the velocity to kilometers per hour. 

> vel2:=convert(vel,units,m/s,km/h)*km/h; 

179:9314425 km 

vel2 := 

h 

Your theoretical speed would be almost 180 km/hr but, as I said, in reality it 

would be somewhat less. The time of descent can now be calculated from 

Z £ 

T = 

p 

(dx=dμ) 2 +(dy=dμ) 2 

dμ; 

2 gy(μ) 

viz:; 

μ=0 

> T:=int(sqrt(diff(x,theta)^2+diff(rhs(Y),theta)^2) 

/sqrt(2*g*rhs(Y)),theta=0..evalf(Theta)); 

T := 5:659009626 

The minimum time of descent is about 5:7 seconds. Again, friction, air resistance, 

and any braking on your part would lengthen this time. Given the 

estimates that we have come up with, do you still want to go through with your 

bizarreescaperoute?" 

\Sure, Mike," Mr. X replies, \and I am counting on you to help attach the 

wire to the appropriate window in the museum roof and leave it open for me." 

\Oh no," Mike groans. \If I get caught, Vectoria's parents are sure to call 

o® our recently announced engagement." 

PROBLEMS: 

Problem 4-41: A di®erent escape route 

Determine Mr. X's escape route if the museum roof slanted at 30 ± to the horizontal, 

the street is 50 m wide, and the vertical section of the museum wall 

adjoining the slanted roof is 25 m tall. Make a plot similar to that in Figure 

4.17. How close to the edge of the roof would Mr. X land? What would be 

his speed at this point? Neglect friction and air resistance. 

Problem 4-42: In search of reality 

Suppose that Mr. X decides that sliding down a steel wire at nearly 180 km/hr 

is a little too insane even for his bizarre tastes. His friend Mike is unavailable for 

technical advice because he is busy with preparations for his upcoming wedding


to Vectoria. So Mr. X comes to you and asks what speed he would reach and 

how long would his slide take if air resistance and friction are included, the same 

wire shape being used. He tells you that his mass is about 60 kg. You may 

assume that the force due to air resistance is 0:5 v 2 newtons, where v is Mr. X's 

speed, and that the coe±cient of kinetic friction is about 0:6 (appropriate for 

steel on steel). Mr. X would also like to know the maximum speed and trip time 

if a braking action is applied. You may assume that the brakes exert a force of 

400 newtons on the wire. Hint: You may ¯nd that using a do loop to compute 

the tangential velocity and position of Mr. X at discrete time intervals is the 

easiest way to attack this problem. 

4.3.4 This Would Be a Great Amusement Park Ride 

To gyre is to go around and round like a gyroscope. 

To gimble is to make holes like a gimlet. 

Lewis Carroll, English writer and mathematician (1832{1898) 

Consider the following possible amusement park ride consisting of a small cage 

of mass m (including the screaming victims inside) being swung around at the 

end of a light, but strong, connecting arm of ¯xed radius r as in Figure 4.18. 

The cage can trace out various spherical surface trajectories depending on the 

y 

z 

θ 

Figure 4.18: Con¯guration of the amusement park ride. 

conditions imposed by the slightly sadistic ride operator. What kind of trajectories 

can the victims be subjected to? 

Our task is to investigate this question using Lagrange's equations of motion, 

an alternative approach to the Newtonian formulation. The Lagrangian L is 

de¯ned as L = T ¡ V ,whereT is the kinetic energy and V is the potential 

energy of the system being studied. Formulating the Lagrangian is often much 

easier than determining all the forces and their components that are required to 

r 

x 

φ 

m


apply Newton's second law. If L depends on the coordinates q1, q2,:::, qi,:::, qN 

and the \velocities" qi _ ´ @qi=@t, it is shown in standard mechanics texts [FC99] 

that the equations of motion are given by Lagrange's equations, 

@L 

¡ 

@qi 

d 

μ 

@L 

=0; (4.18) 

dt @qi _ 

with i =1; 2;:::; N. Since each component equation is of the Euler{Lagrange 

structure, it should come as no surprise that the EulerLagrange command can 

be used to derive the relevant ODEs, given a speci¯ed form of L. Let'snowuse 

this approach for the amusement park ride. 

After we load the plots and VariationalCalculus library packages, 

> restart: with(plots): with(VariationalCalculus): 

the Cartesian coordinates of the mass m are expressed in terms of the spherical 

polar coordinates r (which is ¯xed), μ(t), and Á(t), as in Figure 4.18. 

> x:=r*sin(theta(t))*cos(phi(t)); 

x := r sin(μ(t)) cos(Á(t)) 

> y:=r*sin(theta(t))*sin(phi(t)); 

y := r sin(μ(t)) sin(Á(t)) 

> z:=r*cos(theta(t)); 

z := r cos(μ(t)) 

The kinetic energy of the mass at time t is calculated and simpli¯ed. 

> T:=simplify((m/2)*(diff(x,t)^2+diff(y,t)^2+diff(z,t)^2)); 

T := ¡ 1 

2 mr2 

Ã μ 

d 

¡ 

dt Á(t) 

2 μ 

d 

+ 

dt Á(t) 

2 

cos(μ(t)) 2 μ 

d 

¡ 

dt μ(t) 

! 

2 

The potential energy V = ¡mgr cos(μ(t)), with g the gravitational acceleration, 

is entered and the Lagrangian L = T ¡ V calculated. 

> V:=-m*g*r*cos(theta(t)): L:=T-V; 

L := ¡ 1 

2 mr2 

Ã μ 

d 

¡ 

dt Á(t) 

2 μ 

d 

+ 

dt Á(t) 

2 

cos(μ(t)) 2 μ 

d 

¡ 

dt μ(t) 

! 

2 

+mgrcos(μ(t)) 

The EulerLagrange command is applied to L, the independent variable t being 

speci¯ed, and the two angular coordinates Á(t) andμ(t) being entered as a list. 

The lengthy output is suppressed here in the text. 

> eq:=EulerLagrange(L,t,[phi(t),theta(t)]); 

The ¯rst integral in eq containing the constant K1 is selected. 

> ode1:=select(has,eq,K[1])[]; 

ode1 := ¡ 1 

2 mr2 

μ μ 

d 

¡2 

dt Á(t) 

μ 

d 

+2 

dt Á(t) 

 

cos(μ(t)) 2 

 

= K1


A second ODE relating μ(t) toÁ(t) is required. To this end, we select the 

expression in eq containing d 2 μ(t)=dt 2 and set the result to zero. 

> ode2:=select(has,eq,diff(theta(t),t,t))[]=0; 

ode2 := mr2 μ 

d 

dt Á(t) 

2 

cos(μ(t)) sin(μ(t)) ¡ mgrsin(μ(t)) ¡ mr2 μ 2 d 

μ(t) =0 

dt2 With ode1 and ode2 , we have two coupled nonlinear ODEs. However, we can 

solve the former for dÁ=dt by isolating this term on the lhs of the equation, 

> ode1b:=isolate(ode1,diff(phi(t),t)); 

ode1b := d 

2 K1 

Á(t) =¡ 

dt 

mr2 (¡2+2cos(μ(t)) 2 ) 

and then simplifying with the algebraic substitution cos(μ(t)) 2 =1¡ sin(μ(t)) 2 . 

> ode1c:=algsubs(cos(theta(t))^2=1-sin(theta(t))^2,ode1b); 

ode1c := d 

K1 

Á(t) = 

dt mr2 sin(μ(t)) 2 

Then substituting ode1c and g = r! 2 ,where! is a frequency, into ¡ode2 =(mr2 ) 

and expanding, 

> ode2b:=expand(subs(fode1c,g=r*omega^2g,-ode2/(m*r^2))); 

ode2b := ¡ K1 2 cos(μ(t)) 

m2 r4 sin(μ(t)) 3 + !2 μ 2 d 

sin(μ(t)) + μ(t) =0 

dt2 yields the second-order nonlinear ODE ode2b entirely in terms of μ(t). This 

ODE cannot be solved analytically. Before seeking a numerical solution, it is 

instructive to consider some simpler special cases ¯rst. 

Consider the situation in which the ride operator is feeling in a rare mellow 

mood and allows the cage to swing to and fro at a constant angle Á, so_ Á =0 

and, from ode1c, K1 =0. Thenode2b reduces to the well-known linear ODE for 

the simple undamped plane pendulum with characteristic frequency !. Those 

readers who have gone to an amusement park lately may have seen a \boat" 

ride that behaves as a driven simple pendulum. This ride is much too tame for 

our ride operator and the large teenage market that he is after. 

Another special case corresponds to the cage orbiting in a horizontal circle 

at a ¯xed angle μ(t) = £. (A popular young children's ride does exactly this.) 

We evaluate ode2b at this ¯xed angle. 

> eq2:=eval(ode2b,theta(t)=Theta); 

eq2 := ¡ K1 2 cos(£) 

m 2 r 4 sin(£) 3 + !2 sin(£) = 0 

At the ¯xed angle £, K1 is given from ode1c by the following command line, 

where we have set the angular velocity dÁ(t)=dt =−. 

> K[1]:=Omega*m*r^2*sin(Theta)^2; 

K1 := − mr 2 sin(£) 2


The angular velocity needed to maintain the horizontal circular motion at the 

¯xed angle μ(t) = £ is then obtained by solving eq2 for −. 

> Omega:=solve(eq2,Omega); 

! ! 

−:= p ; ¡ p 

cos(£) cos(£) 

The two angular velocity solutions given above correspond to rotations in the 

opposite sense. 

The two rides described so far are not su±ciently exciting to the teenage 

generation. Taking the nominal values r =1,m =1,and! = 1 for the parameters 

(the reader can experiment with more realistic numbers), can one create 

a more interesting trajectory? For a horizontal circular orbit at an inclination 

to the vertical of £ = 60 ± , the constant K1 is given by 

K1 = !mr2 sin 2 £ 

p cos £ 

= 

r 

9 

¼ 1:06: 

8 

To create a more interesting ride let's take K1 =0:5. The trajectory of the ride 

will be plotted over T = 20 time units. 

> r:=1: m:=1: omega:=1: K[1]:=0.5: T:=20: 

Let's suppose that initially, Á(0) = 0, μ(0) = ¼=3 radians, and _ μ(0) = 0, 

> ic:=phi(0)=0,theta(0)=Pi/3,D(theta)(0)=0: 

and numerically solve ode1c and ode2b subject to this initial condition. 

> sol:=dsolve(fode1c,ode2b,icg,ftheta(t),phi(t)g,numeric): 

To get a feeling for the three-dimensional motion, the spherical surface on which 

the cage can move is created with the sphereplot command. The radius r =1 

of the sphere is speci¯ed as well as the angular ranges μ =0to2¼ and Á =0 

to ¼. Awireframe style is used to represent the spherical surface. 

> gr1:=sphereplot(r,theta=0..2*Pi,phi=0..Pi,style=wireframe): 

The odeplot command is now used to plot the numerically determined trajectory 

of the mass m overthetimeintervalt =0toT . Normal axes are chosen, 

and the minimum number of plotting points is taken to be 500 to obtain a 

smooth curve. 

> gr2:=odeplot(sol,[x,y,z],0..T,axes=normal,numpoints=500, 

thickness=3): 

The display command is used to overlay the two graphs. Axis labels are added, 

and the orientation and size of the 3-dimensional viewing box controlled, as well 

as the number of tickmarks. 

> display(gr1,gr2,labels=["x","y","z"],orientation=[50,-100], 

view=[-1..1,-1..1,-1..1], tickmarks=[2,2,4]); 

Figure 4.19 shows the 3-dimensional trajectory. The viewing box can be rotated 

to observe the trajectory from di®erent angles by dragging the box on the 

computer screen with your mouse.


x 

1 

-1 

-1 

-0.5 

0 

0.5 

z 

1 

Figure 4.19: A wild amusement park ride trajectory on a spherical surface. 

If this trajectory isn't wild enough, feel free to create your own crazy ride. 

You could also try using more realistic values for the parameters instead of the 

nominal values used in the above recipe. 

Haveastomach-churningride! 

PROBLEMS: 

Problem 4-43: A twirling-loop ride 

A vertically oriented circular loop of radius ` rotates with angular velocity ! 

about the z-axis as shown in Figure 4.20. A cage of unit mass (m = 1) is allowed 

to slide along the frictionless loop. If the plane of the loop is oriented along the 

x 

z 

θ 

ω 

m 

y 

y 

1 

-1 

Figure 4.20: Rotating circular loop with cage (mass m) free to slide.


y-axis at t =0,whatarethex-, y-, and z-coordinates of the cage at time t? 

Using the Lagrangian approach, show that the cage's motion is described by 

Äμ + ! 2 1 

0 sin μ ¡ 

2 !2 sin(2 μ) =0; 

with !0 = p g=`. Numerically solve the equation of motion for !0 =1and 

varying values of ! and discuss the behavior to which the cage is subjected. 

Problem 4-44: Ride into the jaws of chaos 

The pivot point O for the simple pendulum is undergoing vertical oscillations 

given by A sin(!t) as indicated in Figure 4.21. 

Asin(ωt) O 

Figure 4.21: An example of parametric excitation. 

Show that the relevant equation of motion is 

Äμ + ! 2 μ 

0 1 ¡ A!2 

 

sin(!t) sin μ =0; 

g 

with !0 = p g=`. This nonlinear ODE with a time-dependent coe±cient is 

referred to in the mathematics literature as an example of parametric excitation. 

Taking !0 =1and! = 1, numerically study the e®ect of changing the ratio 

A=g. Then take this ratio equal to 1, and study the e®ect of changing !. 

Problem 4-45: Horizontally oscillating pivot point 

The pivot point O in the previous problem is undergoing horizontal oscillations 

given by A sin(!t). Derive the relevant equation of motion. Taking !0 =1, 

! =1,andg = 10, numerically study the e®ect of changing the amplitude A. 

θ 

m

Chapter 5 

Linear PDE Models. Part 1 

Because linear partial di®erential equations play such an important role in the 

mathematical description of electromagnetic waves, heat °ow, elastic vibrations, 

and many other scienti¯c phenomena, there is an abundance of wonderful examples 

that can be solved using computer algebra. For this reason, this topic 

is split over two chapters. We begin with examples of checking PDE solutions, 

either obtained by intelligent guessing or quoted, without derivation, in some 

scienti¯c reference. Di®usion and Laplace's equation models are then presented. 

5.1 Checking Solutions 

5.1.1 The Palace of the Governors 

The knowledge of the world is only acquired in the world, 

and not in a closet. 

Phillip Dormer Stanhope, Earl of Chester¯eld (1694{1773) 

While driving back to Phoenix from Los Alamos, where he attended an engineering 

conference at the Los Alamos National Laboratory, Russell stops in 

Sante Fe to have something to eat and to tour the historic Plaza area of town. 

After treating himself to a gourmet lunch, consisting of a tasty rattlesnake 

burger washed down with a Corona beer, he strolls around the Plaza. As an 

engineer, Russell is particularly impressed by the massively thick walls of the 

Palace of the Governors. This long, low adobe structure, which was built in 

1610 by the Spanish, is the oldest continuous seat of government in the United 

States. The thick walls, which is a design feature of many historic buildings in 

the American Southwest, not only o®ered protection from attackers but helped 

to keep the heat of the summer sun at bay as well as excluding the cold breath 

of winter. 

On completing his tour of the Plaza, Russell is intrigued by the question 

of how e®ective a very thick adobe wall is in cutting down the incident solar 

radiation from the summer sun. Pulling out his laptop computer, and ¯nding 

a shady spot, he formulates the following relevant model. For simplicity, he 

207

208 CHAPTER 5. LINEAR PDE MODELS. PART 1 

considers a semi-in¯nite solid medium whose bounding planar surface is periodicallyheatedbythesun. 

Thedirectiontransversetothesurfaceistakento 

be the x-direction, with the surface located at x =0,andpositivextaken to 

be inside the surface. In order to animate the temperature pro¯le inside the 

surface, Russell loads the plots library package. 


From undergraduate thermodynamics, he knows that the time-dependent temperature 

distribution T (x; t) obeys the one-dimensional heat di®usion equation 

@T 

@t = d @2T ; (5.1) 

@x2 with the heat di®usion coe±cient d = K=(½C), where K is the thermal conductivity, 

½ the density, and C the speci¯c heat. Russell enters the heat equation. 

> heateq:=diff(T(x,t),t)-d*diff(T(x,t),x,x)=0; 

³ ´ μ 

heateq := @@tT (x; t) ¡ d @ 2 

 

2 T (x; t) =0 

@x 

To account for the periodic heating of the planar surface due to the sun, he 

takes the temperature variation at x =0tobeT (0;t)=T0 cos(!t). Rather 

than derive the temperature variation T (x; t) forx > 0, Russell decides to 

make an intelligent guess as to its mathematical form and check it by substituting 

the form back into the di®usion equation. Since he is looking for a 

steady-state response and the temperature should decrease as x increases inside 

the surface, Russell conjectures that the solution should be of the structure 

T (x; t) =T0 e ¡®x cos(!t¡ ¯x), with the parameters ® and ¯ as yet undetermined. 

At x = 0, the boundary condition is satis¯ed, and an exponentially 

decaying cosine solution seems reasonable inside the surface. Both ® and ¯ 

must be positive for a waveform propagating in the positive x-direction. 

What forms should ® and ¯ have? From the structure of the heat di®usion 

equation, the units of d are m 2 /s, so 1= p d has units of s 1=2 ¢m ¡1 . Since the 

argument in the exponential function must be dimensionless, ® must have the 

dimensions of m ¡1 . The only other parameter in the problem involving a time 

unit is the frequency ! with units s ¡1 . So, noting that the combination p !=d 

has the units m ¡1 ,Russelltakes® = a p !=d, wherea is a numerical factor, yet 

to be determined. By similar reasoning, the constant ¯ is set equal to b p !=d, 

with b another numerical factor. Both a and b must be positive. 

To check the postulated solution and determine a and b, Russell enters 

T (x; t), which will be automatically substituted into the heat equation. 

> T(x,t):=T[0]*exp(-a*sqrt(omega/d)*x) 

*cos(omega*t-b*sqrt(omega/d)*x); 

T (x; t) :=T0 e (¡a p ! 

d x) cos 

Russell then simpli¯es the heat equation, 

> eq:=simplify(heateq); 

μ 

¡!t+ b 

r ! 

d x

5.1. CHECKING SOLUTIONS 209 

and collects the coe±cients of the cosine and sine terms. 

> collect(eq,fcos,sing); 

¡T0 !e (¡a p ! 

d x) (a2 ¡ b2 μ r 

! 

)cos ¡!t+ b 

d x 

 

¡ T0 !e (¡a p ! 

d x) μ r 

! 

(¡1+2ab)sin ¡!t+ b 

d x 

 

=0 

For the left-hand side of the above output to be equal to zero for arbitrary x 

and t, one must have b = a and ¡1+2ab =0,sothata =1= p 2=b. These 

values of a and b are entered, numerically evaluated, and labeled aa and bb. 

> aa:=evalf(1/sqrt(2)): bb:=aa: 

Adobe brick is made up of a mixture of dried clay and straw, so Russell consults 

a handbook of physical constants that he just happens to have brought along 

with him. He ascertains that for dried clay, the major component of the bricks, 

K =0:4 W/(m¢K), ½ =2000kg/m3 ,andC = 920 J/(kg¢K). Russell then is 

able to calculate the di®usion coe±cient dclay for clay, which he will use as an 

estimate for the adobe brick. 

> K:=0.4: rho:=2000: C:=920: d[clay]:=K/(rho*C); 

dclay := 0:2173913043 10 ¡6 

The rotational period of the earth is about 24 hours, or 24 £ 60 £ 60 = 86;400 

seconds, so its rotational frequency, freq, can be calculated. 

> period:=24*60*60; freq:=evalf(2*Pi/period); 

period := 86400 freq := 0:00007272205218 

Summer temperatures in the American Southwest can vary from daily highs of 

100 ± F or more, to overnight minima in the low to mid-60s. For his calculation, 

Russell takes a mean daily temperature of 80 ± F with a variation of 20 ± on either 

side. Converting to degrees Celsius, this translates into a mean temperature 

of 26:6 ± CandT0 =11 ± for the amplitude of the temperature variation. Now 

T (x; t) is evaluated using the parameter values that have been obtained. 

> T:=eval(T(x,t),fa=aa,b=bb,T[0]=11,d=d[clay],omega=freqg); 

T := 11:e (¡12:93293161 x) cos(¡0:00007272205218 t +12:93293161 x) 

Russell creates a plot of the mean daily temperature, representing the temperature 

by a dashed (linestyle=3) blue line. The spatial range of the plot is 

from x =0to0:5 meters. 

> meantemp:=plot(26.6,x=0..0.5,color=blue,linestyle=3): 

Thetotaltemperaturepro¯leovera0:5-meter spatial range is animated over 

a time interval of one complete period (day). Fifty frames are included in the 

animation. The mean temperature line is included with the background option. 

> animate(plot,[T+26.6,x=0..0.5],t=0..period,frames=50, 

thickness=2,background=meantemp); 

The initial temperature pro¯le in the animation is similar to the solid curve 

shown in Figure 5.1, the horizontal dashed line being the mean temperature.


35 

30 

25 

20 

0 0.1 0.2 0.3 0.4 0.5 

x 

Figure 5.1: Initial temperature pro¯le inside the adobe wall. 

On running the animation, the temperature curve will oscillate up and down 

as the externally imposed temperature on the surface varies during the day. 

However, the temperature variation drops to zero within a penetration depth 

of less than 0:4 m or 40 cm (16 in). Well within the surface, the temperature 

is in thermal equilibrium with the mean exterior temperature. Since the adobe 

walls of many of the historic buildings of Arizona and New Mexico are substantially 

thicker than 16 inches, the interior temperature of these buildings is quite 

insensitive to external temperature variations. 

PROBLEMS: 

Problem 5-1: A house of wood 

In the text recipe, make the following modi¯cations: 

(a) The region x>0 is composed of solid wood for which K =0:15 W/(m¢K), 

½ = 700 kg/m3 ,andC = 1800 J/(kg¢K). 

(b) The mean daily temperature is 30 ± Candtheamplitudeofthetemperature 

variation is 20 ± . 

Run the animation and determine the approximate depth in centimeters at 

which the temperature variation is essentially zero. 

Problem 5-2: Pulsating sphere 

The surface of a sphere of radius r = a, surrounded by an ideal compressible 

°uid, pulsates radially with frequency !. The radial velocity of the surface is 

given by V = U cos(!t). It is stated in an advanced calculus text that the 

steady-state °uid velocity at an arbitrary point r>ais of the form 

V = 

Ua2 (c2 + a2 ! 2 ) r2 £ ¤ 2 2 

(c +ar! )cos(μ)+c!(r ¡ a)sin(μ) ;


where μ ´ ! (r ¡ a)=c ¡ !tand c is the speed of sound in the °uid. 

(a) Check that the solution satis¯es the boundary condition. 

(b) The velocity V is related to the velocity potential © by V = ¡@©=@r. 

Determine the radial dependence of the velocity potential. 

(c) Since © depends only on the distance r from the center of the sphere, it 

satis¯esthewaveequationintheform 

@ 2 (r ©) 

@r 2 

1 

= 

c2 @2 (r ©) 

@t2 : 

Verify that © satis¯es the wave equation, thus ensuring that it is the 

correct solution to the pulsating sphere problem. 

(d) Taking the nominal values U =1,a =1,c =1,and! = 1, animate the 

analytic formula for V in the region outside the spherical surface. 

(e) How far from the surface does the velocity oscillation amplitude drop to 

5% of the value at the surface? 

5.1.2 Play It, Sam 

You just pick a chord, go twang, and you've got music. 

Syd Vicious, British rock musician (1957{1979) 

In a famous scene from the classic movie Casablanca, Humphrey Bogart is 

annoyed by the musical piece that the nightclub pianist is playing and doesn't 

want to hear it. But Ingrid Bergman turns to the piano player, and says \Play 

it, Sam." Humphrey Bogart then echoes her, by saying \If she can stand it, I 

can. Play it." 

To a certain undergraduate physics student watching this old movie, an 

individual who tends to look for deeper understanding rather than simply enjoying 

the movie and the music, the sounds that emanate are of course due to 

the transverse vibrations of the piano strings as each is successively struck by 

a piano hammer. This student is our old friend Vectoria, who is spending a 

lonely evening by herself, since her ¯ance Mike is out of town at a mathematics 

conference. Hoping that Mike will phone before it gets too late, Vectoria 

decides in the meantime to look at the mathematical vibrations of an elastic 

string ¯xed at its ends. 

Not yet having taken the necessary mathematics prerequisites to study the 

topic of vibrating strings in depth, Vectoria consults a physics text that gives 

no derivation but simply the formula for the transverse displacement of a light, 

horizontal, elastic string of length a ¯xed at both ends when struck and given a 

certain initial velocity pro¯le. The information that Vectoria gleans from this 

particular text is as follows.


Neglecting sti®ness, we can model the transverse displacement Ã(x; t) ofa 

light, initially horizontal (Ã(x; 0) = 0) piano string by the wave equation 

@2Ã 1 

¡ 

@x2 c2 @2Ã =0; (5.2) 

@t2 where the wave velocity c equals p T=², T being the tension in the string and 

² the mass per unit length. If a string of length a is held ¯xed at both ends 

(Ã(0;t)=Ã(a; t) = 0) for all times, and is given an initial transverse velocity 

8 

< 4 vx=a; 0


a piecewise function, is slightly harder to verify. One approach is to plot the 

velocity pro¯le predicted by the series solution at t =0andcompareitwiththe 

analytic piecewise form. Adding, say, the ¯rst ¯ve terms in the series solution, 

> psi:=add(psi[n],n=1..5); 

8 va( 

Ã := 

p ³ 

¼x 

´ μ 

¼ct 

2¡1) sin sin 

a a 

¼3 μ μ 

2 ¼x 2 ¼ct 

2 vasin sin 

a 

a 

+ 

c 

¼3 c 

+ 8 

va( 

27 

p μ μ 

3 ¼x 3 ¼ct 

2+1)sin sin 

a 

a 

¼3 c 

+ 8 

va(¡ 

125 

p μ μ 

5 ¼x 5 ¼ct 

2 ¡ 1) sin sin 

a 

a 

¼3 c 

and di®erentiating with respect to time, yields the velocity in the transverse 

direction (output suppressed here). 

> vel:=diff(psi,t); 

To plot the transverse velocity at t = 0, some representative values must be 

substituted for the parameters. Vectoria chooses v =5,a =2,andc =1. 

> t:=0: v:=5: a:=2: c:=1: 

To compare the series representation at t = 0 with the given piecewise initial 

velocity distribution, the latter is entered, 

> V:=piecewise(x


solution isn't too bad, but not great either. Increasing the n value, Vectoria 

observes an increasingly better ¯t. The reader should be the judge of how many 

terms su±ce to give a good ¯t for the parameters chosen. 

Although reasonably satis¯ed that the quoted Fourier series expansion is 

correct, Vectoria still isn't entirely happy until the actual motion predicted 

by the formula is observed. After all, it is not obvious from the series solution 

exactly what the behavior of the string is after being struck. So the displacement 

Ã(x; t) is animated for the time interval t = 0 to 50. First, the time variable t 

must be unassigned. Otherwise, Maple will remember the value t =0usedto 

check the initial velocity pro¯le. 

> unassign('t'): 

Finally, the animation command is given with 100 frames being used. 

> animate(plot,[psi,x=0..a],t=0..50,frames=100,thickness=2); 

On running the animation, Vectoria observes that the wave form begins to grow 

in the region where the string was struck. This makes intuitive sense. Because 

thewaveformiscreatedontheleftsideofthestring,itthenmoveswithwave 

velocity c to the right and re°ects o® the boundary at x = a. On re°ection 

the wave form is inverted, a characteristic feature for a ¯xed-ends boundary 

condition. The wave then propagates to the left boundary at x =0before 

inverting again and repeating the oscillatory behavior. 

Finally, with a feeling of accomplishment, Vectoria is able to appreciate the 

deeper content underlying the simple remarks made by Ingrid Bergman and 

Humphrey Bogart. She will be even more content if Mike phones soon. 

PROBLEMS: 

Problem 5-3: Plucked string 

An elastic string ¯xed between x =0andL, and initially at rest, is \plucked," 

its initial shape being given by the following symmetric triangular pro¯le, 

Ã(x; 0) = 

½ 2 hx=L; 0 · x · L=2; 

2 h (L ¡ x)=L; L=2 · x · L: 

Verify that the motion for t>0 may be described by the Fourier series solution 

8 h 

Ã(x; t) = 

¼2 1X sin(n¼=2) 

n 

n=1 

2 sin(n¼x=L)cos(n¼ct=L) 

and animate the solution for parameter values of your own choosing. 

Problem 5-4: A striking piano hammer 

A piano string ¯xed between x =0 anda is struck by a piano hammer in a 

region of width d centered at x=x0. Its initial velocity distribution is 

½ 

Ã(x; _ v cos(¼ (x ¡ x0)=d); jx ¡ x0j d=2:


Neglecting sti®ness and assuming Ã(x; 0) = 0, 

4 vd 

Ã(x; t) = 

¼2 1X 1 sin(n¼x0=a)cos(n¼d=(2 a)) 

c n (1 ¡ (nd=a) 2 sin(n¼x=a)sin(n¼ct=a) 

) 

n=1 

istheshapeofthestringattimet. Verify that this series solution is correct 

and animate it for parameter values of your own choosing. 

5.1.3 Three Easy Pieces 

I would advise you Sir, to study algebra, if you are not already an 

adept in it: your head would be less muddy ... 

Samuel Johnson, English writer and lexicographer (1709{1784) 

Spurred by her earlier success, and feeling happier after her long phone conversation 

last night with Mike, Vectoria is pursuing another vibrating string 

example, which she has entitled Three Easy Pieces. Thisdoesnotrefertothe 

old Jack Nicholson movie with a similar title (cf. Five Easy Pieces), but to 

the fact that the algebraic manipulations involved in dealing with plane-wave 

propagation along a three-piece string are easy if one uses computer algebra. 

When an intermediate section of a very long horizontal string has a greater 

mass density ² than the remaining two identical portions of the string, a transverse 

plane wave incident on that section will in general experience partial 

re°ection and transmission. Recall that the velocity of the transverse wave is 

given by c = p T =², whereT is the tension in the string. The wave number is 

k = !=c =2¼=¸, where! is the angular frequency and ¸ is the wavelength. 

Since the frequency of the wave and the tension must remain the same in each 

region, the ratio r of wave numbers in two di®erent regions of mass density ²2 

and ²1 is given by r = k2=k1 = p ²2=²1. Vectoria reads in a certain physics 

text that for the case k1 = K, theratior = 3, and the more massive segment 

(labeled 2) stretches from x =0tox = L, the energy transmission (T )and 

re°ection (R) coe±cientsaregivenby, 

9 

8 ¡ 8cos(6KL) 

T = 

; R = 

17 ¡ 8cos(6KL) 17 ¡ 8cos(6KL) : 

Here T and R measure the fraction of the incident power that is transmitted and 

re°ected. The power here is proportional to the square of the string amplitude. 

Vectoria's objective is to verify the cited re°ection and transmission coe±cients 

and plot them for K =1asafunctionofL. Hermethodofattackisto 

write down plane-wave expressions in each region, determine the coe±cients by 

matching the solutions at x =0andx = L, and then calculate T and R. 

She begins by entering r =3,k1 = K, andk2 = rk1. 

> restart: r:=3: k[1]:=K; k[2]:=r*k[1]; 

k1 := K k2 := 3 K


The plane wave is assumed to be traveling in the positive x-direction, its time 

part being e ¡I!t ,whereI = p ¡1. In the ¯rst region, x psi[1]:=exp(I*k[1]*x)+a*exp(-I*k[1]*x); 

Ã1 := e (KxI) (¡I Kx) 

+ ae 

The ¯rst term is the incident plane wave, while the second term is the re°ected 

wave with amplitude a. Since the transmission and re°ection coe±cients involve 

ratios of squared amplitudes, the amplitude of the incident wave has been set 

equal to one without loss of generality. The energy re°ection coe±cient then is 

given by R = jaj2 =1=jaj2 . 

In region 2 (0 · x · L), the wave form must be made up of waves traveling 

in the positive and negative x-directions, viz., Ã2 = beIk2x + ce ¡Ik2x ,with 

undetermined amplitudes b and c. 

> psi[2]:=b*exp(I*k[2]*x)+c*exp(-I*k[2]*x); 

Ã2 := be (3 IKx) (¡3 IKx) 

+ ce 

In the third region, x>L, there will be only a transmitted plane wave, with 

spatial part Ã3 = deIk1x ,thewavenumberbeingk1 = K since the string 

density is the same as in the ¯rst region. The fraction of the energy incident in 

region 1 that is transmitted into region 3 is given by the transmission coe±cient 

T = jdj2 =1=jdj2 . 

> psi[3]:=d*exp(I*k[1]*x); 

Ã3 := de (KxI) 

To evaluate the four unknown coe±cients a, b, c, and d, four independent 

equations are needed. The ¯rst two equations, eq1 and eq2 ,followfromthe 

physical continuity of the string. The string segment in region 2 is joined to 

the segment in region 1 at x =0andtothesegmentinregion3atx = L. 

> eq1:=eval(psi[1]=psi[2],x=0); 

eq1 := 1 + a = b + c 

> eq2:=eval(psi[2]=psi[3],x=L); 

eq2 := be (3 IKL) + ce (¡3 IKL) = de (KLI) 

Since the wave equation, and therefore the second spatial derivative, remains 

¯nite everywhere along the string, the ¯rst derivative of Ã with respect to x 

must be continuous. So, continuity of the slope at x =0andx = L yields the 

third and fourth equations. 

> eq3:=eval(diff(psi[1],x)=diff(psi[2],x),x=0); 

eq3 := KI¡ aKI =3IbK¡ 3 IcK 

> eq4:=eval(diff(psi[2],x)=diff(psi[3],x),x=L); 

eq4 := 3 IbKe (3 IKL) ¡ 3 IcKe (¡3 IKL) = dKe (KLI) I 

The system of four equations is now solved for the four unknown amplitudes, 

and the solution is assigned.


> sol:=solve(feq1,eq2,eq3,eq4g,fa,b,c,dg); assign(sol): 

(¡3 IKL) 2 e 

(3 IKL) e 

sol := fb = ¡ 

e (3 IKL) ;c= ¡ 

¡ 4 e (¡3 IKL) e (3 IKL) ¡ 4 e (¡3 IKL) 

a = ¡ 2(e(3 IKL) ¡ e (¡3 IKL) ) 

e (3 IKL) 3 e 

;d= ¡ 

¡ 4 e (¡3 IKL) (¡3 IKL) (3 IKL) e 

e (KLI) e (3 IKL) g 

¡ 4 e (¡3 IKL) 

Since the amplitudes are complex, the transmission coe±cient is given by T = 

jdj2 = d £ d ¤ , where the asterisk denotes the complex conjugate. The command 

conjugate(d) is used to enter d ¤ . The complex evaluation command, evalc, 

which breaks a complex quantity into real and imaginary parts, is used to 

generate a completely real answer. Finally, the result is simpli¯ed. 

> T:=simplify(evalc(d*conjugate(d))); 

9 

T := ¡ 

¡25+16cos(3KL) 2 

Applying the combine command, with the trig option, produces the desired 

form of the transmission coe±cient. 

> T:=combine(T,trig); 

9 

T := ¡ 

¡17+8cos(6KL) 

Similarly, the re°ection coe±cient is given by R = a £ a ¤ , and the desired form 

follows on using the same command structure as for deriving T . 

> R:=simplify(evalc(a*conjugate(a))); 

R := 16 (¡1+cos(3KL)2 ) 

¡25 + 16 cos(3 KL) 2 

> R:=combine(R,trig); 

R := ¡8+8cos(6KL) 

¡17 + 8 cos(6 KL) 

By energy conservation the sum of the re°ection and transmission coe±cients 

should add up to one. This is checked in the next command line. The value 

K = 1 is also entered for plotting purposes. 

> check:=simplify(R+T); K:=1: 

check := 1 

Finally, Vectoria plots the re°ection and transmission coe±cients in the same 

¯gure over the range L = 0 to 5, using a solid (linestyle=1) red curve for 

R, andadashed(linestyle=3) bluecurveforT . To maintain order, lists are 

used here for the energy coe±cients, the colors, and the line styles. For better 

visualization, thick curves are employed. Labels are also added to the plot, 

entered as strings. 

> plot([R,T],L=0..5,color=[red,blue],linestyle=[1,3], 

thickness=2,labels=["L","R,T"]); 

The resulting picture is shown in Figure 5.3.


1 

0.8 

0.6 

R, T 

0.4 

0.2 

0 

1 2 3 4 5 

L 

Figure 5.3: Re°ection (upper curve) and transmission (lower) coe±cients vs. L. 

When L = n¼=3, with n = 0 or a positive integer, the re°ection coe±cient 

vanishes and there is 100% transmission. Looking back at the expression for 

R, and recalling that k2 =3K here, Vectoria notes that this situation will 

occur whenever k2 L = n¼ or L = n (¸2=2), i.e., when the length L of the thick 

portion exactly accommodates an integer number of half-wavelengths in that 

region. Does this condition make sense to you? 

PROBLEMS: 

Problem 5-5: Variation on the text example 

If ²1 =1,²2 =16,k1 =1,andL = 2, what fraction of the incident plane wave 

energy is transmitted through the middle segment into the third region? 

Problem 5-6: As simple as 1-2-3 

Consider a plane wave of frequency ! traveling from x = ¡1 along an in¯nitely 

long string with three regions of di®erent density, the middle one being located 

between x =0andL. Suppose that the wave number in region 1 (x


(e) What is the maximum value of T ,andforwhatL value does this transmission 

occur? 

Problem 5-7: Five easy pieces (not the movie) 

Consider an in¯nitely long string that has a linear density ²1 =1inregion1 

(x


5.1.4 Complex, Yet Simple 

I adore simple pleasures. They are the last refuge of the complex. 

Oscar Wilde, Anglo-Irish playwright (1854{1900) 

Consider the complex function w = u(x; y)+Iv(x; y) =z+1=z,wherez = x+Iy 

with I = p ¡1. It is stated in a certain mathematical physics text that because 

the real functions u and v satisfy the Cauchy{Riemann conditions, 

@u @v 

= 

@x @y ; 

@v 

@x 

@u 

= ¡ ; (5.5) 

@y 

then both u(x; y) andv(x; y) satisfyLaplace's equation, 

r 2 u = @2u @x2 + @2u @y2 =0; r2v = @2v @x2 + @2v =0; (5.6) 

@y2 and therefore can represent possible real potentials. It is further stated that 

the constant-v curves can be used to represent the equipotentials outside an 

in¯nitely long grounded conducting cylinder of unit radius placed in a previously 

uniform electric ¯eld oriented perpendicular to the cylinder. The electric ¯eld 

lines outside the cylinder are given by the constant-u curves, and the electric 

¯eld vectors by ~ E = ¡rv. 

Our goal in this recipe is to illustrate how simple it is to check these statements 

using Maple. Loading the plots and VectorCalculus packages, 


both z = x + Iyand w = z +1=z are entered. 

> z:=x+I*y; w:=z+1/z; 

1 

z := x + yI w:= x + yI+ 

x + yI 

The complex function w is separated into real and imaginary parts. 

> w:=evalc(w); 

w := x + 

x 

x2 + 

+ y2 μ 

y ¡ 

y 

x 2 + y 2 

 

I 

The term involving I is removed from w, thus yielding u. The term containing 

I is selected and divided by I to produce v. 

> u:=remove(has,w,I); v:=select(has,w,I)/I; 

u := x + 

x 

x 2 + y 2 

v := y ¡ 

y 

x 2 + y 2 

We now check that both Cauchy{Riemann conditions (5.5) are satis¯ed. 

> CR1:=simplify(diff(u,x)-diff(v,y)); 

CR1 := 0 

> CR2:=simplify(diff(v,x)+diff(u,y)); 

CR2 := 0 

A functional operator L is formed to calculate and simplify the Laplacian (r 2 ) 

of a function f expressed in terms of the Cartesian coordinates x and y.


> L:=f->simplify(Laplacian(f,'cartesian'[x,y])): 

Applying L to u and to v yields zero, con¯rming that these functions satisfy 

Laplace's equation and can be regarded as real potentials. 

> L(u); L(v); 

0 0 

To con¯ne our attention to the region outside the cylinder, two piecewise functions, 

pw1 and pw2 , are formed which are equal to zero for x2 + y2 < 1andu 

and v, respectively, outside this circular region. 

> pw1:=piecewise(x^2+y^2=1,u); 

( 2 2 0 x + y < 1 

pw1 := x 

x + 1 · x2 + y2 x 2 + y 2 

> pw2:=piecewise(x^2+y^2=1,v): 

A contour plot operator CP is formed to plot the equipotentials for a given 

potential function f. The color C must also be speci¯ed. The contours are 

drawn for potentials equal to 0:2 n, withn ranging from ¡11 to +11. The grid 

spacing is taken to be 90 £ 90, and constrained scaling is imposed. 

> CP:=(f,C)->contourplot(f,x=-2..2,y=-2..2,contours= 

[seq(0.2*n,n=-11..11)],grid=[90,90],color=C, 

scaling=constrained,thickness=2): 

The curves corresponding to constant u are colored red, those corresponding 

to constant v colored blue. The two sets of curves are superimposed. A blackand-white 

version of the plot is shown in Figure 5.4. 

> display(fCP(pw1,red),CP(pw2,blue)g); 

2 

y 

–2 x 2 

–2 

Figure 5.4: Equipotentials and electric ¯eld lines outside a conducting cylinder.


The horizontal curves are the equipotentials (constant v) and the vertical curves 

are the electric ¯eld lines (constant u). The two sets of curves intersect at 

right angles, as would be expected. The electric ¯eld lines also intersect the 

cylindrical surface perpendicularly, because the conducting surface is also an 

equipotential. 

Instead of showing the electric ¯eld lines, one can plot the electric ¯eld 

vectors. The electric ¯eld ~ E = ¡rv can be calculated in Cartesian coordinates 

using the Gradient command. 

> E:=-Gradient(v,'cartesian'[x,y]); 

E := ¡ 

2 yx 

(x 2 + y 2 ) 2 ex + 

μ 

¡1+ 

1 

x2 ¡ 

+ y2 2 y 2 

(x 2 + y 2 ) 2 

Since we are interested only in the electric ¯eld outside the cylinder, two piecewise 

functions are formed to calculate the x and y components of the electric 

¯eld in the region x 2 + y 2 ¸ 1. 

> Ex:=piecewise(x^2+y^2=1,E[1]): 

> Ey:=piecewise(x^2+y^2=1,E[2]): 

The fieldplot command is used to plot the electric ¯eld vectors as thick red 

arrows, the grid being 12 £ 12. The arrows will point in the direction of ~ E, 

their size being a measure of the magnitude, j ~ Ej. 

> FP:=fieldplot([Ex,Ey],x=-2..2,y=-2..2,arrows=THICK, 

grid=[12,12],color=red): 

The equipotentials and electric ¯eld vectors are superimposed in the same picture, 

a black-and-white rendition being shown in Figure 5.5. 

> display(fCP(pw2,blue),FPg,labels=["x","y"],tickmarks=[2,2]); 

y 

2 

–2 0 x 2 

–2 

Figure 5.5: Equipotentials and electric ¯eld vectors outside the cylinder. 

 

ey

5.2. DIFFUSION AND LAPLACE'S EQUATION MODELS 223 

PROBLEMS: 

Problem 5-9: Intersecting conducting plates 

In the text recipe replace the complex function with w = u + Iv = z2 ,with 

z = x + Iy. Con¯rm that u(x; y) andv(x; y) satisfy the Cauchy{Riemann 

conditions and Laplace's equations. Con¯rm by making a suitable plot that 

the equipotentials and electric ¯eld lines are appropriate to the quarter-space 

bounded by two semi-in¯nite conducting plates intersecting at right angles. 

Make another plot that shows the electric ¯eld vectors and equipotentials. 

Problem 5-10: Fluid °ow around a plate 

Consider the function w = u + iv= p z2 ¡ 1, with z = x + iy. By creating a 

suitable ¯gure in the x-y plane, show that the constant-u curves can represent 

the streamlines (tracks of the °uid particles) for °uid °ow around an in¯nitely 

long plate of ¯nite width, lying between x=¡1 and +1, inserted perpendicular 

to a previously uniform °uid °ow in the y direction. Include the plate and the 

equipotentials (constant v curves) in your ¯gure. 

5.2 Di®usion and Laplace's Equation Models 

In this section, di®usion and Laplace's equation models are solved in Cartesian 

and other common coordinate systems, using some of the standard methods 

of mathematical physics. Of course, we will let Maple do the heavy slogging, 

concentrating on the underlying physics. 

5.2.1 Freeing Excalibur 

A sleeping presence is always a mystery ... seemingly peaceful, 

yet in reality o® on wild adventures in strange landscapes. 

A. Alvarez, British novelist (1929{) 

Russell works as a control systems engineer for an aerospace company in the 

Phoenix area. Earlier in the day, several of his old ASU engineering classmates 

dropped by his workplace and persuaded him to go to the Monastery, a local 

outdoor pub, after work. There they ate chicken wings and pizza, drank pitchers 

of Mexican beer, and swapped stories about their undergraduate days and 

what had happened to each of them since graduating from engineering school. 

Later that night, probably triggered by the day's events, Russell is having 

a wild dream in which he has been transported back to the mythical times of 

Merlin the magician. Merlin has wrapped the fabled sword Excalibur in thin 

insulating material and has embedded it in a large rock with only the tip and 

hilt protruding. The legend is that whoever manages to pull out the sword will 

become ruler of the Kingdom with all its associated wealth. Many have tried,


but all 1 have failed to pull the sword out of the rock. Having taken a course 

in thermodynamics as an engineering undergraduate, Russell speculates as to 

whether the sword could be pulled out by cooling the ends of the sword with 

large buckets of ice, thus causing heat to °ow out of the warmer interior of the 

sword's blade to the ends. If the sword were cooled su±ciently, its diameter 

might shrink slightly, and just possibly he could pull the sword out. 

But remembering the thermodynamics course causes Russell's dream to alter 

direction, and he then dreams of a related problem that appeared on that 

course's ¯nal exam many years ago. A thin 1-m-long rod (the sword's shaft) 

whose lateral surface is insulated to prevent heat °ow through that surface has 

its ends suddenly held at the freezing point of water, 0 ± C(contactwiththe 

buckets of ice). Taking one end of the rod to be at x = 0 and the other at x = 

L = 1, the initial temperature (T ) distribution was T (x; t =0)=100x (1 ¡ x), 

a parabolic pro¯le with a maximum temperature of 25 ± at the midpoint x = 1 

2 . 

In the exam, he was asked to determine the temperature distribution T (x; t) for 

any time t>0. Despite the passage of time, Russell remembers his approach to 

solving this problem very well. His method of attack was to solve the di®usion 

equation 

@T(x; t) 

@t 

by the method of separation of variables, i.e., assume that T (x; t)=S(x) F (t). 

Substituting this assumed form into (5.7) and dividing by T (x; t) yields 

1 dF (t) d d 

= 

F (t) dt S(x) 

2S(x) dx2 : (5.8) 

Since the lhs of (5.8) involves a function of t alone and the rhs involves a function 

of x alone, the only way that it can be generally true is for both sides to be 

equal to a common constant, called the separation constant. The separated 

ODE for S(x) is solved, subject to the boundary conditions, and the complete 

product solution constructed, subject to the initial condition. 

Russell's wild dream is interrupted by the roar of a big jet passing over his 

house on its way into Sky Harbor Airport. Waking up, and unable to get back 

to sleep, he decides to make a cup of instant decaf co®ee and implement an 

animated Maple solution of the heat di®usion problem on his computer. To 

carry out the animation, the plots library package is loaded. Noting that the 

di®usion coe±cient d could be absorbed into either the spatial or time variable, 

Russell sets d = 1 without loss of generality. He also speci¯es the rod's length, 

L = 1, and enters the heat equation (5.7). 

> restart: with(plots): d:=1: L:=1: 

> heateq:=diff(T(x,t),t)=d*diff(T(x,t),x,x); 

= d @2 T (x; t) 

@x 2 ; (5.7) 

heateq := @ 

@2 

T (x; t) = T (x; t) 

@t @x2 A general solution to the heat equation is sought using the PDE solve (pdsolve) 

command, the HINT option explicitly telling Maple to separate variables. 

1 King Arthur has not shown up yet.


> pdsolve(heateq,HINT=S(x)*F(t)); 

·½ 

d 

(T (x; t) =S (x) F (t)) &where 

dt F (t) = c1 F (t); d2 

dx2 S(x) = c1 

¾¸ 

S(x) 

The heat equation has been separated into two ODEs, c1 being the separation 

constant. If the INTEGRATE option is also included, the general solution of each 

ODE will be generated as Russell now illustrates. 

> pdsolve(heateq,HINT=S(x)*F(t),INTEGRATE); 

(T (x; t) =S (x) F (t)) &where 

[ffF (t) = C3 e ( c1 t) g; fS (x) = C1 e ( p c1 x) + C2 e (¡ p c1 x) gg] 

There are three unknown constants in the output, namely C1 , C2 ,and C3 . 

Finally, including the build option produces the general product solution, sol. 

> sol:=pdsolve(heateq,HINT=S(x)*F(t),INTEGRATE,build); 

sol := T (x; t) = C3 e ( c1 t) C1 e (p c1 x) + C3 e ( c1 t) C2 

e (p c1 x) 

Russell then substitutes c1 = ¡k 2 on the rhs of sol and applies the simplify 

command with the symbolic option, which is equivalent to assuming that k>0. 

> temp:=simplify(subs(_c[1]=-k^2,rhs(sol)),symbolic); 

temp := C3 ( C1 e (k (¡tk+xI)) + C2 e (¡k (tk+xI)) ) 

The resulting form is complex in appearance, so the complex evaluation command 

is applied to separate temp into real and imaginary terms. 

> temp:=evalc(temp); 

temp := C3 ( C1 e (¡k2 t) (¡k cos(kx)+ C2 e 2 t) cos(kx)) 

+ C3 ( C1 e (¡k2 t) (¡k sin(kx) ¡ C2 e 2 t) sin(kx)) I 

The temperature has been expressed in terms of cosine, sine, and exponential 

terms, which are now successively collected. 

> temp:=collect(%,[cos,sin,exp]); 

temp := C3 ( C1 + C2) e (¡k2 t) cos(kx)+ C3 ( C1 ¡ C2 ) e (¡k 2 t) sin(kx) I 

Since T (x =0;t) = 0 for all times and cos(kx) does not vanish at x =0,Russell 

removes it from temp. This is the ¯rst boundary condition (bc1). 

> temp:=remove(has,temp,cos); #bc1 

temp := C3 ( C1 ¡ C2 ) e (¡k2 t) sin(kx) I 

Similarly, the temperature must also vanish at x = L =1forallt (the second 

boundary condition). Therefore sin(kL)=0,sok = n¼=Lwith n =1; 2; 3;::: 

The general solution must involve a linear combination of terms involving all 

possible n values. With the coe±cient labeled An, thenth term in the Fourier 

series representation of the temperature will be of the following form. 

> T[n]:=A[n]*subs(k=n*Pi/L,exp(-k^2*t)*sin(k*x)); #bc2 

Tn := An e (¡n2 ¼ 2 t) sin(n¼x)


The complete series solution is T = P1 n=1 Tn, withtheformofAnto be determined 

from the initial condition. At t = 0, one must have 

1X 

T (x; 0) = An sin(n¼x)=100x (1 ¡ x): 

n=1 

If T (x; 0) is multiplied by sin(m¼x) and integrated from x =0tox = L =1, 

the lhs will integrate to zero for every term in the series except for the term 

corresponding to m = n. This is a result of the independence, or orthogonality, 

of the sine terms in the series. The resulting equation can then be solved for 

An. Russell will now apply this approach. He evaluates Tn at t =0,andenters 

the initial temperature pro¯le, T0 . 

> X:=eval(T[n],t=0); T0:=100*x*(1-x); 

X := An sin(n¼x) T0 := 100 x (1 ¡ x) 

Since only the nth term in the series survives, Russell forms T0 ¡ X, multiplies 

this by sin(n¼x), and integrates from x =0to1ineq.Theneq is set equal to 

zero and solved for An. 

> eq:=int((T0-X)*sin(n*Pi*x),x=0..1); 

> A[n]:=solve(eq=0,A[n]); 

200 (¡2+n¼sin(n¼)+2cos(n¼)) 

An := ¡ 

n2 ¼2 (n¼¡ sin(n¼)cos(n¼)) 

Assuming that n is an integer, the nth Fourier term Tn is simpli¯ed, An having 

been automatically substituted. The double colon in assuming is a \type 

match" command. 

> T[n]:=simplify(T[n]) assuming n::integer; 

Tn := ¡ 400 (¡1+(¡1)n ) e (¡n2 ¼ 2 t) sin(n¼x) 

n3 ¼3 Thus, the temperature distribution in the rod satisfying the boundary and 

initial conditions is given by the following in¯nite Fourier series: 

> Temp:=Sum(T[n],n=1..infinity); 

1X 

Ã 

Temp := ¡ 400 (¡1+(¡1)n ) e (¡n2 ¼ 2 t) sin(n¼x) 

n3 ¼3 ! 

n=1 

The sum of the ¯rst ¯ve terms gives a good approximation to T (x; t), because 

the contribution of higher-order terms drops very rapidly with increasing time. 

> TT:=sum(T[n],n=1..5); 

TT := 800 e(¡¼2 t) sin(¼x) 

¼3 + 800 e 

27 

(¡9 ¼2 t) sin(3 ¼x) 

¼3 + 32 e 

5 

(¡25 ¼2 t) sin(5 ¼x) 

¼3 Then TT is animated so its spatial and temporal evolution can be clearly seen. 

> animate(plot,[TT,x=0..1],t=0..1,frames=50); 

Russell is feeling sleepy, so is going back to bed. If you wish to see how the 

initial parabolic temperature pro¯le decays to zero everywhere inside the rod, 

execute the work sheet, click on the plot, and then on the play arrow.


PROBLEMS: 

Problem 5-11: Hot rod 

For the heated rod discussed in the text, at what time is the temperature at 

the center of the rod equal to one-third of its initial value? 

Problem 5-12: The switch 

The temperatures at the ends x =0andx = 100 of a rod (insulated on its 

sides) 100 cm long are held at 0 ± and 100 ± , respectively, until steady state 

is achieved. Then at the instant t = 0, the temperatures of the two ends 

are interchanged. Determine the resultant temperature distribution T (x; t). 

Animate your solution and discuss its behavior. 

5.2.2 Aussie Barbecue 

He who does not mind his belly, will hardly mind anything else. 

Samuel Johnson, English author and lexicographer (1709{1784) 

When coauthor Richard spent a sabbatical leave at the Australian National 

University, in Canberra, he constructed a barbecue consisting of a scrap of 

rectangular iron plate placed on a primitive 2 brickwork support. The myriad 

eucalyptus trees that dot the landscape of the Australian Capital Territory and 

New South Wales constantly shed branches, which were gathered to be used 

as free barbecue fuel. With the iron plate slightly tilted to allow the grease 

from the sizzling lamb chops (accompanied with a cask of wine, the staple of 

the Australian outback barbecue) to drip o® into the ¯re, the °ames on the 

downhill edge of the plate tended to make that edge considerably hotter than 

the other edges of the plate. The temperature distribution in the plate was 

uneven, and vigilance was necessary to prevent the chops from being turned 

into charcoal. The Aussie barbecue is the inspiration for the following twodimensional 

boundary value example, in which the steady-state temperature 

pro¯le in a thin rectangular plate, with prescribed temperatures on the edges, 

is found. 

Consider a uniform solid rectangular plate stretching between x =0 and 

x = L and y =0 and y = h, whereL and h are measured in meters. Since 

what is important is the temperature di®erences across the plate, without loss 

of generality the coldest edge can be set equal to zero. Speci¯cally, we shall 

imagine the plate to have three \cold" edges whose temperatures are set equal to 

zero, i.e., T (0;y)=T (x; 0)=T (x; h)=0. The \hot" edge x=L will be assumed 

to have a parabolic temperature distribution given by T (L; y)=400y (h¡y) ± C. 

If, for example, h=1 m, the temperature along the hot edge will vary from zero 

at the corners to a temperature 100 ± hotter in the middle of that edge. Our goal 

is to determine the steady-state temperature distribution T (x; y) in the plate. 

2 Note: Richard is a theoretician, not an experimentalist.


The answer will require that the two-dimensional form of Laplace's equation, 

@ 2 T (x; y) 

@x 2 

+ @2 T (x; y) 

@y 2 =0; (5.9) 

be solved, subject to the four boundary conditions. After loading the plots 

package, we enter Laplace's equation (LE). 


> LE:=diff(T(x,y),x,x)+diff(T(x,y),y,y)=0; 

μ μ 

2 

2 

@ @ 

LE := T (x; y) + T (x; y) =0 

@x2 @y2 Using the pdsolve commandwiththeHINT=f(x)*g(y),INTEGRATE, andbuild 

options, we generate the general product solution of Laplace's equation. 

> sol:=pdsolve(LE,HINT=f(x)*g(y),INTEGRATE,build); 

sol := T (x; y) = C3 sin( p c1 y) C1 e (p c1 x) + C3 sin( p c1 y) C2 

e (p c1 x) 

+ C4 cos( p c1 y) C1 e (p c1 x) + C4 cos( p c1 y) C2 

e (p c1 x) 

The solution involves one separation constant c1 and four unknown coe±cients. 

For notational convenience, we substitute p c1 = m in the rhs of sol. 

> T:=subs(sqrt(_c[1])=m,rhs(sol)); 

T := C3 sin(my) C1 e (mx) + 

+ C4 cos(my) C2 

e (mx) 

C3 sin(my) C2 

e (mx) 

+ C4 cos(my) C1 e (mx) 

The coe±cients are determined from the four boundary conditions. To satisfy 

T (x; 0) = 0, the cos(my) terms are removed since they don't vanish at y =0. 

> T:=remove(has,T,cos(m*y)); #bc1 

T := C3 sin(my) C1 e (mx) C3 sin(my) C2 

+ 

e (mx) 

Then T is converted completely to trigonometric form. 

> T:=expand(convert(T,trig)); 

T := C3 sin(my) C1 cosh(mx)+ C3 sin(my) C1 sinh(mx) 

C3 sin(my) C2 

+ 

cosh(mx) + sinh(mx) 

The second boundary condition is that T (0;y) = 0. Since cosh(mx)doesn't 

vanish at x = 0, terms involving cosh(mx) are removed from T . 

> T:=remove(has,T,cosh(m*x)); #bc2 

T := C3 sin(my) C1 sinh(mx) 

The third boundary condition, T (x; h) = 0, requires that sin(mh)=0,sothat 

m = n¼=hwith n a positive integer. This relation is substituted into T ,and


the awkward coe±cient combination replaced with An. The resulting form, 

labeled Tn, isthenthterm in the in¯nite Fourier series solution. 

> T[n]:=A[n]*subs(m=n*Pi/h,T)/(_C3*_C1); #bc3 

³ 

n¼y 

´ ³ 

n¼x 

´ 

Tn := An sin sinh 

h 

h 

The remaining boundary condition must be applied along the \hot" edge at 

x = L. ThenTnis evaluated at this value in S, and the temperature distribution 

T0 =400y (h ¡ y) along the hot edge entered. 

> S:=eval(T[n],x=L); T0:=400*y*(h-y); 

³ 

n¼y 

´ μ 

n¼L 

S := An sin sinh 

T0 := 400 y (h ¡ y) 

h 

h 

On the edge x = L, wehave 

1X ³ 

n¼y 

´ μ 

n¼L 

T (L; y) = An sin sinh = T0 = 400 y (h ¡ y): 

h 

h 

n=1 

If T (L; y) is multiplied by sin(m¼y=h) and integrated from y =0toh, thelhs 

will yield zero except for m = n. Thus, the coe±cient An can be determined 

in a manner similar to the last recipe. The next two command lines carry out 

this procedure and calculate An. 

> eq:=int((S-T0)*sin(n*Pi*y/h),y=0..h)=0: #bc4 

> A[n]:=solve(eq,A[n]): 

Assuming that n is an integer, the nth Fourier term is simpli¯ed, An having 

been automatically substituted. 

> T[n]:=simplify(T[n]) assuming n::integer; 

1600 h 

Tn := ¡ 

2 (¡1+(¡1) n ³ 

n¼y 

´ ³ 

n¼x 

´ 

)sin sinh 

h 

h 

n3 μ 

n¼L 

sinh ¼ 

h 

3 

The complete solution is T (x; y) = P1 n=1 Tn. To plot the solution, we will take 

L =1,h = 1, and keep terms in the series up to n =7, 

> L:=1: h:=1: Temp:=sum(T[n],n=1..7); 

Temp := 

3200 sin(¼y) sinh(¼x) 

sinh(¼) ¼ 3 

+ 3200 

27 

sin(3 ¼y) sinh(3 ¼x) 

sinh(3 ¼) ¼ 3 

+ 128 sin(5 ¼y) sinh(5 ¼x) 

5 sinh(5 ¼) ¼3 + 3200 sin(7 ¼y) sinh(7 ¼x) 

343 sinh(7 ¼) ¼3 which is su±ciently accurate here. By itself, the series representation doesn't 

tell us much. To see the steady-state temperature distribution T (x; y) inthe 

plate, a three-dimensional colored contour plot with 20 equally spaced contours 

is created. 

> plot3d(Temp,x=0..L,y=0..h,axes=boxed,style=patchcontour, 

contours=20,orientation=[-132,35],tickmarks=[2,2,2]);


100 

0 

1 

y 

0 

Figure 5.6: Three-dimensional contour plot of the temperature distribution. 

The resulting picture is shown in Figure 5.6. From the ¯gure, the reader can 

get a good feeling for how the temperature distribution drops o® from the hot 

edge. If it is desired to know the temperature at a particular point, a more 

accurate answer can be obtained by substituting the coordinates of the point 

into the series solution. For example, the temperature in the middle (x = L=2, 

y = h=2) of the plate is evaluated, 

> middle:=evalf(eval(Temp,fx=L/2,y=h/2g)); 

and found to be about 20 1 

2 

0 

x 

middle := 20:53145858 

± warmer than on the cold edges. 

PROBLEMS: 

Problem 5-13: Electric potential 

An in¯nitely long hollow conductor has a rectangular cross section with sides L 

and 2 L. One of the longer sides is charged to a potential V =V0 and the other 

three sides are held at zero potential. By solving Laplace's equation, determine 

the potential distribution in the interior region. Taking V0 =1 andL =1, plot 

the equipotential lines. Calculate the electric ¯eld ~ E = ¡rV in the interior 

region. Plot ~ E on the same graph as the equipotential lines. 

Problem 5-14: Barbecue plate 

Suppose that a rectangular barbecue plate has a ¯nite thickness c =0:01 m 

in the z-direction and dimensions a =0:5 mandb =0:5 minthex- and 

y-directions, respectively. Assuming that the bottom of the plate is uniformly 

heated and is 100 ± hotter than the other ¯ve sides, determine the steady-state 

temperature distribution T (x; y; z) inside the plate and make a suitable plot. 

1


5.2.3 Benny's Solution 

A good scientist is a person with original ideas. A good engineer is 

a person who makes a design that works with as few original ideas 

as possible. There are no prima donnas in engineering. 

Freeman Dyson, British-born U.S. physicist (1923{) 

Greg Arious Nerd is currently teaching the mathematical physics course to a 

mixture of future engineers and physicists at Erehwon's most famous academic 

institution, EIT (Erehwon Institute of Technology). The students are being 

instructed in the use of the Elpam computer algebra system in solving their 

mathematical physics problems. As a classroom example, Greg selects a somewhat 

arti¯cial, but pedagogically useful, two-dimensional static potential problem. 

A circular annulus has an angular potential distribution Á(10;μ)=15cosμ 

speci¯ed on the inner radius r1 =10cmandapotentialÁ(20;μ)=30sinμ on 

the outer radius r2 = 20 cm. The question to be answered is, \What is the 

potential distribution in the annular region, and what do the equipotentials 

look like in this region?" 

The following recipe for solving this problem has been submitted by one of 

Greg's engineering students, Benjamin Beetlebrox III. Although, as a descendent 

of one of the founding families on Erehwon, he doesn't like his ¯rst name 

shortened, we shall take Freeman Dyson's words to heart and call him Benny. 

In addition to the plots package needed for plotting the equipotentials, 

Benny loads the VectorCalculus package, because it contains the Laplacian 

command, which will enable him to easily generate Laplace's equation for the 

potential Á(r; μ) in polar coordinates. The radial (r) and angular (μ) polarcoordinates 

are related to the Cartesian coordinates (x; y) through the relations 

x = r cos μ, y = r sin μ. 


Curious about the coordinate systems that the Elpam system supports, Benny 

enters the following command line. On executing this line, a list of the available 

two- and three-dimensional coordinate systems appears in a Help page. 

> ?coords; 

Through the hyperlinks at the bottom of the Help page, Benny is led to the 

coordplot and coordplot3d commands for plotting representative curves and 

surfaces in two and three dimensions, corresponding to holding each coordinate 

equal to a constant value. Closing the Help window, Benny uses coordplot 

to plot the lines r =constantandμ = constant in polar coordinates. The 

grid option is used to control the number of constant values and therefore 

the number of lines drawn. The default is grid=[12,12]. The values of the 

constants are added to the graph by including labeling=true. More detailed 

explanations and other options may be found on the coordplot Help page. 

> coordplot(polar,grid=[5,7],labelling=true, 

scaling=constrained);


Pi 

2/3*Pi 1/3*Pi 

4/3*Pi 

0 

1/4 

1/2 

5/3*Pi 

3/4 

Figure 5.7: Constant r and μ lines in polar coordinates. 

The resulting picture is reproduced in Figure 5.7, circles being produced for 

r =0; 1 1 3 ; ; ; 1 and polar lines for μ =0;¼=3; 2 ¼=3; ¼;:::: 

4 2 4 

Benny now enters Laplace's equation (LE ) in polar coordinates. 

> LE:=expand(Laplacian(phi(r,theta),'polar'[r,theta]))=0; 

LE := 

@ 

Á(r; μ) 

@r 

r 

μ 2 @ 

+ 

Á(r; μ) 

@r2 

+ 

1 

@2 Á(r; μ) 

@μ2 The pdsolve commandwiththeHINT=f(r)*g(theta), INTEGRATE and build 

options is used to ¯nd the general product solution of Laplace's equation. 

> sol:=pdsolve(LE,HINT=f(r)*g(theta),INTEGRATE,build); 

r 2 

=0 

sol := Á(r; μ) = C3 sin( p c1 μ) C1 r (p c1) + C3 sin( p c1 μ) C2 

r (p c1) 

+ C4 cos( p c1 μ) C1 r (p c1) + C4 cos( p c1 μ) C2 

r (p c1) 

Benny notes that the solution must reduce to a cos μ form on one boundary 

and a sin μ form on the other, so he accordingly sets the separation constant 

p c1 = 1 on the rhs of sol. 

> phi:=subs(sqrt(_c[1])=1,rhs(sol)); 

C3 sin(μ) C2 

C4 cos(μ) C2 

Á := C3 sin(μ) C1 r + + C4 cos(μ) C1 r + 

r 

r 

The terms are then grouped by successively collecting cos μ, sinμ, andrterms.


> phi:=collect(phi,[cos(theta),sin(theta),r]); 

μ 

μ 

 

C4 C2 

C3 C2 

Á := C4 C1 r + cos(μ)+ C3 C1 r + sin(μ) 

r 

r 

To simplify the coe±cients in Á, Benny makes the following substitutions. 

> phi:=subs(_C3=a/_C1,_C2=b*_C1/a,_C4=c/_C1,phi); 

μ 

Á := cr+ cb 

μ 

cos(μ)+ ar+ 

ar 

b 

 

sin(μ) 

r 

> phi:=algsubs(c*b=a*d,phi); 

μ 

Á := cos(μ) cr+ d 

μ 

+ ar+ 

r 

b 

 

sin(μ) 

r 

To determine the four unknown coe±cients a, b, c, andd, four boundary conditions 

are required. On the inner boundary, r = 10 cm, the coe±cient of the 

cos μ term must equal 15, while the coe±cient of the sin μ term must equal zero. 

This yields two boundary conditions, labeled bc1 and bc2 . 

> bc1:=eval(coeff(phi,cos(theta)),r=10)=15; 

bc1 := 10 c + d 

10 =15 

> bc2:=eval(coeff(phi,sin(theta)),r=10)=0; 

bc2 := 10 a + b 

10 =0 

On the outer boundary, r =20cm,thecoe±cientofthesinμterm must equal 

30, while the coe±cient of the cos μ term is equal to zero. This yields two more 

boundary conditions, bc3 and bc4 . 

> bc3:=eval(coeff(phi,sin(theta)),r=20)=30; 

bc3 := 20 a + b 

20 =30 

> bc4:=eval(coeff(phi,cos(theta)),r=20)=0;; 

bc4 := 20 c + d 

20 =0 

The four boundary condition equations are solved for a, b, c, andd, 

> coefficients:=solve(fbc1,bc2,bc3,bc4g,fa,b,c,dg); 

½ 

coe±cients := d =200;a=2;c= ¡1 

¾ 

;b= ¡200 

2 

which are assigned to produce the ¯nal solution Á to the potential problem. 

> assign(coefficients): phi:=phi; 

μ 

Á := cos(μ) ¡ r 

μ 

200 

+ + 2 r ¡ 

2 r 

200 

 

sin(μ) 

r 

As requested by Professor Nerd, Benny will now use Á to plot the equipotentials 

in the annular region between r = 10 and 20. He ¯rst converts the potential 

into Cartesian coordinates by substituting r = p x2 + y2 ,cosμ = x= p x2 + y2 , 

and sin μ = y= p x2 + y2 ,intoÁ.


> phi:=subs(fr=sqrt(x^2+y^2),cos(theta)=x/sqrt(x^2+y^2), 

sin(theta)=y/sqrt(x^2+y^2)g,phi): 

He then creates a piecewise potential function © equal to Á for 100 · x2 + y2 · 

400 and zero otherwise. (The © output is suppressed here.) 

> Phi:=piecewise(100implicitplot(fPhi=5*ig,x=-20..20,y=-20..20, 

scaling=constrained,grid=[100,100]): 

The sequence of constant-potential plots is then displayed for i = ¡6 to+6, 

the result being shown in Figure 5.8. 

> display(fseq(F(i),i=-6..6)g); 

20 

y 

10 

–20 –10 10 20 

x 

–10 

–20 

Figure 5.8: Equipotential lines for the circular annulus. 

Although, he hasn't bothered to label the equipotential lines, Benny feels that 

the plot already conveys a better sense of the equipotentials than could be 

gained by staring at the formula, simple as it is. He has left the labeling of the 

equipotentials for you to carry out as a problem. 

PROBLEMS: 

Problem 5-15: Labeling of equipotential lines 

Using the textplot command, add appropriate potential values to Figure 5.8 

so that the equipotential lines are clearly identi¯ed.


Problem 5-16: Split potential 

Along the circumference of a circle of radius r =1,thepotentialÁ(r; μ) has 

the value Á =1fortheangularrange0


5.2.4 Hugo and the Atomic Bomb 

If the radiance of a thousand suns were to burst forth at once in the 

sky, that would be like the splendor of the Mighty One. 

Bhagavad Gita, a philosophical dialogue that is a sacred Hindu text, found in 

the Mahabharata, one of the ancient Sanskrit epics (250 BC{250 AD) 

Hugo, who was formerly a scientist in country X, has emigrated to the New 

World in search of a better life. Unfortunately, he has been forced to temporarily 

drive taxis until he can ¯nd a job more suitable to his educational training. 

While waiting for his next fare, he tries to keep his mind sharp by carrying out 

model calculations on his laptop computer. At this particular moment, Hugo is 

working on the problem of the growth of the neutron density in a nuclear chain 

reaction. Let's eavesdrop on what Hugo is doing. 

Hugo knows that if uranium nuclei are bombarded with neutrons, a given 

nucleus may absorb a neutron, resulting in the splitting of the uranium nucleus 

into two parts with the release of substantial energy as well as two or three of 

the neutrons that were already present in the nucleus. This splitting process 

is called nuclear ¯ssion andisthe¯rststepinachain reaction. Whether the 

reaction will keep on going depends on how many of the released neutrons are 

available to initiate another ¯ssion process. The factor by which the number of 

neutrons increases between one step and the next in the chain reaction is called 

the multiplication factor. In a nuclear reactor, the multiplication factor is kept 

at unity (called the critical condition) by using boron or cadmium control rods 

to \soak up" excess neutrons. In this case, the chain reaction proceeds at a 

constant rate with a steady output of energy. If the multiplication factor is 

greater than unity (the supercritical condition), the chain reaction leads to a 

geometrically increasing number of ¯ssions in a very short time interval with the 

accompanying release of an enormous amount of energy. A nuclear explosion 

takes place | the basis of the atomic bomb. 

Hugo decides that he can learn more about the underlying role that the 

neutrons play in the chain reaction by modeling the time evolution of some 

speci¯ed initial neutron distribution inside a mass of ¯ssionable material. For 

calculational purposes, he takes the mass to be cylindrical in shape with a radius 

r = a, the lower face of the cylinder at z = 0 and the upper face at z = h. 

For simplicity, Hugo takes the neutron density N (number of neutrons per unit 

volume) to be independent of the angular coordinate μ, i.e., N = N(r; z; t). In 

the absence of any production of neutrons by ¯ssion, the neutron density would 

obey the linear di®usion equation. To account for the production of neutrons 

by ¯ssion, Hugo adds a neutron source term ¯N,where¯is a positive rate 

constant, to the di®usion equation, viz., 

@N 

@t = d r2N + ¯N: (5.10) 

In order to use the cylindrical polar form of the Laplacian, he calls up the 

VectorCalculus package.



The di®usion equation is entered with the Laplacian in cylindrical coordinates. 

> de:=diff(N(r,z,t),t)-beta*N(r,z,t) 

=d*Laplacian(N(r,z,t),'cylindrical'[r,theta,z]); 

de := @ 

N (r; z; t) ¡ ¯ N (r; z; t) 

@t 

μ 2 @ 

μ 2 @ 

μμ 

@ 

N (r; z; t) 

@r 

d 

+ r N (r; z; t) + r N (r; z; t) 

@r2 @z2 = 

r 

The di®usion equation de is solved with the pdsolve command, the HINT option 

being omitted, because Maple already assumes a general product solution. After 

the introduction of two separation constants c1 and c2, the PDE is separated 

into three ODEs which are solved and the product formed. 

> sol:=pdsolve(de,INTEGRATE,build); 

sol := N (r; z; t) 

= e (p p c2 z) (td c1) (td c2) (t¯) C5 e e e C1 BesselJ(0; ¡ c1 r) C3 

+ C5 e(td c1) e (td c2) e (t¯) C1 BesselJ(0; p ¡ c1 r) C4 

e (p c2 z) 

+ e (p c2 z) (td c1) (td c2) (t¯) p 

C5 e e e C2 BesselY(0; ¡ c1 r) C3 

+ C5 e(td c1) e (td c2) e (t¯) C2 BesselY(0; p ¡ c1 r) C4 

e (p c2 z) 

The solution sol involves a linear combination of a zeroth-order Bessel function 

of the ¯rst kind (J0( p ¡ c1 r) in math notation) and a zeroth-order Bessel function 

of the second kind (Y0( p ¡ c1 r)). The latter Bessel function diverges at 

r = 0 at arbitrary time, so Hugo imposes the ¯rst boundary condition, removing 

terms involving BesselY from the rhs of the solution. 

> N:=remove(has,rhs(sol),BesselY); #bc1 

N := e (p p c2 z) (td c1) (td c2) (t¯) C5 e e e C1 BesselJ(0; ¡ c1 r) C3 

+ C5 e(td c1) (td c2) (t¯) p 

e e C1 BesselJ(0; ¡ c1 r) C4 

e (p c2 z) 

Hugo simpli¯es N by substituting c1 = ¡¹ 2 and c2 = ¡º 2 and applying the 

simplify command with the symbolic option. 

> N:=simplify(subs(f_c[1]=-mu^2,_c[2]=-nu^2g,N),symbolic); 

N := C5 C1 BesselJ(0; ¹r) e (¡td¹2 ¡tdº 2 +t¯) ( C3 e (ºzI) + C4 e (¡I ºz) ) 

The complex form is split into real and imaginary parts with the complex 

evaluation command. 

> N:=evalc(N);


N := C5 C1 BesselJ(0; ¹r) e (¡td¹2 ¡tdº 2 +t¯) ( C3 cos(ºz)+ C4 cos(ºz)) 

+ C5 C1 BesselJ(0; ¹r) e (¡td¹2 ¡tdº 2 +t¯) ( C3 sin(ºz) ¡ C4 sin(ºz)) I 

Hugo assumes that the neutron density at the surface of the cylindrical mass is 

zero. Thus, since it doesn't go to zero at z =0,thecos(ºz) term is removed 

(a second boundary condition) from N and the result factored. 

> N:=factor(remove(has,N,cos)); #bc2 

N := C5 C1 BesselJ(0; ¹r) e (¡t (¡¯+d¹2 +dº 2 )) sin(ºz)(¡ C4 + C3) I 

In the next command line, Hugo removes the \ugly" Maple coe±cient combination 

from N by forming the product of the 4th, 5th, and 6th operands of N, 

the other operands corresponding to the various constants. 

> N2:=op(4,N)*op(5,N)*op(6,N); 

N2 := BesselJ(0; ¹r) e (¡t (¡¯+d¹2 +dº 2 )) sin(ºz) 

Since the neutron density is zero on the cylindrical surface r = a and at the end 

z = h, two more boundary conditions must be imposed. At r = a, J0(¹a)=0, 

so that ¹ = ¸m=a, where ¸m is the mth zero of J0. These zeros may be 

found with the command BesselJZeros(0,m), wherem =1; 2;:::: At z = h, 

sin(ºh)=0,soº = n¼=h,withn a positive integer. This pair of boundary 

condition relations is substituted into N2. 

> N2:=subs(fmu=BesselJZeros(0,m)/a,nu=n*Pi/hg,N2); #bcs 

μ 

N2 := BesselJ 0; 

 

BesselJZeros(0; m) r 

a 

d BesselJZeros(0;m)2 

e 

(¡t (¡¯ + a2 + dn2 ¼ 2 

h2 ³ 

)) n¼z 

´ 

sin 

h 

By the principle of linear superposition, the complete solution will be 

1X 1X 

1X 1X 

N(r; z; t) = Cm;n N2 ´ Nm;n; 

m=1 n=1 

m=1 n=1 

with the Cm;n determined by the initial neutron density N(r; z; 0) = f. Evaluating 

N2 at t = 0, and labeling the result g, 

> g:=eval(N2,t=0); 

μ 

 


³ 

n¼z 

´ 

g := BesselJ 0; sin 

a 

h 

one has 

1X 1X 

N(r; z; 0) = Cm;n g = f: 

m=1 n=1 

If we multiply both sides of N(r; z; 0) by rJ0(¸m0 r=a)sin(n0¼z=h)andinte grate r from 0 to a and z from 0 to h, the lhs will yield zero unless m = m 0 

and n = n 0 . Solving for Cm 0 ;n 0, and dropping the primes, the coe±cients can 

be calculated from 

R a R h 

0 0 

Cm;n = 

rfgdrdz 

R a R h 

0 0 rg2 dr dz :


For an initial neutron density satisfying the boundary conditions, Hugo takes 

f to be of the following simple structure: 

> f:=(1-r^2/a^2)*sin(Pi*z/h); 

f := 

μ 

1 ¡ r2 

a2 

³ 

¼z 

´ 

sin 

h 

The coe±cients Cm;n are calculated, and then the Fourier term Nm;n. 

> C[m,n]:=int(int(r*f*g,z=0..h),r=0..a) 

/int(int(r*g^2,z=0..h),r=0..a): 

> N[m,n]:=C[m,n]*N2; 

For integer n>1, Nm;n should be equal to zero, since only an n =1term 

occurs in f. Hugo checks that this is the case. 

> simplify(N[m,n]) assuming n::integer,n>1; 

0 

To determine Nm;1 greater care must be exercised, and the limit taken as n ! 1. 

The result then is simpli¯ed with respect to the exponentials. 

> N[m,1]:=simplify(limit(N[m,n],n=1),exp); 

Nm; 1 := 8 sin 

³ 

¼z 

´ μ 

BesselJ 0; 

h 

 


a 

e (¡ (¡¯ a2 h 2 +d BesselJZeros(0;m) 2 h 2 +d¼ 2 a 2 ) t 

a2 h2 ) 

Á 

BesselJ(1; BesselJZeros(0; m)) BesselJZeros(0; m) 3 

The total neutron density at time t then is P 1 

m=1 Nm;1. To plot the density, 

Hugo takes the nominal values a =1,h =1,andd = 1. In the time-dependent 

part of the density, the coe±cient of t is ¯ ¡ d (¸ 2 m =a2 + ¼ 2 =h 2 ). The largest 

possible positive value of this coe±cient occurs when m = 1, corresponding to 

the ¯rst zero of J0. Inthiscase¸1 ¼ 2:405. If 

¯


> Density:=sum(eval(N[m,1],fa=1,h=1,d=1,beta=15.7g),m=1..3); 

Density := 

(0:047209632 t) 

1:108022261 sin(3:141592654 z)BesselJ(0:; 2:404825558 r) e 

(¡24:64086674 t) 

¡ 0:1397775054 sin(3:141592654 z)BesselJ(0:; 5:520078110 r) e 

(¡69:05661119 t) 

+0:04547647069 sin(3:141592654 z)BesselJ(0:; 8:653727913 r) e 

The numerical coe±cient of t in the ¯rst term of the output is slightly positive 

while the coe±cients of t in the other terms are large negative numbers. 

Thus, the ¯rst term in the series will grow with time and all remaining terms 

exponentially decay. If the neutron density grows fast enough, an uncontrolled 

chain reaction, i.e., an explosion, will occur. 

To create a dynamic 3-dimensional picture of the explosive growth in neutron 

density (the \nuclear explosion"), Hugo makes use of the animation command. 

> animate(plot3d,[Density,r=-1..1,z=0..1],t=0..100, 

frames=25,orientation=[65,40],shading=zhue, 

style=patchnogrid,grid=[25,25],axes=framed); 

As Hugo contentedly looks at the \explosion" on his computer screen, running 

in the loop mode and with 200% zoom magni¯cation, he hears a gru® voice. 

\Hey Buddy, if you don't mind, I am running late and have to get to the 

airport." 

Interrupted by harsh reality, Hugo puts his computer away, but hopes that 

one of his job applications will pan out soon. 

PROBLEMS: 

Problem 5-23: Spherical mass 

Carry out a calculation similar to that in the text recipe for a ¯ssionable spherical 

mass of radius r = a andaninitialneutrondensityf(r) =1¡ r2 =a2 . 

Make an animated plot of the neutron density inside the sphere when the mass 

slightly exceeds the critical mass. 

Problem 5-24: Semicircular plate 

A thin semicircular plate of radius r = 1 has its edges held at zero temperature 

and its °at faces insulated. If the initial temperature distribution inside the 

plate is T (r; μ; 0) = 100 r2 cos(2 μ), determine the temperature T (r; μ; t) for 

t>0. Taking the di®usion constant d = 1, animate the temperature pro¯le. 

Problem 5-25: Temperature distribution 

A solid has the shape of an in¯nitely long quarter-cylinder of radius r =1 

and di®usion constant d. The °at sides are insulated, so no heat °ows through 

them, while the curved surface is kept at 100 ± C. Assuming that the temperature 

initially varies as the fourth power of the distance from the axis, ¯nd 

the temperature distribution at any point inside the solid for t>0. Choosing 

nominal parameter values, create an animated plot of the temperature pro¯le. 

Problem 5-26: Cylindrical temperature pro¯le 

A cylinder of unit radius and unit height has its circular ends z =0andz =1


kept at T =0andT = 1, respectively, while the circular surface is kept at 

T = 1. Determine the steady-state temperature pro¯le inside the cylinder and 

plot the contours of equal temperature. 

Problem 5-27: Elliptic cross section 

An in¯nitely long bar of elliptic cross section has its curved surface, x2 +4 y2 =1, 

kept at T =0 ± . Determine the temporal evolution of T in the cylinder if initially 

T =100 ± everywhere inside and the di®usion constant d=1. Animate the result. 

5.2.5 Hugo Prepares for His Job Interview 

I'm notorious for giving a bad interview. I'm an actor and I can't 

help but feel I'm boring when I'm on as myself. 

Rock Hudson, American movie actor (1925{1985) 

Hugo has been invited by International Hydrodynamics Inc. to interview for a 

research position with their company, which specializes in designing streamlined 

hulls for surface vessels as well as underwater craft. As part of his preparation, 

Hugo decides to create some ¯les for the interview that demonstrate his skills 

at simulating liquid °ow around a variety of rigid geometrical shapes. As a 

\warm-up" exercise, he recalls from his undergraduate days a problem involving 

uniform incompressible °uid °ow around a sphere, a problem that can be 

solved analytically. Hugo remembers that the °uid can be characterized by a 

velocity potential U that satis¯es Laplace's equation. The velocity ~ V of a °uid 

element is then given by ~ V = ¡rU. 

Since a spherical boundary of, say, radius a is involved, Hugo chooses to use 

spherical coordinates (r; μ; Á) with the origin at the center of the sphere and the 

physicist's convention that μ is measured from the positive z-axis and Á from 

the positive x-axis. Then, the Cartesian and spherical polar coordinates are 

connected by the relations x = r sin μ cos Á, y = r sin μ sin Á, andz = r cos μ. 

The range of r is from 0 to 1, μ from 0 to ¼, andÁfrom0to2¼. At a distance r far from the sphere, the °uid °ow is assumed to be uniform 

and directed along the z-axis.Theasymptotic(larger) form of the velocity is 

~V = V0 êz where V0 is the undisturbed °uid speed. Thus, since 

~V = ¡ @U 

@x êx ¡ @U 

@y êy ¡ @U 

@z êz = V0 êz; (5.11) 

then, on equating vector components and integrating, the asymptotic velocity 

potential is given (to within an arbitrary constant) by U = ¡V0 z = ¡V0 r cos μ. 

This will serve as one of the boundary conditions on the solution. The other 

boundary condition is at the surface of the sphere. If the surface is idealized to 

be absolutely rigid, it will acquire no momentum from the °uid. This implies 

that the normal component of the °uid velocity vector must vanish at r = a. 

Writing the gradient operator in spherical polar coordinates, this means that 

the velocity potential must satisfy the boundary condition @U=@rjr=a =0.


To solve Laplace's equation, Hugo loads the VectorCalculus package, 


and inputs Laplace's equation in spherical coordinates. He notes that the problem 

has rotational symmetry about the z-axis, the direction of °uid °ow, so 

that there should be no Á dependence in the ¯nal solution. Accordingly, he 

takes U = U(r; μ). 

LE:= 

> LE:=expand(Laplacian(U(r,theta),'spherical'[r,theta,phi]))=0; 

μ 

@ 

2 U (r; μ) 

@r 

r 

μ 2 @ 

+ 

U (r; μ) 

@r2 

+ 

μ 

@ 

cos(μ) U (r; μ) 

@μ 

r 2 sin(μ) 

+ 

@2 U (r; μ) 

@μ2 With no HINT provided, a general product solution is built with the pdsolve 

command, the answer involving one separation constant, c1, and four arbitrary 

constants. The result is then expanded. 

> sol:=expand(pdsolve(LE,INTEGRATE,build)); 

sol := U (r; μ) = 

+ 

+ 

+ 

C1 r (1=2 p μ 

1+4 c1) 1 

C3 LegendreP 

C1 r (1=2 p 1+4 c1) C4 LegendreQ 

p r 

μ 1 

2 

C2 r (¡1=2 p μ 

1+4 c1) 1 

C3 LegendreP 

p r 

C2 r (¡1=2 p μ 

1+4 c1) 1 

C4 LegendreQ 

p r 

2 

p r 

2 

r 2 

=0 

p 

1+4 c1 ¡ 1 

 

; cos(μ) 

2 

p 

1+4 c1 ¡ 1 

 

; cos(μ) 

2 

2 

p 

1+4 c1 ¡ 1 

 

; cos(μ) 

2 

p 

1+4 c1 ¡ 1 

 

; cos(μ) 

2 

The solution is expressed in terms of Legendre functions of the ¯rst kind 

(LegendreP) and of the second kind (LegendreQ). The former are well-behaved 

polynomial functions of cos μ, but the latter diverge at the ends of the angular 

range and therefore must be removed on physical grounds. 

> U:=remove(has,rhs(sol),LegendreQ); 

U := 

C1 r (1=2 p μ 

1+4 c1) 1 

C3 LegendreP 

p r 

C2 r (¡1=2 p μ 

1+4 c1) 1 

C3 LegendreP 

2 

p 

1+4 c1 ¡ 1 

 

; cos(μ) 

2 

p 

1+4 c1 ¡ 1 

 

; cos(μ) 

2 

+ 

p 

r 

2 

In standard math notation, the Legendre polynomials of the ¯rst kind would


be written as Pn(cos μ), where n = 1p 

1+4 c1 ¡ 2 

1. 

To express U in the 

2 

Maple equivalent of the math notation, Hugo substitutes c1 = ¡ 1 

1 +(n + 4 2 )2 , 

and simpli¯es the square root that would otherwise appear by including the 

symbolic option. 

> U:=simplify(subs(_c[1]=-1/4+(n+1/2)^2,U),symbolic); 

U := C3 LegendreP(n; cos(μ)) ( C1 r(n+1=2) + C2 r (¡n¡1=2) ) 

p 

r 

To simplify the notation, and without any loss of generality, Hugo sets C3 =1, 

C1 =A, andC2 =B, inU. On some executions of the recipe the coe±cient 

C3 inU is replaced with C4, so the latter is also set equal to one. 

> U:=subs(f_C3=1,_C4=1,_C1=A,_C2=Bg,U); 

U := LegendreP(n; cos(μ)) (Ar(n+1=2) + Br (¡n¡1=2) ) 

p r 

Only two coe±cients, A and B, remain to be evaluated, so the two boundary 

conditions mentioned earlier are used. The radial derivative of U must be zero 

at r = a, which is now entered, and solved for B in terms of A. 

> bc:=eval(diff(U,r),r=a)=0; 

> B:=simplify(solve(bc,B)); 

n+1) nAa(2 

B := 

n +1 

Now Hugo recalls that the ¯rst few Legendre polynomial functions of the ¯rst 

kind are P0(cos μ) =1,P1(cos μ) =cosμ, andP2(cos μ) = 1 

2 (3 cos2 μ ¡ 1), :::: 

Although the general solution will involve a sum over n from 1 to 1, only the 

n = 1 term has the correct cos μ dependence to match the asymptotic form of 

the velocity potential. The coe±cients of all terms for which n 6= 1mustbe 

equal to zero. As r !1, U ! AP1(cos μ) r = ¡V0 r cos μ, soA = ¡V0. The 

potential U is now evaluated with A = ¡V0 and n = 1, and simpli¯ed. 

> U:=simplify(eval(U,fA=-V[0],n=1g)); 

U := ¡ 1 cos(μ) V0 (2 r 

2 

3 + a3 ) 

r2 So the velocity potential U is now completely known. In standard texts, the 

potential and velocity distributions would probably only be sketched, if drawn 

at all. Hugo knows that this is not good enough for his prospective employers. 

He decides to create a plot of the region outside the sphere that shows the 

equipotential lines as well as the velocity vectors ~ V = ¡rU. For graphing 

purposes, he chooses V0 =1anda = 1, and switches to Cartesian coordinates, 

taking x = 0 so that a two-dimensional plot can be made. 

> U:=subs(fV[0]=1,a=1,cos(theta)=z/sqrt(z^2+y^2), 

r=sqrt(z^2+y^2)g,U): 

Since the region of interest is outside the sphere, Hugo forms a piecewise function 

U2 that is equal to U for z2 + y2 ¸ 1, and zero otherwise. 

> U2:=piecewise(z^2+y^2>=1,U,0);


U2 := 

( 

¡ z (2 (z2 + y 2 ) (3=2) +1) 

2(z 2 + y 2 ) (3=2) 

1 · z 2 + y 2 

0 otherwise 

A functional operator F, whichmakesuseoftheimplicitplot command, is 

formed to plot the equipotentials in steps of 0:25 i, wherei will be taken to be 

an integer. Instead of the 25 £ 25 default grid, which is too coarse here, Hugo 

take the grid to be 100 £ 100. 

> F:=i->implicitplot(U2=.25*i,z=-2..2,y=-2..2,grid=[100,100]): 

Next, Hugo wants to form a graph of the velocity vectors. He calculates the 

velocity in the z-y plane and again forms piecewise functions to give the velocity 

components outside the sphere. 

> Vel:=Gradient(-U,'cartesian'[z,y]); 

> Ve[1]:=piecewise(z^2+y^2>=1,Vel[1],0): 

> Ve[2]:=piecewise(z^2+y^2>=1,Vel[2],0): 

The fieldplot command is used to produce a graph of the velocity vectors. 

> gr:=fieldplot([Ve[1],Ve[2]],z=-2..2,y=-2..2,grid=[18,18], 

arrows=MEDIUM,color=blue): 

The velocity vectors and the equipotential lines corresponding to i = ¡8 to8 

(U = ¡2 to+2instepsof0.25)aredisplayedinFigure5.9. 

> display(fgr,seq(F(i),i=-8..8)g,axes=boxed, 

scaling=constrained); 

Looking at the ¯gure, Hugo can easily \see" the °uid °ow around the sphere, 

with the velocity vectors perpendicular to the equipotentials and varying in 

magnitude near the sphere. 

2 

1 

y 

0 

–1 

–2 

–2 –1 0 1 2 

z 

Figure 5.9: Velocity vectors and equipotentials for °uid °ow around a sphere.


With this simple example under his belt, Hugo feels con¯dent that if he can 

come up with some °uid °ow examples for di®erent hull shapes, he might make 

a good impression on his interviewers and be hired. He realizes that for more 

complicated shapes he will probably have to use Maple's numerical capability. 

So let's leave Hugo to ¯nish his job interview preparation, so he doesn't pull 

aRockHudsonandgiveabadinterview. 

PROBLEMS: 

Problem 5-28: Split potential 

On the surface of a hollow sphere of radius a, the electric potential is 

½ 

+V; 0 · μ

Chapter 6 

Linear PDE Models. Part 2 

In the previous chapter, the computer algebra recipes concentrated on mathematical 

models formulated in terms of the di®usion and Laplace equations. In 

this continuation of our study of linear PDE models, the wave equation and a 

variety of models involving semi-in¯nite and in¯nite domains are presented. 

6.1 Wave Equation Models 

Thewaveequation, 

r 2 Ã(~r; t) = 1 

c2 @2Ã(~r; t) 

@t2 ; (6.1) 

with r2 the Laplace operator, Ã a scalar or vector function of position ~r and 

time t, andcthe wave velocity, applies to many di®erent types of waves, e.g., 

² electromagnetic waves (with Ã a vector); 

² sound waves; 

² tidal waves in an incompressible °uid; 

² elastic waves in a solid; 

² vibrations of elastic strings and membranes. 

In our ¯rst example, Vectoria plucks the humble string. 

6.1.1 Vectoria Encounters Simon Legree 

\We called him Tortoise because he taught us," said the Mock Turtle 

angrily. \Really you are very dull!" 

Lewis Carroll, Alice's Adventures in Wonderland, 1865 

It is some time later in the undergraduate career of Vectoria, the physics student 

who has been featured in several of our earlier stories. Recall that, inspired by 

the movie Casablanca, Vectoria used Maple to check and animate a textbook 

formula for the transverse motion of an initially horizontal piano string that has 

been struck. Since then Vectoria has progressed in her studies and is currently 

enrolled in a mathematical physics course. She has begun to learn how to solve 

wave equation problems using the separation of variables method. 

247


Professor Simon Legree, 1 who is teaching the course, has a reputation for 

assigning large numbers of often di±cult problems, so Vectoria decides once 

again to let Maple help her in deriving the solutions. Although she could do 

the problems by hand, it seems smarter in the long run to develop a computer 

algebra approach to lighten the workload and to avoid mathematical mistakes. 

On talking to the professor, she ¯nds out that, surprisingly, Legree not only 

agrees but suggests that it might be wise to start with a relatively simple 

problem, before tackling the more di±cult ones. Amazed that Legree, despite 

his hard-nosed reputation, has been so helpful, Vectoria decides to follow his 

advice and selects a problem involving the transverse motion of a string that 

has been plucked and released from rest. 

Since she doesn't yet know how to include sti®ness in the string, she opts 

to make use of the linear wave equation that neglects sti®ness. The horizontal 

string is ¯xed at its endpoints, x =0andx = L>0, and is given an initial 

transverse displacement Ã(x; t =0)=2hx=L for 0 · x · L=2 andÃ(x; 0) = 

2 h (L ¡ x)=L for L=2 · x · L. Vectoria recognizes that this is a triangular 

pro¯le with a maximum displacement h at the center (x = L=2) of the string. 

Professor Legree has also stressed that his marking assistant has been instructed 

not to give full credit for a problem solution unless it is accompanied by some 

sort of meaningful plot as well as some pertinent discussion. 

So Vectoria, anticipating that she will animate the vibrational motion of the 

string, begins her recipe with a call to the plots package, and enters the wave 

equation WE for the tranverse displacement Ã(x; t) ofthestringattimet. 


> WE:=diff(psi(x,t),x,x)=(1/c^2)*diff(psi(x,t),t,t); 

WE := @2 

@ 

Ã(x; t) = 

@x2 2 

Ã(x; t) 

@t2 c2 On applying the pdsolve command to WE without any options, 

> pdsolve(WE); 

Ã(x; t) = F1 (ct+ x)+ F2 (ct¡ x) 

Vectoria ¯nds that the solution is given by Ã(x; t) = F 1(ct+ x)+ F 2(ct¡ x), 

where F 1and F 2 are arbitrary functions. This is the well-known general 

solution of the wave equation, which is not too useful for solving the present 

problem with speci¯ed boundary and initial conditions. Clearly, a hint should 

be provided. She could be quite general and enter HINT=f(x)*g(t), butshe 

realizes that sine and cosine functions clearly satisfy the wave equation. Since 

sin(kx), where k is an undetermined constant, is equal to zero at x =0andthus 

satis¯es the boundary condition there, she provides the HINT=sin(k*x)*g(t) 

and integrates and builds up the solution. 

1 Unfortunately, Mrs. Legree, when naming her newborn son, chose Simon as a ¯rst name. 

She was unaware that Simon Legree was the cruel plantation owner in Harriet Beecher Stowe's 

novel, Uncle Tom's Cabin.

6.1. WAVE EQUATION MODELS 249 

> sol:=pdsolve(WE,HINT=sin(k*x)*g(t),INTEGRATE,build); 

sol := Ã(x; t) =sin(kx) C1 sin(ckt)+sin(kx) C2 cos(ckt) 

The string is initially at rest, so the transverse velocity _ Ã(x; 0) is equal to 0. 

Mentally di®erentiating the above output, clearly only the cos(ckt) term should 

be retained, since its derivative vanishes at t = 0, whereas the derivative of the 

sin(ckt) term does not. So Vectoria selects the term on the rhs of the solution 

containing the cosine term. 

> sol2:=select(has,rhs(sol),cos); 

sol2 := sin(kx) C2 cos(ckt) 

The displacement of the string must also vanish at x = L for arbitrary times, so 

onemusthavesin(kL) = 0, which yields k = n¼=L,withnapositive integer. 

She substitutes this result into sol2 , and temporarily removes any coe±cients 

by setting C1 and C2 equal to 1. 

> F[n]:=subs(fk=n*Pi/L,_C1=1,_C2=1g,sol2); 

³ 

n¼x 

´ μ 

cn¼t 

Fn := sin cos 

L 

L 

The general solution satisfying the boundary conditions at x =0andL, and 

with zero initial transverse velocity, then is 

1X 

1X ³ 

n¼x 

´ μ 

cn¼t 

Ã(x; t) = An Fn = An sin cos ; (6.2) 

L 

L 

n=1 

n=1 

where the coe±cients An have to be determined from the initial string pro¯le. 

This pro¯le is given by the piecewise function f. 

> f:=piecewise(xL/2,2*h*(L-x)/L); 

8 

>< 

2 hx 

x< 

f := L 

>: 

L 

2 

2 h (L ¡ x) L 

L 2 0. 

> A[n]:=int(f*sin(n*Pi*x/L),x=0..L)/int(sin(n*Pi*x/L)^2,x=0..L) 

assuming L>0: 

The nth Fourier term in the in¯nite series representing the string displacement 

is then Ãn =An Fn, the result being simpli¯ed by assuming that n is an integer. 

> psi[n]:=simplify(A[n]*F[n]) assuming n::integer; 

³ 

n¼ 

´ ³ 

n¼x 

´ μ 

cn¼t 

8 h sin sin cos 

2 L 

L 

Ãn := 

n2 ¼2 0


The 

P 

transverse displacement of the string at arbitrary time t is Ã(x; t) = 

1 

n=1 Ãn. For animation purposes, Vectoria takes L =10,h =1,andc =1. 

> L:=10: h:=1: c:=1: 

To obtain a good approximation to the initial string pro¯le, Vectoria keeps 

terms in the series up to n = 25, only the partial output being shown here. 

> psi:=sum(psi[n],n=1..25); 

³ 

¼x 

´ μ 

¼t 

8sin cos 

10 10 

Ã := 

¼2 ¡ 8 

μ μ 

3 ¼x 3 ¼t 

sin cos 

10 10 

9 ¼2 + 8 

³ 

¼x 

´ μ 

¼t 

sin cos 

2 2 

25 ¼2 + ¢¢¢¢¢¢+ 8 

μ μ 

21 ¼x 21 ¼t 

sin cos 

10 10 

441 

¼2 ¡ 8 

μ μ 

23 ¼x 23 ¼t 

sin cos 

10 10 

529 

¼2 + 8 

μ μ 

5 ¼x 5 ¼t 

sin cos 

2 2 

625 

¼2 The displacement Ã is animated over the time interval t = 0 to 20, with 100 

equally spaced time frames. 

> animate(plot,[psi,x=0..L],t=0..20,frames=100); 

On running the animation command, Vectoria is at ¯rst surprised by the behavior 

of the string, but on thinking about it decides that it makes sense. After 

executing the recipe, does the behavior make sense to you? What should Vectoria 

write in her explanation of the wave motion? 

PROBLEMS: 

Problem 6-1: Another plucked string 

A light homogeneous horizontal string of length L ¯xed at its ends initially 

has a parabolic shape with height h in the middle. If it is released from rest, 

solve the wave equation to determine its subsequent transverse displacement at 

arbitrary time t. Animate the vibrations for parameters of your own choice and 

choose enough frames to produce a smooth animation. 

Problem 6-2: Struck piano string 

If a horizontal piano string ¯xed at x =0andx = L is struck in such a way 

that its initial displacement Ã(x; 0) is zero and its initial transverse velocity is 

8 

< 4 vx=L; 0


6.1.2 Homer's Jiggle Test 

The fundamental cause of trouble in the world today is that the stupid 

are cocksure while the intelligent are full of doubt. 

Bertrand Russell, British mathematician and philosopher (1872{1970) 

Having successfully used Maple to solve the plucked string problem, Vectoria 

now tackles the ¯rst problem on Professor Legree's wave equation assignment. 

She is asked to determine the transverse displacement Ã(x; y; t) of a light, horizontal, 

rectangular elastic membrane having sides of length a and 2 a that 

is held ¯xed along all four edges. Speci¯cally, the boundary conditions are 

Ã(0;y;t)=Ã(a; y; t) =0for0· y · 2 a and Ã(x; 0;t)=Ã(x; 2 a; t) =0for 

0 · x · a. The membrane is given an initial displacement 

Ã(x; y; 0) ´ f = 4 hx2 (a ¡ x) y3 (2 a ¡ y) 

a7 and is released from rest. Professor Legree also requests that the membrane 

displacement be animated for a =1,h = 1, and wave velocity c =1. 

Vectoria realizes that this membrane problem is mathematically equivalent 

to two ¯xed-ends string problems in the x and y directions. So her \shortcut" 

approach to separating variables in the last recipe can be easily extended. 

After loading the plots package, she enters the wave equation WE in two 

dimensions. 


> WE:=diff(psi(x,y,t),x,x)+diff(psi(x,y,t),y,y) 

=(1/c^2)*diff(psi(x,y,t),t,t); 

μ μ @ 

2 

2 

@ @ 

WE := Ã(x; y; t) + Ã(x; y; t) = 

@x2 @y2 2 

Ã(x; y; t) 

@t2 c2 The pdsolve command is applied to WE with HINT=sin(p*x)*sin(q*y)*T(t), 

where p and q are constants. The assumed form satis¯es the boundary conditions 

along the ¯xed edges x =0andy =0. 

> sol:=pdsolve(WE,HINT=sin(p*x)*sin(q*y)*T(t),INTEGRATE,build); 

sol := Ã(x; y; t) =sin(px)sin(qy) C1 sin(c p p2 + q2 t) 

+sin(px)sin(qy) C2 cos(c p p2 + q2 t) 

The initial transverse velocity is zero, so once again only the term involving 

cos(c p p2 + q2 t)iskept. 

> sol2:=select(has,rhs(sol),cos); 

sol2 := sin(px)sin(qy) C2 cos(c p p 2 + q 2 t) 

At x = a, sin(pa)=0,sop = m¼=a,wherem is a positive integer. Similarly, 

at y =2a, sin(q 2 a) =0,soq = n¼=(2 a), n being a positive integer. Vectoria 

substitutes the p and q relations into sol2 , and temporarily removes all 

coe±cients by setting C1 = C2 =1.


> F[m,n]:=subs(fp=m*Pi/a,q=n*Pi/(2*a),_C1=1,_C2=1g,sol2);; 

Fm; n := sin 

Ã r 

³ 

m¼x 

´ ³ 

n¼y 

´ 

2 2 m ¼ 

sin cos c 

a 2 a 

a2 + n2 ¼2 ! 

t 

4 a2 Since she will need it for determining the coe±cients, Vectoria evaluates Fm;n 

at t = 0, labeling the result gm;n. 

> g[m,n]:=eval(F[m,n],t=0); 

³ 

m¼x 

´ ³ 

n¼y 

´ 

gm;n := sin sin 

a 2 a 

The initial pro¯le of the rectangular membrane is entered. 

> f:=4*h*x^2*(a-x)*y^3*(2*a-y)/a^7; 

f := 4 hx2 (a ¡ x) y3 (2 a ¡ y) 

a7 Now, the general form of the transverse displacement at time t will be given by 

the following linear superposition, 

1X 1X 

1X 1X 

Ã(x; y; t) = Am;n Fm;n ´ Ãm;n; (6.3) 

m=1 n=1 

m=1 n=1 

with the coe±cients Am;n determined by the initial pro¯le f. Setting t =0in 

(6.3) yields 

X1 

1X 

Am;n gm;n = f: 

m=1 n=1 

Multiplying both sides by sin(m 0¼x=a)sin(n0¼y=(2 a)), integrating over x from 

0toa and over y from0to2a, and taking the double sum will yield zero unless 

m = m 0 =andn = n 0 . Removing the primes and solving for Am;n yields 

Z 2 a Z a 

Á Z 2 a Z a 

Am;n = fgm;n dx dy 

g 

0 0 

0 0 

2 m;n dx dy: 

Using this relation, we calculate Am;n in the next command line. 

> A[m,n]:=int(int(f*g[m,n],x=0..a),y=0..2*a)/int(int(g[m,n]^2, 

x=0..a),y=0..2*a): 

Then, the general Fourier term Ãm;n = Am;n Fm;n is determined, the result 

being simpli¯ed assuming that m and n are integers and a>0. 

> psi[m,n]:=simplify(A[m,n]*F[m,n]) assuming m::integer, 

n::integer,a>0; 

Ãm; n := 3072 h ¡ 4+8(¡1) m + ¼2 n2 (¡1) n +2(¡1) (m+n) ¼2 n2 +8 (¡1) (1+m+n) +4(¡1) (1+n)¢ ³ 

m¼x 

´ ³ 

n¼y 

´ 

sin sin 

Ã 

a 2 a 

c¼ 

cos 

p 4 m2 + n2 ! 

t 

. 

(m 

2 a 

3 ¼8 n5 ) 

Given the mathematical form of f, Vectoria is not surprised at how complicated 

the coe±cient turns out to be in Ãm;n.


The parameter values suggested by Professor Legree are entered, 

> a:=1: h:=1: c:=1: 

and the membrane displacement approximated by the ¯nite sum 

Ã = 

X15 

m=1 n=1 

X15 

Ãm;n: 

Vectoria claims that she has kept 15 2 = 225 terms so as to get an accurate 

animation. We will let you be the judge of this, but we have overruled her desire 

to show the lengthy result and advised her to put a colon on the command line. 

> psi:=sum(sum(psi[m,n],m=1..15),n=1..15): 

Finally, a 3-dimensional zhue colored animation of Ã is produced over the time 

interval t = 0 to 5, with 40 frames. 

> animate(plot3d,[psi,x=0..a,y=0..2*a],t=0..5, 

frames=40,axes=frame,shading=zhue,orientation=[-130,50], 

tickmarks=[2,2,3]); 

On executing the animation, Vectoria is reminded of a rerun of an episode 

of a popular cartoon series that she watched recently with her ¯ance, Mike. 

The main character, Homer Simpson, who has a rather fat stomach, went to 

the doctor for a medical checkup. The doctor supposedly measured the fat 

content of Homer's stomach by giving it a jiggle and studying the motion of the 

resulting wave form. The initial shape of the membrane in this animation and 

its subsequent movement somewhat resembles that of Homer's stomach in the 

doctor's jiggle test. Vectoria wonders whether Professor Legree was partially 

motivated in setting up this problem by watching the same cartoon episode. 

Perhaps, underneath his stern exterior, Professor Legree has a sense of humor. 

PROBLEMS: 

Problem 6-3: Vibrations of a square membrane 

A square membrane whose sides are of unit length is given an initial transverse 

displacement Ã(x; y; 0) = xy(1 ¡ x)(1¡ y) and then released. Determine the 

displacement Ã(x; y; t) of the membrane for t>0 and animate the solution for 

nominal values of the parameters. 

Problem 6-4: Free edges 

Consider a rectangular membrane having sides of length a between x =0and 

x = a and sides of length 2 a between y =0andy =2a. The edges at x =0 

and x = a are ¯xed, while those at y =0andy =2a are \free." At a free edge, 

the slope is zero. Explicitly determine Ã(x; y; t) if the membrane is initially at 

rest and has the initial shape 

½ 

2 xh=a; 0 · x · a=2; 

Ã(x; y; 0) = 

(2 h=a)(a ¡ x); a=2 · x · a: 

Animate the solution for a =1,h =1,andc =1.


6.1.3 Vectoria's Second Problem 

Ohyes,thereisavastdi®erencebetweenthesavageandthecivilized 

man, but it is never apparent to their wives until after breakfast. 

Helen Rowland, American journalist, AGuidetoMen,1922 

The second problem on Professor Legree's assignment is to determine the transverse 

displacement Ã(r; μ; t) of a light, horizontal, circular membrane or drum- 

head of radius a ¯xed on its perimeter, given the initial conditions 

μ 

Ã(r; μ; 0) ´ f =20r 1 ¡ r2 

a2 

sin(2 μ); Ã(r; _ μ; 0) = 0: 

The motion of the drumhead is then to be animated for a =1andwavevelocity 

c = 1, and its behavior discussed. 

After loading the plots and VectorCalculus packages, Vectoria uses the 

Laplacian command to enter the wave equation in polar coordinates. 


> WE:=expand(Laplacian(psi(r,theta,t),'polar'[r,theta]) 

=(1/c^2)*diff(psi(r,theta,t),t,t)); 

WE := 

@ 

Ã(r; μ; t) 

@r 

r 

μ 2 @ 

+ 

Ã(r; μ; t) 

@r2 

+ 

@2 Ã(r; μ; t) 

@μ2 r 2 

= 

@2 Ã(r; μ; t) 

@t2 c2 Based on her experience with the ¯rst problem on the assignment, the initial 

condition _ Ã(r; μ; 0) = 0 can be satis¯ed if she assumes a general product solution 

of the form F (r) G(μ) cos(ckt), where the functions F and G and the constant 

k remain to be determined. The assumed form is provided as a HINT in the 

pdsolve command. 

> sol:=pdsolve(WE,HINT=F(r)*G(theta)*cos(c*k*t),INTEGRATE, 

build); 

sol := Ã(r; μ; t) =cos(ckt) C3 sin( p c1 μ) C1 BesselJ( p c1; kr) 

+cos(ckt) C3 sin( p c1 μ) C2 BesselY( p c1; kr) 

+cos(ckt) C4 cos( p c1 μ) C1 BesselJ( p c1; kr) 

+cos(ckt) C4 cos( p c1 μ) C2 BesselY( p c1; kr) 

The Bessel functions of the second kind (Y) diverge at r =0,soareremoved. 

> sol2:=remove(has,rhs(sol),BesselY); 

sol2 := cos(ckt) C3 sin( p c1 μ) C1 BesselJ( p c1; kr) 

+cos(ckt) C4 cos( p c1 μ) C1 BesselJ( p c1; kr) 

Vectoria simpli¯es the separation constant by substituting p c1 = p into sol2 . 

> sol3:=subs(sqrt(_c[1])=p,sol2); 

sol3 := cos(ckt) C3 sin(pμ) C1 BesselJ(p; k r) 

+cos(ckt) C4 cos(pμ) C1 BesselJ(p; k r)


Recognizing that the sines and cosines are independent functions and to be 

consistent with the sin(2 μ) term in the initial displacement, Vectoria removes 

the cos(pμ)termfromsol3 and sets p = 2 in the resulting expression. 

> sol4:=eval(remove(has,sol3,cos(p*theta)),p=2); 

sol4 := cos(ckt) C3 sin(2 μ) C1 BesselJ(2; kr) 

Thedrumheadis¯xedonitsperimeterforallt, i.e.,Ã(a; μ; t) = 0, so that the 

Bessel function J2(ka) equals 0. The allowed k values are the zeros of J2(x). 

The mth zero, km, is entered, 

> k[m]:=BesselJZeros(2,m)/a; 

BesselJZeros(2; m) 

km := 

a 

and substituted into sol4 with all Maple constants removed. 

> F[2,m]:=subs(k=k[m],BesselJ(2,k*r)*sin(2*theta)*cos(c*k*t)); 

μ 

 

μ 


c BesselJZeros(2; m) t 

F2;m:= BesselJ 2; sin(2 μ)cos 

a 

a 

The displacement of the drumhead at time t will be given by 

1X 

1X 

Ã(r; μ; t) = 

Ã2;m; 

m=1 

A2;m F2;m ´ 

with the coe±cients A2;m determined by the initial pro¯le f, whichisentered. 

> f:=20*r*(1-r^2/a^2)*sin(2*theta); 

μ 

f := 20 r 1 ¡ r2 

 

sin(2 μ) 

a 2 

m=1 

Evaluating F2;m at t =0yieldsg2;m. 

> g[2,m]:=eval(F[2,m],t=0); 

μ 

 


g2;m := BesselJ 2; sin(2 μ) 

a 

The mth coe±cient A2;m is given by 

Z a Z 2 ¼ 

Á Z a Z 2 ¼ 

A2;m = rfg2;m dμ dr 

r (g2;m) 2 dμ dr; 

0 

0 

which is now calculated. 

> A[2,m]:=int(int(r*f*g[2,m],theta=0..2*Pi),r=0..a) 

/int(int(r*g[2,m]^2,theta=0..2*Pi),r=0..a): 

Then, the mth Fourier term Ã2;m = A2;m F2;m is determined, the lengthy and 

quite formidable result being suppressed here in the text. 

> psi[2,m]:=A[2,m]*F[2,m]; 

For animation purposes Ã2;m is evaluated at a =1,c =1, 

> psi[2,m]:=eval(psi[2,m],fa=1,c=1g): 

and converted to Cartesian coordinates by noting that sin(2 μ) =2sinμcos μ 

and substituting r = p x2 + y2 and sin(2 μ) =2xy=(x2 + y2 ). 

0 

0


> psi[2,m]:=subs(fr=sqrt(x^2+y^2),sin(2*theta)=2*x*y 

/(x^2+y^2)g,psi[2,m]): 

Vectoria will animate the sum P5 m=1 Ã2;m, which is su±cient to give a good 

animation. Labeling this result as Ã, she numerically evaluates the terms and 

displays them. 

> psi:=evalf(sum(psi[2,m],m=1..5)); 

Ã := 33:78863892 BesselJ(2:; 5:135622302 p x 2 + y 2 ) xycos(5:135622302 t) 

x 2 + y 2 

+ 1:727434530 BesselJ(2:; 8:417244140 p x 2 + y 2 ) xycos(8:417244140 t) 

x 2 + y 2 


x 2 + y 2 


x 2 + y 2 


x 2 + y 2 

Since the displacement Ã applies only to the drumhead, not the region outside, 

Vectoria forms a piecewise function that is equal to Ã for x2 + y2 · 1 and equal 

to zero otherwise. 

> psi2:=piecewise(x^2+y^2 animate(plot3d,[psi2,x=-1..1,y=-1..1],t=0..2,frames=15, 

axes=framed,shading=zhue,style=PATCHCONTOUR, 

orientation=[45,30]); 

On executing the animation, Vectoria observes that in the initial frame of the 

animation the membrane displays two up peaks and two down peaks, located 

radially at r = a= p 3 and separated into four quadrants by perpendicular intersecting 

nodal lines. These nodal lines correspond to sin(2 μ) =0,orμ =0, 

¼=2, ¼, and3¼=2. The up and down peaks correspond to plus and minus signs 

as μ ranges from 0 to 2 ¼ in sin(2 μ). For μ =0to¼=2 and¼to 3 ¼=2, sin(2 μ) 

is positive; for μ = ¼=2 to¼ and 3 ¼=2 to2¼, it is negative. 

As time progresses the up peaks become down peaks and vice versa, the 

\°ipping" of peaks occurring periodically. Having written down her observations, 

she decides that this is enough for the time being, since she has three 

more problems to do on Professor Legree's assignment. She can add more discussion, 

if necessary, when she does a ¯nal review of all her answers. She has a 

date with Mike and would like to ¯nish the assignment before he calls on her.


PROBLEMS: 

Problem 6-5: Circular drumhead 

A circular drumhead of radius r = a and ¯xed on its perimeter is displaced a 

distance h at its center at time t = 0 so that it has a conical shape. Determine 

the displacement of the membrane for t>0. Taking a = h =1andc =1, 

animate the motion of the membrane. 

Problem 6-6: A di®erent initial shape 

A circular drumhead of radius a ¯xed on its outer edge has an initial displacement 

Ã(r; μ; 0) = (1 ¡ r 2 =a 2 )sin(4μ) and its initial velocity is zero. Determine 

the subsequent displacement Ã(r; μ; t), and animate the solution for parameters 

of your own choosing. 

6.1.4 Sound of Music? 

Music is spiritual. The music business is not. 

Van Morrison, Irish rock musician, Times, London, 6 July 1990 

The third problem on Vectoria's assignment, which Professor Legree has worded 

as follows, might be considered to be noteworthy. 

Some musically inclined people like to sing in the shower stall when taking 

a shower. Suppose that a large \shower stall" is empty without the water 

running and consists of a completely enclosed hollow vertical metal cylinder of 

radius a and height h with (approximately) rigid walls and all ¯xtures removed. 

The speed of sound for the air inside the cylinder is c. By solving the scalar 

Helmholtz equation for the spatial part of the velocity potential, determine 

the allowed normal modes inside the cylinder. For rigid walls, the normal 

component of the °uid velocity (or the normal derivative of the potential) must 

vanish at each wall. Taking a=1:83 m, h=3:04 m, and c=344 m/s (assuming 

dry air), determine the three lowest eigenfrequencies. By either consulting a 

musically inclined friend or a music reference book, or going to the Internet, ¯nd 

the closest musical notes on the equal-tempered scale to these eigenfrequencies 

and relate these notes to those found on a piano keyboard. 

Vectoria knows that the velocity potential Á satis¯es the scalar wave equation, 

with c the speed of sound. The velocity of a °uid element is given by 

~v = ¡rÁ. A normal mode of frequency ! is obtained by assuming a solution of 

the form Á(~r; t) =S(~r) cos(!t).Thewaveequationthenreducestothescalar 

Helmholtz equation, 

r 2 S + k 2 S =0; with k = !=c: (6.4) 

Professor Legree in his lectures uses the acronym SHE for the scalar Helmholtz 

equation, and jokingly refers to this equation as \she who must be obeyed". 

Evidently, this refers to a standard phrase uttered by the main character, a 

British barrister, about his wife in a PBS television series (Rumpole of the 

Bailey) thatLegreesawsomeyearsago.


Vectoria begins her solution by loading the VectorCalculus package, 

> restart: with(VectorCalculus): 

and entering SHE in cylindrical coordinates. The z direction is taken along the 

cylinder axis, r is the radial coordinate measured perpendicular to the z-axis, 

and μ is the angular coordinate measured about this axis. 

> SHE:=expand(Laplacian(S(r,theta,z),'cylindrical'[r,theta,z]) 

+k^2*S(r,theta,z))=0; 

SHE := 

@ 

S(r; μ; z) μ 

2 

@r @ 

+ S (r; μ; z) + 

r @r2 @2 S (r; μ; z) 

@μ2 r2 μ 

2 @ 

+ S (r; μ; z) 

@z2 +k 2 S(r; μ; z) =0 

SHE is solved with the pdsolve command, a general product solution of the 

form S(r; μ; z) =R(r)£(μ) Z(z) being assumed. 

> S:=rhs(pdsolve(SHE,HINT=R(r)*Theta(theta)*Z(z),INTEGRATE, 

build)); 

S := e (p c3 z) e ( p c2 μ) C5 C3 C1 %2 + e ( p c3 z) e ( p c2 μ) C5 C3 C2 %1 

+ e(p c3 z) C5 C4 C1 %2 

e (p c2 μ) 

+ e(p c2 μ) C6 C3 C1 %2 

+ C6 C4 C1 %2 

e (p c3 z) 

e (p c3 z) e ( p c2 μ) 

+ e(p c3 z) C5 C4 C2 %1 

e (p c2 μ) 

+ e(p c2 μ) C6 C3 C2 %1 

e (p c3 z) 

C6 C4 C2 %1 

+ 

e (p c3 z) e ( p c2 μ) 

%1 := BesselY( p ¡ c2; p c3 + k2 r) 

%2 := BesselJ( p ¡ c2; p c3 + k2 r) 

The Bessel functions of the second kind diverge at r = 0 and must be removed. 

> S2:=remove(has,S,BesselY); 

S2 := e (p c3 z) e ( p c2 μ) C5 C3 C1 %1 + e (p c3 z) C5 C4 C1 %1 

e (p c2 μ) 

+ e(p c2 μ) C6 C3 C1 %1 

+ C6 C4 C1 %1 

e (p c3 z) 

e (p c3 z) e ( p c2 μ) 

%1 := BesselJ( p ¡ c2; p c3 + k2 r) 

To simplify the notation for matching the boundary conditions, the separation 

constants c2 and c3 are replaced with ¡m2 and ¡q2 in S2 . Vectoria also 

substitutes k2 = ® 2 + q2 , and simpli¯es the result with the symbolic option. 

> S3:=simplify(subs(f_c[2]=-m^2,_c[3]=-q^2,k^2=alpha^2+q^2g, 

S2),symbolic); 

S3 := C1 BesselJ(m; ® r)( C5 C3 e ((qz+mμ) I) ((qz¡mμ) I) 

+ C5 C4 e 

+ C6 C3 e (¡I (qz¡mμ)) + C6 C4 e (¡I (qz+mμ)) )


Then S3 is converted to trig form and the result expanded. 

> S4:=expand(convert(S3,trig)); 

S4 := C1 BesselJ(m; ® r) C5 C3 cos(qz)cos(mμ) 

¡ C1 BesselJ(m; ® r) C5 C3 sin(qz)sin(mμ) 

+ ¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢ 

Approximating the metallic walls of the cylinder as being rigid, the normal 

component of the °uid velocity must vanish at the walls. Since the velocity 

is equal to minus the gradient of the potential, this implies that the normal 

derivative of the potential must vanish at the walls. In the z direction, one 

must have dZ=dz =0atz = 0, so only terms involving cos(qz)areselected. 

> S5:=select(has,S4,cos(q*z)); 

S5 := C1 BesselJ(m; ® r) C5 C3 cos(qz)cos(mμ)+¢¢¢ 

To satisfy the derivative boundary condition on the top face of the cylinder at 

z = h, one must have sin(qh)=0,orq = n¼=h,withn =0; 1; 2;:::: 

For the angular part to remain single-valued as μ increases by 2 ¼, onemust 

have cos(m (μ +2¼)) = cos(μ), which is satis¯ed if m =0; 1; 2; :::: Negative 

integer values of m need not be considered because the minus sign can be 

absorbed in the arbitrary constants. Similar remarks apply to sin(mμ), which 

also appears in S5 . So the allowed Bessel functions are J0; J1; J2; etc. 

To satisfy the derivative condition on the cylindrical wall, one must have 

(d=dr)Jm(®m;s r)jr=a =0,whereslabels the zeros of the derivative of the mthorder 

Bessel function. For later convenience, Vectoria sets ®m;s = ¼pm;s=a, 

the values of pm;s still to be determined. The allowed values of q and ® are 

subsituted into S5 and the result factored. 

> S6:=factor(subs(fq=n*Pi/h,alpha=Pi*p[m,s]/ag,S5)); 

S6 := (¡cos(mμ) C3 I ¡ cos(mμ) C4 I +sin(mμ) C3 ¡ sin(mμ) C4) 

³ 

n¼z 

´ ³ 

cos BesselJ m; 

h 

¼pm; s r 

´ 

C1 ( C5 + C6 ) I 

a 

Replacing the awkward coe±cient combinations with An;m;s and Bn;m;s gives 

normal modes of the following form. For each m value, except m =0,thereare 

actually two modes corresponding to cos(mμ) and sin(mμ). 

> NM:=(A[n,m,s]*cos(m*theta)+B[n,m,s]*sin(m*theta)) 

*select(has,S6,fcos(n*Pi*z/h),BesselJg); 

³ 

n¼z 

´ ³ 

NM := (An; m; s cos(mμ)+Bn; m; s sin(mμ)) cos BesselJ m; 

h 

¼pm; s r 

´ 

a 

Noting that the boundary condition (d=dr)(Jm(¼pm;s r=a)jr=a = 0 may be 

rewritten as (d=dp)(Jm(¼p)jp = pm;s = 0, Vectoria creates a functional operator 

f to apply the derivative boundary condition at the cylinder wall. 

> f:=m->diff(BesselJ(m,Pi*p),p)=0: 

Then using f, she employs a do loop to numerically determine pm;s for m =0 

and 1 and s =0tos =4.


> for m from 0 to 1 do 

> sol[m]:=seq(p[m,s]=fsolve(f(m),p,s..s+1),s=0..4); 

> assign(sol[m]): 

> end do; 

sol 0 := p0; 0 =0:; p0; 1 =1:219669891; p0; 2 =2:233130594; 

p0; 3 =3:238315484;p0; 4 =4:241062864 

sol 1 := p1; 0 =0:5860669999; p1; 1 =1:697050942; p1; 2 =2:717193891; 

p1; 3 =3:726137088; p1; 4 =4:731227206 

Recalling that ! = ck = c p ® 2 + q 2 , the eigenfrequencies are given by 

!n;m;s = ¼c 

r 

³ ´ 

pm;s 

2 

+ 

a 

³ 

n 

´ 2 

h 

radians per second; 

or, on using ! =2¼º, 

ºn;m;s = c 

r 

³pm;s ´ 2 ³ 

n 

´ 2 

+ hertz: 

2 a h 

A functional operator F is formed to calculate the ºn;m;s. 

> F:=(n,m,s)->(c/2)*sqrt((p[m,s]/a)^2+(n/h)^2): 

The given values of the cylinder radius a, cylinder height h, and speed of sound 

c are entered, 

> a:=1.83: h:=3.04: c:=344: 

and the three lowest allowed frequencies calculated using F. 

> nu[0,1,0]:=F(0,1,0); 

º0; 1; 0 := 55:08389289 

> nu[1,0,0]:=F(1,0,0); 

º1; 0; 0 := 56:57894736 

> nu[1,1,0]:=F(1,1,0); 

º1; 1; 0 := 78:96462843 

Consulting The Acoustical Foundations of Music, by John Backus [Bac69], Vectoria 

determines that the closest musical notes are A1 (55.000 Hz) for the ¯rst 

two frequencies and D2 # (77.782 Hz) for the third. Not having much of a 

feeling for these numbers, Vectoria goes to the Internet and ¯nds that the 

lowest-frequency note on a piano is 27.500 Hz and the highest frequency is 

4186.0 Hz. So the allowed frequencies in this \shower stall" example are at the 

low-frequency end of the piano keyboard. 

PROBLEMS: 

Problem 6-7: Acoustical waveguide 

A sound wave of frequency ! is generated at one end (z =0)ofaverylong 

straight cylindrical pipe of radius a having rigid walls.

6.2. SEMI-INFINITE AND INFINITE DOMAINS 261 

(a) Determine and discuss in detail the allowed modes of propagation and 

the cuto® frequency for wave propagation. The cuto® frequency is the 

minimum frequency for propagation of a speci¯c mode. 

(b) If the speed of sound in air is 1100 feet per second and if the frequency of 

the sound wave is 500 Hz, show that only a plane wave will be propagated 

if a


that vanishes at x = 1 is de¯ned as 

r Z 1 

2 

FST (f(x)) ´ F (s) = f(x)sin(sx) dx; (6.5) 

¼ 0 

and the inverse Fourier sine transform of F (s) by 

r Z 1 

2 

f(x) = F (s)sin(sx) ds: (6.6) 

¼ 0 

For the present problem, the temperature distribution will satisfy the onedimensional 

heat di®usion equation (with d the heat di®usion coe±cient), 

@T(x; t) 

= d 

@t 

@2T (x; t) 

@x2 : (6.7) 

Taking the FST of (6.7) with respect to x and integrating the rhs twice by parts, 

and assuming f 0 (x !1) ! 0, will convert the PDE into a ¯rst-order ODE for 

the transformed quantity F (s; t). The ODE will depend on the boundary condition 

T (0;t). The ODE is then solved for F (s; t) making use of the Fourier sine 

transform of the initial condition. Finally, on performing the inverse transform 

with respect to s, the temperature distribution T (x; t) is found. 

Once again, Vectoria disdains doing the problem \the old-fashioned way" 

favored by some instructors, but instead intends to let the computer assist her. 

To this end, she loads the plots and integral transform packages, 

> restart: with(plots): with(inttrans): 

and enters the di®usion equation DE, 

> DE:=diff(T(x,t),t)=d*diff(T(x,t),x,x); 

DE := @ T (x; t) =d 

@t 

μ 

@ 2 

2 T (x; t) 

@x 

and the boundary condition T (0;t)=T0. 

> T(0,t):=T0: 

To keep the recipe general for the moment, Vectoria has not yet speci¯ed the 

value of T0 . She then takes the FST of DE with respect to x, the integration 

by parts and boundary condition substitution being automatically done. 

> FST:=fouriersin(DE,x,s); 

FST := @ 

@t fouriersin(T (x; t);x;s)=ds(p 2 T0 ¡ s fouriersin(T (x; t);x;s) p ¼) 

p 

¼ 

To simplify the notation and facilitate solving the above ODE, she replaces 

fouriersin(T (x; t);x;s)withF (t), temporarily suppressing the argument s. 

> FST:=subs(fouriersin(T(x,t),x,s)=F(t),FST); 

FST := d 

dt F (t) = ds(p2 T0 ¡ s F (t) p ¼) 

p 

¼ 

The Fourier sine transform of the initial condition is zero, so Vectoria solves 

the ¯rst-order linear di®erential equation FST for F (t), subject to F (0) = 0. 

> eq:=dsolve(fFST,F(0)=0g,F(t));


p 

2 T0 

eq := F (t) = 

s p ¼ ¡ e(¡ds2 t) p 2 T0 

s p ¼ 

The Fourier (inverse) sine transform of the rhs of eq with respect to s is taken, 

> T:=fouriersin(rhs(eq),s,x); 

μ 

x 

T := T0 ¡ T0 erf 

2 p 

dt 

yielding the temperature T expressed in terms of the error function. Now T is 

evaluated with the given parameter values (d=100, T0 =100) and simpli¯ed. 

> T:=simplify(eval(T,fd=100,T0=100g)); 

μ 

x 

T := 100 ¡ 100 erf 

20 p 

t 

Choosing the spatial range to be x = 0 to 400 cm, Vectoria animates T over 

the time interval t = 0 to 150 seconds, 50 frames being taken. 

> animate(plot,[T,x=0..400],t=0..150,frames=50,thickness=2); 

On running the animation, Vectoria is pleased with the \sweetness" of the whole 

computer algebra derivation. 

Looking at her watch, she realizes that Mike will be picking her up at 5:00 

p.m., so she had better ¯nish the last problem on Professor Legree's assignment 

before he shows up. She will enjoy her dinner at Gira®es on the waterfront more 

if she has ¯nished her work. 

PROBLEMS: 

Problem 6-9: Fourier sine transform 

Calculate the Fourier sine transform of each f(x) belowandplottheanswers: 

(a) f(x) =e ¡3 jxj ; (b) f(x) =cos(2x); (c) f(x) =sin(x) 2 ; 

(d) f(x) =x=(x2 +1). 

Problem 6-10: A di®erent T (x; 0) 

Use the Fourier sine transform approach to solve the heat conduction problem 

for d =10cm2 /s in a semi-in¯nite rod (0 · x ·1) that has the boundary 

condition T (0;t) = 0 and initial interior temperature distribution T (x; 0) = 

10 x=(x2 + 1). Animate the plot. 

6.2.2 Assignment Complete! 

It is not knowledge, but the act of learning, not possession, but the 

act of getting there, which grants the greatest enjoyment. 

Carl Friedrich Gauss, German mathematician (1777{1855) 

The last problem on Vectoria's assignment is similar to the fourth one, but involves 

a di®erent boundary condition and initial condition. It is now supposed 

that the semi-in¯nite rod (0 · x ·1) has an initial temperature distribution


T (x; 0) = A±(x ¡ a), i.e., is a Dirac delta function 2 of amplitude A located at 

x = a>0. The end x = 0 is perfectly insulated so that no heat can °ow across 

this end. For no heat °ow to occur, the boundary condition is @T=@xjx=0 =0, 

i.e., the gradient of the temperature is zero. The evolution of the temperature 

pro¯le is to be animated over the spatial range x = 0 to 20 for the time interval 

t = 1 to 200, with a =5,A = 5, and a heat di®usion constant d =1. 

In her mathematical physics course, Vectoria has learned that when a gradient 

boundary condition is involved for a semi-in¯nite domain problem the 

Fourier cosine transform should be used. The Fourier cosine transform F (s) of 

a function f(x) forwhichbothf(x) andf 0 (x) ! 0asx !1and its inverse 

transform are 

r 

de¯ned as 

Z 1 

2 

F (s) = f(x) cos(sx) dx; f(x) = 

¼ 

0 

r 2 

¼ 

Z 1 

F (s) cos(sx) ds: 

Guided by her easy conquest of the previous problem, Vectoria realizes that she 

will be done by the time Mike arrives to pick her up. 

She again loads the plots and integral transform library packages, and enters 

the 1-dimensional heat di®usion equation. 


> DE:=diff(T(x,t),t)=d*diff(T(x,t),x,x); 

DE := @ μ 

T (x; t) =d @ 

@t 2 

 

2 T (x; t) 

@x 

The boundary condition bc and the initial condition ic are speci¯ed. The differential 

command D[1](T)(0,t) is used to enter @T(x; t)=@xjx=0, while the 

Dirac delta function is entered with the Dirac command. 

> bc:=D[1](T)(0,t)=0; ic:=A*Dirac(x-a); 

bc := D1(T )(0; t)=0 ic := A Dirac(¡x + a) 

The Fourier cosine transforms of the initial condition (assuming that a>0) and 

the di®usion equation are calculated. The boundary condition is substituted 

into the latter result. 

> F0:=fouriercos(ic,x,s) assuming a>0; 

F0 := A p 2cos(sa) 

p 

¼ 

> FCT:=subs(bc,fouriercos(DE,x,s)); 

FCT := @ 

@t fouriercos(T (x; t); x;s)=¡ds2 fouriercos(T (x; t); x;s) 

As in the previous recipe, fouriercos(T (x; t); x;s) is replaced with F (t)inFCT . 

> FCT:=subs(fouriercos(T(x,t),x,s)=F(t),FCT); 

FCT := d 

dt F (t) =¡ds2 F (t) 

2 The Dirac delta function ±(x ¡ a) is an in¯nitely tall spike located at x = a and is equal 

to zero for x 6= a. One of its most important properties is the sifting property, viz., for a 

smooth function f(x) (not another delta function) and ²>0, R a+² 

f(x) ±(x ¡ a) dx = f(a): 

a¡² 

0


Then, the ordinary di®erential equation FCT is analytically solved for F (t), 

subject to the initial condition F (0) = F0 , 

> eq:=dsolve(fFCT,F(0)=F0g,F(t)); 

eq := F(t) = A p 2cos(sa) e (¡ds2 t) 

p 

¼ 

and the inverse transform performed on the rhs of eq. 

> T:=fouriercos(rhs(eq),s,x); 

r 

¼ 

A 

td 

T := 

e(¡ a2 +x 2 

4 td ) ³ 

ax 

´ 

cosh 

2 td 

¼ 

Looking at the analytic result for T , Vectoria is pleased that she has spent time 

learning how to use the Maple computer algebra system. Once she has set up 

a template for a certain type of problem, then any other problem of that type 

is generally trivial to tackle. In the real world, certain integrals and other steps 

may not be carried out analytically, but traditionally in the world of academia, 

problems are assigned by instructors for which analytic answers exist. 

To ¯nish o® the problem and complete the assignment, Vectoria evaluates 

T with the given parameter values (d =1,a =5,A =5), 

> T:=eval(T,fd=1,a=5,A=5g); 

r μ 

¼ 25+x2 

5 e(¡ 

) 5 x 

4 t cosh 

t 2 t 

T := 

¼ 

and animates the solution over the suggested time interval t = 1 to 200. Clearly, 

the solution cannot be plotted at t = 0, because the initial pro¯le has been 

assumed to be a Dirac delta function. The number of points is controlled to 

product a smooth curve. 

> animate(plot,[T,x=0..20],t=1..200,numpoints=200, 

frames=50,thickness=2); 

As she watches the animation on the computer screen, Mike arrives to whisk 

her o® on their date. 

PROBLEMS: 

Problem 6-11: Fourier cosine transform 

Calculate the Fourier cosine transform of each f(x) belowandplottheanswers 

where possible: 

(a) f(x) =e ¡3 jxj ; (b) f(x) =cos(2x); (c) f(x) =sin(x) 2 ; 

(d) f(x) =x=(x2 +1). 

Problem 6-12: Di®erent initial condition 

Modify the text recipe to ¯nd the temperature distribution inside a semi-in¯nite 

rod (0 · x ·1) that is insulated at x = 0 and has the initial temperature 

distribution T (x >0; 0) = 25 x2 =(x2 +25)andd =1. AnimateT (x; t).


6.2.3 Radioactive Contamination 

The unexamined life is not worth living. 

Socrates, Greek philosopher (470{399 BC) 

When coauthor Richard was a student, he worked one summer as a chemistry 

lab assistant at a uranium mine on Great Bear Lake, which straddles the 

Arctic Circle in Northern Canada. Because it was a summer job, there was a 

period of several weeks when the sun never set, so evening baseball games were 

never called o® because of darkness. The baseball diamond was located on the 

leveled mine tailings, the only relatively °at spot in the small northern min- 

0 

x 

Figure 6.1: The radioactive disposal site. 

ing town of Port Radium, which is perched on almost treeless primordial rock 

formations. The author has always wondered what long-term health problems 

eventually arose among the permanent workers because of working in the mine 

and playing baseball on the radioactive mine tailings. Motivated by these reminiscences, 

we could not resist including a related example involving radioactive 

contamination. 

A radioactive gas is di®using at a steady rate into the atmosphere from 

a leveled contaminated disposal site. The ground and the atmosphere will 

be taken to be semi-in¯nite media with X = 0 at the boundary as shown in 

Figure 6.1. The concentration C(X; T) of radioactive gas in the atmosphere 

obeys the concentration equation (a modi¯ed di®usion equation) 

@C(X; T) 

@T 

= d @2 C(X; T) 

@X 2 ¡ ¸C(X; T); (6.8) 

where d is the di®usion constant and ¸ is the decay rate of the radioactive gas. 

The boundary condition at X =0isgivenbyFick's law, 

¡d @C(0;T) 

= K; (6.9) 

@X


where K is a positive constant with units in kg/(m 2 ¢ s). The coe±cients can be 

eliminated and the two equations cast into dimensionless form by introducing 

the new variables t ´ ¸T, x ´ p ¸=d X, and c = ( p ¸d=K) C. Then the 

concentration equation becomes 

@c(x; t) 

@t 

= @2c(x; t) 

¡ c(x; t); (6.10) 

@x2 with @c=@x = ¡1 as the boundary condition at x =0. 

Assuming that initially c =0forx¸ 0, we want to determine the distribution 

of radioactive gas in the atmosphere for times t ¸ 0 and animate the 

solution. The method of attack will be to make use of the Laplace transform. 

The Laplace transform of a function f(t) isde¯nedas 

Z 1 

L(f(t)) ´ F (s) = f(t) e ¡st dt: (6.11) 

Integrating by parts and assuming that e ¡stf(t)!0 ast!1,then 

μ 

μ 2 df 

d f 

L = sF(s) ¡ f(0); and L 

dt 

dt2 

= s 2 df (0) 

F (s) ¡ sf(0) ¡ : (6.12) 

dt 

To solve the concentration equation (6.10), subject to the boundary and 

initial conditions, the Laplace transform can be applied to the time part of the 

equation, the resultant second-order ODE in x solved, and the inverse Laplace 

transform performed to regain the time dependence. 

This program is now carried out using the laplace command contained in 

the integral transform library package. 


The dimensionless concentration equation is entered, 

> CE:=diff(c(x,t),t)=diff(c(x,t),x,x)-c(x,t); 

CE := @ μ 2 

@ 

c(x; t) = c(x; t) ¡ c(x; t) 

@t @x2 and the initial concentration speci¯ed. 

> c(x,0):=0: 

The Laplace transform of CE is taken with respect to the time variable, and 

the function laplace(c(x; t); t;s) then replaced with f (x) inLT . 

> LT:=laplace(CE,t,s); 

> LT:=subs(laplace(c(x,t),t,s)=f(x),LT); 

μ 2 d 

LT := s f(x) = f (x) ¡ f (x) 

dx2 This second-order ODE is solved for f(x), the following DEtools command line 

being used to obtain exponential solutions, which are convenient here. 

> sol:=DEtools[expsols](LT,f(x)); 

sol := [e (p s+1 x) ;e (¡ p s+1 x) ] 

0


As x !1, the concentration must go to zero, so the negative exponent solution 

is selected and multiplied by an arbitrary coe±cient B to yield f. 

> f:=B*sol[2]; 

f := Be (¡ps+1 x) 

The Laplace-transformed boundary condition bc takes the following form: 

> bc:=eval(diff(f,x),x=0)=laplace(-1,t,s); 

bc := ¡B p s +1=¡ 1 

s 

Then bc is solved for B, and the result BB substituted into f. 

> BB:=solve(bc,B); f:=subs(B=BB,f); 

BB := 

1 

p s +1s 

f := e(¡p s+1 x) 

p s +1s 

To perform the inverse Laplace transform, the substitution s = y ¡ 1is¯rst 

made in f. Looking back at the de¯nition of the Laplace transform, it is then 

necessary to multiply this result by e ¡t , producing f2 . 

> f2:=exp(-t)*subs(s=y-1,f); 

f2 := e(¡t) e (¡p yx) 

p y (y ¡ 1) 

The inverse Laplace transform of f2 with respect to y is carried out assuming 

that x>0. This yields the analytic form c for the concentration, expressed in 

terms of the complementary error function (erfc(z) =1¡ erf(z)). 

> c:=invlaplace(f2,y,t) assuming x>0; 

c := ¡ 1 

2 erfc 

μ 

x +2t 

2 p 

e 

t 

x + 1 

2 erfc 

μ 

¡ ¡x +2t 

2 p 

e 

t 

(¡x) 

The concentration is animated over the spatial region x =0to5andtime 

interval t =0to3. 

> animate(plot,[c,x=0..5],t=0..3,frames=100,thickness=2); 

On running the code and observing the animated concentration pro¯le in the 

region x>0, you will see the radioactive gas di®using into the atmosphere, 

with the concentration attaining a steady-state pro¯le that decreases (approximately) 

exponentially from the contaminated surface as the distance x increases 

from zero. Steady state occurs when there is a balance between the rate at which 

radioactive gas atoms are di®using into the atmosphere and the rate at which 

they are decaying. 

PROBLEMS: 

Problem 6-13: Laplace transform 

Calculate the Laplace transforms of the following functions, simplifying the 

answer where necessary, and identifying any special functions that occur: 

(a) f(t) =e ¡a p t with a>0; (b) f(t) =t cos(at); (c) f(t) = arctan(t);


(d) f(t) =tn ln(t) withn>0; (e) f(t) = tanh(t); (f) f(t) =tanh ¡1 (t); 

(g) f(t) = sin(3 p t) 

; (h) f(t) =J0(t) J1(t). 

t 1=4 

Problem 6-14: Inverse Laplace transform 

Calculate the inverse Laplace transforms of the following functions, identifying 

any special functions that occur in the answer: 

(a) F (s) = 1 

; (b) F (s) = 

s2 a 

s2 ; (c) F (s) = 

+ a2 s2 (s2 + a2 ; 

) 3=2 

1 

(d) F (s) = p 

s2 + a2 ; (e) F (s) =e¡as ; (f) F (s) = sin(as) 

; 

s 

Problem 6-15: Heat °ow in a semi-in¯nite rod 

Consider a semi-in¯nite rod spanning the range x =0tox = 1. The initial 

temperature of the rod is zero. For t>0, the temperature at x =0is 

T (x; 0) = T0. By Laplace transforming the temporal part of the di®usion equation, 

determine the temperature distribution inside the rod for t>0. Animate 

the temperature pro¯le for nominal values of the parameters. 

Problem 6-16: Heat °ow in a bar of varying cross section 

The heat °ow along an insulated semi-in¯nite bar whose cross section varies 

exponentially is described by 

μ 

@ ®x @T ®x @T 

e = e 

@x @x @t : 

If T (x; 0) = 0 for x>0, T (0;t) = 1, and T (1;t) = 0, use the Laplace transform 

approach to show that for t>0, the temperature distribution in the bar is 

T (x; t) = e¡®x 

μ μ 

1 

³ 

erfc xt 

2 2 

¡1=2 ¡ ®t 1=2´ 

+ e ®x μ 

1 

³ 

erfc xt 

2 

¡1=2 + ®t 1=2´ 

: 

Taking ® =1,animateT (x; t) over the range x =0to5fort =0to10. 

Problem 6-17: Convolution theorem 

An important property of the Laplace transform is the convolution theorem. If 

f1(t) andf2(t) are two functions, their convolution is de¯ned to be 

C(T )= 

Z T 

0 

f1(T ¡ t) f2(t) dt: 

If F (s), F1(s), and F2(s) are the Laplace transforms of C(T ), f1(t), and f2(t), 

respectively, the convolution theorem states that 

F (s) =F1(s) F2(s): 

Using the integral transform package, take the Laplace transform of C(t) and 

con¯rm the convolution theorem.


6.2.4 \Play It, Sam" Revisited 

How can you tell Al Gore from a roomful of Secret Service agents? 

He's the sti® one. 

Al Gore, former U.S. Vice President, joking about his reputation for sti®ness, 

New York Times, 14 September 1996 

As pointed out in the subsection, Play It, Sam, the small transverse vibrations 

of a light, horizontal, stretched, elastic string are well modeled by the 

linear wave equation. However, a piano \string," the subject of that section, 

is not actually a °exible elastic string, but rather a wire, possessing a degree 

of sti®ness. To understand the concept of sti®ness, imagine holding a string 

at one end between your ¯ngers. The unsupported end of the string will °op 

vertically downward at the juncture with your ¯ngers. The internal forces of 

thestringareunabletobalancetheshearforceexertedonthestringbyEarth's 

gravitational pull. For a wire, the unsupported end will sag, but not °op verticallydownwardatthejuncturepoint. 

Thewireissaidtohaveadegreeof 

sti®ness. 

For a string under tension, the sti®ness is negligible compared to the tension, 

but for a piano wire sti®ness should be included in the equation of motion. As 

shown in Morse [Mor48], the transverse displacement Ã(x; t) ofahorizontal 

wire is governed by the following fourth-order PDE: 

@2Ã 1 

¡ 

@x2 2 ® 2 

@4Ã 1 

= 

@x4 c2 @2Ã : (6.13) 

@t2 The parameter ® is a measure of the ratio of tension to sti®ness. If ® !1, 

tension predominates and the usual string wave equation results. For intermediate 

values of ®, the full wire equation must be solved. If, on the other hand, 

sti®ness is all important, then ® is very small and the fourth spatial derivative 

term dominates over the second spatial derivative. In this limiting case, the 

equation relevant to a vibrating \bar" results. As a wire is made thicker and 

thicker it becomes a bar. Some common examples of bars are the steel girders 

supporting bridges and used in high-rise construction and railway tracks. 

Labeling a 2 ´ c 2 =(2 ® 2 ), the transverse vibrations of a bar are governed by 

@2Ã @t2 + a2 @4Ã =0: (6.14) 

@x4 Because of the fourth spatial derivative, the vibrations of a bar are substantially 

di®erent from those of a string. 

As a simple example, but one with a complicated answer, suppose that an 

in¯nitely long bar is initially at rest (@Ã(x; 0)=@t = 0) and is given a transverse 

displacement 

Ã(x; 0) = Ae ¡b2x 2 

; 

with A the amplitude. We want to analytically determine the displacement of 

the bar for times t>0 and animate the solution. Our approach will be to make 

use of the \full" Fourier transform and its inverse.


Given a function f(x) that (along with all of its derivatives) vanishes as 

x !§1, the Fourier transform of f(x) and its inverse are de¯ned as 

Z 1 

F (k) = f(x) e ¡Ikx dx; f(x) = 1 

Z 1 

F (k) e 

2 ¼ 

Ikx dk: (6.15) 

¡1 

If the Fourier transform of f(x) (denoted by F[f(x)]) is F (k), then 

· n d f(x) 

F 

dxn ¸ 

=(Ik) n F (k): 

Fourier transforming the spatial part of the bar equation will involve n =4. 

Now let's use the \full" Fourier transform to solve the vibrating bar problem, 

¯rst loading the integral transform package, which contains the necessary 

fourier and invfourier (inverse Fourier) commands. 


The parameters a in the bar equation and b in the initial pro¯le are assumed 

to be positive, as is the time. 

> assume(a>0,b>0,t>0): 

The partial di®erential equation of motion for the bar is entered, using the 

shortcut x$4 to enter the fourth spatial derivative. 

¡1 

> pde:=diff(psi(x,t),t,t)+a^2*diff(psi(x,t),x$4)=0; 

μ 

pde := @ 2 

 

2 Ã(x; t) + a 

@t 2 

μ 

@ 4 

 

4 Ã(x; t) =0 

@x 

The Fourier transform of the spatial part of pde is performed, and the function 

fourier(Ã(x; t); x;k) replaced with F (t). 

> FT:=fourier(pde,x,k); 

> FT:=subs(fourier(psi(x,t),x,k)=F(t),FT); 

FT := a2 k4 μ 2 d 

F (t)+ F (t) =0 

dt2 Since a fourth spatial derivative was involved in the transform, a term (Ik) 4 = 

k4 has resulted in the output. 

The Fourier transform of the initial pro¯le is calculated, 

> F0:=fourier(A*exp(-b^2*x^2),x,k); 

k2 

F0 := Ae 

(¡ 4 b2 ) r ¼ 

b2 and the second-order di®erential equation FT solved for F (t), subject to the 

initial conditions F (0) = F0 , F _ (0) = 0. 

> sol:=dsolve(fFT,F(0)=F0,D(F)(0)=0g,F(t)); 

k2 

Ae(¡ 4 b 

sol := F (t) = 2 ) p ¼ cos(ak2 t) 

b 

To facilitate the calculation of the inverse Fourier transform, the right-hand 

side of sol is converted to exponential form and simpli¯ed.


> F2:=simplify(convert(rhs(sol),exp)); 

F2 := 1 

A 

2 

p μ 

¼ e ( k2 (¡1+4 Iatb 2 ) 

4 b2 b 

We temporarily set ¡1+4Iatb2 = ¡B and1+4Iatb2 = C in F2 . 

) 

+ e 

(¡ k2 (1+4 Iatb 2 ) 

4 b2 ) 

> F3:=subs(f-1+4*I*a*t*b^2=-B,1+4*I*a*t*b^2=Cg,F2); 

F3 := 1 

A 

2 

p ¼ 

μ 

e (¡ k2 B 

4 b2 ) 

+ e 

(¡ k2 C 

4 b2 ) 

To perform the inverse Fourier transform of F3 , it is necessary to make assumptions 

about B and C. Clearly the real part of each is positive and assuming 

Re(A) > 0andRe(B) > 0 will work here. A slightly simpler form results if we 

assume that both B and C are positive, even though they are really complex. 

The inverse Fourier transform of F3 then yields the solution Ã. 

> psi:=invfourier(F3,k,x) assuming B>0,C>0; 

Ã := 1 

μ 

A e 

2 

(¡ x2 b 2 

B ) p C + e (¡ x2 b 2 

C ) p 

B 

p p 

B C 

The original forms of B and C are substituted back into the displacement Ã, 

the lengthy complex output being suppressed. 

> psi:=subs(fB=1-4*I*a*t*b^2,C=1+4*I*a*t*b^2g,psi): 

Since the initial pro¯le was real, Ã should be real as well. The complex evaluation 

command, evalc, is applied and the result simpli¯ed. 

> psi:=simplify(evalc(psi)); 

Ae 

Ã:= 

(¡ b2 x 2 

%1 )μ 

μ 4 2 4 b x at p 

cos 

2 

%1 

p μ 4 2 4 b x at p 

%1+2+sin 

2 

%1 

p 

%1 ¡ 2 

2 p %1 

%1 := 1 + 16 a2 t2 b4 As expected, the displacement Ã is completely real. Noting the subexpression 

%1 (an artifact of exporting the Maple output into the text), the mathematical 

form of Ã is indeed quite complicated, as predicted earlier. 

Taking the nominal values A =1,a =0:1, and b =1inÃ, 

> psi:=eval(psi,fA=1,a=0.1,b=1g): 

the vibrations of the bar are animated over the spatial range x = ¡200 to 200 

and time interval t = 0 to 400. 

> animate(plot,[psi,x=-200..200],t=0..400,frames=150, 

numpoints=500,thickness=2); 

On running the animation, you will observe that the initial Gaussian pro¯le 

of the bar begins to decrease in the vicinity of x = 0, generating oscillations 

that rapidly disperse away from the origin in both directions. The bar does not 

b


overshoot the horizontal equilibrium position near the origin, even for longer 

times. You should experiment with other values of the parameters. 

PROBLEMS: 

Problem 6-18: Fourier transform 

Calculate the Fourier transforms of the following functions and plot the results: 

(a) f(x) =e ¡3 jxj ; (b)f(x) =cos(2x); (c) f(x) =x=(x2 +1). 

Problem 6-19: Inverse Fourier transform 

Calculate the inverse Fourier transforms of the following functions, simplifying 

where necessary, and plot the results: 

(a) f(k) =cos(¼k=2) =(1 ¡ k2 ); 

(b) f(k) =¡2 Ik=(¼ (k2 +2)); 

(c) f(k) =(1= p 2 ¼)(sink=k) 2 . 

Problem 6-20: Bandwidth theorem 

An approximately monochromatic plane wave packet in one dimension has the 

instantaneous form u(x; 0) = f(x) eIk0x ; with f(x) the envelope function and 

k0 the central wave number. Consider the following functions: 

(a) f(x) =2e ¡3 jxj=2 ; 

(b) f(x) =4e ¡x2 =4 ; 

(c) f(x) =5forjxj < 1and0otherwise. 

For each function, perform the following: 

² Calculate the wave number spectrum jA(k)j2 where A(k) is the Fourier 

transform of f(x). 

² Plot the intensities ju(x; 0)j 2 and jA(k)j 2 . 

² Explicitly evaluate the root mean square deviations from the means, ¢x 

and ¢k, de¯ned with respect to the above intensities. 

² Show that in each case the bandwidth theorem (the optical analogue of 

the uncertainty principle) ¢x ¢k ¸ 1 is satis¯ed. 

2 

Problem 6-21: Temperature distribution in an in¯nite rod 

By Fourier transforming the spatial part of the di®usion equation, determine the 

temperature distribution T (x; t) in an in¯nite rod for t>0whenT (x; 0) = T0 

for jxj 0whenT (x; 0) = 

T0 e ¡®2 x 2 

.AnimateT (x; t) for nominal values of the parameters.


6.3 Numerical Simulation of PDEs 

Simulated disorder postulates perfect discipline; simulated fear 

postulates courage; simulated weakness postulates strength. 

Sun Tzu, Chinese general, The Art of War, c.490BC 

To solve some linear and almost all nonlinear PDE models, subject to speci¯ed 

initial and/or boundary conditions, one must resort to numerical means and 

simulate the behavior of the model equation by replacing the spatial and time 

derivatives with ¯nite di®erence approximations and introducing an accurate 

numerical representation of any functional forms. In this section, we will illus- 

t=jk 

k 

j 

0,2 1,2 

0,1 1,1 

0,0 1,0 

h 

i-1,j i, j i+1, j 

P 

i, j-1 

i 

i, j+1 

x=ih 

Figure 6.2: Subdividing the x-t plane with a rectangular computational mesh. 

trate how ¯xed-step explicit schemes are created for a one-spatial-dimensional 

PDE whose dependent variable is some amplitude Ã(x; t), x being the spatial 

coordinate and t the time. The discussion presented here is meant only to give 

you the °avor of a vast and important topic. 

The general approach is to extend the ideas introduced at the end of Chapter 

2 for numerically solving ODEs. Referring to Figure 6.2, the x-t plane can 

be subdivided for computational purposes into a rectangular grid or mesh, with 

each rectangle having sides of length h and k in the x- andt-directions, respectively. 

The coordinates of a representative intersection, or mesh, point P are 

taken to be x = ih, y = jk with i; j =0; 1; 2; :::: In a nonadaptive scheme, 

since h and k are¯xed,theintegersi and j maybeusedtolabelthemesh 

points, the point P being indicated by (i; j). The value of Ã at P is written as 

ÃP = Ã(x = ih;y = jk) ´ Ãi;j. A similar subscript notation may be used for 

the spatial and time derivatives.

6.3. NUMERICAL SIMULATION OF PDES 275 

For example, in the forward di®erence approximation, the ¯rst spatial and 

time derivatives at P are written as 

μ 

@Ã 

= 

@x P 

(Ãi+1;j 

μ 

¡ Ãi;j) @Ã 

; = 

h 

@t P 

(Ãi;j+1 ¡ Ãi;j) 

; (6.16) 

k 

while the \standard" CDAs for the second derivatives at P are 

μ 2 @ Ã 

@x2 

= 

P 

(Ãi+1;j ¡ 2 Ãi;j + Ãi¡1;j) 

h2 ; (6.17) 

μ 2 @ Ã 

@t2 

= 

P 

(Ãi;j+1 ¡ 2 Ãi;j + Ãi;j¡1) 

k2 : (6.18) 

Although rectangular meshes are most commonly employed, diamond-shaped 

grids can also prove to be useful in certain numerical schemes used to model 

wave equations. 

6.3.1 Freeing Excalibur the Numerical Way 

\Are ¯ve nights warmer than one night, then?" 

Alice ventured to ask. \Five times as warm of course." 

\But they could be ¯ve times as cold, by the same rule|" 

\Just so!" cried the Red Queen. 

Lewis Carroll (Charles Lutwidge Dodgson), English writer (1832{1898) 

Recall that Russell, the aerospace engineer, was having a wild dream about 

freeing the sword Excalibur from its stony tomb by cooling its ends with buckets 

of ice water when his dream took a sudden detour and he recalled the 

following related problem from his undergraduate thermodynamics course. 

A thin 1-meter-long rod (the shaft of the sword), whose lateral surface is 

insulated to prevent heat °ow through that surface, has its ends suddenly held 

at the freezing point of water, 0 ± C, by placing them in contact with buckets of 

ice. Taking one end of the rod to be at x = 0 and the other at x = 1, the initial 

temperature distribution was T (x; t =0)=100x (1 ¡ x), a parabolic pro¯le 

with a maximum temperature of 25 ± at the midpoint x = 1 

2 . 

Russell was able to determine the analytic solution to this problem, using 

the separation of variables technique with the aid of Maple. Since it is now 

necessary in his work to numerically solve a system of nonlinear PDEs, and he 

hasn't done any numerical work for a while, he decides to tackle the Excalibur 

problem ¯rst, using an explicit ¯nite-di®erence scheme. 

In the linear di®usion equation (with di®usion coe±cient d =1), Russell 

replaces the time derivative with the forward-di®erence approximation and the 

second spatial derivative with the standard CDA, so that 

@T 

@t = @2T @x2 ) (Ti;j+1 ¡ Ti;j) 

= 

k 

(Ti+1;j ¡ 2 Ti;j + Ti¡1;j) 

h2 : (6.19)


If he sets r ´ k=h 2 and c ´ 1 ¡ 2 r, the explicit scheme is 

Ti;j+1 = rTi¡1;j + cTi;j + rTi+1;j: (6.20) 

The unknown value Ti;j+1 on time step j + 1 is explicitly determined from the 

three known values Ti¡1;j, Ti;j, andTi+1;j on the previous (jth) time step. One 

starts with the bottom row (j = 0) in the numerical mesh shown in Figure 6.2, 

which corresponds to the initial temperature pro¯le here, and calculates T at 

eachinternalmeshpointofthe¯rst(j =1)timerow. Theendmeshpointsof 

each time row are held at zero temperature in accordance with the boundary 

conditions. Once all the Ti;1 values are known on time step 1, the Ti;2 on time 

step 2 are calculated in a similar manner, and so on. 

Since he intends to use a matrix multiplication approach, Russell makes a 

call to the LinearAlgebra library package. He divides the spatial range x =0 

to x = 1 into 12 equal parts, so that there are M = 11 internal mesh points on 

each time row. The value of M is easily increased if necessary. 

> restart: with(LinearAlgebra): M:=11: begin:=time(): 

To display the M £ M =11£ 11 square matrix that will perform the operation 

on the rhs of (6.20), the following interface command is entered. The argument 

rtablesize=infinity allows a matrix with M>10 to be explicitly displayed, 

instead of being represented by a placeholder. 

> interface(rtablesize=infinity): 

Russell introduces an M £ M tridiagonal matrix3 A using the BandMatrix 

command. In the argument, the list [r,c,r] gives the coe±cients on the rhs of 

the explicit scheme (6.20), the number 1 indicates that there is one subdiagonal 

on either side of the main diagonal, and M speci¯es the size of the matrix A. 

> A:=BandMatrix([r,c,r],1,M,M); 

2 

6 

A := 6 

4 

c r 0 0 0 0 0 0 0 0 0 

r c r 0 0 0 0 0 0 0 0 

0 r c r 0 0 0 0 0 0 0 

0 0 r c r 0 0 0 0 0 0 

0 0 0 r c r 0 0 0 0 0 

0 0 0 0 r c r 0 0 0 0 

0 0 0 0 0 r c r 0 0 0 

0 0 0 0 0 0 r c r 0 0 

0 0 0 0 0 0 0 r c r 0 

0 0 0 0 0 0 0 0 r c r 

0 0 0 0 0 0 0 0 0 r c 

Consulting his old numerical analysis text, Russell ¯nds that the ¯xed-step 

explicit scheme for the di®usion equation becomes numerically unstable if the 

ratio r is greater than 1. 

Wanting to carry out a reasonably accurate calcu- 

2 

lation, he takes r =0:05. Inputting the spatial step size h =1=(M +1), he 

3A tridiagonal matrix has nonzero matrix elements only on the central diagonal and two 

adjacent diagonals. 

3 

7 

5


evaluates the time step size k = rh 2 ,aswellastheparameterc. Wanting to 

animate his numerical solution, he decides to create N = 50 plots, with each 

plotseparatedintimebys =20steps. 

> r:=0.05: h:=1.0/(M+1); k:=r*h^2; c:=1-2*r; N:=50: s:=20: 

h := 0:08333333333 k := 0:0003472222222 c := 0:90 

An operator f will be used to input the initial parabolic temperature pro¯le. 

> f:=x->evalf(100*x*(1-x)): 

With the help of f, the initial (j = 0 time row) temperatures at the M internal 

mesh points are calculated, and expressed as a column vector4 T0 so that matrix 

multiplication can be performed. 

> T[0]:=; #input temperatures 

2 

6 

T0 := 6 

4 

7:638888889 

13:88888889 

18:75000000 

22:22222222 

24:30555556 

25:00000000 

24:30555556 

22:22222222 

18:75000000 

13:88888889 

7:638888886 

A graphing operator is formed to plot the temperature pro¯le on the jth step, 

the plotting points being connected by straight lines. The zero temperatures at 

the endpoints are included. 

> gr:=j->plot([[0,0],seq([i/(M+1),T[j][i,1]],i=1..M),[1,0]], 

labels=["x","T"]): 

In the following do loop, the temperature pro¯le is calculated every s=20steps. > for n from 1 to N do 

> T[n]:=A^s . T[n-1]; 

> end do: 

The CPU time on a 3-GHz personal computer to perform the calculation 



is a fraction of a second, hardly worth recording. The plots[display] command 

5 with the insequence=true option is employed to produce an animated 

sequence of the temperature pro¯le in the rod. 

> plots[display]([seq(gr(j),j=0..N)],insequence=true); 

4 Entered with the short-hand notation >. 

5 Ashortcutto¯rstenteringwith(plots) and then display. 

3 

7 

5


To see Russell's animated di®usion equation solution, you will have to execute 

the program and use the animation tool bar. 

PROBLEMS: 

Problem 6-23: Numerical Instability 

In the text recipe, con¯rm that the solution becomes numerically unstable for 

r>0:5. Numerical instability is signaled by the appearance of increasingly 

wild oscillations in the solution as time increases. 

Problem 6-24: Comparison with exact solution 

Explore the change in the percentage error in the numerical mesh values at the 

endoftheruncomparedwiththeexactvaluesasM is increased. 

Problem 6-25: A di®erent pro¯le 

Modify the text recipe to produce an animated numerical solution for the initial 

temperature pro¯le T (x; 0) = 25 sin(¼x). Compare the numerical solution in 

the center of the rod with the exact solution as a function of time. At what 

time is the temperature in the middle equal to one-quarter of the initial value? 

6.3.2 Enjoy the Klein{Gordon Vibes 

The world is never quiet, even its silence eternally resounds with the 

same notes, in vibrations which escape our ears. 

Albert Camus, French-Algerian philosopher, writer (1913{1960) 

After tackling the Excalibur heat-di®usion example, Russell decides to numerically 

investigate the small transverse oscillations of a light stretched horizontal 

string embedded in a stretched vertical elastic membrane. In the absence of 

the membrane, the instantaneous displacement Ã(x; t) of the string satis¯es the 

one-dimensional wave equation. The e®ect of the membrane is to add an additional 

Hooke's law restoring force, proportional to Ã, on the string, which tends 

to speed up the vibrations. The relevant transverse wave equation, called the 

Klein{Gordon equation (KGE), then is 

@2Ã @x2 ¡ @2Ã = aÃ; (6.21) 

@t2 where a is the elastic coe±cient of the membrane and the wave speed has been 

set equal to unity. 

If the string is of unit length and ¯xed at both ends, the boundary conditions 

are Ã(0;t)=Ã(1;t)=0fort ¸ 0. Supposing that the string has the initial 

transverse pro¯le f ´ Ã(x; 0) = x (1 ¡ x) 5 and velocity g ´ _ Ã(x; 0) = x 3 (1¡ x), 

Russell wishes to numerically solve the KGE and animate the oscillations. 

At the internal mesh points, he uses the standard CDAs for the second 

derivatives and approximates the inhomogeneous term with aÃi;j. Thenumer-


ical algorithm for the KGE then is 

(Ãi+1;j ¡ 2 Ãi;j + Ãi¡1;j) 

h 2 

¡ (Ãi;j+1 ¡ 2 Ãi;j + Ã i;j¡1) 

k 2 = aÃi;j; (6.22) 

or, setting r ´ k 2 =h 2 and c ´ 2 ¡ 2 r ¡ k 2 a, and rearranging, 

Ãi;j+1 = rÃi¡1;j + cÃi;j + rÃi+1;j ¡ Ãi;j¡1; (6.23) 

with j = 1; 2; 3; :::: The mesh points involved in this explicit scheme are 

schematically depicted in Figure 6.3. 

ψ i,j+1 

ψ ψ ψ 

i−1,j i,j i+1,j 

ψ i,j−1 

Figure 6.3: Relevant mesh points for numerically solving the KGE. 

In this algorithm, the unknown Ã value on time step (j +1) depends on 

its values on the previous two time steps, j and (j ¡ 1). The second time 

row, corresponding to taking j = 1, will be the ¯rst to be calculated. The Ã 

values are known along the zeroth time row from the initial condition Ã(xi; 0) = 

f(xi). To apply the scheme, the Ã values along the ¯rst time row must also be 

known. These may be determined from the initial transverse velocity. Using 

the forward-di®erence approximation, the condition _ Ã(x; 0) = g(x) yields 6 

Ãi;1 ¡ Ãi;0 

= g(xi); 

k 

or Ãi;1 ´ G = Ãi;0 + kg(xi): (6.24) 

Now Russell programs the explicit scheme (6.23), ¯rst loading the Linear- 

Algebra package so a matrix approach can be used again. 

> restart: with(LinearAlgebra): begin:=time(): 

6 In the problem set, Russell will show us a better approximation than this one.


He divides the range x =0to1into50equalspatialintervals,sothatthe 

number of internal mesh points is M = 49. The size of the time step is taken 

to be k =0:005, the number of time steps is N =200,anda =100. 

> M:=49: k:=0.005: N:=200: a:=100: 

The values of the spatial step size h, theratior, andc are calculated. 

> h:=1.0/(M+1); r:=k^2/h^2; c:=2-2*r-k^2*a; 

h := 0:02000000000 r := 0:06250000000 c := 1:872500000 

Except for the inclusion of the last term on the rhs (and a di®erent de¯nition of 

r), the numerical scheme (6.23) is identical with that for the Excalibur example. 

So, Russell forms a tridiagonal matrix A as before. Because of its large size, the 

full matrix is not explicitly displayed, but given by the following placeholder. 

> A:= BandMatrix([r,c,r],1,M); 

2 

6 

A := 6 

4 

49 x 49 Matrix 

Data Type : anything 

Storage : band[1; 1] 

Shape : band[1; 1] 

Order : Fortran order 

Operators are formed to calculate f(x), g(x), and G(x). 

> f:=x->evalf(x*(1-x)^5): 

> g:=x->evalf(x^3*(1-x)): 

> G:=x->evalf(f(x)+k*g(x)): 

Using the operator f, the initial (zeroth time row) displacements are calculated 

at the M internal mesh points, and put into a column vector format. 

> v[0]:=: 

Similarly, G is used to calculate the displacements at internal mesh points on 

the ¯rst time row and represented as a column vector. 

> v[1]:=: 

A graphing operator is formed to plot the displacement pro¯le on the jth step. 

> gr:=j->plot([[0,0],seq([i/(M+1),v[j][i,1]],i=1..M),[1,0]], 

labels=["x","psi"]): 

The following loop implements the numerical scheme, calculating the displacements 

at the internal mesh points from the second time row to the Nth row. 

> for j from 1 to N do 

> v[j+1]:=A . v[j]-v[j-1]: 

> end do: 

The CPU time is a fraction of a second. 



Russell animates the transverse oscillations by using the display command 

with the insequence=true option. 

3 

7 

5


> plots[display]([seq(gr(j),j=0..N)],insequence =true); 

Once again, you should execute the worksheet to see the oscillations of the 

string. Feel free to change the parameter values or initial conditions. 

Although the KGE can be solved analytically, since it is a linear PDE, 

Russell's numerical approach allows him to tackle nonlinear wave equations 

that cannot be solved analytically. For example, the nonlinear KGE 

@2Ã @x2 ¡ @2Ã @t2 = aÃ+ bÃ3 ; (6.25) 

with a and b positive, might be used to model the symmetric transverse oscillations 

of the string when the membrane is stretched su±ciently that a nonlinear 

correction to Hooke's law should be included. The modi¯cation of Russell's 

numerical scheme is left as a problem for the interested reader. 

PROBLEMS: 

Problem 6-26: a =0 

For a = 0, the KGE reduces to the linear wave equation. Run the code for 

a = 0 and then explore how the results are a®ected by increasing a. 

Problem 6-27: Nonzero initial velocity 

Modify the text recipe for the KGE to handle the initial conditions f(x) =0, 

g(x) =sin(¼x). You may have to adjust the viewing box. 

Problem 6-28: Nonlinear KGE 

Modify the text recipe to numerically simulate the nonlinear KGE (6.25). Take 

the same parameters as in text recipe and explore what happens when increasing 

positive values of b are considered. 

Problem 6-29: Better ¯rst-row approximation 

A better approximation to the ¯rst time row to use for solving the KGE is 

Ãi;1 = f(xi)+kg(xi)+ r 

2 (f(xi+1) ¡ 2 f(xi)+f(xi¡1)) + O(k 3 ): 

Execute the text recipe with this improved approximation. 

6.3.3 Vectoria's Secret 

None are so fond of secrets as those who do not mean to keep them. 

C. C. Colton, English writer (1780{1832) 

Vectoria's secret is out! She and Mike have announced the date of their wedding, 

which will take place in July in a lupine-dappled alpine meadow near 

Mount Baker in the North Cascades. Rumor has it that Mike will throw any 

guest whose cell phone rings into one of the nearby icy Chain Lakes. Since she 

will be too busy making wedding plans, we will not be seeing Vectoria in the 

rest of this text. So, we have asked her to favor us with one last recipe for 

solving the following problem.


In a region of space, the potential V (x; y) satis¯es the Poisson equation, 

@2V @x2 + @2V @y2 = xey ; 0 · x · 2; 0 · y · 1; 

with the four boundary conditions V (0;y)=0, V (2;y)=2ey , V (x; 0) = x, 

and V (x; 1) = ex. Using a central di®erence approximation for each second 

derivative at a mesh point (i, j) and evaluating the rhs of Poisson's equation at 

this point, derive a ¯nite di®erence scheme for solving the equation. Dividing 

the x-interval into 30 steps and the y interval into 15 steps, determine V at 

each mesh point and plot the numerical solution. Here is Vectoria's solution. 

She begins by entering the boundaries and dividing them into the suggested 

intervals in the x- andy-directions. 

> restart: with(plots): begin:=time(): 

In the x-direction, the boundaries are at x=a=0 and x=b=2:0. The interval 

b ¡ a is divided into m=30 steps, and the x step size h=(b ¡ a)=m calculated. 

> a:=0: b:=2.0: m:=30: h:=(b-a)/m; 

h := 0:06666666667 

In the y-direction, the boundaries are at y = c =0 and y = d=1:0. The interval 

d ¡ c is divided into n=15 steps, and the y step size k =(d ¡ c)=n calculated. 

> c:=0: d:=1.0: n:=15: k:=(d-c)/n; 

k := 0:06666666667 

Using central di®erence approximations for the second derivatives, evaluating 

the inhomogeneous term at (i; j), setting r =(h=k) 2 , and rearranging, Poisson's 

equation becomes 

2(1+r) Vi; j ¡ Vi+1;j ¡ Vi¡1;j ¡ r (Vi; j+1 + Vi; j¡1)+h 2 xi e yj =0 (6.26) 

The ratio r is now evaluated. 

> r:=(h/k)^2; 

r := 1:000000000 

The grid coordinates in the x- andy-directions are generated. 

> Xcoords:=seq(x[i]=i*h,i=0..m): 

> Ycoords:=seq(y[j]=j*k,j=0..n): 

The potential V at the grid points along the four bounding edges are generated 

in the following four boundary conditions. 

> bc1:=seq(V[i,0]=i*h,i=0..m): 

> bc2:=seq(V[i,n]=evalf(exp(1))*i*h,i=0..m): 

> bc3:=seq(V[0,j]=0,j=0..n): 

> bc4:=seq(V[m,j]=2*evalf(exp(j*k)),j=0..n): 

Vectoria then assigns Xcoords, Ycoords, and the four boundary conditions. 

> assign(Xcoords,Ycoords,bc1,bc2,bc3,bc4): 

An operator f is formed to calculate equation (6.26) at the grid point (i; j).


> f:=(i,j)->2*(1+r)*V[i,j]-V[i+1,j]-V[i-1,j]-r*(V[i,j+1] 

+V[i,j-1])+h^2*x[i]*exp(y[j])=0; 

f := (i; j) ! 2(1+r) Vi; j ¡ Vi+1;j ¡ Vi¡1;j ¡ r (Vi; j+1 + Vi; j¡1)+h 2 xi e yj 

The mesh equations are determined for all the internal grid points using two 

nested sequences running from i =1tom ¡ 1andfromj =1ton ¡ 1. 

> eqs:=fseq(seq(f(i,j),i=1..m-1),j=1..n-1)g: 

The unknown potentials at the internal grid points are entered as the variables. 

> vars:=fseq(seq(V[i,j],i=1..m-1),j=1..n-1)g: 

The mesh equations are solved (to 6 digits) for the variables, and the solution 

is assigned. 

> sol:=evalf(fsolve(eqs,vars),6): assign(sol): 

Three-dimensional plotting points are formed for all the grid points, including 

those on the boundaries. 

> pts:=seq(seq([x[i],y[j],V[i,j]],i=0..m),j=0..n): 

Using the pointplot3d command, the points representing the potential at each 

grid point are plotted as size-6 blue circles, the result being shown in Figure 6.4. 

> pointplot3d([pts],symbol=circle,symbolsize=6,color=blue, 

axes=boxed,orientation=[135,60],tickmarks=[3,3,3], 

labels=["x","y","V"]); 

4 

V 

2 

0 

2 

1 

x 

0 

0.5 y 

Figure 6.4: Numerical solution of the Poisson equation. 

The CPU time to execute the entire recipe is about 6 seconds. 


cpu := 5:978 

0


PROBLEMS: 

Problem 6-30: Steady-state temperature distribution 

The steady-state temperature distribution T (x; y) in a thin square metal plate 

0:5 m on a side satis¯es Laplace's equation, r2T (x; y) = 0. 

conditions on the edges of the plate are 

The boundary 

T (0;y)=0; T(x; 0) = 0; T(x; 0:5) = 200 x; T (0:5;y)=200y: 

Using the standard CDA for the second derivatives, and choosing a suitable 

mesh spacing, numerically determine the temperature distribution in the plate 

and make a 3-dimensional plot. 

Problem 6-31: Potential distribution 

A square inner conductor 3 cm on a side is held at a potential of 100 V. A 

second square conductor, concentric with the ¯rst and 9 cm long on each of its 

inner sides, is held at 0 V. The potential ©(x; y) in the region between the two 

conductors satis¯es Laplace's equation, r2 ©(x; y) =0. 

(a) Taking the mesh spacing in both directions to be 1 cm, make a mesh 

diagram showing all the interior mesh points for which © is to be found. 

(b) UsingCDAsforthesecondderivatives,writeoutthemeshequationsfor 

the interior points. Make use of symmetry arguments to show that only 

seven interior points need to be used in the calculation of ©. 

(c) Solve the mesh equations and determine © at each interior point. 

(d) Plot © in the region stretching from the inner to the outer conductor.

Part III 

THE DESSERTS 

The way a child discovers the world constantly 

replicates the way science began. You start to 

notice what's around you, and you get very curious 

about how things work. How things interrelate. 

It's as simple as seeing a bug that intrigues you. 

You want to know where it goes at night; 

who its friends are; what it eats. 

David Cronenberg, Canadian ¯lmmaker (1943{) 

It's food too ¯ne for angels; yet come, take 

And eat thy ¯ll! It's Heaven's sugar cake. 

Edward Taylor, English poet (1664{1729) 

285

Chapter 7 

The Hunt for Solitons 

There is no better ... door by which you can enter into the study of 

natural philosophy than by considering the ... physical phenomena of 

acandle. 

Michael Faraday, English physicist (1791{1867) 

Nonlinear PDEs display a rich spectrum of solutions that in most cases must be 

obtained by numerical means. However, there exist special analytic solutions 

to some nonlinear PDEs of physical interest, the best known being soliton solutions 

of nonlinear wave equations. A soliton is a stable solitary wave, which 

is a localized pulse solution that can propagate at some characteristic velocity 

without changing shape despite the \tug of war" between \competing terms" 

in the governing equation of motion. 

A simple physical example [EJMR81] of a solitary wave is provided by the 

°ame of an ordinary lit candle. There exists a dynamic balance between the 

di®usion of the heat from the °ame into the wax and the nonlinear energy 

release as the wax vaporizes. The candle °ame advances into the wax at a 

velocity that just maintains the balance. To check whether a solitary wave 

is stable, i.e., is a soliton, one can subject the solitary wave to some type of 

perturbation and see whether its integrity is preserved. For example, the candle 

°ame may °icker because of an ambient air current, but it tends to preserve its 

shape as the candle burns, so the °ame displays soliton-like behavior. 

Of course, there exist many di®erent possible stability criteria that could be 

invoked to decide whether a solitary wave is a soliton. Historically, however, 

mathematicians have decided that in order for a solitary wave to be deemed 

worthy of the name soliton, it must survive a collision with another solitary-wave 

solution of the same PDE completely unchanged in shape. There are two main 

approaches to applying this collisional stability criterion, either numerically or 

analytically. The numerical simulation approach will be brie°y illustrated in the 

last section. Analytic methods are considerably more complicated to implement 

(see, e.g., [EM00]) and we will be content here only to quote some of the results 

in the form of two-soliton solutions. 

Given a nonlinear PDE, how do we know that it even has the possibility 

287

288 CHAPTER 7. THE HUNT FOR SOLITONS 

of having soliton solutions? This chapter is about the hunt for solitary waves 

(possible solitons), using graphical and analytic approaches, and the analytic 

con¯rmation that some of these solitary waves are indeed solitons. 

First, we should have some idea of what solitary waves look like, and what 

well-known nonlinear PDEs of physical interest are known to have them. To 

keep the discussion simple, let's restrict our attention to wave motion in one 

spatial dimension. Three well-known nonlinear PDEs that describe di®erent 

types of wave motion are the Korteweg{de Vries equation (KdVE), the sine{ 

Gordon equation (SGE), and the nonlinear SchrÄodinger equation (NLSE): 

² @Ã 

@t 

@Ã 

+ ®Ã 

@x + @3Ã 3 =0, KdVE; 

@x 

² @2Ã 1 

2 = 

@x c2 ² i @Ã 

@x 

@ 2 Ã 

2 +sinÃ, SGE; 

@t 

1 @ 

§ 

2 

2 Ã 

@t 2 + jÃj2 Ã =0, NLSE. 

Here x is the spatial coordinate, t is the time, ® is a numerical scale parameter, 

i = p ¡1, and Ã(x; t) is the amplitude. All three equations turn out to be very 

important in nonlinear dynamics because, under suitable approximations, they 

arise in many di®erent contexts. 

The KdVE has been used to describe [SCM73] water waves in shallow canals, 

magnetohydrodynamic waves in plasmas, longitudinal dispersive waves in elastic 

rods, pressure waves in liquid{gas bubble mixtures, and so on. 

The SGE is applicable [BEMS71] to the propagation of magnetic spins in 

ferromagnets, magnetic °ux in Josephson junctions, crystal dislocations, ultrashort 

optical pulses, etc. 

Undoubtedly, the most important application of the NLSE [Has90] is to the 

propagation of optical pulses in glass ¯bers whose refractive index n is of the 

form n = n0 + n1 I, withn0 and n1 positive constants and I the light intensity. 

The light intensity is proportional to jÃj 2 , where Ã is the complex electric 

¯eld amplitude that satis¯es the NLSE. The light intensity is experimentally 

measured rather than the electric ¯eld amplitude. 

The derivation of these three nonlinear wave equations is beyond the scope 

of this text. The reader who is interested in such matters should consult the 

references cited above. As will be demonstrated, all three nonlinear PDEs 

support collisionally stable solitary-wave solutions, i.e., solitons. 

What do solitary waves (solitons) look like? Figure 7.1 shows a sketch of the 

two commonly occurring types. For the peaked variety (called nontopological 

solitary waves by mathematicians), there exists a localized maximum with the 

pulse dropping to zero amplitude at §1. Nontopological solitary waves also 

exist whose pulse amplitude displays a localized dip to zero with the pulse 

increasing to a constant nonzero amplitude at §1. In terms of the intensity, 

the NLSE supports both types of nontopological solitary waves. Those optical 

solitary waves with peaks are called bright solitary waves, while those with the

289 

dip are referred to as black solitary waves. The bright ones occur for the plus 

sign in the NLSE, the black ones for the minus sign. The Korteweg{de Vries 

(KdV) equation also possesses nontopological solitary-wave solutions. 

− ∞ 

U 

peaked kink 

z 

Figure 7.1: Qualitative shapes of two common types of solitary waves. 

For the kink type (referred to as topological solitary waves) in Figure 7.1, the 

pulse amplitude U changes from one constant value (e.g., zero in the ¯gure) at 

¡1 to a larger constant value at + 1. The region in which the change takes 

place is usually quite localized. Antikink solitary waves canalsoexist,forwhich 

the amplitude changes from a constant value at ¡1 to a lower constant value 

at + 1. The SGE displays both kink and antikink solutions. 

Given a nonlinear wave equation, how are these solitary-wave solutions 

found? We know that the one-dimensional linear wave equation has a general 

solution of the structure Ã(x; t) =f(x ¡ ct)+g(x + ct), where f and g 

are arbitrary functions. The function f(x ¡ ct) describes a waveform traveling 

with speed c in the positive x-direction, while the form g(x + ct) describes a 

wave traveling in the negative x-direction. Let us now con¯ne our attention to 

waves traveling in the positive x-direction, the discussion for waves traveling 

in the opposite direction being similar. The linear wave equation can support 

localized solutions Ã(x; t) =f(z = x ¡ ct) such as the peaked solitary waves 

shown in Figure 7.1. These solutions will translate unchanged in shape along 

the positive x-axis. For nonlinear wave equations, we can look for similar types 

of localized solutions. That is to say, we seek solutions of the mathematical form 

Ã(x; t)=Ã(z = x ¡ ct) that have pro¯les qualitatively similar to those shown 

in Figure 7.1. Note that this assumed form reduces the number of independent 

variables from two (x and t) toone(z), thus reducing the PDE to an ODE. 

Borrowing concepts from Chapter 1, we can make a phase-plane portrait for 

the nonlinear ODE. As will be shown in the following section, the solitary-wave 

solutions will correspond to separatrix solutions,aseparatrixbeingatrajectory 

in the phase plane that divides the plane into regions with qualitatively di®erent 

behaviors. The graphical method of hunting for solitons is quite important, 

because it allows for their possible existence to be established, even if analytic 

forms do not exist. 

∞


7.1 The Graphical Hunt for Solitons 

7.1.1 Of Kinks and Antikinks 

If you are idle, be not solitary; if you are solitary, be not idle. 

Samuel Johnson, letter, 27 October 1779, to James Boswell 

In this ¯rst example, a phase-plane portrait will reveal that the SGE has both 

kink and antikink solitary waves. To produce this portrait, the DEtools library 

package must be loaded. In order to implement the assumption Ã(x; t) =Ã(z = 

x ¡ ct), the dchange command will be employed to make a change of variables, 

requiring us also to load PDEtools. The SGE is then entered, 


> SGE:=diff(psi(x,t),x,x)-diff(psi(x,t),t,t)=sin(psi(x,t)); 

μ 

SGE := @ 2 

μ 

2 Ã(x; t) ¡ @ 

@x 2 

 

2 Ã(x; t) =sin(Ã(x; t)) 

@t 

and a variable transformation tr introduced with the \old" variables x, t, and 

Ã(x; t) related to the \new" variables z, ¿, andU(z) byx = z + c¿, t = ¿, and 

Ã(x; t) =U(z), where c is an arbitrary velocity parameter. 

> tr:=fx=z+c*tau,t=tau,psi(x,t)=U(z)g; 

tr := fx = z + c¿; t= ¿; Ã(x; t) =U (z)g 

The variable change (dchange) command applies the transformation to SGE. 

> de1:=dchange(tr,SGE,[z,tau,U(z)]); 

μ 

2 d 

de1 := U (z) ¡ c 

dz2 2 

μ 2 d 

U (z) 

dz2 

=sin(U (z)) 

The SGE has been reduced to an ODE de1 with z as the independent \spatial" 

variable. It can be simpli¯ed by collecting the second derivatives, d 2 U(z)=dz 2 . 

> de2:=collect(de1,diff(U(z),z,z)); 

de2 := (1 ¡ c2 μ 

2 d 

) U (z) =sin(U (z)) 

dz2 The nature of the phase-plane portrait will depend on the parameter c. For 

c>1(so1¡ c2 < 0), de2 is just the undamped plane-pendulum ODE, whose 

portrait was \painted" in Chapter 1. For c1, except that all ¯xed points are shifted by 

¼. Forc =1,sin(U(z)) = 0 and U = 0, or an integer multiple of ¼, for all z. 

The second-order ODE de2 is now rewritten as two ¯rst-order equations by 

setting dU=dz = Y in de3 , and substituting de3 into de2 to produce de4 . 

> de3:=diff(U(z),z)=Y(z); de4:=subs(de3,de2); 

de3 := d 

U (z) =Y (z) 

dz 

de4 := (1 ¡ c2 μ 

d 

) Y (z) =sin(U (z)) 

dz

7.1. THE GRAPHICAL HUNT FOR SOLITONS 291 

Four phase-plane trajectories will be drawn corresponding to the following initial 

conditions, which are chosen to be close to the saddle points that occur for 

c ic:=[[U(0)=0.01,Y(0)=0],[U(0)=-0.01,Y(0)=0], 

[U(0)=-6.27,Y(0)=0],[U(0)=6.27,Y(0)=0]]: 

An operator F is formed to apply the phaseportrait command to the ODE 

system de3 and de4 for, say, c =0:5 for speci¯ed scene parameters A and B. 

Choosing A = U and B = Y will generate the phase-plane portrait, while 

A = z and B = U will produce a plot of U(z). By trial and error, the z range is 

taken to be from 0 to 11. This will produce approximate separatrix trajectories 

that leave from one saddle point and end up at another saddle point. The 

tangent arrows are colored green and are taken (using arrows=MEDIUM) tobe 

\two-headed." The four trajectories are given di®erent colors, so that they are 

easily distinguished on the computer screen. 

> F:=(A,B)->phaseportrait([de3,eval(de4,c=0.5)],[U(z),Y(z)], 

z=0..11,ic,scene=[A,B],U=-6.5..6.5,Y=-6.5..6.5, 

stepsize=0.05,dirgrid=[30,30],color=green,linecolor= 

[blue,red,black,magenta],arrows=MEDIUM): 

Entering F(U,Y)produces the phase-plane portrait shown in Figure 7.2. 

> F(U,Y); 

6 

4 

Y 

2 

–6 –4 –2 0 2 4 6 

U 

–2 

–4 

–6 

Figure 7.2: Separatixes correspond to kink and antikink solitons. 

The solid curves are the separatrixes, two trajectories between the saddle points 

(¡2 ¼;0) and (0,0), and two between (0,0) and (+2 ¼; 0). Inside the separatrixes,


the solutions are periodic as the trajectories cycle around vortex points at (U = 

§ ¼; Y =0)asz increases. The solutions outside the separatrixes are also 

oscillatory, resembling \over-the-top" motion for the undamped pendulum. 

What do the separatrix solutions look like? Qualitatively, the answer is quite 

simple. A trajectory starting at the saddle point (0,0) at z = ¡1 asymptotically 

approaches the right saddle point (2 ¼; 0) as z ! + 1. A similar trajectory 

connects the left saddle point at (¡2 ¼; 0) to the one at the origin as z varies 

from ¡1 to + 1. These are examples of kink solitary waves, whose pro¯les 

U(z) may be seen by entering F(z,U), thus producing Figure 7.3. The curves 

connecting U =2¼ to U =0,andU =0to¡2 ¼, are the antikink solutions. 

> F(z,U); 

U 

6 

4 

2 

0 

–2 

–4 

–6 

2 4 6 8 10 

z 

Figure 7.3: Pro¯les of kink and antikink solitary waves. 

An important physical example of a kink is a so-called Bloch wall between 

two magnetic domains in a ferromagnet as schematically depicted in Figure 7.4. 

The magnetic spins rotate from, say, spin down in one domain to spin up in 

Figure 7.4: Bloch wall between two ferromagnetic domains. 

the adjacent domain. The narrow transition region between down and up spins 

is called a Bloch wall in honor of the theoretical physicist and Nobel laureate


Felix Bloch. Under the in°uence of an applied magnetic ¯eld, the Bloch wall 

(kink soliton) can propagate according to the SGE without changing in shape. 

PROBLEMS: 

Problem 7-1: Variation in velocity 

Explore how the sine{Gordon solitary waves vary in shape as the velocity c is 

altered. 

Problem 7-2: Cosine{Gordon equation 

If the sine term is replaced with a cosine in the SGE, how would the solitarywave 

solutions be a®ected? Con¯rm your reasoning by running the text recipe 

with a cosine present, instead of the sine term. You will have to alter the initial 

conditions to obtain the new separatrixes. 

Problem 7-3: Is there or isn't there? 

Suppose that the nonlinear term in the SGE is replaced with sin 2 Ã. Using 

the phase-plane portrait approach, determine whether there is a solitary-wave 

solution to this modi¯ed SGE. 

7.1.2 In Search of Bright Solitons 

We're all of us sentenced to solitary con¯nement inside our 

own skins, for life! 

Tennessee Williams, American dramatist (1914{1983) 

Our second example illustrates the existence of a bright solitary-wave solution 

to the NLSE for the situation that the equation has the plus sign. 

The DEtools library package is loaded, 


and the NLSE entered. 

> NLSE:=I*diff(E(x,t),x)+(1/2)*diff(E(x,t),t,t) 

+abs(E(x,t))^2*E(x,t)=0; 

μ 

@ 

NLSE := E (x; t) I + 

@x 1 

μ 

2 @ 

E(x; t) + jE (x; t)j 

2 @t2 2 E (x; t) =0 

Because of the complex nature of the equation, a slightly di®erent assumption 

is made here than in the sine{Gordon example. In this case, a solitary-wave 

solution of NLSE of the form E(x; t) =U(t) e ibx is sought, where the parameter 

b, the coordinate x, andU(t) are taken as positive. 

> ode:=eval(NLSE,E(x,t)=U(t)*exp(I*b*x)) 

assuming b>0,x>0,U(t)>0; 

ode := ¡U (t) be (bxI) + 1 

μ 

2 d 

U (t) e 

2 dt2 (bxI) + U (t) 3 e (bxI) =0 

This assumption has reduced the NLSE to an ODE, which is simpli¯ed by 

dividing by eibx and multiplying by 2.


> ode2:=simplify(2*ode/exp(I*b*x)); 

μ 

2 d 

ode2 := ¡2 U (t) b + U (t) +2U (t) 

dt2 3 =0 

The second-order ODE is cast into two ¯rst-order ODEs by setting dU=dt = Y 

in ode3 , and substituting this expression into ode2 . 

> ode3:=diff(U(t),t)=Y(t); ode4:=subs(ode3,ode2); 

ode3 := d 

U (t) =Y(t) 

dt 

μ 

d 

ode4 := ¡2 U (t) b + Y (t) +2U (t) 

dt 3 =0 

Two phase-plane trajectories will be plotted for initial conditions very close to 

the origin. The origin will be revealed to be a saddle point. 

> ic:=[[U(0)=.01,Y(0)=0],[U(0)=-.01,Y(0)=0]]: 

Taking b =1,anoperatorFis formed to apply the phaseportrait command 

to ode3 and ode4 for speci¯ed scene parameters A and B. 

> F:=(A,B)->phaseportrait([ode3,eval(ode4,b=1)],[U(t),Y(t)], 

t=0..9,ic,scene=[A,B],U=-2..2,Y=-2..2,stepsize=0.05, 

dirgrid=[20,20],color=red,linecolor=blue,arrows=MEDIUM): 

The phase-plane portrait results on entering F(U,Y), 

> F(U,Y); 


Y 

2 

1 

–2 –1 0 1 2 

U 

–1 

–2 

Figure 7.5: Separatrixes correspond to bright solitary waves. 

The arrows indicate the direction of increasing t, while the solid curves to the 

left and right of the origin are the two separatrixes. There are vortex points at 

(U = § 1, Y =0)andasaddlepointattheorigin. Theseparatrixlinetothe 

right of the origin starts at U =0 for t !¡1,growstoamaximumpositive


value at intermediate t, and returns to zero as t ! +1. Recalling that the 

intensity is proportional to jÃj 2 = U 2 , we see that the separatrix to the left of 

the origin will produce exactly the same solitary-wave pro¯le for the intensity, 

which is normally what is measured experimentally. Because this solitary-wave 

pro¯le is collisionally stable, it is referred to as a bright soliton. The amplitudes 

U(t) corresponding to the two separatrixes are obtained by entering F(t,U), 

> F(t,U); 


U 

2 

1 

0 

–1 

–2 

2 4 6 8 

t 

Figure 7.6: Bright solitary-wave amplitudes. 

The reader might have been surprised that a di®erent assumed solution was 

used for the NLSE than in the previous section to derive the bright soliton. 

This is because in the underlying derivation of the NLSE, which is beyond the 

scopeofthistext,acoordinatetransformationtoaframemovingatthespeedof 

light has already been made. The bright soliton is stationary (has zero velocity) 

in this moving frame. In the laboratory frame, the bright-soliton solution can 

be interpreted as follows. At a given point in the medium, the bright-soliton 

intensity will be essentially zero until the pulse arrives, then grow to a maximum 

value, and decrease back toward zero as the pulse passes by. 

Bright-soliton solutions have been observed [Has90] experimentally in glass 

¯bers, the time interval over which the light intensity is appreciable at a given 

point in the ¯ber being of the order of a few picoseconds (1 picosecond (ps) = 10 ¡12 

seconds). 

Because of their narrow widths and their stability, solitons are envisioned by 

telecommunications engineers as high-bit-rate carriers of digitized information 

along optical ¯bers, each soliton representing a \one" and a blank space between 

adjacent solitons representing a \zero." 

Blacksolitonsmaybefoundfortheminus-signcaseintheNLSE,thisbeing 

left for you to do as a problem. A few other interesting problems are also 

presented for your intellectual amusement.


PROBLEMS: 

Problem 7-4: Black solitons 

Modify the text recipe to determine the solitary-wave solutions for the minussign 

case in the NLSE. Remembering that the physically observed intensity 

is proportional to jÃj2 , con¯rm that these solutions are black solitary waves. 

These solitary waves are collisionally stable, so are black solitons. 

Problem 7-5: Saturable refractive index 

The NLSE for a dielectric with a saturable refractive index takes the form 

i @Ã 1 

+ 

@x 2 

@2Ã jÃj2 

+ Ã =0; 

@t2 1+a jÃj2 where i = p ¡1andais a positive parameter. Taking a =0:5 and assuming 

a solution of the form Ã(x; t) = U(t) eibx with b = 1, use the phase-plane 

portrait to demonstrate graphically that a solitary-wave solution exists. 

analytic form is not known for this solitary wave. 

Problem 7-6: Burgers' equation 

An 

Burger's equation 

@Ã @Ã 

+ Ã 

@t @x = ¾ @2Ã ; 

@x2 with ¾ a positive parameter, is an example of a nonlinear di®usion equation. 

Graphically show that an antikink solitary-wave solution exists to Burgers' 

equation for a representative value of the di®usion coe±cient ¾. 

Problem 7-7: Boussinesq's equation 

The Boussinesq wave equation, which was ¯rst derived in an attempt to describe 

shallow-water waves (Ã is the surface displacement) propagating in both 

directions, is 

@2Ã @x2 ¡ @2Ã @t2 +6@2 (Ã2 ) 

@x2 + @4Ã =0: 

@x4 Using the phase-portrait option, show that a bright solitary-wave solution exists 

for this equation. 

7.1.3 Can Three Solitons Live Together? 

... for nothing on earth is solitary ... 

Ralph Waldo Emerson, American essayist and philosopher (1803{1882) 

An interesting theoretical problem [ER79] in nonlinear optics is the resonant 

interaction of three collinear waves consisting of two electromagnetic waves labeled 

with subscripts 1 and 2 and a sound wave with subscript 3. The wave 

velocities are, respectively, v1, v2, andv3 ¿ v1;v2. The real amplitudes Á1, Á2, 

and Á3 satisfy the following set of nonlinear PDEs:


@Á1 @Á1 

+ v1 

@t @x = ¡¯1 Á2 Á3; 

@Á2 @Á2 

+ v2 

@t @x = ¯2 Á1 Á3; 

@Á3 @Á3 

+ v3 

@t @x = ¡¯3 Á1 Á2: 

(7.1) 

Here, x is the direction of wave propagation, t is the time, and the coupling 

parameters ¯1, ¯2, and¯3 are real and positive. Our goal is to show graphically 

that there exists a set of three solitary waves, one for each equation, which will 

propagate along together at a common velocity c. 

The DEtools and PDEtools library packages are loaded, 


and the parameter °, which will shortly be introduced, is unprotected from its 

Maple assignment as Euler's constant. To generate the N = 3 PDEs 

> unprotect(gamma): N:=3: 

with a do loop, let's set Á4 = Á1 and Á5 = Á2. 

> phi[4](x,t):=phi[1](x,t): phi[5](x,t):=phi[2](x,t): 

The following do loop then generates the three relevant PDEs. 


> pde[j]:=diff(phi[j](x,t),t)+v[j]*diff(phi[j](x,t),x) 

=(-1)^j*beta[j]*phi[j+1](x,t)*phi[j+2](x,t); 

> end do; 

³ ´ ³ ´ 

pde 1 := @@t Á1(x; t) + @ v1 Á1(x; t) = ¡¯1 Á2(x; t) Á3(x; t) 

@x 

³ ´ ³ ´ 

pde 2 := @@t Á2(x; t) + @ v2 Á2(x; t) = ¯2 Á3(x; t) Á1(x; t) 

@x 

³ ´ ³ ´ 

pde 3 := @@t Á3(x; t) + @ v3 Á3(x; t) = ¡¯3 Á1(x; t) Á2(x; t) 

@x 

Solitary-wave solutions are sought that are functions of the single \new" independent 

\spatial" variable z = x ¡ ct, c being an arbitrary velocity for the 

moment. The relevant variable transformation is entered, with t = ¿ and new 

amplitudes U1(z), U2(z), and U3(z). 

> tr:=fx=z+c*tau,t=tau,phi[1](x,t)=U[1](z), 

phi[2](x,t)=U[2](z),phi[3](x,t)=U[3](z)g; 

tr := fx = z+c¿; t= ¿; Á1(x; t) =U1(z); Á2(x; t) =U2(z); Á3(x; t) =U3(z)g 

and the dchange command applied to each PDE in the following do loop. 


> ode[j]:=dchange(tr,pde[j],[z,tau,U[1](z),U[2](z),U[3](z)]); 

To simplify the output, the substitution ¯j = °j (vj ¡ c) ismadeinthejth 

ODE and each equation divided by (vj ¡ c) and simpli¯ed. 

> ode[j]:=subs(beta[j]=gamma[j]*(v[j]-c),ode[j]):


> ode[j]:=simplify(ode[j]/(v[j]-c)): 

> end do: 

On completion of the do loop, the system sys of three resulting nonlinear ODEs 

is put into a list and displayed. 

> sys:=[ode[1],ode[2],ode[3]]; 

sys := 

· 

d 

dz U1(z) =¡°1 U2(z) U3(z); d 

dz U2(z) =°2 U3(z) U1(z); 

¸ 

d 

dz U3(z) =¡°3 U1(z) U2(z) 

To graphically ¯nd three coexisting solitary-wave solutions, the ° parameters 

are all set equal to the value 1. 

> gamma[1]:=1: gamma[2]:=1: gamma[3]:=1: 

Since the ¯ parameters were all positive, choosing all three ° values to be 

positive implies that c is less than the smallest v value, i.e., smaller than v3. 

Typically, v1 ¼ v2 ¼ 108 m/s and the speed v3 of sound is about 103 m/s. 

So, the electromagnetic solitary waves are very unusual, since they would be 

traveling some ¯ve orders of magnitude more slowly than they normally do. 

Our graphical procedure will produce a three-dimensional viewing box, so 

based on trial and error, it is convenient to choose the following two orientations 

> Orient[1]:=[-90,90]: Orient[2]:=[-90,0]: 

for the viewing box in the following do loop. 


The DEplot3d plotting command will produce solution trajectories but no tangent 

¯eld. The independent variable z is allowed to vary from z = ¡5 toz =20. 

Solitary waves are sought that are kinks (antikinks) or peaked solutions. Peaked 

solutions would start with U =0atz = ¡1, while kinks would start at some 

constant value at this limit. One can't literally start at zero, for example, becausethenitwouldtakeforeverintermsofzto 

generate a solution. In the 

following, U1 and U3 start out with the value 0:01 (close to zero) and peaked 

solutions are sought for these amplitudes. On the other hand, U2 will start with 

the value ¡0:99 (close to ¡1) and a kink solution sought for this amplitude. 

The scene option is taken to be scene=[z,U[1](z),U[2](z)], so that with the 

appropriate orientation, the behavior of U1 and U2 as a function of z can be 

observed. 1 

> DEplot3d(sys,[U[1](z),U[2](z),U[3](z)],z=-5..20, 

[[U[1](0)=0.01,U[2](0)=-0.99,U[3](0)=0.01]], 

scene=[z,U[1](z),U[2](z)],U[1]=0..1.2,U[2]=-1.2..1.2, 

stepsize=0.05,orientation=Orient[i],linecolor=blue); 

> end do; 

On completion of the do loop, the two plots are displayed in Figure 7.7. 

1To see U3 plotted versus z, youcantakescene = [z,U[2](z),U[3](z)] with an orientation 

[-90,90].

7.2. ANALYTIC SOLITON SOLUTIONS 299 

U1 

1 

–5 0 5 z 15 20 

0.5 

U2 

1 

–0.5 

–1 

–5 0 5 z 15 20 

Figure 7.7: Solitary-wave pro¯les for U1 (left) and U2 (right). 

On the left is shown the peaked solitary wave for electromagnetic wave number 

one (U1) and on the right the coexisting kink solitary-wave solution for the 

second electromagnetic wave (U2). The sound-wave amplitude U3 (not shown) 

displays a peaked solitary-wave pro¯le as well. 

David Kaup [Kau76], [KRB79] has extensively investigated the three-wave 

problem and analytically established that the three solitary waves are solitons. 

However, although many years have elapsed since these solitons were predicted, 

there is still no experimental evidence that such solitons can actually be produced 

in the laboratory. 

PROBLEMS: 

Problem 7-8: Solitary sound-wave pro¯le 

Modify the recipe in the text to explicitly graph the solitary sound wave pro¯le 

as a function of z. 

Problem 7-9: Variation with ° 

Discuss how the three solitary-wave pro¯les vary in shape as the values of °1, 

°2, and°3arealtered in the text recipe. For example, try °1 =1,°2 =2, 

and °3 = 3. Support your discussion with the pro¯le plots in each case. Note 

whether any of the pro¯les is still a solitary wave or is a wave train. 

7.2 Analytic Soliton Solutions 

In Chapters 5 and 6, the focus was on solving a wide variety of linear di®usion, 

Laplace, and wave equation models. The coverage was somewhat lengthy not 

only because of the physical importance of these PDEs but also because such 

linear PDEs could be solved analytically, the detailed solutions being readily 

derived with Maple. The mathematical solution of nonlinear di®usion and wave


equation models is much more di±cult, and analytic approaches have consisted 

mainly in ¯nding special solutions, some of which are of physical importance. 

Many of these approaches are quite complicated in nature, but the seeking of 

analytic solitary-wave pro¯les is relatively easy, provided that analytic solutions 

exist. This will now be illustrated in the next two examples. 

7.2.1 Follow That Wave! 

It's just a job ... waves pound the sand. I beat people up. 

Muhammad Ali, American boxer, New York Times, 6April1977 

Probably the ¯rst reported observation of soliton behavior recorded in the scienti¯c 

literature was made by the Scottish engineer and naval architect John 

Scott Russell [Rus44]. In the less formal style of scienti¯c reporting of the day, 

he wrote: 

I was observing the motion of a boat which was rapidly drawn along a narrow 

channel by a pair of horses, when the boat suddenly stopped | not so the 

mass of water in the channel which it had put in motion; it accumulated round 

the prow of the vessel in a state of violent agitation, then suddenly leaving it 

behind, rolled forward with great velocity, assuming the form of a large solitary 

elevation, a rounded smooth and well-de¯ned heap of water, which continued its 

course along the channel apparently without change of form or diminution of 

speed. I followed it on horseback, and overtook it still rolling on at a rate of 

some eight or nine miles an hour, preserving its original ¯gure some thirty feet 

long and a foot to a foot and a half in height. Its height gradually diminished, 

and after a chase of one or two miles I lost it in the windings of the channel. 

Such, in the month of August 1834, was my ¯rst chance interview with that 

singular and beautiful phenomenon :::: 

The \narrow channel" referred to by Russell still exists, being the Union Canal 

linking Edinburgh with Glasgow. Actually, Russell was not observing the 

\rapidly drawn boat" by accident, but was actually carrying out a series of experiments 

to determine the force{velocity characteristics of di®erently shaped 

boat hulls in order to determine design parameters for conversion from horse 

power to steam power. His solitary-wave observations were followed by extensive 

wave-tank experiments in which he established the major properties of 

hydrodynamic solitary waves. [EJMR81] 

The detailed mathematical explanation of Russell's solitary wave had to wait 

50 years until 1895, when the relevant nonlinear Korteweg{de Vries equation, 

@Ã @Ã 

+ ®Ã 

@t @x + @3Ã =0; (7.2) 

@x3 was derived by the Dutch mathematicians Diederik Korteweg and Gustav de 

Vries. In the KdV equation, Ã is the transverse displacement of the horizontal 

water surface, x the spatial coordinate in the direction of wave propagation, t


the time, and ® a numerical factor that can be either scaled out of the equation 

or, alternatively, assigned a convenient numerical value. The KdV equation 

doesn't look like a wave equation, having a ¯rst time derivative and a third 

spatial derivative term. However, it does describe the unidirectional propagation 

of lossless shallow-water waves in a rectangular canal quite well. How a 

peaked solitary wave solution to the KdV equation occurs is rather interesting. 

If the nonlinear term, ®Ã(@Ã=@x), is neglected, the remaining two terms 

produce a dispersive (spreading) e®ect. This is easily understood as follows. A 

localized propagating pulse can be built up out of a Fourier sum of terms of 

the plane wave structure e i(kx+!t) ,wherek is the wave number and ! is the 

frequency. Neglecting the nonlinear term, the dispersion relation ! = k 3 results 

on substituting the plane wave solution. Solving for k, the phase velocity then 

is v = !=k = ! 2=3 . Therefore, high-frequency Fourier components travel faster 

than low-frequency components, i.e., dispersion occurs. 

On the other hand, if the third-derivative term is ignored, it can be analytically 

shown [EM00] that the remaining two terms generate a \shock wave" 

e®ect. A shock wave is characterized by a progressive steepening of the leading 

edge of a propagating localized pulse. The solitary-wave solution corresponds 

to the situation in which the dispersive and shock wave contributions cancel 

exactly. 

To analytically determine the solitary-wave solution, let's follow the procedure 

of the previous section and assume that Ã(x; t) =Ã(z = x ¡ ct), where 

c>0 is the velocity, whose value is not yet unspeci¯ed. The e®ect of this simplifying 

assumption will be to reduce the nonlinear PDE (7.2) to a nonlinear 

ODE. To carry out this reduction explicitly, the PDEtools library is loaded, 


and the KdV equation entered. 

> KdVE:=diff(psi(x,t),t)+alpha*psi(x,t)*diff(psi(x,t),x) 

+diff(psi(x,t),x,x,x)=0; 

μ 

μ μ 

3 

@ 

@ 

@ 

KdVE := Ã(x; t) + ®Ã(x; t) Ã(x; t) + Ã(x; t) =0 

@t @x @x3 The transformation from the \old" independent variables x and t to the new 

variables z and ¿ is entered with x = z + c¿ and t = ¿. The dependent variable 

Ã(x; t) is rewritten as U(z). 

> tr:=fx=z+c*tau,t=tau,psi(x,t)=U(z)g; 

tr := fx = z + c¿; t= ¿; Ã(x; t) =U (z)g 

Then, applying the dchange command to KdVE with the transformation tr, 

> ode1:=dchange(tr,KdVE,[z,tau,U(z)]); 

μ 

μ μ 

3 

d 

d 

d 

ode1 := ¡ U (z) c + ® U (z) U (z) + U (z) =0 

dz dz dz3 yields the third-order nonlinear ODE shown in ode1 . The left-hand side of this 

equation is easily integrated, yielding ode2 .


> ode2:=int(lhs(ode1),z)=0; 

ode2 := ¡c U (z)+ 1 

2 ® U (z)2 μ 

2 d 

+ U (z) =0 

dz2 On the right-hand side of ode2 , we have set the integration constant equal to 

zero by assuming that we are seeking a peaked solitary wave similar to that 

schematically depicted earlier in Figure 7.1. For such a solitary wave, U(z) and 

all of its derivatives must vanish as z !1. 

To integrate the second-order nonlinear ODE ode2 , we can proceed as follows. 

Letting V = dU(z)=dz, then, 

d2U(z) dz2 dV dV dU dV 

= = = V 

dz dU dz dU : 

Along with U(z) =U, this result is substituted into ode2 . 

> ode3:=subs(fdiff(U(z),z,z)=V(U)*diff(V(U),U),U(z)=Ug,ode2); 

ode3 := ¡cU + ®U2 

μ 

d 

+ V (U) V (U) =0 

2 dU 

Since U ! 0andV (U) ´ dU=dz ! 0asjzj !1, we can apply the dsolve 

command to ode3 subject to the condition V (U =0)=0, 

> sol:=dsolve(fode3,V(0)=0g,V(U)); 

p p 

¡3 ®U +9cU 

¡3 ®U +9cU 

sol := V (U) = 

; V (U) =¡ 

3 

3 

yielding positive and negative square root solutions. The positive answer is 

selected and simpli¯ed. 

> sol1:=simplify(sol[1],symbolic); 

p 

¡3 ®U +9cU 

sol1 := V (U) = 

3 

Now, remember that V ´ dU=dz, so from the above output we have 

Z 

3 dU 

z = 

U p 9 c ¡ 3 ®U : 

This integration is now carried out, 

> eq:=z=int(1/rhs(sol1),U); 

μp 

¡3 ®U +9c 

2arctanh 

3 

eq := z = ¡ 

p 

c 

p 

c 

and eq solved for U using the isolate command and the result simpli¯ed. 

> eq2:=simplify(isolate(eq,U)); 

Ã μ p ! 

2 

z c 

3 c tanh ¡ 1 

2 

eq2 := U = ¡ 

® 

Then, substituting z = x ¡ ct into the rhs of eq2 yields the solitary wave Ã, 

which is converted into a more standard form with the sincos option.


> psi:=subs(z=x-c*t,rhs(eq2)); 

Ã μ p ! 

2 

(x ¡ ct) c 

3 c tanh 

¡ 1 

2 

Ã := ¡ 

® 

> psi:=simplify(convert(psi,sincos)); 

3 c 

Ã := μ p 2 

(x ¡ ct) c 

cosh 

® 

2 

At z ´ x¡ct= 0, the solitary-wave solution has its maximum amplitude 3 c=®. 

For a ¯xed value of ®, the maximum amplitude is proportional to the velocity 

c, so by ¯xing this maximum amplitude, the velocity is also ¯xed. Clearly, a 

taller KdV solitary wave will have a larger velocity than a shorter solitary wave. 

On the other hand, the width of the pulse scales inversely with p c,sotaller 

KdV solitary waves are thinner than shorter solitary waves. 

To con¯rm that the solitary-wave solution is indeed localized and travels at 

constant velocity with unchanging shape, the analytic solution can be animated. 

Let's take, for example, ® =1andc = 1 

2 , 

> psi:=eval(psi,falpha=1,c=1/2g); 

Ã := 3 

1 

Ã 

2 

(x ¡ t=2) 

cosh 

p ! 2 

2 

4 

and animate the solitary-wave solution. 

> animate(plot,[psi,x=-20..70],t=0..100,numpoints=200, 

frames=50,axes=frame,thickness=2); 

By running the animation, the reader will see that all the features that have 

been discussed are con¯rmed. Later, by looking at the collision of one solitary 

wave with another, we will con¯rm that the solitary wave is a soliton. 

PROBLEMS: 

Problem 7-10: Modi¯ed KdV equation 

Derive a solitary-wave solution of the modi¯ed KdV equation 

@Ã @Ã 

+ ®Ã2 

@t @x + @3Ã =0; 

@x3 which appears in the theory of double layers in plasmas and as a model of ion 

acoustic solitons in a multicomponent plasma. 

Problem 7-11: Boussinesq's equation 

The Boussinesq water wave equation is 

@2Ã @x2 ¡ @2Ã @2 

+6 

@t2 @x2 ¡ 2 

Ã ¢ + @4Ã =0: 

@x4 Derive the analytic form of the bright solitary-wave solution and animate it.


7.2.2 Looking for a Kinky Solution 

Masterpieces are not single and solitary births; they are the outcome 

of ... thinking by the body of the people, so that the experience of the 

mass is behind the single voice. 

Virginia Woolf, British novelist, ARoomofOne'sOwn,1929 

In our second example, we seek a kink solitary-wave solution of the SGE, 

@2Ã @x2 ¡ @2Ã =sinÃ: (7.3) 

@t2 Recall that the moving kink has been used to model the motion of a magnetic 

domain wall separating two di®erent spin regions in a ferromagnet. 

To ¯nd a kink, the PDEtools package is loaded and the SGE entered. 


> SGE:=diff(psi(x,t),x,x)-diff(psi(x,t),t,t)-sin(psi(x,t))=0; 

μ μ 

2 

2 

@ @ 

SGE := Ã(x; t) ¡ Ã(x; t) ¡ sin(Ã(x; t)) = 0 

@x2 @t2 Then, SGE is reduced to a nonlinear ODE by assuming that Ã(x; t) =U(z) 

with x = z + c¿ and t = ¿, and applying the dchange command. 

> tr:=fx=z+c*tau,t=tau,psi(x,t)=U(z)g: 

> ode1:=dchange(tr,SGE,[z,tau,U(z)]); 

μ μ 

2 

2 

d d 

ode1 := U (z) ¡ U (z) c 

dz2 dz2 2 ¡ sin(U (z)) = 0 

The second derivative terms are collected in ode1 . 

> ode1:=collect(ode1,diff(U(z),z,z)); 

ode1 := (1 ¡ c2 μ 

2 d 

) U (z) ¡ sin(U (z)) = 0 

dz2 For a nontrivial solution, it is necessary that the velocity satisfy c 6= 1. Although 

c could be left as a general parameter, the analytic form is substantially 

simpli¯ed if a speci¯c numerical value is chosen for c, e.g., c = 1 

4 . 

> c:=1/4: 

Paralleling the procedure in the KdV recipe, let's set V = dU=dz, so that 

d2U=dz2 = V (dV=dU), and substitute this result and U(z) =U into ode1 . 

> ode2:=subs(fdiff(U(z),z,z)=V(U)*diff(V(U),U),U(z)=Ug,ode1); 

ode2 := 15 

μ 

d 

V (U) V (U) ¡ sin(U) =0 

16 dU 

For the kink solitary wave, both U and V ´ dU=dz must go to zero as z !¡1. 

So ode2 is solved for V (U), subject to V (0) = 0. 

> sol:=dsolve(fode2,V(0)=0g,V(U)); 

sol := V (U) = 4 p 4 p 

30 ¡ 30 cos(U); V (U) =¡ 30 ¡ 30 cos(U) 

15 

15


The positive square root is chosen (the negative square root yields an antikink) 

and the trig identity cos U =1¡ 2sin 2 (U=2) substituted, 

> sol1:=subs(cos(U)=1-2*sin(U/2)^2,sol[1]); 

sol1 := V (U) = 4 

s 

μ 2 

p U 

60 sin 

15 

2 

and the result simpli¯ed using the square root and symbolic options. 

> sol1:=simplify(sol1,sqrt,symbolic); 

sol1 := V (U) = 8 

μ 

p U 

15 sin 

15 2 

Since V = dU=dz, thenz ´ x ¡ ct is equal to the integral with respect to U of 

the reciprocal of the right-hand side of sol1 . 

> z:=x-c*t=int(1/rhs(sol1),U); 

z := x ¡ t 

μ μ μ 

1 p U U 

= 15 ln csc ¡ cot 

4 4 

2 2 

Then, solving for U and recalling that U(z) ´ Ã(x ¡ ct), 

> solwave:=psi(x,t)=solve(z,U); 

solwave := Ã(x; t) = 

0 

B 

2arctanB 

Ã 

B 

@ 

1+ 

2 e ( (4 x¡t) p 15 

15 

e ( (4 x¡t) p 15 

15 

Ã 

) 

! 2 ; ¡ Ã 

) 

1+ 

e ( (4 x¡t) p 15 

15 

! 2 

) 

e ( (4 x¡t) p 15 

15 

1 

¡ 1 C 

! 2 C 

) A 

yields the analytic solitary-wave solution shown in solwave. The arctan function 

is expressed in terms of two arguments separated by a comma. The term to the 

left (right) of the comma is the numerator (denominator) of arctan. 

The common denominators in the arguments can be removed and the numerator 

of the second argument simpli¯ed with the following assumption: 

> solwave:=simplify(solwave,assume=real); 

solwave := Ã(x; t) =2arctan(2e ( (4 x¡t) p 15 

15 

) 

; ¡e 

( 2(4x¡t) p 15 

) 15 +1) 

If you can't instantly see that this still complicated-appearing result is a kink 

solitary-wave solution of the SGE, you can be excused. However, that it is a 

solution can be con¯rmed by applying the following pdetest command. 

> pdetest(solwave,SGE); 

0 

To see that it is a kink solitary wave, let's animate the rhs of solwave. 

> animate(plot,[rhs(solwave),x=-10..50],t=0..200,frames=50, 

thickness=2,axes=framed); 

On running the animation, we observe that a kink solitary wave travels to the


right, maintaining its initial shape throughout. The kink varies in amplitude 

from 0 at x = ¡1 to 2 ¼ at z =+1. 

PROBLEMS: 

Problem 7-12: Antikink solitary waves 

Show that the choice of the negative square root yields an antikink solution. 

Problem 7-13: Relation between amplitude and velocity 

Is there any relation between the maximum nonzero amplitude and the velocity? 

Is this the same sort of relationship as for the KdV solitary wave or is it 

di®erent? 

Problem 7-14: Burgers' equation 

Burgers' nonlinear di®usion equation is of the form 

@U @U 

+ U 

@t @x = ¾ @2U ; 

@x2 where ¾ is the positive di®usion coe±cient. Analytically derive an antikink 

solitary-wave solution to Burgers' equation and animate it. How do the width 

of the antikink region and the velocity depend on amplitude? 

Problem 7-15: Sine{Gordon breather 

The SGE permits a moving (velocity v) \breather"-mode solution, which is 

localized in space but oscillatory in time, of the form, 

Ã = 4 arctan 

μr m 

1 ¡ m 

sin(° p 1 ¡ m (t ¡ vx)) 

cosh(° p m (x ¡ vt)) 

with ° =1= p 1 ¡ v2 , ¡1


In 1971, Fred Tappert, of Bell Laboratories, found an exact two-soliton 

analytic solution for the KdV equation, this equation now being entered. 


> KdVE:=diff(psi(x,t),t)+psi(x,t)*diff(psi(x,t),x) 

+diff(psi(x,t),x,x,x)=0; 

μ μ μ 

3 

@ 

@ 

@ 

KdVE := Ã(x; t) + Ã(x; t) Ã(x; t) + Ã(x; t) =0 

@t @x @x3 Tappert's two-soliton solution Ã(x; t) isgiven. 

> psi(x,t):=72*(3+4*cosh(2*x-8*t)+cosh(4*x-64*t)) 

/(3*cosh(x-28*t)+cosh(3*x-36*t))^2; 

72 (3 + 4 cosh(2 x ¡ 8 t)+cosh(4x ¡ 64 t)) 

Ã(x; t) := 

(3 cosh(x ¡ 28 t)+cosh(3x ¡ 36 t)) 2 

The lhs of KdVE is extracted and simpli¯ed, Ã(x; t) having been automatically 

substituted. A lengthy expression results, which is suppressed here in the text. 

> check1:=simplify(lhs(KdVE)); 

The combine command, with the trig option, is applied to the numerator of 

check1 , the result being zero, con¯rming that Ã(x; t) is a solution of KdVE . 

> check2:=combine(numer(check1),trig); 

check2 := 0 

The two-soliton solution is now animated. 

> animate(plot,[psi(x,t),x=-20..20],t=-1..1,frames=60, 

numpoints=250,axes=frame,thickness=2); 

In the animation, one initially has a taller, narrower solitary wave to the left of 

a shorter, wider solitary wave. As the animation progresses, both pulses move 

to the right. Having a larger velocity, the taller pulse overtakes the shorter 

pulse and a collision occurs. During the collision, a nonlinear superposition 

takes place, the resultant amplitude being less than the linear sum of the two 

amplitudes. As time progresses, the taller, faster pulse passes through the 

shorter one and emerges unchanged in shape, as does the shorter pulse. The 

solitary waves are indeed solitons. 

Next, we con¯rm that the sine{Gordon equation, 

> SGE:=diff(U(x,t),x,x)-diff(U(x,t),t,t)-sin(U(x,t))=0; 

μ μ 

2 

2 

@ @ 

SGE := U (x; t) ¡ U (x; t) ¡ sin(U (x; t)) = 0 

@x2 @t2 is satis¯ed by the following two-soliton kink{kink solution. [Jac90] 

> U(x,t):=4*arctan(c*sinh(x/sqrt(1-c^2)) 

/cosh(c*t/sqrt(1-c^2))); 

0 μ 1 

x 

B 

c sinh p 

U (x; t) := 4 arctan B 1 ¡ c2 C 

@ 

μ C 

ct A 

cosh p 

1 ¡ c2


The lhs of SGE is simpli¯ed, 

> check3:=simplify(lhs(SGE)); 

and the numerator of check3 expanded and further simpli¯ed. 

> check4:=simplify(expand(numer(check3))); 

check4 := 0 

The result is zero, so U(x; t) satis¯es the sine{Gordon equation. This twosoliton 

kink{kink solution is now animated, with c = 1 

4 . 

> animate(plot,[eval(U(x,t),c=1/4),x=-50..50],t=-100..100, 

frames=50,thickness=2,axes=framed); 

On running the animation, you will observe that the two kinks travel in opposite 

directions, run into each other, and reverse directions after the collision, still 

maintaining their initial shapes. 

PROBLEMS: 

Problem 7-16: Kink{antikink collision 

In the two-soliton kink{kink solution, replace the ¯rst c by 1=c, x by ct,and 

ct by x. Animate the resulting solution and show that it represents a kink{ 

antikink collision. Describe the observed behavior. 

7.3 Simulating Soliton Collisions 

Both authors of this text have spent an academic lifetime jousting with nonlinearities 

in all mathematical shapes and sizes. In this text, we have tried to 

provide a glimpse of the excitement and complexity involved in the study of 

nonlinear dynamics, yet still present the bread-and-butter recipes necessary to 

solve linear ODE and PDE problems, the staple of most undergraduate science 

curricula. Whether the balance of linear and nonlinear recipes is right in our 

computer algebra menu, you will have to be the judge, but we could not resist 

presenting two numerical recipes that simulate soliton collisions. The ¯rst is 

for the Korteweg{de Vries equation, the second for the sine{Gordon equation. 

7.3.1 To Be or Not to Be a Soliton 

There is no means of proving it is preferable to be than not to be. 

E. M. Cioran, French philosopher (1911{1995) 

To prove that solitary-wave solutions are solitons, i.e., whether they survive 

collisions with each other unchanged in shape, is an important area of research 

in nonlinear dynamics. One approach is to numerically collide the solitary 

waves using a ¯nite di®erence scheme to simulate the relevant nonlinear PDE. 

This was done by Norman Zabusky and Martin Kruskal [ZK65] for the KdV

7.3. SIMULATING SOLITON COLLISIONS 309 

equation (taking ® =1), 

@Ã 

@t 

@Ã 

+ Ã 

@x + @3Ã =0: (7.4) 

@x3 They used a CDA for each ¯rst derivative, approximated @ 3 Ã=@x 3 by 

μ 3 @ Ã 

@x3 

=(Ãi+2;j ¡ 2 Ãi+1;j +2Ãi¡1;j ¡ Ãi¡2;j)=(2 h 

P 

3 ); (7.5) 

and averaged Ã in the nonlinear term equally over the three grid points (i+1;j), 

(i; j), and (i ¡ 1;j). Setting r = k=h3 , the Zabusky{Kruskal algorithm is 

Ãi;j+1 = Ãi;j¡1 ¡ rh 2 (Ãi+1;j + Ãi;j + Ãi¡1;j)(Ãi+1;j ¡ Ãi¡1;j)=3 

¡ r (Ãi+2;j ¡ 2 Ãi+1;j +2Ãi¡1;j ¡ Ãi¡2;j); 

(7.6) 

with j =1; 2;:::: This scheme is numerically stable for r


Using the operator F, we add the three time-dependent solitary waves in f, and 

take the time derivative of f in g. 

> f:= add(F(X[i],c[i]),i=1..3): g:=diff(f,t): 

Setting the time to zero, we plot the input pro¯le over the spatial range x =0 

to M, the resulting picture being shown in Figure 7.8. 

> t:=0: plot(f,x=0..M,thickness=2); 

2.5 

2 

1.5 

1 

0.5 

0 50 100 x 150 200 250 

Figure 7.8: Input pro¯le for the three-solitary-wave collision simulation. 

Note that there is a slight overlap of the solitary-wave tails. One could place 

them further apart initially, but this takes more computing time. 

To evaluate f and g at the spatial mesh points, we use the unapply command 

to turn them into operators in terms of the spatial coordinate x. 

> f2:=unapply(f,x): g2:=unapply(g,x): 

Using these two operators, we calculate the input amplitudes Ui;0 ´ f(i; t =0) 

and Ui;1 ´ f(i; t =0)+kg(i; t =0)fori =0toM in the ¯rst and second 

initial conditions, ic1 and ic2 . 

> ic1:=seq(U(i,0)=evalf(f2(i)),i=0..M): 

> ic2:=seq(U(i,1)=evalf(f2(i))+k*g2(i),i=0..M): 

To avoid any possible unknown U values creeping into the double do loop that 

will be used to iterate the numerical algorithm, we \initialize" all U values to 

zero for i =0toM and j =2toN, i.e., for all remaining grid points. These 

zeros will be overwritten as the loop is executed and new U values calculated. 

> init:=seq(seq(U(i,j)=0,i=0..M),j=2..N): 

The two initial conditions and the initialization are assigned. 

> assign(ic1,ic2,init):


An operator G is introduced to calculate the rhs of the Zabusky{Kruskal algorithm 

(7.6) (with Ã replaced with U) for a speci¯ed i and j. 

> G:=(i,j)->U(i,j-1)-r*h^2*(U(i+1,j)+U(i,j)+U(i-1,j)) 

*(U(i+1,j)-U(i-1,j))/3 

-r*(U(i+2,j)-2*U(i+1,j)+2*U(i-1,j)-U(i-2,j)): 

The numerical algorithm is ¯rst iterated from i =2toM ¡ 2foragivenjvalue and then from j =1toN. The spatial index i isstartedat2andendedat 

M ¡ 2 to avoid unknown U values from the two edges of the grid entering into 

the double do loop calculation. 

> for j from 1 to N do; 

> for i from 2 to M-2 do 

> U(i,j+1):=G(i,j): 

> end do: 

> end do: 

A graphing operator is formed to plot the entire pro¯le on the jth time step. 

> gr:=j->plot([seq([i,U(i,j)],i=2..M-2)],thickness=2): 

Using every second graph, the sequence of pictures is now animated. 

> plots[display]([seq(gr(2*j),j=0..N/2)],insequence=true); 

You will have to execute the recipe, click on the resulting plot, and then on the 

start arrow to see the wonderful animation. Despite the initial overlap of the 

solitary-wave tails and the coarse spatial grid used (recall, h = 1), the solitary 

waves are remarkably stable, all three surviving the collision process apparently 

unchanged. After the collision the order of the waves is the reverse of the initial 

ordering, with the smallest pulse on the left and the largest on the right. 


cpu := 29:072 

The CPU time on a 3-GHz PC is about 29 seconds. 

Although we have concentrated on soliton collisions here, the recipe may 

be easily modi¯ed to investigate the behavior of other input pro¯les. For most 

cases, an analytic solution will not exist and the numerical simulation route is 

the only feasible one to take. 

PROBLEMS: 

Problem 7-17: Third derivative 

Using the Taylor expansion, derive the approximation (7.5) to @3Ã=@3x. Problem 7-18: A di®erent scheme 

In the Zabusky{Kruskal ¯nite di®erence scheme for the KdV equation, Ã in the 

nonlinear term Ã (@Ã=@x) was approximated by the average of three Ã terms 

at the grid points (i +1;j), (i; j), and (i ¡ 1;j). Compare the results obtained 

in the text recipe with those you would obtain if U ´ Ã were approximated by 

Ui;j alone. Discuss your result.


Problem 7-19: Ampli¯cation 

Multiply the smallest of the three solitary waves in the text recipe by a factor of 

3 and interpret the resulting behavior when the worksheet is executed. Explore 

the e®ect of multiplying one or more pulses by numerical factors. 

Problem 7-20: Radiative ripples 

In the text recipe, change the sech 2 terms in the input pulses to sech 4 terms, 

then execute the modi¯ed recipe, and discuss the results. 

7.3.2 Are Diamonds a Kink's Best Friend? 

I never hated a man enough to give him diamonds back. 

Zsa Zsa Gabor, Movie actress (1919{) 

Although the sine{Gordon equation, 

@2Ã @x2 ¡ @2Ã =sinÃ; 

@t2 (7.7) 

can be numerically solved with an explicit scheme based on a rectangular mesh, 

it lends itself more naturally to being tackled with a diamond-shaped mesh 

chosen to follow the characteristic directions of the equation. To see how this 

works, let's consider the general PDE 

a @2U @x2 + b @2U @x@y + c @2U + e =0; 

@y2 (7.8) 

where a, b, c, and e are functions of U, @U=@x, @U=@y, but not of higher 

derivatives. For the SGE, U = Ã, y = t, a =1,b =0,c = ¡1, and e = ¡ sin Ã. 

Equation (7.8) can be quite generally solved by the method of characteristics. 

Setting p ´ @U=@x and q ´ @U=@y, equation (7.8) can be written in the form 

a @p @p @q 

+ b + c + e =0: 

@x @y @y 

Since p = p(x; y) andq = q(x; y), then 

(7.9) 

dp @p @p dy 

= + ; 

dx @x @y dx 

dq @q dx @q 

= + : 

dy @x dy @y 

(7.10) 

Substituting @p=@x and @q=@y into (7.9), then multiplying through by dy=dx, 

noting that @q=@x = @p=@y, and rearranging yields 

" μ 2 μ # · 

@p dy dy 

a ¡ b + c ¡ a 

@y dx dx 

dp 

¸ 

dy dq dy 

+ c + e =0: 

dx dx dx dx 

(7.11) 

At this stage, the resulting equation looks like a mathematical mess! However, 

if we choose to work in the characteristic directions whose slopes are given by 

μ 2 μ 

dy dy 

a ¡ b + c =0; (7.12) 

dx dx


then equation (7.11) reduces to 

μ 

dy 

a dp + cdq+ edy =0: (7.13) 

dx 

For the SGE, a =1,b =0,andc = ¡1, so that (7.12) yields (dy=dx) 2 ¡ 1=0, 

or dy=dx = §1. The two characteristic directions have slopes of 45 ± and ¡45 ± , 

respectively. Forming a diamond-shaped mesh with these slopes produces the 

grid illustrated in Figure 7.9. Given the new grid, how is U (or Ã) calculated? 

j=2 

y 

j=1 

P 

L R 

(0,2) (2,2) (4,2) 

(1,1) (3,1) (5,1) 

j=0 

(2,0) (4,0) (6,0) 

i=0 i=1 i=2 

Δx=h 

x x 

min max 

Δy=h 

Figure 7.9: Characteristic directions and labels for solving the SGE. 

Consider the mesh point P in Figure 7.9, where it is desired to calculate the 

unknown UP from the known values UL and UR on the previous time step. The 

subscripts L and R denote advancing \from the left" along the characteristic 

direction dy=dx = 1 and \from the right" along the characteristic direction 

dy=dx = ¡1, respectively. Taking dy=dx = §1 and replacing (7.13) with a 

¯nite di®erence approximation yields the following pair of equations, 

(pP ¡ pL) ¡ (qP ¡ qL) =¡eL (yP ¡ yL); 

¡(pP ¡ pR) ¡ (qP ¡ qR) =¡eR (yP ¡ yR); 

x 

(7.14)


which are easily solved for pP and qP , 

pP = 1 

2 (pR + pL)+ 1 

2 (qR ¡ qL)+ 1 

2 (eR ¡ eL)¢y; 

qP = 1 

2 (pR ¡ pL)+ 1 

2 (qR + qL)+ 1 

2 (eR + eL)¢y; 

(7.15) 

with ¢y = yP ¡ yL = yP ¡ yR =¢x ´ h. 

To obtain UP from the values at L and R, wenotethatU =U(x; y), so that 

dU =(@U=@x) dx +(@U=@y) dy =pdx+ qdy. Replacing this result with a ¯nite 

di®erence approximation along the characteristic direction dy=dx=1 yields 

UP = UL + 1 

2 (pL + pP )(xP ¡ xL)+ 1 

2 (qL + qP )(yP ¡ yL) 

or UP = UL + 1 

2 h (pL + pP + qL + qP ); (7.16) 

where for improved accuracy, the \old" and \new" values of p and q have 

been averaged. Of course, we could also have calculated UP along the other 

characteristic direction dy=dx = ¡1, viz., 

UP = UR + 1 

2 (pR + pP )(xP ¡ xR)+ 1 

2 (qR + qP )(yP ¡ yR); 

or UP = UR + 1 

2 h (¡pR ¡ pP + qR + qP ): (7.17) 

Since the two numerical values of UP will di®er slightly, an equally weighted 

average of the two results for UP is usually taken. 

This method of characteristics scheme is now applied to the collision of a 

sine{Gordon kink with an antikink solitary wave, the input pro¯le taken to be, 

U = 4 arctan 

³ 

e (x ¡ x1 ¡ c1 t)=a ´ ³ 

+ 4 arctan e ¡(x ¡ x2 ¡ c2 t)=b ´ 

; (7.18) 

with a = p 1 ¡ c 2 1 , b = p 1 ¡ c 2 2 , x1 < 0, x2 > 0, c1 > 0, and c2 < 0, and 

the time t = 0. The ¯rst term in the input pro¯le is the kink, 2 the second 

the antikink. Nonzero values of x1 and x2 are used to spatially separate the 

kink and antikink initially. The choice of signs puts the kink to the left of the 

antikink. Since the velocities are of opposite sign, the kink and antikink will 

move toward each other and a collision will ultimately take place. 

So with this preamble behind us, let the ¯nal recipe of this chapter begin! 

Let's consider M =200spatialandN = 100 time steps. 

> restart: begin:=time(): M:=200: N:=100: 

We take x1 = ¡5, x2 =5,c1 =0:8, and c2 = ¡0:8, and calculate a and b. 

> x[1]:=-5: x[2]:=5: c[1]:=0.8: c[2]:=-0.8: 

> a:=sqrt(1-c[1]^2); b:=sqrt(1-c[2]^2); 

a := 0:6000000000 b := 0:6000000000 

2 A more general form than derived earlier.


The spatial range is taken from xmin =3x1 = ¡15 to xmax =3x2 = 15, and 

the step size h =(xmax ¡ xmin)=M calculated. 

> xmin:=3*x[1]: xmax:=3*x[2]: h:=evalf((xmax-xmin)/M); 

h := 0:1500000000 

The kink{antikink pro¯le at time t is entered. 

> U:=4*arctan(exp((x-x[1]-c[1]*t)/a)) 

+4*arctan(exp(-(x-x[2]-c[2]*t)/b)); 

³ 

U := 4 arctan e (1:666666667 x +8:333333335 ¡ 1:333333334 t)´ 

³ 

+ 4 arctan e (¡1:666666667 x +8:333333335 ¡ 1:333333334 t)´ 

The values of p(x; 0) ´ (@U=@x)jt=0 and q(x; 0) ´ (@U=@t)jt=0 are needed, so 

the relevant spatial and time derivatives are calculated and labeled P and Q. 

> P:=diff(U,x): Q:=diff(U,t): t:=0: 

Setting t = 0, the following loop evaluates U, p, andq at each spatial grid point 

on the zeroth time step. Note that the numerical value of U at the ith spatial 

grid point is labeled ui;0 and i is incremented in steps of 2 from 0 to M. 

> for i from 0 to M by 2 do #initial conditions 

> x:=xmin + i*h; 

> u[i,0]:=evalf(U); p[i,0]:=evalf(P); q[i,0]:=evalf(Q); 

> end do: 

The input pro¯le is now plotted and displayed in Figure 7.10. 

> plot([seq([xmin+2*i*h,u[2*i,0]],i=0..M/2)],view= 

[xmin..xmax,-1..15],tickmarks=[3,3],labels=["x","U"]); 

15 

10 

U 

5 

–10 0 

x 10 

Figure 7.10: Input pro¯le for the kink{antikink soliton collision simulation.


The kink is on the left, the antikink on the right, the two pro¯les being joined 

together in the middle. At the x boundaries, one has U ¼ 2 ¼ and @U=@x ¼ 0. 

Since the kink is traveling to the right and the antikink to the left, a collision will 

take place before each reaches the opposite boundary. Provided that we do not 

let the kink and antikink come close to those boundaries, the above boundary 

conditions will prevail for all times in the run. This implies that @U=@t ¼ 0at 

the x boundaries. In the following do loop, the x-boundary grid points (circled 

points in Figure 7.9) are initialized, setting U =2¼ and p = q =0. 

> for j from 2 to N by 2 do #boundary conditions 

> u[0,j]:=2*evalf(Pi): p[0,j]:=0: q[0,j]:=0; 

> u[M,j]:=2*evalf(Pi): p[M,j]:=0: q[M,j]:=0; 

> end do: 

The following double do loop calculates the values of u at the other grid points. 

> for j from 0 to (N-1) do: 

A conditional statement is inserted to start i at i0 =0forj =0; 2; 4; ::: and 

i0 =1forj =1; 3; 5; :::: 

> if j mod 2 = 0 then i0:=0: else i0:=1: end if; 

The following do loop runs over the spatial grid points i, incrementing them 

from i0 to M ¡ 2instepsof2. 

> for i from i0 to (M-2) by 2 do 

Using equation (7.15), we calculate the values of p and q. 

> p[i+1,j+1]:=0.5*(p[i+2,j]+p[i,j]+q[i+2,j]-q[i,j] 

+(-sin(u[i+2,j])+sin(u[i,j]))*h); 

> q[i+1,j+1]:=0.5*(p[i+2,j]-p[i,j]+q[i+2,j]+q[i,j] 

+(-sin(u[i+2,j])-sin(u[i,j]))*h); 

Then U is evaluated using equations (7.16) and (7.17), and the average taken. 

> uP1:=u[i,j]+0.5*h*(p[i,j]+p[i+1,j+1] +q[i,j]+q[i+1,j+1]); 

> uP2:=u[i+2,j]+0.5*h*(-p[i+2,j]-p[i+1,j+1]+q[i+2,j] 

+q[i+1,j+1]); 

> u[i+1,j+1]:=(uP1+uP2)/2; 

> end do: end do: 

The results are plotted, 

> for j from 0 to N by 2 do 

> pl(j):=plot([seq([xmin+2*i*h,u[2*i,j]],i=0..M/2)], 

thickness=3,labels=["x","U"]): 

> end do: 

andanimatedwiththeinsequence=true option. 

> plots[display]([seq(pl(2*j),j=0..N/2)],insequence=true); 

When the work sheet is executed, the kink{antikink hump °ips upside down 

but the shape of the moving pro¯le is identical to the input shape aside from a


phase factor e i¼ = ¡1, indicating that the kink and antikink solitary waves are 

indeed solitons. Finally, the CPU time is calculated. 

> cpu:=(time()-begin)*seconds; 

cpu := 3:024 seconds 

and is about 3 seconds on a 3-GHz PC. 

PROBLEMS: 

Problem 7-21: Ampli¯ed kink{antikink input 

In the text recipe, double the amplitudes of the input kink and antikink solitarywave 

pro¯les. Remembering to also double the value 2¼ in the initialization 

statement to avoid causing an end-e®ect problem, run the ¯le with the ampli¯ed 

input and interpret the outcome. 

Problem 7-22: Kink{kink collision 

Modify the recipe to simulate the collision of a kink solitary wave with another 

kink. Discuss the observed behavior. 

Problem 7-23: Antikink{antikink collision 

Modify the recipe to simulate the collision of an antikink solitary wave with 

another antikink. Discuss the observed behavior. 

Problem 7-24: Rectangular mesh 

Solve the problem of the text recipe using an explicit scheme based on a rectangular 

mesh. 

Problem 7-25: Interacting laser beams 

The interaction of two intense laser pulses of di®erent frequencies as they pass 

through each other in opposite directions in a certain resonant absorbing °uid 

can be described [RE76] by the following normalized PDEs for the laser intensities 

U and V , 

@U @U 

+ 

@x @y = ¡g1 

@V @V 

UV ¡ ®U; ¡ 

@x @y = ¡g2 UV + ®V: 

Here x is the normalized distance inside the °uid medium of length one unit, y 

the normalized time, g1 > 0andg2 > 0arethe\gain"coe±cients,and® ¸ 0 

the absorption coe±cient. The U pulse travels in the positive x direction, while 

the V pulse moves in the negative x direction. 

(a) Find the characteristic directions along which the PDEs reduce to ODEs. 

(b) Devise an explicit numerical scheme that integrates the ODEs along the 

characteristic directions assuming that there are no pulses initially inside 

the °uid (U(x; 0) = V (x; 0) = 0 for 0 < x < 1) and identical ¯niteduration 

U and V pulses are fed in at opposite ends (U(0;y)=V (1;y)= 

f(y) for0· y · Y = 1 and zero for y>Y). 

2 

(c) Numerically solve the equations and animate the results, assuming that 

f(y) =1,g1 =0:4, g2 = 20, and (a) ® =0,(b)® =0:5. 

(d) Discuss the behavior of the two pulses as revealed in the animation.


(e) Repeat the calculation and animation for f(y) =sin(2¼y), the parameter 

values and all boundary and initial conditions remaining the same. 

Compare the results with those obtained for the rectangular pulses.

Chapter 8 

Nonlinear Diagnostic Tools 

In the Appetizers, the reader was introduced to the concept of phase-plane 

analysis of nonlinear ODE models. This involved the creation of phase-plane 

portraits and the location and identi¯cation of the relevant stationary points of 

the ODE system. This graphical approach was extended in the last chapter to 

¯nding solitary wave solutions of physically important nonlinear PDEs. 

Physicists and mathematicians have developed a wide variety of other graphical 

tools for exploring the frontiers of nonlinear dynamics and understanding 

what is observed. In this chapter, a few of the simpler diagnostic tools for 

nonlinear ODEs and nonlinear di®erence equations are presented for the reader 

who craves a ¯nal light but scrumptious intellectual dessert. 

8.1 The Poincare Section 

An important approach to studying the forced motion of nonlinear oscillator 

systems is to create a Poincare section. If the driving frequency is !, onetakesa 

\snapshot" of the phase plane after each period T0 =2¼=! of the driving force. 

After an initial transient time, the ODE system will settle down in steady state 

to either a periodic or a chaotic motion. For the periodic case, the system is said 

to display a period-n response if its period T equals nT0, wheren =1; 2; 3; :::: 

In other words, the frequency response of a period-n solution is !n = !=n. 

If the system evolves to a period-1 solution, with T = T0, the Poincare 

section will consist of a single plot point in the phase plane that is reproduced 

at each multiple of the driving period. On the other hand, if the system evolves 

to a period-2 solution, with T =2T0, thePoincaresectionwillhavetwopoints 

between which the system oscillates as multiples of T0 elapse. And so on. 

In contrast to the periodic situation, for chaotic motion a point is produced 

at a di®erent location at each multiple of T0, and the \sum" of the individual 

snapshots can produce strange, localized patterns (\strange attractors") of plot 

points with complex boundaries in the phase plane. 

In the following recipe, the \period-doubling route to chaos" is explored once 

again for the forced Du±ng oscillator, now from the Poincare section viewpoint. 

319

320 CHAPTER 8. NONLINEAR DIAGNOSTIC TOOLS 

8.1.1 A Rattler Signals Chaos 

Humor is emotional chaos remembered in tranquility. 

James Thurber, American writer, humorist, and cartoonist (1894{1961) 

As an illustrative example of how a Poincare section is produced and how it 

changes character as a control parameter is varied, let's consider Du±ng's ODE 

describing the force oscillations of a nonlinear spring system, 

Äx +2g _x + ®x+ ¯x 3 = F cos(!t): (8.1) 

Here x is the displacement, g the damping coe±cient, ® and ¯ real parameters, 

F the force amplitude, and ! the driving frequency. 

The solutions of Du±ng's equation were examined in some detail in Section 

1.2.1 as the control parameter F was varied, all other parameters being 

held ¯xed. In this recipe, we shall use exactly the same coe±cient values and 

initial conditions as in the earlier treatment, including the same F values. This 

will allow us to compare the results of the Poincare treatmentwiththeconclusions 

reached previously on the response of the forced Du±ng oscillator. 

The number of F values is taken to be N1 = 4, and the maximum number 

of multiples of the driving period T0 =2¼=! considered is N2 =250. 

> restart: with(plots): N1:=4: N2:=250: 

The coe±cient values are entered and the driving period calculated. 

> g:=0.25: alpha:=-1: beta:=1: omega:=1: T[0]:=2*Pi/omega; 

T0 := 2 ¼ 

The force amplitudes are F1 =0:325, F2 =0:35, F3 =0:356, and F4 =0:42. 

> F[1]:=0.325: F[2]:=0.35: F[3]:=0.356: F[4]:=0.42: 

Du±ng's equation can be rewritten as a coupled set of ¯rst-order ODEs, 

_x = y; _y = ¡2 gy¡ ®x¡ ¯x 3 + Fi cos(!t); 

which will be numerically solved with the initial conditions x(0)=0:09, y(0)=0. 

> ic:=x(0)=0.09,y(0)=0: 

The coupled ¯rst-order di®erential equations are entered, an operator being 

used for the second one, the subscript i of the force amplitude Fi to be given. 

> de1:=diff(x(t),t)=y(t): 

> de2:=i->diff(y(t),t)=-2*g*y(t)-alpha*x(t)-beta*x(t)^3 

+F[i]*cos(omega*t): 

An operator sol is introduced to numerically solve de1 and de2 (i), subject to 

the initial conditions, for a given value of i. The option maxfun=0 overrules any 

limit on the maximum number of function evaluations in Maple's numerical 

algorithm. The output is given as a listprocedure. 

> sol:=i->dsolve(fde1,de2(i),icg,fx(t),y(t)g,type=numeric, 

maxfun=0,output=listprocedure): 

Operators X and Y are formed to evaluate x(t) andy(t) 

> X:=i->eval(x(t),sol(i)): Y:=i->eval(y(t),sol(i)):

8.1. THE POINCAR ESECTION 321 

for the ith solution, the evaluations being carried out in the following do loop. 

> for i from 1 to N1 do xx[i]:=X(i); yy[i]:=Y(i); end do: 

An operator Gr is created to graph a phase-plane point, represented by a size-16 

blue cross, at time t = nT0 for the ith amplitude. 

> Gr:=(i,n)->pointplot(f[yy[i](n*T[0]),xx[i](n*T[0])]g 

color=blue,symbol=CROSS,symbolsize=16): 

Using this graphing operator in the following loop generates the Poincare section 

for each of the N1 = 4 amplitude values. To eliminate the transient part of the 

solution, a certain number of initial points must be removed in each plot. This 

number will vary for each numerical run and must usually be determined by 

trial and error. Here, the ¯rst 24 points have been removed. Since N2 =250, 

each plot still contains 225 points. A suitable viewing box is selected, which is 

the same for all four plots, and labels added and the minimum number of tick 

marks controlled. 

> for i from 1 to N1 do 

> display([seq(Gr(i,n),n=25..N2)],axes=boxed,view= 

[-0.4..0.8,-1.4..1.4],labels=["y","x"],tickmarks=[3,3]); 

> end do; 

1 

x 

–1 

0 y 0.5 

1 

x 

–1 

0 y 0.5 

Figure 8.1: Period-1 (left ¯gure) and period-2 (right) Poincare sections. 

The plot on the left of Figure 8.1 corresponds to F1 =0:325, the one on the right 

to F2 =0:35. Despite the fact that 225 points were plotted, we see only a single 

cross in the F1 =0:325 graph, indicating that the system has settled down to 

a period-1 solution. For F2 =0:35, the ODE system oscillates back and forth 

between the two crosses, the Poincare section being characteristic of a period-2 

solution. The observed periodicities agree with the results in Section 1.2.1. 

For F3 =0:356, Figure 8.2 indicates a period-4 solution, again agreeing 

with our earlier conclusion about the periodicity for this forcing amplitude. 

So it is clear that the Poincare section approach is a useful graphical tool for 

interpreting the periodicity of driven oscillator systems.


1 

x 

–1 

0 y 0.5 

Figure 8.2: Period-4 Poincare section. 

Finally, for F4 =0:42, we had previously observed a localized chaotic trajectory, 

characteristic of a strange attractor. The corresponding \strange" 

Poincare section is shown in Figure 8.3. 

x 

1 

0 

–1 

0 y 0.5 

Figure 8.3: Chaotic \rattler." 

Strange attractors are often given colorful or descriptive names. What name 

shallwegivetothechaoticPoincare section obtained above? This depends on 

the reader's experience and imagination. To coauthor Richard, it reminds him 

of an incident that occurred while he was hiking with his family in the Superstition 

Mountains of southern Arizona. While scrambling down a rocky scree 

slope, he had a feeling of impending chaos when his youngest daughter nearly 

stumbled into a coiled rattlesnake. To the author, this particular Poincare sec-


tion resembles that rattler poised to strike. Fortunately, the snake was just as 

frightened as the author's daughter and slithered away without striking. Some 

years later one could joke about the incident, but it was certainly not funny at 

the time. 

PROBLEMS: 

Problem 8-1: A di®erent ¯ value 

Holding all other parameter values as in the text recipe, use the Poincare section 

to determine the response of Du±ngs's ODE for each F value when ¯ =2. 

Problem 8-2: Varying the frequency 

For each of the four Fi,exploretheresponseoftheDu±ngODEasthefrequency 

! is varied, all other parameters being the same as in the text recipe. 

Problem 8-3: Interchanging signs 

Determine the response of the Du±ng ODE for each Fi when all numerical values 

are the same as in the text recipe, but the signs of ® and ¯ are interchanged. 

Problem 8-4: Periodicity? 

Using the Poincare section approach, determine the periodicity of the steadystate 

solution of the following forced oscillator equations: 

(a) Äx +0:7 _x + x3 =0:75 cos t; with x(0) = _x(0) = 0; 

(b) Äx +0:08 _x + x3 =0:2cost; with x(0) = 0:25; _x(0) = 0. 

Problem 8-5: Varying force amplitude 

Using the Poincare section approach, determine the periodicity of the steadystate 

solution of the following ODE for F =0:357 and F =0:35797: 

Äx +0:5 _x ¡ x + x 3 = F cos(t +1); with x(0) = 0:09; _x(0) = 0: 

8.1.2 Hamiltonian Chaos 

Progress everywhere today does seem to come so very heavily 

disguised as Chaos. 

Joyce Grenfell, British actor, writer, Stately as a Galleon, 1978 

In the Hamiltonian formulation of classical mechanics the motion of a single 

particle of unit mass, with coordinates qi and (generalized) momenta pi, moving 

in a conservative potential V (qi) can be described by the Hamiltonian, 

NX 1 

H = 

2 

i 

p2i + V (qi); (8.2) 

where the ¯rst term is the kinetic energy and N is the number of degrees of 

freedom. Hamilton's equations of motion then are given by 

_qi = @H 

; _pi = ¡ 

@pi 

@H 

: (8.3) 

@qi


For N = 1, the motion can be described by a trajectory in the q1 vs. p1 phase 

plane. For N>1, the trajectory is in a 2N-dimensional phase space. 

Originally motivated to study the motion of a star inside a galaxy, Henon 

and Heiles [HH64] introduced a conservative Hamiltonian describing the motion 

of a unit mass in the two-dimensional potential 

V = 1 

2 q2 1 + 1 

2 q2 2 + q 2 1 q2 ¡ 1 

3 q3 2: (8.4) 

The ¯rst two terms in V would generate a paraboloid of revolution characteristic 

of a two-dimensional harmonic oscillator. The force ~ F = ¡rV in this case is 

just the two-dimensional form of Hooke's law. The inclusion of the two cubic 

terms in V distort the shape of the potential away from a paraboloid, add 

nonlinear terms to Hooke's law, and Hamilton's equations generate nonlinear 

ODEs in the 4-dimensional phase space. 

In this recipe, we shall use specialized commands found in the DEtools 

library package to generate these equations, numerically solve them to produce 

the trajectory for a speci¯ed energy and initial conditions, and produce a 

Poincare section. 


Entering the Henon{Heiles potential (8.4), a two-dimensional contour plot is 

generated, the contour lines given by V =0:04 i with i = 0 to 9. To obtain 

smooth curves, the number of plotting points is taken to be 5000. 

> V:=q1^2/2+q2^2/2+q1^2*q2-q2^3/3: 

> contourplot(V,q1=-2..2,q2=-2..2,contours=[seq(0.04*i, 

i=0..9)],numpoints=5000,color=black); 

q2 

2 

1 

–2 –1 0 

1 q1 2 

–1 

–2 

Figure 8.4: Contour plot of the Henon{Heiles potential.


To aid in interpreting the contour plot, let's ¯nd the stationary points where the 

force vanishes. Di®erentiating V with respect to each coordinate and solving 

for the coordinate values that make the force components equal to zero, 

> sol:=solve(fdiff(V,q1),diff(V,q2)g,fq1,q2g); 

sol := fq2 =0; q1 =0g; fq2 =1; q1 =0g; 

fq2 = ¡1 1 

; q1 = 

2 2 RootOf(¡3+ Z 2 ; label = L2 )g 

yields a stationary point at q1 = q2 =0andq1 =0; q2 = 1, as well as others 

at q2 = ¡ 1 and q1 given by the RootOf placeholder. These latter values may 

2 

be found by applying the allvalues command to the third entry in sol. 

> sol3:=allvalues(sol[3]); 

p p 

3 ¡1 ¡1 3 

sol3 := fq1 = ; q2 = g; fq2 = ; q1 = ¡ 

2 2 2 2 g 

1 ; q2 = ¡ 2 2 and at q1 = ¡ p 3; 

q2 = 2 

. The four stationary points are now extracted separately and labeled, 

> s1:=sol[1]; s2:=sol[2]; s3:=sol3[1]; s4:=sol3[2]; 

There are two more ¯xed points at q1 = p 3 

¡ 1 

2 

s1 := fq2 =0; q1 =0g s2 := fq2 =1; q1 =0g 

p p 

3 ¡1 

¡1 3 

s3 := fq1 = ; q2 = g s4 := fq2 = ; q1 = ¡ 

2 2 2 2 g 

and the potential energy at each stationary point determined. 

> U:=v->eval(V,v): V1:=U(s1); V2:=U(s2); V3:=U(s3); V4:=U(s4); 

V1 := 0 V2 := 1 1 1 

V3 := V4 := 

6 6 6 

The stationary point s1 at the origin is the minimum of what would be a 

parabolic potential well if the cubic terms were not present in V . Referring 

to the contour plot, the shape of the contour lines changes as one moves away 

from the origin, the contour lines near the other three stationary points being 

characteristic of saddle points. 1 If the particle has a total energy below the 

potential energy 1 

6 at the saddle point and starts inside the region bounded 

by the three saddle points, it will have a bounded orbit inside this region. If 

E> 1 

6 , the particle could escape through one of the saddle points to in¯nity. 

The Hamiltonian is entered, and the command hamilton eqs used to generate 

Hamilton's equations and a list of the four dependent variables. 

> H:=p1^2/2+p2^2/2+V; 

> hamilton_eqs(H); 

[ d 

d 

p1(t) =¡q1 (t) ¡ 2 q1(t) q2 (t); 

dt dt p2 (t) =¡q2 (t) ¡ q1 (t)2 + q2 (t) 2 ; 

d 

d 

q1 (t) =p1(t); q2 (t) =p2 (t)]; [p1(t); p2 (t); q1 (t); q2 (t)] 

dt dt 

1 You can con¯rm this by making a 3-dimensional contour plot using the plot3d command.


The result is a nonlinear system of four coupled ODEs, which must be solved 

numerically. Let's take as initial conditions q1(t =0)=q10 = ¡0:1, q2(0) = 

q20 = ¡0:2, p2 (0) = p20 = ¡0:05, and a total energy E0 =1=16. This will 

produce a bounded orbit inside the region bounded by the saddle points. 

> q10:=-0.1: q20:=-0.2: p20:=-0.05: E0:=1/16: 

Note that it is not necessary to specify the initial value of p1, since it will be 

determined by energy conservation. In fact, p1(0) can be obtained by entering 

the following generate ic command and asking for one solution. 

> ic:=generate_ic(H,ft=0,p2=p20,q2=q20,q1=q10,energy=E0g,1): 

ic := f[0:; 0:2667708130; ¡0:05; ¡0:1; ¡0:2]g 

The second entry in the output tells us that p1(0) ¼ 0:267. If desired, more initial 

conditions can be generated by stating, e.g., a range for q2(0) and specifying 

the number of p1 (0) values wanted. 

The following poincare command uses a fourth-order Runge{Kutta method 

with three iterations and a step size of 0:05 to produce a trajectory over the 

time interval t =0to300intheq2 vs. p2 vs. q1 phase space. The last number 

speci¯es that a 3-dimensional plot is to be produced. 

> poincare(H,t=0..300,ic,stepsize=.05,iterations=3, 

scene=[q2=-0.4..0.4,p2=-0.4..0.4,q1=-0.4..0.4],3); 

H = :62500000e ¡ 1 ; Initial conditions :; t=0:; 

p1 =0:2667708130; p2 = ¡0:05; q1 = ¡0:1; q2 = ¡0:2; 

Maximum H deviation : :1020000000e ¡ 5 % 

q1 

0.4 

0.2 

0 

–0.2 

–0.4 

–0.4 

Time consumed : 19 seconds 

0 

p2 

H = .6250000000e–1; 

0.4 0.2 0 

q2 

–0.2 

–0.4 

Figure 8.5: Quasiperiodic trajectory for E =1=16.


The output informs us of the input H (energy) value, the initial conditions, 

the maximum percentage deviation from the input H value, and the computer 

time taken to produce a plot of the trajectory, which is shown in Figure 8.5. 

The trajectory executes quasiperiodic motion on the surface of a twisted torus, 

referred to as the Kolmogorov{Moser{Arnold (KAM) torus. 

The Poincare section for the q1 = 0 plane shown in Figure 8.5 can be 

obtained with the following poincare command. In addition to the same information 

as before, the number of points (96 here) in the q2 -p2 plane where 

the trajectory crosses the plane is also given. 

> poincare(H,t=0..300,ic,stepsize=.05,iterations=3, 

scene=[q2,p2]); 

H = :62500000e ¡ 1 ; Initial conditions :; t=0:; 

p1 =0:2667708130; p2 = ¡0:05; q1 = ¡0:1; q2 = ¡0:2 

Number of points found crossing the (q2; p2) plane : 96 

Maximum H deviation : :9800000000e ¡ 6 % 

p2 

0.2 

0.1 

0 

–0.1 

–0.2 

Time consumed : 1 seconds 

H = .6250000000e–1; 

–0.3 –0.2 –0.1 0 0.1 0.2 0.3 

q2 

Figure 8.6: Poincare section for E =1=16. 

The Poincare section can be qualitatively understood. One can think of the 

KAM torus as a twisted \donut," with the trajectories con¯ned to the donut's 

surface. The con¯guration of points in the Poincare section resembles two 

distorted ellipses that one might expect from \slicing" the donut through its 

cross section with the q1 =0plane. 

If the total energy is increased to 1 

8 and the viewing box increased in size, 

the \chaotic" trajectory shown in Figure 8.7 results.


q1 

0.6 

0.4 

0.2 

0 

–0.2 

–0.4 

–0.6 

–0.5 

0 

p2 

H = .1250000000; 

0.6 0.4 0.2 0 –0.2 

q2 

–0.4 

–0.6 

Figure 8.7: Chaotic trajectory for E = 1 

8 . 

A great deal of mathematics research has gone into understanding the onset 

of chaos for the nonlinear Henon{Heiles ODE system. The interested 

reader is referred to the nonlinear dynamics texts by Jackson [Jac90] and by 

Hilborn [Hil94]. 

PROBLEMS: 

Problem 8-6: A di®erent potential 

Replace the potential in the text recipe with 

V = 1 

2 q2 1 

1 + 

2 q2 2 + q4 1 q2 ¡ 1 

4 q3 2 : 

(a) Execute the modi¯ed recipe with E =1=16 and initial conditions as in 

the text recipe. Discuss the resulting plots. 

(b) Explore other initial conditions for the same total energy as in part (a). 

Discuss the results. 

(c) Explore what happens as the energy is increased with the same initial 

conditions as in part (a). Discuss the results. 

(d) Explore other potential energy functions and discuss the results. 

Problem 8-7: Toda potential 

The Toda potential [Jac90] is given by 

V = 1 

μ 

e 

3 

(q2 + p 3 q1) (q2 + e 

¡ p 

3 q1) (¡2 q2) 

+ e ¡ 1: 

(a) Create two- and three-dimensional contour plots of V , choosing suitable 

potential energy contours and viewing ranges.

8.2. THE POWER SPECTRUM 329 

(b) Locate and identify the nature of the stationary points. 

(c) Generate the Hamiltonian equations. 

(d) For E =1:0, p2(0) = ¡0:05, q2(0) = ¡0:2, q1(0) = ¡0:2, determine p1(0). 

(e) Plot the system's trajectory in the q1-q2-p2 space and discuss the result. 

(f) Generate the Poincare section in the q2 vs. p2 plane. Interpret the result. 

(g) Explore the Toda Hamiltonian for other E values and discuss the results. 

8.2 The Power Spectrum 

Still another important diagnostic tool is the power spectrum, which, although 

perhaps better known for its use by engineers in digital signal processing [SK89], 

can be adapted to studying the frequency content of a solution x(t) of a forced 

linear or nonlinear oscillator equation. 

Suppose that in principle, a nonlinear ODE of physical interest has the timedependent 

solution x(t), valid for all t (¡1


evaluate the Fourier transform of x(t) andthenS(f), this procedure would lead 

to an inordinately long computation time. In practice, one tries to obtain an 

accurate power spectrum using a limited number of x points. How this is done 

is now explained in some detail. 

Assume that a sequence of N values of x, viz., xn ´ x(tn = nTs) with 

n =0; 1; 2; :::; N¡ 1 is recorded at evenly spaced time intervals Ts over some 

¯nite time range. As with the Poincare section analysis, one starts recording at 

a su±ciently large time t0, so as to ensure that all transients have died away. 

How large this time must be depends on the nature of the forced oscillator, the 

parameter values, initial conditions, etc., and is determined by trial and error. 

For any choice of the sampling time interval Ts, there is a very special corresponding 

frequency, fNyquist =1=(2 Ts), called the Nyquist frequency. Notethat 

the sampling frequency Fs =1=Ts is twice the Nyquist frequency. Why is the 

Nyquist frequency important? The answer lies in the sampling theorem, dueto 

Nyquist [Nyq28] and Shannon [Sha49], which states: 

If a continuous signal x(t), sampled at an interval Ts, issuchthatitsFourier 

transform X(f) is equal to 0 for all frequencies jfj >fNyquist, thenx(t) is completely 

determined by the sampled values xn. 

In this case, the Nyquist frequency is clearly greater than the maximum frequency 

f max in the signal's frequency spectrum. Thus, Fs > 2 f max. 

In signal processing, engineers ensure that the sampling theorem prevails by 

using a low-pass analog ¯lter on their signal to select f max(< f Nyquist), removing 

all higher frequencies by the process of attenuation. When X(f) is zero outside 

the range ¡f Nyquist to f Nyquist, theyrefertox(t) asbeingbandwidth limited to 

frequencies smaller in magnitude than f Nyquist. 

What happens if x(t) is not bandwidth limited to this frequency range, 

i.e., its Fourier transform X(f) does not vanish outside the range ¡f Nyquist to 

f Nyquist? It turns out that the power outside this frequency range gets \folded 

back" into the range giving an inaccurate power spectrum. This phenomenon is 

called aliasing. To avoid aliasing, one should attempt to make x(t) bandwidth 

limited by taking the sampling frequency Fs > 2 f max. Unfortunately, for forced 

oscillator problems it is usually not known a priori what the maximum frequency 

component is in the signal x(t). On the other hand, suppose that the chosen 

sampling frequency or sampling interval Ts =1=Fs is such that X(f) isnot 

zero at the Nyquist frequency. Then increase Fs to check for possible aliasing. 

Increasing Fs pushes the Nyquist frequency up, allowing us to see whether there 

are indeed higher-frequency components present. 

Keeping these important aspects in mind, let's continue with the formal 

derivation of the power spectrum from the N sampled values xn. It follows from 

elementary mathematics that given these N values, we can generate the Fourier 

transform at only N frequencies. Assuming that x(t) is bandwidth limited, we 

shall take these frequencies to be equally spaced between ¡f Nyquist and f Nyquist, 

viz., the frequencies fk ´ k=(N Ts) withk = ¡N=2; :::;0; :::; N=2. The 

extreme k values generate ¡f Nyquist and f Nyquist. It might seem on counting the


k values that we have N + 1 of them, but due to periodicity of the Fourier 

transform, the extreme k values are not independent, but in fact are equal. 

We now approximate the continuous Fourier transform as follows, 

Z 1 

X(fk) = x(t) e ¡2 ¼Ifk t dt ¼ 

¡1 

N¡1 X 

n=0 

where Xk is the discrete Fourier transform, 

Xk ´ 

N¡1 X 

xn e ¡2 ¼Ifk tn Ts = Ts Xk; 

xn e ¡2 ¼Ikn=N : (8.7) 

n=0 

We can change the k range from ¡N=2; :::; N=2to0;:::;N¡ 1, thus making 

it the same as the n range, by noting that Xk is periodic in k with period N. 

With this standard convention for the range of k, k =0correspondstozero 

frequency, k =1; 2; :::; N=2¡1 to positive frequencies, k = N=2+1; :::; N¡1 

to negative frequencies, and k = N=2 tobothfNyquist and ¡fNyquist. Because of 

symmetry of the power spectrum about k = N=2, it su±ces to plot only the 

positive frequency range, which is what we will do in our recipes. 

In a similar manner, the inverse discrete Fourier transform can be derived: 

xn = 1 

N¡1 X 

N 

k=0 

Xk e 2 ¼Ikn=N : (8.8) 

From the discrete Fourier transform pair, Parseval's theorem then becomes 

N¡1 X 

n=0 

jxnj 2 = 1 

N 

N¡1 X 

k=0 

jXkj 2 ´ 

N¡1 X 

k=0 

SN (k); (8.9) 

where SN (k) =jXkj 2 =N is the power spectrum. 

In calculating the discrete Fourier transform Xk, we will make use of the fast 

Fourier transform (FFT), which is the default numerical algorithm [Bri74] in 

the FourierTransform command found in Maple's DiscreteTransforms library 

package. The FFT is based on the idea of splitting the data set in the discrete 

Fourier transform into even- and odd-labeled points and using the periodicity 

of the exponential function to eliminate redundant operations. A detailed discussion 

of this conversion may be found in standard numerical analysis texts, 

for example in Burden and Faires [BF89] and in Numerical Recipes [PFTV89]. 

Why use this routine? As suggested by the process of eliminating redundant 

calculations, the FFT is signi¯cantly faster than the straightforward evaluation 

of the discrete Fourier transform. 

How much faster? A lot! If N is the number of data points, the discrete 

Fourier transform involves N 2 multiplications, while the FFT turns out 

to involve about N log 2(N) operations. If, for example, N = 10000 as in the 

following recipe, the \normal" discrete Fourier transform requires (10 4 ) 2 =10 8 

computations compared to about 10 4 £ log 2(10 4 ) ¼ 10 5 for the FFT. In this 

case the FFT is about 1000 times faster.


8.2.1 Frank N. Stein's Heartbeat 

The more powerful and original a mind, the more it will incline 

towards the religion of solitude. 

Aldous Huxley, British writer (1894{1963) 

Let's ¯rst illustrate how the power spectrum is calculated for a simple example. 

We are given Frank N. Stein's steady heartbeat described by 

2X 

x = Ai sin(2 ¼fit); i=1 

with fundamental frequency f1 = 1 beat per second (60 beats per minute) and 

a small second harmonic f2 = 2 beats per second. The suitably normalized 

amplitudes are A1 =1andA2 =0:4. Pretending that we do not know the 

frequencies, we will extract them from the power spectrum. (We could be 

dealing with experimental data, not a known analytic form.) 

To calculate the discrete Fourier transform, the DiscreteTransforms package 

must ¯rst be loaded. The number of sampling points is taken to be N = 1000. 

The numerical factor d = 10 will be used to determine the sampling time. 

> restart: with(DiscreteTransforms): N:=1000: d:=10: 

The parameter values are entered and the time T2 =1=f2 is calculated. 

> A[1]:=1: A[2]:=0.4: f[1]:=1: f[2]:=2.0: T[2]:=1/f[2]; 

T2 := 0:5000000000 

An operator is formed to calculate x at a speci¯ed time t. 

> x:=t->add(A[i]*sin(2*Pi*f[i]*t),i=1..2): 

Then, x(t) is plotted over the time interval t =0todT2. 

> plot(x(t),t=0..d*T[2],labels=["t","x"]); 

x 

1 

0.5 

0 

–0.5 

–1 

1 2 3 4 5 

t 

Figure 8.8: Frank N. Stein's heartbeat. 

If one were given only the plot shown in Figure 8.8, it would not be obvious 

exactly what frequencies are contained in Frank's heartbeat. The power spec-


trum will now reveal what they are. The sampling time interval is taken to be 

Ts = T2=d and the sampling frequency Fs =1=Ts is calculated. 

> T[s]:=T[2]/d; F[s]:=1/T[s]; 

Ts := 0:05000000000 Fs := 20:00000000 

The continuous function x(t) is sampled at times t = nTs with n =0toN ¡ 1. 

The sequence of sampled x values is then put into an array format. 

> x:=Array([seq(x(n*T[s]),n=0..N-1)]): 

The (discrete) Fourier transform of x is performed. Maple de¯nes this transform 

such that the kth item in FT is Xk= p N in our notation. 

> FT:=FourierTransform(x): 

The following operator will allow us to calculate S(k) foragivenk value. 

> S:=k->abs(FT[k])^2: 

A plotting point operator pt is introduced that gives the power spectrum S(k) 

at the frequency (k ¡ 1) Fs=N . 

> pt:=k->[F[s]*(k-1)/N,S(k)]: 

Taking the power spectrum to be zero at zero frequency, we produce the points 

in the power spectrum for k ranging up to N=2. 

> pts:=[[0,0],seq(pt(k),k=2..N/2)]: 

The complete power spectrum is plotted, and shown in Figure 8.9. There are 

clearly two frequencies present at 1 and 2 beats per second, as expected. 

> plot(pts,labels=["f","S"],view=[0..5,0..0.3]); 

300 

250 

S 

150 

100 

50 

0 

1 2 3 4 5 

f 

Figure 8.9: Frequencies in Frank N. Stein's heartbeat. 

PROBLEMS: 

Problem 8-8: Third Harmonic 

Suppose that Frank N. Stein's heartbeat also contains the term 0:1 sin(6¼t). 

Plotting p S, show that the presence of the third harmonic is revealed.


8.2.2 The Rattler Returns 

You cannot have power for good without having power for evil too. 

Even mother's milk nourishes murderers as well as heroes. 

George Bernard Shaw, Anglo-Irish playwright, Cusins, in Major Barbara, act3 

To illustrate how the power spectrum is calculated for a nonlinear ODE, let's 

consider the Du±ng oscillator of the previous section with exactly the same parameter 

values and initial conditions. We begin by loading the DiscreteTransforms 

package and taking the number of sample values of x to be N = 10000. 

> restart: with(DiscreteTransforms): N:=10000: 

The Du±ng oscillator ODE is entered, 

> ode:=diff(x(t),t,t)+ 2*g*diff(x(t),t) + alpha*x(t) 

+ beta*x(t)^3=F*cos(omega*t); 

μ μ 2 d d 

ode := x (t) +2g 

dt2 dt x(t) 

 

+ ® x(t)+¯ x (t) 3 = F cos(!t) 

along with the parameter values (taking F =0:325 here) and initial conditions. 

> g:=0.25: alpha:=-1: beta:=1: omega:=1; F:=0.325; #change F 

! := 1 F := 0:325 

> ic:=x(0)=0.09,D(x)(0)=0: 

The force amplitude F will be adjusted to the other values, F =0:35; 0:356; 0:42, 

used previously. The Du±ng oscillator equation is numerically solved, 

> sol:=dsolve(fode,icg,fx(t)g,numeric,maxfun=0, 


and the solution used to evaluate x(t) for arbitrary time t. 

> X:=eval(x(t),sol): 

The driving frequency is ! =1 rad/s, or f =!=(2 ¼)=1=(2 ¼) Hz. The sampling 

frequency fs is taken to be 4 f and the sampling time Ts =1=fs is calculated. 

> f:=omega/(2*Pi): f[s]:=4*f; T[s]:=1/f[s]; #frequencies in Hz 

fs := 2 

Ts := 

¼ 

¼ 

2 

To eliminate the transient solution, the x values are recorded starting at t0 = 

50 ¼ seconds and sampled every Ts seconds up to time (N ¡ 1) Ts. Thesevalues 

are entered with the Array command. 

> x:=Array([seq(X(50*Pi+T[s]*i),i=0..N-1)]): 

The FourierTransform command is applied to the sampled x values. This 

command calculates the discrete Fourier transform (8.7), divided by p N,using 

the fast Fourier transform as the default algorithm. 

> FT:=FourierTransform(x): 

The following operator F enables us to extract the ith term in FT and, since 

the result is generally complex, take the absolute value. Because of Maple's 

de¯nition of the discrete Fourier transform, this is equivalent to taking the


square root of the power spectrum for the ith frequency point. We take the 

square root here because it helps to accentuate smaller peaks in the spectrum. 

> F:=i->abs(FT[i]): #sqrt of S 

An operator pt is introduced to form the ith plotting point for the power 

spectrum, the frequency being expressed in radians per second so a comparison 

can be easily made with the driving frequency ! =1 rad/s. 

> pt:=i->[2*Pi*f[s]*(i-1)/N,F(i)]: 

With the aid of the pt operator, the plotting points are formed into a list of 

lists, with the zero frequency value of the power spectrum taken as zero. Since 

the power spectrum is symmetric about N/2, only the points up to this number 

will be plotted. In the present case, 5000 points will be included. The point, 

e.g., i = 2501 corresponds to an angular frequency of 1 rad/s, for which S ¼ 29. 

> pts:=[[0,0],seq(pt(i),i=2..N/2)]: pts[2501]; 

[1; 28:82963799] 

The points are plotted and labels are added, 

> plot(pts,tickmarks=[3,2],labels=["omega","S"]); 

S 

20 

0 

1 omega 2 

S 

20 

0 

1 omega 2 

Figure 8.10: Power spectrum for period-1 (left) and period-2 (right). 

the power spectrum for F =0:325 being shown in the left plot of Figure 8.10. 

This spectrum shows a single sharp spike located at a frequency of 1 rad/s, i.e., 

exactly at the driving frequency. The period is T =2¼=! =2¼ seconds, so the 

response is period 1, in agreement with the conclusion reached earlier for the 

same Du±ng oscillator on the basis of a Poincare section analysis. 

The plot on the right of Figure 8.10 results on executing the work sheet 

with F =0:35. In addition to the tallest spike at the driving frequency, there 

is a second spike to the left, located at the frequency !=2=0:5 rad/s. This 

corresponds to a period T =2¼=(!=2) = 4 ¼ seconds, i.e., twice as long as 

for F =0:325. The spectrum is characteristic of a period-2 solution. Note 

that there is also a spike in the power spectrum to the right of the driving


frequency, occurring at 1:5 rad/s. This is the third harmonic of the period-2 

frequency. Harmonics are often present in power spectra, so remember that 

only those spikes lying to the left of the driving frequency spike should be 

used to determine the periodicity. It is because these harmonics are present 

that one must be sure to use a su±ciently high sampling frequency so that 

these harmonics do not get \folded" back (aliased) into the frequency range of 

interest. 

On increasing F to 0:356, one sees three spikes to the left of the driving 

frequency in the left plot of Figure 8.11. The spike furthest to the left is located 

at !=4 =0:25 rad/s, corresponding to a period of 8 ¼ seconds. The spectrum 

is characteristic of a period-4 response. Again, harmonics of these frequencies 

appear to the right of the driving frequency. 

S 

20 

0 

1 omega 2 

10 

S 

0 

1 omega 2 

Figure 8.11: Power spectrum for period-4 (left) and chaotic response (right). 

Finally, taking F =0:42 produces the power spectrum shown on the right of 

Figure 8.11. Although a sharp spike is still seen at the driving frequency, the 

spectrum is very chaotic, which is not surprising, since it corresponds to the 

chaotic \rattler" Poincare section seen earlier. 

Recalling Richard's fright when his daughter nearly stumbled and rolled 

over a real rattlesnake, one might fancifully interpret this power spectrum as 

representing the rattling of his teeth or the shaking of his knees at the time of 

the incident. 

PROBLEMS: 

Problem 8-9: A di®erent ¯ value 

Holding all other parameter values as in the text, use the power spectrum to 

determine the response of the forced Du±ng ODE for each F value when ¯ =2. 

Problem 8-10: Varying the frequency 

For each of the four F values, use the power spectrum to explore the response 

of the Du±ng oscillator as ! is varied, all other parameters being unchanged.

8.3. THE BIFURCATION DIAGRAM 337 

Problem 8-11: Interchanging signs 

Use the power spectrum to determine the response of the forced Du±ng oscillator 

for each F value when all numerical values are the same as in the text 

recipe, but the signs of ® and ¯ are interchanged. 

Problem 8-12: Steady-state solution? 

Using the power spectrum approach, determine the nature of the steady-state 

solution for the following forced oscillator equation: 

Äx +0:7 _x + x 3 =0:75 cos t; x(0) = _x(0) = 0: 

Problem 8-13: Solution? 


solution for the following forced Du±ng equation: 

Äx +0:08 _x + x 3 =0:2cost; x(0) = 0:25; _x(0) = 0: 

Problem 8-14: Solutions? 


solution for the following oscillator equation for F =0:357 and F =0:35797: 

Äx +0:5 _x ¡ x + x 3 =0:357 cos(t +1); x(0) = 0:09; _x(0) = 0: 

Problem 8-15: Forced glycolytic oscillator 

The equations describing forced oscillations of the glycolytic oscillator are 

_x = ¡x + ®y+ x 2 y; _y = ¯ ¡ ®y¡ x 2 y + A + F cos(!t): 

Taking ® = ¯ =0,A =0:999, F =0:42, x(0) = 2, and y(0) = 1, determine the 

periodicity of the response using the Poincare sectionapproachfor(a)! =2 

and (b) ! =1:75. Explore the frequency range in between and identify any 

interesting solutions. 

8.3 The Bifurcation Diagram 

Bifurcation diagrams can be generated for both nonlinear ODEs and nonlinear 

di®erence equations by plotting the system \response" versus a \control" parameter 

as the latter is varied. The word bifurcation is derived from the Latin 

word furca for fork. When period doubling occurs from period one to period 

two, the response curve resembles a two-pronged fork, the period-one portion 

being the \handle" and the period-two portion looking like two \prongs." In a 

typical period-doubling scenario, the two prongs then split into four and then 

into eight and so on as the control parmeter is further increased. Period doubling 

is not the only \route" to chaos, so bifurcation diagrams are useful in 

revealing the nature of the route.


8.3.1 Pitchforks and Other Bifurcations 

Though you drive away nature with a pitchfork, she always returns. 

Horace, Roman poet known for his odes (65{8 BC) 

Consider the following di®erence equation, known as the logistic map, 

xn+1 = axn (1 ¡ xn); (8.10) 

where n =0; 1; 2; :::;N and a is allowed to vary between 0 and 4. As a is 

increased over its range, a period-doubling sequence to chaos occurs that can 

be illustrated by plotting xn (at large N to eliminate the transient) versus a. 

The ¯rst part of the following recipe illustrates a period-2 solution of the 

logistic map for a =3:2, N = 119, and initial value x0 =0:1. 

> restart: a:=3.2: N:=119: x[0]:=0.1: 

An operator F is formed to calculate the rhs of the logistic map for a given x. 

> F:=x->a*x*(1-x): 

The logistic map is iterated from n =0toN, 


> x[n+1]:=F(x[n]); 

> end do: 

and the sequence of points [n; xn] plotted, each point being represented by a 

size-12 black circle. The resulting picture is shown in Figure 8.12. 

> plot([seq([n,x[n]],n=0..N)],style=point,symbol=circle, 

symbolsize=12,color=black,labels=["n","x"]); 

0.8 

0.7 

x 

0.5 

0.4 

0.3 

0.2 

0.1 

0 20 40 60 n 100 120 

Figure 8.12: Period-2 solution for the logistic map.


After a short transient interval, the logistic system oscillates back and forth 

between the two branches shown in the ¯gure. This is characteristic of a period- 

2 solution, there being only one branch in steady state for a period-1 solution. 

As a is increased, further bifurcations occur. 

To create a bifurcation diagram for the logistic map, a particular initial 

value of x0 is chosen, and the map iterated for a given a. A certain number of 

the initial points are thrown away to remove the transient part of the solution, 

and the subsequent steady-state points plotted. Then, one increments a by a 

small amount and repeats the process, and so on. The whole process is easily 

automated, as is now illustrated. 


As you may con¯rm by running the ¯rst part of the recipe, a period-1 solution 

prevails until a =3:0, at which point period two begins. In the recipe, the range 

of a is taken from the starting value Sa =2:9 uptothe¯nalavalue Fa =4. If 

this a range is divided into, say, N = 200 equal intervals, 

> Sa:=2.9: Fa:=4: N:=200: stepsize:=(Fa-Sa)/N; x:=0.2; 

stepsize := 0:005500000000 x := 0:2 

the stepsize is ¢a =0:0055. The initial value of x has been taken to be x =0:2. 

A total of 500 iterations will be considered and the ¯rst 100 points ignored 

in order to remove any transients. This leaves 400 points to be plotted for each 

a value. The quantity c is a counter to keep track of the plots, a graph being 

produced for each a value. The counter is initially set equal to zero. 

> totalpts:=500: ignorepts:=100: pts:=totalpts-ignorepts; c:=0: 

pts := 400 

In the following double do loop, the ¯rst, or outer, loop increments a from the 

starting value Sa to the ¯nal a value Fa in incremental steps given by stepsize. 

> for a from Sa to Fa by stepsize do 

The second, or inner, do loop iterates the logistic equation from 1 to totalpts. 

> for n from 1 to totalpts do 

> x:=a*x*(1-x); 

The points for a given a value are formed into a list, 

> pts[n]:=[a,x]; 

> end do; 

and the inner do loop completed. Then, the counter is advanced by one, 

> c:=c+1; 

and a list of lists is made using the ¯nal 400 (presumably) steady-state points. 

> points:=[seq(pts[k],k=ignorepts..totalpts)]: 

Aplotismadeforeachavalue using a point style, 

> Gr[c]:=pointplot(points,symbol=point,color=black): 

> end do:


and the outer do loop ended. The N = 200 plots are superimposed with the 

display command, yielding the bifurcation diagram shown in Figure 8.13. 

x 

> display([seq(Gr[m],m=1..N)],view=[Sa..Fa,0..1], 

1 

0 

labels=["a","x"]); 

3 4 

a 

Figure 8.13: Bifurcation diagram for the logistic map for a =2:9 to4. 

Starting at a =2:9, the reader can observe period 1 occurring up to a =3:0. 

Then the steady-state response undergoes a so-called pitchfork bifurcation to 

period 2, followed by clearly seen bifurcations to period four, period eight, and 

a barely observable period-sixteen solution. At higher a values, the response is 

generally chaotic, but narrow periodic windows also occur. The reader should 

be able to see, for example, a period-3 solution for a ¼ 3:83. In this case, the 

bifurcation to period 3 is an example of a tangent bifurcation. 

Since the periodic windows are often very narrow in terms of the range of a, 

one should really increase the number N, but this leads to a longer computing 

time. A better approach is to leave N unchanged and zoom in on a particular 

range of a by changing the values of Sa and Fa. 

Bifurcation diagrams are very useful diagnostic tools for studying the behavior 

of nonlinear maps as well as forced oscillator ODEs as one or more control 

parameters are varied. 

A complementary graphical approach is to calculate the Lyapunov exponent 

as a function of a as illustrated in the next recipe.


PROBLEMS: 

Problem 8-16: Finer structure 

For the logistic map, produce the bifurcation diagram for the region a =3:54 to 

a =3:6, taking x0 =0:2, and dividing the a interval into 100 steps. Summarize 

the various periodic solutions that you observe. 

Problem 8-17: More ¯ne structure 

Explore the periodic window in the vicinity of a =3:8and report on what 

periodicities you observe at each a value sampled. 

Problem 8-18: Cubic map 

Produce a bifurcation diagram for the cubic map 

xn+1 = axn ¡ x 3 n 

over a suitable range of the parameter a. Determine the values of a in the 

diagram at which the periodicity changes. 

Problem 8-19: Quartic map 

With x0 =0:2, produce the bifurcation diagram for the quartic map 

xn+1 = axn (1 ¡ x 3 n ) 

over the range a =1:5 toa =2:0, taking as small an a step size as you can. 

Summarize the behavior of the map as a varies over the speci¯ed range. 

Problem 8-20: The tent map 

Produce the bifurcation diagram for the tent map 

xn+1 =2axn; 0


8.4 The Lyapunov Exponent 

Named after the Russian mathematician Aleksandr Mikhailovich Lyapunov 

(1857{1918), the Lyapunov exponent ¸ is a measure of the very sensitive dependence 

on initial conditions that is characteristic of chaotic behavior in nonlinear 

di®erence equations (maps) and ODEs. The discussion that follows is for maps, 

a similar analysis applying to ODEs. 

Consider a one-dimensional map 

xn+1 = f(xn); (8.11) 

with f some speci¯ed functional form, and let x0 and y0 be two initial values 

very close to each other. In phase space, they would be represented by two very 

close points. After n iterations, the values of xn and yn will be given by 

xn = f (n) (x0); yn = f (n) (y0); (8.12) 

where f (n) denotes the nth iteration of the map. Because the chaotic regime 

is typically characterized by an extreme sensitivity to initial conditions, for a 

chaotic situation nearby initial points will rapidly separate. On the other hand, 

periodic solutions are insensitive to initial conditions, and nearby initial points 

rapidly converge. 

This suggests that for su±ciently large n, one might assume an approximately 

exponential dependence on n of the separation distance, viz., 

jxn ¡ ynj = jx0 ¡ y0je ¸n ; (8.13) 

with ¸>0 for the chaotic situation and ¸

8.4. THE LYAPUNOV EXPONENT 343 

and the Lyapunov exponent ¸ is given by 

n¡1 ¯ ¯ 

1 X ¯ 

¸ = lim ln ¯ 

df (xk) ¯ 

n!1 n ¯ dxk 

¯ : (8.18) 

k=0 

For periodic solutions, which starting point x0 is chosen doesn't matter, but 

for chaotic trajectories, the precise value of ¸ will depend on x0, i.e., in general 

¸ = ¸(x0). One can, if desired, de¯ne an average ¸, averaged over all starting 

points. Whether this is done or not, ¸>0 should correspond to chaos and 

¸ restart: with(plots): numpts:=300: 

The range of a is taken from the starting value Sa =2:8 to the ¯nal a value 

Fa = 4, with the range divided into N = 480 equal steps. 

> Sa:=2.8: Fa:=4: N:=480: x:=0.2: stepsize:=(Fa-Sa)/N; c:=0: 

stepsize := 0:002500000000 

The stepsize is 0:0025 and the initial value of x is taken to be 0:2. The plots 

counter c has been \initialized" to zero. The outer loop in the following double 

loop increments a in units of stepsize from Sa to Fa. 

> for a from Sa to Fa by stepsize do 

The sum in equation (8.18) is calculated for each value of a, and the sum is 

assigned the name total. Tostarto®,total is set to zero. 

> total:=0; 

The inner do loop runs over the total number of iterations, numpts =300. 

> for j from 1 to numpts do 

The logistic map is entered, and the absolute value of the derivative with respect 

to x formed and labeled d. 

> x:=a*x*(1-x); 

> d:=abs(a*(1-2*x));


If d is not equal to zero, then f =ln(d) iscalculated. Ifd = 0, we would obtain 

f = ¡1. This latter situation is avoided by setting f = 0 in this case. 

> if d0 then f:=ln(d) else f=0 end if; 

To perform the sum in equation (8.18), the total is incremented by the value of 

f, and the inner loop ended. 

> total:=total+f; 

> end do: 

The counter c is incremented by one, 

> c:=c+1; 

and a list of points formed with a as the horizontal coordinate and ¸ = 

total=numpts as the vertical coordinate. This completes the evaluation of ¸ 

in equation (8.18) for a given a value. 

> pts[c]:=[a,total/numpts]: 

> end do: 

The sequence of N = 480 points is plotted, the points being joined using a line 

style. Figure 8.14 shows the Lyapunov exponent ¸ for a =2:8 toa =4. 

> plot([seq(pts[i],i=1..N)],style=line,view=[Sa..Fa,-1.5..1], 

tickmarks=[3,3]); 

1 

0 

-1 

3 3.5 4 

Figure 8.14: Lyapunov exponent (vertical axis) for a =2:8 toa =4. 

If one compares the periodic windows where the Lyapunov exponent goes negative, 

there is good agreement with the bifurcation diagram, Figure 8.13, for 

the logistic map. Because the Lyapunov spikes are sometimes quite narrow, the 

agreement can be improved by zooming in on a particular a region by altering 

the values of Sa and Fa.

8.5. RECONSTRUCTING AN ATTRACTOR 345 

PROBLEMS: 

Problem 8-23: Quartic map 

With x0 =0:2, calculate the Lyapunov exponent for the quartic map 

xn+1 = axn (1 ¡ x 3 n) 

over the range a =1:5 toa =2:0, keeping all other parameters as in the text 

recipe. Over what regions of a do periodic solutions occur? 

Problem 8-24: The tent map 

Plot the Lyapunov exponent ¸ versus a for the tent map 

xn+1 =2axn; 0


so that xn+1 depends only on the previous value xn. If this is the case, then a 

functional form f will exist such that xn+1 = f(xn). 

Assuming that this is the case, one plots pairs of numbers (xn; xn+1) from 

the time series to form the two-dimensional space xn+1 versus xn. Ifthepoints 

appear to lie on a de¯nite geometrical line, this implies that there is an underlying 

attractor and an associated functional form f. If the geometrical shape 

has more structure to it, this could imply that there is an underlying \twodimensional" 

map, namely, 

xn+1 = f(xn)+g(xn¡1); or xn+1 = f(xn)+yn; yn+1 = g(xn): 

If the dimensionality of the underlying map is higher than two, one must increase 

the dimensionality of the space accordingly. To \see" a three-dimensional 

map, for example, one must work in a three-dimensional space. The above procedure 

is then generalized by plotting triplets of numbers. For example, one 

might use (x0; x1; x2), (x1; x2; x3), etc. 

8.5.1 Putting Humpty Dumpty Together Again 

Humpty Dumpty sat on a wall, Humpty Dumpty had a great fall, 

All the king's horses, And all the king's men, 

Couldn't put Humpty Dumpty together again. 


Suppose that we have been given the following lengthy data list x, whereeach 

entry x(n) corresponds to a time t = nts, withn =1; 2; :::;N. 


> x:=[6.24, 9.15, 3.03, 8.24, 5.66, 9.58, 1.56, 5.14, 9.74, .980, 

3.45, 8.81, 4.09, 9.43, 2.10, 6.47, 8.91, 3.79, 9.18, 2.93, 8.07, 

6.07, 9.30, 2.53, 7.37, 7.56, 7.18, 7.89, 6.50, 8.88, 3.89, 9.27, 

2.65, 7.60, 7.12, 8.00, 6.24, 9.15, 3.03, 8.23, 8.23, 5.67, 9.57, 

1.60, 5.23, 9.73, 1.03, 3.59, 8.98, 3.58, 8.97, 3.61, 9.00, 3.51, 

3.86, 9.24, 2.74, 7.76, 6.77, 8.53, 4.89, 9.74, .969, 8.89, 3.41, 

8.77, 4.21, 9.51, 1.82, 5.81, 9.50, 1.87, 5.93, 9.41, 2.16, 6.61, 

8.74, 4.30, 9.56, 1.64, 5.33, 9.71, 1.11, 3.85, 9.23, 2.76, 7.80, 

6.70, 8.62, 4.65, 9.70, 1.13, 3.91, 9.28, 2.59, 7.48, 7.34, 7.61, 

7.10, 8.04, 6.16, 9.23, 2.78, 7.83, 6.62, 8.72, 4.34, 9.58, 1.56, 

5.13, 9.74, .977, 3.44, 8.80, 4.13, 9.45, 2.02, 6.29, 9.10, 3.18, 

8.46, 5.08, 9.75, .960, 3.39, 8.73, 4.31, 9.56, 1.62, 5.30, 9.72, 

1.08, 3.76, 9.15, 3.04, 8.25, 5.63, 9.60, 1.51, 5.00, 9.75, .951, 

3.36, 8.70, 4.43, 9.62, 1.42, 4.75, 9.73, 1.04, 3.63, 9.02, 3.45, 

8.81, 4.09, 9.43, 2.10, 6.47, 8.90, 3.81, 9.20, 2.88, 8.00, 6.23, 

9.16, 3.01, 8.21, 5.73, 9.54, 1.70, 5.49, 9.66, 1.30, 4.41, 9.61, 

1.45, 4.84, 9.74, .989, 3.48, 8.84, 3.98, 9.35, 2.38, 7.07, 8.07, 

6.07, 9.30, 2.52, 7.35, 7.59, 7.14, 7.97, 6.30, 9.09, 3.23, 8.53, 

4.88, 9.74, .972, 3.42]:


Although the data appears to be con¯ned to the approximate range 0 to 10, it 

is not clear at ¯rst glance whether it represents deterministic chaos or is simply 

a \noisy" set of data. Can we con¯rm that it is the former and identify the 

probable identity of the underlying map? Calling this the \Humpty Dumpty 

map," can we then reconstruct his or her mathematical \appearance"? Or, at 

the risk of sounding rather melodramatic, can we put Humpty Dumpty back 

together again? 

The number of operands command reveals that there are 200 entries in x. 

> N:=nops(x); 

N := 200 

First, let us plot x versus n and see whether that reveals any pattern that we 

might have missed. By default, the N points are joined by straight lines. 

> plot([seq([n,x[n]],n=1..N)],labels=["n","x[n]"]); 

x[n] 

8 

6 

4 

2 

0 50 100 150 200 

n 

Figure 8.15: Time series x versus n. 

Although there is a hint of some repetition, it certainly isn't conclusive that the 

series comes from a map producing deterministic chaos rather than just being 

noisy output. Following the procedure mentioned in the introduction, we will 

plot xn+1=10 versus xn=10, the factor of 10 being introduced to reduce both 

vertical and horizontal ranges to 0 to 1 in the graph. The points are represented 

by size-12 black circles. The graph is assigned a name, gr1, soitcan¯rstbe 

displayed and then used to reveal Humpty Dumpty's true identity. 

> gr1:=pointplot([seq([x[n]/10,x[n+1]/10],n=1..N-1)], 

symbol=circle,symbolsize=12,color=black, 

labels=["x[n]","x[n+1]"]): 

Entering gr1 with a command-line-ending semicolon, 

> gr1; #humpty dumpty?


0.8 

0.6 

x[n+1] 

0.4 

0.2 

0.2 0.4 

x[n] 

0.8 

Figure 8.16: xn+1 versus xn for time series. 

produces Figure 8.16. The data points appear to lie on an inverted parabola, 

suggesting that Humpty Dumpty might be a one-dimensional logistic map. Adjusting 

the parameter a by trial and error to provide the best ¯t, the logistic 

curve is now plotted for a =3:9 as a thick red line over the range X =0to1. 

> a:=3.9: #adjust 

> gr2:=plot(a*X*(1-X),X=0..1,color=red,thickness=2): 

The two graphs, gr1 and gr2, are plotted in the same picture, the result being 

showninFigure8.17. 

> display(fgr1,gr2g); #successful reconstruction 

0.8 

0.6 

x[n+1] 

0.4 

0.2 

0 

0.2 0.4 

x[n] 

0.8 1 

Figure 8.17: Best-¯tting logistic curve (solid) and xn+1 vs. xn points.


The points lie on the logistic curve, so we have been successful in identifying and 

reconstructing Humpty Dumpty. We can con¯rm from either of the previous 

two recipes that the logistic map produces deterministic chaos for a =3:9. 

PROBLEMS: 

Problem 8-27: Reconstructing the quartic map 

Taking a =1:97 and x0 =0:9, iterate the quartic map 

xn+1 = axn (1 ¡ x 3 n) 

to produce a time series with 200 points. Plot the time series with a line 

style. Attempt to reconstruct any possible underlying deterministic attractor 

by plotting xn+1 versus xn using a point style. Show that these points lie on 

the curve aX(1 ¡ X 3 )witha =1:97. (This problem is a bit like a dog chasing 

its tail!) 

8.5.2 Random Is Random 

We humans have purpose on the brain. We ¯nd it hard to look at 

anything without wondering what it is \for," what the motive for it 

is, or the purpose behind it. When the obsession with purpose becomes 

pathological it is called paranoia{reading malevolent purpose 

into what is actually random bad luck. But this is just an exaggerated 

form of a nearly universal delusion. Show us almost any object 

or process, and it is hard for us to resist the \Why" question|the 

\What is it for?" question. 

Richard Dawkins, British biologist, author, River Out of Eden, 1995 

In this recipe, we shall use Maple's random-number generator to produce a 

time series and attempt to reconstruct any underlying deterministic attractor. 

The time series will have N =200points. 

> restart: N:=200: 

The command randomize( ) sets the random-number seed to a di®erent value 

based on the computer system clock. 

> randomize(): 

In the following do loop, N random decimal numbers lying between 0 and 1 are 

generated. 


The command rand( ) produces a random positive twelve-digit number. A 

fractional number between 0 and 1 results on dividing by 1012 .Thenumberis 

then put into decimal form by applying the °oating-point evaluation command. 

> x[n]:=evalf(rand()/10^12); 

> end do:


The randomly generated time series is plotted in Figure 8.18. 

> plot([seq([n,x[n]],n=1..N)],labels=["n","x[n]"]); 

1 

0.8 

0.6 

x[n] 

0.4 

0.2 

0 

50 100 n 150 200 

Figure 8.18: Time series for random data. 

If we didn't know how the series was produced, we would probably still 

conclude from the ¯gure that we are dealing with random noise. But one might 

have arrived at the same conclusion in the previous recipe from the time series 

plot. As in that recipe, let's plot xn+1 versus xn using a point style, thus 

producing Figure 8.19. 

> plot([seq([x[n],x[n+1]],n=1..N-1)],style=point,symbol= 

circle,symbolsize=16,labels=["x[n]","x[n+1]"]); 

1 

0.8 

0.6 

x[n+1] 

0.4 

0.2 

0 

0.2 0.4 x[n] 0.8 1 

Figure 8.19: xn+1 versus xn for time series.


There is no discernible pattern in the ¯gure, unlike the situation in the previous 

recipe. By plotting other pairs, or even triplets, of numbers, one can conclude 

(not surprisingly) that there is no underlying chaotic attractor here. 

PROBLEMS: 

Problem 8-28: Rolling a die 

Consulting Maple's Help, produce a random set of 500 numbers from the positive 

integers one to six inclusive. This might simulate the rolling of an honest 

die. Make a plot of the \time series" and xn+1 versus xn to show the randomness. 

Problem 8-29: Literature search 

Apply the techniques of attractor reconstruction to some time series data (e.g., 

the Dow Jones index) extracted from newspapers, magazines, or whatever, and 

see whether a pattern emerges. 

8.5.3 Butter°y Reconstruction 

A people's literature is the great textbook for real knowledge of them. 

Thewritingsofthedayshowthequalityofthepeopleasnohistorical 

reconstruction can. 

Edith Hamilton, American classical scholar (1867{1963) 

Our ¯nal recipe will illustrate how Lorenz's butter°y attractor can be reconstructed 

from time series data extracted from the governing system of three 

coupled nonlinear ODEs. The Lorenz system (which was discussed in Section 

2.2.1) is entered, along with the initial condition x(0) = 2, y(0) = 5, z(0) = 5. 


> sys:=diff(x(t),t)=sigma*(y(t)-x(t)),diff(y(t),t)=-x(t)*z(t) 

+r*x(t)-y(t),diff(z(t),t)=x(t)*y(t)-b*z(t); 

sys := d 

x (t) =¾ (y(t) ¡ x (t)); 

dt 

d 

y(t) =¡x(t) z(t)+r x (t) ¡ y(t); 

dt 

d 

z(t) =x(t) y(t) ¡ b z(t) 

dt 

> ic:=(x(0)=2,y(0)=5,z(0)=5): 

The parameter values are taken to be r =28,b =8=3, and ¾ =10. 

> r:=28: b:=8/3: sigma:=10: 

The total number of entries in the time series will be taken to be N = 2000. 

The total time is T = 50, so the step size ¢ = T=N is 1=40. 

> N:=2000: T:=50: Delta:=T/N;


¢:= 1 

40 

The system of ODEs is solved numerically, the solution being given as a list 

procedure, so that we can create a time series. 

> sol:=dsolve(fsys,icg,fx(t),y(t),z(t)g,numeric,maxfun=0, 


The following line uses the numerical solution to evaluate x(t) atanarbitrary 

time, which must be speci¯ed. 

> S:=eval(x(t),sol): 

Then entering S(n*Delta) will yield x(t) att = n ¢. Using this result, the x 

values are obtained at times t = n ¢forn =0toN. 

> for n from 0 to N do X[n]:=S(n*Delta); end do: 

The time series is created and plotted, the default being to join the points with 

straight lines. The resulting picture is shown in Figure 8.20. 

> plot([seq([n*Delta,X[n]],n=0..N)],labels=["t","x"]); 

x 

15 

10 

5 

0 

–5 

–10 

–15 

10 20 30 40 50 

t 

Figure 8.20: x time series for the Lorenz system. 

To reconstruct the 3-dimensional Lorenz butter°y attractor, it is necessary to 

form triplets of numbers from the time series. By trial and error, we use the 

triplet combination involving n, n +3,andn + 6 to create the plotting points. 

> points:=[seq([X[n],X[n+3],X[n+6]],n=0..(N-6))]: 

Using the spacecurve command with shading=z to color the trajectory, the 

butter°y attractor is revealed. 

> spacecurve(points,style=line,shading=z,axes=framed, 

orientation=[-30,60],tickmarks=[3,3,3], 

labels=["X(n)","X(n+3)","X(n+6)"]); 

A black-and-white version is shown in Figure 8.21. If one compares the picture, 

which can be rotated on the computer screen, with the butter°y picture 

obtained in Chapter 2, the reconstruction of the butter°y is quite good.


X(n+6) 

10 

0 

–10 

–10 

X(n) 

0 

10 

–10 

0 

X(n+3) 

Figure 8.21: Butter°y attractor reconstructed from time series. 

In the preceding recipes, the reconstruction attempt has been tested on 

known examples. So the question is, does the method work for real experimental 

time series data, particularly when the answer is not known? The answer is 

yes! The method has been applied to the Belousov{Zhabotinski chemical oscillator 

reaction (Section 1.2.2) [RSS83] [SWS82], to Taylor{Couette °ow in hydrodynamics 

[AGS84], to ultrasonic cavitation in liquids [LH91], and to measles 

data for the cities of Baltimore and New York [SK86]. 

PROBLEMS: 

Problem 8-30: Butter°y attractor 

Try reconstructing the butter°y attractor from the z(t) time series, and explain 

why you do not obtain two \wings." 

Problem 8-31: RÄossler attractor 

The RÄossler system is given by 

_x = ¡(y + z); _y = x + ay; _z = b + z (x ¡ c): 

Using the time series for x(t) witha =0:2, b =0:2, c =5:7, x(0) = y(0) = 

z(0) = 0:1, construct the RÄossler attractor. Hint: Your picture should resemble 

Figure 1.18. 

10


Epilogue 

Thus grew the tale of Wonderland. 

Thus slowly, one by one, 

Its quant events were hammered out| 

And now the tale is done, 

And home we steer, a merry crew, 

Beneath the setting sun. 


The storybook characters who formed the \merry crew of our \Wonderland," 

having earlier attended Mike and Vectoria's wedding ceremony in a sun-dappled 

alpine meadow, are beginning to slowly drift away from the banquet table to 

dance beneath the setting sun. Eavesdropping on their conversation, it appears 

that everyone was pleased with the entire menu, the appetizers, the entrees, 

and the desserts. The crew also chatted about the new friends that they have 

made. They hope that you the reader, having shared some of their experiences 

and recipes, will include yourself in this group. 

Now your CAS chefs must reluctantly end this tale of Wonderland and 

close this gourmet's guide to some of the advanced mathematical models of 

science. Our guiding principle throughout both volumes of Computer Algebra 

Recipes has been to introduce the reader to what we believe is an important 

educational and scienti¯c innovation, the use of a computer algebra system 

to learn and explore science. This belief is supported by the fact that over 

a million physicists, engineers, mathematicians, chemists, and other groups of 

scientists, are already using one CAS or another to solve technological and 

scienti¯c problems of interest to them. If you have successfully applied the 

computer algebra recipes in our menu to the text problems in both volumes, 

youshouldbeingoodshapetojointheirranks. 

Richard and George, Your CAS chefs

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Index 

acoustical waveguide, 261 

adenosine diphosphate, 98 

aliasing, 330 

Andromeda Strain, 152 

anharmonic force, 34 

annulus, 231 

aorta, 110 

arms race, 105 

arrhythmias, 28 

arrow notation, 52 

arrowheads 

full, 19 

half, 19 

attractor 

butter°y, 83, 351 

chaotic, 345 

Lorenz, 83 

normal, 45 

RÄossler, 353 

reconstruction, 345 

strange, 44 

autonomous system, 32 

baleen whale, 95 

band matrix, 276 

bandwidth limited, 330 

bandwidth theorem, 273 

basketball, falling, 152 

Belousov{Zhabotinski, 40 

Bernoulli equation, 155 

Bernoulli, Daniel, 138 

Bessel functions, 132, 254 

Bessel recurrence relations, 136 

Bessel zeros, 133 

bifurcation, 337 

diagram, 337 

361 

pitchfork, 340 

tangent, 340 

Bloch wall, 292 

blood pressure 

diastolic, 110 

systolic, 110 

blue whale, 160 

boundary condition, 228 

bounded orbit, 325 

Boussinesq's equation, 296, 303 

brachistochrone, 190 

bungee cord, 138 

Burgers' equation, 296, 306 

butter°y reconstruction, 351 

butter°y, Lorenz's, 83 

candle °ame, 287 

capacitor, 29 

Casablanca, 211 

catenary, 194 

Cauchy{Riemann conditions, 220 

CDA, 274 

center, 17 

central di®erence approximation, 90 

chain reaction, 236 

Chandrasekhar, S., 95 

chaos, 33, 38, 336 

characteristic directions, 312 

chemical oscillator, 40 

chemical reaction, 100 

empirical rule, 41 

Oregonator, 41 

classic Maple, 1 

coalescence of roots, 68 

column vector, 85, 277 

competition

362 INDEX 

for food, 27 

warring armies, 57 

complex conjugate, 217 

complex function, 220 

compliance, 110 

computational mesh, 274 

computer algebra system(CAS), 1 

concentration, 41 

conditional statement, 80 

conducting plates, 223 

contour plot, 221 

control parameter, 34, 320 

control rods, 236 

convolution theorem, 269 

coordinates 

bipolar, 235 

bispherical, 235 

Cartesian, 243 

cylindrical, 235, 258 

elliptic, 235 

paraboloidal, 235 

polar, 232, 254 

prolate spheroidal, 235 

spherical, 235 

CPU time, 91 

Crichton,Michael,152 

critical condition, 236 

critical damping, 26, 27 

critical mass, 239 

cross product, 176 

cross section, variable, 269 

crossed ¯elds, 179 

crystal dislocation, 288 

cubic map, 341 

cuto® frequency, 261 

cylinder 

quarter, 240 

temperature distribution, 241 

da Vinci, Leonardo, 58 

damping 

critical, 27 

negative, 29 

over, 26 

under, 26 

Dawkins, Richard, 349 

de Vries, Gustav, 301 

degenerate root, 69 

degrees of freedom, 324 

derivative 

BDA, 90 

CDA, 90 

FDA, 90 

derivative boundary condition, 259 

determinant, 85 

deterministic chaos, 28 

di®erence approximation 

backward, 90 

central, 90 

forward, 90 

fourth derivative, 91 

di®usion coe±cient, 209, 224 

dimensional analysis, 208 

dimensionless variable, 60 

Dirac delta function, 264 

dispersion, 273, 288 

dispersion relation, 301 

do loop, 24 

dot notation, 13 

dot product, 180 

double-well potential, 34 

drag force 

Newton, 161 

Stokes, 24 

driving period, 320 

drumhead, 254 

drumhead, circular, 257 

Du±ng's equation, 34, 149, 320 

hard spring, 33 

inverted, 33 

nonharmonic, 33 

soft spring, 33 

E. coli, 153 

eardrum, 168 

eardrum equation, 66 

Eddington, Arthur, 150 

Edelstein-Keshet, Leah, 152 

eigenfrequency, 257 

eigenfunction, 130

INDEX 363 

eigenvalue, 130 

Eiram Eiruc, 74 

electric ¯eld lines, 222 

elliptic cosine function, 174 

elliptic integral, 8 

complete, 172 

incomplete, 172 

elliptic sine function, 173 

EllipticF, 8 

EllipticK, 172 

energy transmission, 215 

entrainment, 28 

enzyme reaction, 161 

epidemiology, 66 

equipotentials, 222 

Erehwon, 187 

erf, 263 

erfc, 268 

Erlenmeyer °ask, 152 

error function, 263 

error function, complementary, 268 

Euler algorithm, 92 

Euler's constant, 34 

Euler{Lagrange equation, 185 

Euler{Mascheroni constant, 133 

explicit scheme, 91, 275, 279 

falling raindrop, 120 

fast Fourier transform (FFT), 331 

FDA, 274 

Fermat's principle, 190 

ferromagnet, 288 

ferromagnetic domain, 292 

Fick's law, 266, 267 

Field, R. J., 40 

¯nite di®erence approximation, 89 

¯sh harvesting, 100 

¯xed edges, 251 

¯xed point, 15 

°u epidemic, 160 

°uid °ow, 223 

focal/spiral point, 16 

forced sti® ODE, 106 

forward Euler method, 94 

forward-di®erence approximation, 274 

Fourier cosine transform, 264 

Fourier series, 212 

Fourier sine transform, 261 

Fourier transform, 271, 329 

inverse, 271 

fractal dimension, 45 

free body diagram, 181 

free edges, 253 

Frobenius series, 133 

fructose-6-phosphate, 98 

fudge factor, 44 

functional operator, 52 

fur catches, 57 

gamma function, 145 

Gause, 155 

generalized momentum, 324 

geodesic, 191 

glycolysis, 97 

gnus and sung, 73 

golf, 179 

Gompertz's law, 160 

gradient, 222 

great circle, 191 

Greg Arious Nerd, 115 

Henon{Heiles Hamiltonian, 324 

Henon{Heiles potential, 324 

Hamilton's equations, 324 

Hamiltonian chaos, 324 

hard spring, 66, 174 

harmonics, 335 

harvesting, 114 

heat capacity, 209 

heat di®usion, 208 

Helmholtz, 168 

Hermite functions, 145 

higher-order ¯xed point, 49 

hole in one, 184 

Hooke's law, 24 

Hudson's Bay Company, 57 

human °y, 195 

Humpty Dumpty map, 346 

hyperlink, 8 

implicit scheme, 91

364 INDEX 

implicit solution, 169 

incompressible °uid, 241 

inductor, 30 

infectious disease, 66 

in¯nite domain, 261 

initial conditions, 20 

initialization, 310 

Internet, 260 

inverted spring, 34 

ion acoustic soliton, 303 

iron core inductor, 57 

iron sphere temperature, 245 

irreversible reaction, 150 

isoperimetric problem, 191 

Jacobi elliptic function, 173 

Josephson junction, 288 

K}orÄos, E., 40 

KAM torus, 327 

Kaup, David, 299 

KdV equation, 287, 300 

KdV soliton collision, 308 

Kerr medium, 288 

kink, 304 

Kirchho®'s current rule, 29 

Kirchho®'s potential rule, 57 

Klein{Gordon equation, 278 

Korteweg, Diederik, 301 

Kruskal, Martin, 309 

Lagrange's equations, 202 

laminar °ow, 161 

Laplace transform, 121, 267 

Laplace's equation, 220 

Laplace, Pierre-Simon, 121 

laser beam competition, 159 

laser beams, 318 

learning curve, 114 

Legendre polynomials, 243 

limit cycle, 28 

logistic equation, 154 

logistic map, 338 

Lorentz force, 175 

Lorenz equations, 83 

Lorenz's butter°y, 351 

love a®air, 18 

low-pass ¯lter, 330 

Lyapunov exponent, 342, 343 

Lyapunov, Aleksandr M., 342 

magnetic bottle, 175 

magnetic spin, 288, 292 

magnetohydrodynamic waves, 288 

Malthus's equation, 153 

Malthus's law, 152 

Malthus, Thomas, 153 

map 

cubic, 341 

Humpty Dumpty, 346 

logistic, 338 

one-dimensional, 346 

quartic, 341, 345 

sine,341,345 

tent, 341, 345 

Maple 

active derivative, 19 

addition, 4 

aliasing symbols, 122 

animation, 7 

arrow operator, 52 

assignment operator, 4 

assume vs. assuming, 112 

automatic substitution, 19 

classic interface, 1 

clearing internal memory, 4 

colon, 4 

command information, 110 

comment, 5 

conditional statement, 80 

constrained scaling, 6 

copying examples, 8 

CPU time, 91 

curly brackets, 15 

default accuracy, 4 

di®erential operator, 26 

digits control, 103 

ditto operator, 30 

division, 4 

functional operator, 52

INDEX 365 

Help, Full Text Search, 8 

Help, Topic Search, 8 

Help, using, 8 

inert derivative, 19 

inserting comments, 5 

library packages, 7 

list, 6 

list of lists, 6 

list procedure, 127 

menu index ¯le, 2 

multiplication, 4 

placeholder, 280 

plot options, 6 

powers, 4 

prime notation, 126 

procedure, 52 

prompt symbol, 4 

recipe, 1 

rotating viewing box, 31 

semicolon, 4 

set, 15 

square brackets, 6 

standard interface, 1 

string, 6 

superimposing ¯gures, 53 

suppressing output, 4 

text on plot, 42 

toolbar, 6 

type match, 226 

unassignment, 54 

unprotect Maple symbol, 59 

user manuals, 1 

variable change, 29 

warnings, 7 

worksheet, 2 

Maple Command 

^, 4 

¡ >, 52 

*, 4 

+, 4 

-, 4 

., 180 

/, 4 

::, 226 

:=, 4 

:, 4 

;, 4 

>, 277 

< >,176 

?coords, 231 

?inifcns, 132 

?, 8 

Array, 333 

BandMatrix, 276,280 

BesselJZeros, 133, 255 

BesselJ, 132 

BesselY, 133, 237 

CrossProduct, 176 

DEplot3d, 31,45 

DEplot, 24 

DEtools[expsols], 267 

Determinant, 85 

Diff, 19 

Digits, 98 

Dirac, 264 

DotProduct, 180 

D, 26 

Eigenvalues, 85 

EulerLagrange, 188 

FourierTransform, 333 

Gamma, 59 

GenerateMatrix, 85 

Gradient, 222 

HINT, 225 

Hamilton eqs, 325 

INTEGRATE, 225 

Int, 34 

I, 216 

Laplacian('cylindrical'), 258 

Laplacian, 232 

LegendreP, 134 

LegendreQ, 134,242 

Omega, 203 

Order, 128 

Phi, 5 

Pi, 209 

Sum, 226 

SurfaceOfRevolution, 189 

Tangent, 167 

Theta, 60, 203

366 INDEX 

#, 5 

&x, 176 

EnvAllSolutions, 61 

EnvLegendreCut, 135 

abs, 128 

add, 213 

algsubs, 203 

alias, 123 

allvalues, 79, 325 

alpha, 34 

animate, 7 

array, 36 

arrows=MEDIUM, 19 

arrows=THICK, 177 

assign, 22,52 

assume=real, 305 

assume, 172 

assuming, 116 

axes=frame, 42 

axes=none, 140 

axes=normal, 72 

background, 7 

bernoullisol, 155 

beta, 24, 26, 34 

build, 225 

coeff, 233 

collect, 30,68 

color, 20 

combine(trig), 217 

combine, 197 

conjugate, 217 

contourplot, 221, 324 

contours, 221 

convert(exp), 272 

convert(polynom), 89 

convert(radians), 182 

convert(sincos), 163 

convert(units), 5,200 

coordplot3d, 235 

coordplot, 232 

cosh, 177, 228 

cos, 4 

dchange, 30, 60, 290 

declare, 126 

dfieldplot, 19 

diff, 18,59 

dirgrid, 19 

display, 42 

dsolve('series'), 128 

dsolve(classical[foreuler]), 

94 

dsolve(classical[rk4]), 99 

dsolve(method=laplace), 122 

dsolve(numeric), 42,127 

dsolve, 22, 27 

epsilon, 30 

evalc, 124 

evalf, 5 

eval, 5 

expand, 30, 54, 59, 60 

exp, 19 

factor, 30, 238 

fieldplot, 177 

firint, 116 

for...while...do, 24 

fouriercos, 264 

fouriersin, 262 

fourier, 271 

frames, 7 

fsolve, 75, 199 

gamma, 34, 59 

generate ic, 326 

grid, 75 

has, 225 

if...elif... end if, 80 

implicitplot, 75, 234 

infinity, 152 

infolevel[dsolve], 110 

insequence=true, 167 

integer, 229 

interface(imaginaryunit), 66 

intfactor, 116 

int, 34 

invfourier, 272 

invlaplace, 121,268 

isolate, 203 

iterations, 327 

labelling=true, 232 

labels, 34 

lambda, 85

INDEX 367 

laplace, 121,267,268 

lhs, 22 

limit(right), 198 

limit, 152 

linecolor, 20 

linestyle=[DASH,SOLID], 69 

ln, 19 

maxfun=0, 320 

mu, 237 

nops, 78 

numer, 188 

numpoints, 42 

nu, 237 

odeadvisor, 115,131 

odeplot, 42,204 

omega, 24, 31, 34 

op, 140 

orientation, 31, 45 

output=plot, 167 

pdetest, 212,305 

pdsolve, 225 

phaseportrait, 20,35 

phi, 5 

piecewise, 111, 213 

plot3d, 229 

plots[display], 277 

plot, 6 

poincare, 326 

pointplot3d, 93, 178 

pointplot, 7,53 

polar, 232 

psi, 139 

rand( ), 350 

randomize( ), 349 

remove, 84,124 

restart, 4 

rho, 209 

rhs, 19, 22 

riccatisol, 162 

rotate, 140 

round, 151 

rtablesize=infinity, 276 

scaling=constrained, 6 

scene, 21, 31 

select, 124 

seq, 80 

series, 133 

shading=zhue, 45 

showtangent, 167 

sign, 167 

simplify(exp), 239 

simplify(power), 165 

simplify(symbolic), 225 

simplify(trig), 197 

simplify, 22,60 

sin, 4,19 

solve, 5,15 

spacecurve, 98, 104, 352 

sphereplot, 204 

sqrt, 8, 19, 59 

stepsize, 20 

style=line, 53 

style=patchcontour, 229 

style=patchnogrid, 240 

style=point, 62 

subs, 24,30,35 

sum, 226 

symbol=box, 53 

symbol=circle, 7,62 

symbol=cross, 93 

symbolsize, 7 

tau, 30,60 

taylor, 89, 168 

textplot3d, 42 

textplot, 53 

theta, 5,59 

thickness, 42 

tickmarks, 34 

time( ), 98, 276 

title, 42 

unapply, 151 

unassign, 54,63 

unprotect, 34, 59, 297 

with(DEtools), 18, 23, 51 

with(DiscreteTransforms), 332 

with(LinearAlgebra), 83, 276 

with(PDEtools), 29, 59, 126 

with(Student[Calculus1]), 166 

with(VectorCalculus), 175 

with(inttrans), 121,262

368 INDEX 

with(plots), 7,59 

with(plottools, 138 

Mathieu equation, 143 

Mathieu functions, 142 

Mathieu, Emile, 142 

MathieuC, 143 

MathieuS, 143 

matrix, 85 

band, 276 

multiplication, 276 

square, 276 

tridiagonal, 276 

May, Robert, 149 

mean daily temperature, 209 

measles, 353 

membrane 

circular, 254 

free edges, 253 

rectangular, 251 

square, 253 

mesh 

diamond-shaped, 312 

point, 274 

rectangular, 275 

method of characteristics, 312 

Michaelis{Menton equation, 161 

microbiology, 152 

Mike, 23 

mirage, 190 

model 

KdV soliton, 300 

modi¯ed Euler algorithm, 96 

modi¯ed KdV equation, 303 

Mona Lisa, 58 

multiplication factor, 236 

musical notes, 257 

myxamatosis, 160 

National Bureau of Standards, 131 

negative damping, 29 

neutron density, 236 

neutron di®usion, 236 

Newton's drag law, 161 

Newton's second law, 59, 115 

nodal lines, 256 

nodal point, 16 

nonlinear diode, 159 

nonlinear drag, 164 

nonlinear rate equations, 41 

nonlinear SchrÄodinger equation, 287 

nonlinear spring, 72 

normal modes, 257, 259 

Noyes, R. M., 40 

nuclear explosion, 240 

nuclear ¯ssion, 236 

numerical instability, 96, 277 

numerical mesh, 274 

Nyquist frequency, 330 

Nyquist, H., 330 

ODE 

nth-order, 110 

Airy, 23 

analytic solution, 22 

arms race, 105 

autonomous, 13 

baleen whale, 95 

Bernoulli, 155 

Bertalan®y, 161 

Bessel, 131 

blood pressure model, 110 

blue whale, 160 

brachistochrone, 190 

bungee cord, 138 

catenary, 194 

chemical oscillator, 44 

chemical reaction, 100 

competing armies, 57 

con°uent hypergeometric, 23 

dimensionless, 30 

dot notation, 13 

Du±ng, 149, 320 

eardrum, 168 

Euler algorithm, 92 

Euler{Lagrange, 185 

falling basketball, 152 

falling raindrop, 120 

¯nite di®erence approximation, 

89 

¯rst integral, 116

INDEX 369 

¯sh harvesting, 100 

¯xed points, 14 

°u epidemic, 160 

forced sti®, 106 

fox rabies, 102 

Frobenius series, 132 

geodesic, 191 

glycolytic, 96 

gnus and sung, 74 

golf ball, 182 

Gompertz, 160 

growing pendulum, 141 

hard spring, 174 

Hermite, 23 

homogeneous, 110 

inhomogeneous, 110 

integrating factor, 115 

Lagrange, 202 

Laplace transform method, 121 

laser beam, 159 

learning curve, 114 

Legendre, 134 

light ray path, 191 

linear, 14, 110 

logistic, 154 

Lorenz, 83 

Malthus, 153 

Maple odeadvisor, 115 

Mathieu, 142 

Michaelis{Menton, 161 

modi¯ed Euler algorithm, 96 

nonautonomous, 13 

nonlinear, 14 

nonlinear diode, 159 

numerical instability, 96 

numerical solution, 42 

onset of bending, 141 


oscillating pivot, 206 

p-q diagram, 50 

parametric excitation, 206 

pendulum, 23 

phase-plane analysis, 47 

phase-plane portrait, 13 

piecewise, 111 

plotting numerical solution, 42 

population growth, 114 


pursuit, 164 

RÄossler, 45, 88 

rabbits and foxes, 14, 51 

Rayleigh, 23 

Riccati, 162 

RK4 algorithm, 96 

rkf45 method, 42 

RL circuit, 114 

Romeo and Juliet, 18 

rotating circular loop, 206 

SchrÄodinger, 219 

second-order, 14, 118 

Sel'kov, 97 

semi-implicit algorithm, 102 

separable, 150 

series solution, 126 

simple harmonic oscillator, 14 

SIR model, 66 

soft spring, 23 

spherical pendulum, 203 

squid and herring, 82 

standard form, 14 

stationary points, 14 

steady-state solution, 124 

sti®, 105 

sti®ening spring, 141 

Sturm{Liouville, 130 

trajectory, 14 

transient solution, 124 

Van der Pol, 28 

Verhulst, 72 

white dwarf, 95 

yeast model, 154 

optical pulse, 288 

ordinary functions, 130 

ordinary point, 15, 47 


orthogonality, 136, 226 

oscillator 

forced glycolytic, 337 

glycolytic, 96 

tunnel diode, 28

370 INDEX 

overdamping, 26 

p-q diagram, 50 

pacemaker, 28 

Palace of the Governors, 207 

parametric excitation, 206 

parametric plot, 6 

Parseval's theorem, 329 

PDE 

Boussinesq, 296 

Burgers, 296 

cosine{Gordon, 293 

embedded string, 278 

Hamilton, 324 

heat di®usion, 208 

interacting laser beam, 318 

KdV, 287 

Klein{Gordon,278,281 

Laplace, 220 

modi¯ed di®usion, 267 

modi¯ed KdV, 303 

nonlinear, 287 

nonlinear di®usion, 296 

nonlinear Klein{Gordon, 281 

nonlinear SchrÄodinger, 287 

numerical Laplace, 284 

numerical simulation, 274 

Poisson, 282 

scalar Helmholtz, 257 

sine{Gordon, 287 

sound waves, 257 

three-wave, 296 

variable separation, 224 

vibrating bar, 270 

vibrating membrane, 251 

vibrating string, 248 

vibrating wire, 270 

wave, 212, 247 

general solution, 248 

pendulum, growing, 141 

pendulum, rotating, 205 

period doubling, 34, 38 

period eight, 39 

periodfour,38,321,336 

period one, 37 

period two, 38, 319, 335 

period-n response, 319 

periodic windows, 344 

phase plane, 14 

portrait, 14 

trajectory, 14 

phenomenological model, 51 

piano string, 250 

piecewise function, 111 

pitchfork bifurcation, 337 

plane wave, 215 

plate, semicircular, 240 

Poincare section, 319 

Poincare's theorem, 51 

point 

¯xed, 15 

focal/spiral, 16, 49 

higher-order ¯xed, 49 

nodal, 16, 49 

ordinary, 15, 47 

saddle, 16, 17, 50 

saddle-vortex, 71 

simple stationary, 15, 48 

stable equilibrium, 16 

stable focal, 17 

stable nodal, 17, 49 

stationary or ¯xed, 47 

unstable focal, 17 

unstable nodal, 17 

vortex, 16, 17, 67 

Poisson's equation, 282 

polar coordinates, 231 

polynomial force law, 67 

population growth, 153 

potential 

alternating surface, 235 

split, 245 

power spectrum, 329 

chaotic, 336 

predator{prey interaction, 51 


pulsating sphere, 210 

pursuit problem, 164 

quantum tunneling, 219

INDEX 371 

quartic map, 345 

Queen Dido, 191 

RÄossler system, 88 

radian, 59 

radiation, 312 

radioactive contamination, 266 

Rainbow County, 51 

random time series, 350 

random-number generator, 349 

random-number seed, 349 

rectangular barrier, 219 

recurrence formula, 136 

re°ection coe±cient, 215 

resistor, 29 

Riccati equation, 162 

rigid boundary, 241 

rk4 method, 97 

rkf45 algorithm, 42 

RL circuit, 114 

Rocky Mountain, 184 

Romeo and Juliet, 18 

rotational symmetry, 242 

round-o® error, 95 

route to chaos, 34 

Russell, John Scott, 300 

Saccharomyces cerevisiae, 155 

saddle point, 16, 17, 35 

saddle{vortex point, 71 

Sagan, Carl, 343 

sampling frequency, 330, 333, 334 

sampling theorem, 330 

saturable refractive index, 296 

saturation, 155 

scalar Helmholtz equation, 257 

Schizosaccharomyces kephir, 155 

SchrÄodinger equation, 219 

Sel'kov model, 97 

semi-implicit method, 102 

semi-in¯nite domain, 261 

separable ODE, 151 

separation constant, 228 

separation of variables, 224 

separatrix, 17, 289 

Shannon, C. E., 330 

SHE, 257 

shear force, 270 

SHO, 16 

shock wave, 301 

sifting property, 264 

signal processing, 330 

simple ¯xed point 

condition for, 48 

simple harmonic oscillator, 14 

simple pendulum, 58 

simple stationary point, 15, 48 

sine map, 341, 345 

sine{Gordon breather, 306 

sine{Gordon equation, 287 

single-valued, 259 

SIR model, 66 

snowshoe hare, 160 

solitary wave, 287 

antikink, 289 

black, 289 

bright, 289 

kink, 289 

nontopological, 289 

three-wave, 296 

topological, 289 

soliton, 287 

black, 296 

bright, 295 

ion acoustic, 303 

stationary, 295 

sound speed, 257 

special functions, 130 

speci¯c heat, 208 

squid and herring, 81 

stability, collisonal, 287 

stable focal point, 17 

stable limit cycle, 29 

stable nodal point, 17 

static potential, 231 

stationary point, 15 

stationary points 

classi¯cation of, 49 

steady state, 124, 284 

sti® ODE, 105

372 INDEX 

sti®ness,248,270 

Stokes's drag law, 118 

storybook character, 2 

strange attractor, 44, 319 

streamlines, 223 

string 

plucked, 214 

struck, 215 

Strogatz, Steven, 18 

supercritical condition, 236 

superposition 

linear, 238 

surface of revolution, 189 

systemic resistance, 111 

tangent ¯eld, 14 

Tappert, Fred, 307 

Taylor expansion, 48, 89 

Taylor{Couette °ow, 353 

tent map, 341, 345 

terminal velocity, 120 

thermal conductivity, 208 

thermonuclear fusion, 175 

third harmonic, 333 

three-wave problem, 296 

Toda Hamiltonian, 329 

trading records, 57 

transcendental equation, 5 

transcendental function, 61 

transform 

discrete Fourier, 331 

discrete inverse Fourier, 331 

Fourier, 271 

Fourier cosine, 264 

Fourier sine, 261 

inverse Fourier, 271 

Laplace, 267 

transient, 124 

transmission coe±cient, 215 

transverse velocity, 249 

tridiagonal matrix, 276 

tunnel diode, 28 

turbulence, 161 

turning point, 171 

two-soliton solution, 307 

tympanic membrane, 168 

ultrasonic cavitation, 353 

uncertainty principle, 273 

underdamping, 26 

unstable focal point, 17 

unstable nodal point, 17 

Van der Pol equation, 29 

Van der Pol, Balthasar, 28 

variable endpoint, 198 

variable transformation, 290 

variational calculus, 185 

vector 

column, 277 

Vectoria, 23 

velocity potential, 241 

Verhulst, 154 

Verhulst equations, 72 

vibrating cylinder, 261 

vibrating wire, 270 

von Bertalan®y equation, 161 

vortex point, 16, 17, 35, 51, 67 

wave equation, 212 

wave number, 215 

waves 

elastic, 247 

electromagnetic, 247 

longitudinal dispersive, 288 

magnetohydrodynamic, 288 

membrane, 247 

pressure, 288 

shallow water, 288 

sound, 247 

string, 247 

tidal, 247 

weather forecasting, 83 

white dwarf stars, 95 

Whittaker functions, 145 

Woods, Tiger, 183 

yeast, 152 

Zabusky, Norman, 309 

Zabusky{Kruskal algorithm, 309

Computer Algebra Recipes

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