This is an introduction to methods for solving nonlinear partial differential equations (NLPDEs). After the introduction

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*Table of contents : PrefaceAcknowledgmentsNonlinear PDEs are Everywhere Exercises ReferencesCompatibility Charpit's Method Second-Order PDEs Compatibility in (2+1) Dimensions Compatibility for Systems of PDEs Exercises ReferencesDifferential Substitutions Generalized Burgers' Equation KdV-MKdV Connection Generalized KdV Equation Matrix Hopf–Cole Transformation Darboux Transformations Second-Order Darboux Transformations Darboux Transformations Between Two Diffusion Equations Darboux Transformations Between Two Wave Equations Exercises ReferencesPoint and Contact Transformations Contact Transformations Hodograph Transformation Legendre Transformation Ampere Transformation Contact Condition Plateau Problem Linearization Well-known Minimal Surfaces Exercises ReferencesFirst Integrals Quasilinear Second-Order Equations Monge–Ampere Equation The Martin Equation First Integrals and Linearization Hyperbolic MA Equations Parabolic MA Equations Elliptic MA Equations Exercises ReferencesFunctional Separability Exercises ReferencesSolutionsAuthor's BiographyBlank Page*

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Analytical Techniques for Solving Nonlinear Partial Differential Equations Daniel J. Arrigo, University of Central Arkansas This is an introduction to methods for solving nonlinear partial differential equations (NLPDEs). After the introduction of several PDEs drawn from science and engineering, the reader is introduced to techniques used to obtain exact solutions of NPDEs. The chapters include the following topics: Compatibility, Differential Substitutions, Point and Contact Transformations, First Integrals, and Functional Separability. The reader is guided through these chapters and is provided with several detailed examples. Each chapter ends with a series of exercises illustrating the material presented in each chapter. The book can be used as a textbook for a second course in PDEs (typically found in both science and engineering programs) and has been used at the University of Central Arkansas for more than ten years.

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ANALYTICAL TECHNIQUES FOR SOLVING NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS

Series Editor: Steven G. Krantz, Washington University in St. Louis

ARRIGO

Series ISSN: 1938-1743

Analytical Techniques for Solving Nonlinear Partial Differential Equations Daniel J. Arrigo

Analytical Techniques for Solving Nonlinear Partial Diﬀerential Equations

Synthesis Lectures on Mathematics and Statistics Editor Steven G. Krantz, Washington University, St. Louis

Analytical Techniques for Solving Nonlinear Partial Diﬀerential Equations Daniel J. Arrigo 2019

Symmetry Problems. The Navie–Stokes Problem. Alexander G. Ramm 2019

PDE Models for Atherosclerosis Computer Implementation in R William E. Schiesser 2018

An Introduction to Partial Diﬀerential Equations Daniel J. Arrigo 2017

Numerical Integration of Space Fractional Partial Diﬀerential Equations: Vol 2 – Applicatons from Classical Integer PDEs Younes Salehi and William E. Schiesser 2017

Numerical Integration of Space Fractional Partial Diﬀerential Equations: Vol 1 – Introduction to Algorithms and Computer Coding in R Younes Salehi and William E. Schiesser 2017

Aspects of Diﬀerential Geometry III Esteban Calviño-Louzao, Eduardo García-Río, Peter Gilkey, JeongHyeong Park, and Ramón Vázquez-Lorenzo 2017

iv

The Fundamentals of Analysis for Talented Freshmen Peter M. Luthy, Guido L. Weiss, and Steven S. Xiao 2016

Aspects of Diﬀerential Geometry II Peter Gilkey, JeongHyeong Park, Ramón Vázquez-Lorenzo 2015

Aspects of Diﬀerential Geometry I Peter Gilkey, JeongHyeong Park, Ramón Vázquez-Lorenzo 2015

An Easy Path to Convex Analysis and Applications Boris S. Mordukhovich and Nguyen Mau Nam 2013

Applications of Aﬃne and Weyl Geometry Eduardo García-Río, Peter Gilkey, Stana Nikčević, and Ramón Vázquez-Lorenzo 2013

Essentials of Applied Mathematics for Engineers and Scientists, Second Edition Robert G. Watts 2012

Chaotic Maps: Dynamics, Fractals, and Rapid Fluctuations Goong Chen and Yu Huang 2011

Matrices in Engineering Problems Marvin J. Tobias 2011

The Integral: A Crux for Analysis Steven G. Krantz 2011

Statistics is Easy! Second Edition Dennis Shasha and Manda Wilson 2010

Lectures on Financial Mathematics: Discrete Asset Pricing Greg Anderson and Alec N. Kercheval 2010

v

Jordan Canonical Form: Theory and Practice Steven H. Weintraub 2009

The Geometry of Walker Manifolds Miguel Brozos-Vázquez, Eduardo García-Río, Peter Gilkey, Stana Nikčević, and Ramón Vázquez-Lorenzo 2009

An Introduction to Multivariable Mathematics Leon Simon 2008

Jordan Canonical Form: Application to Diﬀerential Equations Steven H. Weintraub 2008

Statistics is Easy! Dennis Shasha and Manda Wilson 2008

A Gyrovector Space Approach to Hyperbolic Geometry Abraham Albert Ungar 2008

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All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means—electronic, mechanical, photocopy, recording, or any other except for brief quotations in printed reviews, without the prior permission of the publisher. Analytical Techniques for Solving Nonlinear Partial Diﬀerential Equations Daniel J. Arrigo www.morganclaypool.com

ISBN: 9781681735337 ISBN: 9781681735344 ISBN: 9781681735351

paperback ebook hardcover

DOI 10.2200/S00907ED1V01Y201903MAS025

A Publication in the Morgan & Claypool Publishers series SYNTHESIS LECTURES ON MATHEMATICS AND STATISTICS Lecture #25 Series Editor: Steven G. Krantz, Washington University, St. Louis Series ISSN Print 1938-1743 Electronic 1938-1751

Analytical Techniques for Solving Nonlinear Partial Diﬀerential Equations

Daniel J. Arrigo University of Central Arkansas

SYNTHESIS LECTURES ON MATHEMATICS AND STATISTICS #25

M &C

Morgan

& cLaypool publishers

ABSTRACT This is an introduction to methods for solving nonlinear partial diﬀerential equations (NLPDEs). After the introduction of several PDEs drawn from science and engineering, the reader is introduced to techniques used to obtain exact solutions of NPDEs. The chapters include the following topics: Compatibility, Diﬀerential Substitutions, Point and Contact Transformations, First Integrals, and Functional Separability. The reader is guided through these chapters and is provided with several detailed examples. Each chapter ends with a series of exercises illustrating the material presented in each chapter. The book can be used as a textbook for a second course in PDEs (typically found in both science and engineering programs) and has been used at the University of Central Arkansas for more than ten years.

KEYWORDS nonlinear partial diﬀerential equations, compatibility, diﬀerential substitutions, point and contact transformations, ﬁrst integrals, functional separability

ix

Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii

1

Nonlinear PDEs are Everywhere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 1.2

2

Compatibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.1 2.2 2.3 2.4 2.5 2.6

3

Charpit’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Second-Order PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Compatibility in .2 C 1/ Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Compatibility for Systems of PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 21 38 41 45 47

Diﬀerential Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.1 3.2 3.3 3.4 3.5

3.6 3.7

4

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Generalized Burgers’ Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . KdV-MKdV Connection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Generalized KdV Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Matrix Hopf–Cole Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Darboux Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Second-Order Darboux Transformations . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Darboux Transformations Between Two Diﬀusion Equations . . . . . . . 3.5.3 Darboux Transformations Between Two Wave Equations . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

53 54 55 58 59 62 65 67 70 71

Point and Contact Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 4.1

Contact Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 4.1.1 Hodograph Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 4.1.2 Legendre Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

x

4.2 4.3

4.4 4.5

5

5.5 5.6

Quasilinear Second-Order Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Monge–Ampere Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 The Martin Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 First Integrals and Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 5.4.1 Hyperbolic MA Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 5.4.2 Parabolic MA Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 5.4.3 Elliptic MA Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

Functional Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 6.1 6.2

7

76 80 82 83 85 87 92

First Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 5.1 5.2 5.3 5.4

6

4.1.3 Ampere Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Contact Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Plateau Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Well-known Minimal Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 Author’s Biography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

xi

Preface This is an introductory book about obtaining exact solutions to nonlinear partial diﬀerential equations (NLPDEs). This book is based, in part, on my lecture notes from a graduate course I have taught at the University of Central Arkansas (UCA) for over a decade. In Chapter 1, I list several NLPDEs (systems of NLPDEs) that appear in science and engineering and provide a springboard into the subject matter. As our world is essentially nonlinear, it is not surprising that the equations that govern most physical phenomena are nonlinear. As Nobel Laureate Werner Heisenberg once said, “The progress of physics will to a large extent depend on the progress of nonlinear mathematics, of methods to solve nonlinear equations.” In Chapter 2, I introduce compatibility. I start with the method of characteristics, a topic from a standard course in PDEs. I then move to Charpit’s method, a method that seeks compatibility between two ﬁrst-order PDEs. From there, I consider the compatibility between ﬁrstand second-order PDEs, particularly, a nonlinear diﬀusion equation, Burgers’ equation, and a linear diﬀusion equation with a nonlinear source term. I then extend the method to the nonlinear diﬀusion equations in .2 C 1/ dimensions and a system of nonlinear PDEs where we consider the Cubic Schrödinger equation. In Chapter 3, I introduce the idea of diﬀerential substitutions. The classic example is the Hopf Cole transformation and Burgers’ equation; it is the example I use to introduce the topic. I extend the results to generalized Burgers’ and KdV equations. I consider Matrix Hopf Cole transformation and use this when I introduce Darboux transformations for linear diﬀusion and wave equations. In Chapter 4, I introduce point and contact transformations. These are a generalization of the usual change of variables one would ﬁnd in an introductory course in PDEs. I introduce three special transformations: the Hodograph transformation, the Legendre transformation, and the Ampere transformation. I also introduce conditions, the contact conditions, which guarantee that a transformation will be a contact transformation. With several examples considered, the chapter ends with the famous Plateau problem from minimal surfaces. In Chapter 5, we consider ﬁrst integrals. Simply put, these are PDEs of a lower order (if they exist) and yield exact solutions of a given equation. I ﬁrst consider second-order quasilinear PDEs in two independent variables and then general Monge-Ampere (MA) equations in two independent variables. Oftentimes, PDEs admit very general classes of ﬁrst integrals. When this happens, it is sometimes possible to transform the given PDE to one that is very simple. I then consider three classes of MA equations: a class of hyperbolic, parabolic, and elliptic MA equations.

xii

PREFACE

The ﬁnal chapter, Chapter 6, is functional separability. This is a generalization of the usual separation of variables that one usually ﬁnds in an introductory course when solving PDEs like the heat equation, the wave equation, or Laplace’s equation. In Chapter 6, I turn the problem into one of compatibility. I consider nonlinear diﬀusion equations, nonlinear dispersion equations in .1 C 1/ dimension, and a three-dimensional linear diﬀusion equation with a source term. It is fair to say that this book does not cover all techniques used to solve nonlinear PDEs but rather represents techniques that I personally have interest in and have successfully used. For example, techniques like the inverse scattering transform and Bäcklund transformations are extremely important and should be added to one’s arsenal. It is also fair to say this is not the only book on techniques for solving PDEs. For example, one should also consider A.R. Forsyth’s A Treatise on Diﬀerential Equations, W. F. Ames’ Nonlinear Partial Diﬀerential Equations in Engineering, Vol. I and II, and L. Debnath’s Nonlinear Partial Diﬀerential Equations for Scientists and Engineers. Daniel J. Arrigo April 2019

xiii

Acknowledgments I ﬁrst would like to thank my wife Peggy. She once again became a book widow. My love and thanks. Second, I would like to thank all of my students who, over the past ten years, volunteered to read the book and gave much-needed input on both the presentation of material and on the complexity of the examples given. Finally, I would like to thank Susanne Filler of Morgan & Claypool Publishers. Once again, she made the process a simple and straightforward one. Daniel J. Arrigo April 2019

1

CHAPTER

1

Nonlinear PDEs are Everywhere Outside of Quantum mechanics, the world around us is modeled by nonlinear partial diﬀerential equations (NLPDEs). Here is just a short list of places that one may ﬁnd NLPDEs. 1. The nonlinear diﬀusion equation (1.1)

u t D .D.u/ux /x

is a NLPDE that models heat transfer in a medium where the thermal conductivity may depend on the temperature. The equation also arises in numerous other ﬁelds such as soil physics, population genetics, ﬂuid dynamics, neurology, combustion theory, and reaction chemistry, to name just a few (see [1] and the references within). 2. The nonlinear wave equation u t t D c.u/2 ux

x

(1.2)

essentially models wave propagation and appears in applications involving one-dimensional gases, shallow water waves, longitudinal threadlines, ﬁnite nonlinear strings, elastic-plastic materials, and transmission lines, to name a few (see [2] and the references within). 3. Burgers’ Equation u t C uux D uxx

(1.3)

is a fundamental partial diﬀerential equation that incorporates both nonlinearity and diﬀusion. It was ﬁrst introduced as a simpliﬁed model for turbulence [3] and appears in various areas of applied mathematics, such as soil-water ﬂow [4], nonlinear acoustics [5], and traﬃc ﬂow [6]. 4. Fisher’s equation u t D uxx C u.1

u/

(1.4)

is a model proposed for the wave of advance of advantageous genes [7] and also has applications in early farming [8], chemical wave propagation [9], nuclear reactors [10], chemical kinetics [11], and theory of combustion [12].

2

1. NONLINEAR PDES ARE EVERYWHERE

5. The Fitzhugh–Nagumo equation u t D uxx C u.1

(1.5)

u/.u C /

models the transmission of nerve impulses [13], [14] and arises in population genetics models [15]. 6. The Korteweg deVries equation (KdV) (1.6)

u t C 6uux C uxxx D 0

describes the evolution of long water waves down a canal of rectangular cross section. It has also been shown to model longitudinal waves propagating in a one-dimensional lattice, ion-acoustic waves in a cold plasma, waves in elastic rods, and used to describe the axial component of velocity in a rotating ﬂuid ﬂow down a tube [16] . 7. The Boussinesq equation 1 u t t C uxx C .2uux /x C uxxxx D 0 3

(1.7)

was introduced by Boussinesq in 1871 [17], [18] to model shallow water waves on long channels. It also arises in other applications such as in one-dimensional nonlinear lattice-waves [19], [20], vibrations in a nonlinear string [21], and ion sound waves in a plasma [22], [23]. 8. The Eikonial equation jruj D F .x/; x 2 Rn

(1.8)

appears in ray optics [24]. 9. The Gross–Pitaevskii equation i

t

D

r2

C V .x/ C j j2

(1.9)

is a model for the single-particle wavefunction in a Bose–Einstein condensate [25], [26]. 10. Plateau’s equation 1 C uy2 uxx

2ux uy uxy C 1 C u2x uyy D 0

(1.10)

arises in the study of minimal surfaces [27]. 11. The Sine–Gordon equation uxy D sin u

(1.11)

1.1. EXERCISES

arises in the study of surfaces of constant negative curvature [28], and in the study of crystal dislocations [29]. 12. Equilibrium equations @xx @xy C C Fx D 0 @x @y @xy @yy C C Fy D 0 @x @y

(1.12)

arise in elasticity. Here, xx ; xy and yy are normal and shear stresses, and Fx and Fy are body forces [30]. These have been used by Cox, Hill, and Thamwattana [31] (see also [32]) to model highly frictional granular materials. 13. The Navier–Stokes equations r uD0 u t C u ru D

rP C r 2 u

(1.13)

describe the velocity ﬁeld and pressure of incompressible ﬂuids. Here is the kinematic viscosity, u is the velocity of the ﬂuid parcel, P is the pressure, and is the ﬂuid density [33].

1.1 1.1.

EXERCISES Show solutions exist for the nonlinear diﬀusion equation u t D .um ux /x ; m 2 R

(1.14)

of the form u D k t p x q for suitable constants k; p; and q . Use these to obtain solutions to u x : (1.15) u t D .uux /x and u t D u x 1.2.

Show that Fisher’s equation u t D uxx C u.1

u/

(1.16)

admit solutions of the form u D f .x ct / where f satisﬁes the ordinary diﬀerential equation (ODE) f 00 C cf 0 C f f 2 D 0: (1.17) Further, show exact solutions can be obtained in the form f D

1 a C be kz

2

(1.18)

3

4

1. NONLINEAR PDES ARE EVERYWHERE

for suitable constants a; b; c; and k [34]. 1.3.

Show that the Fitzhugh–Nagumo equation u t D uxx C u.1

admits solutions of the form u D f .x

u/.u C /

(1.19)

ct / where f satisﬁes the ODE

f 00 C cf 0 C f .1

f /.f C / D 0:

(1.20)

Further, show exact solutions can be obtained in the form f D

1 a C be kz

(1.21)

for suitable constants a; b; c; ; and k . 1.4.

Show that solutions exist of the form uD

x2

ax C bt

(a and b constant) that satisﬁes Burgers’ equation (1.3). 1.5.

Consider the PDE

u t D uxx C 2 sech2 x u:

(1.22)

Even though linear, exact solutions to equations of the form (1.23)

u t D uxx C f .x/u 2

2

can be diﬃcult to ﬁnd. If v D e k t sinh k t or v D e k t cosh kt (k is an arbitrary constant), then u D vx tanh x v satisﬁes (1.22). 1.6.

Show

ˇ ˇ ˇ 2f 0 .x/g 0 .y/ ˇ ˇ; ˇ u D ln ˇ .f .x/ C g.y//2 ˇ

where f .x/ and g.y/ are arbitrary functions satisﬁes Liouville’s equation uxy D e u :

1.1. EXERCISES

1.7.

Show u D 4 tan

1

e axCa

1y

;

where a is an arbitrary nonzero constant satisﬁes the Sine–Gordon equation uxy D sin u:

1.8.

Show that if u D f .x C ct /

satisﬁes the KdV equation (1.6) then f satisﬁes cf 0 C 6ff 0 C f 000 D 0

(1.24)

where prime denotes diﬀerentiation with respect to the argument of f . Show there is one value of c such that f .r/ D 2 sech2 r is a solution of (1.24). 1.9.

The PDE vt

6v 2 vx C vxxx D 0

is known as the modiﬁed Korteweg de Vries (mKdV) equation. Show that if v is a solution of the mKdV, then u D vx v 2 is a solution of the KdV (1.6). 1.10. The Boussinesq equation u t t C uxx

u2

xx

1 uxxxx D 0 3

(1.25)

under the substitution u D 2 .ln F .t; x//xx (and integrating twice) becomes ([35]) FF t t

F t2 C FFxx

Fx2

1 FFxxxx 3

2 4Fx Fxxx C 3Fxx D 0:

(1.26)

Further, show that (1.26) admits solutions of the form F D ax 2 C bt 2 C c for suitable a; b; and c (see [36]).

5

6

REFERENCES

1.2

REFERENCES

[1] B. H. Gilding and R. Kersner, The characterization of reaction-convectiondiﬀusion processes by travelling waves, J. Diﬀer. Equ., 124, pp. 27–79, 1996. DOI: 10.1006/jdeq.1996.0002. 1 [2] W. F. Ames, R. J. Lohner, and E. Adams, Group properties of utt D f .u/ux x , Int. J. Nonlin. Mech., 16(5/6), pp. 439–447, 1981. 1 [3] J. M. Burgers, Mathematical examples illustrating relations occuring in the theory of turbulent ﬂuid motion, KNAW, Amsterdam, Eerste Sectie, Deel XVII, no. 2, pp. 1–53, 1939. DOI: 10.1007/978-94-011-0195-0_10. 1 [4] P. Broadbridge, Burgers’ equation and layered media: Exact solutions and applications to soil-water ﬂow, Math. Comput. Model., 16(11), pp. 163–169, 1992. DOI: 10.1016/08957177(92)90112-x. 1 [5] K. Naugolnykh and L. Ostrovsky, Nonlinear Wave Processes in Acoustics, Cambridge University Press, 1998. DOI: 10.1121/1.429483. 1 [6] R. Haberman, Mathematical Models; Mechanical Vibrations, Population Dynamics, and Trafﬁc Flow: An Introduction to Applied Mathematics, Englewood Cliﬀs, NJ, Prentice Hall, 1977. DOI: 10.1137/1.9781611971156. 1 [7] R. A. Fisher, The wave of advance of advantageous genes, Ann. Eugenic, 7, pp. 353–369, 1937. DOI: 10.1111/j.1469-1809.1937.tb02153.x. 1 [8] A. J. Ammermann and L. L. Cavalli-Sforva, Measuring the rate of spread of early farming, Man, 6, pp. 674–688, 1971. DOI: 10.2307/2799190. 1 [9] R. Arnold, K. Showalter, and J. J. Tyson, Propagation of chemical reactions in space, J. Chem. Educ., 64, pp. 740–742, 1987. DOI: 10.1021/ed064p740. 1 [10] J. Canosa, Diﬀusion in nonlinear multiplicative media, J. Math. Phys., 10, pp. 1862–1868, 1969. DOI: 10.1063/1.1664771. 1 [11] P. C. Fife. Mathematical Aspects of Reacting and Diﬀusing Systems, Lectures in Biomathematics, 28, Springer-Verlag, Berlin, 1979. DOI: 10.1007/978-3-642-93111-6. 1 [12] A. Kolmogorov, I. Petrovsky, and N. Piscunov, A study of the equation of diﬀusion with increase in the quantity of matter and its application to a biological problem, Mosc. Univ. Math. Bull., Ser. Internat., Sec A, 1, pp. 1–25, 1937. 1 [13] R. Fitzhugh, Impulses and physiological states in theoretical models of nerve membrane, Biophys. J., 1, pp. 445–466, 1961. DOI: 10.1016/s0006-3495(61)86902-6. 2

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[14] J. S. Nagumo, S. Arimoto, and S. Yoshizawa, An active pulse transmission line simulating nerve axon, Proc. IRE, 50, pp. 2061–2070, 1962. DOI: 10.1109/jrproc.1962.288235. 2 [15] D. G. Aronson and H. F. Weinberger, Lecture Notes in Mathematics, vol. 446. Partial Differential Equations and Related Topics, Ed., J. A. Goldstein, Springer, Berlin, 1975. DOI: 10.1007/bfb0070592. 2 [16] R. Miura, The Korteweg–deVries equation: A survey of results, SIAM Review, 18, no. 3, pp. 412–459, 1978. DOI: 10.1137/1018076. 2 [17] J. Boussinesq, Theorie de l’intumescence liquide, appelee onde solitaire ou de translation se propagente dans un canal rectangulaire, Comptes Rendus, 72, pp. 755–759, 1871. 2 [18] J. Boussinesq, Theorie des ondes et des remous qui se propagent le long d’un canal rectangulaire horizontal, en communi-quant au liquide contenu dans ce canal des vitesses sensiblemant parielles de la surface au fond, J. Pure Appl. Algebra, 17, pp. 55–108, 1872. 2 [19] M. Toda, Studies of a nonlinear lattice, Phys. Rep., 8, pp. 1–125, 1975. DOI: 10.1016/0370-1573(75)90018-6. 2 [20] N. J. Zabusky, A synergetic approach to problems of nonlinear dispersive wave propagation and interaction, Nonlinear Partial Diﬀerential Equations, pp. 233–258, W. F. Ames, Ed., Academic, New York, 1967. DOI: 10.1016/b978-1-4831-9647-3.50019-4. 2 [21] V. E. Zakharov, On stocastization of one-dimensional chains of nonlinear oscillations, Sov. Phys. JETP-USSR, 38, pp. 108–110, 1974. 2 [22] E. Infeld and G. Rowlands, Nonlinear Waves, Solitons and Chaos, C.U.P., Cambridge, 1990. DOI: 10.1017/cbo9781139171281. 2 [23] A. C. Scott, The application of Backlund transforms to physical problems, Backlund Transformations, R. M. Miura, Ed., Lecture Notes in Mathematics, vol. 515, pp. 80–105, Springer-Verlag, Berlin, 1975. 2 [24] D. D. Holm, Geometric Mechanics Part 1: Dynamics and Symmetry, Imperial College Press, 2011. DOI: 10.1142/p557. 2 [25] E. P. Gross, Structure of a quantized vortex in boson systems, Il Nuovo Cimento, 20(3), pp. 454–457, 1961. DOI: 10.1007/bf02731494. 2 [26] L. P. Pitaevskii, Vortex lines in an imperfect Bose gas, Sov. Phys. JETP-USSR, 13(2), pp. 451–454, 1961. 2 [27] J. C. C. Nitsche, On new results in the theory of minimal surfaces, Bull. Amer. Math. Soc., 71, pp. 195–270, 1965. DOI: 10.1090/s0002-9904-1965-11276-9. 2

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[28] L. P. Eisenhart, A Treatise on the Diﬀerential Geometry of Curves and Surfaces, Dover Publishers, New York, 1960. 3 [29] F. C. Frank and J. H. van der Merwe, One-dimensional dislocations. I. Static theory, Proc. R. Soc. Lond. A, 198, pp. 205–216, 1949. 3 [30] S. Timoshenko and J. N. Goodier, Theory of Elasticity, McGraw-Hill, 1951. 3 [31] G. M. Cox, J. M. Hill, and N. Thamwattana, A formal exact mathematical solution for a sloping rat-hole in a highly frictional granular solid, Acta. Mech., 170, pp. 127–147, 2004. DOI: 10.1007/s00707-004-0118-x. 3 [32] D. J. Arrigo, L. Le, and J. W. Torrence, Exact solutions for a class of ratholes in highly frictional granular solids, DCDIS Series B: Appl. Alg., 19, pp. 497–509, 2012. 3 [33] G. K. Batchelor, An Introduction to Fluid Mechanics, Cambridge University Press, 2000. DOI: 10.1115/1.3601282. 3 [34] M. J. Ablowitz and A. Zeppetella, Explicit solutions of Fisher’s equations for a special wave speed, Bull. Math. Bio., 41, pp. 835–840, 1979. DOI: 10.1016/s0092-8240(79)80020-8. 4 [35] R. Hirota, Exact N-soliton solutions of the wave equation of long waves in shallow-water and in nonlinear lattices, J. Math. Phys., 14, pp. 810–814, 1973. DOI: 10.1063/1.1666400. 5 [36] M. J. Ablowitz and J. Satsuma, Solitons and rational solutions of nonlinear evolution equations, J. Math. Phys., 19, pp. 2180–2186, 1978. DOI: 10.1063/1.523550. 5

9

CHAPTER

2

Compatibility We start our discussion by ﬁrst solving the NPDE uy2 D 2u;

(2.1)

u.x; x/ D 0:

(2.2)

xux

subject to the boundary condition As with most introductory courses in partial diﬀerential equations (see, for example, [37]) we use the method of characteristics. Here, we deﬁne F as F D xp

q2

2u:

(2.3)

The characteristic equations become xs ys us ps qs

D Fp D x; D Fq D 2q; D pFp C qFq D xp 2q 2 ; D Fx pFu D p; D Fy qFu D 2q:

(2.4a) (2.4b) (2.4c) (2.4d) (2.4e)

In order to solve the PDE (2.1) we will need to solve the system (2.4). As (2.1) has a boundary condition (BC), we will create BCs for the system (2.4). In the .x; y/ plane, the line y D x is the boundary where u is deﬁned. To this, we associate a boundary in the .r; s/ plane. Given the ﬂexibility, we can choose s D 0 and connect the two boundaries via x D r . Therefore, we have x D r; y D r; u D 0 when s D 0:

(2.5)

To determine p and q on s D 0, it is necessary to consider the initial condition u.x; x/ D 0: Diﬀerentiating with respect to x gives ux .x; x/ C uy .x; x/ D 0:

(2.6)

uy2 .x; x/ D 0:

(2.7)

From the original PDE (2.1) xux .x; x/

If we denote p0 D ux .x; x/ and q0 D uy .x; x/, then (2.6) and (2.7) become p0 C q0 D 0; rp0

q02 D 0:

(2.8)

10

2. COMPATIBILITY

These are easily solved giving .i / p0 D 0; q0 D 0 .i i / p0 D r; q0 D r:

We will only use case (ii) as case (i) gives rise to the trivial solution u D 0. Solving (2.4a) gives x D a.r/e s ;

(2.9a) (2.9b)

(2.10)

where a is an arbitrary function. From the boundary condition (2.5), we ﬁnd that a.r/ D r gives x D re s :

(2.11)

p D b.r/e s ;

(2.12)

q D c.r/e 2s ;

(2.14)

Solving (2.4d) gives where b is an arbitrary function. From the boundary condition (2.9b), we ﬁnd that b.r/ D r gives p D re s : (2.13) Solving (2.4e) gives

where c is an arbitrary function. From the boundary condition (2.9b), we ﬁnd that c.r/ D r gives q D re 2s : (2.15) To solve (2.4b) we need to use q . Thus,

ys D

2q D 2re 2s

(2.16)

easily integrates, giving y D re 2s C d.r/. Using the BC (2.5) shows that d D 0 gives y D re 2s :

(2.17)

Finally, we see from (2.4b) 2q 2 D r 2 e 2s

us D xp

2r 2 e 4s

(2.18)

which again integrates easily giving u D 12 .r 2 e 2s r 2 e 4s / C e.r/, where e is an arbitrary function. From the boundary condition (2.5), we ﬁnd that e.r/ D 0 gives uD

r 2 e 2s

r 2 e 4s 2

:

(2.19)

Eliminating r and s from (2.11), (2.17), and (2.19) gives uD

x2

y2 2

:

(2.20)

The reader can verify the solution (2.20) does indeed satisfy the PDE (2.1) and the BC (2.2).

2.1. CHARPIT’S METHOD

2.1

11

CHARPIT’S METHOD

Obtaining exact solutions to NLPDEs such as (2.1) can be a diﬃcult task as we are required to solve equations such as (2.4)! The diﬃculty is not so much in solving these characteristic equations but in eliminating the ﬁve arbitrary functions that appear upon integration. So we ask, is it possible to obtain exact solutions another way? Consider the PDE uy D y: (2.21) This integrates to give y2 C f .x/ 2 and substitution into the original PDE (2.1) gives

(2.22)

uD

xf 0 D 2f:

(2.23)

f D cx 2

(2.24)

This ODE is solved giving which from (2.22) leads to

y2 2 and the initial condition (2.2) gives c D 1=2, and leads to the solution given in (2.20). Consider the PDE ux D x: u D cx 2

(2.25)

(2.26)

This integrates to give x2 C g.y/ 2 and substitution into the original PDE (2.1) gives

(2.27)

uD

g 02 D 2g:

(2.28)

This ODE is solved giving (we will omit the trivial solution g D 0) gD

y2 C cy 2

c2 2

(2.29)

which from (2.27) and the initial condition (2.2) leads again to the solution (2.20). As for the ﬁnal example, consider the PDE ux C xuy D x

This integrates to give uDy

1 2 y Ch y 2

(2.30)

xy 1 2 x 2

(2.31)

12

2. COMPATIBILITY

and substitution into the original PDE (2.1) gives 2.

where h D h./ and D y

1 2 x . 2

1/h0

h02

(2.32)

1 D 2h;

This ODE actually has two solutions

h D c

1 1 .c C 1/2 ; h D 2 2 2

;

(2.33)

and leads to the exact solutions uDy

12 Cc y 2

1 u D .x 2 2

1 2 1 x .c C 1/2 ; 2 2 1 2 2 1 2 y /C y x : 2 2

(2.34)

However, only the ﬁrst solution will give rise to (2.20). A question we ask: is it really necessary to solve the reduced ODEs given in (2.23), (2.28), and (2.32)? Consider the solution forms that lead to these ODEs uD

y2 C f .x/ 2

x2 C g.y/ 2 1 2 uDy y Ch y 2 uD

(2.35)

1 2 x : 2

If we impose the initial condition u.x; x/ D 0 in each solution form (2.35), we obtain the solution u D 12 .x 2 y 2 /, but we certainly would have missed obtaining the second exact solution in (2.34). So are there others? For example, both uy D x 2 ; xux C 2yuy D 2u y 2

(2.36)

will lead to exact solutions of the given PDE (2.1), however, the boundary conditions might not be satisﬁed. In fact any PDE of the form ux u2x 2 F ; ; uy C y; xux uy 2u D 0 (2.37) x uy will give rise to an exact solution to (2.1). A number of questions arise. 1. Where did these associated PDEs come from? 2. How do we know that they will lead to a solution that also satisﬁes the BC?

2.1. CHARPIT’S METHOD

13

Before trying to answer such questions, it is important to know that an actual solution exists. Namely, does a solution exist that satisﬁes both the original PDE and second appended PDE? So, in the ﬁrst example, does a solution exist that satisﬁes both uy2 D 2u and uy D

xux

(2.38)

y‹

Here, we use the second in the ﬁrst, and ask: does a solution exist to ux D

y2 2u C ; x x

If so, then they certainly would be compatible, so

uy D

(2.39)

y‹

@ux @uy D . Calculating these gives @y @x

2uy 2y ? C D0 x x

(2.40)

and since uu D y then (2.40) is identically satisﬁed, so the two equations of (2.38) are compatible. For the second example we ask: are the following compatible? xux

uy2 D 2u and ux D x:

(2.41)

We substitute the second into the ﬁrst and then seek compatibility of x2

uy2 D 2u and ux D x:

(2.42)

To show compatibility, we diﬀerentiate the ﬁrst of (2.42) with respect to x giving 2x

(2.43)

2uy uxy D 2ux

and since ux D x then (2.43) is identically satisﬁed then (2.41) are compatible. For the ﬁnal example, we ask: are these compatible? xux

uy2 D 2u and ux C xuy D x

(2.44)

xy:

Deﬁnitely a harder problem to explicitly ﬁnd ux and uy but is that really necessary? If they were compatible, they would have the same second derivatives. If we calculate the x and y derivatives of each we obtain xuxx 2uy uxy xuxy 2uy uyy uxx C xuxy C uy uxy C xuyy

(2.45a) (2.45b) (2.45c) (2.45d)

D ux ; D 2uy ; D 1 y; D x:

Solving (2.45c) and (2.45d) for uxx and uxy gives uxx D x 2 uyy

uy C x 2

y C 1; uxy D

xuyy

x;

(2.46)

14

2. COMPATIBILITY

and the two remaining equations in (2.45) become x.2uy C x 2 /uyy

ux C xuy C x 3 xy C x D 0; .2uy C x 2 /.uyy C 1/ D 0:

(2.47a) (2.47b)

If (2.44) were to be compatible, then (2.47) would be identically satisﬁed. So we see two cases emerge: .i/ 2uy C x 2 D 0; .ii/ 2uy C x 2 ¤ 0:

In the ﬁrst case where uy D 12 x 2 , (2.47a) reduces to 2ux

1 3 x 3

x C xy D 0;

which is compatible with uy D 12 x 2 and and so (2.44) are compatible. In the second case where uyy D 1, (2.47a) reduces to ux C xuy C x.y 1/ D 0 which is identically satisﬁed by virtue of (2.44) and again, (2.44) are compatible. So now we know how to determine when two ﬁrst-order PDEs are compatible. Our next step is to determine how they come about. Consider the compatibility of the following ﬁrst-order PDEs F .x; y; u; p; q/ D 0; G.x; y; u; p; q/ D 0;

(2.48)

where p D ux and q D uy . Calculating the x and y derivatives of (2.48) gives Fx C pFu C uxx Fp C uxy Fq Fy C qFu C uxy Fp C uyy Fq Gx C pGu C uxx Gp C uxy Gq Gy C qGu C uxy Gp C uyy Gq

D D D D

0; 0; 0; 0:

(2.49)

Solving the ﬁrst three (2.49) for uxx , uxy , and uyy gives uxx

D

uxy

D

uyy

D

Fx Gq

p Fu Gq C Fq Gx C p Fq Gu ; Fp Gq Fq Gp Fp Gx p Fp Gu C Fx Gp C p Fu Gp ; Fp Gq Fq Gp Fp2 Gx C p Fp2 Gu Fy Fp Gq q Fu Fp Gq Cq Fu Fq Gp Fx Fp Gp p Fu Fp Gp C Fy Fq Gp .Fp Gq

Fq Gp /Fq

:

2.1. CHARPIT’S METHOD

15

Substitution into the last of (2.49) gives Fp Gx C Fq Gy C .p Fp C q Fq /Gu

.Fx C p Fu /Gp

.Fy C q Fu /Gq D 0;

conveniently written as ˇ ˇ Dx F ˇ ˇ Dx G

ˇ ˇ ˇ D F Fp ˇˇ C ˇˇ y ˇ Gp Dy G

ˇ Fq ˇˇ D 0; Gq ˇ

(2.51)

where Dx F D Fx C p Fu , Dy F D Fy C q Fu and j j is the usual determinant. These are known as the Charpit equations. We also assumed that Fp Gq Fq Gp ¤ 0 and Fq ¤ 0. These cases would need to be considered separately. Example 2.1 Consider xux

uy2 D 2u:

(2.52)

This is the example we considered already; now we will determine all classes of equations that are compatible with this one. Denoting G D xux uy2 2u D xp q 2 2u;

where p D ux and q D uy , then Gx D p; Gy D 0; Gu D

and the Charpit equations are ˇ ˇ Dx F ˇ ˇ p

2; Gp D x; Gq D

ˇ ˇ ˇ D F Fp ˇˇ C ˇˇ y ˇ 2q x

2q;

ˇ Fq ˇˇ D 0; 2q ˇ

or, after expansion, xFx

2qFy C xp

2q 2 Fu C pFp C 2qFq D 0:

Solving this linear PDE by the method of characteristics gives the solution as p p2 2 F ; ; q C y; xp q 2u D 0 x q

(2.53)

which is exactly the one given in (2.37)! Example 2.2 Consider ux uy D 1:

(2.54)

16

2. COMPATIBILITY

Denoting G D ux uy

1 D pq

1;

where p D ux and q D uy , then Gx D 0; Gy D 0; Gu D 0; Gp D q; Gq D p;

and the Charpit equations are ˇ ˇ Dx F ˇ ˇ 0

ˇ ˇ ˇ Dy F Fp ˇˇ ˇ C ˇ ˇ 0 q

ˇ Fq ˇˇ D 0; p ˇ

or, after expansion, qFx C pFy C 2pqFu D 0:

Note that the third term can be replaced by 2Fu , due to the original equation. Solving this linear PDE by the method of characteristics gives the solution as F D F .uq

2y; p; q/

(2.55)

2y; ux ; uy /:

(2.56)

2x; up

or F D F .uuy

2x; uux

For example, if we choose uuy

(2.57)

2x D 0;

then on integrating we obtain u2 D 4xy C f .x/

(2.58)

and substituting into the original PDE (2.54) gives xf 0 D f

(2.59)

u2 D 4xy C cx:

(2.60)

and leads to the exact solution If we choose uux C uy

(2.61)

2y D 0;

then on integrating we obtain u

2

y Cf

yu

x

2 3 y 3

D0

(2.62)

and substituting into the original PDE (2.54) gives ff 02

1D0

(2.63)

2.1. CHARPIT’S METHOD

17

and leads to the exact solution u

3 y C c˙ yu 2 2

x

2 3 y 3

2=3

D 0:

(2.64)

Example 2.3 Consider u2x C uy2 D u2 :

(2.65)

Denoting p D ux and q D uy , then G D u2x C uy2

u2 D p 2 C q 2

u2 :

Thus, Gx D 0; Gy D 0; Gu D

2u; Gp D 2p; Gq D 2q;

and the Charpit equation’s are ˇ ˇ ˇ ˇ ˇ Dx F Fp ˇ ˇ ˇ ˇ ˇ C ˇ Dy F Fq ˇ D 0; ˇ 2pu 2p ˇ ˇ 2qu 2q ˇ or, after expansion, pFx C qFy C p 2 C q 2 Fu C puFp C quFq D 0:

(2.66)

Note that the third term can be replaced by u2 Fu , due to the original equation. Solving (2.66), a linear PDE, by the method of characteristics gives the solution as p q p q F DF x ln u; y ln u; ; : u u u u One particular example is x

p ln u C y u

q ln u D 0; u

or p

If we let u D e

ux C uy D .x C y/ v

u : ln u

, this becomes vx C vy D 2.x C y/;

which, by the method of characteristics, has the solution v D 2xy C f .x

y/:

This, in turn, gives the solution for u as uDe

p 2xyCf .x y/

:

(2.67)

18

2. COMPATIBILITY

Substitution into the original equation (2.65) gives the following ODE: f 02

2f 0

2f C 22 D 0;

y . If we let f D g C 12 2 then we obtain

where f D f ./ and D x

g 02

(2.68)

2g D 0

whose solution is given by .r C c/2 ; g D 0; 2 where c is an arbitrary constant of integration. This, in turn, gives

(2.69)

gD

1 1 f D 2 C c C c 2 ; f D 2 2 2

(2.70)

and substitution into (2.67) gives q

uDe

1 2 x 2 Cy 2 Cc.x y/C 2 c

;

p u D e 2xyC.x

y/2 =2

;

as exact solutions to the original PDE (2.65). It is interesting to note that when we substitute the solution of the compatible equation into the original it reduces to an ODE. A natural question is: does this always happen? This was proven to be true in two independent variables by the author [38]. So now we ask, of the inﬁnite possibilities of compatible equations, can we choose the right one(s) to not only solve a given PDE but also satisfy the given boundary condition? The following example illustrates. Example 2.4 Solve ux uy

xux

(2.71)

yuy D 0

subject to the boundary conditions .i / u.x; 0/ D 0; 1 .i i / u.x; 0/ D x 2 ; 2 .i i i / u.x; x/ D 2x 2 :

(2.72)

Denoting G D pq

xp

yq;

where p D ux and q D uy , then Gx D

p; Gy D

q; Gu D 0; Gp D q

x; Gq D p

y;

2.1. CHARPIT’S METHOD

and the Charpit equations are ˇ ˇ Dx F ˇ ˇ p

Fp q

ˇ ˇ ˇ ˇ ˇ C ˇ Dy F ˇ ˇ x q

Fq p

19

ˇ ˇ ˇ D 0; y ˇ

or, after expansion, .q

x/Fx C .p

y/Fy C .2pq

xp

yq/Fu C pFp C qFq D 0:

(2.73)

Note that the third term can be replaced by pqFu , due to the original PDE. Solving (2.73) by the method of characteristics gives the solution as F D F .q 2

2xq; p 2

2yp; p=q; pq

2u/;

(2.74)

or F D F .uy2

2xuy ; u2x

2yux ; ux =uy ; ux uy

2u/:

(2.75)

So how do we incorporate the boundary conditions? We will look at each separately. Boundary Condition (i) In this case u.x; 0/ D 0, and diﬀerentiating with respect to x gives ux .x; 0/ D 0 on the boundary. From the original PDE (2.71) we then have uy .x; 0/ D 0. Substituting these into (2.75) gives F D F .0; 0; ‹; 0/: (2.76)

Note that we have included a ? in the third argument of F since we have 00 . So how do we now use (2.76)? If we choose any of the arguments in (2.76) that are zero, then the boundary conditions are satisﬁed by that particular PDE. So, if we choose the ﬁrst for example, we have uy2

2xuy D 0

(2.77)

and we know that this is compatible with the original PDE (2.71) and satisﬁes the BC (2.72) and will give rise to the desired solution. As there are two cases (a) uy D 0 and (b) uy 2x D 0, we consider each separately. Case (a) If uy D 0 then ux D 0 from the original PDE, giving u D c and the BC u.x; 0/ D 0 gives c D 0, so the solution is u 0. Case (b) In the second case where uy 2x D 0, integrating gives u D 2xy C g.x/ and substituting into the original PDE gives xf 0 D 0 so f D c . Thus, we have the exact solution u D 2xy C c . The BC u.x; 0/ D 0 gives that c D 0 and so the solution is u D 2xy: 1 Boundary Condition (ii) In this case u.x; 0/ D x 2 , and diﬀerentiating with respect to x gives 2 ux .x; 0/ D x on the boundary. From the original PDE we then have xuy .x; 0/ x 2 D 0 or uy .x; 0/ D x . Substituting these into (2.75) gives F D F . x 2 ; x 2 ; 1; 0/:

(2.78)

20

2. COMPATIBILITY

So how do we use (2.78)? Again, if we choose any combination of the arguments that is zero, then the boundary conditions are satisﬁed by that particular PDE. Thus, if we choose the sum of the ﬁrst two arguments, i.e., uy2 2xuy C u2x 2yux D 0, then the solution of this will satisfy the BC u.x; y/ D 21 x 2 . However, this PDE is nonlinear. We would like to solve a linear problem if we can so we will choose a diﬀerent combination. Another choice would be choosing the third argument in (2.78) equals 1, i.e., ux =uy D 1. So we are to solve ux

(2.79)

uy D 0:

This is easily solved giving u D f .x C y/ and substitution into the original PDE gives f 02

f 0 D 0 or f 0 D 0; f 0

D 0;

(2.80)

where D x C y . Only the second gives rise to a correct solution and we ﬁnd that f D 12 2 1 and leads to the exact solution u D .x C y/2 . 2 Boundary Condition (iii) In this case u.x; x/ D 2x 2 , and diﬀerentiating with respect to x gives ux .x; x/ C uy .x; y/ D 4x on the boundary. From the original PDE we then have ux .x; x/uy .x; x/ xux .x; x/ yy .x; x/ D 0. Solving for ux .x; x/ and uy .x; x/ gives ux .x; x/ D uy .x; x/ D 2x . Substituting these into (2.75) gives F D F .0; 0; 1; 0/:

(2.81)

So now we have lots of possibilities again: if we choose any combination of the arguments that is zero, then the boundary conditions are satisﬁed by that particular PDE. For example, .a/ uy2 2xuy D 0; .b/ u2x 2yux D 0; .c/ ux =uy 1 D 0;

(2.82)

will all leads to solutions that satisfy the BC. PDE (a) leads to u D 2xy , PDE (b) leads to the same solution, and PDE (c) leads to u D 21 .x C y/2 . As another possibility, consider 1 b F .a; b; c; d / D d : 2 c This leads to yuy D u

(2.83)

which is easily solved, giving u D yf .x/: Substitution into the original PDE (2.71) gives ff 0

xf 0

f D 0:

(2.84)

2.2. SECOND-ORDER PDES

21

Now (2.84) is nonlinear and can be integrated. Sometimes the solution form and boundary condition give us a hint of what solution to look for. Since u.x; x/ D 2x 2 , using u D yf .x/ gives xf .x/ D 2x 2 or f .x/ D 2x , which in fact does satisfy the ODE (2.84)! Up to this point, we have considered the compatibility of two ﬁrst-order PDEs. We now extend this technique to higher-order PDEs.

2.2

SECOND-ORDER PDES

Consider the following pairs of PDEs u t C ux D x;

u t D uxx

(2.85)

u t C ux D t;

u t D uxx :

(2.86)

and We ask: does a common solution exist for each pair, (2.85) and (2.86)? We certainly could solve each of the ﬁrst-order PDEs in (2.85) and (2.86) and substitute the solutions into the second equation of (2.85) and (2.86) and determine whether solutions exist, but instead we will ask if they are compatible. Compatible in the sense that all higher order derivatives would be the same. Let us consider the pair (2.85). If we isolate the derivatives u t and uxx , then ut D x

ux ; uxx D x

(2.87)

ux :

If these are to be compatible, then .u t /xx D .uxx / t . Using (2.87), this compatibility restraint becomes .x ux /xx D .x ux / t or uxxx D u tx (2.88) and, by virtue of the diﬀusion equation in (2.85), (2.88) is identically satisﬁed and, thus, the pair (2.85) are compatible. For the second pair, (2.86), isolating the derivatives u t and uxx gives ut D t

ux ; uxx D t

(2.89)

ux :

If these are to be compatible, then .u t /xx D .uxx / t . Therefore, .t

ux /xx D .t

ux / t

or

uxxx D 1

u tx ;

(2.90)

and, by virtue of the diﬀusion equation in (2.86), gives 0 D 1 which clearly is not true, so (2.86) are not compatible. As a matter of exercise, let us determine functions f .t; x/ such that the following are compatible u t C ux D f .t; x/; u t D uxx : (2.91) As we did previously, we isolate the derivatives u t and uxx . This gives u t D f .t; x/

ux ; uxx D f .t; x/

ux :

(2.92)

22

2. COMPATIBILITY

If these are to be compatible, then .u t /xx D .uxx / t . This gives .f .t; x/

ux /xx D .f .t; x/

ux / t

or fxx

uxxx D f t

u tx

(2.93)

and, by virtue of the diﬀusion equation in (2.91), gives fxx D f t . Thus, if f is any solution of the diﬀusion equation, then (2.91) will be compatible. So how complicated can this get? The next example illustrates. Determine functions f and g such that the following are compatible: u t C f .t; x/ux D g.t; x/u;

(2.94)

u t D uxx :

We ﬁrst isolate the derivatives u t and uxx . This gives u t D g.t; x/u

f .t; x/ux ; uxx D g.t; x/u

(2.95)

f .t; x/ux :

Imposing .u t /xx D .uxx / t gives gxx u C 2gx ux C guxx

fxx ux

2fx uxx

f uxxx D g t u C gu t

f t ux

f u tx :

(2.96)

Using (2.95) to eliminate the derivatives u t ; u tx ; uxx and uxxx gives .gxx

2gfx

g t / u C .2gx

fxx C 2ffx C f t / ux D 0:

(2.97)

For (2.94) to be compatible, then (2.97) this would have to be identically zero and, since u can vary, then the coeﬃcients of u and ux in (2.97) must be zero. This means that f and g must satisfy g t C 2gfx f t C 2ffx C 2gx

gxx D 0; fxx D 0:

(2.98a) (2.98b)

So, the problem of determining compatibility for the pair (2.94) turns to a very complicated problem, as we now have to solve a coupled nonlinear system of PDEs for f and g . The good news is, we can actually solve this system (see Section 3.5.1). In light of our discussion so far, we have been seeking compatible equations with the diﬀusion equation that are linear. Must we assume this on the onset? The answer is no, as the next example illustrates. Determine functions f .u/ such that u t D uxx and u t D f .u/u2x

(2.99)

are compatible. As we have done previously, we rewrite (2.99) as u t D f .u/u2x ;

uxx D f .u/u2x ;

(2.100)

2.2. SECOND-ORDER PDES

and the compatibility of (2.100) requires that .u t /xx (2.100) that u tx

23

.uxx / t D 0: It follows from the ﬁrst of

D 2f ux uxx C f 0 u3x D 2f 2 u3x C f 0 u3x D f 0 C 2f 2 u3x :

Note that the second of (2.100) has been used. Further, u txx D f 00 C 4ff 0 u4x C 3 f 0 C 2f 2 u2x uxx D f 00 C 4ff 0 u4x C 3 f 0 C 2f 2 f u4x D f 00 C 7ff 0 C 6f 3 u4x :

(2.101)

(2.102)

Similarly, uxxt

D 2f ux uxt C f 0 u tu2x 2 u3x C f 0 f u4x D 2f ux f 0 C 2f D 3ff 0 C 4f 3 u4x :

(2.103)

Therefore, compatibility gives rise to f 00 C 7ff 0 C 6f 3 u4x

or and since ux ¤ 0, then

3ff 0 C 4f 3 u4x D 0

f 00 C 4ff 0 C 2f 3 u4x D 0 f 00 C 4ff 0 C 2f 3 D 0:

(2.104) 1 The solution of (2.104) is rather complicated but it does admit the simple solution f .u/ D u and, thus, u2 u t D x ; u t D uxx (2.105) u are compatible. Up to this point, we have been determining compatible equations for the diﬀusion equation—a linear second-order PDE. We would really like to harness the power of this technique to hopefully gain some insight into NLPDEs. The next example illustrates this very nicely. Example 2.5 Seek compatibility with the nonlinear diﬀusion equation u t D .um ux /x

(2.106)

u t C A.t; x/ux D B.t; x/u:

(2.107)

.m ¤ 0/ and the linear ﬁrst-order PDE

24

2. COMPATIBILITY

Isolating u t and uxx from (2.106) and (2.107) gives mu2x ; um u gives (using (2.108) where appropriate)

u t D B.t; x/u

A.t; x/ux ; uxx D

B.t; x/u

A.t; x/ux

(2.108)

and requiring compatibility .uxx / t D .u t /xx ux .A t C 2AAx mAB/ m C .2.m C 1/Bx Axx / ux u mB 2 B t 2BAx CBxx u C D 0: um 1 Since A and B are independent of u and ux , we obtain A t C 2AAx mAB Axx 2.m C 1/Bx B t C 2BAx mB 2 Bxx

D 0; D 0; D 0; D 0:

(2.109a) (2.109b) (2.109c) (2.109d)

At this point we impose compatibility for A and B , i.e., .Axx / t D .A t /xx and .Bxx / t D .B t /xx . This gives .3m C 4/Bx ..m C 1/B Ax / D 0; .3m C 4/Bx2 D 0;

(2.110a) (2.110b)

and from (2.110b) we see two cases emerge: (i) m D (ii) m ¤

4=3; 4=3 .so Bx D 0/:

Case (i) m D 4=3 In this case we solve (2.109b) and (2.109d) for A and B giving 1 a.t/x 2 C c.t /x C d.t /; B D a.t /x C b.t/; (2.111) 3 where a d are arbitrary functions of integration. Further, substitution into the remaining equations of (2.109) gives 1 4 4 4 2 2 aP C 2ac C ab x C cP C 2c C bc x C dP C 2cd C bd D 0; (2.112a) 3 3 3 3 4 4 (2.112b) aP C 2ac C ab x C bP C 2bc C b 2 D 0; 3 3 AD

and since a

d are independent of x , this leads to 4 4 aP C 2ac C ab D 0; bP C 2bc C b 2 D 0; 3 3 4 4 2 P cP C 2c C bc D 0; d C 2cd C cd D 0: 3 3

(2.113a) (2.113b)

2.2. SECOND-ORDER PDES

25

We will assume that a ¤ 0 as this is contained in case (ii). Eliminating the nonlinear terms in (2.113) gives b aP abP D 0; c aP acP D 0; d aP adP D 0; (2.114) and are easily solved giving b D k1 a; c D k2 a;

(2.115)

d D k3 a;

where k1 ; k2 and k3 are arbitrary constants. With these assignments, (2.113a) becomes 4 aP C 2k2 C k1 a2 D 0; (2.116) 3 and is easily solved giving 1 ; aD 4 2k2 C k1 t C k4 3

(2.117)

where k4 is an additional constant. Thus, from (2.115) and (2.117) we obtain the forms for a; b; c and d , and through (2.111), the forms of A and B . Thus, (2.107) is compatible with equations of the form 1 2 x C k2 x C k3 x C k2 ut C 3 ux D u: (2.118) 4 4 2k2 C k1 t C k4 2k2 C k1 t C k4 3 3 Case (ii) m ¤ 4=3 In this case, we have Bx D 0. We solve (2.109b) and (2.109d) giving A D c.t/x C d.t /;

(2.119)

B D b.t/;

where b; c; and d are arbitrary functions of integration. Further, substitution into the remaining equations of (2.109) gives cP C 2c 2 mbc x C dP C 2cd mbd D 0; (2.120a) Pb C 2bc mb 2 D 0; (2.120b) and since b; c; and d are independent of x , this leads to bP C 2bc

mb 2 D 0; cP C 2c 2

mbc D 0; dP C 2cd

These can be solved as in the previous case. Their solution is: If b D 0 then c0 d0 cD ; dD ; 2c0 t C k 2c0 t C k

mbc D 0:

(2.121a)

(2.122)

26

2. COMPATIBILITY

giving (2.107) as 2c0 x C d0 ux D 0: 2c0 t C k

(2.123)

1 c0 d0 ; cD ; dD ; m/t C k .2k1 m/t C k .2k1 m/t C k

(2.124)

ut C

If b ¤ 0 then bD

.2k1

where b0 ; c0 ; d0 and k are all arbitrary constants. The result from (2.107) is ut C

c0 x C d0 ux D .2k1 m/b0 t C k .2k1

b0 u: m/b0 t C k

(2.125)

We now would like to use these results to ﬁnd exact solutions of the original PDE (2.106). In the case of m D 4=3, we use (2.118). We will consider the case where k1 D k2 D k3 D 0 and k4 D 1. So (2.118) becomes 1 2 ut x ux D x u: (2.126) 3 Solving gives 3 u D x 3F t ; (2.127) x and substituting into (2.106) gives

F0 F D9 F 4=3 0

0

:

(2.128)

Integrating once gives

F0 C c1 : (2.129) F 4=3 Even though (2.129) can be solved leading to an implicit solution, we will suppress the constant c1 . This leads to the exact solution 3=4 27 1 F D ; (2.130) 4 C 0 F D9

and through (2.127) gives

0

13=4

1 B 27 C (2.131) @ A : 4 3 t C 0 x Next, we will use (2.123). We will consider the case where k D d0 D 0. So (2.123) becomes uDx

3

ut C

x ux D 0: 2t

(2.132)

2.2. SECOND-ORDER PDES

Solving gives uDF

x p

t

27

(2.133)

;

and substituting into (2.106) gives 1 0 F D F m F 00 C mF m 2

1

p F 02 ; D x= t:

(2.134)

Now any solution of this ODE will give rise to an exact solution of the original PDE (2.106). Example 2.6 Seek compatibility of Burgers’ equation u t C uux D uxx

(2.135)

u t C X.t; x; u/ux D U.t; x; u/:

(2.136)

and the quasilinear ﬁrst-order PDE

Isolating u t and uxx from (2.135) and (2.136) gives u t D U.t; x; u/

X.t; x; u/ux ; uxx D U.t; x; u/ C .u

X.t; x; u//ux ;

(2.137)

and requiring compatibility .uxx / t D .u t /xx (using (2.137) where appropriate) gives .Xxx

Xuu u3x C .2Xxu Uuu C 2uXu 2XXu/ u2x 2Uxu X t C uXx 2XXx C 2UXu C U / ux CU t C 2UXx C uUx Uxx D 0:

(2.138)

Since X and U are independent of ux , for (2.138) to be identically satisﬁed, the coeﬃcients of um x , m D 0; 1; 2 and 3 must be zero. This gives

Xxx

2Uxu

Xuu D 0; 2Xxu Uuu C 2uXu 2XXu D 0; X t C uXx 2XXx C 2UXu C U D 0; U t C 2UXx C uUx Uxx D 0:

(2.139a) (2.139b) (2.139c) (2.139d)

We can easily solve the ﬁrst two equations in (2.139). This gives rise to X D Ku C A; 1 U D K.1 K/u3 C .Kx 3

(2.140a) AK/2 C Bu C C;

(2.140b)

where K; A; B and C are arbitrary functions of t and x . Substituting (2.140) into (2.139c) and (2.139d) and isolating coeﬃcients with respect to u gives

28

2. COMPATIBILITY

A.Kxx 1 .1 3

C t C 2CAx Cxx B t C Cx C 2BAx C 2CKx Bxx K t / C K tx Kxxx C Bx C K.Axx 2AAx A t / C .4Ax C 2B/Kx 8 1 2 2K/K t .2K C 1/AKx C Kx2 C .K C 1/Kxx .2K 2 C K/Ax 3 3 3 1 .2K 2 1/Kx 3

D 0; (2.141a) D 0; (2.141b) D 0; (2.141c) D 0; (2.141d) D 0; (2.141e)

and Kt

Axx 2AAx C C A t C 2KC 2Bx D 0; 3Kxx C 2KB C B C 2KAx C 2AKx C Ax D 0; K .4Kx 2KA A/ D 0; 1 K.2K C 1/.K 1/ D 0: 3

(2.142a) (2.142b) (2.142c) (2.142d)

The systems (2.141) and (2.142) represent an overdetermined system of equations to determine the form of A; B; C; and K . From (2.142d) we see that three cases emerge: .i/ K D 0; .i i / K D 1; .i i i / K D 1=2:

Each will be considered separately. Case (i) K D 0 In this case, the systems (2.141) and (2.142) become Cxx D 0; Bxx D 0; Bx D 0;

(2.143a) (2.143b) (2.143c)

2AAx C C D 0; B C Ax D 0:

(2.144a) (2.144b)

C t C 2CAx B t C Cx C 2BAx

and Axx

At

2Bx

From (2.143c), (2.144a), and (2.144b) we ﬁnd A D a.t /x C b.t/;

BD

a.t /;

C D aP C 2a2 x C bP C 2ab;

(2.145)

where a and b are arbitrary functions of t . Here we ﬁnd that (2.143b) is automatically satisﬁed while (2.143a) becomes aR C 6aaP C 4a3 x C bR C 2b aP C 4a2 b D 0; (2.146)

2.2. SECOND-ORDER PDES

29

which leads to aR C 6aaP C 4a3 D 0; bR C 2b aP C 4a2 b D 0:

(2.147a) (2.147b)

1 cP 1 dP ; bD ; 2c 2c

(2.148)

If we let aD

then (2.147) becomes « c D 0; c

d« D 0; c

(2.149)

which readily integrates, giving c D c2 t 2 C c1 t C c0 ; d D c3 t 2 C c4 t C c5 ;

where c0

(2.150)

c5 are arbitrary constants leading to a and b as aD

1 2c2 t C c1 2c3 t C c4 1 ; bD : 2 c2 t 2 C c1 t C c0 2 c2 t 2 C c1 t C c0

(2.151)

This gives compatible equations to Burgers’ equation in the form 1 .2c2 t C c1 /x C 2c3 t C c4 2c2 t C c1 c2 x C c3 1 ut C uC : ux D 2 2 2 2 c2 t C c1 t C c0 2 c2 t C c1 t C c0 c2 t C c1 t C c0 (2.152)

Case (ii) K D 1 In this case, the systems (2.141) and (2.142) become C t C 2CAx Cxx D 0; B t C Cx C 2BAx Bxx D 0; A t 2AAx C Bx C Axx D 0; Ax D 0

(2.153a) (2.153b) (2.153c) (2.153d)

and Axx

At

2Bx

2AAx C 3C D 0; 3B C 3Ax D 0; 3A D 0:

(2.154a) (2.154b) (2.154c)

From (2.154) we readily see that A D B D C D 0 and that (2.153) is automatically satisﬁed. From (2.140) we obtain X D u and U D 0. From (2.135) and (2.136) we obtain u t C uux D 0; uxx D 0;

(2.155)

30

2. COMPATIBILITY

which has the common solution uD

where c1

c3 are arbitrary constants.

c1 x C c3 ; c1 t C c2

(2.156)

Case (iii) K D 1=2 In this case, the systems (2.141) and (2.142) become C t C 2CAx B t C Cx C 2BAx A t C 2AAx C 2Bx

Cxx D 0; Bxx D 0; Axx D 0:

(2.157)

It is interesting to note that in this last case, we are required to solve the nonlinear system of PDEs (2.157). Recall the section where we sought compatibility with the diﬀusion equation where the system (2.98) was derived and reference was made to Section 3.5.1. In this case, the system (2.157) also appears in Section 3.5.1. Example 2.7 Seek compatibility with the linear diﬀusion equation with a source term u t D uxx C Q.u/;

Q00 ¤ 0;

(2.158)

and general quasilinear ﬁrst-order PDE (2.159)

u t C X.t; x; u/ux D U.t; x; u/:

Isolating u t and uxx from (2.158) and (2.159) gives u t D U.t; x; u/

X.t; x; u/ux and uxx D U.t; x; u/

X.t; x; u/ux

Q.u/;

(2.160)

and requiring compatibility .uxx / t D .u t /xx (using (2.160) where appropriate) gives C .Xxx CU t

Xuu u3x C .2Xxu Uuu 2XXu / u2x 2Uxu C 2UXu 3QXu 2XXx X t / ux Uxx C 2UXx C QUu 2QXx UQ0 D 0:

(2.161)

Since X and U are independent of ux we obtain

Xxx Ut

Xuu 2Xxu Uuu 2XXu 2Uxu C 2UXu 3QXu 2XXx X t Uxx C 2UXx C QUu 2QXx UQ0

D 0; D 0; D 0; D 0:

(2.162a) (2.162b) (2.162c) (2.162d)

Integrating (2.162a) and (2.162b) gives X D P u C A; U D

1 2 3 P u C .Px 3

AP / u2 C Bu C C;

(2.163)

2.2. SECOND-ORDER PDES

31

where A; B; C and P are arbitrary functions of t and x . Substituting further into (2.162c) gives 2 3PQ C P 3 u3 2P .2Px AP / u2 3 C .P t C 3Pxx 2PAx 2APx 2BP / u A t Axx C 2AAx C 2Bx 2CP D 0:

(2.164)

From (2.164) we ﬁnd that if P ¤ 0 then we have a preliminary from of Q. Thus, two cases emerge: .i/ P ¤ 0; .i i / P D 0:

Case 1 P ¤ 0 Dividing (2.164) by 3P and isolating Q gives P t C 3Pxx 2PAx 2APx 2BP 2 2 3 2 2 QD P u C .2Px AP / u u 9 3 3P A t Axx C 2AAx C 2Bx 2CP : 3P

(2.165)

As Q D Q.u/ only, from (2.165) we deduce that P D p , A D a, B D b and C D c , all constant. Thus, Q has the form 2 2 3 p u 9

QD

2 2 2 apu2 C bu C c: 3 3 3

(2.166)

With this choice of Q we ﬁnd the remaining equation in (2.162) is automatically satisﬁed. Thus, we ﬁnd X D pu C a;

U D

1 2 3 p u 3

pau2 C bu C c;

(2.167)

and the following two are compatible: 2 2 2 2 3 2 p u apu2 C bu C c; 9 3 3 3 1 2 3 2 u t C .pu C a/ux D p u pau C bu C c: 3

u t D uxx

(2.168a) (2.168b)

Case 2 P D 0 In this case, (2.163) reduces to X D A; U D Bu C C;

(2.169)

while (2.162c) and (2.162d) become A t C 2AAx C 2Bx

Axx D 0;

(2.170)

32

2. COMPATIBILITY

and .Bu C C /Q0 C .2Ax

B/Q D .B t C 2BAx

Bxx /u C C t C 2CAx

Cxx :

(2.171)

Since (2.171) should only be an ODE for Q, four cases arise: (i) (ii) (iii) (iv)

B B B B

D 0, D 0, ¤ 0, ¤ 0,

C C C C

D 0, ¤ 0, D 0, ¤ 0.

Each will be considered separately. Subcase (i) B D 0, C D 0 In this case, (2.170) and (2.171) become A t C 2AAx

Axx D 0;

2Ax Q.u/ D 0;

(2.172)

from which we deduce that A D a, a constant, since Q ¤ 0. Thus, for arbitrary Q.u/, the following are compatible: u t D uxx C Q.u/; u t C aux D 0: (2.173) Subcase (ii) B D 0, C ¤ 0 In this case, equation (2.171) becomes CQ0 C 2Ax Q D C t C 2CAx

(2.174)

Cxx ;

and dividing through by C gives Q0 .u/ C

2Ax C t C 2CAx Q.u/ D C C

Cxx

:

(2.175)

Since (2.175) should be only an equation involving u, then

C t C 2CAx C

2Ax D m; C Cxx D k1 ;

(2.176a) (2.176b)

where m and k1 are arbitrary constants. With these assignments, (2.175) becomes Q0 .u/

mQ.u/ D k1 :

(2.177)

We note that m ¤ 0, as m D 0 gives Q00 D 0 which is inadmissible. From (2.176a), we solve for C giving 2Ax C D ; (2.178) m

2.2. SECOND-ORDER PDES

33

and with this choice (2.176b) becomes Axxx C 2A2x

A tx

(2.179)

k1 Ax D 0:

Within this subcase (2.170) becomes A t C 2AAx

(2.180)

Axx D 0:

It is left as an exercise to the reader to show that if (2.179) and (2.180) are consistent, then k1 Ax D 0; and since Ax D 0 gives C D 0, then we require that k1 D 0. We can readily solve (2.179) and (2.180) for A giving c1 x C c2 AD ; (2.181) 2c1 t C c0 where c0

c2 are arbitrary constants while solving (2.177) gives Q.u/ D k2 e mu ;

(2.182)

where k2 is an additional arbitrary constant. Thus, the following are compatible: u t D uxx C k2 e mu ;

ut C

c1 x C c2 ux D 2c1 t C c0

2c1 : m.2c1 t C c0 /

(2.183)

Subcase (iii) B ¤ 0, C D 0 In this case, equation (2.171) becomes BuQ0 .u/ C .2Ax

B/ Q.u/ D .B t C 2BAx

Bxx / u;

(2.184)

and dividing through by B gives uQ0 .u/ C

2Ax B B t C 2BAx Q.u/ D B B

Bxx

u:

(2.185)

Since (2.185) should be only an equation involving u, then 2Ax B D m; B B t C 2BAx Bxx D k1 ; B

(2.186a) (2.186b)

where m and k1 are arbitrary constants and (2.185) becomes uQ0 .u/

mQ.u/ D k1 u:

(2.187)

We can solve (2.186a) for B explicitly provided that m ¤ 1. As this is a special case, we consider it ﬁrst.

34

2. COMPATIBILITY

Special Case: m D 1 If m D 1, then from (2.187)

uQ0 .u/

(2.188)

Q.u/ D k1 u;

whose solution is Q.u/ D .k1 ln u C k2 / u;

(2.189)

where k2 is a constant of integration. For this case, (2.186a) becomes (2.190)

Ax D 0;

which must be solved in conjunction with (2.170) and (2.186b). These are easily solved giving A D c1 e k1 t C c2 ; B D c3 e k1 t

1 c1 k1 e k1 t x; 2

(2.191)

where c1 ; c2 and c3 are arbitrary constants. Thus, the following are compatible: 1 k1 t k1 t k1 t c1 k1 e x u: (2.192) u t D uxx C .k1 ln u C k2 / u; u t C c1 e C c2 ux D c3 e 2 In the case where m ¤ 1, we solve (2.186a) for B giving BD

2Ax : 1 m

(2.193)

Thus, (2.170) and (2.186b) become A t C 2AAx C

3Cm Axx D 0; A tx C 2A2x 1 m

Axxx D k1 Ax :

(2.194)

These are compatible provided that either k1 D 0 or m D 3. Of course there are other choices but we require that B ¤ 0 which eliminate those like A D 0 and Ax D 0 . The solution for A in each case is c1 x C c3 S0 AD (2.195) .k1 D 0/; A D 3 .m D 3/; 2c1 t C c0 S k1 S D 0. For each case B is given through (2.193) and Q is 4 obtained through (2.187). Our ﬁnal results:

where S D S.x/ satisﬁes S 00 C

u t D uxx C k2 um ; u t C

and 3

u t D uxx C k2 u

c1 x C c3 ux D 2c1 t C c0 .1

1 k1 u; u t 2

2c1 u ; m/.2c1 t C c0 /

0 0 S0 S 3 ux D 3 u; S S

(2.196)

(2.197)

2.2. SECOND-ORDER PDES

35

are compatible. Case (iv) B ¤ 0, C ¤ 0 In this case, dividing (2.187) by B gives C .2Ax B/ .B t C 2BAx uC Q0 C QD B B B

Bxx /

uC

C t C 2CAx B

Cxx

:

(2.198)

Since (2.198) should be only an equation involving u, then C D a; B B/ D m;

.2Ax B .B t C 2BAx Bxx / D k1 ; B C t C 2CAx Cxx D k2 ; B

(2.199a) (2.199b) (2.199c) (2.199d)

where a, m, k1 and k2 are arbitrary constants. From (2.199) we deduce k2 D ak1 and from (2.198) that Q satisﬁes .u C a/Q0 .u/ mQ.u/ D k1 .u C a/; (2.200)

which is (2.187) with a translation in the argument u. Since we can translate u in the original PDE without loss of generality, we can set a D 0 without loss of generality, and thus we are led back to case (iii). Exact Solutions One of the main goals of deriving compatible equations is to use them to construct exact solutions of a given PDE. As cubic source terms appear to be special we will focus on these and consider PDEs of the form u t D uxx C q1 u3 C q2 u (2.201)

where q1 and q2 are constant. This PDE is know as the Newel-Whitehead-Segel equation and was introduced by Newel and Whitehead [40] and Segel [41] to model various phenomena in ﬂuid mechanics. As cubic source terms arose in two places in our study, we consider each separately. We ﬁrst consider (2.168). Here we will set a D 0; b D 3k and c D 0, where k is constant, and for convenience we choose p D 3. In this case, (2.168) becomes u t C 3uux D 3u3 3ku; u t D uxx 2u3 2ku:

(2.202a) (2.202b)

uxx C 3uux C u3 C ku D 0

(2.203)

Eliminating u t in (2.202) gives

36

2. COMPATIBILITY

which involves the second member of the Riccati hierarchy. Under the substitution u D Sx =S , (2.203) becomes Sxxx C kSx D 0: (2.204) The solution of (2.204) diﬀers depending on the sign of k and is given as: S.t / D s2 .t/e !x C s1 .t/e !x C s0 .t / .k D ! 2 /; S.t / D s2 .t/x 2 C s1 .t /x C s0 .t / .k D 0/; S.t / D s2 .t/ sin.!x/ C s1 .t/ cos.!x/ C s0 .t/ .k D ! 2 /;

(2.205a) (2.205b) (2.205c)

where s1 ; s2 and s3 are arbitrary functions of integration. Substitution of any of these into (2.202a) leads to the following ODEs for s1 ; s2 and s3 : s2 s00

s1 s20 s2 s10 D 0; s0 s20 3ks0 s2 D 0;

(2.206a) (2.206b)

s1 s20 s2 s10 D 0; s0 s20 6s22 D 0;

(2.207a) (2.207b)

for k ¤ 0 and s2 s00

for k D 0. These are easily solved, giving S.t / D c1 e !x C c2 e !x C c3 e

3! 2 t

s.t/ .k D

! 2 /;

S.t/ D c1 .x 2 C 6t / C c2 x C c3 s.t / .k D 0/; 2 S.t/ D c1 sin.!x/ C c2 cos.!x/ C c3 e 3! t s.t/ .k D ! 2 /;

where c1

(2.208a) (2.208b) (2.208c)

c3 are constant and via u D Sx =S gives the exact solutions ! .c1 e !x

c2 e

!x

/

.k D ! 2 /; C c2 e C c3 e 3! 2 t 2c1 x C c2 uD .k D 0/; 2 c1 .x C 6t / C c2 x C c3 ! .c1 cos.!x/ c2 sin.!x// uD .k D ! 2 /: 2t 3! c1 cos.!x/ C c2 sin.!x/ C c3 e uD

c1

e !x

!x

(2.209)

The second place we saw cubic source terms was in (2.197). We will set k1 D 4k and k2 D q for convenience, so (2.197) becomes 0 0 S S0 u; u t 3 ux D 3 (2.210a) S S (2.210b) u t D uxx C qu3 2ku;

2.2. SECOND-ORDER PDES

37

00

where S satisﬁes S C kS D 0. We will distinguish between two diﬀerent cases: .i/ k D 0; .i i / k ¤ 0:

In the case of k D 0, then S D c1 x C c2 . However, we can set c1 D 1 and c2 = 0 without loss of generality. In this case, (2.210a) becomes 3 ux D x

ut

3 u; x2

(2.211)

which is solved giving u D xF .x 2 C 6t /;

(2.212)

where F is an arbitrary function of its argument. Substitution of (2.212) into (2.210b) (with k D 0) gives q F 00 C F 3 D 0; (2.213) 4 where F D F ./; D x 2 C 6t . The solution of (2.213) can be expressed in terms of the Jacobi elliptic functions so 2 F D p cn ; p1 2 q

2 and F D p

q

nc ; p1 2

depending on the sign of q . In the case of k ¤ 0, we integrate (2.210a), giving uD

kS.x/ F 3k t S 0 .x/

ln.S 0 / ;

(2.214)

where F is again an arbitrary function of its argument; substitution into (2.210b) gives rise to the ODE F 00 C 3F 0 C 2F C qF 3 D 0; (2.215) where F D F ./; D 3kt

ln.S 0 /. If we introduce the change of variables D

ln ;

F D G;

(2.216)

where G D G./, then (2.215) becomes G 00 C qG 3 D 0;

(2.217)

which is essentially (2.213). If we impose the change of variables (2.216) on the solution (2.214), then we obtain u D kS.x/e 3kt G S 0 .x/e 3kt ; (2.218)

38

2. COMPATIBILITY

and substitution of (2.218) into (2.210b) gives (2.217) directly! Furthermore, as kS D S”, we can write (2.218) u D S 00 .x/e 3kt G S 0 .x/e 3kt (2.219) or u D SN 0 .x/e

3kt

G SN .x/e

3kt

;

(2.220)

where SN satisﬁes the same equation as S . Thus, we again obtain solutions in terms of Jacobi ElSN 0 .x/e 3k t SN 0 .x/e 3kt 3kt N liptic functions u D cn.; p1 / and u D nc.; p1 /, D S.x/e dep p 2 2 q q pending on the sign of q . We note that these exact solutions appear in EqWorld [42].

2.3

COMPATIBILITY IN .2 C 1/ DIMENSIONS

Up until now, we have considered compatibility of PDEs in .1 C 1/ dimensions. We now extend this idea and consider PDEs in .2 C 1/ dimensions. In this section we consider the compatibility between the .2 C 1/ dimensional reaction–diﬀusion equation u t D uxx C uyy C Q.u; ux ; uy /;

(2.221)

and the ﬁrst-order partial diﬀerential equation u t D F t; x; y; u; ux ; uy :

(2.222)

This section is based on the work of Arrigo and Suazo [39]. We will assume that F in (2.222) is nonlinear in the ﬁrst derivatives ux and uy . The case where F is linear in the ﬁrst derivatives ux and uy , is left as an exercise to the reader. Compatibility between (2.221) and (2.222) gives rise to the compatibility equation constraints Fpp C Fqq D 0; Fxp Fyq C pFup qFuq C .F Q/Fpp D 0; Fxq C Fyp C qFup C pFuq C .F Q/Fpq D 0; F t C Fxx C Fyy C 2pFxu C 2qFyu C 2.F Q/Fyq C p 2 C q 2 Fuu C2q.F Q/Fuq C .F Q/2 Fqq C Qp Fx C Qq Fy C pQp C qQq Q Fu pQu Fp qQu Fq C FQu D 0:

(2.223a) (2.223b) (2.223c)

(2.223d)

Eliminating the x and y derivatives in (2.223b) and (2.223c) by (i ) cross diﬀerentiation and (i i ) imposing (2.223a) gives 2Fup C .Fp 2Fuq C .Fp

Qp /Fpp C .Fq Qp /Fpq C .Fq

Qq /Fpq D 0; Qq /Fqq D 0:

(2.224a) (2.224b)

2.3. COMPATIBILITY IN .2 C 1/ DIMENSIONS

39

Further, eliminating Fup and Fuq by again (i ) cross diﬀerentiation and (i i ) imposing (2.223a) gives rise to .2Fpp

Qpp C Qqq /Fpp C 2.Fpq Qpq /Fpq D 0; .Qpp Qqq /Fpq C 2Qpq Fqq D 0:

(2.225a) (2.225b)

Solving (2.223a), (2.225a) and (2.225b) for Fpp , Fpq and Fqq gives rise to two cases: .i/ Fpp D Fpq D Fqq D 0; 1 1 .i i / Fpp D .Qpp Qqq /; Fpq D Qpq ; Fqq D .Qqq 2 2

(2.226a) Qpp /:

(2.226b)

As we are primarily interested in compatible equations that are more general than quasilinear, we omit the ﬁrst case. If we require that the three equations in (2.226b) be compatible, then to within equivalence transformations of the original equation, Q satisﬁes (2.227)

Qpp C Qqq D 0:

Using (2.227), we ﬁnd that (2.226b) becomes Fpp D Qpp ; Fpq D Qpq ; Fqq D Qqq ;

(2.228)

F D Q.u; p; q/ C X.t; x; y; u/p C Y.t; x; y; u/q C U.t; x; y; u/;

(2.229)

from which we ﬁnd that

where X , Y and U are arbitrary functions. Substituting (2.229) into (2.224a) and (2.224b) gives 2Qup C XQpp C YQpq C 2Xu D 0; 2Quq C XQpq C YQqq C 2Yu D 0;

(2.230a) (2.230b)

while (2.223b) and (2.223c) become (using (2.227) and (2.230)) .Xp C Yq C 2U / Qpp C .Xq Yp/ Qpq C 2 Xx Yy D 0; .Xq Yp/ Qpp .Xp C Yq C 2U / Qpq 2 Xy C Yx D 0:

(2.231a) (2.231b)

If we diﬀerentiate (2.230a) and (2.230b) with respect to x and y , we obtain Xx Qpp C Yx Qpq C 2Xxu D 0; Xx Qpq C Yx Qqq C 2Yxu D 0; Xy Qpp C Yy Qpq C 2Xyu D 0; Xy Qpq C Yy Qqq C 2Yyu D 0:

If Xx2 C Yx2 ¤ 0, then solving (2.227) and (2.232a) for Qpp , Qpq and Qqq gives Qpp D

Qqq D

2 .Yx Yxu Xx Xxu / ; Xx2 C Yx2

Qpq D

2 .Xx Yxu C Yx Xxu / : Xx2 C Yx2

(2.232a) (2.232b)

40

2. COMPATIBILITY

If Xy2 C Yy2 ¤ 0, then solving (2.227) and (2.232b) for Qpp , Qpq and Qqq gives 2 Yy Yyu Xy Xyu 2 Xy Yyu C Yy Xyu Qpp D Qqq D ; Qpq D : Xy2 C Yy2 Xy2 C Yy2 In any case, this shows that Qpp , Qpq and Qqq are, at most, functions of u only. Thus, if we let Qpp D

Qqq D 2g1 .u/; Qpq D g2 .u/;

for arbitrary functions g1 and g2 , then Q has the form Q D g1 .u/ p 2 q 2 C g2 .u/p q C g3 .u/p C g4 .u/q C g5 .u/; where g3

g5 are further arbitrary functions. Substituting (2.233) into (2.231) gives 2 .Xp C Yq C 2U / g1 C .Xq Yp/ g2 C 2 Xx Yy D 0; 2 .Xq Yp/ g1 .Xp C Yq C 2U / g2 2 Xy C Yx D 0:

(2.233)

(2.234)

Since both equations in (2.234) must be satisﬁed for all p and q , this requires that each coeﬃcient of p and q must vanish. This leads to 2g1 X g2 Y D 0; g2 X C 2g1 Y D 0;

(2.235)

2g1 U C Xx Yy D 0; g2 U C Xy C Yx D 0:

(2.236)

and

From (2.235) we see that either g1 D g2 D 0 or X D Y D 0. If g1 D g2 D 0, then Q is quasilinear giving that F is quasilinear, which violates our non-quasilinearity condition. If X D Y D 0, we are led to a contradiction, as we imposed Xx2 C Yx2 ¤ 0 or Xy2 C Yy2 ¤ 0. Thus, it follows that Xx2 C Yx2 D 0;

Xy2 C Yy2 D 0;

or Xx D 0; Xy D 0;

Yx D 0; Yy D 0:

Furthermore, from (2.236) we obtain U D 0. Since Q is not quasilinear then from (2.231) we deduce that .Xp C Y q/2 C .Xq Yp/2 D 0; from which we obtain X D Y D 0. With this assignment, we see from (2.229) that F D Q and from (2.230) that Q satisﬁes Qup D 0; Quq D 0; (2.237)

2.4. COMPATIBILITY FOR SYSTEMS OF PDES

41

which has the solution (2.238)

Q D G.p; q/ C H.u/

for arbitrary functions G and H . From (2.227), we ﬁnd that G satisﬁes Gpp C Gqq D 0; from (2.223d), that H satisﬁes H 00 D 0 giving that H D cu where c is an arbitrary constant, noting that we have suppressed the second constant of integration due to translational freedom. This leads to our main result. Equations of the form u t D uxx C uyy C cu C G ux ; uy

are compatible with the ﬁrst-order equations u t D cu C G ux ; uy ;

where c is an arbitrary constant and G.p; q/, a function satisfying Gpp C Gqq D 0.

2.4

COMPATIBILITY FOR SYSTEMS OF PDES

We now extend the idea of compatibility to systems of PDEs. In this section we consider the Cubic Schrödinger equation i

If we assume that

t

C

xx

C kj j2

D 0:

(2.239)

D u C iv then (2.239) becomes v t C uxx C ku.u2 C v 2 / D 0; u t C vxx C kv.u2 C v 2 / D 0:

(2.240)

Here, we seek compatibility with the pair of ﬁrst-order PDEs u t C A.t; x; u; v/ux C B.t; x; u; v/vx D U.t; x; u; v/; u t C C.t; x; u; v/ux C D.t; x; u; v/vx D V .t; x; u; v/;

(2.241)

for some functions A; B; C; D; U and V to be determined. We solve (2.240) and (2.241) for u t ; v t ; uxx and vxx and require compatibility, that is .u t /xx D .uxx / t

and .v t /xx D .vxx / t :

(2.242)

Isolating the coeﬃcients of ux and vx gives rise to two sets of determining equations for the unknowns A; B; C; D; U and V . Each set contains ten determining equations. We only list six of each ten as they are the only ones needed in our preliminary analysis. Also, they are the smaller

42

2. COMPATIBILITY

equations. These are:

Uvv

Auu D 0; 2Auv C Buu D 0; Avv C 2Buv D 0; Bvv D 0; Uuu 2Axu C .2C B/Au AAv 2ABu DCu C Cv D 0; 2Bxv C 2DAv C DBu .C C 4B/Bv BDu C .A 2D/Dv D 0; 2Uuv 2Axv 2Bxu C 3DAu C .C 2B/Av 3BBu 3ABv BCu C .A 2D/Cv DDu CDv D 0;

(2.243a) (2.243b) (2.243c) (2.243d) (2.243e) (2.243f )

Cuu D 0; 2Cuv C Duu D 0; Cvv C 2Duv D 0; Dvv D 0; 2Cxu C .2A D/Au C CAv C .B C 4C /Cu ACv 2ADu D 0; Vvv 2Dxv C BBu C ABv C 2DCv C DDu C .C 2B/Dv D 0; 2Vuv 2Cxv 2Dxu C BAu C AAv C .2A D/Bu C CBv C3DCu C 3C Cv C .2C B/Du 3ADv D 0:

(2.244a) (2.244b) (2.244c) (2.244d) (2.244e) (2.244f )

(2.243g)

and

Vuu

(2.244g)

We easily solve (2.243a)–(2.243d) and (2.244a)–(2.244d) giving A D A1 .t; x/uv B1 .t; x/v 2 C A2 .t; x/u C A3 .t; x/v C A4 .t; x/; B D A1 .t; x/u2 C B1 .t; x/uv C B2 .t; x/u C B3 .t; x/v C B4 .t; x/; C D C1 .t; x/uv D1 .t; x/v 2 C C2 .t; x/u C C3 .t; x/v C C4 .t; x/; D D C1 .t; x/u2 C D1 .t; x/uv C D2 .t; x/u C D3 .t; x/v C D4 .t; x/;

(2.245a) (2.245b) (2.245c) (2.245d)

where Ai ; Bi ; Ci and Di ; i D 1; 2; 3; 4 are arbitrary functions of integration. We impose (2.245) on the remaining equations in (2.243) and (2.244) and then require the resulting equations to be compatible in the sense that .Uuu /v D .Uuv /u ; .Uuv /v D .Uvv /u ; .Vuu /v D .Vuv /u ; and .Vuv /v D .Vvv /u :

(2.246)

We performing this calculation and set the coeﬃcients with respect to u; v; u2 ; uv and v 2 to zero. The coeﬃcients of u2 ; uv and v 2 are presented below: A21 D 0; 2A1 B1 A1 C1 2C1 D1 D 0; 3A1 D1 B1 C1 C B12 C 4D12 D 0;

(2.247a) (2.247b) (2.247c)

2.4. COMPATIBILITY FOR SYSTEMS OF PDES

(2.248a) (2.248b) (2.248c)

B1 D1 D 0; 6A1 B1 3A1 C1 C 4C1 D1 D 0; 3B12 C 2D12 A1 D1 3C1 B1 D 0;

2A21

43

(2.249a) (2.249b) (2.249c)

A1 C1 D 0; 6C1 D1 3B1 D1 C 4A1 B1 D 0; 3C12 C A1 D1 C 3B1 C1 D 0 D 0;

and D12 D 0; 2A1 B1 C B1 D1 2C1 D1 D 0; 2 4A1 C C12 C 3A1 D1 B1 C1 D 0:

(2.250a) (2.250b) (2.250c)

From (2.247)–(2.250) we see that A1 D D1 D 0 and B1 .B1

C1 / D 0; C1 .B1

(2.251)

C1 / D 0;

from which we deduce that C1 D B1 . We now return to (2.246) and isolate coeﬃcients with respect to u and v . These are: (2.252a) (2.252b) (2.252c) (2.252d) (2.252e) (2.252f ) (2.252g) (2.252h)

B1 . 7A3 11B2 3C2 C 3D3 / D 0; B1 . 3A2 C 2C3 C D2 / D 0; B1 .5B3 C3 C 2D2 / D 0; B1 .5A3 C 9B2 C C2 9D3 / D 0; B1 .9A2 B3 9C3 5D2 / D 0; B1 . 2A3 C B2 5C2 / D 0; B1 . A3 2B2 C 3D3 / D 0; B1 . 3A2 C 3B3 C 11C3 C 7D2 / D 0;

from which we see two cases emerge: (i) B1 ¤ 0 and (ii) B1 D 0. Case (i) B1 ¤ 0 Solving (2.252) gives A2 D 0; A3 D

2B2 ; C1 D B1 ; C2 D B2 ; C3 D B3 ; D2 D

2B3 ; D3 D 0;

(2.253a)

and from the remaining compatibility conditions (2.246), we obtain B4 D

B2 B3 B1

3 B1x B2 B3 3 B1x ; C4 D C ; 2 B1 B1 2 B1

D4 D A4 C

B22

B32 B1

:

(2.254)

44

2. COMPATIBILITY

Now that we have consistency, we solve (2.243e)–(2.243g) and (2.244e)–(2.244g) for U and V . As these are fairly complicated (with 20 terms each), we suppress the output. We now return to (2.242) and consider the remaining 4 and 4 terms. As we are able to isolate terms involving u and v , we do so. This gives rise to a total of 114 new determining equations. It is left as an exercise to the read to show that the ﬁnal forms of A; B; C; D; U; and V are: AD

c1 v 2 C 2c2 t C 2c3 ; B D C D c1 uv; D D c1 u2 C 2c2 t C 2c3 ; U D c1 .c2 t C c3 / .u2 C v 2 /v .c2 x C c4 /v; V D c1 .c2 t C c3 / .u2 C v 2 /u C .c2 x C c4 /u;

(2.255)

giving rise to the following compatible equations: u t C . c1 v 2 C 2c2 t C 2c3 / ux C c1 uvvx D c1 .c2 t C c3 /v.u2 C v 2 / .c2 x C c4 /v; v t C c1 uvux C . c1 u2 C 2c2 t C 2c3 /vx D c1 .c2 t C c3 /u.u2 C v 2 / C .c2 x C c4 /u; (2.256)

where c1

c4 are arbitrary constants.

Case (ii) B1 D 0 In order to facilitate ﬁnding a solution to (2.243) and (2.244) with the forms of A; B; C; and D given in (2.245) (with A1 D B1 D C1 D D1 D 0), we note that the original system (2.240) is invariant under the transformation u!

v;

(2.257)

v ! u:

We would expect the augmented pair (2.241) to also be invariant under the same transformation. In this case system (2.241) is u t C .A2 u C A3 v C A4 /ux C .B2 u C B3 v C B4 /vx D U.t; x; u; v/; v t C .C2 u C C3 v C C4 /ux C .D2 u C D3 v C D4 /vx D V .t; x; u; v/;

(2.258)

which becomes under (2.257) u t C .D3 u D2 v C D4 /ux C . C3 u C C2 v C4 /vx D V .t; x; v; u/; v t C . B3 u C B2 v B4 /ux C .A3 u A2 v C A4 /vx D U.t; x; v; u/:

(2.259)

Comparing (2.258) and (2.259) gives A2 D D3 ; A3 D D2 ; A4 D D4 ; B2 D C3 ; B3 D C2 ; B4 D C4 ; C2 D B3 ; C3 D B2 ; C4 D B4 ; D2 D A3 ; D3 D A2 ; D4 D A4 ;

(2.260)

and from (2.260) we have A2 D B2 D C2 D D2 D A3 D B3 D C3 D D3 D 0; and C4 D

B4 ; D4 D A4 :

(2.261)

2.5. EXERCISES

45

With these assignments (2.243e)–(2.243g) and (2.244e)–(2.244g) reduce considerably and the entire set of determining equations (10 C 10) can be fully integrated, eventually leading to the compatible equations c1 u .c2 x c4 /v c1 x C 2c2 t C c3 ux D ; 2c1 t C c0 2c1 t C c0 c1 x C 2c2 t C c3 .c2 x c4 /u c1 v vt C vx D ; 2c1 t C c0 2c1 t C c0

ut C

where c0

(2.262)

c4 are arbitrary constants.

This chapter has considered the compatibility between PDEs: either single PDEs or systems of PDEs. It is interesting to note that several authors (see, for example, Pucci and Saccomandi [43], Arrigo and Beckham[44], and Niu et al. [45]) have shown that compatibility of a given PDE and a ﬁrst-order quasilinear PDE is equivalent to the nonclassical method in the symmetry analysis of diﬀerentials (Bluman and Cole [46]). Symmetry analysis of diﬀerential equations, ﬁrst introduced by Lie [47], plays a fundamental role in the construction of exact solutions to nonlinear partial diﬀerential equations and provides a uniﬁed explanation for the seemingly diverse and ad hoc integration methods used to solve ordinary diﬀerential equations. At the present time, there is extensive literature on the subject, and we refer the reader to the books by Arrigo [48], Bluman and Kumei [49], Cherniha et al. [50], and Olver [51] to name just a few.

2.5 2.1.

EXERCISES Solve the following PDEs by the method of characteristics: .i/ u2x 3uy2 u D 0; .i i / u t C u2x C u D 0;

.i i i / .iv/

2.2.

u2x C uy2 D 1; u2x u uy D 0;

u.x; 0/ D x 2 ; u.x; 0/ D x; p u.x; 1/ D x 2 C 1; u.x; y/ D 1 along y D 1

x:

Use Charpit’s method to ﬁnd compatible ﬁrst-order PDEs for the following: .i/ u2x C uy2 .i i / u t C uu2x

D 2x; D 0:

Use any one of the compatible equations derived above to obtain an exact solution of the PDEs given. 2.3.

Show the PDE u t D .uux /x

46

2. COMPATIBILITY

is compatible with (2.265)

uxxx D 0:

Further show that it admits solutions of the form u D a.t /x 2 C b.t /x C c.t/:

2.4.

Find the value of k such that the PDE u t D uuxx C ku2x

is compatible with .i / uxxxx D 0; .i i / uxxxxx D 0:

Further show that the original PDE admits solutions of the form .i/ u D a.t /x 3 C b.t /x 2 C c.t/x C d.t /; .i i / u D a.t /x 4 C b.t /x 3 C c.t/x 2 C d.t /x C e.t/:

2.5.

Show that u t D uxx C u ln u;

u tx D

u t ux ; u

are compatible. Further show that the ﬁrst PDE admits solutions of the form u D T .t /X.x/. 2.6.

Show that u t D uxx C u2x C u2 ;

uxxx C ux D 0;

are compatible. Further show that the ﬁrst PDE admits solutions of the form u D a.t/ cos x C b.t / [52]. 2.7.

Consider the compatibility between u t D D.u/uxx and u t D F .u/unx :

(2.266)

Can you identify a relationship between D.u/ and F .u/ and for what powers n, such that these two PDEs are compatible?

2.6. REFERENCES

2.6

47

REFERENCES

[37] D. J. Arrigo, An Introduction to Partial Diﬀerential Equations, Morgan & Claypool, 2017. 9 [38] D. J. Arrigo, Nonclassical contact symmetries and Charpit’s method of compatibility, J. Nonlinear Math. Phys., 12(3), pp. 321–329, 2005. DOI: 10.2991/jnmp.2005.12.3.1. 18 [39] D. J. Arrigo and L. R. Suazo, First-order compatibility for a (2 C 1)-dimensional diffusion equation, J. Phys. A: Math. Gen., 41, no. 025001, 2008. DOI: 10.1088/17518113/41/2/025001. 38 [40] A. C. Newell and J. A. Whitehead, Finite bandwidth, ﬁnite amplitude convection, J. Fluid Mech., 38(2), pp. 279–303, 1969. DOI: 10.1017/s0022112069000176. 35 [41] L. A. Segel, Distant side-walls cause slow amplitude modulation of cellular convection, J. Fluid Mech., 38, pp. 203–224, 1969. DOI: 10.1017/s0022112069000127. 35 [42] A. D. Polyanin and V. F. Zaitsev, Exact solutions of the Newell–Whitehead Equation. http://eqworld.ipmnet.ru 38 [43] E. Pucci and G. Saccomandi, On the weak symmetry groups of partial diﬀerential equations, J. Math. Anal. Appl., 163, pp. 588–598, 1992. DOI: 10.1016/0022-247x(92)90269-j. 45 [44] D. J. Arrigo and J. R. Beckham, Nonclassical symmetries of evolutionary partial differential equations and compatibility, J. Math. Anal. Appl., 289, pp. 55–65, 2004. DOI: 10.1016/j.jmaa.2003.08.015. 45 [45] X. Niu, L. Huang, and Z. Pan, The determining equations for the nonclassical method of the nonlinear diﬀerential equation(s) with arbitrary order can be obtained through the compatibility, J. Math. Anal. Appl., 320, pp. 499–509, 2006. DOI: 10.1016/j.jmaa.2005.06.058. 45 [46] G. W. Bluman and J. D. Cole, The general similarity solution of the heat equation, J. Math. Mech., 18, pp. 1025–1042, 1969. 45 [47] S. Lie, Klassiﬁkation und Integration von gewohnlichen Diﬀerentialgleichen zwischen x , y die eine Gruppe von Transformationen gestatten, Mathematische Annalen, 32, pp. 213– 281, 1888. DOI: 10.1007/bf01444068. 45 [48] D. J. Arrigo, Symmetries Analysis of Diﬀerential Equations—An Introduction, Wiley, Hoboken, NJ, 2015. 45 [49] G. Bluman, and S. Kumei, Symmetries and Diﬀerential Equations, Springer, Berlin, Germany, 1989. DOI: 10.1007/978-1-4757-4307-4. 45

48

REFERENCES

[50] R. Cherniha, S. Mykola, and O. Pliukhin, Nonlinear Reaction-Diﬀusion-Convection Equations: Lie and Conditional Symmetry, Exact Solutions and their Applications, CRC Press, Boca Raton, FL, 2018. DOI: 10.1201/9781315154848. 45 [51] P. J. Olver, Applications of Lie Groups to Diﬀerential Equations, 2nd ed., Springer, Berlin, Germany, 1993. DOI: 10.1007/978-1-4612-4350-2. 45 [52] V. A. Galaktionov, On new exact blow-up solutions for nonlinear heat conduction equations with source and applications, International Journal of Diﬀerential Equations, 3, pp. 863–874, 1990. 46

49

CHAPTER

3

Diﬀerential Substitutions One of the simplest NLPDEs is Burgers’ equation u t C uux D uxx :

(3.1)

The PDE was introduced by Burgers in 1934 as a rough model for turbulence. This equation has the steepening eﬀect of the NPDE u t C uux D 0;

(3.2)

and the diﬀusive eﬀect of the heat equation u t D uxx :

(3.3)

Here, we ﬁrst consider a related PDE–potential Burgers’ equation. If we let u D vx ;

(3.4)

1 v t C vx2 D vxx : 2

(3.5)

then (3.1) can be integrated once, giving

Note that the function of integration can be set to zero with outloss of generality. A remarkable fact is that this nonlinear PDE (3.5) can be linearized. If we let v D f .w/;

(3.6)

1 f 0 .w/w t C f 02 .w/wx2 D f 0 .w/wxx C f 00 .w/wx2 : 2

(3.7)

then (3.5) becomes

Choosing 1 02 f 2

(3.8)

w t D wxx Š

(3.9)

f 00 D

gives (3.7) as the heat equation

50

3. DIFFERENTIAL SUBSTITUTIONS

Solving (3.8) gives f D

2 ln jw C c1 j C c2 ;

(3.10)

where c1 and c2 are constants of integration. Setting c1 D c2 D 0 then f D 2 ln jwj and from (3.6) v D 2 ln jwj; (3.11) transforming the potential Burgers’ equation (3.5) to (3.9), the heat equation. Now we return to the actual Burgers’ equation (3.1). Combining (3.4) and (3.11) tells us that Burgers’ equation and the heat equation are related via wx uD 2 : (3.12) w This transformation is known as the Hopf–Cole transformation, introduced independently by Hopf [53] and Cole [54], but appeared earlier in Forsythe [55]. If we start with Burgers’ equation (3.1), we can see that it is transformed to the heat equation through the Hope–Cole transformation (3.12). The following illustrates: 1 2 u ut D ux 2 x 2 ! w wx2 1 wxx w x x 2 D 2 C2 2 2 w t w w 2 w x w w xx x D (3.13) ww x ww t xx t D w x w w t D wxx C c 0 .t/w; where c 0 .t/ is arbitrary. However, we can set c 0 .t/ D 0 as introducing the new variable wQ D e c.t/ w , gives the last of (3.13) with c 0 .t/ D 0 and further leaves (3.12) unchanged. We now consider two examples. Example 3.1 Solve the initial value problem u t C uux D uxx ; 0 < x < ; u.0; t / D u.; t / D 0; u.x; 0/ D sin x:

(3.14)

Passing the PDE, the boundary conditions, and the initial condition through the Hopf–Cole transformation (3.12) gives w t D wxx wx .0; t / D wx .; t / D 0 1 cos x : w.t; 0/ D exp 2

(3.15)

3. DIFFERENTIAL SUBSTITUTIONS

51

With the usual separation of variables, we ﬁnd the solution of (3.15) 1 X 1 a0 C an e 2 nD1

wD

where

2 an D

Z

0

n2 t

cos nx;

(3.16)

1 1 exp cos x cos nx dx D 2In 2 2

(3.17)

and In .x/ is the modiﬁed Bessel function of the ﬁrst kind. Passing (3.16) through the Hopf– Cole transformation (3.12) gives 1 X 1 2 4 nIn e n t sin nx 2 nD1 uD : (3.18) 1 X 1 n2 t 1 I0 . 2 / C 2 In e cos nx 2 nD1 If we solve the heat equation with the same boundary and initial conditions as given in (3.14), we obtain the simple solution u D e t sin x: (3.19) Figure 3.1 shows a comparison of the solutions (3.18) and (3.19) at a variety of times. Example 3.2 Solve the initial value problem u t C uux D uxx ;

1 < x < 1; u.t; 0/ D f .x/

Passing the initial condition through the Hopf–Cole transformation gives Z 1 x w.t; 0/ D W .x/ D exp f ./ d : 2 0 The solution of the heat equation (3.9) with this initial condition is Z 1 1 2 w.x; t/ D p W ./ e .x / =4t d 4 t 1 from which we obtain wx .x; t/ D

p

Z

1

1

4 t

.x

/ 2t

1

W ./e

(3.20)

(3.21)

(3.22)

.x /2 =4t

d :

(3.23)

From (3.12) we ﬁnd the following: Z u.x; t / D

1 1

Z

.x

/ 2t

1 1

W ./e

W ./e

.x /2 =4=t

d :

.x /2 =4=t

d

(3.24)

52

3. DIFFERENTIAL SUBSTITUTIONS

t = 0.25

t = 0.5

t = 0.75

t = 1.0

Figure 3.1: A comparison of the solutions (3.18) (red) and (3.19) (blue) at times t = 0.25, 0.5, 0.75 and 1.

3.1. GENERALIZED BURGERS’ EQUATION

53

As a particular example, we consider 8 ˆ 1;

where k is a positive constant. The initial condition (3.21) becomes 8 k x< 1 ˆ 1: With this initial condition, (3.22) becomes e k ek xC1 x 1 C 1 erf 1 C erf w.x; t / D p p 2 2 2 t 2 t k 2 t kx e 2kt x 1 2kt x C 1 erf erf ; p p 2 2 t 2 t

(3.26)

(3.27)

and through the Hopf–Cole transformation (3.12) we obtain the solution to (3.20), subject to the initial condition (3.25).

3.1

GENERALIZED BURGERS’ EQUATION

Here we consider a generalization of the Hopf–Cole transformation. Can we connect the following NLPDEs u t C A.u/ux D uxx ; v t C B.v/vx D vxx ;

(3.28a) (3.28b)

via the substitution u D F .v; vx /;

Fvx ¤ 0‹

(3.29)

Substituting (3.29) into (3.28a) and imposing (3.28b) leads to the following determining equations: Fpp D 0; A.F /Fp B.v/Fp 2pFvp D 0; A.F /Fv B.v/Fp pB 0 Fp pFvv D 0;

(3.30a) (3.30b) (3.30c)

where p D vx . Diﬀerentiating (3.30b) with respect to p twice (using (3.30a)) gives Fp3 A00 .F / D 0:

(3.31)

54

3. DIFFERENTIAL SUBSTITUTIONS

From (3.31) and (3.30a) we ﬁnd A.F / D c1 F C c2 ;

(3.32)

F D F1 .v/p C F2 .v/;

where c1 and c2 are arbitrary constants and F1 and F2 are arbitrary functions. With these assignments, returning to (3.30b) and isolating coeﬃcients with respect to p gives 2F10 c1 F12 D 0; F1 .B c1 F2 c2 / D 0;

(3.33a) (3.33b)

which we conveniently solve as F1 D

2 ; B D c1 F C c2 ; c1 v C c3

(3.34)

where c3 is an additional arbitrary constant. The remaining equation in (3.30c) becomes F200 D 0;

(3.35)

F2 D c4 v C c5 :

(3.36)

which we solve as With appropriate translation and scaling of variables we can set the following: c1 D 1, c2 D c4 D c5 D 0, and suppress the subscript in c3 . Thus, we have the following: solutions of u t C u ux D uxx

(3.37)

vx Ccv v

(3.38)

can be obtained via uD

2

where v satisﬁes v t C c v vx D vxx :

(3.39)

If we set c D 0 we get the Hopf–Cole transformation. It is interesting to note that if we set c D 1, we get a transformation which gives rise to solutions of the same equation.

3.2

KDV-MKDV CONNECTION

The Korteweg–deVries equation (KdV) u t C 6uux C uxxx D 0;

(3.40)

ﬁrst introduced by Korteweg and DeVries [56] to model shallow water waves, is a remarkable NLPDE. It has a number of applications and possesses a number of special properties (see, for example, Miura [57]). In 1968, Robert Miura found this remarkable transformation [58]. He

3.3. GENERALIZED KDV EQUATION

55

found that solutions of the KdV equation (3.40) can be found using solutions of the modiﬁed Korteweg–deVries (MKdV) equation 6v 2 vx C vxxx D 0;

vt

(3.41)

via the transformation u D vx

v2;

(3.42)

which today is known as the Miura transformation. We ask whether it’s possible to connect two general KdV-type equations.

3.3

GENERALIZED KDV EQUATION

In this section we connect the following general NLPDEs: u t C A.u/ux C uxxx D 0; v t C B.v/vx C vxxx D 0;

(3.43a) (3.43b)

u D F .v; vx /; Fvx ¤ 0:

(3.44)

via the substitution Substituting (3.44) into (3.43a) and imposing (3.43b) leads to the following determining equations:

A.F /Fv

A.F /Fp B.v/Fp

Fpp Fvp B.v/Fp C 3pFvv pB 0 Fp C p 2 Fvvv

D 0; D 0; D 0; D 0;

(3.45a) (3.45b) (3.45c) (3.45d)

where p D vx . Diﬀerentiating (3.45c) with respect to p twice (using (3.45a) and (3.45b)) gives Fp3 A00 .F / D 0:

(3.46)

From (3.46), (3.45a) and (3.45b) we see A.F / D c1 F C c2 ;

F D c3 p C F1 .v/;

(3.47)

where c0 c3 are arbitrary constants and F1 an arbitrary function. With these assignments, returning to (3.45c) and isolating coeﬃcients with respect to p gives 3F100 C c1 c32 D 0; B c1 F1 c2 D 0;

which we solve as F1 D

c1 c32 2 v C c4 v C c5 ; 6

B D c1 F1 C c2 ;

(3.48a) (3.48b)

(3.49)

56

3. DIFFERENTIAL SUBSTITUTIONS

(c4 and c5 additional constants) and shows that (3.45d) is automatically satisﬁed. As we have the ﬂexibility of translation and scaling in the variables, without loss of generality we can set c0 D 0; c1 D 6; c2 D 1; c3 D a; c4 D b . Thus, solutions of u t C 6uux C uxxx D 0

(3.50)

v 2 C av C b;

(3.51)

v 2 C av C b vx C vxxx D 0:

(3.52)

can be obtained via u D vx

where v satisﬁes vt C 6

Setting a D b D 0 gives the Muira transformation (3.42); the usual KdV-MKdV connection, setting b D 0, gives a connection between the KdV and what is known as the Gardner equation [59]. Example 3.3 We consider the NLPDE uxx C 2uuxy C u2 uyy D 0:

(3.53)

This particular PDE arises in the study of highly frictional granular materials [60]. We consider a diﬀerential substitution u D F .vx ; vy /; (3.54) and the target PDE (3.55)

vxx C A.v; vx ; vy /vxy C B.v; vx ; vy /vyy C C.v; vx ; vy / D 0:

Substitution of (3.54) into (3.53) and imposing (3.55), on isolating the coeﬃcients of vxy ; vyy ; vxyy ; vyyy and various products gives rise to the following determining equations: C 2 Fpp C .C Cp pCv C qACv 2qF Cv /Fp qCv Fq 2C.A F /Fpp 2CFpq .Cq C qAv /Fq C .qAAv pAv 2F Cp 2qFAv Cq C CAp C 2ACp /Fp 2BCFpp 2F CFpq .Cq C qBv /Fq C .ACq 2F Cq pBv C CBp 2qFBv C BCp C qABv /Fp .F A/2 Fpp C 2.F A/Fpq C Fqq C .2AAp 2FAp Aq /Fp Ap Fq 2B.A F /Fpp C 2.F 2 AF B/Fpq C 2FFqq .Aq C Bp /Fq C .BAp Bq 2FBp 2FAq C AAq C 2ABp /Fp B 2 Fpp 2FBFpq C F 2 Fqq C .ABq 2FBq C BBp /Fp Bq Fq .F 2 C A2 B 2AF /Fp C .2F A/Fq B.A 2F /Fp C .F 2 B/Fq

D 0; (3.56a) D 0; (3.56b) D 0; (3.56c) D 0; (3.56d) D 0; D 0; D 0; D 0:

(3.56e) (3.56f ) (3.56g) (3.56h)

3.3. GENERALIZED KDV EQUATION

57

Eliminating Fp from (3.56g) and (3.56h) gives B D AF

F 2;

(3.57)

and returning to either (3.56g) and (3.56h) gives .A

A/Fp C Fq D 0;

2F / .F

(3.58)

FFp C Fq , but each case ultimately leads to the Fp same result and thus, we will only pursue the ﬁrst case. On setting A D 2F in equations (3.56d), (3.56e) and (3.56f), we obtain

leading to two cases: (i) A D 2F and (ii) A D

F 2 Fpp 2FFpq C Fqq C 4FFp2 4Fp Fq D 0; F Fpp 2F 2 Fpq C FFqq C 3F 2 Fp2 2FFp Fq Fq2 D 0; F 4 Fpp 2F 3 Fpa C F 2 Fqq C 2F 3 Fp2 2FFq2 D 0; 3

(3.59a) (3.59b) (3.59c)

and on the elimination of the second-order derivatives we obtain Fq

(3.60)

FFp D 0;

and with this shows that (3.59) is identically satisﬁed. As F satisﬁes (3.60), both (3.56b) and (3.56c) reduce Cq F Cp D 0: (3.61) With F and C satisfying (3.59) and (3.61), and the last equation in (3.56), (3.56a) becomes C Cp pCv qF Cv Fp C C 2 Fpp D 0: (3.62) If we perform the operation @q

F @p on (3.62) and use (3.60) and (3.61), we obtain

2CFp Cp

.p C qF /Fp Cv C 3C 2 Fpp D 0:

(3.63)

From (3.62) and (3.63), (upon subtraction), we have C Fp Cp C 2CFpp D 0:

(3.64)

If C D 0 then we have u D F .p; q/; vxx C 2F vxy C F 2 vyy D 0;

Fq

FFp D 0:

(3.65)

If C ¤ 0 then the compatibility of Fp Cp C 2CFpp D 0 with (3.60) and (3.61) leads to Fpp D 0;

(3.66)

58

3. DIFFERENTIAL SUBSTITUTIONS

and from (3.61) and (3.64) gives Cp D Cq D 0, and from (3.62) gives .p C qF /Cv D 0, which leads to two cases. If F D p=q , then uD

vx ; vy2 vxx vy

2vx vy vxy C vx2 vyy C c.v/vy2 D 0;

(3.67)

where c.v/ is an arbitrary function. If Cv D 0, then vx C c1 ; vy C c2 C c3 .vy C c2 /2 D 0; uD

.vy C c2 /2 vxx

2.vx C c1 /.vy C c2 /vxy C .vx C c1 /2 vyy

(3.68)

where c1 and c2 are arbitrary constants. However, we can set c1 D c2 D 0 in (3.68) without loss of generality showing (3.68) to be a special case of (3.67). A natural question one might ask is “what was the point?” We have shown that we are able to express solutions of one NLPDE, (3.53), in terms of another NLPDE, (3.65) or (3.67)—a rather esoteric exercise. However, as it turns out, these PDEs, ((3.65) and (3.67)) are linearizable! This is a topic covered in the next chapter.

3.4

MATRIX HOPF–COLE TRANSFORMATION

In this section we extend the Hopf–Cole transformation to matrices. Before doing so, we must establish the derivative of inverse matrices. Consider an n n invertible matrix ˆ with variable entries. Then ˆˆ 1 D I (3.69) where I is the usual identity matrix. Diﬀerentiating with respect to x gives ˆx ˆ

1

C ˆ ˆx 1 D 0;

(3.70)

from which we obtain noting that ˆx 1 D ˆ

1

x

ˆx 1 D

ˆ

ˆt 1 D

ˆ

1

ˆx ˆ

1

;

(3.71)

and not ˆx 1 D .ˆx / 1 . Similar for the t derivative, 1

ˆt ˆ

1

:

(3.72)

We know consider the Matrix Burgers’ equation t C 2x D xx ;

(3.73)

and matrix Hopf–Cole transformation D

ˆx ˆ

1

:

(3.74)

3.5. DARBOUX TRANSFORMATIONS

Substituting (3.74) into (3.73) (using (3.71) and (3.72) appropriately) gives ˆ tx ˆ 1 C ˆx ˆ 1 ˆ t ˆ 1 2 ˆxx ˆ 1 C ˆx ˆ 1 ˆx ˆ 1 ˆx ˆ 1 D ˆxxx ˆ 1 C 2ˆxx ˆ 1 ˆx ˆ 1 C ˆx ˆ 1 ˆxx ˆ 1 2ˆx ˆ 1 ˆx ˆ 1 ˆx ˆ 1 :

59

(3.75)

Expanding and canceling appropriately gives ˆ tx ˆ

1

C ˆx ˆ

1

1

ˆt ˆ

D

ˆxxx ˆ

1

C ˆx ˆ

1

ˆxx ˆ

which becomes on multiplying on the left by ˆ 1 and on the right by ˆ ˆ 1 ˆ t x D ˆ 1 ˆxx x :

1

;

(3.76)

(3.77)

Integrating with respect to x gives the linear heat matrix equation ˆ t D ˆxx :

(3.78)

Note that, like the scalar case, the matrix function of integration can be set to zero without loss of generality.

3.5

DARBOUX TRANSFORMATIONS

In 1882, Darboux considered the following problem: is it possible to ﬁnd functions f .x/ such that solutions of y 00 C f .x/y D 0; z 00 C kz D 0 (3.79) can be connected through y D z 0 C A.x/z‹

(3.80)

A direct substitution of (3.80) into the ﬁrst of (3.79) gives z 000 C Az 00 C 2A0 z 0 C A00 z C f .x/ z 0 C Az D 0;

(3.81)

where prime denoted diﬀerentiation with respect to their argument. Imposing the second of (3.79) gives f C 2A0 k z 0 C A00 C fA kA z D 0; (3.82) and since this must be true for all z leads to f C 2A0 k D 0; A C fA kA D 0: 00

If f D k

(3.83a) (3.83b)

2A0 then (3.83b) becomes A00

2AA0 D 0;

(3.84)

60

3. DIFFERENTIAL SUBSTITUTIONS

which integrates once to become A0

A2 D c

(3.85)

where c is an arbitrary constant. If we let A D 0 = where D .x/, (3.85) becomes 00 C c D 0:

(3.86)

Composing the result gives the solution of y 00 C 2 .ln /00 C k y D 0;

(3.87)

obtained via the Darboux transformation 0 z;

y D z0

(3.88)

where and z satisfy 00 C c D 0; z 00 C kz D 0;

(3.89)

respectively. For example, if we choose solutions D cosh x and z D ax C b (a; b constant) as solutions of (3.89), then via (3.87) and (3.88) solutions of 2y

y 00 C

cosh2 x

D0

(3.90)

are given by yDa

.ax C b/ tanh x:

(3.91)

A natural question is: does this method apply to PDEs? The answer: Yes! A very important class of PDEs is the .1 C 1/ dimensional Schrödinger equation i

„ 2m

C V .x/

(3.92)

u t D uxx C f .t; x/u;

(3.93)

t

D

xx

from Quantum Mechanics. Here, we will consider

and seek Darboux transformations of the form u D vx C A.t; x/v;

(3.94)

where v satisﬁes the standard heat equation v t D vxx :

(3.95)

3.5. DARBOUX TRANSFORMATIONS

61

Substituting (3.94) into (3.93) and imposing (3.95) gives .A t

Axx

.2Ax C f / v D 0:

fA/ vx

(3.96)

Since (3.96) must be true for all v , we obtain At

2Ax C f D 0; Axx fA D 0:

(3.97a) (3.97b)

From (3.97a), we ﬁnd f D 2Ax , and eliminating f in (3.97b) gives A t C 2AAx

Axx D 0:

(3.98)

As we saw previously, this is Burgers’ equation which was linearized via the Hopf–Cole transformation A D x =: (3.99) This, in turn, gives f D 2 .ln /xx . Thus, we have the following: solutions of u t D uxx C 2 .ln /xx u

(3.100)

are given by x v;

u D vx

(3.101)

where both and v satisfy the heat equation. We now consider a few examples, choosing various simple solutions of the heat equation. Example 3.4 If we choose D x; then solutions of u t D uxx

2u x2

(3.102)

are obtained by

v ; x where here, and in the examples that follow, v is any solution of the heat equation. u D vx

Example 3.5

(3.103)

If we choose D x 2 C 2t; then solutions of u t D uxx

.x 2 2t / u .x 2 C 2t /2

(3.104)

2x v: C 2t

(3.105)

are obtained by u D vx

x2

62

3. DIFFERENTIAL SUBSTITUTIONS

Example 3.6 If we choose D e t cosh x; then solutions of u t D uxx C

u

(3.106)

cosh2 x

are obtained by tanh xv:

u D vx

(3.107)

An interesting question arises: is it possible to use Darboux transformations for any time independent potential? See the exercise section for further examination.

3.5.1 SECOND-ORDER DARBOUX TRANSFORMATIONS A natural extension is to consider a transformation that includes second derivatives, i.e., (3.108)

u D vxx C A.t; x/vx C B.t; x/v:

Substituting (3.108) into into (3.93) and imposing (3.95) gives .2Ax C f / vxx C .A t

2Bx

Axx

fA/ vx C .B t

Bxx

f B/ v D 0:

(3.109)

As this must be satisﬁed for all v we obtain

At

2Bx Bt

2Ax C f D 0; Axx fA D 0; Bxx f B D 0:

(3.110a) (3.110b) (3.110c)

We see again f D 2Ax , and the two remaining equations in (3.110) are A t C 2AAx 2Bx B t C 2BAx

Axx D 0; Bxx D 0:

(3.111a) (3.111b)

It is interesting to note that in order to ﬁnd a second-order Darboux transformation to solve a linear PDE, we must solve a system of nonlinear systems of PDEs —a task that might seem hopeless. However, we see that by introducing the matrix 0 1 D ; (3.112) B A Burgers’ system (3.111) can be written as t C 2x

xx D 0

(3.113)

3.5. DARBOUX TRANSFORMATIONS

63

and, as we saw in the section on the matrix Burgers’ equation, can be linearized via a Matrix Hopf–Cole transformation. Thus, if we let D

where ˆD

11 21

12 22

;

ˆx ˆ

1

(3.114)

ij D ij .t; x/; i; j D 1; 2;

(3.115)

ˆxx D 0:

(3.116)

then (3.113) becomes ˆt

Thus, each element of ˆ satisﬁes the heat equation. From (3.112), (3.114) and (3.115), we have (3.117)

ˆx D ˆ;

which becomes

11 x 21 x

12 x 22 x

D

0 B

1 A

11 21

12 22

:

(3.118)

Expanding (3.118) and equating elements gives the following: 11 x D 21 ; 12 x D 22 ;

21 x D B11 C A21 ; 22 x D B12 C A22 :

(3.119)

If we let 11 D !1 and 12 D !2 , then 21 D !1x and 22 D !2 then from (3.119) B!1 C A!1x D B!2 C A!2x D

(3.120a) (3.120b)

!1xx ; !2xx ;

a system of two equations in the two unknowns A and B . Solving gives ˇ ˇ ˇ ˇ ˇ !1x ˇ !1 !2 ˇˇ !2x ˇˇ ˇ ˇ ˇ !1xx !2xx ˇ ˇ !1xx !2xx ˇ ˇ ; BD ˇ ˇ : ˇ AD ˇ !1 ˇ !1 !2 ˇˇ !2 ˇˇ ˇ ˇ ˇ !1x !2x ˇ ˇ !1x !2x ˇ

(3.121)

Furthermore, f D 2Ax which can be written as f D 2 .ln W2 /xx

where W2 D !1 !2x

Substituting A and B from (3.121) into (3.108) gives ˇ ˇ ˇ !1 !2 ˇˇ ˇ ˇ !1xx !2xx ˇ ˇ vx C ˇ u D vxx ˇ ˇ !1 ! 2 ˇ ˇ ˇ !1x !2x ˇ

ˇ ˇ !1x ˇ ˇ !1xx ˇ ˇ !1 ˇ ˇ !1x

!2 !1x : ˇ !2x ˇˇ !2xx ˇ ˇ v !2 ˇˇ !2x ˇ

(3.122)

(3.123)

64

3. DIFFERENTIAL SUBSTITUTIONS

which can conveniently be written as ˇ ˇ !1 !2 v ˇ ˇ !1x ! v 2x x ˇ ˇ !1xx !2xx vxx ˇ ˇ uD ˇ !1 !2 ˇˇ ˇ ˇ !1x !2x ˇ

or uD

where W is the usual Wronskian. Example 3.7

ˇ ˇ ˇ ˇ ˇ ˇ

;

jW .!1 ; !2 ; v/j ; jW .!1 ; !2 /j

(3.124)

(3.125)

If we choose !1 D x; and !2 D x 3 C 6xt , then from (3.121) and (3.122) 3 3 ; B D 2; x x

6 ; AD x2

f D

and solutions of

(3.126)

6u x2

(3.127)

3vx 3v C 2: x x

(3.128)

u t D uxx

are obtained by u D vxx 2

2

Example 3.8 If we choose !1 D e a t cosh ax; and !2 D e b t sinh bx; where a and b are constants, then from (3.121) and (3.122) f

D

A D B D

.a2 b 2 /.b 2 cosh2 ax C a2 cosh2 bx/ ; .a sinh ax sinh bx b cosh ax cosh bx/2 .a2 b 2 / cosh ax sinh bx ; a sinh ax sinh bx b cosh ax cosh bx ab.a cosh ax cosh bx b sinh ax sinh bx/ ; a sinh ax sinh bx b cosh ax cosh bx

2

(3.129)

and solutions of u t D uxx

2

.a2 b 2 /.b 2 cosh2 ax C a2 cosh2 bx/ u .a sinh ax sinh bx b cosh ax cosh bx/2

(3.130)

can be obtained using (3.108) with A and B given in (3.129). For example, choosing a D 1 and b D 2 shows that PDEs of the form u t D uxx C

6u cosh2 x

;

(3.131)

3.5. DARBOUX TRANSFORMATIONS

65

admit a Darboux transformation of the form 3 sinh x 2 cosh2 x 3 vx C v: cosh x cosh2 x

u D vxx

(3.132)

These results generalize to nth order Darboux transformations; we refer the reader to Arrigo and Hickling [61] for futher details.

3.5.2

DARBOUX TRANSFORMATIONS BETWEEN TWO DIFFUSION EQUATIONS In this section we wish to extend the previous section’s results and now have v satisfy v t D vxx C g.t; x/v:

(3.133)

First-Order Darboux Transformations

We will start with ﬁrst-order Darboux transformations as we did earlier and seek solutions of u t D uxx C f .t; x/u

(3.134)

via the Darboux transformations of the form u D vx C A.t; x/v;

(3.135)

where v satisﬁes (3.133). As we did previously, substituting (3.135) into (3.134), imposing (3.133) and isolating coeﬃcients with respect to v and vx gives At

2Ax C f g D 0; fA C gx C gA D 0:

Axx

(3.136a) (3.136b)

From (3.136a) we see that f Dg

2Ax ;

(3.137)

which in turn gives (3.136b) as A t C 2AAx

Axx C gx D 0:

(3.138)

The Hopf–Cole transformation (3.99), i.e., AD

x =

(3.139)

works well here, where substitution and integrating (3.138) gives t

xx C g.t; x/ D 0:

(3.140)

Note that the function of integration is omitted without loss of generality. Combining the results, solutions of u t D uxx C .g.t; x/ C 2.ln /xx /u (3.141)

66

3. DIFFERENTIAL SUBSTITUTIONS

are obtained via the Darboux transformation u D vx

x v;

(3.142)

where v and satisfy (3.133) and (3.140), respectively. Second-Order Darboux Transformations

We now extend the Darboux transformation to second order and seek solutions of (3.143)

u t D uxx C f .t; x/u

via the Darboux transformations of the form (3.144)

u D vxx C A.t; x/vx C B.t; x/v;

where v satisﬁes (3.133). As we did previously, substituting (3.144) into (3.143), imposing (3.133) and isolating coeﬃcients with respect to v , vx , and vxx , gives 2Ax C f g D 0; A t Axx 2Bx fA C 2gx C gA D 0; B t Bxx C gB f B C Agx C gxx D 0:

(3.145a) (3.145b) (3.145c)

From (3.145a) we again see that f Dg

(3.146)

2Ax ;

which, in turn, gives (3.145b) and (3.145c) as A t C 2AAx B t C 2BAx

Axx 2Bx C 2gx D 0: Bxx C gxx C Agx D 0:

(3.147a) (3.147b)

If we introduce the matrices D

0 B

1 A

; GD

g gx

0 g

;

(3.148)

the system of equations (3.147) conveniently becomes t C 2x

xx C G

G C Gx D 0:

(3.149)

Again, the Matrix Hopf–Cole transformation D

ˆx ˆ

1

(3.150)

works well here, leading to the linear Matrix equation ˆt

ˆxx

Gˆ D 0:

(3.151)

3.5. DARBOUX TRANSFORMATIONS

67

As the process is now identical to that presented in (3.117)–(3.124), we simply state our results. Solutions of u t D uxx C .g.t; x/ C 2.ln /xx / u (3.152) can be obtained via the Darboux transformation ˇ ˇ ˇ !1 !2 ˇˇ ˇ ˇ !1xx !2xx ˇ ˇ ˇ vx u D vxx ˇ !1 !2 ˇˇ ˇ ˇ !1x !2x ˇ where D !1 !20

ˇ ˇ !1x ˇ ˇ !1xx C ˇ ˇ !1 ˇ ˇ !1x

ˇ !2x ˇˇ !2xx ˇ ˇ v; !2 ˇˇ !2x ˇ

(3.153)

!2 !10 , !1 , !2 , and v are all independent solutions of w t D wxx C g.t; x/w:

(3.154)

These results also generalize to nth order Darboux transformations; we refer the reader to Arrigo and Hickling [62] for futher details.

3.5.3

DARBOUX TRANSFORMATIONS BETWEEN TWO WAVE EQUATIONS In this section we wish to connect solutions to two diﬀerent wave equations. In particular, ﬁnd wave equations of the form u t t D uxx C f .x/u; (3.155) that admit Darboux transformations where v satisﬁes v t t D vxx :

(3.156)

Again, we will consider ﬁrst- and second-order transformations First-Order Darboux Transformations

We will start with ﬁrst-order Darboux transformations as we did earlier and seek solutions of (3.155) via the Darboux transformations of the form u D vx C A.x/v;

(3.157)

where v satisﬁes (3.156). As we did previously, substituting (3.157) into (3.155), imposing (3.156) and isolating coeﬃcients with respect to v and vx , gives 2A0 C f D 0; A00 C fA D 0;

(3.158a) (3.158b)

where prime denotes diﬀerentiation with respect to the argument. From (3.158a) we see that f D

2A0 ;

(3.159)

68

3. DIFFERENTIAL SUBSTITUTIONS

which in turn gives (3.158b) as A00

2AA0 D 0:

(3.160)

A2 D c;

(3.161)

This can be integrated once giving A0

where c is a constant of integration. If we introduce a Hopf–Cole transformation 0 =;

(3.162)

00 C c D 0:

(3.163)

u t t D uxx C 2.ln /00 u

(3.164)

AD

where D .x/; then (3.161) becomes Combining the results, solutions of

are obtained via the Darboux transformation u D vx

0 v;

(3.165)

where v is a solution of the wave equation (3.156) and a solution of (3.163). Second-Order Darboux Transformations

We now extend the Darboux transformation to second order and seek solutions of u t D uxx C f .x/u

(3.166)

via the Darboux transformations of the form u D vxx C A.x/vx C B.x/v;

(3.167)

where v satisﬁes (3.156). As we did previously, substituting (3.167) into (3.166), imposing (3.156) and isolating coeﬃcients with respect to v , vx , and vxx , gives 2A0 C f D 0; A00 C 2B 0 C fA D 0: B 00 C f B D 0:

(3.168a) (3.168b) (3.168c)

From (3.168a) we again see that 2A0 ;

(3.169)

2AA0 C 2B 0 D 0; B 00 2BA0 D 0:

(3.170a) (3.170b)

f D

which, in turn, gives (3.168b) and (3.168c) as A00

3.5. DARBOUX TRANSFORMATIONS

If we introduce the matrix D

0 B

1 A

69

(3.171)

;

the system of equations (3.170) conveniently becomes 00

20 D 0:

(3.172)

Again, the Matrix Hopf–Cole tranformation D

ˆ0 ˆ

1

(3.173)

works well here leading to the linear Matrix equation 1

ˆ

ˆxx

0

(3.174)

D 0;

which we integrate giving ˆxx D ˆ c;

(3.175)

where c is a 2 by 2 matrix of arbitrary constants, namely cD

c11 c21

c12 c22

(3.176)

:

As the process is now identical to that presented in (3.173)–(3.124), we simply state our results. Solutions of u t t D uxx C 2.ln /00 u

are obtained via the Darboux transformation ˇ ˇ !1 !2 ˇ ˇ !1 00 !2 00 ˇ u D vxx ˇ !1 !2 ˇ ˇ !1 0 !2 0 where D !1 !20

ˇ ˇ ˇ ˇ !1 0 ˇ ˇ ˇ !1 0 ˇ ˇ vx C ˇ ˇ !1 ˇ ˇ ˇ ˇ ˇ !1 0

(3.177) ˇ !2 0 ˇˇ !2 0 ˇ ˇ v; !2 ˇˇ !2 0 ˇ

(3.178)

!2 !10 , with !1 , !2 and v satisfying the ODEs !100 D c11 !1 C c21 !2 ; !200 D c12 !1 C c22 !2 ;

and the wave equation (3.156).

(3.179)

70

3. DIFFERENTIAL SUBSTITUTIONS

3.6 3.1.

EXERCISES Find a Darboux transformation in the form y D z 0 C A.x/z connecting y 00 C f .x/y D 0; z 00 C g.x/z D 0:

3.2.

Darboux transformations for (3.93) naturally extend to nth order where the nth order Darboux transformation is given by uD

W .!1 ; !2 ; ; !n ; v/ W .!1 ; !2 ; ; !n /

(3.180)

applicable for f D 2 .ln W /xx where W D W .!1 ; !2 ; ; !n / :

(3.181)

.a/ Show that by picking the following polynomial solutions of the heat equation !1 !2 !3 !4 !5

D D D D D

x; x 3 C 6xt; x 5 C 20x 3 t C 60xt 2 ; x 7 C 42x 5 t C 420x 3 t 2 C 840xt 3 ; x 9 C 72x 7 t C 1512x 5 t 2 C 10080x 3 t 3 C 15120xt 4 ;

that f in the case of n D 3; 4 and 5, becomes 12 ; f D x2

f D

20 ; f D x2

30 : x2

(3.182)

Conjecture the general form of f is the case of arbitrary n. .b/ Show that by picking the following solutions of the heat equation !n D

( 2 e n t cosh nx e

n2 t

for n odd for n even

sinh nx

(3.183)

that for n D 3; 4 and 5, that f becomes f D

12 2

cosh x

; f D

20 2

cosh x

; f D

30 cosh2 x

Conjecture the general form of f is the case of arbitrary n.

:

(3.184)

3.7. REFERENCES

3.3.

71

Show by a suitable transformation u D F .v/, the Thomas equation uxy C aux uy C bux C cuy D 0

(a; b and c constant) can be linearized. 3.4.

Show that the forced Burgers’ equation u t C 2uux D uxx C f .x/

can be linearized via the Hopf–Cole transformation.

3.7

REFERENCES

[53] E. Hopf, The partial diﬀerential equation u t C uux D uxx , Comm. Pure Appl. Math., 3, pp. 201–230, 1950. DOI: 10.1002/cpa.3160030302. 50 [54] J. D. Cole, On a quasi-linear parabolic equation occurring in aerodynamics, Q. Appl. Math., 9, pp. 225–236, 1951. DOI: 10.1090/qam/42889. 50 [55] A. R. Forsyth, Theory of Diﬀerential Equations, vol. VI, p. 102, Ex. 3, Cambridge University Press, 1906. 50 [56] D. J. Korteweg and F. de Vries, On the change of form of long waves advancing in a rectangular canal, and on a new type of long stationary waves, Philos. Mag., 39, pp. 422– 443, 1895. DOI: 10.1080/14786449508620739. 54 [57] R. M. Miura, The Korteweg–deVries Equation: A survey of results, SIAM Review, 18(3), pp. 412–459, 1978. DOI: 10.1137/1018076. 54 [58] R. M. Miura, Korteweg–de Vries Equation and generalizations. I. A remarkable explicit nonlinear transformation, J. Math. Phys., 9, pp. 1202–1204, 1968. DOI: 10.1063/1.1664700. 54 [59] R. M. Miura, C. S. Gardner, and M. D. Kruskal, Korteweg–de Vries Equation and generalizations. II. Existence of conservation laws and constants of motion, J. Math. Phys., 9, pp. 1204–1209, 1968. DOI: 10.1063/1.1664701. 56 [60] G. M. Cox, J. M. Hill, and N. Thamwattana, A formal exact mathematical solution for a sloping rat-hole in a highly frictional granular solid, Acta Mech., 170, pp. 127–147, 2004. DOI: 10.1007/s00707-004-0118-x. 56 [61] D. J. Arrigo and F. Hickling, A Darboux transformation for a class of linear parabolic partial diﬀerential equation, J. Phys. A: Math. Gen., 35, pp. 389–399, 2002. DOI: 10.1088/0305-4470/35/28/101. 65

72

REFERENCES

[62] D. J. Arrigo and F. Hickling, An nth order Darboux transformation for the one dimensional time dependent Schrödinger equation, J. Phys. A: Math. Gen., 36, pp. 1615–1621, 2003. DOI: 10.1088/0305-4470/36/6/307. 67

73

CHAPTER

4

Point and Contact Transformations In an introductory course in PDEs we found that by introducing new variables x D f .r; s/;

@.x; y/ ¤ 0; @.r; s/

y D g.r; s/;

(4.1)

we were able to transform PDEs of the form Auxx C Buxy C C uyy C l.o.t.s D 0;

(4.2)

where A; B and C are functions of .x; y/ and l.o.t.s are lower-order terms to standard form, i.e.,

urr

uss C l.o.t.s D 0 parabolic; urs C l.o.t.s D 0 modiﬁed hyperbolic; C uss C l.o.t.s D 0 elliptic:

(4.3)

We now ask, can we generalize these transformation? For example, can we consider transformations of the form x D F .X; Y; U /; y D G.X; Y; U /;

u D H.X; Y; U /;

(4.4)

where U D U.X; Y /? Transformations of this type are referred to as point transformations. Consider, for example, the transformation x D X; y D U;

(4.5)

u D Y:

In order to see the eﬀect on a PDE, we will need to calculate how derivatives transform. The easiest way is using Jacobians. For example, the derivative ux transforms as @.u; y/ @.Y; U / @.u; y/ @.X; Y / @.X; Y / ux D D D D @.x; y/ @.X; U / @.x; y/ @.X; Y / @.X; Y /

UX ; UY

(4.6)

noting that we have used the transformation (4.5) in (4.6). Similar for uy , @.x; u/ @.X; Y / @.x; u/ 1 @.X; Y / @.X; Y / uy D D D D : @.x; y/ @.X; U / @.x; y/ UY @.X; Y / @.X; Y /

(4.7)

74

4. POINT AND CONTACT TRANSFORMATIONS

The process easily extends to second-order derivatives. For example, uxx transforms as

uxx

D D

UY @. U ;U/ @.ux ; y/ X @.ux ; y/ @.X; Y / @.X; Y / D D @.x; y/ @.X; U / @.x; y/ @.X; Y / @.X; Y / UY2 UXX 2UX UY UX Y C UX2 UY Y ; UY3

(4.8)

while uxy and uyy transform as

uxy

D D

@. U1Y ; U / @.uy ; y/ @.uy ; y/ @.X; Y / @.X; Y / D D @.x; y/ @.X; U / @.x; y/ @.X; Y / @.X; Y / UX UY Y UY UXY ; UY3

(4.9)

and

uyy

D D

@.X; U1Y / @.x; uy / @.x; uy / @.X; Y / @.X; Y / D D @.x; y/ @.X; U / @.x; y/ @.X; Y / @.X; Y / UY Y : UY3

(4.10)

Thus, under the transformation (4.5), the ﬁrst- and second-order derivatives transform as

uxx D

UY2 UXX

UX ; UY C UX2 UY Y

ux D 2UX UY UXY UY3

1 ; UY UX UY Y D

(4.11)

uy D ; uxy

UY UXY UY3

; uyy D

UY Y : UY3

We now ask—why would this be useful? Consider, for example, the heat equation uxx D uy :

Under (4.5) (using the derivatives obtained in (4.11)), (4.12) would become UY2 UXX

2UX UY UX Y C UX2 UY Y 1 D ; 3 UY UY

(4.12)

4.1. CONTACT TRANSFORMATIONS

75

or UY2 UXX

2UX UY UXY C UX2 UY Y C UY2 D 0:

So one may ask—what was the point? We clearly made the problem more diﬃcult. However, suppose we started with the NLPDE uy2 uxx

2ux uy uxy C ux2 uyy C uy2 D 0;

(4.13)

a PDE given in (3.67) (with c.v/ D 1), then under (4.5), (4.13) becomes UXX D UY ;

the linear heat equation!

4.1

CONTACT TRANSFORMATIONS

We now extend transformations to include ﬁrst derivatives x D F .X; Y; U; P; Q/; y D G.X; Y; U; P; Q/;

u D H.X; Y; U; P; Q/;

(4.14)

where U D U.X; Y /, P D UX and Q D UY . We also need the relation that ux D M.X; Y; U; P; Q/;

uy D N.X; Y; U; P; Q/;

(4.15)

and impose the condition that (4.15) cannot have higher order derivatives. Before talking about condition which will ensure this, we consider three very famous contact transformations: 1. the Hodograph transformation, 2. the Legendre transformation, and 3. the Ampere transformation.

4.1.1 HODOGRAPH TRANSFORMATION These transformations are of the form u D Y;

(4.16)

x D U; y D Y; u D X:

(4.17)

x D X; y D U;

or The ﬁrst transformation (4.16) was previously given in (4.5). For the second transformation (4.5), ﬁrst-order derivatives transform as ux D

1 ; uy D UX

UY ; UX

(4.18)

76

4. POINT AND CONTACT TRANSFORMATIONS

and the second-order derivatives transform as uxx D uxy D

UXX ; UX3 UY UXX

UX UX Y UX3

UY2 UXX

uyy D

(4.19)

;

2UX UY UXY C UX2 UY Y : UX3

4.1.2 LEGENDRE TRANSFORMATION These transformations are of the form x D UX ;

y D UY ; u D X UX C Y UY

U:

(4.20)

Using Jacobians as we did for point transformations, we ﬁnd the ﬁrst derivatives transform as ux D X;

(4.21)

uy D Y;

and the second derivatives transform as uxx D

UY Y ; uxy D 2 UXX UY Y UXY

UXY ; 2 UXX UY Y UXY

uyy D

UXX : 2 UXX UY Y UXY

(4.22)

4.1.3 AMPERE TRANSFORMATION These transformations are of the form x D UX ; y D Y; u D XUX

U:

(4.23)

Using Jacobians, we ﬁnd the ﬁrst derivatives transform as ux D X;

uy D

(4.24)

UY ;

and the second derivatives transform as uxx D

1 UXX

;

uxy D

UXY ; uyy D UXX

2 UXX UY Y UXY : UXX

(4.25)

Note that we could easily have chosen x D X; y D UY ;

u D Y UY

and the derivatives would have transformed similarly. We now consider some examples.

U;

(4.26)

4.1. CONTACT TRANSFORMATIONS

77

Example 4.1 Consider uy2 uxx

2ux uy uxy C ux uuu C uy3 D 0:

(4.27)

Under the Hodograph transformation (4.16), Equation (4.27) becomes UXX

(4.28)

1 D 0;

which integrates giving 1 2 X C F .Y /X C G.Y /; 2 where F and G are arbitrary functions; via (4.16) we obtain the exact solution U D

yD

1 2 x C F .u/x C G.u/: 2

(4.29)

(4.30)

Example 4.2 Consider the nonlinear diﬀusion equation ut D

uxx : u2x

(4.31)

Under the Hodograph transformation t D T; x D U; u D X;

(4.32)

UT D UXX ;

(4.33)

Equation (4.31) becomes the heat equation! Example 4.3 Consider uxx

x uxx uyy

u2xy D 0:

(4.34)

Under the Legendre transformation (4.20), Eq. (4.34) becomes 2 UXY UY Y UY Y UXX U D 0; X 2 2 2 2 2 UXX UY Y UXY UXX UY Y UXY UXX UY Y UXY .UXX UY Y UXY / (4.35) which, after simpliﬁcation, becomes UY Y

(4.36)

UX D 0;

the heat equation. Example 4.4 Consider uxx

uyy

ux

uy

uxx uyy

u2xy D 0:

(4.37)

78

4. POINT AND CONTACT TRANSFORMATIONS

Under the Legendre transformation (4.20), Eq. (4.37) becomes 2 UY Y UXX UXY UY Y UXX .X Y / D 0; 2 UXX UY Y UXY .UXX UY Y UX2Y /2

(4.38)

which, after simpliﬁcation becomes UXX

UY Y D X

(4.39)

Y:

A particular solution of (4.39) is U D .X 3 C Y 3 /=6; with this, we can transform (4.39) to the standard wave equation VXX VY Y D 0; (4.40) via

1 1 U D V C X 3 C Y 3: 6 6

(4.41)

The solution of (4.40) is V D F .X C Y / C G.X

Y /:

(4.42)

Composing (4.42), (4.41) and (4.20) gives 1 Y / C X 2; 2 1 y D F 0 .X C Y / G 0 .X Y / C Y 2 ; 2 u D .X C Y /F 0 .X C Y / C .X Y /G 0 .X Y / 1 1 F .X C Y / G.X C Y / C X 3 C Y 3 ; 3 3 x

D F 0 .X C Y / C G 0 .X

(4.43)

the exact solution of (4.37) Example 4.5 Consider uxx uyy

u2xy D 1:

(4.44)

Under the Ampere transformation (4.23), (4.44) becomes 1 UXX UY Y UX2Y UXX UXX

UX2Y D 1; 2 UXX

(4.45)

which, after simpliﬁcation, becomes UXX C UY Y D 0;

Laplace’s equation!

(4.46)

4.1. CONTACT TRANSFORMATIONS

79

It is possible to combine several transformations as the following example demonstrates. Example 4.6 We consider the PDE uxx uyy

u2xy D ux2 C uy2

2

(4.47)

:

This PDE appears in elasticity [64]. Under the Ampere transformation, x D X;

y D UY ; u D Y U Y

U;

and (4.47) becomes UXX C UX2 C Y 2

2

ux D

UX ; uy D Y;

(4.48)

(4.49)

UY Y D 0:

If we let U D Y V , (4.49) becomes

2 VXX C Y 3 VX2 C 1 .Y VY Y C 2VY / D 0:

(4.50)

Introducing the new variable Y D 1=S , the Y derivatives transform as VY D

so that (4.50) becomes

S 2 VS ; VY Y D S 4 VS S C 2S 3 VS ; 2 VXX C VX2 C 1 VSS D 0:

(4.51)

The transformation (4.48) under the change of variables so far is x D X; y D V

SVS ; u D

VS ; ux D

VS ; S

uy D

1 : S

(4.52)

Next, we perform a Legendre transformation on (4.51) X D W ; S D W ; V D W C W

giving

W; VX D ; VS D ;

(4.53)

2 W C 2 C 1 W D 0:

(4.54)

D tan ; W D sec Q;

(4.55)

Q

C Q C Q D 0:

(4.56)

If we let then (4.54) becomes Composing all the transformations into one (with new variables) gives x D cos XUX C sin XU;

y D sin XUX

cos XU;

u D Y;

(4.57)

80

4. POINT AND CONTACT TRANSFORMATIONS

and (4.47) is transformed to the Helmholtz equation (4.58)

UXX C UY Y C U D 0:

These results are presented in Arrigo and Hill [64]. It is important to realize that not all transformations of the form (4.14) are contact transformations. Consider, for example, x D X; y D U;

(4.59)

u D UX :

First derivatives transform as ux D

UY UXX

UX UX Y UY

; uy D

UX Y ; UY

(4.60)

showing that they possess second order terms. Thus, we wish to establish conditions to guarantee that a transformation is, in fact, a contact transformation.

4.2

CONTACT CONDITION

We ﬁrst consider the problem with ODEs before moving to PDEs. Suppose we have a transformation X D X.x; y; p/; Y D Y .x; y; p/; P D P .x; y; p/; (4.61) where p D

dY dY dy and P D . We wish to calculate . From (4.61) we have dx dX dX dy d 2y C Y p dY dx dx 2 D P; D dX dy d 2y Xx C Xy C Xp 2 dx dx Yx C Yy

or Yx C Yy

dy d 2y C Yp 2 D P dx dx

As X and Y are independent of

dy d 2y Xx C Xy C Xp 2 : dx dx

d 2y , then from (4.63) we have dx 2 Yx C pYy D P Xx C pXy ; Yp D PXp :

(4.62)

(4.63)

(4.64a) (4.64b)

The set of equations (4.64) is know as the contact conditions. In 1872, Lie [63] gave the following deﬁnition of a contact transformation: if X; Y; P are independent functions of x; y; p such that dY

P dX D .dy

pdx/

(4.65)

4.2. CONTACT CONDITION

81

for some D .x; y; p/, then X D X.x; y; p/; Y D Y .x; y; p/; P D P .x; y; p/ is a contact transformation. Inserting the appropriate diﬀerentials into (4.65) gives Yx dx C Yy dy C Yp dp

P .Xx dx C Xy dy C Xp dp/ D .dy

pdx/:

(4.66)

Isolating coeﬃcients with respect to dx , dy and dp gives Yx Yy Yp

(4.67a) (4.67b) (4.67c)

PXx D p; PXy D ; PXp D 0;

and it is a simple matter to show that eliminating in (4.67) gives (4.64). With Lie’s deﬁnition in hand, we can extend this to PDEs. If d U P dX Qd Y D .du pdx qdy/ ; (4.68) then under the transformations X D X.x; y; u; p; q/; Y D Y.x; y; u; p; q/;

U D U.x; y; u; p; q/;

(4.69)

(4.68) becomes Ux dx C Uy dy C Uu du C Up dp C Uq dq P Xx dx C Xy dy C Xu du C Xp dp C Xq dq Q Yx dx C Yy dy C Yu du C Yp dp C Yq dq D .d U UX dX UY d Y / :

(4.70) (4.71) (4.72) (4.73)

Expanding and equating diﬀerentials gives Up Uq Uu Ux Uy

PXp PXq PXu PXx PXy

QYp QYq QYu QYx QYy

D 0; D 0; D ; D p; D q:

(4.74)

The relations (4.74) are the contact conditions. Example 4.7 Consider the Hodograph transformation X D x;

Y D u U D y:

(4.75)

Calculating the ﬁrst derivatives, we ﬁnd P D

p=q;

Q D 1=q:

(4.76)

82

4. POINT AND CONTACT TRANSFORMATIONS

Substituting (4.75) and (4.76) into (4.74) show these are satisﬁed if D 1; hence, (4.75) is a contact transformation. Example 4.8 Consider the Legendre transformation X D p;

Y D q;

U D xp C yq

u:

(4.77)

Calculating the ﬁrst derivatives, we ﬁnd P D X;

(4.78)

Q D Y:

Substituting (4.77) and (4.78) into (4.74) shows these are satisﬁed if D 1; hence, (4.77) is a contact transformation. Example 4.9 Consider the Ampere transformation X D x;

Y D q;

U D yq

u:

(4.79)

Calculating the ﬁrst derivatives, we ﬁnd P D

p;

Q D y:

(4.80)

Substituting (4.79) and (4.80) into (4.74) shows these are satisﬁed if D 1; thus, (4.79) is a contact transformation. Example 4.10 Consider X D x C q; Y D y C p;

U D u C pq:

(4.81)

Calculating the ﬁrst derivatives we ﬁnd P D p;

Q D q:

(4.82)

Substituting (4.81) and (4.82) into (4.74) shows these are satisﬁed if D 1; thus, (4.81) is a contact transformation.

4.3

PLATEAU PROBLEM

Consider the surface area of u D u.x; y/ on some region R ZZ q SA D 1 C u2x C uy2 dA:

(4.83)

R

For this surface to be a minimum, then the Euler-Lagrange equation from the calculus of variations is @ @L @ @L @L C D0 (4.84) @x @ux @y @uy @u

with L D

q

4.3. PLATEAU PROBLEM

83

1 C u2x C uy2 . This gives 1 C uy2 uxx

2ux uy uxy C 1 C u2x uyy D 0:

(4.85)

It is well known that solutions of (4.85) can be generated by the Enneper–Weierstrass formulas given by Z x D Re f .1 g 2 /dz; Z y D Re if .1 C g 2 /dz; (4.86) Z u D Re 2 f g dz: For example, choosing f D 1 and g D z gives

z3 ; D Re z 3 3 z y D Re i z C ; 3 u D Re z 2 ; x

(4.87)

and denoting z D p C i q gives rise to p3 C pq 2 ; 3 q3 y D q p2q C ; 3 u D p2 q2; x

D p

(4.88)

commonly referred to as Enneper’s minimal surface. Here we present a new way of generating a solution to (4.85). We will show that (4.85) is linearizable and furthermore, we can represent its solution parametrically in terms of solutions of Laplace’s equation.

4.3.1 LINEARIZATION Under the Legendre transformation x D vr ; y D vs u D rvr C svs

v;

(4.89)

ﬁrst-order derivatives transform as ux D r; uy D s;

(4.90)

84

4. POINT AND CONTACT TRANSFORMATIONS

while second-order derivatives transform as vss vrs vrr uxx D ; uxy D ; uyy D ; 2 2 2 vrr vss vrs vrr vss vrs vrr vss vrs thus transforming (4.85) to the linear PDE 1 C r 2 vrr C 2rsvrs C 1 C s 2 vss D 0:

(4.91)

(4.92)

If we denote A D 1 C r 2 , B D 2rs and C D 1 C s 2 then B 2 4AC < 0 so that the PDE (4.92) is elliptic. We now show that (4.92) is transformable to Laplace’s equation. Introducing polar coordinates r D cos b; s D sin b (4.93) transforms (4.92) to

4 C 2 v C vbb C v D 0:

Further, if we let D

1 ; sinh a

(4.94) (4.95)

then (4.94) becomes

2va D 0: sinh a cosh a Next we transform the dependent variable v as vaa C vbb C

vD

w ; tanh a

(4.96)

(4.97)

leading to waa C wbb C

2w

D 0: cosh2 a Finally, if we introduce the ﬁrst-order Darboux transformation w D Wa

tanh a W;

(4.98)

(4.99)

then equation (4.98) becomes @ r 2W @a

tanh ar 2 W D 0;

(4.100)

where r 2 W D 0 is Laplace’s equation. Integrating (4.100) gives r 2 W D F .b/ cosh a;

(4.101)

where F .b/ is arbitrary, but we can set F .b/ D 0 without loss of generality. The rationale is as follows. If F .b/ is arbitrary, then we can set F D G 00 C G so r 2 W D G 00 .b/ C G.b/ cosh a: (4.102)

4.3. PLATEAU PROBLEM

85

If e C G.b/ cosh a; W DW

(4.103)

e D 0: r 2W

(4.104)

then (4.102) gives rise to Laplace’s equation

Further substitution into (4.99) shows it is left unchanged. In order to connect solutions of the original equation (4.85) to Laplace’s equation, it is necessary to compose the transformations (4.89), (4.93), (4.95), (4.97) and (4.99). This gives rise to the following result. Exact solutions to the minimal surface equation 1 C uy2 uxx 2ux uy uxy C 1 C ux2 uyy D 0

(4.105)

are given parametrically by x D cosh a cos b Wa C sinh a sin b Wb sinh a cos b Waa cosh a sin b Wab ; y D cosh a sin b Wa sinh a cos b Wb C cosh a cos b Wab C sinh a sin b Wbb ;(4.106) u D W Waa ;

where W satisﬁes Laplace’s equation Waa C Wbb D 0:

(4.107)

In the next section we will show that several well-known minimal surfaces may be obtained with simple solutions of Laplace’s equation.

4.3.2

WELL-KNOWN MINIMAL SURFACES

Catenoid and Helicoid

Probably the simplest non-zero solution to Laplace’s equation, (4.107) is given by W D a sin ˛ C b cos ˛;

(4.108)

where ˛ is constant. Substituting this into (4.106) gives rise to x D cosh a cos b sin ˛ C sinh a sin b cos ˛; y D cosh a sin b sin ˛ sinh a cos b cos ˛; u D a sin ˛ C b cos ˛:

(4.109)

x D sinh a sin b; y D sinh a cos b; u D b;

(4.110)

Setting ˛ D 0 gives

86

4. POINT AND CONTACT TRANSFORMATIONS

the surface of a helicoid, while setting ˛ D

2

gives

x D cosh a cos b; y D cosh a sin b; u D a;

(4.111)

the surface of a catenoid [68]. Enneper’s Surface

By choosing 1 .sinh 2a C cosh 2a/ cos 2b; 3 as a solution of Laplace’s equation gives, (4.106) gives W D

1 e a cos b C e 3a cos3 b C e 3a sin2 b cos b; 3 1 3a 3 a y D e sin b C e sin b C e 3a cos2 b sin b; 3 u D e 2a cos2 b sin2 b : x

(4.112)

D

(4.113)

Identifying that e a cos b;

pD

qD

e a sin b;

(4.114)

gives Enneper’s minimal surface [68] 1 3 p C pq 2 ; 3 1 y D q C q 3 p 2 q; 3 u D p2 q2: x

D p

(4.115)

Henneberg’s Surface

By choosing 2 cosh 2a cos 2b; 3 as a solution of Laplace’s equation gives, (4.106) gives W D

2 sinh 3a cos 3b 2 sinh a cos b; 3 2 y D sinh 3a sin 3b C 2 sinh a sin b; 3 u D 2 cosh 2a cos 2b; x

(4.116)

D

known as Henneberg’s minimal surface [68].

(4.117)

4.4. EXERCISES

87

Catalan’s Surface

By choosing W D 4 sinh a sin b

2a cosh a sin b

2b sinh a cos b;

(4.118)

as a solution of Laplace’s equation, (4.106) gives x D 2b cosh 2a sin 2b; y D 1 cosh 2a cos 2b; u D 4 sinh a sin b;

(4.119)

known as Catalan’s minimal surface [68]. Bour’s Surface

By choosing 1 .cosh 3a C sinh 3a/ cos 3b; 2 as a solution of Laplace’s equation gives, from (4.106) W D

e 4a cos 4b; 2 4a e e 2a sin 2b sin 4b; 2 4 3a e cos 3b: 3

x

D e 2a cos 2b

y

D

u D

(4.120)

(4.121)

Identifying that r D e 2a and b D =2 gives rise to r2 cos 2; 2 r2 r sin sin 2; 2 4 4=3 3 r cos ; 3 2

x

D r cos

y

D

u D

(4.122)

the minimal surface known as Bour’s surface [68].

4.4 4.1.

EXERCISES Show the following are contact transformations: .i / x D X C Y C UX ; y D X C 3UX ; u D 2XUX .i i / x D e Y ; y D U; u D X 2 e Y ; U2 2UX .i i i / x D Y C ; y D U; u D e X X2 ; UY UY 1 U XUX Y UY .iv/ x D UX ; y D ; uD : UY UY

2U C X 2 C Y 2 ;

88

4. POINT AND CONTACT TRANSFORMATIONS

4.2.

Given X; Y; P and Q, show a U exists such that the contact conditions are satisﬁed; further, ﬁnd U xp x ; P D ; Q D x; q q .i i / X D x 2p; Y D y; P D p; Q D q; u u q .i i i / X D q; Y D y ; P D ; QD ; q pq p 1 1 .iv/ X D x C ; Y D y C ; P D A.p; q/; Q D B.p; q/: p q .i / X D u; Y D y

In part (iv), you will need to determine the forms A and B . 4.3.

Show that the following is a contact transformation: t D 2T; x D

2U

.T C X/uX U .T C X/UX ; uD : 2.T C X / 2.T C X/2

(4.123)

Further, show that the Hunter–Saxon equation [65] 1 u tx C uuxx C u2x D 0; 2

(4.124)

is transformed to [66] UTX D

4.4.

4U : .T C X/2

2.UT C UX / T CX

(4.125)

The boundary Layer equations from ﬂuid mechanics are (4.126a) (4.126b)

ux C vy D 0; uux C vuy D uyy : 4 .i / Show that (4.126b) can be written as u2 x C uv

uy

y

(4.127)

D 0:

4 .i i / Show that with the introduction of the stream functions .x; y/ such that uD

y;

vD

x;

u2 D y ;

uv

uy D

D

.x; y/ and D

x ;

(4.128)

4.4. EXERCISES

(4.126a) and (4.127) are automatically satisﬁed. Furthermore, from (4.128), satisfy 2 y

C

yy

x

y

89

and (4.129a) (4.129b)

D y ; D x :

4 .i i i / Show that under the Hodograph transformation, x D X; y D ‰;

D Y;

D ˆ;

(4.130)

system (4.129) becomes ˆY D

1 ; ‰Y

‰Y Y ; ‰Y3

(4.131)

2‰Y2 Y ‰Y Y C 2 D 0; 3 ‰Y ‰Y

(4.132)

ˆX D

which, on eliminating ˆ, gives ‰Y ‰XY

or

‰X ‰Y Y

1 ‰Y

X

D

‰Y Y ‰Y3

:

(4.133)

Y

Under the Hodograph transformation (4.130), u becomes uD

1 ; ‰Y

(4.134)

giving (4.133) as (4.135)

uX D .uuY /Y

a nonlinear diﬀusion equation! This has been noted in Ames [67]. 4 .iv/ Show under the Hodograph transformation, x D X; y D ˆ;

D ‰; D Y;

(4.136)

system (4.129) becomes ‰Y2 ˆ2Y

1 D 0; ˆY

‰Y Y ˆ2Y

‰Y ˆY Y ‰X ‰Y C 3 ˆY ˆY

‰Y2 ˆX ˆX C D 0; 2 ˆY ˆY

(4.137)

which on simplifying gives ‰Y2

ˆY D 0; ‰X

‰Y Y D 0: ‰Y3

(4.138)

90

4. POINT AND CONTACT TRANSFORMATIONS

Diﬀerentiating the second in (4.138) gives .‰Y /X D

‰Y Y ‰Y3

(4.139)

:

Y

Show that under the Hodograph transformation (4.136) u becomes uD

giving (4.139) as

u

X

D

1 ; ‰Y

u Y u 3

(4.140)

(4.141)

;

Y

where u D u 1 , another nonlinear diﬀusion equation! Rogers and Shadwick have shown that diﬀerent nonlinear diﬀusion equations can be linked via reciprocal transformations [71]. 4.5.

The Harry–Dym equation is given by u t D u3 uxxx :

Show that by introducing the change of variable u D v within scaling of t ) v x v t D 3=2 ; v xx which, under the substitution v D wx , becomes ! wxx wt D : wx3=2 x

1=2

, this equation becomes (to

(4.142)

Show under a Hodograph transformation, (4.142) remains unchanged. 4.6.

The equations for irrotational steady two-dimensional ﬂows in ﬂuid mechanics are given by .u/x C .v/y D 0; uux C vuy D uvx C vvy D r vE D 0:

px ; py ;

(4.143a) (4.143b) (4.143c) (4.143d)

4.4. EXERCISES

(i) Show that for 2 - D ﬂows, (4.143d) is uy introducing a potential u D x and v D y .

91

vx D 0, and is identically satisﬁed by

(ii) Assuming that p D c 2 where c is a constant speed of sound, show that the system reduces to the single equation x2

c 2 xx C 2x y xy C y2

c 2 yy D 0:

(4.144)

(iii) Show that under a Legendre transformation this equation linearizes. 4.7.

The governing equations for highly frictional granular materials is given by @xx C @x @xy C @x xx yy

@xy D 0; @y @yy D g; @y 2 xy D 0;

(4.145a) (4.145b) (4.145c)

where xx , yy and xy are normal and shear stresses, respectively, the material density, and g , the gravity. In (4.145), (4.145a) and (4.145b) are the equilibrium equations, whereas (4.145c) is the constitutive relation for the material [69]. (i) Show that by introducing a potential u, such that u2x ; uy

(4.146)

2ux uy uxy C u2x uyy C guy2 D 0:

(4.147)

xx D uy ; xy D

ux ; then yy D

Eq. (4.145) reduces to uy2 uxx

(ii) Show that under a Hodograph transformation, this linearizes. 4.8.

The shallow water wave equations are given by u t C uux C gx D 0; t C Œu. h.x//x D 0;

(4.148a) (4.148b)

where u D u.x; t/ the velocity, D .x; t / the free surface, h D h.x/ the bottom surface, and g is gravity [70].

92

REFERENCES

(i) Introducing a potential function where D x C h.x/;

uD

t ; x

(4.149)

shows that the shallow water equations become x2 t t

gx3 xx

2 t x tx C t2

gh0 .x/u3x D 0:

(4.150)

(ii) Show that under a modiﬁed Hodograph transformation, this equation for h.x/ D ˛x can be transformed to UX3 UT T

gUXX D 0:

(4.151)

Further show that under a Legendre transformation, this equation can be linearized. Compose the two transformations.

4.5

REFERENCES

[63] S. Lie, Göttinger Nachrichten, p. 480, 1872. 80 [64] D. J. Arrigo and J. M. Hill, Transformations and equation reductions in ﬁnite elasticity. I. Plane strain deformations, Math. Mech. Solids, 1(2), pp. 155–175, 1996. DOI: 10.1177/108128659600100201. 79, 80 [65] J. K. Hunter and R. Saxton, Dynamics of director ﬁelds, SIAM J. Appl. Math., 51(6), pp. 1498–1521, 1991 DOI: 10.1137/0151075. 88 [66] O. L. Morozov, Contact equivalence of the generalized hunter—Saxton Equation and the Euler-Poisson Equation, ArXiv:math-ph/0406016v2. 88 [67] W. F. Ames, Nonlinear Partial Diﬀerential Equations in Engineering, vol. I, 1968. DOI: 10.1063/1.3034120. 89 [68] E. W. Weisstein, Minimal Surfaces, From MathWorld–A Wolfram Web Resource. http: //mathworld.wolfram.com 86, 87 [69] D. J. Arrigo, L. Le, and J. W. Torrence, Exact solutions for a class of ratholes in highly frictional granular materials, DCDIS Series B: Appl. Alg., 19, pp. 497–509, 2010. 91 [70] J. J. Stoker, Water Waves—The Mathematical Theory with Applications, Wiley-Interscience, New York, Interscience, 1958. DOI: 10.1002/9781118033159. 91 [71] C. Rogers and W. F. Shadwick, Backlünd Transformations and their Applications, Academic Press, 1982. 90

93

CHAPTER

5

First Integrals In an introductory course in PDEs, the wave equation u t t D c 2 uxx ;

(5.1)

where c > 0 is a constant wave speed, is introduced. The general solution of (5.1) is u D F .x

ct / C G.x C ct/;

(5.2)

where F and G are arbitrary functions of their arguments. In this solution, there are two waves; one is traveling right (F .x ct/) and one is traveling left (G.x C ct/). Each of these solutions can be obtained from the following ﬁrst-order PDEs u t C cux D 0; u t cux D 0:

(5.3a) (5.3b)

PDE (5.3a) gives rise to u D F .x ct /, whereas PDE (5.3b) gives rise to u D G.x C ct/. We question whether it is possible to show this directly—whether the solutions of (5.3a) and/or (5.3b) will give rise to solutions of (5.1). For example, diﬀerentiating (5.3a) with respect to t and x gives u t t C cu tx D 0; u tx C cuxx D 0:

(5.4)

It is a simple matter to show that upon elimination of u tx in (5.4) we obtain (5.1). The same follows from (5.3b). In this example, we considered the linear wave equation, where the solution of a lower-order PDE gave rise to solutions of a higher order PDE. Does this idea apply for NLPDEs? The following example illustrates the concept. Consider the following pair of PDEs: uxx

ux uy D 1; uy 4 uyy D 0:

(5.5a) (5.5b)

We ask, will solutions of (5.5a) give rise to solutions of (5.5b)? For example, one exact solution p of (5.5a) is u D 2 xy . Calculating all necessary derivatives gives p p p p y y x x ux D p ; uy D p ; uxx D (5.6) p ; uyy D p ; y 2y y x 2x x

94

5. FIRST INTEGRALS

and substituting into (5.5b) shows it is identically satisﬁed. The reader can also verify that u D x C y also satisﬁes both PDEs. We naturally ask, will all of the solutions of (5.5a) give rise to solutions (5.5b)? It’s a simple matter to calculate the derivatives of (5.5a) ux D

1 uy

uxx D

uxy uy2

uxy D

uyy uy2

(5.7)

and eliminating uxy from (5.7) gives (5.5b). If a PDE of lower-order F .x; y; u; ux ; uy / D 0

(5.8)

exists such that diﬀerential consequences of (5.8) gives rise to the original (higher-order) PDE, then the lower-order PDE is called a ﬁrst integral. We see that (5.5a) is one ﬁrst integral of (5.5b). Are there others and, if so, how do we ﬁnd them? If we assume a general form as in (5.8), then diﬀerential consequences give Fx C Fu ux C Fp uxx C Fq uxy D 0; Fy C Fu uy C Fp uxy C Fq uyy D 0;

(5.9a) (5.9b)

where p D ux and q D uy . Eliminating uxy from (5.9) gives Fp2 uxx

Fq2 uyy C Fq .Fx C pFu /

Fp Fy C qFu :

(5.10)

In order to obtain our targeted PDE (5.5b), we need to solve the following system of PDE for F Fp2 q 4 Fq2 D 0; (5.11a) Fq .Fx C pFu / Fp Fy C qFu D 0: (5.11b) Since (5.11a) factors Fp q 2 Fq Fp C q 2 Fq D 0, there are two cases to consider. We consider the ﬁrst case here and leave it to the reader to consider the second case. Solving Fp

gives

q 2 Fq D 0

F D G x; y; u; p

(5.12) 1 ; q

(5.13)

where G is an arbitrary function of its arguments. Substituting (5.13) into (5.11b) and letting D p 1=q gives .Gx C Gu / q 2 Gy D 0: (5.14) Since G D G.x; y; u; / and is now independent of q , then from (5.14) we have both Gx C Gu D 0; Gy D 0;

(5.15)

5.1. QUASILINEAR SECOND-ORDER EQUATIONS

which are easily solved, giving G D G.; x u/, and the ﬁnal form of F as 1 1 F D G1 p ;x p u : q q The second case is solved the same way and gives rise to 1 1 F D G2 p C ; x p C q q

u :

95

(5.16)

(5.17)

As we can see from (5.16) and (5.17), there is actually an enormous class of ﬁrst integrals to (5.8). As we will see later, PDEs admitting general classes of ﬁrst integrals sometimes have important properties.

5.1

QUASILINEAR SECOND-ORDER EQUATIONS

In this section we consider general second-order quasilinear PDEs in two independent variables Ar C Bs C C t D E;

(5.18)

where r D uxx , s D uxy , t D uyy and A; B; C and E are functions of x; y; u; ux and uy . We assume a ﬁrst integral of the form (5.8). Diﬀerentiating with respect to x and y (see (5.9)) and isolating uxx and uyy gives rD

Fx C pFu C sFq ; tD Fp

Fy C qFu C sFp ; Fq

(5.19)

where we are assuming that Fp Fq ¤ 0. Substituting (5.19) into (5.18) and simplifying gives AFq2 BFp Fq C CFp2 s C AFq .Fx C pFu / C CFp Fy C qFu C EFp Fq D 0 (5.20) and, since F is independent of s , this leads to AFq2 BFp Fq C CFp2 D 0; AFq .Fx C pFu / C CFp Fy C qFu C EFp Fq D 0:

(5.21a) (5.21b)

Note that (5.21a) is quadratic in Fq =Fp and can be solved depending on whether B 2 4AC > 0, B 2 4AC D 0 or B 2 4AC < 0. We have assumed that Fp Fq ¤ 0. Cases where either Fp D 0 or Fq D 0 would be considered special cases and would need to be considered separately. Example 5.1 Obtain a ﬁrst integral to uy2 uxx

2ux uy uxy C u2x uyy D uy3 :

(5.22)

96

5. FIRST INTEGRALS

Identifying A D q 2 , B D 2pq , C D p 2 , and E D q 3 , from the ﬁrst integral equations (5.21) gives q 2 Fq2 C 2pqFp Fq C p 2 Fp2 D 0; q 2 Fq .Fx C pFu / C p 2 Fp Fy C qFu C q 3 Fp Fq D 0:

(5.23a) (5.23b)

We ﬁnd (5.23a) a perfect square, which gives (5.24)

pFp C qFq D 0;

which simpliﬁes (5.23b) to

pFy C q 2 Fp D 0;

qFq qFx

(5.25)

or qFx

pFy C q 2 Fp D 0:

(5.26)

Note that we have excluded the case where Fq D 0 as this would give Fp D 0. The solution of (5.24) is p F D F x; y; u; ; (5.27) q and substituting into (5.26) and letting p D q gives Fx

whose solution is

(5.28)

Fy C F D 0;

F D F u; x

Thus, (5.22) admits ﬁrst integral of the form ux F u; x ;y uy

; y

1 2 x C x : 2

1 2 xux x C 2 uy

D 0:

(5.29)

(5.30)

For example, if we choose ux

xuy D 0;

(5.31)

then we obtain the exact solution u D f .x 2 C 2y/ of the original PDE. Other choices of F in (5.30) could lead to exact solutions. For example, choosing 2xux C 2y x 2 uy D 0; (5.32) y x would lead to the exact solution u D F C . As a ﬁnal possibility, we choose x 2 ux x D f .u/: (5.33) uy

5.1. QUASILINEAR SECOND-ORDER EQUATIONS

97

This is very much like (5.31) involving an arbitrary function f .u/ and leads to the exact solution 1 y C x2 D 2

f .u/x C g.u/;

(5.34)

involving now two arbitrary functions f and g . Example 5.2 Obtain a ﬁrst integral to uy2 C uy uxx 2ux uy C ux uxy C u2x uyy D 0: Here A D q 2 C q , B D 2pq (5.21)

(5.35)

p , C D p 2 , and E D 0, giving from the ﬁrst integral equations

q 2 C q Fq2 C .2pq C p/Fp Fq C p 2 Fp2 D 0; q 2 C q Fq .Fx C pFu / C p 2 Fp Fy C qFu D 0:

From (5.36a), we ﬁnd pFp C qFq

pFp C .1 C q/Fq D 0;

(5.36a) (5.36b) (5.37)

showing that there are two cases to consider. Case 1: If pFp C qFq D 0;

(5.38)

then from (5.36b) we obtain the second equation .q C 1/Fx

pFy C pFu D 0;

and the solution of the over-determined system (5.38) and (5.39) is p F D F y C u; : q

(5.39)

(5.40)

Case 2: If pFp C .q C 1/Fq D 0;

(5.41)

then from (5.36b) we obtain the second equation qFx

pFy D 0;

and the solution of the over-determined system (5.41) and (5.42) is p : F D F u; qC1

(5.42)

(5.43)

98

5. FIRST INTEGRALS

From either (5.40) or (5.43) we can ﬁnd the general solution of (5.35). For example, if we choose from (5.40) p D f .u C y/; (5.44) q where f is arbitrary, if we let f D 1= 0 , the solution of (5.44) is x D .y C u/ C

where and

.u/;

(5.45)

are arbitrary functions of their arguments.

Example 5.3 Consider 1 C uy2 uxx

2ux uy uxy C u2x uyy D 0:

(5.46)

q˙i x; y; u; ; p

(5.48)

Here A D 1 C q 2 ; B D 2pq; C D p 2 ; E D 0 and our ﬁrst integral equations are 1 C q 2 Fq2 C 2pqFp Fq C p 2 Fp2 D 0; (5.47a) 2 2 1 C q Fq .Fx C pFu / C p Fp .Fy C qFu / D 0: (5.47b) We solve (5.47a), giving

where i is the usual i D

p

F DF

1. Requiring that (5.48) satisﬁes (5.49) gives the ﬁnal form q˙i F D F u iy; : (5.49) p

One will notice that the ﬁrst integrals presented in (5.49) are complex. However, it is possible to construct real solutions to (5.46). For example, if we choose qCi D u iy; p q i D u C iy; p

(5.50a) (5.50b)

then we can split (5.50) into real and imaginary parts giving (5.51a) (5.51b)

q D up; yp D 1;

leading to the common solution uD

c

x y

:

In this example, c is an arbitrary constant and one can verify that (5.52) satisﬁes (5.46).

(5.52)

5.1. QUASILINEAR SECOND-ORDER EQUATIONS

99

If we choose qCi 1 D ; p u iy q i 1 D ; p u C iy

(5.53a) (5.53b)

then we can split (5.53) into real and imaginary parts, giving uq C y D p; u yq D 1;

(5.54a) (5.54b)

u D tan.x C c/y:

(5.55)

leading to the solution Finally, choosing qCi D .u iy/2 ; p q i D .u C i/2 ; p

then splitting (5.56) into real and imaginary parts gives q D u2 y 2 p; 2yup D 1;

(5.56a) (5.56b)

(5.57a) (5.57b)

leading to the solution u2 D

y2 2

x C c: y

(5.58)

The reader can verify that (5.55) and (5.58) are both solutions of (5.46). Example 5.4 Obtain a ﬁrst integral to uxx C 2uuxy C u2 uyy D 0:

(5.59)

Identifying that A D 1, B D 2u, C D u2 , and E D 0, gives from the ﬁrst integral equations (5.21) Fq2 2uFp Fq C u2 Fp2 D 0; Fq .Fx C pFu / C u2 Fp Fy C qFu D 0:

(5.60a) (5.60b)

From (5.60a) we ﬁnd Fq

uFp D 0;

(5.61)

100

5. FIRST INTEGRALS

which simpliﬁes (5.60b) to uFp Fx C uFy C .p C uq/ Fu D 0

(5.62)

Fx C uFy C .p C uq/ Fu D 0:

(5.63)

or Note that we have excluded the case where Fp D 0, as this would gives Fq D 0. The solution of (5.61) is F D G .x; y; u; p C uq/ : (5.64) Substituting this into (5.63) and letting D p C uq gives

Gx C uGy C Gu C qG D 0:

(5.65)

Since G is independent of q , this gives G D 0:

(5.66)

Normally we would say that G D 0 which would then give that F is independent of p and q , i.e., no ﬁrst integral. But we also have the choice that D 0, and one can show that, in fact, D p C uq D 0 is a ﬁrst integral.

5.2

MONGE–AMPERE EQUATION

In this section we consider Monge–Ampere equations. These are second-order PDEs in two independent variables in the form Ar C Bs C C t C D rt s 2 D E; (5.67) where in addition to the forms of A; B; C; and E in the previous section, we also assume that D ¤ 0 and is a function of x; y; u; ux ; and uy . We again assume a ﬁrst integral of the form (5.8) and on isolating the derivatives r and t and substituting into (5.67) and isolating coeﬃcients with respect to s gives AFq2 BFp Fq C CFp2 D D Fx Fp C Fy Fq C pFu Fp C qFu Fq ; (5.68a) AFq .Fx C pFu / C CFp Fy C qFu C EFp Fq D D Fx Fy C pFy Fu C qFx Fu C pqFu2 : (5.68b) In an attempt to solve (5.68), we assume there exists a .x; y; u; p; q/, such that AFq2 BFp Fq C CFp2 D Fx Fp C Fy Fq C pFu Fp C qFu Fq CAFq .Fx C pFu / C CFp Fy C qFu C EFp Fq D Fx Fy C pFy Fu C qFx Fu C pqFu2 D A1 Fx C B1 Fu C C1 Fp C D1 Fq A2 Fy C B2 Fu C C2 Fp C D2 Fq :

(5.69)

5.2. MONGE–AMPERE EQUATION

101

Expanding (5.69) and equating coeﬃcients of the product of Fx ; Fy ; Fu ; Fp ; Fq gives A1 A2 D D; A2 B1 D pD; A1 B2 D qD; B1 B2 D pqD; A1 C2 D D; D1 A2 D D; A2 C1 D C; A1 D2 D A; C1 C2 D C; D1 D2 D A; B1 D2 C D1 B2 D pA qD; B1 C2 C C1 B2 D qC pD; C1 D2 C D1 C2 D B C E;

(5.70)

from which we ﬁnd the following: B1 D pA1 ;

A2 D

qD ; A1

D ; B2 D A1

A1 C ; D1 D A1 ; D D A C2 D ; D2 D ; A1 A1

C1 D

(5.71)

where satisﬁes D 2 2

BD C AC C DE D 0:

(5.72)

Solving (5.72) for gives rise to the following possibilities: (i) two real distinct roots, (ii) two real repeated roots, and (iii) two complex roots. In the case of two distinct roots, say 1 and 2 , (5.69) can be factored as DFx C pDFu CFp C 1 DFq DFy C qDFu C 1 DFp AFq D 0; DFx C pDFu CFp C 2 DFq DFy C qDFu C 2 DFp AFq D 0:

(5.73a) (5.73b)

There are four systems to be considered, however, two of these give 1 D 2 ; as we have assumed that 1 ¤ 2 , these are inadmissible. Thus, we shall only consider the systems D .Fx C pFu/ CFp C 1 DFq D 0; D Fy C qFu C 2 DFp AFq D 0;

(5.74a) (5.74b)

D .Fx C pFu/ CFp C 2 DFq D 0; D Fy C qFu C 1 DFp AFq D 0:

(5.75a) (5.75b)

and

In the case of a repeated root, we only have one set of equations for F to consider, as (5.74) and (5.75) coalesce into one. The following examples illustrate.

102

5. FIRST INTEGRALS

Example 5.5 Obtain ﬁrst integral to s 2 D 1:

r C 3s C t C rt

(5.76)

Identifying that A D 1, B D 2, C D 1, D D 1, and E D 1, then (5.72) gives 2

3 C 2 D 0;

(5.77)

leading to D 1; 2. Thus, we have two cases to consider. Case 1: From (5.74) we have Fx C pFu Fp C Fq D 0; Fy C qFu C 2Fp Fq D 0:

Using the method of characteristic, we solve the ﬁrst, giving 1 F D F x C p; y; u C p 2 ; x q : 2

(5.78a) (5.78b)

(5.79)

Substituting into the second gives Fy C .x

/Fˇ C 2 F˛ C .˛

x/Fˇ C F D 0;

(5.80)

where ˛ D x C p , ˇ D u C 12 p 2 , and D x q . With these variables, F is independent of x ; this leads to Fˇ D 0; Fy C 2F˛ C F D 0; (5.81) from which we obtain the ﬁnal form of ﬁrst integral F .x

2y C p; y

x C q/ D 0:

(5.82)

Case 2: From (5.75) we have Fx C pFu Fp C 2Fq D 0; Fy C qFu C Fp Fq D 0:

Using the method of characteristic, we solve the ﬁrst, giving 1 2 F D F x C p; y; u C p ; 2x q : 2

(5.83a) (5.83b)

(5.84)

Substituting into the second gives Fy C .2x

/Fˇ C F˛ C .˛

x/Fˇ C F D 0;

(5.85)

5.2. MONGE–AMPERE EQUATION

where ˛ D x C p , ˇ D u C this leads to

1 2 p , 2

and D 2x

103

q . With these variables, F is independent of x ;

(5.86)

Fˇ D 0; Fy C F˛ C F D 0;

from which we obtain the ﬁnal form of ﬁrst integral F .x

y C p; y

(5.87)

2x C q/ D 0:

Any choice of F in (5.82) or (5.87) will lead to an exact solution of the original PDE. For example, from (5.87), choosing a.x

y C p/ C b.y

(5.88)

2x C q/ D 0;

where a and b are arbitrary constants and f is an arbitrary function, gives uD

b2

ab a2 2 2a b x C xy C f .bx 2a2 a

ay/;

(5.89)

pq:

(5.90)

an exact solution to (5.76). Example 5.6 Obtain ﬁrst integral to xqr

.x C y/s C ypt C xy rt

s2 D 1

Identifying that A D xq , B D .x C y/, C D yp , D D xy , and E D 1 pq , then from (5.72) gives x 2 y 2 2 C xy.x C y/ C xypq C xy.1 pq/ D 0; (5.91) 1 1 ; . Thus, we have two cases to consider. leading to D x y Case 1: From (5.74) we have xFx C xpFu yFy C yqFu

(5.92a) (5.92b)

pFp Fq D 0; Fp qFq D 0:

Using the method of characteristic, we solve the ﬁrst, giving F D F .xp; y; u

xp ln x; ln x C q/ :

(5.93)

Substituting into the second gives Fy C y.

ln x/Fˇ

xF˛ C x ln xFˇ C .ln x

/ F D 0;

(5.94)

yFˇ C x ln xFˇ D 0;

(5.95)

or, re-arranging gives Fy C y Fˇ

F

xF˛ C ln x F

104

5. FIRST INTEGRALS

where ˛ D x C p , ˇ D u C 12 p 2 , and D x q . With these variables, F is independent of x and thus leads to Fy D 0; F˛ D 0; Fˇ D 0; F D 0; (5.96) gives that F is constant. So in this case, there is no ﬁrst integral. Case 2: From (5.75) we have xyFx C xypFu xyFy C xyqFu

(5.97a) (5.97b)

ypFp xFq D 0; yFp xqFq D 0:

Using the method of characteristic, we solve the ﬁrst, giving F D F .xp; y; u C xp ln x; x C yq/ :

(5.98)

Substituting (5.98) into (5.97b) and re-arranging gives Fy yF˛ Fˇ x C Fˇ x 2 xy ln xFˇ D 0;

(5.99)

where ˛ D xp , ˇ D u C xp ln x , and D x C yq . With these variables, F is independent of x ; this leads to Fˇ D 0; Fy F˛ D 0; (5.100) from which we obtain F D F . ; y C ˛/ and the ﬁnal form of ﬁrst integral (5.101)

F .xp C y; yq C x/ D 0:

Any choice of F in (5.101) will lead to an exact solution of the original PDE. For example, choosing xp C yq C x C y D 0 (5.102) leads to the exact solution uD

(5.103)

.x C y/ C f .x=y/;

where f is arbitrary as an exact solution to (5.90). Example 5.7 Obtain ﬁrst integral to yqr C 2xys C y 2 t C y 2 rt

s2 D x2

Identifying that A D yq , B D 2xy , C D y 2 , D D y 2 , and E D x 2 y 4 2

2xy 3 C x 2 y 2 D 0;

yq:

(5.104)

yq , from (5.72) we get

(5.105)

5.2. MONGE–AMPERE EQUATION

105

leading to the repeated root D x=y . Thus, we have only a single case to consider. From (5.74) we have yFx C ypFu yFp C xFq D 0; yFy C yqFu C xFp yFq D 0:

(5.106a) (5.106b)

Using the method of characteristic, we solve the ﬁrst, giving F D F x C p; y; 2u C p 2 ; 2yq

x2 :

(5.107)

Substituting into the second gives yFy C Fˇ C F˛ C 2˛Fˇ x

Fˇ x 2 D 0;

(5.108)

where ˛ D x C p , ˇ D u C 12 p 2 , and D 12 x 2 yq . With these variables, F is independent of x ; this leads to Fˇ D 0; F˛ C 2˛Fˇ D 0; yFy C Fˇ D 0; (5.109) from which we obtain F D F . /, leading to the ﬁnal form of ﬁrst integral F 2yq x 2 D 0:

(5.110)

For example, if we choose 2yuy

x2 D 0

(5.111)

we integrate, giving 1 2 x ln jyj C C.x/; (5.112) 2 where C.x/ is an arbitrary function and substitution into (5.104) shows it is exactly satisﬁed. uD

Example 5.8 Consider u2xy D 0:

uxx C uyy C 2 uxx uyy

(5.113)

Here A D 1; B D 0; C D 1;

D D 2;

E D 0;

(5.114)

and from (5.72) we have 42 C 1 D 0;

(5.115)

giving D ˙i=2. Thus, (5.74) and (5.75) become 2Fx C 2pFu Fp iFq D 0; 2Fy C 2qFu C iFp Fq D 0;

(5.116a) (5.116b)

106

5. FIRST INTEGRALS

and 2Fx C 2pFu 2Fy C 2qFu

Fp C iFq D 0; iFp Fq D 0:

(5.117a) (5.117b)

The solutions of each are F1 .x C 2ux iy; ux C i uy / D 0; F2 .x C 2ux C iy; ux iuy / D 0:

(5.118)

Again, one will notice that the ﬁrst integrals in (5.118) are complex. However, it is again possible to construct real solutions to (5.113). For example, if we choose x C 2ux iy C ux C i uy D 0; x C 2ux C iy C ux i uy D 0;

(5.119a) (5.119b)

then we can split into real and imaginary parts, giving x C 3ux D 0; y uy D 0;

(5.120a) (5.120b)

which leads to the solution uD

y2 x2 C C c; 6 2

(5.121)

where c is an arbitrary constant. If we choose .x C 2ux iy/2 C ux C iuy D 0; .x C 2ux C iy/2 C ux iuy D 0;

(5.122a) (5.122b)

then we can split into real and imaginary parts, giving 4yux uy C 2xy D 0; 4u2x C .4x C 1/ux C x 2 y 2 D 0;

(5.123a) (5.123b)

which leads to the common solution uD

2x 2 C 2y 2 C x .1 C 8x C 16y 2 /3=2 ˙ C c; 8 96

(5.124)

where c is an arbitrary constant. The reader can verify that (5.121) and (5.124) do indeed satisfy (5.113).

5.3. THE MARTIN EQUATION

5.3

107

THE MARTIN EQUATION

The governing equations of an inviscid, one-dimensional nonsteady gas, neglecting heat condution and heat radiation are (5.125a)

t C .u/x D 0; Px u t C uux D ;

(5.125b)

or, in conservative form t C .u/x D 0; .u/ C u2 C P x D 0:

(5.126a) (5.126b)

Normally, a third equation is given, conservation of energy, or an equation of state, but here we will leave the system underdetermined. If we introduce stream functions and N such that D x; u D 2 N u D x ; u C P D

t; N t ;

(5.127a) (5.127b)

then (5.126a) and (5.126b) are automatically satisﬁed. If we introduce a new variable such that N D

(5.128)

tP;

then (5.127b) becomes u D x

tPx ; u2 D t C tP t :

(5.129)

Following Martin [74], we choose new independent variables . ; P /, so that (5.127a) and (5.129) become tP D (5.130a) ; J xP u D ; (5.130b) J P t tP t t u D ; (5.130c) J P x xP tx u2 D ; (5.130d) J where J D t xP tP x . From which (5.130a) and (5.130b) we deduce xP uD : (5.131) tP Eliminating and u from (5.130c) and (5.130d) using (5.130a) and (5.131) gives xP D t t C tP

t P ;

(5.132a)

D tx C xP

x P ;

(5.132b)

xP2 tP

108

5. FIRST INTEGRALS

from which we deduce .t

P / t xP

tp x

D 0:

(5.133)

Thus, (5.134)

t D P ;

as we are assuming that t xP

tp x ¤ 0. From (5.132) we obtain xP D PP ;

(5.135)

and from (5.134) and (5.135) gives (5.131) as (5.136)

uD :

We return to (5.130a) and with (5.134) this becomes x D P

C ;

(5.137)

where D 1 . Eliminating x from (5.135) and (5.137) gives the Monge–Ampere equation

PP

2 P D P :

(5.138)

With a renaming of variables, Martin [75] asked which forms of the Monge–Ampere equation u2xy C 2 D 0;

uxx uyy

D X.x/Y .y/;

(5.139)

admit ﬁrst integrals. He was able to deduce the following forms: D Y.y/;

D

xm 1 ; y mC1

(5.140)

where m is an arbitrary constant and Y.y/ an arbitrary function of its argument and identiﬁed the special cases D y 2; D x 1y 1; D ex ey : (5.141) We now consider this same problem, however, we will remove the assumption that is separable. From the section on ﬁrst integrals, if a ﬁrst integral F .x; y; u; p; q/ exists, then F satisﬁes Fx C pFu .x; y/Fq D 0; Fy C qFu C .x; y/Fp D 0:

(5.142a) (5.142b)

Requiring that these two be compatible gives rise to the additional equation Fu

1 x Fp 2

1 y Fq D 0: 2

(5.143)

5.3. THE MARTIN EQUATION

If we further require that (5.142) and (5.143) be compatible, we obtain xx 3 x xy 3 y Fp C Fq D 0; 2 2 x x xy 3 x yy 3 y Fp C Fq D 0: y 2 y 2

109

(5.144a) (5.144b)

Since Fp Fq ¤ 0, from (5.144) we obtain xx 3 x yy 3 y xy 3 y xy 3 x D 0; x 2 y 2 x 2 y 2 or 2 2xx 32x 2yy 3y2 2xy 3x y D 0:

(5.145) (5.146)

Melshenko [76] also obtains (5.146) and states that this equation admits solutions of the form x C c1 D H .c1 x C c2 y/ and D .y C c2 / 2 H ; (5.147) y C c2 for arbitrary constants c1 ; c2 and arbitrary function H . Remarkably, under the substitution D

1 ; !2

(5.148)

Eq. (5.146) becomes the homogeneous Monge–Ampere equation !xx !yy

2 !xy D 0;

(5.149)

which is know to admit the solutions ([77]) ! D f .c1 x C c2 y/ C c3 x C c4 y C c5 ; c5 and arbitrary function f , c4 x C c5 y C c6 ! D .c1 x C c2 y C c3 / f C c7 x C c8 y C c9 ; c1 x C c2 y C c3

(5.150)

for arbitrary constants c1

with arbitrary constants c1

(5.151)

c9 and arbitrary function f , and the parametric solutions

! D tx C f .t /y C g.t/;

x C f 0 .t /y C g 0 .t / D 0;

(5.152)

with arbitrary functions f and g noting that (5.147) are special cases of (5.150) and (5.151). One can also show that (5.149) also admits solutions of the form c4 x C c5 y C c6 C c10 x C c11 y C c12 ; (5.153) ! D .c1 x C c2 y C c3 / f c7 x C c8 y C c9 with arbitrary constants c1 c12 and arbitrary function f , but with the constraint that c1 c5 c9 C c2 c6 c7 C c3 c4 c8 c1 c6 c8 c2 c4 c9 c3 c5 c7 D 0.

110

5. FIRST INTEGRALS

5.4

FIRST INTEGRALS AND LINEARIZATION

As we’ve seen in the preceding section, some PDEs admit very general classes of ﬁrst integrals. A natural question is: can they be used to simplify the form of a PDE?

5.4.1 HYPERBOLIC MA EQUATIONS Lie [72] and [73] considered PDEs of the form s 2 D E;

(5.154)

F2 .˛2 ; ˇ2 / D 0:

(5.155)

Ar C Bs C C t C D rt

that admitted two general ﬁrst integrals, say F1 .˛1 ; ˇ1 / D 0;

In (5.155), it is assumed that F1 and F2 are arbitrary functions of ˛i and ˇi that are functions of x; y; u; p; and q . Lie was able to show that when these ﬁrst integrals exist, it is possible to transform (5.154) to UXY D 0: We deﬁne X D ˛1 ; Y D ˛2 ; P D ˇ1 ; Q D ˇ2 (5.156) and we’ll assume that U exists, such that the contact conditions are satisﬁed. We consider UXY D 0:

This can be rewritten as @.X; P / @.X; P / @.X; Y / @P D D = D 0; @Y @.X; Y / @.x; y/ @.x; y/

giving @.X; P / D0 @.x; y/

(5.157)

or Xx Py

Xy Px D 0:

We now bring in (5.156) and suppress subscripts. Thus, (5.158) becomes ˛x C p˛u C r˛p C s˛q ˇy C qˇu C sˇp C tˇq ˛y C q˛u C s˛p C t˛q ˇx C pˇu C rˇp C sˇq D 0:

(5.158)

(5.159)

Expanding (5.159) gives ˛ ˇ C qˇ ˇ ˛ C q˛ p y u p y u r C ˇq .˛x C p˛u / C ˛q ˇy C qˇu ˇq ˛y C q˛u ˛p .ˇx C pˇu / s (5.160) C ˇq .˛x C p˛u/ ˛q .ˇx C pˇu / t 2 C ˛p ˇq ˛qˇp rt s D ˛y C q˛u .ˇx C pˇu / .˛x C p˛u / ˇy C qˇu :

5.4. FIRST INTEGRALS AND LINEARIZATION

111

As ˛ and ˇ are ﬁrst integrals, they satisfy the following: D .Fx C pFu/ CFp C 1 DFq D 0; D Fy C qFu C 2 DFp AFq D 0;

(5.161a) (5.161b)

where D 2 2

BD C AC C DE D 0:

(5.162)

As we are assuming that D ¤ 0, then using (5.161), we eliminate the terms ˛x C p˛u ; ˛y C q˛u ; ˇx C pˇu ; ˇy C qˇu

(5.163)

from (5.160). Simplifying gives A ˛p ˇq D C C ˛p ˇq D D

1 2

ˇp ˛q r C .1 C 2 / ˛p ˇq ˇp ˛q s ˇp ˛q t C ˛p ˇq ˛q ˇp rt s 2 AC ˛p ˇq ˇp ˛q : 2 D

As we have assumed that ˛p ˇq

ˇp ˛q ¤ 0, this term cancels, giving C AC A 2 r C .1 C 2 / s C t C rt s D 1 2 : D D D2

From (5.162), we deduce that 1 C 2 D B=D and 1 2 D .AC C DE/=D 2 , giving B C AC C DE AC A 2 rC sC t C rt s D : D D D U2 D2

(5.164)

Simplifying gives rise to (5.154). Recall the contact conditions from the previous chapter. They are: Up Uq Uu Ux Uy

PXp PXq PXu PXx PXy

QYp QYq QYu QYx QYy

D 0; D 0; D ; D p; D q:

(5.165)

s 2 C 9 D 0:

(5.166)

These will be needed for the following examples. Example 5.9

Consider 3r C s C t C rt

112

5. FIRST INTEGRALS

This PDE admits the ﬁrst integrals F1 .x C 2y C p; 3x

3y

q/ D 0;

F2 .x

3y C p; 2x C 3y C q/ D 0:

(5.167)

Here we try (5.168a) (5.168b)

X D x C 2y C p; Y D x 3y C p; P D 3x 3y q; Q D 2x C 3y C q:

At this point, we need to ﬁnd U such that the contact conditions (5.165) are satisﬁed, namely Up Uq Uu Ux Uy

D D D D D

5x; 0; ; 5x p; 15y 5q

(5.169) q:

Cross diﬀerentiation of (5.169) shows they are consistent; thus, U (and ) exist. Integrating (5.169) gives 15 2 5 y 5u; (5.170) U D 5xp C x 2 2 2 and calculating UXY using (5.168) and (5.170) gives UXY D

Example 5.10

3uxx C uxy C uyy C uxx uyy 5 .uxx C 1/

u2xy C 9

D 0:

(5.171)

Consider uy2 uxx

3ux uy uxy C 2u2x uyy D ux uy2 :

This PDE admits the ﬁrst integrals p F1 y C 2 ; e q

xp

q

D 0;

F2 u; e

x

p q2

(5.172)

D 0:

(5.173)

Here we try p X D y C 2 ; Y D u; q p xp P De ; Q D e x 2: q q

(5.174a) (5.174b)

5.4. FIRST INTEGRALS AND LINEARIZATION

113

At this point, we need to ﬁnd U such that the contact conditions (5.165) are satisﬁed, namely p ; q2 p2 D 2e x 3 ; q x p De C ; q2 D p; p De x q: q

Up D 2e Uq Uu Ux Uy

x

(5.175a) (5.175b) (5.175c) (5.175d) (5.175e)

Cross diﬀerentiation of (5.175) shows they are not consistent, so we modify our choice. We now pick p X D y C 2 ; Y D u; q xp ; Q D be P D ae q

(5.176a) x

p ; q2

(5.176b)

with hopefully suitably chosen constants a and b . The contact conditions become p ; q2 p2 D 2ae x 3 ; q x p C ; D be q2 D p; p D ae x q: q

Up D 2ae Uq Uu Ux Uy

x

(5.177a) (5.177b) (5.177c) (5.177d) (5.177e)

Compatibility between these gives 1 q2 x p pp C 2ae q2 1 qp ae x q p q 2be x 3 q p C be

x

D 0;

(5.178a)

D 0;

(5.178b)

D 0;

(5.178c)

D 0;

(5.178d) (5.178e)

114

5. FIRST INTEGRALS

pq C 2ae

2 xp 3 q

p q2 p x C pu be x 2 q y C qu p qx py C ae x q qq C C ae

x

D 0;

(5.178f )

D 0;

(5.178g)

D 0;

(5.178h)

D 0;

(5.178i) (5.178j)

D 0:

From (5.178a) and (5.178c) we see that b D a, and further from (5.178a) and (5.178b) we get p D ae x 2 : q This assignment shows that the entire system (5.178) is satisﬁed; thus, U exists satisfying the contact conditions. Integrating (5.177) (with a D 1) gives U De

2 xp ; q2

(5.179)

and under the contact transformation p X D y C 2 ; Y D u; U D e q

2 xp ; q2

(5.180)

gives UXY as UXY D

Example 5.11 Consider uy uy C 1 uxx

e

x

uy2 uxx

uy 2zy2 zxx

3ux uy uxy C 2u2x uyy 4zx zy zxy C 2zx2 zyy

ux uy2 D 0: zx u2z

2ux uy C ux C uy C 1 uxy C ux .ux C 1/ uyy D 0:

This PDE admits the ﬁrst integrals qC1 pC1 F1 x C u; D 0; F2 y C u; D 0: p q

(5.181)

(5.182)

(5.183)

Here we try X D x C u; Y D y C q; qC1 pC1 P D ; QD : p q

(5.184a) (5.184b)

5.4. FIRST INTEGRALS AND LINEARIZATION

115

At this point, we need to ﬁnd U such that the contact conditions (5.165) are satisﬁed, namely Up Uq Uu Ux Uy

D 0; D 0; pC1 D q qC1 D p pC1 D q

qC1 C ; p

(5.185)

p; q:

However, cross diﬀerentiation of (5.185) shows they are not consistent; thus, another choice will be required. Next we try X D x C u; Y D y C u; qC1 pC1 P DA ; QDB ; p q

(5.186a) (5.186b)

and determine whether the functions A and B can be chosen such that the contact conditions can be satisﬁed. With (5.186), the contact conditions (5.165) become Up Uq Uu Ux Uy

D 0; D 0; qC1 pC1 D A B C ; q p qC1 D A p; p pC1 q; D B q

or, after eliminating , the latter two of (5.187) become qC1 pC1 Ux C pUu D .p C 1/A C pB ; p q qC1 pC1 Uy C qUu D qA C .q C 1/B : p q Diﬀerentiating (5.188a) with respect to q (or (5.188b) with respect to p ) gives qC1 pC1 q 2 A0 p2B 0 D 0: p q If we let pD

sC1 r C1 ; qD ; rs 1 rs 1

(5.187)

(5.188a) (5.188b)

(5.189)

(5.190)

116

5. FIRST INTEGRALS

then (5.189) becomes .r C 1/2 A0 .r/

.s C 1/2 B 0 .s/ D 0;

(5.191)

from which we deduce that .r C 1/2 A0 .r/ D .s C 1/2 B 0 .s/ D

c1

(5.192)

for some constant c1 . Each can be integrated giving A .r/ D

c1 C c2 ; r C1

B .s/ D

Returning to (5.188) and simplifying gives

c1 C c3 : sC1

Ux C pUu D .c1 C c2 C c3 / p C c2 ; Uy C qUu D .c1 C c2 C c3 / q C c3 :

(5.193)

(5.194a) (5.194b)

We choose c1 D 1 and c2 D c3 D 0. Solving (5.194) subject to Up D Uq D 0 gives U D u and thus, the contact transformation is X D x C u; Y D y C u; U D u:

Under this contact (point) transformation, we have 2ux uy C ux C uy C 1 uxy C ux C u2x uyy uy C uy2 uxx UXY D D 0: 3 1 C ux C uy

(5.195)

(5.196)

5.4.2 PARABOLIC MA EQUATIONS As we’ve seen in the previous section, hyperbolic Monge–Ampere equations admitting four ﬁrst integrals can be linearized. In this section we consider parabolic Monge–Ampere equations admitting three ﬁrst integrals. Suppose Ar C Bs C C t C D rt

s 2 D E;

(5.197)

which admits three general ﬁrst integrals F .˛; ˇ; / D 0;

(5.198)

where F is an arbitrary function of ˛ , ˇ and that are functions of x; y; u; p; and q . Then it is possible to transform (5.197) to UXX D 0: We deﬁne Y D ˛; P D ˇ; (5.199) and we will assume for the moment that X; U and Q exist such that these, in addition to (5.199), satisfy the contact conditions (5.165). Consider UXX D 0:

5.4. FIRST INTEGRALS AND LINEARIZATION

117

This can be written as @P @.P; Y / @.P; Y / @.X; Y / D D = D 0; @X @.X; Y / @.x; y/ @.x; y/

giving @.P; Y / D 0; @.x; y/

(5.200)

or Px Yy

Py Yx D 0:

(5.201)

The rest follows as in the hyperbolic case (steps (5.159)–(5.164)) and we refer the reader there for details. However, unlike the hyperbolic case where there is a set of two independent ﬁrst integrals allowing one to choose X , Y , P , and Q, and use these to ﬁnd U , in the parabolic case we only have three ﬁrst integrals and we have already used two to deﬁne Y and P . We show how to use the third ﬁrst integral to obtain X; U , and Q. Recall the contact conditions: Up Uq Uu Ux Uy

PXp PXq PXu PXx PXy

QYp QYq QYu QYx QYy

D 0; D 0; D ; D p; D q:

(5.202)

We deﬁne U as (5.203)

U D XP C W;

so that (5.202), after eliminating , becomes Wp C XPp QYp D 0; Wq C XPq QYq D 0; Wx C pWu C X .Px C pPu / Q .Yx C pYu/ D 0; Wy C qWu C X Py C qPu Q Yy C qYu D 0:

(5.204a) (5.204b) (5.204c) (5.204d)

Since Y and P are independent, then it is possible to choose a pair of equations from (5.204) and solve for X and Q. To demonstrate, let us suppose for the sake of argument that Eqs. (5.204b) and (5.204c). From these two, we solve for X and Q XD

.Yx C pYu /Wq .Yx C pYu /Pq

noting that .Yx C pYu /Pq tions in (5.204) gives

.Wx C pWu /Pq .Yx C pYu /Pq

.Px C pPu /Wq ; .Px C pPu /Yq (5.205) .Px C pPu /Yq ¤ 0. Substituting (5.205) into the remaining equa.Wx C pWu /Yq ; .Px C pPu /Yq

QD

@.Y; P; W / D 0; @.x; N p; q/

@.Y; P; W / D 0: @.x; N y; N q/

(5.206)

118

5. FIRST INTEGRALS

In (5.206) we have the usual Jacobian notation, and partial derivatives with respect to xN and yN are deﬁned as @ @ @ @ @ @ D Cp ; D Cq : (5.207) @xN @x @u @yN @y @u As both Y and P are ﬁrst integrals to (5.197), they both satisfy the equations C Fp C D B Fy C qFu C Fp 2D

Fx C pFu

B Fq D 0; 2D A Fq D 0: D

(5.208a) (5.208b)

Using these in (5.206) gives rise to two equations for W , which is precisely (5.208). As we have a third independent solution of (5.208) (i.e., the third ﬁrst integral), we use this for W . The following examples illustrate this method. Example 5.12 Consider s2 D

r C 2s C 2t C rt

(5.209)

1:

This equation admits the ﬁrst integral F 2x

y C p; y

x C q; u

xp

yq

x 2 C xy

y 2 =2 D 0:

(5.210)

Here, we will choose Y Dy

x C q;

P D 2x

y C p:

(5.211)

The system (5.204) becomes Wp C X D 0; Wq Q D 0; Wx C pWu C 2X C Q D 0; Wy C qWu X Q D 0:

(5.212a) (5.212b) (5.212c) (5.212d)

We solve (5.212a) and (5.212b) for X and Q. This gives XD

Wp ;

(5.213)

Q D Wq ;

giving the remaining two equations in (5.212) as Wx C pWu 2Wp C Wq D 0; Wy C qWu C Wp Wq D 0:

(5.214a) (5.214b)

These are easily solved, giving (5.210). If we choose W Du

xp

yq

x 2 C xy

y 2 =2;

(5.215)

5.4. FIRST INTEGRALS AND LINEARIZATION

119

then from (5.213), (5.211), and (5.203) we have X D x;

Y Dy

x C uy ;

yuy C x 2

U Du

y2 ; 2

(5.216)

and under this contact transformation UXX D

u2xy C 1

uxx C 2uxy C uyy C uxx uyy uyy C 1

D 0:

(5.217)

We also note that X D x;

Y Dy

x C uy ;

yuy C x 2

U Du

y2 ; 2

(5.218)

will transform (5.209) to UXX D 0. Example 5.13 Consider xuy2 uxx

2xux uy uxy C xu2x uyy D ux uy2 C xu2x uy :

This equation admits the ﬁrst integral 2xq xq F u; x 2 C ; e p p Here, we will choose Y D u; P D x 2 C

y

D 0:

2xq : p

(5.219)

(5.220)

(5.221)

The system (5.204) becomes 2xq X D 0; p 2x Wq C X D 0; p q Wx C pWu C 2 x C X pQ D 0; p Wy C qWu qQ D 0: Wp

(5.222a) (5.222b) (5.222c) (5.222d)

We solve (5.222b) and (5.222d) for X and Q. This gives XD

pWq ; 2x

QD

Wy C qWu : q

(5.223)

Thus, the remaining two equations in (5.222) become Wx

p Wy q

pW C qW D 0; p q q pC Wq D 0: x

(5.224a) (5.224b)

120

5. FIRST INTEGRALS

These are easily solved, giving (5.220). If we choose 2xqe p

W D

y

(5.225)

;

then from (5.223), (5.221), and (5.203) we have X De

y

; Y D u; U D x 2 e

y

(5.226)

;

and under this contact transformation UXX D

xuy2 uxx

2xux uy uxy C xu2x uyy e

ux uy2

xu2x uy

y u3 x

D 0:

(5.227)

With the solution of (5.227) as U D A.Y /X C B.Y /, with A and B arbitrary, through the contact transformation (5.226) we obtain the general solution of the original PDE as x 2 D A.u/ C B.u/e y :

(5.228)

Example 5.14 Consider u2xy D u2x uy2 :

2uux uy uxy C u2 uxx uyy

This equation admits the ﬁrst integral F x

u ;y p

Here, we will choose Y Dy

u u ; q pq

u ; q

P D

D 0:

u : pq

(5.229)

(5.230)

(5.231)

The system (5.204) becomes u Wp X D 0; p2q u u X Q D 0; Wq 2 pq q2 p 1 Wx C pWu C X C Q D 0; q q 1 Wy C qWu C X D 0: p

(5.232a) (5.232b) (5.232c) (5.232d)

We solve (5.232a) and (5.232b) for X and Q. This gives XD

p 2 qWp ; u

QD

q.pWp qWq ; u

(5.233)

5.4. FIRST INTEGRALS AND LINEARIZATION

121

and the two remaining equations in (5.232) as pq Wq D 0; u pq Wy C qWu C Wp D 0: u

(5.234a)

Wx C pWu C

(5.234b)

These are easily solved giving (5.230). If we choose W Dx

u ; p

(5.235)

then from (5.233), (5.231), and (5.203) we have u ; U D x; q

X D q; Y D y

P D

u ; pq

QD

q ; p

(5.236)

D 0:

(5.237)

and under this contact transformation UXX D

u2xy

2uux uy uxy C u2 uxx uyy

u2x uy2

u3x uy2 uyy

We note that through (5.236), we are able to explicitly solve for x; y; and u, giving x D U;

yDY

XP ; uD Q

X 2P ; Q

(5.238)

and one can show that under this transformation, (5.229) becomes UXX D 0.

5.4.3 ELLIPTIC MA EQUATIONS We now extend our results to elliptic Monge–Ampere equations. We again consider Ar C Bs C C t C D rt

s 2 D E;

(5.239)

and suppose that it admits two general ﬁrst integrals, say F .˛1 ˙ iˇ1 ; ˛2 ˙ iˇ2 / D 0;

(5.240)

where F is an arbitrary function of ˛i ˙ iˇi ; i D 1; 2; and ˛i ; ˇi are real functions of x; y; u; p , and q . We will show it is possible to transform (5.239) to UXX C UY Y D 0. We deﬁne X D ˛1 ; Y D ˇ1 ; P D ˇ2 ; Q D ˛2 (5.241) and we’ll assume that U exists such that the contact conditions are satisﬁed. Consider UXX C UY Y D 0:

122

5. FIRST INTEGRALS

This can be written as @P @Q @.P; Y / @.X; Q/ C D C @X @Y @.X; Y / @.X; Y / @.P; Y / @.X; Y / @.X; Q/ @.X; Y / D = C = D 0; @.x; y/ @.x; y/ @.x; y/ @.x; y/

(5.242)

giving @.P; Y / @.X; Q/ C D 0; @.x; y/ @.x; y/

or Px Yy

Py Yx C Xx Qy

(5.243)

Xy Qx D 0:

We now bring in (5.241), so (5.243) becomes ˇ2 x C pˇ2 u C rˇ2p C sˇ2 q ˇ1y C qˇ1 u C sˇ1p C tˇ1 q ˇ1 x C pˇ1 u C rˇ1p C sˇ1 q ˇ2y C qˇ2 u C sˇ2p C tˇ2 q C ˛1x C p˛1u C r˛1p C s˛1q ˛2y C q˛2u C s˛2p C t˛2q ˛2x C p˛2u C r˛2p C s˛2q ˛1y C q˛1u C s˛1p C t˛1q D 0:

(5.244)

As F in (5.240) are ﬁrst integrals, they satisfy the following (assuming that D ¤ 0): ˛1x

iˇ1 x C p .˛1u

˛1y

iˇ1y C q .˛1u

˛2x

iˇ2 x C p .˛2u

˛2y

iˇ2y C q .˛2u

C ˛1p iˇ1p C .˛1 iˇ1 / ˛1q D A i ˇ1 u / C .˛1 C i ˇ1 / ˛1p i ˇ1p ˛1q D C iˇ2 u / ˛2p iˇ2p C .˛2 iˇ2 / ˛2q D A i ˇ2 u / C .˛2 C i ˇ2 / ˛2p i ˇ2p ˛2q D iˇ1 u /

iˇ1 q D 0; iˇ1 q D 0; iˇ2 q D 0; iˇ2 q D 0;

(5.245a) (5.245b) (5.245c) (5.245d)

where D 2 2

BD C AC C DE D 0; D a ˙ ib:

(5.246)

We isolate real and imaginary parts of (5.245) and solve for ˛i x ; ˛i y ; ˇi x ; ˇi y and eliminate these in (5.244). This leads to A C r C 2a s C t C rt D D

s 2 D a2 C b 2

AC : D2

(5.247)

From (5.246), we have that a2

b2 D2

B 2aD D 0; aBD C AC C DE D 0;

(5.248a) (5.248b)

2

from which we deduce a C b

2

5.4. FIRST INTEGRALS AND LINEARIZATION

123

2

D D AC C DE . Eliminating a and b in (5.247) gives s 2 D E;

Ar C Bs C C t C D rt

which is (5.239). The next two examples illustrate these results. Example 5.15

Consider .1 C uy2 /uxx

2ux uy uxy C u2x uyy D 0:

(5.249)

As was shown in Example 5.3, this PDE admits the ﬁrst integrals qi F u ˙ iy; D 0: p

(5.250)

Here we try 1 q ; QD : p p

X D u; Y D y; P D

(5.251)

Unfortunately, the contact conditions are not satisﬁed, but with the slight adjustment, X D u; Y D y; P D

1 ; QD p

q ; p

(5.252)

they lead to U D x to which we add to (5.252) and under X D u; Y D y;

U D x;

P D

or x D U; y D Y; u D X;

pD

1 ; p

QD

1 ; qD UX

q : p

(5.253)

UY UX

(5.254)

the PDE (5.249) is transformed to UXX C UY Y D 0. Example 5.16

Consider uxx C uyy C 2uxx uyy

2u2xy D 0:

(5.255)

As was shown in Example 5.8, this PDE admits the ﬁrst integrals F .x C 2p ˙ iy; q ˙ ip/ D 0:

(5.256)

Here we try X D x C 2p; Y D y;

P D p;

Q D q:

(5.257)

124

5. FIRST INTEGRALS

We ﬁnd the contact conditions are satisﬁed and lead to U D u C p 2 , and from xDX

2UX ; y D Y; u D U

UX2 p D

1 ; qD UX

UY ; UX

(5.258)

the PDE (5.255) is transformed to UXX C UY Y D 0. Example 5.17

Consider uy uxx

uyy C 1 C uy

uxx uyy

u2xy D uy :

(5.259)

This PDE admits the ﬁrst integrals F .p

x ˙ i q; y C q C ln q ˙ ix/ D 0:

(5.260)

Here we try X Dp

x;

Y D q;

P D x;

Q D y C q C ln q:

(5.261)

We ﬁnd that the contact conditions (5.165) are satisﬁed, leading to U D xp C yq

u

1 2 q C q ln q 2

(5.262)

q:

From (5.261) and (5.262), x D UX ; y D UY

Y

ln Y; u D XUX C Y UY

1 U C UX2 2

1 2 Y 2

Y

(5.263)

transforms (5.259) to UXX C UY Y D 0:

5.5 5.1.

EXERCISES Find a ﬁrst integral for the following: .i/ .i i / .i i i / .iv/

5.2.

(5.264)

q 2 r 2pqs C p 2 t D 0; qr .p C q C 1/s C .p C 1/t D 0; rt s 2 C 1 D 0; yr ps C t C y.rt s 2 / C 1 D 0:

The equations that model one-dimensional gas ﬂow are t C .u/x D 0; Px u t C uux D ;

(5.265a) (5.265b)

5.6. REFERENCES

where P D P ./. If we introduce a stream function D

x;

u D

125

such that (5.266)

t;

then (5.265a) is identically satisﬁed, whereas (5.265b) becomes 2 x

tt

2

t

x

tx

C

2 t

xx

2 0 xP

.

x/

xx

D 0:

(5.267)

Determine the forms of P ./ such that (5.267) admits a ﬁrst integral. 5.3.

Show the PDE 2xuy uxy C uy2 uyy C x 2 uxx uyy

2 uxy D uy2

(5.268)

admits the ﬁrst integral F

q ;u x

xp

1 2 q ;p 2

yq C q 2 x

D 0:

(5.269)

Use this to ﬁnd a contact transformation that linearizes the PDE. 5.4.

Show the PDE 2xuy uxy C x 2 uxx uyy

admits the ﬁrst integral

F

1 ˙ iq ; y ˙ i.xp x

u2xy D 1 C uy2 u/ D 0:

(5.270)

(5.271)

Use this to ﬁnd a contact transformation that linearizes the PDE.

5.6

REFERENCES

[72] S. Lie, Neue integrations methode der Monge–Ampèreshen gleichung, Arch. Math. Kristiania, 2, pp. 1–9, 1877. 110 [73] A. R. Forsyth, Theory of Diﬀerential Equations, vol. 6, Cambridge University Press, Cambridge, 1906. 110 [74] M. H. Martin, The propagation of a plane shock into a quiet atmosphere, Can. J. Math., 5, pp. 37–39, 1953. DOI: 10.4153/CJM-1953-004-2. 107 [75] M. H. Martin, The Monge–Ampere partial diﬀerential equation rt Math., 3, pp. 165–187, 1953. DOI: 10.2140/pjm.1953.3.165. 108

s 2 C 2 D 0, Pac. J.

[76] S. V. Meleshko, Methods for Constructing Exact Solutions of Partial Diﬀerential Equations, Springer, 2005. 109 [77] A. D. Polyanin and V. F. Zaitsev, Handbook of Nonlinear Partial Diﬀerential Equations, Chapman and Hall/CRC, Boca Raton, FL, 2004. DOI: 10.1201/9780203489659. 109

127

CHAPTER

6

Functional Separability In a standard course in PDEs, one encounters the heat equation u t D uxx ;

(6.1)

and solutions in the form u D T .t/X.x/, often referred to as separable solutions. A natural question is: do nonlinear PDEs admit separable solutions? In general, the answer is no but sometimes they admit some sort of a variation of separable solution. For example, u t D .uux /x

(6.2)

does admit solutions of the form u D TX , leading to .XX 0 / T0 D : T2 X

(6.3)

u D a.t/x 2 C b.t /x C c.t /:

(6.4)

It also admits solutions of the form

Substitution of (6.4) into (6.2) and isolating coeﬃcients of x gives a0 b0 c0

D 6a2 ; D 6ab; D 2ac C b 2 :

(6.5) (6.6)

uxx 1 C u2x

(6.7)

However, ut D

doesn’t admit separable solutions of the form u D T .t /X.x/, since substitution of this form into (6.7) leads to TX 00 T 0X D (6.8) 1 C T 2 X 02 which doesn’t separate. However, Eq. (6.7) does admit separable solutions of the form u D T .t / C X.x/. Sometimes it is not at all obvious that a particular PDE admits some form of a separable solution. The following two examples illustrate this.

128

6. FUNCTIONAL SEPARABILITY

Example 6.1 Consider u t t D uxx C e u

(6.9)

u D ln.T .t / C X.x//:

(6.10)

and separable solutions of the form

Substitution of (6.10) into (6.9) gives .T C X/T 00 T 02 .T C X/X 00 X 02 D C .T C X/: .T C X/2 .T C X/2

(6.11)

.T C X/T 00

(6.12)

Simplifying gives T 02 D .T C X /X 00

X 02 C .T C X/3 ;

and diﬀerentiating (6.12) with respect to t and x gives X 0 T 000 D T 0 X 000 C 6a.T C X/T 0 X 0 ;

(6.13)

X 000 T 000 D 0 C 6.T C X/: 0 T X

(6.14)

or

This leads to the two equations T 000

6T T 0 D kT 0 ; X 000 C 6XX 0 D kX 0 ;

(6.15)

where k is an arbitrary constant. Integrating each twice gives 1 02 T 2

T3 D

1 2 kT C c1 T C c2 ; 2

1 02 1 X C X 3 D kX 2 C c3 X C c4 ; 2 2

(6.16)

where c1 c4 are constants of integration. Finally, substituting (6.16) back into (6.12) shows that it is identically satisﬁed if c1 C c3 D 0 and c2 c4 D 0. Thus, any solution of T 02 D 2T 3 C kT 2

2c1 T C 2c2 ; X 02 D

2X 3 C kX 2 C 2c1 X C 2c2 ;

(6.17)

when substituted into (6.10), will give rise to an exact solution of the PDE (6.9). Example 6.2 Consider ut t

uxx D sin u;

(6.18)

X.x/ : T .t /

(6.19)

and solutions of the form u D 4 tan

1

6. FUNCTIONAL SEPARABILITY

129

Substituting (6.19) into (6.18) and re-arranging terms gives 2T 02

T T 00

X2

T 00 C 2X 02 T

T2

X 00 X

XX 00

T2

X 2 D 0:

Diﬀerentiating (6.20) twice with respect to t and x gives 00 0 00 0 1 T 1 X C D 0; 0 0 TT T XX X

(6.20)

(6.21)

leading to the equations 1 TT0

T 00 T

0

1 XX 0

D k;

X 00 X

0

D

k;

(6.22)

where k is again a constant. Each can be integrated twice, leading to X 02 D c1 X 4 C c2 X 2 C c3 ; T 02 D

c1 T 4 C c4 T 2 C c5 :

(6.23)

Substituting (6.23) into (6.20) leads to the conditions c4 c2 D 1, c3 C c5 D 0. Any solution of the Jacobi elliptic equations (6.23) with these constant restrictions will lead to an exact solution of (6.18) in the form of (6.19). A natural question is: how did one know that separable solutions of u t t D uxx C e u ;

(6.24)

u D ln.T .t / C X.x//;

(6.25)

were of the form or separable solutions of ut t

uxx D sin u;

(6.26)

were of the form u D 4 tan

1

.X.x/=T .t //‹

(6.27)

We also ask: are there others? If an equation admits functional separable solutions, then it would be of the form f .u/ D T .t/X.x/; or f .u/ D T .t / C X.x/: However, these are equivalent, as we can take natural logarithms of both sides of the ﬁrst and reset f; T and X . Thus, we only consider f .u/ D T .t / C X.x/:

Diﬀerentiating (6.28) with respect to t and x gives f 0 u tx C f 00 u t ux D 0;

(6.28)

130

6. FUNCTIONAL SEPARABILITY

or (6.29)

u tx D F .u/u t ux ;

where we set F D f 00 =f 0 . If a PDE admits solutions of the form (6.28) then it will be compatible with (6.29). As an example, we will reconsider (6.9). Taking t and x derivatives of both (6.9) and (6.29) gives ut t t u t tx

u txx uxxx u t tx u txx

D euut ; D e u ux ; D F 0 u2t ux C F u t t ux C F u t u tx ; D F 0 u t u2x C F u tx ux C F u t uxx :

(6.30a) (6.30b) (6.30c) (6.30d)

Solving (6.30) for the third-order derivatives u t t t ; u t tx ; u txx ; and uxxx gives ut t t u t tx u txx uxxx

D F ux u tx C F u t uxx C F 0 u t u2x C e u u t D F ux u t t C F u t u tx C F 0 u2t ux D F ux u tx C F u t uxx C F 0 u t u2x D F ux u t t C F u t u tx C F 0 u2t ux e u ux :

(6.31a) (6.31b) (6.31c) (6.31d)

Requiring compatibility .u t tx / t D .u t t t /x ;

.u txx / t D .u t tx /x ;

.uxxx / t D .u txx /x ;

and using both (6.9) and (6.29) gives F 00 C 2FF 0 u3t ux C u3x C 3F 0 C 2F 2 C F

1 e u u t ux D 0;

(6.32)

(6.33)

from which we obtain two equations for F F 00 C 2FF 0 D 0; 3F C 2F 2 C F 1 D 0:

(6.34a) (6.34b)

0

Requiring that (6.34) be compatible gives F D

1; F D

1 : 2

(6.35)

Since F D f 00 =f 0 , upon integrating we get f D c1 e u C c2 ;

f D c1 e

u=2

C c2 ;

(6.36)

where c1 and c2 are arbitrary constants. Thus, u t t D uxx C e u

(6.37)

6. FUNCTIONAL SEPARABILITY

131

admits separable solutions of the form c1 e u C c2 D T .t / C X.x/; c1 e u=2 C c2 D T .t / C X.x/:

(6.38a) (6.38b)

Note that we can set c1 D 1 and c2 D 0 without loss of generality. Solving for u, we obtain u D ln jT .t/ C X.x/j; u D 2 ln jT .t / C X.x/j;

(6.39a) (6.39b)

and the ﬁrst solution in (6.39) we saw earlier. Example 6.3 Seek functional separable solutions to the nonlinear diﬀusion equation u x ut D : u2 x

(6.40)

We now wish that (6.40) and u tx D F .u/u t ux .noting that F D

f 00 =f 0 /

(6.41)

are compatible. From our original PDE (6.40), we have uxx D u2 u t C

2u2x : u

(6.42)

From cross diﬀerentiation .uxx / t D .u tx / t and (6.42) and (6.41), we obtain u2 u t t C 2uu2t C

4ux u tx u

2u2x u t D F 0 u t u2x C F ux u tx C F u t uxx : u2

(6.43)

Using (6.42) and (6.41) to eliminate derivatives u tx and uxx , we obtain ut t D

F 0 u t u2x u2

2u2t 2u t u2x F 2 u t u2x C C C F u2t u u4 u2

2F u t u2x : u3

We now further require that (6.41) and (6.44) be compatible. This leads to 00 3 F 2FF 0 2F 2 2F 3F 0 2 2 C 2 u t ux C C C 3 u t u3x D 0; 2 F CF u u u2 u2 u u4

(6.44)

(6.45)

from which we obtain two equations for F 3 3F C 2 D 0; u u F 00 2FF 0 2F 2 2F C C 3 D 0: u2 u2 u u4 F0 C F2

(6.46a) (6.46b)

132

6. FUNCTIONAL SEPARABILITY

Requiring that (6.46) be compatible gives F D

1 3 ; F D : u u

Each case will be considered separately. Case 1: F D u 1 . Since F D f 00 =f 0 , this gives f 00 D f0

1 ; u

and integrating twice gives f .u/ D c1 C c2 ln u:

Thus, our equation admits a separable solution of the form ln u D T .t / C X.x/:

Note that we set c1 D 1 and c2 D 0 without loss of generality. With a resetting of variables, this is just u D T .t /X.x/. Case 2: F D u 3 : Since F D f 00 =f 0 , this gives f 00 D f0

3 ; u

and integrating twice gives f .u/ D c1 C

c2 : u2

Thus, our equation admits a separable solution of the form 1 uD p : T .t / C X.x/

(6.47)

Note that we set c1 D 1 and c2 D 0 without loss of generality. Substituting (6.47) into our original PDE shows that 2XX 00 X 02 C 2TX 00 2T 0 D 0; which eventually leads to the exact solution uD p

1 c2 x 2 C c1 x C c0 C c3 e 2c2 t

:

6. FUNCTIONAL SEPARABILITY

133

GENERAL TRAVELING WAVE SOLUTIONS We now wish to consider a slightly diﬀerent form of solutions, namely f .u/ D A.t/x C B.t /;

(6.48)

which are known as general traveling wave solutions. As we did previously, we diﬀerentiate (6.48), but this time with respect to x twice, giving f 0 uxx C f 00 u2x D 0;

(6.49)

uxx D F .u/u2x ;

(6.50)

or where F D f 00 =f 0 . If our given equation is compatible with (6.50), then it will admit solutions in the form (6.48). The following examples illustrate this. Example 6.4 Consider the diﬀusion equation with a nonlinear source u t D uxx C Q.u/:

(6.51)

u t D F .u/u2x C Q.u/:

(6.52)

Using (6.50), (6.51) becomes Imposing compatibility between (6.50) and (6.52) gives Q00

FQ0

QF 00 C F 00 C 4FF 0 C 2F 3 u2x D 0;

(6.53)

from which we obtain the determining equations Q00 FQ0 QF ” D 0; F 00 C 4FF 0 C 2F 3 D 0:

(6.54a) (6.54b)

Since F D f 00 =f 0 , then (6.54a) integrates readily, giving QD

c1 f C c2 ; f0

(6.55)

where c1 and c2 are arbitrary constants. The remaining equation in (6.54) becomes f .4/ f0

7f 00 f 000 8f 003 C D 0: f 02 f 03

(6.56)

Fortunately, (6.56) admits several integrating factors 1 ; f 02

f ; f 02

f2 ; f 02

ln f 0 ; f 02

(6.57)

134

6. FUNCTIONAL SEPARABILITY

and can be used to integrate (6.56). The ﬁrst integrating factor leads to f 000 f 03

the integrating factor f 0 leads to

f 002 D c3 ; f 04

(6.58)

f 00 D c3 f C c4 ; f 02

(6.59)

2

and the integrating factor f 0 leads to ln jf 0 j D

1 c3 f 2 C c4 f C c5 ; 2

or f 0 D e c3 f

2 Cc

4f

Cc5

;

(6.60) (6.61)

where c3 c5 are further arbitrary constant. Note that we have absorbed the factor of 1=2 into c3 . For example, if we choose c3 D 0; c4 D 1; c5 D 0 then (6.61) integrates to give f D ln u;

(6.62)

where we have suppressed the constant of integration. In this case, Q in (6.55) becomes Q D .c1 ln u C c2 /u;

(6.63)

u t D uxx C .c1 ln u C c2 /u

(6.64)

u D e A.t /xCB.t / :

(6.65)

and PDEs of the form admits solutions of the form The reader can verify that substituting (6.65) into (6.64) shows that A and B satisfy the ODE A0 D c1 A; B 0 D

2A2 C c1 B C c2 :

(6.66)

Example 6.5 Consider the nonlinear diﬀusion equation u t D .D.u/ux /x :

(6.67)

Expanding (6.67) and using (6.50) gives u t D .D 0 C DF /u2x :

(6.68)

If we set G D D 0 C DF , then (6.68) becomes u t D Gu2x :

(6.69)

6. FUNCTIONAL SEPARABILITY

135

Requiring that (6.50) and (6.69) be compatible gives G 00 C 3F G 0 C .F 0 C 2F 2 /G D 0;

and since F D f 00 =f 0 , then (6.70) becomes 000 f f 00 G 00 C 3 0 G 0 f f0

3

f 002 f 02

G D 0:

(6.70)

(6.71)

Equation (6.71) can be explicitly solved, giving G D .c1 f C c2 / f 0 ;

(6.72)

where c1 and c2 are constants of integration. Further, since G D D 0 C FD , then D0

f 00 D D .c1 f C c2 / f 0 : f0

(6.73)

Since f .u/ D A.t /x C B.t/ we have the freedom to set the constants c1 and c2 without loss of generality. If c1 D 0, then we can set c2 D 1 without loss of generality. If c1 ¤ 0, we can set c1 D ˙1 and c2 D 0. In the ﬁrst case, where c1 D 0, we can solve (6.73) giving Z D.u/ du D A.t/x C B.t /; (6.74) f .u/ D uCd (d is an arbitrary constant) and substitution of (6.74) into (6.67) gives A0 D 0; B 0 D A2 :

(6.75)

These easily integrate, giving Z

D.u/ du D c1 x C c12 t C c2 ; uCd

(6.76)

is an exact solution to the NPDE u t D .D.u/ux /x :

(6.77)

Of course, (6.76) is nothing more than the usual traveling solution. In the second case we consider c1 D 1, (6.73) becomes D0

f 00 DD f0

ff 0 :

(6.78)

Substitution of f .u/ D A.t/x C B.t / into (6.67), with D and F satisfying (6.78), leads to A0 D

A3 ; B 0 D

A2 B:

(6.79)

136

6. FUNCTIONAL SEPARABILITY

These easily integrate AD p

1 2.t C t0 /

;

BDp

x0 2.t C t0 /

;

(6.80)

and we set t0 D x0 D 0 without loss of generality. Thus, we obtain solutions of the form x f .u/ D p : 2t

(6.81)

Now (6.78) would need to be integrated for a given D . However, this is a nonlinear ODE for F . Instead, for a given F , we will deduce a form of D , where exact solutions (6.67) can be obtained. Integrating (6.78) gives Z D.u/ D d f .u/ du f 0 .u/; (6.82) where d is a constant of integration. Table 6.1 contains several examples where exact solutions of the nonlinear diﬀusion equation (6.67) are given for a variety of diﬀusivities. Table 6.1: Exact solutions of the nonlinear diﬀusion equation (6.67) Solution

f(u) u

x u= √2t

u2

u=±

1 u

u=

eu

u = 1n

tan u

D(u) d–

x √2t

2u d –

1 3 u 3

1n |u| – d u2

√2t x x √2t

u = tan–1

1 2 u 2

(d – eu)eu x √2t

(d + 1n | cos u|) sec2 u

Example 6.6 Consider the nonlinear dispersion equation u t C k.u/ux C uxxx D 0;

(6.83)

where k.u/ is an arbitrary function of u. Our goal is to determine forms of k such that (6.83) admits separable solutions. Expanding (6.83) and using (6.50) gives ut D

k.u/ux

F 0 C 2F 2 u3x :

(6.84)

6. FUNCTIONAL SEPARABILITY

137

Requiring that (6.84) and (6.50) be compatible gives rise to the following set of determining equations: F

000

00

C 9FF C 6F

02

k 00 C F k 0 D 0; C 30F 2 F 0 D 0:

(6.85a) (6.85b)

Since F D f 00 =f 0 , then (6.85a) integrates to give (6.86)

k D c1 f C c2 ;

where c1 and c2 are arbitrary constants. Furthermore, (6.85b) becomes f .5/ 13f 00 f .4/ 9f 0002 C C f0 f 02 f 02

81f 002 f 000 72f 004 C D 0; f 03 f 04

(6.87)

which possesses the integrating factor f 0 3 , which allows us to integrate, giving f .4/ f 04

9f 000 f 00 12f 003 C C c3 D 0: f 05 f 06

(6.88)

Equation (6.88) possesses the integrating factor f 0 leading to 3f 002 C c3 f C c4 D 0; f 04

f 000 f 03

(6.89)

where c3 and c4 are additional arbitrary constants. If we perform a Hodograph transformation on (6.89) (interchanging the roles of f and u) we get the linear ODE ufff

(6.90)

.c3 f C c4 / uf D 0:

If if c3 ¤ 0 this can be solved in terms of integrated Airy functions. If c3 D 0, then (6.89) can be integrated twice, giving 1 f0 D p : (6.91) 2 c4 u C c5 u C c6 It can further be integrated, depending on the sign of c4 . Here we consider the cases where c4 D 1, c4 D 0; and c4 D 1. We will also set c1 D 1, c2 D 0. Case 1. c4 D 1 In this case, (6.91) integrates to give f D tan

1

p

u

1 c 2 5

u2 C c5 u C c6

!

C c7 :

(6.92)

We will set c2 D c5 D c7 D 0 and c1 D c6 D 1. From (6.83), (6.86), and (6.48), PDEs of the form u u t C tan 1 p ux C uxxx D 0 (6.93) 1 u2

138

6. FUNCTIONAL SEPARABILITY

admit separable solutions of the form u D sin .A.t/x C B.t// ;

(6.94)

where A and B satisfy A0 C A2 D 0;

B 0 C AB

A3 D 0:

(6.95)

1 ; .t C a0 /2

(6.96)

These are easily solved, giving AD

1 b0 ; BD t C a0 t C a0

a0 ; b0 are constant, and via (6.94) (with a0 D b0 D 0) gives xt 1 u D sin : t2

Case 2. c4 D 0 In this case (6.91) integrates to give p 2 c5 u C c6 f D C c7 : c5

(6.97)

(6.98)

We will set c1 D 1; c2 D 0; c5 D 4; c6 D 0; c7 D 0. From (6.83), (6.86), and (6.48), PDEs of the form p u t C uux C uxxx D 0 (6.99) admit separable solutions of the form u D .A.t/x C B.t //2 ;

(6.100)

A0 C A2 D 0; B 0 C AB D 0:

(6.101)

where A and B satisfy These are easily solved, giving AD

1 ; t C a0

BD

b0 ; t C a0

(6.102)

a0 ; b0 are constant, and via (6.100) (with a0 D b0 D 0) gives uD

x 2 t

:

(6.103)

6. FUNCTIONAL SEPARABILITY

Case 3. c4 D 1 In this case, (6.91) integrates to give ˇ ˇ p ˇ1 ˇ 2 ˇ f D ln ˇ c5 C u C u C c5 u C c6 ˇˇ C c7 : 2

139

(6.104)

We will set c2 D c5 D c7 D 0 and c1 D c6 D 1. From (6.83), (6.86), and (6.48), PDEs of the form p u t C ln ju C u2 C 1jux C uxxx D 0 (6.105) admit separable solutions of the form u D sinh .A.t/x C B.t // ;

(6.106)

A0 C A2 D 0; B 0 C AB C A3 D 0:

(6.107)

where A and B satisfy These are easily solved, giving AD

1 b0 1 ; ; BD C t C a0 t C a0 .t C a0 /2

a0 ; b0 are constant, and via (6.106) (with a0 D b0 D 0) gives xt C 1 u D sinh : t2

If we were to chose c6 D 1, we would obtain the exact solution xt C 1 u D cosh t2 for the PDE u t C ln ju C

p

u2

1jux C uxxx D 0:

(6.108)

(6.109)

(6.110)

(6.111)

HIGHER-DIMENSIONAL FUNCTIONAL SEPARABLE SOLUTIONS In this section we extend our results to PDEs in three dimensions, and, in particular, we consider the nonlinear elliptic equation uxx C uyy C uzz D Q.u/:

(6.112)

We will seek separable solutions of the form f .u/ D A.x/ C B.y/ C C.z/:

(6.113)

140

6. FUNCTIONAL SEPARABILITY

The reader can verify that (6.113) arises on solving uxy D F .u/ux uy ; uxz D F .u/ux uz ; uyz D F .u/uy uz ;

(6.114)

where F D f 00 =f 0 . We wish to make (6.112) and (6.114) compatible. By diﬀerentiating (6.112) and (6.114) with respect to x; y; and z , we calculate all third-order derivatives. Requiring that these be compatible with each other gives rise to 2 F 0 C F 2 uxx F u2x F 00 C 2FF 0 u2x C uy2 C u2z C Q00 FQ0 3F 0 C 2F 2 Q D 0; (6.115a) 2 F 0 C F 2 uyy F uy2 F 00 C 2FF 0 u2x C uy2 C u2z C Q00 FQ0 3F 0 C 2F 2 Q D 0; (6.115b) 2 F 0 C F 2 uzz F u2z F 00 C 2FF 0 u2x C uy2 C u2z C Q00 FQ0 3F 0 C 2F 2 Q D 0; (6.115c) and the ﬁrst-order condition 3F 00 C 8FF 0 C 2F 3 u2x C uy2 C u2z

3Q00 C 3FQ0 C 7F 0 C 4F 2 Q D 0:

(6.116)

We can solve (6.115) for uxx ; uyy ; and uzz , provided that F 0 C F 2 ¤ 0. As F 0 C F 2 D 0 is a special case, we consider this ﬁrst. This gives F D u 1 , with the constant of integration omitted. Equation (6.115) reduce to u2 Q00 uQ0 C Q D 0: (6.117) This integrates to Q D q1 u C q2 u ln juj;

(6.118)

where q1 and q2 are arbitrary constants, with q2 ¤ 0 as we are interested in NLPDEs. However, as we can scale u in our original PDE (6.112) with this particular Q, we can set q1 D 0 without loss of generality. Since F D f 00 =f 0 , then from (6.113) we obtain u D e ACBCC :

Substitution of (6.119) into (6.112) with (6.118) gives A00 C A02 C B 00 C B 02 C C 00 C C 02 e ACBCC D q2 e ACBCC .A C B C C / ;

(6.119)

(6.120)

which we see separates into A00 C A02 D q2 A C a;

B 00 C B 02 D q2 B C b; C 00 C C 02 D q2 C C c;

(6.121)

where a; b and c are constants such that a C b C c D 0. We can further transfer the constants a; b; and c to the solution (6.119), allowing us to set a D b D c D 0 and the ODEs in (6.121)

6. FUNCTIONAL SEPARABILITY 00

141

02

become the single ODE w C w D q2 w; w D w./ which can be solved to quadature. If the ﬁrst constant of integration is omitted, we obtain the explicit solution wD

1 q2 C . 2 4

w 0 /2 ;

(6.122)

where w0 is a constant of integration. Thus, the forms of A; B , and C are AD

1 q2 C .x 2 4

x0 /2 ; B D

1 q2 C .y 2 4

y 0 /2 ; C D

1 q2 C .z 2 4

z 0 /2 :

(6.123)

As we have the ﬂexibility of translational and scaling symmetry, we arrive at our ﬁnal solution u D e q .x

2 Cy 2 Cz 2

/C3=2

(6.124)

as an exact solution to the PDE uxx C uyy C uzz D 4qu ln juj:

Note that we set q2 D 4q . We now assume that F 0 C F 2 ¤ 0. We now can solve for From (6.115) we obtain .F 00 C 2FF 0 / u2x C uy2 C u2z C Q00 FQ0 2 uxx D F ux C 2 .F0 C F 2 / 00 0 2 2 .F C 2FF / ux C uy C u2z C Q00 FQ0 uyy D F uy2 C 2 .F0 C F 2 / 2 00 0 2 .F C 2FF / ux C uy C u2z C Q00 FQ0 uzz D F u2z C 2 .F 0 C F 2 /

(6.125) all second-order derivatives. 3F 0 C 2F 2 Q 3F 0 C 2F 3F 0 C 2F

2

Q

2

Q

;

(6.126a)

;

(6.126b)

:

(6.126c)

to which we add those given in (6.114). Requiring (6.114) and (6.126) be compatible gives F 000 C 3FF 00 C 2F 02 C 2F 2 F 0 u2x C uy2 C u2z Q000 C 2FQ00 C 4F 0 C F 2 Q0 C 3F 00 C FF 0 2F 3 Q D 0: (6.127) Eliminating u2x C uy2 C u2z between (6.116) and (6.127) gives rise to our ﬁrst constraint on F and Q. 3F 00 C 8FF 0 C 2F 3 Q000 C 2FQ00 C .3F 0 C F 2 /Q0 C .3F 00 C FF 0 2F 3 /Q F 000 C 3FF 00 C 2F 02 C 2F 2 F 0 3Q00 C 3FQ0 C .7F 0 C 4F 2 /Q D 0: (6.128) If we diﬀerentiate (6.116) with respect to either x; y , or z and use (6.114), (6.116), and (6.126) we obtain a second constraint 3 3F 00 C 8FF 0 C 2F 3 Q000 9F 000 C 45FF 00 C 24F02 C 74F 2 F 0 Q00 C 9FF 000 30F 0 F 00 C 24F 2 F 00 56FF 02 2F 3 F 0 Q0 C 21F 0 C 12F 2 F 000 27F 002 C 26F 3 28FF 0 F 00 C 56F 03 C 42F 2 F 02 C 48F 4 F 0 C 8F 6 D 0: (6.129)

142

6. FUNCTIONAL SEPARABILITY

Eliminating Q000 between (6.128) and (6.129) gives two factors: one is F 0 C F 2 , which we assume is not zero; the second is 3F 0 C F 2 Q00 3F 00 C 11FF 0 C 3F 3 Q0 C 5FF 00 7F 02 C 7F 2 F 0 C 2F 4 Q D 0: (6.130) It is left as an exercise to the reader to show that if (6.130) is satisﬁed, then (6.128) and (6.129) are automatically satisﬁed. To solve (6.130) for a given Q would seem like a daunting task— solving a highly nonlinear second order ODE for F . However, the ODE is linear for Q and it is this road we will take. We will ﬁrst introduce the substitution F D 3S 0 =S . The motivation for this is in the ﬁrst term 3F 0 C F 2 . This gives S 2 S 00 Q00 S 2 S 000 C 8SS 0 S 00 Q0 C 5S S 0 S 000 7SS 002 C 20S 02 S 00 Q D 0: (6.131) One exact solution of (6.131) would then allow us to reduce the order of the ODE. Seeking a solution of the form Q D Q.S; S 0 /, we are led to the solution Q D S 5 S 0 ; letting Q D S 5 S 0 P where P D P .u/ gives SS 0 S 00 P 00 SS 0 S 000 2SS 002 2S 02 S 00 P 0 D 0; (6.132) from which we integrate, giving P D p1 C p2

Z

S 00 S 2 S 002

(6.133)

du;

where p1 and p2 are constant. Now we will relate this to f . Since F D S3 D

1 , which allows (6.133) to be integrated, giving f0 02 f P D p1 C p2 3 00 2f : f

Q D q1

f 00 C q2 f 03

2ff 00 f 03

3 f0

f 00 , then f0

(6.134)

We note that f 00 ¤ 0 as f 00 D 0 gives Q D 0. This, in turn, gives 02 p1 f 00 p2 f 00 f QD 3 2f ; 3 f 03 3 f 03 f 00 or

3S 0 D S

;

(6.135)

(6.136)

where q1 and q2 are constant. Now that we know the form of Q, we go after the forms of A; B; and C . Recall that f .u/ D A.x/ C B.y/ C C.z/, so ux D

A0 B0 ; u D ; y f0 f0

uz D

C0 ; f0

(6.137)

6. FUNCTIONAL SEPARABILITY

and uxx D

A00 f0

f 00 A02 B 00 ; u D yy f 03 f0

f 00 B 02 C 00 ; u D zz f 03 f0

f 00 C 02 : f 03

Substitution of these into (6.112) with Q given in (6.136) gives A02 C B 02 C C 02 f 00 A00 C B 00 C C 00 2ff 00 f 00 D q C q 1 2 f0 f 03 f 03 f 03 or

3 f0

143

(6.138)

;

(6.139)

A00 C B 00 C C 00 C 3q2 D A02 C B 02 C C 02 C q1 C 2q2 .A C B C C / g;

(6.140)

where g D f 00 =f 02 . If we diﬀerentiate (6.140) with respect to x , y; and z , we obtain g 0 A0 A000 D 2A0 A00 C 2q2 A0 g C A02 C B 02 C C 02 C q1 C 2q2 .A C B C C / ; f0 0 0 gB ; B 000 D 2B 0 B 00 C 2q2 B 0 g C A02 C B 02 C C 02 C q1 C 2q2 .A C B C C / f0 g0 C 0 C 000 D 2C 0 C 00 C 2q2 C 0 g C A02 C B 02 C C 02 C q1 C 2q2 .A C B C C / : f0

(6.141a) (6.141b) (6.141c)

Further cross diﬀerentiation of (6.141) and canceling the common term A0 B 0 =f 0 gives 0 g0 0 00 00 02 02 02 D 0; (6.142a) 2 A C B C 2q2 g C A C B C C C q1 C 2q2 .A C B C C / 0 f 0 0 g D 0; (6.142b) 2 A00 C C 00 C 2q2 g 0 C A02 C B 02 C C 02 C q1 C 2q2 .A C B C C / 0 f g0 0 2 B 00 C C 00 C 2q2 g 0 C A02 C B 02 C C 02 C q1 C 2q2 .A C B C C / D 0: (6.142c) f0 Upon subtraction, (6.142) gives rise to A00 B 00 g 0 D 0; A00

C 00 g 0 D 0;

B 00

C 00 g 0 D 0:

(6.143)

This leads to A00 D B 00 D C 00 D 2 (constant);

(6.144)

f .u/ D ax C by C cz C d;

(6.145)

as g 0 D 0 gives that F 0 C F 2 D 0 which we assume to be nonzero. If D 0 then A, B; and C are linear, so from (6.113) solutions are of the form

where a d are constant and it’s relatively straightforward to deduce that (6.112) will admit solutions of this type, provided that Q is of the form QD

a2 C b 2 C c 2

f 00 : f 03

(6.146)

144

6. FUNCTIONAL SEPARABILITY

If ¤ 0 then A, B , and C are at most quadratic, so from (6.144) and (6.113), solutions are of the form1 f .u/ D x 2 C y 2 C z 2 C k; (6.147) k is constant and it’s relatively straightforward to deduce that (6.112) will admit solutions of this type, provided that Q is of the form QD

6.1 6.1.

6.2.

4kf 00 f 03

2.2ff 00 3f 02 / f 03

(6.148)

EXERCISES Seek separable solutions for the following: ux .i / ut D p u x uy ux C p .i i / ut D p u x u y

u D .a.x/t C b.x//2 u D .a.x; y/t C b.x; y//2

Determine the form of Q such that uxx C uyy D Q.u/

(6.149)

admits functional separable solutions of the form ([78], [79], and [80]) f .u/ D A.t/ C B.x/:

6.3.

(6.150)

Determine the form of Q such that u t t D uxx C Q.u; u t ; ux /

(6.151)

admits solutions of the form [83] u D A.t/ C B.x/:

6.4.

(6.152)

Determine the form of Q such that u t D uxx C Q.u; ux/

(6.153)

f .u/ D A.t/ C B.x/:

(6.154)

admits solutions of the form

1 we

have omitted the linear terms in x , y and z as the original PDE admits translation in these variables.

6.2. REFERENCES

6.5.

145

Find conditions on D.u/ and Q.u/ (in terms of diﬀerential equations) such that the following admit functional separable solutions (see [81] and [82]): .i/ u t D .D.u/ux /x C Q.u/; .i i / u t t D .D.u/ux /x C Q.u/:

Can any of these be solved explicitly? 6.6.

Find separable solutions to

(6.155)

f .u/ D A.t/ C B.x/ C C.y/:

(6.156)

u t D .D.u/ux /x C D.u/uy

y

of the form 6.7.

The stream function formulation of the boundary layer equations is y

xy

Find separable solutions of the form

6.2

x

yy

D

yyy

C U.x/:

(6.157)

D A.y/x C B.y/ (Polyanin [84]).

REFERENCES

[78] A. M. Grundland and E. Infeld, A family of nonlinear Klein–Gordon equations J. Math. Phys., 33(7), pp. 2498–2503, 1992. DOI: 10.1063/1.529620. 144 [79] W. Miller and L. A. Rubel, Functional separation of variables for Laplace equations in two dimensions, J. Phys. A: Math. Gen., 26, pp. 1901–1913, 1993. DOI: 10.1088/03054470/26/8/017. 144 [80] R. Z. Zhdanov, Separation of variables in the nonlinear wave equation, J. Phys. A: Math. Gen., 27, pp. L291–297, 1994. DOI: 10.1088/0305-4470/27/9/009. 144 [81] C. Qu, S. Zhang, and R. Liu, Separation of variables and exact solutions to quasilinear diﬀusion equations with nonlinear source, Physica D, 144, pp. 97–123, 2000. DOI: 10.1016/s0167-2789(00)00069-5. 145 [82] P. G. Estevez and C. Z. Qu, Separation of variables in nonlinear wave equation with a variable wave speed, Theor. Math. Phys., 133(2), pp. 1490–1497, 2002. DOI: 10.1023/A:1021190509331. 145 [83] C. Z. Qu, W. He, and J. Dou, Separation of variables and exact solutions of generalized nonlinear Klein–Gordon equations, Prog. Theor. Phys., 105, pp. 379–398, 2001. 144 [84] A. D. Polyanin, Exact solutions and transformations of the equations of a stationary laminar boundary layer, Theoretical Foundations of Chemical Engineering, 35, pp. 319–328, 2001. DOI: 10.1023/A:1010462116343. 145

147

CHAPTER

7

Solutions CHAPTER 1 x2 2t ; u D 2. 6t x 2. .a 1/.a C 1/ D 0; 4k 2 3. .a 1/.a C 1/ D 0; ck 4. a D 4; b D 2 . 5. a D b D c:

1. u D

2ck C 1 D 0 k 2 C ck 1 D 0. k 2 D 0 cka C ak 2 2a C

CHAPTER 2 .2x C y/2 .2x y/2 and u D ; 4 4 t 2t (ii) u D .x p 1/e C e ; p (iii) u D x 2 C y 2 and u D x 2 C .2 y/2 ; 1 (iv) u D e xCy 2 : p 2 3 2. (i) F D F x; q; pq y; u 2xp C 3 p , 2 1 p (ii) F D F t; ; pu; 2t uq x : 2q 2 q 3. a; b; and c satisfy a0 D 6a2 ; b 0 D 6ab; c 0 D 2ac C b 2 : 4 (i) For all k , (ii) k D 2=3, (iii) k D 3=4. 5. TP D T ln T C kT; XR kX X ln X . 6. aP D 2ab a; bP D a2 C b 2 : 2F 3 F 2 D 0 4FF 0 7. n D 2; F 00 C C 2 D 0. D D D2

1. (i) u D

CHAPTER 3 1. A D 2. f D

0 ; f D g C 2 .ln /00 ; 00 C .g.x/ C / D 0: n.n C 1/ n.n C 1/ ; f D . 2 x cosh2 x

1 D 0.

148

7. SOLUTIONS

3. F D

1 ln v . a

CHAPTER 4 1. (i)

p D 3X Y C U Y; q D Y X U Y;. 2.XP C 2Q/ 2Xe Y (ii) p D ; qD ; P P e XP e XP (iii) p D ; qD ; Q Q2 X (iv) p D ; p D U XP: Q

x2p C c; q (ii) U D u p 2 C c; (iii) U D x C c; (iv) A D c1 p C c2 ; B D c1 q C c3 ; 1 1 U D c1 .u ln p ln q/ C c2 x C C c3 y C C c4 ; p q

2. (i) U D

CHAPTER 5 1. (i)

F DF

p xp u; ; Cy , q q

pC1 q/ ; F D F2 x C y; , q .x q; y C p/; F D F2 .x C q; y p/ ; (iii) F D F1 1 2 (iv) F D F1 y C q p ; F D F2 .x C yp; y C q/ : 2 2 2. P 0 ./ D 4 .a C b/ ˙p C abq C b 2 pq ap.a C bp/x C .abpq C a2 q p 2 /y C bp.a C bp/u F DF ; . bp.a C bp/ abpq C a2 q p 2 1 Y U YQ 1 Y2 3. x D ; y D Q; u D C : X X X X 2 X2 YQ U 1 : 4. x D ; y D Q; u D X X

(ii) F D F1 .x C u; p

CHAPTER 6 1. (i) a00 D a2 ; b 00 D ab; (ii) axx C ayy D a2 ; bxx C byy D ab:

7. SOLUTIONS 00

0

2. (a) F C 2FF D 0; Q

00

FQ

0

0

2

.3F C 2F /Q D 0; f 00 (b) f .u/ D ax C by C c; Q D .a2 C b 2 / 03 for all f .u/; f 0 ¤ 0; f 00 f ff 00 f 02 2 2 for all f .u/; f 0 ¤ 0: (c) f .u/ D .x C y / C k; Q D 4k 03 4 f f 03 3. (a) Q D A.u t / C B.ux /; A; B arbitrary; 00 2 2 (b) Q D F .u/ u t ux C G.u/ where F C 2FF 0 D 0; G 00 C 2F 0 G D 0: f 00 p 2 C c1 f C C.f 0 p/ 4. Q.u; p/ D where c1 and C is an arbitrary constant and function. f0

149

151

Author’s Biography DANIEL J. ARRIGO Daniel J. Arrigo earned his Ph.D. from the Georgia Institute of Technology in 1991. He has been on staﬀ in the Department of Mathematics at the University of Central Arkansas since 1999 and is currently Professor of Mathematics. He has published over 30 journal articles and 2 books. His research interests include the construction of exact solutions of PDEs; symmetry analysis of nonlinear PDEs; and solutions to physically important equations, such as nonlinear heat equations and governing equations modeling of granular materials and nonlinear elasticity. In 2008, Dr. Arrigo received the Oklahoma-Arkansas Section of the Mathematical Association of America’s Award for Distinguished Teaching of College or University Mathematics.