Multiplicative Partial Differential Equations 9781032575032, 9781032576046, 9781003440116

Citation preview

Multiplicative Partial Differential Equations This book presents an introduction to the theory of multiplicative partial differential equations (MPDEs). It is suitable for all types of basic courses on MPDEs. The authors’ aim is to present a clear and well-organized treatment of the concept behind the development of mathematics and solution techniques. The text material of this book is presented in a highly readable, mathematically solid format. Many practical problems are illustrated displaying a wide variety of solution techniques. Key features: •  The book includes new classification and canonical forms of second-order MPDEs. • Authors propose a new technique to solving the multiplicative wave equation such as the method of separation of variables and the energy method. • The proposed technique in the book can be used to give the basic properties of multiplicative elliptic problems, fundamental solutions, multiplicative integral representation of multiplicative harmonic functions, mean-value formulas, strong principle of maximum, multiplicative Poisson equation, multiplicative Green functions, method of separation of variables and theorems of Liouville and Harnack. Svetlin G. Georgiev was born in 1974 in Rouse, Bulgaria. He is a mathematician who has worked in various areas of mathematics. He currently focuses on harmonic analysis, functional analysis, partial differential equations, ordinary differential equations, Clifford and quaternion analysis, integral equations and dynamic calculus on timescales. Khaled Zennir was born in Skikda, Algeria in 1982. He received his PhD in mathematics in 2013 from Sidi Bel Abbès University, Algeria (assist. professor). He obtained his highest diploma in Algeria (habilitation, mathematics) from Constantine University, Algeria in 2015. He is now an associate professor at Qassim University, KSA. His research interests lie in nonlinear hyperbolic partial differential equations—global existence, blow-up and longtime behavior.

Advances in Applied Mathematics Series Editor: Daniel Zwillinger Introduction to Quantum Control and Dynamics Domenico D’Alessandro Handbook of Radar Signal Analysis Bassem R. Mahafza, Scott C. Winton, Atef Z. Elsherbeni Separation of Variables and Exact Solutions to Nonlinear PDEs Andrei D. Polyanin, Alexei I. Zhurov Boundary Value Problems on Time Scales, Volume I Svetlin Georgiev, Khaled Zennir Boundary Value Problems on Time Scales, Volume II Svetlin Georgiev, Khaled Zennir Observability and Mathematics Fluid Mechanics, Solutions of Navier-Stokes Equations, and Modeling Boris Khots Handbook of Differential Equations, Fourth Edition Daniel Zwillinger, Vladimir Dobrushkin Experimental Statistics and Data Analysis for Mechanical and Aerospace Engineers James Middleton Advanced Engineering Mathematics with MATLAB®, Fifth Edition Dean G. Duffy Handbook of Fractional Calculus for Engineering and Science Harendra Singh, H. M. Srivastava, Juan J. Nieto Advanced Engineering Mathematics A Second Course with MATLAB• Dean G. Duffy Quantum Computation Helmut Bez and Tony Croft Computational Mathematics An Introduction to Numerical Analysis and Scientific Computing with Python Dimitrios Mitsotakis Delay Ordinary and Partial Differential Equations Andrei D. Polyanin, Vsevolod G. Sorkin, Alexi I. Zhurov Clean Numerical Simulation Shijun Liao Multiplicative Partial Differential Equations Svetlin G. Georgiev and Khaled Zennir https://www.routledge.com/Advances-in-Applied-Mathematics/book-series/CRCADVAPP MTH?pd=published,forthcoming&pg=1&pp=12&so=pub&view=list

Multiplicative Partial Differential Equations

Svetlin G. Georgiev Khaled Zennir

First edition published 2024 by CRC Press 2385 NW Executive Center Drive, Suite 320, Boca Raton FL 33431 and by CRC Press 4 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN CRC Press is an imprint of Taylor & Francis Group, LLC © 2024 Svetlin G. Georgiev and Khaled Zennir Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. ISBN: 978-1-032-57503-2 (hbk) ISBN: 978-1-032-57604-6 (pbk) ISBN: 978-1-003-44011-6 (ebk) DOI: 10.1201/9781003440116 Typeset in CMR10 font by KnowledgeWorks Global Ltd.

Contents

Preface

vii

1 General Introduction

1

2 Classification of Second-Order Multiplicative Partial Differential Equations

3

3 Classification and Canonical Forms 3.1 Multiplicative Linear MPDEs . . . . . . . . . . . . . . . . . 3.1.1 The multiplicative hyperbolic case . . . . . . . . . . . 3.1.2 The multiplicative elliptic case . . . . . . . . . . . . . 3.1.3 The parabolic case . . . . . . . . . . . . . . . . . . . . 3.2 Classification and Canonical Form of Multiplicative Quasilinear MPDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 The multiplicative hyperbolic case . . . . . . . . . . . 3.2.2 The multiplicative elliptic case . . . . . . . . . . . . . 3.2.3 The multiplicative parabolic case . . . . . . . . . . . . 3.3 n Multiplicative Independent Variables . . . . . . . . . . . . 3.4 Classification of First-Order Systems with Two Multiplicative Independent Variables . . . . . . . . . . . . . 3.5 Advanced Practical Exercises . . . . . . . . . . . . . . . . . . 4 The Multiplicative Wave Equation 4.1 The One-Dimensional Multiplicative Wave Equation . . . . 4.1.1 The Cauchy problem and D’Alambert formula . . . 4.1.2 The Cauchy problem for the nonhomogeneous wave equation . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.3 Separation of multiplicative variables . . . . . . . . . 4.1.4 The energy method: uniqueness . . . . . . . . . . . . 4.2 The Multiplicative Wave Equation in R3∗ . . . . . . . . . . 4.2.1 Radially symmetric solutions . . . . . . . . . . . . . 4.2.2 The Cauchy problem . . . . . . . . . . . . . . . . . . 4.3 The Two-Dimensional Wave Equation . . . . . . . . . . . . 4.4 The (2n + 1)-Dimensional Wave Equation . . . . . . . . . . 4.5 The (2n) − 1-Dimensional Wave Equation . . . . . . . . . . 4.6 The Cauchy Problem for a Nonlinear Hyperbolic Equation

8 8 22 49 61 73 79 82 83 84 94 99

. .

104 104 104

. . . . . . . . . .

109 119 129 131 131 132 141 144 154 155

v

vi

Contents 4.7 4.8

5 The 5.1 5.2 5.3

5.4 5.5 5.6 5.7 5.8

The Riemann Function . . . . . . . . . . . . . . . . . . . . . Advanced Practical Exercises . . . . . . . . . . . . . . . . . .

161 165

Heat Equation The Weak Maximum Principle . . . . . . . . . . The Strong Maximum Principle . . . . . . . . . The Cauchy Problem . . . . . . . . . . . . . . . 5.3.1 The fundamental solution . . . . . . . . . 5.3.2 The Cauchy problem . . . . . . . . . . . . The Mean Value Formula . . . . . . . . . . . . . The Maximum Principle for the Cauchy Problem The Method of Separation of Variables . . . . . The Energy Method: Uniqueness . . . . . . . . . Advanced Practical Exercises . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

168 168 171 175 175 177 187 192 195 201 202

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

. . . . . . . . . . . . .

205 205 208 209 212 214 215 218 223 227 230 230 235 239

Cauchy–Kovalevskaya Theorem Analytic Functions of One Variable . . . . . . . . Autonomous Multiplicative Differential Equations Systems of Ordinary Differential Equations . . . . Partial Differential Equations . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

242 242 248 250 253

6 The 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10

Laplace Equation Basic Properties of Elliptic Problems . . . . . The Fundamental Solution . . . . . . . . . . . Integral Representation of Harmonic Functions Mean Value Formulas . . . . . . . . . . . . . . Strong Maximum Principle: Uniqueness . . . . The Poisson Equation . . . . . . . . . . . . . . The Green Function of the Dirichlet Problem . The Poisson Formula for a Multiplicative Ball Theorems of Liouville and Harnack . . . . . . Separation of Variables . . . . . . . . . . . . . 6.10.1 Multiplicative rectangles . . . . . . . . . 6.10.2 Multiplicative circular domains . . . . . 6.11 Advanced Practical Exercises . . . . . . . . . .

7 The 7.1 7.2 7.3 7.4

. . . . . . . . . . . . .

. . . . . .

References

257

Index

259

Preface

This book presents an introduction to the theory of multiplicative partial differential equations (MPDEs). The book is suitable for all types of basic courses on MPDEs. In Chapters 2 and 3, the classification and canonical forms of second-order MPDEs are considered. Chapter 4 is concerned with the multiplicative wave equation. They are investigated by even and odd dimensional multiplicative wave equations, method of separation of variables and energy method. It is introduced as the Riemann functions. Chapter 5 deals with the multiplicative heat equation. They are considered the weak and strong maximum principles, the Cauchy problem, the mean value formula, the method of separation of variables and the energy method. The multiplicative Laplace equation is introduced in Chapter 6. They are given the basic properties of multiplicative elliptic problems, the fundamental solutions, multiplicative integral representation of multiplicative harmonic functions, mean value formulas, strong principle of maximum, the multiplicative Poisson equation, multiplicative Green functions, method of separation of variables and theorems of Liouville and Harnack. Chapter 7 is focused on a multiplicative analogues of the classical Cauchy–Kovalevskaya Theorem. It is considered in the case of both MDEs and MPDEs. The aim of this book is to present a clear and well-organized treatment of the concept behind the development of mathematics and solution techniques. The text of this book is presented in a highly readable, mathematically solid format. Many practical problems are illustrated displaying a wide variety of solution techniques. Svetlin G. Georgiev and Khaled Zennir Paris, France

vii

1 General Introduction

A multiplicative partial differential equation (MPDE) describes a relation between an unknown function and its multiplicative partial derivatives. The general form of a MPDE for a function u(x1 , . . . , xn ) is F (x1 , . . . , xn , u, u∗x1 , . . . , u∗xn , . . .) = 0∗ , or

 F

x1

x1 , . . . , x n , e

ux 1 u

,...,e

uxn

uxn u

 ...,

= 1,

where xi , 1 ≤ i ≤ n, are multiplicative independent multiplicative variables, u is the unknown variable, u∗xi denote the multiplicative partial derivative of u with respect to xi , uxi denote the classical partial derivative of u with respect to xi , i ∈ {1, . . . , n}. The equation is, in general, supplemented by additional conditions such as initial conditions or boundary conditions. The analysis of MPDEs has many facets. The classical approach is to develop methods for finding explicit solutions. The aim is to discover some of the solution properties before computing it, and sometimes even without a complete solution. There exist many equations that cannot be solved. All we can do in these cases is to obtain qualitative information on the solution. Furthermore, it is desired in many cases that the solution will be unique, and that it will be stable under small perturbations of the data. A theoretical understanding of the equation enables us to check whether these conditions are satisfied. As we will see in what follows, there are many ways to solve MPDEs, each way applicable to a certain class of MPDEs. Therefore, it is important to have an analysis of the equation before or during solving it. The fundamental theoretical question is whether the problem consisting of the equation and its associated side conditions is well-posed. Definition 1.1 A problem is called well-posed if it satisfies the following criteria. 1. Existence: The problem has a solution. 2. Uniqueness: There is no more than one solution. 3. Stability: A small change in the equation or in the side conditions gives rise to a small change in the solution.

DOI: 10.1201/9781003440116-1

1

2

General Introduction

If one or more conditions above does not hold, we say that the problem is ill-posed. MPDEs are often classified into different types. In fact, there exist several such classifications. The first classification is according to the order of the equation. Definition 1.2 The order is defined to be the order of the highest multiplicative derivative in the equation. If the highest multiplicative derivative is of order m, then the equation is said to be of order m. Other important classifications will be given in the next chapters. Definition 1.3 A function in the set C∗m that satisfies a MPDE of order m will be called a solution. Definition 1.4 If each of the functions u1 , . . . , ul satisfies a MPDE and every multiplicative linear combination of them satisfies that equation too, this property is called multiplicative superposition principle. It allows the construction of complex solutions through multiplicative combinations of simple solutions. In addition, we will use the superposition principle to obtain uniqueness of solutions of some PDEs.

2 Classification of Second-Order Multiplicative Partial Differential Equations

Definition 2.1 By a second-order multiplicative partial differential equation in n multiplicative variables x1 , . . . , xn , we mean any equation of the form ∗∗ ∗∗ F (x1 , . . . , xn , u, u∗x1 , . . . , u∗xn , u∗∗ x1 x1 , ux1 x2 , . . . , uxn xn ) = 0∗

(2.1)

or  ux uxn 1 F x1 , . . . , xn , ex1 u , . . . , exn u ,  x1

e

ux1 u

+x1

 xn

...,e

ux1 x1 u−u2 x1 u2

uxn u

+xn

 x1 x2

u

,e

uxn xn u−u2 xn u2

x1 x2 u−ux1 ux2 u2



,



= 1.

Example 2.1 The equation n X

−∗

u∗∗ xi xi = f (u)

∗i=1

will be called a multiplicative nonlinear Poisson equation. It is a second-order MPDE. In fact, we can rewrite it in the following manner: f (u)

= −∗

n X

ux i u

xi

e

+xi

ux x u−u2 xi i i u2

!

∗i=1 n P

xi

= −∗ ei=1 −

= e

DOI: 10.1201/9781003440116-2

n P i=1

xi

ux i u

ux i u

+xi

+xi

ux x u−u2 xi i i u2

ux x u−u2 xi i i u2

!

!

.

3

4 Classification of Second-Order Multiplicative Partial Differential Equations Example 2.2 The equation u∗xn −∗

n−1 X

∗∗

(uγ∗ )xi xi = 0∗

∗i=1

will be called multiplicative porous medium equation, where γ > 0 is a constant. It is a second-order MPDE. It can be rewritten in the form 0∗

= 1 γ



= exn

γ

γ

γ

2



(e(log u) )xi (e(log u) )xi xi (e(log u) )−(e(log u) )xi   n−1 x +x i i γ X (e(log u) ) (e(log u)2γ ) −∗ e

uxn u

∗i=1 n−1 P

= exn

xn

uxn u

uxn u

−∗ e

−

= e

n−1 P i=1

i=1

 xi 

 xi 

γ

γ

γ

γ

2



(e(log u) )xi (e(log u) )xi xi (e(log u) )−(e(log u) )xi  +x i γ (e(log u) ) (e(log u)2γ ) γ

γ

γ

γ

2



(e(log u) )xi (e(log u) )xi xi (e(log u) )−(e(log u) )xi  +xi γ (log u) (e ) (e(log u)2γ )

,

or  
γ
γ
γ
2 (log u)γ n−1 e(log u) xi xi e(log u) − e(log u) xi uxn X  e xi  = 0.
+ xi
xn − xi (log u)γ (log u)2γ u e e i=1 Definition 2.2 If equation (2.1) can be written in the form n X

aij (x1 , . . . , xn , u, u∗x1 , . . . , u∗xn ) ·∗ u∗∗ x i x j +∗

∗i,j=1

n X

bi (x1 , . . . , xn , u) ·∗ u∗xi

∗i=1

= f (x1 , . . . , xn , u, u∗x1 , . . . , u∗xn ), then we say that the equation is multiplicative quasilinear. Example 2.3 The equation ∗∗ 2∗ u∗x2 ·∗ u∗∗ = x21∗ u2∗ x1 x1 −∗ ux1 x2 ·∗ u

is a multiplicative quasilinear second-order MPDE. Definition 2.3 If equation (2.1) can be written in the form n X ∗i,j=1

aij (x1 , . . . , xn ) ·∗ u∗∗ xi xj +

n X

bi (x1 , . . . , xn ) ·∗ u∗xi

∗i=1

= f (x1 , . . . , xn , u, u∗x1 , . . . , u∗xn ), then we say that the equation is multiplicative semilinear.

Classification of Second-Order Multiplicative Partial Differential Equations 5 Example 2.4 The equation 2∗ ∗∗ ∗∗ 4∗ u∗∗ x1 x1 −∗ ux1 x2 + ux3 x3 = x1 ·∗ u

is a multiplicative semiilinear second-order MPDE. It can be written in the form x21∗ ·∗ u4∗ 2

= e(log x1 ) ·∗ e(log u) 2

= e(log x1 ) 

= e

x1

ux 1 u



+∗ e

x3



= e

x1

ux 1 u

4

(log u)2

+x1

ux3 u

ux x u−u2 x1 1 1 u2

+x3

+x1



−∗ e

ux3 x3 u−u2 x3 u2

ux x u−u2 x1 1 1 u2

x1 x2

u

x1 x2 u−ux1 ux2 u2





 −x1 x2

u

x1 x2 u−ux1 ux2 u2



 +x3

ux 3 u

+x3

ux x u−u2 x3 3 3 u2



,

or (log x1 )2 (log u)4    ux1 x1 u − u2x1 ux1 ux1 x2 u − ux1 ux2 = x1 + x1 − x1 x2 u u2 u2  2  ux x u − u ux3 +x3 + x3 3 3 2 x3 . u u 

Definition 2.4 If equation (2.1) can be written in the form n X

aij (x1 , . . . , xn ) ·∗ u∗∗ xi xj +∗

∗i,j=1

n X

bi (x1 , . . . , xn ) ·∗ u∗xi = f (x1 , . . . , xn ),

∗i=1

then we say that the equation is multiplicative linear. Moreover, if f (x1 , . . . , xn ) = 0, then the equation is said to be multiplicative linear homogeneous second-order MPDE. Otherwise, the equation is said to be multiplicative linear nonhomogeneous second-order MPDE. Example 2.5 The equation n X ∗i=1

u∗∗ xi xi = 0∗

6 Classification of Second-Order Multiplicative Partial Differential Equations is called multiplicative Laplace equation. It is a multiplicative linear homogeneous second-order MPDE. It can be rewritten in the form 0∗

=

1 n X

=

∗i=1 n P

e

ux i u

xi

= ei=1 or

n X

 xi

i=1

ux i u

xi

+xi

+xi

ux x u−u2 xi i i u2

ux x u−u2 xi i i u2

ux x u − u2 uxi + xi i i 2 xi u u

!

!

,

 = 0.

Example 2.6 The equation u∗∗ x1 x1 −∗

n X

u∗∗ xi xi = 0∗

∗i=2

will be called multiplicative wave equation. It is a multiplicative linear homogeneous second-order MPDE. It can be rewritten in the form 0∗

=

1 

= e

x1

ux 1 u

ux x u−u2 x1 +x1 1 1u2



ux 1 u

ux x u−u2 x1 1 1 u2



n X

−∗

xi

e

ux i u

+xi

ux x u−u2 xi i i u2

∗i=2  x1

= e

+x1

−

n P

xi

i=2

ux i u

+xi

ux x u−u2 xi i i u2

!

!

,

or  x1

ux x u − u2 ux1 + x1 1 1 2 x1 u u

 −

n X

 xi

i=2

ux x u − u2 uxi + xi i i 2 xi u u

 = 0.

Example 2.7 The equation n X

u∗x1 −

u∗∗ xi xi = 0∗

∗i=2

will be called multiplicative heat (multiplicative diffusion) equation. It is a multiplicative linear homogeneous second-order MPDE. It can be rewritten in the form 0∗

=

1

= e

ux x1 u1

= ex1

ux 1 u

−∗ −

n X

∗i=2 n X

xi

e

xi

i=2



ux i u

+xi

ux x u−u2 xi i i u2

!

ux x u − u2 uxi + xi i i 2 xi u u

 ,

Classification of Second-Order Multiplicative Partial Differential Equations 7 or

n

ux1 X + xi u i=2

x1



ux x u − u2 uxi + xi i i 2 xi u u

 = 0.

Example 2.8 The equation u∗x1 −∗

n X

aij ·∗ u∗∗ xi xj +∗

∗i,j=2

n X

bi ·∗ u∗xi = x21∗ +∗ · · · +∗ x2n∗ ,

∗i=2

will be called multiplicative Kolmogorov equation, where aij , bi , 1 ≤ i, j ≤ n, are positive constants. It is a multiplicative linear nonhomogeneous secondorder MPDE. It can be rewritten in the form 2

2

x21∗ +∗ · · · +∗ x2n∗ = e(log x1 ) +∗ · · ·∗ +∗ e(log xn ) = e(log x1 )

= e

ux x1 u1

2

+···+(log xn )2



n X

−∗

e

xi xj

ux x u−ux ux i j i j u2



∗i,j=2,i6=j

−∗

n X

xi

e

ux i u

+xi

ux x u−u2 xi i i u2

!

∗i=2 x1

= e

ux1 u

−

n P

 xi xj

i,j=2,i6=j

ux x u−ux ux i j i j u2

 −

n P i=2

xi

ux i u

+xi

ux x u−u2 xi i i u2

!

,

or (log x1 )2 + · · · + (log xn )2

= x1

ux1 − u

n X i,j=2,i6=j

  X   n uxi xj u − uxi uxj uxi xi u − u2xi uxi xi xj − x + x . i i u2 u u2 i=2

Definition 2.5 A second-order MPDE that is not multiplicative linear is said to be multiplicative nonlinear. Example 2.9 The equation 2∗ ∗∗ 3∗ ∗ u2∗ ·∗ u∗3 ·∗ u∗∗ −∗ u∗x1 x 1 +∗ u x1 x2 +∗ u ·∗ ux2 x2 = u

is a multiplicative nonlinear second-order MPDE.

3 Classification and Canonical Forms for Multiplicative Linear and Multiplicative Quasilinear Second-Order Multiplicative Partial Differential Equations

3.1

Classification and Canonical Forms for Multiplicative Linear Second-Order Multiplicative Partial Differential Equations in Two Multiplicative Independent Multiplicative Variables

Let U be a domain in R2∗ . Definition 3.1 A multiplicative linear differential operator of second order for the function u = u(x1 , x2 ) is given by 2 ∗∗ ∗∗ L(u) = a ·∗ u∗∗ x1 x1 +∗ e ·∗ b ·∗ ux1 x2 +∗ c ·∗ ux2 x2 ,

where the coefficients a, b and c are supposed to be nonnegative continuously multiplicative differentiable and not simultaneously multiplicative vanishing functions of x1 and x2 in the domain U. We consider the multiplicative differential operator ˜ L(u) = L(u) +∗ g(x1 , x2 , u, u∗x1 , u∗x2 ) := L(u) +∗ · · · ,

(3.1)

where g is not necessarily multiplicative linear and does not contain secondorder multiplicative derivatives. ˜ Definition 3.2 The operator L is called the principal part of the operator L. Our aim is to transform the operator (3.1) or the corresponding equation L(u) +∗ · · · = 0∗ into a simple form, called the canonical form, by introducing new multiplicative independent multiplicative variables.

DOI: 10.1201/9781003440116-3

8

Multiplicative Linear MPDEs

9

Let ξ1 and ξ2 be new multiplicative independent multiplicative variables which are connected with x1 and x2 in the following ways: ξ1 = φ1 (x1 , x2 ), ξ2 = φ2 (x1 , x2 ), where φ1 , φ2 ∈ C∗2 (U ). We will denote with u(ξ1 , ξ2 ) the transformed function u(x1 , x2 ) into the multiplicative variables ξ1 and ξ2 . We have the following relations: u∗x1

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x , 1 1

u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x , 2 2

u∗∗ x1 x1


∗

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

=

∗ u∗ξ1 ·∗ ξ1x 1

=

u∗ξ1

=


∗ ∗∗ ∗ ∗ ∗ ∗∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1 ·∗ ξ1x1 +∗ uξ1 ·∗ ξ1x1 x1


∗

· x1 ∗


∗ x1

x1

∗ +∗ u∗ξ2 ·∗ ξ2x 1


∗ x1

∗ ∗∗ ξ1x +∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 1 1 x1


∗

· x1 ∗

∗ ∗∗ ξ2x +∗ u∗ξ2 ·∗ ξ2x 1 1 x1


∗ ∗∗ ∗ ∗ ∗ ∗∗ + u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1 ·∗ ξ2x1 +∗ uξ2 ·∗ ξ2x1 x1 =

∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1


2∗

∗ ∗ ∗ ∗∗ +∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 ·∗ ξ2x1 +∗ uξ1 ·∗ ξ1x1 x1

∗ ∗ ∗∗ ∗ +u∗∗ ξ1 ξ2 ·∗ ξ1x1 ·∗ ξ2x1 +∗ uξ2 ξ2 ·∗ ξ2x1

=

∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1


2∗


2∗

∗∗ +∗ u∗ξ2 ·∗ ξ2x 1 x1

∗ ∗ ∗∗ ∗ +∗ e2 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 ·∗ ξ2x1 +∗ uξ2 ξ2 ·∗ ξ2x1

∗∗ ∗∗ +u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 x1 1 x1

=

∗2∗ ∗2∗ 2 ∗∗ ∗ ∗ ∗∗ u∗∗ ξ1 ξ1 ·∗ φ1x1 +∗ e ·∗ uξ1 ξ2 ·∗ φ1x1 ·∗ φ2x1 +∗ uξ2 ξ2 ·∗ φ2x1 ∗ ∗∗ +∗ u∗ξ1 ·∗ φ∗∗ 1x1 x1 +∗ uξ2 ·∗ φ2x1 x1 ,


2∗

10 ux2 x2

Classification and Canonical Forms
∗

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

=

∗ u∗ξ1 ·∗ ξ1x 2

=

u∗ξ1

=


∗ ∗∗ ∗ ∗ ∗ ∗∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2 ·∗ ξ1x2 +∗ uξ1 ·∗ ξ1x2 x2


∗

· x2 ∗


∗ x2

x2

∗ +∗ u∗ξ2 ·∗ ξ2x 2


∗ x2

∗ ∗∗ ξ1x +∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 2 2 x2


∗

· x2 ∗

∗ ∗∗ ξ2x +∗ u∗ξ2 ·∗ ξ2x 2 2 x2


∗ ∗∗ ∗ ∗ ∗ ∗∗ + u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2 ·∗ ξ2x2 +∗ uξ2 ·∗ ξ2x2 x2 =

∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2


2∗

∗ ∗ ∗ ∗∗ +∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 ·∗ ξ2x2 +∗ uξ1 ·∗ ξ1x2 x2

∗ ∗ ∗∗ ∗ +u∗∗ ξ1 ξ2 ·∗ ξ1x2 ·∗ ξ2x2 +∗ uξ2 ξ2 ·∗ ξ2x2

=

∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2


2∗


2∗

∗∗ +∗ u∗ξ2 ·∗ ξ2x 2 x2

∗ ∗ ∗∗ ∗ +∗ e2 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 ·∗ ξ2x2 +∗ uξ2 ξ2 ·∗ ξ2x2


2∗

∗∗ ∗∗ +∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 x2 2 x2

=

∗2∗ ∗2∗ 2 ∗∗ ∗ ∗ ∗∗ u∗∗ ξ1 ξ1 ·∗ φ1x2 +∗ e ·∗ uξ1 ξ2 ·∗ φ1x2 ·∗ φ2x2 +∗ uξ2 ξ2 ·∗ φ2x2 ∗ ∗∗ +u∗ξ1 ·∗ φ∗∗ 1x2 x2 +∗ uξ2 ·∗ φ2x2 x2 ,

ux1 x2


∗

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

=

∗ u∗ξ1 ·∗ ξ1x 1

=

u∗ξ1

=


∗ ∗∗ ∗ ∗ ∗ ∗∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2 ·∗ ξ1x1 +∗ uξ1 ·∗ ξ1x1 x2


∗

· x2 ∗


∗ x2

x2

∗ +∗ u∗ξ2 ·∗ ξ2x 1


∗ x2

∗ ∗∗ ξ1x +∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 1 1 x2


∗

· x2 ∗

∗ ∗∗ ξ2x +∗ u∗ξ2 ·∗ ξ2x 1 1 x2


∗ ∗∗ ∗ ∗ ∗ ∗∗ + u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2 ·∗ ξ2x1 +∗ uξ2 ·∗ ξ2x1 x2 =

∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ1x1 ·∗ ξ2x2 +∗ uξ1 ·∗ ξ1x1 x2 ∗ ∗ ∗∗ ∗ ∗ ∗ ∗∗ +u∗∗ ξ1 ξ2 ·∗ ξ1x2 ·∗ ξ2x1 +∗ uξ2 ξ2 ·∗ ξ2x1 ·∗ ξ2x2 +∗ uξ2 ·∗ ξ2x1 x2

Multiplicative Linear MPDEs

11

∗ ∗ ∗∗ ∗ ∗ ∗ ∗ = u∗∗ ξ1 ξ1 ·∗ φ1x1 ·∗ φ1x2 +∗ uξ1 ξ2 ·∗ φ1x1 ·∗ φ2x2 +∗ φ1x2 ·∗ φ2x1




∗ ∗ ∗ ∗∗ ∗ ∗∗ +u∗∗ ξ2 ξ2 ·∗ φ2x1 ·∗ φ2x2 +∗ uξ1 ·∗ φ1x1 x2 +∗ uξ2 ·∗ φ2x1 x2 .

From here, 2 ∗∗ ∗∗ L(u) = a ·∗ u∗∗ x1 x1 +∗ e ·∗ b ·∗ ux1 x2 +∗ c ·∗ ux2 x2

 ∗2∗ ∗2∗ 2 ∗∗ ∗ ∗ ∗∗ = a ·∗ u∗∗ ξ1 ξ1 ·∗ φ1x1 +∗ e ·∗ uξ1 ξ2 ·∗ φ1x1 ·∗ φ2x1 +∗ uξ2 ξ2 ·∗ φ2x1 ∗ ∗∗ +∗ u∗ξ1 ·∗ φ∗∗ 1x1 x1 +∗ uξ2 ·∗ φ2x1 x1



 ∗ ∗ ∗∗ +∗ e2 ·∗ b ·∗ u∗∗ ξ1 ξ1 ·∗ φ1x1 ·∗ φ1x2 +∗ uξ1 ξ2 ·∗ φ∗1x1 ·∗ φ∗2x2 +∗ φ∗1x2 ·∗ φ∗2x1




∗ ∗ ∗ ∗∗ ∗ ∗∗ +∗ u∗∗ ξ2 ξ2 ·∗ φ2x1 ·∗ φ2x2 +∗ uξ1 ·∗ φ1x1 x2 +∗ uξ2 ·∗ φ2x1 x2



 ∗2∗ ∗2∗ 2 ∗∗ ∗ ∗ ∗∗ +∗ c ·∗ u∗∗ ξ1 ξ1 ·∗ φ1x2 +∗ e ·∗ uξ1 ξ2 ·∗ φ1x2 ·∗ φ2x2 +∗ uξ2 ξ2 ·∗ φ2x2 ∗ ∗∗ +∗ u∗ξ1 ·∗ φ∗∗ 1x2 x2 +∗ uξ2 ·∗ φ2x2 x2



∗2∗ ∗2∗ 2 ∗ ∗ = u∗∗ ξ1 ξ1 ·∗ a ·∗ φ1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2




∗ ∗ ∗ ∗ ∗ ∗ +∗ e2 ·∗ u∗∗ ξ1 ξ2 ·∗ a ·∗ φ1x1 ·∗ φ2x1 +∗ b ·∗ φ1x2 ·∗ φ2x1 +∗ φ1x1 ·∗ φ2x2

+∗ c ·∗ φ∗1x2 ·∗ φ∗2x2




∗2∗ ∗2∗ 2 ∗ ∗ +∗ u∗∗ ξ2 ξ2 ·∗ a ·∗ φ2x1 +∗ e ·∗ b ·∗ φ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2

2 ∗∗ ∗∗ +∗ u∗ξ1 ·∗ a ·∗ φ∗∗ 1x1 x1 +∗ e ·∗ b ·∗ φ1x1 x2 +∗ c ·∗ φ1x2 x2





2 ∗∗ ∗∗ +∗ u∗ξ2 ·∗ a ·∗ φ∗∗ 2x1 x1 +∗ e ·∗ b ·∗ φ2x1 x2 +∗ c ·∗ φ2x2 x2 .







12

Classification and Canonical Forms

Let ∗2∗ 2 ∗ ∗ ∗ α = a ·∗ φ∗2 1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2 ,


β = a ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ b ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2 +∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 , ∗2∗ 2 ∗ ∗ ∗ γ = a ·∗ φ∗2 2x1 +∗ e ·∗ b ·∗ φ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2 ,

2 ∗∗ ∗∗ α1 = a ·∗ φ∗∗ 1x1 x1 +∗ e ·∗ b ·∗ φ1x1 x2 +∗ c ·∗ φ1x2 x2 , 2 ∗∗ ∗∗ γ1 = a ·∗ φ∗∗ 2x1 x1 +∗ e ·∗ b ·∗ φ2x1 x2 +∗ c ·∗ φ2x2 x2 .

Then the multiplicative differential operator L assumes the following form: 2 ∗∗ ∗∗ ∗ ∗ L(u) = α ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ β ·∗ uξ1 ξ2 +∗ γ ·∗ uξ2 ξ2 +∗ α1 ·∗ uξ1 +∗ γ1 ·∗ uξ2 ,

which is called the canonical form of the operator L. We set 2 ∗∗ ∗∗ Λ(u) = α ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ β ·∗ uξ1 ξ2 +∗ γ ·∗ uξ2 ξ2 .

Proposition 3.1 a, b, c and α, β, γ are related as follows: α ·∗ γ −∗ β 2∗ = (a ·∗ c −∗ b2∗ ) ·∗ φ∗1x1 ·∗ φ∗2x2 −∗ φ∗1x2 ·∗ φ∗2x1


2∗

(3.2)

Proof We have ∗2∗ 2 ∗ ∗ ∗ α ·∗ γ −∗ β 2∗ = a ·∗ φ∗2 1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2 ∗2∗ 2 ∗ ∗ ∗ ·∗ a ·∗ φ∗2 2x1 +∗ e ·∗ b ·∗ φ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2







−∗ a ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ b ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2 +∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 =





2∗

∗2∗ ∗2∗ 2 ∗ ∗ ∗ a2∗ ·∗ φ∗2 1x1 ·∗ φ2x1 +∗ e ·∗ a ·∗ b ·∗ φ1x1 ·∗ φ2x1 ·∗ φ2x2 ∗2∗ ∗2∗ 2 ∗ ∗ ∗ +∗ a ·∗ c ·∗ φ∗2 1x1 ·∗ φ2x2 +∗ e ·∗ a ·∗ b ·∗ φ1x1 ·∗ φ1x2 ·∗ φ2x1

+∗ e4 ·∗ b2∗ ·∗ φ∗1x1 ·∗ φ∗1x2 ·∗ φ∗2x1 ·∗ φ∗2x2 +∗ e2 ·∗ b ·∗ c ·∗ φ∗1x1 ∗2∗ ∗2∗ ∗2∗ 2 ∗ ∗ ∗ ·∗ φ∗1x2 ·∗ φ∗2 2x2 +∗ a ·∗ c ·∗ φ1x2 ·∗ φ2x1 +∗ e ·∗ b ·∗ c ·∗ φ1x2 ·∗ φ2x1 ·∗ φ2x2

Multiplicative Linear MPDEs

13

∗2∗ ∗2∗ 2∗ 2∗ ∗ ∗ +∗ c2∗ ·∗ φ∗2 ·∗ φ∗2 ·∗ 1x2 ·∗ φ2x2 −∗ a 1x1 ·∗ φ2x1 −∗ b ∗2∗ ∗2∗ ∗2∗ 2 ∗ ∗ ∗ ∗ ∗ φ∗2 1x2 ·∗ φ2x1 +∗ e ·∗ φ1x1 ·∗ φ1x2 ·∗ φ2x1 ·∗ φ2x2 ·∗ +∗ φ1x1 ·∗ φ2x2




∗2∗ 2 ∗ −∗ c2∗ ·∗ φ∗2 1x2 ·∗ φ2x2 −∗ e ·∗ a ·∗ b ·∗ ∗2∗ ∗ ∗ ∗ φ∗1x1 ·∗ φ∗1x2 ·∗ φ∗2 2x1 +∗ φ1x1 ·∗ φ2x1 ·∗ φ2x2




−∗ e2 ·∗ a ·∗ c ·∗ φ∗1x1 ·∗ φ∗1x2 ·∗ φ∗2x1 ·∗ φ∗2x2 −∗ e2 ·∗ b ·∗ c ·∗ ∗2∗ ∗ ∗ ∗ ∗ ∗ φ∗2 1x2 ·∗ φ2x1 ·∗ φ2x2 +∗ φ1x1 ·∗ φ1x2 ·∗ φ2x2




∗2∗ ∗2∗ ∗2∗ 2 ∗ ∗ ∗ ∗ ∗ = a ·∗ c ·∗ φ∗2 1x1 ·∗ φ2x2 +∗ φ1x2 ·∗ φ2x1 −∗ e ·∗ φ1x1 ·∗ φ1x2 ·∗ φ2x1 ·∗ φ2x2




∗2∗ 2 ∗ ∗ ∗ ∗ ∗ −∗ b2∗ ·∗ φ∗2 1x2 ·∗ φ2x1 ·∗ −∗ e ·∗ φ1x1 ·∗ φ1x2 ·∗ φ2x1 ·∗ φ2x2 ∗2∗ ∗ +∗ φ∗2 1x1 ·∗ φ2x2




= a ·∗ c ·∗ φ∗1x1 ·∗ φ∗2x2 −∗ φ∗1x2 ·∗ φ∗2x1


2 ∗

−∗ b2∗ ·∗ φ∗1x1 ·∗ φ∗2x2 −∗ φ∗1x2 ·∗ φ∗2x1


2 ∗

= (a ·∗ c −∗ b2∗ ) ·∗ φ∗1x1 ·∗ φ∗2x2 −∗ φ∗1x2 ·∗ φ∗2x1


2∗

,

which completes the proof. Proposition 3.2 (Identity for the Characteristic Multiplicative Quadratic Form) We have a ·∗ l2∗ +∗ e2 ·∗ b ·∗ l ·∗ m +∗ c ·∗ m2∗ = α ·∗ λ2∗ +∗ e2 ·∗ β ·∗ λ ·∗ µ +∗ γ ·∗ µ2∗ , where l

= λ ·∗ φ1x1 j +∗ µ ·∗ φ∗1x2 ,

m

= λ ·∗ φ∗2x1 +∗ µ ·∗ φ∗2x2 .

Proof We have a ·∗ l2∗ +∗ e2 ·∗ b ·∗ l ·∗ m +∗ c ·∗ m2∗ = a ·∗ λ ·∗ φ∗1x1 +∗ µ ·∗ φ∗1x2


2∗



+∗ e2 ·∗ b ·∗ λ ·∗ φ∗1x1 +∗ µ ·∗ φ∗1x2 ·∗ λ ·∗ φ∗2x1 +∗ µ ·∗ φ∗2x2 +∗ c ·∗ λ ·∗ φ∗2x1 +∗ µ ·∗ φ∗2x2


2∗

14 =

Classification and Canonical Forms
∗ +∗ e2 ·∗ λ ·∗ µ ·∗ φ∗1x1 ·∗ φ∗1x2 +∗ µ2∗ ·∗ φ∗2 1x2

∗ a ·∗ λ∗2∗ ·∗ φ∗2 1x1

 +∗ e2 ·∗ b ·∗ λ2∗ ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ λ ·∗ µ ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ λ ·∗ µ ·∗ φ∗1x1 ·∗ φ∗2x2 +∗ µ2∗ ·∗ φ∗1x2 ·∗ φ∗2x2




2 ∗ ∗ 2∗ ∗ ∗ +∗ c ·∗ λ2∗ ·∗ φ∗2 ·∗ φ∗2 2x1 +∗ e ·∗ λ ·∗ µ ·∗ φ2x1 ·∗ φ2x2 +∗ µ 2x2 . Also, α ·∗ λ2∗ +∗ e2 ·∗ β ·∗ λ ·∗ µ +∗ γ ·∗ µ2∗ =

∗2∗ 2 ∗ ∗ ∗ λ2∗ ·∗ a ·∗ φ∗2 1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2





+∗ e2 ·∗ a ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ b ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2  +∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 ·∗ λ ·∗ µ
∗2∗ 2 ∗ ∗ 2∗ ∗ +∗ a ·∗ φ∗2 2x1 +∗ e ·∗ b ·∗ φ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2 ·∗ µ =

2 ∗ ∗ 2∗ ∗ ∗ a ·∗ λ2∗ ·∗ φ∗2 ·∗ φ∗2 1x1 +∗ e ·∗ λ ·∗ µ ·∗ φ1x1 ·∗ φ2x1 +∗ µ 2x1




+∗ e2 ·∗ b ·∗ λ2∗ ·∗ φ∗1x1 ·∗ φ∗1x2 +∗ λ ·∗ µ ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ2x2 ·∗ +∗ µ2∗ ·∗ φ∗2x1 ·∗ φ∗2x2








2 ∗ ∗ 2∗ ∗ ∗ +∗ c ·∗ λ2∗ ·∗ φ∗2 ·∗ φ∗2 1x2 +∗ e ·∗ λ ·∗ µ ·∗ φ1x2 ·∗ φ2x2 +∗ µ 2x2 , which completes the proof. We will impose two conditions on the transformed coefficients α, β and γ so that to obtain a simple canonical form Λ(u). We consider the following cases: 1. α = −∗ γ, β = 0∗ or α = γ = 0∗ . 2. α = γ, β = 0∗ . 3. β = γ = 0∗ . The transformations φ1 and φ2 satisfy one of the above cases. This depends on the algebraic character of the characteristic quadratic form: Q(l, m)

= a ·∗ l2∗ +∗ e2 ·∗ b ·∗ l ·∗ m +∗ c ·∗ m2∗ = α ·∗ λ2∗ +∗ e2 ·∗ β ·∗ λ ·∗ µ +∗ γ ·∗ µ2∗ .

Multiplicative Linear MPDEs

15

Geometrically speaking, this depends on the character of the quadratic curve in the l, m-multiplicative plane, i.e., for fixed x1 and x2 such that Q(l, m) = e; this curve may be a multiplicative hyperbola, multiplicative ellipse or multiplicative parabola. From here and equation (3.2), we get to the following definition. Definition 3.3 At a point (x1 , x2 ), the operator L(u) will be called as follows: 1. Multiplicative hyperbolic, if a ·∗ c −∗ b2∗ < 1 or 1

>

elog a log c −∗ e(log b)

2

2

= elog a log c−log b , or log a log c − (log b)2 < 0. 2. Multiplicative elliptic, if a ·∗ c −∗ b2∗ > 1, or log a log c − (log b)2 > 0. 3. Multiplicative parabolic, if a ·∗ c −∗ b2∗ = 1, or log a log c − (log b)2 = 0.

Remark 3.1 Note that there are cases in which a second-order multiplicative linear multiplicative differential operator is multiplicative parabolic and at the same time the corresponding classical analogue is hyperbolic. We will see this in the following example. Example 3.1 Consider the operator 2 ∗∗ ∗∗ L(u)(x1 , x2 ) = u∗∗ x1 x1 −∗ e ·∗ ux1 x2 +∗ ux2 x2 ,

(x1 , x2 ) ∈ R2∗ .

(3.3)

Its classical analogue is L1 (u) = ux1 x1 − e2 ux1 x2 + ux2 x2 ,

(x1 , x2 ) ∈ R2 .

Let a(x1 , x2 )

= e,

b(x1 , x2 )

= e,

c(x1 , x2 )

= e,

(x1 , x2 ) ∈ R2∗ ,

(3.4)

16

Classification and Canonical Forms

and a1 (x1 , x2 )

=

1,

b1 (x1 , x2 )

=

e2 , 2

c1 (x1 , x2 )

=

1,

(x1 , x2 ) ∈ R2 .

Then, log(a(x1 , x2 )) log(c(x1 , x2 )) − (log(b(x1 , x2 ))2

=

log e log e − (log e)2

=

1−1

=

0,

(x1 , x2 ) ∈ R2∗ ,

and a1 (x1 , x2 )c1 (x1 , x2 ) − (b1 (x1 , x2 ))

2

 =

1·1−

=

1−

e2 2

2

e4 4 (x1 , x2 ) ∈ R2 .

< 0,

Thus, L is multiplicative parabolic in R2∗ and L1 is hyperbolic in R2 . Example 3.2 Let 2

4 −∗ (x1 +∗ x2 )/∗ e L(u) = e−∗ x1 ·∗ u∗∗ ·∗ u∗∗ x1 x1 +∗ e ·∗ e x1 x2 ∗ ∗ +∗ e−∗ x2 ·∗ u∗∗ x2 x2 +∗ ux1 +∗ ux2 ,

(x1 , x2 ) ∈ R2∗ . Here a(x1 , x2 )

b(x1 , x2 )

=

e−∗ x1

=

e x1 ,

=

e4 ·∗ e−∗ (x1 +∗ x2 )/∗ e

=

e4 ·∗ e−∗ ((x1 x2 )/∗ e

1

2

2

Multiplicative Linear MPDEs

c(x1 , x2 )

17

=

e4 ·∗ e−∗ e

=

e4 ·∗ e−∗ e

=

e4 ·∗ ee

=

elog e

=

e4e

=

e−∗ x2

=

e x2 ,

−

4

−

log(x1 x2 ) log e2

log(x1 x2 ) 2

log(x1 x2 ) 2

−

log ee

log(x1 x2 ) 2

1

log(x1 x2 ) 2

,

(x1 , x2 ) ∈ R2∗ .

Then, log(a(x1 , x2 )) log(c(x1 , x2 )) − (log(b(x1 , x2 )))2  1  1    − log(x1 x2 )  2 x1 log e x2 − log e4e = log e =

 2 log(x1 x2 ) 1 − 4e− 2 x1 x2

=

log(x1 x2 ) 1 − 16e−2 2 x1 x2

=

1 − 16e− log(x1 x2 ) x1 x2

=

1 16 − x1 x2 x1 x2

= − < 0,

15 x1 x2 (x1 , x2 ) ∈ R2∗ ,

because x1 > 1, x2 > 1. Thus the considered operator is multiplicative hyperbolic. Example 3.3 Let ∗∗ 4 ∗∗ L(u) = u∗∗ x1 x1 +∗ sin∗ x1 ·∗ cos∗ x2 ·∗ ux1 x2 +∗ e ·∗ ux2 x2 ,

18

Classification and Canonical Forms

(x1 , x2 ) ∈ R2∗ . Here a(x1 , x2 )

=

e,

b(x1 , x2 )

=

e−2 ·∗ sin∗ x1 ·∗ cos∗ x2

=

e−2 ·∗ esin(log x1 ) ·∗ ecos(log x2 )

=

−2 sin(log x1 ) ) log(elog(cos x2 ) ) elog(e ) log(e

=

e

=

e4 ,

c(x1 , x2 )

sin(log x1 ) cos(log x2 ) 2

,

(x1 , x2 ) ∈ R2∗ .

Then, log(a(x1 , x2 )) log(c(x1 , x2 )) − (log(b(x1 , x2 )))2 =


  sin(log x1 ) cos(log x2 ) 2 2 log e log e4 − log e

=

4−

2 sin(log x1 ) cos(log x2 ) 2 (sin(log x1 ))2 (cos(log x2 ))2 4− 4 

=

> 0,

(x1 , x2 ) ∈ R2∗ .

Hence, the operator L is multiplicative elliptic. Example 3.4 Let 2∗ 2 ∗∗ ∗∗ L(u) = x21∗ ·∗ u∗∗ x1 x1 +∗ e ·∗ x1 ·∗ x2 ·∗ ux1 x2 +∗ x2 ·∗ ux2 x2

+∗ e4 ·∗ u∗x1 −∗ u∗x2 , (x1 , x2 ) ∈ R2∗ . Here a(x1 , x2 )

b(x1 , x2 )

c(x1 , x2 )

=

x21∗

=

e(log x1 ) ,

=

x1 ·∗ x2

=

elog x1 log x2 ,

=

x22∗

=

e(log x2 ) ,

2

2

(x1 , x2 ) ∈ R2∗ .

Multiplicative Linear MPDEs

19

Then, log(a(x1 , x2 )) log(c(x1 , x2 )) − (log(b(x1 , x2 )))2    

2 2 2 = log e(log x1 ) log e(log x2 ) − log elog x1 log x2 = (log x1 )2 (log x2 )2 − (log x1 log x2 )2 = 0,

(x1 , x2 ) ∈ R2∗ .

Therefore, the operator L is multiplicative parabolic. Exercise 3.1 Determine the operator L: 3 ∗∗ 4 ∗∗ ∗ 2 ∗ 1. L(u) = e2 ·∗ u∗∗ x1 x1 −∗ e ·∗ ux1 x2 +∗ e ·∗ ux2 x2 +∗ ux1 −∗ e ·∗ ux2 , 2 (x1 , x2 ) ∈ R∗ . 2 ∗∗ 3 ∗∗ 2 2. L(u) = u∗∗ x1 x1 −∗ e ·∗ ux1 x2 +∗ e ·∗ ux2 x2 + u, (x1 , x2 ) ∈ R∗ . 4 ∗∗ ∗∗ 2 3. L(u) = u∗∗ x1 x1 +∗ e ·∗ ux1 x2 +∗ ux2 x2 , (x1 , x2 ) ∈ R∗ . ∗∗ 3 ∗∗ 2 4. L(u) = e−2 ·∗ u∗∗ x1 x1 +∗ ux1 x2 +∗ e ·∗ ux2 x2 , (x1 , x2 ) ∈ R∗ . 2 ∗∗ 4 ∗∗ ∗ 5. L(u) = e−3 ·∗ u∗∗ x1 x1 +∗ e ·∗ ux1 x2 −∗ e ·∗ ux2 x2 −∗ ux1 , 2 (x1 , x2 ) ∈ R∗ . ∗∗ ∗∗ 2 6. L(u) = u∗∗ x1 x1 −∗ ux1 x2 +∗ ux2 x2 , (x1 , x2 ) ∈ R∗ . 8 ∗∗ 8 ∗∗ 2 7. L(u) = e2 ·∗ u∗∗ x1 x1 +∗ e ·∗ ux1 x2 +∗ e ·∗ ux2 x2 , (x1 , x2 ) ∈ R∗ .

Answer

1. Multiplicative elliptic.

2. Multiplicative elliptic. 3. Multiplicative hyperbolic. 4. Multiplicative hyperbolic. 5. Multiplicative elliptic. 6. Multiplicative elliptic. 7. Multiplicative parabolic. Example 3.5 Let us consider the operator 2 ∗∗ ∗∗ ∗ L(u) = x1 ·∗ u∗∗ x1 x1 −∗ e ·∗ ux1 x2 +∗ x2 ·∗ ux2 x2 +∗ ux1 ,

Here a(x1 , x2 )

= x1 ,

b(x1 , x2 )

= e−1 ,

c(x1 , x2 )

= x2 ,

(x1 , x2 ) ∈ R2∗ .

(x1 , x2 ) ∈ R2∗ .

20

Classification and Canonical Forms

Then, log(a(x1 , x2 )) log(c(x1 , x2 )) − (log(b(x1 , x2 )))2 = log x1 log x2 − (log(e−1 ))2 = log x1 log x2 − (−1)2 = log x1 log x2 − 1,

(x1 , x2 ) ∈ R2∗ .

Thus, we have the following cases: 1. If log x1 log x2 < 1,

(x1 , x2 ) ∈ R2∗ ,

then the considered equation is multiplicative hyperbolic. 2. If log x1 log x2 = 1,

(x1 , x2 ) ∈ R2∗ ,

then the considered equation is multiplicative parabolic. 3. If log x1 log x2 > 1,

(x1 , x2 ) ∈ R2∗ ,

then the considered equation is multiplicative elliptic. Example 3.6 Consider the multiplicative analogue of the classical Tricomi operator: ∗∗ L(u) = u∗∗ (x1 , x2 ) ∈ R2∗ . x1 x1 +∗ x1 ·∗ ux2 x2 , Here a(x1 , x2 )

= e,

b(x1 , x2 )

=

c(x1 , x2 )

= x1 ,

1, (x1 , x2 ) ∈ R2∗ .

Then, log(a(x1 , x2 )) log(c(x1 , x2 )) − (log(b(x1 , x2 )))2

=

log e log x1 − (log 1)2

=

log x1 ,

(x1 , x2 ) ∈ R2∗ .

Thus, we have the following cases: 1. If x1 < e, then log x1 < 1 and the considered equation is multiplicative hyperbolic.

Multiplicative Linear MPDEs

21

2. If x1 = e, then log x1 = 1 and the considered equation is multiplicative parabolic. 3. If x1 > e, then log x1 > 1 and the considered equation is multiplicative elliptic. The corresponding canonical forms of the multiplicative differential operator L(u) are as follows: 1. α = −∗ γ,

β = 0∗ . ∗∗ Λ(u) +∗ · · · = α(u∗∗ ξ1 ξ1 −∗ uξ2 ξ2 ).

When α = γ = 0∗ , we have Λ(u) +∗ · · · = e2 ·∗ β ·∗ u∗∗ ξ 1 ξ2 . 2. α = γ,

β = 0∗ . ∗∗ Λ(u) +∗ · · · = α ·∗ (u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 ).

3. β = γ = 0∗ .

Λ(u) +∗ · · · = α ·∗ u∗∗ ξ 1 ξ1 .

The corresponding canonical forms of the considered multiplicative differential equations are as follows: 1. α = −∗ γ,

β = 0∗ . ∗∗ u∗∗ ξ1 ξ1 −∗ uξ2 ξ2 +∗ · · · = 0∗ .

When α = γ = 0∗ , we have u∗∗ ξ1 ξ2 +∗ · · · = 0∗ . 2. α = γ,

β = 0∗ . ∗∗ u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 +∗ · · · = 0∗ .

3. β = γ = 0∗ .

u∗∗ ξ1 ξ1 +∗ · · · = 0∗ .

22

Classification and Canonical Forms

For fixed (x1 , x2 ) ∈ R2∗ , such a canonical form can be obtained by the multiplicative linear transformation which takes Q into the corresponding canonical form. If we assume that the operator L is of the same type in every point of the domain U ⊆∈ R2∗ , we will search functions φ1 and φ2 which will transform L(u) into a canonical form at every point of U. To find such functions, it depends on whether certain first-order systems of multiplicative linear multiplicative partial differential equations can be solved. Without loss of generality, we suppose that a 6= 0∗ everywhere in the domain U.

3.1.1

The multiplicative hyperbolic case

We suppose that L(u) is multiplicative hyperbolic in U and α = γ = 0∗ = 1. Then, using the definitions for α and γ, we obtain the following system: ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2 = 0∗ , ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 2x1 +∗ e ·∗ b ·∗ φ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2 = 0∗ ;

or a ·∗ (φ∗1x1 )/∗ (φ∗1x2 )


2∗

+∗ e2 ·∗ b ·∗ ((φ∗1x1 )/∗ (φ∗1x2 )) +∗ c = 0∗ ,

a ·∗ (φ∗2x1 )/∗ (φ∗2x2 )


2∗

+∗ e2 ·∗ b ·∗ ((φ∗2x1 )/∗ (φ∗2x2 )) +∗ c = 0∗ .

From the above system we see that if there exist such functions φ1 and φ2 , then ((φ∗1x1 )/∗ (φ∗1x2 )) and (φ∗2x1 )/∗ (φ∗2x2 ) satisfy the multiplicative quadratic equation: a ·∗ p2∗ +∗ e2 ·∗ b ·∗ p +∗ c = 0∗ , (3.5) or 0∗

= 1 = e0 2

= a ·∗ e(log p) +∗ e2 ·∗ elog b log p +∗ c = elog a log e

(log p)2

+∗ elog e

2

log elog b log p

+∗ elog c

2

= elog a(log p) +∗ e2 log b log p +∗ elog c 2

= elog a(log p)

+2 log b log p+log c

,

or log a(log p)2 + 2 log b log p + log c = 0,

Multiplicative Linear MPDEs

23

where p is unknown. Since L(u) is multiplicative hyperbolic in U, log a log c − (log b)2 < 0 and then equation (3.5) has two roots p1 and p2 . Thus in the multiplicative hyperbolic case, we obtain the canonical form e2 ·∗ β ·∗ u∗∗ ξ1 ξ2 +∗ · · · by determining the functions φ1 and φ2 from the multiplicative differential equations: φ∗1x1 −∗ p1 ·∗ φ∗1x2 = 0∗ , φ∗2x1 −∗ p2 ·∗ φ∗2x2 = 0∗ . These two first-order multiplicative linear homogeneous multiplicative partial differential equations yield two families of multiplicative curves: φ1 = const,

φ2 = const.

These two families can be defined as the families of solutions of the multiplicative ordinary differential equations: (d∗ x2 )/∗ (d∗ x1 )

=

−∗ p1 ,

(d∗ x2 )/∗ (d∗ x1 )

=

−∗ p2 ,

and since p1 and p2 are roots of equation (3.5), we have a ·∗ ((d∗ x2 )/∗ (d∗ x1 ))

∗2∗

−∗ e2 ·∗ b ·∗ ((d∗ x2 )/∗ (d∗ x1 )) +∗ c = 0∗ .

(3.6)

Here x2 is considered as a function of x1 along the multiplicative curves of the family. We have that 1 p1,2 = (b ±∗ (b2∗ −∗ a ·∗ c) 2 ∗ )/∗ a and let us set 1

p1

=

(b +∗ (b2∗ −∗ a ·∗ c) 2 ∗ )/∗ a,

p2

=

(b −∗ (b2∗ −∗ a ·∗ c) 2 ∗ /∗ a.

1

Then, Z x 2 +∗ Z∗ x 2 +∗

1

(b +∗ (b2∗ −∗ a ·∗ c) 2 ∗ )/∗ a ·∗ d∗ x1

=

const,

(b −∗ (b2∗ −∗ a ·∗ c) 2 ∗ )/∗ a ·∗ d∗ x2

=

const,

p1 −∗ p2

=

(e2 ·∗ (b2∗ −∗ a ·∗ c))/∗ a.

1

∗

24

Classification and Canonical Forms

Definition 3.4 The multiplicative curves   Z ξ1 = φ1 ·∗ x2 +∗ (b +∗ (b2∗ −∗ a ·∗ c))/∗ a ·∗ d∗ x1 , ∗

(3.7) 

 Z

ξ2 = φ2 ·∗ x2 +∗

(b −∗ (b2∗ −∗ a ·∗ c))/∗ a ·∗ d∗ x1 ,

∗

are called the characteristic multiplicative curves of the multiplicative linear multiplicative hyperbolic differential operator L(u). Remark 3.2 For convenience, in practice, we very often take Z
ξ1 = x2 +∗ (b +∗ (b2∗ −∗ a ·∗ c))/∗ a ·∗ d∗ x1 , ∗

Z ξ2

=

x 2 +∗


(b −∗ (b2∗ −∗ a ·∗ c))/∗ a ·∗ d∗ x1 .

∗

Definition 3.5 The multiplicative curves ξ1 = φ1 (x1 , x2 ) = const,

(x1 , x2 ) ∈ R2∗ ,

ξ2 = φ2 (x1 , x2 ) = const,

(x1 , x2 ) ∈ R2∗ ,

which satisfy the system (3.7), are called the characteristic multiplicative curves of the multiplicative linear multiplicative hyperbolic operator L(u). In the case when α = −∗ γ, β = 0∗ , the functions φ1 and φ2 satisfy the system: ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2 ∗2∗ 2 ∗ ∗ ∗ = −∗ (a ·∗ φ∗2 2x1 +∗ e ·∗ b ·∗ φ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2 )

a ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ b ·∗ (φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x2 ·∗ φ∗2x1 ) +∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 = 0∗ or ∗2∗ 2 ∗ ∗ ∗ ∗ ∗ a ·∗ (φ∗2 1x1 +∗ φ2x1 ) +∗ e ·∗ b ·∗ (φ1x1 ·∗ φ1x2 +∗ φ2x1 ·∗ φ2x2 ) ∗2∗ ∗2∗ +∗ c ·∗ (φ1x2 +∗ φ2x2 ) = 0∗ ,

a ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ b ·∗ (φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2 ) +∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 = 0∗ . Thus we have the canonical form: ∗∗ α ·∗ (u∗∗ ξ1 ξ1 −∗ uξ2 ξ2 ) +∗ · · · = 0∗ .

(3.8)

Multiplicative Linear MPDEs

25

Definition 3.6 Equation (3.6) is called the characteristic equation. Now we will simplify equation (3.8). Let us assume that φ3 , φ4 ∈ C∗1 (U ) and φ 1 = φ 3 +∗ φ 4 , φ 2 = φ 3 −∗ φ 4 . Then, φ∗1x1 = φ∗3x1 +∗ φ∗4x1 , φ∗1x2 = φ∗3x2 +∗ φ∗4x2 , φ∗2x1 = φ∗3x1 −∗ φ∗4x1 , φ∗2x2 = φ∗3x2 −∗ φ∗4x2 , and ∗2∗ ∗ φ∗2 1x1 +∗ φ2x1

=

φ∗3x1 +∗ φ∗4x1


2 ∗

+∗ φ∗3x1 −∗ φ∗4x1


2 ∗

∗2∗ ∗2∗ 2 ∗ ∗ ∗ = φ∗2 3x1 +∗ e ·∗ φ3x1 ·∗ φ4x1 +∗ φ4x1 +∗ φ3x1 ∗ −∗ e2 ·∗ φ∗3x1 ·∗ φ∗4x1 +∗ φ∗2 4x1

= ∗2∗ ∗ φ∗2 1x2 +∗ φ2x2

= =


∗2∗ ∗ e2 ·∗ φ∗2 3x1 +∗ φ4x1 , φ∗3x2 +∗ φ∗4x2


2 ∗

+∗ φ∗3x2 −∗ φ∗4x2


2 ∗

∗2∗ ∗2∗ 2 ∗ ∗ ∗ φ∗2 3x2 +∗ e ·∗ φ3x2 ·∗ φ4x2 +∗ φ4x2 +∗ φ3x2 ∗ −∗ e2 ·∗ φ∗3x2 ·∗ φ∗4x2 +∗ φ∗2 4x2

=


∗2∗ ∗ e2 ·∗ φ∗2 3x2 +∗ φ4x2 ,

and φ∗1x1 ·∗ φ∗1x2 +∗ φ∗2x1 ·∗ φ∗2x2

=



φ∗3x1 +∗ φ∗4x1 ·∗ φ∗3x2 +∗ φ∗4x2

+∗ φ∗3x1 −∗ φ∗4x1 ·∗ φ∗3x2 −∗ φ∗4x2

= φ∗3x1 ·∗ φ∗3x2 +∗ φ∗3x1 ·∗ φ∗4x2 +∗ φ∗4x1 ·∗ φ∗3x2 +∗ φ∗4x1 ·∗ φ∗4x2 +∗ φ∗3x1 ·∗ φ∗3x2 −∗ φ∗3x1 ·∗ φ∗4x2 −∗ φ∗4x1 ·∗ φ∗3x2 +∗ φ∗4x1 ·∗ φ∗4x2
= e2 ·∗ φ∗3x1 ·∗ φ∗3x2 +∗ φ∗4x1 ·∗ φ∗4x2 ;

26

Classification and Canonical Forms

and

∗2∗ 2 ∗ ∗ ∗ ∗ a ·∗ φ∗2 1x1 +∗ φ2x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ φ2x1 ·∗ φ2x2
∗2∗ ∗ +∗ c ·∗ φ∗2 1x2 +∗ φ2x2

∗2∗ 4 ∗ ∗ ∗ ∗ ∗ = e2 ·∗ a ·∗ φ∗2 3x1 +∗ φ4x1 +∗ e ·∗ b ·∗ φ3x1 ·∗ φ3x2 +∗ φ4x1 ·∗ φ4x2
∗2∗ ∗ +∗ e2 ·∗ c ·∗ φ∗2 3x2 +∗ φ4x2 , from where

∗2∗ 2 ∗ ∗ ∗ ∗ ∗ a ·∗ φ∗2 3x1 +∗ φ4x1 +∗ e ·∗ b ·∗ φ3x1 ·∗ φ3x2 +∗ φ4x1 ·∗ φ4x2
∗2∗ ∗ +∗ c ·∗ φ∗2 3x2 +∗ φ4x2 = 0∗ .

(3.9)

Moreover, φ∗1x1 ·∗ φ∗2x1

= =

φ∗1x2 ·∗ φ∗2x2

= =

φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2

=



φ∗3x1 +∗ φ∗4x1 ·∗ φ∗3x1 −∗ φ∗4x1 ∗2∗ ∗ φ∗2 3x1 −∗ φ4x1 ,



φ∗3x2 +∗ φ∗4x2 ·∗ φ∗3x2 −∗ φ∗4x2 ∗2∗ ∗ φ∗2 3x2 −∗ φ4x2 ,



φ∗3x2 +∗ φ∗4x2 ·∗ φ∗3x1 −∗ φ∗4x1

+∗ φ∗3x1 +∗ φ∗4x1 ·∗ φ∗3x2 −∗ φ∗4x2

=

φ∗3x1 ·∗ φ∗3x2 +∗ φ∗3x1 ·∗ φ∗4x2 −∗ φ∗3x2 ·∗ φ∗4x1 −∗ φ∗4x1 ·∗ φ∗4x2 +∗ φ∗3x1 ·∗ φ∗3x2 −∗ φ∗3x1 ·∗ φ∗4x2 +∗ φ∗4x1 ·∗ φ∗3x2 −∗ φ∗4x1 ·∗ φ∗4x2


= e2 ·∗ φ∗3x1 ·∗ φ∗3x2 −∗ φ∗4x1 ·∗ φ∗4x2 , and
a ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ b ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2 +∗ c ·∗ φ∗1x2 ·∗ φ∗2x2

∗2∗ 2 ∗ ∗ ∗ ∗ ∗ = a ·∗ φ∗2 3x1 −∗ φ4x1 +∗ e ·∗ b ·∗ φ3x1 ·∗ φ3x2 −∗ φ4x1 ·∗ φ4x2
∗2∗ ∗ +c ·∗ φ∗2 3x2 −∗ φ4x2 .

Multiplicative Linear MPDEs

27

Using the last identity and equations (3.8), (3.9), we obtain

∗2∗ 2 ∗ ∗ ∗ ∗ ∗ a ·∗ φ∗2 3x1 +∗ φ4x1 +∗ e ·∗ b ·∗ φ3x1 ·∗ φ3x2 +∗ φ4x1 ·∗ φ4x2
∗2−∗ ∗ +∗ c ·∗ φ∗2 = 0∗ 3x2 +∗ φ4x2

2∗ 2 ∗ ∗ ∗ ∗ ∗ a ·∗ φ∗2 3x1 −∗ φ4x1 +∗ e ·∗ b ·∗ φ3x1 ·∗ φ3x2 −∗ φ4x1 ·∗ φ4x2
∗ +∗ c ·∗ φ23x∗ 2 −∗ φ∗2 4x2 = 0∗ , from where ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 3x1 +∗ e ·∗ b ·∗ φ3x1 ·∗ φ3x2 +∗ c ·∗ φ3x2

=

0∗

∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 4x1 +∗ e ·∗ b ·∗ φ4x1 ·∗ φ4x2 +∗ c ·∗ φ4x2

=

0∗ .

In this way, we have reduced the case α = −∗ γ, β = 0∗ to the case α = γ = 0∗ . Example 3.7 We will find the canonical form of the multiplicative analogue of the classical Tricomi equation: ∗∗ u∗∗ x1 x1 +∗ x1 ·∗ ux2 x2 = 0∗ ,

x1 < 0∗ .

The characteristic equation is 2∗

(d∗ x2 )

+∗ x1 ·∗ (d∗ x1 )

2∗

= 0∗ ,

x1 < 0∗ ,

whereupon 2∗

((d∗ x2 )/∗ (d∗ x1 ))

+∗ x1 = 0∗ ,

or

1

(d∗ x2 )/∗ (d∗ x1 ) = ±∗ (−∗ x1 ) 2 ∗ , and

1

d∗ x2 = ±∗ (−∗ x1 ) 2 ∗ ·∗ d∗ x1 , and

Z

Z d∗ x2 = ±∗

∗

x1 < 1, x1 < 1, x1 < 1,

1

(−∗ x1 ) 2 ∗ ·∗ d∗ x1 +∗ c,

x1 < 1,

∗

or 3

x2 = ∓∗ ((−∗ x1 ) 2 ∗ /∗ ( or

 x2 ±∗

2 3∗



3 )) +∗ c, 2∗ 3

x1 < 1,

·∗ (−∗ x1 ) 2 ∗ = c.

28

Classification and Canonical Forms We set  ξ1

=

x 2 +∗

ξ2

=

x 2 −∗



2 3∗



2 3∗



3

·∗ (−∗ x1 ) 2 ∗ , 3

·∗ (−∗ x1 ) 2 ∗ ,

x1 < 1.

Then, 1

∗ ξ1x 1

=

−∗ (−∗ x1 ) 2 ∗ ,

∗ ξ1x 2

=

e,

∗ ξ2x 1

=

(−∗ x1 ) 2 ∗ ,

∗ ξ2x 2

=

e,

1

x1 < 1.

Hence, u∗x1

∗ = u∗ξ1 ·∗ xi∗1x1 +∗ u∗ξ2 ·∗ ξ2x 1 1

1

= −∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ1 +∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ2 , u∗∗ x1 x1

=

  1 1 (e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ) ·∗ u∗ξ1 −∗ (−∗ x1 ) 2 ∗ ·∗ ∗ ∗∗ ∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1




   1 1 −∗ e/∗ e2 ·∗ (−∗ x1 )∗2 ·∗ u∗ξ2 +∗ (−∗ x1 ) 2 ∗ ·∗ ∗ ∗∗ ∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1

=






     3 3 e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ1 −∗ e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ2

  1 1 1 2 ∗ · u∗∗ −∗ (−∗ x1 ) 2 ∗ ·∗ −∗ (−∗ x1 ) 2 ∗ ·∗ u∗∗ ∗ ξ1 ξ 2 ξ1 ξ1 +∗ (−∗ x1 )   1 1 1 2 ∗ · u∗∗ +∗ (−∗ x1 ) 2 ∗ −∗ (−∗ x1 ) 2 ∗ ·∗ u∗∗ + (− x ) ∗ ∗ 1 ∗ ξ1 ξ 2 ξ2 ξ2 =



     3 1 e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ1 −∗ e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ2

∗∗ ∗∗ ∗∗ −∗ x1 ·∗ u∗∗ ξ1 ξ1 +∗ x1 ·∗ uξ1 ξ2 +∗ x1 ·∗ uξ1 ξ2 −∗ x1 ·∗ uξ2 ξ2

Multiplicative Linear MPDEs

=



29

     1 1 e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ1 −∗ e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ2

2 ∗∗ ∗∗ −∗ x1 ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x1 ·∗ uξ1 ξ2 −∗ x1 ·∗ uξ2 ξ2 ,

u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= u∗ξ1 +∗ u∗ξ2 , u∗∗ x2 x2

∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ = u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2 +∗ uξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2 2 ∗∗ ∗∗ = u∗∗ ξ1 ξ1 +∗ e ·∗ uξ1 ξ2 +∗ uξ2 ξ2 .

From here,    1 ∗∗ 2 2∗ u∗∗ ·∗ u∗ξ1 x1 x1 +∗ x1 ·∗ ux2 x2 = e/∗ e ·∗ (−∗ x1 )    1 −∗ e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ2 −∗ x1 ·∗ u∗∗ ξ1 ξ 1 ∗∗ ∗∗ 2 ∗∗ +∗ e2 ·∗ x1 ·∗ u∗∗ ξ1 ξ2 −∗ x1 ·∗ uξ2 ξ2 +∗ x1 ·∗ uξ1 ξ1 +∗ e ·∗ x1 ·∗ uξ1 ξ2

+∗ x1 ·∗ u∗∗ ξ 2 ξ2 =



     1 1 e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ1 −∗ e/∗ e2 ·∗ (−∗ x1 ) 2 ∗

·∗ u∗ξ2 +∗ e4 ·∗ x1 ·∗ u∗∗ ξ1 ξ2 ,

x1 < 1.

Therefore, 0∗

∗∗ = u∗∗ x1 x1 +∗ x1 ·∗ ux2 x2

=



     1 1 e/∗ e2 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗ξ1 −∗ e/∗ e2 ·∗ (−∗ x1 ) 2 ∗

·∗ u∗ξ2 +∗ e4 ·∗ x1 ·∗ u∗∗ ξ1 ξ2 3

∗ ∗ = e−8 ·∗ (−∗ x1 ) 2 ∗ ·∗ u∗∗ ξ1 ξ2 +∗ uξ1 −∗ uξ2 ∗ ∗ = e6 ·∗ (ξ2 −∗ ξ1 ) ·∗ u∗∗ ξ1 ξ2 +∗ uξ1 −∗ uξ2 ,

30

Classification and Canonical Forms

i.e., ∗ ∗ e6 ·∗ (ξ2 −∗ ξ1 ) ·∗ u∗∗ ξ1 ξ2 +∗ uξ1 −∗ uξ2 = 0∗

is the canonical form of the considered equation. Example 3.8 We will find the canonical form of the equation: ∗∗ 2 ∗ x52∗ ·∗ u∗∗ x1 x1 −∗ x2 ·∗ ux2 x2 +∗ e ·∗ ux2 = 0∗ ,

x2 6= 0∗ .

The characteristic equation is x52∗ ·∗ (d∗ x2 )2∗ −∗ x2 ·∗ (d∗ x1 )2∗ = 0∗ ,

x2 6= 0∗ ,

whereupon x22∗ ·∗ d∗ x2 = ±∗ d∗ x1 . Hence, Z

x22∗ ·∗ d∗ x2 = ±∗

∗

Z d∗ x1 +∗ c,

x2 6= 0∗ ,

∗

or
x32∗ /∗ e3 = ±∗ x1 +∗ c,

x2 6= 0∗ ,

or x32∗ ±∗ e3 ·∗ x1 = c,

x2 6= 0∗ .

We set ξ1

=

x32∗ +∗ e3 ·∗ x1 ,

ξ2

=

x32∗ −∗ e3 ·∗ x1 ,

x2 6= 0∗ .

Then, ∗ ξ1x (x1 , x2 ) 1

= e3 ,

∗ ξ1x (x1 , x2 ) 2

= e3 ·∗ x22∗ ,

∗ ξ2x (x1 , x2 ) 1

= e−3 ,

∗ ξ2x (x1 , x2 ) 2

= e3 ·∗ x22∗ ,

From here, u∗x1

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

=

e3 ·∗ u∗ξ1 −∗ e3 ·∗ u∗ξ2 ,

x2 6= 0∗ .

Multiplicative Linear MPDEs u∗∗ x1 x1

31

∗ ∗∗ ∗ = e3 ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1




∗ ∗∗ ∗ −∗ e3 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x2

=






3 3 3 ∗∗ 3 ∗∗ e3 ·∗ e3 ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ uξ1 ξ2 −∗ e ·∗ e ·∗ uξ1 ξ2 −∗ e ·∗ uξ2 ξ2

18 9 ∗∗ = e9 ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ1 − ∗ e ξ1 ξ2 +∗ e ·∗ uξ2 ξ2 ,

u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= e3 ·∗ x22∗ ·∗ u∗ξ1 +∗ e3 ·∗ x22∗ ·∗ u∗ξ2 , u∗∗ x2 x2

2∗ 3 ∗∗ ∗ ∗∗ ∗ = e6 ·∗ x2 ·∗ u∗∗ ξ1 +∗ e ·∗ x2 ·∗ uξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2




∗ ∗∗ ∗ +∗ e6 ·∗ x2 ·∗ u∗ξ2 +∗ e3 ·∗ x22∗ ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2

= e6 ·∗ x2 ·∗ u∗ξ1 +∗ e6 ·∗ x2 ·∗ u∗ξ2 +∗ e3 ·∗ x22∗ ·∗ 2∗ 3 ∗∗ e3 ·∗ x22∗ ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x2 ·∗ uξ1 ξ2




2∗ 3 ∗∗ +∗ e3 ·∗ x22∗ ·∗ e3 ·∗ x22∗ ·∗ u∗∗ ξ1 ξ2 +∗ e ·∗ x2 ·∗ uξ2 ξ2




= e6 ·∗ x2 ·∗ u∗ξ1 +∗ e6 ·∗ x2 ·∗ u∗ξ2 +∗ e9 ·∗ x42∗ ·∗ u∗∗ ξ1 ξ1 4∗ 9 ∗∗ +∗ e18 ·∗ x42∗ ·∗ u∗∗ ξ1 ξ2 +∗ e ·∗ x2 ·∗ uξ2 ξ2 .

Therefore, 0∗

=

∗∗ 2 ∗ x52∗ ·∗ u∗∗ x1 x1 −∗ x2 ·∗ ux2 x2 +∗ e ·? ux2

=

5∗ 18 9 ∗∗ e9 ·∗ x52∗ ·∗ u∗∗ ·∗ x52∗ ·∗ u∗∗ ξ1 ξ 1 − ∗ e ξ1 ξ2 +∗ e ·∗ x2 ·∗ uξ2 ξ2

−∗ e6 ·∗ x22∗ ·∗ u∗ξ1 −∗ e6 ·∗ x22∗ ·∗ u∗ξ2 −∗ e9 ·∗ x52∗ ·∗ u∗∗ ξ 1 ξ1 5∗ 9 ∗∗ −∗ e18 ·∗ x52∗ ·∗ u∗∗ ξ1 ξ2 −∗ e ·∗ x2 ·∗ uξ2 ξ2

+∗ e6 ·∗ x22∗ ·∗ u∗ξ1 +∗ e6 ·∗ x22∗ ·∗ u∗ξ2 =

−∗ e36 ·∗ x52∗ ·∗ u∗∗ ξ1 ξ 2 ,

whereupon u∗∗ ξ1 ξ2 = 0∗ is the canonical form of the considered equation.




32

Classification and Canonical Forms

Example 3.9 We will find the canonical form of the equation: 2 ∗∗ 2∗ ∗ u∗∗ ·∗ u∗∗ x1 x1 −∗ e ·∗ sin∗ x1 ·∗ ux1 x2 −∗ (cos∗ x1 ) x2 x2 −∗ cos∗ x1 ·∗ ux2 = 0∗ .

The characteristic equation is (d∗ x2 )2∗ +∗ e2 ·∗ sin∗ x1 ·∗ d∗ x1 ·∗ d∗ x2 −∗ (cos∗ x1 )2∗ ·∗ (d∗ x1 )2∗ = 0∗ , whereupon ((d∗ x2 )/∗ (d∗ x1 ))

2∗

+∗ e2 ·∗ sin∗ x1 ·∗ ((d∗ x2 )/∗ (d∗ x1 )) −∗ (cos∗ x1 )2∗ = 0∗ .

Hence, ((d∗ x2 )/∗ (d∗ x1 ))1,2

−∗ sin∗ x1 ±∗ (sin∗ x1 )2∗ +∗ (cos∗ x1 )2∗

=

= −∗ sin∗ x1 ±∗ e, or d∗ x2 = (−∗ sin∗ x1 ±∗ e) ·∗ d∗ x1 , from where Z

Z d∗ x2

(−∗ sin∗ x1 ±∗ e) ·∗ d∗ x1 +∗ c

= ∗

= cos∗ x1 ±∗ x1 +∗ c and x2 −∗ cos∗ x1 ∓∗ x1

= c.

We set ξ1

= x2 −∗ cos∗ x1 −∗ x1 ,

ξ2

= x2 −∗ cos∗ x1 +∗ x1 .

Then, ∗ ξ1x = sin∗ x1 −∗ e, 1 ∗ ξ1x = e, 2 ∗ ξ2x = sin∗ x1 +∗ e, 1 ∗ ξ2x = 1. 2

From here, u∗x1

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

= (sin∗ x1 −∗ e) ·∗ u∗ξ1 +∗ (sin∗ x1 +∗ 1)u∗ξ2 ,






/∗ e

Multiplicative Linear MPDEs

u∗∗ x1 x1

33

= cos∗ x1 ·∗ u∗ξ1 +∗ cos∗ x1 ·∗ u∗ξ2 +∗ (sin∗ x1 −∗ e) ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1




∗ ∗∗ ∗ +∗ (sin∗ x1 +∗ e) ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1




= cos∗ x1 ·∗ u∗ξ1 +∗ cos∗ x1 ·∗ u∗ξ2 ∗∗ +∗ (sin∗ x1 −∗ e) ·∗ (sin∗ x1 −∗ e) ·∗ u∗∗ ξ1 ξ1 +∗ (sin∗ x1 +∗ e) ·∗ uξ1 ξ2




∗∗ +∗ (sin∗ x1 +∗ e) ·∗ (sin∗ x1 −∗ e) ·∗ u∗∗ ξ1 ξ2 +∗ (sin∗ x1 +∗ e) ·∗ uξ2 ξ2




= cos∗ x1 ·∗ u∗ξ1 +∗ cos∗ x1 ·∗ u∗ξ2 +∗ (sin∗ x1 −∗ e)2∗ ·∗ u∗∗ ξ1 ξ1 2∗ +∗ e2 ·∗ ((sin∗ x1 )2∗ −∗ e) ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ2 +∗ (sin∗ x1 +∗ e) ξ 2 ξ2 ,

u∗∗ x1 x2

∗ ∗∗ ∗ = (sin∗ x1 −∗ e) ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2




∗ ∗∗ ∗ +∗ (sin∗ x1 +∗ e) ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2




∗∗ = (sin∗ x1 −∗ e) ·∗ (u∗∗ ξ1 ξ1 +∗ uξ1 ξ2 ) ∗∗ +∗ (sin∗ x1 +∗ e) ·∗ u∗∗ ξ1 ξ2 +∗ uξ2 ξ2




2 ∗∗ = (sin∗ x1 −∗ e) ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ sin∗ x1 uξ1 ξ2

+∗ (sin∗ x1 +∗ e) ·∗ u∗∗ ξ2 ξ2 , u∗x2

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= u∗ξ1 +∗ u∗ξ2 , u∗∗ x2 x2

∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ = u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1 +∗ uξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1 2 ∗∗ ∗∗ = u∗∗ ξ1 ξ1 +∗ e ·∗ uξ1 ξ2 +∗ uξ2 ξ2 .

34

Classification and Canonical Forms

Therefore, 2 ∗∗ 2∗ ∗ = u∗∗ ·∗ u∗∗ x1 x1 −∗ e sin∗ x1 ·∗ ux1 x2 −∗ (cos∗ x1 ) x2 x2 −∗ cos∗ x1 ·∗ ux2

0∗

= cos∗ x1 ·∗ u∗ξ1 +∗ cos∗ x1 ·∗ u∗ξ2 +∗ (sin∗ x1 −∗ e)2∗ ·∗ u∗∗ ξ 1 ξ1 2∗ +∗ e2 ·∗ ((sin∗ x1 )2∗ −∗ e) ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ2 +∗ (sin∗ x1 +∗ e) ξ 2 ξ2 4 2∗ −∗ e2 ·∗ sin∗ x1 ·∗ (sin∗ x1 −∗ e) ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ (sin∗ x1 ) ξ1 ξ2 2∗ −∗ e2 ·∗ sin∗ x1 ·∗ (sin∗ x1 +∗ e) ·∗ u∗∗ ·∗ u∗∗ ξ2 ξ2 −∗ (cos∗ x1 ) ξ1 ξ1 2∗ ∗ −∗ e2 ·∗ (cos∗ x1 )2∗ ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ2 −∗ (cos∗ x1 ) ξ2 ξ2 −∗ cos∗ x1 uξ1

−∗ cos∗ x1 u∗ξ2 −∗ e4 ·∗ u∗∗ ξ 1 ξ2 , = u∗∗ ξ1 ξ2 , i.e., u∗∗ ξ1 ξ2 = 0∗ is the canonical form of the considered equation. Exercise 3.2 Find the canonical form of the following equations: 5 ∗∗ 6 ∗∗ 1. u∗∗ x1 x1 −∗ e ·∗ ux1 x2 +∗ e ·∗ ux2 x2 = 0∗ . 5 ∗∗ 4 ∗∗ 2. u∗∗ x1 x1 +∗ e ·∗ ux1 x2 +∗ e ·∗ ux2 x2 = 0∗ . 2∗ ∗∗ 2 ∗ 3. x21∗ ·∗ u∗∗ x1 x1 −∗ x2 ·∗ ux2 x2 −∗ e ·∗ x2 ·∗ ux2 = 0∗ , x2 6= 0.

Answer

1. u∗∗ ξ 1 ξ2

=

0∗ ,

x2 +∗ e3 ·∗ x1

=

ξ1 ,

x2 +∗ e2 ·∗ x1

=

ξ2 .

u∗∗ ξ 1 ξ2

=

0∗ ,

x2 +∗ e4 ·∗ x1

=

ξ1 ,

x2 +∗ x1

=

ξ2 .

2.

x1 6= 0,

Multiplicative Linear MPDEs

35

3. 2 u∗∗ ξ1 ξ2 +∗ e/∗ e ·∗ ξ1

·∗ u∗ξ2

=

0,

x2 /∗ x1

=

ξ1 ,

x1 ·∗ x2

=

ξ2 .





Exercise 3.3 Consider the equation ∗∗ 2 ∗ 2 ∗ u∗∗ x1 x1 −∗ ux2 x2 +∗ e ·∗ ux1 +∗ e ·∗ ux2 = 0∗ .

1. Find the canonical form. 2. Find the general solution. 3. Find the solution for which u(1, x2 ) u∗x1 (1, x2 ) Solution

= e, = −∗ e.

1. The characteristic equation is (d∗ x2 )2∗ −∗ (d∗ x1 )2∗ = 0∗ ,

whereupon d∗ x2 = ±∗ d∗ x1 and x2 ±∗ x1 = c. We set

Then, ∗ ξ1x 1

= e,

∗ ξ1x 2

= e,

∗ ξ2x 1

= −∗ e,

ξ1

= x 2 +∗ x 1 ,

ξ2

= x 2 −∗ x 1 .

36

Classification and Canonical Forms ∗ ξ2x 2

u∗x1

= e, ∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

= u∗ξ1 −∗ u∗ξ2 , u∗x1 x1

∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ = u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1 −∗ uξ1 ξ2 ·∗ ξ1x1 −∗ uξ2 ξ2 ·∗ ξ2x1 ∗∗ ∗∗ ∗∗ = u∗∗ ξ1 ξ1 −∗ uξ1 ξ2 −∗ uξ1 ξ2 +∗ uξ2 ξ2 2 ∗∗ ∗∗ = u∗∗ ξ1 ξ1 −∗ e ·∗ uξ1 ξ2 +∗ uξ2 ξ2 ,

u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= u∗ξ1 +∗ u∗ξ2 , u∗∗ x2 x2

∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ = u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2 +∗ uξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2 ∗∗ ∗∗ ∗∗ = u∗∗ ξ1 ξ1 +∗ uξ1 ξ2 +∗ uξ1 ξ2 +∗ uξ2 ξ2 2 ∗∗ ∗∗ = u∗∗ ξ1 ξ1 +∗ e ·∗ uξ1 ξ2 +∗ uξ2 ξ2 .

Hence, 0∗

=

∗∗ 2 ∗ 2 ∗ u∗∗ x1 x1 −∗ ux2 x2 +∗ e ·∗ ux1 +∗ e ·∗ ux2

=

2 ∗∗ ∗∗ ∗∗ u∗∗ ξ1 ξ1 −∗ e ·∗ uξ1 ξ2 +∗ uξ2 ξ2 −∗ uξ1 ξ1 ∗∗ 2 ∗ −∗ e2 ·∗ u∗∗ ξ1 ξ2 −∗ uξ2 ξ2 +∗ e ·∗ uξ1

−∗ e2 ·∗ u∗ξ2 +∗ e2 ·∗ u∗ξ1 +∗ e2 ·∗ u∗ξ2 4 ∗ −∗ e4 ·∗ u∗∗ ξ1 ξ2 +∗ e ·∗ uξ1 ,

whereupon ∗ u∗∗ ξ1 ξ2 −∗ uξ1 = 0∗

is the canonical form. 2. Fix ξ1 and set

v = u∗ξ1 .

(3.10)

Multiplicative Linear MPDEs

37

Then equation (3.10) takes the form vξ2 −∗ v = 0∗ , from where v = f1 (ξ1 ) ·∗ eξ2 . Hence, u∗ξ1 = f1 (ξ1 ) ·∗ eξ2 and u = eξ2 ·∗ f (ξ1 ) +∗ g(ξ2 ), where f and g are C∗2 -functions. Therefore, u(x1 , x2 ) = f (x1 +∗ x2 ) ·∗ ex2 −∗ x1 +∗ g(x2 −∗ x1 )

(3.11)

is the general solution. 3. Using equation (3.11), we get u(1, x2 ) u∗x1 (x1 , x2 )

= f (x2 ) ·∗ ex2 +∗ g(x2 ), = f ∗ (x1 +∗ x2 ) ·∗ ex2 −∗ x1 −∗ f (x1 +∗ x2 ) ·∗ ex2 −∗ x1 −∗ g ∗ (x2 −∗ x1 ),

u∗x1 (1, x2 )

= f ∗ (x2 ) ·∗ ex2 −∗ f (x2 ) ·∗ ex2 −∗ g ∗ (x2 ),

i.e., we obtain the system f (x2 ) ·∗ ex2 +∗ g(x2 ) = e, (3.12) f ∗ (x2 ) ·∗ ex2 −∗ f (x2 ) ·∗ ex2 −∗ g ∗ (x2 ) = −∗ e. We multiplicative differentiate the first equation of the last system with respect to x2 and we obtain f ∗ (x2 ) ·∗ ex2 +∗ f (x2 ) ·∗ ex2 +∗ g ∗ (x2 ) = 0∗ , f ∗ (x2 ) ·∗ ex2 −∗ f (x2 ) ·∗ ex2 −∗ g ∗ (x2 ) = −∗ e. whereupon e2 ·∗ f ∗ (x2 ) ·∗ ex2 = −∗ e or f ∗ (x2 ) = −∗ (e/∗ e2 ) ·∗ e−∗ x2 or

1

f (x2 ) = e 2 ·∗ e−∗ x2 +∗ c.

38

Classification and Canonical Forms From here and from the first equation of (3.12), we find g(x2 )

=

e −∗ f (x2 ) ·∗ ex2  1  e −∗ e 2 ·∗ e−∗ x2 +∗ c ·∗ ex2

=

e 2 −∗ c ·∗ ex2 .

=

1

Therefore, u(x1 , x2 )

=



 1 1 e 2 ·∗ e−∗ x1 −∗ x2 +∗ c ·∗ ex2 −∗ x1 +∗ e 2 −∗ c ·∗ ex2 −∗ x1

1

= e 2 ·∗ e−∗ e

2

1

·∗ x1

+∗ e 2 .

Exercise 3.4 Consider the equation x21∗ ·∗ u∗x1 x1 −∗ e2 ·∗ x1 ·∗ x2 ·∗ u∗x1 x2 −∗ e3 ·∗ x22∗ ·∗ u∗x2 x2 = 0∗ ,

x1 > 0∗ ,

x2 > 0∗ .

1. Find the canonical form. 2. Find the general solution. 3. Find the solution u(x1 , x2 ) for which

Solution

u(x1 , e)

=

1,

u∗x2 (x1 , e)

=

(e4 /∗ (e3 ·∗ x14 ∗ ).

5

1. The characteristic equation is x21∗ ·∗ (d∗ x2 )2 +∗ e2 ·∗ x1 ·∗ x2 ·∗ d∗ x1 ·∗ d∗ x2 −∗ e3 ·∗ x22∗ ·∗ (d∗ x1 )2∗ = 0∗ ,

whereupon x21∗ ·∗ ((d∗ x2 )/∗ (d∗ x1 ))

2∗

+∗ e2 ·∗ x1 ·∗ x2 ·∗ ((d∗ x2 )/∗ (d∗ x1 ))

−∗ e3 ·∗ x22∗ = 0∗ and ((d∗ x2 )/∗ (d∗ x1 ))1,2

(d∗ x2 )/∗ (d∗ x1 )

=

−∗ x1 ·∗ x2 ±∗ (x21∗ ·∗ x22∗  1 +∗ e3 ·∗ x21∗ ·∗ x22∗ ) 2 ∗ /∗ (x21∗ )

=

(−∗ x1 ·∗ x2 ±∗ e2 ·∗ x1 ·∗ x2 )/∗ (x21∗ ),

= x2 / ∗ x1

Multiplicative Linear MPDEs

39

and (d∗ x2 )/∗ (d∗ x1 ) = −∗ e3 (x2 /∗ x1 ). Hence, x2 /∗ x1

=

c,

x31∗ ·∗ x2

=

c.

We set ξ1

=

x2 /∗ x1 ,

ξ2

=

x31∗ ·∗ x2 .

Then, ∗ ξ1x 1

=

−∗ (x2 /∗ (x21∗ )),

∗ ξ1x 2

=

e/∗ x1 ,

∗ ξ2x 1

=

e3 ·∗ x21∗ ·∗ x2 ,

∗ ξ2x 2

=

x31∗ ,

u∗x1

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

= −∗ (x2 /∗ (x21∗ )) ·∗ u∗ξ1 +∗ e3 ·∗ x21∗ ·∗ x2 ·∗ u∗ξ2 , u∗∗ x1 x1

=

((e2 ·∗ x2 )/∗ (x31∗ )) ·∗ u∗ξ1 −∗ (x2 /∗ (x21∗ )) ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1




+∗ e6 ·∗ x1 ·∗ x2 ·∗ u∗ξ2 +∗ e3 ·∗ x21∗ ·∗ x2 ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1

=




((e2 ·∗ x2 )/∗ (x31∗ )) ·∗ u∗ξ1 −∗ (x2 /∗ (x21∗ )) ·∗ 2∗ 3 ∗∗ −∗ (x2 /∗ (x21∗ )) ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x1 ·∗ x2 ·∗ uξ1 ξ2




40

Classification and Canonical Forms

+∗ e6 ·∗ x1 ·∗ x2 ·∗ u∗ξ2 +∗ e3 ·∗ x21∗ ·∗ x2 ·∗ −∗ (x2 /∗ (x21∗ )) ·∗ u∗∗ ξ1 ξ 2 +∗ e3 ·∗ x21∗ ·∗ x2 ·∗ u∗∗ ξ2 ξ 2 =




((e2 ·∗ x2 )/∗ (x31∗ )) ·∗ u∗ξ1 +∗ e6 ·∗ x1 ·∗ x2 ·∗ u∗ξ2 +∗ ((x22∗ )/∗ (x41∗ )) ·∗ u∗∗ ξ1 ξ1 4∗ 2∗ 9 ∗∗ −∗ e6 ·∗ x22∗ ·∗ u∗∗ ξ1 ξ2 +∗ e ·∗ x1 ·∗ x2 ·∗ uξ2 ξ2 ,

u∗∗ x1 x2

=

−∗ (e/∗ (x21∗ )) ·∗ u∗ξ1 −∗ (x2 /∗ (x21∗ )) ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2




+∗ e3 ·∗ x21∗ ·∗ u∗ξ2 +∗ e3 ·∗ x21∗ ·∗ x2 ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2




= −∗ (e/∗ (x21∗ )) ·∗ u∗ξ1 +∗ e3 ·∗ x21∗ ·∗ u∗ξ2 −∗ (x2 /∗ (x21∗ )) 3∗ ∗∗ ·∗ (e/∗ x1 ) ·∗ u∗∗ ξ1 ξ1 +∗ x1 ·∗ uξ1 ξ2




3∗ ∗∗ +∗ e3 ·∗ x21∗ ·∗ x2 ·∗ (e/∗ x1 ) ·∗ u∗∗ ξ1 ξ2 +∗ x1 ·∗ uξ2 ξ2




= −∗ (e/∗ (x21∗ )) ·∗ u∗ξ1 +∗ e3 ·∗ x21∗ ·∗ u∗ξ2 −∗ (x2 /∗ (x31∗ )) 2 ∗∗ ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x1 ·∗ x2 ·∗ uξ1 ξ2

+∗ e3 ·∗ x51∗ ·∗ x2 ·∗ u∗∗ ξ 2 ξ2 , u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

=

(e/∗ x1 ) ·∗ u∗ξ1 +∗ x31∗ ·∗ u∗ξ2 ,

Multiplicative Linear MPDEs

u∗x2 x2

=

41

∗ ∗∗ ∗ (e/∗ x1 ) ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2 ∗ ∗∗ ∗ +∗ x31∗ ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2

=







3∗ ∗∗ (e/∗ x1 ) ·∗ (e/∗ x1 ) ·∗ u∗∗ ξ1 ξ1 +∗ x1 ·∗ uξ1 ξ2 3∗ ∗∗ +∗ x31∗ ·∗ (e/∗ x1 ) ·∗ u∗∗ ξ1 ξ2 +∗ x1 ·∗ uξ2 ξ2

=







2∗ 6∗ 2 ∗∗ ∗∗ (e/∗ (x21∗ )) ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x1 ·∗ uξ1 ξ2 +∗ x1 ·∗ uξ2 ξ2 .

Therefore, 0∗

2 ∗∗ 3 2∗ ∗∗ = x21∗ ·∗ u∗∗ x1 x1 −∗ e ·∗ x1 ·∗ x2 ·∗ ux1 x2 −∗ e x2 ·∗ ux2 x2

= x21∗ ·∗

 ((e2 ·∗ x2 )/∗ (x31∗ )) ·∗ u∗ξ1 +∗ e6 ·∗ x1 ·∗ x2 ·∗ u∗ξ2

2∗ 6 +∗ ((x22∗ )/∗ (x41∗ )) ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ x2

·∗ u∗∗ ξ1 ξ 2

+∗ e

9

·∗ x41∗ ·∗

x22∗ ·∗

u∗∗ ξ2 ξ2



−∗ e2 ·∗ x1 ·∗ x2 ·∗

 −∗ (e/∗ (x21∗ )) ·∗ u∗ξ1 +∗ e3 ·∗ x21∗ ·∗ u∗ξ2 −∗ (x2 /∗ (x31∗ )) ·∗ u∗∗ ξ 1 ξ1 +∗ e

2

·∗ x1 ·∗ x2 ·∗ u∗∗ ξ1 ξ 2

−∗ e

3

·∗ x22∗ ·∗



+∗ e

3

·∗ x51∗ ·∗

x2 ·∗ u∗∗ ξ2 ξ2



(e/∗ (x21∗ )) ·∗ u∗∗ ξ1 ξ1

6∗ ∗∗ +∗ e2 ·∗ x21∗ ·∗ u∗∗ ξ1 ξ2 +∗ x1 ·∗ uξ2 ξ2



= e2 ·∗ (x2 /∗ x1 ) ·∗ u∗ξ1 +∗ e6 ·∗ x31∗ ·∗ x2 ·∗ u∗ξ2 +∗ ((x22∗ )/∗ (x21∗ )) 2∗ 2∗ 6 ∗∗ ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ x1 ·∗ x2 ·∗ uξ1 ξ2 2 ∗ +∗ e9 ·∗ x61∗ ·∗ x22∗ ·∗ u∗∗ ξ2 ξ2 +∗ e ·∗ (x2 /∗ x1 ) ·∗ uξ1

−∗ e6 ·∗ x31∗ ·∗ x2 ·∗ u∗ξ2 +∗ e2 ·∗ ((x22∗ )/∗ (x21∗ )) ·∗ u∗∗ ξ1 ξ 1

42

Classification and Canonical Forms

6∗ 2∗ 6 ∗∗ −∗ e4 ·∗ x21∗ ·∗ x22∗ u∗∗ ξ1 ξ2 −∗ e ·∗ x1 ·∗ x2 ·∗ uξ2 ξ2 2∗ 2∗ 6 ∗∗ −∗ e3 ·∗ ((x22∗ )/∗ (x21∗ )) ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ x1 ·∗ x2 ·∗ uξ1 ξ2

−∗ e3 ·∗ x61∗ ·∗ x22∗ ·∗ u∗∗ ξ2 ξ2 = e4 ·∗ (x2 /∗ x1 ) ·∗ u∗ξ1 −∗ e16 ·∗ x21∗ ·∗ x22∗ ·∗ u∗∗ ξ1 ξ 2 3∗ ∗ = e4 ·∗ u∗∗ ξ1 ξ2 −∗ (e/∗ (x1 ·∗ x2 )) ·∗ uξ1 ,

i.e., ∗ e4 ·∗ u∗∗ ξ1 ξ2 −∗ (e/∗ ξ2 ) ·∗ uξ1 = 0∗

(3.13)

is the canonical form of the considered equation. 2. Fix ξ1 and set

v = u∗ξ1 .

Then equation (3.13) takes the form e4 ·∗ vξ∗2 −∗ (e/∗ ξ2 ) ·∗ v = 0∗ or e4 ·∗ ((vξ∗2 )/∗ v) = (e/∗ ξ2 ) or

1

v = x24 ∗ ·∗ f1∗ (ξ1 ), and

1

u∗ξ1 = ξ24 ∗ ·∗ f1∗ (ξ1 ) and

1

u = ξ24 ∗ ·∗ f (ξ1 ) +∗ g(ξ2 ), i.e., 1

u(x1 , x2 ) = (x31∗ ·∗ x2 ) 4 ∗ ·∗ f (x2 /∗ x1 ) +∗ g(x31∗ ·∗ x2 )

(3.14)

is the general solution of the considered equation, where f and g are C∗2 -functions. 3. From equation (3.14), we get u(x1 , 1) u(x1 , x2 )

=

4

(x31∗ ) 3 ∗ ·∗ f (e/∗ x1 ) +∗ g(x31∗ ), 1

1

= e 4 ·∗ ((x31∗ )/∗ (x32∗ )) ·∗ f (x2 /∗ x1 ) +∗ (x2 /∗ x1 ) 4 ∗ ·∗ f ∗ (x2 /∗ x1 ) +∗ x31∗ ·∗ g ∗ (x31∗ ·∗ x2 ),

Multiplicative Linear MPDEs

u(x1 , e)

1

43 1

1

= e 4 ·∗ (x31∗ ) 4 ∗ ·∗ f (e/∗ x1 ) +∗ (e/∗ (x1 ) 4 ∗ ) ·∗ f ∗ (e/∗ x1 ) +∗ x31∗ ·∗ g ∗ (x31∗ ).

In this way we get the system 1

(x31∗ ) 4 ∗ ·∗ f (e/∗ x1 ) +∗ g(x31∗ ) = 0∗ 1

1

1

e 4 ·∗ (x31∗ ) 4 ∗ ·∗ f (e/∗ x1 ) +∗ (e/∗ (x1 ) 4 ∗ ) ·∗ f ∗ (e/∗ x1 )

(3.15)

1

+∗ x31∗ ·∗ g ∗ (x31∗ ) = (e4 /∗ (e3 ·∗ x1 4 ∗ ·∗ x1 ). We multiplicative differentiate the first equation of the last system and we get 5

1

(e3 /∗ (e4 ·∗ x1 4 ∗ ) ·∗ f (e/∗ x1 ) −∗ (e/∗ (x14 ∗ ) ·∗ f ∗ (e/∗ x1 ) +∗ e3 ·∗ x21∗ ·∗ g ∗ (x31∗ ) = 0∗ 1

1

1

e 4 ·∗ (x31∗ ) 4 ∗ ·∗ f (e/∗ x1 ) +∗ (e/∗ (x1 4 ∗ ) ·∗ f ∗ (e/∗ x1 ) 1

+∗ x31∗ ·∗ g ∗ (x31∗ ) = (e4 /∗ (e3 ·∗ x1 4 ∗ ·∗ x1 ), whereupon 3

1

1

e 4 ·∗ (x31∗ ) 4 ∗ ·∗ f (e/∗ x1 ) −∗ (e/∗ (x1 4 ∗ )) ·∗ f ∗ (e/∗ x1 ) +∗ e3 ·∗ x31∗ ·∗ g ∗ (x31∗ ) = 0∗ , 1

1

1

e 4 ·∗ (x31∗ ) 4 ∗ ·∗ f (e/∗ x1 ) +∗ (e/∗ (x14 ∗ )) ·∗ f ∗ (e/∗ x1 ) 1

+∗ x31∗ ·∗ g ∗ (x31∗ ) = (e4 /∗ (e3 ·∗ x1 4 ∗ ·∗ x1 )) and f ∗ (e/∗ x1 ) = e/∗ x1 , i.e., f ∗ (z) = z. Therefore,

1

f (z) = e 2 ·∗ z 2∗ +∗ c. Hence, from the first equation of (3.15), we obtain
1 (x31∗ ) 4 ∗ ·∗ e/∗ (e2 ·∗ x21∗ ) +∗ c +∗ g(x31∗ ) = 0∗ .

44

Classification and Canonical Forms From here, 1

5

g(z) = −∗ c ·∗ z 4 ∗ −∗ (e/∗ (e2 ·∗ z 12 ∗ )). Consequently, u(x1 , x2 )

1

=

(x31∗ ·∗ x2 ) 4 ∗ ·∗ f (x2 /∗ x1 ) +∗ g(x31∗ ·∗ x2 )

=

 1  1 (x31∗ ·∗ x2 ) 4 ∗ ·∗ e 2 ·∗ ((x22∗ )/∗ (x21∗ )) +∗ c 1

−∗ c ·∗ (x31∗ ·∗ x2 ) 4 ∗ −∗ ( 9

1

5 5 e 2 (e ·∗ x14 ∗ ·∗ x212 ∗ ) /∗

5

5

5

= e 2 ·∗ ((x24 ∗ )/∗ (x14 ∗ ) −∗ (e/∗ (e2 ·∗ x14 ∗ ·∗ x212 ∗ ). Exercise 3.5 Consider the equation 6 ∗∗ 16 u∗∗ ·∗ u∗∗ x1 x1 +∗ e ·∗ ux1 x2 −∗ e x2 x2 = 0∗ .

1. Find the canonical form. 2. Find the general solution. 3. Find the solution u(x1 , x2 ) for which u(−∗ x1 , e2 ·∗ x1 ) u(x1 , 1) Solution

= x1 , =

e2 ·∗ x1 .

1. The characteristic equation is (d∗ x2 )2∗ −∗ e6 ·∗ d∗ x1 ·∗ d∗ x2 −∗ e16 ·∗ (d∗ x1 )2 = 0∗ ,

whereupon 2∗

((d∗ x2 )/∗ (d∗ x1 ))

−∗ e6 ·∗ ((d∗ x2 )/∗ (d∗ x1 )) −∗ e16 = 0∗

and (d∗ x2 )/∗ (d∗ x1 )

=

1

(e3 ±∗ (e9 +∗ e16 ) 2 ∗ )/∗ e

= e3 ±∗ e5 , d∗ x2

= e8 ·∗ d∗ x1 ,

d∗ x2

= −∗ e2 ·∗ d∗ x1 .

Multiplicative Linear MPDEs

45

Therefore, x2 −∗ e8 ·∗ x1

= c,

x2 +∗ e2 ·∗ x1

= c.

We set ξ1

=

x2 −∗ e8 ·∗ x1 ,

ξ2

=

x2 +∗ e2 ·∗ x1 .

Then, ∗ ξ1x 1

= −∗ e 8 ,

∗ ξ1x 2

= e,

∗ ξ2x 1

= e2 ,

∗ ξ2x 2

= e,

u∗x1

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

= −∗ e8 ·∗ u∗ξ1 +∗ e2 ·∗ u∗ξ2 , u∗∗ x1 x1

∗ ∗∗ ∗ = −∗ e8 ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1




∗ ∗∗ ∗ +∗ e2 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1




2 ∗∗ = −∗ e8 ·∗ −∗ e8 ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ uξ1 ξ2




2 ∗∗ +∗ e2 ·∗ −∗ e8 ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ uξ2 ξ2




32 4 ∗∗ = e64 ·∗ u∗∗ ·∗ u∗∗ ξ 1 ξ1 − ∗ e ξ1 ξ2 +∗ e ·∗ uξ2 ξ2 ,

u∗∗ x1 x2

∗ ∗∗ ∗ = −∗ e8 ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2




∗ ∗∗ ∗ +∗ e2 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2




∗∗ = −∗ e8 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 ξ2




∗∗ +∗ e2 ·∗ u∗∗ ξ1 ξ2 +∗ uξ2 ξ2




6 ∗∗ 2 ∗∗ = −∗ e8 ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ uξ1 ξ2 +∗ e ·∗ uξ2 ξ2 ,

46

Classification and Canonical Forms u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= u∗ξ1 +∗ u∗ξ2 , u∗x2 x2

∗ ∗∗ ∗ ∗∗ ∗ ∗∗ ∗ = u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2 +∗ uξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2 ∗∗ ∗∗ ∗∗ = u∗∗ ξ1 ξ1 +∗ uξ1 ξ2 +∗ uξ1 ξ2 +∗ uξ2 ξ2 2 ∗∗ ∗∗ = u∗∗ ξ1 ξ1 +∗ e ·∗ uξ1 ξ2 +∗ uξ2 ξ2 .

Therefore, 0∗

6 ∗∗ 16 = u∗∗ ·∗ u∗∗ x1 x1 +∗ e ·∗ ux1 x2 −∗ e x2 x2 32 4 ∗∗ = e64 ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ 1 − ∗ e ξ1 ξ2 +∗ e ·∗ uξ2 ξ2 36 12 −∗ e48 ·∗ u∗∗ ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ 1 − ∗ e ξ 1 ξ2 + ∗ e ξ2 ξ2 32 16 = −∗ e16 ·∗ u∗∗ ·∗ u∗∗ ·∗ u∗ξ2 ξ2 ξ1 ξ 1 − ∗ e ξ1 ξ2 −∗ e

= −∗ e100 ·∗ u∗∗ ξ1 ξ2 , i.e., u∗∗ ξ1 ξ2 = 0∗

(3.16)

is the canonical form of the considered equation. 2. Fix ξ2 and set

v = u∗ξ2 .

Then equation (3.16) takes the form vξ∗1 = 0∗ , from where v = f1 (ξ2 ) and u∗ξ2 = f1 (ξ2 ). Hence, u = f (ξ2 ) +∗ g(ξ1 ), i.e., u(x1 , x2 ) = f (x2 +∗ e2 ·∗ x1 ) +∗ g(x2 −∗ e8 ·∗ x1 )

(3.17)

is the general solution of the considered equation, where f and g are C∗2 -functions.

Multiplicative Linear MPDEs

47

3. From equation (3.17), we get u(−∗ x1 , 2x1 )

= f (e2 ·∗ x1 −∗ e2 ·∗ x1 ) +∗ g(e2 ·∗ x1 +∗ e8 ·∗ x1 ) = f (0∗ ) +∗ g(e10 ·∗ x1 ),

u(x1 , 1)

= f (e2 ·∗ x1 ) +∗ g(−∗ e8 ·∗ x1 ).

In this way we get the system f (1) +∗ g(e10 ·∗ x1 ) = x1 (3.18) f (e2 ·∗ x1 ) +∗ g(−∗ e8 ·∗ x1 ) = e2 ·∗ x1 , whereupon g(e10 ·∗ x1 ) and

=

x1 −∗ f (0∗ )

1

g(z) = e 10 ·∗ z −∗ f (0∗ ). From here and from the second equation of (3.18), we find f (e2 ·∗ x1 ) = e2 ·∗ x1 −∗ g(−∗ e8 ·∗ x1 ) = e2 ·∗ x1 +∗ (e8 ·∗ x1 )/∗ e10 +∗ f (0∗ ) 14

= e 5 ·∗ x1 +∗ f (0∗ ), i.e.,

7

f (z) = e 5 ·∗ z +∗ f (0∗ ). Consequently, u(x1 , x2 )

=

7

e 5 ·∗ (x2 +∗ e2 ·∗ x1 ) +∗ f (0∗ ) +∗ (x2 −∗ e8 ·∗ x1 )/∗ e10 −∗ f (0∗ ) 3

= e 2 ·∗ x2 +∗ e2 ·∗ x1 . Exercise 3.6 Consider the equation 2 ∗∗ 3 ∗∗ u∗∗ x1 x1 +∗ e ·∗ ux1 x2 −∗ e ·∗ ux2 x2 = 0∗ .

1. Find the canonical form. 2. Find the general solution.

48

Classification and Canonical Forms 3. Find the solution u(x1 , x2 ) for which = e3 ·∗ x21∗ ,

u(x1 , 1) u∗x2 (x1 , 1) Answer

=

1.

1. u∗∗ ξ1 ξ 2

=

0∗ ,

ξ1

= x2 −∗ e3 ·∗ x1 ,

ξ2

= x 2 +∗ x 1 .

2. u(x1 , x2 ) = f (x2 −∗ e3 ·∗ x1 ) +∗ g(x2 +∗ x1 ), where f and g are C∗2 -functions. 3. u(x1 , x2 ) = e3 ·∗ x21∗ +∗ x22∗ . Exercise 3.7 Consider the equation ∗∗ ∗∗ x1 ·∗ u∗∗ x1 x1 +∗ (x1 +∗ x2 ) ·∗ ux1 x2 +∗ x2 ·∗ ux2 x2 = 0∗ ,

x1 > 0∗ , x2 > 0.

1. Find the canonical form. 2. Find the general solution. 3. Find a solution u(x1 , x2 ) for which   1 u x1 , = x31∗ , x1 u∗x1 (x1 , (e/∗ x1 )) Answer

= e2 ·∗ x21∗ .

1. ∗ −∗ ξ1 ·∗ u∗∗ ξ1 ξ2 +∗ uξ2

=

0∗ ,

ξ1

= x 2 −∗ x 1 ,

ξ2

= x2 / ∗ x1 .

2. u(x1 , x2 ) = (x2 −∗ x1 ) ·∗ f (x2 /∗ x1 ) +∗ g(x2 −∗ x1 ), where f and g are C∗2 -functions, 3. u(x1 , x2 ) = (x21∗ )/∗ x2 .

Multiplicative Linear MPDEs

3.1.2

49

The multiplicative elliptic case

We suppose that L(u) is multiplicative elliptic in U. In this case, we have a ·∗ c −∗ b2∗ > 0∗ . Also, α=γ and β = 0∗ . Then the multiplicative linear independent variables ξ1 = φ1 (x1 , x2 ) and ξ2 = φ2 (x1 , x2 ) satisfy the system ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2 ∗2∗ 2 ∗ ∗ ∗ = a ·∗ φ∗2 2x1 +∗ e ·∗ b ·∗ φ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2

a ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ b ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2




+∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 = 0∗ or



a ·∗ φ∗1x1 −∗ φ∗2x1 ·∗ φ∗1x1 +∗ φ∗2x1 +∗ e2 ·∗ b ·∗ φ∗1x1 ·∗ φ∗1x2 −∗ φ∗2x1 ·∗ φ∗2x2






+∗ c ·∗ φ∗1x2 −∗ φ∗2x2 ·∗ φ∗1x2 +∗ φ∗2x2 = 0∗ , a ·∗ φ∗1x1 φ∗2x1 +∗ b ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2

(3.19)


+∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 = 0∗ . Next, we rewrite equation (3.19) in the following form:

∗ 2∗ ∗ a ·∗ φ∗2 +∗ e2 ·∗ b ·∗ φ∗1x1 ·∗ φ∗1x2 −∗ φ∗2x1 ·∗ φ∗2x2 1x1 +∗ (i ·∗ φ2x1 )
∗2∗ ∗ +∗ c ·∗ φ∗2 = 0∗ 1x2 +∗ (i ·∗ φ2x2 ) e2 ·∗ i ·∗ a ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ e2 ·∗ i ·∗ b ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2 +∗ e2 ·∗ i ·∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 = 0∗




50

Classification and Canonical Forms

or a ·∗ φ∗1x1 +∗ i ·∗ φ∗2x1


∗2∗



+∗ e2 ·∗ b ·∗ φ∗1x1 +∗ i ·∗ φ∗2x1 ·∗ φ∗1x2 +∗ i ·∗ φ∗2x2

+∗ c ·∗ φ∗x2 +∗ i ·∗ φ∗2x2


2∗

= 0∗ . (3.20)

Let φ3

=

φ∗1x1 +∗ i ·∗ φ∗2x1 ,

φ4

=

φ∗1x2 +∗ i ·∗ φ∗2x2 .

Then, using equation (3.20), we have a ·∗ φ23∗ +∗ e2 ·∗ b ·∗ φ3 ·∗ φ4 +∗ c ·∗ φ24∗ = 0∗ and 2∗

a ·∗ (φ3 /∗ φ4 )

+∗ e2 ·∗ b(φ3 /∗ φ4 ) +∗ c = 0∗ .

Consequently, 1

(φ3 /∗ φ4 )1,2 = (−∗ b ±∗ i ·∗ (a ·∗ c −∗ b2∗ ) 2 ∗ )/∗ a or
φ3 = −∗ (b/∗ a) ±∗ i ·∗ ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ φ4 , or   1 φ∗1x1 +∗ i ·∗ φ∗2x1 = −∗ (b/∗ a) ±∗ i ·∗ ((a ·∗ c −∗ b2∗ ) 2 ∗ )/∗ a
·∗ φ∗1x2 +∗ i ·∗ φ∗2x2 . Let φ∗1x1



+∗ i ·∗ φ∗2x1 = −∗ (b/∗ a) +∗ i ·∗ ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ φ∗1x2 +∗ i ·∗ φ∗2x2 .

Then, φ∗1x1 +∗ i ·∗ φ∗2x1

=

−∗ (b/∗ a) ·∗ φ∗1x2 +∗ i ·∗ ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ φ∗1x2 −∗ (b/∗ a) ·∗ i ·∗ φ∗2x2 −∗ ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ φ∗2x2 .

From here, φ∗1x1

=

−∗ (b/∗ a) ·∗ φ∗1x2 −∗ ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ φ∗2x2 ,

φ∗2x1

=

((a ·∗ c −∗ b2∗ )/∗ a) ·∗ φ∗1x2 −∗ (b/∗ a) ·∗ φ∗2x2 .

Multiplicative Linear MPDEs

51

From the first equation of the last system, we get φ∗2x2 = −∗ ((a ·∗ φ∗1x1 +∗ b ·∗ φ∗1x2 )/∗ (a ·∗ c −∗ b2∗ )). Then, using its second equation, we find φ∗2x1

= ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ φ∗1x2   1 −∗ (b/∗ a) ·∗ −∗ ((a ·∗ φ∗1x1 +∗ b ·∗ φ∗1x2 )/∗ (a ·∗ c −∗ b2∗ ) 2 ∗ ) =

1

(((a ·∗ c −∗ b2∗ ) 2 ∗ )/∗ a) ·∗ φ∗1x2 +∗ ((·∗ a ·∗ b ·∗ φ∗1x1 +∗ b2∗ ·∗ φ∗1x2 )/∗ 1

(a ·∗ (a ·∗ c −∗ b2∗ ) 2 ∗ )) =

(a ·∗ c −∗ b2∗ +∗ b2∗ ·∗ φ∗1x2 +∗ a ·∗ b ·∗ φ∗1x1 )/∗ 1

(a ·∗ ((a ·∗ c −∗ b2∗ ) 2 ∗ )) 1

=

(a ·∗ c ·∗ φ∗1x2 +∗ a ·∗ b ·∗ φ1 x∗1 )/∗ (a ·∗ (a ·∗ c −∗ b2∗ ) 2 ∗ )

=

(b ·∗ φ∗1x1 +∗ c ·∗ φ∗1x2 )/∗ ((a ·∗ c −∗ b2∗ ) 2 ∗ ),

1

i.e., we obtain the system 1

φ∗2x1 = (b ·∗ φ∗1x1 +∗ c ·∗ φ∗1x2 )/∗ ((a ·∗ c −∗ b2∗ ) 2 ∗ ) 1

φ∗2x2 = −∗ ((a ·∗ φ∗1x1 +∗ b ·∗ φ∗1x2 )/∗ ((a ·∗ c −∗ b2∗ ) 2 ∗ )).

(3.21)

Definition 3.7 Equation (3.21) will be called multiplicative Beltrami differential equations. From these multiplicative Beltrami differential equations, by eliminating one of the unknowns, for instance, φ2 , we get   1 (∂∗ /∗ (∂∗ x1 )) ·∗ (a ·∗ φ∗1x1 +∗ b ·∗ φ∗1x2 )/∗ (a ·∗ c −∗ b2∗ ) 2 ∗   1 +∗ (∂∗ /∗ (∂∗ x2 )) ·∗ (b ·∗ φ∗1x1 +∗ c ·∗ φ∗1x2 )/∗ ((a ·∗ c −∗ b2∗ ) 2 ∗ ) = 0∗ . (3.22) Using equation (3.21), we find φ∗2x1 ·∗ φ∗1x2 −∗ φ∗1x1 ·∗ φ∗2x2 = ((b ·∗ φ∗1x1 +∗ c ·∗ φ∗1x2 ) 1

/∗ ((a ·∗ c −∗ b2∗ ) 2 ∗ ))/∗ (φ∗1x2 ) 1

+∗ ((a ·∗ φ∗1x1 +∗ b ·∗ φ∗1x2 )/∗ ((a ·∗ c −∗ b2∗ ) 2 ∗ )) ·∗ φ∗1x1 =


1 ∗2∗ 2 ∗ ∗ ∗ (e/∗ ((a ·∗ c −∗ b2∗ ) 2 ∗ )) ·∗ a ·∗ φ∗2 1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2 .

52

Classification and Canonical Forms

If we assume that ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2 = 0∗ ,

then from the first equation of (3.19), we get ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 2x1 +∗ e ·∗ b ·∗ φ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2 = 0∗ .

Therefore, (φ∗1x1 )/∗ (φ∗1x2 ) and (φ∗2x1 )/∗ (φ∗2x2 ) are the roots of the quadratic equation: a ·∗ ν 2∗ +∗ e2 ·∗ b ·∗ ν +∗ c = 0∗ . From here, (φ∗1x1 )/∗ (φ∗1x2 ) +∗ ((φ∗2x1 )/∗ (φ∗2x2 ))

= −∗ ((e2 ·∗ b)/∗ a),

((φ∗1x1 )/∗ (φ∗1x2 )) ·∗ ((φ∗2x1 )/∗ (φ∗2x2 ))

= c/∗ a.

From the last relations and from the second equation of (3.19), we obtain 0∗

=

a ·∗ ((φ∗1x1 )/∗ (φ∗1x2 )) ·∗ ((φ∗2x1 )/∗ (φ∗2x2 ))
+∗ b ·∗ ((φ∗1x1 )/∗ (φ∗2x1 )) +∗ ((φ∗1x2 )/∗ (φ∗2x2 )) +∗ c

=


a ·∗ (c/∗ a) +∗ b −∗ ((e2 ·∗ b)/∗ a) +∗ c

=

e2 ·∗ c −∗ ((e2 ·∗ b2∗ )/∗ a)

=

e2 ·∗ ((a ·∗ c −∗ b2∗ )/∗ a)

6=

0∗ .

Consequently, ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 1x1 +∗ e ·∗ b ·∗ φ1x1 ·∗ φ1x2 +∗ c ·∗ φ1x2 6= 0∗ , ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 2x1 +∗ e ·∗ b ·∗ φ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2 6= 0∗ ,

and φ∗2x1 ·∗ φ∗1x2 −∗ φ∗1x1 ·∗ φ∗2x2 6= 0.

(3.23)

In other words, the transformation of the differential equation to the canonical form
∗∗ α ·∗ u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 +∗ · · · = 0∗

Multiplicative Linear MPDEs

53

in a neighborhood of a point is given by any pair of functions satisfying equation (3.21) and having multiplicative nonvanishing multiplicative Jacobian (3.23). Such functions are determined once we have a solution of equation (3.22) with multiplicative nonvanishing multiplicative gradient. If a, b, c ∈ C∗2 (U ), such a solution always exists, at least locally, and hence the system φ1 , φ2 may be introduced in a neighborhood of any point. Definition 3.8 The multiplicative curves ξ1 = φ1 (x1 , x2 ) = const and ξ2 = φ2 (x1 , x2 ) = const are called the characteristic multiplicative curves of the multiplicative linear multiplicative elliptic differential operator L(u). Let φ = φ 1 +∗ i · ∗ φ 2 . Then, φ∗x1

= φ∗1x1 +∗ i ·∗ φ∗2x1 ,

φ∗x2

= φ∗1x2 +∗ i ·∗ φ∗2x2 .

Hence, from equation (3.20), we get 2 ∗ ∗ ∗2∗ ∗ a ·∗ φ∗2 x1 +∗ e ·∗ b ·∗ φx1 ·∗ φx2 +∗ c ·∗ φx2 = 0∗ ,

whereupon
2∗ a ·∗ (φ∗x1 )/∗ (φ∗x2 ) +∗ e2 ·∗ b ·∗ ((φ∗x1 )/∗ (φ∗x2 )) +∗ c = 0∗ and

1

(φ∗x1 )/∗ (φ∗x2 ) = (−∗ b ±∗ i ·∗ (a ·∗ c −∗ b2∗ ) 2 ∗ )/∗ a or

  1 a ·∗ φ∗x1 +∗ b ±∗ i ·∗ (a ·∗ c −∗ b2∗ ) 2 ∗ ·∗ φ∗x2 = 0∗ .

Therefore, 1

(d∗ x1 )/∗ a = (d∗ x2 )/∗ (b ±∗ i ·∗ (a ·∗ c −∗ b2∗ ) 2 ∗ ) or 1 (d∗ x2 )/∗ (d∗ x1 ) = (b ±∗ i ·∗ (a ·∗ c −∗ b2∗ ) 2 ∗ )/∗ a,

(3.24)

54

Classification and Canonical Forms

and a ·∗ (d∗ x2 )2∗ −∗ e2 ·∗ b ·∗ d∗ x1 ·∗ d∗ x2 +∗ c ·∗ (d∗ x1 )2∗  2∗ 1 = a ·∗ (b ±∗ i ·∗ (a ·∗ c −∗ b2∗ ) 2 ∗ )/∗ a ·∗ (d∗ x1 )2∗ 1

−∗ e2 ·∗ b ·∗ ((b ±∗ i ·∗ (a ·∗ c −∗ b2∗ ) 2 ∗ )/∗ a) ·∗ (d∗ x1 )2∗ +∗ c ·∗ (d∗ x1 )2∗ =

0∗ .

Definition 3.9 The equation a ·∗ (d∗ x2 )2∗ −∗ e2 ·∗ b ·∗ d∗ x1 ·∗ d∗ x2 +∗ c ·∗ (d∗ x1 )2∗ = 0∗ is called the characteristic equation of the multiplicative elliptic operator L(u). Note that the general multiplicative integrals of equation (3.24) are given by φ5 (x1 , x2 ) ±∗ iφ6 (x1 , x2 ) = const, where φ5 and φ6 are multiplicative real-valued functions. In practice, for convenience, we very often take ξ1

= φ5 (x1 , x2 ),

ξ2

= φ6 (x1 , x2 ).

Example 3.10 Consider the equation 2∗ 2 ∗∗ 2 ∗∗ x22∗ ·∗ u∗∗ x1 x1 +∗ e ·∗ x1 ·∗ x2 ·∗ ux1 x2 +∗ e ·∗ x1 ·∗ ux2 x2

+∗ x2 ·∗ u∗x2 = 0∗ ,

x1 , x2 6= 0∗ .

Here a(x1 , x2 )

=

x22∗ ,

b(x1 , x2 )

=

x1 ·∗ x2 ,

c(x1 , x2 )

=

e2 ·∗ x21∗ ,

x1 , x2 6= 0∗ .

Then, a ·∗ c −∗ b2∗

= e2 ·∗ x21∗ ·∗ x22∗ −∗ x21∗ ·∗ x22∗ = x21∗ ·∗ x22∗ > 0∗ .

Multiplicative Linear MPDEs

55

Therefore, the considered equation is a multiplicative elliptic equation. The characteristic equation is x22∗ ·∗ (d∗ x2 )2∗ −∗ e2 ·∗ x1 ·∗ x2 ·∗ d∗ x1 ·∗ d∗ x2 +∗ e2 ·∗ x21∗ ·∗ (d∗ x1 )2∗ = 0∗ , whereupon x22∗ ·∗ ((d∗ x2 )/∗ (d∗ x1 ))

2∗

−∗ e2 ·∗ x1 ·∗ x2 ·∗ ((d∗ x2 )/∗ (d∗ x1 )) +∗ e2 ·∗ x21∗ = 0∗ .

Hence, ((d∗ x2 )/∗ (d∗ x1 ))1,2 = ((x1 ·∗ x2 ±∗ i ·∗ x1 ·∗ x2 )/∗ (x22∗ )). Consider (d∗ x2 )/∗ (d∗ x1 ) = (x1 +∗ i ·∗ x1 )/∗ x2 . Then, x2 ·∗ d∗ x2 = (x1 +∗ i ·∗ x1 ) ·∗ d∗ x1 and x22∗ = x21∗ +∗ i ·∗ x21∗ +∗ c. We set ξ1

=

x21∗ −∗ x22∗ ,

ξ2

=

x21∗ .

Then, ∗ ξ1x 1

=

e2 ·∗ x1 ,

∗ ξ1x 2

=

−∗ e2 ·∗ x2 ,

∗ ξ2x 1

=

e2 ·∗ x1 ,

∗ ξ2x 2

= 0∗ ,

u∗x1

u∗∗ x1 x1

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

=

e2 ·∗ x1 ·∗ u∗ξ1 +∗ e2 ·∗ x1 ·∗ u∗ξ2 ,

=

∗ ∗∗ ∗ e2 ·∗ u∗ξ1 +∗ e2 ·∗ x1 ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1




∗ ∗∗ ∗ +∗ e2 ·∗ u∗ξ2 +∗ e2 ·∗ x1 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1




56

Classification and Canonical Forms = e2 ·∗ u∗ξ1 +∗ e2 ·∗ u∗ξ2 +∗ e2 ·∗ x1 ·∗ 2 ∗∗ e2 ·∗ x1 ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x1 ·∗ uξ1 ξ2




2 ∗∗ +∗ e2 ·∗ x1 ·∗ e2 ·∗ x1 ·∗ u∗∗ ξ1 ξ2 +∗ e ·∗ x1 ·∗ uξ2 ξ2




= e2 ·∗ u∗ξ1 +∗ e2 ·∗ u∗ξ2 +∗ e4 ·∗ x21∗ ·∗ u∗∗ ξ1 ξ 1 2∗ 4 ∗∗ +∗ e8 ·∗ x21∗ ·∗ u∗∗ ξ1 ξ2 +∗ e ·∗ x1 ·∗ uξ2 ξ2 ,

u∗∗ x1 x2

∗ ∗∗ ∗ = e2 ·∗ x1 ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2




∗ ∗∗ ∗ +∗ e2 ·∗ x1 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2






2 2 ∗∗ = e2 ·∗ x1 ·∗ −∗ e2 ·∗ x2 ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x1 ·∗ −∗ e ·∗ x2 ·∗ uξ1 ξ2 4 ∗∗ = −∗ e4 ·∗ x1 ·∗ x2 ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ x1 ·∗ x2 ·∗ uξ1 ξ2 ,

u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= −∗ e2 ·∗ x2 ·∗ u∗ξ1 , u∗∗ x2 x2

∗ ∗∗ ∗ = −∗ e2 ·∗ u∗ξ1 −∗ e2 ·∗ x2 ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2




= −∗ e2 ·∗ u∗ξ1 +∗ e4 ·∗ x22∗ ·∗ u∗∗ ξ1 ξ1 . Hence, 0∗

2∗ 2 ∗∗ 2 ∗∗ ∗ = x22∗ ·∗ u∗∗ x1 x1 +∗ e ·∗ x1 ·∗ x2 ·∗ ux1 x2 +∗ e ·∗ x1 ·∗ ux2 x2 +∗ x2 ·∗ ux2

 = x22∗ e2 ·∗ u∗ξ1 +∗ e2 ·∗ u∗ξ2 +∗ e4 ·∗ x21∗ ·∗ u∗∗ ξ1 ξ 1 +∗ e

8

·∗ x21∗ ·∗

u∗∗ ξ1 ξ2

+∗ e

4

·∗ x21∗ ·∗

u∗∗ ξ2 ξ 2



4 ∗∗ +∗ e2 ·∗ x1 ·∗ x2 ·∗ −∗ e4 ·∗ x1 ·∗ x2 ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ x1 ·∗ x2 ·∗ uξ1 ξ2

+∗ e2 ·∗ x21∗ ·∗ −∗ e2 ·∗ u∗ξ1 +∗ e4 ·∗ x22∗ ·∗ u∗∗ ξ1 ξ 1 +∗ x2 ·∗ −∗ e2 ·∗ x2 ·∗ u∗ξ1










Multiplicative Linear MPDEs

57

= −∗ e4 ·∗ x21∗ ·∗ u∗ξ1 +∗ e2 ·∗ x22∗ ·∗ u∗ξ2 2∗ 2∗ 4 ∗∗ +∗ e4 ·∗ x21∗ ·∗ x22∗ ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x1 ·∗ x2 ·∗ uξ2 ξ2 2∗ 2∗ ∗∗ ∗ 2 ∗ = u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 −∗ (e/∗ (x2 )) ·∗ uξ1 +∗ (e/∗ (e ·∗ x1 )) ·∗ uξ2 ∗∗ ∗ 2 ∗ = u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 −∗ (e/∗ (ξ2 −∗ ξ1 )) ·∗ uξ1 +∗ (e/∗ (e ·∗ ξ2 )) ·∗ uξ2

is the canonical form of the considered equation. Example 3.11 Consider the equation ∗∗ u∗∗ x1 x1 +∗ x1 ·∗ ux2 x2 = 0∗ ,

x1 > 0∗ .

Here a(x1 , x2 )

=

e,

b(x1 , x2 )

=

0∗ ,

c(x1 , x2 )

=

x1 ,

x1 > 0∗ .

Then, a(x1 , x2 ) ·∗ c(x1 , x2 ) −∗ (b(x1 , x2 )2∗

=

x1

>

0∗ ,

i.e., the considered equation is a multiplicative elliptic equation. The characteristic equation is (d∗ x2 )2∗ +∗ x1 ·∗ (d∗ x1 )2∗ = 0∗ and d∗ x2

1

= ±∗ i ·∗ x1 2 ∗ ·∗ d∗ x1 ,

and

3

2

x2 = ±∗ e 3 ·∗ i ·∗ x12 ∗ +∗ c. We set xi1

=

x2 ,

ξ2

=

e 3 ·∗ x12 ∗ .

2

3

58

Classification and Canonical Forms

Then, ∗ ξ1x 1

= 0∗ ,

∗ ξ1x 2

= e,

∗ ξ2x 1

= x12 ∗ ,

∗ ξ2x 2

= 0∗ ,

u∗x1

1

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1 1

= x12 ∗ ·∗ u∗ξ2 , u∗∗ x1 x1

1

1

1

∗ ∗∗ ∗ = e 2 ·∗ (e/∗ (x12 ∗ )) ·∗ u∗ξ2 +∗ x12 ∗ ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1

=

 1  1 1 (e/∗ (e2 ·∗ x12 ∗ )) ·∗ u∗ξ2 +∗ x12 ∗ ·∗ x12 ∗ ·∗ u∗∗ ξ2 ξ 2

=

(e/∗ (e2 ·∗ x12 ∗ )) ·∗ u∗ξ2 +∗ x1 ·∗ u∗∗ ξ2 ξ 2 ,

1

u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= u∗ξ1 , u∗∗ x2 x2

∗ ∗∗ ∗ = u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2

= u∗∗ ξ1 ξ1 . Hence, 0∗

∗∗ = u∗∗ x1 x1 +∗ x1 ·∗ ux2 x2 1

=

∗∗ (e/∗ (e2 ·∗ x12 ∗ ) ·∗ u∗ξ2 +∗ x1 ·∗ u∗∗ ξ2 ξ2 +∗ x1 ·∗ uξ1 ξ1

=

u∗∗ ξ 1 ξ1

+∗ u∗∗ ξ 2 ξ2

 +∗



2

3 2∗

e/∗ e ·∗ x1



·∗ u∗ξ2

∗∗ 3 ∗ = u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 +∗ (e/∗ (e ·∗ ξ2 )) ·∗ uξ2

is the canonical form of the considered equation.




Multiplicative Linear MPDEs

59

Example 3.12 Consider the equation 2∗ ∗∗ ∗ ∗ (e +∗ x21∗ ) ·∗ u∗∗ x1 x1 +∗ (e +∗ x2 ) ·∗ ux2 x2 +∗ x1 ·∗ ux1 +∗ x2 ·∗ ux2 = 0∗ .

Here a(x1 , x2 )

= e +∗ x21∗ ,

b(x1 , x2 )

=

c(x1 , x2 )

= e +∗ x22∗ .

0∗ ,

Then, a(x1 , x2 ) ·∗ c(x1 , x2 ) −∗ (b(x1 , x2 ))2∗

=

(e +∗ x21∗ )(e +∗ x22∗ )

>

0∗ ,

i.e., the considered equation is a multiplicative elliptic equation. The characteristic equation is (e +∗ x21∗ ) ·∗ (d∗ x2 )2∗ +∗ (e +∗ x22∗ ) ·∗ (d∗ x1 )2∗ = 0∗ or

1

1

(e +∗ x21∗ ) 2 ∗ ·∗ d∗ x2 = ±∗ i ·∗ (e +∗ x22∗ ) 2 ∗ ·∗ d∗ x1 , or

1

1

((d∗ x2 )0 /∗ (e +∗ x22∗ ) 2 ∗ ) = ±∗ i ·∗ ((d∗ x1 )/∗ (e +∗ x21∗ ) 2 ∗ ) or   1 1 log∗ x2 +∗ (e +∗ x22∗ ) 2 ∗ = ±∗ i ·∗ log∗ (x1 +∗ (e +∗ x21∗ ) 2 ∗ ) +∗ c. We set ξ1

=

  1 log∗ x2 +∗ (e +∗ x22∗ ) 2 ∗ ,

ξ2

=

log∗ (x1 +∗ (e +∗ x21∗ ) 2 ∗ ).

1

Then, ∗ ξ1x 1

=

0∗ ,

∗ ξ1x 2

=

(e/∗ (e +∗ x22∗ ) 2 ∗ ),

∗ ξ2x 1

=

(e/∗ (e +∗ x21∗ ) 2 ∗ ,

1

1

60

Classification and Canonical Forms ∗ ξ2x 2

u∗x1

= 0∗ , ∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1 1

= (e/∗ (e +∗ x21∗ ) 2 ∗ ) ·∗ u∗ξ2 , u∗∗ x1 x1

3

1

= −∗ (x1 /∗ (e +∗ x21∗ ) 2 ∗ ) ·∗ u∗ξ2 +∗ (e/∗ (e +∗ x21∗ ) 2 ∗ ) ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1




3

= −∗ (x1 /∗ (e +∗ x21∗ ) 2 ∗ ) ·∗ u∗ξ2 +∗ (e/∗ (e +∗ x21∗ )) ·∗ u∗∗ ξ 2 ξ2 , u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= u∗∗ x2 x2

1

(e/∗ (e +∗ x22∗ ) 2 ∗ ) ·∗ u∗ξ1 , 3

1

= −∗ (x2 /∗ (e +∗ x22∗ ) 2 ∗ ) ·∗ u∗ξ1 +∗ (e/∗ (e +∗ x22∗ ) 2 ∗ ) ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2




3

= −∗ (x2 /∗ (e +∗ x22∗ ) 2 ∗ ) ·∗ u∗ξ1 +∗ (e/∗ (e +∗ x22∗ )) ·∗ u∗∗ ξ 1 ξ1 . Hence, 2∗ ∗∗ ∗ ∗ 0∗ = (e +∗ x21∗ ) ·∗ u∗∗ x1 x1 +∗ (e +∗ x2 ) ·∗ ux2 x2 +∗ x1 ·∗ ux1 +∗ x2 ·∗ ux2

  3 = (e +∗ x21∗ ) ·∗ −∗ (x1 /∗ (e +∗ x21∗ ) 2 ∗ ) ·∗ u∗ξ2 +∗ (e/∗ (e +∗ x21∗ )) ·∗ u∗∗ ξ 2 ξ2   3 +∗ (e +∗ x22∗ ) ·∗ −∗ (x2 /∗ (e +∗ x21∗ ) 2 ∗ ) ·∗ u∗ξ1 +∗ (e/∗ (e +∗ x22∗ )) ·∗ u∗∗ ξ 1 ξ1 1

1

+∗ (x1 /∗ (e +∗ x21∗ ) 2 ∗ ) ·∗ u∗ξ2 +∗ (x2 /∗ (e +∗ x22∗ ) 2 ∗ ) ·∗ u∗ξ1 ∗∗ = u∗∗ ξ1 ξ1 +∗ uξ2 ξ2

is the canonical form of the considered equation.

Multiplicative Linear MPDEs

61

Exercise 3.8 Find the canonical form of the equation ∗∗ x2 ·∗ u∗∗ x1 x1 +∗ x1 ·∗ ux2 x2 = 0∗ ,

x1 , x2 > 0∗ .

Answer 3

ξ1

= x22 ∗ ,

ξ2

= x12 ∗ ,

3

and ∗∗ 3 ∗ 3 ∗ u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 +∗ (e/∗ (e ·∗ ξ1 )) ·∗ uξ1 +∗ (e/∗ (e ·∗ ξ2 )) ·∗ uξ2 = 0∗ .

3.1.3

The parabolic case

In this case, we have a ·∗ c −∗ b2∗ = 0∗ and β = γ = 0∗ . Then,
a ·∗ φ∗1x1 ·∗ φ∗2x1 +∗ b ·∗ φ∗1x2 ·∗ φ∗2x1 +∗ φ∗1x1 ·∗ φ∗2x2 +∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 = 0∗ ∗2∗ 2 ∗ ∗ ∗ a ·∗ φ∗2 2x1 +∗ e ·∗ bφ2x1 ·∗ φ2x2 +∗ c ·∗ φ2x2 = 0∗ . (3.25)

Let λ1 = ((φ∗2x1 )/∗ (φ∗2x2 )). Then, from the second equation of (3.25), we get a ·∗ λ21∗ +∗ e2 ·∗ b ·∗ λ1 +∗ c = 0∗ , whereupon (λ1 )1,2

=

1

(−∗ b ±∗ (b2∗ −∗ a ·∗ c) 2 ∗ )/∗ a

= −∗ (b/∗ a), or ((φ∗2x1 )/∗ (φ∗2x2 )) = −∗ (b/∗ a), or a ·∗ φ∗2x1 +∗ b ·∗ φ∗2x2 = 0∗ .

(3.26)

Hence, from the first equation of (3.25), we find

a ·∗ φ∗1x1 ·∗ −∗ (b/∗ a) ·∗ φ∗2x2 +∗ b ·∗ φ∗1x2 ·∗ −∗ (b/∗ a) ·∗ φ∗2x2
+∗ φ∗1x1 ·∗ φ∗2x2 +∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 = 0∗ ,

62

Classification and Canonical Forms

or −∗ b ·∗ φ∗1x1 ·∗ φ∗2x2 −∗ ((b2∗ )/∗ a) ·∗ φ∗1x2 ·∗ φ∗2x2 +∗ b ·∗ φ∗1x1 ·∗ φ∗2x2 +∗ c ·∗ φ∗1x2 ·∗ φ∗2x2 = 0∗ , or ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ φ∗1x2 ·∗ φ∗2x2 = 0∗ . Therefore, the function φ2 can be determined by equation (3.26) and the function φ1 is arbitrarily chosen C∗1 -function in U. Equation (3.26) gives a family of solutions of the multiplicative ordinary differential equation: ((d∗ x2 )/∗ (d∗ x1 )) = b/∗ a,

(3.27)

where x2 is considered as a function of x1 along the multiplicative curves of the family. Definition 3.10 The curves ξ1

= φ1 (x1 , x2 ) = const,

ξ2

= φ2 (x1 , x2 ) = const

are called the characteristic multiplicative curves of the multiplicative linear multiplicative parabolic operator L(u). Using equation (3.27), we get a ·∗ ((d∗ x2 )/∗ (d∗ x1 ))

2∗

−∗ e2 ·∗ b ·∗ ((d∗ x2 )/∗ (d∗ x1 )) +∗ c

= (b∗ )/∗ a +∗ e2 ·∗ ((b2∗ )/∗ a) +∗ c = −∗ ((b2∗ )/∗ a) +∗ c = 0∗ . Definition 3.11 The equation a ·∗ (d∗ x2 )2∗ −∗ e2 ·∗ b ·∗ d∗ x1 ·∗ d∗ x2 +∗ c ·∗ (d∗ x1 )2∗ = 0∗ is called the characteristic equation of the multiplicative parabolic operator L(u). Remark 3.3 If φ(x1 , x2 ) = const is the general solution of equation (3.27); in practice, we take ξ1

=

φ(x1 , x2 ),

ξ2

=

x1 ,

Multiplicative Linear MPDEs

63

or ξ2

= x2 ,

ξ1

= x1

or

ξ1

= x2

and

ξ2

= φ(x1 , x2 ).

or

Example 3.13 Consider the equation 2∗ 2 ∗∗ ∗∗ ∗ x21∗ ·∗ u∗∗ x1 x1 −∗ e ·∗ x1 ·∗ x2 ·∗ ux1 x2 +∗ x2 ·∗ ux2 x2 +∗ x1 ·∗ ux1

+∗ x2 ·∗ u∗x2 = 0∗ ,

x1 , x2 > 0∗ .

Here a(x1 , x2 )

= x21∗ ,

b(x1 , x2 )

= −∗ x 1 · ∗ x 2 ,

c(x1 , x2 )

=

x22∗ ,

x1 , x2 > 0∗ .

Then, a(x1 , x2 ) ·∗ c(x1 , x2 ) −∗ (b(x1 , x2 ))2∗

= x21∗ ·∗ x22∗ −∗ x21∗ ·∗ x22∗ =

0∗ .

Therefore, the considered equation is multiplicative parabolic. The characteristic equation is x21∗ ·∗ (d∗ x2 )2∗ +∗ e2 ·∗ x1 ·∗ x2 ·∗ d∗ x1 ·∗ d∗ x2 +∗ x22∗ ·∗ (d∗ x1 )2∗ = 0∗ , whereupon (x1 ·∗ d∗ x2 +∗ x2 ·∗ d∗ x1 )

2∗

= 0∗

and x1 ·∗ d∗ x2 = −∗ x2 ·∗ d∗ x1 , and (d∗ x2 )/∗ x2 = −∗ ((d∗ x1 )/∗ x1 ), and x1 ·∗ x2 = c.

64

Classification and Canonical Forms We set ξ1

= x1 ,

ξ2

= x1 ·∗ x2 .

Then, u∗x1

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

= u∗ξ1 +∗ x2 ·∗ u∗ξ2 , u∗∗ x1 x1

∗ ∗∗ ∗ = u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1 ∗ ∗∗ ∗ +∗ x2 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1




∗∗ ∗∗ ∗∗ = u∗∗ ξ1 ξ1 +∗ x2 ·∗ uξ1 ξ2 +∗ x2 ·∗ uξ1 ξ2 +∗ x2 ·∗ uξ2 ξ2

= u∗∗ x1 x2




2∗ 2 ∗∗ ∗∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x2 ·∗ uξ1 ξ2 +∗ x2 ·∗ uξ2 ξ2 ,

∗ ∗∗ ∗ ∗ = u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2 +∗ uξ2 ∗ ∗∗ ∗ +∗ x2 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2




∗ ∗∗ = x1 ·∗ u∗∗ ξ1 ξ2 +∗ uξ2 +∗ x1 ·∗ x2 ·∗ uξ2 ξ2 ,

u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= x1 ·∗ u∗ξ2 , u∗∗ x2 x2

∗ ∗∗ ∗ = x1 ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2




= x21∗ ·∗ u∗∗ ξ2 ξ2 . Hence, 0∗

=

2∗ 2 ∗∗ ∗∗ ∗ x21∗ ·∗ u∗∗ x1 x1 −∗ e ·∗ x1 ·∗ x2 ·∗ ux1 x2 +∗ x2 ·∗ ux2 x2 +∗ x1 ·∗ ux1

+∗ x2 ·∗ u∗x2

Multiplicative Linear MPDEs

=

65

2∗ 2 ∗∗ ∗∗ x21∗ ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x2 ·∗ uξ1 ξ2 +∗ x2 ·∗ uξ2 ξ2




∗∗ ∗ −∗ e2 ·∗ x1 ·∗ x2 ·∗ x1 ·∗ u∗∗ ξ1 ξ2 +∗ x1 ·∗ x2 ·∗ uξ2 ξ2 +∗ uξ2





∗ ∗ ∗ +∗ x21∗ ·∗ x22∗ ·∗ u∗∗ ξ2 ξ2 +∗ x1 ·∗ uξ1 +∗ x2 ·∗ uξ2 +∗ x1 ·∗ x2 ·∗ uξ2 =

∗ ξ1 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1

is the canonical form of the considered equation. Hence,
∗ ξ1 ·∗ u∗ξ1 ξ = 0∗ 1

and ξ1 ·∗ u∗ξ1 = f (ξ2 ); u∗ξ1 = (e/∗ ξ1 ) ·∗ f (ξ2 ) and u = log∗ (ξ1 ) ·∗ f (ξ2 ) +∗ g(ξ2 ); and u(x1 , x2 ) = log∗ (x1 ) ·∗ f (x1 ·∗ x2 ) +∗ g(x1 ·∗ x2 ) is the general solution of the considered equation, where f and g are C∗2 functions. Example 3.14 Consider the equation 2 ∗∗ ∗∗ 2 ∗ ∗ x21∗ ·∗ u∗∗ x1 x1 −∗ e ·∗ x1 ·∗ ux1 x2 +∗ ux2 x2 +∗ e ·∗ ux2 −∗ x1 ·∗ ux1 = 0∗ .

Here a(x1 , x2 )

= x21∗ ,

b(x1 , x2 )

= −∗ x 1 ,

c(x1 , x2 )

= e.

Then, a(x1 , x2 ) ·∗ c −∗ (b(x1 , x2 ))2∗

= x21∗ −∗ x21∗ =

0∗ .

Therefore, the considered equation is multiplicative parabolic. The characteristic equation is x21∗ ·∗ (d∗ x2 )2 +∗ e2 ·∗ x1 ·∗ d∗ x1 ·∗ d∗ x2 +∗ (d∗ x1 )2∗ = 0∗ ,

66

Classification and Canonical Forms

whereupon x1 ·∗ d∗ x2 +∗ d∗ x1 = 0∗ and x1 ·∗ ex2 = c. We set ξ1

=

x1 ·∗ ex2 ,

ξ2

=

x2 .

Then, ∗ ξ1x = ex2 , 1 ∗ ξ1x = x1 ·∗ ex2 , 2 ∗ ξ2x = 0∗ , 1 ∗ ξ2x = e, 2 ∗ ∗ u∗x1 = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

= ex2 ·∗ u∗ξ1 , x2 ∗ ∗∗ ∗ u∗∗ ·∗ u∗∗ x1 x1 = e ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1

= ee

2

·∗ x2




·∗ u∗∗ ξ1 ξ 1 ,

x2 ∗ ∗∗ ∗ u∗∗ ·∗ u∗ξ1 +∗ ex2 ·∗ u∗∗ x1 x2 = e ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2

∗∗ = ex2 ·∗ u∗ξ1 +∗ ex2 ·∗ x1 ·∗ ex2 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 ξ2

= ex2 ·∗ u∗ξ1 +∗ x1 ·∗ ee

2

·∗ x2

∗ ∗ u∗x2 = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= x1 ·∗ ex2 ·∗ u∗ξ1 +∗ u∗ξ2 ,







x2 ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ 1 + ∗ e ξ1 ξ2 ,

Multiplicative Linear MPDEs

67

x2 ∗ ∗∗ ∗ u∗∗ ·∗ u∗ξ1 +∗ x1 ·∗ ex2 ·∗ u∗∗ x2 x2 = x1 ·∗ e ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2




∗ ∗∗ ∗ +∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2

∗∗ = x1 ·∗ ex2 ·∗ u∗ξ1 +∗ x1 ·∗ ex2 ·∗ x1 ·∗ ex2 ·∗ u∗∗ ξ1 ξ 1 + ∗ u ξ1 ξ 2




∗∗ +∗ x1 ·∗ ex2 ·∗ u∗∗ ξ1 ξ2 +∗ uξ2 ξ2

= x1 ·∗ ex2 ·∗ u∗ξ1 +∗ x21∗ ·∗ ee

2

·∗ x2

2 x2 ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x1 ·∗ e ξ1 ξ2

+∗ u∗∗ ξ2 ξ2 . Hence, 2 ∗∗ ∗∗ 2 ∗ ∗ 0∗ = x21∗ ·∗ u∗∗ x1 x1 −∗ e ·∗ x1 ·∗ ux1 x2 +∗ ux2 x2 +∗ e ·∗ ux2 −∗ x1 ·∗ ux1

= x21∗ ·∗ ee 

2

·∗ x2

2 ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ x1 ·∗

ex2 ·∗ u∗ξ1 +∗ x1 ·∗ ee

2

·∗ x2

x2 ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ1 + ∗ e ξ1 ξ2

+∗ x1 ·∗ ex2 ·∗ u∗ξ1 +∗ x21∗ ·∗ ee

2

·∗ x2



·∗ u∗∗ ξ1 ξ 1

∗∗ +∗ e2 ·∗ x1 ·∗ ex2 ·∗ u∗∗ ξ1 ξ2 +∗ uξ2 ξ2

+∗ e2 ·∗ x1 ·∗ ex2 ·∗ u∗ξ1 +∗ e2 ·∗ u∗ξ2 −∗ x1 ·∗ ex2 ·∗ u∗ξ1 2 ∗ = u∗∗ ξ2 ξ2 +∗ e ·∗ uξ2

is the canonical form of the considered equation. Let v = u∗ξ2 . Then, vξ∗2 = −∗ e2 ·∗ v and v = e −∗ e

2

·∗ ξ2

and u∗ξ2 = e−∗ e

2

·∗ f (ξ1 ),

·∗ ξ2

·∗ f1 (ξ1 )

68

Classification and Canonical Forms

and u = f (ξ1 ) ·∗ e−∗ e

2

·∗ ξ2

+∗ g(ξ1 ),

i.e., u(x1 , x2 ) = e−∗ e

2

·∗ x2

·∗ f (x1 ·∗ ex2 ) +∗ g (x1 ·∗ ex2 )

is the general solution of the considered equation, where f and g are C∗2 functions. Example 3.15 Consider the equation 2 ∗∗ 2∗ u∗∗ ·∗ u∗∗ x1 x1 −∗ e ·∗ sin∗ x1 ux1 x2 +∗ (sin∗ x1 ) x2 x2 = 0∗ .

Here a(x1 , x2 )

= e,

b(x1 , x2 )

= −∗ sin∗ x1 ,

c(x1 , x2 )

=

(sin∗ x1 )2∗ .

Then, a(x1 , x2 ) ·∗ c(x1 , x2 ) −∗ (b(x1 , x2 ))2∗

=

(sin∗ x1 )2∗ −∗ (sin∗ x1 )2∗

=

0∗ .

Therefore, the considered equation is multiplicative parabolic. The characteristic equation is (d∗ x2 )2∗ +∗ 2 sin∗ x1 ·∗ d∗ x1 ·∗ d∗ x2 +∗ (sin∗ x1 )2∗ ·∗ (d∗ x1 )2∗ = 0∗ , whereupon 2

(d∗ x2 +∗ sin∗ x1 ·∗ d∗ x1 ) = 0∗ and d∗ x2 +∗ sin∗ x1 ·∗ d∗ x1 = 0∗ , and x2 −∗ cos∗ x1 = c. We set ξ1

= x2 −∗ cos∗ x1 ,

ξ2

= x1 .

Multiplicative Linear MPDEs

69

Then, ∗ ξ1x 1

= sin∗ x1 ,

∗ ξ1x 2

=

e,

∗ ξ2x 1

=

e,

∗ ξ2x 2

= 0∗ ,

u∗x1

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

= sin∗ x1 ·∗ u∗ξ1 +∗ u∗ξ2 , u∗x1 x1

∗ ∗∗ ∗ = cos∗ x1 ·∗ u∗ξ1 +∗ sin∗ x1 ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1




∗ ∗∗ ∗ +∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1 ∗∗ = cos∗ x1 ·∗ u∗ξ1 +∗ sin∗ x1 ·∗ sin∗ x1 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 ξ2




∗∗ +∗ sin∗ x1 ·∗ u∗∗ ξ1 ξ2 +∗ uξ2 ξ2 2 ∗∗ = cos∗ x1 ·∗ u∗ξ1 +∗ (sin∗ x1 )2∗ ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ sin∗ x1 ·∗ uξ1 ξ2

+∗ u∗∗ ξ2 ξ 2 , u∗∗ x1 x2

∗ ∗∗ ∗ = sin∗ x1 ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2 ∗ ∗∗ ∗ +∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2 ∗∗ = sin∗ x1 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 ξ2 ,

u∗x2

u∗∗ x2 x2

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

=

u∗ξ1 ,

=

∗ ∗∗ ∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2

=

u∗∗ ξ1 ξ1 .




70

Classification and Canonical Forms

Hence, 0∗

=

2 ∗∗ 2∗ u∗∗ ·∗ u∗∗ x1 x1 −∗ e ·∗ sin∗ x1 ·∗ ux1 x2 +∗ (sin∗ x1 ) x2 x2

2 ∗∗ ∗∗ = cos∗ x1 u∗ξ1 +∗ (sin∗ x1 )2∗ ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ sin∗ x1 ·∗ uξ1 ξ2 +∗ uξ2 ξ2


∗∗ 2∗ −∗ e2 ·∗ sin∗ x1 ·∗ sin∗ x1 ·∗ u∗∗ ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 ξ2 +∗ (sin∗ x1 ) ξ 1 ξ1 = cos∗ x1 ·∗ u∗ξ1 +∗ u∗∗ ξ2 ξ2 = cos∗ (ξ2 ) ·∗ u∗ξ1 +∗ u∗∗ ξ 2 ξ2 is the canonical form of the considered equation. Exercise 3.9 Find the canonical form of the equation 9 ∗∗ ∗ ∗ u∗x1 x1 +∗ e6 ·∗ u∗∗ x1 x2 +∗ e ·∗ ux2 x2 +∗ ux1 +∗ ux2 = 0∗ .

Answer ξ1

= x1 ,

ξ2

= x2 −∗ e3 ·∗ x1 ,

and ∗ 2 ∗ u∗∗ ξ1 ξ1 +∗ uξ1 −∗ e ·∗ uξ2 = 0∗ .

Exercise 3.10 Consider the equation 4 ∗∗ ∗∗ 2 ∗ e4 ·∗ x22∗ ·∗ u∗∗ x1 x1 +∗ e ·∗ x2 ·∗ ux1 x2 +∗ ux2 x2 +∗ e ·∗ ux1 = 0∗ .

1. Find the canonical form. 2. Find the general solution. 3. Find the solution u(x1 , x2 ) for which u(x1 , 1) u∗x2 (x1 , 1) Solution

= e9 ·∗ sin∗ x1 , = ex1 .

1. Here a(x1 , x2 )

=

e4 ·∗ x22∗ ,

b(x1 , x2 )

=

e2 ·∗ x2 ,

c(x1 , x2 )

=

e.

Multiplicative Linear MPDEs

71

Then, (b(x1 , x2 ))2∗ −∗ a(x1 , x2 ) ·∗ c(x1 , x2 )

=

e4 ·∗ x22∗ −∗ e4 ·∗ x22∗

=

0∗ .

Therefore, the considered equation is multiplicative parabolic. The characteristic equation is e4 ·∗ x22∗ ·∗ (d∗ x2 )2∗ −∗ e4 ·∗ x2 ·∗ d∗ x1 ·∗ d∗ x2 +∗ (d∗ x1 )2∗ = 0∗ , whereupon (e2 ·∗ x2 ·∗ d∗ x2 −∗ d∗ x1 )2∗ = 0∗ and x2 ·∗ d∗ x2 −∗ d∗ x1 = 0∗ and x22∗ −∗ x1 = c. We set ξ1

=

x22∗ −∗ x1 ,

ξ2

=

x2 .

Then, ∗ ξ1x 1

= −∗ e,

∗ ξ1x 2

= e2 ·∗ x2 ,

∗ ξ2x 1

= 1,

∗ ξ2x 2

= e,

u∗x1

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

= −∗ u∗ξ1 , u∗x1 x1

∗ ∗∗ ∗ = −∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1

= u∗∗ ξ1 ξ1 ,




72 u∗∗ x1 x2

∗ = −∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2

Classification and Canonical Forms
∗ +∗ u∗∗ ξ1 ξ2 ·∗ ξ2x2

∗∗ = −∗ e2 ·∗ x2 ·∗ u∗∗ ξ1 ξ1 −∗ uξ1 ξ2 ,

u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= e2 ·∗ x2 ·∗ u∗ξ1 +∗ u∗ξ2 , u∗∗ x2 x2

∗ ∗∗ ∗ = e2 ·∗ u∗ξ1 +∗ e2 ·∗ x2 ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2




∗ ∗∗ ∗ +∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2 ∗∗ = e2 ·∗ u∗ξ1 +∗ e2 ·∗ x2 ·∗ e2 ·∗ x2 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 ξ2




∗∗ +∗ e2 ·∗ x2 ·∗ u∗∗ ξ1 ξ2 +∗ uξ2 ξ2 4 ∗∗ ∗∗ = e2 ·∗ u∗ξ1 +∗ e4 ·∗ x22∗ ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x2 ·∗ uξ1 ξ2 +∗ uξ2 ξ2 .

Hence, 0∗

4 ∗∗ ∗∗ 2 ∗ = e4 ·∗ x22∗ ·∗ u∗∗ x1 x1 +∗ e ·∗ x2 ·∗ ux1 x2 +∗ ux2 x2 +∗ e ·∗ ux1 4 2 ∗∗ ∗∗ = e4 ·∗ x22∗ ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ x2 ·∗ −∗ e ·∗ x2 ·∗ uξ1 ξ1 −∗ uξ1 ξ2

+∗ e2 ·∗ u∗ξ1 +∗ e4 ·∗ x22∗ ·∗ u∗∗ ξ1 ξ 1 ∗∗ 2 ∗∗ +∗ e4 ·∗ x2 ·∗ u∗∗ ξ1 ξ2 +∗ uξ2 ξ2 −∗ e ·∗ uξ1

= u∗ξ2 ξ2 is the canonical form of the considered equation. 2. Let v = u∗ξ2 . Then, vξ2 = 0∗ and v = f (ξ1 ); u∗ξ2 = f (ξ1 ), and u = ξ2 ·∗ f (ξ1 ) +∗ g(ξ1 ); and u(x1 , x2 ) = x2 ·∗ f (x22∗ −∗ x1 ) +∗ g(x22∗ −∗ x1 )




Classification and Canonical Form of Multiplicative Quasilinear MPDE

73

is the general solution of the considered equation, where f and g are C∗2 -functions. 3. We have u(x1 , 1) u∗x2 (x1 , x2 )

=

g(−∗ x1 ),

= f (x22∗ −∗ x1 ) +∗ e2 ·∗ x22∗ ·∗ f ∗ (x2 −∗ x1 ) +∗ e2 ·∗ x2 ·∗ g ∗ ·∗ (x22∗ −∗ x1 ),

u∗x2 (x1 , 1)

=

f (−∗ x1 ).

In this way we obtain the system f (−∗ x1 )

=

ex1

g(−∗ x1 )

=

e9 ·∗ sin∗ x1 ,

whereupon f (x1 )

= e−∗ x1

g(x1 )

= −∗ e9 ·∗ sin∗ x1 .

Consequently,
2∗ u(x1 , x2 ) = x2 ·∗ ex1 −∗ x2 −∗ e9 ·∗ sin∗ x22∗ −∗ x1 .

3.2

Classification and Canonical Form of Multiplicative Quasilinear Second-Order Multiplicative Partial Differential Equations in Two Independent Variables

Here we consider the equation 2 ∗∗ ∗∗ L(u) = a ·∗ u∗∗ x1 x1 +∗ e ·∗ b ·∗ ux1 x2 +∗ c ·∗ ux2 x2 +∗ d = 0∗

where a

= a(x1 , x2 , u, u∗x1 , u∗x2 ),

b = b(x1 , x2 , u, u∗x1 , u∗x2 ), c = c(x1 , x2 , u, u∗x1 , u∗x2 ), d = d(x1 , x2 , u, u∗x1 , u∗x2 ) are given functions and u is unknown.

in U,

(3.28)

74

Classification and Canonical Forms

Definition 3.12 The operator L(u) will be called as follows: 1. Multiplicative hyperbolic at the point (x01 , x02 ) ∈ U , if a ·∗ c −∗ b2∗ < 0∗ . 2. Multiplicative elliptic at the point (x01 , x02 ) ∈ U , if a ·∗ c −∗ b2∗ > 0∗ . 3. Multiplicative parabolic at the point (x01 , x02 ) ∈ U , if a ·∗ c −∗ b2∗ = 0∗ .

Definition 3.13 The operator L(u) will be called as follows: 1. Multiplicative hyperbolic in U , if it is multiplicative hyperbolic at any point of U . 2. Multiplicative elliptic in U , if it is multiplicative elliptic at any point of U . 3. Multiplicative parabolic in U , if it is multiplicative parabolic at any point of U . However, since a, b, c and d depend on the unknown u and its derivatives u∗x1 , u∗x2 , then the character of the operator L(u) at a point (x1 , x2 ) also depends on u, u∗x1 , u∗x2 . We introduce the characteristic parameters: ξ1

= φ1 (x1 , x2 ),

ξ2

= φ2 (x1 , x2 ).

We consider u, x1 and x2 as functions of ξ1 and ξ2 : u = u(ξ1 , ξ2 ), x1

= x1 (ξ1 , ξ2 ),

x2

= x2 (ξ1 , ξ2 ).

Now we multiplicative differentiate with respect to x1 and x2 the second and the third equations of the last system and we get e

∗ ∗ = x∗1ξ1 ·∗ ξ1x +∗ x∗1ξ2 ·∗ ξ2x , 1 1

1

∗ ∗ = x∗1ξ1 ·∗ ξ1x +∗ x∗1ξ2 ·∗ ξ2x , 2 2

Classification and Canonical Form of Multiplicative Quasilinear MPDE 1 = e

75

∗ ∗ x∗2ξ1 ·∗ ξ1x +∗ x∗2ξ2 ·∗ ξ2x , 1 1

∗ ∗ = x∗2ξ1 ·∗ ξ1x +∗ x∗2ξ2 ·∗ ξ2x . 2 2

Hence, setting D = e/∗ (x∗1ξ1 ·∗ x∗2ξ2 −∗ x∗1ξ2 ·∗ x∗2ξ1 ), we go to the system ∗ ξ1x = x∗2ξ2 ·∗ D, 1 ∗ ξ1x = −∗ x∗1ξ2 ·∗ D, 2

(3.29) ∗ ξ2x = −∗ x∗2ξ1 ·∗ D, 1 ∗ ξ2x = x∗1ξ1 ·∗ D. 2

Next, we multiplicative differentiate u with respect to ξ1 and ξ2 and we get u∗ξ1 = u∗x1 ·∗ x∗1ξ1 +∗ u∗x2 ·∗ x∗2ξ1 , (3.30) u∗ξ2 = u∗x1 ·∗ x∗1ξ2 +∗ u∗x2 ·∗ x∗2ξ2 , whereupon u∗x1 = u∗ξ1 ·∗ x∗2ξ2 ·∗ D −∗ u∗ξ2 ·∗ x∗2ξ1 ·∗ D, (3.31) u∗x2 = −∗ u∗ξ1 ·∗ x∗1ξ2 ·∗ D +∗ u∗ξ2 ·∗ x∗1ξ1 ·∗ D. Now we multiplicative differentiate equation (3.30) with respect to ξ1 and ξ2 and we get ∗2∗ ∗2∗ 2 ∗∗ ∗ ∗ ∗∗ u∗∗ x1 x1 ·∗ x1ξ1 +∗ e ·∗ ux1 x2 ·∗ x1ξ1 ·∗ x2ξ1 +∗ ux2 x2 ·∗ x2ξ1 ∗ ∗∗ ∗ ∗∗ = u∗∗ ξ1 ξ1 −∗ ux1 ·∗ x1ξ1 ξ1 −∗ ux2 ·∗ x2ξ1 ξ1 , ∗ ∗ ∗∗ ∗ ∗ ∗ ∗ u∗∗ x1 x1 ·∗ x1ξ1 ·∗ x1ξ2 +∗ ux1 x2 ·∗ x1ξ1 ·∗ x2ξ2 +∗ x1ξ2 ·∗ x2ξ1 ∗ ∗ +∗ u∗∗ x2 x2 ·∗ x2ξ1 ·∗ x2ξ2 ∗ ∗∗ ∗ ∗∗ = u∗∗ ξ1 ξ2 −∗ ux1 ·∗ x1ξ1 ξ2 −∗ ux2 ·∗ x2ξ1 ξ2 , ∗2∗ ∗2∗ 2 ∗∗ ∗ ∗ ∗∗ u∗∗ x1 x1 ·∗ x1ξ2 +∗ e ·∗ ux1 x2 ·∗ x1ξ2 ·∗ x2ξ2 +∗ ux2 x2 ·∗ x2ξ2 ∗ ∗∗ ∗ ∗∗ = u∗∗ ξ2 ξ2 −∗ ux1 ·∗ x1ξ2 ξ2 −∗ ux2 ·∗ x2ξ2 ξ2 .




76

Classification and Canonical Forms Let ∗ x∗2 1ξ1



A

  ∗  x1ξ ·∗ x∗1ξ 1 2 =    ∗  x∗2 1ξ2 

B

∗ x∗2 2ξ1

e2 ·∗ x∗1ξ1 ·∗ x∗2ξ1 x∗1ξ1 ·∗

x∗2ξ2

+∗ x∗1ξ2 ·∗

x∗2ξ1

x∗2ξ1 ·∗

x∗2ξ2

∗ x∗2 2ξ2

e2 ·∗ x∗1ξ2 ·∗ x∗2ξ2

∗ ∗∗ ∗ ∗∗ u∗∗ ξ1 ξ1 −∗ ux1 ·∗ x1ξ1 ξ1 −∗ ux2 ·∗ x2ξ1 ξ1



  ∗∗ ∗ ∗∗  uξ ξ −∗ u∗x ·∗ x∗∗ 1ξ1 ξ2 −∗ ux2 ·∗ x2ξ1 ξ2 1 2 1 =    ∗∗ ∗ ∗∗  uξ ξ −∗ u∗x ·∗ x∗∗ 1ξ2 ξ2 −∗ ux2 ·∗ x2ξ2 ξ2 2 2 1

   .   

We have that detA = e/∗ (D3∗ ). The multiplicative cofactors of the matrix A are as follows: ∗ x1ξ ·∗ x∗2ξ +∗ x∗1ξ ·∗ x∗2ξ x∗2ξ ·∗ x∗2ξ 1 2 2 1 1 2 a11 = ∗2∗ 2 ∗ ∗ e · x · x x ∗ 1ξ2 ∗ 2ξ2 2ξ2

∗

∗ ·∗ x∗2 2ξ2 ,

a12

= (e/∗ D) ∗ x1ξ ·∗ x∗1ξ 1 2 = −∗ ∗ x∗2 1ξ2

x∗2ξ1 ·∗ x∗2ξ2 ∗2∗ x2ξ2

∗

= −∗ (e/∗ D) ·∗ x∗1ξ2 ·∗ x∗2ξ2 ,

a13

=

∗ x1ξ ·∗ x∗1ξ 1 2 ∗ x∗2 1ξ2

x∗1ξ1 ·∗ x∗2ξ2 +∗ x∗1ξ2 ·∗ x∗2ξ1 2 ∗ ∗ e ·∗ x1ξ2 ·∗ x2ξ2

∗ = (e/∗ D) ·∗ x∗2 1ξ2 ,

a21

=

=

2 e ·∗ x∗1ξ ·∗ x∗2ξ 1 1 −∗ 2 ∗ ∗ e ·∗ x1ξ2 ·∗ x2ξ2

∗ x∗2 2ξ1 ∗2∗ x2ξ2

−∗ ((e2 )/∗ D) ·∗ x∗2ξ1 ·∗ x∗2ξ2 ,

∗

∗

    ,   

Classification and Canonical Form of Multiplicative Quasilinear MPDE ∗2∗ ∗ x1ξ x∗2 2ξ1 1 a22 = ∗2∗ ∗2∗ x1ξ2 x2ξ2 ∗
= (e/∗ D) ·∗ x∗1ξ1 ·∗ x∗2ξ2 +∗ x∗1ξ2 ·∗ x∗2ξ1 , ∗2∗ x1ξ e2 ·∗ x∗1ξ1 ·∗ x∗2ξ1 1 a23 = −∗ ∗2∗ 2 ∗ ∗ x e · x · x ∗ 1ξ2 ∗ 2ξ2 1ξ2

77

∗

·∗ x∗1ξ1 ·∗

2

a31

x∗1ξ2 ,

= −∗ ((e )/∗ D) e2 ·∗ x∗1ξ1 ·∗ x∗2ξ1 = ∗ ∗ ∗ ∗ x1ξ1 ·∗ x2ξ2 +∗ x1ξ2 ·∗ x2ξ1

∗ x∗2 2ξ1

x∗2ξ1 ·∗ x∗2ξ2



∗



∗

∗ ·∗ x∗2 2ξ1 ,

a32

a33

= (e/∗ D) ∗ x∗2 1ξ1 ·∗ = −∗ ∗ ∗ x1ξ1 ·∗ x1ξ2

∗ x∗2 2ξ1

x∗2ξ1 ·∗ x∗2ξ2



∗

= −∗ (e/∗ D) ·∗ x∗1ξ1 ·∗ x∗2ξ1 , ∗ x∗2 e2 ·∗ x∗1ξ1 ·∗ x∗2ξ1 1ξ1 = ∗ ∗ ∗ ∗ ∗ ∗ x1ξ1 ·∗ x1ξ2 x1ξ1 ·∗ x2ξ2 +∗ x1ξ2 ·∗ x2ξ1 ∗ = (e/∗ D) ·∗ x∗2 1ξ1 .

Hence,  A− ∗ 1

∗ x∗2 2ξ2

−∗ e2 ·∗ x∗2ξ1 ·∗ x∗2ξ2

   −∗ x∗1ξ ·∗ x∗2ξ 2 2 = D2∗ ·∗     x∗2∗

x∗1ξ1 ·∗

1ξ2

x∗2ξ2

+∗ x∗1ξ2 ·∗

−∗ e2 ·∗ x∗1ξ1 ·∗ x∗1ξ2

Therefore, 

u∗∗ x1 x1

x∗2ξ1



 ∗∗   ux1 x2  = A−∗ 1 ·∗ B   u∗∗ x2 x2

∗ x∗2 2ξ1

−∗ x∗1ξ1 ·∗ ∗ x∗2 1ξ1

 x∗2ξ1

   .   

78

Classification and Canonical Forms

and the canonical form of the operator L(u) is as follows:  2∗ 2 ∗ ∗ ∗ L(u) = D ·∗ a ·∗ x∗2 2ξ2 −∗ e ·∗ b ·∗ x1ξ2 ·∗ x2ξ2  ∗2∗ +∗ c ·∗ x1ξ2 ·∗ u∗∗ ξ 1 ξ1 
+∗ −∗ e2 ·∗ a ·∗ x∗2ξ1 ·∗ x∗2ξ2 +∗ e2 ·∗ b x∗1ξ1 ·∗ x∗2ξ2 +∗ x∗1ξ2 ·∗ x∗2ξ1  −∗ e2 ·∗ c ·∗ x∗1ξ1 ·∗ x∗1ξ2 ·∗ u∗∗ ξ1 ξ2   ∗2∗ 2 ∗ ∗ ∗∗ ∗ +∗ a ·∗ x∗2 2ξ1 −∗ e ·∗ b ·∗ x1ξ1 ·∗ x2ξ1 +∗ c ·∗ x1ξ1 ·∗ uξ2 ξ2   ∗ ∗∗ 2 ∗ ∗ ∗∗ ∗ −∗ ux1 ·∗ a ·∗ x∗2 2ξ2 ·∗ x1ξ1 ξ1 −∗ e ·∗ x2ξ1 ·∗ x2ξ2 ·∗ x1ξ1 ξ2  ∗2∗ ∗∗ +∗ x2ξ1 ·∗ x1ξ2 ξ2  2 +∗ e ·∗ b ·∗ −∗ x∗1ξ2 ·∗ x∗2ξ2 ·∗ x∗∗ 1ξ1 ξ1
∗ ∗ ∗ ∗ +∗ x∗∗ 1ξ1 ξ2 ·∗ x1ξ1 ·∗ x2ξ2 +∗ x1ξ2 ·∗ x2ξ1  ∗ ∗ ∗∗ −∗ x1ξ1 ·∗ x2ξ1 ·∗ x1ξ2 ξ2  ∗∗ 2 ∗ ∗ ∗∗ ∗ +∗ c · ∗ x∗2 1ξ2 ·∗ x1ξ1 ξ1 −∗ e ·∗ x1ξ1 ·∗ x1ξ2 ·∗ x1ξ1 ξ2  ∗2∗ ∗∗ +∗ x1ξ1 ·∗ x1ξ2 ξ2   ∗2∗ ∗ 2 ∗ ∗ ∗∗ −∗ ux2 ·∗ a ·∗ x∗∗ 2ξ1 ξ1 ·∗ x2ξ2 −∗ e ·∗ x2ξ1 ·∗ x2ξ2 ·∗ x2ξ1 ξ2  ∗2∗ ∗∗ +∗ x2ξ1 ·∗ x2ξ2 ξ2  2 +∗ e ·∗ b ·∗ −∗ x∗1ξ2 ·∗ x∗2ξ2 ·∗ x∗∗ 2ξ1 ξ1 
+∗ x∗1ξ1 · ∗ x∗2ξ2 +∗ x∗1ξ2 ·∗ x∗2ξ1 ·∗ x∗∗ 2ξ1 ξ2  −∗ x∗1ξ1 ·∗ x∗2ξ1 ·∗ x∗∗ 2ξ2 ξ2  ∗∗ 2 ∗ ∗ ∗∗ ∗ +∗ c ·∗ x∗2 1ξ2 ·∗ x2ξ1 ξ1 −∗ e ·∗ x1ξ1 ·∗ x1ξ2 ·∗ x2ξ1 ξ2  ∗∗ ∗ +∗ x∗2 · · x +∗ d 1ξ1 ∗ ∗ 2ξ2 ξ2 = 0∗ . (3.32)

Classification and Canonical Form of Multiplicative Quasilinear MPDE

3.2.1

79

The multiplicative hyperbolic case

In this case, we have a ·∗ c −∗ b2 < 0∗ . The characteristic parameters ξ1

= φ1 (x1 , x2 ),

ξ2

= φ2 (x1 , x2 ),

satisfy, as in the multiplicative linear case, the equations φ∗1x1 −∗ λ1 ·∗ φ∗1x2 = 0∗ , (3.33) φ∗2x1 −∗ λ2 ·∗ φ∗2x2 = 0∗ , where λ1 and λ2 are the roots of the multiplicative quadratic equation (3.5). By equation (3.33), we find ∗ ∗ ξ1x = λ1 ·∗ ξ1x , 1 2 ∗ ∗ ξ2x = λ2 ·∗ ξ2x . 1 2

Hence, from equation (3.29), we get x∗2ξ2 = −∗ λ1 ·∗ x∗1ξ2 , (3.34) x∗2ξ1 = −∗ λ2 ·∗ x∗1ξ1 . In the multiplicative hyperbolic case, we have α = γ = 0∗ or

∗2∗ ∗2∗ ∗ a ·∗ ξ1x +∗ e2 ·∗ b ·∗ ξ1x · ξ ∗ +∗ c ·∗ ξ1x = 0∗ 1 ∗ 1x2 1 2 ∗2∗ ∗2∗ ∗ a ·∗ ξ2x +∗ e2 ·∗ b ·∗ ξ2x · ξ ∗ +∗ c ·∗ ξ2x = 0∗ . 1 ∗ 2x2 1 2

Hence, applying equation (3.29), we get ∗2∗ 2 ∗ ∗ ∗ a ·∗ x∗2 2ξ2 −∗ e ·∗ b ·∗ x2ξ2 ·∗ x1ξ2 +∗ c ·∗ x1ξ2 = 0∗ , ∗2∗ 2 ∗ ∗ ∗ a ·∗ x∗2 2ξ1 −∗ e ·∗ b ·∗ x2ξ1 ·∗ x1ξ1 +∗ c ·∗ x1ξ1 = 0∗ .

80

Classification and Canonical Forms

Then the canonical form of the operator L(u) is n L(u) = D2∗ ·∗ −∗ e2 ·∗ a ·∗ x∗2ξ1 ·∗ x∗2ξ2 +∗ e2 ·∗ b·∗ 
x∗1ξ1 ·∗ x∗2ξ2 +∗ x∗1ξ2 ·∗ x∗2ξ1 −∗ e2 ·∗ c ·∗ x∗1ξ1 ·∗ x∗1ξ2 ·∗ u∗∗ ξ 1 ξ2   ∗∗ 2 ∗ ∗ ∗∗ ∗ −∗ u∗x1 ·∗ a ·∗ x∗2 2ξ2 ·∗ x1ξ1 ξ1 −∗ e ·∗ x2ξ1 ·∗ x2ξ2 ·∗ x1ξ1 ξ2   ∗∗ 2 ∗ +∗ x∗2 · x + e · b · −∗ x∗1ξ2 ·∗ x∗2ξ2 ·∗ x∗∗ ∗ ∗ ∗ 1ξ1 ξ1 2ξ1 ∗ 1ξ2 ξ2 ∗ ∗ ∗ ∗ +∗ x∗∗ 1ξ1 ξ2 ·∗ x1ξ1 ·∗ x2ξ2 +∗ x1ξ2 ·∗ x2ξ1  ∗ ∗ ∗∗ −∗ x1ξ1 ·∗ x2ξ1 ·∗ x1ξ2 ξ2




  ∗2∗ ∗∗ 2 ∗ ∗ ∗∗ ∗∗ ∗ +∗ c ·∗ x∗2 1ξ2 ·∗ x1ξ1 ξ1 −∗ e ·∗ x1ξ1 ·∗ x1ξ2 ·∗ x1ξ1 ξ2 +∗ x1ξ1 ·∗ x1ξ2 ξ2   ∗2∗ 2 ∗ ∗ ∗∗ −∗ u∗x2 ·∗ a ·∗ x∗∗ 2ξ1 ξ1 ·∗ x2ξ2 −∗ e ·∗ x2ξ1 ·∗ x2ξ2 ·∗ x2ξ1 ξ2   ∗∗ 2 ∗ +∗ x∗2 · x + e · b · −∗ x∗1ξ2 ·∗ x∗2ξ2 ·∗ x∗∗ ∗ ∗ ∗ 2ξ1 ξ1 2ξ1 ∗ 2ξ2 ξ2
+∗ x∗1ξ1 ·∗ x∗2ξ2 +∗ x∗1ξ2 ·∗ x∗2ξ1 ·∗ x∗∗ 2ξ1 ξ2  −∗ x∗1ξ1 ·∗ x∗2ξ1 ·∗ x∗∗ 2ξ2 ξ2  o ∗2∗ ∗∗ 2 ∗ ∗ ∗∗ ∗∗ ∗ +∗ c ·∗ x∗2 1ξ2 ·∗ x2ξ1 ξ1 −∗ e ·∗ x1ξ1 ·∗ x1ξ2 ·∗ x2ξ1 ξ2 +∗ x1ξ1 ·∗ x2ξ2 ξ2 +∗ d. Now, using equation (3.34), we get n  L(u) = D2∗ ·∗ −∗ e2 ·∗ a ·∗ λ1 ·∗ λ2 −∗ e2 ·∗ b ·∗ (λ1 +∗ λ2 ) −∗ e2 ·∗ c ·∗ x∗1ξ1 ·∗ x∗1ξ2 ·∗ u∗∗ ξ 1 ξ2 
∗∗ ∗ −∗ u∗x1 ·∗ a ·∗ λ21∗ +∗ e2 ·∗ b ·∗ λ1 +∗ c ·∗ x∗2 1ξ2 ·∗ x1ξ1 ξ1
+∗ −∗ e2 ·∗ a ·∗ λ1 ·∗ λ2 −∗ e2 ·∗ b ·∗ (λ1 +∗ λ2 ) −∗ e2 ·∗ c ·∗ x∗1ξ1 ·∗ x∗1ξ2 ·∗ x∗∗ 1ξ1 ξ2

Classification and Canonical Form of Multiplicative Quasilinear MPDE 
∗∗ ∗ +∗ a ·∗ λ22∗ +∗ e2 ·∗ b ·∗ λ2 +∗ c ·∗ x∗2 1ξ1 ·∗ x1ξ1 ξ2 −∗ u∗x2 ·∗




∗∗ ∗ a ·∗ λ21∗ +∗ e2 ·∗ b ·∗ λ1 +∗ c ·∗ x∗2 1ξ2 ·∗ x2ξ1 ξ1


+∗ −∗ e2 ·∗ a ·∗ λ1 ·∗ λ2 −∗ e2 ·∗ b ·∗ (λ1 +∗ λ2 ) −∗ e2 ·∗ c ·∗ x∗1ξ1 ·∗ x∗1ξ2 ·∗ x∗∗ 2ξ1 ξ2 o
∗∗ ∗ +∗ a ·∗ λ22∗ +∗ e2 ·∗ b ·∗ λ2 +∗ c ·∗ x∗2 +∗ d 1ξ1 ·∗ x2ξ2 ξ2 = 0∗ . Next, using a ·∗ λ21∗ +∗ e2 ·∗ b ·∗ λ1 +∗ c =

0∗ ,

a ·∗ λ22∗ +∗ e2 ·∗ b ·∗ λ2 +∗ c =

0∗ ,

and λ 1 +∗ λ 2 λ1 ·∗ λ2

= −∗ ((e2 ·∗ b)/∗ a), = c/∗ a,

we get  D2∗ ·∗ −∗ e4 ·∗ ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ x∗1ξ1 ·∗ x∗1ξ2 ·∗ u∗∗ ξ 1 ξ2 ∗ +∗ e4 ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ x∗1ξ1 ·∗ x∗1ξ2 ·∗ x∗∗ 1ξ1 ξ2 ·∗ ux1

 ∗ +∗ e4 ·∗ ((a ·∗ c −∗ b2∗ )/∗ a) ·∗ x∗1ξ1 ·∗ x∗1ξ2 ·∗ x∗∗ 1ξ1 ξ2 ·∗ ux2 +∗ d = 0∗ or

 ∗∗ ∗ ∗ x∗1ξ1 ·∗ x∗1ξ2 ·∗ u∗∗ ξ1 ξ2 −∗ x1ξ1 ξ2 ·∗ D ·∗ −∗ λ1 ·∗ uξ1 ·∗ x1ξ2 +∗ λ2 ·∗ u∗ξ2 ·∗ x∗1ξ1




∗ ∗ ∗ ∗ −∗ x∗∗ 1ξ1 ξ2 ·∗ D ·∗ −∗ uξ1 ·∗ x1ξ2 +∗ uξ2 ·∗ x1ξ1

= ((a ·∗ d)/∗ (e4 ·∗ (a ·∗ c −∗ b2∗ ))) ·∗ D2∗ is the canonical form of L(u).




81

82

Classification and Canonical Forms

3.2.2

The multiplicative elliptic case

In this case, we have a ·∗ c −∗ b2∗ > 0∗ and α = γ, β = 0∗ . Then, ∗2∗ ∗2∗ ∗ a ·∗ ξ1x +∗ e2 ·∗ b ·∗ ξ1x · ξ ∗ +∗ c ·∗ ξ1x 1 ∗ 1x2 1 2 ∗2∗ ∗2∗ ∗ = a ·∗ ξ2x +∗ e2 ·∗ b ·∗ ξ2x · ξ ∗ +∗ c ·∗ ξ2x 1 ∗ 2x2 1 2 ∗ ∗ ∗ a ·∗ ξ1x · ξ ∗ +∗ b ·∗ ξ1x · ξ ∗ +∗ ξ1x · ξ∗ 1 ∗ 2x1 2 ∗ 2x1 1 ∗ 2x2




∗ +∗ c ·∗ ξ1x · ξ∗ 2 ∗ 2x2

= 0∗ . Hence, from equation (3.29), we obtain ∗2∗ 2 ∗ ∗ ∗ a ·∗ x∗2 2ξ2 −∗ e ·∗ b ·∗ x2ξ2 ·∗ x1ξ2 +∗ c ·∗ x1ξ2 ∗2∗ 2 ∗ ∗ ∗ = a ·∗ x∗2 2ξ1 −∗ e ·∗ bx2ξ1 ·∗ x1ξ1 +∗ c ·∗ x1ξ1


−∗ a ·∗ x∗2ξ1 ·∗ x∗2ξ2 +∗ b ·∗ x∗1ξ2 ·∗ x∗2ξ1 +∗ x∗1ξ1 ·∗ x∗2ξ2 −∗ c ·∗ x∗1ξ1 ·∗ x∗1ξ2 = 0∗ . Consequently, the canonical form of the operator L(u) is as follows:   ∗2∗ 2 ∗ ∗ ∗ L(u) = D2∗ ·∗ a ·∗ x∗2 − e · b · x · x + c · x ∗ ∗ ∗ ∗ ∗ ∗ 1ξ2 2ξ2 2ξ2 1ξ2 ∗∗ ·∗ u∗∗ ξ1 ξ1 +∗ uξ2 ξ2




  ∗2∗ 2 ∗ ∗ ∗ −∗ u∗x1 ·∗ a ·∗ x∗2 2ξ2 −∗ e ·∗ b ·∗ x2ξ2 ·∗ x1ξ2 +∗ c ·∗ x1ξ2 ∗∗ ·∗ x∗∗ 1ξ1 ξ1 +∗ x1ξ2 ξ2




  ∗2∗ 2 ∗ ∗ ∗ −∗ u∗x2 ·∗ a ·∗ x∗2 − e · b · x · x + c · x ∗ ∗ ∗ ∗ ∗ ∗ 2ξ2 1ξ2 2ξ2 1ξ2 ∗∗ ·∗ x∗∗ 2ξ1 ξ1 +∗ x2ξ2 ξ2

= 0∗




+∗ d

Classification and Canonical Form of Multiplicative Quasilinear MPDE

83

or   ∗2∗ 2 ∗ ∗ ∗ D2∗ ·∗ a ·∗ x∗2 2ξ2 −∗ e ·∗ b ·∗ x1ξ2 ·∗ x2ξ2 +∗ c ·∗ x1ξ2
·∗ ∆∗ u −∗ u∗x1 ·∗ ∆∗ x1 −∗ u∗x2 ·∗ ∆∗ x2 +∗ d = 0∗ , where u∗x1 and u∗x2 are determined with equation (3.31).

3.2.3

The multiplicative parabolic case

In this case, we have a ·∗ c −∗ b2∗ = 0∗ and β = γ = 0∗ . Then, ∗ ∗ ∗ a ·∗ ξ1x · ξ ∗ +∗ b ·∗ ξ1x · ξ ∗ +∗ ξ1x · ξ∗ 1 ∗ 2x1 2 ∗ 2x1 1 ∗ 2x2




∗ +∗ c ·∗ ξ1x · ξ∗ 2 ∗ 2x2

= 0∗ ∗2∗ ∗2∗ ∗ a ·∗ ξ2x +∗ e2 ·∗ b ·∗ ξ2x · ξ ∗ +∗ c ·∗ ξ2x = 0∗ . 1 ∗ 2x2 1 2

Hence, from equation (3.29), we obtain a ·∗ x∗2ξ1 ·∗ x∗2ξ2 −∗ b ·∗ x∗1ξ2 ·∗ x∗2ξ1 +∗ x∗1ξ1 ·∗ x∗2ξ2




+∗ c ·∗ x∗1ξ1 ·∗ x∗1ξ2 = 0∗ a

∗ ·∗ x∗2 2ξ1

∗ −∗ e2 ·∗ b ·∗ x∗2ξ1 ·∗ x∗1ξ1 +∗ c ·∗ x∗2 1ξ1 = 0∗ .

Therefore, the canonical form of the operator L(u) is as follows:   ∗2∗ 2 ∗ ∗ ∗∗ ∗ L(u) = D2∗ ·∗ a ·∗ x∗2 − e · b · x · x + c · x ∗ ∗ ∗ 1ξ2 ∗ 2ξ2 ∗ ∗ 1ξ2 ·∗ uξ1 ξ1 2ξ2   ∗2∗ 2 ∗ ∗ ∗∗ ∗ −∗ u∗x1 ·∗ a ·∗ x∗2 − e · b · x · x + c · x ∗ ∗ ∗ 1ξ2 ∗ 2ξ2 ∗ ∗ 1ξ2 ·∗ x1ξ1 ξ1 2ξ2    ∗2∗ 2 ∗ ∗ ∗∗ ∗ −∗ u∗x2 ·∗ a ·∗ x∗2 − e · b · x · x + c · x · x ∗ ∗ ∗ 1ξ2 ∗ 2ξ2 ∗ ∗ 1ξ2 ∗ 2ξ1 ξ1 2ξ2 +∗ d = 0∗ , where u∗x1 and u∗x2 are determined with equation (3.31).

84

Classification and Canonical Forms

3.3

Classification and Canonical Form of Second-Order Linear Partial Differential Equations in n Multiplicative Independent Variables

Consider a general second-order multiplicative linear multiplicative partial differential equation in n multiplicative independent variables: n X

aij ·∗ u∗∗ xi xj +∗

∗i,j=1

n X

bi ·∗ u∗xi +∗ c ·∗ u +∗ d = 0∗ ,

(3.35)

∗i=1

where the coefficients aij , bi , 1 ≤ i, j ≤ n, c, d and the unknown u are functions of x = (x1 , . . . , xn ). Let   a11 a12 · · · a1n  a21 a22 · · · a2n    A =  . ,  ..  an1

an2

...

ann

b = (b1 , · · · , bn ). Assume that the matrix A is symmetric. Otherwise, we can always find a symmetric matrix: aij

1

= e 2 ·∗ (aij +∗ aji ) 1

= e 2 ·∗ (aij aji ) = e = e

 1 log e 2 log(aij aji )

log(aij )+log(aji ) 2

such that equation (3.35) can be rewritten in the form n X

aij ·∗ u∗∗ xi xj

+∗

∗i,j=1

n X

bi ·∗ u∗xi +∗ c ·∗ u +∗ d = 0∗ .

∗i=1

Now we consider the transformation ξ = Q ·∗ x, where ξ = (ξ1 , . . . , ξn ) and Q = (qij ) is an n × n matrix. We have that ξi∗

=

n X ∗j=1

qij ·∗ xj .

n Multiplicative Independent Variables

85

Using the chain rule, we have u∗xi =

n X

∗ u∗ξk∗ ·∗ ξkx i

∗k=1

=

n X

u∗ξk∗ ·∗ qki ,

∗k=1

u∗∗ xi xj =

n X

∗ ∗ u∗∗ ∗ ξ ∗ ·∗ ξkx ·∗ ξlx ξk i j l

∗k,l=1

=

n X

u∗∗ ∗ ξ ∗ ·∗ qki ·∗ qlj . ξk l

∗k,l=1

This allows equation (3.35) to be expressed as   n n X X  qki ·∗ aij ·∗ qlj  ·∗ u∗∗ ξk ξl + ∗ lower ∗k,l=1

order

terms = 0∗ .

∗i,j=1

The coefficient matrix of the terms u∗∗ ξk ξl in this transformed expression is equal to (qki ·∗ aij ·∗ qlj ) = QT ·∗ A ·∗ Q. We know that for any real symmetric matrix A, there is an associative multiplicative orthogonal matrix Q1 such that QT1 ·∗ A ·∗ Q1 = Λ. Here Q1 is called diagonalizing matrix of the matrix A and Λ is a diagonal matrix whose elements are the multiplicative eigenvalues λi of the matrix A and the columns of Q1 are multiplicative linearly independent eigenvectors of A, qi = (q1i , . . . , qni ), 1

2∗ 2∗ 2 ∗ |qi |∗ = (q1i +∗ · · · +∗ qni ) = e.

So, we have Q =

(qij ),

Λ

(λi ·∗ δij ) ,

=

i, j = 1, . . . , n,

where δij are the multiplicative Kronecker delta. Now, if Q is taken to be a diagonalizing matrix of A, it follows that QT ·∗ A ·∗ Q =

Λ 

  =  



λ1 λ2 ..

  . 

. λn

86

Classification and Canonical Forms

The members λ1 , λ2 , . . . , λn are real numbers because the matrix A is symmetric. Definition 3.14 Equation (3.35) is called as follows: 1. Multiplicative elliptic, if all multiplicative eigenvalues λi of A are multiplicative nonzero and have the same sign. 2. Multiplicative hyperbolic, if all multiplicative eigenvalues λi of A are multiplicative nonzero and have the same sign except for one of the multiplicative eigenvalues. 3. Multiplicative parabolic, if the multiplicative eigenvalues λi of A are all multiplicative positive or all multiplicative negative, save one that is multiplicative zero. 4. Multiplicative ultrahyperbolic, if there is more than one multiplicative positive multiplicative eigenvalue and more than one multiplicative negative multiplicative eigenvalue, and there are multiplicative nonzero multiplicative eigenvalues. The multiplicative nonsingular transformation x = QT ·∗ ξ transforms equation (3.35) in the canonical form: n X

λi ·∗ u∗∗ ξi ξi +∗ lower

order

terms = 1.

∗i=1

Example 3.16 Consider the equation ∗∗ ∗∗ u∗∗ x1 x1 −∗ ux1 x2 +∗ ux2 x2 = f (x1 , x2 ).

We can rewrite this equation in the following form: 1

1

∗∗ ∗∗ ∗∗ 2 2 u∗∗ x1 x1 −∗ e ·∗ ux1 x2 −∗ e ·∗ ux2 x1 +∗ ux2 x2 = f (x1 , x2 ).

Then,   A= 

e e

− 12

1

e− 2



e

 . 

We will find the multiplicative eigenvalues of the matrix A. We have 1 e −∗ λ e− 2 det∗ (A −∗ λ ·∗ I∗ ) = − 1 e −∗ λ e 2 ∗

= 0∗

n Multiplicative Independent Variables

87

if and only if

1

(λ −∗ e)2∗ −∗ e 4 = 1, if and only if

3

λ2∗ −∗ e2 ·∗ λ +∗ e 4 = 1, if and only if 3

λ1

= e2 ,

λ2

= e2 .

1

Since λ1 , λ2 > 0∗ , the considered equation is a multiplicative elliptic equation. Now we will find the matrix Q. Let A1 = (ea , eb ) be an eigenvector of the 3 matrix A corresponding to the eigenvalue λ1 = e 2 . Then,   a    1 1  e −∗ e 2 −∗ e 2 1            , ·∗ = 1 b  − e 12 e2   e   1  ∗ i.e., ea +∗ eb = 1. We take q1 = e a

1 √ 2

·∗ (e, −∗ e).

b

Let A2 = (e , e ) be an eigenvector of the matrix 1 eigenvalue λ2 = e 2 . Then,   a  1 1  e e2 −∗ e 2       ·∗  b  = 1  − e 12    2 e e ∗

A corresponding to the



ea = eb . q2 = e

1 √ 2

·∗ (e, e).

Therefore, Q =

q1T , q2T

= e

1 √ 2




 ·∗

e −∗ e

e e



   ,  1 

i.e., We take

1

 .

88

Classification and Canonical Forms

Let ξ

= Q ·∗ x   1 1 √ √   e 2 e 2 x1   =  1 1  ·∗ √ √ x2 −∗ e 2 e 2   1 √ e 2 ·∗ (x1 +∗ x2 )     =  √1 .  e 2 ·∗ (−∗ x1 +∗ x2 ) 

Then, 1 √ 2

∗ ξ1x 1

= e

∗ ξ1x 2

= e

∗ ξ2x 1

= −∗ e

∗ ξ2x 2

= e

u∗x1

1 √ 2

, 1 √ 2

,

,

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 1 1

= e u∗∗ x1 x1

1 √ 2

,

= e

1 √ 2

·∗ u∗ξ1 −∗ e

1 √ 2

1 √ 2

∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1

·∗ u∗ξ2 ,


1 √ 2


∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x1 +∗ uξ2 ξ2 ·∗ ξ2x1  1  1 1 √ √ √ 2 · u∗∗ = e 2 ·∗ e 2 ·∗ u∗∗ − e ∗ ∗ ξ1 ξ1 ξ1 ξ2  1  1 1 √ √ √ 2 · u∗∗ −∗ e 2 ·∗ e 2 ·∗ u∗∗ − e ∗ ∗ ξ 1 ξ2 ξ 2 ξ2 −∗ e

1

1

∗∗ ∗∗ 2 = e 2 ·∗ u∗∗ ξ1 ξ1 −∗ uξ1 ξ2 +∗ e ·∗ uξ2 ξ2 ,

u∗∗ x1 x2

= e

1 √ 2

∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2 1 √ 2





∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2  1  1 1 √ √ √ 2 · u∗∗ = e 2 ·∗ e 2 ·∗ u∗∗ + e ∗ ∗ ξ1 ξ1 ξ1 ξ2  1  1 1 √ √ √ 2 · u∗∗ −∗ e 2 ·∗ e 2 ·∗ u∗∗ + e ∗ ∗ ξ 1 ξ2 ξ 2 ξ2 −∗ e

1

1

= e 2 ·∗ u∗ξ1 ξ1 −∗ e 2 ·∗ u∗ξ2 ξ2 ,

n Multiplicative Independent Variables

u∗x2

∗ ∗ = u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x 2 2

= e u∗∗ x2 x2

89

= e

1 √ 2

·∗ u∗ξ1 +∗ e

1 √ 2

1 √ 2

∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x2 +∗ uξ1 ξ2 ·∗ ξ2x2

·∗ u∗ξ2 ,


1 √ 2


∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2  1  1 1 √ √ √ 2 · u∗∗ = e 2 ·∗ e 2 ·∗ u∗∗ ∗ ξ1 ξ2 ξ1 ξ1 +∗ e  1  1 1 √ √ √ 2 · u∗∗ +∗ e 2 ·∗ e 2 ·∗ u∗∗ ∗ ξ 2 ξ2 ξ 1 ξ2 + ∗ e +∗ e

1

1

∗∗ ∗∗ 2 = e 2 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 ξ2 +∗ e ·∗ uξ2 ξ2 .

Therefore, ∗∗ ∗∗ u∗∗ x1 x1 −∗ ux1 x2 +∗ ux2 x2

1

1

∗∗ ∗∗ 2 = e 2 ·∗ u∗∗ ξ1 ξ1 −∗ uξ1 ξ2 +∗ e ·∗ uξ2 ξ2 1

1

∗∗ 2 −∗ e 2 ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ uξ2 ξ2 1

1

∗∗ ∗∗ 2 +∗ e 2 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 ξ2 +∗ e ·∗ uξ2 ξ2 1

3

∗∗ 2 = e 2 ·∗ u∗∗ ξ1 ξ1 +∗ e ·∗ uξ2 ξ2 .

Hence, the canonical form of the considered equation is  1  1 1 3 √ √ 2 · u∗∗ 2 · (ξ − ξ ), e 2 · (ξ + ξ ) e 2 ·∗ u∗∗ + e = f e . ∗ ∗ ∗ 2 ∗ 2 ∗ 1 ∗ 2 ξ1 ξ1 ξ 2 ξ2 Example 3.17 Consider the equation 2 ∗∗ ∗∗ ∗∗ u∗∗ x1 x1 +∗ e ·∗ ux1 x3 +∗ ux2 x2 +∗ ux3 x3 = 0∗ .

We can rewrite it in the following form: ∗∗ ∗∗ ∗∗ ∗∗ u∗∗ x1 x1 +∗ ux1 x3 +∗ ux2 x2 +∗ ux3 x1 +∗ ux3 x3 = 0∗ .

Then, 

e A= 1 e

1 e 1

 e 1 . e

We will find the multiplicative eigenvalues of the matrix A. We have e −∗ λ 1 e 1 e −∗ λ 1 det∗ (A −∗ λ ·∗ I∗ ) = e 1 e −∗ λ ∗ =1

90

Classification and Canonical Forms

if and only if −∗ (λ −∗ e)3∗ +∗ λ −∗ e = 1, if and only if
−∗ (λ −∗ e) ·∗ (λ −∗ e)2∗ −∗ e = 1, if and only if (λ −∗ e) ·∗ (λ −∗ e2 ) ·∗ λ = 1, if and only if λ1

=

1,

λ2

= e,

λ3

= e2 .

Since λ1 = 0∗ and λ2 , λ3 > 0∗ , the considered equation is a multiplicative parabolic equation. Let A1 = (a, b, c) be a multiplicative eigenvector of the matrix A corresponding to the multiplicative eigenvalue λ1 = 0∗ . Then,       e 1 e a 1  1 e 1  ·∗  b  =  1  , e 1 e c 1 i.e., a +∗ c

=

1,

b

=

1.

We take q1 = e

1 √ 2

·∗ (e, 1, −∗ e).

Let A2 = (a, b, c) be a multiplicative eigenvector of the matrix A corresponding to the multiplicative eigenvalue λ2 = e. Then,       1 1 e a 1  1 1 1  ·∗  b  =  1  , e 1 1 c 1 i.e., a

=

1,

c

=

1.

We take q2 = (1, e, 1).

n Multiplicative Independent Variables

91

Let A3 = (a, b, c) be a multiplicative eigenvector of the matrix A corresponding to the multiplicative eigenvalue λ3 = e2 . Then,       −∗ e 1 e a 1  1 −∗ e 1  ·∗  b  =  1  , e 1 −∗ e c 1 i.e., a

= c,

b

=

We take q3 = e

1 √ 2

1.

·∗ (e, 1, e).

Hence, Q =

q1T , q2T , q3T 

 = 

e

1 √ 2




1 e 1

1 1 √ −∗ e 2

e e

1 √ 2



1

 

1 √ 2

and 



 ξ1  ξ2  ξ3

=

 

Then, 1 √ 2

∗ ξ1x 1

=

e

∗ ξ1x 2

=

1,

∗ ξ1x 3

=

−∗ e

∗ ξ2x 1

=

1,

,

1 √ 2

,

1 √ 2

1 1 √ −∗ e 2

1 e 1

e

1 √ 2

1 e

1 √ 2





 x1   x2   ·∗ x3

 1 √ ·∗ x1 +∗ e 2 ·∗ x3   x2  . 1 1 √ √ −∗ e 2 ·∗ x1 +∗ e 2 ·∗ x3 

=

e

e

1 √ 2

92

Classification and Canonical Forms ∗ ξ2x 2

=

∗ ξ2x 3

= 1,

∗ ξ3x 1

=

e

∗ ξ3x 2

=

1,

∗ ξ3x 3

=

e

u∗x1

=

∗ ∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x +∗ u∗ξ3 ·∗ ξ3x 1 1 1

=

e

=

e

u∗∗ x1 x1

e,

1 √ 2

1 √ 2

,

,

1 √ 2

·∗ u∗ξ1 −∗ e

1 √ 2

∗ ∗∗ ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x1 +∗ uξ1 ξ2 ·∗ ξ2x1 +∗ uξ1 ξ3 ·∗ ξ3x1

−∗ e

1 √ 2

e

1 √ 2

−∗ e

u∗∗ x1 x3

·∗ u∗ξ3 ,



 1  1 √ √ 2 · u∗∗ ·∗ e 2 ·∗ u∗∗ ∗ ξ1 ξ3 ξ1 ξ 1 − ∗ e 1 √ 2

 1  1 √ √ 2 · u∗∗ ·∗ e 2 ·∗ u∗∗ − e ∗ ∗ ξ1 ξ 3 ξ3 ξ 3

1

1

=

∗∗ ∗∗ 2 e 2 ·∗ u∗∗ ξ1 ξ1 −∗ uξ1 ξ3 +∗ e ·∗ uξ3 ξ3 ,

=

e

1 √ 2

−∗ e =

e

1 √ 2



=

e

1




 ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ3 ·∗ ξ1x1 +∗ uξ2 ξ3 ·∗ ξ2x1

∗ +∗ u∗∗ ξ3 ξ3 ·∗ ξ3x1

=

1 √ 2

∗ ∗∗ ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x3 +∗ uξ1 ξ2 ·∗ ξ2x3 +∗ uξ1 ξ3 ·∗ ξ3x3 1 √ 2

∗ ∗∗ ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ3 ·∗ ξ1x3 +∗ uξ2 ξ3 ·∗ ξ2x3 +∗ uξ3 ξ3 ·∗ ξ3x3

  1 1 1 √ √ √ 2 · u∗∗ 2 · ·∗ e 2 ·∗ u∗∗ + e ∗ ∗ ∗ ξ1 ξ 1 ξ1 ξ3 − ∗ e

1 √ 2




·∗ u∗∗ ξ1 ξ 3 + ∗ e 1

1 √ 2

·∗ u∗∗ ξ3 ξ 3

∗∗ 2 e 2 ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ uξ3 ξ3 ,






n Multiplicative Independent Variables u∗x2

u∗∗ x2 x2

u∗x3

u∗∗ x3 x3

93

=

∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x +∗ u∗∗ ξ3 ξ3x2 2 2

=

u∗ξ2 ,

=

∗ ∗∗ ∗ ∗∗ ∗ u∗∗ ξ1 ξ2 ·∗ ξ1x2 +∗ uξ2 ξ2 ·∗ ξ2x2 +∗ uξ2 ξ3 ·∗ ξ3x2

=

u∗∗ ξ2 ξ 2 ,

=

∗ ∗ ∗ u∗ξ1 ·∗ ξ1x +∗ u∗ξ2 ·∗ ξ2x +∗ u∗ξ3 ·∗ ξ3x 3 3 3

=

e

=

e

1 √ 2

·∗ u∗ξ1 +∗ e

1 √ 2

∗ ∗∗ ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ1 ·∗ ξ1x3 +∗ uξ1 ξ2 ·∗ ξ2x3 +∗ uξ1 ξ3 ·∗ ξ3x1

+∗ e =

e

1 √ 2

+∗ e =

1

1 √ 2

1 √ 2

·∗ u∗ξ3 ,


∗ ∗∗ ∗ ∗∗ ∗ ·∗ u∗∗ ξ1 ξ3 ·∗ ξ1x3 +∗ uξ2 ξ3 ·∗ ξ2x3 +∗ uξ3 ξ3 ·∗ ξ3x3




 1  1 √ √ 2 · u∗∗ ·∗ e 2 ·∗ u∗∗ + e ∗ ∗ ξ1 ξ 1 ξ1 ξ3 1 √ 2

  1 √ 2 · u∗∗ ·∗ u∗∗ ∗ ξ3 ξ 3 ξ1 ξ 3 + ∗ e 1

∗∗ ∗∗ 2 e 2 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 ξ3 +∗ e ·∗ uξ3 ξ3 .

From here, 1

1

2 ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ ∗∗ 2 2 u∗∗ x1 x1 +∗ e ·∗ ux1 x3 +∗ ux2 x2 +∗ ux3 x3 = e ·∗ uξ1 ξ1 −∗ uξ1 ξ3 +∗ e ·∗ uξ3 ξ3

 1  1 ∗∗ 2 +∗ e2 ·∗ e 2 ·∗ u∗∗ ξ1 ξ1 −∗ e ·∗ uξ1 ξ3 1

∗∗ ∗∗ 2 +∗ u∗∗ ξ2 ξ2 +∗ e ·∗ uξ1 ξ1 +∗ uξ1 ξ3

1

+∗ e 2 ·∗ u∗∗ ξ3 ξ 3 ∗∗ = e2 ·∗ u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 .

Therefore, the canonical form of the considered equation is ∗∗ e2 ·∗ u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 = 0∗ .

94

Classification and Canonical Forms

Exercise 3.11 Classify the three-dimensional multiplicative wave equation:
c2 ∗∗ ∗∗ ∗∗ u∗∗ x1 x1 −∗ e ·∗ ux2 x2 +∗ ux3 x3 +∗ ux4 x4 = 1, where c is a constant, c 6= 0. Answer Multiplicative hyperbolic.

3.4

Classification of First-Order Systems with Two Multiplicative Independent Variables

Consider the system A ·∗ u∗x1 +∗ B ·∗ u∗x2 = F,

(3.36)

where  A

a11 a21 .. .

a12 a22

... ...

an1

an2

...

b11 b12  b21 b22  =  .  .. bn1 bn2   f1  f2    =  . ,  .. 

... ...

  =   

B

F

...

 a1n a2n   ,  ann  b1n b2n   ,  bnn

fn    u =  

u∗1 u∗2 .. . u∗n

   , 

   u∗xi =  

u∗1xi u∗2xi .. . u∗nxi

   , 

i = 1, 2,

aij , bij , fi , i, j ∈ {1, 2, . . . , n}, are given functions, u is unknown. Now we consider the transformation v = P −1∗ u,

i.e., u = P ·∗ v,

where P is a multiplicative nonsingular n × n-matrix. Then, u∗xi

=

(P ·∗ v)xi

=

Px∗i ·∗ v +∗ P ·∗ vx∗i ,

i = 1, 2,

Classification of First-Order Systems

95

and equation (3.36) takes the following form:

A ·∗ Px∗1 ·∗ v +∗ P ·∗ vx∗1 +∗ B ·∗ Px∗2 ·∗ v +∗ P ·∗ vx∗2 = F or
A ·∗ P ·∗ vx∗1 +∗ B ·∗ P ·∗ vx∗2 = F −∗ A ·∗ Px∗1 +∗ B ·∗ Px∗2 ·∗ v = G. Assuming that A is multiplicative nonsingular, we multiplicative multiply the above equation by (A ·∗ P )−1∗ to obtain (A ·∗ P )−1∗ ·∗ A ·∗ P ·∗ vx∗1 +∗ (A ·∗ P )−1∗ ·∗ B ·∗ P ·∗ vx∗2 = (A ·∗ P )−1∗ ·∗ G or vx∗1 +∗ P −1∗ ·∗ A−1∗ ·∗ B ·∗ P ·∗ vx∗2 = (A ·∗ P )−1∗ ·∗ G.

(3.37)

We set D

= A−1∗ ·∗ B,

H

=

(A ·∗ P )−1∗ ·∗ G.

Then, vx∗1 +∗ P −1∗ ·∗ D ·∗ P vx∗2 = H. If P is taken to be the diagonalizing matrix of D and Λ is a diagonal matrix whose elements are the multiplicative eigenvalues λi of D and the columns of P are multiplicative linearly independent multiplicative eigenvectors of D, pi = (p1i , p2i , . . . , pni ), |pi | = 1, i ∈ {1, 2, . . . , n}. So, P

=

(pij ),

Λ

=

(λj δij ),

i, j ∈ {1, 2, . . . , n},

where δij is the multiplicative Kronecker delta. It follows that P −1∗ ·∗ D ·∗ P = Λ. Thus, we can write equation (3.37) in the following form: vx∗1 +∗ Λ ·∗ vx∗2 = H or vx∗1 +∗ λi ·∗ vx∗2 = hi ,

i ∈ {1, 2, . . . , n},

with n characteristics given by (d∗ x2 )/∗ (d∗ x1 ) = λi , i ∈ {1, 2, . . . , n}.

(3.38)

96

Classification and Canonical Forms

Definition 3.15 Equation (3.38) is said to be the canonical form of equation (3.36). The classification of equation (3.36) is done basing on the nature of the multiplicative eigenvalues λi of D. Definition 3.16 1. If all the n multiplicative eigenvalues of D are real and distinct, then equation (3.36) is said to be multiplicative hyperbolic. 2. If all the n multiplicative eigenvalues of D are complex, then equation (3.36) is said to be multiplicative elliptic. 3. If some of the n multiplicative eigenvalues of D are real and other complex, equation (3.36) is considered as hybrid of multiplicative elliptic-hyperbolic type. 4. If all the n multiplicative eigenvalues of D are real and some of them are repeated, then equation (3.36) is said to be multiplicative parabolic. Example 3.18 Consider the system u∗1x1 +∗ α ·∗ u∗1x2 +∗ β ·∗ u∗2x2

=

0∗

u∗2x1 +∗ γ ·∗ u∗1x2 +∗ δ ·∗ u∗2x2

=

0∗ .

Then,  A

= 

B

=

e 1

1 e



α γ

β δ



, .

The matrix A is multiplicative nonsingular and D = B. We will find the multiplicative eigenvalues of the matrix D. We have α −∗ λ β det∗ (D −∗ λ ·∗ I∗ ) = γ δ −∗ λ ∗ =

0∗

if and only if (λ −∗ α) ·∗ (λ −∗ δ) −∗ β ·∗ γ = 0∗ , if and only if λ2∗ −∗ (α +∗ δ) ·∗ λ +∗ α ·∗ δ −∗ β ·∗ γ = 0∗ .

Classification of First-Order Systems

97

Therefore, 1. If (α −∗ δ)2∗ +∗ e4 ·∗ β ·∗ γ > 0∗ , then the considered system is multiplicative hyperbolic. 2. If (α −∗ δ)2∗ +∗ e4 ·∗ β ·∗ γ = 0∗ , then the considered system is multiplicative parabolic. 3. If (α −∗ δ)2∗ +∗ e4 ·∗ β ·∗ γ < 0∗ , then the considered system is multiplicative elliptic. Example 3.19 Consider the system u∗1x1 −∗ u∗2x2

=

0∗ ,

u∗1x2 −∗ u∗2x1

=

0∗ .

e 1

1 −∗ e



1 e

−∗ e 1



Here  A =  B

=

, .

We have detA = −∗ e 6= 0∗ , i.e., the matrix A is multiplicative nonsingular. Also,   e 1 A−1∗ = , 1 −∗ e D

= A−1∗ ·∗ B   1 −∗ e = . −∗ e 1

We will find the multiplicative eigenvalues of the matrix D. We have −∗ λ −∗ e det(D −∗ λ ·∗ I∗ ) = −∗ e −∗ λ ∗ =

0∗

if and only if λ2∗ −∗ e = 1,

98

Classification and Canonical Forms

if and only if λ1

=

1,

λ2

= −∗ e.

Therefore, the considered system is multiplicative hyperbolic. Example 3.20 Consider the system u∗1x2 −∗ u∗2x1 ∗

∗

(ρ ·∗ u1 )x1 +∗ (ρ ·∗ u2 )x2 where ρ = e +∗ u21∗ +∗ u22∗

=

1

=

0∗ ,


σ ∗

and σ is a multiplicative constant. We have
σ∗ −∗ e
ρ∗xi = σ ·∗ e +∗ u21∗ +∗ u22∗ ·∗ e2 ·∗ u1 u∗1xi +∗ e2 ·∗ u2 u∗2xi , ∗

(ρ ·∗ uj )xi

=

ρ∗xi ·∗ uj +∗ ρ ·∗ u∗jxi ,

i, j = 1, 2.

Therefore, we can rewrite the considered system we can rewrite in this form: u∗1x2 −∗ u∗2x1 ρ ·∗ u∗1x1 +∗ ρ ·∗ u∗2x2

=

0∗

= −∗ ρ∗x1 ·∗ u1 −∗ ρ(x2 ·∗ u2 .

Here A

=

B

=



1 ρ



e 1

 −∗ e , 1  1 . ρ

Since det∗ A = ρ 6= 0∗ , the matrix A is multiplicative nonsingular and   1 e/∗ ρ −1∗ A = , −∗ e 1 D

= A−1∗ ·∗ B   1 e = . −∗ e 1

We will find the multiplicative eigenvalues of the matrix D. We have −∗ λ e det∗ (D −∗ λ ·∗ I∗ ) = −∗ e −∗ λ ∗ =

0∗

Advanced Practical Exercises

99

if and only if λ2∗ +∗ e =

1,

if and only if λ1,2 = ±∗ i. Therefore, the considered system is multiplicative elliptic. Exercise 3.12 Prove that the system u∗1x2 −∗ u∗2x1

=

0∗ ,

u∗1x1 −∗ e9 ·∗ u∗2x2

=

0∗

is a multiplicative hyperbolic system.

3.5

Advanced Practical Exercises

Problem 3.1 Determine the type of the operator L, where 1.

√

L(u) = e

3

√

4 ·∗ u∗∗ x 1 x 1 −∗ e

2.

√

2 L(u) = −∗ u∗∗ x 1 x 1 −∗ e

2

3

∗∗ ·∗ u∗∗ x1 x2 +∗ ux2 x2 +∗ u.

∗∗ ∗ ·∗ u∗∗ x1 x2 +∗ ux2 x2 +∗ ux2 .

3. ∗∗ L(u) = u∗∗ x1 x1 +∗ ux2 x2 +∗ u.

4. ∗∗ ∗ L(u) = u∗∗ x1 x1 −∗ ux2 x2 +∗ ux1 .

5. ∗ ∗ L(u) = e2 ·∗ u∗∗ x1 x2 −∗ ux1 −∗ ux2 .

6. 2 L(u) = u∗∗ x 1 x 1 +∗ e

√

3

3 ∗∗ ·∗ u∗∗ x1 x2 +∗ e ·∗ ux2 x2 .

7. ∗ L(u) = u∗x1 −∗ u∗∗ x2 x2 +∗ ux2 .

Answer

1. Multiplicative hyperbolic.

2. Multiplicative hyperbolic. 3. Multiplicative elliptic.

100

Classification and Canonical Forms 4. Multiplicative hyperbolic. 5. Multiplicative hyperbolic. 6. Multiplicative parabolic. 7. Multiplicative parabolic.

Problem 3.2 Find the canonical form of the following equations: 1. 3 ∗∗ 2 ∗∗ u∗∗ x1 x1 −∗ e ·∗ ux1 x2 +∗ e ·∗ ux2 x2 = 0∗ .

2. ∗∗ 2 ∗∗ u∗∗ x1 x1 −∗ ux1 x2 −∗ e ·∗ ux2 x2 = 0∗ .

3. ∗∗ 2 3 ∗ u∗∗ x2 x2 −∗ (x1 /∗ x2 ) ·∗ ux1 x2 +∗ (e /∗ (e ·∗ x2 )) ·∗ ux2 = 0∗ .

Answer

1. u∗∗ ξ1 ξ2

=

0∗ ,

ξ1

= x 1 +∗ x 2 ,

ξ2

= x2 +∗ e2 ·∗ x1 .

2. u∗∗ ξ1 ξ 2

=

0∗ ,

ξ1

= x2 +∗ e2 ·∗ x1 ,

ξ2

= x 2 −∗ x 1 .

3. 3 ∗ u∗∗ ξ1 ξ2 +∗ (e/∗ (e ·∗ ξ2 )) ·∗ uξ1

=

0,

ξ1

= x1 ·∗ x2 ,

ξ2

= x1 .

Problem 3.3 Consider the equation ∗∗ u∗∗ x1 x1 −∗ ux2 x2 = e.

Advanced Practical Exercises

101

1. Find the canonical form. 2. Find the general solution. 3. Find the solution u(x1 , x2 ) for which u(x1 , 1) u∗x2 (x1 , 1)

=

φ1 (x1 ),

φ1 ∈ C∗ 1 (R1∗ ).

where φ0 ∈ C∗2 (R1∗ ), Answer

= φ0 (x1 ),

1. e4 ·∗ u∗∗ ξ1 ξ2

= e,

ξ1

= x1 +∗ x2 ,

ξ2

= x 1 −∗ x 2 .

2. u(x1 , x2 ) = (x21∗ −∗ x22∗ )/∗ e4 +∗ f (x1 +∗ x2 ) +∗ g(x1 −∗ x2 ), where f and g are C∗2 -functions. 3. u(x1 , x2 ) = −∗ (x22∗ /∗ e2 ) +∗ (φ0 (x1 +∗ x2 ) +∗ φ0 (x1 −∗ x2 ))/∗ e2

1 2

x1Z+∗ x2

+∗ e ·∗

φ1 (x) ·∗ d∗ x. ∗x1 −∗ x2

Problem 3.4 Consider the equation 4 ∗∗ 4 ∗ u∗∗ x1 x1 +∗ e ·∗ ux1 x2 −∗ e ·∗ ux1 = 0∗ .

1. Find the canonical form. 2. Find the general solution. 3. Find a solution u(x1 , x2 ) such that u(x1 , e8 ·∗ x1 )

=

1,

u∗x1 (x1 , e8 ·∗ x1 )

=

−∗ e32 ·∗ x1 .

102

Classification and Canonical Forms

Answer

1. ∗ u∗∗ ξ1 ξ2 −∗ uξ2

=

0∗ ,

ξ1

=

x2 ,

ξ2

=

x2 −∗ e4 ·∗ x1 .

2. u(x1 , x2 ) = ex2 ·∗ f (x2 −∗ e4 ·∗ x1 ) +∗ g(x2 ), where f is a C∗2 -function and g is a C∗1 -function, 3.   8 1 u(x1 , x2 ) = −∗ x2 −∗ e4 ·∗ x1 +∗ e 2 ·∗ e−∗ x2 +∗ e ·∗ x1 1

+∗ (x2 /∗ e2 ) +∗ e 2 . Problem 3.5 Find the canonical form of the equation ∗∗ u∗∗ x1 x1 +∗ x2 ·∗ ux2 x2 = 0∗ ,

x2 > 0∗ .

Answer 1

ξ1

=

e2 ·∗ (x2 ) 2 ∗ ,

ξ2

=

x1

and ∗∗ ∗ u∗∗ ξ1 ξ1 +∗ uξ2 ξ2 −∗ (e/∗ ξ1 ) ·∗ uξ1 = 0∗ .

Problem 3.6 Find the canonical form of the equation 4 ∗∗ 4 ∗∗ 2 ∗ ∗ u∗∗ x1 x1 +∗ e ·∗ ux1 x2 +∗ e ·∗ ux2 x2 −∗ e ·∗ ux1 +∗ ux2 = 0∗ .

Answer ξ1

= x2 ,

ξ2

= x2 −∗ e2 ·∗ x1

and ∗ 5 ∗ e4 ·∗ u∗∗ ξ1 ξ1 +∗ uξ1 +∗ e ·∗ uξ2 = 0∗ .

Problem 3.7 Consider the equation 2 ∗∗ ∗∗ ∗ x21∗ ·∗ u∗∗ x1 x1 −∗ e ·∗ x1 ·∗ ux1 x2 +∗ ux2 x2 +∗ x1 ·∗ ux1 = 0∗ ,

x1 > 0∗ .

Advanced Practical Exercises

103

1. Find the canonical form. 2. Find the general solution. 3. Find the solution u(x1 , x2 ) for which

Answer

2

·∗ x2

u(e, x2 )

=

ee

u∗x1 (e, x2 )

=

1.

ξ1

=

x1 ·∗ ex2 ,

ξ2

=

x2 ,

,

1.

u∗∗ ξ2 ξ2 = 0∗ . 2. u(x1 , x2 ) = x2 ·∗ g (x1 ·∗ ex2 ) +∗ f (x1 ·∗ ex2 ) , where f and g are C∗2 -functions. 3.


2 u(x1 , x2 ) = e −∗ e2 ·∗ log∗ x1 ·∗ x21∗ ·∗ ee ·∗ x2 .

Problem 3.8 Classify the three-dimensional multiplicative heat equation:
∗∗ ∗∗ u∗x1 −∗ α2∗ ·∗ u∗∗ x2 x2 +∗ ux3 x3 +∗ ux4 x4 = 0∗ , where α is a multiplicative constant, α 6= 0∗ . Answer Multiplicative parabolic. Problem 3.9 Prove that the system u∗1x2 −∗ u∗2x1

=

0∗ ,

u∗1x1 +∗ e2 ·∗ u∗1x2 +∗ e4 ·∗ u∗2x2

=

0∗

is a multiplicative elliptic system.

4 The Multiplicative Wave Equation

4.1 4.1.1

The One-Dimensional Multiplicative Wave Equation The Cauchy problem and D’Alambert formula

The multiplicative homogeneous wave equation in one (spatial) dimension has the form 2∗ u∗∗ ·∗ u∗∗ tt −∗ c xx = 0∗ ,

−∗ ∞ ≤ a < x < b ≤ ∞,

b > 0∗ ,

(4.1)

where c ∈ R∗ is called the multiplicative wave speed. Introducing the new multiplicative variables ξ1

= x +∗ c ·∗ t,

ξ2

= x −∗ c ·∗ t,

we get the canonical form of equation (4.1): 2∗ u∗∗ ·∗ u∗∗ tt −∗ c xx

= −∗ e4 ·∗ c2∗ ·∗ u∗∗ ξ1 ξ2 =

0∗ .

Therefore, its general solution is given by u(x, t) = f (x +∗ c ·∗ t) +∗ g(x −∗ c ·∗ t),

(4.2)

where f, g ∈ C∗2 (R∗ ) are two arbitrary functions. Conversely, any two functions f, g ∈ C∗2 (R∗ ) define a solution of the multiplicative wave equation (4.1) via equation (4.2). The function g(x −∗ c ·∗ t) represents a multiplicative wave moving to the right with multiplicative velocity c and it is called a multiplicative forward wave. The function f (x +∗ c ·∗ t) is a multiplicative wave traveling to the left with the same multiplicative speed, and it is called a multiplicative backward wave. Equation (4.2) shows the fact that any solution of the multiplicative wave equation is the multiplicative sum of two such multiplicative traveling waves. Since for any two piecewise continuous functions f and g, equation (4.2) defines a piecewise continuous function u that is a DOI: 10.1201/9781003440116-4

104

The One-Dimensional Multiplicative Wave Equation

105

multiplicative superposition of a multiplicative backward and a multiplicative forward wave traveling in multiplicative opposite directions with multiplicative speed c. Let {fn (s)}n∈N and {gn (s)}n∈N be sequences of multiplicative smooth functions converging at any point t to f and g, respectively, which converge multiplicative uniformly to these functions in any bounded and closed interval that does not contain points of discontinuity. The function un (x, t) = fn (x +∗ c ·∗ t) +∗ gn (x −∗ c ·∗ t) is a proper solution of the wave equation, but the limiting function u(x, t) = f (x +∗ c ·∗ t) +∗ g(x −∗ c ·∗ t) is not necessarily twice multiplicative differentiable, therefore it might be not a solution of (4.1). Definition 4.1 We call a function u that satisfies (4.2) with piecewise continuous functions f and g a generalized solution of the multiplicative wave equation (4.1). The Cauchy problem for the one-dimensional multiplicative homogeneous wave equation is given by 2∗ u∗∗ ·∗ u∗∗ tt −∗ c xx = 0∗ ,

−∗ ∞ < x < ∞,

t > 0∗ ,

(4.3)

u(x, 1) = φ(x), (4.4) u∗t (x, 1) = ψ(x),

−∗ ∞ < x < ∞,

where φ ∈ C∗2 (R∗ ) and ψ ∈ C∗1 (R∗ ). Definition 4.2 A classical (proper) solution of the Cauchy problem (4.3), (4.4) is a function u that is 1. continuously twice multiplicative differentiable for all t > 0∗ , 2. u and u∗t are continuous in the half-space t ≥ 0∗ and such that (4.3), (4.4) are satisfied. Recall that the general solution of the multiplicative wave equation is of the form (4.2). We will find the functions f and g using the initial condition (4.4). Substituting t = 0∗ in (4.2), we obtain u(x, 1) = f (x) +∗ g(x) (4.5) = φ(x). Multiplicative differentiating (4.2) with respect to t and substituting t = 0∗ , we get u∗t (x, 1)

=

c ·∗ f ∗ (x) −∗ c ·∗ g ∗ (x)

=

ψ(x).

106

The Multiplicative Wave Equation

Multiplicative integrating the last equation over [1, x], we get Zx f (x) −∗ g(x) = (e/∗ c) ·∗

ψ(s) ·∗ d∗ s +∗ C,

(4.6)

∗1

where C = f (1) −∗ g(1). Equations (4.5) and (4.6) are two multiplicative linear algebraic equations for f and g. The solution of this system of equations is given by f (x)

1 2

= e ·∗ φ(x) +∗ e

1 2c

Zx

C

·∗

ψ(s) ·∗ d∗ s +∗ e 2 , ∗1

g(x)

1 2

= e ·∗ φ(x) −∗ e

1 2c

Zx

C

·∗

ψ(s) ·∗ d∗ s −∗ e 2 . ∗1

Substituting these expressions for f and g into the general solution (4.2), we obtain 2

u(x, t) = (φ(x +∗ c ·∗ t) +∗ φ(x −∗ c ·∗ t))/∗ e +∗ e

1 2c

x+ Z∗ c·∗ t

·∗

ψ(s) ·∗ d∗ s, ∗x−∗ c·∗ t

(4.7) which is called multiplicative d’Alambert’s formula, shortly d’Alambert’s formula. Example 4.1 Consider the following Cauchy problem: 9 ∗∗ u∗∗ tt −∗ e ·∗ uxx

=

0∗ ,

u(x, 1)

=

u∗t (x, 1) = x2∗ ,

0∗ < x < ∞,

t > 0∗ ,

0∗ < x < ∞.

Here c =

3,

φ(x)

= x2∗ ,

ψ(x)

= x2 .

Then, using d’Alambert’s formula, we get u(x, t)

=

(((x +∗ e3 ·∗ t)2∗ +∗ (x −∗ e3 ·∗ t)2∗ )/∗ e2 )

1 6

3 x+Z ∗ e ·∗ t

+∗ e ·∗ ∗x−∗ e3 ·∗ t

s2∗ ·∗ d∗ s

The One-Dimensional Multiplicative Wave Equation 3

107

·∗ t

=

s=x+∗ e 1 x2∗ +∗ e9 ·∗ t2∗ +∗ e 18 ·∗ s3∗

=

x2∗ +∗ e9 ·∗ t2∗ +∗ e 18 ·∗ (x +∗ e3 ·∗ t)3∗ −∗ (x −∗ e3 ·∗ t)3∗

=

x2∗ +∗ e9 ·∗ t2∗ +∗ x2∗ ·∗ t +∗ e3 ·∗ t3∗ .

s=x−∗ e3 ·∗ t

1

Example 4.2 Consider the following Cauchy problem: 4 ∗∗ u∗∗ tt −∗ e ·∗ uxx

=

0∗ ,

u(x, 1)

=

φ(x) ( e −∗ x2∗

=

−∗ ∞ < x < ∞,

t > 0,

|x|∗ ≤ e

1 otherwise, u∗t (x, 1)

= =

ψ(x) ( (x −∗ e) ·∗ (x −∗ e2 ) e ≤ x ≤ e2 1 otherwise.

We will find u(e, e). Using d’Alambert’s formula, we have

u(e, e)

=

3

2

Ze

1 4

3

((φ(e ) +∗ φ(−∗ e))/∗ e ) +∗ e ·∗

ψ(s) ·∗ d∗ s ∗−∗ e

2

1

Ze

1

∗e 2 Ze

= e 4 ·∗

(s −∗ e) ·∗ (s −∗ e2 ) ·∗ d∗ s

(s2∗ −∗ e3 ·∗ s +∗ e2 ) ·∗ d∗ s

= e 4 ·∗ ∗e

 s=e2 s=e2 s=e2  1 3 3 2 2 ∗ ∗ = e ·∗ e 3 ·∗ s −∗ e 2 · ∗ s +∗ e ·∗ s 1 4

s=e

s=e

s=e

1 24

= −∗ e . Exercise 4.1 Solve the Cauchy problem ∗∗ u∗∗ tt −∗ uxx

=

u(x, 1) = u∗t (x, 1) =

0∗ ,

−∗ ∞ < x < ∞,

t > 0∗ ,

x, cos∗ x,

−∗ ∞ < x < ∞.

Answer u(x, t) = x +∗ sin∗ t ·∗ cos∗ x.




108

The Multiplicative Wave Equation

Theorem 4.1 Fix T > 0∗ . The Cauchy problem (4.3), (4.4) in the domain 0 < x < ∞, 1 ≤ t ≤ T, is well-posed for φ ∈ C∗2 (R∗ ), ψ ∈ C∗1 (R∗ ). Proof D’Alambert’s formula provides us with a solution, and we have shown that any solution of the Cauchy problem (4.3), (4.4) is necessarily equal to d’Alambert solution. Since φ ∈ C∗2 (R∗ ), ψ ∈ C∗1 (R∗ ), we have that u ∈ C∗2 (R∗ × (1, ∞)) ∩ C∗1 (R∗ × [1, ∞)). Therefore, the d’Alambert solution is a classical solution. Now we will prove the stability of the Cauchy problem. Let  > 0∗ be arbitrarily chosen. We take 1 < δ < (/∗ (e +∗ T )). Let also u1 and u2 be solutions of the Cauchy problem with initial conditions φ1 , ψ1 and φ2 , ψ2 , respectively, such that |φ1 (x) −∗ φ2 (x)|∗


1. x∈R∗

4.1.2

The Cauchy problem for the nonhomogeneous wave equation

Consider the following Cauchy problem: 2∗ u∗∗ ·∗ u∗∗ tt −∗ c xx = f (x, t),

0∗ < x < ∞,

t > 1,

(4.8)

u(x, 1) = φ(x), (4.9) u∗t (x, 1) = ψ(x),

0∗ < x < ∞,

where f ∈ C∗ (R∗ × (1, ∞)), φ ∈ C∗2 (R∗ ), and ψ ∈ C∗1 (R∗ ) are given functions. Theorem 4.2 The Cauchy problem (4.8), (4.9) admit at most one solution. Proof Suppose that u1 and u2 are solutions to the problem (4.8), (4.9). Then the function u = u1 −∗ u2 is a solution to the homogeneous problem 2∗ u∗∗ ·∗ u∗∗ tt −∗ c xx = 0∗ ,

1 < x < ∞,

t > 1,

(4.10)

u(x, 1) = 1, (4.11) u∗t (x, 1) = 1,

1 < x < ∞.

Note that v = 1 is also a solution to the homogeneous problem (4.10), (4.11). Hence from Theorem 4.1, we conclude that u = v = 1,

110

The Multiplicative Wave Equation

i.e., u1 = u2 . This completes the proof. Theorem 4.3 Let f, fx∗ ∈ C∗ (R∗ × [1, ∞)), φ ∈ C∗2 (R∗ ), ψ ∈ C∗1 (R∗ ). Then the Cauchy problem (4.8), (4.9) has a solution given by u(x, t) = ((φ(x +∗ c ·∗ t) +∗ φ(x −∗ c ·∗ t))/∗ e2 )

+∗ e

1 2c

x+ Z∗ c·∗ t

·∗

ψ(s) ·∗ d∗ s (4.12)

∗x−∗ c·∗ t

+∗ e

1 2c

Zt

x+∗ c·Z∗ (t−∗ τ )

·∗

f (ξ, τ ) ·∗ d∗ ξ ·∗ d∗ τ. ∗1 ∗x−∗ c·∗ (t−∗ τ )

Definition 4.3 Equation (4.12) is also called d’Alambert’s formula. Remark 4.3 Note that for f = 1 both d’Alambert’s formulas (4.12) and (4.7) coincide. Proof Let (x0 , t0 ) ∈ R∗ × (1, ∞) be arbitrarily chosen. Let 4 be the multiplicative triangle with edges the points (x0 , t0 ), (x0 −∗ c ·∗ t0 , 1) and (x0 +∗ c ·∗ t0 , 1). Multiplicative integrating both sides of equation (4.8) over the multiplicative triangle 4, we get Z Z  2  ∗∗ ec ·∗ u∗∗ (x, t) − u (x, t) ·∗ d∗ x ·∗ d∗ t ∗ tt xx ∗4

Z Z = −∗

f (x, t) ·∗ d∗ x ·∗ d∗ t. ∗4

Using the multiplicative Green formula, we obtain Z Z I
−∗ f (x, t) ·∗ d∗ x ·∗ d∗ t = u∗t (x, t) ·∗ d∗ x +∗ c2∗ ·∗ u∗x (x, t) ·∗ d∗ t ∗∆

∗∂4

(4.13) ∗(x0 −Z∗ c·∗ t0 ,1)

= (x0 ,t0 )

u∗t (x, t) ·∗ d∗ x +∗ c2∗ ·∗ u∗x (x, t) ·∗ d∗ t




The One-Dimensional Multiplicative Wave Equation (x0 +Z ∗ c·∗ t0 ,1)

+∗

111

u∗t (x, t) ·∗ ·∗ d∗ x +∗ c2∗ ·∗ u∗x (x, t) ·∗ d∗ t




∗(x0 −∗ c·∗ t0 ,1) (xZ0 ,t0 )

+∗


u∗t (x, t) ·∗ d∗ x +∗ c2∗ ·∗ u∗x (x, t) ·∗ d∗ t .

∗(x0 +∗ c·∗ t0 ,1)

Note that (x0 −Z ∗ c·∗ t0 ,0)

u∗t (x, t) ·∗ d∗ x +∗ c2∗ ·∗ u∗x (x, t) ·∗ d∗ t




∗(x0 ,t0 ) (x0 −Z ∗ c·∗ t0 ,1)

=

c ·∗ u∗t (x, t) ·∗ d∗ t

∗(x0 ,t0 )

+∗ c

·∗ u∗x (x, t) ·∗



(4.14)

d∗ x

(x0 −Z ∗ c·∗ t0 ,1)

= c ·∗

d∗ u ∗(x0 ,t0 )

= c ·∗ (u(x0 −∗ c ·∗ t0 , 1) −∗ u(x0 , t0 )) = c ·∗ (φ(x0 −∗ c ·∗ t0 ) −∗ u(x0 , t0 )) , (x0 +Z ∗ c·∗ t0 ,1)

u∗t (x, t) ·∗ d∗ x +∗ c2∗ ·∗ u∗x (x, t) ·∗ d∗ t




∗(x0 −∗ c·∗ t0 ,1) x0 + Z∗ c·∗ t0

=

u∗t (x, 0∗ ) ·∗ d∗ x

∗x0 −∗ c·∗ t0 x0 + Z∗ c·∗ t0

ψ(x) ·∗ d∗ x,

= ∗x0 −∗ c·∗ t0

(4.15)

112

The Multiplicative Wave Equation

and (xZ0 ,t0 )

u∗t (x, t) ·∗ d∗ x +∗ c2∗ ·∗ u∗x (x, t) ·∗ d∗ t




∗(x0 +∗ c·∗ t0 ,1) (xZ0 ,t0 )



=

−∗ c ·∗ u∗t (x, t) ·∗ d∗ t

∗(x0 +∗ c·∗ t0 ,1)

−∗ c ·∗ u∗x (x, t) ·∗ d∗ x



(4.16)

(xZ0 ,t0 )

= −∗ c · ∗

·∗ d∗ u ∗(x0 +∗ c·∗ t0 ,1)

= −∗ c ·∗ (u(x0 , t0 ) −∗ u(x0 +∗ c ·∗ t0 , 1)) = −∗ c ·∗ (u(x0 , t0 ) −∗ φ(x0 +∗ c ·∗ t0 )) . We substitute (4.14), (4.15) and (4.16) into (4.13) and we find Z Z −∗ f (x, t) ·∗ d∗ x ·∗ d∗ t ∗4 x0 + Z∗ c·∗ t0

= c ·∗ (φ(x0 −∗ c ·∗ t0 ) −∗ u(x0 , t0 )) +∗

ψ(x) ·∗ d∗ x ∗x0 −∗ c·∗ t0

−∗ c ·∗ (u(x0 , t0 ) −∗ φ(x0 +∗ c ·∗ t0 )) , or u(x0 , t0 )

= ((φ(x0 +∗ c ·∗ t0 ) +∗ φ(x0 −∗ c ·∗ t0 ))/∗ e2 )

+∗ e

1 2c

x0 + Z∗ c·∗ t0

·∗

ψ(x) ·∗ d∗ x ∗x0 −∗ c·∗ t0

+∗ e

1 2c

Z Z ·∗

f (x, t) ·∗ d∗ x ·∗ d∗ t. ∗4

Because (x0 , t0 ) ∈ R∗ × (1, ∞) was arbitrarily chosen, we finally obtain an explicit formula for the solutions of the Cauchy problem (4.8), (4.9) at an arbitrary point (x, t) given by (4.12). Now we will prove that the function

The One-Dimensional Multiplicative Wave Equation

113

u(x, t) given by the formula (4.12) is indeed a solution to the Cauchy problem (4.8), (4.9). We have u(x, 0) u∗t (x, t)

=

φ(x),

= (c ·∗ φ∗ (x +∗ c ·∗ t) −∗ c ·∗ φ∗ (x −∗ c ·∗ t))/∗ e2 +∗ (ψ(x +∗ c ·∗ t) +∗ ψ(x −∗ c ·∗ t))/∗ e2 Zt

1 2

+∗ e ·∗

(f (x +∗ c ·∗ (t −∗ τ ), τ ) ∗1

+∗ f (x −∗ c ·∗ (t −∗ τ ), τ )) ·∗ d∗ τ, u∗t (x, 1)

=

ψ(x),

u∗∗ tt (x, t)

=

c2∗ ·∗ (φ∗∗ (x +∗ c ·∗ t) +∗ φ∗∗ (x −∗ c ·∗ t))/∗ e2 +∗ (c ·∗ ψ ∗ (x +∗ c ·∗ t) −∗ c ·∗ ψ ∗ (x −∗ c ·∗ t))/∗ e2 +∗ f (x, t) Zt

1 2

+∗ e ·∗

(c ·∗ fx∗ (x +∗ c ·∗ (t −∗ τ ), τ )

∗1

−∗ c ·∗ fx (x −∗ c ·∗ (t −∗ τ ), τ )) ·∗ d∗ τ u∗x (x, t)

=

(φ∗ (x +∗ c ·∗ t) +∗ φ∗ (x −∗ c ·∗ t))/∗ e2 +∗ (ψ(x +∗ c ·∗ t) −∗ ψ(x −∗ c ·∗ t))/∗ ee

+∗ e

1 2c

2

·∗ c

Zt ·∗

(f (x +∗ c ·∗ (t −∗ τ ), τ ) ∗1

−∗ f (x −∗ c ·∗ (t −∗ τ ), τ )) ·∗ d∗ τ, u∗xx (x, t)

=

(φ∗∗ (x +∗ c ·∗ t) +∗ φ∗∗ (x −∗ c ·∗ t))/∗ e2 +∗ (ψ ∗ (x +∗ c ·∗ t) −∗ ψ ∗ (x −∗ c ·∗ t))/∗ ee

+∗ e

1 2c

Zt ·∗

(fx∗ (x +∗ c ·∗ (t −∗ τ ), τ )

∗1

−∗ fx∗ (x

−∗ c ·∗ (t −∗ τ ), τ )) ·∗ d∗ τ,

2

·∗ c

114

The Multiplicative Wave Equation

whereupon 2∗ u∗∗ ·∗ u∗∗ tt (x, t) −∗ c xx (x, t) = f (x, t).

Therefore, u is a solution to the Cauchy problem (4.8), (4.9). This completes the proof. Example 4.3 Consider the following Cauchy problem: 4 ∗∗ u∗∗ tt −∗ e ·∗ uxx

=

ex +∗ sin∗ t,

u(x, 1)

=

1,

u∗t (x, 1)

=

e/∗ (= e +∗ x2∗ ),

1 < x < ∞,

t > 1,

1 < x < ∞.

Here c = e2 , f (x, t) φ(x)

= ex +∗ sin∗ t, =

ψ(x) = e/∗ (e +∗ x2∗ ).

1,

Then, using d’Alanbert’s formula (4.12), we get

u(x, t)

2 x+Z ∗ e ·∗ t

1 4

(e/∗ (e +∗ s2∗ )) ·∗ d∗ s

= e ·∗ ∗x−∗ e2 ·∗ t

1 4

Zt

x+∗ e2Z·∗ (t−∗ τ )

+∗ e · ∗


eξ +∗ sin∗ τ ·∗ d∗ τ

∗1 ∗x−∗ e2 ·∗ (t−∗ τ )

s=x+∗ e2 ·∗ t 1 1 = e 4 ·∗ arctan∗ s +∗ e 4 · ∗ 2 s=x−∗ e ·∗ t

Zt


ξ=x+∗ e2 ·∗ (t−∗ τ ) eξ +∗ ξ ·∗ sin∗ τ ·∗ d∗ τ 2 ξ=x−∗ e ·∗ (t−∗ τ )

∗1

=

(arctan∗ (x +∗ e2 ·∗ t) −∗ arctan(x −∗ e2 ·∗ t))/∗ e4

1

+∗ e 4 · ∗

Zt 

ex+∗ e

2

·∗ t−∗ e2 ·∗ τ

−∗ ex−∗ e

∗1


+∗ e4 ·∗ (t −∗ τ ) ·∗ sin∗ τ ·∗ d∗ τ

2

·∗ t+∗ e2 ·∗ τ

The One-Dimensional Multiplicative Wave Equation

115

1 4

= e ·∗ arctan∗ (x +∗ e2 ·∗ t) −∗ arctan∗ (x −∗ e2 ·∗ t) −∗ ex
+∗ ex ·∗ cosh∗ (e2 ·∗ t) +∗ t −∗ sin∗ t. Example 4.4 Consider the following Cauchy problem: ∗∗ u∗∗ tt −∗ uxx

=

cos∗ (x +∗ t),

u(x, 1)

=

x,

u∗t (x, 1)

=

sin∗ x,

1 < x < ∞,

t > 1,

1 < x < ∞.

Here c =

e,

f (x, t)

=

cos∗ (x +∗ t),

φ(x)

=

x,

ψ(x)

=

sin∗ x.

Then, using d’Alambert’s formula (4.12), we have

u(x, t)

=

2

x+ Z ∗t

1 2

(x +∗ t +∗ x −∗ t)/∗ e +∗ e ·∗

sin∗ s ·∗ d∗ s ∗x−∗ t

Zt

1 2

x+∗Z(t−∗ τ )

+∗ e ·∗

cos∗ (ξ +∗ τ ) ·∗ d∗ ξ ·∗ d∗ τ ∗1 x−∗ (t−∗ τ )

=

Zt s=x+∗ t ξ=x+∗ (t−∗ τ ) 1 2 x −∗ e ·∗ cos∗ s +∗ e ·∗ sin∗ (ξ +∗ τ ) ·∗ d∗ τ 1 2

s=x−∗ t

ξ=x−∗ (t−∗ τ )

∗1 1

= x −∗ e 2 ·∗ (cos∗ (x +∗ t) −∗ cos∗ (x −∗ t)) Zt

1 2

+∗ e ·∗


sin∗ (x +∗ t) −∗ sin∗ (x −∗ t +∗ e2 ·∗ τ ) ·∗ d∗ τ

∗1 1 2

1

= x +∗ e ·∗ sin∗ x ·∗ sin∗ t +∗ e 2 ·∗ t ·∗ sin∗ (x +∗ t).

116

The Multiplicative Wave Equation

Exercise 4.2 Solve the Cauchy problem ∗∗ u∗∗ tt −∗ uxx

= x ·∗ t,

u(x, 1) =

1 < x < ∞,

t > 1,

1,

u∗t (x, 1) = ex ,

1 < x < ∞.

Answer

1

u(x, t) = ex ·∗ sinh∗ t +∗ e 6 ·∗ x ·∗ t3∗ . Theorem 4.4 Let T > 1 be fixed and f, fx∗ ∈ C∗ (R∗ × [1, ∞)), φ ∈ C∗2 (R∗ ), ψ ∈ C∗1 (R∗ ). Then the Cauchy problem (4.8), (4.9) is well-posed in the domain 1 < x < ∞, 1 ≤ t ≤ T. Proof Let  > 1 be arbitrarily chosen and 1 < δ < /∗ (e +∗ T +∗ ((T 2∗ )/∗ ∗ e )). Let also fi , φi and ψi , i = 1, 2, be such that fi , fix ∈ C∗ (R∗ × [1, ∞)), 2 1 φi ∈ C∗ (R∗ ), ψi ∈ C∗ (R∗ ), and 2

|f1 (t, x) −∗ f2 (t, x)|∗


1.

120

The Multiplicative Wave Equation

Since u is a nontrivial solution, it follows that X ∗ (1) = X ∗ (eL ) = 1. Therefore, X(x) should be a solution of the eigenvalue problem: X ∗∗ +∗ λ ·∗ X = 1,

1 < x < eL ,

(4.22)

X ∗ (1) = X ∗ (eL ) = 1.

(4.23)

For the general solution of the first equation of (4.21) we have the following: 1. 1

1

X(x) = c1 ·∗ cosh∗ ((−∗ λ) 2 ∗ ·∗ x) +∗ c2 ·∗ sinh∗ ((−∗ λ ·∗ x)) 2 ∗ for λ < 1. 2. X(x) = c1 +∗ c2 ·∗ x for λ = 1. 3.

1

1

X(x) = c1 ·∗ cos∗ ((λ) 2 ∗ ·∗ x) +∗ c2 ·∗ sin∗ ((λ) 2 ∗ x) for λ > 1, where c1 and c2 are arbitrary real constants. 1. When λ < 1, using (4.23), we find c1 = c2 = 1. Therefore, X(x) ≡ 1 and the multiplicative eigenvalue problem (4.22), (4.23) does not admit multiplicative negative eigenvalues. 2. When λ = 1, using (4.23), we find X(x) = c1 . 3. When λ > 1, then c2 = 1 and 1

1

c1 ·∗ (λ) 2 ∗ ·∗ sin∗ (L(λ) 2 ∗ ) = 1. If c1 = 1, then X(x) ≡ 1 and the multiplicative eigenvalue problem (4.22), (4.23) does not admit multiplicative positive eigenvalues. Thus 1

(λ) 2 ∗ ·∗ eL = enπ ,

n ∈ N,

The One-Dimensional Multiplicative Wave Equation or λ = en

2

π2

2

/∗ (eL ),

121

n ∈ N.

The associated multiplicative eigenfunction is X(x) = cos∗ ((enπ ·∗ x)/∗ eL ), which is uniquely determined up to a multiplicative factor. Hence, the solution of the eigenvalue problem (4.22), (4.23) is an infinite sequence of multiplicative nonnegative simple multiplicative eigenvalues and their associated multiplicative eigenfunctions will be denoted by Xn (x)

=

cos∗ ((enπ ·∗ x)/∗ eL ),

λn

=



enπ

2

L

2∗

,

n ∈ N0 .

Consider the second equation of (4.21) for λ = λn . The solutions are T0 (t) = α0 +∗ β0 ·∗ t, Tn (t) = αn ·∗ cos∗ (c ·∗ e

πn L

·∗ t) +∗ βn ·∗ sin∗ (c ·∗ e

πn L

·∗ t),

n ∈ N,

where αn , βn , n ∈ N0 are multiplicative real constants. Then the multiplicative product solutions of the initial boundary value problem (4.17)–(4.19) are given by u0 (x, t) = X0 (x) ·∗ T0 (t) = A0 +∗ B0 ·∗ t, un (x, t) = cos∗ (e

nπ L

·∗ x) ·∗ An ·∗ cos∗ (c ·∗ e

+∗ Bn ·∗ sin∗ (c ·∗ e

πn L


·∗ t) ,

πn L

·∗ t)

n ∈ N.

Applying the multiplicative superposition principle, the function u(x, t)

=

A0 + ∗ B 0 · ∗ t + ∗

∞  X

An ·∗ cos∗ (c ·∗ e

πn L

·∗ t)

∗n=1

+∗ Bn ·∗ sin∗ (c ·∗ e

πn L

 πn ·∗ t) ·∗ cos∗ (c ·∗ e L ·∗ x)

is a generalized (formal) solution of the problem (4.17)–(4.19). To find the constants An and Bn , n ∈ N0 , we will use the initial condition (4.19). Assume that the initial data φ and ψ can be expanded into multiplicative generalized Fourier series with respect to the sequence of the multiplicative eigenfunctions

122

The Multiplicative Wave Equation

of the problem (4.22), (4.23) and these series are multiplicative uniformly convergent, ∞ X πn φ(x) = a0 +∗ an ·∗ cos∗ (c ·∗ e L ·∗ x), ∗n=1

ψ(x) = b0 +∗

∞ X

bn ·∗ cos∗ (c ·∗ e

πn L

·∗ x).

∗n=1

We have that Ze

1 L

L

a0 = e ·∗

φ(x) ·∗ d∗ x, ∗1

Ze

1 L

L

b0 = e · ∗

ψ(x) ·∗ d∗ x, ∗1

Ze

2 L

L

am = e ·∗

cos∗ (c ·∗ e

πm L

·∗ x) ·∗ φ(x) ·∗ d∗ x,

∗1

Ze

2 L

L

bm = e ·∗

cos∗ (c ·∗ e

πm L

·∗ x) ·∗ ψ(x) ·∗ d∗ x,

m ∈ N.

∗1

Hence, u(x, 1)

=

∞ X

A0 + ∗

An ·∗ cos∗ (c ·∗ e

πn L

·∗ x)

∗n=1

=

∞ X

a0 +∗

an ·∗ cos∗ (c ·∗ e

πn L

·∗ x),

∗n=1

whereupon an = An ,

n ∈ N0 .

Also, u∗t (x, 1)

=

B 0 +∗

∞ X

Bn ·∗ c ·∗ e

πn L

·∗ cos∗ (c ·∗ e

∗n=1

= b0 + ∗

∞ X ∗n=1

bn ·∗ cos∗ (c ·∗ e

πn L

·∗ x).

πn L

·∗ x)

The One-Dimensional Multiplicative Wave Equation

123

Therefore, B0

=

b0 ,

Bn

= (eL ·∗ bn )/∗ c ·∗ e

πn L

n ∈ N.

,

Thus the problem (4.17)–(4.19) is formally solved. Example 4.5 We will find a formal solution to the following problem: ∗ u∗∗ tt −∗ uxx

=

1 < x < eπ ,

1,

u∗x (1, t)

= u∗x (eπ , t) = 1,

u(x, 1)

=

(cos∗ x)2∗ ,

u∗t (x, 1)

=

(sin∗ x)2∗ ,

t > 1,

t ≥ 1,

1 ≤ x ≤ eπ .

Here, L = π, φ(x)

=

(cos∗ x)2∗ ,

ψ(x)

=

(sin∗ x)2∗ .

Then, Xn (x) λn a0

= cos∗ (en ·∗ x), 2

= en , Ze

π

π

= (e/∗ e ) ·∗

(cos∗ x)2∗ ·∗ d∗ x

∗1 1

= e2 ,

b0

Ze

1 π

π

(sin∗ x)2∗ ·∗ d∗ x

= e ·∗ ∗1 1

= e2 ,

am

Ze

2 π

π

= e ·∗

cos∗ (em ·∗ x)(cos∗ x)2∗ ·∗ d∗ x

∗1

=

  1 

m 6= 2 1

e2

m = 2,

124

The Multiplicative Wave Equation bm

Ze

2 π

π

cos∗ (em ·∗ x)(sin∗ x)2∗ ·∗ d∗ x

= e ·∗ ∗1

  1

=



m 6= 2 1

−∗ e 2

m = 2.

Therefore, A0 = e, B0 = e, 1

A2 = e 2 , 1

B2 = −∗ e 4 , Am = B m = 1,

m 6= 2,

m ∈ N,

and  1  1 t 1 u(x, t) = e 2 +∗ e 2 +∗ e 2 ·∗ cos∗ (e2 ·∗ t) −∗ e 4 ·∗ sin∗ (e2 ·∗ t) ·∗ cos∗ (e2 ·∗ x). Exercise 4.5 Find a formal solution to the following problem: ∗∗ u∗∗ tt −∗ uxx

=

1,

u∗x (1, t)

=

u∗x (eπ , t) = 1,

u(x, 1)

=

1,

u∗t (x, 1)

=

(sin∗ x)3∗ ,

1 < x < eπ ,

t > 1,

t ≥ 1,

1 ≤ x ≤ eπ .

Answer u(x, t)

= e

4 3π

· ∗ t +∗

∞ X

12

e πn(4n2 −1)(4n2 −9) ·∗ sin∗ (e2n ·∗ t) ·∗ cos∗ (e2n ·∗ x).

∗n=1

Now we consider the nonhomogeneous initial boundary value problem: 2

c ∗∗ u∗∗ tt −∗ e uxx = f (x, t),

1 < x < eL ,

u∗x (1, t) = u∗x (eL , t) = 1,

t ≥ 1,

t > 1,

(4.24) (4.25)

u(x, 1) = φ(x), (4.26) u∗t (x, 1) = ψ(x),

1 ≤ x ≤ eL ,

The One-Dimensional Multiplicative Wave Equation

125

where f (x, t) = e

C0

+∗ e

D0

∞  X

· ∗ t +∗

eCn ·∗ cos∗ (e

cπn L

·∗ t)

∗n=1

+∗ e

Dn

·∗ sin∗ (e ∞ X

φ(x) = ea0 +∗

cπn L

 ·∗ t) ·∗ cos∗ (e

ean ·∗ cos∗ (e

πn L

πn L

1 ≤ x ≤ eL ,

·∗ x),

t ≥ 1,

·∗ x),

∗n=1 ∞ X

ψ(x) = eb0 +∗

ebn ·∗ cos∗ (e

πn L

1 ≤ x ≤ eL ,

·∗ x),

t ≥ 1,

∗n=1

φ∗ (1) = φ∗ (eL ) = ψ ∗ (1) = ψ ∗ (eL ) = 1, Cn , Dn , an and bn , n ∈ N0 are given constants. Let u(x, t) be a solution to the problem (4.24)–(4.26) in the following form: ∞ X

u(x, t) = T0 (t) +∗

Tn (t) ·∗ cos∗ (e

πn L

·∗ x).

(4.27)

∗n=1

Substituting (4.27) in (4.24), we get T0∗∗ +∗

∞  X

Tn∗∗ +∗ e

c2 π 2 n 2 L2

 πn ·∗ Tn ·∗ cos∗ (e L ·∗ x)

∗n=1

= e

C0

+∗ e

D0

· ∗ t +∗

∞  X

eCn ·∗ cos∗ (e

cπn L

·∗ t)

∗n=1

+∗ eDn ·∗ sin∗ (e

cπn L

 πn ·∗ t) ·∗ cos∗ (e L ·∗ x).

Hence, T0∗∗ = eC0 +∗ eD0 ·∗ t, Tn∗∗ +∗ e

c2 π 2 n 2 L2

·∗ Tn = eCn ·∗ cos∗ (e

cπn L

·∗ t) +∗ eDn ·∗ sin∗ (e

cπn L

·∗ t).

126

The Multiplicative Wave Equation

Therefore, C0 2

·∗ t2∗ +∗ e

T0 (t)

= e

Tn (t)

= eαn ·∗ cos∗ (e

D0 6

cπn L

Dn L

·∗ t3∗ +∗ eβ0 ·∗ t +∗ eα0 ,

·∗ t) +∗ eβn ·∗ sin∗ (e

−∗ e 2cπn ·∗ t cos∗ (e

cπn L

·∗ t)

Cn L

cπn L

·∗ t) +∗ e 2cπn ·∗ t ·∗ sin∗ (e

cπn L

·∗ t),

n ∈ N.

We will find the constants eα0 , eαn and eβn , n ∈ N, using the initial condition (4.26). We have u(x, 1)

=

∞ X

eα0 +∗

eαn ·∗ cos∗ (e

πn L

·∗ x)

∗n=1 ∞ X

= ea0 +∗

ean ·∗ cos∗ (e

πn L

·∗ x),

∗n=1

whereupon αn = an , n ∈ N0 . Also, u∗t (x, 1)

eβ0 +∗

=

∞  X

eβn ·∗ e

cπn L

 Dn L ( πn −∗ e 2cπn ·∗ cos∗ e L ·∗ x)

∗n=1

= e b 0 +∗

∞ X

ebn ·∗ cos∗ (e

πn L

·∗ x),

∗n=1

whereupon β0

=

b0 ,

βn

=

L cπn

 bn +

Dn L 2cπn

 .

Therefore, u(x, t)

=

e

C0 2

+∗

·∗ t2∗ +∗ e ∞  X

D0 6

·∗ t3∗ +∗ eb0 ·∗ t +∗ ea0 ·∗

ean ·∗ cos∗ (e

cπn L

·∗ t)

∗n=1

  Dn L L cπn +∗ e cπn ·∗ ebn +∗ e 2cπn ·∗ sin∗ (e L ·∗ t) Dn L

−∗ e 2cπn ·∗ t∗ cos∗ (e ·∗ cos∗ (e

πn L

·∗ x).

cπn L

Cn L

·∗ t) +∗ e 2cπn ·∗ t ·∗ sin∗ (e

cπn L

·∗ t)



The One-Dimensional Multiplicative Wave Equation

127

Example 4.6 We will find a formal solution to the following problem: ∗∗ 2π u∗∗ ·∗ x) ·∗ cos∗ (e2π ·∗ t), tt −∗ uxx = cos∗ (e

1 < x < e, u∗x (1, t) = u∗x (e, t) = 1, u(x, 1) = (cos∗ (πx))2∗ ,

t > 1,

t ≥ 1,

u∗t (x, 1) = e2 ·∗ cos∗ (e2π ·∗ x),

1 ≤ x ≤ e.

Here, c =

e,

f (x, t)

=

cos∗ (e2π ·∗ x) ·∗ cos∗ (e2π ·∗ t),

φ(x)

=

e 2 +∗ e 2 ·∗ cos∗ (e2π ·∗ x),

ψ(x)

= e2 ·∗ cos∗ (e2π ·∗ x),

1

1 ≤ x ≤ e,

1

1 ≤ x ≤ e.

Then, C0

= D0 =

0,

C1

=

0,

C2

=

1,

Cn

=

0,

n ∈ N,

Dn

=

0,

n ∈ N,

a0

=

1 , 2

a1

=

0,

a2

=

1 , 2

an

=

0,

b0

=

0,

n ∈ N,

n ≥ 3,

n ≥ 3,

t ≥ 1,

128

The Multiplicative Wave Equation

b1

=

0,

b2

=

2,

bn

=

0,

n ∈ N,

n ≥ 3.

Therefore,  1  1 u(x, t) =e 2 +∗ e 2 ·∗ cos∗ (e2π ·∗ t) +∗ ((t +∗ e4 )/∗ e4π ) ·∗ sin∗ (e2π ·∗ t) ·∗ cos∗ (e2π ·∗ x). Exercise 4.6 Find a formal solution to the following problem: 2

c ∗∗ u∗∗ tt −∗ e ·∗ uxx

=

1,

1 < x < eL ,

u(1, t) = u(eL , t)

=

1,

t ≥ 1,

u(x, 1)

=

φ(x),

u∗t (x, 1)

=

ψ(x),

t > 1,

1 ≤ x ≤ eL ,

where φ(1) φ(eL )

= φ∗∗ (1) = ψ(1) = 1, = φ∗∗ (eL ) = ψ(eL ) = 1.

Answer u(x, t) =

∞ X

eAn ·∗ cos∗ (e

cπn L

·∗ t) +∗ eBn ·∗ sin∗ (e

cπn L


nπ ·∗ t) ·∗ sin∗ (e L ·∗ x),

∗n=1

2 L

eAn = e ·∗

Ze

L

·∗ φ(x) ·∗ sin∗ (e

nπ L

·∗ x) ·∗ d∗ x,

∗1

Bn = e

2 cnπ

Ze

L

·∗

ψ(x) ·∗ sin∗ (e ∗1

nπ L

·∗ x) ·∗ d∗ x,

n ∈ N.

The One-Dimensional Multiplicative Wave Equation

4.1.4

129

The energy method: uniqueness

The energy method is a fundamental tool in the theory of MPDEs for proving the uniqueness of the solutions of initial boundary value problems. Consider the problem 2

c ∗∗ u∗∗ tt −∗ e ·∗ uxx = f (x, t), L

u(1, t) = g(t), u(x, 1) = φ(x), L

where f ∈ C∗ ([1, e ] × [1, ∞)),

g (1) = ψ(1),

t > 1,

(4.28)

t ≥ 1,

u(e , t) = h(t),

u∗t (x, 1) = ψ(x), g, h ∈ C∗2 ([1, ∞)),

∗

g(1) = φ(1),

1 < x < eL ,

(4.29) L

1≤x≤e , φ∈

∗

C∗2 (R∗ ),

L

h (1) = ψ(e ),

ψ∈

(4.30) C∗1 (R∗ )

and

L

h(1) = φ(e ).

Definition 4.5 The condition (4.29) is called Dirichlet boundary condition. Let u1 and u2 be two solutions of the problem (4.28)–(4.30). We set v(x, t) = u1 (x, t) −∗ u2 (x, t). Then, 2

∗∗ ∗∗ vtt −∗ ec ·∗ vxx

=

1,

1 < x < eL ,

v(1, t) = v(eL , t)

=

1,

t ≥ 1,

v(x, 1) = vt (x, 1)

=

1,

1 ≤ x ≤ eL .

t > 1,

Define the total energy L

1 2

E(t) = e ·∗

Ze 

 2 (vt∗ (x, t))2∗ +∗ ec (vx∗ (x, t))2∗ ·∗ d∗ x.

∗1

Here, Ze

1 2

L

(vt∗ (x, t))2∗ d∗ x

e ·∗ ∗1

is the multiplicative total kinetic energy, while e

c2 2

Ze

L

·∗

(vx∗ (x, t))2∗ d∗ x

∗1

is the multiplicative total potential energy. We have L

∗

E (t) =

Ze  1

 2 ∗∗ ∗∗ vt∗ (x, t) ·∗ vtt (x, t) +∗ ec ·∗ vx∗ (x, t) ·∗ vxt (x, t) ·∗ d∗ x.

(4.31)

130

The Multiplicative Wave Equation

Note that 2

∗∗ ec ·∗ vx∗ (x, t) ·∗ vxt (x, t) 2

∗∗ = ec ·∗ ((vx∗ (x, t) ·∗ vt∗ (x, t))x −∗ vxx (x, t) ·∗ vt∗ (x, t)) 2

2

∗∗ = ec ·∗ (vx∗ (x, t) ·∗ vt∗ (x, t))x −∗ ec ·∗ vxx (x, t) ·∗ vt∗ (x, t) 2

∗∗ = ec ·∗ (vx∗ (x, t) ·∗ vt∗ (x, t))x −∗ vtt (x, t) ·∗ vt∗ (x, t).

Hence and by (4.31), we get ∗

E (t)

= e

c2

Ze

L

·∗

(vx∗ (x, t) ·∗ vt (x, t))x ·∗ d∗ x

∗1

x=eL 2 = ec ·∗ vx∗ (x, t) ·∗ vt∗ (x, t) x=1

= 1, where we have used that vt∗ (1, t) = vt∗ (eL , t) = 1. Therefore, E(t) = const. By the initial conditions, we have vt∗ (x, 1) = vx∗ (x, 1) = 1, 1 ≤ x ≤ eL . Therefore, E(1) = 1 and E(t) ≡ 1 on [1, ∞). Consequently, 2

(vt∗ (x, t))2∗ +∗ ec ·∗ (vx∗ (x, t))2∗ = 1 on [1, eL ] × [1, ∞). Hence, vt∗ (x, t) = vx∗ (x, t) = 1 on [1, eL ] × [1, ∞). From here, v(x, t) = const on [1, eL ] × [1, ∞). Using v(x, 1) = 1 on [1, eL ], we conclude that v(x, t) = 1 on [1, L] × [1, ∞), from where u1 (x, t) = u2 (x, t)

on

[1, eL ] × [1, ∞).

The Multiplicative Wave Equation in R3∗

131

Exercise 4.7 Prove that the problem 2

c ∗∗ u∗∗ tt −∗ e ·∗ uxx

u∗∗ x (1, t) = g(t),

=

f (x, t),

u∗x (eL , t)

=

h(t),

u∗t (x, 1)

=

ψ(x),

u(x, 1) = φ(x),

1 < x < eL ,

t > 1,

t ≥ 1, 1 ≤ x ≤ eL ,

where f ∈ C∗ ([1, eL ] × [1, ∞)), h, g ∈ C∗1 ([1, ∞)), φ ∈ C∗2 (R∗ ), ψ ∈ C∗1 (R∗ ) and h(1)

=

φ∗ (L),

g(1)

=

φ∗ (1),

h∗ (1)

=

ψ ∗ (eL ),

g ∗ (1)

=

ψ ∗ (1),

has unique solution.

4.2 4.2.1

The Multiplicative Wave Equation in R3∗ Radially symmetric solutions

We seek solutions u(x1 , x2 , x3 , t) to the multiplicative wave equation u∗tt −∗ c2 ∆u = 0∗ ,

(x1 , x2 , x3 ) ∈ R3 ,

−∗ ∞ < t < ∞,

(4.32)

2∗ ∗∗ ∗∗ ∆∗ u = u∗∗ x1 x1 +∗ ux2 x2 +∗ ux3 x3 , that are of the form u = u(r, t), r = (x1 2∗ 2∗ 12 ∗ +∗ x2 +∗ x3 ) . Thus

u∗xi u∗∗ xi xi

= u∗r ·∗ (xi /∗ r), = u∗rr ·∗ ((x2i ∗ )/∗ (r2∗ )) +∗ u∗r ·∗ ((r2∗ −∗ x2i ∗ )/∗ (r3∗ )),

i = 1, 2, 3,

and ∗∗ ∗∗ ∗∗ 2 ∗ u∗∗ x1 x1 +∗ ux2 x2 +∗ ux3 xz3 = urr +∗ (e /∗ r) ·∗ ur ,

i.e., the function u(r, t) satisfies the equation
c2 ∗∗ 2 ∗ u∗∗ tt −∗ e ·∗ urr +∗ (e /∗ r) ·∗ ur = 0∗ .

(4.33)

132

The Multiplicative Wave Equation

Defining v(r, t) = r ·∗ u(r, t), we see that v satisfies the equation 2

∗∗ ∗∗ vtt −∗ ec ·∗ vrr = 0∗ .

This is exactly the one-dimensional multiplicative wave equation. Therefore, the general radial solution for equation (4.33) is given by u(r, t) = (e/∗ r) ·∗ (f (r +∗ ec ·∗ t) +∗ g(r −∗ ec ·∗ t)) , where f, g ∈ C∗2 (R∗ ). We can solve the Cauchy problem that consists of (4.33) for t > 1 and the initial conditions u(r, 1) = φ(r), (4.34) u∗t (r, 1) = ψ(r),

1 ≤ r ≤ ∞.

Note that the initial conditions are only given along the ray r ≥ 1 and not for all r. If we suppose that φ, ψ ∈ C∗1 ([1, ∞)) and φ∗ (1) = ψ ∗ (1) = 1, then we can extend φ and ψ to the whole line 0 < r < ∞ by defining them to be multiplicative even extensions of the given φ and ψ. Hence, the initial conditions for v are multiplicative odd functions, and therefore v(r, t) is multiplicative odd, which implies that u(r, t) is a multiplicative even function. ˜ reIf we denote the multiplicative even extensions of φ and ψ by φ˜ and ψ, spectively, we thus obtain the following radially symmetric solution for the three-dimensional radial multiplicative wave equation:  ˜ +∗ ec ·∗ t) +∗ (r −∗ ec ·∗ t) u(r, t) = (e/∗ (e2 ·∗ r)) ·∗ (r +∗ ec ·∗ t)φ(r  ˜ −∗ ec ·∗ t) φ(r c r+ Z∗ e t

+∗ (e/∗ (e2c ·∗ ) ·∗ r ·∗

˜ ·∗ d∗ s. s ·∗ ψ(s)

∗r−∗ ec t

4.2.2

The Cauchy problem

Let (x, t) = (x1 , x2 , x3 , t) ∈ R3∗ × [1, ∞) be arbitrarily chosen. With S we will denote the sphere |y −∗ x|2∗∗ = t2∗ ,

The Multiplicative Wave Equation in R3∗

133

where y = (y1 , y2 , y3 ), |x −∗ y|∗ is the distance between the points x and y. Let also µ be an arbitrary real-valued twice continuously multiplicative differentiable function defined on S. We will prove that the function Z u(x, t) = (e/∗ t) ·∗ µ(y1 , y2 , y3 ) ·∗ d∗ sy (4.35) ∗S

is a solution to equation (4.32) in the case when c = 1. If c 6= 1, then we make the change t = (e/∗ ec ) ·∗ τ. Really, the change of variables yi −∗ xi = t ·∗ ξi ,

i = 1, 2, 3,

brings the expression (4.35) in the form Z u(x, t) = t ·∗ µ(x1 +∗ t ·∗ ξ1 , x2 +∗ t ·∗ ξ2 , x3 +∗ t ·∗ ξ3 ) ·∗ d∗ σξ ,

(4.36)

∗σ

where σ is the unit sphere |ξ|∗ = e and d∗ σξ = (d∗ sy )/∗ (t2∗ ) is an element of the unit sphere. By (4.36), we get Z X 3

∆∗ u(x, t) = t ·∗

µ∗∗ yi yi (x1 +∗ t ·∗ ξ1 , x2 +∗ t ·∗ ξ2 , x3 +∗ t ·∗ ξ3 ) ·∗ d∗ σξ

∗σ ∗i=1

and u∗t (x, t) Z = µ(x1 +∗ t ·∗ ξ1 , x2 +∗ t ·∗ ξ2 , x3 +∗ t ·∗ ξ3 ) ·∗ d∗ σξ ∗σ

+∗ t · ∗

Z X 3

ξi ·∗ µ∗yi (x1 +∗ t ·∗ ξ1 , x2 +∗ t ·∗ ξ2 , x3 +∗ t ·∗ ξ3 ) ·∗ d∗ σξ

∗σ ∗i=1

= (u(x, t)/∗ t) +∗ t ·∗

Z X 3 ∗σ ∗i=1

·∗ d∗ σξ .

ξi ·∗ µ∗yi (x1 +∗ t ·∗ ξ1 , x2 +∗ t ·∗ ξ2 , x3 +∗ t ·∗ ξ3 )

134

The Multiplicative Wave Equation We set

Z X 3

I=

µ∗yi (y) ·∗ νi (y) ·∗ d∗ sy ,

∗S ∗i=1

where ν(y) = (ν1 (y), ν2 (y), ν3 (y)) is the outer multiplicative normal vector to S at the point y. Then, t ·∗

Z X 3

ξi ·∗ µ∗yi (x +∗ t ·∗ ξ) ·∗ d∗ σξ = (e/∗ t2∗ ) ·∗ I

∗σ ∗i=1

and u∗t (x, t) = u(x, t)/∗ t +∗ (e/∗ t) ·∗ I. We multiplicative differentiate the last equation with respect to t and we find ∗ 2∗ 2∗ ∗ u∗∗ tt (x, t) = ut (x, t)/∗ t −∗ ((u(x, t))/∗ (t )) −∗ (e/∗ (t )) ·∗ I +∗ (e/∗ t) ·∗ It

= (e/∗ t) ·∗ (u(x, t)/∗ t +∗ (e/∗ t) ·∗ I) −∗ (e/∗ (t2∗ )) ·∗ u(x, t) −∗ (e/∗ (t2∗ )) ·∗ I +∗ (e/∗ t) ·∗ It∗ = (e/∗ t) ·∗ It∗ . (4.37) By the Gauss–Ostrogradsky formula, we have that π

2π

Zt Ze Ze I=

∆∗ µ ·∗ ρ2∗ ·∗ sin∗ θ ·∗ d∗ φ ·∗ d∗ θ ·∗ d∗ ρ.

∗1 ∗1 ∗1

Hence, π

It∗

= t

2

2π

Ze Ze

∆∗ µ ·∗ sin∗ θ ·∗ d∗ φ ·∗ d∗ θ ∗1 ∗1

= t

2∗

Z ·∗

∆∗ µ ·∗ d∗ σξ ∗σ

= t ·∗ ∆∗ u. We substitute the last formula in (4.37) and we obtain that u satisfies equation (4.32). Example 4.7 Consider the function Z u(x, t) = (e/∗ t) ·∗ (y1 +∗ y2 +∗ y3 ) ·∗ d∗ sy . ∗S

The Multiplicative Wave Equation in R3∗

135

Then, 2π π Ze Ze 

u(x, t)

= (e/∗ t) ·∗

x1 +∗ x2 +∗ x3 +∗ t ·∗ cos∗ φ ·∗ sin∗ θ ∗1 ∗1



+∗ t ·∗ sin∗ φ ·∗ sin∗ θ +∗ t ·∗ cos∗ θ ·∗ t2∗ ·∗ sin∗ θ ·∗ d∗ θ ·∗ d∗ φ e4π ·∗ t ·∗ (x1 +∗ x2 +∗ x3 )

= and

∗∗ ∗∗ ∗∗ u∗∗ tt −∗ ux1 x1 −∗ ux2 x2 −∗ ux3 x3 = 0∗ .

Exercise 4.8 Check that the function Z u(x, t) = (e/∗ t) ·∗ (y1 −∗ e2 ·∗ y2 +∗ e3 ·∗ y3 )d∗ sy ∗S

satisfies equation (4.32). We set

Z

M (µ) = (e/∗ t2∗ ) ·∗

µ(y1 , y2 , y3 ) ·∗ d∗ sy . ∗S

Then,

u(x, t) = t ·∗ M (µ)

is a solution to equation (4.32) for c = 1. Now we consider the following Cauchy problem: u∗∗ (x1 , x2 , x3 ) ∈ R3∗ , t > 1, (4.38) tt −∗ ∆∗ u = 1, u(x1 , x2 , x3 , 1) = φ(x1 , x2 , x3 ), (4.39) u∗t (x1 , x2 , x3 , 1) = ψ(x1 , x2 , x3 ),

(x1 , x2 , x3 ) ∈ R3∗ ,

where φ ∈ C∗3 (R3∗ ), ψ ∈ C∗2 (R3∗ ). We will prove that 1

1

u(x, t) = e 4π ·∗ t ·∗ M (ψ) +∗ e 4π ·∗ (∂∗ /∗ (∂∗ t)) (t ·∗ M (φ))

(4.40)

is its solution. Definition 4.6 The equality (4.40) is called the multiplicative Kirchhoff formula, shortly the Kirchoff formula. We have that the function (4.40) satisfies equation (4.38). We will prove that it satisfies the initial condition (4.39). Indeed, 1 u(x1 , x2 , x3 , 0) = e 4π ·∗ M (φ) t=1 Z 1 = e 4π ·∗ φ(x1 , x2 , x3 ) ·∗ d∗ σξ |ξ|=e

=

φ(x1 , x2 , x3 ),

136

The Multiplicative Wave Equation u∗t (x1 , x2 , x3 , t)

 = (∂∗ /∗ ∂∗ t) ·∗ +∗ e =

1 4π

1

e 4π ·∗ t ·∗ M (ψ) 

·∗ (∂∗ /∗ (∂∗ t)) ·∗ (t ·∗ M (φ))

1

1

e 4π ·∗ M (ψ) +∗ e 4π ·∗ t ·∗ (∂∗ /∗ (∂∗ t)) ·∗ M (ψ) 1

+∗ e 4π ·∗ (∂∗2 /∗ (∂∗ t2∗ )) ·∗ (t ·∗ M (φ)) =

1

1

e 4π ·∗ M (ψ) +∗ e 4π ·∗ t ·∗ (∂∗ /∗ (∂∗ t)) ·∗ M (ψ) 1

+∗ e 4π ·∗ t ·∗ ∆∗ M (φ), u∗t (x1 , x2 , x3 , 1)

= =

1 e 4π ·∗ M (ψ) 1

t=1

Z

e 4π ·∗

ψ(x1 , x2 , x3 ) ·∗ d∗ σξ ∗|ξ|=e

=

ψ(x1 , x2 , x3 ).

Example 4.8 Consider the following Cauchy problem: ∗∗ ∗∗ ∗∗ u∗∗ tt −∗ ux1 x1 −∗ ux2 x2 −∗ ux3 x3 = 1,

(x1 , x2 , x3 ) ∈ R3∗ ,

u(x1 , x2 , x3 , 1) = x1 , u∗t (x1 , x2 , x3 , 1) = x3 ,

(x1 , x2 , x3 ) ∈ R3∗ .

Here, φ(x1 , x2 , x3 )

= x1 ,

ψ(x1 , x2 , x3 )

= x3 .

Hence, Z φ(y1 , y2 , y3 )dsy ∗|x−∗ y|∗ =t

Z y1 ·∗ d∗ sy

= ∗|x−∗ y|∗ =t

t > 1,

The Multiplicative Wave Equation in R3∗ 2π

Ze

Ze

π

·∗

=

137

(x1 +∗ t ·∗ cos∗ φ ·∗ sin∗ θ) ·∗ t2∗ ·∗ sin∗ θ ·∗ d∗ θ ·∗ d∗ φ

∗1

1

= e4π ·∗ t2∗ ·∗ x1 , Z ψ(y1 , y2 , y3 ) ·∗ d∗ sy ∗|x−∗ y|∗ =t

Z y3 ·∗ d∗ sy

= ∗|x−∗ y|∗ =t 2π

Ze Ze

π

=

(x3 +∗ t ·∗ cos∗ θ) ·∗ t2∗ ·∗ sin∗ θ ·∗ d∗ θ ·∗ d∗ φ

∗1 ∗1

= e4π ·∗ t2∗ ·∗ x3 . Then, using the Kirchhoff formula, we get u(x1 , x2 , x3 , t)


1 1 = e 4π ·∗ (e/∗ t) e4π ·∗ t2∗ ·∗ x3 +∗ e 4π ·∗ (∂∗ /∗ (∂∗ t)) ·∗ (e/∗ t) ·∗ e4π ·∗ t2∗ ·∗ x1




= t ·∗ x3 +∗ x1 . Exercise 4.9 Find a solution to the Cauchy problem: ∗∗ ∗∗ ∗∗ u∗∗ tt −∗ ux1 x1 −∗ ux2 x2 −∗ ux3 x3 = 1,

(x1 , x2 , x3 ) ∈ R3∗ ,

1

t > 1,

1

u(x1 , x2 , x3 , 1) = e 2 ·∗ x21∗ +∗ e 2 ·∗ x22∗ +∗ x23∗ , u∗t (x1 , x2 , x3 , 1) = 1,

(x1 , x2 , x3 ) ∈ R3∗ .

Answer 1

1

u(x1 , x2 , x3 , t) = e 2 ·∗ x21∗ +∗ e 2 ·∗ x22∗ +∗ x23∗ +∗ e2 ·∗ t2∗ . Next, we consider the following Cauchy problem: u∗∗ tt −∗ ∆∗ u = f (x1 , x2 , x3 , t),

(x1 , x2 , x3 ) ∈ R3∗ ,

u(x1 , x2 , x3 , 1) = φ(x1 , x2 , x3 ), u∗t (x1 , x2 , x3 , 1) = ψ(x1 , x2 , x3 ),

t > 1, (4.41)

(x1 , x2 , x3 ) ∈ R3∗ ,

where φ ∈ C∗3 (R3∗ ), ψ ∈ C∗2 (R3∗ ), f ∈ C∗2 (R3∗ × [1, ∞)).

138

The Multiplicative Wave Equation We set v(x1 , x2 , x3 , t) = u(x1 , x2 , x3 , t) −∗ φ(x1 , x2 , x3 ) −∗ t ·∗ ψ(x1 , x2 , x3 ).

Then, v(x1 , x2 , x3 , 1) = =

u(x1 , x2 , x3 , 0) −∗ φ(x1 , x2 , x3 ) 1,

vt∗ (x1 , x2 , x3 , t)

= u∗t (x1 , x2 , x3 , t) −∗ ψ(x1 , x2 , x3 ),

vt∗ (x1 , x2 , x3 , 1)

= u∗t (x1 , x2 , x3 , 1) −∗ ψ(x1 , x2 , x3 ) =

∗∗ vtt (x1 , x2 , x3 , t)

vx∗∗i xi (x1 , x2 , x3 , t)

1,

= u∗∗ tt (x1 , x2 , x3 , t), ∗∗ = u∗∗ xi xi (x1 , x2 , x3 , t) −∗ φxi xi (x1 , x2 , x3 )

−∗ t ·∗ ψx∗∗i xi (x1 , x2 , x3 ), ∗∗ vtt −∗ ∆ ∗ v

i = 1, 2, 3,

= u∗tt −∗ ∆∗ u +∗ ∆∗ φ +∗ t ·∗ ∆∗ ψ = f +∗ ∆∗ φ +∗ t ·∗ ∆∗ ψ.

Therefore, v(x1 , x2 , x3 , t) satisfies the Cauchy problem: ∗∗ vtt −∗ ∆∗ v = g(x1 , x2 , x3 , t),

(x1 , x2 , x3 ) ∈ R3∗ ,

t > 1,

v(x1 , x2 , x3 , 1) = vt (x1 , x2 , x3 , 1) = 1,

(4.42)

(x1 , x2 , x3 ) ∈ R3∗ ,

where g(x1 , x2 , x3 , t) = f (x1 , x2 , x3 , t) +∗ ∆∗ φ(x1 , x2 , x3 ) +∗ t ·∗ ∆∗ ψ(x1 , x2 , x3 ). Let τ > 1 be arbitrarily chosen. Consider the following Cauchy problem: ∗∗ wtt −∗ ∆∗ w = 1,

(x1 , x2 , x3 ) ∈ R3∗ ,

w(x1 , x2 , x3 , τ ) = 1, wt∗ (x1 , x2 , x3 , τ ) = g(x1 , x2 , x3 , τ ).

t > τ, (4.43)

The Multiplicative Wave Equation in R3∗

139

Let z(x1 , x2 , x3 , t, τ ) be the solution of the problem (4.43) which is supposed to be continued as identically multiplicative zero for t ≤ τ. We set Zt h(x1 , x2 , x3 , t) = z(x1 , x2 , x3 , t, τ ) ·∗ d∗ τ. ∗1

Then, h(x1 , x2 , x3 , 1)

=

1, Zt

ht (x1 , x2 , x3 , t)

=

zt (x1 , x2 , x3 , t, τ ) ·∗ d∗ τ

z(x1 , x2 , x3 , t, t) +∗ ∗1

Zt =

zt∗ (x1 , x2 , x3 , t, τ ) ·∗ d∗ τ,

∗1

h∗t (x1 , x2 , x3 , 1)

h∗∗ tt (x1 , x2 , x3 , t)

=

1,

=

zt∗ (x1 , x2 , x3 , t, t)

Zt +∗

∗∗ ztt (x1 , x2 , x3 , t, τ ) ·∗ d∗ τ

∗1

Zt =

g(x1 , x2 , x3 , t) +∗

∗∗ ztt (x1 , x2 , x3 , t, τ ) ·∗ d∗ τ

∗1

Zt =

∆z(x1 , x2 , x3 , t, τ ) ·∗ d∗ τ

g(x1 , x2 , x3 , t) +∗ 0

=

g(x1 , x2 , x3 , t) +∗ ∆∗ h(x1 , x2 , x3 , t),

i.e., h(x1 , x2 , x3 , t) is a solution to the Cauchy problem (4.42). Now we apply the Kirchhoff formula and we obtain 1

∗ z(x1 , x2 , x3 , t, τ ) = e 4π ·∗ (t −∗ τ ) ·∗ Mt− (g(x1 , x2 , x3 , τ )), ∗τ

whereupon v(x1 , x2 , x3 , t) = e

1 4π

Zt ·∗ ∗1

∗ (t −∗ τ )Mt− (g(x1 , x2 , x3 , τ )) ·∗ d∗ τ ∗τ

140

The Multiplicative Wave Equation

and u(x1 , x2 , x3 , t) = φ(x1 , x2 , x3 ) +∗ t ·∗ ψ(x1 , x2 , x3 ) 1

Zt

+∗ e 4π ·∗

 ∗ (t −∗ τ ) ·∗ Mt− f (x1 , x2 , x3 , τ ) ∗τ

∗1

 +∗ ∆∗ φ(x1 , x2 , x3 ) +∗ τ ·∗ ∆∗ ψ(x1 , x2 , x3 ) ·∗ d∗ τ is a solution to the Cauchy problem (4.41). Example 4.9 Consider the following Cauchy problem: u∗∗ tt −∗ ∆∗ u = x1 ,

(x1 , x2 , x3 ) ∈ R3∗ ,

t > 1,

u(x1 , x2 , x3 , 1) = x2 , u∗t (x1 , x2 , x3 , 1) = x3 ,

(x1 , x2 , x3 ) ∈ R3∗ .

Here, f (x1 , x2 , x3 , t)

=

x1 ,

φ(x1 , x2 , x3 )

=

x2 ,

ψ(x1 , x2 , x3 )

=

x3 .

Then, using (4.44), u(x1 , x2 , x3 , t)

= x 2 +∗ t · ∗ x 3 +∗ e

1 4π

Zt ·∗

(e/∗ (t −∗ τ )) ·∗ ∗1

Z y1 ·∗ d∗ sy ·∗ d∗ τ ∗|x−∗ y|=t−∗ τ

= x2 +∗ t ·∗ x3

+∗ e

1 4π

Zt ·∗

(e/∗ (t −∗ τ )) ·∗ ∗1

2π

Ze Ze

π

(x1 +∗ (t −∗ τ ) ·∗ cos∗ φ ·∗ sin∗ θ) ∗1 ∗1

·∗ (t −∗ τ )2∗ ·∗ sin∗ θ ·∗ d∗ θ ·∗ d∗ φ ·∗ d∗ τ 1

= x2 +∗ t ·∗ x3 +∗ e 2 ·∗ x1 ·∗ t2∗ .

(4.44)

The Two-Dimensional Wave Equation

141

Exercise 4.10 Solve the Cauchy problem: 4 6 u∗∗ tt −∗ ∆∗ u = −∗ e ·∗ x1 −∗ e ·∗ x2 ,

(x1 , x2 , x3 ) ∈ R3∗ ,

t > 1,

u(x1 , x2 , x3 , 1) = x31∗ +∗ x32∗ , u∗t (x1 , x2 , x3 , 1) = 1,

(x1 , x2 , x3 ) ∈ R3∗ .

Answer u(x1 , x2 , x3 , t) = x31∗ +∗ x32∗ +∗ t2∗ ·∗ x1 .

4.3

The Two-Dimensional Wave Equation

We consider the Cauchy problem for the wave equation with two spatial variables: u∗tt −∗ u∗x1 x1 −∗ u∗x2 x2 = 0∗ , (x1 , x2 ) ∈ R2 , t > 0∗ , (4.45) u(x1 , x2 , 1) = φ(x1 , x2 ), (4.46) u∗t (x1 , x2 , 1) = ψ(x1 , x2 ),

(x1 , x2 ) ∈ R2∗ ,

where φ ∈ C∗3 (R2∗ ), ψ ∈ C∗2 (R2∗ ). The solution u(x1 , x2 , t) of the problem (4.45), (4.46) can be derived using Kirchhoff’s formula. Then, Z u(x1 , x2 , t) = (e/∗ (e4π ·∗ t)) ·∗ ψ(x1 +∗ y1 , x2 +∗ y2 ) ·∗ d∗ sy |y|∗ =t





1  +∗ e 4π ·∗ (∂∗ /∗ ∂∗ t) (e/∗ t) ·∗

Z  φ(x1 +∗ y1 , x2 +∗ y2 ) ·∗ d∗ sy  , |y|∗ =t

(4.47) which is independent of x3 and satisfies (4.45), (4.46). Note that the projection d∗ y1 ·∗ d∗ y2 of the element of the arc d∗ sy of the sphere |y|∗ = t on the circle y12∗ +∗ y22∗ ≤ t2∗ is expressed by the following formula d∗ y1 ·∗ d∗ y2 = (|y3 |∗ /∗ t) ·∗ d∗ sy . To compute the integrals on the right-hand side of the formula (4.47), we should project on the multiplicative circle y12∗ +∗ y22∗ ≤ t2∗

142

The Multiplicative Wave Equation

both the upper hemisphere y3 > 1 and the lower hemisphere y3 < 1 of the sphere |y|∗ = t. Therefore, Z 1 1 2π u(x1 , x2 , t) = e ·∗ (ψ(y1 , y2 )/∗ (t2∗ −∗ (y1 −∗ x1 )2∗ −∗ (y2 −∗ x2 )2∗ ) 2 ∗ ) ∗B

·∗ d∗ y1 ·∗ d∗ y2 1

+∗ e 2π ·∗ (∂∗ /∗ ∂∗ t) Z 1 ·∗ ((φ(y1 , y2 ))/∗ (t2∗ −∗ (y1 −∗ x1 )2∗ −∗ (y2 −∗ x2 )2∗ ) 2 ∗ ) ∗B

·∗ d∗ y1 ·∗ d∗ y2 , (4.48) where B is the multiplicative circle (y1 −∗ x1 )2∗ +∗ (y2 −∗ x2 )2∗ ≤ t2∗ . Definition 4.7 The equality (4.48) is called Poisson’s formula. Example 4.10 Consider the following Cauchy problem: ∗∗ ∗∗ u∗∗ tt −∗ ux1 x1 −∗ ux2 x2 = 0∗ ,

(x1 , x2 ) ∈ R2∗ ,

t > 0∗ ,

u(x1 , x2 , 1) = x1 , u∗t (x1 , x2 , 1) = x2 ,

(x1 , x2 ) ∈ R2∗ .

Then, using Poisson’s formula, we have Z 1 u(x1 , x2 , t) = e 2π ·∗ (y2 /∗ (t2∗ −∗ (y1 −∗ x1 )2∗ −∗ (y2 −∗ x2 )2∗ )) ∗B

·∗ d∗ y1 ·∗ d∗ y2 Z

1

+∗ e 2π (∂∗ /∗ ∂∗ t) ·∗

(y1 /∗ (t2∗ −∗ (y1 −∗ x1 )2∗

∗B 2∗

−∗ (y2 −∗ x2 ) )

1 2∗

) ·∗ d∗ y1 ·∗ d∗ y2 ,

B : (y1 −∗ x1 )2∗ +∗ (y2 −∗ x2 )2∗ ≤ t2∗ . Let y1

= x1 +∗ r ·∗ cos∗ φ,

y2

= x2 +∗ r ·∗ sin∗ φ,

r ∈ [1, t],

φ ∈ [1, e2π ].

The Two-Dimensional Wave Equation

143

Hence, 2π

u(x1 , x2 , t)

=

e

1 2π

Zt Ze ·∗

1

((x2 +∗ r ·∗ sin∗ φ)/∗ (t2∗ −∗ r2∗ ) 2 ∗ )

∗1 ∗1

·∗ r ·∗ d∗ φ ·∗ d∗ r 2π

+∗ e

1 2π

Zt Ze ·∗ (∂∗ /∗ ∂∗ t) ·∗

1

((x1 +∗ r ·∗ cos∗ φ)/∗ (t2∗ −∗ r2∗ ) 2 ∗

∗1 ∗1

·∗ r ·∗ d∗ φ ·∗ d∗ r =

x2 ·∗ t +∗ x1 .

Exercise 4.11 Solve the Cauchy problem: ∗∗ ∗∗∗ u∗∗ tt −∗ ux1 x1 −∗ ux2 x2 = 1,

(x1 , x2 ) ∈ R2∗ ,

t > 1,

u(x1 , x2 , 1) = x1 −∗ x2 , u∗t (x1 , x2 , 1) = x2 ,

(x1 , x2 ) ∈ R2∗ .

Answer u(x1 , x2 , t) = x1 +∗ x2 ·∗ (t −∗ e). Next, we consider the following Cauchy problem: ∗∗ ∗∗ u∗∗ tt −∗ ux1 x1 −∗ ux2 x2 = f (x1 , x2 , t),

(x1 , x2 ) ∈ R2∗ ,

t > 1,

u(x1 , x2 , 1) = φ(x1 , x2 ), u∗t (x1 , x2 , 1) = ψ(x1 , x2 ),

(x1 , x2 ) ∈ R2∗ ,

where φ ∈ C∗3 (R2∗ ), ψ ∈ C∗2 (R2∗ ) and f ∈ C∗2 (R2∗ × [1, ∞)). Using the formula (4.44), as in above, for its solution u(x1 , x2 , t), we have the following representation: u(x1 , x2 , t) = φ(x1 , x2 ) +∗ t ·∗ ψ(x1 , x2 )

+∗ e

1 2π

Zt Z ·∗

(f (y1 , y2 , τ ) +∗ ∆∗ φ(y1 , y2 ) +∗ τ ·∗ ∆∗ (ψ(y1 , y2 ))) ∗1 Bt−∗ τ 1

/∗ ((t −∗ τ )2∗ −∗ (y1 −∗ x1 )2∗ −∗ (y2 −∗ x2 )2∗ ) 2 ∗ ·∗ d∗ y1 ·∗ d∗ y2 ·∗ dτ, where Bt−∗ τ : (y1 −∗ x1 )2∗ +∗ (y2 −∗ x2 )2∗ ≤ (t −∗ τ )2∗ .

144

The Multiplicative Wave Equation

Exercise 4.12 Solve the Cauchy problem: 2∗ ∗∗ ∗∗ u∗∗ tt −∗ ux1 x1 −∗ ux2 x2 = x2 ,

(x1 , x2 ) ∈ R2∗ ,

t > 1,

u(x1 , x2 , 1) = e, u∗t (x1 , x2 , 1) = x1 ,

(x1 , x2 ) ∈ R2∗ .

Answer 1

1

u(x1 , x2 , t) = e +∗ t ·∗ x1 +∗ e 2 ·∗ x22∗ ·∗ t2∗ +∗ e 12 ·∗ t4∗ .

4.4

The (2n + 1)-Dimensional Wave Equation

We will start this section with the following useful lemma: Lemma 4.1 Let φ : R∗ → R be C∗k+1 -function. Then for k ∈ N we have 1. (d2∗∗ /∗ d∗ r2∗ ) ·∗ ((e/∗ r) ·∗ (d∗ /∗ d∗ r)) = ((e/∗ r) ·∗ (d∗ /∗ d∗ r))

 ·∗ r(2k)∗

k∗

2. (k−1)∗

((e/∗ r) ·∗ (d∗ /∗ (d∗ r))) =

k−1 X



  ·∗ r(2k−1)∗ ·∗ φ(r) (4.49)  ·∗ (d∗ φ/∗ d∗ r)(r) ,

(k−1)∗

 r(2k−1)∗ ·∗ φ(r)

k

eβj ·∗ r(j+1)∗ ·∗ ((dj∗∗ φ)/∗ (d∗ rj∗ ))(r),

(4.50)

∗j=0

where βjk =



k −∗ 1 j

 (2k − 1) . . . (k + 1 + j),

Proof 1. We will use induction. (a) For k = 1, we have ((d2∗∗ )/∗ (d∗ r2∗ )) ·∗ (r ·∗ φ(r)) =

(d∗ /∗ d∗ r) ·∗ (φ(r) +∗ r ·∗ φ∗ (r))

= e2 ·∗ φ∗ (r) +∗ r ·∗ φ∗∗ (r),

j = 0, 1, . . . , k − 1.

The (2n + 1)-Dimensional Wave Equation ((e/∗ r) ·∗ (d∗ /∗ d∗ r)) r2∗ ·∗ φ∗ (r) =

145



(e/∗ r) ·∗ e2 ·∗ r ·∗ φ∗ (r) +∗ r2∗ ·∗ φ∗∗ (r)

= e2 ·∗ φ∗ (r) +∗ r ·∗ φ∗∗ (r). Therefore, the assertion is valid for k = 1. (b) Assume that (4.49) holds for some k ∈ N. (c) We will prove that   k ((d2∗∗ )/∗ (d∗ r2∗ )) ((e/∗ r) ·∗ (d∗ /∗ d∗ r)) ∗ ·∗ r(2k+1)∗ ·∗ φ(r)   (k+1)∗ = ((e/∗ r) ·∗ (d∗ /∗ d∗ r)) r(2k+2)∗ ·∗ (d∗ φ/∗ d∗ r)(r) . (4.51) Indeed,  
k (d2∗∗ /∗ d∗ r2∗ ) ·∗ ((e/∗ r) ·∗ (d∗ /∗ d∗ r)) ∗ ·∗ r(2k+1)∗ ·∗ φ(r)
(k−1)∗ = (d2∗∗ /∗ d∗ r2∗ ) ·∗ ((e/∗ r) ·∗ (d∗ /∗ d∗ r))   ·∗ ((e/∗ r) ·∗ (d∗ /∗ d∗ r)) ·∗ r(2k+1)∗ ·∗ φ(r)
(k−1)∗ = (d2∗∗ /∗ d∗ r2∗ ) ·∗ ((e/∗ r)(d∗ /∗ d∗ r)) ·∗ ((e/∗ r))   ·∗ e(2k+1) ·∗ r(2k)∗ ·∗ φ(r) +∗ r(2k+1)∗ ·∗ φ∗ (r)
(k−1)∗ = (d2∗∗ /∗ d∗ r2∗ ) ·∗ ((e/∗ r) ·∗ (d∗ /∗ d∗ r))   ·∗ e2k+1 ·∗ r(2k−1)∗ ·∗ φ(r) +∗ r(2k)∗ ·∗ φ∗ (r)
(k−1)∗ = d2∗∗ /∗ d∗ r2∗ ·∗ ((e/∗ r) ·∗ (d∗ /∗ d∗ r)) 
 ·∗ r(2k−1)∗ e2k+1 ·∗ φ(r) +∗ r ·∗ φ∗ (r) k

= ((e/∗ r) ·∗ (d∗ /∗ d∗ r)) ∗ 
 ·∗ r(2k)∗ ·∗ e2k+2 ·∗ φ∗ (r) +∗ r ·∗ φ∗∗ (r) (4.52) and ((e/∗ r) ·∗ (d∗ /∗ d∗ r)) =

(k+1)∗

  ·∗ r(2k+2)∗ ·∗ (d∗ φ/∗ d∗ r)(r) k

((e/∗ r) ·∗ (d∗ /∗ d∗ r)) ∗ ·∗ ((e/∗ r) ·∗ (d∗ /∗ d∗ r))   ·∗ r(2k+2)∗ ·∗ φ∗ (r)

146

The Multiplicative Wave Equation

k

=

((e/∗ r) ·∗ (d∗ /∗ d∗ r)) ∗ ·∗ ((e/∗ r))   ·∗ e2k+2 ·∗ r(2k+1)∗ ·∗ φ∗ (r) +∗ r(2k+2)∗ ·∗ φ∗∗ (r)

=

((e/∗ r) ·∗ (d∗ /∗ d∗ r)) ∗ 
 ·∗ r(2k)∗ ·∗ e2k+2 ·∗ φ∗ (r) +∗ r ·∗ φ∗∗ (r) .

k

Hence from (4.52), we get (4.51). 2. We have ((e/∗ r) ·∗ (d∗ /∗ d∗ r))

(k−1)∗

  ·∗ r(2k−1)∗ ·∗ φ(r)

= (e/∗ r(k−1)∗ ) ·∗ (d∗ /∗ d∗ r)

= (e/∗ r

(k−1)∗

) ·∗

k−1 X



k−1 j

∗j=0

  ·∗ r(2k−1)∗ ·∗ φ(r)

(k−1)∗



(k−1−j)∗

·∗ (d∗ /∗ d∗ r) ∗

  ·∗ r(2k−1)∗ ·∗ ((dj∗∗ φ)/∗ (d∗ rj∗ ))(r)

= (e/∗ (r

(k−1)∗

)) ·∗

k−1 X



∗j=0



k−1 j

∗

·∗ (2k − 1)∗ · · · (2k − 1 − k + 1 + j − 1)∗ ·∗ r(2k−1−k+1+j)∗ ·∗ ((dj∗∗ φ)/∗ (d∗ rj∗ ))(r) = (e/∗ r

(k−1)∗

) ·∗

k−1 X



∗j=0

k−1 j

 ·∗ (2k − 1)∗ · · · (k + 1 + j)∗ ∗

·∗ r(k+j)∗ ·∗ ((dj∗∗ φ)/∗ (d∗ rj∗ ))(r) =

k−1 X ∗j=0



k−1 j



·∗ (2k − 1)∗ · · · (k + 1 + j)∗ ·∗ r(j+1)∗ ∗

·∗ ((dj∗∗ φ)/∗ (d∗ rj∗ ))(r), which completes the proof.

The (2n + 1)-Dimensional Wave Equation

147

Now we consider the multiplicative Cauchy problem: u∗∗ tt −∗ ∆∗ u = f (x, t),

(x, t) ∈ R2n+1 × (1, ∞), ∗

u(x, 1) = φ(x),

(4.53)

u∗t (x, 1) = ψ(x), where ∆∗ u =

2n+1 X

x ∈ R2n+1 , ∗

∗ m 2n+1 u∗∗ ), ψ ∈ C∗m−1 xi xi , x = (x1 , x2 , . . . , x2n+1 ), φ ∈ C∗ (R

∗i=1

(R2n+1 ), f ∈ C∗m−2 (R2n+1 × [1, ∞)), m ≥ 2. We denote ∗ α(n) =

n

((e2(2π) )/∗ (e1···3···(2n−1) )),

β(n) = e

2n!(4π)n (2n+1)!

1

= e 2n+1 ·∗ α(n). Then, if B(x, r) is a ball with centerpoint x and radius r, we have µ∗ (B(x, r)) µ∗ (∂∗ B(x, r))

= β(n) ·∗ r(2n+1)∗ , = α(n) ·∗ r(2n)∗ ,

where µ∗ (B(x, r)) and µ∗ (∂∗ B(x, r)) are the multiplicative volume of the multiplicative ball B(x, r) and the multiplicative area of the multiplicative sphere ∂∗ B(x, r), respectively. Let u ∈ C∗m (R2n+1 × [1, ∞)) be a solution to the prob∗ lem (4.53). For x ∈ R2n+1 , t > 1, r > 1, we define Z U (x, r, t) = (e/∗ (α(n) ·∗ r(2n)∗ )) ·∗ u(y, t) ·∗ d∗ sy , ·∗ ∂∗ B(x,r)

F (x, r, t)

=

(e/∗ (α(n) ·∗ r(2n)∗ )) ·∗

Z f (y, t) ·∗ d∗ sy , ∗∂∗ B(x,r)

Φ(x, r)

=

(e/∗ (α(n) ·∗ r

(2n)∗

Z )) ·∗

φ(y) ·∗ d∗ sy , ∗∂∗ B(x,r)

Ψ(x, r)

=

(e/∗ (α(n) ·∗ r

(2n)∗

Z )) ·∗

ψ(y) ·∗ d∗ sy . ∗∂B(x,r)

With D∗ u(x, t), we denote the multiplicative vector (u∗x1 (x, t), . . . , u∗x2n+1 (x, t)).

148

The Multiplicative Wave Equation

Theorem 4.6 We have U ∈ C∗m (R∗ × [1, ∞)) and ∗∗ Utt∗∗ −∗ Urr −∗ (e2n /∗ r) ·∗ Ur∗ = F (x, r, t), (r, t) ∈ R∗ × (1, ∞),

(4.54)

U (r, 1) = Φ(r), (4.55) Ut∗ (r, 1) = Ψ(r),

r ∈ R∗ ,

where R∗ = (1, ∞). Proof We have Z U (x, r, t) = (e/∗ α(n)) ·∗

u(x +∗ r ·∗ z, t) ·∗ d∗ sz . ∗∂∗ B(1,e)

Then, lim U (x, r, t) = u(x, t)

r→1+∗

and Ur∗ (x, r, t)

Z = (e/∗ α(n)) ·∗

D∗ u(x +∗ r ·∗ z, t) ·∗ z ·∗ d∗ sz ∗∂∗ B(1,e)

Z

= (e/∗ (α(n) ·∗ r(2n)∗ )) ·∗

D∗ u(y, t) ·∗ ((y −∗ x)/∗ r) ·∗ d∗ sy ∗∂∗ B(x,r)

Z

= (e/∗ (α(n) ·∗ r(2n)∗ )) ·∗

(∂∗ u/∗ ∂∗ ν) ·∗ d∗ sy ∗∂∗ B(x,r)

Z

= (e/∗ (α(n) ·∗ r(2n)∗ )) ·∗

∆∗ u(y, t) ·∗ ∆∗ y ∗B(x,r)

Z

= (r/∗ ((2n + 1) ·∗ β(n) ·∗ r(2n+1)∗ )) ·∗

∆∗ u(y, t) ·∗ d∗ y ∗B(x,r)

Z = (r/∗ ((2n + 1) ·∗ β(n))) ·∗

∆∗ u(x +∗ r ·∗ z, t) ·∗ d∗ z. ∗B(0,1)

Therefore, lim Ur∗ (x, r, t) = 1

r→1+∗

The (2n + 1)-Dimensional Wave Equation

149

and ∗∗ Urr (x, r, t) = (e/∗ ((2n + 1)∗ ·∗ β(n) ·∗ r(2n+1)+∗ )) ·∗

Z ∆∗ u(y, t) ·∗ d∗ y ∗B(x,r)

Z

−∗ (e/∗ (β(n) ·∗ r(2n+1)∗ )) ·∗

∆∗ u(y, t) ·∗ d∗ y ∗B(x,r)

Z

+∗ (e/∗ ((2n + 1)∗ ·∗ β(n) ·∗ r(2n)∗ )) ·∗

∆∗ u(y, t) ·∗ d∗ sy ∗∂∗ B(x,r)

= ((e/∗ (2n + 1)∗ ) −∗ e) ·∗ (e/∗ (β(n) ·∗ r(2n+1)∗ )) Z ·∗

∆∗ u(y, t) ·∗ d∗ y

∗B(x,r)

+∗ (e/∗ (α(n) ·∗ r(2n)∗ )) ·∗

Z ∆∗ u(y, t) ·∗ d∗ sy ∗∂∗ B(x,r)

= ((e/∗ (2n + 1)∗ ) −∗ e) ·∗ (e/∗ β(n)) Z ·∗

∆∗ u(x +∗ r ·∗ z, t) ·∗ d∗ z

B(0,1)

Z +∗ (e/∗ α(n)) ·∗

∆∗ u(x +∗ r ·∗ z, t) ·∗ d∗ sz . ∗∂∗ B(0,1)

(4.56) Consequently, ∗∗ lim Urr (x, r, t) = (e/∗ (2n + 1)∗ ) ·∗ ∆∗ u(x, t).

r→1+∗

∗∗∗ Using (4.56), we can compute Urrr (x, r, t) etc. So, U ∈ C∗m (R∗ × [1, ∞)). Also, we have Z r(2n)∗ ·∗ Ur∗ (x, r, t) = (e/∗ ((2n + 1)∗ ·∗ β(n))) ·∗ ∆∗ u(y, t) ·∗ d∗ y ∗B(x,r)

Z = (e/∗ ((2n + 1)∗ ·∗ β(n))) ·∗ B(x,r)

(u∗∗ tt (y, t) −∗ f (y, t)) ·∗ d∗ y

150

The Multiplicative Wave Equation

and so   r(2n)∗ ·∗ Ur∗ (x, r, t)

= r

(e/∗ ((2n + 1)∗ ·∗ β(n))) Z ·∗

(u∗∗ tt (y, t) −∗ f (y, t)) ·∗ d∗ sy

∂∗ B(x,r)

=

((r(2n)∗ )/∗ (α(n) ·∗ r(2n)∗ )) Z

(u∗∗ tt (y, t) −∗ f (y, t)) ·∗ d∗ sy

·∗ ∗∂∗ B(x,r)

=

r(2n)∗ ·∗ Utt∗∗ (x, r, t) −∗ r(2n)∗ ·∗ F (x, r, t),

i.e., U (x, r, t) satisfies equation (4.54). Moreover, U (x, r, 1)

= (e/∗ (α(n) ·∗ r

(2n)∗

Z )) ·∗

u(y, 1) ·∗ d∗ sy ∗∂∗ B(x,r)

=

(e/∗ (α(n) ·∗ r

(2n)∗

Z )) ·∗

φ(y) ·∗ d∗ sy ∗∂∗ B(x,r)

= Ut∗ (x, r, t)

Φ(x, r),

= (e/∗ (α(n) ·∗ r(2n)∗ )) ·∗

Z

u∗t (y, t) ·∗ d∗ sy ,

∗∂B(x,r)

Ut∗ (x, r, 1)

Z

= (e/∗ (α(n) ·∗ r(2n)∗ )) ·∗

u∗t (y, 1)dsy

∗∂∗ B(x,r)

=

(e/∗ (α(n) ·∗ r

(2n)∗

Z )) ·∗

ψ(y) ·∗ d∗ sy ∗∂∗ B(x,r)

=

Ψ(x, r),

i.e., U (x, r, t) satisfies the initial condition (4.55).

The (2n + 1)-Dimensional Wave Equation

151

Definition 4.8 Equation (4.54) is called multiplicative Euler1 –Poisson2 – Darboux3 equation. Let ˜ (r, t) U

= ((e/∗ r) ·∗ (∂∗ /∗ ∂∗ r))

(n−1)∗

  ·∗ r(2n−1)∗ ·∗ U (x, r, t) ,

F˜ (r, t)

= ((e/∗ r) ·∗ (∂∗ /∗ ∂∗ r))

(n−1)∗

  ·∗ r(2n−1)∗ ·∗ F (x, r, t) ,

˜ Φ(r)

= ((e/∗ r) ·∗ (∂∗ /∗ ∂∗ r))

(n−1)∗

  ·∗ r(2n−1)∗ ·∗ Φ(x, r) ,

˜ Ψ(r)

= ((e/∗ r) ·∗ (∂∗ /∗ ∂∗ r))

(n−1)∗

  ·∗ r(2n−1)∗ ·∗ Ψ(x, r) .

Then, ˜ (r, 1) U

=

˜ Φ(r),

˜t∗ (r, 1) U

=

˜ Ψ(r).

Theorem 4.7 We have ˜ ∗∗ −∗ U ˜ ∗∗ = F˜ (r, t), U tt rr

(r, t) ∈ R∗ × (1, ∞),

˜ (r, 1) = Φ(r), ˜ U ˜t (r, 1) = Ψ(r), ˜ U ˜ (1, t) = 1, U

r ∈ R∗ ,

t ∈ (1, ∞).

Proof If r > 0, applying Lemma 4.1, we have ∗∗ ˜rr U (r, t)

=

((∂∗2∗ )/∗ (∂∗ r2∗ )) ((e/∗ r) ·∗ (∂∗ /∗ ∂∗ r))

(n−1)∗

  r(2n−1)∗ ·∗ U (x, r, t) = ((e/∗ r) ·∗ (∂∗ /∗ ∂∗ r))

n∗

  ·∗ r(2n)∗ ·∗ Ur∗ (x, r, t)

1 Leonhard Euler (April 15, 1707–September 18, 1783) was a Swiss mathematician, physicist, astronomer, logician and engineer who made important and influential discoveries in many branches of mathematics. 2 Simeon Denis Poisson (June 21, 1781–April 25, 1840) was a French mathematician, geometer and physicist. 3 Jean Gaston Darboux (August 14, 1842–February 23, 1917) was a French mathematician, who made several important contributions to geometry and mathematical analysis.

152

The Multiplicative Wave Equation (n−1)∗

= ((e/∗ r) ·∗ (∂∗ /∗ (∂∗ r)))

  ∗∗ ·∗ r(2n−1)∗ ·∗ Urr (x, r, t) +∗ e2n ·∗ r(2n−2)∗ ·∗ Ur∗ (x, r, t) = ((e/∗ r) ·∗ (∂∗ /∗ ∂∗ r)) 

(n−1)∗


 ∗∗ r(2n−1)∗ ·∗ Urr (x, r, t) +∗ (e2n /∗ r) ·∗ Ur∗ (x, r, t)

= ((e/∗ r) ·∗ (∂∗ /∗ (∂∗ r))) 

(n−1)∗

 r(2n−1)∗ ·∗ Utt∗∗ (x, r, t) −∗ r(2n−1)∗ ·∗ F (x, r, t)

˜tt∗∗ (r, t) −∗ F˜ (r, t). = U Again applying Lemma 4.1, we have n−1 X

˜ (r, t) = U

n

eβj ·∗ r(j+1)∗ ·∗ ((∂∗j∗ )/∗ (∂∗ rj∗ )) ·∗ U (x, r, t).

∗j=0

Hence from Theorem 4.6, we find that ˜ (r, t) = 1, lim U

r→1+∗

which completes the proof. By Theorem 4.7, we have ˜ (r, t) U

1 2

e ·∗

=



t+ Z ∗r  1 ˜ ˜ ˜ 2 Φ(r +∗ t) −∗ Φ(t −∗ r) +∗ e ·∗ Ψ(y) ·∗ d∗ y ∗t−∗ r

1 2

Zt

(t−Z ∗ τ )+∗ r

+∗ e ·∗

F˜ (ξ, τ ) ·∗ dξ ·∗ dτ

∗1 ∗(t−∗ τ )−∗ r

for all r ∈ R∗ , t ≥ 1. Because u(x, t) = lim U (x, r, t) r→1

and ˜ (r, t) U

=

=

((e/∗ r) ·∗ (∂∗ /∗ ∂∗ r)) n−1 X ∗j=0

n

(n−1)∗

  ·∗ r(2n−1)∗ ·∗ U (x, r, t)

eβj r(j+1)∗ ·∗ (∂∗j /∗ ∂∗ rj∗ )U (x, r, t),

The (2n + 1)-Dimensional Wave Equation

153

we get n

˜ (r, t)/∗ (eβ0 ·∗ r)) lim (U

r→1+∗

=

lim U (x, r, t)

r→1

= u(x, t). Therefore,  n ˜ +∗ r) −∗ Φ(t ˜ −∗ r))/∗ (e2 ·∗ r) = (e/∗ eβ0 ) ·∗ lim (Φ(t

u(x, t)

r→1

t+ Z ∗r

2

+∗ (e/∗ (e ·∗ r)) ·∗

˜ Ψ(y) ·∗ d∗ y

∗t−∗ r (t−Z ∗ τ )+∗ r

Zt

2

+∗ (e/∗ (e ·∗ r)) ·∗

F˜ (ξ, τ ) ·∗ d∗ ξ ·∗ d∗ τ



∗1 ∗(t−∗ τ )−∗ r

 =

n

˜ ∗ (t) +∗ Ψ(t) ˜ +∗ (e/∗ eβ0 ) ·∗ Φ

Zt

 F˜ (t −∗ τ, τ ) ·∗ d∗ τ  ,

∗1

whereupon n

u(x, t) = (e/∗ eβ0 ) ·∗

 ((∂∗ /∗ ∂∗ t)) ·∗ ((e/∗ t) ·∗ (∂∗ /∗ ∂∗ t))



(n−1)∗

 Z

 ·∗ (e/∗ (α(n) ·∗ t)) ·∗

 φ(y) ·∗ d∗ sy  ∂∗ B(x,t) (n−1)∗

+∗ ((e/∗ t) ·∗ (∂∗ /∗ (∂∗ t))) 

 Z

 ·∗ (e/∗ (α(n) ·∗ t)) ·∗

 ψ(y) ·∗ d∗ sy  ∂∗ B(x,t)

Zt ((e/∗ (t −∗ τ )) ·∗ (∂∗ /∗ (∂∗ (t −∗ τ ))))

+∗

(n−1)∗

∗1



 Z

 ·∗ (e/∗ (α(n) ·∗ (t −∗ τ ))) ·∗

  f (y, τ ) ·∗ d∗ sy  ·∗ d∗ τ .

∗∂∗ B(x,t−∗ τ )

(4.57)

154

The Multiplicative Wave Equation

Exercise 4.13 Prove that the function u(x, t) defined by (4.57) satisfies the problem (4.53).

4.5

The (2n) − 1-Dimensional Wave Equation

Now we consider the following Cauchy problem: u∗tt −∗ ∆u = f (x, t),

(x, t) ∈ R2n ∗ × (1, ∞),

u(x, 1) = φ(x), u∗t (x, 1) = ψ(x), where ∆∗ u =

2n X

(4.58) x ∈ R2n ∗ ,

m 2n m−1 u∗∗ (R2n xi xi , x = (x1 , . . . , x2n ), φ ∈ C∗ (R ), ψ ∈ C∗ ∗ ) and

∗i=1

f ∈ C∗m−2 (R2n ∗ × [1, ∞)). As we have deducted the solution for the two-dimensional wave equation using the solution of the three-dimensional wave equation, one can get  n (n−1)∗ u(x, t) = (e/∗ eβ0 ) ·∗ (∂∗ /∗ ∂∗ t) ((e/∗ t) ·∗ (∂∗ /∗ ∂∗ t)) 



 2 (e /∗ α(n)) ·∗

Z

1  (φ(y)/∗ (t2∗ −∗ |y −∗ x|2∗ ) 2 ∗ dy 

∗B(x,t)

+∗ ((e/∗ t) ·∗ (∂∗ /∗ ∂∗ t))

(n−1)∗





 2 (e /∗ α(n)) ·∗

Z

1  (ψ(y)/∗ (t2∗ −∗ |y −∗ x|2∗ ) 2 ∗ ) ·∗ d∗ y 

∗B(x,t)

Zt

(n−1)∗

((e/∗ (t −∗ τ )) ·∗ (∂∗ /∗ (∂∗ (t −∗ τ ))))

+∗ ∗1





 2 (e /∗ α(n)) ·∗

Z

 1  (f (y, τ )/∗ ((t −∗ τ )2∗ −∗ |y −∗ x|2∗ ) 2 ∗ ) ·∗ dy ·∗ dτ .

∗B(x,t−∗ τ )

(4.59) Exercise 4.14 Prove that the function u(x, t) defined by (4.59) satisfies the problem (4.58).

The Cauchy Problem for a Nonlinear Hyperbolic Equation

4.6

155

The Cauchy Problem for a Nonlinear Hyperbolic Equation

Consider the equation ∗ ∗ u∗∗ x1 x2 = f (x1 , x2 , u, ux1 , ux2 ),

(4.60)

where f is a given function. Suppose that the curve l is expressed in the form x2 = g(x1 ), where g is a C∗1 -function and g ∗ (x1 ) > 1. The Cauchy data for equation (4.60) are given by u(x1 , g(x1 ))

= φ(x1 ),

u∗x1 (x1 , g(x1 ))

= ψ(x1 ),

u∗x2 (x1 , g(x1 ))

= κ(x1 ),

where φ, ψ, κ are C∗1 -functions satisfying the consistency condition: φ∗ (x1 ) = ψ(x1 ) +∗ κ(x1 ) ·∗ g ∗ (x1 ). Without loss of generality, we assume that the functions φ, ψ and κ vanish identically. Otherwise, we introduce a new dependent variable: v = u −∗ φ(x1 ) −∗ (x2 −∗ g(x1 )) ·∗ κ(x1 ). Then, v(x1 , g(x1 )) = u(x1 , g(x1 )) −∗ φ(x1 ) = 1, vx1 (x1 , x2 ) = u∗x1 −∗ φ∗ (x1 ) +∗ g ∗ (x1 ) ·∗ κ(x1 ) −∗ (x2 −∗ g(x1 )) ·∗ κ∗ (x1 ), vx∗1 (x1 , g(x1 )) = u∗x1 (x1 , g(x1 )) −∗ φ∗ (x1 ) +∗ g ∗ (x1 ) ·∗ κ(x1 ) = ψ(x1 ) −∗ φ∗ (x1 ) +∗ g ∗ (x1 ) ·∗ κ(x1 ) = −∗ κ(x1 ) ·∗ g ∗ (x1 ) +∗ κ(x1 ) ·∗ g ∗ (x1 ) = 1,

156

The Multiplicative Wave Equation vx∗2 (x1 , x2 ) = u∗x2 (x1 , x2 ) −∗ κ(x1 ),

vx∗2 (x1 , g(x1 )) = u∗x2 (x1 , g(x1 )) −∗ κ(x1 ) = κ(x1 ) −∗ κ(x1 ) = 1, vx∗∗1 x2 (x1 , x2 ) = u∗x1 x2 (x1 , x2 ) −∗ κ∗ (x1 ). Equation (4.60) is converted to a similar equation: vx∗∗1 x2 = κ∗ (x1 ) +∗ f (x1 , x2 , v +∗ φ(x1 ) +∗ (x2 −∗ g)κ, vx1 +∗ φ∗ −∗ g ∗ ·∗ κ +∗ (x2 −∗ g) ·∗ κ∗ , vx∗2 +∗ κ). Therefore, we consider the problem ∗ ∗ u∗∗ x1 x2 = f (x1 , x2 , u, ux1 , ux2 )

u(x1 , g(x1 )) = u∗x1 (x1 , g(x1 )) = u∗x2 (x1 , g(x1 ))

(4.61)

= 1. Theorem 4.8 Let f (x1 , x2 , u, u∗x1 , u∗x2 ) be a continuous function of its variables in a neighborhood N of the point (x∗10 , x∗20 , 0∗ , 0∗ , 0∗ ), where (x∗10 , x∗20 ) lies on a multiplicative smooth curve l whose multiplicative tangent is never multiplicative horizontal or multiplicative vertical. Let also f satisfy in N a multiplicative Lipschitz condition with respect to each of the variables u, u∗x1 and u∗x2 . Then, in a sufficiently small neighborhood of (x∗10 , x∗20 ), there exists a unique solution of the problem (4.61). Proof Let δ > 0∗ be chosen so small that the region S defined by the inequalities |x1 −∗ x∗10 |∗

≤ δ,

|x2 −∗ x∗20 |∗

≤ δ,

|u|∗

≤ δ,

|u∗x1 |∗

≤ δ,

|u∗x2 |∗

≤ δ,

The Cauchy Problem for a Nonlinear Hyperbolic Equation

157

lies entirely in N. We denote m = max |f |∗ and let k1 , k2 and k3 denote the S

multiplicative Lipschitz constants associated with u, u∗x1 , and u∗x2 , respectively. We set M = max{m, k1 , k2 , k3 }. Now, we choose a multiplicative positive number p so small that M ·∗ p2∗


1,

u(1, t) = u∗x (eL , t) = 1,

t ≥ 1,

u(x, 1) = φ(x), u∗t (x, 1) = ψ(x), where φ ∈ C∗2 ([1, eL ]), ψ ∈ C∗1 ([1, eL ]) and φ(1) = ψ(1) = φ∗ (eL ) = ψ ∗ (eL ) = 1. Answer u(x, t) =

∞  X

An ·∗ cos∗ ((e(2n+1)cπ ·∗ t)/∗ e2L ) +∗ Bn

∗n=0

 ·∗ sin∗ ((e(2n+1)cπ ·∗ t)/∗ e2L ) ·∗ sin∗ ((e(2n+1)π ·∗ x)/∗ e2L ),

2 L

Ze

L

An = e · ∗

ψ(x) ·∗ sin∗ ((e(2n+1)π ·∗ x)/∗ e2L ) ·∗ d∗ x,

∗1

Bn = e

4 (2n+1)πc

Ze

L

·∗

ψ(x) ·∗ sin∗ ((e(2n+∗ 1)π ·∗ x)/∗ e2L ) ·∗ d∗ x,

n ∈ N0 .

∗1

Problem 4.5 Check that the function Z u(x, t) = (e/∗ t) ·∗ (y12∗ −∗ e2 ·∗ y1 ·∗ y2 +∗ y3 ) ·∗ d∗ sy , ∗S

where S : |x −∗ y|2∗∗ = t2∗ satisfies equation (4.32). Problem 4.6 Find a solution to the Cauchy problem: ∗∗ ∗∗ ∗∗ u∗∗ tt −∗ ux1 x1 −∗ ux2 x2 −∗ ux3 x3 = 1,

(x1 , x2 , x3 ) ∈ R3∗ ,

t > 1,

u(x1 , x2 , x3 , 1) = e2 ·∗ x1 −∗ e3 ·∗ x2 +∗ e4 ·∗ x3 , u∗t (x1 , x2 , x3 , 1) = e3 ,

(x1 , x2 , x3 ) ∈ R3∗ .

Advanced Practical Exercises

167

Answer u(x1 , x2 , x3 , t) = e2 ·∗ x1 −∗ e3 ·∗ x2 +∗ e4 ·∗ x3 +∗ e3 ·∗ t,

(x1 , x2 , x3 ) ∈ R3∗ ,

t > 1.

Problem 4.7 Solve the Cauchy problem: u∗tt −∗ u∗x1 x1 −∗ u∗x2 x2 −∗ u∗x3 x3 = −∗ e4 ,

(x1 , x2 , x3 ) ∈ R3∗ ,

t > 1,

u(x1 , x2 , x3 , 1) = x21∗ +∗ x22∗ +∗ x23∗ , u∗t (x1 , x2 , x3 , 1) = 1,

(x1 , x2 , x3 ) ∈ R3∗ .

Answer u(x1 , x2 , x3 , t) = x21∗ +∗ x22∗ +∗ x23∗ +∗ t2∗ ,

(x1 , x2 , x3 ) ∈ R3∗ ,

t > 1.

Problem 4.8 Solve the Cauchy problem: u∗tt −∗ u∗x1 x1 −∗ u∗x2 x2 = 1,

(x1 , x2 ) ∈ R2∗ ,

t > 1,

u(x1 , x2 , 1) = e2 ·∗ x1 −∗ e3 ·∗ x2 , u∗t (x1 , x2 , 1) = e3 ·∗ x1 +∗ x2 ,

(x1 , x2 ) ∈ R2∗ .

Answer u(x1 , x2 , t) = (e2 +∗ e3 ·∗ t) ·∗ x1 +∗ (t −∗ e3 ) ·∗ x2 ,

(x1 , x2 ) ∈ R2∗ ,

t > 1.

Problem 4.9 Solve the Cauchy problem: u∗tt −∗ u∗x1 x1 −∗ u∗x2 x2 = t,

(x1 , x2 ) ∈ R2∗ ,

t > 1,

u(x1 , x2 , 1) = x21∗ , u∗t (x1 , x2 , 1) = x22∗ ,

(x1 , x2 ) ∈ R2∗ .

Answer 1

u(x1 , x2 , t) = x21∗ +∗ t ·∗ x22∗ +∗ e 2 ·∗ t3∗ +∗ t2∗ .

(x1 , x2 ) ∈ R2∗ ,

Problem 4.10 Let L = L∗ . Prove that RL (x1 , x2 , ξ1 , ξ2 ) = RL∗ (ξ1 , ξ2 , x1 , x2 ).

t > 1.

5 The Heat Equation

5.1

The Weak Maximum Principle

Consider the multiplicative heat equation for a function u(x1 , . . . , xn , t) in an n-dimensional bounded domain D, u∗t = ek ·∗ ∆∗ u, where ∆u =

n X

(x1 , . . . , xn ) ∈ D,

t > 1,

(5.1)

u∗∗ xi xi , k is a positive constant. Define the domain

∗i=1

QT = {(x1 , . . . , xn , t) : (x1 , . . . , xn ) ∈ D,

1 < t ≤ eT },

where T > 1 is arbitrarily chosen. Define the parabolic boundary of QT as follows: ∂∗p QT = {D × {1}} ∪ {∂∗ D × [1, eT ]}. Let C∗QT be the class of functions that are twice multiplicative continuously differentiable in QT with respect to (x1 , . . . , xn ) and once multiplicative continuously differentiable with respect to t in QT , and continuous in QT . Theorem 5.1 Let u ∈ C∗QT and u∗t −∗ ek ·∗ ∆∗ u < 0∗ in QT . Then, u has no local maximum in QT and u achieves its maximum in ∂∗p QT . Proof Assume that u has a local maximum at some point (x, t) ∈ QT . Then, u∗t (x, t) = 1 and ∆∗ u(x, t) ≤ 1, which is a contradiction. Since u is continuous in QT , then its maximum is achieved somewhere in ∂QT . If the maximum is achieved at a point (x0 , T ) ∈ D × {T }, then u∗t (x, T ) = 1 and ∆∗ u(x, T ) ≤ 1, which is a contradiction. Exercise 5.1 Let u ∈ C∗QT and u∗t −∗ ek ·∗ ∆∗ u > 0∗ in QT . Prove that u has no local minimum in QT and u achieves its minimum in ∂p QT . DOI: 10.1201/9781003440116-5

168

The Weak Maximum Principle

169

Theorem 5.2 (Weak Maximum Principle for the Heat Equation) Let u ∈ C∗QT be a solution to the multiplicative heat equation (5.1). Then u achieves its maximum (minimum) on ∂∗p QT . Proof Let  > 0 be arbitrarily chosen and eM = max u. ∂∗p QT

Define the function v(x, t) = u(x, t) −∗ e ·∗ t. We have max v ≤ eM .

∂∗p QT

Hence, vt∗ −∗ ∆∗ v

=

u∗t −∗ e −∗ ∆∗ u

=

−∗ e


0 was arbitrarily chosen, we conclude that u ≤ eM in QT , which completes the proof. Corollary 5.1 Let u ∈ C∗QT be a solution to the heat equation (5.1). If u(x, t) ≥ (≤)0∗ on ∂∗p QT , then u(x, t) ≥ (≤)0∗ on QT .

170

The Heat Equation

Theorem 5.3 Let u ∈ C∗QT be a solution to the equation u∗t −∗ ek ·∗ ∆∗ u = f (x, t),

0∗ < t ≤ eT ,

(x1 , . . . , xn ) ∈ D,

(5.2)

where f ∈ C∗ (QT ) and |f |∗ ≤ eN on QT . Let also |u(x, t)|∗ ≤ em on ∂∗p QT . Then, |u(x, t)|∗ ≤ eN ·∗ t +∗ em

on

QT .

(5.3)

Proof Let  > 0 be arbitrarily chosen and w±∗ (x, t) = (eN +∗ e ) ·∗ t +∗ em ±∗ u(x, t). Note that w±∗ (x, t) ≥ 1 on ∂∗p QT . Also, ∗

(w±∗ )t −∗ ek ·∗ ∆∗ (w±∗ )

=

eN +∗ e ±∗ u∗t ∓∗ ek ·∗ ∆∗ u


= eN +∗ e ±∗ u∗t −∗ ek ·∗ ∆∗ u = e N +∗ e  ±∗ f > 0∗

on QT .

Hence from Exercise 5.1, we conclude that w±∗ achieves its maximum on QT . Because w±∗ ≥ 1 on ∂∗p QT , using Corollary 5.1, we obtain w±∗ ≥ 1 on QT , i.e., ±∗ u(x, t) ≤ (eN +∗ e ) ·∗ t +∗ m on QT . Because  > 0 was arbitrarily chosen, we get (5.3). Theorem 5.4 Let u1 , u2 ∈ C∗QT be solutions to equation (5.2) with initial condition ui (x, 1) = φi (x), x ∈ D, i = 1, 2, and boundary condition ui (x, t) = ψi (x, t),

0∗ ≤ t ≤ eT ,

x ∈ ∂D,

i = 1, 2,

respectively, where φi ∈ C∗2 (D), ψi ∈ C∗2 (QT ), i = 1, 2. Set δ = max |φ1 −∗ φ2 |∗ +∗ D

max |ψ1 −∗ ψ2 |∗ .

∂∗ D×{1}

Then |u1 −∗ u2 |∗ ≤ δ,

on

QT .

(5.4)

The Strong Maximum Principle

171

Proof Let w = u1 −∗ u2 . Then w is a solution to the problem wt∗ −∗ ek ·+ ∗∆∗ w = 1 on

D × (1, eT ],

w(x, 1) = φ1 (x) −∗ φ2 (x),

x ∈ D,

w(x, t) = ψ1 (x, t) −∗ ψ2 (x, t),

x ∈ ∂∗ D,

1 ≤ x ≤ eT .

Hence from Theorem 5.2, we obtain min w(x, t) ≤ w(x, t) ≤ max w(x, t)

∂∗p QT

∂∗p QT

in QT ,

whereupon we get (5.4).

5.2

The Strong Maximum Principle

Theorem 5.5 (The Stong Maximum Principle for the Multiplicative Heat Equation) Let u ∈ C∗QT be a solution to equation (5.1). Let also u(x0 , t0 ) = max u(x, t) = m > 0∗ QT

in some point (x0 , t0 ) ∈ QT \∂∗p QT . Then, u(x, t) = m for all (x, t) ∈ Qt0 for which there exists a continuous curve that connects (x, t) and (x0 , t0 ) and lies in Qt0 . Proof Suppose the contrary. Then there exists a point (x1 , t1 ) so that 0 x ∈ D, 0∗ ≤ t1 < et + and u(x1 , t1 ) < em1 < em . Let Q1 be the cylinder:   ! 12 ∗ n   X Q1 = (x, t) : (xi −∗ x1i )2∗ ≤ eρ , t1 ≤ t ≤ t2 ,   1

∗i=1

where 1 > ρ > 0 and t1 < t2 < t0 are chosen so that Q1 ⊂ Qt0 and ! 12 ∗ n X 1 m1 1 2∗ u(x, t ) < e for all x for which (xi −∗ xi ) ≤ eρ . ∗i=1

Let α > 0 be chosen so that 4k 2 (n + 2)2 − 8kαρ2 < 0.

172

The Heat Equation

We consider the function m

w(x, t) = e −∗ e

(m−m1 )

·∗

e

ρ2

!2∗

n X

−∗ (xi −∗ x1i )2∗ ∗i=1 1

·∗ e−∗ α·∗ (t−∗ t ) −∗ u(x, t). Then,  2

 eρ −∗

!2∗ ∗

n X

(xi −∗ x1i )2∗



∗i=1

=

=

2

e ·∗

xj

e

ρ2

4

−∗ e ·∗

 2

 eρ −∗

!

n X

−∗ (xi −∗ x1i )2∗ ∗i=1 e

ρ2

!

n X

−∗ (xi −∗ x1i )2∗ ∗i=1 !2 ∗∗

(xi −∗ x1i )2∗

 xj xj


−∗ e4 ·∗ −∗ e2 ·∗ (xj −∗ x1j ) ·∗ (xj −∗ x1j ) 4

−∗ e ·∗

=

·∗ (xj −∗ x1j ),

∗

n X ∗i=1

=


·∗ −∗ e2 ·∗ (xj −∗ x1j )

e

4

−∗ e ·∗



e

n X

2

∆ ∗  e ρ −∗

ρ2

ρ2

n X

!

n X

!

−∗ (xi −∗ x1i )2∗ ∗i=1 −∗ (xi −∗ x1i )2∗ ∗i=1

+∗ e8 ·∗ (xj −∗ x1j )2∗ ,

!2∗  (xi −∗ x1i )2∗



∗i=1

=

−∗ e

4n

·∗

+∗ e8 ·∗

e

n X

ρ2

!

n X

−∗ (xi −∗ x1i )2∗ ∗i=1

(xi −∗ x1i )2∗ ,

∗i=1

from where  ∆∗ w(x, t)

=

n X

2

−∗ e(m−m1 ) ·∗ ∆∗  eρ −∗

∗i=1

·∗ e−∗ e

α

1

·∗ (t−∗ t )

−∗ ∆∗ u(x, t)

!2∗  (xi −∗ x1i )2∗



The Strong Maximum Principle

=

e

4n(m−m1 )

−∗ e

−∗ e ·∗ ∆∗ w(x, t)

−∗ e

=

·∗

8(m−m1 )

·∗ e−∗ e k

173

α

e

!

n X

ρ2

−∗ (xi −∗ x1i )2∗ ∗i=1 !

n X

·∗ (xi −∗ x1i )2∗ ∗i=1

·∗ (t−∗ t1 )

−∗ ∆∗ u(x, t),

4kn(m−m1 )

+∗ e8k(m−m1 ) ·∗

·∗

−∗ (xi −∗ x1i )2∗ ∗i=1

e

n X

!

n X

ρ2

! (xi −∗ x1i )2∗

∗i=1

·∗ e−∗ e

wt∗ (x, t)

=

e

α

·∗ (t−∗ t1 )

α(m−m1 )

e−∗ e

α

·∗

e

+∗ ek ·∗ ∆∗ u(x, t)

ρ2

!2∗

n X

−∗ (xi −∗ x1i )2∗ ∗i=1

·∗ (t−∗ t1 )

−∗ u∗t

(m−m1 )



and k

wt (x, t) −∗ e ·∗ ∆∗ w(x, t) = e

2

−∗ e4kn ·∗

eρ −∗

α

·∗ e ·∗

e

ρ2

−∗ (xi −∗ x1i )2∗ ∗i=1

!

n X

(xi −∗ x1i )2∗

−∗ e8k ·∗

+∗ e8kρ

=e

(m−m1 )

−∗ e

·∗ e−∗ e

α

e ρ −∗

·∗

e

in Q1 .

(xi −∗ x1i )2∗

∗i=1

ρ

e

2

ρ2

−∗

!2∗

n X

−∗ (xi −∗ x1i )2∗ ∗i=1 n X

∗i=1

≥1

!

·∗ (t−∗ t1 )

 ·∗ eα ·∗

4k(n+2)

n X

2

∗i=1

 2

!2∗

n X

! (xi −∗ x1i )2∗

+∗ e

8kρ2



e −∗ e

α

·∗ (t−∗ t1 )

174

The Heat Equation When

! 12 ∗

n X

(xi −∗ x1i )2∗

= eρ

∗i=1 1

2

and t ≤ t ≤ t , we have w(x, t)

=

em −∗ u(x, t)

≥

0∗ .

When t = t1 and ! 12 ∗

n X

(xi −∗ x1i )2∗ ∗i=1

≤ eρ ,

we have 2

= em −∗ e(m−m1 ) ·∗

w(x, t)

e ρ −∗

n X

!2∗ (xi −∗ x1i )2∗

−∗ u(x, t)

∗i=1 4

≥ em −∗ e(m−m1 )ρ −∗ em 1 >

em−(m−m1 )−m1

=

1.

Hence from Corollary 5.1, we conclude that w(x, t) ≥ 1 in Q1 . Let (x2 , t2 ) ∈ Q1 and n X 2 (x2i −∗ x1i )2∗ < eρ . ∗i=1 2

2

Then, w(x , t ) ≥ 0∗ and 2

2

w(x , t )

m

= e −∗ e ·∗ e−∗ e ≥

α

(m−m1 )

·∗

·∗ (t2 −∗ t1 )

e

ρ2

n X

−∗ (x2i ∗i=1 2 2

!2∗ −∗ x1i )2∗

−∗ u(x , t )

1,

whereupon 2

2

u(x , t ) ≤
1.

We will seek a solution u(x, t) of (5.5) that has the special structure:   1 −∗ e α −∗ e 2 u(x, t) = t ·∗ v t ·∗ |x|∗ ,

(5.5)

(5.6)

where α is a constant and v is a function that must be found, |x|∗ =

! 12 ∗

n X

x2i ∗

.

∗i=1

We set

1

y = t−∗ e 2 ·∗ x and insert (5.6) into (5.5). We get 1

eα ·∗ v(|y|∗ ) +∗ e 2 ·∗ |y|∗ ·∗ v ∗ (|y|∗ ) +∗ (en−1 /∗ |y|∗ ) ·∗ v ∗ (|y|∗ ) +∗ v ∗∗ (|y|∗ ) = 1. We take α=

n 2

and set r1 = |y|∗ .

176

The Heat Equation

Then, n

1

e 2 ·∗ v(r1 ) +∗ e 2 ·∗ r1 ·∗ v ∗ (r1 ) +∗ (en−1 /∗ r1 ) ·∗ v ∗ (r1 ) +∗ v ∗∗ (r1 ) = 1, (n−1)∗

which we multiply by r1 1

and we find 1

(n−1)∗

(n−1)∗

·∗ v ∗ (r1 ) +∗ r1 whereupon 

(n−2)∗

·∗ v(r1 ) +∗ e 2 ·∗ r1n∗ ·∗ v ∗ (r1 ) +∗ en−1 ·∗ r1

e 2 n ·∗ r1

(n−1)∗

r1

·∗ v ∗

·∗ v ∗ ∗ (r1 ) = 1,

∗

1

∗

+∗ e 2 ·∗ (r1n∗ ·∗ v) = 1.

Thus, (n−1)∗

r1

1

·∗ v ∗ +∗ e 2 ·∗ r1n∗ ·∗ v = eC1

for some constant C1 . Assuming lim v(r1 ), v ∗ (r1 ) = 1,

r1 →∞

we get C1 = 1 and (n−1)∗

r1

1

·∗ v ∗ +∗ e 2 ·∗ r1n∗ ·∗ v = 1,

whence

1

v ∗ = −∗ e 2 ·∗ r1 ·∗ v and

2∗ )/

v = eC ·∗ e−∗ ((r1

∗e

4

)

,

C > 0.

From here,   2∗ 4 n 1 n u(x, t) = (e/∗ t 2 ∗ ) ·∗ v (|x|∗ /∗ t 2 ∗ ) = (eC /∗ t 2 ∗ ) ·∗ e−∗ (|x| /∗ (e ·∗ t)) solves the multiplicative heat equation (5.5). Definition 5.1 The function  2∗ 4 1   ·∗ e−∗ (|x|∗ )/∗ (e ·∗ t) , n  4π 2 (e ·∗ t) ∗ Φ(x, t) =    1, x ∈ Rn∗ , t < 1,

x ∈ Rn∗ ,

is called the fundamental solution of the heat equation. Theorem 5.6 For each t > 1 we have Z Φ(x, t) ·∗ d∗ x = e. ∗Rn ∗

t > 1,

The Cauchy Problem Proof We have Z Φ(x, t) ·∗ d∗ x

177

= (e/∗ (e

4π

·∗ t)

n 2∗

Z

2∗

e−∗ (|x|∗

) ·∗

∗Rn ∗

/∗ (e4 ·∗ t))

·∗ d∗ x

∗Rn ∗

Z

n

=

(e/∗ e

π2

) ·∗

2∗

e−∗ |z|∗ ·∗ d∗ z

∗Rn ∗

= e

1 n π2

·∗

n Z∞ Y

2∗

e−∗ zi ·∗ d∗ zi

i=1− ∞ ∗

= e.

5.3.2

The Cauchy problem

Consider the following Cauchy problem u∗t −∗ ∆∗ u = 1 in

Rn∗ × (1, ∞), (5.7)

u=φ

on Rn∗ × {t = 1}.

Theorem 5.7 Assume φ ∈ C∗ (Rn∗ ) and |φ(x)|∗ ≤ eM for all x ∈ Rn∗ and for some M > 1. Let u(x, t) be defined by Z 2∗ 4 n 4π 2 e−∗ ((|x−∗ y|∗ )/∗ (e ·∗ t)) u(x, t) = (e/∗ (e ·∗ t) ∗ ) ·∗ ∗Rn ∗

·∗ φ(y) ·∗ d∗ y,

x ∈ Rn∗ ,

t > 1.

Then 1. u ∈ C∗∞ (Rn∗ × (1, ∞)), 2. u∗t (x, t) −∗ ∆∗ u(x, t) = 1, 3.

x ∈ Rn∗ ,

t > 1,

0

lim u(x, t) = φ(x ) for each x0 ∈ Rn∗ . (x, t) → (x0 , 1) x ∈ Rn∗ , t > 1

Proof n

2∗

4

1. Let t1 > 1 be arbitrarily chosen. Since (e/∗ t 2 ∗ ) ·∗ e−∗ (|x|∗ /∗ (e ·∗ t)) is infinetely many multiplicative differentiable with uniformly bounded multiplicative derivatives of all orders on Rn∗ × [t1 , ∞) and t1 > 1 was arbitrarily chosen, we conclude that u ∈ C∗∞ (Rn∗ × (1, ∞)).

178

The Heat Equation 2. Note that u∗t (x, t)

Z −∗ ∆∗ u(x, t)

=

(Φ∗t −∗ ∆∗x Φ) (x −∗ y, t) ·∗ φ(y) ·∗ d∗ y

∗Rn ∗

=

1,

because Φt (x −∗ y, t)

=

−∗ n 2

n

−∗ e 2 (4π)

2∗ )/

n

·∗ t−∗ 2 ∗ −∗ e ·∗ e−∗ (|x−∗ y|∗

n

n

+∗ e(4π) 2 ·∗ t−∗ 2 ∗ −∗ e

2∗ )/

=

1

−∗ e 2 (4π)

−∗ n 2

1

−∗ n 2

1

−∗ n 2

−∗ e 2 (4π) +∗ e 4 (4π)

∆∗x Φ(x −∗ y, t)

=

n

−∗ n 2

1

−∗ n 2

−∗ e 2 (4π) +∗ e 4 (4π)

2∗

·∗ e−∗ (|x−∗ y|∗

4

·∗ t)

,

∗ (e

4

·∗ t)

, 2∗ )/

n

·∗ t−∗ 2 ∗ −∗ e e−∗ (|x−∗ y|∗

∗ (e

4

·∗ t)

n

·∗ t−∗ 2 ∗ −∗ e ·∗ (xj −∗ yj )2∗

2∗ )/

·∗ e−∗ (|x−∗ y|∗

∗ (e

n

2∗ )/

=

∗ (e

4

·∗ t)

, 2∗ )/

n

·∗ t−∗ 2 ∗ −∗ e e−∗ (|x−∗ y|∗

∗ (e

4

·∗ t)

n

·∗ t−∗ 2 ∗ −∗ e ·∗ |x −∗ y|2∗∗ )/∗ (e4 ·∗ t)

,

i.e., Φ itself solves the heat equation. 3. Fix x0 ∈ Rn∗ and  > 0. Then there exists δ = δ() > 0 such that |φ(y) −∗ φ(x0 )|∗ < e Let

if |y −∗ x0 |∗ < eδ ,

y ∈ Rn∗ .

δ

|x −∗ x0 |∗ ≤ e 2 . Then, Z
0 0 |u(x, t) −∗ φ(x )|∗ = Φ(x −∗ y, t) ·∗ φ(y) −∗ φ(x ) ·∗ d∗ y Rn∗ Z ≤ ∗Rn ∗

·∗ t)

·∗ t−∗ 2 ∗ −∗ e ·∗ (xj −∗ yj )

·∗ e−∗ (|x−∗ y|∗ Φ∗∗ xj xj (x −∗ y, t)

4

2

·∗ |x −∗ y|2∗∗ e−∗ (|x−∗ y|∗ Φ∗xj (x −∗ y, t)

∗ (e

|Φ(y −∗ x, t)|∗ ·∗ |φ(y) −∗ φ(x0 )|∗ d∗ y

∗

The Cauchy Problem

179 Z

Φ(x −∗ y, t) ·∗ |φ(y) −∗ φ(x0 )|∗ ·∗ d∗ y

= ∗B(x0 ,δ)

Z Φ(x −∗ y, t) ·∗ |φ(y)

+∗ 0 ∗Rn ∗ \B(x ,δ)

−∗ φ(x0 )|∗ ·∗ d∗ y =: I1 +∗ I2 . Note that Z

≤ e ·∗

I1

Φ(x −∗ y, t) ·∗ d∗ y ∗B(x0 ,δ)

≤ e ·∗

Z Φ(x −∗ y, t) ·∗ d∗ y ∗Rn ∗

= e and Z I2


Φ(x −∗ y, t) ·∗ |φ(y)|∗ +∗ |φ(x0 )|∗ ·∗ d∗ y

≤ 0 ∗Rn ∗ \B(x ,δ)

Z

≤ e2M ·∗

Φ(x −∗ y, t) ·∗ d∗ y 0 ∗Rn ∗ \B(x ,δ)

Z

n

= (e2M /∗ (e4π ·∗ t) 2 ∗ ) ·∗

2∗ )/∗ (e4 ·∗ t))

e−( (|x−∗ y|∗

·∗ d∗ y

0 ∗Rn ∗ \B(x ,δ)



≤

(e

1

|x −∗ y|∗ ≥ e 2 ·∗ |y −∗ x0 |∗

2M

/∗ (e

4π

·∗ t)

n 2∗



Z

e−∗ (|x

) ·∗

0

−∗ y|2∗∗ /∗ (e16 ·∗ t))

·∗ d∗ y

0 ∗Rn ∗ \B(x ,δ)

≤

(e

C1

/∗ t

n 2∗

Z∞ ) ·∗

e−∗ (r

2∗ /

∗ (e

16

·∗ t))

·∗ r(n−1)∗ ·∗ d∗ r → 1

∗eδ

as

t → 1+∗ ,

where C1 is a multiplicative constant independent of t. Therefore, if δ |x −∗ x0 |∗ ≤ e 2

180

The Heat Equation and t > 1 is small enough, we have I2 ≤ e and |u(x, t) −∗ φ(x0 )|∗ ≤ e2 , which completes the proof.

Example 5.1 Consider the following Cauchy problem: u∗t −∗ u∗∗ xx = 0∗

(−∗ ∞, ∞) × (1, ∞),

in

u(x, 1) = x

(−∗ ∞, ∞).

in

Then, u(x, t)

= (e/∗ (e

√ 2 π

·∗ t

1 2∗

Z∞ )) ·∗

e−∗ ((x−∗ y)

2∗ /

∗ (e

4

·∗ t))

·∗ y ·∗ d∗ y

∗−∗ ∞

=

(e/∗ (e

√ 2 π

·∗ t

1 2∗

Z∞ )) ·∗

e−∗ (z

2∗ /

∗ (e

4

·∗ t))

·∗ (x −∗ z) ·∗ d∗ z

∗−∗ ∞ √

=

(e/∗ (e2

π

Z∞

 x ·∗

1

·∗ t 2 ∗ )) ·∗

e−∗ (z

2∗ /

∗ (e

4

·∗ t))

Z∞ ·∗ d∗ z −∗

∗−∗ ∞

 e−∗ (z

2∗ /

∗ (e

4

·∗ t))

·∗ z ·∗ d∗ z

∗−∗ ∞

= x. Example 5.2 Consider the following Cauchy problem: ∗∗ u∗t −∗ u∗∗ x1 x1 −∗ ux2 x2 = 1 in

R2∗ × (1, ∞),

u(x1 , x2 , 1) = x21∗ +∗ x22∗

in R2∗ .

We have u(x1 , x2 , t) =

(e/∗ (e4π ·∗ t)) ·∗

Z

2∗ /

e−∗ (|x−∗ y|

∗ (e

4

·∗ t))

·∗ (y12∗ +∗ y22∗ ) ·∗ d∗ y1 ·∗ d∗ y2

∗R2∗

=

(e/∗ (e4π ·∗ t)) ·∗ Z ∗R2∗

2∗

e−∗ ((z1

+∗ z22∗ )/∗ (e4 ·∗ t))


·∗ (x1 −∗ z1 )2∗ +∗ (x2 −∗ z2 )2∗ ·∗ d∗ z1 ·∗ d∗ z2

The Cauchy Problem

181

= (x21∗ +∗ x22∗ ) ·∗ (e/∗ (e4π ·∗ t)) ·∗ Z∞

Z∞

2∗ +∗ z22∗ )/∗ (e4 ·∗ t))

e−∗ ((z1

·∗ d∗ z1 ·∗ d∗ z2

∗−∗ ∞ ∗−∗ ∞

−∗ (x1 /∗ (e

2π

Z∞

Z∞

·∗ t)) ·∗

2∗ +∗ z22∗ )/∗ (e4 ·∗ t))

e−∗ ((z1

·∗ z1 ·∗ d∗ z1 ·∗ d∗ z2

∗−∗ ∞ ∗−∗ ∞

−∗ (x2 /∗ (e

2π

Z∞

Z∞

·∗ t)) ·∗

2∗ +∗ z22∗ )/∗ (e4 ·∗ t))

e−∗ ((z1

·∗ z2 ·∗ d∗ z1 ·∗ d∗ z2

∗−∗ ∞ ∗−∗ ∞

+∗ (e/∗ (e4π ·∗ t)) ·∗ Z∞

Z∞

2∗ +∗ z22∗ )/∗ (e4 ·∗ t))

e−∗ ((z1

·∗ (z12∗ +∗ z22∗ ) ·∗ d∗ z1 ·∗ d∗ z2

∗−∗ ∞ ∗−∗ ∞

=

x21∗ +∗ x22∗ +∗ e4 ·∗ t.

Exercise 5.2 Find a solution to the Cauchy problem: ∗∗ ∗∗ u∗t −∗ u∗∗ x1 x1 −∗ ux2 x2 −∗ ux3 x3 = 1 in

R3∗ × (1, ∞)

u(x1 , x2 , x3 , 1) = e2 ·∗ x1 −∗ x2 +∗ e3 ·∗ x3

in R3∗ .

Answer u(x1 , x2 , x3 , t) = e2 ·∗ x1 −∗ x2 +∗ e3 ·∗ x3 . Next, we consider the following Cauchy problem: u∗t −∗ ∆∗ u = f (x, t)

in Rn∗ × (1, ∞) (5.8)

u = 0 on

Rn∗ × {t = 1}.

Theorem 5.8 Let f ∈ C∗2 (Rn∗ , C∗1 ([1, ∞))), |ft∗ |∗ , |fx∗i |∗ , |fx∗∗i xi |∗ ≤ eM , i = 1, . . . , n, in Rn∗ × [1, ∞) for some positive constant eM . Let also Zt Z Φ(x −∗ y, t −∗ s) ·∗ f (y, s) ·∗ d∗ y ·∗ d∗ s.

u(x, t) = ∗1 Rn ∗

182

The Heat Equation

Then 1. u ∈ C∗2 (Rn∗ , C∗1 ((1, ∞))), 2. u∗t −∗ ∆∗ u = f (x, t) in Rn∗ × (1, ∞), 3.

lim u(x, t) = 1 for each x0 ∈ Rn∗ . 0 (x, t) → (x , 1) x ∈ Rn∗ , t > 1

Proof 1. We change the variables and we get Zt Z Φ(y, s) ·∗ f (x −∗ y, t −∗ s) ·∗ d∗ y ·∗ d∗ s.

u(x, t) = ∗1

Rn ∗

Since f ∈ C∗2 (Rn∗ , C∗1 ((1, ∞))) and |f |∗ , |ft∗ |∗ , |fx∗i |∗ , |fx∗∗i xi |∗ ≤ eM , i = 1, . . . , n, we conclude that u ∈ C∗2 (Rn∗ , C∗1 ((1, ∞))). 2. We have u∗t (x, t)

Zt Z =

Φ(y, s) ·∗ ft∗ (x −∗ y, t −∗ s) ·∗ d∗ y ·∗ d∗ s

∗1 ∗Rn ∗

Z Φ(y, t) ·∗ f (x −∗ y, 1) ·∗ d∗ y

+∗ ∗Rn ∗

Zt Z = −∗

Φ(y, s) ·∗ fs∗ (x −∗ y, t −∗ s) ·∗ d∗ y ·∗ d∗ s

∗1 ∗Rn ∗

Z +∗

Φ(y, t) ·∗ f (x −∗ y, 1) ·∗ d∗ y,

∗Rn ∗

Zt Z ∆∗ u(x, t)

Φ(y, s) ·∗ ∆∗x f (x −∗ y, t −∗ s) ·∗ d∗ y ·∗ d∗ s

= ∗1

∗Rn ∗

Zt Z Φ(y, s) ·∗ ∆∗y f (x −∗ y, t −∗ s) ·∗ d∗ y ·∗ d∗ s.

= ∗1 ∗Rn ∗

The Cauchy Problem

183

Therefore, u∗t (x, t) −∗ ∆∗ u(x, t) Zt Z

Φ(y, s) (−∗ fs∗ (x −∗ y, t −∗ s) −∗ ∆∗y f (x −∗ y, t −∗ s))

= ∗1 ∗Rn ∗

·∗ d∗ y ·∗ d∗ s Z Φ(y, t) ·∗ f (x −∗ y, 1) ·∗ d∗ y

+∗ ∗Rn

Zt Z

Φ(y, s) ·∗ (−∗ fs∗ (x −∗ y, t −∗ s) −∗ ∆∗y f (x −∗ y, t −∗ s))

= ∗e ∗Rn ∗

·∗ d∗ y ·∗ d∗ s 

Ze Z Φ(y, s) ·∗ (−∗ fs (x −∗ y, t −∗ s) −∗ ∆∗y f (x −∗ y, t −∗ s))

+∗ ∗1 ∗Rn ∗

·∗ d∗ y ·∗ d∗ s Z Φ(y, t) ·∗ f (x −∗ y, 1) ·∗ d∗ y

+∗ ∗Rn ∗

= J1 +∗ J2 +∗ J3 . Note that Zt Z Φ(y, s) ·∗ (−∗ fs (x −∗ y, t −∗ s) −∗ ∆∗y f (x −∗ y, t −∗ s))

J1 = e ∗Rn ∗

·∗ d∗ y ·∗ d∗ s Zt Z (Φs (y, s) −∗ ∆∗y Φ(y, s)) ·∗ f (x −∗ y, t −∗ s) ·∗ d∗ y ·∗ d∗ s

= e ∗ ∗Rn

184

The Heat Equation Z

Φ(y, e ) ·∗ f (x −∗ y, t −∗ e ) ·∗ d∗ y

+∗ ∗Rn ∗

Z −∗

Φ(y, t) ·∗ f (x −∗ y, 1) ·∗ d∗ y ∗Rn ∗

Z =

Φ(y, e ) ·∗ f (x −∗ y, t −∗ e ) ·∗ d∗ y −∗ J3 .

∗Rn

Hence, Z

u∗t (x, t) −∗ ∆∗ u(x, t) =

Φ(y, e ) ·∗ f (x −∗ y, t −∗ e ) ·∗ d∗ y −∗ J3

∗Rn ∗ 

Ze Z Φ(y, s) ·∗ (−∗ fs (x −∗ y, t −∗ s) −∗ ∆∗y f (x −∗ y, t −∗ s))

+∗ ∗1 ∗Rn ∗

·∗ d∗ y ·∗ d∗ s +∗ J3 Z

Φ(y, e ) ·∗ f (x −∗ y, t −∗ e ) ·∗ d∗ y

= ∗Rn



Ze Z +∗

Φ(y, s) ·∗ (−∗ fs (x −∗ y, t −∗ s) −∗ ∆∗y f (x −∗ y, t −∗ s))

∗1 ∗Rn ∗

·∗ d∗ y ·∗ d∗ s and u∗t (x, t) −∗ ∆∗ u(x, t) = lim

Z

e →0∗ ∗Rn ∗

Φ(y, e ) ·∗ f (x −∗ y, t −∗ e ) ·∗ d∗ y.

We observe that Z   Φ(y, e ) ·∗ f (x −∗ y, t −∗ e ) ·∗ d∗ y −∗ f (x, t) ∗Rn∗ ∗ Z ≤ Φ(y, e ) ·∗ |f (x −∗ y, t −∗ e ) −∗ f (x, t)|∗ ·∗ d∗ y ∗Rn ∗

The Cauchy Problem Z = Φ(y, e ) ·∗ |f (x −∗ y, t −∗ e ) −∗ f (x, t)|∗ ·∗ d∗ y

185

∗B(x,δ)

Z

Φ(y, e ) ·∗ |f (x −∗ y, t −∗ ) −∗ f (x, t)|∗ ·∗ d∗ y

+∗ ∗Rn ∗ \B(x,δ)

J4

=

J4 +∗ J5 ,

≤

e ·∗

Z

Φ(y, e ) ·∗ d∗ y

∗Rn ∗

= J5

≤

e , (e

2M

/∗ (e

4π

)

n 2∗

Z

2∗ )/

e−∗ (|y|∗

) ·∗

∗ (e

4

)

·∗ d∗ y

∗Rn ∗ \B(x,δ)

≤

e

C2 n 2

Z∞ ·∗

e−∗ ((r

2∗ )/

∗ (e

4

))

·∗ r(n−1)∗ ·∗ d∗ r

eδ

≤

C3 

e

,

where C2 and C3 are positive constants independent of 0 <  < 1. Therefore, Z Φ(y, e ) ·∗ f (x −∗ y, t −∗ e ) ·∗ d∗ y −∗ f (x, t) ≤ e(1+C3 ) . ∗Rn∗ ∗

From here, Z lim 

e →0∗ ∗Rn ∗

Φ(y, e ) ·∗ f (x −∗ y, t −∗ e ) ·∗ d∗ y = f (x, t)

and u∗t (x, t) −∗ ∆∗ u(x, t) = 1 in

Rn∗ × (1, ∞).

3. From the definition of the function u, we have Z t Z |u(x, t)|∗ = Φ(y, s) ·∗ f (x −∗ y, t −∗ s) ·∗ d∗ y ·∗ d∗ s ∗1 ∗Rn∗

∗

Zt Z ≤

Φ(y, s) ·∗ |f (x −∗ y, t −∗ s)|∗ ·∗ d∗ y ·∗ d∗ s ∗1 ∗Rn ∗

186

The Heat Equation ≤ e

M

Zt Z ·∗

Φ(y, s) ·∗ d∗ y ·∗ d∗ s ∗1

∗Rn ∗

= eM ·∗ t. Consequently, lim u(x, t) = 1 for (x, t) → (x0 , 1) x ∈ Rn∗ , t > 1

each x0 ∈ Rn∗ .

Example 5.3 Consider the following Cauchy problem: ∗∗ u∗t −∗ u∗∗ x1 x1 −∗ ux2 x2 = x1 +∗ x2

in R2∗ × (1, ∞)

u(x1 , x2 , 1) = 1 in R2∗ . We have Zt u(x1 , x2 , t)

=

Z

(e/∗ (e4π ·∗ (t −∗ s))) ·∗

∗1

2∗ )/∗ (e4 ·∗ (t−∗ s)))

e−∗ ((|x−∗ y|

∗R2∗

·∗ (y1 +∗ y2 ) ·∗ d∗ y1 ·∗ d∗ y2 ·∗ d∗ s Zt =

(e/∗ (e4π ·∗ (t −∗ s)))

∗1

Z∞ Z∞ ·∗

2∗ +∗ z22∗ )/∗ (e4 ·∗ (t−∗ s)))

e−∗ ((z1

−∗ ∞ −∗ ∞

·∗ (x1 +∗ x2 −∗ z1 −∗ z2 ) ·∗ d∗ z1 ·∗ d∗ z2 ·∗ d∗ s = t ·∗ (x1 +∗ x2 ). Exercise 5.3 Find a solution to the Cauchy problem: ∗∗ ∗∗ 2 3 u∗t −∗ u∗∗ x1 x1 −∗ ux2 x2 −∗ ux3 x3 = e ·∗ t ·∗ (x1 −∗ x2 +∗ x3 ) in R∗ × (1, ∞)

u(x1 , x2 , x3 , 1) = 1 in R3∗ . Answer u(x1 , x2 , x3 , t) = t2∗ ·∗ (x1 −∗ x2 +∗ x3 ).

The Mean Value Formula

187

Remark 5.1 If u1 is a solution to the problem (5.7) and u2 is a solution to the problem (5.8), then by Theorem 5.7 and Theorem 5.8, it follows that u(x, t)

=

u1 (x, t) +∗ u2 (x, t) Zt Z

Z Φ(x −∗ y, t) ·∗ φ(y) ·∗ d∗ y +∗

=

Φ(x −∗ y, t −∗ s) ·∗ f (y, s) ∗1

∗Rn ∗

∗Rn ∗

·∗ d∗ y ·∗ d∗ s, where f and φ satisfy the hypotheses above, is a solution to the problem: u∗t −∗ ∆∗ u = f u=φ

in Rn∗ × (1, ∞) on Rn∗ × {t = 1}.

Exercise 5.4 Find a solution to the Cauchy problem: ∗∗ 2 2 3∗ u∗t −∗ u∗∗ +∗ x1 +∗ e3 ·∗ t2∗ ·∗ x22∗ x1 x1 −∗ ux2 x2 = e ·∗ t −∗ e ·∗ t

in R2∗ × (1, ∞) u(x1 , x2 , 1) = x2

in R2∗ .

Answer u(x1 , x2 , t) = t2∗ +∗ t ·∗ x1 +∗ t3∗ ·∗ x22∗ +∗ x2 .

5.4

The Mean Value Formula

For fixed x ∈ Rn∗ , t ∈ R∗ , we define  W (x, t, r) = (y, s) ∈ Rn+1 : s ≤ t, ∗

Φ(x −∗ y, t −∗ s) ≥ (e/∗ rn∗ ) .

We set W (r) = W (1, 1, r). Example 5.4 For n = 2 we will compute Z Z (|y|2∗∗ /∗ s2∗ ) ·∗ d∗ y ·∗ d∗ s. ∗W (1)

We have n 2∗ 4 W (e) = (y, s) ∈ R3∗ : −∗ (e/∗ (e4π ·∗ s)) ·∗ e(|y|∗ /∗ (e ·∗ s)) ≥ e,

o s≤1 .

188

The Heat Equation

Hence, n 1 W (e) = (y, s) ∈ R3∗ : |y|∗ ≤ e2 ·∗ (s ·∗ log∗ (−∗ e4π ·∗ s)) 2 ∗ , Therefore, Z Z

o 1 −∗ e 4π ≤ s ≤ 1 .

(|y|2∗ /∗ s2∗ ) ·∗ d∗ y ·∗ d∗ s

∗W (e)

Z1

Z

(|y|2∗ /∗ s2∗ ) ·∗ d∗ y ·∗ d∗ s

= 1

1

∗−∗ e 4π ∗|y|∗ ≤e2 ·∗ (s·∗ log∗ (−∗ e4π ·∗ s)) 2 ∗

= e

2π

Z1

2∗

·∗

Ze

(e/∗ s ) ·∗

2 1

(s ·∗ log∗ (−∗ e4π ·∗ s)) 2 ∗ ·∗ r3∗ ·∗ d∗ r ·∗ d∗ s

1

1

∗−∗ e 4π

= e4 . Exercise 5.5 Consider the case n = 3 and compute Z Z (|y|2∗ /∗ s2∗ ) ·∗ d∗ y ·∗ d∗ s. ∗W (e)

Generalize the result for arbitrary n. Answer e4 . Theorem 5.9 (The Mean Value Formula) Let u ∈ C∗QT solve the heat equation: u∗t −∗ ∆∗ u = 1. Then, u(x, t) = (e/∗ (e4 ·∗ rn∗ )) ·∗

Z

Z

u(y, s)(|x −∗ y|2∗∗ /∗ (t −∗ s)2∗ ) ·∗ d∗ y ·∗ d∗ s.

∗W (x,t,r)

Proof Without loss of generality, we will prove the formula for x = 1, t = 1. Otherwise, we translate the space and time coordinates so that x = 1, t = 1. Let Z Z φ(r) = u(r ·∗ z, r2∗ ·∗ τ ) ·∗ (|z|2∗∗ /∗ τ 2∗ ) ·∗ d∗ z ·∗ d∗ τ. ∗W (e)

Then, ∗

φ (r)

Z

Z

= ∗W (e)

n X ∗i=1

·∗ d∗ z ·∗ d∗ τ

! u∗yi ·∗

2∗

2∗

2

zi (|z| /∗ τ ) +∗ e ·∗ r

·∗ u∗s ·∗

2∗

(|z| /∗ τ )

The Mean Value Formula

189

= (e/∗ r(n+1)∗ ) ·∗

Z

n X

Z

∗i=1

∗W (r)

Z

+∗ (e2 /∗ r(n+1)∗ ) ·∗

u∗yi ·∗ yi ·∗ (|y|2∗ /∗ s2∗ ) ·∗ d∗ y ·∗ d∗ s

Z

u∗s ·∗ (|y|2∗ /∗ s) ·∗ d∗ y ·∗ d∗ s

∗W (r)

= A +∗ B. We introduce the function n

ψ(y, r, s) = −∗ e 2 ·∗ log(−∗ e4π ·∗ s) +∗ (|y|2∗ /∗ (e4 ·∗ s)) +∗ en ·∗ log∗ r. We have n

(e/∗ (−∗ e4π ·∗ s) 2 ∗ ) ·∗ e(|y|

2∗ /

∗ (e

4

·∗ s))

= (e/∗ rn∗ ) on ∂∗ W (r).

Hence, ψ = 1 on ∂∗ W (r). Also, ψy∗i

n X

(yi /∗ (e2 ·∗ s)),

=

yi ·∗ ψy∗i

∗i=1 n X

(e/∗ (e2 ·∗ s)) ·∗

=

yi2∗

∗i=1

(|y|2∗ /∗ (e2 ·∗ s)),

= whereupon

(|y|2∗ /∗ s) = e2 ·∗

n X

yi ·∗ ψy∗i .

∗i=1

Therefore, B

=

(e4 /∗ r(n+1)∗ ) ·∗

Z

Z

u∗s ·∗

(e4 /∗ r(n+1)∗ ) ·∗

Z

Z ∗W (r)

=

4

(e /∗ r

(n+1)∗

Z

Z

) ·∗

yi ·∗ ψy∗i ·∗ d∗ y ·∗ d∗ s

∗i=1

∗W (r)

=

n X

u∗s ·∗

n X


(yi ·∗ ψ)∗yi −∗ ψ ·∗ d∗ y ·∗ d∗ s

∗i=1

−∗ nu∗s ·∗

ψ

∗W (r)

·∗ d∗ y ·∗ d∗ s (ψ = 1 on ∂∗ W (r))

+∗ u∗s ·∗

n X

(yi ·∗ ψ)∗yi ∗i=1

!

190 = (e4 /∗ r(n+1)∗ ) ·∗

Z

Z

n X

−∗ nu∗s ·∗ ψ −∗

The Heat Equation ! u∗∗ syi ·∗ yi ·∗ ψ

∗i=1

∗W (r)

·∗ d∗ y ·∗ d∗ s. Now, we multiplicative integrate by parts with respect to s and we get ! Z Z n X B = (e4 /∗ r(n+1)∗ ) ·∗ −∗ nu∗s ·∗ ψ +∗ u∗yi ·∗ yi ·∗ ψs ∗i=1

∗W (r)

·∗ d∗ y ·∗ d∗ s = (e4 /∗ r(n+1)∗ ) ·∗

Z

Z

−∗ nu∗s ·∗ ψ +∗

n X

u∗yi

∗i=1

∗W (r)

! n

2

2∗

4

2∗

·∗ yi −∗ (e /∗ (e ·∗ s)) −∗ (|y| /∗ (e ·∗ s )) =

·∗ d∗ y ·∗ d∗ s

−∗ (e4 /∗ r(n+1)∗ ) Z

Z

nu∗s ·∗ ψ +∗ (en /∗ (e2 ·∗ s)) ·∗

·∗

n X

! u∗yi ·∗ yi

·∗ d∗ y ·∗ d∗ s

∗i=1

∗W (r)

−∗ (e/∗ r(n+1)∗ ) ·∗

Z

Z ∗W (r)

=




n X

u∗yi ·∗ yi ·∗ (|y|2∗ /∗ s2∗ ) ·∗ d∗ y ·∗ d∗ s

∗i=1

−∗ (e4 /∗ r(n+1)∗ ) Z

Z

nu∗s ·∗

·∗

n

2

ψ +∗ (e /∗ (e ·∗ s)) ·∗

n X

! u∗yi ·∗

yi

∗i=1

∗W (r)

·∗ d∗ y ·∗ d∗ s −∗ A. Consequently, φ∗ (r)

= −∗ (e4 /∗ r(n+1)∗ ) Z

Z

·∗ ∗W (r)

νs∗ ·∗

n

2

ψ +∗ (e /∗ (e ·∗ s)) ·∗

n X ∗i=1

! u∗yi ·∗

yi

·∗ d∗ y ·∗ d∗ s

The Mean Value Formula

191

= −∗ (e4 /∗ r(n+1)∗ ) Z

Z

n

·∗

n

2

e ·∗ ∆∗ u ·∗ ψ +∗ (e /∗ (e ·∗ s)) ·∗

n X

! u∗yi ·∗

yi

∗i=1

∗W (r)

·∗ d∗ y ·∗ d∗ s (ψ = 1

on ∂∗ W (r))

= −∗ (e4 /∗ r(n+1)∗ ) Z

Z

n

·∗

−∗ e ·∗

n X

u∗yi ·∗

n

ψyi +∗ (e /∗ (e ·∗ s)) ·∗

∗i=1

∗W (r)

n X

2

! u∗yi ·∗

∗i=1

·∗ d∗ y ·∗ d∗ s = −∗ (e4 /∗ r(n+1)∗ ) ·∗

Z

Z

−∗ (en /∗ (e2 ·∗ s)) ·∗

n X

2

+∗ (e /∗ (e ·∗ s)) ·∗

u∗yi ·∗ yi

∗i=1

∗W (r)

n

n X

! u∗yi ·∗

yi

·∗ d∗ y ·∗ d∗ s

∗i=1

= 1. Thus φ is a multiplicative constant. Hence, φ(r)

=

lim φ(t)

t→1

Z

Z

(|y|2∗ /∗ s2∗ ) ·∗ d∗ y ·∗ d∗ s

= u(1, 1) ·∗ ∗W (1)

= e4 ·∗ u(1, 1), whereupon u(1, 1) = (e/∗ (e4 ·∗ rn∗ )) ·∗

Z

Z ∗W (r)

u(y, s) ·∗ (|y|2∗ /∗ s2∗ ) ·∗ d∗ y ·∗ d∗ s.

yi

192

The Heat Equation

5.5

The Maximum Principle for the Cauchy Problem

Here we consider the following Cauchy problem: u∗t −∗ ∆∗ u = 1 in

Rn∗ × (1, eT ), (5.9)

on Rn∗ × {t = 1},

u=φ where T > 0 is fixed.

Theorem 5.10 (The Maximum Principle for the Cauchy Problem) Suppose that φ ∈ C∗ (Rn∗ ) and u ∈ C∗2 (Rn∗ , C∗1 ((1, eT ])) ∩ C∗ (Rn∗ × [1, eT ]) solves the Cauchy problem (5.9) and satisfies the growth estimate u(x, t) ≤ eA ·∗ ee

a

·∗ |x|2∗∗

x ∈ Rn∗ ,

,

1 ≤ t ≤ eT ,

for some constants A, a > 0. Then, sup

u = sup φ. Rn ∗

T Rn ∗ ×[1,e ]

Proof 1. Let e4aT < e. Then there are  > 0 and γ > 0 such that 1

e4a(T +) < e and e 4(T +) = ea+γ . We fix y ∈ Rn∗ and µ > 0. Define the function n

v(x, t) = u(x, t) −∗ (eµ /∗ (eT +∗ e −∗ t) 2 ∗ ) 2∗ /

·∗ e(|x−∗ y|∗

4 T  ∗ (e ·∗ (e +∗ e −∗ t)))

x ∈ Rn∗ , 1 < t ≤ eT .

,

We have vt∗ (x, t)

=

n

n

u∗t (x, t) −∗ e 2 µ ·∗ (eT +∗ e −∗ t)−∗ 2 ∗ −∗ e 2∗ /

·∗ e(|x−∗ y|∗

∗ (e

4

·∗ (eT +∗ e −∗ t)))

µ

n

−∗ e 4 ·∗ (eT +∗ e −∗ t)−∗ ( 2 −2)∗ 2∗ /∗ (e4 ·∗ (eT +∗ e −∗ t)))

·∗ |x −∗ y|2∗∗ e(|x−∗ y|∗ vx∗i (x, t)

=

µ

,

n

u∗xi (x, t) −∗ e 2 ·∗ (eT +∗ e −∗ t)−∗ ( 2 −1)∗ 2∗

·∗ (xi −∗ yi ) ·∗ e(|x−∗ y|∗

/∗ (e4 ·∗ (eT +∗ e −∗ t)))

,

The Maximum Principle for the Cauchy Problem vx∗∗i xi (x, t)

=

193

µ 2

T  u∗∗ xi xi (x, t) −∗ e ·∗ (e +∗ e −∗ t) 2∗ /

·∗ e(|x−∗ y|∗

∗ (e

4

−∗ ( n 2 −1)∗

·∗ (eT +∗ e −∗ t)))

µ

n

−∗ e 4 ·∗ (eT +∗ e −∗ t)−∗ ( 2 +2)∗ ·∗ (xi −∗ yi )2∗ 2∗ /

·∗ e(|x−∗ y|∗ ∆∗ v(x, t)

=

∗ (e

4

·∗ (eT +∗ e −∗ t)))

,

i = 1, . . . , n,

n

∆∗ u(x, t) −∗ e 2 µ 2∗ /

n

·∗ (eT +∗ e −∗ t)−∗ 2 ∗ e(|x−∗ y|∗ µ

4 T  ∗ (e ·∗ (e +∗ e −∗ t)))

n

−∗ e 4 ·∗ (eT +∗ e −∗ t)−∗ ( 2 +2)∗ 2∗ /∗ (e4 ·∗ (eT +∗ e −∗ t)))

·∗ |x −∗ y|2∗∗ e(|x−∗ y|∗

,

whence vt∗ (x, t) −∗ ∆∗ v(x, t) = 0∗

in Rn∗ × (1, eT ].

Let D = B(y, r). Then, by Theorem 5.2, we get max v = max v. ∂∗p QT

QT

If x ∈ Rn∗ , then µ

v(x, 1)

n

2∗

u(x, 1) −∗ e (T +) 2 ·∗ e(|x−∗ y|∗

=

/∗ (e4 ·∗ (eT +∗ e )))

≤ u(x, 1) = φ(x). If |x −∗ y|∗ = r, 1 ≤ t ≤ eT , then n

v(x, t) = u(x, t) −∗ (eµ /∗ (eT +∗ e −∗ t) 2 ∗ er ≤ eA ·∗ ee ·∗ e(r

a

2∗ /

·∗ |x|2∗∗

∗ (e

4

2∗ /

∗ (e

n

4

·∗ (eT +∗ e −∗ t))

−∗ (eµ /∗ (eT +∗ e −∗ t) 2 ∗ )

·∗ (eT +∗ e −∗ t)))

194

The Heat Equation ≤ eA ·∗ ee ·∗ e(r

a

2∗ /

= eA ·∗ ee

a

·∗ (|y|∗ +∗

∗ (e

4

r)2∗

n

−∗ (eµ /∗ (eT +∗ e ) 2 ∗ )

·∗ (eT +∗ e )))

·∗ (|y|∗ +∗ r)2∗

n

−∗ (eµ /∗ (e4 ·∗ (ea +∗ eγ )) 2 ∗ ·∗ e(e

a

+∗ eγ )·∗ r 2∗

≤ sup φ Rn ∗

for r > 0∗ is sufficiently small. Consequently, v(x, t) ≤ sup φ Rn ∗

for all x ∈ Rn∗ , t ∈ [1, eT ]. Let µ → 0. Then, u(x, t) ≤ sup φ Rn ∗

for all x ∈ Rn∗ and 1 ≤ t ≤ eT . T1 T1 2T1 2. If 4aT ≥ 1, then we  apply  the above result on [1, e ], [e , e ], 1 . . . , for some T1 ∈ 0, . 4a

Exercise 5.6 Suppose that φ ∈ C∗ (Rn∗ ) and u ∈ C∗2 (Rn∗ , C∗1 ((1, eT ])) ∩ C∗ (Rn∗ × [1, eT ]) solves the Cauchy problem (5.9) and satisfies the growth estimate u(x, t) ≥ −∗ eA ·∗ ee

a

·∗ |x|2∗∗

,

x ∈ Rn∗ ,

1 ≤ t ≤ eT ,

for some constants A, a > 0. Prove that inf

T Rn ∗ ×[1,e ]

u = inf φ. n R∗

Hint 5.1 Use the function n

2∗ )/∗ (e4 ·∗ (eT +∗ e −∗ t))

v(x, t) = u(x, t) +∗ (eµ /∗ (eT +∗ e −∗ t) 2 ∗ ·∗ e(|x−∗ y|∗

,

x ∈ Rn∗ , 1 < t ≤ eT , for some µ > 0. Theorem 5.11 Let φ ∈ C∗ (Rn∗ ), f ∈ C∗ (Rn∗ × [1, eT ]). Then there exists at most one solution u ∈ C∗2 (Rn∗ , C∗1 ((1, eT ])) ∩ C∗ (Rn∗ × [1, eT ]) of the Cauchy problem: u∗t −∗ ∆∗ u = f (x, t) in Rn∗ × (1, eT ], (5.10) u = φ on Rn∗ × {t = 1},

The Method of Separation of Variables

195

satisfying the growth estimate |u(x, t)|∗ ≤ eA ·∗ ee

a

·∗ |x|2∗

x ∈ Rn∗ ,

,

1 ≤ t ≤ eT ,

(5.11)

for some constants A, a > 0. Proof Assume that u1 , u2 ∈ C∗2 (Rn∗ , C∗1 ((1, eT ])) ∩ C∗ (Rn∗ × [1, eT ]) satisfy (5.10), (5.11). Then u1 −∗ u2 satisfies u∗t −∗ ∆∗ u = 0 in u = 1 on

Rn∗ × (1, eT ], Rn∗ × {t = 1},

and (5.11). Hence and Theorem 5.10, applied for ±∗ (u1 −∗ u2 ), we get sup

|u1 −∗ u2 |∗ = 1,

T Rn ∗ ×[1,e ]

which completes the proof.

5.6

The Method of Separation of Variables

Consider the following initial boundary value problem: u∗t −∗ ek ·∗ u∗∗ xx = 1,

1 < x < eL ,

t > 1,

(5.12)

u(1, t) = 1, (5.13) u(eL , t) = 1, u(x, 1) = φ(x),

t ≥ 1, 1 ≤ x ≤ eL ,

(5.14)

where φ is a given initial condition and k > 0 is a given constant. Assume the compatibility condition φ(1) = φ(eL ) = 1. We seek solutions of the problem (5.12)–(5.14) that have the following special form: u(x, t) = X(x) ·∗ T (t). (5.15) We are not interested in the zero solution u(x, t) = 1. Therefore, we seek functions X and T that do not vanish identically. We substitute (5.15) into (5.12) and we find (T ∗ (t)/∗ (ek ·∗ T (t))) = (X ∗∗ (x)/∗ X(x)).

196

The Heat Equation

Since x and t are independent variables, from the last equality, it follows that there exists a constant λ, which is called the separation constant, such that (T ∗ (t)/∗ (ek ·∗ T (t))) = (X ∗∗ (x)/∗ X(x)) = −∗ eλ . Using the boundary condition (5.13), because u is not the multiplicative trivial solution u = 1, it follows that X(1) = X(eL ) = 1. Thus, the function X should be a solution to the boundary problem: X ∗∗ +∗ eλ ·∗ X = 1,

1 < x < eL ,

X(1) = X(eL ) = 1. As in (4.1.3), we find X(x)

=

eλ

=

sin∗ ((enπ ·∗ x)/∗ eL ), e

nπ L


2∗

,

n ∈ N.

For convenience, we use the notation Xn (x)

=

eλ n

=

sin∗ ((enπ ·∗ x)/∗ eL ), e

nπ L


2 ∗

,

n ∈ N.

Hence,  nπ 2∗ −∗ e k e L ·∗ t

Tn (t) = eBn ·∗ e

,

n ∈ N.

Thus, we obtain the following sequence of separated solutions: un (x, t)

= Xn (x) ·∗ Tn (t) = e

Bn

 nπ 2∗ −∗ ek ·∗ e L ·∗ t

·∗ e

·∗ sin∗ ((enπ ·∗ x)/∗ eL ),

n ∈ N,

where Bn , n ∈ N, are constants. The superposition principle implies that u(x, t) =

∞ X

eBn ·∗ sin∗ ((enπ ·∗ x)/∗ eL ) ·∗ e

 nπ 2∗ −∗ ek ·∗ e L ·∗ t

∗n=1

is a formal solution of the problem (5.12)–(5.13). We will find the constants Bn using the initial condition (5.14). We have u(x, 1)

= =

φ(x) ∞ X ∗n=1

eBn ·∗ sin∗ ((enπ ·∗ x)/∗ eL ).

The Method of Separation of Variables

197

Fix m ∈ N and multiply by sin∗ ((emπ ·∗ x)/∗ eL ) the last equality. Then integrating it term-by-term over [1, eL ], we find ∞ X

e

Bn

Ze

L

·∗

∗n=1

sin∗ ((enπ ·∗ x)/∗ eL ) ·∗ sin∗ ((emπ ·∗ x)/∗ eL ) ·∗ d∗ x

∗1

L

Ze =

sin∗ ((emπ ·∗ x)/∗ eL ) ·∗ φ(x) ·∗ d∗ x,

∗1

whereupon

e

Bm

Ze

2 L

L

sin∗ ((emπ ·∗ x)/∗ eL ) ·∗ φ(x) ·∗ d∗ x.

= e ·∗ ∗1

Example 5.5 Consider the initial boundary value problem: u∗t −∗ u∗∗ xx = 1,

1 < x < eπ ,

t > 1,

u(1, t) = 1, u(eπ , t) = 1,

t ≥ 1,

u(x, 1) = x2∗ ,

1 ≤ x ≤ eπ .

Here L = π, φ(x) = x2∗ , k = e. Then,

e

Bm

Ze

2 π

π

= e ·∗

sin∗ (em ·∗ x)x2∗ ·∗ d∗ x

∗1

= e

2 mπ

4

(−∗ e)(m+1)∗ +∗ e m3 π ·∗ ((−∗ e)m∗ −∗ e) .

Then, u(x, t) =

∞  X

 4 2 e n π (−∗ e)(n+1)∗ +∗ e n3 π ·∗ ((−∗ e)n∗ −∗ e)

∗n=1

·∗ sin∗ (en ·∗ x)e−∗ n

2∗ ·∗ t

is a formal solution of the considered problem.

198

The Heat Equation

Exercise 5.7 Find a formal solution to the problem: u∗t −∗ u∗∗ xx = 1,

1 < x < eπ ,

t > 1,

u(1, t) = 1, u(π, t) = 1,

t ≥ 1,

u(x, 1) = x +∗ cos∗ x,

1 ≤ x ≤ eπ .

Answer ∞  X

u(x, t) = e2 ·∗ sin∗ x ·∗ e−∗ t +∗

 2n 2 e n (−∗ e)(n+1)∗ +∗ e π(n2 −1) ((−∗ e)n +∗ e)

∗n=2

·∗ sin∗ (en ·∗ x) ·∗ e−∗ e

n2

·∗ t

.

Next, we consider the following problem: u∗t −∗ ek ·∗ u∗∗ xx = f (x, t),

1 < x < eL ,

t > 1,

(5.16)

u(1, t) = 1, u(eL , t) = 1,

t ≥ 1,

u(x, 1) = φ(x),

1 ≤ x ≤ eL ,

(5.17)

(5.18)

where f (x, t) φ(x)

= =

∞ X ∗n=1 ∞ X

fn (t) ·∗ sin∗ ((enπ ·∗ x)/∗ eL ), eAn sin∗ ((enπ ·∗ x)/∗ eL ),

∗n=1

fn are given continuous functions and An are given constants, n ∈ N. Let u(x, t) be a formal solution to the problem (5.16)–(5.18) that has the form ∞ X u(x, t) = Tn (t) ·∗ sin∗ ((enπ ·∗ x)/∗ eL .) (5.19) ∗n=1

Substituting (5.19) into (5.16), we get ∞  X

Tn∗ (t) +∗ ek ·∗ e

n2 π 2 L2

 ·∗ Tn (t) ·∗ sin∗ ((enπ ·∗ x)/∗ eL )

∗n=1

=

∞ X ∗n=1

fn (t) ·∗ sin∗ ((enπ ·∗ x)/∗ eL ),

The Method of Separation of Variables

199

whereupon Tn∗ +∗ ek ·∗ e

n2 π 2 L2

·∗ Tn = fn

and Tn (t) = e

2 2 kn π L2

−∗ e

·∗ t

·∗

e

Z

Bn

+∗

fn (t)e

e

2 2 kn π L2

! ·∗ t

dt ,

t > 1,

∗

where Bn , n ∈ N, are constants which will be determined below. We set Z 2 2 kn π 2 gn (t) = fn (t) ·∗ ee L ·∗ t ·∗ d∗ t. ∗

Therefore, u(x, t) =

∞ X

e−∗ e

2 2 kn π L2 ·∗ t


·∗ eBn +∗ gn (t) ·∗ sin∗ ((enπ ·∗ x)/∗ eL ).

∗n=1

We will find the constants Bn using the initial condition (5.18). We have u(x, 1)

∞ X

= =

∗n=1 ∞ X


eBn +∗ gn (1) ·∗ sin∗ ((enπ ·∗ x)/∗ eL ) eAn ·+ ∗ sin∗ ((enπ ·∗ x)/∗ eL ).

∗n=1

Hence, eBn = eAn −∗ gn (1) and u(x, t) =

∞ X

e −∗ e

2 2 kn π L2

·∗ t


·∗ eAn −∗ gn (0) +∗ gn (t) ·∗ sin∗ ((enπ ·∗ x)/∗ eL ).

∗n=1

Example 5.6 Consider the following problem: u∗t −∗ u∗∗ xx = t ·∗ x,

1 < x < eπ ,

u(1, t) = 1, u(eπ , t) = 1,

t ≥ 1,

u(x, 1) = x2∗ ,

1 ≤ x ≤ eπ .

Here, f (x, t) φ(x)

= t ·∗ x, = x2∗ ,

t > 1,

200

The Heat Equation k

=

1,

L = π. Note that Ze

π

x ·∗ sin∗ (en ·∗ x) ·∗ d∗ x =

π

e n ·∗ (−∗ e)(n+1)∗ ,

∗1

Ze

π

x2∗ ·∗ sin∗ (en ·∗ x) ·∗ d∗ x =

π2 n

e

2

·∗ (−∗ e)(n+1)∗ +∗ e n3 ·∗ ((−∗ e)n∗ −∗ e).

∗1

Then, f (x, t)

= t ·∗

∞ X

2

e n ·∗ (−∗ e)(n+1)∗ ·∗ sin∗ (en ·∗ x),

∗n=1 2 n

fn (t)

= e ·∗ (−∗ e)(n+1)∗ ·∗ t,

φ(x)

=

∞  X

 4 2π e n ·∗ (−∗ e)(n+1)∗ +∗ e n3 π ·∗ ((−∗ e)n∗ −∗ e)

∗n=1

·∗ sin∗ (en ·∗ x), eAn gn (t)

4

2π

= e n ·∗ (−e)(n+1)∗ +∗ e n3 π ·∗ ((−∗ e)n∗ −∗ e), 2

= e n ·∗ (−∗ e)(n+1)∗ ·∗

Z

t ·∗ ee

n2

·∗ t

·∗ d∗ t

∗

  2 1 n2 = e n3 ·∗ (−∗ e)(n+1)∗ ·∗ t −∗ e n2 ·∗ ee ·∗ t , gn (1) eBn

2

= e n5 ·∗ (−∗ e)n∗ , 4

2π

2

= e n ·∗ (−∗ e)(n+1)∗ +∗ e n3 π ·∗ ((−∗ e)n∗ −∗ e) −∗ e n5 ·∗ (−∗ e)n∗ .

Consequently, u(x, t)

=

∞ X

e −∗ e

n2

·∗ t

 2π 4 ·∗ e n ·∗ (−∗ e)(n+1)∗ +∗ e n3 π ·∗ ((−∗ e)n∗ −∗ e)

∗n=1 2

−∗ e n5 ·∗ (−∗ e)n∗    2 1 n2 +∗ e n3 ·∗ (−∗ e)(n+1)∗ ·∗ t −∗ e n2 ·∗ ee ·∗ t ·∗ sin∗ (en ·∗ x).

The Energy Method: Uniqueness

201

Exercise 5.8 Find a formal solution to the problem: 3 u∗t −∗ u∗∗ xx = sin∗ x ·∗ cos∗ (e ·∗ x),

1 < x < eπ ,

t > 1,

u(1, t) = 1, u(eπ , t) = 1,

t ≥ 1,

u(x, 1) = x,

1 ≤ x ≤ eπ .

Answer u(x, t)

∞ X

= ∗

5.7

2

e n ·∗ (−∗ e)(n+1)∗ ·∗ e−∗ e

n2

·∗ t

·∗ sin∗ (en ·∗ x)

n=1 n 6= 2, 4

+∗ e −∗ e

4

+∗ e−∗ e

16

·∗ t



·∗ t

7

1

−∗ e 8 −∗ e 8 ·∗ ee

4

·∗ t



·∗ sin∗ (e2 ·∗ x)

  16 17 1 ·∗ −∗ e 32 +∗ e 32 ·∗ ee ·∗ t ·∗ sin∗ (e4 ·∗ x).

The Energy Method: Uniqueness

Consider the following Cauchy problem: u∗t −∗ ∆∗ u = f (x, t)

in QT , (5.20)

u = φ on D × {t = 1}, where f ∈ C∗ (QT ), φ ∈ C∗ (D). Theorem 5.12 There exists at most one solution u ∈ C∗QT of the problem (5.20). Proof Suppose that u1 , u2 ∈ C∗QT are solutions of the problem (5.20). Let v = u1 −∗ u2 . Then v satisfies the problem vt∗ −∗ ∆∗ v = 1 in v = 1 on Set

Z E(t) = ∗D

QT D × {t = 1}.

(v(x, t))2∗ ·∗ d∗ x,

1 ≤ t ≤ eT .

202

The Heat Equation

Then, ∗

E (t)

2

Z

= e ·∗

v(x, t) ·∗ vt∗ (x, t) ·∗ d∗ x

∗D

= e2 ·∗

Z v(x, t) ·∗ ∆∗ v(x, t) ·∗ d∗ x ∗D

= −∗ e2 ·∗

Z

|∇v(x, t)|2∗∗ ·∗ d∗ x

∗D

≤ 1,

1 ≤ t ≤ eT .

Hence, E(t) ≤ E(1) = 1,

1 ≤ t ≤ eT .

Consequently, u1 = u2 in QT .

5.8

Advanced Practical Exercises

Problem 5.1 Find a solution to the following Cauchy problem: ∗∗ ∗∗ u∗t −∗ u∗∗ x1 x1 −∗ ux2 x2 −∗ ux3 x3 = 1 in

R3∗ × (1, ∞),

u(x1 , x2 , x3 , 1) = e2 ·∗ x21∗ −∗ e2 ·∗ x22∗ +∗ e3 ·∗ x23∗

in R3∗ .

Answer u(x1 , x2 , x3 , t) = e2 ·∗ x21∗ −∗ e2 ·∗ x22∗ +∗ e3 ·∗ x23∗ +∗ e6 ·∗ t. Problem 5.2 Find a solution to the following Cauchy problem: ∗∗ ∗∗ 2 6 2∗ u∗t −∗ u∗∗ ·∗ x3 x1 x1 −∗ ux2 x2 −∗ ux3 x3 = e ·∗ t ·∗ x1 +∗ x2 +∗ e ·∗ t

in R3∗ × (1, ∞), u(x1 , x2 , x3 , 1) = 1 in R3∗ . Answer u(x1 , x2 , x3 , t) = t2∗ ·∗ x1 +∗ t ·∗ x2 +∗ e2 ·∗ t3∗ ·∗ x3 . Problem 5.3 Find a solution to the following Cauchy problem: 2∗ 2∗ 2∗ ∗∗ ∗∗ 2 6 2∗ u∗t −∗ u∗∗ x1 x1 −∗ ux2 x2 −∗ ux3 x3 = e ·∗ t ·∗ (x1 +∗ x2 +∗ x3 ) −∗ e ·∗ t

in R3∗ × (1, ∞), u(x1 , x2 , x3 , 1) = x1

in R3∗ .

Advanced Practical Exercises

203

Answer u(x1 , x2 , x3 , t) = t2∗ ·∗ (x21∗ +∗ x22∗ +∗ x23∗ ) +∗ x1 . Problem 5.4 Find a formal solution to the following problem: u∗t −∗ u∗∗ xx = 1,

1 < x < eπ ,

t > 1,

u(1, t) = 1, u(eπ , t) = 1, ( u(x, 1) =

t ≥ 1, π

x 1 ≤ x ≤ e2 π

e π −∗ x

e 2 ≤ x ≤ eπ .

Answer 4

u(x, t) = e π ·∗

∞ X

(−1)n+1

e (2n−1)2 ·∗ sin∗ (e(2n−1) ·∗ x) ·∗ e−∗ e

(2n−1)2

·∗ t

.

∗n=1

Problem 5.5 Find a formal solution to the following problem: 2 u∗t −∗ u∗∗ xx = sin∗ (x/∗ e ),

1 < x < eπ ,

t > 1,

u(1, t) = 1, u(eπ , t) = 1,

t ≥ 1,

u(x, 1) = x,

1 ≤ x ≤ eπ .

Answer u(x, t)

=

∞ X

e −∗ e

n2

·∗ t

 2 8 (−1)n+1 +(−1)n n(4n28−1)π (−1)n−1 ·∗ e n +∗ e n(4n2 −1)π

∗n=1

·∗ ee

n2

·∗ t



·∗ sin∗ (en ·∗ x).

Problem 5.6 Find a formal solution to the following problem: u∗t −∗ ek ·∗ u∗∗ xx = 1, u∗x (1, t) u∗x (eL , t) = 1,

1 < x < eL ,

t > 1,

= 1,

t ≥ 1, u(x, 1) = φ(x),

1 ≤ x ≤ eL ,

where φ ∈ C∗2 ([1, eL ]), φ∗ (1) = φ∗ (eL ) = 1. Answer u(x, t) =

∞ X ∗n=0

eBn ·∗ e−∗ e

2 2 kn π L2 ·∗ t

·∗ cos∗ ((enπ ·∗ x)/∗ eL ),

204

The Heat Equation

where Bn

2 L

=

ZL φ(x) cos

nπx dx, L

n ∈ N,

0

B0

1 L

=

ZL φ(x)dx. 0

Problem 5.7 Find a formal solution to the following problem: u∗t −∗ ek ·∗ u∗∗ xx = 1,

1 < x < eL ,

t > 1,

u(1, t) = 1, u∗x (eL , t) = 1,

t ≥ 1,

u(x, 1) = φ(x),

1 ≤ x ≤ eL ,

where φ ∈ C∗2 ([1, eL ]), φ(1) = φ∗ (eL ) = 1. Answer u(x, t)

=

∞ X

eBn e−∗ e

k

∗n=0 ZL

Bn

=

2 L

φ(x) sin

(2n+1)2 π 2 4L2

·∗ t

·∗ sin∗ (e

(2n + 1)πx dx, 2L

(2n+1) 2πL

) ·∗ x),

n ∈ N0 .

0

Problem 5.8 Find a formal solution to the following problem: u∗t −∗ ek ·∗ u∗∗ xx = 1, u∗x (1, t) L

u(e , t) = 1,

1 < x < eL ,

t > 1,

= 1,

t ≥ 1, u(x, 1) = φ(x),

1 ≤ x ≤ eL ,

where φ ∈ C∗2 ([1, eL ]), φ∗ (1) = φ(eL ) = 1. Answer u(x, t)

=

∞ X

eBn ·∗ e−∗ e

∗n=0 ZL

Bn

=

2 L

φ(x) cos 0

k

(2n+1)2 π 2 4L2

·∗ t

cos∗ (e

(2n + 1)πx dx, 2L

(2n+1)π 2l

n ∈ N0 .

·∗ x),

6 The Laplace Equation

6.1

Basic Properties of Elliptic Problems

Let D be a domain in Rn∗ , n ≥ 2, with sufficiently multiplicative smooth boundary ∂∗ D. Consider the multiplicative Laplace1 equation, shortly Laplace equation, n X ∆∗ u = u∗∗ x = (x1 , x2 , . . . , xn ) ∈ D. (6.1) xi xi = 1, ∗i=1

Definition 6.1 A C∗2 -function u satisfying (6.1) is called a multiplicative harmonic function, shortly harmonic equation. The Laplace equation is a special case of a more general equation ∆∗ u = f (x),

x ∈ D.

(6.2)

where f is a given function. Definition 6.2 Equation (6.2) is called multiplicative Poisson equation, shortly Poisson equation. Definition 6.3 The problem defined by Poisson equation and the Dirichlet boundary condition u(x) = φ(x), x ∈ ∂D, for a given function φ is called the Dirichlet problem. Definition 6.4 The problem defined by the Poisson equation and the Neumann boundary condition ∂∗ν u(x) = φ(x),

x ∈ ∂D,

where φ is a given function, ν = (ν1 , . . . , νn ) denotes the unit outward multiplicative normal to ∂∗ D, and ∂∗ν denotes a multiplicative differentiation in the multiplicative direction of ν, i.e., ∂∗ν = ν ·∗ ∇∗ , is called the Neumann problem. 1 Pierre-Simon Laplace (23 March 1749- March 1827) was an influental French scholar whose work was important to the development of mathematics, statistcs, physics, and astronomy. He was one of the first scientists to postulate the existence of black holes and the notion of gravitational collapse.

DOI: 10.1201/9781003440116-6

205

206

The Laplace Equation

Definition 6.5 The problem defined by the Poisson equation and the Robin2 boundary condition (the boundary condition of the third kind) u(x) +∗ α(x) ·∗ ∂∗ν u(x) = φ(x),

x ∈ ∂∗ D,

where α and φ are given functions, is called the Robin problem (problem of the third kind). Theorem 6.1 A necessary condition for the existence of a solution to the Neumann problem ∆∗ u = f (x), x ∈ D, x ∈ ∂∗ D,

∂∗ν u(x) = φ(x), is

Z

Z φ(x(s)) ·∗ d∗ s =

f (y) ·∗ d∗ y,

(6.3)

∗D

∗∂∗ D

where x(s) is a multiplicative parametrization of ∂∗ D. Proof Note that ∆∗ u = ∇∗ ·∗ ∇∗ u. Therefore, we can rewrite the Poisson equation as follows: ∇∗ ·∗ ∇∗ u = f (x), which we multiplicative integrate over D and using the multiplicative Gauss theorem, we find Z Z ∇∗ u ·∗ ν ·∗ d∗ s = f (y) ·∗ d∗ y, ∗D

∗∂∗ D

i.e., equation (6.3) holds. Remark 6.1 It is useful to observe that for harmonic functions we have Z ∂∗ν u ·∗ d∗ s = 1 (6.4) ∗Γ

for any closed curve Γ that is fully contained in D. Theorem 6.2 Let u and v be harmonic functions defined in D. Then, Z
v(y) ·∗ ∂∗ν(y) u(y) −∗ u(y) ·∗ ∂∗ν(y) v(y) ·∗ d∗ sy = 1. (6.5) ∗∂∗ D 2

Victor Gustave Robin (1855–1897) was a French mathematical analyst and applied mathematician who lectured in mathematical physics at the Sorbonne in Paris and also worked in the area of thermodynamics. He is known especially for the Robin boundary condition.

Basic Properties of Elliptic Problems

207

Proof Note that X

v ·∗ u∗xi


∗

=

xi

∗i=1

n X

vx∗i ·∗ u∗xi +∗

∗i=1

=

n X

n X

v ·∗ u∗∗ xi xi

∗i=1

u∗xi ·∗ vx∗i ,

∗i=1 n X

u ·∗ vx∗i




=

xi

∗i=1

n X

u∗xi ·∗ vxi .

∗i=1

Therefore, n  X

v ·∗ u∗xi


∗ xi

−∗ u ·∗ vx∗i


∗  xi

= 1,

∗i=1

which we multiplicative integrate over D and using the Gauss theorem, we obtain the desired result (6.5). This completes the proof. Example 6.1 (The Hadamard3 Example) Consider the Laplace equation in the domain −∗ ∞ < x1 < ∞, x2 > 1, under the Cauchy conditions: un (x1 , 1) = 1, (6.6) n n un∗ x2 (x1 , 1) = sin∗ (e ·∗ x1 )/∗ e ,

−∗ ∞ < x1 < ∞,

where n ∈ N. Let 1

un (x1 , x2 ) = e n2 ·∗ sin∗ (en ·∗ x1 ) ·∗ sinh∗ (en ·∗ x2 ). Then, un∗∗ x1 x1

=

−∗ sin∗ (en ·∗ x1 ) ·∗ sinh∗ (en ·∗ x2 ),

un∗∗ x2 x2

=

sin∗ (en ·∗ x1 ) ·∗ sinh∗ (en ·∗ x2 ),

whereupon ∆∗ un (x1 , x2 ) = 1, i.e., un (x1 , x2 ) is a harmonic function satisfying (6.6). When n is large enough, the initial condition (6.6) describes an arbitrary small perturbation of the trivial solution u = 0. On the other hand, sup |un (x1 , x2 )|∗ grows multiplicative exponentially fast as n → ∞ for any x1 ∈Rn ∗

x2 > 1. Therefore, the Cauchy problem for the Laplace equation is not stable and hence it is not well-posed with respect to the initial condition (6.6). 3

Jacques Salomon Hadamard (December 8, 1865–October 17, 1963) was a French mathematician who made major contributions in number theory, complex function theory, differential geometry and partial differential equations.

208

The Laplace Equation

Definition 6.6 We define a harmonic polynomial of degree m to be a harmonic function Pm (x1 , . . . , xn ) of the form X Pm (x1 , . . . , xn ) = eai1 ...in ·∗ xi11∗ ·∗ · · · ·∗ xinn∗ , ∗0≤i1 +···+in ≤m

where ai1 ...in are constants. Example 6.2 The polynomial P1 (x1 , . . . , xn ) = x1 +∗ · · · +∗ xn is a harmonic polynomial of degree 1. Example 6.3 The polynomial P3 (x1 , x2 ) = x31∗ −∗ e3 ·∗ x1 ·∗ x22∗ −∗ x2 is a harmonic polynomial of degree 3. Definition 6.7 The harmonic polynomials X h Pm (x1 , . . . , xn ) = eai1 ...in ·∗ xi11∗ . . . xinn∗ ∗i1 +···+in =m

are called homogeneous harmonic polynomials of order m.

6.2

The Fundamental Solution

We will find a solution u of the Laplace equation (6.1) in D = Rn∗ that has the form u(x) = v(r), where

1

r = |x|∗ = (x21∗ +∗ · · · +∗ x2n∗ ) 2 ∗ . We have u∗xi

=

v ∗ (r) ·∗ (xi /∗ r),

u∗∗ xi xi

=

v ∗∗ (r) ·∗ (x2i ∗ /∗ r2∗ ) +∗ v ∗ (r) ·∗ ((r2∗ −∗ x2i ∗ )/∗ r3∗ ),

∆∗ u = = =

n X ∗i=1 n X ∗i=1 ∗∗

i = 1, . . . , n,

u∗∗ xi xi
v ∗∗ (r) ·∗ (x2i ∗ /∗ r2∗ ) +∗ v ∗ (r) ·∗ ((r2∗ −∗ x2i ∗ )/∗ r3∗ )

v (r) +∗ (en−1 /∗ r) ·∗ v ∗ (r).

Integral Representation of Harmonic Functions

209

Then, ∆∗ u = 1 if and only if v ∗∗ +∗ (en−1 /∗ r) ·∗ v ∗ = 1. 1. Let n = 2. Then, v ∗∗ +∗ (e/∗ r) ·∗ v ∗ = 1, whence v(r) = ea ·∗ log∗ r +∗ eb ,

a, b = const.

2. Let n ≥ 3. Then, v(r) = (ea /∗ r(n−2)∗ ) +∗ eb ,

a, b = const.

Definition 6.8 The function, defined by  1  −∗ e 2π ·∗ log∗ |x|∗ n = 2 Φ(x) =  (n−2)∗ e/∗ (en(n−2)κ(n) ·∗ |x|∗ ) n ≥ 3, defined for x ∈ Rn∗ , x 6= 1, is called the fundamental solution of the Laplace equation. Here κ(n) denotes the volume of the unit ball in Rn , n π2
. κ(n) = Γ n2 +∗ 1

6.3

Integral Representation of Harmonic Functions

Theorem 6.3 Let u be multiplicative harmonic in D and continuous in D ∪ ∂∗ D with its multiplicative partial derivatives of first order. Then Z u(x) = Φ(x −∗ y) ·∗ ∂∗ν∗y u(y) ·∗ d∗ sy ∗∂∗ D

(6.7) Z −∗

u(y) ·∗ ∂∗ν∗y Φ(x −∗ y) ·∗ d∗ sy ,

x ∈ D.

∗∂∗ D

Proof Let x ∈ D be arbitrarily chosen. We take  > 0 so that B(x, ) ⊂ D. Denote D = D\B(x, ). We have ∂∗ D = ∂∗ D ∪ ∂∗ B(x, ).

210

The Laplace Equation

Then, using Theorem 6.2, we have Z
0 = Φ(x −∗ y) ·∗ ∂∗ν∗y u(y) −∗ u(y) ·∗ ∂∗ν∗y Φ(x −∗ y) ·∗ d∗ sy ∗∂∗ D

Z


Φ(x −∗ y) ·∗ ∂∗ν∗y u(y) −∗ u(y) ·∗ ∂∗ν∗y Φ(x −∗ y) ·∗ d∗ sy

= ∗∂∗ D

Z


Φ(x −∗ y) ·∗ ∂∗ν∗y u(y) −∗ u(y) ·∗ ∂∗ν∗y Φ(x −∗ y) ·∗ d∗ sy .

−∗ ∗∂∗ B(x,)

Hence, Z


Φ(x −∗ y) ·∗ ∂∗ν∗y u(y) −∗ u(y) ·∗ ∂∗ν∗y Φ(x −∗ y) ·∗ d∗ sy

∗∂∗ D

Z


Φ(x −∗ y) ·∗ ∂∗ν∗y u(y) −∗ u(y) ·∗ ∂∗ν∗y Φ(x −∗ y) ·∗ d∗ sy

= ∗∂∗ B(x,)

Z Φ(x −∗ y) ·∗ ∂∗ν∗y u(y) ·∗ d∗ sy

= ∗∂∗ B(x,)

Z −∗

(u(y) −∗ u(x)) ·∗ ∂∗ν∗y Φ(x −∗ y) ·∗ d∗ sy

∗∂∗ B(x,)

Z −∗

u(x) ·∗ ∂∗ν∗y Φ(x −∗ y) ·∗ d∗ sy .

∗∂∗ B(x,)

(6.8) Note that Φ(x −∗ y)

=

 1  −∗ e 2π ·∗ log∗ |x −∗ y|∗ 

∂∗ν∗y Φ(x −∗ y)

= ∂∗ B(x,)

n=2 (n−2)∗

(e/∗ (en(n−2)κ(n) ·∗ |x −∗ y|∗

 1   −∗ e 2π

n=2

1   − e nκ(n) n−1 ∗

n ≥ 3.

)) n ≥ 3,

Integral Representation of Harmonic Functions

211

Therefore, Z (u(y) −∗ u(x)) ·∗ ∂ν∗y Φ(x −∗ y) ·∗ d∗ sy ∗∂∗ B(x,)

=

Z  1  2π −∗ e ·∗ (u(y) −∗ u(x)) ·∗ d∗ sy       ∗∂∗ B(x,)  1   nκ(n)n−1 ·  − e  ∗ ∗  

n=2

Z (u(y) −∗ u(x)) ·∗ d∗ sy

n≥3 (6.9)

∗∂∗ B(x,)

Z

=

 1  −∗ e 2π ·∗ (u(x +∗  ·∗ z) −∗ u(x)) ·∗ d∗ sz       ∗∂∗ B(0,1)   1   −∗ e nκ(n) ·∗   

→1

as

n=2

Z (u(x +∗  ·∗ z) −∗ u(x)) ·∗ d∗ sz

n≥3

∗∂∗ B(0,1)

 → 1, Z u(x) ·∗ ∂∗ν∗y Φ(x −∗ y) ·∗ d∗ sy

∗∂∗ B(x,)

=

Z  1  2π · u(x) · − e d∗ sy  ∗ ∗ ∗      ∗∂∗ B(x,)  1    −∗ e nκ(n)n−1 ·∗ u(x) ·∗   

=

  −∗ u(x) n = 2 

−∗ u(x) n ≥ 3,

n=2 (6.10)

Z d∗ sy ∗∂∗ B(x,)

n≥3

212

The Laplace Equation Z Φ(x −∗ y) ·∗ ∂∗ν∗y u(y) ·∗ d∗ sy ∗∂∗ B(x,)

∗

Z

≤

  eC ·∗ | log∗ e |∗ ·∗ d∗ sy       ∗∂∗ B(x,)   C   e n−2 ·∗   

≤

eC1 

→1

(6.11)

Z d∗ sy

n≥3

∗∂∗ B(x,)

 C  e 1 ·∗ | log∗ e |∗ 

n=2

n=2

n≥3

as

 → 1.

Here C1 and C are constants independent of . From (6.8)–(6.11), we obtain the desired result (6.7). This completes the proof.

6.4

Mean Value Formulas

Theorem 6.4 If u ∈ C∗2 (D) is multiplicative harmonic, then Z u(x) = (e/∗ (enκ(n) ·∗ r(n−1)∗ )) ·∗ u(y) ·∗ d∗ sy ∗∂∗ B(x,r)

(6.12) = (e/∗ (e

κ(n)

Z

n∗

·∗ r )) ·∗

u(y) ·∗ d∗ y ∗B(x,r)

for each ball B(x, r) ⊂ D. Proof Note that Φ(x −∗ y) ·∗

∂∗ν∗y Φ(x −∗ y)

= ∂∗ B(x,r)

∗∂∗ B(x,r)

 1  −∗ e 2π ·∗ log∗ r 

=

n=2

(e/∗ (en(n−2)κ(n) ·∗ r(n−2)∗ ))

n ≥ 3,

 2π  −∗ (e/∗ (e ·∗ r)) n = 2 

−∗ (e/∗ (enκ(n) ·∗ r(n−1)∗ )) n ≥ 3.

Mean Value Formulas

213

Hence from (6.7), we get Z  1  2π · log r ·  − e ∂∗ν∗y u(y) ·∗ d∗ sy +∗ (e/∗ (e2π ·∗ r)) ∗ ∗ ∗ ∗      ∗∂∗ B(x,r)      Z     · u(y) ·∗ d∗ sy n = 2  ∗     ∗∂∗ B(x,r)      u(x) = (e/∗ (en(n−2)κ(n) ·∗ r(n−2)∗ ))     Z     · ∂∗ν∗y u(y) ·∗ d∗ sy +∗ (e/∗ (enκ(n) ·∗ r(n−1)∗ ))  ∗     ∗∂∗ B(x,r)      Z      · u(y) ·∗ d∗ sy n ≥ 3.  ∗   ∗∂∗ B(x,r)

Hence from (6.4), we obtain (6.12). This completes the proof. Definition 6.9 The formulas in (6.12) are known as multiplicative mean value formulas for multiplicative harmonic functions, for a multiplicative sphere and for a multiplicative ball, respectively. Theorem 6.5 (Converse to Multiplicative Mean Value Formula) If u ∈ C∗2 (D) satisfies (6.12) for each ball B(x, r) ⊂ D, then u is multiplicative harmonic in D. Proof Suppose that ∆∗ u 6≡ 1 in D. Then there exists some multiplicative ball B(x, r) ⊂ D such that, without loss of generality, ∆∗ u > 1 within B(x, r). By (6.12), we have Z u(x) = (e/∗ (enκ(n) ·∗ r(n−1)∗ )) ·∗ u(y) ·∗ d∗ sy ∗∂∗ B(x,r)

Z

1

= e nκ(n) ·∗

u(x +∗ r ·∗ z) ·∗ d∗ sz , ∗∂B(0,1)

whereupon 1

= e

1 nκ(n)

Z ·∗

∇∗ u(x +∗ r ·∗ z) ·∗ z ·∗ d∗ sz ∗∂∗ B(0,1)

=

(e/∗ (e

nκ(n)

·∗ r

(n−1)∗

Z )) ·∗

∇∗ u(y) ·∗ ((y −∗ x)/∗ r) ·∗ d∗ sy ∗∂∗ B(x,r)

214

The Laplace Equation

= (e/∗ (e

nκ(n)

·∗ r

(n−1)∗

Z )) ·∗

∆∗ u(y) ·∗ d∗ y ∗B(x,r)

>

1,

which is a contradiction. This completes the proof.

6.5

Strong Maximum Principle: Uniqueness

Theorem 6.6 Suppose u ∈ C∗2 (D) ∩ C∗ (D) is multiplicative harmonic in D. 1. If there exists a point x0 ∈ D such that u(x0 ) = max u, D

then u = const

within

D.

2. We have max u = max u. D

∗∂∗ D

(6.13)

Proof 1. Let eM = max u. D

Suppose u(x0 ) = eM for some x0 ∈ D. Let 1 < e < edist(x0 ,∂∗ D) be arbitrarily chosen. Assume that there exists y ∈ B(x0 , ) so that u(y) < eM . Since u ∈ C∗ (D), there is a δ > 0 so that B(y, δ) ⊂ B(x0 , ) and u(ξ) < eM for all ξ ∈ B(y, δ). Also, using (6.12), we have Z 1 eM = u(x0 ) = e κ(n)n ·∗ u(y) ·∗ d∗ y < eM , ∗B(x0 ,)

which is a contradiction. Therefore, u(y) = eM for any y ∈ B(x0 , ). Let x ∈ D be arbitrarily chosen and l be a continuous curve lying within D and joining the points x and x0 . We take 1 < e1 < edist(l,∂∗ D) . Note that if y ∈ l, we have u(η) = eM for any η ∈ B(y, 1 ). Therefore, u(x) = eM . Because x ∈ D was arbitrarily chosen, we conclude that u(x) = eM for all x ∈ D.

The Poisson Equation

215

2. If there is x0 ∈ D so that u(x0 ) = max u(x), then u(x) = u(x0 ) for D

any x ∈ D, whereupon we get (6.13). This completes the proof. Theorem 6.7 The Dirichlet problem ∆∗ u = f

in

D

u=φ

on

∂∗ D,

(6.14)

where f ∈ C∗ (D), φ ∈ C∗ (∂∗ D) has at most one solution in C∗2 (D) ∩ C∗ (D). Proof Let u1 , u2 ∈ C∗2 (D) ∩ C∗ (D) be two solutions of the Dirichlet problem (6.14). Then, v = u1 −∗ u2 ∈ C∗2 (D) ∩ C∗ (D) satisfies the Dirichlet problem ∆∗ v = 1 in v = 1 on

D ∂∗ D.

Hence from Theorem 6.6, we conclude that v = 1 in D, i.e., u1 = u2 in D. This completes the proof.

6.6

The Poisson Equation

Theorem 6.8 Let f ∈ C∗2 (Rn∗ ) be with compact support. Let also Z u(x) = −∗ Φ(x −∗ y) ·∗ f (y) ·∗ d∗ y, x ∈ Rn∗ . ∗Rn ∗

Then, 1. u ∈ C∗2 (Rn∗ ), 2. ∆∗ u = f

in

Rn∗ .

Proof 1. We have

Z u(x) = −∗

Φ(y) ·∗ f (x −∗ y) ·∗ d∗ y.

∗Rn ∗

If ej = (1, . . . , 1, e, 1, . . . , 1), j = 1, . . . , n, the e in the jth slot, then ((u(x +∗ h ·∗ ej ) −∗ u(x))/∗ h Z = −∗ ∗Rn ∗

Φ(y)((f (x +∗ h ·∗ ej −∗ y) −∗ f (x −∗ y))/∗ h) ·∗ d∗ y.

216

The Laplace Equation Because lim (f (x +∗ h ·∗ ej −∗ y) −∗ f (x−y ))/∗ h) = fx∗j ·∗ (x −∗ y)

h→1

uniformly on Rn∗ , we get u∗xj (x) = −∗

Z

Φ(y) ·∗ fx∗j (x −∗ y) ·∗ d∗ y.

(6.15)

∗Rn ∗

As in above, u∗∗ xi xj (x) = −∗

Z

Φ(y) ·∗ fx∗∗i xj (x −∗ y) ·∗ d∗ y.

(6.16)

∗Rn ∗

As the expressions on the right-hand side of (6.15) and (6.16) are continuous in the variable x, we conclude that u ∈ C∗2 (Rn∗ ). 2. Let x ∈ Rn∗ and  > 0 be arbitrarily chosen and fixed. Then, Z ∆∗ u(x) = −∗ Φ(y) ·∗ ∆∗x f (x −∗ y) ·∗ d∗ y Rn ∗

Z = −∗

Φ(y) ·∗ ∆∗x f (x −∗ y) ·∗ d∗ y

(6.17)

∗B(1,e )

Z −∗

Φ(y) ·∗ ∆∗x f (x −∗ y) ·∗ d∗ y.

 ∗Rn ∗ \B(1,e )

We have Z Φ(y) ·∗ ∆∗x f (x −∗ y) ·∗ d∗ y ∗B(1,e )

∗

Z ≤

|Φ(y)|∗ ·∗ |∆∗x f (x −∗ y)|∗ ·∗ d∗ y ∗B(1,e )

Z ≤ sup |∆∗ f |∗ ·∗ Rn ∗

≤

|Φ(y)|∗ ·∗ d∗ y ∗B(1,e )

 C2 ·∗ | log∗ e |∗  e 

eC

2

→ 1 as

n≥3  → 0.

n=2

(6.18)

The Poisson Equation

217

Here C is a constant independent of . Also, Z −∗ Φ(y) ·∗ ∆∗x f (x −∗ y) ·∗ d∗ y  ∗Rn ∗ \B(1,e )

Z = −∗

Φ(y) ·∗ ∆∗y f (x −∗ y) ·∗ d∗ y

 ∗Rn ∗ \B(1,e )

(6.19)

Z ∇∗ Φ(y) ·∗ ∇∗y f (x −∗ y) ·∗ d∗ y

=  ∗Rn ∗ \B(1,e )

Z −∗

Φ(y) ·∗ fν∗y (x −∗ y) ·∗ d∗ sy ,

∗∂∗

B(1,e )

Z Φ(y) ·∗ fν∗y (x −∗ y) ·∗ d∗ sy ∗∂∗ B(1,e ) ∗ Z ≤ eC ·∗ |Φ(y)|∗ ·∗ d∗ sy ∗∂∗ B(1,e )

( ≤

eC ·∗ | log∗ e |∗ e

C

n=2

n≥3

→ 1 as

 → 0,

where C is a constant independent of . Note that 1

∇∗ Φ(y)

=

−∗ e nκ(n) ·∗ (y/∗ (|y|n∗ ∗ )),

Φν∗y (y)

=

e nκ(n)n−1

1

on ∂∗ B(1, e ).

Therefore, Z ∇∗ Φ(y) ·∗ ∇∗y f (x −∗ y) ·∗ d∗ y  ∗Rn ∗ \B(1,e )

Z =

−∗

∆∗ Φ(y) ·∗ f (x −∗ y) ·∗ d∗ y

 ∗Rn ∗ \B(1,e )

Z Φν∗y (y) ·∗ f (x −∗ y) ·∗ d∗ sy

+∗ ∗∂∗ B(1,e )

Z Φν∗y (y) ·∗ f (x −∗ y) ·∗ d∗ sy

= ∗∂∗ B(1,e )

(6.20)

218

The Laplace Equation Z

1

=

e nκ(n)n−1 ·∗

f (x −∗ y) ·∗ d∗ sy ∗∂∗

=

B(1,e )

Z

1

e nκ(n) ·∗

f (x −∗ z) ·∗ d∗ sz ∗∂∗ B(1,e)

→ f (x)

as

 → 0.

From here and (6.19) and (6.20), we obtain Z −∗ Φ(y) ·∗ ∆∗x f (x −∗ y) ·∗ d∗ y → f (x) as

 → 0.

 ∗Rn ∗ \B(1,e )

Hence from (6.17) and (6.18), letting  → 0, we get that ∆∗ u(x) = f (x), which completes the proof.

6.7

The Green Function of the Dirichlet Problem

Definition 6.10 The Green4 function of the Dirichlet problem for the Laplace equation in a domain D is a function G(x, y) depending on two points x, y ∈ D which possesses the following properties: 1. G(x, y) has the form G(x, y) = Φ(x −∗ y) +∗ g(x, y), where g(x, y) is a multiplicative harmonic function with respect to x, y ∈ D. 2. When x ∈ ∂∗ D or y ∈ ∂∗ D, the equality G(x, y) = 1 is fulfilled.

Theorem 6.9 We have G(x, y) ≥ 1

in

D.

4 George Green (July 14, 1793–May 31, 1841) was a British mathematical physicist who introduced several important concepts, among them a theorem similar to the modern Green theorem, the idea of potential functions and the concept of what are now called Green functions.

The Green Function of the Dirichlet Problem

219

Proof Let y ∈ D be arbitrarily chosen. Let also δ > 0 be sufficiently small so that B(y, δ) ⊂ D. Denote Dδ = D\B(y, eδ ). Since lim G(x, y) = ∞, then x→y

we must have, for sufficiently small δ > 0 that G(x, y) > 1 when x ∈ B(y, eδ ). Therefore, G(x, y) ≥ 1 on ∂∗ Dδ . Hence from the maximum principle, we conclude that G(x, y) ≥ 1 for all x ∈ Dδ . Consequently, G(x, y) ≥ 1 in D. This completes the proof. Theorem 6.10 (Symmetry Property of the Green Function) We have G(x, y) = G(y, x)

for

any

x, y ∈ D.

Proof Let x, y ∈ D, x 6= y, be arbitrarily chosen and fixed. Take  > 0 small enough so that B(x, e ) ⊂ D, B(y, e ) ⊂ D and B(x, e ) ∩ B(y, e ) = Ø. Denote D = D\ (B(x, e ) ∪ B(y, e )) . Note that G(z, y) is multiplicative harmonic in D\B(y, e ) and G(z, x) is multiplicative harmonic in D\B(x, e ). Applying (6.5) to the domain D for G(z, x) and G(z, y), we get Z 1 = (G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x) ·∗ Gν∗z (z, y)) ·∗ d∗ sz ∗∂∗ D

Z (G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x) ·∗ Gν∗z (z, y)) ·∗ d∗ sz

= ∗∂∗ D

Z −∗

(G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x) ·∗ Gν∗z (z, y)) ·∗ d∗ sz

∗∂∗ B(y,e )

Z −∗

(G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x) ·∗ Gν∗z (z, y)) ·∗ d∗ sz ,

∗∂∗ B(x,e )

whereupon Z (G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x) ·∗ Gν∗z (z, y)) ·∗ d∗ sz ∗∂∗ D

Z (G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x)Gν∗z (z, y)) ·∗ d∗ sz

= ∗∂∗ B(y,e )

Z +∗

(G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x) ·∗ Gν∗z (z, y)) ·∗ d∗ sz .

∗∂∗ B(x,e )

Because G(z, y) = G(z, x) = 1 for z ∈ ∂D,

220

The Laplace Equation

we get Z (G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x) ·∗ Gν∗z (z, y)) ·∗ d∗ sz ∗∂∗

B(x,e )

Z (−∗ G(z, y) ·∗ Gν∗z (z, x) +∗ G(z, x) ·∗ Gν∗z (z, y)) ·∗ d∗ sz .

= ∗∂∗

B(y,e )

(6.21) Note that Z (G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x) ·∗ Gν∗z (z, y)) ·∗ d∗ sz ∗∂∗ B(x,e )

Z

    Φ(z −∗ y) +∗ g(z, y) ·∗ Φν∗z (z −∗ x) +∗ gν∗z (z, x)

= ∗∂∗ B(x,e )

    −∗ Φ(z −∗ x) +∗ g(z, x) ·∗ Φν∗z (z −∗ y) +∗ gν∗z (z, y) ·∗ d∗ sz Z (Φ(z −∗ y) ·∗ Φν∗z (z −∗ x) −∗ Φ(z −∗ x)Φν∗z (z −∗ y)) ·∗ d∗ sz

= ∗∂∗

B(x,e )

Z (g(z, y) ·∗ Φν∗z (z −∗ x) −∗ gν∗z (z, y) ·∗ Φ(z −∗ x)) ·∗ d∗ sz

+∗ ∗∂∗

B(x,e )

Z (Φ(z −∗ y) ·∗ gν∗z (z, x) −∗ g(z, x) ·∗ Φν∗z (z −∗ y)) ·∗ d∗ sz

+∗ ∗∂∗

B(x,e )

Z +∗

(g(z, y) ·∗ gν∗z (z, x) −∗ g(z, x) ·∗ gν∗z (z, y)) ·∗ d∗ sz .

∗∂∗ B(x,e )

(6.22) Since g(z, y) and g(z, x) are multiplicative harmonic with respect to z in B(x, ), using Theorem 6.2, we get Z (g(z, y) ·∗ gν∗z (z, x) −∗ g(z, x) ·∗ gν∗z (z, y)) ·∗ d∗ sz = 1. (6.23) ∗∂∗ B(x,e )

The Green Function of the Dirichlet Problem

221

Because Φ(z −∗ y) and g(z, x) are multiplicative harmonic with respect to z in B(x, e ), using Theorem 6.2, we obtain Z (Φ(z −∗ y) ·∗ gν∗z (z, x) −∗ g(z, x) ·∗ Φν∗z (z −∗ y)) ·∗ d∗ sz = . ∗∂∗ B(x,e )

(6.24) Since Φ(z −∗ y) is multiplicative harmonic with respect to z in B(x, ), using Theorem 6.3, we have Z (Φ(z −∗ y) ·∗ Φν∗z (z −∗ x) −∗ Φ(z −∗ x) ·∗ Φν∗z (z −∗ y)) (6.25)

∗∂∗ B(x,e )

·∗ d∗ sz = Φ(x −∗ y). Because g(z, y) is multiplicative harmonic with respect to z in B(x, ), using Theorem 6.3, we obtain Z (g(z, y) ·∗ Φν∗z (z −∗ x) −∗ gν∗z (z, y) ·∗ Φ(z −∗ x)) ·∗ d∗ sz = g(x, y). ∗∂∗ B(x,e )

(6.26) From (6.22)–(6.26), we obtain Z (G(z, y) ·∗ Gν∗z (z, x) −∗ G(z, x) ·∗ Gν∗z (z, y)) (6.27)

∗∂∗ B(x,e )

·∗ d∗ sz → Φ(x −∗ y) +∗ g(x, y) = G(x, y) as  → 0. Similarly, Z (G(z, x) ·∗ Gν∗z (z, y) −∗ G(z, y)Gν∗z (z, x)) ·∗ d∗ sz → G(y, x) ∗∂∗ B(y,e )

(6.28) as  → 0. From (6.21), (6.27) and (6.28), letting  → 0, we get G(x, y) = G(y, x). This completes the proof. Now we suppose that u is multiplicative harmonic in D and satisfies the boundary condition u = φ on ∂∗ D. (6.29)

222

The Laplace Equation

Then, by Theorem 6.3, we get Z u(x) = Φ(x −∗ y) ·∗ ∂ν∗y u(y) ·∗ d∗ sy ∗∂∗ D

Z −∗

u(y) ·∗ ∂∗ν∗y ·∗ Φ(x −∗ y) ·∗ d∗ sy . ∗∂∗ D

Hence, using Theorem 6.2, Φ(x −∗ y) = G(x, y) −∗ g(x, y) and the boundary condition (6.29), we obtain Z u(x) = (G(x, y) −∗ g(x, y)) ·∗ ∂∗ν∗y u(y) ·∗ d∗ sy ∗∂∗ D

Z


u(y) ·∗ Gν∗y (x, y) −∗ gν∗y (x, y) ·∗ d∗ sy

−∗ ∗∂∗ D

Z G(x, y) ·∗ ∂∗ν∗y u(y) ·∗ d∗ sy

= ∗∂∗ D

Z −∗

u(y) ·∗ Gν∗y (x, y) ·∗ d∗ sy

∗∂∗ D

Z



+∗

 u(y) ·∗ gν∗y (x, y) −∗ g(x, y) ·∗ u∗ν∗y (y) ·∗ d∗ sy

∗∂∗ D

Z = −∗

u(y) ·∗ Gν∗y (x, y) ·∗ d∗ sy

∗∂∗ D

Z = −∗

φ(y) ·∗ Gν∗y (x, y) ·∗ d∗ sy ,

∗∂∗ D

i.e., Z u(x) = −∗

φ(y) ·∗ Gν∗y (x, y) ·∗ d∗ sy ,

x ∈ D.

(6.30)

∗∂∗ D

The multiplicative harmonicity of the function u expressed by the formula (6.30) follows from the fact that the Green function G(x, y) is multiplicative harmonic with respect to x for x 6= y. The fact that this function satisfies the boundary condition (6.29) requires special proof.

The Poisson Formula for a Multiplicative Ball

6.8

223

The Poisson Formula for a Multiplicative Ball

Let D be the multiplicative ball B(0∗ , e) and let x, y be two interior points of that multiplicative ball. Theorem 6.11 The function G(x, y) = Φ(x −∗ y) −∗ Φ (|x|∗ ·∗ y −∗ (x/∗ |x|∗ )) is the Green function for the multiplicative ball D. Proof Here g(x, y) = −∗ Φ (|x|∗ ·∗ y −∗ (x/∗ |x|∗ )) . Note that if y = x/∗ (|x|2∗ ), then |y|∗

=

(e/∗ |x|∗

>

e,

i.e., |x|∗ ·∗ y −∗ (x/∗ |x|∗ 6= 1) for any x, y ∈ B(0∗ , e). Therefore, g(x, y) is multiplicative harmonic with respect to x and y in B(1, e). If |y|∗ = e, then |y −∗ x|∗

=


1 |x|2∗∗ −∗ e2 ·∗ x ·∗ y +∗ e 2 ∗

=

||y|∗ ·∗ x −∗ (y/∗ |y|∗ )|∗

=

||x|∗ ·∗ y −∗ (x/∗ |x|∗ )|∗ ,

i.e., Φ(x −∗ y) = Φ (|x|∗ ·∗ y −∗ (x/∗ |x|∗ )) and G(x, y) = 1. Similarly, if |x|∗ = e, then G(x, y) = 1, which completes the proof. For |y|∗ = e, we have 1

Gνy∗ (x, y) = −∗ e nκ(n) ·∗

n X

(((yi ·∗ (yi −∗ xi ))/∗ |y −∗ x|n∗ ∗ )

∗i=1 n

−∗ |x|∗ ((yi ·∗ (|x|∗ ·∗ yi −∗ (xi /∗ |x|∗ )))/∗ ||x|∗ ·∗ y −∗ (x/∗ |x|∗ )|∗ ∗ ))

224

The Laplace Equation = −∗ e ·∗

1 nκ(n)

n X

(((yi ·∗ (yi −∗ xi ))/∗ |y −∗ x|n∗ ∗ ) −∗ |x|∗

∗i=1

·∗ ((yi ·∗ (|x|∗ yi −∗ (xi /∗ |x|∗ )))/∗ |y −∗ x|n∗ ∗ )) 1

= −∗ e nκ(n) ·∗

n X

((yi2∗ −∗ xi ·∗ yi −∗ yi2∗

∗i=1

·∗ |x|2∗ +∗ xi ·∗ yi )/∗ |y −∗ x|n∗ ∗ ) 1

= −∗ e nκ(n) ·∗

n X

((yi2∗ ·∗ (e −∗ |x|2∗∗ ))/∗ |y −∗ x|n∗ ∗ )

∗i=1 1

= −∗ e nκ(n) ·∗ ((|y|2∗∗ ·∗ (e −∗ |x|2∗∗ ))/∗ |y −∗ x|n∗ ∗ ) 1

=e nκ(n) ·∗ ((|x|2∗∗ −∗ e)/∗ |y −∗ x|n∗ ∗ ). Then, applying (6.30), we obtain Z 1 nκ(n) u(x) = e ·∗ ((e −∗ |x|2∗∗ )/∗ |y −∗ x|n∗ ∗ ) ·∗ φ(y) ·∗ d∗ sy ,

(6.31)

∗|y|∗ =e

where φ ∈ C∗ (∂∗ B(1, e)). Definition 6.11 Equation (6.31) is known as Poisson’s formula. The Poisson formula for the multiplicative ball B(x0 , er ) is Z 1 2 u(x) = e rnκ(n) ·∗ ((er −∗ |x −∗ x0 |2∗∗ )/∗ |y −∗ x|n∗ ∗ ) ·∗ φ(y) ·∗ d∗ sy , ∗|y−∗ x0 |∗ =er

where φ ∈ C∗ (∂∗ B(x0 , er )). Exercise 6.1 Prove that Z 1 e nκ(n) ·∗ ((e −∗ |x|2∗∗ )/∗ |y −∗ x|n∗ ∗ ) ·∗ d∗ sy = e

for x ∈ B(1, e).

∗∂∗ B(0,1)

Let x0 ∈ ∂B(1, e) be arbitrarily chosen and fixed. Now, we will prove that lim u(x) = φ(x0 ), x → x0 x ∈ B(1, e)

The Poisson Formula for a Multiplicative Ball

225

where u(x) is defined by (6.31). We have Z 1 nκ(n) u(x) −∗ φ(x0 ) =e ·∗ ((e −∗ |x|2∗∗ )/∗ |y −∗ x|n∗ ∗ ) ·∗ (φ(y) −∗ φ(x0 )) ∗|y|∗ =e

·∗ d∗ sy . Let  > 0 be arbitrarily chosen. Then there exists δ = δ() > 0 so that |φ(y) −∗ φ(x0 )|∗ < e

for y ∈ B(x0 , eδ ) ∩ ∂∗ B(1, e).

Set
S = ∂∗ B(1, e)\ B(x0 , eδ ) ∩ ∂∗ B(1, e) . We choose δ > 0 small enough so that e −∗ |x|2∗∗ ≤ e

nκ(n) 4M Q

for x ∈ B(1, e),

δ

|x −∗ x0 |∗ ≤ e 2 ,

where Z Q =

(e/∗ |y −∗ x0 |n∗ ∗ ) ·∗ d∗ sy ,

∗S

M

sup |φ|∗ .

=

∂∗ B(1,e)

Then, for δ

|x −∗ x0 |∗ ≤ e 2 , x ∈ B(1, e), we have |u(x) −∗ φ(x0 )|∗ ≤e

1 nκ(n)

Z

((e −∗ |x|2∗∗ )/∗ |y −∗ x|n∗ ∗ ) ·∗ |φ(y) −∗ φ(x0 )|∗ ·∗ d∗ sy

·∗ ∗∂∗ B(1,e)

=e

1 nκ(n)

Z ·∗

((e −∗ |x|2∗∗ )/∗ |y −∗ x|n∗ ∗ )

∗B(x0 ,eδ )∩∂∗ B(1,e)

·∗ |φ(y) −∗ φ(x0 )|∗ ·∗ d∗ sy +∗ e

1 nκ(n)

Z ·∗

((e −∗ |x|2∗∗ )/∗ |y −∗ x|n∗ ∗ ) ·∗ |φ(y) −∗ φ(x0 )|∗ d∗ sy

∗S

= I1 +∗ I2 . (6.32)

226

The Laplace Equation

Note that Z

1

I1 ≤ e nκ(n) ·∗

((e −∗ |x|2∗∗ )/∗ |x −∗ y|n∗ ∗ ) ·∗ d∗ sy

∗B(x0 ,eδ )∩∂∗ B(1,e)

Z

1

≤ e nκ(n) ·∗

((e −∗ |x|2∗∗ )/∗ |x −∗ y|n∗ ∗ ) ·∗ d∗ sy

∗∂∗ B(1,e)

= e

δ

for |x −∗ x0 |∗ ≤ e 2 ,

For

x ∈ B(1, e). δ

|x −∗ x0 |∗ ≤ e 2 , x ∈ B(1, e), and |y −∗ x0 |∗ ≥ eδ , y ∈ ∂∗ B(1, e), we have |y −∗ x0 |∗

≤

|y −∗ x|∗ +∗ |x −∗ x0 |∗

≤

|y −∗ x|∗ +∗ e 2

≤

|y −∗ x|∗ +∗ e 2 ·∗ |y −∗ x0 |∗ ,

δ

1

i.e., if δ

|x −∗ x0 |∗ ≤ e 2 , x ∈ B(1, e), and |y −∗ x0 |∗ ≥ eδ , y ∈ ∂B(1, e), we have 1

|y −∗ x|∗ ≥ e 2 ·∗ |y −∗ x0 |∗ . Hence, for δ

|x −∗ x0 |∗ ≤ e 2 , x ∈ B(1, e), we have I2

Z

2M

≤ e nκ(n) ·∗

((e −∗ |x|2∗∗ )/∗ |x −∗ y|n∗ ∗ ) ·∗ d∗ sy

∗S

≤ e

4M nκ(n)

Z ·∗

((e −∗ |x|2∗∗ )/∗ |y −∗ x0 |n∗ ∗ ) ·∗ d∗ sy

∗S

≤ e . From the last estimate and (6.32) and (6.33), we get |u(x) −∗ φ(x0 )|∗ ≤ e2 δ

for |x −∗ x0 |∗ ≤ e 2 , x ∈ B(1, e).

(6.33)

Theorems of Liouville and Harnack

227

Exercise 6.2 Let D = {x ∈ Rn∗ : xn > 1}. 1. Prove that

G(x, y) = Φ(x −∗ y) −∗ Φ(x −∗ y 0 ),

where x, y ∈ D, y = (y1 , . . . , yn−1 , yn ), y 0 = (y1 , . . . , yn−1 , −∗ yn ) is the Green function. 2. Let φ ∈ C∗ (∂∗ D) ∩ L∞ ∗ (∂∗ D). Prove that lim ((e2 ·∗ xn )/∗ (enκ(n) )) x → x0 x∈D Z ·∗

(φ(y)/∗ |x −∗ y|n∗ ∗ ) ·∗ d∗ sy = φ(x0 ),

x0 ∈ ∂D.

∗∂∗ D

6.9

Theorems of Liouville and Harnack

Theorem 6.12 Let u be multiplicative harmonic throughout the space Rn∗ and u(x) ≥ (≤)1, x ∈ Rn∗ . Then u is identically equal to a constant in Rn∗ . Proof Suppose that u(x) ≥ 1, x ∈ Rn∗ . Let R > 0 be arbitrarily chosen. Then by (6.31), we have Z 1 2 u(x) = e nκ(n)R ·∗ ((eR −∗ |x|2∗∗ )/∗ |y −∗ x|n∗ ∗ ) ·∗ φ(y) ·∗ d∗ sy , |x|∗ < eR , ∗|y|∗ =eR

(6.34) for φ ∈ C∗ (∂∗ B(1, eR )), and u(x) → φ(x0 ) as x → x0 , x ∈ B(1, eR ), x0 ∈ ∂∗ B(1, eR ). Hence, Z 1 2 nκ(n)R u(1) = e ·∗ (eR /∗ |y|n∗ ∗ ) ·∗ φ(y) ·∗ d∗ sy ∗|y|∗ =eR

= e

1 nκ(n)Rn−1

Z ·∗

φ(y) ·∗ d∗ sy . ∗|y|∗

=eR

Since eR −∗ |x|∗

≤

|y|∗ −∗ |x|∗

≤

|y −∗ x|∗

≤

|y|∗ +∗ |x|∗

≤ eR +∗ |x|∗

228

The Laplace Equation

for |y|∗ = eR , |x|∗ < eR , we get Z 1 2 u(x) ≥ e nκ(n)R ·∗ ((eR −∗ |x|2∗∗ )/∗ (eR +∗ |x|∗ )n∗ ) ·∗ φ(y) ·∗ d∗ sy |y|∗ =eR

= ((eR

n−2

1

2

·∗ (eR −∗ |x|2∗∗ ))/∗ (eR +∗ |x|∗ )n∗ )e nκ(n)Rn−1

Z ·∗

φ(y) ·∗ d∗ sy

|y|∗ =eR

((eR

=

n−2

2

·∗ (eR −∗ |x|2∗∗ ))/∗ (eR +∗ |x|∗ )n∗ ) ·∗ u(1),

and u(x) ≤ e

1 nκ(n)R

Z

2

((eR −∗ |x|2∗∗ )/∗ (en R −∗ |x|∗ )n∗ ) ·∗ φ(y) ·∗ d∗ sy

·∗ ∗|y|∗ =eR

=

((eR

n−2

1

2

·∗ (eR −∗ |x|2∗∗ ))/∗ (eR −∗ |x|∗ )n∗ ) ·∗ e nκ(n)Rn−1

Z ·∗

φ(y) ·∗ d∗ sy

|y|∗ =eR

=

((eR

n−2

2

·∗ (eR −∗ |x|2∗∗ ))/∗ (eR −∗ |x|∗ )n∗ ) ·∗ u(1),

|x|∗ < eR .

Therefore, ((eR

n−2

2

·∗ (eR −∗ |x|2∗∗ ))/∗ (eR +∗ |x|∗ )n∗ ) ·∗ u(1) ≤ u(x)

≤ ((eR

n−2

2

·∗ (eR −∗ |x|2∗∗ ))/∗ (eR −∗ |x|∗ )n∗ ) ·∗ u(1),

|x|∗ < eR .

Making R tend to ∞ we get u(x) = u(1) for any x ∈ Rn∗ . Theorem 6.13 (The Liouville5 Theorem) Let u be multiplicative harmonic throughout Rn∗ and it is bounded above (below) in Rn∗ . Then u is identically equal to a constant in Rn∗ . Proof Let u(x) ≤ eM , x ∈ Rn∗ . Set v(x) = eM −∗ u(x),

x ∈ Rn∗ .

Then v is multiplicative harmonic throughout Rn∗ and it is multiplicative nonnegative in Rn∗ . Hence from Theorem 6.12, we conclude that eM −∗ u(x) = eM −∗ u(1), 5

x ∈ Rn∗ ,

Joseph Liouville (March 24, 1809–September 8, 1882) was a French mathematician who worked in a number of different fields in mathematics, including number theory, complex analysis, differential geometry, topology, mathematical physics and astronomy.

Theorems of Liouville and Harnack

229

whereupon u(x) = u(1)

x ∈ Rn∗ .

Exercise 6.3 Prove that the Dirichlet problem for the half space xn > 1 cannot have more than one solution in the class of bounded functions. Theorem 6.14 (The Harnack6 Theorem) Let um (x), m ∈ N, be multiplicative harmonic functions in a domain D which are continuous in D and the ∞ X series um (x) is uniformly convergent on the boundary ∂∗ D. Then this ∗m=1

series is uniformly convergent in D and its sum u(x) =

∞ X

um (x) is a mul-

∗m=1

tiplicative harmonic function in D. Proof Let  > 0 be arbitrarily chosen. Since

∞ X

um (x) is uniformly con-

∗m=1

vergent on ∂∗ D, there exists an index N = N () such that p X uN +i (y) < e ∗i=1

holds for all p ≥ 1 and y ∈ ∂∗ D. Note that

∗

p X

uN +i (x) is multiplicative har-

∗i=1

monic in D and continuous in D. Hence from the maximum principle, we conclude that p X uN +i (x) < e for all x ∈ D. ∗i=1

Therefore,

∞ X

∗

um (x) is uniformly convergent in D. Let now x0 ∈ D be an

∗m=1

arbitrary point. We take R > 0 so that B(x0 , eR ) ⊂ D. Then, Z 1 2 um (x) =e nκ(n)R ·∗ ((eR −∗ |x −∗ x0 |2∗∗ )/∗ |y −∗ x|n∗ ∗ ) ·∗ um (y) ∂∗ B(x0 ,eR )

·∗ d∗ sy ,

m ∈ N,

|x −∗ x0 |∗ < eR .

6 Carl Gustav Axel von Harnack (May 7, 1851–April 3, 1888) was a German mathematician who contributed to potential theory Harnack inequality applied to harmonic functions. He also worked on the real algebraic geometry of plane curves, proving Harnack’s curve theorem for real plane algebraic curves.

230

The Laplace Equation

Hence, using

∞ X

um (x) that is uniformly convergent in D, we get

∗m=1

u(x)

=

∞ X

um (x)

∗m=1

Z

1

2

((eR −∗ |x −∗ x0 |2∗∗ )/∗ |y −∗ x|n∗ ∗ )

= e nκ(n)R ·∗ ∗∂∗ B(x0 ,eR )

·∗

∞ X

um (y) ·∗ d∗ sy

∗m=1

= e

1 nκ(n)R

Z

2

((eR −∗ |x −∗ x0 |2∗∗ )/∗ |y −∗ x|n∗ ∗ )

·∗ ∗∂∗ B(x0 ,eR )

·∗ u(y) ·∗ d∗ sy ,

|x −∗ x0 |∗ < eR .

Therefore, u is multiplicative harmonic in |x −∗ x0 |∗ < eR . Because x0 ∈ D was arbitrarily chosen, we conclude that u is multiplicative harmonic everywhere in D.

6.10 6.10.1

Separation of Variables Multiplicative rectangles

Let u be the solution to the Dirichlet problem in multiplicative rectangular domain ∆∗ u = e, ea < x1 < eb , ec < x2 < ed , (6.35) with the boundary conditions u(ea , x2 ) = φ1 (x2 ), u(eb , x2 ) = φ2 (x2 ),

ec ≤ x2 ≤ ed , (6.36)

u(x1 , ec ) = ψ1 (x1 ), u(x1 , ed ) = ψ2 (x1 ),

ea ≤ x1 ≤ eb ,

Separation of Variables

231

where φ1 , φ2 ∈ C∗ ([ec , ed ]), ψ1 , ψ2 ∈ C∗ ([ea , eb ]), and φ1 (ec ) = ψ1 (ea ), φ1 (ed ) = ψ2 (ea ), φ2 (ec ) = ψ1 (eb ), φ2 (ed ) = ψ2 (eb ). We split u into the form u = u1 +∗ u2 , where u1 solves ∆∗ u1 = 1,

ea < x1 < eb ,

ec < x2 < ed ,

u1 (ea , x2 ) = φ1 (x2 ), u1 (eb , x2 ) = φ2 (x2 ),

ec ≤ x2 ≤ ed ,

(6.37)

u1 (x1 , ec ) = u1 (x1 , ed ) = 1,

ea ≤ x1 ≤ eb ,

∆∗ u2 = 1,

ea < x1 < eb ,

and u2 satisfies ec < x2 < ed ,

u2 (ea , x2 ) = u2 (eb , x2 ) = 1,

ec ≤ x2 ≤ ed ,

(6.38)

u2 (x1 , ec ) = ψ1 (x1 ), u2 (x1 , ed ) = ψ2 (x1 ),

ea ≤ x1 ≤ eb ,

under the compatibility condition φ1 (ec ) = ψ1 (ea ) = φ1 (ed ) = ψ2 (ea ) = φ2 (ec ) = ψ1 (eb ) = φ2 (ed ) = ψ2 (eb ) = 1. We will seek a solution u1 = u1 (x1 , x2 ) of the problem (6.37) in the form u1 (x1 , x2 ) = X 1 (x1 )Y 1 (x2 ). Substituting such solution into the multiplicative Laplace equation, we obtain ∗∗ X 1 (x1 ) −∗ eλ ·∗ X 1 (x1 ) = 1, ea < x1 < eb , (6.39) ∗∗

Y 1 (x2 ) +∗ eλ ·∗ Y 1 (x2 ) = 1,

ec < x2 < ed .

(6.40)

232

The Laplace Equation

Also, we get Y 1 (ec ) = Y 1 (ed ) = 1. Thus, using (6.40), we obtain the Sturm7 –Liouville problem for Y 1 (x2 ): ∗∗

Y 1 (x2 ) +∗ eλ ·∗ Y 1 (x2 ) = 1 ec < x2 < ed , (6.41) Y 1 (ec ) = Y 1 (ed ) = 1. 1

Solving (6.41), as in (4.1.3), we derive a sequence of eigenvalues eλn and 1 a sequence of eigenfunctions Yn1 (x2 ). Then we substitute the sequence eλn into (6.39) and we obtain an associated sequence Xn1 (x1 ). The formal solution u1 (x1 , x2 ) of the problem (6.37) is written as follows: u1 (x1 , x2 ) =

∞ X

Xn1 (x1 ) ·∗ Yn1 (x2 ).

∗n=1

The remaining boundary conditions for u1 , u1 (ea , x2 ) = φ1 (x2 ), u1 (eb , x2 ) = φ2 (x2 ) will be used to eliminate the two free parameters associated with Xn1 (x1 ) for each n ∈ N, as is done in (4.1.3). Now, we seek a solution u2 (x1 , x2 ) of the problem (6.38) in the form u2 (x1 , x2 ) = X 2 (x1 ) ·∗ Y 2 (x2 ). Substituting it in ∆∗ u2 = 0, we get

∗∗

X 2 (x1 ) −∗ eλ ·∗ X 2 (x1 ) = 1, Y

2 ∗∗

(x2 ) +∗ eλ ·∗ Y 2 (x2 ) = 1,

ea < x1 < eb ,

(6.42)

ec < x2 < ed .

(6.43)

Using that u2 (ea , x2 ) = u2 (eb , x2 ) = 1, ec ≤ x2 ≤ ed , we get X 2 (ea ) = X 2 (eb ) = 1. Hence from (6.42), we obtain the Sturm–Liouville problem for X 2 (x1 ). ∗∗

X 2 (x1 ) −∗ eλ ·∗ X 2 (x1 ) = 0,

ea < x1 < eb , (6.44)

2

a

2

b

X (e ) = X (e ) = 1. For the problem (6.44), as in (4.1.3), we get a sequence of eigenvalues 2 2 eλn and a sequence of eigenfunctions Xn2 (x1 ). Substituting eλn into (6.43), we 7 Jacques Charles Francois Sturm (September 29, 1803–December 15, 1855) was a French mathematician who discovered the theorem which bears his name and which concerns the determination of the number and the localization of the real roots of a polynomial equation included between given limits.

Separation of Variables

233

obtain an associated sequence Yn2 (x2 ). The formal solution u2 (x1 , x2 ) of the problem (6.38) is written as follows: u2 (x1 , x2 ) =

∞ X

Xn2 (x1 ) ·∗ Yn2 (x2 ).

∗n=1

We find the two free parameters in Yn2 (x2 ), n ∈ N, using the boundary conditions u2 (x1 , ec ) = ψ1 (x1 ), u2 (x1 , ed ) = ψ2 (x1 ). The formal solution u(x1 , x2 ) of the problem (6.35), (6.36) is written in the following way: u(x1 , x2 ) =

∞ X


Xn1 (x1 ) ·∗ Yn1 (x2 ) +∗ Xn2 (x1 ) ·∗ Yn2 (x2 ) .

∗n=1

Example 6.4 Consider the problem ∆∗ u = 1 in

1 < x1 , x2 < eπ ,

u(1, x2 ) = e, u(eπ , x2 ) = u(x1 , 1) = u(x2 , eπ ) = 1,

1 ≤ x1 , x2 ≤ eπ .

We seek a nontrivial formal solution u(x1 , x2 ) = X(x1 ) ·∗ Y (x2 ) Then, (X ∗∗ (x1 )/∗ X(x1 )) = −∗ (Y ∗∗ (x2 )/∗ Y (x2 )) = eλ , where λ is a constant. Since u is nontrivial and u(x1 , 1)

=

u(x1 , eπ )

=

1,

we obtain Y (1) = Y (eπ ) = 0. Thus we get the Sturm–Liouville problem: Y ∗∗ (x2 ) +∗ eλ ·∗ Y (x2 ) = 1, Y (1) = Y (eπ ) = 1.

1 < x2 < eπ ,

234

The Laplace Equation

Hence, Yn (x2 ) = sin∗ (en ·∗ x2 ), and

λn = n2 ,

2

Xn∗∗ (x1 ) −∗ en ·∗ Xn (x1 ) = 1, Therefore, Xn (x1 ) = eAn ·∗ ee

n

·∗ x1

ea < x1 < eb .

+∗ eBn ·∗ e−∗ e

n

·∗ x1

and u(x1 , x2 ) =

∞  X

eAn ·∗ ee

n

·∗ x1

+∗ eBn e−∗ e

n

·∗ x1



·∗ sin∗ (en ·∗ x2 ),

∗n=1

where An and Bn are constants which we will determine using the boundary conditions u(1, x2 ) = e and u(eπ , x2 ) = 1, 1 ≤ x2 ≤ eπ . We get u(1, x2 )

=

∞ X

n (eAn +∗B ) ·∗ sin∗ (en ·∗ x2 ) e

∗n=1

= e, whereupon multiplying by sin∗ (en x2 ) the last equality and integrating over [1, eπ ], we obtain (eAn +∗ eBn ) ·∗

Ze

π

π

(sin∗ (en ·∗ x2 ))2∗ ·∗ d∗ x2 =

∗1

or

Ze

sin∗ (en ·∗ x2 ) ·∗ d∗ x2

∗1 2

eAn +∗ eBn = −∗ e nπ ((−1)

n

−1)

.

(6.45)

Also, u(eπ , x2 ) =

∞ X


eAn ·∗ enπ +∗ eBn ·∗ e−∗ nπ ·∗ sin∗ (en ·∗ x2 ) = 1,

∗n=1

whence eAn ·∗ enπ +∗ eBn ·∗ e−∗ nπ = 1. From the last equality and from (6.45), we go to the system 2

n

eAn +∗ eBn = −∗ e nπ ((−1) eAn ·∗ enπ +∗ eBn ·∗ e−∗ nπ = 1.

−1)

Separation of Variables

235

For its solution, we have An =

2 1 ((−1)n − 1) 2nπ nπ e −1

Bn = −

2 e2nπ ((−1)n − 1) 2nπ . nπ e −1

Therefore, 4

eAn ·∗ en·∗ x1 +∗ eBn ·∗ e−∗ n·∗ x1 = e nπ

1 ((−1)n −1) e2nπ enπ −1

·∗ sinh∗ (e·∗ n(x1 −∗ eπ ))

and u(x1 , x2 ) =

∞ X

4

e nπ

1 ((−1)n −1) e2nπ enπ −1

·∗ sinh∗ (en (x1 −∗ eπ )) ·∗ sin∗ (en ·∗ x2 ).

∗n=1

Exercise 6.4 Find a formal solution to the following problem: ∆∗ u = 1,

1 < x1 , x2 < e,

u(1, x2 ) = x2 ·∗ (e −∗ x2 ), u(e, x2 ) = 1,

1 ≤ x2 ≤ e,

u(x1 , 1) = sin∗ (eπ ·∗ x1 ), u(x1 , e) = 1,

1 ≤ x1 ≤ e.

Answer u(x1 , x2 )

=

((sinh∗ (eπ ·∗ (e −∗ x2 )))/∗ (sinh∗ eπ )) ·∗ sin∗ (eπ ·∗ x1 ) 8

+∗ e π3 ·∗

∞ X

3

((sinh∗ ((e(2n+1)π ·∗ (x1 −∗ e))))/∗ e(2n+1) )

∗n=0

·∗ ((sin∗ (e(2n+1)π ·∗ x2 ))/∗ (sinh∗ (e(2n+1)π ))).

6.10.2

Multiplicative circular domains

Consider the Dirichlet problem: ∆∗ u = 1,

(x1 , x2 ) ∈ B(1, ea ), (6.46)

u(x1 , x2 ) = φ(x1 , x2 ),

(x1 , x2 ) ∈ ∂B(1, ea ).

Introduce multiplicative polar coordinates x1

=

r ·∗ cos∗ θ,

x2

=

r ·∗ sin∗ θ,

r > 1,

θ ∈ [1, e2π ].

236

The Laplace Equation

Then,  B(1, ea ) = (r, θ) : 1 ≤ r ≤ ea , φ(x1 , x2 )

∂B(1,ea )

1 ≤ θ ≤ e2π ,

= φ(ea ·∗ cos∗ θ, ea ·∗ sin∗ θ) = h(θ).

Thus the problem (6.46) takes the following form: ∗ 2∗ ∗∗ u∗∗ rr +∗ (e/∗ r) ·∗ ur +∗ (e/∗ r ) ·∗ uθθ = 1 in

B(1, ea )

u(ea , θ) = h(θ), lim u(r, θ)

(6.47)

exists

r→1

and

is

finite.

We will search a formal solution of the problem (6.47) in the form u(r, θ) = R(r) ·∗ Θ(θ). Substituting this function in (6.47) and using the arguments in (4.1.3), we find r2∗ ·∗ R∗∗ (r) +∗ r ·∗ R∗ (r) −∗ eλ ·∗ R(r) = 1,

1 < r < ea ,

Θ∗∗ (θ) +∗ eλ ·∗ Θ(θ) = 1,

1 < θ < e2π ,

Θ∗ (1) = Θ∗ (e2π )

Θ(1) = Θ(e2π ),

R(ea ) ·∗ Θ(θ) = h(θ), lim u(r, θ)

r→1

1 ≤ θ ≤ e2π ,

exists

and

is

finite.

(6.48) Since we search a solution u(r, θ) of the class C∗2 , we need to impose the periodicity conditions: Θ(1)

=

Θ(e2π ),

Θ∗ (1)

=

Θ∗ (e2π ).

Hence from the second equation of (6.48), we get Θn (θ) λn

= eAn ·∗ cos∗ (en ·∗ θ) +∗ eBn ·∗ sin∗ (en ·∗ θ), = n2 ,

n ∈ N0 .

Separation of Variables

237

Substituting the eigenvalues λn into the first equation of (6.48), we find 2

r2∗ ·∗ R∗∗ (r) +∗ r ·∗ R∗ (r) −∗ en ·∗ R(r) = 1, whereupon Rn (r) = eCn ·∗ rn∗ +∗ eDn ·∗ r−∗ n∗ ,

n ∈ N,

R0 (r) = C0 +∗ D0 ·∗ log∗ r. Since we want lim u(r, θ) to exist and and to be finite, we get Dn = 0, n ∈ N0 . r→1

Therefore, we obtain a formal solution as follows: u(r, θ) =

∞ X

Rn (r) ·∗ Θn (θ)

∗n=0

=e

α0 2

∞ X

+∗


rn∗ ·∗ eαn ·∗ cos∗ (en ·∗ θ) +∗ eβn ·∗ sin∗ (en ·∗ θ) .

∗n=1

(6.49) Formally differentiating this series term-by-term, we verify that (6.49) is indeed multiplicative harmonic. Imposing the boundary condition u(ea , θ) = h(θ), 1 ≤ θ ≤ e2π , we obtain 2π

eα 0

1 π

Ze

= e ·∗

h(θ) ·∗ d∗ θ, ∗1 2π

eαn

= e

1 πan

Ze ·∗

·∗ h(θ) ·∗ cos∗ (en ·∗ θ) ·∗ d∗ θ,

∗1 2π

e

βn

= e

1 πan

Ze ·∗

h(θ) ·∗ sin∗ (en θ) ·∗ d∗ θ,

n ∈ N.

∗1

Example 6.5 Consider the Dirichlet problem: ∆∗ u = 1,

x21∗ +∗ x22∗ < e,

u(x1 , x2 ) = x22∗

on x21∗ +∗ x22∗ = e.

Introducing polar coordinates x1

=

r ·∗ cos∗ θ,

x2

=

r ·∗ sin∗ θ,

r > 1,

θ ∈ [1, e2π ],

238

The Laplace Equation

we get the problem ∗ 2∗ ∗∗ u∗∗ rr +∗ (e/∗ r) ·∗ ur +∗ (e/∗ r ) ·∗ uθθ = 1

in

B(0, 1),

u(cos∗ θ, sin∗ θ) = (sin∗ θ)2∗ ,

θ ∈ [1, e2π ].

Here, h(θ) = (sin∗ θ)2∗ . Then, 2π

eα0

Ze

1 π

= e ·∗

(sin∗ θ)2∗ ·∗ d∗ θ

∗1

= e, 2π

eαn

Ze

1 π

= e ·∗

(sin∗ θ)2∗ ·∗ cos∗ (en ·∗ θ) ·∗ d∗ θ

∗1

=

 1  e −∗ 2 

n=2

n 6= 2,

1

n ∈ N,

2π

eβn

1 π

Ze

= e ·∗

(sin∗ θ)2∗ ·∗ sin∗ (en ·∗ θ) ·∗ d∗ θ

∗1

=

1,

n ∈ N.

Consequently, u(x1 , x2 )

= u(r, θ) 1

= e 2 −∗ (r2∗ /∗ e2 ) ·∗ cos∗ (e2 ·∗ θ) 1

= e 2 −∗ ((r2∗ ·∗ (cos∗ θ)2∗ −∗ r2∗ ·∗ (sin∗ θ)2∗ )/∗ e2 1

= e 2 ·∗ (e −∗ x21∗ +∗ x22∗ ). Exercise 6.5 Find a formal solution to the following problem: ∆∗ u = 1 in u(x1 , x2 ) = x32∗

x21∗ +∗ x22∗ < e, on x21∗ +∗ x22∗ = e.

Advanced Practical Exercises

239

Answer u(r, θ) = (r/∗ e4 ) ·∗ (e3 ·∗ sin∗ θ −∗ r2∗ ·∗ sin∗ (e3 ·∗ θ)) +∗ (r/∗ e4 ) ·∗ (e3 ·∗ cos∗ θ −∗ r2∗ ·∗ cos∗ (e3 ·∗ θ)). Exercise 6.6 Find a formal solution to the Dirichlet problem: ∆∗ u = 1 in

R2∗ \B(1, ea ),

u(x1 , x2 ) = φ(x1 , x2 )

on ∂∗ B(1, ea ),

where φ ∈ C∗ (∂∗ B(1, ea )). Answer u(r, θ) = e

α0 2

+∗

∞ X

n∗

(ea /∗ r)


·∗ eαn ·∗ cos∗ (en ·∗ θ) +∗ eβn ·∗ sin∗ (en ·∗ θ) ,

∗n=1

where 2π

e

α0

=

1 π

Ze

e ·∗

h(θ) ·∗ d∗ θ, ∗1 2π

eαn

=

1 π

Ze

h(θ) ·∗ cos∗ (en ·∗ θ) ·∗ d∗ θ,

e ·∗ ∗1 2π

eβn

=

1 π

Ze

e ·∗

h(θ) ·∗ sin∗ (en ·∗ θ) ·∗ d∗ θ,

n ∈ N,

∗1

h(θ)

6.11

=

φ(ea ·∗ cos∗ θ, ea ·∗ sin∗ θ).

Advanced Practical Exercises

Problem 6.1 Let D = {x = (x1 , . . . , xn ) ∈ Rn∗ : 1 < xn < ea }, where a is a positive constant. Prove that G(x, y) =

∞ X

m
Φ(x, y m ) −∗ Φ(x, y 0 ) ,

∗m=−∞

where y = (y1 , . . . , yn−1 , yn ), y m = (y1 , . . . , yn−1 , e2ma +∗ yn ), y 0 yn−1 , e2ma −∗ yn ) is a Green function.

m

= (y1 , . . . ,

240

The Laplace Equation

Problem 6.2 Find a formal solution to the following problem: ∆∗ u = 1,

1 < x1 , x2 < e,

u(x1 , 1) = e +∗ sin∗ (eπ ·∗ x1 ), u(x1 , e) = e2 ,

1 ≤ x1 ≤ e,

u(1, x2 ) = u(e, x2 ) = e +∗ x 2 ,

1 ≤ x2 ≤ e.

Answer u(x1 , x2 ) = sin∗ (eπ x1 ) ·∗ ((sinh∗ (eπ ·∗ (e −∗ x2 )))/∗ (sinh∗ eπ )) +∗ e +∗ x2 . Problem 6.3 Find a formal solution to the following problem: ∆∗ u = 1,

1 < x2 < e2 ,

1 < x1 < e,

u(x1 , 1) = 1, u(x1 , e2 ) = e,

1 ≤ x1 ≤ e,

u∗x1 (1, x2 ) = 1, u∗x1 (e, x2 ) = sin∗ (e2π ·∗ x2 ),

1 ≤ x2 ≤ e2 .

Answer u(x1 , x2 ) = x2 /∗ e2 +∗ ((cosh∗ (e2π ·∗ x1 ) ·∗ sin∗ (e2π ·∗ x2 ))/∗ (e2π ·∗ sinh∗ (e2π ))). Problem 6.4 Find a bounded formal solution to the following problem: ∆∗ u = 1,

1 < x1 < e,

x2 > 1,

u(x1 , 1) = e, u(1, x2 ) = u(e, x2 ) = 1. Answer 4

u(x1 , x2 ) = e π ·∗

∞ X ∗n=0

1

(2n+1)π

e 2n+1 e−∗ e

·∗ x2

·∗ sin∗ (e(2n+1)π ·∗ x1 ).

Advanced Practical Exercises

241

Problem 6.5 Find a formal solution to the following problem: ∆∗ u = 1,

1 < x1 , x2 < eπ ,

u∗x1 (1, x2 ) = u∗x1 (eπ , x2 ) = 1,

1 ≤ x2 ≤ eπ ,

u∗x2 (x1 , 1) = 1, π

u∗x2 (x1 , eπ ) = x1 −∗ e 2 ,

1 ≤ x1 ≤ eπ .

Answer 4

u(x1 , x2 ) = eA0 −∗ e π ·∗

∞ X

((cos∗ (e(2n−1) ·∗ x1 ) ·∗ cosh∗ (e(2n−∗ 1) ·∗ x2 ))/∗

∗n=0 2

(e(2n−1) ·∗ sinh∗ (e(2n−1)π ))), where A0 is a constant. Problem 6.6 Find a formal solution to the following problem: ∆∗ u = 1 in u(x1 , x2 ) = x21∗ Answer

x21∗ +∗ x22∗ < e, on x21∗ +∗ x22∗ = e.

1

u(x1 , x2 ) = e 2 ·∗ (e +∗ x21∗ −∗ x22∗ ). Problem 6.7 Let u(x) be a multiplicative harmonic function in D. Prove that u ∈ C∗∞ (D).

7 The Cauchy–Kovalevskaya Theorem

7.1

Analytic Functions of One Variable

Definition 7.1 A power series is an expression of the form ∞ X

f (z) =

ean ·∗ (z −∗ ec )n∗ ,

(7.1)

∗n=0

with coefficients an ∈ C and centre c ∈ C. Assume that the power series (7.1) is convergent. Then, |ean |∗ ·∗ |z −∗ ec |n∗ ∗ → 1

as n → ∞.

Therefore, there exists a constant M > 0 such that |ean |∗ ·∗ |z −∗ ec |n∗ ∗ ≤ eM

for any

n ∈ N,

whereupon |z −∗ ec |∗ ≤ eR (7.2) = sup{r ≥ 1 :

sup |ean |∗ ·∗ rn∗ < ∞}.

Definition 7.2 The above defined eR ∈ [1, ∞] is called the convergence radius of the power series (7.1). We introduce the notation DR (ec ) = {z ∈ C∗ : |z −∗ ec |∗ < eR },

DR = DR (1).

Note that if |z −∗ ec | > eR , then the power series (7.1) is divergent. Theorem 7.1 Let eR be defined by (7.2). Then the power series (7.1) converges absolutely uniformly on each compact subset of the open disk DR (ec ) and diverges at every z ∈ C∗ \DR (ec ). Moreover, eR can be determined by the Cauchy–Hadamard formula 1

e/∗ eR = lim sup |ean |∗n ∗ n→∞

DOI: 10.1201/9781003440116-7

242

Analytic Functions of One Variable

243

with the conventions e/∗ ∞ = 0 and e/∗ 1 = ∞, and, furthermore, provided that ean = 1 for only finitely many n, one can estimate eR by the ratio test lim inf |ean |∗ /∗ |ean+1 |∗ n→∞

≤

eR

≤

lim sup |ean |∗ /∗ |ean+1 |∗ . n→∞

Proof Without loss of generality, we suppose that c = 0. Above we have ∞ X demonstrated the divergence of ean ·∗ z n∗ at every z ∈ C∗ \DR . Take ∗n=1

r < R. Let z ∈ Dr . Then, for any ρ ∈ (r, R), we have |ean ·∗ z n∗ |∗ < |ean |∗ ·∗ rn∗ = |ean |∗ ρn∗ ·∗ (r/∗ ρ)

n∗

n∗

≤ M ·∗ (r/∗ ρ)

for some constant M > 0, M < ∞. Since r/∗ ρ < 1, we have that

∞ X

e a n · ∗ z n∗

∗n=1

converges uniformly in Dr . Because any z ∈ DR is in some Dr , r < R, the series converges absolutely uniformly on each compact subset of DR . Let now, ρ be defined with 1

e/∗ ρ = lim sup |ean |∗n ∗ . n→∞

We will prove that ρ = R. For any  ∈ (0, 1), we have |ean |∗ ·∗ ρn∗ ≥ (e −∗ )n∗ for infinitely many n. Also, there is n such that |ean |∗ ·∗ ρn∗ ≤ (e +∗ )n∗ for any n > n . Thus, if |z|∗ > ρ, then |ean ·∗ z n∗ |∗ > |ean |∗ ·∗ ρn∗ . Hence, |ean ·∗ z n∗ |∗ > e for infinitely many n provided that |z|∗ > ρ. Therefore, the ∞ X series ean ·∗ z n∗ diverges, which implies that ρ ≥ R. On the other hand, ∗n=1

if |z|∗ < ρ, then for any  > 0 we have |ean ·∗ z n |∗ = |ean |∗ ·∗ ρn∗ ·∗ (|z|n∗ ∗ /∗ ρn∗ ) ≤ (e +∗ )n∗ (|z|n∗ ∗ /∗ ρn∗ = k n∗ for any n > n . By choosing  > 0 small enough, one can ensure that k ∈ [0, 1), ∞ X and so ean ·∗ z n∗ converges. Therefore, ρ ≤ R. Consequently, ρ = R. Let ∗n=1

α = lim inf (|ean |∗ /∗ |ean+1 |∗ ). n→∞

Suppose that |z|∗ < α. Then for any  > 0, we have |ean |∗ ≥ (α −∗ ) ·∗ |ean+1 |∗ for all sufficiently large n. This gives |ean ·∗ z n∗ |∗ ≤ C ·∗ (|z|∗ /∗ (α −∗ ))

n∗

244

The Cauchy–Kovalevskaya Theorem

for all sufficiently large n, with some constant C > 0. By choosing  > 0 small ∞ X enough, we show the convergence of ean ·∗ z n∗ . Therefore, α ≤ R. Let ∗n=1

now

an

β = lim sup(|e |∗ /∗ |ean+1 |∗ . n→∞

Suppose that |z|∗ > β and  = |z|∗ −∗ β > 1. Then, |ean |∗ ≤ (β +∗ ) ·∗ |ean+1 |∗ for sufficiently large n. So |ean ·∗ z n∗ |∗ ≥ C ·∗ (|z|n∗ ∗ )/∗ (β +∗ )n∗ ≥ C for some constant C > 0, and the series diverges. Therefore, R ≤ β. Let Ω denotes an open set in C∗ . Definition 7.3 A multiplicative complex-valued function f : Ω → C∗ is called multiplicative analytic at z ∈ Ω, if there are coefficients ean ∈ C and a radius er > 1 such that ∞ X f (z +∗ h) = ean ·∗ hn∗ ∗n=0

for all h ∈ Dr . Moreover, f is said to be multiplicative analytic on Ω if it is analytic at each z ∈ Ω. The set of multiplicative analytic functions on Ω is denoted by C∗ω (Ω). Theorem 7.2 Let eR > 1 be the convergence radius of the power series f (z) =

∞ X

ean ·∗ (z −∗ ec )n∗ .

∗n=0

Let also d ∈ DR (c). Then    ∞ ∞  X X n  ean ·∗ (d −∗ ec )(n−j)∗  ·∗ (z −∗ d)j∗ , f (z) = j ∗j=0

∗n=j

where the convergence radius of the power series is at least eR−∗ |d−∗ c|∗ . In particular, we have f ∈ C∗ω (DR (ec )), and the convergence radius of a rearranged power series depends continuously on its centre. Proof We have (z −∗ ec )n

= (z −∗ d +∗ d −∗ ec )n

=

 n  X n (z −∗ d)j∗ ·∗ (d −∗ ec )(n−j)∗ . j

∗j=0

Analytic Functions of One Variable

245

Hence, ∞ X

f (z) =

ean ·∗ (z −∗ ec )n∗

∗n=0 ∞ X

=



  n  X n ·∗ (z −∗ d)j∗ ·∗ (d −∗ ec )(n−j)∗  . j

ean

∗n=0

∗j=0

c

Let |z −∗ d|∗ ≤ eρ−∗ (d−∗ e ) with ρ < R. Then,  n  X n (n−j)∗ n |z −∗ d|j∗∗ ·∗ |d −∗ ec |∗ = (|z −∗ d|∗ +∗ |d −∗ ec |∗ ) ∗ ≤ ρn∗ j ∗j=0

 ρ  n∗ n and since ean ·∗ ρn∗ = ean R ·∗ e R , we obtain that ∞ X

 ean

∗n=0

  n  X n ·∗ (z −∗ d)j∗ ·∗ (d −∗ ec )(n−j)∗  j ∗j=0

is absolutely convergent on each compact subset of Dr (d). Therefore,    ∞ n  X X n ean (z −∗ d)j∗ ·∗ (d −∗ ec )(n−j)∗  j ∗n=0

∗j=0

=

 ∞ X ∞  X n ean ·∗ (z −∗ d)j∗ ·∗ (d −∗ ec )(n−j)∗ . j

∗j=0 ∗n=j 0

Let eR denote the convergence radius of the rearranged series centred at 0 c 0 R ed . Then eR ≥ eR−∗ |d−∗ e |∗ or eR−∗ R ≤ |d −∗ ec |∗ . If |d −∗ ec |∗ < e 2 , and ec ∈ DR0 (ed ), which means that the above reasoning can be applied with the 0 roles of the two power series interchanged, giving eR −∗ R ≤ |ec −∗ ed |∗ . Theorem 7.3 Let eR be the convergence radius of the power series (7.1). Then both ∞ X g(z) = nean ·∗ (z −∗ ec )(n−1)∗ and ∗n=0

F (z) =

∞ X

(ean /∗ en+1 ) ·∗ (z −∗ ec )(n+1)∗

∗n=0

have convergence radii equal to eR and f∗ = g

and

F∗ = f

in

DR (ec ).

246

The Cauchy–Kovalevskaya Theorem 0

Proof Let eR be the convergence radius of g. For z 6= ec , we have g(z) = (e/∗ (z −∗ ec )) ·∗

∞ X

enan ·∗ (z −∗ ec )n∗ .

∗n=0

Therefore, 0

(e/∗ eR ) = lim sup e

√ n

nan

≥ lim sup e

n→∞

√ n

an

1

= eR ,

n→∞

0

i.e., eR ≤ eR . Let er < eR . Then for any e > 1, there is a constant C > such that en·∗ |an |∗ ·∗ rn∗ ≤ eC ·∗ (e +∗ e )n∗ ·∗ |an |∗ rn∗ = eC ·∗ (e +∗ e )n∗ ·∗ e( R ) r

n

·∗ |an |∗

·∗ Rn∗ . 0

Hence, choosing e > 1 small enough, we see that er ≤ eR . Consequently, R R0 e ≤ e , whence R = R0 . Now, we will prove that f ∗ = g in DR (ec ), i.e., for each z ∈ DR (ec ), we have f (z +∗ h) = f (z) +∗ g(z) ·∗ h +∗ o(|h|∗ ). Note that f (z +∗ h) −∗ f (z)

=

∞ X

ean ·∗ ((z +∗ h)n∗ −∗ z n∗ )

∗n=0

= h ·∗

∞ X

ean ·∗

∗n=0

n−1 X

(z +∗ h)j∗ ·∗ z (n−1−j)∗

∗j=0

= h ·∗ λz (h). Let r < R be such that |z|∗ < er , and consider all h satisfying |z +∗ h|∗ ≤ er . Then ∞ X ∗n=0

|ean |∗

n−1 X

(n−1−j)∗

|z +∗ h|j∗∗ ·∗ |z|∗

≤

∞ X

rn∗ ·∗ en·∗ |an |∗ < ∞,

∗n=0

∗j=0

so λz (h) converges locally uniformly in a neighborhood of the multiplicative origin. Hence, λz is continuous at 1. Moreover, λz (1) = g(z). Therefore, λz (h) = g(z) +∗ o(e), with o(e) → 1 as |h|∗ → ∞, i.e., f (z +∗ h) = f (z) +∗ g(z) ·∗ h +∗ o(|h|∗ ). The statements about F follow from the above if we replace f with F and g with f.

Analytic Functions of One Variable

247

Definition 7.4 By repeatedly applying Theorem 7.3, we see that the coefficients of the power series f about ec ∈ Ω are given by ean = ((f ∗(n) (ec ))/∗ n!∗ . Therefore, if f ∈ C∗ω (Ω) and ec ∈ Ω, then the following Taylor series converges in a neighborhood of ec . f (z) =

∞ X

((f ∗(n) (ec ))/∗ n!∗ ) ·∗ (z −∗ ec )n∗ .

∗n=0

Definition 7.5 A multiplicative accumulation point of a set D ⊂ C∗ is a point z ∈ C∗ such that any neighborhood of z contains a point w 6= z from D. Definition 7.6 We say that z ∈ D is a multiplicative isolated point if it is not a multiplicative accumulation point of D. Definition 7.7 If all points of D are multiplicative isolated, then D is called multiplicative discrete. Theorem 7.4 (Identity Theorem) Let Ω be a multiplicative connected open set in C∗ and f ∈ C∗ω (Ω). Let also at least one of the following statements hold. 1. There is eb ∈ Ω such that f ∗(n) (eb ) = 1 for all n ∈ N. 2. The multiplicative zero set of f has a multiplicative accumulation point in Ω. Then, f ≡ 1 in Ω. Proof Observe that each of the sets Σn = {z ∈ Ω : f ∗(n) (z) = 1} \ is relatively closed in Ω. Then Σ = Σn is also closed. Since z ∈ Σ implies n

that f ≡ 1 in a small disk centred at z by a Taylor series argument, we have that Σ is open. 1. Suppose that there is eb ∈ Ω such that f ∗(n) (b) = 1 for any n. Since eb ∈ Σ, we have that Σ is nonempty. Therefore, Σ = Ω. 2. Suppose that the zero set of f has a multiplicative accumulation point ec . If ec ∈ Σ, then Σ = Ω. Let ec ∈ / Σ. Then there is an n such that f ∗(n) (ec ) 6= 1. So, we have that f (z) = (z −∗ ec )n∗ ·∗ g(z) for some continuous function g such that g(ec ) 6= 1. Hence, there is a neighborhood of ec where f has at most one multiplicative zero, which is a contradiction because ec is a multiplicative accumulation point of the multiplicative zero set of f. Therefore, ec ∈ Σ and then Σ = Ω.

248

7.2

The Cauchy–Kovalevskaya Theorem

Autonomous Multiplicative Differential Equations

Here we will give the MDE case of the Cauchy–Kovalevskaya1 theorem which is simpler and contains half of the main ideas. Consider the initial value problem: u∗ = f (u), (7.3) u(1) = ee , where f : C∗ → C∗ is a given function multiplicative analytic at 1, u is the unknown function. Theorem 7.5 The initial value problem (7.3) has a unique solution u that is multiplicative analytic at 1. Proof Without loss of generality, we suppose that ee = 1. Otherwise, we change u with u −∗ ee and f with f (ee +∗ ·). We repeatedly differentiate the first equation of (7.3) and we get u∗∗

=

(f (u))∗

= f ∗ (u) ·∗ u∗ , u∗∗∗

=

(f (u))∗∗

= f ∗∗ (u) ·∗ (u∗ )2∗ +∗ f ∗ (u) ·∗ u∗∗ , .. . u∗(k)

=

(f (u))∗(k−1)

= qk (f (u), . . . , f ∗(k−1) (u), u∗ , . . . , u∗(k−1) ), where qk is a multivariate polynomial with multiplicative nonnegative integer coefficients. We evaluate this at 1 and we get u∗(k) (1)

= qk (f (1), . . . , f ∗(k−1) (1), u∗ (1), . . . , u∗(k−1) (1)).

1 Sofia Kovalevskaya (January 15, 1850–February 10, 1891) was the first major Russian female mathematician and responsible for important original contributions to analysis, partial differential equations and mechanics.

Autonomous Multiplicative Differential Equations

249

Now repeatedly applying the same formula with k having values k − 1, k − 2 etc., to eliminate u∗(m) (1) from the right-hand side, we obtain u∗(k) (1) = Qk (f (1), . . . , f ∗(k−1) (1)),

(7.4)

where Qk is another multivariate polynomial having multiplicative nonnegative integer coefficients. This proves uniqueness of multiplicative analytic solutions of (7.3), because (7.4) fixes their Maclaurin series coefficients at 1. Provided that the Maclaurin series ∞ X u(z) = ((u∗(n) (1))/∗ n!∗ ) ·∗ z n∗ , (7.5) ∗n=0

with u∗(n) (1) given by (7.4) converges in a neighborhood of 1, the function v = u∗ −∗ f (u) is multiplicative analytic at 1 and its Maclaurin series is identically multiplicative zero. Hence, by the Identity Theorem, v must vanish wherever it is defined. Now we will prove that (7.5) converges in a neighborhood of 0. Since the right-hand side f is multiplicative analytic at 1, there exist constants M > 0 and r > 0 such that ((|f ∗(k) (1)|∗ )/∗ (k!∗ )) ≤ (eM /∗ rk∗ ),

k = 0, 1, . . ..

Then the function: F (z)

= (eM /∗ (e −∗ z/∗ er )) M

M

= eM +∗ e r ·∗ z +∗ · · · +∗ e rk ·∗ z k∗ +∗ · · · majorizes f at 1. Consider the initial value problem: U ∗ = F (U ), (7.6) U (1) = 1. Then, using (7.5), we have U ∗(k) (1) = Qk (F (1), . . . , F ∗(k−1) (1)) and |u∗(k) (1)|∗

= |Qk (f (1), . . . , f ∗(k−1) (1))|∗ ≤ Qk (|f (1)|∗ , . . . , |f ∗(k−1) (1)|∗ ) ≤ Qk (|F (1)|∗ , . . . , |F ∗(k−1) (1)|∗ ) = U ∗(k) (1).

250

The Cauchy–Kovalevskaya Theorem

Therefore, the solution u of (7.3) is majorized by the solution U of (7.6) at 1. Note that (7.6) is solvable with   1 U (z) = r e −∗ (e −∗ (eM ·∗ z)/∗ r) 2 ∗ = eM ·∗ (z/∗ e2 ) +∗ · · · , whose Taylor series around 1 has multiplicative nonnegative coefficients. Consequently, u is multiplicative analytic at 1.

7.3

Systems of Ordinary Differential Equations

In this section, we will consider the Cauchy–Kovalevskaya theorem for the system u∗j u∗j (1)

= fj (z1 , . . . , zn , u1 , . . . , um ), =

1,

j = 1, . . . , m.

We could have eliminated the dependence of f on z1 , . . . , zn by introducing the new variables u∗m+1 = z1 , . . . , u∗m+n = zn with the equations u∗m+k = e, k = 1, . . . , n. The above system can be written in the following manner: u∗ = f (z, u), (7.7) u(1) = 1, n where u = (u1 , . . . , um ) has values in Cm ∗ , z = (z1 , . . . , zn ). In C∗ , a power series is an expression of the form

f (z) =

∞ X ∗α1 =0

...

∞ X

aα1 ...αn ·∗ (z1 −∗ ec1 )α1∗ ·∗ · · · ·∗ (zn −∗ ecn )αn∗ ,

∗αn =0

with coefficients aα1 ...αn ∈ C∗ , and centre ec ∈ Cn∗ . Introduce the multi-index α = (α1 , . . . , αn ) ∈ N0 and the conventions |α| = α1 + · · · + αn

α

α

and z α∗ = z1 1∗ ·∗ · · · ·∗ zn n∗

for

z ∈ Cn∗ .

Then the above series can also be written as follows: X f (z) = aα ·∗ (z −∗ ec )α∗ .

(7.8)

∗|α|≥0

If the series (7.8) converges for some z, then there is a constant 0 < M < ∞ such that α

1∗ |aα |∗ ·∗ |z1 −∗ ec1 |α . . . |zn −∗ ecn |∗ n∗ ≤ eM ∗

for

all α.

Systems of Ordinary Differential Equations

251

In particular, if this series converges in a neighborhood of ec , then there are −∗ |α| constants 0 < M < ∞ and r > 0, such that |aα |∗ ≤ eM r for all α. On the α 1 other hand, if r ∈ Rn∗ and M < ∞ satisfy |aα |∗ ·∗ r1 ∗ . . . rnαn∗ ≤ eM for all α, then the series converges absolutely for all z ∈ Cn∗ satisfying |zi −∗ eci |∗ < ri for each i ∈ {1, . . . , n}. Definition 7.8 Let Ω be an open subset of Cn∗ . A complex-valued function f : Ω → C∗ is called multiplicative analytic at ec ∈ Ω if there are constants aα ∈ C∗ , α ∈ Nn0 , such that the power series (7.8) converges in a neighborhood of ec . Moreover, f is said to be multiplicative analytic in Ω if it is multiplicative analytic at each ec ∈ Ω. The set of multiplicative analytic functions on Ω is denoted by C∗ω (Ω). As in the single-variable case, one can show that if f is multiplicative analytic at ec , then the series (7.8) is its multivariate Taylor series, i.e., the coefficients are given by aα

= (∂∗α f (ec )/∗ α!∗ ) α1 αn = (∂∗1 . . . ∂∗n f (ec ))/∗ (α1 !∗ . . . αn !∗ ),

where we have introduced the conventions α!∗ = α1 !∗ ·∗ . . . ·∗ αn !∗ and α α ∂∗jj f = ∂∗zjj f, j = 1, . . . , n. Theorem 7.6 (Identity Theorem) Let f ∈ C∗ω (Ω) with Ω a multiplicative connected open set in Cn∗ , and with some eb ∈ Ω, let ∂∗α f (eb ) = 1 for all α. Then f ≡ 1 in Ω. Proof Let Then Σα

Σα = {z ∈ Ω : ∂∗α f (z) = 1}. \ is relatively closed in Ω. Hence, Σ = Σα is also closed. Since α

z ∈ Σ implies that f ≡ 1 in a neighborhood of z by a Taylor series argument, we have that Σ is also open. Because eb ∈ Σ, we have that Σ is nonempty. Consequently, Σ = Ω. This completes the proof. Theorem 7.7 Consider the following Cauchy problem: u∗j = fj (z, u), u∗j (1) = e∗j ,

j = 1, . . . , m,

where fj : Cn+m → C∗ are multiplicative analytic at 1 for each j = 1, . . . , m. ∗ Then there exists a unique solution u that is multiplicative analytic at 1.

252

The Cauchy–Kovalevskaya Theorem

Proof Without loss of generality, we suppose that e∗j = 1, j = 1, . . . , m. First of all, we will determine the higher derivatives of u. For this aim, we multiplicative differentiate the equation u∗j = fj (z, u) and we get u∗∗ j

= (fj (z, u))

= ∂∗z fj +∗

∗

m X

∂∗ui fj ·∗ u∗i ,

∗i=1

u∗∗∗ j

= (fj (z, u))

=

2 ∂∗z f j +∗

∗∗

m X m X

∂∗ui ∂∗ul fj · u∗i ·∗ u∗l +∗

∗l=1 ∗i=1

n X

∂∗ui fj ·∗ u∗i ,

∗i=1

.. . ∗(k)

uj

  = qk ∂∗β fj (z, u), u∗(l) ,

where qk is a multivariate polynomial with multiplicative nonnegative coefficients and the arguments of qk are all ∂∗β fj (z, u) with |β| ≤ k − 1, and all (l) components of all u∗ with l ≤ k − 1. We evaluate this at z = 1, and we use u(1) = 1, to get   ∗(k) uj (1) = qk ∂∗β fj (1), u∗(l) (1)
= Qjk ∂∗β f (1) ,   where Qjk ∂jβ f (1) is a multivariate polynomial having multiplicative nonnegative coefficients. Note that the arguments of Qjk are all components of all ∂∗β f (1) with |β| ≤ k − 1. Since f is componentwise multiplicative analytic at 0, there exist constants M > 0 and r > 0 such that (|∂∗α fj (1)|∗ )/∗ α!∗ ≤ (eM /∗ r|α|∗ )

for

all α,

and

all j.

Note that Fj (z, u) = eM /∗ (e −∗ z/∗ r) (e −∗ (u∗1 +∗ · · · +∗ u∗m )/∗ r) majorizes fj at 1, j = 1, . . . , m. Consider the system Uj∗ = Fj (z, U ), (7.9) Uj∗ (1) = 1,

j = 1, . . . , m.

Partial Differential Equations

253

Using the multiplicative positivity of the coefficients of Qjk , we get
|∂∗α u∗j (1)|∗ = Qjk ∂∗β f (1) ∗ ≤ Qjk |∂∗β f (1)|∗





≤ Qjk ∂∗β F (1) = ∂∗α Uj∗ (1), i.e., Uj∗ majorizes u∗j at 1. Observe that   r 1 ∗ U1∗ (z) = ·∗ = Um (z) = e m ·∗ e −∗ (e +∗ e2mM ·∗ log∗ (e −∗ (z/∗ r))) 2 ∗ solves (7.9), which is multiplicative analytic at 0. Consequently, u is multiplicative analytic at 1.

7.4

Partial Differential Equations

Theorem 7.8 (Cauchy–Kovalevskaya Theorem) Consider the following Cauchy problem: ∂∗n u∗j = fj (z, u, ∂∗1 u, . . . , ∂∗n−1 u),

u∗j (ζ, 1) = 1,

ζ ∈ Cn−1 , ∗

j = 1, . . . , m. (7.10) n+m+(n−1)×m Let fj ∈ C∗ → C∗ be multiplicative analytic at 1 for all j = 1, . . . , m. Then there exists a unique solution u that is multiplicative analytic at 1. Proof Without loss of generality, we can assume that fj (1) = 1 for all j = 1, . . . , m. Otherwise, we replace u(z) with u(z) −∗ zn ·∗ ∂∗n u(1). With pik we denote the multiplicative slot of fj that takes ∂i u∗k as its arguments, i.e, (n−1)×m fj = fj (z, u, p) with z ∈ Cn∗ , u ∈ Cm , j = 1, . . . , m. Because ∗ , p ∈ C∗ ∗ n−1 uj (ζ, 1) = 1, ζ ∈ C∗ , j = 1, . . . , m, we have ∂∗α u(1) = 1 if αn = 0. The multiplicative derivatives ∂∗α u with αn > 0 can be found by multiplicative differentiating equation (7.10). We get ∂∗k ∂∗n uj = ∂∗zk fj +∗

m X ∗q=1

∂∗uq fj ·∗ ∂∗zk uq +∗

n−1 X

m X

∗i=1 ∗q=1

∂∗pi q fj ·∗ ∂∗zi ∂∗zk uq ,

254

The Cauchy–Kovalevskaya Theorem

in general, for α with αn > 0, we have
∂∗α uj = qα ∂∗β fj , ∂∗γ u ,

j = 1, . . . , m,

where qα is a polynomial with multiplicative nonnegative coefficients, depending on ∂∗β fj with |β| ≤ |α| − 1 and ∂∗γ u with |γ| ≤ |α| and γn ≤ αn − 1. As in the previous sections, we can eliminate the terms ∂∗γ u and evaluate the result at 1 to get

∂∗α uj (1) = qα ∂∗β fj (1), ∂∗γ u(1) = Qjα ∂∗β f (1) , j = 1, . . . , m, where Qjα is a polynomial with multiplicative nonnegative coefficients depending on ∂∗β fj with |β| ≤ |α| − 1. Since f is componentwise multiplicative analytic at 1, there exist constants M > 0 and r > 0 such that M

|∂∗α fj (1)|∗ /∗ α!∗ ≤ e r|α|

for

all α

and

all j.

Note that Fj (z, u, p) = eM /∗ (e −∗ (z1 +∗ · · · +∗ zn−∗ 1 +∗ (zn /∗ ρ)  +∗ u∗1 +∗ · · · +∗ u∗m )/∗ r) ·∗ e −∗ (e/∗ r)

 X

pik  −∗ eM ,

∗i,k

where ρ ∈ (1, e] is a constant which will be determined below, majorizes fj at 1, j = 1, . . . , m. Consider the system Uj∗ = Fj (z, U ), (7.11) Uj∗ (1) = 1,

j = 1, . . . , m.

Using the positivity of the coefficients of Qjα , we get
|∂∗α uj (1)|∗ = Qjα ∂∗β f (1) ∗ ≤


Qjα ∂∗β f (1) ∗

≤


Qjα ∂∗β F (1) ∗

=

∂∗α Uj (1),

i.e., Uj majorizes uj at 1, j = 1, . . . , m. Now we will prove that (7.11) has a solution that is multiplicative analytic at 0. Put s = z1 +∗ · · · +∗ zn−1 , t = zn , v

∗ = U1∗ = · · · = Um ,

Partial Differential Equations

255

to get ∂∗t v = eM /∗ (e −∗ ((s +∗ t)/∗ ρ +∗ m ·∗ v)/∗ r)   ·∗ e −∗ (e(n−1)mρ/∗ r ) ·∗ ∂∗s v −∗ eM . We define the new variable σ = t +∗ ρ ·∗ s and assume that v depends only on σ. Then,   (n−1)mρ ∂∗σ v = eM /∗ (e −∗ ((eσ /∗ eρ ) +∗ m ·∗ v)/∗ r) ·∗ e −∗ e r ·∗ ∂∗σ v −∗ eM , whereupon 

e −∗ e

(n−∗ 1)mM ρ r



·∗ ∂∗σ v −∗ e

(n−1)mρ r

2∗

·∗ (∂∗σ v)

(7.12) =e

M e−∗ (σ/∗ ρ+∗ m·∗ v)/∗ r

M

−∗ e .

We choose ρ ∈ (0, 1] so small that ee−∗ e

((n−1)mM ρ)/∗ r

> 1.

Then equation (7.12) can be solved for ∂∗σ v in the power series   σ 2∗  σ ∂∗σ v = ec1 ·∗ e ρ +∗ m ·∗ v +∗ ec2 ·∗ e ρ +∗ m ·∗ v +∗ · · · σ

convergent for some e ρ +∗ m ·∗ v 6= 1, with multiplicative nonnegative coefficients eck . In other words, there is a function g multiplicative analytic at 1, with multiplicative nonnegative Maclaurin series coefficients and with g(1) = 1, such that   σ

∂∗σ v = g ·∗ e ρ +∗ m ·∗ v .

(7.13)

Now, we can apply the Cauchy–Kovalevskaya theorem for multiplicative analytic MDEs from the previous sections of this chapter to conclude that equation (7.13) has a solution v multiplicative analytic at 1, satisfying v(1) = 1, whose Maclaurin series coefficients are multiplicative nonnegative. Hence, the vector function U with components Uj (z) = v ·∗ (ρ ·∗ (z1 +∗ · · · +∗ zn−1 ) +∗ zn ) ,

j = 1, . . . , m,

solves (7.11). Since the Maclaurin series coefficients of v are multiplicative nonnegative, the same holds for Uj implying that Uj majorizes 1 at 1. zn =1

References

[1] Arsenin, V. Ya. Basic Equations and Special Functions of Mathematical Physics, Iliffe, 1968. [2] Asmar, N. Partial Differential Equations with Fourier Series and Boundary Value Problems, Pearson Education, Inc., 2005. [3] Bitsadse, A. Equations of Mathematical Physics, Mir Publishers, Moscow, 1980. (in English). [4] Copson, E. Partial Differential Equations, Cambridge University Press, Cambridge, 1975. [5] Courant, R. Methods of Mathematical Physics, Vol. II, Wiley-Interscience, New York, 1962. [6] Courant, R. Course of Differential and Integral Calculus, Vols. I–III, Pergamon, New York, 1965. [7] Courant, R. Calculus of Variations, New York, 1962. [8] Epstein, B. Partial Differential Equations: An Introduction, Robert E. Krieger Publishing Company, 1975. [9] Evans, L. Partial Differential Equations, AMS, 1997. [10] Folland, G. Introduction to Partial Differential Equations, Princeton University Press, Princeton, 1995. [11] Jeffrey, A. Applied Partial Differential Equations: An Introduction, Academic Press, 2003. [12] John, F. Partial Differential Equations, reprint of the fourth edition, Applied Mathematical Sciences Vol. 1. Springer, Berlin, 1991. [13] Jost, J. Partial Differential Equations, Springer, New York, 2002. [14] Miranda, K. Lectures on Partial Differential Equations, WileyInterscience, New York, 1954. [15] Petrovsky, I. G. Lectures on Partial Differential Equations, Dover Publications Inc, 1991.

257

258

References

[16] Pinchover, Y., J. Rubinstein. An Introduction to Partial Differential Equations, Cambridge University Press, Cambridge, 2005. [17] Rauch, J. Partial Differential Equations, Springer, 1996. [18] Renardy, M., R. Rogers. An Introduction to Partial Differential Equations, Springer, New York, 2004. [19] Vladimirov, V. Equations of Mathematical Physics, Marcel Dekker, Inc., New York, 1971. [20] Webster, A. Partial Differential Equations of Mathematical Physics, Dover Publications, Inc., 1955.

Index

Boundary Condition of the Third Kind, 206

Homogeneous Harmonic Polynomial, 208

Canonical Form of First-Order System, 96 Characteristic Equation, 25 Characteristic Equation of Multiplicative Linear Multiplicative Parabolic Operator, 62 Characteristic Multiplicative Curves of Multiplicative Linear Multiplicative Hyperbolic Differential Operator, 24 Characteristic Multiplicative Curves of Multiplicative Linear Multiplicative Parabolic Operator, 62 Classical Solution, 105 Convergence Radius, 242

Ill-Posed Problem, 1

D’Alambert Formula, 110 Dirichlet Boundary Conditions, 129 Dirichlet Problem, 205 Fundamental Solution of Laplace Equation, 209 Fundamental Solution of the Multiplicative Heat Equation, 176 Green Function, 218 Hadamard Example, 207 Harmonic Function, 205 Harmonic Polynomial, 208 Harnack Theorem, 229

Kirchhoff formula, 135 Liouville Theorem, 228 Multiplicative Accumulation Point, 247 Multiplicative Adjoint Operator of a Multiplicative Hyperbolic Differential Operator, 162 Multiplicative Analytic Function, 244, 251 Multiplicative Backward Wave, 104 Multiplicative Beltrami Differential Equations, 51 Multiplicative Diffusion Equation, 6 Multiplicative Discrete Set, 247 Multiplicative Elliptic First-Order System, 96 Multiplicative Elliptic Multiplicative Linear Multiplicative Differential Operator, 15 Multiplicative Elliptic Multiplicative Quasilinear Differential Operator, 74 Multiplicative Elliptic Second Order Equation, 86 Multiplicative Elliptic-Hyperbolic First-Order System, 96 Multiplicative Forward Wave, 104 Multiplicative Generalized Solution, 105 Multiplicative Heat Equation, 6

259

260 Multiplicative Hyperbolic First-Order System, 96 Multiplicative Hyperbolic Multiplicative Linear Multiplicative Differential Operator, 15 Multiplicative Hyperbolic Multiplicative Quasilinear Differential Operator, 74 Multiplicative Hyperbolic Second Order Equation, 86 Multiplicative Isolated Point, 247 Multiplicative Kolmogorov Equation, 7 Multiplicative Laplace Equation, 5 Multiplicative Linear Differential Operator, 8 Multiplicative Linear Homogeneous Second Order MPDE, 5 Multiplicative Linear Nonhomogeneous Second Order MPDE, 5 Multiplicative Linear Second Order MPDE, 5 Multiplicative Mean-Value Formula for Multiplicative Harmonic Functions for a Multiplicative Sphere, 213 Multiplicative Mean-Value Formula for Multiplicative Harmonic Functions for a Multiplicative Ball, 213 Multiplicative Nonlinear Poisson Equation, 3 Multiplicative Nonlinear Second Order MPDE, 7 Multiplicative Parabolic First-Order System, 96 Multiplicative Parabolic Multiplicative Linear Multiplicative Differential Operator, 15

Index Multiplicative Parabolic Multiplicative Quasilinear Differential Operator, 74 Multiplicative Parabolic Second-Order Equation, 86 Multiplicative Quasilinear Second Order MPDE, 4 Multiplicative Semiilinear Second Order MPDE, 4 Multiplicative Superposition Principle, 2 Multiplicative Ultrahyperbolic Second Order Equation, 86 Multiplicative Wave Equation, 6 Neumann Boundary Conditions, 119 Neumann Problem, 205 Order of Equation, 2 Poisson Equation, 205 Poisson Formula, 142, 224 Power Series, 242 Problem of the Third Kind, 206 Proper Solution, 105 Riemann Function, 164 Robin Boundary Condition, 206 Robin Problem, 206 Second Order Multiplicative Partial Differential Equation, 3 Separation Constant, 196 Solution, 2 Symmetry Property of the Green Function, 219 Taylor Series, 247 The Mean-Value Formula, 188 The Srong Maximum Principle for the Wave Equation, 171 Weak Maximum Principle for the Heat Equation, 169 Well-Posed Problem, 1