K. Sankara Rao

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Introduction

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pastern E.ononY Edition

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PARTIAT

DIFFERENTIAL

EquATr0Ns

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I

0. Partial Differential Equations of First Order

Exerckes 77

2. Elliptic Differential Equations

'.4.t Occunence of the Laplace and Poisson Equations 79 2.1.1 Derivation of Laplace Equation 79

. 2.1.2 Derivation of Poisson Equation 8/ 2.2 Botndary Value Problems (BVPs) :82 .,tl Some Important Mathematical Tools 82

C/ Properties of Harmonic Functions 84 2.4.1 The Spherical Mean 85

v 0 . 1

,,91

w,6.3 wg4

-09

\9.7

y.{

'0.9

\y.1,0

,,xrr

u

t-46

79-146

Introduction I

Surfaces and Normals 2 Curves and their Tangents 4

Formation of Partial Differential Equation 6

Solution of Partial Differential Equations of First Order ,10

Integral Surfaces Passing through a Given Curve ,{Z

The Cauchy Problem for First Order Equations 20

Surfaces Orthogonal to a Given System of Surfaces 21

First Order Non-linear Equations 22

0.9.1 Cauchy Method of Characteristics 2.1

Compatible Systems of First Order Equations 30

Charpit's Method 34

0.ll.l Special Tlpes of First Order Equations 38

Exercises 43

1. Fundamental Concepts

47:78

1.1 Introduction 47

''./2 Classification of Second Order PDE 47

-14 Canonical Forms 48

1.3.1 Canonical Form for Hyperbolic Equation 49 1.3.2 Canonical Form for Parabolic Equation 51 1.3.3 Canonical Form for Elliptic Equation 53 I .4 Adjoint Operators 62

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vt CoNTENTS

2.4.2 Mean Value Theorem for Harmonic Functions g6

,2.4.3 Maximum-Minimum principle and Consequences gZ

4 Separation of Variables 9-l

,;tK Dirichlet Problem for a Rectangle 94

Z1 The Neumann Problem for a Rectanple 92

-ZA Interior Dirichler Problem for a CircL 9g

_29 Exterior Dirichler Problem for a Circle 102

-.f l0 lnterior Neumann Problem for a Circle 106

._V1,1 Solution of Laplace Equation in Cylindrical Coordinates /0g

91 2 Solution of Laplace Equation in Spherical _{Coordinates 1/J}

,2. l3 Miscelianeous Examples /22

Exercises I44

3. Parabolic Differential Equations

_{147_lgg}

3.1 Occurrence of the Diffusion Equation 147 3.2 Boundary Conditions 149

3.3 Elementary Solutions _{of the Diffusion Equarion 150}

3.4 Dirac Delta Function /54

3.5 Separation of Variables Method /j9

3.6 Solution of Diffusion Equation in Cylindrical Coordinares 121

3.7 Solution of Diffusion Equation in Spherical Coordinares _124

3.8 Maximum-Minimum Principle and Consequences _1Zl

1 . 9 M i s c e l l a n e o u s E x a m p l e s / 2 9 Exercises I86

4. Hyperbolic Differential Equations

_{lgg_232}

4.1 Occurrence _{of the Wave Equation 1g9}

4.2 Derivation of One-dimensional _{Wave Equation /g9}

4.3 solution of one-dimensional _{wave Equation by canonical Reduction /92}

4.4 The Initial Value Problem; _{D'Alembert's Solution 196}

4.5 Vibrating Srring-Vadables Separable _{Solution 200}

4.6 Forced Vibrations-Solution of Non_homogeneous _{Equation ?0g}

4.7 Boundary and Initial Value problem for Two-dimensional Wave

Equarions*Merhod of Eigenfunction 2I0

4.8 Periodic Solution of One-dimensional _{Wave Equation in Cylindrical}

Coordinates 213

4.9 Periodic Solution of one-dimensional _{wave Equarion}_{in Sphericar}polar

Coordinares 21J

4.10 Vibration of a Circular Membrane 2/Z

4.ll Uniqueness _{of the Solution for the Wave Equation 2/9}

4.12 Duhamel's Principle 220

4.13 Miscellaneous Examples 222

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I

i CoNTENTS

I

5. Green's

Function

233-21

I t . , l n r r o d u c r i o n z J J

5.2 Green's Function for Laplace Equation 239 5.3 The Methods of Images 245

5.4 The Eigenfuncrion Method 2jZ

5.5 Green's Function for the Wave Equation_Helmh oltz Theorcm 254 5.6 Green's Function for the Diffusion Equation 259

Exercises 263

6. Laplace Transform Methods _Z6S-J1

6.1 Introducrion 265

6.2 Transform of Some Elementary Functions 26g 6.3 Propenies of Laplace Transform 270

6.4 Transform of a Periodic Function 228 6.5 Transform of Error Function 280 6.6 Transform of Bessel's Function 28J 6.7 Transform of Dirac Delta Function 2g5 6.8 Inverse Transform 285

6.9 Convolution Theorem (Faltung Theorem) 292 6.10 Transform of Unit Step Function 296

6.ll Complex Inversion Formula (Mellin-Fourier Integral) 299 6.12 Solution of Ordinary Differential Equations 302

6.13 Solution of Partial Differenrial Equations 302

6.13.1 Solurion of Diffusion Equarion 308 6.13.2 Solution of Wave Equarion j13

6.14 Miscellaneous Examples 321

Exercises 329

7. Fourier Transform Methods

_333-3g

7.1 Introduction 333

7.2 Foti,er Integral Representations -tj.t 7.2.1 Fourier Integral Theorem 335

7.2.2 Sine and Cosine Integral Representations jjg 7.3 Fourier Transform Pairs -139

7.4 Transform of Elementary Functions 340

7.5 Properties of Fourier Trasnform -i4j

7.6 Convolution Theorem (Faltung Theorem) -i56

7.7 Parseval's Relation -t58

7.8 Transform of Dirac Delta Function 3i9 7.9 Multiple Fourier Transforms .li9 7.10 Finite Fourier Transforms 360

7.10.1 Finite Sine Transform i61 7.10.2 Finire Cosine Transform 362

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CONTENTS

7.ll Solution of Diffusion Equation 363 7.12 Solution of Wave Equation 367 7.13 Solution of Laplace Equation 371 7.14 Miscellaneous Examples J73

Exercises 384

Ansvers and Keys to Eterches

Bibliagaphy

Index

388-423

425

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PREFACE

'\'ith the remarkable advances made in various branches of science, engineering and technology,

:rday, more than ever before, the study of partial differential equations has become essential. For,

:.: have an indepth understanding _{of subjects like fluid dynamics and heat transfer, aerodynamics,}

:lasticity, waves, and electromagnetics, the knowledge of finding solutions to partial differential

:quations is absolutely necessary.

This book on Partial Differential Equations is the outcome of a series of lectures delivered

ry me, over several years, to the postgraduate students of Applied Mathematics at Anna

university, chennai. It is written mainly to acquaint the reader with various well-known

:nathematical techniques, _{namely, the variables separable method, integral transform techniques,}

and Green's function approach, so as to solve various boundary value problems involving

parabolic, _{elliptic and hyperbolic partial differenrial equations,}_{which arise in many physical}

situations. _{In fact, the Laplace equation, the heat conduction equation and the wave equation}

have been derived by taking inro account certain physical problems.

The book has been organized in a logical order and the topics are discussed in a systematic

manner In chapter 0, partial differential equations of first order are dealt with. In chapter I, the classification of second order partial differential equations, and their canonical forms are given. The concept of adjoint operators is introduced and illustrated through examples, and Riemann's method of solving cauchy's problem described. chapter 2 deals with elliptic differential

equations. Also, basic mathematical _{tools as well as various DroDerties}_{of harmonic functions are}

discussed. Further, the Dirichlet and Neumann boundary value prtblems are solved using variables

separable method in cartesian, _{cylindrical and spherical coordinate systems. chapter 3 is devoted}

to a discussion _{on the solution of boundary value problems describing the parabolic or diffusion}

equation in various coordinate systems using the variables separable method. Elementary

solutions are also given. Besides, the maximum-minimum principle is discussed, and the concept

of Dirac delta function is introduced along with a few properties. chapter 4 provides a detailed study of the wave equation representing the hyperbolic partial differential equation, and gives

D'Alembert's solution.

In addition, the chapter presents _{problems like vibrating string, vibration of a circular}

membrane, and periodic solutions of wave equation, shows the uniqueness of the solutions, and

illustrates Duhamel's principle. chapter 5 introduces the basic concepts in the construction of

Green's function for various boundary value problems using the eigenfunction method and the method of images. chapter 6 on Laplace transform method is self-contained since the subject matter has been developed from the basic definition. various properties of the transform and inverse transform are described and detailed proofs are given, besides presenting the convolution theorem and complex inversion formula. Further, the Laplace transform methods are applied to solve several initial value, boundary value and initial boundary value problems. Finally in

Chapter 7, the theory of Fourier transform is discussed in detail. Finite Fourier transforms are also

introduced, and their applications to diffusion, wave and Laolace equations have been analvzed.

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CHAPTER O

PARTIAL DIFFERENTIAL

EQUATIONS OF FIRST ORDER

: . 1 I N T R O D U C T I O N

::rial differential equations _{offirst order occur in many practical}_situatrons_{such as Brownian}

motion.

r:3 theofy of stochastic processes, radioactive disintigration, noise in communi.utron ,yrt.,rr,

:rDulation growth and in many problems dealing wiih telephone traffic, traffic flow along a

-.ghway and gas dynamics and so on. In fact, their study is essential to understand the nat-ure

:: solutions _{and forms a guide to find the solutions of hijher order partial differentiar equations.}

A first order partial differentiar equation (usually denoted by pDE) in two independent variabres

,:. _{.r, and one unknown z, also called dependent}

viriable, is an equation of the iorm

o ( r r,,,(.*)=o

\ dx dy)

( 0 .

t )

I n r r o d u c i n g t h e n o l a t i o n O Z d 7

P = ^ ,

_ox

s =

-dy

Equation _{(0.1) can be written in symbolic form as}

F t x , y , z . p , q ) = 0 . _{( 0 3 )}

A solution _{of Eq. (0.1) in some domain O of IR2 is a function z = f(x, y) defined and is of C,, in}

Q should satisfy the following two conditions:

( i ) F o r e v e r y (x , y ) e O, the point (r,y,", p,q) is in the domain of the function F. ( i i ) W h e n z = f(x,y) is substituted i n t o E q . ( 0 . 1 ) , i t s h o u l d r e d u c e t o a n i d e n r i t y i n x , y f o r

all (x, y) e Q.

we classify the PDE of first order depending _{upon the form of the function F. An equation}

of the form

S , ) .

P1x.

_{y. ztli + Qtx. y. zt+ = R\x. y. z)}

ox d y

is a quasi-linear _{PDE of first order, if tr,e derivatives}_{dz/dx and dzr0y tr-tatappear}

in the function

-E are Iihear' while the coefficients p,

e and R depend on the indepe;dent variabres _{x, I,and arso}

on the dependent _{variable z. Similarly, an equation of the form}

( 0 . 2 )

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INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

) , s ,

P t x , _{y t l + Qtx. y\f = R(x.}_{y . z )}

ox dy

( 0 . s )

is called almost linear PDE of first order, if the coefficients _{p and o are functions of the}

inde-pendent variables only. An equation of the fonn

; ? s "

a lx . y t d + b l x . y \ f i + c t x . lt t : = d \ r . i

is cafled a finear PDE of first order, if the function F is linear in 0zl0x, dzldy and z, while the

coefficients a, b, c and d depend only on the independent variables x and y. An equation which

does not fit into any of the above categories is called non-linear. For example,

0z dz

\ t , x 1 + Y . = n z

ox oy

is a linear PDE of first order.

dz dz )

l t t ) x 1 +Y a

=z-ox oy

is an almost linear PDE of first order.

s , 2 ,

( i i i )

P { z ) = + - = 0

ox oy

is a quasi-linear PDE of first order.

, a , Z / . \ z

(iv)

_{l+ I .l ?l =t}

\ d x ) \ d y )

is a non-linear PDE of first order.

Before discussing various methods for finding the solutions of the first order pDEs, we shall

review some of the basic definitions and concepts needed from calculus.

0,2 SURFACES

AND NORMALS

Let O be a domain

in three-dimensional

space

IRr and suppose

F(x,y,z) is a function

in the

class C'(O), then the vector valued function grad F can be wrrrten as

. ^ ( a p a F a F \

g r a o _{f = l . . - " . . - l}

\ o x o y d z )

( 0 . 6 )

If we assume that the partial derivatives of F do not vanish sirnultaneously at any point then the

set of points (x, y, z) in Q, satisfying the equation

F(x, y, z) = (

( 0 . 7 )

( 0 . 8 ₎

is a surface in o for some constant c. This surface denoted by s6 is called a level surface ofF.

lf (,16, y6, z6) is a given point in Q, then by taking F(x6,y|,zi=(, we get an equation bf the

form

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER 3

rlich represents _{a surface in f), passing through the point (r0,J0,zo). Here, Eq. (0.9) represents}

. oe-paraneter family of surface in o. The value of grad F is a vector, normal to the level rrfrce- Now, one may ask, if it is possible to solve Eq. (0.8) for z in terms of ;r and y. To-answer G qnestion, let us consider a set of relations of the fomr

a = f r ( u , v ) , y = f 2 Q t , v ) , 2 = f i ( u , v )

z = f(x, y) F . . = f ( x , y ) - z = 0 .

( 0 . 1 0 )

llere for every pair of values of a and v, we will have three numbers ;r, Jr and z, which represetts r point in space. However, it may be noted that, every point in space need not correspond to a peir u and v. But, if the Jacobian

u!{'r"

_*,

o \u, v) ( 0 . l 1 )

len, the first two equations of (0.10) can be solved and z and y. can be exDressed as functions

ofr and y like

u = f,(x. y), v = p(x, y).

Thus, r and v are obtained once x and y are known, and the third relation of Eq. (0.10) gives the ralue of z in the form

z = fift(x, y), p(x, y)l

( 0 . 1 2 )

This is, of course, a functional relation between the coordinates x, y and z as in Eq. (0.g). Hence, ry point (x, y, z) obtained from Eq. (0.10) always lie on a fixed surface. Equations (0.10) are also called paranetric equations of a surface. It may be noted that the parametric equation of a srface need not be unique, which can be seen in the following example:

The parametric equations

- r = r s r n d c o s d , y = r s i n A s i n d , _{z = r c o s 0}

and

" = , ( 1 - o ' )

" o " e ,

, = r Q - Q 1 - ) " r n r .

, =

r r o ,

( 1 + 0 ' )

( r + 0 " )

t + 0 '

both represent the same surface x2 + y2 + z2 = 12 which is a sphere, where r is a constant. If the equation of the surface is of the form

( 0 . 1 3 )

( 0 . 1 4 )

Then

; Differentiating partially with respect _{to .r and .y, we obtain} d F .0F dz ^ d F d F d z

= - f - - : - - - = v , - + - - = { l

d x .dz dx 0 y 0 z 0 y

from which we get

0z dFldx dF

-- = --:-::T- = -- (usiue 0.14)

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4 o r

INTRODUCTION TO PARTIAL D]FFERENTIAL EOUATIONS

o " = n dx Similarly, we obtain d l . ) E , - = q _{a n d : 1 = _ l} dv ' _{d_}

Hence, the direction cosines of the normal to the surface _{at a point (x, y, z) are given as}

( 0 . 1 5 )

( 0 . l 7 )

where 1is some interval on the real axis. In component _{form, Eq. (0.17) can be written as}

x = I ( 1 ) . y = f 2 G ) , z = f r ( t \

( 0 .

r 8 )

( 0 . l e )

Now, returning to the level surface given by Eq. (0.g), it is easy to write the equation of the tangent plane to the surface ,.t" at a point (rp, y6, z6) as

l a F

I

_{. l a r}

_J

_{l a r}

_l

( x - x 0 ) l

, _

( x 0 . l o . z o )

l + ( l - r o ) l l- r x s . y 6 . z 6 ) l + r r - r s t l f r 1 6 . y n . - - o r l = 0 .

{ 0 . t b r

L a x ) _{, - ,} _s _{L d :} " l

0.3 CURVES AND THEIR TANGENTS

A c u r v e in t h r e e - d i m e n s i o n a l s p a c e I R I c a n b e d e s c r i b e d i n t e r m s o f p a r a m e t r i c e o u a t i o n s

Suppose i denotes _{the position vector of a point on a curve c, then the vector equation ol C mav}

b e w r i h e n a s

v = F l t l t ^ r l c I

where _{i = (r, /, z) and F(t) =[fi(t), f2(), hO] and the funcrions}

fr, f2 and /3 betongs to C,(1). Further, we assume that

W'ryT)*,0'o'0,

This non-vanishing vecror is tangenr _{ro the curve C at the poinr (x, y, z) or at [J;O, f20, f](t))}

of the curve C.

Another way of describing _{a curve in three-dimensional}_{space IRI is by using rhe fact that}

t h e i n t e r s e c l i o n o f t w o s u r f a c e s _{g i v e s ri s e lo a curve.} Let \(x, y, z) = C1 | F2@, y, z) = Crl| \ ( 0.20 ₎

are two surfaces. Their intersecrion. _{ir nor empty. is always a curve, provided grad F, and grad}

F: are not collinear at any poinr of ct in IRj. In other words, the intersection of surfices siven

b 1 E q . ( 0 . 2 0 ) is a c u r v e if

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PARTIAL DIFFERENTIAL EOUATIONS OF FIRST ORDER

etad 4Q, y,z)xglad F2@, y, z) + (0,0,0) (0.2 :;:* every _{@,y,2)eQ. For various}values of C1 and C2, Eq. (0.20) desuibes different curve ::r€ totality ofthese curves is called a two parameter family of curves. Here, c1 and c2 are referr( 15 parameters _{of this family. Thus, if we have two surfaces}_denoted_{by s] and 52 whose}_equatior r: in the form

rJ

F ( x .

y ,

z ) = o l

G(x, y, z) = 0l

( 0 . 2 "

l::en, the equation _{of the tangent plane to,Sl at a point p(xs, ys, z()) is}

) F ) F ) F

( x - x o ) i + ( y - y o t ; + ( z - 2 6 ; = 0

!:nilarly, the equation of the tangent plane to,S2 at the point p(x6,y6,26) is

.

.aG

dG

aG

( . x - x0)=- + (/ - yn)^- +\z - z =-=l).

o x d y - d z

(0.2'

iere, the partial derivatives dFllx, dGldx,etc. are evaluated at p(xx,yx,zs). The intersection r :Fse two tangent planes is the tangent line I at P to the curve C, which is the int3rsection (

::e surfaces _{s' and .92. The equation of the tangent line z to the curue c at (xo, yo,}

"i is obtaine

::r'm Eqs. (0.23) and (0.24) as ( x - , r o )

_ ( y - y o \ _ ( z - z o l

AF-AG aF E

-AF-AI -aF

aG-a y aG-a r - aG-a , aG-a , aG-a" dr-E aG-a, E n- n aG-a,

( 0 . 2 :

(0.2.

( x - x s ) _{( y - y i} _{\ z - z o )}

e@a=a(F,q=Ztr.gl

0 (y, ,) 0 (2, x) 0 (r, y) 'l}rerefore,

the direction cosines of I are proportional to

l a 1 r , c 7 a @ , G )

d ( F , q 1

Lao,d'

aea' a@,r\l

( 0 . 2 (

( 0 . 2 1

aor illustration, let us consider the following examples:

EX4MPLE 0.1 Findlhe tangent vector at (0,1, nl2) ro the helix described by the equation

x = c o s t , - y = s i n r , z = t , 1 € 1 i n l R ' .

Solution _{The tangent vector to the helix at (r, y, :) is}

( dx dv d:\

(12)

7 t

6 nnnoDUcTIoN To PARTTAL DIFFERENTIAI. EQUATIoNS

we observe that the point (0,1, tr/2) corresponds _{to t = ttl2. At this point (0,1, n/2), the tangent} vector to the given helix is (-1,0, l).

EXAMPLE 0,2 Find the equation of the tangent line to the space circle

f + y 2 + 2 2 = 1 , x + y + z = o

0Fl0u d" 1 arlA, a" 1 ^

ELa,' yr y a,la,*

yt l='

(0.2e)

d r l d u . d , l a r l a u d v f

a " L a r * E q

_{I d , L a y + E s}

_{) = u}

at the point

(1/Jt4,

2tJ14,

-3tJ:,4).

Solation The space circle is described as

F ( x , Y, z) = Y2 +Y2 +t2 -l=0 G ( x , Y , z ) = s a _{a Y a 2 = g}

Recalling Eq. (0.25), the equation of rhe tangent _{plane at (l(i4, 2lJA, 4lJw} can be wrirten

x -rilta

_{v -zlJu}

; l " \ 7 \

-,"h-,1+")

,(#)-,(#)

z +3/Jl4

, [ t ) - , f z )

-\J'4, -l.Ji?l

x-rtJ14 y-2/.64 z+3ktl4

0.4 FORMATION OF PARTIAL DIFFERENTTAL EQUATION

Suppose u and y are any two given functions _{of x, y and z. Let F be an arbitrary function of z} and v of the form

F ( u , v ) = 6 _{( 0 . 2 8 )}

we can form a differential equation by eliminating the arbitrary function F. For, we differentiate Eq. (0.28) partially with respect _{to r and l, to get}

and

(13)

PARTIAL DIFFERENTIAL EQUATIONS OF FIRS-I' ORDER

\-.rv. eliminatin 0Fldu and 0Fl0v fton Eqs. (0.29) and (0.30), we obtain

;1ich simpliltes to

fhis is a linear PDE of the t)?e

'* hele du du dv 0v - + - D - + - D d x d z ' d x d z ' 0u 0u dv dv - + - o - + - o d), dz dy d2 d ( u , r ) 0 1 u , v ) 0 \ u , v )

P

_{a o A " q a e $ = a l r , y )}

P P + Q q = R , ^ d \ u . v \ ^ O ( u , v ) ^ O \ u , v \

'

0 0 . 2 ) '

'

a e , . \ ) '

"

d \ x , y )

( 0 . 3 _{I )} ( 0 . 3 2 ) ( 0 . 1 3 ₎

:quation (0.32) is called Lagrange's PDE of first order. The following examples illustrate the idea

:.: formation of PDE.

EXAMPLE 0.3 Form the PDE by eliminating the arbitrary function from (t) z = f (x + it) + g(r - tr), where t=J-1

( 1 i ) f ( x + y + 2 , * 2 + y 2 + t 2 1 = 0 . . Solution

(i) Given z = f Qc +it) + g(x -it)

Differentiating Eq. (1) twice partially with respect to r and ,, we get

( r )

( 1 )

s "

: = f ' l x + i t ) + g ' ( x - i t l ox - ) " . ; = f " ( x + i t ) + s " t r - i t ) . ( 2 )

dx--iere, /' indicates derivative of /with respect to (-r+tt) and g' indicates derivative of g with :.spect to (x-il). Also, we have

) ,

: _{= if ' (x + it) - iC' lx - it)}

dt

- )

o^-

1 =

-1" t* * ,,\ - g" rx - it).

clt-i:om Eqs. (2) and (3), we at once, find that

;hich is the required PDE.

(14)

(ii) The given relation is of the form

Q ( u ' v ) = o ' wherc u = x + y + t, u = 12 + y2 +

"2 Hence, the required PDE is of the form

pp+ eq = R, (Lagrange equation) ( l ) where

l a u A r l

, = a ^ ( u ' ' )

= l a , a v l = l t 2 v l

d \ y . z ) ldu

a v l I

z , l = z f " - , 1

IE EI

I d "

0 r l

,=#3=1fr

_frl=li

_1";,,,-,,

t o x d x l

l?! a'l

p = ! ( r . r \ _ l d *

d x t l t

2 r l

"- a*r)=l+

_{+l=l' zrl=2{t-')}

I dy dy I

Hence, the required PDE is

2 ( z - y ) p + 2 ( x - z ) q = 2 ( y - x) o r

( z - y ) p + ( x - z ) q = y - x .

EXAMPLE 0.4 Eliminate the arbitrary _{function from the following and hence,}_{obtain the corresponding}

partial differeniial equation: 1 i 1 z = x y + f ( x 2 + y 2 )

(ii) z = f(xylz). Solution

( r ) U r v e n z = x y + J \ x - + y ' )

Differentiating _{Eq. (l) partially with respect}_{to .r and J.,, we obtarn}

) -! 1 = y+2xf'\x2 * yr1= p dz a y = x + z Y f ' ( x ' + Y ' ) = q ( l ) (2) (3)

(15)

PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

Eininating /' from Eqs. (2) and (3) we get

W - ) K I = y 2 - x 2 ,

*ich is the required PDE.

(ii) Given 2= f(ry/z)

Differentiating partially Eq. (l) with respect to r and /, we get ) ,

:: = !- f,(xylz) = p o x z

2 .

+ =: I,kytz\=

_{oy z}

s

Etuinating /' from Eqs. (2) and (3), we find x p - v q = o

J

( 1 )

w=(]v

(4)

lich is the required PDE.

E/IMPLE 0.5 Form the partial differential equation by eliminating the constants from

z = e + b y + a b .

Solution Given z = ax+ by + ab

Differentiating Eq. (l) partially witi respect to.x and / we obtain dz

- - = a = p ox dz

a v = ' =

q

9$stituting p and q for a and b in Eq. (l), we get the requifed PDE as z = p x + q y + p q

EGMPLE 0,6 Find the partial differential equation of the family of planes, the sum of whose ! r'. ? intercepts is equal to unity.

Solution Let * +1, +1=l be the equation ofthe plane in intercept form, s o rhal a+b+c=1.

a D c Thus, we have x y z -- + -- + -- = l a b l - a - b (4) ( 2 ) (3) ( l ) (2) (3) ( l )

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7

IO NTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

Differentiating Eq. (l) with respect to x and y, we have

l * _P = n p I a l - a - b _{l - r _ b = - ;} and I ' _Q _{- n} _q ₁ b t - a - b - " " ' _{t _ " 4 = -} _b

From Eqs. (2) and (3), we get

! = !

q a

Also, from Eqs. (2) and (4), we ger

p a = a + b - 1 = a + ! a - l q ( 2 ) ( 3 ) ( 4 )

a l t + L - p l = 1 .

\ s )

Therefore, a = q / ( p + q _ p q ) Sirnilarly, from Eqs. (3) and (4), we find

b = p t ( p + q _ p q )

Substituting _{the values of a and 6 from Eqs. (5) and (6) respectively}_{to Eq. (r), we have}

p + q _ p q x + p + q _ p q _{, * P + Q _} P Q _{z = l} q p _pq or ! + l - ' = | q p p q p + q - p q That is, P x + q Y - 2 = - - - ! ! - , 0 ) p + q - p q

which is the required PDE.

0.5 SOLUTION OF PARTIAL DIFFERENTTAL EOUATIONS _{OF FIRST ORDER} In Section 0.4, we have observed that relations of the form

F(x, y, z, a, b) = 0 (0.34)

( 5 )

(17)

. . - :ise to PDE of first order of the form

f ( x , y , z , p , q ) = 9 .

I t

( 0 . 3 _{5 )}

- -,:. _{any relation of the form (0.34) containing}

two arbitrary constants a and 6 is a solution of

'. ?DE of the form (0.35) and is called a complete solurion or complete integral. fonsider a first order PDE of the form

) , ) , P \ x . _{y . z ) - + Q(x,}y , z)"+ = R\x. y . z l ox dy ' r . n p l y ( 0 . 3 6 ) P p + Q q = R , ( 0 . 3 _{7 )}

"-::. -r ?rd.;,, are independent variables. The solution of Eq. (0.37) is a surface S lying in the

...--)-space, called an integral surface. If we are given that z=-f(x,y) is an integral surface

.' ::: PDE (0.37). Then, the normal to this surface will have direction cosines proportional to

: - : x . dzldy, -l) or (p, g,-l). Therefore, t h e d i r e c t i o n o f t h e n o r m a l i s g i v e n b y i = \ p , q , - 1 \ .

:'.': the PDE (0.37), we observe that the normal ii is perpendicular to the direction defined by

- : .:ctor / =\P, Q, R) (see Fig. 0.1).

Fig. 0.1 Integral surlace z=f(x,y).

--=:efore,

any integral surface must be tangential to a vector with components _{lP, Q, Rj, and}

-:-:e. we will never leave the integral surface or solutions surface. Also, the total differential d: - : r e n b y

a , = ( a , * ( a y

dx dy

: - : r Eqs. (0.37) and (0.38), we find

lP, Q, Rl =ldx, dy, dz\1

'.:.i. rhe soiution to Eq. (0.37) can be obtained using the following theorem:

( 0 . 3 8 )

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12 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

Theorem 0.I The general solution ofthe linear pDE P P + Q q = R

can be written in the form F(u,v\=Q, where I'is an arbitrary function, and u(x,y,z)=Ct and v(x, y, z) = C2 form a solution of the equation

d t _ d y

p(x, y, z) e@, y, _") R(x, y, z)

Proof We observe that Eq. (0.a0) consists of a set of two independent ordinary equations, that is, a two parameter family of curves in space, one such set can be

dv OG. v. z\

a=;G;;

which is referred to as "characferistic crrrve". In quasiJinear until z(r,/) is known. Recalling Eqs. (0.37) and (0.38), nomllon as dz

L

lax the

o l t a z D x \ ( R \

- l l t = l I _{t 0 . 4 2 )} dy)\dz/dy) _\dz)

integral surface. For the existence of finite solutions of

( 0 . 4 0 ) differential written as

(0.4

_{r )}

case, Eq. (0.41) cannot be evaluated

we may recast them using matrix

o l

; t = 0

a y l

( 0 . 4 3 ₎

Both the equations must hold on

Eq. (0.42), we must have

D / 1 1 | D

a x _{a y l ldx}

R I IR

dzl ldz

on expanding the determinants, we have

dx dy

which are called auxiliary ln order to complete

generated by the integral

Let

P (x, y, z) Q@, y, z) 99!4j!ens for a given PDE.

the proof of the theorem, curves of Eq. (0.44) has an

dz R(x, y, z)

we have yet to

equation of the

(0.44)

show that any surface form F(r.i, v) = 0.

u ( x , y , z ) = ( , a n d v ( x , y , z ) = C , ( 0 . 4 5 ) be two indeperident integrals of the ordinary differential equations (0.44). If Eqs. (0.45) satisfy Eq. (0.44). then. we have

du du du ; d x + = - d y + ; - d z = d u = 0 ox dy dz d v , 0 v d v . a x + - d y + - & = d v = U . ox dy dz and

(19)

Solving these equations, we find

dx dv

_;_-____;_____=_

==_j...-du dv ==_j...-du dv 0u dv ==_j...-dufr

at at- a" a, E Ar- arE

dz du 0v du 0v'

a, ay- ay dr

l dx dy dz ---;---i = ---t---- = --i-:---:-o \ u , v ) d \ u , v ' t d \ u , v l 0 ( y , z ) d ( z , x ) 0 ( r , y )

:iow, we may recall from Section 0.4 that the relation F(u, $ = A, where F is an arbitrary function, bds to the partial differentiA',equation

d(u.v\ A(u$ _ [email protected])

P ; . , + q

-'

d(y, z) ' d(2, x) d(x, y) ( 0 . 4 7 )

Br virtue of Eqs. (0.37) and (0.47), Eq. (0.46) can be written as d x

= d y _ d z P Q R

T b e s o l u t i o n o f t h e S e e q u a t i o n s _{a r e k n o w n t o b e u ( x , y , z ) - _ C 1 and v(x,y,z)=C2. Hence,}

f(2. v) = g is the required solution of Eq. (0.37), if u and v are given by Eq. (0.a5), We shall illustrate this method through following examples:

EX4MPLE 0.7 Find the general integal of the following linear partial differential equations: ( i ) y 2 p - x y q = y ( 2 - 2y1

( i i ) ( y + z x ) p - ( x + y z ) q = 7 s 2 - 12. Solulion

(i) The integral surface of the given PDE is generated by the integral curves ofthe auxiliary eouatlon

dx dy dz

y' -ry x(z -2y)

fhe first two members of the above equation give us

: = = o r x d x = - y d y ,

rhich on integration results in

_2

t 3

rfiich can be rewritten as

( 0 . 4 6 )

( 1 )

- . 2

L + C _{o r x ' + v ' = C ,}

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( t )

( 2 )

l 4 The

INTRODUCTION _{TO PARTIAL}_DIFFERENTIAL_EQUATIONS

last two members of Eq. (l) give d Y = t u ^

o r z d y - 2 y d y = _ y d z - y z - z y

That is,

2 y d y = y d z + z d y ,

which on integration yields

y2 = yz +C2 or y2 - yz =C2 (J)

Hence, the curves given by Eqs. (2) and (3) generate _{the required integral surface as}

F \ x z + y 2 , y 2 - l z ) = 0 .

(ii) The integral surface _{ofthe given pDE is generated}_{by the integral}_{curves ofthe auxiliary}

equation

d x _ d y dz

y + z x - ( x + _{l z )} _{* 2 - y ,}

To get the first integral curve, let us consider the first combination as

x d x + y d y d z

;;;7 _;;fr=? _

yz

or x d x + y d y _ d z That is, " (r2 - yz) ,2 - y2 x d x + y d y = 2 f u . On integration, we get t2 ,'2 ,2 _ + " 2 V - t = L o r x ' + Y ' - z ' = C t

Similarly, _{for getting the second integral curve, let us consider the combination such as}

y d x + x d y _{d z}

7;;;7__y=7_t

o r y d x + x d y + d z = 0 , w h i c h o n i n t e g r a t i o n r e s u l t s i n x Y + z = C 2

Thus, the curves given by Eqs. (2) and (3) generate _{the required integral surface as}

F ( x 2 + y 2 - z 2 , x y + z ) = g .

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a dz i_ox EGIMPLE 0.8 Use PARTIAL DIFFERENTIAL Lagrange's method to EeuATioNS oF FrRsr oRDER

solve the equation

,'T

I

f

r l - ^

1 l o z l - l I

o y l

l 5 t h e r e z = z ( x , y ) . Solution The

The corresponding auxiliary. equations are

dx

Ihrefore,

rhich on integration yields

Similarly, using multipliers a,

given PDE can be written as

t ^ a z l I a , 1

| d , o z f

x l - p - y ; - l - y l - a - y - ^ _{l + z l d = - - l t - l = t )} L oyJ L dxJ L dy dx) 2 . '

-( y y - Pz\1i+@z-ygi!= Bx-ay

( t )

dx dy dy _{( 2 )} ( y y - Fz) (az-yx) (fx-ay) Lsing multipliers x, y, and z we find that each fraction is

_ x d x + y d y + z d z 0

x d x + y d y + z d z = 0 ,

, 2 + y 2 + " 2 = c ,

f , nd y, we find from Eq. (2) that a d x + B d y + y d z = 0 ,

riich on integration gives

a x + p y + y z = C z Thus, the general solution of the given equation is found to be

'

F ( t z + y 2 + 2 2 , a x + B y + y z ) = 0

EXAMPLE 0.9 Find the general integrals of the following linear pDEs: ( i ) pz-qz=22 +(x+y)2

each fraction is equal to

( 3 )

( 4 )

(22)

t 6

Solution

(i) The integral equation

surface of the given PDE is generated by the integral curves of the auxiliary

The first two members of Eq.

Now, considering Eq. (2) and the first

o r dx dy dz z _ z z2 +(x+ y)2 ( 1 ) g i v e d r + d Y - 9 , , 2 - * y x + Y = C l

and last members of Eq. (l), we obtain

, z d z

z ' + c i

2z dz z ' + C ; l n ( 2 2 + C l ) = 2 x + C 2 ( l ) ( 2 )

O I

l n l z " + ( x + y ) " ) - 2 x = C ,

Thus, the curves given by Eqs. (2) and (3) generates the integral surface for the given PDE ( 3 )

as

Equation (1) can be rewritten as

d x - d y _{_} d y - d z _{_}

by the integral

dz

d z - d x

curves of the auxiliary

( l )

( 2 )

( x - y ) ( x + y + z ) ( y - z ) ( x + y + z ) ( z - x ) ( x + y + z )

Considering the trst two terms of Eq. (2) and integrating, we get ln (r - ]') = ln (),- z) + ln C'r

O I

- v l

F(x + y, log \x2 + y2 + 22 + 2xyj -2x) = 0 (ii) The integral surface of the given PDE is given

equation

* _{= 0 ,} , z - t n y 2 - " ,

(23)

S.nilarly, considering the last two terms of Eq. (2) and integrating, we obtain

fi=c,

rus, the integral curves given by Eqs. (3) and (4) generate the integral surface / \

-

( y - ' ' , - x ) " '

3,6 INTEGRAL SURFACES PASSING THROUGH A GTVEN CURVE the previous section, we have seen how a general solution for a given

:rained. Now, we shall make use of this general solution to find an integral

. :iven curve as explained below:

Suppose, we have obtained two integral curves described by

u(x, Y, z) = (t ) t t v(x, Y, z) = C, ) l 7 l i n e a r P D E . c a n b e s u r f a c e c o n t a i n i n g ( 0 . 4 8 ) can be written

(0.4e)

C described by ( 0 . 5 _{0 )}

( 0 . 5

r )

obtain a relation ( 4 )

' rm the auxiliary equations of a given PDE. Then, the solution of the given PDE

- the form

F ( u , v ) = Q

Suppose, we wish to determine an integral surface, containing a given curve

-.3 parametric equations of the form

x = x ( t ) , y = y ( t ) , _{z = z ( t ) ,}

.:.ere 1is a parameter. Then, the particular solution (0.48) must be like

u{x(t),

y(t),

r0)} = cr

_I

I vIx(t), y(t), z(t) = C2)

:.us, we have two relations, from which we can eliminate the parameter / to

: : r h e t y p e

.:ich leads to the solution : uple of examples. t.Y4MPLE 0.10 Find t\e

- : n t a i n i r r g t h e s t r a i g h t l i n e x + y = g , s = 1 .

Solution The auxiliary equations for the given

___1

x \ y 2 + z ) - y t x z - : 1

F ( C . , C ) = 0 ( 0 5 2 )

g i v e n b y E q . ( 0 . a 9 ) . F o r i l l u s t r a t i o n , l e t u s c o n s i d e r t h e f o l l o w i n g

integral surface of the linear PDE

x ( y 2 + z ) p - y ( x 2 + z ) q = 1 x 2 - y 2 1 z

PDE are dz (r' - yt),

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18 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUAIIONS

Using the multiplier xyz, we have

y z d x + z x d y + x y d z = 0 . On integration, we get

xtz = Ct Q)

Suppose, we use the multipliers x, y and z. Then find that each fraction in Eq. (l) is equal to

x d x + y d y + z d z = 0 ,

* 2 + y 2 + " 2 = C , ( 3 )

For the curve in question, we have the equations in parametric form as

x = t , y = - t , z = l

Substituting these values in Eqs. (2) and.(3), we obtain

- t 2 = C , )

_{' f}

( 4 )

zt2

+t = Cz)

Eliminating

the parameter

t' we find

1_zcr=c2

or

2 C y + C 2 - l = 0 Hence, the required integral surface is

t 2 + y 2 + 1 2 + 2 r y 2 - 1 = 0 .

EXAMPLE 0.ll Find the integral surface of the linear PDE

which contains the circle defined by

x . P + Y q = z

t 2 + y 2 + 1 2 = 4 , x + y + z = 2 .

Solution The integral surface of the given PDE is generated by the integral curves of the

auxiliary equation

d x _dy =dz x y z

( l )

I Integration of the first two members of Eq. (l)gives

l n .r = l n / + I n C

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PARTIAL D]FFERENTIAL EQUATIONS OF FIRST ORDER

i:milarly, integration of the last t\ryo members of Eq. (l) yields

v

, = C z

z ::3nce, the integral surface of the given pDE rs

o[:, zl=o

_s)

\ y z )

-: this integral surface also contains the given circle, then we have to find a relation betwaen : 1' and ylz.

The equation of the circle is

t 9

( 3 )

* 2 + y 2 + " 2 = 4 x + y + z = 2 i:om Eqs. (2) and (3), we have

! = x l C v z = y l C r = a 1 g r g ,

S.rbstituting _{these values ofy and: in Eqs. (5) and (6), we find}

z x 2 t 2 " ( r r )

x ' + - + - : -

_{, = 4 , o I . * 2 l l + - ! + - - l "}

_{l = a}

L l L r L 2 | C i C i C ; ) x x ( r t \ x + - + - = 2 . _{o r x l l + ' +} ' _{l = 2}

q C r c z

^ 1 " c , - c E ) - '

i:om Eqs. (7) and (8) we observe

1 * { * -l -=f,t t l'

ci cici l. cr crc2J'

-hich on simplification gives us

) 1 1

_ + _ + - - = 0

Cr CrCz c r'C,

l:rat is,

C r C 2 r C t + t = 0 .

\ow, repfacing Cl by xly and C2 by ylz, we get ihe required integral surface as

x v x - - + _ + t = 0 , y z y

(s)

(6) ( 7 ) ( 8 )

(e)

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20 I N T R O D U C T I O N T O P A R T I A L D I F F E R E N T I A L E O U A T I O N S

z y

Y l , r w ' t l ' ? - n

0.7 THE CAUCHY PROBLEM FOR FIRST ORDER EQUATIONS

Consider an interval 1on the real line. If xo(s), y6(s) and zs(s) are three arbitrary functions of

a single variable se1 such that they are continuous in the interval lwith their first derivatives.

Then, the Cauchy problem for a first order PDE of the form

F(x, y, z, p, q) = Q ( 0 . 5 3 ₎

is to find a region IR in (r, y), i.e. the space containing (xo(s),-},o(s)) for all s e 1, and a solution

z = dQ, y) of the PDE (0.53) such that

Z [ x 6 ( s ) , y 6 ( s ) ] = Z o ( 5 )

and QQ,y) together with its partial derivatives with respecl _{to n and ), are continuous}functions

of x and y in the region IR.

Geometrically, there exists a surface z=QG,y) which passes through the curve f, called

d a l u m c u r v e . w h o s e p a r a m e t r i c e q u a l i o n s a r e

x = , x 6 ( s ) , y = y o ( s ) , z = z s ( s )

and at every point of which the direction (p, q,-1) of the normal is such that F(x, Y, z, P, q) = Q

This is only one form of the problem of Cauchy.

In order to prove the existence of a solution of Eq. (0.53) containing the curve f, we have

to make further assumptions about the lorm of the function F and the nature of f. Based on these

assumptions, we have a whole class of existence theorems which is beyond the scope of this

book. However, we shall quote one form of the existence theorem without proof, which is due

to Kowalewski (see Senddon, 1986). T h e o r e m 0 . 2 I f

(i) SCy) and all of its derivatives are continuous for y y6l<d,

( i i ) x 6 i s a g i v e n n u m b e r _{a n d ; 6 = g ( y 6 ) , q 0 = C ' O o ) a n d f(x,y,z,q) and all of its partial}

derivatives are continuous in a region S defined by

l x x p < d , y - y o l < 5 , c l - a o l < 3 , then, there exists a unique function /(x,y) such that

(.i) d!,y) and all of its partial derivatives are continuous in a region IR defined by

(27)

(ii) For all (.r, y) in IR, z = Q$, y) is a solution of the equation

F(x, y, z)= (

z = d @ , y )

- rhe surface which cuts each of the given system orthogonally (see Fig. 0.2).

21

o z .( d z \ - a = J l x . y , z , = - l _{a n d}

o x _\ _{d y )}

(iii) For all values of y in the interval _{ll _ tsl < 6b Leo, y) = S0).}

0.8 SURFACES ORTHOGONAL _{TO A GIVEN SYSTEM OF SURFACES}

one of the useful applications _{of the theory of linear first order pDE is to find the sysrem of} surfaces orthogonal _{to a given system of surfaces.}_{Let a one-parameter}_{famiry of surfaces}_is described by the equation

Then, the task is to determine _{the system of surfaces}_{which cut each of the given surfaces} onhogonally..Let _{(x, y, z\ be a point on the surface}_{given by Eq. (0.54), where thJnormal to the} surface _{will have direction ratios ()Fldx,7F/dy, drn4 *ni.n ,nuy be denoted by p,}

e, R. ( 0 . 5 4 )

(0.5s)

( 0 . 5 6 )

Fig.0.2 Orthogonal surtace to a given system ot surfaces.

l::n, its normal atlhe point (x, y, z.) will have direction ratios (dz/dx, dz/O.y, _l) which, of course,

':'r be perpendicurar to the normar _{to the surfaces}_{characterized}_{by Eq. (0.54). As u.onr.qu.n..}

'.: have a relation

r(*e!-n=o

ox dy

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22 rNrRoDUcrroN ro PARIAL DTFFERENnAL EQUAnoNs

I which is a linear PDE of Lagranges type, and can be recast into

I

#x.Hx=#

rosol

_I

Thus, any solution of the linear first order PDE of the type given by either Eq. (0.57) or (0'58) |

is o*hogonat to every surface of the system described by Eq. (0.5a). In other words, the surfaces I

orthogonal to the system (0.54) are the surfaces genelated by the integral curves of the auxiliary | equations

rl

I

#^=h=#""

(o5e)

o.e FrBSr

oRDEB

NoN-LINEAR

EouArloNs

I

In rhis section, we will discuss the problem of finding the solution of first order non-linear partial I

differentiar

equations

(PDEs)

in."";fl::,T;:T:

(0.60t1

where

,=*, ,=fr

|

We also assume that the function possesses continuous second order derivatives with respect tol its arguments over a domain Q of (x,y,z, p, q)-space, and either Fp or Fq is not zero at every _I

Pont

such

that

.l

't*?r

r,

_I

I

Fig.0.3 cone of normals to the integral surtace

l

The PDE (0.60) establishes the fact that at every point (x; y, z) of the region, there exists a I

relation between the numbers _{p and q such that 0@;q)=0' which defines the direction of the I}

normal fi = \p, Q, - l) to the desired integral surface z = z(x, y) of Eq. (0.60). Thus, the direction I

of the normal to the desired integral surface at certain point (x, y, :) is not defined uniquely. i

However, a certain cone ofadmissable directions ofthe normals exist satisling the relation ((p,q)=0:

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PARTIAL DJFFERENTIAL EQUATIONS OF FIRST ORDER 23

-. herefore, the problem of finding the solution of Eq. (0.60) reduces to finding an integral

':':.e z - z(x, y). the normals at every point of which are directed along one of the permissible - -:-:ions of the cone of normals at that point.

Ihus, the integral or the solution of Eq. (0.60) essentially depends on two arbitrary constants

: fonn

J G , y , z , a , b ) = 0 , ( 0 . 6 1 )

:r is called a complete integral. Hence, we get a two-parameter family of integral surfaces

- .gh the same Polnt.

- -q,1 Cauchy's Method of Characteristics

: ntegral surface z=z(x,y) ofEq. (0.60) that passes through _{a given curye ro=x6(s), )0=,)'0(r),}

: -:x(s) may be visualized as consisting of points lying on a certain one-parameter famlly 01'

. : s r = x ( l , s ) , y = ) , ( / , s ) , z = z ( t , s ) , w h e r e s i s a p a r a m e t e r o f t h e f a m i l y c a l l e d c l r a r a c t e r i s t i c s .

:lere, we shall discuss the Cauchy's method for solving Eq. (0.60), which is based on geometrical

. ieralions. Let z=z(x,y) represents an integral surface S of Eq. (0.60) in (r, y.:)-space.

. - . . , , p , q , - l \ a r e th e d i r e c t i o n r a t i o s o f t h e n o r m a l t o S . N o w , th e d i f f e r e n t i a l e q u a t i o n ( 0 . 6 0 ) ' . i s that at a given point P(xx,y6,z6) on S, the relationship between p0 and 90, tllat

. rr.1 _{. _r,s,}_{:6 . po , qo ), need not be necessarily}linear. Hence, all the tangent planes to possible

:ral surfaces through P form a family of planes enveloping a conical surface called Monge

:- \\'ith P as its vertex. In other words, the problem of solvlng the PDE (0.60) is to find

. ,:es which touch the Monge cone at each point along a generator. F o r e x a m p l e . l e t u s

. i e r the non-linear P D E

P 2 - q 2 = 1 '

.,ery point of the r1,;-space, the relation (0.62) can be expressed parametrically as

p = c o s h p , q = s i n h l , - * < p < *

.-:. the equation of the tangent planes at (ro,_yo,"o) can be written as

( x - x 0 ) c o s h p + ( y - y s ) s i n h 4 - ( z - : . ) = Q

.,rr0..1,6,:0) be the vertex and QQ,y,z) be any point on the generator. Then, the direction

. of the generator are (x x6), (,r, ya),G-zo). Now, the direction ratios of the axis of the

: \ \ h i c h is p a r a l l e l t o x - a x i s a r e (1 , 0 , 0 ) ( s e e F i g . 0 . 4 ) . L e t t h e s e m i - v e r t i c l e a n g l e o f t h e c o n e - l. Then. 7r ,1 ( 0 . 6 2 ) ( 0 . 6 3 ₎ ( 0.61) ( r - x u ) l + { . } -) u ) 0 + ( z - z u i 0 l .J2 J t . - * 6 t 2 + ( r - ' t , 6 1 2 + 1 : - 2 , , 1 2

(.r rx )2 + (-t, y0)2 + (z ziz = 2 (,t .ro )2

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24 _{IN'TRODUCTION TO PARTIAL DII.FERENTIAL EQUATIONS}

Thus, we see that the Monge cone of the pDE (0.62) is given by Eq. (0.65). This is a right

circular cone with semi-vertical _{angel 4 whose axis is the straight line passing through (..b, _yo,}_zo)

and parallel to t-axis.

Fig,0.4 Monge cone.

since an integral surface is touched by a Monge cone along its generator, we must have a method to determine the generator of the Monge cone of the pDE (0.60) which is explained below:

It nray be noted that the equation of the tangent plane to the integral surface z = z (x, ),) at the

point (,re,1:e,;s) is given by _\,

p ( x - x o ) + q ( y - y o ) = k - z _.

Now, the given nonJinear PDE (0.60) can be recasted into an equivalent fonn as q = q ( x o , y o , z o , p )

indicating _{that p and ? are not independent}_{at (xo,yo,zo). At each point of the surface s, there}

exists a Monge cone which touches the surface along the generator of the cone. The lines of contact between the tangent planes of the integral surface and the corresponding cones, that is

the generators _{along which the surface is touched, define a direction field on the surface S. These}

directions are called the characteristic _{directions, also called Monge directions on S and lie along}

the generators _{of the Monge cone. The integral cunr'es of this field of directions on the intesral}

surface .l define a family of curves called characteristic curves as shown in Fig. 0.5. The Monge cone can be obtained by eliminating p from the followins equations:

<----+---+---

<---+--+--+--

<----+---+---) ( 0 . 6 6 ) ( 0 . 6 7 ) l(.\,,. r,,. --r )

(31)

p(x - xi + q@0, _{yo, zo, p) (y - yo) = Q - zo)}

25 ( 0 . 6 _{8 )} (0.6e) (0.70) ( 0 . 7 1 ) anl

o-fserving that 4 is a function

.1-( r - x 0 ) + .1-( ) / - ) . , 0 ) : 1 = 0 . ap of p and differentiating Eq.

a.r4 df dI< da

- , = - ; - + = - - = u .

ap dp dq dp

(0.60) with respect to p, we

\.-'\\i eliminating (dqldp) from Eqs. (0.69) and (0.70), we obtain

dF dF (x - x6)

a p - A vyi='

= l r - r n l - l n F F ' p ' ( l

l*:.erefore, the equations describing _{the Monge cone are given by} q = q(xo, _Yo,zo, p), (x - xs)p+ (y - ysl q = Q - zs)

(0.72)

I - - X n l - 9 n ' P - q

l:: second and third of Eqs. (0.72) define the generator of the Monge cone. Solving them for : - r o ) ( y - j l o ) a n d ( z - z o ) , w e g e t

x - x n l - V n z - z ^

Fp Fq pFo + qFn

F :alfy, replacing (x- xo),Q - yi and (z - zs) by dx, dy and dz respectively, which corresponds r infinitesimal movement from (ro,ys,z6) along the generator, Eq. (0.73) becomes

dx dy dz

Fp Fq pF, + qFo'

D:roting the ratios in Eq. (0.7a) by dt, we observe that the characteristic curves on s can be c,::ained by solving the ordinary differential equations

4*= Fo{x,

_t,

(32)

I

dz

dt

26 and

"! = F"|x, y, z(x, y'), p(x, y),q(x. y)l.

at

Also, we note that

Therefore, 0z dx dz dv dx dv = - ; - - + ; - - = _{p _ + q} -dx at dy aI at aI d: - = D t - + a f ^ dt

Along the characteristic curve, p is a function of r, so that dD dp dx 0p dv

a=;,a.6,a

Now, using Eqs. (0.75) and (0.76), the above equation becomes dp 0o 0F dp dF

- - . . - = - : - + 3 - _

dt 0x 0p dy dq

Since z,, =zyic or Py =q,, we have

dD dD dI1 d0 dr

: = +

-dt 0x dp dx dq

Similatly, we can show that

( 0 . 7 6 )

( 0 . 7 ? ₎

Arso,

differentiating

Eq.

(0.60);;^

;Tec,;:

j"

)irr,

: - + - : - P + ^ - : * - : - - = U

d x d z ' d p d x d q d x

Using Eq. (0.79), Eq' (0.78) becomes

: = -\r\ + prz)

_ : a = - ( F v + q F : )

AT

Thus, given an integral surface, we have shown that there exists a family of characterislic curves along which x, !, z, p and g vary according to Eqs. (0.75), (0.76), (0'77)' (0'80) and (0.81). Collecting these results together, we may write

( 0 . 7 8 )

( 0 . 7 9 )

( 0 . 8 0 )

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2 7

PARTIAL DIFFERENTIAL EOUATIONS OF FIRST ORDER

dx dt dz - = _{P t a} D + q r o , dt = = - ( f , + p F , t a n d A T = = - ( F , + q F - ) . d l

:.:: equations are known as characteristic equations of the given PDE (0.60). The last three

:'- ...:rons of(0.82) are also called compatibility conditions. Without knowing the solution z = z(x' y)

c' r: PDE (0.60), it is possible to find the functions x(t),y(t),2(t), p(t),q(r) from Eqs. (0.82)

:--: rs. we can find the curves r=r(t),/= y(t),2 -- z(t) called characteristics and at each point

,: ' - :haracteristic, we can find the numbers p = p(t) and q = q(t) that determine the direction ol

p ( X - x ) + q ( Y - y)= (Z - z). ( 0 . 8 3 )

T-, rharacteristics, together with the plane (0.83) refened to each of its points is called a

c - : : ' : : i e r i s t i c s t r i p . T h e s o l u t i o n x = x ( / ) , / = y ( t ) , 2 = z ( t ) , p = p ( t ) , q = S Q ) o f t h e c h a r a c t e r i s t i c .:-,:::Jns (0.82) satisfy the strip condition

F

dy (tt

( 0 . 8 2 )

fi

= rr,t!,

* a,lL

(0

84)

[ -.. b. noted that not every set of ltve functions can be interpreted as a strip. A strip should

l. . rhat the planes with normals (P. q. -l) be tangential to the charactertstic curve. That is' thel

h .rrisfy the strip condition (0.84) and the normals should vary continuously along the curve.

[ .r rmportant consequence of the Cauchy's method of characteristic is stated in the following

F .

- - .

h= . r..- 0.3 AJong every strip (characteristic strip) of the PDE: F(.r. .1 . --. p. g) = 0. the function

l " . . : . P ' g l i s c o n s t a n t

I eroo| Along the characteristic strip, we have

|

,,

_{0F dx . dF (b. aF (t: +rydp *dj ,tq}

|

; r t y ( ' ) , ) . ( r ) . 2 ( r ) . p ( r ) . q a \ r = ; ; + i i * * A - a o i - A , t ,

t using the results listed in Eq. (0.82), the righrhand side of the above equation becomes T

I 4 F p + F , F q + F , ( p F r + q F o ) - F o ( F , + p F , ) - F o ( F r + q F , ) = 0 .

| '... tt'" function .F(x, y, z, p, 4) is constant along the strip of the characteristic equations of F

- r e

d e f i n e d b y E q . ( 0 . 6 0 ) .

| . .: illusrration. we consider the following examples:

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EXAMPLE 0,12 Find the characteristics ofthe equation pg = z and determine the integral surface

which passes through the straight line x=l,z= y.

Solution If the initial data curve is given in parametric form as

,o(r) = I, yo(s) = s, z6(s) = s,

then ordinarily the solution is sought in parametric form as : = x(r, s), y = y(t, s), z = z(t, s). Thus, using the given data, the differential equation becomes

P 6 ( s ) q 6 ( s ) - s = 0 = r ' and the strip condition gives

I = pe(0) + q6(l) or Qo = l. Therefore,

qo =1, _{Po = r} (unique initial striP)

Now, the characteristic equations for the given PDE are d x d y d z ^ d D d s

A = q ' A = e ' a =

z P q '

; = e ' ; = q

On integration, we get

p = cl exp (r), q = c2 exp (t), x = c2 exp (r) + ca I

y=cl exp(t)+c4, z =2cpz exp (2t) + cs _'J

Now taking into account the initial conditions

x o = 1 , y o = J r z o = s , p o = s , q o = l

we can determine the constants of integration and obtain (since c2 = l, ct = 0) P = s e x P ( t ) , 4 = e x P ( t ) , x = e x P ( ) l

. f

/ = s e x p ( / ) , z = s e x p ( Z t ) j Consequently, the required integral surface is obtained from Eq. (7) as

. ^ t

-EXAMPLE T.lJ Find the characteristics ofthe equation pq=z and hence, determine the integral

surface which passes through the parabola x=0, y'=2.

Solution The initial data curve is

r o ( s ) = 0 , /o(s)=s, zo(s)=s'. Using this data, the given PDE becomes

p o ( s ) q o ( s ) - s" =0=F _{6 )} ( l ) ( 2 ) ( 3 ) (4) ( 5 ) (6) ( 7 )

(35)

The strip condition gives

2s= ps(o)+qs(l)

_{or qs_2s=o}

₍₂₎

Therefore,

Qo = 2s and po = zolQo = ,212, = t B\

2 ' - ,

Now, the characteristic equations of the given pDE are given by

*

=

r' * =,, *=roo,

*

=,,

#

=,

On integration, we obtain

p=qexp(t), q = c2 exp (t), x =c2 exp(l)+c, _I . y = c t e x p O + c a , z = c1c2 e x p ( 2 t ) + c 5 J Taking into account the initial conditions

t o = 0 , .Io =s, zO=s2, ps=s/2, qt=2s, h = s/2, cz =2s, ct = Qs, ca = 3l), gt =Q

r =| exp

1r;, q =2s exp

(t),

x = 2s [exp

(r) - l], .y =

;texp

(r) + tl

z = s2 exp

(2t)

(4) ( 5 ) u'e find Therefore, we have (6)

Elirninating .r and t from the last three equations of (6), we get

167 = (4y + x)2 . This is the required integral surface.

EYAMPLE 0.11 Find the characte stics of the PDE

P 2 + q 2 = 2

md determine the inte$al surface which passes through x=0,2=y. Solution The initial data curve is

r o ( r ) = 0 , y o ( s ) = s , z 6 ( s ) = s . Lsing this data, the given PDE becomes

(36)

D = X l . o = 1 . x = ! 2 t . )

I

Y = 2 t + s . z = 4 t + s . )

The last-three equations of(6) are parametric equations ofthe desired integral surface. Elimi

the parameters s and r, we get

z = y ! x .

. O.1O COMPATIBLE SYSTEMS OF FIRST ORDER EQUATIONS

Two first order PDEs are said to be compatible, if they have a common solution. We shal

derive the necessary and sufficient conditions for the two partial differential equations

f(x, Y, z, P, q) = s and

and the strip condition gives

Hence,

to be compatible. Let

INTRODUCTION TO PARTIAL DIFFERENTIAL EOUATIONS

1 = p o ( 0 ) + q e ( l ) o r q o - l = 0

4 = r r , 4 = rr . * = r o '

_{d r d r d r t}

+ z q 2

= +l

!=0.41=o

I

d t d t l

8 o = l ' P o = l l

Now, the characteristic equations for the given PDE are given by

P = q , q = c ) , x = 2 c t l + q l

'l

y = 2 c 2 t + c 4 , _{z = 4 t + c s )} Taking into account the initial conditions

x o = 0 , lo = s , z o = r , p o = ! 1 , q o = 1 , we find

( 0 . 8 7 )

Since Eqs. (0.85) and (0.86) have common solution, we can solve them and obtain explicit expressions

for u and a in the form

g(x, y, z, p, q) = 0

, - d ( f , s \ , n

d (p, q)

( 0 . 8 _{8 )}

(37)

and then, the differential relation

P d x + q d Y = 7 7 o r

Q @, y,2) dx + ry \x, y, z) dy = dz (0.89) should be integrable, for which the necessary condition is

- i . c u r l * = 0 v h e r e X = I d , V , - l ) . T h a t is ,

li

i

il

<di

+,yi

-iqlanx

dDy dtdzl=s

lo

v

rl

o r

uhich can be rewritten as

dGv,) +v(Q")

= v, - Qy

V t , + 0 V " = Q y + V @ ,

3 l

(0.

e0

₎

)JoW differentiating Eq. (0.85) with respect to x and z, we get

1yofi;nfr=o

t.r,ffi.r,ffi=o

But, from Eq. (0.89), we have

d p =dd o q =dv and so on.

L:sins

these

,.rutt..

tt'" ,0"* .ii"t'li t'""Jt 'lJu'"'"

f*+ fp!,+ fqv/,=o

ano

. f " + f p f , + f q V , = 0 .

\lultiplying the second one ofthe above pair by Q and adding to the firsl one. we readily obtain

(f, + 0 f,) + f p@, + 00,) + .fq(V,

+ 0v

_")

= 0

S i m i l a r l y . f r o m E q . ( 0 . 8 6 ) w e c a n d e d u c e t h a t

(38)

32

Sotving the above pair

of equations for (V,/x _{+ 0rl/z), we have}

( V , + O V / , ) _{_} I foG, + QE,) - s r(f, + 0"f,) .fnq o - go.f,

O I

v, + Ov" = jUtot, - s of) + Q(f os" - s of))

= t . l o _ \ t . g ) _{+ 6 d _ \ l}, g t _| _{1 O . o t )} t l d 1 x , p ) ' A Q , p ) )

where J is defined in Eq. (0.87). Similarly, differentiating Eq. (0.85) with respect to y and z and using Eq. (0.88), we can show that

6"

'

* 16,

= - ll a+4 *, al|. stl

r L d t y , q ) d \ z , q t )

( 0 . e 2 )

Finally, substituting the values _{of Vt + QV, and (, +ry/, from Eqs.}(0.91) and (0.92) into Eq. (0.90), we obtain

atf

_{.il *rd^tf}

.g)

_{=_11!+.ra:l.etl}

d ( x . p \ d ( 2 . p ) _{l A ( y , S )} _{d t t . S ) )}

In view of Eqs. (0.88), we can replace _{$ and y by p and q, respectively}to get

a C ' s )

+ p d - ( f

' s ) +a_(f

. s )

* o d , t f

. d _o

( 0 . 9 3 )

0 ( t , p ) ' 0 ( 2 . p ) d ( y , q ) ' , d ( r , q ) - "

This is the desired compatibility condition. For illustration, let us consider the following examples: EXAMPLE 0./5 Show that the following PDES

x p - y q = x a n d , 2 p + q = r "

are compatible and hence, find their solution.

Solutioh Suppose, we have

. f = x p - y q - x = 0 . g = x 2 p + q - x z = 0 . ( l ) Then, A ^ a \ J , g ) l t p - t ) x d(x, p) l(2xp- z) ,2

a ( f ,

_{s ) _l o}

_{x l_ _z}

a . , - l r l - ' t o \ 2 , p t l - x x . I = pr2 - 12 -2x2p+ xz = xz - x2p- xz , ( 2 )

(39)

aU, e\ l-q -vl

= * = l - ' i = - o . d \ y , q ) l 0 l l

a j . i l

_{| o}

- v l

-:-:--- = | t=-xv, d ( z , q ) | -:r I I ard we find

a j , c ) . -d(f,d , aj,d

a ( f , d

₎

. . , ^ + p - - + - ; ! J + q l z e ! = _{r " - r ' p - 1 2} _{+ p x 2 - q - q x y} d \ x , p ) d \ 2 , p ) d ( y , q ) ' d ( z , q ) = o _ q _ q r y _ r , = x z - q , x ( q y + x ) = o - q _ x 2 p = 0

Hence, the given PDEs are. compatible.

Now, solving Eqs. (1) and (2) for p and g, we obtain P = q - l

r q z + x x3+x2z x+x2y from which we get

x(1+ vz\ l+ r,z

p=

,G$= u,

and x 2 ( z - x \ x ( z - x )

c =

4 r - ; a 9 =

r . ,

ln order to get the solution of the given system, we have to integrate Eq. (0.89), that is

. ( 1 + u z \ . x ( z - x \ d z = ' ' ' & + ' d v ( r l

l + r y

OT v ( z - x \ - x ( z - x \ 1z - dk = !...:...-:- _{dx * j} _: o, o r d z - & _ y d x + x d y z - x l + x y On integration, we get ln (z-I) = ln (l + r1,) + ln c. Ihat is, z - x = c ( l + x y ) 33

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Hence, the solution of the given system is found to be z = x + c ( \ + x y ) ,

which is of one-parameter family.

0.11 CHARPIT'S METHOD

In this section, we will discuss a general method for finding the complete

solution of a noirlinear PDE of first order of the form

integral or complete

a r - . , - - ^ \ - n

J \ ^ . ) ' . . , y a t t t - v ,

(0.

e4

)

This method is known as Charpit's method. The basic idea in Charpit's rnethod is the introduction

of another PDE of first order of the form

( 4 )

g ( x ' v ' z , p , q ) = o

and then, solve Eqs. (0.94) and (0.95) for p and q and substitute in

/7 = p(x, y, z, a)dx + q(x, y, z, a)dy.

Now, the solution of Eq. (0.96) if it exists is the complete integral of Eq. (0 94).

The main task is the determination of the second equation (0.95) which is already discussed in the previous section. Now, what is required, is to seek an equation of the form

g ( x , _{v ' z ' p ' q ) = o} compatible with the given equation

f ( x , y , z ' p ' q ) = o for which the necessary and sufficient condition is

On expansion, we have

( a f d s _aI ae\,"(at ds -af as\

\ a * a p o p d x )'

' \ a . ' dp ap a= )

a \ f . g \ + ,dtf ,g, dU. g\ od(J,t, _0.

d ( * . p ) ' d t z - p \ - d l y , q ) ' a ( z . q J

.t44-4+).,(++-++)=,

\ d Y d q d q d Y ) \ d z d q d q o : ) d\ dy dz dp

fp

fq rf, + q1'n -(f, + Pf,)

_ d q

-U' + q1"1

which can be recast into

( 0 . e 8 )

This is a linear PDE, from which we can determine g. The auxiliary equations of (0.98) are

( 0.95 ₎

( 0 . e 6 ₎

( 0 . q 7 )

r, fr . t, fi * 1rf

n +

o1:01ff

- u, * pf

)n- (f

, * nf,)#

=

o

(41)

-Ihese

equations are called Charpit's equations. Any integral of Eq. (0.99) involving p or q or both

:an be laken as the second relation (0.95). Then, the integration of Eq. (0.96) gives the complete

:regral as desired. It may b6 noted that all charpits .quutibn, need nor be ur.I, bu, it is enough

:r choose the simplest of them. This method is illustrated through the following examples:

EXAMPLE 0.16 Find the complete integral of

6? a q2yy = qz Solution Suppose

:hen, we have

o r

From Eqs. ( l). and (3),

and 7 = 1 p 2 + q 2 ) y - q z = 0

p'L+qj-1zr= o

dx dy dz dp dq

fp fq Pfr+q1t, -(fr+ p7,1 -(fr+q1,7

( r )

/ * = u , J r = p - + q " J , = - s

f ^ = 2 p v . f - = 2 a v - 2 .

\ow, the charpits auxiliary equations are given by

I hat ls.

dx dy

2 p y 2 q y - t 2 p 2 y + 2 q z y - q z

elp dq

P q - l ( p 2 * q 2 ) - q 2 l

From the last two members of Eq. (2), we have

d P = d q Pq -p2 P d P + q d q = g P 2 + q 2 = 4 ( c o n s t a n t ) we obtain a y - q z = 0 _{o r Q = q , / z} (2) ( 3 ₎ ( a v " - \ ; 1 o t 2 - az y21ttz ( 1 )

(42)

Substituting these values of p and q in we get

O I

which can be rewritten as

. On int€gration, we find

or

Hence, the complete integral is

That is,

From Eq. (2), it follows that

dz= pdx + q dy,

o,=Ja

-77

ax+zay

r---i-_---:'--; z dz - ay dy = I m' - o' y' & d 1*2 - a2 y2 7tt2 \x+ b)- = \z-/ a) - y-. . - t ) 1 . l x + b ) - + y - = z ' l a .

EXAMPLE 0.17 Find the complete integral of the PDE: ,2 = pq ry. Solution In this example, given

f =

"2 - pqxy. Then, we have

'fx = - P4/' fY = -PA*' f" =22

f o = - e W ' f q = - P x Y ' Now, the Charpit's auxiliary equations bre given by

( l )

dt _dy = d" =

dp

₌

aq

fp

fq pf, + q1:n -(f, + pf") -(f, + q1'"1

dx dy dz dp ds

-qxy _{- pxy -2pqry} _pqy-zpz _{pqx -Zqz}

dplp dqlq dxlx dvlv

qy-22 px-22 -qy -px