K. Sankara Rao

86
Introduction z E o |JJ o z E IJ |JJ m pastern E.ononY Edition to PARTIAT DIFFERENTIA EquATr0Ns

Transcript of K. Sankara Rao

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Introduction

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PARTIATDIFFERENTIAL

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I

CONTENTS

P reface

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 82C/ Properties of Harmonic Functions 84

2.4.1 The Spherical Mean 85v

0 .1,,91w,6.3

wg4

-09\9.7y.{'0.9

\y.1,0,,xrr

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t-46

79-146

Introduction ISurfaces and Normals 2Curves and their Tangents 4Formation of Partial Differential Equation 6Solution of Partial Differential Equations of First Order ,10Integral Surfaces Passing through a Given Curve ,{ZThe Cauchy Problem for First Order Equations 20Surfaces Orthogonal to a Given System of Surfaces 21First Order Non-linear Equations 220.9.1 Cauchy Method of Characteristics 2.1Compatible Systems of First Order Equations 30Charpit's Method 340.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 491.3.2 Canonical Form for Parabolic Equation 511.3.3 Canonical Form for Elliptic Equation 53

I .4 Adjoint Operators 62I .5 Riemann's Method 64

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vt CoNTENTS

2.4.2 Mean Value Theorem for Harmonic Functions g6

,2.4.3 Maximum-Minimum principle and Consequences gZ4 Separation of Variables 9-l,;tK Dirichlet Problem for a Rectangle 94Z1 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 /0g91 2 Solution of Laplace Equation in Spherical Coordinates 1/J,2. l3 Miscelianeous Examples /22

Exercises I44

3. Parabolic Differential Equations 147_lgg3.1 Occurrence of the Diffusion Equation 1473.2 Boundary Conditions 1493.3 Elementary Solutions of the Diffusion Equarion 1503.4 Dirac Delta Function /543.5 Separation of Variables Method /j93.6 Solution of Diffusion Equation in Cylindrical Coordinares 1213.7 Solution of Diffusion Equation in Spherical Coordinares _1243.8 Maximum-Minimum Principle and Consequences 1Zl1.9 Miscel laneous Examples /29

Exercises I86

4. Hyperbolic Differential Equations lgg_2324.1 Occurrence of the Wave Equation 1g94.2 Derivation of One-dimensional Wave Equation /g94.3 solution of one-dimensional wave Equation by canonical Reduction /924.4 The Initial Value Problem; D'Alembert's Solution 1964.5 Vibrating Srring-Vadables Separable Solution 2004.6 Forced Vibrations-Solution of Non_homogeneous Equation ?0g4.7 Boundary and Initial Value problem for Two-dimensional Wave

Equarions*Merhod of Eigenfunction 2I04.8 Periodic Solution of One-dimensional Wave Equation in Cylindrical

Coordinates 2134.9 Periodic Solution of one-dimensional wave Equarion in Sphericar polar

Coordinares 21J4.10 Vibration of a Circular Membrane 2/Z4.ll Uniqueness of the Solution for the Wave Equation 2/94.12 Duhamel's Principle 2204.13 Miscellaneous Examples 222

Exercises 230

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Ii CoNTENTS

II 5. Green's Function 233-21I t . , lnr roducr ion zJJ

5.2 Green's Function for Laplace Equation 2395.3 The Methods of Images 2455.4 The Eigenfuncrion Method 2jZ5.5 Green's Function for the Wave Equation_Helmh oltz Theorcm 2545.6 Green's Function for the Diffusion Equation 259

Exercises 263

6. Laplace Transform Methods Z6S-J1

6.1 Introducrion 2656.2 Transform of Some Elementary Functions 26g6.3 Propenies of Laplace Transform 2706.4 Transform of a Periodic Function 2286.5 Transform of Error Function 2806.6 Transform of Bessel's Function 28J6.7 Transform of Dirac Delta Function 2g56.8 Inverse Transform 2856.9 Convolution Theorem (Faltung Theorem) 2926.10 Transform of Unit Step Function 2966.ll Complex Inversion Formula (Mellin-Fourier Integral) 2996.12 Solution of Ordinary Differential Equations 3026.13 Solution of Partial Differenrial Equations 302

6.13.1 Solurion of Diffusion Equarion 3086.13.2 Solution of Wave Equarion j13

6.14 Miscellaneous Examples 321Exercises 329

7. Fourier Transform Methods 333-3g7.1 Introduction 3337.2 Foti,er Integral Representations -tj.t

7.2.1 Fourier Integral Theorem 3357.2.2 Sine and Cosine Integral Representations jjg

7.3 Fourier Transform Pairs -1397.4 Transform of Elementary Functions 3407.5 Properties of Fourier Trasnform -i4j7.6 Convolution Theorem (Faltung Theorem) -i567.7 Parseval's Relation -t587.8 Transform of Dirac Delta Function 3i97.9 Multiple Fourier Transforms .li97.10 Finite Fourier Transforms 360

7.10.1 Finite Sine Transform i617.10.2 Finire Cosine Transform 362

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CONTENTS

7.ll Solution of Diffusion Equation 3637.12 Solution of Wave Equation 3677.13 Solution of Laplace Equation 3717.14 Miscellaneous Examples J73

Exercises 384

Ansvers and Keys to Eterches

Bibliagaphy

Index

388-423

425

427-430

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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 deliveredry me, over several years, to the postgraduate students of Applied Mathematics at Annauniversity, 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 involvingparabolic, elliptic and hyperbolic partial differenrial equations, which arise in many physicalsituations. In fact, the Laplace equation, the heat conduction equation and the wave equationhave 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 systematicmanner In chapter 0, partial differential equations of first order are dealt with. In chapter I, theclassification 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'smethod of solving cauchy's problem described. chapter 2 deals with elliptic differentialequations. Also, basic mathematical tools as well as various DroDerties of harmonic functions arediscussed. Further, the Dirichlet and Neumann boundary value prtblems are solved using variablesseparable method in cartesian, cylindrical and spherical coordinate systems. chapter 3 is devotedto a discussion on the solution of boundary value problems describing the parabolic or diffusionequation in various coordinate systems using the variables separable method. Elementarysolutions are also given. Besides, the maximum-minimum principle is discussed, and the conceptof Dirac delta function is introduced along with a few properties. chapter 4 provides a detailedstudy of the wave equation representing the hyperbolic partial differential equation, and givesD'Alembert's solution.

In addition, the chapter presents problems like vibrating string, vibration of a circularmembrane, and periodic solutions of wave equation, shows the uniqueness of the solutions, andillustrates Duhamel's principle. chapter 5 introduces the basic concepts in the construction ofGreen's function for various boundary value problems using the eigenfunction method and themethod of images. chapter 6 on Laplace transform method is self-contained since the subjectmatter has been developed from the basic definition. various properties of the transform andinverse transform are described and detailed proofs are given, besides presenting the convolutiontheorem and complex inversion formula. Further, the Laplace transform methods are applied tosolve several initial value, boundary value and initial boundary value problems. Finally inChapter 7, the theory of Fourier transform is discussed in detail. Finite Fourier transforms are alsointroduced, and their applications to diffusion, wave and Laolace equations have been analvzed.

The text is inr.erspersed with solved examplesl also. miscellaneous examples are given in

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CHAPTER O

PARTIAL DIFFERENTIALEQUATIONS OF FIRST ORDER

: .1 INTRODUCTION

::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 )

Inr roducing the nolat ion

O Z d 7

P=^ , s=-ox 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,, inQ should satisfy the following two conditions:

( i ) For every (x ,y)e O, the point ( r ,y , " , p ,q) is in the domain of the funct ion F.( i i ) When z= f (x ,y) is subst i tu ted in to Eq. (0.1) , i t should reduce to an idenr i ty inx,y for

all (x, y) e Q.

we classify the PDE of first order depending upon the form of the function F. An equationof 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 arsoon the dependent variable z. Similarly, an equation of the form

(0 .2 )

( 0 .4 )

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INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

) , s ,Ptx, yt 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 l x . y td+ b l x . y \ f i + c t x . l t t : = d \ r . i

is cafled a finear PDE of first order, if the function F is linear in 0zl0x, dzldy and z, while thecoefficients a, b, c and d depend only on the independent variables x and y. An equation whichdoes 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 theclass C'(O), then the vector valued function grad F can be wrrrten as

. ^ (ap aF aF\g r a o f = l . . - " . . - l\ox oy dz )

( 0 .6 )

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

F(x, y, z) = F(xo, y6, zs), (0 e)

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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 levelrrfrce- Now, one may ask, if it is possible to solve Eq. (0.8) for z in terms of ;r and y. To-answerG qnestion, let us consider a set of relations of the fomr

a= f r (u , v ) , y= f2Qt , v ) , 2= f i (u , v )

z = f(x, y)

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

(0.10)

llere for every pair of values of a and v, we will have three numbers ;r, Jr and z, which represettsr point in space. However, it may be noted that, every point in space need not correspond to apeir 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 functionsofr 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 theralue of z in the form

z = f i f t(x, y), p(x, y) l (0 .12 )

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) arealso called paranetric equations of a surface. It may be noted that the parametric equation of asrface 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 0and

"=, (1-o ' ) "o"e, ,=rQ-Q1-)" rnr . ,= r ro,(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 .13 )

(0.14)

Then

; Differentiating partially with respect to .r and .y, we obtaindF .0F dz ^ dF dF dz= - f - - : - - - = v , - + - - = { ldx .dz dx 0y 0z 0y

from which we get

0z dFldx dF-- = --:-::T- = -- (usiue 0.14).tx df ldz dx

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4

or

INTRODUCTION TO PARTIAL D]FFERENTIAL EOUATIONS

o" = n

dxSimilarly, we obtain

d l . )E, - = q a n d : 1 = _ ldv '

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= f2G) , z= f r ( t \ (0 . r8 )

(0 . l e )

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

laF I . lar J lar l( x -x0 ) l ,_ (x0 . lo .zo ) l+ ( l - ro ) l l - r xs .y6 .z6 ) l+ r r - rs t l f r16 .yn . - -o r l=0 . {0 . tb rL a x ) , - , s L d : " l

0.3 CURVES AND THEIR TANGENTS

A curve in three-d imensional space IRI can be descr ibed in terms of parametr ic eouat ionsSuppose i denotes the position vector of a point on a curve c, then the vector equation ol C mavbe wr ihen as

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 thatthe in tersecl ion of two sur faces g ives r ise 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 gradF: are not collinear at any poinr of ct in IRj. In other words, the intersection of surfices sivenb1 Eq. (0.20) is a curve i f

-

and

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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 equatiorr: in the form

rJF(x. y, z)=ol

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 -xo) i + (y - yo t ;+ (z - 26 ; =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 lAF-AG aF E

-AF-AI -aF aG-

ay ar- a, a, a" dr-E a, E n- n a,

(0 .2 :

(0.2.

( x -xs ) ( y - y i \ z - zo )e@a=a(F,q=Ztr.gl0 (y, ,) 0 (2, x) 0 (r, y)

'l}rerefore, the direction cosines of I are proportional to

la1r ,c7 a@,G) d(F,q1Lao,d' aea' a@,r\l

(0.2(

(0.21

aor il lustration, 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:\\a 'a ' i )={ - "n

t ' cos l ' l ) '

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7t

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 tangentvector to the given helix is (-1,0, l).

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

f +y2 +22 =1 , x+y+z=o

0Fl0u d" 1 arlA, a" 1 ^ELa,' yr y a, la,* yt l=' (0.2e)

dr ldu.d, l ar lau dvfa"Lar*Eq I d,Lay+Es )=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 ) = sa a Ya2=gRecalling 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 ) - , fz)-\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 zand v of the form

F(u,v) =6 (0.28)

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

and

(0.30)

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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 - + - Dd x d z ' d x d z '

0u 0u dv dv- + - o - + - od), dz dy d2

d ( u , r ) 0 1 u , v ) 0 \ u , v )P aoA"q ae$= a l r ,y )

P P + Q q = R ,

^ d \ u . v \ ^ O ( u , v ) ^ O \ u , v \' 00 .2 ) ' ' ae , . \ ) ' " d \x ,y )

(0 .3 I )

(0.32)

(0 .13 )

: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 lox

-)" . ; = 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.

(4)

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INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

(ii) The given relation is of the form

Q(u'v) =o '

wherc u = x + y + t, u = 12 + y2 + "2Hence, the required PDE is of the form

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

lau Arl, =a^(u' ' ) =la, avl=l t 2vl

d\y .z ) ldu av l I z , l=z f " - ,1IE EII d" 0r l,=#3=1fr frl=li 1";,,,-,,t o x d x l

l?! a' lp=! ( r . r \ _ ld * dx t l t 2 r l"- a*r)=l+ +l=l' zrl=2{t-')

I dy dy IHence, the required PDE is

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

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

EXAMPLE 0.4 Eliminate the arbitrary function from the following and hence, obtain the correspondingpartial differeniial equation:

1 i1 z=xy+ f (x2 +y2)

(ii) z = f(xylz).

Solution( r ) U rven z=xy+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)

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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) = po x z

2.+ =: I,kytz\= soy z

Etuinating /' from Eqs. (2) and (3), we find

xp -vq=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 = pox

dz

av='= 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 -- + - + - = la 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 Ia l - a - b l - r _ b = - ;

and

I ' Q - n q 1b 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 - lq

(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 = lq p _pq

or

! + l - ' = |q p pq p+q- pq

That is,

Px+qY-2=--- ! ! - , 0)p+q -pq

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)

(6)

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

. . - :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 . np 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 . dz ldy, - l ) or (p, g , - l ) . Therefore, the d i rect ion of the normal isg ivenby i=\p, q , -1\ .

:'. ': the PDE (0.37), we observe that the normal ii is perpendicular to the direction defined by- : . :c tor / = \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 en by

a,=(a,*(aydx dy

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

lP, Q, Rl =ldx, dy, dz\1'.:.i. rhe soiution to Eq. (0.37) can be obtained using the following theorem:

(0.3 8)

(0.3 e)

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12 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

Theorem 0.I The general solution ofthe linear pDE

PP+Qq=R

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

d t _ d yp(x, y, z) e@, y,

") R(x, y, z)

Proof We observe that Eq. (0.a0) consists of a set of two independent ordinaryequations, 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 quasiJinearuntil z(r,/) is known. Recalling Eqs. (0.37) and (0.38),nomllon as

dz

Llax

the

o l tazDx \ ( R \- l l t = l I t 0 . 4 2 )dy)\dz/dy) \dz)

integral surface. For the existence of finite solutions of

(0.40)

differentialwritten as

(0.4 r )

case, Eq. (0.41) cannot be evaluatedwe may recast them using matrix

ol; t=0ayl

(0.43 )

Both the equations must hold onEq. (0.42), we must have

D / 1 1 | D

ax ay l ldx

RI IRdzl ldz

on expanding the determinants, we have

dx dy

which are called auxiliaryln order to complete

generated by the integralLet

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 toequation of the

(0.44)

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

u(x , y , z )= ( , and v (x , y , z )=C, (0 .45 )

be two indeperident integrals of the ordinary differential equations (0.44). If Eqs. (0.45) satisfyEq. (0.44). then. we have

du du du;dx+=-dy+ ; -dz=du=0ox dy dz

dv , 0v dv. ax+ -dy+-&=dv=U.

ox dy dz

and

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

Solving these equations, we find

dx dv_;_-____;_____=_

=__=___j......-

du dv du dv 0u dv dufrat at- a" a, E Ar- arE

dzdu 0v du 0v'a, ay- ay dr

l

dx dy dz---;-------i = ---t---- = --i-:---:-o \ u , v ) d \ u , v ' t d \ u , v l0(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)(0.47)

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

dx =dy _dzP Q R

Tbe solut ion of theSe equat ions are known to be u(x,y ,z) -_C1 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 ) y2 p - xy q = y(2 - 2y1

( i i ) (y+zx)p-(x+yz)q=7s2 - 12.

Solulion

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

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

t3

rfiich can be rewritten as

(0.46)

( 1 )

- .2L + C o r x ' + v ' = C ,2

(2)

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( t )

(2)

l 4

The

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

last two members of Eq. (l) give

dY= tu^ o r zdy -2ydy=_ydz-y z -zy

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 auxiliaryequation

dx_dy dzy+zx - (x+ lz) *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=? _ yzor

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 asy d x + x d y d z

7;;;7__y=7_tor

y d x + x d y + d z = 0 ,which on in tegrat ion resul ts in

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 .

(3 )

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a

dziox

EGIMPLE 0.8 Use

PARTIAL DIFFERENTIAL

Lagrange's method to

EeuATioNS oF FrRsr oRDER

solve the equation

, 'TI

f r l -^1 lo z l - l Ioyl

l 5

there 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^az l Ia ,1 |d ,ozfx l - p - y ; - l - y l - a - y - ^ l + z l d= - - l t - l = t )L oyJ L dxJ L dy dx)

2. ' -(yy - Pz \1 i+@z-yg i != Bx-ay ( t )

dx dy

dy (2)(yy- 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 z0

xdx+ydy+zdz=0 ,

, 2+y2+"2=c ,

f , nd y, we find from Eq. (2) that

adx+Bdy+ydz=0 ,

riich on integration gives

ax+ py+yz=Cz

Thus, the general solution of the given equation is found to be' F ( t z +y2 +22 , ax+ By+yz )=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)

(ii l (x2 - lz) p +(y2 - u)q = ,2 - 'y.

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t 6

Solution

(i) The integralequation

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

surface of the given PDE is generated by the integralcurves of the auxiliary

The first two members of Eq.

which on integration yields

Now, considering Eq. (2) and the first

or

dx dy dz

z _z z2 +(x+ y)2

(1) g ive

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 ;

ln (22 + Cl) =2x + C2

( l )

(2)

which on integration yields

OI

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

dx-dy _ dy-dz _

by the integral

dz

dz-dx

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

OI

- v l

F(x + y, log \x2 + y2 + 22 + 2xyj -2x) = 0

(ii) The integral surface of the given PDE is givenequation

* =0 ,,z - tn y2 -" ,

(3)

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

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 )tt

v(x, Y, z) = C, )

l 7

l inear PDE. can besur face conta in ing

( 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 II

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

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

.:ich leads to the solution: uple of examples.

t.Y4MPLE 0.10 Find t\e

- :nta in i r rg the st ra ight l ine x+y=g, s=1.

Solution The auxiliary equations for the given

___1x \ y 2 + z ) - y t x z - : 1

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

g iven by Eq. (0.a9) . For i l lust rat ion, le t us consider the fo l lowing

integral surface of the linear PDE

x(y2 + z)p - y(x2 +z)q=1x2 - y21z

PDE are

dz

(r' - yt),( l )

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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 ,

which on integration yields

* 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 theauxiliary equation

dx _dy =dzx y z

( l )

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

ln . r = ln/+ InC

(2)

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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, =Czz

::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 lCv z=ylCr=a1grg,

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=aL l L r L 2 | C i C i C ; )

x x ( r t \x+ -+ -=2 . o r x l l + ' + '

l =2q Crcz ^1"c,-cE)- '

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

1*{* - l -=f,*t* t l 'c i cici l . cr crc2J'-hich on simplification gives us

) 1 1_ + _ + - - = 0Cr 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 INTRODUCTION TO PARTIAL DIFFERENTIAL EOUATIONS

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

Z[x6(s) , y6(s) ]= Zo(5)

and QQ,y) together with its partial derivatives with respecl to n and ), are continuous functionsof x and y in the region IR.

Geometrically, there exists a surface z=QG,y) which passes through the curve f, calleddalum curve. whose parametr ic equal ions are

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 theseassumptions, we have a whole class of existence theorems which is beyond the scope of thisbook. However, we shall quote one form of the existence theorem without proof, which is dueto Kowalewski (see Senddon, 1986).

Theorem 0.2 I f

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

( i i ) x6 is a g iven number and;6 =g(y6) , q0=C'Oo) and f (x ,y ,z ,q) and a l l o f i ts par t ia lderivatives 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

l x - : 1 6 l < d 1 , ! - l o < 5 2 ,

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

(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

oz .( dz\- a =J l x . y , z ,= - l andox \ dy )

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

0.8 SURFACES ORTHOGONAL TO A GIVEN SYSTEM OF SURFACESone of the useful applications of the theory of linear first order pDE is to find the sysrem ofsurfaces orthogonal to a given system of surfaces. Let a one-parameter famiry of surfaces isdescribed by the equation

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

(0.54)

(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=oox dy

Pp+Qq = R (0. s7)

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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=# rosolI

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 Iorthogonal 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 IIn rhis section, we will discuss the problem of finding the solution of first order non-linear partial Idifferentiar 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 IPont such that

.l 't*?r

r, I

IFig.0.3 cone of normals to the integral surtace

lThe 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 Inormal fi = \p, Q, - l) to the desired integral surface z = z(x, y) of Eq. (0.60). Thus, the direction Iof the normal to the desired integral surface at certain point (x, y, :) is not defined uniquely. iHowever, a certain cone ofadmissable directions ofthe normals exist satisling the relation ((p,q)=0:

(see Fig. 0.3).

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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) , wheres is a parameter of the fami ly ca l led c l raracter is t ics.

: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 \ are the d i rect ion rat ios of the normal to S. Now, the d i f ferent ia l equat ion (0.60)' . is that at a g iven point P(xx,y6,z6) on S, the re lat ionship between p0 and 90, t l la t. 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 a long a generator . For example. le t us. ier the non- l inear PDE

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 ich is para l le l to x-ax is are (1, 0 ,0) (see Fig. 0 .4) . Let the semi-ver t ic le angle of the cone- l . Then.

7r,1

( 0 .62 )

(0 .63 )

( 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

1 r r e ) 2 - ( - r , - l o ) 2 - ( z - z i 2 - o ( 0 . 6 5 )

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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 rightcircular 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 methodto 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 thepoint (,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 (xo, yo, zo, p)

indicating that p and ? are not independent at (xo,yo,zo). At each point of the surface s, thereexists a Monge cone which touches the surface along the generator of the cone. The lines ofcontact between the tangent planes of the integral surface and the corresponding cones, that isthe generators along which the surface is touched, define a direction field on the surface S. Thesedirections are called the characteristic directions, also called Monge directions on S and lie alongthe generators of the Monge cone. The integral cunr'es of this field of directions on the intesralsurface .l define a family of curves called characteristic curves as shown in Fig. 0.5. The Mongecone can be obtained by eliminating p from the followins equations:

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

)

(0.66)

( 0 .6 7)

l(.\,,. r,,. --r )

Fig. 0.5 Characteristic directions on an integral. surface.

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

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

25

(0.6 8 )

(0.6e)

(0.70)

(0 .71 )

anl

o-fserving that 4 is a function

.1-(r-x0)+() /-) . ,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)ap-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: - ro)(y- j lo) and (z-zo) , we get

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 correspondsr infinitesimal movement from (ro,ys,z6) along the generator, Eq. (0.73) becomes

dx dy dzFp Fq pF, + qFo'

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

4*= Fo{x, t, "@, y), p(x, y),q(x, y)} (0.75)

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I

dz

dt

26

and

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

"! = 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 dva=;,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

(0.76)

(0.7? )

Arso, differentiating Eq. (0.60);;^ ;Tec,;:

j" )irr,: - + - : - P + ^ - : * - : - - = Udx dz ' dp dx dq dx

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 curvesalong 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 .78 )

(0 .79 )

(0.80 )

(0 .81 )

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27PARTIAL DIFFERENTIAL EOUATIONS OF FIRST ORDER

dx

dt

dz-= P ta D + q ro ,dt

==-( f , + pF, t andAT

==- (F ,+qF- ) .dl

:.:: 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 .83 )

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

c- : : ' : : ier is t ic s t r ip . The solut ion x=x( / ) , / = y( t ) ,2=z( t ) ,p= p( t ) ,q=SQ) of the character is t ic

.:-,:::Jns (0.82) satisfy the strip condition

F

dy(tt

(0.82)

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 is constant

I eroo| Along the characteristic strip, we have

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

| ; r ty( ' ) , ) . ( r ) .2(r ) .p(r ) .qa\r=; ;+ i i * *A- aoi -A, t ,

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

I 4Fp+F ,Fq+F , (pFr+qFo) -Fo (F ,+pF , ) -Fo (F r+qF , )=0 .

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

F -re def ined by Eq. (0.60).

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

I

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28 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

EXAMPLE 0,12 Find the characteristics ofthe equation pg = z and determine the integral surfacewhich 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= zPq '

;= e ' ;=qOn 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 exP(t ) , 4=exP ( t ) , x=exP() l. f

/ =sexp ( / ) , z=sexp (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 integralsurface which passes through the parabola x=0, y'=2.

Solution The initial data curve is

ro(s) = 0, /o(s)=s, zo(s)=s ' .

Using this data, the given PDE becomes

po(s) qo(s) - s" =0=F 6)

( l )

(2)

(3)

(4)

(5)

(6)

(7)

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER 29

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=ctexpO+ca, z = c1c2 exp (2t) + c5 J

Taking into account the initial conditions

to =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

P2 +q2 =2

md determine the inte$al surface which passes through x=0,2=y.

Solution The initial data curve is

ro(r) = 0, yo(s)=s, z6(s)=s.Lsing this data, the given PDE becomes

p 'o+q6 -2=0=F ( l )

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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. Elimithe 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 shalderive 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=po(0 )+qe( l ) o r qo - l=0

4 =rr, 4 = rr . * = ro' +zq2 = +ld rd rd r t

!=0.41=o Id t d t l

8 o = l ' P o = l l

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

On integration, we get

P = q , q = c ) , x = 2 c t l + q l' ly=2c2 t+c4 , z=4 t+cs )

Taking into account the initial conditions

xo=0 , l o =s , zo = r , po= !1 , qo=1 ,

we find

(0 .87 )

Since Eqs. (0.85) and (0.86) have common solution, we can solve them and obtain explicit expressionsfor u and a in the form

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

, -d ( f , s \ ,nd (p, q)

( 0 . 8 8 )P = OG, Y' z) , q =t / / (x, v, z)

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

and then, the differential relation

P d x + q d Y = 7 7or

Q @, y,2) dx + ry \x, y, z) dy = dz (0.89)

should be integrable, for which the necessary condition is

- i . cu r l *=0

vhere X =Id,V,- l ) . That is,

l i i i l<di +,yi -iqlanx dDy dtdzl=s

lo v rlor

uhich can be rewritten as

dGv,) +v(Q") = v, - Qy

Vt ,+0V"=Qy+V@,

3 l

(0. e0 )

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

1*yofi*;nfr=o

t.r,ffi.r,ffi=oBut, from Eq. (0.89), we have

dp =dd oq =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

Simi lar ly . f rom Eq. (0.86) we can deduce that

G, + ds,) + s p(Q, + 00,) + cc(V, + Qv ") = o

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32

Sotving the above pair

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

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

(V,+OV/,) _ I

foG, + QE,) - s r(f, + 0"f,) .fnq o - go.f,

OI

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

= t . l o_ \ t . g ) + 6d_ \ l

, g t | 1O .o t )t l d1x , p ) ' AQ, p ) )

where J is defined in Eq. (0.87). Similarly, differentiating Eq. (0.85) with respect to y and z andusing 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 .e2)

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

atf . i l *rd^tf .g) =_11!+.ra:l .et ld(x . p \ d (2 . p ) lA (y ,S ) d t t .S ) )

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

aC's) + pd-( f 's ) +a_( f .s ) *od, t f .d _o (0 .93)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 =xp -yq -x=0 .

g=x2p+q -xz=0 .

( l )

Then,A ^a \J ,g ) l t p - t ) xd(x, p) l(2xp- z) ,2

a(f , s) _l o x l_ _za . , - l r l - ' to \ 2 , p t l - x x . I

= pr2 - 12 -2x2p+ xz = xz - x2p- xz ,

(2)

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

aU, e\ l-q -vl= * = l

- ' i = - o .

d \y ,q ) l 0 l l

aj . i l | o -v l-:-:------- = | t=-xv,d(z,q) | - : r I I

ard we find

aj ,c) . -d( f ,d , a j ,d a( f ,d ). . ,^+ p--+- ; !J+qlze!= r " - r 'p-12 + px2 -q-qxyd\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 - lrqz+x x3+x2z x+x2y

from which we get

x(1+ vz\ l+ r,zp= ,G$= u,and

x 2 ( z - x \ x ( z - x )c= 4r-;a9= 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 + ry l+ ry

OT

v(z - x \ - x(z-x \1z - dk = !...:.........-:- dx * j

_: o,

or

d z - & _ y d x + x d yz - 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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34 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

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 formintegral or complete

a r - . , - - ^ \ - nJ \ ^ . ) ' . . , 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

(af ds _aI ae\,"(at ds -af as\\a* ap op dx ) '

' \a . ' dp ap a= )

a\ f .g \ + ,d t f ,g , *dU. g \ *od(J , t , _0 .d ( * . p ) ' d t z - p \ - d l y , q ) ' a ( z . q J

.t44-4+).,(++-++)=,\dY dq dq dY ) \dz dq dq o : )

d\ dy dz dp

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

_ d q-U' + q1"1

which can be recast into

(0 .e8)

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

( 0.95 )

(0. e6 )

( 0 . q 7 )

r, fr . t, fi * 1rf n + o1:01ff - u, * pf )n- (f , * nf,)# = o

(0 .ee)

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER 35

-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 il lustrated through the following examples:

EXAMPLE 0.16 Find the complete integral of

6? a q2yy = qz

Solution Suppose

:hen, we have

or

On integration, we get

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

and

7=1p2+q2 )y -qz=0p'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

2py 2qy - t 2p2y+2qzy -qz

elp dq

Pq - l ( p2 *q2 ) -q2 l

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

d P = d q

Pq -p2

P d P + q d q = g

P2 +q2 =4(constant)

we obtain

a y - q z = 0 o r Q = q , / z

(2)

(3 )

(av

" - \ ; 1ot2 - az y21t tz

( 1 )

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36 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

Substituting these values of p and q in

we get

OI

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 .lx+b)- +y- = z ' la .

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

fo=-eW' fq=-PxY'

Now, the Charpit's auxiliary equations bre given by

( l )

dt _dy = d" = dp = aqfp fq pf, + q1:n -(f, + pf") -(f, + q1'"1

dx dy dz dp ds-qxy - pxy -2pqry pqy-zpz pqx -Zqz

dplp dqlq dxlx dvlvqy-22 px-22 -qy -px

(2)

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

.irich can be rewritten as37

ft integation, we find

Fmm the given PDE, we have

rtich gives

dplp - dqlq _-dxlx+dytyqy-px qy- px

dp dq dy dx

P q

D X- - = c(constant)qv

p = cwlx

shere a = l/J7.Hence,

P = zlax'9rbstituting these values of p and q in

dz= pdx+ q dy,get

is the complete integral of the given pDE.

0.18 Find the complete integral of

,2 = pqry=rq2y2

q2 ="2/"y2 or q=zl^fcy=azly,

a z = - d x + - d !a x y

d z l & , d yz a x y

Il n z = : l n x + a l n y + l n b

a '

z = brrro yo

Charpit's method.,2 p2 +y2q2 -4=o

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38 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

Solution The Charpit's equations for the given PDE can be written as

dx__ dy _ dz _ dP

;7;- ti i- 'a2 P2 + Y2q21- -2'P2

--!c- (l)-2 yq'

Considering the first and last but one of Eq. (1), we have

- !+-1 or * *42=o2x'p -Zxp' x P

On integration, we gel

ln (xp)=1Y1s or xP=a (2)

From the given PDE and using the result (2)' we get

y 2 q 2 = 4 _ o 2 ( 3 )

Substituting one set of p and q values from Eqs (2) and (3) in

dz= pdr + q dy,

we find that

4, = o& *.J;I dY .

x y

On integration, the complete integral of the given PDE is found to be

. = o t n r * r [ 4 J h y * h .

0.11.1 Speciat Types of First Order Equations

Type I Equations Involving p and q only'

That is, equations of the tYPe

f(p, q) = 0. (0. I 00)

Let z=al+by+c=0 is a so lut ion of the g iven PDE, descr ibed by f ( .p 'q)=o, 16gn

1 t .

p=?=o.q= : .=b.1x uf

Substituting these values of p and q in the given PDE' we get

f ( a . b ) = 0 ( 0 l 0 l t

Solv ing for b. we get , 6=/(d) . say Then'

z - - a x + A @ ) . v + c ( 0 l 0 l r

is the complete integral of the given PDE

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PARTIAL DIFFERENTIAL EQUA]IONS OF FIRST ORDER

E,\-IMPLE 0.19 Find a complete integral of the equation

" lp*"G=tSolution The given PDE is of the font f(p,q)=g. Therefore, let us assume the solution

-: :he form

z = a x + b y + c';:-:rg

"G+Jt '= t o r b=( r -J ; )2

Hirce, the complete integral is found to be

z = ax + (1- nla)2 y+c.

EX{MPLE 0.20 Find the complete integral of the PDE

Ps= l

Solution Since the given PDE is of the torm f(p,q)=O, we assume the solulion in the:t;:n z=ax+by-lc, where ab=l or b=lla- Hence, the complete integral is

Iz=ax+ -y+c .

I!

Tlpe Il Equations Not Involving the Independent Variables.

-,ar is, equations of the type

f (2 , p ,d=o (o . lo3)

-ai a trial solution, let us assume that z is a function of u = )c + ay, where a is an arbitrary constant.

z = f(u) = f(x + ay) (0.104)

dz dz du dz'

dx du 0x du

dz dz 0u dzt= s r= * ,=a ^

S -:stituting these values ofp and g in the given PDE, we get

f ) - , - \

I l" .a.,+l=o (o.ros)\ a u a u )

r::ch is an ordinary differential equation of first ordetSolving Eq. (0.105) for dz/du, we obtain

] = Q k , a ) \ s a v )

i :

39

dz

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PARTIAL DIFFERENTIAL EQUA-I IONS OF FiRST ORDER

EI{-I{PLE 0.19 Find a complete integral of the equation

Ji * Jq =t.

Solulion The given PDE is of the form /(p, g) = g. Therefore, let us assume the solution: :--: form

z = a r + b y + c

, ! a + . l b = l o r b = \ l - J t t ) t

l-l:::e. the complete integral is found to be

z = a x + Q - J i ) 2 y + c .

E.\${PLE 0.20 Find the complete integral of the pDE

P q = t '

Solution Since tlie given PDE is of the form f(p,q)=O, we assume the solution in the- : - , z=ar+W*c, where ab=1 or b=11a. Hence, the complete in tegra l is

1z = a x + - y + c .

a

Tlpe II Equations Not Involving the Independent Variables.

.l-:: is, equations of the type

f(2, p,q)=o (0 .103 ):-. : rial solution, let us assume that z is a function of u = x+ ay, where a is an arbitrary constant.

z= f (u )= f ( x+ay )

0z dz 0u dz' 0x du dx du

o=+=++=,+oy du dy du

S,::rituting these values ofp and q in the given PDE, we get

f(,.+. "!)=o\ au du)

r::ch is an ordinary differential equation of first order.Solving Eq. (0.105) for dz/du, we obtain

dz

^ = Q Q , a ) ( s a v )

dz---:"-----a = au.Q G , a )

39

(0. 104)

(0. r 05)

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40 INTRODUCTION TO PARTIAL DIFFEREN1IAL EQUATIONS

On integration, we find

I t ! '=u*"J 9\2' a t

That is,

F (z ,a )=u+s=Y lay t t .

which is the complete integral of the given PDE.

EXAMPLE 0.21 Find the complete integral of

P(1+ q) = qz

Solution Let us assume the solution in the form

7= f (u )=a+6y

Then,dz dz

P=A ' q=a du

Substituting these values in the given PDE, we get

dz ( . dz \ dzal'*oa)=o'd".

That is,dz dz

oa r=* - ' o r a - ' -=4u

On integration, we findln (az - l )=u+c=x+aY+c

which is the required complete integral.

EXAMPLE 0.22 Find the complete integal of the PDE:

P2 "2 +qz =1 '

Sotation Let us assume lhat z = f(u)= x+ ay is a solution of the given

dz dze= du ' c=a du

Substituting these values ofp and g in the given PDE, we obtain

l d z l 1 l l C E l

la") '- +a-\du) =l

That is,

d z l

du tlr2 +o2

iPDE. Then,

(*\ <,, *ort=r or\du )

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER 4 l

ar lnte$ation, we get

'[7*t a"=au

r-i-----i t f r-------''lz l z - +a - a - . I z+ !z - +a -

t = t r u v ' r D2 2 l a I

-:.;h is the required complete integral of the given PDE.

Ty pe lll Separable Equations

r.- eOuation in which z is absent and the terms containing x and p can be separated from those

::-::iningy and 4 is called a separable equation.

Tlat is, equations of the tYPe

f(x, p)= F(v'q)

: trial solution, let us assume that

f(x 'P)=F(Y'q)=a (saY)

.. solving them for p and q, we obtainp=0@)' c=v0)

(0. r 07)

(0. 108)

(0 .10e)

a"=* a" * *ay= pdx+qdydx dy

dz =O@dx+v(;)dy

r ::egration, we ge1 the complete integral in the form

"=lO<Oa'*JyQ)dY+b

TUPLE 0,23 Find the complete integral of the PDE:

pz y (1+ x2'1= qx2

Solution The given PDE is of separable type and can be rewritten as

o2 ( l + ,2 \.--. .-=9=o (say), an arbitrary constant.x ' y

J-o*e=;67' q=aY

r:r rring these values of P and q in

42=pdx+qdf ,

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42

we get

Type Mlairaut's Form

A first order PDE is said to be of clairaut's form if it can be written as

z= px+W+f @,q )

The corresponding Charpit's equations are

&dv

"+ fr y+.fq px+qy+pfr+q1fo

2 = :(x + a)"' + | (y + a)'', + b.5 J

( 0 . l l l )p - p q - q

The integration of the last two equations of (0.lll) gives us

p = a , q = b

Substituting these values ofp and q in the given PDE, we get the required complete integral inthe form

INTRODUCTION TO PARTIAL DIFIERENTIAL EQUATIONS

' {otu= Jl;7ar+aYry

On inlegation, we obtain

'=Ji [*t *1y'*bwhich is the complete integral of the given PDE.

EXAMPLE 0.21 Find the complete integral of

P2+q2= '+Y

Solution The given PDE is of separable type and can be rewritten as

p2 - t= y-q2 =a (say)

Then,

P =J'+i,

Now, substituting these values ofp and 4 in

q=Jy+a .

dz=p&+qdy

we find

dz

dp dq

dz =.lx + a dx+,JW dy,

On integration, the complete integral is found to be

z=ax+by+ f (a ,b ) (0. r l2)

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PARTIAL DIFFERENTIAL EQUATIONS OF NRST ORDER

E\-lrtPLE 0.25 Find the complete integral of the equation

,= p* *qy* r l t * p \ t ,

Solation The given PDE is in the Clairaut's form. Hence, its complete integral is

t=* * ry * r [ * ] *F .

Find the complete integral of( p + q ) ( z - x p - y q ) = I

The given PDE can be rewritten as

Iz=xp+yq+ -p+q

:s in the Clairaut's form,

EXERCISES

. Eliminate the arbitrary function in the following and hence obtain the corresponding pDE

z=x+y+ f (A ) .

Form the PDE from the following by eliminating the constants

z=(x2 + a) (y2 +b) .

Find the integral surface (general solution) of the differential equarion

"dz "dzx - ; -+y '7 -=$+y )2 .ox oy

Find the general integrals of the following linear pDEs:,)

i ) - P + x z q = y -

, i i ) (y + l )p+(x +1)q = s .

Find the integral surface of the linear PDE

x P - Y q = z

;h ich conta ins the c i rc le x2 + y2 =1, 2=1.

43

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INTRODUC'TION TO PARTIAL DIFFERENTIAL EQUATIONS

6. Find the equation of the integral surface of the PDE

2y(z -3)p + (2x - z)q = y(2x - 3)

whiclr contains the circle x2 + y2 =2a, 2 =g.

7. Find the general integral of the PDE

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

which contains the circle *2 + y2 =1, "

=1.

8. Find the solution of the equation

, = L1p2 + q21 + 1p _ x\ (q _ y)2

which passes through the x-axis.

9. Find the characteristics of the equation

pq=ry

and determine the integral surface which passes through the curve z=x,y=0.

10. Determine the characteristics of the equation

" = p z - q 2

and find the integral surface which passes through the parabola 4z+x2 =0, y=0.

rll. Show that the PDEs

xp= yq and z(xp+ yq)=Zxy

are compatible and hence find its solution.

12. Show that the equations

p2 + q2 =! arrd (p' + q')x = pz

are compatible and hence find its solution.

13. Find the complete integral of the equation

(p2 +q27x= pz

where P=AzDx' q=0zl0Y.

14. Find the complete integrals of the equations

( i ) pxs - 4q3 x2 +6x22-2=o

(r i ) 2(z+xP+Yq)=YP2.

Find the complete integral of the equation

p + q = p q .

Find the complete integrals of the following equations:

( l ) z p q = p + q

(lr) p2 q2 + *t ut = *'qt (" + y').

15.

16.

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PARTIAL DIFFERENTIAL EQUATIONS OF FIRST

17. Find the complete integral of the PDE

2= px+ qy _sin(pq)

Find the complete integrals of the rollowing PDEs:

(\) tp3 q2 + yp2q3 +1p3 +q3)-"p2q2 =o

( i1) pqz= p2(xq + p2)+q2(yp+q2).

Find the surface which intersects the surfaces of the system

z(x+Y1=s132a11

orthogonally and passes through the circle

x 2 + Y 2 = 1 , 7 = 1 '

Find the complete integral of the equation

1p2 +q21x= pz

t l

. 0202where p=2, Q=2, (GATE-Maths, 1996)

Find the integral surface of th€ linear PDE

s-D*-G-v+4!=,dx dy

which contains z-1 and *2 +y2 =1. (GATE-Maths, 1999)

the correct answer in the following questions:

Using the transformation u =Wly in the PDE xux = u + yu, the transformed equation

has a solution of the form l/=

@) ax-by+(ab-2 +ba)1

(D) ax + by -(ab-2 +ba-2).

45

r8 .

1 9 .

10.

(A) f(x/y) (B) /(x + y)

(C) f (x - y) (D) .f (,y). (cATE-Maths, l e97)

The complete integral ofthe partial differential equation ,p3 q2 + yp2q3 +1p3 +q31-tp2q2 =O

l s z =

(A) ax + by + (ab-2 + baal

(C) -ax + by + (bo-2 - ob-z)

(GATE-Maths, 1997)

The partial differential equation of the family of surfaces z=(x+ y)+ A(ry) is

(A ) xp - yq=g (B ) xP-Yq=Y-Y

(C) xp+yq=x+y (D) xP+Yq=o .(GATE-Maths, 1998)

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INTRODUCTION TO PANTIAL DIFFERETMAL EQUATIONS IThe complete integral of the ppt 7=psaqy-sin(pq) is I(A) z=at+by+sin(ab) @) z=u.+by-sin(ab) I(C) z=ar+y1sh16; (D) z=x+6y-s in (a) . I

IIIIII{

l

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CHAPTER T

FUNDAMENTAL CONCEPTS

. INTBODUCTION

!::. practical problems in science and engineering, when formulated mathematically, give rise tor--:l differential equations (often

_ refened to as pDE). In order to understand the physical-r'. iou' ofthe mathematical model, it is necessary to have some knowledge about the mathematical.::;ter, properties, and the sorution of the governing pDE. An equation which invorves several::Tndent variables (usually denoted by x, y, z, r, ...), a dependeni function a ofthese variables,: -re partial derivatives of the dependent function a with respect to the independent variables

F ( x , y , 2 , t , . . . , u , u y , u z , h , . . . , u r x , u r , . . . , u r , . . - ) = 0

:.:.led a partial differential equation. A few well-krown examDles are:

u, = k(uo +uw + urz)

u * + u , ^ . + u - - = O' ) .

u = c- (u$ + u)ry + uzz)

q + u u r = p u x r

Iinear three-dimensional heat equation]

[Laplace equation in three dimensions]

Iinear three-dimensional wave equation]

[nonlinear one-dimensional Burger equation].

( l . l )

:. these examples, z is the dependent function and the subscripts denote partial differentiation--: respect to these variables.

inition l.l rhe order ofthe partial differential equation is the order of the highest derivative-:ring in the equation. Thus the above examples are partiar differentiar equations of second

:::. whereas

q = u u r n + s t n x:--. example for third order partial differential equation.

CLASSIFICATION OF SECOND ORDEF PDE

. nost general linear second order pDE, with one dependent functron u on a domain e of- : X = ( \ , x2 , . . . , xn ) . n > l . i sn n

S s1 -z.t ri1urixJ *

L b,'r' + F(u) = Qi , i= t t= l

classification of a PDE depends only on the highest order derivatives present.47

(1 .2 )

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48 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

The classification of PDE is motivated by the classification of the quadratic equation of theform

Ax2 + Bxy + Cy2 + Dx + Ey + F =0 ( 1 .3 )

which is elliptic, parabolic, or hlperbolic according as the discrimina Bz -4AC is negative.zero or positive. Thus, we have the following second order linear PDE in two variables ;r and y':

Auo + Bu,y + Curr + Dux + Eur + Fu = G ( 1 . 4 )

where the coefficients A, B, C, ... may be functions of .r and y, however, for the sake of simplicitywe assume them to be constants. Equation (1.4) is elliptic, parabolic or hyperbolic at a point(.x6, y6) according as the discriminant

Bz (xo, y) - 4 A(xo, ro ) C (;ro , rb )

is negative, zero or positive. If this is true at all points in a domain Q, then Eq. (1.4) is said to

be elliptic, parabolic or hyperbolic in that domain. If the number of independent variables is twoor three, a transformation can always be found to reduce the given PDE to a canonical form (also

called normal form). In general, when the number of independent variables is greater than 3, itis not always possible to find such a transformation except in certain special cases. The idea of

reducing the given PDE to a canonical form is that the transformed equation assumes a simple

form so that the subsequent analysis of solving the equation is made easy.

1.3 CANONICAL FORMS

Consider the most general transformation of the independent variables x and y of Eq. (1.4) to newvariables 1,q, wherc

5=1@,y) , n=q@,y)such that the functions { and r7 arc continuously differentiable and the Jacobian

( 1 . 5 )

(1 6 )

in the domain O where Eq. (l.a) holds. Using the chain rule of partial differentiation, the partial

derivatives become

ux = u|1x +uTna

uy = ut'j +unn)

' =#3=,;,',1= o*, -€,n,) *o

u* = uE461. +2u4r€,4, + rnrr\! + rg ,to + urno

uo = ugl*4y + utT(4,ryy + 1yry") +urrn,n, + u|q n, +u44n

ur, =u6gC,? +2u6r1yey +rrrrt r? +u,{r, +urr1r^

Substituting these expressions into the original differential equation (1.4), we get

(1 7 )

( I 8 )Au"" + Bu,- + Cu-- + Du, + Eu- + Fu = G9 9 S ' l ' t ' I 9 t I

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FUNDAMENTAL CONCEPTS

- :te

2=A1l+Bt" t . .+Ct?

E =2A1,t1, + B((,q, + (rr7) +2C(rr,,

e = 4l + nr7,r1, +Cr72,

E = Anxr + Bry,y + Cnw + Dttx + Enr

| = 1 , ( - ' = U

-: nay be noted that the transformed equation (1.8) has the same form as that of the original:::ation (1.4) under the general transformation (1.5).

Since the classification of Eq. (1.4) depends on the coefficients A, B and C, we can also:: *rite the equation in the form

Au* + Bu^. +Cu.^. = H (x. v. u.u-. u-. \ ( l . l 0 )

-: can be shown easily that under the transformation (1.5), Eq. (1.10) takes one of the following'-l:ee canonical forms:

( i ) u lq -u r ,=0 (6 , r y ,u ,%,ua ) ( l . l l a )

: :

uq=Q,(6,n,u,uq,ao) in the hyperbol ic case

( i i ) ury+urr=0(6,4,u,ue ,ua) in the el l ipt ic case ( l . l lb)

( i i t u$ = tg ,n ,u ,u€ ,u? ) ( l . l l c )

: l

uaa = Q(6'n,u'u1, zo) in the parabolic case

*e shall discuss in detail each of these cases separately.Using Eq. (1.9) it can also be verified that

Ez _ qlc = G,?ry _ tr4)2 {a, _ +,1c1

-d therefore we conclude that the transformation of the independent variables does not modifyr-e type of PDE.

1.3.1 Canonical Form for Hyperbolic Equation

:;nce the discr iminanr E2-qAe>0 for hyperbol ic case, we set 7=0 and f=0 in Eq. (1.9).;hich will give us the coordinates € and ry that reduce the given PDE to a canonical form in'*hich the coefficients of u$,u44 €re zero. Thus we have

,4 = A4', + B€,6y + c6; =o

e = .tql + Bq,q, +cryj = o

49

( l .e )

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50 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUAT]ONS

which, on rewriting, become

Solving these equations forr-..-

4x _-B+,1 B" -4AC, a t

F--;-ry , _ -B- l B ' -4AC

4y 2A

The condition B' >4AC implies that the slopes of the curves ( (x, y) = C1, r\Q, y) = Cz are real.Thus, if B' > 4AC, then at any point (x, _y), there exists two real directions given by the two roots(1.12) along which the PDE (1.4) reduces to the canonical form. These are called characterislicequations. Though there are two solutions for each quadratic, we have considered only onesolution for each. Otherwise we will end up with the same two coordinates.

Afong the curve ( (x ,y)=cr , we have

d ( = ( , d x + ( r d y = 0

Hence.

( r . r3)

Similarf y, af ong the atwe q(x, y) = c2, we have

( l . l 4 )

Integrating Eqs. (1.13) and (1.14), we obtain the equations of family of characteristics €(r, y)=ctand ryQ,y)=c2, which are ca l led the character is t ics of the PDE (1.4) . Now to obta in thecanonicai form for the given PDE, we substitute the expressions of{ and ? into Eq. (1.8) whichreduces to Eq. ( l . l I a) .

To make the ideas clearer, let

Comparing with the standard PDEthe given equation is a hyperbolic

( r \ 2 ( r \al ! l +nl i l+c=o

\qv ) \{r i

,fr'l'.uiz,l.. =o\ t t y ) \ay )

GJ11,) and (n'/ny)' we get

( r .12 )

4=-lL1dx \1, )

4=-luldx \,tn )

us consider the following example:

3u* + l ]u- . + 3t . - . = 0

(1.4), we have A =3, B =10, C =3, 82 - 4AC = 64 > 0. HencePDE. The corresponding characteristics are:

dy ( c, \ ( -a*, [F-+rc) 1dx \6 r ) \ 2n ) 3

Page 58: K. Sankara Rao

| ,,NDAMENTAL coNcEprs slI| +=-(u)=-(-a-' l- 'a-qAc)-.|

* \ 4 ' ) \ 24 ) -

|

-: tina ( and ry, we first solve for 7 by integrating the above equations. Thus, we get

| '=3x+c1 ' ,= ! ' * ' ,

J :h give the constants as

I ,=Y-3x. c r=y-x /3

|

: - c r o r e '

F = v - 1 . - - I

| €=y-3x=c1, n=t - ! ,=c ,

| - -::e are.the characterislic lines for the given hyperbolic equation. ln this example, the characteristics

| ,,=,'r#

to be strarght lines in the (x, y)-plane along which the inirial data, impulses will

J To find rhe canonical equation, we substitute the expressions for f and 7 into Eq. ( I .9) to get

| ,= .e ( |+B l ( r+cq j= t1_ t12+10(_3) ( r )+3=0

I I = ZAi,ry, + Bli,Uy + qyry,t +2c6rny

I

I =,(3,er(-;).,ofr-:xrr.r(-J)]+z(:)(r)(r)II

| =o=ro(_fJ+e =,,_+=_+

I .=o D=0. E =0. F=o

lF: :e. the required canonical lorm is

' * " ' =o o r ' ' : n=o

f- :rlegmtion, we obtain

| ,G.n\= "f c)+ sot)

f

---:/and g are arbitrary. Going back to the original variabres, the generar sorution is

|

, ( x , y )= f ( y_3x )+s (y_x t3 )

f 3.2 Canonical Form for parabotic Equation

f " :eparabof ic equat ion. rhe disci iminant E2-q,qe =0, which can be true i f B=0 and i or C

F :j.rat to zero. Suppose we set first l=0 in Eq. (1.9). Then we obrain

|

, =A { }+B( , ( , +cs , ,=0

I

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INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

which givesr--;-

t - -B!,1 B' - 4AC;= , ," y

Using the condition for parabolic case, we get

9 r

{ , = - 2A

Hence, to find the function 6=1Q,y) which satisfies Eq.(1.15), we set

) , , r p

A=-7.= rA

and get the implicit solution

6Q,v)=ctIn fact, one can verify that 7=O implies B=0 as follows:

B = 2AS,t7, + B (6,tl y + 6yn x\ + 2C1yry'

Since 82 -4AC=0, the above relation reduces to

( r \ 2 ( r \A l : z | + B l : : r l +C=011, ) \6, )

<t

or

(1 .15 )

Howevel

B =2A€r4, +2.[ 'tc 1q,r7, + qyq,)+2cEyn,

= 2 6fAq, + Je 6 ; (-.aq, + J- ry n1

q, =- B =-z . !AC =-4y 2A 2A

Hence,

E = z1J-e6 -.,F.t6"11J-.t t, + Jcrl)=sWe therefore choose f in such a way that both 7 and E are zero. Then 17 can be chosen in anyway we like as long as it is not parallel to the f- coordinale. ln other words, we choose r7 suchthat the Jacobian of the transformation is not zero. Thus we can write the canonical equation forparabolic case by simply substituting { and 7 into Eq. (1.8) which reduces to either ofthe forms( l . l l c ) .

To illustrate the procedure, we consider the following example:

x2uo -2ryu, + y2uo = er

The discriminant 82 - 4AC = 4x2 y2 - 4x2 y2 =g, and hence the given PDE is parabolic everywhere.

The characteristic equation is

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FUNDAMENTAL CONCEPTS

dv t -B2xvv-----L = - !-

dx €t 2A 2x2 x

- .-.:eSTation, we havex l = c

: :.ence ( = xy will satisfy the characteristic equation and we can choose 17 = y. To find the:::cal equation, we substitute the expressions forf and ? into Eq. (1.9) to get

A= Ayz + Bxy+cx2 =t2y ' -2t2y2 +y2x2 =0

53

r=0, e=y2,E=0, F=0,

-:.. the transformed equation is

D = -2xy

G=e"

: :rronical form is, therefore,

lTur, - LxyuE = e'

42unt=26uq+e1n

21 | r,"ur, = -7 u1 +1e, '

i 3 Canonical Form for Elliptic Equation

-:. :he discriminant 82-4AC<0, for elliptic case, the characteristic equations

dx

dy_B+ lB " -4ACdx 24

--. complex conjugate coordinates, say { and ri. Now, we make another transformation from- :o (d, p) so that

o=€+n. s=€-n2 2 i

:- give us the required canonical equation in the form (l.l lb).-:

;l lustrate the procedure, we consider the following example:

: rriminant Bz - 4AC = -4x2. ,. "#J;",iiu"r,oro,

o e iptic. The characreristic equations

-'[472

B-dy=dx

= -tX

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54 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

Integration of these equations yields

i y+ l=s , , - i r+L="") " 4

Hence, we may assume that

t = ! x 2 + i v . , = ! 1 2 - i u' 22 '

Now, introducing the second transformationF + n F - n

a= - ,p= - ;

we obtain

12a=; , F=v

The canonical form can now be obtained by computing

]=Aa l+ fa ,ay+ca j=y2

E =2Aa,f ,+ B(a*F, + arfr) +Zc(arg) = 0

e=eB | +BB,B I+cp2r=x2

D = Aao + Bary + cayy + Ddx + Eat = |

E = A9* + BBxy + cByy + DB, + E Bn = 0

F=0, c=oThus the required canonical equation is

,2uoo + ,2u OO +uo =O

or

uoo+upp=- f i

EXAMPLE 1,1 Classify and reduce the relation

y2 uo - Zxyu, * *' u o = t u, * t u,xy

to a canonical form and solve it.

Solution The discriminant of the given PDE is

82 -4AC =4x2yz -4x2y2 =g

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FUNDAMENTAL CONCEPTS 55

-:-:3 rhe given equation is ofa parabolic type. The characteristic equation is

dy __1, _ B _-2ry __xdx 1r 2A Zy2 y

L- .::ation gives x2 +y2 =cr. Therefore, €=12 +y2 satisfies the characteristic equation. The-- : :.rrdinate can be chosen arbitrarily so that it is not parallel to {, i.e. the Jacobian of the trans-i. -::ion is not zero. Thus we choose

6 = 1 2 + y 2 ,

: ::d the canonical equation, we compute

7 = ,lEl + nqg, + G1l, =a*2y2 -Bx2y2 + 4x2y2 =0

B=0, e =4r2y2, D=E=F =G=oF:-::. the required canonical equation is

4x2Yzur, =O orunn=0

::.re this equation, we integrate it twice with respect to ry to getur=fG), u=f(€)rt+sG)

^:-: .fG) and SG) are arbitrary functions off. Now, going back to the original independenri:::-es. the required solution is

u= y2 f (x2 + y21+ g l*2 + y21,-tltPLE 1.2 Reduce the following equation to a canonical form:

( l + x 2 1 u o + ( + y 2 \ u y y + x u x + y u y = 0

:,tlution The discriminant of the given PDE is

82 _ 4AC = _4(1+ r211t+ y21<o

:: rhe given PDE is an elliptic type. The characteristic equations are

B - G4o+?nrt5

- - , ' 2

2A

dx

dy

dx

2 ( l + x 2 )

E;V' ! t+ r2

- -::gration, we get

B + J Br:4AC

6 = ln ( t+. , [2 + t .y- i tn 1y+ n[2 + t t =. ,

n = tn (, * JP * D + i tn 1y +,[1]i1 = g,second transformation

F + n n - fa = 2 _ _ : J . B = ' ' '

22 i

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56

we obtain

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

o=tn ( r *J r ' * t )

P = tn1y+, [1 ] + t )

Then the canonical form can be obtained by computing

f =^ l a?+Ba ,ay+Ca2 r=1 , E=0 , C=1 , D=E=F=G=O

Thus the canonical equation for the given PDE is

udd+upg=0

EruMPLE 1.J Reduce the following equation to a canonical form and hence solve it:

uo-2sinxu*-cos2 rrr - cos:ay =0

Solution Computing with the general second order PDE (1.4), we have

A=1 , B=-2s inx , C=-cos2x ,D=0 , t = - cos r , F=0 , G=0

The discriminate 82 - 4.,q,C = + (sin2 x + cos2 r) = 4 > 3. Hence the given pDE is hyperbolic. Therelevant characteristic eouations are

B-JF -qACdy=dx

dy_dx

2A= - s i n r - l

= l - s i n xB+J* -+rcZA

On integation, we get

y = cos r -.r+ cl, y = cos x+ x + c2

Thus, we choose the characteristic lines as

I = x+ y -cos x=q , r y - - - x+ y - cos x=c2

In order to find the canonical equation, we compute

2= Al l + M"e. ,+c{3 =o

E = 2A1,ry. + B (6,11 y + I yn ) + 2cqyq y= 2 ( s in r+ l ) ( s i n x -1 ) -4 s in2 x -2 cos2 r = -4

C =0 , D =0 , I ' - n

ur- =0

Thus, the required canonical equation is

F = O G=0

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FUNDAMENTAL CONCEPTS

:::!raling with respect to 6, we obtain

un = f(ry)...3re / is arbitrary. Integrating once again with respect to ?, we have

u = J f@) dtl+ cG)

e(6) isPDE is

u=V(rt)+cG)another arbitrary function. Retuming to the old variables r, lr, the solution of the

u(x, y) = ry (y - x -cos x)+ g(y +x - cos x)

E\4.\IPLE 1.r' Reduce the Tricomi equation

u f t + x u y y = 0 , x * 0'" z.i x, y to canonical form.

solution The discriminant 82 - 4AC = -4x. Hence the given pDE is of mixed type: hyperbolic' - . < 0 a n d e l l i p t i c f o r x > 0 .

,_-: I ln the half-plane x<0, the characteristic equations are

dv n-

dx n-.

t*'[F -+,qc

, =? g*yr/2 + r,

, = -? 943/2 + "2

)re, the new coordinates are

( (x, f i=1y-1f i ' f =, ,2

4{x ,y )= }y+1 .1 -a13 =g ,z -

are cubic parabolas.order to find the canonical equation, we compute

I = l t ? + B r F +cF1= -9 r *g *2 r=g" 5 . x 5 ) / . " 5 1 4 . .

, . 4 ^

E=9x, e =0, p=-1g4) t2 =-E,

2A

F - C - n

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5E INTRODUCTION TO PARTIAL D1FFERENTIAL EOUATIONS

On integration, we have

dy&

On integration, we get

/ = . r - s i n r f c l , y = x + s i n x + c 2

Thus, the characteristic equations are

E = y - x + s i n x , r y = y - x - s i n x

EXAMPLE 1.6 Reduce the following equation to a canonical form and hence solve it:

Thus, the required canonical equation is

9ru,- -2(-t\-tt2u, +1Gx\-tt2 u- =oe ' t 4 '

' t 4

OI

Iu h = - l u q - u a )

Case II ln the half-plane x > 0, the charactedstic equations are given by

dv . - dv7=iJx . a=- iJ ictx ctx

((x. y)= | y - i tJx)" , 4(x. y) = | y + i (Jx\ '2 ' Z '

Introducing the second transformation

t+n ^ t -nd= i . p=- ;

we obtain3 ^ . t- .1o=ry , p=-6 txr

The corresponding normal or canonical form is1^

uooluBB+---- ' :up =U

EXAMPLE 1.5 Find the characteristics of the equation

uxr + 2ury + sinz1x1uo + u, = 0

when it is of hyperbolic type.

Solution The discriminant 82 - 4AC = 4 - 4 sin2 x = 4 cos2 x. Hence for alI x*(2n-l)212,the given PDE is of hyperbolic type. The characteristic equations are

B+JF -qAC= l+cos . r

yut + (x+ y)ury + xuw = 0

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FUNDAMENTAI- CONCEPTS 59

Solution Thediscriminant

8 2 - 4 A C = ( x + y 1 2 - 4 r y = ( x _ y ) 2 > 0-:-:e the given PDE is hyperbolic everywhere except along the line y:x; whereas on the line

= r. it is parabolic. When y*x, the characteristic equations are/--;-

d y _ B + r l B . _ 4 A C _ ( x + y \ + ( x _ y )

dx 2A 2y

,- - rregration, we obtain

y=x+q ,

the characteristic equations are

1=v-x ,

dy

dx4=td t y

y2 =x2 +cz

t t=y2 -x2

straight lines and rectangular hyperbolas. The canonical form can be obtained by

7=, tg ]+n4qn+C( l=y-x-y+x=0, E=_z( ,_y)2 ,

c=0. D=0, E=2(x-y \ F=C=ocanonical equation for the given PDE is

-2(x - y)2 ug + 2(x - y1u, = 0

6u€4 + un

+2( -Ou,=0

d I - d u l -=; l E " l=u05\ on )

E*=r<ntorywe obtain

l ru=EJ f t t t \dry+c,Gl

-t2uh

::::ion yields

:. - :rtegrating with respect to ?.

i:neral solution.

I tu= - .1 f ( y ' - x " )d t y ' - x ' ) +g1y -x )

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60 INTRODUCTION TO PARTIAL D]FFERENTIAL EQUATIONS

EXAMPLE 1.7 Classiff and transform the following equation to a canonical form:

sin2(x)rio + sin (2x)u, I cos2 (x1u o .= x

Solution The discriminant of the given pDE is

82 - 4AC = sin2 2x - 4 sin2 x cos2 r = o

Hence, the given equation is of parabolic type. The characteristic equation is

d v B

4 t = ; = c o t x

Integation gives

) r = l n s i n x + c tHence, the characteristic equations are:

6 = / - l n s i n n , r y = y

4 is chosen in such a way that the Jacobian of the transformation is nonzero. Now the canonicalform can be obtained by computing

A=0, a =0,

E=0. F=0,Hence. the canonical equation is

cos2 (x) ur, + u, = x

1l- e2(n-1t1uro -- sin-t pn-|1- u,

EXAMPLE 1.8 Show that the eouation

uo+ryu,=Lu,,x a '

where 1y' and a are constants, is hyperbolic and obtain its canonical form.

Solution Comparing with the general PDE (1.4) and replacingy by /, we have A=1, B=0,C = - 1 / a ' , D = 2 N l x , a n d E = F = G = 0 . T h e d i s c r i m i n a n t 8 2 - 4 A C = 4 1 a 2 > , J . H e n c e . t h e s i v e nPDE is hyperbolic. The characteristic equations are

f = "or2

r, D =t,

dt

dxI=+-CI

P r

Therefore,

On integation, we get

d t l

d x a

d t l

[ a a

t = - !+c r , t = !+c .a a '

82 - 4AC ,,14/a'

2

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Page 69: K. Sankara Rao

1.4 ADJorNr oPERAroRs

I

where r is a differential operator given or tu=' (t'16)

It= aolxylL + ar<'){\ +...+ o,{')

|One way of introducing the adjoint differential operator Z* associated with I is to form the Iproduct vZu and integrate it over the interval of interest. Let

I

I tn"uar l fn+lB r t *uar (1.11 |

rvhich is obtained after repeated integration by parts. Here, Z* is the operator adjoint to t, where Ithe functions u and y are completely arbitrary except lhat Lu and.ttv should exist.

IEXAMPLE 1.10 Let Lu = a(x) (d2uldx2-S + n14 g"ldr'1+c(.r)r/; construct its adjoint Z+.

ISolution Consider the equation

It'**= -t.,',:!:;'*,.' :,':..*),',,1,* ^,,. I= J n

(av)---= ax + I o @v) *dxJ n Gv) u dx

Page 70: K. Sankara Rao

:: -1 ,\ ever,

FUNDAMENTAL CONCEPTS

r B r ! 2 " ? 8 S

| 1av)Td;r= | to")| tu' \a*JA dX- JA dX

=1u'val,^ - l" {ou)'u'a,

=[u'av]l -tu(av)'ll + tu u@v)"dx

lB vtu dx =7u'(av) - u(av)' + u (bv)lBn +

JB ul(av)" - (bv)' + (cv)) dx

equation with Eq. (1.17), we get

1 * y = (av)" - (bv)' + (cv) = 6y" a 12o' - b) v, + (a,, - b, + c) v

-b)!+{a" -b'+c)dx

:::scribed over the

( l . 18 )

functions of x and y. In

s,a-J

I 'ooffa. =tu(bv)l i -[u ubi 'a*

L(u) = ( l . l e )

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64 TNTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

Its adjoint operator is defined by

r-(u)= ). j. ,n,rr-f.!@,u)*"u (r.zo)i j=1 dx'ox t E dxi

Here it is assumed that A4 eCQ) and B, ec(l). For any pair of functions u,reCQ\, it can beshown that

- r t f , | , ' - . \ / , - r r \ lvL(u\-uL*(v\=\*lI r, l ,?-,+1.*lt,-2?ll (r.2r)

, : d x , l - t r

' \ d x , d x i ) \ i a r , 1 1 \ " - ' l

This is known as Lagrange's identity.

EXAMPLE 1.11 Construct an adjoint to the Laplace operator given by

L ( u ) = \ $ + u D (1.22)

Solution Comparing Eq. (1.22) with the general linear PDE (1.19), we have All=l,lrr=1.From Eq. (1.20), the adjoint of (1.22) is given by

L+O)=+O)+{{ i=uo * , " "dx- dy-

Therefore,

L * ( u ) = u o + u ,

Hence, the Laplace operator is a self-adjoint operator.

EXAMPLE 1.12 Find the adjoint of the differential operator

L ( u ) = u o - r t ' / l t 1 \

Solution Comparing Eq. (1.23) with the general second order PDE (1.19), we have 4r =1,Br =-1. From Eq. (1.20), the adjoint of (1.23) is given by

L*O)= 4;0\-4Cn)=u**,dx' oI

Therefore,

L* (u) = uo t, Y,

It may be noted that the diffusion operator is not a self-adjoint operator.

1.5 RIEMANN'S METHOD

In Section 1.2, we have noted with interest that a linear second order PDE

L(u) = 61'' 11is classified as hlperbolic if 82 > 4AC, and it has two families of real characteristic curves in thexy-plane whose equations are

Page 72: K. Sankara Rao

FUNDAMENTAL CONCEPTS

1=f i@,y )=c t , 4= fz@,y)=cz:re' (|'TD are the natural coordinates for the hyperboric system. In the -r,r'-prane, the curves:;:),::i:\:1:::!)i"^r:,::!","haracteristics of the given pDe as shown i"'nig. i.ifale in the 6?-plane, the curves 6=ct and e=cz are furn'il i". oi.truight lines parallel to the; as shown in Fig. l.l(b).

A linear second order partial differential equation in two variables, once classified as a hyperboricrrion, can always be reduced to the canonical form

2'2,/1 z, z-, 1L )

d+ dy

consider an eguation which is alteady reduced fo its canonicd form in te vaiab)es

d-u du . duL(u) = - + a- + b-; + cu = F(x. y)

ox oy ox oy

r Z is a linear differential operator and a, b, c, F are functions of r and y only and arerntiable in some domain IR.

(a) (b)

Fig. 1.1 Families of characteristic lines.

d v(x, y) be an arbitrary function having continuous second order partial derivatives. Let us

ler the adjoint operator Z* of Z defined by

(1.24)

( t .2s))2,, ) )Z * (v) = -:-:- - -:- (av) - + (bv) + cv

ox oy ox oy

u =o*-u4, N =buv+vfu- ov' dx

we introduce

(1.26)

Page 73: K. Sankara Rao
Page 74: K. Sankara Rao

66

then

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

M, + N, = ur(ov) + u(av), - u rv, - w, + u r(bv) + u(bv), + v ru, + vu,

Adding and subtracting cuv, we get

T - ) ' l r - 1 ' r du .du lu,+N,=-,1 - l - ' " - ! wr- levy+cvl*ul ! .^ *o - *o - *rul'

Loxoy ox dy ) ldxdy dx dy l

l e.

y L u _ u L * v = M t r N y ( 1 . : - '

This is known as Lagrange identity which will be used in the subsequent discussion. The operarrZ is a self-adjoint if and only if L=L*. Now we shall attempt to solve Cauchy's problem whiciis described as follows: Let

L@)= 16,r ' , ( 1 . 1 8 r

with the condition (Cauchy data)

(i) u= f(x) on f, a curve in the xy-plane;

) , ,( i i ) : : = s( x) on r.

onThis is a, and its normal derivatives are prescribed on a curve I which is not a characteristic lina

Let f be a smooth initial curve which is also continuous as shown in Fig. 1.2. Since Eq. (1.ltlis in canonical form, .r and y are the characteristic coordinates. We also assume that the tangetto f is nowhere parallel to the coordinate axes.

Flg. 1.2 Cauchy data.

LeI P(6,D be a point at which the solution to the Cauchy problem is sought. Let us drarthe characteristics PQ and PR through P to meet the curve f at Q and R. We assume that ru,, uy ate prescribed along f. Let ?lR be a closed contour PpRP bounding lR. Since Eq. (l.ltris already in canbnical form, the characteristics are lines parallel to x and y axes. Using Greel':theorem, we have

f f f

l l ( M , + N , ) d x d y = Q - ( M d y - N d x )J J " J l ' R 'R

P(1, tt)

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FUNDAMENTAL CONCEPTS

dlR is the boundary of lR. Applying this theorem to the surface integral ofEq. (1.27), we

[ ̂ _q ay-Na,1=![tu4o)-ut*{D]a,at (1.30)J ' R

o .

other words,

| <*or- rua 'y* [ (Mdy-Ndx)+ l rvar-N, / " )= f [ pL@)-uL*(v) )dxt ryJ r ' J a ' J r o ' ' - ' '

J J

using the fact that dy=o on PQ and rk=o on PR, we have

l_1u ay-Na'1+l .u ay- [-^ l, 'a"=[[ 1"21u)-uL+ (v)]dx dyJr ' ' JRp ' .1 pe , i i . . , -

Eq. (1.26), we find that

Irn'*=1n u"a*+[o * 'd*

by parts the second term on the right-hand side and grouping, the above equation

l,n u a, =[uv)an + la u@v - v,1 dx

this result into Eq. (1.31), we obtain

luvl, = luvlg + J, u (bv - v,) dx -J u

u@v - v r) dl

( 1 .31 )

( 1 .35 )

- [ , w ar - u a4 *l! lu4u) _ ,L* {u)]a, at

IR

.u,* choose v(x,y:6,q) to be a solution of the adjoint equation

? t* 1,1= g

r de same time satisfy the following conditions:

vt=bv when y=q, i .e . , on PQyy = av when "r = 4, i.e., on PR

v = l w h e n r = 6 , y = q

rzi rhis function v(x,y;€,D as the Riemann function ot the Riemann-Green function. Since=F-Eq. (1.32) reduces to

( 1 .32 )

( 1 .33 )

( 1.34a)

(1.34b)

( 1.3ac)

lul p = luv)e - [, [u (av - v r) trt, - v (bu + u -) dl + [ I eF) ttx dy

= :alled the Riemann-Green solution for the Cauchy problem described by Eq. (1.28) when;r- are prescribed on l. Equalion (1.35) can also be written as

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68 INTRoDUCTI0N To PARTIAL DIFFERENTIAL EQUAI.IONS

I f r rl u l p = l u v l o _ l - u v \ a d y _ b d x ) + | ( u v , d y _ v u , d x ) + l l ( v F ) d x r t v { 1 . 3 6 1- . t f J r J d

This relation gives us the value of z at a point p when u and u, arc prescribed on f. But whenu and u, are prescribed on f, we obtain

t t , r rl u l p = L u v l R - J r u v \ d

d y - b d x ) - J r ( / v r d x + w ) . d y ) + J J

( v F ) d x d y ( t . 1 7 )JR

By adding Eqs. (1.36) and (1.37), the value of u at p is given by

l . - f l rl u l , = - { l uv l " + [av lp l - l _uv (s dy -bd r ) - ^ l ue ,d r - v , ,dy )2 r r 2J r "

I f r r+ - l v (u ,d r -uudy \+ l l 1v f1dx ay ( t . i 8 )2J r ^ " ' d

Tlrus, we can see that the solution to the Cauchy problem at a pornt (q,q) depends only onthe cauchy data on f. The knowledge of the Riemann-Green function therefore enables us tosolve Eq. (1.28) with the Cauchy data prescribed on a noncharacteristic curve_

EXAMPLE 1.1J Obtain the Riemann solution for the equation-)

d x d Y - * - " '

grven

( t ) u = f ( x ) o n f

),,(u) - = g(x) on I

otx

where f is the curve y=x.

Solution Here, the given PDE is

^ t

L@) = -:--L. = r1x. y1 (t.3e)ox oy

We construct the adjoint La of L as follows: setting

M = quv-uyy, N =buv+vu,

and comparing the given equation (1.39) with the standard canonical form of hyperbolic equation( I .24.1, we have

a = b = c = 0

Therefore,

M =-uvy, N =yux (1.40)

Page 77: K. Sankara Rao

FUNDAMENTAL CONCEPTS

M x + Ny = vury - ur) ry = vl(u) - uLi (v)

^7

l * (v)=__:__:dx dy

self-adjoint operator. Using Green's theorem

l l w,+ N,)d, dt = | ._w ay - u a,;"fi '

" dt<

t f

ll lvL(u) - uL* tv)ldi dy = J a{v

ay - u a')I R '

f l Ur -,1. ri la* ay = | ̂ _ g ay - u a"t'ril rdt<

69

1-:

F:::. Z = Z* and is a

l ro{u at, - N a4 =l r<u at, - w aO * ! n, {u ar - u ai + ! r^ {u at - N a,)

r : 1 a v e

J, w at, - N a,t = [,( -"!.0, - "!*), r \ d y d x )

1.3, we have on f, r=/. Therefore, dx=dy. Hence

J, tv at, - u a,t = [,[-, t, -, ' i)*

( 1 . 4 1 )

( 1 . 4 2 )

( l .4 r )

( 1 . 4 4 )

Flg. 1.3 An il lustration ol Exampte 1.13.

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70 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

since on qP, y = constant. Therefore, dy = 0. Thus,

I tuar-ua* l=f -Nat=f - ,La* ( r .4s)JQP' . JQP JQP Ox

Similarly, on PR, x = constant. Hence, dx = 0. Thus

I say-xa*t=[--r+=[ - ,?+ (r.46)J p R ' ' J p R ' J p R d y

'

Substituting Eqs. (1.44)-(1.46) into Eq. (1.43), we obtain from Eq. Q.al, the relation

! ! w r - r * o ty a* + = J,l, ̂ L, a, -' * *)* I ",-" fi a, * J, r-, fi o,n " \ oY ox ) r v r

But

t 0, . -P r 0vJpp- ' 2ra '=F 'u lo

* Joo"E*Therefore,

t t - - t ' 1

" )

)l vr - "L* <uila, at = [.] -"ff,a, -,7* |n " \

o Y o x )

D r ) v c d v+l-vulb+),pufrdx+)ro-ufidt 0.41)

Now choosing v(x,y;6,il as the solution of the adjoint equation such that

(i) Z*v=0 throughout the.ry-plane

dv( i i )

^=t t when y=4, i .e . , on QP

Av( i i i ; | =0 when x = 6, i .e . , on PR

oy

( iv) v=l a t P(( ,4) .

Equation (1.47) becomes

t t t ( ov ou . \| {vDax a t = | - l

-u -ax-v ;dx l+(uv \p- (u )p

n , , \ oy ox )

or

(u)p=(uv)e*1-( - , !a , - ' ** l - f i (vF)dxdt ( r .48)- , r \ oy dx J , t

Page 79: K. Sankara Rao

FUNDAMENTAL CONCEPTS

f t f a 2 1(uv)q - tuv)p = | _ d(uvt= l - l i1n)ax + l twlay IJ t lox dy I

f= J,

(urv dx + uv, dx + urv dy + uv, dy\

'. .r Eq. (1.48) can be rewritten as

(u)p = (uvt p = J, (ur, dy + *, ay1 - lJ pr; axayIR

: -:il1', adding Eqs. (1.48) and (1.49), we get

I l r .(u)p =11@v)p+{*)n)+

2 ) r (uv, dx - vu, dx)

* )[ ,{ur, at *,', an - II eFt dx d.yIR

at-l.ltPlE 1.14 Yerify that the Green function for the equation

d2u * 2 ( du * ! ! \ =ndx dy x+y \dx dy )

-

. , : : c r t o u=0 ,0u ldx=3x2 on y= r , i s g i ven by

(x + y) l2xy + 1( - rl.) g - y) + 2(r7), r ^ , t , t , ' r t -T

i-: :ltain the solution of the equation in the form

y=1x -y1 (2x2 - xy +2y2 )

Solution In the given problem,

,,=#k.h*.h*=, (, so)-::ring this equation with the standard canonical hyperbolic equation (1.24), we have

o=b= 2

, C=0 , F=ox+ y

:: lint equation is Z+(v)=Q, \vhs1g

L.(D= -! ' : - a= (Zl-!(JL).d x d y d x \ x + y ) d y \ x + y )

'-:at

Lrv=0 throughout the r/-plane

1 l

( l .4e)

( l . s l )

Page 80: K. Sankara Rao

nt

, , ^ dv z(l l l - = -v

dx x+ y

(iii) +=---:--voy x+y

( i v ) v= l

If v is defined by

INTRODUCTION TO PARTTAL DIFFERENTIAL EQUATIONS

on PQ, i,e., on y=q

on PR, i.e., on -r = f

at P(4,t1).

/ - r . , \v (x, y; (,4"1= :-:--!J-lzxy + (( _ r) (x _ y) + 2qryl

\e +D"Then

(1rs2)

( l .s3)

( 1.54)

(1.s5)

0.s6)

and

#=ffi,,,.o-^v|*ut.#*l

*= A#*, + 2Y2 + 2x (( - 4) + 2(41

dzv _4 (x+y )0x 0y G + tD3

dv = -)- .gry +zx2 -2y\q _ D +z6tt)a l 11+ r ' f ' ' " ' ' ' ^ ' t \e t t '

Using the results described by Eqs. (1.53f{1.56), Eq. (t.51) becomes

' *P1= !:-- ' (? * ' :)* o'dxdy x+ y\dx 0y) 1x+ y)2

=1.! r* ! )_ , ,?._ , r l4xy+z(xz +y2) lG+d" (x+y) ( (+4) r "

or

z*1uy= 4( '+Y)- 4(x+v) =sG +q)" (6 +d'

Hence condition (i) of Eq. (1.52) is satisfied. Also, on y=7.

* =

A#*' + 2t12 + 2x (( - 11) + 2(41

Page 81: K. Sankara Rao

FUNDAMENTAL CONCEPTS 73

*1,=,= dipq2 + zx (( +'D + z(qt

2v/(x + y) at y =r7 is given by

2v2x+y x+q ff irr*r*n

-Do-d+z6q)

I= . t lznz +2x(( +r)+2{41G+?D '

Eqs. (1.57) and (1.58), we get

0v2- : -= -Y a t V=ndx x+ y

propertyatx=1,

(ii) in Eq. (1.52) has been verified. Similarly, property (iii) can also be verified.v=q '

G+i lG+D2G + ril3

( t .s7)

( l . s8 )

F + nv = -2-+124n +G _ tD' +26ry1=

\q+nr(iv) in Eq. (1.52) has also been verified.(1.50) and (1.51), we have

vl(u\ - uL * (v) = v !+ -, !+. +( 4). 4( 2L\d x d y d x d y d x \ x + y ) d y \ x + y )

d(0u \ a ( av \ d (2vu \ 0 (2vu \= - l ! - : - l - - l r = - l + - l - l += - l - |dy \ dx ) dx \ dy ) dx \ x+y ) dy \ x+y )

d ( zvu dv) 0 ( 2vu du\= - - l - - u - - l + ^ l - +v : - |dx \ x+y dy ) dy \x+y dx . )

AM dN=-+ -dx dy

M=2* -uaru , N= 2 ' " *u7ux + y d y x + y d x

-.ing Green's theorem, we have

tr c r.O

)) lvL(u\ - uL* (v\l dx dy --JaRW dy- N d') = J;

(M dy- N dx\IR

* !' <u ay- u ai+!R g ay - r a'1 ( l .5e)

Page 82: K. Sankara Rao

74

(see

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

Fig. 1.3) on QP, y = C. Hence, dy = 0; on PR, x = C. Therefore, dx = 0

pl l zuv lv l . | 2u, a^ . f= | | 1 - - u - ^ - l d y - \ - + v - = - l d x lJR L t r+ / dy l ' l r +y d r ) . l

rP l I zuu d" ]1 . ra l [ 2u, 0 , ]1 ._ t t1_+v___f ldY+ l | 1 :_r j l ldyJ? L t ;+ / dx ) ) JP L t - r+ / dY t . t

However,

( ( Z u v . a u \ . r P 2 u v p c P i v

J? \x+v '*)*=Jn

f i ,x*t* ' 'n- ln ' ^a-Now, using the condition u=0 on y=x, Eq. (1.59) becomes

!! t uut - uL* qu d' dv = I: ( * -,*)* - 11 | * -, ' i" l*R

J / ( \ x + ) , d y ) . , n \ x + . y d x t

- l ' 2u ' d t - (u r \o+ (uv rn - l ' r \ a ,

r Q x + y " J Q d x

.l',#*-tl("#,)*Also, using conditions (iiF(iv) of Eq. (1.52), the above equation simplifies to

(u)p=(uv)q-Ji"**

Now using the given condition, viz.

* = 3 t 2 o n n Q

we obtain

(u) p = (uv)e - 3 J: 4?##4)e=-+-l 'st *,3q41a,

\€+ry r "E

=- *-r lir,

-$*lenrn, -s^tl

= ffi,4/

* 62 r72 + rta 1 + 3qrt 1t, * rl, ))

=G-, ie€z -6n+zn2)

Page 83: K. Sankara Rao

FUNDAMENTAL CONCEPTS

u(x, y) = (x - y) (2x2 - xy + 2y21

Show that the Green's function for the equation

d-u- ; ; + a = 0ox oy

75'--erefore,

:ce the result.

.+VPLE 1.15

v(x, y|€,D = J0

J0 denotes Bessel's function of the first kind of order zero.

Solution Comparing with the standard canonical hyperbolic equation (1.24), we have

a = b = 0 . c = l

: 3 self-adjoint equation and, therefore, the Green's function v can be obtained from

d-v;--;- +Y =Uox t7v

d v ^- " -=udx

0v- ; -=Uoy

on Y=q

on x=1

at x=6, y=7t

0k =a(x - i l \ -n )

dv 0v d(dx dQ 0x

kQHl=oo- ,1)dx

1=io"rty-nlo x K

Page 84: K. Sankara Rao

76

Thus,

INTRODUCTION TO PARTI{ DIFFERENTIAI EQUATIONS

0 v d v a . t - r . -; - = - ; ; ; Q " ' \ y - 4 )a x d g E

. t ^ T ^d ' v d l d v a . t - L I

a,ay= ayLaie' 'v-i l)

=110"* 2+rr-nto-k ?u a0,. . - , , r r-r , . - ,a '1, ag-]k L dQ aQ ay\Y-ry)*e v-nt

N ay)

However,

fr=io"-<'-aTherefore,

+=Ap'o!*1r-r16-k(,-q)o-n)ld'o**d' ' )2un' ' ' Idxdy kL do ' k ' ap ' --u-ntf i i0 '-r tx-t t ]

Hence,

ffi*"=ogives

zlo! 6ro-r, a1*{11_e14 *r,r -Lf*,=okLk ' dO' k ' 'd0 '

aO)or

o-( ,r-r

d2: * rr* 4 l* u = ok'\ d0" d0 )

OT

^ t 2Q ' v " + Q v ' + L Q K v = 0

Let k=2, a=4. Then the above equation reduces to

Q2v" + (v, +Q2v=0=v,,+lv,+v (Bessel,s equation)0

Its solution is known to be of the form

u = Jo@) = JoQJG - g U - rt)

which is the desired Green's function.

Page 85: K. Sankara Rao

l r

FTJNDAMENTAL CONCEPTS

EXERCISES

I' Find the region in the xy-prane in which the following equation is hyperbolic:

l(x- y)2 - lluo + 2u, +f(x - y)2 _tlu, =g

2. Find the families of characteristics of the pDE

(l- x2)uo - u, =o

in the elliptic and hyperbolic cases.3. Reduce the following PDE to a canonical form

uB + t"yurry =0

{. Classifr and reduce the following equations to a canonical form:

(a) y2uo-x2uo=0, r>0, y>0.

(b) uo +2u, +\0 =0.

(c) e 'uo*eYuo=u.

(d) x2uo t2ryu, + y2u, = 0.

(e) 4uo+5u, +\ry +ux +uy =2.

5. Reduce the following equation to a canonical form and hence solve it:

3uo+ l \u r+3u ) ry=O

6. lf L(u)=c2yo-ur, then show that its adjoint operator is given by

L* = c2vo -vx

-. Determine the adjoint operator ,* corresponding to

L(u) = Ayo I pu, +Curry + Du, * Eu, t Fu

wherc A, B, C, D, E and F are functions of .r and y only.i. Find the solution of the following Cauchy problem

u, = F(x, y)

glven

u = f (x), fi= sUl on the line y = I

using Riemann's method which is of the form

I t " v ^u(xo, yi =

ilf @o> + t<yr)]+ iJ, sel e - J) F@, y) dx dy

Page 86: K. Sankara Rao

INTRODUCTION TO PARTIAL DIFFERENTIAL EOUATIONS

where IR is the triangular region in the xy-plane bounded by the line y = I and the linesx = x6, ! = lo through (-16, y6).

9. The characteristics of the partial differential equation

d 2 " . ̂ d 2 t , 1 2 " a - a --;-= - ,;-i + cssz vj all + 3! = 0dx' ox oy dv- dx dy

when it is of hyperbolic type are... and... (CATE-Maths, 1997)

10. Using ry = x + y as one of the transformation variable, obtain the canonical form of

dzu ^02u . d2u- : - ; - z ; - - - f - . = -=u .

dx' ox oy dy'

Choose the correct answer in the following qu€stions:

11. The PDE

y3uo - (r2 -1)u, =O is

(A) parabol ic in {(x,y):x<0}

@) hyperbolic in {(x, y) : y > 0}

i (C) elliptic in IR2(D) parabolic in {(r, y):x > 0}.

12. The equation

(B) r /+0

(D) r/ > 0.

(GATE-Maths, 1998)

(GATE-Maths, 1998)

x21y - l1zo - x7y2 - t)rry+y(yz - l )zo+ z,=0

is hyperbolic in the entire x/-plane except along(A) x-axis @) y-axis(C) A line parallel to y-axis (D) A line parallel to r-axis.

(GATE-Maths, 2000)13. The characteristic curves of the equation

,2uo - Y2,o = ,2 Y2 +r, x>0 are

(A) rectangular hyperbola (B) parabola(C) circle (D) straight line.

(GATE-Maths, 2000)14. Pick the region in which the following PDE is.hyperbolic:

yuo+2xyar+xu)ry=ux+uy

(A) xy +l

(C) xy>l

(GATE-Maths, 2003)