K. Sankara Rao

z E

pastern E.ononY Edition

o |JJ

o

z E

IJ

Introduction to

|JJ

m

PARTIAT DIFFERENTIA EquATr0Ns

CONTENTS u

Preface

0. Partial Differential Equationsof First Order

t-46

0 . 1 Introduction

I Surfacesand Normals 2 w,6.3 Curves and their Tangents 4 Formation of Partial Differential Equation 6 wg4 Solution of Partial Differential Equationsof First Order ,10 -09 Integral SurfacesPassingthrough a Given Curve ,{Z \9.7 The Cauchy Problem for First Order Equations 20 '0.9 SurfacesOrthogonalto a Given Systemof Surfaces 21 First Order Non-linear Equations 22 0.9.1 Cauchy Method of Characteristics 2.1 \y.1,0 CompatibleSystemsof First Order Equations 30 Charpit's Method 34 0.ll.l SpecialTlpes of First Order Equations 38

,,91

y.{

I

,,xrr

Exercises 43

1. Fundamental Concepts

47:78

1.1 Introduction 47 ''./2 Classificationof SecondOrder PDE 47 Canonical Forms 48 -14 1.3.1 CanonicalForm for Hyperbolic Equation 49 1.3.2 CanonicalForm for Parabolic Equation 51 1.3.3 CanonicalForm for Elliptic Equation 53 Adjoint I .4 Operators 62 I .5 Riemann's Method 64

Exerckes 77

2. Elliptic Differential Equations '.4.t . 2.2 .,tl C/

Occunenceof the Laplaceand PoissonEquations 79 2.1.1 Derivation of LaplaceEquation 79 2.1.2 Derivation of Poisson Equation 8/ BotndaryValueProblems(BVPs) :82 SomeImportantMathematical Tools 82 Properties of HarmonicFunctions 84 2.4.1 The SphericalMean 85 v

79-146

vt

CoNTENTS

2.4.2 Mean ValueTheoremfor HarmonicFunctions g6 Maximum-Minimumprinciple and ConsequencesgZ ,2.4.3 Separationof Variables 9-l 4 Dirichlet Problem for a Rectangle 94 ,;tK The NeumannProblemfor a Rectanple 92 Z1 InteriorDirichlerProblemfor a CircL 9g -ZA Exterior Dirichler Problem for a Circle 102 _29 -.f l0 lnterior NeumannProblem for a Circle 106 ._V1,1 Solutionof LaplaceEquationin CylindricalCoordinates /0g 91 2 Solutionof LaplaceEquationin SphericalCoordinates 1/J Examples /22 ,2. l3 Miscelianeous Exercises I44

3. Parabolic Differential Equations 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 1.9

4. Hyperbolic Differential Equations 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.ll 4.12 4.13

147_lgg

Occurrenceof the Diffusion Equation 147 Boundary Conditions 149 ElementarySolutionsof the Diffusion Equarion 150 Dirac Delta Function /54 Separationof VariablesMethod /j9 Solutionof Diffusion Equationin CylindricalCoordinares 121 Solutionof Diffusion Equationin SphericalCoordinares _124 Maximum-MinimumPrincipleand Consequences1Zl M i s c e l l a n e o uEsx a m p l e s / 2 9 Exercises I86

Occurrenceof the Wave Equation 1g9 Derivationof One-dimensional Wave Equation /g9 solution of one-dimensionalwave Equationby canonicalReduction /92 The Initial ValueProblem;D'Alembert'sSolution 196 Vibrating Srring-Vadables SeparableSolution 200 ForcedVibrations-Solutionof Non_homogeneous Equation ?0g Boundaryand Initial Value problemfor Two-dimensional Wave Equarions*Merhodof Eigenfunction 2I0 PeriodicSolutionof One-dimensional Wave Equationin Cylindrical Coordinates 213 PeriodicSolutionof one-dimensional wave Equarionin Sphericarpolar Coordinares 21J Vibrationof a CircularMembrane 2/Z Uniquenessof the Solutionfor the Wave Equation 2/9 Duhamel'sPrinciple 220 Miscellaneous Examples 222 Exercises 230

lgg_232

Ii

I I I

CoNTENTS

5. Green'sFunction t., 5.2 5.3 5.4 5.5 5.6

233-2

l n r r o d u c r i o nz J J Green'sFunctionfor LaplaceEquation 239 The Methods of Images 245 The EigenfuncrionMethod 2jZ Green's Function for the Wave Equation_Helmholtz Theorcm 254 Green'sFunctionfor the Diffusion Equation 259 Exercises 263

6. Laplace Transform

Methods

Z6S-J1

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.ll 6.12 6.13

Introducrion 265 Transformof Some ElementaryFunctions 26g Propeniesof LaplaceTransform 270 Transform of a Periodic Function 228 Transformof Error Function 280 Transform of Bessel'sFunction 28J Transform of Dirac Delta Function 2g5 InverseTransform 285 Convolution Theorem (Faltung Theorem) 292 Transform of Unit Step Function 296 Complex Inversion Formula (Mellin-Fourier Integral) 299 Solution of Ordinary Differential Equations 302 Solutionof PartialDifferenrialEquations 302 6.13.1 Solurionof Diffusion Equarion 308 6.13.2 Solutionof Wave Equarion j13 6.14 Miscellaneous Examples 321 Exercises 329

7. Fourier Transform Methods 7.1 7.2

7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10

Introduction 333 Foti,er Integral Representations -tj.t 7.2.1 Fourier Integral Theorem 335 7.2.2 Sine and Cosine Integral Representationsjjg Fourier TransformPairs -139 Transform of ElementaryFunctions 340 Propertiesof Fourier Trasnform -i4j Convolution Theorem (Faltung Theorem) -i56 Parseval'sRelation -t58 Transform of Dirac Delta Function 3i9 Multiple Fourier Transforms .li9 Finite FourierTransforms 360 7.10.1Finite Sine Transform i61 7.10.2Finire CosineTransform 362

333-3g

CONTENTS

7.ll Solutionof DiffusionEquation 363 7.12 Solutionof Wave Equation 367 7.13 Solutionof LaplaceEquation 371 Examples J73 7.14 Miscellaneous Exercises 384 Ansversand Keysto Eterches Bibliagaphy Index

388-423 425 427-430

PREFACE

'\'ith the remarkableadvancesmade in various branchesof science,engineeringand technology :rday, more than ever before, the study of partial differential equationshas becomeessential.For, :.: have an indepth understandingof subjectslike fluid dynamics and heat transfer,aerodynamics :lasticity, waves, and electromagnetics,the knowledge of finding solutions to partial differentia :quationsis absolutelynecessary. This book on Partial Differential Equationsis the outcome of a seriesof lecturesdelivered me, over several years, to the postgraduatestudentsof Applied Mathematicsat Anna ry university, chennai. It is written mainly to acquaint the reader with various well-known :nathematicaltechniques,namely, the variables separablemethod, integral transform techniques and Green's function approach,so as to solve various boundary value problems involving parabolic,elliptic and hyperbolicpartial differenrialequations,which arise in many physica situations.In fact, the Laplaceequation,the heat conductionequationand the wave equation have been derived by taking inro account certain physical problems. The book has been organizedin a logical order and the topics are discussedin a systemati manner In chapter 0, partial differential equationsof first order are dealt with. In chapter I, the classificationof secondorder partial differential equations,and their canonical forms are given. The concept of adjoint operatorsis introduced and illustrated through examples,and Riemann's method of solving cauchy's problem described. chapter 2 deals with elliptic differential equations.Also, basic mathematicaltools as well as various DroDertiesof harmonic functions are discussed. Further,the Dirichletand Neumannboundaryvalueprtblemsare solvedusingvariable separablemethod in cartesian,cylindrical and sphericalcoordinatesystems.chapter 3 is devoted to a discussionon the solution of boundary value problemsdescribingthe parabolicor diffusion equation in various coordinate systems using the variables separablemethod. Elementar solutionsare also given. Besides,the maximum-minimumprinciple is discussed,and the concept of Dirac delta function is introduced along with a few properties.chapter 4 provides a detailed study of the wave equation representingthe hyperbolic partial differential equation, and gives D'Alembert'ssolution. In addition, the chapterpresentsproblemslike vibrating string, vibration of a circular membrane,and periodic solutions of wave equation,shows the uniquenessof the solutions,and illustratesDuhamel'sprinciple.chapter 5 introducesthe basic conceptsin the constructionof Green's function for various boundary value problems using the eigenfunction method and the method of images. chapter 6 on Laplace transform method is self-containedsince the subject matter has been developed from the basic definition. various properties of the transform and inverse transformare describedand detailedproofs are given, besidespresentingthe convolution theorem and complex inversion formula. Further, the Laplace transform methodsare applied to solve severalinitial value, boundary value and initial boundary value problems.Finally in Chapter7, the theory of Fourier transformis discussedin detail. Finite Fourier transformsare also introduced,and their applicationsto diffusion, wave and Laolace equationshave been analvzed The text is inr.erspersed with solvedexampleslalso. miscellaneous examplesare given in

CHAPTER O

PARTIAL DIFFERENTIAL EQUATIONSOF FIRST ORDER :.1 INTRODUCTION

::rial differentialequations offirst orderoccurin manypracticalsituatrons suchas Brownianmotion r:3 theofy of stochasticprocesses, radioactivedisintigration,noise in communi.utron,yrt.,rr, :rDulation growth and in many problemsdealing wiih telephonetraffic, traffic flow along a -.ghwayand gas dynamicsand so on. In fact, their study is essentialto understand the nat-u :: solutionsand forms a guide to find the solutionsof hijher order partial differentiarequation A first orderpartialdifferentiarequation(usuallydenotedby pDE) in two independent variabre ,:..r,and one unknownz, also called dependentviriable, is an equationof the iorm

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

\

dy)

( 0 .t )

I n r r o d u c i ntgh e n o l a t i o n OZ

P=^,

ox

d7

s=-

dy

(0.2

Equation(0.1) can be written in symbolicform as F t x ,y , z . p , q ) = 0 .

(03)

A solutionof Eq. (0.1) in somedomain O of IR2 is a function z = f(x, y) definedand is of C,, in Q shouldsatisfythe following two conditions: ( i ) F o r e v e r y( x , y ) e O , t h e p o i n t( r , y , " , p , q ) i s i n t h e d o m a i no f t h e f u n c t i o nF . ( i i ) W h e nz = f ( x , y ) i s s u b s t i t u t eidn t o E q . ( 0 . 1 ) ,i t s h o u l d r e d u c et o a n i d e n r i t yi n x , y f o r all (x, y) e Q.

we classifythe PDE of first order dependingupon the form of the function F. An equation of the form S,

).

P1x.y. ztli + Qtx.y. zt+ = R\x.y. z) ox dy

(0 . 4 )

is a quasi-linear PDE of first order,if tr,ederivativesdz/dx and dzr0y tr-tatappear in the function -Eare Iihear'while the coefficientsp, e andR dependon the indepe;dentvariabresx, I,and arso on the dependentvariablez. Similarly,an equationof the form

INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS

), s, Ptx,ytl + Qtx.y\f = R(x.y. z) ox

dy

(0.s)

is calledalmost linear PDE of first order,if the coefficientsp and o are functionsof the independentvariablesonly. An equationof the fonn ;?

s"

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

(0 . 6 )

is cafleda finearPDE of first order,if the functionF is linear in 0zl0x, dzldy andz, while the coefficientsa, b, c and d dependonly on the independent variablesx and y. An equationwhich doesnot fit into any of the abovecategoriesis called non-linear.For example, 0z dz \t, x 1 +Y. =nz ox oy is a linear PDE of first order. dz dz ) ltt) x 1 +Y a =zox 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 \dx ) \dy )

is a non-linearPDE of first order. Beforediscussingvariousmethodsfor finding the solutionsof the first order pDEs, we shall review some of the basic definitionsand conceptsneededfrom calculus.

AND NORMALS 0,2 SURFACES LetO be a domainin three-dimensional IRr andsuppose space F(x,y,z) is a functionin the class C'(O), then the vector valued function grad F can be wrrrten as

(ap aF aF\

. ^ g r a of = l . . - " . . - l \ox oy

dz )

(0.7)

If we assumethat the partial derivativesof F do not vanishsirnultaneously at any point then the set of points (x, y, z) in Q, satisfyingthe equation F(x, y, z) = (

( 0 . 8)

is a surfacein o for someconstantc. This surfacedenotedby s6 is calleda level surfaceofF. lf (,16,y6, z6) is a given point in Q, then by taking F(x6,y|,zi=(, we get an equationbf the form F(x, y, z) = F(xo, y6, zs),

(0e)

PARTIALDIFFERENTIAL EQUATIONS OF FIRSTORDER

3

rlich representsa surfacein f), passingthrough the point (r0,J0,zo). Here, Eq. (0.9) represent . oe-paraneter family of surface in o. The value of grad F is a vector, normal to the level rrfrce- Now, one may ask, if it is possibleto solve Eq. (0.8) for z in terms of ;r and y. To-answer G qnestion, let us consider a set of relations of the fomr

a=fr(u,v), y= f2Qt,v), 2=fi(u,v)

(0.10

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

u!{'r"*, o \u, v)

( 0 .l 1 )

len, the first two equationsof (0.10) can be solved and z and y. can be exDressedas functions ofr and y like u = f,(x. y),

v = p(x, y).

Thus, r and v are obtainedoncex and y are known, and the third relation of Eq. (0.10) givesthe ralue of z in the form

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

(0.12

This is, of course,a functional relation betweenthe coordinatesx, y andz as in Eq. (0.g). Hence, ry point (x, y, z) obtained from Eq. (0.10) always lie on a fixed surface.Equations (0.10) are also called paranetric equationsof a surface.It may be noted that the parametricequation of a srface need not be unique, which can be seen in the following example: The parametric equations -r=rsrndcosd, y=rsinAsind,

z=rcos0

and

rro, , = , = r Q Q 1 ) " r n r . " = , ( 1 -+o0' ")' )o " e , (r+0") t+0' both representthe same surfacex2 + y2 + z2 = 12 which is a sphere,where r is a constant. If the equation of the surface is of the form

z = f(x, y)

(0.13

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

(0.14)

Then

;

partiallywith respectto .r and.y, we obtain Differentiating dF .0F - - : - - dz -=v, ^

=-f

dx .dz dx

dF

dF dz

0y

0z 0y

-+--={l

from which we get 0z --:-::TdFldx dF -= = -- (usiue 0.14) .tx df ldz dx

4

INTRODUCTION TO PARTIALD]FFERENTIAL EOUATIONS

or o"

dx

= n

we obtain Similarly, dl.

,-=q

)E

and

'

:1

=_l

dv d_ Hence,the directioncosinesof the normalto the surfaceat a point(x, y, z) are givenas (0.15) Now, returningto the level surfacegiven by Eq. (0.g), it is easy to write the equationof the tangent plane to the surface ,.t" at a point (rp, y6, z6) as

.lar J l a r r 1 6 . y n . -l- o r l = 0 . ( x - x 0 )l a l F( x 0 . l o . zIo ) {0.tbr , _ l + ( l r o ) l l r x s . y 6 . z 6 ) l + r r r s tlf "l Lax ) ,-,

0.3

s

Ld:

CURVES AND THEIR TANGENTS

A c u r v e i n t h r e e - d i m e n s i o nsapl a c eI R I c a n b e d e s c r i b e di n t e r m s o f p a r a m e t r i ce o u a t i o n s Supposei denotesthe positionvectorof a point on a curvec, then the vectorequationol C mav b e w r i h e na s v=Fltl

t^r

lcI

( 0 .l 7 )

where 1is some interval on the real axis. In componentform, Eq. (0.17) can be written as

x=I(1).

y=f2G), z=fr(t\

( 0 .r 8 )

wherei = (r, /, z) andF(t) =[fi(t), f2(), hO] andthefuncrions to C,(1). fr, f2 and/3 betongs Further,we assumethat

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

( 0 .l e )

This non-vanishing vecroris tangenrro the curve C at the poinr (x, y, z) or at [J;O, f20, f](t)) of the curve C. Anotherway of describinga curve in three-dimensional spaceIRI is by using rhe fact that t h e i n t e r s e c l i oonf t w o s u r f a c e sg i v e sr i s e l o a c u r v e . Let and

\(x, y, z) = C1 | F2@,y, z) = Crl|

(0.20)

\ are two surfaces. Their intersecrion. ir nor empty.is alwaysa curve, providedgrad F, and grad F: are not collinearat any poinr of ct in IRj. In other words,the intersectionof surfices siven b 1 E q . ( 0 . 2 0 )i s a c u r v e i f

PARTIALDIFFERENTIAL EOUATIONS OF FIRSTORDER

etad 4Q, y,z)xglad F2@,y, z) + (0,0,0)

(0.2

:;:* every@,y,2)eQ. For variousvaluesof C1 andC2,Eq. (0.20)desuibesdifferentcurv ::r€ totalityofthesecurvesis calleda two parameter familyof curves.Here,c1 andc2 arereferr 15parameters of this family.Thus,if we havetwo surfacesdenotedby s] and52 whoseequatio r: in the form

F ( xy. ,z ) = o l

rJ

(0.2

G(x,y, z) = 0l l::en, the equationof the tangentplaneto,Sl at a point p(xs, ys,z()) is )F

)F

)F

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

=0

(0.2

!:nilarly, the equationof the tangentplaneto,S2at the point p(x6,y6,26) is

dG aG . .aG ( . x- x 0 ) = - + ( / - y n ) ^ - + \ z - z = - = l ) . oxdy-dz

(0.2

iere, the partial derivatives dFllx, dGldx,etc. are evaluatedat p(xx,yx,zs). The intersectionr :Fse two tangent planes is the tangent line I at P to the curve C, which is the int3rsection( ::e surfacess' and .92.The equationof the tangentline z to the curue c at (xo, yo, is obtaine "i ::r'm Eqs. (0.23) and (0.24) as (x-,ro)

AF-AG aF E

_

(y- yo\

(z-zol _ -aF -AF-AI

aG-

(0.2

ay ar- a, a, a" dr-E a, E n- n a, (x-xs)

(y- yi

\z-zo)

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

(0.2

0 (y, ,) 0 (2,x) 0 (r, y) 'l}rerefore, the directioncosinesof I are proportionalto

l a 1 r , c 7a @ , G )d ( F , q 1

Lao,d'aea' a@,r\l

(0.2

aor illustration,let us considerthe following examples: EX4MPLE 0.1 Findlhe tangentvector at (0,1,nl2) ro the helix describedby the equation x=cost, Solution

- y = s i nr ,

z=t,

1€1in lR'.

The tangent vector to the helix at (r, y, :) is ( dx dv d:\

\a'a'i

)={-"n

t ' c o s ll') '

7t

6

nnnoDUcTIoN To PARTTAL DIFFERENTIAI. EQUATIoNS

we observethat the point (0,1,tr/2) corresponds to t = ttl2. At this point (0,1,n/2), the tangent vectorto the givenhelix is (-1,0,l). EXAMPLE 0,2 Find the equationof the tangentline to the spacecircle f + y 2 + 2 2= 1 , x + y + z = o

at thepoint(1/Jt4,2tJ14,-3tJ:,4). Solation

The spacecircle is describedas F(x,Y,z) = Y2+Y2+t2 -l=0

G(xY , ,z )= s a Y a 2 = g RecallingEq. (0.25),the equationof rhe tangentplaneat (l(i4,

x -rilta

2lJA, 4lJw

can be wrirten

v-zlJu

,"h-,1+") ,(#)-,(#)

-;---l"\--7----\-

z+3/Jl4

,[t)-,fz) -\J'4, -l.Ji?l

x-rtJ14 y-2/.64

0.4

z+3ktl4

FORMATIONOF PARTIAL DIFFERENTTALEQUATION

Supposeu andy are any two given functionsof x, y and z. Let F be an arbitraryfunctionof z andv of the form F(u,v)=6

(0.28)

we can form a differentialequationby eliminatingthe arbitraryfunctionF. For, we differentiate Eq.(0.28)partiallywith respectto r andl, to get

0Fl0u d" 1 arlA, a" 1 ^

ELa,' yr y a,la,*yt l='

(0.2e)

and

drldu.d, l arlau dvf

a"Lar*Eq I d,Lay+Es )=u

(0.30)

PARTIAL DIFFERENTIAL EQUATIONS OF FIRS-I' ORDER

\-.rv. eliminatin 0Fldu and 0Fl0v fton Eqs. (0.29) and (0.30), we obtain -du + - Ddu dx dz'

-dv + - D0v dx dz'

-0u + - o 0u

-dv + - o dv

d),

dy

dz

d2

;1ich simpliltesto d(u,r)

0 1 u , v ) 0 \ u ,v )

Pa o A " q a e $ = a l r , y )

( 0 . 3I )

fhis is a linear PDE of the t)?e

(0.32)

PP+Qq=R, '* hele

'^

d\u.v\

00.2)'

^

'

O(u,v)

ae,.\)'

^

"

O\u,v\

d\x,y)

( 0 . 1)3

:quation (0.32) is called Lagrange'sPDE of first order.The following examplesillustrate 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 (1i) f(x+y+2, *2 +y2 +t21=0. . Solution (i) Given z = f Qc+it) + g(x -it) Differentiating Eq. (1) twice partially with respectto r and ,, we get

(1)

s"

:= f ' l x + i t )+ g ' ( x i t l ox -) ".;= f"(x+it)+ s"tr-it). dx-iere, /' indicates derivative of /with :.spect to (x-il). Also, we have

(2)

respect to (-r+tt) and g' indicates derivative of g with

), : = if ' (x + it) - iC' - it) lx dt

-)

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

(r)

clt-

i:om Eqs. (2) and (3), we at once, find that (4) ;hich is the requiredPDE.

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 + "2 Hence, the required PDE is of the form pp+

eq = R, (Lagrangeequation)

(l)

where

lau Arl =la, avl=lt 2vl , =a^(u'') d\y.z) ldu

avl I

IE

EI

Id"

0rl

tox

dxl

z,l=zf"-,1

,=#3=1fr 1";,,,-,, frl=li l?!

"- a*r)=l+ p=!(r.r\ _ld* I dy

a'l dxt lt

2rl

zrl=2{t-') +l=l' dy I

Hence,the requiredPDE is 2(z- y)p +2(x - z)q =2(y - x) or (z-y)p+(x-z)q=y-x. EXAMPLE 0.4 Eliminatethe arbitraryfunctionfrom the following andhence,obtainthe corresponding partial differeniial equation: 1i1 z=xy+ f(x2 +y2)

(ii) z = f(xylz). Solution ( r ) U r v e nz = x y + J \ x - + y ' )

(l)

DifferentiatingEq. (l) partiallywith respectto .r and we obtarn J.,, )! 1 = y + 2 x f ' \ x 2 * y r 1 =p dz ay=x+zYf'(x'+Y')=q

(2) (3)

PARTIAL DIFFERENTIALEQUATIONSOF FIRST ORDER

Eininating /'

from Eqs. (2) and (3) we get W-)KI=y2-x2,

*ich is the required PDE. (ii) Given 2= f(ry/z) Differentiating partially Eq. (l) with respectto r and /, we get

(4)

(1)

), :: = !f,(xylz) = p oxz

(2)

2.

(3)

=: I,kytz\=s + oy z Etuinating /' from Eqs. (2) and (3), we find xp-vq=o J

w=(]v lich

(4)

is the required PDE.

E/IMPLE

0.5 Form the partial differential equation by eliminating the constantsfrom z=e+by+ab.

Solution

Given z = ax+ by + ab

(l)

Differentiating Eq. (l) partially witi respectto.x and / we obtain -dz -=a= ox

p

(2)

dz q av='=

(3)

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

EGMPLE 0,6 Find the partial differential equationof the family of planes,the sum of whose ! r'. ? interceptsis equal to unity. Solution

Let

*

+1, +1=l

be theequationofthe planein interceptform,so rhala+b+c=1.

aDc

Thus, we have -x+y-z+- - = l a b l-a-b

(l)

7 IO

NTRODUCTIONTO PARTIALDIFFERENTIAL EQUATIONS

Differentiating Eq. (l) with respectto x and y, we have l* a

P l-a-b

I' b

Q -n t-a-b-"

p

=n

I

(2)

1 b

(3)

l-r_b=-;

and "'

q t_"4=-

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

!=!

(4)

qa

Also, from Eqs.(2) and (4), we ger pa=a+b-1=a+!a-l q

alt+L-pl=1.

\s)

Therefore, a=q/(p+q_ pq)

(5)

Sirnilarly, from Eqs. (3) and (4), we find b=pt(p+q_pq)

(6)

Substitutingthe valuesof a and 6 from Eqs. (5) and (6) respectivelyto Eq. (r), we have p+ q_ pq q

x+p+q_ p

pq , * P + Q_pq _

P Qz = l

or

| !+l-'= q p pq p+q- pq That is, Px+qY-2=---!!-, p+q-pq

0)

whichis the requiredPDE. 0.5 SOLUTIONOF PARTIALDIFFERENTTAL EOUATIONSOF FIRST ORDER In Section0.4, we have observedthat relationsof the form F(x, y, z, a, b) = 0

(0.34)

PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

It

. . - :ise to PDE of first order of the form ( 0 . 35) f ( x , y , z ,p , q ) = 9 . - -,:. any relationof the form (0.34) containingtwo arbitraryconstantsa and 6 is a solutionof '. ?DE of the form (0.35) and is called a complete solurionor completeintegral. fonsider a first order PDE of the form

), ), P\x.y. z)- + Q(x,y, z)"+ = R\x.y. zl

( 0 . 36 )

Pp+Qq=R,

(0.37)

ox

dy

'r.nply

"-::. -r ?rd.;,,are independentvariables.The solutionof Eq. (0.37) is a surfaceS lying in the ...--)-space,called an integralsurface.If we are given that z=-f(x,y) is an integralsurface .' ::: PDE (0.37).Then, the normal to this surfacewill have directioncosinesproportionalto : - : x . d z l d y , - l ) o r ( p , g , - l ) . T h e r e f o r et h, e d i r e c t i o n o f t h e n o r m a li s g i v e n b y i = \ p , q , - 1 \ . :'.': the PDE (0.37),we observethat the normal ii is perpendicular to the directiondefinedby -: .:ctor/ =\P, ( s e e R ) F i g . 0 . 1 ) . Q,

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

any integralsurfacemust be tangentialto a vector with componentslP, Q, Rj, and -:-:e. we will never leavethe integralsurfaceor solutionssurface.Also, the total differentiald: -:ren by

a,=(a,*(ay dx

:-:r

dy

(0.38)

E q s .( 0 . 3 7 )a n d ( 0 . 3 8 ) ,w e f i n d lP, Q, Rl =ldx, dy, dz\1

'.:.i. rhe soiutionto Eq. (0.37) can be obtainedusingthe following theorem:

(0.3e)

12

INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS

Theorem0.I The generalsolutionofthe linearpDE PP+Qq=R canbe writtenin the form F(u,v\=Q, whereI'is an arbitraryfunction,andu(x,y,z)=Ct and v(x, y, z) = C2 form a solutionof the equation dt_dy

dz

p(x, y, z)

e@, y, ")

R(x,y, z)

(0.40)

Proof We observethat Eq. (0.a0)consistsof a set of two independent ordinarydifferential equations,that is, a two parameterfamily of curvesin space,one such set can be written as dv OG. v. z\

(0.4r)

a=;G;;

which is referredto as "characferisticcrrrve". In quasiJinearcase,Eq. (0.41) cannotbe evaluated until z(r,/) is known. Recalling Eqs. (0.37) and (0.38), we may recast them using matrix nomllon as

o - l ll t a z D x \ ( R \ t=l I dy)\dz/dy) \dz)

L lax

t0.42)

Both the equationsmust hold on the integralsurface.For the existenceof finite solutionsof Eq. (0.42), we must have D

ax

/11

| D

ayl ldx

RI IR

ol ; t=0 ayl

dzl ldz

(0.43 )

on expandingthe determinants, we have

dx P (x, y, z)

dy

dz

Q@,y, z)

R(x, y, z)

(0.44)

which are called auxiliary99!4j!ensfor a givenPDE. ln order to complete the proof of the theorem,we have yet to show that any surface generatedby the integralcurvesof Eq. (0.44)hasan equationof the form F(r.i,v) = 0. Let

u ( x ,y , z ) = ( ,

and v(x,y,z)=C,

(0.45)

be two indeperident integralsof the ordinarydifferentialequations(0.44).If Eqs.(0.45) satisfy (0.44). we Eq. then. have du du du ;ox dx+=-dy+;-dz=du=0 dy

dz

and

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

dz

PARTIAL DIFFERENTIALEQUATIONSOF FIRST ORDER

t3

Solving these equations,we find _;_-____;_____=_

dx

dv

du dv du dv

=__=___j......-

0u dv dufr

at at- a" a, E Ar- arE dz du 0v

du 0v'

a, ay- ay dr rfiich can be rewritten as

l

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

---;-------i

(0.46)

:iow, we may recall from Section0.4 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 _dz PQR

T b e s o l u t i o n o f t h e S ee q u a t i o n sa r e k n o w n t o b e u ( x , y , z ) - _ C 1a n dv ( x , y , z ) = C 2 . H e n c e f(2. v) = g is the required solution of Eq. (0.37), if u andv are given by Eq. (0.a5), We shall illustrate this method through following examples: EX4MPLE 0.7 Find the general integal of the following linear partial differential equations: ( i ) y 2p - x y q = y ( 2 - 2 y 1 (ii) (y+zx)p-(x+yz)q=7s2 - 12. Solulion (i) The integral surface of the given PDE is generatedby the integral curves ofthe auxiliary eouatlon dx y'

dy -ry

dz x(z -2y)

(1)

fhe first two members of the above equation give us or

:==

xdx=-ydy,

rhich on integration results in _2

-.2 L+C 2

or

x'+ v'=C,

(2)

l4

INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS

The

lasttwo members of Eq. (l) give dY= tu^ -y z-zy

or zdy-2ydy=_ydz

That is, 2ydy=ydz+zdy, which on integrationyields y2 = yz +C2

or

y2 - yz =C2

(J)

Hence,the curvesgiven by Eqs. (2) and (3) generatethe requiredintegralsurfaceas F\xz+y2, y2 - lz)=0. (ii) The integralsurfaceofthe given pDE is generated by the integralcurvesofthe auxiliary equation

dx_dy y+zx

-(x+

dz *2 -y,

lz)

(t)

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

dz

;;;7 _;;fr=? _yz or xdx+ydy _ dz (r2 - yz) ,2 - y2 "

That is,

xdx+ydy=2fu. On integration,we get ,'2 ,2 _t2+ " =L 2 V-t

or

x'+Y'-z'=Ct

(2)

Similarly,for gettingthe secondintegralcurve, let us considerthe combinationsuch as ydx+xdy dz

7;;;7__y=7_t or ydx+xdy+dz=0, w h i c h o n i n t e g r a t i orne s u l t si n xY+z=C2 Thus, the curvesgiven by Eqs. (2) and (3) generatethe requiredintegralsurface as F ( x 2+ y 2 - z 2 , x y + z ) = g .

(3)

PARTIAL DIFFERENTIALEeuATioNS oF FrRsroRDER

EGIMPLE

l5

0.8 Use Lagrange'smethod to solve the equation

,'T a

f

dz iox

ozl

I rl-^

1l

oyl

-l I

there z=z(x,y). Solution

The given PDE can be written as

x tl -^pa- yz;l-Ila - y, l1- a - y - ^ oyJ L L

|d,ozf

l +z ld = - - l t - l = t ) dxJ L dy dx)

2.

'-

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

(t)

dy

The correspondingauxiliary. equationsare dx (yy- Fz)

dy (az-yx)

(2) (fx-ay)

Lsing multipliers x, y, and z we find that each fraction is _xdx+ydy+zdz 0

Ihrefore, xdx+ydy+zdz=0, rhich on integrationyields ,2+y2+"2=c,

(3)

Similarly,usingmultipliersa, f , nd y, we find from Eq. (2) that each fraction is equal to adx+Bdy+ydz=0, riich on integrationgives ax+py+yz=Cz Thus,the generalsolutionof the given equationis foundto be ' F ( t z + y 2 + 2 2 ,a x + B y + y z ) = 0 EXAMPLE0.9 Findthe generalintegrals of the followinglinearpDEs: (i) pz-qz=22 +(x+y)2 (iil (x2 - lz) p +(y2 - u)q = ,2 - 'y.

(4)

t6

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

Solution (i) The integral surfaceof the given PDE is generatedby the integral curvesof the auxiliary equation dx z

dy _z

dz

(l)

z 2+ ( x + y ) 2

The first two members of Eq. ( 1 ) g i v e dr+dY-9, which on integrationyields (2)

x+Y=Cl Now, consideringEq. (2) and the first and last membersof Eq. (l), we obtain zdz

,

z'+ci or 2z dz z'+C; which on integrationyields

ln (22+ Cl) =2x + C2 OI

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

(3)

Thus,the curvesgiven by Eqs.(2) and (3) generates the integralsurfacefor the given PDE as F(x + y, log \x2 + y2 + 22 + 2xyj -2x) = 0 (ii) The integralsurfaceof the given PDE is given by the integralcurvesof the auxiliary equation

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

dz

(l)

,2 -*y

Equation(1) can be rewrittenas dx-dy dy-dz dz-dx _ _ (x- y)(x+ y+ z) (y-z)(x+y+z) (z-x)(x+y+z)

(2)

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

-vl

(3)

PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

l7

S.nilarly,consideringthe last two terms of Eq. (2) and integrating,we obtain (4)

fi=c, rus, the integral curves given by Eqs. (3) and (4) generatethe integral surface /\ "' (y-'' ,-x) 3,6

INTEGRAL SURFACES PASSING THROUGH A GTVENCURVE

the previoussection,we have seenhow a generalsolution for a given l i n e a r P D E .c a n b e :rained.Now, we shall make use of this generalsolutionto find an integrals u r f a c ec o n t a i n i n . :iven curve as explainedbelow: Suppose,we have obtainedtwo integralcurvesdescribedby u(x, Y, z) = (t ) t t v(x, Y, z) = C, )

(0.48)

' rm the auxiliaryequationsof a given PDE.Then,the solutionof the given PDE can be written - the form F(u,v)= Q

(0.4e

Suppose,we wish to determinean integral surface,containing a given curve C describedby -.3 parametricequationsof the form x=x(t),

y=y(t),

z=z(t),

(0.5 0)

.:.ere1is a parameter.Then, the particularsolution(0.48) must be like

u{x(t),y(t),r0)} = cr I

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

(0.5 r)

:.us,we have two relations,from which we can eliminatethe parameter/ to obtain a relation :: rhetype F(C.,C)=0

(0 52)

.:ich leadsto the solution g i v e n b y E q . ( 0 . a 9 ) .F o r i l l u s t r a t i o nl ,e t u s c o n s i d e trh e f o l l o w i n g : uple of examples. t.Y4MPLE 0.10 Find t\e integralsurfaceof the linear PDE x(y2 + z)p - y(x2 +z)q=1x2 - y21z -:ntainirrg t h e s t r a i g h lti n e x + y = g , s = 1 . Solution

The auxiliary equationsfor the given PDE are ___1 x\y2+ z)

- y t x z- : 1

dz (r' - yt),

(l)

INTRODUCTION TO PARTIALDIFFERENTIAL EQUAIIONS

18

Using the multiplier xyz, we have yzdx+zxdy+xydz=0. On integration,we get xtz = Ct

Q)

Suppose,we use the multipliersx, y and z. Then find that each fraction in Eq. (l) is equalto xdx+ydy+zdz=0, which on integrationyields *2 +y2 +"2 =C,

(3)

For the curve in question, we have the equationsin parametricform as x=t,

y=-t,

z=l

Substitutingthesevaluesin Eqs. (2) and.(3), we obtain - t 2= C , )

'f

(4)

zt2+t = Cz) theparameter t' we find Eliminating 1_zcr=c2 or 2Cy+C2-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 integralsurfaceof the linear PDE x.P+Yq=z

which contains the circle defined by t2+y2+12=4,

x+y+z=2.

The integral surface of the given PDE is generatedby the integral curves of the Solution auxiliary equation dx _dy =dz xyz I

(l)

Integrationof the first two membersof Eq. (l)gives

ln .r= ln/+ InC

(2)

PARTIALD]FFERENTIAL EQUATIONS OF FIRSTORDER

t9

i:milarly, integrationof the last t\ryomembersof Eq. (l) yields

,v = C z

(3)

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

o[:,zl=o z)

s)

\y

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

(s)

*2+y2 +"2 =4 x+y+z=2

(6)

i:om Eqs. (2) and (3), we have !=xlCv

z=ylCr=a1grg,

thesevaluesofy and: in Eqs. (5) and (6), we find S.rbstituting t2"(rr) xz' +x 2- + -:, =4, oI. *2ll+-!+--l" Ll

LrL2

|

xx(rt\

x+-+-=2.

q Crcz

Ci

CiC;)

l=a

' o r x^1"c,-cE)-' ll+'+ l=2

(7)

(8)

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

1*{* -l -=f,*t* t l' ci cici l. cr crc2J' -hich on simplificationgives us )11

_+_+--=0

Cr CrCz c r'C, l:rat is, C r C 2 r C t+ t = 0 . \ow, repfacingCl by xly and C2 by ylz, we get ihe required integral surface as -x-v+ _ +xt = 0 , yz y

(e)

20

I N T R O D U C T I O NT O P A R T I A LD I F F E R E N T I A LE O U A T I O N S

zy Yl,rw'tl'?-n

0.7

THE CAUCHY PROBLEM FOR FIRST ORDER EQUATIONS

Consideran interval1on the real line. If xo(s),y6(s) and zs(s) are three arbitraryfunctionsof a singlevariablese1 such that they are continuousin the intervallwith their first derivatives. Then, the Cauchy problem for a first order PDE of the form F(x, y, z, p, q) = Q

( 0 . 53)

is to find a region IR in (r, y), i.e. the spacecontaining(xo(s),-},o(s)) for all s e 1, and a solution z = dQ, y) of the PDE (0.53) such that Z [ x 6 ( s ) y, 6 ( s ) ] =Z o ( 5 ) and QQ,y) togetherwith its partialderivativeswith respeclto n and ), are continuousfunctions of x and y in the region IR. Geometrically,there exists a surfacez=QG,y) which passesthrough the curve f, called c q u a l i o n sa r e d a l u mc u r v e .w h o s ep a r a m e t r i e x=,x6(s), y=yo(s),

z=zs(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 problemof Cauchy. In order to prove the existenceof a solutionof Eq. (0.53) containingthe curve f, we have to make further assumptionsabout the lorm of the function F and the nature of f. Basedon these we have a whole class of existencetheoremswhich is beyond the scope of this assumptions, book. However,we shall quote one form of the existencetheoremwithout proof, which is due to Kowalewski(see Senddon,1986). T h e o r e m0 . 2 I f (i) SCy)and all of its derivativesare continuousfor y y6l 0 . ,_-: I

ln the half-planex 4AC, and it has two families of real characteristiccurves in the xy-plane whose equationsare

FUNDAMENTAL CONCEPTS

1=fi@,y)=ct,

4=fz@,y)=cz

:re' (|'TD are the natural coordinates for the hyperboric system. In the -r,r'-prane,the curves

of the givenpDe as showni"'nig. i.ifa

:;:),::i:\:1:::!)i"^r:,::!","haracteristics le in the the curves

6?-plane, 6=ct and e=cz are furn'ili". oi.truight lines parallelto the ; as shown in Fig. l.l(b). A linearsecondorder partial differentialequationin two variables, onceclassifiedas a hyperboric rrion, can always be reduced to the canonical form 2'2, /1 z, z-,

d+ dy

1L )

consideran eguationwhich is alteadyreducedfo its canonicd form in te vaiab)es d-u du . du L(u) = + a- + b-; + cu = F(x. y) ox oy ox oy

(1.24)

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

(a)

(b) lines. Fig. 1.1 Familiesof characteristic

Let us d v(x, y) be an arbitraryfunctionhavingcontinuoussecondorderpartialderivatives. ler the adjoint operatorZ* of Z definedby )2,,

)

)

Z * (v) = -:-:- - -:- (av)- + (bv)+ cv ox oy ox oy

(t.2s)

u = o * - u- 4 , N = b u v + v f u dx ov'

(1.26)

we introduce

66

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

then M, + N, = ur(ov) + u(av), - urv, - w, + u r(bv) + u(bv), + vru, + vu,

Addingand subtractingcuv, we get T -)

'l

r -1'r

l -l-'"- ! wr-levy+ cvl*ul!.^ *od-u *o. d-u *rul u,+N,=-,1 ' ox dy dx dy Loxoy

)

ldxdy

l

l e.

yLu_uL*v=MtrNy

(1.:-'

This is known as Lagrange identity which will be used in the subsequentdiscussion.The operarr Z is a self-adjointif and only if L=L*. Now we shall attemptto solve Cauchy'sproblemwhici is describedas follows: Let

L@)=16,r',

(1.18r

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

(ii) :: = s(x) on r.

on This is a, and its normal derivativesare prescribedon a curve I which is not a characteristiclina Let f be a smoothinitial curvewhich is alsocontinuousas shownin Fig. 1.2.SinceEq. (1.ltl is in canonical form, .r and y are the characteristiccoordinates.We also assumethat the tanget to f is nowhere parallel to the coordinate axes.

P(1,tt)

Flg. 1.2 Cauchy data. LeI P(6,D be a point at which the solution to the Cauchy problem is sought. Let us drar the characteristicsPQ and PR through P to meet the curve f at Q and R. We assumethat r u,, uy ate prescribedalong f. Let ?lR be a closed contour PpRP bounding lR. Since Eq. (l.ltr is already in canbnical form, the characteristicsare lines parallel to x and y axes. Using Greel': theorem, we have ff

f

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

JJ R

FUNDAMENTALCONCEPTS

dlR is the boundaryof lR. Applying this theoremto the surface integral ofEq. (1.27), we

[ ^_q ay-Na,1=![tu4o)-ut*{D]a,at

J'R

(1.30)

o.

other words,

< * o r - r u a ' y *J[a ' ( M d y - N d x )J+r lo 'r v a' r - N ,-/' "' ) =JfJ[ p L @ ) - u L * ( v ) ) d x t r y

J| r '

using the fact that dy=o on PQ and rk=o

on PR, we have

- (v)]dx dy ' ' .[-^ Jl_1u r ' ay-Na'1+l J R.u p ay1p el,'a"=[[ , i i 1"21u)-uL+ ..,

(1 . 3 1 )

Eq. (1.26), we find that

*'d* Irn'*=1nu"a*+[o by parts the second term on the right-hand side and grouping, the above equation

l,n

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

this result into Eq. (1.31), we obtain

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

-

u

u@v- vr) dl

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

(1.32)

IR

.u,*choosev(x,y:6,q)

to be a solution of the adjoint equation

g t* 1,1=

?

(1 . 3 3 )

r de same time satisfy the following conditions: vt=bv yy = av v=l

w h e ny = q , i . e . , o n P Q when "r = 4, i.e., on PR w h e nr = 6 ,

y=q

( 1.34a) (1.34b) ( 1.3ac)

rzi rhis function v(x,y;€,D as the Riemannfunction ot the Riemann-Greenfunction. Since = F - E q . ( 1 . 3 2 )r e d u c e st o

lulp = luv)e- [, [u(av- vr) trt,- v(bu+ u-) dl + eF)ttxdy [I

(1 . 3 5 )

= :alled the Riemann-Greensolution for the Cauchy problem describedby Eq. (1.28) when ;r- are prescribedon l. Equalion (1.35) can also be written as

68

INTRoDUCTI0N To PARTIAL DIFFERENTIAL EQUAI.IONS

I f rr l 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 xr t v .tf Jr Jd

{1.361

This relation gives us the value of z at a point p when u and u, arc prescribedon f. But when u and u, are prescribed on f, we obtain t t , rr d y - b d x ) - J r ( / v r d x + w ) . d y ) + ( v F ) d xd y lulp=LuvlRJruv\d JJ

(t.17)

JR

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

-

f

lr

l u l , = - { l u v l " + [ a v l p l -l _ u v ( sd y - b d r ) - ^ l u e , d r - v , , d y ) rr 2 2Jr " I f

+ - l v ( u^ , d r - u u d y \ r+r l 1l v f 1 daxy "'d 2Jr

(t.i8)

Tlrus, we can see that the solution to the Cauchy problem at a pornt (q,q) dependsonly on the cauchy data on f. The knowledge of the Riemann-Greenfunction therefore enablesus to solve Eq. (1.28) with the Cauchy data prescribed on a noncharacteristiccurve_ EXAMPLE 1.1J Obtain the Riemann solution for the equation -) - *-"' dxdY grven (t) u=f(x) ),, (u) - = g(x) otx

onf on I

where f is the curve y=x. Solution

Here, the given PDE is ^t

L@) = -:--L. = r1x. y1 ox oy

(t.3e)

We construct the adjoint La of L as follows: setting M = quv-uyy,

N =buv+vu,

and comparingthe given equation(1.39) with the standardcanonicalform of hyperbolic equation ( I .24.1,we have a=b=c=0 Therefore, M =-uvy,

N =yux

(1.40)

FUNDAMENTAL CONCEPTS

69

1-:

M x + Ny = vury- ur) = vl(u) - uLi (v) ry ^7

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

(1.41)

F:::. Z = Z* and is a self-adjoint operator.Using Green's theorem

dt= | ._w ay- u a,; ll w,+ N,)d, ' dt<

"fi

"

r:1ave tf = ll lvL(u) uL*tv)ldidy Ja{v ay u a')

IR'

fl 'ril

Ur -,1. rila*ay=rdt< | ^_g ay- u a"t

=l l ro{uat, Na4 r0,

(b) uo +2u, +\0 =0. (c) e'uo*eYuo=u. (d) x2uo t2ryu, + y2u, = 0. (e) 4uo+5u, +\ry +ux +uy =2. 5. Reducethe following equationto a canonicalform and hencesolve it: 3uo+l\ur+3u)ry=O 6. lf L(u)=c2yo-ur,

then show that its adjoint operatoris given by L* = c2vo -vx

-. Determine the adjoint operator,* corresponding to L(u) = Ayo I pu, +Curry+ Du, * Eu, t Fu whercA, B, C, D, E and F are functions of .r and y only. i. Find the solutionof the following Cauchyproblem u, = F(x, y) glven

u=f (x),

= fi= sUl ontheliney I

using Riemann's method which is of the form I t yi = @o> u(xo, +t