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Introduction to

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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
!-lo

<52,

PARTIAL DIFFERENTIALEQUATIONSOF FIRST ORDER

21

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

-oz a = J.(l x . y , z , =dz\ -l ox

dy)

\

and

(iii) For all valuesof y in the intervalll _ tsl < 6b Leo, y) = S0). 0.8 SURFACESORTHOGONALTO A GIVENSYSTEMOF SURFACES one of the usefulapplicationsof the theoryof linear first order pDE is to find the sysrem of surfacesorthogonalto a given systemof surfaces.Let a one-parameter famiry of surfacesis described by the equation F(x,y, z)= (

(0.54)

Then,the task is to determinethe systemof surfaceswhich cut each of the given surfaces onhogonally..Let (x, y, z\ be a point on the surfacegiven by Eq. (0.54),wherethJnormal to the surfacewill havedirectionratios()Fldx,7F/dy, drn4 *ni.n ,nuy be denotedby p, e, R.

z=d@,y) -

(0.5s)

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

Fig.0.2 Orthogonal surtace to a givensystem ot surfaces. l::n, its normal atlhe point (x, y, z.) will havedirection ratios(dz/dx, dz/O.y,_l) which, of course, ':'r be perpendicurar to the normarto the surfacescharacterized by Eq. (0.54).As u.onr.qu.n.. '.: havea relation

r(*e!-n=o ox dy

(0.56)

Pp+Qq= R

(0.s7)

22

EQUAnoNs DTFFERENnAL ro PARIAL rNrRoDUcrroN

type, and can be recastinto which is a linear PDE of Lagranges

#x.Hx=#

I

ros I

I

Thus, any solution of the linear first order PDE of the type given by either Eq. (0.57) or (0'58) | is o*hogonat to every surface of the systemdescribedby Eq. (0.5a). In other words, the surfacesI orthogonalto the system(0.54) are the surfacesgenelatedby the integral curves of the auxiliary | equations

#^=h=#"" EouArloNs o.e FrBSroRDEBNoN-LINEAR

rl I

(o5e I

partial the problemof findingthe solutionof first ordernon-linear we will discuss In rhissection, I

(PDEs) in."";fl::,T;:T: equations differentiar

(0.60t

where ,=*, ,=fr

|

continuoussecondorderderivativeswith respecttol We also assumethat the functionpossesses over a domainQ of (x,y,z, p, q)-space,and eitherFp or Fq is not zero at everyI its arguments

that such Pont

.l

't*?r r,

surtace to theintegral Fig.0.3 coneof normals

I

I

l The PDE (0.60) establishesthe fact that at every point (x; y, z) of the region, there exists a I relationbetweenthe numbersp and q such that 0@;q)=0' which definesthe directionof the I normalfi = \p, Q,- l) to the desiredintegralsurfacez = z(x, y) of Eq. (0.60).Thus, the directionI of the normal to the desired integral surface at certain point (x, y, :) is not defined uniquely.i ofthe normalsexistsatislingthe relation((p,q)=0: directions a certainconeofadmissable However, (see Fig. 0.3).

PARTIAL DJFFERENTIAL EQUATIONS OF FIRST ORDER

23

-.herefore,the problemof finding the solutionof Eq. (0.60) reducesto finding an integral ':':.e z - z(x, y). the normalsat every point of which are directedalongone of the permissible - -:-:ions of the cone of normalsat that point. Ihus, the integralor the solutionof Eq. (0.60)essentiallydependson two arbitraryconstants : fonn (0.61)

JG,y,z,a,b)=0,

family of integralsurfaces :r is called a completeintegral.Hence,we get a two-parameter - .gh the same Polnt. - -q,1 Cauchy's Method of Characteristics througha givencurye ro=x6(s), )0=,)'0(r), : ntegralsurfacez=z(x,y) ofEq. (0.60)thatpasses famlly 01' : -:x(s) may be visualizedas consistingof points lying on a certainone-parameter . : s r = x ( l , s ) , y = ) , ( / , s ) , z = z ( t , s ) , w h e r e si s a p a r a m e t eorf t h e f a m i l yc a l l e dc l r a r a c t e r i s t i c s . :lere,we shalldiscussthe Cauchy'smethodfor solvingEq. (0.60),which is basedon geometrical . ieralions. Let z=z(x,y) representsan integralsurfaceS of Eq. (0.60) in (r, y.:)-space. . - . . , , p , q , - l \ a r e t h e d i r e c t i o nr a t i o so f t h e n o r m a lt o S . N o w , t h e d i f f e r e n t i ael q u a t i o n( 0 . 6 0 ) ' . i s t h a t a t a g i v e n p o i n t P ( x x , y 6 , z 6 )o n S , t h e r e l a t i o n s h i pb e t w e e np 0 a n d 90, tllat . rr.1 linear.Hence,all the tangentplanesto possible :6 . po, qo), need not be necessarily . _r,s, :ral surfacesthroughP form a family of planesenvelopinga conical surfacecalled Monge :- \\'ith P as its vertex. In other words, the problem of solvlng the PDE (0.60) is to find . , : e s w h i c h t o u c h t h e M o n g e c o n e a t e a c h p o i n t a l o n g a g e n e r a t o rF. o r e x a m p l e .l e t u s . i e r t h e n o n - l i n e aPr D E P2

(0 . 6 2 )

-q2 =1'

.,ery point of the r1,;-space,the relation(0.62) can be expressedparametricallyas -*
(0.61)

.,rr0..1,6,:0)be the vertex and QQ,y,z) be any point on the generator.Then, the direction . of the generatorare (x x6),(,r, ya),G-zo). Now, the directionratios of the axis of the l en g l eo f t h e c o n e : \ \ h i c hi s p a r a l l etlo x - a x i sa r e ( 1 , 0 , 0 ) ( s e eF i g .0 . 4 ) .L e t t h e s e m i - v e r t i c a - l. Then. 7r ,1

( r - x u ) l + { . }- ) u ) 0 + ( z - z u i 0

l

- ' t , 61 2 + 1 : - 2 , , 1 2 .J2 Jt .-*6t2 +(r

(.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

(0.65)

24

IN'TRODUCTION TO PARTIAL DII.FERENTIAL EQUATIONS

Thus, we see that the Monge cone of the pDE (0.62) is given by Eq. (0.65). This is a right circular conewith semi-verticalangel 4 whoseaxis is the straightline passingthrough (..b, _yo,zo) and parallel to t-axis. l(.\,,.r,,.--r)

Fig,0.4 Monge cone. since an integral surfaceis touchedby a Monge cone along its generator,we must have a method to determinethe generatorof the Monge cone of the pDE (0.60) which is explained below: It nray be noted that the equationof the tangentplane to the integral surface z = z (x, ),) at the point (,re,1:e,;s)is given by \, p(x-xo)+q(y-yo)=k-z

.

(0.66)

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

(0 . 67 )

indicatingthat p and ? are not independentat (xo,yo,zo).At eachpoint of the surfaces, there exists a Monge cone which touches the surface along the generatorof the cone. The lines of contact between the tangent planes of the integral surface and the correspondingcones, that is the generatorsalong which the surfaceis touched,define a direction field on the surfaceS. These directions are called the characteristicdirections,also called Monge directions on S and lie along the generatorsof the Monge cone. The integral cunr'esof this field of directions on the intesral surface.l define a family of curves called characteristiccurves as shown in Fig. 0.5. The Monge cone can be obtained by eliminating p from the followins equations:

<----+---+--<---+--+--+-<----+---+--) Fig. 0.5 Characteristic directionson an integral. surface.

PARTIAL DIFFERENTIALEQUATIONSOF FIRST ORDER

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

25 (0.68)

anl .1-

(r-x0)+()/-).,0):1=0. ap

(0.6e)

o-fservingthat 4 is a function of p and differentiatingEq. (0.60) with respectto p, we

-a.r4 , = -df; - + =dI<- -da = u . ap

dp

dq dp

(0.70)

\.-'\\i eliminating (dqldp) from Eqs. (0.69) and (0.70), we obtain dF dF (x - x6)

ap-A vyi=' r-rn

l-

'FpF

ln

(0.71)

'(l

the equationsdescribingthe Mongeconeare given by l*:.erefore, q = q(xo,Yo,zo,p), (x - xs)p+ (y - yslq = Q - zs) (0.72)

=l I--Xn

l-9n

'P

-q

l:: secondand third of Eqs. (0.72) define the generatorof the Monge cone. Solving them for :-ro)(y-jlo) and (z-zo), we get x-xn

l-Vn

z-z^

Fp

Fq

pFo + qFn

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

dy

dz

Fp

Fq

pF, + qFo'

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

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

(0.75)

26

TO PARTIALDIFFERENTIAL EQUATIONS INTRODUCTION

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

(0.76)

at

Also, we note that dz dt

dv = -0z ; - -dx + ; - -dz = dv p dx _ + q aI- dx at dy aI at

Therefore, -d:= D t - + a f ^ dt

(0.7? )

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

a=;,a.6,a Now, using Eqs. (0.75) and (0.76),the aboveequationbecomes dp 0o 0F dt 0x 0p

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

dp dF dy dq

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

-dD: = - -dD+ - dI1 dt 0x dp

Eq.(0.60);;^ differentiating Arso, ;Tec,;: :-+-:-P+ dx dz'

d0

dr

(0.78) I

dx dq

j"

)irr, = U ^ -:*-:dp dx dq dx

(0.79)

Using Eq. (0.79), Eq' (0.78) becomes = -\r\ + prz)

(0.80)

_:a=-(Fv+qF:)

( 0 . 8)1

: Similatly, we can show that AT

curves Thus,givenan integralsurface,we haveshownthat thereexistsa family of characterislic (0.81). (0'80) (0.76), (0'77)' and to Eqs.(0.75), alongwhichx, !, z, p and g vary according Collectingtheseresultstogether,we may write

27

PARTIAL DIFFERENTIAL EOUATIONS OF FIRST ORDER

F

dx dt dy (tt dz

-= dt

(0.82)

P t aD + q r o ,

==-(f,

+ pF,t and

AT

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

equationsof the given PDE (0.60).The last three equationsare known as characteristic :.:: :'- ...:ronsof(0.82) are also calledcompatibilityconditions.Without knowing the solution z = z(x' y) c' r: PDE (0.60),it is possibleto find the functionsx(t),y(t),2(t), p(t),q(r) from Eqs. (0.82) and at eachpoint y(t),2 -- z(t) calledcharacteristics :--: rs. we can find the curvesr=r(t),/= ' = = the direction ol q q(t) determine p p(t) and that ,: - :haracteristic,we can find the numbers

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

(0.83)

togetherwith the plane (0.83) refened to each of its points is called a T-, rharacteristics, c - : : ' : : i e r i s t i cs t r i p .T h e s o l u t i o nx = x ( / ) , / = y ( t ) , 2 = z ( t ) , p = p ( t ) , q = S Q ) o f t h e c h a r a c t e r i s t i c .:-,:::Jns (0.82) satisfythe strip condition

(084)

= * a,lL fi rr,t!, [ l. h [

F.

-.. b. notedthat not every set of ltve functionscan be interpretedas a strip. A strip should curve.Thatis' thel . rhatthe planeswith normals(P. q. -l) be tangentialto the charactertstic .rrisfy the strip condition (0.84) and the normals should vary continuouslyalong the curve. is statedin the following of the Cauchy'smethodof characteristic .r rmportantconsequence

--.

strip)of the PDE: F(.r..1. --.p. g) = 0. the function h= . r..- 0.3 AJongeverystrip (characteristic . .:. P'gl is constant l" strip, we have eroo| Along the characteristic I

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

|

,,

|

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

t T

aoi-A,t,

becomes usingthe resultslistedin Eq. (0.82),the righrhandsideof the aboveequation

4Fp+F,Fq+F,(pFr+qFo)-Fo(F,+pF,)-Fo(Fr+qF,)=0. I '... tt'" function.F(x,y, z, p, 4) is constant of equations alongthe stripof the characteristic | -re (0.60). by Eq. defined F we considerthe followingexamples: . .: illusrration. |

I

INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS

28

EXAMPLE 0,12 Find the characteristicsofthe equation pg = z and determinethe integralsurface which passesthroughthe straightline x=l,z= y. Solution

If the initial data curve is given in parametricform as ,o(r) = I,

yo(s)= s,

z6(s)= s,

then ordinarily the solution is sought in parametricform as : = x(r, s),

y = y(t, s),

z = z(t, s).

Thus, using the given data, the differential equation becomes (l)

P6(s)q6(s)-s=0=r' and the strip condition gives I = pe(0)+ q6(l)

or

(2)

Qo= l.

Therefore, qo =1,

Po = r

(3)

(uniqueinitial striP)

Now, the characteristicequationsfor the given PDE are dxdydz^dDds

A=q'

A= e'

a=

zPq'

;= e' ;=q

(4)

On integration,we get p = cl exp (r), q = c2 exp (t), x = c2 exp (r) + ca I y=cl exp(t)+c4, z =2cpz exp(2t) + cs

(5)

'J

Now taking into account the initial conditions x o = 1 , y o = J r z o= s , p o = s , q o = l

(6)

we can determinethe constantsof integration and obtain (since c2 = l, ct = 0) P = s e x P ( t ) ,4 = e x P ( t ) , x = e x P ( ).l f

/ = s e x p ( / ) ,z = s e x p ( Z t )

(7)

j

the requiredintegralsurfaceis obtainedfrom Eq. (7) as Consequently, .-^t-

EXAMPLE T.lJ Find the characteristicsofthe equation pq=z surfacewhich passesthrough the parabolax=0, y'=2. Solution

and hence,determinethe integral

The initial data curve is

r o ( s )= 0 , / o ( s ) = s , z o ( s ) = s ' . Using this data, the given PDE becomes p o ( s )q o ( s )- s " = 0 = F

6)

PARTIAL DIFFERENTIALEQUATIONSOF FIRST ORDER

29

The strip condition gives

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

qs_2s=o

(2)

Therefore, t po = zolQo= ,212, = 2 pDE Now, the characteristicequations of the given are given by Qo= 2s

and

B\ '-,

= * r' * =,,*=roo, * =,,# =,

(4)

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

. Takinginto accountthe initial conditions

(5)

to =0, .Io=s, zO=s2,ps=s/2, qt=2s, u'e find h = s/2, cz =2s, ct = Qs, ca= 3l), gt =Q Therefore,we have

r =| exp1r;,q =2s exp(t), x = 2s[exp(r)- l], .y=

;texp

(r)+ tl

(6)

z = s2exp(2t) Elirninating.r and t from the last three equationsof (6), we get

167= (4y + x)2. This is the requiredintegralsurface. EYAMPLE 0.11 Find the characte stics of the PDE P 2+ q 2= 2 md determinethe inte$al surfacewhich passesthroughx=0,2=y. Solution

The initial datacurveis ro(r)= 0, yo(s)=s,z6(s)=s.

Lsing this data,the given PDE becomes p'o+q6-2=0=F

(l)

INTRODUCTION TO PARTIAL DIFFERENTIAL EOUATIONS

and the strip conditiongives

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

qo-l=0

Hence, 8 o = l ' P o= l l Now, the characteristicequations for the given PDE are given by

=rr. * = ro'+zq2 =+l 4d r d=rr, r d r t4

!=0.41=o dtdtl

I

On integration,we get

'l

P=q, q=c), x=2ctl+ql

y = 2 c 2 t + c 4z, = 4 t + c s ) Takinginto accountthe initial conditions x o = 0 ,l o = s , z o= r , p o = ! 1 , q o = 1 , we find D=Xl. o =1.x=!2t.)

I

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

equationsof(6) are parametricequationsofthe desiredintegralsurface.Elimi The last-three s and r, we get the parameters z=y!x. . O.1O COMPATIBLE SYSTEMS OF FIRST ORDER EQUATIONS Two first order PDEs are said to be compatible,if they have a commonsolution.We shal derivethe necessaryand sufficientconditionsfor the two partial differentialequations f(x, Y, z, P, q) = s and g(x, y, z, p, q) = 0 to be compatible. Let

,-d(f,s\,n

(0.87)

d (p, q)

SinceEqs.(0.85)and(0.86)havecommonsolution,we cansolvethemandobtainexplicitexpressions for u and a in the form

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

( 0 . 88)

PARTIAL DIFFERENTIAL EQUATIONS OF FIRST ORDER

3l

andthen, the differential relation Pdx+qdY=77 or

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

(0.89)

for whichthe necessary conditionis shouldbe integrable, - i . c u r l* = 0 vhereX =Id,V,-l). That is,

li


lo

i

dDy

v

il

dtdzl=s

rl

or

dGv,)+v(Q")= v, - Qy uhich can be rewrittenas

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

(0.e0)

Eq. (0.85)with respectto x andz, we get )JoWdifferentiating

1*yofi*;nfr=o

t.r,ffi.r,ffi=o But,from Eq. (0.89),we have dp =dd

oq =dv and so on.

tt'",0"* .ii"t'li t'""Jt 'lJu'"'" ,.rutt.. these L:sins f*+ fp!,+ fqv/,=o ano .f"+fpf,+fqV,=0. \lultiplying the secondone ofthe abovepair by Q andaddingto the firsl one. we readilyobtain

(f, + 0f,) + f p@,+ 00,)+ .fq(V,+ 0v = 0 ") S i m i l a r l yf.r o m E q . ( 0 . 8 6 )w e c a n d e d u c et h a t

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

32

INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

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

(V,+OV/,) I _ foG, + QE,) s r(f, + 0"f,) .fnqo go.f, OI

v, +Ov"= jUtot, - sof) +Q(fos"- sof)) ) ,gt = t.lo_\t.g+ 6 ' d_\l | t l d 1 x ,p ) A Q ,p ) )

1O.ot)

whereJ is definedin Eq. (0.87).Similarly,differentiating Eq. (0.85)with respectto y andz and usingEq. (0.88),we can showthat

=- ll a+4 *, al|.stl 6" ' * 16, r d\z,qt Ldty,q)

(0.e2)

)

thevaluesof Vt + QV, and(, +ry/, fromEqs.(0.91)and(0.92)intoEq.(0.90), Finally, substituting we obtain

atf.il *rd^tf .g)=_11!+.ra:l.etl d ( x .p \

d ( 2 .p )

d t t . S )) lA(y,S) In view of Eqs. (0.88),we can replace$ andy by p and q, respectivelyto get

a C ' s ) p d - ( 'fs ) + ' 0 ( 2 . p ) + a _ (. fs )* o d , t f. d _ o 0(t,p)

d(y,q)',

(0.93)

d(r,q)-"

This is the desiredcompatibility condition. For illustration, let us considerthe following examples: EXAMPLE 0./5

Show that the following PDES xp-yq=x

and ,2p+q=r"

are compatible and hence, find their solution. Solutioh

Suppose,we have

.f=xp-yq-x=0.

(l)

g=x2p+q-xz=0.

(2)

Then, A^

a\J,g) ltp-t) d(x, p) l(2xp- z)

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

a(f,s)_l o

x l__z

o \ 2 ,p t

x. I

a.

,-l

l-x

rl-'t

PARTIAL DIFFERENTIALEQUATIONSOF FIRST ORDER

-vl 'i=-o.

aU,e\ l-q-

=*=l

d\y,q) l0

ll

o

-vl

d(z,q) | -:r

I I

aj.il = | -:-:------|

33

t=-xv,

ard we find

a j , c ) . -pd- -(+f-,; !dJ +,qal zje,!d= ..,^+ d\x, p)

d\2, p)

d(y,q)

'

a(f,d

)

r"-r'p-12

d(z,q)

+ p x 2- q - q x y

=o_q_qry_r, =xz-q,x(qy+x) =o-q_x2p =0

Hence,the given PDEsare.compatible. Now, solvingEqs. (1) and (2) for p and g, we obtain P=q-l rqz+x x3+x2z x+x2y from which we get

p=x(1+ vz\ l+ r,z ,G$= u, and x2(z-x\

c=

x(z-x)

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 v dz='

l+ry

l+ry

OT

v(z- x\ - x(z-x\ 1z- dk = !...:.........-:dx * j o, _: or dz-& _ydx+xdy z-x l+xy On integration,we get ln (z-I)

= ln (l + r1,)+ ln c.

Ihat is, z-x=c(l+xy)

(rl

EQUATIONS TO PARTIALDIFFERENTIAL INTRODUCTION

34

Hence, the solution of the given system is found to be (4)

z=x+c(\+xy), family. which is of one-parameter 0.11 CHARPIT'S METHOD

In this section,we will discussa generalmethodfor finding the completeintegralor complete solutionof a noirlinearPDE of first order of the form

(0.e4)

ar-.,--^\-n J\^.)'..,yattt-v,

This methodis known as Charpit'smethod.The basicidea in Charpit'srnethodis the introduction of anotherPDE of first order of the form

(0.95)

g(x'v'z,p,q)=o and then, solve Eqs. (0.94) and (0.95) for p and q and substitutein /7 = p(x, y, z, a)dx + q(x, y, z, a)dy.

(0.e6)

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

(0.95)whichis alreadydiscussed of the secondequation The maintaskis the determination of the form an equation is to seek what is required, Now, in the previoussection. g ( x ,v ' z ' p ' q ) = o with the givenequation compatible f ( x , y , z 'p ' q ) = o and sufficientconditionis for whichthe necessary

a \ f . g \ + , d t f , g ,* d U . g \* o d ( J , t , _ 0 . 'a(z.qJ 'dtz-p\-dly,q)

(0.q7)

d(*. p)

On expansion,we have

(af ds_aI ae\,"(atds-af as\ '\a.' dp apa= ) \ a * a p o p d x) '

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

dq dY )

\dz

dq o: )

dq

which can be recast into

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

(0.e8)

(0.98) are This is a linear PDE, from which we can determineg. The auxiliary equationsof d\

dy

fp

fq _dq

dz

dp

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

-U' + q1"1

(0.ee)

PARTIALDIFFERENTIAL EQUATIONS OF FIRSTORDER

35

-Ihese equationsare calledCharpit'sequations. Any integralof Eq. (0.99) involvingp or q or both :an be laken as the secondrelation(0.95).Then,the integrationof Eq. (0.96) givesthe complete :regralas desired.It may b6 notedthat all charpits.quutibn, neednor be ur.I, bu, it is enough :r choosethe simplestof them. This methodis illustratedthroughthe following examples: EXAMPLE 0.16 Find the complete integral of

6? a q2yy= qz Solution

Suppose

(r)

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

p'L+qj-1zr= o

:hen,we have

/*=u, Jr=p-+q" f^ =2pv.

J,=-s

f- =2av-2.

\ow, the charpitsauxiliary equationsare given by dx

fp

dy

fq

dz

dp

dq

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

I hat ls.

dx

dy

2py 2qy-t

2p2y+2qzy-qz

elp

dq l ( p 2* q 2 ) - q 2 l Pq

(2)

of Eq. (2), we have Fromthe lasttwo members dP=dq Pq -p2 or PdP+qdq=g On integration,we get P2+q2 =4(constant)

( 3)

From Eqs. ( l). and (3), we obtain ay-qz=0

or

Q=q,/z

and

(1) (av "-\;

1 o t 2- a zy 2 1 t t z

36

INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS

Substitutingthese values of p and q in dz= pdx + q dy, we get

-77 o,=Ja ax+zay OI r---i-_---:'--;

z dz- ay dy = I m' - o' y' & which can be rewritten as d 1*2 - a2y2 7tt2

. On int€gration,we find

or \x+ b)- = \z-/a) - y-

Hence,the completeintegralis .

.-t

)

1.

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

EXAMPLE 0.17 Find the completeintegralof the PDE: ,2 = pq ry. Solution

In this example,given f =

"2

- pqxy.

(l)

Then,we have 'fx = - P4/'

fY = -PA*'

f" =22

fo=-eW' fq=-PxY' Now, the Charpit'sauxiliaryequationsbre given by

dt _dy = d" dp aq = = -(f, -(f, pf, q1:n + + pf") + q1'"1 fp fq That is, dx -qxy

dy - pxy

dz -2pqry

dp pqy-zpz

ds pqx -Zqz

From Eq. (2), it follows that dplp qy-22

dqlq px-22

dxlx -qy

dvlv -px

(2)

PARTIAL DIFFERENTIALEQUATIONSOF

FIRST ORDER

.irich can be rewritten as dplp - dqlq _-dxlx+dyty

qy-px dp

qy- px dq

dy

dx

Pq ft

integation, we find

-D X- = c(constant) qv p = cwlx Fmm the given PDE, we have ,2 = pqry=rq2y2 rtich gives q2 ="2/"y2 or q=zl^fcy=azly,

sherea = l/J7. Hence,

P = zlax' 9rbstitutingthesevaluesof p andq in dz= pdx+ q dy, get az=-dx+-d! axy

dz

l&,

dy

z

ax

y

I

lnz=:lnx+alny+lnb a'

z = brrroyo

is the complete integral of the given pDE. 0.18 Find the complete integral of ,2p2 +y2q2 -4=o Charpit's method.

37

EQUATIONS TO PARTIALDIFFERENTIAL INTRODUCTION

38

Solution

The Charpit's equationsfor the given PDE can be written as dx__

dy _

;7;- tii-

dz

_

dP

-2'P2 'a2P2+Y2q21(l)

--!c-

-2 yq'

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

-!+-1 2x'p

-Zxp'

or *x *42=o P

On integration,we gel ln (xp)=1Y1s or

xP=a

(2)

From the given PDE and using the result (2)' we get (3)

y 2q 2 = 4 _ o 2 (3) in Substitutingone set of p and q valuesfrom Eqs (2) and dz= pdr + q dy, we find that

dY .

4, = o& *.J;I xy

found to be On integration,the completeintegralof the given PDE is .=otnr*r[4Jhy*h. 0.11.1 Speciat Types of First Order Equations EquationsInvolving p and q only' That is, equationsof the tYPe

Type I

(0. I 00)

f(p, q) = 0. L e tz = a l + b y + c = 0

by f(.p'q)=o, 16gn i s a s o l u t i o no f t h e g i v e nP D E , d e s c r i b e d 1

t.

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

uf

we get Substitutingthesevaluesof p and q in the given PDE' f ( a . b )= 0

(0 l0lt

S o l v i n gf o r b . w e g e t ,6 = / ( d ) . s a y T h e n ' z--ax+A@).v+c

(0 l0lr

is the completeintegralof the given PDE

P A R T I A L D I F F E R E N T I A LE Q U A ] I O N S O F F I R S T O R D E R

39

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

"lp*"G=t Solution -: :he form

The given PDE is of the font f(p,q)=g.

Therefore,let us assumethe solution

z=ax+by+c ';:-:rg

"G+Jt'=t

or b=(r-J;)2

Hirce, the completeintegralis foundto be z = ax+ (1- nla)2 y+c. EX{MPLE 0.20 Find the completeintegralof the PDE Ps=l Solution Sincethe given PDE is of the torm f(p,q)=O, we assumethe solulionin the :t;:n z=ax+by-lc, whereab=l or b=lla- Hence,the completeintegralis I z=ax+-y+c.

I! Tlpe Il -,ar

Equations Not Involving the Independent Variables.

is, equations of the type

f ( 2 ,p , d = o

(o.lo3)

-ai a trial solution, let us assumethat z is a function of u = )c+ ay, where a is an arbitrary constant. z = f(u) = f(x + ay) dz

dz du

dx

du 0x

dz

dz 0u

'

(0.104)

dz du dz

t= sr= * ,=a ^

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

)-

,-\

Il".a.,+l=o au) \ au r::ch is an ordinary differential equation of first ordet Solving Eq. (0.105) for dz/du, we obtain ]=Qk,a) i:

\sav)

(o.ros)

PARTIAL DIFFERENTIAL EQUA-IIONS OF FiRST ORDER

39

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

Ji * Jq=t. Solulion : :--: form

The given PDE is of the form /(p, g) = g. Therefore, let us assumethe solution z=ar+by+c ,!a+.lb=l

or

b=\l-Jtt)t

l-l:::e. the complete integral is found to be z=ax+Q-Ji)2y+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 assumethe solution in the - : - , z = a r + W * c , w h e r ea b = 1 o r b = 1 1 a .H e n c e ,t h e c o m p l e t ei n t e g r a li s 1 z=ax+-y+c. a Tlpe II

EquationsNot Involving the IndependentVariables.

.l-:: is, equationsof the type

f(2, p,q)=o

(0.103)

:-. : rial solution, let us assumethat z is a function of u = x+ ay, where a is an arbitrary constant.

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

0z 0x

dz 0u du dx

(0.104)

dz du

o=+=++=,+ oy

du dy

du

S,::rituting thesevaluesofp and q in the given PDE, we get

f(,.+. du) \ a u "!)=o r::ch is an ordinary differential equation of first order. Solving Eq. (0.105) for dz/du, we obtain dz ^=QQ,a)

(sav)

dz ---:"-----a = au. QG,a)

(0.r 05)

EQUATIONS INTRODUCTION TO PARTIALDIFFEREN1IAL

40

On integration,we find

I t! '=u*"

J 9\2' a t

That is, F(z,a)=u+s=Ylaytt. which is the completeintegralof the given PDE. EXAMPLE 0.21 Find the completeintegral of P(1+q) = qz Solution

Let us assumethe solution in the form 7=f(u)=a+6y

Then, dz P=A'

q=a

dz du

thesevaluesin the given PDE, we get Substituting dz(.

dz\

dz

al'*oa)=o'd". That is, dz oar=*-'

dz or a-'-=4u

On integration,we find ln(az-l)=u+c=x+aY+c which is the requiredcompleteintegral. EXAMPLE 0.22 Find the completeintegal of the PDE: P 2" 2 + q z= 1 ' Sotation Let us assumelhat z = f(u)= x+ ay is a solutionof the given PDE.Then, dz dz c=a du du' given PDE,we obtain thesevaluesofp and g in the Substituting e=

ldzl

la")

llCEl =l '-1 +a-\du)

That is,

(*\ \du )

<,,*ort=r or

dzl du

tlr2 +o2

i

PARTIAL DIFFERENTIALEQUATIONSOF FIRST ORDER

4l

'[7*t a"=au ar lnte$ation, we get

r-i-----i zlz- +a-

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

2l

2

a

t=truv'rD

I

-:.;h is the requiredcompleteintegralof the given PDE. Equations Type lll Separable from those r.- eOuation in which z is absentand the termscontainingx andp canbe separated equation. ::-::iningy and4 is calleda separable tYPe of the Tlat is, equations (0.r07) f(x, p)= F(v'q) : trial solution,let us assumethat f(x'P)=F(Y'q)=a (saY)

(0.108)

.. solvingthem for p and q, we obtain

p = 0 @ ) 'c = v 0 ) a"=* a"**ay= pdx+qdy dx

dy

dz=O@dx+v(;)dy integralin the form we ge1the complete r ::egration,

"=lO
The given PDE is of separabletype and can be rewritten as o 2( l + , 2 \ .--. .-=9=o x'y

(say),an arbitrary constant.

J-o*

e=;67'

q=aY

r:r rring these values of P and q in 42=pdx+qdf,

(0.10e)

42

INTRODUCTIONTO PARTIAL DIFIERENTIAL EQUATIONS

we get ' u=

{ot

Jl;7ar+aYry

On inlegation, we obtain

'=Ji[*t

*1y'*b

which is the completeintegralof the given PDE. EXAMPLE 0.21 Find the completeintegral of P2+q2='+Y Solution

The given PDE is of separabletype and can be rewritten as p2 -t= y-q2 =a (say)

Then,

P =J'+i,

q=Jy+a.

Now, substitutingthesevaluesofp and 4 in dz=p&+qdy we find dz=.lx + a dx+,JW

dy,

On integration,the complete integral is found to be 2 = :(x + a)"' + | (y + a)'', + b. 5J

Form Type Mlairaut's A first orderPDE is said to be of clairaut'sform if it can be written as z= px+W+f @,q) Charpit'sequationsare The corresponding

&dv dz fr y+.fq px+qy+pfr+q1fo "+ dp p-p

dq

The integration of the last two equationsof (0.lll) p=a,

(0.lll)

q-q gives us

q=b

Substitutingthese values ofp and q in the given PDE, we get the required complete integral in the form

z=ax+by+f(a,b)

(0.r l2)

PARTIAL DIFFERENTIALEQUATIONSOF NRST ORDER

E\-lrtPLE

43

0.25 Find the complete integral of the equation

,= p**qy*rlt*p\ t, Solation ThegivenPDEis in the Clairaut's form.Hence,its complete integralis t=**ry*r[*]

*F.

Find the complete integral of (p+q)(z-xp-yq)=I The given PDE can be rewritten as

I z=xp+yq+p+q :s in the Clairaut's form,

EXERCISES .

pDE Eliminatethe arbitraryfunctionin the followingandhenceobtainthe corresponding 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

x"-d; -z+ y ' 7" d- =z $ + y ) 2 . ox oy Find the general integrals of the following linear pDEs: ,) yi) -P+xzq= ,i i )

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

Find the integral surface of the linear PDE xP-Yq=z ; h i c h c o n t a i n st h e c i r c l ex 2 + y 2 = 1 , 2 = 1 .

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 containsthe circle x2 + y2 =2a, 2 =g. 7. Find the generalintegral of the PDE (x-y)p+(y-x-z)q=z which containsthe circle *2 + y2 =1, =1. " 8. Find the solution of the equation , = L1p2 + q21+ 1p _ x\ (q _ y) 2

which passesthroughthe x-axis. 9. Find the characteristicsof the equation

pq=ry and determinethe integral surface which passesthrough the curve z=x,y=0. 10. Determine the characteristicsof the equation pz -q2 "= and find the integral surface which passesthrough the parabola4z+x2 =0, y=0. rll. Show that the PDEs and z(xp+ yq)=Zxy

xp= yq

are compatibleand 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 ) p x s- 4 q 3 x2+6x22-2=o (ri) 2(z+xP+Yq)=YP2. 1 5 . Find the complete integral of the equation p+q=pq. 1 6 . Find the complete integrals of the following equations: (l) zpq=p+q (lr) p2q2 + *t ut = *'qt (" + y').

45

EQUATIONS OF FIRST PARTIALDIFFERENTIAL

17. Find the complete integral of the PDE 2= px+ qy _sin(pq)

r 8 . Find the complete integralsof

the rollowing PDEs:

(\) tp3q2 + yp2q3+1p3+q3)-"p2q2 =o ( i 1 ) p q z = p 2 ( x q+ p 2 ) + q 2 ( y p + q 2 ) . 1 9 . Find the surface which intersectsthe surfaces of the system

z(x+Y1=s132a11 orthogonally and passesthrough the circle x 2 + Y 2= 1 , 7 = 1 '

10. Find the complete integral of the equation 1 p 2+ q 2 1 x = p z

.0202 wherep=2, tl

(GATE-Maths, 1996)

Q=2,

Find the integralsurfaceof th€ linearPDE

s-D*-G-v+4!=, dy dx which containsz-1

and *2 +y2 =1.

(GATE-Maths,1999)

the correct answer in the following questions: Using the transformationu =Wly in the PDE xux = u + yu, the transformedequation has a solution of the form l/=

(A) f(x/y) (C) f (x - y)

(B) /(x + y) (D) .f (,y).

(cATE-Maths,l e97) ofthepartialdifferential equation,p3q2 + yp2q3+1p3+q31-tp2q2 =O integral Thecomplete ls z=

(A) ax+ by+ (ab-2+ baal (C) -ax + by + (bo-2- ob-z)

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

(GATE-Maths,1997) The partialdifferentialequationof the family of surfacesz=(x+ y)+ A(ry) is (B) xP-Yq=Y-Y (A) xp- yq=g (C) xp+yq=x+y

(D) xP+Yq=o. (GATE-Maths, 1998)

INTRODUCTION TOPANTIAL DIFFERETMAL EQUATIONS

Thecomplete integral of the ppt 7=psaqy-sin(pq) is (A) z=at+by+sin(ab) @) z=u.+by-sin(ab) (C) z=ar+y1sh16; (D) z=x+6y-sin(a).

I

I I I

II I

I II {

l

CHAPTER T

FUNDAMENTALCONCEPTS . INTBODUCTION !::. practicalproblemsin scienceand engineering,when formulatedmathematically,give rise to r--:l differential equations(often refened to as pDE). In order to understandthe physical _ -r'. iou' ofthe mathematical model,it is necessaryto havesomeknowledgeaboutthe mathematical .::;ter, properties,and the sorutionof the governingpDE. An equationwhich invorves several ::Tndent variables(usuallydenotedby x, y, z, r, ...), a dependenifunctiona ofthese variables, : -re partial derivatives of the dependentfunction a with respectto the independentvariables F ( x ,y , 2 , t , . . . , u , u y ,u z ,h , . . . , u r x , u r , . . . , u r , . . - )= 0

(l.l)

:.:.leda partialdifferentialequation.A few well-krown examDlesare: u, = k(uo +uw + urz) Iinear three-dimensional heatequation] u* +u,^+ . u - -= O '). u = c- (u$ + u)ry+ uzz)

[Laplaceequationin threedimensions] Iinear three-dimensional waveequation]

q+uur=puxr

[nonlinearone-dimensional Burgerequation].

:. theseexamples,z is the dependentfunction and the subscriptsdenote partial differentiation --: respectto these variables. inition l.l rhe order ofthe partialdifferentialequationis the order of the highest derivative -:ring in the equation. Thus the above examplesare partiar differentiar equations of second :::. whereas q=uurn+stnx :--.examplefor third order partial differential equation.

CLASSIFICATION OF SECONDORDEF PDE . nost generallinear secondorder pDE, with one dependentfunctronu on a domaine of - : X = ( \ , x 2 , . . . x, n )n. > l . i s nn s1 S ri1urixJ * b,'r' + F(u) = Q z.t L i ,i = t t=l

classification of a PDE dependsonly on the highestorderderivativespresent. 47

(1.2)

48

INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS

The classificationof PDE is motivated by the classificationof the quadraticequation of the form Ax2 + Bxy+ Cy2 + Dx + Ey + F =0

(1.3)

which is elliptic, parabolic,or hlperbolic accordingas the discrimina Bz -4AC is negative. zero or positive. Thus, we have the following secondorder linear PDE in two variables;r and y': Auo + Bu,y + Curr + Dux + Eur + Fu = G

(1.4)

where the coefficientsA, B, C, ... may be functions of .r and y, however,for the sakeof simplicity we assumethem to be constants.Equation (1.4) is elliptic, parabolic or hyperbolic at a point (.x6,y6) according as the discriminant Bz(xo, y) - 4A(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 independentvariablesis two or three, a transformationcan always be found to reducethe given PDE to a canonicalform (also called normal form). In general,when the number of independentvariables is greaterthan 3, it is not always possible to find such a transformationexcept in certain special cases.The idea of reducing the given PDE to a canonical form is that the transformedequation assumesa simple form so that the subsequentanalysisof solving the equation is made easy. 1.3

CANONICAL FORMS

Considerthe most generaltransformationof the independentvariablesx andy of Eq. (1.4) to new variables 1,q, wherc (1.5) 5=1@,y), n=q@,y) such that the functions { and r7 arc continuously differentiable and the Jacobian

*o ' =#3=,;,',1= o*,-€,n,)

(16)

in the domain O where Eq. (l.a) holds. Using the chain rule of partial differentiation,the partial derivativesbecome ux = u|1x +uTna uy = ut'j +unn)

+2u4r€,4,+ rnrr\! + rg,to + urno u* = uE461. uo = ugl*4y + utT(4,ryy + 1yry")+urrn,n, + u|qn, +u44n ur, =u6gC,? +2u6r1yey +rrrrt r?+u,{r, +urr1r^

(17)

into the originaldifferentialequation(1.4), we get theseexpressions Substituting + Bu,Au"" + Cu-+ Du,9 + Eu+ Fu = G 't'I tI 99 S'l

(I 8)

FUNDAMENTAL CONCEPTS

49

- :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

(l.e)

-: nay be notedthat the transformedequation(1.8) hasthe sameform as that of the original (1.5). :::ation (1.4) underthe generaltransformation Sincethe classificationof Eq. (1.4) dependson the coefficientsA, B and C, we can also :: *rite the equationin the form Au* + Bu^.+Cu.^.= H (x. v.u.u-.u-.\

(l.l0)

(1.5),Eq. (1.10)takesone of the following -: canbe showneasilythat underthe transformation '-l:eecanonicalforms: (i) ulq-ur,=0(6,ry,u,%,ua) (l.lla) :: uq=Q,(6,n,u,uq,ao)in the hyperbolic case (ii) ury+urr=0(6,4,u,ue,ua) in theellipticcase u$ =tg,n,u,u€,u?) (iit

(l.llb) (l.llc)

:l

uaa= Q(6'n,u'u1,zo) in the parabolic case *e shalldiscussin detail eachof thesecasesseparately. UsingEq. (1.9) it can also be verifiedthat

_ tr4)2 {a, _ +,1c1 Ez_ qlc = G,?ry -d thereforewe concludethat the transformation of the independent variablesdoesnot modify r-e type of PDE. 1.3.1 Canonical Form for Hyperbolic Equation E2-qAe>0 for hyperbolic :;ncethe discriminanr case,we set 7=0 andf=0 in Eq. (1.9). ;hich will give us the coordinates and that reduce the given PDE to a canonicalform in € ry '*hich the coefficientsof u$,u44 €re zero.Thus we have ,4= A4', + B€,6y+ c6; =o

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

50

INTRODUCTION TO PARTIALDIFFERENTIAL EQUAT]ONS

which, on rewriting, become

(r

al!l

(r\

\2

\qv)

+nli l+c=o \{r i

=o ,fr'l'.uiz,l.. \tty )

Solvingtheseequationsfor

\ay )

GJ11,)and(n'/ny)'we get r-..-

4x_-B+,1B" -4AC ,

at

F--;-

r y ,_ - B - l B ' - 4 A C 2A 4y

(r.12)

The conditionB' >4AC implies that the slopesof the curves( (x, y) = C1,r\Q, y) = Cz are real. Thus, if B' > 4AC, then at any point (x, _y),there existstwo real directions given by the two roots (1.12) along which the PDE (1.4) reducesto the canonicalform. Theseare called characterislic equations.Though there are two solutionsfor each quadratic,we have consideredonly one solutionfor each. Otherwisewe will end up with the sametwo coordinates. A f o n g t h e c u r v e( ( x , y ) = c r , w e h a v e d( =(,dx+(rdy=0 Hence.

4=-lL1 dx

(r.r3)

\1, )

Similarfy, afong the atwe q(x, y) = c2, we have

4=-lul dx

(l.l4)

\,tn )

Eqs.(1.13)and (1.14),we obtainthe equationsof family of characteristics Integrating €(r, y)=ct a n d r y Q , y ) = c 2 , w h i c h a r e c a l l e d t h e c h a r a c t e r i s t i cosf t h e P D E ( 1 . 4 ) . N o w t o o b t a i n t h e canonicaiform for the given PDE, we substitutethe expressions of{ and ? into Eq. (1.8) which r e d u c e st o E q . ( l . l I a ) . To make the ideas clearer, let us considerthe following example: 3 u * + l ] u - . + 3 t . - .= 0 Comparing with the standard PDE (1.4), we haveA =3, B =10, C =3, 82 - 4AC = 64 > 0. Hence the given equationis a hyperbolicPDE. The correspondingcharacteristicsare:

dy (c,\ dx

\6r)

(-a*,[F-+rc)

1

2n

3

\

)

,,NDAMENTALcoNcEprs

|

I | | -: |

|

+=-(u)=-(-a-'l-'a-qAc)-. * 24

\4') \ )tina ( andry, we first solve for 7 by integratingthe aboveequations.Thus,we get

I

,=!'*',

,=Y-3x.

cr=y-x/3

:-crore'

|

|

'=3x+c1' :h give the constantsas

J

|

sl

F €==vy- 1- .3- -x = c 1 ,

I

n=t-!,=c,

- -::e

are.thecharacterislic linesfor thegivenhyperbolicequation.ln thisexample,the characteristics to be strarghtlines in the (x, y)-plane along which the inirial data, impulseswill ,,=,'r#

|

To find rhe canonicalequation,we substitute the expressions for

J

f and 7 into Eq. ( I .9) to get

,=.e(|+Bl(r+cqj=t1_t12+10(_3)(r)+3=0 |

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

I

D=0. E=0. F=o

:e. therequired canonical lormis

lF: ' f-

| f |

*"'=o

or '':n=o

:rlegmtion, we obtain

,G.n\="fc)+ sot)

---:/and g arearbitrary. Goingbackto the originalvariabres, the generar sorutionis ,(x,y)=f(y_3x)+s(y_xt3)

3.2 Canonical Form for parabotic Equation ":eparabof ic equation. rhedisciiminant E2-q,qe=0, whichcanbetrueif B=0 andi or C f :j.rat to zero.Suppose we set first l=0 in Eq. (1.9).Thenwe obrain F ,=A{}+B(,(,+cs,,=0 f

|

I


INTRODUCTIONTO PARTIAL DIFFERENTIALEQUATIONS

or (r\ (r\2 Al:z | +Bl::r l+C=0

11,)

\6, )

which gives t;=

-B!,1 r--;B' - 4AC ,,

"y

Using the condition for parabolic case, we get 9r

(1.15)

{,=- 2A Eq.(1.15),we set Hence,to find the function6=1Q,y) whichsatisfies ),,rp

A=-7.= rA andget the implicit solution

6Q,v)=ct In fact,one can verify that 7=O impliesB=0 as follows: + B(6,tly + 6ynx\ + 2C1yry' B = 2AS,t7, to Since82 -4AC=0, the aboverelationreduces

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

= 26fAq,+ Je 6; (-.aq,+J- ryn1 Howevel

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

-.,F.t6"11J-.t E= z1J-e6 t, +Jcrl)=s Wethereforechoosef in sucha way that both 7 andE arezero.Then 17can be chosenin any way we like as long as it is not parallelto the f- coordinale.ln otherwords,we chooser7 such is not zero.Thuswe canwrite the canonicalequationfor that the Jacobianof the transformation to eitherofthe forms parabolic caseby simplysubstituting { and7 intoEq.(1.8)whichreduces (l.llc). we consider the followingexample: the procedure, To illustrate x2uo -2ryu,

+ y2uo = er

The discriminant82 - 4AC = 4x2y2 - 4x2y2 =g, andhencethe givenPDE is paraboliceverywhere. The characteristicequation is

53

FUNDAMENTAL CONCEPTS

dvt-B2xvv dx -

-----L

2A

€t

=

-

!-

x

2x2

we have .-.:eSTation, xl=c : :.ence ( = xy will satisfy the characteristic equation and we can choose 17= y. To find the

forf :::cal equation,we substitutethe expressions

and ? into Eq. (1.9) to get

A = A y z + B x y + c x 2= t 2 y ' - 2 t 2 y 2 + y 2 x 2 = 0

r=0, E=0,

e=y2, F=0,

D = -2xy G=e"

-:.. the transformedequationis lTur, - LxyuE= e' 42unt=26uq+e1n : :rronicalform is, therefore, 21 | ur, = -7 u1+1e, r,"' i 3 Canonical Form for Elliptic Equation -:. :he discriminant 82-4AC<0, for ellipticcase,the characteristic equations

dx

dy_B+lB"-4AC dx

24

--. complex conjugatecoordinates,say { and ri. Now, we make anothertransformationfrom - :o (d, p) so that

o=€+n. s=€-n 22i

:- give us the requiredcanonicalequationin the form (l.llb). -: ;llustratethe procedure,we considerthe following example:

: rriminantBz- 4AC= -4x2.,.

o e iptic.Thecharacreristic equations

"#J;",iiu"r,oro, dy= Bdx

-'[47 2

= -tX

54

INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS

Integrationof these equationsyields

iy+l=s,,

-ir+L=""

t' = 2!2x'2 + i v .

,=!12 -iu

)"4

Hence, we may assumethat

the secondtransformation Now,introducing F+n

F-n

a=-,p=-; we obtain 12

a=;,

F=v

The canonicalform can now be obtainedby computing ]=Aal+fa,ay+caj=y2 E =2Aa,f ,+ B(a*F, + arfr) +Zc(arg) = 0 e=eB| +BB,BI+cp2r=x2 D = Aao + Bary + cayy+ Ddx + Eat = | E = A9* + BBxy+ cByy+ DB, + EBn = 0

F=0,

c=o

Thusthe requiredcanonicalequationis ,2uoo+ ,2uOO+uo =O or

uoo+upp=-fi EXAMPLE1,1 Classifyandreducethe relation y2uo - Zxyu, * *' uo = t u, * t u, xy

to a canonicalform and solveit. Solution

The discriminantof the given PDE is 82 -4AC =4x2yz-4x2y2=g

FUNDAMENTALCONCEPTS

55

rhe given equation is ofa parabolic type. The characteristicequation is

-:-:3

d y_ _ 1 , _ B _ - 2 r y _ _ x dx 1r 2A Zy2 y L- .::ation givesx2 +y2 =cr. Therefore, equation.The €=12 +y2 satisfiesthe characteristic -- : :.rrdinatecanbe chosenarbitrarilysothat it is not parallelto {, i.e. the Jacobianof the transi. -::ion is not zero.Thus we choose 6=12 +y2, : ::d the canonicalequation,we compute

--

,'2

-Bx2y2 7 = ,lEl + nqg, + G1l,=a*2y2 =0 + 4x2y2 e =4r2y2,

B=0, F:-::.

D = E = F= G = o

the requiredcanonicalequationis

=O orunn=0 4x2Yzur, ::.re this equation,we integrateit twice with respectto ry to get

ur=fG),

u=f(€)rt+sG)

^:-:

.fG) and SG) are arbitraryfunctionsoff. Now, going back to the original independenr i:::-es. the requiredsolutionis u = y 2f ( x 2 + y 2 1 +g l * 2 + y 2 1 ,-tltPLE 1.2 Reduce the following equationto 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
G4o+?nrt5 ' ! E;V

Bdx

-

2(l+ x2)

dy

B+ J Br:4AC

dx

2A

-::gration,we get

6 = ln (t+.,[2 + t.y-i tn1y+n[2 + t t =.,

n = tn(, * JP * D+i tn1y+,[1]i1 = g, second transformation F +n

a =2__:J.

22i

n-f

B=''

'

t+r2

56

INTRODUCTIONTO PARTIAL DIFFERENTIALEQUATIONS

we obtain

o=tn(r*Jr'*t) P = t n 1 y + , [ 1+] t ) Thenthe canonicalform can be obtainedby computing f = ^ l a ? + B a , a y + C a 2 r = 1 ,E = 0 ,

C=1, D=E=F=G=O

Thusthe canonicalequationfor the given PDE is udd+upg=0 EruMPLE 1.J Reducethe following equationto a canonicalform and hencesolveit: uo-2sinxu*-cos2 rrr - cos:ay=0 Solution

Computingwith the generalsecondorderPDE (1.4), we have A=1,

B=-2sinx, C=-cos2x,

D=0,

t=-cosr,

F=0,

G=0

= + (sin2x + cos2r) = 4 > 3. Hencethe given pDE is hyperbolic.The The discriminate82 - 4.,q,C characteristic relevant eouationsare dy= B-JF -qAC =-sinr-l dx 2A d y _ B+J* -+rc =l-sinx dx ZA On integation, we get y = cosr -.r+ cl,

y = cosx+ x + c2

Thus,we choosethe characteristic lines as I = x + y - c o sx = q ,

r y - - - x +y - c o sx = c 2

In orderto find the canonicalequation,we compute

2 = A l l +M " e . , + c { = 3o E = 2A1,ry. + B(6,11 y +I yn) +2cqyq y = 2 ( s i nr + l ) ( s i nx - 1 ) - 4 s i n 2x - 2 c o s r2 = - 4 C =0,

D =0,

I'-n

Thus,the requiredcanonicalequationis ur- =0

F =O

G=0

FUNDAMENTALCONCEPTS

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

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

u = f@) dtl+ cG) J u=V(rt)+cG) to the old variables e(6)is anotherarbitraryfunction.Retuming r, lr, the solutionof the

PDEis

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

E\4.\IPLE 1.r' Reducethe Tricomi equation uft+xuyy=0, '"

x*0

z.i x, y to canonicalform.

The discriminant82 - 4AC = -4x. Hencethe givenpDE is of mixed type:hyperbolic solution ' - . < 0 a n d e l l i p t i cf o r x > 0 . ,_-: I

ln the half-planex<0, the characteristic equationsare

dv dx

nn-.

t*'[F -+,qc 2A

, =? g*yr/2+ r, , = -? 943/2+

"2

)re, the new coordinatesare

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

4 { x , y ) = } y + 1 . 1 - a 1=3g , zarecubicparabolas. orderto find the canonicalequation,we compute I = l t ? + "B5 .rxF5 ) / +. "c5F1 1 = - 9 r * g *, .24 ^r = g 4..

E=9x,

e =0,

p=-1g4)t2 =-E,

F-C-n

INTRODUCTIONTO PARTIAL D1FFERENTIALEOUATIONS

5E

Thus,the requiredcanonicalequationis -2(-t\-tt2u, +1Gx\-tt2 u- =o 9ru,' e't t 4'

4

OI

I uh=-luq-ua) Case II

ln the half-plane x > 0, the charactedsticequationsare given by

dv .-

dv

7 =iJx. ctx

a =-iJi ctx

On integration,we have

((x. y)= | y - itJx)", 2'

4(x.y) = | y + i(Jx\' Z'

the secondtransformation Introducing t+n d=i.

p^= -t ;- n

we obtain 3

o=ry,

p^ = - 6. tt-.1 xr

normalor canonicalform is The corresponding 1 ^ =U uooluBB+----':up of the equation EXAMPLE 1.5 Find the characteristics uxr+ 2ury+ sinz1x1uo+ u, = 0 when it is of hyperbolictype. Solution The discriminant82 - 4AC= 4 - 4 sin2x = 4 cos2x. Hencefor alI x*(2n-l)212, equationsare the given PDE is of hyperbolictype. The characteristic dy

&

B+JF -qAC= l + c o s . r

On integration, we get /=.r-sinrfcl,

y=x+sinx+c2

Thus, the characteristicequations are E=y-x+sinx,

ry=y-x-sinx

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

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

FUNDAMENTAI-CONCEPTS

Solution

59

Thediscriminant 8 2 - 4 A C = ( x + y 1 2 - 4 r y = ( x _ y ) 2> 0

-:-:e the given PDE is hyperboliceverywhereexceptalong the line y:x; = r. it is parabolic.When y*x, the characteristic equationsare /--;dy _B+rlB. _4AC _(x+y\+(x_y) dx 2A 2y dy dx

whereason the line

4=t dty

,- - rregration,we obtain

y=x+q,

y2 =x2 +cz

the characteristic equationsare

1=v-x,

tt=y2 -x2

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

7=,tg]+n4qn+C(l=y-x-y+x=0, c=0.

D=0,

E=2(x-y\

E=_z(,_y)2,

F=C=o

canonical equation for the given PDE is

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

-t2uh + 2 ( - O u , = 0 d I -dul

-

6u€4+ un= ; l E l = u " ) 05\ on ::::ion yields

E*=r
u=EJ fttt\dry+c,Gl

It

u=-.1 i:neral solution.

f (y'-x")dty' -x')+g1y-x)

INTRODUCTION TO PARTIALD]FFERENTIAL EQUATIONS

60

EXAMPLE 1.7 Classiff and transform the following equation to a canonical form: sin2(x)rio + sin(2x)u, I cos2 (x1u .=x o Solution

The discriminant of the given pDE is 82 - 4AC = sin22x - 4 sin2x cos2r = o

Hence, the given equation is of parabolic type. The characteristicequation is dvB

c o tx

4t=;= Integation gives

)r=lnsinx+ct Hence, the characteristicequationsare: 6=/-lnsinn,

ry=y

4 is chosenin such a way that the Jacobianof the transformationis nonzero.Now the canonical form can be obtained by computing

a =0, F=0,

A=0, E=0.

f =

D =t,

r, "or2

Hence.the canonicalequationis

cos2(x) ur, + u, = x -- sin-tpn-|1- u, 1l- e2(n-1t1uro EXAMPLE 1.8 Show that the eouation

uo+ryu,=Lu,, xa'

where 1y'and a are constants,is hyperbolic and obtain its canonical form. Solution Comparingwith the generalPDE (1.4) and replacingyby /, we have A=1, B=0, C = - 1 / a ' , D = 2 N l x , a n d E = F = G = 0 . T h ed i s c r i m i n a 8n 2t - 4 A C = 4 1 a 2 > , J .H e n c et.h es i v e n PDE is hyperbolic.The characteristicequationsare dt

Pr

82 - 4AC

,,14/a'

dx

2

Therefore, dtl

dtl

dxa

[aa

On integation, we get

t=-!+cr, aa'

t=!+c.

= + ICI

oPERAroRs 1.4 ADJorNr tu=' givenor where r is a differential operator

I

(t'16)

I

t= aolxylL+ ar<'){\ +...+o,{') |

One way of introducingthe adjoint differentialoperatorZ* associatedwith I is to form the I productvZu and integrateit over the intervalof interest.Let I

Itn"uarlfn+lB rt*uar

(1.11 |

rvhichis obtainedafterrepeatedintegrationby parts.Here,Z* is the operatoradjointto t, where I the functionsu and y arecompletelyarbitraryexceptlhat Lu and.ttv shouldexist. I EXAMPLE 1.10 Let Lu = a(x) (d2uldx2-S + n14g"ldr'1+c(.r)r/;constructits adjoint Z+. I Solution Considerthe equation I

t'**= :,':..*),',,1, -t.,',:!:;'*,.' ^,,. I = (av)---=ax+ @v) Gv)u dx Jn Io *dxJ n

FUNDAMENTAL CONCEPTS

:: -1,\ ever, rB

r!2"

?8

S

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

dX

- {ou)'u'a, =1u'val,^ l" =[u'av]l-tu(av)'ll +

tu

u@v)"dx

-[uubi'a* I'ooffa.=tu(bv)li

lB

vtu dx=7u'(av)- u(av)'+ u (bv)lBn + ul(av)"- (bv)'+ (cv))dx JB

equationwith Eq. (1.17), we get 1 * y = (av)" - (bv)' + (cv) = 6y" a 12o'- b) v, + (a,, - b, + c) v

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

(l . 1 8 ) functions of x and y. In

:::scribed over the

L(u) =

s

,a-J

(l.le)

TNTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS

64

Its adjoint operator is defined by

r-(u)=). j. ,n,rr-f.!@,u)*"u dx'ox dxi ij=1

t

(r.zo)

E

Here it is assumedthat A4 eCQ) and B, ec(l). For any pair of functionsu,reCQ\, it can be shown that -r

tf,

|

,'-

.

\

,

/

-rr \l

vL(u\-uL*(v\=\*lI r,l,?-,+1.*lt,-2?ll '\ dx, dx, dxi) ar, ,:

\

l-tr

i

11

(r.2r \"-'l

This is known as Lagrange'sidentity. EXAMPLE 1.11 Construct an adjoint to the Laplace operator given by L(u)=\$+uD

(1.22)

Solution ComparingEq. (1.22)with the generallinearPDE (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)=uo+u, Hence, the Laplace operator is a self-adjoint operator. EXAMPLE 1.12 Find the adjoint of the differential operator L(u)=uo-rt

'/l t1\

Solution ComparingEq. (1.23)with the generalsecondorder 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

R I E M A N N ' SM E T H O D

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

L(u)= 61'' 11 is classifiedas hlperbolic if 82 > 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 r
(1.42)

(l.4r)

- Na,t= -"!.0,J,w at, ,[,( r\ dy "!*) dx ) 1.3, we have on f, r=/.

Therefore,dx=dy. Hence

- ua,t=[,[-, -,'i)* J,tvat, t,

Flg.1.3 An illustration ol Exampte 1.13.

(1.44)

INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS

70

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

f -,La* I t u a. r - u a * l =JQP f - N a t =JQP

JQP'

Ox

(r . 4 s )

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

-,?+' I say-xa*t=[--r+=[ ' ' JpR JpR dy

JpR'

(r.46)

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

a,-' *o x*)* a,*J,r-,fi o, !n! wr - r *otya*+=J,l, v r fi ^L, o Y ) rI",-" "\ But 0,

0v r * J p p -2' r a ' = F ' u l o J o o " E * t

.

-P

Therefore, tt-

)ln

vr -

t'

1

) " =[.] -"ff,a,-,7* at
"\

Dr)vcdv

+l-vulb+),pufrdx+)ro-ufidt Now choosingv(x,y;6,il (i) Z*v=0 (ii)

dv ^=tt

Av (iii; | =0 oy (iv) v=l

0.41)

as the solution of the adjoint equation such that

throughoutthe.ry-plane w h e ny = 4 , i . e . , o n Q P w h e nx = 6 , i . e . , o n P R at P((,4).

Equation(1.47) becomes

ov ou.\ tt at= t( -u x-v;dxl+(uv\p-(u)p | { v D a x ,|,-\l o- a y ox ) n

or

( u ) p = ( u- v ) e * 1 - ( - , ! a , - ' * * l - f i ( v F ) d x d t ,r\ oy

dx

J ,t

(r.48)

FUNDAMENTAL

1l

CONCEPTS

f tf a (uv)q-tuv)p = | _ d(uvt= l-li1n)ax Jt

2 + ltwlay dy

lox

=

f

J, '. .r Eq. (1.48)can be rewrittenas

1 I I

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

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

( l .4e)

IR

: -:il1', addingEqs. (1.48) and (1.49), we get I

(u)p=11@v)p+{*)n)+

*

lr.

2) r

(uv, dx- vu, dx)

at*,', an- II eFtdxd.y )[ ,{ur, IR

at-l.ltPlE 1.14 Yerify that the Greenfunctionfor the equation d2u 2 (du*!!\=n * d xd y x + y \ d x d y ) . , : : c r t o u = 0 , 0 u l d x = 3 x 2o n y = r , i s g i v e nb y (x + y) l2xy + 1( - rl.)g - y) + 2(r7) ,r^,t,t,'rt-T i-: :ltain the solutionof the equationin the form - x y+ 2 y 2 ) y=1x-y1(2x2 Solution

In the given problem,

,,=#k.h*.h*=,

(,so)

-::ring this equationwith the standardcanonicalhyperbolicequation(1.24),we have o=b=

2 , x+ y

C=0,

F=o

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

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

'-:at Lrv=0

throughout the r/-plane

dy\x+y)

(l.sl)

nt

INTRODUCTIONTO PARTTALDIFFERENTIALEQUATIONS

,,^ (lll

-d v = -v z dx x+ y

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

(iii) +=---:--v oy x+y

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

(iv) v=l

at P(4,t1).

(1rs2)

If v is definedby /-r

.,\

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

\e+D"

( l .s3)

Then

#=ffi,,,.o-^v|*ut.#*l + 2Y2 +2x(( - 4)+2(41

*= A#*,

( 1.54)

and

dzv _4(x+y)

0x 0y

(1.s5)

G + tD3

dv = -)' t \ e _ tDt '+z6tt) .gry +zx2-2y\q a l 1 1 +r ' f ' ' " ' ' ' ^ Usingtheresultsdescribed by Eqs.(1.53f{1.56), Eq.(t.51)becomes

'*P1=!:-dxdy

' (? *':)* x+ y\dx 0y)

= 1 . ! r * ! ) _, G+d"

o' 1x+y)2

, ? . _ , r l 4 x y + z (+xyz2 ) l

(x+y)((+4)r"

or

z * 1 u y4=( ' + Y ) -4 ( x + v=) s G +q)" (6+d' Hencecondition(i) of Eq. (1.52) is satisfied.Also, on y=7.

*

= A#*'

+ 2t12 +2x(( - 11) +2(41

0.s6)

73

FUNDAMENTAL CONCEPTS

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

(t.s7)

2v/(x+ y) at y =r7 is given by 2v2 x+y x+q =

ffirr*r*n

-Do-d+z6q)

I

(l.s8)

.tlznz+2x(( +r)+2{41 G+?D'

Eqs.(1.57)and (1.58),we get -0: v- =2 - Y dx x+ y

at V=n

property(ii) in Eq. (1.52)hasbeenverified.Similarly,property(iii) can also be verified. atx=1, v = q ' F+n G+ilG+D2 v= -2-+124n +G _ tD' +26ry1= \q+nr G + ril3

(iv) in Eq. (1.52) has also beenverified. (1.50)and (1.51),we have

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

vu\ 0(2vu\ = -d (l0! -u: -\ l - -a (l r =a- v l\+ -d (l 2 l+=-l| dy\ dx) dx\ dy) dx\x+y) dy\x+y) ( = -d- l - - zvu u-dx\x+y = -AM +dx

dv) 0 ( 2vu l+ ^ l-+v:dy) dy\x+y

du\ | dx.)

dN dy

M=2* -uaru, N= 2'" *u7u x+y

dy

x+y

dx

-.ing Green's theorem, we have tr c r.O - uL* (v\l dxdy --JaRW (M dy- N dx\ dy- N d') = )) lvL(u\ J; IR

* !'
( l .5e)

74

INTRODUCTION TO PARTIALDIFFERENTIAL EQUATIONS

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

p l l z u v- u - ^l v- l ld y.-

= |

J R| 1 Ltr+/

| 2 u ,+ v -a= ^- l d .xf

\-

dyl'

lr+y

dr)

l

.l

z u u d " ] 1 . r a l[ 2 u , 0 , ] 1 . _ tr Pt l1I _ + v _ _ _ f l d Y +l | 1 : _ r j l l d y J?

Lt;+/

J PL t - r + /

dx))

dYt.t

However, p cP iv au\ . rP 2uv ((Zuv J?\x+v. '*)*=Jn fi,x*t*''n-ln' ^aNow, using the conditionu=0 on y=x, Eq. (1.59) becomes

qud'dv=JI:/ ( (* -,*)* - 11 -,'i"l* !!R t uut- uL* . , n |* dy) dx t \x+), \x+.y 2u'

-l'

rQx+y

dt-(ur\o+(uvrn-l' r\a, "

JQ

dx

.l',#*-tl("#,)* (iiF(iv) of Eq. (1.52),the aboveequationsimplifiesto Also,usingconditions

(u)p=(uv)q-Ji"** Now using the given condition,viz. *=3t2

on nQ

we obtain

- 3J: (u) p=(uv)e 4?##4)e =-+-l'st \€+ryr"E

*,3q41a,

-$*lenrn, =-s^tl *-r lir, = ffi,4/

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

= G - , i e € z- 6 n + z n 2 )

FUNDAMENTAL CONCEPTS

'--erefore,

u(x, y) = (x - y) (2x2- xy + 2y21 :ce the result. .+VPLE 1.15 Showthat the Green'sfunctionfor the equation d-u -;;+a=0 ox oy

v(x,y|€,D = J0 J0 denotesBessel'sfunctionof the first kind of order zero. Solution

Comparing with the standardcanonical 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 =U ox t7v

-d"v-^= u

o nY = q

dx

-0v ;-=U

o nx = 1

oy

a tx = 6 , y = 7 t 0 k= a ( x - i l \ - n ) dv dx

0v d( dQ 0x

kQHl=oo-,1) dx

1=io"rty-nl oxK

75

76

INTRODUCTIONTO PARTI{ DIFFERENTIAI EQUATIONS

Thus, 0v dv a .t-r. ;-=-;;;Q"'\y-4) ax dgE .t d'v

-

^T^ dldva.t-L

'v-il)I a,ay= ayLaie' ag-] =110"* ?ua0,..-, , rr-r,. 2+rr-nto-k v - n-,a'1, t k dQ aQ ay\Y-ry)*e

L

N ay)

However,

fr=io"-<'-a Therefore,

)2un''' I +=Ap'o!*1r-r16-k(,-q)o-n)ld'o**d'' ' '--u-ntfii0'-rtx-tt] dxdykL do k' ap Hence,

ffi*"=o

gives

a1*{11_e14 zlo! 6ro-r, ' ' d 0*r,r ' -Lf*,=o kLk' dO' k aO ) or

o-( d2: * rr* 4 l* u= o ,r-r d0) k'\ d0" OT

^ t2 Q ' v "+ Q v ' + L Q K v = 0 Let k=2, a=4. Then the above equation reducesto Q2v"+ (v, +Q2v=0=v,,+lv,+v

0

(Bessel,sequation)

Its solution is known to be of the form

u= Jo@)= JoQJG- g U - rt) which is the desiredGreen'sfunction.

lr

FTJNDAMENTALCONCEPTS

EXERCISES I' Find the regionin the xy-pranein which the following equationis hyperbolic: l(x- y)2 - lluo + 2u, +f(x - y)2 _tlu, =g 2. Find the familiesof characteristics of the pDE (l- x2)uo- u, =o in the elliptic and hyperboliccases. 3. Reducethe following PDE to a canonicalform =0 uB + t"yurry {. Classifrand reducethe following equationsto a canonicalform: (a) y2uo-x2uo=0,

y>0.

r>0,

(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
INTRODUCTION TO PARTIALDIFFERENTIAL EOUATIONS

where IR is the triangular region in the xy-plane boundedby the line y = I and the lines x = x6, ! = lo through (-16,y6). 9. The characteristicsof the partial differential equation , 12" aad2" . ^ d2t -;-= - ,;-i + csszvj all + 3! = 0 ox oy dx dy dx' dvwhen it is of hyperbolictype are... and... (CATE-Maths,1997) = y 10. Using ry x + as one of the transformationvariable, obtain the canonical form of -d: -z; -uz ; - ^ - -0 f -2 . =u- = u.. d 2 u

dx'

ox oy

dy' (GATE-Maths,1998)

Choosethe correct answer in the following qu€stions: 11. The PDE y3uo - (r2 -1)u, =O is

i

(A) parabolic in {(x,y):x<0} @) hyperbolicin {(x, y) : y > 0} (C) ellipticin IR2 (D) parabolic in {(r, y):x > 0}. 12. The equation x21y-l1zo - x7y2-t)rry+y(yz -l)zo+ z,=0

(GATE-Maths, 1998)

is hyperbolicin the entirex/-planeexceptalong (A) x-axis @) y-axis (C) A line parallelto y-axis (D) A line parallelto r-axis. (GATE-Maths,2000) 13. The characteristic curvesof the equation ,2uo - Y2,o = ,2Y2+r, (A) rectangular hyperbola (C) circle

x>0 are (B) parabola (D) straightline. (GATE-Maths, 2000)

14. Pickthe regionin whichthe followingPDE is.hyperbolic: yuo+2xyar+xu)ry=ux+uy

(A) xy +l

(B) r/+0

(C) xy>l

(D) r/ > 0. (GATE-Maths,2003)