Victor P.Pikulin Stanislav 1.Pohozaev Equations in Mathematica Physics A practicalcourse TranslatedfromtheRussian by AndreiIacob Springer BaselAG Authors: Victor P.Pikulin Department of Higher Mathematics Moscow Power EngineeringInstitute (MPEI) Krasnokazarmenaja str.14 111250 Moscow Russia 2000Mathematical Subject ClassificationM16053 Stanislav I.POhozaev Steklov Institute 01Mathematics Gubkina str.8 117 966 Moscow Russia Originally publishedinRussianby Izdatel'skaya firma"Physics-Matematicheskaya Literatura",Aussian Academy of Sciences A CIP cataloguerecord forthisbook is available Irom the Library of Congress,WashingtonD.C.,USA Deutsche Bibliothek Cataloging-in-Publication Data Pikulin,Viktor P.: Equations inmathematical physics: a practical course / Victor P.Pikulin / Stanislav I. Pohozaev.- Basel: Boston: Berlin: Birkhuser,2001 ISBN978-3-7643-6501-1ISBN 978-3-0348-8285-9(eBook) 10.1007/978-3-0348-8285-9 Thisworkissubjectto copyright.Allrightsarereserved,whetherthewholeor part01thematerialis concerned,specilicallytherights01translation,reprinting,re-use01illustrations,recitation,broadca-sting,reproduction onmicrofilms or in other ways,andstorage in data banks.For anykind of use per-missionof the copyright owner must be obtained. 2001SpringerBasel AG Originallypublished by Birkhuser Verlagin2001 Softcover reprint of the hardcover 1st edition 2001 Cover design:Micha Lotrovsky,4106 Therwil,Switzerland Printedon acid-free paper produced 01chlorine-Iree pulp.TCF00 ISBN978-3-7643-6501-1 987654321 DOI Contents Preface.................................................... ........... .Vll Introduction......................................................... .1 Chapter1.Ellipticproblems............................................ .7 1.1TheDirichletproblemfortheLaplaceequationinanannulus........7 1.2Examplesof Dirichletproblemsinanannulus.......................10 1.3TheinteriorandexteriorDirichletproblems.........................11 1.4ThePoissonintegralforthe disc.Complexform. Solutionof theDirichletproblemwhentheboundar.ycondition isarationalfunctionR( "in 'P,cos 'P).............................14 1.5The interiorandexteriorDirichletproblems.........................18 1.6BoundaryvalueproblemsforthePoissonequation inadiscandinanannulus..........................................20 1. 7Boundary valueproblemsfortheLaplaceand Poissonequationsinarectangle.....................................19 1.8BoundaryvalueproblemsfortheLaplaceand Poissonequationsinaboundedcylinder.............................27 1.9BoundaryvalueproblemsfortheLaplaceand Pois"oneqnationsinaball..................................... .33 1.10Boundary val1H'problemsfortheHelmholt7,equations...........43 1.11Boundary valueproblemfortheHelrnoltzequationinacvliuder.....44 1.12I30undaryvalueproblemsfortheHdmolt7,equationilladisc........46 1.1:\BoundaryvalueproblemsfortheHelmoltzt'CjnatiOlIin(\hall... . 1.14Guidedelectromagneticwaves.................................. . 1.15Themethodof conformalmappings(forthesolutioll of boundaryvalueproblemsintheplane)...... . 1. HiTheGreenfune! ionmethod.................. . 1.17Othermethods................................ . 1.18Problemsforindependentstudy................................ . 1.19Answers...................................................... . Chapter2.Hyperbolicproblems......................................... . 2.1 2.2 Thetravelling-wavelllethod.................................... . Themethodof selectionof particularsolutions................ . 49 54 55 60 68 73 76 81 81 93 2.3TheFourierintegraltransformmethod..........................96 2.4The Laplaceintegraltransformmethod............... .........III 2.5TheHankelintegraltransformmethod...........................115 2.6The methodof standing waves.Oscillationsof Clh01ll1 0, 11=Clonthelateralsurfaceonx(0. f). u=f(.]:,g,z)onthelowerbase(t= 0)of the(,ylinder. 4INTRODUCTION Inaddition to physicalphenomena that evolveinspace and time,there exist phenomena which do not change with time.Most of these phenomena are described byellipticboundary valueproblems.Incontrasttothehyperbolicwaveequation and the parabolic heat equation, elliptic boundary value problems require no initial conditions.For them only boundary conditions needtobe imposed.The following threetypesof boundary conditionsarethemostimportantones: (1)boundary conditionof the firstkind(Dirichletcondition); (2)boundary conditionof the secondkind(Neumanncondition); (3)boundary conditionof the thirdkind(Robincondition). For example, the boundary value problem with the condition of the firstkind (theDirichletboundary valueproblem)fortheLaplaceequationisformulatedas follows:findthe solution of the equation D.u=0 in some domain of space(or of the plane)whichtakesgivenvaluesontheboundary.Asaconcretephysicalexample one can give the problem of determining the steady temperature distribution inside adomain0,ifthetemperatureonitsboundary00isgiven.Anotherexample: findthedistributionofelectricpotentialinsideadomainifthepotentialonits boundary isknown.The mathematicalmodelof both phenomena is { D.u= 0 u=tp wheretpisagivenfunction inthe domain0, ontheboundary 00, Theboundaryvalueproblemwithboundaryconditionofthesecondkind (theNeumannboundaryvalueproblem)isposedasfollows:findthesolutionof the givenequation insomedomain0of space(orof the plane)assumingthat the outward normal derivative ou/on (which isproportional to the heat ormassflux) isgiven on 00.This generalboundary valueproblem,forthe heat equation orfor theequationof electrostatics,withthefluxgivenontheboundary,iswrittenas follows: {~ ~ = :on inthe domain0, ontheboundary on. Unlike the Dirichlet problem forthe Laplace equation, the Neumann problem hasameaningonlyinthecasewhenthetotalfluxthroughtheboundary00 is equalto zero,i.e.,Jan~ ~ ds=O.Forexample,the interiorNeumannproblemfor the unitdisc, { D. u=00 ::;p < 1,0::;tp::;27f, ou op (1, tp)=1,0::;tp::;27f, doesnothaveaphysicalmeaningbecauseaconstantunitfluxinsidethedomain cannotensurethat thesolutionisstationary. INTRODUCTION The DirichletandNeumannboundaryvalueproblemsforthePoissonequa-tion1:::.1L= fareformulatedinasimilarmanner.Letusmentiononlythatforthe Neumannboundary valueproblem {1:::.11.. =f OlL - =cp on inthedomainn. ontheboundaryan. tohaveasolutionitisnecessaryandsufficientthat /"f d:r=/"cpds. JnJaIl AnotherpeculiarityoftheNeumannproblemforthePoissonequation.which distinguishesitfromtheotherboundaryvalueproblemsconsideredhere.isthat itssolutionisnotunique. Theboundary valueproblem with boundary condition of the third kind(the Robinboundary valueproblem)for,say,the Poisson equation,isposedasfollows: findasolution u( M)of the equation insomedomainn of space(orof theplane), whichsatisfiesontheboundaryantheconditionau./nll+ (JU-=cpowh 0). For n= 0 equation(1.4)has twosolutions:1 and lnp.Thus,wenowhavean infinitesetof functions("atom"solutions) 1,lnp,l' cos(n R.() 0of thecomplex(-plane (Figurel. 7),in suchamannert.hat.: t.heposit.ive.r:-sclIliaxisismappedintothepositivprealtheposit.ivey-semiaxii-lismappedintot.henegatiVE'realThus,wearriveatthe followingconclusion: Boundaryvalueproblem intheplane(x, y) BoundaryvaluE'problelll intheplane 1/)

u.=0,.:r> 0,y> 0, ul:r=o=O.y::;'0, /LIFO=8(.J:- 1),x::;'O. { =O.'I ill '1=1)={ O. if>1. if 0,M1(xo,YO,-zo)isthepointsymmetric toMo(xo, Yo,zo)withrespecttotheplanez=0,andM(x, y, z)isanarbitrary pointof thehalf-planez> 0. Physically the Greenfunctioncan beinterpreted asthe potential of the field producedbypoint-likechargesplacedatthepointMo (overthegroundedplane z= 0)and thepointM1(Figure1.11). x FIGURE1.11. 1.16.THEGREENFUNCTIONMETHOD z Mo(xo, Yo,zo) ...-R/""......,M(x. y.z) MMoI II I IRMM1 I y V Ml (XO,Yo,-Zo) The potentialatthepointM (x, y, z)equals G(MM) =1_1 047rRMMO47rRMM] (z=0 isagroundedconductingplane) 63 Noticethatinthecaseofahalf-plane(y>0)theGreenfunctionhasthe form(Figurel.12) 1111 G(M,Mo)=-In-- - -In--. 21fRMMo21fR'vlf1,h y Mo(xo, Yo) f-._ I- ......,M(x,y) I I O r ~ ~ ~ ~.r I FIGURE1.12.ThepotentialatthepointM(x.y)equals G(M. Mo)=-21 In-R1- -21 In-I"1 7rllil1\I O7TlAIMI 64CHAPTER1.ELLIPTICPROBLEMS Examplesof problemssolvedbymeansof theGreenfunction.Supposewewant to solvethe Dirichletproblemforthe Laplace equation in ahalf-plane {flu= 0,-00 < x< 00,y> 0, u(x,O)=J(x),-00 < x< 00. The solution of thisproblem is u(x,y) =-JL1OJ(8)8GId 'if-00at t=O8 (weputMo= Mo(x, y),M= M(8, t)),where 11 G(xy' 8t)=- ~ = : = ; = = = = = : =,"2. /()2()2 'ifyX-8+y-t 11 2'ifJ(x - 8)2+ (y + t)2 Calculating 8G/8tl t =o,weobtain y 100 J(8) u(x,y) =- ()22 d8. 'if-008 - X+ y (1.55) Example1[3,no.244].Findafunctionu(x, y),harmonicinthe half-planey> 0, if itisknown that x u(x,O)=-2-1 . x+ Solution.Wemustcalculate the integral y100 8 u(x, y)=- (2)[()22]d8. 'if-001 + 88 - X+ y Apparently,theeasiestwaytodothisistousethemethodof residues,namely, the followingformula: 1008 (2)[()22]d8=2'ifi[res[J(Z)]z=i+ res[J(z)]z=x+iy, -001 + 88- X+ y whereJ(z)= z/ ((1+ z2)[(Z- X)2+ y2]). Since 1 res[J]z=i=2[(i_X)2+ y2]' itfollowsthat x+iy res[J(z)]z=x+iy=2iy[1+ (x+ iy)2] , JL 100 8d8=iy+x + iy= 'if-00(1+ 82)[(8- x)2+ y2][(i- X)2+ y2][1+ (x+ iy)2] iyx + iy [(i- x) + iy][(i- x)- iy]+ (x + iy - l)(x + iy + 1) I.Hi.THEGREENFUNCTIONMETHOD = [i - .1'1- -iy- i- + iY]+[:1'+ /y- 1+ .r+ //J+ 1 ] = [i (1- - .1'- i (1+- .1'+; (;1/- +.l'1- i (l ++ :1' l = [i (1+ :y)+ J'- i (1+ - ./.] 1[.1'- i (1+ y).1'+ i (1+ y)].1' =:2.r2+ (1+ y)'2+.1'2+ (1+ y)2=.1'2+ (l + .1/)2' Therefore.t hesolutionoftheproblelIlisgivellby .r u(;r,y)='2(.r .1'+l+y-Remark2.The solutionoftheproblelllconsideredahoyc.

'II=0,-x 0): { 611=0,-00 < x< 00..If> O. 1Ily=o=R(.r).-,00< Y' 0, uly=o- 1 + x2 'k= const,-00 < x< 00. Solution.Usingformula(1.57),wehave [ k]kk(y + 1) u(z)= -2 Reres(1+ (2)(( _z)(=-i = -2Re 2i(z + i)=-,x2;:-+--'-=-:-(y-+-----'---,1)=2 1.17.Othermethods Inthissectionwewillconsidermethodsforsolvingboundary valueproblemsfor thebiharmonicequationandtheequations~ u =i, aswellasboundaryvalue problemdfortheLaplaceandPoissonequations(withoutemployingtheGreen function). Biharmonicequation. Example1.Solvethefollowingboundaryvalueprobleminthedisc{(p, cp):0:0(isafundamental solution),i.e., 1 ~ =0. y'x2 + y2+ (z + 1)2 Nowletusdifferentiateboth sidesof thisequality with respecttoz.Weget ~ z+l=0 [x2 + y2+ (z + 1)2]3/2. Differentiatingonemoretimewith respecttozwehave x2 + y2- 2(z + 1)2 ~ =0 [X2+ y2+ (z+ 1)2]5/2. This suggeststo considerthe function x2 + y2- 2(z + 1)2 u(x,y,Z)=[22()2]5/2' x+y+z+1 Thisfunctionisharmonicin the wholehalf-spacez> 0(since~ u = 0,aswejust showed),and forz = 0 wehave x2 + y2- 2 ulz=o=(1+ x2 + y2)5/2 ' whichprovesthat u(x, y, z)isthe soughtsolution. l.1R.PROBLEMSFORINDEPENDENTSTUDY73 1.18.Problemsforindependentstudy 1.Findthedistributionofthepotentialofaninfinitelylong(-::x:;l.-oc 0, { L II.(X, O)- 0,11/(.r, O)- 0,-x < .1'< X. thenusing theDuhamel '8methodorprinciple(orthemethodof Impulsesonecan obtain the solutioniIItheform 1IIj.l+o.(t-T) II(:]', t)=-;- dTf(C. T)dC. 2a.():r-o.(t-T) Example1[4,Ch.II,no.57].Supposethatawaveoftheforlll.p(.r- af)travels alonganinfinitestring.Takingthiswaveastheinitialconditiollattimet=0, findthestate of the string attimet> O. Solution. In our problem the travelling waveat time t=0 characterized by nonzero "initial"deviationsandvelocities 11(.1'.0)='P(J:).11.t ( x, 0)=- a'P' (.1' ) ,-x O. II/(0. t)+ 211,(0.t)=-4t.I ?n. (2.1 ) i.e.,thefunctionI1(J:.t)issoughtinthefirstquadrant.r> O.t> 0of the(x. t)-plane(Figure2.4). Solution.Firstlet.usfindaparticularsolutionoftheeqnation11tt=V:r:r- 6. withoutpayingattention totheinitialandboundary conditiolls.\Veshallassume thatthisparticular solutionuldependsonlyon:r.i.e' ..111=1I1(:r).Thm in order 84 t ....., I II FIGURE2.4. CHAPTER2.HYPERBOLICPROBLEMS t DU=-6 x o x FIGURE2.5 tofinditwemustsolvetheequationd2ud dx2 = 6,whichyieldsUl (x)= 3x2. Nowletus consider the new function v(x, t)= u(x, t) -Ul(X). Then fromequation (2.1)weobtain the equation Vtt=Vxxand the problem from which one determines v(x, t): { Vtt= vxx ,x> 0,t> 0, v(x,O)= _2X2,Vt(x,O)= 0,x::::0, Vt(O, t)+ 2vx(0, t)=-4t,t::::0, (2.2) Clearly,fort< xthe solution v(x, t)can be writtenvia d'Alembertformula in the form(seeFigure2.5) V(x, t)=-(x - t)2- (x + t)2=-2(x2 + t2). For t> xwewillseek the solution of problem(2.2)inthe form V(x, t)=f(x + t)+ g(t - x). Tofindthefunctionsf(x)andg(x)weusethecontinuityofthesolution v(x, t)on the characteristic t= x and the boundary condition Vt(O, t) + 2vx(0, t)= -4t.Indeed,onthecharacteristict=xwehavef(2t)+ g(O)=-4t2,whence f(t)= _t2 - g(O).Further,fromtheboundary condition in(2.2)itfollowsthat g'(t) + f'(t)+ 2f'(t) - 2g'(t)=-4t, whichafterintegration yields 3f(t) - g(t)= _2t2 + 3f(0) - g(O). Clearly,f(O)+ g(O)=O.Thus,wearrivedatasystemoftwoequationsforthe twounknown functionsf(t)and g(t): { f(t)=_t2 + f(O), 3f(t) - g(t)= _2t2 + 4f(0). 2.1.THETRAVELLING-WAVEMETHOD85 Thisgivesg(t)=-t2 - frO).Thus,fort> ;r, v(x, t)=-(t + :x:)2+ f(O)- (t- .1:)2- f(O)=-2(.x:2 + t2). Weconcludethatthe soughtsolutionisgivenfort> O.. X:> 0by 22 l1.(x. t)=v(x, t)+ U1(X)=J'- 2t. Example3[4,Ch.II,no.78].Consideraninfiniteelasticbealllconstructedby joining in thepointJ:= 0twohomogeneoussemi-infinitebeams.For.r< 0[resp., x> 0]themassdensit.y,t.heelasticit.ymodulus,andthespeedofpropagationof smalltransversalperturbations areequalto PI,E1,andu 1[r('sp ..{J2.E2.and a2]. AssumethatfromthedomainJ:< 0awaveUj (x, t)=f (t- .r / Ill)travelsalong thebeam.Studythebehaviourof solutionwhenE2 0andE2 x. Solution.Thedeviationsof thepointsof thebeamarcgovernedbytheequations -00 < J:< O.I> O. o < x< 00.t> n, (2.3) andtheinitialconditions { Uj (x:,0):f (-:r / (1 d '. t ( x, 0)=f' (-:r / a d. 11.2(.1:,0)- O.IL2tCr.0)- 0,0 0 (Ll' if t+ .I:..< o. al Next,solvingthe systems(2.8)wefind = 0,z> 0, whence Nowletususetheconjugationconditionstofindthefunctionsh(s)and g2 (8).Fromthe equalityUI (0, t)= U2 (0, t)itfollowsthat f(t) + g1 (t)= h(t). Similarly,fromthe equalityE1 Ulx(O, t)= E2U2x(0, t)itfollowsthat Sinceal=VEl/PIanda2=VE2/P2,weobtain VE1Pl + (t)]=VE2 P2, whichafterintegration yields VE1Pl+ g] (t)]= VE2P2 2.1.THETRAVELLING WAVEMETHOD87 Therefore, wehavethe following system of equatiolls forthe desiredfUllctions 12(t)andg1(t): { f(t)+ 91(t)= h(t), [-f(t) + 91(t)l=VE2(i2[-f:zU)] Solvingthissystemweget gl(t)= - JE;P;f(t). yfE;p; + VE2 (i2 h(t) =,v f(l). Weconcludethatthe solutionof om problem isrqm'sent cdbythefollowing pair of functions: { f(t,- +- ve;p:;f(t+ . III(II ILl(.1',t)=V + Vf(t- , (11 ift+> O. (JI ift+< O. (II { 2 jE;p;f(t- . U'2(.r.t)=vE2(i2((2 ift -> O. II .f(((2). ift- .1< O. II....' The term VEIPI- VE'2P2f(.t) It+-VEIfh+ VE2P2.01 inIlJ (I,.x:)fort+ .r/a1>0isareflectedwave.Obviousl.".thiswaveic.itbsentif E1 P1=E2 P2If =0,thenthercflectedwaveisf (t+.r / 111):ifE2-=x. then thereflectedwaY('is_.- f (t + :1' /ad Therefractedwaveis/I2(:r. f).If E2= O.thmtlwamplitudeof thiswaveis twicethatof theincidelltwave:if '2=,x.thenthereic.llOrcfractedW 0, and the initial conditions u(x, y, 0)= 'P(x, y),Ut (x,y, 0)= 1jJ(x,y),00< x, y< 00, i.e.,thefunctionu(x, y, t)issoughtintheupperhalf-spacet > 0(Figure2.6). Thesolutionofthisproblemexists,isunique,andisgivenbythePoisson formula u(x,y,t)=_1_[ ~ j r 'P((,T))d(dT)+j"r1jJ((,T))d(dr/+ 21TaatJ Katva2t2 - p2J Katva2t2_p2 +rtdTfrrf((,T),T)d(d] Jo.JKa(t_T)va2(t-T)2-p2T), wherep2=((- X)2+ (T)- y)2and K"sdenotesthediscof radiusascenteredat thepoint(x, y)(Figure2.7). 2.1.THETRAVELLING-WAVEl\lETHOD89 FIGCRE2.6. r} o FIGURE2.7. Example5[6.Cll.IV,12.37(2)].SolvetheCauchyproblem {11ft.. =I.'rr+ +-..IX.< :r .1.1 n. U(:r.y. 0)=.r- y.IJ.{(;1'.y. 0)=JY.--0(,< Y.y< iX. Solution.LetllSapplythePoissonformula,setting 22 r.p(J:,y)=.r- y, y)=:ry,1(J', /I. t)=f).ryt. 90CHAPTER2.HYPERBOLICPROBLEMS Wehave 1a Jh(2- 7]2 u(x,y,t) =-- d(d7] + 27ratK,y't2 - (( - x)2- (7]_y)2 +Jr r(7]d( d7]+ 27rJK,y't2 - (( - x)2- (7]- y)2 +rt dTJr r6(7]Td( d7]. 27rJoJK'_Ty'(t - T)2- (( - x)2- (7]- y)2 The integration iscarriedoutoverthe disksKt andKt-r.Wecan write { ( = x + ~ s cp, 7]=y+psmcp, { t, where0 < P < s= - - t- T, Letuscalculate separately eachof the integralswritten above: J 1a12K1t(x+pcoscp)2_(y+psincp)2d 1=-- pdpcp= 27rat00y't2 _p2 1a12K 1tx2 - y21a1t2xl dp12K =- - pdpdcp+ - - cos cpdcp-27rat00y't2 - p227rat0y't2 - p20 1ait2yp2 dp12K1aitp2dp12K --- cos cpdcp + -- cos(2cp)dcp= 27rat0Jt2 - p2027rat0Jt2 - p20 =x2 - y2!!.- (27rrtp2dp)= x2 _y2. 27ratJo y't2_p2' h=~ r2K rt(x + pcoscp)(y+ psincp)pdpdcp= 27rJo JoJt2 - p2 11txyp12K11txp212K =- dpdcp+ - dpsincp dcp + 27r0y't2 - p2027r0y't2 - p20 11typ212K +- ~ d p cos cpdcp + 27r0Y t2 - p20 11tp212K +- ~ dpsincpcoscpdcp=xyt. 27r0Y t2 - p20 Finally, J2 =~ t dTr2K rt-r 6T(X + pcoscp)(y + psincp)pdpdcp = 27rJoJoJoy'(t - T)2- p2 2.1.THETRAVELLING-WAVEIvIETHOD91 31' It-TXYPj.LK =- TdTripdc;= 7r().()vi (t- T)2- p2. II = 6:ryj.t T riTII-T Pdp=6:ry.I' (t- T)T lIT=.f!)1 (l . . I() 22' ,().()Vt- T-- P. II Therefore,thesoughtsolutionis 2:1:1 u(:r:.y,t)=h+ J2+ .l:J=,ryt(l + t) +.r- Y. Thethree-dimensionalcase(threespacevariables).B.\'definition.theCauchy problemforthethree-dimensional11Iaveequationistheproblclllof findingafunc-tionu(:r. y. z. t)thatsatisfiesthe equation 1111=OL 611+ f(.r,y, z, t),-rx;< .r, y . .::< 'X,t > O. andthe initialconditions 11(:r,y. z, 0)=cp(:r,y.z).I1t (:r,y,z, 0) y . .:;).-x < .f. y .. < 'X. The solutionof thisproblem exists,isunique,andisgivenbyKirchhoff'sformula 1[()Jf rio()ff' u(x, y,'::. t)=--2-:-).dO'+ 47r(l(tSuit.. + l ,[T /[ "., /((,T),[,,], whereSapdenotesthesphereofradiusapcenteredatthepoint(.I: . .I/ . .:;)(Figure 2.8). ( __________________rl FrCCRE2.1'\. 92CHAPTER2.HYPERBOLICPROBLEMS Example6[6,Ch.IV,12.38(1)].SolvetheCauchyproblem { Utt= Uxx+Uyy+Uzz+2xyz,-00 < x, y, z< 00,t> 0, u(x, y, Z, 0= x2 + y2- 2z2,Ut(x, y, z, 0)=1,-00 < x, y, z< 00. Solution.Letusapply the Kirchhoff formula,setting cp(x,y, z)= x2 + y2- 2z2,?jJ(x,y, z)=1,f(x, y, z, t)= 2xyz,a=1. Weobtain 1aJe r e + T)2- 2(2 u(x,y,z,t)=4Jrat1St tdir+ +Je rdu+(dT Jr r du. 4Jr1St t4Jr10ls'_Tt-T Here the integration iscarried out over the spheres Stand St-T ofradii t and t - T, respectively,centered at the point(x, y, z)(seeFigure 2.8).If rl,()ESt, then clearlywecanwrite = x+t sin 0 cos cp, T)= y +t sin 0 sin cp, (= z +t cos 0, where0:Scp:S2Jr,0:S0:SJr.Letuscalculateseparatelyeachoftheabove integrals: J1 r27r r [(x+tsinecoscp)2+(y+tsinesincp)2_ 4Jrat 1010t 2(X+tcoSO)2]2lldlld - tSllluucp= t 1a 127l'17l' = --a(x2 +2tx sin 0 c:os cp+t2 sin2 Ocos2 cp)t2 sinOdOdcp+ 4Jrtoo +r27l'r (y2+2ty sin 0 sin cp+t2 sin2 0 sin2 cp)t2 sin 0 dOdcp+ 4Jrat1010 + aar27l'r (-2z2- 4tz cos 0 - 2t2 c:os2 0)t2 sin 0 dOdcp= 4Jrt 1010 2.2.METHODOFPARTICULARSOJXTIONS 1 'J1"l'21e(,2222 - -;- 2t;; 1)1Il()COSfJdOd'P= J'+.IJ -hr111.(). [) 1(jj.L"llT1:2 J)=-- -t1)1IlfJcosfJdfJd;;=t. ()t.Il.()t and 1j.tj'L"1" 2(.l:+ (t- T)sinO cos'P)(Y+ (I- T)c;in()Sill;;) h=- (iTX ,0.().()t- T '2 X(:::+ (t.- T)('Oi'i0) (t- T)sin ()d() Ii;;= 1/112"j'lTj'l =-;-- (I- T)dT.ryzsinOdOd'P =2J',1F(I- T)liT= ('-.1',1;:::' 27f,().()Il.II TllPrefore.thesolutiolltoomCauchyproblem :1'2,2:2 u(.J'.y.:::,t)=.r+y-2z+1+1.1'.11:::. 2.2.Themethodof selectionof particularsolutions Sometimes it is convenient. instead of using the general forlllulathat giv('i'ithe solu-tion of the Cauchy probleIll fort.hewave equation, to take advHnt age of t he specific formof theright-handsideof theequationandof theinitial('onditiolli'i.Indeed, thed' AlcmbcrtianDu.=CPl1/ at:2- 6v. definedOIltwicc-differentia blpfunctioni'i: takes,forexample.afunctionofthpformIl(.r.y.z.t)=(ulsill(/)U)cos(I':)Pm(t). whereP",(I)isapoIVllomialof degrecITI.illtoafUllctiollof tll('sallieforlll.Using thii'iobservation.itii'ieasyto choose the requisite parti('ular solut ion of t heCauchy problem. Example1.SolvetheCauchyproblem { 1111=611+ t/)l sin(3y) -:)G< .1',y ..:< x.t > o. Ult=()= Ullt=o=- c:IIJ+42sin(S.r).--x< .r.,If.:< x. SolutionLetusdecompost'theproblelIlintotwosllhprohlcllli'iili'ifollows: (a) { = :" +cos(4z),.-x_ 'd>... v21l'-00 Theoperationofpassingfromf (x)to1(>..)iscalledtheFourierintegral transformation,whereasthatofpassingfrom1(>..)tof(x)iscalledtheinverse Fouriertransformation. If the functionf(x)is given on the half-line 0 < x< 00,then one can consider itsFouriercosinetransform jc(>..)=if 100 f(x) cos(>..x) dx. TopassfromJc(>..)to the original functionf(x)oneusesthe formula fi (Xlf(x)=V:;;:Jor(>..) cos(>..x)d>... Similarly,onedefinestheFouriersinetransform if 100 r(>..)=- f(x) sin(>..x) dx, 1l'0 andonehasthe inversionformula fi roof(x)=v:;;: Jof"(>..)sin(>.x) d>.. 2.:).INTEGHALTHANSFORI\I\JETHOD97 Herearetwobasicpropertiesof theFouriertransformatioll. (1)Lillearity: !'l,/J+ ('2.1'2=clh +1''2/2(where ('1and ('2art' arbitrary (,()llstants): (2)Ruleoftransformationofpartialderivatives:ifII=- It(,/'. I)andtheFouricr transformatioIlistakewithrespecttothe,f'variable,tlH'n {I,=i AU,117,.=(iA )2U,, .. =(i A) IIii, '-v-' (thisisestablishedbyintegrationbyparts),Thet ransforlllsoft hI'partial derivativeswithrespect.totarcgivenby (J2 if Uti=-:-:1'' ",vt I (N' . , '-v-' iJlIu iii II (Ilndertheasslll1lptiollthattheindicatedpartialcieri vilt i \'('Swi t.1lrpspectto Isatisfyc(']'tainconditiolls), \VesccthatundertheactionoftheFouriertrallsforlllat iOlltheoperation ofdifferentiationwithrespe('(to,];becoIl](,stheoperat iOlloflIlultipliciltioll(by iA).ThisimportantfactisIlscdillsolving \'all1('pmbklllsforpartial differentialequations, TheFouriertransforlllatiollelljoysmallYotherint prest illgprop('rtieswhich willnotbeciis(,llsscdhere, Examples of FouriertransformsaregiveninTable:2.1. Table2.1Fouriertransformsof severalfunctions no,FUllctionf (,],)Functionf(A)

{ 1.ifI,rlS(J,/:2sin (A(J) 1 V; ---0, ifI,rl> o. A " 1 ,2 2 (- (f1 ._f -10 v:2(1

(,-(11,,1 '20 V; ._--(12+ A2 1 1c: /1 I '_'(-li!'\ +'!.V '!. I (OJ) I-(A);) - r-(I(I 6I(;r- 0) (I 0),1/(a2 + XL),etc. Remark2.Tosolvetheproblemonthehalf-linewiththeboundarycondition v.(O.t)=0wemustusctheFouriersinetransform.Indeed.if U'(A. t)=.II (exou((. t) sin(A() de. V;: J() then n(.I:. t)=II j'exoU'(A. t) sill(A,r) riA. Y;:,o whenceu(O,t)=O. Nowlet.usconsideramorecomplicatedproblemillwhi('ht heFouriertrami-formationapplies. Example2[4,Cli.II.no.176].SolvetheCauchyproblelll { 'J:l lItt=u-u:r:r+ CU+ f(:r,f), 11(r, 0)= O.11/(1:.:,0)= 0, -x T. Then the solution ute, t)of t.heoriginal(Fourier-transformed)problemisgivenby theformula where Thisyields lt 11((, t)=v((, t, T)dT. () { ~ =.C1cos(v. a'(' - c' T)+ C, Sin (Va'(' .. c' T). f((. T)=-C] Va2(2- (;2sin (Va2(2- ('2T) + + C2va2(2- c2 cos(Va2(2- ('2T). 102CHAPTER2.HYPERBOLICPROBLEMS Denoteva2(2- C2 =b.Then whence and I COS(bT)0If(CT) C2=.)1(,7)=-b- COS(bT). -sm(bT-b-Therefore, i.e., V(Ct,T)=_f((;T)sin(bT)cos(bt)+ !((;T)cos(bT)sin(bt)= =f( C T)sin b( t_T)=f( C T)sin ( va2(2- c2 (t- T)), bva2(2- c2 lot~ sin (va2(2- c2(t- T)) u(C t)=f(C T)VdT. oa2(2- c2 Nowapplying the formulafortheinverseFouriertransform weget u(x, t)=_1_ JOOu(C t)ei(x d( = J21f-00 1lot ~ sin(Va2(2-c2(t-T)). =- dTf(C T)e"(xd(= J21f0=va2(2- c2 1lotJOOjDOsin(la2i2-c2(t-T)) =-- dTd(f(A, T)V\22ei(x-).,)dA. J21f0-00-00va (- c Inthe theoryof Besselfunctiononehastheformula sinr=~ (ITJo(rsincpsinB)eircos'PcosesinBdB. r2 Jo Nowletuschangeof variableinthisidentity accordingtotherule rcoscp=-a((t - T), Then r2=(t - T)2(a2(2- c2). r sin cp= ic( t- T). 2.:3.FOURIERINTEGRALTRANSFORMrvlETHOI) It followsthat sin(Ja2(2- c2(t-T))1111" --r=::=::==::::::::=c.-'-- =- Jo(ic(t - T)sin (1)c-W((I-T) cos H t;in (1. Ja2(2_(,2(t-T)20 Next,letusmake the change of variable cos e =/i / (a( 1- T))inthis integral.Then -sinede = drJ/[a.(t- T)]and 1[11". =2:Jo(ic(t- T)sin e)e-1(l((t-T) (()SI)sin (1= ,0 1ja(t-T)()('-I(i1 =- .'.10iC(t-T)1- 2()2--d{J= 2a-a(t-T)at- Tt-- T 1!(l(I-T)(fP)(-I(i1 =- .10c(t- T)2- 2--d,d. 2a,--u(I-T)0.I- T Consequently 111jX U(:l', t)=- dT.dCx 47r(L(J-x !x[j(lU-T)(J2) xf (A, T)10c(t- T) 2- , -=-(l(t-T) Letusset Then J0(t-T)(1)2) .. ioC -a(I-T)(L /'0(1T) df3=y(jje'((C-1) d,d. ,-0.(1T) BytheFourierintegralformula, - d(g([-J)ei((r->--i1) df3=g(,/'- A). 1I'x;j'= 27r, -x-ex; Thisyields(bythedefinitionof thefunctiong(;3)) Wefinallyconcludethat if x- a(t- T) 0, Solving this newproblem byDuhamel's formula(method of variation of con-stants),weobtain itsin (a(t- T)))..2+ jI.2) u()..,jI.,t)=f()..,jI., t))dT. oa)..2+ jI.2 Next,applyingthe inverseFouriertransformation wehave 2,:1,FOURIERINTEGRALTRANSFOR!,ll\TETHOD105 Insertingherethe valueoff(>"M, t)and denotingv=VAL+ p2,We'arrivpat the relation 1if 1L(;r:,y, t)=-()2dTX 27f() -x Tocalculatethequadruple indefinite integral weintroducepolar coordinates viatherelations { (- J'= p ip, II-Y=PSlIlip, Thisimmediatelygives {A =[) cos 0, II=[) sin fJ, A(( - ,r)+ p(TJ- y)=pvcos(O- ip)=()[JCOSt' wherelj'=e - ip,Itfollowsthat ex .fffl f((, TJ,T) - T) ei[),(:I-()+/L(Y-I/J]d( dll riAdll= -'XC =I'x! rip(exdvflc dipj'2Kf((,TJ, T)sin(av(t - T)c-il"nos 'I'PlJdu', , ()./0./0()W) Nowobservethat the followingrdation(integralrepresentation of t heBessel functions)holds: Inparticular, J( ,)-i,rcost!'+illl,'/., (')11jK , ".r- CII;, 27f-K 1j,271' ,}o(:1')=27f() COSIii lit" or(considering.rreal) It followsthat 1j'LK ,}o(.z;)=27f0 COSI/'dc', ,.SIllav- T,.,IJI'('(" j'X)j''XI'LITj'LK,,((/))' lipdlJdipj((./I,T)(f'(JlJr//;'= ,II,II'II()(J./) ')I'xj'XiLK =_7fripdlJf((,rl, T)sin(alJ(t ([,II'I IIl 106CHAPTER2.HYPERBOLICPROBLEMS Weshallalsousethe factthat 00{0, 1 Jo(pv) sin(av(t - T))dv=1 oJa2(t-T)2-p2' ifa(t-T) 0, Solution.ApplyingtheFouriertransformationwithrespecttothevariablesx,y, zto the equation andinitialconditions,weobtain theCauchyproblem { ~ t t \ j.L,v, t)+ a2 (>.2+2 + v)U(>.,IL,v, t)= 1(>., j.L,v, t),t> 0, U(>.,j.L,v, t)lt=o= 0,Ut(>',IL, v, t)lt=o= 0, where 00 ~ \ t)- 1Jf] -i(.\x+py+vz).(t) ddd u/\,j.L,V,- (21f)3/2eux,y,z,.xyz, -00 and 00 j~ \t)- 1Jlf -i(,\x+py+vz)j(t) ddd /\,j.L,V,- (21fr>/2..ex,y,z,xyz. -00 :2.:1.FOUHIERINTEGRALTRANSFORlII'dETHO])107 TIl('solutionof thisproblem(b.\'analog.\'withtheprecedillgOllt')isgiwll the integral - 1'1 sin(o(l(1-- T)) H(A,ILlJ,t)=f(A,II,I/,t)rIT. ,IIII f! wherepL=A"+pL+1/'-Bytheformulafort IteinverseFouriertransfornl (lJ', ift< !l.r. 11,5 Example2[4,no.839].Solvethefollowinginitial-boundaryvalueproblemonthe half-line: 0< x\)iscalledtheFourier-Bessel-Hankeltransformorimageof the functionf(p),whichinturn iscalledtheoriginal. Clearly,it isadvisable to usethe Hankel transformation inthe case when the Laplaceoperator!1u(ortheoperator!12u)iswritteninpolarcoordinatesand thepolarradiusrangesfromto00.Letusgivesomerelevantexamples. Example1[4,Ch.4,no.109].Solvetheboundary valueproblem { Utt= a2 +,< p < 00, u(p,O)=A,Ut(p,O) = 0,0::;p < 00, VI + p2/b2 0< t< 00, wherea2,A,and b2 are someconstants. Solution.ToderivetheequationfortheimageU()., t),letusapplytheHankel transformation (of order zero)with respectto the variable p to the given equation. Wehave U()..,t)= 100 pu(p, t)Jo()..p) dp,andalsou(p, t)= 100 )..U()., t)Jo()..p) d)". The left-handsideof our equation transformsas Utt(p, t)f-+Utt()., t). The right-hand sideof the equation transformsasfollows(here weuseintegration byparts): 100 :p (ppJo()..p) dp= 100 :p (pJo()..p)dp= (p Jo()"p))[-100 )..p dp=-)..100 p dp= =-)..+ )..100 u dp= )..100 + Jg()..p)p)..)udp= 100 + )"JM)..p))udp, where the integrated terms vanish thanks to the condition u(oo, t)= up(oo, t)= 0. Bydefinition,theBesselfunctionJo(x)satisfiesthe equation whence )..2 pJg ()..p)+ =_p)..2 Jo()..p). P 2.5.HANKELINTEGRALTRANSFORr-.!METHOD117 Therefore, JOG10(OU)2jOC' - -:--)P-;:;- pJO()..p)dp=-)..pu.Jo()..(I)dp=-)..u()..,t). oP (p0(10 Weseethatthegivenpartialdifferentialequationistransformedintothe ordinarydifferentialequation I> O. Now let us calculate the Fourier- Bessel-Hankel image of the function u(p, 0)= AI VI + p2 Ib2.Wehave Ii()..,0)=ICC(lA,JU((lA)rip. oVI + p21b2 It isknownthat whence I'Xe-w.\1 ,()-A- AJo(pA) dA=vw2 + p2 Therefore,bytheinversionformula, j'X1e-w''\ V 22(lJo(pA)ri(l=-\- . ,0w+p/\ whichgives I"XAbe11.\ rip=Ab-. .0Vb2 + p2A Wethusarriveatt heCauchyproblem t> O. /LIIA,t+ aAUA, t=,0,, { ()2 ) Abb.\ ( 11/\,0=Te- ,iLlA.O)=O. The solutionof thisproblemis )Ab11.\..(\/) Il(A.t=TEe(OS(1/\. Toobtain the solutioll of the original problem.itremainsto applyt heinverse HankeltranSfOrIllatioIl: u(p. t)= IX AU(A,t)Jo(Ap) liA=Ab(K riA. ().10 118CHAPTER2.HYPERBOLICPROBLEMS Since cos(a>.t)= Ree-ia..\t, weget u(p,t)=Ab.RetOe-..\(b+iat)Jo(>.p)d>'=Ab'Re1. J()Vp2+ (b+iat)2 Further, 11111 Re=Re - .=- Re -----=== Vp2+(b+iat)2b bJo+i{3' wherewedenote Wehave (3=2at b. 11 Re=Re J 0+ i(3V02 + ,82 (cos i+ i sin i) (here wechoose the firstbranch of Jo + i(3,corresponding to k =0 in the expres-sionJo + i(3=v02 + (32ei('P+27rk)/2,wherek = 0,1).Obviously, with and Re1=cos(r.p/2) Jo + i,8(3 tanr.p=-o cos(r.p/2)==- VI + cosr.p=- 1 +. v1+eo,,,11ja 2J2J2V02 + (32 Therefore, 11 Re=-In + i(3J2 10 V02 + ,82 + 02 + ,82 . Thus,thesolutionof ourproblemis u(p, t) ;2[ 2.5.HANKELINTEGRALTRANSFORl\!\IETHOD119 Example2 [4,eh.VI,no.110].Find t h(' radially symmetri(' transvprsal oscillations ofallinfiniteplate,as:mmingthatitsinitialpositionisgiven(andciclJencisonly ontheradius)andtheinitialvelocityissetequaltozero. Solution.It isknownthatthetransversaloscillationsofallillfilliteplat(,arede-scrihedbythee(jllation/Lit+ /;2 t::,.2u=0wit hprescrihedinitial(ollditiollS.IIIthe presentcaseitiscOIlvenienttopasstopolar ("oordinat('sand write the llwthemat-icalformulationof theproblemilltheform { Utt+ b" i J ) 2 ~ +; ~ ) )"211=O.0< p< x. (Ipup u(p,O)=f(p),u,(p.O)= O.()Sp < X. l1(OC.t)=Ilfl(OC.t)=UfiP(OC.t)=IlflPfi(X.t)=O. O'p) dp= >.4Jou(p, t)pJo(>'p) dp=>.4U(>., t), and soour original problems istransformed into the followingCauchy problem for an ordinary differential equation: {Utt+b2>.4U(>',t)=0,t>o, U(>',O)= 1(>.),Ut(>',O)= 0, where 1(>.)= Jooo J(p)pJo(>'p) dp. Clearly,the solutionof thelatter isgivenby U(>.,t)= 1(>.) COS(b>.2t). The solution of the original problem isrecoveredusing the inversion formula: u(p, t)= 100 >.1(>.) COS(b>.2t)JO(>'p) d>'= = 100 >.cos(b>.2t)JO(>'p)(100 p}(fJ) Jo(>'IL)dfJ)d>'= = 1C fJJ(fJ)(1C >.cos(b>.2t)JO(>.p)Jo(>'IL) d>.)dfJ The integralinsideparenthesesisfurthertransformedasfollows: 2.6.STANDINGWAVES.BOUNDEDSTRINe121 _R1.(p"+II")/(1lhl) J(Pfl",)_ ~ ~ '()~ ~- '-2ibt'.-'2ibt-1(.f/2+ IL2.p2+ p'2)( PI}) =HeICOS+ SillIII~2bt4bt4bt2bt 1.p2+ /L2(PII) =2btSill4btI()2bt. Thus,thesolutionhastheform 1f=p2+ 112((1/1) alp, t)=2btIlf l/1) sin4blII)2btel11. . () 2.6.The method of standing waves.Oscillations of a bounded string WealreadyknowhowtosolvethewaveequationI1tt=(I2U r ,rOilthewholeline -00 < J'< DC(Cauchyproblem).Thed'Alcmbertformulaexpn'ss('sthcsolution asasum of twotravellingwavesthatpropagateilloppositedirections.If wenow consider the same equation onabounded interval O. d.1'.O)=O.l't(.r.O)=:r - 2.(J'S.I'S2. 126CHAPTER2.HYPERBOLICPROBLEMS Asafirststepletussolvethefollowingproblem:findasolutionoftheequa-tionVtt=Vxx+ v,notidenticallyequaltozero,whichsatisfiesthehomoge-neousboundaryconditionsv(O, t)=v(2, t)=0andisrepresentableintheform v(x, t)=X(x)T(t).Wegetthe equation Til X= X"T + XT, or,upondividingboth sidesbyXT, TilX" --1=-TX The left- [resp.,right-]hand side of this equation depends only on t[resp.,x].Since tandxareindependentvariables, TilX" - -1 =- =-A TX' whereA isthe separation constant.This yieldstwoordinary differential equations: X" + AX = 0,(2.14) and T"+(A-1)T=0.(2.15) Fromtheboundaryconditionsv(O, t)=v(2, t)=0andtheexpressionof v(t, x)itfollowsthatX(O)T(t)=0andX(2)T(t)= 0,whenceX(O)= X(2)= O. Thus,wearrivedattheSturm-Liouvilleproblem {X" + AX =0,0< x< 2, X(O)=X(2)= O. The eigenvalues and corresponding eigenfunctions of thisproblem areAn=(1727r) 2 andXn(x)= sin ( ~ 7 r x),n=1,2, .... Next,fromequation(2.15)itfollowsthat whereAnandBnarearbitraryconstantsandfLn=J('';) 2- 1,n=1,2, ... Clearly,the functions v,,(x, t)=[AnCOS(fLnt)+ Bn sin(fLnt)] sin (n21Tx) areparticularsolutionsof theequationVtt=Vxx+ vthatsatisfynullboundary conditions. and 2.7.MIXEDPROBLEMSFORSTRING Nowletussolvethe followingtwoproblems (I) { Vtt=v,: + v, v(x,O)- 0, v(O, t)=0, 0< x< 2,t> 0, Vt(x,O)=x - 2,0::;x::;2, v(2, t)=0,t2'0, {Vtt=VXX+V+t(2-X),0 0, ulx=o= ul,r=7r= uly=o= uIY=7r= 0,t 2"0, u(x, y, 0)= 0,Ut(x, y. 0)= 0,0::;x, y::;7r. 18.Solvethefollowingmixedproblem onthe oscillationsof amembrane: Utt=C2(uxx+Uyy),O O. u(a, tp,t)= Asin(2tp),0 -:;tp-:;Jr,t2:0, u(p, tp,0)= 0,Ut(p,tp,0)= 0,0-:;p -:;a,0-:;tp-:;2Jr. 30.Find the oscillations of acircular membrane(0-:;p-:;a),with null boundary conditions,generatedby the motionof itsboundary accordingto thelaw u(a, tp,t)=Asin(wt),A=const. 2.12.PROBLEMSFORINDEPENDENTSTUDY153 31.Solve the followingmixedproblem forthe waveequation inabounded cylin-der: o < (!< (1,0< ;:;< h. t > 0, { 2(ILl).(2'll') Utt.= 6.u + Uo Psmh_Z, ulz=o- 0,Ulz=h- 0,ulp=n- 0, t;::.0, ull=o= 0,Utlt=o= 0, whereILlisthe firstpositiverootof the equationJo(J:)= O. 32.Find the oscillations of a gas in a spherical container, generated by oscillations of its wallsthat start at t= O.Itisassumed that thpvelocit,vof the particles aredirectedalongtheradiiofthecontainerandhavemagnitudeequalto t2 Pn (cos ()),whereP" (x)istheLegendrepolynomialof degreen. 33.Showthatif thefunctionsf(x),uo(x)andU1(:1:)areharmonicinthewhole Euclidean space R" and g(t)is a continuously differentiable function fort;::.0, then the solutionof theCauchyproblem {Utt=C26.U+f(X)g(t),::r:ER".t>O. ult=o= uo(x),Utll=o= U1(:1-)..rER". isgivenbythe formula u(x, t)=uo(x) + tUl(X)+ f(x) 1t (t- T)g(T) dT. o 34.Find the oscillations of an infinite circular cylinder of radius(1,generatedby the motion of its lateral surface according to the lawA sin(....:t) cos (1/ 0, Utlt=o= O..1'ER". isgivenbytheformulau(x, t)=8uj(x, t)/8t. 36.Solvethe nonlinearCauchyproblem { Uft=a26.u + (\7U)2- u;, u(:r,O)=0,Ut(x,O)=f(x), .rER:l,t>0, .1'ER:l. 154CHAPTER2.HYPERBOLICPROBLEMS 37.Showthat thefunctionu(x, t)isthe solution of theCauchyproblem {Utt+6.2U=0,xERn, ult=o=f(x),utlt=o= 0, t> 0, xERn. if and onlyif the function V(x, t)= u(x, t)+ irt 6u(x, T)dT Jo isthe solution of the Cauchy problem forthe homogeneous Schrodinger equa-tion { Vt= i6v,xERn,t> 0, vlt=o=f(x),xERn. 38.Letthe functionu(x, t)bethe solution of theCauchyproblem forthe homo-geneousSchrodingerequation { Ut=i6u, ult=o=f(x), xERn,t> 0, xERn, wheref(x)is a real-valued function.Find the solution ofthe Cauchy problem { Vtt+ 6v =0, vlt=o=f(x), xERn, vtlt=o=0, t> 0, :1':ERn. 39.Letthefunctionf(x, t)bebiharmonicinthevariablexforeachfixedt2:: (i.e.,62 f= 0).Find the solution of theCauchyproblem { Utt+ 62u=f(x, t), ult=o= 0,Utlt=o= 0, 40.Find the solution of theCauchyproblem xER",t > 0, xERn. { Utt+ 62 U= 0,xERn,t> 0, ult=o= Uo(x),Ut It=o= Ul (x),xERTl, whereuo(x)and Ul(X)arebiharmonicfunctions(i.e.,62uo= 0,62ul= 0). 2.1:3.ANSWERS 2.13.Answers 1 1.11=-20 ) sinh(2nlt). 2.II=p_(r2+tL)[.rcosh(2:rt)- tsinh(2:rt)]. 3.11=f- 1 +(-I +sin(.!'+f). 4.1/=.r2+ t'2+ {SillY. '2:z28:;14L2Ii 5.II=Y+I.:+81+:3 1+12t.1'+45f. f(r+ af)+ f(r- af)1j"c+(J/(C)d 6.II=+ - q 0aretherootsof the equationJO(J:)=O. 29.71=sin(2:p)tAnh (II;'p)cost)+ A sin(2:p),where rn=l andJim> 0aretherootsof the equationh (x)=(). ( uJ)xc Jo - p""'(/l'TI)( ('/1 n) 30.71=A sin(wt)-An.Io-psin-{ ,where .10 a() (11- I 2Awl,a(W)( I'll) An=.uJ2pJo-:-f!Jo-(Jrip. Cf1naJO(7a) [JdJin)].()(1/ cistheconstantfromtheequationVtt=c2611and/1/1> ()aret 1]('rootsof the equation.10 Cr). 158CHAPTER2.HYPERBOLICPROBLEMS (f-ll).(27r) xJo-;;PsmhZ . pn Pn(COS(})22Pn(cos(})L:oo 1 32.U=t- --x 2npn-la2npn-3((n))2 o0k=lf-lk wherethe coefficientsAkare givenby the formula 1ipO( (n)) A n+3/2Jf-lkd k=Pn+l/2-pp, .cl[1_n(n+l)]J2((n))0Po 2(!Lkn))2n+1/2f-lk Poisthe radiusof the container,are thepositiverootsof the equation and a isthe constantin the equation Utt= a2 !:lu. 33.Hint.Considerthe twoproblems and { Utt= a2!:lu + f(x)g(t), ult=o= uo(x),utlt=o= O. J(W)00 34.u=A J: sin(wt) cos (n () ILl> () FIGI'RE:3.2 J' Letus mention that several problems of electrodynamics abo reduce to equa-tionsof parabolictype. 3.1.TheFourierintegraltransformmethod WehavealreadyseenhowtheFouriertransformatiollcallheul)('dto solvehyper-bolicproblems.Inthissectionweconsider the applicatioll of t heFouriert ransfor-mationtosolvingparabolicproblems. Problemsonthelineandhalf-line. Example1[4,eh. III,no.[)Sj.ApplyingtheFourieriut cgraltransformat iOIl,solve thefollowingboundary valueproblem: { lit+ f).. -DO< .r< x. u(.1, 0)- 0,oc< .r< oc. t>O. 162CHAPTER3.PARABOLICPROBLEMS Solution.Denotebyt)theFouriertransform(withrespecttothe variablex) of the functionu(x, t),i.e., t)= /00u(x)e-iExdx, v 21T-00 or,inasimplerform,t)=F[u(x, t)],whereFistheFouriertransformation operator.It iswellknownthat F[uxx(x, t)]= (i02 F[u(x, t)], i.e., F[uxx(x, t)]=t). Further, F[u(x, 0)]= 0)= O. Also,considerF[J(x, t)]=t).ThentheheatequationinFouriertrans-formsreads This isafirst-orderordinary differentialequation inthe variable t(with playing the roleof aparameter).Therefore,wehavethe followingCauchyproblem: { t) t)= t), 0)- O. t > 0, WewillsolvethisCauchyproblemusingDuhamel'smethod.Tothisendlet ussolvefirstthe auxiliaryproblem { 22 Vt ( t)+a(t)= 0,t> 0, v(C t)lt=T=T). Wehave Then,asweknow, trt t)= Jot;T)dT=Jof(C T)e-ea2(t-T)dT. It followsthat 1100 c1it100 c2 2(t)- c u(x, t)=- =- dT a.-Tf(C T)e"'X= j21T-00j21T0-00 3.1.FOURIERTRANSFORMMETHOD163 1ITJoc(A_x)2~=- dTf(>-',T)d>-.e - 4a2(t-T):2= 27f()_ oca(t - T) (>.._.r):2 1itJoce - ~=r:;;.dTf(>-',T).;t=Td>-.. 2ay 7f0-x;t- T Inthiscalculationwehaveusedthe well-knownexpressionfort heFouriertrans-formof the functione-b:r2 (b> 0),namely Weconcludethatthe solution of our problem is (..\._.1')2 u(x, t)=1 r:;;.r dT Jocf()...,T) e;;;;; d>-.. 2ay7fio-oct-T Example2[4,Ch.III,no.57].Applying the Fourier integral transformation,solve the boundary valueproblem { Ut= a2u:ra,0< x, t< 'Xi, u,,(O,t)=0,0 < t< iX. u(x,O)=f(x),0< x< iX. Solution.Letusapply theFouriercosinetransformatioll u'(>-.,t)=filC u(x,t)cos(>-..r)d.r. y-;() Recallthatthe inverseFouriercosim'transformationisgivenh ~ the rule 11(;1",t)=fi r=u'()., t) cos(>...r)d>-.. Y-; io Fromthisequalityitreadilyfollowsthat 164CHAPTER3.PARABOLICPROBLEMS whenceux(O, t)= 0,Le.,theboundary conditioninourproblem issatisfied.Fur-ther,observethat =(_>.)2 I!100 u(x,t) cos(>.x) dx and jc(>.)=I!100 f(x) cos(>.x) dx. InthiswayweobtainthefollowingCauchyproblemforanordinarydiffer-ential equation: { t) + a: >.2UC(>.,t)=0, UC(>.,0)= r(>.), where>.isregardedasaparameter. The solution of thisproblem has the form t> 0, Applying the inverseFouriercosinetransformation weobtain f2roo- 22 u(x, t)=V;: ior(>')e-a >.tcos (>.x) d>'= = 100 (100 cos(>.x)e-a2>.2td>'= = 100 [100 e-a2 >.2tcos >.(x d>'+ 100 e-a2 >.2tcos >.(x- d>.]= =--+ e- 4at1100 [(x+o2

2a,,;;t0 In this calculation weusedthe factthat 10022Jif{32 e-a Ycos(f3y) dy=- o2a (a,f3>O). Example 3[4,Ch. III,no.113].Find the temperature distribution along an infinite rod composedof twohomogeneousrodsin contactatthepointx=0,withchar-acteristics aI, kland a2,k2,respectively(hereaI, a2are the constants in the heat :3.1.FOURIERTRANSFORMMETHOD165 equation and kl' k2are the heat conductivity coefficients).The initial temperature IS if x< O. if x> O. Solution.From thePoissonformula whichgivesthetemperaturedistributionalonganinfiniterodundertheinitial condition u(x, 0).itfollowsthatthe solutionof theheatequation with initialcondition isgivenbythefunction -00 < x< 00,t> o. { T*,if x> 0, u(x,O)= T2,if x< 0 T+ T*T*- T().) u(x,t)=+~ . 222avt wherethefunction(::r:)=ftc ~ ; r e-y2 dyisthe so-callederrorintegral. Tosolvetheoriginalproblemwewillargueasfollows.Firstweextendthe leftrod(with temperature Tl)to the rightto obtain an infinitehomogeneousrod. N extwefindthetemperatureof theresultinginfiniterodundert heassumption that itsinitialtemperature is where Ttissomeconstant. if x< O. if x> O. Wethen proceed ina similar manner with the rightsemi-infinite rod.assum-ingthat { T2,if x> 0, U2(X,0)= T5.,if x< O. The constants Ttand T5.arefoundfromthe conjugation(matching)conditions: (1)continuityof thetemperature attht'point.r= 0: (2)continuityof theheatfluxatthepointx=0: aUlaU2 klax(0, t)=k2ax(0, t). 166CHAPTER3.PARABOLICPROBLEMS Fromthe formulagivenaboveitfollowsthat () _Tl+ T{T{- Tl in(_X_) UlX, t- 2+2'i'fi' 2al vt () __ T2+ T2'T2- T2'in(_X_) U2X, t-- +2'i'fi' 22a2Vt Hence,theconjugationconditionsyieldthefollowingsystemofequationswith unknownsT{and T2': Solving this system,weobtain Therefore, where Ul (x, t)= To+ (To- Tt} 0, ult=o- 0,-00 < x, y, z< 00. Solution.WeusetheFourier transformation withrespectto the spacevariables: 00 -( \t)- 1Je {{(t)-i('>"x+l-lY+vz)ddd UA,/-l,l/,- (2n)3/2)} UX,y,Z,exyZ, -00 3.1.FOURIERTRANSFORMMETHOD 00 g()..,j.L,v, t)=JJJg(X, y, z, t)e-i(>.:r+llil+VZ)d:rdy dz. -00 Weobtain aCauchyproblemforthe Fouriertransforms: { - 2(222)- -+ aA+ j.L+ vu= g(A, j.L,v, t). ult=o= O. t> O. Applying hereDuhamel'smethodwehave NowusingtheinverseFourier transformationweobtain But 00 (t) - 1fr!f-(\t)i(>':r+I'!J+vo)d\Id-uX,Y.X,- (21frl/2 JuuA,j.L, v,eA( Itv--00 xei(AX+11Y+VZ)dAdj.Ldv= x fr!f222 Xr e-(A+1'+VdAdJ.Ldv. -x x d\Id-(P,v--:xc =a:l(t- rp/2 (' Weconclude that 167 168CHAPTER3.PARABOLICPROBLEMS Remark1.Suppose that g(x, y, z, t)doesnotdepend on t,i.e.,wearedealing with theCauchy problem { Ut=+ + uzz )+ g(x, y, z), ult=o- 0,00< x, y, z< 00. Then the solutionisgivenby where 2r2 (a)=yHJoe-P dp, Indeed,letusdenote W=2aF' Then -00 < x, y, z< '00,t> 0, =_1 J"7J'(yH - (2a''vte-w2 dW)] = 47fa2 rryH2Jo -00 =_1[1- (_r )]47fa2 J.r2a/t -00 Remark2.Suppose g(x, y,z, t)doesnotdepend onz,i.e.,weare dealing with the Cauchyproblem { Ut= a2(uxx+ Uyy ) + g(x, y, t), ult=o= 0,00< x,y < 00. -00 < x, y< 00,t> 0, Then clearlyin thiscasethesolutionisgivenby 3.1.FOURIERTRANSFORMMETHOD169 Example5[4,Ch.V,no.67].Findthetemperaturedist ributioTlintheupper half-spaceif theinitialtemperatureinthehalf-spaceisL;Cro,thetemperaturein theplanez=()is?;ero,andinthehalf-planeitself thereisadistributionof heat sourceswithdensityfCr,y, z. t),i.e.,solvetheCauchyproblem 2 { 7Lt=a_(UI:r +. . ..+ f(X.,y, z, t),. -00 < .f: .I! O.f> 0, alz=o- 0,Cle< .1,lJ< Cle,t;::,0, I1lt=o=0,-x < J:,lJ< 00,z;::'O. Solution.InthepresentcasewewilltaketheFouriertransformatiollint heform x U(A.lt.V.t)=1/21:)/2JIrXu(x.y.z.t)e-i(AI+11IJisin(v:)d:rdyd.:. 2'if..10 -"XC Then ex: 11(."':,y, z, f)=21.I.! ,LX u( A, 11,v,t )ei(A.r+llI/) sill (1/.:)riAdp. dl; -x (notethatthisguaranteesthatthecondition nlz=o=0 issat.isfied),The ordinary differentialequation(withrespecttof)fortheFouriprt rallCifonllsis {ill+ a:'(A'2+ /12+ v'2Fi=1(\ /1,v. fl. ull=o= 0, The solutionof thisproblemisgivenby Therefore. x I> n, - 1fj' l'xt))(A;dlll/"(')1\1i-u"'.y,f)- ,1/2:1/2/1A, /1,v . .L:-ill!IJ,(A (p.l /J-2'if,,0 ->C 170CHAPTER3.PARABOLICPROBLEMS OG =dT/flOG 27f()() -OG 00 XJJ100 sin(v() sin(vz) d)"dJi.dv. -OG Letuscompute the secondtripleintegral: OG JJe-a2(..\2+112)(t-r)d)"dJi1OGe -a2v2(t-r) sin(v() sin(vz) dv = -00 7f(x_02+(y_,,)2 -::--,-------,--e4a2(t-T)X a2(t- T) [100 e-a2v2(t-r)cosv((_z)dv_lC e-aV(t-r)cosv((+Z)dV] Weconclude that the solution of the originalproblem isgivenby the formula 1itdT100 u(x, y, z, t)=(J7TP(P/2d(x 2a7f()t- T0 -oc 3.2.TheLaplaceintegraltransformmethod Wehavealready madeacquaintance with theapplication of the Laplacetransfor-mationtohyperbolicequations.InthissectionwewillusetheLaplacetransfor-mationtosolveseveralparabolicequations.Letusnotethattheblockdiagram forsolvingproblemsbythismethodisidenticalto thatinthehyperboliccase. Example1[3,Ch.5,no.832].Solvetheboundary valueproblem { Ut=u,: + U- f(x), _0< x, t< 00, u(O,t)- t,ur(O,t)- 0, t< 00. (3.1 ) :\.2.LAPLACETRANSFORf..IMETHOD171 Solution.WeshallusetheLaplacetransformation\villlrespectt ()thevariablef. Usingitsproperties,wecanwrite or 11 ( 1:, t)==U (p, t). vd.r, t)~ UI(p,t), IIr(J:,t)~ pU(p, t)- t. I(;)")~ 1'"(p). ApplyingtheLaplacetransformationtobothsidesof (',.pand denoting Y(:1') I:r=v'>,p=R(p), weobtain R'=...f>,.y', R" = AY", where now the prime denotes differentiation with respectto .r.Therefore.wehave the equation x2y"+ xy' + x2y = 0, whichisrecognizedtobe theBesselequation of indexzero.Itssolutionis where Jo (x)and Yo (x)are the Bessel functions of index zero of the firstand second kind,respectively,andC1,C2 arearbitraryconstants.Clearly.C2 =0,because thesolutionofourproblemmustbefiniteandthecenterofthedisc.Hence, y(x)=CJo(x),or,takingC=1, R(p)=Jo(...f>,.p). By the boundary conditions,R(b)=0,i.e., Recallingthatthepositiverootsof theBesselfunction.10aredenotedby /JOn,we findthatthe eigenvaluesof ourproblemareAn= 2.wheren=1.2 ....The correspondingeigenfunctionsare Rn(P)= Jo p),n= 1. 2 .... 178CHAPTER3.PARABOLICPROBLEMS Next,wefindthe corresponding solutionsTn(t)of equation(2),namely, ("ona)2 Tn(t)= Cne- -b- t. Therefore,the solution the originalproblem isgivenbyaseries ~ ("ona )2t(f.1.on) u(p, t)= ~ Cne- -b- Jo -b-P . n=l Usingthe initial conditiononereadilyseesthatC3 =1 andCn =0 if nf- 3.We concludethatthe solution is ( "03 a)2(f.1.03) u(p,t)=e--b- tJo bP. Thecaseof arectangle. Example 4.Solve the following problem of heat propagation in a rectangle(Figure 3.4). Ut=a2(uxx+~ y , 0 < x< b1,t> 0, .(27rX).(7rY) ult=o= smb";smb;, ulr = 0,t2::0, whererdenotestheboundary of the rectangle. FIGURE3.4. Heatpropagation in arectangle Wewillseekthe solution of problem(3.6),(3.8)in the form u(x, y, t)= T(t)V(x, y). Substituting this expression in the heatequation weobtain or T'=Vxx+ Vyy=-,\ a2TV. (3.6) (3.7) (3.8) :U.FOURIERI\IETHOD.SEPAR.ATIONOF\ARIABLES179 Thisyieldsthefollowingtwoequations: (:3.9) (3.10) Notethat(:3.9)isapartialdifferentialequation,whereas(:3.10)isanordinary differelltialequat ion. Firstletussolve(:3.9).FromthcboundaryconditiollILk=()isfollowsthat V II'=O.Letusconsidertheeigellvalucprohlelll { ;!l + VIJIJ+ AV=0, tI,'=() andseekitssolutionintheform V(x, y)=X(.r)Y(y). Substitutingthisexpressioninequatioll(:3.11)weget or or,equivalently, X"y + Xy" + AXY =O. X"Y" - + - +A =0, XY yllX" -+A=--=P, YX Thisyieldstwoordinary differentialequat.ions: X" + liX =0, Y"+(A-p)Y=O. (3.11 ) (3,12) From theboundary conditionV k =()weobtain thebOlllHlarycOllditionsX (0)= X(btl=0forc 0, U(2) It=o= e-(X-y)2,-00 < x, y< 00. Thesolutionof problem(1)hastheformu(l)(X,y,t)=it.Letusfindthe solution of problem(2).Weshallassumethat U(2) (x, y, t)= v(x - y; t). Then denotingx- y= zand observing that 8U(2)8v 8z8v 8x8z 8x8z' 82u(2)82v82u(2)82v 8x2 8z2'8y28z2' weobtain theCauchy problem {182V Vt=4:8z2 '-00 < z0, I _z2 Vt=O=e,-00 < z< 00. The solution of this problem isgivenbythe Poisson formula(with a=1/2): a=(l+t (( _ _ z). Vtt+1 :3.4.MODIFIEDSEPARATION-OF-VARIABLESl\[ETHOD195 Therefore, .)1(.1_.11)'2 11(-1 (.r..I). t)=Jl + {-l+T and thesolutionof th('originalproblemis 11(.r-YJ" n(r. y. t)= -t +.8v1+t Example5[6.Ch.IV.1:3.7(4)].SolvetIl('Cauchyproblelll { 'ILt=tll1+ Y + ,:;),-oc < .r..I/.: