MATH Greens functions, integral equations and applications

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1 MATH 3432 Greens functions, integrl equtions nd pplictions Willim J. Prnell Spring 213 1

2 Contents Pge 1 Introduction nd motivtion 6 2 Green s functions in 1D Ordinry Differentil Equtions: review Generl forcing nd the influence (Green s) function Liner differentil opertors Sturm-Liouville (S-L) eigenvlue problems Existence nd uniqueness of BVPs for ODEs: The Fredholm Alterntive Wht is Green s function? Green s functions for Regulr S-L problems vi eigenfunction expnsions Green s functions for Regulr S-L problems using direct pproch Green s functions for the wve eqution with time hrmonic forcing The djoint Green s function Green s functions for non S-A BVPs Inhomogeneous boundry conditions Existence of zero eigenvlue - modified Green s functions Revision checklist Green s functions in 2 nd 3D Self-djointness An eigenvlue problem on rectngulr domin Eigenvlue problem for the Lplcin opertor Multidimensionl Dirc Delt Function Green s functions for the Lplce nd Poisson eqution Applictions of Poisson s eqution Helmholtz eqution in two sptil dimensions Where next? Revision checklist Theory of integrl equtions nd some exmples in 1D Liner integrl opertors Wht is n integrl eqution? Volterr integrl equtions govern IVPs Fredholm integrl equtions govern BVPs

3 4.5 Seprble (degenerte) kernels Neumnn series solution Wve propgtion in heterogeneous medi Revision checklist A Some helpful stuff (which you should know!) 1 B Exmple sheets 12 3

4 Syllbus Section 1: Introduction nd motivtion. Wht use re Green s functions nd integrl equtions? Some exmple pplictions. (.5 lecture) Section 2: Green s functions in 1D. Ordinry differentil equtions review, influence function, Liner differentil opertors, Green s identity, djoint nd self-djoint opertors, Sturm-Liouville eigenvlue ODE problems, Fredholm Alterntive, Green s functions s eigenfunction expnsions, dirc delt function nd generlized functions, direct pproch for determining Green s functions vi method of vrition of prmeters, the wve eqution, djoint Green s function, non Sturm-Liouville problems, modified Green s function nd inhomogeneous boundry conditions. (9.5 lectures). Section 3: Green s functions in 2 nd 3D. Sturm-Liouville problems in 2 nd 3D, Green s identity, Multidimensionl eigenvlue problems ssocited with the Lplcin opertor nd eigenfunction expnsions, bsics of Bessel functions, Green s function for Lplce s eqution in 2 nd 3D (unbounded nd simple bounded domins) nd ssocited pplictions, Green s function for Helmholtz eqution in 2D (unbounded nd simple bounded domins) nd ssocited wve scttering nd cloking problems. (7 lectures). Section 4: Integrl equtions in 1D. Liner integrl opertors nd integrl equtions in 1D, Volterr integrl equtions govern initil vlue problems, Fredholm integrl equtions govern boundry vlue problems, seprble (degenerte) kernels, Neumnn series solutions nd iterted kernels, pplictions to scttering. (5 lectures) 4

5 Course lecturer The course lecturer is Dr. Willim Prnell My office is in the School of Mthemtics, Aln Turing building. If you hve ny questions plese use either emil or preferbly sk me questions directly fter the lectures. You will hve plenty of time to discuss further spects in the exmples clsses. Course rrngements There will be two lectures per week in weeks 1-11 nd one exmples clss per week in weeks exmple sheets will be set, distributed ppropritely between weeks Sheet 1 is minly revision mteril - ensure you know it! Exmples clsses will be held in weeks There is no clss in week 1. If you cnnot do the mteril on Sheet 1, look bck t your MT1121 nd MT241 notes but of course sk me if in the end you re still hving problems. Students should work on the exmples sheets before the Exmple clss so tht they cn flg up ny difficulties. Some hours in week 12 will be set side for revision s should be expected. Lectures re held on Mondys, in the Schuster Moseley Lecture thetre nd Fridys in Aln Turing, G.17. The exmples clss follows the Fridy clss, lso in G.17. The purpose of this is for you to work through some of the problems on the exmples sheet tht you hve lredy looked t nd sk for help if you need it. The end of semester 2 hour exmintion ccounts for 8 % nd mid-term 5 minute test on the Fridy of week 7 ccounts for 2 %. The test will be on mteril from Section 2 only nd the ccompnying Exmple sheets 1-5 of the course. This test will help you with revision nd it is good to get it out of the wy before Ester. A note bout the notes These notes re pretty comprehensive. You should not relly need to look t ny other books s result of this. You my lso hve to look bck t your notes from MT1121 nd MT241 from time to time. There re plenty of exmples provided both in the notes nd on the Exmples sheets. In the lectures I will go through most of the notes but not lwys ll of the detils. The notes ccompny the lectures nd you should certinly still ttend nd listen crefully even though I provide these notes. I certinly will not necessrily write ll of the text on the bord lthough I will mention nd describe ll of the relted mthemticl ides. It is up to you to red the notes crefully. In lectures I will minly focus on the mthemtics, the theory nd model exmples to id understnding. Sometimes I will sk you to work through some of the exmples in the notes in your own time. And remember you need to spend gret del of your own time reding through the notes to understnd them! You will notice tht t the end of ech section I provide revision check-list. This should help you to understnd wht you do nd do not understnd t the end of the section with the id of the notes nd the relted exmples sheets. I urge you to look t the exmples sheets before the exmples clss. Otherwise you will not mke the most of the help vilble in the session nd you my fll behind. 5

6 Willim J. Prnell: MT3432. Section 1: Motivtion 6 1 Introduction nd motivtion In this course nd these notes we will discuss the solution to brod clss of problems in pplied mthemtics. We will lrgely focus on solving ordinry differentil equtions (ODEs) nd prtil differentil equtions (PDEs). These will tke the form Lu(x) = f(x) for ODEs nd we re interested in boundry vlue problems where x [, b] for some rel nd b with boundry conditions prescribed on nd b. In the end we wnt to solve for the field vrible u(x). We cn lso nlyse initil vlue problems where initil conditions re specified t x = but we only hve 22 lectures! Here L is known s n ordinry differentil opertor, e.g. L = d 2 /dx 2. The function f(x) is forcing function. PDEs will tke the form Lu = Q(x) (1.1) on some domin x D where x = (x, y, z) in three dimensionl problems. Here L is prtil differentil opertor, e.g. L = 2 / x / y / z 2, the Lplcin. Note tht for resons of clrity nd time restrictions we do not consider problems with explicit time dependence or forcing, or rther we consider certin types of time dependent problems, e.g. exp( iωt) for wve problems 1 which yield problems in the form (1.1) where (1.1) is the stedy-stte forcing. We lso restrict ttention to sclr problems so tht u is sclr field (temperture, pressure, etc.). As result of the bove, the PDEs tht we consider in this course re ll elliptic. This therefore includes the stedy stte het eqution (Lplce s eqution 2 u = Q(x)) nd the time hrmonic wve eqution (Helmholtz eqution 2 u + k 2 u = Q(x)). We ve done ll this before you my sy. Well you hve done some of it, but we will be lerning bout specil technique to solve inhomogeneous PDEs, i.e. when the forcing terms f(x) nd Q(x) bove re non-zero. This technique is the method of Green s functions 2. It trnspires tht the solution to the problem cn (in generl) be written s weighted integrl of the forcing over the domin, where the weighting is the Green s function. This is topic tht hs been nd is still of gret interest s reserch topic in pplied mthemtics. Green s functions hve pervded mny res of mthemtics, science, engineering nd computtion, often in surprising wys. In prticulr, Green s functions cn be used in order to re-write the differentil eqution forms of the problems in integrl eqution form. The subject of boundry element methods, n re of gret interest for solving problems numericlly, stems from this development. In ddition to the fct tht they re of gret use, they re lso very interesting mthemticlly. We will be ble to discuss vrious ides nd theoreticl spects pertining to the theory of ordinry nd prtil differentil equtions. As n exmple of the use of Green s functions, consider the simple ordinry differentil eqution of the form d 2 u = f(x) (1.2) dx2 1 In some of the Exmple Sheets we do consider smll subset of time dependent PDEs. The reson for doing this is to see the context in which our problems without time dependence reside. 2 nmed fter the brillint pplied mthemticin nd Nottinghm Miller George Green ( ) who developed them s tool in the 183s

7 Willim J. Prnell: MT3432. Section 1: Motivtion 7 where f is some forcing function, on domin x [, L] with homogeneous boundry conditions e.g. u() =, du/dx(l) =. This corresponds to the stedy stte het eqution in one dimension with het source term f(x) nd with fixed temperture t x = nd n insulted boundry t x = L (no het flux cross the boundry). This problem is of course boundry vlue problem, i.e. n ODE governing some function u (the temperture) with corresponding boundry conditions t the edge of the domin. It trnspires tht solution of the problem cn be written in the form u(x) = L G(x, x )f(x ) dx (1.3) where G(x, x ) is the corresponding Green s function which stisfies n ssocited boundry vlue problem. We will not describe this here but will of course in detil in lter chpters. Note tht (1.3) is strictly n integrl eqution, lthough it does not hve to be solved so it cn be sid to be n integrl expression for the function u(x). In two nd three dimensions, the corresponding solution cn be written 3 u(x) = G(x,x )f(x ) dx (1.4) V where D is the two/three dimensionl domin nd G is the corresponding Green s function. We will describe the theory behind the bove nlysis nd describe in prticulr some pplictions in the context of het conduction nd wve propgtion. In prticulr for problems involving inhomogeneous medi (think of solid body with n inclusion embedded inside it) we re ble to write down integrl equtions which govern the sclr field u(x). We shll describe methods to solve these interesting problems. Indeed in lter chpters we will mke links to some modern reserch topics. These include coustic scttering theory i.e. how sound wves re scttered from obstcles, coustic cloking theory i.e. how we cn try to mke objects invisible to sound nd the study of composite mterils, lthough we probbly will not hve the time to consider ll of these pplictions. I will of course mke it cler wht is nd is not exminble. Here re some brief detils of the ppliction res described bove. Acoustic scttering theory Suppose tht we hve uniform medium nd within this domin we embed n inclusion, it could hve rbitrry shpe. Imgine tht sound (coustic) wves re incident on the inclusion. This cuses the wves to be scttered. How do we solve for this scttered field? One exmple is shown in figure 1. We shll describe how we do this for simple geometries in section 3 vi Green s functions. In section 5 we describe more generl cse nd describe how the problem cn be reformulted in terms of integrl equtions. We describe technique tht cn be implemented in order to predict the scttered field. 3 In hrder problems this not the cse - we will consider some of these in sections 4 nd (??) when we discuss integrl equtions.

8 Willim J. Prnell: MT3432. Section 1: Motivtion 8 Figure 1: An coustic (sound) field is generted by forcing t the point in the white circle. Outgoing circulr wves re generted. These outgoing wves re subsequently scttered by the circulr blck region. Becuse in this instnce the wvelength is commensurte with the size of the circulr region, scttering is strong: we see cler shdow region nd bckscttered field. The field is time hrmonic so tht we re showing the mplitude of the wve field t single instnt in time. Acoustic cloking theory Suppose tht we did not wnt the field to be scttered from the circulr region bove. How could we enble this to hppen? The development of the two nd three dimensionl Green s function enbles us to esily describe the concept of coustic cloking. This is topic of gret interest presently. The ide is to design n coustic mteril which possesses properties in order to guide the coustic wves round region of interest. See figure 2. This is of interest in number of pplictions minly due to the fct tht outside the clok region, one cnnot tell t ll tht there is circulr region or nything inside it. We will describe how this concept of cloking cn be chieved theoreticlly in section 3. Composite mterils Suppose tht we hve mteril which consists of lots of smll inclusions embedded inside n otherwise uniform host medium (see figure 3). This type of so-clled composite mteril is used in thousnds of pplictions in engineering, medicl science, the utomotive nd defence industries nd erospce sector mongst mny others. If the inclusions nd host medium hve different therml conductivities, how do we theoreticlly predict

9 Willim J. Prnell: MT3432. Section 1: Motivtion 9 Figure 2: A mteril with specil mteril properties is wrpped round the blck circulr region. These properties guide the incoming coustic (sound) wve, generted t the point just to the right of the imge, round the region. The region is therefore cloked nd nything inside will not be seen in the fr-field. The field is time hrmonic so tht we re showing the mplitude of the wve field t single instnt in time. wht the so-clled overll (or effective) therml conductivity is nd how it depends on the volume frction (reltive quntities of the different constituents), conductivities nd shpe of the constituents of the mteril in question? In section 5 we will use integrl equtions in order to motivte one pproch to solving this problem. It trnspires tht we cn introduce smll mount of the inclusion mteril in order to significntly influence (nd improve) the overll (or effective) therml conductivity of the mteril. This cn ssist in decresing the cost, improving the effectiveness, etc. of the mteril. Interesting mthemtics underlies these pplictions! The three pplictions bove will be considered in this course but note tht bove ll we will be interested in the interesting mthemtics tht sits underneth nd describes these importnt phenomen. Understnding the mthemtics is key to getting sensible predictions in these ppliction res. These reserch res re of gret current interest nd mny scientists re currently undertking relted mthemticl reserch with ssocited pplictions in physics, mterils science, chemistry, medicl imging nd dignostics, medicl implnts, non destructive evlution of components in industry nd mny more.

10 Willim J. Prnell: MT3432. Section 1: Motivtion 1 Figure 3: We show composite mteril which consists of mny smll inclusions distributed throughout uniform host mteril. The question is how do we predict the overll mteril properties from knowledge of the constituent mterils?

11 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 11 2 Green s functions in 1D We now come on to the introduction of the concept of Green s function nd we shll strt in one dimension, i.e. with ordinry differentil equtions (ODEs). We will usully be interested in solutions of second order (highest derivtive is two) ODEs. This includes mny problems tht re of interest in prctice, for exmple the (stedy stte) het eqution nd the wve eqution t fixed frequency. 2.1 Ordinry Differentil Equtions: review You hve seen the mteril here before (MT1121). We will review it briefly but look bck t your notes to ensure tht you know it thoroughly! Let us consider second order Ordinry Differentil Equtions (ODEs) of the form p(x)u (x) + r(x)u (x) + q(x)u(x) = f(x) (2.1) where p(x), r(x), q(x) nd f(x) re rel functions. Two type of problems cn be considered: Boundry Vlue Problems (BVPs) nd Initil Vlue Problems. For BVPs, x is sptil vrible e.g. x [, b] nd we require ssocited boundry conditions (BCs) e.g. B = {u() =, u(b) = 1}, etc. For IVPs, x is time so x [, ) nd we require ssocited initil conditions (ICs) e.g. I = {u() =, u () = 1}. If the BCs or ICs hve zero right hnd side they re known s homogeneous. Otherwise they re known s inhomogeneous. We will consider exclusively BVPs in this section. We will consider IVPs in section 4 (integrl equtions in 1D). Note tht often we cn divide through by p(x) in order to give unit coefficient of u (x). However in generl we hve to be creful with this. Some singulr problems (tht re physicl) do not llow us to do this. The generl solution of the ODE is in generl written in the form u(x) = u c (x) + u p (x) (2.2) where u c (x) is known s the complementry function nd is the solution to the homogeneous ODE p(x)u c(x) + r(x)u c(x) + q(x)u c (x) = (2.3) wheres u p (x) is known s the prticulr solution nd is the solution to the inhomogeneous ODE p(x)u p(x) + r(x)u p(x) + q(x)u p (x) = f(x). (2.4) Once we hve determined (2.2) it will hve some undetermined constnts (these re lwys in the complementry function) which re then determined by imposing the BCs or ICs on the generl solution. How do we determine the complementry function nd prticulr solution? Let us discuss this now. We note tht in prticulr we re interested in two types of ODEs: Constnt coefficient ODEs nd those of Euler type since these my be solved nlyticlly. ODEs tht cnnot be solved nlyticlly cn of course be treted by numericl methods but this is outside the scope of this course.

12 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Homogeneous ODEs: The complementry function For constnt coefficient ODES, with r, q C we cn write u c(x) + ru c(x) + qu c (x) =. (2.5) Here we relly cn tke the coefficient of u c (x) to be unity since we cn divide through by the constnt p. We know tht since the ODE is second order there will be two fundmentl solutions sy u 1 (x) nd u 2 (x) tht contribute to the complementry function nd it cn be written s u c (x) = c 1 u 1 (x) + c 2 u 2 (x) for some rel constnts c 1, c 2 R. To find u 1 nd u 2, seek solutions of the form exp(mx) where m R nd find the λ tht ensure solutions from m 2 + rm + q =. There will either be two rel, two complex conjugte or repeted roots. In the cse of the ltter one of these solutions must be multiplied by x in order to obtin the second linerly independent solution (see question 4 of Exmple sheet 1). Exmple 2.1 Find the solution of u c(x) + u c(x) 2u c (x) =. (2.6) Seeking solutions in the form exp(mx) gives m 2 + m 2 = (m + 2)(m 1) = so tht m = 2, 1. The solution is therefore for some constnts c 1, c 2. Exmple 2.2 Find the solution of u c (x) = c 1 exp( 2x) + c 2 exp(x) (2.7) u c(x) + 2u c(x) + u c (x) =. (2.8) Seeking solutions in the form exp(mx) gives m 2 + 2m + 1 = (m + 1) 2 = so tht λ = 1 (repeted). The solution is therefore for some constnts c 1, c 2. Euler equtions re of the form u c (x) = c 1 exp( x) + c 2 x exp( x) (2.9) x 2 u c (x) + rxu c (x) + qu c(x) = (2.1) for some r, q R nd x. Solutions re then sought in the form x m. Exmple 2.3 Find the solution of the Euler ODE x 2 u c (x) + 2xu c (x) 6u c(x) =. (2.11) Seeking solutions in the form x m gives m(m 1)+2m 6 = m 2 +m 6 = (m+3)(m 2) so tht m = 2 nd m = 3. The solution is therefore for some constnts c 1, c 2. u c (x) = c 1 x 2 + c 2 x 3 (2.12)

13 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Inhomogeneous ODEs Let us now consider how we find the prticulr solution u p (x). We cn obtin this by two lterntive techniques: the method of undetermined coefficients nd the method of vrition of prmeters. Inhomogeneous ODEs: Method of undetermined coefficients Consider gin the generl second-order ODE of the form p(x)u (x) + r(x)u (x) + q(x)u(x) = f(x). (2.13) We must seek prticulr solutions u p (x) in order to tke cre of the inhomogeneous term f(x) on the right hnd side. A simple method is known s the method of undetermined coefficients. This is sometimes lso clled the method of intelligent guessing! Exmple 2.4 Find the prticulr solution for the ODE u (x) + u (x) 2u(x) = 1 exp(3x) (2.14) We note tht exp(3x) is not one of the fundmentl solutions (you cn check this). Therefore pose prticulr solution in the form u p (x) = exp(3x) for some R to be determined. Substituting this into the ODE we find tht (9 exp(3x) + 3 exp(3x) 2 exp(3x)) = 1 exp(3x) (2.15) nd so for consistency we note tht we require = 1. If the right hnd side of the ODE is one of the fundmentl solutions we multiply our choice by x (note the specil cse of n Euler ODE with fundmentl solution 1/x with forcing term 1/x would hve u p (x) = (/x) ln x). Clerly this method cn sometimes be difficult to pply becuse we re using our judgement s to wht we should choose s cndidte solution. It would be preferble if we could derive more lgorithmic pproch. Inhomogeneous ODEs: Method of vrition of prmeters We cnnot lwys use the method of undetermined coefficients. Sometimes we just cnnot see the prticulr solution. Consider gin the generl second-order ODE of the form p(x)u (x) + r(x)u (x) + q(x)u(x) = f(x). (2.16) We will now briefly describe the method of vrition of prmeters. In order to pply this method we need to know the complementry function. This is impertive (remember tht this ws not the cse with the method of undetermined coefficients). We know from section tht the complementry function hs the form u c (x) = c 1 u 1 (x) + c 2 u 2 (x). (2.17)

14 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 14 We will pose prticulr solution of the form u p (x) = v 1 (x)u 1 (x) + v 2 (x)u 2 (x) (2.18) nd so we need to determine the two unknown functions v 1 (x) nd v 2 (x). Let us differentite u p (x): nd mke the ssumption tht Differentite u p (x) gin u p(x) = v 1(x)u 1 (x) + v 1 (x)u 1(x) + v 2(x)u 2 (x) + v 2 (x)u 2(x) (2.19) v 1(x)u 1 (x) + v 2(x)u 2 (x) =. (2.2) u p(x) = v 1(x)u 1(x) + v 2(x)u 2(x) + v 1 (x)u 1(x) + v 2 (x)u 2(x). (2.21) Substituting u p (x) nd its derivtives into the governing ODE nd rerrnging we find p(x)[v 1(x)u 1(x) + v 2(x)u 2(x)] + v 1 (x)[p(x)u 1(x) + r(x)u 1(x) + q(x)u 1 (x)] + v 2 (x)[p(x)u 2 (x) + r(x)u 2 (x) + q(x)u 2(x)] = f(x). (2.22) Of course in the second nd third terms on the left hnd side, the terms in squre brckets re zero. Therefore p(x)(v 1(x)u 1(x) + v 2(x)u 2(x)) = f(x). (2.23) This together with the ssumption (2.2) gives us two equtions to solve for v 1 (x) nd v 2(x). We solve to find v 1 (x) = u 2 (x)f(x) p(x)(u 1 (x)u 2 (x) u 2(x)u 1 (x)), v 2 (x) = u 1 (x)f(x) p(x)(u 1 (x)u 2 (x) u (2.24) 2(x)u 1 (x)). We note tht since u 1 (x) nd u 2 (x) re fundmentl solutions the Wronskin is non-zero: W(x) = u 1 (x)u 2 (x) u 2(x)u 1 (x). (2.25) So, we cn integrte in ech of (2.24) between nd x to find v 1 (x) = x u 2(x )f(x ) p(x )W(x ) dx + v 1 (), v 2 (x) = x u 1 (x )f(x ) p(x )W(x ) dx + v 2 (). (2.26) We cn set v 1 () = v 2 () = becuse from (2.18) these merely generte dditionl terms tht re of the form of the complementry function. Therefore v 1 (x) = x u 2(x )f(x ) p(x )W(x ) dx, v 2 (x) = x Therefore we cn ssert tht the generl solution to the ODE is u 1 (x )f(x ) p(x )W(x ) dx. (2.27) u(x) = u c (x) + u p (x) (2.28) = (c 1 + v 1 (x))u 1 (x) + (c 2 + v 2 (x))u 2 (x) (2.29)

15 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Generl forcing nd the influence (Green s) function In order to give full description of Green s functions, wht they re nd why they re useful we need lot more ODE theory some (most?) of which you will not hve come cross before. We will come on to this in moment but let us consider simple problem here first in order to motivte the ide of Green s function. In prticulr we should sk if we cn obtin solution form for n ODE with n rbitrry forcing term f(x) on the right hnd side? In order to nswer this question let us consider cnonicl problem nd one tht hs very importnt ppliction. Consider the simple eqution d 2 u/dx 2 = u (x) = f(x) (2.3) on the domin x [, L] subject to homogeneous boundry conditions B = {u() =, u(l) = }. This problem is in fct the stedy stte het eqution. I.e. the het eqution without ny time dependence 4. Temperture is fixed to be zero on the boundries. In order to solve this problem, we note tht the complementry function stisfies u c (x) = (2.31) nd by direct integrtion, the fundmentl solutions re 1 nd x. However it turns out to be very convenient to hve fundmentl solutions one of which stisfies one of the homogeneous boundry conditions nd one of which stisfies the other. Therefore we choose liner combintions, to obtin u 1 (x) = x, u 2 (x) = L x (2.32) stisfying the left nd right boundry condition respectively. Using (2.27), since W = u 1 u 2 u 2 u 1 = x( 1) (L x)(1) = L, we find tht The full solution is therefore v 1 (x) = 1 L v 2 (x) = 1 L x x f(x )(L x ) dx (2.33) f(x )x dx (2.34) u(x) = (c 1 + v 1 (x))x + (c 2 + v 2 (x))(l x) (2.35) so finlly let us pply the BCs. Setting x = mens tht c 2 = nd for x = L we find so tht c 1 = v 1 (L). We then note tht = (c 1 + v 1 (L))L (2.36) c 1 + v 1 (x) = v 1 (L) + v 1 (x) (2.37) = 1 L = 1 L L L x f(x )(L x ) dx + 1 L x f(x )(L x ) dx (2.38) f(x )(L x ) dx. (2.39) 4 In relity ll problems hve to hve some time dependence of course. Wht usully hppens is tht fter some initil trnsients hve decyed we re left with stedy stte solution which my or my not be the trivil one u =.

16 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 16 We cn therefore write u(x) = x L L x (x L)f(x ) dx + (x L) L Finlly this mens we cn write the solution in the form where u(x) = L x x f(x ) dx G(x, x )f(x ) dx (2.4) x G(x, x ) = L (x L), x x, x L (x L), x x L. (2.41) The function G(x, x ) cn be thought of s n influence function. It is in fct the Green s function for this problem nd we will sy more bout this lter on. Note tht G(x, x ) = G(x, x) = G(x, x) here, i.e. it is symmetric (the overline or br denotes the complex conjugte, recll z = + ib, z = ib). The Green s function does not lwys possess this full symmetry; it only occurs for specil types of boundry vlue problems. In prticulr G(x, x ) = G(x, x) lwys occurs for specil clss of problems clled self-djoint opertor problems (which we will consider shortly). Note tht we my write (2.41) in the form G(x, x ) = x L (x L)H(x x) + x L (x L)H(x x ) (2.42) which lso illustrtes the symmetry, where { 1, x >, H(x) =, x < is the so-clled Heviside step function. When determining Green s function lter, I would lwys encourge you to write them in this form. It helps gret del, especilly when integrting them! Finlly we note tht by directly integrting twice we could in fct obtin the solution in the form (see question 5 on Exmple Sheet 1) u(x) = x x f(x 1 ) dx 1 dx + c 1 x + c 2. (2.43) You re sked to show tht this is equivlent to (2.4) in question 5 on Exmple Sheet 1.

17 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Liner differentil opertors It turns out to be very useful to define the nottion L to men liner opertor, which mens tht L(c 1 u 1 + c 2 u 2 ) = c 1 Lu 1 + c 2 Lu 2. for (possibly complex) constnts c j. In this chpter it will be ssocited with second order ordinry differentil opertor, e.g. L = d 2 /dx 2. In the next chpter it will be ssocited with prtil differentition. Remember tht in generl n opertor will tke function nd turn it into nother function. The functions in generl will belong to some function spce which possess some specific properties, i.e. L 2 [, b] which mens tht they re squre integrble on [, b], (i.e. f L 2 [, b] mens f(x) 2 dx < ) etc. We re interested in the liner BVP Lu = p(x) d2 u dx + r(x)du + q(x)u = f(x) (2.44) 2 dx where for now we do not mke ny restrictions on the functions p(x), r(x), q(x) nd f(x) but they cn be complex functions nd we usully consider them s continuous. The (rel) domin on which the ODE holds is x [, b] nd it is of course subject to BCs on x =, b which we shll denote s B. We will restrict ttention to homogeneous BCs nd for now these could be of ny form, e.g. B = {u() =, u(b) = }, Dirichlet (2.45) B = {u () =, u(b) = }, Dirichlet-Neumnn, (2.46) B = {u () =, u (b) = }, Neumnn, (2.47) B = {u() + hu () =, u(b) = }, Robin-Dirichlet, (2.48) B = {u() = u(b), u () = u (b)}, Periodic, (2.49) B = {u() + hu (b) =, u(b) = }, Mixed-Dirichlet. (2.5) Extension to the cse of inhomogeneous BCs is not too difficult - we shll discuss this in section The BVP therefore consists of the eqution Lu = f(x) nd the BCs B Inner products The function spces to which the functions tht we re interested in belong, re endowed with n inner product. This mens tht they re inner product spces. This bsiclly mens tht they possess nice properties such s Cuchy-Schwrz nd the tringle inequlity. We do not worry too much bout this here, usully ssuming tht the functions we re interested in re in L 2 [, b]. The notion nd nottion of n inner product is useful. We define the usul inner product s f, g = f(x)g(x) dx (2.51) where we note tht f(x) denotes the complex conjugte of the function f, i.e. we hve defined this inner product over the set of complex vlued functions (this includes the set of rel functions of course).

18 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 18 We hve the importnt properties of inner product spces tht f, g = g, f, (2.52) f, αg 1 + βg 2 = α f, g 1 + β f, g 2, (2.53) f, f with equlity if nd only if f =. (2.54) αg 1 + βg 2, f = α g 1, f + β g 2, f. (2.55) The djoint opertor It is useful to define so-clled djoint BVP ssocited with the originl BVP bove. This djoint problem consists of n djoint opertor L nd ssocited djoint BCs B. These re defined by v, Lw = L v, w noting tht this prescribes both n opertor nd BCs nd in generl L L nd B B. Exmple 2.5 Assuming tht u, v L 2 [, b] (i.e. they re squre integrble on [, b]), find the djoint opertor nd BCs for the following problems (i) (ii) (iii) (iv) (v) (vi) L = d2 dx2, B = {u() =, u(1) = }, (2.56) L = d2 dx + d + 1, 2 dx B = {u() =, u(1) = }, (2.57) L = d2 dx 2 + 1, B = {u() =, u(1) = u ()}, (2.58) L = d2 dx 2 + 1, B = {u() = u(1), u () = u (1)}, (2.59) L = d2 dx + i d + 1, 2 dx B = {u() =, u(1) = }, (2.6) L = x 2 d2 dx + x d 2 dx + 1, B = {u(1) =, u (2) = }, (2.61) (vii) L = d2 dx + 2 k2, B = {u (x) ± iku(x) s x ± }, (2.62) The trick is to use integrtion by prts to interchnge the order of integrtion onto the other function. (i) Let us follow through the rgument, using integrtion by prts: 1 v, Lu = v d2 u dx dx 2 [ = v du ] 1 1 du dv dx dx dx dx [ = v du ] [ 1 [ u dv ] 1 dx dx [ = v du ] 1 1 dx udv + dx [ = v du ] 1 dx 1 d 2 v dx2u dx ] u d2 v dx dx 2 + L u, v (2.63)

19 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 19 where L = d 2 /dx 2 nd in the lst step we hve imposed the BCs on u. In order to ensure tht the term in brckets is zero we must choose B = {v() = v(1) = } but this is equivlent to hving v() = v(1) = (If v is complex function then it being zero mens both its rel nd imginry prts must be zero nd hence these conditions re equivlent). We see tht L = L nd the djoint BCs re the sme s the originl BCs, i.e. B = B. (ii) The first term of the opertor is identicl with tht in (i) so we cn use tht result. 1 ( d 2 u v, Lu = v dx + du ) 2 dx + u dx [ = v du ] 1 1 dx udv d 2 v 1 + dx dx 2u dx + [uv]1 u dv 1 dx dx + vu dx [ = v du ] 1 1 ( dx udv dx + uv d2 v + dx dv ) 2 dx + v u dx [ = v du ] 1 + L v, u (2.64) dx nd we note here tht L L due to the first derivtive term. The djoint BCs re unchnged however, B = B. (iii) Using (i) bove it is esily shown tht 1 ( ) d 2 u v, Lu = v dx + u dx 2 [ = v du ] 1 1 ( ) dx udv d2 v + dx dx + v u dx 2 [ = v du ] 1 dx udv + L v, u. (2.65) dx so tht L = L. Let us now determine the djoint BCs, B. We need [ v du ] 1 dx udv = (2.66) dx nd using the BCs u() = nd u(1) = u () we see tht [ v du ] 1 dx udv = (v(1)u (1) u(1)v dx (1)) (v()u () u()v ()), = v(1)u (1) u(1)v (1) v()u (), which implies tht we require the djoint BCs to be = v(1)u (1) (v (1) + v())u () (2.67) v(1) =, v (1) = v(). (2.68) Note in prticulr tht in this exmple, lthough L = L the djoint BCs re different from the originl BCs, B B. (iv)-(vii) See question 5 on Exmple Sheet 2.

20 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 2 In question 6 of Exmple Sheet 2 you re sked to show tht the djoint opertor ssocited with the generl ODE (2.44) is ( L = p(x) d2 dx + 2 dp ) ( d 2 dx r dx + d2 p dx dr ) 2 dx + q. (2.69) Lgrnge s 5 identity Lgrnge derived very useful identity. This is: [ p vlu L vu = d dx ( du dx v udv dx ) + ( r dp ) ] uv dx (2.7) You re sked to prove this in question 7 on Exmple Sheet 2. Green s 6 second identity We cn integrte both sides of Lgrnge s identity (2.7) between x = nd x = b to get vlu L vu dx = [ ( p v du ) ( dx udv + r dp ) ] b uv. (2.71) dx dx Note tht this generl identity is very useful in order to determine the djoint BCs B required bove. With inner product nottion we note tht we cn write (2.71) s v, Lu L v, u = [ ( p v du ) ( dx udv + r dp ) ] b uv. (2.72) dx dx I hd sentence here which referred to rel function spces ; plese delete nd ignore - it ws very confusing nd did not dd nything! Apologies Self-djoint opertors Self-djoint (S-A) opertors re specil opertors with the property tht the djoint problem is identicl to the originl problem, i.e. both the djoint opertor nd the djoint BCs re the sme s the originl physicl BVP. I.e. L = L nd B = B. E.g. Exmple 2.5(i) bove. It is sometimes the cse tht the differentil opertor is the sme, i.e. L = L but the boundry conditions re not, e.g. Exmple 2.5(iii) bove. In this cse the opertor is sid to be formlly self-djoint. 5 Joseph-Louis Lgrnge ( ) ws brillint Itlin-born French mthemticin nd stronomer. He mde significnt contributions in mny brnches of science, in prticulr to nlysis, number theory, nd clssicl nd celestil mechnics. Note tht Frnce hs n incredible history in mthemtics nd engineering - if you re ever in Pris, go to the Eiffel Tower nd look t the nmes engrved on ech side of the lower prt of the tower. You cn lso see this on wiki: of the 72 nmes on the Eiffel Tower nd note tht Lgrnge is present! 6 We hve lredy mentioned Green - he ws the Nottinghm miller!

21 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 21 Exmple 2.6 Referring to Exmple 2.5 bove, determine which of (i)-(vi) re self djoint. (i) L = L nd B = B. So self-djoint. (ii) L L so not self-djoint. (iii) Although L = L, B B so not self-djoint. (This is clled only formlly selfdjoint) (iv)-(vii) See Question 5 on Exmple Sheet 2. In generl, mixed BCs do not led to self-djoint opertors, lthough if p(x) =constnt, then periodic BCs (which re mixed) do yield self-djoint opertors. For complex liner opertors (i.e. where p, r nd q re complex functions, the conditions for self-djointness re complicted. They re in fct tht p(x) hs to be rel function with p = Re(r) nd 2Im(q) = (Im(r)) (here Re nd Im denote the rel nd imginry prts of the function respectively. See Question 8 on Exmple Sheet 2. For simplicity let us restrict ttention from now on to rel opertors, so tht p, q nd r re rel functions. Of course the functions u nd v could still be complex. We see then from the form of the generl djoint opertor in (2.69) tht necessry condition for second order differentil opertor to be formlly self-djoint (i.e. L = L) is tht r(x) = p (x). The opertor cn then be written s Lu = d dx ( p(x) du dx In this cse Green s second identity simplifies to v, Lu Lv, u = ) + q(x)u(x). (2.73) [ ] b p(x)(v(x)u (x) v (x)u(x)) (2.74) nd in order to derive the djoint BCs B stisfied by v we choose them such tht the right hnd side of (2.74) is zero. For given B stisfied by u this defines the conditions B stisfied by v. This lso shows tht even if L = L, we my not hve B = B. We therefore reiterte here tht the property of self-djointness requires properties of BCs, not just the opertor itself. In prticulr it could be tht the opertor is formlly self-djoint so tht L = L but the required djoint BCs in order to ensure tht (2.74) is stisfied re not the sme s the originl BCs Forcing forml self-djointness In fct we cn use wht we know bout first order ODEs in order to write ll second order ODEs in forml self-djoint form s we show in section However, even though we cn do this, we note tht the BCs my not led to fully self-djoint opertor. Let us now consider very specil type of BVP, the so-clled Sturm-Liouville problems.

22 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Sturm-Liouville (S-L) eigenvlue problems The problems tht we will be concerned with in this section re the so-clled Sturm- Liouville 7 ODE BVPs which tke the form of n opertor in S-A form, i.e. Lu = d ( p(x) du ) + q(x)u(x) (2.75) dx dx with x [, b]: this could lso be the whole rel line or the semi-infinite domin, e.g. x [, ). The functions p, q nd µ re rel nd continuous. In generl p is nonnegtive (nd usully positive lmost everywhere) nd µ is positive. We will ssocite some homogeneous BCs with this ODE shortly. If the opertor if NOT in the form (2.75), the problem is NOT S-L problem. Nturlly rising problems in the physicl sciences often led to the eqution Lφ(x) + λµ(x)φ(x) = (2.76) where µ(x) rises vi the physics in the derivtion of the governing equtions. This is ccompnied by boundry conditions. Solutions to this problem exist only for prticulr vlues of λ sy λ k (the eigenvlues), for k = 1, 2, 3,..., with ssocited solution φ k (x) (the eigenfunctions). The eigenvlues nd eigenfunctions re usully of gret physicl interest nd significnce. Regulr S-L problem The regulr Sturm-Liouville eigenvlue problem is defined by the ODE Lφ(x) + λµ(x)φ(x) = (2.77) with L s defined in (2.75) nd homogeneous boundry conditions of the form dφ B = {α 1 φ() + α 2 dx () =, β dφ 1φ(b) + β 2 (b) = }, (2.78) dx where α n, β n re rel, x [, b] ( finite intervl), the functions p(x), q(x) nd µ(x) re rel nd continuous, p (x) exists nd is continuous, nd p(x), µ(x) re positive. We note tht the BCs here re not mixed. This is importnt s we shll see lter. Also, note tht the fct tht α n nd β n re rel ensures the self-djointness (i.e. full S-A not just forml) of the problem: Regulr S-L problems re fully self-djoint! (but note the mny conditions required for regulrity!) Singulr S-L problem We sometimes wnt to relx the conditions bove since physicl problems re often not quite s constrined. We will not be too prescriptive here bout the type of non-regulr 7 nmed fter the French mthemticins Jcques Chrles Frncois Sturm ( ) nd Joseph Liouville ( ) who studied these in the erly 19th century. This work ws very influentil for the theory of ODEs.

23 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 23 S-L problem we consider but will occsionlly refer to them s we proceed. Wht often hppens in singulr S-L problems is tht e.g. p(x) vnishes t one of the end points of the intervl [, b] or e.g. the boundry conditions re not quite of the form in (2.78), e.g. periodic conditions with p =constnt Theorems ssocited with Regulr S-L problems for ODEs For regulr S-L ODE problem we hve the following importnt theorems: 1. All eigenvlues λ re rel 2. There re n infinite number of eigenvlues λ 1 < λ 2 <... < λ n < λ n+1 <... (2.79) There is smllest eigenvlue λ 1 but no lrgest eigenvlue: λ n s n. 3. Corresponding to ech eigenvlue λ n there is n eigenfunction sy φ n (x) which is unique to within n rbitrry multiplictive constnt. φ n (x) hs n 1 zeros for x (, b). 4. The eigenfunctions form complete set. This mens tht ny piecewise smooth function g(x) cn be represented in the form g(x) = n φ n (x) (2.8) n=1 Importntly this series is convergent, converging to (g(x+)+g(x ))/2 where x+ nd x denote pproching x from bove nd below respectively. Thus for continuous functions this series converges to g(x). 5. Eigenfunctions ssocited with different eigenvlues re orthogonl reltive to the weight function µ(x). I.e. if λ m λ n (m n) µ(x)φ m (x)φ n (x) = (2.81) If the S-L problem is singulr, these theorems my still hold, but not necessrily. Since this is course on Green s functions rther thn ODEs, we do not go into the detils of these theorems too much. Although let us discuss simple exmple to illustrte their usefulness in simple importnt cse.

24 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D A model exmple to illustrte the theorems Exmple 2.7 We set p = 1, q = in (2.75) nd thus consider the ssocited eigenvlue problem for the Lplcin opertor in one dimension, with the weighting µ(x) = 1. These eigenfunctions re therefore pproprite for the het eqution nd wve problems in one spce dimension s you will hve seen in MT241. The eigenfunction eqution is φ (x) + λφ(x) = (2.82) for x [, L]. Let us consider the cse when B = {φ() =, φ(l) = }. This is therefore regulr S-L problem. The solutions of this problem tke the form (see question 2 on Exmple Sheet 3) φ n (x) = sin ( nπx ) ( nπ ) 2, λ n = (2.83) L L with n = 1, 2,... nd therefore the solution is of the form of Fourier sine series: u(x) = n φ n (x) (2.84) n=1 for some rel coefficients n (i.e. u(x) is rel function). Rel eigenvlues In determining this result you usully ssume rel eigenvlues. Seeking complex ones cn be hrd! This theorem tells us tht once we hve found ll of the rel eigenvlues we cn stop s there re no complex ones! Eigenvlue ordering We see tht indeed we hve n infinite number of eigenvlues λ n = (nπ/l) 2 nd tht indeed we hve smllest: (π/l) 2, but no lrgest. Zeros of eigenfunctions Eigenfunctions φ n (x) = sin ( ) nπx L should hve n 1 zeros inside (, b). This is clerly true. Eigenfunction convergence The eigenfunction expnsion (2.84) is Fourier Sine series nd we know (from MT241) vi Fourier s convergence theorem tht ny piecewise smooth function cn be represented s so. Remember tht this helped in MT241 s we could use seprtion of vribles successfully in mny cses.

25 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 25 Eigenfunction orthogonlity The weight function µ(x) here is simply unity. We cn use the inner product nottion nd we know tht if m n φ n, φ m = L sin(nπx/l) sin(mπx/l) =. Orthogonlity of the eigenfunctions enbles the coefficients n to be determined in strightforwrd mnner s n == u(x), φ L n φ m, φ m = u(x)φ n dx. (x) dx Proofs of S-L Theorems 1. nd 5. L φ2 m Some of the Theorems 1-5 bove relting to S-L problems re difficult to prove. Two of them re reltively simple however: Theorem 1 pertining to rel eigenvlues nd Theorem 5 pertining to orthogonl eigenfunctions. For resons tht will become cler shortly, we will prove Theorem 5 first. Theorem 5 - A modified inner product nd orthogonl eigenfunctions Tke two eigenfunctions, φ nd ψ (ssocited with regulr S-L opertor L) corresponding to distinct eigenvlues λ nd ν sy, so tht Lφ = λµ(x)φ(x), Lψ = νµ(x)ψ(x). (2.85) Since the opertor is regulr S-L, it is S-A so tht = Lφ, ψ φ, Lψ, (2.86) = λ µφ, ψ + ν φ, µψ, (2.87) = λ µφ, ψ + ν φ, µψ, (2.88) = (ν λ) µ(x)φ(x)ψ(x) dx (2.89) nd therefore since the eigenvlues re distinct, using stndrd inner product nottion µ(x)φ(x)ψ(x) dx = φ, µψ =. I.e. the weighted eigenfunctions re orthogonl with respect to the usul inner product defined in (2.51). Given the bove however, it is convenient to define modified inner product f, g = µ(x)f(x)g(x) dx. (2.9) Then the eigenfunctions themselves re orthogonl with respect to this newly defined inner product. Unless otherwise stted, we ssume tht the weighting µ(x) = 1.

26 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 26 Theorem 1 - Rel eigenvlues Tke the eigenvlue λ corresponding to the eigenfunction φ(x) ssocited with regulr S-L opertor L. We hve, working with the modified inner product (2.9) bove, Also we hve Therefore, since problem is S-A, Lφ, φ = λφ, φ, (2.91) = λ φ, φ. (2.92) φ, Lφ = φ, λφ, (2.93) = λ φ, φ. (2.94) = Lφ, φ φ, Lφ, (2.95) = (λ λ) φ, φ (2.96) so tht λ = λ nd therefore the eigenvlues must be rel. It trnspires tht this result holds for regulr S-L problems, singulr S-L problems in the sense tht p(x) = t n end point, nd lso if the BCs re periodic.

27 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Existence nd uniqueness of BVPs for ODEs: The Fredholm Alterntive Recll the following theorem for Initil Vlue Problems ssocited with ODEs: Theorem 2.1 Given the ODE u (t) + p(t)u (t) + q(t)u(t) = f(t) subject to ICs u(t ) = x, u (t ) = v, if p(t), q(t) nd f(t) re continuous on the intervl [, b] contining t, the solution of the IVP exists nd is unique. Unfortuntely the sitution is not s simple for BVPs. It cn be the cse tht BVPs hve (i) no solution, (ii) unique solution or (iii) infinitely mny solutions! Let us first stte the following theorem which gurntees the existence of two fundmentl solutions to homogeneous ODE: Theorem 2.2 Given the homogeneous ODE p(x)u (x) + r(x)u (x) + q(x)u(x) =, with p, r nd q continuous nd p never zero on the domin of interest, there lwys exist two fundmentl solutions u 1 (x) nd u 2 (x) which generte the generl solution u(x) = c 1 u 1 (x) + c 2 u 2 (x). Therefore whether solution exists or not depends on the BCs. As very simple exmple to illustrte tht BVPs cn hve unique solution, no solution or infinitely mny solutions, let us consider the following problem. Exmple 2.8 Consider the homogeneous ODE subject to inhomogeneous BCs u (x) + u(x) = (i) u() = 1, u(π) = 1, (ii) u() = 1, u(π/2) = 1, (iii) u() = 1, u(2π) = 1. The fundmentl solutions re cosx nd sin x so tht u(x) = c 1 cosx + c 2 sin x. The BCs in (i) re inconsistent nd therefore there is no solution. The BCs in (ii) yield the unique solution u(x) = cosx + sin x. The BCs in (iii) yield the infinite fmily of solutions u(x) = cosx + c 2 sin x where c 2 is rbitrry.

28 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 28 Let us now consider the cse of n inhomogeneous ODE subject to homogeneous BCs (recll tht this is the min thrust of our enquiries in this course). We re ble to stte rther generl theorem regrding existence nd uniqueness of solutions to this problem. We consider the dditionl effect of inhomogeneous BCs in section We shll consider n exmple which illustrtes the min issues tht rise. Exmple 2.9 Find the solution to the ODE subject to u() = u(l) =. u (x) + u(x) = f(x) Fundmentl solutions of the homogeneous ODE re sin x nd cos x but remember tht the generl solution cn be ny liner combintion of these nd it is convenient to use u 1 (x) = sin x nd u 2 (x) = sin(x L) (since sin(x L) = sin x cosl cos L sin x). This is convenient since they stisfy the left nd right BCs respectively. Let us therefore write the solution to the homogeneous problem s u(x) = c 1 sin x + c 2 sin(x L). We then know from (2.29) tht the solution to the inhomogeneous problem cn be written where v 1 (x) = v 2 (x) = u(x) = (c 1 + v 1 (x)) sin x + (c 2 + v 2 (x)) sin(x L) (2.97) x u 2(x )f(x ) p(x )W(x ) dx = u 1 (x )f(x ) p(x )W(x ) dx = x x x sin(x L)f(x ) sin L sin x f(x ) sin L dx, (2.98) dx. (2.99) noting tht W(x) = sin x cos(x L) sin(x L) cos x = sin(x (x L)) = sin L is constnt. Now impose boundry conditions, with u() = giving whilst u(l) = gives c 1 sin L = L c 2 =, (2.1) sin(x L)f(x ) dx. (2.11) This lst eqution giving c 1 is perfectly vlid, unless L = nπ which knocks out the left hnd side! In tht cse there is then only solution if nπ nπ f(x ) sin(x nπ) dx = ( 1) n f(x ) sin x dx =. Tke, e.g. L = π. Even if this condition is stisfied then there re infinitely mny solutions becuse we cn dd on ny multiple of sin x to the solution, i.e. u(x) = u PS (x) + c sin x Often the only solution to the homogeneous BVP is the zero solution. When L = π bove we see tht sin x is non trivil solution to the homogeneous BVP. This corresponds to n existence of so-clled zero eigenvlue. Interestingly, this tells us something very specil bout the existence nd uniqueness of the solution to the inhomogeneous problem s we now describe vi generl theorem. We will return to the exmple bove fter we hve stted the theorem to see how it ligns with the theorem.

29 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 29 The Fredholm Alterntive for ODE BVPs We cn stte the following theorem Theorem 2.3 We introduce the BVP consisting of the liner ODE Lu = p(x)u (x) + r(x)u (x) + q(x)u(x) = f(x) subject to homogeneous BCs B with p(x), r(x), q(x) nd f(x) rel nd continuous on the intervl [, b], with p(x) on [, b]. Consider lso the ssocited homogeneous djoint problem L v = with ssocited homogeneous BCs B. Then EITHER 1. If the only solution to the homogeneous djoint problem is the trivil solution v(x) = then the solution to the inhomogeneous problem u(x) exists nd is unique OR 2. If there re non-trivil solutions to the homogeneous djoint problem v(x) then either There re infinitely mny solutions if v(x)f(x) =, or There is no solution if v(x)f(x). See question 1 of Exmple Sheet 4 for some more detils of this theorem. Clerly if the problem is self-djoint then L = L nd B = B nd so the djoint homogeneous problem is simply the homogeneous version of the originl BVP. Exmple 2.1 How is the Fredholm Alterntive Theorem consistent with exmple 2.9? Firstly the BVP is S-A nd so the djoint problem is merely the homogeneous version of the originl problem. It therefore hs solution v(x) = d 1 sin x + d 2 cos x. Imposing v() = yields d 2 = nd v(l) = gives d 1 sin L = (2.12) which mens tht if L nπ we need d 1 = nd therefore the only solution to this problem is the trivil one v(x) =. From the Fredholm Alterntive Theorem, the solution u(x) to the originl problem is unique. If L = nπ then (2.12) is trivilly stisfied for ny d 1. Therefore non-trivil solution to the homogeneous djoint problem is v(x) = sin x

30 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 3 which from the Fredholm Theorem mens tht if L sin x f(x ) dx = there re infinitely mny solution to the originl problem, wheres if L sin x f(x ) dx there re no solutions. This corresponds exctly to the Exmple bove. The existence of non-trivil solution to the homogeneous problem corresponds to the existence of zero eigenvlue. We will see lter in section 2.13 tht when this hppens, the stndrd Green s function (s we will define shortly) does not exist nd modified form hs to be considered. One finl point. This theorem llows us to sy gret del bout the existence nd uniqueness of solutions to inhomogeneous ODEs without ctully hving to solve the problems! We illustrte this with n exmple. Exmple 2.11 For the following ODE/BC pirings use the Fredholm Alterntive to stte if solution exists nd if so if it is unique (note tht you do not solve the inhomogeneous BVP in order to show this!). u (x) + ψu(x) = sin x with () ψ = 1, B = {u() =, u(π) = } (b) ψ = 1, B = {u () =, u (π) = } (c) ψ = 1, B = {u() =, u(π) = } (d) ψ = 2, B = {u() =, u(π) = } All problems re self-djoint. () A non-trivil solution to the homogeneous problem is v(x) = sin x. But we note tht π sin 2 x dx so therefore solution does not exist. (Verify this yourself by trying to solve the inhomogeneous problem). Prts (b)-(d) re considered in question 2 on Exmple Sheet 4.

31 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 31 The Fredholm Alterntive for Liner Systems As perhps should be expected, the Fredholm Alterntive is fr more generl thn just governing ODEs. Theorem 2.4 We introduce the liner system Lu = f where L is n m n mtrix nd u nd f re 1 n vectors where f is given nd u is unknown. Consider the homogeneous djoint (trnspose) problem L T v = where superscript T denotes the trnspose of the mtrix. Then EITHER 1. If the only solution to the homogeneous djoint problem is the trivil solution u = then the solution to the inhomogeneous problem u exists nd is unique OR 2. If there re non-trivil solutions to the homogeneous djoint problem v then either There re infinitely mny solutions if v f =, or There is no solution if v f. It trnspires tht this theorem is useful for liner integrl equtions in lter sections.

32 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Wht is Green s function? Hving ddressed mny spects of ODE theory, let us now focus on the min issue of this course - defining nd using Green s functions. The method of Green s functions is simply method in order to solve inhomogeneous BVPs. One of the interesting spects of Green s functions is tht they enble the solution to be written down in very generl form for vriety of forcing functions. The Green s function lso often corresponds to something physiclly importnt. We hve lredy seen one exmple where the Green s function enbles the solution to be written in generl form in section 2.2. At tht time we did not think of it s Green s function, it ws considered merely s n influence function for the inhomogeneous forcing term f(x). 2.7 Green s functions for Regulr S-L problems vi eigenfunction expnsions Consider gin the regulr S-L problem of the form Lu = f(x) (2.13) with L given by (2.75), x [, b] nd u is subject to two homogeneous BCs of the form (2.78). Also consider the relted eigenvlue problem Lu = λµ(x)u (2.14) with some ppropritely chosen µ(x). We cn solve (2.13) by posing n eigenfunction expnsion of the form (see Exmple Sheet 3) u(x) = n φ n (x). (2.15) This cn be differentited term-by-term (see MT241) so tht, pplying L we find Lu(x) = n=1 n λ n µ(x)φ n (x) = f(x). (2.16) n=1 Let us multiply by φ m (x) nd integrte over the domin x [, b]. The orthogonlity of the eigenfunctions (with respect to the weight µ(x)) llows us to then show tht Therefore u(x) = n λ n = f(x ) nd so we recognize tht we cn write u(x) = f(x)φ n(x) dx (2.17) (x)µ(x) dx. φ2 n ( ) φ n (x)φ n (x ) λ n φ2 n µ(x 1) dx 1 n=1 dx (2.18) f(x )G(x, x ) dx (2.19)

33 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 33 where G(x, x ) = ( ) φ n (x)φ n (x ) λ n φ2 n(x 1 )µ(x 1 ) dx 1 n=1 (2.11) which is therefore n eigenfunction expnsion of the Green s function. Note tht G(x, x ) = G(x, x) in this setting. We note tht the definition (2.11) would run into difficulty if one of the eigenvlues is zero (i.e. if there is non-trivil solution to the homogeneous djoint problem!). We return to this point lter on in section Exmple 2.12 Let us return to the fmilir exmple with u() = u(l) = nd the relted eigenvlue problem Lu = d2 u = f(x) (2.111) dx2 d 2 φ = λφ (2.112) dx2 with φ() = φ(l) =. We lredy know from exmple 2.7 tht λ n = (nπ/l) 2 nd φ n (x) = sin(nπx/l) with n = 1, 2, 3... Therefore with reference to the theory bove, u(x) is given by where u(x) = = G(x, x ) = 2 L n φ n (x), (2.113) n=1 L n=1 f(x )G(x, x ) dx (2.114) sin(nπx/l) sin(nπx /L) (nπ/l) 2. (2.115) Finlly we sk, how is this representtion of the Green s function in terms of eigenfunctions relted to the form derived in (2.41) or (2.42) bove. They must be equivlent! We discuss this in question 4 on Exmple Sheet 4.

34 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Green s functions for Regulr S-L problems using direct pproch For problems of regulr (nd some singulr) S-L type we hve shown bove in equtions (2.19)-(2.11) tht the eqution hs the solution u(x) = Lu = f(x) (2.116) G(x, x )f(x ) dx (2.117) for some ppropritely defined function G(x, x ) which we hve termed the Green s function. We hve n eigenfunction representtion for the Green s function defined in (2.11). This pproch shows tht the Green s function exists provided tht there is no zero eigenvlue, see section We cn obtin the Green s function for S-L using vrition of prmeters. We will describe this shortly but first we need some discussion of few rther unusul functions The Dirc delt function The representtion of the solution in the form (2.117) shows tht the source term f(x) represents forcing t ll of the points t which it is non-zero. We cn isolte the effect of ech point in the following mnner. First we tke function f(x) nd consider splitting it up in order to tke into ccount the seprte contributions from intervls of width x i such s we do when crrying out the process of Riemnn integrtion, see figure 4. Consider decomposing the function f(x) into liner combintion of unit pulses strting t the points x i nd being of width x i, see figure 5. y x i x Figure 4: Figure depicting the prtition of function f(x) into liner contributions of unit pulses, similrly to the process of Riemnn integrtion.

35 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 35 So we would write f(x) i f(x i ) (unit pulse strting t x = x i ). (2.118) nd we know tht this is only good pproximtion if the intervls re smll (infinitesiml in fct!) Indeed this is very similr to something like n integrl. Only the x i is missing! Let us now introduce this nd limiting process in the following mnner: f(x) = lim x i = lim x i (unit pulse) f(x i ) x i (2.119) x i i f(x i )(Dirc pulse) x i. (2.12) i We now pper to hve introduced strnge object - wht we hve termed here the Dirc pulse. It hs height 1/ x i nd width x i. We picture this in figure 6. Note tht this pulse hs unit re. In the limit s x i this pulse represents concentrted pulse of infinite mplitude locted t single point. It is not relly function but is often termed generlized function. We will cll this object the Dirc Delt function 8, which when locted t the point x = x i is written s δ(x x i ). It cnnot be written down in the form δ(x x i ) =... We think of this object s concentrted source or impulsive force, nd ccording to (2.12), in the limit, we hve the definition f(x) = f(x i )δ(x x i ) dx i. (2.121) The intervl of integrtion here is ll x i. The property in (2.121) is known s the sifting property of the Dirc delt function. The dirc delt function cn be thought of s the limit of the sequence of vrious different functions, not only the rectngulr type depicted bove. 1 x i x i + x Figure 5: A unit pulse. We now note some importnt properties of the function. Firstly, we note tht with f(x) = 1 1 = δ(x x i ) dx i. (2.122) 8 Nmed fter the brillint twentieth century mthemticl physicist Pul Dirc ( )

36 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 36 x 1 x x i x i + x Figure 6: A Dirc pulse. The function is even, δ(x x i ) = δ(x i x). Furthermore it is strongly linked with the Heviside function H(x x i ) which we hve lredy defined bove, but repet here for completeness, s { 1, x > x i, H(x x i ) = (2.123), x < x i vi the expression H(x x i ) = x δ(y x i ) dy. (2.124) The Heviside function is not defined t x = x i - we hve freedom to choose its vlue there. Usully the most convenient is to choose its vlue s 1/2. This is the verge of the limit from both sides of x i of course. Finlly we note tht δ[c(x x i )] = 1 c δ(x x i) (2.125) for some constnt c. For proofs of the lst two properties see question 6 on Exmple Sheet 4. The introduction of this function now llows us to determine n eqution governing the Green s function Reltionship between the Dirc delt function nd the Green s function Given the solution (2.117), we note tht the Green s function G(x, x ) is n influence function for the source f(x). As n exmple, let us suppose tht f(x) is now concentrted source t x = s, i.e. f(x) = δ(x s) with < s < b. This then gives u(x) = δ(x s)g(x, x ) dx = G(x, s) (2.126) by the sifting property (2.121). We therefore obtin the fundmentl interprettion of the Green s function: it is the response t x due to concentrted source t x : L x G(x, x ) = δ(x x ) (2.127)

37 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 37 where the subscript x on the opertor ensures tht we know tht the derivtives re with respect to x. The source position x is prmeter in the problem. We cn check tht (2.117) stisfies (2.116) vi the definition of the Green s function (2.127) by operting on ech side of (2.117) with L x to give L x u = f(x )L x [G(x x )] dx = vi the sifting property (2.121). f(x )δ(x x ) dx = f(x) (2.128) Boundry conditions for the Green s function BVP If we tke (2.127) together with pproprite homogeneous boundry conditions s n independent definition of the Green s function (which we shll!) then we lso wnt to derive the solution strting with this independent definition. Strt with Green s identity in 1D for ODEs with opertors in S-L form (2.75), which turns out to be v, Lu Lv, u = [ ( p(x) v du )] b dx udv. (2.129) dx Let v = G(x, x ). The right hnd side vnishes s long s we choose the homogeneous BCs for the Green s function to be the sme s those for the originl problem ssocited with u. Then G(x, x ), Lu(x) LG(x, x ), u(x) =. (2.13) Using the definition of the Dirc delt function nd interchnging vribles we obtin u(x) = f(x )G(x, x) dy. (2.131) For regulr S-L opertors the Green s function is (Hermitin) symmetric (G(x, x) = G(x, x )) s we shll show shortly, so tht u(x) = f(x )G(x, x ) dx. (2.132) Reciprocity nd symmetry of the Green s function for fully S-A problems. Let us suppose tht the BVP is fully self-djoint (e.g. regulr S-L problem). Let us once gin use (2.129) nd let u = G(x, x 1 ) nd v = G(x, x 2 ). Both stisfy homogeneous boundry conditions of the form (2.78). Furthermore since L x u = δ(x x 1 ) we use Green s second identity to find [ ] G(x, x 2 )δ(x x 1 ) LG(x, x 2 )G(x, x 1 ) dx =. (2.133)

38 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 38 Therefore from the sifting property of the Dirc function, G(x 1, x 2 ) = = = LG(x, x 2 )G(x, x 1 ) dx (2.134) LG(x, x 2 )G(x, x 1 ) dx (2.135) δ(x x 2 )G(x, x 1 ) dx (2.136) = G(x 2, x 1 ) (2.137) = G(x 2, x 1 ) (2.138) Note tht this is ll relint on the fct tht the opertor is fully self-djoint. If it is not, much of the bove theory hs to be modified s we shll see in section 2.1. Physiclly, the property (2.138) sys tht the response t x 1 due to concentrted source t x 2 is the sme s the response t x 2 due to concentrted source t x 1. This is not immeditely physiclly obvious! Jump conditions t x = x The Green s function cn be determined from the governing eqution (2.127). For x < x, G(x, x ) stisfies this eqution with homogeneous BC t x =. Similrly for x > x with homogeneous BC t x = b. Wht hppens t the point x = x? We need to consider the type of singulrity tht rises in (2.127) with reference to the property (2.124). Suppose firstly tht G(x, x ) hs jump discontinuity t x = x ( property shred by the Heviside function H(x x )). Then dg(x, x )/dx would hve delt function singulrity nd so d 2 G(x, x )/dx 2 would be more singulr thn the ctul right hnd side of (2.127). Therefore we conclude tht G(x, x ) must be continuous t x = x which we denote by [G(x, x )] x=x+ x=x = (2.139) where x + nd x denote pproching x = x from bove nd below respectively, e.g. x + = lim ǫ x +ǫ, x = lim ǫ x ǫ with ǫ >. On the other hnd dg(x, x )/dx does hve jump discontinuity t x = x. In order to illustrte this for S-L problems, integrte Lu = f(x) where L is the S-L opertor (2.75), between x = x nd x = x+ re continuous t x = x ) Since p(x) is continuous function this then gives to give (since q nd G [ p(x) dg ] x=x + = 1. (2.14) dx x=x [ ] x=x + dg = 1 dx x=x p(x ). (2.141)

39 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Summry: Green s function for regulr S-L problems Given regulr S-L problem of the form L x u = d dx ( p(x) du dx ) + q(x)u(x) = f(x) (2.142) together with homogeneous boundry conditions B t x =, b, the corresponding Green s function will be defined by L x G(x, x ) = δ(x x ) (2.143) together with the sme homogeneous boundry conditions B t x =, b, nd the following conditions t x = x : [G(x, x )] x=x+ x=x = (2.144) nd [ ] x=x + dg = 1 dx x=x p(x ). (2.145) Let us use these steps to construct the Green s function for simple exmple.

40 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 4 Exmple 2.13 Consider gin the stedy stte het eqution d 2 u = f(x) (2.146) dx2 with u() =, u(l) =. We cn write the solution to this problem in the form u(x) = L where the Green s function G(x, x ) stisfies with G(, x ) = nd G(L, x ) =. f(x )G(x, x ) dx (2.147) d 2 G(x, x ) dx 2 = δ(x x ) (2.148) Now tht we hve governing eqution for the Green s function we cn esily obtin its solution for x x : { + bx, x < x, G(x, x ) = (2.149) c + dx, x > x but we note tht the constnts could be different for different x - the source loction. The BC t x = pplies for x < x nd imposing this gives =. Similrly G(L, x ) = gives c + dl =. Therefore we hve { bx, x < x, G(x, x ) = (2.15) d(x L), x > x. We lso know from the discussion bove tht G(x, x ) is continuous t x = x. This gives bx = d(x L). (2.151) The jump condition on the derivtive t x = x, gives (since p = 1) Solve (2.151) nd (2.152) to obtin d b = 1. (2.152) b = (x L), d = x L L noting in prticulr the dependence on x here. This gives { x L G(x, x ) = (x L), x x, x L (x L), x x L, (2.153) (2.154) which grees with wht we found in (2.41). In fct, for regulr (nd some singulr) S-L problems we cn derive the Green s function explicitly vi the method of vrition of prmeters s we describe in these steps:

41 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Explicit solution for the Green s function for regulr S-L problems 1. Find the two independent solutions of the homogeneous eqution (2.142) (i.e. the complementry function u c ) sy u 1 (x) nd u 2 (x). 2. Tke liner combintions of these solutions in order to find solution which stisfies the left (t x = ) nd right (t x = b) homogeneous boundry conditions. Cll these u L (x) nd u R (x) respectively. 3. Write the Green s function s G(x, x ) = { c L (x )u L (x), x x, c R (x )u R (x), x x b (2.155) where we hve noted the explicit dependence of c L nd c R on the source loction x. Note tht we re ble to put here becuse the Green s function is continuous t x = x. 4. Enforce the condition on G(x, x ) t x = x : 5. Enforce the condition on dg/dx t x = x : c L (x )u L (x ) = c R (x )u R (x ) (2.156) c R (x ) du R dx (x ) c L (x ) du L dx (x ) = 1 p(x ). (2.157) 6. Finlly we cn solve (2.156) nd (2.157) for c L (x ) nd c R (x ) to get c L (x ) = u R(x ) p(x )W(x ), c R(x ) = u L(x ) p(x )W(x ), (2.158) where we hve defined the ssocited Wronskin W(x ) = u L(x ) u R (x ) u L (x ) u R (x ). (2.159) And therefore the Green s function is known immeditely once u L nd u R hve been determined. Although such n explicit form is lso vilble when the problem is not of Sturm-Liouville type it needs little more justifiction, see section Let us first consider some exmples which do conform to S-L type. Exmple 2.14 Let us reconsider the stedy stte het eqution from Exmple 2.13 where we determined the Green s function by using direct method, insted of the vrition of prmeters procedure bove. We note tht p(x) = 1 nd we hve the two homogeneous solutions u L (x) = x nd u R (x) = x L. Therefore since from (2.159) we hve W(x ) = x (x L) = L, from (2.155) nd (2.158) we see tht G(x, x ) = { 1 x(x L L), x x, 1 L (x L), x x L (2.16) which grees with (2.154) which we note in turn greed with the lterntive derivtion of (2.41). Note how quickly we could derive the Green s function with the procedure bove!

42 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 42 Exmple 2.15 Let us consider once gin the Green s function ssocited with the stedy stte het eqution d 2 u = f(x) (2.161) dx2 but now with boundry conditions u() =, u (L) =. The Green s function G(x, x ) stisfies with G(, x ) = nd dg/dx(l, x ) =. d 2 G(x, x ) dx 2 = δ(x x ) (2.162) By the rgument bove, ll we hve to do is to find the homogeneous solutions which stisfy seprtely the left nd right boundry conditions nd we cn immeditely write down the Green s function. These re respectively u L (x) = x nd u R (x) = 1. Therefore W(x ) = 1 nd { x, x x, G(x, x ) = x, x x L = xh(x x) x H(x x ) (2.163) The next exmple is n exmple of singulr S-L problem showing tht the method lso works in tht cse. This is the exmple tht ws inconsistent in the lectures. Note tht now I hve chnged the forcing f(x) so tht the problem IS consistent. Note tht the ctul construction of the Green s function hs not chnged. Exmple 2.16 Construct the Green s function ssocited with the BVP consisting of the ODE x 2 u (x) + 2xu (x) 2u(x) = f(x) (2.164) together with the BCs u() = nd u(1) =. Use the Green s function to find the solution when f(x) = x 2 nd confirm this by finding the solution vi stndrd techniques. First we note tht the ODE is of S-L type since it my be written ( d x 2du ) 2u(x) = f(x). (2.165) dx dx Therefore we cn ppel to ll of the theory bove. Since the ODE is of Euler type we seek solutions in the form u(x) = x m (2.166) nd therefore we require m 2 +m 2 = so tht (m+2)(m 1) = nd therefore m = 2 or m = 1 nd therefore we hve u 1 (x) = 1/x 2 nd u 2 (x) = x. these cn be combined to yield the solutions u L (x) = x, u R (x) = x 1 x2. (2.167)

43 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 43 We find tht p(x )W(x ) = 3 nd therefore so tht c L (x ) = 1 3 u R(x ) = (x3 1), (2.168) 3x 2 c R (x ) = 1 3 u L(x ) = x 3 x 2 (2.169) ( ) x x 3 1, x x 3 x G(x, x ) = 2, ( ) x x 3 1, x 3 x 2 x 1, = x ( ) x 3 1 H(x x) + x ( ) x 3 1 H(x x 3 3 x 2 ) (2.17) The solution is therefore u(x) = 1 G(x, x )f(x )dx nd for us since f(x ) = x 2, using the (Heviside form of the) Green s function bove, 1 x u(x) = x ( x ) dx + (x3 1) x 3x 2 = x [ ] x4 x + 1 (x 1x ) x 4 x 12 2 = x ( ) 4 x4 + x + x5 12 x2 12 x 3 dx = 1 x(x 1). (2.171) 4 This cn be confirmed using the stndrd solution form u(x) = u c (x) + u p (x) where u c (x) = c 1 x + c 2 x 2. It is esily verified tht u p = x 2 /4. BCs require c 2 = nd c 1 = 1/4 which recovers (2.171).

44 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Green s functions for the wve eqution with time hrmonic forcing Consider the wve eqution in one sptil dimension, 2 U ξ = 1 2 U + F(ξ, t) 2 c 2 t2 where c is the wvespeed nd F(ξ, t) is some forcing term. This pplies to number of physicl systems, e.g. trnsverse wves on string. Suppose tht the forcing is time hrmonic F(ξ, t) = g(ξ)e iωt so tht the wves re lso time hrmonic, i.e. nd then we hve U(ξ, t) = u (ξ)e iωt u (ξ) + k2 u (ξ) = g(ξ) where k = ω/c is the wvenumber (dimensions 1/L)nd ω is the frequency (dimensions 1/T). Note tht using complex exponentil is for convenience. It is much neter mthemticlly thn using sine or cosine function. After ll nlysis the ide is to simply tke the rel prt s clerly in relity we need rel solution. We cn non-dimensionlize (scle) ξ on the wvenumber, introducing x = kξ nd u(x) = ku so tht where f(x) = 1 k g(x/k). u (x) + u(x) = f(x) (2.172) We hve lredy determined the existence nd uniqueness properties of this problem for the BCs u() = u(l) = in Exmple 2.9. This corresponds to the sitution when the ends of the string re fixed - common scenrio! Let us ssume, for now, tht L nπ so tht there re no existence nd uniqueness issues. We then know the solution to the problem, it is (combining nd simplifying ll of the terms in Exmple 2.9) u(x) = L x sin(x L) sin x x f(x ) dx + sin L nd we note tht we cn therefore write this s sin(x L) sin x f(x ) dx (2.173) sin L where u(x) = L G(x, x )f(x ) dx (2.174) { G(x, x ) = 1 sin(x L) sin x, x x, sin L sin x sin(x L), x x L (2.175) nd we note tht G(x, x ) = G(x, x). In prticulr it is helpful to write G(x, x ) = 1 sin L (sin(x L) sin xh(x x) + sin x sin(x L)H(x x )).

45 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 45 In question 1 on Exmple Sheet 5 we show tht this Green s function cn be derived directly by following the steps in the explicit solution rther thn the method of vrition of prmeters bove. Let us consider slightly different problem now: n infinite domin. Exmple 2.17 Consider string of infinite length. Wht this mens in relity is tht it is so long tht the end conditions re never importnt in the problem. Wht is the ssocited Green s function for forcing t the origin? Since we re interested in physicl problems, let us work in the physicl domin so let us solve G (x, ) + k 2 G(x, ) = δ(x) where k = ω/c is the wvenumber defined bove. This is subject to the usul continuity conditions t x = x =. Since the string is of infinite extent we do not hve ny BCs! So wht do we use s dditionl conditions?! Well we know tht if we re forcing the string hrmoniclly t the point x = then wves must be moving wy from tht point to infinity. They clerly cnnot be coming in from infinity. So this pplies for x ± which gives us nother two conditions. It is convenient in this cse to write our fundmentl solutions s u 1 (x) = exp(ix) nd u 2 (x) = exp( ix). In order for the solution to be outgoing s x ± it is cler tht we need to tke { c L () exp( ikx), x, G(x, ) = (2.176) c R () exp(ikx), x. We hve retined the generl form with rgument x = here since we will generlize to rbitrry x lter on. Why did we choose this form? Well we hve time dependence of exp( iωt) so tht for x >, we hve solution of the form exp(i(kx ωt)). Tht this corresponds to wve moving in the positive x direction is cler since when we increse time, if we wnt to sty t the sme point on the wve we hve to increse x. Similrly exp( i(kx+ωt)) corresponds to left propgting wve nd so is vlid for x <. Another wy of sying this is to impose tht G (x, ) ikg(x, ) s x, (2.177) G (x, ) + ikg(x, ) s x. (2.178) Continuity t x = gives c L () = c R (). And the jump condition on the derivtive gives so tht c L () = 1/(2ik) nd [ ] x= + dg = ikc R () + ikc L () = 2ikc L () = 1 (2.179) dx x= G(x, ) = 1 exp(ik x ). (2.18) 2ik

46 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 46 It is not difficult to show tht (see question 2 of Exmple Sheet 4) for generl point of forcing x = x, G(x, x ) = 1 2ik exp(ik x x ). (2.181) Note tht G(x, x ) = G(x, x). This is interesting becuse this is not Hermitin symmetry nd therefore this problem cnnot be self-djoint. But this is strnge. Why not?! It looks like it should be! Well, this is subtle: note tht the coefficients in the boundry (rdition) conditions (2.177), (2.178) re complex nd this mens tht we re not gurnteed selfdjointness. See prt (ii) of question 2 on Exmple Sheet 4. In prt (iii) of tht question we lso consider the problem of wves on semi-infinite string forced hrmoniclly. Why work with complex solutions exp(±ikx) rther trigonometric functions sin kx nd cos kx? We do this for wves problems minly becuse it is convenient in terms of lgebr: when we tke products of exponentils we cn combine terms dditively in the exponent. A good exmple is the time hrmonic dependence: exp(ikx) exp( iωt) = exp(i(kx ωt)). Of course t the end we wnt REAL solutions so we hve to tke the rel prt of whtever solution we obtin in prctice.

47 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D The djoint Green s function We derived the result (2.132) only for S-A problems nd in these cses we note tht the Green s function is lso symmetric. Wht if the problem is not S-A? Cn we still define Green s function nd if so, wht does the solution form for u(x) look like? We will del with this here, lthough we note tht physiclly mny of the equtions we del with re of the S-A form. First we introduce the governing eqution L x u = f(x) (2.182) where now L x is not S-A. Homogeneous BCs B ccompny (2.182). We note tht we cn introduce the corresponding Green s function in the usul mnner with equivlent homogeneous BCs to the BVP. L x G(x, x ) = δ(x x ) (2.183) Theorem 2.5 For non S-A problems with homogeneous BCs, the solution cn still be written u(x) = G(x, x )f(x ) dx (2.184) The proof of this is esy. We cn perform L x on both sides of (2.184) s we did in (2.128). However, how is this consistent with wht ws derived bove using Green s identity, becuse there we hd to use the symmetry properties of the Green s function? In fct wht we cn do is use the djoint opertor nd define the so-clled djoint Green s function G which stisfies L xg = δ(x x ) (2.185) where L is the djoint opertor together with the necessry djoint BCs B. We use the definition of the djoint opertor, v, Lu = L v, u (2.186) Let us choose u = G(x, x 1 ) nd v = G (x, x 2 ), noting tht G stisfies the sme BCs s the originl problem wheres G stisfies the djoint BCs (note lso here tht we hve seen few exmples of cses where L = L but the djoint BCs re different nd therefore this would clerly give G (x, y) G(x, y)). This ll gives G (x, x 2 ), LG(x, x 1 ) = L G (x, x 2 ), G(x, x 1 ). (2.187) Using the definitions of these functions we find tht G (x, x 2 )δ(x x 1 ) dx = L G (x, x 2 )G(x, x 1 ) dx (2.188)

48 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 48 nd the sifting property of the delt function gives G (x 1, x 2 ) = So we hve proved the following result: = L G (x, x 2 )G(x, x 1 ) dx δ(x x 2 )G(x, x 1 ) dx = G(x 2, x 1 ) (2.189) Theorem 2.6 For opertors tht re not fully S-A, the Green s function is not Hermitin symmetric. But there is symmetry reltion relting the Green s function nd its djoint: G(x, x ) = G (x, x). (2.19) Next, choose u to stisfy the originl inhomogeneous problem nd v = G in (2.186) so tht we obtin G, L x u u, L G =. (2.191) Once gin, using the sifting property of the delt function nd interchnging vribles we obtin u(x) = Finlly, we use Theorem 2.6 to get u(x) = G (x, x)f(x ) dx. (2.192) G(x, x )f(x ) dx. (2.193) Wht the theory just presented regrding the djoint Green s function tells us is tht we do not hve to ever worry bout constructing the djoint Green s function! We cn lwys just construct the Green s function (if it exists), stisfying the sme BCs s the originl problem, even when it is not S-A nd still write the solution in the form This is convenient! u(x) = G(x, x )f(x ) dx. In the next section we will construct Green s function for non S-A opertor.

49 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Green s functions for non S-A BVPs The theory in this section (in blue below) is non-exminble. Wht you should tke from this section is tht for Non-SA problems, you cn use the sme explicit method to derive the Green s function s we derived for regulr S-L problems BUT ONLY WHEN BCs re NOT MIXED! I WILL NOT SET ANY PROBLEMS ON THE EXAM INVOLVING FINDING THE GREENS FUNCTION WITH MIXED BCs! Above we showed explicitly how to determine the Green s function when the problem is of regulr S-L type. Wht if it is not of this type? Three different cses rise. These re: (i) When the ODE is in S-A form, with B B (but BCs re not mixed) (ii) When the ODE is not in S-A form with non-mixed BCs. (iii) When ny BC is of mixed type. In cse (i) we cn follow the explicit pproch described bove for regulr S-L problems with no problem. In cse (ii) we now show tht we cn trnsform the ODE into S-A form nd hence there is no issue. One cn pproch the Green s function construction in exctly the mnner described for regulr S-L problems bove. We cn force the ODE into S-A form s now explin. Tke the usul ODE p(x) d2 G dx + r(x)dg 2 dx + q(x)g = δ(x x ). (2.194) Divide by p(x), generte the integrting fctor ( ) I(x) = exp r(x)/p(x) dx nd then multiply through by this to get I(x) d2 G dx + r(x) 2 p(x) I(x)dG I(x) + q(x)i(x)g = dx p(x) δ(x x ). nd now re-write the left hnd side in the form ( d I(x) dg ) + q(x)i(x)g = I(x) dx dx p(x) δ(x x ). (2.195) It my not be beneficil to do this in order to solve the problem. Indeed solving (2.194) my be esier (but they must give the sme solution!). But we cn now integrte (2.195) between x = x nd x = x + to get the jump condition s before [ dg dx ] x + x = 1 p(x ). (2.196) Hence we cn in fct follow exctly the sme procedure s for regulr S-L problems in order to generte the Green s function. The only difference is tht the Green s function will not be Hermitin symmetric. Insted we hve the lterntive symmetry reltion G(x, x) = G (x, x ) s proven bove.

50 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 5 Exmple 2.18 Solve the BVP consisting of the ODE x 2 u + 4xu + 2u = f(x) nd the BCs u(1) = u(2) = by finding the Green s function nd write down the explicit solution in the cse when f(x) = x. Confirm tht this is wht one would expect by solving vi direct methods for this specific f(x). Fundmentl solutions to the homogeneous problem re given by solving x 2 u + 4xu + 2u = which is n Euler eqution so we seek solutions of the form u(x) = x m which gives so tht u(x) = x 2 nd u(x) = x 1. m 2 + 3m + 2 = (m + 2)(m + 1) = Combintion of these solutions stisfying the left nd right BCs re u L (x) = 1 x 1 x 2, u R(x) = 1 x 2 x 2 (2.197) The ssocited Wronskin is (check this!) W(x ) = u L (x )u R (x ) u R (x )u L (x ) = 1 x 4 so tht p(x )W(x ) = 1/x 2. Therefore c L (x ) = u R(x ) p(x )W(x ) = x 2, c R (x ) = u L(x ) p(x )W(x ) = x 1. (2.198) nd the Green s function is { c L (x )u L (x), 1 x x, G(x, x ) = c R (x )u R (x), x x 2. { 1 (x x = 2 2)(x 1), 1 x x, 1 (x x 2 1)(x 2), x x 2. The solution of the problem is = 1 x 2(x 2)(x 1)H(x x) + 1 x 2(x 1)(x 2)H(x x ). u(x) = 2 1 f(x )G(x, x ) dx. With f(x) = x, crrying out the integrtions we obtin (check this!) u(x) = 1 x 2 7x 6 + x 6 In question 4 on Exmple Sheet 5 you re sked to show tht the sme result would hve been obtined if you hd used the djoint Green s function.

51 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 51 THE REST OF SECTION 2.11 IS NON-EXAMINABLE. If BCs re of mixed type (i.e. (iii) bove), then strictly there is no left or right BC to stisfy. Wht this mens is tht we cnnot solve the problem in the sme mnner s the explicit method bove. In prticulr it mens tht the x nd x dependence in the Green s function is NOT seprble. Once we hve determined fundmentl solutions u 1 nd u 2 wht we must do is to pose solution in the form { c L (x )u 1 (x) + d L (x )u 2 (x), x x, G(x, x ) = (2.199) c R (x )u 1 (x) + d R (x )u 2 (x), x x b nd then determine c L, d L, c R, d R from the two boundry conditions nd continuity conditions imposed t x = x. Let us illustrte with n exmple. Exmple 2.19 Determine the Greens function for the ODE u (x) + 3u (x) + 2u(x) = subject to the mixed BCs u() = u (1) nd u(1) = 2u() + u (). Both BCs re mixed so there is no left or right BC. However we cn esily solve the homogeneous ODE with u 1 (x) = e x, u 2 (x) = e 2x so let us tke the Green s function in the form bove: { c L (x )e x + d L (x )e 2x, x x, G(x, x ) = c R (x )e x + d R (x )e 2x, x x 1 (2.2) Imposing the BCs leds to the conditions 1 c L + d L = c R e 2d R e, 2 c R e + d R e = c 2 L together with the stndrd continuity conditions t x = x (which give n ddition two conditions) we cn then determine c L, d L, c R, d R. (see question 8 on Exmple Sheet 5) Inhomogeneous boundry conditions In generl we would like to solve problems which re not restricted to boundry conditions tht re homogeneous. There re two pproches to solving such problems. Consider the following problem LU = f(x) (2.21)

52 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 52 subject to U() = α, U(b) = β. Since the problem is liner we cn decompose the solution in the form where u nd v stisfy U(x) = u(x) + v(x) (2.22) Lu = f(x), Lv = (2.23) subject to u() =, u(b) = nd v() = α, v(b) = β. The problem for v is simply to find liner combintion of the fundmentl solutions such tht the boundry conditions re met. However note tht from Exmple (2.8) we re not lwys gurnteed tht such solution will exist! The problem for u is equivlent to the homogeneous boundry condition problems bove with ssocited Green s function tht lso stisfies homogeneous BCs. Thus if both v(x) nd the Green s function exists, the solution will be U(x) = G(x, x )f(x ) dx + v(x) (2.24) Alterntively we cn derive the solution directly vi the Green s function nd ppliction of Lgrnge s identity. For conciseness let us consider the fully self-djoint cse. Then [ GLU ULG dx = G du dx U dg ] b dx = U() dg dx (, x ) U(b) dg dx (b, x ) Therefore simplifying the left hnd side, the solution is U(x) = = α dg dx (, x ) β dg dx (b, x ) (2.25) G(x, x )f(x ) dx + β dg dx The two solutions (2.24) nd (2.26) re equivlent of course. (b, x) αdg(, x). (2.26) dx An exmple ssocited with non-homogeneous boundry conditions cn be found in question 9 on Exmple Sheet 5.

53 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 53 THIS LAST SECTION IS NOT EXAMINABLE. BUT: You should ensure you know wht goes wrong when there is zero eigenvlue, i.e. the norml green s function cnnot be defined nd therefore you need modified Green s function. But you won t be sked to construct ny or know the theory of them, etc Existence of zero eigenvlue - modified Green s functions As is cler from the eigenfunction expnsion of the Green s function in (2.11), there is clerly problem when there exists zero eigenvlue. In this instnce the Green s function does not exist! As we know, zero eigenvlue corresponds to non-trivil solution of the homogeneous BVP, nd this exists for the fixed ends string problem bove, when L = nπ so let us consider tht problem. Try to determine the Green s function using the method bove - it will not work! In this cse s we know from the Fredholm Alterntive, there will only exist solutions to the inhomogeneous BVP if L sin x f(x ) dx =, so we ssume tht this holds. If n is even, ny even function f(x ) will ensure this. The reson tht the Green s function does not exist is becuse, once gin by the Fredholm Alterntive, for its existence (it is itself defined by n inhomogeneous BVP) we require L sin x δ(x x ) dx = sin x = for ll x [, L] but this clerly does not hold. Therefore insted of using tht Green s function s defined clssiclly, wht we should do is introduce different comprison problem. We define the so-clled modified Green s function G m (x, x ) vi the governing eqution LG m (x, x ) = δ(x x ) + cφ(x) where c R nd φ(x) is the non-trivil eigenfunction corresponding to the zero eigenvlue (here it is φ(x) = sin x). All other conditions on the Green s function remin the sme. In the generl cse then, we must choose c so tht the right hnd side is orthogonl to the eigenfunction corresponding to the zero eigenvlue. I.e. here we choose so tht (δ(z x ) + cφ(z))φ(z) dz = c = φ(x ) φ2 (z) dz where x is the loction of the source. Therefore the modified Green s function G m (x, x ) is defined by the eqution LG m (x, x ) = δ(x x ) φ(x)φ(x ) φ2 (z) dz.

54 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 54 subject to the usul homogeneous BCs. Unfortuntely since the right hnd side is orthogonl to φ(x), by the Fredholm Alterntive there re n infinite number of solutions so tht the modified Green s function is not uniquely defined. It trnspires tht the prticulr solution of the BVP cn be written s usul in terms of the modified Green s function G m (x, x ) (which cn be chosen to be symmetric) in the form u(x) = G m (x, x )f(x ) dx. Exmple 2.2 Derive the modified Green s function for the problem subject to u () =, u (L) =. u (x) = f(x) As we hve discussed bove, constnt c is homogeneous solution (n eigenfunction corresponding to zero eigenvlue). We note tht by the Fredholm Alterntive, for solution to exist therefore we must hve L f(x) dx =. Let us ssume tht we hve such n f(x). In tht cse the modified Green s function stisfies d 2 G m dx 2 = δ(x x ) + c. subject to G m (, x ) =, G m (L, x ) =. For modified Green s function to exist, gin by the Fredholm Alterntive the right hnd side hs to be orthogonl to constnt, i.e. so tht c = 1/L. Thus for x y, we must hve L δ(x x ) + c dx = By direct integrtion, dg m dx = d 2 G m dx 2 = 1 L. x L + c 1, x x, x L + c 2, x x L. If we choose c 1 nd c 2 to stisfy the BCs t x =, L we find c 1 =, c 2 = 1. The jump condition on dg m /dx is lredy stisfied. Integrting once gin we find x2 G m (x, x ) = 2L + d 1, x x, x2 2L + x + d 2, x x L.

55 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D 55 Imposing continuity of the Green s function t x = x we get d 1 = x + d 2. So we find x2 G m (x, x ) = 2L + x + d 2, x x, x2 2L + x + d 2, x x L. which illustrtes tht it is not unique. Imposing symmetry of the Green s function, i.e. we find tht (see question?? on Exmple Sheet 5) where β is constnt, independent of x. Therefore we hve, G m (x, x ) = nd solution of the BVP is u(x) = d 2 = 1 x 2 L 2 + β x2 2L + x + β, x x, x2 L 2L + x + β, x x L. G m (x, x )f(x ) dx. since f(x ) is orthogonl to constnt, we cn tke β = without loss of generlity.

56 Willim J. Prnell: MT3432. Section 2: Green s functions in 1D Revision checklist The following is guide to wht you should know. Red ech point nd sk yourself if you understnd wht it mens! Also, remember tht ssocited theory from the relevnt sections is exminble. For constnt coefficient nd Euler ODEs: be ble to find the complementry function u c nd be ble to find the prticulr solution u p by method of undetermined coefficients nd vrition of prmeters. Use integrtion by prts with inner products to derive the djoint opertor nd BCs. Know Lgrnge s nd Green s identities nd be ble to use them. Be ble to identify self djoint opertors - this requires both L = L nd B = B. Be ble to identify Sturm-Liouville problem nd know the difference between regulr nd singulr Sturm-Liouville problems. Know the 5 theorems ssocited with regulr S-L problems nd be ble to see how they relte to model problems. Understnd the Fredholm Alterntive theorem nd be ble to pply it to determine if solutions to inhomogeneous ODEs (with homogeneous BCs) exist nd re unique. Be ble to derive the Green s function vi eigenfunction expnsions Understnd some bsic properties of the Dirc delt function nd its reltionship to the Green s function Understnd the conditions tht define the Green s function (governing BVP, conditions t x = x ) Be ble to derive the Green s function for regulr S-L problems by vrition of prmeters (i.e. pplying v.o.p. to the originl ODE), nd the direct pproch (explicit solution - section 2.8.7) Be ble to derive Green s functions for the wve eqution t fixed frequency (time hrmonic) Understnd the problem tht rises if zero eigenvlue exists, understnd its reltionship with the Fredholm Alterntive nd the fct tht Green s function does not exist in this cse Be ble to define nd derive the djoint Green s function nd its reltionship to the Green s function Be ble to derive Green s functions for non S-L problems (WHEN BCs re NOT MIXED). Understnd liner superposition in order to derive solutions to inhomogeneous ODEs with inhomogeneous BCs

57 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 57 3 Green s functions in 2 nd 3D Unlike the one dimensionl cse where Green s functions cn be found explicitly for number of different problems, the two nd three dimensionl cse is somewht more difficult. Explicit Green s functions re rther difficult to determine especilly for bounded domins. However some specil cses turn out to be extremely importnt nd we cn do gret del with them. We shll denote two or three dimensionl domin by D nd its boundry by D. It will be cler from the problem whether we re in 2 or 3 dimensions or lterntively, the benefit of this unified nottion is tht it llows us to write down equtions which pply to both cses. For exmple, if D denotes ll spce (two or three dimensionl spce) then we cn prove some rther generl theorems ssocited with the sclr prtil differentil eqution (p(x) φ(x)) + q(x)φ(x) + λµ(x)φ(x) = (3.1) on the two or three dimensionl domin D subject to some homogeneous BCs on D. Note tht this eqution hs gret similrity to tht discussed in the Sturm-Liouville (one dimensionl) context in section (2.4). We do not hve the time to consider this in detil, lthough some spects re discussed. Insted let us consider the eigenvlue problem ssocited with the Lplcin opertor L = 2 in two nd three dimensions which mounts to the considertion of the problem subject to the homogeneous BC 2 φ + λφ = (3.2) α(x)φ + β(x) φ n =, x D. (3.3) Eqution (3.2) is known s Helmholtz eqution 9. Note tht this eigenvlue problem is importnt in (t lest) two physicl contexts. The first is the het eqution governing the temperture field u(x) which is (k(x) U) = ρ(x)c(x) U t + Q(x, t) (3.4) where k(x) is the conductivity, ρ(x) the mss density nd c(x) the specific het. The term +Q(x, t) is het source. The second is the wve eqution (E(x) U) = ρ(x) 2 U + Q(x, t) (3.5) t2 where E(x) is some physicl property to be specified nd ρ(x) is gin, the mss density. The term Q(x, t) is ny pplied forcing. The physicl interprettion of u(x) depends on the context. 9 Hermnn Ludwig Ferdinnd von Helmholtz ( ) ws Germn physicin nd physicist who mde significnt contributions to severl widely vried res of modern science. He is most known in the wves community for Helmholtz eqution, often lso known s the reduced wve eqution hving ssumed time hrmonic behviour

58 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 58 When the medium is homogeneous (mteril prmeters do not depend on x) these equtions reduce to nd k 2 U = ρc U t 2 U = 1 c 2 p + Q(x, t) (3.6) 2 U + Q(x, t) (3.7) t2 where c p is the wve speed. If you remember bck to MT241 you will hve determined eigenfunction expnsions for two nd three dimensionl problems for the Lplcin opertor. It trnspires tht you cn find such eigenfunction expnsions explicitly in number of cses, but the seprbility depends on the geometry (coordinte systems). Things work out nice nd clenly in number of systems but in prticulr in rectngles nd circles s we shll recll. Prtil differentil opertors in cylindricl nd sphericl coordinte systems (gin see MT241) re reviewed in Appendix A. We consider problems for (3.8) when Q is independent of time nd the resulting solution is lso independent of time, U(x, t) = u(x) so tht 2 u = 1 Q(x) (3.8) k nd (3.9) when Q is time-hrmonic, Q(x, t) = Q(x)e iωt so tht U(x, t) = u(x)e iωt nd so where K = ω/c p is known s the wvenumber. Both of these re very importnt problems in prctice. 2 u + K 2 u = Q(x) (3.9) 3.1 Self-djointness Assuming tht L is now some prtil differentil opertor, we cn define the djoint opertor in exctly the sme mnner s we did in one dimension. Tking n inner product v, u = vu dx which holds in two or three dimensions, we cn define the djoint opertor L vi D v, Lu = L v, u but ctully determining L is often difficult! See however question 2 on Exmple sheet 6. In this section we will consider only the Lplcin nd Helmholtz opertors. As we will show shortly in this cse L = L.

59 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D Lgrnge s identity nd Green s second identity First, note tht (question 1 on Exmple sheet 6) for two functions f(x), g(x), show tht (f g) = f g + f 2 g. (3.1) In the cse when L = 2 + K 2 (the Helmholtz opertor) with K R ( very useful cse!), we cn write down multidimensionl Lgrnge s identity s introduced in one dimension in section 2.3: v( 2 u + K 2 u) u( 2 v + K 2 v) = (u v v u). (3.11) And it follows then tht Green s second identity lso hs nlogues in two nd three dimensions: [ ] v( 2 u + K 2 u) u( 2 v + K 2 v) = (u v v u) nds (3.12) D D where s denotes prmetriztion of the boundry D of D. The nottion n denotes n outwrd pointing norml to D nd D is trversed in counter-clockwise mnner. With the homogeneous boundry condition B: α(x)u + β(x) u n = with α, β rel functions, choosing B = B (from the right hnd side of (3.12)) we hve L = L (left hnd side of (3.12)). Tht is [ ] vlu Lvu =. (3.13) D 3.2 An eigenvlue problem on rectngulr domin Exmple 3.1 Consider the two dimensionl eigenvlue problem ssocited with either het distribution or vibrting rectngulr membrne on rectngulr domin. The full problems re considered in question 3 on Exmple sheet 6. In tht question it is shown tht the relevnt two dimensionl eigenvlue problem is 2 φ x + 2 φ + λφ = (3.14) 2 y2 subject to the BCs which for simplicity here we shll tke s zero temperture (het) or pinned (membrne), i.e. for x [, L] nd y [, H]. φ(x, ) =, φ(x, H) =, (3.15) φ(l, y) =, φ(l, y) = (3.16) In tht question you lso show, by seprting vribles, tht the two dimensionl eigenfunctions re ( nπx ) ( mπy ) φ mn (x, y) = sin sin (3.17) L H for m, n = 1, 2, 3,..., with ssocited eigenvlues ( mπ ) 2 ( nπ ) 2 λ mn = +. (3.18) H L

60 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 6 Circulr domins led to eigenvlue expnsions in terms of Bessel functions which you my hve seen before. These cn be rther messy but we del with one of the simplest Bessel functions lter when we discuss the free spce Green s function for Helmholtz eqution. 3.3 Eigenvlue problem for the Lplcin opertor As in the one dimensionl cse, we cn mke some rther generl sttements bout certin eigenvlue problems. In prticulr let us consider the eigenvlue problem ssocited with the Lplcin opertor: subject with homogeneous BC 2 φ + λφ = (3.19) α(x)φ + β(x) φ n = (3.2) where α, β re rel functions. As we see in question 2 on Exmple sheet 6, this is self-djoint BVP. Associted with the bove eigenvlue problem we hve the following theorems: 1. All eigenvlues re rel 2. There re n infinite number of eigenvlues. There is smllest eigenvlue but no lrgest. 3. Corresponding to ny eigenvlue there my be mny eigenfunctions (note tht this ws not the cse for the 1D Sturm-Liouville problems) 4. The eigenfunctions form complete set, i.e. ny piecewise smooth function cn be represented by generlized Fourier Series in terms of the eigenfunctions. 5. Eigenfunctions ssocited with different eigenvlues re orthogonl, reltive to weight µ(x) with µ(x) = 1 here. [Furthermore, even different eigenfunctions tht correspond to the sme eigenvlue cn be mde orthogonl vi Grm-Schmidt orthogonliztion (but we will not consider this here)]. 3.4 Multidimensionl Dirc Delt Function The Dirc delt function extends nturlly to higher dimensions. For Crtesin coordintes nd two nd three dimensions we hve δ(x x ) = δ(x x )δ(y y ), (3.21) δ(x x ) = δ(x x )δ(y y )δ(z z ), (3.22) respectively. There re lso versions in cylindricl n sphericl coordintes but we do not need them here. We use the nottion x = (x, y, z) nd x = (x, y, z ). The filtering property of the Dirc delt function in higher dimensions is written s { f(x) x D, f(x )δ(x x ) dx = (3.23), x / D. D

61 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D Green s functions for the Lplce nd Poisson eqution Consider the eqution 2 u = Q(x) (3.24) on domin D in either two or three dimensions. It will lso be subject to some homogeneous BCs usully of the form u = or u n = or αu + β u n = for some rel functions α, β. We note tht with Q = corresponds to the cse of Lplce s eqution. For simplicity we will ssume tht ll functions nd prmeters, including the Green s function ssocited with the Lplcin problem re rel. This mens we don t hve to worry bout conjugtes, etc. in this subsection. Motivted by the one dimensionl cse, the ssocited Green s function is therefore defined by the eqution with equivlent homogeneous BCs, i.e. 2 G(x,x ) = δ(x x ) (3.25) αg(x,x ) + β G(x,x ) n =, x D. This is often known s Green s function for Lplce s or Poisson s eqution. It is perhps more ccurte to describe it s Green s function ssocited with the Lplcin opertor. We note tht eqution (3.24) is the forced stedy stte (independent of time) het eqution (3.8), i.e. it is tht temperture field tht persists fter ll trnsients hve decyed Symmetry for the Lplcin opertor Use Green s second identity with L = 2 in (3.13) (i.e. tke K = ) nd with v(x) = G(x,x 1 ) nd u(x) = G(x,x 2 ) so tht (remembering everything is rel!) (G(x,x 1 ) 2 G(x,x 2 ) 2 G(x,x 1 )G(x,x 2 )) dx =, (3.26) D which cn be written G(x,x 1 )δ(x x 2 ) dx = D = = D D D 2 G(x,x 1 )G(x,x 2 )) dx (3.27) 2 G(x,x 1 )G(x,x 2 ) dx (3.28) δ(x x 1 )G(x,x 2 ) dx (3.29) nd therefore using the filtering property of the Dirc Delt function (3.23), when x 1,x 2 D, G(x 2,x 1 ) = G(x 1,x 2 ) (3.3) nd s with the one dimensionl cse, this in fct holds for ll self-djoint opertors.

62 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D Solution representtion for the Lplcin opertor Use Green s second identity with u(x) stisfying the boundry vlue problem bove nd v(x) = G(x,x ) to get (gin remembering we ssume everything is rel) G(x,x ) 2 u(x) dx = 2 G(x,x )u(x) dx, (3.31) so tht D D G(x,x )Q(x) dx = = D D D 2 G(x,x )u(x) dx (3.32) δ(x x )u(x) dx (3.33) = u(x ). (3.34) Interchnge x nd x nd use (3.3) so tht we get u(x) = G(x,x )Q(x ) dx. (3.35) D So, the solution representtion in higher dimensions is entirely nlogous to the one dimensionl cse. Now let us consider finding some Green s functions ssocited with the Lplcin opertor Multidimensionl eigenfunction expnsions Let us use eigenfunctions to construct the Green s function for Lplce s eqution. We note here tht this pproch my not be idel - it my be very slowly convergent! Write G(x,x ) = λ λ φ λ (x) where φ λ is the eigenfunction corresponding to the eigenvlue λ. Then pply the Lplcin opertor so tht 2 G(x,x ) = λ 2 φ λ (x) = λ λφ λ (x) λ λ Since the left hnd side is δ(x x ), by orthogonlity nd the filtering property of the Dirc delt function we hve D λ λ = φ λ(x)δ(x x ) dx φ λ (x ) D φ2 λ (x) dx = (x) dx. D φ2 λ Exmple 3.2 Use the two dimensionl eigenfunction from exmple 3.1 to construct the ssocited Green s function, solving subject to homogeneous BCs on x =, L nd y =, H. 2 G(x,y) = δ(x y) (3.36)

63 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 63 From Exmple 3.1 we know tht the corresponding eigenvlues re ( nπ ) 2 ( mπ ) 2 λ mn = + L H with m = 1, 2, 3,... nd n = 1, 2, 3,... The corresponding eigenfunctions re nd in this cse φ mn (x, y) = sin (nπx/l) sin (mπy/h) D φ 2 mn (x, y) dx = (L/2)(H/2). Therefore the eigenfunction expnsion of the Green s function is G(x,x ) = 4 sin(nπx/l) sin(mπy/h) sin(nπx /L) sin(mπy /H). LH (nπ/l) 2 + (mπ/h) 2 n=1 m= Fredholm Alterntive As in the one dimensionl context, if there is zero eigenvlue (λ = ) then there re problems in terms of defining Green s function. We do not worry too much bout this here s we do not hve time! Let us ssume from now on tht λ will never be zero nd therefore the Green s function lwys exists Nonhomogeneous boundry conditions We cn lso cter for nonhomogeneous boundry conditions in firly simple wy s in the one dimensionl cse. We use Green s identity. For exmple if we hve u(x) = h(x) on the boundry for the Lplcin problem, we still retin the homogeneity of the Greens function BCs which enbles us to write, using (3.12) with v = G (nd remembering everything is rel!) G(x,x ) 2 u 2 G(x,x )u dx = (G u u G) n ds (3.37) D D = h(x)n x G(x,x ) ds (3.38) using the boundry conditions G = nd u = h on D. Therefore, given G we know the right hnd side of this eqution. Exploiting the forms of the left hnd side we find u(x ) = Q(x)G(x,x ) dx + h(x)n x G(x,x ) ds. D Finlly, interchnging x nd x nd using (3.3) (in the first integrl only!) we find u(x) = G(x,x )Q(x ) dx + h(x )n x G(x,x) ds. (3.39) D Note tht we hve to be very creful with the lst surfce integrl term. Plese consider this crefully nd look t the relevnt Questions on Exmple Sheet 7 nd lso the exmples tht we will consider lter in Section 3.6 to help you with your understnding!. D D D

64 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D Conditions stisfied by the Green s function Thus fr, by nlogy with the one dimensionl cse we hve sid tht the Green s function stisfies subject to the boundry condition 2 G(x,x ) = δ(x x ) (3.4) αg(x,x ) + βn x G(x,x ) = (3.41) for x D nd α, β cn be rel functions of x in generl. In order to derive the GF directly s in the one dimensionl cse we need to understnd wht hppens ner the source point x = x. In one dimension, remember tht the Green s function ws continuous but its derivtive ws discontinuous. To determine wht hppens in two nd three dimensions, let us first define locl polr coordinte system in two nd three dimensions (with reference to figure 7): x = x + r (3.42) where r = { r(cos θ, sin θ) in two dimensions, r(cosθ sin φ, sinθ sin φ, cosφ) in three dimensions (3.43) noting tht r = x x, θ [, 2π] nd φ [, π]. In both cses we note tht r = x x. This is just the system of cylindricl nd sphericl polr coordintes remember. D y r θ x D x Figure 7: Figure illustrting the locl coordinte system introduce round the source t x = x (in two dimensions). The three dimensionl cse is nlogous but hrder to drw! Let us now introduce these coordintes, nd integrte (3.4) over smll circle(sphere) S ǫ = x : x x ǫ in two(three) dimensions, of rdius ǫ with its centre on the source

65 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 65 point, with the intention of letting ǫ. Note tht we could lso define S ǫ with respect to the locl coordinte system r vi S ǫ = r : r = r ǫ. Therefore 2 G(x,x )dx = δ(x x ) dx (3.44) S ǫ S ǫ = δ(r) dr (3.45) S ǫ = 1 (3.46) where we hve used the filtering property of the delt function on the right hnd side. On the left hnd side we use Green s theorem so tht S ǫ G(x,x ) nds = 1. (3.47) This dicttes the type of singulrity tht the Green s function must possess. We note in prticulr tht this implies tht the Green s function is not continuous s it is in the one dimensionl context. And remember we re interested in the limiting cse when ǫ to understnd the singulr nture of the Green s function t the source loction. We cn esily see the type of singulrity required in two nd three dimensions s we now show. In two dimensions S ǫ is circle, n = e r, ds = rdθ so tht (3.47) sys tht we need 2π ( ) G lim ǫ r r dθ = 1 r=ǫ nd therefore we need the leding order behviour of the Green s function to be G r 1 2πr so tht upon integrtion for x close to x, i.e. r 1, G 1 2π ln r = 1 2π ln x x. Importntly, note tht this is the locl behviour of the Green s function ner the source point x = x. It is not the ctul Green s function globlly. In three dimensions S ǫ is sphere, n = e r, ds = r 2 sin φdθ so we need π 2π ( ) G lim ǫ r r2 sin φdφdθ = 1 r=ǫ nd so if we get the correct behviour. φ= θ= G 1 4πr = 1 4π x x The behviour exhibited by the Green s functions in the cses bove close to the source point mens tht the Green s function is clerly not continuous (it is singulr) but it possesses type of singulrity tht is integrble, i.e. G(x,x ) dx D

66 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 66 exists for ll x. We stress gin, s we stted bove tht this is just the locl behviour close to the source point. The Green s function could behve quite differently wy from the source point, depending upon the geometry nd boundry conditions of the problem. However there is specil (very importnt) cse where the bove llows us to determine the Green s function immeditely s we now describe Free-spce Green s function Wht hppens if we do not hve ny boundries? I.e. we seek Green s functions for infinite spce, or s it is often known, the free-spce Green s function. Well wht do we require for the Green s function for the Lplcin? It must solve Lplce s eqution nd lso hve the correct singulr behviour ner the source. So immeditely we re ble to sy tht the free spce Green s function for Lplce s eqution in two nd three dimensions re respectively G 2 (x,x ) = 1 2π ln x x, 1 G 3 (x,x ) = 4π x x (3.48) where the subscripts n refers to n dimensions nd free spce. Why is this? Well, clerly both possess the correct singulr behviour - they re exctly the singulr functions tht we required bove! But importntly they lso stisfy Lplce s eqution - you re sked to check this in question 4 on Exmple Sheet 6. We must be creful however when we use the free spce Green s functions. In order to derive the solution representtion for physicl problems, we usully pply Green s identity but this requires condition on the boundry. Here the boundry is t infinity. In order to understnd wht hppens we must pply Green s identity to finite domin nd then let the outer boundry tend to infinity, ensuring tht the contribution to the solution from tht boundry term does not diverge in the limit. Note tht this is not mysterious - it just substitutes for boundry condition tht would be present in finite problems. Let us focus on the solution of 2 u = Q(x) for free spce problems. We cn pply Green s identity from (3.12) with K = on finite domin D to obtin (remembering gin tht everything is rel!) (G(x,x ) 2 u(x) 2 G(x,x )u(x)) dx = (G u u G) ds. (3.49) D Let us tke D to be circle(sphere) of rdius R, centred on x = x, the source loction. Crrying out the usul steps on the left hnd side we obtin u(x ) = Q(x)G(x,x ) dx + (G(x,x ) x u(x) u(x) x G(x,x )) n ds. (3.5) D D Interchnging x nd x nd using the symmetry of the Green s function (except in the very lst term where we cnnot!) we find u(x) = Q(x )G(x,x ) dx + (G(x,x ) x u(x ) u(x ) x G(x,x)) n ds. D D D (3.51)

67 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 67 We could require the integrl t infinity (second term on the right hnd side) to do number of things. It could be chosen to give certin behviour to the solution t infinity (e.g. it could be constnt), or lterntively it could be chosen so tht it decys (is zero in the limit). In the ltter cse, using the forms of the Green s functions in two nd three dimensions, we require ( lim R u r ln r u ) r ( u + r u ) r lim R r=r r=r You re sked to confirm this in question 5 on Exmple Sheet 6. =, (3.52) =. (3.53) Bounded domins In fct, it turns out tht we my use the free spce solution to solve for Green s functions on domins tht do hve boundries. In prticulr the nice thing is tht the free spce solution contins the correct singulr behviour nd so we cn write the solution for new problem s G(x,x ) = G n (x,x ) + V (x,x 1 ) (3.54) where V (x,x 1 ) is chosen to stisfy the correct boundry conditions on the boundries. We use the so-clled method of imges which you my hve come cross in fluid dynmics for potentil flow problems. As usul let us restrict ttention to homogeneous boundry conditions here. Exmple 3.3 Determine the two dimensionl Green s function for Lplce s eqution on the semi-infinite domin D = { < x <, y < }, (upper hlf-spce) subject to G = on the boundry y =. Let us seek the solution in the form (3.54), with G(x,x ) = 1 2π ln x x + V (x,x 1 ) (3.55) so tht the first term tkes cre of the singulrity t the source. We need the function V to ensure tht the boundry condition is stisfied. Let us choose V (x,x 1 ) = 1 2π ln x x 1 (3.56) where x 1 = (x, y ). By symmetry we see tht the response should be zero t y = nd we will verify this shortly. We think of x 1 s n imge source. Thus the Green s function is G(x,x ) = 1 2π (ln x x ln x x 1 ) = 1 ( ) (x 4π ln x ) 2 + (y y ) 2. (3.57) (x x ) 2 + (y + y ) 2

68 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 68 Evluting this on y = we find G(x,x ) = 1 ( ) (x y= 4π ln x ) 2 + y 2 (x x ) 2 + y 2 s required. = 1 ln 1 = (3.58) 4π

69 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D Applictions of Poisson s eqution The beuty of Green s functions is tht they re useful! Let us consider n exmple here where we determine the stedy stte het distribution on two dimensionl domin. In question 5 of Exmple Sheet 6 you re sked to consider three dimensionl problem relting to the distribution of electricl chrge. This ws the problem for which Green originlly devised Green s functions! How gret is tht?! Exmple 3.4 Suppose tht we hve very lrge two dimensionl domin nd suppose tht it is so lrge tht boundry effects cn be ignored, i.e. the boundry is so fr wy tht ny het source genertes het distribution which hs decyed to zero long time nd distnce before the boundry. We use het source to het circulr region C of rdius uniformly with mgnitude q nd this source is mintined for ll times. Wht is the stedy stte temperture field tht results? You my use without proof the result { 2π ln(β 2, β 2 < 1, + 1 2β cosψ) dψ = (3.59) 2πβ 2, β 2 > 1. Since we cn ignore boundry effects we cn ssume tht the circulr domin C hs its centre t the origin of our coordinte system, i.e. t x =. The problem is two dimensionl nd the field tht results (fter trnsients hve decyed) is simply u(x) = G(x,x )q dx C nd we pproximte the Green s function s the free-spce one since the question sys we cn ignore boundry effects. So we hve u(x) = q ln( x x ) dx. 2π Introduce the two polr coordinte systems C x = r cosθ, y = r sin θ, (3.6) x = r cosθ, y = r sin θ, (3.61) nd thus, upon simplifying the rgument we find u(r, θ) = q 4π = q 4π = q 4π 2π 2π 2π = q2 2 ln r + q 4π ln(r 2 + r 2 2rr cos(θ θ)) r dr dθ ( ) ln r r2 r 2r 2 r cos(θ θ) r dr dθ [ ( )] ln r 2 + ln 1 + r2 r 2r 2 r cos(θ θ) r dr dθ (3.62) 2π ( ) ln 1 + r2 r 2r 2 r cos(θ θ) r dr dθ (3.63) Let β = r /r nd ψ = θ θ so tht (dr = rdβ nd dψ = dθ ) u(r, θ) = q2 2 ln r + qr2 4π /r β 2π ln[β β cosψ] dβdψ (3.64)

70 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 7 where we note tht the limits of the θ integrl simplified becuse we could write 2π θ θ = θ 2π 2π + 2π θ nd since the integrnd is periodic with period 2π, the first nd lst terms on the right hnd side cncel. We now use the result given in the question, so tht if r >, the β integrl runs from to /r < 1 nd therefore is zero. Therefore for r >, u(r, θ) = (q 2 /2) lnr. If r <, the β integrl runs from to /r > 1 so we write it s qr 2 4π ( 1 + /r 1 ) β 2π ln[β β cosψ] dψdβ. The first term (integrl between nd 1) is zero (using the result (3.59)), the second, gin using the result (3.59) gin qr 2 /r 1 [ β ln β dβ = qr 2 β ] /r 2 β2 ln β 1 = q2 4 + q2 qr2 [ln ln r] (3.65) where we hve evluted this using integrtion by prts. Finlly when combined with the first term in (3.64), this reduces to nd therefore we hve q 4 (r2 2 ) + q2 2 ln q u(r, θ) = 4 (r2 2 ) + q2 ln, r, 2 q 2 (3.66) 2 ln r, r. which we note is continuous t r = on the boundry of the forcing disc C. Finlly we just hve to check tht the integrl t infinity from Green s identity does not diverge s described bove. If it diverges then the solution (3.4) is not the correct one nd our ssumptions from the outset re wrong. So we must see if (3.52) holds. Well it clerly does since with u = (q 2 )/2 lnr, lim R ( u r ln r u r ) r=r ( q 2 = lim lnr r ln r 1 R 2 r) r=r (3.67) =. (3.68) so the boundry term is not just zero in the limit, it is identiclly zero. We plot the solution in figure 8

71 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 71 Figure 8: Plot of the stedy stte temperture distribution u(x, y) in domin where boundries my be neglected with the het source inside the circulr domin x 2 + y 2 < 1. Here we hve = q = 1. Exmple 3.5 Consider the stedy stte het distribution governed by 2 u = on the semi-infinite domin D = { < x <, y } subject to the inhomogeneous boundry condition u(x, ) = h(x). Use the semi-infinite Green s function to determine the solution to the problem in integrl form nd determine the solution explicitly when { qx, x, h(x) = (3.69), otherwise for, q R, with >. Since we hve semi-infinite domin let us employ the semi-infinite Green s function tht we determined in (3.57) bove. We cn use the expression (3.39) derived in section with Q(x) =, i.e. u(x) = h(x )n x G(x,x) ds. (3.7) D

72 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 72 Firstly, we note tht the dummy vrible of integrtion is x. This is importnt! The boundry should be prtitioned s D = D D, i.e. one long the boundry y =, D nd one from infinity, D. See figure 9. We ssume tht the boundry integrl long D, i.e. the boundry contribution from infinity is zero. But we will check this lter! The orienttion of the integrl is nti-clockwise with the norml pointing outwrds. Therefore, since lso h(x ) = outside x, nd it is qx inside tht intervl, we hve u(x) = qx n x G(x,x) dx. (3.71) y = where we note tht s = x on D nd this integrl is long y = (figure 9. Next, we note tht n = (, 1) so tht x G(x,x) = y G(x,x) Next we use the Green s function tht we determined in Exmple 3.3, in the form G(x,x ) = 1 2π (ln x x ln x x 1 ) where x = (x, y ) with y >, nd x 1 = (x, y ). We hve G(x,x) = G(x,x ). Therefore G(x,x) = 1 ( ) y y y 2π (x x ) 2 + (y + y ) y + y 2 (x x ) 2 + (y y ) 2 nd we hve to evlute this on y =. G(x,x) y = 1 ( ) y y = 2π (x x ) 2 + y y 2 (x x ) 2 + y 2 y = π((x x ) 2 + y 2 ) Finlly then u(x) = qy π x ((x x ) 2 + y 2 ) ds. (3.72) where we hve tken qy outside the integrl. You will hopefully remember how to evlute integrls of this form from the first yer! Wht we hve to do is to re-write it by using the stndrd trick of dding zero, i.e. u(x) = qy π = qy π = qy π x ((x x ) 2 + y 2 ) dx. (3.73) x x x ((x x ) 2 + y 2 ) dx. (3.74) x x ((x x ) 2 + y 2 ) dx + qxy 1 π ((x x ) 2 + y 2 ) dx. (3.75)

73 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 73 y D R n D x Figure 9: Figure illustrting the decomposition of the boundry D into contribution from D, i.e. from infinity (s R ) nd one on y =, i.e. D. The first integrl is qy π nd let x x ((x x ) 2 + y 2 ) dx = qy [ ln((x x ) 2 + y 2 ) ] 2π The second integrl is qxy π = qy 2π = qy 2π ln 1 ((x x ) 2 + y 2 ) dx = qx πy (x + ) 2 + y 2, (3.76) [ ln((x ) 2 + y 2 ) ln((x + ) 2 + y 2 ) ], (3.77) ( ) (x ) 2 + y 2. (3.78) 1 (1 + (x x) 2 /y 2 ) dx p = (x x) y so tht dx = y dp nd the lower nd upper limits become ( x)/y nd ( x)/y respectively: qxy π ( x)/y 1 ((x x ) 2 + y 2 ) dx = qx 1 dp (3.79) π ( x)/y (1 + p 2 ) = qx [ ] ( x)/y rctn(p) (3.8) π ( x)/y = qx [ ( ) ( )] ( x) x rctn rctn (3.81) π y y = qx [ ( ) ( )] (x + ) x rctn rctn (3.82) π y y where the lst step used the fct tht rctn is n odd function. So combing (3.78) nd (3.82), the solution is u(x) = qy ( ) ((x ) 2 2π ln + y 2 ) + qx [ ( ) ( )] (x + ) x rctn rctn ((x + ) 2 + y 2 ) π y y (3.83)

74 Willim J. Prnell: MT3432. Section 3: Green s functions in 2 nd 3D 74 Figure 1: Plot of the stedy stte temperture distribution u(x, y) in the semi-infinite domin y with boundry distribution h(x) given by (3.69). Let us check tht this recovers the right behviour s y. The first term tends to zero so we eliminte tht stright wy from our investigtions. The behviour of the second term is little more subtle. Let us fix q = = 1 s in Figure 1. We hve to consider three different domins for x. We study F(x, y) = x [ ( ) ( )] (x + 1) x 1 rctn rctn π y y in the limit s y + (i.e. from bove). We should find tht lim F(x, y) = h(x) (3.84) y + First consider x < 1. In this cse the rguments of rctn become (x + 1) lim, y + y (3.85) (x 1) lim, y + y (3.86) so both rctns yield π/2 nd the respective contribution cncels, s is required for h(x).