Boundary-value problems

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Chpter 10 Boundry-vlue problems The initil-vlue problem is chrcterized by the imposition of uxiliry dt t single point: if the eqution is of the nth order, the n otherwise rbitrry constnts in its solution re determined by prescribing the function nd its first n 1 derivtives ll t single vlue of the independent vrible. Boundry-vlue problems provide the uxiliry dt by providing informtion, not t single point, but t two points 1, usully t the endpoints (or boundry) of n intervl. They rise often in the solution of prtil-differentil equtions. Exmple 10.0.2 Consider the temperture u (x, t) in wll of thickness L whose interior, t x = 0, is mintined t temperture A nd whose exterior, t x = L, is mintined t temperture B. Initilly, t t = 0, the distribution of temperture is u = u 0 (x). Its subsequent evolution is governed by the prtil-differentil eqution (the het eqution) ρ (x) u t = x ( p (x) u x ). (10.1) Here ρ nd p re known functions depending on the construction of the wll, nd both my be ssumed positive. The problem of finding solution of eqution (10.1) together with the imposed initil dt nd boundry dt is n exmple of n initil-boundry-vlue problem. There is simple, time-independent solution of this eqution stisfying the boundry conditions: x 0 U (x) = A + (B A) p (s) 1 ds L 0 p (s) 1 ds. (10.2) 1 Or possibly more thn two, but the commonly rising cse is tht of two points. 227

228 If we write u = U (x) + v in eqution (10.1), we find tht v stisfies exctly the sme eqution (10.1), but its boundry conditions now tke the form v = 0 t x = 0 nd t x = L. Its initil vlue is v = v 0 (x) = u 0 (x) U (x). The method of seprtion of vribles is n ttempt to find solution of eqution (10.1) for v, stisfying the boundry conditions, in the form v (x, t) = X (x) T (t) of product of function of x only with function of t only. Substitution of this expression into eqution (10.1) results in the eqution ρ (x) X (x) T (t) = T (t) (p (x) X (x)). If one further ssumes tht the functions X nd T re not zero except possibly t isolted points, then by dividing either side of this lst eqution by ρxt one obtins T (t) T (t) = (p (x) X (x)) ρ (x) X (x). Since the left-hnd side depends on t only nd the right-hnd side on x only, they must be equl to the sme constnt, λ sy. This requires of these functions tht nd ( d p (x) dx ) λρ (x) X = 0 (10.3) dx dx dt dt = λt, (10.4) where λ is some s yet undetermined constnt. This will provide solution v = X (x) T (t) of the het eqution if eqution (10.3) hs solution stisfying the boundry dt X (0) = 0, X (L) = 0. This combintion of the differentil eqution (10.3) with boundry dt is clled boundry-vlue problem, to distinguish it from the initil-vlue problem. Unlike initil-vlue problems, boundry-vlue problems do not lwys hve solutions, s we shll see below (cf. Exmple 10.1.1). Even if the boundry-vlue problem consisting of eqution (10.3) does hve solution stisfying the boundry dt, it will not in generl provide solution to the originl initil-boundry-vlue problem, since the initil vlue T (0) X (x) will not in generl gree with v 0 (x). These deficiencies will be delt with below. Eqution (10.3) of the preceding exmple is liner, homogenous nd of the second order. We will mostly confine the discussion to this cse. We begin with certin lgebric properties of liner opertors.

229 10.1 Opertors nd their djoints Consider the generl, liner, homogeneous opertor M of the second order defined by the formul Mu f (x) u + g (x) u + h (x) u (10.5) on the intervl (, b). The coefficients f, g, h re ssumed to be C 2 (i.e, continuous together with their first two derivtives). If we multiply M u by n rbitrry C 2 function v, integrte over the intervl, nd use integrtion by prts to move derivtives of u to derivtives of v, we find where vmu dx = { fvu (fv) u + guv } b + (M v) u dx, M v (fv) (gv) + hv = fv + ( 2f g ) v + ( f g + h ) v. (10.6) The opertor M is sid to be djoint to the opertor M. between them cn be summrized s follows: The reltion (vmu um v) dx = { f ( vu uv ) + ( g f ) uv } b. (10.7) The form on the right, which is biliner in the vlues of u, u, v, v t the endpoints, is clled the biliner concomitnt. A cse of specil importnce is tht of the self-djoint opertor, M = M. For this to be true we must hve 2f g = g nd f g + h = h. This in turn holds precisely when g = f ; in this cse the opertor M tkes the form Mu d ( f (x) du ) + h (x) u. (10.8) dx dx This is the form of the opertor ppering on the left-hnd side of eqution (10.3), nd is of very common occurrence in problems of mthemticl physics, which is the origin of its specil importnce. For the self-djoint cse, eqution (10.7) cn be written (vmu umv) dx = { f ( vu uv )} b. (10.9)

230 The eqution (10.3) of the preceding exmple hs the form Mu = λρu where the opertor of eqution (10.8) is defined by f = p nd h = 0. The constnt λ tht rose in the course of seprting vribles my pper rbitrry, unrelted to the problem to be solved, but this is not so. It cnnot be ssigned rbitrrily, or the problem of solving eqution (10.3) together with the boundry conditions X (0) = 0 nd X (L) = 0 will hve only the trivil solution X 0. Exmple 10.1.1 In the Exmple 10.0.2, tke ρ (x) = p (x) = 1. Then eqution (10.1) tkes the form u t = u xx nd eqution (10.3) tkes the form X = λx. Solutions my be written X = c sin ( λx ) + d cos ( λx ), where we hve ssumed tht λ 0; this will be verified subsequently. Imposition of the boundry condition X (0) = 0 then implies tht d = 0, so imposition of the remining condition t x = L implies c sin ( L λ ) = 0. If c = 0, the solution X 0. To hve nonzero solution we must hve L λ = π, 2π,..., i.e., λ cnnot be rbitry but must tke one or nother of the discrete set of vlues λ n = n 2 π 2 /L 2. There re therefore infinitely mny seprtion solutions of the form v n (x, t) = e n2 π 2 t/l 2 sin (nπx/l). (10.10) Ech stisfies the eqution nd the boundry conditions. Since the eqution nd the boundry conditions re liner nd homogeneous, ny liner combintion with rbitrry coefficients {c n }, v (x, t) = c n v n (x, t), lso stisfies the eqution nd the boundry conditions. If this sum cn be mde to stisfy the initil condition v = v 0 (x), complete solution will be chieved. The ltter condition is written v 0 (x) = c n sin (nπx/l). If the summtion is llowed to run from 1 to, this represents the expnsion of the function v 0 in Fourier sine series. Under pproprite conditions on the function v 0, this is indeed possible by choosing the coefficients {c n } to be the Fourier coefficients of v 0, lthough the convergence of the series for v (x, t) requires further considertion.

231 In the preceding exmple, the vlues of the prmeter λ for which nontrivil solutions exist re clled eigenvlues, nd the corresponding functions sin ( λx ) re clled eigenfunctions. Eigenvlues nd their eigenfunctions ply centrl role in boundry-vlue problems. We therefore consider them in greter generlity. 10.2 Eigenvlue problems Consider the self-djoint differentil opertor on the intervl [, b], which we now write in the form Mu ( p (x) u ) + q (x) u = λρ (x) u, (10.11) where the coefficients stisfy the following conditions: p C 1 nd p > 0 on [, b]; q, ρ C on [, b] nd ρ > 0 on (, b). (10.12) This form of the opertor is generl enough to pply to problems rising in diverse pplictions. Tht the coefficient p is continuous nd positive on the closed intervl [, b] implies tht it is bounded wy from zero (i.e., tht p (x) δ for some δ > 0 for ech x [, b]), nd therefore the equtions Mu λρu = 0 nd Mu λρu = r cn be put into the stndrd form to which the theory of Chpter 2 pplies. The minus sign hs been introduced for lter convenience : pplied to Exmple 10.1.1, the new sign convention of eqution (10.11) would mke the eigenvlues turn out positive rther thn negtive. The integrl reltion (10.9) now tkes the form (vmu umv) dx = { p ( uv vu )} b. (10.13) The biliner concomitnt hs been reduced to the product of the Wronskin of the two functions multiplied by the coefficient p. The eqution (10.11) must be supplied with boundry conditions. These we will tke to be homogeneous boundry conditions generlizing those of Exmples 10.0.2 nd 10.1.1. The generl structure of homogeneous boundry conditions is: 11 u () + 12 u () + b 11 u (b) + b 12 u (b) = 0, nd 21 u () + 22 u () + b 21 u (b) + b 22 u (b) = 0. (10.14) Here the numbers { ij, b ij }, i = 1, 2, j = 1, 2 re prescribed: they determine the boundry dt. Rrely is one clled on to consider boundry dt

232 of such generlity. In the exmples bove, 11 nd b 21 re nonzero, nd ll the others vnish. But before pssing on to more specific cses, we mke couple of observtions tht follow from the liner nd homogeneous nture of both the eqution nd the boundry conditions. One is tht u 0 is lwys solution of eqution (10.11) together with the uxiliry dt (10.14). Another is tht if u is solution so lso is cu for ny rbitrry constnt c nd, more generlly, if u nd v re solutions so lso is ny liner combintion cu + dv with rbitrry constnts c nd d. Given n opertor M, we know how to construct the djoint opertor M, s in eqution (10.6) bove. We my consider, long with the boundryvlue problem Mu = λρu, < x < b nd conditions 10.14 (10.15) the relted boundry-vlue problem M v = µρv, < x < b nd conditions 10.17, (10.16) where the boundry conditions for this relted problem, 11v () + 12v () + b 11v (b) + b 12v (b) = 0, nd 21v () + 22v () + b 21v (b) + b 22v (b) = 0, (10.17) re similr in structure { to } equtions (10.14) but with possibly different vlues for the coefficients ij. Suppose now tht we cn choose these coefficients such tht, whenever v stisfies the conditions (10.17) while u stisfies the conditions (10.14), the biliner concomitnt vnishes, i.e., by eqution (10.7) bove, vmu dx = um v dx, (10.18) for ll C 2 functions u, v stisfying conditions (10.14), (10.17), respectively. In this cse, the eigenvlue problem (10.16) will be referred to s the djoint problem to the eigenvlue problem (10.14). Tht the djoint boundry conditions cn in fct be found such tht the biliner comcomitnt vnishes my be seen s follows. Let the four-component vector U be defined by the equtions U = (u(), u (), u(b), u (b)) = (u 1, u 2, u 3, u 4 ),

233 nd similrly for V : V = (v(), v (), v(b), v (b)) = (v 1, v 2, v 3, v 4 ). The pir of boundry conditions (10.14) my be written A 1 U = 0 nd A 2 U = 0, (10.19) where A 1 nd A 2 re ppropritely chosen four-component vectors. These must be linerly independent if the two equtions (10.14) re to represent two distinct boundry conditions, nd we my therefore, without loss of generlity, suppose they re orthogonl: A 1 A 2 = 0. We cn extend these to n orthogonl bsis A 1, A 2, A 3, A 4 for R 4, nd observe tht U = αa 3 + βa 4 by virtue of the boundry conditions (10.19). The biliner concomitnt, the right-hnd side of eqution (10.7), is biliner in U nd V. If we cll it C, we my write C = V KU where K is four-by-four mtrix whose entries cn be red off eqution (10.7). Therefore C = α (V KA 3 ) + β (V KA 4 ). The boundry conditions on u re stisfied for ny choices of α nd β so the vnishing of the biliner concomitnt C requires tht V KA 3 = 0 nd V KA 4 = 0. This implies there re t lest two linerly independent choices of the vector V tht result in the vnishing of the biliner comcomitnt, proving the clim. If the mtrix K is nonsingulr, there re precisely two, since the vectors KA 3 nd KA 4 re then linerly independent: reding K from eqution (10.7) shows tht it is indeed nonsingulr, so there re precisely two. We shll, beginning in the next section, restrict considertion to the cse when one of the boundry conditions refers to the left-hnd endpoint only, the other to the right-hnd endpoint only, but we provide here simple exmples of boundry conditions tht ech involve both endpoints. Exmple 10.2.1 Boundry-vlue problems re by no mens confined to equtions of the second order. A simple exmple is the first-order eqution u = λu on the intervl [0, 2π] with the single endpoint condition u(0) = 2u(2π). If the opertor L is defined by Lu = u then the djoint opertor is

234 defined by L v = v. It is esy to see tht the djoint boundry conditions is v(0) = (1/2)v(2π). The solution is u = u 0 exp (λx) nd the eigenvlue reltion is exp (2πλ) = 1/2. The permissible vlues of λ re therefore (ln 2/2π)+ik, where k tkes on integer vlues. Exmple 10.2.2 Consider the eqution u + λu = 0 on the intervl [0, 2π], with the boundry vlues u (0) u (2π) = 0 nd u (0) u (2π) = 0. This problem is self-djoint nd the djoint boundry conditions re the sme s those bove for u. The following ssertions cn be verified explicitly: The solution is identiclly zero unless λ is equl to one or nother of the eigenvlues 0, 1, 4,... k 2,.... If so, the corresponding solutions (the eigenfunctions) re 1; cos x, sin x; cos 2x, sin 2x;... ; cos kx, sin kx;..., where to ech eigenvlue k 2 except for k = 0 there correspond the pir of eigenfunctions cos (kx), sin (kx). By n eigenfunction we men nontrivil solution of homogeneous boundry-vlue problem. This does not exist for generl vlues of the prmeter λ: it requires tht λ hve one of specil set of vlues, i.e. n eigenvlue. The problem of finding these eigenvlues nd eigenfunctions we refer to s n eigenvlue problem. The definitions of the terms djoint nd self-djoint tht we hve used re used in functionl nlysis in mnner tht is relted but not identicl. There the notion of djointness is introduced in connection with tht of liner functionl on liner spce. A typicl such functionl my be written < v, u > where v lbels the functionl nd u rnges over the liner spce B. If M : B 0 B is liner opertor on the spce B 0, then < v, Mw > for w B 0 is gin liner functionl on B 0, which is denoted by < M v, w >. This defines the djoint mpping M : B B0: < v, Mu >=< M v, u >. The connection of this with our discussion bove is s follows. Suppose tht the functions w(x) we wish to consider re in the spce of C 2 functions on [, b] which stisfy the boundry conditions

235 (10.14). This is liner spce; cll it L. Now, for fixed function v(x) s yet unspecified, consider f v [u] = v(x)u(x)dx = (v, u) where u L, the spce of continuous functions on [, b]. This is esily seen to be liner functionl on L, for ny fixed choice of v. Then, noting tht M : L L for the opertor of eqution (10.5), put u = Mw If we further restrict v to be in C 2 nd to stisfy the djoint boundry conditions (10.17), then, with our definitions of M nd M, we derive (v, Mu) = (M v, u). Among the differences in the two pproches is one of generlity: we do not consider ll liner functionls on L, but only those defined by integrtion ginst certin specil clss of functions v. 10.3 The Sturm-Liouville problem We shll emphsize the fmily of boundry-vlue problems in which the endpoint conditions re seprted: i.e., there is one condition t ech endpoint, so the conditions (10.14) tke the form αu () + α u () = 0 nd βu (b) + β u (b) = 0, (10.20) where it is ssumed tht the coefficients α nd α re not both zero, nd likewise for β, β. The ssumption (10.20) excludes certin cses for which the endpoint conditions re not seprted but re mixed, s in the more generl conditions (10.14), or Exmples 10.2.2 nd 10.2.1 bove. The Sturm-Liouville problem is the eigenvlue problem consisting of the self-djoint eqution (10.11), under conditions on the coefficients given by eqution (10.12), together with the seprted endpoint conditions (10.20). It is the bsic eigenvlue problem of ordinry differentil equtions, nd tkes its nme from two of the mthemticins who developed its theory. We shll study the theory in greter depth in Chpter 11. In this section we estblish some of its bsic lgebric properties. A simple but useful observtion is the following. Lemm 10.3.1 Suppose two functions u nd v both stisfy the endpoint condition (10.20) t x =. Then their Wronskin vnishes there.

236 Proof: The ssumption mens tht αu () + α u () = 0, αv () + α v () = 0. This is system of two liner equtions for α nd α. The determinnt of this system is the Wronskin of the functions evluted t x =. If it were nonzero, we would rrive t the contrdiction tht α nd α both vnish. Remrks: A similr conclusion holds for the other endpoint x = b. If the functions u nd v re llowed to be complex-vlued functions of the rel vrible x, this conclusion is unchnged. We now esily find: Theorem 10.3.1 For the eigenvlue problem (10.11), (10.20) with seprted endpoint conditions, there is t most one linerly independent eigenfunction corresponding to single vlue of λ. Proof: Suppose there were two linerly independent solutions u nd v. They would both hve to stisfy the endpoint condition t x =, so their Wronskin would hve to vnish there. But they re both solutions of the eqution (10.11) nd if they were linerly independent their Wronskin could not vnish nywhere, hence they cnnot be linerly independent. Remrks: Exmple 10.2.2 bove, for which the endpoint conditions re not seprted, demonstrtes tht there my be two linerly independent eigenfunctions corresponding to single eigenvlue in tht cse. We infer tht for n eigenvlue problem with seprted endpoint conditions, ny eigenfunction is determined uniquely up to multipliction by n rbitrry constnt. Exmple 10.2.1 shows tht for some boundry-vlue problems, even though the coefficients in the eqution nd in the boundry conditions re rel, the eigenvlues nd eigenfunctions my be complex. We shll show first tht, for the Sturm-Liouville boundry-vlue problem, the eigenvlues re rel. To estblish this we must llow for the possibility tht they be complex. We continue to restrict considertion to rel

237 vlues of the independent vrible x, nd to rel vlues of the coefficient functions p, q, ρ nd of the prmeters α, α nd β, β of the boundry conditions. If λ is complex, the solution u of eqution (10.11) is necessrily lso complex: u (x) = u 1 (x) + iu 2 (x) where u 1 nd u 2 re rel. If we write u = u 1 iu 2 for the complex conjugte, we find Mu = λρu, wheres if u stisfies the boundry conditions (10.20) so lso does u. We therefore hve the two equtions Mu = λρu, Mu = λρu where both u nd u stisfy the boundry conditions (10.20). Applying the integrl condition (10.13) with v = u we find (umu umu) dx = ( λ λ ) ρ u 2 dx = { p ( uu uu )} b. The lst term vnishes in view of Lemm 10.3.1 nd we therefore find ( ) λ λ ρ u 2 dx = 0. If the integrl vnishes, then u 0 on [, b]: the integrnd is non-negtive nd continuous, so if it should be nonzero t ny point, there would be neighborhood of tht point throughout which it is positive, nd this neighborhood would mke positive contribution to the integrl, which would therefore be nonzero. Since u is n eigenfunction, it cnnot vnish identiclly nd we conclude tht the integrl is nonzero, nd therefore λ = λ, i.e., λ is rel. We hve proved Theorem 10.3.2 Any eigenvlue of the Sturm-Liouville problem is rel. In investigting eigenvlues, we cn now restrict our ttention to rel vlues of λ. We cn then lso restrict ttention to rel-vlued solutions u, since ny complex-vlued solution is obtined by multiplying rel-vlued solution by complex constnt. The integrl uv dx of the product of the functions u nd v represents n nlog of the inner product (or sclr product, or dot product) of vectors. It is n nlogy tht cn be pushed quite fr. It is therefore nturl to refer to two functions for which this integrl vnishes s orthogonl. Definition 10.3.1 The functions u nd v re orthogonl on (, b) with respect to the weight function ρ if u (x) v (x) ρ (x) dx = 0. (10.21)

238 Theorem 10.3.3 Eigenfunctions of the Sturm-Liouville problem belonging to different eigenvlues re orthogonl with respect to the weight function ρ. Proof: Let u nd v be eigenfunctions belonging to different eigenvlues λ nd µ respectively. Then (vmu umv) dx = (λ µ) uvρ dx = { p ( uv vu )} b = 0. The conclusion now follows since λ µ. In Exmple 10.1.1 bove, n importnt element ws the bility to represent n essentilly rbitrry function s n infinite sum of sine functions. More generlly it is likewise importnt to be ble to represent n essentilly rbitrry function f on (, b) in the form f (x) = c n u n (x) n=0 where {u n } is sequence of eigenfunctions of Sturm-Liouville problem. If we ssume such development is possible nd tht we cn interchnge summtion nd integrtion, then the orthogonlity of the eigenfunctions provides the coefficients in this sum: c n = f (x) u n (x) ρ (x) dx u n (x) 2 ρ (x) dx. PROBLEM SET 10.3.1 1. Refer to Exmple 10.0.2. Verify tht the function U (x) given there is indeed solution of eqution (10.1), tht if u = U + v then v indeed stisfies (10.1) if u does nd tht X (x) T (t) is indeed solution of (10.1) if X nd T re solutions of equtions (10.3) nd (10.4) respectively. 2. Work out (M ). 3. For the opertor M given by the formul (10.5), find function I (x) such tht IM is self-djoint. Wht do you need to ssume bout the coefficients of M?

239 4. In Exmple 10.1.1 there ppers to be n implicit ssumption tht λ < 0. Assume insted tht λ = µ 2, where µ is rel number. Cn you chieve nontrivil solution of the boundry-vlue problem of tht exmple? 5. Prove the conclusions sserted in Exmple 10.2.2. 6. Consider the mixed-endpoint boundry-vlue problem d ( p (t) du ) + q (t) u = λu, u (0) = u (T ), u (0) = u (T ), dt dt (10.22) where the coefficients p nd q re defined nd continuous for ll rel t nd periodic with period T : p (t + T ) = p (t) nd q (t + T ) = q (t); in ddition ssume tht p is positive nd C 1. Show tht ny solution stisfying the given dt is likewise periodic with period T. 7. Suppose λ = λ 1 + iλ 2 nd u (x) = u 1 (x) + iu 2 (x) represent complex eigenvlue nd eigenfunction, respectively, of the Sturm-Liouville problem M u = λρu together with the boundry conditions (10.20). Write this problem s pir of rel differentil equtions for u 1 nd u 2 nd obtin Theorem 10.3.2 in this wy. 8. Prove the sttement in the text following Theorem 10.3.2 tht if the eigenvlue λ is rel then ny complex eigenfunction is complex multiple of rel eigenfunction. 9. The eigenfunctions of Exmple 10.1.1 re u n (x) = sin (nπx/l). Show by explicit integrtion tht these re orthogonl on (0, L) with weight function ρ(x) 1. 10. Consider the fourth-order opertor Mu p 0 u (4) + p 1 u + p 2 u + p 3 u + p 4 u. Assume tht the coefficients p i, i = 0,..., 4, re sufficiently differentible nd tht p 0 does not vnish. Find the djoint opertor M by successive integrtions by prts. Find the conditions on the coefficients for M to be self-djoint, nd re-express the self-djoint opertor in the form ( ) Mu = d2 dx 2 P d2 u dx 2 + d ( Q du ) + Ru, dx dx

240 identifying the coefficients P, Q, R in terms of the originl coefficients {p i }. Show tht if the boundry dt re such tht the corresponding biliner concomitnt vnishes, i.e., if (vmu umv) dx = 0, then the eigenvlues of ny such self-djoint opertor re rel. 11. For Exmple 10.2.1 the opertor is M = d/dx nd M = M (verify this); M is sid to be skew-djoint. For the third-order liner opertor Mu p 0 u + p 1 u + p 2 u + p 3 u form the djoint opertor by successive integrtions by prts, nd show tht the opertor M cnnot be self-djoint, but cn be skewdjoint under pproprite conditions on the coefficients. Show further tht if the boundry dt re such tht the corresponding biliner concomitnt vnishes, i.e., if (vmu + umv) dx = 0, then the eigenvlues of ny such skew djoint opertor re pure-imginry. 12. Let {P n (x)} 0 be sequence of polynomils on [, b], with P n of degree n, nd suppose they re orthogonl with respect to positive weight function ρ. Show tht P n hs n zeros in [, b]. 2 10.4 Green s Functions Consider the generl problem of solving the nth-order, inhomogeneous, differentil eqution Lu = r, < x < b (10.23) where Lu p 0 (x) u (n) + p 1 (x) u (n 1) + + p n (x) u nd the coefficients r nd {p i } re given, continuous functions on [, b] nd p 0 does not vnish there. Now, however, we suppose the dditionl conditions 2 This describes the Legendre polynomils, which indeed stisfy differentil eqution of self-djoint form, but the proof should be independent of the theory of differentil equtions

241 on the solution u nd its derivtives to be given in the form of homogeneous boundry dt like those of eqution (10.14) rther thn initil dt. In prticulr, we suppose the boundry conditions to hve the form n ij u (j 1) () + b ij u (j 1) (b) = 0, i = 1, 2,..., n. (10.24) j=1 We shll find tht this boundry-vlue problem my fil to hve solution. However, if it does hve solution, it cn be expressed in the form u (x) = G (x, ξ) r (ξ) dξ, (10.25) where the function G (x, ξ) ppering in the integrnd depends on the opertor L nd the boundry dt, but not on the function r. The function G, clled the Green s function for this boundry-vlue problem, therefore plys similr role to tht plyed by the influence function for initil-vlue problems (cf. eqution 2.26). We outline the principl result in 10.4.2. It tkes its simplest form in the (importnt) cse of the Sturm-Liouville boundry-vlue problem, so we consider this in detil. 10.4.1 The Sturm-Liouville Cse Consider now the inhomogeneous eqution Mu λρu = r (10.26) with the sme definition of the opertor M nd the sme seprted boundry conditions s in 10.3; in prticulr, M is given by eqution (10.11). If λ is n eigenvlue for the homogeneous problem nd v the corresponding eigenfunction, then vr dx = v (Mu λρu) dx = { p ( uv vu )} b + u (Mv λρv) dx = 0, since the integrl vnishes becuse of the ssumptions on v nd the integrted prt vnishes becuse u nd v both stisfy the boundry conditions. We conclude tht eqution (10.26) cn fil to hve solution: if λ is n eigenvlue nd r is chosen so tht vr dx 0 (e.g., r = v), there cn be no solution. We then inquire whether there is lwys, i.e., for every continuous function r on [, b], solution of this boundry-vlue problem provided λ is

242 not n eigenvlue of the homogeneous problem. Here the nswer will be ffirmtive, nd we cn express the solution in the form u (x) = G (x, ξ) r (ξ) dξ. (10.27) The function G (x, ξ) is clled the Green s function, nd is the counterprt for boundry-vlue problems to the influence function nd the vrition-ofprmeters formul obtined erlier (Chpter 2) in the discussion of liner initil-vlue problems. In fct, our derivtion of the Green s function, to which we now proceed, will be mde to depend on tht formul. Let u 1 (x, λ) nd u 2 (x, λ) be ny bsis of solutions of the homogenous differentil eqution Mu λρu = 0. (10.28) Then there is lwys prticulr integrl of eqution (10.26) which my be written x ( ) u2 (ξ) u 1 (x) u 2 (x) u 1 (ξ) U (x) = r (ξ) dξ, (10.29) p (ξ) W (ξ) which is esily checked on writing the eqution (10.26) in the stndrd form u + p p u + λρ q p = r p. The solutions u 1 nd u 2 depend on the prmeter λ but since λ will not chnge in wht follows, we suppress this dependence in eqution (10.29) nd in other reltions below. Since the most generl solution of eqution (10.26) cn be written in the form u (x) = U (x) + c 1 u 1 (x) + c 2 u 2 (x) (10.30) for some choice of constnts c 1 nd c 2, this must lso be true for the solution of the boundry-vlue problem (10.26) with the boundry conditions (10.20). It is just mtter of clculting c 1 nd c 2. The clcultion is simplified somewht if we choose the bsis u 1, u 2 ppropritely. In prticulr, we ll choose u 1 to stisfy the left-hnd boundry condition nd u 2 to stisfy the right-hnd boundry condition. This cn lwys be done: for exmple, choose u 1 to solve the initil-vlue problem consisting of the differentil eqution (10.28) together with the initil dt u 1 () = α, u 1 () = α, nd similrly for u 2: it cn be chosen to be solution of eqution (10.28) with initil dt u 2 (b) = β, u 2 (b) = β. These

243 solutions re linerly independent: if they were linerly dependent we would hve u 1 (x) = Cu 2 (x) where C is constnt, nd then u 1 would stisfy both sets of boundry conditions nd hence be n eigenfunction, contrdicting the ssumption tht λ is not n eigenvlue. It s convenient to introduce shorthnd nottion for the boundry conditions. For ny C 1 function v on [, b] define A[v] = αv () + α v (), B[v] = βv (b) + β v (b). (10.31) In terms of these liner opertors A nd B the bsis functions u 1, u 2 stisfy A[u 1 ] = 0, B[u 2 ] = 0; A[u 2 ] 0, B[u 1 ] 0. Note lso tht our choice of prticulr integrl U is such tht U nd U both vnish t x = nd, s result A[U] = 0. Now pply the boundry conditions A[u] = 0 nd B[u] = 0 to the generl solution (10.30) bove. We find A[u] = c 2 A[u 2 ] = 0 nd B[u] = B[U] + c 1 B[u 1 ] = 0. This implies tht c 2 = 0 nd c 1 = B[U]/B[u 1 ]. But it is strightforwrd clcultion using the explicit form of U in eqution (10.29) tht u 2 (ξ) B[U] = B[u 1 ] r (ξ) dξ. (10.32) p (ξ) W (ξ) This gives for the solution u the formul x u (x) = identifying G (x, ξ) = u 1 (ξ) u 2 (x) b r (ξ) dξ p (ξ) W (ξ) x { u2 (x) u 1 (ξ) /p (ξ) W (ξ) if ξ < x u 2 (ξ) u 1 (x) /p (ξ) W (ξ) if ξ > x u 1 (x) u 2 (ξ) r (ξ) dξ, (10.33) p (ξ) W (ξ) }. (10.34) One cn now verify tht the formul (10.33) stisfies both the eqution (10.26) nd the boundry conditions (10.20). We now hve the following Theorem 10.4.1 Suppose λ is not n eigenvlue of the Sturm-Liouville problem (10.11), (10.20). Then, for ny continuous function r on [, b], there is unique solution of equtions (10.26), (10.20).

244 Proof: One solution u is provided by the formul (10.33). To see tht it is unique, suppose there were second v. Then the difference u v would stisfy equtions (10.11), (10.20) of the Sturm-Liouville problem. Unless u v is the zero solution, the ltter hs n eigenfunction, i.e., λ is n eigenvlue. This is contrdiction. There is nother chrcteriztion of the Green s function for this problem tht is often useful. Fix ξ in the intervl (, b). Then G is the unique, continuous function stisfying the boundry conditions t ech endpoint, stisfying the differentil eqution MG = 0 t ech point of the intervl except t x = ξ, nd stisfying lso the jump-discontinuity condition [ ] G = 1/p (ξ). x Here the brcket nottion mens, for function f (x), ξ [f] ξ = lim x ξ + f (x) lim x ξ f (x), i.e., the difference of the limiting vlues pproched from the right nd from the left. It is not difficult to check tht the explicit expression (10.34) stisfies these conditions. 10.4.2 The Generl Cse The ide of the Green s function cn be generlized beyond the Sturm- Liouville boundry-vlue problem, nd indeed beyond the theory of ordinry differentil equtions. In this brief section we merely outline the generliztion to the boundry-vlue problem consisting of the differentil eqution (10.23) together with the liner, homogeneous boundry dt (10.24). Note tht the sign convention for the opertor L of eqution (10.23) differs from tht of the preceding subsection. We suppose tht the coefficient p 0 does not vnish on the intervl [, b]. We suppose further tht the homogeneous problem Lu = 0 together with the boundry dt (10.24) hs only the trivil solution u 0 on [, b]. Under these conditions, one cn define function G (x, ξ) s follows: Definition 10.4.1 For fixed ξ in (, b), G stisfies the boundry dt (10.24), it stisfies the differentil eqution Lu = 0 except t x = ξ nd, t x = ξ [ k ] { G 0 if k = 0, 1,..., n 2 x k = 1/p 0 (ξ) if k = n 1. ξ

245 This function is the Green s function for the boundry-vlue problem, i.e., the (unique) solution of the ltter is given by the formul u (x) = PROBLEM SET 10.4.1 G (x, ξ) r (ξ) dξ. (10.35) 1. Consider the differentil eqution Mu λρu = 0 on [, b], but with the inhomogeneous boundry dt αu () + α u () = A, βu (b) + β u (b) = B. Convert this to n inhomogeneous differentil eqution of the form (10.26) with homogeneous boundry dt. 2. Verify eqution (10.32). 3. Verify tht the formul (10.33) provides solution to equtions (10.26) nd (10.20). 4. Show tht the denomintor p (ξ) W (ξ) in eqution (10.34) is constnt (i.e., independent of ξ). Under wht conditions does it vnish? 5. Find the Green s function for the eqution u + λu = r (x) on the intervl [0, π] with boundry conditions u (0) = 0 nd u (π) = 0. For wht vlues of λ does it become singulr? 6. Show tht the Green s function (10.34) is symmetric: G (x, ξ) = G (ξ, x). 7. Suppose tht λ is n eigenvlue with eigenfunction u 1 (x), nd suppose the condition u 1r dx = 0 is stisfied. Show tht if in eqution (10.33) the function u 1 is this eigenfunction nd u 2 is ny linerly independent solution of the differentil eqution, the inhomogeneous boundry-vlue problem (10.26), (10.20) is stisfied. 8. Verify tht the expression (10.35) solves the boundry-vlue problem (10.23), (10.24); show tht the solution is unique. 9. Consider the liner opertor Lu = u nd the single boundry condition αu() + βu(b) = 0. Under wht conditions on α nd β is it true tht the boundry-vlue problem Lu = 0 hs only the trivil solution? Under this condition, work out the Green s function for the solution of the inhomogeneous problem (10.23) with the sme boundry condition.

246 10. The sme s problem (9) but with Lu = u + u.

314

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