STURM-LIOUVILLE PROBLEMS

STURM-LIOUVILLE PROBLEMS Mrch 8, 24 We hve seen tht in the process of solving certin liner evolution equtions such s the het or wve equtions we re led in very nturl wy to n eigenvlue problem for second order liner differentil opertor with two boundry conditions. In this chpter we shll study collection of such eigenvlue problems. Let (, b) be n intervl on the rel line, nd consider the following problem: Lu := (p(x)u ) + q(x)u = λ w(x)u, < x < b, (1) B u := A 1 u() + A 2 u () + 1 u(b) + 2 u (b) =, (2) B b u := b 1 u() + b 2 u () + B 1 u(b) + B 2 u (b) =. (3) We note tht L, B nd B b re liner opertors nd hence stisfy the Superposition Principle. This is n importnt nd useful fct. Next we impose some restrictions on the intervl (, b) nd the coefficients occurring in the opertors L, B, nd B b. Assumptions. The intervl (, b) is bounded, i.e. < < b <. All coefficients re rel. Moreover, the functions p, p, q, nd w re continuous on [, b] while p(x) > nd w(x) > on [, b]. We lso ssume tht the two boundry conditions re linerly independent, tht is to sy tht B is not multiple of B b nor is B b multiple of B. Equivlently, the vectors (A 1, A 2, 1, 2 ) nd (b 1, b 2, B 1, B 2 ) re linerly independent. When these ssumptions re stisfied then we sy tht equtions (1)-(3) constitute regulr Sturm- Liouville problem (RSLP). When the boundry conditions hve the form: B u = A 1 u() + A 2 u () =, (4) B b u = B 1 u(b) + B 2 u (b) =, (5) i.e. the coefficients denoted with lower cse letters re ll zero, then we sy tht the boundry conditions re seprted boundry conditions. When the boundry conditions hve the form B u = u() u(b) =, (6) B b u = u () u (b) =, (7) 1

we sy tht the boundry conditions re periodic boundry conditions. The reson for this terminology is tht continuously differentible function on the rel line tht is periodic with period b stisfies (6)-(7). The following is simple but useful little result. Lemm. (Green s first nd second identities) If u C 2 (, b) C 1 [, b] then: (I) v (Lu) dx = p u v b + {p u v + q u v} dx (II) [v (Lu) u (Lv)] dx = [p u v p u v b. Proof. The proof essentilly consists of nothing more thn integrting by prts. v (Lu) dx = Now, by symmetry we see tht [ v (pu ) + v q u] dx = p u v b + {p u v + q u v} dx u (Lv) dx = [ u (pv ) + u q v] dx = p v u b + {p v u + q v u} dx Subtrcting this lst eqution from (I) we obtin (II). Remrk. Note tht the requirement f C 2 (, b) sys nothing bout u t the end points, while u C 1 [, b] implies tht u () nd u (b) re defined s one-sided derivtives. Definition. The RSLP (1)-(3) is sid to be selfdjoint if whenever u, v C 2 (, b) C 1 [, b] nd stisfy the boundry conditions (2) nd (3) then [v (Lu) u (Lv)] dx =. By the second Green s identity this is equivlent to sying tht whenever u nd v re two functions tht belong to C 2 (, b) C 1 [, b] nd stisfy the boundry conditions B u = B v = B b u = B b v =, then [p u v p u v] b =. 2

Lemm. A RSLP with seprted boundry conditions is selfdjoint. periodic boundry conditions is selfdjoint. A RSLP with p() = p(b) nd with Proof. We must show tht [p u v p u v] b =. This is obvious if u nd v stisfy periodic boundry conditions, so from now on let us suppose tht u nd v stisfy seprted boundry conditions: nd A 1 u() + A 2 u () = (8) B 1 u(b) + B 2 u (b) = (9) A 1 v() + A 2 v () = (1) B 1 v(b) + B 2 v (b) =. (11) It is cler tht [p u v p u v] b = if u v u v = t both endpoints nd b. Regrding eqution (8) nd (1) s homogeneous system of liner lgebric equtions for A 1 nd A 2 we see tht since A 1 nd A 2 re not both zero it must be true tht the determinnt of the coefficient mtrix must be zero. But this determinnt is u() v () u () v(). A similr rgument for equtions (9) nd (11) shows tht u(b) v (b) u (b) v(b), must be zero. Definition. A sclr λ is sid to be n eigenvlue for the Sturm-Liouville problem (1)-(3) if there exists function u C 2 (, b) C 1 [, b] such tht u is not identiclly zero nd u stisfies (1)-(3). The function u is clled n eigenfunction nd the pir (λ, u) is clled n eigenpir. The next theorem proves couple of fundmentl results (prts (i) nd (ii) ) bout eigenpirs for the RSLP. The third nd fourth prts of the theorem re not very importnt but cn occsionlly sve us little bit of work. Theorem 1. Suppose tht (1)-(3) is regulr selfdjoint Sturm-Liouville problem, then: (i) If (λ, u) is n eigenpir then λ is rel. (ii) If (λ, u) nd (µ, v) re two eigenpirs with λ µ then < u, v > w =. (iii) If the boundry conditions re seprted nd A 1 A 2 nd B 1 B 2 then for ny eigenpir (λ, u) we hve min{q(x) : x b} λ mx{w(x) : x b}. (12) (iv) Under the hypotheses of (iii) the inequlity is strict (>) unless ll of the following hold: q constnt, w constnt, nd A 1 = B 1 = (Neumnn problem). 3

Proof. (i) Since ll the coefficients re rel we get upon tking the complex conjugte of (1)-(3): Lu = λ w u, B u, B b u =. This mens tht (λ, u) is lso n eigenpir. By Green s second identity we hve [ulu ulu] dx = ( λ λ ) wuu dx. By selfdjointness the left hnd side is zero while the right hnd side is (λ λ) u 2 w. But since u is not identiclly zero this mens tht λ = λ. (ii) First we note tht vlu = λwvu nd ulv = µwuv. Tking the difference of these two equtions nd integrting we obtin [vlu ulv] dx = (λ µ) wvu dx = (λ µ) < v, u > w. Agin, by virtue of selfdjointness the left hnd side is zero, implying the orthogonlity of u nd v. iii) By Green s first identity we hve If we cn show tht nd λ wuu dx = then λ w u 2 dx q u 2 dx, so tht u (Lu) dx = p u u b + {p u u + q u u} dx p()u ()u() p(b)u (b)u(b) + λ q u 2 dx w u 2 dx q u 2 dx. u()u () (13) u(b)u (b), (14) min{q(x) : x b} mx{w(x) : x b} nd we re done. It remins to prove (13) nd (14). We will prove (13) nd leve (14) s n exercise for the reder. From the boundry condition t we hve A 1 u() + A 2 u () =, (15) A 1 u() u() + A 2 u () u() =, (16) If A 2 = then A 1 nd we hve from (15) tht u() = (since both coefficients cn not be zero). Therefore u () u() = nd hence (13) is stisfied. On the other hnd, if A 2 then from (16) it follows tht u () u() = (A 1 /A 2 ) u() 2, where the lst inequlity follows from the ssumption A 1 A 2. 4

Exercise 2. Complete the proof of (iii) by showing tht u(b)u (b). Exercise 2b. Prove prt (iv) of theorem 1. Hint: Trce crefully through the proof of (12). Exmple 1. Let us use the bove theorem to help us find ll the eigenvlues nd eigenfunctions for the problem u = λu, < x < L, (17) u() =, u (L) =. (18) First we observe tht this is regulr Sturm-Liouville problem. Since the problem hs seprted boundry conditions it is lso selfdjoint. By theorem 1 ll eigenvlues must therefore be rel. Moreover, since in this cse A 1 = 1, A 2 =, B 1 =, nd B 1 = 1, we hve A 1 A 2 = nd B 1 B 2 =, so tht we my conclude tht ll eigenvlues stisfy (12). In the present cse we hve q(x) 1 nd w(x) 1, so we see tht ll eigenvlues re nonnegtive (ctully we my use exercise 2b to conclude tht ll eigenvlues re positive). However, let us suppose tht ll we know is tht the eigenvlues re rel. In tht cse we must look t three cses. Cse λ < : We my write λ = ω 2 with ω > so tht the differentil eqution becomes u = ω 2 u. This hs the generl solution α e ωx + β e ωx or equivlently α cosh(ωx) + β sinh(ωx). In this prticulr cse the second form is more convenient. We esily see tht the boundry condition t requires α =. Hence we my ssume tht u = sinh(ωx) - once we know α = it is not necessry to crry long the remining coefficient since we know tht if u is n eigenfunction then so is ny multiple of u. Imposing the boundry condition t L we obtin ω cosh(ωl) =. But this is impossible since neither fctor ω or cosh(ωl) cn be zero. Cse λ = : Now the differentil eqution is simply u = which hs generl solution u = α + βx. The two boundry conditions yield α = nd β =, nd once gin we hve no eigenfunction. Cse λ > : We my write λ = ω 2 with ω >. The differentil eqution u = ω 2 u hs generl solution u = α cos(ωx) + β sin(ωx). Imposing the boundry condition t tells us tht α =, so tht we my ssume tht u = sin(ωx). Imposing the second boundry condition yields ω cos(ωl) =. This requires tht ω = (2n 1)π/2L for some integer n >. We denote these llowed vlues of ω by ω n : ω n = (2n 1)π/2, n = 1, 2,, λ n = ω 2 n, n = 1, 2,. The eigenfunctions re functions of the form β n sin(ω n x). However the coefficients re rbitrry nd do not crry ny useful informtion, so we my consider just the eigenfunctions φ n (x) = sin(ω n x), n = 1, 2,. Exercise 3. Show by direct integrtion tht the eigenfunctions in the bove exmple re mutully orthogonl, i.e. show tht whenever m n. L sin(ω n x) sin(ω m x) dx = In the bove exmple we sw gin tht the eigenvlues form n infinite discrete collection. The next theorem tells us tht this is typicl. But first nother definition: We sy tht set is countbly infinite if 5

its elements cn be put in one-to-one correspondence with subset of the nturl numbers {1, 2, 3,...}. This is equivlent to sying tht the elements of the set my be indexed by nturl numbers. Theorem 2. The eigenvlues of the RSLP (1)-(3) with seprted boundry conditions form countbly infinite set of rel numbers tht my be rrnged s n incresing sequence: λ < λ 2 < with lim λ n = +. n For ech λ n there is unique (up to sclr multiple) eigenfunction. The eigenfunctions my be tken to be rel nd they form complete orthogonl fmily (i.e. bsis) in the spce L 2 w(, b). The proof of this theorem is beyond the scope of these notes. Its consequences, however, re crucilly importnt to much of wht follows. We see tht corresponding to ech RSLP with seprted boundry conditions we cn find n orthonorml bsis for the spce L 2 w(, b). Such bsis my then be used to form corresponding (generlized) Fourier series. For exmple the usul Fourier sine series on (, π) corresponds to the RSLP u = λu, < x < π with seprted boundry conditions u() = u(π) =. In generl, corresponding to the selfdjoint RSLP (1)-(3) we hve the generlized Fourier series on (, b): f(x) c n φ n (x), n=1 where c n = f(x) φ n(x) dx φ n(x) 2 dx For exmple, for the selfdjoint RSLP (17)-(18) we hve the generlized Fourier series on (, L) given by f(x) c n cos(ω n x) n=1 where c n = L L f(x) cos(ω nx) dx = 2 cos2 (ω n x) dx L L f(x) cos(ω n x) dx Remrk on the cse of periodic boundry conditions. Wht bout RSLP s which do not hve seprted boundry conditions but re selfdjoint (like the cse of periodic boundry conditions)? We do know, from theorem 1, tht the eigenvlues re rel nd tht eigenfunctions corresponding to different eigenvlues re orthogonl. However it my hppen tht there re two linerly independent eigenfunctions corresponding to the sme eigenvlue. Since the Sturm-Liouville problem is second order we know tht it is impossible to hve more thn two linerly independent solutions irrespective of the boundry conditions! When we hve two linerly independent solutions u 1 nd u 2 corresponding to the sme eigenvlue λ nd 6

both u 1 nd u 2 stisfy the boundry conditions B u 1 = B b u 1 = B u 2 = B b u 2 =, then we cn find two mutully orthogonl eigenfunctions v 1 nd v 2 corresponding to the eigenvlue λ. We do this by the so-clled Grm-Schmidt process: ( ) < u2, u 1 > w v 1 = u 1, v 2 = u 2 u 1, < u 1, u 1 > w where <, > w denotes the L 2 inner product with weight w, i.e. < φ, ψ > w = φ(x)ψ(x) w(x) dx. This mens tht corresponding to ech eigenvlue we hve either one linerly independent eigenfunction or we hve two mutully orthogonl eigenfunctions. The totlity of ll these eigenfunctions forms n orthogonl fmily in L 2 w(, b) nd it cn be shown tht they form complete set of eigenfunctions (i.e. n orthogonl bsis) for L 2 w(, b). The eigenvlues cn gin be shown to form countbly infinite set which tends to infinity. In summry: ll sttements of theorem 2 re still vlid for the cse of generl selfdjoint RSLP s except the sttement bout there being unique linerly independent eigenfunction corresponding to ech eigenvlue. Tht sttement becomes the sttement tht corresponding to ech eigenvlue there is either one linerly independent eigenfunction or there re two mutully orthogonl eigenfunctions. The clssicl exmple of this sitution is the RSLP u (x) = λu(x), π < x < π, u( π) = u(π), u ( π) = u (π). In this cse the eigenvlues re λ n = n 2, n =, 1, 2,. Corresponding to the eigenvlue we hve only one linerly independent eigenfunction ( e.g. φ (x) 1, nd corresponding to the other eigenvlues λ n, n >, we cn find two mutully orthogonl eigenfunctions (e.g. sin(nx) nd cos(nx)). The totlity of these eigenfunctions form n orthogonl bsis for L 2 ( π, π). Note tht within the two-dimensionl eigenspce corresponding to ny positive eigenfunction we my, of course, prefer to choose the mutully orthogonl eigenfunctions e inx nd e inx in which cse our bsis becomes { e inx : n =, ±1, ±2,... } In this cse it must nturlly be ssumed tht we del with the liner spce of complex squre-integrble functions over the field of complex sclrs. We therefore see tht the theory of Sturm-Liouville equtions give us substntil generliztion of Fourier series. We cll the series expnsion in terms of the eigenfunctions of n RSLP generlized Fourier series or eigenfunction expnsions. Exercise 4. Prove tht v 1 nd v 2 s given bove, re two mutully orthogonl eigenfunctions. Given N linerly independent functions, {u 1, u 2,, u N } tht re squre integrble with respect to weight function w, design procedure to find N mutully orthonorml functions {v 1, v 2,, v N } such tht v k is liner combintion of {u 1, u 2,, u k }. The next question tht rises nturlly concerns pointwise nd uniform convergence of generlized Fourier series. Concerning this mtter we will stte, without proof, the following, rther comprehensive theorem. 7

Theorem 3. Let {(λ n, φ n )} n=1 be mximl collection of linerly independent eigenpirs for the RSLP (1)-(3) with seprted boundry conditions. Let f L 2 w(, b) nd let c n denote the n th generlized Fourier coefficient of f, nd consider the generlized Fourier series c n = < f, φ n > w < φ n, φ n > w, c n φ n (x). n=1 (i) This series converges to f in the men-squre with weight w, i.e. in the norm u w := < u, u > w. (ii)this series converges to 1 [f(x+) + f(x )] 2 for ny x (, b) provided tht f is piecewise smooth. (iii) If f C[, b] nd f is piecewise continuous nd if f stisfies both boundry conditions of the Sturm- Liouville problem, then the series converges uniformly on [, b]. (iv) If f is piecewise smooth nd continuous on the subintervl [α, β] (, b), then the series converges uniformly on the subintervl [α, β]. The study of Sturm-Liouville problems thus provides substntil generliztion of ordinry Fourier series. We conclude with severl exmples. Exmple 2. Consider the RSLP (xu ) = λ x 1 u, 1 < x < e, u(1) =, u (e) =. ) Find ll eigenpirs. b) Expnd the constnt function f(x) 1 in terms of the eigenfunctions. c) Discuss the convergence of the series obtined in b). d) Use b) nd c) to determine the vlue of 1 + 1/3 1/5 1/7 + 1/9 + 1/11 1/13 1/15 + 1/17 +. (19) Solution: ) Let us first observe tht by theorem 1 ll eigenvlues re rel nd nonnegtive. This mens tht we my represent ech eigenvlue λ s λ = ω 2 with ω : x 2 u + xu + ω 2 u =. This is the Euler eqution (lso sometimes clled the equidimensionl eqution). It is known to possess solutions of the form u = x r. Substituting this into the eqution we obtin x 2 r(r 1)x r 2 + xrx r 1 + ω 2 x r =. Upon simplifying this becomes (r 2 + ω 2 )x r =, 8

so tht we must hve r = ±iω, nd the corresponding solutions re Consequently the generl solution is given by u = x ±iω = e ±iω ln x = cos(ω ln x) ± i sin(ω ln x). u = A cos(ω ln x) + B sin(ω ln x). Next we need to stisfy the boundry conditions. At x = 1 we hve = u(1) = A nd t x = e we hve = u (e) = ω B e 1 cos(ω ln e). This cn only be true if either ω = or if ω is n odd multiple of π/2. But if ω = then u which cnnot be n eigenfunction. We therefore rrive t the following collection of eigenpirs: where b) f (λ k, φ k ) = (ω 2 k, sin(ω k ln x)), ω k = k=1 (2k 1)π, k = 1, 2,. 2 c k φ k, where c k = < f, φ k > w < φ k, φ k > w. In order to compute the generlized Fourier coefficient we obtin < φ k, φ k > w = e If we mke the substitution z = ω k ln x this integrl becomes Therefore ωk ω 1 k c k = e 1 ωk sin 2 z dz = (2ω k ) 1 [1 cos(2z)] dz = 1 x 1 sin 2 (ω k ln x) dx. 1 (2k 1)π ωk x 1 sin(ω k ln x) dx = 2ω 1 4 k sin z dz = (2k 1)π So we hve the generlized Fourier series 1 f k=1 ( 4 (2k 1)π (2k 1) π sin 2 [z 12 ] (2k 1)π/2 sin(2z) = 1/2. [ cos z)](2k 1)π/2 = ) ln x. 4 (2k 1)π. c) Since the function f 1 does not stisfy the boundry conditions we do not hve uniform convergence on the intervl [1, e], however we do hve pointwise convergence to 1 for ll x on the intervl 1 < x < e. Also, we hve uniform convergence on ny closed subintervl [α, β] with 1 < α < β < e. d) Since 1 < e < e, we see tht the bove series converges to 1 t x = e. This tells us tht 1 = k=1 ( 4 (2k 1)π (2k 1) π sin 4 9 ),

or π 4 = ( 1 (2k 1)π (2k 1) sin 4 k=1 Hence the series (19) sums to π 2/4. ) = 1 2 [1 + 1/3 1/5 1/7 + 1/9 +....]. Although the bove exmple provides nice illustrtion of Sturm-Liouville problem with vrible coefficients, most of the Sturm-Liouville problems we will encounter re those for which the differentil eqution is simply u = λu. However this does not men tht the problem is trivil, s the following exmples show. Exmple 3. Find ll the eigenpirs for the problem u = λu, < x < π, u() u () =, u(π) + bu (π) =, where we ssume tht > nd b >. From theorem 1 it follows tht the eigenvlues re nonnegtive. This mens tht we cn let λ = ω 2, with ω, nd the eqution becomes u = ω 2 u, which hs the generl solution u = A cos(ωx) + B sin(ωx), ω >, u = A + Bx, if ω =. First let us exmine the cse ω =. In this cse the boundry conditions imply tht A B = nd A + Bπ + Bb =. Substituting into the second eqution we get ( + b + π)b = nd therefore B = nd A = nd we hve no eigenfunction. When ω >, u (x) = ωa sin(ωx) + Bω cos(ωx) so tht the boundry conditions red A ωb =, A cos(ωπ) + B sin(ωπ) + b[ ωa sin(ωπ) + ωb cos(ωπ)] =, which my be written s ( 1 ω [cos(ωπ) bω sin(ωπ)] [sin(ωπ) + bω cos(ωπ)] ) ( A B In order for this eqution to hve nontrivil solution (i.e. solution besides the solution A = B = ) the determinnt of the coefficient mtrix must be zero: ) = ( sin(ωπ) + bω cos(ωπ) + ω cos(ωπ) bω 2 sin(ωπ). (2) Let us ssume tht cos(ωπ) nd let us divide the bove eqution by cos(ωπ). After rerrnging terms we then obtin ( + b) ω tn(ωπ) = bω 2 1. (21) Let us define ω = (b) 1/2 The solutions ω > of eqution (21) correspond to the intersections of the curves y = tn(πω) nd y = (+) ω/(bω 2 1) in the hlf plne ω >, < y < (see figure below). Plese note tht the second curve hs n symptote ω = ω. We lbel the first coordintes of the intersections with subscripts: ω 1 < ω 2 <. These vlues my be obtined by some numericl method such s Newton s Method. As fr s we re concerned we will consider the problem of finding the eigenvlues completed t this point. We hve λ k = ωk 2. ). 1

Fig. 6.1 Loction of the ω k s for exmple 3. The corresponding eigenfunctions hve the form φ k (x) = A cos(ω k x)+b sin(ω k x), however we sw from our boundry conditions tht A = ω k B so tht we my write φ k (x) = B [ ω k cos(ω k x) + sin(ω k x)]. Finlly, since we only need one eigenfunction per eigenvlue, we my set B = 1, so tht φ k (x) = ω k cos(ω k x) + sin(ω k x). There is one loose end to tke cre of. The previous rguments re bsed on the ssumption tht cos(ωπ). Is it possible to hve solutions ω of eqution (2) for which cos(ωπ) =? Clerly this would imply ωπ = (2m 1)π/2 for some positive integer m. Eqution (2) then implies tht sin(ωπ) = bω 2 sin(ωπ), where sin(ωπ) = ±1, nd hence my be cncelled to yield bω 2 = 1, so tht lso ω = ω. To summrize, in the exceptionl cse tht (b) 1/2 = m 1 2 for some positive integer m there will be nother eigenpir (λ, φ ) where λ = ω 2 = (m 1 2 )2, φ (x) = ω cos(ω x) + sin(ω x). Exmple 4. Let us find ll the eigenvlues nd eigenfunctions for the Sturm-Liouville problem u (x) = λu(x), < x < 1, u () =, u (1) = γu(1). Since the boundry conditions re seprted the problem is selfdjoint. In this cse the hypotheses of theorem 1-(iii) re not stisfied. Indeed, we hve B 1 = γ, B 2 = 1, so tht B 1 B 2 = γ <. This indictes we my expect negtive eigenvlue, nd we must consider ll three cses λ >, λ =, nd λ <. Let us begin with the esiest cse. Fig. 6.2 Loction of ω for exmple 4. The cse λ = : The solution is of the form A + Bx. When we impose the two boundry conditions we see tht B = nd B = γ(a + B). But this mens tht both A nd B must be zero. The cse λ < : Let λ = ω 2, with ω >, nd the differentil eqution becomes u = ω 2 u. The generl solution is u = A cosh(ωx) + B sinh(ωx), or equivlently u = e ωx + b e ωx. 11

Either form my be used, but in this cse the first form is little more convenient. Since u (x) = ωa sinh(ωx) + ωb cosh(ωx), the boundry condition t x = tells us tht B =, so tht u = A cosh(ωx). The boundry condition t x = 1 yields ω sinh(ωx) = γ cosh(ωx) evluted t x = 1. This yields tnh(ω) = γ/ω, ω >. Solutions of this correspond to intersections of the curves y = tnh ω nd y = γ/ω in the hlf plne ω >, < y <. As we see from the bove grph, there is single solution which we shll denote by ω. This yields the single eigenpir λ = ω, 2 φ (x) = cosh(ω x). Cse λ > : We set λ = ω 2 with ω >. The generl solution to the differentil eqution is now nd its derivtive is given by u(x) = A cos(ωx) + B sin(ωx), u (x) = ωa sin(ωx) + ωb cos(ωx). The boundry condition t x = implies B =. The boundry condition t x = 1 then yields the informtion ωa sin ω = γa cos ω or tn ω = γ/ω. As before, the positive solutions of this eqution correspond to the intersections of the curves y = tn ω nd y = γ/ω. We gin lbel these solutions with positive integers: < ω 1 < ω 2 <. This yields the eigenpirs (λ k, φ k ) with λ k = ωk 2, φ k(x) = cos(ω k x), where k = 1, 2, 3,. In this prticulr cse it ws more convenient to index the set of linerly independent eigenfunctions by the nonnegtive integers rther thn by the positive integers. We note Fig 6.3 Loction of the ω k s for exmple 4. tht if f L 2 (, 1) then we cn find the generlized Fourier series where the Fourier coefficients re given by f c cosh(ω x) + c = N 1 c k = N 1 k 1 1 c k cos(ω k x), k=1 f(x) cosh(ω x) dx, f(x) cos(ω k x) dx, k >, 12

where N = 1 cosh 2 (ω x) dx, N k = 1 cos 2 (ω k x) dx, k >. Wrning. These integrls cnnot be evluted without knowing the numericl vlues of ω k. N k is not simply 1/2 s one might hve guessed from our previous exmples! The integrl of cos 2 (ωx) over n intervl (, b) is equl to hlf its length, (b )/2, if the intervl contins precisely n integrl number of qurter-cycles, i.e. if 2ω(b )/π is positive integer. This is not true in generl. Exercise 5. Show tht N k = 1 2 [1 γ/(ω2 k + γ2 )]. Exmple 5. As our finl exmple we consider the Sturm-Liouville problem u (x) = λu(x), < x < 1 u() =, u (1) = γ u(1), γ > 1. Fig 6.4 Loction of ω for exmple 5. Agin we cnnot pply theorem 1-(iii) since B 1 B 2 <, nd so we hve to consider three seprte cses. Cse λ = : In this cse u = A + Bx. Imposing the boundry conditions leds to A = B =, so tht we do not hve n eigenfunction. Cse λ < : Now we cn set λ = ω 2 nd the differentil eqution becomes u = ω 2 u. Clerly, in view of the fct tht we must hve u() = the solution must hve the form u = B sinh(ωx). The second boundry condition tells us tht ω must be positive solution of tnh ω = ω/γ. Since γ > 1 this eqution hs exctly one solution ω (see figure bove). We therefore hve, s in the previous exmple, one linerly independent eigenfunction corresponding to negtive eigenvlue: λ = ω 2, φ = sinh(ω x). Exercise 6. Complete the bove exmple by finding ll the eigenpirs corresponding to positive eigenvlues. Exercise 7. Consider the eigenvlue problem u = λu, < x < L, u () + αu() =, u (L) + βu(l) =. 13

Show tht under pproprite restrictions on L, α, nd β it is possible to hve two negtive eigenvlues, but no more. We conclude this chpter with one very importnt observtion. Note tht in ll eigenvlue problems tht we hve looked t in this chpter, the boundry conditions hve been homogeneous. This is typiclly so nd is n essentil prerequisite for the vlidity of the Superposition Principle. 14