STURMLIOUVILLE BOUNDARY VALUE PROBLEMS


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1 STURMLIOUVILLE BOUNDARY VALUE PROBLEMS Throughout, we let [, b] be bounded intervl in R. C 2 ([, b]) denotes the spce of functions with derivtives of second order continuous up to the endpoints. Cc 2 ([, b]) is the subspce of functions tht vnish ner the endpoints. Let L denote second order differentil opertor of the form (1) Lu(x) = r(x)u (x) + r (x)u (x) + q(x)u(x) = d ( r(x) du ) + q(x)u(x). dx dx We ssume tht r C 1 ([, b]) nd q C 0 ([, b]) re rel, nd tht r(x) c for some c > 0. The opertor L is the most generl second order rel ODE which is formlly selfdjoint on L 2 (dx), in tht (Lu) v dx = u (Lv) dx u, v C 2 c ([, b]). The condition u, v C 2 c ([, b]) implies tht when integrting by prts the boundry terms vnish. Since L hs rel coefficients, conjugting v or not does not ffect the definition. For generl u, v C 2 ([, b]), (2) (Lu)v u(lv) dx = r ( u v u v ) b nd we need to impose first order conditions on u, v t the endpoints to mke the right hnd side vnish. A boundry condition B is n expression of the form Bu = αu() + βu(b) + γu () + δu (b) for rel constnts α, β, γ, δ. We will impose two conditions B 1 u = 0 nd B 2 u = 0 where B 1 nd B 2 re independent (i.e. the corresponding vectors (α, β, γ, δ) re independent), chosen to gurntee tht the right hnd side of (2) vnish. Definition 1. The boundry conditions B 1, B 2 re selfdjoint for L if, for ll u, v C 2 ([, b]) which stisfy B 1 u = B 2 u = B 1 v = B 2 v = 0, then (Lu) v dx = u (Lv) dx. In other words, the vnishing of B j u nd B j v implies the righthnd side of (2) vnishes. 1
2 2 526/556 LECTURE NOTES Dirichlet conditions: B 1 u = u(), B 2 u = u(b). Neumnn conditions: B 1 u = u (), B 2 u = u (b). Robin conditions: B 1 u = u () αu(), B 2 u = u (b)+βu(b), α, β > 0. The bove re seprted boundry conditions, in tht B 1 is condition t nd B 2 is condition t b. Any pir of seprted conditions is selfdjoint for generl L. The most common nonseprted condition is Periodic conditions: B 1 u = u(b) u(), B 2 u = u (b) u (). These re selfdjoint for L if r(b) = r(). Another wy to stte selfdjointness is to consider the subspce C 2 B([, b]) = { u C 2 ([, b]) : B 1 u = B 2 u = 0 }. Then (L, B 1, B 2 ) is selfdjoint provided tht Lu, v = u, Lv for u, v CB 2 ([, b]), where u, v = u v dx. We next fix positive weight function ρ(x) C 2 ([, b]), so ρ(x) c > 0 for x [, b], nd consider the SturmLiouville eigenvlue problem Lu = λ ρ u, B 1 u = B 2 u = 0. We sy tht the number λ is n eigenvlue if there is nonzero solution u C 2 ([, b]) to this eqution, nd cll u n eigenfunction. Lemm 2. Let (L, B 1, B 2, ρ) be selfdjoint SturmLiouville system.. The ssocited eigenvlues re ll rel numbers. b. Eigenfunctions ssocited to different eigenvlues re orthogonl in the inner product b u, v ρ = u(x) v(x) ρ(x) dx. c. The dimension of ech eigenspce is t most 2; if the boundry conditions re seprted then it is exctly 1. Proof. Note tht n eigenfunction u is n eigenvector for the opertor ρ 1 L, i.e. ρ 1 Lu = λu, nd tht ρ 1 Lu, v ρ = u, ρ 1 Lv ρ, so tht ρ 1 L is selfdjoint on the domin C 2 B ([, b]) with respect to the inner product, ρ. The proof of nd b then follow exctly s for finite dimensionl opertors. For c, we note tht the spce of solutions to (L λρ)u = 0 is 2dimensionl subspce of C 2 ([, b]). If one imposes seprted condition B 1 u = 0, this restricts the initil conditions ( u(), u () ) to 1dimensionl spce, hence there is t most 1dimensionl spce of solutions to (L λρ)u = 0 with the boundry conditions imposed. Theorem 3. Given selfdjoint SturmLiouville system s bove, there is n orthonorml bsis for the spce L 2 ρ([, b]) consisting of eigenfunctions for the SturmLiouville problem. The eigenvlues stisfy λ n.
3 526/556 LECTURE NOTES 3 Here, L 2 ρ([, b]) is the spce of mesurble u on [, b] such tht u 2 L = u(x) 2 ρ(x) dx <. 2 ρ Since ρ(x) is bounded bove nd below, this is the sme spce of functions s L 2 ([, b]), but the norm nd inner product, re different. The mp ρ u ρ 1 2 u is esily seen to be unitry mp of L 2 ρ onto L 2 : ρ 1 2 u L 2 = u L 2 ρ. In prticulr, {u j } j=1 is n orthonorml bsis for L2 ρ iff {ρ 1 2 u j } j=1 is n orthonorml bsis for L 2. We will prove Theorem 3 in the cse of seprted boundry conditions for simplicity, but it holds for generl selfdjoint boundry conditions. We strt the proof by reducing to the cse where ρ = 1. Consider the opertor ( ) Lu = ρ 1 2 L(ρ 1 d r du 2 u) = + qu, q = ρ 1 2 L(ρ 1 2 ), dx ρ dx which is formlly selfdjoint on L 2 (dx), nd Lu = λρu iff L(ρ 1 2 u) = λ ρ 1 2 u. We lso define boundry conditions B j (u) = B j (ρ 1 2 u); it follows esily tht B 1, B 2 re selfdjoint for L. We conclude there is orthonorml bsis for L 2 (ρdx) of eigenfunctions for ρ 1 L stisfying B j iff there is n orthonorml bsis for L 2 (dx) consisting of eigenfunctions for L stisfying B j, where the bses re relted by multiplying by ρ 1 2. We thus ssume ρ = 1, nd consider the eigenfunction problem Lu = λu, where Lu = (ru ) + qu, nd we impose selfdjoint conditions B 1 u = B 2 u = 0. Lemm 4. If λ is n eigenvlue for Lu = λu, then λ C for some constnt C depending on (L, B 1, B 2 ). Proof. Integrting by prts we hve λ u 2 dx = (Lu)ū dx = r u 2 + q u 2 dx + r(x)u(x)u (x) x=b x=. For Dirichlet or Neumnn conditions, the boundry terms vnish, nd (3) λ u 2 dx q u 2 dx ( ) mx q [,b] u 2 dx. For Robin conditions u () = αu(), u (b) = βu(b), we get λ u 2 dx = r u 2 + q u 2 dx αr() u() 2 βr(b) u(b) 2.
4 4 526/556 LECTURE NOTES In the physiclly relistic cse α, β 0, then (3) still pplies. If one or both is negtive, we need more work. We bound mx [,b] u2 min [,b] u2 d dx u2 dx = 2 u u dx ɛ Tking ɛ smll, for C sufficiently lrge we hve nd then λ α r() u() 2 + β r(b) u(b) 2 u 2 dx (q + C ) u 2 dx u 2 + 2ɛ 1 u 2 dx. r u 2 + C u 2 dx ( ) C + mx q [,b] u 2 dx. We replce L by L λ 0 with λ 0 = 1+C, which hs the sme eigenfunctions, but with eigenvlues shifted by λ 0. We my thus ssume ll eigenvlues stisfy λ 1, nd in prticulr (4) Lu = 0, B 1 u = B 2 u = 0, implies u = 0. We produce eigenfunctions for L by finding eigenfunctions for the opertor L 1, which we express s n integrl kernel. Thus, we seek to express the solution to in the form Lu = f, B 1 u = B 2 u = 0, where f C([, b]), (5) u(x) = G(x, y) f(y) dy. The function G(x, y) is clled Green s kernel for the problem (L, B 1, B 2 ). We will pply vrition of prmeters: let u 1 nd u 2 be nonzero rel solutions to Lu 1 = Lu 2 = 0, B 1 u 1 = 0, B 2 u 2 = 0. Since B 1 nd B 2 re seprted, then u 1 nd u 2 re determined up to constnt multiple. Furthermore, u 1 nd u 2 re linerly independent; otherwise they re both solutions to (4). Thus [ ] u1 (x) u 2 (x) W (x) = det u 1 (x) u 2 (x) 0. By Abel s theorem, rw + r W = 0, so rw = constnt. The vrition of prmeters method sttes tht if [ ] [ ] [ ] u1 (x) u 2 (x) c 1 (x) 0 u 1 (x) u 2 (x) c 2 (x) = f/r
5 526/556 LECTURE NOTES 5 then u = c 1 u 1 + c 2 u 2 solves Lu = f. This hs solution c 1 = u 2f rw, c 2 = u 1f rw. Since B 1 is first order opertor t, then B 1 u = c 1 ()B 1 u 1 + c 2 ()B 1 u 2, nd this vnishes if c 2 () = 0. Similrly B 2 u = 0 if c 1 (b) = 0. Thus we set c 1 (x) = Then (5) holds, where x u 2 (y) rw f(y) dy, c 2(x) = G(x, y) = x u 1 (y) rw { (rw ) 1 u 1 (x) u 2 (y), y x, (rw ) 1 u 2 (x) u 1 (y), x y. f(y) dy. Note tht G(x, y) = G(y, x), nd tht G(x, y) is continuous on [, b] [, b]. Furthermore, G is rel since u 1 nd u 2 re. The kernel G is lso left inverse for L, in tht if v CB 2 ([, b]), then v(x) = G(x, y) (Lv)(y) dy. This follows by uniqueness of solutions (4). In prticulr, if one considers the mps L : C 2 B([, b]) C([, b]), G : C([, b]) C 2 B([, b]), then these mps re respectively the inverse of ech other. It follows tht G is 11 on the spce of continuous functions, but we need stronger result for the digonliztion rgument. Lemm 5. Suppose tht f L 2 ([, b]), nd tht G(x, y) f(y) dy = 0 for ll x. Then f(y) = 0.e. Proof. We will show tht f(y)φ(y) dy = 0 for ll φ Cc 2 ([, b]), nd the result follows by density of Cc 2 in L 2. We then write ( ) f(y) φ(y) dy = f(y) G(y, x) (Lφ)(x) dx dy using Fubini s theorem. = ( G(x, y) f(y) dy ) (Lφ)(x) dx The opertor T f(x) = G(x, y)f(y) dy is then selfdjoint, compct opertor on L 2 ([, b]), nd 0 is not n eigenvlue of T. There thus exists n orthonorml bsis {u j } j=1 for L2 ([, b]), where G(x, y) u j (y) dy = ν j u j (x), ν j 0. Since the left hnd side is continuous in x so is u j (x), nd thus the left hnd side belongs to CB 2 ([, b]), nd so u j CB 2 ([, b]). We then hve Lu j = λ j u j, λ j = νj 1,
6 6 526/556 LECTURE NOTES nd by Lemm 2 ech eigenspce is 1dimensionl (seprted boundry conditions). We hve rrnged tht λ j 1, which mens we cn order them so tht λ j in decresing mnner. The eigenvlues for the originl problem re λ j + λ 0, which still decrese to, but there my be finitely mny positive eigenvlues, depending on q nd the boundry conditions.
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