advertisement

Spectral Properties of Elliptic Operators In previous work we have replaced the strong version of an elliptic boundary value problem LßuÝxÞà = fÝxÞ BCßuÝxÞà = gÝxÞ x5U [email protected] with the weak problem find u 5 V such that Bßu, và = fÝvÞ -v 5 V. The connection between theweak problem and the strong problem, (and in particular the boundary conditions), was established by making use of the abstract Green’s formula Bßu, và = ÝLßuà, vÞ H + ÖT 0 v, T 1 u× -u, v 5 V. If the bilinear form B is V ? elliptic, Bßu, uà ³ c 0 || u|| 2V -u 5 V. then it follows immediately from the Lax-Milgram lemma that the weak problem has a unique solution for every f. This is the same thing as saying that L is an isomorphism from V onto H. On the other hand, if B is only V ? H ? coercive, Bßu, uà ³ c 0 || u|| 2V ? c 1 ||u|| 2H -u 5 V, then we are not able to apply this lemma directly to the weak problem. Instead, we replace the weak problem with find u 5 V such that B W ßu, và = fÝvÞ + WÝu, vÞ H -v 5 V. For W sufficiently large, B W is V-elliptic and the Lax Milgram lemma implies the existence of a solution to this perturbed problem. Then L W is an isomorphism from V onto H, hence there is a bounded inverse, L ?1 W from H to V. Then, since the embedding of V into H is compact by the Rellich lemma, we have that i E L ?1 W : H ¸ H is compact. That is, L W u = f + Wu ?1 é u = L ?1 W f + WL W u é ÝI ? KÞu = F ?1 where Ku = WL ?1 W u, and F = L W f. The operator I ? K is an operator of Fredholm type for which there is a theorem, known as the Fredholm alternative theorem, stating conditions under which there is a unique solution to ÝI ? KÞu = F. By translating this theorem into the notation and terminology of the weak boundary value problem, we obtain a similar existence-uniqueness theorem for the weak boundary value problem. Now we are going to apply a similar approach in considering the eigenvalue problem for an elliptic operator. We suppose that LßuÝxÞà = ?divÝAÝxÞ 4uÝxÞÞ + cÝxÞ uÝxÞ x5U 1 is uniformly elliptic on U, and we will consider the eigenvalue problem of finding values of the scalar V, such that LßuÝxÞà = V uÝxÞ uÝxÞ = 0 x5U [email protected] has nontrivial solutions. Note that in the abscence of first order terms, -d, f 5 C K0 ÝUÞ ÝLßdà, fÞ 0 = Ýd, LßfàÞ 0 and we say that L is symmetric. Then it follows that the associated bilinear form B is also symmetric, Bßu, và = Bßv, uà -u, v 5 V = H 10 ÝUÞ. Unless cÝxÞ ³ 0, the bilinear form B is not V-elliptic, but it is V-H-coercive, which is to say, B W is V-elliptic, for W sufficiently large. Then we recall that L W : H 10 ÝUÞ ¸ H 0 ÝUÞ is an isomorphism for W sufficiently large. Then, instead of considering the eigenvalue problem for L, we will consider the eigenvalue problem for L ?1 W which is a compact operator for which there is a classical theorem. By translating the conclusions of the classical theorem into the terminology of our elliptic operator, we obtain the following theorem. Theorem Under the assumptions on L, there is a countable collection áV k â of scalars such that LßuÝxÞà = V uÝxÞ uÝxÞ = 0 x5U [email protected] has nontrivial solutions. Moreover, (a) the V k are all real and L < V 1 ² V 2 ² ` ¸ K (b) Let N k = áu 5 H 10 ÝUÞ : Lßuà = V k uâ. Then dim N k < K all k and Ýu, vÞ 0 = 0 if u 5 Nk, v 5 Nj (c) The weak solutions ád k â , for j®k LßdÝxÞà = V dÝxÞ dÝxÞ = 0 ||d|| 0 = 1, x5U [email protected] form an orthonormal basis for H 0 ÝUÞ, and at the same time, an orthogonal (but not orthonormal) basis for H 10 ÝUÞ. 2 Proof- Let L denote the smallest value such that L L : H 10 ÝUÞ ¸ H 0 ÝUÞ is an isomorphism. Then for any W ³ L, there is a bounded linear operator S W : H 0 ÝUÞ ¸ H 10 ÝUÞ, which is inverse to L W ; i.e., L W ÝS W fÞ = f for all f 5 H 0 ÝUÞ. Let K = i E S W : H 0 ÝUÞ ¸ H 10 ÝUÞ Ð H 0 ÝUÞ. Then K is a compact linear operator on H 0 ÝUÞ. Note that SWf = u B W ßu, và = Ýf, vÞ 0 -v 5 H 10 ÝUÞ é é LWu = f For arbitrary f, g 5 H 0 ÝUÞ, let Kf = u, Kg = v. Then Ýf, KgÞ 0 = Ýf, vÞ 0 = B W ßu, và = B W ßv, uà = Ýg, uÞ 0 = Ýu, gÞ 0 = ÝKf, gÞ 0 . Thus the symmetry of the bilinear form implies that K is symmetric, and since K is bounded, in fact compact, it follows that K is self adjoint. In addition, ÝKf, fÞ 0 = Ýu, fÞ 0 = B W ßu, uà ³ c 0 ||u|| 21 , so K is a compact self adjoint, positive operator on the Hilbert space H = H 0 ÝUÞ. We have the following theorem about such operators, Theorem (Fredholm-Riesz-Schauder) For K a compact self adjoint, positive operator on the Hilbert space H, (a) K has a countable collection of positive eigenvalues, áS k â, accumulating at zero; i.e., ||K|| LÝHÞ ³ S 1 ³ S 2 ³ ` ¸ 0 (b) Let N k = áu 5 H : Kßuà = S k uâ. Then dim N k < K all k and Ýu, vÞ H = 0 if u 5 Nk, v 5 Nj j®k (c) The normalized eigenfunctions for K form an orthonormal basis for H. The compact self adjoint, positive operators on the Hilbert space H are the analogues of symmetric positive definite matrices on R n . Now é L W ßSdà = d Kd = S d é B W ßSd, và = Ýd, vÞ 0 -v 5 V = H 10 ÝUÞ hence L W ßSdà = d é L W ßdà = 1 d S so the eigenfunctions of K are the eigenfunctions of L W but the eigenvalues of L W are the reciprocals of the eigenvalues of K. Thus 3 ||K|| LÝHÞ ³ S 1 ³ S 2 ³ ` ¸ 0 implies 0< 1 < J 1 ² J 2 ² ` ¸ K. ||K|| LÝHÞ L W ßd k à = J k d k é Lßd k à = ÝJ k ? WÞ d k = V k d k In addition, with 0< 1 ? W < V 1 ² V 2 ² ` ¸ K. ||K|| LÝHÞ If ád k â denote the weak solutions of L W ßdÝxÞà = J dÝxÞ dÝxÞ = 0 ||d|| 0 = 1, x5U [email protected] u = > k>0 Ýu, d k Þ 0 d k = > k>0 u k d k then for u 5 V = H 10 ÝUÞ, write Define ÝÝu, vÞÞ V = B W ßu, và + Ýu, vÞ H Note that ÝÝu, vÞÞ V = B W > k>0 u k d k , > j>0 v j d j + > k>0 u k d k , > j>0 v j d j -u, v 5 V. H = > k > j u k v j B W ßd j , d k à + > k u k v k . B W ßd j , d k à = ÝL W d j , d k Þ H = J j Ýd j , d k Þ H = V j N jk But ÝÝu, vÞÞ V = > k ÝJ k + 1Þu k v k . hence and, in particular, ÝÝd j , d k ÞÞ V = ÝJ k + 1Þ N jk . To see that the eigenfunctions form a complete family for the inner product on V, suppose ÝÝu, d k ÞÞ V = 0 Then -k. 0 = ÝÝu, d k ÞÞ V = B W ßu, d k à + Ýu, d k Þ H = J k Ýu, d k Þ H + Ýu, d k Þ H i.e., ÝJ k + 1ÞÝu, d k Þ H = 0 But Jk ³ hence 1 >0 ||K|| LÝHÞ Ýu, d k Þ H = 0 -k. -k -k and since the family is complete in H, u = 0.n 4 u 5 V = H 10 ÝUÞ iff ||u|| 2V = ÝÝu, uÞÞ V = > k ÝJ k + 1Þu 2k < K Note that u 5 H = H 0 ÝUÞ iff ||u|| 2H = Ýu, uÞ 0 = > k u 2k < K Similarly, For F 5 D v ÝUÞ, f 5 DÝUÞ Ð H, FÝfÞ = FÝ> k f k d k Þ = > k f k FÝd k Þ = > k f k F k and we have F 5 V v iff |FÝfÞ| = > k f k F k = > k Ý1 + J k Þ 1/2 f k Ý1 + J k Þ ?1/2 F k ² > k Ý1 + J k Þ ?1 F 2k In addition, u 5 H iff ||f|| 2V Ý1 + J k Þ ?1/2 F k F 5 V v iff i.e., 1/2 5 §2. áu k â 5 § 2 . Ý1 + J k Þ 1/2 u k u 5 V iff 5 §2. and we can define, more generally, u 5 H s ÝUÞ iff Ý1 + J k Þ s/2 u k 5 §2, and we have, with this definition, ÝH s ÝUÞÞ v = H ?s ÝUÞ. This scale of Hilbert spaces, each with the inner product ÝÝu, vÞÞ s = > k Ý1 + J k Þ s u k v k is completely analogous to the Fourier spaces H s ÝR n Þ we have defined previously. Consider now the following examples. 1) LßuÝxÞà = ?u”ÝxÞ, uÝ0Þ = uÝ^Þ = 0 0<x<^ Here cÝxÞ = 0 so W = 0 and J k = V k . For this operator, it is easy to compute the eigenvalues and eigenfunctions d k ÝxÞ = 2 2 ^ sinÝkxÞ and V k = k k = 1, 2, u 5 Then the function 2x ^ x 2 1? ^ uÝxÞ = has u k = Ýu, d k Þ 0 = ? 4 2 sin ^k Ý1 + V k Þ 1/2 ~ k Now i.e. so k^ 2 0 < x < ^/2 ^/2 < x < ^ . Ý1 + V k Þ s/2 u k ~ k s?2 5 § 2 for s ² 1; u 5 H 1 = H 10 ÝUÞ. vÝxÞ = 1, 0 < x < ^, On the other hand, for the function vk = we find 1 ? Ý?1Þ k 5 §2 k but Ý1 + V k Þ s/2 v k ~ k s?1 5 § 2 i.e., v 5 H 0 = L 2 ÝUÞ. wÝxÞ = NÝx ? x 0 Þ Finally, for wk = and wk 6 §2 for s > 1/2; N 5 H ?1 ÝUÞ. LßuÝx, yÞà = ?4 2 uÝx, yÞ u=0 2) x 0 5 Ý0, ^Þ 2 ^ sinÝkx 0 Þ 5 § K Ý1 + V k Þ ?s/2 w k ~ k ?s 5 § 2 In particular, for s ² 0; in U = á0 < x, y < ^â on @ Then, again, c = 0 so J = V, and 2 sinÝjxÞ sinÝkyÞ d j,k Ýx, yÞ = ^ In this case, for V j,k = Ýj 2 + k 2 Þ, j, k = 1, 2, u wÝx, yÞ = NÝx ? x 0 Þ NÝy ? y 0 Þ Ýx 0 , y 0 Þ 5 U we have 2 sinÝjx 0 Þ sinÝky 0 Þ w j,k = ^ and Ý1 + V j,k Þ ?s/2 w k ~ Ý1 + j 2 + k 2 Þ ?s/2 5 § 2 for s > 1 6 Then w 6 H ?1 ÝUÞ. For the function uÝx, yÞ = we have u j,k = 0 < x < ^/2, 0 < y < ^ ^/2 < x < ^, 0 < y < ^ c j,k 5 §2 j k2 Ý1 + j 2 + k 2 Þ s/2 u j,k ~ but 2x ^ x 2 1? ^ Ý1 + j 2 + k 2 Þ s/2 6 § 2 if s = 1 j k2 Then u 5 H but u 6 V. . 7