Sums of Reciprocal Eigenvalues

Transcription

1 Sums of Reciprocal Eigenvalues Bodo Dittmar Martin Luther niversity Halle-Wittenberg (Germany) Queen Dido Conference Carthage, Tunisia 200 Contens.Introduction 2.Membrane problems - isoperimetric inequalities 2. Fixed membrane 2.2 Free membrane 3. Sums of all reciprocal eigenvalues 3. Fixed membrane 3.2 Free membrane

2 Introduction The eigenvalue problem of the fixed membrane u + λu = 0 in D, the eigenvalue problem of the free membrane u = 0 on D, () v + µv = 0 in D, v n = 0 on D (2) n stands for the normal to D, λ and µ for the eigenvalue parameters. It is well-known that there exists an infinity of eigenvalues with finite multiplicity 0 < λ λ 2 λ 3..., 0 = µ < µ 2 µ Furthermore the method mentioned works also for the Stekloff problems and u = 0 in D, u n = νu on D, (3) u = 0 in D, u n = σu on C, (4) u = 0 on C 2, where D = C = C C 2. There is also an infinity of eigenvalues with finite multiplicity 0 < σ σ 2 σ 3..., 0 = ν < ν 2 ν

3 2 Membrane problems 2. Fixed membrane Theorem (Pólya, Schiffer 954) For any n ṙ 2 i= λ i i= λ (o) i where λ (o) i unit disk. denotes the eigenvalues of the unit disk. Equality holds if and only if D is the 3

4 = {z : z < } u + λu f (z) 2 = 0 in, u = 0, u i u j f (z) 2 da z = δ ij, i, j =, 2,..., u j (ζ) = λ j G(z, ζ) f (z) 2 u j (z)da z. G(z, ζ) f (z) f (ζ) = j= u j (z) f (z) u j (ζ) f (ζ) λ j 4

5 Lemma For the eigenvalues of the fixed membrane problem holds = max λ j L n G (z, ζ) f (z) 2 f (ζ) 2 v j (z)v j (ζ)da z da ζ, where f (z) 2 v i (z)v j (z)da z = δ i,j, i, j =, 2,..., n, v j L 2 () and v j is a basis of the space L n. v j = j i= c ji u (o) i, c jj 0, k= λ k k= 2 u(o) k f (z) 2 λ (o) k 5

6 Theorem 2 Let u (o) k be the eigenfunctions of the fixed membrane problem in the unit disk, λ k the corresponding eigenvalues and let f(z) = z + a 2 z be a conformal mapping of the unit disk onto D. Then, for any n 2 we have k= λ k k= λ (o) k + k= λ (o) k j 2 a j 2 j=2 u (o) 2 k r 2j 2 da Corollary λ 2 k= k k= λ (o) k f (z) 2 G 2 (z, ζ)da ζ da z G 2 (z, ζ)da z = π πρ2 + π ρ 2n+2 n(n + ) π ρ 2n n 2, ρ = ζ 2 G 2 (z, ζ)da z da ζ = k= λ (o) k 2 6

7 2.2 Free membrane Lemma 2 Let N f (z, ζ) be the following symmetric function depending on an univalent conformal map f where and A = f (z) 2 da z <. Then N f (z, ζ) = A N(z, ζ) + H f (z) + H f (ζ), z, ζ, H f (z) = f (ζ) 2 N(z, ζ)da ζ, z z N f (z, ζ) = f (z) 2, z ζ, z, N f (z, ζ) n z = 0 on, H f (z) = 4 f(z) 2 + h(z), on for a suffi- where h(z) is a harmonic function in with h f 2 n = A/(2π) /4 n ciently smooth f(z) and n is the outward pointing normal. In particular H f z (z) = 4 z 2. 7

8 v j (ζ) = µ j N f (z, ζ)v j (z) f (z) 2 da z, j = 2, 3,..., A where A = f (z) 2 da < and the kernel N f (z, ζ) f (z) f (ζ) with the eigenfunctions v j (z) f (z) with the eigenvalues µ j /A in the space V f. Lemma 3 A = max (N f (z, ζ) AC) f (z) h i (z) f (ζ) h i (ζ)da z da ζ, µ j=2 j L n i=2 where {h i } n i=2 is a basis of L n satisfying the orthonormality conditions h ih j da = δ ij. 8

9 Lemma 4 Let v m (o) be the eigenfunctions of the free membrane problem in the unit disk and let f(z) = z + a 2 z be a conformal mapping of the unit disk onto D. For a radial eigenfunction we have for any conformal mapping A ( 2 v m (o) 2 f (z) 2 da v m (o) f (z) da) 2 A, where A is the area of the domain D. Let v m (o) (z) and v (o) m+ (z) be the normalized eigenfunctions of the unit disk belonging to the same eigenvalue µ (o) m, such that (v m (o) (z)) 2 +(v (o) m+ (z))2 is radial. Then A (v m (o) 2 (o) + v m+2 ) f (z) 2 da ( 2 ( 2 v m (o) f (z) da) 2 v (o) m+ f (z) da) 2 2A. Equality occurs in both inequalities if and only if f(z) = z. 9

10 Theorem 3 Let D be a simply connected domain in the plane with area A < and maximal conformal radius. Then, for any n 2 we have µ j µ (o) j, where µ (o) j are the free membrane eigenvalues of the unit disk. Equality occurs if and only if D is the unit disk. 0

11 2.3 Sums of all reciprocal eigenvalues 2.3. Fixed membrane Theorem 4 Let f be a conformal mapping from the unit disk onto the domain D with the area A, then it holds G 2 (z, ζ) f (z) 2 f (ζ) 2 da z da ζ = where G(z, ζ) denotes Green s function of the unit disk., λ 2 j= j Theorem 5 λ 2 j= j = (A m,l + B m,l )a 0,m a 0,l + m=0 l=0 (C k,m,l + D k,m,l + E k,m,l, k= m= l= k=2 m= l= where the coefficients A, B, C, D, E are known and f (r, ϕ) 2 = a 0,n r n (a m,n cos mϕ + b m,n sin mϕ)r m n=0 m= n=

12 Examples. Disk j= λ 2 j = n= 8 4(2n+4)(4n+4)(2n+2) = π Remark Let λ j (n) be zeros of the Bessel function J n with the same order n (Rayleigh, Scientific papers, 899). 2. Kardioide j= λ j (n) 2 = 6(n + ) 2 (n + 2). j= λ 2 j = 3 64 π Similar results are obtained for the image of the unit disk by f n (z) = z+ n zn, n = 2, 3,... 2

13 3. Regular n-gone The conformal mapping f n of the unit disc onto a regular n-gone is well-known f n (z) = z dζ ( ζ n, n 3, ) 2/n The following table contains some numerical results n j= λ 2 j An open problem is: Prove that among all n-gones with the same maximal conformal radius the regular n- gone has the least value for the sum above. 3

14 2.3.2 Free membrane Theorem 6 If D is a simply connected sufficiently smooth bounded domain with the area A = f (z) 2 da z <. Then for the eigenvalues of the free membrane it holds ( ) 2 f N f (z, ζ) AC (z) 2 f (ζ) 2 da z da ζ = A 2 j=2 where C = A 2 N(z, ζ) f (z) 2 f (ζ) 2 da z da ζ., µ 2 j 4