Sums of Reciprocal Eigenvalues
Bodo Dittmar
Martin Luther University Halle-Wittenberg (Germany)
E-mail: bodo.dittmar@mathematik.uni-halle.de
Queen Dido Conference
Carthage, Tunisia 2010
Contens
1.Introduction
2.Membrane problems - isoperimetric inequalities
2.1 Fixed membrane
2.2 Free membrane
3. Sums of all reciprocal eigenvalues
3.1 Fixed membrane
3.2 Free membrane
1
1
Introduction
The eigenvalue problem of the fixed membrane
∆u + λu = 0 in D,
u = 0 on ∂D,
(1)
the eigenvalue problem of the free membrane
∆v + µv = 0 in D,
∂v
= 0 on ∂D
∂n
(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 < λ1 ≤ λ2 ≤ λ3 ...,
0 = µ1 < µ2 ≤ µ3 ... .
Furthermore the method mentioned works also for the Stekloff problems
∆u = 0 in D,
∂u
= νu on ∂D,
∂n
(3)
and
∆u = 0 in D,
∂u
= σu on C1 ,
∂n
u = 0 on C2 ,
(4)
where ∂D = C = C1 ∪ C2 . There is also an infinity of eigenvalues with finite multiplicity
0 < σ1 ≤ σ2 ≤ σ3 ≤ ...,
0 = ν1 < ν2 ≤ ν3 ≤ ... .
2
2
2.1
Membrane problems
Fixed membrane
Theorem 1 (Pólya, Schiffer 1954) For any n
n
n
X
1
1
1 X
≥
2
(o)
ṙ i=1 λi
i=1 λ
i
(o)
where λi denotes the eigenvalues of the unit disk. Equality holds if and only if D is the
unit disk.
3
U = {z : |z| < 1}
∆u + λu|f 0 (z)|2 = 0 in U,
u|∂U = 0,
Z
ui uj |f 0 (z)|2 dAz = δij , i, j = 1, 2, ... ,
U
Z
uj (ζ) = λj
G(z, ζ)|f 0 (z)|2 uj (z)dAz .
U
G(z, ζ)|f 0 (z)||f 0 (ζ)| =
∞
X
uj (z)|f 0 (z)|uj (ζ)|f 0 (ζ)|
j=1
4
λj
Lemma 1 For the eigenvalues of the fixed membrane problem holds
n
X
1
1
λj
= max
Ln−1
n Z Z
X
1
U
U
GU (z, ζ)|f 0 (z)|2 |f 0 (ζ)|2 vj (z)vj (ζ)dAz dAζ ,
where U |f 0 (z)|2 vi (z)vj (z)dAz = δi,j , i, j = 1, 2, ..., n, vj ∈ L2 (U ) and vj is a basis of the
space Ln−1 .
R
vj =
j
X
(o)
cji ui , cjj 6= 0,
i=1
n
X
1
k=1
λk
≥
n
X
R
U
(o) 2
uk
|f 0 (z)|2
(o)
λk
k=1
5
(o)
Theorem 2 Let uk be the eigenfunctions of the fixed membrane problem in the unit disk,
λk the corresponding eigenvalues and let f (z) = z + a2 z 2 + ... be a conformal mapping of
the unit disk onto D. Then, for any n ≥ 2 we have
n
X
1
k=1
λk
≥
n
X
1
(o)
k=1
+
λk
Z
n
∞
X
1 X
(o) 2
2
2
j
|a
|
uk r2j−2 dA
j
(o)
k=1
λk
U
j=2
Corollary 1
n
X
1
λ2
k=1 k
Z
U
≥−
G2U (z, ζ)dAz
n
X
1
(o) 2
k=1 λk
Z
|f 0 (z)|2
+2
U
Z
U
G2U (z, ζ)dAζ dAz
∞
∞
X
X
ρ2n+2
ρ2n
π 3 2
= − πρ + π
−π
, ρ = |ζ|
2 4
n(n + 1)
n2 − 1
1
2
Z Z
U
U
G2U (z, ζ)dAz dAζ =
n
X
k=1
6
1
(o) 2
λk
2.2
Free membrane
Lemma 2 Let Nf (z, ζ) be the following symmetric function depending on an univalent
conformal map f
Nf (z, ζ) = A N (z, ζ) + Hf (z) + Hf (ζ), z, ζ ∈ U,
where
Hf (z) = −
and A =
R
U
Z
U
|f 0 (ζ)|2 N (z, ζ)dAζ , z ∈ U
|f 0 (z)|2 dAz < ∞. Then
∆z Nf (z, ζ) = |f 0 (z)|2 , z 6= ζ, z ∈ U,
∂Nf (z, ζ)
= 0 on ∂U,
∂nz
1
Hf (z) =
|f (z)|2 + h(z),
4
2
|
∂h
where h(z) is a harmonic function in U with ∂n
= A/(2π) − 1/4 ∂|f
∂n on ∂U for a sufficiently smooth f (z) and n is the outward pointing normal.
In particular
1
Hf ≡z (z) = |z|2 .
4
7
vj (ζ) =
µj
A
Z
U
Nf (z, ζ)vj (z)|f 0 (z)|2 dAz , j = 2, 3, ... ,
where A = U |f 0 (z)|2 dA < ∞ and the kernel Nf (z, ζ)|f 0 (z)||f 0 (ζ)| with the eigenfunctions
vj (z)|f 0 (z)| with the eigenvalues µj /A in the space Vf .
R
Lemma 3
n
X
A
µ
j=2 j
= max
Ln−1
n Z Z
X
i=2 U
U
(Nf (z, ζ) − AC)|f 0 (z)|hi (z)|f 0 (ζ)|hi (ζ)dAz dAζ ,
where {hi }ni=2 is a basis of Ln−1 satisfying the orthonormality conditions
8
R
U
hi hj dA = δij .
(o)
Lemma 4 Let vm be the eigenfunctions of the free membrane problem in the unit disk
and let f (z) = z + a2 z 2 + ... be a conformal mapping of the unit disk onto D. For a radial
eigenfunction we have for any conformal mapping
Z
A
U
(o) 2 0
vm
|f (z)|2 dA
2
Z
−
U
(o)
(o) 0
vm
|f (z)|2 dA
≥ A,
(o)
where A is the area of the domain D. Let vm (z) and vm+1 (z) be the normalized eigenfunc(o)
(o)
(o)
tions of the unit disk belonging to the same eigenvalue µm , such that (vm (z))2 +(vm+1 (z))2
is radial. Then
Z
(o) 2
(o) 2
A (vm
+ vm+1 )|f 0 (z)|2 dA
U
Z
−
U
2
(o) 0
vm
|f (z)|2 dA
−
Z
U
Equality occurs in both inequalities if and only if f (z) = z.
9
2
(o)
vm+1 |f 0 (z)|2 dA
≥ 2A.
Theorem 3 Let D be a simply connected domain in the plane with area A < ∞ and
maximal conformal radius 1. Then, for any n ≥ 2 we have
n
X
1
1
µj
≥
n
X
1
(o)
1
(o)
,
µj
where µj are the free membrane eigenvalues of the unit disk. Equality occurs if and only
if D is the unit disk.
10
2.3
Sums of all reciprocal eigenvalues
2.3.1
Fixed membrane
Theorem 4 Let f be a conformal mapping from the unit disk U1 onto the domain D with
the area A, then it holds
Z
U1
Z
U1
G2 (z, ζ)|f 0 (z)|2 |f 0 (ζ)|2 dAz dAζ =
∞
X
1
j=1
λ2j
,
where G(z, ζ) denotes Green’s function of the unit disk.
Theorem 5
∞
X
1
j=1
λ2j
=
∞ X
∞
X
∞ X
∞ X
∞
X
m=0 l=0
k=1 m=1 l=1
(Am,l + Bm,l )a0,m a0,l +
(Ck,m,l + Dk,m,l +
∞
∞ X
∞ X
X
k=2 m=1 l=1
where the coefficients A, B, C, D, E are known and
|f 0 (r, ϕ)|2 =
∞
X
n=0
a0,n rn
∞ X
∞
X
(am,n cos mϕ + bm,n sin mϕ)rm
m=1 n=1
11
Ek,m,l ,
Examples
1. Disk
P∞
1
j=1 λ2
j
=
4
2·43
+
P∞
8
n=1 4(2n+4)(4n+4)(2n+2)
=
π2
48
−
5
32
Remark 1 Let λj (n) be zeros of the Bessel function Jn with the same order n (Rayleigh,
Scientific papers, 1899).
∞
X
1
1
=
.
2
λ (n)
16(n + 1)2 (n + 2)
j=1 j
2. Kardioide
P∞
1
j=1 λ2
j
=
3 2
64 π
−
551
1536
Similar results are obtained for the image of the unit disk by fn (z) = z+ n1 z n , n = 2, 3, ...
.
12
3. Regular n-gone
The conformal mapping fn of the unit disc onto a regular n-gone is well-known
Z z
fn (z) =
dζ
, n ≥ 3,
(1 − ζ n )2/n
The following table contains some numerical results
n
4
5
6
7
8
−2
j=1 λj
P∞
0.0514
0.0498
0.0496
0.0494
0.0494
An open problem is:
Prove that among all n-gones with the same maximal conformal radius the regular ngone has the least value for the sum above.
13
2.3.2
Free membrane
Theorem
6 If D is a simply connected sufficiently smooth bounded domain with the area
R
A = U |f 0 (z)|2 dAz < ∞. Then for the eigenvalues of the free membrane it holds
Z Z U
where C = − A12
U
Nf (z, ζ) − AC
2
|f 0 (z)|2 |f 0 (ζ)|2 dAz dAζ = A2
∞
X
1
j=2
R R
U U
N (z, ζ)|f 0 (z)|2 |f 0 (ζ)|2 dAz dAζ .
14
µ2j
,