Resonances and balls in obstacle scattering with Neumann boundary conditions

0.13 MB·English

by  T. J. Christiansen

RESONANCES AND BALLS IN OBSTACLE SCATTERING WITH
NEUMANN BOUNDARY CONDITIONS
8 T.J.CHRISTIANSEN
0
0
2 Abstract. We consider scattering by an obstacle in d, d ≥ 3 odd. We show that for the
R
Neumann Laplacianifanobstacle has the sameresonances as the ballof radiusρdoes, then
n theobstacleisaballofradiusρ.Wegiverelatedresultsforobstacleswhicharedisjointunions
a ofseveralballsofthesameradius.
J

4

]
h 1. Introduction
p
The purpose of this is note is to show that for obstacle scattering in d, d ≥ 3 odd, a ball is
- R
h uniquelydeterminedbyitsresonancesfortheLaplacianwithNeumannboundaryconditions.We
t actually show a somewhat stronger result: If O and O have the same (Neumann) resonances
a 1 2
m and O is the disjoint union of m balls, each of radius ρ, then so is O .
1 2
Wefixsomenotation.LetO ⊂ d beacompactsetofdimensiondwithsmoothboundary∂O.
[
Let ∆ denote the Laplacian Rwith Neumann boundary conditions on d \O. Set R (λ) =
d\O O
1 (∆ R−λ2)−1, with the conventionthat R (λ) is bounded on L2( d\O)Rwhen ℑλ>0.Then,
0v forRodd\Odd,itisknownthatRO(λ):L2comp(RdO\O)→L2loc(Rd\O)hasRameromorphiccontinuation
6 to . Set
6 C
R ={λ ∈ : R (λ)has a pole at λ , repeated according to multiplicity}.
0 O 0 O 0
C
1. The multiplicity can be defined via
0
8 mO(λ)=rank RO(z)dz, 0<ǫ≪1.
0 I|z−λ|=ǫ
v: Set B(ρ) to be the closed ball of radius ρ centered at the origin.
i
X Theorem 1.1. If d≥3 is odd, and O ⊂ d is a smooth compact set with R =R , then O
O B(ρ)
R
r is a translate of B(ρ).
a
We note that a simple scaling argument shows that if ρ 6= ρ then R 6= R so that
1 2 B(ρ1) B(ρ2)
B(ρ) is determined (up to translation) by its Neumann resonances.
Hassell and Zworski [7] proved that for the Dirichlet Laplacian in 3, if a connected set O
R
has the same resonances as B(ρ), then O is a translate of B(ρ). It seems that their argument
also works for Neumann boundary conditions, again in dimension 3 with the restriction that
O is connected. That we can prove this result for all odd dimensions for Neumann boundary
conditions follows partly from the fact that for the Neumann case the resonances determine the
PartiallysupportedbyNSFgrantDMS0500267.
1
2 T.J.CHRISTIANSEN
determinant of the scattering matrix up to one unknown constant, while for the Dirichlet case
there are two unknown constants– compare Lemma 2.1 to [7].
We shall actually prove the following theorem, from which Theorem 1.1 follows immediately.
Theorem 1.2. Let d ≥ 3 be odd. Let O ⊂ d be a disjoint union of m closed balls each of
1
R
the same radius ρ, and let O ⊂ d be a smooth compact set. If R = R , then O is also a
2 R O1 O2 2
disjoint union of m closed balls of radius ρ.
The proofofTheorem1.2usesheatcoefficients.These arecloselyrelatedto the singularityat
t=0 of the distribution
(1) u (t)=Tr cos t ∆ −cos t ∆ .
O d\O d
R R
This is rather informal. One way t(cid:16)o m(cid:16)akqe precise(cid:17)sense o(cid:16)fp(1) is(cid:17)a(cid:17)s follows. With ρ′ chosen
sufficiently large that O⊂B(ρ′),
(2)
u (t)=Tr cos t ∆ − cos t ∆ +Tr cos t ∆
O d\O d\B(ρ′) d d\B(ρ′) B(ρ′) d B(ρ′)
R 1R R 1R 1 R 1
where is t(cid:16)he c(cid:16)haqracteristi(cid:17)c function of th(cid:16)e pset E,(cid:17)compare [(cid:17)15]. T(cid:16)he Poisson(cid:16)fpormula(cid:17) for (cid:17)
E
1
resonances in odd dimensions is
u (t)= eiλj|t|, t6=0
O
λjX∈RO
see [3, 8, 9, 14], and [13] for an application to the existence of resonances. This shows that any
singularitiesofthedistributionu (t)atnonzerotimearedeterminedbytheresonancesof∆ .
O d\O
WhenOisnotconnected,weexpectthatthedistributionof(2)hassingularitiesatnonzerotRimes,
see [10]. In particular, if O consists of two disjoint convex obstacles, the distribution (2) has a
singularity at twice the distance between them (e.g. [6] or [10, Section 6.4]). From this and
Theorem 1.2 we have
Corollary 1.3. Let d ≥ 3 be odd, and let O , O ⊂ d be smooth compact sets. Suppose O
1 2 1
R
is the disjoint union of 2 balls of radius ρ a distance δ > 0 apart. If R = R , then O is
O1 O2 2
obtained from O by a rigid motion.
1
This paper was inspired by [7], and we use some of the same notation.
2. Proof of Theorem 1.2
Let S (λ) be the scattering matrix for the Neumann Laplacian on d\O, and set s (λ) =
O O
R
detS (λ). Set
O
d zj
E(z)=(1−z)exp .
j
!
1
X
Inthelemmabelowwehavenotmadeanyassumptionsontheconnectivityof d\O.If d\O
R R
has a bounded component, then R contains the squareroots,and the negative squareroots,of
O
eigenvalues of the Neumann Laplacian on the bounded component(s). These do not cause poles
of the determinant of the scattering matrix. We remark that it is possible to have 0∈R here,
O
if d\O has a bounded component, but not if d\O is connected.
R R
RESONANCES AND BALLS 3
Lemma 2.1. Let d≥3 be odd and O⊂ d be a smooth compact set. Then
R
E(−λ/λ )
s (λ)=eicOλd j
O
E(λ/λ )
j
λj∈RYO,λj6=0
for some real constant c .
O
For a large number of more general self-adjoint operators (for example, for the Laplacian
with different boundary conditions), the resonances determine the determinant of the scattering
matrix up to two unknown constants, e.g. [7].
Proof. It follows from [14] that
E(−λ/λ )
s (λ)=eigO(λ) j
O
E(λ/λ )
j
λj∈RYO,λj6=0
where g (λ) is a polynomial of degree at most d. It remains only to show that g (λ)=c λd.
O O O
We now use the expression of [11] for the scattering matrix. Choose ρ′ > 0 so that O is
contained in the ball of radius ρ′ centered at the origin. For ψ ∈C∞( d), let
c R
ψ(λ):L2( d)→L2( d−1)
E± R S
be the operator with Schwartz kernel ψ(x)exp(±iλhx,ωi). For i=1, 2, 3, choose χ ∈C∞( d)
i c R
so that χ ≡1 if |x|<4+ρ′ and χ ≡1 on the support of χ , i=1, 2. Set
i i+1 i
(3) A (λ)=c˜ λd−2 χ3(λ)[∆ ,χ ]R (λ)[∆ ,χ ]t χ3(λ).
O d E+ Rd 1 O Rd 2 E−
Here c˜ = iπ(2π)−d and t χ3 denotes the transpose of χ3. Then the scattering matrix S (λ)
d E− E− O
associated to P is given by
(4) S (λ)=I+A (λ).
O O
Next, note that for the Neumann Laplacian,
R (0)[∆ ,χ ]t χ3(0)=χ
O Rd 2 E− 2
since [∆ ,χ ]t χ3(0)=∆ χ , and χ satisfies the Neumann boundary conditions. But then
Rd 2 E− Rd 2 2
χ3(0)[∆ ,χ ]χ = ∆ χ =0
E+ Rd 1 2 Z d Rd 1
R
so that A (λ) vanishes to order at least d−1 at λ=0. Thus
O
(5) s (λ)−1=det(I +A (λ))−1=O(|λ|d−1)
O O
at λ = 0. Using the fact that E(λ/λj) = O(|λ|d) at λ = 0, we see that g(j)(λ) = 0 for j =
0,...,d−2. It is also well known that for λ∈ , s (λ)s (−λ)=1, implying that g(d−1)(0)=0.
O O
Finally,sincefor λ∈ s (λ)=s (−λ), wehRavethatg(d)(0)isreal.This finishes the proof. (cid:3)
O O
R
We fix some notation. For a smooth compact set O ⊂ d with boundary ∂O of dimension
R
d−1,letκ ...,κ denotetheprincipalcurvaturesof∂O.Weusethenormalizationthat
1,∂O d−1,∂O
the mean curvature H is H = d−1κ .
∂O ∂O 1 j,∂O
P
4 T.J.CHRISTIANSEN
Lemma 2.2. Let d ≥ 3 be odd and let O ⊂ d be a smooth compact set with boundary ∂O of
R
dimension d−1. Then R determines
O
d−1
Vol(∂O), H , and 13H2 +2 κ2 .
∂O ∂O j,∂O
Z∂O Z∂O 1 !
X
Proof. As t↓0,
∞
(6) Tr(e−t∆Rd\O −e−t∆Rd)≃t−d/2 tn/2an.
n=0
X
The tracein (6) canbe made precise in exactlythe same manner that (1) is made precise in (2).
Explicit expressions for the first few a can be found in [4]. For Neumann boundary conditions
n
in d, we have
R
a =α Vol(O)
0 0
a =α Vol(∂O)
1 1
a =α H
2 2 ∂O
Z∂O
d−1
a =α 13H2 +2 κ2
3 3 ∂O j,∂O
Z∂O 1 !
X
where the α are nonzero constants which depend on the dimension. Using Lemma 2.1, (6),
i
Tr(e−t∆Rd\O −e−t∆Rd)= 21πi ∞ e−tλ2ddλlogsO(λ)dλ+ 12 e−tλ2j
Z−∞ λj∈XRO∩R
(e.g. [12, Corollary 2.10] for the case of d\O connected, or [5]) and the expressions for the a
i
above, we prove the lemma. R (cid:3)
We are now ready for the proof of our theorem.
Proof of Theorem 1.2. SupposeO ,O areasinthestatementofthetheorem,withR =R .
1 2 O1 O2
Then by Lemma 2.2,
(7) Vol(∂O )=Vol(∂O ), H = H and
1 2 ∂O1 ∂O2
Z∂O1 Z∂O2
d−1 d−1
13H2 +2 κ2 = 13H2 +2 κ2 .
∂O1 j,∂O1 ∂O2 j,∂O2
Z∂O1 1 ! Z∂O2 1 !
X X
By the Cauchy-Schwarz inequality, for i=1, 2,
2
H
∂Oi ∂Oi ≤ H2
(cid:16)RVol(∂Oi)(cid:17) Z∂Oi ∂Oi
2
withequalityifandonlyifH isconstant.Likewise, d−1κ2 ≥ 1 d−1κ ,with
∂Oi 1 j,∂Oi d−1 1 j,∂Oi
equality if and only if κ1,∂Oi = κ2,∂Oi = ··· = κd−1,∂OPi. Since equality ho(cid:16)lPds for O1, th(cid:17)e three
RESONANCES AND BALLS 5
equalities of (7) mean that H is a constantand the principal curvatures of ∂O are all equal.
∂O2 2
Thus O must be the disjointn union of l balls of fixed radius ρ′. Again using the first two
2
equalities of (7), we see that we must have l =m and ρ′ =ρ. (cid:3)
We remark that the proof of this theorem is a bit delicate, in the sense that it is important
for us that in the expression for a the coefficients of H2 and d−1κ2 have the same sign.
3 ∂O 1 j,∂O
However, if R = R and we know a priori that O is a (d-dimensional) convex set, then we
O B(ρ) P
do notneedto use the coefficienta to provethatO is,upto translation,B(ρ).Instead,onecan
3
use first two equalities of (7) and the fact that in the Alexandrov-Fenchel inequality
Vol(∂O) 1/(d−1) ≤ ∂OHO 1/(d−2)
Vol(∂B(ρ)) H
(cid:18) (cid:19) ∂BR(ρ) B(ρ)!
equality holds if and only if O is a ball [2, ChapterR4, Section 9] or [1].
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Department of Mathematics
University of Missouri
Columbia, Missouri 65211
e-mail: [email protected]