Faber-Krahn Type Inequalities in Inverse Scattering Theory David - - PowerPoint PPT Presentation

Faber-Krahn Type Inequalities in Inverse Scattering Theory

David Colton

colton@math.udel.edu

Department of Mathematical Sciences, University of Delaware, USA

AIP , Vienna July 20-24, 2009 – p. 1/18

Scattering by a Perfect Conductor

u

s

D

ν

i

u

∆u + k2u = 0

in R2 \ D

u = us + eikx·d

in R2 \ D

u = 0

• n ∂D

lim

r→∞

√r ∂us ∂r − ikus

• = 0

The scattered field us has the asymptotic behavior

us(x) = eikr √r u∞(ˆ x, d) + O

• r−3/2

as r → ∞, where ˆ

x = x/|x|, |d| = 1, r = |x| and k > 0 is the wave

• number. u∞(ˆ

x, d) is called the far field pattern of the scattered field us.

AIP , Vienna July 20-24, 2009 – p. 2/18

Scattering by a Perfect Conductor

We define the far field operator F : L2(S) → L2(S) by

(Fg)(ˆ x) :=

• S

u∞(ˆ x, d)g(d)ds(d), S is the unit circle.

The far field equation is

(Fg)(ˆ x) = Φ∞(ˆ x, z), g ∈ L2(S)

where

Φ∞(ˆ x, z) = eiπ/4 √ 8πk e−ikˆ

x·z

is the far field pattern of the fundamental solution

Φ(x, z) := i 4H(1) (k|x − z|) .

AIP , Vienna July 20-24, 2009 – p. 3/18

The Linear Sampling Method

The linear sampling method for determining D from u∞(ˆ

x, d) attempts

to use regularization methods to solve the far field equation

(Fg)(ˆ x) = Φ∞(ˆ x, z), z ∈ R2

then looks for points z where the norm of g is relatively large. This method can only be justified if k2 is not a Dirichlet eigenvalue.

AIP , Vienna July 20-24, 2009 – p. 4/18

Dirichlet Eigenvalues

What happens when k2 is equal to a Dirichlet eigenvalue? Example: Let D be the unit disk and set z = 0. Then

Φ∞ = γ := eiπ/4/ √ 8πk and Fg = γ can be solved explicitly to give g(d) := γH(1)

0 (k)

2πJ0(k) .

Let k01 be the first zero of J0(k). Then k2

01 is the first Dirichlet

eigenvalue λ1, for D and

lim

k2→λ1 g = ∞.

In general the far field equation is not solvable but for z ∈ D it can be shown using a result of Tilo Arens that the regularized solution of

Fg = Φ∞ has relatively large norm for k2 near λ1.

AIP , Vienna July 20-24, 2009 – p. 5/18

Dirichlet Eigenvalues

Let the scattering object D be the rectangle [−0.4, 0.4] × [−0.5, 0.5].

2 4 6 8 10 12 14 16 18 20 2 4 6 8 10 12 14 16 18 20 Norm of g Dirichlet b.c. k

Graph of g versus k Faber-Krahn inequality

λ1 ≥ πk2

01

area D From the Faber-Krahn inequality, a lower bound for the area of D can be determined from a knowledge of the smallest eigenvalue λ1.

AIP , Vienna July 20-24, 2009 – p. 6/18

Scattering by a Dielectric

u

s

D

ν

i

u

∆u + k2n(x)u = 0

in R2

u = us + ui

in R2

lim

r→∞

√r ∂us ∂r − ikus

• = 0

where ui(x) := eikx·d. We assume that n − 1 has compact support D, n(x) > 0 for x ∈ D and n is in L∞(D). us again has the asymptotic behavior

us(x) = eikr √r u∞(ˆ x, d) + O

• r−3/2

AIP , Vienna July 20-24, 2009 – p. 7/18

Transmission Eigenvalues

We again consider the far field operator

(Fg)(ˆ x) :=

• S

u∞(ˆ x, d)g(d)ds(d)

and the corresponding far field equation

(Fg)(ˆ x) = Φ∞(ˆ x, z).

Instead of g becoming large near a Dirichlet eigenvalue, g now becomes large when k is a transmission eigenvalue. The linear sampling method for determining D from u∞(ˆ

x, d) only

works if k is not a transmission eigenvalue.

AIP , Vienna July 20-24, 2009 – p. 8/18

Transmission Eigenvalues

Definition: k > 0 is a transmission eigenvalue if there exists a nontrivial solution v ∈ L2(D), w ∈ L2(D), v − w ∈ H2(D) of the interior transmission problem

∆w + k2n(x)w = 0

in

D ∆v + k2v = 0

in

D w = v

• n

∂D ∂w ∂ν = ∂v ∂ν

• n

∂D

Remark: Note that if n = 1 the interior transmission problem is degenerate.

AIP , Vienna July 20-24, 2009 – p. 9/18

Transmission Eigenvalues

Theorem: Let 0 < n ∈ L∞(D) and either n − 1 ≥ δ > 0 or

1 − n ≥ δ > 0 in D. Then the set of transmission eigenvalues is a

discrete set. Colton-Kirsch-P¨ aiv¨ arinta (1989), Rynne-Sleeman (1991). Theorem: Let 0 < n ∈ L∞(D) and either n − 1 ≥ δ > 0 or

1 − n ≥ δ > 0 in D. Then there exist an infinite number of transmission

eigenvalues. P¨ aiv¨ arinta and Sylvester (2008) Cakoni, Gintides, Haddar (to appear) Theorem: Let k be a transmission eigenvalue. If supD|n(x) − 1| → 0 then k → ∞. Cakoni-Colton-Monk (2007)

AIP , Vienna July 20-24, 2009 – p. 10/18

Numerical Examples

Transmission eigenvalues can be determined from the far field data, i.e. values of k for wich the regularized solution of Fg = Φ∞ becomes relatively large:

−1.5 −1 −0.5 0.5 1 1.5 −1.5 −1 −0.5 0.5 1 1.5

The scatterer D

0.5 1 1.5 2 2.5 1 2 3 4 5 6 7 8 9

Wave number k Norm of g

n = 16

AIP , Vienna July 20-24, 2009 – p. 11/18

Transmission Eigenvalues

We now obtain an analog of the Faber-Krahn inequality for the first transmission eigenvalue k1. We again denote the first Dirichlet eigenvalue for D by λ1. Theorem: Suppose n − 1 ≥ δ > 0 for D. Then

k2

1 >

λ1

supDn Colton-P¨ aiv¨ arinta-Sylvester (2007) Open problem: If 1 − n ≥ δ > 0 all that be said is k2

1 > λ1.

In our previous example where D is an L shaped obstacle and n = 16 we have that k1 ≈ 1.09. Since λ1 ≈ 9.65 the above inequality now give the lower bound n > 8.1. Upper bounds for n and improved lower bounds have recently, been obtained by Cakoni and Gintides.

AIP , Vienna July 20-24, 2009 – p. 12/18

Transmission Eigenvalues

D D D

• n = 1 in D0

n − 1 ≥ δ > 0 in D \ D0

The case when there are regions in

D where n = 1 (i.e. cavities) is more

delicate. Let λ1 be again the first eigenvalue of

−∆ in D.

Theorem: Suppose n − 1 ≥ δ > 0 for x ∈ D \ D0 and n = 1 in D0. Then

k2

1 >

λ1

supD\D0n. Cakoni-Colton-Haddar (to appear)

AIP , Vienna July 20-24, 2009 – p. 13/18

Transmission Eigenvalues

Theorem: Let k be a transmission eigenvalue. Then if either a) supD\D0(n(x) − 1) = 0 or b) area(D \ D0) → 0 then k → ∞. Cakoni-Cay¨

• ren-Colton (2008)

−1 −0.5 0.5 1 −1 −0.5 0.5 1

(a)

−1 −0.5 0.5 1 −1 −0.5 0.5 1

(b)

−1 −0.5 0.5 1 −1 −0.5 0.5 1

(c)

(a) is dielectric with n = 16, (b) and (c) show medium with a circular void

• f radius 0.1

AIP , Vienna July 20-24, 2009 – p. 14/18

Transmission Eigenvalues

1.5 2 2.5 3 3.5 4 4.5 10 20 30 40 50 60 70 80

X: 1.73 Y: 41.91 X: 1.86 Y: 26.3 X: 3.44 Y: 26.95 X: 3.71 Y: 20.42 X: 4.17 Y: 74.78 X: 1.83 Y: 24.89

k ||g(.)||

Black dashed line corresponds to square without cavity (a), black solid line corresponds to the square with the cavity (b), and gray solid line corresponds to the square with the cavity (c).

AIP , Vienna July 20-24, 2009 – p. 15/18

Near Field Data

D

Ω Λ D δΩ

O

∆u + k2n(x)u = 0

in R2 \ {x0}

u = Φ(·, x0) + us

for x ∈ R2

lim

r→∞

√r ∂us ∂r − ikus

• = 0

Here the initial field is a point source located at x0 ∈ Λ and the Cauchy data of the total field is measured on ∂Ω, n − 1 has support D and there may be cavities D0 ⊂ D where n = 1.

AIP , Vienna July 20-24, 2009 – p. 16/18

Near Field Data

Define

H(Ω) :=

• v ∈ H1(Ω) : ∆v + k2v = 0
• The reciprocity gap operator R : H(Ω) → L2(Λ) is defined by

(Rv)(x0) :=

• Ω
• u(·, x0)∂v

∂ν − v∂u(· , x0) ∂ν

• ds

Theorem: R : H(Ω) → L2(Λ) is injective if k is not a transmission eigenvalue for D. Cakoni-Cay¨

• ren-Colton (2008)

AIP , Vienna July 20-24, 2009 – p. 17/18

Near Field Data

Instead of the whole space H(Ω) we can consider the dense set of H consisting of all Herglotz wave functions

vg(x) :=

• S

eikx·dg(d) ds(d), g ∈ L2(S).

Assuming that z ∈ D and defining the near field operator

N : L2(S) → L2(Λ) by Ng := Rvg we expect that the regularized

solution of

(Ng)(x0) = (RΦ(· z))(x0), x0 ∈ Λ

will be large if k is a transmission eigenvalue. Remark: Other dense sets of H(Ω) can be used yielding different near field operators.

AIP , Vienna July 20-24, 2009 – p. 18/18