Inverse Scattering for Materials with a
Conductive Boundary
Isaac Harris
Texas A&M University, Department of Mathematics
College Station, Texas 77843-3368
iharris@math.tamu.edu
Joint work with: O. Bondarenko and A. Kleefeld
March 2016
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Formulation of the Problem
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The Direct Scattering Problem
We consider the time-harmonic scattering problem in R2 and R3
1. D ⊂ Rm be a bounded open region with ∂D-Smooth
2. The function n ∈ L∞ (D) is the refractive index
3. The function η ∈ L∞ (∂D) is conductivity parameter
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The Direct Scattering Problem
We consider the scattering by an inhomogeneous material with a
1 (R3 )
conductive boundary. The total field u = u s + u i ∈ Hloc
satisfies:
−
u =u
+
in
R3 \ D
∆u + k 2 n(x)u = 0
in
D
on
∂D
−
+
+
∂ν u = ∂ν u + ηu
∂r u − iku = o r −1
as r → ∞.
s
and
∆u + k 2 u = 0
s
Let u i = eikx·d then it is known that the scattering problem is
well-posed for that radiating scattered fields u s (x, d; k) which
depends on the incident direction d and the wave number k.
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The Inverse Scattering Problem
The Inverse Problem
Given the Cauchy Data (u s , ∂ν u s ) on the collection curve C for all
d on the unit sphere and a range of wave numbers k ∈ [kmin , kmax ]
can we obtain information about the scatterer D and it’s material
parameters n and η?
F. Cakoni, David Colton and Peter Monk,
“The inverse electromagnetic scattering problem for a partially coated
dielectric”, J. Comput. Appl. Math. 204, 256-267 (2007).
O. Bondarenko and X. Liu
“The factorization method for inverse obstacle scattering with conductive
boundary condition”, Inverse Problems 29 (2013) 095021.
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The Far Field Pattern
The scattered field, has the following asymptotic expansion
eik|x|
u (x, d; k) =
4π|x|
s
∞
u (x̂, d; k) + O
1
|x|
as |x| → ∞.
Using Green’s Representation theorem one can show that the far
field pattern is given by
Z
u ∞ (x̂, d; k) =
u s (y , d; k)∂ν e−ik x̂·y − ∂ν u s (y , d; k)e−ik x̂·y ds(y )
C
where x̂, d ∈ S = {x ∈ R3 : |x| = 1}.
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Reconstruction via the Far Field Operator
We now define the far field operator as F : L2 (S) 7−→ L2 (S)
Z
(F g )(x̂) :=
u ∞ (x̂, d; k)g (d) ds(d)
S
The inverse problem now reads: given the far field operator F for
a range of wave numbers k ∈ [kmin , kmax ] obtain information
about the scatterer D and it’s material parameters n and η.
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Solution to the Inverse
Scattering Problem
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Reconstruction Methods
1. Optimization Methods: Solve nonlinear model using
iterative scheme to reconstruct the material parameters, which
requires a priori information, and can be computationally
expensive. For matrix valued coefficient the inverse problem
lacks uniqueness.
2. Qualitative Methods: Using nonlinear model to reconstruct
limited information with no a priori information, such as the
support of the region in a computationally simple manner (i.e.
solving a linear integral equations).
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Reconstruction of D by the LSM
Let Φ∞ (x̂, z) = e−ikz·x̂ be the far field pattern for the fundamental
solution of the Helmholtz equation.
The Linear Sampling Method (LSM)
The Far Field equation in R3 is given by (fix the wave number k)
(F g )(x̂) = Φ∞ (x̂, z)
for a z ∈ R3 ,
then the regularized solution of the far-field equation gz,δ satisfies
I
kgz,δ kL2 (S) is bounded as δ → 0 provided that z ∈ D,
kgz,δ kL2 (S) is unbounded as δ → 0 provided that z ∈
/ D.
∞
Assuming that F gz,δ − Φ (·, z) 2 → 0 as δ → 0.
I
L (S)
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Numerical Reconstruction i.e. plot z 7→ 1/kgz,δ k
F. Cakoni, David Colton and Peter Monk,
“The inverse electromagnetic scattering problem for a partially coated
dielectric”, J. Comput. Appl. Math. 204, 256-267 (2007).
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Reconstruction of D by the FM
Provided the far field operator has a suitable factorization, we
define the positive self-adjoint compact operator
F] := < F + = F : L2 (S) 7−→ L2 (S).
The Factorization Method (FM)
Let the pair (λi , ψi ) ∈ R+ × L2 (S) be an orthonormal eigensystem
of F] . Then
z ∈ D ⇐⇒ W (z) =
∞
X
|(Φ∞ (x̂, z), ψi )|2
i=1
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λi
< ∞.
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Numerical Reconstruction i.e. plot z 7→ 1/W (z)
O. Bondarenko and X. Liu
“The factorization method for inverse obstacle scattering with conductive
boundary condition”, Inverse Problems 29 (2013) 095021.
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TE-Problem for a Material
with a Conductive Boundary
O. Bondarenko, I. Harris, and A. Kleefeld
“The interior transmission eigenvalue problem for an inhomogeneous
media with a conductive boundary”
Submitted to Applicable Analysis - (arXiv:1510.01762).
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Motivation
I
The Transmission eigenvalue problem is non-selfadjoint and
non-linear
I
The linear sampling method fails if k is a transmission
eigenvalue
I
The Transmission eigenvalues can be used to
determine/estimate the material properties.
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Injectivity of the Far Field Operator
The associated Herglotz function is of the form
Z
vg (x) := e ikx·d g (d) ds(d).
S
The far-field operator F is injective with dense range if and only if
the only pair (w , vg ) that satisfies
∆w + k 2 nw = 0
in
D
∆vg + k vg = 0
in
D
w = vg and ∂ν w = ∂ν vg + ηvg
on
∂D
2
is given by (w , vg ) = (0, 0)
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Reconstructing the Real TEs
Recall that the far field equation is given by
(F gz )(x̂) = Φ∞ (x̂, z).
Determination of TEs from scattering data:
Let z ∈ D and gz,δ regularized solution to the Far Field equation
I
if k is not a TE then kgz,δ kL2 (S) is bounded as δ → 0
I
if k is a TE then kgz,δ kL2 (S) is unbounded as δ → 0.
Since the Transmission eigenvalues can be reconstructed from the
data the next question one can ask is what information do they
hold about the coefficients.
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Numerical Examples i.e. plot k 7→ ||gz,δ ||
Figure: Plot of the average ||gz,δ ||L2 (S) for 25 points z ∈ D
I. Harris, F. Cakoni, and J. Sun
”Transmission eigenvalues and non-destructive testing of anisotropic
magnetic materials with voids” Inverse Problems, 30 035016 (2014).
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The Transmission Eigenvalues
Values of k ∈ C where nontrivial (w , v ) ∈ L2 (D) × L2 (D) such
that w − v ∈ H 2 (D) ∩ H01 (D) satisfies
∆w + k 2 nw = 0
w −v =0
and
∆v + k 2 v = 0
and ∂ν w − ∂ν v = ηv
in D
on ∂D.
Where we assume that n > 0 and η > 0. Equivalently there exists
u = w − v ∈ H 2 (D) ∩ H01 (D) satisfies
1
(∆ + k 2 n)u = 0
n−1
η
and ∂ν u = − 2
(∆ + k 2 n)u
k (n − 1)
(∆ + k 2 )
u=0
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in D
on ∂D.
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The Variational Formulation
By Green’s 2nd Theorem
1
(∆ + k 2 n)u = 0
n−1
η
(∆ + k 2 n)u
and ∂ν u = − 2
k (n − 1)
(∆ + k 2 )
u=0
in D
on ∂D
has the variational form
Z 2
Z
k
1
∂ν u ∂ν ϕ ds +
(∆u + k 2 nu)(∆ϕ + k 2 ϕ) dx = 0,
η
n−1
∂D
D
for all ϕ ∈ H 2 (D) ∩ H01 (D).
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The transmission eigenvalue problem can be written as
Ak u − k 2 Bu = 0
Z
(Ak u, ϕ)H 2 (D) =
D
where
1
(∆u + k 2 u)(∆ϕ + k 2 ϕ) + k 4 uϕ dx
n−1
Z 2
k
+
∂ν u ∂ν ϕ ds
η
∂D
and
Z
∇u · ∇ϕ dx
(Bu, ϕ)H 2 (D) =
D
Assume that either n > 1 a.e. in D and that η > 0 on ∂D then
1. the operator B is positive, compact, and self-adjoint.
2. the mapping k 7→ Ak is continuous from (0, ∞) to the set of
self-adjoint coercive operators on H 2 (D) ∩ H01 (D).
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The Key Ingredient
F. Cakoni and H. Haddar
“On the existence of transmission eigenvalues in an inhomogenous
medium”, Applicable Analysis 88, no 4, 475-493 (2009).
We recall this key result in the following lemma.
Lemma
If k 7→ Ak is continuous from (0, ∞) to the set of self-adjoint positive definite
bounded operators and B is a self-adjoint non-negative compact operator.
1. Ak0 − k02 B is a positive operator,
2. Ak1 − k12 B is non-positive on a m dimensional subspace
then each of the equations λj (k) − k 2 = 0 for j = 1, . . . , m has at least one
solution in [k0 , k1 ] where λj (k) is such that Ak − λj (k)B has a non-trivial kernel.
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Notice that we have the following estimates
Ak u − k 2 Bu, u 2
≥ αkuk2H 2 (D) − k 2 k∇uk2L2 (D)
H (D)
≥ αkuk2H 2 (D) − k 2 kuk2H 2 (D)
≥ (α − k 2 )kuk2H 2 (D) ,
where α is the coercivity constant for the operator Ak (that is
independent of k). Hence, for all k 2 sufficiently small we have that
Inverse Scattering for Conductive Boundary
Ak u − k 2 Bu, u
H 2 (D)
> 0.
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Let xi ∈ D where Bi = B(xi , ε) := {x ∈ Rm : |x − xi | < ε} with
ε > 0, sufficiently small. Also assume n(x) ≥ nmin and find
∆ + kε2
1
∆ + kε2 nmin ui = 0
nmin − 1
in Bi
where ui ∈ H02 (Bi ).
Extend ui to H02 (D) by zero, then one can then show that
Akε u − kε2 Bu, u 2
≤ 0, for all u ∈ Span{u1 , u2 , · · · }.
H (D)
Theorem (O. Bondarenko, I. H, and A. Kleefeld)
Assume that either n > 1 or 0 < n < 1 a.e. in D and η > 0, then
there exists infinitely many real transmission eigenvalues. Moreover
the set of real transmission eigenvalues is discrete.
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It is known that for a every fixed k ∈ (0, ∞) there exists an
increasing sequence λj (k; n, η) of positive generalized eigenvalues
Ak u − λj (k)Bu = 0
that satisfy
(Ak u, u)H 2 (D)
U∈Uj u∈U\{0} (Bu, u)H 2 (D)
λj (k; n, η) = min max
where Uj is the set of all j-dimensional subspaces of
H02 (D) ∩ H01 (D).
Recall that the real transmission eigenvalues satisfy
λj (k; n, η) − k 2 (n, η) = 0.
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Monotonicity Results
Theorem (O. Bondarenko, I. H, and A. Kleefeld)
Assume that 0 < n with 0 < η and that kj is a transmission
eigenvalue such that λj (k) − k 2 = 0,
1. if 1 < n, then kj is decreasing with respect to n.
2. if 1 < n, then kj is decreasing with respect to η.
3. if 0 < n < 1, then kj is increasing with respect to n.
4. if 0 < n < 1, then kj is increasing with respect to η.
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An Inverse Spectral Result
Corollary (O. Bondarenko, I. H, and A. Kleefeld)
1. If it is known that n > 1 or 0 < n < 1 is a constant refractive
index with η known and fixed, then n is uniquely determined
by the first transmission eigenvalue.
2. If n > 1 or 0 < n < 1 is known and fixed with η a constant,
then the first transmission eigenvalue uniquely determines η.
Since the Transmission eigenvalues can be reconstructed from the
data this uniqueness result can be used for non-destructive testing.
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Let D =Unit Sphere with n and η constant. Using separation of
variables we have that k is a transmission eigenvalue if and only if
√ −jm (k)
jm (k n)
√
√
= 0.
det
0 (k) − ηj (k)
0 (k n)
−kjm
k njm
m
Here, jm denotes the spherical Bessel function of the first kind.
η
1
2
3
10
1.
2.798 386
2.458 714
2.204 525
1.743 402
2.
3.029 807
2.914 716
2.809 294
2.467 800
3.
3.141 593
3.141 593
3.141 593
3.138 749
4.
3.601 813
3.514 484
3.435 429
3.141 593
Table: The first four interior transmission eigenvalues for a unit sphere
using the index of refraction n = 4 and various choices of η.
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Let n and η be constants, therefore by Green’s representation
w (x) = SLk √n (∂ν w )(x) − DLk √n (w )(x) ,
v (x) = SLk (∂ν v )(x) − DLk (v )(x) ,
x ∈ D,
x ∈D,
where
Z
SLk (f )(x) =
Φk (x, y )f (y ) ds(y ) ,
x ∈D,
Z∂D
DLk (f )(x) =
∂ν(y ) Φk (x, y )f (y ) ds(y ) ,
x ∈D,
∂D
are the single and double layer potentials that are defined for
points in the domain D, respectively.
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Boundary Integral Operators
Z
Sk (f )(x) =
Φk (x, y )f (y ) ds(y ) ,
x ∈ ∂D ,
Z∂D
Dk (f )(x) =
∂ν(y ) Φk (x, y )f (y ) ds(y ) ,
x ∈ ∂D ,
∂D
are the single and double layer boundary integral operators,
respectively.
Z
Kk (f )(x) =
∂ν(x) Φk (x, y )f (y ) ds(y ) ,
x ∈ ∂D ,
∂D Z
Tk (f )(x) = ∂ν(x)
∂ν(y ) Φk (x, y )f (y ) ds(y ) ,
x ∈ ∂D ,
∂D
are the normal derivative of the single and double layer boundary
integral operators, respectively.
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Boundary Integral Operators
Using the notation ψ1 = ∂ν v (x) and ψ2 = v (x), x ∈ ∂D,
!
Sk √n − Sk
−Dk √n + Dk + ηSk √n ψ1 0 =
.
ψ2
0
Kk √n − Kk −Tk √n + Tk + η Kk √n − 21 I
The boundary integral operator Z depending on the wave number
k is given by
!
−Dk √n + Dk + ηSk √n Sk √n − Sk
Z(k) =
.
Kk √n − Kk −Tk √n + Tk + η Kk √n − 12 I
Abstractly k is a transmission eigenvalue provided that
Z(k) has a non-trivial kernel.
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Numerical Results
For a cushion-shaped obstacle with index of refractions n = 1/2
and n = 4 for η = 1/2, η = 1, and η = 3.
(n, η)
(1/2,1/2)
(1/2,1)
(1/2,3)
(4,1/2)
(4,1)
(4,3)
1.
1.481
1.889
2.482
2.754
2.678
2.391
2.
1.754
2.245
2.947
2.987
2.930
2.723
3.
2.080
2.713
3.640
3.460
3.404
3.196
W.J. Beyn
“An integral method for solving nonlinear eigenvalue problems”, Linear
Algebra Appl. 436 (2012) 3839-3863.
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Numerical Results
The monotonicity of the first interior transmission eigenvalues for
the peanut-shaped obstacle using n = 4 for increasing η.
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Convergence as η → 0+
We conjecture that as η → 0+ the transmission eigenvalues
converge to the transmission eigenvalues for η = 0.
η
1
1/2
1/4
1/8
1/16
1/32
1/64
ERROR
1.118· 10−1
5.468· 10−2
2.695· 10−2
1.337· 10−2
6.659· 10−3
3.323· 10−3
1.660· 10−3
EOC
1.032
1.021
1.011
1.006
1.003
1.001
ERROR
9.063· 10−2
4.535· 10−2
2.264· 10−2
1.130· 10−2
5.647· 10−3
2.822· 10−3
1.411· 10−3
EOC
0.999
1.002
1.003
1.001
1.001
1.000
Table: The estimated order of convergence for the 2nd and 4th
transmission eigenvalue for a unit sphere using n = 4 as η → 0+ .
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Some Answered Questions:
I
Discreteness of the transmission eigenvalues
I
Existence of transmission eigenvalues
I
Monotonicity of the transmission eigenvalues
I
Reconstruction from far field pattern
Some ‘Unanswered’ Questions:
I
Existence of Complex eigenvalues
I
Inverse Spectral Analysis
I
Convergence Analysis as η → 0+
I
Asymptotic Analysis as η → 0+
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Figure: Questions?
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