The Factorization Method for Imaging Defects in
Anisotropic Materials
Isaac Harris
Texas A&M University, Department of Mathematics
College Station, Texas 77843-3368
iharris@math.tamu.edu
Joint work with: F. Cakoni
Research Funded by: NSF Grant DMS-1106972
and University of Delaware Graduate Student Fellowship
May 2016
The Factorization Method for Defects
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The Physical Model
We consider the time harmonic Acoustic scattering in R3 or
Electromagnetic scattering in R2 (scalar) TE-polarization case.
Inverse Problem
Reconstruct the support of the defective region D0 in a known
Anisotropic Material D from the measured scattered field.
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Reconstruction Methods
1. Optimization Methods: Reconstruct all material parameters
which require a priori information, and can be computationally
expensive. Moreover our problem lacks uniqueness.
2. Qualitative Methods: Reconstruct limited information, such
as the support of the scattering object in a computationally
simple manner, usually by constructing an indicator function.
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The Scattered Field usb
We consider the scattering by a healthy material where we find the
1 (Rm ) with u i = exp(ikx · d).
scattered field usb ∈ Hloc
∇ · Ã∇ubs + k 2 ñubs = ∇ · (I − Ã)∇u i + k 2 (1 − ñ)u i
+ SRC in Rm
D ⊂ Rm be a bounded simply connected open region
1. Ã := AχD + I(1 − χD ) and ñ := nχD + 1(1 − χD )
2. The matrix A ∈ Lip(D, Rm×m ) is symmetric-positive definite
3. The function n ∈ L∞ (D) such that n(x) > 0 for a.e. x ∈ D.
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The Scattered Field us0
We consider the scattering by a defective material where we find
1 (Rm ) with u i = exp(ikx · d).
the scattered field us0 ∈ Hloc
∇·Ã0 ∇u0s +k 2 ñ0 u0s = ∇·(I−Ã0 )∇u i +k 2 (1−ñ0 )u i
+ SRC in Rm
Now D0 ⊂ D be possibly multiple connected open set
1. Ã0 := A0 χD0 + Ã(1 − χD0 ) and ñ0 := n0 χD0 + ñ(1 − χD0 )
2. The matrix A0 ∈ Lip(D0 , Cm×m ) is symmetric
3. The function (possibly complex valued) n0 ∈ L∞ (D0 ).
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Far-Field Patterns
It is known that the radiating scattered fields which depends on
the direction d, has the following asymptotic expansion
1
e ik|x|
∞
s
ub (x̂, d) + O
as |x| → ∞
ub (x, d) =
m−1
|x|
|x| 2
us0 (x, d)
e ik|x|
=
|x|
m−1
2
1
∞
u0 (x̂, d) + O
as |x| → ∞
|x|
where the observation direction x̂ and incident direction d are in
the the unit circle or sphere S = {x ∈ Rm : |x| = 1}.
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Some Related Work
O. Bondarenko, A. Kirsch and X. Liu
The factorization method for inverse acoustic scattering in a layered
medium
Inverse Problems 29 (2013) 045010.
In this paper the FM is developed to reconstruct impenetrable sound
soft(or hard) defects in a layer media.
Y. Grisel, V. Mouysset, P.A. Mazet and J.P. Raymond
Determining the shape of defects in non-absorbing inhomogeneous media
from far-field measurements
Inverse Problems 28 (2012) 055003.
In this paper the FM is developed to reconstruct penetrable defects
in an inhomogeneous isotropic media.
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The Scattering Operator
Just as in the two other applications of the factorization method
for imaging inside of an object we see that the Scattering operator
for the ‘healthy’ material will be used in the factorization.
The Operator S
S : L2 (S) 7−→ L2 (S) for the healthy background where the
e iπ/4
1
constant γ, is given by γ = √
in R2 and γ = 4π
in R3
8πk
Z
S = I + 2ikγFb
where
(Fb g )(x̂) :=
u∞
b (x̂, d)g (d) ds(d).
S
Since A and n are real valued, the scattering operator is unitary,
i.e. SS ∗ = S ∗ S = I.
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The Relative Far-Field Operator
We will give a factorization for the relative far-field operator
defined by F := F0 − Fb . It can be shown that the operator F is
the far-field operator corresponding to only the defective region.
Z
(Fg )(x̂) :=
∞
u0 (x̂, d) − u∞
b (x̂, d) g (d) ds(d)
S
The Far-Field Data
1. u∞
0 (x̂, d) - is known from physical measurements
2. u∞
b (x̂, d) - can be computed directly since we assume A(x)
and n(x) along with D are known
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Modified Herglotz Operator
1 (Rm \ {z}) be the radiating Green’s function
Let G(·, ·) ∈ Hloc
of the background media,
∇ · Ã∇G(·, z) + k 2 ñG(·, z) = −δ( · − z) in Rm \ {z}
+ SRC
Define the operator H : L2 (S) 7−→ H 1 (D0 ), by
Z
s
Hg :=
ub (x, d) + exp(ikx · d) g (d) ds(d).
S
Theorem
The operator H ∗ : H 1 (D0 ) 7−→ L2 (S) satisfies the following:
1. H ∗ is compact with dense range.
2. S ∗ G∞ (·, z) ∈ Range(H ∗ ) if and only if z ∈ D0 .
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A Factorization Emerges
1 (Rm ) that satisfies
Given v ∈ H 1 (D0 ) we can construct u ∈ Hloc
∇ · Ã∇u + k 2 ñu = ∇ · (Ã − Ã0 )∇v + k 2 (ñ − ñ0 )v + SRC in Rm
Define the operator T : H 1 (D0 ) 7−→ H 1 (D0 ) such that
Z
(Tv , ϕ)H 1 (D0 ) = − (A − A0 )∇(v + u) · ∇ϕ − k 2 (n − n0 )(v + u)ϕ dx.
D0
Theorem
The far field operator F : L2 (S) 7−→ L2 (S) associated with the
defect can be factorized as F = −γSH∗ TH.
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Analysis of the Middle Operator
We define the real and imaginary part of an operators by
<(T ) =
T + T∗
2
and =(T ) =
T − T∗
.
2i
For simplicity lets assume we have that A0 and n0 are real valued.
Theorem
The operator T : H 1 (D0 ) 7−→ H 1 (D0 ) satisfies the following:
1. ±<(T ) is the compact perturbation of a coercive operator
2. =(T ) is non-positive and compact
Provided that A − A0 is either positive or negative definite D0 .
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Main Result (for Non-Absorbing Defects)
We define the selfadjoint compact operator
F] := < γ −1 S ∗ F + = γ −1 S ∗ F : L2 (S) 7−→ L2 (S).
Theorem
If A − A0 is either positive or negative definite in D0 , and let
(λi , ψi ) ∈ R+ × L2 (S) be an orthonormal eigensystem of F] . Then
"
z ∈ D0 ⇐⇒ W (z) =
∞
X
|(S ∗ G∞ (x̂, z), ψi )|2
i=1
λi
#−1
> 0.
For the results on absorbing defects see:
F. Cakoni and I. Harris
The factorization method for a defective region in an anisotropic material.
Inverse Problems 31 (2015) 025002
(arXiv:1410.3737).
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Numerical Reconstructions i.e. plot z 7−→ W (z)
Example 1. We consider D = [−2, 2]2 where the defective region
is a void D0 (i.e A0 = I and n0 = 1 in D0 ) embedded in isotropic
media. The coefficients in D are given by A = 0.5I and n = 3.
Figure: On the left is the reconstruction of the 2 circular voids. While on
the right is the reconstruction of the a circular void of radius 1. Where
the wavenumber is k = 1. Dashed line: exact boundaries of the scatterer
and void(s).
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Example 2. For this example we now reconstruct voids in an
anisotropic square scatterer D = [−2, 2]2 .The coefficients in D are
chosen to be given by n = 3 and
0.6022 0.1591
A=
0.1591 0.7478
Figure: On the left is the reconstruction of the 2 circular voids. While on
the right is the reconstruction of the a circular void of radius 1. Where
the wavenumber is k = 1. Dashed line: exact boundaries of the scatterer
and void(s).
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Figure: Questions?
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