Identification of surface impedance of thin dielectric objects from far-field data This article has been downloaded from IOPscience. Please scroll down to see the full text article. 2011 Inverse Problems 27 025011 (http://iopscience.iop.org/0266-5611/27/2/025011) Download details: IP Address: 128.114.163.7 The article was downloaded on 02/03/2013 at 09:51 Please note that terms and conditions apply. View the table of contents for this issue, or go to the journal homepage for more Home Search Collections Journals About Contact us My IOPscience
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Identification of surface impedance of thin dielectric objects from far-field data
This article has been downloaded from IOPscience. Please scroll down to see the full text article.
2011 Inverse Problems 27 025011
(http://iopscience.iop.org/0266-5611/27/2/025011)
Download details:
IP Address: 128.114.163.7
The article was downloaded on 02/03/2013 at 09:51
Please note that terms and conditions apply.
View the table of contents for this issue, or go to the journal homepage for more
Home Search Collections Journals About Contact us My IOPscience
Received 13 October 2010, in final form 2 December 2010Published 25 January 2011Online at stacks.iop.org/IP/27/025011
AbstractWe consider the inverse scattering problem of determining pointwise thesurface impedance of a thin dielectric infinite cylinder having an open arcas cross section from the knowledge of the TM-polarized scattered far-fieldelectromagnetic field at a fixed frequency. In this work, we show that the surfaceimpedance is uniquely determined by the far-field data. We derive a Newton-type method to reconstruct the surface impedance. Numerical examples aregiven showing the efficaciousness of our algorithms.
(Some figures in this article are in colour only in the electronic version)
1. Introduction
In this paper, we investigate the inverse problem of using time-harmonic electromagneticfar-field measurements to determine information about the thickness and physical propertiesof thin dielectric films. This work needs prior detection of the shape and location of dielectricsfrom far-field measurements.
The authors have developed, in [11], qualitative methods to solve inverse scatteringproblems to determine the shape and location of complex thin dielectrics. In particular, theauthors were able to detect them by applying the linear sampling method. This type ofqualitative method was first introduced in [1] for the case of an obstacle with empty interior,and adapted in [2] for the case of thin dielectrics.
In addition to the detection of cracks, the authors developed a technique to reconstructthe surface impedance for the case when it is constant along the dielectric. In this paper, wedevelop the theoretical framework for the inverse problem, prove that the surface impedancecan be uniquely determined from the knowledge of the far-field data and develop a Newton-type method to determine the surface impedance pointwise for the case when it is a continuousfunction along the crack, denoted by �.
Considerable work has been done on inversion techniques using Newton-type methods.We refer the reader to [9], [7] and [8] for more information on these type of optimization
methods. These methods are known for their high-quality reconstruction from a fairly smallamount of measured data. The Newton-type methods are based on solving a nonlinear ill-posedequation of the form
G(λ) = u∞ (1)
where G : Cλc([−1, 1]) → L2(�) maps a continuous surface impedance λ into the far-field
pattern u∞ (to be stated more precisely later) corresponding to λ, the dielectric � and a fixedwavenumber k. The operator G is called the impedance to the measurement operator. Wedenote � = [0, 2π ] × [0, 2π ], G(λ) = Gλ and
Cλc([−1, 1]) := {f ∈ C([−1, 1]) : f (x) > λc > 0 for x ∈ [−1, 1]}.
Newton-type methods are based on solving the linearized version of equation (1):
G′λh + Gλ = u∞ (2)
for small h; then, the current parameterized function λ is replaced by λ+h. The differentiabilityof the operator Gλ will be discussed later.
The main concern of this paper is to solve the inverse problem of determining λ pointwisefrom the far-field pattern. In the next section, we briefly describe the direct problem. Insection 3, we investigate the uniqueness results for this inverse problem. In section 4, wedefine the operator Gλ and study its differentiability. Finally, in section 5, we show thenumerical implementation of our algorithm to solve the ill-posed inverse problem. Numericalexamples are provided to show the efficaciousness of this method.
2. The direct problem
In this work, we assume that the obstacle is a thin dielectric right cylinder with an open arc inR
2 as cross section whose properties depend only on the cross section of the cylinder and theincident electromagnetic field is E-polarized. This leads to the exterior boundary problem
�u + k2n(x)u = 0
outside the cylinder, where n(x) is the index of refraction of the background medium whichsatisfies Re n > 0 and Im n � 0, and u = ui + us is the total field. Note that us is the scatteredfield due to an incident field ui by the cross section of the cylinder.
We refer the reader to [5], [6] and [4] for information on the asymptotic analysis involvedin the derivation of the boundary conditions. For the sake of simplicity, we assume thatthe medium is homogeneous, i.e. n(x) = 1 outside the cylinder. We also assume that� ⊂ R
2 is a simple piecewise smooth arc, i.e. � = {ρ(s) : s ∈ [s0, s1]}, where the mappingρ : [s0, s1] → R
2 is one to one, continuous and piecewise smooth. Hence, we arrive at thefollowing boundary value problem for the scattered field us due to an incident field ui scatteredby the crack �:
�us + k2us = 0 in R2 \ � (3)[
∂(us + ui)
∂ν
]= 0 on � (4)
[(us + ui)] − iλ∂(us + ui)+
∂ν= 0 on � (5)
limr→∞
√r
(∂us
∂r− ikus
)= 0 (6)
2
Inverse Problems 27 (2011) 025011 N Zeev
where u±(x) = limh→0+
u(x ± hν) and ∂u±∂ν
(x) = limh→0+
ν · ∇u(x ± hν) for x ∈ �, [u] := u+ − u−
and[
∂u∂ν
]:= ∂u+
∂ν− ∂u−
∂νare the respective jumps across �, and the dimensionless positive
continuous function λ > λc > 0 involves electric permittivity and magnetic permeability ofthe dielectric medium and the background as well as the thickness ht and frequency ω. TheSommerfeld radiation condition (6) is satisfied uniformly in x = x/|x| with r = |x|. Here,we assume that k is the wavenumber in the air. In this study, we use the incident field as aplane wave and the normal vector ν pointing to the right side of �.
In order to formulate the previous system in more general form, we let λ be a piecewisesmooth function on � such that λ(x) > λc > 0. Given f ∈ H− 1
2 (�) and h ∈ H− 12 (�), find
v ∈ H 1loc(R
2 \ �) satisfying
�v + k2v = 0 in R2 \ � (7)[
∂v
∂ν
]= f on � (8)
[v] − iλ∂v+
∂ν= h on � (9)
limr→∞
√r
(∂v
∂r− ikv
)= 0. (10)
It is shown in [11] that the well posedness of the direct problem by using the integralequation method, i.e. the integral equation( i
λIc + T�
)[v] = i
λh +
(K ′
� − I
2
)f, (11)
has a unique solution [v] ∈ H12 (�) that is continuously dependent on the data f and h,
where Ic is the identity compact embedding operator from H 1/2(�) to H−1/2(�), the operatorsK ′
� : H−1/2(�) → H−1/2(�) and T� : H 1/2(�) → H−1/2(�) are defined by
(K ′�ψ)(x) :=
∫�
ψ(y)∂
∂νx
�(x, y)dsy for x ∈ �,
(T�ψ)(x) := ∂
∂νx
∫�
ψ(y)∂
∂νy
�(x, y) dsy for x ∈ �,
respectively, where �(x, y) = i/4H(1)0 (k |x − y|) is the fundamental solution of the Helmholtz
equation with H(1)0 being a Hankel function of the first kind of order zero.
The scattered field can then be computed,
us(x) :=∫
�
[v](y)∂
∂νx
�(x, y) dsy for x ∈ �,
and solved using boundary element methods.A set of multistatic data was computed in [11], i.e. the corresponding far field measured at
many observation directions on a subset of the unit circle with incident directions on a subsetof the unit circle.
3. Uniqueness of the inverse problem
The inverse problem we investigate is based on detecting the surface impedance λ ∈Cλc
([−1, 1]) (parameterized along �) pointwise from the knowledge of the location and
3
Inverse Problems 27 (2011) 025011 N Zeev
shape of � (reconstructed) and the far-field data u∞(x, d), where x and d are on the unit circleS := {x ∈ R
2 : |x| = 1} and λc > 0.In this section, we assume that the incident field is a time harmonic plane wave given by
ui := eikx·d for x ∈ R2, where the unit vector d ∈ S is the incident direction. In this setting, the
scattered field us satisfies (7)–(10), f := −[∂eikx·d
∂ν
] = 0, and h := −[eikx·d ]+iλ∂eikx·d∂ν
= iλ∂eikx·d∂ν
.It is shown in [3] that the scattered field, which also depends on d now, has the asymptoticbehavior
u(x) = eikr
√ru∞(x, d) + O(r−3/2) (12)
where u∞ is the far-field pattern of the scattered wave x = x/|x| and r = |x|.Uniqueness results for the detection of � and constant surface impedance λ were proved
by the authors ( see lemma 2.2.4) in [10].
Lemma 3.1. Assume that �1 and �2 are two scattering obstacles with corresponding constantsurface impedances λ1 and λ2, such that for a fixed wavenumber the far-field patterns coincidefor all incident directions d, and then �1 = �2 and λ1 = λ2.
Proving that the scatterers �1 and �2 coincide, do not use the assumption that λ1 and λ2
are constant.In this work, we develop the theoretical framework and a numerical method to uniquely
detect λ ∈ Cλc([−1, 1]). In order to do that, we recall that a Herglotz wavefunction is a
solution of the Helmholtz equation in R2 of the form
vig(x) :=
∫S
g(d) eikx·d ds(d) (13)
where g ∈ L2(S) is the kernel of vig and S is the unit circle. Let F : L2(S) → L2(S) be the
corresponding far-field operator defined by
(Fg)(θ) =∫
S
u∞(θ, φ)g(φ) dφ.
By superposition we have the following relation:
(Fg) = B(iλHg)
where H : L2(S) → H− 12 (�) is defined by
Hg := ∂vig
∂ν(14)
and B : H− 12 (�) → L2(S) takes h ∈ H− 1
2 (�) to the far-field pattern u∞ of the solution to(7)–(10) with n(x) = 1, f := 0 and h.
For β ∈ H12 (�), we construct the double layer potential
D(β)(x) :=∫
�
β(y)∂
∂νy
�(x, y) ds(y)
which has as a far-field pattern γFβ, where
Fβ :=∫
�
β(y)∂ e−ikx·y
∂νy
ds(y)
and γ = eiπ/4√8πk
.In order to show that λ is uniquely determined by the far-field pattern u∞, we first show
the following theorem.
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Inverse Problems 27 (2011) 025011 N Zeev
Theorem 3.2. Let vig be the Herglotz wavefunction with kernel g and vg = vi
g + vsg be the
solution of (7)–(10) with incident wave vig . Then,∫
�
λ(x)
∣∣∣∣∂vg
∂ν
∣∣∣∣2
ds(x) = −k ‖Fg‖2 +√
8πk Im(eiπ/4(Fg, g)).
Proof. Let D be any bounded domain with a piecewise smooth boundary extending �. Letvs and ws be two radiating solutions of the Helmholtz equation with far fields v∞ and w∞. Byboundary conditions, we get∫
∂D
([v]
∂w
∂ν− [w]
∂v
∂ν
)ds = 2i
∫�
λ(x)∂v
∂ν
∂w
∂νds(x).
By the Green’s second identity in D,∫∂D
([v]
∂w
∂ν− [w]
∂v
∂ν
)ds =
∫∂D
(v+ ∂w
∂ν− w+ ∂v
∂ν
)ds
=∫
∂D
(vs ∂ws
∂ν− ws ∂vs
∂ν
)ds +
∫∂D
(vs ∂wi
∂ν− wi ∂vs
∂ν
)ds
+∫
∂D
(vi ∂ws
∂ν− ws ∂vi
∂ν
)ds +
∫∂D
(vi ∂wi
∂ν− wi ∂vi
∂ν
)ds.
Let �R be a ball of radius R containing D. Using the Green’s second identity in �R \ D
and the asymptotic behavior of vs and ws as R → ∞, we get∫∂D
(vs ∂ws
∂ν− ws
∂vs
∂ν
)ds = −2ik
∫S
v∞w∞ ds.
Let wih be a Herglotz wavefunction with kernel h:
wih(x) =
∫S
eikx·dh(d) ds(d).
From the far-field pattern identity,
v∞(x) = eiπ/4
√8πk
∫∂D
(vs(y)
∂ e−ikx·y
∂ν− e−ikx·y ∂vs(y)
∂ν
)ds(y),
interchanging the order of integrations, we have∫∂D
(vs(y)
∂wih(y)
∂ν− wi
h(y)∂vs(y)
∂ν
)ds
=∫
S
h(x)
∫∂D
(vs(y)
∂ e−ikx·y
∂ν− e−ikx·y ∂vs(y)
∂ν
)ds(y) ds(x)
=√
8πk e−iπ/4∫
S
h(x)v∞(x) ds(x).
Let vg = vig + vs
g be the solution to (7)–(10). Then, from the above results we get
2i∫
�
λ(x)
∣∣∣∣∂vg
∂ν
∣∣∣∣2
ds(x) =∫
∂D
([vg]
∂vg
∂ν− [vg]
∂vg
∂ν
)ds
=∫
∂D
(vs
g
∂vgs
∂ν− vg
s∂vs
g
∂ν
)ds +
∫∂D
(vs
g
∂vgi
∂ν− vg
i∂vs
g
∂ν
)ds
+∫
∂D
(vi
g
∂vgs
∂ν− vg
s∂vi
g
∂ν
)ds +
∫∂D
(vi
g
∂vgi
∂ν− vg
i∂vi
g
∂ν
)ds
= −2ik ‖Fg‖2 +√
8πk e−iπ/4(Fg, g) −√
8πk eiπ/4(g, Fg).
This proves the theorem. �
5
Inverse Problems 27 (2011) 025011 N Zeev
We proceed by showing the denseness property of the Herglotz wavefunction.
Lemma 3.3. Let
W :={f ∈ L2(�) : f = ∂vg
∂ν|� for all g ∈ L2(S)
}where vg is defined as in the previous theorem. Then, W is dense in L2(�).
Proof. Let D be any bounded domain with a piecewise smooth boundary extending �. Inorder to prove the denseness of W , it is sufficient to show that if ψ ∈ L2(�) satisfies∫
�
ψ∂vg
∂νds = 0 for all f = ∂vg
∂ν|� ∈ W,
then ψ = 0.Let w ∈ H 1
loc(R2 \ �) be the solution of
�w + k2n(x)w = 0 in R2 \ � (15)[
∂w
∂ν
]= 0 on � (16)
[w] − iλ∂w+
∂ν= ψ on � (17)
limr→∞
√r
(∂w
∂r− ikw
)= 0. (18)
Using boundary conditions, we get
0 =∫
�
ψ∂vg
∂νds =
∫�
∂vg
∂ν
([w] − iλ
∂w
∂ν
)ds
=∫
∂D
(∂vg
∂ν[w] − [vg]
∂w
∂ν
)ds.
By same arguments as in the previous theorem, we get∫S
vg,∞(x, d)w∞(x, d) ds(x) = 0
for all g ∈ L2(S), and all d ∈ S. Since the far-field operator vg,∞ is injective and has a denserange, we get
w∞ = 0 for x, d ∈ S.
Hence, w = 0 in R2 \ �. Therefore, ψ = 0.
�
Theorem 3.4. Assume that �1 and �2 are two cracks with surface impedances λ1 and λ2
such that the far-field patterns u1∞(x, d) and u2
∞(x, d) coincide for all x, d ∈ S; then, λ1 =λ2.
Proof. By lemma 2.1, since u1∞(x, d) = u2
∞(x, d) for all x, d ∈ S, then �1 = �2, and thecorresponding scattered fields coincide us
1 = us2 (see lemma 2.2.4 in [10]) in R
2 \ � for alld ∈ S. Hence, if v
jg is the solution to (7)–(10) corresponding to u
j∞ with eikx·d replaced by vi
g ,then by the previous theorem, we have∫
�
(λ1(x) − λ2(x))
∣∣∣∣∂vg
∂ν
∣∣∣∣2
ds(x) = 0.
By the previous lemma, we have λ1 = λ2 almost everywhere on � which proves the uniquenessof λ. �
6
Inverse Problems 27 (2011) 025011 N Zeev
4. Frechet differentiability of the impedance to the measurement operator
In this section, we want to define the impedance to the measurement operator G :Cλc
([−1, 1]) → L2(�), where � = [0, 2π ] × [0, 2π ] and study its differentiability. Werecall that the scattered field can be written as
us = K(Ic − iλT�)−1Rui (19)
where Rui = iλ∂ui
∂νand K is the double layer potential
Kψ(x) =∫
�
ψ(y)∂
∂νy
�(x, y) ds(y) on R2 \ �.
Then, the corresponding far-field pattern can be written as
u∞ = K∞(Ic − iλT�)−1Rui (20)
where
K∞ψ(x) = γ
∫�
ψ(y)∂ e−ikx·y
∂νy
ds(y).
Let
Gλ(θ, φ) = u∞(θ, φ)
where θ and φ represent the direction of the incident wave ui and the direction of measurement,respectively. From the above definition, it is easy to see that if λ ∈ Cλc
([−1, 1]) is a functionparameterized on �, then the operator Gλ = K∞(Ic − iλT�)−1Rui is differentiable withrespect to λ. Furthermore, we have the following theorem.
Theorem 4.1. The Frechet derivative of the far-field operator Gλ : Cλc([−1, 1]) → L2(�)
with respect to λ is given by
G′λh = K∞ψh
where
(Ic − iλT�)ψh = −ihT�(Ic − iλT�)−1Rui + ih∂ui
∂ν.
Proof. The proof comes from differentiating (20) with respect to λ. Note that in order to dothat, we use the fact that (Ic − iλT�) is invertible and Frechet differentiable; hence (see [9])the Frechet derivative of the inverse is of the form
(Ic − iλT�)−1(Ic − iλT�)′(Ic − iλT�)−1. �
The following theorem shows that the Frechet derivative G′λ is well defined.
Theorem 4.2. The operator G′λ is injective.
Proof. Assume G′λq = 0. Then, the solution to (7)–(10) with f = 0 and h replaced by
−iqT�(Ic − iλT�)−1Rui + iq ∂ui
∂νhas a vanishing far-field pattern. By Rellich’s lemma, v = 0
in R2 \ �. Therefore,
−iqT�(Ic − iλT�)−1Rui + iq∂ui
∂ν= 0.
Since (Ic − iλT�)−1Rui = [u] and the fact that
[u] − iλ∂u+
∂ν= iλ
∂eikx·d
∂ν,
we have q = 0.We also note that the operator Gλ as defined above is compact. Hence, equation (2) is
ill-posed. Therefore, we need to use regularization theory to get its solution. �
7
Inverse Problems 27 (2011) 025011 N Zeev
5. Numerical examples
In this section, we shall briefly describe the reconstruction of the crack � and then use theNewton-type method to solve the inverse problem of reconstructing λ from the knowledgeof noisy synthetic far-field data u∞ ∈ L2(�), the shape of the reconstructed � and a fixedwavenumber k.
We note that the scattered field
us =∫
�
(Ic − iλT�)−1Rui(y)∂
∂νy
�(x, y) ds(y)
has the far field
u∞(x) = γ
∫�
(Ic − iλT�)−1Rui(y)∂
∂νy
e−ikx·y ds(y) (21)
where γ = eiπ/4/√
8πk.In our numerical experiments, the wavenumber k = 5 and the scatterer � is be described
by ρ : [−1, 1] → R2 of the form
ρ :={(
s, 2 sin
(3π
2s
)+ 2, sin
(3π
2s + π
)): −1 � s � 1
}and
ρ := {(s − 2, 2s) : −1 � s � 1} .
In all choices of �, the distance between its endpoints is about one wavelength. Weparameterize λ along the crack by l : [−1, 1] → Cλc
([−1, 1]) of the form
(l ◦ g) := {0.25s + 0.55 : −1 � s � 1}and
(l ◦ g) :={
0.5 cosπs
2+ 0.2 sin
πs
2− 0.1 cos
3πs
2: −1 � s � 1
},
with g : [−1, 1] → ρ(s) being a function parameterizing the crack in the interval [−1, 1].Using time harmonic plane waves as the incident field ui(x) = eikx·d , we computed the
approximate far-field pattern (21) for p equidistant points x in the unit circle S,
u∞ =N∑
n=−N
u∞,n exp(inθ)
where d is a unit vector representing the direction of incidence.Adding random noise, we write
u∞,a =N∑
n=−N
u∞,a,n exp(inθ)
for each incident direction d, and u∞,a,n = u∞,n(1+εχn) with χn a random variable in [−1, 1](ε = 0.05 or ε = 0.1 in our examples).
It is shown in [11] (theorem 3.3), using regularization theory, that for an arbitrary arcL ⊂ � and βL ∈ H
12 (L), we can find a bounded solution g ∈ L2(S) that satisfy∫
S
u∞(x, d)g(d) ds(d) = γ
∫L
βL(y)∂
∂νy
e−ikx·y dsy (22)
with discrepancy ε. In addition, all approximate solutions to (22) are unbounded for L �⊂ �.
8
Inverse Problems 27 (2011) 025011 N Zeev
(a) (b)
Figure 1. Panels (a) and (b) show the original crack (dashed red curve) and the interpolationcurve of the reconstructed points along the crack (blue curve). The reconstruction was obtained asdescribed above using the linear sampling method with 5% noise, and the wavenumber k = 5.
We then sample a region containing � in its interior. Let L be a small segment centeredat a sampling point z with the unit normal vector nz and βL a sequence that converges to δ(z).Thus, in the limiting case, (22) is replaced by∫ π
−π
u∞(x, θ)gz,nz(θ) dθ = −ikγ nz · x e−ikx·z x ∈ S, z ∈ R
2 and nz ∈ S.
We are able to detect � (see figure 5.2, 5.3 and 5.4 in [11]) using the solution gz,nzas our
indicator function, since if z ∈ � and nz coincides with the normal vector to � at z, then wecan find a bounded gz,nz
∈ L2(S) that approximate the above equation. Otherwise all suchgz,nz
∈ L2(S) are unbounded.We are now interested in obtaining pointwise information of the position and slopes of
the reconstructed crack �. Hence, we use the Matlab command ‘contour’ to get informationon the level curves of the matrix
A =⎛⎝∑
nz∈S
1
‖gz,nz‖
⎞⎠
i,j
,
where z = (xi, yj ) is a point at the sampled space. In our numerical experiments, we usexi, yi ∈ [−5, 5].
It is now possible to find good approximations of the reconstructed crack. In particular,since the level curves enclose the desired curve �, in our numerical experiments, we averagepoints on both sides of the same level curves reasonably close to each other.
After getting a set of points approximating the location of the crack, we obtain a piecewisecubic interpolation using the Matlab command ‘pp=interp(x,y,‘cubic’,‘pp’)’. Having theinterpolating curve, we are able to get pointwise information of the curve as well as its slopeusing the Matlab command ‘y=ppval(pp,x)’ (see figure 1).
Note that the accuracy of the interpolation depends on the discretization in the linearsampling method. We now proceed to obtain equally distributed values of points along thereconstructed crack and use them to reconstruct λ.
Having � reconstructed, we can now go back to equation (2):
G′λh + Gλ = u∞.
9
Inverse Problems 27 (2011) 025011 N Zeev
(a) (b)
Figure 2. Panel (a) shows a plot of λ(x) (in red) and the initial guess λ0(x) (in blue) chosen alongthe crack. Panel (b) shows a plot of λ(x) (in red) and lambda reconstructed (in blue) after teniterations with Newton-type methods. One incident field was used, k = 5, 5% noise and N = 3.
We choose h(x) = ∑Nk=0 akfk(x); hence, we want to solve
N∑k=0
akG′λfk + Gλ = u∞
for real ak. We considered Chebyshev polynomials fk(x) = cos(k cos−1(x)), and to havelocal effects we added radial basis functions fk(x) = e−c|x−ζk |2 , with x ∈ [−1, 1], c > 0 andζk ∈ [−1, 1].
To avoid an inverse crime, we choose a different number of quadrature points in thecomputations of the direct (n = 64 points) and inverse problems (n = 32 points). We usea collocation method with respect to m equidistant points x1 · · · xm in the unit circle, i.e. wesolve
for real ak.Even though we have computed the far-field data corresponding to multiple incident
directions for the reconstructed �, we only need few incident directions for the reconstructionof λ. In particular, our numerical experiments were produced using a single incident direction.
We applied the Tikhonov regularization to deal with the ill-posedness of (23) and sincein general (23) is over-determined (m > N + 1), we consider the minimization the followingequivalent problem:
C(p, a0, . . . , aN) :=p∑
k=0
⎧⎨⎩
m∑i=1
|bijkaj − cik|2 + α
N∑j=0
θjka2j
⎫⎬⎭ (24)
for real values ak, where bijk = (G′λfj )(xi; dk), cik = u∞(xi; dk) − Gλ(xi; dk), α is the
regularization parameter, and θjk are positive weights.We solve the linear system
p∑k=1
N∑l=0
m∑i=1
bijkbilkal + α
p∑k=1
θjkaj =p∑
k=1
m∑i=1
bijkcik, j = 0, . . . , N. (25)
10
Inverse Problems 27 (2011) 025011 N Zeev
(a) (b)
Figure 3. Panel (a) shows a plot of λ(x) and the initial guess λ0(x) chosen along the crack. Panel(b) shows a plot of λ(x) and lambda reconstructed after nine iterations with Newton-type methods.One incident field was used, k = 5, 5% noise and N = 3.
(a) (b)
Figure 4. Panel (a) shows a plot of λ(x) and the initial guess λ0(x) chosen along the crack.Panel (b) shows a plot of λ(x) and lambda reconstructed after 12 iterations with Newton-typemethods. One incident field was used, k = 5, 5% noise and N = 3.
In our numerical tests, we use α = 2−j , where j is the number of iteration, and p = 1 (oneincident direction). We solve (24) by using a variation of the Levenberg–Marquardt choosingθj,k = ∑m
i=1 |bi,j,k|2.We solve (25) on each iteration of the numerical examples. We iterate until the impedance
to the measurement operator Gλ satisfies‖Gλ − u∞‖
‖u∞‖ < δ
for small enough δ (δ = O(10−2) in our experiments).The initial guesses chosen in the numerical experiments were
We are able to reconstruct the surface impedance pointwise after few iterations of theNewton-type method (see figures 2–4). Figure 4 suggests that this method is robust andproduce good reconstructions even when the initial guess is far from the real value.
11
Inverse Problems 27 (2011) 025011 N Zeev
Acknowledgments
The author is very grateful to Professor Dr F Cakoni for her interest and for useful discussionson the topic of this paper. The research of the author is supported by the ODU SummerResearch Fellowship. The support is gratefully acknowledged.
References
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absorbing obstacles: the scalar case Math. Models Methods Appl. Sci. 15 1273–300[6] Haddar H, Joly P and Nguyen H M 2004 Asymptotic models for scattering by highly conducting bodies INRIA
POEMS Technical Reports[7] Kress R 1995 Frechet differentiability of the far field operator for scattering from a crack J. Inverse Ill-Posed
Problems 3 305–13[8] Monch L 1997 On the inverse acoustic scattering problem by an open arc: the sound-hard case Inverse
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