The inverse electromagnetic scattering problem by a penetrable cylinder at oblique incidence

1.8 MB·

by  Drossos Gintides

The inverse electromagnetic scattering problem by a
penetrable cylinder at oblique incidence
Drossos Gintides∗1 and Leonidas Mindrinos†2
1
Department of Mathematics, National Technical University of Athens, Greece.
2Computational Science Center, University of Vienna, Austria.
7
1
0
2

n Abstract
a
J In this work we consider the method of non-linear boundary integral equation for solving
numerically the inverse scattering problem of obliquely incident electromagnetic waves by a
7
penetrable homogeneous cylinder in three dimensions. We consider the indirect method and

] simple representations for the electric and the magnetic fields in order to derive a system
A of five integral equations, four on the boundary of the cylinder and one on the unit circle
N wherewemeasurethefar-fieldpatternofthescatteredwave. Wesolvethesystemiteratively
by linearizing only the far-field equation. Numerical results illustrate the feasibility of the
.
h proposed scheme.
t
a Keywordsinverseelectromagneticscattering,obliqueincidence,integralequationmethod
m
[
1 Introduction

1
v
The inverse obstacle scattering problem is to image the scattering object, i.e. find its shape
9
5 and location, from the knowledge of the far-field pattern of the scattered wave. The medium
8 is illuminated by light at given direction and polarization. Then, Maxwell’s equations are used
1 to model the propagation of the light through the medium, see [6, 7] for an overview. Due to
0
the complexity of the combined system of equations for the electric and the magnetic fields, it
.
1 is common to impose additional assumptions on the incident illumination and the nature of the
0 scatterer.
7
Weconsidertime-harmonicincidentelectromagneticplanewavethatduetothelinearityofthe
1
problem will result to a time-independent system of equations. In addition, the penetrable object
:
v
is considered as an infinitely long homogeneous cylinder. Then, it is characterized by constant
i
X permittivityandpermeability. Theproblemisfurthersimplifiedifweimposeobliqueincidencefor
r the incident wave.
a The three-dimensional scattering problem modeled by Maxwell’s equations is then equivalent
to a pair of two-dimensional Helmholtz equations for two scalar fields (the third components
of the electric and the magnetic fields). This approach reduces the difficulty of the problem
but results to more complicated boundary conditions. The transmission conditions now contain
also the tangential derivatives of the electric and magnetic fields. In [10] we showed that the
corresponding direct problem is well-posed and we constructed a unique solution using the direct
integral equation method. A similar problem has been considered for an impedance cylinder
embedded in a homogeneous [24], and in an inhomogeneous medium [22]. A numerical solution
∗[email protected]
†[email protected]
1
of the direct problem has been also proposed using the finite element method [4], the Galerkin
method [20], and the method of auxiliary sources [23].
On the other hand, the inverse problem is non-linear and ill-posed. The non-linearity is due to
the dependence of the solution of the scattering problem on the unknown boundary curve. The
smoothness of the mapping from the boundary to the far-field pattern reflects the ill-posedness of
the inverse problem. The unique solvability of the inverse problem is still an open problem. The
first and only, to our knowledge, uniqueness result was presented recently in [21] for the case of an
impedance cylinder using the Lax-Phillips method.
In this work, we solve the inverse problem by formulating an equivalent system of non-linear
integralequationsthatissolvedusingaregularizediterativescheme. Thismethodwasintroduced
by Kress and Rundell [18] and then considered in many different problems, in acoustic scattering
problems [12, 13], in elasticity [5, 9] and in electrical impedance problem [8]. We propose an
iterative scheme that is based on the idea of Johansson and Sleeman [14] applied to the inverse
problem of recovering a perfect conducting cylinder. See [1, 5] for some recent applications. We
assumeintegralrepresentationsforthesolutionsthatresultstoasystemconsistingoffourintegral
equationsontheunknownboundary(consideringthetransmissionconditions)andoneontheunit
circle (taking into account the asymptotic expansion of the solutions).
We solve this system in two steps. First, given an initial guess for the boundary curve we solve
the well-posed subsystem (equations on the boundary) to obtain the corresponding densities and
then we solve the linearized (with respect to the boundary) ill-posed far-field equation to update
the initial approximation of the radial function. We consider Tikhonov regularization and the
normal equations are solved by the conjugate gradient method.
The paper is organized as follows: in Section 2 we present the direct scattering problem, the
elastic potentials and the equivalent system of integral equations that provide us with the far-
field data. The inverse problem is stated in Section 3 where we construct an equivalent system of
integralequationusingtheindirectintegralequationmethod. InSection4thetwo-stepmethodfor
theparametrizedformofthesystemandthenecessaryFr´echetderivativeoftheintegraloperators
arepresented. Thenumericalexamplesgivesatisfactoryresultsandjustifytheapplicabilityofthe
proposed iterative scheme.
2 The direct problem
We consider the scattering of an electromagnetic wave by a penetrable cylinder in R3. Let x =
(x,y,z) ∈ R3. We denote by Ω = {x : (x,y) ∈ Ω,z ∈ R} the cylinder, where Ω is a bounded
int
domain in R2 with smooth boundary Γ. The cylinder Ω is oriented parallel to the z-axis and
int
Ω is its horizontal cross section. We assume constant permittivity (cid:15) and permeability µ for the
0 0
exterior domain Ω := R3 \Ω . The interior domain Ω is also characterized by constant
ext int int
parameters (cid:15) and µ .
1 1
We define the exterior magnetic Hext(x,t) and electric field Eext(x,t) for x∈Ω , t∈R and
ext
theinteriorfieldsHint(x,t)andEint(x,t)forx∈Ω , t∈R,thatsatisfytheMaxwell’sequations
int
∂Hext ∂Eext
∇×Eext+µ =0, ∇×Hext−(cid:15) =0, x∈Ω ,
0 0 ext
∂t ∂t
(1)
∂Hint ∂Eint
∇×Eint+µ =0, ∇×Hint−(cid:15) =0, x∈Ω .
1 1 int
∂t ∂t
and the transmission conditions
nˆ×Eint =nˆ×Eext, nˆ×Hint =nˆ×Hext, x∈Γ, (2)
where nˆ is the outward normal vector, directed into Ω .
ext
2
z
Einc
Ωext Hinc
(cid:15)0,µ0
dˆ
Ωint
θ
Ω
O
(cid:15)1,µ1
y
φ
x
Figure 1. Thegeometryofthescatteringproblem.
We illuminate the cylinder with an incident electromagnetic plane wave at oblique incidence,
meaningtransversemagnetic(TM)polarizedwave. Wedefinebyθ theincidentanglewithrespect
tothenegativezaxisandbyφthepolarangleoftheincidentdirectiondˆ(insphericalcoordinates),
see Figure 1. Then, dˆ= (sinθcosφ,sinθsinφ,−cosθ) and the polarization vector is given by
pˆ=(cosθcosφ,cosθsinφ,sinθ), satisfying dˆ⊥pˆand assuming that θ ∈(0,π/2)∪(π/2,π).
In the following, due to the linearity of the problem we suppress the time-dependence of the
fields and because of the cylindrical symmetry of the medium we express the incident fields as
separable functions of x:=(x,y) and z.
√
Let ω >0 be the frequency and k =ω µ (cid:15) the wave number in Ω . We define β =k cosθ
0 0 0 ext 0
(cid:112)
and κ = k2−β2 =k sinθ and it follows that the incident fields can be decomposed to [10]
0 0 0
Einc(x;dˆ,pˆ)=einc(x)e−iβz, Hinc(x;dˆ,pˆ)=hinc(x)e−iβz, (3)
where
1 1
einc(x)= √ pˆeiκ0(xcosφ+ysinφ), hinc(x)= √ (sinφ,−cosφ,0)eiκ0(xcosφ+ysinφ).
(cid:15) µ
0 0
After some calculations, we can reformulate Maxwell’s euqations (1) as a system of equations
only for the z-component of the electric and magnetic fields [10]. The interior fields eint(x) and
3
hint(x), x ∈ Ω := Ω and the exterior fields eext(x) and hext(x), x ∈ Ω := R2 \Ω satisfy the
3 1 3 3 0
Helmholtz equations
∆eint+κ2eint =0, ∆hint+κ2hint =0, x∈Ω ,
3 1 3 3 1 3 1
(4)
∆eext+κ2eext =0, ∆hext+κ2hext =0, x∈Ω ,
3 0 3 3 0 3 0
3
where κ2 =µ (cid:15) ω2−β2. Here, we assume µ (cid:15) >µ (cid:15) cos2θ in order to have κ2 >0.
1 1 1 1 1 0 0 1
The transmission conditions (2) can also be written only for the z-component of the fields.
Let (nˆ,τˆ) be a local coordinate system, where nˆ = (n ,n ) is the outward normal vector and
1 2
τˆ = (−n ,n ) the outward tangent vector on Γ. We define ∂ = nˆ ·∇ , ∂ = τˆ ·∇ , where
2 1 ∂n t ∂τ t
∇ = e ∂ + e ∂ and e ,e denote the unit vectors in R2. Then, we rewrite the boundary
t 1∂x 2∂y 1 2
conditions as [10]
eint =eext, x∈Γ,
3 3
∂hint ∂eint ∂hext ∂eext
µ˜ ω 3 +β 3 =µ˜ ω 3 +β 3 , x∈Γ,
1 1 0 0
∂n ∂τ ∂n ∂τ
(5)
hint =hext, x∈Γ,
3 3
∂eint ∂hint ∂eext ∂hext
(cid:15)˜ ω 3 −β 3 =(cid:15)˜ ω 3 −β 3 , x∈Γ,
1 1 0 0
∂n ∂τ ∂n ∂τ
where µ˜ = µ /κ2, (cid:15)˜ = (cid:15) /κ2, β = β/κ2, for j = 0,1. The exterior fields are decomposed to
j j j j j j j j
eext = esc + einc and hext = hsc + hinc, where esc and hsc denote the scattered electric and
3 3 3 3 3 3 3 3
magnetic field, respectively. From (3) we see that
1
einc(x)= √ sinθeiκ0(xcosφ+ysinφ), hinc(x)=0. (6)
3 (cid:15) 3
0
Toensurethatthescatteredfieldsareoutgoing,weimposeinadditiontheradiationconditions
in R2 :
√ (cid:18)∂esc (cid:19) √ (cid:18)∂hsc (cid:19)
lim r 3 −iκ esc =0, lim r 3 −iκ hsc =0, (7)
r→∞ ∂r 0 3 r→∞ ∂r 0 3
where r =|x|, uniformly over all directions.
Now we are in position to formulate the direct transmission problem for oblique incident wave:
Find the fields hint,hsc,eint and esc that satisfy the Helmholtz equations (4), the transmission
3 3 3 3
conditions (5) and the radiation conditions (7).
Theorem 2.1 If κ2 is not an interior Dirichlet eigenvalue and κ2 is not an interior Dirichlet and
1 0
Neumann eigenvalue, then the direct transmission problem (4) – (7) admits a unique solution.
Proof: The proof is based on the integral representation of the solution resulting to a Fredholm
type system of boundary integral equations. For more details see [10, Theorem 3.2]. (cid:3)
In the following, j = 0,1 counts for the exterior (x ∈ Ω ) and interior domain (x ∈ Ω ),
0 1
respectively. We introduce the single- and double-layer potentials defined by
(cid:90)
(S f)(x)= Φ (x,y)f(y)ds(y), x∈Ω ,
j j j
(cid:90)Γ (8)
∂Φ
(D f)(x)= j (x,y)f(y)ds(y), x∈Ω ,
j j
∂n(y)
Γ
where Φ is the fundamental solution of the Helmholtz equation in R2 :
j
i
Φ (x,y)= H(1)(κ |x−y|), x,y ∈Ω , x(cid:54)=y, (9)
j 4 0 j j
4
and H(1) is the Hankel function of the first kind and zero order. We define also the integral
0
operators
(cid:90)
(S f)(x)= Φ (x,y)f(y)ds(y), x∈Γ,
j j
Γ
(cid:90)
∂Φ
(D f)(x)= j (x,y)f(y)ds(y), x∈Γ,
j
∂n(y)
Γ
(cid:90)
∂Φ
(NS f)(x)= j (x,y)f(y)ds(y), x∈Γ,
j
∂n(x)
Γ
(cid:90) ∂2Φ (10)
(ND f)(x)= j (x,y)f(y)ds(y), x∈Γ,
j
∂n(x)∂n(y)
Γ
(cid:90)
∂Φ
(TS f)(x)= j (x,y)f(y)ds(y), x∈Γ,
j
∂τ(x)
Γ
(cid:90) ∂2Φ
(TD f)(x)= j (x,y)f(y)ds(y), x∈Γ.
j
∂τ(x)∂n(y)
Γ
The following theorem was proven in [10].
Theorem 2.2 Let the assumptions of Theorem 2.1 still hold. Then, the potentials
eint(x)=−(D φ )(x)+(S η )(x), x∈Ω ,
3 1 1 1 1 1
hint(x)=−(D ψ )(x)+(S ξ )(x), x∈Ω ,
3 1 1 1 1 1
(11)
eext(x)=(D φ )(x)−(S η )(x), x∈Ω ,
3 0 0 0 0 0
hext(x)=(D ψ )(x)−(S ξ )(x), x∈Ω ,
3 0 0 0 0 0
solve the direct transmission problem (4) – (7) provided that the densities φ ∈ H1/2(Γ) and
0
ψ ∈H1/2(Γ) satisfy the system of integral equations
0
(cid:18) (cid:19)
φ
(D +K ) 0 =b , (12)
0 0 ψ 0
0
where
D =(cid:18)D0− 21I 0 (cid:19), K =(cid:32) −(cid:15)(cid:15)˜˜01S0K1 −(cid:15)˜01ωS0(β1L1+β0L0)(cid:33),
0 0 D − 1I 0 1 S (β L +β L ) −µ˜1S K
0 2 µ˜0ω 0 1 1 0 0 µ˜0 0 1
(cid:18) −S ∂ + (cid:15)˜1S K (cid:19)
b0 = − 1 S0 (ηβ ∂(cid:15)˜0+0β 1L ) ei3nc,
µ˜0ω 0 0 τ 1 1
and K := (cid:0)NS ± 1I(cid:1)−1ND , L := 2(TD −TS K ). The rest of the densities satisfy φ =
j j 2 j j j j j 1
φ +einc, ψ =ψ , η =K φ and ξ =K ψ .
0 3 1 0 j j j j j j
The solutions esc and hsc of (4) – (7) have the asymptotic behavior
3 3
eiκ0r eiκ0r
esc(x)= √ e∞(xˆ)+O(r−3/2), hsc(x)= √ h∞(xˆ)+O(r−3/2), (13)
3 r 3 r
where xˆ=x/|x|. The pair (e∞,h∞) is called the far-field pattern corresponding to the scattering
problem(4)–(7). Itsknowledgeisessentialfortheinverseproblemandusing(11)wecancompute
it by
e∞(xˆ)=(D∞φ )(xˆ)−(S∞η )(xˆ), xˆ∈S,
0 0
(14)
h∞(xˆ)=(D∞ψ )(xˆ)−(S∞ξ )(xˆ), xˆ∈S,
0 0
5
where S is the unit ball. The far-field operators are given by
(cid:90)
(S∞f)(xˆ)= Φ∞(xˆ,y)f(y)ds(y), xˆ∈S,
(cid:90)Γ ∂Φ∞ (15)
(D∞f)(xˆ)= (xˆ,y)f(y)ds(y), xˆ∈S,
∂n(y)
Γ
where Φ∞ is the far-field of the Green function Φ, given by
eiπ/4
Φ∞(xˆ,y)= √ e−iκ0xˆ·y.
8πκ
0
3 The inverse problem
The inverse scattering problem, we address here, reads: Find the shape and the position of the
inclusion Ω, meaning reconstruct its boundary Γ, given the far-field patterns (e∞(xˆ),h∞(xˆ)), for
all xˆ∈S, for one or few incident fields (6).
3.1 The integral equation method
Tosolvetheinverseproblemweapplythemethodofnonlinearboundaryintegralequations,which
in our case results to a system of four integral equations on the unknown boundary and one on
the unit circle where the far-field data are defined. This method was first introduced in [18] and
further considered in various inverse problems, see for instance [2, 3, 5, 9, 13]. Since the direct
problem was solved with the direct method (Green’s formulas), in order to obtain our numerical
data, here we adopt a different approach based on the indirect integral equation method, using
simple representations for the fields.
Weassumeadouble-layerrepresentationfortheinteriorfieldsandasingle-layerrepresentation
for the exterior fields. Thus, we set
1 1
eint(x)= (D φ )(x), hint(x)= (D φ )(x), x∈Ω ,
3 (cid:15)˜ 1 e 3 µ˜ 1 h 1
1 1 (16)
1 1
esc(x)= (S ψ )(x), hsc(x)= (S ψ )(x), x∈Ω .
3 (cid:15)˜ 0 e 3 µ˜ 0 h 0
0 0
Substituting the above representations in the transmission conditions (5) and considering the
well-known jump relations, we get the system of integral equations
(cid:18) (cid:19)
1 1 1
D − φ − S ψ =einc, on Γ,
(cid:15)˜ 1 2 e (cid:15)˜ 0 e 3
1 0
β (cid:18) 1 ∂ (cid:19) (cid:18) 1(cid:19) β ∂einc
ωND φ + 1 TD − φ −ω NS − ψ − 0TS ψ =β 3 , on Γ,
1 h 1 e 0 h 0 e 0
(cid:15)˜ 2∂τ 2 (cid:15)˜ ∂τ
1 0
(17)
(cid:18) (cid:19)
1 1 1
D − φ − S ψ =0, on Γ,
1 h 0 h
µ˜ 2 µ˜
1 0
β (cid:18) 1 ∂ (cid:19) (cid:18) 1(cid:19) β ∂einc
ωND φ − 1 TD − φ −ω NS − ψ + 0TS ψ =(cid:15)˜ ω 3 , on Γ.
1 e 1 h 0 e 0 h 0
µ˜ 2∂τ 2 µ˜ ∂n
1 0
In addition, given the far-field operators (15) and the representations (16) of the exterior fields
weseethattheunknownboundaryΓandthedensitiesψ andψ satisfyalsothefar-fieldequations
e h
6
1
S∞ψ =e∞, on S, (18a)
e
(cid:15)˜
0
1
S∞ψ =h∞, on S, (18b)
h
µ˜
0
wheretheright-handsidesaretheknownfar-fieldpatternsfromthedirectproblem. Theequation
(17) in matrix form reads
(T+K)φ=b, (19)
where
ω β   β β 
1 ∂ 0 0 −ωNS − 1TD 0TS ωND
τ 0 1 0 1
2 2µ˜1   µ˜1 µ˜0 
   
 1   1 1 
0 − 0 0   0 D − S 0 
 2µ˜   µ˜ 1 µ˜ 0 
T= 1 , K= 1 0 ,
0 0 ω −β1 ∂τ −β0TS0 ωND1 −ωNS0 β1TD1
 2 2(cid:15)˜   (cid:15)˜ (cid:15)˜ 
 1   0 1 
 1   1 1 
0 0 0 − − S 0 0 D
0 1
2(cid:15)˜ (cid:15)˜ (cid:15)˜
1 0 1
   
ψ (cid:15)˜ ω∂
e 0 n
φ=φh, b= 0 einc.
ψh β0∂τ  3
φ 1
e
The matrix T due to its special form and the boundness of ∂ : H1/2(Γ) → H−1/2(Γ) has a
τ
bounded inverse given by
2 2β 
1
∂ 0 0
τ
ω ω 
 
0 −2µ˜ 0 0 
T−1 = 1 . (20)
 2 2β 
0 0 − 1∂ 
 ω ω τ
 
0 0 0 −2(cid:15)˜
1
Then, equation (19) takes the form
(I+C)φ=g, (21)
where now I is the identity matrix and
 2 
−2NS 0 (β −β )TS 2ND
0 0 1 0 1
ωµ˜
 0 
 µ˜ 
 0 −2D 2 1S 0 
C=T−1K= 1 µ˜0 0 ,
 2 
− (β0−β1)TS0 2ND1 −2NS0 0 
 ω(cid:15)˜0 
 (cid:15)˜ 
2 1S 0 0 −2D
0 1
(cid:15)˜
0
 
2(cid:15)˜ ∂
0 n
 0 
g=T−1b=2 einc.
 (β −β )∂  3
 0 1 τ
ω
−2(cid:15)˜
1
7
Using the mapping properties of the integral operators [17], we see that the operator C :
(H−1/2(Γ)×H1/2(Γ))2 →(H−3/2(Γ)×H−1/2(Γ))2 is compact.
We observe that we have six equations (21) and (18) for the five unknowns: Γ and the four
densities. Thus, we consider the linear combination (cid:15)˜ ·(18a) + µ˜ ·(18b) as a replacement for the
0 0
far-field equations in order to state the following theorem as a formulation of the inverse problem.
Theorem 3.1 Giventheincidentfield (6)andthefar-fieldpatterns(e∞(xˆ),h∞(xˆ)),forallxˆ∈S,
if the boundary Γ and the densities ψ , φ , ψ and φ satisfy the system of equations
e h h e
2
ψ −2NS ψ + (β −β )TS ψ +2ND φ =2(cid:15)˜ ∂ einc, (22a)
e 0 e ωµ˜ 0 1 0 h 1 e 0 n 3
0
µ˜
φ −2D φ +2 1S ψ =0, (22b)
h 1 h 0 h
µ˜
0
2 2
− (β −β )TS ψ +2ND φ +ψ −2NS ψ = (β −β )∂ einc, (22c)
ω(cid:15)˜ 0 1 0 e 1 h h 0 h ω 0 1 τ 3
0
(cid:15)˜
2 1S ψ +φ −2D φ =−2(cid:15)˜ einc, (22d)
(cid:15)˜ 0 e e 1 e 1 3
0
S∞ψ +S∞ψ =(cid:15)˜ e∞+µ˜ h∞, (22e)
e h 0 0
then, Γ solves the inverse problem.
Theintegraloperatorsin(22)arelinearwithrespecttothedensitiesbutnon-linearwithrespect
to the unknown boundary Γ. The smoothness of the kernels in the far-field equation (22e) reflects
the ill-posedness of the inverse problem.
To solve the above system of equations, we consider the method first introduced in [14] and
then applied in different problems, see for instance [1, 5, 19]. More precisely, given an initial
approximationfortheboundaryΓ,wesolvethesubsystem(22a)-(22d)forthedensitiesψ , φ , ψ
e h h
and φ . Then, keeping the densities ψ and ψ fixed we linearize the far-field equation (22e) with
e e h
respect to the boundary. The linearized equation is solved to obtain the update for the boundary.
ThelinearizationisperformedusingFr´echetderivativesoftheoperatorsandwealsoregularizethe
ill-posed last equation.
To present the proposed method in details, we consider the following parametrization for the
boundary
Γ={z(t)=r(t)(cost, sint):t∈[0,2π]},
where z :R→R2 is a C2-smooth, 2π-periodic, injective in [0,2π), meaning that z(cid:48)(t)(cid:54)=0, for all
t∈[0,2π]. The non-negative function r represents the radial distance of Γ from the origin. Then,
we define
ζ (t)=ψ (z(t)), ζ (t)=ψ (z(t)), ξ (t)=φ (z(t)), ξ (t)=φ (z(t)), t∈[0,2π]
e e h h e e h h
and the parametrized form of (22) is given by
         
A B C D F
1 1 1 1 1
A2 B2 C2 D2 F2
         
A3(r;ζe)+B3(r;ξh)+C3(r;ζh)+D3(r;ξe)=F3, (23)
         
A4 B4 C4 D4 F4
A B C D F
5 5 5 5 5
8
with the parametrized operators
(cid:90) 2π
(A (r;ζ))(t)=(C (r;ζ))(t)=ζ(t)−2 MNS0(t,s)ζ(s)ds,
1 3
0
µ˜ 2 (cid:90) 2π
(A (r;ζ))(t)=− 0(C (r;ζ))(t)=− (β −β ) MTS0(t,s)ζ(s)ds,
3 1 0 1
(cid:15)˜ ω(cid:15)˜
0 0 0
µ˜ (cid:15)˜ (cid:15)˜ (cid:90) 2π
(A (r;ζ))(t)= 0 1(C (r;ζ))(t)=2 1 MS0(t,s)ζ(s)ds,
4 2
µ˜ (cid:15)˜ (cid:15)˜
1 0 0 0
(cid:90) 2π
(A (r;ζ))(t)=(C (r;ζ))(t)= Φ∞(zˆ(t),z(s))ζ(s)|z(cid:48)(s)|ds,
5 5
0
(cid:90) 2π
(B (r;ξ))(t)=(D (r;ξ))(t)=ξ(t)−2 MD1(t,s)ξ(s)ds,
2 4
0
(cid:90) 2π
(B (r;ξ))(t)=(D (r;ξ))(t)=2 MND1(t,s)ξ(s)ds,
3 1
0
and the right-hand side
2
(F (r))(t)=2(cid:15)˜ ∂ einc(z(t)), (F (r))(t)= (β −β )∂ einc(z(t)),
1 0 n 3 3 ω 0 1 τ 3
(F (r))(t)=−2(cid:15)˜ einc(z(t)), (F )(t)=(cid:15)˜ e∞(zˆ(t))+µ˜ h∞(zˆ(t)).
4 1 3 5 0 0
In addition, we set A2 =B1 =B4 =B5 =C4 =D2 =D3 =D5 =F2 =0. The matrix MKj
denotesthediscretizedkerneloftheoperatorK .Theexplicitformsofthekernelscanbefoundfor
j
example in [10, Equation 4.3]. The operators A , B , C , D , k =1,2,3,4,5 act on the densities
k k k k
and the first variable r shows the dependence on the unknown parametrization of the boundary.
Only F is independent of the radial function.
5
Let the function q stand for the radial function of the perturbed boundary
Γ ={q(t)=q(t)(cost, sint):t∈[0,2π]}.
q
Then the iterative method reads:
Iterative Scheme 3.2 Let r(0) be an initial approximation of the radial function. Then, in the
kth iteration step:
i. We assume that we know r(k−1) and we solve the subsystem
         
A B C D F
1 1 1 1 1
AA23(r(k−1);ζe)+BB23(r(k−1);ξh)+CC23(r(k−1);ζh)+DD23(r(k−1);ξe)=FF23,
A B C D F
4 4 4 4 4
(24)
to obtain the densities ζ(k), ξ(k), ζ(k) and ξ(k).
e h h e
ii. Keeping the densities ζ and ζ fixed, we linearize the fifth equation of (23), namely
e h
A (r(k−1);ζ(k))+(A(cid:48)(r(k−1);ζ(k)))(q)+C (r(k−1);ζ(k))+(C(cid:48)(r(k−1);ζ(k)))(q)=F . (25)
5 e 5 e 5 h 5 h 5
We solve this equation for q and we update the radial function r(k) =r(k−1)+q.
The iteration stops when a suitable stopping criterion is satisfied.
9
Remark 3.3 In order to take advantage of the available measurement data, we can also keep the
overdetermined system (17) and (18) instead of (22e) and replace equation (25) with
(cid:32)A(cid:48)(r(k−1);ζ(k))(cid:33) (cid:18)F (cid:19) (cid:32)A (r(k−1);ζ(k))(cid:33)
5 e q = e − 5 e , (26)
A(cid:48)5(r(k−1);ζh(k)) Fh A5(r(k−1);ζh(k))
where now F =(cid:15)˜ e∞ and F =µ˜ h∞.
e 0 h 0
The Fr´echet derivatives of the operators are calculated by formally differentiating their kernels
with respect to r
eiπ/4 (cid:90) 2π (cid:18) z(cid:48)(s)·q(cid:48)(s)(cid:19)
((A(cid:48)(r;ζ))(q))(t)= √ e−iκ0zˆ(t)·z(s) −iκ zˆ(t)·q(s)|z(cid:48)(s)|+ ζ(s)ds.
5 8πκ 0 |z(cid:48)(s)|
0 0
(27)
Recall that A =C =S∞. If κ2 is not an interior Neumann eigenvalue, then the operator A(cid:48) is
5 5 0 5
injective[13]. Usingsimilarargumentsasin[1,11]wecanrelatetheaboveiterativeschemetothe
classical Newton’s method.
The iterative scheme 3.2 can also be generalized to the case of multiple illuminations einc, l=
l
1,...,L.
Iterative Scheme 3.4 (Multiple illuminations) Letr(0) beaninitialapproximationofthera-
dial function. Then, in the kth iteration step:
i. We assume that we know r(k−1) and we solve the L subsystems
     
A B C
1 1 1
AA23(r(k−1);ζe,l)+BB23(r(k−1);ξh,l)+CC23(r(k−1);ζh,l)
A B C
4 4 4
   
D F
1 1,l
+DD23(r(k−1);ξe,l)=FF23,,ll, l=1,...,L (28)
D F
4 4,l
to obtain the densities ζ(k), ξ(k), ζ(k) and ξ(k).
e,l h,l h,l e,l
ii. Then, keeping the densities fixed, we solve the overdetermined version of the linearized fifth
equation of (23)
A(cid:48)(r(k−1);ζ(k)+ζ(k)) F −A (r(k−1);ζ(k)+ζ(k))
5 e,1 h,1 5,1 5 e,1 h,1
A(cid:48)(r(k−1);ζ(k)+ζ(k)) F −A (r(k−1);ζ(k)+ζ(k))
 5 e,2 h,2   5,2 5 e,2 h,2 
 . q = . 
 .   . 
. .
   
A(cid:48)(r(k−1);ζ(k)+ζ(k)) F −A (r(k−1);ζ(k)+ζ(k))
5 e,l h,l 5,L 5 e,l h,l
for q and we update the radial function r(k) =r(k−1)+q.
The iteration stops when a suitable stopping criterion is satisfied.
4 Numerical implementation
In this section, we present numerical examples that illustrate the applicability of the proposed
method. We use quadrature rules for integrating the singularities considering trigonometric inter-
polation. The convergence and error analyses are given in [15, 16]. Then, the system of integral
10