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Inverse Problems and Imaging
Volume 4, No. 3, 2010, 1–XX
RECONSTRUCTING ELECTROMAGNETIC OBSTACLES BY
THE ENCLOSURE METHOD
Ting Zhou
Department of Mathematics,
University of Washington,
Seattle, WA 98105, USA
Abstract. We show that one can determine Perfectly Magnetic Conductor obstacles, Perfectly Electric Conductor obstacles and obstacles satisfying impedance boundary condition, embedded in a known electromagnetic
medium, by making electromagnetic measurements at the boundary of the
medium. The boundary measurements are encoded in the impedance map
that sends the tangential component of the electric field to the tangential component of the magnetic field. We do this by probing the medium with complex
geometrical optics solutions to the corresponding Maxwell’s equations and extend the enclosure method to this case.
1. Introduction. We consider in this paper an inverse problem in electromagnetism. We probe a medium in which there is an unknown obstacle with electromagnetic waves. We would like to determine the obstacle from electromagnetic
measurements made only at the boundary of the medium. We consider first Perfectly Magnetic Conductor (PMC) obstacles which means that the tangential component of the magnetic field vanishes at the boundary of the obstacle. We also
consider Perfectly Electric Conductor (PEC) obstacles which means that the tangential component of the electric field vanishes at the boundary of the obstacle and
also obstacles satisfying more general impedance conditions at the boundary. In
all these cases we show that under some conditions one can recover the obstacle
by making electromagnetic measurements at its boundary. We use the enclosure
method [11, 12] of Ikehata that has been applied successfully for the case of electrostatics and acoustics. This method uses as probing waves the so-called complex
geometrical optics (CGO) solutions to the conductivity equation, in the case of electrostatics, and Helmholtz equations in the case of acoustics. The case of Maxwell’s
equations presents new challenges since it is considerable more difficult in this case
to construct CGO solutions. Moreover it is also more difficult to establish the
bounds needed for the indicator function in Ikehata’s method. We now describe
more precisely the mathematical problem and some historical background.
Let Ω ⊂ R3 be a bounded domain with smooth boundary, filled with an isotropic
electromagnetic material, characterized by three parameters: the electrical permittivity ε(x), the conductivity σ(x) and magnetic permeability µ(x). In particular,
2000 Mathematics Subject Classification. Primary: 35R30, 35Q61.
Key words and phrases. Maxwell’s equations, enclosure methods, complex geometrical optics
soltuions.
The author is supported by NSF grant 0758357.
1
c
2010
American Institute of Mathematical Sciences
2
Ting Zhou
we consider non-dissipative materials for which σ = 0. A PMC obstacle is a subdomain D such that the electric-magnetic field (E, H) satisfies Maxwell’s equations
with tangential component of the magnetic field vanishing at the boundary of D,

 ∇ ∧ E = iωµH, ∇ ∧ H = −iωεE in Ω \ D,
(1)
ν ∧ E|∂Ω = f,

ν ∧ H|∂D = 0
where ν is the unit outer normal vector to the boundary ∂Ω ∪ ∂D. A PEC obstacle
is such that the tangential component of the electric field
ν ∧ E|∂D = 0.
Both PMC (λ → 0) and PEC (λ → ∞) conditions are two extreme cases of the
general impedance boundary condition
iν ∧ H − λ(ν ∧ E) ∧ ν|∂D = 0
where λ ∈ L∞ (∂D) is a complex-valued function, see [9]. Suppose we are given the
impedance map which takes the tangential component of the electric field ν ∧ E|∂Ω
to the tangential component of the magnetic field ν ∧ H|∂Ω . This map encodes
all the possible electromagnetic measurements that can be made at the boundary
of the medium. Our goal is to retrieve information of the shape of D from the
impedance map.
For the case of electrostatics, the well-known Calderón’s problem [6] is to determine the conductivity of a medium by making voltage and current measurements
of the boundary. The information is encoded in the Dirichlet-to-Neumann map for
the conductivity equation ∇ · (γ∇u) = 0. In [25], Sylvester and Uhlmann constructed complex geometrical optics (CGO) solutions for the conductivity equation
and Schrödinger operator ∆ − q and proved unique identifiability of C 2 isotropic
conductivities in dimensions n ≥ 3 from the Dirichlet-to-Neumann map. Further
developments including improved regularity assumption, 2D problems and partial
data problems, were obtained based on this idea. See [27] for a survey of recent
developments.
As another application of CGO solutions, Ikehata [11, 12] introduced the enclosure method to reconstruct obstacles embedded in an electric conductor (conductivity equations) or acoustic medium (Helmholtz equations). Intuitively, one can
enclose the obstacle by hyperplanes. Then one constructs CGO solutions that decay
on one side of the hyperplane and grow on the other side. In [16], CGO solutions
with nonlinear radial weights are constructed and called complex spherical waves.
Using these the enclosure method was generalized to enclose non-convex obstacles,
see [15] and [21]. In addition of the conductivity equation and Helmholtz equation,
other types of systems are considered. For example, in [28], CGO solutions were
constructed for systems with Laplacian as the leading order term, e.g., for the elasticity equations, and used to recover inclusions.
For the case of Maxwell’s equations, the uniqueness of parameters in a domain
Ω from the impedance map was obtained in [24], [22] and [23]. Particularly, in [23],
the Maxwell’s operator was reduced to a matrix Schrödinger operator, and vector
CGO solutions were constructed to recover electromagnetic parameters. With these
solutions at hand, the enclosure method is applicable using the following steps: one
first defines an indicator function Iρ (τ, t) for each direction ρ ∈ S2 ; by adjusting
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
3
t, the hyperplane moves along ρ; for each ρ and t, the asymptotic behavior of
Iρ (τ, t) as τ 1 produces the support function of the convex hull of D. One of the
main difficulties to determine obstacles in electromagnetic materials is that, unlike
the Schrödinger equation, the CGO solutions for Maxwell’s equations don’t behave
as small perturbations (w.r.t. τ ) of Calderón’s solutions. However, this can be
overcome by carefully choosing appropriate probing fields (see section 3). We will
focus on showing the asymptotic behavior of the indicator function of these CGO
solutions for large τ. (see section 4)
For the reconstruction of the non-convex part of the obstacles, we apply an inversion transformation with respect to a sphere to the domain. Since this transformation maps spheres to planes, by enclosing the image obstacle using hyperplanes, we
actually enclose the original obstacle by spheres. The argument relies on the invariance of Maxwell’s equations under this transformation, and a relation between the
impedance maps for the original system and the image systems. Throughout the
process of reconstruction, one actually constructs the CSW for the original system,
which is the pull-back of the CGO solution of the image system by the inversion
transformation. See section 5 for more details. In another direction we mention
the work of Ammari et al. [2, 3, 4] to determine asymptotically inhomogeneities of
small volume embedded in an homogeneous medium, and [1] where efficient numerical algorithms were derived in this context.
The rest of the paper is organized as following. In Section 2, we formulate the
forward problem for a Perfect Magnetic Conductor (PMC) obstacle and define rigorously the impedance map. Then we construct the appropriate CGO solutions
for Maxwell’s equations in Section 3. The main reconstruction scheme for PMC
obstacles is developed in Section 4. Section 5 is devoted to introducing the enclosure method to recover the non-convex part of the obstacle based on inversion. In
Section 6, we show that the scheme also applies to PEC obstacles and obstacles
with impedance boundary conditions. In Section 7, we present our conclusions and
propose some open problems related to this paper. We point out that PEC and
impedance boundary conditions are more natural boundary conditions in practice.
However, this paper presents a detailed proof in the PMC setting, which requires
the well-posedness of a mixed boundary value problem for Maxwell’s equations. For
completeness, we include a proof in Appendix A.
2. Direct Problems and the main result. Let Ω be a bounded domain in
R3 with smooth boundary and connected complement R3 \ Ω. We consider the
electric permittivity ε(x), conductivity σ(x) and magnetic permeability µ(x) of the
background medium as globally defined functions with the following properties:
there are positive constants εm , εM , µm , µM , εc and µc such that for all x ∈ Ω
εm ≤ ε(x) ≤ εM , µm ≤ µ(x) ≤ εM , σ(x) = 0
and ε − εc , µ − µc ∈ C03 (Ω).
An obstacle D is a Perfect Magnetic Conductor (PMC) if it is a subset of Ω such
that Ω \ D is connected. Moreover, the electric field E and the magnetic field H in
Ω\D satisfy the following boundary value problem for the time-harmonic Maxwell’s
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
4
Ting Zhou
equations
(2)
∇ ∧ E = iωµH, ∇ ∧ H = −iωεE in Ω \ D,
ν ∧ E = f ∈ T H 1/2 (∂Ω) on ∂Ω,
and the boundary condition on the interface ∂D is given by
(ν ∧ H)|∂D = 0,
(3)
where ν is the unit outer normal vector on ∂Ω ∪ ∂D.
Notations. If F is a function space on ∂Ω, the subspace of all those f ∈ F 3 which
3
are tangent to ∂Ω is denoted by T F . For example, for u ∈ H s (∂Ω) , we have the
decomposition u = ut + uν ν, where the tangential component ut = −ν ∧ (ν ∧ u) ∈
T H s (∂Ω) and the normal component uν = u · ν ∈ H s (∂Ω). In addition, we define
the weighted L2 space in R3 :
Z
L2δ = f ∈ L2loc (R3 ) : kf k2L2 = (1 + |x|2 )δ |f (x)|2 dx < ∞ .
δ
Admissibility. It can be shown (see Appendix A.) that for f ∈ T H 1/2 (∂Ω) and
g ∈ T H 1/2 (∂D), the boundary value problem for Maxwell’s equations

 ∇ ∧ E = iωµH, ∇ ∧ H = −iωεE in Ω \ D̄,
ν ∧ E|∂Ω = f
(4)

ν ∧ H|∂D = g,
3
3
has a unique solution (E, H) ∈ H 1 (Ω \ D) × H 1 (Ω \ D) , except for a discrete set
of magnetic resonance frequencies {ωn }. It satisfies
kEkH 1 (Ω\D) + kHkH 1 (Ω\D) ≤ C kf kT H 1/2 (∂Ω) + kgkT H 1/2 (∂D) .
Denote by (E0 , H0 ) the solution of Maxwell’s equations in the domain Ω without
an obstacle, namely, the solution of
∇ ∧ E0 = iωµH0 , ∇ ∧ H0 = −iωεE0 , in Ω,
ν ∧ E0 |∂Ω = f.
Main result. Now we are in the position to define the impedance map for nonresonant frequencies,
ΛD (ν ∧ E|∂Ω ) = ν ∧ H|∂Ω .
and it can be shown that
ΛD : T H 1/2 (∂Ω) → T H 1/2 (∂Ω)
is bounded. If ω is a resonance frequency, one can replace the impedance map by
the set of Cauchy data
Cω = {(ν ∧ E|∂Ω , ν ∧ H|∂Ω ) | (E, H) satisfies (2) and (3)}
⊂ T H 1/2 (∂Ω) × T H 1/2 (∂Ω).
Denote by Λ∅ the impedance map for the domain without an obstacle. Then the
main result of the presenting work is to show
Theorem 2.1. There exists a reconstruction scheme to determine the convex hull
of the obstacle D from the impedance map ΛD .
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
5
3. Construction of Complex Geometric Optics (CGO) solutions. In [23],
Maxwell’s operator was reduced to an 8 × 8 second order Schrödinger matrix operator by introducing the generalized Sommerfeld potentials. A vector CGO-solution
was constructed to simplify the proof in [22]. Similar techniques also appeared in
[8] when dealing with the partial data problems for Maxwell’s equations. For completeness, we include the construction of the solution in this work (see [23] for more
details). Moreover, to obtain CGO solutions for our use in the enclosure method, a
special ”incoming” field will be chosen.
Define the scalar fields Φ and Ψ by
i
i
(5)
Φ = ∇ · (εE), Ψ = ∇ · (µH).
ω
ω
Under appropriate assumptions on Φ and Ψ, Maxwell’s equations are equivalent to
1
1
1
1
(6)
∇∧E− ∇
Ψ − iωµH = 0, ∇ ∧ H + ∇
Φ + iωεE = 0.
ε
µ
µ
ε
Moreover, in this case, Φ and Ψ vanish, leading to a solution of Maxwell’s equations.
Let X = (ϕ, e, h, ψ)T ∈ (D0 )8 with
1
1
e = ε1/2 E, h = µ1/2 H, ϕ = 1/2 Φ, ψ = 1/2 Ψ.
εµ
ε µ
Then (5) and (6) read
(P (i∇) − k + V )X = 0, in Ω
where
(7)

∇·
0
0
∇∧
−∇∧
0
0
∇·

0
0 

∇ 
0

∇·
0
0 −∇∧
∇∧
0
0
∇·
 
0

0 
 D D−1
∇  
0
0
 ∇

P (i∇) = 
0
0
and
0
 ∇

V = (k − κ)18 + 
 0
0
are matrix operators and
D = diag(µ1/2 , ε1/2 13 , µ1/2 13 , ε1/2 ), κ = ω(εµ)1/2 , k = ω(ε0 µ0 )1/2 .
An important property of this operator is that it allows to reduce Maxwell’s equations to a Schrödinger matrix equation by observing that
(P (i∇) − k + V )(P (i∇) + k − V T ) = −(∆ + k 2 )18 + Q,
where
Q = V P (i∇) − P (i∇)V T + k(V + V T ) − V V T
is a zeroth-order matrix multiplier. Hence, by writing an ansatz for X, one can
define the generalized Sommerfeld potential Y by
X = (P (i∇) + k − V T )Y.
So it satisfies the Schrödinger equation
(8)
Inverse Problems and Imaging
(−∆ − k 2 + Q)Y = 0.
Volume 4, No. 3 (2010), 1–XX
6
Ting Zhou
The following CGO-solution is constructed using Faddeev’s kernel. Let ζ ∈ C3 be a
vector with ζ · ζ = k 2 . Suppose y0,ζ ∈ C8 is a constant vector with respect to x and
bounded with respect to ζ. We refer eix·ζ y0,ζ as the ”incoming” field. Then there
exists a unique solution to (8) of the form
Yζ (x) = eix·ζ (y0,ζ − vζ (x)) ,
where vζ (x) ∈ (L2δ+1 )8 satisfying
kvζ kL2δ+1 ≤ C/|ζ|
for δ ∈ (−1, 0). Moreover, one can show that vζ ∈ (H s (Ω))8 for 0 ≤ s ≤ 2, e.g., see
[5], and
kvζ (x)kH s (Ω) ≤ C|ζ|s−1 .
(9)
Lemma 3.1 in [23] states that if we choose y0,ζ such that the first and the last
components of (P (ζ)−k)y0,ζ vanish, where P (ζ) is the matrix obtained by replacing
i∇ by ζ in (7), then for large |ζ|, Xζ provides the solution to the original Maxwell’s
equations. We proceed to provide a more specific choice of y0,ζ such that the CGO
solution for Maxwell’s equations has special properties.
As in [23], choose
y0,ζ =
1
(ζ · a, ka, kb, ζ · b)T ,
|ζ|
where
p
ζ = −iτ ρ + τ 2 + k 2 ρ⊥ ,
√
with ρ, ρ⊥ ∈ S2 and ρ · ρ⊥ = 0. τ > 0 is used to control the size of |ζ| = 2τ 2 + k 2 .
Then we obtain


0
2

1 
 −(ζ · a)ζ − kζ ∧ b + k2 a 
x0,ζ := (P (−ζ) + k)y0,ζ =


kζ
∧
a
−
(ζ
·
b)ζ
+
k
b
|ζ|
0
satisfying the condition in Lemma 3.1 in [23]. Taking τ → ∞, we have
ζ
1
→ ζ̂ = √ (−iρ + ρ⊥ ).
|ζ|
2
In this article, for the enclosure method, we expect different asymptotic behavior
of the electric and magnetic fields with respect to the phase growth. Therefore we
choose a and b, and correspondingly y0,ζ , such that
ζ̂ · b = 1,
ζ̂ · a = 0.
This is satisfied, for example, by taking a ∈ R3 and b ∈ C3 satisfying
a ⊥ ρ, a ⊥ ρ⊥ , b = ζ̂.
Given these choices, It’s easy to see that as τ → ∞
η := (x0,ζ )2 → −k ζ̂ ∧ b = ikρ ∧ ρ⊥ = O(1),
θ := (x0,ζ )3 = O(τ ).
Then Xζ = (P (i∇) + k − V T )Yζ is written in the form
√
Xζ = eτ (x·ρ)+i
Inverse Problems and Imaging
τ 2 +k2 x·ρ⊥
(x0,ζ + rζ (x))
Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
7
where
(10)
rζ = P (−ζ)vζ + P (i∇)vζ − V T y0,ζ + kvζ − V T vζ
satisfying
krζ kL2 (Ω) ≤ C
for C > 0 independent of ζ. Summing up, we obtain the following
Proposition 1. Let ρ, ρ⊥ ∈ S2 with ρ · ρ⊥ = 0. Assume ω is not a resonant
frequency. Given θ, η ∈ C3 as above, then for τ > 0 large enough, there exists a
3
3
unique complex geometric optics solution (E0 , H0 ) ∈ H 1 (Ω) ×H 1 (Ω) of Maxwell’s
equations
∇ ∧ E0 = iωµH0 ∇ ∧ H0 = −iωεE0 in Ω
of the form
E0 = ε−1/2 eτ (x·ρ)+i
√
τ 2 +k2 x·ρ⊥
(η + R)
√
−1/2 τ (x·ρ)+i τ 2 +k2 x·ρ⊥
H0 = µ
e
(θ + Q).
Moreover, we have
η = O(1),
θ = O(τ ) for τ 1,
and R = (rζ )2 , Q = (rζ )3 are bounded in (L2 (Ω))3 for τ 1.
For reconstruction, one computes directly the boundary tangential CGO-fields
(ν ∧ E0 |∂Ω , ν ∧ H0 |∂Ω ). On the other hand, given unknown background parameters, one can recover tangential CGO fields from the impedance map Λ∅ by solving
appropriate boundary integral equations, see [23, 22]. This allows detection of an
obstacle by measurements taken before and after its appearance without invading
the background area.
4. Reconstruction Scheme. Adding a parameter t > 0 into the CGO-solution
in Proposition 1, we use
(11)
E0 = ε−1/2 eτ (x·ρ−t)+i
√
H0 = µ−1/2 eτ (x·ρ−t)+i
τ 2 +k2 x·ρ⊥
√
(η + R)
τ 2 +k2 x·ρ⊥
(θ + Q)
to define an indicator function which physically measures the differences between
the energies required to keep the same boundary CGO electric field for the domain
Ω with and without the obstacle D.
Definition 4.1. For ρ ∈ S2 , τ > 0 and t > 0 we define the indicator function
Z
Iρ (τ, t) := iω
(ν ∧ E0 ) · (ΛD − Λ∅ )(ν ∧ E0 ) ∧ ν dS
∂Ω
where E0 is a CGO solution of Maxwell’s equations given by (11).
The enclosure method’s aim is to recover the convex hull ch(D) of D by reconstructing the following support function.
Definition 4.2. For ρ ∈ S2 , we define the support function of D by
hD (ρ) := sup x · ρ.
x∈D
Then, the reconstruction scheme in Theorem 2.1 is
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
8
Ting Zhou
Theorem 4.3. There is a subset Σ ⊂ S2 of measure zero such that when ρ ∈ S2 \ Σ,
the support function hD (ρ) can be recovered by
hD (ρ) = inf{t ∈ R | lim Iρ (τ, t) = 0}.
τ →∞
Moreover, if D is strictly convex, one can reconstruct D.
Remark 1. The proof mainly consists of showing the following:
lim Iρ (τ, t) = 0 if t > hD (ρ),
τ →∞
lim inf Iρ (τ, t) =
τ →∞
C>0
∞
if t = hD (ρ),
if t < hD (ρ).
Remark 2. Gaussian curvature at a point on the boundary is defined to be the
product of the two principal curvatures, which measure how the surface bends by
different amounts in different directions at that point. A surface with positive
Gaussian curvature at a point is locally convex. In the assumption, the set Σ
consists of the directions ρΣ such that some point in {x ∈ ∂D | x · ρΣ = hD (ρΣ )}
has Gaussian curvature 0. For a smooth boundary, one can show that the measure
of Σ is zero.
To show the limits in Remark 1, we need the following equality for Maxwell’s
equations.
3
3
Lemma 4.4. Assume (E, H) ∈ H 1 (Ω \ D) × H 1 (Ω \ D) is a solution of (2) and
(3) with
ν ∧ E|∂Ω = ν ∧ E0 |∂Ω .
Denote the reflected solution in Ω \ D by Ẽ = E − E0 and define
Z
h
i
I := iω
(ν ∧ E0 ) · (ΛD − Λ∅ )(ν ∧ E0 ) ∧ ν dS.
∂Ω
Then we have
Z
(12)
I=
µ−1 |∇ ∧ Ẽ|2 − ω 2 ε|Ẽ|2 dx +
Ω\D̄
Z
µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx.
D
Proof. First using integration by parts and by boundary conditions, we have
Z
µ−1 (∇ ∧ E) · (∇ ∧ Ẽ) − ω 2 εE · (Ẽ)dx
Ω\D̄
Z
=−
Z
−
∂Ω
iω(ν ∧ H) · (Ẽ)dS = 0.
∂D
Adding this to the following equality
Z
I=
(ν ∧ E0 ) · (−iωH + iωH0 )dS
∂Ω
Z
=
−µ−1 (∇ ∧ E0 ) · (∇ ∧ E) + ω 2 εE0 · Edx
Ω\D̄
Z
Z
+
µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx +
(ν ∧ E0 ) · (−iωH)dS,
Ω
∂D
since the last term vanishes due to the zero-boundary condition on the interface,
we obtain (12).
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
9
Proof of Theorem 4.3. We proceed to show the first limit in Remark 1
(13)
lim Iρ (τ, t) = 0 if t > hD (ρ)
τ →∞
by proposing an upper bound of the indicator function.
3
The reflected solution Ẽ ∈ H 1 (Ω \ D) satisfies

 ∇ ∧ (µ−1 ∇ ∧ Ẽ) − sε∇(∇ · εẼ) − ω 2 εẼ = 0 in Ω \ D,
ν ∧ Ẽ|∂Ω = 0,

ν ∧ (µ−1 ∇ ∧ Ẽ)|∂D = −iων ∧ H0 |∂D ∈ T H 1/2 (∂D).
Moreover, we have by (45)
(14)
kẼk2H 1 (Ω\D) ≤ Ckν ∧ H0 k2H −1/2 (∂D) ≤ C(k∇ ∧ E0 k2L2 (D) + kE0 k2L2 (D) ),
where the second inequality is valid since
iωhν ∧ H0 , Fi∂D = (µ−1 ∇ ∧ E0 , ∇ ∧ F)D − (iωεE0 , F)D
3
F ∈ H 1 (D) .
Therefore, by (12) and (14), we have
Iρ (τ, t) ≤ C(kE0 k2L2 (D) + k∇ ∧ E0 k2L2 (D) ) ≤ C(kE0 k2L2 (D) + kH0 k2L2 (D) ).
For the CGO-solution (11), we have the following estimates:
kE0 k2L2 (D) ≤ Ce2τ (hD (ρ)−t) kη + Rk2L2 (D)3 ∼ e2τ (hD (ρ)−t)
(15)
kH0 k2L2 (D) ≤ Ce2τ (hD (ρ)−t) kθ + Qk2L2 (D)3 ∼ τ 2 e2τ (hD (ρ)−t)
as τ 1. Therefore, we obtain
Iρ (τ, t) ≤ Cτ 2 e2τ (hD (ρ)−t)
for τ large enough, proving the first limit (13).
To show that when t = hD (ρ),
(16)
lim inf Iρ (τ, hD (ρ)) = C > 0,
τ →∞
by (12), it’s sufficient to show the following Lemmas 4.5, 4.6.
As for the case when t < hD (ρ), the infinity limit is obtained by observing
Iρ (τ, t) = Iρ (τ, hD (ρ))e2τ (hD (ρ)−t) .
This completes the proof of the theorem.
Lemma 4.5. If t = hD (ρ), then for some positive constant C,
Z
lim inf
µ−1 |∇ ∧ E0 |2 dx = C.
τ →∞
D
Proof. Thanks to the special choice of H0 in Proposition 1,
Z
Z
−1
2
2
µ |∇ ∧ E0 | dx ≥ CkH0 kL2 (D)3 ≥ C
τ 2 e2τ (x·ρ−hD (ρ)) dx ≥ C
D
D
3
for τ 1. We denote by x0 a point in {x ∈ R | x · ρ = hD (ρ)} ∩ ∂D. The last
inequality is due to the accumulation of the integrand in a small cone at x0 . A
detailed proof is provided in [13].
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
10
Ting Zhou
Lemma 4.6. If t = hD (ρ), then as τ → ∞,
R
R
ω 2 Ω\D ε|Ẽ|2 dx + D ε|E0 |2 dx
R
(17)
→ 0.
µ−1 |∇ ∧ E0 |2 dx
D
Proof. The proof follows a similar scheme in [11] for Helmholtz equations.
Noticing that for τ 1
R
R
ω 2 D ε|E0 |2 dx
ε|E0 |2 dx
R
RD
=
= O(τ −2 ),
−1 |∇ ∧ E |2 dx
2 dx
µ
µ|H
|
0
0
D
D
it suffices to show
ω2
(18)
lim R
τ →∞
D
R
Ω\D
µ−1 |∇
ε|Ẽ|2 dx
ε|Ẽ|2 dx
RΩ\D
=
lim
= 0.
∧ E0 |2 dx τ →∞ D µ|H0 |2 dx
R
To estimate the numerator, we consider the auxiliary boundary value problem:

 ∇ ∧ P = iωµQ, ∇ ∧ Q = −iωεP − iε
ω Ẽ in Ω \ D,
ν ∧ P|∂Ω = 0,

ν ∧ Q|∂D = 0.
This is equivalent to the BVP

 ∇ ∧ (µ−1 ∇ ∧ P) − sε∇(∇ · εP) − ω 2 εP = εẼ in Ω \ D,
(19)
ν ∧ P|∂Ω = 0,

ν ∧ (∇ ∧ P)|∂D = 0,
since
∇ · εP = 0,
in Ω \ D.
3
When ω is admissible, (19) is well-posed since εẼ ∈ H 1 (Ω \ D) and divergence
free. Moreover, by Proposition 3, one has P ∈ H 2 (Ω \ D)3 satisfying
kPkH 2 (Ω\D) ≤ CkẼkL2 (Ω\D) .
By the Sobolev embedding theorem, we have
|P(x) − P(y)| ≤ C|x − y|1/2 kẼkL2 (Ω\D) for x, y ∈ Ω \ D,
sup |P(x)| ≤ CkẼkL2 (Ω\D) .
x∈Ω\D
Now we are in the position to estimate the numerator in (18)
Z
Z
ε|Ẽ|2 dx =
Ẽ · (∇ ∧ (µ−1 ∇ ∧ P) − ω 2 εP)dx
Ω\D
Ω\D
Z
=
∇ ∧ (µ−1 ∇ ∧ Ẽ) · P − ω 2 εẼ · Pdx
Ω\D
(20)
Z
Z −
ν ∧ (µ−1 ∇ ∧ Ẽ) · PdS
−
∂Ω
Z
=−
∂D
ν ∧ (µ−1 ∇ ∧ E0 ) · PdS.
∂D
Suppose {x ∈ ∂D | x · ρ = hD (ρ)} consists of N points {xj }N
j=1 . Without loss of
generality, we take N = 2. We decompose D as D1 ∪ D2 ∪ Γ where Dj are two
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
11
disjoint subdomains of D with xj ∈ Γj = ∂Dj \ Γ (j = 1, 2). Then ∂D = Γ1 ∪ Γ2 .
Expanding the RHS of (20)at xj , one has
Z
ε|Ẽ|2 dx
Ω\D
=iω
(Z
2
X
j=1
)
Z
(P(xj ) − P(x)) · (ν ∧ H0 )dS −
Γj
P(xj ) · (ν ∧ H0 )dS
∂Dj
Z
+ iω (P(x1 ) − P(x2 )) · (ν ∧ H0 )dS
(ΓZ
Z
2
X
=
iω
(P(xj ) − P(x)) · (ν ∧ H0 )dS −
Γj
j=1
)
2
ω εP(xj ) · E0 dS
Dj
Z
+ iω (P(x1 ) − P(x2 )) · (ν ∧ H0 )dS

 Γ
Z
Z
2 Z

X
|x − xj |1/2 |ν ∧ H0 |dS +
|E0 |dx + |ν ∧ H0 |dS kẼkL2 (Ω\D)
≤C


D
Γ
j=1 Γj

Z
2 Z
X
τ |x − xj |1/2 eτ (x·ρ−hD (ρ)) dS +
eτ (x·ρ−hD (ρ)) dx
≤C

Γ
D
j
j=1
Z
τ (x·ρ−hD (ρ))
+
τe
dS kẼkL2 (Ω\D) .
Γ
This yields
kẼk2L2 (Ω\D) dx
≤C

2
X

1/2 τ (x·ρ−hD (ρ))
e
dS
Γj
Z
e
D
|x − xj |
τ
j=1
+
!2
Z
τ (x·ρ−hD (ρ))
2 Z
2 )
τ (x·ρ−hD (ρ))
dx +
τe
dS
.
Γ
R
It’s easy to see that the second term can be absorbed by D µ|H0 |2 dx and the third
term decays for τ 1 since points on Γ are a positive distance away from xj .
Therefore, we only need to show that
Z
lim τ
|x − xj |1/2 eτ (x·ρ−hD (ρ)) dS = 0, j = 1, 2.
τ →∞
Γj
This is shown in [11], where nonzero Gaussian curvature of ∂D at xj is required,
which is satisfied by the choice of ρ.
5. Complex Spherical Waves and Reconstruction of Non-convex Obstacles. For various applications of CGO solutions, it is natural to consider using CGO
solutions with non-linear limiting Carleman weights (LCW). In [16], a full characterization of possible LCWs in Euclidean spaces is given. Among those, to tackle
the partial data problems for the Schrödinger equations in [17], a radial weight
ln |x − x0 | (x0 ∈ R3 \ Ω) was chosen to construct CGO solutions, known as complex
spherical waves (CSW). For Helmholtz equations, they are used in the enclosure
method to probe non-convex acoustic obstacles in [21] and [28]. A key ingredient
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
12
Ting Zhou
in the construction of CSW are Carleman estimates.
Instead, in this work we construct CSW with LCW of the form
2 x − x0
(21)
ϕ(x) = R
+ x0 · ρ
|x − x0 |2
for R > 0, x0 ∈ R3 and ρ ∈ R3 \ {0}. The main tool in the construction is the
inversion transformation with respect to the sphere S(x0 , R):
x − x0
+ x0 := y.
(22)
Tx0 ,R : x 7→ R2
|x − x0 |2
We have that Tx20 ,R = Id. This transformation preserves angles and maps generalized spheres (spheres and hyperplanes) into generalized spheres. Geometrically,
fixing a reference circle S(x0 , R), enclosing D with spheres passing through x0 (CSW
with LCW ϕ) corresponds to enclosing D̂x0 ,R = Tx0 ,R (D) with hyperplanes that
do not pass through x0 , where the reconstruction scheme in Section 4 applies, see
Figure 1.
Figure 1. Enclosing D with CSW
A rigorous argument consists of describing the behavior of the electromagnetic
waves in the image domain Ω̂x0 ,R = Tx0 ,R (Ω). To simplify the notation, throughout
the following, we drop the subscript (x0 , R).
To see more clearly the invariance of Maxwell’s equations under the inversion
transformation, we consider electromagnetic fields written as differential 1-forms
(Einstein summation notation)
E = Ej dxj ,
H = Hj dxj .
The push-forward of E by T in Ω̂, denoted by Ê = T∗ E, is defined by
Ê(y) = Êj dy j = (DT −1 )kj (y)Ek (T −1 (y)) dy j ,
y = T (x).
Similarly, for the magnetic field H, we define Ĥ = T∗ H. Suppose (E, H) solves
Maxwell’s equations in Ω \ D
∇ ∧ E = iωµH,
Inverse Problems and Imaging
∇ ∧ H = −iωεE.
Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
13
Then we have (Ê, Ĥ) satisfies in Ω̂ \ D̂
∇ ∧ Ê = iω µ̂Ĥ,
∇ ∧ Ĥ = −iω ε̂Ê,
where the material parameter µ̂ in Ω̂ is defined by
(DT )T ◦ µ ◦ (DT )
(23)
µ̂ = T∗ µ :=
◦ T −1 ,
|det(DT )|
and ε̂ = T∗ ε is defined similarly.
Now we are at the point of checking the one-to-one correspondence between
boundary measurements. Since the outer normal unit ν̂ of ∂ Ω̂ is the normalization
of T∗ ν, and conversely ν is the normalization of (T −1 )∗ ν̂, one has
(ν ∧ E)|∂Ω (x) = K1 (x)(ν̂ ∧ Ê)|∂ Ω̂ ◦ T (x),
(ν̂ ∧ Ĥ)|∂ Ω̂ (y) = K2 (y)(ν ∧ H)|∂Ω ◦ T −1 (y),
where K1 and K2 are defined by
det(DT (x))DT −1 (y) ,
|DT (x)T ν̂(y)|
y=T (x)
K2 (y) := det(DT −1 (y)) DT (x)T ν̂(y) DT (x) x=T −1 (y) .
K1 (x) :=
Hence, knowing the impedance map ΛD for the original domain, one can compute
the map Λ̂D̂ for the image domain by
h
i
(24)
Λ̂D̂ (ν̂ ∧ Ê)|∂ Ω̂ = K2 (y) ΛD K1 (x)(ν̂ ∧ Ê)|∂ Ω̂ ◦ T (x) ◦ T −1 (y).
We compute explicitly
(25)
R2
DT −1 (z) = DT (z) =
|z − x0 |2
z − x0
I −2
|z − x0 |
z − x0
|z − x0 |
T !
,
and
3
R2
det(DT (z)) = det(DT (z)) =
(1 − 2|z − x0 |2 ).
|z − x0 |2
According to (25), the push-forward of the isotropic parameters µ̂ and ε̂ given by
(23) remain isotropic and satisfy the hypothesis in the previous sections when x0
is outside Ω. Therefore, by Section 3, one can construct CGO solutions with linear
phase in Ω̂, denoted by (Ê0 , Ĥ0 ), depending on τ and t. Note that the pull-back of
(Ê0 , Ĥ0 ) is a CSW with weight (21) of the original Maxwell system. Compute the
impedance map Λ̂D̂ using (24), and define the indicator function for ρ ∈ S2
Z
Iˆρ (τ, t) = iω
(ν̂ ∧ Ê0 ) · [(Λ̂D̂ − Λ̂∅ )(ν̂ ∧ Ê0 ) ∧ ν̂]dS.
−1
∂ Ω̂
The support function of D̂
hD̂ (ρ) = sup y · ρ,
for each ρ ∈ S2 \ Σ
y∈D̂
is recovered using Theorem 4.3. Passing back to the original domain by applying the
inversion transformation, we recover T (ch(D̂)). The corresponding radial support
function of D is
x − x0
2
h̃D (ρ) = sup R
· ρ + x0 · ρ .
|x − x0 |2
x∈D
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
14
Ting Zhou
Summing up, we have shown the following theorem.
Theorem 5.1. Given x0 ∈ R3 \ Ω and R > 0 such that Ω ⊂ B(x0 , R), there is a
zero measure subset Σ of S2 , s.t., when ρ ∈ S2 \ Σ, we have
h̃D (ρ) = inf{t ∈ R | lim Iˆρ (τ, t)}.
τ →∞
To conclude this section, we give a remark describing the shape we reconstructed,
i.e., T (ch(D̂)), using this type of CGO solutions.
Suppose x0 ∈ R3 \Ω and R > 0 such that Ω ⊂ Bx0 ,R . In Figure 1, when enclosing
D̂, hyperplanes passing through x0 (e.g., l4 , l5 ) have as pre-images (generalized
spheres C4 , C5 ) under T themselves. Not only does the cone at x0 , formed by
these hyperplanes, narrow the searching area of D, but also its intersection with
∂D separates ∂D into two subsets: the front Fx0 that faces x0 and the back Bx0 =
∂D \ Fx0 (see Figure 2). We observe that in Figure 1, some CSWs (e.g., C1 , C2 )
Figure 2. Front and back of ∂D
probe the front Fx0 using their exterior while some (e.g., C3 ) probe the back Bx0
using their interior.
Hence in practice, one can expand a sphere passing through x0 , by moving its
center along one direction (perpendicular to ρ), until the indicator function is no
longer vanishing as τ → ∞, suggesting that the sphere intersects the front Fx0 .
In this way, some non-convex part of D from the front side can be recovered. By
moving x0 ∈ R3 \ Ω, since the spheres passes x0 , we expect that this reconstruction
scheme will recover more non-convex part of the obstacle than the one using CSWs
with logarithm weight introduced in [17], whose level surfaces are spheres centered
at x0 .
6. Enclosing other obstacles. In this section, we consider reconstruction of other
types of obstacles, including a Perfect Electric Conducting (PEC) obstacle and an
obstacle with impedance boundary condition. For the acoustic setting, we mention
papers [12] and [13] for similar results. In both cases, we suppose our domain Ω,
the obstacle D and all the electromagnetic parameters in the background satisfy
the same hypothesis as in the PMC case.
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
15
6.1. PEC obstacles. We consider the electromagnetic field (E, H) satisfying Maxwell’s
equations (2) with the boundary condition
ν ∧ E|∂D = 0.
(26)
Then the reconstruction scheme Theorem 4.3 is valid simply by a similar equality
as Lemma 4.4.
3
3
Lemma 6.1. Suppose (E, H) ∈ H 1 (Ω) × H 1 (Ω) satisfies (2) and (26) with f =
ν ∧ E0 |∂Ω . Then we have
Z
Z
(27)
−I =
µ−1 |∇ ∧ Ẽ|2 − ω 2 ε|Ẽ|2 dx +
µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx.
Ω\D̄
D
Therefore, the proof of Theorem 4.3 essentially follows the PMC case, except
that when showing
R
ε|Ẽ|2 dx
Ω\D
= 0,
lim R
τ →∞
µ|H0 |2 dx
D
one needs to apply the Sobolev embedding theorem to the solution of

 ∇ ∧ P = iωµQ, ∇ ∧ Q = −iωεP − ωi Ẽ in Ω \ D,
ν ∧ P|∂Ω = 0,

ν ∧ P|∂D = 0.
A regularity result for this BVP was studied in the appendix of [16].
6.2. Obstacles with impedance boundary conditions. In this section, we consider a more general interface condition
(iν ∧ H − λ(ν ∧ E) ∧ ν)|∂D = 0,
(28)
where λ ∈ L∞ (∂D) and Im λ(x) ≥ C > 0. This condition applies when the obstacle D is not perfectly conducting but does not allow the electromagnetic wave to
penetrate deeply into D, where λ denotes the electromagnetic impedance of D.
First, we state that the following BVP is well-posed by using a similar argument
to the one in Appendix A.

 ∇ ∧ E = iωµH, ∇ ∧ H = −iωεE, in Ω \ D,
(29)
(iν ∧ H − λ(ν ∧ E) ∧ ν)|∂D = g ∈ T H 1/2 (∂D),

ν ∧ E|∂Ω = f ∈ T H 1/2 (∂Ω).
Moreover, one has
kEkH 1 (Ω\D) + kHkH 1 (Ω\D) ≤ C kf kH 1/2 (∂Ω) + kgkH 1/2 (∂D) .
A generalization of the equalities in Lemma 4.4 and Lemma 6.1 can be obtained
as following, analogous to Proposition 2.1 in [14],
Lemma 6.2. Suppose (E, H) satisfies (29) with
g = 0,
Inverse Problems and Imaging
and f = ν ∧ E0 |∂Ω .
Volume 4, No. 3 (2010), 1–XX
16
Ting Zhou
Then we have
Z
Z
I=
µ−1 |∇ ∧ Ẽ|2 − ω 2 ε|Ẽ|2 dx +
µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx
Ω\D
D
Z
2
(30)
ω
−iImλ |(ν ∧ E) ∧ ν| dS + Reλ |(ν ∧ E0 ) ∧ ν|2 − |(ν ∧ Ẽ) ∧ ν|2
∂D
h
i
− 2Imλ Im (ν ∧ Ẽ) ∧ ν · (ν ∧ E0 ) ∧ ν dS.
Proof. First we have with boundary terms
Z
I=
−µ−1 (∇ ∧ E0 ) · (∇ ∧ E) + ω 2 εE0 · Edx
Ω\D
Z
Z
+
µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx + ω
E0 · i(ν ∧ H)dS.
Ω
∂D
Adding
Z
µ−1 (∇ ∧ E) · (∇ ∧ Ẽ) − ω 2 εE · Ẽdx − ω
0=
Z
i(ν ∧ H) · ẼdS,
∂D
Ω\D
we get
Z
I=
µ−1 |∇ ∧ Ẽ|2 − ω 2 ε|Ẽ|2 dx +
Z
µ−1 |∇ ∧ E0 |2 − ω 2 ε|E0 |2 dx + I1
D
Ω\D
where the boundary term
Z
I1 =ω
E0 · i(ν ∧ H) + i(ν ∧ H) · E0 − i(ν ∧ H) · EdS
Z∂D
=ω
λ ((ν ∧ E0 ) ∧ ν) · (ν ∧ E) ∧ ν + λ ((ν ∧ E) ∧ ν) · (ν ∧ E0 ) ∧ ν
∂D
− λ|(ν ∧ E) ∧ ν|2 dS
Z
=ω
Reλ |(ν ∧ E0 ) ∧ ν|2 − |(ν ∧ Ẽ) ∧ ν|2 − iImλ|(ν ∧ E) ∧ ν|2
∂D
− iImλ((ν ∧ E0 ) ∧ ν) · (ν ∧ E) ∧ ν + iImλ((ν ∧ E0 ) ∧ ν) · (ν ∧ E0 ) ∧ ν
+ iImλ((ν ∧ E) ∧ ν) · (ν ∧ E0 ) ∧ ν − iImλ((ν ∧ E0 ) ∧ ν) · (ν ∧ E0 ) ∧ νdS.
Then we have (30).
To deal with the boundary terms, we need the following lemma
Lemma 6.3. Let Ω̃ be a bounded domain in Rn with a smooth boundary Γ. Then
there exists a positive constant KΩ̃ depending on Ω̃ such that
Z
(31)
|(ν ∧ F) ∧ ν|2 dS = kν ∧ Fk2L2 (Γ) ≤ KΩ̃ kFk2H 1 (Ω̃) ,
Γ
Z
(32)
Γ
|(ν ∧ F) ∧ ν|2 dS ≤ KΩ̃ kFk2H 1 (Ω̃) + −1 kFk2L2 (Ω̃)
3
for all F ∈ H 1 (Ω̃) and ∈ (0, 1).
Proof. The proof of (31) is a direct corollary of a standard trace theorem. The proof
of (32) is a variation of Theorem 1.5.1.10 in [10] for vector cases. For completeness,
we include the proof here.
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
17
Given Γ smooth, one can extend the outer normal unit vector function ν(x) on Γ
to a (C 1 (Ω̃))3 function, denoted by νext . Applying integration by parts, we obtain
Z
Z
2
|(ν ∧ F) ∧ ν| dS = |ν ∧ F|2 dS
Γ
ZΓ
= (νext ∧ F) · (∇ ∧ F) − ∇ ∧ (νext ∧ F) · Fdx
ZΩ̃
= (νext ∧ F) · (∇ ∧ F) − (∇ · F)(νext · F) − [(ν · ∇)νext ] · F
Ω̃
+ (∇ · νext )|F|2 + [(νext · ∇)F] · Fdx
Z
Z
≤C
|∇F||F|dx +
|F|2 dx
Ω̃
Ω̃
≤C k∇FkL2 (Ω̃) kFkL2 (Ω̃) + kFk2L2 (Ω̃) .
Finally, one has
Z
|(ν ∧ F) ∧ ν|2 dS ≤ C k∇Fk2L2 (Ω̃) + −1 kFk2L2 (Ω̃) + kFk2L2 (Ω̃) .
Γ
This implies (32) for ∈ (0, 1), where KΩ̃ is the final general constant C, depending
on ν hence on the domain.
The reflected solution Ẽ = E − E0 ∈ H 1 (Ω \ D) is a solution of (29) with f = 0
and g = (−iν ∧ H0 + λ(ν ∧ E0 ) ∧ ν)|∂D ∈ T H 1/2 (∂D). Therefore,
kẼkH 1 (Ω\D) ≤Ck − iν ∧ H0 + λ(ν ∧ E0 ) ∧ νkH 1/2 (∂D)
≤C(kH0 kH 1 (D) + kE0 kH 1 (D) ).
Combined with Lemma 6.2 and Lemma 6.3 one can propose an upper bound
|Re(I)| ≤ C kH0 k2H 1 (D) + kE0 k2H 1 (D) ,
which proves the decaying property
(33)
lim |Re(Iρ (τ, t))| = 0,
τ →∞
for t > hD (ρ).
Notice that although the H 1 norm of (E0 , H0 ) introduces one order of τ to the L2
norm, it doesn’t cancel the exponential decay.
To show the non-vanishing property at t = hD (ρ), we first present several estimates.
Since Ẽ satisfies (37) with f = 0 and
g = ω λ(ν ∧ Ẽ) ∧ ν + λ(ν ∧ E0 ) ∧ ν − iν ∧ H0 ∈ T H 1/2 (∂D) ⊂ T H −1/2 (∂D),
∂D
using (45), we have
kẼkH 1 (Ω\D) ≤C kgkH −1/2 (∂D)
n
o
≤C k∇ ∧ ẼkL2 (Ω\D) + kẼkL2 (Ω\D) + k∇ ∧ E0 kL2 (D) + kE0 kL2 (D)
n
o
(34)
≤C k∇ ∧ ẼkL2 (Ω\D) + kẼkL2 (Ω\D) + kH0 kL2 (D) + kE0 kL2 (D)
Inverse Problems and Imaging
Volume 4, No. 3 (2010), 1–XX
18
Ting Zhou
3
The second inequality is valid since for F ∈ H 1 (Ω \ D) , one has
Z
Z
(∇ ∧ Ẽ) · F − Ẽ · (∇ ∧ F)dx,
((ν ∧ Ẽ) ∧ ν) · (ν ∧ F)dS =
∂D
Ω\D
the same estimate holds for k(ν ∧ E0 ) ∧ νkH −1/2 (∂D) on D and the second part of
(14) is true.
In addition, by the well-posedness of Maxwell’s equations satisfied by E0 in D
with boundary condition
ν ∧ µ−1 (∇ ∧ E0 )|∂D = iω(ν ∧ H0 )|∂D ∈ T H −1/2 (∂D),
one obtains the estimate
(35)
kE0 kH 1 (D) ≤ Ckν ∧ H0 kH −1/2 (∂D) ≤ C k∇ ∧ E0 kL2 (D) + kE0 kL2 (D) .
Combining Lemma 6.2, Lemma 6.3, (34) and (35), we obtain
!
|Re(I)| ≥ min{µ−1 } − L k∇ ∧ Ẽk2L2 (Ω\D) − C kẼk2L2 (Ω\D)
Ω\D
+ min{µ−1 } − L k∇ ∧ E0 k2L2 (D) − C kE0 k2L2 (D)
D
where L > 0 and C > 0. Choosing small enough such that
K := min{µ−1 } − L > 0
Ω
and dropping the first term, we obtain the lower bound
|Re(I)| ≥ −C kẼk2L2 (Ω\D) + Kk∇ ∧ E0 k2L2 (D) − C kE0 k2L2 (D) .
The remainder part of the proof is essentially the same as the PMC case. We
propose the auxiliary problem

 ∇ ∧ (µ−1 ∇ ∧ P) − sε∇(∇ · εP) − ω 2 εP = εẼ in Ω \ D,
ν ∧ P|∂Ω = 0,

ν ∧ (µ−1 ∇ ∧ P) − ωλ(ν ∧ P) ∧ ν|∂D = 0.
Then it can be verified that
Z
Z
2
ε|Ẽ| dx = −
[ν ∧ µ−1 (∇ ∧ E0 )] · P − λ[(ν ∧ E0 ) ∧ ν] · PdS
Ω\D
∂D
Z
Z
= − iω
(ν ∧ H0 ) · PdS +
λ[(ν ∧ E0 ) ∧ ν] · PdS.
∂D
∂D
We have shown the first term can be absorbed by kH0 k2L2 (D) for large τ using the
regularity of P and Sobolev embedding theorem. It is not hard to see the same
holds for the second term since
Z
Z
λ[(ν ∧ E0 ) ∧ ν] · PdS ≤ C
eτ (x·ρ−hD (ρ)) dS kẼkL2 (Ω\D) .
∂D
∂D
This proves
(36)
lim inf |Re(Iρ (τ, t))| =
τ →∞
C>0
∞
t = hD (ρ),
t < hD (ρ).
Summing up, by (33) and (36), we obtain the main reconstruction theorem for
obstacles with impedance interface conditions as following
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Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
19
Theorem 6.4. There is a subset Σ ⊂ S2 of measure zero such that when ρ ∈ S2 \ Σ,
the support function hD (ρ) can be recovered by
hD (ρ) = inf{t ∈ R | lim |Re (Iρ (τ, t))| = 0}.
τ →∞
7. Conclusion and Open problems. Since the first introduction of the enclosure
method in the context of electrostatics, it has been generalized to Helmholtz equations, elasticity equations and some systems with Laplacian as the leading term,
the heat equation and so on. In this work, for the first time, we generalized it to
Maxwell’s equations. On this direction of research, there are several open problems
and further development that we believe can be done in the future.
In the acoustic case, [19] recently introduced the enclosure method for recovering
penetrable obstacles, namely, coefficients bear a jump discontinuity on the obstacle but allow propagation of waves through the interface. There is loss of elliptic
regularity for equations with L∞ coefficients and the Sobolev embedding theorem
is not valid. This was tackled by developing some Hölder type of estimates for the
equations. It would be interesting to generalize this to the electromagnetic setting.
A more general and also much more difficult problem is to explore the method
for the full dissipative Maxwell’s equations, in which the conductivity is nonzero.
The difficulty is that the complex coefficients will compromise the lower bounds for
the indicator function.
Another interesting open problem is whether the depth stability analysis in [20]
could be extended to Maxwell’s equations.
A. Well-posedness of a mixed boundary value problem. Let Ω ⊂ R3 be a
bounded domain with smooth boundary and D ⊂ Ω, Consider the boundary value
problem for Maxwell’s equations

 ∇ ∧ E = iωµH ∇ ∧ H = −iωγE + J in Ω \ D,
(37)
ν ∧ E|∂Ω = f,

ν ∧ (µ−1 ∇ ∧ E)|∂D = iων ∧ H|∂D = g.
Here µ and γ are complex-valued functions in C k (Ω \ D) with positive real parts
and ω ∈ C.
Theorem A.1. There is a discrete subset Σ of C such that for ω not in Σ, there
3
3
exists a unique solution (E, H) ∈ H k (Ω \ D) × H k (Ω \ D) of (37) given any
f ∈ T H k−1/2 (∂Ω), g ∈ T H k−1/2 (∂D) and J ∈ H k−1 (Ω \ D) with ∇ · J = 0. The
solution satisfies
(38)
kEkH k (Ω\D) + kHkH k (Ω\D) ≤C kf kT H k−1/2 (∂Ω) + kgkT H k−1/2 (∂D)
+kJkH k−1 (Ω\D)
with C > 0.
Here we proceed to prove the theorem by modifying the variational method in
[16], [7] and [18]. From the Maxwell’s equations (37), the electric field E satisfies
the second order equation
∇ ∧ (µ−1 ∇ ∧ E) − ω 2 γE = iωJ in Ω \ D,
and
(39)
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∇ · γE = 0 in Ω \ D.
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20
Ting Zhou
Therefore, we consider
∇ ∧ (µ−1 ∇ ∧ E) − sγ∇(∇ · γE) − ω 2 γE = iωJ in Ω \ D
(40)
where s is a positive real number. The equation (39) will follow from
s∇ · (γ∇(∇ · γE)) + ω 2 ∇ · γE = 0
which is obtained by taking divergence of (40). Denoting the L2 (Ω \ D) inner
product by (·, ·) and L2 (Γ) inner products by h·, ·iΓ (where Γ = ∂Ω or ∂D), we
define the bilinear form associated with the elliptic system (40):
B(E, F) := (µ−1 ∇ ∧ E, ∇ ∧ F) + s(∇ · γE, ∇ · γF)
(41)
for E, F ∈ X where
X = {F ∈ (H 1 (Ω \ D))3 | ν ∧ F|∂Ω = 0, ν · γF|∂D = 0}.
By Green’s formulae, we have that B is related to the differential operator
P = ∇ ∧ (µ−1 ∇∧) − sγ∇(∇ · γ)
by
(42)
B(E, F) = (P E, F) − hν ∧ (µ−1 ∇ ∧ E), Fi∂(Ω\D) + hs∇ · γE, ν · γFi∂(Ω\D)
for E, F ∈ (H 1 (Ω \ D))3 . Then for f˜ ∈ X 0 the weak formulation of the mixed
boundary value problem

in Ω \ D,
 P E = f˜
ν ∧ E|∂Ω = 0,

ν ∧ (µ−1 ∇ ∧ E)|∂D = 0.
is: find E ∈ X such that
B(E, F) = (f˜, F)
for all F ∈ X.
By (42), this implies the natural boundary condition
(43)
ν ∧ (µ−1 ∇ ∧ E)|∂D = 0,
∇ · γE|∂Ω = 0.
To show the theorem, one first has for the homogeneous boundary conditions,
Proposition 2. Suppose γ and µ are complex functions in C 1 (Ω \ D) with positive
real parts, and let s be a positive real number. There is a discrete set Σs ⊂ C such
that if ω is outside this set, then for any f ∈ X 0 there exists a unique solution
E ∈ X of
(44)
∇ ∧ (µ−1 ∇ ∧ E) − sγ∇(∇ · γE) − ω 2 γE = f˜
satisfying
kEkH 1 (Ω\D) ≤ Ckf˜kX 0 .
Proof. It is sufficient to show that B is bounded and coercive on X. It is clear that
B is bounded,
|B(E, F)| ≤ CkEkH 1 (Ω\D) kFkH 1 (Ω\D) .
To show the coercivity, first we have
|B(E, E)| ≥ c k∇ ∧ Ek2L2 (Ω\D) + k∇ · Ek2L2 (Ω\D) − CkEk2L2 (Ω\D) .
It can be shown that there is a Poincaré inequality for 1-forms in X similar to that
in [26]
n
o
kEk2H 1 (Ω\D) ≤ C kEk2L2 (Ω\D) + k∇ ∧ Ek2L2 (Ω\D) + k∇ · Ek2L2 (Ω\D) ,
E ∈ X.
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Volume 4, No. 3 (2010), 1–XX
Reconstructing electromagnetic obstacles
21
We have then
B(E, E) ≥ C1 kEk2H 1 (Ω\D) − C2 kEk2L2 (Ω\D) .
To show the higher order regularity of solutions, we define for k ≥ 2,
X k := F ∈ (H k (Ω \ D))3 |
ν ∧ F|∂Ω = ∇ · γF|∂Ω = ν · γF|∂D = ν ∧ (µ−1 ∇ ∧ F)|∂D = 0 .
Then we have
Proposition 3. Let γ and µ be functions in C k (Ω \ D), k ≥ 2, with positive real
parts, and let s > 0. Suppose ω ∈
/ Σs , then for any f˜ ∈ (H k−2 (Ω \ D))3 the equation
(44) has a unique solution E ∈ X k and
kEk k
≤ Ckf˜k k−2
.
H (Ω\D)
H
(Ω\D)
This can be proved by the same techniques in Section 5.9 of [26] for the Hodge
Laplacian.
Proof of Theorem A.1. As in [16], we take Σ to be the set Σ1 in previous propositions, then s can be chosen such that ω ∈
/ Σs (for more details see [16]).
3
3
To show uniqueness, suppose (E, H) ∈ H 1 (Ω \ D) ×H 1 (Ω \ D) solves (37) with
f = g = 0 and J = 0. One has
∇ ∧ (µ−1 ∇ ∧ E) = ω 2 γE,
∇ · γE = 0.
It follows that E ∈ X (by the natural boundary conditions) is a solution of (44)
with f˜ = 0, which implies E = H = 0 by Proposition 2.
For existence, given f ∈ T H k−1/2 (∂Ω), g ∈ T H k−1/2 (∂D) ⊂ T H k−3/2 (∂D) and
J ∈ H k−1 (Ω \ D), we can find E0 ∈ H k (Ω \ D) with
ν ∧ E0 |∂Ω = f,
ν ∧ (µ−1 ∇ ∧ E0 )|∂D = g
such that the extension is bounded, namely
kE0 kH k (Ω\D) ≤ C(kf kH k−1/2 (∂Ω) + kgkH k−3/2 (∂D) )
Suppose Ẽ ∈ X k is a solution, given by Proposition 3, of (44) with
f˜ = −∇ ∧ (µ−1 ∇ ∧ E0 ) + sγ∇(∇ · γE0 ) + ω 2 γE0 + iωJ ∈ (H k−2 (Ω \ D))3 .
Notice that
kf˜kH k−2 (Ω\D) ≤C kE0 kH k (Ω\D) + kJkH k−2 (Ω\D)
≤C kE0 kH k (Ω\D) + kJkH k−1 (Ω\D) .
Then E = E0 + Ẽ ∈ (H k (Ω \ D))3 satisfies
∇ ∧ (µ−1 ∇ ∧ E) − sγ∇(∇ · γE) − ω 2 γE = iωJ.
This implies ∇ · γE = 0 by an earlier argument and particular choice of s. If
1
we define H = iωµ
∇ ∧ E ∈ (H k−1 (Ω \ D))3 , then we have (E, H) the solution of
Maxwell’s equations.
Applying the same argument to H, which satisfies a second order elliptic system
∇ ∧ (γ −1 ∇ ∧ H) − ω 2 µH = ∇ ∧ (γ −1 J) ∈ H k−2 (Ω \ D)
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Volume 4, No. 3 (2010), 1–XX
22
Ting Zhou
by eliminating E from the original Maxwell’s equations. Notice that
∇ · (∇ ∧ (γ −1 J)) = 0.
By uniqueness, one has H ∈ (H k (Ω \ D))3 .
Finally, the estimate (38) is derived from
(45)
kEkH k (Ω\D) ≤kE0 kH k (Ω\D) + kẼkH k (Ω\D)
≤C kE0 kH k (Ω\D) + kf˜kH k−2 (Ω\D)
≤C kE0 kH k (Ω\D) + kJkH k−1 (Ω\D)
≤C kf kH k−1/2 (∂Ω) + kgkH k−3/2 (∂D) + kJkH k−1 (Ω\D)
≤C kf kH k−1/2 (∂Ω) + kgkH k−1/2 (∂D) + kJkH k−1 (Ω\D) .
The same computation applies to H,
kHkH k (Ω\D) ≤C kf kH k−3/2 (∂Ω) + kgkH k−1/2 (∂D) + k∇ ∧ (γ −1 J)kH k−2 (Ω\D)
≤C kf kH k−1/2 (∂Ω) + kgkH k−1/2 (∂D) + kJkH k−1 (Ω\D) .
Acknowledgments. I would like to express my deepest gratitude to my advisor
Gunther Uhlmann for his encouragement and help. We would like to thank Matti
Lassas and Mikko Salo for helpful conversations.
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Received xxxx 20xx; revised xxxx 20xx.
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Volume 4, No. 3 (2010), 1–XX