The Factorization Method for EIT in the Case of
Mixed Inclusions
Susanne Schmitt
Institut fuer Algebra und Geometrie, Universitaet Karlsruhe (TH), 76128 Karlsruhe,
Germany
E-mail: [email protected]
Abstract.
The inverse problem of electrical impedance tomography (EIT) is to recover the
conductivity inside an investigated subject from boundary measurements of current
and voltage. In this work we deal with the simpler case in which the background
conductivity is known a priori and we wish to locate inclusions, i.e. domains inside
the investigated subject having a different conductivity than the background. A very
successful method for solving this problem is the factorization method.
However, for the factorization method in EIT it is usually assumed that there is
only one type of inclusions present, e.g. only inclusions with a higher or with a lower
conductivity than the background. The question whether the factorization method also
works in the mixed case is still an open problem. Here we present a modified version of
the method to cope with this case that makes use of some a priori information about
where the inclusions are roughly located. We show that under this assumption we can
even detect inclusions in the mixed case.
AMS classification scheme numbers: 35R30, 35Q60
Detecting mixed inclusions in EIT
2
1. Introduction
In electrical impedance tomography (EIT) we inject a current f into the surface of the
investigated subject B and measure the induced potential at the surface. The potential
u solves the boundary value problem
div(γ∇u) = 0 in B,
∂γ u = f on ∂B,
where ∂γ u := γ ∂ν u denotes the conormal derivative at the boundary ∂B. From data
sets of this kind we wish to extract information about the conductivity γ inside the
subject. This work is based on the knowledge of all possible measurement data sets
{f, u|∂B }, i.e. we assume to know the Neumann-to-Dirichlet map Λ that maps each
current pattern f to the corresponding boundary potential u|∂B . This idealized model
is called the continuum model.
The inverse problem that we deal with is the detection of anomalies inside B from the
knowledge of the Neumann-to-Dirichlet map Λ, i.e. the location domains in which the
conductivity is different from the background conductivity γ0 which is assumed to be
known a priori. Thus we suppose that the conductivity has the form γ = γ0 + χΩ γ1 ,
where χΩ is the characteristic function of the inclusion Ω.
A very successful method for solving this problem is the factorization method that
has first been suggested by A. Kirsch in [1] for scattering problems. The factorization
method for electrical impedance tomography has been derived in [2], [3], [4] and has
been extended and improved continuously since then: In [5] the case of complex and
anisotropic conductivities is considered, while in [6], [7] the complete electrode model is
applied which represents a much more realistic model than the continuum model.
The proof that the factorization method works is based on the result that the ranges of
of certain operators coincide. For the proof of this range identity it is crucial to assume
that all the inclusions are of the same type, i.e. for real isotropic conductivities that they
all have either a higher or a lower conductivity than the background medium. However,
in some applications this is not the case, e.g. in imaging of the human thorax. Also in
geoelectrical imaging one might expect a mixed conductivity distribution.
Numerical examples suggest that the factorization method even works for the mixed case
(cf. figure 7 in [8]), but to our knowledge there is no proof and it is still an open problem.
Thus we apply a method that was originally developed by N. Grinberg and A. Kirsch
for scattering from different types of obstacles (cf. [9], [10], [11]) to the EIT problem.
This method is based on the idea of covering one of the obstacle types in order to obtain
a modified factorization for which one can prove the corresponding range identity and
thus that the factorization method works.
For an extensive description of important contributions to the inverse problem of
electrical impedance tomography we refer to the review paper [12] and the references
therein. There are several recent works dealing with related problems of weakening
Detecting mixed inclusions in EIT
3
assumptions on the problem setting for the factorization method. In the article [13] it
is shown that the assumption on the contrast between the background conductivity and
the inclusion conductivity can be weakened but it is still assumed that the conductivity
inside all inclusions is either higher or lower than in the background. In [14] a different
partial differential equation is considered and it is shown that the factorization method
locates inhomogeneities both in the leading coefficient and in the coefficient of 0-th
order. Again, the contrasts in these coefficients need to have the same direction for all
inclusions.
This work is organized as follows: In the next section we consider the direct problem
of EIT and state the main assumptions on the problem setting. Moreover, we show the
well-known factorization for the case of two disjoint inclusions and give an outline of
the factorization method in EIT. Finally, we derive a representation for the operator
that appears in the middle of this factorization and that this representation gives a
connection to the middle operator for the case of only one inclusion.
In section 3 we state the required additional a priori assumptions and use the results
from section 2 to derive our new method. We obtain two slight modifications of the
original factorization and show that these can be used to detect inclusions in the mixed
case. In the last section 4 we show some numerical examples using our new method.
2. The factorization method for two disjoint inclusions
We start by explaining the problem setting and by providing an overview over the
original factorization method for impedance tomography. In order to investigate the
direct problem we need to state some basic assumptions on the underlying geometry of
the body considered and the inclusions as well as the conductivity distribution. Since
our new method is aimed at the detection of mixed inclusion types, we restrict ourselves
to the case of at least two disjoint inclusions.
Assumption 2.1. Let B ⊂ Rn (n = 2, 3) be a bounded domain and ∂B ∈ C 2 . Let
Ω1 , Ω2 be subdomains of B with C 2 -boundaries, Ω1 , Ω2 ⊂ B and Ω1 ∩ Ω2 = ∅. By Ω we
denote the union of both inclusions: Ω = Ω1 ∪ Ω2 . Let B \ Ω be connected.
Additionally, we allow isotropic and real-valued conductivities that have properties as
stated below:
Assumption 2.2. Let the conductivity be as follows: γ : B → R and


x ∈ B \ Ω,

γ0 (x),
γ(x) = γ0 (x) + γ1 (x), x ∈ Ω1 ,


γ (x) + γ (x), x ∈ Ω .
0
2
2
Let the background conductivity satisfy γ0 ∈ C 2,α (B) for some α > 0 and let γ be
strictly positive, i.e. there exists c > 0 with
γ(x) ≥ c > 0 for all x ∈ B.
Detecting mixed inclusions in EIT
4
We assume that γj ∈ L∞ (Ωj ) (j = 1, 2) and allow two different types of inclusions:
Either there exists cj > 0 such that
γ0 (x) − γ(x) ≥ cj > 0
for almost all x ∈ Ωj , or there exists cj > 0 such that
γ(x) − γ0 (x) ≥ cj > 0
for almost all x ∈ Ωj (j = 1, 2). In the first case we call Ωj an inclusion of type 1 and
in the second case an inclusion of type 2.
This means that type 1 inclusions have a lower conductivity than the background, while
type 2 inclusions have a higher conductivity than the background.
Throughout this work we will deal with the following function spaces:
L2⋄ (∂B) = {g ∈ L2 (∂B) : hg, 1i = 0}
1
1
H⋄2 (∂B) = {g ∈ H 2 (∂B) : hg, 1i = 0},
−1
1
H⋄ 2 (∂B) = {g ∈ H − 2 (∂B) : hg, 1i = 0},
1
H⋄1 (B) = {u ∈ H 1 (B) : u|∂B ∈ H⋄2 (∂B)},
where h·, ·i denotes the scalar product in L2 (∂B) as well as the dual pairing between
−1
1
H⋄ 2 (∂B) and H⋄2 (∂B).
−1
Now we can formulate the direct problem. For a given current pattern f ∈ H⋄ 2 (∂B)
find a weak solution u ∈ H⋄1 (B) to the boundary value problem
div(γ∇u) = 0 in B,
∂γ u = f on ∂B.
Using Green’s formula we obtain the weak formulation for this problem:
ZZ
Z
γ∇u · ∇ϕ dx = f ϕ ds for all ϕ ∈ H⋄1 (B).
B
(1)
∂B
By the Lax-Milgram theorem (cf. [15]) this problem is uniquely solvable for every
−1
f ∈ H⋄ 2 (∂B). Using the trace theorem we obtain the Neumann-to-Dirichlet Operator
−1
1
Λ : H⋄ 2 (∂B) → H⋄2 (∂B), f 7→ u|∂B ,
where u solves (1). Λ0 denotes the Neumann-to-Dirichlet Operator for the homogeneous
case, i.e. γ ≡ γ0 in B, respectively. Thus Λ, Λ0 are well-defined and bounded operators.
We proceed by deriving the factorization method as in [5] but for the special case of
two disjoint inclusions. At first we define the operators appearing in the factorization
of Λ − Λ0 . Consider the operator
Detecting mixed inclusions in EIT
5
−1
−1
1
G : H⋄ 2 (∂Ω1 ) × H⋄ 2 (∂Ω2 ) → H⋄2 (∂B), (ψ1 , ψ2 )⊤ 7→ v|∂B .
1
Here v ∈ H 1 (B \ Ω) with v|∂B ∈ H⋄2 (∂B) solves the following boundary value problem
in the weak sense:
div(γ0 ∇v) = 0 in B \ Ω,
∂γ0 v = ψj on ∂Ωj (j = 1, 2),
∂γ0 v = 0 on ∂B.
1
As shown in [5], G is compact, one-to-one, and R(G) is dense in H⋄2 (∂B). Additionally,
the following operator will appear:
1
1
−1
−1
T : H⋄2 (∂Ω1 ) × H⋄2 (∂Ω2 ) → H⋄ 2 (∂Ω1 ) × H⋄ 2 (∂Ω2 ),
(h1 , h2 )⊤ 7→ (∂γ0 w|+,1, ∂γ0 w|+,2)⊤
1
Here, w ∈ H 1 (B \∂Ω) with w|∂B ∈ H⋄2 (∂B) is the solution to the transmission boundary
value problem
div(γ∇w) = 0 in B \ ∂Ω,
∂γ0 w = 0 on ∂B,
∂γ0 w|+,j − ∂γ w|−,j = 0 on ∂Ωj (j = 1, 2),
(2)
w|+,j − w|−,j = hj on ∂Ωj (j = 1, 2).
The notation w±,j indicates the trace of w from the exterior and interior of Ωj ,
respectively. The corresponding operator for the homogenous case γ ≡ γ0 is denoted by
T0 .
A weak formulation for this transmission boundary value problem is to find w ∈
1
H 1 (B \ ∂Ω), w|∂B ∈ H⋄2 (∂B) satisfying the jump conditions w|+,j − w|−,j = hj (j = 1, 2)
and
ZZ
γ∇w · ∇ϕ dx = 0 for all ϕ ∈ H⋄1 (B).
B
Since this weak formulation with an additional jump condition is not suitable for our
purpose, we derive an equivalent formulation.
1
For hj ∈ H⋄2 (∂Ωj ) choose ŵ (j) ∈ H 1 (Ωj ) such that ŵ (j) |∂Ωj = hj (j = 1, 2). Then by
setting w̃ = w + ŵ (1) + ŵ (2) we get the formulation: find w̃ ∈ H⋄1 (B) satisfying
ZZ
ZZ
ZZ
(1)
γ∇w̃ · ∇ϕ dx =
γ∇ŵ · ∇ϕ dx +
γ∇ŵ (2) · ∇ϕ dx
(3)
B
Ω1
Ω2
Detecting mixed inclusions in EIT
6
for all ϕ ∈ H⋄1 (B). Here we can also show unique solvability by the Lax-MilgramTheorem, and from the trace theorem we have well-definedness and boundedness of
T, T0 .
During the remainder of this section we recall the main results of the factorization
method for EIT. For the corresponding proofs we refer to [5]. First we state the wellknown factorization that can be proven independently of γ1 and γ2 ,
Λ − Λ0 = G(T − T0 )G∗ .
(4)
Additionally, T and T0 are self-adjoint. Under the assumption that only one inclusion
type is present inside B the middle operator T − T0 satisfies the following coercivity
relations:
Lemma 2.3. (a) If Ω1 and Ω2 are both of type 1, then h(T − T0 )h, hi∂Ω ≥ ckhk2
1
H 2 (∂Ω)
1
2
for all h ∈ H⋄ (∂Ω),
(b) If Ω1 and Ω2 are both of type 2, then h(T0 − T )h, hi∂Ω ≥ ckhk2
1
H 2 (∂Ω)
for all
1
2
h ∈ H⋄ (∂Ω).
An immediate conclusion of lemma 2.3 is that the operator T − T0 is one-to-one. Later
we will see that these coercivity results are not valid in the mixed case. However, we
are still able to show the following lemma, which we will need in section 3.
Lemma 2.4. Assume that Ω1 is of type 1 while Ω2 is of type 2. Then the operator
T − T0 is one-to-one.
1
1
Proof. Let h = (h1 , h2 )⊤ ∈ H⋄2 (∂Ω1 ) × H⋄2 (∂Ω2 ) be such that (T − T0 )h = 0. Then
∂γ0 w|+,1 = ∂γ0 w0 |+,1 , ∂γ0 w|+,2 = ∂γ0 w0 |+,2 ,
where w, w0 are the solutions tho the transmission boundary value problem (2) for the
inhomogeneous and the homogeneous case, respectively. By uniqueness of the Neumann
problem in B \ Ω we know that w ≡ w0 in B \ Ω, from which we can conclude using the
jump conditions at ∂Ω1 , ∂Ω2 that
∂γ w|−,1 = ∂γ0 w0 |−,1 , ∂γ w|−,2 = ∂γ0 w0 |−,2 ,
w|−,1 = w0 |−,1 , w|−,2 = w0 |−,2.
Using Green’s formula, the fact that Ω1 is a type 1 inclusion and the Dirichlet principle
we can estimate
ZZ
ZZ
2
γ|∇w| dx =
γ0 |∇w0 |2 dx
Ω1
Ω
≥
Z 1Z
2
γ|∇w0 | dx + c1
Ω
≥
Z 1Z
Ω1
ZZ
|∇w0 |2 dx
Ω
γ|∇w|2 dx + c1
Z Z1
Ω1
|∇w0 |2 dx.
Detecting mixed inclusions in EIT
7
Now it follows that ∇w0 = 0 and ∇w = 0 in Ω1 , and thus w, w0 are constant in Ω1 .
This means that the normal derivatives of w, w0 at ∂Ω1 from both sides are zero, which
1
implies that w = w0 = 0 in B \ Ω. Now h1 has to be constant, and since h1 ∈ H⋄2 (∂Ω1 )
we have that h1 = 0. Now we have to apply the very same arguments to inclusion Ω2
using its type 2 property to obtain h2 = 0.
The coercivity results from lemma 2.3 play a crucial role in the proof of the range
identity
R |Λ − Λ0 |
1
2
= R(G),
(5)
Now this range identity can be used in the following way: It is a well-known result that
y ∈ Ω ⇔ ϕy ∈ R(G),
where ϕy is the trace at ∂B of the potential of a dipole located in y, i.e. ϕy (x) =
γ0 â · ∇y N(x, y) for x ∈ B where N is the Neumann function for the domain B and the
background conductivity γ0 while â is an arbitrary unit vector.
Together with the range identity (5) and the Picard criterion we deduce the equivalence
y∈Ω⇔
∞
X
hϕy , ψk i2
k=1
λk
<∞
with an orthonormal eigensystem {λj , ψj } of |Λ − Λ0 |.
This binary criterion can be used to decide if a point y lies inside the inclusion or not.
However, we have to assume that there is only one inclusion type present, which is a
severe restriction.
In order to investigate the factorization method in the special case where at least two
disjoint inclusions are present we now take a closer look at the middle operator. Our
aim is to write T − T0 in the form
!
(11)
(12)
T (11) − T0
T (12) − T0
T − T0 =
(21)
(22)
T (21) − T0
T (22) − T0
(ij)
where T (ij) , T0
it follows that
−1
1
1
1
: H⋄2 (∂Ωj ) → H⋄ 2 (∂Ωi ) Then for h = (h1 , h2 )⊤ ∈ H⋄2 (∂Ω1 ) ×H⋄2 (∂Ω2 )
T (11) − T0
and
(11)
(12)
h1 + T (12) − T0
h2 = ((T − T0 )h)|∂Ω1 ,
T (21) − T0
respectively.
(21)
(22)
h1 + T (22) − T0
h2 = ((T − T0 )h)|∂Ω2 ,
Detecting mixed inclusions in EIT
8
Additionally, we wish to find a connection to the cases where only one of the two
inclusions is present in order to make use of the coercivity results from lemma 2.3. The
subsequent conclusions are based on the weak formulation (3).
1
1
Let (h1 , h2 )⊤ ∈ H⋄2 (∂Ω1 ) ×H⋄2 (∂Ω2 ) be given. For these densities, choose ŵ (j) ∈ H 1 (Ωj )
such that ŵ (j) |∂Ωj = hj (j = 1, 2).
(1)
Now consider the following problems: On the one hand find w̃0 ∈ H⋄1 (B) such that
ZZ
ZZ
(1)
(1)
γ0 ∇w̃0 · ∇ϕ dx =
γ0 ∇ŵ0 · ∇ϕ dx for all ϕ ∈ H⋄1 (B).
(6)
Ω1
B
(2)
w̃0
On the other hand find
∈ H⋄1 (B) such that
ZZ
ZZ
(2)
(2)
γ0 ∇w̃0 · ∇ϕ dx =
γ0 ∇ŵ0 · ∇ϕ dx for all ϕ ∈ H⋄1 (B).
(7)
Ω2
B
(1)
(2)
By addition of (6) and (7) it is easy to see that w̃0 := w̃0 + w̃0 is solution to (3) for
the homogeneous case γ ≡ γ0 . Now we define the components of T0 :
(11)
T0
(12)
T0
(21)
T0
(22)
T0
1
−1
(1)
(1)
−1
(2)
(2)
−1
(1)
(1)
−1
(2)
(2)
: H⋄2 (∂Ω1 ) → H⋄ 2 (∂Ω1 ), h1 7→ ∂γ0 w̃0 |+,1 , w̃0 solves (6),
1
: H⋄2 (∂Ω2 ) → H⋄ 2 (∂Ω1 ), h2 7→ ∂γ0 w̃0 |+,1 , w̃0 solves (7),
1
: H⋄2 (∂Ω1 ) → H⋄ 2 (∂Ω2 ), h1 7→ ∂γ0 w̃0 |+,2 , w̃0 solves (6),
1
: H⋄2 (∂Ω2 ) → H⋄ 2 (∂Ω2 ), h2 7→ ∂γ0 w̃0 |+,2 , w̃0 solves (7).
(11)
It is easy to see that T0 corresponds to T0 in the case where only the inclusion Ω1
(22)
is present, and that T0 corresponds to T0 in the case where Ω2 is the only inclusion,
respectively.
(12)
Lemma 2.5. T0
(21)
and T0
are compact operators. Thus T0 can be represented by
!
(11)
T0
0
T0 =
+ K0
(22)
0
T0
where K0 is a compact operator.
(12)
Proof. We only prove the assertion for T0
1
2
(21)
; for T0
the arguments are analogous.
Let (h2,j )j∈N be a bounded sequence in H⋄ (∂Ω2 ). Then the corresponding sequence
(2,j)
of solutions w̃0
of (7) is also bounded in H 1 (B \ Ω2 ). Now let U ⊂ B \ Ω2
j∈N
be a subdomain
U ⊂ B \ Ω2 and ∂Ω1 ⊂ U. Following Theorem. 8.8 in [16]
with
(2,j)
the sequence w̃0
is even bounded in H 2 (U) and by the trace theorem we get:
j∈N
(2,j)
k∂γ0 w̃0 kH 12 (∂Ω )
1
−1
1
≤ c j ∈ N. By compactness of the embedding J : H⋄2 (∂Ω1 ) →
H⋄ 2 (∂Ω1 ) we get the assertion.
Detecting mixed inclusions in EIT
9
For T the partitioning is not as simple as for T0 , since the different conductivities γ1 , γ2
appear in equation (3). However, we start as before and divide T into parts contributed
by jump conditions at the individual inclusion boundaries.
1
1
As before, for (h1 , h2 )⊤ ∈ H⋄2 (∂Ω1 ) × H⋄2 (∂Ω2 ) choose ŵ (j) ∈ H 1 (Ωj ) such that
ŵ (j) |∂Ωj = hj (j = 1, 2) and consider the following problems: find w̃ (j) ∈ H⋄1 (B)
(j = 1, 2) satisfying
ZZ
γ0 ∇w̃
(1)
· ∇ϕ dx +
ZZ
γ1 ∇w̃
(1)
· ∇ϕ dx +
Ω
B
=
Z1Z
ZZ
γ2 ∇w̃ (1) · ∇ϕ dx
ZZ
γ2 ∇w̃ (2) · ∇ϕ dx
Ω2
γ∇ŵ
(1)
(8)
· ∇ϕ dx,
Ω1
ZZ
γ0 ∇w̃
(2)
· ∇ϕ dx +
ZZ
γ1 ∇w̃
(2)
· ∇ϕ dx +
Ω
B
=
Z1Z
Ω2
(9)
γ∇ŵ (2) · ∇ϕ dx
Ω2
for all ϕ ∈ H⋄1 (B). One can easily see that w̃ := w̃ (1) + w̃ (2) is a solution to (3). As
before, we define component operators:
1
−1
1
−1
1
−1
1
−1
T (11) : H⋄2 (∂Ω1 ) → H⋄ 2 (∂Ω1 ), h1 7→ ∂γ0 w̃ (1) |+,1 , w̃ (1) solves (8),
T (12) : H⋄2 (∂Ω2 ) → H⋄ 2 (∂Ω1 ), h2 7→ ∂γ0 w̃ (2) |+,1 , w̃ (2) solves (9),
T (21) : H⋄2 (∂Ω1 ) → H⋄ 2 (∂Ω2 ), h1 7→ ∂γ0 w̃ (1) |+,2 , w̃ (1) solves (8),
T (22) : H⋄2 (∂Ω2 ) → H⋄ 2 (∂Ω2 ), h2 7→ ∂γ0 w̃ (2) |+,2 , w̃ (2) solves (9).
All four partial operators are well-defined and bounded. It can be shown that T (12)
and T (21) are compact operators for which we only give an outline of a proof: first we
have to perform a decomposition of T (12) and T (21) as in the proof of theorem 2.2 in
[5] and show compactness of one of the decomposed operators using compactness of the
imbeddings between sobolev spaces as is done in lemma 2.5.
Here T (11) is not identical to T in the case where Ω1 is the only inclusion. The same
holds for T (22) . Therefore we carry out one more decomposition of these two operators.
1
−1
Define T (1) : H⋄2 (∂Ω1 ) → H⋄ 2 (∂Ω1 ) by h1 7→ ∂γ0 w̃ (11) |+,1 , where w̃ (11) solves the weak
problem
ZZ
ZZ
ZZ
(11)
(11)
γ0 ∇w̃
· ∇ϕ dx +
γ1 ∇w̃
· ∇ϕ dx =
γ∇ŵ (1) · ∇ϕ dx
(10)
B
Ω1
Ω1
Detecting mixed inclusions in EIT
10
for all ϕ ∈ H⋄1 (B). T (1) is obviously equivalent to T in the case where the inclusion Ω2
−1
1
is not present. Analogously, define T (2) : H⋄2 (∂Ω2 ) → H⋄ 2 (∂Ω2 ) by h2 7→ ∂γ0 w̃ (22) |+,2 ,
where w̃ (22) solves
ZZ
ZZ
ZZ
(22)
(22)
γ0 ∇w̃
· ∇ϕ dx +
γ2 ∇w̃
· ∇ϕ dx =
γ∇ŵ (2) · ∇ϕ dx
(11)
Ω2
B
Ω2
for all ϕ ∈ H⋄1 (B). T (2) is equal to T in the case where Ω2 is the only inclusion. Now
we continue by investigating the difference between T (11) and T (1) (between T (22) and
T (2) , respectively) and show that it is a compact operator.
Lemma 2.6. The following identities hold:
T (11) = T (1) + S (1) , T (22) = T (2) + S (2) ,
(12)
where S (1) , S (2) are compact operators.
Proof. We start by defining S (1) , S (2) and show that the representations in (12) are
−1
1
valid. Let S (1) : H⋄2 (∂Ω1 ) → H⋄ 2 (∂Ω1 ), h1 7→ ∂γ0 ṽ (1) |+,1 where ṽ (1) ∈ H 1 (B) solves
ZZ
γ0 ∇ṽ
(1)
· ∇ϕ dx+
ZZ
γ1 ∇ṽ
(1)
· ∇ϕ dx +
Ω
B
=−
Z 1Z
ZZ
γ2 ∇ṽ (1) · ∇ϕ dx
Ω2
γ2 ∇w̃
(11)
(13)
· ∇ϕ dx
Ω2
for all ϕ ∈ H 1 (B), where w̃ (11) is solution to problem (10).
1
2
H⋄ (∂Ω2 ) →
−1
H⋄ 2 (∂Ω2 ),
ZZ
γ0 ∇ṽ
(2)
Analogously S (2) :
h2 7→ ∂γ0 ṽ (2) |+,2 where ṽ (2) ∈ H 1 (B) is solution to
· ∇ϕ dx+
ZZ
γ1 ∇ṽ
(2)
· ∇ϕ dx +
Ω
B
=−
Z 1Z
ZZ
Ω2
γ1 ∇w̃
(22)
γ2 ∇ṽ (2) · ∇ϕ dx
(14)
· ∇ϕ dx
Ω1
for all ϕ ∈ H 1 (B) where w̃ (22) is solution to (11).
By addition of (10) and (13) (and of (11) and (14), respectively) we realize that
w̃ (1) := w̃ (11) + ṽ (1) solves (8) (and that w̃ (2) := w̃ (22) + ṽ (2) solves (9), respectively).
Thus it follows that T (1) + S (1) = T (11) and T (2) + S (2) = T (22) .
Now we have to show that S (1) and S (2) are compact operators.
We only prove the assertion for S1 and decompose S (1) by S (1) = S̃˜ ◦ S̃. Here
1
1
1
S̃ : H⋄2 (∂Ω1 ) → H⋄2 (∂Ω2 ), h1 7→ w̃ (11) |−,2, where w̃ (11) solves (10) and S̃˜ : H⋄2 (∂Ω2 ) →
−1
H⋄ 2 (∂Ω1 ), h̃2 7→ ∂γ0 ṽ|+,1 , where ṽ solves
Detecting mixed inclusions in EIT
ZZ
γ0 ∇ṽ · ∇ϕ dx +
11
ZZ
γ1 ∇ṽ · ∇ϕ dx +
Ω1
B
=−
ZZ
ZZ
γ2 ∇ṽ · ∇ϕ dx
Ω2
γ2 ∇ŵ (2) · ∇ϕ dx
Ω2
for all ϕ ∈ H⋄1 (B), and ŵ (2) ∈ H 1 (Ω2 ) was chosen such that ŵ (2) |∂Ω2 = h̃2 . As in
the proof of theorem 2.2 of [5] it can be shown that S̃˜ is compact. Furthermore S̃ is
bounded, which yields the assertion.
Altogether we have now derived the representation
!
(1)
T
0
+K
T =
0 T (2)
with a compact operator K and by combination with lemma 2.5
!
(11)
(1)
T − T0
0
+ K̃
T − T0 =
(22)
(2)
0
T − T0
(15)
with a compact operator K̃.
3. The covering method
Now we restrict ourselves to the mixed case, i.e. there is an inclusion Ω1 of type 1 as well
as an inclusion Ω2 of type 2. In this section we derive the factorization method for slight
modifications of Λ − Λ0 under some additional a priori assumptions on the inclusions’
locations. We show that using these we can reconstruct inclusions even in the case of
different types of inclusions. This setting will be defined in the next assumption.
Assumption 3.1. We assume that we know C 2 -domains Ω̃1 , Ω̃2 ⊂ B where Ω̃1 ∩ Ω̃2 = ∅,
Ω̃1 ∪ Ω̃2 ⊂ B, and Ωj ⊂ Ω̃j (j = 1, 2) holds and B \ (Ω̃1 ∪ Ω̃2 ) is connected.
This configuration is illustrated in figure (3). The a priori knowledge of Ω̃1 , Ω̃2 is –at
least in some applications– no strong restriction. For example, in medical applications
one knows roughly where the different tissues lie and which conductivities they usually
have.
Corresponding to lemma 2.3 the following coercivity assertions hold for the operators
on the diagonal in the representation (15) of T − T0 .
(11)
h(T (1) − T0
(22)
h(T0
−T
1
2 (∂Ω1 )
for all h1 ∈ H⋄2 (∂Ω1 ),
)h2 , h2 i∂Ω2 ≥ ckh2 k2
1
2 (∂Ω2 )
for all h2 ∈ H⋄2 (∂Ω2 ).
H
(2)
1
)h1 , h1 i∂Ω1 ≥ ckh1 k2
H
1
Detecting mixed inclusions in EIT
12
Ω1
Ω̃1
Ω2
Ω̃2
Ω1
B
Figure 1. Sketch of the inclusions and the a priori known domains Ω̃1 , Ω̃2
Now it is easy to see from (15) that the operator T −T0 is neither positively nor negatively
coercive, so that we are not able to apply the factorization method immediately. In order
to derive a factorization for a slight modification of Λ−Λ0 , we now define some auxiliary
operators.
−1
−1
Qj : H⋄ 2 (∂Ωj ) → H⋄ 2 (∂ Ω̃j ), ϕ 7→ ∂γ0 w|+ on ∂ Ω̃j
1
Here w ∈ H 1 (B \Ωj ) with w|∂B ∈ H⋄2 (∂B) (j = 1, 2) solves the boundary value problem:
div(γ0 ∇w) = 0 in B \ Ωj ,
(16)
∂γ0 w = ϕ on ∂Ωj ,
∂γ0 w = 0 on ∂B.
As in the proof of lemma 2.5 we can show that Q1 , Q2 are compact, one-to-one and have
dense range. The same holds for the adjoints Q∗1 , Q∗2 .
Furthermore, we make use of the composed operators
Q1
I
!
I
Q2
!
−1
−1
−1
−1
: H⋄ 2 (∂Ω1 ) × H⋄ 2 (∂Ω2 ) → H⋄ 2 (∂ Ω̃1 ) × H⋄ 2 (∂Ω2 )
:
−1
H⋄ 2 (∂Ω1 )
×
−1
H⋄ 2 (∂Ω2 )
→
−1
H⋄ 2 (∂Ω1 )
×
(17)
−1
H⋄ 2 (∂ Ω̃2 )
Q1
These are defined correspondingly to Q1 , Q2 :
is defined by (ϕ1 , ϕ2 )⊤ 7→
I
(∂γ0 w|+,1, ϕ2 )⊤ , where w solves an slight modification of the boundary value problem
Detecting mixed inclusions in EIT
13
(16): w is solution to the conductivity equation in B \ Ω and has Neumann boundary
values ϕ1 , ϕ2 at the boundaries ∂Ω1 , ∂Ω2 .
⊤
⊤
I
Q2 is defined by (ϕ1 , ϕ2 ) 7→ (ϕ1 , ∂γ0 w|+,2 ) , where w solves the same boundary value
problem in B \ Ω.
−1
1
Additionally, we make use of the operators G̃j : H⋄ 2 (∂ Ω̃j ) → H⋄2 (∂B) (j = 1, 2)
G1 that are defined just asnG but
with
Ω
replaced
by
Ω̃
j . We formally write G as
G2 .
o
Analogously, we define
G̃1
G2
just the same as G, but here Ω1 is replaced by Ω̃1 . The
n o
1
corresponding definition holds for G
.
G̃2
Now it is easy to check that the following equalities are valid:
Lemma 3.2. (a)
( ) ( )
!
( )
! ( )∗
G1
G̃1
Q1
G̃
I
0
G̃1
1
=
, G̃1 G̃∗1 =
G2
G2
I
G2
0 0
G2
(b)
( ) ( )
!
( )
! ( )∗
G1
G1
I
G1
0 0
G1
=
, G̃2 G̃∗2 =
G2
G̃2
Q2
G̃2
0 I
G̃2
n o n o
1
1
It is quite clear that G̃1 , G̃2 , G̃
, G
have the same properties as G concerning
G2
G̃2
1 ∗
compactness, injectivity and denseness of the range. The adjoint G
maps from
G2
−1
1
1
(∂B) tonH⋄2 o
(∂Ω1 ) × H⋄2 (∂Ω1 ) and the corresponding mapping property holds for
H⋄ 2 o
n
∗
∗
G1
G̃1
and
G2
G̃2
Now choose ρ1 , ρ2 ∈ C such that Re{ρj } > 0 (j = 1, 2) and consider the modified
Neumann-to-Dirichlet maps Λ̃1 = Λ − Λ0 − ρ1 G̃1 G̃∗1 and Λ̃2 = Λ − Λ0 + ρ2 G̃2 G̃∗2 . First,
we remark that for the acquisition of both maps we only need measured data (i.e. Λ)
and information known a priori. Using lemma 3.2 and the well-known factorization (4)
we can derive factorizations of Λ̃1 and Λ̃2 :
( )
( )∗
G1
G1
Λ̃1 =
(T − T0 )
− ρ1 G̃1 G̃∗1
G2
G2
( )"
!
!∗
G̃1
Q1
Q1
=
(T − T0 )
−
G2
I
I
ρ1 I 0
0 0
( )
( )∗
G1
G1
Λ̃2 =
(T − T0 )
+ ρ2 G̃2 G̃∗2
G2
G2
( )"
!
!∗
G1
I
I
=
(T − T0 )
+
G̃2
Q2
Q2
0 0
0 ρ2 I
!# ( )∗
G̃1
,
G2
!# ( )∗
G1
.
G̃2
Detecting mixed inclusions in EIT
14
For Λ̃1 we choose ρ1 such that Im{ρ1 } < 0 while for Λ̃2 we choose Im{ρ2 } > 0. In
order to apply the factorization method we have to show some required properties of
the middle operator S defined by
!
Q
1
T − T0
I
!
I
Q2
S1 =
Q1
I
S2 =
I
Q2
or
T − T0
!∗
!∗
!
(18)
!
(19)
−
ρ1 I 0
0 0
+
0 0
0 ρ2 I
.
1
1
Lemma 3.3. (a) ImhS1 h, hi ≥ 0 for all h = (h1 , h2 )⊤ ∈ H⋄2 (∂ Ω̃1 ) × H⋄2 (∂Ω2 ) and
1
1
ImhS2 h, hi ≥ 0 for all (h1 , h2 )⊤ ∈ H⋄2 (∂Ω1 ) × H⋄2 (∂ Ω̃2 ).
(b) If (h1 , h2 )⊤ 6= 0 such that (ReSj )h = 0, then ImhSj h, hi > 0 (j = 1, 2).
Proof. We only consider the first factorization (18). For (19) the arguments are the same.
1
1
−1
−1
We denote X = H⋄2 (∂ Ω̃1 ) × H⋄2 (∂Ω2 ) and its dual space X ∗ = H⋄ 2 (∂ Ω̃1 ) × H⋄ 2 (∂Ω2 ).
Then the middle operator is
Q1
I
!
Q
1
S1 =
T − T0
I
!
ρ1 I 0
=: H −
.
0 0
!∗
+
−ρ1 I 0
0
0
!
∗
By lemma 2.4 and the injectivity of QI1 and QI1 we can conclude that H : X → X ∗
is such that ReH is injective and ImhHh, hi = 0 for all h ∈ X . Thus
ImhS1 h, hi = −Im{ρ1 }kh1 k2 ≥ 0,
and the first assertion is shown.
Let h = h02 ∈ X such that (ReS1 )h = 0. Then (ReS1 )h = (ReH)h = 0. From the
injectivity of ReH it follows that h2 = 0. This means that for h = hh12 ∈ X where
h 6= 0 and (ReS1 )h = 0 we have that h1 6= 0. Now assume that such an h exists, then:
ImhS1 h, hi = −Im{ρ1 }kh1 k2 > 0,
which proves the second assertion.
We derive one more representation of Λ̃1 and Λ̃2 that is used to prove that the middle
operator is a compact perturbation of a coercive operator. For this we make use of the
Detecting mixed inclusions in EIT
15
represention (15), lemma 3.2 and the fact that
!∗
!
(11)
(1)
Q∗1
0
Q1 T − T0
Q1
Q1
+ K1 ,
=
T − T0
(22)
I
I
0
T (2) − T0
!
!
!
(11)
I ∗
T (1) − T0
0
I + K2
=
T − T0
(22)
Q2
Q2
Q∗2
0
Q2 T (2) − T0
!
(20)
where K1 , K2 are compact operators. The representations (20) can be shown similarly
to the derivations in section 2. Now all preparations for the derivation of the new
representations for Λ̃1 , Λ̃2 are met an we can write
!
# ( )∗
( )"
(11)
T (1) − T0
0
G1
G1
− ρ1 G̃1 G̃∗1
+
K̃
Λ̃1 =
(22)
(2)
G
G2
0
T − T0
2
!
( )"
−ρ1 I
0
G̃1
=
(22)
(2)
G2
0
T − T0
!
# ( )∗
(11)
(1)
∗
G̃1
Q1 (T − T0 )Q1 0
+
+ K̃
G2
0
0
( )"
!
# ( )∗
−ρ1 I
0
G̃1
G̃1
˜
=
+ K̃
,
(22)
(2)
G2
G2
0
T − T0
(21)
!
# ( )∗
( )"
(11)
G1
G1
T (1) − T0
0
+ K̃
Λ̃2 =
+ ρ2 G̃2 G̃∗2
(22)
(2)
G2
G2
0
T − T0
( )"
!
(11)
G1
T (1) − T0
0
=
G̃2
0
ρ2 I
!
# ( )∗
0
0
G1
+
+ K̃
(22)
(2)
∗
G̃2
0 Q2 (T − T0 )Q2
( )"
!
# ( )∗
(11)
G1
G1
T (1) − T0
0
˜
=
+ K̃
.
G̃2
G̃2
0
ρ2 I
(22)
Regarding the factorizations (21), (22) and lemma 3.3 we can conclude that in both
cases we have a factorization of the form
F = GSG ∗
satisfying the following properties:
(a) F : L2⋄ (∂B) → L2⋄ (∂B) is a bounded operator.
Detecting mixed inclusions in EIT
−1
16
−1
1
1
(b) In (21) G : H⋄ 2 (∂ Ω̃1 ) × H⋄ 2 (∂Ω2 ) → L2⋄ (∂B) and S : H⋄2 (∂ Ω̃1 ) × H⋄2 (∂Ω2 ) →
−1
−1
H⋄ 2 (∂ Ω̃1 ) × H⋄ 2 (∂Ω2 ) are bounded operators.
−1
1
−1
1
(c) In (22) G : H⋄ 2 (∂Ω1 ) × H⋄ 2 (∂ Ω̃2 ) → L2⋄ (∂B) and S : H⋄2 (∂Ω1 ) × H⋄2 (∂ Ω̃2 ) →
−1
−1
H⋄ 2 (∂Ω1 ) × H⋄ 2 (∂ Ω̃2 ) are bounded operators.
1
(d) G is compact and R(G) = H⋄2 (∂B).
(e) ReG = C + K where K is compact and C is self-adjoint. In (21) −C is coercive, and
in (22) C is coercive.
1
1
(f) ImhSh, hi ≥ 0 for all h = (h1 , h2 )⊤ ∈ H⋄2 (∂ Ω̃1 ) × H⋄2 (∂Ω2 ) (for all h = (h1 , h2 )⊤ ∈
1
1
H⋄2 (∂Ω1 ) × H⋄2 (∂ Ω̃2 ), respectively).
(g) ImS is positive on the nullspace of ReS.
Thus we are able to apply the F# -factorization method in the form of the following
theorem which is cited from [17].
Theorem 3.4. Let X ⊂ U ⊂ X ∗ be a Gelfand triple with a Hilbert space U and a
reflexive Banach space X such that the imbedding is dense. Let H be a second Hilbert
space and let F : H → H, G : X ∗ → H and S : X → X ∗ be linear and bounded operators
with F = GSG ∗ and
(a) G is one-to-one and compact with dense range in H,
(b) ReS = C + K where C (or −C) is coercive and K is compact,
(c) ImS is non-negative on X,
(d) S is injective or ImS is positive on the finite-dimensional nullspace of ReS.
Then F# := |ReF | + ImF is positive definite and the ranges of G : X ∗ → H and
1
F#2 : H → H coincide.
As a result, we can conclude for our factorizations:
Corollary 3.5. The operators Λ#,1, Λ#,2 : L2⋄ (∂B) → L2⋄ (∂B) defined by
Λ#,1 = |Λ − Λ0 − Re{ρ1 }G̃1 G̃∗1 | − Im{ρ1 }G̃1 G̃∗1 ,
Λ#,2 = |Λ − Λ0 + Re{ρ2 }G̃2 G̃∗2 | + Im{ρ2 }G̃2 G̃∗2 ,
are positive definite, and the ranges of
1
2
coincide.
and Λ#,2
n
G̃1
G2
o
1
2
as well as the ranges of
and Λ#,1
n
G1
G̃2
o
By means of the test functions ϕy from section 2 we can again derive a binary criterion
to decide whether a point y lies inside an inclusion or not. As above we conclude
n o
1
.
Lemma 3.6. (a) Let y ∈ Ω̃1 . Then: y ∈ Ω1 ⇔ ϕy ∈ R G
G̃2
n o
1
(b) Let y ∈ Ω̃2 . Then: y ∈ Ω2 ⇔ ϕy ∈ R G̃
.
G2
Detecting mixed inclusions in EIT
17
Proof. We only prove assertion a). Here we can state y ∈ Ωn1 ⇔oy ∈ Ω1 ∪ Ω̃2 , since Ω̃1
1
and Ω̃2 are disjoint. Additionally, y ∈ Ω1 ∪ Ω̃2 ⇔ ϕy ∈ R G
as is already known
G̃2
from the original factorization method.
Now by combination of lemma 3.6 and theorem 3.5 we obtain the following equivalences
using the Picard criterion
Corollary 3.7. (a) Let y ∈ Ω̃1 . Then
y ∈ Ω1 ⇔
Here
∞
X
k=1
D
(2)
ϕy , ψk
(2)
λk
E2
< ∞.
(23)
o
n
(2)
(2)
λk , ψk : k ∈ N is an eigensystem of Λ#,2 .
(b) Let y ∈ Ω̃2 . Then
y ∈ Ω2 ⇔
∞
X
k=1
D
(1)
ϕy , ψk
(1)
λk
E2
< ∞.
(24)
o
n
(1)
(1)
Here
λk , ψk : k ∈ N is an eigensystem of Λ#,1 .
These two binary criteria can now be used to reconstruct inclusions. However, in contrast
to the original factorization method we now have to perform two steps in order to identify
both inclusions. In the first step we use criterion (23) and reconstruct Ω1 , while in the
second step we use criterion (24) to reconstruct Ω2 .
The new method can now be interpreted as follows: For the reconstruction of Ω1 we use
the parameter ρ2 to synthesize a type 1 inclusion Ω̃2 that covers the disturbing type 2
inclusion Ω2 . On the other hand, we cover Ω1 by an artificial type 2 inclusion that is
produced by ρ1 in oder to identify Ω2 .
4. Numerical Experiments
In this section we give some results of numerical experiments with our new method. In
all our examples B is the unit circle in R2 , and the background conductivity is γ0 (x) ≡ 1.
Thus the dipole test function ϕy can explicitly be represented by
ϕy (x) =
1 (y − x) · â
,
2π |y − x|2
where â is a unit vector that represents the dipole axis. As a basis for the current
patterns as well as the boundary potentials we use the trigonometric functions
1
1
√ cos(kθ), √ sin(kθ), k = 1, 2, . . . ,
π
π
where θ denotes the argument of boundary points in polar coordinates.
Detecting mixed inclusions in EIT
18
In order to obtain approximations of Λ and Λ0 we solve the homogeneous and the
inhomogeneous forward problem for the basis current patterns up to an end index N ∈ N
using a standard finite element method.
In our example there are two inclusions: Ω1 has a kite shape, lies within Ω̃1 and
has conductivity γ = 0.5, while in the lower covering domain Ω2 is an ellipse with
conductivity γ = 2. The inclusion boundaries are marked by white lines in our
illustrations.
In order to reconstruct the inclusions, we evaluate the functions

E2 −1
D
(j)
N

X ϕy , ψk
Wj (y) = 
 (j = 1, 2)
(j)
λk
k=1
on a mesh of points y. As stated in theorem 3.7, we obtain the reconstruction of Ω1 by
(2)
(2)
evaluation of W2 in points inside Ω̃1 , where {(λk , ψk ) : k ∈ N } is an eigensystem of
Λ#,2 . On the other hand, we can reconstruct Ω2 by evaluating W1 on a mesh inside Ω̃2 ,
(1)
(1)
where {(λk , ψk ) : k ∈ N } is an eigensystem of Λ#,1 .
The experience from scattering theory (cf. [10]) suggests that the original factorization
method produces better reconstructions that the covering method although it is not
proven to work in the case of mixed obstacles.
What is more, the larger the absolute value of the parameters ρ1 , ρ2 get the more
they might influence the reconstruction as we can see in our theory: The larger |ρj | is
(j = 1, 2) the larger the contrast of the artificial covering inclusion to the background
gets, and the factorization method emphasizes this synthetic inclusion rather than the
one that we wish to identify.
This can also be seen in the representations (18), (19) of the middle operator. The
information about the desired inclusion lies inside the first part
Q1
I
!
while the second part
Q
1
T − T0
I
!∗
or
!
−ρ1 I 0
or
0
0
I
Q2
!
T − T0
0 0
0 ρ2 I
I
Q2
!∗
,
!
only contains information about the covering domains Ω̃1 , Ω̃2 , but carries more weight
as |ρ| gets larger.
In the following two numerical examples we verify these expectations for our new method
where we performed both reconstructions using different values of the parameters ρ1 ,
ρ2 .
Now the covering domains Ω̃1 and Ω̃2 are ellipses and defined as:
Detecting mixed inclusions in EIT
19
(a) ρ2 = 0
(b) ρ2 = 10−3 (1 + i)
(c) ρ2 = 10−2 (1 + i)
(d) ρ2 = 10−1 (1 + i)
Figure 2. Reconstruction type 1 inclusion; no noise
x22
1
,
Ω̃1 = x ∈ B : (x1 − 0.5) +
<
4
9
1
x22
2
.
<
Ω̃2 = x ∈ B : (x1 + 0.5) +
4
9
2
At first we consider the case of no artificial noise. In figure 2 we can see the
reconstruction of the type 1 inclusion, i.e. the kite shape for ρ2 = 0, ρ2 = 10−3 (1 + i),
ρ2 = 10−2 (1 + i) and ρ2 = 10−1 (1 + i). Here we can see that for ρ2 = 0 we get quite
a good reconstruction, while for growing absolute value of ρ2 the reconstructions get
worse.
The reconstruction of the ellipse is illustrated in figure 3 for the values ρ1 = 0,
ρ1 = 10−3 (1 − i), ρ1 = 10−2 (1 − i) and ρ1 = 10−1 (1 − i). Again, for ρ1 = 0 the
reconstruction is best, and for growing |ρ1 | the reconstructions deteriorate.
Figures 4 and 5 contain the same reconstructions for 0.5% white noise added to Λ,
while the observations concerning ρ1 , ρ2 are the same as for the noiseless case. We thus
recommend to choose ρ1 , ρ2 such that their absolute values are very small compared to
the the first few eigenvalues of |Λ − Λ0 |.
Detecting mixed inclusions in EIT
20
(a) ρ1 = 0
(b) ρ1 = 10−3 (1 − i)
(c) ρ1 = 10−2 (1 − i)
(d) ρ1 = 10−1 (1 − i)
Figure 3. Reconstruction of type 2 inclusion; no noise
In our next experiment we used smaller coverings Ω̃1 , Ω̃2 in order to improve the
reconstructions:
Ω̃1 =
Ω̃2 =
2
x ∈ B : (x1 − 0.5) +
x22
1
<
9
,
1
x ∈ B : (x1 + 0.5) + (x2 − 0.4) <
16
2
2
.
For ρ1 we used the values 10−3 (1 − i) and 10−2 (1 − i) while for ρ2 we used 10−3 (1 + i)
and 10−2 (1 + i).
Figure 6 illustrates the corresponding reconstructions. We observe that they are
noticeably better than the corresponding reconstructions for the same values of ρ1 ,
ρ2 and the previous larger covering domains. Since the synthetic inclusion caused by
the covering method is smaller than before it is less disturbing for the location of the
desired inclusion. It is thus recommended to choose the coverings as small as possible.
Our last experiment deals with the case in which the covering domains do not cover the
inclusions properly. Here we chose
Detecting mixed inclusions in EIT
21
(a) ρ2 = 0
(b) ρ2 = 10−3 (1 + i)
(c) ρ2 = 10−2 (1 + i)
(d) ρ2 = 10−1 (1 + i)
Figure 4. Reconstruction type 1 inclusion; 0.5% noise
1
,
Ω̃1 = x ∈ B : (x1 − 0.5) + (x2 − 0.2) <
9
1
2
2
Ω̃2 = x ∈ B : (x1 + 0.5) + (x2 − 0.2) <
,
16
2
2
while the values for ρ1 , ρ2 are the same as before.
The reconstructions for this case are shown in figure 7. They are slightly worse than the
reconstructions in figure 6 but still have an edge on the first ones where the coverings
were larger. This observation is in some sense plausible since the unions Ω̃j ∪Ωj (j = 1, 2)
are only slightly larger than the coverings in figure 6 but still much smaller than the
coverings in the first example. However, we have to emphasize that this case is not
covered by our theory.
Conclusions
We have shown that under some additional a priori information about the inclusions’
positions we can derive the factorization method for slight modifications of the operator
Detecting mixed inclusions in EIT
22
(a) ρ1 = 0
(b) ρ1 = 10−3 (1 − i)
(c) ρ1 = 10−2 (1 − i)
(d) ρ1 = 10−1 (1 − i)
Figure 5. Reconstruction of type 2 inclusion; 0.5% noise
Λ − Λ0 . Our numerical examples show that this new method can be used to reconstruct
inclusions even in the mixed case. However, the reconstructions are not as good as for
ρ = 0, i.e. for the standard version of the factorization method which is not proven
to work in the mixed case. This observation is the same as for this method applied to
scattering problems (cf. [10]).
We are confident that our method also gives some further insight into the factorization
method which could lead to future extensions of the currently required assumptions on
the problem setting.
Acknowledgments
This work was supported by the German Ministry of Education and Research in
the BMBF-project ‘Regularization techniques for electrical impedance tomography in
medical and geological sciences’
Detecting mixed inclusions in EIT
23
(a) ρ2 = 10−3 (1 + i)
(b) ρ2 = 10−2 (1 + i)
(c) ρ1 = 10−3 (1 − i)
(d) ρ1 = 10−2 (1 − i)
Figure 6. Reconstruction of both inclusions; no noise
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