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Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 93, pp. 1–30.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
EXISTENCE OF SOLUTION FOR A SEGMENTATION
APPROACH TO THE IMPEDANCE TOMOGRAPHY PROBLEM
RENIER MENDOZA, STEPHEN KEELING
Abstract. In electrical impedance tomography (EIT), image reconstruction
of the conductivity distribution of a body can be calculated using measured
voltages at the boundary. This is done by solving an inverse problem for
an elliptic partial differential equation (PDE). In this work, we present some
sensitivity results arising from the solution of the PDE. We use these to show
that a segmentation approach to the EIT inverse problem has a unique solution
in a suitable space using a fixed point theorem.
1. Introduction
Electrical impedance tomography (EIT) is an imaging technique proposed by
Calderon [6] in recovering the spatial distribution of the conductivities in the interior
of a body Ω based on the voltage and current measurements from electrodes placed
around its boundary ∂Ω. EIT is a non-invasive imaging technique with a wide range
of applications. We can refer to the following works [12, 13, 22, 23, 24, 29, 30, 38].
The EIT consists of two sub-problems: the forward problem and the inverse
problem. Suppose Ω ⊆Rnis a bounded domain with a sufficiently smooth bound-
ary. In the forward EIT problem, given the boundary currents f∈L2(∂Ω) and
the conductivity distribution σ∈L∞(Ω) satisfying σ(x)≥σ > 0, for all x∈Ω,
the electric potential φin Ω and the boundary voltage V=φ∂Ωare solved. These
electrical measurements satisfy a generalized Laplace equation:
∇ · (σ∇φ) = 0 in Ω,
σ∂φ
∂n =fon ∂Ω,(1.1)
where nis the outward normal direction at ∂Ω. The boundary currents are chosen
so that R∂Ωf dS = 0. This condition is imposed to satisfy the conservation of
charge. Furthermore, the electric potential φmust satisfy R∂Ωφ dS = 0. This
amounts to choosing the reference voltage. The equation (1.1) can be viewed as a
generalized Ohm’s law and is a well-posed boundary value problem with a unique
solution (up to a constant) φ∈H1(Ω). The partial differential equation (PDE)
(1.1) is called the continuum model of EIT. We focus our study on this model.
Other EIT models are discussed in [5, 8, 34].
2010 Mathematics Subject Classification. 35J20, 47H10, 35R30.
Key words and phrases. Electrical impedance tomography problem;
two-phase segmentation algorithm; fixed point theorem.
c
2020 Texas State University.
Submitted October 23, 2019. Published September 16, 2020.
1
2 R. MENDOZA, S. KEELING EJDE-2020/93
The inverse EIT problem or the conductivity reconstruction problem is the re-
covery of σinside Ω given Vand fin ∂Ω. Denote
˜
L2(∂Ω) := f∈L2(∂Ω) : Z∂Ω
f dS = 0
and define Λσ:˜
L2(∂Ω) →˜
L2(∂Ω) by
Λσ(f) = φ∂Ω,(1.2)
where φ∈H1(Ω) satisfies (1.1) and R∂Ωφ dS = 0. The inverse EIT problem is the
recovery of σgiven Λσ. Although the reconstruction problem is severely ill-posed, a
unique solution exists. Physically, this makes sense but to show this mathematically
is not trivial. For the discussion of this result, we refer the readers to [36, 35] for
the case n≥3 and to [2, 5, 25] for the case n= 2.
Because of its ill-posedness, the inverse EIT problem is an active research area.
Hence, several approaches have been proposed to solve this problem. Different
techniques are discussed in [5, 8, 20, 28, 37]. In this work, we focus on a technique
proposed by Mendoza and Keeling in [27]. We assume that the conductivity σis
piecewise constant. This assumption is based on the fact that the conductivities
of healthy tissues show great contrast [3, 17]. By assuming that σis piecewise-
constant, the inverse problem is treated as a segmentation problem. A segmentation
technique called “Multi-Phase Segmentation”, proposed by F¨urtinger in [15], is
explored in [27]. Moreover, it is assumed that the desired conductivity can be
expressed in terms of Mphases, i.e., of the form
σ(x) =
M
X
m=1
σm(x)χm(x),(1.3)
where for the mth phase χmis the characteristic function of a subdomain Ωm⊂Ω
and σmis globally smooth. In [27], the number of phases is fixed to 2, hence the
method is referred to as a two-phase segmentation approach. This is possible if the
subdomain Ω1has disjoint non-adjacent components. The subdomains Ω1and Ω2
form a disjoint partition of Ω, i.e., Ω1∩Ω2=∅and Ω = Ω1∪Ω2. The conductivity
σ2in Ω2is assumed to be known and χ2= 1 −χ1. Therefore, the inverse EIT
problem becomes a problem of identifying σ1and χ1. It is shown in [27] that σ1
can be expressed in terms of χ1. Given an initial guess for χ1, an iterative algorithm
is proposed. The main goal of this paper is to show that this iterative process has
a unique solution given an initial guess for χ1.
In the next section, we briefly discuss the two-phase segmentation algorithm.
Then an analysis of the algorithm is carried out. We show that the algorithm can
be expressed as a fixed point iteration. Finally, the existence of a fixed point in a
suitable space is presented.
2. Two-phase segmentation algorithm
Let us fix f∈˜
L2(∂Ω) and define the function F:L2(Ω) →˜
L(∂Ω) by
F(σ) = φ∂Ω,(2.1)
where φis the solution of (1.1) given σand f. Let σ?be the actual conductivity
distribution in Ω. The inverse problem is to recover σ?from Λσ?. Suppose f∈
˜
L2(∂Ω) and let V?= Λσ?(f) = F(σ?) be the exact boundary voltage. Moreover,
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 3
let ˜
V≈V?be the measured boundary voltage. Let ˜σ1be an estimate of σ1. To
solve the EIT inverse problem, our aim is to minimize
˜
J(σ1, χ1) = Z∂Ω
|F(σ)−˜
V|2dS +ZΩ
α|∇σ1|2(χ1+) + λ(σ1−˜σ1)2dV (2.2)
with σ=σ1χ1+σ2(1 −χ1) and σ2is given. The first term of the integral is the
fidelity term, the second term provides smoothness on σ1on Ω1and the third term
comes from Tikohonov regularization.
Before we proceed, we first need to define the forward solution and the adjoint
solution of the EIT problem.
Definition 2.1. The forward solution φis the solution of (1.1) given f∈˜
L2(∂Ω)
and σ∈L∞(Ω). Moreover, let ˜
Vbe the measured boundary voltage. We define
the adjoint solution φ∗as the solution of
∇ · (σ∇φ∗) = 0 in Ω,
σ∂φ∗
∂n =F(σ)−˜
Von ∂Ω.(2.3)
Both φand φ∗satisfy R∂Ωφ dS = 0 and R∂Ωφ∗dS = 0.
Using (1.1), (2.3), and (1.3), the variational formulations of the forward and
adjoint problems are
ZΩ
(σ1χ1+σ2(1 −χ1))∇φ· ∇v dV =Z∂Ω
fv dS, (2.4)
ZΩ
(σ1χ1+σ2(1 −χ1)∇φ∗· ∇v dV =Z∂Ω
(F(σ1χ1+σ2(1 −χ1)) −˜
V)v dS, (2.5)
for all v∈H1(Ω). Formulations (2.4) and (2.5) have unique solutions [27] in
˜
H1(Ω) := {v∈H1(Ω)|R∂Ωv dS = 0}. Because of (1.3), the forward and adjoint
solutions φand φ∗are dependent on σ1and χ1alone. Thus, we have the following
definition.
Definition 2.2. We define Φ(σ1, χ1) and Φ∗(σ1, χ1) to be the operators that map
any given σ1∈L∞(Ω) and characteristic function χ1to the respective solutions
φand φ∗of (2.4) and (2.5), respectively. Equivalently, Φ : (σ1, χ1)→φand
Φ∗: (σ1, χ1)→φ∗.
The computation of the derivative of ˜
Jis necessary to express σ1in terms of χ1.
For a fixed χ1, the variational derivative of ˜
Jin (2.2) with respect to σ1∈H1(Ω)
in the direction of δσ1∈H1(Ω) is given by
δ˜
J
δσ1
(σ1, χ1;δσ1) = −ZΩ
2χ1δσ1∇φ· ∇φ∗dV +ZΩ
2α(χ1+)∇(δσ1)· ∇σ1dV
+ZΩ
2λ(σ1−˜σ1)δσ1dV.
If we equate the above expression to 0, we conclude that the following optimality
condition must be satisfied,
ZΩ
[α(χ1+)∇σ1· ∇v+λ(σ1−˜σ1)v]dV =ZΩ
(χ1∇φ· ∇φ∗)v dV, (2.6)
for all v∈H1(Ω). Given χ1and an estimate ˜σ1, the quantity σ1is calculated via
(2.6). We formalize this in the following definition.
4 R. MENDOZA, S. KEELING EJDE-2020/93
Definition 2.3. We define the operator Σ1:χ1→σ1that maps an element
χ1∈L∞(Ω) to an element σ1∈H1(Ω) via (2.6) and the the operator Σ : χ1→σ
via σ(χ1) = Σ1(χ1)χ1+σ2(1 −χ1).
Definitions 2.2 and 2.3 are used to replace ˜σ1and σ1in (2.6) with σk
1and
σk+1
1= Σ1(χ1), respectively. Hence,
ZΩ
α(χ1+)∇Σ1(χ1)· ∇v dV +ZΩ
λ(Σ1(χ1)−σk
1)v dV
=ZΩ
χ1∇Φ(σk
1, χ1)· ∇Φ∗(σk
1, χ1)v dV.
(2.7)
Furthermore, the following operators are defined for the global conductivity,
Σk(χ1) := σk
1χ1+σ2(1 −χ1),(2.8)
Σk+1(χ1) := Σ1(χ1)χ1+σ2(1 −χ1),(with Σ1(χ1) = σk+1
1).(2.9)
Under some assumptions, we will show that (2.7) admits a unique solution (see
Lemma 3.25). Observe that the functional ˜
J(σ1, χ1) in (2.2) can be written as a
functional ˜
J(Σ1(χ1), χ1) depending only on χ1. To determine χ1, we add a Total
Variation (TV) regularization to (2.2) to penalize oscillations. For discussions of
TV-regularization, one can refer to [31, 32, 10, 21]. Given a (sufficiently smooth)
function f, its total variation is given by
T V (f) := ZΩ
|∇f|dV ≈ZΩp|∇f|2+β2dV,
for some 0 < β 1 (compare, e.g., [7, 11]). To determine the optimal χ1, our aim
is to minimize the TV-regularized functional
J(χ1) = Z∂Ω
|F(Σ(χ1)) −˜
V|2dS +ZΩ
α|∇Σ1(χ1)|2(χ1+)
+ZΩ
λ(Σ1(χ1)−˜σ1)2+γp|∇χ1|2+β2dV,
(2.10)
for α, λ, γ > 0 and , β ∈(0,1). Thus we find an update for χ1that reduces the cost
J. This update can be obtained using the method of steepest descent [33], which is
given in weak form for Jas follows,
ZΩ
χk+1
1v dV =ZΩ
χk
1v dV −ωδJ
δχ1
(χk
1;v),(2.11)
for all v∈H1(Ω), where ω∈(0,1) is the step size and k∈N. Let χk
1, δχk
1∈L2(Ω)
and suppose
δΣ1
δχk
1
(χk
1;δχk
1)∈H1(Ω),
then the variational derivative of Jin (2.10) is
δJ
δχk
1
(χk
1;δχ1) = ZΩ
−2(Σ1(χk
1)−σ2)δχk
1∇Φ(σk
1, χk
1)· ∇Φ∗(σk
1, χk
1)dV
+ZΩ
α|∇Σ1(χk
1)|2δχk
1dV +γZΩ
∇(δχk
1)· ∇χk
1
p|∇χk
1|2+β2dV.
(2.12)
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 5
Remark 2.4. Observe that (2.12) requires the calculation of ∇χk
1but since χk
1is
binary, a smooth approximation of χk
1is necessary. This will be discussed later. The
assumption that δΣ1
δχk
1(χk
1;δχk
1)∈H1(Ω) can be shown if χk
1is sufficiently smooth
(see Theorem 3.28).
Instead of performing the iteration (2.11) by evaluating δJ/δχ1(χk
1;v) explicitly
in terms of χk
1, the iteration may be performed semi-implicitly by evaluating part
of the variational derivative of Jin (2.12) at χk+1
1as follows:
ZΩvχk+1
1+ωγ ∇v· ∇χk+1
1
p|∇χk
1|2+β2dV =ZΩ
vG(χk
1)dV, (2.13)
for all v∈H1(Ω), where
G(χ1) = χ1−ωα|∇Σ1(χ1)|2+ 2ω(Σ1(χ1)−σ2)∇Φ(˜σ1, χ1)· ∇Φ∗(˜σ1, χ1).(2.14)
We show later that (2.13) admits a unique solution (see Lemma 4.3).
As mentioned in Remark 2.4, a smooth approximation of χ1is necessary. To
do this, we introduce the kernel function ξδ(x) = 1
4πδ e−x2
4δ, for some δ > 0. We
approximate χk
1using the convolution of χk
1and ξδ, i.e., we let
˜χk
1:= χk
1∗ξδ=ZR2
ξδ(x−y)χk
1(y)dy. (2.15)
The following result gives the regularity and continuity of the above mollification.
Theorem 2.5. Let k∈N, then ˜χk
1is a real analytic function on Ωand ˜χk
1→χk
1
almost everywhere as δ→0. Furthermore, suppose 0≤χk
1≤1, for all x∈Ω.
Then 0≤˜χk
1≤1 [15].
Using (2.15), we approximate (2.13) by
ZΩ
ωγ ∇˜χk+1
1· ∇v
p|∇˜χk
1|2+β2+ ˜χk+1
1v dV =ZΩ
G(˜χk
1)v dV, ∀v∈H1(Ω).(2.16)
Definition 2.6. We define the operator Θ : L2(Ω) →L2(Ω) to be the solution
˜χk+1
1∈L2(Ω) of (2.16) for a given ˜χk
1∈L2(Ω).
Because the solution of (2.16) is not binary, the update for χis obtained by
performing a thresholding step.
We summarize the two-phase segmentation method in the algorithm below. The
analysis of the numerical solution of the PDEs arising from this algorithm can be
found in [26].
Two-phase segmentation algorithm.
(1) Given fand ˜
V. Choose parameters , δ, λ, γ, β 1, α1, and ζ, ω ∈
(0,1). Select the maximum number of iterations Kand the tolerance ρ.
Set k= 1 and choose the initial σk
1. Select an initial guess χk
1. The value
of σ2is given and χk
2= 1 −χk
1.
(2) Take ˜χk
1=χk
1∗ξδand ˜χk+1
1= Θ(˜χk
1). Then, χ1is updated by
χk+1
1(x) = (1,if ˜χk+1
1(x)≥ζ,
0,otherwise.
Set χk+1
2= 1 −χk+1
1.
6 R. MENDOZA, S. KEELING EJDE-2020/93
(3) If k=Kor kχk+1
1−χk
1kL2(Ω) < ρ, the algorithm terminates. Otherwise,
k←k+ 1 and go back to step 2.
3. Analysis of the algorithm
In this section, we analyze the two-phase segmentation algorithm. We start
with results obtained by assuming that σ∈L∞(Ω) and that χ1is a characteris-
tic function. Because the characteristic function χ1is binary, we use its smooth
approximation (2.15) instead. This is necessary because some of the essential re-
sults require χ1to have a higher regularity. This might seem like a deviation from
our proposed method but we will argue that these modifications can be justified.
We will then introduce a modification of the two-phase segmentation algorithm to
adapt with the mollification of χ1. In the next section, we prove that the modified
version of two-phase segmentation algorithm has a fixed point via Schauder’s Fixed
Point Theorem.
3.1. Preliminaries. We already emphasized that φand φ∗are solved using σk. In
this section, we try to understand how a perturbation on σkaffects φand φ∗. Recall
that σkdepends on χ1. Therefore, φand φ∗depend on χ1as well. Working under
the assumption that σ∈L∞(Ω) and χ1is a characteristic function, we show that
φand φ∗depend continuously on σkand on χ1. We begin this section by showing
that the variational forward problem and the variational adjoint problem both
have unique solutions in H1(Ω) under stated assumptions on σ. In the succeeding
sections, we study the behavior of φand φ∗when we require additional regularity
of σk.
Theorem 3.1. Let σk∈L∞(Ω) such that 0< σ ≤σk(x)for all x∈Ωand
f∈˜
L2(∂Ω). Then ( the variational forward EIT problem)
ZΩ
σk∇φ· ∇v dV =Z∂Ω
σk∂φ
∂n v dS, ∀v∈H1(Ω) (3.1)
has a unique solution φ∈H1(Ω) with R∂Ωφ dS = 0. Similarly, let ˜
V∈˜
L2(∂Ω) be
the known boundary voltage. Then ( the variational adjoint problem)
ZΩ
σk∇φ∗· ∇v dV =Z∂Ω
(φ−˜
V)v dS, ∀v∈H1(Ω) (3.2)
has a unique solution φ∗∈H1(Ω) with R∂Ωφ∗dS = 0.
Proof. We define I:= {u∈H1(Ω) : R∂Ωu dS = 0},a(u, v) := RΩσk∇u·∇v dV , and
b(v) := R∂Ωfv dS. Clearly, ais bilinear and bis linear. Using the Cauchy-Schwarz
identity, H¨older’s inequality, and the definition of the H1norm, ais bounded, i.e.,
|a(u, v)| ≤ kσkkL∞(Ω)kukH1(Ω)kvkH1(Ω) .
Because u∈I, then R∂Ωu dS = 0. Therefore, using the lower bound of σ, and the
generalized Friedrich’s inequality [4], we obtain
|a(u, u)| ≥ σk∇uk2
L2(Ω)
=σ
2k∇uk2
L2(Ω) +σ
2k∇uk2
L2(Ω)
≥σ
21
Ckuk2
L2(Ω) −Z∂Ω
u dS2+σ
2k∇uk2
L2(Ω)
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 7
=σ
2Ckuk2
L2(Ω) +σ
2k∇uk2
L2(Ω)
≥σmin 1
C,1kuk2
H1(Ω),
for some C > 0. Finally, by the Trace Theorem [14], bis bounded. Hence, by
the Lax-Milgram Theorem ∃!φ∈H1(Ω) satisfying (3.1). Similarly, there exists a
unique φ∗∈Isatisfying (3.2).
Corollary 3.2. Let φand φ∗satisfy the variational forward and the variational
adjoint problem stated in the previous theorem. We have the following estimates:
kφkH1(Ω) ≤C1kfk˜
L2(∂Ω) (3.3)
kφ∗kH1(Ω) ≤C2k(φ−˜
V)k˜
L2(∂Ω) (3.4)
for some C1, C2>0.
Note that using the Trace Theorem and the triangle inequality, φ∗can be further
estimated by
kφ∗kH1(Ω) ≤C3kfk˜
L2(∂Ω) +k˜
Vk˜
L2(∂Ω)(3.5)
for some C3>0. Throughout this work, we use the following notation.
Definition 3.3. We let δσkand δχ1denote perturbations of σk∈L∞(Ω) and
χ1∈L∞(Ω), respectively. In Definition (2.2), Φ and Φ∗are operators that map
(σk
1, χ1) to φand φ∗, respectively. But because σk=σk
1χ1+σ2(1 −χ1), we can
make the identifications
Φ(σk) = Φ(σk
1, χ1),Φ∗(σk)=Φ∗(σk
1, χ1).
Hence, Φ : σk→φand Φ∗:σk→φ∗.
Remark 3.4. Given a perturbation δσk∈L∞(Ω), how can we choose η > 0 so that
the forward and the adjoint problems have unique solutions if we use σk+ηδσk? We
know that the forward and adjoint problems have unique solutions given σk∈L∞
if σk(x)≥σ > 0 for all x∈Ω. To make sure that Φ(σk+ηδσk) is unique, we can
simply select ηsufficiently small so that (σk+ηδσk)(x)≥στ>0 for all x∈Ω
and η∈(0, τ ) for some τ > 0. This is possible because σk(x)≥σ > 0. Thus,
the coercivity of the bilinear functional in the variational formulations of both the
forward and the adjoint problems is guaranteed and the solvability of these problems
is assured. Consequently, by (3.3) and (3.5) there exist C1, C2>0 such that
kΦ(σk+ηδσk)kH1(Ω) ≤C1kfk˜
L2(∂Ω),(3.6)
kΦ∗(σk+ηδσk)kH1(Ω) ≤C2(kfk˜
L2(∂Ω) +k˜
Vk˜
L2(∂Ω)),(3.7)
for any η∈(0, τ).
The result below shows how a perturbation δσkaffects φand φ∗.
Theorem 3.5. Let σk, δσk∈L∞(Ω). Then there exist C1, C2>0such that
kΦ(σk+ηδσk)−Φ(σk)kH1(Ω) ≤C1ηkδσkkL∞(Ω),(3.8)
kΦ∗(σk+ηδσk)−Φ∗(σk)kH1(Ω) ≤C2ηkδσkkL∞(Ω),(3.9)
for any η∈(0, τ ), where τis chosen according to Remark 3.4.
8 R. MENDOZA, S. KEELING EJDE-2020/93
Proof. From (3.1), we have
ZΩ
σk∇Φ(σk)· ∇v dV =ZΩ
fv dV. (3.10)
Similarly, for σk+ηδσk,
ZΩ
(σk+ηδσk)∇(Φ(σk) + δφ)· ∇v dV =ZΩ
fv dV, (3.11)
where we denote δφ := Φ(σk+ηδσk)−Φ(σk). Subtracting (3.10) from (3.11), we
obtain ZΩ
σk∇δφ · ∇v dV =−ZΩ
ηδσk∇Φ(σk+ηδσk)· ∇v dV. (3.12)
We define a(u, v) := RΩσk∇u·∇v dV and b1(v) := −RΩηδσk∇Φ(σk+ηδσk)·∇v dV .
Clearly, aand b1are bilinear and linear, respectively. Recall from Theorem (3.1)
that for any u, v ∈I,a(u, v) is coercive and continuous. By the Cauchy-Schwarz
inequality and (3.6), b1is bounded. Thus, if we take u=v=δφ, use the previous
inequality, and the coercivity of a(u, v) we obtain
kδφkH1(Ω) ≤1
¯
C¯
C1ηkfk˜
L2(∂Ω)kδσkkL∞(Ω),(3.13)
where ¯
C > 0 is the coercivity constant and ¯
C1is the constant from (3.6). This
proves the first inequality. Using similar arguments, one can show (3.9).
Definition 3.6. Recall that from (2.8), Σk(χ1) := σk
1χ1+σ2(1 −χ1) = σk. There-
fore, φand φ∗depend on χ1for a fixed σk
1. Hence, for brevity we denote
Φ(χ1) := Φ(Σk(χ1)) and Φ∗(χ1) := Φ∗(Σk(χ1)).(3.14)
From here onwards, it is assumed that φand φ∗are solved using σk
1χ1+σ2(1 −χ1).
We now show that φand φ∗depend continuously on χ1.
Corollary 3.7. Let χ1∈L∞(Ω), then ∃¯
C1,¯
C2>0such that
kΦ(χ1+ηδχ1)−Φ(χ1)kH1(Ω) ≤¯
C1ηkδχ1kL∞(Ω) ,(3.15)
kΦ∗(χ1+ηδχ1)−Φ∗(χ1)kH1(Ω) ≤¯
C2ηkδχ1kL∞(Ω) ,(3.16)
for any η∈(0, τ ), where τis chosen according to Remark 3.4.
Proof. We use the decomposition
Σk(χ1) = σk
1χ1+σ2(1 −χ1).(3.17)
Let η∈(0, τ ). If we use χ1+ηδχ1instead of χ1, we have
Σk(χ1) + δσk= Σk(χ1) + η(σk
1−σ2)δχ1,(3.18)
where δσkis the associated change in σkgiven a ηδχ1perturbation of χ1. Sub-
tracting (3.17) from (3.18), we obtain
δσk=η(σk
1−σ2)δχ1.(3.19)
The inequalities we need to show directly follow from Theorem (3.5) and
kδσkkL∞(Ω) ≤ηkσk
1−σ2kL∞(Ω)kδχ1kL∞(Ω).
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 9
3.2. Smooth approximation of χ1.From (2.7), the quantity σk+1
1:= Σ1(χ1)
can be obtained via
ZΩ
α(χ1+)∇σk+1
1· ∇v dV +ZΩ
λ(σk+1
1−σk
1)v dV
=ZΩ
χ1∇Φ(χ1)· ∇Φ∗(χ1)v dV.
(3.20)
This equation can be interpreted as
a(σk+1
1, v) = b(v),∀v∈H1(Ω),(3.21)
where
a(σk+1
1, v) := ZΩ
α(χ1+)∇σk+1
1· ∇v dV +ZΩ
λσk+1
1v dV,
b(v) := ZΩ
λσk
1v dV +ZΩ
χ1∇Φ(χ1)· ∇Φ∗(χ1)v dV.
To guarantee solvability of (3.21), b(v) must be bounded. To show this, it is
necessary that χ1∇Φ(χ1)· ∇Φ∗(χ1) be in L2(Ω). If either ∇Φ(χ1) or ∇Φ∗(χ1) is in
L∞(Ω), then ∇Φ(χ1)·∇Φ∗(χ1)∈L2(Ω). In [9], it was shown that k∇Φ(χ1)kL∞(Ω0)
can be bounded by k∇Φ(χ1)kL2(Ω) for some Ω0compactly embedded in Ω. This was
proven under the assumption that σk∈C1(¯
Ω). Recall that σk=σk
1χ1+σ2(1−χ1).
Clearly, σkis not necessarily in C1(¯
Ω) because χ1is a characteristic function. We
have introduced a mollification χδ
1of χ1in (2.15) to resolve this. Thus, σk
1∈C∞(¯
Ω).
Moreover, σkis not just in C1(¯
Ω) but in C∞(¯
Ω) as well. This might seem like
a deviation from our proposed method but technically, we can choose δto be
extremely close to 0 so that χδ
1is a good approximation of χ1. We show that the
mollification affects the corresponding φand φ∗to a very small extent as long as
the distance between χδ
1and χ1is small enough. But first, we need the following
results.
Lemma 3.8. Let 1≤p < ∞and take 1≤r≤ ∞ such that 1
r+ 1 −1
p∈[0,1].
Define the operator from Lp(Ω) to Lr(Ω) by
Tδ(g) := g∗ξδ.(3.22)
Then Tδis continuous and injective [15].
Lemma 3.9. For any g∈Lp(Ω),1≤p < ∞,∂ν(g∗ξδ)=(∂νg)∗ξδ, for |ν| ≤ 1.
Moreover, ∂ν(g∗ξδ)→∂νgalmost everywhere as δ→0.
The proof of the above lemma is rather straightforward and is omitted. The
second assertion follows from Lemma (2.5) (compare with [14]). For brevity, from
here onwards we let
χδ
1:= χ1∗ξδ
denote the mollification of χ1. We now show how the perturbation of χ1affects the
solution of the forward and adjoint problems.
Theorem 3.10. For a fixed δ, the solution of the forward problem given χ1∈
L∞(Ω) and the solution of the forward problem given χδ
1satisfy
kΦ(χδ
1)−Φ(χ1)kH1(Ω) ≤Cδ
1kχδ
1−χ1kL∞(Ω) (3.23)
for some C1>0. The same applies to the adjoint problem
kΦ∗(χδ
1)−Φ∗(χ1)kH1(Ω) ≤Cδ
2kχδ
1−χ1kL∞(Ω) (3.24)
10 R. MENDOZA, S. KEELING EJDE-2020/93
for some C2>0.
Proof. We proved in (3.15) that φdepends continuously on χ1. Note that we proved
this given the assumption that χ1∈L∞(Ω), which implies that χ1∈L2(Ω) as well.
By Lemma (3.8), we can infer that χδ
1∈L∞(Ω) by choosing p= 2 and r=∞.
Finally, Theorem (3.5) proves the rest of our claim.
Remark 3.11. The superscript δin Cδ
1and Cδ
2from (3.23) and (3.24) are used to
emphasize the dependence of the inequality constants on the mollification parameter
δ. From here onwards, we use the same notation for all constants dependent on δ.
We know that χδ
1converges to χ1pointwise [15]. Although it does not guarantee
that kχδ
1−χ1kL∞(Ω) converges to 0, this can still be a gauge to measure the distance
between the solution of the forward problem using χ1and the solution using χδ
1.
To make our modifications consistent, we find a new thresholding approach to
adapt with the mollification of χ1. First, we define a space that will be important
in our succeeding computations. Let B(Ω) be the Borel σ−algebra over Ω and let
µ(·) denote the Lebesgue measure. For any A1, A2∈ B(Ω), we define the symmetric
difference of A1and A2to be
A14A2:= (A1\A2)∪(A2\A1).
Definition 3.12. Let the distance d:B(Ω) × B(Ω) →R∪ {∞} be d(A1, A2) =
µ(A14A2). We now define M(Ω) = (B(Ω), d)/ker(d).
The space M(Ω) is, in fact, a metric space [19, 15]. In our algorithm, we did a
thresholding on Θ in order to get an update for χ1. In the following definition, we
modify this thresholding to make it coherent with the mollification of χ1.
Definition 3.13. Let z∈L2(Ω)\H1(Ω). We define H:L2(Ω) → M(Ω) by
H(g) = {x∈Ω : ((gδ−ζ+δz)∗ξδ)(x)≥0},(3.25)
for some ζ∈(0,1) and gδ=g∗ξδ. Moreover, define M:M(Ω) →L2(Ω) as the
map that assigns ω∈ M(Ω) to its characteristic function χω, that is,
M(ω) = χω.(3.26)
Observe that if we let δ→0, then
(gδ−ζ+δz)∗ξδ→g−ζ. (3.27)
See [15]. This means that as δ→0, H(g) becomes the set of x∈Ω for which
g(x)≥ζ. Using the above-mentioned functions Tδ,H,M, and Θ (Definition 2.6),
we can modify the two-phase segmentation algorithm into
χk+1
1= (Tδ◦M◦H◦Θ◦Tδ)(χk
1).(3.28)
The main goal of this article is to show that (3.28) has a fixed point.
Remark 3.14. It is again emphasized that the introduction of z∈L2(Ω)\H1(Ω)
in (3.25) and the modification of Algorithm (1) as a result of the mollification of χ1
are purely technical devices and are used only for theoretical purposes. This might
seem like a deviation from Algorithm (1) but as shown in (3.23), (3.24) and (3.27),
these changes are justified.
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 11
3.3. Gradient of the functional J.This section is devoted to showing the va-
lidity of the explicit formulation of the gradient of Jin (2.10). As shown in the
computation of (2.12), it is sufficient to show that
δΣ1
δχ1
(χ1;δχ1)∈H1(Ω).
This is not necessarily true for an arbitrary characteristic function χ1. Hence, we
use χδ
1instead of χ1so that we are dealing with a smooth function rather than a
characteristic function. Thus, we are now solving σk+1
1using the equation
ZΩ
α(χδ
1+)∇σk+1
1· ∇v dV +ZΩ
λ(σk+1
1−σk
1)v dV =ZΩ
χδ
1∇φ· ∇φ∗v dV, (3.29)
for all v∈H1(Ω). Before we start our calculations, we first make few assumptions.
These will be used throughout our analysis.
Let α, σ, λ, δ > 0. We assume that σk
1∈C∞(¯
Ω) such that
Σk(χδ
1)≥σ > 0. We also assume that ∂Ω is sufficiently smooth. (3.30)
The assumption that σk
1∈C∞(¯
Ω) might seem like a strong assumption but if we
can show that σk+1
1∈C∞(¯
Ω) as well, then this assumption makes sense. Moreover,
the initial guess for σ1in our algorithm can be chosen to be constant throughout
Ω so that it is in C∞(¯
Ω).
We now investigate what happens to φand φ∗if we use the above assumption.
Lemma 3.15. Under assumption (3.30),
k∇φkL∞(Ω) ≤CδkφkH1(Ω),(3.31)
for some Cδ>0(compare with [9]). In fact, φ∈C∞(¯
Ω).
Proof. By assumption (3.30) and Lemma (2.5), σk∈C∞(¯
Ω). Let l≥1. Obviously,
σk∈Cl(¯
Ω). Therefore, using standard regularity estimates (see, e.g., [16]), we
obtain
kφkHl+2(Ω) ≤C1kφkH1(Ω) ,(3.32)
for some C1>0. Furthermore, by the Sobolev imbedding theorem [14], we have
kφkCl,γ (¯
Ω) ≤C2kφkHl+2(Ω) .(3.33)
By the definition of k·kCl,γ (¯
Ω), the embedding
C1,γ (¯
Ω) →C1(¯
Ω) (3.34)
is continuous (see [18]). If we compare (3.32), (3.33), and (3.34), we can deduce
that there exists C > 0 such that
k∇φkL∞(Ω) ≤Ck∇φkH1(Ω),(3.35)
which completes the first part of the proof. Moreover, because l≥1, it follows that
φ(χδ
1)∈C∞(¯
Ω).
Lemma 3.16. Under assumption (3.30), there exists Cδ>0such that
k∇φ∗kL∞(Ω) ≤Cδkφ∗kH1(Ω).(3.36)
Furthermore, φ∗∈C∞(¯
Ω).
12 R. MENDOZA, S. KEELING EJDE-2020/93
The proof of this theorem is similar to that of the previous theorem. We now
analyze the dependence of φand φ∗on χδ
1. From here onwards, we denote
δχδ
1:= δχ1∗ξδ
so that (χ1+ηδχ1)∗ξδ=χδ
1+ηδχδ
1.
Suppose we replace χδ
1with χδ
1+ηδχδ
1. Again by Remark 3.4, ηshould be taken
from the set (0, τ ) where τis chosen so that σk+ηδσk≥στ>0. Therefore, from
(3.6), (3.7), (3.14), (3.31), and (3.36), the following estimates hold
k∇Φ(χδ
1+ηδχδ
1)kL∞(Ω) ≤C1kfk˜
L2(∂Ω),(3.37)
k∇Φ∗(χδ
1+ηδχδ
1)kL∞(Ω) ≤C2kfk˜
L2(∂Ω) +k˜
Vk˜
L2(∂Ω),(3.38)
for some C1, C2>0 and for all η∈(0, τ ).
In Corollary 3.7, we have shown that φand φ∗depend continuously on χ1.
Observe that this theorem holds with any χ1whose value is between 0 and 1.
Therefore, this also holds when we use χδ
1instead because 0 ≤χδ
1≤1 as proven in
Lemma (2.5). We state this in the following lemma.
Lemma 3.17. Under assumption (3.30), there exists Cδ
1, Cδ
2>0such that
kΦ(χδ
1+ηδχδ
1)−Φ(χδ
1)kH1(Ω) ≤Cδ
1ηkδχ1kL2(Ω) ,(3.39)
kΦ∗(χδ
1+ηδχδ
1)−Φ∗(χδ
1)kH1(Ω) ≤Cδ
2ηkδχ1kL2(Ω) ,(3.40)
for any η∈(0, τ ), where τis chosen according to Remark 3.4.
We define
ψ(χδ
1) := ∇Φ(χδ
1)· ∇Φ∗(χδ
1).(3.41)
Observe that the right-hand side of (3.29) includes ψ(χδ
1). In the next lemma,
we show that ψdepends continuously on χ1. This will be necessary when we
analyze the solution of (3.29). Note that ∇Φ∗(χδ
1)∈L2(Ω) by Corollary 3.2 and
∇Φ(χδ
1)∈L∞(Ω) by (3.31). Therefore, by H¨older’s inequality,
ψ(χδ
1)∈L2(Ω).(3.42)
This means that ψis a map from χ1∈L2(Ω) to ∇Φ(χδ
1)· ∇Φ∗(χδ
1)∈L2(Ω). In
the following lemma, we prove that this mapping is continuous.
Lemma 3.18. Under assumption (3.30), there exists Cδ≥0such that
kψ(χδ
1+ηδχδ
1)−ψ(χδ
1)kL2(Ω) ≤Cδηkδχ1kL2(Ω) ,(3.43)
for any η∈(0, τ ), where τis chosen according to Remark 3.4.
Proof. Adding and subtracting ∇Φ(χδ
1+ηδχδ
1)· ∇Φ∗(χδ
1) to ψ(χδ
1+ηδχδ
1)−ψ(χδ
1),
we obtain
ψ(χδ
1+ηδχδ
1)−ψ(χδ
1) =: A1(χδ
1;δχδ
1) + A2(χδ
1;δχδ
1),
where
A1(χδ
1;δχδ
1) = ∇Φ(χδ
1+ηδχδ
1)· ∇Φ∗(χδ
1+ηδχδ
1)− ∇Φ(χδ
1+ηδχδ
1)· ∇Φ∗(χδ
1),
A2(χδ
1;δχδ
1) = ∇Φ(χδ
1+ηδχδ
1)· ∇Φ∗(χδ
1)− ∇Φ(χδ
1)· ∇Φ∗(χδ
1).
Thus
kψ(χδ
1+ηδχδ
1)−ψ(χδ
1)kL2(Ω)
≤ kA1(χ1;δχ1)kL2(Ω) +kA2(χ1;δχ1)kL2(Ω).(3.44)
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 13
We can estimate A1(χδ
1;δχδ
1) using the H¨older’s inequality, (3.37), and (3.40). Thus,
kA1(χδ
1;δχδ
1)kL2(Ω) ≤C1ηkfk˜
L2(∂Ω)kδχ1kL2(Ω),
for some C1>0. Similarly, using the H¨older’s inequality, (3.36), and (3.39), we
obtain
kA2(χδ
1;δχδ
1)kL2(Ω) ≤C2ηk∇Φ∗(χδ
1)kL∞(Ω)kδχ1kL2(Ω),
for some C2>0. The estimates for A1(χδ
1;δχδ
1) and A2(χδ
1;δχδ
1), together with
(3.44), complete the proof.
Recall that the goal of this section is to validate the explicit formulation of the
gradient of J(χ1) by showing that δΣ1
δχ1(χδ
1;δχδ
1)∈H1(Ω). To accomplish this, we
first need to show that the derivatives of both Φ and Φ∗with respect to χ1converge
in H1(Ω). We start by looking for candidates for the derivatives.
Lemma 3.19. Under assumption (3.30), there exists Dφ(χ1;δχ1)∈H1(Ω) satis-
fying
ZΩ
σk∇Dφ(χ1;δχ1)· ∇v dV =ZΩ
(σk
1−σ2)δχδ
1∇Φ(χδ
1)· ∇v dV (3.45)
with R∂ΩDφ(χ1;δχ1)dS = 0, for all v∈H1(Ω), such that R∂Ωv dS = 0.
Proof. Let u, v ∈H1(Ω) such that R∂Ωu dS =R∂Ωv dS = 0. We define
a(u, v) := ZΩ
σk∇u· ∇v dV, b(v) := ZΩ
(σk
1−σ2)δχδ
1∇Φ(χδ
1)· ∇v dV.
From the proof of Theorem (3.1), ais bilinear, coercive and bounded. Obviously,
bis linear. We wish to employ the Lax-Milgram theorem so it is sufficient to show
that bis bounded. Indeed, using the Cauchy-Schwarz inequality and the H¨older’s
inequality we obtain
|b(v)|≤k(σk
1−σ2)δχδ
1kL∞(Ω)k∇Φ(χδ
1)kL2(Ω)kvkL2(Ω) .
The right-hand side of the last inequality is bounded because of Corollary (3.2) and
the fact that (σk
1−σ2)δχδ
1∈C∞(¯
Ω).
Lemma 3.20. Under assumption (3.30), there exists Dφ∗(χ1;δχ1)∈H1(Ω) satis-
fying
ZΩ
σk∇Dφ∗(χ1;δχ1)· ∇v dV
=ZΩ
(σk
1−σ2)δχδ
1∇Φ∗(χδ
1)· ∇v dV +Z∂Ω
Dφ(χ1;δχ1)v dV,
(3.46)
with R∂ΩDφ∗(χ1;δχ1)dS = 0, for all v∈H1(Ω), such that R∂Ωv dS = 0.
The proof of this lemma is similar to the proof of the previous lemma; we omit
it. Now we prove that Dφ(χ1;δχ1) and Dφ∗(χ1;δχ1) in the last two lemmas are
the derivatives of Φ and Φ∗at χ1in the direction of δχ1, respectively. We state
and prove this in the following lemma.
Lemma 3.21. Under assumption (3.30), we have
lim
η→0
Φ(χδ
1+ηδχδ
1)−Φ(χδ
1)
η−Dφ(χδ
1;δχδ
1)
H1(Ω) = 0,(3.47)
14 R. MENDOZA, S. KEELING EJDE-2020/93
lim
η→0
Φ∗(χδ
1+ηδχδ
1)−Φ∗(χδ
1)
η−Dφ∗(χδ
1;δχδ
1)
H1(Ω) = 0.(3.48)
Thus, we can make the identifications
δΦ
δχ1
(χδ
1;δχδ
1) = Dφ(χ1;δχ1),δΦ∗
δχ1
(χδ
1;δχδ
1) = Dφ∗(χ1;δχ1).
Furthermore, because Dφ(χ1;δχ1), Dφ∗(χ1;δχ1)∈H1(Ω) we have
δΦ
δχ1
(χδ
1;δχδ
1),δΦ∗
δχ1
(χδ
1;δχδ
1)∈H1(Ω),
with
Z∂Ω
δΦ
δχ1
(χδ
1;δχδ
1)dS =Z∂Ω
δΦ∗
δχ1
(χδ
1;δχδ
1)dS = 0.
Proof. From (3.1), we have RΩσk∇Φ(χδ
1)· ∇v dV =RΩfv dV , where σk= Σk(χδ
i).
Then we obtain
ZΩ
σk∇Φ(χδ
1)· ∇v dV =ZΩ
fv dV. (3.49)
Similarly, for χδ
1+ηδχδ
1,
ZΩ
[σk+η(σk
1−σ2)δχδ
1]∇Φ(χδ
1+ηδχδ
1)· ∇v dV =ZΩ
fv dV. (3.50)
Subtracting (3.49) from (3.50), we obtain
ZΩ
σk∇(Φ(χδ
1+ηδχδ
1)−Φ(χδ
1)) · ∇v dV
=ZΩ
η(σk
1−σ2)δχδ
1∇Φ(χδ
1+ηδχδ
1)· ∇v dV.
(3.51)
Dividing by η, we have
ZΩ
σk∇Φ(χδ
1+ηδχδ
1)−Φ(χδ
1)
η· ∇v dV
=ZΩ
(σk
1−σ2)δχδ
1∇Φ(χδ
1+ηδχδ
1)· ∇v dV.
(3.52)
Recall from (3.45) that
ZΩ
σk∇Dφ(χδ
1;δχδ
1)· ∇v dV =ZΩ
(σk
1−σ2)δχδ
1∇Φ(χδ
1)· ∇v dV. (3.53)
Subtracting (3.52) and (3.53), we obtain
ZΩ
σk∇Φ(χδ
1+ηδχδ
1)−Φ(χδ
1)
η−Dφ(χδ
1;δχδ
1)· ∇v dV =: A(v) (3.54)
where
A(v) = ZΩ
(σk
1−σ2)δχδ
1∇[Φ(χδ
1+ηδχδ
1)−Φ(χδ
1)] · ∇v dV,
which can be estimated using the Cauchy-Schwarz inequality and (3.15):
|A(v)| ≤ C1ηk(σk
1−σ2)δχδ
1kL∞(Ω)kδχ1kL2(Ω)kvkH1(Ω) .(3.55)
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 15
It is worth noting that (σk
1−σ2)δχδ
1∈L∞(Ω) and δχ1∈L2(Ω) so that the right-
hand side of the above inequality is bounded. Now observe that a(u, v) := RΩσk∇u·
∇v dV is coercive as demonstrated in the proof of Theorem (3.1), i.e.,
|a(u, u)| ≥ ¯
Ckuk2
H1(Ω),(3.56)
for some ¯
C > 0, for any u∈H1(Ω) such that R∂Ωu dS = 0. Hence, the left hand
side of (3.54) is bounded from above if we set
u=Φ(χδ
1+ηδχδ
1)−Φ(χδ
1)
η−Dφ(χδ
1;δχδ
1).
Using this fact and comparing (3.54) and (3.55), we obtain the estimate
¯
CkΦ(χδ
1+ηδχδ
1)−Φ(χδ
1)
η−Dφ(χδ
1;δχδ
1)kH1(Ω)
≤C1ηk(σk
1−σ2)δχδ
1kL∞(Ω)kδχ1kL2(Ω).
Therefore, taking the limit of the last equality as η→0, we obtain
lim
η→0kΦ(χδ
1+ηδχδ
1)−Φ(χδ
1)
η−Dφ(χδ
1;δχδ
1)kH1(Ω) = 0.(3.57)
The rest of the proof is similar to show the convergence of the derivative of Φ∗
with respect to χ1in H1(Ω). The last statements of the lemma can be inferred
directly from the last two lemmas.
Now that we have shown that the derivatives of both Φ and Φ∗converge in
H1(Ω), the next step is to show that ψin (3.41) has a derivative with respect to
χ1which converges in L2(Ω). We first prove the following lemma.
Lemma 3.22. Under assumption (3.30), we let
δφ := Φ(χδ
1+ηδχδ
1)−Φ(χδ
1).(3.58)
Then there exists Cδ>0such that
k∇δφkL∞(Ω)
≤Cδk∇δφkL2(Ω) +ηk∇ · ((σk
1−σ2)δχδ
1∇φ(χδ
1+ηδχδ
1))kH1(Ω),(3.59)
for any η∈(0, τ ), where τis chosen according to Remark 3.4.
Proof. Set
δσk= Σk(χδ
1+ηδχδ
1)−Σk(χδ
1).(3.60)
Then δφ and δσ satisfy
∇ · (Σk(χδ
1)∇(δφ)) = −∇ · (δσk∇Φ(χδ
1+ηδχδ
1)) on Ω,
σ∂(δφ)
∂n = 0 on ∂Ω.
By assumption (3.30) and since δχδ
1is a mollification of δχ1, we deduce that ∇ ·
(δσk∇Φ(χδ
1+ηδχδ
1)) ∈H1(Ω). Because χδ
1, σk
1∈C∞(¯
Ω), then Σk(χδ
1)∈C∞(¯
Ω).
Thus, σk(χδ
1)∈C1(¯
Ω). Using standard regularity estimate (see, e.g., [16]),
kδφkH3(Ω) ≤C1(kδφkH1(Ω) +k∇ · (δσk∇Φ(χδ
1+ηδχδ
1))kH1(Ω)),(3.61)
for some C1>0. Furthermore, by the Sobolev imbedding theorem, we have
kδφkC1,γ (¯
Ω) ≤C2kδφkH3(Ω) .(3.62)
16 R. MENDOZA, S. KEELING EJDE-2020/93
By the definition of k·kCl,γ (¯
Ω), the embedding
C1,γ (¯
Ω) →C1(¯
Ω) (3.63)
is continuous (see, e.g., [18]). If we compare (3.61), (3.62), and (3.63), we can
deduce that ∃¯
C > 0 such that
k∇δφkL∞(Ω) ≤C2kδφkH1(Ω) +k∇ · (δσk∇Φ(χδ
1+ηδχδ
1))kH1(Ω).(3.64)
Observe that δσk=η(σk
1−σ2)δχδ
1. Therefore,
k∇ · (δσk∇Φ(χδ
1+ηδχδ
1))kH1(Ω)
=ηk∇ · ((σk
1−σ2)δχδ
1∇Φ(χδ
1+ηδχδ
1))kH1(Ω).(3.65)
Recall that all solutions of the forward problem have zero boundary integral. Thus,
Z∂Ω
δφ dS =Z∂Ω
Φ(χδ
1+ηδχδ
1)−Φ(χδ
1)dS = 0 −0=0.
Using this and the generalized Friedrich’s inequality, we obtain
k∇δφk2
L2(Ω) =1
2k∇δφk2
L2(Ω) +1
2k∇δφk2
L2(Ω)
≥C3
2kδφk2
L2(Ω) −1
2Z∂Ω
δφ dS 2+1
2k∇δφk2
L2(Ω)
=C3
2kδφk2
L2(Ω) +1
2k∇δφk2
L2(Ω)
≥min C3
2,1
2kδφk2
H1(Ω)
(3.66)
for some C3>0. Using (3.66) and (3.65), (3.64) becomes
k∇δφkL∞(Ω) ≤C2kδφkH1(Ω) +k∇ · (δσk∇Φ(χδ
1+ηδχδ
1))kH1(Ω)
≤C2n1
qmin{C3
2,1
2}
k∇δφkL2(Ω)
+ηk∇ · ((σk
1−σ2)δχδ
1∇Φ(χδ
1+ηδχδ
1))kH1(Ω)o
≤2C2max n1
qmin{C3
2,1
2}
,1onk∇δφkL2(Ω)
+ηk∇ · ((σk
1−σ2)δχδ
1∇Φ(χδ
1+ηδχδ
1))kH1(Ω)o.
Because of (3.58) and (3.60), our claim immediately follows from the above inequal-
ity.
Note that
∇Φ(χδ
1)∈L∞(¯
Ω),∇Φ∗(χδ
1)∈L∞(Ω),∇δΦ
δχδ
1
(χδ
1;δχδ
1),∇δΦ∗
δχδ
1
(χδ
1;δχδ
1)∈L2(Ω).
Therefore, ∇Φ(χδ
1)· ∇δΦ∗
δχ1(χδ
1;δχδ
1) + ∇δΦ
δχ1(χδ
1;δχδ
1)· ∇Φ∗(χδ
1)∈L2(Ω). We show
in the next lemma that this is in fact the derivative of ψdefined in (3.41).
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 17
Lemma 3.23. Under assumption (3.30), we have
lim
η→0kψ(χδ
1+ηδχδ
1)−ψ(χδ
1)
η−δψ
δχ1
(χδ
1;δχδ
1)kL2(Ω) = 0 (3.67)
with
δψ
δχ1
(χδ
1;δχδ
1) := ∇Φ(χδ
1)· ∇δΦ∗
δχ1
(χδ
1;δχδ
1) + ∇δΦ
δχ1
(χδ
1;δχδ
1)· ∇Φ∗(χδ
1).
Proof. For any perturbation δχ1of χ1, we have
ψ(χδ
1+ηδχδ
1)−ψ(χδ
1)
η
=∇Φ(χδ
1+ηδχδ
1)· ∇Φ∗(χδ
1+ηδχδ
1)
η−∇Φ(χδ
1)· ∇Φ∗(χδ
1)
η
=∇Φ(χδ
1+ηδχδ
1)·∇Φ∗(χδ
1+ηδχδ
1)− ∇Φ∗(χδ
1)
η
+∇Φ(χδ
1+ηδχδ
1)− ∇Φ(χδ
1)
η· ∇Φ∗(χδ
1).
(3.68)
To continue with our proof, we first perform some convenient calculations. By
adding and subtracting a term, the following is obtained:
∇Φ(χδ
1+ηδχδ
1)·∇Φ∗(χδ
1+ηδχδ
1)− ∇Φ∗(χδ
1)
η
− ∇Φ(χδ
1)· ∇δΦ∗
δχδ
1
(χδ
1;δχδ
1) := A1+A2,
(3.69)
where
A1=∇Φ(χδ
1+ηδχδ
1)·∇Φ∗(χδ
1+ηδχδ
1)− ∇Φ∗(χδ
1)
η
− ∇Φ(χδ
1+ηδχδ
1)· ∇δΦ∗
δχ1
(χδ
1;δχδ
1),
A2=∇Φ(χδ
1+ηδχδ
1)·δ∇Φ∗
δχ1
(χδ
1;δχδ
1)− ∇Φ(χδ
1)· ∇δΦ∗
δχ1
(χδ
1;δχδ
1).
Using the H¨older’s inequality and (3.37), we obtain
kA1kL2(Ω) ≤C1kfk˜
L2(∂Ω)k∇Φ∗(χ1+ηδχ1)− ∇Φ∗(χ1)
η− ∇δΦ∗
δχ1
(χδ
1;δχδ
1)kL2(Ω),
for some C1>0. Recall from (3.39) that
kΦ(χδ
1+ηδχδ
1)−Φ(χδ
1)kH1(Ω) ≤C2ηkδχ1kL2(Ω) ,(3.70)
for some C2>0. Using the Cauchy-Schwarz inequality, (3.48), and (3.59), we
obtain
kA2kL2(Ω) ≤C3ηC4kδχ1kL2(Ω) +k∇ · ([σk
1−σ2]δχδ
1∇Φ(χδ
1+ηδχδ
1))kH1(Ω)
× k∇δΦ∗
δχ1
(χδ
1;δχδ
1)kH1(Ω),
for some C3, C4>0. Define
A3:= ∇Φ(χ1+ηδχ1)− ∇Φ(χ1)
η· ∇Φ∗(χ1)− ∇ δΦ
δχ1
(χ1;δχ1)· ∇Φ∗(χ1).(3.71)
18 R. MENDOZA, S. KEELING EJDE-2020/93
Hence,
ψ(χδ
1+ηδχδ
1)−ψ(χδ
1)
η=
3
X
i=1
Ai.
Using H¨older’s inequality and (3.36),
kA3kL2(Ω)
≤ k∇Φ(χδ
1+ηδχδ
1)− ∇Φ(χδ
1)
η− ∇ δΦ
δχ1
(χδ
1;δχδ
1)kL2(Ω)k∇Φ∗(χδ
1)kL∞(Ω).
Comparing (3.68), (3.69) and (3.71) we obtain
kψ(χδ
1+ηδχδ
1)−ψ(χδ
1)
η−δψ
δχ1
(χδ
1;δχδ
1)kL2(Ω) ≤
3
X
i=1
kAik,
using the triangle inequality. Now we only need to show that all the terms on the
right-hand side of this inequality converge to 0 as ηgoes to 0. From the estimate of
the L2-norm of A2above, we can see that kA2kL2(Ω) →0. From (3.47) and (3.48),
we can deduce that
lim
η→0k∇Φ(χδ
1+ηδχδ
1)− ∇Φ(χδ
1)
η− ∇ δΦ
δχδ
1
(χδ
1;δχδ
1)kL2(Ω) = 0,
lim
η→0k∇Φ∗(χδ
1+ηδχδ
1)− ∇∗Φ(χδ
1)
η− ∇δΦ∗
δχδ
1
(χδ
1;δχδ
1)kL2(Ω) = 0.
These imply that kA1kL2(Ω),kA3kL2(Ω) →0.
We now use our results on Φ and Φ∗to study Σ1. We first show that under
assumption (3.30), (3.29) has a unique solution σk+1
1. We then proceed with finding
the regularity of the said solution. Observe that σk+1
1depends on φand φ∗. Hence,
we can investigate how the mollification of χ1affects σk+1
1. We show that σk+1
1
continuously depends on χ1. Furthermore, we prove that δΣ1
δχ1(χδ
1;δχδ
1)∈H1(Ω).
We start by equipping H1(Ω) with a suitable norm.
Proposition 3.24. Under assumption (3.30) we define
|v|2
H1(Ω) := αZΩ
(χ+)|∇v|2dV +θλ Zv2dV,
where χ(x)∈[0,1] for all x∈Ω, and let k·kH1(Ω) be the standard H1(Ω) norm.
Then |·|H1(Ω) and k·kH1(Ω) are equivalent.
Proof. Observe that
min{α, λ}kvk2
H1(Ω) ≤αZΩ
|∇v|2dV +θZΩ
|v|2dV
≤αZΩ
(χ+)|∇v|2dV +λZΩ
|v|2dV =|v|2
H1(Ω).
On the other hand,
|v|2
H1(Ω) =αZΩ
(χ+)|∇v|2dV +λZΩ
|v|2dV
≤αZΩ
(1 + )|∇v|2dV +λZΩ
|v|2dV
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 19
≤max {α(1 + ), λ} kvk2
H1(Ω).
We now establish the regularity of σk+1
1.
Lemma 3.25. Under assumption (3.30), the variational formulation
ZΩ
α(χδ
1+)∇σk+1
1· ∇v dV +ZΩ
λ(σk+1
1−σk
1)v dV
=ZΩ
χδ
1∇Φ(χδ
1)· ∇Φ∗(χδ
1)v dV
(3.72)
for all v∈H1(Ω), has a unique solution σk+1
1∈H1(Ω). Furthermore,
kσk+1
1kH1(Ω) ≤2pµ(Ω) max{Cδ
1, Cδ
2}
min{α, λ}(3.73)
where Cδ
1=k∇Φ(χδ
1)kL∞(Ω)k∇Φ∗(χδ
1)kL∞(Ω) and Cδ
2=λkσk
1kL∞(Ω).
Proof. Let u, v ∈H1(Ω) and define a(u, v) = RΩα(χδ
1+)∇u· ∇v dV +RΩλuv dV
and b(v) = RΩ(χδ
1)∇Φ(χδ
1)· ∇Φ∗(χδ
1)v dV +RΩλσk
1v dV . It is obvious that ais
bilinear and bis linear. Using the Cauchy-Schwarz inequality, one can prove that
a(u, v) is continuous. We can also easily show that a(u, v) is coercive using the
previous proposition:
|a(u, u)|=ZΩ
α(χδ
1+)|∇u|2V+ZΩ
λu2dV
=|u|2
H1(Ω) ≥min{α, θ}kuk2
H1(Ω).
Furthermore, the continuity of b(v) can be proven using the Cauchy-Schwarz in-
equality, the bounds in (3.31), and (3.36). Hence, by Lax-Milgram Theorem there
is a unique σk
1∈H1(Ω) satisfying (3.72) for all v∈H1(Ω). The H1(Ω) bound for
σk+1
1directly follows.
Remark 3.26. Given any perturbation δχδ
1and η > 0, we have to make sure
that the quantity Σ1(χδ
1+ηδχδ
1) is well-defined. As shown in the proof of Lemma
(3.25), replacing χδ
1with χδ
1+ηδχδ
1cannot be done with just any η. To make sure
that the bilinear functional ais coercive, χδ
1+ηδχδ
1+must be positive. Since
χδ
1+ > 0, we can choose ηsmall enough so that χδ
1+ηδχδ
1+ > 0 is satisfied.
Hence, similar to Remark 3.4, we take ηfrom the set (0,¯τ) for ¯τsufficiently small
so that Σ1(χδ
1+ηδχδ
1) makes sense. Therefore, combining (3.37), (3.38) and (3.73),
there exists C > 0 such that
kΣ1(χδ
1+ηδχδ
1)kH1(Ω) <∞,(3.74)
for any η∈(0,ˆτ), where ˆτ= min {τ , ¯τ}.
Because χδ
1is a mollification of χ1, it is real analytic. Then χδ
1+∈C∞(¯
Ω).
Moreover, from Lemma (3.15), ∇Φ(χδ
1),∇Φ∗(χδ
1)∈C∞(¯
Ω). This means that
χδ
1∇Φ(χδ
1)· ∇Φ∗(χδ
1), χδ
1+,λare all in C∞(¯
Ω). Then using standard regularity
estimates [16] on
−α∇ · [(χδ
1+)∇σk+1
1] + λ(σk+1
1−σk
1) = χδ
1∇Φ(χδ
1)· ∇Φ∗(χδ
1) on Ω,
∂σk+1
1
∂n = 0 on ∂Ω,
(3.75)
20 R. MENDOZA, S. KEELING EJDE-2020/93
gives us σk+1
1∈C∞(¯
Ω). Consequently, k∇Σ1(χδ
1)kL∞(Ω) <∞. Furthermore,
k∇Σ1(χδ
1+ηδχδ
1)kL∞(Ω) <∞,(3.76)
for any η∈(0,ˆτ), where ˆτ= min {τ, ¯τ},τand ¯τare chosen according to Remark
3.4 and Remark 3.26, respectively. Before we can show that δΣ1
δχ1(χδ
1;δχδ
1)∈H1(Ω),
we first need to find a candidate derivative.
Lemma 3.27. Under assumption (3.30), there exists Dσ(χ1;δχ1)∈H1(Ω) such
that
ZΩ
α(χδ
1+)∇Dσ(χδ
1;δχδ
1)· ∇v dV +ZΩ
λDσ(χδ
1;δχδ
1)v dV
=ZΩ
χδ
1
δψ
δχ1
(χδ
1;δχδ
1)v+δχδ
1∇Φ(χδ
1)· ∇Φ∗(χδ
1)v−αδχδ
1∇(Σ1(χδ
1)) · ∇v dV,
(3.77)
for all v∈H1(Ω).
The proof of the above lemma is similar to that of Lemma 3.19; we omit it. We
now have all the necessary tools to show that δΣ1
δχ1(χδ
1;δχδ
1)∈H1(Ω). We prove
that this is exactly Dσ(χδ
1;δχδ
1) computed in the previous lemma.
Theorem 3.28. Under assumption (3.30), there exists Cδ>0such that
kΣ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)kH1(Ω) ≤Cδηkδχ1kL2(Ω) (3.78)
for any η∈(0,ˆτ), where ˆτ= min {τ, ¯τ},τand ¯τare both chosen according to
Remark 3.4 and Remark 3.26, respectively. Also,
lim
η→0kΣ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)
η−Dσ(χδ
1;δχδ
1)k= 0.(3.79)
Furthermore, we make the identification δΣ1
δχ1(χδ
1;δχδ
1) = Dσ(χδ
1;δχδ
1)∈H1(Ω).
Proof. Let v∈H1(Ω) and η > 0. From (3.72), we have
ZΩ
α(χδ
1+)∇Σ1(χδ
1)· ∇v+λ(Σ1(χδ
1)−σk
1)v dV
=ZΩ
χδ
1∇Φ(χδ
1)· ∇Φ∗(χδ
1)v dV.
(3.80)
Similarly, if we replace χδ
1in (3.72) with χδ
1+ηδχδ
1, we have
ZΩ
α(χδ
1+ηδχδ
1+)∇(Σ1(χδ
1+ηδχδ
1)) · ∇v dV
+ZΩ
λ(Σ1(χδ
1+ηδχδ
1)−σk
1)v dV =: A(v),
(3.81)
where
A(v) = ZΩ
(χδ
1+ηδχδ
1)∇Φ(χδ
1+ηδχδ
1)· ∇Φ∗(χδ
1+δχδ
1)v dV.
Subtracting (3.80) from (3.81), we obtain
A1(v) + A2(v) = B1(v) + B2(v) + B3(v),(3.82)
where
A1(v) := ZΩ
α(χδ
1+)∇[Σ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)] · ∇v dV,
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 21
A2(v) := ZΩ
λ(Σ1(χδ
1+ηδχδ
1)−Σ1(χδ
1))v dV,
B1(v) := ZΩ
χδ
1[∇Φ(χδ
1+ηδχδ
1)· ∇Φ∗(χδ
1+ηδχδ
1)− ∇Φ(χδ
1)· ∇Φ∗(χδ
1)]v dV,
B2(v) := ZΩ
ηδχδ
1∇Φ(χδ
1+ηδχδ
1)· ∇Φ∗(χδ
1+ηδχδ
1)v dV,
B3(v) := −ZΩ
αηδχδ
1∇(Σ1(χδ
1+ηδχδ
1)) · ∇v dV,
for all v∈H1(Ω). We define a(Σ1(χδ
1+ηδχδ
1)−Σ1(χδ
1), v) := A1(v) + A2(v) and
b(v) := B1(v) + B2(v) + B3(v), for all v∈H1(Ω). In Lemma (3.25), we have shown
already that ais bilinear, coercive, and continuous. Clearly, bis linear. Now we only
need to show that bis continuous. So we need to estimate B1(v), B2(v), and B3(v).
For all these, we use the Cauchy-Schwarz inequality. To show continuity of B1(v),
we also use (3.43). For B2(v) and B3(v), one uses the Cauchy-Schwarz inequality,
H¨older’s inequality, and Young’s inequality for convolutions to show continuity.
It is worth noting that k∇Φ(χδ
1+ηδχδ
1)kL∞(Ω),k∇Φ∗(χδ
1+ηδχδ
1)kL∞(Ω), and the
quantity kΣ1(χδ
1+ηδχδ
1)kH1(Ω) are independent of ηas shown in (3.37), (3.38), and
(3.74). Combining these implies that b(v) is bounded.
We make the substitution u=v= Σ1(χδ
1+ηδχδ
1)−Σ1(χδ
1). Furthermore, using
the coercivity of aand the boundedness of b, we conclude that
kΣ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)kH1(Ω) ≤Cηkδχ1kL2(Ω) .(3.83)
Hence, the proof of our first statement is complete.
Subtracting (3.77) from (3.82), we obtain
D1(v) + D2(v) = E1(v) + E2(v) + E3(v),(3.84)
where
D1(v) = ZΩ
α(χδ
1+)∇Σ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)
η−Dσ(χδ
1;δχδ
1)· ∇v dV,
D2(v) = ZΩ
λΣ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)
η−Dσ(χδ
1;δχδ
1)v dV,
E1(v) = ZΩ
χδ
1ψ(χδ
1+ηδχδ
1)−ψ(χδ
1)
η−δψ
δχδ
1
(χδ
1;δχδ
1)v dV,
E2(v) = ZΩ
δχδ
1ψ(χδ
1+ηδχδ
1)−ψ(χδ
1)v dV,
E3(v) = −ZΩ
αδχδ
1∇Σ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)· ∇v dV,
and ψis the function defined in (3.41).
Using the Cauchy-Schwarz inequality, H¨older’s inequality, (3.43), and (3.83), we
can estimate E1(v), E2(v), and E3(v). Indeed,
|E1(v)|≤kψ(χδ
1+ηδχδ
1)−ψ(χδ
1)
η−δψ
δχ1
(χδ
1;δχδ
1)kL2(Ω)kvkL2(Ω) .
Similarly,
|E2(v)| ≤ C2ηkδχδ
1kL∞(Ω)kδχ1kL2(Ω)kvkH1(Ω) ,
22 R. MENDOZA, S. KEELING EJDE-2020/93
for some C2>0. Finally,
|E3(v)| ≤ αkδχδ
1kL∞(Ω)k∇[Σ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)]kL2(Ω)k∇vkL2(Ω)
≤Cηkδχδ
1kL∞(Ω)kδχ1kL2(Ω)kvkH1(Ω) .
If we make the substitution
v=Σ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)
η−Dσ(χδ
1;δχδ
1),
then by Proposition 3.24, we obtain
|D1(v) + D2(v)|=
Σ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)
η−Dσ(χδ
1;δχδ
1)
2
H1(Ω)
≥min{α, λ}kΣ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)
η−Dσ(χδ
1;δχδ
1)k2
H1(Ω).
Hence, the above inequality and (3.84) imply
min{α, λ}kΣ1(χδ
1+ηδχδ
1)−Σ1(χδ
1)
η−Dσ(χδ
1;δχδ
1)kH1(Ω)
≤ |E1(v)|+|E2(v)|+|E3(v)|.
Taking the limit as η→0, it is clear that |E2(v)|+|E3(v)| → 0.
Lastly, from (3.67), we have |E1(v)| → 0, which then implies our second state-
ment. The last statement follows immediately from the previous lemma.
Now that we have established that δΣ1
δχ1(χδ
1;δχδ
1)∈H1(Ω), the gradient of the
functional Jshown in (2.14) is justified.
4. Existence of a fixed point
In (3.28), the update for χ1was introduced. In this section, we show that this
update has a fixed point. In other words, we show that
Υ(χ1) := (Tδ◦M◦H◦Θ◦G◦Tδ)(χ1) (4.1)
has a fixed point on some suitable space. We use the following fixed point theorem
to prove this.
Theorem 4.1 (Schauder fixed point).Let Kbe a convex subset of L2(Ω) and
suppose Υ : K→L2(Ω) is continuous. Suppose Υ(K)is a compact subset of K.
Then Υhas a fixed point in K(see [15]).
Thus, it is necessary to show that Υ is continuous on a convex subset Kof
L2(Ω) and that Υ(K) is compact in K. The previous section justified the calculated
formulation of the function Gdefined in (2.14). We now show that
G(χ1) = χ1−ω[−2(Σ1(χ1)−σ2)ψ(χ1) + α|∇Σ1(χ1)|2] (4.2)
is continuous.
Lemma 4.2. Under assumption (3.30),
lim
η→0kG(χδ
1+ηδχδ
1)−G(χδ
1)kL2(Ω) = 0.(4.3)
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 23
Proof. Denote
A1(χδ
1) := 2ω[Σ1(χδ
1+ηδχδ
1)−σ2]ψ(χδ
1),
A2(χδ
1) := −αω|∇Σ1(χδ
1)|2.
Thus, using (4.2), the triangle inequality, and Young’s inequality for a convolution,
we obtain
kG(χδ
1+ηδχδ
1)−G(χδ
1)kL2(Ω)
≤ηkξδkL1(Ω)kδχ1kL2(Ω) +kA1(χδ
1+ηδχδ
1)−A1(χδ
1)kL2(Ω)
+kA2(χδ
1+ηδχδ
1)−A2(χδ
1)kL2(Ω).
(4.4)
Adding and subtracting 2ω[Σ1(χδ
1+ηδχδ
1)−σ2]ψ(χδ
1) to A1(χδ
1+ηδχδ
1)−A1(χδ
1)
and using the triangle inequality, we obtain
kA1(χδ
1+ηδχδ
1)−A1(χδ
1)kL2(Ω) ≤ kB1(χ1;δχ1)kL2(Ω) +kB2(χ1;δχ1)kL2(Ω),
where
B1(χδ
1;δχδ
1) = −2ω[Σ1(χδ
1+ηδχδ
1)−σ2]ψ(χδ
1+ηδχδ
1)
+ 2ω[Σ1(χδ
1+ηδχδ
1)−σ2]ψ(χδ
1),
B2(χδ
1;δχδ
1) = −2ω[Σ1(χδ
1+ηδχδ
1)−σ2]ψ(χδ
1)+2ω[Σ1(χδ
1)−σ2]ψ(χδ
1).
By using H¨older’s inequality, (3.76), (3.43), (3.31), (3.36), and (3.78), we can esti-
mate B1(χδ
1;δχδ
1) and B2(χδ
1;δχδ
1) as follows:
kB1(χδ
1;δχδ
1)kL2(Ω) ≤2C1ωηk[Σ1(χδ
1+ηδχδ
1)−σ2]kL∞(Ω)kδχ1kL2(Ω),(4.5)
for some C1>0 and
kB2(χδ
1;δχδ
1)kL1(Ω) ≤C2ηkδχ1kL2(Ω) k∇Φ(χδ
1)kL∞(Ω)k∇Φ∗(χδ
1)kL2(Ω),(4.6)
for some C2>0. Also, A2(χδ
1+ηδχδ
1)−A2(χδ
1) can be estimated using (3.76) and
(3.78):
kA2(χδ
1+ηδχδ
1)−A2(χδ
1)kL2(Ω)
≤αωC3ηkδχ1kL2(Ω)kΣ1(χδ
1ηδχδ
1)kL∞(Ω) +kΣ1(χδ
1)kL∞(Ω),(4.7)
for some C3>0. Comparing (4.4), (4.5), (4.6), and (4.7) implies the existence of a
C > 0 such that
kG(χδ
1+ηδχδ
1)−G(χδ
1)kL2(Ω) ≤Cηkδχ1kL2(Ω) ,
for any η∈(0,ˆτ), where ˆτ= min {τ, ¯τ},τand ¯τare both chosen according to
Remark 3.4 and Remark 3.26, respectively. Taking the limit of the above inequality
as η→0 gives us our desired result.
Now that we have shown the continuity of the function G, we prove the continuity
of the operator Θ (see Definition 2.6). Before proving continuity, we first show that
given χδ
1, (2.16) has a solution in H1(Ω). Note that because χδ
1∈C∞(Ω), ∇χδ
1is
bounded in Ω for a fixed δ. Hence,
q|∇χδ
1|2+β2≤qk∇χδ
1k2
L∞(Ω) +β2=: ¯
K < ∞.
Or equivalently, 1
p|∇χδ
1|2+β2≥1
¯
K.(4.8)
24 R. MENDOZA, S. KEELING EJDE-2020/93
Obviously, for any β > 0, p|∇χδ
1|2+β2≥β, which can be expressed as
1
p|∇χδ
1|2+β2≤1
β.(4.9)
For u, v ∈H1(Ω), we define
a(u, v) = ZΩ
ωγ ∇u· ∇v
p|∇χδ
1|2+β2+uv dV, (4.10)
b(v) = ZΩ
G(χδ
1)v dV.
Clearly, aand bare bilinear and linear, respectively. Observe that using the Cauchy-
Schwarz inequality and (4.9), we obtain
|a(u, v)| ≤ 2 max{ωγ
β,1}kukH1(Ω)kvkH1(Ω) ,
for all v∈H1(Ω), which makes acontinuous. It can also be shown using the
Cauchy-Schwarz inequality that b(v) is also continuous. Using (4.9), acan be
proven to be coercive. Therefore, using the Lax-Milgram theorem, there exists a
unique θ∈H1(Ω) satisfying a(θ, v) = b(v)] for all v∈H1(Ω).
Lemma 4.3. Under assumption (3.30), there exists a unique θ∈H1(Ω) satisfying
ZΩ
ωγ ∇θ· ∇v
p|∇χδ
1|2+β2+θv dV =ZΩ
G(χδ
1)v dV, (4.11)
for any v∈H1(Ω) provided that ∂ v
∂n = 0 on ∂Ω. Furthermore, there exists C > 0
such that
kθkH1(Ω) ≤1
min{ωγ
¯
K,1}kG(χδ
1)kL2(Ω).
The above lemma tells us that given χ1, a mollification can be performed to
obtain a unique solution θ∈H1(Ω) to (4.11). Hence, we can think of Θ(χδ
1) as a
function that maps an element χ1∈L2(Ω) to an element θ∈H1(Ω). Note that
given a perturbation δχδ
1of χδ
1, Θ(χδ
1+ηδχδ
1) is well-defined for any 0 < η < ∞
because for coercivity we just need qk∇(χδ
1+ηδχδ
1)k2
L∞(Ω) +β2to be finite. Since
χδ
1+ηδχδ
1∈C∞(¯
Ω), this is not a problem. From the definition of Gand the
inequalities (3.37), (3.38), and (3.76), we can infer that
kΘ(χδ
1+ηδχδ
1)kH1(Ω) ≤CkG(χδ
1+ηδχδ
1)kL2(Ω) <∞,(4.12)
for some C > 0. We now prove that this map is continuous.
Lemma 4.4. Under assumption (3.30),
lim
η→0kΘ(χδ
1+ηδχδ
1)−Θ(χδ
1)kH1(Ω) = 0.
Proof. By the previous lemma, note that Θ(χδ
1) satisfies
ZΩ
ωγ ∇Θ(χδ
1)· ∇v
p|∇χδ
1|2+β2+ Θ(χδ
1)v dV =ZΩ
G(χδ
1)v dV. (4.13)
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 25
Similarly, Θ(χδ
1+ηδχδ
1) satisfies
ZΩ
ωγ ∇Θ(χδ
1+ηδχδ
1)· ∇v
p|∇(χδ
1+ηδχδ
1)|2+β2+ Θ(χδ
1+ηδχδ
1)v dV
=ZΩ
G(χδ
1+ηδχδ
1)v dV.
(4.14)
Subtracting (4.13) from (4.14), we obtain
A1(v) + A2(v) = B(v),(4.15)
with
A1(v) := ZΩ
ωγh∇Θ(χδ
1+ηδχδ
1)· ∇v
p|∇χδ
1+ηδχδ
1|2+β2−∇Θ(χδ
1)· ∇v
p|∇χδ
1|2+β2idV,
A2(v) := ZΩ
[Θ(χδ
1+ηδχδ
1)−Θ(χδ
1)]v dV,
B(v) := ZΩ
[G(χδ
1+ηδχδ
1)−G(χδ
1)]v dV.
We subtract and add the term
ZΩ
ωγ ∇Θ(χδ
1+ηδχδ
1)· ∇v
p|∇χδ
1|2+β2dV
to A1(v) to obtain
A1(v) = A3(v) + A4(v),(4.16)
with
A4(v) := ZΩ
ωγD(η)∇Θ(χδ
1+ηδχδ
1)· ∇v
p|∇(χδ
1)|2+β2,
A3(v) := ZΩ
ωγ [∇Θ(χδ
1+ηδχδ
1)− ∇Θ(χδ
1)] · ∇v
p|∇χδ
1|2+β2dV,
D(η) := 1
p|∇(χδ
1+ηδχδ
1)|2+β2−1
p|∇χδ
1|2+β2.
From (4.15) and (4.16), we have
A3(v) + A2(v) = B(v)−A4(v).(4.17)
From the definition of the bilinear functional ain (4.10), we deduce that
A3(v) + A2(v) = a(Θ(χδ
1+ηδχδ
1)−Θ(χδ
1), v).
From the coercivity of awe can show that
|(A3+A2)(Θ(χδ
1+ηδχδ
1)−Θ(χδ
1))|
≥min(ωγ
¯
K,1)kΘ(χδ
1+ηδχδ
1)−Θ(χδ
1)k2
H1(Ω).(4.18)
On the other hand, using the Cauchy-Schwarz inequality, we obtain
|B(v)|≤kG(χδ
1+ηδχδ
1)−G(χδ
1)kL2(Ω)kvkH1(Ω) .(4.19)
Moreover, using the Cauchy-Schwarz inequality, H¨older’s inequality, and (4.9), we
have
|A4(v)| ≤ ωγ
βkD(η)kL∞(Ω)kΘ(χδ
1+ηδχδ
1)kH1(Ω)kvkH1(Ω) .(4.20)
26 R. MENDOZA, S. KEELING EJDE-2020/93
Comparing (4.17), (4.18), (4.19), and (4.20), we obtain
min{ωγ
¯
K,1}kΘ(χδ
1+ηδχδ
1)−Θ(χδ
1)k2
H1(Ω)
≤ kG(χδ
1+ηδχδ
1)−G(χδ
1)kL2(Ω) +ωγ
βkD(η)kL∞(Ω)kΘ(χδ
1+ηδχδ
1)kH1(Ω).
From (4.12), the right-hand side of the above inequality is bounded for any
η∈(0,ˆτ), where ˆτ= min {τ, ¯τ},τand ¯τare both chosen according to Remark 3.4
and Remark 3.26, respectively. Obviously, limη→0kD(η)kL∞(Ω) = 0. The above
inequality, together with (4.3), establishes our claim.
We now establish that the function Hdefined in (3.25) is continuous. The details
of the proof can be found in [15].
Lemma 4.5. Let z∈L2(Ω)\H1(Ω) and suppose {gn}∞
n=1 ⊂L2(Ω) converge to g
in L2(Ω). Then
lim
n→∞ µ(H(g)4H(gn)) = 0,(4.21)
where
H(g) = {x∈Ω : ((gδ−ζ+δz)∗ξδ)(x)≥0},
for some ζ∈(0,1) and gδ=g∗ξδ. In other words, His continuous in L2(Ω) (see
[15]).
In our next computations, we prove the continuity of the function Mdefined in
(3.26). Recall that Mmaps elements of M(Ω) to their corresponding characteristic
functions. We now try to find a suitable space for these characteristic functions.
Intuitively, convergence of these characteristic functions is dependent upon the
convergence of their associated supports. We choose L2(Ω) to be the space of
the characteristic functions and select M(Ω) to be the space of their associated
supports. Recall that M(Ω) is a metric space equipped with the measure of the
symmetric difference. The following lemma proves how these two spaces are related.
Lemma 4.6. Let ˆχand χbe characteristic functions on Ωwhose supports are given
by Ωˆχand Ωχ, respectively. Then
µ(Ωˆχ4Ωχ) = kˆχ−χk2
L2(Ω).
Proof. Because χand ˆχare characteristic functions, we have
Ωˆχ\Ωχ={x:x∈Ωˆχ∧x /∈Ωχ}={x: ˆχ(x)=1∧χ(x)=0}.
Similarly,
Ωχ\Ωˆχ={x:χ(x)=1∧ˆχ(x)=0}.
Thus, from the definition of symmetric difference and the fact that Ωˆχ\Ωχand
Ωχ\Ωˆχare disjoint sets, we obtain
µ(Ωˆχ4Ωχ) = µ((Ω ˆχ\Ωχ)∪(Ωχ\Ωˆχ))
=µ({x: ˆχ(x)=1∧χ(x)=0}) + µ({x:χ(x) = 1 ∧ˆχ(x)=0})
=ZΩ
ˆχ(1 −χ)dV +ZΩ
χ(1 −ˆχ)dV
=kˆχ−χk2
L2(Ω).
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 27
Now that we have established a mode of convergence for the characteristic func-
tions and their associated sets, we can prove that Mis continuous.
Lemma 4.7. Suppose {ωn}∞
n=1 ⊂ M(Ω) such that ωn→ωin M(Ω), that is,
limn→∞ µ(ωn4ω) = 0. Then
lim
n→∞ kM(ωn)−M(ω)kL2(Ω) = 0,
where M:M(Ω) →L2(Ω) is a function that maps ωto its corresponding charac-
teristic function, that is, M(ω) = χω. In other words, Mis continuous on M(Ω).
Proof. We denote
M(ωn) =: χnand M(ω) := χω.
Then by Lemma (4.6),
lim
n→∞ kM(ωn)−M(ω)k2
L2(Ω) = lim
n→∞ kχn−χωk2
L2(Ω) = lim
n→∞ µ(ωn4ω) = 0.
We have shown continuity of G, Θ, Hand M. Finally, we can prove that Υ has
a fixed point.
Theorem 4.8. Under assumption (3.30) we let z∈L2(Ω)\H1(Ω). Then the
function Υ : L2(Ω) →L2(Ω) defined by
Υ(χ1) := (Tδ◦M◦H◦Θ◦G◦Tδ)(χ1)
has a fixed point in the set
K:= {χ1∈L2(Ω) : 0 ≤χ1≤1a.e. Ω}.(4.22)
Proof. We employ the Schauder Fixed Point Theorem. We first need to show that K
is a convex set in L2(Ω). Let χ1,¯χ1∈Kand λ∈(0,1). Obviously, λχ1+(1−λ) ¯χ1∈
L2(Ω). We only need to show that 0 ≤λχ1+ (1 −λ) ¯χ1≤1. Because λ, 1−λ > 0
then λχ1+ (1 −λ)¯χ1≥0. Furthermore, λχ1+ (1 −λ)¯χ1≤λ+ 1 −λ= 1. Thus,
λχ1+ (1 −λ)¯χ1∈Kand Kis convex in L2(Ω).
We show next that Υ is continuous. The functions G, Θ, H, and Mare continu-
ous as proven in Lemma 4.2, Lemma 4.4, Lemma 4.5, and Lemma 4.7, respectively.
From Lemma 3.8, Tδis continuous as well by choosing p= 2, r= 2, and q= 1.
Because composition of continuous functions is continuous, Υ is continuous.
We only need to show that Υ(K)⊂Kand that Υ(K) is compact in K. Recall
that (M◦H◦Θ◦G)(χδ
1) is a characteristic function. Thus,
0≤(M◦H◦Θ◦G)(χδ
1)≤1.
By Theorem 2.5,
0≤(Tδ◦M◦H◦Θ◦G)(χδ
1)≤1,
and so Υ(χ1)∈K. Let ¯χ1be an arbitrary element of K. Let us denote ω:=
(H◦Θ◦G)(¯χδ
1), χω:= M(ω), and χω
δ:= Tδ(χω). By Lemma 3.9, H¨older’s inequality,
and the Cauchy-Schwarz inequality, we obtain
|∇χω
δ(x)|=|ZΩ
∇ξδ(x−y)χω(y)dy|
≤ kχωkL∞(Ω) ZΩ
|∇ξδ(x−y)|dy
≤pµ(Ω)k∇ξδkL2(Ω).
28 R. MENDOZA, S. KEELING EJDE-2020/93
Hence,
k∇χω
δk2
L2(Ω) =ZΩ
|ZΩ
∇ξδ(x−y)χω(y)dy|2dx ≤µ(Ω)2k∇ξδk2
L2(Ω).
From Lemma 2.5, χω
δis real analytic and so χω
δ∈H1(Ω). For a fixed δ, we compute
the H1(Ω) norm of χω
δusing Young’s inequality for convolutions and the H¨older’s
inequality:
kΥ(¯χ1)k2
H1(Ω) =kχω
δk2
H1(Ω)
=kχω
δk2
L2(Ω) +k∇χω
δk2
L2(Ω)
≤ kχω∗ξδk2
L2(Ω) +µ(Ω)2k∇ξδk2
L2(Ω)
≤ kχωk2
L1(Ω)kξδk2
L2(Ω) +µ(Ω)2k∇ξδk2
L2(Ω)
≤ kχωk2
L∞(Ω)µ(Ω)2kξδk2
L2(Ω) +µ(Ω)2k∇ξδk2
L2(Ω)
≤µ(Ω)2kξδk2
L2(Ω) +k∇ξδk2
L2(Ω).
Since ¯χ1is arbitrary, any sequence {Υ(χn
1)}∞
n=1 is bounded in the H1(Ω) norm
for a fixed δ. Because Ω is bounded, H1(Ω) is compactly embedded in L2(Ω) and
{Υ(χn
1)}∞
n=1 has a convergent subsequence [1]. Therefore Υ(K) is compact. Using
Schauder Fixed Point theorem, Υ1has a fixed point on K.
The fixed point is attained given any arbitrary χ1∈L2(Ω) such that 0 ≤χ1≤1
and ¯σ1in C∞(Ω), which can be chosen to be a constant. The introduction of a
mollifier was used to guarantee the existence of the fixed point. Note that a fixed
point is guaranteed for any arbitrary δ > 0.
5. Conclusion
The EIT problem is the image reconstruction of the conductivity distribution of
a body Ω given current and electrical potential data on the boundary ∂Ω. In [27],
a two-phase segmentation algorithm was proposed in reconstructing conductivity
distribution in EIT. The algorithm arised from the minimization of a functional
which depends on the conductivity distribution σ=σ1χ1+σ2(1 −χ1). The value
of σ2is fixed and known while σ1is expressed in terms of χ1. Hence, the functional
depends on χ1alone. An iterative algorithm using the method of steepest descent
is then explored. Moreover, the algorithm is summarized using a composition of
several functions of χ1. By introducing a mollification on χ1, continuity of these
functions was shown. Finally, the existence of a fixed point of the proposed method
was proved using the Schauder Fixed Point theorem.
Acknowledgments. R. Mendoza acknowledges the Office of the Chancellor of the
University of the Philippines Diliman, through the Office of the Vice Chancellor
for Research and Development, for funding support through the Ph.D. Incentive
Award.
References
[1] R. Adams, J. Fournier; Sobolev spaces (second edition), Elsevier, 2003.
[2] K. Astala, L. P¨aiv¨arinta; Calderon’s inverse conductivity problem in the plane, Annals of
Mathematics ,163 (2006), 265–299.
[3] D. C. Barber, B. H. Brown; Progress in electrical impedance tomography, SIAM Inverse
Problems in Partial Differential Equations, (1990), 151–164.
EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 29
[4] P. Bochev, R. B. Lehoucq; On the finite element solution of the pure Neumann problem,
SIAM Review 47 (2005), 50–66.
[5] L. Borcea; Electrical impedance tomography, IOP Publishing Inverse Problems ,18 (2002),
R99–R136.
[6] A. P. Calderon; On an inverse boundary value problem, Seminar on Numerical Analysis and
its Applications to Continuum Physics (Rio de Janeiro), (1980), 65–73.
[7] T. F. Chan, H. M. Zhou, R. H. Chan; Continuation method for total variation denoising
problems, Proceeding of SPIE Annual Meeting, 5(1995).
[8] M. Cheney, D. Isaacson, J. Newell; Electrical impedance tomography, SIAM Review 42 (1999),
no. 1, 85–101.
[9] D. Dobson; Convergence of a reconstruction method for the inverse conductivity problem,
SIAM Journal of Applied Mathematics, 52 (1992), 442–458.
[10] D. Dobson, F. Santosa; An image enhancement technique for electrical impedance tomogra-
phy, IOP Publishing Inverse Problems 10 (1994), no. 2, 317.
[11] D. Dobson, C. Vogel; Continuation method for total variation denoising problems, SIAM
Journal on Numerical Analysis, 34 (1997), no. 5, 1779–1791.
[12] P. Drabek, J. Milota; Methods of nonlinear analysis, Birkh¨auser, 2007.
[13] M. R. Eggleston, R. J. Schwabe, D. Isaacson, L.F. Coffin; The application of electric cur-
rent computed tomography to defect imaging in metals, Review of Progress in Quantitative
Nondestructive Evaluation (1990), 455–462.
[14] L. Evans; Partial differential equations, American Mathematical Society, 2010 Second Edi-
tion.
[15] S. F¨urtinger; An approach to computing binary edge maps for the purpose of registering
intensity modulated images, Ph.D. thesis, Karl-Franzens University Graz, 2012.
[16] D. Gilbarg, N. Trudinger; Elliptic partial differential equations of second order, Springer,
Reprint of the 1998 Edition.
[17] G. A. Gray; A variational study of the electrical impedance tomography problem, Ph.D. thesis,
Rice University, 2002.
[18] C. Großmann, H-G Roos, M. Stynes; Numerical treatment of partial differential equations
(universitext), Springer, 2007.
[19] P. Halmos; Measure theory, Springer, 1974.
[20] M. Hinterm¨uller, A. Laurain; Electrical impedance tomography: from topology to shape, Con-
trol and Cybernetics 37 (2008), no. 4.
[21] David Holder (ed.); Electrical impedance tomography: Methods, history and applications
(series in medical physics and biomedical engineering), IOP Publishing, 2004.
[22] D. Isaacson, J. Mueller, S. Siltanen; Biomedical applications of electrical impedance tomog-
raphy, IOP Publishing Physiological Measurement 24 (2003), no. 2.
[23] D. Isaacson, J. Newell, J. C. Goble, M. Cheney; Thoracic impedance images during venti-
lation, Proceedings of the Twelfth Annual International Conference of the IEEE (1-4 Nov,
1990), 106–107.
[24] J. Jordana, M. Gasulla, R. Pallas-Areny; Electrical resistance tomography to detect leaks
from buried pipes, IOP Publishing Measurement Science and Technology 12 (2001), no. 8.
[25] H. Kang; A uniqueness theorem for an inverse boundary value problem in two dimensions,
Journal of Mathematical Analysis and Applications, 270 (2002), no. 1, 291–302.
[26] R. Mendoza, S. Keeling; FEM convergence of a segmentation approach to the electrical
impedance tomography problem, AIP Conference Proceedings 1707 (2016), no. 1, 050009.
[27] R. Mendoza, S. Keeling; A two-phase segmentation approach to the impedance tomography
problem, Inverse Problems, 33 (2017), no. 1, 015001.
[28] R. Mendoza, J. E. Lope; Reconstructing images in electrical impedance tomography problem
using hybrid genetic algorithms, Science Diliman, 242 (2012), no. 2, 50–66.
[29] R. Parker; The inverse problem of resistivity sounding, Geophysics, 49 (1984), 2143–2158.
[30] A. Ramirez, W. Daily, D. Labrecque, E. Owen, D. Chestnut; Monitoring an underground
steam injection process using electrical resistance tomography, Water Resources Research 29
(1993), no. 1, 73–87.
[31] L. I. Rudin; Images, numerical analysis of singularities and shock filters, Ph.D. thesis, Cali-
fornia Institute of Technology, 1987.
[32] L. I. Rudin, S. Osher, E. Fatemi; Nonlinear total variation bases noise removal algorithms,
Physica D 60 (1992), 259–268.
30 R. MENDOZA, S. KEELING EJDE-2020/93
[33] J. Snyman; Practical mathematical optimization: An introduction to basic optimization the-
ory and classical and new gradient-based algorithms, Springer, 2005.
[34] E. Somersalo, M. Cheney, D. Isaacson; Existence and uniqueness for electrode models for
electric current computed tomography, SIAM Journal on Applied Mathematics, 52 (1992),
no. 4, 1023–1040.
[35] J. Sylvester, G. Uhlmann; A global uniqueness theorem for an inverse boundary value prob-
lem, Annals of Mathematics 125 (1987), no. 1, 153–169.
[36] G. Uhlmann; Electrical impedance tomography and calderon’s problem, IOP, 25 (2009).
[37] A. C. Velasco, M. Darbas, R. Mendoza, M. Bacon, J. C. de Leon; Comparative study of
heuristic algorithms for electrical impedance tomography, Philippine Journal of Science 149
(2020), no. 3a, 747–761.
[38] Y. Zuo, Z. Guo; A review of electrical impedance techniques for breast cancer detection,
Elsevier Medical Engineering & Physics, 25 (2003), no. 2, 79–90.
Renier Mendoza
Institute of Mathematics, University of the Philippines, Diliman, Quezon City, Philip-
pines
Email address:rmendoza@math.upd.edu.ph
Stephen Keeling
Institute for Mathematics and Scientific Computing, Karl-Franzens University of Graz,
Austria
Email address:stephen.keeling@uni-graz.at
