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 ⊆ Rn is 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 φ in Ω and the boundary voltage V = φ ∣∣ ∂Ω are solved. These electrical measurements satisfy a generalized Laplace equation: ∇ · (σ∇φ) = 0 in Ω, σ ∂φ∂n = f on ∂Ω, (1.1) where n is the outward normal direction at ∂Ω. The boundary currents are chosen so that ∫ ∂Ω f dS = 0. This condition is imposed to satisfy the conservation of charge. Furthermore, the electric potential φ must satisfy ∫ ∂Ω φdS = 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 σ inside Ω given V and f in ∂Ω. Denote L̃2(∂Ω) := { f ∈ L2(∂Ω) : ∫ ∂Ω f dS = 0 } and define Λσ : L̃2(∂Ω)→ L̃2(∂Ω) by Λσ(f) = φ ∣∣ ∂Ω , (1.2) where φ ∈ H1(Ω) satisfies (1.1) and ∫ ∂Ω φdS = 0. The inverse EIT problem is the recovery of σ given 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 σ is piecewise constant. This assumption is based on the fact that the conductivities of healthy tissues show great contrast [3, 17]. By assuming that σ is piecewise- constant, the inverse problem is treated as a segmentation problem. A segmentation technique called “Multi-Phase Segmentation”, proposed by Fürtinger in [15], is explored in [27]. Moreover, it is assumed that the desired conductivity can be expressed in terms of M phases, i.e., of the form σ(x) = M∑ m=1 σm(x)χm(x), (1.3) where for the mth phase χm is the characteristic function of a subdomain Ωm ⊂ Ω and σm is 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 Ω1 has disjoint non-adjacent components. The subdomains Ω1 and Ω2 form a disjoint partition of Ω, i.e., Ω1 ∩Ω2 = ∅ and Ω = Ω1 ∪Ω2. The conductivity σ2 in Ω2 is assumed to be known and χ2 = 1 − χ1. Therefore, the inverse EIT problem becomes a problem of identifying σ1 and 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 ∈ L̃2(∂Ω) and define the function F : L2(Ω)→ L̃(∂Ω) by F (σ) = φ ∣∣ ∂Ω , (2.1) where φ is the solution of (1.1) given σ and f . Let σ? be the actual conductivity distribution in Ω. The inverse problem is to recover σ? from Λσ? . Suppose f ∈ L̃2(∂Ω) and let V ? = Λσ?(f) = F (σ?) be the exact boundary voltage. Moreover, EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 3 let Ṽ ≈ V ? be the measured boundary voltage. Let σ̃1 be an estimate of σ1. To solve the EIT inverse problem, our aim is to minimize J̃(σ1, χ1) = ∫ ∂Ω |F (σ)− Ṽ |2 dS + ∫ Ω α|∇σ1|2(χ1 + ε) + λ(σ1 − σ̃1)2 dV (2.2) with σ = σ1χ1 + σ2(1 − χ1) and σ2 is given. The first term of the integral is the fidelity term, the second term provides smoothness on σ1 on Ω1 and 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 φ is the solution of (1.1) given f ∈ L̃2(∂Ω) and σ ∈ L∞(Ω). Moreover, let Ṽ be the measured boundary voltage. We define the adjoint solution φ∗ as the solution of ∇ · (σ∇φ∗) = 0 in Ω, σ ∂φ∗ ∂n = F (σ)− Ṽ on ∂Ω. (2.3) Both φ and φ∗ satisfy ∫ ∂Ω φdS = 0 and ∫ ∂Ω φ∗ dS = 0. Using (1.1), (2.3), and (1.3), the variational formulations of the forward and adjoint problems are∫ Ω (σ1χ1 + σ2(1− χ1))∇φ · ∇v dV = ∫ ∂Ω fv dS, (2.4)∫ Ω (σ1χ1 + σ2(1− χ1)∇φ∗ · ∇v dV = ∫ ∂Ω (F (σ1χ1 + σ2(1− χ1))− Ṽ )v dS, (2.5) for all v ∈ H1(Ω). Formulations (2.4) and (2.5) have unique solutions [27] in H̃1(Ω) := {v ∈ H1(Ω)| ∫ ∂Ω v dS = 0}. Because of (1.3), the forward and adjoint solutions φ and φ∗ are dependent on σ1 and χ1 alone. 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 χ1 to the respective solutions φ and φ∗ of (2.4) and (2.5), respectively. Equivalently, Φ : (σ1, χ1) → φ and Φ∗ : (σ1, χ1)→ φ∗. The computation of the derivative of J̃ is necessary to express σ1 in terms of χ1. For a fixed χ1, the variational derivative of J̃ in (2.2) with respect to σ1 ∈ H1(Ω) in the direction of δσ1 ∈ H1(Ω) is given by δJ̃ δσ1 (σ1, χ1; δσ1) = − ∫ Ω 2χ1δσ1∇φ · ∇φ∗ dV + ∫ Ω 2α(χ1 + ε)∇(δσ1) · ∇σ1 dV + ∫ Ω 2λ(σ1 − σ̃1)δσ1 dV. If we equate the above expression to 0, we conclude that the following optimality condition must be satisfied,∫ Ω [α(χ1 + ε)∇σ1 · ∇v + λ(σ1 − σ̃1)v] dV = ∫ Ω (χ1∇φ · ∇φ∗)v dV, (2.6) for all v ∈ H1(Ω). Given χ1 and an estimate σ̃1, the quantity σ1 is 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 → σ1 that 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 σ̃1 and σ1 in (2.6) with σk1 and σk+1 1 = Σ1(χ1), respectively. Hence,∫ Ω α(χ1 + ε)∇Σ1(χ1) · ∇v dV + ∫ Ω λ(Σ1(χ1)− σk1 )v dV = ∫ Ω χ1∇Φ(σk1 , χ1) · ∇Φ∗(σk1 , χ1)v dV. (2.7) Furthermore, the following operators are defined for the global conductivity, Σk(χ1) := σk1χ1 + 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 J̃(σ1, χ1) in (2.2) can be written as a functional J̃(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 TV (f) := ∫ Ω |∇f | dV ≈ ∫ Ω √ |∇f |2 + β2 dV, 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) = ∫ ∂Ω |F (Σ(χ1))− Ṽ |2 dS + ∫ Ω α|∇Σ1(χ1)|2(χ1 + ε) + ∫ Ω λ(Σ1(χ1)− σ̃1)2 + γ √ |∇χ1|2 + β2 dV, (2.10) for α, λ, γ > 0 and ε, β ∈ (0, 1). Thus we find an update for χ1 that reduces the cost J . This update can be obtained using the method of steepest descent [33], which is given in weak form for J as follows,∫ Ω χk+1 1 v dV = ∫ Ω χk1v dV − ω δJ δχ1 (χk1 ; v), (2.11) for all v ∈ H1(Ω), where ω ∈ (0, 1) is the step size and k ∈ N. Let χk1 , δχ k 1 ∈ L2(Ω) and suppose δΣ1 δχk1 (χk1 ; δχk1) ∈ H1(Ω), then the variational derivative of J in (2.10) is δJ δχk1 (χk1 ; δχ1) = ∫ Ω −2(Σ1(χk1)− σ2)δχk1∇Φ(σk1 , χ k 1) · ∇Φ∗(σk1 , χ k 1) dV + ∫ Ω α|∇Σ1(χk1)|2δχk1 dV + γ ∫ Ω ∇(δχk1) · ∇χk1√ |∇χk1 |2 + β2 dV. (2.12) EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 5 Remark 2.4. Observe that (2.12) requires the calculation of ∇χk1 but since χk1 is binary, a smooth approximation of χk1 is necessary. This will be discussed later. The assumption that δΣ1 δχk1 (χk1 ; δχk1) ∈ H1(Ω) can be shown if χk1 is sufficiently smooth (see Theorem 3.28). Instead of performing the iteration (2.11) by evaluating δJ/δχ1(χk1 ; v) explicitly in terms of χk1 , the iteration may be performed semi-implicitly by evaluating part of the variational derivative of J in (2.12) at χk+1 1 as follows:∫ Ω [ vχk+1 1 + ωγ ∇v · ∇χk+1 1√ |∇χk1 |2 + β2 ] dV = ∫ Ω vG(χk1) dV, (2.13) for all v ∈ H1(Ω), where G(χ1) = χ1−ωα|∇Σ1(χ1)|2 + 2ω [ (Σ1(χ1)−σ2)∇Φ(σ̃1, χ1) ·∇Φ∗(σ̃1, 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 χ1 is necessary. To do this, we introduce the kernel function ξδ(x) = 1 4πδ e − x2 4δ , for some δ > 0. We approximate χk1 using the convolution of χk1 and ξδ, i.e., we let χ̃k1 := χk1 ∗ ξδ = ∫ R2 ξδ(x− y)χk1(y)dy. (2.15) The following result gives the regularity and continuity of the above mollification. Theorem 2.5. Let k ∈ N, then χ̃k1 is a real analytic function on Ω and χ̃k1 → χk1 almost everywhere as δ → 0. Furthermore, suppose 0 ≤ χk1 ≤ 1, for all x ∈ Ω. Then 0 ≤ χ̃k1 ≤ 1 [15]. Using (2.15), we approximate (2.13) by∫ Ω ωγ ∇χ̃k+1 1 · ∇v√ |∇χ̃k1 |2 + β2 + χ̃k+1 1 v dV = ∫ Ω G(χ̃k1)v dV, ∀v ∈ H1(Ω). (2.16) Definition 2.6. We define the operator Θ : L2(Ω) → L2(Ω) to be the solution χ̃k+1 1 ∈ L2(Ω) of (2.16) for a given χ̃k1 ∈ L2(Ω). Because the solution of (2.16) is not binary, the update for χ is 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 f and Ṽ . Choose parameters ε, δ, λ, γ, β � 1, α � 1, and ζ, ω ∈ (0, 1). Select the maximum number of iterations K and the tolerance ρ. Set k = 1 and choose the initial σk1 . Select an initial guess χk1 . The value of σ2 is given and χk2 = 1− χk1 . (2) Take χ̃k1 = χk1 ∗ ξδ and χ̃k+1 1 = Θ(χ̃k1). Then, χ1 is updated by χk+1 1 (x) = { 1, if χ̃k+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 = K or ‖χk+1 1 − χk1‖L2(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 χ1 is a characteris- tic function. Because the characteristic function χ1 is binary, we use its smooth approximation (2.15) instead. This is necessary because some of the essential re- sults require χ1 to 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 φ and φ∗ are solved using σk. In this section, we try to understand how a perturbation on σk affects φ and φ∗. Recall that σk depends on χ1. Therefore, φ and φ∗ depend on χ1 as well. Working under the assumption that σ ∈ L∞(Ω) and χ1 is a characteristic function, we show that φ and φ∗ depend continuously on σk and 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 φ and φ∗ when we require additional regularity of σk. Theorem 3.1. Let σk ∈ L∞(Ω) such that 0 < σ ≤ σk(x) for all x ∈ Ω and f ∈ L̃2(∂Ω). Then ( the variational forward EIT problem)∫ Ω σk∇φ · ∇v dV = ∫ ∂Ω σk ∂φ ∂n v dS, ∀v ∈ H1(Ω) (3.1) has a unique solution φ ∈ H1(Ω) with ∫ ∂Ω φdS = 0. Similarly, let Ṽ ∈ L̃2(∂Ω) be the known boundary voltage. Then ( the variational adjoint problem)∫ Ω σk∇φ∗ · ∇v dV = ∫ ∂Ω (φ− Ṽ )v dS, ∀v ∈ H1(Ω) (3.2) has a unique solution φ∗ ∈ H1(Ω) with ∫ ∂Ω φ∗ dS = 0. Proof. We define I := {u ∈ H1(Ω) : ∫ ∂Ω u dS = 0}, a(u, v) := ∫ Ω σk∇u ·∇v dV , and b(v) := ∫ ∂Ω fv dS. Clearly, a is bilinear and b is linear. Using the Cauchy-Schwarz identity, Hölder’s inequality, and the definition of the H1 norm, a is bounded, i.e., |a(u, v)| ≤ ‖σk‖L∞(Ω)‖u‖H1(Ω)‖v‖H1(Ω). Because u ∈ I, then ∫ ∂Ω u dS = 0. Therefore, using the lower bound of σ, and the generalized Friedrich’s inequality [4], we obtain |a(u, u)| ≥ σ‖∇u‖2L2(Ω) = σ 2 ‖∇u‖2L2(Ω) + σ 2 ‖∇u‖2L2(Ω) ≥ σ 2 ( 1 C ‖u‖2L2(Ω) − (∫ ∂Ω u dS )2) + σ 2 ‖∇u‖2L2(Ω) EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 7 = σ 2C ‖u‖2L2(Ω) + σ 2 ‖∇u‖2L2(Ω) ≥ σmin { 1 C , 1 } ‖u‖2H1(Ω), for some C > 0. Finally, by the Trace Theorem [14], b is bounded. Hence, by the Lax-Milgram Theorem ∃! φ ∈ H1(Ω) satisfying (3.1). Similarly, there exists a unique φ∗ ∈ I satisfying (3.2). � Corollary 3.2. Let φ and φ∗ satisfy the variational forward and the variational adjoint problem stated in the previous theorem. We have the following estimates: ‖φ‖H1(Ω) ≤ C1‖f‖L̃2(∂Ω) (3.3) ‖φ∗‖H1(Ω) ≤ C2‖(φ− Ṽ )‖L̃2(∂Ω) (3.4) for some C1, C2 > 0. Note that using the Trace Theorem and the triangle inequality, φ∗ can be further estimated by ‖φ∗‖H1(Ω) ≤ C3 ( ‖f‖L̃2(∂Ω) + ‖Ṽ ‖L̃2(∂Ω) ) (3.5) for some C3 > 0. Throughout this work, we use the following notation. Definition 3.3. We let δσk and δχ1 denote perturbations of σk ∈ L∞(Ω) and χ1 ∈ L∞(Ω), respectively. In Definition (2.2), Φ and Φ∗ are operators that map (σk1 , χ1) to φ and φ∗, respectively. But because σk = σk1χ1 + σ2(1 − χ1), we can make the identifications Φ(σk) = Φ(σk1 , χ1), Φ∗(σk) = Φ∗(σk1 , χ1). Hence, Φ : σk → φ and 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 η sufficiently 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 ‖Φ(σk + ηδσk)‖H1(Ω) ≤ C1‖f‖L̃2(∂Ω), (3.6) ‖Φ∗(σk + ηδσk)‖H1(Ω) ≤ C2(‖f‖L̃2(∂Ω) + ‖Ṽ ‖L̃2(∂Ω)), (3.7) for any η ∈ (0, τ). The result below shows how a perturbation δσk affects φ and φ∗. Theorem 3.5. Let σk, δσk ∈ L∞(Ω). Then there exist C1, C2 > 0 such that ‖Φ(σk + ηδσk)− Φ(σk)‖H1(Ω) ≤ C1η‖δσk‖L∞(Ω), (3.8) ‖Φ∗(σk + ηδσk)− Φ∗(σk)‖H1(Ω) ≤ C2η‖δσk‖L∞(Ω), (3.9) for any η ∈ (0, τ), where τ is chosen according to Remark 3.4. 8 R. MENDOZA, S. KEELING EJDE-2020/93 Proof. From (3.1), we have∫ Ω σk∇Φ(σk) · ∇v dV = ∫ Ω fv dV. (3.10) Similarly, for σk + ηδσk,∫ Ω (σk + ηδσk)∇(Φ(σk) + δφ) · ∇v dV = ∫ Ω fv dV, (3.11) where we denote δφ := Φ(σk + ηδσk)− Φ(σk). Subtracting (3.10) from (3.11), we obtain ∫ Ω σk∇δφ · ∇v dV = − ∫ Ω ηδσk∇Φ(σk + ηδσk) · ∇v dV. (3.12) We define a(u, v) := ∫ Ω σk∇u·∇v dV and b1(v) := − ∫ Ω ηδσk∇Φ(σk+ηδσk)·∇v dV . Clearly, a and b1 are 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), b1 is bounded. Thus, if we take u = v = δφ, use the previous inequality, and the coercivity of a(u, v) we obtain ‖δφ‖H1(Ω) ≤ 1 C̄ C̄1η‖f‖L̃2(∂Ω)‖δσ k‖L∞(Ω), (3.13) where C̄ > 0 is the coercivity constant and C̄1 is 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) := σk1χ1 +σ2(1−χ1) = σk. There- fore, φ and φ∗ depend on χ1 for a fixed σk1 . 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 φ and φ∗ are solved using σk1χ1 +σ2(1−χ1). We now show that φ and φ∗ depend continuously on χ1. Corollary 3.7. Let χ1 ∈ L∞(Ω), then ∃C̄1, C̄2 > 0 such that ‖Φ(χ1 + ηδχ1)− Φ(χ1)‖H1(Ω) ≤ C̄1η‖δχ1‖L∞(Ω), (3.15) ‖Φ∗(χ1 + ηδχ1)− Φ∗(χ1)‖H1(Ω) ≤ C̄2η‖δχ1‖L∞(Ω), (3.16) for any η ∈ (0, τ), where τ is chosen according to Remark 3.4. Proof. We use the decomposition Σk(χ1) = σk1χ1 + σ2(1− χ1). (3.17) Let η ∈ (0, τ). If we use χ1 + ηδχ1 instead of χ1, we have Σk(χ1) + δσk = Σk(χ1) + η(σk1 − σ2)δχ1, (3.18) where δσk is the associated change in σk given a ηδχ1 perturbation of χ1. Sub- tracting (3.17) from (3.18), we obtain δσk = η(σk1 − σ2)δχ1. (3.19) The inequalities we need to show directly follow from Theorem (3.5) and ‖δσk‖L∞(Ω) ≤ η‖σk1 − σ2‖L∞(Ω)‖δχ1‖L∞(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∫ Ω α(χ1 + ε)∇σk+1 1 · ∇v dV + ∫ Ω λ(σk+1 1 − σk1 )v dV = ∫ Ω 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) := ∫ Ω α(χ1 + ε)∇σk+1 1 · ∇v dV + ∫ Ω λσk+1 1 v dV, b(v) := ∫ Ω λσk1v dV + ∫ Ω 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 ‖∇Φ(χ1)‖L∞(Ω′) can be bounded by ‖∇Φ(χ1)‖L2(Ω) for some Ω′ compactly embedded in Ω. This was proven under the assumption that σk ∈ C1(Ω̄). Recall that σk = σk1χ1 +σ2(1−χ1). Clearly, σk is not necessarily in C1(Ω̄) because χ1 is a characteristic function. We have introduced a mollification χδ1 of χ1 in (2.15) to resolve this. Thus, σk1 ∈ C∞(Ω̄). Moreover, σk is not just in C1(Ω̄) but in C∞(Ω̄) as well. This might seem like a deviation from our proposed method but technically, we can choose δ to be extremely close to 0 so that χδ1 is a good approximation of χ1. We show that the mollification affects the corresponding φ and φ∗ to a very small extent as long as the distance between χδ1 and χ1 is 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 Tδ is 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)→ ∂νg almost 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 χδ1 := χ1 ∗ ξδ denote the mollification of χ1. We now show how the perturbation of χ1 affects 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 χδ1 satisfy ‖Φ(χδ1)− Φ(χ1)‖H1(Ω) ≤ Cδ1‖χδ1 − χ1‖L∞(Ω) (3.23) for some C1 > 0. The same applies to the adjoint problem ‖Φ∗(χδ1)− Φ∗(χ1)‖H1(Ω) ≤ Cδ2‖χδ1 − χ1‖L∞(Ω) (3.24) 10 R. MENDOZA, S. KEELING EJDE-2020/93 for some C2 > 0. Proof. We proved in (3.15) that φ depends 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 χδ1 ∈ L∞(Ω) by choosing p = 2 and r = ∞. Finally, Theorem (3.5) proves the rest of our claim. � Remark 3.11. The superscript δ in Cδ1 and Cδ2 from (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 χδ1 converges to χ1 pointwise [15]. Although it does not guarantee that ‖χδ1−χ1‖L∞(Ω) converges to 0, this can still be a gauge to measure the distance between the solution of the forward problem using χ1 and the solution using χδ1. 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 A1 and A2 to be A1 4A2 := (A1\A2) ∪ (A2\A1). Definition 3.12. Let the distance d : B(Ω) × B(Ω) → R ∪ {∞} be d(A1, A2) = µ(A1 4A2). 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)(χk1). (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 J in (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 χδ1 instead of χ1 so that we are dealing with a smooth function rather than a characteristic function. Thus, we are now solving σk+1 1 using the equation∫ Ω α(χδ1 + ε)∇σk+1 1 · ∇v dV + ∫ Ω λ(σk+1 1 − σk1 )v dV = ∫ Ω χδ1∇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 σk1 ∈ C∞(Ω̄) such that Σk(χδ1) ≥ σ > 0. We also assume that ∂Ω is sufficiently smooth. (3.30) The assumption that σk1 ∈ C∞(Ω̄) might seem like a strong assumption but if we can show that σk+1 1 ∈ C∞(Ω̄) as well, then this assumption makes sense. Moreover, the initial guess for σ1 in our algorithm can be chosen to be constant throughout Ω so that it is in C∞(Ω̄). We now investigate what happens to φ and φ∗ if we use the above assumption. Lemma 3.15. Under assumption (3.30), ‖∇φ‖L∞(Ω) ≤ Cδ‖φ‖H1(Ω), (3.31) for some Cδ > 0 (compare with [9]). In fact, φ ∈ C∞(Ω̄). Proof. By assumption (3.30) and Lemma (2.5), σk ∈ C∞(Ω̄). Let l ≥ 1. Obviously, σk ∈ Cl(Ω̄). Therefore, using standard regularity estimates (see, e.g., [16]), we obtain ‖φ‖Hl+2(Ω) ≤ C1‖φ‖H1(Ω), (3.32) for some C1 > 0. Furthermore, by the Sobolev imbedding theorem [14], we have ‖φ‖Cl,γ(Ω̄) ≤ C2‖φ‖Hl+2(Ω). (3.33) By the definition of ‖ · ‖Cl,γ(Ω̄), the embedding C1,γ(Ω̄) ↪→ C1(Ω̄) (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 ‖∇φ‖L∞(Ω) ≤ C‖∇φ‖H1(Ω), (3.35) which completes the first part of the proof. Moreover, because l ≥ 1, it follows that φ(χδ1) ∈ C∞(Ω̄). � Lemma 3.16. Under assumption (3.30), there exists Cδ > 0 such that ‖∇φ∗‖L∞(Ω) ≤ Cδ‖φ∗‖H1(Ω). (3.36) Furthermore, φ∗ ∈ C∞(Ω̄). 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 φ and φ∗ on χδ1. From here onwards, we denote δχδ1 := δχ1 ∗ ξδ so that (χ1 + ηδχ1) ∗ ξδ = χδ1 + ηδχδ1. Suppose we replace χδ1 with χδ1 + ηδχδ1. Again by Remark 3.4, η should be taken from the set (0, τ) where τ is chosen so that σk + ηδσk ≥ στ > 0. Therefore, from (3.6), (3.7), (3.14), (3.31), and (3.36), the following estimates hold ‖∇Φ(χδ1 + ηδχδ1)‖L∞(Ω) ≤ C1‖f‖L̃2(∂Ω), (3.37) ‖∇Φ∗(χδ1 + ηδχδ1)‖L∞(Ω) ≤ C2 ( ‖f‖L̃2(∂Ω) + ‖Ṽ ‖L̃2(∂Ω) ) , (3.38) for some C1, C2 > 0 and for all η ∈ (0, τ). In Corollary 3.7, we have shown that φ and φ∗ depend continuously on χ1. Observe that this theorem holds with any χ1 whose value is between 0 and 1. Therefore, this also holds when we use χδ1 instead because 0 ≤ χδ1 ≤ 1 as proven in Lemma (2.5). We state this in the following lemma. Lemma 3.17. Under assumption (3.30), there exists Cδ1 , C δ 2 > 0 such that ‖Φ(χδ1 + ηδχδ1)− Φ(χδ1)‖H1(Ω) ≤ Cδ1η‖δχ1‖L2(Ω), (3.39) ‖Φ∗(χδ1 + ηδχδ1)− Φ∗(χδ1)‖H1(Ω) ≤ Cδ2η‖δχ1‖L2(Ω), (3.40) for any η ∈ (0, τ), where τ is chosen according to Remark 3.4. We define ψ(χδ1) := ∇Φ(χδ1) · ∇Φ∗(χδ1). (3.41) Observe that the right-hand side of (3.29) includes ψ(χδ1). In the next lemma, we show that ψ depends continuously on χ1. This will be necessary when we analyze the solution of (3.29). Note that ∇Φ∗(χδ1) ∈ L2(Ω) by Corollary 3.2 and ∇Φ(χδ1) ∈ L∞(Ω) by (3.31). Therefore, by Hölder’s inequality, ψ(χδ1) ∈ L2(Ω). (3.42) This means that ψ is a map from χ1 ∈ L2(Ω) to ∇Φ(χδ1) · ∇Φ∗(χδ1) ∈ L2(Ω). In the following lemma, we prove that this mapping is continuous. Lemma 3.18. Under assumption (3.30), there exists Cδ ≥ 0 such that ‖ψ(χδ1 + ηδχδ1)− ψ(χδ1)‖L2(Ω) ≤ Cδη‖δχ1‖L2(Ω), (3.43) for any η ∈ (0, τ), where τ is chosen according to Remark 3.4. Proof. Adding and subtracting ∇Φ(χδ1 + ηδχδ1) ·∇Φ∗(χδ1) to ψ(χδ1 + ηδχδ1)−ψ(χδ1), we obtain ψ(χδ1 + ηδχδ1)− ψ(χδ1) =: A1(χδ1; δχδ1) +A2(χδ1; δχδ1), where A1(χδ1; δχδ1) = ∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1 + ηδχδ1)−∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1), A2(χδ1; δχδ1) = ∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1)−∇Φ(χδ1) · ∇Φ∗(χδ1). Thus ‖ψ(χδ1 + ηδχδ1)− ψ(χδ1)‖L2(Ω) ≤ ‖A1(χ1; δχ1)‖L2(Ω) + ‖A2(χ1; δχ1)‖L2(Ω). (3.44) EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 13 We can estimate A1(χδ1; δχδ1) using the Hölder’s inequality, (3.37), and (3.40). Thus, ‖A1(χδ1; δχδ1)‖L2(Ω) ≤ C1η‖f‖L̃2(∂Ω)‖δχ1‖L2(Ω), for some C1 > 0. Similarly, using the Hölder’s inequality, (3.36), and (3.39), we obtain ‖A2(χδ1; δχδ1)‖L2(Ω) ≤ C2η‖∇Φ∗(χδ1)‖L∞(Ω)‖δχ1‖L2(Ω), for some C2 > 0. The estimates for A1(χδ1; δχδ1) and A2(χδ1; δχδ1), 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 (χδ1; δχδ1) ∈ H1(Ω). To accomplish this, we first need to show that the derivatives of both Φ and Φ∗ with respect to χ1 converge 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 ∫ Ω σk∇Dφ(χ1; δχ1) · ∇v dV = ∫ Ω (σk1 − σ2)δχδ1∇Φ(χδ1) · ∇v dV (3.45) with ∫ ∂Ω Dφ(χ1; δχ1) dS = 0, for all v ∈ H1(Ω), such that ∫ ∂Ω v dS = 0. Proof. Let u, v ∈ H1(Ω) such that ∫ ∂Ω u dS = ∫ ∂Ω v dS = 0. We define a(u, v) := ∫ Ω σk∇u · ∇v dV, b(v) := ∫ Ω (σk1 − σ2)δχδ1∇Φ(χδ1) · ∇v dV. From the proof of Theorem (3.1), a is bilinear, coercive and bounded. Obviously, b is linear. We wish to employ the Lax-Milgram theorem so it is sufficient to show that b is bounded. Indeed, using the Cauchy-Schwarz inequality and the Hölder’s inequality we obtain |b(v)| ≤ ‖(σk1 − σ2)δχδ1‖L∞(Ω)‖∇Φ(χδ1)‖L2(Ω)‖v‖L2(Ω). The right-hand side of the last inequality is bounded because of Corollary (3.2) and the fact that (σk1 − σ2)δχδ1 ∈ C∞(Ω̄). � Lemma 3.20. Under assumption (3.30), there exists Dφ∗(χ1; δχ1) ∈ H1(Ω) satis- fying ∫ Ω σk∇Dφ∗(χ1; δχ1) · ∇v dV = ∫ Ω (σk1 − σ2)δχδ1∇Φ∗(χδ1) · ∇v dV + ∫ ∂Ω Dφ(χ1; δχ1)v dV, (3.46) with ∫ ∂Ω Dφ∗(χ1; δχ1) dS = 0, for all v ∈ H1(Ω), such that ∫ ∂Ω v 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 χ1 in 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(χδ1 + ηδχδ1)− Φ(χδ1) η −Dφ(χδ1; δχδ1) ∥∥ H1(Ω) = 0, (3.47) 14 R. MENDOZA, S. KEELING EJDE-2020/93 lim η→0 ∥∥Φ∗(χδ1 + ηδχδ1)− Φ∗(χδ1) η −Dφ∗(χ δ 1; δχδ1) ∥∥ H1(Ω) = 0. (3.48) Thus, we can make the identifications δΦ δχ1 (χδ1; δχδ1) = Dφ(χ1; δχ1), δΦ∗ δχ1 (χδ1; δχδ1) = Dφ∗(χ1; δχ1). Furthermore, because Dφ(χ1; δχ1), Dφ∗(χ1; δχ1) ∈ H1(Ω) we have δΦ δχ1 (χδ1; δχδ1), δΦ∗ δχ1 (χδ1; δχδ1) ∈ H1(Ω), with ∫ ∂Ω δΦ δχ1 (χδ1; δχδ1) dS = ∫ ∂Ω δΦ∗ δχ1 (χδ1; δχδ1) dS = 0. Proof. From (3.1), we have ∫ Ω σk∇Φ(χδ1) · ∇v dV = ∫ Ω fv dV , where σk = Σk(χδi ). Then we obtain ∫ Ω σk∇Φ(χδ1) · ∇v dV = ∫ Ω fv dV. (3.49) Similarly, for χδ1 + ηδχδ1,∫ Ω [σk + η(σk1 − σ2)δχδ1]∇Φ(χδ1 + ηδχδ1) · ∇v dV = ∫ Ω fv dV. (3.50) Subtracting (3.49) from (3.50), we obtain∫ Ω σk∇(Φ(χδ1 + ηδχδ1)− Φ(χδ1)) · ∇v dV = ∫ Ω η(σk1 − σ2)δχδ1∇Φ(χδ1 + ηδχδ1) · ∇v dV. (3.51) Dividing by η, we have∫ Ω σk∇ (Φ(χδ1 + ηδχδ1)− Φ(χδ1) η ) · ∇v dV = ∫ Ω (σk1 − σ2)δχδ1∇Φ(χδ1 + ηδχδ1) · ∇v dV. (3.52) Recall from (3.45) that∫ Ω σk∇Dφ(χδ1; δχδ1) · ∇v dV = ∫ Ω (σk1 − σ2)δχδ1∇Φ(χδ1) · ∇v dV. (3.53) Subtracting (3.52) and (3.53), we obtain∫ Ω σk∇ (Φ(χδ1 + ηδχδ1)− Φ(χδ1) η −Dφ(χδ1; δχδ1) ) · ∇v dV =: A(v) (3.54) where A(v) = ∫ Ω (σk1 − σ2)δχδ1∇[Φ(χδ1 + ηδχδ1)− Φ(χδ1)] · ∇v dV, which can be estimated using the Cauchy-Schwarz inequality and (3.15): |A(v)| ≤ C1η‖(σk1 − σ2)δχδ1‖L∞(Ω)‖δχ1‖L2(Ω)‖v‖H1(Ω). (3.55) EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 15 It is worth noting that (σk1 − σ2)δχδ1 ∈ L∞(Ω) and δχ1 ∈ L2(Ω) so that the right- hand side of the above inequality is bounded. Now observe that a(u, v) := ∫ Ω σk∇u· ∇v dV is coercive as demonstrated in the proof of Theorem (3.1), i.e., |a(u, u)| ≥ C̄‖u‖2H1(Ω), (3.56) for some C̄ > 0, for any u ∈ H1(Ω) such that ∫ ∂Ω u dS = 0. Hence, the left hand side of (3.54) is bounded from above if we set u = Φ(χδ1 + ηδχδ1)− Φ(χδ1) η −Dφ(χδ1; δχδ1). Using this fact and comparing (3.54) and (3.55), we obtain the estimate C̄‖Φ(χδ1 + ηδχδ1)− Φ(χδ1) η −Dφ(χδ1; δχδ1)‖H1(Ω) ≤ C1η‖(σk1 − σ2)δχδ1‖L∞(Ω)‖δχ1‖L2(Ω). Therefore, taking the limit of the last equality as η → 0, we obtain lim η→0 ‖Φ(χδ1 + ηδχδ1)− Φ(χδ1) η −Dφ(χδ1; δχδ1)‖H1(Ω) = 0. (3.57) The rest of the proof is similar to show the convergence of the derivative of Φ∗ with respect to χ1 in 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 ψ in (3.41) has a derivative with respect to χ1 which converges in L2(Ω). We first prove the following lemma. Lemma 3.22. Under assumption (3.30), we let δφ := Φ(χδ1 + ηδχδ1)− Φ(χδ1). (3.58) Then there exists Cδ > 0 such that ‖∇δφ‖L∞(Ω) ≤ Cδ { ‖∇δφ‖L2(Ω) + η‖∇ · ((σk1 − σ2)δχδ1∇φ(χδ1 + ηδχδ1))‖H1(Ω) } , (3.59) for any η ∈ (0, τ), where τ is chosen according to Remark 3.4. Proof. Set δσk = Σk(χδ1 + ηδχδ1)− Σk(χδ1). (3.60) Then δφ and δσ satisfy ∇ · (Σk(χδ1)∇(δφ)) = −∇ · (δσk∇Φ(χδ1 + ηδχδ1)) on Ω, σ ∂(δφ) ∂n = 0 on ∂Ω. By assumption (3.30) and since δχδ1 is a mollification of δχ1, we deduce that ∇ · (δσk∇Φ(χδ1 + ηδχδ1)) ∈ H1(Ω). Because χδ1, σ k 1 ∈ C∞(Ω̄), then Σk(χδ1) ∈ C∞(Ω̄). Thus, σk(χδ1) ∈ C1(Ω̄). Using standard regularity estimate (see, e.g., [16]), ‖δφ‖H3(Ω) ≤ C1(‖δφ‖H1(Ω) + ‖∇ · (δσk∇Φ(χδ1 + ηδχδ1))‖H1(Ω)), (3.61) for some C1 > 0. Furthermore, by the Sobolev imbedding theorem, we have ‖δφ‖C1,γ(Ω̄) ≤ C2‖δφ‖H3(Ω). (3.62) 16 R. MENDOZA, S. KEELING EJDE-2020/93 By the definition of ‖ · ‖Cl,γ(Ω̄), the embedding C1,γ(Ω̄) ↪→ C1(Ω̄) (3.63) is continuous (see, e.g., [18]). If we compare (3.61), (3.62), and (3.63), we can deduce that ∃C̄ > 0 such that ‖∇δφ‖L∞(Ω) ≤ C2 ( ‖δφ‖H1(Ω) + ‖∇ · (δσk∇Φ(χδ1 + ηδχδ1))‖H1(Ω) ) . (3.64) Observe that δσk = η(σk1 − σ2)δχδ1. Therefore, ‖∇ · (δσk∇Φ(χδ1 + ηδχδ1))‖H1(Ω) = η‖∇ · ((σk1 − σ2)δχδ1∇Φ(χδ1 + ηδχδ1))‖H1(Ω). (3.65) Recall that all solutions of the forward problem have zero boundary integral. Thus,∫ ∂Ω δφ dS = ∫ ∂Ω Φ(χδ1 + ηδχδ1)− Φ(χδ1) dS = 0− 0 = 0. Using this and the generalized Friedrich’s inequality, we obtain ‖∇δφ‖2L2(Ω) = 1 2 ‖∇δφ‖2L2(Ω) + 1 2 ‖∇δφ‖2L2(Ω) ≥ C3 2 ‖δφ‖2L2(Ω) − 1 2 (∫ ∂Ω δφ dS )2 + 1 2 ‖∇δφ‖2L2(Ω) = C3 2 ‖δφ‖2L2(Ω) + 1 2 ‖∇δφ‖2L2(Ω) ≥ min {C3 2 , 1 2 } ‖δφ‖2H1(Ω) (3.66) for some C3 > 0. Using (3.66) and (3.65), (3.64) becomes ‖∇δφ‖L∞(Ω) ≤ C2 { ‖δφ‖H1(Ω) + ‖∇ · (δσk∇Φ(χδ1 + ηδχδ1))‖H1(Ω) } ≤ C2 { 1√ min{C3 2 , 1 2} ‖∇δφ‖L2(Ω) + η‖∇ · ((σk1 − σ2)δχδ1∇Φ(χδ1 + ηδχδ1))‖H1(Ω) } ≤ 2C2 max { 1√ min{C3 2 , 1 2} , 1 }{ ‖∇δφ‖L2(Ω) + η‖∇ · ((σk1 − σ2)δχδ1∇Φ(χδ1 + ηδχδ1))‖H1(Ω) } . Because of (3.58) and (3.60), our claim immediately follows from the above inequal- ity. � Note that ∇Φ(χδ1) ∈ L∞(Ω̄),∇Φ∗(χδ1) ∈ L∞(Ω), ∇ δΦ δχδ1 (χδ1; δχδ1),∇δΦ ∗ δχδ1 (χδ1; δχδ1) ∈ L2(Ω). Therefore, ∇Φ(χδ1) · ∇ δΦ∗ δχ1 (χδ1; δχδ1) +∇ δΦ δχ1 (χδ1; δχδ1) · ∇Φ∗(χδ1) ∈ L2(Ω). We show in the next lemma that this is in fact the derivative of ψ defined in (3.41). EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 17 Lemma 3.23. Under assumption (3.30), we have lim η→0 ‖ψ(χδ1 + ηδχδ1)− ψ(χδ1) η − δψ δχ1 (χδ1; δχδ1)‖L2(Ω) = 0 (3.67) with δψ δχ1 (χδ1; δχδ1) := ∇Φ(χδ1) · ∇δΦ ∗ δχ1 (χδ1; δχδ1) +∇ δΦ δχ1 (χδ1; δχδ1) · ∇Φ∗(χδ1). Proof. For any perturbation δχ1 of χ1, we have ψ(χδ1 + ηδχδ1)− ψ(χδ1) η = ∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1 + ηδχδ1) η − ∇Φ(χδ1) · ∇Φ∗(χδ1) η = ∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1 + ηδχδ1)−∇Φ∗(χδ1) η + ∇Φ(χδ1 + ηδχδ1)−∇Φ(χδ1) η · ∇Φ∗(χδ1). (3.68) To continue with our proof, we first perform some convenient calculations. By adding and subtracting a term, the following is obtained: ∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1 + ηδχδ1)−∇Φ∗(χδ1) η −∇Φ(χδ1) · ∇δΦ ∗ δχδ1 (χδ1; δχδ1) := A1 +A2, (3.69) where A1 = ∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1 + ηδχδ1)−∇Φ∗(χδ1) η −∇Φ(χδ1 + ηδχδ1) · ∇δΦ ∗ δχ1 (χδ1; δχδ1), A2 = ∇Φ(χδ1 + ηδχδ1) · δ∇Φ∗ δχ1 (χδ1; δχδ1)−∇Φ(χδ1) · ∇δΦ ∗ δχ1 (χδ1; δχδ1). Using the Hölder’s inequality and (3.37), we obtain ‖A1‖L2(Ω) ≤ C1‖f‖L̃2(∂Ω)‖ ∇Φ∗(χ1 + ηδχ1)−∇Φ∗(χ1) η −∇δΦ ∗ δχ1 (χδ1; δχδ1)‖L2(Ω), for some C1 > 0. Recall from (3.39) that ‖Φ(χδ1 + ηδχδ1)− Φ(χδ1)‖H1(Ω) ≤ C2η‖δχ1‖L2(Ω), (3.70) for some C2 > 0. Using the Cauchy-Schwarz inequality, (3.48), and (3.59), we obtain ‖A2‖L2(Ω) ≤ C3η { C4‖δχ1‖L2(Ω) + ‖∇ · ([σk1 − σ2]δχδ1∇Φ(χδ1 + ηδχδ1))‖H1(Ω) } × ‖∇δΦ ∗ δχ1 (χδ1; δχδ1)‖H1(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(χδ1 + ηδχδ1)− ψ(χδ1) η = 3∑ i=1 Ai. Using Hölder’s inequality and (3.36), ‖A3‖L2(Ω) ≤ ‖∇Φ(χδ1 + ηδχδ1)−∇Φ(χδ1) η −∇ δΦ δχ1 (χδ1; δχδ1)‖L2(Ω)‖∇Φ∗(χδ1)‖L∞(Ω). Comparing (3.68), (3.69) and (3.71) we obtain ‖ψ(χδ1 + ηδχδ1)− ψ(χδ1) η − δψ δχ1 (χδ1; δχδ1)‖L2(Ω) ≤ 3∑ i=1 ‖Ai‖, 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 η goes to 0. From the estimate of the L2-norm of A2 above, we can see that ‖A2‖L2(Ω) → 0. From (3.47) and (3.48), we can deduce that lim η→0 ‖∇Φ(χδ1 + ηδχδ1)−∇Φ(χδ1) η −∇ δΦ δχδ1 (χδ1; δχδ1)‖L2(Ω) = 0, lim η→0 ‖∇Φ∗(χδ1 + ηδχδ1)−∇∗Φ(χδ1) η −∇δΦ ∗ δχδ1 (χδ1; δχδ1)‖L2(Ω) = 0. These imply that ‖A1‖L2(Ω), ‖A3‖L2(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 1 depends on φ and φ∗. Hence, we can investigate how the mollification of χ1 affects σk+1 1 . We show that σk+1 1 continuously depends on χ1. Furthermore, we prove that δΣ1 δχ1 (χδ1; δχδ1) ∈ H1(Ω). We start by equipping H1(Ω) with a suitable norm. Proposition 3.24. Under assumption (3.30) we define |v|2H1(Ω) := α ∫ Ω (χ+ ε)|∇v|2 dV + θλ ∫ v2 dV, where χ(x) ∈ [0, 1] for all x ∈ Ω, and let ‖ · ‖H1(Ω) be the standard H1(Ω) norm. Then | · |H1(Ω) and ‖ · ‖H1(Ω) are equivalent. Proof. Observe that min{αε, λ}‖v‖2H1(Ω) ≤ α ∫ Ω ε|∇v|2 dV + θ ∫ Ω |v|2 dV ≤ α ∫ Ω (χ+ ε)|∇v|2 dV + λ ∫ Ω |v|2 dV = |v|2H1(Ω). On the other hand, |v|2H1(Ω) = α ∫ Ω (χ+ ε)|∇v|2 dV + λ ∫ Ω |v|2 dV ≤ α ∫ Ω (1 + ε)|∇v|2 dV + λ ∫ Ω |v|2 dV EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 19 ≤ max {α(1 + ε), λ} ‖v‖2H1(Ω). � We now establish the regularity of σk+1 1 . Lemma 3.25. Under assumption (3.30), the variational formulation∫ Ω α(χδ1 + ε)∇σk+1 1 · ∇v dV + ∫ Ω λ(σk+1 1 − σk1 )v dV = ∫ Ω χδ1∇Φ(χδ1) · ∇Φ∗(χδ1)v dV (3.72) for all v ∈ H1(Ω), has a unique solution σk+1 1 ∈ H1(Ω). Furthermore, ‖σk+1 1 ‖H1(Ω) ≤ 2 √ µ(Ω) max{Cδ1 , Cδ2} min{αε, λ} (3.73) where Cδ1 = ‖∇Φ(χδ1)‖L∞(Ω)‖∇Φ∗(χδ1)‖L∞(Ω) and Cδ2 = λ‖σk1‖L∞(Ω). Proof. Let u, v ∈ H1(Ω) and define a(u, v) = ∫ Ω α(χδ1 + ε)∇u · ∇v dV + ∫ Ω λuv dV and b(v) = ∫ Ω (χδ1)∇Φ(χδ1) · ∇Φ∗(χδ1)v dV + ∫ Ω λσk1v dV . It is obvious that a is bilinear and b is 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)| = ∫ Ω α(χδ1 + ε)|∇u|2V + ∫ Ω λu2 dV = |u|2H1(Ω) ≥ min{αε, θ}‖u‖2H1(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 σk1 ∈ H1(Ω) satisfying (3.72) for all v ∈ H1(Ω). The H1(Ω) bound for σk+1 1 directly follows. � Remark 3.26. Given any perturbation δχδ1 and η > 0, we have to make sure that the quantity Σ1(χδ1 + ηδχδ1) is well-defined. As shown in the proof of Lemma (3.25), replacing χδ1 with χδ1 + ηδχδ1 cannot be done with just any η. To make sure that the bilinear functional a is coercive, χδ1 + ηδχδ1 + ε must be positive. Since χδ1 + ε > 0, we can choose η small enough so that χδ1 + ηδχδ1 + ε > 0 is satisfied. Hence, similar to Remark 3.4, we take η from the set (0, τ̄) for τ̄ sufficiently small so that Σ1(χδ1 +ηδχδ1) makes sense. Therefore, combining (3.37), (3.38) and (3.73), there exists C > 0 such that ‖Σ1(χδ1 + ηδχδ1)‖H1(Ω) <∞, (3.74) for any η ∈ (0, τ̂), where τ̂ = min {τ, τ̄}. Because χδ1 is a mollification of χ1, it is real analytic. Then χδ1 + ε ∈ C∞(Ω̄). Moreover, from Lemma (3.15), ∇Φ(χδ1),∇Φ∗(χδ1) ∈ C∞(Ω̄). This means that χδ1∇Φ(χδ1) · ∇Φ∗(χδ1), χδ1 + ε, λ are all in C∞(Ω̄). Then using standard regularity estimates [16] on −α∇ · [(χδ1 + ε)∇σk+1 1 ] + λ(σk+1 1 − σk1 ) = χδ1∇Φ(χδ1) · ∇Φ∗(χδ1) 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∞(Ω̄). Consequently, ‖∇Σ1(χδ1)‖L∞(Ω) <∞. Furthermore, ‖∇Σ1(χδ1 + ηδχδ1)‖L∞(Ω) <∞, (3.76) for any η ∈ (0, τ̂), where τ̂ = min {τ, τ̄}, τ and τ̄ are chosen according to Remark 3.4 and Remark 3.26, respectively. Before we can show that δΣ1 δχ1 (χδ1; δχδ1) ∈ 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∫ Ω α(χδ1 + ε)∇Dσ(χδ1; δχδ1) · ∇v dV + ∫ Ω λDσ(χδ1; δχδ1)v dV = ∫ Ω χδ1 δψ δχ1 (χδ1; δχδ1)v + δχδ1∇Φ(χδ1) · ∇Φ∗(χδ1)v − αδχδ1∇(Σ1(χδ1)) · ∇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 (χδ1; δχδ1) ∈ H1(Ω). We prove that this is exactly Dσ(χδ1; δχδ1) computed in the previous lemma. Theorem 3.28. Under assumption (3.30), there exists Cδ > 0 such that ‖Σ1(χδ1 + ηδχδ1)− Σ1(χδ1)‖H1(Ω) ≤ Cδη‖δχ1‖L2(Ω) (3.78) for any η ∈ (0, τ̂), where τ̂ = min {τ, τ̄}, τ and τ̄ are both chosen according to Remark 3.4 and Remark 3.26, respectively. Also, lim η→0 ‖Σ1(χδ1 + ηδχδ1)− Σ1(χδ1) η −Dσ(χδ1; δχδ1)‖ = 0. (3.79) Furthermore, we make the identification δΣ1 δχ1 (χδ1; δχδ1) = Dσ(χδ1; δχδ1) ∈ H1(Ω). Proof. Let v ∈ H1(Ω) and η > 0. From (3.72), we have∫ Ω α(χδ1 + ε)∇Σ1(χδ1) · ∇v + λ(Σ1(χδ1)− σk1 )v dV = ∫ Ω χδ1∇Φ(χδ1) · ∇Φ∗(χδ1)v dV. (3.80) Similarly, if we replace χδ1 in (3.72) with χδ1 + ηδχδ1, we have∫ Ω α(χδ1 + ηδχδ1 + ε)∇(Σ1(χδ1 + ηδχδ1)) · ∇v dV + ∫ Ω λ(Σ1(χδ1 + ηδχδ1)− σk1 )v dV =: A(v), (3.81) where A(v) = ∫ Ω (χδ1 + ηδχδ1)∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1 + δχδ1)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) := ∫ Ω α(χδ1 + ε)∇[Σ1(χδ1 + ηδχδ1)− Σ1(χδ1)] · ∇v dV, EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 21 A2(v) := ∫ Ω λ(Σ1(χδ1 + ηδχδ1)− Σ1(χδ1))v dV, B1(v) := ∫ Ω χδ1[∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1 + ηδχδ1)−∇Φ(χδ1) · ∇Φ∗(χδ1)]v dV, B2(v) := ∫ Ω ηδχδ1∇Φ(χδ1 + ηδχδ1) · ∇Φ∗(χδ1 + ηδχδ1)v dV, B3(v) := − ∫ Ω αηδχδ1∇(Σ1(χδ1 + ηδχδ1)) · ∇v dV, for all v ∈ H1(Ω). We define a(Σ1(χδ1 + ηδχδ1) − Σ1(χδ1), 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 a is bilinear, coercive, and continuous. Clearly, b is linear. Now we only need to show that b is 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, Hölder’s inequality, and Young’s inequality for convolutions to show continuity. It is worth noting that ‖∇Φ(χδ1 + ηδχδ1)‖L∞(Ω), ‖∇Φ∗(χδ1 + ηδχδ1)‖L∞(Ω), and the quantity ‖Σ1(χδ1 +ηδχδ1)‖H1(Ω) are independent of η as shown in (3.37), (3.38), and (3.74). Combining these implies that b(v) is bounded. We make the substitution u = v = Σ1(χδ1 + ηδχδ1)−Σ1(χδ1). Furthermore, using the coercivity of a and the boundedness of b, we conclude that ‖Σ1(χδ1 + ηδχδ1)− Σ1(χδ1)‖H1(Ω) ≤ Cη‖δχ1‖L2(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) = ∫ Ω α(χδ1 + ε)∇ [Σ1(χδ1 + ηδχδ1)− Σ1(χδ1) η −Dσ(χδ1; δχδ1) ] · ∇v dV, D2(v) = ∫ Ω λ [Σ1(χδ1 + ηδχδ1)− Σ1(χδ1) η −Dσ(χδ1; δχδ1) ] v dV, E1(v) = ∫ Ω χδ1 [ψ(χδ1 + ηδχδ1)− ψ(χδ1) η − δψ δχδ1 (χδ1; δχδ1) ] v dV, E2(v) = ∫ Ω δχδ1 [ ψ(χδ1 + ηδχδ1)− ψ(χδ1) ] v dV, E3(v) = − ∫ Ω αδχδ1∇ [ Σ1(χδ1 + ηδχδ1)− Σ1(χδ1) ] · ∇v dV, and ψ is the function defined in (3.41). Using the Cauchy-Schwarz inequality, Hölder’s inequality, (3.43), and (3.83), we can estimate E1(v), E2(v), and E3(v). Indeed, |E1(v)| ≤ ‖ψ(χδ1 + ηδχδ1)− ψ(χδ1) η − δψ δχ1 (χδ1; δχδ1)‖L2(Ω)‖v‖L2(Ω). Similarly, |E2(v)| ≤ C2η‖δχδ1‖L∞(Ω)‖δχ1‖L2(Ω)‖v‖H1(Ω), 22 R. MENDOZA, S. KEELING EJDE-2020/93 for some C2 > 0. Finally, |E3(v)| ≤ α‖δχδ1‖L∞(Ω)‖∇[Σ1(χδ1 + ηδχδ1)− Σ1(χδ1)]‖L2(Ω)‖∇v‖L2(Ω) ≤ Cη‖δχδ1‖L∞(Ω)‖δχ1‖L2(Ω)‖v‖H1(Ω). If we make the substitution v = Σ1(χδ1 + ηδχδ1)− Σ1(χδ1) η −Dσ(χδ1; δχδ1), then by Proposition 3.24, we obtain |D1(v) +D2(v)| = ∣∣Σ1(χδ1 + ηδχδ1)− Σ1(χδ1) η −Dσ(χδ1; δχδ1) ∣∣2 H1(Ω) ≥ min{αε, λ}‖Σ1(χδ1 + ηδχδ1)− Σ1(χδ1) η −Dσ(χδ1; δχδ1)‖2H1(Ω). Hence, the above inequality and (3.84) imply min{αε, λ}‖Σ1(χδ1 + ηδχδ1)− Σ1(χδ1) η −Dσ(χδ1; δχδ1)‖H1(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 (χδ1; δχδ1) ∈ H1(Ω), the gradient of the functional J shown in (2.14) is justified. 4. Existence of a fixed point In (3.28), the update for χ1 was 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 K be a convex subset of L2(Ω) and suppose Υ : K → L2(Ω) is continuous. Suppose Υ(K) is a compact subset of K. Then Υ has a fixed point in K (see [15]). Thus, it is necessary to show that Υ is continuous on a convex subset K of L2(Ω) and that Υ(K) is compact in K. The previous section justified the calculated formulation of the function G defined 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→0 ‖G(χδ1 + ηδχδ1)−G(χδ1)‖L2(Ω) = 0. (4.3) EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 23 Proof. Denote A1(χδ1) := 2ω[Σ1(χδ1 + ηδχδ1)− σ2]ψ(χδ1), A2(χδ1) := −αω|∇Σ1(χδ1)|2. Thus, using (4.2), the triangle inequality, and Young’s inequality for a convolution, we obtain ‖G(χδ1 + ηδχδ1)−G(χδ1)‖L2(Ω) ≤ η‖ξδ‖L1(Ω)‖δχ1‖L2(Ω) + ‖A1(χδ1 + ηδχδ1)−A1(χδ1)‖L2(Ω) + ‖A2(χδ1 + ηδχδ1)−A2(χδ1)‖L2(Ω). (4.4) Adding and subtracting 2ω[Σ1(χδ1 + ηδχδ1) − σ2]ψ(χδ1) to A1(χδ1 + ηδχδ1) − A1(χδ1) and using the triangle inequality, we obtain ‖A1(χδ1 + ηδχδ1)−A1(χδ1)‖L2(Ω) ≤ ‖B1(χ1; δχ1)‖L2(Ω) + ‖B2(χ1; δχ1)‖L2(Ω), where B1(χδ1; δχδ1) = −2ω[Σ1(χδ1 + ηδχδ1)− σ2]ψ(χδ1 + ηδχδ1) + 2ω[Σ1(χδ1 + ηδχδ1)− σ2]ψ(χδ1), B2(χδ1; δχδ1) = −2ω[Σ1(χδ1 + ηδχδ1)− σ2]ψ(χδ1) + 2ω[Σ1(χδ1)− σ2]ψ(χδ1). By using Hölder’s inequality, (3.76), (3.43), (3.31), (3.36), and (3.78), we can esti- mate B1(χδ1; δχδ1) and B2(χδ1; δχδ1) as follows: ‖B1(χδ1; δχδ1)‖L2(Ω) ≤ 2C1ωη‖[Σ1(χδ1 + ηδχδ1)− σ2]‖L∞(Ω)‖δχ1‖L2(Ω), (4.5) for some C1 > 0 and ‖B2(χδ1; δχδ1)‖L1(Ω) ≤ C2η‖δχ1‖L2(Ω)‖∇Φ(χδ1)‖L∞(Ω)‖∇Φ∗(χδ1)‖L2(Ω), (4.6) for some C2 > 0. Also, A2(χδ1 + ηδχδ1)−A2(χδ1) can be estimated using (3.76) and (3.78): ‖A2(χδ1 + ηδχδ1)−A2(χδ1)‖L2(Ω) ≤ αωC3η‖δχ1‖L2(Ω) { ‖Σ1(χδ1ηδχ δ 1)‖L∞(Ω) + ‖Σ1(χδ1)‖L∞(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 ‖G(χδ1 + ηδχδ1)−G(χδ1)‖L2(Ω) ≤ Cη‖δχ1‖L2(Ω), for any η ∈ (0, τ̂), where τ̂ = min {τ, τ̄}, τ and τ̄ are 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 χδ1, (2.16) has a solution in H1(Ω). Note that because χδ1 ∈ C∞(Ω), ∇χδ1 is bounded in Ω for a fixed δ. Hence,√ |∇χδ1|2 + β2 ≤ √ ‖∇χδ1‖2L∞(Ω) + β2 =: K̄ <∞. Or equivalently, 1√ |∇χδ1|2 + β2 ≥ 1 K̄ . (4.8) 24 R. MENDOZA, S. KEELING EJDE-2020/93 Obviously, for any β > 0, √ |∇χδ1|2 + β2 ≥ β, which can be expressed as 1√ |∇χδ1|2 + β2 ≤ 1 β . (4.9) For u, v ∈ H1(Ω), we define a(u, v) = ∫ Ω ωγ ∇u · ∇v√ |∇χδ1|2 + β2 + uv dV, (4.10) b(v) = ∫ Ω G(χδ1)v dV. Clearly, a and b are bilinear and linear, respectively. Observe that using the Cauchy- Schwarz inequality and (4.9), we obtain |a(u, v)| ≤ 2 max{ωγ β , 1}‖u‖H1(Ω)‖v‖H1(Ω), for all v ∈ H1(Ω), which makes a continuous. It can also be shown using the Cauchy-Schwarz inequality that b(v) is also continuous. Using (4.9), a can 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∫ Ω ωγ ∇θ · ∇v√ |∇χδ1|2 + β2 + θv dV = ∫ Ω G(χδ1)v dV, (4.11) for any v ∈ H1(Ω) provided that ∂v ∂n = 0 on ∂Ω. Furthermore, there exists C > 0 such that ‖θ‖H1(Ω) ≤ 1 min{ωγ K̄ , 1} ‖G(χδ1)‖L2(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(χδ1) as a function that maps an element χ1 ∈ L2(Ω) to an element θ ∈ H1(Ω). Note that given a perturbation δχδ1 of χδ1, Θ(χδ1 + ηδχδ1) is well-defined for any 0 < η < ∞ because for coercivity we just need √ ‖∇(χδ1 + ηδχδ1)‖2L∞(Ω) + β2 to be finite. Since χδ1 + ηδχδ1 ∈ C∞(Ω̄), this is not a problem. From the definition of G and the inequalities (3.37), (3.38), and (3.76), we can infer that ‖Θ(χδ1 + ηδχδ1)‖H1(Ω) ≤ C‖G(χδ1 + ηδχδ1)‖L2(Ω) <∞, (4.12) for some C > 0. We now prove that this map is continuous. Lemma 4.4. Under assumption (3.30), lim η→0 ‖Θ(χδ1 + ηδχδ1)−Θ(χδ1)‖H1(Ω) = 0. Proof. By the previous lemma, note that Θ(χδ1) satisfies∫ Ω ωγ ∇Θ(χδ1) · ∇v√ |∇χδ1|2 + β2 + Θ(χδ1)v dV = ∫ Ω G(χδ1)v dV. (4.13) EJDE-2020/93 IMPEDANCE TOMOGRAPHY PROBLEM 25 Similarly, Θ(χδ1 + ηδχδ1) satisfies∫ Ω ωγ ∇Θ(χδ1 + ηδχδ1) · ∇v√ |∇(χδ1 + ηδχδ1)|2 + β2 + Θ(χδ1 + ηδχδ1)v dV = ∫ Ω G(χδ1 + ηδχδ1)v dV. (4.14) Subtracting (4.13) from (4.14), we obtain A1(v) +A2(v) = B(v), (4.15) with A1(v) := ∫ Ω ωγ [ ∇Θ(χδ1 + ηδχδ1) · ∇v√ |∇χδ1 + ηδχδ1|2 + β2 − ∇Θ(χδ1) · ∇v√ |∇χδ1|2 + β2 ] dV, A2(v) := ∫ Ω [Θ(χδ1 + ηδχδ1)−Θ(χδ1)]v dV, B(v) := ∫ Ω [G(χδ1 + ηδχδ1)−G(χδ1)]v dV. We subtract and add the term∫ Ω ωγ ∇Θ(χδ1 + ηδχδ1) · ∇v√ |∇χδ1|2 + β2 dV to A1(v) to obtain A1(v) = A3(v) +A4(v), (4.16) with A4(v) := ∫ Ω ωγD(η) ∇Θ(χδ1 + ηδχδ1) · ∇v√ |∇(χδ1)|2 + β2 , A3(v) := ∫ Ω ωγ [∇Θ(χδ1 + ηδχδ1)−∇Θ(χδ1)] · ∇v√ |∇χδ1|2 + β2 dV, D(η) := 1√ |∇(χδ1 + ηδχδ1)|2 + β2 − 1√ |∇χδ1|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 a in (4.10), we deduce that A3(v) +A2(v) = a(Θ(χδ1 + ηδχδ1)−Θ(χδ1), v). From the coercivity of a we can show that |(A3 +A2)(Θ(χδ1 + ηδχδ1)−Θ(χδ1))| ≥ min( ωγ K̄ , 1)‖Θ(χδ1 + ηδχδ1)−Θ(χδ1)‖2H1(Ω). (4.18) On the other hand, using the Cauchy-Schwarz inequality, we obtain |B(v)| ≤ ‖G(χδ1 + ηδχδ1)−G(χδ1)‖L2(Ω)‖v‖H1(Ω). (4.19) Moreover, using the Cauchy-Schwarz inequality, Hölder’s inequality, and (4.9), we have |A4(v)| ≤ ωγ β ‖D(η)‖L∞(Ω)‖Θ(χδ1 + ηδχδ1)‖H1(Ω)‖v‖H1(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 K̄ , 1}‖Θ(χδ1 + ηδχδ1)−Θ(χδ1)‖2H1(Ω) ≤ ‖G(χδ1 + ηδχδ1)−G(χδ1)‖L2(Ω) + ωγ β ‖D(η)‖L∞(Ω)‖Θ(χδ1 + ηδχδ1)‖H1(Ω). From (4.12), the right-hand side of the above inequality is bounded for any η ∈ (0, τ̂), where τ̂ = min {τ, τ̄}, τ and τ̄ are both chosen according to Remark 3.4 and Remark 3.26, respectively. Obviously, limη→0 ‖D(η)‖L∞(Ω) = 0. The above inequality, together with (4.3), establishes our claim. � We now establish that the function H defined 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, H is continuous in L2(Ω) (see [15]). In our next computations, we prove the continuity of the function M defined in (3.26). Recall that M maps elements ofM(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 χ̂ and χ be characteristic functions on Ω whose supports are given by Ωχ̂ and Ωχ, respectively. Then µ(Ωχ̂ 4 Ωχ) = ‖χ̂− χ‖2L2(Ω). Proof. Because χ and χ̂ are characteristic functions, we have Ωχ̂\Ωχ = {x : x ∈ Ωχ̂ ∧ x /∈ Ωχ} = {x : χ̂(x) = 1 ∧ χ(x) = 0}. Similarly, Ωχ\Ωχ̂ = {x : χ(x) = 1 ∧ χ̂(x) = 0}. Thus, from the definition of symmetric difference and the fact that Ωχ̂\Ωχ and Ωχ\Ωχ̂ are disjoint sets, we obtain µ(Ωχ̂ 4 Ωχ) = µ((Ωχ̂\Ωχ) ∪ (Ωχ\Ωχ̂)) = µ({x : χ̂(x) = 1 ∧ χ(x) = 0}) + µ({x : χ(x) = 1 ∧ χ̂(x) = 0}) = ∫ Ω χ̂(1− χ) dV + ∫ Ω χ(1− χ̂) dV = ‖χ̂− χ‖2L2(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 M is continuous. Lemma 4.7. Suppose {ωn}∞n=1 ⊂ M(Ω) such that ωn → ω in M(Ω), that is, limn→∞ µ(ωn 4 ω) = 0. Then lim n→∞ ‖M(ωn)−M(ω)‖L2(Ω) = 0, where M :M(Ω) → L2(Ω) is a function that maps ω to its corresponding charac- teristic function, that is, M(ω) = χω. In other words, M is continuous on M(Ω). Proof. We denote M(ωn) =: χn and M(ω) := χω. Then by Lemma (4.6), lim n→∞ ‖M(ωn)−M(ω)‖2L2(Ω) = lim n→∞ ‖χn − χω‖2L2(Ω) = lim n→∞ µ(ωn 4 ω) = 0. � We have shown continuity of G, Θ, H and 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 ≤ 1 a.e. Ω}. (4.22) Proof. We employ the Schauder Fixed Point Theorem. We first need to show thatK is a convex set in L2(Ω). Let χ1, χ̄1 ∈ K and λ ∈ (0, 1). Obviously, λχ1+(1−λ)χ̄1 ∈ L2(Ω). We only need to show that 0 ≤ λχ1 + (1− λ)χ̄1 ≤ 1. Because λ, 1− λ > 0 then λχ1 + (1 − λ)χ̄1 ≥ 0. Furthermore, λχ1 + (1 − λ)χ̄1 ≤ λ + 1 − λ = 1. Thus, λχ1 + (1− λ)χ̄1 ∈ K and K is convex in L2(Ω). We show next that Υ is continuous. The functions G, Θ, H, and M are continu- ous as proven in Lemma 4.2, Lemma 4.4, Lemma 4.5, and Lemma 4.7, respectively. From Lemma 3.8, Tδ is 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) ⊂ K and that Υ(K) is compact in K. Recall that (M ◦H ◦Θ ◦G)(χδ1) is a characteristic function. Thus, 0 ≤ (M ◦H ◦Θ ◦G)(χδ1) ≤ 1. By Theorem 2.5, 0 ≤ (Tδ ◦M ◦H ◦Θ ◦G)(χδ1) ≤ 1, and so Υ(χ1) ∈ K. Let χ̄1 be an arbitrary element of K. Let us denote ω := (H◦Θ◦G)(χ̄δ1), χω := M(ω), and χωδ := Tδ(χ ω). By Lemma 3.9, Hölder’s inequality, and the Cauchy-Schwarz inequality, we obtain |∇χωδ (x)| = | ∫ Ω ∇ξδ(x− y)χω(y)dy| ≤ ‖χω‖L∞(Ω) ∫ Ω |∇ξδ(x− y)|dy ≤ √ µ(Ω)‖∇ξδ‖L2(Ω). 28 R. MENDOZA, S. KEELING EJDE-2020/93 Hence, ‖∇χωδ ‖2L2(Ω) = ∫ Ω | ∫ Ω ∇ξδ(x− y)χω(y)dy|2dx ≤ µ(Ω)2‖∇ξδ‖2L2(Ω). From Lemma 2.5, χωδ is real analytic and so χωδ ∈ H1(Ω). For a fixed δ, we compute the H1(Ω) norm of χωδ using Young’s inequality for convolutions and the Hölder’s inequality: ‖Υ(χ̄1)‖2H1(Ω) = ‖χωδ ‖2H1(Ω) = ‖χωδ ‖2L2(Ω) + ‖∇χωδ ‖2L2(Ω) ≤ ‖χω ∗ ξδ‖2L2(Ω) + µ(Ω)2‖∇ξδ‖2L2(Ω) ≤ ‖χω‖2L1(Ω)‖ξδ‖ 2 L2(Ω) + µ(Ω)2‖∇ξδ‖2L2(Ω) ≤ ‖χω‖2L∞(Ω)µ(Ω)2‖ξδ‖2L2(Ω) + µ(Ω)2‖∇ξδ‖2L2(Ω) ≤ µ(Ω)2 ( ‖ξδ‖2L2(Ω) + ‖∇ξδ‖2L2(Ω) ) . Since χ̄1 is arbitrary, any sequence {Υ(χn1 )}∞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(χn1 )}∞n=1 has a convergent subsequence [1]. Therefore Υ(K) is compact. Using Schauder Fixed Point theorem, Υ1 has a fixed point on K. � The fixed point is attained given any arbitrary χ1 ∈ L2(Ω) such that 0 ≤ χ1 ≤ 1 and σ̄1 in 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 σ2 is fixed and known while σ1 is expressed in terms of χ1. Hence, the functional depends on χ1 alone. 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. 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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 1. Introduction 2. Two-phase segmentation algorithm Two-phase segmentation algorithm 3. Analysis of the algorithm 3.1. Preliminaries 3.2. Smooth approximation of 1 3.3. Gradient of the functional J 4. Existence of a fixed point 5. Conclusion Acknowledgments References