Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 19 (2000), No. 1, 109–120

Domain Identification

for Semilinear Elliptic Equations in the Plane: the Zero Flux Case

D. D. Trong and D. D. Ang

Abstract. We consider the problem of identifying the domain Ω ⊂ ℝ^{2} of a semilinear elliptic equation subject to given Cauchy data on part of the known outer boundary Γ and to the zero flux condition on the unknown inner boundary γ, where it is assumed that Γ is a piecewise C^{1} curve and that γ is the boundary of a finite disjoint union of simply connected domains, each bounded by a piecewise C^{1} Jordan curve. It is shown that, under appropriate smoothness conditions, the domain Ω is uniquely determined. The problem of existence of solution for given data is not considered since it is usually of lesser importance in view of measurement errors giving data for which no solution exists.

Keywords: Domain identification, semilinear elliptic equations, finitely many holes, zero flux AMS subject classification: 35 R 30, 35 J 60

1. Introduction

Let Ω be a domain in ℝ^{2} bounded by a known outer boundary Γ and containing finitely many internal holes represented by simply connected domains. Let γ be the (unknown) inner boundary of Ω. Consider the equation

A(x, u)u(x) = 0 (x ∈ Ω)

where A(x, u) is a (possibly semilinear) elliptic differential operator.

We address the problem of identifying the domain Ω of the equation A(x, u)u = 0, subject to given Cauchy conditions on an open portion Γ_{0} of Γ and to the condition of zero A(x, u)-conormal derivative on γ. We note at once that the problem of existence of a solution is not considered here. In fact, as is often the case with inverse problems, the question of existence for given data is less important than that of uniqueness. Indeed, data given by measurements are usually affected with errors leading to a problem without solution. As a consequence, one works with a problem that may not have a solution and hence has to resort to a regularization. The particular case that A(x, u) is the Laplacian is of special interest. The corresponding problem is related to the detection of cracks by the electric method and has been considered in [6] for the special case of one internal hole, the boundary of which is star-shaped and with the condition of zero normal derivative prescribed on γ. In physical terms, the function u is the electric potential, the normal derivative \frac{∂u}{∂n} on Γ_{0} is the electric flux through Γ_{0} (the conductivity being assumed to be 1), and the condition \frac{∂u}{∂n} = 0 on γ means that the interior of the crack is filled with a non-conducting material (like air). The use of the electric method in crack detection was initiated in the pioneering work of Friedman and Vogelius [10]. The method was applied by Kubo [13], McIver [11], Alessandrini [1, 2] and others (see also [14]). It is noted that a variant of the problem corresponding to the condition u = 0 on γ is considered in [9] and in [17]. It was the case of the Laplace equation that motivated the present study since except for our recent paper [6], the published literature on crack detection by the electric method, to our knowledge, has been limited to infinitely thin cracks. However, the full strength our result is not utilized. In fact the result could find applications in heat conduction with temperature dependent conductivity and other problems in the nonlinear realm. These and related problems will be the object of a future study.

Both authors: Hochiminh City Nat. Univ., Dept. Math. & Comp. Sci., 227 Nguyen Van Cu, Q5, Hochiminh City, Vietnam ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

Let Ω be as above and let the internal holes in Ω be represented by the simply connected domains ω_{1}, ..., ω_{m} with disjoint closures, i.e.,

ω̄_{i} ∩ ω̄_{j} = ∅  (i ≠ j).  (1)

By our definition of γ, we have

γ = ⋃_{i=1}^{m} ∂ω_{i}.

It is assumed that γ is piecewise of C^{1} type. We have in mind cracks in a solid, and thus, the discontinuities of ∂ω_{i} would correspond to crack edges.

Before giving a precise formulation of the problem, several remarks are in order.

First, the assumption of simply connected cracks excludes rectilinear cracks for which uniqueness does not hold with one boundary measurement (although it does hold with two boundary measurements).

Second, the condition of disjoint closures (1) can be weakened. We still need the ω′_{i}s to be mutually disjoint, while, two distinct ∂ω_{i} and ∂ω_{j} can have an intersection consisting of a finite set of points. Note that if ∂ω_{i} ∩ ∂ω_{j} (i ≠ j) contains a segment, then it can be shown that there is no uniqueness with one measurement.

As a third remark, we note that the present approach does not apply to the 3dimensional case. For the latter case, we require the crack surfaces to be piecewise analytic (cf. Ang, Mennicken and Trong [5]).

Finally, we remark that in the case of elastic solids, stresses and displacements measured on the outer boundary uniquely determine the locations and shapes of internal cracks (Ang, Trong and Yamamoto [7, 8]). We also refer to the paper of Andrieux, Abda and Bui [4] dealing with the problem of rectilinear or planar cracks in elastic bodies from boundary measurements in terms of a functional introduced by the authors.

We now give a precise formulation of our problem. Let A(x, u) be a differential operator of the form

A(x, u)u ≡ ∑_{i,j=1}^{2} D_{j}(a_{ij}(x, u)D_{i}u) − c(x)F (u)  (2)

where F ∈ C^{1}(ℝ), a_{ij} ∈ C^{1}(ℝ^{2} × ℝ) and (a_{ij}) satisfies the ellipticity condition, i.e. there exists a constant c_{0} > 0 such that

∑_{i,j=1}^{2} a_{ij}(x, σ)ξ_{i}ξ_{j} ≥ c_{0} (ξ_{1}^{2} + ξ_{2}^{2})

for all ξ = (ξ_{1}, ξ_{2}), (x, σ) ∈ ℝ^{2} × ℝ. Consider the equation

A(x, u)u = 0  (3)

subject to the boundary conditions

u|_{Γ_{0}} = f  (4)
∑_{i,j=1}^{2} a_{ij}(x, u(x)) n_{j}(x)D_{i}u(x) = g(x) (x ∈ Γ_{0})  (5)

where Γ_{0} is an open subset of Γ, n(x) = (n_{1}(x), n_{2}(x)) is the unit vector normal to Γ∪γ at x and

∑_{i,j=1}^{2} a_{ij}(x, u(x)) n_{j}(x)D_{i}u(x) = 0  (x ∈ γ_{∗} = γ \ {y_{1}, ..., y_{k}}),  (6)

where {y_{1}, ..., y_{k}} is a finite subset of γ such that γ_{∗} = γ \ {y_{1}, ..., y_{k}} is of C^{1} type. The functions c and F are assumed to satisfy

c ∈ C(ℝ^{2}), c(x) > 0 a.e. or c ≡ 0 in Ω  (7)
f, g ∈ C(Γ̄_{0}), F (0) = 0, F′(v) > 0 for all v ≠ 0.  (8)

The remainder of the paper consists of two sections. In Section 2, we state the main result of our paper and give a counterexample. Section 3 is devoted to the proof of the main result.

2. Main result, counterexample, and preliminary lemmas

We assume that Ω is in a family of plane domains with outer boundary Γ, and containing finitely many holes ω_{i} (i = 1, ..., m) such that

Γ, ∂ω_{i} are piecewise C^{1} Jordan curves in ℝ^{2}  (9)
Γ ∩ ∂ω_{i} = ∅, ω̄_{i} ∩ ω̄_{j} = ∅ (i ≠ j).  (10)

Then we have

Theorem 1. Let (1), (2) and (7) − (10) hold. If we have either f ≢ const or g ≢ 0, then there exists at most one pair (Ω, u), u ∈ C(Ω̄) ∩ C^{2}(Ω ∪ Γ_{0} ∪ γ_{∗}) ∩ H^{1}(Ω) for which (3) − (6) hold.

We shall give a counterexample to show that condition (7) on the coefficient c is essential. Indeed, concerning the coefficient c in equation (3), it would intituitively seem that the condition c(x) ≥ 0 a.e. alone would ensure uniqueness. However, it is a classical result that this is not true. For the reader’s convenience, we nevertheless include a counterexample.

Counterexample. Let c̃ ∈ C(ℝ) be a function satisfying

c̃(x_{1}) \{ \begin{matrix}> 0 & for all x_{1} > 0 \\ = 0 & for all x_{1} ≤ 0, \end{matrix}  (11)

put c(x_{1}, x_{2}) = c̃(x_{1}) for all (x_{1}, x_{2}) ∈ ℝ^{2} and let z be the solution of the Volterra equation

z(x_{1}) = α + ∫_{−1}^{x_{1}}∫_{−1}^{ξ} c̃(τ)z(τ) dτdξ  (α > 0).  (12)

Then clearly z ∈ C^{2}(ℝ^{2}) and in view of (11), (12) we get

z(x_{1}) = α (x_{1} ≤ 0)  (13)
z”(x_{1}) = c̃(x_{1})z(x_{1}).  (14)

Put v(x_{1}, x_{2}) = z(x_{1}) and c(x_{1}, x_{2}) = c̃(x_{1}). In view of (13) and (14) we have

Δv − cv = 0  for all x ∈ ℝ^{2}  (15)
v(x_{1}, x_{2}) = α  for all x_{1} ≤ 0.  (16)

Now, let ω be any simply connected domain satisfying

(i) ω̄ ⊂ (−1, 0) × (−1, 1)

(ii) γ = ∂ω is a C^{1} Jordan curve.

Put

\begin{aligned}Ω_{γ} & = (−1, 1) × (−1, 1) \ ω̄ \\ Γ_{0} & = \{(x_{1}, 1) : −1 ≤ x_{1} ≤ 1\}. \end{aligned}

By (15) and (16), v satisfies

\begin{matrix}Δv − cv & = 0 on Ω_{γ} \\ v|_{Γ_{0}} = z(x_{1}), \frac{∂v}{∂n}|_{Γ_{0}} & = 0 \\ \frac{∂v}{∂n}|_{γ} & = 0. \end{matrix} \}

So (Ω_{γ}, v) is a solution of problem (3) - (6), but clearly uniqueness fails in this case.

We now state some lemmas that will be used in the proof of Theorem 1.

Lemma 1. Let C be a piecewise C^{1} Jordan curve. Then there is a homeomorphism h : ℝ^{2} ⟶ ℝ^{2} such that h(C) = ∂B(0, 1), h(U_{1}) = B(0, 1) and h(U_{2}) = ℝ^{2} \ \overline{B(0, 1)} where U_{1} is the simply connected domain of ℝ^{2} bounded by C, U_{2} = ℝ^{2} \ Ū_{1} and B(0,r) is the ball of radius r > 0 centered at 0 ∈ ℝ^{2}.

Figure 1

Lemma 2. Let C, U_{1}, U_{2} be as in Lemma 1, C_{1} an arc of C and G an open neighborhood of \overline{C}_{1} in ℝ^{2}. Then there exists a neighborhood U of \overline{C}_{1} in ℝ^{2}, U ⊂ G, such that U is of the form (see Figure 1)

U = V_{1} ∪ V_{2} ∪ S

where V_{1} = U ∩U_{1}, V_{2} = U ∩U_{2}, S = U ∩C are connected and furthermore S ⊂ ∂V_{1}∩∂V_{2}.

The proof of Lemma 1 which relies on the Riemann mapping theorem is omitted. Lemma 2 is a direct consequence of Lemma 1.

3. Proof of Theorem 1

Let (Ω^{1}, u^{1}) and (Ω^{2}, u^{2}) be solutions of problem (3) - (6). We claim that Ω^{1} = Ω^{2} and u^{1} = u^{2}.

We first set some notations. Let ω_{1}^{i}, ..., ω_{m_{i}}^{i} (i = 1, 2) be the holes in Ω^{i} and put

γ^{i} = ⋃_{j=1}^{m_{i}} ∂ω_{j}^{i} (i = 1, 2).

By assumption, γ^{i} is C^{1}-smooth except at a finite number of points {y_{1}^{i}, ..., y_{k_{i}}^{i}} ⊂ γ^{i}. Put

γ_{∗}^{i} = γ^{i} \ {y_{1}^{i}, ..., y_{k_{i}}^{i}}.  (17)

We denote by W the (connected) component of Ω^{1} ∩ Ω^{2} such that Γ ⊂ ∂W. For x ∈ γ^{2} ∩ Ω^{1}, let m_{x} be the index in the set {1, …, m_{2}} such that x ∈ ∂ω_{m_{x}}^{2} and let C_{x} be the maximal arc of ∂ω_{m_{x}}^{2} such that x ∈ C_{x} and C_{x} ⊂ ∂ω_{m_{x}}^{2} ∩ Ω^{1}.

With these notations, we begin the proof of Theorem 1. The proof is by contradiction. If Ω^{1} \ Ω̄^{2} = ∅ and Ω^{2} \ Ω̄^{1} = ∅, then Ω^{1} = Ω^{2}. Thus we can assume, for example, that Ω^{1} \ Ω̄^{2} ≠ ∅. The proof is divided into several steps. The crucial one is Step 3 which establishes the existence of an open subset U_{0} of Ω^{1} such that ∂U_{0} is piecewise of C^{1} type and that ∂U_{0} ⊂ γ^{1} ∪ (∂W \ Γ). Note that

∂(Ω^{1} \ \overline{W}) \ γ^{1} = ∂W \ (γ^{1} ∪ Γ).  (18)

Step 1. u^{1} = u^{2} on W.

Proof. We have

∑_{i,j=1}^{2} D_{j}(a_{ij}(x, u^{1})D_{i}u^{1}) − c(x)F (u^{1}) = 0  (19)

∑_{i,j=1}^{2} D_{j}(a_{ij}(x, u^{2})D_{i}u^{2}) − c(x)F (u^{2}) = 0.  (20)

Letting v ≡ u^{2} − u^{1}, one has

\begin{aligned}a_{ij}(x, u^{2}) − a_{ij}(x, u^{1}) & = v ∫_{0}^{1} \frac{∂a_{ij}}{∂u} (x, u^{1} + t(u^{2} − u^{1})) dt \\ F (u^{2}) − F (u^{1}) & = v ∫_{0}^{1} F′(u^{1} + t(u^{2} − u^{1})) dt. \end{aligned}

Substracting (19) from (20) and using the latter relations, we get after some computations

∑_{i,j=1}^{2} D_{j}(a_{ij}(x, u^{1})D_{i}v) − ∑_{i,j=1}^{2} D_{j}(v b_{ij}) − c(x)c̃(x)v = 0  (21)

where

\begin{matrix}b_{ij}(x) & = b_{ij}(x, u_{1}(x), u_{2}(x)) \\ c̃(x) & = c̃(x, u_{1}(x), u_{2}(x)). \end{matrix} \}

On the other hand, on Γ_{0} one has

v|_{Γ_{0}} = 0  (22)
∑_{i,j=1}^{2} a_{ij}(x, u^{1}(x))n_{j}(x)D_{i}v(x) = 0 (x ∈ Γ_{0})  (23)

where we have used the condition u^{1} = u^{2} on Γ_{0}. By uniqueness of the Cauchy problem for elliptic equations [15] we have v = 0 on W, i.e. u^{1} = u^{2} on W, which completes the proof of Step 1

Step 2. Let x ∈ ∂W \ (γ^{1} ∪ Γ). Then x ∈ γ^{2} ∩ Ω^{1} and C_{x} ⊂ ∂W \ (γ^{1} ∪ Γ) = ∂(Ω^{1} \ \overline{W}) \ γ^{1}.

Proof. Noting that ∂Ω^{1} = Γ∪γ^{1} and W ⊂ Ω^{1}, we have ∂W \(γ^{1} ∪Γ) ⊂ \overline{Ω}^{1} \∂Ω^{1} = Ω^{1}. On the other hand, since W is a component of Ω^{1} ∩ Ω^{2}, we have

∂W ⊂ ∂(Ω^{1} ∩ Ω^{2}) ⊂ ∂Ω^{1} ∪ ∂Ω^{2} ⊂ γ^{1} ∪ γ^{2} ∪ Γ.  (24)

It follows that ∂W \ (γ^{1} ∪ Γ) ⊂ γ^{2}. Hence ∂W \ (γ^{1} ∪ Γ) ⊂ γ^{2} ∩ Ω^{1}.

We now show that

C_{x} ⊂ ∂W \ (γ^{1} ∪ Γ)  for x ∈ ∂W \ (γ^{1} ∪ Γ).  (25)

Let C_{1} be a subarc of C_{x} satisfying \overline{C}_{1} ⊂ C_{x}. By Lemma 2, there exists a neighborhood ω of C_{1} such that

ω ⊂ Ω^{1}  and  ω ∩ ( ⋃_{j=1}^{m_{2}} ω_{j}^{2} \ ω_{m_{x}}^{2}) = ∅,

ω being of the form

ω = U_{1} ∪ U_{2} ∪ S, S ⊂ ∂U_{1} ∩ ∂U_{2} (26)

where U_{1} ≡ ω ∩ ω_{m_{x}}^{2}, U_{2} ≡ ω ∩ Ω^{2}, S ≡ ω ∩ C_{x} are connected.

Figure 2

From (24) one has ω ∩ (∂Ω^{1} ∪ ∂W) ≠ ∅. The relation ω ⊂ Ω^{1} implies ω ∩ ∂Ω^{1} = ∅. Hence ω ∩ ∂W ≠ ∅, i.e.

ω ∩ W ≠ ∅.  (27)

On the other hand, we have

U_{1} = ω ∩ ω_{m_{x}}^{2} ⊂ Ω^{1} \ Ω̄^{2} ⊂ Ω^{1} \ \overline{W} .  (28)

Hence (26) and (27) together imply that U_{2} ∩ W ≠ ∅. The connectedness of U_{2} and the maximality of W as a connected subset of Ω^{1} ∩ Ω^{2} imply that

U_{2} ⊂ W.  (29)

From (26) and (29) we get C_{1} ⊂ ∂W \ (γ^{1} ∪ Γ). Since C_{1} is any arc of C_{x} such that \overline{C}_{1} ⊂ C_{x}, one has S = ω ∩C_{x} ⊂ ∂W \(γ^{1} ∪Γ). This establishes (25) and thus completes the proof of Step 2

Step 3. There is an open subset U_{0} of Ω^{1} such that ∂U_{0} ⊂ γ^{1} ∪ (∂W \ Γ) and that ∂U_{0} is piecewise of C^{1} type.

Proof. We have the following two cases.

Case (i): There is an x_{0} ∈ ∂W \ (Γ ∪ γ^{1}) such that ∂ω_{m_{x_{0}}}^{2} ∩ γ^{1} = ∅ (see Figure 3), i.e. ∂ω_{m_{x_{0}}}^{2} ∩ ∂ω_{i}^{1} = ∅ (i = 1, ..., m_{1}).

Figures 3 and 4

Then put U_{0} = ω_{m_{x_{0}}}^{2} ∩ Ω^{1}. We have ∂U_{0} = ∂ω_{m_{x_{0}}}^{2} ∪ (∪_{i∈I}∂ω_{i}^{1}) where I is the set of the i’s in {1, ..., m_{1}} satisfying ω_{i}^{1} ⊂ ω_{m_{x_{0}}}^{2} . Hence U_{0} is the required set.

Case (ii): ∂ω_{m_{x}}^{2} ∩ γ^{1} ≠ ∅ for all x ∈ ∂W \ (Γ ∪ γ^{1}).

Then for each x ∈ ∂W \ (Γ ∪ γ^{1}), the Jordan arc C_{x} (cf. Step 2) has its edge points (i.e. the points in C̄_{x} \ C_{x}) in ∂ω_{m_{x}^{1}}^{1} and ∂ω_{m_{x}^{2}}^{1} where ω_{m_{x}^{1}}^{1} and ω_{m_{x}^{2}}^{1} are holes in Ω^{1}. To fix ideas, we assume that m_{x}^{1} ≤ m_{x}^{2}. The following three situations are to be considered:

(ii)_{1} m_{x_{0}}^{1} = m_{x_{0}}^{2} for some x_{0} ∈ ∂W \ (Γ ∪ γ^{1}).

(ii)_{2} m_{x}^{1} ≠ m_{x}^{2} for all x ∈ ∂W \ (Γ ∪ γ^{1}) and there exist y_{0}, z_{0} ∈ ∂W \ (Γ ∪ γ^{1}) such

that (m_{y_{0}}^{1}, m_{y_{0}}^{2}) = (m_{z_{0}}^{1}, m_{z_{0}}^{2}) and C_{y_{0}} ≠ C_{z_{0}}.

(ii)_{3} m_{x}^{1} ≠ m_{x}^{2} and C_{y} = C_{z} or (m_{y}^{1}, m_{y}^{2}) ≠ (m_{z}^{1}, m_{z}^{2}) for all x, y, z ∈ ∂W \ (Γ ∪ γ^{1}).

If (ii)_{1} holds, the set of end points of C_{x_{0}} has at most two points and, moreover, by the maximality of C_{x_{0}}, these points are in ∂ω_{m_{x_{0}}^{1}}^{1} ⊂ γ_{1}. If the end points of C_{x_{0}} are

distinct (see Figure 4), then ∂ω_{m_{x_{0}}^{1}}^{1} is the union of two Jordan arcs l_{α} and l_{β} such that l_{α} ∩ l_{β} = {α, β}. Note that both sets C_{x_{0}} ∪ l_{α} and C_{x_{0}} ∪ l_{β} are Jordan curves and that one of them, say γ_{0} = C_{x_{0}} ∪ l_{α} does not contain the other in its interior. Call U′_{0} the domain interior to γ_{0} and put U_{0} = U′_{0} ∩ Ω^{1}. Then clearly U_{0} is the desired open set. This proves Step 3 in the case α ≠ β. If α = β, i.e. if C̄_{x_{0}} is a Jordan curve, the proof is similar as (in fact easier than) for the case of distinct end points.

If (ii)_{2} holds (see Figure 5), we denote the end points of C_{y_{0}} and C_{z_{0}} by α, β and α′, β′, respectively. As in the proof for the case (ii)_{1}, we choose U′_{0} as the domain interior to the Jordan curve C_{y_{0}} ∪ l_{1} ∪ C_{z_{0}} ∪ l_{2} (cf. Figure 5). Put U_{0} = U′_{0} ∩ Ω^{1}. Just as for the case (ii)_{1} we get that ∂U_{0} is piecewise of C^{1} type. This proves Step 3 for the case (ii)_{2}.

Figure 5

Finally, we consider the case (ii)_{3}. We shall prove that Step 3 holds for U_{0} = Ω^{1} \\overline{W}. Letting x ∈ ∂W \ (Γ ∪ γ^{1}), Step 2 gives C_{x} ⊂ ∂W \ (Γ ∪ γ^{1}) = ∂(Ω^{1} \ \overline{W}) \ γ^{1}. Hence

∂(Ω^{1} \ \overline{W}) \ γ^{1} = ⋃_{x∈∂(Ω^{1}\\overline{W})\γ^{1}} C_{x}.  (30)

For each pair (p, q) with p, q ∈ {1, ..., m_{1}} and p ≤ q, by assumptions (ii)_{3} there is at most one C_{x} such that p = m_{x}^{1} and q = m_{x}^{2}. Hence, there are at most m_{0} arcs C_{z_{1}}, ..., C_{z_{m_{0}}} in ∂(Ω^{1} \ \overline{W}) \ γ^{1} such that

⋃_{x∈∂(Ω^{1}\\overline{W})\γ^{1}} C_{x} = ⋃_{i=1}^{m_{0}} C_{z_{i}}  (31)

where m_{0} ≤ \frac{1}{2}m_{1}(m_{1} + 1) (= the number of pairs (p, q) above). Combining (30) and (31) gives

∂(Ω^{1} \ \overline{W}) \ γ^{1} = ⋃_{i=1}^{m_{0}} C_{z_{i}}.  (32)

Let z_{i}^{1} and z_{i}^{2} be the end points of C_{z_{i}}, i.e. {z_{i}^{1}, z_{i}^{2}} = C̄_{z_{i}} \ C_{z_{i}} (i = 1, ..., m_{0}), and put B = \{z_{i}^{1}, z_{i}^{2} : 1 ≤ i ≤ m_{0}\}. For x ∈ (∂(Ω^{1} \ \overline{W}) ∩ γ^{1}) \ B we denote by D_{x} the maximal arc of γ^{1} \ B such that x ∈ D_{x}. By the same arguments as for Step 2, we get

D_{x} ⊂ ∂(Ω^{1} \ \overline{W}) ∩ γ^{1}  for x ∈ ∂(Ω^{1} \ \overline{W}) ∩ γ^{1} \ B.  (33)

Since (33) holds, the set ∂(Ω^{1} \ \overline{W}) ∩ γ^{1} \ B can be divided into equivalence classes with respect to the equivalence relation

x ∼ y  ⟺  D_{x} = D_{y}

(note that the maximality of D_{x} gives D_{x} = D_{y} or D_{x}∩D_{y} = ∅ for all x, y ∈ ∂(Ω^{1}\\overline{W})∩ γ^{1} \ B). We claim that the number of ∼-equivalence classes is finite. We first note that, for a couple x, y in a Jordan curve C, there are only two Jordan arcs of C having x, y as its edges points. From the maximality of D_{x} (x ∈ ∂(Ω^{1} \ \overline{W}) ∩ γ^{1} \ B), its edge points (i.e. the points in \overline{D}_{x} \ D_{x}) are in B. Hence for z, z′ ∈ B there are at most two point x_{1}, x_{2} ∈ ∂(Ω^{1} \ \overline{W}) ∩ γ^{1} \ B such that D_{x_{1}} ≠ D_{x_{2}} and \overline{D}_{x_{1}} \ D_{x_{1}} = \{z, z′\} = \overline{D}_{x_{2}} \ D_{x_{2}}. On the other hand, B is finite. Hence the set of ∼-equivalence classes is finite.

From the foregoing arguments, we can find y_{1}, ..., y_{r} ∈ ∂(Ω^{1} \ \overline{W}) ∩ γ^{1} \ B such that

∂(Ω^{1} \ \overline{W}) ∩ γ^{1} \ B = ⋃_{j=1}^{r} D_{y_{j}}.  (34)

From (32) and (34) one has

∂(Ω^{1} \ \overline{W}) = B ∪ (∂U_{0} \ γ^{1}) ∪ ((∂U_{0} ∩ γ^{1}) \ B) = B ∪ ( ⋃_{i=1}^{m_{0}} C_{z_{i}}) ∪ ( ⋃_{j=1}^{r} D_{y_{j}}).

Here we recall U_{0} = (Ω^{1} \ \overline{W}). This completes the proof of Step 3

Summarizing, we have shown that in our proof of Theorem 1 by contradiction, the assumption Ω^{1} \ Ω̄^{2} ≠ ∅ led to the existence of an open set U_{0} as in Step 3.

Step 4. Using the latter result we will show in this final step that u^{1} = const on Ω^{1}.

Proof. By Step 3, there is an open set U_{0} ⊂ Ω^{1} such that ∂U_{0} is piecewise of C^{1} type and that ∂U_{0} ⊂ γ^{1} ∪ (∂W \ Γ). Hence, there is a finite set B_{1} ⊂ γ^{1} ∪ γ^{2} such that ∂U_{0} \ B_{1} is a finite union of C^{1}-smooth Jordan arcs and that {y_{i}^{1}, y_{j}^{2} : 1 ≤ i ≤ k_{1}, 1 ≤ j ≤ k_{2}} ⊂ B_{1} where y_{i}^{1} and y_{j}^{2} are as in (17). One has ∂U_{0} = B_{1} ∪ (∂U_{0} \ B_{1}). For x ∈ ∂U_{0} \ B_{1}, the following two cases are to be considered:

(a) x ∈ γ_{∗}^{1} = γ^{1} \ {y_{1}^{1}, ..., y_{k_{1}}^{1}}.

(b) x ∈ ∂W \ (γ^{1} ∪ Γ) ⊂ γ^{2}.

For the case (a), we have

∑_{i,j=1}^{2} a_{ij}(x, u^{1}(x))n_{j}(x)D_{i}u^{1}(x) = 0.  (35)

For the case (b), since x ∈ ∂W \ (γ^{1} ∪ Γ) ⊂ γ^{2}, (6) gives

∑_{i,j=1}^{2} a_{ij}(x, u^{1}(x))n_{j}(x)D_{i}u^{1}(x) = ∑_{i,j=1}^{2} a_{ij}(x, u^{2}(x))n_{j}(x)D_{i}u^{2}(x) = 0.  (36)

From (35) and (36) we conclude that (35) holds for all x ∈ ∂U_{0} \ B_{1}. Multiplying (3) by u^{1}, integrating on U_{0} and using the divergence theorem we get in view of (35) and (36)

∫_{U_{0}} ∑_{i,j=1}^{2} a_{ij}(x, u^{1}(x))D_{i}u^{1}(x)D_{j}u^{1}(x) dx + ∫_{U_{0}} c(x)F (u^{1}(x))u^{1}(x) dx = 0.  (37)

Domain Identification for Elliptic Equations 119

Now, if c(x) > 0 a.e., then (37) implies F (u^{1}(x))u^{1}(x) = 0 a.e. on U_{0}. From (8) it follows that u^{1}(x) = 0 for all x ∈ U_{0}. By Pederson’s theorem on the uniqueness of elliptic continuation [15], the above equality gives u^{1} = 0 on Ω^{1}, which implies f = g ≡ 0, which is a contradiction.

If c(x) = 0 for all x ∈ ℝ^{2}, then we get in view of (37) and the ellipticity of the operator A(x, u) (cf. (2)) that ∇u^{1}(x) = 0 a.e. on U_{0}. If ω_{0} is a component of U_{0}, then the latter equality gives u^{1} = c_{1} (constant) on ω_{0}. By uniqueness of elliptic continuation, we have u^{1} = c_{1} on Ω^{1}. Hence f = c_{1} and g = 0, which is a contradiction. This completes the proof of Theorem 1

Acknowledgements. This work was started while the authors were visiting Ecole Polytechnique, Paris, in the Summer of 1996. We wish to express our heart-felt thanks and appreciation to Professor Alain Damlamian for his warm hospitality and for his most valuable suggestions. The authors wish to thank the referees for their kind comments and most valuable suggestions.

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Received 15.03.1999; in revised form 02.08.1999