UDC 517.917 : 947.42

ON THE UNIQUENESS OF THE SOLUTION OF THE EXTERIOR INVERSE PROBLEM FOR A NEWTONIAN POTENTIAL

A. I. PRILEPKO

The paper investigates the question of uniqueness of the solution of the exterior inverse problem for a Newtonian potential, the formulation of which is given below.

1. We adopt the following notation. \(E^n\) is Euclidean space of \(n\) dimensions; \(x=(x_1,x_2,\ldots,x_n)\), \(y=(y_1,y_2,\ldots,y_n)\) are points of this space; \(r_{xy}=|y-x|\) is the distance between \(x\) and \(y\); \(d_x S\) or \(d_x\sigma\) is the element of area of an \((n-1)\)-dimensional surface at the point \(x\); \(\Gamma A\) is the boundary of an open set \(A\) in \(E^n\); \(\overline{K(\delta_x)}\) is the closed \(n\)-dimensional ball with center at the point \(x\) and radius \(\delta\); \(K(x,y)\) is the fundamental solution of Laplace’s equation

\[ \sum_{k=1}^{n}\frac{\partial^2 u}{\partial x_k^2}\equiv \Delta u=0 \tag{0.1} \]

(for example, \(K(x,y)=\dfrac{1}{4\pi}\dfrac{1}{r_{xy}}\) for \(n=3\), \(K(x,y)=\dfrac{1}{2\pi}\ln\dfrac{1}{r_{xy}}\) for \(n=2\)).

We denote by

\[ V_T(x)=\int_T \mu(y)K(x,y)\,dy \tag{0.2} \]

the Newtonian potential in the space \(E^n\), \(n\ge 2\). Everywhere in what follows, unless otherwise stated, we shall assume that the domain \(T\) is, generally speaking, multiply connected and, moreover, that \(\mu(y)\ne 0\) almost everywhere for \(y\in T\).

Problem. Let \(T_\alpha\) \((\alpha=1,2)\) be finite domains (or open sets); let \(\mu(y)\) be a function defined in the whole space \(E^n\) \((n\ge 2)\); and let \(V_{T_\alpha}(x)\) be the potential determined by the body \(T_\alpha\) with the given density \(\mu(y)\). It is required to determine the mutual arrangement of the domains \(T_\alpha\) under the condition that the exterior potentials are equal, i.e.,

\[ V_{T_1}(x)=V_{T_2}(x)\quad \text{for } x\in E^n\setminus(\overline{T}_1\cup \overline{T}_2). \tag{0.3} \]

This problem belongs to the ill-posed problems of mathematical physics, since, in the formulation given, generally speaking, it has not a unique-

natural solution. The clarification of additional conditions ensuring the uniqueness of the solution of the indicated problem is of great importance in modern geophysics.

For the case when the set \(E^n\setminus(\overline{T}_1\cup\overline{T}_2)\) has one component under various restrictions on the domains \(T_\alpha\), this problem has been considered by many authors. The first proof of the uniqueness theorem for the exterior problem of a Newtonian potential in the class of stellar bodies of constant density is due to P. S. Novikov [16]. For constant density this problem was also investigated in the works of L. N. Sretenskii [23], I. M. Rapoport [19], M. M. Lavrent’ev [13], V. K. Ivanov [4–5], V. P. Simonov [20], Yu. A. Shashkin [27–29], A. Gel’mins [11], I. T. Todorov and D. Zidorov [26], L. E. Kazakova [8], R. A. Smith [21], and others. The question of the uniqueness of the solution of the problem with variable density for the logarithmic potential was studied by methods of the theory of functions of a complex variable in the works of Yu. A. Shashkin [27–29] and V. P. Simonov [20]. The two-dimensional case of this problem was studied in the works of V. K. Ivanov [2–3] and M. M. Lavrent’ev [14]. For some domains \(T_\alpha\), for which the set \(E^n\setminus(\overline{T}_1\cup\overline{T}_2)\) consists of more than one component, the problem in the formulation given above was considered in the works of I. T. Todorov and D. Zidorov [26] and Yu. A. Shashkin [29]. For the case when a potential of the form \(V_A(x)\) is considered, where \(A\) is an open set, in the simplest cases the problem was considered in the works of A. A. Zamorev [1] and R. A. Smith [21]. In a somewhat different formulation the same problem was considered in the work of A. Gel’mins [11], where examples of non-uniqueness of the solution are indicated. We note that, in our formulation, the examples of [11] lose their force, since the potentials \(V_{A_1}(x)\) and \(V_{A_2}(x)\) in the indicated examples do not coincide identically in at least one of the components of the set

\[ E^n\setminus(\overline{A}_1\cup\overline{A}_2). \]

The question of uniqueness of the solution of the problem is closely connected with the question of the stability of exterior inverse problems.

A. N. Tikhonov [24] formulated a general criterion of stability in any compact class of uniqueness. Questions of stability and concrete estimates are found in the works of I. M. Rapoport [19], V. K. Ivanov [5], M. M. Lavrent’ev [12], Yu. A. Shashkin [27–29], L. E. Kazakova [8]. Methods for solving ill-posed problems were proposed in [6, 25, 12, 15] and others.

The present paper contains a number of new uniqueness theorems for the solution of the problem under various restrictions on the domains. First (Theorem 1), for the case of one class of variable densities, a uniqueness theorem is proved for the problem under consideration for domains \(T_\alpha\) for which the set \(\overline{T}_1\cap\overline{T}_2\) is star-shaped with respect to some point \(O\subset\overline{T}_1\cap\overline{T}_2\). In Theorem 2 non-stellar domains are indicated for the case of the logarithmic potential of constant density, for which uniqueness of the solution of the problem holds. From Theorems 1 and 2 there follow, as corollaries, the results of [16, 20]. Next, Theorems 3 and 4 are proved for constant density in \(n\)-dimensional space for various multiply connected domains and open sets. From the indicated theorems there follow the results of [1, 21, 23, 26].

A brief announcement of part of the results of this article was published in [17].

§ 1. DEFINITIONS AND BASIC LEMMAS

1°. Let \(A\) be an open set, and let \(\mu(y)\) be a bounded summable function defined in the whole space \(E^n\). Consider the Newtonian potential

\[ V_A(x)=\int_A \mu(y)K(x,y)\,dy, \tag{1.1} \]

where \(K(x,y)\) is the fundamental solution of the Laplace equation in the space \(E^n\) \((n\geqslant 2)\).

Lemma 1. If the density \(\mu(y)\) is such that, for the set \(A\), the potential \(V_A(x)\) is identically equal to zero outside \(A\), i.e.

\[ V_A(x)\equiv 0 \quad \text{for } x\in (E^n\setminus \overline{A}), \tag{1.2} \]

then for any harmonic function \(U(y)\) defined in a domain \(D\supset \overline{A}\), the equality

\[ \int_A \mu(y)U(y)\,dy=0 \tag{1.3} \]

holds.

Proof. Construct two domains \(D_k\) \((k=1,2)\) with analytic boundaries \(S_k\), and such that \(D\supset \overline{D}_1\supset D_1\supset D_2\supset \overline{A}\). By the known formula, for any harmonic function in the domain \(D\) the following relation holds at points of the domain \(D_1\):

\[ U(x)=\int_{S_1} K(x,y)\frac{\partial U(y)}{\partial n_y}\,d_yS -\int_{S_1} U(y)\frac{\partial}{\partial n_y}K(x,y)\,d_yS . \tag{1.4} \]

Multiplying expression (1.4) by \(\mu(x)\) and integrating over \(A\), we obtain

\[ \int_A \mu(x)U(x)\,dx = \int_A \mu(x)\left[ \int_{S_1} K(x,y)\frac{\partial U(y)}{\partial n_y}\,d_yS - U(y)\frac{\partial}{\partial n_y}K(x,y)\,d_yS \right]dx . \]

Interchanging the order of differentiation and integration, we obtain

\[ \int_A \mu(x)U(x)\,dx = \int_{S_1}\left[ V_A(y)\frac{\partial U(y)}{\partial n_y} - U(y)\frac{\partial V_A(y)}{\partial n_y} \right]d_yS . \tag{1.5} \]

Since from condition (1.2) of the lemma we have \(V_A(y)=0\), \(\dfrac{\partial V_A(y)}{\partial n_y}=0\) on \(S_1\), and the functions \(U(y)\), \(\dfrac{\partial U(y)}{\partial n_y}\) are bounded on \(S_1\), it follows from (1.5) that (1.3) holds.

2°. Before formulating Lemma 2, let us introduce the definition of a generalized solution of the Dirichlet problem for the Laplace equation, as is done in works [9–10]. We assume henceforth that the boundary \(S\) of the domain \(T\) (generally speaking, multiply connected) has no interior points, i.e. we consider such domains that, in a neighborhood of any point of their boundary, there exist points of the set complementary to \(T\). If in the domain \(T\) the solution of the Dirichlet problem for the Laplace equation arises—

possible for arbitrary continuous values, then we call it a normal domain. Let \(f(x)\) be a continuous function given on the boundary \(S\) of an arbitrarily connected and bounded domain \(T\). Construct a continuous function \(\varphi(x)\) in the whole space, coinciding with \(f(x)\) on the boundary \(S\).

Consider a sequence of normal domains

\[ D_1,\ D_2,\ \ldots,\ D_m,\ \ldots \]

with boundaries

\[ S_1,\ S_2,\ \ldots,\ S_m,\ \ldots, \]

containing the closed domain \(T\cup S\) and converging to the domain \(T\) in such a way that every closed set belonging to \(E^n\setminus \overline{T}\), starting from some number \(m_0\), remains outside the domain \(D_{m_0}\). We may assume that the components of the boundary \(S_m\) are analytic, and moreover \(\overline{D}_{m+1}\subset D_m\) for each \(m\). Denote by \(U_{m\varphi}\) the solution of the Dirichlet problem for the Laplace equation in the domain \(D_m\) with boundary data \(\varphi|_{S_m}\). As is proved in [9—10], the sequence of functions

\[ U_{1\varphi},\ U_{2\varphi},\ \ldots,\ U_{m\varphi},\ \ldots \]

converges in the closed domain \(\overline{T}\), and \(U_{m\varphi}\) converges uniformly in every closed domain \(\overline{T}'\subset T\). The limiting function \(U_f(x)\) satisfies the Laplace equation and does not depend on the special choice of the domains \(D_m\) or on the construction of the function \(\varphi(x)\).

Definition. The function \(U_f(x)\) constructed in the above manner will be called the generalized solution of the Laplace equation for the Dirichlet problem in the domain \(T\) with boundary data \(f(x)\), continuous on the boundary \(S\).

\(3^\circ\). Let \(T\) be a bounded domain with boundary \(S\), satisfying the conditions of item \(2^\circ\), and let \(\mu(y)\) be a summable bounded function in \(E^n\).

Lemma 2. If the density \(\mu(y)\) is such that the potential \(V_T(x)\) of the domain \(T\) is identically equal to zero outside \(T\), i.e.

\[ V_T(x)\equiv 0 \quad \text{for } x\in (E^n\setminus \overline{T}), \]

then, for any generalized solution \(U_f\) of the Dirichlet problem in the domain \(T\), the equality

\[ \int_T \mu(y)U_f(y)\,dy=0 \tag{1.6} \]

holds.

Proof. Construct a sequence of domains

\[ D_1,\ D_2,\ \ldots,\ D_m,\ \ldots\quad (\overline{D}_{m+1}\subset D_m), \]

containing \(T\cup S\), with analytic boundaries \(S_m\), and by the method indicated in item \(2^\circ\) construct the sequence of solutions of the Dirichlet problem for the Laplace equation for the function \(f\), continuous on \(S\):

\[ \{U_{m\varphi}\}\quad (m=1,2,\ldots),\qquad U_{m\varphi}\to U_f \quad \text{as } m\to\infty . \]

For the function \(U_{m\varphi}\), harmonic in the domain \(\overline{D}_{m+1}\), by Lemma 1 we have

\[ \int_T \mu(y)U_{m\varphi}(y)\,dy=0. \tag{1.7} \]

By virtue of the boundedness of \(f(x)\), the extended function \(\varphi(x)\) can be chosen so that \(|f|\le M\), \(|\varphi|\le M=\mathrm{const}\) throughout the whole space. From the maximum principle for a harmonic function it follows that \(|U_{m\varphi}|\), \(|U_f|\le M\) uniformly for all \(m\). Then, according to Lebesgue’s dominated convergence theorem, we shall have

\[ \lim_{m\to\infty}\int_T \mu(y)U_{m\varphi}(y)\,dy = \int_T \mu(y)U_f(y)\,dy . \tag{1.8} \]

From equality (1.7) it follows that the left-hand side of (1.8) is equal to zero; therefore equality (1.6) holds, and the lemma is proved.

§ 2. UNIQUENESS THEOREMS

In this section uniqueness theorems are proved for the solution of the problem posed, under certain restrictions on the domains. First the proof is carried out for finite domains. Then one theorem is extended to open bounded sets.

\(1^\circ\). Let \(T_\alpha\) \((\alpha=1,2)\) be finite domains bounded by surfaces \(S_\alpha\), where \(S_\alpha\) is the boundary of \(E^n\setminus \overline{T}_\alpha\). Denote by \(S^e\) the boundary of the set \(\overline{T}_1\cup \overline{T}_2\). We also introduce notation. If \(T_1\ne T_2\), set

\[ S_1^i=S_1\cap(\overline{T}_1\cap\overline{T}_2),\qquad S_1^e=S_1\setminus S_1^i, \tag{2.1} \]

\[ S_2^e=S_2\cap S^e,\qquad S_2^i=S_2\setminus S_2^e . \tag{2.2} \]

It is not difficult to show that \(S^e=S_1^e\cup S_2^e\). If \(T_1=T_2\), we put \(S_\alpha^e=S_\alpha\) \((\alpha=1,2)\). We note that some of the sets \(S_\alpha^e, S_\alpha^i\) may be empty. In addition, throughout what follows we always assume that each boundary \(S_\alpha\) \((\alpha=1,2)\) belongs to the class \(A^{(1,\lambda)}\). Denote by \((\rho,\theta)\) the spherical coordinates of the point \(x\) of the space \(E^n\) \((n\ge 2)\); by \((\mathbf R,\mathbf n_y)\) denote the scalar product of the vector \(\mathbf R\) with the vector \(\mathbf n_y\).

\(2^\circ\). Theorem 1. If there exists at least one point \(0\in(\overline{T}_1\cap\overline{T}_2)\) such that

1) for the radius vector \(\mathbf R_y=\mathbf 0_y\) (with origin at the point \(0\)) and the unit vector \(\mathbf n_y\) of the exterior normal to the surface \(S_\alpha\) at the point \(y\), the relation

\[ (\mathbf R_y,\mathbf n_y)>0 \quad \text{for } \quad y\in S_1^i,\; S_2^i, \tag{2.3} \]

holds;

2) for the positive function \(\mu(y)\in C^1\) in \(\overline{T}_1\cup\overline{T}_2\), the condition

\[ \frac{\partial}{\partial\rho}\left(\rho^n\mu\right)>0,\quad \rho\ne 0,\quad y\in \overline{T}_1\cup\overline{T}_2 \tag{2.4} \]

\[ (n\ge 2), \]

is satisfied;

3) for the domains \(T_\alpha\), with the given density, the equality

\[ V_{T_1}(x)=V_{T_2}(x) \quad \text{for } \quad x\in E^n\setminus(\overline{T}_1\cup\overline{T}_2), \tag{2.5} \]

holds, then \(T_1=T_2\).

Proof. Denote by \(\overline{T}_0=\overline{T}_1\cap\overline{T}_2\) (one may assume that \(\operatorname{mes} T_0\ne 0\)), and by \(T^e\) the domain bounded by \(S^e\). Introduce the function

\[ \mu^*(y)= \begin{cases} \mu(y), & \text{for } y\in (T_1\setminus T_0),\\ -\mu(y), & \text{for } y\in (T_2\setminus T_0),\\ 0, & \text{for } y\in T_0 . \end{cases} \tag{2.6} \]

Then condition (2.5) of the theorem is written in the form

\[ \int_{T^e} \mu^*(y)K(x,y)\,dy \equiv V_{T^e}(x)=0 \quad \text{for } x\in (E^n\setminus \overline{T^e}). \]

By Lemma 1, for any function \(U\) harmonic in a domain \(D\supset (T_1\cup T_2)\), we have

\[ \int_{T^e} \mu^*(y)U(y)\,dy=0. \]

Or, taking relation (2.6) into account, we rewrite the last equality in the form

\[ \int_{T_1} \mu(y)U(y)\,dy=\int_{T_2} \mu(y)U(y)\,dy. \tag{2.7} \]

We shall prove the theorem by contradiction, assuming that \(T_1\ne T_2\). First of all, note the following: if in (2.7) we put \(U\equiv 1\), then we obtain that the masses of the bodies \(T_1\) and \(T_2\) are equal, and therefore neither of the domains \(T_\alpha\) lies strictly inside the other. In the case when the domains \(T_1\) and \(T_2\) have different connectivity, from the fact that one domain is not contained inside the other and from condition (2.3) we conclude that each of the sets \(S_\alpha^e\) \((\alpha=1,2)\) is nonempty. Let \(H(y)\) be a function harmonic in the domain \(D\); then the function

\[ U(y)=\sum_{k=1}^{n} y_k\,\frac{\partial H(y)}{\partial y_k} \tag{2.8} \]

is also harmonic in the domain \(D\). Substituting the function (2.8) into equality (2.7), we obtain

\[ \int_{T_1} \mu(y)\left[\sum_{k=1}^{n} y_k\,\frac{\partial H}{\partial y_k}\right]\,dy - \int_{T_2} \mu(y)\times \]

\[ \times \left[\sum_{k=1}^{n} y_k\,\frac{\partial H}{\partial y_k}\right]\,dy=0. \]

We rewrite this equality in the form

\[ \int_{T_1}\left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k H)\right]\,dy - \int_{T_2}\left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k H)\right]\,dy - \tag{2.9} \]

\[ -\int_{T_1} H\left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k)\right]\,dy + \int_{T_2} H\left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k)\right]\,dy=0. \]

Transforming, in the first two terms of (2.9), the volume integrals into surface integrals, we obtain

\[ \int_{S_1} H \cdot \mu \left[ \sum_{k=1}^{n} y_k \cos \left(\widehat{y_k \cdot n_y}\right) \right] d_y S - \]

\[ -\int_{S_2} H \cdot \mu \left[ \sum_{k=1}^{n} y_k \cos \left(\widehat{y_k \cdot n_y}\right) \right] d_y S - \]

\[ -\int_{T_2} H \left[ \sum_{k=1}^{n} \frac{\partial}{\partial y_k}(\mu y_k) \right] dy + \]

\[ +\int_{T_2} H \left[ \sum_{k=1}^{n} \frac{\partial}{\partial y_k}(\mu y_k) \right] dy =0. \]

We shall write the last equality in the form

\[ J(H)=0, \tag{2.10} \]

where

\[ J(H)= \int_{S_1} H\mu \cdot (R_y,n_y)\,d_yS - \int_{S_2} H\mu \cdot (R_y,n_y)\,d_yS - \]

\[ -\int_{T_1 \times T_0} H \left[ \sum_{k=1}^{n} \frac{\partial}{\partial y_k}(\mu y_k) \right] dy + \int_{T_2 \times T_0} H \left[ \sum_{k=1}^{n} \frac{\partial}{\partial y_k}(\mu y_k) \right] dy. \]

Define on the surface \(S^e\) the function \(f(y)\) in the following way:

\[ f(y)= \begin{cases} 1 & \text{for } y\in S_1^e,\\ 0 & y\in S_2^e . \end{cases} \tag{2.11} \]

(As was noted earlier, for each of the sets \(\operatorname{mes} S_\alpha^e \ne 0\) \((\alpha=1,2)\).)

We now extend equality (2.10) to a function \(H_f\) harmonic in the domain \(T^e\), which on the boundary of the surface \(S^e\) assumes the values \(f(y)\) everywhere except at points of surface measure zero.

For this purpose we construct a sequence of pieces of surfaces \((S_1^e)_k\), with

\[ \overline{(S_1^e)_k}\subset (S_1^e)_{k+1}, \tag{2.12} \]

and the surface measure
\[ \operatorname{mes}\,[S_1^e\setminus (S_1^e)_k]=\varepsilon_k \]
is such that \(\varepsilon_k\to 0\) as \(k\to\infty\).

Construct a monotonically increasing sequence of continuous functions \(\{f_k(y)\}\) on the surface \(S^e\), \(f_{k+1}(y)\ge f_k(y)\) for points \(y\in S^e\), by setting

\[ f_k(y)= \begin{cases} 1 & \text{for } y\in (S_1^e)_k,\\ 0 & \text{for } y\in S_2^e,\; S_1^e\setminus (S_1^e)_{k+1},\\ \gamma_k(y) & \text{for } y\in (S_1^e)_{k+1}\setminus (S_1^e)_k, \end{cases} \tag{2.13} \]

where \(\gamma_k(y)\) is a continuous function lying between zero and one. For each function \(f_k(y)\), continuous on \(S^e\), we construct the generalized solution \(H_{f_k}(y)\) of the Dirichlet problem. Since \(f_{k+1}(y) \geq f_k(y)\), \(y \in S^e\), it follows, on the basis of the maximum principle in the domain \(\overline{T}^{\,e}\), that

\[ H_{f_{k+1}}(y) \geq H_{f_k}(y) \quad \text{for} \quad y \in (\overline{T}_1 \cup \overline{T}_2), \tag{2.14} \]

and \(|H_{f_k}| < 1\). Consequently, \(\{H_{f_k}\}\) is an increasing sequence of bounded functions in \(\overline{T}^{\,e}\) and harmonic in the domain \(T^e\); by Harnack’s theorem this sequence converges uniformly inside \(T^e\) to a function \(H_f\), where \(H_f\) is harmonic in the domain \(T^e\), and the sequence \(H_{f_k}(y) \to H_f(y)\) for all \(y \in (\overline{T}_1 \cup \overline{T}_2)\).

By known results [9, 10], \(H_{f_k}\) assumes on the boundary \(S^e\) the values \(f_k(y)\) at all regular points of the boundary \(S^e\). Therefore, from the construction of the function \(f_k(y)\) by formula (2.7), it follows that the limiting function \(H_f(y)\) assumes on the boundary \(S^e\) the values of the function \(f(y)\), except for a set of measure zero. Since, for fixed \(k\) (see § 1, item \(2^\circ\)), the sequence \(H_{m\varphi_k}\), which determines the function \(H_{f_k}\), converges in the closed domain \(\overline{T}^{\,e}\) and \(|H_{m\varphi_k}| < 1\) \((m=1,2,\ldots)\), then, according to Lebesgue’s dominated convergence theorem, we have

\[ \lim_{m \to \infty} J(H_{m\varphi_k}) = J(H_{f_k}) \]

for each fixed \(k\). Since, in turn, we have

\[ |H_{f_k}| < 1 \quad (k=1,2,\ldots) \]

and \(H_{f_k}\) converges to \(H_f\) in the closed domain \(\overline{T}^{\,e}\), applying Lebesgue’s theorem again, we obtain

\[ \lim_{k \to \infty} J(H_{f_k}) = J(H_f). \]

In view of (2.10), \(J(H_{m\varphi_k})=0\) for fixed \(k\) and \((m=1,2,\ldots)\); hence \(J(H_{f_k})=0\) for every \(k\). Consequently,

\[ J(H_f)=0, \tag{2.15} \]

where \(H_f\) is a function harmonic in the domain \(T^e\), assuming on the boundary \(S^e\) the values \(f(y)\) almost everywhere, and moreover

\[ 0 < H_f < 1. \tag{2.16} \]

Let us rewrite \(J(H_f)\) in the following form:

\[ J(H_f)= \int_{S_1^e+S_1^i} H_f \mu \cdot (R_y,n_y)\,d_y S - \int_{S_2^e+S_2^i} H_f \mu \cdot (R_y,n_y)\,d_y S - \tag{2.17} \]

\[ - \int_{T_1 \setminus T_0} H_f \left[ \sum_{k=1}^{n} \frac{\partial}{\partial y_k}(\mu y_k) \right]\,dy + \int_{T_2 \setminus T_0} H_f \left[ \sum_{k=1}^{n} \frac{\partial}{\partial y_k}(\mu y_k) \right]\,dy . \]

Since

\[ \frac{\partial}{\partial \rho}(\rho^n \mu) = \rho^{\,n-1} \left( n\mu+\rho\frac{\partial \mu}{\partial \rho} \right) = \rho^{\,n-1} \sum_{k=1}^{n} \frac{\partial}{\partial y_k}(\mu y_k), \tag{2.18} \]

then from condition (2.4) of the theorem it follows that

\[ \sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k)>0 \quad\text{for } \rho\ne 0 \text{ at points } y\in(\overline{T}_1\cup \overline{T}_2). \tag{2.19} \]

Let us estimate the value of the integral \(J(H_f)\). From (2.16), (2.17), and (2.19) it follows that

\[ \begin{aligned} J(H_f)>& \int_{S_1^e+S_1^i} H_f\mu\cdot(R_y,\mathbf n_y)\,d_yS - \int_{S_2^e+S_2^i} H_f\mu\cdot(R_y,\mathbf n_y)\,d_yS \\ &- \int_{T_1\setminus T_0} 1\cdot \left[ \sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k) \right]\,dy + \int_{T_2\setminus T_0} H_f \left[ \sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k) \right]\,dy . \end{aligned} \]

Using the fact that

\[ \int_{T_1\setminus T_0} \left[ \sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k) \right]\,dy = \int_{S_1^e}\mu\cdot(R_y,\mathbf n_y)\,d_yS - \int_{S_2^i}\mu\cdot(R_y,\mathbf n_y)\,d_yS, \]

where \(\mathbf n_y\) is the exterior normal to the surface \(S_\alpha\) \((\alpha=1,2)\), from the inequality for \(J(H_f)\) we obtain

\[ \begin{aligned} J(H_f)>& \int_{S_1^e+S_1^i} H_f\mu\cdot(R_y,\mathbf n_y)\,d_yS - \int_{S_2^e+S_2^i} H_f\mu\cdot(R_y,\mathbf n_y)\,d_yS \\ &- \int_{S_1^e}\mu\cdot(R_y,\mathbf n_y)\,d_yS + \int_{S_2^i}\mu\cdot(R_y,\mathbf n_y)\,d_yS \\ &+ \int_{T_2\setminus T_0} H_f \left[ \sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k) \right]\,dy . \end{aligned} \]

As was already noted earlier, on the boundary \(S^e=S_1^e\cup S_2^e\) the function \(H_f\) assumes the values \(f(y)\) almost everywhere; therefore, substituting the data from (2.11) into the last inequality, we obtain the following inequality:

\[ \begin{aligned} J(H_f)>& \int_{S_1^i} H_f\mu\cdot(R_y,\mathbf n_y)\,d_yS + \int_{S_2^i} (1-H_f)\mu\cdot(R_y,\mathbf n_y)\,d_yS \\ &+ \int_{T_2\setminus T_0} H_f \left[ \sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k) \right]\,dy . \end{aligned} \tag{2.20} \]

From (2.15) and (2.19) it follows that

\[ \int_{T_2\setminus T_0} H_f \left[ \sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\mu y_k) \right]\,dy>0 . \]

From (2.15), in view of the fact that \(\mu(y)>0\) for \(y\in \overline{T}_1\cup \overline{T}_2\), and from the second condition of the theorem, it follows that

\[ \int_{S_1^i} H_f\cdot \mu\cdot(\mathbf R_y,\mathbf n_y)\,d_yS \geq 0,\quad \int_{S_2^i} (1-H_f)\cdot \mu\cdot(\mathbf R_y,\mathbf n_y)\,d_yS \geq 0. \tag{2.21} \]

Finally we obtain \(J(H_f)>0\), which contradicts (2.15). The theorem is proved.

1. For the case \((n=2)\) of the logarithmic potential with constant density we shall prove the theorem for domains constructed below.

Let \(\hat G\) be an arbitrary bounded simply connected domain of the plane \(w=\xi_1+i\xi_2\), and let \(f(z)\) \((z=\rho e^{i\theta})\), \(f(0)=0\), \(f'(0)=1\), be a function holomorphic and univalent in the disk \(|z|<R\) (\(R\) an arbitrary positive number), mapping the disk \(|z|<R\) onto the domain \(\hat G\) of the \(w\)-plane. Denote by \(K_{\rho_0}\) the disk \(|z|<\rho_0\), where \(\rho_0=\mathrm{const}\) and \(0<\rho_0<(2-\sqrt3)R\). Let \(T_\alpha\) \((\alpha=1,2)\) be connected bounded domains, with \(T_\alpha\subset K_{\rho_0}\), and suppose that there exists at least one point \(0\in(\overline T_1\cap \overline T_2)\), such that \((\mathbf R_y,\mathbf n_y)\geq 0\) for \(y\in S_1^i,S_2^i\). Denote by \(G_\alpha\) \((\alpha=1,2)\) the connected domains of the \(w\)-plane that are the images of the domains \(T_\alpha\) under the mapping of the disk \(|z|<R\) by the above function onto the domain \(\hat G\).

Theorem 2. If, under the assumptions made, in the domains \(G_\alpha\) \((\alpha=1,2)\) the equality

\[ \int_{G_1}\ln\frac{1}{r_{\xi\eta}}\,d\eta = \int_{G_2}\ln\frac{1}{r_{\xi\eta}}\,d\eta \quad \text{for } \xi\in E^2\setminus(\overline G_1\cup \overline G_2), \tag{2.22} \]

holds, then \(G_1=G_2\).

Proof. First of all, note that from Koebe’s distortion theorem the following assertion is valid [20]. If \(f(z)\), \(f(0)=0\), \(f'(0)=1\) \((z=\rho e^{i\theta})\), is a function holomorphic and univalent in the disk \(|z|<R\), then

\[ \frac{\partial}{\partial\rho}\,|zf'(z)|>0 \quad \text{for } 0<|z|<\rho_0,\ \text{where } 0<\rho_0<(2-\sqrt3)R. \tag{2.23} \]

Indeed, by the distortion theorem we have

\[ \frac{R-|z|}{(R+|z|)^3}\leq |f'(z)|\leq \frac{R+|z|}{(R-|z|)^3}, \]

or, rewriting this otherwise,

\[ \frac{2|z|-4R}{R^2-|z|^2}\leq \frac{\partial}{\partial |z|}\ln |f'(z)|\leq \frac{2|z|+4R}{R^2-|z|^2}. \tag{2.24} \]

Since

\[ \frac{\partial}{\partial |z|}\ln |z|=\frac{1}{|z|}, \]

then, adding this equality to the left-hand side of inequality (2.24), we obtain

\[ \frac{\partial}{\partial |z|}\ln |zf'(z)| \geq \frac{z^2-4R|z|+R^2}{|z|(R^2-|z|^2)}, \]

whence, for \(0<|z|=\rho<\rho'\), where \(\rho'=(2-\sqrt3)R\), it follows that

\[ \frac{\partial}{\partial \rho}\bigl(|zf'(z)|\bigr)>0 \quad \text{for } \rho\ne 0. \]

We pass to the proof of the theorem by the method of contradiction. From the assertion of Lemma 1 we obtain

\[ \int_{G_1} V\,d\xi=\int_{G_2} V\,d\xi \quad (d\xi=d\xi_1\,d\xi_2), \tag{2.25} \]

where \(V(\xi)\) is a function harmonic in the domain \(G^*\supset (G_1\cup \overline{G}_2)\). We make the change of variables \(w=f(z)\), \(w=\xi_1+i\xi_2\), \(z=x_1+ix_2\), where \(f(z)\) is the function defining the domains \(G_\alpha\); then we obtain

\[ \int_{T_1} \mu(x)U(x)\,dx=\int_{T_2}\mu(x)U(x)\,dx, \tag{2.26} \]

where \(\mu(x)=|f'(z)|^2\), and \(U(x)\) is a function harmonic in the domain \(T^*\supset(\overline{T}_1\cup \overline{T}_2)\). Since

\[ \frac{\partial}{\partial \rho}\bigl(\rho^2\mu\bigr) = \frac{\partial}{\partial \rho}\bigl[\rho f'(z)\bigr]^2 = 2\rho |f'(z)|\,\frac{\partial}{\partial \rho}\,|zf'(z)|, \]

it follows from (2.23) that \(\dfrac{\partial}{\partial \rho}\bigl(\rho^2\mu\bigr)>0\) for \(\rho\ne 0\). Further, to complete the proof of Theorem 2 the construction of the proof of Theorem 1 is repeated.

Let us note that, for the logarithmic potential in the case when the domains \(T_1\) and \(T_2\) are star-shaped with respect to a common interior point, Theorems 1 and 2 were proved in the work of V. P. Simonov [20].

\(4^\circ\). For the case of constant density in the space \(E^n\) (\(n\ge 2\)) the following theorem is proved. Let \(\gamma\) and \(\beta\) be two arbitrary numbers, with \(\gamma^2+\beta^2\ne 0\), and let \(\mathbf q=(q_1,\ldots,q_n)\) be a constant vector.

Theorem 3. If there exists a point \(0\), numbers \(\gamma,\beta\), and a vector \(\mathbf q\) such that: 1) for the surfaces \(S_\alpha\) (\(\alpha=1,2\)) the relation holds

\[ \int_{S_1^i} |\Phi(y)|\,d_yS+\int_{S_2^i}|\Phi(y)|\,d_yS < \int_{S_1^e}|\Phi(y)|\,d_yS+\int_{S_2^e}|\Phi(y)|\,d_yS, \tag{2.27} \]

where the expression

\[ \Phi(y)=(\gamma R_y+\beta \mathbf q,\mathbf n_y) \tag{2.28} \]

is the scalar product of the vector \(\gamma R_y+\beta \mathbf q\) with the unit vector of the exterior normal \(\mathbf n_y\) to the surface \(S_\alpha\) (\(\alpha=1,2\));

2) for the domains \(T_\alpha\), with density \(\mu\equiv 1\), the equality

\[ V_{T_1}(x)=V_{T_2}(x)\quad \text{for } x\in E^n\setminus(\overline{T}_1\cup \overline{T}_2), \tag{2.29} \]

holds, then \(T_1=T_2\).

Proof. Suppose the contrary, that \(T_1\ne T_2\). In view of condition (2.29) and Lemma 1, it follows that

\[ \int_{T_1} U\,dy=\int_{T_2} U\,dy \tag{2.30} \]

for a function \(U(y)\) harmonic in the domain \(D\supset(\overline{T}_1\cup \overline{T}_2)\). Substituting into (2.30) the function

\[ U(y)=\sum_{k=1}^{n}\frac{\partial}{\partial y_k}\bigl[(\gamma y_k+\beta q_k)H(y)\bigr], \]

where \(H(y)\) is a harmonic function in the domain \(D \supset (\overline{T}_1 \cup \overline{T}_2)\), we obtain

\[ \int_{T_1}\left\{\sum_{k=1}^{n}\frac{\partial}{\partial y_k}\left[(\gamma y_k+\beta q_k)H(y)\right]\right\}dy = \int_{T_2}\left\{\sum_{k=1}^{n}\frac{\partial}{\partial y_k}\left[(\gamma y_k+\beta q_k)H(y)\right]\right\}dy . \tag{2.31} \]

Transforming the volume integrals into surface integrals, we obtain

\[ \int_{S_1} H(y)(\gamma R_y+\beta q,n_y)\,d_yS = \int_{S_2} H(y)(\gamma R_y+\beta q,n_y)\,d_yS . \tag{2.32} \]

Using the notation (2.28), we rewrite equality (2.32) in the form

\[ \int_{S_1^e} H(y)\overline{\Phi}(y)\,d_yS - \int_{S_2^e} H(y)\overline{\Phi}(y)\,d_yS = \]

\[ = \int_{S_2^i} H(y)\Phi(y)\,d_yS - \int_{S_1^i} H(y)\Phi(y)\,d_yS . \tag{2.33} \]

Define on the surface \(S^e\) the function \(f(y)\) as follows:

\[ f(y)= \begin{cases} -\operatorname{sign}\Phi(y), & \text{for } y\in S_1^e,\\ \operatorname{sign}\Phi(y), & \text{for } y\in S_2^e . \end{cases} \tag{2.34} \]

As in the proof of Theorem 1, we extend equality (2.33) to the function \(H_f\), harmonic in the domain \(T^e\), which on the boundary \(S^e\) almost everywhere takes the values \(f(y)\). Indeed, the function \(f(y)\) can also be written in the form

\[ f(y)= \begin{cases} 1, & \text{for } y\in S_*^e,\\ -1, & \text{for } y\in (S^e\setminus S_*^e). \end{cases} \]

Define on the surface \(S^e\) the functions

\[ f^1(y)= \begin{cases} 1, & \text{for } y\in S_*^e,\\ 0, & \text{for } y\in (S^e\setminus S_*^e); \end{cases} \]

\[ f^2(y)= \begin{cases} 0, & \text{for } y\in S_*^e,\\ 1, & \text{for } y\in (S^e\setminus S_*^e). \end{cases} \]

Analogously to subsection \(2^\circ\), from the functions \(f^1\) and \(f^2\) we construct generalized solutions \(H_{f^1}\), \(H_{f^2}\), for which

\[ H_{f_k^1}\to H_{f^1},\quad H_{f_k^2}\to H_{f^2}, \]

and then

\[ H_{f_k^1}-H_{f_k^2}\to H_{f^1}-H_{f^2}\quad \text{in } \overline{T}^{\,e}. \]

Therefore

\[ H_f=H_{f^1}-H_{f^2}. \tag{2.35} \]

there is the sought function, harmonic in the domain \(T^e\), which on the boundary \(S^e\) almost everywhere assumes the values \(f(y)\). By Lebesgue’s theorem on passage to the limit and by transformations as in § 2, from (2.33) we obtain

\[ \int_{S_1^e} H_f \Phi(y)\,d_yS-\int_{S_2^e} H_f \Phi(y)\,d_yS = \]

\[ = \int_{S_2^i} H_f \Phi(y)\,d_yS-\int_{S_1^i} H_f \Phi(y)\,d_yS . \tag{2.36} \]

Substituting in (2.36), in place of \(H_f\), the values \(f(y)\) for points \(y\in S^e\), we have

\[ \int_{S_1^e} |\Phi(y)|\,d_yS+\int_{S_2^e} |\Phi(y)|\,d_yS = \]

\[ = \int_{S_2^i} H_f \Phi(y)\,d_yS-\int_{S_1^i} H_f \Phi(y)\,d_yS . \tag{2.37} \]

Since, by the maximum principle, \(|H_f|\leq 1\), it follows from (2.37) that

\[ \int_{S_1^e} |\Phi(y)|\,d_yS+\int_{S_2^e} |\Phi(y)|\,d_yS \leq \]

\[ \leq \int_{S_2^i} |\Phi(y)|\,d_yS+\int_{S_1^i} |\Phi(y)|\,d_yS, \]

which contradicts the condition of the theorem (2.27). The theorem is proved.

Remark 1. If \(\gamma=1\) and \(\beta=0\), then \(\Phi(y)=(R_y,n_y)\), and the condition of the theorem (2.27) is replaced by the condition

\[ \int_{S_1^i} |(R_y,n_y)|\,d_yS+\int_{S_2^i} |(R_y,n_y)|\,d_yS < \]

\[ < \int_{S_1^e} |(R_y,n_y)|\,d_yS+\int_{S_2^e} |(R_y,n_y)|\,d_yS, \tag{2.27'} \]

which refines the formulation of Theorem 1 of [26]. We also note that the assertion of Theorem 3 remains valid if, instead of condition (2.27′), one requires the condition

\[ \int_{S_1^i} |(R_y,n_y)|\,d_yS+\int_{S_2^i} |(R_y,n_y)|\,d_yS < \]

\[ < \int_{S_1^e} (R_y,n_y)\,d_yS+\int_{S_2^e} (R_y,n_y)\,d_yS. \tag{2.27''} \]

Obviously, (2.27′) follows from (2.27″). Let us note that condition (2.27′) is satisfied, in particular, when \(\overline{T}_1\cap \overline{T}_2\) is star-shaped with respect to some point \(0\in \overline{T}_1\cap \overline{T}_2\) and each of the sets \(S_\alpha^e\) \((\alpha=1,2)\) is nonempty, while \(\operatorname{mes}(S_1^i\cap S_2^i)=0\).

Indeed, let there be a finite domain \(D\) of the space \(E^n\), bounded by a surface \(S\). Then the volume of the domain \(D\) is represented in the form

\[ \begin{aligned} \operatorname{mes} D &= \int_D dy = \frac{1}{n}\int_D n\,dy \\ &= \frac{1}{n}\int_D \left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}y_k\right]dy = \frac{1}{n}\int_S (\mathbf R_y,\mathbf n_y)\,d_yS . \end{aligned} \tag{2.38} \]

Consequently, we have the formula

\[ \operatorname{mes} D=\frac{1}{n}\int_S(\mathbf R_y,\mathbf n_y)\,d_yS . \tag{2.39} \]

In particular, if the set \(\overline T_1\cap \overline T_2\) is star-shaped with respect to at least one point \(0\in \overline T_1\cap \overline T_2\) and \(\operatorname{mes}(S_a^e)\ne0,\operatorname{mes}(S_1^i\cap S_2^i)=0\), then \((\mathbf R_y,\mathbf n_y)>0\) on \(S_a^i\) \((a=1,2)\), and from (2.39) we obtain that condition \((2.27')\) means

\[ \operatorname{mes}(T_1\cup T_2)>\operatorname{mes}(T_1\cap T_2). \]

Remark 2. If \(\gamma=0\) and \(\beta=1\), then \(\Phi(y)=(\mathbf q,\mathbf n_y)\), and condition (2.27) of Theorem 3 is replaced by the condition

\[ \int_{S_1^i}|(\mathbf q,\mathbf n_y)|\,d_yS + \int_{S_2^i}|(\mathbf q,\mathbf n_y)|\,d_yS < \int_{S_1^e}|(\mathbf q,\mathbf n_y)|\,d_yS + \int_{S_2^e}|(\mathbf q,\mathbf n_y)|\,d_yS, \tag{2.27'''} \]

which refines Theorem 2 of [26]. In particular, if \(\operatorname{mes}(S_1^i\cap S_2^i)=0\) and a straight line parallel to the vector \(\mathbf q\) intersects \(S_1^i\cup S_2^i\) in no more than two points or segments, then condition \((2.27''')\) holds. We shall prove this assertion when justifying Remark 3.

Remark 3. Conditions \((2.27')\) and \((2.27''')\) can be given a certain geometric interpretation.

Let there be some piece of surface

\[ \widehat S=\widehat S^1+\widehat S^2+\ldots+\widehat S^m, \]

where a ray issuing from the point \(0\) intersects \(\widehat S^1,\ldots,\widehat S^m\) at one point or along one segment. Denote

\[ \|V_{K(\widehat S)}\| = |V_{K(\widehat S^1)}| + |V_{K(\widehat S^2)}| +\ldots+ |V_{K(\widehat S^m)}|, \tag{2.40} \]

where \(|V_{K(\widehat S^l)}|\), \(l=1,\ldots,m\), denotes the absolute value of the volume of the cone with vertex at the point \(0\), constructed on the piece of surface \(\widehat S^l\) \((l=1,\ldots,m)\). In this notation condition \((2.27')\) is written in the form

\[ \|V_{K(S_1^i)}\|+\|V_{K(S_2^i)}\| < \|V_{K(S_1^e)}\|+\|V_{K(S_2^e)}\|. \tag{2.41} \]

Similarly, let the piece of surface

\[ \widehat S=\widehat S^1+\widehat S^2+\ldots+\widehat S^m \]

such that every line parallel to the vector \(\mathbf q\) projects each of the pieces \(\hat S^1,\ldots,\hat S^m\) in a one-to-one manner onto the plane \(N\), perpendicular to the vector \(\mathbf q\). Then denote

\[ \|\sigma_{\Pi(\hat S)}\|=|\sigma_{\Pi(\hat S^1)}|+|\sigma_{\Pi(\hat S^2)}|+\cdots+|\sigma_{\Pi(\hat S^m)}|, \tag{2.42} \]

where \(|\sigma_{\Pi(\hat S^l)}|\), \((l=1,\ldots,m)\), denotes the absolute value of the \((n-1)\)-dimensional area of the projection \(\Pi(\hat S^l)\) of the piece of surface \(\hat S^l\) onto the plane \(N\). In this notation condition \((2.27''')\) is replaced by the condition

\[ \|\sigma_{\Pi(\hat S_1^i)}\|+\|\sigma_{\Pi(\hat S_2^i)}\| < \|\sigma_{\Pi(\hat S_1^e)}\|+\|\sigma_{\Pi(\hat S_2^e)}\|. \tag{2.43} \]

Indeed, from formula (2.39) we have that the volume of the cone constructed on the piece of surface \(\hat S^l\) with center at the point \(0\), taking account of sign, is written in the form

\[ V_{K(\hat S^l)}=\frac1n\int_{\hat S^l}|(\mathbf R_y,\mathbf n_y)|\,d_yS, \]

since on the conical surface \((\mathbf R_y,\mathbf n_y)=0\). Therefore, the absolute value of this same cone is

\[ |V_{K(\hat S^l)}|=\frac1n\int_{\hat S^l}|(\mathbf R_y,\mathbf n_y)|\,d_yS. \]

Therefore, from the notation (2.40), condition \((2.27')\) is rewritten in the form (2.41). Similarly, for condition (2.42) note that

\[ d\sigma^m=|(\mathbf q,\mathbf n_y)|\,d_y\hat S^m, \]

where \(d\sigma^m\) is an element of \((n-1)\)-dimensional area of the projection \(\Pi(\hat S^m)\) of the piece of surface \(\hat S^m\) onto the plane \(N\). Therefore

\[ |\sigma_{\Pi(\hat S^m)}|=\int_{\hat S^m}|(\mathbf q,\mathbf n_y)|\,d_yS. \tag{2.44} \]

From the notation (2.42) and \((2.27''')\) follows (2.43). In particular, from (2.44) follows the assertion of Remark 2.

\(5^\circ\). Let \(A_1\) and \(A_2\) be open bounded sets, each of which consists of a finite number of domains

\[ A_1=\bigcup_{j=1}^{j_1}T_1^j,\qquad A_2=\bigcup_{j=1}^{j_2}T_2^j, \]

where \(j_1,j_2\) are fixed numbers, and moreover \(S_1^j,S_2^j\in A^{(1,\lambda)}\). Denote by \(\Gamma^e\) the boundary of the set \(\overline{A_1\cup A_2}\). If \(A_1\ne A_2\), then denote

\[ \begin{gathered} \Gamma_1^i=\Gamma\overline{A_1}\cap(\overline{A_1}\cap\overline{A_2}),\qquad \Gamma_1^e=\Gamma A_1\setminus\Gamma_1^i,\\ \Gamma_2^e=\Gamma A_2\cap\Gamma^e,\qquad \Gamma_2^i=\Gamma A_2\setminus\Gamma_2^e. \end{gathered} \tag{2.45} \]

If \(A_1=A_2\), then we set \(\Gamma_\alpha^e=\Gamma A_\alpha\), \((\alpha=1,2)\).

Theorem 4. If there exist a point \(0\), numbers \(\gamma,\beta\) \((\gamma^2+\beta^2\ne 0)\), and a constant vector \(\mathbf q\) such that:

1)

\[ \int_{\Gamma_1^i} |\Phi(y)|\,d_yS+\int_{\Gamma_2^i}|\Phi(y)|\,d_yS< \int_{\Gamma_1^e}|\Phi(y)|\,d_yS+\int_{\Gamma_2^e}|\Phi(y)|\,d_yS, \tag{2.46} \]

where

\[ \Phi(y)=(\gamma,\mathbf R_y+\beta \mathbf q,\mathbf n_y); \]

2) for constant density \(\mu=1\) we have

\[ V_{A_1}(x)=V_{A_2}(x)\quad \text{for } x\in E^n\setminus(\overline{A}_1\cup\overline{A}_2), \tag{2.47} \]

then \(A_1=A_2\).

Proof. Consider the set \(\overline{A}_1\cup\overline{A}_2\). If \((\overline{A}_1\cup\overline{A}_2)\) is not a connected set, then construct a set \(A^*\subset [E^n\setminus(\overline{A}_1\cup\overline{A}_2)]\) such that the set \(\overline{A}^*\cap(\overline{A}_1\cup\overline{A}_2)\) consists of a finite number of points. We may assume that \(A^*\) consists of the union of domains whose boundaries, with the exception of a finite number of points, belong to \(A^{(1,\lambda)}\); moreover, by construction the set \(A^*\) is chosen so that \(\overline{A}_1\cup\overline{A}_2\cup\overline{A}^*\) is a connected set. Denote

\[ \overline{B}_1=\overline{A}_1\cup\overline{A}^*,\qquad \overline{B}_2=\overline{A}_2\cup\overline{A}^* . \tag{2.48} \]

Consider the potential of density \(\mu=1\) of the set \(A^*\), which we denote by \(V_{A^*}(x)\). From condition (2.47) of the theorem it follows that

\[ V_{A_1}(x)+V_{A^*}(x)=V_{A_2}(x)+V_{A^*}(x) \quad \text{for } x\in E^n\setminus(\overline{A}_1\cup\overline{A}_2\cup\overline{A}^*), \]

i.e.

\[ V_{B_1}(x)=V_{B_2}(x)\quad \text{for } x\in E^n\setminus(\overline{B}_1\cup\overline{B}_2). \tag{2.49} \]

Since \(\overline{B}_1\cup\overline{B}_2\) is a connected set, on the basis of Lemma 1 we have

\[ \int_{B_1} U\,dy=\int_{B_2} U\,dy \tag{2.50} \]

for any function \(U(y)\) harmonic in a domain \(D\supset(\overline{B}_1\cup\overline{B}_2)\). The last equality, on the basis of (2.48), can be rewritten in the form

\[ \int_{A_1} U(y)\,dy=\int_{A_2} U(y)\,dy \tag{2.51} \]

for any harmonic function \(U\) defined in a domain \(D\supset(\overline{B}_1\cup\overline{B}_2)\).

Next, as in the proof of Theorem 3, define on \(\Gamma_1^e\cup\Gamma_2^e\cup\Gamma A^*\) a function of the form

\[ f(y)= \begin{cases} -\operatorname{sign}\Phi(y), & \text{for } y\in\Gamma_1^e,\\ \operatorname{sign}\Phi(y), & \text{for } y\in\Gamma_2^e,\\ 0, & \text{for } y\in\Gamma A^* . \end{cases} \tag{2.52} \]

Then, taking as \(U\) a function of the form

\[ U=\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\gamma y_k+\beta q_k)H \]

and repeating the arguments used in the proof of Theorem 3, we obtain that

\[ \int_{\Gamma_1} H_f \Phi(y)\,d_yS-\int_{\Gamma_2} H_f \Phi(y)\,d_yS=0. \]

We write this equality in the form

\[ \int_{\Gamma_1^e} H_f \Phi(y)\,d_yS-\int_{\Gamma_2^e} H_f \Phi(y)\,d_yS = \int_{\Gamma_2^i} H_f \Phi(y)\,d_yS-\int_{\Gamma_1^i} H_f \Phi(y)\,d_yS. \tag{2.53} \]

Since \(|H_f|\leq 1\) on the set \(\overline{A}_1\cup\overline{A}_2\), using the boundary data (2.52), from (2.53) we obtain

\[ \int_{\Gamma_1^e}|\Phi(y)|\,d_yS+\int_{\Gamma_2^e}|\Phi(y)|\,d_yS \leq \int_{\Gamma_1^i}|\Phi(y)|\,d_yS+\int_{\Gamma_2^i}|\Phi(y)|\,d_yS, \]

which contradicts condition (2.46). The theorem is proved.

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Received by the editors
September 7, 1965

Institute of Mathematics, Siberian Branch, Academy of Sciences of the USSR