UDC 517.946. 42:947.42

ON INVERSE PROBLEMS OF POTENTIAL THEORY

A. I. PRILEPKO

The paper investigates two inverse problems of potential theory: Problem 1—the determination of the density of a given body from its external potential; Problem 2—the determination of the shape of a body of given density from its external potential. Problems 1 and 2 are equivalent to the study of a class of linear and nonlinear integral equations of the first kind. Both of these problems have non-unique solutions and belong to problems that are ill-posed in the classical sense, methods for solving which are developed in works [1, 2, 5, 6, 8—10, 17, 18] and others. The main content of the paper is devoted to the uniqueness of the solution of the indicated problems. The uniqueness of the solution of Problem 1 for the volume potential was studied in [12, b)]. The same problem in the case of the Newtonian potential of a star-shaped domain of one class of densities was considered in [1, a)], and for the potential of a simple layer in [7]. Under certain restrictions, approximate methods for solving this kind of problem are available in works on geophysics [16] and others. The uniqueness of the solution of Problem 2 for variable density in the two-dimensional case by methods of the theory of functions of a complex variable was studied in [13, 21], and in the general case in [14] and [12, b)], where the literature is indicated in more detail. The mixed inverse problem is considered in [3].

In § 1 of the present paper, uniqueness theorems are proved for the solution of Problem 1 for more general classes of densities than in [12, b)], and also for certain classes of densities and bodies the solution is written out or a method for finding it is indicated. In § 2, uniqueness theorems are proved for the solution of Problem 2 for classes of variable densities, generally speaking, not of constant sign, in contrast to works [11, 12, a), 13—15, 19, 21], where positive density is considered. The solution of Problem 2 with variable density not of constant sign is closely connected with the study of the inverse problem of electrical prospecting for variable densities [20].

Let us note that the study of Problems 1 and 2 is carried out for metaharmonic \((\varkappa \geqslant 0)\) and Newtonian \((\varkappa = 0)\) potentials. In particular, for the case \((\varkappa = 0)\) the results obtained are new also for the Newtonian potential.

\(1^\circ\). Let \(A_\alpha(\alpha = 1, 2)\) be open bounded sets of the space \(E^n(n \geqslant 2)\). We denote the metaharmonic \((\varkappa \geqslant 0)\) potential [12, b)] of the set \(A_\alpha\) with density \(\mu_\alpha(\alpha = 1, 2)\), almost everywhere nonzero for points \(y \in A_\alpha\), by

\[ V(x; A_\alpha, \mu_\alpha)=\int_{A_\alpha}\mu_\alpha(y)K(x,y)\,dy, \tag{0.1} \]

where \(x=(x_1,\ldots,x_n)\), \(y=(y_1,\ldots,y_n)\), \(K(x,y)\) is the fundamental solution of the metaharmonic equation

\[ \Delta u-\varkappa^{2}u=0 \quad (\varkappa=\mathrm{const}\geqslant 0) \tag{0.2} \]

with the corresponding decay at infinity.

§ 1. On determining the density of a given body from the external potential

Let \(A=A_1=A_2\) be an open set consisting of a finite number of bounded domains \(T_j\), whose boundaries \(S_j\) belong to the class \(A^{(1,\lambda)}\). Denote by

\[ V(x; A,\mu_\alpha)=V(x; A_\alpha,\mu_\alpha)\quad (\alpha=1,2) \tag{1.1} \]

the potential of the given set \(A\) with density \(\mu_\alpha(y)\ne 0\) almost everywhere for \(y\in A\).

\(2^\circ\). Consider the class of functions \(\mu_\alpha(y)\) \((\alpha=1,2)\) satisfying the conditions

\[ \mu_\alpha(y)=\eta(y)v_\alpha(y), \tag{1.2} \]

where

a) \(v_\alpha(y)\) are bounded functions for \(y\in \overline A\), and moreover
\[ \frac{\partial v_\alpha}{\partial y_{k_1}}=0 \]
(\(k_1\) is a fixed number);

b) \(\eta(y)>0\), and either
\[ \frac{\partial}{\partial y_{k_1}}\eta(y)\geqslant 0 \quad \text{for } y\in \overline A, \]
or
\[ \frac{\partial}{\partial y_{k_1}}\eta(y)\leqslant 0 \quad \text{for } y\in \overline A. \]

Theorem 1 (case \(\varkappa\geqslant 0\)). If, for functions \(\mu_\alpha(y)\) of the class (1.2), the external potentials of the given set \(A\) satisfy the condition

\[ V(x; A,\mu_1)=V(x; A,\mu_2)\quad \text{for } x\in E^n\setminus \overline A, \tag{1.3} \]

then \(\mu_1(y)=\mu_2(y)\) for \(y\in A\).

Proof. Suppose that the set \(A\) consists of one finitely connected domain \(T\), bounded by a surface \(S\) of class \(A^{(1,\lambda)}\). The proof of the theorem in the general case is carried out analogously. From condition (1.3) it follows that for any metaharmonic function \(U(y)\) in a domain \(D\supset \overline T\) the equality \([12,\ \mathrm{в}]\) holds:

\[ \int_T \mu_1(y)U(y)\,dy=\int_T \mu_2(y)U(y)\,dy. \tag{1.4} \]

Denote \(\mu(y)=\mu_1(y)-\mu_2(y)\). We carry out the proof of the theorem by contradiction, assuming that \(\mu(y)\ne 0\) for \(y\in T\). From condition (1.2, a) we have that the function \(v(y)=v_1(y)-v_2(y)\) does not have constant sign on \(S\). Indeed, if, for example, the sign of the function \(v(y)\) were positive on the boundary \(S\), then the function \(\mu(y)>0\) for \(y\in T\). Choosing in the equality (1.4) as \(U(y)\) a positive solution of equation (0.1) in the domain \(D\), we obtain a contradiction. Consequently, \(\operatorname{sign}\mu(y)\ne \mathrm{const}\) for \(y\in S\). Suppose \(v(y)\ne 0\). Taking in equality (1.4)

\[ U(y)=-\frac{\partial H(y)}{\partial y_{k_1}}, \]

where \(H(y)\) is a metaharmonic function in the domain \(D\supset T\), we obtain the equality

\[ \int_S H(y)\mu(y)(q,n_y)\,d_yS-\int_T H(y)\frac{\partial \mu(y)}{\partial y_{k_1}}\,dy=0, \tag{1.5} \]

where \((\mathbf q,\mathbf n_y)\) is the scalar product of the unit vector \(\mathbf q\) (the direction of the vector \(\mathbf q\) coincides with the direction of the axis \(Oy_{k_1}\)) and \(\mathbf n_y\), the unit vector of the exterior normal at the point \(y \in S\).

1. Consider the case
\[ \frac{\partial \eta(y)}{\partial y_{k_1}} > 0 \]
for \(y \in \overline A\). Define on the surface \(S\) the function \(f(y)\) as follows:
\[ f(y)=\operatorname{sign}\mu(y)\quad \text{for } y\in S. \tag{1.6} \]

We note that \(f(y)\not\equiv \mathrm{const}\) for \(y\in S\). Extending the equality (1.5) to the function \(H_f(y)\), which is metaharmonic in the domain \(T\) and on the boundary \(S\) almost everywhere takes the values (1.6), we obtain the equality
\[ J(H_f)=0, \tag{1.7} \]
where
\[ J(H_f)=\int_S |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS -\int_{T'} H_f(y)\frac{\partial \mu(y)}{\partial y_{k_1}}\,dy, \tag{1.8} \]
and
\[ T'=\left\{\,y\in T,\ \frac{\partial}{\partial y_{k_1}}\eta(y)>0\,\right\}, \]
i.e. \(T'\) denotes the set of points \(y\in T\) for which
\[ \frac{\partial}{\partial y_{k_1}}\eta(y)>0. \]

For the purpose of simplifying the proof of the theorem, we assume that \(T'\) is a domain, and that the boundary \(S'\) of the domain \(T'\) belongs to the class \(A^{(1,\lambda)}\); moreover, \(\operatorname{mes} T'\ne 0\). Since \(|H_f(y)|<1\) for \(y\in T\), it follows from the definitions of \(T'\) and the property of the function \(\nu(y)\) (1.2, a) that
\[ \int_{T'} H_f(y)\frac{\partial\mu(y)}{\partial y_{k_1}}\,dy < \int_{S'} |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS. \tag{1.9} \]

If \(T'=T\), then from (1.8) and (1.9) we obtain \(J(H_f)>0\).

If \(T'\subset T\), and a straight line parallel to the vector \(\mathbf q\) intersects each of the boundaries \(S'\) and \(S\) in no more than two points, then in this case from (1.8) and (1.9) we obtain
\[ \begin{aligned} J(H_f) &> \int_S |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS -\int_{S'} |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS \\ &= \left[ \int_{S^+} |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS - \int_{S'{}^+} |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS \right] \\ &\quad+ \left[ -\int_{S'{}^-} |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS + \int_{S^-} |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS \right], \end{aligned} \tag{1.10} \]
where it is denoted
\[ S^+=\{y\in S,\ (\mathbf q,\mathbf n_y)>0\},\qquad S^-=\{y\in S,\ (\mathbf q,\mathbf n_y)\le 0\}, \]
\[ S'{}^+=\{y\in S',\ (\mathbf q,\mathbf n_y)>0\},\qquad S'{}^-=\{y\in S',\ (\mathbf q,\mathbf n_y)\le 0\}. \]

From the notation and condition (1.2, a), we observe that each expression in square brackets on the right-hand side of inequality (1.10) is nonnegative, and at least one of them is strictly greater than zero; therefore \(J(H_f)>0\). In the case,

if one does not require additional conditions on the structure of the boundaries \(S'\) and \(S\), then, by a slight modification of the partitions, we find that

\[ \int\limits_{\dot S} |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS - \int\limits_{\dot S'} |\mu(y)|(\mathbf q,\mathbf n_y)\,d_yS>0. \]

Consequently \(J(H_f)>0\). Carrying out similar arguments in all other cases of the structure of the set \(T'\), we find that \(J(H_f)>0\); this contradicts (1.7).

The theorem for case 1 is proved.

1. Consider the case

\[ \frac{\partial \eta(y)}{\partial y_{k_1}}\leqslant 0 \quad \text{for } y\in \bar A . \]

Introduce the notation

\[ \begin{aligned} S^+&=\{\,y\in S,\ \mu(y)(\mathbf q,\mathbf n_y)>0\,\},\qquad S^-=\{\,y\in S,\ \mu(y)(\mathbf q,\mathbf n_y)\leqslant 0\,\},\\ T_{\nu+}&=\{\,y\in T,\ \nu(y)\geqslant 0\,\},\qquad T_{\nu-}=\{\,y\in T,\ \nu(y)<0\,\}. \end{aligned} \tag{1.11} \]

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

\[ f(y)= \begin{cases} 1, & \text{if } y\in S^+,\\ 0, & \text{if } y\in S^-. \end{cases} \tag{1.12} \]

(Note that in the case where \(f(y)=\mathrm{const}\) for \(y\in S\), then \(\nu\equiv 0\).) Extending equality (1.5) to the function \(H_f(y)\), metaharmonic in the domain \(T\) and taking on the boundary \(S\), almost everywhere, the values (1.12), we obtain

\[ J(H_f)=0, \tag{1.13} \]

where

\[ \begin{aligned} J(H_f) &= \int\limits_{\dot S^+} \mu(y)(\mathbf q,\mathbf n_y)\,d_yS - \int\limits_{\dot T_{\nu+}} H_f(y)\,\frac{\partial \mu(y)}{\partial y_{k_1}}\,dy \\ &\qquad - \int\limits_{\dot T_{\nu-}} H_f(y)\,\frac{\partial \mu(y)}{\partial y_{k_1}}\,dy . \end{aligned} \tag{1.14} \]

Consider the case \(T_{\nu-}=T'_{\nu-}\), where

\[ T'_{\nu-} = \left\{\,y\in T_{\nu-},\ -\frac{\partial \eta(y)}{\partial y_{k_1}}<0\,\right\}. \tag{1.15} \]

(In all other cases the arguments are carried out analogously.) From the notation (1.11) and (1.15), and also from the properties of the function \(\mu(y)\), we have

\[ \frac{\partial \mu(y)}{\partial y_{k_1}}>0 \quad \text{for } y\in T_{\nu-};\qquad \frac{\partial \mu(y)}{\partial y_{k_1}}\leqslant 0 \quad \text{for } y\in T_{\nu+}. \tag{1.16} \]

Denote by \(S_{\nu-}\) the boundary of the set \(T_{\nu-}\); also denote

\[ S_{\nu-}^{+}=\{\,y\in S_{\nu-},\ (\mathbf q,\mathbf n_y)<0\,\};\qquad S_{\nu-}^{-}=\{\,y\in S_{\nu-},\ (\mathbf q,\mathbf n_y)\geqslant 0\,\}. \]

Since \(0<H_f(y)<1\) for \(y\in T\), it follows that

\[ -\int\limits_{\dot T_{\nu-}} H_f(y)\,\frac{\partial \mu(y)}{\partial y_{k_1}}\,dy > -\int\limits_{S_{\nu-}} \mu(y)(\mathbf q,\mathbf n_y)\,d_yS = \]

\[ = - \int_{S_v^{+}} \mu(y)(\mathbf q,\mathbf n_y)\,d_yS -\int_{S_{v-}^{-}}\mu(y)(\mathbf q,\mathbf n_y)\,d_yS . \tag{1.17} \]

From (1.14) and (1.17) we have

\[ \begin{aligned} J(H_f) \geq& \left[ \int_{S^{+}}\mu(y)(\mathbf q,\mathbf n_y)\,d_yS -\int_{S_{v-}^{-}}\mu(y)(\mathbf q,\mathbf n_y)\,d_yS \right] +\\ &+\left[-\int_{S_{v-}^{+}}\mu(y)(\mathbf q,\mathbf n_y)\,d_yS\right] +\left[-\int_{T_{v+}^{+}} H_f(y)\frac{\partial \mu(y)}{\partial y_{k_1}}\,dy\right]. \end{aligned} \tag{1.18} \]

From (1.16), (1.11), and the properties of the functions \(v(y)\), \(\eta(y)\), it follows that on the right-hand side of inequality (1.18) each of the three terms enclosed in square brackets is nonnegative, and at least one of them is strictly greater than zero. Consequently, we obtain \(J(H_f)>0\), which contradicts (1.13).

The theorem is proved.

3°. Let \((\rho,\Theta)\) denote the spherical coordinates of the point \(x\) of the space \(E^n\) \((n\geq 2)\). Consider the class of functions \(\mu_\alpha(y)\) \((\alpha=1,2)\) satisfying the conditions

\[ \mu_\alpha(y)=\xi(y)\delta_\alpha(y), \tag{1.19} \]

where

a) \(\delta_\alpha(y)\) \((\alpha=1,2)\) are bounded functions for \(y\in \bar A\), and moreover
\[ \frac{\partial \delta_\alpha}{\partial \rho}=0; \]

b) \(\xi(y)>0\), and either
\[ \frac{\partial}{\partial \rho}(\rho^n\xi)\geq 0, \]
or
\[ \frac{\partial}{\partial \rho}(\rho^n\xi)\leq 0 \]
for \(y\in \bar A\)

(\(\rho\) is the length of the radius vector of the point \(y\) with origin at the point \(0\in E^n\setminus \bar A\)). For case b), when the origin of coordinates \(O\in\bar A\), we assume that the function \(\xi(y)>0\), and
\[ \frac{\partial}{\partial \rho}(\rho^n\xi)>0 \]
for \(\rho<\rho_0\),
\[ \frac{\partial}{\partial \rho}(\rho^n\xi)\geq 0 \]
for \(\rho\geq\rho_0\), where \(\rho_0\) is a sufficiently small positive fixed number.

Theorem 2 (the case \(\chi=0\)). If, for functions \(\mu_\alpha(y)\) of the class (1.19), the exterior potentials of a given set \(A\) satisfy the condition

\[ V(x;A,\mu_1)=V(x;A,\mu_2)\quad \text{for } x\in E^n\setminus \bar A, \tag{1.20} \]

then \(\mu_1(y)=\mu_2(y)\) for \(y\in A\).

The proof of this theorem is carried out according to a scheme analogous to the proof of Theorem 1.

Consider the case \(\xi(y)>0\),
\[ \frac{\partial}{\partial \rho}(\rho^n\xi)>0 \]
for \(y\in \bar A\). From condition (1.20), for any function \(H(y)\) harmonic in the domain \(D\supset \bar T\), we obtain the equality (see [12, c)])

\[ J(H)=0, \tag{1.21} \]

where

\[ J(H)= \int_S H(y)\mu(y)(\mathbf R_y,\mathbf n_y)\,d_yS -\int_T H(y)\left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(y_k\mu)\right]\,dy, \tag{1.22} \]

and \((\mathbf R_y,\mathbf n_y)\) is the scalar product of the radius vector \(\mathbf R_y\) with the vector \(\mathbf n_y\). Denote \(\mu(y)=\mu_1(y)-\mu_2(y)\). Note that \(\mu(y)\not\equiv \operatorname{const}\), \(y\in S\), if \(\mu\not\equiv 0\) for \(y\in A\).

Define on the boundary \(S\) the function \(f(y)\) as follows:

\[ f(y)=\operatorname{sign}\mu(y)\quad \text{for } y\in S. \tag{1.23} \]

Extending the equality (1.23) to the function \(H_f(y)\), harmonic in the domain \(T\) and assuming on the boundary \(S\) the values (1.23) almost everywhere, we obtain

\[ J(H_f)=0, \tag{1.24} \]

where

\[ J(H_f)=\int_S |\mu(y)|(\mathbf R_y,\mathbf n_y)\,d_yS -\int_T H_f(y)\left[\sum_{k=1}^n \frac{\partial}{\partial y_k}(\mu\, y_k)\right]\,dy. \tag{1.25} \]

Since \(|H_f(y)|<1\), under our assumptions on the function \(\mu(y)\) we have from (1.25) \(J(H_f)>0\), which contradicts (1.24). The proof of Theorem 2 in all the remaining cases is carried out in the same way as in Theorem 1, if everywhere we replace the scalar product \((\mathbf q,\mathbf n_y)\) by \((\mathbf R_y,\mathbf n_y)\), and replace the quantity

\[ \frac{\partial \mu(y)}{\partial y_{k_1}} \]

by the expression

\[ \sum_{k=1}^n \frac{\partial}{\partial y_k}(\mu\, y_k). \]

\(4^\circ\). Consider the class of functions \(\mu_\alpha(y)\) \((\alpha=1,2)\) satisfying the conditions

\[ \mu_\alpha(y)=\gamma \xi(y)\delta_\alpha(y)+\beta\eta(y)\nu_\alpha(y), \tag{1.26} \]

where \(\gamma,\beta\) are nonnegative constants, with \(\gamma^2+\beta^2\ne0\), and the functions \(\delta_\alpha(y)\) and \(\nu_\alpha(y)\) satisfy conditions (1.2, a) and (1.19, a), respectively, while for the functions \(\eta(y)\) and \(\xi(y)\) the conditions

\[ \eta(y)>0,\quad \frac{\partial}{\partial y_{k_1}}\eta(y)>0;\quad \xi(y)>0,\quad \frac{\partial}{\partial \rho}(\rho^n\xi)>0 \quad \text{for } y\in A \]

are fulfilled.

From the analysis of the proofs of Theorems 1 and 2 it follows

Theorem 3 (case \(\varkappa=0\)). If for the functions \(\mu_\alpha(y)\) of the class (1.26) the external potentials of a given set \(A=\displaystyle\bigcup_{j=1}^{j_1} T_j\) satisfy the condition

\[ V(x;A,\mu_1)=V(x;A,\mu_2)\quad \text{for } x\in E^n\setminus \overline A, \]

then the function \(\mu(y)=\mu_1(y)-\mu_2(y)\) has constant sign on \(S_j\), the boundary of \(T_j\).

\(5^\circ\). Let the function \(\mu(y)\in C^{(1,\lambda)}(\overline T)\) be different from zero almost everywhere for \(y\in \overline T\), and let the domain \(T\), bounded by the surface \(S\), belong to the class \(A^{(1,\lambda)}\).

Consider the Newtonian potential \((\varkappa=0)\)

\[ V(x)=\int_T \mu(y)K(x,y)\,dy, \tag{1.27} \]

where

\[ K(x,y)= \begin{cases} \dfrac{1}{(n-2)\sigma_n r_{xy}}, & \text{for } n>2,\\[6pt] \dfrac{1}{2\pi}\ln\dfrac{1}{r_{xy}}, & \text{for } n=2. \end{cases} \]

\[ r=|x-y|=\sqrt{\sum_{k=1}^{n}(x_k-y_k)^2},\quad \sigma_n \]
is the area of the surface of the unit sphere in \(n\)-dimensional space.

Consider the following expression:
\[ W(x)= \begin{cases} \gamma(n-1)V(x)-\displaystyle\sum_{k=1}^{n}(\gamma x_k+\beta q_k)\dfrac{\partial}{\partial x_k}V(x), & \text{for } n>2,\\[1.2em] 2\gamma V(x)-\gamma-\displaystyle\sum_{k=1}^{n}(\gamma x_k+\beta q_k)\dfrac{\partial}{\partial x_k}V(x), & \text{for } n=2, \end{cases} \tag{1.28} \]
where \(\beta^2+\gamma^2\ne0\), \(\beta\) and \(\gamma\) are constants, \(\mathbf q=\{q_1,\ldots,q_n\}\) is a unit vector directed along the axis \(Oy_k\). Using the equality
\[ \sum_{k=1}^{n}\frac{\partial}{\partial y_k}(\gamma y_k+\beta q_k)K(x,y)= \]
\[ = \begin{cases} \gamma(n-1)K(x,y)-\displaystyle\sum_{k=1}^{n}(\gamma x_k+\beta q_k)\dfrac{\partial}{\partial x_k}K(x,y), & \text{for } n>2,\\[1.2em] 2\gamma K(x,y)-\gamma-\displaystyle\sum_{k=1}^{2}(\gamma x_k+\beta q_k)\dfrac{\partial}{\partial x_k}K(x,y), & \text{for } n=2, \end{cases} \]
we transform the function \(W(x)\) as follows:
\[ W(x)=\int_S \mu(y)\Phi(y)K(x,y)\,d_yS- \]
\[ -\int_T\left[\sum_{k=1}^{n}(\gamma y_k+\beta q_k)\frac{\partial\mu}{\partial y_k}\right]K(x,y)\,dy. \tag{1.29} \]
Here \(\Phi(y)=(\gamma R_y+\beta\mathbf q,\mathbf n_y)\) is the scalar product of the vector \(\gamma R_y+\beta\mathbf q\) with the vector \(\mathbf n_y\). Denote by \(\mathbf n_{x_0}\) the direction of the exterior normal to the surface \(S\) at the point \(x_0\), and by \((\mathbf n_{x_0},\mathbf r_{xy})\) the angle between the vectors \(\mathbf n_{x_0}\) and \(\mathbf r_{xy}\), where the point \(x\) is taken on the normal \(\mathbf n_{x_0}\), \(x\in E^n\setminus T\). We have
\[ \frac{\partial W(x)}{\partial n_{x_0}} = \int_S \frac{\cos(\widehat{\mathbf n_{x_0},\mathbf r_{xy}})} {(n-2)\sigma_n r_{xy}^{\,n-1}} \mu(y)\Phi(y)\,d_yS - \]
\[ -\int_T \frac{\cos(\widehat{\mathbf n_{x_0},\mathbf r_{xy}})} {(n-2)\sigma_n r_{xy}^{\,n-1}} \left[\sum_{k=1}^{n}(\gamma y_k+\beta q_k)\frac{\partial\mu}{\partial y_k}\right]dy. \]

From the formula for the jump of the normal derivative of the simple-layer potential and from the continuity of the first derivatives of the potential (1.27) in the whole space, we obtain the equation
\[ A(\mu\Phi)=f+P(\mu)\quad (n>2), \tag{1.30} \]

where

\[ A(\mu \Phi)=\frac{1}{2}\mu(x)\Phi(x)-\int_S \frac{\cos\widehat{(\mathbf n_x,\mathbf r_{xy})}}{(n-2)\sigma_n r_{xy}^{\,n-1}} \,\mu(y)\Phi(y)\,d_yS,\quad x\in S, \tag{1.31} \]

\[ P(\mu)=-\int_T \frac{\cos\widehat{(\mathbf n_x,\mathbf r_{xy})}}{(n-2)\sigma_n r_{xy}^{\,n-1}} \left[\sum_{k=1}^n(\gamma y_k+\beta q_k)\frac{\partial\mu(y)}{\partial y_k}\right]dy, \quad x\in S, \tag{1.32} \]

\[ f(x)=\left.\frac{\partial W}{\partial n_x}\right|_{-} =\lim_{x'\to x}\frac{\partial W(x')}{\partial n_x}, \tag{1.33} \]

where in (1.33) the point \(x'\in E^n\setminus \overline T\) is taken on the normal \(\mathbf n_x\) at the point \(x\in S\). We thus arrive at the following assertion.

Lemma 1. If the potential \(V(x)\) (1.27) is given for \(x\in E^n\setminus \overline T\), then its density \(\mu(x)\) satisfies the integro-differential equation (1.30).

From Lemma 1 one can obtain a number of corollaries for various classes of densities.

Consider the class of functions \(\mu(y)\) satisfying the condition

\[ \frac{\partial\mu}{\partial\rho}=0\quad ((\rho,\Theta)=y,\ y\in \overline T). \tag{1.34} \]

For functions of the class (1.34), putting \(\gamma=1,\ \beta=0\) in (1.30)—(1.33), we obtain

\[ A(\mu\Phi)=f, \tag{1.35} \]

where \(\Phi(x)=(\mathbf R_x,\mathbf n_x)\) for \(x\in S\).

In various function spaces, equation (1.30), as well as equation (1.35), can be transformed. For example, consider the space \(R_2\) [6, a), 12, a)], whose norm is equivalent to the norm in the space \(C^{(1,\lambda)}\). If we assume that the domain \(T\) belongs to the class \(A^{(2,\lambda)}\), and that the function \(\mu(y)\), satisfying condition (1.34), belongs to the class \(C^{(1,\lambda)}(S)\), then all the terms in (1.35) belong to the space \(R_2\), which follows from the properties of potentials. The operator \(A\) in the space \(R_2\) has a bounded inverse \(A^{-1}\), which follows from the uniqueness of the solution of the exterior Neumann problem for Laplace’s equation and the well-known Banach theorem on the inverse operator. Therefore equation (1.35) is replaced by the equivalent equation

\[ \mu\Phi=A^{-1}(f). \tag{1.36} \]

From equation (1.36) it follows that

Corollary 1. If for the potential (1.27) with density \(\mu(y)\) of the class (1.34) the condition \(V(x)=0\) is fulfilled for \(x\in E^n\setminus \overline T\), then \(\mu(y)=0\) for \(y\in T\).

Corollary 2. If the domain \(T\) is star-shaped with respect to an interior point and the function \(\mu(y)\) satisfies condition (1.34), then the values of the function \(\mu(x)\) for \(x\in S\) are expressed in terms of the exterior potential \(V(x)\) in the form

\[ \mu(x)=A^{-1}(f)\Phi^{-1}(x), \]

where \(\Phi(x)=(\mathbf R_x,\mathbf n_x)>0\) for \(x\in S\), and the function \(f(x)\) is expressed in terms of the values of the exterior potential \(V(x)\) (1.27) by formulas (1.33) and (1.28).

Let us note that, for the case of a star-shaped domain and densities of the class (1.34), in [1, a)] the values of the function \(\mu\) are written in another form in terms of its potential.

Consider the class of functions \(\mu(y)\) satisfying the conditions

\[ \frac{\partial \mu(y)}{\partial y_{k_1}}=0 \quad \text{for } y\in \overline{T}. \tag{1.37} \]

For functions of the class (1.37), putting in (1.30)—(1.33) \(\beta=1,\ \gamma=0\), we obtain the equation

\[ A(\mu\Phi)=f, \tag{1.38} \]

where \(\Phi(x)=(q,\mathbf n_x),\ x\in S\). From equation (1.38) it follows that

Corollary 3. If, for the potential (1.27) with density \(\mu(y)\) from the class (1.37), the condition \(V(x)=0\) for \(x\in E^n\setminus \overline{T}\) is satisfied, then \(\mu(y)=0\) for \(y\in T\).

6°. To determine the density \(\mu(y)\) in the domain \(T\), if the potential \(V(x)\) (1.27) is known for \(x\in E^n\setminus \overline{T}\), one can indicate one more method.

Consider the class of functions \(\mu(y)\) that are solutions of the equation

\[ L\mu=0, \tag{1.39} \]

where \(L\) is a prescribed operator for points \(y\in T\). Under appropriate smoothness requirements on the function \(\mu(y)\) and on the boundary \(S\), by virtue of the properties of the potential \(V(x)\) we have

\[ L\Delta V(x)=0 \quad \text{for } x\in T. \tag{1.40} \]

Define on the surface \(S\) the functions \(\varphi(x_0)\) and \(\psi(x_0)\) by the formulas

\[ \varphi(x_0)=\lim_{x\to x_0} V(x); \quad (x\in E^n\setminus \overline{T},\ x_0\in S), \tag{1.41} \]

\[ \psi(x_0)=\lim_{x\to x_0}\sum_{k=1}^{n} a_k \frac{\partial V(x)}{\partial x_k}; \quad (x\in E^n\setminus \overline{T},\ x_0\in S), \tag{1.42} \]

where \(a_k=a_k(x_0)\) are prescribed functions on the surface \(S\).

Consider the following problem: for points \(x\in T\), find a function \(U(x)\) that is a solution of the equation

\[ L\Delta U(x)=0 \tag{1.43} \]

with the following boundary data:

\[ U\big|_{x_0\in S}=\varphi(x_0), \tag{1.44} \]

\[ \sum_{k=1}^{n} a_k \frac{\partial U(x)}{\partial x_k}\bigg|_{x_0\in S}=\psi(x_0). \tag{1.45} \]

Suppose that the solution of the problem (1.43)—(1.45) has been found; then

\[ \mu(x)=\Delta U(x), \quad \text{for } x\in T. \]

For example, consider the class of functions \(\mu(y)\) that are solutions of the Laplace equation for points \(x\in T\):

\[ \Delta\mu=0. \tag{1.46} \]

Then the function \(U(x)\) for \(x\in T\) is found from the following data:

\[ \Delta^2 U=0, \tag{1.47} \]

\[ U\big|_{x \in S}=V(x)\big|_{x \in S}, \tag{1.48} \]

\[ \frac{\partial U}{\partial n}\bigg|_{x \in S} = \frac{\partial V}{\partial n}\bigg|_{x \in S}. \tag{1.49} \]

Solving problem (1.47)—(1.49), we find the function \(u(x)\) for \(x \in T\); moreover, by virtue of the uniqueness of the solution of problem (1.47)—(1.49), we have

Corollary 4. If for the potential (1.27) with density \(\mu(y)\) of class (1.46) the condition \(V(x)=0\) for \(x \in E^n \setminus \overline{T}\) is fulfilled, then \(\mu(y)=0\) for \(y \in T\).

Let us note that this corollary easily follows from Lemmas 1, 2 of [12, c)]. If we consider the class of functions \(\mu(y)\) satisfying conditions (1.37), then Corollary 3 can be proved by the method described above, putting

\[ L=\frac{\partial}{\partial x_{k_1}}. \]

Indeed, in this case the density \(\mu(y)\) belongs to the class (1.37), and therefore the potential \(V(x)\) (1.27) satisfies the equation (see [4])

\[ \frac{\partial}{\partial x_{k_1}}\Delta V=0 \quad \text{for } x \in T. \tag{1.50} \]

Denote \(Z(x)=\dfrac{\partial}{\partial x_{k_1}}V(x)\); then from (1.50) we obtain

\[ \Delta Z(x)=0 \quad \text{for } x \in T, \]

and, by virtue of the equality to zero of the exterior potential \(V(x)\), we have

\[ Z(x)\big|_{x \in S}=0. \]

Consequently, \(Z(x)=0\) for \(x \in T\). Therefore

\[ \frac{\partial V(x)}{\partial x_{k_1}}=0 \quad \text{for } x \in T, \]

and from the condition of equality to zero of the exterior potential we have

\[ V(x)\big|_{x \in S}=0. \]

Thus, we conclude that \(V(x)=0\) for \(x \in T\), and then \(\mu(x)=0\) for \(x \in T\).

§ 2. ON DETERMINING THE SHAPE OF A BODY OF GIVEN DENSITY FROM THE VALUES OF THE EXTERIOR POTENTIAL

\(7^\circ\). Let \(A_1\) and \(A_2\) be open bounded sets of the space \(E^n\), 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 the surfaces \(S_1^j\) and \(S_2^j\) belong to the class \(A^{(1,\lambda)}\).

Introduce the notation (see [12, c)]). Let \(\Gamma^e\) be the boundary of the set \(\overline{A}_1 \cup \overline{A}_2\). If \(A_1 \ne A_2\), denote

\[ \Gamma_1^i=\Gamma_1 \cap (\overline{A}_1 \cap \overline{A}_2), \qquad \Gamma_1^e=\Gamma_1 \setminus \Gamma_1^i, \]

\[ \Gamma_2^e=\Gamma_2 \cap \Gamma^e, \qquad \Gamma_2^i=\Gamma_2 \setminus \Gamma_2^e. \tag{2.1} \]

If \(A_1=A_2\), we put \(\Gamma_\alpha^e=\Gamma_\alpha\) \((\alpha=1,2)\).

Theorem 4 (the case \(\varkappa \gg 0\)). If there exists at least one constant vector \(\mathbf q\) such that the following conditions are satisfied:

1) each of the sets \((S_1^\varepsilon)^j,\ j=1,2,\ldots,j_1;\ (S_2^\varepsilon)^j,\ j=1,2,\ldots,j_2\) is nonempty, and a straight line parallel to the vector \(\mathbf q\) intersects \(\Gamma_1^i \cup \Gamma_2^i\) in no more than two points (or two segments);

2) for the function \(\mu(y)\) the condition
\[ \frac{\partial \mu(y)}{\partial y_{k_1}}=0 \]
is satisfied (the direction of the axis \(Oy_{k_1}\) coincides with the direction of the unit vector \(\mathbf q\); the function \(\mu(y)\) is, generally speaking, not of constant sign for \(y\in \overline A_1\cup \overline A_2\));

3) for the sets \(A_\alpha\), for the given density, the equality
\[ V(x; A_1,\mu)=V(x; A_2,\mu)\quad \text{for } x\in E^n\setminus(\overline A_1\cup \overline A_2) \]
holds,

then \(A_1=A_2\).

Proof. For simplicity, assume that \(A_\alpha=T_\alpha\). From condition 3) it follows that, for any metaharmonic function \(H(y)\) in a domain \(D,\ D\supset(\overline T_1\cup \overline T_2)\), the equality \(J(H)=0\) holds, where
\[ J(H)=\int_{S_1}\mu(y)H(y)(\mathbf q,\mathbf n_y)\,d_yS -\int_{S_2}\mu(y)H(y)(\mathbf q,\mathbf n_y)\,d_yS . \]

Introduce the notation:
\[ S_1^+=\{y\in S_1,\ [\mu(y)(\mathbf q,\mathbf n_y)]>0\};\quad S_1^-=S_1\setminus S_1^+; \]
\[ S_2^-=\{y\in S_2,\ [\mu(y)(\mathbf q,\mathbf n_y)]<0\};\quad S_2^+=S_2\setminus S_2^- . \]

Define on the surface \(S^\varepsilon\) the function \(f(y)\) as follows:
\[ f(y)= \begin{cases} 1 & \text{for } y\in S_1^{\varepsilon+},\ S_2^{\varepsilon-},\\ 0 & \text{for } y\in S_1^{\varepsilon-},\ S_2^{\varepsilon+}. \end{cases} \]

Further, repeating the reasoning used in the proof of Theorem 2 of the author’s paper [12, c)], pp. 198–202, we obtain the contradictory assertion
\[ J(H_f)>0. \]

The theorem is proved. From Theorem 4, as a corollary, follows

Theorem \(4'\) (the case \(\varkappa \gg 0\)). If the potentials \(V(x;A_\alpha,\mu)\) \((\alpha=1,2)\) are such that

1) every straight line parallel to the axis \(Oy_{k_1}\) intersects \(\Gamma_1\) and \(\Gamma_2\) in no more than two points (or two segments);

2) the density \(\mu(y)\) satisfies the condition
\[ \frac{\partial \mu(y)}{\partial y_{k_1}}=0 \quad \text{for } y\in(\overline A_1\cup \overline A_2); \]

3) \(V(x;A_1,\mu)=V(x;A_2,\mu)\) for \(x\in E^n\setminus(\overline A_1\cup \overline A_2)\), then \(A_1=A_2\).

8°. Let the function \(\mu(y)\) satisfy the conditions
\[ \mu(y)=\xi(y)\delta(y), \tag{2.2} \]
where

a) the function \(\delta(y)\), generally speaking not of constant sign, satisfies the condition
\[ \frac{\partial \delta}{\partial \rho}=0, \]

b) the function \(\xi(y)>0\) and
\[ \frac{\partial}{\partial \rho}\bigl(\rho^\eta \xi\bigr)>0 \quad \text{for } y\in \overline A_\alpha . \]

Let \(T_\alpha\) \((\alpha=1,2)\) be finite domains of the space \(E^n\), for which the notation (2.1) is preserved when \(A_\alpha=T_\alpha\).

Theorem 5 (case \(\varkappa=0\)). If there exists at least one point \(O \in (\overline{T}_1 \cap \overline{T}_2)\) such that

1) for the radius vector \(\mathbf{R}_y\) (with origin at the point \(O\)) the relation

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

holds;

2) for the domains \(T_\alpha\) \((\alpha=1,2)\), for a given density of class (2.2), the equality

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

holds, then \(T_1=T_2\).

We prove the theorem by contradiction. From the second condition of the theorem, for any harmonic function \(U(y)\) in a domain \(D \supset (\overline{T}_1\cup \overline{T}_2)\) the equality

\[ \int_{T_1}\mu(y)U(y)\,dy-\int_{T_2}\mu(y)U(y)\,dy=0 \]

holds.

Putting

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

where \(H(y)\) is a harmonic function in the domain \(D\), we obtain the equality

\[ J(H)=0, \tag{2.4} \]

where

\[ \begin{aligned} J(H)=&\int_{S_1} H(y)\mu(y)(\mathbf{R}_y,\mathbf{n}_y)\,d_yS -\int_{S_2} H(y)\mu(y)(\mathbf{R}_y,\mathbf{n}_y)\,d_yS \\ &-\int_{(T_1^\circ\setminus T_0)} H(y)\left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(y_k\mu)\right]\,dy +\int_{(T_2^\circ\setminus T_0)} H(y)\left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(y_k\mu)\right]\,dy. \end{aligned} \tag{2.5} \]

Write the left-hand side of equality (2.4) in the form

\[ J(H)=J_{\mu+}(H)+J_{\mu-}(H), \]

where \(J_{\mu+}(H)\) denotes the integrals on the right-hand side of expression (2.5), taken respectively over the surfaces \(S_\alpha\) and the sets \((T_\alpha\setminus T_0)\), for which the function \(\mu(y)>0\) when \(y\in(\overline{T}_1\cup\overline{T}_2)\); and \(J_{\mu-}(H)\) denotes the integrals of expression (2.5) taken respectively where \(\mu(y)\leq 0\) when \(y\in\overline{T}_1\cup\overline{T}_2\).

Introduce the notation:

\[ S_{\mu+}=\{y\in S,\ \mu(y)>0\}; \qquad S_{\mu-}=\{y\in S,\ \mu(y)\leq 0\}; \]

\[ T_{\mu+}=\{y\in T,\ \mu(y)>0\}; \qquad T_{\mu-}=\{y\in T,\ \mu(y)\leq 0\}. \tag{2.6} \]

Taking into account the notation (2.1) and (2.6), we have a number of new notations, for example,

\[ S_{\alpha\mu+}^{e}=\{y\in S_{\alpha}^{e},\ \mu(y)>0\}. \]

Define on the surface \(S^e=S_1^e\cup S_2^e\) the function \(f(y)\) as follows:

\[ f(y)= \begin{cases} 1 & \text{for } y\in S_{1\mu+}^{e},\ S_{2\mu-}^{e},\\ 0 & \text{for } y\in S_{1\mu-}^{e},\ S_{2\mu+}^{e}. \end{cases} \tag{2.7} \]

Note that from the conditions of the theorem it follows that \(f(y)\ne \mathrm{const}\) for \(y\in S^{e}\). As in the proof of Theorem 1 of the author’s paper [12, b)], we extend equality (2.4) to the solution \(H_f(y)\) of the Dirichlet problem for the Laplace equation, for which the function \(H_f(y)\) on the boundary \(S^{e}\) almost everywhere takes the values \(f(y)\) from (2.7). We obtain the equality

\[ J(H_f)=0, \tag{2.8} \]

where

\[ J(H_f)=J_{\mu+}(H_f)+J_{\mu-}(H_f), \]

and

\[ \begin{aligned} J_{\mu+}(H_f)=& \int_{S^{e}_{1\mu+}} 1\mu(y)(R_y,n_y)\,d_yS +\int_{S^{i}_{1\mu+}} H_f(y)\mu(y)(R_y,n_y)\,d_yS \\ &-\int_{S^{i}_{2\mu+}} H_f(y)\mu(y)(R_y,n_y)\,d_yS -\int_{(T_1\setminus T_0)_{\mu+}} H_f(y) \left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(y_k\mu)\right]\,dy \\ &+\int_{(T_2\setminus T_0)_{\mu+}} H_f(y) \left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(y_k\mu)\right]\,dy; \end{aligned} \tag{2.9} \]

\[ \begin{aligned} J_{\mu-}(H_f)=& \int_{S^{i}_{1\mu-}} H_f(y)\mu(y)(R_y,n_y)\,d_yS -\int_{S^{e}_{2\mu-}} 1\cdot\mu(y)(R_y,n_y)\,d_yS \\ &-\int_{S^{i}_{2\mu-}} H_f(y)\mu(y)(R_y,n_y)\,d_yS -\int_{(T_1\setminus T_0)_{\mu-}} H_f(y) \left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(y_k\mu)\right]\,dy \\ &+\int_{(T_2\setminus T_0)_{\mu-}} H_f(y) \left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(y_k\mu)\right]\,dy. \end{aligned} \tag{2.10} \]

From conditions (2.2) and the notation (2.6) we have

\[ \left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(y_k\mu)\right]>0 \quad \text{for } y\in (T_a\setminus T_0)_{\mu+}, \tag{2.11} \]

\[ \left[\sum_{k=1}^{n}\frac{\partial}{\partial y_k}(y_k\mu)\right]<0 \quad \text{for } y\in (T_a\setminus T_0)_{\mu-}. \tag{2.12} \]

Using the properties of the function \(H_f\) \((0<H_f(y)<1\) for \(y\in T^{e})\), from (2.2), (2.11), and (2.12) we obtain the estimates:

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

\[ -\int_{S^{e}_{1\mu+}} \mu(y)(R_y,n_y)\,d_yS +\int_{S^{i}_{2\mu+}} \mu(y)(R_y,n_y)\,d_yS, \tag{2.13} \]

\[ \int_{(T_2 \dot{\times} T_0)_{\mu-}} H_f(y)\left[\sum_{k=1}^n \frac{\partial}{\partial y_k}(y_k\mu)\right]\,dy \geq \]

\[ \geq \int_{S^e_{2\mu-}} \mu(y)(R_y,\mathbf n_y)\,d_yS -\int_{S^i_{1\mu-}} \mu(y)(R_y,\mathbf n_y)\,d_yS . \tag{2.14} \]

From (2.11)—(2.14) we obtain

\[ J_{\mu+}(H_f)\geq \int_{S^i_{1\mu+}} H_f(y)\mu(y)(R_y,\mathbf n_y)\,d_yS +\int_{S^i_{2\mu+}}(1-H_f)\mu(y)(R_y,\mathbf n_y)\,d_yS+ \]

\[ +\int_{(T_2\dot{\times}T_0)_{\mu+}} H_f(y)\left[\sum_{k=1}^n \frac{\partial}{\partial y_k}(y_k\mu)\right]\,dy, \tag{2.15} \]

\[ J_{\mu-}(H_f)\geq \left[\int_{S^i_{1\mu-}}(H_f-1)\mu(y)(R_y,\mathbf n_y)\,d_yS\right]+ \]

\[ +\left[-\int_{S^i_{2\mu-}} H_f(y)\mu(y)(R_y,\mathbf n_y)\,d_yS\right]+ \]

\[ +\left[-\int_{(T_1\dot{\times}T_0)_{\mu-}} H_f(y)\left[\sum_{k=1}^n \frac{\partial}{\partial y_k}(y_k\mu)\right]\right]\,dy . \tag{2.16} \]

From the conditions of the theorem (2.3), the properties of the function \(\mu(y)\) satisfying (2.2), and also the properties of the function \(H_f(y)\) determined by the data (2.7), we obtain that every term on the right-hand side enclosed in square brackets in inequalities (2.15) and (2.16) is nonnegative and at least one of them is strictly greater than zero. Consequently, we obtain \(J(H_f)>0\), which contradicts (2.8). The theorem is proved.

Remark 1. Condition (2.3) of Theorem 5 is satisfied if the set \(\overline T_1\cap \overline T_2\) is star-shaped with respect to a point \(O\in(\overline T_1\cap \overline T_2)\), and the set \(E^n\setminus(\overline T_1\cap \overline T_2)\) consists of one component. In particular, if the sets \(\overline T_1\) and \(\overline T_2\) are star-shaped with respect to a common point \(O\), then condition (2.3) holds.

Remark 2. The assertion of Theorem 5 is also valid in the case when, for the density \(\mu(y)\) of class (2.2), the function \(\xi(y)\) satisfies the condition \(\xi(y)>0\), \(\dfrac{\partial}{\partial \rho}(\rho^n\xi)>0\) for \(\rho<\rho_0\), \(\dfrac{\partial}{\partial \rho}(\rho^n\xi)\geq 0\) for \(\rho\geq \rho_0\).

Remark 3. If for the potentials \(V(x;A_\alpha,\mu)\) \((\varkappa=0)\) with density \(\mu(y)\), satisfying condition (2.2), the equality \(V(x;A_1,\mu)=V(x,A_2,\mu)\) holds for \(x\in E^n\setminus(\overline A_1\cup \overline A_2)\), then \(\overline A_1\cap \overline A_2\ne 0\).

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Received by the editors
April 10, 1966

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