Solvability and Invertibility of the Problems of Logarithmic Potential

Abstract

In the inverse problems of logarithmic potential with various initial conditions, there is usually a unique solution. However, sometimes a solution to a problem of simple layer potential is not unique. Thus, it is necessary to separate these two cases with the help of some definitions. In the Introduction, we formulate the direct and the inverse problem of logarithmic potential (direct A-problem and inverse A-problem, respectively) and set similar problems of logarithmic simple layer potential (direct B-problem and inverse B-problem, respectively). Then we give the definitions of the unique and non-unique solvability and invertibility of these problems. In §1, we study the connection between the gradient of logarithmic potential in the inverse A-problem and the mapping function in the direct A-problem for domains with smooth boundaries. Wherein special attention is paid to the cases of expansion of the gradient in finite or infinite series in a neighborhood of infinity. The cases when the gradient has finite number of algebraic branching points are also considered. The conclusions about the solvability and invertibility of these A-problems are drawn. In §2, the indicated A-problems are solved for domains with corner points formed by the intersection of line segments and circular arcs in various combinations. The respective examples are given together with the conclusions about the solvability and invertibility. In item 3.1 from §3, we give an example of the inverse B-problem which is non-uniquely solvable. Wherein its respective direct B-problem is non-uniquely invertible. These results are presented in the form of Th. 3.1 and its corollaries. In item 3.2, we solve the direct B-problem for a symmetric lune bounded by circular arcs, and compare the result to the similar direct A-problem. The passage to the limit from lunes to the unit circle is studied.

INTRODUCTION

This article continues the authors’ research made in [1–6]. Let us recall the setting of the problem of logarithmic potential.

Let $L$ be a closed piecewise analytic contour without self-intersections which encloses a bounded simply connected domain $D$ ($0\in D$). It is filled with a homogeneous gravitating mass with constant density $\mu>0$ (equal 1, for simplicity). In addition, there is the gradient of a logarithmic potential $u(z)$. Except for a finite set of branching points, $u(z)$ is an analytic function in the domain $D^{-}={\textbf{C}}-D$ with normalization $u(\infty)=0$.

In the direct A-problem, given domain $D$, one has to find function $u(z)$ in explicit form. In the inverse A-problem, given the function

$$u(z)=\iint\limits_{D}\frac{d\sigma(\tau)}{\tau-z}=\sum_{k=0}^{\infty}\frac{c_{k}}{z^{k+1}},$$

(1)

where $\tau\in D$, $d\sigma(\tau)$ is the area element, one has to find univalent domain $D$ (or its boundary $L$). We will search for domain $D$ as the image of the open unit disk $E=\{t:|t|<1\}$ under the mapping of univalent analytic function $z(t)$ with normalization $z(0)=0$, $z^{\prime}(0)>0$.

Simultaneously with these problems (we will call them ‘‘domain’’ problems), consider the problem of logarithmic simple layer potential which was first set in [7]. The potential gradient in it is given by the formula

$$u(z)=\int\limits_{L}\frac{|dt|}{t-z},z\in D^{-}.$$

(2)

By the direct B-problem we will call the problem of finding a function $u(z)$ defined for the given contour $L$ by formula (2). By the inverse B-problem we will call the problem of finding a contour $L$ satisfying formula (2) for the given function $u(z)$ represented in a neighborhood of infinity by the Laurent series. The contour in these problems is sought as the image of the unit circle $\{|\zeta|=1\}$ under the mapping $z=f(\zeta)$ with the former normalization.

In connection with these problems, we give the following definitions.

Definition 1. If the solution to the inverse A-problem (inverse B-problem), domain $D$ (contour $L$), is unique, we call this inverse A-problem (inverse B-problem) uniquely solvable.

Definition 2. If the solution to the inverse A-problem (inverse B-problem) consists of two or more contours, we call this inverse A-problem (inverse B-problem) non-uniquely solvable.

Consider a uniquely solvable inverse A-problem (inverse B-problem) with the given gradient $u(z)$. And, given the obtained domain $D$ (contour $L$), solve the inverse A-problem (inverse B-problem) finding gradient $\tilde{u}(z)$.

Definition 3. If function $\tilde{u}(z)$ exists, is unique, and $\tilde{u}(z)\equiv u(z)$, we call the original inverse A-problem (inverse B-problem) uniquely invertible.

Definition 4. If function $\tilde{u}(z)$ exists, is unique, but $\tilde{u}(z)\neq u(z)$, we call the original inverse A-problem (inverse B-problem) non-invertible.

Consider a direct A-problem (direct B-problem) with the given domain $D$ (contour $L$). And, given the obtained gradient $u(z)$, solve the inverse A-problem (inverse B-problem) finding domain $\tilde{D}$ (con- tour $\tilde{L}$).

Definition 5. If domain $\tilde{D}$ exists, is unique, and $\tilde{D}=D$ (contour $\tilde{L}$ exists, is unique, and $\tilde{L}=L$), we call the original direct A-problem (direct B-problem) uniquely invertible.

Definition 6. If domain $\tilde{D}$ exists, is unique, but $\tilde{D}\neq D$ (contour $\tilde{L}$ exists, is unique, but $\tilde{L}\neq L$), we call the original direct A-problem (direct B-problem) non-invertible.

Definition 7. If there exist several solutions $\tilde{D_{1}}$,.., $\tilde{D_{n}}$ ($\tilde{L_{1}}$,.., $\tilde{L_{n}}$) ($n\geq 2$), and there exists $k$ such that $\tilde{D_{k}}=D$ ($\tilde{L_{k}}=L$), we call the original direct A-problem (direct B-problem) non-uniquely invertible.

A theorem on the uniqueness of solution to the inverse A-problem with constant density $\mu>0$ is proved for star-like domains in [9]. This theorem and our definitions above imply

Theorem 1. If the solution to the inverse A-problem is a star-like domain $D$, this inverse A-problem is uniquely solvable and uniquely invertible.

1 CONNECTION BETWEEN FUNCTIONS $u(z)$ AND $z(t)$ IN ‘‘DOMAIN’’ PROBLEMS WITH SMOOTH BOUNDARIES

There are many works devoted to this topic (see [2, 3, 8, 10–12]). We will list the possible cases and give the respective examples.

1.1. Consider the inverse A-problem for $u(z)=\sum_{k=0}^{n}\frac{c_{k}}{z^{k+1}}$. Then, $z(t)=\sum_{k=1}^{n+1}a_{k}t^{k}$ [11]. In order to find unknown coefficients $a_{k}$ given known coefficients $c_{k}$ ($k=1,...,n$), one constructs a system of $n$ complex or $2n$ real equations. The case $n=2$ is completely studied [2, Th. 3]: the inverse A-problem with function (2) is uniquely solvable, and due to Th. 0.1, is uniquely invertible.

Consider in more detail the A-problem for a domain $D(\alpha,\beta)$ which is the image of $E$ under the mapping by the function $z(t)=\alpha(t+\beta t^{2})$ with known $\alpha>0$, $\beta\in R$, $0<|\beta|\leq 1/2$. For such $\alpha$, $\beta$, domain $D(\alpha,\beta)$ is star-like. The solution to the given direct A-problem has the form [3]

$$u(z)=\frac{2\alpha^{2}(1+2\beta^{2})}{z}+\frac{2\alpha^{3}\beta}{z^{2}}.$$

For the inverse A-problem with $u(z)=\frac{c_{0}}{z}+\frac{c_{1}}{z^{2}}$, it was proved in [3] that it has a unique solution in the class of star-like functions. Let us show now that this solution coincides with $D(\alpha,\beta)$. Let $z(t)=a_{1}t+a_{2}t^{2}$. The system for finding $a_{1}$, $a_{2}$ has the form

$$a_{1}^{2}+2a_{2}^{2}=\frac{c_{0}}{2},\quad a_{1}^{2}a_{2}=\frac{c_{1}}{2}.$$

Substitute $c_{0}=2(\alpha^{2}+2\alpha^{2}\beta^{2})$, $c_{1}=2\alpha^{3}\beta$ into it. Direct calculations show that $a_{1}=\alpha$, $a_{2}=\alpha\beta$ is the solution to this system. We have proved the unique solvability and invertibility of the inverse A-problem. Since

$$|c_{0}|=\iint\limits_{D}d\sigma(\tau)=S(D),$$

(3)

where $S(D)$ is the area of domain $D$, we have $S(D(\alpha,\beta))=2\alpha^{2}(1+2\beta^{2})$.

1.2. Consider the inverse A-problem for the function

$$u(z)=\sum\limits_{k=0}^{\infty}\frac{c_{k}}{z^{k+1}}.$$

(4)

Denote by $z_{k}$ $(k=\overline{1,s})$ the singular points of $u(z)$. Given the function $z(t)=\sum_{k=1}^{\infty}a_{k}t^{k}$, construct the new analytic function $z^{*}(t)=\overline{z(1/\bar{t})}$ defined in the domain $E^{-}=\{t:|t|>1\}$. It is clear that the system for finding $a_{k}$ given $c_{k}$ will consist of infinite number of equations. A system of the form $\{z(t_{k})=z_{k}\}$ for finding unknown points $t_{k}$ serves as its analogue. In one special case, the solution to such system was obtained in [2, Ex. 1].

Consider one more example. Given the inverse A-problem with the potential gradient in the form

$$u(z)=\frac{A_{1}}{z-b}+\frac{A_{2}}{z+b},$$

(5)

where $A_{1}$, $A_{2}$ are real, $b>0$. Due to [12, Th. 4.5],

$$z(t)=\frac{\alpha_{1}}{t-1/a}+\frac{\alpha_{2}}{t+1/a},$$

where $\pm a$ are the preimages of points $\pm b$. Denote $c=1/a$, then we have $z(t)=[(\alpha_{1}+\alpha_{2})t+c(\alpha_{1}-\alpha_{2})]/(t^{2}-c^{2})$, and the system $\{z(\pm a)=\pm b\}$ obtains the form

$$\frac{(\alpha_{1}+\alpha_{2})a+c(\alpha_{1}-\alpha_{2})}{a^{2}-c^{2}}=b,\quad\frac{-(\alpha_{1}+\alpha_{2})a+c(\alpha_{1}-\alpha_{2})}{a^{2}-c^{2}}=-b.$$

Whence $\alpha_{1}=\alpha_{2}=\alpha$ and $z(t)=2\alpha t/(t^{2}-c^{2})$. The equality $z(a)=b$ gives the first equation for finding $\alpha$ and $a$: $2\alpha a^{3}/(a^{4}-1)=b$. To obtain the second equation, we use Ivanov’s integral equation [10]

$$z^{*}(t)=-\frac{1}{2\pi i}\int\limits_{|t|=1}\frac{u(z(\tau))}{\tau-t}d\tau,\quad|t|>1.$$

(6)

Since $z(t)$ is a conformal map, $z(t)-b=z^{\prime}(a)(t-a)+O(t-a)$ in a neighborhood of point $a$, and $z(t)+b=z^{\prime}(-a)(t+a)+O(t+a)$ in a neighborhood of point $-a$. Hence,

$$z^{\prime}(a)=z^{\prime}(-a)=-\frac{2\alpha(a^{4}+1)}{(a^{4}-1)^{2}},\ u(z(\tau))=\frac{A}{z^{\prime}(a)(\tau-a)}+\frac{A}{z^{\prime}(-a)(\tau+a)}+u_{1}(\tau),$$

where $u_{1}(\tau)$ is a function analytic in the unit disk $E$.

Since

$$z^{*}(t)=-\frac{2\alpha a^{2}t}{t^{2}-a^{2}}=-\left(\frac{\alpha a^{2}}{t-a}+\frac{\alpha a^{2}}{t+a}\right),$$

formula (6) implies

$$-\left(\frac{\alpha a^{2}}{t-a}+\frac{\alpha a^{2}}{t+a}\right)=-\left(res_{a}\frac{u(z(\tau))}{\tau-t}+res_{-a}\frac{u(z(\tau))}{\tau-t}\right)=\frac{A_{1}}{z^{\prime}(a)(\tau-a)}+\frac{A_{2}}{z^{\prime}(-a)(\tau+a)}.$$

Hence,

$$A_{1}=A_{2}=A=-\alpha a^{2}z^{\prime}(a)=\frac{2\alpha^{2}a^{2}(a^{4}+1)}{(a^{4}-1)^{2}}.$$

Finally, we obtain the system

$$\frac{2\alpha a^{3}(a^{4}-1)}{a^{4}-1}=b,\quad\frac{2\alpha^{2}a^{2}(a^{4}+1)}{(a^{4}-1)^{2}}=A.$$

Under the condition $A>b^{2}$, this system has the unique solution

$$a=\frac{\sqrt{A-\sqrt{A^{2}-b^{4}}}}{b^{2}},\quad\alpha=\frac{b(a^{4}-1)}{2a^{3}}.$$

Thus, the function $z(t)=2\alpha t/(t^{2}-c^{2})$ is uniquely defined, and it maps $E$ onto the interior of a hippopede [12]. A solution to the direct A-problem for the function

$$z(t)=\frac{2\alpha t}{t^{2}-c^{2}},\quad\alpha>0,\quad a\in R,\quad|a|>1,$$

was found in [3]. It coincides with function $u(z)$ defined by formula (5). Thus, we have proved the following lemma.

Lemma 1.2. The inverse A-problem with function $u(z)$ defined by formula (5) is uniquely solvable and uniquely invertible.

Due to formula (3), the area of the hippopede equals $4\alpha^{2}(a^{4}+1)/(a^{4}-1)^{2}$.

1.3. Consider the cases when function $u(z)$ has finite number of algebraic branching points. The simplest example of such function is

$$u(z,A,c)=A(\sqrt{z^{2}-c^{2}}-z),$$

(7)

where $A,c>0$. Function $z(t)$ has the form [12]

$$z(t)=\alpha t+\frac{\alpha_{1}}{t},\quad\alpha,\alpha_{1}\in R,\quad|\alpha_{1}|<|\alpha|,$$

and it maps the unit circle onto the ellipsis $\frac{x^{2}}{(\alpha+\alpha_{1})^{2}}+\frac{y^{2}}{(\alpha-\alpha_{1})^{2}}=1$. In order to find $a=\alpha+\alpha_{1}$, $b=\alpha-\alpha_{1}$, we get the system $a^{2}-b^{2}=c^{2},\ 2ab=Ac^{2},$ which has the unique solution

$$a=c\sqrt{\frac{\sqrt{A^{2}+1}+1}{2}},\quad b=c\sqrt{\frac{\sqrt{A^{2}+1}-1}{2}}.$$

Now, consider the inverse A-problem for the exterior of the ellipsis $x^{2}/a^{2}+y^{2}/b^{2}=1$. It was shown in [3] that

$$\tilde{u}(z)=\frac{2ab}{c^{2}}\left(\sqrt{z^{2}-c^{2}}-z\right).$$

And since $\tilde{u}(z)\equiv u(z)$, the following lemma holds true.

Lemma 1.3. The inverse A-problem with function $u(z)$ defined by formula (7) is uniquely solvable and uniquely invertible.

A more complex example of solving the inverse A-problem was considered in [12, p. 61]. Let the sought domain $D$ be enclosed by a contour $L$ containing points $z_{1}=c$, $z_{2}=b$ $(b>c>0)$. Points $z_{3}=-c$, $z_{4}=-b$ do not lie in $D$. Function $u(z)$ is defined by the formula

$$u(z)=A\sqrt{\frac{z^{2}-b^{2}}{z^{2}-c^{2}}}-\Phi(z),$$

(8)

where

$$\Phi(z)=\frac{A}{2\pi i}\int\limits_{L}\sqrt{\frac{\zeta^{2}-b^{2}}{\zeta^{2}-c^{2}}}\frac{d\zeta}{\zeta-z},\ z\in D,\ A>0.$$

The author of the cited paper claims that $u(z)$ is analytic in $D^{-}$, i.e., it does not have singularities at $z_{3}$ and $z_{4}$. This is, however, not true. The Cauchy-type integral $\Phi(z)$ is an analytic function both in $D$ and $D^{-}$, so it cannot eliminate singularities $z_{3}$ and $z_{4}$ of the function $A\sqrt{\frac{z^{2}-b^{2}}{z^{2}-c^{2}}}$. Moreover, $u(z)$ does not belong to the class of functions which vanish at infinity, since $\lim\limits_{z\rightarrow\infty}u(z)=A\neq 0$. Next, this author suggests searching for the mapping function $z(t)$ in two various forms, but the explicit form of function $u(z)$ has not been obtained.

Let us analyze both forms as direct A-problems, and show that they lead to two different functions $u(z)$ which do not coincide with function (8). To simplify the calculations, we discard the condition of normalization $z(0)=0$.

a) $z_{1}(t)=\alpha\sqrt{t+a}$, $\alpha>0$, $a>1$. Obtain the equation for the boundary of domain $D_{1}=z_{1}(E)$ in Cartesian coordinates:

$$(x+iy)^{2}=\alpha^{2}t+\alpha^{2}a\ \Leftrightarrow\ (x^{2}-y^{2}-\alpha^{2})+4x^{2}y^{2}=\alpha^{4}.$$

This is a fourth order algebraic curve splitting into two convex ovals: the left one $L_{1}$ and the right one $L_{2}$. They intersect $Ox$ at points $-\alpha\sqrt{a+1}$, $-\alpha\sqrt{a-1}$ and $\alpha\sqrt{a-1}$, $\alpha\sqrt{a+1}$, respectively. The right oval corresponds to function $z_{1}(t)$. To calculate $u(z)$, we use the method of the Schwarz function [12]. First, from the equation of the curve of the form $F(z,\bar{z})=0$ we find $\bar{z}=S(z)$ ($S(z)$ is called the Schwarz function for curve $L$). Then, we apply the complex Green formula and obtain for the potential gradient the expression

$$u(z)=\iint\limits_{D}\frac{d\sigma(\tau)}{\tau-z}=\frac{1}{2i}\int\limits_{L}\frac{d\tau}{\tau-z}.$$

It follows from the equality $z(t)=\alpha\sqrt{t+a}$ that

$$z^{2}\bar{z}^{2}=\alpha^{4}(t+a)(\bar{t}+a)=\alpha^{4}(1+a^{2})+\alpha^{4}a(t+\bar{t}).$$

Since $t=z^{2}/\alpha^{2}-a$, $\bar{t}=\bar{z}^{2}/\alpha^{2}-a$, we have $z^{2}\bar{z}^{2}=\alpha^{2}az^{2}+\alpha^{2}a\bar{z}^{2}+\alpha^{4}(1-a^{2})$, and then

$$\bar{z}=\pm\alpha\sqrt{\frac{az^{2}+\alpha^{2}(1-a^{2})}{z^{2}-\alpha^{2}a}}.$$

The $+$ sign corresponds to curve $L_{2}$, hence, its Schwarz function has the form

$$S(z)=\alpha\sqrt{\frac{az^{2}+\alpha^{2}(1-a^{2})}{z^{2}-\alpha^{2}a}}.$$

Then,

$$u_{1}(z)=\frac{\alpha}{2i}\int\limits_{L_{2}}\sqrt{\frac{a\tau^{2}+\alpha^{2}(1-a^{2})}{\tau^{2}-\alpha^{2}a}}\frac{d\tau}{\tau-z}.$$

(9)

Function $S(z)$ has four algebraic branching points $\pm\alpha\sqrt{a}$, $\pm\alpha\sqrt{(a^{2}-1)/a}$ —two inside $L_{2}$, and two outside. According to [3], we obtain the third variant of the classification for $S(z)$, hence, integral (9) cannot be calculated in elementary functions.

b) $z_{2}(t)=\alpha/(\sqrt{t+a})$, $\alpha>0$, $a>1$.

The equation of curve $z_{2}(\partial E)$ in Cartesian coordinates has the form $(1-a^{2})(x^{2}+y^{2})+2a\alpha^{2}(x^{2}-y^{2})-\alpha^{4}=0$. This is also a fourth order algebraic curve splitting into two convex ovals: the left one $L_{3}$ and the right one $L_{4}=z_{2}(\partial E)$. The Schwarz function for $L_{4}$ has the form

$$\bar{z}=\alpha\sqrt{\frac{az^{2}-\alpha^{2}}{(a^{2}-1)z^{2}-a\alpha^{2}}}.$$

Then,

$$u_{2}(z)=\frac{\alpha}{2i}\int\limits_{L_{4}}\sqrt{\frac{a\tau^{2}-\alpha^{2}}{(a^{2}-1)\tau^{2}-a\alpha^{2}}}\frac{d\tau}{\tau-z}.$$

(10)

By the same reasoning as in item a), this integral is not calculated in elementary functions.

Note that, so far, the inverse A-problems with functions of the form (9), (10) have not been considered in the articles on this topic.

2 CONNECTION BETWEEN FUNCTIONS $u(z)$AND $z(t)$ IN ‘‘DOMAIN’’ PROBLEMS WITH BOUNDARY CORNER POINTS

2.1 Corner Points Formed by the Intersection of Line Segments

In this case, the boundary of such domain $D_{n}$ is an $n$-gon, and this variant is studied in [3]. Consider the direct A-problem, and denote the vertices and the interior angles of domain $D_{n}$ by $a_{k}$ and $\alpha_{k}\in[0,2\pi]$, respectively ($k=\overline{1,n}$). Then, the behavior of function $u(z,D_{n})$ in a neighborhood of $a_{k}$ has the form

$$u(z,D_{n})\sim(e^{-2i\gamma_{k-1}}-e^{-2i\gamma_{k}})(z-a_{k})\ln(a_{k}-z),$$

where $\gamma_{k}=\arg(a_{k+1}-a_{k})$. Function $u(z,D_{n})$ itself is defined by the equality

$$u(z,D_{n})=\sum_{k=1}^{n}e^{-i(\gamma_{k+1}-\gamma_{k})}(z-a_{k})\ln(a_{k}-z).$$

(11)

Denote by $A_{k}$ the preimages of points $a_{k}$ on the unit circle $\{|t|=1\}$. Then, the mapping function $z(t)$ is the Schwarz–Christoffel integral

$$z(t)=C\int\limits_{t_{0}}^{t}\prod_{k=1}^{n}(\zeta-A_{k})^{\alpha_{k}/\pi-1}d\zeta+C_{1}.$$

Now, consider the inverse A-problem with function (11). If $D_{n}=z(E)$ is a star-like domain, then, due to Th. 0.1, this problem is uniquely solvable and uniquely invertible.

2.2 Corner Points Formed by a Line Segment and a Circular Arc

Consider the direct A-problem with the domain in the form of the sector $D(R,\alpha)=\{z:|z|<R,0,\arg z<\alpha\}$. It was shown in [1] that the potential gradient has the form

$$u(z,R,\alpha)=\frac{\alpha R^{2}}{2z}+\frac{i}{2}\left[R-Re^{-i\alpha}+\left(z-\frac{R^{2}}{z}\right)\ln\left(1-\frac{R}{z}\right)-\left(ze^{-2i\alpha}-\frac{R^{2}}{z}\right)\ln\left(1-\frac{Re^{i\alpha}}{z}\right)\right].$$

In a neighborhood of point $z=R$, the gradient behaves as follows: $u(z,R,\alpha)\sim i(z-R)\ln\left(1-\frac{R}{z}\right)+O(z-R)$. In a neighborhood of $z=Re^{i\alpha}$, the behavior of the gradient is of the form $u(z)\sim ie^{-2i\alpha}(z-Re^{i\alpha})\ln\left(1-\frac{Re^{i\alpha}}{z}\right)+O(z-Re^{i\alpha})$. The mapping function $z(t)$ is written in the form

$$z(t)=R\left(\frac{\sqrt{1-t}-\sqrt{ia(1+t)}}{\sqrt{1-t}+\sqrt{ia(1+t)}}\right)^{-\alpha},\ a>0.$$

2.3 Corner Points Formed by the Intersection of Circular Arcs

Such domains have not been considered so far, hence, we study this case in more detail. We start with the direct A-problem where domain $D(a,R)$ contains $0$ and is formed by the intersection of the circular arcs $L_{-a}=\{z:|z+a|<R\}$ and $L_{a}=\{z:|z-a|<R\}$, $R>a>0$. From two possible variants of lunes for $D(a,R)$, we choose the largest by area. The corner points here are $\pm i\sqrt{R^{2}-a^{2}}=\pm ib$, $b>0$. Denote $\alpha=\arg(a-ib)<0$. By the Green formula, we obtain

$$u(z,a,R)=\iint\limits_{D(a,R)}\frac{d\sigma}{\tau-z}=\frac{1}{2i}\left(\int\limits_{L_{-a}}\frac{\bar{\tau}d\tau}{\tau-z}+\int\limits_{L_{a}}\frac{\bar{\tau}d\tau}{\tau-z}\right).$$

For $\tau\in L_{-a}$, we have $|\tau+a|=R\ \Leftrightarrow\ (\tau+a)(\bar{\tau}+a)=R^{2}\ \Leftrightarrow\ \bar{\tau}=(b^{2}-a\tau)/(\tau+a)$, hence,

$$\int\limits_{L_{-a}}\frac{\bar{\tau}d\tau}{\tau-z}=\int\limits_{ib}^{-ib}\frac{b^{2}-a\tau}{(\tau+a)(\tau-z)}d\tau=-\frac{2i\alpha R^{2}}{z+a}+\frac{az-b^{2}}{z+a}(\ln(z-ib)-\ln(z+ib)).$$

For $\tau\in L_{a}$, we have $|\tau-a|=R\ \Leftrightarrow\ (\tau-a)(\bar{\tau}-a)=R^{2}\ \Leftrightarrow\ \bar{\tau}=(b^{2}+a\tau)/(\tau-a)$, and hence,

$$\int\limits_{L_{a}}\frac{\bar{\tau}d\tau}{\tau-z}=\int\limits_{-ib}^{ib}\frac{b^{2}+a\tau}{(\tau-a)(\tau-z)}d\tau=-\frac{2i\alpha R^{2}}{z-a}+\frac{az+b^{2}}{z-a}(\ln(z-ib)-\ln(z+ib)).$$

Then,

$$u(z,a,R)=\frac{1}{2i}\left(-\frac{2i\alpha R^{2}}{z+a}-\frac{2i\alpha R^{2}}{z-a}+\frac{2a(z-ib)(z+ib)}{(z-a)(z+a)}\ln\frac{z-ib}{z+ib}\right).$$

Note that $\lim\limits_{z\to\pm ib}u(z,a,R)<\infty$.

Consider passing to the limit as $a\to 0$. Then, in the limit, $D(a,R)$ coincides with the open disk $D(0,R)=\{z:|z|<R\}$, $\alpha\to-\pi/2$, and

$$\lim_{a\to 0}u(z,a,R)=\frac{1}{2i}\left(-\frac{4i(-\pi/2)R^{2}}{z}\right)=\frac{\pi R^{2}}{z}.$$

We have obtained the known logarithmic potential for $D(0,R)$ [3, Th. 2]. Denote the angle between arcs $L_{-a}$ and $L_{a}$ by $\gamma\in(0,\pi/2)$, then the inverse function $t(z)$ to the mapping function $z(t)$ is expressed by the formula

$$t(z)=\frac{(z+ib)^{\pi/\gamma}+(z-ib)^{\pi/\gamma}}{(z+ib)^{\pi/\gamma}-(z-ib)^{\pi/\gamma}},$$

it has branching points when $\pi/\gamma$ is not an integer.

Due to the star-likeness of the domains from items 2.2 and 2.3, the inverse A-problems for functions $u(z,R,\alpha)$ and $u(z,a,R)$ are uniquely solvable and uniquely invertible.

3 PROBLEMS OF SIMPLE LAYER POTENTIAL FOR DOMAINS WITH SMOOTH AND PIECEWISE SMOOTH BOUNDARIES

3.1. Such problems were studied in [5]. Denote by $z=f(\zeta)$ a function which maps the unit circle $|\zeta|=1$ onto the sought contour. This function has the form

$$f(\zeta)=\int\limits_{0}^{\zeta}P_{n}(\omega)d\omega=\zeta Q_{2n}(\zeta).$$

Function $f(\zeta)$ is a polynomial of degree $2n+1$ univalent in the closed unit disk. The gradient of the simple layer potential is given by the formula

$$u(z)=\int\limits_{L}\frac{|dt|}{t-z}=\sum_{k=1}^{n+1}\frac{A_{k}}{z^{k}}.$$

(12)

Wherein $|A_{1}|$ is the length of contour $L$. If $n=1$, we have

$$f(\zeta)=b\int\limits_{0}^{\zeta}(1+a\omega)^{2}d\omega,\quad u(z)=\frac{A_{1}}{z}+\frac{A_{2}}{z^{2}},\quad A_{1},A_{2}<0.$$

The connection between coefficients $a$, $b$ and $A_{1}$, $A_{2}$ are given by the system

$$A_{1}=-2\pi b(a^{2}+1),\quad A_{2}=-2\pi b^{2}a.$$

(13)

The direct B-problem is to find $A_{1}$ and $A_{2}$ given $a$ and $b$. The inverse B-problem is to find $a$ and $b$ given $A_{1}$ and $A_{2}$. System (13) implies $2\pi b^{3}A_{1}+A^{2}_{2}=-4\pi^{2}b^{4}$, wherein there are three possible cases [5]:

a) for $4\pi^{2}A_{2}^{2}>\frac{27}{256}A_{1}^{4}$, system (13) (and hence, the inverse B-problem) does not have a solution (is inconsistent);

b) for $4\pi^{2}A_{2}^{2}=\frac{27}{256}A_{1}^{4}$, system (13) (and hence, the inverse B-problem) has a unique solution;

c) for $4\pi^{2}A_{2}^{2}<\frac{27}{256}A_{1}^{4}$, system (13) (and hence, the inverse B-problem) has two different solutions.

In case c) for $A_{1}^{0}=-130\pi/19$, $A_{2}^{0}=-12\pi\sqrt{114}/19$, we have two star-like functions

$$f_{1}(\zeta)=2\int\limits_{0}^{\zeta}\left(1+\frac{3\sqrt{3}}{\sqrt{38}}\omega\right)^{2}d\omega,\ f_{2}(\zeta)=3\int\limits_{0}^{\zeta}\left(1+\frac{2\sqrt{2}}{\sqrt{57}}\omega\right)^{2}d\omega.$$

Now, two curves $f_{1}(\partial E)$ and $f_{2}(\partial E)$ are star-like, moreover, $f_{2}(\partial E)$ encompasses $f_{1}(\partial E)$. If we consider the direct B-problem, first with function $f_{1}(\zeta)$, and then with function $f_{2}(\zeta)$, we will obtain the same solution

$$u(z)=\frac{A_{1}^{0}}{z}+\frac{A_{2}^{0}}{z^{2}}.$$

(14)

Let us formulate the obtained result.

Theorem 2. There exist direct B-problems where two given different star-like mapping functions in the form of a third order polynomial are corresponded by the same simple layer potential (14), and vice versa, the solution to the inverse B-problem with potential (14) is a pair of different star-like functions.

Corollary. 1) The inverse B-problem with potential (14) is non-uniquely solvable according to Def. 2.

2) Two direct B-problems with functions $f_{1}(\zeta)$ and $f_{2}(\zeta)$ from case c) are non-uniquely invertible according to Def. 7. Potential (14) is their mutual solution, it generates two different contours one of which coincides with the original contour.

However, if we restrict ourselves to the class of convex functions $z=f(\zeta)$, the solvable inverse B-problem for functions (12) has a unique solution [6, Th. 2].

3.2. In comparison to a ‘‘domain’’ problem, consider the direct B-problem for the domain from item 2.3. Using the notation introduced there we obtain (letting $b=1$)

$$u(z)=\int\limits_{\partial D(a,R)}\frac{|dt|}{t-z}=\int\limits_{L_{a}}\frac{ds}{t-z}+\int\limits_{L_{-a}}\frac{ds}{t-z}=J_{1}+J_{2}.$$

Denote $\alpha={\textrm{arccotan}}a$. If $t\in L_{-a}$, $t=a+\sqrt{a^{2}+1}e^{i\theta}$. The arc length equals $ds=|dt|=\sqrt{a^{2}+1}d\theta$, $-(\pi-\alpha)\leq\theta\leq\pi-\alpha$. Then,

$$J_{1}=\int\limits_{-(\pi-\alpha)}^{\pi-\alpha}\frac{\sqrt{a^{2}+1}e^{i\theta}d\theta}{e^{i\theta}(a+\sqrt{a^{2}+1}e^{i\theta}-z)}=\frac{\sqrt{a^{2}+1}}{z-a}\left[-i\ln(a+\sqrt{a^{2}+1}e^{i\theta}-z)|_{-(\pi-\alpha)}^{\pi-\alpha}-2(\pi-\alpha)\right]$$

$$=\frac{\sqrt{a^{2}+1}}{z-a}\left[-i\ln\frac{z-i}{z+i}-2(\pi-\alpha)\right].$$

If $t\in L_{a}$, $t=-a+\sqrt{a^{2}+1}e^{i\theta}$, $\alpha\leq\theta\leq 2\pi-\alpha$, and

$$J_{2}=\frac{\sqrt{a^{2}+1}}{z+a}\left[-i\ln(-a+\sqrt{a^{2}+1}e^{i\theta}-z)|_{\alpha}^{2\pi-\alpha}-2(\pi-\alpha)\right]=\frac{\sqrt{a^{2}+1}}{z-a}\left[-i\ln\frac{z+i}{z-i}-2(\pi-\alpha)\right].$$

The gradient of the simple layer potential for domain $D(a,R)$ has the form

$$u(z)=\sqrt{a^{2}+1}\left\{\frac{1}{z-a}\left[-i\ln\frac{z-i}{z+i}-2(\pi-\alpha)\right]+\frac{1}{z+a}\left[-i\ln\frac{z+i}{z-i}-2(\pi-\alpha)\right]\right\}.$$

Consider the behavior of function $u(z)$ at the corner points $\pm i$. It is straightforward to show that

$$u(z)\sim i\sqrt{a^{2}+1}\left[-\frac{1}{z-a}\ln(z-i)+\frac{1}{z+a}\ln(z-i)\right]+O(z-i)$$

in a neighborhood of point $z=i$, and

$$u(z)\sim i\sqrt{a^{2}+1}\left[\frac{1}{z-a}\ln(z+i)-\frac{1}{z+a}\ln(z+i)\right]+O(z+i)$$

in a neighborhood of point $z=-i$. Comparing to item 2.3, here we have $\lim_{z\to\pm i}u(z)=\infty$.

Consider passing to the limit as $a\to 0$. For this, represent $u(z)$ in the form

$$u(z)=\sqrt{a^{2}+1}\left(\frac{1}{z-a}-\frac{1}{z}\right)\left[-i\ln(a+\sqrt{a^{2}+1}e^{i\theta}-z)|_{-(\pi-\alpha)}^{\pi-\alpha}-2(\pi-\alpha)\right]$$

$${}+\sqrt{a^{2}+1}\left(\frac{1}{z+a}-\frac{1}{z}\right)\left[-i\ln(-a+\sqrt{a^{2}+1}e^{i\theta}-z)|_{\alpha}^{2\pi-\alpha}-2(\pi-\alpha)\right]$$

$${}+\frac{\sqrt{a^{2}+1}}{z}\left\{-i\ln(a+\sqrt{a^{2}+1}e^{i\theta}-z)|_{-(\pi-\alpha)}^{\pi-\alpha}--i\ln(-a+\sqrt{a^{2}+1}e^{i\theta}-z)|_{\alpha}^{2\pi-\alpha}\right\}$$

$${}-\frac{4\sqrt{a^{2}+1}(\pi-\alpha)}{z}.$$

Denote by $D^{+}(a,R)$ and $D^{-}(a,R)$ the interior and the exterior of the lune, respectively. For $z\in D^{+}(a,R)$ and $a\to 0$, we have $\alpha\to\frac{\pi}{2}$, $\{\cdot\}\to 2\pi$. Calculate the limit value of the potential gradient

$$\lim_{a\to 0}u^{+}(z)=\frac{2\pi}{z}-\frac{2\pi}{z}=0.$$

For $z\in D^{-}(a,R)$ and $a\to 0$, we have $\alpha\to\frac{\pi}{2}$, $\{\cdot\}\to 0$, and

$$\lim_{a\to 0}u^{-}(z)=-\frac{2\pi}{z}.$$

These results coincide with those obtained in [5].