MATHEMATICS

L. G. MIKHAILOV

INVESTIGATION OF ONE NEW TYPE OF TWO-DIMENSIONAL INTEGRAL EQUATIONS

(Presented by Academician M. A. Lavrent’ev, 16 X 1957)

Consider the equation*

\[ f(z)+\lambda \frac{1}{\pi z}\iint_D \frac{K(\zeta)}{\zeta-z}\, f(\zeta)\,ds=g(z), \tag{1} \]

where \(z=x+iy;\ \zeta=\xi+i\eta;\ ds=d\xi\,d\eta;\ K(z)\) and \(g(z)\) are given, and \(f(z)\) is the sought, complex-valued function of real variables; \(\lambda\) is a complex parameter; \(D\) is a bounded domain whose boundary consists of a finite number of simple closed nonintersecting rectifiable lines; the origin lies inside the domain \(D\).

Let \(\gamma_\varepsilon\) be the circle \(|z|\leq \varepsilon\) and \(D_\varepsilon=D-\gamma_\varepsilon\). The integral in (1), and everywhere below, will be understood in the sense of the principal value, i.e., as

\[ \lim_{\varepsilon\to 0}\iint_{D_\varepsilon}, \]

if

\[ \iint_{D_\varepsilon} \]

exists in the Lebesgue sense for every \(\varepsilon>0\). The class of functions integrable in the indicated sense will be denoted by \(L^*(D)\), while retaining the notation \(L_p(D)\) for functions summable with power \(p\geq 1\).

1. Homogeneous equation. Define reciprocal classes of kernels and solutions by the conditions

\[ \begin{aligned} \text{A.}\quad & K(z)f(z)\in L_p(D_\varepsilon),\quad p>2 \quad \text{for every } \varepsilon>0;\\ \text{B.}\quad & K(z)f(z)\in L^*(D). \end{aligned} \tag{2} \]

Differentiating the equation with respect to \(\bar z\) in the sense of S. L. Sobolev \((^2)\), and then integrating the resulting differential equation \((^{1,3,7})\), we shall have

\[ f(z)=\frac{\Phi(z)}{z^2}e^{\lambda\omega(z)}, \tag{3} \]

where \(\Phi(z)\) is some holomorphic function and

\[ \omega(z)=-\frac{1}{\pi}\iint_D \frac{K(\zeta)}{\zeta(\zeta-z)}\,ds. \tag{4} \]

If \(K(z)/z\) is not integrable, then an analytic regularizer can be introduced \((^3)\).

Thanks to formula (3), the solution of the integral equation is reduced to finding the holomorphic function \(\Phi(z)\). Under conditions (2), from the very

* The problem was posed by I. N. Vekua; in the general formulation, under the integral sign, instead of \(f(\zeta)\) there stands \(\overline{f(\zeta)}\).

it follows from the equation \((^{1,2,7})\) that \(\Phi(z)\) is continuous and holomorphic in the whole plane, except for the point \(z=0\), and has the expansion

\[ \Phi(z)=a_0+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots \tag{5} \]

To determine \(\Phi(z)\), we substitute (3) into the integral equation and bring it to the form

\[ \frac{\Phi(z)}{z}e^{\lambda\omega(z)} = -\frac{1}{\pi}\iint_D \frac{\partial}{\partial \bar\zeta} \left[ \frac{\Phi(\zeta)}{\zeta}e^{\lambda\omega(z)} \right] \frac{ds}{\zeta-z}. \tag{6} \]

Here we have the condition that \(\frac{\Phi(z)}{z}e^{\lambda\omega(z)}\) coincides with the integral of its derivative. It can be shown that \(\Phi(z)\) satisfies condition (6) if and only if the equalities

\[ \lim_{\varepsilon\to 0}\int_{l_\varepsilon}\Phi(t)e^{\lambda\omega(t)}t^k\,dt=0,\qquad k=-1,0,1,2,\ldots, \tag{7} \]

hold, where \(l_\varepsilon\) is the circle \(|z|=\varepsilon\). Thus, the following theorem has been obtained:

Theorem 1. In order that the homogeneous integral equation (1) have a solution, it is necessary and sufficient that there exist an entire analytic function (5) satisfying conditions (7).

2. Passing to the nonhomogeneous equation, set \(K(z)g(z)\in L_p(D)\), \(p>2\). By the substitution of the unknown function \(f=f_1+g\) we obtain a new free term

\[ g_1(z)=-\frac{1}{\pi}\iint \frac{K(\zeta)g(\zeta)}{\zeta-z}\,ds, \]

which is continuous and differentiable in the sense of S. L. Sobolev, as we shall assume.

Differentiating (1) with respect to \(\bar z\) and then integrating the resulting nonhomogeneous differential equation as in (7), we shall have

\[ f(z)=g(z)+\lambda\frac{e^{\lambda\omega(z)}}{z}\,[\Omega(z)+\Phi(z)], \tag{8} \]

where \(\omega(z)\) and \(\Phi(z)\) are given by formulas (4) and (5), and

\[ \Omega(z)=-\frac{1}{\pi}\iint \frac{K(\zeta)e^{-\lambda\omega(\zeta)}}{\zeta-z}g(\zeta)\,ds. \]

It may happen that \(Ke^{-\lambda\omega}g\) is not integrable, and then it is necessary to introduce an analytic regularizer.

With the help of (8), equation (1) is transformed into the equality

\[ [\Omega(z)+\Phi(z)]e^{\lambda\omega(z)} = -\frac{1}{\pi}\iint_D \frac{\partial}{\partial \bar\zeta} [\Omega(\zeta)+\Phi(\zeta)]e^{\lambda\omega(\zeta)} \frac{ds}{\zeta-z}, \]

which, like (6), expresses that the function \((\Omega+\Phi)e^{\lambda\omega}\) coincides with the integral of its derivative, and this reduces to the conditions

\[ \lim_{\varepsilon\to 0}\int_{l_\varepsilon} [\Omega(t)+\Phi(t)]e^{\lambda\omega(t)}t^k\,dt=0,\qquad k=0,1,2,\ldots. \tag{9} \]

Theorem 2. In order that the nonhomogeneous integral equation (1) have a solution, it is necessary and sufficient that there exist an entire analytic function (5) satisfying conditions (9).

3. On the basis of Theorems 1 and 2 one can investigate various concrete cases. In Theorem 3 the discussion will concern the simplest solutions of the homogeneous equation

\[ z^{-m}e^{\lambda\omega(z)},\qquad m=2,3,\ldots, \tag{10} \]

which are obtained from (3), if for \(\Phi(z)\) one takes successively \(1, z^{-1}, z^{-2}, \ldots\)

Theorem 3. If \(K(z)\) satisfies the Hölder condition at the point \(z=0\), then for every \(\lambda\) the homogeneous equation has no solutions of the form (10), while the nonhomogeneous equation has, moreover, a unique solution.

Proof. We shall verify conditions (7) for the functions (10).

For all \(m=2,3,\ldots\), the condition
\[ \lim_{\varepsilon\to 0}\int_{l_\varepsilon} e^{\lambda\omega(t)}t^{-1}\,dt=0 \]
is fulfilled, or
\[ \lim_{r\to 0}\int_0^{2\pi} e^{\lambda\omega(r,\varphi)}\,d\varphi=0. \tag{11} \]

If \(K(0)=0\), then \(\omega(z)\) is continuous. Passing to the limit under the integral sign in (11) gives
\[ \lim_{r\to 0}\int_0^{2\pi} e^{\lambda\omega(r,\varphi)}\,d\varphi=e^{\lambda\omega(0)}\ne 0. \]
This shows that no function (10) can serve as a solution.

For \(K(0)\ne 0\), the integral (4) can be represented in the form
\[ \omega(z)=a(z)+K(0)\frac{\bar z}{z}, \]
where \(a(z)\) is a continuous function. Substituting this expression for \(\omega(z)\) in (11), we again see that the condition is not fulfilled. The first part of the theorem is proved.

The solution of the nonhomogeneous equation is given by formula (8) for \(\Phi(z)\equiv 0\). It is easy to verify that conditions (9) are fulfilled, taking into account that \(\omega(z)\) is bounded and \(\Omega(z)\) is continuous. The theorem is proved.

Theorem 4. Let \(\omega(z)\) depend only on \(r=|z|\). Then, if
\[ \lim_{r\to 0} e^{\lambda\omega(r)}\ne 0 \]
or does not exist, the homogeneous equation has no solutions different from the trivial one; but if
\[ \lim_{r\to 0} e^{\lambda\omega(r)}=0, \]
then it has an infinite number of linearly independent solutions (10).

Example. Let \(D\) be the disk \(|z|\le R<1\) and
\[ K(z)=\frac{1}{2\ln r}\,\frac{z}{\bar z}. \]
Here \(K(z)\) is continuous (but does not satisfy the Hölder condition) and
\[ \omega(z)=\ln\ln\frac{1}{|z|}-\ln\ln\frac{1}{R}. \]
The nonhomogeneous equation is solvable for every \(\lambda\); the homogeneous equation has an infinite number of solutions in the half-plane \(\operatorname{Re}\lambda>0\) and has no solutions for \(\operatorname{Re}\lambda\le 0\). This shows that the principle of contraction mappings is not applicable to the operator (1).

If, following (6), one calls the index \(x\) the difference between the numbers of solutions of the given and of the transposed equations, then in our example \(x\) takes the values \(-\infty, 0, +\infty\). The equation is neither Fredholm nor Noetherian.

Theorem 5. If \(K(z)\) is continuous at the point \(z=0\), then the nonhomogeneous equation is solvable for every \(\lambda\).

Theorem 6. If \(K(z)\) is continuous at the point \(z=0\), then the nonhomogeneous equation is solvable for
\[ |\lambda|<\frac{1}{2M}, \]
where
\[ M=\lim_{z\to 0}|K(z)|. \]

Let us estimate the behavior of \(\omega(z)\) at the point \(z=0\). If \(K(z)\) is continuous at the point \(z=0\), then
\[ |\omega(z)|<\varepsilon\ln\frac{1}{|z|}+N(\varepsilon), \]
where \(\varepsilon>0\) is an arbitrarily small quantity and \(N(\varepsilon)\) is a constant depending on it. If \(K(z)\) is bounded at the point \(z=0\) and
\[ M=\lim_{z\to 0}|K(z)|, \]
then
\[ |\omega(z)|<2(M+\varepsilon)\ln\frac{1}{|z|}+N_\varepsilon. \]
Substituting these estimates into (9) for \(\Phi(z)\equiv 0\), we obtain the proofs of Theorems 5 and 6.

It is appropriate to make a comparison with one-dimensional equations. It is known that if \(K(x,t)\) satisfies the Hölder condition, then the kernel
\[ \frac{K(x,t)}{x-t} \]
is singular (not Fredholm) (4), while the kernel
\[ \frac{K(x,t)}{x-c} \]
can be assigned to the Fredholm—

...with (5). Of interest is the question whether the kernel

\[ \frac{K(x,t)}{\sqrt{(x-t)(x-c)}} \]

is Fredholm or singular. In the two-dimensional case the kernel

\[ \frac{K(\zeta,z)}{(\zeta-z)^2} \]

is likewise singular, whereas

\[ \frac{K(\zeta,z)}{(z-c)^2} \]

is Fredholm. The “intermediate” case

\[ \frac{K(\zeta)}{z(\zeta-z)} \]

was studied above. Theorem 3 shows that, under a Hölder condition on \(K(z)\), this kernel is Fredholm. The example shows that already for continuous \(K(z)\) it becomes singular.

In conclusion I express my gratitude to Corresponding Member of the Academy of Sciences of the USSR I. N. Vekua for posing the problem and for valuable suggestions.

Tajik State
University

Received
15 X 1957

REFERENCES

¹ I. N. Vekua, Matem. sborn., 31 (73), 2 (1952).
² I. N. Vekua, DAN, 89, No. 5 (1953).
³ I. N. Vekua, DAN, 98, No. 2 (1954).
⁴ N. I. Muskhelishvili, Singular Integral Equations, M.—L., 1946.
⁵ G. Fubini, Atti d. Accad. Naz. d. Lincei, Rend. 21, (1912).
⁶ I. Ts. Gokhberg, DAN, 78, No. 4 (1951).
⁷ L. G. Mikhailov, Uch. zap. Tajik. univ., 10 (1957).