Georgian Mathematical Journal Volume 9 (2002), Number 1, 179–196

SOLUTION OF A LINEAR INTEGRAL EQUATION OF THIRD KIND

D. SHULAIA

Abstract. The aim of this paper is to study, in the class of Hölder functions, a nonhomogeneous linear integral equation with coefficient cos x. Necessary and sufficient conditions for the solvability of this equation are given under some assumptions on its kernel. The solution is constructed analytically, using the Fredholm theory and the theory of singular integral equations.

2000 Mathematics Subject Classification: 45B05, 45E05.

Key words and phrases: Integral equations of third kind, eigenfunctions, singular integral operator, parameter, boundary values, homogeneous equation.

1. Introduction

In this paper, a method is described for solving an integral equation which frequently occurs when studying many important problems of mathematical physics, in particular, in problems of the multidimensional transfer theory. An equation is written in the form

cos xφ(x) = ∫_{0}^{2π}K(x, y)φ(y)dy + f(x), x ∈ (0, 2π).  (1)

Such equations are frequently called equation of third kind. After the early works of Hilbert [1] and Picard [2], in the course of the last century there appeared a lot of papers on equations of third kind though attention given to them can hardly be compared with attention given to equations of first and second kind. Most of standard studies of integral equations touch very briefly upon equations of third kind. The investigation should certainly begin by the question what we must mean by a solution of equation (1). Various answers are suggested in literature. Following the classical definition of a solution which is natural and typical of some frequently occurring in the applications of third kind equations, we assume that the kernel K ∈ H (i.e., satisfies the Hölder condition H), while a free term belongs to the Muskhelishvili class H^{∗} (see [3]). Therefore we look for a solution φ of equation (1) which belongs to the class H^{∗}.

Our investigation is based on the ideas of the theory of spectral expansion and the approach proposed by Hilbert and Schmidt for an integral equation of second

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

kind. This paper is the continuation of [4]. The paper is organized as follows. In Section 2, using the initial equation, we introduce integral operators and their corresponding integral equations which depend on the parameter. Some of their properties, which will play an important role in further considerations, are introduced. In Section 3, singular integral operators, which are closely connected with the introduced operators and equations, are defined and studied. The main result states that an arbitrary function from H^{∗} can be represented through singular operators and the eigenfunctions of the parameter-dependent operator. In the last section, a solution of the considered problem is given as an application of the results obtained in Section 2.

2. Notation and the Background

Let Ω_{z} denote the linear integral operator defined by the formula

Ω_{z}(g(z, ·))(x) := g(z, x) + ∫_{0}^{2π} \frac{K(x, y)}{z − cos y} g(z, y)dy, x ∈ [0, 2π], z ∉ [−1, 1], (2)

where the real-valued function K satisfies the condition H with respect to both variables.

Recall that a function φ(t) is said to satisfy the condition H (Hölder condition) on a certain smooth line Γ, if for any two points t_{1}, t_{2} of Γ

|φ(t_{1}) − φ(t_{2})| ≤ A|t_{1} − t_{2}|^{γ},

where A and γ are some positive constants (see, e.g., [3]).

If a function φ(t), given on Γ satisfies the condition H in each closed part of Γ not containing the ends, and near any ends c, φ(t) behaves as

φ(t) = \frac{φ^{∗}(t)}{|t − c|^{δ}} , 0 ≤ δ < 1,

where φ^{∗}(t) belongs to the class H, then it is said φ(t) to belong to the class H^{∗} on Γ.

Recall also that [3] if Φ(z) is a function holomorphic in each finite domain not containing points of a certain smooth line Γ and the function Φ(z) is continuous both from the left and from the right on Γ, with a possible exception for the ends, but near the ends satisfies the condition

|Φ(z)| ≤ \frac{C}{|z − c|^{α}} ,

where c is the corresponding end, C and α are certain real constants, α < 1, then Φ(z) is called piecewise holomorphic function with the line of discontinuity Γ.

The operator Ω_{z}(g(z, ·))(x), operating on any function g(z, x) piecewise holomorphic in z with a cut on the real axis [−1, 1] and satisfying the condition H in x, will gives a piecewise holomorphic function with a cut on the [−1, 1].

Using the Plemelj–Sokhotskii’s formulas (see, e.g.,[5], p. 86), we can calculate the boundary values of Ω_{z}(g) as

Ω_{ζ}^{±}(g^{±}(ζ, ·))(x) := g^{±}(ζ, x) + ∫_{0}^{2π} \frac{K(x, y)}{ζ − cos y} g^{±}(ζ, y)dy

±\frac{iπ}{\sqrt{1 − ζ^{2}}} (K(x, ζ_{a})g^{±}(ζ, ζ_{a}) + K(x, ζ̄_{a})g^{±}(ζ, ζ̄_{a})), ζ ∈ (−1, +1), (3)

where ζ_{a} = arccos ζ, ζ̄_{a} = 2π − arccos ζ. Here the integration with respect to y is understood in the Cauchy principal value sense.

Theorem (Tamarkin). Suppose the kernel K(x, ξ, λ) of the integral equation

u(x) = f(x) + ∫_{a}^{b} K(x, ξ, λ)u(ξ)dξ

is analytic in λ on an open domain A of the λ-plane and for almost all points (x, ξ) of the square

a ≤ x ≤ b; a ≤ ξ ≤ b.

Suppose that the integrals

∫_{a}^{b} |K(x, ξ, λ)|^{2}dξ,  ∫_{a}^{b} |K(x, ξ, λ)|^{2}dx

exist for almost all values of x and ξ, respectively, and that, on every closed subdomain A_{0} of A we have

∫_{a}^{b} |K(x, ξ, λ)|^{2}dξ ≤ F_{0}(x),  ∫_{a}^{b} |K(x, ξ, λ)|^{2}dx ≤ F_{0}(ξ),

where F_{0}(x) is a positive function which depends only on A_{0} and which is integrable on (a, b). Then the reciprocal Ω(x, ξ, λ) of the kernel K(x, ξ, λ) either is meromorphic on A for almost all values of x, ξ or does not exist at all.

For a proof, see [6].

Let κ denote the set of all values of z when the homogeneous equation

Ω_{z}(g) = 0  (4)

admits non-zero solutions. Such values of z are called eigenvalues of K(x, y) or of Ω_{z}. Since the kernel of the operator Ω_{z} is a function piecewise analytic in z, satisfying the condition H and vanishing as z → ∞, the Tamarkin theorem tells us κ is at most countable in the plane z with a cut on [−1, 1]. Obviously, if g is a solution of the latter equation, then the function

φ_{z_{k}}(x) = \frac{g(z_{k}, x)}{cos x − z_{k}} ,

where z_{k} ∈ κ, is also a solution of the equation

(cos x − z)φ_{z}(x) = ∫_{0}^{2π}K(x, y)φ_{z}(y)dy, x ∈ [0, 2π],

when z = z_{k} and vice versa. Moreover, sets of the eigenvalues of the operators Ω_{z} and

Ω_{z}^{∗}(h(z, ·))(x) := h(z, x) + ∫_{0}^{2π} \frac{K(y, x)}{z − cos y} h(z, y)dy, x ∈ [0, 2π],

are identical. Clearly, if z ≠ z′, then

∫_{0}^{2π}φ_{z}^{∗}(x)φ_{z′}(x)dx = 0,  (5)

where φ_{z}^{∗} is a solution of the equation

(cos x − z)φ_{z}^{∗}(x) = ∫_{0}^{2π}K(y, x)φ_{z}^{∗}(y)dy, x ∈ [0, 2π].  (6)

It is not difficult to show that if K(x, y) = K(y, x), then all eigenvalues, if they exist, are real.

The functions φ_{z_{k}}(x) and φ_{z_{k}}^{∗}(x), z_{k} ∈ κ, are called the eigenfunctions of the kernel K(x, y) and K(y, x), respectively. Moreover, in the sequel the operators and functions determined by K(y, x) just in the same way as by K(x, y), will be provided with the superscript ^{∗}.

Define

α(t, y) := (1 − σ(t))(1 − σ(y)) + σ(t)σ(y), t, y ∈ [0, 2π],

where σ(t) is the characteristic function of the segment [0, π] and consider the following integral equation of second kind

M_{0}(t, x) = ∫_{0}^{2π}K̃(t, x, y)M_{0}(t, y)dy, x ∈ [0, 2π],  (7)

where

K̃(t, x, y) = \frac{K(x, y) − K(x, t)}{cos y − cos t}α(t, y) + \frac{K(x, y) − K(x, t̄)}{cos y − cos t}α(t̄, y)

and t̄= 2π − t. Here t ∈ [0, π] is a parameter. We see that the kernel K̃ of this equation does not belong to the type which is usually called regular. In spite of this to the equation (7) we may apply the basic Fredholm theorems. Indeed, it is possible to reduce such equations to the Fredholm equation with bounded kernel (see, e.g., [3], Section 111 and [7], Section 16).

Using this fact we have

Theorem 2.1. Let for some value of the parameter t = t′ ∈ [0, π] the homogeneous equation (7) have only a trivial solution. Then z′ = cos t′ ∉ κ.

Proof. The proof in a completely analogous to that of Theorem 1 from [4].

In order to eliminate additional arguments, in the sequel we shall assume that:

R_{1}. κ is a finite set.

R_{2}. The homogeneous equations (7) and

M_{0}^{∗}(t, x) = ∫_{0}^{2π}K̃^{∗}(t, x, y)M_{0}^{∗}(t, y)dy, x ∈ [0, π],

where

K̃^{∗}(t, x, y) = \frac{K(y, x) − K(t, x)}{cos y − cos t}α(t, y) + \frac{K(y, x) − K(t̄, x)}{cos y − cos t}α(t̄, y),

admit only trivial solutions for any values of the parameter t ∈ [0, π].

Note that these is an important restriction, although it is satisfied for a sufficiently wide class of kernels K(x, y), in particular, for kernels from the problems of the transport theory [8]–[11].

In view of Theorem 2.1 the assumption R_{2} implies:

A_{1}. κ ∩ [−1, +1] = ∅.

A_{2}. The nonhomogeneous integral equations

M(t, x) = ∫_{0}^{2π}K̃(t, x, y)M(t, y)dy + | sin t|K(x, t), t, x ∈ [0, 2π],  (8)

and

M^{∗}(t, x) = ∫_{0}^{2π}K̃^{∗}(t, x, y)M^{∗}(t, y)dy + | sin t|K(t, x), t, x ∈ [0, 2π], (9)

admit only unique solutions satisfying the H condition with respect to t and x.

Remark 2.2. If a kernel K ∈ H admits an expansion into a uniformly convergent series of the form

K(x, y) = ∑_{n}(2n + 1)g_{n}| sin x|P_{n}(cos x)P_{n}(cos y),

where g_{n} is a real number and P_{n} is the nth order Legendre polynomial, then equations (8) and (9) have unique solutions as functions expressed by uniformly convergent series of the form

M(t, x) = ∑_{n}(2n + 1)g_{n}| sin x|P_{n}(cos x)h_{n}(t)

and

M^{∗}(t, x) = ∑_{n}(2n + 1)g_{n}| sin t|P_{n}(cos x)h_{n}(t),

respectively, where h_{n} is defined from the recurrent relation

(n + 1)h_{n+1}(t) + nh_{n−1}(t) = (2n + 1)(cos t − 4g_{n})h_{n}(t),
h_{0}(t) = | sin t|, (n = 0, 1, ⋯).

In our further consideration we will also need the following identity, which immediately can be obtained by using the Bertrand–Poincaré formula [5]:

∫_{0}^{2π} \frac{M^{∗}(t_{0}, x)}{cos x − cos t_{0}} dx ∫_{0}^{2π} \frac{M(t, x)}{cos x − cos t}u(t)dt

= ∫_{0}^{2π} \frac{u(t)}{cos t_{0} − cos t}dt( ∫_{0}^{2π} \frac{M^{∗}(t_{0}, x)M(t, x)}{cos x − cos t_{0}}dx − ∫_{0}^{2π} \frac{M^{∗}(t_{0}, x)M(t, x)}{cos x − cos t}dx)

π((M̄^{∗}(t_{0}, t_{0})M̄(t_{0}, t_{0}) + M̄^{∗}(t_{0}, t̄_{0})M̄(t_{0}, t̄_{0}))u(t_{0})
+(M̄^{∗}(t_{0}, t_{0})M̄(t̄_{0}, t_{0}) + M̄^{∗}(t_{0}, t̄_{0})M̄(t̄_{0}, t̄_{0}))u(t̄_{0}), t_{0} ∈ (0, 2π), (10)

where u ∈ H^{∗}, M̄(t, x) and M̄^{∗}(t, x) are unique solutions of the nonhomogeneous integral equations

M̄(t, x) = ∫_{0}^{2π}K̃(t, x, y)M̄(t, y)dy + K(x, t), t, x ∈ [0, 2π],

and

M̄^{∗}(t, x) = ∫_{0}^{2π}K̃^{∗}(t, x, y)M̄^{∗}(t, y)dy + K(t, x), t, x ∈ [0, 2π],

respectively. It is obvious that

M(t, x) = | sin t|M̄(t, x) and M^{∗}(t, x) = | sin t|M̄^{∗}(t, x).

3. Representation Theorem for a Function of the Class H^{∗}

As is known that in the Hilbert–Schmidt theory for integral equations of second kind an important role is played by one property of eigenfunctions, which in our case corresponds to the following equality:

(cos x − z)φ_{z_{k}}(x) − ∫_{0}^{2π}K(x, y)φ_{z_{k}}(y)dy

= (z_{k} − z)φ_{z_{k}}(x), x ∈ [0, 2π], z_{k} ∈ κ.  (11)

But in contrast to this theory, in our case the set of eigenfunctions, is not a complete system even in the class of solutions of the initial equation. Therefore, based on the spectral expansion theory, for this aim we try to find such a singular operator which has a similar property.

Theorem 3.1. Let M ∈ H be a solution of (8) and let u be function such that

L(u(·))(x) := A(x)u(x) + B(x̄)u(x̄) + ∫_{0}^{2π} \frac{M(t, x)}{cos x − cos t}u(t)dt, x ∈ (0, 2π), (12)

where x̄ = 2π − x,

A(x) = | sin x| + ∫_{0}^{2π} \frac{α(x, y)M(x, y)}{cos x − cos y}dy, B(x̄) = ∫_{0}^{2π} \frac{α(x, y)M(x̄, y)}{cos x − cos y}dy,

is meaningful and, moreover, integrable. Then the equality

(cos x − z)L(u(·))(x) − ∫_{0}^{2π}K(x, y)L(u(·))(y)dy

= L((cos(·) − z)u(·))(x), x ∈ (0, 2π),

holds.

Proof. Indeed, in view of (8) and (12) straightforward calculations shows that

∫_{0}^{2π}K(x, y)L(u(·))(y)dy = ∫_{0}^{2π}M(t, x)u(t)dt.

By (8) we obtain

L((cos x − cos(·))u(·))(x) = ∫_{0}^{2π}M(t, x)u(t)dt

and the result follows.

The latter result gives a motive to study the operator L more thoroughly. Before beginning our systematic study we have to prove

Lemma 3.2. For every t ∈ (0, π) the system of algebraic equations

\begin{matrix}A(t)v(t) + B(t̄)w(t) = 0, \\ B(t)v(t) + A(t̄)w(t) = 0 \end{matrix}  (13)

with respect to (v, w) admits only a trivial solution.

Proof. Let us assume the contrary. Suppose for some t = t′ ∈ (0, π) this system has a non-zero solution (v(t′), w(t′)). Then by virtue of (8) and (13) the function

M(t′, x)v(t′) + M(t̄′, x)w(t′)

is a non-zero solution of equation (4) when z′ = cos t′, which contradicts the assumption A_{1}.

The basic result for L is the following Theorem 3.3. Let f ∈ H^{∗}. Then provided that the conditions

∫_{0}^{2π}fφ_{z_{k}}^{∗}dx = 0, z_{k} ∈ κ,  (14)

are fulfilled, there exists a unique solution u of the class H^{∗}, for which

L(u) = f.  (15)

Proof. Suppose that u ∈ H^{∗} satisfies (15) and introduce into consideration the function

Ψ(z, x) = \frac{1}{2πi} ∫_{0}^{2π} \frac{M(t, x)}{cos t − z} u(t)dt, x ∈ [0, 2π], z ∉ [−1, 1].

This function possesses the following properties:

P_{1}. In the plane with a cut [−1, +1] it is piecewise holomorphic with respect to the variable z, while for the variable x it satisfies the H condition.

P_{2}. As z → ∞ it vanishes uniformly in x.

P_{3}. By the Plemelj–Sokhotskii formulas for boundary values, we have

Ψ^{±}(ζ, x) = \frac{1}{2πi} ∫_{0}^{2π} \frac{M(t, x)}{cos t − ζ} u(t)dt ± ±\frac{1}{2\sqrt{1 − ζ^{2}}} (M(ζ_{a}, x)u(ζ_{a})
+ M(ζ̄_{a}, x)u(ζ̄_{a})), ζ ∈ (−1, +1), x ∈ [0, 2π].

Combining (3) with the latter equality we get

Ω_{ζ}^{+}(Ψ^{+}(ζ, ·))(x) − Ω_{ζ}^{−}(Ψ^{−}(ζ, ·))(x)

= \frac{1}{\sqrt{1 − ζ^{2}}} (M(ζ_{a}, x)u(ζ_{a}) + M(ζ̄_{a}, x)u(ζ̄_{a}))

+\frac{1}{\sqrt{1 − ζ^{2}}} ∫_{0}^{2π} \frac{K(x; y)}{ζ − cos y} (M(ζ_{a}, y)u(ζ_{a}) + M(ζ̄_{a}, y)u(ζ̄_{a}))dy

+\frac{1}{\sqrt{1 − ζ^{2}}} (K(x, ζ_{a}) ∫_{0}^{2π} \frac{M(t; ζ_{a})}{ζ − cos t}u(t)dt + K(x, ζ̄_{a}) ∫_{0}^{2π} \frac{M(t; ζ̄_{a})}{ζ − cos t}u(t)dt),

ζ ∈ (−1, +1), x ∈ [0, 2π].

Recall that M(t, x) satisfies (8) and then by straightforward calculation and using (15) we obtain

Ω_{ζ}^{+}(Ψ^{+})(x) − Ω_{ζ}^{−}(Ψ^{−})(x) = \frac{1}{\sqrt{1 − ζ^{2}}} (K(x, ζ_{a})f(ζ_{a}) + K(x, ζ̄_{a})f(ζ̄_{a})),

ζ ∈ (−1, +1), x ∈ [0, 2π].

Now it is evident that if we consider the function

F (z, x) = Ω_{z}(Ψ)(x),

then we will see that it is also piecewise holomorphic in z, vanishing as z → ∞ and its boundary values satisfy the condition

F^{+}(ζ, x) − F^{−}(ζ, x) = \frac{1}{\sqrt{1 − ζ^{2}}} (K(x, ζ_{a})f(ζ_{a}) + K(x, ζ̄_{a})f(ζ̄_{a})),

ζ ∈ (−1, +1), x ∈ [0, 2π].

Consequently, using the Plemelj–Sokhotskii formulas we get

F (z, x) = \frac{1}{2πi} ∫_{−1}^{+1}\frac{1}{\sqrt{1 − ξ^{2}}} \frac{K(x, ξ_{a})f(ξ_{a}) + K(x, ξ̄_{a})f(ξ̄_{a})}{ξ − z}dξ.  (16)

Here ξ_{a} = arccos ξ, ξ̄_{a} = 2π − arccos ξ. After some transformation of the righthand side of (16) we conclude that the function Ψ(z, x) satisfies the integral equation

Ω_{z}(Ψ)(x) = \frac{1}{2πi} ∫_{0}^{2π} \frac{K(x, t)}{cos t − z} f(t)dt, x ∈ [0, 2π].  (17)

However the condition of the solubility of the integral equation (17) is that its free term be orthogonal to the eigenfunctions of the kernel K(y, x), that is,

∫_{0}^{2π}φ_{z_{k}}^{∗}(x) ∫_{0}^{2π} \frac{K(x, t)}{cos t − z_{k}} f(t)dt dx = 0, z_{k} ∈ κ.

Now, using (6) we immediately come to conditions (14).

Now suppose that f ∈ H^{∗} satisfies conditions (14). Then by virtue of Tamarkin’s Theorems it follows that there is a unique solution Ψ of (17) which is a piecewise holomorphic function in z. Moreover, using (3) from (17) we obtain

Ψ̃(t_{0}, x) − ∫_{0}^{2π}K(x, y)Ψ̃(t_{0}, y)dy + K(x, t_{0})( ∫_{0}^{π} \frac{Ψ̃(t, t_{0})}{cos t − cos t_{0}} dt + f(t_{0}))

+K(x, t̄_{0})( ∫_{0}^{π} \frac{Ψ̃(t, t̄_{0})}{cos t − cos t_{0}} dt + f(t̄_{0})) = 0, t_{0} ∈ (0, 2π), x ∈ [0, 2π], (18)

where

Ψ̃(t_{0}, x) = | sin t_{0}|(Ψ^{+}(cos t_{0}, x) − Ψ^{−}(cos t_{0}, x)).

Now consider the nonhomogeneous system of equations:

A(t_{0})u(t_{0}) + B(t̄_{0})v(t_{0}) = ∫_{0}^{π} \frac{Ψ̃(t, t_{0})}{cos t − cost_{0}} dt + f(t_{0}),

B(t_{0})u(t_{0}) + A(t̄_{0})v(t_{0}) = ∫_{0}^{π} \frac{Ψ̃(t, t̄_{0})}{cos t − cost_{0}} dt + f(t̄_{0}), t_{0} ∈ (0, π),

(19) with respect to (u, v). Lemma 3.2 shows us that there is a unique solution of (19) and, moreover, it is easy to see that v(t_{0}) = u(t̄_{0}).

Let

M̃(t_{0}, x) = Ψ̃(t_{0}, x) − M(t_{0}, x)u(t_{0}) − M(t̄_{0}, x)u(t̄_{0}).

A quick calculation using (18) and (8) shows that M̃(t_{0}, x) is a solution of the homogeneous equation (4) when z = cos t_{0}. However, since K has no the eigenvalues on [−1, +1], we obtain

Ψ̃(t_{0}, x) = M(t_{0}, x)u(t_{0}) + M(t̄_{0}, x)u(t̄_{0}), t_{0} ∈ (0, 2π), x ∈ [0, 2π].

Combining this with (19), we have

A(t_{0})u(t_{0}) + B(t̄_{0})u(t̄_{0}) = f(t_{0}) + ∫_{0}^{π} \frac{M(t, t_{0})}{cos t − cos t_{0}} u(t)dt

+ ∫_{π}^{2π} \frac{M(t, t_{0})}{cos t − cos t_{0}} u(t)dt, t_{0} ∈ (0, 2π), x ∈ [0, 2π].

Consequently (15) holds and the proof is completed.

Now a major goal of ours is to invert the operator L. To do this, we need to introduce, in the class H^{∗}, the integral operators

S(v(·))(t) := A(t)v(t) + B(t)v(t̄) + ∫_{0}^{2π} \frac{M(t, y)}{cos y − cos t}v(y)dy, t ∈ (0, 2π), (20)

and S^{∗} introduced similar to S according to the rule mentioned above. The operator (20) can be rewritten as

S(v(·))(t) := | sin t|v(t)

+ ∫_{0}^{2π}(\frac{v(y) − v(t)}{cos y − cos t}α(t, y) + \frac{v(y) − v(t̄)}{cos y − cos t}α(t̄, y))M(t, y)dy, t ∈ (0, 2π);

the operator S^{∗} can be rewritten similarly.

It is easy to see that for any u and v from H^{∗}

∫_{0}^{2π}u(t)S(v(·))(t)dt = ∫_{0}^{2π}v(x)L(u(·))(x)dx.

Hence, if there exists u such that (15) is fulfilled, then it is necessary that

∫_{0}^{2π}v(x)f(x)dx = 0,  (21)

where v is an arbitrary solution of the homogeneous equation

S(v) = 0.  (22)

SOLUTION OF A LINEAR INTEGRAL EQUATION OF THIRD KIND 189

The converse statement is also true. Before proving this, we want to give some properties of the operators S and S^{∗}.

The following lemma is an immediate consequence of the construction of S, together with equation (8).

Lemma 3.4. The equality

S(K(x, ·))(t) = M(t, x), x, t ∈ [0, 2π],

holds.

The same type of arguments leads to

Lemma 3.5. The equality

S^{∗}(K(·, x))(t) = M^{∗}(t, x), x, t ∈ [0, 2π],

holds.

Lemma 3.6. The equality

S^{∗}(M(t_{0}, ·))(t) = S(M^{∗}(t, ·))(t_{0}), t_{0}, t ∈ [0, 2π],  (23)

holds. Proof. In view of (8), by Lemma 3.5 we conclude that

S^{∗}(M(t_{0}, ·))(t) = | sin t_{0}|M^{∗}(t, t_{0})

+ ∫_{0}^{2π}(\frac{M^{∗}(t, y) − M^{∗}(t, t_{0})}{cos y − cos t_{0}}α(t_{0}, y) + \frac{M^{∗}(t, y) − M^{∗}(t, t̄_{0})}{cos y − cos t_{0}}α(t̄_{0}, y))M(t_{0}, y)dy

= S(M^{∗}(t, ·))(t_{0}).

We can put these pieces together to show a very important property of our operators.

Let us start with some notation:

\begin{matrix}A_{M}(t_{0}) = Ã(t_{0}) − π^{2}(M̄^{∗}(t_{0}, t_{0})M̄(t_{0}, t_{0}) + M̄^{∗}(t_{0}, t̄_{0})M̄(t_{0}, t̄_{0})), \\ B_{M}(t̄_{0}) = B̃(t̄_{0}) − π^{2}(M̄^{∗}(t_{0}, t_{0})M̄(t̄_{0}, t_{0}) + M̄^{∗}(t_{0}, t̄_{0})M̄(t̄_{0}, t̄_{0})), \end{matrix} (24)

where t̄_{0} = 2π − t_{0} and

Ã(t_{0}) = A^{∗}(t_{0})A(t_{0}) + B^{∗}(t̄_{0})B(t̄_{0}),
B̃(t_{0}) = A^{∗}(t_{0})B(t̄_{0}) + B^{∗}(t_{0})A(t̄_{0}).

Theorem 3.7. The composition S^{∗}L contains no singular parts and the equality

S^{∗}(L(u))(t_{0}) = A_{M}(t_{0})u(t_{0}) + B_{M}(t̄_{0})u(t̄_{0}), t_{0} ∈ (0, 2π),

holds.

Proof. Performing the operations indicated on the left-hand side of (24) and using identity (10), by the repeated integration we obtain

S^{∗}(L(u))(t_{0}) = A_{M}(t_{0})u(t_{0}) + B_{M}(t̄_{0})u(t̄_{0})

+ ∫_{0}^{2π} \frac{u(t)}{cos t_{0} − cos t} (S^{∗}(M(t, ·))(t_{0}) − S(M^{∗}(t_{0}, ·)(t))dt.

Lemma 3.6 now completes the proof.

The following is an immediate consequence of Lemma 3.6 if in (23) we successively take: a) t_{0} = t = s, b) t_{0} = 2π − s, t = s, c) t_{0} = s, t = 2π − s,

d) t_{0} = t = 2π − s.

Corollary 3.8. The equalities

a) A^{∗}(s)M(s, s) + B^{∗}(s̄)M(s, s̄) = A(s)M^{∗}(s, s) + B(s̄)M^{∗}(s, s̄),
b) A^{∗}(s̄)M(s, s̄) + B^{∗}(s)M(s, s) = A(s)M^{∗}(s̄, s) + B(s̄)M^{∗}(s̄, s̄),
c) A^{∗}(s)M(s̄, s) + B^{∗}(s̄)M(s̄, s̄) = A(s̄)M^{∗}(s, s̄) + B(s)M^{∗}(s, s),
d) A^{∗}(s̄)M(s̄, s̄) + B^{∗}(s̄)M(s̄, s) = A(s̄)M^{∗}(s̄, s̄) + B(s)M^{∗}(s̄, s),
s ∈ (0, 2π),

where s̄ = 2π − s, hold.

Our next task is to examine the system of equations

\begin{matrix}A_{M}(t_{0})v(t_{0}) + B_{M}(t̄_{0})w(t_{0}) = 0, \\ B_{M}(t_{0})v(t_{0}) + A_{M}(t̄_{0})w(t_{0}) = 0. \end{matrix}  (25)

Let

C(t_{0}) = ( \begin{matrix}A(t_{0}) & B(t̄_{0}) \\ B(t_{0}) & A(t̄_{0}) \end{matrix} ) ,
M̄(t_{0}) = ( \begin{matrix}M̄(t_{0}, t_{0}) & M̄(t̄_{0}, t_{0}) \\ M̄(t_{0}, t̄_{0}) & M̄(t̄_{0}, t̄_{0}) \end{matrix} ) ,
C_{M}(t_{0}) = ( \begin{matrix}A_{M}(t_{0}) & B_{M}(t̄_{0}) \\ B_{M}(t_{0}) & A_{M}(t̄_{0}) \end{matrix} ) .

Defines the operators C^{∗} and M̄^{∗} similarly. By Corollary 3.8 a simple calculation shows that

C^{∗′} M̄ = M̄^{∗′}C,

where C^{∗′} and M̄^{∗′} are the transposed operators of C^{∗} and M̄^{∗}, respectively. As a result of this equality we have

Lemma 3.9. There exists a factorization

C_{M} = (C^{∗′} + iπM̄^{∗′})(C − iπM̄).

The next lemma can be considered as a corollary of Lemmas 3.2 and 3.9.

SOLUTION OF A LINEAR INTEGRAL EQUATION OF THIRD KIND 191

Lemma 3.10. For every t_{0} ∈ (0, π), the homogeneous system of algebraic equations (25) admits only a trivial solution.

Denote

T (v(·))(t_{0}) : = \frac{1}{Δ(t_{0})}(A_{M}(t̄_{0})S^{∗}(v(·))(t_{0})
− B_{M}(t_{0})S^{∗}(v(·))(t̄_{0})), t_{0} ∈ (0, 2π),  (26)

where

Δ(t_{0}) = det C_{M}(t_{0}),

moreover, here Δ(t̄_{0}) = Δ(t_{0}).

By using the Theorem 3.7 we have

Theorem 3.11. The equality

T (L(u)) = u  (27)

holds.

We shall now consider a relationship between the eigenfunctions and the above introduced operators.

Theorem 3.12. The eigenfunctions of the kernel K(x, y) are solutions of the homogeneous equation (22). Proof. By (20) and (6) we obtain

S(φ_{z_{k}}^{∗}(·))(t) = A(t)φ_{z_{k}}^{∗}(t) + B(t)φ_{z_{k}}^{∗}(t̄)

+ ∫_{0}^{2π} \frac{M(t, x)}{cos x − cos t} \frac{∫_{0}^{2π}K(y, x)φ_{z_{k}}^{∗}(y)dy}{cos x − z_{k}} dx, t ∈ (0, 2π).

It is easy to see that

\frac{1}{cos x − cos t} \frac{1}{cos x − z_{k}} = (\frac{1}{cos x − cos t} − \frac{1}{cos x − z_{k}} ) \frac{1}{cos t − z_{k}} .

Therefore

S(φ_{z_{k}}^{∗}(·))(t) = A(t)φ_{z_{k}}^{∗}(t) + B(t)φ_{z_{k}}^{∗}(t̄)

+ ∫_{0}^{2π} \frac{M(t, x)}{cos x − cos t} \frac{1}{cos t − z_{k}} dx ∫_{0}^{2π}K(y, x)φ_{z_{k}}^{∗}(y)dy − ∫_{0}^{2π} \frac{M(t, x)}{cos t − z_{k}} φ_{z_{k}}^{∗}(x)dx.

But according to (8) we have

∫_{0}^{2π} \frac{φ_{z_{k}}^{∗}(x)}{cos t − z_{k}} M(t, x)dx = A(t)φ_{z_{k}}^{∗}(t) + B(t)φ_{z_{k}}^{∗}(t̄)

+ ∫_{0}^{2π} \frac{M(t, x)}{cos x − cos t} \frac{1}{cos t − z_{k}} dx ∫_{0}^{2π}K(y, x)φ_{z_{k}}^{∗}(y)dy.

Hence we have proved the theorem.

Analogously we obtain

Theorem 3.13. The equalities

S^{∗}(φ_{z_{k}}) = 0, z_{k} ∈ κ,

hold.

Corollary 3.14. The equalities

T (φ_{z_{k}}) = 0, z_{k} ∈ κ,

hold.

Now we shall prove one important property of the eigenfunctions

Theorem 3.15. The systems of eigenvalues {φ_{z_{k}}} and {φ_{z_{k}}^{∗}} are biorthogonal systems.

Proof. Owing to equality (5) it remains to prove that the numbers

N_{z_{k}} = ∫_{0}^{2π}φ_{z_{k}}φ_{z_{k}}^{∗}dx, z_{k} ∈ κ,

are different from zero. Let us assume the contrary, i.e., that N_{z_{p}} = 0 holds for some z_{p}. Then it is obvious that φ_{z_{p}} satisfies the conditions of Theorem 3.3 and the integral equation

L(u) = φ_{z_{p}}

has a unique solution. It follows from Theorem 3.11 and Corollary 3.14 that φ_{z_{p}} = 0. Thus we get a contradiction and the theorem is proved.

Remark 3.16. This results implies:

(i) Only the functions φ_{z_{k}}^{∗}, z_{k} ∈ κ, and their linear combination are solutions of the homogeneous equation (22).

(ii) Condition (21) is also sufficient for the solvability of equation (15).

The main result of this section is summarized in the following theorem.

Theorem 3.17. Let ψ ∈ H^{∗}. Then

ψ = ∑_{k} a_{z_{k}}φ_{z_{k}} + L(u),  (28)

where

a_{z_{k}} = \frac{1}{N_{z_{k}}} ∫_{0}^{2π}ψφ_{z_{k}}^{∗}dx, u = T (ψ).

Moreover, a_{z_{k}} and u are defined uniquely.

Proof. In view of Theorem 3.15 it is evident that the function

ψ_{0} = ψ − ∑_{k} a_{z_{k}}φ_{z_{k}}

has the property

∫_{0}^{2π}ψ_{0}φ_{z_{k}}^{∗}dx = 0, z_{k} ∈ κ.

By Theorem 3.3 this yields (28). The uniqueness of a_{z_{k}} and u is also obvious.

4. Main Results

The results obtained in the preceding sections can be successfully applied to solve of the nonhomogeneous integral equation

(cos x − z)φ̃_{z}(x) = ∫_{0}^{2π}K(x; y)φ̃_{z}(y)dy + f(x), x ∈ (0, 2π),  (29)

where f ∈ H^{∗}.

Theorem 4.1. Let f ∈ H^{∗} and let z ∉ [−1, +1] ∪ κ. Then equation (29) has one and only one solution φ̃_{z} ∈ H^{∗} expressed by the formula

φ̃_{z}(x) = ∑_{k} \frac{φ_{z_{k}}(x)}{z_{k} − z} \frac{1}{N_{z_{k}}} ∫_{0}^{2π}f(y)φ_{z_{k}}^{∗}(y)dy + L(\frac{1}{cos(·) − z} T (f)(·))(x). (30)

Proof. Let φ̃_{z} ∈ H^{∗} be a solution of equation (29). By virtue of Theorem 3.17 the function φ̃_{z} can be written in the form

φ̃_{z} = ∑_{k} ã_{z_{k}}φ_{z_{k}} + L(ũ).  (31)

To find the coefficients ã_{z_{k}} and the function ũ, we proceed in the following way. Putting (31) into equation (29) and using relation (11) and Theorem 3.1, we get

∑_{k} ã_{z_{k}}(z_{k} − z)φ_{z_{k}} + L((cos(·) − z)ũ(·)) = f.

From this, by Theorem 3.17 we obtain

ã_{z_{k}}(z_{k} − z) = \frac{1}{N_{z_{k}}} ∫_{0}^{2π}fφ_{z_{k}}^{∗}dy, z_{k} ∈ κ,

(cos t − z)u(t) = T (f)(t), t ∈ (0, 2π).

Now let we show that the function φ̃_{z} from (30) satisfies equation (29). Substituting (30) into (29) and using again the relation (11) and Theorem 3.1, we obtain

∑_{k} φ_{z_{k}} \frac{1}{N_{z_{k}}} ∫_{0}^{2π}fφ_{z_{k}}^{∗}dy + L(T (f)) = f.

According to Theorem 3.17 last equality is true.

Theorem 4.2. Let z = z_{1} be r-tiple eigenvalue of the kernel K. Then the solution of equation (29) exists if and only if the conditions

∫_{0}^{2π}fφ_{z_{k}}^{∗}dx = 0, k ≤ r  (32)

are fulfilled. In this case equation (29) has in the class H^{∗} infinitely many solutions represented by the formula

φ̃_{z_{1}} = ∑_{k≤r} c_{k}φ_{z_{k}} + ∑_{k>r} \frac{φ_{z_{k}}(x)}{z_{k} − z_{1}} \frac{1}{N_{z_{k}}} ∫_{0}^{2π}f(y)φ_{z_{k}}^{∗}(y)dy + L(\frac{1}{cos(·) − z_{1}} T (f)(·)), (33)

where c_{k} are arbitrary constant numbers.

Proof. Let φ̃_{z_{1}} ∈ H^{∗} be a solution of (29). Then multiplying (29) by φ_{z_{k}}^{∗} and integrating with respect to x from 0 to 2π, and using, in addition, equality (11), we get

(z_{k} − z_{1}) ∫_{0}^{2π}φ̃_{z_{1}}(x)φ_{z_{k}}^{∗}(x)dx = ∫_{0}^{2π}f(x)φ_{z_{k}}^{∗}(x)dx.

Since z_{k} = z_{1} for k ≤ r,we obtain

∫_{0}^{2π}f(x)φ_{z_{k}}^{∗}(x)dx = 0, k ≤ r.

Thus we have r necessary conditions for the solvability of equation (29).

Before proving that the function represented by formula (33) satisfies equation (29), we show that this equation satisfies the function

φ̄_{z_{1}} = L(\frac{1}{cos(·) − z_{1}} T (f)(·)) + ∑_{k>r} \frac{φ_{z_{k}}}{z_{k} − z_{1}} \frac{1}{N_{z_{k}}} ∫_{0}^{2π}fφ_{z_{k}}^{∗}dx.

Just as in Theorem 4.1 we find

∑_{k>r} φ_{z_{k}} \frac{1}{N_{z_{k}}} ∫_{0}^{2π}fφ_{z_{k}}^{∗}dx + L(T (f)) = f.

But by (32) and Theorem 3.17 we conclude that latter equality is true. Now it remains to observe that the function

φ̃_{z_{1}} = ∑_{k≤r} c_{k}φ_{z_{k}} + φ̄_{z_{1}}

satisfies equation (29).

Theorem 4.3. Let z = cos t_{0}, where t_{0} ∈ (0, π). In order that equation (29) be solvable in the class H^{∗}; it is necessary and sufficient that its free term satisfy the conditions

T (f)(t_{0}) = T (f)(t̄_{0}) = 0.  (34)

Then the unique solution φ̄_{t_{0}} ∈ H^{∗} can be written by formula (30).

Proof. Just as in Theorem 4.1, if a solution of (29) is exists, then

(cos t − cos t_{0})u(t) = T (f)(t)

whence for t = t_{0} and t = t̄_{0} there follows equality (34). This in fact is the necessary condition which is satisfied by the function f(x) in order that equation (29) would hold.

We can show that the function φ_{z} ∈ H^{∗}, where z = cos t_{0}, defined by formula (30) satisfies equation (29) the condition (34) is fulfilled. Thus the theorem is proved.

Corollary 4.4. Let z ∉ κ and let

f(x) = ∑_{k} a_{z_{k}}φ_{z_{k}}(x).  (35)

Then equation (29) has one and only one solution which is expressed by the formula

φ̃_{z}(x) = ∑_{k} \frac{a_{z_{k}}}{z_{k} − z} φ_{z_{k}}(x).

Corollary 4.5. Let z = z_{1} be an r-tiple eigenvalue of the kernel K and let the function f(x) has form (35). Then the solution of equation (29) exists if and only if the conditions a_{z_{k}} = 0, k ≤ r, are fulfilled. In this case equation (29) has infinitely many solutions represented by the formula

φ̃_{z_{1}} = ∑_{k≤r} c_{k}φ_{z_{k}} + ∑_{k>r} \frac{a_{z_{k}}}{z_{k} − z_{1}} φ_{z_{k}}(x),

where c_{k} are arbitrary constant numbers.

Obviously as a particular case of the above stated results we obtain the following theorem which answers the question posed for equation (1) in the Introduction. Note that in this connection we shall use (26).

Theorem 4.6. Let K ∈ H be such that the assumptions R_{1} and R_{2} are fulfilled. Then the solution of equation (1) exists in the class H^{∗} if and only if f(x) ∈ H^{∗} satisfies the conditions

A_{M} (\frac{3π}{2} )S^{∗}(f)(\frac{π}{2} ) = B_{M} (\frac{π}{2} )S^{∗}(f))(\frac{3π}{2} ),

A_{M} (\frac{π}{2} )S^{∗}(f)(\frac{3π}{2} ) = B_{M} (\frac{3π}{2} )S^{∗}(f)(\frac{π}{2} ),

where A_{M} and B_{M} are defined by (24). Moreover, if the latter conditions are fulfilled, then the solution is unique and can be expressed by the formula

φ(x) = ∑_{k} φ_{z_{k}}(x)\frac{z_{k}^{−1}}{N_{z_{k}}} ∫_{0}^{2π}f(y)φ_{z_{k}}^{∗}(y)dy + L(\frac{1}{cos(·)}T (f)(·))(x).

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(Received 1.05.2001)

Author’s address: I.Vekua Institute of Applied Mathematics I. Javakhishvili Tbilisi State University 2, University St., Tbilisi 380043 Georgia