Memoirs on Differential Equations and Mathematical Physics

Volume 1. 1994

Tariel Kiguradze

SOME BOUNDARY VALUE PROBLEMS FOR SYSTEMS OF LINEAR PARTIAL DIFFERENTIAL EQUATIONS OF HYPERBOLIC TYPE

Abstract. The linear hyperbolic system

\frac{∂^{2}u}{∂x∂y} = P_{0}(x, y)u + P_{1}(x, y)\frac{∂u}{∂x} + P_{2}(x, y)\frac{∂u}{∂y} + q(x, y)  (1)

is considered, where P_{0}, P_{1}, P_{2} and q are respectively the n×n matrices and the n-dimensional vector whose components are measurable and essentially bounded functions in the rectangle D_{ab} = [0, a] × [0, b] or in the strip D_{b} = = ℝ × [0, b].

For system (1) problems with general functional boundary conditions are investigated in the rectangle D_{ab} and problems on bounded, almost-periodic and periodic solutions in the strip D_{b}.

Optimal in a certain sense conditions are established, guaranteeing the unique solvability of the problems and the stability of their solutions with respect to small perturbations of the coefficients of system (1) and of the boundary conditions.

Mathematics Subject Classification. 35L55

Key words and phrases. System of partial differential equations of hyperbolic type, boundary value problem in the rectangle, classical solution, absolutely continuous solution, generalized solution, periodic solution, almostperiodic solution, bounded solution.

�������� ���������� ������ ����������� �������

\frac{∂^{2}u}{∂x∂y} = P_{0}(x, y)u + P_{1}(x, y)\frac{∂u}{∂x} + P_{2}(x, y)\frac{∂u}{∂y} + q(x, y),  (1)

����� P_{0}, P_{1}, P_{2} �� q, ���������� n×n ��������� �� n- �������������� ��������� ������� ������������ D_{ab} = [0, a] × [0, b] ������������ �� D_{b} = ℝ×[0, b] ����� ������������ ������ �� ��������� ������������� �����������

D_{ab} ������������ (1) ������������� ������������ ��������� ������ ������������� ��������� ���������� ���� D_{b} ����� —��������� �������������� ������� ��������� �� ��������� ������������ ��������

���������� ��������� ����� ���������� ��������� �������� ������ �������� �������� ��������� �������� ������������ �� ���������� ���������� (1) �������� ��������������� �� ��������� �������������� ����� ����������� ������� PREFACE

0.1. Statement of the main problems. Beginning from the 60ies, problems on periodic solutions in a strip or in the large as well as problems with boundary conditions connecting the values of an unknown solution in various characteristics have been intensively studied for partial differential equations of hyperbolic type (see, e.g., [1-4, 6-10, 16, 17, 29, 31-36, 38-47,51]). These problems naturally lead us to boundary value problems in a rectangle with general functional boundary conditions, and also to problems on bounded and almost-periodic in a strip solutions. This work deals just with such a class of problems for the linear hyperbolic system

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q(x, y),  (0.1)

where P_{0}, P_{1}, P_{2} and q are, respectively, n×n matrices and an n-dimensional vector whose components are real measurable and essentially bounded functions given in the rectangle

D_{ab} = [0, a] × [0, b]

or in the strip

D_{b} = ℝ × [0, b]

(a and b are positive numbers, ℝ is the set of real numbers).

We start with the following two definitions.

Definition 0.1. A vector function u : D_{ab} → ℝ^{n} (u : D_{b} → ℝ^{n}) is called a solution of system (0.1) if it is absolutely continuous in D_{ab} ^{1} (in every rectangle contained in D_{b}) and satisfies system (0.1) almost everywhere in D_{ab} (in D_{b}).

Definition 0.2. A solution u : D_{ab} → ℝ^{n} (u : D_{b} → ℝ^{n}) of system (0.1) is called classical if it has the continuous partial derivatives \frac{∂u}{∂x}, \frac{∂u}{∂y} and \frac{∂^{2}u}{∂x∂y} in D_{ab} (in D_{b}).

1According to the well-known definition of an absolutely continuous function of many variables (see [5, §570], [11] and [49]), a vector function u : D_{ab} → ℝ^{n} is absolutely continuous if and only if it admits the representation

u(x, y) = v_{1}(x) + v_{2}(y) + ∫_{0}^{x} ∫_{0}^{y} v(s, t)ds dt for (x, y) ∈ D_{ab}, where v_{1} : [0, a] → ℝ^{n} and v_{2} : [0, b] → ℝ^{n} are absolutely continuous and v : D_{ab} → ℝ^{n} is summable.

Definition 0.1 was earlier used when investigating initial value problems for hyperbolic systems with discontinuous coefficients [11-13,49,50]. As for boundary value problems for system (0.1), they were studied, as a rule, in terms of the concept of a classical solution for P_{i} (i = 0, 1, 2) and q being continuous or even smooth. In fact, the concepts either of a classical solution or of an absolutely continuous one, do not appear to be sufficient for investigating even very simple boundary value problems. ^{2} Therefore we have to extend the concept of a solution on the basis of Picone’s canonic representation of system (0.1) [37].

For an arbitrary x ∈ [0, a] (y ∈ [0, b]) by Z_{1}(x, ·) (Z_{2}(·, y)) we denote the fundamental matrix of the system of ordinary differential equations

\frac{dz(x, y)}{dy} = P_{1}(x, y)z(x, y) (\frac{dz(x, y)}{dx} = P_{2}(x, y)z(x, y)),

satisfying the initial condition

Z_{1}(x, 0) = E (Z_{2}(0, y) = E),

where E is the unit n × n matrix.

According to Lemma 3.2 below, if the matrix function P_{2}(x, ·) is absolutely continuous, then a solution u : D_{ab} → ℝ^{n} of system (0.1) is also a solution of the system

\frac{∂}{∂y} [Z_{1}^{−1}(x, y)Z_{2}(x, y)\frac{∂}{∂x}(Z_{2}^{−1}(x, y)u(x, y))] =

= Z_{1}^{−1}(x, y)(P(x, y)u(x, y) + q(x, y)),  (0.1′)

where

P(x, y) = P_{0}(x, y) + P_{1}(x, y)P_{2}(x, y) − \frac{∂P_{2}(x, y)}{∂y}

and vice versa. System (0.1′) is called the Picone canonic form of system (0.1).

Definition 0.3. A vector function u : D_{ab} → ℝ^{n} (u : D_{b} → ℝ^{n}) is called a generalized solution of system (0.1) if: i) u admits the representation

u(x, y) = Z_{2}(x, y)[v_{0}(x, y) + v_{1}(y)],

where v_{0} : D_{ab} → ℝ^{n} (v_{0} : D_{b} → ℝ^{n}) is absolutely continuous (locally absolutely continuous) and v_{1} : [0, b] → ℝ^{n} is summable; ii) equality (0.1′) holds almost everywhere in D_{ab} (in D_{b}).

2For the periodic boundary value problem this fact was taken into account in [51].

In the rectangle D_{ab} for system (0.1) we study boundary value problems of four types

u(x, 0) = φ_{0}(x), h(\frac{∂u(·, y)}{∂y})(y) = φ_{1}(y);  (0.2)
\frac{∂u(x, 0)}{∂x} − P_{2}(x, 0)u(x, 0) = ψ_{0}(x), h(\frac{∂(·, y)}{∂y})(y) = ψ_{1}(y);  (0.3)
lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), h(u(·, y))(y) = ψ_{1}(y); (0.4)

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), \\ h(\frac{∂}{∂y} (u(·, y) − Z_{2}(·, y)u(0, y)))(y) = ψ_{1}(y), \end{matrix}  (0.5)

where φ_{0} : [0, a] → ℝ^{n} is absolutely continuous, ψ_{0} : [0, a] → ℝ^{n}, φ_{1} and ψ_{1} : [0, b] → ℝ^{n} are summable and h is a linear continuous operator acting from the space of absolutely continuous in [0, a] n-dimensional vector functions to the space of measurable and essentially bounded in [0, b] ndimensional vector functions. We are especially interested in the case when h(v)(y) ≡ v(a)−v(0), i.e. when boundary conditions (0.2)-(0.5) are periodic,

u(x, 0) = φ_{0}(x), \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y} + φ_{1}(y);  (0.2_{1})

\begin{matrix}\frac{∂u(x, 0)}{∂x} − P_{2}(x, 0)u(x, 0) = ψ_{0}(x), \\ \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y} + ψ_{1}(y); \end{matrix}  (0.3_{1})

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), \\ u(a, y) = u(0, y) + ψ_{1}(y); \end{matrix}  (0.4_{1})

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), \\ \frac{∂}{∂y} (u(a, y) − u(0, y)) = ψ_{1}(y). \end{matrix}  (0.5_{1})

For any k ∈ {2, 3, 4, 5} under a solution (a classical solution, a generalized solution) of problem (0.1),(0.k) we understand a solution (a classical solution, a generalized solution) of system (0.1) satisfying boundary conditions (0.k) almost everywhere in D_{ab}.

The behaviour of the matrix function M_{0} : [0, b] → ℝ^{n×n} induced by the operator h and the fundamental matrix Z_{2}, ^{3} affects essentially the solvability and correctness of boundary value problems of type (0.1),(0.k) and differential properties of solutions of these problems. For example: 1) if M_{0} is singular at isolated points, then despite the smoothness of P_{i} (i = = 0, 1, 2) and q and the unique solvability of problem (0.1),(0.2) the latter’s solution may be non-classical (see Remark 4.2); 2) if M_{0} is singular in the set of positive measure, then problem (0.1),(0.2) may be non-Fredholmian (see Theorems 4.1 and 4.1′) or may have the unique classical solution and an infinite set of absolutely continuous solutions (see Remark 4.3); 3) if M_{0}(y) ≡ Θ, where Θ is the zero n × n matrix, then problem (0.1),(0.4) may have a unique generalized solution which is not absolutely continuous (see Remark 4.13).

3I.e., M_{0} is the matrix function satisfying the equality h(Z_{2}(·, y)c) = M_{0}(y)c almost for all y ∈ [0, b] and an arbitrary c ∈ ℝ^{n}.

Two fundamentally different cases

(i) det(M_{0}(y)) ≢ 0 and

(ii) M_{0}(y) ≡ Θ. will be treated by us separately.

The optimal in a certain sense conditions for the existence and uniqueness of classical, absolutely continuous and generalized solutions of problems (0.1),(0.2) and (0.1),(0.4) (problems (0.1),(0.3) and (0.1),(0.5)) and for the stability of these solutions with respect to small, in an integral sense, perturbations of coefficients of system (0.1) are obtained in case (i) (case (ii)). Moreover, the effective methods for constructing a solution of problem (0.1),(0.2) are developed.

The sufficient conditions for the unique solvability of problems (0.1),(0.2) and (0.1),(0.4) are obtained also in case (ii). In this case, however, the latter problems are ill-posed, since for their solvability it is necessary that certain integral equalities (see equalities (4.49) and (4.84)) be fulfilled; these equalities can be violated at arbitrarily small perturbations of the coefficients of system (0.1) or the vector functions φ_{k} and ψ_{k} (k = 0, 1).

The above results and their particular cases for problems (0.1),(0.k_{1}) (k = 2, 3, 4, 5) are stated in Chapter II (§§4-6). They are proved by the unified method that consists in reducing problems under consideration to the modified characteristic initial value problem with either of the two conditions below

\begin{matrix}u(x, 0) = φ_{0}(x), \frac{∂u(0, y)}{∂y} = ψ(y) + \\ + ∫_{0}^{a} [Q_{0}(s, y)u(s, y) + γ^{−1}(s)Q_{1}(s, y)\frac{∂u(s, y)}{∂s}]ds \end{matrix}  (0.6)

and

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = φ(x), \\ u(0, y) = ψ(y) + ∫_{0}^{y} ∫_{0}^{a} q(y, s, t)u(s, t)ds dt. \end{matrix}  (0.7)

The material required to realize the method in question is stated in Chapter I. More exactly, in this chapter: (a) the sufficient conditions for the existence and uniqueness of a solution of a system of special type operator equations are established, on whose basis the unique solvability and correctness of auxiliary problems (0.1),(0.6) and (0.1),(0.7) are proved (see §§1 and 3); (b) new conditions for the unique solvability of the general linear boundary value problem for a system of ordinary differential equations with a parameter are obtained and the properties of matrix functions induced by this problem are studied (see §2).

The problem

u(x, 0) = φ(x), ess sup_{(x,y)∈D_{b}} (∥\frac{∂u(x, y)}{∂x}∥ + ∥\frac{∂u(x, y)}{∂y}∥) < +∞ (0.8)

in the strip D_{b} is considered for system (0.1) in Chapter III, where the sufficient conditions for the existence and uniqueness of its solution are also given (see §7). When P_{i} (i = 0, 1, 2) and q are periodic in the first argument with a period a and φ(x + a) ≡ φ(x), this solution is also a solution of the periodic problem

u(x, 0) = φ(x), u(x + a, y) = u(x, y).  (0.9)

Other conditions for the unique solvability of problem (0.1),(0.9) follow from the results of Chapter II that deals with problems (0.1),(0.k_{1}) (k = 2, 3, 4, 5) (see §8). In the last §9 the problem of the almost-periodicity of a solution of problem (0.1),(0.8) is considered assuming that φ is almost periodic, while P_{i} (i = 0, 1, 2) and q are almost-periodic in the first argument. Theorem 9.1 proved here is an analogue of Favard’s well-known theorem [16]. Applying this theorem, one can obtain from the results of §7 the effective conditions ensuring the almost periodicity in the first argument of a solution of problem (0.1),(0.8).

0.2. Basic notation and terms. ℝ = (−∞, +∞), ℝ_{+} = [0, +∞).

ℝ^{m×n} the space of m × n matrices X = (x_{ij}) with real components x_{ij} (i = 1, …, m; j = 1, …, n) and the norm

∥X∥ = ∑_{i=1}^{m} ∑_{j=1}^{n} |x_{ij}|.

ℝ^{n} = ℝ^{n×1}.

(x_{ij})_{i,j=1}^{n} is the square matrix with components x_{ij} (i, j = 1, …, n) and (x_{i})_{i=1}^{n} is the n-dimensional column vector with components x_{i} (i = = 1, …, n).

By an absolute value of the matrix X = (x_{ij}) ∈ ℝ^{m×n} we understand the matrix |X| = (|x_{ij}|) ∈ ℝ^{m×n} with components |x_{ij}| (i = 1, …, m; j = = 1, …, n).

A matrix X = (x_{ij}) ∈ ℝ^{m×n} is called non-negative if x_{ij} ≥ 0 (i = = 1, …, m; j = 1, …, n).

The inequalities between the matrices X = (x_{ij}) and Y = (y_{ij}) ∈ ℝ^{m×n} are understood componentwise, i.e.,

X ≤ Y ⇔ x_{ij} ≤ y_{ij} (i = 1, …, m; j = 1, …, n).

If X_{k} = (x_{ijk}) ∈ ℝ^{m×n} (k = 1, …, k_{0}), then

max_{1≤k≤k_{0}} X_{k} = ( max_{1≤k≤k_{0}} x_{ijk}).

det(X) is the determinant of the matrix X ∈ ℝ^{n×n}.

r(X) is the spectral radius of the matrix X ∈ ℝ^{n×n}.

X^{−1} is the matrix reciprocal to X ∈ ℝ^{n×n}.

diag(x_{1}, …, x_{n}) is the diagonal n × n matrix with diagonal elements x_{1}, …, x_{n}.

E is the unit matrix.

Θ is the zero matrix.

A matrix function Z : D → ℝ^{m×n} is called measurable, summable, continuous, etc., if its components are such.

Let D be a k-dimensional segment, k_{0} ∈ {1, …, k}, D^{0} be the projection of D onto ℝ^{k_{0}} and M_{0} : D^{0} → ℝ^{m×m}. A matrix function Z : D → ℝ^{m×m} is called M_{0}-continuous (M_{0}-summable) if it admits in D the representation

Z(x_{1}, …, x_{k}) = M_{0}(x_{1}, …, x_{k_{0}})Z_{0}(x_{1}, …, x_{k}),

where Z_{0} : D → ℝ^{m×n} is continuous (summable).

Let M_{0} : [0, b] → ℝ^{m×m}. We say that a matrix function Z : D_{ab} → → ℝ^{m×n} satisfies the Carathéodory condition with M_{0} weight if it admits the reprezentation

Z(x, y) = M_{0}(y)Z_{0}(x, y)

in D_{ab}, where Z_{0}(·, y) : [0, a] → ℝ^{m×n} is measurable for all y ∈ [0, b], Z_{0}(x, ·) : [0, b] → ℝ^{m×n} is continuous almost for all x ∈ [0, a] and max_{0≤y≤b} ∥Z_{0}(·, y)∥ is summable in [0, a].

ℂ(D; ℝ^{m×n}), L_{∞}(D; ℝ^{m×n}) and L(D; ℝ^{m×n}) are the spaces of continuous, measurable and essentially bounded and summable functions Z : D → → ℝ^{m×n} with the norms

∥Z∥_{ℂ} = max_{(x_{1},... ,x_{k})∈D} ∥Z(x_{1}, …, x_{k})∥,

∥Z∥_{L_{∞}} = ess sup_{(x_{1},... ,x_{k})∈D} ∥Z(x_{1}, …, x_{k})∥,

∥Z∥_{L} = ∫ ⋯ ∫_{D} ∥Z(x_{1}, …, x_{k})∥dx_{1} ⋯ dx_{k}.

If Z = (z_{ij}) ∈ L_{∞}(D; ℝ^{m×n}), then

ess sup_{(x_{1},... ,x_{k})∈D} Z(x_{1}, …, x_{k}) = ( ess sup_{(x_{1},... ,x_{k})∈D} z_{ij}(x_{1}, …, x_{k})).

If Z ∈ L(D_{ab}; ℝ^{m×n}), then

∥Z∥_{L}^{(0)} = ess sup_{(x,y)∈D_{ab}} ∥ ∫_{0}^{y} ∫_{0}^{x} Z(s, t)dsdt∥,

∥Z∥_{L}^{(1)} = max_{(x,y)∈D_{ab}} [ ∫_{0}^{y} ∥ ∫_{0}^{x} Z(s, t)ds∥dt + ∫_{0}^{x} ∥ ∫_{0}^{y} Z(s, t)dt∥ds],

∥Z∥_{L}^{(2)} = ess sup_{(x,y)∈D_{ab}} [∥ ∫_{0}^{x} Z(s, y)ds∥ + ∥ ∫_{0}^{y} Z(x, t)dt∥].

ℂ^{k}([0, a]; ℝ^{m×n}) is the space of k times continuously differentiable matrix functions Z : [0, a] → ℝ^{m×n} with the norm

∥Z∥_{ℂ^{k}} = max_{x∈[0,a]} ∑_{i=0}^{k} ∥Z^{(i)}(x)∥.

ℂ̃^{k−1}([0, a]; ℝ^{m×n}) is the space of matrix functions Z : [0, a] → ℝ^{m×n} which are absolutely continuous together with their derivatives up to the k − 1 order inclusive, with the norm

∥Z∥_{ℂ^{k−1}} = ∑_{i=1}^{k−1} ∥Z^{(i)}(0)∥ + ∫_{0}^{a} ∥Z^{(k)}(x)∥dx.

ℂ̃_{∞}(ℝ; ℝ^{m×n}) is the space of bounded and Lipschitz continuous matrix functions Z : ℝ → ℝ^{m×n}.

ℂ̃_{∞}^{k−1}([0, a]; ℝ^{m×n}) is the space of matrix functions Z : [0, a] → ℝ^{m×n} which are Lipschitz continuous together with their derivatives up to the k − 1 order inclusive, with the norm

∥Z∥_{ℂ_{∞}^{k−1}} = ∑_{i=1}^{k−1} ∥Z^{(i)}(0)∥ + ess sup_{x∈[0,a]} ∥Z^{(k)}(x)∥dx.

ℂ̃([0, a]; ℝ^{m×n})=ℂ̃^{0}([0, a]; ℝ^{m×n}), ℂ̃_{∞}([0, a]; ℝ^{m×n})=ℂ̃_{∞}^{0}([0, a]; ℝ^{m×n}). ℂ̃(D_{ab}; ℝ^{m×n}) is the space of absolutely continuous matrix functions Z :

D_{ab} → ℝ^{m×n} with the norm

∥Z∥_{ℂ} = ∥Z(0, 0)∥ + ∫_{0}^{a} ∥\frac{∂Z(x, 0)}{∂x}∥dx + ∫_{0}^{b} ∥\frac{∂Z(0, y)}{∂y}∥dy +

∫_{0}^{a} ∫_{0}^{b} ∥\frac{∂^{2}Z(x, y)}{∂x∂y}∥dx dy.

ℂ̃_{loc}(D_{b}; ℝ^{m×n}) is the space of locally absolutely continuous matrix functions Z : D_{b} → ℝ^{m×n}.

If Z ∈ ℂ̃(D_{ab}; ℝ^{m×n}), then

∥Z∥(1) ; _{ℂ} = ∥Z(0, 0)∥ + max_{(x,y)∈D_{ab}} ( ∫_{0}^{x} ∥\frac{∂Z(s, y)}{∂s}∥ds + ∫_{0}^{y} ∥\frac{∂Z(x, t)}{∂t}∥dt),

∥Z∥(2) ; _{ℂ} = ∥Z(0, 0)∥ + ess sup_{(x,y)∈D_{ab}} (∥\frac{∂Z(x, y)}{∂x}∥ + ∥\frac{∂Z(x, y)}{∂y}∥).

ℂ̃_{∞}^{(−1,k−1)}(D_{ab}; ℝ^{m×n}) is the space of measurable matrix functions Z : D_{ab} → ℝ^{m×n} such that Z(x, ·) ∈ ℂ̃_{∞}^{k−1}([0, b]; ℝ^{m×n}) almost for every x ∈ ∈ [0, a], and

ess sup_{(x,y)∈D_{ab}} ∑_{i=0}^{k} ∥\frac{∂^{i}Z(x, y)}{∂y^{i}} ∥ < +∞.

ℂ̃_{∞}^{(k−1,−1)}(D_{ab}; ℝ^{m×n}) is the space of measurable matrix functions Z : D_{ab} → ℝ^{m×n} such that Z(·, y) ∈ ℂ̃_{∞}^{k−1}([0, a]; ℝ^{m×n}) almost for every y ∈ ∈ [0, b], and

ess sup_{(x,y)∈D_{ab}} ∑_{i=0}^{k} ∥\frac{∂^{i}Z(x, y)}{∂x^{i}} ∥ < +∞.

CHAPTER I AUXILIARY STATEMENTS

§ 1. On a System of Operator Equations

Let n, n_{1} and n_{2} be natural numbers, I be a finite or an infinite interval of the real axis,

D = I × [0, b]

and Λ_{1},Λ_{2} and Λ be some nonempty closed subsets of the spaces L_{∞}(D;ℝ^{n_{1}}), L_{∞}(D; ℝ^{n_{2}}) and L_{∞}(D; ℝ^{n}). The boundary value problems considered in this work are reduced to a system of operator equations of the type

z_{i}(x, y) = g_{i}(z_{1}, z_{2})(x, y) (i = 1, 2)  (1.1)

or to the equation

z(x, y) = g(z)(x, y),  (1.2)

where g_{i} : Λ_{1} × Λ_{2} → Λ_{i} (i = 1, 2) and g : Λ → Λ are continuous operators.

By a solution of system (1.1) (equation (1.2)) we shall understand a pair of vector functions (z_{i})_{i=1}^{2} ∈ Λ_{1} ×Λ_{2} (vector function z ∈ Λ), which satisfies system (1.1) (equation (1.2)) almost everywhere in D.

For arbitrary y ∈ [0, b] and z ∈ L_{∞}(D, ℝ^{k}) assume

|z|_{I,y} = ess sup_{(x,t)∈I×[0,y]} |z(x, t)|, |z(·, y)|_{I} = ess sup_{x∈I} |z(x, y)|,

∥z∥_{I,y} = ess sup_{(x,t)∈I×[0,y]} ∥z(x, t)∥, ∥z(·, y)∥_{I} = ess sup_{x∈I} ∥z(x, y)∥.

Let Λ_{0} be a subspace of the space L_{∞}(I; ℝ^{n_{2}}) satisfying the following conditions:

1) if v(x) = c ∈ ℝ^{n_{2}} for x ∈ I , then v ∈ Λ_{0};

2) if v_{1} and v_{2} ∈ Λ_{0} and v(x) = max_{1≤i≤2} v_{i}(x), then v ∈ Λ_{0};

3) if ζ_{0} and ζ ∈ Λ_{2}, then |ζ(·, y) − ζ_{0}(·, y)| ∈ Λ_{0} almost for all y ∈ [0, b]. An operator l : Λ_{0} → ℝ^{n_{2}} is called non-negative if for any non-negative

vector function v ∈ Λ_{0} the vector l(v) is non-negative.

If A ∈ L_{∞}(I; ℝ^{n_{2}×n_{2}}) and A(·)c ∈ Λ_{0} for c ∈ ℝ^{n_{2}}, then by l(A) we shall denote a matrix such that

l(A(·)c) = l(A)c for c ∈ ℝ^{n_{2}}.

Lemma 1.1. Let for any ζ_{i} and ζ̄_{i} ∈ Λ_{i} (i = 1, 2) the inequalities

∥g_{1}(ζ_{1}, ζ_{2})(x, y) − g_{1}(ζ̄_{1}, ζ̄_{2})(x, y)∥ ≤
≤ ∫_{0}^{y} g_{0}(t)(∥ζ_{1} − ζ̄_{1}∥_{I,t} + ∥ζ_{2} − ζ̄_{2}∥_{I,t}) dt  (1.3)

and

|g_{2}(ζ_{1}, ζ_{2})(x, y) − g_{2}(ζ̄_{1}, ζ̄_{2})(x, y)| ≤
≤ A_{01}|ζ_{1} − ζ̄_{1}|_{I,y} + ∫_{0}^{y} A_{02}(t)|ζ_{2} − ζ̄_{2}|_{I,t} dt +
+A_{1}(x, y)|ζ_{2}(·, y) − ζ̄_{2}(·, y)|_{I} + A_{2}(x, y)l(|ζ_{2}(·, y) − ζ̄_{2}(·, y)|) (1.4)

hold almost evrywhere in D_{ab} where g_{0} : [0, b] → ℝ_{+} is a summable function, A_{01} is a non-negative constant n_{2} × n_{1} matrix, A_{02} ∈ L([0, b]; ℝ^{n_{2}×n_{2}}) and A_{i} ∈ L_{∞}(D; ℝ^{n_{2}×n_{2}}) (i = 1, 2) are non-negative matrix functions,

A_{i}(·, y)c ∈ Λ_{0} for y ∈ [0, b], c ∈ ℝ^{n_{2}} (i = 1, 2)

and l : Λ_{0} → ℝ^{n_{2}} is a non-negative linear operator. Besides, let

ess sup_{0≤y≤b} r(l(A_{2}(·, y))) < 1  (1.5)

and

ess sup_{0≤y≤b} r(A(y)) < 1,  (1.6)

where

A(y) = ess sup_{x∈I} [A_{1}(x, y) + A_{2}(x, y)(E − l(A_{2}(·, y)))^{−1}l(A_{1}(·, y))].(1.7)

Then system (1.1) has the unique solution (z_{i})_{i=1}^{2} ∈ Λ_{1} × Λ_{2}. Moreover, for an arbitrary (z_{i0})_{i=1}^{2} ∈ Λ_{1} × Λ_{2} we have

z_{im}(x, y) ⇉ z_{i}(x, y) for m → +∞ (i = 1, 2),  (1.8)

where

z_{im}(x, y) = g_{i}(z_{1 m−1}, z_{2 m−1})(x, y) (i = 1, 2; m = 1, 2, …). (1.9)

To prove this lemma we need

Lemma 1.2. Let B_{01} be a non-negative constant n_{2} × n_{1} matrix, B_{02} ∈ ∈ L([0, b]; ℝ^{n_{2}×n_{2}}) and B_{i} ∈ L_{∞}(D; ℝ^{n_{2}×n_{2}}) (i = 1, 2) be non-negative matrix functions,

B_{i}(·, y)c ∈ Λ_{0} for y ∈ [0, b], c ∈ ℝ^{n_{2}} (i = 1, 2)

and l : Λ_{0} → ℝ^{n_{2}} be a non-negative linear operator. Moreover, let

ess sup_{0≤y≤b} r(l(B_{2}(·, y))) < 1  (1.10)

and

ess sup_{0≤y≤b} r(B(y)) < 1,  (1.11)

where

B(y) = ess sup_{x∈I} [B_{1}(x, y) + B_{2}(x, y)(E − l(B_{2}(·, y)))^{−1}l(B_{1}(·, y))]. (1.12)

Then there exists a positive number ρ such that for vector functions w_{i} ∈ L_{∞}(D; ℝ^{n_{i}}) (i = 1, 2) satisfying the condition

|w_{2}(·, y)| ∈ Λ_{0}

almost everywhere in [0, b] and the inequalities

∥w_{1}(x, y)∥ ≤ δ + ∫_{0}^{y} g_{0}(t)(∥w_{1}∥_{I,t} + ∥w_{2}∥_{I,t}) dt  (1.13)

and

|w_{2}(x, y)| ≤ η + B_{01}|w_{1}|_{I,y} + ∫_{0}^{y} B_{02}(t)|w_{2}|_{I,t} dt +
+ B_{1}(x, y)|w_{2}(·, y)|_{I} + B_{2}(x, y)l(|w_{2}(·, y)|),  (1.14)

almost everywhere in D, the estimates

∥w_{i}∥_{I,b} ≤ ρ exp (ρ ∫_{0}^{b} g_{0}(t) dt)(δ + ∥η∥) (i = 1, 2)  (1.15)

are valid for any δ ∈ ℝ_{+}, η ∈ ℝ^{n_{2}} and g_{0} : [0, b] → ℝ_{+}.

Proof. Because of the non-negativity of l and the restrictions imposed on the space Λ_{0}, from (1.14) we have

l(|w_{2}(·, y)|) ≤ l(η + B_{01}|w_{1}|_{I,y} + ∫_{0}^{y} B_{02}(t)|w_{2}|_{I,t} dt) +
+ l(B_{1}(·, y))|w_{2}(·, y)|_{I} + l(B_{2}(·, y))l(|w_{2}(·, y)|)

and hence

[E − l(B_{2}(·, y))]l(|w_{2}(·, y)|) ≤ l(η + B_{01}|w_{1}|_{I,y} +
+ ∫_{0}^{y} B_{02}(t)|w_{2}|_{I,t} dt) + l(B_{1}(·, y))|w_{2}(·, y)|_{I},

which by virtue of condition (1.10) implies

l(|w_{2}(·, y)|) ≤ (E − l(B_{2}(·, y)))^{−1}l(η + B_{01}|w_{1}|_{I,y} +
+ ∫_{0}^{y} B_{02}(t)|w_{2}|_{I,t} dt) + (E − l(B_{2}(·, y)))^{−1}l(B_{1}(·, y))|w_{2}(·, y)|_{I}.

According to the latter estimate and equality (1.12) we find from (1.14) that

|w_{2}(·, y)|_{I} ≤ B_{0}(η + B_{01}|w_{1}|_{I,y} + ∫_{0}^{y} B_{02}(t)|w_{2}|_{I,t} dt) +
+ B(y)|w_{2}(·, y)|_{I},

where B_{0} is the non-negative constant n_{2} × n_{2} matrix depending on l and B_{2} only. From this and taking into account (1.11), we get

|w_{2}(·, y)|_{I} ≤ (E − B(y))^{−1}B_{0}(η + B_{01}|w_{1}|_{I,y} + ∫_{0}^{y} B_{02}(t)|w_{2}|_{I,t} dt)

and

∥w_{2}∥_{I,y} ≤ ρ_{1}(∥η∥ + ∥w_{1}∥_{I,y} + ∫_{0}^{y} ∥B_{02}(t)∥ ∥w_{2}∥_{I,t} dt),

where

ρ_{1} = (1 + ∥B_{01}∥)∥B_{0}∥ ess sup_{0≤y≤b} ∥(E − B(y))^{−1}∥.

By Gronwall’s lemma ([19],p.37) from the latter inequality it follows that

∥w_{2}∥_{I,y} ≤ ρ_{2}(∥η∥ + ∥w_{1}∥_{I,y}),  (1.16)

where

ρ_{2} = ρ_{1} exp (ρ_{1} ∫_{0}^{b} ∥B_{02}(t)∥ dt).

In view of (1.13) and (1.16)

∥w_{1}∥_{I,y} ≤ δ + ρ_{2}∥η∥ ∫_{0}^{b} g_{0}(t) dt + (1 + ρ_{2}) ∫_{0}^{y} g_{0}(t)∥w_{1}∥_{I,t} dt.

From this again by Gronwall’s lemma we obtain

∥w_{1}∥_{I,y} ≤ (δ + ρ_{2}∥η∥ ∫_{0}^{b} g_{0}(t) dt) exp ((1 + ρ_{2}) ∫_{0}^{y} g_{0}(t) dt) ≤
≤ (1 + ρ_{2}) exp ((2 + ρ_{2}) ∫_{0}^{b} g_{0}(t) dt)(δ + ∥η∥).

Estimates (1.15) follow from the latter inequality and (1.16), where ρ = = 2(1 + ρ_{2})^{2} is the number depending on B_{0i}, B_{i} (i = 1, 2) and l only. ■

Proof of Lemma 1.1. In view of (1.5)-(1.7) there exists γ ∈ (0, 1) such that the matrix functions

B_{i}(x, y) = γ^{−1}A_{i}(x, y) (i = 1, 2)  (1.17)

satisfy conditions (1.10) and (1.11), where B is the matrix given by (1.12). First let us prove the existence of a solution of system (1.1).

Let us choose arbitrarily (z_{i0})_{i=1}^{2} ∈ Λ_{1} × Λ_{2}, compose the sequence (z_{im})_{i=1}^{2} (m = 1, 2, …) by formulas (1.9) and show that this sequence is fundamental in L_{∞}(D; ℝ^{n_{1}}) × L_{∞}(D; ℝ^{n_{2}}). Put

v_{im}(x, y) = γ^{−m}[z_{im}(x, y) − z_{i m−1}(x, y)] (i = 1, 2; m = 1, 2, …).

Then in view of (1.3),(1.4) and (1.17) we obtain

\begin{matrix}∥v_{1m}(x, y)∥ ≤ γ^{−1} ∫_{0}^{y} g_{0}(t)(∥v_{1 m−1}∥_{I,t} + ∥v_{2 m−1}∥_{I,t})dt \\ (m = 2, 3, …) \end{matrix} (1.18)

and

|v_{2m}(x, y)| ≤ B_{01}|v_{1 m−1}|_{I,y} + ∫_{0}^{y} B_{02}(t)|v_{2 m−1}|_{I,t} dt +
+B_{1}(x, y)|v_{2 m−1}(·, y)|_{I} + B_{2}(x, y)l(|v_{2 m−1}(·, y)|) (m = 2, 3, …), (1.19)

where

B_{01} = γ^{−1}A_{01}, B_{02}(t) = γ^{−1}A_{02}(t).  (1.20)

Introducing the notation

w_{im}(x, y) = max_{1≤k≤m} |v_{ik}(x, y)| (i = 1, 2; m = 1, 2, …),

δ = ∥v_{11}∥_{I,b}, η = |v_{21}|_{I,b}, \overline{g}_{0}(t) = n_{1}γ^{−1}g_{0}(t),

from (1.18) and (1.19) we get

∥w_{1m}(x, y)∥ ≤ δ + ∫_{0}^{y} \overline{g}_{0}(t)(∥w_{1m}∥_{I,t} + ∥w_{2m}∥_{I,t}) dt (m = 1, 2, …),
|w_{2m}(x, y)| ≤ η + B_{01}|w_{1m}|_{I,y} + ∫_{0}^{y} B_{02}(t)|w_{2m}|_{I,t} dt +

+ B_{1}(x, y)|w_{2m}(·, y)|_{I} + B_{2}(x, y)l(|w_{2m}(·, y)|) (m = 1, 2, …),

whence according to Lemma 1.2 it follows that

∥w_{im}∥_{I,b} ≤ ρ_{0} (i = 1, 2; m = 1, 2, …),

where ρ_{0} is a positive constant independent of m.

Therefore

∥z_{im} − z_{i m−1}∥_{I,b} ≤ ρ_{0}γ^{m} (i = 1, 2; m = 1, 2, …).

Consequently, ((z_{im})_{i=1}^{2})_{m=1}^{+∞} is a fundamental sequence in L_{∞}(D; ℝ^{n_{1}})× ×L_{∞}(D; ℝ^{n_{2}}). Put

z_{i}(x, y) = lim_{m→+∞} z_{im}(x, y) (i = 1, 2).

Since the sets Λ_{i} (i = 1, 2) are complete and the operators g_{i} : Λ_{1} ×Λ_{2} → Λ_{i} (i = 1, 2) are continuous, it becomes obvious that (z_{i})_{i=1}^{2} ∈ Λ_{1} × Λ_{2} is a solution of system (1.1).

To complete the proof of the lemma we have to show that system (1.1) has no solution different from (z_{i})_{i=1}^{2}. Let (z̄_{i})_{i=1}^{2} be an arbitrary solution of this system. Then in view of (1.3) and (1.4)

∥z_{1}(x, y) − z̄_{1}(x, y)∥ ≤ ∫_{0}^{y} g_{0}(t)(∥z_{1} − z̄_{1}∥_{I,t} + ∥z_{2} − z̄_{2}∥_{I,t})dt

and

|z_{2}(x, y) − z̄_{2}(x, y)| ≤ A_{01}|z_{1} − z̄_{1}|_{I,y} + ∫_{0}^{y} A_{02}(t)|z_{2} − z̄_{2}|_{I,t}dt +
+A_{1}(x, y)|z_{2}(·, y) − z̄_{2}(·, y)|_{I} + A_{2}(x, y)l(|z_{2}(·, y) − z̄_{2}(·, y)),

whence according to conditions (1.5) and (1.6) and Lemma 1.2 it follows that

∥z_{i} − z̄_{i}∥_{I,b} = 0 (i = 1, 2). ■

When l is the zero operator Lemma 1.1 takes the form of

Lemma 1.3. Let for any ζ_{i} and ζ̄_{i} ∈ Λ_{i} (i = 1, 2) inequality (1.3) hold almost everywhere in D and

|g_{2}(ζ_{1}, ζ_{2})(x, y) − g_{2}(ζ̄_{1}, ζ̄_{2})(x, y)| ≤ A_{01}|ζ_{1} − ζ̄_{1}|_{I,y} +
+ ∫_{0}^{y} A_{02}(t)|ζ_{2} − ζ̄_{2}|_{I,t}dt + A(y)|ζ_{2}(·, y) − ζ̄_{2}(·, y)|,  (1.21)

where g_{0} : [0, b] → ℝ_{+} is a summable function, A_{01} is a non-negative constant n_{2} × n_{1} matrix, A_{02} ∈ L([0, b]; ℝ^{n_{2} × n_{2}}) and A ∈ L_{∞}(D; ℝ^{n_{2}×n_{2}}) are non-negative matrix functions. Moreover, it is assumed that A satisfies condition (1.6). Then system (1.1) has the unique solution (z_{i})_{i=1}^{2} ∈ Λ_{1}×Λ_{2} and for arbitrary (z_{i0})_{i=1}^{2} ∈ Λ_{1} × Λ_{2} conditions (1.8) hold, where

z_{im}(x, y) = g_{i}(z_{1 m−1}, z_{2 m−1})(x, y) (i = 1, 2; m = 1, 2, …).

Equation (1.2) is equivalent to system (1.1), where

n_{1} = n, n_{2} = 1, Λ_{1} = Λ, Λ_{2} = L_{∞}(D; ℝ),
g_{1}(z_{1}, z_{2})(x, y) ≡ g(z_{1})(x, y), g_{2}(z_{1}, z_{2})(x, y) ≡ 0.

Therefore from Lemma 1.3 immediately follows

Lemma 1.4. Let for any ζ and ζ̄ ∈ Λ the inequality

∥g(ζ)(x, y) − g(ζ̄)(x, y)∥ ≤ ∫_{0}^{y} g_{0}(t)∥ζ − ζ̄∥_{I,t}dt  (1.22)

hold almost everywhere in D, where g_{0} : [0, b] → ℝ_{+} is a summable function. Then equation (1.2) has the unique solution z ∈ Λ and for arbitrary z_{0} ∈ Λ we have

z_{m}(x, y) ⇉ z(x, y) for m → +∞,  (1.23)

where

z_{m}(x, y) = g(z_{m−1})(x, y) (m = 1, 2, …).  (1.24)

Remark 1.1. From the above arguments it becomes clear that if conditions of Lemma 1.1 take place, then

∥z_{i} − z_{im}∥_{I,b} ≤ ρ_{0}γ^{m} (m = 1, 2, …),

where γ ∈ (0, 1) and ρ_{0} > 0 are constants independent of m. If conditions of Lemma 1.4 hold, then

∥z − z_{m}∥ ≤ \frac{ρ_{0}^{m}}{m!} (m = 1, 2, …).

§ 2. On Boundary Value Problems for Systems of Linear Ordinary Differential Equations with a Parameter

In this section for a system of linear differential equations

\frac{dz(x, y)}{dx} = A(x, y)z(x, y) + c(x, y)  (2.1)

depending on a parameter y ∈ [0, b], we investigate boundary value problems of the type

h(z(·, y))(y) = φ(y),  (2.2)

where

A(·, y) = (a_{ij}(·, y)) ∈ L_{∞}([0, a]; ℝ^{n×n}) for y ∈ [0, b],

γ = sup_{(x,y)∈D_{ab}} ∥A(x, y)∥ < +∞,

c(·, y) ∈ L_{∞}([0, a]; ℝ^{n}), φ(y) ∈ ℝ^{n} for y ∈ [0, b]

(2.3)

and h is a linear continuous operator acting from the space ℂ̃([0, a]; ℝ^{n}) to a subspace of the space L_{∞}([0, b]; ℝ^{n}).

The operator h uniquely defines matrix functions

H_{0} ∈ L_{∞}([0, b]; ℝ^{n×n}), H ∈ L_{∞}(D_{ab}; ℝ^{n×n})  (2.4)

such that

\begin{matrix}h(u)(y) = H_{0}(y)u(0) + ∫_{0}^{a} H(s, y)u′(s) ds \\ for u ∈ ℂ̃([0, a]; ℝ^{n}), y ∈ [0, b].^{4} \end{matrix}  (2.5)

Along with (2.1),(2.2) let us consider the corresponding homogeneous boundary value problem

\frac{dz(x, y)}{dx} = A(x, y)z(x, y),  (2.1_{0})
h(z(·, y))(y) = 0.  (2.2_{0})

By Z_{0}(·, y) will be meant a fundamental matrix of system (2.1_{0}), satisfying the initial condition

Z_{0}(0, y) = E.

By M_{0} and M we denote matrix functions given by the equalities

\begin{aligned}M_{0}(y) & = H_{0}(y) + ∫_{0}^{a} H(s, y)\frac{∂Z_{0}(s, y)}{∂s} ds, \\ M(x, y) & = H(x, y)Z_{0}(x, y) + ∫_{x}^{a} H(s, y)\frac{∂Z_{0}(s, y)}{∂s} ds. \end{aligned}  (2.6)

Moreover, we shall use the notation

I_{M_{0}} = {y ∈ [0, b] : det M_{0}(y) ≠ 0}.

In view of (2.5) and (2.6),

\begin{matrix}h(Z_{0}(·, y)u(·))(y) = M_{0}(y)u(0) + ∫_{0}^{a} M(s, y)u′(s) ds \\ for u ∈ ℂ̃([0, a]; ℝ^{n}), y ∈ [0, b]. \end{matrix}  (2.7)

It is easily seen that the condition

y ∈ I_{M_{0}}  (2.8)

is necessary and sufficient for problem (2.1_{0}),(2.2_{0}) to have only the trivial solution.

4See Lemma 2.1.

2.1. Some comments on the integral representation of the operator h. In this subsection we shall give some results of Dunford-Pettis’ theorem ([20],Ch.XI, §1, Theorem 6) on the integral representation of the operator h which will be necessary in the sequel.

Lemma 2.1_{1}.h is the linear continuous operator acting from ℂ̃([0, a]; ℝ^{n}) to L_{∞}([0, b]; ℝ^{n}) if and only if representation (2.5) is valid. H_{0} and H here are the matrix functions satisfying conditions (2.4).

Proof. It is easy to see that if the matrix functions H_{0} and H satisfy conditions (2.4), then the operator h given by equality (2.5) is a linear continuous operator acting from ℂ̃([0, a]; ℝ^{n}) to L_{∞}([0, b]; ℝ^{n}). Let us now show the converse: if h : ℂ̃([0, a]; ℝ^{n}) → L_{∞}([0, b]; ℝ^{n}) is a linear continuous operator, then representation (2.5) is valid. For an arbitrary vector function v ∈ L([0, a]; ℝ^{n}) assume that

l(v)(x) = ∫_{0}^{x} v(s) ds

and

h̄(v)(y) = h(l(v))(y).

It follows from the continuity of operators l : L([0, a]; ℝ^{n}) → ℂ̃([0, a]; ℝ^{n}) and h : ℂ̃([0, a]; ℝ^{n}) → L_{∞}([0, b]; ℝ^{n}) that h̄ is a linear continuous operator acting from L([0, a]; ℝ^{n}) to L_{∞}([0, b]; ℝ^{n}). Therefore, according to the above-mentioned Dunford-Pettis theorem, there exists a unique matrix function H ∈ L_{∞}(D_{ab}; ℝ^{n×n}) such that

h̄(v)(y) = ∫_{0}^{a} H(s, y)v(s) ds for v ∈ L([0, a]; ℝ^{n}), y ∈ [0, b]. (2.9)

On the other hand, since c → h(c)(y) is a linear continuous operator acting from ℝ^{n} to L_{∞}([0, b]; ℝ^{n}), there exists a unique matrix function H_{0} ∈ Ł_{∞}([0, b]; ℝ^{n×n}) such that

h(c)(y) = H_{0}(y)c for c ∈ ℝ^{n}, y ∈ [0, b].  (2.10)

Let u ∈ ℂ̃([0, a]; ℝ^{n}) be an arbitrary vector function. Then

u(x) = u(0) + l(u′)(x) for x ∈ [0, a].

Therefore, in view of (2.9) and (2.10), we have

h(u)(y) = h(u(0))(y) + h̄(u′)(y) =
= H_{0}(y)u(0) + ∫_{0}^{a} H(s, y)u′(s) ds for y ∈ [0, b].

Consequently, representation (2.5) is valid, where H_{0} and H are the matrix functions satisfying conditions (2.4). ■

Lemma 2.1_{2}.^{5} h is a linear continuous operator acting from ℂ̃([0, a]; ℝ^{n}) to ℂ([0, b]; ℝ^{n}) if and only if representation (2.5) is valid, where

H_{0} ∈ ℂ([0, b]; ℝ^{n×n}), H ∈ L_{∞}(D_{ab}; ℝ^{n×n})  (2.11)

and

∫_{0}^{x} H(s, ·) ds ∈ ℂ([0, b]; ℝ^{n×n}) for x ∈ [0, a].  (2.12)

Proof. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ([0, b]; ℝ^{n}) be a linear continuous operator. Then, according to Lemma 2.1_{1} representation (2.5) is valid, where H_{0} and H are matrix functions satisfying conditions (2.4). On the other hand,

H_{0}(·)(c) = h(c) ∈ ℂ([0, b]; ℝ^{n}) for c ∈ ℝ^{n}

and

( ∫_{0}^{x} H(s, ·) ds)c = h(χ_{c,x}) ∈ ℂ([0, b]; ℝ^{n}) for c ∈ ℝ^{n}, x ∈ [0, a],

where

χ_{c,x}(s) = \begin{cases}sc & for 0 ≤ s ≤ x \\ xc & for s > x \end{cases} .

Consequently conditions (2.11) and (2.12) hold. Thus the first part of the lemma is proved.

Assume now that H_{0} and H are the matrix functions satisfying conditions (2.11) and (2.12), and h is an operator admitting representation (2.5). First of all let us show that h acts from ℂ̃([0, a]; ℝ^{n}) to ℂ([0, b]; ℝ^{n}). To do this, it is enough to show that for an arbitrary v ∈ L([0, a]; ℝ^{n}) the condition

w(·) = ∫_{0}^{a} H(s, ·)v(s) ds ∈ ℂ([0, b]; ℝ^{n})  (2.13)

holds.

Without loss of generality we may assume that

γ_{0} = sup_{(x,y)∈D_{ab}} ∥H(x, y)∥ < +∞.  (2.14)

Let v_{k} : [0, a] → ℝ^{n} (k = 1, 2, …) be a sequence of continuously differentiable vector functions satisfying conditions

∫_{0}^{a} ∥v_{k}(s) − v(s)∥ ds → 0 for k → +∞.  (2.15)

Put

w_{k}(y) = ∫_{0}^{a} H(s, y)v_{k}(s) ds.

5This lemma can also be obtained from B.Z.Vulikh’s theorem ([48], Theorem 2).

Then

w_{k}(y) = H_{1}(a, y)v_{k}(a) − ∫_{0}^{a} H_{1}(s, y)v′_{k}(s) ds,

where

H_{1}(x, y) = ∫_{0}^{x} H(s, y) ds,

whence, in view of (2.11) and (2.12), it follows that

w_{k} ∈ ℂ([0, b]; ℝ^{n}) (k = 1, 2, …).

On the other hand, in view of (2.14) and (2.15),

sup_{y∈[0,b]} ∥w_{k}(y) − w(y)∥ ≤ γ_{0} ∫_{0}^{a} ∥v_{k}(s) − v(s)∥ ds → 0 for k → +∞.

Now, applying the well-known Weierstrass theorem, the validity of condition (2.13) becomes evident. Hence h is a continuous operator acting from ℂ̃([0, a]; ℝ^{n}) to ℂ([0, b]; ℝ^{n}). ■

Lemma 2.1_{3}. Let k be an arbitrary positive integer. h is a linear continuous operator acting from ℂ̃([0, a]; ℝ^{n}) to ℂ̃_{∞}^{k−1}([0, b]; ℝ^{n}) if and only if condition (2.5) holds, where

H_{0} ∈ ℂ̃_{∞}^{k−1}([0, b]; ℝ^{n×n})  (2.16)

and

H ∈ ℂ̃_{∞}^{(−1,k−1)}(D_{ab}; ℝ^{n×n}).  (2.17)

Proof. It is evident that if the matrix functions H_{0} and H satisfy conditions (2.16) and (2.17), respectively, then operator the h, given by the equality (2.5), transforms ℂ̃([0, a]; ℝ^{n}) into ℂ̃_{∞}^{k−1}([0, b]; ℝ^{n}) and is linear and continuous. Let us show the converse: if h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}^{k−1}([0, b]; ℝ^{n}) is a linear continuous operator, then (2.5) is valid and H_{0} and H satisfy conditions (2.16) and (2.17), respectively.

For an arbitrary vector function u ∈ ℂ̃([0, a]; ℝ^{n}) we put

h^{(i)}(u)(y) = \frac{d^{i}}{dy^{i}} h(u)(y) (i = 0, …, k).

It is evident that h^{(k)} and h^{(i)}(·)(0) (i = 0, …, k − 1) are linear continuous operators acting from ℂ̃([0, a]; ℝ^{n}) respectively to L_{∞}([0, b]; ℝ^{n}) and ℝ^{n}. According to Lemma 2.1_{1},

h^{(k)}(u)(y) = H_{0k}(y)u(0) + ∫_{0}^{a} H_{k}(s, y)u′(s) ds
for u ∈ ℂ̃([0, a]; ℝ^{n}), y ∈ [0, b],

and

h^{(i)}(u)(0) = H_{0i}u(0) + ∫_{0}^{a} H_{i}(s)u′(s) ds (i = 0, 1, …, k − 1),

where

H_{0k} ∈ L_{∞}([0, b]; ℝ^{n×n}), H_{k} ∈ L_{∞}(D_{ab}; ℝ^{n×n}),
H_{0i} ∈ ℝ^{n×n}, H_{i} ∈ L_{∞}([0, a]; ℝ^{n×n}) (i = 0, …, k − 1).

Therefore from the identity

h(u)(y) = ∑_{i=0}^{k−1} \frac{y^{i}}{i!} h^{(i)}(u)(0) + \frac{1}{(k − 1)!} ∫_{0}^{y}(y − t)^{k−1}h^{(k)}(u)(t) dt

there follows representation (2.5), where

H_{0}(y) = ∑_{i=0}^{k−1} \frac{y^{i}}{i!} H_{0i} + \frac{1}{(k − 1)!} ∫_{0}^{y}(y − t)^{k−1}H_{0k}(t)dt,

and

H(x, y) = ∑_{i=0}^{k−1} \frac{y^{i}}{i!} H_{i}(x) + \frac{1}{(k − 1)!} ∫_{0}^{y}(y − t)^{k−1}H_{k}(x, t) dt

are the matrix functions satisfying conditions (2.16) and (2.17). ■

Lemma 2.1_{4}. Let k be an arbitrary positive integer. h is a linear continuous operator acting from ℂ̃([0, a]; ℝ^{n}) to ℂ^{k}([a, b]; ℝ^{n}) if and only if (2.5) is valid, where

H_{0} ∈ ℂ^{k}([0, b]; ℝ^{n×n}),

the matrix function H satisfies condition (2.17), and

∫_{0}^{x} H(s, ·) ds ∈ ℂ^{k}([0, b]; ℝ^{n×n}) for x ∈ [0, a].

This lemma can be proved similarly to Lemma 2.1_{3}, with the only difference that instead of Lemma 2.1_{1} we use Lemma 2.1_{2}.

2.2. Characteristics of the matrix functions Z_{0}, M_{0} and M.

Lemma 2.2_{1}. If

A ∈ L_{∞}(D_{ab}; ℝ^{n×n}),  (2.18)

then

Z_{0} ∈ L_{∞}(D_{ab}; ℝ^{n×n})  (2.19)

and

\frac{∂Z_{0}}{∂x} ∈ L_{∞}(D_{ab}; ℝ^{n×n}).  (2.20)

In the case when

∫_{0}^{x} A(s, ·) ds ∈ ℂ([0, b]; ℝ^{n×n}) for x ∈ [0, a],  (2.21)

then instead of (2.19) we have

Z_{0} ∈ ℂ(D_{ab}; ℝ^{n×n}).  (2.22)

Proof. In view of (2.3) and (2.18) it is clear that the matrix function Z_{0} is measurable and satisfies inequalities

∥\frac{∂Z_{0}(x, y)}{∂x}∥ ≤ γ∥Z_{0}(x, y)∥

and

∥Z_{0}(x, y)∥ ≤ n + γ ∫_{0}^{x} ∥Z_{0}(s, y)∥ ds

almost everywhere in D_{ab}. From this, using Gronwall’s lemma, we obtain

∥Z_{0}(x, y)∥ ≤ n exp(γx), ∥\frac{∂Z_{0}(x, y)}{∂x}∥ ≤ nγ exp(γx).  (2.23)

Hence Z_{0} satisfies conditions (2.19) and (2.20).

Assume now that (2.21) holds. Then, in view of (2.3), the matrix function

A_{1}(x, y) = ∫_{0}^{x} A(s, y) ds

is continuous in D_{ab}. Therefore it is clear that the function

ω(δ) = max_{\begin{matrix}0≤x≤a, 0≤y,ȳ≤b \\ |y−ȳ|≤δ \end{matrix}} ∥A_{1}(x, y) − A_{1}(x, ȳ)∥  (2.24)

satisfies the condition

lim_{δ→0} ω(δ) = 0.  (2.25)

In view of (2.3), (2.23) and (2.24), from the equality

Z_{0}(x, y) − Z_{0}(x, ȳ) = ∫_{0}^{x}[A(s, y) − A(s, ȳ)]Z_{0}(s, y) ds + + ∫_{0}^{x} A(s, ȳ)[Z_{0}(s, y) − Z_{0}(s, ȳ)] ds = [A_{1}(x, y) − A_{1}(x, ȳ)]Z_{0}(x, y) −

−∫_{0}^{x}[A_{1}(s, y)−A_{1}(s, ȳ)]A(s, y)Z_{0}(s, y) ds+∫_{0}^{x} A(s, ȳ)[Z_{0}(s, y)−Z_{0}(s, ȳ)]ds it follows that

∥Z_{0}(x, y) − Z_{0}(x, ȳ)∥ ≤ γ_{1}ω(|y − ȳ|) + γ ∫_{0}^{x} ∥Z_{0}(s, y) − Z_{0}(s, ȳ)∥ ds
for 0 ≤ x ≤ a, 0 ≤ y, ȳ ≤ b,

where

γ_{1} = n(1 + aγ) exp(γa),

whence, according to Gronwall’s lemma, we get

∥Z_{0}(x, y) − Z_{0}(x, ȳ)∥ ≤ γ_{0}ω(|y − ȳ|) for 0 ≤ x ≤ a, 0 ≤ y, ȳ ≤ b, (2.26)

where γ_{0} = γ_{1} exp(γa). The validity of (2.22) becomes evident if we take into account the above estimate and conditions (2.20) and (2.25). ■

Lemma 2.2 _{2}. If

A ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}),  (2.27)

then

Z_{0} ∈ ℂ̃(D_{ab}; ℝ^{n×n}),  (2.28)
\frac{∂Z_{0}}{∂x} , \frac{∂Z_{0}}{∂y} and \frac{∂^{2}Z_{0}}{∂x∂y} ∈ L_{∞}(D_{ab}; ℝ^{n×n}).  (2.29)

When

A and \frac{∂A}{∂x} ∈ ℂ(D_{ab}; ℝ^{n×n}),  (2.30)

then instead of (2.29) we have

\frac{∂Z_{0}}{∂x} , \frac{∂Z_{0}}{∂y} and \frac{∂^{2}Z_{0}}{∂x∂y} ∈ ℂ(D_{ab}; ℝ^{n×n}).  (2.31)

Proof. In view of (2.27) there exists a positive number l_{0} such that the function ω given by (2.24) satisfies the inequality

ω(δ) ≤ l_{0}δ for δ ≥ 0.

Therefore from (2.26) we get

∥Z_{0}(x, y) − Z_{0}(x, ȳ)∥ ≤ γ_{0}l_{0}|y − ȳ| for 0 ≤ x ≤ a, 0 ≤ y, ȳ ≤ b. (2.32)

Taking into account the above arguments and conditions (2.3),(2.23) and (2.27), it follows from the identity

\frac{∂Z_{0}(x, y)}{∂x} = A(x, y)Z_{0}(x, y)  (2.33)

that

\frac{∂Z_{0}}{∂x} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}),  (2.34)
\frac{∂^{2}Z_{0}(x, y)}{∂x∂y} = A(x, y)\frac{∂Z_{0}(x, y)}{∂y} + \frac{∂A(x, y)}{∂y}Z_{0}(x, y)  (2.35)

and

\frac{∂Z_{0}(x, y)}{∂y} = Z_{0}(x, y) ∫_{0}^{x} Z_{0}^{−1}(s, y)\frac{∂A(s, y)}{∂y}Z_{0}(s, y) ds.  (2.36)

The validity of conditions (2.28) and (2.29) becomes evident if we take into consideration (2.32)-(2.35) together with (2.3),(2.20) and (2.27).

In the case when conditions (2.30) hold, conditions (2.31) follow from equalities (2.33),(2.35) and (2.36). ^{6} ■

By virtue of Lemmas 2.1_{1} - 2.1_{4}, 2.2_{1} and 2.2_{2}, from (2.6) we get the following propositions:

Lemma 2.3_{1}. If h : ℂ̃([0, a]; ℝ^{n}) → L_{∞}([0, b]; ℝ^{n}) is a linear continuous operator and the matrix function A satisfies condition (2.18), then

M_{0} ∈ L_{∞}([0, b]; ℝ^{n×n}), M ∈ L_{∞}(D_{ab}; ℝ^{n×n}).

Lemma 2.3_{2}. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ([0, b]; ℝ^{n}) be a linear continuous operator and let the matrix function A satisfy conditions (2.18),(2.21). Then

M_{0} ∈ ℂ([0, b]; ℝ^{n×n}), M ∈ L_{∞}(D_{ab}; ℝ^{n×n})

and

∫_{0}^{x} M(s, ·) ds ∈ ℂ([0, b]; ℝ^{n×n}) for x ∈ [0, a].

Lemma 2.3_{3}. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) be a linear continuous operator and let the matrix function A satisfy condition (2.27). Then

M_{0} ∈ ℂ̃_{∞}([0, b]; ℝ^{n×n})

and

M ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}).  (2.37)

6This fact is a consequence of the well-known Peano theorem on the differentiability of a solution of the Cauchy problem with respect to a parameter ([19], Theorem 3.1).

Lemma 2.3_{4}. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ^{1}([0, b]; ℝ^{n}) be a linear continuous operator and let the matrix function A be continuous and have a continuous partial derivative in the second argument. Then

M_{0} ∈ ℂ^{1}([0, b]; ℝ^{n×n}),

the matrix function M satisfies condition (2.37) and

∫_{0}^{x} M(s, ·) ds ∈ ℂ^{1}([0, b]; ℝ^{n×n}) for x ∈ [0, a].

2.3. Criterion for the existence and uniqueness of a solution of problem (2.1),(2.2). Properties of a solution.

Lemma 2.4. Let the parameter y ∈ [0, b] be such that

\begin{matrix}M(x, y)Z_{0}^{−1}(x, y)c(x, y) = M_{0}(y)c_{0}(x, y) for 0 ≤ x ≤ a, \\ φ(y) = M_{0}(y)φ_{0}(y), \end{matrix} (2.38)

where

c(·, y) ∈ L([0, a]; ℝ^{n}) and φ_{0}(y) ∈ ℝ^{n}.

Then the vector function

z(x, y) = Z_{0}(x, y)[φ_{0}(y) − ∫_{0}^{a} c_{0}(s, y) ds + ∫_{0}^{x} Z_{0}^{−1}(s, y)c(s, y) ds] (2.39)

is a solution of problem (2.1),(2.2), which is unique if and only if condition (2.8) holds.

Proof. According to the Cauchy formula, ^{7} the vector function

z(x, y) = Z_{0}(x, y)[α + ∫_{0}^{x} Z_{0}^{−1}(s, y)c(s, y) ds]  (2.40)

is a solution of system (2.1) for every α ∈ ℝ^{n} and, vice versa, for every solution of this system there exists α ∈ ℝ^{n} such that representation (2.40) is valid.

If we substitute (2.40) into (2.2), then with regard to (2.7) and (2.38) we shall find

M_{0}(y)α + ∫_{0}^{a} M(s, y)Z_{0}^{−1}(s, y)c(s, y) ds = φ(y)

and

M_{0}(y)α = M_{0}(y)[φ_{0}(y) − ∫_{0}^{a} c_{0}(s, y) ds].  (2.41)

7See [19], Ch.4, §2, Corollary 2.1.

Consequently the vector function (2.40) is a solution of problem (2.1),(2.2) if and only if α is a solution of the system of linear algebraic equations (2.41). It is obvious that

α = φ_{0}(y) − ∫_{0}^{a} c_{0}(s, y) ds

is a solution of this system which is unique if and only if condition (2.8) holds. ■

According to Lemmas 2.2_{1}, 2.2_{2} and 2.3_{1} -2.3_{4}, from Lemma 2.4 we get the following assertions.

Lemma 2.5_{1}. Let h : ℂ̃([0, a]; ℝ^{n}) → L_{∞}([0, b]; ℝ^{n}) be a linear continuous operator, the matrix function A satisfy condition (2.18) and

mes I_{M_{0}} = b.

Moreover, let

c ∈ L_{∞}(D_{ab}; ℝ^{n}), φ ∈ L([0, b]; ℝ^{n}), η ∈ L([0, b]; ℝ^{n}),

where

η(y) = ∥M_{0}^{−1}(y)∥ + ∥M_{0}^{−1}(y)φ(y)∥.

Then for almost every y ∈ [0, b] problem (2.1),(2.2)has the unique solution z(·, y), and

ess sup_{(x,y)∈D_{ab}} (∥z(x, y)∥ + ∥\frac{∂z(x, y)}{∂x}∥)/η(y) < +∞.

Lemma 2.5_{2}. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ([0, b]; ℝ^{n}) be a linear continuous operator, A ∈ ℂ(D_{ab}; ℝ^{n×n}) and I_{M_{0}} = [0, b]. Moreover, let

c ∈ ℂ(D_{ab}; ℝ^{n}), φ ∈ ℂ([0, b]; ℝ^{n}).

Then for every y ∈ [0, b] problem (2.1),(2.2) has the unique solution z(·, y), and

z and \frac{∂z}{∂x} ∈ ℂ(D_{ab}; ℝ^{n}).

Lemma 2.5_{3}. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) be a linear continuous operator, the matrix function A satisfy condition (2.27), and I_{M_{0}} = = [0, b]. Let, in addition,

c ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}), φ ∈ ℂ̃([0, b]; ℝ^{n}).

Then for every y ∈ [0, b] problem (2.1),(2.2) has the unique solution z(·, y) and

z ∈ ℂ̃(D_{ab}; ℝ^{n}).

Lemma 2.5_{4}. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ^{1}([0, b]; ℝ^{n}) be a linear continuous operator, A : D_{ab} → ℝ^{n×n} and c : D_{ab} → ℝ^{n} be continuous and have the continuous partial derivatives \frac{∂A}{∂y} and \frac{∂c}{∂y}, φ : [0, b] → ℝ^{n} be continuously differentiable and I_{M_{0}} = [0, b]. Then for every y ∈ [0, b] problem (2.1),(2.2) has the unique solution z(·, y), where z : D_{ab} → ℝ^{n} is continuous and has the continuous partial derivatives \frac{∂z}{∂x}, \frac{∂z}{∂y} and \frac{∂^{2}z}{∂x∂y}.

2.4. Lemmas on the unique solvability of p0oblem (2.1),(2.2) and on estimation of the matrix M_{0}^{−1}(y). For every k ∈ {1, …, n} let e_{k} denote an n-dimensional column vector whose k-th component is unity and the remainder are zero. We have

Lemma 2.6. If I_{M_{0}} ≠ ∅, then

∥M_{0}^{−1}(y)∥ = ∑_{k=1}^{n} ∥z_{k}(0, y)∥ for y ∈ I_{M_{0}},  (2.42)

where every z_{k}(·, y) is a solution of system (2.1_{0}) satisfying the condition

h(z_{k}(·, y))(y) = e_{k}.  (2.43)

Proof. According to Lemma 2.4, problem (2.1_{0}),(2.43) has the unique solution z_{k}(·, y) for every k ∈ {1, …, n} and y ∈ I_{M_{0}}. Let Z(x, y) be a matrix with columns z_{1}(x, y), …, z_{n}(x, y). Then

Z(x, y) = Z_{0}(x, y)Z(0, y).

Hence, in view of (2.7) and (2.43), we have

E = M_{0}(y)Z(0, y) for y ∈ I_{M_{0}}.

Consequently, equality (2.42) is valid. ■

Let us introduce the notation

\begin{matrix}A_{0}(s, x, y) = Θ, A_{1}(s, x, y) = A(s, y), \\ A_{j+1}(s, x, y) = ∫_{s}^{x} A(ξ, y)A_{j}(s, ξ, y) dξ (j = 1, 2, …), \end{matrix}  (2.44)
M_{00}(y) = H_{0}(y), M_{0j}(y) = H_{0}(y) +
+∫_{0}^{a} H(s, y)A(s, y)[E+∑_{i=0}^{j−1} ∫_{0}^{s} A_{i}(ξ, s, y) dξ] ds (j = 1, 2, …). (2.45)

If the inequality

det M_{0 k−1}(y) ≠ 0

holds for any y ∈ [0, b] and natural k, then for every natural m we put

B_{1m}(s, x, y)=[E+ ∑_{i=0}^{m−1} ∫_{0}^{x} A_{i}(ξ, x, y) dξ]M_{00}^{−1}(y)H(s, y)A(s, y), (2.46)

B_{km}(s, x, y) = [E + ∑_{i=0}^{m−1} ∫_{0}^{x} A_{i}(ξ, x, y) dξ]M_{0 k−1}^{−1}(y) ×

× ∫_{s}^{a} H(ξ, y)A(ξ, y)A_{k−1}(s, ξ, y) dξ for k > 1  (2.47)

and

B_{km}^{0}(y) = max_{0≤x≤a} [ ∫_{0}^{x} |A_{m}(s, x, y) − B_{km}(s, x, y)| ds +

+ ∫_{x}^{a} |B_{km}(s, x, y)| ds].  (2.48)

Lemma 2.7. Let there exist natural k and m and a nonempty set I_{0} ⊂ [0, b] such that

det M_{0 k−1}(y) ≠ 0 for y ∈ I_{0}  (2.49)

and

r(B_{km}^{0}(y)) < 1 for y ∈ I_{0}.  (2.50)

Then I_{0} ⊂ I_{M_{0}} and

∥M_{0}^{−1}(y)∥ ≤ ρ_{0}∥(E − B_{km}^{0}(y))^{−1}∥ ∥M_{0 k−1}^{−1}(y)∥ for y ∈ I_{0}, (2.51)

where ρ_{0} is a positive constant.

Proof. Let z(·, y) be an arbitrary solution of system (2.1_{0}). Then for any natural j we have

z(x, y)=[E + ∑_{i=0}^{j−1} ∫_{0}^{x} A_{i}(s, x, y) ds]z(0, y)+∫_{0}^{x} A_{j}(s, x, y)z(s, y) ds.(2.52_{j})

In view of (2.5),(2.44) and (2.45), from (2.52_{k}) we find that

h(z(·, y))(y) = M_{00}(y)z(0, y) + ∫_{0}^{a} H(s, y)A(s, y)z(s, y) ds for k = 1

and

h(z(·, y))(y) = M_{0 k−1}(y)z(0, y) +
+ ∫_{0}^{a} ( ∫_{s}^{a} H(ξ, y)A(ξ, y)A_{k−1}(s, ξ, y) dξ)z(s, y) ds for k > 1.

Hence, by virtue of (2.49) it follows that

z(0, y) = M_{00}^{−1}(y)h(z(·, y))(y) − M_{00}^{−1}(y) ∫_{0}^{a} H(s, y)A(s, y)z(s, y) ds
for k = 1, y ∈ I_{0}

and

z(0, y)=M_{0 k−1}^{−1}(y)h(z(·, y))(y)−M_{0 k−1}^{−1}(y) ∫_{0}^{a} ( ∫_{s}^{a} H(ξ, y)×
×A(ξ, y)A_{k−1}(s, ξ, y) dξ)z(s, y)ds for k > 1, y ∈ I_{0}.

If we substitute the value z(0, y) into the formula (2.52_{m}), then, in view of the notation (2.46) and (2.47), we shall get

z(x, y) = [E + ∑_{i=0}^{m−1} ∫_{0}^{x} A_{i}(s, x, y) ds]M_{0 k−1}^{−1}(y)h(z(·, y))(y) +

+∫_{0}^{x}[A_{m}(s, x, y)−B_{km}(s, x, y)]z(s, y) ds−∫_{x}^{a}B_{km}(s, x, y)z(s, y) ds for y ∈ I_{0},

whence, with regard to (2.48), we obtain

z̄(y) ≤ B_{km}^{0}(y)z̄(y) + A_{m}^{0}(y)|M_{0 k−1}^{−1}(y)||h(z(·, y))(y)| for y ∈ I_{0},

where

z̄(y) = max_{0≤x≤a} |z(x, y)|

and

A_{m}^{0}(y) = max_{0≤x≤a} |E + ∑_{i=0}^{m−1} ∫_{0}^{x} A_{i}(ξ, x, y) dξ|.

By condition (2.50) it follows that

z̄(y) ≤ (E − B_{km}^{0}(y))^{−1}A_{m}^{0}(y)|M_{0 k−1}^{−1}(y)||h(z(·, y))(y)| for y ∈ I_{0},

and, consequently,

∥z(x, y)∥ ≤ ∥A_{m}^{0}(y)∥ ∥(E − B_{km}^{0}(y))^{−1}∥ ∥M_{0 k−1}^{−1}(y)∥ ∥h(z(·, y))(y)∥ for 0 ≤ x ≤ a, y ∈ I_{0}.

However, in view of (2.3),

ρ_{0} = n sup_{0≤y≤b} ∥A_{m}^{0}(y)∥ < +∞.

Thus we have proved that an arbitrary solution z(·, y) of system (2.1_{0}) admits the estimate

\begin{matrix}∥z(x, y)∥ ≤ \frac{ρ_{0}}{n} ∥(E − B_{km}^{0}(y))^{−1}∥ ∥M_{0 k−1}^{−1}(y)∥∥h(z(·, y))(y)∥ \\ for 0 ≤ x ≤ a, y ∈ I_{0}, \end{matrix} (2.53)

where ρ_{0} is independent of I_{0} and z(·, y). It is easily seen that for every y ∈ I_{0} problem (2.1_{0}),(2.2_{0}) has only the trivial solution, i.e. I_{0} ⊂ I_{M_{0}}.

By virtue of Lemma 2.4, for every y ∈ I_{0} and k ∈ {1, …, n} problem (2.1_{0}),(2.43) has the unique solution z_{k}(·, y). On the other hand, in view of (2.53)

∥z_{k}(x, y)∥ ≤ \frac{ρ_{0}}{n} ∥(E − B_{km}^{0}(y))^{−1}∥ ∥M_{0 k−1}^{−1}(y)∥

for 0 ≤ x ≤ a, y ∈ I_{0} (k = 1, …, n).

With regard to these inequalities, from (2.42) we get (2.51). ■

Remark 2.1. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ([0, b]; ℝ^{n}) be a linear continuous operator, the matrix function A satisfy condition (2.21) and let I_{0} ⊂ [0, b] be a closed set. Then, as follows from Lemmas 2.2_{1} and 2.3_{2}, for conditions (2.49) and (2.50) to be fulfilled for some k and m, it is not only sufficient but also necessary that

I_{0} ⊂ I_{M_{0}}.

Lemma 2.8. Let

h(u)(y) ≡ (u_{i}(a_{i}(y)))_{i=1}^{n},  (2.54)

where a_{i} : [0, b] → [0, a] (i = 1, …, n) are measurable functions. Moreover, let there exist a nonempty set I_{0} ⊂ [0, b] such that the matrix function A is bounded in [0, a] × I_{0} and

r(A^{0}(y)) < \frac{π}{2a} for y ∈ I_{0},  (2.55)

where

A^{0}(y) = ess sup_{0≤x≤a} |A(x, y)|.

Then I_{0} ⊂ I_{M_{0}} and

∥M_{0}^{−1}(y)∥ ≤ ρ_{0}∥(E − \frac{2a}{π}A^{0}(y))^{−1}∥ for y ∈ I_{0},  (2.56)

where ρ_{0} is a positive constant.

Proof. Let z(·, y) = (z_{i}(·, y))_{i=1}^{n} be an arbitrary solution of (2.1_{0}). Then

z_{i}(x, y) = z_{i}(a_{i}(y), y) + ∑_{j=1}^{n} ∫_{a_{i}(y)}^{x} a_{ij}(s, y)z_{j}(s, y) ds (i = 1, …, n)

and

\begin{matrix}|z_{i}(x, y)|≤|z_{i}(a_{i}(y), y)|+∑_{j=1}^{n} a_{ij}^{0}(y)| ∫_{a_{i}(y)}^{x} |z_{j}(s, y)| ds| \\ (i = 1, …, n), \end{matrix}  (2.57)

where

a_{ij}^{0}(y) = ess sup_{0≤x≤a} |a_{ij}(x, y)|.

By Minkovsky’s inequality it follows that

( ∫_{0}^{a} z_{i}^{2}(x, y) dx)^{1/2} ≤ a^{1/2}|z_{i}(a_{i}(y), y)| +
+ ∑_{j=1}^{n} a_{ij}^{0}(y)( ∫_{0}^{a} | ∫_{a_{i}(y)}^{x} |z_{j}(s, y)| ds|^{2}dx)^{1/2} (i = 1, …, n). (2.58)

However, by Virtinger’s inequality, ^{8}

∫_{0}^{a} | ∫_{a_{i}(y)}^{x} |z_{j}(s, y)| ds|^{2}dx ≤ (\frac{2a}{π})^{2} ∫_{0}^{a} z_{j}^{2}(x, y) dx (i, j = 1, …, n).

Taking into account the above inequalities and condition (2.54), we get from (2.58)

z̄(y) ≤ a^{1/2}|h(z(·, y))(y)| + \frac{2a}{π}A^{0}(y)z̄(y),

where

z̄(y) = (( ∫_{0}^{a} z_{i}^{2}(x, y) dx)^{1/2})_{i=1}^{n}.

From this inequality, in view of condition (2.55), one arrives at the following estimate

z̄(y) ≤ a^{1/2}(E − \frac{2a}{π}A^{0}(y))^{−1}|h(z(·, y))(y)| for y ∈ I_{0},

by which means from (2.57) we get

|z(x, y)| ≤ |h(z(·, y))(y)| + a^{1/2}A^{0}(y)z̄(y) ≤

≤ |h(z(·, y))(y)| + aA^{0}(y)(E − \frac{2a}{π}A^{0}(y))^{−1}|h(z(·, y))(y)| =

= (E + \frac{(π − 2)a}{π}A^{0}(y))(E − \frac{2a}{π}A^{0}(y))^{−1}|h(z(·, y))(y)|.

The fact that A^{0} is bounded in the set I_{0} results in

∥z(x, y)∥ ≤ \frac{1}{n}ρ_{0}∥(E − \frac{2a}{π}A^{0}(y))^{−1}∥ ∥h(z(·, y))(y)∥ for y ∈ I_{0}, (2.59)

where

ρ_{0} = n sup_{y∈I_{0}} ∥E + \frac{(π − 2)a}{π}A^{0}(y)∥

is a constant independent of z(·, y).

By virtue of (2.59) it follows from Lemmas 2.4 and 2.6 that I_{0} ⊂ I_{M_{0}} and estimate (2.56) is valid. ■

8See [18], Ch.VI, Theorem 256.

If problem (2.1_{0}),(2.2_{0}) is periodic, i.e.

h(z(·, y))(y) = z(a, y) − z(0, y),  (2.60)

then H_{0}(y) ≡ 0, H(x, y) ≡ E and, consequently,

M_{0}(y) ≡ Z_{0}(a, y) − E.

The lemmas below have something to do with the considered case.

Lemma 2.9. Let condition (2.60) hold and there exist a nonempty set I_{0} ⊂ [0, b], a diagonal matrix function

A_{0}(x, y) = diag[a_{01}(x, y), …, a_{0n}(x, y)]  (2.61)

and a non-negative matrix function B : I_{0} → ℝ^{n×n} such that for every y ∈ I_{0} a_{0i}(·, y) : [0, a] → ℝ (i = 1, …, n) are summable and of constant sign,

sup_{y∈I_{0}} ∫_{0}^{a} |a_{0i}(s, y)| ds < +∞ (i = 1, …, n),  (2.62)
∫_{0}^{a} a_{0i}(s, y) ds ≠ 0 for y ∈ I_{0} (i = 1, …, n),  (2.63)
r(B(y)) < 1 for y ∈ I_{0}  (2.64)

and

|A(x, y) − A_{0}(x, y)| ≤ |A_{0}(x, y)|B(y) for 0 ≤ x ≤ a, y ∈ I_{0}. (2.65)

Then I_{0} ⊂ I_{M_{0}} and

∥M_{0}^{−1}(y)∥ ≤ ρ_{0}∥(E − B(y))^{−1}( ∫_{0}^{a} A_{0}(s, y) ds)^{−1}∥ for y ∈ I_{0}, (2.66)

where ρ_{0} is a positive constant.

Proof. First of all let us note that in view of conditions (2.62),

ρ_{1} = sup_{y∈I_{0}} ∥ exp ( ∫_{0}^{a} |A_{0}(s, y)| ds)∥ < +∞,

ρ_{0} = sup_{0<|u|≤ρ_{1}} \frac{n|u|}{| exp(u) − 1|} < +∞.

Let y ∈ I_{0} and z(·, y) be an arbitrary solution of (2.1_{0}). Then

\frac{dz(x, y)}{dx} = A_{0}(x, y)z(x, y) + [A(x, y) − A_{0}(x, y)]z(x, y).

Therefore, according to conditions (2.61),(2.63) and Lemma 2.4, we have

z(x, y)=exp ( ∫_{0}^{x} A_{0}(ξ, y) dξ)[ exp ( ∫_{0}^{a} A_{0}(s, y) ds)−E]^{−1}h(z(·, y))(y)+

+[E−exp ( ∫_{0}^{a} A_{0}(s, y) ds)]^{−1} ∫_{0}^{x} exp ( ∫_{s}^{x} A_{0}(ξ, y) dξ)[A(s, y)−

−A_{0}(s, y)]z(s, y) ds+exp ( ∫_{0}^{a} A_{0}(s, y) ds)[E−exp ( ∫_{0}^{a} A_{0}(s, y) ds)]^{−1}×

×∫_{x}^{a} exp ( ∫_{s}^{x} A_{0}(ξ, y) dξ)[A(s, y)−A_{0}(s, y)]z(s, y) ds. (2.67)

If we assume

z̄(y) = max_{0≤x≤a} |z(x, y)|,

then, in view of conditions (2.61), (2.63) and (2.65) and taking into account the fact that functions a_{i}(·, y) (i = 1, …, n) are of constant sign, we obtain from (2.67) that

|z(x, y)| ≤ \frac{ρ_{0}}{n} | ∫_{0}^{a} A_{0}(s, y) ds|^{−1}|h(z(·, y))(y)| +

+|E − exp ( ∫_{0}^{a} A_{0}(s, y) ds)|^{−1}[| ∫_{0}^{x} exp ( ∫_{s}^{x} A_{0}(ξ, y) dξ)A_{0}(s, y) ds| +

+ exp ( ∫_{0}^{a} A_{0}(s, y) ds)| ∫_{x}^{a} exp ( ∫_{s}^{x} A_{0}(ξ, y) dξ)A_{0}(s, y) ds|]B(y)z̄(y) =

= \frac{ρ_{0}}{n} | ∫_{0}^{a} A_{0}(s, y) ds|^{−1}|h(z(·, y))(y)| +

+|E − exp ( ∫_{0}^{a} A_{0}(s, y) ds)|^{−1}[|E − exp ( ∫_{0}^{x} A_{0}(ξ, y) dξ)| +

+| exp ( ∫_{0}^{x} A_{0}(ξ, y) dξ) − exp ( ∫_{0}^{a} A_{0}(ξ, y) dξ)|]B(y)z̄(y) =

= \frac{ρ_{0}}{n} | ∫_{0}^{a} A_{0}(s, y) ds|^{−1}|h(z(·, y))(y)| + B(y)z̄(y)

and

z̄(y) ≤ B(y)z̄(y) + \frac{ρ_{0}}{n} | ∫_{0}^{a} A_{0}(s, y) ds|^{−1}|h(z(·, y))(y)|.

By virtue of inequality (2.64), we have

|z̄(y)| ≤ \frac{1}{n}ρ_{0}|(E − B(y))^{−1}( ∫_{0}^{a} A_{0}(s, y) ds)^{−1}| |h(z(·, y))(y)|.

Consequently,

∥z(x, y)∥ ≤ \frac{ρ_{0}}{n} ∥(E − B(y))^{−1}( ∫_{0}^{a} A_{0}(s, y) ds)^{−1}∥∥h(z(·, y))(y)∥

for 0 ≤ x ≤ a, y ∈ I_{0}.

According to this estimate, it follows from Lemmas 2.4 and 2.6 that I_{0} ⊂ I_{M_{0}} and estimate (2.66) is valid. ■

Lemma 2.10. Let condition (2.60) hold and there exist a nonempty set I_{0} ⊂ [0, b] and functions σ_{i} : I_{0} → {−1, 1} (i = 1, …, n) such that

ā_{ii}(y) = ess sup_{0≤x≤a} (σ_{i}(y)a_{ii}(x, y)) < +∞,

ā_{ij}(y)=ess sup_{0≤x≤a} |a_{ij}(x, y)|<+∞ (i≠j, i, j = 1, …, n) for y ∈ I_{0} (2.68)

and for every y ∈ I_{0} the real parts of eigenvalues of the matrix

Ā(y) = (ā_{ij}(y))_{i,j=1}^{n}

are negative. Then I_{0} ⊂ I_{M_{0}} and

∥M_{0}^{−1}(y)∥ ≤ ρ∥Ā^{−1}(y)∥ for y ∈ I_{0},  (2.69)

where ρ > 0 is a constant.

Proof. From the asymptotic stability of the matrix Ā(y) (the real parts of eigenvalues of Ā(y) are negative for y ∈ I_{0}) we have

ā_{ii}(y) < 0 for y ∈ I_{0} (i = 1, …, n)  (2.70)

and the matrix

B(y) = ((1 − δ_{ij})\frac{ā_{ij}(y)}{|ā_{ii}(y)|})_{i,j=1}^{n},

where δ_{ij}, Kronecker’s symbol, satisfies condition (2.64).

Assume

A_{0}(x, y) = diag[a_{11}(x, y), …, a_{nn}(x, y)], Ā_{0}(y) = diag[ā_{11}(y), …, ā_{nn}(y)].

Then by (2.68) and (2.70),

|Ā_{0}(y)| ≤ |A_{0}(x, y)| for 0 ≤ x ≤ a, y ∈ I_{0}

and

|A(x, y) − A_{0}(x, y)| ≤ ((1 − δ_{ij})ā_{ij}(y))_{i,j=1}^{n} = |Ā_{0}(y)|B(y) ≤
≤ |A_{0}(x, y)|B(y) for 0 ≤ x ≤ a, y ∈ I_{0}.

Consequently, all conditions of Lemma 2.9 are fulfilled. Therefore I_{0} ⊂ I_{M_{0}} and estimate (2.66), where ρ_{0} > 0 is a constant, takes place. On the other hand,

(E − B(y))^{−1} = Ā^{−1}(y)Ā_{0}(y)

and

∥Ā_{0}(y)( ∫_{0}^{a} A_{0}(s, y) ds)^{−1}∥=∑_{i=1}^{n} |ā_{ii}(y)|| ∫_{0}^{a} a_{ii}(s, y) ds|^{−1} ≤ \frac{n}{a} for y ∈ I.

According to these conditions, from (2.66) there follows estimate (2.69), where ρ = \frac{n}{a}ρ_{0}. ■

§ 3. On Two Auxiliary Problems with the Integral Conditions for Linear Hyperbolic Systems

In this section for the linear hyperbolic system

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q(x, y)  (3.1)

we shall investigate two modifications of the characteristic initial value problem

\begin{matrix}u(x, 0) = φ_{0}(x),  \frac{∂u(0, y)}{∂y} = ψ(y) + \\ + ∫_{0}^{a} (Q_{0}(s, y)u(s, y) + γ_{1}^{−1}(s)Q_{1}(s, y)\frac{∂u(s, y)}{∂s})ds \end{matrix}  (3.2)

and

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = φ(x), \\ u(0, y) = ψ(y) + ∫_{0}^{y} ∫_{0}^{a} Q(y, s, t)u(s, t)dsdt, \end{matrix}  (3.3)

where

\begin{matrix}P_{i} ∈ L_{∞}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ L_{∞}(D_{ab}; ℝ^{n}), \\ φ_{0} ∈ ℂ̃([0, a]; ℝ^{n}), γ_{1}(x) = 1 + ∥φ′_{0}(x)∥, \end{matrix}  (3.4)

and the vector and matrix functions φ : [0, a] → ℝ^{n}, ψ : [0, b] → ℝ^{n} and Q_{i} : D_{ab} → ℝ^{n×n} (i = 0, 1), Q : [0, b] × D_{ab} → ℝ^{n×n} are summable. Moreover, there exists a summable function η : [0, b] → [0, +∞) such that

∥Q(y, s, t)∥ ≤ η(y) for y ∈ [0, b], (s, t) ∈ D_{ab}.  (3.5)

For an arbitrary x ∈ [0, a] (y ∈ [0, b]) by Z_{1}(x, ·) (Z_{2}(·, y)) we mean the fundamental matrix of the system of ordinary differential equations

\frac{dz}{dy} = P_{1}(x, y)z (\frac{dz}{dx} = P_{2}(x, y)z),

which satisfies the initial condition

Z_{1}(x, 0) = E (Z_{2}(0, y) = E).

Before we proceed to formulating the conditions of the unique solvability of problems (3.1),(3.2) and (3.1),(3.3), we have to establish some properties of solutions of system (3.1).

3.1. Boundedness of the first order partial derivatives of solutions of system (3.1).

Lemma 3.1. Every solution u of system (3.1) satisfies the conditions

ess sup_{(x,y)∈D_{ab}} [∥\frac{∂u(x, y)}{∂x}∥/(1 + ∥\frac{∂u(x, 0)}{∂x}∥)] < +∞  (3.6)

and

ess sup_{(x,y)∈D_{ab}} [∥\frac{∂u(x, y)}{∂y}∥/(1 + ∥\frac{∂u(0, y)}{∂y}∥)] < +∞.  (3.7)

Proof. According to conditions (3.4) and due to the absolute continuity of u, there exists a positive number γ such that

∥P_{1}(x, y)∥ ≤ γ

and

∫_{0}^{y} ∥P_{0}(x, t)u(x, t) + P_{2}(x, t)\frac{∂u(x, t)}{∂t} + q(x, t)∥dt ≤ γ

almost for every (x, y) ∈ D_{ab}. Therefore from the equality

\frac{∂u(x, y)}{∂x} = \frac{∂u(x, 0)}{∂x} + ∫_{0}^{y} [P_{0}(x, t)u(x, t) +

+ P_{1}(x, t)\frac{∂u(x, t)}{∂x} + P_{2}(x, t)\frac{∂u(x, t)}{∂t} + q(x, t)]dt

we get

∥\frac{∂u(x, y)}{∂x}∥ ≤ (1 + γ)(1 + ∥\frac{∂u(x, 0)}{∂x}∥) + γ ∫_{0}^{y} ∥\frac{∂u(x, t)}{∂x}∥dt.

It follows from the above inequality and Gronwall’s lemma that

∥\frac{∂u(x, y)}{∂x}∥ ≤ (1 + γ)(1 + ∥\frac{∂u(x, 0)}{∂x}∥) exp(γb).

Hence, inequality (3.6) holds. The validity of inequality (3.7) can be proved similarly. ■

3.2. Picone canonic form of linear hyperbolic systems.

Lemma 3.2. If P_{2}(x, ·) : [0, b] → ℝ^{n×n} is absolutely continuous almost for every x ∈ [0, a], then every solution of (3.1) is a solution of the system

\frac{∂}{∂y} [Z_{1}^{−1}(x, y)Z_{2}(x, y)\frac{∂}{∂x}(Z_{2}^{−1}(x, y)u)] =

= Z_{1}^{−1}(x, y)(P(x, y)u(x, y) + q(x, y)),  (3.8)

where

P(x, y) = P_{0}(x, y) + P_{1}(x, y)P_{2}(x, y) − \frac{∂P_{2}(x, y)}{∂y},  (3.9)

and, vice versa, every solution of (3.8) is a solution of (3.1).

Proof. Let u : D_{ab} → ℝ^{n} be an arbitrary absolutely continuous vector function. Then, by virtue of the equalities

\frac{∂Z_{1}^{−1}(x, y)}{∂y} = −Z_{1}^{−1}(x, y)P_{1}(x, y)

and

\frac{∂Z_{2}^{−1}(x, y)}{∂x} = −Z_{2}^{−1}(x, y)P_{2}(x, y)

we have

\frac{∂}{∂x}(Z_{2}^{−1}(x, y)u(x, y)) = −Z_{2}^{−1}(x, y)P_{2}(x, y)u(x, y) + Z_{2}^{−1}(x, y)\frac{∂u(x, y)}{∂x}

and

\frac{∂}{∂y} [Z_{1}^{−1}(x, y)Z_{2}(x, y)\frac{∂}{∂x}(Z_{2}^{−1}(x, y)u(x, y))] =

= \frac{∂}{∂y} [Z_{1}^{−1}(x, y)(\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y))] =

= Z_{1}^{−1}(x, y)[\frac{∂^{2}u(x, y)}{∂x∂y} + (P_{1}(x, y)P_{2}(x, y) − \frac{∂P_{2}(x, y)}{∂y})u(x, y) −

−P_{1}(x, y)\frac{∂u(x, y)}{∂x} − P_{2}(x, y)\frac{∂u(x, y)}{∂y}].

The latter equality shows that u is a solution of system (3.1) if and only if it is a solution of system (3.8). ■

3.3. On the integral representation of a generalized solution of system (3.1).

Lemma 3.3. If

P_{2} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}),  (3.10)

then an arbitrary generalized solution u of system (3.1) admits the representation

u(x, y) = Z_{2}(x, y)[v_{0}(y) + ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)v_{1}(s)ds] +

+Z_{2}(x, y) ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)(P(s, t)u(s, t) + q(s, t))ds dt, (3.11)

where

v_{0}(y) = u(0, y), v_{1}(x) = lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) (3.12)

and, vice versa, whatever summable vector functions v_{0} : [0, b] → ℝ^{n} and v_{1} : [0, a] → ℝ^{n} might be, the summable vector function u : D_{ab} → ℝ^{n}, admitting representation (3.11), is a generalized solution of system (3.1) satisfying conditions (3.12).

Proof. Let u be an arbitrary generalized solution of (3.1). Then, in view of conditions (3.4),(3.9) and (3.10), the vector function

Z_{1}^{−1}(·, ·)(P(·, ·)u(·, ·) + q(·, ·)) : D_{ab} → ℝ^{n}

as well as the vector functions v_{0} : [0, b] → ℝ^{n} and v_{1} : [0, a] → ℝ^{n} defined by (3.12), is summable. As for the vector function

Z_{1}^{−1}(x, ·)Z_{2}(x, ·)\frac{∂}{∂x}(Z_{2}^{−1}(x, ·)u(x, ·)) : [0, b] → ℝ^{n},

it is absolutely continuous almost for all x ∈ [0, a] according to Lemma 2.2_{2}. Moreover,

Z_{2}(x, y)\frac{∂}{∂x}(Z_{2}^{−1}(x, y)u(x, y)) = \frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y). (3.13)

Therefore, the integration of equality (3.8) from 0 to y yields

Z_{1}^{−1}(x, y)Z_{2}(x, y)\frac{∂}{∂x}(Z_{2}^{−1}(x, y)u(x, y)) =

= v_{1}(x) + ∫_{0}^{y} Z_{1}^{−1}(x, t)(P(x, t)u(x, t) + q(x, t))dt

and, consequently,

\frac{∂}{∂x}(Z_{2}^{−1}(x, y)u(x, y)) = Z_{2}^{−1}(x, y)Z_{1}(x, y)v_{1}(x) +

+ ∫_{0}^{y} Z_{2}^{−1}(x, y)Z_{1}(x, y)Z_{1}^{−1}(x, t)(P(x, t)u(x, t) + q(x, t))dt.

Then, because of the fact that the vector function

Z_{2}^{−1}(·, y)u(·, y) : [0, a] → ℝ^{n}

is absolutely continuous almost for every y ∈ [0, b], we get equality (3.11).

Assume now that v_{0} : [0, b] → ℝ^{n}, v_{1} : [0, a] → ℝ^{n} and u : D_{ab} → ℝ^{n} are arbitrary summable vector functions. Moreover, u admits representation (3.11), i.e.

u(x, y) = Z_{2}(x, y)[v_{0}(y) + v(x, y)],

where

v(x, y) = ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)v_{1}(s)ds +
+ ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)(P(s, t)u(s, t) + q(s, t))ds dt.

Then by Lemmas 2.2_{1} and 2.2_{2} and conditions (3.4),(3.9) and (3.10), the vector function v : D_{ab} → ℝ^{n} is absolutely continuous. It is also evident that the vector function u satisfies system (3.8) and conditions (3.12) almost everywhere in D_{ab}. Consequently, u is a generalized solution of system (3.1). ■

3.4. Conditions for the unique solvability of problems (3.1), (3.2) and (3.1), (3.3).

Lemma 3.4. Problem (3.1),(3.2) has one and only one solution u, and

\begin{matrix}u_{k}(x, y) ⇉ u(x, y), γ_{1}^{−1}(x)\frac{∂u_{k}(x, y)}{∂x} ⇉ γ_{1}^{−1}(x) \frac{∂u(x, y)}{∂x}, \\ γ_{2}^{−1}(y)\frac{∂u_{k}(x, y)}{∂y} ⇉ γ_{2}^{−1}(y)\frac{∂u(x, y)}{∂y} for k → +∞, \end{matrix} (3.14)

where

γ_{2}(y) = 1 + ∥ψ(y)∥ + ∫_{0}^{a}[∥Q_{0}(s, y)∥ + ∥Q_{1}(s, y)∥]ds,  (3.15)

u_{0}(x, y) ≡ 0 and

u_{k}(x, y) = φ_{0}(x) + ∫_{0}^{y} ψ(t)dt +

+ ∫_{0}^{y} ∫_{0}^{a} [Q_{0}(s, t)u_{k−1}(s, t) + γ_{1}^{−1}(s)Q_{1}(s, t)\frac{∂u_{k−1}(s, t)}{∂s}]ds dt +

+ ∫_{0}^{y} ∫_{0}^{x} [P_{0}(s, t)u_{k−1}(s, t) + P_{1}(s, t)\frac{∂u_{k−1}(s, t)}{∂s} +

+P_{2}(s, t)\frac{∂u_{k−1}(s, t)}{∂t} + q(s, t)]ds dt.  (3.16)

Proof. Let n_{1} = 2n, n_{2} = n and γ_{0} > 0 be large such that

r(A) < 1, where A ≡ γ_{0}^{−1} ess sup_{(x,y)∈D_{ab}} |P_{2}(x, y)|.  (3.17)

For arbitrary vector functions z_{1} = (z_{1}^{i})_{i=0}^{1} ∈ L_{∞}(D_{ab}; ℝ^{n_{1}}), where z_{1}^{i} ∈ ∈ L_{∞}(D_{ab}; ℝ^{n}) (i = 0, 1) and z_{2} ∈ L_{∞}(D_{ab}; ℝ^{n_{2}}), assume that

g_{1}^{0}(z_{1}, z_{2})(x, y) = φ_{0}(x) + ∫_{0}^{y} ψ(t)dt +
+ ∫_{0}^{y} ∫_{0}^{a}[Q_{0}(s, t)z_{1}^{0}(s, t) + Q_{1}(s, t)z_{1}^{1}(s, t)]ds dt +
+ ∫_{0}^{y} ∫_{0}^{x}[P_{0}(s, t)z_{1}^{0}(s, t) + γ_{1}(s)P_{1}(s, t)z_{1}^{1}(s, t) +
+γ_{2}(t) exp(γ_{0}s)P_{2}(s, t)z_{2}(s, t) + q(s, t)]ds dt;  (3.18)
\begin{matrix}g_{1}^{1}(z_{1}, z_{2})(x, y) = γ_{1}^{−1}(x)\frac{∂}{∂x}g_{1}^{0}(z_{1}, z_{2})(x, y), \\ g_{1}(z_{1}, z_{2})(x, y) = (g_{1}^{i}(z_{1}, z_{2})(x, y))_{i=0}^{1}, \end{matrix}  (3.19)

and

g_{2}(z_{1}, z_{2})(x, y) = γ_{2}^{−1}(y) exp(−γ_{0}x)\frac{∂}{∂y} g_{1}^{0}(z_{1}, z_{2})(x, y).  (3.20)

Let us show that problem (3.1),(3.2) is equivalent to the system of functional equations (1.1).

Indeed, let problem (3.1),(3.2) have a solution u. Introduce the notation

\begin{matrix}z_{1}^{0}(x, y) = u(x, y), z_{1}^{1}(x, y) = γ_{1}^{−1}(x)\frac{∂u(x, y)}{∂x}, \\ z_{1}(x, y) = (z_{1}^{i}(x, y))_{i=1}^{1}, z_{2}(x, y) = γ_{2}^{−1}(y) exp(−γ_{0}x)\frac{∂u(x, y)}{∂y}. \end{matrix} (3.21)

According to Lemma 3.1,

(z_{i})_{i=1}^{2} ∈ L_{∞}(D_{ab}; ℝ^{n_{1}}) × L_{∞}(D_{ab}; ℝ^{n_{2}}).

On the other hand, it becomes evident from equalities (3.18)-(3.21) that (z_{i})_{i=1}^{2} is a solution of system (1.1).

The converse is obvious: if z_{1}^{i} ∈ L_{∞}(D_{ab}; ℝ^{n}) (i = 1, 2), z_{1} = (z_{1}^{i})_{i=0}^{1}, z_{2} ∈ L_{∞}(D_{ab}; ℝ^{n_{2}}) and (z_{i})_{i=1}^{2}is a solution of system (1.1), then u(·, ·) = z_{1}^{0}(·, ·) is a solution of problem (3.1),(3.2), and equalities (3.21) are valid.

To complete the proof of this lemma, it remains to show that system (1.1) in the space L_{∞}(D_{ab}; ℝ^{n_{1}}) × L_{∞}(D_{ab}; ℝ^{n_{2}}) has the unique solution (z_{i})_{i=1}^{2} and conditions (1.8) hold, where z_{i0}(x, y) ≡ 0 (i = 1, 2) and (z_{ik})_{k=1}^{+∞} (i = 1, 2) are the sequences given by (1.9).

According to conditions (3.4) and the restrictions imposed on the vector and matrix functions φ_{0}, ψ and Q_{i} (i = 1, 2), it follows from (3.18)-(3.20) that operators g_{i} : L_{∞}(D_{ab}; ℝ^{n_{1}})×L_{∞}(D_{ab}; ℝ^{n_{2}}) → L_{∞}(D_{ab}; ℝ^{n_{i}}) (i = 1, 2) satisfy, for arbitrary ζ_{i} and ζ̄_{i} ∈ L_{∞}(D_{ab}; ℝ^{n_{i}}) (i = 1, 2), inequalities (1.3) and (1.21) almost everywhere in D_{ab}, where I = [0, a],

g_{0}(t) = γ_{2}(t)[1 + (1 + ∫_{0}^{a} γ_{1}(s)ds) ess sup_{(x,y)∈D_{ab}} (∥P_{0}(x, y)∥ +

+∥P_{1}(x, y)∥ + exp(γ_{0}x)∥P_{2}(x, y)∥)],

A_{01} = ess sup_{0≤y≤b} [γ_{2}^{−1}(y) ∫_{0}^{a}(|Q_{0}(s, y)| + |Q_{1}(s, y)|)ds] +

+ ess sup_{0≤y≤b} ∫_{0}^{a}(|P_{0}(s, y)| + γ_{1}(s)|P_{1}(s, y)|)ds, A_{02}(y) ≡ Θ

and

A(y) ≡ A.

The validity of Lemma 3.4 becomes evident if we take condition (3.17) into account and apply Lemma 1.3. ■

Lemma 3.4′. If P_{i} : D_{ab} → ℝ^{n×n} (i = 0, 1, 2), q : D_{ab} → ℝ^{n}, ψ : [0, b] → ℝ^{n} are continuous, φ_{0} : [0, a] → ℝ^{n} is continuously differentiable and Q_{i} : D_{ab} → ℝ^{n} (i = 0, 1) satisfy the Carathéodory conditions, ^{9} then problem (3.1),(3.2) is uniquely solvable, its solution u is classical and

\begin{matrix}u_{k}(x, y) ⇉ u(x, y), \frac{∂u_{k}(x, y)}{∂x} ⇉ \frac{∂u(x, y)}{∂x}, \\ \frac{∂u_{k}(x, y)}{∂y} ⇉ \frac{∂u(x, y)}{∂y} for k → ∞, \end{matrix}  (3.22)

where u_{0}(x, y) = 0 and for an arbitrary natural k the vector function u_{k} is given by equality (3.16).

Proof. By Lemma 3.4, problem (3.1),(3.2) has the unique solution u and conditions (3.14) hold. On the other hand, according to the restrictions imposed on P_{i} (i = 0, 1, 2), ψ, φ_{0} and Q_{i} (i = 0, 1), it becomes clear from (3.16) that

u_{k}, \frac{∂u_{k}}{∂x} and \frac{∂u_{k}}{∂y} ∈ ℂ(D_{ab}; ℝ^{n}) (k = 1, 2, …),

γ_{1} ∈ ℂ([0, a]; (1, +∞)), γ_{2} ∈ ℂ([0, b]; (1, +∞)).

Therefore, by virtue of the well-known Weierstrass theorem, it follows from (3.14) that u is a classical solution and conditions (3.22) take place. ■

Lemma 3.5. If condition (3.10) holds, then problem (3.1),(3.3) has one and only one generalized solution u and

\begin{matrix}γ^{−1}(y)u_{k}(x, y) ⇉ γ^{−1}(y)u(x, y), \frac{∂u_{k}(x, y)}{∂x} − P_{2}(x, y)u_{k}(x, y) ⇉ \\ ⇉ \frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y) for k → +∞, \end{matrix}(3.23)

where γ(y) = 1 + ∥ψ(y)∥ + η(y), u_{0}(x, y) = 0 and

u_{k}(x, y) = Z_{2}(x, y)[ψ(y) + ∫_{0}^{y} ∫_{0}^{a} Q(y, s, t)u_{k−1}(s, t)ds dt] +

+Z_{2}(x, y) ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)φ(s)ds + Z_{2}(x, y) ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y) ×

×Z_{1}(s, y)Z_{1}^{−1}(s, t)[P(s, t)u_{k−1}(s, t) + q(s, t)]dsdt.  (3.24)

Proof. According to Lemma 3.3, an arbitrary generalized solution of problem (3.1), (3.3) is a solution of the system of integral equations

u(x, y) = Z_{2}(x, y)[ψ(y) + ∫_{0}^{y} ∫_{0}^{a} Q(y, s, t)u(s, t)ds dt] +

9I.e., Q_{i}(·, y) : [0, a] → ℝ^{n×n} are measurable for all y ∈ [0, b], Q_{i}(x, ·) : [0, b] → ℝ^{n×n} are continuous almost for all x ∈ [0, a] and there exists a summable function γ : [0, a] → → [0, +∞) such that ∥Q_{i}(x, y)∥ ≤ γ(x) almost for all (x, y) ∈ D_{ab} (i = 0, 1).

+Z_{2}(x, y) ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)φ(s)ds + Z_{2}(x, y) ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y) ×

×Z_{1}(s, y)Z_{1}^{−1}(s, t)[P(s, t)u(s, t) + q(s, t)]ds dt  (3.25)

and, vice versa, an arbitrary summable solution of this system is a generalized solution of problem (3.1),(3.3 ). On the other hand, in view of conditions (3.4)-(3.6), it becomes clear that a summable function u : D_{ab} → ℝ^{n} is a solution of system (3.25) if and only if the vector function

z(x, y) = γ^{−1}(y)u(x, y)

belongs to L_{∞}(D_{ab}; ℝ^{n}) and is a solution of functional equation (1.2), where

g(z)(x, y) = γ^{−1}(y)Z_{2}(x, y)[ψ(y) + ∫_{0}^{y} ∫_{0}^{a} Q(y, s, t)γ(t)z(s, t)ds dt] +

+γ^{−1}(y)Z_{2}(x, y) ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)φ(s)ds + γ^{−1}(y)Z_{2}(x, y) ×

× ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)[P(s, t)γ(t)z(s, t) + q(s, t)]ds dt.

It follows from conditions (3.4)-(3.6) that the operator g transforms the space L_{∞}(D_{ab}; ℝ^{n}) into itself and for every ζ and ζ̄ ∈ L_{∞}(D_{ab}; ℝ^{n}) satisfies inequality (1.22), where I = [0, a], g_{0}(t) = c_{0}η(t) and c_{0} is a positive constant. Therefore, by virtue of Lemma 1.4,system (1.2) has the unique solution and condition (1.23) holds, where z_{0}(x, y) ≡ 0 and (z_{k})_{k=1}^{∞} is a sequence given by (1.24). It is clear from the above arguments that problem (3.1),(3.3) has the unique solution

u(x, y) = γ(y)z(x, y)

and conditions (3.23) take place. ■

Remark 3.1. From the proof of Lemma 3.5 it is clear that the following assertions are valid:

a) if ψ ∈ L_{∞}([0, b]; ℝ^{n}) and Q ∈ L_{∞}([0, b]×D_{ab}; ℝ^{n×n}), then the generalized solution u of problem (3.1),(3.3) belongs to L_{∞}(D_{ab}; ℝ^{n}), and condition (3.23) takes the form

\begin{matrix}u_{k}(x, y) ⇉ u(x, y), \frac{∂u_{k}(x, y)}{∂x} − P_{2}(x, y)u_{k}(x, y) ⇉ \\ ⇉ \frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y) for k → +∞; \end{matrix}  (3.26)

b) if ψ ∈ ℂ([0, b]; ℝ^{n}), Q ∈ L_{∞}([0, b] × D_{ab}; ℝ^{n×n}) and Q(·, s, t) : [0, b] → → ℝ^{n×n} is continuous almost for every (s, t) ∈ D_{ab}, then the generalized solution u of problem (3.1),(3.3) is continuous, and condition (3.23) takes the form of (3.26).

Lemma 3.5′. Let the matrix function P_{2} satisfy condition (3.10),

ψ ∈ ℂ̃([0, b]; ℝ^{n}), Q ∈ L_{∞}([0, b] × D_{ab}; ℝ^{n×n}),

Q(·, s, t) : [0, b] → ℝ^{n×n} be absolutely continuous almost for every (s, t) ∈ ∈ D_{ab} and the inequality

∥\frac{∂Q(y, s, t)}{∂y}∥ ≤ η(y)η_{0}(s, t)

hold in [0, b] × D_{ab}, where η : [0, b] → ℝ_{+} and η_{0} : D_{ab} → ℝ_{+} are summable functions. Then problem (3.1),(3.3) has the unique generalized solution u which is absolutely continuous, and

\begin{matrix}u_{k}(x, y) ⇉ u(x, y), \frac{∂u_{k}(x, y)}{∂x} ⇉ \frac{∂u(x, y)}{∂x}, \\ γ^{−1}(y)\frac{∂u_{k}(x, y)}{∂y} ⇉ γ^{−1}(y)\frac{∂u(x, y)}{∂y} for k → +∞, \end{matrix}  (3.27)

where γ(y) = 1 + ∥ψ′(y)∥ + η(y), u_{0}(x, y) = 0 and the vector function u_{k} is given by equality (3.24) for an arbitrary natural k.

From Lemmas 2.2_{2} and 3.5′ there immediately follows

Lemma 3.5′′. Let P_{i} : D_{ab} → ℝ^{n×n} (i = 0, 1, 2), q : D_{ab} → ℝ^{n} and φ : [0, a] → ℝ^{n} be continuous and P_{2} have a continuous partial derivative in the second argument. Moreover, let

ψ ∈ ℂ^{1}([0, b]; ℝ^{n}), Q ∈ L_{∞}([0, b] × D_{ab}; ℝ^{n×n}),

Q(·, s, t) : [0, b] → ℝ^{n×n} be absolutely continuous almost for all (s, t) ∈ D_{ab},

Q(·, s, ·) : [0, b] → ℝ^{n×n} be continuous almost for all s ∈ [0, a], ∫_{0}^{x} Q(·, s, t)ds :

[0, b] → ℝ^{n×n} be continuously differentiable for x ∈ [0, a] and almost for all t ∈ [0, b], and the inequality

∥\frac{∂Q(y, s, t)}{∂y}∥ ≤ η_{0}(s, t)

hold in [0, b] × D_{ab}, where η_{0} : D_{ab} → [0, ∞) is a summable function. Then problem (3.1),(3.3) is uniquely solvable, its solution u is classical and condition (3.22) takes place, where u_{0}(x, y) = 0 and the vector function u_{k} is given by equality (3.24) for an arbitrary natural k.

3.5. A priori estimates of solutions of problems (3.1),(3.2)and(3.1),(3.3)

Lemma 3.6. For any α > 0 there exists β > 0 such that if

\begin{matrix}∥P_{0}∥_{L} ≤ α, ∥P_{i}∥_{L_{∞}} ≤ α (i = 1, 2), \\ ∥Q_{0}∥_{L} ≤ α, ∥γ_{1}^{−1}Q_{1}∥_{L_{∞}} ≤ α, \end{matrix}  (3.28)

then the solution u of problem (3.1),(3.2) admits the estimate

∥u∥_{ℂ} ≤ β(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}).  (3.29)

Proof. In view of (3.1) and (3.2)

\frac{∂u(x, y)}{∂x} = φ′_{0}(x) + ∫_{0}^{y} (P_{0}(x, t)u(x, t) +

+P_{1}(x, t)\frac{∂u(x, t)}{∂x} + P_{2}(x, t)\frac{∂u(x, t)}{∂t} + q(x, t))dt,  (3.30)

\frac{∂u(x, y)}{∂y} = Z_{2}(x, y)[ψ(y) + ∫_{0}^{a} (Q_{0}(s, y)u(s, y) +

+γ_{1}^{−1}(s)Q_{1}(s, y)\frac{∂u(s, y)}{∂s})ds] + Z_{2}(x, y) ∫_{0}^{x} Z_{2}^{−1}(s, y) ×

×[P_{0}(s, y)u(s, y) + P_{1}(s, y)\frac{∂u(s, y)}{∂s} + q(s, y)]ds  (3.31)

and

u(x, y) = φ_{0}(x) + ∫_{0}^{y} \frac{∂u(x, t)}{∂t}dt.  (3.32)

Moreover, as follows from (3.28),

∥Z_{2}(x, y)∥ ≤ α_{1}, ∥Z_{2}^{−1}(x, y)∥ ≤ α_{1},  (3.33)

where α_{1} = n exp(aα).

If we assume

ρ_{0}(y) = max_{0≤x≤a} ∥u(x, y)∥, ρ(y) = ∫_{0}^{a} ∥\frac{∂u(s, y)}{∂s}∥ds,

then taking into account (3.28) and (3.33), from (3.31) and (3.32) we obtain

∥\frac{∂u(x, y)}{∂y}∥ ≤ α_{1}∥ψ(y)∥ + [ ∫_{0}^{a}(α_{1}∥Q_{0}(s, y)∥ +

+α_{1}^{2}∥P_{0}(s, y)∥)ds]ρ_{0}(y) + α(α_{1} + α_{1}^{2})ρ(y) + α_{1}^{2} ∫_{0}^{a} ∥q(s, y)∥ds (3.34)

and

ρ_{0}(y) ≤ ∥φ_{0}∥_{ℂ} + α_{1}∥ψ∥_{L} + ∫_{0}^{y} [ ∫_{0}^{a} (α_{1}∥Q_{0}(s, t)∥ +

+α_{1}^{2}∥P_{0}(s, t)∥)ds]ρ_{0}(t)dt + α(α_{1} + α_{1}^{2}) ∫_{0}^{y} ρ(t)dt + α_{1}^{2}∥q∥_{L}, whence according to Gronwall’s lemma and inequalities (3.28)

ρ_{0}(y) ≤ [∥φ_{0}∥_{ℂ} + α_{1}∥ψ∥_{L} + α_{1}^{2}∥q∥_{L} + α(α_{1} + α_{1}^{2}) ∫_{0}^{y} ρ(t)dt] ×

×exp (α_{1}∥Q_{0}∥_{L}+α_{1}^{2}∥P_{0}∥_{L})≤α_{2}[∥φ_{0}∥_{ℂ}+∥ψ∥_{L}+∥q∥_{L}+∫_{0}^{y} ρ(t)dt], (3.35)

where

α_{2} = (1 + α)(α_{1} + α_{1}^{2}) exp(α(α_{1} + α_{1}^{2})).

In view of (3.28),(3.34) and (3.35) from (3.30) we get

ρ(y) ≤ ∥φ_{0}∥_{ℂ} + ∫_{0}^{y} ( ∫_{0}^{a} ∥P_{0}(s, t)∥ds)ρ_{0}(t)dt +

+α ∫_{0}^{y} ρ(t)dt + α ∫_{0}^{y} ∫_{0}^{a} ∥\frac{∂u(s, t)}{∂t}∥dsdt + ∥q∥_{L} ≤

≤ ∥φ_{0}∥_{ℂ} + αα_{2}[∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L} + ∫_{0}^{y} ρ(t)dt] + α ∫_{0}^{y} ρ(t)dt +

+aαα_{1}∥ψ∥_{L} + aα^{2}(α_{1} + α_{1}^{2})[∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L} + ∫_{0}^{y} ρ(t)dt] +

+aα^{2}(α_{1} + α_{1}^{2}) ∫_{0}^{y} ρ(t)dt + aαα_{1}^{2}∥q∥_{L} + ∥q∥_{L} ≤

≤ α_{3}[∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L} + ∫_{0}^{y} ρ(t)dt],  (3.36)

where

α_{3} = αα_{2} + αα_{1}α_{2}(α_{1} + α_{1}^{2}) + α_{1}(1 + α)(α_{1} + α_{1}^{2}).

Applying again Gronwall’s lemma, from (3.36) we obtain

ρ(y) ≤ α_{4}(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}),  (3.37)

where α_{4} = α_{3} exp(bα_{3}).

By (3.34),(3.35) and (3.37) we have

∫_{0}^{a} ∥\frac{∂u(s, y)}{∂s}∥ds ≤ α_{4}(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}),

∥u(x, y)∥ ≤ α_{5}(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}),

∫_{0}^{b} ∥\frac{∂u(x, t)}{∂t}∥dt ≤ α_{6}(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}),

where α_{5} = α_{2}(1 + bα_{4}),

α_{6} = α_{1}^{2} + α_{4}(α_{1} + α_{1}^{2})(α_{4} + α_{5}).

With regard to the latter estimates and conditions (3.28) we get

∥u∥_{ℂ} = ∥φ_{0}∥_{ℂ} + ∫_{0}^{b} ∥\frac{∂u(0, t)}{∂t}∥dt + ∫_{0}^{a} ∫_{0}^{b} ∥\frac{∂^{2}u(s, t)}{∂s∂t}∥dsdt ≤

≤ ∥φ_{0}∥_{ℂ} + α_{6}(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}) + ∫_{0}^{a} ∫_{0}^{b} [∥P_{0}(s, t)∥∥u(s, t)∥ +

+∥P_{1}(s, t)∥∥\frac{∂u(s, t)}{∂s}∥ + ∥P_{2}(s, t)∥∥\frac{∂u(s, t)}{∂t}∥]dsdt + ∥q∥_{L} ≤

≤ (1 + α_{6} + αα_{5} + bαα_{4} + aαα_{6})(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}).

Consequently, estimate (3.29) is valid, where β = 1+α_{6}+αα_{5}+bαα_{4}+aαα_{6} depends on a, b and α only. ■

Lemma 3.7. For any α > 0 there exists β > 0 such that if

\begin{matrix}∥P_{0}∥_{L}^{(2)} ≤ α, ∥P_{i}∥_{L_{∞}} ≤ α (i = 1, 2), \\ ∥Q_{0}∥_{L}^{(2)} ≤ α, ∥γ_{1}^{−1}Q_{1}∥_{L_{∞}} ≤ α, \end{matrix}  (3.38)

then the solution u of problem (3.1),(3.2) admits the estimate

∥u∥(1) ; _{ℂ} ≤ β(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}^{(1)}),  (3.39)

but for φ_{0} ∈ ℂ̃_{∞}([0, a]; ℝ^{n}) and ψ ∈ L_{∞}([0, b]; ℝ^{n}), it admits the estimate

∥u∥(2) ; _{ℂ} ≤ β(∥φ_{0}∥_{ℂ_{∞}} + ∥ψ∥_{L_{∞}} + ∥q∥_{L}^{(2)}).  (3.40)

Proof. Put

ρ_{1}(x, y) = ∥φ_{0}(0)∥ + max_{(s,t)∈D_{xy}} ( ∫_{0}^{s} ∥\frac{∂u(ξ, t)}{∂ξ}∥dξ + ∫_{0}^{t} ∥\frac{∂u(s, τ)}{∂τ}∥dτ).

It is clear that ρ_{1} : D_{ab} → ℝ_{+} is continuous and

ρ_{1}(a, b) = ∥u∥(1) ; _{ℂ}.

From (3.1) and (3.2) we obtain

∥φ_{0}(0)∥ + ∫_{0}^{x} ∥\frac{∂u(s, y)}{∂s}∥ds + ∫_{0}^{y} ∥\frac{∂u(x, t)}{∂t}∥dt ≤ ∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} +

+ ∫_{0}^{y} ∥ ∫_{0}^{a} [Q_{0}(s, t)u(s, t) + γ_{1}^{−1}(s)Q_{1}(s, t)\frac{∂u(s, t)}{∂s}]ds∥dt +

+ ∫_{0}^{x} ∥ ∫_{0}^{y} P_{0}(s, t)u(s, t)dt∥ds + ∫_{0}^{y} ∥ ∫_{0}^{x} P_{0}(s, t)u(s, t)ds∥dt +

+ ∫_{0}^{x} ∥ ∫_{0}^{y} P_{1}(s, t)\frac{∂u(s, t)}{∂s}dt∥ds + ∫_{0}^{y} ∥ ∫_{0}^{x} P_{1}(s, t)\frac{∂u(s, t)}{∂s}ds∥dt +

+ ∫_{0}^{x} ∥ ∫_{0}^{y} P_{2}(s, t)\frac{∂u(s, t)}{∂t}dt∥ds +

+ ∫_{0}^{y} ∥ ∫_{0}^{x} P_{2}(s, t)\frac{∂u(s, t)}{∂t}ds∥dt + ∥q∥_{L}^{(1)}.  (3.41)

But

∫_{0}^{a} Q_{0}(s, t)u(s, t)ds = ( ∫_{0}^{a} Q_{0}(s, t)ds)[φ_{0}(a) + ∫_{0}^{t} \frac{∂u(a, τ)}{∂τ}dτ] −

− ∫_{0}^{a} ( ∫_{0}^{s} Q_{0}(ξ, t)dξ)\frac{∂u(s, t)}{∂s}ds,  (3.42)

∫_{0}^{x} P_{0}(s, t)u(s, t)ds = ( ∫_{0}^{x} P_{0}(s, t)ds)[φ_{0}(x) + ∫_{0}^{t} \frac{∂u(x, τ)}{∂τ}dτ] −

− ∫_{0}^{x} ( ∫_{0}^{s} P_{0}(ξ, t)dξ)\frac{∂u(s, t)}{∂s}ds,  (3.43)

∫_{0}^{y} P_{0}(s, t)u(s, t)dt = ( ∫_{0}^{y} P_{0}(s, t)dt)[φ_{0}(s) + ∫_{0}^{y} \frac{∂u(s, t)}{∂t}dt] −

− ∫_{0}^{y} ( ∫_{0}^{t} P_{0}(s, τ)dτ)\frac{∂u(s, t)}{∂t}dt.  (3.44)

If along with these inequalities we take into account conditions (3.38), then from (3.41) we shall get

∥φ_{0}(0)∥ + ∫_{0}^{x} ∥\frac{∂u(s, y)}{∂s}∥ds + ∫_{0}^{y} ∥\frac{∂u(x, t)}{∂t}∥dt ≤ ∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} +

+∥q∥_{L}^{(1)} + α ∫_{0}^{y} [∥φ_{0}(a)∥ + ∫_{0}^{t} ∥\frac{∂u(a, τ)}{∂τ}∥dτ + 2 ∫_{0}^{a} ∥\frac{∂u(s, t)}{∂s}∥ds]dt +

+α ∫_{0}^{x} [∥φ_{0}(s)∥ + 2 ∫_{0}^{y} ∥\frac{∂u(s, t)}{∂s}∥dt]ds +

+α ∫_{0}^{y} [∥φ_{0}(x)∥ + ∫_{0}^{t} ∥\frac{∂u(x, τ)}{∂τ}∥dτ + ∫_{0}^{x} ∥\frac{∂u(s, t)}{∂s}∥ds]dt +

+2α ∫_{0}^{y} ∫_{0}^{x} ∥\frac{∂u(s, t)}{∂s}∥dsdt + 2α ∫_{0}^{x} ∫_{0}^{y} ∥\frac{∂u(s, t)}{∂t}∥dtds ≤

≤ (1 + 2bα + aα)∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}^{(1)} + 2α ∫_{0}^{y} ρ_{1}(a, t)dt +

+3α ∫_{0}^{y} ρ_{1}(x, t)dt + 4α ∫_{0}^{x} ρ_{1}(s, y)ds

and, consequently,

ρ_{1}(x, y) ≤ (1 + 2bα + aα)(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}^{(1)}) +

+5α ∫_{0}^{y} ρ_{1}(a, t)dt + 4α ∫_{0}^{x} ρ_{1}(s, y)ds.

Applying twice Gronwall’s lemma, we find from the latter inequality that

ρ_{1}(x, y) ≤ (1 + 2bα + aα) exp(4aα)(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}^{(1)}) +

+5α exp(4aα) ∫_{0}^{y} ρ_{1}(a, t)dt,

ρ_{1}(a, y) ≤ β_{0}(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}^{(1)})

and, consequently,

∥u∥(1) ; _{ℂ} ≤ β_{0}(∥φ_{0}∥_{ℂ} + ∥ψ∥_{L} + ∥q∥_{L}^{(1)}),  (3.45)

where

β_{0} = (1 + 2bα + aα) exp[4aα + 5bα exp(4aα)].

Now consider the case when

φ_{0} ∈ ℂ̃_{∞}([0, a]; ℝ^{n}), ψ ∈ L_{∞}([0, b]; ℝ^{n}).

In this case according to Lemma 3.1 and conditions (3.2),(3.38) and (3.42), the vector functions \frac{∂u(x,y)}{∂x} and \frac{∂u(x,y)}{∂y} are essentially bounded. Assume

ρ_{2}(y) = max_{0≤x≤a} ∥\frac{∂u(x, y)}{∂y}∥.

Taking into account conditions (3.38) and (3.42)-(3.44) we obtain from (3.1) and (3.2)

∥\frac{∂u(x, y)}{∂x}∥ ≤ ∥φ′_{0}(x)∥ + ∥ ∫_{0}^{y} P_{0}(x, t)u(x, t)dt∥ +

+∥ ∫_{0}^{y} P_{1}(x, t)\frac{∂u(x, t)}{∂x}dt∥ + ∥ ∫_{0}^{y} P_{2}(x, t)\frac{∂u(x, t)}{∂t}dt∥ + ∥q∥_{L}^{(2)} ≤

≤ ∥φ_{0}∥_{ℂ_{∞}} + ∥q∥_{L}^{(2)} + α[∥φ_{0}(x)∥ + 2 ∫_{0}^{y} ∥\frac{∂u(x, t)}{∂t}∥dt] +

+α ∫_{0}^{y} ∥\frac{∂u(x, t)}{∂x}∥dt + α ∫_{0}^{y} ∥\frac{∂u(x, t)}{∂t}∥dt ≤ (1 + α + aα)∥φ_{0}∥_{ℂ_{∞}} +

+∥q∥_{L}^{(2)} + 3α ∫_{0}^{y} ρ_{2}(t)dt + α ∫_{0}^{y} ∥\frac{∂u(x, t)}{∂x}∥dt ≤

≤ (1 + 3α + aα)[∥φ_{0}∥_{ℂ_{∞}} + ∥q∥_{L}^{(2)} + ∫_{0}^{y} ρ_{2}(t)dt] + α ∫_{0}^{y} ∥\frac{∂u(x, t)}{∂x}∥dt

and

∥\frac{∂u(x, y)}{∂y}∥ ≤ ∥ψ∥_{L_{∞}} + ∥ ∫_{0}^{a} (Q_{0}(s, y)u(s, y) +

+γ_{1}^{−1}(s)Q_{1}(s, y)\frac{∂u(s, y)}{∂s})ds∥ + ∥ ∫_{0}^{x} P_{0}(s, y)u(s, y)ds∥ +

+∥ ∫_{0}^{x} P_{1}(s, y)\frac{∂u(s, y)}{∂s}ds∥ + ∥ ∫_{0}^{x} P_{2}(s, y)\frac{∂u(s, y)}{∂y}ds∥ +

+∥q∥_{L}^{(2)} ≤ ∥ψ∥_{L_{∞}} + ∥q∥_{L}^{(2)} + α[∥φ_{0}(a)∥ + ∫_{0}^{y} ∥\frac{∂u(a, t)}{∂t}∥dt +

+2 ∫_{0}^{a} ∥\frac{∂u(s, y)}{∂s}∥ds]+α[∥φ_{0}(x)∥+∫_{0}^{y} ∥\frac{∂u(x, t)}{∂t}∥dt +

+ ∫_{0}^{x} ∥\frac{∂u(s, y)}{∂s}∥ds] + α ∫_{0}^{x} ∥\frac{∂u(s, y)}{∂s}∥ds + α ∫_{0}^{x} ∥\frac{∂u(s, y)}{∂y}∥ds ≤

≤ (2α + 2aα)∥φ_{0}∥_{ℂ_{∞}} + ∥ψ∥_{L_{∞}} + ∥q∥_{L^{(2)}} + 4α ∫_{0}^{a} ∥\frac{∂u(s, y)}{∂s}∥ds +

+2α ∫_{0}^{y} ρ_{2}(t)dt + α ∫_{0}^{x} ∥\frac{∂u(s, y)}{∂y}∥ds ≤ (1 + 4α + 2aα)[∥φ_{0}∥_{ℂ_{∞}} +

+∥ψ∥_{L_{∞}} + ∥q∥_{L}^{(2)} + ∫_{0}^{y} ρ_{2}(t)dt + ∫_{0}^{a} ∥\frac{∂u(s, y)}{∂s}∥ds] + α ∫_{0}^{x} ∥\frac{∂u(s, y)}{∂y}∥ds,

whence by Gronwall’s lemma

∥\frac{∂u(x, y)}{∂x}∥ ≤ β_{1}[∥φ_{0}∥_{ℂ_{∞}} + ∥ψ∥_{L_{∞}} + ∥q∥_{L}^{(2)} + ∫_{0}^{y} ρ_{2}(t)dt] (3.46)

and

∥\frac{∂u(x, y)}{∂y}∥ ≤ β_{2}[∥φ_{0}∥_{ℂ_{∞}} + ∥ψ∥_{L_{∞}} + ∥q∥_{L}^{(2)} +

+ ∫_{0}^{y} ρ_{2}(t)dt + ∫_{0}^{a} ∥\frac{∂u(s, y)}{∂s}∥ds],  (3.47)

where

β_{1} = (1 + 3α + aα) exp(bα), β_{2} = (1 + 4α + 2aα) exp(aα).

By virtue of (3.46) and (3.47) we get

ρ_{2}(y) ≤ (1 + β_{1}a)β_{2}[∥φ_{0}∥_{ℂ_{∞}} + ∥ψ∥_{L_{∞}} + ∥q∥_{L}^{(2)} + ∫_{0}^{y} ρ_{2}(t)dt].

If we apply again Gronwall’s lemma, then from the latter inequality we obtain

ρ_{2}(y) ≤ β_{3}[∥φ_{0}∥_{ℂ_{∞}} + ∥ψ∥_{L_{∞}} + ∥q∥_{L}^{(2)}],  (3.48)

where β_{3} = (1 + β_{1}a)β_{2} exp(β_{2}b + β_{1}β_{2}ab).

The estimate

∥u∥(2) ; _{ℂ} ≤ β_{4}(∥φ_{0}∥_{ℂ_{∞}} + ∥ψ∥_{L_{∞}} + ∥q∥_{L}^{(2)})  (3.49)

follows from (3.46) and (3.48), where

β_{4} = β_{1}(1 + β_{3}b) + β_{3} + 1.

According to (3.45) and (3.49) it is clear that estimates (3.39) and (3.40) are valid, where β = β_{0} + β_{4} depends on a, b and α only. ■

Lemma 3.8. For any α > 0, P_{1} ∈ ℂ̃_{∞}^{(0,−1)}(D_{ab}; ℝ^{n×n}) and P_{2} ∈ ℂ̃_{∞}^{(−1,0)} (D_{ab}; ℝ^{n×n}) there exists β >0 such that if ψ ∈L_{∞}([0, b]; ℝ^{n}), Q ∈ L_{∞}([0, b]× ×D_{ab}; ℝ^{n}),

∥P_{0}∥_{L} ≤ α, ∥Q∥_{L_{∞}} ≤ α,  (3.50)

then the generalized solution u of problem (3.1),(3.3) admits the estimate

∥u∥_{L_{∞}} ≤ β(∥φ∥_{L} + ∥ψ∥_{L_{∞}} + ∥q∥_{L}^{(0)}).  (3.51)

Proof. By Lemma 3.3 the generalized solution u of problem (3.1),(3.3) admits the representation

u(x, y) = ∫_{0}^{x} Z(x, y, s, y,)Z_{1}(s, y)φ(s)ds +
+Z_{2}(x, y)[ψ(y) + ∫_{0}^{y} ∫_{0}^{a} Q(y, s, t)u(s, t)dsdt] +
+ ∫_{0}^{y} ∫_{0}^{x} Z(x, y, s, t)[P(s, t)u(s, t) + q(s, t)]dsdt,  (3.52)

where

Z(x, y, s, t) = Z_{2}(x, y)Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t),

and P is a matrix function given by (3.9).

However,

∫_{0}^{y} ∫_{0}^{x} Z(x, y, s, t)q(s, t)dsdt = ∫_{0}^{y} ∫_{0}^{x} q(s, t)dsdt −

− ∫_{0}^{y} \frac{∂Z(x, y, x, t)}{∂t}( ∫_{0}^{t} ∫_{0}^{x} q(s, τ)dsdτ)dt −

− ∫_{0}^{x} \frac{∂Z(x, y, s, y)}{∂s}( ∫_{0}^{y} ∫_{0}^{s} q(ξ, t)dξdt)ds +

+ ∫_{0}^{y} ∫_{0}^{x} [\frac{∂^{2}Z(x, y, s, t)}{∂s∂t} ∫_{0}^{t} ∫_{0}^{s} q(ξ, τ)dξdτ]dsdt.

On the other hand, by virtue of Lemma 2.2_{2} and the restrictions imposed on the matrix functions P_{1} and P_{2}, there exists a positive number α_{0} such that the inequalities

∥P_{1}(x, y)P_{2}(x, y)∥ + ∥\frac{∂P_{2}(x, y)}{∂y}∥ ≤ α_{0}, ∥Z_{i}(x, y)∥ ≤ α_{0} (i = 1, 2) (3.53)

and

∥Z(x, y, s, t)∥ + ∥\frac{∂Z(x, y, s, t)}{∂s}∥ + ∥\frac{∂Z(x, y, s, t,)}{∂t}∥ +

+∥\frac{∂^{2}Z(x, y, s, t)}{∂s∂t}∥ ≤ α_{0}  (3.54)

hold almost everywhere in D_{ab} and D_{ab} × D_{ab}. Therefore

∥ ∫_{0}^{x} ∫_{0}^{y} Z(x, y, s, t)q(s, t)dsdt∥ ≤ [1 + (a + b + ab)α_{0}]∥q∥_{L}^{(0)} (3.55)

and

∥P(x, y)∥ ≤ ∥P_{0}(x, y)∥ + α_{0}.  (3.56)

Put

ρ(y) = max_{0≤x≤a} ∥u(x, y)∥.

Then with regard to inequalities (3.50) and (3.53)-(3.56) we obtain from (3.52)

ρ(y) ≤ α_{0}^{2}∥φ∥_{L} + α_{0}[∥ψ∥_{L_{∞}} + aα ∫_{0}^{y} ρ(t)dt] +

+α_{0} ∫_{0}^{y} [ ∫_{0}^{a} ∥P_{0}(s, t)∥ds + α_{0}a]ρ(t)dt + [1 + (a + b + ab)α_{0}]∥q∥_{L}^{(0)} ≤

≤ (1 + α_{0} + α_{0}^{2})(1 + a + b + ab)(∥φ∥_{L} + ∥ψ∥_{L_{∞}} + ∥q∥_{L^{(0)}}) +
+α_{0} ∫_{0}^{y} [aα + α_{0}a + ∫_{0}^{a} ∥P_{0}(s, t)∥ds]ρ(t)dt,

which by Gronwall’s lemma and (3.50) implies estimate (3.51), where β = (1 + α_{0} + α_{0}^{2})(1 + a + b + ab) exp (α_{0}α + abα_{0}(α + α_{0})) is a constant depending on P_{1}, P_{2}, a, b and α only. ■

3.6. Correctness of problems (3.1),(3.2) and (3.1),(3.3). Alongside with problems (3.1),(3.2) and (3.1),(3.3), for any natural k let us consider the problems

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0k}(x, y)u(x, y) + P_{1k}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2k}(x, y)\frac{∂u(x, y)}{∂y} + q_{k}(x, y),  (3.57)

u(x, 0) = φ_{0k}(x), \frac{∂u(0, y)}{∂y} = ψ_{k}(y) +

+ ∫_{0}^{a} [Q_{0k}(s, y)u(s, y) + γ_{1k}^{−1}(s)Q_{1k}(s, y)\frac{∂u(s, y)}{∂s}]ds  (3.58)

and

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0k}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q_{k}(x, y),  (3.59)

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = φ_{k}(x), \\ u(0, y) = ψ_{k}(y) + ∫_{0}^{y} ∫_{0}^{a} Q_{k}(y, s, t)u(s, t)dsdt, \end{matrix}  (3.60)

where

P_{ik} ∈ L_{∞}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q_{k} ∈ L_{∞}(D_{ab}; ℝ^{n}),
φ_{0k} ∈ ℂ̃([0, a]; ℝ^{n}), γ_{1k}(x) = 1 + ∥φ′_{0k}(x)∥,
φ_{k} ∈ L([0, a]; ℝ^{n}), ψ_{k} ∈ L([0, b]; ℝ^{n}),
Q_{ik} ∈ L(D_{ab}; ℝ^{n×n}) (i = 0, 1)

and

Q_{k} ∈ L_{∞}([0, b] × D_{ab}; ℝ^{n×n}).

As above, by γ_{1} and γ_{2} we imply the functions γ_{1}(x) = 1+∥φ′_{0}(x)∥, γ_{2}(y) = 1+∥ψ(y)∥+∫_{0}^{a} [∥Q_{0}(s, y)∥+∥Q_{1}(s, y)∥]ds.

Lemma 3.9. Let

sup_{k≥1} ∥P_{ik}∥_{L_{∞}} < +∞ (i = 1, 2), sup_{k≥1} ∥γ_{1k}^{−1}Q_{1k}∥_{L_{∞}} < +∞, (3.61)
lim_{k→+∞} ∥P_{ik} − P_{i}∥_{L} = 0 (i = 0, 1, 2), lim_{k→+∞} ∥q_{k} − q∥_{L} = 0 (3.62)

and

lim_{k→+∞} ∥φ_{0k} − φ_{0}∥_{ℂ} = 0, lim_{k→+∞} ∥ψ_{k} − ψ∥_{L} = 0,

lim_{k→+∞} ∥Q_{0k} − Q_{0}∥_{L} = 0 lim_{k→+∞} ∥γ_{1}γ_{1k}^{−1}Q_{1k} − Q_{1}∥_{L} = 0. (3.63)

Then

lim_{k→+∞} ∥u_{k} − u∥_{ℂ} = 0,  (3.64)

where u and u_{k} are, respectively, solutions of problems (3.1),(3.2) and (3.57),(3.58). Proof. For an arbitrary natural k the vector function

v(x, y) = u_{k}(x, y) − u(x, y)

is a solution of the problem

\frac{∂^{2}v(x, y)}{∂x∂y} = P_{0k}(x, y)v(x, y) + P_{1k}(x, y)\frac{∂v(x, y)}{∂x} +

+P_{2k}(x, y)\frac{∂v(x, y)}{∂y} + q̃_{k}(x, y),  (3.65)

v(x, 0) = φ̃_{0k}(x), \frac{∂v(0, y)}{∂y} = ψ̃_{k}(y) +

+ ∫_{0}^{a} [Q_{0k}(s, y)v(s, y) + γ_{1k}^{−1}(s)Q_{1k}(s, y)\frac{∂v(s, y)}{∂s}]ds,  (3.66)

where

q̃_{k}(x, y) = [P_{0k}(x, y) − P_{0}(x, y)]u(x, y) + [P_{1k}(x, y) − P_{1}(x, y)] ×

×\frac{∂u(x, y)}{∂x} + [P_{2k}(x, y) − P_{2}(x, y)]\frac{∂u(x, y)}{∂y} + q_{k}(x, y) − q(x, y) (3.67)

and

φ̃_{0k}(x) = φ_{0k}(x) − φ_{0}(x), ψ̃_{k}(y) = ψ_{k}(y) − ψ(y) +
+ ∫_{0}^{a} [(Q_{0k}(s, y) − Q_{0}(s, y))u(s, y) +
+(γ_{1k}^{−1}(s)Q_{1k}(s, y) − γ_{1}^{−1}(s)Q_{1}(s, y))\frac{∂u(s, y)}{∂s}]ds.  (3.68)

In view of conditions (3.61)-(3.63) there exists α > 0 such that

\begin{matrix}∥P_{0k}∥_{L} ≤ α, ∥P_{ik}∥_{L_{∞}} ≤ α (i = 1, 2), \\ ∥Q_{0k}∥_{L} ≤ α, ∥γ_{1k}Q_{1k}∥_{L_{∞}} ≤ α (k = 1, 2, …). \end{matrix}  (3.69)

By virtue of Lemma 3.1 we may consider without loss of generality that

∥u(x, y)∥ ≤ α, ∥\frac{∂u(x, y)}{∂x}∥ ≤ αγ_{1}(x), ∥\frac{∂u(x, y)}{∂y}∥ ≤ αγ_{2}(y). (3.70)

By Lemma 3.6 conditions (3.69) guarantee the existence of a positive constant β such that

∥u_{k} − u∥_{ℂ} ≤ β(∥φ̃_{0k}∥_{ℂ} + ∥ψ̃_{k}∥_{L} + ∥q̃_{k}∥_{L}) (k = 1, 2, …).  (3.71)

However, in view of (3.67),(3.68) and (3.70),

\begin{matrix}∥q̃_{k}∥_{L} ≤ α[∥P_{0k} − P_{0}∥_{L} + ∑_{i=1}^{2} ∥γ_{i}(P_{ik} − P_{i})∥_{L} + ∥q_{k} − q∥_{L}] \\ (k = 1, 2, …) \end{matrix} (3.72)

and

\begin{matrix}∥ψ̃_{k}∥_{L} ≤ ∥ψ_{k} − ψ∥_{L} + a[∥Q_{0k} − Q_{0}∥_{L} + ∥γ_{1}γ_{1k}^{−1}Q_{1k} − Q_{1}∥_{L}] \\ (k = 1, 2, …). \end{matrix} (3.73)

According to conditions (3.61)-(3.63), from estimates (3.71)-(3.73) we get equality (3.64). ■

Lemmas 3.10 and 3.11 below alsi deal with the correctness of problem (3.1),(3.2).

Lemma 3.10. Let conditions (3.61) hold,

sup_{k≥1} ∥P_{0k}∥_{L}^{(2)} < +∞, sup_{k≥1} ∥Q_{0k}∥_{L}^{(2)} < +∞,  (3.74)
lim_{k→+∞} ∥P_{ik} − P_{i}∥_{L}^{(1)} = 0 (i = 0, 1, 2), lim_{k→+∞} ∥q_{k} − q∥_{L}^{(1)} = 0, (3.75)

lim_{k→+∞} ∥φ_{0k} − φ_{0}∥_{ℂ} = 0, lim_{k→+∞} ∥ψ_{k} − ψ∥_{L} = 0,

lim_{k→+∞} ∥Q_{0k} − Q_{0}∥_{L}^{(1)} = 0, lim_{k→+∞} ∥γ_{1}γ_{1k}^{−1}Q_{1k} − Q_{1}∥_{L}^{(1)} = 0. (3.76)

Then

lim_{k→+∞} ∥u_{k} − u∥(1) ; _{ℂ} = 0,  (3.77)

where u and u_{k} are, respectively, the solutions of problems (3.1),(3.2) and (3.57),(3.58).

Lemma 3.11. Let conditions (3.61) take place, φ_{0} and φ_{0k} ∈ ℂ̃_{∞}([0, a]; ℝ^{n}), ψ and ψ_{k} ∈ L_{∞}([0, b]; ℝ^{n}),

\begin{matrix}lim_{k→+∞} ∥P_{ik} − P_{i}∥_{L}^{(2)} = 0 (i = 0, 1, 2), \\ lim_{k→+∞} ∥q_{k} − q∥_{L}^{(2)} = 0, \end{matrix}  (3.78)

lim_{k→+∞} ∥φ_{0k} − φ_{0}∥_{ℂ_{∞}} = 0, lim_{k→+∞} ∥ψ_{k} − ψ∥_{L_{∞}} = 0,

lim_{k→+∞} ∥Q_{0k} − Q_{0}∥_{L}^{(2)} = 0, lim_{k→+∞} ∥Q_{1k} − Q_{1}∥_{L}^{(2)} = 0. (3.79)

Then

lim_{k→+∞} ∥u_{k} − u∥(2) ; _{ℂ} = 0.  (3.80)

To prove the above stated lemmas we need three auxiliary assertions.

Lemma 3.12. Let p_{k} : D_{ab} → ℝ (k = 1, 2, …) be a sequence of summable functions satisfying the condition

lim_{k→+∞} ∥p_{k}∥_{L}^{(i)} = 0  (3.81)

for some i ∈ {1, 2} and let the function z : D_{ab} → ℝ be such that z(·, y) and z(x, ·) are absolutely continuous almost for any y ∈ [0, b] and x ∈ [0, a], respectively, and the inequalities

|\frac{∂z(x, y)}{∂x}| ≤ z_{1}(x), |\frac{∂z(x, y)}{∂y}| ≤ z_{2}(y)  (3.82)

take place almost everywhere in D_{ab}, where z_{1} : [0, a] → ℝ_{+} and z_{2} : [0, b] → → ℝ_{+} are summable functions. Then

lim_{k→+∞} ∥p_{k}z∥_{L}^{(i)} = 0.  (3.83)

Proof. In view of (3.82), from the equalities

∫_{0}^{x} p_{k}(s, y)z(s, y)ds = z(x, y) ∫_{0}^{x} p_{k}(s, y)ds −

− ∫_{0}^{x} ( ∫_{0}^{s} p_{k}(ξ, y)dξ)\frac{∂z(s, y)}{∂s}ds

and

∫_{0}^{y} p_{k}(x, t)z(x, t)dt = z(x, y) ∫_{0}^{y} p_{k}(x, t)dt −
− ∫_{0}^{y} ( ∫_{0}^{t} p_{k}(x, τ)dτ)\frac{∂z(x, t)}{∂t}dt  (3.84)

we have

∥p_{k}z∥_{L}^{(i)} ≤ [ max_{(x,y)∈D_{ab}} ∥z(x, y)∥ + ∫_{0}^{a} z_{1}(s)ds + ∫_{0}^{b} z_{2}(t)dt]∥p_{k}∥_{L}^{(i)},

whence according to (3.81) we have equality (3.83). ■

Lemma 3.13. Let p_{k} : D_{ab} → ℝ (k = 1, 2, …) be a sequence of measurable and essentially bounded functions satisfying conditions (3.81) for some i ∈ {0, 1} and

α = sup_{k≥1} ∥p_{k}∥_{L_{∞}} < +∞.  (3.85)

Then equality (3.83) is valid for any summable function z : D_{ab} → ℝ.

Proof. Since z is summable, there exists a sequence of functions z_{m} : D_{ab} → → ℝ (m = 1, 2, …) such that for any m ≥ 1, j and l ∈ {1, …, m} the function z_{m} is constant in the rectangle (\frac{j−1}{m}a, \frac{j}{m}a) × (\frac{l−1}{m}b, \frac{l}{m}b) and

lim_{m→+∞} ∥z − z_{m}∥_{L} = 0.  (3.86)

By (3.81) and (3.85)

∥p_{k}z∥_{L}^{(i)} ≤ ∥p_{k}(z − z_{m})∥_{L}^{(i)} + ∥p_{k}z_{m}∥_{L}^{(i)} ≤ α∥z − z_{m}∥_{L}^{(i)} + ∥p_{k}z_{m}∥_{L}^{(i)}

and

lim_{k→+∞} ∥p_{k}z_{m}∥_{L}^{(i)} = 0 (m = 1, 2, …).

Therefore

lim sup_{k→+∞} ∥p_{k}z∥_{L}^{(i)} ≤ α∥z − z_{m}∥_{L}^{(i)} (m = 1, 2, …),

from which with regard to (3.86) we get equality (3.83). ■

Lemma 3.14. Let p_{k} : D_{ab} → ℝ (k = 1, 2, …) be a sequence of measurable essentially bounded functions satisfying condition (3.85) and

lim_{k→+∞} ∥p_{k}∥_{L}^{(2)} = 0.  (3.87)

Then for any function z belonging to ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ) or ℂ̃_{∞}^{(0,−1)}(D_{ab}; ℝ) we have

lim_{k→+∞} ∥p_{k}z∥_{L}^{(2)} = 0.  (3.88)

Proof. To be more precise, we assume that z ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ) since the case when z ∈ ℂ̃_{∞}^{(0,−1)}(D_{ab}; ℝ) is considered similarly.

Choose β > 0 such that the inequalities

|z(x, y)| ≤ β, |\frac{∂z(x, y)}{∂y}| ≤ β  (3.89)

hold almost everywhere in D_{ab}. Then from (3.84) we find that

| ∫_{0}^{y} p_{k}(x, t)z(x, t)dt| ≤ (1 + b)β∥p_{k}∥_{L}^{(2)}.  (3.90)

Let ε be an arbitrarily small positive number. Choose a natural m such that

\frac{aαβ}{m} < \frac{ε}{2}.

For every j ∈ {1, …, m} we put

z_{0j}(x) = z(x, \frac{jb}{m})

and choose a step-function z_{j} : [0, a] → ℝ such that

α ∫_{0}^{a} |z_{0j}(s) − z_{j}(s)|ds < \frac{ε}{2}.

Then, taking into account (3.85) and (3.89),

| ∫_{0}^{x} p_{k}(s, y)z(s, y)ds| ≤ ∫_{0}^{x} |p_{k}(s, y)||z(s, y) − z(s, \frac{jb}{m})|ds +

+ ∫_{0}^{x} |p_{k}(s, y)||z_{0j}(s) − z_{j}(s)|ds + | ∫_{0}^{x} p_{k}(s, y)z_{j}(s)ds| ≤

≤ \frac{aαβ}{m} + α ∫_{0}^{a} |z_{0j}(s) − z_{j}(s)|ds + ∥p_{k}z_{j}∥_{L}^{(2)} ≤

≤ ε + ∥p_{k}z_{j}∥_{L}^{(2)} for \frac{j − 1}{m}b ≤ y ≤ \frac{jb}{m}.  (3.91)

It follows from (3.90) and (3.91) that

∥p_{k}z∥_{L}^{(2)} ≤ ε + (1 + b)β∥p_{k}∥_{L}^{(2)} + max_{1≤j≤m} ∥p_{k}z_{j}∥_{L}^{(2)}.

On the other hand, in view of (3.87) we have

lim_{k→+∞} ∥p_{k}z_{j}∥_{L}^{(2)} = 0 (j = 1, …, m).

Therefore

lim sup_{k→+∞} ∥p_{k}z∥_{L}^{(2)} ≤ ε.

From this and in view of the arbitrariness of ε, we get equality (3.88). ■

Proof of Lemma 3.10. For every natural k the vector function v(x, y) = = u_{k}(x, y) − u(x, y) is a solution of problem (3.65),(3.66), where q̃_{k}, φ̃_{0k} and ψ̃_{k} are the vector functions given by equalities (3.67) and (3.68).

According to conditions (3.61),(3.74) and Lemma 3.7 there exists a positive number β such that

∥u_{k} − u∥(1) ; _{ℂ} ≤ βε_{k} (k = 1, 2, …),  (3.92)

where ε_{k} = ∥q̃_{k}∥_{L}^{(1)} + ∥φ_{0k}∥_{ℂ} + ∥ψ̃_{k}∥_{L}. On the other hand, by virtue of Lemma 3.1 inequalities (3.70) take place for some α > 0.

In view of (3.67) and (3.68)

ε_{k} ≤ ∥(P_{0k} − P_{0})u∥_{L}^{(1)} + ∥(P_{1k} − P_{1})w_{1}∥_{L}^{(1)} + ∥(P_{2k} − P_{2})w_{2}∥_{L}^{(1)} +

+∥q_{k} − q∥_{L}^{(1)} + ∥φ_{0k} − φ_{0}∥_{ℂ} + ∥ψ_{k} − ψ∥_{L} +

+∥(Q_{0k} − Q_{0})u∥_{L}^{(1)} + ∥(γ_{1}γ_{1k}^{−1}Q_{1k} − Q_{0})γ_{1}^{−1}w_{1}∥_{L}^{(1)},

where

w_{1}(x, y) = \frac{∂u(x, y)}{∂x}, w_{2}(x, y) = \frac{∂u(x, y)}{∂y},

from which by virtue of conditions (3.61) and (3.74)-(3.76) and Lemmas 3.12 and 3.13 we have

lim_{k→+∞} ε_{k} = 0.

Then taking into account estimate (3.92), we obtain equality (3.77). ■

Proof of Lemma 3.11 proceeds analogously to that of Lemma 3.10. The only difference is that instead of Lemma 3.13 we use Lemma 3.14.

Lemma 3.15. Let P_{1} ∈ ℂ̃_{∞}^{(0,−1)}(D_{ab}; ℝ^{n×n})

sup_{k≥1} ∥P_{0k}∥_{L_{∞}} < +∞, sup_{k≥1} ∥Q_{k}∥_{L_{∞}} < +∞,  (3.93)
lim_{k→+∞} ∥P_{0k} − P_{0}∥_{L}^{(0)} = 0, lim_{k→+∞} ∥q_{k} − q∥_{L}^{(0)} = 0,  (3.94)
lim_{k→+∞} ∥φ_{k} − φ∥_{L} = 0, lim_{k→+∞} ∥ψ_{k} − ψ∥_{L_{∞}} = 0  (3.95)

and

lim_{k→+∞} ess sup_{\begin{matrix}0≤y_{1}<y_{2}≤b, \\ (x,y)∈D_{ab} \end{matrix}} ∥ ∫_{y_{1}}^{y_{2}} ∫_{0}^{x} [Q_{k}(y, s, t) − Q(y, s, t)]dsdt∥ = 0 (3.96)

Then

lim_{k→+∞} ∥u_{k} − u∥_{L_{∞}} = 0,  (3.97)

where u and u_{k} are, respectively, the generalized solutions of problems (3.1), (3.3) and (3.59),(3.60).

Proof. For every natural k the vector function v(x, y) = u_{k}(x, y) − u(x, y) is a generalized solution of the problem

\frac{∂^{2}v(x, y)}{∂x∂y} = P_{0k}(x, y)v(x, y) + P_{1}(x, y)\frac{∂v(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂v(x, y)}{∂y} + q̃_{k}(x, y),

lim_{y→0} (\frac{∂v(x, y)}{∂x} − P_{2}(x, y)v(x, y)) = φ̃_{k}(x),

v(0, y) = ψ̃_{k}(y) + ∫_{0}^{y} ∫_{0}^{a} Q_{k}(y, s, t)v(s, t)dsdt,

where

q̃_{k}(x, y) = [P_{0k}(x, y) − P_{0}(x, y)]u(x, y) + q_{k}(x, y) − q(x, y),
φ̃_{k}(x) = φ_{k}(x) − φ(x), ψ̃_{k}(y) = ψ_{k}(y) − ψ(y) +
+ ∫_{0}^{y} ∫_{0}^{a}[Q_{k}(y, s, t) − Q(y, s, t)]u(s, t)dsdt.

By conditions (3.93) and Lemma 3.8 there exists β > 0 such that

∥u_{k} − u∥_{L_{∞}} ≤ β(∥φ̃_{k}∥_{L} + ∥ψ_{k}∥_{L_{∞}} + ∥q̃_{k}∥_{L}^{(0)}) (k = 1, 2, …). (3.98)

By virtue of conditions (3.93),(3.94) and Lemma 3.13

lim_{k→+∞} ∥q̃_{k}∥_{L}^{(0)} = 0.  (3.99)

On the other hand, in view of (3.93) and (3.96) we have

lim_{k→+∞} sup_{0≤y≤b} ∥ ∫_{0}^{y} ∫_{0}^{a} [Q_{k}(y, s, t) − Q(y, s, t)]u(s, t)dsdt∥ = 0 (3.100)

Taking into account conditions (3.95),(3.99) and (3.100), we obatin equality (3.97) from (3.98). ■

CHAPTER II

BOUNDARY VALUE PROBLEMS IN A RECTANGLE

§ 4. Existence and Uniqueness Theorems

In this chapter for the linear hyperbolic system

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q(x, y)  (4.1)

we study boundary value problems of four types

u(x, 0) = φ_{0}(x), h(\frac{∂u(·, y)}{∂y})(y) = φ_{1}(y),  (4.2)
\frac{∂u(x, 0)}{∂x} − P_{2}(x, 0)u(x, 0) = ψ_{0}(x), h(\frac{∂u(·, y)}{∂y})(y) = ψ_{1}(y), (4.3)
lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), h(u(·, y))(y) = ψ_{1}(y) (4.4)

and

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), \\ h(\frac{∂}{∂y} (u(·, y))(y) − Z_{2}(·, y)u(0, y)))(y) = ψ_{1}(y), \end{matrix}  (4.5)

whose special cases are periodic boundary value problems

u(x, 0) = φ_{0}(x), \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y} + φ_{1}(y),  (4.2_{1})
\frac{∂u(x, 0)}{∂x} − P_{2}(x, 0)u(x, 0) = ψ_{0}(x), \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y} + ψ_{1}(y), (4.3_{1})
\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), \\ u(a, y) = u(0, y) + ψ_{1}(y) \end{matrix}  (4.4_{1})

and

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), \\ \frac{∂}{∂y} (u(a, y) − u(0, y)) = ψ_{1}(y). \end{matrix}  (4.5_{1})

Here and everywhere below, unless stated otherwise, we assume

\begin{matrix}P_{i} ∈ L_{∞}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ L_{∞}(D_{ab}; ℝ^{n}), \\ φ_{0} ∈ ℂ̃([0, a]; ℝ^{n}), ; φ_{1} ∈ L([0, b]; ℝ^{n}), \\ ψ_{0} ∈ L([0, a]; ℝ^{n}), ψ_{1} ∈ L([0, b]; ℝ^{n}) \end{matrix}  (4.6)

and h : ℂ̃([0, a]; ℝ^{n}) → L_{∞}([0, b]; ℝ^{n}) is a linear continuous operator. Additional restrictions on the coefficients of system (4.1) and on the boundary conditions will be given in the theorems formulated below.

By Z_{1} : D_{ab} → ℝ^{n×n} and Z_{2} : D_{ab} → ℝ^{n×n} will be meant solutions of the matrix differential equations

\frac{∂Z_{1}(x, y)}{∂y} = P_{1}(x, y)Z_{1}(x, y)

and

\frac{∂Z_{2}(x, y)}{∂x} = P_{2}(x, y)Z_{2}(x, y),

satisfying the initial conditions

Z_{1}(x, 0) = E for 0 ≤ x ≤ a

and

Z_{2}(0, y) = E for 0 ≤ y ≤ b.

According to Lemmas 2.1_{1} and 2.3_{1},

\begin{matrix}h(v)(y) = H_{0}(y)v(0) + ∫_{0}^{a} H(s, y)v′(s)ds \\ for v ∈ ℂ̃([0, a]; ℝ^{n}), y ∈ [0, b] \end{matrix}  (4.7)

and

\begin{matrix}h(Z_{2}(·, y)v(·))(y) = M_{0}(y)v(0) + ∫_{0}^{a} M(s, y)v′(s)ds \\ for v ∈ ℂ̃([0, a]; ℝ^{n}), y ∈ [0, b], \end{matrix}  (4.8)

where

H_{0} ∈ L_{∞}([0, b]; ℝ^{n×n}), H ∈ L_{∞}(D_{ab}; ℝ^{n×n}),  (4.9)

\begin{matrix}M_{0}(y) = H_{0}(y) + ∫_{0}^{a} H(s, y)\frac{∂Z_{2}(s, y)}{∂s}ds, \\ M(x, y) = H(x, y)Z_{2}(x, y) + ∫_{x}^{a} H(s, y)\frac{∂Z_{2}(s, y)}{∂s}ds \end{matrix}  (4.10)

and

M_{0} ∈ L_{∞}([0, b]; ℝ^{n×n}), M ∈ L_{∞}(D_{ab}; ℝ^{n×n}).  (4.11)

Assume

I_{M_{0}} = {y ∈ [0, b] : det M_{0}(y) ≠ 0}.

4.1. Problem (4.1),(4.2) (case M_{0}(y)≢Θ). Alongside with (4.1),(4.2) we have to consider the homogeneous problem

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} + P_{2}(x, y)\frac{∂u(x, y)}{∂y}, (4.1_{0})
u(x, 0) = 0, h(\frac{∂u(·, y)}{∂y})(y) = 0.  (4.2_{0})

Theorem 4.1. Let the vector and the matrix functions

φ_{1}, MZ_{2}^{−1}q, MZ_{2}^{−1}P_{0}, (1 + ∥φ′_{0}∥)MZ_{2}^{−1}P_{1}  (4.12)

be M_{0}-summable. Then problem (4.1),(4.2) is solvable and its solution is unique if and only if

mes I_{M_{0}} = b.  (4.13)

Moreover, if condition (4.13) is violated, then the space of solutions of homogeneous problem (4.1_{0}),(4.2_{0}) is infinite dimensional.

Proof. According to the M_{0}-summability of vector and matrix functions (4.12), there exist

φ_{10} ∈ L([0, b]; ℝ^{n}), q_{0} ∈ L(D_{ab}; ℝ^{n}), Q_{i} ∈ L(D_{ab}; ℝ^{n×n}) (i = 0, 1)

such that

\begin{matrix}φ_{1}(y) = M_{0}(y)φ_{10}(y), M(x, y)Z_{2}^{−1}(x, y)q(x, y) = −M_{0}(y)q_{0}(x, y), \\ M(x, y)Z_{2}^{−1}(x, y)P_{0}(x, y) = −M_{0}(y)Q_{0}(x, y), \\ M(x, y)Z_{2}^{−1}(x, y)P_{1}(x, y) = −γ_{1}^{−1}(x)M_{0}(y)Q_{1}(x, y), \end{matrix} (4.14)

where γ_{1}(x) = 1 + ∥φ′_{0}(x)∥.

Let u be an arbitrary solution of system (4.1). Then

\frac{∂u(x, y)}{∂y} = Z_{2}(x, y)(\frac{∂u(0, y)}{∂y} + ∫_{0}^{x} Z_{2}^{−1}(s, y)[P_{0}(s, y)u(s, y) +

+P_{1}(s, y)\frac{∂u(s, y)}{∂s} + q(s, y)]ds),  (4.15)

whence, in view of representation (4.8), it is clear that the boundary condition

h(\frac{∂u(·, y)}{∂y})(y) = φ_{1}(y)  (4.16)

holds if and only if

φ_{1}(y) = M_{0}(y)\frac{∂u(0, y)}{∂y} + ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)(P_{0}(s, y)u(s, y) +

+P_{1}(s, y)\frac{∂u(x, y)}{∂s} + q(s, y))ds for 0 ≤ y ≤ b.

By (4.14) from the above equality we have

M_{0}(y)\frac{∂u(0, y)}{∂y} = M_{0}(y)(ψ(y) +

+ ∫_{0}^{a} [Q_{0}(s, y)u(s, y) + γ_{1}^{−1}(s)Q_{1}(s, y)\frac{∂u(s, y)}{∂s}]ds),

where

ψ(y) = φ_{10}(y) + ∫_{0}^{a} q_{0}(s, y)ds.

Consequently, for conditions (4.2) to be fulfilled, it is sufficient, and when (4.13) holds, it is necessary that

\begin{matrix}u(x, 0) = φ_{0}(x), \frac{∂u(0, y)}{∂y} = ψ(y) + \\ + ∫_{0}^{a} [Q_{0}(s, y)u(s, y) + γ_{1}^{−1}(s)Q_{1}(s, y)\frac{∂u(s, y)}{∂s}]ds. \end{matrix}  (4.17)

However, according to Lemma 3.4, problem (4.1),(4.17) has the unique solution. Thus we have proved that problem (4.1),(4.2) is solvable and its solution is unique in the case if condition (4.13) is fulfilled.

For completion of the proof it remains to show that if

mes I_{M_{0}} < b,  (4.18)

then homogeneous problem (4.1_{0}),(4.2_{0}) has an infinite dimensional space of solutions.

In view of (4.18) there exists a measurable function c_{0} : [0, b] → ℝ^{n} such that

∥c_{0}(y)∥ = 1 for y ∉ I_{M_{0}}, c_{0}(y) = 0 for y ∈ I_{M_{0}}  (4.19)

and

M_{0}(y)c_{0}(y) = 0 for y ∈ [0, b].

As shown above, the solution u of system (4.1_{0}) satisfies the condition

h(\frac{∂u(·, y)}{∂y})(y) = 0

if and only if

M_{0}(y)\frac{∂u(0, y)}{∂y} = M_{0}(y) ∫_{0}^{a} [Q_{0}(s, y)u(s, y) + γ_{1}^{−1}(s)Q_{1}(s, y)\frac{∂u(s, y)}{∂s}]ds.

Therefore it is clear that for all natural k the solution of system (4.1_{0}) satisfying boundary conditions

\begin{matrix}u(x, 0) = 0, \frac{∂u(0, y)}{∂y} = y^{k−1}c_{0}(y) + ∫_{0}^{a} [Q_{0}(s, y)u(s, y) + \\ +γ_{1}^{−1}(s)Q_{1}(s, y)\frac{∂u(s, y)}{∂s}]ds \end{matrix} (4.20)

is the solution of problem (4.1_{0}), (4.2_{0}).

According to Lemma 3.4, problem (4.1_{0}),(4.20) has the unique solution which we denote by u_{k}. It follows immediately from (4.18) and (4.19) that functions u_{k} (k = 1, 2, …) are linearly independent. ■

Corollary 4.1. If condition (4.13) holds and

∫_{0}^{a} ∫_{0}^{b} ∥M_{0}^{−1}(y)∥(∥P_{0}(x, y)∥ + (1 + ∥φ′_{0}(x)∥)∥P_{1}(x, y)∥ +
+∥q(x, y)∥ + ∥φ_{1}(y)∥)dxdy < +∞,  (4.21)

then problem (4.1),(4.2) has one and only one solution.

Proof. According to conditions (4.6),(4.11) and Lemma 2.2_{1},

MZ_{2}^{−1} ∈ L_{∞}(D_{ab}; ℝ^{n}).

If we take this and conditions (4.21) into consideration, then it becomes evident that

M_{0}^{−1}φ_{1} ∈ L([0, b]; ℝ^{n}), M_{0}^{−1}MZ_{2}^{−1}q ∈ L(D_{ab}; ℝ^{n}),
M_{0}^{−1}MZ_{2}^{−1}P_{0} ∈ L(D_{ab}; ℝ^{n×n}),
(1 + ∥φ′_{0}∥)M_{0}^{−1}MZ_{2}^{−1}P_{1} ∈ L(D_{ab}; ℝ^{n×n}).

Thus, the vector and the matrix functions (4.12) are M_{0}-summable. ■

Remark 4.1. In the above proven corollary condition (4.21) is essential and we cannot neglect it. To convince ourselves that is so consider the problem

\frac{∂^{2}u(x, y)}{∂x∂y} = y \frac{∂u(x, y)}{∂y} + 1,

u(x, 0) = 0, \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y}

for which all conditions of Theorem 4.1, except of (4.21), are fulfilled because

M_{0}(y) = exp(ay) − 1.

Let us show that this problem has no solution. In fact, otherwise we should have

\frac{∂u(x, y)}{∂y} = exp(xy)\frac{∂u(0, y)}{∂y} + \frac{exp(xy) − 1}{y} for 0 < y ≤ b,

from which according to the condition

\frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y}

it follows that

\frac{∂u(0, y)}{∂y} = −\frac{1}{y} for 0 < y ≤ b

and

\frac{∂u(x, y)}{∂y} = −\frac{1}{y} for 0 < y ≤ b.

But this contradicts the absolute continuity of u.

Theorem 4.1′. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ([0, b]; ℝ^{n}) be a linear continuous operator,

\begin{matrix}P_{i} ∈ ℂ(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ ℂ(D_{ab}; ℝ^{n}), \\ φ_{0} ∈ ℂ^{1}([0, a]; ℝ^{n}), φ_{1} ∈ ℂ([0, b]; ℝ^{n}). \end{matrix}  (4.6′)

Moreover, let the vector and the matrix functions MZ_{2}^{−1}q and MZ_{2}^{−1}P_{i}(i = = 0, 1), respectively, satisfy the Carathéodory condition with M_{0}-weight, and let φ_{1} be M_{0}-continuous. Then problem (4.1),(4.2) has at least one classical solution but for its uniqueness it is necessary and sufficient that

Ī_{M_{0}} = [0, b].  (4.13′)

When condition (4.13′) is violated the space of classical solutions of problem (4.1_{0}),(4.2_{0}) is infinite dimensional.

Proof. According to Lemma 2.3_{2} and the restrictions imposed on φ_{1}, MZ_{2}^{−1}q and MZ_{2}^{−1}P_{i} (i = 0, 1),

M_{0} ∈ ℂ([0, b]; ℝ^{n×n})  (4.22)

and the representations

φ_{1}(y) = M_{0}(y)φ_{10}(y), M(x, y)Z_{2}^{−1}(x, y)q(x, y) = −M_{0}(y)q_{0}(x, y),
M(x, y)Z_{2}^{−1}(x, y)P_{i}(x, y) = −M_{0}(y)Q_{i}(x, y) (i = 0, 1)

are valid, where φ_{10} ∈ ℂ([0, b]; ℝ^{n}), while q_{0} ∈ L(D_{ab}; ℝ^{n}) and Q_{i} ∈ ∈ L(D_{ab}; ℝ^{n×n}) (i = 0, 1) satisfy the Carathéodory condition.

Let u be an arbitrary classical solution of system (4.1). It satisfies condition (4.16) if and only if

M_{0}(y)\frac{∂u(0, y)}{∂y} = M_{0}(y)(ψ(y) +

+ ∫_{0}^{a} [Q_{0}(s, y)u(s, y) + Q_{1}(s, y)\frac{∂u(s, y)}{∂s}]ds),

where

ψ(y) = φ_{10}(y) + ∫_{0}^{a} q_{0}(s, y)ds,

whence, in view of the continuity of \frac{∂u(0,y)}{∂y}, it becomes clear that for boundary conditions (4.2) to be fulfilled, it is sufficient, and when condition (4.13′) holds, it is necessary that

\begin{matrix}u(x, 0) = φ_{0}(x), \\ \frac{∂u(0, y)}{∂y} = ψ(y) + ∫_{0}^{a} [Q_{0}(s, y)u(s, y) + Q_{1}(s, y)\frac{∂u(s, y)}{∂s}]ds. \end{matrix} (4.23)

But according to Lemma 3.4′, problem (4.1),(4.23) has the unique solution u and this solution is classical. Thus, problem (4.1),(4.2) has at least one classical solution and when (4.13′) holds the solution is unique.

Suppose that condition (4.13′) is not fulfilled. Then, according to (4.22), there exist an interval [b_{1}, b_{2}] ⊂ (0, b) and a continuous vector function c_{0} : [0, b] → ℝ^{n} such that

∥c_{0}(y)∥ > 0 for b_{1} < y < b_{2}, c_{0}(y) = 0 for y ∉ (b_{1}, b_{2})

and

M_{0}(y)c_{0}(y) = 0 for y ∈ [0, b].

Now, repeating the same arguments as in the proof of the second part of Theorem 4.1 and applying Lemma 3.4′ instead of Lemma 3.4, we convince ourselves that homogeneous problem (4.1_{0}),(4.2_{0}) has a countable system of classical solutions. ■

Remark 4.2. In Theorem 4.1′ the requirements for φ_{1} to be M_{0}-continuo-us and for the vector and the matrix functions MZ_{2}^{−1}q and MZ_{2}^{−1}P_{i} (i=0,1) to satisfy the Carathéodory conditions with M_{0} weight are the most essential ones. The examples below show that if at least one of these requirements is not fulfilled, then problem (4.1),(4.2) has no classical solution despite the fact that the coefficients of system (4.1) and the functions given in boundary conditions (4.2) are smooth.

Consider the boundary value problems

\frac{∂^{2}u(x, y)}{∂x∂y} = |y − b_{0}|^{2k−1} \frac{∂u(x, y)}{∂y} − x(y − b_{0})^{2k−1}|y − b_{0}|^{2k−1}, (4.24_{1})
u(x, 0) = b_{0} + \frac{x}{2k} b_{0}^{2k}, \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y} + a(y − b_{0})^{2k−1}; (4.25_{1})
\frac{∂^{2}u(x, y)}{∂x∂y} = |y − b_{0}|^{2k−1} \frac{∂u(x, y)}{∂y} + (y − b_{0})^{2k−1},  (4.24_{2})
u(x, 0) = 0, \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y}  (4.25_{2})

\begin{matrix}\frac{∂^{2}u(x, y)}{∂x∂y} = |y − b_{0}|^{2k−1} \frac{∂u(x, y)}{∂y} + \frac{1}{b_{0}} (y − b_{0})^{2k−1}u(x, y) + \\ +\frac{1}{b_{0}} |y − b_{0}|(y − b_{0})^{2k−1}, \end{matrix} (4.24_{3})

u(x, 0) = 0, \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y}  (4.25_{3})

and

\frac{∂^{2}u(x, y)}{∂x∂y} = |y − b_{0}|^{2k−1} \frac{∂u}{∂y} − exp(−x)(y − b_{0})^{2k−1} \frac{∂u}{∂x}, (4.24_{4})
u(x, 0) = exp(x) + b_{0}, \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y},  (4.25_{4})

where b_{0} ∈ (0, b) and k is a natural number. For each of these problems (4.24i), (4.25_{i}) (i = 1, 2, 3, 4) we have

h(v)(y) ≡ v(a) − v(0), Z_{2}(x, y) = exp(x|y − b_{0}|^{2k−1}),
M_{0}(y) = exp(a|y − b_{0}|^{2k−1}) − 1,
M(x, y) = exp(a|y − b_{0}|^{2k−1}), I_{M_{0}} = [0, b]\{b_{0}}.

Proceeding from this, it is easy to convince ourselves that for these problems all conditions of Theorem 4.1 are fulfilled and therefore they are uniquely solvable. The following functions

u_{1}(x, y) = |y − b_{0}| + \frac{x}{2k} (y − b_{0})^{2k}, u_{2}(x, y) = b_{0} − |y − b_{0}|,

u_{3}(x, y) = b_{0} − |y − b_{0}|, u_{4}(x, y) = exp(x) + |y − b_{0}|

are respectively the solutions of problems (4.24_{i}), (4.25_{i}) (i = 1, 2, 3, 4). But none of these solutions is classical, because they have no partial derivative in the second argument when y = b_{0}. This case is due to the fact that for each of these problems one (and only one) of the conditions of Theorem 4.1′ is violated.

More precisely, for problem (4.24_{1}),(4.25_{1}) the function

φ_{1}(y) = a(y − b_{0})^{2k−1}

is not M_{0}-continuous, for problem (4.24_{2}),(4.25_{2}) the function

M(x, y)Z_{2}^{−1}(x, y)q(x, y) = exp ((a − x)|y − b_{0}|^{2k−1})(y − b_{0})^{2k−1}

does not satisfy the Carathéodory condition with M_{0} weight, for problem (4.24_{3}), (4.25_{3}) the function

M(x, y)Z_{2}^{−1}(x, y)P_{0}(x, y) = \frac{1}{b_{0}} exp ((a − x)|y − b_{0}|^{2k−1})(y − b_{0})^{2k−1}

and for problem (4.24_{4}),(4.25_{4}) the function

M(x, y)Z_{2}^{−1}(x, y)P_{1}(x, y) = exp ((a − x)|y − b_{0}|^{2k−1} − x)(y − b_{0})^{2k−1} behaves similarly.

Remark 4.3. Let all conditions of Theorem 4.1′ hold. If, in addition, condition (4.13) takes place, then problem (4.1),(4.2) has the unique solution and it is classical. If condition (4.13) is violated, then this problem has the unique classical solution and an infinite set of absolutely continuous solutions.

As an example let us consider the problem

\frac{∂^{2}u(x, y)}{∂x∂y} = p(y)\frac{∂u(x, y)}{∂y}  (4.26)
u(x, 0) = 0, \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y},  (4.27)

where p : [0, b] → [0, +∞) is a continuous function with the set of zeros J_{p} which is nowhere dense in [0, b] and has a positive measure. In this case

M_{0}(y) = exp(xp(y)) − 1

and

I_{M_{0}} = [0, b]\J_{p}.

Consequently, for problem (4.26),(4.27) all conditions of Theorem 4.1′ are fulfilled, but condition (4.13) is violated. Therefore u_{0}(x, y) ≡ 0 is the unique classical solution of the problem under consideration. On the other hand, for every summable function c : [0, b] → ℝ the function

u_{c}(x, y) = ∫_{0}^{y} δ(t)c(t) exp(xp(t))dt,

where

δ(t) = \begin{cases}1 & for t ∈ J_{p} \\ 0 & for t ∉ J_{p} \end{cases}

is an absolutely continuous solution of problem (4.26),(4.27).

Corollary 4.1′. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) be a linear continuous operator, P_{j} (i = 0, 1, 2), q and φ_{i} (i = 0, 1) satisfy conditions (4.6′) and

I_{M_{0}} = [0, b].  (4.13′′)

Then problem (4.1),(4.2) has one and only one solution u and it is classical.

Proof. By virtue of Lemmas 2.1_{3} and 2.2_{1},

H_{0} ∈ ℂ̃_{∞}([0, b]; ℝ^{n×n}), H ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}),

Z_{2} and \frac{∂Z_{2}}{∂x} ∈ ℂ(D_{ab}; ℝ^{n×n}),

whence by equalities (4.10) and conditions (4.6′) and (4.13′′) it follows that φ_{1} is M_{0}-continuous and MZ_{2}^{−1}q and MZ_{2}^{−1}P_{i} (i = 0, 1) satisfy the Carathéodory condition with M_{0} weight. Consequently, all conditions of Theorem 4.1′ take place. Moreover, instead of (4.13′) we have stronger restriction (4.13′′). Therefore, by virtue of Theorems 4.1 and 4.1′, problem (4.1),(4.2) has the unique solution and it is classical. ■

Remark 4.4. When the matrix function P_{2} satisfies the Lappo-Danilevsky condition in the first argument, i.e. when

\begin{matrix}P_{2}(x, y)( ∫_{0}^{x} P_{2}(s, y)ds) = ( ∫_{0}^{x} P_{2}(s, y)ds)P_{2}(x, y) \\ for (x, y) ∈ D_{ab}, \end{matrix}  (4.28)

we have

Z_{2}(x, y) = exp ( ∫_{0}^{x} P_{2}(s, y)ds).

Therefore the matrix function M_{0} is calculated explicitly and conditions (4.13) and (4.21) may be verified more or less effectively. But if condition (4.28) is violated, then to verify these conditions effectively, we shall have to apply Lemmas 2.7-2.10.

Let us introduce the notation

A_{0}(s, x, y) = Θ, A_{1}(s, x, y) = P_{2}(s, y),
A_{j+1}(s, x, y) = ∫_{s}^{x} P_{2}(ξ, y)A_{j}(s, ξ, y)dξ (j = 1, 2, …),
M_{00}(y) = H_{0}(y), M_{0j}(y) = H_{0}(y) +

+ ∫_{0}^{a} H(s, y)P_{2}(s, y)[E + ∑_{i=0}^{j−1} ∫_{0}^{s} A_{i}(ξ, s, y)dξ]ds (j = 1, 2, …).

If the inequality

det M_{0k−1}(y) ≠ 0

holds for some y ∈ [0, b] and natural k, then for every natural m we assume

B_{1m}(s, x, y) = [E + ∑_{i=0}^{m−1} ∫_{0}^{x} A_{i}(ξ, x, y)dξ] ×

× H_{0}^{−1}(y)H(s, y)P_{2}(s, y),

B_{km}(s, x, y) = [E + ∑_{i=0}^{m−1} ∫_{0}^{x} A_{i}(ξ, x, y)dξ]M_{0k−1}^{−1}(y) ×

× ∫_{s}^{a} H(ξ, y)P_{2}(ξ, y)A_{k−1}(s, ξ, y)dξ for k > 1

and

B_{km}^{0}(y) = max_{0≤x≤a} [ ∫_{0}^{x} |A_{m}(s, x, y) − B_{km}(s, x, y)|ds +

+ ∫_{x}^{a} |B_{km}(s, x, y)|ds].

From Lemma 2.7 and Corollaries 4.1 and 4.1′ there follow the following assertions.

Corollary 4.2. Let there exist natural k and m such that the inequalities

det M_{0k−1}(y) ≠ 1, r(B_{km}^{0}(y)) < 1  (4.29)

hold almost everywhere in [0, b] and

∫_{0}^{a} ∫_{0}^{b} ∥(E − B_{km}^{0}(y))^{−1}∥ ∥M_{0k−1}^{−1}(y)∥(∥P_{0}(x, y)∥ +
+(1 + ∥φ′_{0}(x)∥)∥P_{1}(x, y)∥ + ∥q(x, y)∥ + ∥φ_{1}(y)∥)dxdy < +∞.

Then Problem (4.1),(4.2) has one and only one solution.

Corollary 4.2′. Let h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) be a linear continuous operator and φ_{i}, P_{i} (i = 0, 1) and q satisfy conditions (4.6′). Moreover, let inequality (4.29) hold everywhere in [0, b]. Then problem (4.1),(4.2) has one and only one solution and this solution is classical.

From Corollaries 4.1 and 4.1′ according to Lemmas 2.8, 2.9 and 2.10 we obtain respectively Corollaries 4.3 and 4.3′, 4.4 and 4.4′ and 4.5 and 4.5′.

Corollary 4.3. Let

h(v)(y) = (v_{i}(a_{i}(y)))_{i=1}^{n},  (4.30)

where a_{i} : [0, b] → [0, a] (i = 1, …, n) are measurable functions. Moreover, let the matrix function

A(y) = ess sup_{0≤x≤a} |P_{2}(x, y)|

satisfy the inequality

r(A(y)) < \frac{π}{2a}  (4.31)

almost everywhere in [0, b], and

∫_{0}^{a} ∫_{0}^{b} ∥(E − \frac{2a}{π}A(y))^{−1}∥(∥P_{0}(x, y)∥ +

+(1 + ∥φ′_{0}(x)∥)∥P_{1}(x, y)∥ + ∥q(x, y)∥ + ∥φ_{1}(y)∥)dxdy < +∞. (4.32)

Then problem (4.1),(4.2) has one and only one solution.

Corollary 4.3′. Let conditions (4.6′) and (4.30) hold, where a_{i} : [0, b] → → [0, a] (i = 1, …, n) are continuous functions. Let, moreover, the matrix function

A(y) = max_{0≤x≤a} |P_{(}x, y)|

be such that

r(A(y)) < \frac{π}{2a} for 0 ≤ y ≤ b.  (4.33)

Then problem (4.1),(4.2) has one and only one solution and this solution is classical.

Remark 4.5. Condition (4.33) is optimal in the sense that it is impossible to replace it by the requirement that inequality (4.31) be fulfilled evrywhere in [0, b] except at some finite number of points and that integral (4.32) be convergent. To convince ourselves that this is so consider the problem

\begin{matrix}\frac{∂^{2}u_{1}(x, y)}{∂x∂y} = \frac{∂u_{2}(x, y)}{∂y}, \\ \frac{∂^{2}u_{2}(x, y)}{∂x∂y} = −ω_{α}^{2}(y)\frac{∂u_{1}(x, y)}{∂y} + xω_{α}^{2}(y), \end{matrix}  (4.34)

u_{i}(x, 0) = 0 (i = 1, 2), \frac{∂u_{1}(0, y)}{∂y} = \frac{∂u_{2}(a, y)}{∂y} = 0,  (4.35)

where

ω_{α}(y) = \frac{π}{2a}[1 − (\frac{y}{b} )^{α}], α ∈ (0, 1].

In that case

P_{2}(x, y) = ( \begin{matrix}0 & 1 \\ −ω_{α}^{2}(y) & 0 \end{matrix} ) , A(y) = ( \begin{matrix}0 & 1 \\ ω_{α}^{2}(y) & 0 \end{matrix} )

and

r(A(y)) = ω_{α}(y).

Consequently, all conditions in Corollary 4.3′ except (4.33) hold. Instead of condition (4.33) we have

r(A(y)) < \frac{π}{2a} for 0 < y ≤ b, r(A(0)) = \frac{π}{2a}.

In addition,

(E − \frac{2a}{π}A(y))^{−1} = (\frac{y}{b} )^{−α}[2 − (\frac{y}{b} )^{α}]^{−1} ( \begin{matrix}1 & \frac{2a}{π} \\ \frac{2a}{π}ω_{α}^{2}(y) & 1 \end{matrix} )

for 0 < y ≤ b.

From the above it becomes evident that condition (4.32) holds for α ∈ (0, 1) and is not fulfilled for α = 1. According to Corollary 4.3, problem (4.1),(4.2) has the unique solution (u_{i})_{i=1}^{2} for every α ∈ (0, 1). By immediate verification we convince ourselves that

u_{1}(x, y) = − ∫_{0}^{y} \frac{sin(ω_{α}(t)x)}{ω_{α}(t) cos(aω_{α}(t))}dt + xy,
u_{2}(x, y) = − ∫_{0}^{y} \frac{cos(xω_{α}(t))}{cos(aω_{α}(t))}dt + y,

and this solution is not classical because

lim_{y→0} \frac{∂u_{i}(x, y)}{∂y} = −∞ for 0 < x < a (i = 1, 2).

Remark 4.6. As been admitted above, for problem (4.34),(4.35) all conditions of Corollary 4.3 except (4.32) take place for α = 1. Let us show that in this case problem (4.34),(4.35) has no solution. Suppose the converse holds, i.e. the problem does have a solution (u_{i})_{i=1}^{2}. Assume

z(x, y) = \frac{∂u_{1}(x, y)}{∂y}.

Then for every y ∈ (0, b) we shall have

\frac{∂^{2}z(x, y)}{∂x^{2}} = −ω_{1}^{2}(y)z(x, y) + xω_{1}^{2}(y),

z(0, y) = 0, \frac{∂z(x, y)}{∂x}|_{x=a} = 0.

Therefore

z(x, y) = −\frac{1}{ω_{1}(y) cos(aω_{1}(y))} sin(ω_{1}(y)x) + x

and

\frac{∂u_{1}(x, y)}{∂y} = −\frac{1}{ω_{1}(y) sin (\frac{πy}{2b} )} sin(ω_{1}(y)x) + x.

But this is impossible because for every x ∈ (0, a] the function in the righthand side is not summable in the second argument on the interval [0, b]. The obtained contradiction proves that problem (4.34),(4.35) has no solution. Thus it is impossible to omit condition (4.32) from Corollary 4.3.

Remark 4.7. When α = 1 problem (4.34),(4.35) is an example of a problem of type (4.1),(4.2) which is unsolvable although the corresponding homogeneous problem has only the trivial solution.

Corollary 4.4. Let there exist a diagonal matrix function A_{0} ∈ L_{∞}(D_{ab}; ℝ^{n×n}) such that

det (A_{0}(x, y)) ≠ 0  (4.36)

almost for every (x, y) ∈ D_{ab} and the inequality

r(A(y)) < 1  (4.37)

holds almost everywhere in [0, b], where

A(y) = ess sup_{0≤x≤a} |A_{0}^{−1}(x, y)P_{2}(x, y) − E|.

Let, moreover, every diagonal element of the matrix A_{0}(·, y) be a function of constant sign almost for every y ∈ [0, b] and

∫_{0}^{a} ∫_{0}^{b} ∥(E − A(y))^{−1}( ∫_{0}^{a} A_{0}(s, y)ds)^{−1}∥(∥P_{0}(x, y)∥ +
+(1 + ∥φ′_{0}(x)∥)∥P_{1}(x, y)∥ + ∥q(x, y)∥ + ∥φ_{1}(y)∥)dxdy < +∞. (4.38)

Then problem, (4.1), (4.2_{1}) has one and only one solution.

Corollary 4.4′. Let conditions (4.6′) hold and there exist a diagonal matrix function A_{0} ∈ ℂ(D_{ab}; ℝ^{n×n}) such that

det (A_{0}(x, y)) ≠ 0 for (x, y) ∈ D_{ab}  (4.39)

and

r(A(y)) < 1 for 0 ≤ y ≤ b,  (4.40)

where

A(y) = max_{0≤x≤a} |A_{0}^{−1}(x, y)P_{2}(x, y) − E|.

Then problem (4.1),(4.2_{1}) has one and only one solution and this solution is classical.

Remark 4.8. The example considered in Remark 4.1 shows that it is impossible to omit condition (4.38) from Corollary 4.4 and to replace condition (4.39) by the requirement for inequality (4.36) be fulfilled everywhere in D_{ab} except for one segment {(x, y_{0}) : 0 ≤ x ≤ a} for some y_{0} ∈ [0, b].

Remark 4.9. Condition (4.40) is optimal in the sense that if it is violated even at one point while all other conditions of Corollary 4.4′ and condition (4.38) hold, problem (4.1),(4.2_{1}) may have no classical solution. As an example verifying this fact, consider the problem

\begin{matrix}\frac{∂^{2}u_{1}(x, y)}{∂x∂y} = \frac{∂u_{1}(x, y)}{∂y} + (1 − ε_{α}(y))^{2} \frac{∂u_{2}(x, y)}{∂y} − ε_{α}(y), \\ \frac{∂^{2}u_{2}(x, y)}{∂x∂y} = \frac{∂u_{1}(x, y)}{∂y} + \frac{∂u_{2}(x, y)}{∂y} + 1, \end{matrix} (4.41)

u_{i}(x, 0) = 0, \frac{∂u_{i}(0, y)}{∂y} = \frac{∂u_{i}(a, y)}{∂y} (i = 1, 2),  (4.42)

where

ε_{α}(y) = (1 − \frac{y}{b} )^{α}, α ∈ (0, +∞).

In that case

P_{2}(x, y) = ( \begin{matrix}1 & (1 − ε_{α}(y))^{2} \\ 1 & 1 \end{matrix} ) , A_{0}(x, y) = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) ,

A(y) = ( \begin{matrix}0 & (1 − ε_{α}(y))^{2} \\ 1 & 0 \end{matrix} )

and

r(A(y)) = 1 − ε_{α}(y),

whence it is clear that all conditions of Corollary 4.4′ except (4.40) take place. Instead of (4.40) we have

r(A(y)) < 1 for 0 ≤ y < b, r(A(b)) = 1.

Besides,

(E − A(y))^{−1} = \frac{1}{ε_{α}(y)(2 + ε_{α}(y))} ( \begin{matrix}1 & (1 − ε_{α}(y))^{2} \\ 1 & 1 \end{matrix} ) .

Consequently, condition (4.38) holds for α ∈ (0, 1) while this condition is violated for α ≥ 1. Let us show that for α ∈ (0, 1) the solution of problem (4.41),(4.42), whose existence and uniqueness follow from Corollary 4.4, is not classical and for α ≥ 1 this problem has no solution whatsoever. Indeed, let (u_{i})_{i=1}^{2} be the solution of problem (4.41),(4.42). Assume

z(x, y) = \frac{∂u_{1}(x, y)}{∂y} + (ε_{α}(y) − 1)\frac{∂u_{2}(x, y)}{∂y}.

Then

\frac{∂z(x, y)}{∂x} = ε_{α}(y)z(x, y) − 1,

z(0, y) = z(a, y).

Therefore

z(x, y) = \frac{1}{ε_{α}(y)} for 0 ≤ x ≤ a, 0 ≤ y < b

and

lim_{y→b} z(x, y) = +∞ for 0 ≤ x ≤ a.

Besides, for α ≥ 1 the function z is not summable in D_{ab}, which is impossible due to the absolute continuity of the functions u_{i} (i = 1, 2).

Corollary 4.5. Let P_{2}(x, y) = (p_{2ij}(x, y))_{i,j=1}^{n} and there exist functions σ_{i} : [0, b] → {−1, 1} (i = 1, …, n) such that the real parts of eigenvalues of the matrix

A(y) = (a_{ij}(y))_{i,j=1}^{n},

where

a_{ii}(y) = ess sup_{0≤x≤a}(σ_{i}(y)p_{2ii}(x, y)), a_{ij}(y) = ess sup_{0≤x≤a} |p_{2ij}(x, y)|

for (i ≠ j; i, j = 1, …, n)

are negative almost for all y ∈ [0, b] and

∫_{0}^{b} ∥A^{−1}(y)∥(1 + ∥φ_{1}(y)∥)dy < +∞.  (4.43)

Then problem (4.1),(4.2_{1}) has one and only one solution.

Corollary 4.5′. Let conditions (4.6′) hold, P_{2}(x, y) = (p_{2ij}(x, y))_{i,j=1}^{n} and there exist numbers σ_{i} ∈ {−1, 1} (i = 1, …, n) such that the real parts of eigenvalues of the matrix

A(y) = (a_{ij}(y))_{i,j}^{n},  (4.44)

where

a_{ii}(y) = max_{0≤x≤a}(σ_{i}p_{2ii}(x, y)), a_{ij}(y) = max_{0≤x≤a} |p_{2ij}(x, y)|

for (i ≠ j; i, j = 1, …, n)

are negative for all y ∈ [0, b]. Then problem (4.1),(4.2_{1}) has one and only one solution and this solution is classical.

Remark 4.10. The restriction in Corollary 4.5′ imposed on the eigenvalues of matrix (4.44) is optimal and cannot be weakened. As an example let us consider problem (4.41),(4.42). If σ_{1} = σ_{2} = −1, then matrix (4.44) for system (4.41) has the form

A(y) = ( \begin{matrix}−1 & (1 − ε_{α}(y))^{2} \\ 1 & −1 \end{matrix} )

and its eigenvalues

λ_{1}(y) = ε_{α}(y) − 2, λ_{2}(y) = −ε_{α}(y)

satisfy the conditions

λ_{1}(y) < 0 for 0 ≤ y ≤ b, λ_{2}(y) = −ε_{α}(y) 0 ≤ y < b, λ_{2}(b) = 0

and condition (4.43) also holds for α ∈ (0, 1). Nevertheless, problem (4.41), (4.42) has no classical solution for any α > 0.

4.2. Problems (4.1),(4.2) and (4.1),(4.3) (case M_{0}(y) ≡ Θ).

Theorem 4.2. Let

P_{i} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}), (4.45)
φ_{0} ∈ ℂ̃([0, a]; ℝ^{n}), φ_{1} ∈ ℂ̃([0, b]; ℝ^{n}),  (4.46)

and h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) be a linear continuous operator such that

M_{0}(y) = Θ for 0 ≤ y ≤ b  (4.47)

and

\begin{matrix}det ( ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)[P_{0}(s, y) + P_{1}(s, y)P_{2}(s, y)]Z_{2}(s, y)ds)≠0 \\ for 0 ≤ y ≤ b. \end{matrix}(4.48)

Then for the unique solvability of problem (4.1),(4.2) it is necessary and sufficient that

∫_{0}^{a} M(s, 0)Z_{2}^{−1}(s, 0)[P_{0}(s, 0)φ_{0}(s) + P_{1}(s, 0)φ′_{0}(s) +
+q(s, 0)]ds = φ_{1}(0).  (4.49)

Proof. Let u be an arbitrary solution of system (4.1). Then by Lemma 3.3,

u(x, y) = Z_{2}(x, y)[u(0, y) + ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)φ(s)ds] + Z_{2}(x, y) ×

× ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)(P(s, t)u(s, t) + q(s, t))ds dt (4.50)

and

\frac{∂}{∂x}(Z_{2}^{−1}(x, y)u(x, y)) = Z_{2}^{−1}(x, y)Z_{1}(x, y) ×

×[φ(x) + ∫_{0}^{y} Z_{1}^{−1}(x, t)(P(x, t)u(x, t) + q(x, t))dt],  (4.51)

where

φ(x) = \frac{∂u(x, 0)}{∂x} − P_{2}(x, 0)u(x, 0)  (4.52)

and

P(x, y) = P_{0}(x, y) + P_{1}(x, y)P_{2}(x, y) − \frac{∂P_{2}(x, y)}{∂y}.

In view of (4.50), from the obvious equality

Z_{2}^{−1}(x, y)\frac{∂u(x, y)}{∂y} = \frac{∂}{∂y} (Z_{2}^{−1}(x, y)u(x, y)) − \frac{∂Z_{2}^{−1}(x, y)}{∂y}u(x, y)

we obtain

Z_{2}^{−1}(x, y)\frac{∂u(x, y)}{∂y} = \frac{∂u(0, y)}{∂y} + ∫_{0}^{x} Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)φ(s)ds +

+ ∫_{0}^{x} Z_{2}^{−1}(s, y)(P̄(s, y)u(s, y) + q(s, y))ds + ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y)P_{1}(s, y) ×

×Z_{1}(s, y)Z_{1}^{−1}(s, t)(P(s, t)u(s, t) + q(s, t))ds dt + v(x, y), (4.53)

where

P̄(x, y) = P_{0}(x, y) + P_{1}(x, y)P_{2}(x, y)  (4.54)

and

v(x, y) = − ∫_{0}^{x} Z_{2}^{−1}(s, y)\frac{∂P_{2}(s, y)}{∂y}u(s, y)ds + ∫_{0}^{x} \frac{∂Z_{2}^{−1}(s, y)}{∂y}Z_{1}(s, y) ×

×[φ(s) + ∫_{0}^{y} Z_{1}^{−1}(s, t)(P(s, t)u(s, t) + q(s, t))dt]ds − \frac{∂Z_{2}^{−1}(x, y)}{∂y}u(x, y).

But

Z_{2}^{−1}(x, y)\frac{∂P_{2}(x, y)}{∂y} =

= Z_{2}^{−1}(x, y)[\frac{∂}{∂y} (P_{2}(x, y)Z_{2}(x, y)) − P_{2}(x, y)\frac{∂Z_{2}(x, y)}{∂y}]Z_{2}^{−1}(x, y) =

= Z_{2}^{−1}(x, y)[\frac{∂^{2}Z_{2}(x, y)}{∂x∂y} − \frac{∂Z_{2}(x, y)}{∂x}Z_{2}^{−1}(x, y)\frac{∂Z_{2}(x, y)}{∂y}]Z_{2}^{−1}(x, y) =

= [Z_{2}^{−1}(x, y)\frac{∂}{∂x}(\frac{∂Z_{2}(x, y)}{∂y}) + \frac{∂Z_{2}^{−1}(x, y)}{∂x} · \frac{∂Z_{2}(x, y)}{∂y}]Z_{2}^{−1}(x, y) =

= \frac{∂}{∂x}(Z_{2}^{−1}(x, y)\frac{∂Z_{2}(x, y)}{∂y})Z_{2}^{−1}(x, y) =

= −\frac{∂}{∂x}(\frac{∂Z_{2}^{−1}(x, y)}{∂y}Z_{2}(x, y))Z_{2}^{−1}(x, y).

If we take into account the above equality and identity (4.51), we shall have

v(x, y) = ∫_{0}^{x} \frac{∂}{∂s}(\frac{∂Z_{2}^{−1}(s, y)}{∂y}Z_{2}(s, y))Z_{2}^{−1}(s, y)u(s, y)ds +

+ ∫_{0}^{x} \frac{∂Z_{2}^{−1}(s, y)}{∂y}Z_{2}(s, y)\frac{∂}{∂s}(Z_{2}^{−1}(s, y)u(s, y))ds − \frac{∂Z_{2}^{−1}(x, y)}{∂y}u(x, y) =

= ∫_{0}^{x} \frac{∂}{∂s}(\frac{∂Z_{2}^{−1}(s, y)}{∂y}u(s, y))ds − \frac{∂Z_{2}^{−1}(x, y)}{∂y}u(x, y) =

= \frac{∂Z_{2}^{−1}(0, y)}{∂y}u(0, y) = 0.  (4.55)

Therefore from (4.53) we get

Z_{2}^{−1}(x, y)\frac{∂u(x, y)}{∂y} = \frac{∂u(0, y)}{∂y} + ∫_{0}^{x} Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)φ(s)ds +

+ ∫_{0}^{x} Z_{2}^{−1}(s, y)(P̄(s, y)u(s, y) + q(s, y))ds + ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y) ×

×P_{1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)(P(s, t)u(s, t) + q(s, t))ds dt. (4.56)

According to representations (4.8) and (4.56) and condition (4.47), u satisfies boundary condition (4.2) if and only if

φ(x) = φ′_{0}(x) − P_{2}(x, 0)φ_{0}(x),  (4.57)
u(0, 0) = φ_{0}(0)  (4.58)

and

φ_{1}(y) = ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)(P_{1}(s, y)Z_{1}(s, y)φ(s) + q(s, y))ds +

+ ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P̄(s, y)u(s, y)ds +
+ ∫_{0}^{y} ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y) ×
×Z_{1}^{−1}(s, t)(P(s, t)u(s, t) + q(s, t))ds dt for 0 ≤ y ≤ b. (4.59)

On the other hand, if conditions (4.57) and (4.59) hold, then by (4.52) we have

u(x, 0) = φ_{0}(x) + Z_{2}(x, 0)(u(0, 0) − φ_{0}(0))

and

φ_{1}(0) = ∫_{0}^{a} M(s, 0)Z_{2}^{−1}(s, 0)[P_{0}(s, 0)φ_{0}(s) + P_{1}(s, 0)φ′_{0}(s) + q(s, 0)]ds +

+( ∫_{0}^{a} M(s, 0)Z_{2}^{−1}(s, 0)P̄(s, 0)Z_{2}(s, 0)ds)(u(0, 0) − φ_{0}(0)),

whence it is clear that condition (4.49) is necessary and sufficient for equality (4.58) to be fulfilled.

Thus we have proved that condition (4.49) is necessary for problem (4.1),(4.2) to be solvable. Moreover, if this condition is fulfilled, then the solution u of system (4.1) satisfies boundary conditions (4.2) if and only if

\frac{∂u(x, 0)}{∂x} − P_{2}(x, 0)u(x, 0) = φ(x) for 0 ≤ x ≤ a  (4.60_{1})

and equality (4.59) holds, where φ is the vector function given by (4.57).

Thus to complete the proof we have to show that system (4.1) has one and only one solution satisfying (4.59) and (4.60_{1}).

According to (4.50),

∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P̄(s, y)u(s, y)ds = Q_{0}(y)u(0, y) + + ∫_{0}^{y} ( ∫_{0}^{a} ∫_{0}^{x} Q_{1}(x, y, s)Z_{1}^{−1}(s, t)P(s, t)u(s, t)ds dx)dt + φ_{2}(y),

where

Q_{0}(y) = ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P̄(s, y)Z_{2}(s, y)ds,
Q_{1}(x, y, s) = M(x, y)Z_{2}^{−1}(x, y)P̄(x, y)Z_{2}(x, y)Z_{2}^{−1}(s, y)Z_{1}(s, y)

and

φ_{2}(y) = ∫_{0}^{a} ∫_{0}^{x} Q_{1}(x, y, s)[φ(s) + ∫_{0}^{y} Z_{1}^{−1}(s, t)q(s, t)dt]ds dx.

But

∫_{0}^{a} ∫_{0}^{x} Q_{1}(x, y, s)Z_{1}^{−1}(s, t)P(s, t)u(s, t)ds dx = ∫_{0}^{a} Q_{2}(y, s, t)u(s, t)ds, where

Q_{2}(y, s, t) = ( ∫_{s}^{a} Q_{1}(x, y, s)dx)Z_{1}^{−1}(s, t)P(s, t).

Therefore

∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P̄(s, y)u(s, y)ds =
= Q_{0}(y)u(0, y) + ∫_{0}^{y} ∫_{0}^{a} Q_{2}(y, s, t)u(s, t)ds dt + φ_{2}(y).

If we take onto account the formulas above and condition (4.48), then equality (4.59) will take the form

u(0, y) = ψ(y) + ∫_{0}^{y} ∫_{0}^{a} Q(y, s, t)u(s, t)ds dt for 0 ≤ y ≤ b, (4.60_{2})

where

ψ(y) = Q_{0}^{−1}(y)[φ_{1}(y) − φ_{2}(y) −
− ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)(P_{1}(s, y)Z_{1}(s, y)φ(s) + q(s, y))ds] −

−Q_{0}^{−1}(y) ∫_{0}^{y} ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)q(s, t)ds dt and

Q(y, s, t) = −Q_{0}^{−1}(y)[Q_{2}(y, s, t) +
+M(s, y)Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)P(s, t)].

By conditions (4.45),(4.46) and (4.48) and Lemmas 2.2_{2} and 2.3_{2}, the vector and matrix functions φ, ψ and Q satisfy the conditions of Lemma 3.5′. Therefore problem (4.1),(4.60_{1}),(4.60_{2}) has one and only one solution. ■

Theorem 4.2′. Let P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument, φ_{0} and φ_{1} be continuously differentiable and h : ℂ̃([0, a]; ℝ^{n}) → ℂ^{1}([0, b]; ℝ^{n}) be a linear continuous operator satisfying conditions (4.47) and (4.48). Then the fulfilment of (4.49) is the necessary and sufficient condition for problem (4.1),(4.2) to be uniquely solvable and to have a classical solution.

This theorem is proved using the same arguments as in proving Theorem 4.2, but instead of Lemmas 2.3_{3} and 3.5′ we apply Lemmas 2.3_{4} and 3.5′′.

Consider the case when boundary conditions (4.2) have the form

u(x, 0) = φ_{0}(x), ∑_{k=1}^{m} Φ_{k}(y)\frac{∂u(a_{k}, y)}{∂y} = φ_{1}(y).  (4.61)

In that case

M_{0}(y) = ∑_{k=1}^{m} Φ_{k}(y)Z_{2}(a_{k}, y)

and

M(x, y) = ∑_{k=1}^{m} Φ_{k}(y)Z_{2}(a_{k}, y)χ_{k}(x),

where

χ_{k}(x) = \begin{cases}1 & for 0 ≤ x ≤ a_{k} \\ 0 & for a_{k} < x ≤ a \end{cases} .

Therefore Theorems 4.2 and 4.2′ imply several assertions.

Corollary 4.6. Let conditions (4.45) and (4.46) be fulfilled and Φ_{k} ∈ ∈ ℂ̃_{∞}([0, b]; ℝ^{n×n}) and a_{k} ∈ [0, a] (k = 1, …, m) be such that

∑_{k=1}^{m} Φ_{k}(y)Z_{2}(a_{k}, y) = Θ for 0 ≤ y ≤ b  (4.62)

and

det ( ∑_{k=1}^{m} Φ_{k}(y)Z_{2}(a_{k}, y) ∫_{0}^{a_{k}} Z_{2}^{−1}(s, y)[P_{0}(s, y) +

+P_{1}(s, y)P_{2}(s, y)]Z_{2}(s, y)ds) ≠ 0 for 0 ≤ y ≤ b.  (4.63)

Then problem (4.1), (4.61) is uniquely soluvable if and only if

∑_{k=1}^{m} Φ_{k}(0)Z_{2}(a_{k}, 0) ∫_{0}^{a_{k}} Z_{2}^{−1}(s, 0) ×

×[P_{0}(s, 0)φ_{0}(s) + P_{1}(s, 0)φ′_{0}(s) + q(s, 0)]ds = φ_{1}(0).  (4.64)

Corollary 4.6′. Let P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument, φ_{0}, φ_{1} and Φ_{k} (k = = 1, …, m) be continuously differentiable, a_{k} ∈ [0, a] (k = 1, …, m), and conditions (4.62) and (4.63) hold. Then then fulfilment of equality (4.64) is necessary and sufficient for problem (4.1),(4.61) to be uniquely solvable and to have the classical solution.

Corollaries 4.6 and 4.6′ for problem (4.1),(4.2_{1}) take the form of

Corollary 4.7. Let conditions (4.45),(4.46) hold,

Z_{2}(a, y) = E for 0 ≤ y ≤ b  (4.65)

and

\begin{matrix}det ( ∫_{0}^{a} Z_{2}^{−1}(s, y)[P_{0}(s, y) + P_{1}(s, y)P_{2}(s, y)]Z_{2}(s, y)ds) ≠ 0 \\ for 0 ≤ y ≤ b. \end{matrix} (4.66)

Then problem (4.1),(4.2_{1}) is uniquely solvable if and only if

∫_{0}^{a} Z_{2}^{−1}(s, 0)[P_{0}(s, 0)φ_{0}(s) + P_{1}(s, 0)φ′_{0}(s) + q(s, 0)]ds = φ_{1}(0). (4.67)

Corollary 4.7′. Let P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument, φ_{0} and φ_{1} be continuously differentiable and conditions (4.65) and (4.66) hold. Then problem (4.1),(4.2_{1}) is uniquely solvable and its solution is classical if and only if equality (4.64) takes place.

Problem (4.1),(4.2) is ill-posed under conditions (4.47) and (4.48) due to the fact that its solvability may be violated at arbitrarily small perturbations either of coefficients of the system under consideration or of boundary conditions. But problem (4.1),(4.3) is free from such a deficiency. Namely, the following theorem is valid.

Theorem 4.3. Let

P_{i} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}),

ψ_{0} be summable, ψ_{1} absolutely continuous and h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) be a linear continuous operator satisfying conditions (4.47) and (4.48). Then problem (4.1),(4.3) has one and only one solution.

Proof. First we assume that problem (4.1),(4.3) has a solution u. Then u is a solution of problem (4.1),(4.2), where

φ_{0}(x) = Z_{2}(x, y)u(0, 0) + z_{0}(x), φ_{1}(y) = ψ_{1}(y)

and z_{0} is the solution of the system of ordinary differential equations

\frac{dz(x)}{dx} = P_{2}(x, 0)z(x) + ψ_{0}(x)

with the initial condition

z(0) = 0.

According to Theorem 4.2,

Q_{0}(0)u(0, 0) = c_{0},

where

Q_{0}(0) = ∫_{0}^{a} M(s, 0)Z_{2}^{−1}(s, 0)[P_{0}(s, 0) + P_{1}(s, 0)P_{2}(s, 0)]Z_{2}(s, 0)ds,

c_{0} = ψ_{1}(0) − ∫_{0}^{a} M(s, 0)Z_{2}^{−1}(s, 0)[P_{0}(s, 0)z_{0}(s) + P_{1}(s, 0)z′_{0}(s) + q(s, 0)]ds.

However, by condition (4.48),

det Q_{0}(0) ≠ 0.

Therefore

u(0, 0) = Q_{0}^{−1}(0)c_{0}.

Thus we have proved that every solution of problem (4.1),(4.3) is a solution of problem (4.1),(4.2), where

φ_{0}(x) = Z_{2}(x, 0)Q_{0}^{−1}(0)c_{0} + z_{0}(x), φ_{1}(y) = ψ_{1}(y).  (4.68)

The converse statement can be easily verified. If equalities (4.68) take place, then every solution of problem (4.1),(4.2) is a solution of problem (4.1),(4.3). Consequently, problem (4.1),(4.3) is equivalent to problem (4.1),(4.2), where φ_{0} and φ_{1} are given by equalities (4.68). But from (4.68) there follows equality (4.49). Therefore by Theorem 4.2 we conclude that in that case problem (4.1),(4.2) has one and only one solution. ■

In the same way we can prove

Theorem 4.3′. Let P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument, ψ_{0} be continuous, ψ_{1} be continuously differentiable and h : ℂ̃([0, a]; ℝ^{n}) → ℂ^{1}([0, b]; ℝ^{n}) be a linear continuous operator satisfying conditions (4.47) and (4.48). Then problem (4.1),(4.3) has one and only one solution and this solution is classical.

When boundary conditions (4.3) have the form

\frac{∂u(x, 0)}{∂x} − P_{0}(x, 0)u(x, 0) = ψ_{0}(x), ∑_{k=1}^{m} Φ_{k}(y)\frac{∂u(a_{k}, y)}{∂y} = φ_{1}(y), (4.69)

Theorems 4.3 and 4.3′ result in the following statements.

Corollary 4.8. Let

P_{i} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}),

ψ_{0} be summable, ψ_{1} absolutely continuous and Φ_{k} ∈ ℂ̃_{∞}([0, b]; ℝ^{n×n}) and a_{k} ∈ [0, a] (k = 1, …, m) be such that conditions (4.62) and (4.63) hold. Then problem (4.1),(4.69) has one and only one solution.

Corollary 4.8′. Let P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument. Let ψ_{0} be continuous, ψ_{1} continuously differentiable and Φ_{k} ∈ ℂ^{1}([0, b]; ℝ^{n×n}) and a_{k} ∈ [0, a] (k = = 1, …, m) be such that conditions (4.62) and (4.63) take place. Then problem (4.1),(4.69) has one and only one solution and this solution is classical.

Corollary 4.9. Let

P_{i} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}),

ψ_{0} be summable, ψ_{1} be absolutely continuous and conditions (4.65) and (4.66) hold. Then problem (4.1),(4.3_{1}) has one and only one solution.

Corollary 4.9′. Let P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument, ψ_{0} be continuous, ψ_{1} continuously differentiable and conditions (4.65),(4.66) take place. Then problem (4.1),(4.3_{1}) has one and only one solution and this solution is calssical.

Condition (4.48), appearing in Theorem 4.3 and its corollaries, is optimal in the sense that it cannot be weakened. As an example, let us consider problem (4.1),(4.3_{1}) for P_{2}(x, y) ≡ 0, i.e. when it has the form

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} + q(x, y), (4.70)
\frac{∂u(x, 0)}{∂x} = ψ_{0}(x), \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y} + ψ_{1}(y)  (4.71)

For this problem we have

Corollary 4.10. If

P_{0} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}),

then fulfilment of the inequality

det ( ∫_{0}^{a} P_{0}(s, y)ds) ≠ 0 for 0 ≤ y ≤ b

is necessary and sufficient for problem (4.70),(4.71) to be uniquely solvable for every P_{1}, q, ψ_{0} and ψ_{1} satisfying the conditions

\begin{matrix}P_{1} ∈ ℂ̃^{(−1,0)}(D_{ab}; ℝ^{n×n}), q ∈ ℂ̃^{(−1,0)}(D_{ab}; ℝ^{n}), \\ ψ_{0} ∈ L([0, a]; ℝ^{n}), ψ_{1} ∈ ℂ̃([0, b]; ℝ^{n}). \end{matrix}  (4.72)

Proof. The sufficiency follows from Corollary 4.9. Thus we have to show that if for any y_{0} ∈ [0, b]

det ( ∫_{0}^{a} P_{0}(s, y_{0})ds) = 0,

then there exist matrix and vector functions which satisfy conditions (4.72) and for which problem (4.70),(4.71) has no solution. Indeed, choose c ∈ ℝ^{n} such that the system of algebraic equations

( ∫_{0}^{a} P_{0}(s, y_{0})ds)z = c  (4.73)

is unsolvable and assume

P_{1}(x, y) = ∫_{0}^{x} P_{0}(s, y)ds, q(x, y) = −\frac{c}{a}, ψ_{0}(x) = 0, ψ_{1}(y) = 0. (4.74)

Suppose that for such P_{1}, q, ψ_{0}, and ψ_{1} problem (4.70),(4.71) has the solution u. Then

\frac{∂}{∂x}(\frac{∂u(x, y)}{∂y}) = \frac{∂}{∂x}( ∫_{0}^{x} P_{0}(s, y)ds)u(x, y)) − \frac{c}{a}.

Integrating this identity with respect to x from 0 to a and taking into account conditions (4.71) and (4.74), we obtain

( ∫_{0}^{a} P_{0}(s, y_{0})ds)u(a, y_{0}) = c.

But this is impossible because of the fact that system (4.73) is unsolvable. The obtained contradiction shows that if P_{1}, q, ψ_{0} and ψ_{1} are given by equalities (4.74), then problem (4.70),(4.71) has no solution despite the fact that conditions (4.72) hold. ■

4.3. Problems (4.1),(4.4) and (4.1),(4.5).

Theorem 4.4. Let

P_{2} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}),
mes I_{M_{0}} = b, ∫_{0}^{b}(1 + ∥ψ_{1}(y)∥)∥M_{0}^{−1}(y)∥dy < +∞.  (4.75)

Then problem (4.1),(4.4) has one and only one generalized solution.

Proof. Let u be an arbitrary generalized solution of system (4.1). Then, according to Lemma 3.3,

u(x, y) = Z_{2}(x, y)[u(0, y) + ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)v_{1}(s)ds] + Z_{2}(x, y) ×

× ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)(P(s, t)u(s, t) + q(s, t))dsdt, (4.76)

where

P(x, y) = P_{0}(x, y) + P_{1}(x, y)P_{2}(x, y) − \frac{∂P_{2}(x, y)}{∂y},

v_{1}(x) = lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)).

From this and representation (4.8) it becomes clear that boundary conditions (4.4) are fulfilled if and only if

v_{1}(x) = ψ_{0}(x)

and

ψ_{1}(y) = M_{0}(y)u(0, y) + ψ_{2}(y) + ∫_{0}^{y} ∫_{0}^{a} Q_{1}(y, s, t)u(s, t)ds dt,

ψ_{2}(y) = ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)Z_{1}(s, y)[ψ_{0}(s) + ∫_{0}^{y} Z_{1}^{−1}(s, t)q(s, t)dt]ds,

Q_{1}(y, s, t) = M(s, y)Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)P(s, t).

Moreover, as follows from Lemmas 2.2_{2} and 2.3_{1},

ψ_{2} ∈ L_{∞}([0, b]; ℝ^{n×n}), Q_{1} ∈ L_{∞}([0, b] × D_{ab}; ℝ^{n×n}).  (4.77)

Obviously, problem (4.1),(4.4) is equivalent to the problem of finding the generalized solution of system (4.1), satisfying the boundary conditions

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂y} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), \\ u(0, y) = ψ(y) + ∫_{0}^{y} ∫_{0}^{a} Q(y, s, t)u(s, t)ds dt, \end{matrix}  (4.78)

where

ψ(y) = M_{0}^{−1}(y)(ψ_{1}(y) − ψ_{2}(y)),
Q(y, s, t) = −M_{0}^{−1}(y)Q_{1}(y, s, t).

However, by conditions (4.75) and (4.77),

ψ ∈ L([0, b]; ℝ^{n}), Q ∈ L([0, b] × D_{ab}; ℝ^{n×n}),
∥Q(y, s, t)∥ ≤ η(y) for y ∈ [0, b], (s, t) ∈ D_{ab}

and

η ∈ L([0, b]; ℝ).

Therefore, by Lemma 3.5 problem (4.1),(4.78) has one and only one generalized solution. ■

Theorem 4.4′. Let

P_{2} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}), ψ_{1} ∈ ℂ̃([0, b]; ℝ^{n}),

and h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) be a linear continuous operator such that

I_{M_{0}} = [0, b].  (4.79)

Then problem (4.1),(4.4) has the unique generalized solution and this solution is absolutely continuous.

Theorem 4.4′′. Let P_{i} (i = 0, 1, 2) and q be continuous, P_{2} have a continuous partial derivative in the second argument, ψ_{0} be continuous, ψ_{1} be continuously differentiable and h : ℂ̃([0, a]; ℝ^{n}) → ℂ^{1}([0, b]; ℝ^{n}) be a linear continuous operator, satisfying condition (4.79). Then problem (4.1),(4.4) has a unique generalized solution which is classical.

Theorem 4.4′ (Theorem 4.4′′) can be proved similarly to Theorem 4.4 but in that case instead of Lemmas 2.3_{1} and 3.5 we apply Lemmas 2.3_{3} and 3.5′ (Lemmas 2.3_{4} and 3.5′′).

Remark 4.11. The effective conditions guaranteeing the fulfilment of condition (4.75) (Condition (4.79)) are given in the above proven Corollaries 4.2-4.5 (Corollaries 4.2′-4.5′).

Remark 4.12. The restrictions imposed on the operator h in Theorem 4.4′ are optimal in the sense that they cannot be weakened. As an example, consider the problem

\frac{∂^{2}u(x, y)}{∂x∂y} = 0,  (4.80)
lim_{y→0} \frac{∂u(x, y)}{∂x} = \frac{1}{a}, u(a, y)=(1+y^{α})u(0, y),  (4.81)

where α ∈ (0, 1). In that case h(v)(y) = v(a) − (1 + y^{α})v(0) and

M_{0}(y) = −y^{α}.

Consequently, the operator h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃([0, b]; ℝ^{n}) is continuous and satisfies condition (4.75). On the other hand, problem (4.80),(4.81) has the unique generalized solution

u(x, y) = y^{−α} + \frac{x}{a},

which is not absolutely continuous.

Finally, letus pass tothe investigation of problem (4.1),(4.4) when M_{0}(y)≡ ≡ Θ and h acts from ℂ̃([0, a]; ℝ^{n}) to ℂ̃_{∞}([0, b]; ℝ^{n}). Let us introduce the matrix functions

P̄(x, y) = P_{0}(x, y) + P_{1}(x, y)P_{2}(x, y)

and

M̄(y) = H′_{0}(y) + ∫_{0}^{a} [\frac{∂H(s, y)}{∂y} · \frac{∂Z_{2}(s, y)}{∂s} +

+M(s, y)Z_{2}^{−1}(s, y)P̄(s, y)Z_{2}(s, y)]ds.  (4.82)

Theorem 4.5. Let

P_{2} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}), ψ_{1} ∈ ℂ̃([0, b]; ℝ^{n})

and h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) be a linear continuous operator such that

\begin{matrix}M_{0}(y) = Θ for 0 ≤ y ≤ b, \\ mes IM̄ = b, ∫_{0}^{b}(1 + ∥ψ′_{1}(y)∥)∥M̄^{−1}(y)∥dy < +∞. \end{matrix}  (4.83)

Then for the existence and uniqueness of a generalized solution of problem (4.1), (4.4) it is necessary and sufficient that

ψ_{1}(0) = ∫_{0}^{a} M(s, 0)Z_{2}^{−1}(s, 0)ψ_{0}(s)ds.  (4.84)

Proof. In view of representation (4.10) and condition M_{0}(y) ≡ Θ, we have

H_{0}(y) + ∫_{0}^{a} H(s, y)\frac{∂Z_{2}(s, y)}{∂s} ds = Θ for 0 ≤ y ≤ b.  (4.85)

If we differentiate this identity, then by virtue of Lemmas 2.1_{3} and 2.2_{2} we obtain

H′_{0}(y) + ∫_{0}^{a} \frac{∂H(s, y)}{∂y} · \frac{∂Z_{2}(s, y)}{∂s}da = − ∫_{0}^{a} H(s, y)\frac{∂^{2}Z_{2}(s, y)}{∂s∂y}ds. (4.86)

Let u be an arbitrary generalized solution of system (4.1) and

ū(x, y) = u(x, y) − Z_{2}(x, y)u(0, y).

Then, according to Lemma 3.3, representation (4.76) is valid and, consequently,

Z_{2}^{−1}(x, y)ū(x, y) = ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)v_{1}(s)ds +
+ ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)(P(s, t)u(s, t) + q(s, t))ds dt, (4.87)

where

v_{1}(x) = lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)),

P(x, y) = P_{0}(x, y) + P_{1}(x, y)P_{2}(x, y) − \frac{∂P_{2}(x, y)}{∂y}.

Proceeding from this representation in a way which is similar to that of proving equality (4.56), we can show that

Z_{2}^{−1}(x, y)\frac{∂ū(x, y)}{∂y} = ∫_{0}^{x} Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)v_{1}(s)ds +

+ ∫_{0}^{x} Z_{2}^{−1}(s, y)(P̄(s, y)u(s, y) + q(s, y))ds +

+ ∫_{0}^{x} ∫_{0}^{y} Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)(P(s, t)u(s, t) +

+q(s, t))ds dt − Z_{2}^{−1}(x, y)\frac{∂Z_{2}(x, y)}{∂y}u(0, y).  (4.88)

According to representations (4.7),(4.10) and condition (4.85), we obtain

h(u(·, y))(y) = h(ū(·, y))(y) = ∫_{0}^{a} H(s, y)\frac{∂ū(s, y)}{∂s}ds =

= ∫_{0}^{a} M(s, y)\frac{∂}{∂s}(Z_{2}^{−1}(s, y)u(s, y))ds.

Therefore

h(u(·, 0))(0) = ∫_{0}^{a} M(s, 0)\frac{∂}{∂s}(Z_{2}^{−1}(s, 0)u(s, 0))ds =

= ∫_{0}^{a} M(s, 0)Z_{2}^{−1}(s, 0)v_{1}(s)ds

and

\frac{d}{dy} h(u(·, y))(y) = ∫_{0}^{a} \frac{∂H(s, y)}{∂y} · \frac{∂ū(s, y)}{∂s}ds + h(\frac{∂ū(·, y)}{∂y})(y),

whence it is clear that boundary conditions (4.4) are fulfilled if and only if

v_{1}(x) = ψ_{0}(x),  (4.89)
∫_{0}^{a} \frac{∂H(s, y)}{∂y} · \frac{∂ū(s, y)}{∂s}ds + h(\frac{∂ū(·, y)}{∂y})(y) = ψ′_{1}(y)  (4.90)

and equality (4.84) holds. In view of (4.8),(4.86) and (4.89) from (4.88) we obtain

h(\frac{∂ū(·, y)}{∂y})(y) = ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)ψ_{0}(s)ds +

+ ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)(P̄(s, y)u(s, y) + q(s, y))ds +

+ ∫_{0}^{y} ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)(P(s, t)u(s, t) +

+q(s, t))ds dt − ( ∫_{0}^{a} H(s, y)\frac{∂^{2}Z_{2}(s, y)}{∂s∂y}ds)u(0, y) = ψ_{2}(y) +

+ ∫_{0}^{y} ∫_{0}^{a} Q_{1}(y, s, t)u(s, t)ds dt + ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y) × P̄(s, y) ×
×u(s, y)ds + [H′_{0}(y) + ∫_{0}^{a} \frac{∂H(s, y)}{∂y} · \frac{∂Z_{2}(s, y)}{∂s}ds]u(0, y), (4.91)

where

ψ_{2}(y) = ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)[P_{1}(s, y)Z_{1}(s, y)ψ_{0}(s) + q(s, y)]ds +

+ ∫_{0}^{y} ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)q(s, t)ds dt,

Q_{1}(y, s, t) = M(s, y)Z_{2}^{−1}(s, y)P_{1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)P(s, t).

But according to (4.76),

∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P̄(s, y)u(s, y)ds =

= [ ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)P̄(s, y)Z_{2}(s, y)ds]u(0, y) + ψ_{3}(y) +

+ ∫_{0}^{y} ∫_{0}^{a} ( ∫_{s}^{a} Q_{2}(x, y, s)dx)Z_{1}^{−1}(s, t)P(s, t)u(s, t)ds dt, (4.92)

where

Q_{2}(x, y, s) = M(x, y)Z_{2}^{−1}(x, y)P̄(x, y)Z_{2}(x, y)Z_{2}^{−1}(s, y)Z_{1}(s, y),

ψ_{3}(y) = ∫_{0}^{a} ∫_{0}^{x} Q_{2}(x, y, s)[ψ_{0}(s) + ∫_{0}^{y} Z_{1}^{−1}(s, t)q(s, t)dt]ds dx.

On the other hand, in view of (4.87),

∫_{0}^{a} \frac{∂H(s, y)}{∂y} · \frac{∂ū(s, y)}{∂s}ds = ψ_{4}(y) + ∫_{0}^{y} ∫_{0}^{a} Q_{3}(y, s, t)u(s, t)ds dt, (4.93)

where

ψ_{4}(y) = ∫_{0}^{a} \frac{∂H(x, y)}{∂y}[Z_{1}(x, y)ψ_{0}(x) +

+\frac{∂Z_{2}(x, y)}{∂x} ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)ψ_{0}(s)ds]dx +

+ ∫_{0}^{y} ∫_{0}^{a} \frac{∂H(x, y)}{∂y}[Z_{1}(x, y)Z_{1}^{−1}(x, t)q(x, t) +

+\frac{∂Z_{2}(x, y)}{∂x} ∫_{0}^{x} Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)q(s, t)ds]dx dt,

Q_{3}(y, s, t) = [ ∫_{s}^{a} \frac{∂H(x, y)}{∂y} · \frac{∂Z_{2}(x, y)}{∂x}Z_{2}^{−1}(s, t)dx + \frac{∂H(s, y)}{∂y}] ×

×Z_{1}(s, y)Z_{1}^{−1}(s, t)P(s, t).

From (4.89)-(4.93) we obtain equalities (4.78), where

ψ(y) = M̄^{−1}(y)[ψ′_{1}(y) − ψ_{2}(y) − ψ_{3}(y) − ψ_{4}(y)],
Q(y, s, t) = −M̄^{−1}(y)[Q_{1}(y, s, t) +
+( ∫_{s}^{a} Q_{2}(x, y, s)dx)Z_{1}^{−1}(s, t)P(s, t) + Q_{3}(y, s, t)].

Thus we have proved that condition (4.84) is necessary for the solvability of problem (4.1),(4.4). Moreover, if it is fulfilled, then problem (4.1),(4.4) is equivalent to problem (4.1),(4.78). On the other hand, by virtue of Lemmas 2.1_{3}, 2.2_{2} and 2.3_{3} and condition (4.83), the vector and matrix functins, ψ and Q respectively, satisfy conditions of Lemma 3.5. Therefore problem (4.1),(4.78) has one and only one generalized solution. ■

If instead of Lemma 3.5 we apply Lemma 3.5′ (Lemma 3.5′′), we shall be able to convince ourselves that under specific additional restrictions on P_{i}(i = 0, 1, 2), q, h, ψ_{0} and ψ_{1} the generalized solution of problem (4.1),(4.4) is absolutely continuous (classical). Namely, the following assertions are valid.

Theorem 4.5′. Let

P_{i} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}),
ψ_{1} ∈ ℂ̃^{1}([0, b]; ℝ^{n})

and h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}^{1}([0, b]; ℝ^{n}) be a linear continuous operator such that

M_{0}(y) = Θ for 0 ≤ y ≤ b, IM̄ = [0, b].  (4.94)

Then the fulfilment of equality (4.84) is necessary and sufficient for problem (4.1),(4.4) to have the unique generalized solution and for this solution to be absolutely continuous.

Theorem 4.5′′. Let P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument, ψ_{0} be continuous, ψ_{1} twice continuously differentiable and h : ℂ̃([0, a]; ℝ^{n}) → ℂ^{2}([0, b]; ℝ^{n}) be a linear continuous operator satsfying conditions (4.94). Then the fulfilment of equality (4.84) is necessary and sufficient for problem (4.1),(4.4) to have the unique generalized solution and for this solution to be classical.

Remark 4.13. The restrictions imposed on P_{i} (i = 0, 1, 2) and q in Theorems 4.5′ and 4.5′′ are optimal and they cannot be weakened. As an example, consider the problems

\frac{∂^{2}u(x, y)}{∂x∂y} = p_{0}(y)u(x, y) − 1,

lim_{y→0} \frac{∂u(x, y)}{∂x} = 0, u(a, y) = u(0, y),

\frac{∂^{2}u(x, y)}{∂x∂y} = u(x, y) − p_{1}(y)\frac{∂u}{∂x} − x,

lim_{y→0} \frac{∂u(x, y)}{∂x} = 1, u(a, y) = u(0, y) + a

and

\frac{∂^{2}u(x, y)}{∂x∂y} = u(x, y) − q(y),

lim_{y→0} \frac{∂u(x, y)}{∂x} = 0, u(a, y) = u(0, y),

where p_{0} : [0, b] → (0, +∞), p_{1} : [0, b] → ℝ and q : [0, b] → ℝ are continuous functions. All conditions of Theorem 4.5 for these problems are fulfilled. Therefore each of these problems has the unique generalized solution

u(x, y) = \frac{1}{p_{0}(y)}, u(x, y) = x + p_{1}(y) and u(x, y) = q(y).

Moreover these solutions are absolutely continuous (classical) if and only if p_{0}, p_{1} and q are absolutely continuous (classical).

Similarly to Theorem 4.5 we can prove

Theorem 4.6. Let

P_{2} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}),
M_{0}(y) = Θ for 0 ≤ y ≤ b

and

mes IM̄ = b, ∫_{0}^{b}(1 + ∥ψ_{1}(y)∥)∥M̄^{−1}(y)∥dy < +∞.

Then problem (4.1),(4.5) has one and only one generalized solution.

Theorem 4.6′. Let

P_{i} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2),
q ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}), ψ_{1} ∈ b̃c([0, b]; ℝ^{n})

and h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) be a linear continuous operator satisfying conditions (4.94). Then problem (4.1),(4.5) has one and only one generalized solution and this solution is absolutely continuous.

Theorem 4.6′′. Let P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument, ψ_{0} be continuous, ψ_{1} continuously differentiable and h : ℂ̃([0, a]; ℝ^{n}) → ℂ^{1}([0, b]; ℝ^{n}) be a linear continuous operator satisfying conditions (4.94). Then problem (4.1),(4.5) has one and only one generalized solution and this solution is classical.

In conclusion, let us admit that the conditions of unique solvability of problems (4.1), (4.4_{1}) and (4.1),(4.5_{1}) are sufficiently transparent because for these problems

M_{0}(y) = Z_{2}(a, y) − E

and

M̄(y) = ∫_{0}^{a} Z_{2}^{−1}(s, y)P̄(s, y)Z_{2}(s, y)ds.

§ 5. Ways of Constructing Solutions

Consider the boundary value problem

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q(x, y),  (5.1)
u(x, 0) = φ_{0}(x), h(\frac{∂u(·, y)}{∂y})(y) = φ_{1}(y),  (5.2)

where

P_{i} ∈ L_{∞}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ L_{∞}(D_{ab}; ℝ^{n}),
φ_{0} ∈ ℂ̃_{∞}([0, a]; ℝ^{n}), φ_{1} ∈ L_{∞}(D_{ab}; ℝ^{n})

and h : ℂ̃([0, a]; ℝ^{n}) → L_{∞}([0, b]; ℝ^{n}) is a linear continuous operator.

As in Section 4, by Z_{2} we shall mean the solution of the matrix differential equation

\frac{∂Z_{2}(x, y)}{∂x} = P_{2}(x, y)Z_{2}(x, y)

with the initial condition

Z_{2}(0, y) = E.

According to Lemma 2.3_{1}, the operator h admits the representation

h(Z_{2}(·, y)v(·))(y) = M_{0}(y)v(0) +
+ ∫_{0}^{a} M(s, y)v′(s)ds for v ∈ ℂ̃([0, a]; ℝ^{n}),

where

M_{0} ∈ L_{∞}([0, b]; ℝ^{n} × n), M ∈ L_{∞}(D_{ab}; ℝ^{n×n}).

Analyzing the proof of Theorem 4.1, we can see that the following statement is valid.

Theorem 5.1. Let

ess inf_{0≤y≤b} | det M_{0}(y)| > 0.  (5.3)

Then problem (5.1),(5.2) has the unique solution u and

\begin{matrix}u_{m}(x, y) ⇉ u(x, y), \frac{∂u_{m}(x, y)}{∂x} ⇉ \frac{∂u(x, y)}{∂x}, \\ \frac{∂u_{m}(x, y)}{∂y} ⇉ \frac{∂u(x, y)}{∂y} for m → +∞, \end{matrix}  (5.4)

where u_{0}(x, y) ≡ 0 and for an arbitrary m the vector function u_{m} is the solution of the problem

\frac{∂^{2}u_{m}(x, y)}{∂x∂y} = P_{2}(x, y)\frac{∂u_{m}(x, y)}{∂y} + P_{0}(x, y)u_{m−1}(x, y) +

+P_{1}(x, y)\frac{∂u_{m−1}(x, y)}{∂x} + q(x, y),  (5.5)
u_{m}(x, 0) = φ_{0}(x), h(\frac{∂u_{m}(·, y)}{∂y})(y) = φ_{1}(y).  (5.6)

To construct u_{m} almost for all y ∈ [0, b] we have to solve the system of ordinary differential equations

\frac{∂z(x, y)}{∂x} = P_{2}(x, y)z(x, y) + P_{0}(x, y)u_{m−1}(x, y) +

+P_{1}(x, y)\frac{∂u_{m−1}(x, y)}{∂x} + q(x, y)  (5.7)

with the boundary condition

h(z)(y) = φ_{1}(y).  (5.8)

In view of (5.3) problem (5.7),(5.8) has the unique solution z_{m}(·, y) and

u_{m}(x, y) = φ_{0}(x) + ∫_{0}^{y} z_{m}(x, t)dt.

However, if n > 1, then the solution z_{m}(·, y) can be effectively constructed only in exceptional cases. Consequently, the method of constructing the solution of problem (5.1),(5.2) described above fails in the general case.

We shall consider the case when P_{2} and h admit the representation

P_{2}(x, y) = P_{20}(x, y) + P_{21}(x, y)

and

h(v)(y) = h_{0}(v)(y) − ∑_{j=1}^{ν} H_{j}v(s_{j}),

where

P_{20} and P_{21} ∈ L_{∞}(D_{ab}; ℝ^{n×n}),

s_{j} ∈ [0, a], H_{j} ∈ ℝ^{n×n} (j = 1, …, ν),

and the problem

\frac{∂z(x, y)}{∂x} = P_{20}(x, y)z(x, y), h_{0}(z)(y) = 0  (5.9)

has only the trivial solution almost for all y ∈ [0, b].

In that case problem (5.1),(5.2) takes the form

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+[P_{20}(x, y) + P_{21}(x, y)]\frac{∂u(x, y)}{∂y} + q(x, y),  (5.10)
u(x, 0) = φ_{0}(x), h_{0}(\frac{∂u(·, y)}{∂y})(y) = ∑_{j=1}^{ν} H_{j} \frac{∂u(s_{j}, y)}{∂y} + φ_{1}(y). (5.11)

Let u_{0}(x, y) ≡ 0. For every natural m by u_{m} we denote the solution of the problem

\frac{∂^{2}u_{m}(x, y)}{∂x∂y} = P_{20}(x, y)\frac{∂u_{m}(x, y)}{∂y} + P_{0}(x, y)u_{m−1}(x, y) +

+P_{1}(x, y)\frac{∂u_{m−1}(x, y)}{∂x} + P_{21}(x, y)\frac{∂u_{m−1}(x, y)}{∂y} + q(x, y), (5.12)

\begin{matrix}u_{m}(x, 0) = φ_{0}(x), \\ h_{0}(\frac{∂u_{m}(·, y)}{∂y})(y) = ∑_{j=1}^{ν} H_{j} \frac{∂u_{m−1}(s_{j}, y)}{∂y} + φ_{1}(y). \end{matrix}  (5.13)

Below we determine the conditions whose fulfilment guarantees the fulfilment of conditions (5.4).

By Z_{0} will be meant the solution of the matrix differential equation

\frac{∂Z_{0}(x, y)}{∂x} = P_{20}(x, y)Z_{0}(x, y)

with the initial condition

Z_{0}(0, y) = E,

and by G(·, ·, y) the Green’s matrix of problem (5.9).

By Lemma 2.3_{1} operator h_{0} admits the representation

h_{0}(Z_{0}(·, y)v(·))(y) = M_{10}(y)v(0) +
+ ∫_{0}^{a} M_{1}(s, y)v′(s)ds for v ∈ ℂ̃([0, a]; ℝ^{n}),  (5.14)

where

M_{10} ∈ L_{∞}([0, b]; ℝ^{n×n}), M_{1} ∈ L_{∞}(D_{ab}; ℝ^{n} × n).

Therefore

G(x, s, y) = \begin{cases}Z_{0}(x, y)[E − M_{10}^{−1}(y)M_{1}(s, y)]Z_{0}^{−1}(s, y) & for s ≤ x \\ −Z_{0}(x, y)M_{10}^{−1}(y)M_{1}(s, y)Z_{0}^{−1}(s, y) & for s > x. \end{cases} (5.15)

We have

Theorem 5.2. Let

ess inf_{0≤y≤b} | det M_{10}(y)| > 0  (5.16)

and there exist a matrix function A ∈ L_{∞}([0, b]; ℝ^{n×n}) such that

ess sup_{0≤y≤b} r(A(y)) < 1  (5.17)

and the inequality

∑_{j=1}^{ν} |Z_{0}(x, y)M_{10}^{−1}(y)H_{j}| + ∫_{0}^{a} |G(x, s, y)P_{21}(s, y)|ds ≤ A(y) (5.18)

holds almost everywhere in D_{ab}. Then problem (5.10),(5.11) has the unique solution u and condition (5.4) takes place, where u_{0}(x, y) ≡ 0 and for every natural m the vector function u_{m} is a solution of problem (5.12),(5.13).

Proof. Let n_{1} = 2n, n_{2} = n. For arbitrary z_{1} = (z_{1}^{i})_{i=0}^{1} ∈ L_{∞}(D_{ab}; ℝ^{n_{1}}), where z_{1}^{i} ∈ L_{∞}(D_{ab}; ℝ^{n}) (i = 0, 1) and z_{2} ∈ L_{∞}(D_{ab}; ℝ^{n_{2}}), assume

g_{1}^{0}(z_{1}, z_{2})(x, y) = φ_{0}(x) +

+ ∫_{0}^{y} Z_{0}(x, t)M_{10}^{−1}(t)[ ∑_{j=1}^{ν} H_{j}z_{2}(s_{j}, t) + φ_{1}(t)]dt +

+ ∫_{0}^{y} ∫_{0}^{a} G(x, s, t)[P_{0}(s, t)z_{1}^{0}(s, t) + P_{1}(s, t)z_{1}^{1}(s, t) +

+P_{21}(s, t)z_{2}(s, t) + q(s, t)]ds dt,  (5.19)

\begin{matrix}g_{1}^{1}(z_{1}, z_{2})(x, y) = \frac{∂g_{1}^{0}(z_{1}, z_{2})(x, y)}{∂x}, g_{1}(z_{1}, z_{2})(x, y) = \\ = (g_{1}^{i}(z_{1}, z_{2})(x, y))_{i=0}^{1}, g_{2}(z_{1}, z_{2})(x, y) = \frac{∂g_{1}^{0}(z_{1}, z_{2})(x, y)}{∂y} \end{matrix} (5.20)

and show that problem (5.10),(5.11) is equivalent to the system of operator equations

z_{i}(x, y) = g_{i}(z_{1}, z_{2})(x, y) (i = 1, 2).  (5.21)

Indeed, let problem (5.10),(5.11) have a solution u. Put

z_{1}^{0}(x, y) = u(x, y), z_{1}^{1}(x, y) = \frac{∂u(x, y)}{∂x}, z_{2}(x, y) = \frac{∂u(x, y)}{∂y}.

By Lemma 3.1 and equalities (5.19) and (5.20)

(z_{i})_{i=1}^{2} ∈ L_{∞}(D_{ab}; ℝ^{n_{1}}) × L_{∞}(D_{ab}; ℝ^{n_{2}})

is the solution of system (5.21). Taking into account (5.15) and (5.16), we can easily show that if z_{1}^{i} ∈ L_{∞}(D_{ab}; ℝ^{n}) (i = 0, 1), z_{1} = (z_{1}^{i})_{i=0}^{1}, z_{2} ∈ ∈ L_{∞}(D_{ab}; ℝ^{n}) and (z_{i})_{i=1}^{2} is a solution of system (5.21), then u(·, ·) = = z_{1}^{0}(·, ·) is a solution of problem (5.10),(5.11) and the equalities

z_{1}^{1}(x, y) = \frac{∂u(x, y)}{∂x}, z_{2}(x, y) = \frac{∂u(x, y)}{∂y}

are valid.

In view of (5.15)-(5.20), for any ζ_{i} and ζ̄ ∈ L_{∞}(D_{ab}; ℝ^{n_{i}}) (i = 1, 2) the operators g_{i} : L_{∞}(D_{ab}; ℝ^{n_{1}}) × L_{∞}(D_{ab}; ℝ^{n_{2}}) → L_{∞}(D_{ab}; ℝ^{n_{i}}) (i = 1, 2) satisfy inequalities (1.3) and (1.21) almost everywhere in D_{ab}, where I = = [0, a], g_{0}(t) ≡ const, A_{01} is a non-negative constant n_{2} × n_{1} matrix and A_{02} is the zero matrix. The validity of the theorem becomes evident by applying Lemma 1.3. ■

Theorem 5.3. Let inequality (5.16) hold,

ess sup_{0≤y≤b} r( ∑_{j=1}^{ν} |H_{j}| |Z_{0}(s_{j}, y)M_{10}^{−1}(y)|) < 1  (5.22)

and there exist a matrix function A ∈ L_{∞}([0, b]; ℝ^{n×n}) satisfying condition (5.17) such that the inequality

|Z_{0}(x, y)M_{10}^{−1}(y)|[E − ∑_{j=1}^{ν} |H_{j}| |Z_{0}(s_{j}, y)M_{10}^{−1}(y)|]^{−1} ×

× ∑_{j=1}^{ν} ∫_{0}^{a} |H_{j}| |G(s_{j}, s, y)P_{21}(s, y)|ds +

+ ∫_{0}^{a} |G(x, s, y)P_{21}(s, y)|ds ≤ A(y)  (5.23)

holds almost everywhere in D_{ab}. Then problem (5.10),(5.11) has the unique solution u and condition (5.4) takes place, where u_{0}(x, y) ≡ 0 and for any natural m the vector function u_{m} is a solution of problem (5.12),(5.13).

Proof. Let n_{1} = 2n, n_{2} = n, Λ_{1} = L_{∞}(D_{ab}; ℝ^{n_{1}}), Λ_{2} be a set of all ζ ∈ ∈ L_{∞}(D_{ab}; ℝ^{n_{2}}) such that ζ(·, y) ∈ ℂ([0, a]; ℝ^{n}) almost for all y ∈ [0, b] and Λ_{0} = ℂ([0, a]; ℝ^{n}), while g_{i} : Λ_{1} × Λ_{2} → Λ_{i} (i = 1, 2) be the operators given by equalities (5.19),(5.20).

As shown above, problem (5.1),(5.2) is equivalent to the system of operator equations (5.21) due to condition (5.16).

In view of (5.15),(5.16),(5.19) and (5.20) for any ζ_{i} and ζ̄_{i} ∈ Λ_{i} (i = 1, 2) the operators g_{i} (i = 1, 2) satisfy inequalities (1.3),(1.4) almost everywhere in D_{ab}, where I = [0, a], g_{0}(t) ≡ const, A_{01} is a non-negative constant n_{2} × n_{1} matrix, A_{02} is the zero matrix,

A_{1}(x, y) = ∫_{0}^{a} |G(x, s, y)P_{21}(s, y)|ds, A_{2}(x, y) = |Z_{0}(x, y)M_{10}^{−1}(y)|

and l : Λ_{0} → ℝ^{n_{2}} is a non-negative linear operator given by the equality

l(v) = ∑_{j=1}^{ν} |H_{j}|v(s_{j}).

In view of (5.22), condition (1.5) takes place and in view of (5.23) we may assume without loss of generality that equality (1.7) holds. Besides, since condition (1.6) is also fulfilled, by virtue of Lemma 1.1 system (5.21) has the unique solution (z_{i})_{i=1}^{2} and conditions (1.8) take place. The validity of Theorem 5.3 immediately follows from the above arguments. ■

Let us give two corollaries of the above proven theorems for the periodic boundary value problem. For the convenience we rewrite system (5.1) and the boundary condition in the scalar form

\frac{∂^{2}u_{i}(x, y)}{∂x∂y} = ∑_{k=1}^{n} (p_{0ik}(x, y)u_{k}(x, y) + p_{1ik}(x, y)\frac{∂u_{k}(x, y)}{∂x} +

+p_{2ik}(x, y)\frac{∂u_{k}(x, y)}{∂y}) + q_{i}(x, y) (i = 1, …, n),  (5.24)
u_{i}(x, 0) = φ_{0i}(x), \frac{∂u_{i}(a, y)}{∂y} = \frac{∂u_{i}(0, y)}{∂y} + φ_{1i}(y) (i = 1, …, n). (5.25)

Moreover, as above we assume that

p_{0ik}, p_{1ik}, p_{2ik} and q_{i} ∈ L_{∞}(D_{ab}; ℝ) (i, k = 1, …, n),
φ_{0i} ∈ ℂ̃_{∞}([0, a]; ℝ), φ_{1i} ∈ L_{∞}([0, b]; ℝ) (i = 1, …, n).

Corollary 5.1. Let there exist σ_{i} ∈ {−1, 1} (i = 1, …, n) and δ > 0 such that the real parts of eigenvalues of matrix S(y) = (s_{ij}(y))_{i,j=1}^{n}, where

s_{ii}(y) = ess sup_{0≤x≤a}{σ_{i}p_{2ii}(x, y)},

s_{ij}(y) = ess sup_{0≤x≤a} |p_{2ij}(x, y)| for (i ≠ j; i, j = 1, ⋯, n) (5.26)

are less than −δ almost for all y ∈ [0, b]. Then problem (5.24),(5.25) has the unique solution (u_{i})_{i=1}^{n} and

\begin{matrix}u_{im}(x, y) ⇉ u_{i}(x, y), \frac{∂u_{im}(x, y)}{∂x} ⇉ \frac{∂u_{i}(x, y)}{∂x}, \\ \frac{∂u_{im}(x, y)}{∂y} ⇉ \frac{∂u_{i}(x, y)}{∂y} for m → +∞ (i = 1, …, n), \end{matrix} (5.27)

where u_{i0}(x, y) ≡ 0 (i = 1, …, n) and for any natural m and i ∈ {1, …, n} the function u_{im} is a solution of the equation

\frac{∂^{2}u_{im}(x, y)}{∂x∂y} = p_{2ii}(x, y)\frac{∂u_{im}(x, y)}{∂y} +

+ ∑_{i=1}^{n}(1 − δ_{ij})p_{2ij}(x, y)\frac{∂u_{jm−1}(x, y)}{∂y} +

+ ∑_{j=1}^{n} (p_{0ij}(x, y)u_{jm−1}(x, y) + p_{1ij}(x, y)\frac{∂u_{jm−1}(x, y)}{∂x}) + q_{i}(x, y) (5.28)

with the boundary conditions

u_{im}(x, 0) = φ_{0}(x), \frac{∂u_{im}(a, y)}{∂y} = \frac{∂u_{im}(0, y)}{∂y} + φ_{1i}(y). (5.29)

Proof. From the restrictions imposed on the matrix function S it follows that

s_{ii}(y) ≤ −δ almost for all y ∈ [0, b]  (5.30)

and

A(y) = ((1 − δ_{ij})\frac{s_{ij}(y)}{|s_{ii}(y)|})_{i,j=1}^{n}  (5.31)

satisfies condition (5.17).

Assume

\begin{matrix}P_{20}(x, y) = (δ_{ij}p_{2ij}(x, y))_{i,j=1}^{n}, \\ P_{21}(x, y) = ((1 − δ_{ij})p_{2ij}(x, y))_{i,j=1}^{n}, \\ h_{0}(v)(y) = (v_{i}(a) − v_{i}(0))_{i=1}^{n}, H_{j} = Θ (j = 1, …, ν). \end{matrix} (5.32)

Then

M_{10}(y) = diag (γ_{1}(y), …, γ_{n}(y)),

where

γ_{i}(y) = exp ( ∫_{0}^{a} p_{2ii}(s, y)ds) − 1 (i = 1, …, n).

In view of (5.26) and (5.30)

|γ_{i}(y)| ≥ |1 − exp(s_{ii}(y)a)| ≥ |1 − exp(−aδ)| > 0 (i = 1, …, n), (5.33)

whence it is clear that M_{10} satisfies condition (5.16). According to Theorem 5.2, to complete the proof it suffices to show that inequality (5.18) holds almost everywhere in D_{ab}.

In view of (5.32) and (5.33) Green’s matrix of problem (5.9) has the form

G(x, s, y) = diag(g_{1}(x, s, y), …, g_{n}(x, s, y)),

where

g_{i}(x, s, y) = \{ \begin{matrix}γ_{i}^{−1}(y) exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ) for s ≤ x \\ γ_{i}^{−1}(y) exp ( ∫_{0}^{a} p_{2ii}(ξ, y)dξ + ∫_{s}^{x} p_{2ii}(ξ, y)dξ) for s > x \end{matrix} .

Therefore

∑_{j=1}^{ν} |Z_{0}(x, y)M_{10}^{−1}(y)H_{j}| + ∫_{0}^{a} |G(x, s, y)P_{21}(s, y)|ds =

= ((1 − δ_{ij}) ∫_{0}^{a} |g_{i}(x, s, y)p_{2ij}(s, y)|ds)_{i,j=1}^{n}.  (5.34)

According to (5.26) and (5.30s), for σ_{i} = 1 we have

∫_{0}^{a} |g_{i}(x, s, y)p_{2ij}(s, y)|ds ≤ |γ_{i}^{−1}(y)|\frac{s_{ij}(y)}{|s_{ii}(y)|} ×

× ∫_{0}^{x} exp ( − ∫_{s}^{x} |p_{2ii}(ξ, y)|dξ)|p_{2ii}(s, y)|ds +

+|γ_{i}^{−1}(y)|\frac{s_{ij}(y)}{|s_{ii}(y)|} exp ( − ∫_{0}^{x} |p_{2ii}(ξ, y)|dξ) ×

× ∫_{x}^{a} exp ( − ∫_{s}^{a} |p_{2ii}(ξ, y)|dξ)|p_{2ii}(s, y)|ds = \frac{s_{ij}(y)}{|s_{ii}(y)|}.

Similarly, we can show that for σ_{i} = −1, the estimate

∫_{0}^{a} |g_{i}(x, s, y)p_{2ij}(s, y)|ds ≤ \frac{s_{ij}(y)}{|s_{ii}(y)|}

is also valid.

By virtue of this estimate, inequality (5.18) follows from equalities (5.31) and (5.34). ■

Corollary 5.2. Let the conditions of Corollary 5.1 take place. Then problem (5.24), (5.25) has the unique solution (u_{i})_{i=1}^{n} and condition (5.27) holds, where u_{i0}(x, y) ≡ 0 (i = 1, …, n) and for any natural m and i ∈ ∈ {1, …, n} the function u_{im} is a solution of equation (5.28) with the boundary conditions

u_{im}(x, 0) = φ_{0i}(x), \frac{∂u_{im}(x_{i}, y)}{∂y} = \frac{∂u_{im−1}(s_{i}, y)}{∂y} + φ_{1i}(y), (5.35)

where s_{i} = \frac{1+σ_{i}}{2} a and x_{i} = \frac{1−σ_{i}}{2} a.

Proof. To prove the corollary it suffices to find that if ν = n,

P_{20}(x, y) = (δ_{ik}p_{2ik}(x, y))_{i,k=1}^{n}, P_{21}(x, y) = ((1 − δ_{ik})p_{2ik}(x, y))_{i,k=1}^{n},

h_{0}(v)(y) = (v_{i}(x_{i}))_{i=1}^{n}, H_{j} = (δ_{ij}δ_{ik})_{i,k=1}^{n},

then all conditions of Theorem 5.3 are fulfilled. In our case Z_{0}(x, y) = diag ( exp ( ∫_{0}^{x} p_{211}(s, y)ds), …, exp ( ∫_{0}^{x} p_{2nn}(s, y)ds)), M_{10} = diag ( exp ( ∫_{0}^{x_{1}} p_{211}(s, y)ds), …, exp ( ∫_{0}^{x_{n}} p_{2nn}(s, y)ds))

and Green’s matrix of problem (5.9) has the form

G(x, s, y) = diag(g_{1}(x, s, y), …, g_{n}(x, s, y)),

where

g_{i}(x, s, y) = \{ \begin{matrix}exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ) sign(x − x_{i}) for (x−x_{i})(s−x_{i})≥0, \\ |s − x_{i}| ≤ |x − x_{i}| \\ 0 for (x − x_{i})(s − x_{i}) ≥ 0, |s − x_{i}| > |x − x_{i}| \\ 0 for (x − x_{i})(s − x_{i}) < 0 \end{matrix} ,

whence in view of (5.26) and (5.30) we obtain

M_{10}(y) ≥ E,

∑_{j=1}^{ν} |H_{j}| |Z_{0}(s_{j}, y)M_{10}^{−1}(y)| =

= diag ( exp ( ∫_{x_{1}}^{s_{1}} p_{211}(s, y)ds), …, exp ( ∫_{x_{n}}^{s_{n}} p_{2nn}(s, y)ds)) = = diag ( exp ( − ∫_{0}^{a} |p_{211}(s, y)|ds), …, exp ( − ∫_{0}^{a} |p_{2nn}(s, y)|ds)) ≤

≤ diag(exp(−δa), …, exp(−δa))

and

|Z_{0}(x, y)M_{10}^{−1}(y)|[E − ∑_{j=1}^{ν} |H_{j}| |Z_{0}(s_{j}, y)M_{10}^{−1}(y)|]^{−1} ×

× ∑_{j=1}^{ν} ∫_{0}^{a} |H_{j}| |G(s_{j}, s, y)P_{21}(s, y)|ds + ∫_{0}^{a} |G(x, s, y)P_{21}(s, y)|ds =

= ((1 − δ_{ij})[γ_{i}^{−1}(y) exp ( ∫_{x_{i}}^{x} p_{2ii}(s, y)ds) ∫_{0}^{a} |g_{i}(s_{i}, s, y)| ×

×|p_{2ij}(s, y)|ds + ∫_{0}^{a} |g_{i}(x, s, y)||p_{2ij}(s, y)|ds])_{i,j=1}^{n},  (5.36)

where

\overline{γ}_{i}(y) = 1 − exp ( − ∫_{0}^{a} |p_{2ii}(s, y)|ds) (i = 1, …, n).

Consequently, inequalities (5.16) and (5.22) are fulfilled. On the other hand,

∫_{0}^{a} g_{i}(s_{i}, s, y)|p_{2ij}(s, y)|ds =

= | ∫_{x_{i}}^{s_{i}} exp ( ∫_{s}^{s_{i}} p_{2ii}(ξ, y)dξ)|p_{2ij}(s, y)|ds| ≤

≤ \frac{s_{ij}(y)}{|s_{ii}(y)|}| ∫_{0}^{a} exp ( ∫_{s}^{s_{i}} p_{2ii}(ξ, y)dξ)p_{2ii}(s, y)ds| =

= \overline{γ}_{i}(y)\frac{s_{ij}(y)}{|s_{ii}(y)|} (i = 1, …, n)

and

∫_{0}^{a} g_{i}(x, s, y)|p_{2ij}(s, y)|ds ≤

≤ \frac{s_{ij}(y)}{|s_{ii}(y)|}| ∫_{x_{i}}^{x} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)p_{2ii}(s, y)ds| =

= \frac{s_{ij}(y)}{|s_{ii}(y)|}|1 − exp ( ∫_{x_{i}}^{x} p_{2ii}(ξ, y)dξ)|.

Therefore from equalities (5.31) and (5.36) there follows estimate (5.23) and, as admitted above, A satisfies condition (5.17). ■

Remark 5.1. If in boundary condition (5.35) we put s_{i} = \frac{1−σ_{i}}{2} a and x_{i} = \frac{1+σ_{i}}{2} a for any i ∈ \{1, …, n\}, then the conclusion of Corollary 5.2 becomes invalid. Indeed, consider the problem

\frac{∂^{2}u(x, y)}{∂x∂y} = \frac{∂u(x, y)}{∂y} + 1,  (5.37)

u(x, 0) = 0, \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y}

for which all conditions of Corollary 5.2 are fulfilled. In that case n = 1 and σ_{1} = −1. If we assume that x_{1} = \frac{1+σ_{1}}{2} = 0 and s_{1} = a, then (5.35) takes the form

u_{m}(x, 0) = 0, \frac{∂u_{m}(0, y)}{∂y} = \frac{∂u_{m−1}(a, y)}{∂y}.  (5.38)

For any natural m problem (5.37),(5.38) has the unique solution u_{m} and

\frac{∂u_{m}(x, y)}{∂y} = (\frac{∂u_{m−1}(a, y)}{∂y} + 1) exp(x) − 1.

Since u_{1}(a, y) = 0, from the last equality we obtain by induction

\frac{∂u_{m}(x, y)}{∂y} = ( exp((m − 1)a) + 1) exp(x) − 1 → +∞ for m → +∞.

§ 6. Dependence of Solutions on Coefficients of the System and on Boundary Conditions

In this section for the hyperbolic system

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q(x, y)  (6.1)

we consider the boundary value problems

u(x, 0) = φ_{0}(x), h(\frac{∂u(·, y)}{∂y})(y) = φ_{1}(y);  (6.2)
lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), h(u(·, y))(y) = ψ_{1}(y) (6.3)

and establish the conditions for the stability of their solutions with respect both to small perturbations of coefficients of the system and boundary data. As above, unless otherwise stated, it is assumed that

P_{i} ∈ L_{∞}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q ∈ L_{∞}(D_{ab}; ℝ^{n}),
φ_{0} ∈ ℂ̃([0, a]; ℝ^{n}), φ_{1} ∈ L([0, b]; ℝ^{n}),
ψ_{0} ∈ L([0, a]; ℝ^{n}), ψ_{1} ∈ L([0, b]; ℝ^{n}),

h : ℂ̃([0, a]; ℝ^{n}) → ℂ̃_{∞}([0, b]; ℝ^{n}) is a linear continuous operator and Z_{2} is the solution of the matrix differential equation

\frac{∂Z(x, y)}{∂x} = P_{2}(x, y)Z(x, y)

satisfying the initial condition Z(0, y) = E.

6.1. Problem (6.1),(6.2). Alongside with (6.1),(6.2), for any natural k, consider the problem

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0k}(x, y)u(x, y) + P_{1k}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2k}(x, y)\frac{∂u(x, y)}{∂y} + q_{k}(x, y),  (6.4)
u(x, 0) = φ_{0k}(x), h_{k}(\frac{∂u(·, y)}{∂y})(y) = φ_{1k}(x),  (6.5)

where

P_{ik} ∈ L_{∞}(D_{ab}; ℝ^{n×n}) (i = 0, 1, 2), q_{k} ∈ L_{∞}(D_{ab}; ℝ^{n}),
φ_{0k} ∈ ℂ̃([0, a]; ℝ^{n}), φ_{1k} ∈ L([0, b]; ℝ^{n}),

and h_{k} : ℂ̃([0, a]; ℝ^{n}) → L_{∞}([0, b]; ℝ^{n}) is a linear continuous operator.

By Lemmas 2.1_{1} and 2.3_{1} for an arbitrary v ∈ ℂ̃([0, a]; ℝ^{n}) we have

h(v)(y) = H_{0}(y)v(0) + ∫_{0}^{a} H(s, y)v′(s)ds,  (6.6)
h_{k}(v)(y) = H_{0k}(y)v(0) + ∫_{0}^{a} H_{k}(s, y)v′(s)ds,  (6.7)

where

H_{0} ∈ ℂ̃_{∞}([0, b]; ℝ^{n} × n) and H_{0k} ∈ L_{∞}([0, b]; ℝ^{n×n}),
H ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) and H_{k} ∈ L_{∞}(D_{ab}; ℝ^{n×n}).

Put

\begin{matrix}M_{0}(y) = H_{0}(y) + ∫_{0}^{a} H(s, y)\frac{∂Z_{2}(s, y)}{∂s}ds, \\ M(x, y) = H(x, y)Z_{2}(x, y) + ∫_{x}^{a} H(s, y)\frac{∂Z_{2}(s, y)}{∂s}ds \end{matrix}  (6.8)

and

\begin{matrix}M_{0k}(y) = H_{0k}(y) + ∫_{0}^{a} H_{k}(s, y)\frac{∂Z_{2k}(s, y)}{∂s}ds, \\ M_{k}(x, y) = H_{k}(x, y)Z_{2k}(x, y) + ∫_{x}^{a} H_{k}(s, y)\frac{∂Z_{2k}(s, y)}{∂s}ds, \end{matrix} (6.9)

where Z_{2k} is the solution of the matrix differential equation

\frac{∂Z(x, y)}{∂x} = P_{2k}(x, y)Z(x, y)

satisfying the initial condition Z(0, y) = E.

Theorem 6.1. Let

ess inf_{0≤y≤b} | det(M_{0}(y))| > 0,  (6.10)
sup ∥P_{ik}∥_{L_{∞}} < +∞ (i = 1, 2),

ess sup_{(x,y)∈D_{ab}} ∥ ∫_{0}^{x} [P_{2k}(s, y) − P_{2}(s, y)]ds∥ → 0 for k → +∞, (6.11)

lim_{k→+∞} ∥P_{ik} − P_{i}∥_{L} = 0 (i = 0, 1, 2), lim_{k→+∞} ∥q_{k} − q∥_{L} = 0 (6.12)

and

lim_{k→+∞} ∥φ_{0k} − φ_{0}∥_{ℂ} = 0, lim_{k→+∞} ∥φ_{1k} − φ_{1}∥_{L} = 0,

lim_{k→+∞} |||h − h_{k}|||^{10} = 0 (6.13)

Then, starting from some k_{0}, problem (6.4), (6.5) has the unique solution u_{k} and

lim_{k→+∞} ∥u_{k} − u_{0}∥_{ℂ} = 0.  (6.14)

where u_{0} is the solution of problem (6.1), (6.2). Proof. By Lemmas 2.2_{1} and 2.3_{3}

Z_{2} and Z_{2}^{−1} ∈ ℂ̃_{∞}^{(0,−1)}(D_{ab}; ℝ^{n×n}),  (6.15)
M_{0} ∈ ℂ̃_{∞}([0, b]; ℝ^{n×n}), M ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}).  (6.16)

Let us show that

lim_{k→+∞} ∥Z_{2k} − Z_{2}∥_{L_{∞}} = 0, lim_{k→+∞} ∥Z_{2k}^{−1} − Z_{2}^{−1}∥_{L_{∞}} = 0. (6.17)

Put

Z_{0k}(x, y) = Z_{2k}(x, y) − Z_{2}(x, y).

Then

Z_{0k}(x, y) = ∫_{0}^{x} [P_{2k}(s, y)Z_{2k}(s, y) − P_{2}(s, y)Z_{2}(s, y)]ds = = ∫_{0}^{x} P_{2k}(s, y)Z_{0k}(s, y)ds + ∫_{0}^{x} [P_{2k}(s, y) − P_{2}(s, y)]Z_{2}(s, y)ds.

Therefore

∥Z_{0k}(x, y)∥ ≤ α ∫_{0}^{x} ∥Z_{0k}(s, y)∥ds + ε_{k},

where

ε_{k} = ess sup_{(x,y)∈D_{ab}} ∥ ∫_{0}^{x} [P_{2k}(s, y) − P_{2}(s, y)]Z_{2}(s, y)ds∥,

α = sup_{k≥1} ∥P_{2k}∥_{L_{∞}},

whence by Gronwall’s lemma we have

∥Z_{0k}∥_{L_{∞}} ≤ ε_{k} exp(aα).

However, according to conditions (6.11) and (6.15)

lim_{k→+∞} ε_{k} = 0.^{11}

Therefore

lim_{k→+∞} ∥Z_{2k} − Z_{2}∥_{L_{∞}} = 0.

10Here and everywhere below by |||h||| is understood the norm of operator h.

11See the proof of Lemma 3.14.

According to representations (6.6) and (6.7), it follows from the convergence by the norm of a sequence of operators h_{k} (k = 1, 2, …) to the operator h that

lim_{k→+∞} ∥H_{0k} − H_{0}∥_{L_{∞}} = 0, lim_{k→+∞} ∥H_{k} − H∥_{L_{∞}} = 0.  (6.18)

In view of (6.8) and (6.9),

M_{0k}(y) − M_{0}(y) = H_{0k}(y) − H_{0}(y) + ∫_{0}^{a} [H_{k}(s, y) −

−H(s, y)]P_{2k}(s, y)Z_{2k}(s, y)ds + ∫_{0}^{a} H(s, y)P_{2k}(s, y)[Z_{2k}(s, y) −

−Z_{2}(s, y)]ds + ∫_{0}^{a} H(s, y)[P_{2k}(s, y) − P_{2}(s, y)]Z_{2}(s, y)ds,
M_{k}(x, y) − M(x, y) = H_{k}(x, y)Z_{2k}(x, y) − H(x, y)Z_{2}(x, y) +

+ ∫_{x}^{a} [H_{k}(s, y) − H(s, y)]P_{2k}(s, y)Z_{2k}(s, y)ds + ∫_{x}^{a} H(s, y)P_{2k}(s, y) ×

×[Z_{2k}(s, y) − Z_{2}(s, y)]ds + ∫_{x}^{a} H(s, y)[P_{2k}(s, y) − P_{2}(s, y)]Z_{2}(s, y)ds.

Therefore

∥M_{0k} − M_{0}∥_{L_{∞}} ≤ ∥H_{0k} − H_{0}∥_{L_{∞}} +
+α_{0}(∥H_{k} − H∥_{L_{∞}} + ∥Z_{2k} − Z_{2}∥_{L_{∞}}) + ε_{0k},  (6.19)
∥M_{k} − M∥_{L_{∞}} ≤ ∥H_{k}Z_{2k} − HZ_{2}∥_{L_{∞}} +
+α_{0}(∥H_{k} − H∥_{L_{∞}} + ∥Z_{2k} − Z_{2}∥_{L_{∞}}) + ε_{0k},  (6.20)

where

α_{0} = sup_{k≥1}(a∥P_{2k}Z_{2k}∥_{L_{∞}} + a∥HP_{2k}∥_{L_{∞}}),

ε_{0k} = ess sup_{(x,y)∈D_{ab}} ∥ ∫_{x}^{a} H(s, y)[P_{2k}(s, y) − P_{2}(s, y)]Z_{2}(s, y)ds∥.

Because of the fact that H ∈ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) and Z_{2} ∈ℂ̃_{∞}^{(0,−1)}(D_{ab};ℝ^{n×n}) and taking into account condition (6.11),

lim_{k→+∞} ε_{0k} = 0.

If together with this we take into consideration conditions (6.17) and (6.18), then from (6.19) and (6.20) we get

lim_{k→+∞} ∥M_{0k} − M_{0}∥_{L_{∞}} = 0, lim_{k→+∞} ∥M_{k} − M∥_{L_{∞}} = 0. (6.21)

According to conditions (6.10) and (6.21), there exist a natural k_{0} and a positive number δ such that

ess inf_{0≤y≤b} | det(M_{0k}(y))| > δ for k ≥ k_{0}  (6.22)

and

lim_{k→+∞} ∥M_{0k}^{−1} − M_{0}^{−1}∥_{L_{∞}} = 0.  (6.23)

By Theorem 4.1 condition (6.10) guarantees the unique solvability of problem (6.1),(6.2), while condition (6.22) ensures the unique solvability of problem (6.4),(6.5) for an arbitrary k ≥ k_{0}. Let us denote the solutions of these problems by u_{0} and u_{k}. In proving Theorem 4.1 we showed that u = u_{0} and u = u_{k} satisfy, respectively, the boundary conditions

u(x, 0) = φ_{0}(x), \frac{∂u(0, y)}{∂y} = ψ(y) +

+ ∫_{0}^{a} [Q_{0}(s, y)u(s, y) + γ_{1}^{−1}(s)Q_{1}(s, y)\frac{∂u(s, y)}{∂s}]ds  (6.24)

and

u(x, 0) = φ_{0k}(x), \frac{∂u(0, y)}{∂y} = ψ_{k}(y) +

+ ∫_{0}^{a} [Q_{0k}(s, y)u(s, y) + γ_{1k}^{−1}(s)Q_{1k}(s, y)\frac{∂u(s, y)}{∂s}]ds,  (6.25)

where

\begin{matrix}γ_{1}(x) = 1 + ∥φ′_{0}(x)∥, \\ ψ(y) = M_{0}^{−1}(y)[φ_{1}(y) − ∫_{0}^{a} M(s, y)Z_{2}^{−1}(s, y)q(s, y)ds], \\ Q_{0}(x, y) = −M_{0}^{−1}(y)M(x, y)Z_{2}^{−1}(x, y)P_{0}(x, y), \\ Q_{1}(x, y) = −γ_{1}(x)M_{0}^{−1}(y)M(x, y)Z_{2}^{−1}(x, y)P_{1}(x, y) \end{matrix} (6.26)

and

\begin{matrix}γ_{1k}(x) = 1 + ∥φ′_{0k}(x)∥, \\ ψ_{k}(y)=M_{0k−1}^{−1}(y)[φ_{1k}(y)−∫_{0}^{a} M_{k}(s, y)Z_{2k}^{−1}(s, y)q_{k}(s, y)ds], \\ Q_{0k}(x, y) = −M_{0k}^{−1}(y)M_{k}(x, y)Z_{2k}^{−1}(x, y)P_{0k}(x, y), \\ Q_{1k}(x, y) = −γ_{1k}(x)M_{0k}^{−1}(y)M_{k}(x, y)Z_{2k}^{−1}(x, y)P_{1k}(x, y). \end{matrix} (6.27)

More exactly, problem (6.1),(6.2) is equivalent to problem (6.4),(6.24) and for any k ≥ k_{0} problem (6.4),(6.5) is equivalent to problem (6.4),(6.25).

By conditions (6.15)-(6.17) and (6.21)-(6.23) we have

ε_{1k} = ∥M_{0k}^{−1} − M_{0}^{−1}∥_{L_{∞}} +

+ ∥M_{0k}^{−1}M_{k}Z_{2k}^{−1} − M_{0}^{−1}MZ_{2}^{−1}∥_{L_{∞}} → 0 for k → +∞

and

β = ∥M_{0}^{−1}∥_{L_{∞}} + ∥M_{0}^{−1}MZ_{2}^{−1}∥_{L_{∞}} < +∞.

If along with this we take into account conditions (6.11)-(6.13), then from equalities (6.26) and (6.27) we find

sup_{k≥1} ∥γ_{1k}^{−1}Q_{1k}∥_{L_{∞}} ≤ β sup_{k≥1} ∥P_{1k}∥_{L_{∞}} < +∞,

∥Q_{0k} − Q_{0}∥_{L} ≤ ∥M_{0k}^{−1}M_{k}Z_{2k}^{−1} − M_{0}^{−1}MZ_{2}^{−1}∥_{L_{∞}}∥P_{0k}∥_{L} +
+∥M_{0}^{−1}MZ_{2}^{−1}∥_{L_{∞}}∥P_{0k} − P_{0}∥_{L} ≤
≤ ε_{1k}∥P_{0k}∥_{L} + β∥P_{0k} − P_{0}∥_{L} → 0 for k → +∞,
∥γ_{1}γ_{1k}^{−1}Q_{1k} − Q_{1}∥_{L} ≤ ε_{1k}∥γ_{1}P_{1k}∥_{L} + β∥γ_{1}(P_{1k} − P_{1})∥_{L} → 0
for k → +∞,

∥ψ_{k} − ψ∥ ≤ ε_{1k}∥φ_{1k}∥_{L} + β(∥φ_{1k} − φ_{1}∥_{L} + ∥q_{k} − q∥_{L}) → for k → +∞.

Consequently, all conditions of Lemma 3.9 are fulfilled and by virtue of this lemma equality (6.14) is valid. ■

Based on Lemmas 3.10 and 3.11, similarly to the above reasoning, we prove the following theorems.

Theorem 6.2. Let conditions (6.10) and (6.13) hold,

sup ∥P_{ik}∥_{L_{∞}} < +∞ (i = 0, 1, 2), sup_{k≥1} ∥q_{k}∥_{L_{∞}} < +∞, (6.28)

ess sup_{(x,y)∈D_{ab}} ∥ ∫_{0}^{x} [P_{2k}(s, y) − P_{2}(s, y)]ds∥ → 0 for k → +∞

and

lim_{k→+∞} ∥P_{ik} − P_{i}∥_{L}^{(1)} = 0 (i = 0, 1, 2), lim_{k→+∞} ∥q_{k} − q∥_{L}^{(1)} = 0.

Then, starting from some k_{0}, problem (6.4),(6.5) has the unique solution u_{k} and

lim_{k→+∞} ∥u_{k} − u_{0}∥(1) ; _{ℂ} = 0,  (6.29)

where u_{0} is the solution of problem (6.1),(6.2).

Theorem 6.3. Let conditions (6.10) and (6.28) hold and, besides, φ_{0} and φ_{0k} ∈ ℂ̃_{∞}([0, a]; ℝ^{n}), φ_{1} and φ_{1k} ∈ L_{∞}([0, b]; ℝ^{n}),

lim_{k→+∞} ∥P_{ik} − P_{i}∥_{L}^{(2)} = 0 (i = 0, 1, 2), lim_{k→+∞} ∥q_{k} − q∥_{L}^{(2)} = 0

and

lim_{k→+∞} ∥φ_{0k} − φ_{0}∥_{ℂ_{∞}} = 0, lim_{k→+∞} ∥φ_{1k} − φ_{1}∥_{L_{∞}} = 0, lim_{k→+∞} |||h − h_{k}||| = 0.

Then, starting from some k_{0}, problem (6.4),(6.5) has the unique solution u_{k} and

lim_{k→+∞} ∥u_{k} − u_{0}∥(2) ; _{ℂ} = 0,  (6.30)

where u_{0} is the solution of problem (6.1),(6.2).

Remark 6.1. In Theorems 6.2 and 6.3 the condition

sup_{k≥1} ∥P_{1k}∥_{L_{∞}} < +∞, sup_{k≥1} ∥q_{k}∥ < +∞  (6.31)

are essential and cannot be neglected. As an example, consider the problems

\frac{∂^{2}u(x, y)}{∂x∂y} = \frac{∂u(x, y)}{∂y},  (6.32)
u(x, 0) = 0, u(0, y) − u(a, y) = 0  (6.33)

and

\frac{∂^{2}u(x, y)}{∂x∂y} = −[k cos k^{2}(x + y)]\frac{∂u(x, y)}{∂x} + \frac{∂u(x, y)}{∂y} + k sin k^{2}(x + y), (6.34)
u(x, 0) = 0, u(0, y) − u(a, y) = 0,  (6.35)

for which all conditions of Theorems 6.2 and 6.3, except (6.31), are fulfilled. Since M_{0k}(y) ≡ M_{0}(y) = exp(ay) − 1, by Corollary 4.1 problem (6.32), (6.33) has only the trivial solution u_{0}(x, y) ≡ 0 and problem (6.34), (6.35) has the unique solution u_{k} for any natural k. Moreover,

\frac{∂u_{k}(x, y)}{∂x} = exp ( − \frac{sin k^{2}(x + y)}{k}) ×

× ∫_{0}^{y} exp (\frac{sin k^{2}(x + t)}{k})(\frac{∂u_{k}(x, t)}{∂t} + k sin k^{2}(x + t))dt.

The assumption for conditions (6.29) or (6.30) to be valid leads us to the false equality

lim_{k→+∞} k ∫_{0}^{x} | ∫_{0}^{y} exp (\frac{sin k^{2}(s + t)}{k}) sin k^{2}(s + t)dt|ds = 0,

because

k ∫_{0}^{y} exp (\frac{sin k^{2}t}{k}) sin k^{2}tdt = \frac{1}{k} − \frac{cos k^{2}y}{k} exp (\frac{sin k^{2}y}{k} ) +

+\frac{1}{2} ∫_{0}^{y} exp (\frac{sin k^{2}t}{k})(1 + cos 2k^{2}t)dt ⇉ \frac{y}{2} for k → +∞.

6.2. Problem (6.1),(6.3). For any natural k consider the hyperbolic system

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0k}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q_{k}(x, y)  (6.36)

with the boundary conditions

lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0k}(x), h_{k}(u(·, y))(y) = ψ_{1k}(y), (6.37)

where

P_{0k} ∈ L_{∞}(D_{ab}; ℝ^{n×n}), q_{k} ∈ L_{∞}(D_{ab}; ℝ^{n}),
ψ_{0k} ∈ L([0, a]; ℝ^{n}), ψ_{1k} ∈ L_{∞}([0, b]; ℝ^{n})

and h_{k} : ℂ̃([0, a]; ℝ^{n}) → L_{∞}([0, b]; ℝ^{n}) is a linear continuous operator.

As above, we use representations (6.6)-(6.8) and by M_{0k} and M_{k} are meant the matrix functions given by the equalities

M_{0k}(y) = H_{0k}(y) + ∫_{0}^{a} H_{k}(s, y)\frac{∂Z_{2}(s, y)}{∂s}ds,

M_{k}(x, y) = H_{k}(x, y)Z_{2}(x, y) + ∫_{x}^{a} H_{k}(s, y)\frac{∂Z_{2}(s, y)}{∂s}ds.

Theorem 6.4. Let P_{1} ∈ ℂ̃_{∞}^{(0,−1)}(D_{ab}; ℝ^{n×n}), P_{2} ∈ ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}),

ess inf_{0≤y≤b} | det(M_{0}(y))| > 0,
sup_{k≥1} ∥P_{0k}∥_{L_{∞}} < +∞, sup_{k≥1} ∥q_{k}∥_{L_{∞}} < +∞,  (6.38)
lim_{k→+∞} ∥P_{0k} − P_{0}∥_{L}^{(0)} = 0, lim_{k→+∞} ∥q_{k} − q∥_{L}^{(0)} = 0  (6.39)

and

lim_{k→+∞} ∥ψ_{0k} − ψ_{0}∥_{L} = 0, lim_{k→+∞} ∥ψ_{1k} − ψ_{1}∥_{L_{∞}} = 0,

lim_{k→+∞} |||h_{k} − h||| = 0. (6.40)

Then, starting from some k_{0}, problem (6.36),(6.37) has the unique generalized solution u_{k} and

lim_{k→+∞} ∥u_{k} − u_{0}∥_{L_{∞}} = 0,  (6.41)

where u_{0} is the generalized solution of problem (6.1),(6.3).

Proof. As shown above, conditions (6.21)-(6.23) follow from the convergence by the norm of the sequence of operators h_{k} (k = 1, 2, ⋯) to h and the restrictions imposed on det(M_{0}(y)), where k_{0} is a natural number and δ is a positive constant independent of k.

The unique solvability of problem (6.1),(6.3) as well as of problem (6.36), (6.37) for any k ≥ k_{0} follows from Theorem 4.4. In proving this theorem it was admitted that u = u_{0} and u = u_{k} satisfy, respectively, the boundary conditions

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ_{0}(x), \\ u(0, y) = ψ(y) + ∫_{0}^{y} ∫_{0}^{a} Q(y, s, t)u(s, t)dsdt \end{matrix}  (6.42)

and

\begin{matrix}lim_{y→0} (\frac{∂u(x, y)}{∂y} − P_{2}(x, y)u(x, y)) = ψ_{0k}(x), \\ u(0, y) = ψ_{k}(y) + ∫_{0}^{y} ∫_{0}^{a} Q_{k}(y, s, t)u(s, t)dsdt, \end{matrix}  (6.43)

where

ψ(y) = M_{0}^{−1}(y)ψ_{1}(y) − M_{0}^{−1}(y) ∫_{0}^{a} M(x, y)Z_{2}^{−1}(x, y)Z_{1}(x, y) ×
×[ψ_{0}(x) + ∫_{0}^{y} Z_{1}^{−1}(x, t)q(x, t)dt]dx,
Q(y, s, t) = −M_{0}^{−1}(y)M(s, y)Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)P(s, t),

P(x, y) = P_{0}(x, y) + P_{1}(x, y)P_{2}(x, y) − \frac{∂P_{2}(x, y)}{∂y}

and

ψ_{k}(y) = M_{0k}^{−1}(y)ψ_{1k}(y) − M_{0k}^{−1}(y) ∫_{0}^{a} M_{k}(x, y)Z_{2}^{−1}(x, y)Z_{1}(x, y) ×

×[ψ_{0k}(x) + ∫_{0}^{y} Z_{1}^{−1}(x, t)q_{k}(x, t)dt]dx,
Q_{k}(y, s, t) = −M_{0k}^{−1}(y)M_{k}(s, y)Z_{2}^{−1}(s, y)Z_{1}(s, y)Z_{1}^{−1}(s, t)P_{k}(s, t),

P_{k}(x, y) = P_{0k}(x, y) + P_{1}(x, y)P_{2}(x, y) − \frac{∂P_{2}(x, y)}{∂y}.

More exactly, problem (6.1),(6.3) is equivalent to problem (6.1),(6.42) and problem (6.36),(6.37) to problem (6.36),(6.43).

It follows from conditions (6.21)-(6.23) and (6.38)-(6.40) that

sup_{k≥1} ∥Q_{k}∥_{L_{∞}} < +∞, lim_{k→+∞} ∥ψ_{k} − ψ∥_{L_{∞}} = 0

and

ess sup_{\begin{matrix}0≤y_{1}<y_{2}≤b \\ (x,y)∈D_{ab} \end{matrix}} ∥ ∫_{y_{1}}^{y_{2}} ∫_{0}^{x} [Q_{k}(y, s, t) − Q(y, s, t)]dsdt∥ → 0 for k → +∞.

Consequently, all conditions of Lemma 3.15 hold and by virtue of this lemma equality (6.41) is valid. ■

Remark 6.2. If h_{k} = h (k = 1, 2, ⋯), the operator h being such that

H ∈ ℂ̃_{∞}^{(0,−1)}(D_{ab}; ℝ^{n×n}),  (6.44)

then the assumption

sup_{k≥1} ∥q_{k}∥_{L_{∞}} < +∞  (6.45)

in Theorem 6.4 becomes unnecessary. If, however, (6.44) is violated, then restriction (6.45) is essential and cannot be neglected. As an example, consider the problems

\frac{∂^{2}u(x, y)}{∂x∂y} = 0  (6.46)
\frac{∂u(x, 0)}{∂x} = 0, u(0, y) + ∫_{0}^{a} H(s)\frac{∂u(s, y)}{∂s}ds = 0  (6.47)

and

\frac{∂^{2}u(x, y)}{∂x∂y} = −k^{3} sin k^{8}x,  (6.48)
\frac{∂u(x, 0)}{∂x} = 0, u(0, y) + ∫_{0}^{a} H(s)\frac{∂u(s, y)}{∂s}ds = 0,  (6.49)

where

H(x) = ∑_{m=1}^{+∞} \frac{2}{m^{2}} sin m^{8}x.

For these problems all conditions of Theorem 6.4, except (6.44), are fulfilled. On the other hand, problem (6.46),(6.47) has the unique solution u_{0}(x, y) ≡ ≡ 0 and problem (6.48),(6.49) has the unique solution

u_{k}(x, y) = k^{3}y ∫_{0}^{a} H(s) sin k^{8}sds + \frac{y}{k^{5}} (cos k^{8}x − 1)

for any natural k.

However,

k^{3} ∫_{0}^{a} H(s) sin k^{8}sds = 2k ∫_{0}^{a} sin^{2} k^{8}sds +

+ ∑_{m≠k,m=1}^{+∞} \frac{2k^{3}}{m^{2}} ∫_{0}^{a} sin m^{8}s sin k^{8}sds = k ∫_{0}^{a}(1 − cos 2k^{8}s)ds +

+ ∑_{m≠k,m=1}^{+∞} \frac{k^{3}}{m^{2}} ∫_{0}^{a}[cos(k^{8} − m^{8})s − cos(k^{8} + m^{8})s]ds ≥

≥ ak − \frac{1}{2k^{7}} − ∑_{m≠k,m=1}^{+∞} \frac{k^{3}}{m^{2}} (\frac{1}{|k^{8} − m^{8}|} + \frac{1}{k^{8} + m^{8}} ) > ak − α,

where

α = ∑_{m=1}^{+∞} \frac{1}{m^{2}} .

Therefore

∥u_{k} − u_{0}∥_{L_{∞}} > abk − αb − 2b → +∞ for k → +∞,

i.e. (6.41) is violated. ■

CHAPTER III

BOUNDARY VALUE PROBLEMS IN A STRIP

§ 7. Bounded Solutions

This section deals with the problem on the existence and uniqueness in a strip D_{b} of a solution u of the linear hyperbolic system

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q(x, y),  (7.1)

satisfying the conditions

u(x, 0) = φ(x), ess sup_{(x,y)∈D_{b}} (∥\frac{∂u(x, y)}{∂x}∥ + ∥\frac{∂u(x, y)}{∂y}∥) < +∞, (7.2)

where

\begin{matrix}P_{j} = (p_{jik})_{j,k=1}^{n} ∈ L_{∞}(D_{b}; ℝ^{n×n}) (j = 0, 1, 2), \\ q = (q_{i})_{i=1}^{n} ∈ L_{∞}(D_{b}; ℝ^{n}), φ = (φ_{i})_{i=1}^{n} ∈ ℂ̃_{∞}(ℝ, ℝ^{n}). \end{matrix}  (7.3)

For an arbitrary function z ∈ L_{∞}(D_{b}; ℝ) the following notation will be used:

I_{+}(z) = {y ∈ [0, b] : sup_{x≥0} ∫_{0}^{x} z(s, y)ds = +∞},

I_{−}(z)={y ∈[0, b] : sup_{x≥0} ∫_{0}^{x} z(s, y)ds<+∞ and sup_{x≤0} ∫_{0}^{x} z(s, y)ds=+∞},

I_{0}(z) = {y ∈ [0, b] : sup_{x∈ℝ} ∫_{0}^{x} z(s, y)ds < +∞},

χ(z)(y) = \begin{cases}+∞ & for y ∈ I_{+}(z) \\ 0 & for y ∈ I_{0}(z) \\ −∞ & for y ∈ I_{−}(z) \end{cases} .

For an arbitrary set I ⊂ ℝ we denote by mes I its Lebesque measure and by Ī its closure.

Theorem 7.1. Let there exist constants α ∈ (0, 1), β > 0 and essentially bounded measurable functions a_{ik} : [0, b] → [0, +∞) (i ≠ k; i, k = 1, …, n) such that the spectral radius of the matrix A(y) = (a_{ik}(y))_{i,k=1}^{n}, where a_{ii}(y) ≡ 0 (i = 1, …, n), is less than α almost for all y ∈ [0, b] and the inequalities

| ∫_{x}^{χ(p_{2ii})(y)}exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ) ×

×(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| + |q_{i}(s, y)|)ds| ≤ β (i, k = 1, …, n) (7.4)

and

\begin{matrix}| ∫_{x}^{χ(p_{2ii})(y)}exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds| ≤ a_{ik}(y) \\ (i ≠ k; i, k = 1, …, n) \end{matrix}  (7.5)

hold almost everywhere in D_{b}. Then problem (7.1),(7.2) is solvable; moreover, the solution is unique if and only if

mes I_{0}(p_{2ii}) = 0 (i = 1, …, n).  (7.6)

Proof. In view of (7.3), without loss of generality it can be assumed that the inequality

∥P_{0}(x, y)∥ + b∥P_{1}(x, y)∥ + ∥P_{2}(x, y)∥ ≤ β  (7.7)

is fulfilled in D_{b}. First we prove the solvability of the problem under consideration on the basis of Lemma 1.3.

Let

ψ_{i}(y) = \begin{cases}1 & for y ∈ I_{0}(p_{2ii}) \\ 0 & for y ∉ I_{0}(p_{2ii}) \end{cases} ,  (7.8)
ψ(y) = (ψ_{i}(y))_{i=1}^{n},

and let c : [0, b] → ℝ be an arbitrary continuous function. For any z_{j} = = (z_{ji})_{i=1}^{n} ∈ L_{∞}(D_{b}; ℝ^{n}) (j = 1, 2) and i ∈ {1, …, n} we put

f_{i}(z_{1}, z_{2})(x, y) = ∑_{k=1}^{n} p_{0ik}(x, y)[φ_{i}(x) + ∫_{0}^{y} z_{2i}(x, t)dt] +

+ ∑_{k=1, k≠i}^{n} p_{2ik}(x, y)z_{2k}(x, y) + ∑_{k=1}^{n} p_{1ik}(x, y)z_{1k}(x, y) + q_{i}(x, y), (7.9)
g_{1i}(z_{1}, z_{2})(x, y) = φ′_{i}(x) +
+ ∫_{0}^{y} [p_{2ii}(x, t)z_{2i}(x, t) + f_{i}(z_{1}, z_{2})(x, t)]dt,  (7.10)

g_{2i}(z_{1}, z_{2})(x, y) = exp ( ∫_{0}^{x} p_{2ii}(ξ, y)dξ)c(y)ψ_{i}(y) +

+ ∫_{χ(p_{2ii})(y)}^{x} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)f_{i}(z_{1}, z_{2})(s, y)ds  (7.11)

and

\begin{matrix}g_{1}(z_{1}, z_{2})(x, y) = (g_{1i}(z_{1}, z_{2})(x, y))_{i=1}^{n}, \\ g_{2}(z_{1}, z_{2})(x, y) = (g_{2i}(z_{1}, z_{2})(x, y))_{i=1}^{n}. \end{matrix}  (7.12)

By virtue of conditions (7.3)-(7.5) and (7.7)-(7.12) the operators g_{1} and g_{2} transform the space L_{∞}(D_{b}; ℝ^{n}) × L_{∞}(D_{b}; ℝ^{n}) into L_{∞}(D_{b}; ℝ^{n}) and satisfy conditions (1.3) and (1.21), where g_{0}(t) ≡ nβ, while A_{01} and A_{02} are constant n × n matrices whose all elements equal to β.

Thus, all conditions of Lemma 1.3 are satisfied and therefore system (1.1) has the unique solution (z_{1}, z_{2}). Assume

u(x, y) = φ(x) + ∫_{0}^{y} z_{2}(x, t)dt.

On account of conditions (7.3)-(7.5) equalities (7.9)-(7.11) yield that u is locally absolutely continuous,

\frac{∂^{2}u(x, y)}{∂x∂y} = \frac{∂z_{2}(x, y)}{∂x} = P_{0}(x, y)u(x, y) + P_{1}(x, y)z_{1}(x, y) +

+P_{2}(x, y)z_{2}(x, y) + q(x, y),

\frac{∂u(x, y)}{∂x} = φ′(x) + ∫_{0}^{y} \frac{∂z_{2}(x, t)}{∂x}dt = φ′(x) +

+ ∫_{0}^{y}[P_{0}(x, t)u(x, t) + P_{1}(x, t)z_{1}(x, t) +

+P_{2}(x, t)z_{2}(x, t) + q(x, t)]dt = z_{1}(x, y)

and

\frac{∂u(0, y)}{∂y} = c(y)ψ(y).  (7.13)

Consequently, u is the solution of problem (7.1),(7.2). On the other hand, if mes I_{0}(p_{2ii}) > 0 for any i ∈ {1, …, n}, then in view of equalities (7.8) and (7.13) and an arbitrary choice of c it is evident that the problem under consideration has an infinite dimensional set of solutions.

To complete the proof we have to show that (7.1),(7.2) is uniquely solvable when (7.6) is fulfilled.

In view of (7.6) we get from (7.8) and (7.11) that

\begin{matrix}g_{2i}(z_{1}, z_{2})(x, y) = ∫_{χ(p_{2ii})(y)}^{x} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)f_{i}(z_{1}, z_{2})(s, y)ds \\ (i = 1, …, n). \end{matrix} (7.14)

As proved above, system (1.1) has the unique solution (z_{1}^{0}, z_{2}^{0}); moreover, u^{0}(x, y) = φ(x) + ∫_{0}^{y} z_{2}^{0}(x, t)dt is the solution of problem (7.1),(7.2). Let u be an arbitrary solution of this problem. Assume

z_{1}(x, y) = \frac{∂u(x, y)}{∂x}, z_{2}(x, y) = \frac{∂u(x, y)}{∂y}.

Then

ess sup_{y∈[0,b]} ( sup_{x∈ℝ} ∥z_{2}(x, y)∥) < +∞,  (7.15)

and for any x_{0i} ∈ ℝ (i = 1, …, n) we have

\begin{matrix}z_{1}(x, y) = g_{1}(z_{1}, z_{2})(x, y), \\ z_{2i}(x, y) = exp ( ∫_{x_{0i}}^{x} p_{2ii}(ξ, y)dξ)z_{2i}(x_{0i}, y) + \\ + ∫_{x_{0i}}^{x} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)f_{i}(z_{1}, z_{2})(s, y)ds (i = 1, …, n). \end{matrix} (7.16)

On the other hand, in view of (7.6), we have

lim inf_{x_{0i}→χ(p_{2ii})(y)} ∫_{x_{0i}}^{x} p_{2ii}(ξ, y)dξ = −∞ for x ∈ ℝ  (7.17)

almost for every y ∈ [0, b]. By virtue of conditions (7.14),(7.15) and (7.17) it follows from (7.16) that (z_{1}, z_{2}) is a solution of system (1.1). Thus z_{i}(x, y) ≡ ≡ z_{i}^{0}(x, y) (i = 1, 2) and, consequently,

\frac{∂u(x, y)}{∂x} ≡ z_{1}^{0}(x, y), \frac{∂u(x, y)}{∂y} ≡ z_{2}^{0}(x, y).

Then with regard to (7.2) we get

u(x, y) ≡ u^{0}(x, y). ■

The following theorem can be proved similarly to Theorem 7.1.

Theorem 7.1′. Let P_{j} (j = 0, 1, 2) and q be continuous and bounded, φ be continuously differentiable and bounded together with its derivative and for every i ∈ {1, …, n} either Ī_{+}(p_{2ii}) = [0, b] or Ī_{−}(p_{2ii}) = [0, b]. Moreover,let the integrals

∫_{0}^{χ_{i}} exp ( ∫_{s}^{0} p_{2ii}(ξ, y)dξ)(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| +
+(1 − δ_{ik})|p_{2ik}(s, y)| + |q_{i}(s, y)|)ds (k = 1, …, n)  (7.18)

where χ_{i} = +∞ and χ_{i} = −∞ for Ī_{+}(p_{2ii}) = [0, b] and Ī_{−}(p_{2ii}) = [0, b], respectively, converge uniformly with respect to y ∈ [0, b]. Let, besides, there exist a constant β > 0 and continuous functions a_{ik} : [0, b] → [0, +∞) (i ≠ k; i, k = 1, …, n) such that the spectral radius of the matrix A(y) = = (a_{ik}(y))_{i,k=1}^{n}, where a_{ii}(y) ≡ 0 (i = 1, …, n), is less than unity for every y ∈ [0, b] and the inequalities

\begin{matrix}| ∫_{x}^{χ_{i}} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| + \\ +|q_{i}(s, y)|)ds| ≤ β (k = 1, …, n) \end{matrix}  (7.19)

and

\begin{matrix}| ∫_{x}^{χ_{i}} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds| ≤ a_{ik}(y) \\ (i ≠ k; i, k = 1, …, n) \end{matrix}  (7.20)

hold in D_{b}. Then problem (7.1), (7.2) has the unique classical solution.

Remark 7.1. If for any i ∈ {1, …, n} integral (7.18) does not converge uniformly, then problem (7.1),(7.2) may have no classical solutions. Indeed in ℂ̃_{loc}(D_{b}; ℝ) the problem

\frac{∂^{2}u(x, y)}{∂x∂y} = |y − \frac{b}{2}|^{3} \frac{∂u(x, y)}{∂y} + (y − \frac{b}{2})^{3},

u(x, 0) = 0, ess sup_{(x,y)∈D_{b}} (|\frac{∂u(x, y)}{∂x}| + |\frac{∂u(x, y)}{∂y}|) < +∞ (7.21)

has the unique solution

u(x, y) = \frac{b}{2} − |y − \frac{b}{2}|,

which is not classical despite the fact that all conditions of Theorem 7.1′ except those of the uniform convergence of integral (7.18) which in this case takes the form

∫_{0}^{+∞} exp ( − |y − \frac{b}{2}|^{3}s)|y − \frac{b}{2}|^{3}ds,

are fulfilled.

Remark 7.2. The conditions of Theorem 7.1′ ensure the uniqueness only of a classical but not of an absolutely continuous solution. In fact, let p_{2} : [0, b] → [0, +∞) be a continuous function with a nowhere dense set of zeros of positive measure. It is clear that

I_{0}(p_{2}) = {y ∈ [0, b] : p_{2}(y) = 0},
I_{+}(p_{2}) = [0, b]\I_{0}(p_{2}), Ī_{+}(p_{2}) = [0, b].

For an arbitrary continuous c : [0, b] → ℝ the function

u_{c}(x, y) = ∫_{0}^{y} exp(p_{2}(t)x)c(t)ψ(t)dt,

where

ψ(y) = \begin{cases}1 & for y ∈ I_{0}(p_{2}) \\ 0 & for y ∈ I_{+}(p_{2}) \end{cases}

belongs to ℂ̃_{loc}(D_{b}; ℝ) and this function is a solution of the equation

\frac{∂^{2}u(x, y)}{∂x∂y} = p_{2}(y)\frac{∂u(x, y)}{∂y},  (7.22)

satisfying condition (7.21). On the other hand, problem (7.22),(7.21), for which all conditions of Theorem 7.1′ hold, has the unique classical solution u_{0}(x, y) ≡ 0.

Corollary 7.1. Let there exist constants a > 0, α ∈ (0, 1), β_{0} > 0 and essentially bounded measurable functions a_{ik} : [0, b] → ℝ_{+} (i ≠ k, i, k = = 1, …, n), σ_{i} : [0, b] → {−1, 1} and γ_{i} : [0, b] → [0, +∞) (i = 1, …, n) such that the spectral radius of the matrix A(y) = (a_{ik}(y))_{i,k=1}^{n}, where a_{ii}(y) ≡ 0 (i = 1, …, n), is less than α almost for all y ∈ [0, b] and the inequalities

σ_{i}(y) ∫_{x}^{x+a}p_{2ii}(ξ, y)dξ ≤ −γ_{i}(y) (i = 1, …, n),  (7.23)

∫_{x}^{x+a}(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| + |q_{i}(s, y)|)ds ≤

≤ β_{0}γ_{i}(y) (i, k = 1, …, n)  (7.24)

and

| ∫_{x−σ_{i}(y)a}^{x} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds| ≤

≤ a_{ik}(y)(1 − exp(−γ_{i}(y))) (i ≠ k; i, k = 1, …, n)  (7.25)

hold almost everywhere in D_{b}. Then problem (7.1),(7.2) is solvable, and for the uniqueness of a solution it is sufficient that

mes{y ∈ [0, b] : γ_{i}(y) = 0} = 0 (i = 1, …, n).  (7.26)

Proof. Put

β_{1} = ∥P_{2}∥_{L_{∞}} < +∞, β = β_{0} exp(aβ_{1}),

I_{i}^{0} = {y ∈ [0, b] : γ_{i}(y) = 0}, I_{i}^{+} = {y ∈ [0, b] : γ_{i}(y) > 0, σ_{i}(y) = −1},

I_{i}^{−} = {y ∈ [0, b] : γ_{i}(y) > 0, σ_{i}(y) = 1}, (i = 1, …, n).

Then by virtue of (7.23),

[0, b] = I_{i}^{+} ∪ I_{i}^{0} ∪ I_{i}^{−}, I_{+}(p_{2ii}) ⊃ I_{i}^{+}, I_{−}(p_{2ii}) ⊃ I_{i}^{−}

and

I_{i}^{0} ⊃ I_{0}(p_{2ii}) (i = 1, …, n).  (7.27)

If y ∈ I_{i}^{0}, then in view of (7.24) without loss of generality we may assume that

p_{0ik}(x, y) ≡ p_{1ik}(x, y) ≡ q_{i}(x, y) ≡ 0 for x ∈ ℝ, k ∈ {1, …, n}.

Therefore it is obvious that inequalities (7.4) and (7.5) are fulfilled in ℝ×I_{i}^{0} for every i ∈ {1, …, n}.

If y ∈ I_{i}^{+}, then χ(p_{2ii})(y) = +∞ and in view of inequalities (7.23)–(7.25) we have

∫_{x}^{+∞} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| + |q_{i}(s, y)|)ds =

= ∑_{m=1}^{+∞} exp ( − ∑_{l=1}^{m} ∫_{x+(l−1)a}^{x+la} p_{2ii}(ξ, y)dξ) ×

× ∫_{x+(m−1)a}^{x+ma} exp ( ∫_{s}^{x+ma}p_{2ii}(ξ, y)dξ)(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| +

+|q_{i}(s, y)|)ds ≤ β_{0}γ_{i}(y) exp(aβ_{1}) ∑_{m=1}^{+∞} exp(−mγ_{i}(y)) =

= βγ_{i}(y)( exp(γ_{i}(y)) − 1)^{−1} ≤ β (k = 1, …, n)

and

∫_{x}^{+∞} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds =

= ∑_{m=1}^{+∞} exp ( ∫_{x}^{x+(m−1)a}p_{2ii}(ξ, y)dξ) ×

× ∫_{x+(m−1)a}^{x+ma} exp ( ∫_{s}^{x+(m−1)a}p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds ≤

≤ a_{ik}(y)(1 − exp(−γ_{i}(y))) ∑_{m=1}^{+∞} exp(−(m − 1)γ_{i}(y)) = a_{ik}(y)

(k ≠ i; i, k = 1, …, n).

Therefore, inequalities (7.4) and (7.5) are fulfilled in ℝ × I_{i}^{+}. Similarly, we can prove that these inequalities are also fulfilled in ℝ × I_{i}^{−} (i = 1, …, n).

As for (7.6), these equalities are fulfilled by virtue of (7.27) if (7.26) holds. The validity of Corollary 7.1 becomes evident by applying Theorem 7.1. ■

Corollary 7.1′. Let P_{j} (j = 0, 1, 2) and q be continuous and bounded, φ be continuously differentiable and bounded together with its derivative. Let, besides, there exist constants a > 0, β_{0} > 0, ε > 0, σ_{i} ∈ {−1, 1} (i = 1, …, n) and continuous functions γ_{i} : [0, b] → [0, +∞) (i = 1, …, n) and a_{ik} : [0, b] → [0, +∞) such that the set of zeros of γ_{i} is nowhere dense in D_{b} (i = 1, …, n), the spectral radius of the matrix A(y) = (a_{ik}(y))_{i,k=1}^{n}, where a_{ii}(y) ≡ 0 (i = 1, …, n), is less than unity for every y ∈ [0, b] and the inequalities

σ_{i} ∫_{x}^{x+a}p_{2ii}(ξ, y)dξ ≤ −γ_{i}(y) (i = 1, …, n),  (7.28)

∫_{x}^{x+a}(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| + (1 − δ_{ik})|p_{2ik}(s, y)| + |q_{i}(s, y)|)ds ≤

≤ β_{0}γ_{i}^{1+ε}(y) (i = 1, …, n)  (7.29)

and

| ∫_{x−σ_{i}a}^{x} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds| ≤

≤ a_{ik}(y)(1 − exp(−γ_{i}(y))) (i ≠ k; i, k = 1, …, n)  (7.30)

holds in D_{b}. Then problem (7.1), (7.2) has one and only one classical solution.

Proof. In view of (7.28),

\begin{matrix}σ_{i}( ∫_{0}^{x} p_{2ii}(ξ, y)dξ) sign x ≤ −(\frac{|x|}{a} + 1)γ_{i}(y) + 2aβ_{1} \\ (i = 1, …, n), \end{matrix}  (7.31)

where

β_{1} = sup_{(x,y)∈D_{b}} ∥P_{2}(x, y)∥.

Taking into account the fact that the set of zeros of γ_{i} is nowhere dense in [0, b] (i = 1, …, n), we obtain from (7.31) that

Ī_{+}(p_{2ii}) = [0, b] for σ_{i} = −1, Ī_{−}(p_{2ii}) = [0, b] for σ_{i} = 1. (7.32)

Let us show that for every i ∈ {1, …, n} integrals (7.18) converge uniformly with respect to y ∈ [0, b]. In view of (7.32) we have χ_{i} = +∞ and χ_{i} = −∞ for σ_{i} = −1 and σ_{i} = 1, respectively. First we consider the case when σ_{i} = −1. By (7.29) and (7.31) for any x > 0 and y ∈ [0, b] we have

∫_{x}^{+∞} exp ( ∫_{s}^{0} p_{2ii}(ξ, y)dξ)(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| +

+(1 − δ_{ik})|p_{2ik}(s, y)| + |q_{i}(s, y)|)ds =

= ∑_{m=1}^{+∞} ∫_{x+(m−1)a}^{x+ma} exp ( − ∫_{s}^{0} p_{2ii}(ξ, y)dξ)(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| +

+(1 − δ_{ik})|p_{2ik}(s, y)| + |q_{i}(s, y)|)ds ≤

≤ β_{0} exp(aβ_{1})γ_{i}^{1+ε}(y) ∑_{m=1}^{+∞} exp ( − (\frac{x}{a} + m)γ_{i}(y)) =

= β_{0} exp(aβ_{1})γ_{i}^{1+ε}(y)( exp(γ_{i}(y)) − 1)^{−1} exp ( − \frac{x}{a}γ_{i}(y)) ≤

≤ β_{0} exp(aβ_{1})γ_{i}^{ε}(y) exp ( − \frac{x}{a}γ_{i}(y)) ≤ β_{2}x^{−ε} (k = 1, …, n),

where β_{2} = β_{0} exp(aβ_{1} −ε)(aε)^{ε}. Consequently, for σ_{i} = −1 integrals (7.18) converge uniformly. Analogously we can prove that integrals (7.18) converge uniformly for σ_{i} = 1 as well.

Similarly to the prove of Corollary 7.1, we can show that conditions (7.28)-(7.30) imply conditions (7.19) and (7.20).

Applying now Theorem 7.1′, the validity of Corollary 7.1′ becomes obvious. ■

Corollary 7.2. Let the inequalities

\begin{matrix}σ_{i}(y)p_{2ii}(x, y) ≤ l_{ii} (i = 1, …, n), \\ |p_{2ik}(x, y)| ≤ l_{ik} (i ≠ k; i, k = 1, …, n), \end{matrix}  (7.33)

where σ_{i} : [0, b] → {−1, 1} (i = 1, …, n) are measurable functions and l_{ik} (i, k = 1, …, n) are constants such that the real parts of eigenvalues of the matrix (l_{ik})_{i,k=1}^{n} are negative. Then problem (7.1),(7.2) has one and only one solution.

Proof. Since l_{ik} (i ≠ k; i, k = 1, …, n) are non-negative and eigenvalues of the matrix (l_{ik})_{i,k=1}^{n} are negative, we have

l_{ii} < 0 (i = 1, …, n)  (7.34)

and the spectral radius of the matrix (a_{ik})_{i,k=1}^{n}, where

a_{ii} = 0 (i = 1, …, n), a_{ik} = \frac{l_{ik}}{|l_{ii}|} (i ≠ k; i, k = 1, …, n),

is less than unity.

In view of (7.3),(7.33) and (7.34), inequalities (7.23) and (7.24) hold almost everywhere in D_{b}, where a = 1, γ_{i}(y) ≡ |l_{ii}|,

β_{0} = (∥P_{0}∥_{L_{∞}} + ∥P_{1}∥_{L_{∞}} + ∥q∥_{L_{∞}})/|l_{ii}|.

Besides,

| ∫_{x−σ_{i}(y)}^{x} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds| ≤

≤ l_{ik}| ∫_{x−σ_{i}(y)}^{x} exp (σ_{i}(y)(s − x)|l_{ii}|)ds| =

= \frac{l_{ik}}{|l_{ii}|}(1 − exp(−|l_{ii}|)) (i ≠ k; i, k = 1, …, n),

i.e. inequalities (7.25) also hold. Now applying Corollary 7.1, the unique solvability of problem (7.1),(7.2) becomes obvious. ■

Based on Corollary 7.1′, similarly to Corollary 7.2, we prove the validity of

Corollary 7.2′. Let P_{j} (j = 0, 1, 2) and q be continuous and bounded, φ be continuously differentiable and bounded together with its derivative and the inequalities

σ_{i}p_{2ii}(x, y) ≤ l_{ii} (i = 1, …, n),
|p_{2ik}(x, y)| ≤ l_{ik} (i ≠ k; i, k = 1, …, n),

hold in D_{b}, where σ_{i} ∈ {−1, 1} and l_{ik} (i, k = 1, …) are constants and the real parts of eigenvalues of matrix (l_{ik})_{i,k=1}^{n} are negative. Then problem (7.1), (7.2) has one and only one solution and this solution is classical.

§ 8. Periodic Solutions

In this section for the system

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q(x, y)  (8.1)

the conditions of the existence and uniqueness of a solution (generalized, absolutely continuous and classical) defined in the strip D_{b} and satisfying one of the following two boundary conditions

u(x, 0) = φ(x), u(x + a, y) = u(x, y)  (8.2)

and

lim_{y→0} (\frac{∂u(x, y)}{∂x} − P_{2}(x, y)u(x, y)) = ψ(x), u(x + a, y) = u(x, y) (8.3)

are established. Besides, we investigate the problem of relation between problem (8.1),(8.2) and the problem of bounded solutions

u(x, 0) = φ(x), ess sup_{(x,y)∈D_{b}} (∥\frac{∂u(x, y)}{∂x}∥ + ∥\frac{∂u(x, y)}{∂y}∥).  (8.4)

Throughout the remainder of this section it is assumed that a is a positive number, P_{j} = (p_{jik})_{i,k=1}^{n} : D_{b} → ℝ^{n×n} (j = 0, 1, 2) and q = (q_{i})_{i=1}^{n} : D_{b} → ℝ^{n} are measurable and essentially bounded, φ : ℝ → ℝ^{n} is absolutely continuous and ψ : ℝ → ℝ^{n} is locally summable,

\begin{matrix}P_{i}(x + a, y) ≡ P_{i}(x, y) (i = 0, 1, 2), q(x + a, y) ≡ q(x, y), \\ φ(x + a) ≡ φ(x), ψ(x + a) ≡ ψ(x). \end{matrix} (8.5)

Moreover, use is made of the notation

N_{0}(y) = Z_{2}(a, y) − E,
N(y) = ∫_{0}^{a} Z_{2}^{−1}(s, y)[P_{0}(s, y) + P_{1}(s, y)P_{2}(s, y)]Z_{2}(s, y)ds.

Let u be a solution of problem (8.1),(8.2). Then its restriction on D_{ab} satisfies the boundary conditions

u(x, 0) = φ(x), \frac{∂u(a, y)}{∂y} = \frac{∂u(0, y)}{∂y}.  (8.6)

Assume now that u is a solution of problem (8.1),(8.6). Then in view of the periodicity of φ we shall have u(a, y) = u(0, y). Let \overline{u} be the extension of u in D_{b} satisfying the condition

\overline{u}(x + a, y) ≡ \overline{u}(x, y).

In view of (8.5) it is evident that \overline{u} is a solution of problem (8.1),(8.2). Consequently, problems (8.1),(8.2) and (8.1),(8.6) are equivalent when (8.5) holds. But (8.1),(8.6) is the special case of problem (4.1),(4.2) for h(v)(y) = = v(a) − v(0) and φ_{1}(y) ≡ 0. Now from representations (4.7) and (4.10) it is clear that H_{0} ≡ Θ, H(x, y) ≡ E,

M_{0}(y) = N_{0}(y), M(x, y) = Z_{2}(a, y).

Therefore Theorem 4.1 and Corollaries 4.1 and 4.1′ yield the following assertions.

Theorem 8.1. Let the vector and the matrix functions Z_{2}^{−1}q, Z_{2}^{−1}P_{0}, (1+ ∥φ′_{0}∥)×Z_{2}^{−1}P_{1} be N_{0}-summable in D_{ab}. Then problem (8.1),(8.2) is solvable and the non-singularity of N_{0} almost everywhere in [0, b] is necessary and sufficient for the solution to be unique. If N_{0} is singular in the set with a positive measure, then the homogeneous problem corresponding to (8.1), (8.2) has an infinite dimensional set of solutions.

Corollary 8.1. If N_{0} is non-singular almost everywhere in [0, b] and

∫_{0}^{a} ∫_{0}^{b} ∥N_{0}^{−1}(y)∥(∥P_{0}(x, y)∥ +
+(1 + ∥φ′(x)∥)∥P_{1}(x, y)∥ + ∥q(x, y)∥)dxdy < +∞,

then problem (8.1), (8.2) has one and only one solution.

Corollary 8.1′. If P_{i} (i = 0, 1, 2) and q are continuous, φ_{0} is continuously differentiable and N_{0} is non-singular in [0, b], then problem (8.1), (8.2) has the unique solution and this solution is classical.

Assume

B_{0}(y) = ∫_{0}^{a} P_{2}(s, y)ds,

and when B_{0}(y) is nonsingular

B(y) = max_{0≤x≤a} [ ∫_{0}^{x} |B_{0}^{−1}(y) ∫_{0}^{s} P_{2}(ξ, y)dξP_{2}(s, y)|ds +

+ ∫_{x}^{a} |B_{0}^{−1}(y) ∫_{s}^{a} P_{2}(ξ, y)dξP_{2}(s, y)|ds].

For k = 2 and m = 1 the following assertions follow from Corollaries 4.2 and 4.2′.

Corollary 8.2. Let the inequalities

det(B_{0}(y)) ≠ 0, r(B(y)) < 1  (8.7)

hold almost everywhere in [0, b] and

∫_{0}^{b} ∥(E − B(y))^{−1}∥∥B_{0}^{−1}(y)∥dy < +∞.

. Then problem (8.1),(8.2) has one and only one solution.

Collolary 8.2′. If P_{i} (i = 0, 1, 2) and q are continuous, φ is continuously differentiable and inequalities (8.7) hold in [0, b], then problem (8.1),(8.2) has one and only one solution and this solution is classical.

The conditions for the unique solvability of problem (8.1),(8.2) are given also in [2,4,7,42], where the periodic boundary value problem is investigsted for quasilinear hyperbolic equations and systems with continuous right sides.

For example, L.Cesari [7] proved the solvability of problem (8.1),(8.2) under the assumptions that det(N_{0}(0)) ≠ 0, b is sufficiently small and

2a∥P_{2}(x, 0)∥ < 1 for x ∈ ℝ.  (8.8)

It is obvious that this result follows from Corollary 8.1′; moreover, condition (8.8) is unnecessary.

A.K.Aziz and S.L.Brodsky [2] considered a system with a small parameter ε > 0

\frac{∂^{2}u(x, y)}{∂x∂y} = ε[P_{0}(x, y)u(x, y)+P_{1}(x, y)\frac{∂u(x, y)}{∂x}+P_{2}(x, y)\frac{∂u(x, y)}{∂y}+q(x, y)]

and proved the unique solvability of problem (8.1),(8.2) for sufficiently small b and ε assuming that

det ( ∫_{0}^{a} P_{2}(s, 0)ds) ≠ 0.

This result also follows from Corollary 8.1′.

A.K.Aziz and A.M.Meyers [4] considered problem for n = 1 and proved its unique solvability under assumptions that P_{1} has a continuous partial derivative in the first argument,

P_{2}(x, y) ≠ 0 for (x, y) ∈ D_{b}  (8.9)

and

l[\frac{1}{l_{∗}} + \frac{exp(l^{∗}a) − 1}{l^{∗}(exp(l_{∗}a) − 1)}] < 1,  (8.10)

where

l_{∗} = min_{(x,y)∈D_{b}} |P_{2}(x, y)|, l^{∗} = max_{(x,y)∈D_{b}} |P_{2}(x, y)|,

l = max_{(x,y)∈D_{b}} |P_{0}(x, y) + P_{1}(x, y)P_{2}(x, y) − \frac{∂P_{1}(x, y)}{∂x}|.

But, in view of Corollary 8.1′, for n = 1 it is sufficient to have instead of condition (8.9) a more weaker condition

∫_{0}^{a} P_{2}(s, y)ds ≠ 0 for 0 ≤ y ≤ b.

As for condition (8.10) and the requirement for \frac{∂P_{1}}{∂x} to exist, they are unnecessary.

B.P.Tkach [42] proved the unique solvability of problem (8.1),(8.2) under the assumptions that

det(B_{0}(y)) ≠ 0, r(\overline{B}(y)) < 1 for 0 ≤ y ≤ b,

where

\overline{B}(y) = \frac{a^{2}}{2} |B_{0}^{−1}(y)|[ max_{(x,y)∈D_{b}} |P_{2}(x, y)|]^{2}.

This result is the special case of Corollary 8.2 because

B(y) ≤ \overline{B}(y) for 0 ≤ y ≤ b.

The following assertions follow from Theorems 4.2, 4.2′, 4.5 and 4.5′.

Theorem 8.2. Let the restrictions of P_{i} (i = 0, 1, 2) and q on D_{ab} belong, respectively, to ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) and ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}) (P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument), φ_{0} be absolutely continuous (continuously differentiable) and

N_{0}(y) = Θ, det(N(y)) ≠ 0 for 0 ≤ y ≤ b.

Then problem (8.1),(8.2) is uniquely solvable (and its solution is classical) if and only if

∫_{0}^{a} Z_{2}^{−1}(s, 0)[P_{0}(s, 0)φ(s) + P_{1}(s, 0)φ′(s) + q(s, 0)]ds = 0.

Theorem 8.3. Let the restriction of P_{2} on D_{ab} belong to ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}),

N_{0}(y) = Θ, det(N(y)) ≠ 0 almost everywhere in [0, b]

and

∫_{0}^{b} ∥N^{−1}(y)∥dy < +∞.

Then problem (8.1),(8.3) has one and only one generalized solution if and only if

∫_{0}^{a} Z_{2}(s, 0)φ(s)ds = 0.  (8.11)

Theorem 8.3′. Let the restriction of P_{i} (i = 0, 1, 2) and q on D_{ab} belong, respectively, to ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) and ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}) (P_{i} (i = 0, 1, 2) and q be continuous and have a continuous partial derivative in the second argument, ψ be continuous) and

N_{0}(y) = Θ, det(N(y)) ≠ 0 for 0 ≤ y ≤ b.

Then the fulfilment of condition (8.11) is necessary and sufficient for problem (8.1), (8.3) to have the unique generalized solution which is absolutely continuous (classical).

For P_{2}(x, y) ≡ Θ the result similar to Theorem 8.3 was obtained by S.V.Žestkov [51].

In Theorem 8.3′ the requirement for N to be non-singular in [0, b] is optimal and cannot be weakened. The following corollary concerning to the problem

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} + q(x, y), (8.12)
lim_{y→0} \frac{∂u(x, y)}{∂x} = 0, u(x + a, y) = u(x, y)  (8.13)

verifies this assertion.

Corollary 8.3. Let the restriction of P_{0} in D_{ab} belong to ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) (P_{0} be continuous and have a continuous partial derivative in the second argument). Then for an arbitrary P_{1} and q whose restrictions in D_{ab} belong repectively, to ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n×n}) and ℂ̃_{∞}^{(−1,0)}(D_{ab}; ℝ^{n}) (are continuous and have a continuous partial derivative in the second argument), problem (8.12), (8.13) has the unique generalized solution and this solution is absolutely continuous (classical) if and only if

det ( ∫_{0}^{a} P_{0}(s, y)ds) ≠ 0 for 0 ≤ y ≤ b.

This corollary follows from Corollary 4.10 and Theorem 8.3′.

Now consider the problem of relation between problems (8.1),(8.2) and (8.1),(8.4).

Theorem 8.4. Let φ_{0}′ be essentially bounded and almost for every y ∈ ∈ [0, b]

γ_{0i}(y) ≡ ∫_{0}^{a} p_{2ii}(s, y)ds ≠ 0 (i = 1, …, n).  (8.14)

Let, besides, there exist constants α ∈ (0, 1),β_{0} > 0 and essentially bounded measurable functions a_{ik} : [0, b] → ℝ_{+} (i ≠ k, i, k = 1, …, n), such that the spectral radius of the matrix A(y) = (a_{ik}(y))_{i,k=1}^{n}, where a_{ii}(y) ≡ 0 (i = 1, …, n), is less than α almost for all y ∈ [0, b] and the inequalities

\begin{matrix}∫_{0}^{a} (|p_{0ik}(s, y)| + |p_{1ik}(s, y)| + |q_{i}(s, y)|)ds ≤ β_{0}|γ_{0i}(y)| \\ (i, k = 1, …, n) \end{matrix} (8.15)

and

∫_{x}^{x+a} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds ≤
≤ a_{ik}(y)(1 − exp(−γ_{0i}(y))) (i ≠ k; i, k = 1, …, n)  (8.16)

hold almost everywhere in D_{ab}. Then problems (8.1),(8.2) and (8.1),(8.4) are uniquely solvable and their solutions coincide.

Proof. In view of (8.5), conditions (8.14)-(8.16) yield conditions (7.23)- (7.26), where

γ_{i}(y) = |γ_{0i}(y)|, σ_{i}(y) = \begin{cases}−1 & for γ_{0i}(y) ≥ 0 \\ 1 & for γ_{0i}(y) < 0 \end{cases} (i = 1, …, n).

Therefore, by virtue of Corollary 7.1 problem (8.1),(8.4) has the unique solution u_{0}. Consider the vector function

\overline{u}(x, y) = u_{0}(x + a, y).

According to (8.5), \overline{u} is also a solution of problem (8.1),(8.4) and in view of the unique solvability of the latter, we have \overline{u}(x, y) ≡ u_{0}(x, y). Consequently, u_{0} is a solution of problem (8.1),(8.2). Theorem will be proved if we show that an arbitrary solution u of problem (8.1),(8.2) is also a solution of problem (8.1),(8.4), i.e. u satisfies the condition

ess sup_{(x,y)∈D_{b}} (∥\frac{∂u(x, y)}{∂x}∥ + ∥\frac{∂u(x, y)}{∂y}∥) < +∞.

By virtue of Lemma 3.1 and the essential boundedness of φ′, we have

ess sup_{(x,y)∈D_{b}} ∥\frac{∂u(x, y)}{∂x}∥ < +∞.  (8.17)

Consequently, it remains to show that

ess sup_{(x,y)∈D_{b}} ∥\frac{∂u(x, y)}{∂y}∥ < +∞.  (8.18)

Put

\frac{∂u(x, y)}{∂y} = (z_{i}(x, y))_{i=1}^{n},

P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} + q(x, y) = (q_{0i}(x, y))_{i=1}^{n}.

Then

\frac{∂z_{i}(x, y)}{∂x} = ∑_{k=1}^{n} p_{2ik}(x, y)z_{k}(x, y) + q_{0i}(x, y) (i = 1, …, n) (8.19)
z_{i}(0, y) = z_{i}(a, y) (i = 1, …, n).  (8.20)

On the other hand, in view of conditions (8.5),(8.15),(8.17) and the essential boundedness of γ_{i} (i = 1, …, n), there exists a constant β such that the inequalities

∫_{x}^{x+a} |q_{0i}(s, y)|ds ≤ β|1 − exp(−γ_{0i}(y))| (i = 1, …, n) (8.21)

hold almost everywhere in D_{ab}. In view of (8.14) from (8.19) and (8.20), we have

z_{i}(x, y) = [ exp(−γ_{0i}(y)) − 1]^{−1} ∑_{k=1}^{n}(1 − δ_{ik}) ×

× ∫_{x}^{x+a} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)p_{2ik}(s, y)z_{k}(s, y)ds +

+[ exp(−γ_{0i}(y)) − 1]^{−1} ∫_{x}^{x+a} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)q_{0i}(s, y)ds (8.22)
(i = 1, …, n).

If we assume

\overline{z}(y) = ( max_{0≤x≤a} |z_{i}(x, y)|)_{i=1}^{n}, \overline{q} = β( exp(a∥p_{2ii}∥_{L_{∞}}))_{i=1}^{n},

then with regard to (8.16) and (8.21), from (8.22) we find

\overline{z}(y) ≤ A(y)\overline{z}(y) + \overline{q},

whence in view of the condition

r(A(y)) < α < 1

and the essential boundedness of A it follows that

z(y) ≤ [E − A(y)]^{−1}\overline{q}

and

ess sup_{0≤y≤b} ∥\overline{z}(y)∥ < +∞.

Consequently, condition (8.18) holds. ■

The following assertions follow from Theorem 8.4 and Corollary 7.1′. Theorem 8.4′. Let P_{j} and q be continuous, φ be continuously differen-

tiable and

γ_{0i}(y) ≡ ∫_{0}^{a} p_{2ii}(s, y)ds ≠ 0 for 0 ≤ y ≤ b (i = 1, …, n).

Let, besides, there exist continuous functions a_{ik} : [0, b] → ℝ_{+} (i ≠ k, i, k = = 1, …, n), such that the spectral radius of the matrix A(y) = (a_{ik}(y))_{i,k=1}^{n}, where a_{ii}(y) ≡ 0 (i = 1, …, n), is less than unity for every y ∈ [0, b] and inequalities (8.16) hold in D_{b}. Then problems (8.1),(8.2) and (8.1),(8.4) are uniquely solvable and they have one and the same solution which is classical.

Corollary 8.4. Let the inequalities

σ_{i}(y)p_{2ii}(x, y) ≤ l_{ii} (i = 1, …, n),
|p_{2ik}(x, y)| ≤ l_{ik} (i ≠ k; i, k = 1, …, n),

where σ_{i} : [0, b] → {−1, 1} (i = 1, …, n) are measurable functions and l_{ik} (i, k = 1, …, n) constants such that the real parts of eigenvalues of the matrix (l_{ik})_{i,k=1}^{n} are negative. Then problems (8.1),(8.2) and (8.1),(8.4) are uniquely solvable and their solutions coincide.

Corollary 8.4′. Let P_{j} (j = 0, 1, 2) and q be continuous, φ be continuously differentiable and the inequalities

σ_{i}p_{2ii}(x, y) ≤ l_{ii} (i = 1, …, n),
|p_{2ik}(x, y)| ≤ l_{ik} (i ≠ k; i, k = 1, …, n),

hold in D_{b}, where σ_{i} ∈ {−1, 1} and l_{ik} (i, k = 1, …) are constants and the real parts of eigenvalues of matrix (l_{ik})_{i,k=1}^{n} are negative. Then problems (8.1),(8.2) and (8.1),(8.4) are uniquely solvable and they have one and the same solution which is classical.

§ 9. Almost-Periodic Solutions

In this section we investigate the problem of almost-periodicity in the first argument of a solution of the problem

\frac{∂^{2}u(x, y)}{∂x∂y} = P_{0}(x, y)u(x, y) + P_{1}(x, y)\frac{∂u(x, y)}{∂x} +

+P_{2}(x, y)\frac{∂u(x, y)}{∂y} + q(x, y),  (9.1)
u(x, 0) = φ(x), ess sup_{(x,y)∈D_{b}} (∥\frac{∂u(x, y)}{∂x}∥ + ∥\frac{∂u(x, y)}{∂y}∥) < +∞. (9.2)

In addition, it is assumed everywhere that

P_{j} = (p_{jik})_{i,k=1}^{n} ∈ L_{∞}(D_{b}; ℝ^{n×n}) (j = 0, 1, 2),
q = (q_{i})_{i=1}^{n} ∈ L_{∞}(D_{b}; ℝ^{n}), φ = (φ_{i})_{i=1}^{n} ∈ ℂ̃_{∞}(ℝ; ℝ^{n}).

The concepts of almost-periodicity and S-almost-periodicity in the first argument of a matrix function of two variables which are introduced below, are the modification of Bohr’s and Stepanov’s concepts of almost-periodicity of a function of one variable ([30], Ch.1,§1 and Ch.5,§2).

Definition 9.1. A continuous matrix function Z : D_{b} → ℝ^{m×n} is called almost- periodic in the first argument if for an arbitrary ε > 0 there exists l > 0 such that an arbitrary segment [x_{0}, x_{0}+l] contains at least one number τ for which the inequality

sup_{(x,y)∈D_{b}} ∥Z(x + τ, y) − Z(x, y)∥ < ε

takes place.

Definition 9.2. A locally summable matrix function Z : D_{b} → ℝ^{m×n} is called S-almost-periodic in the first argument if for an arbitrary ε > 0 there exists l > 0 such that an arbitrary segment [x_{0}, x_{0} + l] contains at least one number τ for which the inequality

sup_{x∈ℝ} ∫_{0}^{b} ∫_{x}^{x+1} ∥Z(s + τ, t) − Z(s, t)∥dsdt < ε

takes place.

For any locally summable matrix function Z : D_{b} → ℝ^{m×n} we put

∥Z∥_{S} = sup_{x∈ℝ} ∫_{0}^{b} ∫_{x}^{x+1} ∥Z(s, t)∥dsdt.

A sequence of locally summable matrix functions Z_{k} : D_{b} → ℝ^{m×n} (k = = 1, 2, …) is called S-convergent to Z if

lim_{k→+∞} ∥Z_{k} − Z∥_{S} = 0.

Definition 9.3. A continuous (locally summable) matrix function Z:D_{b} → → ℝ^{m×n} is called normal (S-normal) if for any sequence of real numbers (λ_{k})_{k=1}^{+∞} the sequence of matrix functions (Z_{k})_{k=1}^{+∞}, where Z_{k}(x, y) = Z(x + +λ_{k}, y), contains uniformly convergent (S-convergent) subsequence.

Lemma 9.1. A continuous (locally summable) matrix function Z : D_{b} → → ℝ^{m×n} is almost-periodic (S-almost-periodic) in the first argument if and only if it is normal (S-normal).

This lemma is an analogue of Bochner’s well-known theorem ([30], Theorem 5.4.2) for matrix functions of two variables and can be proved in the same way as Bochner’s theorem.

Definition 9.4. Let \overline{P}_{i} ∈ L_{∞}(D_{b}; ℝ^{n×n}) (i = 0, 1, 2) and there exists a sequence of real numbers (λ_{k})_{k=1}^{+∞} such that

lim_{k→+∞} ∥P_{ik} − \overline{P}_{i}∥_{S} = 0,

where

P_{ik}(x, y) = P_{i}(x + λ_{k}, y) (i = 0, 1, 2; k = 1, 2, …).

Then we shall say that the triple of matrix functions (\overline{P}_{0}, \overline{P}_{1}, \overline{P}_{2}) belong to the class H(P_{0}, P_{1}, P_{2}).

Theorem 9.1. Let P_{i} (i = 0, 1, 2) be S-almost-periodic in the first argument and for any (\overline{P}_{0}, \overline{P}_{1}, \overline{P}_{2}) ∈ H(P_{0}, P_{1}, P_{2}) the homogeneous problem

\frac{∂^{2}u(x, y)}{∂x∂y} = \overline{P}_{0}(x, y)u(x, y) + \overline{P}_{1}(x, y)\frac{∂u(x, y)}{∂x} + \overline{P}_{2}(x, y)\frac{∂u(x, y)}{∂y}, (9.3)
u(x, 0) = 0, ess sup_{(x,y)∈D_{b}} (∥\frac{∂u(x, y)}{∂x}∥ + ∥\frac{∂u(x, y)}{∂y}∥) < +∞  (9.4)

have only the trivial solution. Then for any almost-periodic φ : ℝ → ℝ^{n} and S-almost-periodic in the first argument q : D_{b} → ℝ^{n} a solution of problem (9.1), (9.2), if it exists, is almost-periodic in the first argument.

Proof. Let for an almost-periodic φ : ℝ → ℝ^{n} and S-almost-periodic in the first argument q : D_{b} → ℝ^{n} problem (9.1),(9.2) have a solution u. According to Lemma 9.1, to prove the theorem it suffices to show that u is normal.

Assume on the contrary that u is not normal. Then by Lemma 9.1 there exist sequences of real numbers (λ_{k})_{k=1}^{+∞}, of natural numbers (k_{l})_{l=1}^{+∞} and (j_{l})_{l=1}^{+∞} and a positive number δ such that the sequences of matrix and vector functions

(P_{ik})_{k=1}^{+∞} (i = 0, 1, 2), (q_{k})_{k=1}^{+∞}  (9.5)

and

(φ_{k})_{k=1}^{+∞},  (9.6)

where

P_{ik}(x, y) = P_{i}(x + λ_{k}, y) (i = 0, 1, 2),
q_{k}(x, y) = q(x + λ_{k}, y) φ_{k}(x) = φ(x + λ_{k})

are, respectively, S-convergent and uniformly convergent and

sup_{(x,y)∈D_{b}} ∥u(x + λ_{k_{l}}, y) − u(x + λ_{j_{l}}, y)∥ > δ (l = 1, 2, …),

from which the existence of sequences x_{l} ∈ [0, a] and y_{l} ∈ [0, b] (l = 1, 2, …) such that

∥u(x_{l} + λ_{k_{l}}, y_{l}) − u(x_{l} + λ_{j_{l}}, y_{l})∥ > δ (l = 1, 2, …)  (9.7)

becomes evident.

Assume

u_{l}(x, y) = u(x + x_{l} + λ_{k_{l}}, y) − u(x + x_{l} + λ_{j_{l}}, y),
\overline{φ}_{l} = φ_{k_{l}}(x + x_{l}) − φ_{j_{l}}(x + x_{l}),
\overline{P}_{il}(x, y) = P_{i}(x + x_{l} + λ_{k_{l}}, y) (i = 0, 1, 2)

and

\overline{q}_{l}(x, y) = [P_{0}(x + x_{l} + λ_{k_{l}}, y) − P_{0}(x + x_{l} + λ_{j_{l}}, y)]u(x + x_{l} + λ_{j_{l}}, y) +

+[P_{1}(x + x_{l} + λ_{k_{l}}, y) − P_{1}(x + x_{l} + λ_{j_{l}}, y)]\frac{∂u(x + x_{l} + λ_{j_{l}}, y)}{∂x} +

+[P_{2}(x + x_{l} + λ_{k_{l}}, y) − P_{1}(x + x_{l} + λ_{j_{l}}, y)]\frac{∂u(x + x_{l} + λ_{j_{l}}, y)}{∂y} +

+q(x + x_{l} + λ_{k_{l}}, y) − q(x + x_{l} + λ_{j_{l}}, y).  (9.8)

Then, in view of (9.1) and (9.2), for any natural l we have

\frac{∂^{2}u_{l}(x, y)}{∂x∂y} = \overline{P}_{0_{l}}(x, y)u_{l}(x, y) + \overline{P}_{1_{l}}(x, y)\frac{∂u_{l}(x, y)}{∂x} +

+\overline{P}_{2_{l}}(x, y)\frac{∂u_{l}(x, y)}{∂y} + \overline{q}_{l}(x, y),  (9.9)
u_{l}(x, 0) = \overline{φ}_{l}(x)  (9.10)

and

sup_{(x,y)∈D_{b}} ∥u_{l}(x, y)∥ ≤ γ, ess sup_{(x,y)∈D_{b}} (∥\frac{∂u_{l}(x, y)}{∂x}∥ + ∥\frac{∂u_{l}(x, y)}{∂y}∥) < γ, (9.11)

where γ is a positive constant independent of l. On the other hand, in view of (9.7)

∥u_{l}(0, y_{l})∥ > δ (l = 1, 2, …).  (9.12)

Because of the S-convergence of sequences (9.5) and the uniform convergence of sequence (9.6) we have

∥\overline{q}_{l}∥_{s} ≤ γ ∑_{j=0}^{2} ∥P_{ik_{l}} − P_{ij_{l}}∥_{S} + ∥q_{k_{l}} − q_{j_{l}}∥_{S} → 0 for l → +∞ (9.13)

and

lim_{l→+∞} \overline{φ}_{l}(x) = 0 uniformly in ℝ.  (9.14)

Taking into account Lemma 9.1 and an essential boundedness of P_{i} (i = = 0, 1, 2), without loss of generality we may assume that the sequences (\overline{P}_{il})_{l=1}^{+∞} (i = 0, 1, 2) are S-convergent to some matrix functions \overline{P}_{i} ∈ ∈ L_{∞}(D_{b}; ℝ^{n×n}) (i = 0, 1, 2), i.e.

lim_{l→+∞} ∥\overline{P}_{il} − \overline{P}_{i}∥_{S} = 0.  (9.15)

On the other hand, by virtue of the Arzela-Ascoli lemma and conditions (9.11), again we may assume without loss of generality that the sequence (u_{l})_{l+1}^{+∞} is uniformly convergent in the rectangle D_{ab} for an arbitrary a ∈ ℝ.

Put

u_{0}(x, y) = lim_{l→+∞} u_{l}(x, y).  (9.16)

From (9.11),(9.12) and (9.14) it is clear that u_{0} : D_{b} → ℝ^{n} is Lipschitz continuous,

u_{0}(x, 0) = 0, ess sup_{(x,y)∈D_{b}} (∥\frac{∂u_{0}(x, y)}{∂x}∥ + ∥\frac{∂u_{0}(x, y)}{∂y}∥) ≤ γ (9.17)

and

max_{0≤y≤b} ∥u_{0}(0, y)∥ > 0.  (9.18)

Besides, for an arbitrary a ∈ ℝ we have

lim_{l→+∞} ∫_{0}^{x} \frac{∂u_{l}(s, y)}{∂s}ds = ∫_{0}^{x} \frac{∂u_{0}(s, y)}{∂s}ds, lim_{l→+∞} ∫_{0}^{y} \frac{∂u_{l}(x, t)}{∂t}dt =

= ∫_{0}^{y} \frac{∂u_{0}(x, t)}{∂t}dt uniformly in D_{ab}.  (9.19)

It follows from (9.9) and (9.10) that

u_{l}(x, y) = \overline{φ}_{l}(x) − φ_{l}(0) + u_{l}(0, y) + ∫_{0}^{x} ∫_{0}^{y} [\overline{P}_{0l}(s, t)u_{l}(s, t) +

+\overline{P}_{1l}(s, t)\frac{∂u_{l}(s, t)}{∂s} + \overline{P}_{2l}(s, t)\frac{∂u_{l}(s, t)}{∂t} + q_{l}(s, t)]dsdt.  (9.20)

However, in view of conditions (9.15),(9.16) and (9.19) and Lemma 3.13

lim_{l→+∞} ∫_{0}^{x} ∫_{0}^{y} \overline{P}_{0l}(s, t)u_{l}(s, t)dsdt = ∫_{0}^{x} ∫_{0}^{y} \overline{P}_{0}(s, t)u_{0}(s, t)dsdt,

lim_{l→+∞} ∫_{0}^{x} ∫_{0}^{y} \overline{P}_{1l}(s, t)\frac{∂u_{l}(s, t)}{∂s}dsdt = ∫_{0}^{x} ∫_{0}^{y} \overline{P}_{1}(s, t)\frac{∂u_{0}(s, t)}{∂s}dsdt,

lim_{l→+∞} ∫_{0}^{x} ∫_{0}^{y} \overline{P}_{2l}(s, t)\frac{∂u_{l}(s, t)}{∂t}dsdt = ∫_{0}^{x} ∫_{0}^{y} \overline{P}_{2}(s, t)\frac{∂u_{0}(s, t)}{∂t}dsdt

uniformly in D_{ab}.

If, alongside with the latter three equalities, we take into account conditions (9.13) and (9.14), then from (9.20) we obtain

u_{0}(x, y) = u_{0}(0, y) +

+ ∫_{0}^{x} ∫_{0}^{y} [\overline{P}_{0}(s, t)u_{0}(s, t) + \overline{P}_{1}(s, t)\frac{∂u_{0}(s, t)}{∂s} + \overline{P}_{2}(s, t)\frac{∂u_{0}(s, t)}{∂t}]dsdt.

From this and condition (9.17) it follows that u_{0} is a solution of problem (9.3),(9.4). But, according to the conditions of the theorem, problem (9.3),(9.4) has only the trivial solution. Consequently, u_{0}(x, y) ≡ 0. But this contadicts inequality (9.18). The obtained contradiction proves the theorem. ■

The theorem proved above is an analogue of J.Favard’s well-known theorem [15] for system (9.1).

Theorem 9.2. Let φ be almost-periodic, P_{i} (i = 0, 1, 2) and q be S-almost-periodic in the first argument and there exist constants α ∈ (0, 1), β > 0 and essentially bounded measurable functions a_{ik} : [0, b] → [0, +∞) (i ≠ k; i, k = 1, …, n), a_{i} : [0, b] → (0, +∞) and σ_{i} : [0, b] → {−1, 1} (i = 1, …, n) such that the spectral radius of the matrix (a_{ik}(y))_{i,k=1}^{n}, where a_{ii}(y) ≡ 0 (i = 1, …, n), is less than α almost for all y ∈ [0, b] and the inequalities

σ_{i}(y)p_{2ii}(x, y) ≤ −a_{i}(y) (i = 1, …, n),  (9.21)

\begin{matrix}| ∫_{x}^{χ_{i}(y)} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)(|p_{0ik}(s, y)| + |p_{1ik}(s, y)| + \\ +|q_{i}(s, y)|)ds| ≤ β (i, k = 1, …, n) \end{matrix} (9.22)

and

\begin{matrix}| ∫_{x}^{χ_{i}(y)} exp ( ∫_{s}^{x} p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds| ≤ a_{ik}(y) \\ (i ≠ k; i, k = 1, …, n) \end{matrix}  (9.23)

take place almost everywhere in D_{b}, where χ_{i}(y) = +∞ for σ_{i}(y) = −1 and χ_{i}(y) = −∞ for σ_{i}(y) = 1, respectively. Then problem (9.1),(9.2) has the unique solution and this solution is almost-periodic in the first argument.

Proof. Let I_{0}(p_{2ii}) and χ(p_{2ii}) (i = 1, …, n) be, respectively, the sets and the functions appearing in Theorem 7.1. Because the function a_{i} (i = = 1, …, n) is positive, from inequalities (9.21) we have

χ(p_{2ii})(y) = χ_{i}(y), mes I_{0}(p_{2ii}) = 0.

Therefore inequalities (7.4) and (7.5) follow from equalities (9.22) and (9.23). Consequently, all conditions of Theorem 7.1 hold, which guarantees the unique solvability of problem (9.1),(9.2). It remains to show that the solution of the problem is almost-periodic in the first argument. By virtue of Theorems 7.1 and 9.1 it suffices to show that if

\overline{P}_{j}(x, y) = (\overline{p}_{jik}(x, y))_{i,k=1}^{n} (j = 0, 1, 2)

and

(\overline{P}_{0}, \overline{P}_{1}, \overline{P}_{2}) ∈ H(P_{0}, P_{1}, P_{2})

then the inequalities

σ_{i}(y)\overline{p}_{2ii}(x, y) ≤ −a_{i}(y) (i = 1, …, n),  (9.24)

\begin{matrix}| ∫_{x}^{χ_{i}(y)} exp ( ∫_{s}^{x} \overline{p}_{2ii}(ξ, y)dξ)(|\overline{p}_{0ik}(s, y)| + |\overline{p}_{1ik}(s, y)|)ds| ≤ β \\ (i, k = 1, …, n) \end{matrix} (9.25)

and

\begin{matrix}| ∫_{x}^{χ_{i}(y)} exp ( ∫_{s}^{x} \overline{p}_{2ii}(ξ, y)dξ)|\overline{p}_{2ik}(s, y)|ds| ≤ a_{ik}(y) \\ (i ≠ k; i, k = 1, …, n) \end{matrix}  (9.26)

hold almost everywhere in D_{b}. According to the definition of the class H there exists a sequence of real numbers (λ_{l})_{l=1}^{+∞} such that

ε_{l} = max_{\begin{matrix}1≤i,k≤n \\ 0≤j≤2 \end{matrix}} [ sup_{x∈ℝ} ∫_{0}^{b} ∫_{x}^{x+1} |\overline{p}_{jik}(s, t) − p_{jik}(s + λ_{l}, t)|dsdt| → 0

for l → +∞.

(9.27)

By (9.21)

∫_{y}^{t} ∫_{x}^{s} σ_{i}(τ)p_{2ii}(ξ + λ_{l}, τ)dξdτ ≤ −(s − x) ∫_{y}^{t} a_{i}(t)dt
for 0 ≤ y < t ≤ b, x < s (i = 1, …, n),

whence, in view of (9.27), we have that

∫_{y}^{t} ∫_{x}^{s} σ_{i}(τ)\overline{p}_{2ii}(ξ, τ)dξdτ ≤ −(s − x) ∫_{y}^{t} a_{i}(τ)dτ
for 0 ≤ y < t ≤ b, x < s (i = 1, …, n).

It is clear from these estimates that inequalities (9.24) hold almost everywhere in D_{b}. Inequalities (9.25) and (9.26) can be proved similarly. We shall give here the proof of inequality (9.26) only. For any natural m we put

χ_{im}(y) = \begin{cases}m & for σ_{i}(y) = −1 \\ −m & for σ_{i}(y) = 1 \end{cases} .

| ∫_{x}^{x+χ_{im}(y)}exp ( ∫_{s}^{x} p_{2ii}(ξ + λ_{l}, y)dξ)|p_{2ik}(s + λ_{l}, y)|ds| =

= | ∫_{x+λ_{l}}^{x+λ_{l}+χ_{im}(y)}exp ( ∫_{s}^{x+λ_{l}}p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds| ≤

≤ | ∫_{x+λ_{l}}^{x+χ_{i}(y)}exp ( ∫_{s}^{x+λ_{l}}p_{2ii}(ξ, y)dξ)|p_{2ik}(s, y)|ds| ≤ a_{ik}(y)

(i ≠ k; i, k = 1, …, n).

Therefore

| ∫_{y}^{t} ∫_{x}^{x+χ_{im}(y)}exp ( ∫_{s}^{x} \overline{p}_{2ii}(ξ, τ)dξ)|\overline{p}_{2ik}(s, τ)|dsdτ| ≤

≤ ∫_{y}^{t} | ∫_{x}^{x+χ_{im}(y)}exp ( ∫_{s}^{x} p_{2ii}(ξ + λ_{l}, τ)dξ)|p_{2ik}(s + λ_{l}, τ)| ×

× exp ( ∫_{s}^{x}[\overline{p}_{2ii}(ξ, τ) − p_{2ii}(ξ + λ_{l}, τ)]dξ)ds|dτ +

+ ∫_{y}^{t} | ∫_{x}^{x+χ_{im}(y)}|\overline{p}_{2ik}(s, τ) − p_{2ik}(s + λ_{l}, τ)| exp ( ∫_{s}^{x} \overline{p}_{2ii}(ξ, τ)dξ)ds|dτ ≤

≤ ∫_{y}^{t} exp (| ∫_{x}^{x+χ_{im}(y)}|\overline{p}_{2ii}(ξ, τ) − p_{2ii}(ξ + λ_{l}, τ)|dτ|)a_{ik}(τ)dτ +

+ exp(c_{0}m) ∫_{y}^{t} | ∫_{x}^{x+χ_{im}(y)}|\overline{p}_{2ik}(s, τ) − p_{2ik}(s + λ_{l}, τ)|ds|dτ (9.28)
for 0 ≤ y ≤ t (i ≠ k; i, k = 1, …, n),

where

c_{0} = max_{1≤i≤n} ∥p_{2ii}∥_{L_{∞}}.

But,

exp (| ∫_{x}^{x+χ_{im}(y)}|\overline{p}_{2ii}(ξ, τ) − p_{2ii}(ξ + λ_{l}, τ)|ds|) ≤

≤ 1 + | ∫_{x}^{x+χ_{im}(y)}|\overline{p}_{2ii}(ξ, τ) − p_{2ii}(ξ + λ_{m}, τ)|ds| exp(2c_{0}m)

and

∫_{y}^{t} | ∫_{x}^{x+χ_{im}(y)}|\overline{p}_{2ik}(s, τ) − p_{2ik}(s + λ_{l}, τ)|ds|dτ ≤ mε_{l}.

According to the above arguments from (9.28) we obatin

∫_{y}^{t} | ∫_{x}^{x+χ_{im}(y)}exp ( ∫_{s}^{x} \overline{p}_{2ii}(ξ, y)dξ)|\overline{p}_{2ik}(s, τ)|ds|dτ ≤

≤ ∫_{y}^{t} a_{ik}(τ)dτ + m(1 + ∥a_{ik}∥_{L_{∞}}) exp(2c_{0}m)ε_{l}
for 0 ≤ y ≤ t ≤ b (i ≠ k; i, k = 1, …, n).

Passing in these inequalities to the limit first for l → +∞ and then for m → +∞, on account of (9.27), we find

∫_{y}^{t} | ∫_{x}^{x+χ_{i}(y)}exp ( ∫_{s}^{x} \overline{p}_{2ii}(ξ, τ)dξ)|\overline{p}_{2ik}(s, τ)|ds|dτ ≤ ∫_{y}^{t} a_{ik}(τ)dτ

for 0 ≤ y ≤ t ≤ b (i ≠ k; i, k = 1, …, n).

from which it is clear that inequalities (9.26) hold almost everywhere in D_{b}. ■

Similarly to Corollary 7.2 we prove

Corollary 9.1. Let φ_{0} be almost-periodic and P_{i} (i = 0, 1, 2) and q be S-almost- periodic in the first argument. Let, besides, inequalities

σ_{i}(y)p_{2ii}(x, y) ≤ l_{ii} (i = 1, …, n),
|p_{2ik}(x, y)| ≤ l_{ik} (i ≠ k, i, k = 1, …, n)

hold almost everywhere in D_{b}, where σ_{i} : [0, b] → {−1, 1} (i = 1, …, n) are measurable functions and l_{ik} (i, k = 1, …, n) are constants such that the real parts of eigenvalues of the matrix (l_{ik})_{i,k=1}^{n} are negative. Then problem (9.1),(9.2) has the unique solution and this solution is almost-periodic in the first argument.

REFERENCES

1. A.K.Aziz,Periodic solutions of hyperbolic partial differential equations. Proc. Amer. Math.Soc. 17(1966), No 3, 557-566.

2. A.K.Aziz and S.L.Brodsky, Periodic solutions of a class of weakly nonlinear hyperbolic partial differential equations. SIAM J. Math. Anal. 3(1972), No 2, 300-313.

3. A.K.Aziz and M.G.Horak, Periodic solutions of hyperbolic partial differential equations in the large. SIAM J. Math. Anal. 3(1972), No 1, 176-182.

4. A.K.Aziz and A.M.Meyers, Periodic solutions of hyperbolic partial differential equations in a strip. Trans. Amer. Math. Soc. 146(1969), 167-178.

5. C.Carathéodory, Vorlesungen über reelle Functionen. Chelsea Publ. Comp., New York, 1968.

6. L.Cesari, Periodic solutions of hyperbolic partial differential equations. Proc. Internat. Sympos. Non-linear Vibrations (Kiev 1961), vol. 2, Izd. Akad. Nauk Ukrain. SSR, Kiev, 1963, 440-457.

7. L.Cesari, A criterion for the existence in a strip of periodic solutions of hyperbolic partial differential equations. Rend. Circ. Mat. Palermo (2) 14(1965), 95-118.

8. L.Cesari, Existence in the large of periodic solutions of hyperbolic partial differential equations. Arch. Rational Mech. Anal. 20(1965), No 2, 170-190.

9. L.Cesari, A boundary value problem for quasilinear hyperbolic systems in Shauder’s canonic form. Ann. Scuola norm. super. Pisa 1(1974), No 3-4, 311-358.

10. L.Cesari, Un problema ai limiti per sistemi di equazioni iperboliche quasi lineari nella forma canonica di Shauder. Rend. Accad. Naz. Lincei. Cl. Sc. fis. mat. e natur. 57(1974), No 5, 303-307.

11. K.Deimling, A Carathéodory theory for systems of integral equations. Ann. mat. pura ed appl. 86(1970), 217-260.

12. K.Deimling, On the set of solutions of a system of Volterra integral equations.Ann. mat. pura ed appl. 85(1970), 369-382.

13. K.Deimling, Absolutely continuous solutions of Cauchy problem for u_{xy} = f(x, y, u, u_{x}, u_{y}). Ann. mat. pura ed appl. 89(1971), 381-391.

14. B.P.Demidovich, Lectures on the mathematical theory of stability. (Russian) Nauka, Moscow, 1967.

15. J.Favard, Sur les équations differentialles á coefficients presquepériodiques. Acta Math. 51(1927), 31-81.

16. M.P.Grigolia, On a generalized characteristic boundary value problem for hyperbolic systems. (Russian)Differentsial’nye Uravneniya21(1985), No 4, 678-686.

17. J.K.Hale, Periodic solutions of a class of hyperbolic equations containing a small parameter. Arch. Rat. Mech. Anal. 23(1967), No 5, 380-398.

18. G.H.Hardy, J.E.Littlewood and G.Pólya. Inequalities. Cambridge Univ. Press, Cambridge, 1934.

19. P.Hartman, Ordinary differential equations. John Wiley, New York, 1964.

20. L.V.Kantorovich and G.P.Akilov, Functional analysis. (Russian) Nauka, Moscow, 1977.

21. I.T.Kiguradze, Boundary value problems for systems of ordinary differential equations. (Russian) Modern problems in mathematics. The latest achievements (Itogi Nauki i tekhniki. VINITI Acad. Sci. USSR), Moscow, 1987, V. 30, 3-103.

22. T.I.Kiguradze, On existence and uniqueness of solutions of periodic boundary value problems for linear hyperbolic equations. Bull. Acad. Sci. of Georgia 146(1992), No 3, 501-505.

23. T.I.Kiguradze, On the correctness of periodic boundary value problems for linear hyperbolic equations. Bull. Acad. sci. of Georgia 147(1993), No 1, 26-30.

24. T.I.Kiguradze, On a boundary value problem for quasilinear hyperbolic systems. (Russian) Doklady Acad. Nauk 328(1993), No 2, 135-138.

25. T.I.Kiguradze, On the periodic boundary value problems for linear hypebolic equations. I. (Russian) Differentsial’nye Uravneniya 29(1993), No 2, 281-297.

26. T.I.Kiguradze, On the periodic boundary value problems for linear hyperbolic equations. II. (Russian) Differentsial’nye Uravneniya 29(1993), No 4, 637-645.

27. M.A.Krasnoselsky and S.G.Krein, On the average principle in the nonlinear mechanics. (Russian) Uspekhi Mat. Nauk 10(1955), No 3, 147-152.

28. J.Kurzweil and Z.Vorel, On the continuous dependence of solutions of differential equations on a parameter. (Russian) Czechosl. Math. J. 7(1957), No 4, 568-583.

29. V.Lakshmikantham and S.G.Pandit, Periodic solutions of hyperbolic partial differential equations. Comput. and Math. 11(1985), No 1-3, 249-259.

30. B.M.Levitan, Almost-periodic functions. (Russian) Gostekhizdat, Moscow, 1953.

31. Liu Baoping, The integral operator method for finding almostperiodic solutions of nonlinear wave equations. Nonlinear Anal., Theory, Meth. and Appl. 11(1987), No 5, 553-564.

32. Yu.A.Mitropolsky and G.P.Khoma, On periodic solutions of the second order wave equations. I. (Russian) Ukrain. Mat. Z. 38(1986), No 5. 593-600.

33. Yu.A.Mitropolsky and G.P.Khoma, On periodic solutions of the second order wave equations. II. (Russian) Ukrain. Mat. Z. 38(1986), No 6, 733-739.

34. Yu.A.Mitropolsky and G.P.Khoma, On periodic solutions of the second order wave equations. III. (Russian) Ukrain. Mat. Z. 39(1987), No 3, 347-353.

35. Yu.A.Mitropolsky and B.P.Tkach, Periodic solutions of nonlinear systems of partial differential equations of neutral type. (Russian) Ukrain. Mat. Z. 21(1969), No 4, 475-486.

36. Yu.A.Mitropolsky and L.B.Urmancheva, On two-point boundary value problem for systems of hyperbolic equations. (Russian) Ukrain Mat. Z. 42(1990), No 12, 1657-1663.

37. M.Picone, Sulle equationi alle derivate parziali del second ordine del tipo iperbolico in due variabili independenti. Rend. Circ. Mat. Palermo 30(1910), 349-376.

38. B.I.Ptashnyck, Incorrect boundary value problems for partial differential equations. (Russian) Naukova Dumka, Kiev, 1964.

39. P.H.Rabinowitz, Periodic solutions of nonlinear hyperbolic partial differential equations. Comm. Pure Appl. Math. 20(1967), 145-205.

40. P.H.Rabinowitz, Periodic solutions of nonlinear hyperbolic partial differential equations. II. Comm. Pure Appl. Math. 22(1969), 15-39.

41. P.H.Rabinowitz, Time periodic solutions of nonlinear wave equations. Manuscripta Math. 5(1971), 165-194.

42. B.P.Tkach, Periodic solutions of systems of partial differential equations of neutral type. (Russian) Differentsial’nye Uravneniya 5(1969), No 4, 735-748.

43. B.P.Tkach, Periodic and quasiperiodic solutions of systems with distributed parameters. (Russian) Doctoral thesis, Institute of Mathematics of the Ukrainean Academy of Sciences, Kiev, 1991.

44. M.T.Tsutsunava, On a boundary value problem for systems of hyperbolic equations. (Russian) Kraevye Zadachi. Perm Polytechn. Inst. Press, Perm, 1979, 58-60.

45. M.T.Tsutsunava, On a boundary value problem for systems of linear partial differential equations of hyperbolic type. Bull. Acad. Sci. Georgian SSR 114(1984), No 3, 465-467.

46. O.Vejvoda, Periodic solutions of a linear and weakly nonlinear wave equation in one dimension. I. Czechosl. Math. J. 14(1964), No 3, 341-382.

47. A.N.Vitjuck, On the solvability of a boundary value problem for systems of hyperbolic equations. (Russian) General theory of boundary value problems. Kiev, 1983, 257.

48. B.Vulich, Sur les formes générales de certaines opérations linéaires. Mat. Sb. 2(1937), No 2, 275-305.

49. S.Walczak, Absolutely continuous functions of several variables and their application to differential equations. Bull. Pol. Acad. Sci. Math. 35(1987), No 11-12, 733-744.

50. W.Walter, Differential and integral inequalities. Springer-Verlag, Berlin, Heidelberg, New York, 1970.

51. S.V.Z̆estkov, On periodic solutions of nonlinear hyperbolic systems in a strip. (Russian) Differentsial’nye Uravneniya 25(1989), No 10, 1806-1807.

(Received 15.10.1993)

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

CONTENTS

T. Kiguradze. Some Boundary Value Problems for Systems of Linear Partial Differential Equations of Hyperbolic Type. . . . . . . . . . . . . . . . . . . . . . . . . . .1

Preface 0.1. Statement of the main problems . . . . . . . . . . . . . . . . . . . . . . . . . . 3 0.2. Basic notation and terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Chapter I. Auxiliary Statements

§1. On a System of Operator Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .11 §2. On Boundary Value Problems for Systems of Linear Ordinary

Differential Equations with a Parameter . . . . . . . . . . . . . . . . . . . . . . .17 2.1. Some comments on the integral representation

of the operator h . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.2. Characteristics of the matrix functions Z_{0}, M_{0} and M . . 22 2.3. Criterion for the existence and uniqueness of a solution

of problem (2.1),(2.2). Properties of a solution . . . . . . . . . . 26 2.4. Lemmas on the unique solvability of problem

(2.1),(2.2) and on estimation of the matrix M_{0}^{−1} . . . . . . . . 28

§3. On Two Auxiliary Problems with the Integral Conditions for Linear Hyperbolic Systems . . . . . . . . . . . . . . . . . . . . . 36 3.1. Boundedness of the first order partial derivatives

of solutions of system (3.1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.2. Picone’s canonic form of linear hyperbolic systems . . . . . . 37 3.3. On the integral representation of

a generalized solution of system (3.1) . . . . . . . . . . . . . . . . . . . . 38 3.4. Conditions for the unique solvability

of problems (3.1),(3.2) and (3.1),(3.3) . . . . . . . . . . . . . . . . . . . 40 3.5. A priori estimates of solutions

of problems (3.1),(3.2) and (3.1),(3.3) . . . . . . . . . . . . . . . . . . . 44 3.6. Correctness of problems (3.1),(3.2) and (3.1),(3.3) . . . . . . . 52

Chapter II. Boundary Value Problems in a Rectangle

§4. Existence and Uniqueness Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . 60 4.1. Problem (4.1),(4.2) (case M_{0}(y) ≢ Θ). . . . . . . . . . . . . . . . . . . .62 4.2. Problems (4.1),(4.2) and (4.1),(4.3) (case M_{0}(y) ≡ Θ)….76 4.3. Problems (4.1),(4.4) and (4.1),(4.5) . . . . . . . . . . . . . . . . . . . . . 84

§5. Ways of Constructing Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 §6. Dependence of Solutions on Coefficients of the System

and on Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 6.1. Problem (6.1),(6.2). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .102 6.2. Problem (6.1),(6.3). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .109 Chapter III. Boundary Value Problems in a Strip §7. Bounded Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 §8. Periodic Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 §9. Almost-Periodic solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .130 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139