ON SOME BOUNDARY VALUE PROBLEMS FOR SYSTEMS OF LINEAR FUNCTIONAL DIFFERENTIAL EQUATIONS

ROBERT HAKL^{∗}

Dept. of Math. Analysis, Masaryk University, Brno, Czech Republic

1991 Mathematics Subject Classification. 34K10

Key words and phrases. System of linear functional differential equations, linear differential system with deviating arguments, boundary value problem

1. Introduction

1.1. Statement of the Problem. On the segment I = [a, b] consider the system of linear functional differential equations

x′_{i}(t) = ∑_{k=1}^{n} ℓ_{ik}(x_{k})(t) + q_{i}(t) (i = 1, …, n)  (1)

and its particular case

x′_{i}(t) = ∑_{k=1}^{n} p_{ik}(t)x_{k}(τ_{ik}(t)) + q_{i}(t) (i = 1, …, n)  (1′)

with the boundary conditions

∫_{a}^{b} x_{i}(t)dφ_{i}(t) = c_{i} (i = 1, …, n).  (2)

Here ℓ_{ik} : C(I; ℝ) → L(I; ℝ) are linear bounded operators, p_{ik} and q_{i} ∈ L(I; ℝ), c_{i} ∈ ℝ (i, k = 1, …, n), φ_{i} : I → ℝ (i = 1, …, n) are the functions with bounded variations, and τ_{ik} : I → I (i, k = 1, …, n) are measurable functions.

The simple but important particular case of the conditions (2) are the two–point boundary conditions

x_{i}(b) = λ_{i}x_{i}(a) + c_{i} (i = 1, …, n),  (3)

∗ e-mail: hakl@math.muni.cz.

the periodic boundary conditions

x_{i}(b) = x_{i}(a) + c_{i} (i = 1, …, n),  (4)

and the initial conditions

x_{i}(t_{0}) = c_{i} (i = 1, …, n),  (5)

where t_{0} ∈ I and λ_{i} ∈ ℝ (i = 1, …, n).

By a solution of the system (1) (of the system (1′)) we understand an absolutely continuous vector function (x_{i})_{i=1}^{n} : I → ℝ which satisfies the system (1) (the system (1′)) almost everywhere on I. A solution of the system (1) (of the system (1′)) which satisfies the condition (j), where j ∈ {2, 3, 4, 5}, is said to be a solution of the problem (1), (j).

As for the differential system

x′_{i}(t) = ∑_{k=1}^{n} p_{ik}(t)x_{k}(t) + q_{i}(t) (i = 1, …, n),

the boundary value problems have been studied in detail (see [4,5,8–10] and references therein). There are also a lot of interesting results concerning the problems (1), (k) and (1′), (k) (k = 2, 3, 4, 5) (see [2,3,6,7,11–13]). In this paper, the optimal conditions for the unique solvability of the problems (1), (2) and (1′), (2) are established which are different from the previous results.

1.2. Basic Notation. Throughout this paper the following notation and terms are used:

I = [a, b], ℝ =] − ∞, +∞[, ℝ_{+} = [0, +∞[;

δ_{ik} is the Kronecker’s symbol, i.e.,

δ_{ik} = \begin{cases}1 & for i = k, \\ 0 & for i ≠ k; \end{cases}

ℝ^{n} is the space of n-dimensional column vectors x = (x_{i})_{i=1}^{n} with the components x_{i} ∈ ℝ (i = 1, …, n) and the norm

∥x∥ = ∑_{i=1}^{n} |x_{i}|;

ℝ^{n×n} is the space of n × n-matrices X = (x_{ik})_{i,k=1}^{n} with the components x_{ik} ∈ ℝ (i, k = 1, …, n) and the norm

∥X∥ = ∑_{i,k=1}^{n} |x_{ik}|;

ℝ_{+}^{n} = {(x_{i})_{i=1}^{n} ∈ ℝ^{n} : x_{i} ≥ 0, i = 1, …, n}; ℝ_{+}^{n×n} = {(x_{ik})_{i,k=1}^{n} ∈ ℝ^{n×n} : x_{ik} ≥ 0, i, k = 1, …, n}; the inequalities between vectors x and y ∈ ℝ^{n}, and between matrices X and

Y ∈ ℝ^{n×n} are considered componentwise, i.e.,

x ≤ y ⇔ (y − x) ∈ ℝ_{+}^{n}, X ≤ Y ⇔ (Y − X) ∈ ℝ_{+}^{n×n};

r(X) is the spectral radius of the matrix X ∈ ℝ^{n×n}; X^{−1} is the inverse matrix to X ∈ ℝ^{n×n}; E is the unit matrix; C(I; ℝ^{n}) is the space of continuous^{1} vector functions x : I → ℝ^{n} with the norm

∥x∥_{C} = sup{∥x(t)∥ : t ∈ I};

C(I; ℝ_{+}^{n}) = {x ∈ C(I; ℝ^{n}) : x(t) ∈ ℝ_{+}^{n} for t ∈ I}; L(I; ℝ^{n}) is the space of summable vector functions x : I → ℝ^{n} with the norm

∥x∥_{L} = ∫_{I} ∥x(t)∥dt;

L(I; ℝ_{+}^{n}) = {x ∈ L(I; ℝ^{n}) : x(t) ∈ ℝ_{+}^{n} for almost all t ∈ I}; C̃(I; ℝ^{n}) is the space of absolutely continuous vector functions x : I → ℝ^{n} with

the norm

∥x∥C̃ = ∥x∥_{C} + ∥x′∥_{L};

P_{I} is the set of linear operators ℓ : C(I; ℝ) → L(I; ℝ) mappings C(I; ℝ_{+}) into L(I; ℝ_{+});

L_{I} is the set of linear continuous operators ℓ : C(I; ℝ) → L(I; ℝ), for each of them there exists an operator \overline{ℓ} ∈ P_{I} such that for any u ∈ C(I; ℝ) the inequalities

|ℓ(u)(t)| ≤ \overline{ℓ}(|u|)(t)

holds almost everywhere on I; for any u ∈ L(I; ℝ)

η(u)(t, s) = ∫_{t}^{s} u(ξ)dξ.

1The vector function x = (x_{i})_{i=1}^{n} : I → ℝ^{n} is said to be continuous, bounded, summable, etc., if the components x_{i} : I → ℝ (i = 1, …, n) have such a property.

1.3. Criterion on the Unique Solvability of the Problem (1), (2). The results in general theory of boundary value problems (see [12], Theorems 1.1 and 1.4) yield the following

Theorem 1. If ℓ_{ik} ∈ L_{I} (i, k = 1, …, n), then the boundary value problem (1), (2) with arbitrary c_{i} ∈ ℝ and q_{i} ∈ L(I; ℝ) (i = 1, …, n) is uniquely solvable if and only if the corresponding homogeneous problem

x′_{i}(t) = ∑_{k=1}^{n} ℓ_{ik}(x_{k})(t)  (i = 1, …, n),  (1_{0})

∫_{a}^{b} x_{i}(s)dφ_{i}(s) = 0  (i = 1, …, n)  (2_{0})

has only the trivial solution. If the latter condition is fulfilled, then the solution of the problem (1), (2) admits the representation

x_{i}(t) = ∑_{k=1}^{n} y_{ik}(t)c_{k} + g_{i}(q_{1}, …, q_{n})(t)  (i = 1, …, n),  (6)

where y_{ik} ∈ C̃(I; ℝ) (i, k = 1, …, n), and g_{i} : L(I; ℝ^{n}) → C̃(I; ℝ) (i = 1, …, n) are

linear continuous operators such that the vector function (∑_{k=1}^{n} y_{ik}c_{k})_{i=1}^{n} is the solution

of the problem (1_{0}), (2), and the vector function (g_{i}(q_{1}, …, q_{n}))_{i=1}^{n} is the solution of the problem (1), (2_{0}).

Remark 1. The operator (g_{i})_{i=1}^{n} : L(I; ℝ^{n}) → C̃(I; ℝ^{n}) is called the Green’s operator of the problem (1_{0}), (2_{0}). According to Danford–Pettis Theorem (see [1], Ch. XI, §1, Theorem 6), there exists the unique matrix function G = (g_{ik})_{i,k=1}^{n} : I × I → ℝ^{n×n} with the essentially bounded components g_{ik} : I × I → ℝ (i, k = 1, …, n) such that

g_{i}(q_{1}, …, q_{n})(t) ≡ ∑_{k=1}^{n} ∫_{a}^{b} g_{ik}(t, s)q_{k}(s)ds (i = 1, …, n).

Consequently, the formula (6) can be rewritten as follows:

x_{i}(t) = ∑_{k=1}^{n} y_{ik}(t)c_{k} + ∑_{k=1}^{n} ∫_{a}^{b} g_{ik}(t, s)q_{k}(s)ds (i = 1, …, n).  (6′)

This formula is called the Green’s formula for the problem (1), (2), and the matrix G is called the Green’s matrix of the problem (1_{0}), (2_{0}).

The aim of the following is to find effective criteria for the unique solvability of the above formulated problems. With a view to achieve this goal, we will need one lemma which is proved in Section 2.

2. Lemma on Boundary Value Problem for the System of Functional Differential Equations

Consider the system of differential inequalities

|y′_{i}(t) − ℓ_{i}(y_{i})(t)| ≤ ∑_{k=1}^{n} h_{ik}(t)∥y_{k}∥_{C}  (i = 1, …, n)  (7)

with the boundary conditions

∫_{a}^{b} y_{i}(s)dφ_{i}(s) = 0 (i = 1, …, n),  (8)

where

ℓ_{i} ∈ L_{I}, h_{ik} ∈ L(I; ℝ_{+}) (i, k = 1, …, n),

c_{i} ∈ ℝ (i = 1, …, n), and φ_{i} : I → ℝ (i = 1, …, n) are functions with bounded variations.

Along with (7), (8) for every i ∈ {1, …, n} consider the homogeneous problem

y′(t) = ℓ_{i}(y)(t),  ∫_{a}^{b} y(s)dφ_{i}(s) = 0.  (9_{i})

Lemma 1. Let for every i ∈ {1, …, n} the homogeneous problem (9_{i}) have only the trivial solution and there exist a matrix A = (a_{ik})_{i,k=1}^{n} ∈ ℝ_{+}^{n×n} such that

r(A) < 1  (10)

and

∫_{a}^{b} |g_{i}(t, s)|h_{ik}(s)ds ≤ a_{ik}  for t ∈ I (i, k = 1, …, n),  (11)

where g_{i} is the Green’s function of the problem (9_{i}). Then the problem (7), (8) has only the trivial solution.

Proof. Let (y_{i})_{i=1}^{n} be a solution of (7), (8). Then for every i ∈ {1, …, n}, the function y_{i} is the solution of the problem

y′(t) − ℓ_{i}(y)(t) = q_{i}(t),  ∫_{a}^{b} y(s)dφ_{i}(s) = 0,  (12)

where

q_{i}(t) =^{def} y′_{i}(t) − ℓ_{i}(y_{i})(t).  (13)

By the Green’s formula we have

y_{i}(t) = ∫_{a}^{b} g_{i}(t, s)q_{i}(s)ds for t ∈ I (i = 1, …, n),  (14)

Due to (7) and (13),

∫_{a}^{b} |g_{i}(t, s)||q_{i}(s)|ds ≤ ∑_{k=1}^{n} ∫_{a}^{b} |g_{i}(t, s)|h_{ik}(s)∥y_{k}∥_{C}ds for t ∈ I (i = 1, …, n).

In view of (11) and the last inequalities from (14) we obtain

|y_{i}(t)| ≤ ∑_{k=1}^{n} a_{ik}∥y_{k}∥_{C}  for t ∈ I (i = 1, …, n).  (15)

Consequently, (15) yields

(E − A)(∥y_{i}∥_{C})_{i=1}^{n} ≤ 0.  (16)

Since A is a nonnegative matrix satisfying (10), there exists the nonnegative inverse matrix (E − A)^{−1}. Then by (16) we obtain y_{i}(t) ≡ 0 (i = 1, …, n). □

3. Existence and Uniqueness Theorems

Throughout the following we will assume that ℓ_{ik} ∈ L_{I} (i, k = 1, …, n) and for any u ∈ C(I; ℝ) the inequalities

|ℓ_{ik}(u)(t)| ≤ \overline{ℓ}_{ik}(|u|)(t) (i, k = 1, …, n)

hold almost everywhere on I, where \overline{ℓ}_{ik} ∈ P_{I} (i, k = 1, …, n).

Theorem 2. Let there exist operators ℓ_{i}, ℓ̃_{ik} ∈ L_{I} (i, k = 1, …, n), functions h_{ik} ∈ L(I; ℝ_{+}) (i, k = 1, …, n), and a matrix A = (a_{ik})_{i,k=1}^{n} ∈ ℝ_{+}^{n×n} satisfying (10) such that:

(i) any solution of the system (1_{0}) is also a solution of the system

x′_{i}(t) = ℓ_{i}(x_{i})(t) + ∑_{k=1}^{n} ℓ̃_{ik}(x_{k})(t)  (i = 1, …, n);  (17)

(ii) for any y ∈ C̃(I; ℝ), the inequalities

|ℓ̃_{ik}(y)(t)| ≤ h_{ik}(t)∥y∥_{C}  (i, k = 1, …, n)  (18)

holds almost everywhere on I;

(iii) for every i ∈ {1, …, n} the problem (9_{i}) has only the trivial solution and the inequalities (11) are fulfilled, where g_{i} is the Green’s function of the problem (9_{i}). Then the problem (1), (2) has a unique solution.

Proof. Let (y_{i})_{i=1}^{n} be a solution of the problem (1_{0}), (2_{0}). Then by (17) and (18) it is also a solution of the problem (7), (8). Now it is obvious that all the assumptions of Lemma 1 are fulfilled. Therefore y_{i}(t) ≡ 0 (i = 1, …, n). Thus the homogeneous problem (1_{0}), (2_{0}) has only the trivial solution and consequently, by Theorem 1, the problem (1), (2) has a unique solution. □

Corollary 1. Let there exist operators ℓ_{i} ∈ L_{I} (i = 1, …, n), functions h_{ik} ∈ L(I; ℝ_{+}) (i, k = 1, …, n), and a matrix A = (a_{ik})_{i,k=1}^{n} ∈ ℝ_{+}^{n×n} satisfying (10) such that:

(i) for any y ∈ C̃(I; ℝ), the inequalities

|ℓ_{ii}(y)(t) − ℓ_{i}(y)(t)| ≤ h_{ii}(t)∥y∥_{C} (i = 1, …, n),
|ℓ_{ik}(y)(t)| ≤ h_{ik}(t)∥y∥_{C} (i ≠ k; i, k = 1, …, n)

holds almost everywhere on I;

(ii) for every i ∈ {1, …, n} the problem (9_{i}) has only the trivial solution and the inequalities (11) are fulfilled, where g_{i} is the Green’s function of the problem (9_{i}). Then the problem (1), (2) has a unique solution.

Proof. Put

ℓ̃_{ii}(y)(t) ≡ ℓ_{ii}(y)(t) − ℓ_{i}(y)(t) (i = 1, …, n),
ℓ̃_{ik}(y)(t) ≡ ℓ_{ik}(y)(t) (i ≠ k; i, k = 1, …, n).

Then the assumptions of Theorem 2 are fulfilled. Consequently, the problem (1), (2) has a unique solution. □

Corollary 2. Let

∫_{a}^{b} exp (∫_{a}^{s} ℓ_{ii}(1)(ξ)dξ) dφ_{i}(s) ≠ 0  (i = 1, …, n)  (19)

and there exist a matrix A = (a_{ik})_{i,k=1}^{n} ∈ ℝ_{+}^{n×n} satisfying (10) such that (11) is fulfilled, where g_{i} is the Green’s function of the problem

y′(t) = ℓ_{ii}(1)(t)y(t),  ∫_{a}^{b} y(s)dφ_{i}(s) = 0  (20_{i})

and

h_{ik}(t) = \overline{ℓ}_{ii}(|η(\overline{ℓ}_{ik}(1))(t, ·)|)(t) + (1 − δ_{ik})\overline{ℓ}_{ik}(1)(t) (i, k = 1, …, n). (21)

Then the problem (1), (2) has a unique solution.

Proof. The condition (19) is necessary and sufficient for the problem (20_{i}) to have only the trivial solution for every i ∈ {1, …, n}.

On the other hand, every solution (x_{i})_{i=1}^{n} of the system (1_{0}) satisfies

x′_{i}(t) = ℓ_{ii}(1)(t)x_{i}(t) + ℓ_{ii}(x_{i}(·) − x_{i}(t))(t) + ∑_{k=1}^{n}(1 − δ_{ik})ℓ_{ik}(x_{k})(t) =  (22)

= ℓ_{ii}(1)(t)x_{i}(t) + ∑_{k=1}^{n} [ℓ_{ii}(|η(ℓ_{ik}(x_{k}))(t, ·)|)(t) + (1 − δ_{ik})ℓ_{ik}(x_{k})(t)] (i = 1, …, n).

Put

ℓ_{i}(y)(t) ≡ ℓ_{ii}(1)(t)y(t) (i = 1, …, n),
ℓ̃_{ik}(y)(t) ≡ ℓ_{ii}(|η(ℓ_{ik}(y))(t, ·)|)(t) + (1 − δ_{ik})ℓ_{ik}(y)(t) (i, k = 1, …, n).

Then any solution of the system (1_{0}) is also a solution of the system (17). Now it is obvious that all the assumptions of Theorem 2 are fulfilled. Consequently, the problem (1), (2) has a unique solution. □

If

ℓ_{ik}(y)(t) ≡ p_{ik}y(τ_{ik}(t)) (i, k = 1, …, n),

then the system (1) has the form (1′). In that case

\overline{ℓ}_{ik}(y)(t) ≡ |p_{ik}|y(τ_{ik}(t)) (i, k = 1, …, n),

\overline{ℓ}_{ii}(|η(\overline{ℓ}_{ik}(1))(t, ·)|)(t) ≡ |p_{ii}(t) ∫_{t}^{τ_{ii}(t)}|p_{ik}(s)|ds| .

Therefore from Corollary 2 it follows Corollary 2′. Let

∫_{a}^{b} exp (∫_{a}^{s} p_{ii}(ξ)dξ) dφ_{i}(s) ≠ 0 (i = 1, …, n)

and there exist a matrix A = (a_{ik})_{i,k=1}^{n} ∈ ℝ_{+}^{n×n} satisfying (10) such that (11) is fulfilled, where g_{i} is the Green’s function of the problem

y′(t) = p_{ii}(t)y(t),  ∫_{a}^{b} y(s)dφ_{i}(s) = 0

and

h_{ik}(t) = |p_{ii}(t) ∫_{t}^{τ_{ii}(t)}|p_{ik}(s)|ds| + (1 − δ_{ik})|p_{ik}(t)|  (21′)
(i, k = 1, …, n).

Then the problem (1′), (2) has a unique solution.

Corollary 3. Let λ_{i} ≠ 1, μ_{i} = max{1, |λ_{i}|} (i = 1, …, n) and the matrix

A = (\frac{μ_{i}}{|1 − λ_{i}|} ∫_{a}^{b} \overline{ℓ}_{ik}(1)(s)ds)_{i,k=1}^{n}

satisfies (10). Then the problem (1), (3) has a unique solution.

Proof. Since λ_{i} ≠ 1 (i = 1, …, n), for every i ∈ {1, …, n} the problem

y′(t) = 0,  y(b) = λ_{i}y(a)  (23_{i})

has only the trivial solution. Moreover, the Green’s function of (23_{i}) is of the form

g_{i}(t, s) = \begin{cases}\frac{λ_{i}}{λ_{i}−1} & for a ≤ s ≤ t ≤ b, \\ \frac{1}{λ_{i}−1} & for a ≤ t < s ≤ b. \end{cases}

Put

ℓ_{i}(y)(t) ≡ 0,  ℓ̃_{ik}(y)(t) ≡ ℓ_{ik}(y)(t),  h_{ik}(t) ≡ \overline{ℓ}_{ik}(1)(t) (i, k = 1, …, n).

Then all the assumptions of Theorem 2 are fulfilled. Consequently, the problem (1),

(3) has a unique solution □ Corollary 3′. Let λ_{i} ≠ 1, μ_{i} = max{1, |λ_{i}|} (i = 1, …, n) and the matrix

A = (\frac{μ_{i}}{|1 − λ_{i}|} ∫_{a}^{b} |p_{ik}(s)|ds)_{i,k=1}^{n}

satisfies (10). Then the problem (1′), (3) has a unique solution.

Corollary 4. Let

∫_{a}^{b} ℓ_{ii}(1)(s)ds ≠ 0  (i = 1, …, n)  (24)

and there exist a matrix A = (a_{ik})_{i,k=1}^{n} ∈ ℝ_{+}^{n×n} satisfying (10) such that (11) is fulfilled, where g_{i} is the Green’s function of the problem

y′(t) = ℓ_{ii}(1)(t)y(t),  y(b) = y(a),  (25_{i})

and h_{ik} is defined by (21). Then the problem (1), (4) has a unique solution.

Proof. The condition (24) is necessary and sufficient for the problem (25_{i}) to have only the trivial solution for every i ∈ {1, …, n} and its Green’s function is of the form

g_{i}(t, s) = \begin{cases}(1 − exp(∫_{a}^{b} ℓ_{ii}(1)(ξ)dξ))^{−1} exp(∫_{s}^{t} ℓ_{ii}(1)(ξ)dξ) \\ for a ≤ s ≤ t ≤ b, \\ ( exp(− ∫_{a}^{b} ℓ_{ii}(1)(ξ)dξ) − 1)^{−1} exp(∫_{s}^{t} ℓ_{ii}(1)(ξ)dξ) \\ for a ≤ t < s ≤ b. \end{cases}  (26_{i})

Now it is obvious that all the assumptions of Corollary 2 are fulfilled. Consequently, the problem (1), (4) has a unique solution. □

Corollary 4′. Let

∫_{a}^{b} p_{ii}(s)ds ≠ 0  (i = 1, …, n)  (24′)

and there exist a matrix A = (a_{ik})_{i,k=1}^{n} ∈ ℝ_{+}^{n×n} satisfying (10) such that (11) is fulfilled, where g_{i} is the Green’s function of the problem

y′(t) = p_{ii}(t)y(t), y(b) = y(a),

and h_{ik} is defined by (21′). Then the problem (1′), (4) has a unique solution.

Corollary 5. Let (24) be fulfilled and there exist σ_{i} ∈ {−1, 1}, α_{i} ∈]0, +∞[, α_{ik} ∈ [0, +∞[ (i, k = 1, …, n) such that the real part of every eigenvalue of the matrix A = (−δ_{ik}α_{i} + α_{ik})_{i,k=1}^{n} is negative and the inequalities

σ_{i}ℓ_{ii}(1)(t) ≤ −α_{i}  (i = 1, …, n),  (27)
\overline{ℓ}_{ii}(|η(\overline{ℓ}_{ik}(1))(t, ·)|)(t) + (1 − δ_{ik})\overline{ℓ}_{ik}(1)(t) ≤ α_{ik}  (28)
(i, k = 1, …, n)

hold almost everywhere on I. Then the problem (1), (4) has a unique solution.

Proof. At first note that according to Theorems 1.13 and 1.18 in [10] the negativeness of real parts of the eigenvalues of the matrix A yields the inequality

r(\overline{A}) < 1,  (29)

where

\overline{A} = (\frac{α_{ik}}{α_{i}} )_{i,k=1}^{n} .

On the other hand, from (27) it follows that for every i ∈ {1, …, n} the problem (25_{i}) has only the trivial solution and its Green’s function g_{i} is given by (26_{i}). Put

Δ_{i}(s, t) = exp(−σ_{i}α_{i}(t − s)) (i = 1, …, n).

Then for every i ∈ {1, …, n} from (26_{i}) and (27) we obtain

|g_{i}(t, s)| ≤ \begin{cases}[σ_{i}(1 − Δ_{i}(a, b))]^{−1}Δ_{i}(s, t) & for a ≤ s ≤ t ≤ b, \\ [σ_{i}(1 − Δ_{i}(a, b))]^{−1}Δ_{i}(a, b)Δ_{i}(s, t) & for a ≤ t < s ≤ b. \end{cases}  (30)

Define the functions h_{ik} by (21). Then from (28) and (30) we get

∫_{a}^{b} |g_{i}(t, s)|h_{ik}(s)ds ≤  (31)

≤ α_{ik}[σ_{i}(1 − Δ_{i}(a, b))]^{−1} (∫_{a}^{t} Δ_{i}(s, t)ds + Δ_{i}(a, b) ∫_{t}^{b} Δ_{i}(s, t)ds) =

= \frac{α_{ik}}{α_{i}} [1 − Δ_{i}(a, b)]^{−1}(1 − Δ_{i}(a, t) + Δ_{i}(a, b)Δ_{i}(b, t) − Δ_{i}(a, b)) =
= \frac{α_{ik}}{α_{i}}  (i, k = 1, …, n).

Taking into account (29) we conclude that all the assumptions of Corollary 2 are fulfilled. Consequently, the problem (1), (4) has a unique solution. □

Corollary 5′. Let (24′) be fulfilled and there exist σ_{i} ∈ {−1, 1}, α_{i} ∈]0, +∞[, α_{ik} ∈ [0, +∞[ (i, k = 1, …, n) such that the real part of every eigenvalue of the matrix A = (−δ_{ik}α_{i} + α_{ik})_{i,k=1}^{n} is negative and the inequalities

σ_{i}p_{ii}(t) ≤ −α_{i} (i = 1, …, n),

|p_{ii}(t) ∫_{t}^{τ_{ii}(t)}|p_{ik}(s)|ds| + (1 − δ_{ik})|p_{ik}(t)| ≤ α_{ik}

(i, k = 1, …, n)

hold almost everywhere on I. Then the problem (1′), (4) has a unique solution.

The last two corollaries concern with the Cauchy problems (1), (5) and (1′), (5).

Corollary 6. Let t_{0} ∈ I and there exist a nonnegative integer m_{0}, a natural number m > m_{0}, and α ∈]0, 1[ such that

ρ_{im}(t) ≤ αρ_{im_{0}}(t)  for t ∈ I  (i = 1, …, n),  (32)

where

ρ_{i0}(t) ≡ 1 (i = 1, …, n),

ρ_{ij}(t) = ∑_{k=1}^{n} |∫_{t_{0}}^{t} \overline{ℓ}_{ik}(ρ_{kj−1})(s)ds| (i = 1, …, n; j = 1, 2, …).

Then the problem (1), (5) has a unique solution.

Proof. For every i ∈ {1, …, n} we define the following sequences of operators ρ_{ij} : C(I; ℝ^{n}) → C(I; ℝ):

ρ_{i0}(u_{1}, …, u_{n})(t) =^{def} u_{i}(t),

ρ_{ij}(u_{1}, …, u_{n})(t) =^{def} ∑_{k=1}^{n} |∫_{t_{0}}^{t} \overline{ℓ}_{ik}(ρ_{kj−1}(u_{1}, …, u_{n}))(s)ds| (j = 1, 2, …).

Then for any (u_{i})_{i=1}^{n} ∈ C(I; ℝ^{n}),

ρ_{ij}(u_{1}, …, u_{n})(t) = ρ_{ij−j_{0}}(ρ_{1j_{0}}(u_{1}, …, u_{n}), …, ρ_{nj_{0}}(u_{1}, …, u_{n}))(t)  (33)
(i = 1, …, n; j ≥ j_{0}; j, j_{0} = 0, 1, 2, …)

and

ρ_{ij}(1, …, 1)(t) = ρ_{ij}(t) (i = 1, …, n; j = 0, 1, 2, …).  (34)

Now let (y_{i})_{i=1}^{n} be a solution of (1_{0}) satisfying the initial conditions

y_{i}(t_{0}) = 0 (i = 1, …, n).

Then for every nonnegative integer j,

|y_{i}(t)| ≤ ρ_{ij}(|y_{1}|, …, |y_{n}|)(t) (i = 1, …, n).  (35_{j})

By (34) from (35_{m_{0}}) we find

|y_{i}(t)| ≤ ρ_{im_{0}}(t) ∑_{k=1}^{n} ∥y_{k}∥_{C}  (i = 1, …, n).  (36)

Let for every i ∈ {1, …, n},

v_{i}(t) = \begin{cases}0 & if ρ_{im_{0}}(t) = 0, \\ \frac{|y_{i}(t)|}{ρ_{im_{0}}(t)} & if ρ_{im_{0}}(t) ≠ 0. \end{cases}

Then (36) yields

γ_{i} = ess sup{v_{i}(t) : t ∈ I} < +∞ (i = 1, …, n)

and

|y_{i}(t)| ≤ γ_{i}ρ_{im_{0}}(t) = γ_{i}ρ_{im_{0}}(1)(t) (i = 1, …, n),

whence by (32), (33) and (35_{m−m_{0}}) for every i ∈ {1, …, n} we get

|y_{i}(t)| ≤ ρ_{im−m_{0}}(|y_{1}|, …, |y_{n}|)(t) ≤
≤ γρ_{im−m_{0}}(ρ_{1m_{0}}(1), …, ρ_{nm_{0}}(1))(t) =
= γρ_{im}(1)(t) = γρ_{im}(t) ≤ γαρ_{im_{0}}(t),

where γ = max{γ_{1}, …, γ_{n}}. Hence we obtain

v_{i}(t) ≤ αγ (i = 1, …, n)

and, consequently,

γ ≤ αγ.

Since α ∈]0, 1[, we have γ = 0, which implies y_{i}(t) ≡ 0 (i = 1, …, n). Consequently, the problem (1), (5) has a unique solution. □

If m = 2, m_{0} = 1, then Corollary 6 yields the following result for the problem (1′), (5):

Corollary 6′. Let t_{0} ∈ I and α ∈]0, 1[ be such that

∑_{k=1}^{n} |∫_{t_{0}}^{t} |p_{ik}(s)| ∑_{j=1}^{n} | ∫_{t_{0}}^{τ_{ik}(s)}|p_{kj}(ξ)|dξ| ds| ≤ α ∑_{k=1}^{n} |∫_{t_{0}}^{t} |p_{ik}(s)|ds| for t ∈ I

(i = 1, …, n).

Then the problem (1′), (5) has a unique solution.

At the end of this subsection we give the examples verifying the optimality of the above formulated conditions in the existence and uniqueness theorems.

Example 1. Let n = 2, λ_{1} ∈ [−1, 1[, λ_{2} ∈] − ∞, −1[∪]1, +∞[. On the segment I = [0, 1] consider the system (1′) with the boundary conditions (3), where

p_{1k}(t) = \begin{cases}δ_{1k}(λ_{1} − 1) & for 0 ≤ t ≤ \frac{1}{2} \\ (1 − δ_{1k})(λ_{1} − 1) & for \frac{1}{2} < t ≤ 1 \end{cases} (k = 1, 2),

p_{2k}(t) = \begin{cases}δ_{2k}(\frac{λ_{2}−1}{λ_{2}}) & for 0 ≤ t ≤ \frac{1}{2} \\ (1 − δ_{2k})(\frac{λ_{2}−1}{λ_{2}}) & for \frac{1}{2} < t ≤ 1 \end{cases} (k = 1, 2),

τ_{11}(t) ≡ τ_{21}(t) ≡ 0, τ_{12}(t) ≡ τ_{22}(t) ≡ 1,

q_{i} ∈ L(I; ℝ), and c_{i} ∈ ℝ (i = 1, 2). Then all the assumptions of Corollary 3′ with μ_{1} = 1, μ_{2} = |λ_{2}|,

A = (\begin{matrix}1/2 & 1/2 \\ 1/2 & 1/2 \end{matrix})

are fulfilled except the condition (10) instead of which we have

r(A) = 1.  (37)

On the other hand, the homogeneous problem

x′_{i}(t) = ∑_{k=1}^{2} p_{ik}(t)x_{k}(τ_{ik}(t)) (i = 1, 2),  (38)

x_{1}(1) = λ_{1}x_{1}(0),  x_{2}(1) = λ_{2}x_{2}(0)  (39)

has the nontrivial solution

x_{1}(t) = (λ_{1} − 1)t + 1, x_{2}(t) = \frac{λ_{2} − 1}{λ_{2}}t + \frac{1}{λ_{2}} .

This example shows that the strict inequality (10) in Corollaries 3 and 3′ cannot be replaced by the nonstrict one.

Example 2. On the segment I = [a, b] consider the problem (1′), (4) with n ≥ 2, constant coefficients p_{ii}(t) ≡ p_{ii} = −1, p_{ik}(t) ≡ p_{ik} = \frac{1}{n−1} (i ≠ k; i, k = 1, …, n), and τ_{ik} : I → I (i, k = 1, …, n) arbitrary measurable functions. Then the vector γ = (γ_{i})_{i=1}^{n} ∈ ℝ^{n}, where γ_{1} = γ_{2} = ⋯ = γ_{n} ≠ 0 satisfies the equality

P γ = 0,

where P = (p_{ik})_{i,k=1}^{n}, i.e., P has a zero eigenvalue. Thus all the assumptions of Corollary 5′ are fulfilled with σ_{i} = 1, α_{i} = |p_{ii}|, α_{ik} = (1 − δ_{ik})p_{ik}, (i, k = 1, …, n), i.e., A = P, except the negativeness of real part of every eigenvalue of the matrix A.

On the other hand, the vector (γ_{i})_{i=1}^{n} is a nontrivial solution of the homogeneous problem

x′_{i}(t) = ∑_{k=1}^{n} p_{ik}x_{k}(τ_{ik}(t)), x_{i}(b) = x_{i}(a).

This example shows that in Corollaries 5 and 5′ the requirement on the negativeness of the real part of every eigenvalue of the matrix A cannot be weakened.

Example 3. Let I = [0, 1], t_{0} = 0, τ_{ik}(t) ≡ 1 (i, k = 1, …, n)

p_{ik}(t) = \begin{cases}1 & for t ∈ [\frac{k−1}{n}, \frac{k}{n}[ \\ 0 & for t ∈ I \ [\frac{k−1}{n}, \frac{k}{n}[ \end{cases} (i, k = 1, …, n),

and consider the problem (1′), (5). Put

ρ_{i0}(t) ≡ 1, ρ_{ij}(t) = ∑_{k=1}^{n} ∫_{0}^{t} p_{ik}(s)ρ_{ij−1}(τ_{ik}(s))ds

(i = 1, …, n; j = 1, 2, …).

Then

ρ_{ij}(t) = t (i = 1, …, n; j = 1, 2, …)

and for every nonnegative integer m_{0} and every natural number m > m_{0} we have

ρ_{im}(t) ≤ ρ_{im_{0}}(t) for t ∈ I (i = 1, …, n).

On the other hand,

x_{i}(t) = t (i = 1, …, n)

is a nontrivial solution of the homogeneous problem

x′_{i}(t) = ∑_{k=1}^{n} p_{ik}(t)x_{k}(τ_{ik}(t)), x_{i}(t_{0}) = 0.

The last example shows that in Corollaries 6 and 6′ we cannot choose α = 1.

Acknowledgement

This work was supported by the Ministry of Education of the Czech Republic under the Research Intention No. J07/98:143 100 001.

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