ON A NON-LOCAL BOUNDARY VALUE PROBLEM FOR LINEAR FUNCTIONAL DIFFERENTIAL EQUATIONS

where ` : C([a, b];R) → L([a, b];R) and h : C([a, b];R) → R are linear bounded operators, q ∈ L([a, b];R), and c ∈ R. By a solution to the problem (1.1), (1.2) we understand an absolutely continuous function u : [a, b] → R satisfying the equation (1.1) almost everywhere on the interval [a, b] and verifying also the boundary condition (1.2). The question on the solvability of various types of boundary value problems for functional differential equations and their systems is a classical topic in the theory of differential equations (see, e.g., [1,3–5,7–9,11–14] and references therein). Many particular cases of the boundary condition (1.2) are studied in detail (namely, periodic, anti-periodic and multi-point conditions), but only a few efficient conditions is known in the case, where a general non-local boundary condition is considered. In the present paper, new efficient conditions are found sufficient for the unique solvability of the problem (1.1), (1.2). It is clear that the ordinary differential equation u′ = p(t)u + q(t), (1.3)


Introduction
On the interval [a, b], we consider the problem on the existence and uniqueness of a solution to the equation The question on the solvability of various types of boundary value problems for functional differential equations and their systems is a classical topic in the theory of differential equations (see, e.g., [1, 3-5, 7-9, 11-14] and references therein). Many particular cases of the boundary condition (1.2) are studied in detail (namely, periodic, anti-periodic and multi-point conditions), but only a few efficient conditions is known in the case, where a general non-local boundary condition is considered. In the present paper, new efficient conditions are found sufficient for the unique solvability of the problem (1.1), (1.2). It is clear that the ordinary differential equation is satisfied. Below, we establish new solvability conditions for the problem (1.1), (1.2) in terms of norms of the operators appearing in (1.1) and (1.2) (see Theorems 2.1-2.4). Moreover, we apply these results to the differential equation with an argument deviation ] is a measurable function (see Theorems 2.5 and 2.6), and we show that the assumptions of the statements obtained reduce to the condition (1.4) in the case, where the equation (1.5) is the ordinary one (see Remark 2.6). All the main results are formulated in Section 2, their proofs are given in Section 3.
The following notation is used throughout the paper: (1) R is the set of all real numbers,

Main Results
In theorems stated below, we assume that the operator admits the representation = 0 − 1 with 0 , 1 ∈ P ab . This is equivalent to the fact that is not only bounded, but it is strongly bounded (see, e.g., [6, Ch.VII, §1.2]), i.e., that there exists a function η ∈ L([a, b]; R + ) such that the condition is satisfied.
We first consider the case, where the boundary condition (1.2) is understood as a non-local perturbation of a two-point condition of an anti-periodic type. More precisely, we consider the boundary condition where λ ≥ 0, h 0 , h 1 ∈ P F ab , and c ∈ R. We should mention that there is no loss of generality in assuming this, because an arbitrary functional h can be represented in the form Note also that we have studied the problem (1.1), (2.1) with λ < 0 in the paper [10]. EJQTDE, 2009 No. 36, p. 2 and either the conditions be satisfied, or the conditions hold. Then the problem (1.1), (2.1) has a unique solution.    Using the transformation described in the previous remark, we can immediately derive from Theorem 2.1 the following statement.
In the case, where λ = 0 in (2.1), we consider the problem and from Theorem 2.1 we get be satisfied, or the conditions hold. Then the problem (2.13) has a unique solution.
Now we give two statements dealing with the unique solvability of the problem (1.1), (1.2). We assume in Theorems 2. 3

and 2.4 that
There is no loss of generality in assuming this, because every linear bounded functional h : C([a, b]) → R can be expressed in such a form.
h + (1) 0 + 1 < h(1) be fulfilled. Then the problem (1.1), (1.2) has a unique solution. (1) In what follows, we establish two theorems, which can be also derived from Theorems 2.3 and 2.4, and which require that the deviation τ (t) − t is "small" enough. In order to simplify formulation of statements, we put Moreover, having h + , h − ∈ P F ab , we denote

Proofs
It is well-known that the linear problem has the Fredholm property, i. e., the following assertion holds (see, e. g., [2,4]; in the case, where the operator is strongly bounded, see also [1,14]). has only the trivial solution.
Proof of Theorem 2.1. According to Lemma 3.1, to prove the theorem it is sufficient to show that the homogeneous problem has only the trivial solution. Assume that, on the contrary, u is a nontrivial solution to the problem (3.2), (3.3). First suppose that u changes its sign. Put It is clear that (3.6) We can assume without loss of generality that t M < t m . The integration of the equality (3.2) from t M to t m , from a to t M , and from t m to b, in view of (3.4), (3.5), and the assumption 0 , 1 ∈ P ab , yields On the other hand, from the boundary condition (3.3), in view of the relations (3.5), (3.6) and the assumption h 0 , h 1 ∈ P F ab , we get Hence, it follows from the relation (3.8) that We first assume that 0 ≥ 1. Then the conditions (2.6) and (2.7) are supposed to be satisfied. It is clear that the inequality (2.7) implies λ > 0 and 1 < 1 − 1 λ h 0 (1) and thus B 1 < 1, Using these inequalities and the relations (3.6), from (3.7) and (3.10) we obtain which yields that Obviously, On the other hand, by virtue of (2.2), it follows from the inequality (2.7) that and thus we obtain Now, from (3.11), (3.12), and (3.13) we get which contradicts the inequality (2.7). Now assume that 0 < 1. Then, in view of the relations (3.6), the inequalities (3.7) and (3.9) yield and thus we get 1 ≥ B 1 > 1 and (3.14) Obviously, If 0 ≥ 1−h 0 (1)− λ+h 1 (1) 2 then the conditions (2.6) and (2.7) are supposed to be satisfied. Therefore, we obtain from the inequality (2.6) that 1 ≤ 1 + λ + h 1 (1) and thus it is easy to verify that (3.16) Now, it follows from (3.14), (3.15), and(3.16) that which contradicts the inequality (2.6).
then, taking the above-mentioned condition 1 > 1 and the obvious inequality into account, from the relations (3.14) and (3.15) we get which contradicts the inequality (2.4).
Now suppose that u does not change its sign. Then, without loss of generality, we can assume that and choose t M0 , t m0 ∈ [a, b] such that It is clear that (3.22) Notice that if the assumptions of the theorem are fulfilled, then both inequalities The integration of the equality (3.2) from a to t M0 and from t M0 to b, in view of the relations (3.17), (3.18), and (3.19) and the assumption 0 , 1 ∈ P ab , yields

The last two inequalities yield
and thus, using (3.3), (3.18), and the assumption h 0 , h 1 ∈ P F ab , we get (3.25) First suppose that (3.21) holds. The integration of the equality (3.2) from t m0 to t M0 , in view of (3.17), (3.18), and (3.19) and the assumption 0 , 1 ∈ P ab , results in which contradicts the inequality (3.24). Now assume that (3.22) holds. The integration of the equality (3.2) from t M0 to t m0 , in view of (3.17), (3.18), and (3.19) and the assumption 0 , 1 ∈ P ab , yields The last inequality, together with (3.25), results in By virtue of Corollary 2.2, the problem (1.1), (1.2) is uniquely solvable under the assumptions Moreover, using the transformation described in Remark 2.2, it is not difficult to verify that the problem (1.1), (1.2) is uniquely solvable also under the assumptions Combining these two cases we obtain the required assertion. The operator can be expressed in the form = 0 − 1 , where 0 , 1 ∈ P ab are such that 0 (1) ≡ p 0 and 1 (1) ≡ p 1 and, moreover, the functionalh admits the representationh =h + −h − in whichh + ,h − ∈ P F ab are such thath + (1) = µ 0 and h − (1) = µ 1 . Consequently, by virtue of Theorem 2.3, the problem (3.27) has only the trivial solution and thus u ≡ 0. This means that the problem (3.26) has only the trivial solution.
Proof of Theorem 2.6. The proof is analogous to those of Theorem 2.5, only Theorem 2.4 must be used instead of Theorem 2.3.