A Two-Point Boundary Value Problem with Reflection of the Argument

Abstract

We consider the following two-point boundary value problems in and in by setting and being a Caratheodory function. When , for a.e. with strict inequality on a positive measurable subset of , and for a.e. as well as sufficiently large , several existence theorems will be obtained, with or without a sign condition.

1. Introduction

Let us study the existence of solutions of the two-point boundary value problems with reflection of the argument: by setting and being a Caratheodory function. That means (i) is continuous in for a.e.,(ii) is measurable in for all and(iii)for each , there exists an such thatfor a.e. and for all . Concerning the nonlinear growth of , let us make an assumption :

(H) There exists a constant , for and for a.e. with strict inequality on a positive measurable subset of , such that (i)for a.e. and for all as well as (ii)for a.e. and for all both hold.

On the other hand, either with or without a Landesman-Lazer condition (see (15) below), solvability of the resonance problem has been extensively studied under the condition that the nonlinearity of is assumed to have either the following: (i)linear growth in as (see [1–11])(ii)superlinear growth in in one of the directions or as well as may be bounded in the opposite direction (see [12, 13])

Similar study on (1) in addition to a new order term under a different assumption has been done by [14]. The research on (2) has been first studied by [15] when is bounded, while [16] focused on the nonresonance case by allowing to grow linearly in as . However, research on the boundary value problems (1) and (2) has been studied but not thoroughly enough.

The purpose of this paper is to establish solvability theorems for (1) and (2) when is satisfied. Based on the well-known Leray-Schauder continuation method (see [17, 18]) some new solvability results will be obtained, with or without a sign condition (that is in with , and ).

We shall make use of real Banach spaces , , as well as Sobolev spaces and in the following procedure. The norms of , , , and are denoted by , , , and , respectively. By saying “a solution of (1),” we mean that with and satisfies the differential equation in (1), a.e.

2. Existence Theorems

For each with , we write and . To obtain the main results of this paper with reflection of the argument, we need to deduce the following two lemmas which are extensions of [7], Lemma 1.

Lemma 1. Let be a nonnegative -function such that for a.e., with strict inequality on a positive measurable subset of . Then, there exists a constant such that whenever with for a.e., and with .

Proof. Just as in the proof in [7], Lemma 1, there exists a constant such that whenever with for a.e. and with . Since and we have therefore Furthermore, since and , we have , and Just as in the proof of Lemma 1, we can apply the equality to obtain the next lemma when is replaced by

Lemma 2. Let be a nonnegative -function such that for a.e., with strict inequality on a positive measurable subset of . Then, there exists a constant such that whenever with for a.e., and with .

Proof. for some constant independent of with for a.e., and with .

Theorem 3. Let be a Caratheodory function satisfying . Then for each the problem (1) has a solution , provided that holds, where and .

Proof. Given a fixed , Consider the boundary value problems for , which becomes the original problem when . Since , (16) has only a trivial solution when by Lemma 1. To apply the Leray-Schauder continuation method, it suffices to show first that solutions to (16) for have a priori bound in . To this end, let be a continuous function such that for , and for . Define , and . Then, are Caratheodory functions, such that for a.e. and and for a.e. and If is a possible solution to (16) for some , then by using (18), (19), and Lemma 1, we have which implies that for some constant independent of . It remains to show that solutions of (16) for have an a priori bound in . We will show this by contradiction. Suppose that there exists a sequence and a corresponding sequence in such that is a solution of (16) with and for all . Let , then for all , and by (21), we have as . Since and for all , we may assume without loss of generality that converges to in with for some , and is pointwise bounded by an -function independent of . In particular, converges uniformly on , which implies that converges to in . Now, let us consider only the case , for the case can be treated similarly. Using the elementary inequality for all as well as with , and the fact that converges to 0 uniformly on , we have on for sufficiently large . Multiplying each side of (16) by , and integrating them over when and , we find for sufficiently large . It follows from (3) and that is bounded below by a function in independent of . By applying Fatou’s lemma to the inequality , we find , which contradicts the second inequality in (15). Therefore, the theorem is proven.

Theorem 4. Let be a Caratheodory function satisfying with for a.e. and for all . Then for each the problem (1) has a solution , provided that .
By modifying the proof of Theorem 3 slightly, we obtain the next solvability theorem, where the nonlinearity of satisfies the following condition:
(F) There exist constants and such that for a.e. and for all Also for a.e. and for all At the same time, condition (15) may be replaced by the inequality: where and

Theorem 5. Let be a Caratheodory function satisfying and . Then for each problem (1) has a solution , provided that (26) holds.

Proof. In the process of showing Theorem 3, (15) is used only to see the contradiction in the end. Thus, we may follow exactly the same process as in the proof of Theorem 3, to the point where on is considered and (23) holds. By and (22), we find that for a.e. with and for all By (3) and (22), we also have for a.e. with and for all . Combining (27) and (28), we find that is bounded below by a function in independent of . By applying Fatou’s lemma to the left hand side of the following inequality we find or equivalently which contradicts the second inequality in (26). Thus, we have proven the theorem.

Again, by modifying the proofs of Theorem 3, Theorem 4, and Theorem 5 slightly, we may use Lemma 2 to obtain the following solvability results for (2).

Theorem 6. Let be a Caratheodory function satisfying . Then for each , the problem (2) has a solution , provided that (15) holds.

Theorem 7. Let be a Caratheodory function satisfying with for a.e. and for all . Then for each , the problem (2) has a solution , provided that .

Theorem 8. Let be a Caratheodory function satisfying and . Then for each , the problem (2) has a solution , provided that (26) holds.

Remark 9. The conclusions of Theorem 5 and Theorem 8 both remain true if , but to use the condition instead of in .

Also by applying Lemma 2, the next result may be obtained when is replaced by .

Remark 10. Our results remain valid when problems (1) and (2) are replaced by respectively.

Data Availability

Data will be made available upon request.

Conflicts of Interest

The authors declare that they have no conflicts of interest.