Nonlocal boundary value problems with resonant or non-resonant conditions
© Wang and Yang; licensee Springer. 2013
Received: 12 March 2013
Accepted: 9 October 2013
Published: 9 November 2013
We study solvability of nonlocal boundary value problems for second-order differential equations with resonance or non-resonance. The method of proof relies on Schauder’s fixed point theorem. Some examples are presented to illustrate the main results.
Keywordsnonlocal boundary value problems resonance non-resonance Schauder’s fixed point theorem
where , for , , , .
where is continuous, is continuous and for , , .
Theorem 1.1 
Suppose that there are two constants such that
(A1) for any , ;
Then (1.2) has at least one solution.
For (1.1), condition (A2) in Theorem 1.1 implies that for . It follows that (1.1) has infinitely many solutions ( is the solution of (1.1)). At this point, Theorem 1.1 has little significance for (1.1). Moreover, there are few papers considering multiple results at resonance. For the case with non-resonance, there is an extensive literature; see [14–17] and the references therein.
The main purpose of this article is to discuss the existence of solutions of equation (1.1) by means of Schauder’s fixed point theorem. We only need to consider the behavior of g and f on some closed sets. Consequently, information on the location of the solution is obtained and multiple results are obtained if g and f satisfy the given conditions on distinct regions. Our approach is valid for the cases at resonance or non-resonance. In addition, some of our conditions are easily certified (see Corollaries 3.1 and 3.2).
The paper is organized as follows. Section 2 introduces an important lemma. Section 3 is devoted to the existence results of (1.1). In Section 4 we extend some results of Section 3 to the general boundary conditions.
where , , for , and .
Boundary value problem (2.1) has a unique solution .
If on J and , then on J.
If for all , then on J; if for all , then on J.
If () on J, then on J.
Define an operator by , where ; then A is completely continuous.
The conclusion is obvious.
Here we only prove the case of . We consider two cases.
which implies that is nonincreasing on J. Noting , we obtain that is nonincreasing. Thus .
Since is continuous, by the intermediate value theorem, there exists such that .
which implies that . Hence, on J.
If , then , which implies that . Hence, on J.
Case 3.2 There exist such that and . We assume that . Otherwise, for all . Similar to Case 3.1, one can show that for , which is impossible.
which is a contradiction.
- (4)Since , using the conclusion of (3), we have
If , then and . Thus on J.
- (5)Let in . For any , there exists such that
which implies that is equicontinuous. It follows that is relatively compact in and A is a completely continuous operator. The proof is complete. □
The following well-known Schauder fixed point theorem is crucial in our arguments.
Lemma 2.2 
Let X be a Banach space and be closed and convex. Assume that is a completely continuous map; then T has a fixed point in D.
3 Main results
The following theorem is the main result of the paper.
Then (1.1) has at least one solution x with .
Also, the fact that A is completely continuous and H is continuous gives that is a continuous, compact map. By Lemma 2.2, has at least one fixed point in Ω. The proof is complete. □
Corollary 3.1 Assume that and the following condition holds:
Then (1.1) has at least one solution x with .
Proof Since , there exists such that the function is nondecreasing in . When , . We directly apply Theorem 3.1 and this ends the proof. □
Corollary 3.2 Assume that and the following condition holds:
Then (1.1) has at least one solution x with .
Proof Since , there exists such that the function is nondecreasing in . Condition (3.1) is satisfied if (3.4) holds. The proof is complete. □
Hence, by Corollary 3.1, (3.5) has a solution . Since n is an arbitrary positive integer, (3.5) has infinitely many solutions.
Using Corollary 3.2, we obtain that (3.6) has at least a nonnegative solution . Moreover, it is not difficult to show that for any .
where and for , , , .
where is sufficiently large, . We introduce the following assumptions.
(P1) The condition implies that boundary value problem (4.2) has a unique solution .
(P2) The condition implies that on J.
(P3) The condition () implies that on J.
(P4) The condition () implies that on J.
We say that the boundary condition satisfies P123 if (4.2) satisfies conditions (P1), (P2), (P3), and P134 if (4.2) satisfies conditions (P1), (P3), (P4).
Assume that the boundary condition satisfies P123 and (H1) holds. Then (4.1) has at least one solution x with .
Assume that the boundary condition satisfies P134 and (H2) holds. Then (4.1) has at least one solution x with .
The proof of Theorem 4.1 is similar to that of Theorem 3.1 and we omit it.
Then the solution obtained is nonnegative and nontrivial.
Remark 4.2 The boundary condition satisfies P134 if it satisfies P123.
One can easily check that boundary conditions (4.3), (4.4) satisfy P123, and conditions (4.5), (4.6), (4.7) satisfy P134.
where α, β, λ, μ, (), () are constants and , , .
If , , then the boundary condition satisfies P123.
If , or , , then the boundary condition satisfies P134.
has a unique solution .
By the maximum principle, we obtain that , a contradiction.
If and , , then .
Now suppose that on J and .
If , from the boundary conditions, we obtain that , , which implies that . The case has been discussed. If , from the boundary conditions, we obtain that , , which implies that . The case has also been discussed. The proof is complete. □
where , are constants.
By Theorems 4.1 and 4.2, (4.10) has a solution . Hence, (4.10) has infinitely many solutions.
where , , are constants.
Equation (4.11) has a solution and , for all . Set , then for all . By Theorems 4.1 and 4.2, (4.11) has a solution . Now we show that for . Assume that there exists with . Since , is minimum value and , . On the other hand, , a contradiction. If , then . This is impossible.
Equation (4.11) has infinitely many solutions for . Set , , where is an integer. Since for all , (4.11) has a solution . Hence, (4.11) has infinitely many solutions.
The work is supported by the NNSF of China (11171085) and Hunan Provincial Natural Science Foundation of China.
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