Table of Contents
Chinese Journal of Mathematics
Volume 2017, Article ID 7838102, 8 pages
https://doi.org/10.1155/2017/7838102
Research Article

Problems with Mixed Boundary Conditions in Banach Spaces

Department of Mathematics, IME-USP, Cidade Universitária, 05508-090 São Paulo, SP, Brazil

Correspondence should be addressed to Dionicio Pastor Dallos Santos; rb.psu.emi@oicinoid

Received 22 November 2016; Accepted 22 February 2017; Published 15 March 2017

Academic Editor: Zhilin Yang

Copyright © 2017 Dionicio Pastor Dallos Santos. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Using Leray-Schauder degree or degree for -condensing maps we obtain the existence of at least one solution for the boundary value problem of the following type: ,   where is a homeomorphism with reverse Lipschitz constant such that , is a continuous function, is a positive real number, and is a real Banach space.

1. Introduction

The purpose of this article is to obtain some existence results for the nonlinear boundary value problem of the formwhere is a homeomorphism such that and is Lipschitz constant, is a continuous function, is a positive real number, and is a real Banach space. Of course, a solution of this problem is a function of class such that the function is continuously differentiable, satisfying the boundary conditions, and for all .

The existence of solutions for second-order boundary value problems has been studied by many authors using various methods (see [16]).

In particular, the authors in [3] have studied the following boundary value problem:where , and . They obtained the existence of solutions of (2) using Darbo fixed point theorem and properties of the measure of noncompactness.

Recently, Zhou and Peng [5] have studied the following boundary value problem:where is a continuous function and is a Banach space. They obtained the existence of solutions of (3), where the main tools used in the study are Sadovskii fixed point theorem and precise computation of measure of noncompactness.

Inspired by these results, the main aim of this paper is to study the existence of at least one solution for the boundary value problem (1) using Leray-Schauder degree or degree for -condensing maps. For this, we reduce the nonlinear boundary value problem to some fixed points problem. Next, we shall essentially consider two types of regularity assumptions for . In Theorem 10 we suppose that is completely continuous, which allows us to prove that the associated fixed point operator is completely continuous required by a Leray-Schauder approach. In Theorem 11 we only assume some regularity conditions expressed in terms of the measure of noncompactness, which allows us to apply the methods of topological degree theory for -condensing maps.

The paper is organized as follows. In Section 2, we establish the notation, terminology, and various lemmas which will be used throughout this paper. In Section 3, we formulate the fixed point operator equivalent to problem (1). In Section 4, we give main results in this paper. In Section 5, we study the existence of at least one solution for (1) in Hilbert spaces. For these results, we adapt the ideas of [79] to the present situation.

2. Notations and Preliminary Results

We first introduce some notation. For fixed , we denote the usual norm in by . Let denote the Banach space of continuous functions from into , endowed with the uniform norm and let denote the Banach space of continuously differentiable functions from into , equipped with the usual norm .

We introduce the following applications:

The Nemytskii operator isthe integration operator isand the continuous linear application is

Throughout this paper, we denote by a real Banach space and . For , we use the notation

Definition 1. Let be a Banach space and let be the family of bounded subsets of . The Kuratowski measure of noncompactness is the map defined for

Properties(a) iff is compact.(b) then .(c).(d).(e), where and .(f), where .(g).

The details of and its properties can be found in [10].

Definition 2 (see [11]). Assuming that the mapping is said to be a condensing operator if is continuous and bounded (sends bounded sets into bounded sets), and, for any nonrelatively compact and bounded set ,

The following lemmas are of great importance in the proof of our main results. The proofs can be found in [11].

In the following, we denote and as the noncompactness measure in and , respectively.

Lemma 3. Let be a bounded subset of real numbers and a bounded subset of . Then where .

Lemma 4. Let be bounded subsets of Banach spaces and , respectively, with Then

Lemma 5. If is bounded and equicontinuous, then we have the following:
(1) .
(2) .

Lemma 6. If is a bounded set in , then
(i) ,
(ii) .

Lemma 7. If is a bounded set in and equicontinuous, then

3. Fixed Point Formulations

Let us consider the operator Here is understood as the operator defined for . It is clear that is continuous and sends bounded sets into bounded sets.

Lemma 8. is a solution of (1) if and only if is a fixed point of the operator .

Proof. Let be a solution of (1). This implies that Integrating of to and using the fact that , we deduce that Applying and to both of its members and using that , we have thatConversely, since, by definition of the mapping , it is a simple matter to see that if is such that then is a solution to (1).

Using the theorem of Arzelà-Ascoli we show that the operator is completely continuous.

Lemma 9. If is completely continuous, then the operator is completely continuous.

Proof. Let be a bounded set. Then, if , there exists a constant such thatNext, we show that is a compact set. Let be a sequence in , and let be a sequence in such that . Using (19), we have that there exists a constant such that, for all , which implies thatHence the sequence is bounded in . Moreover, for and for all , we have that which implies that is equicontinuous.
On the other hand, for ,where Recalling that the convex hull of a set is given by it follows thatwhich implies thatUsing the fact that is completely continuous, we deduce that . Hence, is a relatively compact set in . Thus, by the Arzelà-Ascoli theorem there is a subsequence of , which we call , which is convergent in . Using the fact that is continuous it follows from that the sequence is convergent in and hence is convergent in . Finally, let be a sequence in . Let be such thatLet be a subsequence of such that it converges to . It follows that and converge to . This concludes the proof.

In order to apply Leray-Schauder degree to the operator , we introduced a family of problems depending on a parameter . For , we consider the family of boundary value problemsNotice that (30) coincide, for , with (1). So, for each , the operator associated with (30) for Lemma 8 is the operator , where is defined on by Using the same arguments as in the proof of Lemma 9 we show that the operator is completely continuous. Moreover, using the same reasoning as above, system (30) (see Lemma 8) is equivalent to the problem

4. Main Results

In this section, we present and prove our main results.

Theorem 10. Let be a Banach space and a homeomorphism with Lipschitz constant . Suppose that is completely continuous and that there exist two numbers such that Then problem (1) has at least one solution.

Proof. Let be such that . Using (32) we have that is solution of (30), which implies that Using the fact that is a homeomorphism with Lipschitz constant , we deduce that By Gronwall’s Inequality, we have Hence, . Because is such that , we have that and hence Using that is completely continuous we deduce that, for each , the Leray-Schauder degree is well-defined for any , and by the homotopy invariance we have that Hence, . This, in turn, implies that there exists such that , which is a solution for (1).

Theorem 11. Let be a Banach space and a homeomorphism with Lipschitz constant . Assume that is continuous and satisfies the following conditions: (1)There exist two numbers such that (2)For all bounded subsets in , Then problem (1) has at least one solution.

Proof. Observe that maps bounded sets into bounded sets. Furthermore, its continuity follows by the continuity of the operators which compose . We show that the operator is condensing (-condensing). In fact, for a bounded set in , there exists a constant such that For we have that which means is equicontinuous. Applying Lemma 7 there exists or with or Let us consider the first case. Using the properties of , we see that Using the fact that is a homeomorphism with Lipschitz constant , we deduce that Applying Lemma 3 and again the properties of , we obtain that Using (41), we have thatThis implies, by Lemma 6, that Consider the alternative case. Proceeding as before, we obtain Therefore, in either case, we obtain By (41), we get , and therefore is -condensing.
Let us consider the function Let be such that . Using the fact that is a homeomorphism with Lipschitz constant and Gronwall’s Inequality, we deduce that there exists a constant such that .
Finally, we show the existence of at least one solution of (1) using the homotopy invariance of the degree for -condensing maps. Let be bounded in . Then Then we have that, for each , the degree is well-defined and, by the properties of that degree, that Then, from the existence property of degree, there exists such that , which is a solution for (1).

Remark 12. In [3], the nonlinear term is bounded; in our result, the nonlinear term may possess no more than linear growth.

Now we consider an example to illustrate our results.

Example 13. Consider the boundary value problem in with norm . Now we can regard this problem as a problem of form (1), where and with It is clear that is a continuous function and that On the other hand, for any bounded subsets we have and henceSo, by Theorem 11 we get one solution.

5. Boundary Value Problems in Hilbert Spaces

Throughout this section, let denote a real Hilbert space. Assume that satisfies the following conditions:

(1) is a homeomorphism with Lipschitz constant .

(2) For any ,

Lemma 14. Let be such thatfor all . If is such that , then there exists such that .

Proof. Let be such that . Using (32), we have that is solution of (30), which implies that where, for all , we obtain On the other hand, because is a homeomorphism such thatfor all , then and hence Using the integration-by-parts formula and the boundary conditions, we deduce that Since and is solution of (30) we have that Hence,It follows that there exists such that . Because is such that , we deduce that and hence Finally, if , then , so the proof is complete.

Now we show the existence of at least one solution for problem (1) by means of Leray-Schauder degree.

Theorem 15. Let be completely continuous. Assume that satisfies the conditions of Lemma 14. Then (1) has at least one solution.

Proof. Let and be a possible fixed point of . Then, using Lemma 14 we deduce that Therefore, if , it follows from the homotopy invariance of Leray-Schauder degree that is independent of so that if we notice that , Hence there exists such that it is a solution for (1).

Using a proof similar to that of Theorem 11, we obtain the following existence result.

Theorem 16. Let be continuous. Assume that satisfies the following conditions: (1)There exists such that for all .(2) sends bounded sets into bounded sets.(3)For any bounded set in , Then problem (1) has at least one solution.

Proof. Let be bounded in . Applying Lemma 7 there exists or with or Proceeding as Theorem 11, we obtain in either case Using the homotopy invariance of the degree for -condensing maps, we obtain Then, from the existence property of degree, there exists such that , which is a solution for (1).

The following corollary is concerned with the existence of one solution for (1).

Corollary 17. Assume that satisfies the following conditions: (1)Suppose that for any the mapping is bounded and uniformly continuous in , where .(2)There exists such that for all .(3)There exists a constant with such that Then problem (1) has at least one solution.

Proof. Let be a bounded set in . Let us consider Cleary, , is bounded and equicontinuous. Thus, by using the conclusion of Lemma 5, we have Using (84), we obtain By using the arguments of Theorem 16, we can obtain the conclusion of Corollary 17.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this article.

Acknowledgments

This research was supported by CAPES and CNPq/Brazil.

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