Discrete Dynamics in Nature and SocietyVolume 2012 (2012), Article ID 786404, 16 pagesdoi:10.1155/2012/786404

Research Article

Existence of Solution to a Second-Order Boundary Value Problem via Noncompactness Measures

Wen-Xue Zhou1,2 and Jigen Peng¹

¹College of Sciences, Xi'an Jiaotong University, Xi'an 710049, China²Department of Mathematics, Lanzhou Jiaotong University, Lanzhou 730070, China

Received 26 December 2011; Revised 16 February 2012; Accepted 20 February 2012

Academic Editor: Guang Zhang

Copyright © 2012 Wen-Xue Zhou and Jigen Peng. 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.

The existence and uniqueness of the solutions to the Dirichlet boundary value problem in the Banach spaces is discussed by using the fixed point theory of condensing mapping, doing precise computation of measure of noncompactness, and calculating the spectral radius of linear operator.

1. Introduction

This paper is mainly concerned with the following second-order Dirichlet boundary value problem: in a Banach space , where , is the zero element of .

In the last several decades, there has been much attention focused on the boundary value problems for various nonlinear ordinary differential equations, difference equations, and functional differential equations, see [1–20] and the references therein. The existence of solutions for Neumann boundary value problems has been considerably investigated in many publications such as [2–5, 8–10]. Dirichlet boundary value problems have deserved the attention of many researchers, see [11–20] and the references therein.

In particular, the authors in [11] have studied the following two-point boundary value problem: where and . They obtained the existence of solutions by means of the Darbo fixed point theorem and properties of the measure of noncompactness.

We would like to mention the results due to [11]. First, we point out that many authors applied the famous Sadovskii’s fixed point theorem to investigate similar problems and used the following hypothesis with respect to the Kuratowski measure of noncompactness : there exists a constant such that for any bounded and equicontinuous set and , . What is more, they required a stronger condition, that is, for and the constant satisfies (see Remarks 3.2–3.6).

The authors in [15, 18] have studied the following boundary value problem: where is a real Banach space, , is continuous, , and for . They obtained the existence of solutions by means of Sadovskii’s fixed point theorem and properties of the measure of noncompactness.

Motivated by the above-mentioned work [11, 15, 18], the main aim of this paper is to study the existence and uniqueness of solutions for the problem (1.1) under the new conditions. The main new features presented in this paper are as follows First, the existence and uniqueness of solutions to Banach space’s Dirichlet boundary value problem is proved precisely calculating the spectral radius of linear operation. Second, the conditions imposed on the BVP (1.1) are weak. Third, the main tools used in the analysis are Sadovskii’s fixed point theorem and precise computation of measure of noncompactess. Our results can be seen as a supplement of the results in [11] (see Remarks 3.2–3.6).

This paper is organized as follows. In Section 2, we provide some basic definitions, preliminaries facts, and various lemmas which will be used throughout this paper. In Section 3, we give main results in this paper.

2. Preliminaries and Lemmas

Let be a real Banach space and be a cone in which defines a partial ordering in by if and only if · is said to be normal if there exists a positive constant such that implies , where denotes the zero element of , and the smallest is called the normal constant of (it is clear, ). If and , we write . For details on cone theory, see the monograph [7].

Let . By we denote the Banach space of all continuous functions from into with the norm

Definition 2.1 (see [7]). Assume that is a bounded set in . Let be expressed as the union of a finite number of sets with diameter . is said to be the Kuratowski measure of noncompactness and is called the noncompactness measure for short. For details and properties of the noncompactness measure see [7].

Definition 2.2 (see [7]). The mapping is said to be a condensing operator if is continuous, bounded, and for any nonrelatively compact and bounded set , The following lemmas are of great importance in the proof of our main results.

Lemma 2.3. Suppose that . Then for any , the linear boundary value problem has a unique solution , and is bounded linear operator.

Proof. Note that , which assures second-order boundary value problem has only a zero solution. To obtain a solution of the problem (2.3), we require a mapping whose kernel is the Green’s function of the boundary value problem (2.4). Let , we consider three cases.Case 1. if , we have Case 2. if , we have Case 3. if , , we have After direct computations, it is easy to see that is continuously differentiable, and is a solution of (2.3).We now claim that solution of the boundary value problem (2.3) is unique. The proof is as follows. If possible, suppose that is another solution of the problem (2.3). For any ( denotes the dual space of ), let , thus we obtain , . By (2.3), we have that is, is a solution of the boundary value problem (2.4). However, on the other hand, problem (2.4) has only a zero solution, therefore we have . Thus, we get , hence in , which implies that solution of the problem (2.3) is unique, say, , and .It is easy to see that is bounded linear operator. This completes the proof.

Remark 2.4. If , it is easy to see that .

Lemma 2.5. Assume that , and is given by (2.8). Then (1)the spectral radius ;(2)If is an ordered Banach space, then is a positive operator, that is, if , then .

Proof. (1) Define operator by where . By Lemma 2.3, we have that is bounded invertible operator of , and if , then has a bounded invertible operator, thus .Let , is a eigenvalue of . For any , since is eigenvector of , then the spectrum of operator is . By the spectral mapping theorem [21], we get so .(2) If , by definition of and , then we get , so by (2.8), we have , that is, is a positive operator. This achieves the proof.

Remark 2.6. In particular. If , , then by (1) of Lemma 2.5, we get , where is an operator in : and . In fact: This means that , therefore . However, on the other hand, . As a result, we obtain .

Lemma 2.7. Let , , . Then where , is closed convex hull of .

Proof. If , then (2.14) is true. We suppose that , and take a partition of : Let , , by definition of Riemann integral, we get We take sufficiently large, such that , then we get This finishes the proof.

Lemma 2.8. Suppose that is a bounded set in , then there exists a countable subset of , such that

Proof. Let , . For , take , then is not a cover of . Take , then is not a cover of . Continuing this process, take , then is not a cover of .Set , then we get , where denote the distance between two points and of . Thus it follows that .Setting , choose sufficiently large such that , that is, . The proof is completed.

Lemma 2.9. If is a bounded set in , . Then

Proof. For any , there exists a partition such that for . Choose . Since is uniformly continuous on , there exists , such that , and , we have Let be a partition of , and . Set , . Clearly, we have For any and , it follows from (2.20) and (2.21) that So, Thus it follows that . Therefore, by using the arbitrariness of , we have . The lemma is proved.

Lemma 2.10 (see [7]). Assume that is bounded and equicontinuous. Then is continuous on and

Lemma 2.11 (see [7]). Suppose that is a countable family of strongly measurable functions . If there exists a function such that for a.e. , then and

Lemma 2.12. Assume that is equicontinuous in . Then is equicontinuous.

Proof. For any , it follows from the equicontinuity of that there exists such that implies for all and .For any , by virtue of definition of , we have Thus, we get Hence, is equicontinuous. This finishes the proof.

Lemma 2.13 (see [7]). If is bounded and equicontinuous, then is continuous in and .

Lemma 2.14 (see [7] (Sadovskii’s Theorem)). Assume that is a nonempty bounded, closed, and convex set. If a mapping is condensing, then has a fixed point in .

3. Main Results

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

Theorem 3.1. Let be a Banach space. Suppose that and the following conditions hold: there exist two positive numbers and , such that for any a bounded set in , there exists a constant such that there exist two positive numbers and with , . Then problem (1.1) has at least one solution. Proof. Define the integral operator Then is continuous, and it is clear that is a solution of the problem (1.1) if and only if is a fixed point of .We now show that is a condensing operator. Let be bounded in , by , we claim that is bounded. Since , we know that is bounded, this means that is equicontinuous. Therefore, it follows from Lemma 2.13 that .For any , , from Lemma 2.7, we have Hence, Using the properties of the noncompactness measure together with , we obtain By Lemma 2.9, we have Hence, By , we get , therefore is condensing. Let , we will prove that , for sufficiently large. By means of the homotopy invariance theorem, we have . By virtue of the solvability of Kronecker [6], we know that has a fixed point in , and the fixed point of is a solution of the problem (1.1).Indeed. If there exists a constant , such that , then satisfies Let , and is an operator in defined by Then by Lemma 2.5 we have .By (3.9) and , we have Thus, continuing this process, by induction, we obtain which implies that By the Gelfand theorem [22], we have Set , then we have . By (3.15), there exists an integer , such that as , that is, , this means that . In view of series converges, we know that also converges.Denote . By (3.14), we get , which implies that , hence .Take , then we have , . Thus problem (1.1) has at least one solution. This proves the theorem.

Remark 3.2. In [11], the nonlinear term is bounded, if , in our result, the nonlinear term may no more than a linear growth.

Remark 3.3. In [11], if , the growth restriction of for is . However, in our result, satisfies .

Theorem 3.4. Let be a Banach space, and . Assume that condition and the following conditions hold: for all , for any a bounded set in , there exists a constant such that there exist two positive numbers and with , . Then problem (1.1) has at least one solution.

Proof. Assume that the operator is defined the same as in Theorem 3.1. We show that the operator is condensing. In fact, for a bounded set , there exists a countable subset , such that . However, on the other hand, we have By Lemma 2.11 and , we obtain Thus, By , we get , therefore is condensing. By using the same arguments of Theorem 3.1, we can obtain the conclusion of Theorem 3.4. The detailed proof is omitted here. The proof is achieved.

Next, we establish a uniqueness of solution for the problem (1.1).

Theorem 3.5. Let be a Banach space. Suppose that and that there exists a constant with such that Then problem (1.1) has a unique solution.

Proof. Assume that operator is defined the same as in Theorem 3.1, and the fixed point of is a solution of the problem (1.1).We will prove that for sufficiently large the operator is a contraction operator. Indeed, by the definition of and (3.20), we have the estimate By induction, we have Thus, Moreover, we can choose to be sufficiently large such that tends to .Further, take , there exists an integer , such that as . By (3.23), we obtain which implies that is a contraction mapping by . By the contraction mapping principle, we conclude that there exists a unique fixed point for , this proves that problem (1.1) has a unique solution. This completes the proof.

Remark 3.6. By the direct application of the Banach contraction mapping principle, the conclusion of Theorem 3.5 holds true under the condition . However, we require the condition , here is optimum.The following theorem is concerned with the existence of positive solutions for problem (1.1).

Theorem 3.7. Let be an ordered Banach space, be a normal cone with positive elements. Suppose that satisfy the following conditions: (P₁) there exists a constant with , and , such that (P₂) for any a bounded set in , there exists a constant with such that Then problem (1.1) has at least one positive solution.

Proof. Consider the linear boundary value problem By Lemma 2.3, the linear boundary value problem (3.27) has unique a positive solution .Set , then is bounded and convex closed set in . For any , by , we have Multiply by and integrate from 0 to 1, we obtain that is, , so . By the proof of Theorem 3.1, it follows that is equicontinuous. Thus, by Lemma 2.12, we know that   is equicontinuous.Next we show that is condensing. For any , then is bounded and equicontinuous, therefore is bounded and equicontinuous. By Lemma 2.13, we have However, on the other hand, we have Thus, For any , , we acquire Hence, Further, we obtain This means that By , we have , so is condensing. Applying Lemma 2.14, we conclude that has a fixed point which is a solution of problem (1.1). The proof of the theorem is completed.

Acknowledgments

Wen-Xue Zhou's work was supported by NNSF of China (10901075), Program for New Century Excellent Talents in University (NCET-10-0022), the Key Project of Chinese Ministry of Education (210226), and NSF of Gansu Province of China (1107RJZA091).

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