`Abstract and Applied AnalysisVolume 2012, Article ID 401923, 14 pageshttp://dx.doi.org/10.1155/2012/401923`
Research Article

## Positive Solutions for Neumann Boundary Value Problems of Second-Order Impulsive Differential Equations in Banach Spaces

Department of Mathematics, Northwest Normal University, Lanzhou 730070, China

Received 8 September 2011; Revised 28 December 2011; Accepted 6 January 2012

Copyright © 2012 Xiaoya Liu and Yongxiang Li. 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 of positive solutions for Neumann boundary value problem of second-order impulsive differential equations , , , , , , in an ordered Banach space was discussed by employing the fixed point index theory of condensing mapping, where is a constant, , , , , and is the cone of positive elements in . Moreover, an application is given to illustrate the main result.

#### 1. Introduction

The theory of impulsive differential equations is a new and important branch of differential equation theory, which has an extensive physical, chemical, biological, engineering background and realistic mathematical model, and hence has been emerging as an important area of investigation in the last few decades; see [1]. Correspondingly, boundary value problems of second-order impulsive differential equations have been considered by many authors, and some basic results have been obtained; see [27]. But many of them obtained extremal solutions by monotone iterative technique coupled with the method of upper and lower solutions; see [24]. The research on positive solutions is seldom and most in real space ; see [57].

In this paper, we consider the existence of positive solutions to the second-order impulsive differential equation Neumann boundary value problem in an ordered Banach space : where is a constant, ,  ;;   is an impulsive function, .    denotes the jump of at that is, , where and represent the right and left limits of at , respectively.

In the special case where ,    ,   , NBVP (1.1) has been proved to have positive solutions; see [8, 9]. Motivated by the aforementioned facts, our aim is to study the positive solutions for NBVP (1.1) in a Banach space by fixed point index theory of condensing mapping. Moreover, an application is given to illustrate the main result. As far as we know, no work has been done for the existence of positive solutions for NBVP (1.1) in Banach spaces.

#### 2. Preliminaries

Let be an ordered Banach space with the norm and partial order ≤, whose positive cone is normal with normal constant . Let ;   , , . Let is continuous at , and left continuous at , and exists, . Evidently, is a Banach space with the norm , where ; see [2]. An abstract function is called a solution of NBVP (1.1) if satisfies all the equalities of (1.1).

Let denote the Banach space of all continuous -value functions on interval with the norm . Let denote the Kuratowski measure of noncompactness of the bounded set. For the details of the definition and properties of the measure of noncompactness, see [10, 11]. For any and , set . If is bounded in , then is bounded in , and .

Now, we first give the following lemmas in order to prove our main results.

Lemma 2.1 (see [12]). Let be equicontinuous. Then is continuous on , and

Lemma 2.2 (see [13]). Let be a bounded and countable set. Then is Lebesgue integral on , and

Lemma 2.3 (see [14]). Let be bounded. Then there exists a countable set , such that .

To prove our main results, for any , we consider the Neumann boundary value problem (NBVP) of linear impulsive differential equation in : where .

Lemma 2.4. For any , and , the linear NBVP (2.3) has a unique solution given by where

Proof. Suppose that is a solution of (2.3); then
Let ; then Integrating (2.7) from 0 to , we have Again, integrating (2.7) from to , where , then Repeating the aforementioned procession, for , we have Hence, For , integrating (2.11) from 0 to , we have Notice that ; thus, for , we have In view of that , we have Substituting (2.15) into (2.13), for , we obtain
Inversely, we can verify directly that the function defined by (2.4) is a solution of the linear NBVP (2.3). Therefore, the conclusion of Lemma 2.4 holds.

By (2.5), it is easy to verify that has the following property:

Evidently, is also an ordered Banach space with the partial order ≤ reduced by the positive cone . is also normal with the same normal constant .

Define an operator as follows: Clearly, is continuous, and the positive solution of NBVP (1.1) is the nontrivial fixed point of operator . However, the integral operator is noncompactness in general Banach space. In order to employ the topological degree theory and the fixed point theory of condensing mapping, there demands that the nonlinear and impulsive function satisfy some noncompactness measure condition. Thus, we suppose the following.

For any and   are bounded and where is arbitrarily countable set, and are counsants and satisfy .

Lemma 2.5. Suppose that condition is satisfied; then is condensing.

Proof. Since is bounded and equicontinuous for any bounded and nonrelative compact set , by Lemma 2.3, there exists a countable set , such that By assumption and Lemma 2.1, Since is equicontinuous, by Lemma 2.1, we have Combining (2.20) and , we have Hence, is condensing.

Let be a cone in defined by where .

Lemma 2.6. For any ,.

Proof. For any , by (2.18) and the second inequality of (2.17), we have By this, (2.18), and the first inequality of (2.17), we have Hence, .

Thus, for any is condensing mapping; the positive solution of NBVP (1.1) is equivalent to the nontrivial fixed point of in . For , let , and , which is the relative boundary bound of in . Denote that ; then the fixed point of in is the positive solution of NBVP (1.1). We will use the fixed point theory of condensing mapping to find the fixed point of in

Let be a Banach space and let be a cone in . Assume that is a bounded open subset of and let be its bound. Let be a condensing mapping. If for every , then the fixed point index is defined. If , then has a fixed point in . As the fixed point index theory of completely continuous mapping, see [10, 11], we have the following lemmas that are needed in our argument for condensing mapping.

Lemma 2.7. Let be condensing mapping; if then .

Lemma 2.8. Let be condensing mapping; if there exists , such that then .

#### 3. Main Results

(i) There exist ,   , such that for all and ,  ,  , and .(ii) There exist ,  , and , such that for all and ,  ,  , and . (i) There exist ,  , such that for all and ,  ,  , and .(ii) There exist ,  , and , such that for all and ,  ,  , and .

Theorem 3.1. Let be an ordered Banach space, whose positive cone is normal, , and . Suppose that conditions and or are satisfied; then the NBVP (1.1) has at least one positive solution.

Proof. We show, respectively, that the operator defined by (2.18) has a nontrivial fixed point in two cases that is satisfied and is satisfied.
Case 1. Assume that is satisfied; let , where is the constant in condition , to prove that satisfies If (3.1) is not true, then there exist and , such that ; by the definition of , satisfies Integrating (3.2) from 0 to 1, using of assumption , we have Since , for any , by the definition of , we have , , and thus that is; . So we obtain that in , which contracts with . Hence (3.1) is satisfied; by Lemma 2.7, we have Let ,  , and obviously . We show that if is large enough, then In fact, if there exist ,   such that , then ; by the definition of satisfies By of assumption , we have Integrating on and using of assumption , we have If , for any , by the definition of , we have ;  ; thus By and the normality of cone , we have If , then for any , by the definition of , we have ,  ; thus By , and the normality of , we have
Let ; then (3.6) is satisfied; by Lemma 2.8, we have Combining (3.5), (3.14), and the additivity of fixed point index, we have Therefore has a fixed point in , which is the positive solution of NBVP (1.1).

Case 2. Assume that is satisfied; let , where is the constant in condition , to proof that satisfies where . In fact, if there exists and , such that , then satisfies (3.7) and of condition , and we have Integrating on and using of assumption , we have
If , for any , by the definition of , we have , for all, and thus By , we obtain that , which contracts with . Hence (3.16) is satisfied.
If , for any , by the definition of , we have ,  , for all , and thus By , we obtain that , which contracts with . Hence (3.16) is satisfied.
Hence, by Lemma 2.8, we have
Next, we show that if is large enough, then In fact, if there exists and such that , then satisfies (3.2). Integrating (3.2) on , and using of , we have For any , by the definition of , we have ,  , for all, and thus By , we have By the normality of , we have Thus
Let ; then (3.22) is satisfied; by Lemma 2.7, we have
Combining (3.21), (3.28), and the additivity of fixed index, we have
Therefore has a fixed point in , which is the positive solution of NBVP (1.1).

Remark 3.2. The conditions and are a natural extension of suplinear condition and sublinear condition in Banach space . Hence if , then Theorem 3.1 improves and extends the main results in [8, 9].

#### 4. Example

We provide an example to illustrate our main result.

Example 4.1. Consider the following problem: where ,,, and    is a constant.

Conclusion
Problem (4.1) has at least one positive solution.

Proof. Let and ; then is a Banach space with norm , and   is a positive cone of . Let ; then the problem (4.1) can be transformed into the form of NBVP (1.1), where ,  , and  . Evidently is a Banach space with norm , and is positive cone of . Let , where ; then is a cone in , and for any , we have .
Next, we will verify that the conditions and in Theorem 3.1 are satisfied.
It is easy to verify that for any ,   and   are bounded. Let ; then is completely continuous. So for any countable bounded set , we have and for ,  and  , by simple calculations, . So condition is satisfied.
Let ; then for , we have Let , ; by simple calculations, we have . So of condition is satisfied.
Let , for ; we have . For ,  , we have . Hence, let ,   ; then for any , we have Let , and   ; by simple calculations, we have , so of condition is satisfied.
By Theorem 3.1, Problem (4.1) has at least one positive solution.

#### Acknowledgment

This work was supported by NNSF of China (no. 10871160, 11061031).

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