Research Article | Open Access

Volume 2011 |Article ID 989401 | https://doi.org/10.5402/2011/989401

Wanjun Li, "Eigenvalue of Coupled Systems for Hammerstein Integral Equation with Two Parameters", International Scholarly Research Notices, vol. 2011, Article ID 989401, 11 pages, 2011. https://doi.org/10.5402/2011/989401

# Eigenvalue of Coupled Systems for Hammerstein Integral Equation with Two Parameters

Accepted24 Feb 2011
Published11 Oct 2011

#### Abstract

By using the fixed-point index theory, we discuss the existence, multiplicity, and nonexistence of positive solutions for the coupled systems of Hammerstein integral equation with parameters.

#### 1. Introduction

In recent years, the study of solutions for Hammerstein integral equations has been an interesting topic, since the solution of some boundary value problems for differential equations are usually equivalent to solutions of Hammerstein integral equations . And many results concerning the existence of solutions for Hammerstein integral equations have been obtained by many authors [1, 79]. For example, in  the Hammerstein integral equation: was considered, where is a bounded domain. When the nonlinear term is of the form or , , some existence results of nonnegative solutions in for (1.1) were obtained; when the nonlinear term is a general , some multiple results for (1.1) in space were derived. In , by means of the decomposition of the operator and the critical point theory, the existence of infinitely many solutions for (1.1) was considered. In  the integral equation was studied, where is a parameter. Using the Leggett-Williams fixed-point theorem, when belongs to some intervals, the existence of triple positive solutions for (1.2) was proved.

More recently, in  the integral equation was studied and obtained that there exists a such that (1.3) has at least two, one and no positive solutions for , , and , respectively.

Motivated by the papers mentioned above, we consider the coupled systems of Hammerstein integral equation (1.4) in this paper, where , , . Under some new assumptions, we show that there exists a continuous curve separating into two disjoint subsets and such that problem (1.4) has at least two, one and no positive solutions for , and , respectively. Proofs of our results are mainly based on the fixed-point index theory, for this type of results see .

The rest of the paper is organized as follows. In Section 2, we present some preliminaries and lemmas that will be used to prove our main results. In Section 3, we discuss the existence, multiplicity, and nonexistence of positive solution of the systems (1.4).

The vector is said to be a positive solution of problem (1.4) if and only if satisfies problem (1.4) and , or for any .

#### 2. Preliminaries and Lemmas

In the rest of the paper, we always suppose the following assumptions hold:; ; there exist , , , such that for and ; are nondecreasing on for , that is, , whenever , where the inequality on can be understood componentwise and or for all ; there exist constants , such that for all ; uniformly for .

We will consider the Banach space equipped with the standard norm

Define It is easy to see that is a cone in .

We define the operators and by

Obvious, the existence of a positive solution of problem (1.4) is equivalent to the existence of a nontrivial fixed point of in . It is easy to prove that the following lemma is true.

Lemma 2.1. Assume that hold. Then and is completely continuous.

Finally we list two lemmas, which are crucial to prove our main results.

Lemma 2.2 (see ). Let be a Banach space, a cone and an open bounded subset of . Let and be a completely continuous mapping. Suppose that for all and all . Then .

Lemma 2.3 (see ). Let be a Banach space and a cone in . Assume that is a compact map such that for . If for all , then .

#### 3. Main Results

In this section, we consider the existence of positive solutions for (1.4) in .

Lemma 3.1. Assume that hold, and let be a compact subset of . Then there exists a constant such that for all and all possible positive solutions of (1.4) at , one has .

Proof. Suppose by contradiction that there is a sequence of positive solutions of (1.4) at such that for all and . Then and thus Since is compact, the sequence has a convergent subsequence which we denote without loss of generality still by such that , , and at least one or , hence for sufficiently large, we have .
Then by , there exists such that where satisfies
Thus for and , by using (3.1) and (3.2), we get for all sufficiently large. This is a contraction.

Lemma 3.2. Assume that hold, and let (1.4) have a positive solution at . Then the problem also has a positive solution at for all .

Proof. Let be a positive solution of (1.4) at , and let with . First, we assume , and , then Set and let , , , , , , . Then where , . By the compactness of the operator , , Lemma 2.1, and the Lebesgue dominated convergence theorem, the sequence , and converges to and , respectively. It is clear that and is a solution (1.4) at . Proof of the case , and or , and can be done similarly. The proof is complete.

Lemma 3.3. Assume that hold. Then there exists such that (1.4) has a positive solution for all .

Proof. Let , , then . Take , . Then and at , we get This shows that and for . Set for . Then We conclude the proof similarly to Lemma 3.2.

Define (1.4) has a positive solution at ; then by Lemma 3.3, , and it is easy to see that is a partially ordered set.

Lemma 3.4. Assume that hold. Then is bounded above.

Proof. Suppose to the contrary that there exists a fixed-point sequence of at such that . Considering a subsequence if necessary, we assume . The proof for the case can be shown by an analogous way. Then there are two cases to be considered:(i)there exists a constant such that ;(ii)there exists a subsequence such that ,which is impossible by Lemma 3.1. So we only consider (i). In view of we can choose such that or and further or for . Thus for , we know Now we will distinguish two cases.Case 1. If and , we have where , which implies that or , which is a contradiction.Case 2. If that one of and is satisfied, without loss of generality, we assume . The proof for the case can be shown by an analogous way. By we have where ; which implies that or , which is a contradiction. The proof is complete.

Lemma 3.5. Assume that hold. Then every chain in has a unique supremum in .

Proof. Let be a chain in . Since is a partially ordered set, it is enough to show that has an upper bound in . Without loss of generality, we may choose a distinct sequence such that . By Lemma 3.4, two sequences and converge to, say, and , respectively. If , then the proof is done. Since the sequence is bounded above, we may assume that the sequence belongs to a compact rectangle in and Lemma 3.1 implies that the corresponding solutions are uniformly bounded in . By the compactness of the integral operators and , the sequence has a subsequence converging to, say, . We can easily show, by the Lebesgue convergence theorem, that is a solution of (1.4) at . Thus and this completes the proof.

Lemma 3.6. Assume that hold. Then there exists such that (1.4) has a positive solution at for all , and no solution at for all . Similarly, there exists such that (1.4) has a positive solution at for all , and no solution at for all , where is upper bound of .

Proof. We know by Lemma 3.3 that (1.4) has a positive solution at for all . Thus is a nonempty chain in and by Lemma 3.5, it has a unique supremum of the form in . The proof of the second part is similar.

Lemma 3.7. Assume that hold. Then there exists a continuous curve separating into two disjoint subsets and such that problem (1.4) has at least one positive solution for and no solution for .

Proof. We first construct the curve on . Define At , we know by Lemma 3.6, . Thus is a nonempty chain in and Lemma 3.5 implies that the chain has a unique supremum. We show . Indeed, otherwise, we may choose such that and (1.4) has a solution at . Thus by Lemma 3.2, (1.4) has a solution at and this contradicts Lemma 3.6. Similarly, we get .
For , we know by Lemmas 3.3, 3.5, and 3.6 that is a nonempty chain in and thus the chain also has a unique supremum.
We notice that , for or . Now for , let us define Then is well defined and and . Similar to [10, Theorem  3.1], it is easy to prove that is continuous on .
Consequently, the curve separates into two disjoint subsets and , where is bounded and is unbounded. It is obvious that (1.4) has a positive solution at for all . If and so if , then and . Thus by Lemma 3.2, . On the other hand, if and if , then either is not defined when or when . We get for both cases and the proof is done.

Now, we show the existence of the second positive solution for . Let , then we may choose such that . We know by Lemma 3.7 that (1.4) has a positive solution at , so let be the solution at and let us denote . Then obviously and we have the following theorem.

Lemma 3.8. Let . Then there exists and for all , such that where and .

Proof. From , there exists constant such that Then by the uniform continuity of on a compact set, there exists small enough such that for all , , Furthermore, we have for all The inequalities for can be shown similarly and the proof is done.

We now state and prove our main result in this section.

Theorem 3.9. Assume . Then there exists a continuous curve separating into two disjoint subsets and such that problem (1.4) has at least two positive solution on , at least one positive solution on and no solution on .

Proof. By Lemma 3.7, it is enough to show the existence of the second positive solution of (1.4) for . Let . Let and be a positive solution of (1.4) at and let , where is given in Lemma 3.8. Let then is bounded open in , and be a completely continuous mapping, since it is completely continuous. Let . Then there exists such that either or . Suppose that . Then by and Lemma 3.8, for all . Similarly, for the case , we can get for all . Thus , for all and all and by Lemma 2.2, Modifying (3.2) for , that is, there exists such that where satisfies Let , where is given in Lemma 3.1 with a compact rectangle in containing . Let . Then for , by Lemma 3.1. Furthermore, if , then Thus by (3.24), , for all and Therefore and by Lemma 2.3, Consequently by the additivity of the fixed-point index, Since and thus has a fixed point on and another on , and this completes the proof.

#### Acknowledgments

The author is very grateful to the referee for her/his important comments and suggestions. This work is sponsored by the Tutorial Scientific Research Program Foundation of Education Department of Gansu Province, China (0810-03).

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