Abstract and Applied Analysis

Volume 2013 (2013), Article ID 296038, 7 pages

http://dx.doi.org/10.1155/2013/296038

Research Article

Multiple Solutions for Boundary Value Problems of th-Order Nonlinear Integrodifferential Equations in Banach Spaces

Yanlai Chen

School of Economics, Shandong University, Jinan, Shandong 250100, China

Received 23 June 2013; Revised 18 September 2013; Accepted 20 September 2013

Academic Editor: Pavel Kurasov

Copyright © 2013 Yanlai Chen. 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 boundary value problems of a class of th-order nonlinear integrodifferential equations of mixed type in Banach space are considered, and the existence of three solutions is obtained by using the fixed-point index theory.

Guo [1] considered the initial value problems of a class of integrodifferential equations of Volterra type and obtained the existence of maximal and minimal solutions by establishing a comparison result. In [2], the author and Qin investigated a first-order impulsive singular integrodifferential equation on the half line in a Banach space and proved the existence of two positive solutions by means of the fixed-point theorem of cone expansion and compression with norm type. For other results related to integrodifferential equations in Banach spaces please see also [3–6] and the references therein. It is worth pointing out that the nonlinear terms involved in the equations they considered are either sublinear or superlinear globally.

In this paper, by using fixed-point index theory (for details please see [7]), we consider the th-order integrodifferential equations with nonlinear terms neither sublinear nor superlinear globally and prove the existence of three solutions.

Let be a real Banach space and 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 . If and , we write . is said to be solid if its interior is not empty; that is, . In case of , we write . For details on cone theory, please see [8].

We consider the following boundary value problem (BVP for short) in :

where , , denotes the zero element of , and

with , , , and the set of all nonnegative numbers. Let

Denote that is a map from into and is continuous on . It is clear that is a Banach space with norm defined by

Let

It is obvious that and are two cones in and , respectively.

Lemma 1. is the solution of problem (1) if and only if is the fixed point of operator defined by

Proof. For , Taylor's formula with the integral remainder term gives

Taking , we have

Substituting

into (8), we get

Let be the solution of BVP (1). Then (10) implies

Comparing this with (6), we have , which means that is the fixed point of the operator in .

On the other hand, let be the fixed point of the operator . By (6),

where . It follows by taking and in (12) that

It is also clear from (12) that

Hence, . Then (13)–(14) imply that is the solution for BVP (1) in .

To continue, let us formulate some conditions.

Let be bounded and uniformly continuous in on , . There exist nonnegative constants  such that

 where , denotes the Kuratowski measure of noncompactness, and . Assume that

 where

 , , and is defined by . There exist , , and such that

 for , , , and .

Remark 2. By and , one can see that is neither sublinear nor superlinear globally.

Lemma 3 (see [8]). Let be a bounded set of . Then

where .

Lemma 4 (see [8]). Let be a bounded set of . Suppose that is equicontinuous. Then

where is defined by Lemma 3 and .

Lemma 5. Let hold. Then operator defined by (6) is a strict set contraction from into .

Proof. It is easy to see that and is a bounded operator by (6), (12), and .

Now we check that operator is continuous from into . Let , , and

For any , by (6),

Then the Lebesgue dominated convergence theorem gives

Hence,

Similarly, in view of (12), we get

Then

Consequently, the continuity of operator is proved.

Let be bounded. Then is bounded. We prove that is equicontinuous on . In fact, , by (12),

According to the absolute continuity of Lebesgue integral, is equicontinuous on . Therefore, Lemma 4 implies that

where ( is fixed, ). By (6), we see that

where ,

It follows from (31) and that

which implies, according to Lemma 3, that

where in view of (15).

Similarly, we have

Thus, we get by (34) and (35). Noticing that is bounded and continuous, the conclusion follows.

Theorem 6. Let be a normal solid cone and let , , and hold. Then BVP (1) has at least three solutions in .

Proof. Condition implies that there exist and , such that, for ,

Choose . Let

For , we have , , and . So, it follows from (6), (12), and (36) that

Hence, . Thus, we have shown that

Similarly, by (18), it is easy to get that there is a number such that and

where   and is the normal constant of .

Let

It is easy to see that , , and are all nonempty bounded open convex sets of , and

As the proof of (38), for , by ,

On the other hand, according to , for , , , and , we get by (12) that

Condition also implies that

Consequently, in view of (43) and (45), we have shown that

It follows from (39), (40), (42), (46), and Lemma 5 that

where denotes the fixed-point index [7]. Therefore, has three fixed points , , and. By Lemma 1, BVP (1) has at least three solutions in .

An application of Theorem 6 is as follows.

Example 7. Consider

where and .

Obviously, () is the trivial solution of BVP (48).

Conclusion. BVP (48) has at least two nontrivial nonnegative solutions.

Proof. Let , -dimensional space, with norm and

Then is a normal and solid cone in and (48) can be regarded as a BVP of the form (1), where

and with

Obviously, and is satisfied for since is finite-dimensional.

One can see that

Then (51) implies that

Therefore,

Hence,

On the other hand, it is easy to see that

Thus, (55) and (56) imply that is satisfied.

Now, we check . Let , and , . Obviously, and, for , , , , , and (i.e., , , , , , , , ). Then (51) implies that

where . So, we have . Hence, is satisfied. And, finally, the conclusion follows from Theorem 6.

Acknowledgment

The author is grateful to Professor Guo Dajun and two anonymous referees for their valuable suggestions and comments.

References

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