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Journal of Mathematics
Volume 2013, Article ID 316576, 8 pages
http://dx.doi.org/10.1155/2013/316576
Research Article

Positive Solutions for a Fourth-Order Boundary Value Problem

Department of Mathematics, Qingdao Technological University, Fushun Road No. 11, Quingdao, Shandong 266033, China

Received 24 August 2012; Accepted 29 October 2012

Academic Editor: Andrea Pascucci

Copyright © 2013 Kun Wang and Zhilin Yang. 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

This paper deals with the existence and multiplicity of positive solutions for the fourth-order boundary value problem . Here . We use fixed point index theory to establish our main results based on a priori estimates achieved by utilizing some integral identities and integral inequalities.

1. Introduction

The purpose of this paper is to study the existence and multiplicity of positive solutions for the fourth-order boundary value problem where . By a positive solution of (1), we mean a function that solves (1) and satisfies for all .

In recent years, fourth-order boundary value problems for nonlinear ordinary differential equations have been extensively studied by using diverse methods, including fixed point theorems on cones [16], the method of lower and upper solutions [710], the iterative method [1114], critical point theory [15], and the shooting method [16]. It should be remarked that none of the results cited above involves derivatives of all orders in their nonlinearities. In [9], Ma and Yang studied the fourth-order four-point boundary value problem where , . The authors used the method of upper and lower solutions and a new maximum principle to establish their existence results. In [14], Pei and Chang studied a class of fourth-order boundary value problem By using the monotone iterative technique, the authors proved that problem (3) has at least one symmetric positive solution under certain conditions. In [17], Pang et al. studied the existence and multiplicity of nontrivial solutions for the fourth-order boundary value problem where . Making use of the theory of Leray-Schauder degree, under appropriate conditions on the nonlinearity , the authors proved that problem (4) has at least six different nontrivial solutions, including two positive ones and two negative ones.

In [18], Yang and Sun studied the fourth-order boundary value problem where . By using fixed point index theory based on a priori estimates, the authors established the main results on the existence, multiplicity and uniqueness of positive solutions for (5).

Motivated by [18], in this paper, we will discuss the existence and multiplicity of positive solutions for problem (1). As in [18], we first use the method of order reduction to transform (1) into a boundary value problem for a second-order integro-differential equation and then seek the existence and multiplicity of positive solutions for the resultant problem. To overcome the difficulties stemming from the presence of derivatives of all orders and the difference between (1) and (5) in their boundary value conditions, we have to prove that the maximum of every nonnegative concave function can be dominated by the integrals (see Lemmas 5 and 6 below for more details). Based on a priori estimates achieved by utilizing some integral identities and integral inequalities, we use fixed point index theory to prove the existence and multiplicity of positive solutions for (1).

This paper is organized as follows. In Section 2, we transform (1) into a boundary value problem for a second-order integro-differential equation and then establish some basic integral identities and integral inequalities that are useful in deriving the priori estimates in the next section. Our main results, namely Theorems 1012, are stated and proved in Section 3.

2. Preliminaries

Let and , where . Furthermore, let Clearly is a real Banach space and is a cone in .

Let

Define the linear integral operators and by Substituting into (1), we transform (1) into the following second-order boundary value problem for the integro-differential equation which is equivalent to the nonlinear integral equation

Define the operator by

Now implies that is a completely continuous operator. In our setting, the existence of positive solutions for (1) is equivalent to that of positive fixed points of .

To establish the priori estimates of positive solutions for some problems associated with (11), we need several integral identities and integral inequalities below.

Lemma 1. If , then

Proof. Integrating by parts and using , we have from which (13) follows. This completes the proof.

Lemma 2. If , then

Proof. Notice . Integrating by parts, we obtain so that Notice and . Integrating the right hand of the above by parts again gives (16). This completes the proof.

If both and are continuous and decreasing on , then Chebyshev's inequality for and is given by In addition, if both and are continuous, with one of them decreasing on and the other increasing on , then Chebyshev's inequality for and becomes

Lemma 3. If , and is decreasing on , then

Proof. Notice and . Since is increasing on and is decreasing on , it follows from Chebyshev's inequality that
Notice that is increasing on . Chebyshev's inequality again implies
Note that (15) is guaranteed by . By virtue of (15), we have
Combining this with (22) yields (21). This complete the proof.

Lemma 4. If , then

Proof. By (15), we have This completes the proof.

Lemma 5. If and , then

Proof. It is easy to see that . The condition implies . By the concavity of , we have This completes the proof.

Lemma 6. If and , then we have

Proof. By the concavity of and , we have This completes the proof.

Lemma 7 (see [19]). Let be a real Banach space and a cone in . Suppose that is a bounded open set and that is a completely continuous operator. If there exists such that then , where indicates the fixed point index on .

Lemma 8 (see [19]). Let be a real Banach space and a cone in . Suppose that is a bounded open set with and that is a completely continuous operator. If then .

3. Existence and Multiplicity of Positive Solutions for (1)

For simplicity, we denote by and for in this section. Now we list our hypotheses.(H1) . (H2) There exist , , , such that , and holds for all .(H3) There exists a function such that for all and where and are given as in (H2).(H4) There exist , , such that , and holds for all .(H5) There exist , , such that , and holds for all .(H6) There are nonnegative constants , , , such that , and holds for all .(H7) There is a constant such that is increasing on in , and

Remark 9. is said to be increasing in if holds for every pair with and for all , where the partial ordering in is understood componentwise.

We denote for in the sequel.

Theorem 10. If (H1)–(H4) hold, then (1) has at least one positive solution.

Proof. Let where . We are now going to prove that is bounded. Indeed, if , then there exist such that , which can be written in the form which is equivalent to By (H2), we have Note (27) and (29). Multiply the above by , integrate over , and use Lemmas 3, 5, and 6 to obtain so that Now Lemma 4 implies where is given in (H3). Furthermore, this estimate leads to for all . Let Now (46) implies that for all . By (H3), there is a function such that Thus Combining the last inequality with (35), we see that there exists such that for all . This means that is bounded. Taking , we have
Now Lemma 7 yields
Let
Now we want to prove that . In fact, if , then and there is such that , that is,
which can be written in the form
By (H4), we have
Note (27) and (29). Multiply (56) by , integrate over , and use Lemmas 5 and 6 to obtain so that , whence and , as required. A consequence of this is Now Lemma 8 yields This together with (52) implies Therefore has at least one fixed point on and thus (1) has at least one positive solution. This completes the proof.

Theorem 11. If (H1), (H5), and (H6) hold, then (1) has at least one positive solution.

Proof. Let
We now assert that is bounded. Indeed, if , then there is such that , which is equivalent to
Now (H5) implies
Note (27) and (29). Multiply the above by , integrate over , and use Lemmas 5 and 6 to obtain so that for all , where . By Lemma 4, we obtain
Furthermore, this estimate leads to for all . By (63), we have
Note and . Integrating the above over , we obtain
This proves the boundedness of . Taking , we have
Now Lemma 8 yields On the other hand, let where is given by . Now we want to prove that . Indeed, if , then there exists such that , which can be written as By (H6), we have Note (27) and (29). Multiply the above by , integrate over , and use Lemmas 3, 5, and 6 to obtain so that , whence . This proves , as required. As a result of this, we obtain
Lemma 7 implies
Now (70) and (76) combined lead to
Consequently, has at least one fixed point on . Thus (1) has at least one positive solution. This completes the proof.

Theorem 12. If (H1), (H2), (H3), (H6), and (H7) hold, then (1) has at least two positive solutions.

Proof. If and , then we have
By (H7), we have
Consequently, for all , we have The preceding inequalities imply and thus Now Lemma 5 yields By (H2), (H3), and (H6), we know that (52) and (76) hold (see the proofs of Theorems 10 and 11). Note that we can choose in (52) and (76). Combining (52), (76), and (84), we obtain Therefore, has at least two fixed points, with one on and the other on . Hence (1) has at least two positive solutions. This completes the proof.

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