Chinese Journal of Mathematics

Volume 2014 (2014), Article ID 717290, 11 pages

http://dx.doi.org/10.1155/2014/717290

Research Article

Existence and Multiplicity of Positive Solutions for a System of Fourth-Order Boundary Value Problems

Shoucheng Yu and Zhilin Yang

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

Received 26 February 2014; Accepted 15 April 2014; Published 18 June 2014

Academic Editor: Salvatore A. Marano

Copyright © 2014 Shoucheng Yu 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.

We study the existence and multiplicity of positive solutions for the system of fourth-order boundary value problems , and where . We use fixed point index theory to establish our main results based on a priori estimates achieved by utilizing some integral identities and inequalities and -monotone matrices.

1. Introduction

In this paper we study the existence and multiplicity of positive solutions for the system of fourth-order boundary value problems:

where . It is well known that the deflection of elastic beams can be described by some fourth-order boundary value problems; for example, see [1, 2]. Consequently, fourth-order boundary value problems play a very important role in both theory and applications. Boundary value problems for systems of nonlinear ordinary differential equations have been also studied by several authors; for example, see [3–11] and the references therein. In [3], Li et al. use fixed point theorems on cones to establish the existence of positive solutions for a system of third-order boundary value problems:

In [4], Lü et al. study the existence of positive solutions for a system of boundary value problems:

where . In [11], the authors investigate the existence and multiplicity of positive solutions for the system

The hypotheses imposed on the nonlinearities and are formulated in terms of two linear functions and . The main results in [11] are established by using fixed point index theory based on a priori estimates of positive solutions achieved by utilizing new integral inequalities and nonnegative matrices.

This paper is organized as follows. In Section 2, we use the method of order reduction to transform (1) into a system of boundary value problems for second-order integrodifferential equations. Also, in this section, we develop some basic integral identities and inequalities that are useful in deriving a priori estimates in Section 3. Our main results, namely, Theorems 11–14, are stated and proved in Section 3. Finally, three examples that illustrate our main results are presented in Section 4.

2. Preliminaries

Let , , . Then is a real Banach space and is a solid cone in . Let

Define the linear integral operator by

It is well known that is a completely continuous, positive, linear operator and

where . Substituting , into (1), we transform (1) into the following system of seconed-order boundary value problems for the integrodifferential equations:

which is equivalent to the system of nonlinear integral equations:

Define the operators and by

If , then and are completely continuous operators. Clearly, the existence of positive solutions for (1) is equivalent to that of positive fixed points of .

Lemmas 1–4 below are cited from [12].

Lemma 1. If , , then

Lemma 2. If , then

Lemma 3. If , , and is decreasing on , then

Lemma 4. If , , then

Lemma 5. Suppose is not identically vanishing on , and is a concave function. Let . Then

Proof. By the concavity of and the nonnegativity of , we have

This completes the proof.

Lemma 6 (see [13]). 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 7 (see [13]). 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 .

Lemma 8 (see [14, Lemma 2.4]). If is concave on , with , then is increasing on and

for all .

3. Main Results

Let . Let for . We list our hypotheses as follows.

(H1).(H2)There exist a nonnegative matrix and a constant that

holds for all , with the matrix

being an -monotone matrix.(H3)There exist and such that (1) is concave; (2) there are two constants and such that

(H4)There exist two functions , such that

for all , and

where , with being specified in Theorem 11.(H5)There exist a nonnegative matrix and a constant that

holds for all , with the matrix

being an -monotone matrix.(H6)There exist a nonnegative matrix and a constant such that

holds for all , with the matrix

being an -monotone matrix.(H7)There exist a nonnegative matrix and a constant that

holds for all , with the matrix

 being an -monotone matrix. (H8) and are increasing in , and there is a constant such that

Remark 9. A real matrix is said to be nonnegative if all elements of are nonnegative.

Remark 10 (see [15, page 112]). A real square matrix is called -monotone if, for any column vector , .

Theorem 11. If (H1), (H2), (H4), and (H5) hold, then (1) has at least one positive solution.

Proof. Let

where . We will prove that is bounded. Indeed, if , then there exists such that , which can be written in the form

Taking differentiation of the preceding equations twice, we obtain

and thus we have, by (H2),

Multiply the above by and integrate over and use (11) and to obtain (12)

It is readily seen that, for any , and are decreasing on . By (13), we obtain

so that

Since is an -monotone matrix, we have

Now Lemma 4 implies

Furthermore, these estimates lead to

for all . Let

Equation (42) implies that . Let . By (H4) again, there are two functions , for all , such that

Thus we have

This implies, for all

and, in turn,

By (H4) again, there exists a constant such that

This means that is bounded. Taking , we have

Now Lemma 6 yields

Let

where is given in (H5). We are in the position to prove . Indeed, if , then for some , which can be written componentwise as

Taking differentiation of the preceding equations twice, we obtain

By (H5), we have

Multiply the above by and use (11) and (12) to obtain

so that

Since is an -monotone matrix, it follows that and , so that . This proves , as required. Consequently,

Now Lemma 7 yields

Combining (51) and (59) we arrive at

Therefore has at least one fixed point on . Thus (1) has at least one positive solution, which completes the proof.

Theorem 12. If (H1), (H3), (H4), and (H5) are satisfied, then (1) has at least one positive solution.

Proof. By (H3), we obtain

for all , . We claim that the set

is bounded, where . Indeed, if , then there exists a constant such that , which can be written in the form

By (H3) and (61), we have

for all . The nonnegativity and concavity of imply . Now Lemma 8 and Jensen’s inequality imply

This, together with (64) and (H3), implies

Multiply the last inequality by and integrate over and use (7) twice to obtain

so that

By Lemma 5, we have

Multiply the first inequality of (64) by , integrate over , and use (7) to obtain

This, along with (68), implies

By Lemma 5, we have

so that

(H3) implies that is strictly increasing and (see Lemma 8). Consequently, there exists such that

Let . Then

This establishes the a priori bound of for . Now it remains to derive the a priori bound of for , which is similar to the derivation of the a priori bound of for in Theorem 11. This means that is bounded. Taking , we have

Now Lemma 6 yields

Note that (H1) and (H5) imply (59) holds (see the proof of Theorem 11). Combining (77) and (59) we arrive at

Therefore has at least one fixed point on . Thus (1) has at least one positive solution, which completes the proof.

Theorem 13. If (H1), (H6), and (H7) are satisfied, then (1) has at least one positive solution.

Proof. Let

We will prove that is bounded. Indeed, if , then and , for some , which can be written componentwise as

Taking differentiation of the preceding equations twice, we obtain

By (H6), we have

Multiply the above by and use (11) and (12) to obtain:

so that

Since is an -monotone matrix, we have

Now Lemma 4 implies

Furthermore, these estimates lead to

Therefore, for all , we have

Applying Lemma 3 in [16], we may establish the boundedness of of . Taking , we have

Lemma 7 yields

Let

where and is given in (H7). Next we will prove . Indeed, if , then there exists such that , which can be written in the form

Taking differentiation of the preceding equations twice, we obtain

By (H7), we have

Multiply the above by and integrate over and use (11) and (13). Notice that are decreasing on ; we obtain

so that

Since is an -monotone matrix, it follows that and , so that . This proves , as required. Consequently,

Now Lemma 6 yields

Combining (90) and (98), we arrive at

Therefore has at least one fixed point on . Hence (1) has at least one positive solution, which completes the proof.

Theorem 14. If (H1), (H2), (H4), (H7), and (H8) hold, then (1) has at least two positive solutions.

Proof. By (H8), we have

for all and all . Consequently, we have, for all ,

The preceding inequalities imply

thus

Now Lemma 7 yields

By (H2), (H4), and (H7), we find that (51) and (98) hold. Note that we can choose in (51) and (98) (see the proofs of Theorems 11 and 13). Combining (51), (98), and (104), 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.

4. Examples

Example 15 . Let

where , , and . Then (H1), (H2), (H4), and (H5) hold. By Theorem 11, (1) has at least one positive solution.

Example 16. Let

where , , and . Then (H1), (H6), and (H7) hold. By Theorem 13, (1) has at least one positive solution.

Example 17. Let

where , , , , and ; is sufficiently small. Then (H1), (H2), (H4), (H7), and (H8) hold. By Theorem 14, (1) has at least two positive solutions.

Conflict of Interests

The authors declare that there is no conflict of interests for any of them.

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