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

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.