#### Abstract

By using the method of order reduction and the fixed point index, the existence of positive solutions for a fourth-order boundary value problem is studied. We provide conditions under which the existence results hold. Such conditions are related to the first eigenvalue corresponding to the relevant linear differential equation with dependence on the derivatives of unknown function.

#### 1. Introduction

In this paper, we consider the following fourth-order ordinary differential equation boundary value problem (BVP):where is continuous.

BVP (1) is used to model such phenomena as the deformations of an elastic beam in equilibrium state, whose one end-point is fixed and the other is freed. In mechanics, BVP (1) is called cantilever beam equation. Owing to its importance in mechanics, the existence of solutions to this problem has been studied by many authors; see [1–7] and references therein.

Very recently, Li [3] used the theory of the fixed point index to discuss the existence of solutions of BVP (1) when is superlinear or sublinear growth on . It should be remarked that two constants and , where is the first eigenvalue of the linear eigenvalue problem corresponding to BVP (1), play an important role in the discussion. The first eigenvalue principles were also used in [8–24].

However, none of these known results use the first eigenvalues of the corresponding linear differential equation that contains the derivative of the unknown function. This is because the presence of derivatives in linear eigenvalue problem will make the study extremely difficult. To overcome this difficulty, we employ the method of order reduction to develop spectral properties of associated linear differential equation that contains the derivative of the unknown function. Then, we use the fixed point index theory to investigate the existence results of positive solutions for BVP (1) under some conditions concerning the first eigenvalue corresponding to the relevant linear eigenvalue problem that contains the derivative of the unknown function. It should be noted that the method of order reduction was also used in [7, 25] to transform higher order boundary value problem to a lower order integrodifferential equation.

#### 2. Main Results

Let and . We use the classical Banach space with the norm , respectively. Let denote the cone of all nonnegative functions in .

We assume the following hypothesis holds throughout this paper.

Firstly, we consider the existence of solutions to BVP (1) when the nonlinearity satisfies the sublinear growth property. To this end, we use the method of order reduction to transform BVP (1). For , we setwhereThen we have . From the above expressions, we easily see that are three completely continuous linear operators, and and have the following properties: Hence we conclude that

Using the above transformations , BVP (1) can be converted into the equivalent nonlinear integral equation:

Define an operator by Then the existence of a positive solution of BVP (1) is equivalent to the existence of a nontrivial fixed point of on . Now condition implies that is a completely continuous operator.

For with , we define linear operator byClearly, is a completely continuous linear operator.

Lemma 1. *For the operator defined by (8), the spectral radius and had unique positive unit eigenfunction corresponding to its first eigenvalue .*

*Proof. *Take ; clearly ; by (3) we have So we have Thus there exists a constant such that , . From Lemma in [22], we know that the spectral radius and had a positive eigenfunction corresponding to its first eigenvalue . Hence Lemma 1 holds.

It follows from the definition of operator that the function , corresponding to the first eigenvalue of the operator , belongs to and satisfies the equation In particular, for , the positive eigenfunction of the operator corresponding to the first eigenvalue belongs to and satisfies the equation

Theorem 2. *If satisfies () and the following conditions, then BVP (1) has at least one positive solution.**There are with and such that and **There are with and such that and *

*Proof. *Let . We will prove that is a bounded set. In fact, if , by the definition of for some . Hence we have By , we have , which can be rewritten in the form , where . The condition ensures that has a bounded inverse operator which is given by . It follows from that . So we have and is bounded. Select ; we have , where . Now Lemma .3.1 in [26] impliesLet be the positive eigenvalue function of (with replacement of by in Lemma 1). We may suppose that has no fixed point on (otherwise, the proof is finished). Now we shall show thatwhere . In fact, if (17) is not valid, there exist and such that . Hence we have . Since , by the definitions of and (), we have Hence from , we obtain thatfrom which it follows thatSet . It is easy to see that and . It follows from that . Therefore, by (20), which contradicts the definition of . Hence (17) is true and we have from Corollary .3.1 in [26] thatBy (16) and (22), we have thatThen has at least one fixed point in which means that BVP (1) has a positive solution. The proof of Theorem 2 is completed.

Next, we consider the existence of solutions to BVP (1) when the nonlinearity satisfies the superlinear growth condition. In this case, we assume that is independent of and ; that is, we consider the following simplified BVP:As in [3], a positive solution of BVP (24) is equivalent to a nontrivial fixed point of defined bywhere is Green’s function of the homogeneous linear problem , which is explicitly given byFrom (26) we can easily verify that has the following properties:For with (we also use denoting ). Define a linear integral operator by

In what follows we work on the Banach space and utilize the cone

Lemma 3 (see [3, 6]). *If , then and .*

Lemma 4 (see [3, 6]). * are two completely continuous operators.*

Based on the definition of operator , (12) can be rewritten in the following form:which means that the operator has an eigenvalue , and its related eigenfunction can be taken by . In fact, we have from equivalent differential equation (12).

Theorem 5. *If satisfies () and the following conditions, then BVP (24) has at least one positive solution.**There are with and such that and **There are with and such that and *

*Proof. *Set such that . Take . For every , it follows from (27) and Lemma 3 thatWe may suppose that has no fixed point on (otherwise, the proof is finished). Now we show thatwhere is the positive eigenfunction of operator related to its first eigenvalue , and follows from (5).

In fact, if (34) is not valid, there exist and such thatHence we have . Set . It is easy to see that and . Since , by (33), we obtain thatTherefore by (30), , which contradicts the definition of . Hence (34) is true and we have from Corollary .3.1 in [26] thatNext, we prove thatIf this is false, then there exist and such that . It follows from Lemma 3 that . Then by assumption , we have Notice that ; we obtain , which is a contradiction with . Hence, (38) holds. Then by Lemma .3.1 in [26], we haveBy (37) and (39), we have thatThen has at least one fixed point in which means that BVP (24) has a positive solution. The proof of Theorem 5 is completed.

#### Conflicts of Interest

The author declares that they have no conflicts of interest.

#### Acknowledgments

The project is supported by NNSF of China (11371221; 11571207).