Research Article | Open Access

# Positive Solution of Fourth-Order Integral Boundary Value Problem with Two Parameters

**Academic Editor:**D. Anderson

#### Abstract

The author investigates the fourth-order integral boundary value problem with two parameters , where nonlinear term function is allowed to change sign. Applying the fixed point index theorem on cone together with the operator spectrum theorem, some results on the existence of positive solution are obtained.

#### 1. Introduction

The theory of boundary value problems with integral boundary conditions for ordinary differential equations arises in different areas of applied mathematics and physics. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and plasma physics can all be reduced to nonlocal problems with integral boundary conditions (see, e.g., [1–3]). For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to the papers by Gallardo [4], Karakostas and Tsamatos [5], and Lomtatidze and Malaguti [6] and the references therein. For more information about the general theory of integral equations and their relation to boundary value problems, we refer to the books of Corduneanu [7] and Agarwal and O'Regan [8].

Moreover, boundary value problems with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multipoints and nonlocal boundary value problems as special cases. The existence and multiplicity of positive solutions for such problems have received a great deal of attention. To identify a few, we refer the reader to [9–15] and the references therein.

In the recent literature, several sorts of boundary value problems with integral boundary conditions have been studied further, see [16–20]. Especially, Ruyun Ma and Yulian An [18] investigated the global structure of positive solutions for nonlocal boundary value problems by using global bifurcation techniques, where . In [19], Jiqiang Jiang et al. investigated the existence of positive solution for second-order singular Sturm-Liouville integral boundary value problems by using the fixed point theory in cones, where .

On the other hand, the fourth-order boundary value problem describe the deformations of an elastic beam in equilibrium state. Owing to its importance in physics, the existence of solutions to this problem has been studied by many authors; see, for example, [21–24] and references therein. Especially, in [22], Li studied existence of positive solution for fourth-order boundary value problem by using the fixed point index theorem, where .

Motivated by the above-mentioned works [18, 19, 22], in this paper, we study the following fourth-order integral boundary value problem (for short BVP in the sequel) with two parameters: where nonlinear term function is allowed to change sign. To the best of our knowledge, BVP has not been investigated up to now. In the literature such as above-mentioned paper [18, 19, 22], the nonnegativity on is a usual assumption. In the present paper, since the function is not assumed to be nonnegative, the corresponding integral operator doesn't map the cone into cone, and so, there exists difficulty in applying the cone fixed point theorem. On the other hand, owing to the occurrence of parameter in this boundary value problem including integral boundary conditions, it is not easy to transform the BVP (1.4) into an integral equation directly. To overcome these difficulties, we first introduce operator spectrum method combined with some analysis technique, next apply the fixed point index theorem, and establish existence of positive solution to BVP (1.4).

Let us begin with listing the following assumption conditions, which will be used in the sequel:

Let .(H1)(H2).Let be the roots of the polynomial ; namely, By (H2), it is to see that .

Let . Then (H2) implies . Let be the real Banach space equipped with the norm . Denote by the set in .

#### 2. Preliminaries

In this section, we shall give some important preliminary lemmas, which will be used in proving of our main results.

Lemma 2.1 (see [22, 23]). * Suppose that (H2) holds, then there exist unique satisfying
**
respectively, and on , where is as in (1.6). Moreover, have the expression
**
where .*

Let be the Green function of the linear boundary value problem By [22, 23], can be expressed by the formula where

Lemma 2.2 (see [22, 23]). * have the following properties: *(i)*.
*(ii)*.
*(iii)*, where*

Put . Set , where is described as before. We need also the following assumptions in the sequel.

(H3) Functions , satisfy .

Let , consider the following BVP: By papers [22, 23], BVP (2.7) has a unique solution expressed by

Let . Since , by Lemma 2.2, it is easy to verify that .

Let where is as in (2. 1). By Lemmas 2.1 and 2.2, we have and On the other hand, satisfies the following relation: So, from (2.10)–(2.11), it follows that Now, we make the following decomposition: So by (2.10), (2.12)-(2.13), it follows that Similarly, by setting we have

For any , define as Obviously, for any .

Let ; consider the BVP with integral boundary conditions

Denote operator B on by It is easy to see that maps into .

Define operator L: as follows:

We need the following Lemma.

Lemma 2.3. *Let (H2) holds. Assume that and . Then is a solution of (2.18) if and only if is a solution of operator equation in .*

*Proof. *(1) Assume is a solution of (2.18). By (2.14)–(2.20), we have
Let . Then ,. Thus, by (2.7)-(2.8), we have , and so .

(2) Inversely, assume satisfies . Then . By (2.7), (2.8),(2.14)–(2.20), we have
Consequently,
Hence, is a solution of (2.18). The proof is complete.

We have also the following lemma.

Lemma 2.4. *Suppose (H3) holds. Then is a bounded operator with and .*

*Proof. *In view of Lemma 2.2 (ii), by (2.9),(2.15),(2.19) and (H3), noticing that , for any and , we have
Thus, , and so .

On the other hand, from , we have . So, Lemma 2.4 is true.

By (2.7)-(2.8), it follows from that

For any , let and . Under conditions (H1)–(H3), consider the following auxiliary BVP:
Notice that satisfies (2.25), it is easy to see that is a solution of (2.26) if and only if is a solution of the following BVP:
Thus, if and only if , then is a solution of BVP (1.4).

Now, by Lemma 2.3, is a solution of (2.26) if is a fixed point of the operator . So, we only need focusing our attention on the existence of the fixed point of .

For the remainder of this section, we give the definition of positive solution.

By a positive solution of BVP (1.4), we mean a function such that , , and satisfies (1.4).

#### 3. Main Results

We introduce now some notations, which will be used in the sequel.

Let , , and be as described in Lemma 2.2 and (H3), respectively. We also set We also need the following assumption.

(H4) There exists a number , and such that

We are now in a position to state and prove our main results on the existence.

Theorem 3.1. *Suppose that (H1)–(H4) hold. If , then BVP (1.4) has a positive solution.*

*Proof. *By Lemma 2.4 together with (H3), we have (<1). By operator spectrum theorem, we know that exists and is bounded. Furthermore, by Neumann expression, can be expressed by
Noticing that and from (3.3), we have
Thus, from the reversibility of , we have
The following proof will be divided into five steps.*Step 1. *We will show that is completely continuous.(1) maps into .For any , it follows from (H1) that , and so . By (H1)-(H2) together with Lemma 2.2, for any , we have
where .From the continuity of , it is easy to see that , and so .(2) is a compact operator on .Assume that is a arbitrary bounded set in . Then there exists a such that for all . Also, we have for all since . Consequently,
where . That means is a uniformly bounded set in .On the other hand, the continuity of on yields that for every , there exists such that for any with , the following inequality
holds for all , and so,
for any , where . That is, is equicontinuous.

Hence, in view of Arzela-Ascoli theorem, we know that the operator is compact on .(3)Now, we show that the operator is continuous.Indeed, for any sequence in with and any , we have
Thus, , and, by Lemma 2.2, it follows from the continuity of that
By (1)–(3) we obtain that is completely continuous.

Now, from (3.4), we have is continuous, and so, is completely continuous.

Now we set
where are described in Lemma 2.2. Set
Obviously, is a cone in .*Step 2. *.

In fact, for any and every in , by Lemma 2.2, we have
Thus, we have
Since , by (3.4) together with (3.16) for every , we have
On the other hand, since , by (3.5), we have
Inequality (3.17) together with (3.18) implies for every
namely, . Thus, we obtain that maps into .*Step 3. *We shall deduce that for any and , the following inequality holds:
where .

In fact, in view of Lemma 2.2 and the symmetry of , we have
Thus, keeping in mind that , it follows from that
On the other hand, from , it follows that
Thus, by (3.22)-(3.23), we have
and so,
where .*Step 4. *By (H4), we have
Let . By (2.7)-(2.8), we easily know that is a positive eigenfunction of operator with respect to positive eigenvalue , that is, .

Now, we show that , that is, . We discuss it in three different cases.(1). In this case, , and .(i)If , then . By Jordan's inequality, we have
(ii)If , by setting , we have . Then from (3. 12), it follows that
Thus, by (i)-(ii) above, we have
(2). In this case, , and ,.(i)If , by setting , we have
From , it follows that . Keeping in mind that for all , it follows immediately that
(ii)If , by setting , we have . From (2)(i) above, it follows that
Hence, by (2)(i)-(ii) above, we have
On the other hand, by (3.27)-(3.28), we have
Thus, we have immediately
It is easy to verity that . Hence, .

(3). In this case, , and , .(i)If , then . Thus,
(ii)If , from (i), by letting , then we have , and
Thus, (3)(i)-(ii) above implies that
Summing up (1)–(3) keeping in mind that , we have
that is, .

Now, set . We claim that
Indeed, if not, then exists a and with . Without loss of generality, assume that (otherwise, by proving later on, we will know that the theorem is true). By , we have , and so, it follows from (3.25) that