Existence of Positive Solutions for Nonlocal Fourth-Order Boundary Value Problem with Variable Parameter

Abstract

By using the Krasnoselskii's fixed point theorem and operator spectral theorem, the existence of positive solutions for the nonlocal fourth-order boundary value problem with variable parameter , , , is considered, where is a parameter, and , .

1. Introduction

The existence of positive solutions for nonlinear fourth-order multipoint boundary value problems has been studied by many authors using nonlinear alternatives of Leray-Schauder, the fixed point theory, and the method of upper and lower solutions (see, e.g., [1–15] and references therein). The multipoint boundary value problem is in fact a special case of the boundary value problem with integral boundary conditions.

Recently, Bai [16] studied the existence of positive solutions of nonlocal fourth-order boundary value problem

(1.1)

under the assumption:

(A1) and ,

(A2), , , , , .

In this paper, we study the above generalizing form with variable parameters BVP

(1.2)

where , is a parameter.

Obviously, BVP(1.1) can be regarded as the special case of BVP(1.2) with . Since the parameters is variable, we cannot expect to transform directly BVP(1.2) into an integral equation as in [16]. We will apply the cone fixed point theory, combining with the operator spectra theorem to establish the existence of positive solutions of BVP(1.2). Our results generalize the main result in [16].

Let , and we assume that the following conditions hold throughout the paper:

(H1) and ,

(H2), , , and , .

2. The Preliminary Lemmas

Set , and

(2.1)

By (H1), (H2), we get , . Denote by the Green's function of the problem

(2.2)

and the Green's function of the problem

(2.3)

Then, carefully calculation yield

(2.4)

Lemma 2.1 (see [16]).

Suppose that (A1), (A2) hold. Then, for any , solves the problem

(2.5)

if and only if

Let , and , for . , , , , .

It is easy to show that are norms on .

Lemma 2.2 (see [16]).

and () is a Banach space.

Lemma 2.3 (see [5]).

Assume that (A1), (A2) hold. Then,

(i), for , ; , for , ,

(ii), for , ,

where , ; , .

Denote

(2.6)

Computations yield the following results.

Lemma 2.4 (see [3]).

(i)when , ,

(ii)when , ,

(iii)when , .

Lemma 2.5 (see [16]).

Suppose that (A1), (A2) hold and , , are given as above. Then,

(i),

(ii), .

By Lemmas 2.4 and 2.5, .

Take , by Lemma 2.5, .

Define

(2.7)

Lemma 2.6.

is completely continuous, and .

Proof.

It is similar to Lemma 6 of [3], so we omit it.

Lemma 2.7 (see [17]).

Let E be a Banach space, a cone, and , be two bounded open sets of with . Suppose that is a completely continuous operator such that either

(i) and , , or

(ii) and ,

holds. Then, has a fixed point in .

3. The Main Results

Suppose that , , , , , , and , are defined as in Section 2, we introduce some notations as follows:

(3.1)

Theorem 3.1.

Assume that (H1), (H2) hold and . Then BVP(1.2) has at least one positive solution if one of the following cases holds:

(i), ,

(ii), .

Proof.

For any , consider the following BVP:

(3.2)

It is easy to see that the above question is equivalent to the following question:

(3.3)

For any , let . Obviously, the operator is linear. By Lemma 2.2, for all , , . Hence , and so . On the other hand, is a solution of (3.3) if and only if satisfies , that is,

(3.4)

Owing to and , the operator maps into . From (by Lemma 2.6) together with and condition , applying operator spectral theorem, we have that the exists and is bounded. Let , then (3.4) is equivalent to . By the Neumann expansion formula, can be expressed by

(3.5)

The complete continuity of with the continuity of yields that the operator is completely continuous. For all , let , then , and . So, we have, . Hence,

(3.6)

and so , .

Assume that for all , , , let , by (3.6) we have , and so , . Thus by induction, it follows that, for all , , . By (3.5), for all , we have

(3.7)

and so .

On the other hand, for all , we have

(3.8)

(3.9)

(3.10)

For any , define . By (H1) and (H2), we have that is continuous. It is easy to see that being a positive solution of BVP(1.2) is equivalent to being a nonzero solution equation as follows:

(3.11)

Let . Obviously, is completely continuous. We next show that the operator has a nonzero fixed point in . Let

(3.12)

It is easy to know that is a cone in , . Now, we show .

For , by (2.7), there is ,. Hence, by (3.7), ,,. By proof of Lemma 2.5 in [16],

(3.13)

By (3.7) and (3.10),

(3.14)

Thus .

• (i) Since , by the definition of , there exists such that

  

  (3.15)

Let , one has

(3.16)

So, by (3.10), we get

(3.17)

Hence, for ,

(3.18)

On the other hand, since , there exists such that

(3.19)

Choose , let . For ,, there is . Thus,

(3.20)

Hence, for ,

(3.21)

By the use of the Krasnoselskii's fixed point theorem, we know there exists such that , namely, is a solution of (1.2) and satisfied , , .

• (ii) The proof is similar to (i), so we omit it.

Corollary 3.2.

Assume that (H1), (H2) hold, and . Then that (1.2) has at least two positive solution, if satisfy

(i), ,

(ii)There exists such that , for , .

Proof.

By the proof of Theorem 3.1, we know that (1) from the condition , there exists , such that , , (2) from the condition , there exists , , such that , , (3) from the condition (ii), there exists , , such that , . By the use of Krasnoselskii's fixed point theorem, it is easy to know that (1.2) has at least two positive solutions.

Corollary 3.3.

Assume (H1), (H2) hold, and . Then problem (1.2) has at least two positive solution, if satisfy

(i), ,

(ii)There exists such that , for , .

Proof.

The proof is similar to Corollary 3.2, so we omit it.

Example 3.4.

Consider the following boundary value problem

(3.22)

In this problem, we know that ,,, , then we can get , , , , , , . Further more, we obtain , , then , , so

(3.23)

Thus, ,,, and satisfy the conditions of Theorem 3.1, and there exists at least a positive solution of the above problem.