## Boundary Value Problems

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# Bifurcation from interval and positive solutions for a class of fourth-order two-point boundary value problem

Boundary Value Problems20132013:170

DOI: 10.1186/1687-2770-2013-170

Accepted: 8 July 2013

Published: 22 July 2013

## Abstract

We consider the fourth-order two-point boundary value problem , , , which is not necessarily linearizable. We give conditions on the parameters k, l and that guarantee the existence of positive solutions. The proof of our main result is based upon topological degree theory and global bifurcation techniques.

MSC:34B15.

### Keywords

topological degree fourth-order ordinary differential equation bifurcation positive solution eigenvalue

## 1 Introduction

The deformations of an elastic beam in an equilibrium state with fixed both endpoints can be described by the fourth-order boundary value problem
(1.1)

where is continuous, is a parameter and l is a given constant. Since problem (1.1) cannot transform into a system of second-order equations, the treatment method of the second-order system does not apply to it. Thus, the existing literature on problem (1.1) is limited. When , the existence of positive solutions of problem (1.1) has been studied by several authors, see [15]. Especially, when , Xu and Han [6] studied the existence of nodal solutions of problem (1.1) by applying disconjugate operator theory and bifurcation techniques.

Recently, motivated by [6], when k, l satisfy (A1), Shen [7] studied the existence of nodal solutions of a general fourth-order boundary value problem by applying disconjugate operator theory [8, 9] and Rabinowitz’s global bifurcation theorem
(1.2)

where

(A1) one of following conditions holds:
1. (i)
k, l satisfying are given constants with
(1.3)

2. (ii)
k, l satisfying are given constants with
(1.4)

In this paper, we consider bifurcation from interval and positive solutions for problem (1.2). In order to prove our main result, condition (A1) and the following weaker conditions are satisfied throughout this paper:

(H1) is continuous and there exist functions , , , and such that
(1.5)
for some functions , defined on with
(1.6)
uniformly for , and
(1.7)
for some functions , defined on with
(1.8)

uniformly for .

(H2) for and .

(H3) There exists a function with in any subinterval of such that
(1.9)

It is the purpose of this paper to study the existence of positive solutions of (1.2) under conditions (A1), (H1), (H2) and (H3). The main tool we use is the following global bifurcation theorem for the problem which is not necessarily linearizable.

Theorem A (Rabinowitz [10])

Let V be a real reflexive Banach space. Let be completely continuous such that , . Let () be such that is an isolated solution of the following equation:
(1.10)
for and , where , are not bifurcation points of (1.10). Furthermore, assume that
(1.11)
where is an isolating neighborhood of the trivial solution. Let
Then there exists a continuum (i.e., a closed connected set) of containing , and either
1. (i)

is unbounded in , or

2. (ii)

.

Remark 1.1 For other results on the existence and multiplicity of positive solutions and nodal solutions for boundary value problems of fourth-order ordinary differential equations based on bifurcation techniques, see [1120].

## 2 Hypotheses and lemmas

Let
(2.1)

Theorem 2.1 (see [[7], Theorem 2.4])

Let (A1) hold. Then
1. (i)
is disconjugate on , and has a factorization
(2.2)

where with ();
1. (ii)
if and only if
(2.3)

where
(2.4)

Theorem 2.2 (see [[7], Theorem 2.7])

Let (A1) hold and with on any subinterval of . Then
1. (i)
the problem
(2.5)

has an infinite sequence of positive eigenvalues
(2.6)
1. (ii)

as ;

2. (iii)

to each eigenvalue , there corresponds an essential unique eigenfunction which has exactly simple zeros in and is positive near 0;

3. (iv)

given an arbitrary subinterval of , an eigenfunction that belongs to a sufficiently large eigenvalue changes its sign in that subinterval;

4. (v)

for each , the algebraic multiplicity of is 1.

Theorem 2.3 (see [[7], Theorem 2.8]) (Maximum principle)

Let (A1) hold. Let with on and in . If satisfies
(2.7)

then on .

Let with the norm . Let with its usual norm . By a positive solution of (1.2), we mean x is a solution of (1.2) with (i.e., in and ).

Let with the inner product and the norm . Further, define the linear operator
(2.8)
with
(2.9)

Then is a closed operator and is completely continuous.

Lemma 2.4 Let be the first eigenfunction of (2.5). Then, for all , we get
(2.10)
Proof Obviously, , we have
Integrating by parts, we obtain

□

Let be the closure of the set of positive solutions of the problem
(2.11)
We extend the function f to a continuous function defined on by
(2.12)
Then for . For , let x be an arbitrary solution of the problem
(2.13)

Since for , we have for . Thus x is a nonnegative solution of (2.11), and the closure of the set of nontrivial solutions of (2.13) in is exactly Σ.

Let be the Nemytskii operator associated with the function
(2.14)
Then (2.13), with , is equivalent to the operator equation
(2.15)
In the following, we shall apply the Leray-Schauder degree theory, mainly to the mapping ,
(2.16)

For , let , and let denote the degree of on with respect to 0.

Lemma 2.5 Let be a compact interval with . Then there exists a number with the property
(2.17)

Proof Suppose to the contrary that there exist sequences and in , in E, such that for all , then in .

Set . Then and . Now, from condition (H1), we have the following:
(2.18)
and, accordingly,
(2.19)
Let and denote the nonnegative eigenfunctions corresponding to and , respectively. Then we have, from the first inequality in (2.19),
(2.20)
From Lemma 2.4, we have
(2.21)
Since in E, from (1.6) we have
(2.22)
By the fact that , we conclude that in E. Thus,
(2.23)
Combining this and (2.21) and letting in (2.20), we get
(2.24)
and consequently
(2.25)
Similarly, we deduce from the second inequality in (2.19) that
(2.26)

Thus, . This contradicts . □

Corollary 2.6 For and , .

Proof Lemma 2.5, applied to the interval , guarantees the existence of such that for ,
(2.27)
Hence, for any ,
(2.28)

which ends the proof. □

Lemma 2.7 Suppose . Then there exists such that with , ,
(2.29)

where is the nonnegative eigenfunction corresponding to .

Proof We assume to the contrary that there exist and a sequence , with and in E, such that for all . As
(2.30)
and in , it follows that
(2.31)
Notice that has a unique decomposition
(2.32)

where and . Since on and , we have from (2.32) that .

Choose such that
(2.33)
By (H1), there exists such that
(2.34)
Therefore, for , ,
(2.35)
Since , there exists such that
(2.36)
and consequently
(2.37)
Applying Lemma 2.4 and (2.37), it follows that
(2.38)
(2.39)
Thus,
(2.40)

Corollary 2.8 For and , .

Proof Let , where is the number asserted in Lemma 2.7. As is bounded in , there exists such that for all . By Lemma 2.7, one has
(2.41)
Hence
(2.42)

□

Now, using Theorem A, we may prove the following.

Proposition 2.9 is a bifurcation interval from the trivial solution for (2.15). There exists an unbounded component C of a positive solution of (2.15), which meets . Moreover,
(2.43)
Proof For fixed with , let us take that , and . It is easy to check that for , all of the conditions of Theorem A are satisfied. So, there exists a connected component of solutions of (2.15) containing , and either
1. (i)

is unbounded, or

2. (ii)

.

By Lemma 2.5, the case (ii) cannot occur. Thus is unbounded bifurcated from in . Furthermore, we have from Lemma 2.5 that for any closed interval , if , then in E is impossible. So, must be bifurcated from in . □

## 3 Main results

Theorem 3.1 Let (A1), (H1), (H2), (H3) hold. Assume that either
(3.1)
or
(3.2)

then problem (1.2) has at least one positive solution.

Proof of Theorem 3.1 It is clear that any solution of (2.15) of the form yields a solution x of (1.2). We will show that C crosses the hyperplane in . To do this, it is enough to show that C joins to . Let satisfy
(3.3)

We note that for all since is the only solution of (2.15) for and .

Case 1. .

In this case, we show that

We divide the proof into two steps.

Step 1. We show that is bounded.

Since , . From (H3), we have
(3.4)

Let denote the nonnegative eigenfunction corresponding to .

From (3.4), we have
(3.5)
By Lemma 2.4, we have
(3.6)
Thus
(3.7)

Step 2. We show that C joins to .

From (3.3) and (3.7), we have that . Notice that (2.15) is equivalent to the integral equation
(3.8)
which implies that
(3.9)
We divide both of (3.9) by and set . Since is bounded in E, there exists a subsequence of and , with and on , such that
(3.10)
relabeling if necessary. Thus, (3.9) yields that
(3.11)
which implies that
(3.12)
Let and denote the nonnegative eigenfunction corresponding to and , respectively. Then we have, from the first inequality in (3.12),
From Lemma 2.4, integrating by parts, we obtain that
and consequently
(3.13)
Similarly, we deduce from the second inequality in (3.12) that
(3.14)
Thus
(3.15)

So, C joins to .

Case 2. .

In this case, if is such that
and
then
(3.16)
and, moreover,
(3.17)
Assume that is bounded; applying a similar argument to that used in Step 2 of Case 1, after taking a subsequence and relabeling if necessary, we obtain
(3.18)

Again C joins to and the result follows. □

## Declarations

### Acknowledgements

This work is supported by the NSF of Gansu Province (No. 1114-04).

## Authors’ Affiliations

(1)
Department of Basic Courses, Lanzhou Institute of Technology

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