Abstract

By using the Krein-Rutman theorem and bifurcation methods, we discuss the existence of positive solutions for the boundary value problems of a sixth-order ordinary differential equation.

1. Introduction

In recent years, the following boundary value problems for sixth-order ordinary differential equations have been studied extensively (see, e.g., [1–7] and the references therein): where , , and are some given real constants and is a continuous function on . The boundary value problems were motivated by the study for stationary solutions of the sixth-order parabolic differential equations: This equation arose in the formation of the spatial periodic patterns in bistable systems and is also a model for describing the behavior of phase fronts in materials that are undergoing a transition between the liquid and solid state. When , it was studied by [4, 5].

In [2], the existence and multiplicity results of nontrivial solutions for (1.1) were proved using a minimization theorem and Clarks theorem [6], respectively, when and . The authors studied also the homoclinic solutions for (1.1) when and , where is a positive periodic function and is a positive constant, by the mountain-pass theorem and concentration-compactness arguments. In [3], by variational tools, including Brezis-Nirenbergs linking theorems, Gyulov et al. studied also the existence and multiplicity of nontrivial solutions of BVP (1.1). In [7], using the fixed point index theory of cone mapping, the authors gave some results for existence and multiplicity of positive solutions of BVP (1.1).

On the other hand, in [8, 9], by the Krein-Rutman theorem and the global bifurcation techniques, Ma et al. were concerned with the existence of positive solutions of the following fourth-order boundary value problem: where is continuous and satisfies some conditions.

Inspired by the works of the above papers, in this paper, we consider the following nonlinear sixth-order boundary value problem: under the following assumptions on the nonlinear term.() is continuous and there exist functions with and on such that uniformly for , and uniformly for . Here .() for and .() There exist constants satisfying and for .

The existence of positive solution for (1.4) is proved using Krein-Rutman theorem [10] and the Global Bifurcation Theory [11]. The idea of this work comes from [8, 9].

The rest of the paper is organized as follows. In Section 2, we present some preliminaries and lemmas that will be used to prove our main results. In Section 3, we discuss the existence of positive solution of the problem (1.4).

2. Preliminaries

In this section, we will make some preliminaries which are needed to show our main results. Let us assume that() with or or with .

Definition 2.1. We say is a generalized eigenvalue of the linear problem if (2.1) has nontrivial solutions.

Let and be the Banach space equipped with the norm . is defined as

For any , we have where Based on (2.3) and , we come to Since , the norm of can be defined as It is not difficult to verify that is a Banach space. Let Then the cone is normal and has a nonempty interior .

Lemma 2.2. For , then .

Proof. (1) By , there is a such that , and so . Hence . By , there is a , which makes , thereby we come to . And we have .
(2) Because of , we come to . Correspondingly, and we can obtain for the same reason.
(3) By the definition of , we know , that is, . Correspondingly, we come to .
Based on the combination of (1), (2) and (3), the conclusion can be reached and the lemma is thus proved.

For any , define a linear operator by

Theorem 2.3. Assume that holds, let be the spectral radius of , then Problem (2.1) has an algebraically simple eigenvalue, , with a positive eigenfunction . Moreover, there is no other eigenvalue with a positive eigenfunction.

Proof. It is easy to check that Problem (2.1) is equivalent to the integral equation .
We define .
In fact, for , we have Combining this with the fact , it can be concluded that where Thus, there exist corresponding constants , which make Consequently, we obtain , thus . The assertion is proved.
If , then , and correspondingly, So, , and correspondingly .
Because , and is compactly embedded in , thus we obtain that is completely continuous.
Next, we will prove that is strongly positive.
(1) For any , if on , then there exists a constant such that It is easy to verify that there exists , such that on . Thus .
For any , if on , then there exists a constant such that Then there exists , which makes on . Thus, .
For any , if on , similarly, we can verify that there exists , which makes on .
Hence we obtain , for all , in which .
(2) Thanks to the definition of , we come to For any , if on , there exist and such that For any , if on , there exist and such that To , if on , there exist and such that Hence we obtain , for all , where .
(3) It is easy to come to for all , if on , there exist constants and such that for all , if on , there exist constants and such that for all , if on , there exist constants and such that Hence we obtain , for all , where .
By (1), (2), and (3), we have .
According to Krein-Rutman theorem, we know that has a single algebraic eigenvalue which corresponds to the eigenvector . Furthermore, there is no other eigenvalues with corresponding positive eigenfunctions. Correspondingly, is an algebraic single eigenvalue of Problem (2.1) with a corresponding positive eigenvector , and there is no any other eigenvalues which have corresponding positive feature vector. The theorem is thus proved.

3. Main Results

The main result of this paper is as follows.

Theorem 3.1. Let , , and hold. Assume that either Then (1.4) has at least one positive solution.

Proof. Define that by , , where It is easy to verify that is compact.
Let , and satisfy Obviously, by the condition , we have Let It is easy to see the fact that is monotone, not decreasing and
Let us consider as a bifurcation problem from the trivial solution .
It is easy to verify that (3.7) is equivalent to equation From the proof of Theorem 2.3, we know that is strong positive and compact: Define by then by (3.4) and Lemma 2.2, we know that when , Based on Theorem  2 in literature [11], we come to the following conclusion.
There exists an unbounded connected subset for the following set: such that .
Next, we will verify the result of this theorem.
Obviously, any solution of (3.7), such as , is the solution of problem (1.4). If we want to verify passing through hyperplane , we only need to verify that connects and .
Let and satisfy Since is the only solution of (3.7) when and , we have for all .
Case 1 (). If we want to verify that (1.4) have at least one positive solution, we only need to verify that connects and , that is, to verify The proof can be divided into the following two steps.Step 1. If we can verify that there exists a constant such that , where , then this connects and .
By (3.13), when , , and we divide the two sides of equation with at the same time. Let , then is bounded in , and we have already known that is compact and the bounded set is mapped as bicompact set, so there exists convergent subsequence in we might as well still mark it as ; and satisfy Furthermore, because according to (3.6) and Lemma 2.2, we obtain Hence there is where . Hence . Combined with Theorem 2.3, it is easy to obtain To sum up, connects with .
Step 2. To verify the fact that arbitrary , there exists such that .
Thanks to the Lemma  2.1 in [8], we only need to verify that nonlinear operator has linear function , and there exists such that and . It follows from that there exist such that , and To , let Then is the linear function of .
Again, as that is, by the Lemma  2.1 in [8], we obtain The conclusion is thus proved.Case 2 (). If , which satisfies , and , then we obtain hence we have .
Similar to Case 1, the verification of Case 2 can also be divided into two steps with the conclusion that connects and . And we come to the conclusion that passes through hyperplane on , hence (1.4) have at least one positive solution.