Table of Contents
ISRN Applied Mathematics
Volume 2012, Article ID 926952, 12 pages
http://dx.doi.org/10.5402/2012/926952
Research Article

The Existence of Positive Solutions for a Nonlinear Sixth-Order Boundary Value Problem

1Department of Mathematics, Longdong University, Gansu, Qingyang 745000, China
2Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, China

Received 19 February 2012; Accepted 13 March 2012

Academic Editors: G. Kyriacou and F. Tadeo

Copyright © 2012 Wanjun Li et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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., [17] and the references therein):𝑢(6)+𝐴𝑢(4)+𝐵𝑢+𝐶𝑢𝑓(𝑡,𝑢)=0,0<𝑥<𝐿,𝑢(0)=𝑢(𝐿)=𝑢(0)=𝑢(𝐿)=𝑢(4)(0)=𝑢(4)(𝐿)=0,(1.1) where 𝐴, 𝐵, and 𝐶 are some given real constants and 𝑓(𝑥,𝑢) is a continuous function on 𝑅2. The boundary value problems were motivated by the study for stationary solutions of the sixth-order parabolic differential equations:𝜕𝑢=𝜕𝜕𝑡6𝑢𝜕𝑥6𝜕+𝐴4𝑢𝜕𝑥4𝜕+𝐵2𝑢𝜕𝑥2+𝑓(𝑥,𝑢).(1.2) 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 𝑓(𝑥,𝑢)=𝑢𝑢3, 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 𝐶=1 and 𝑓(𝑥,𝑢)=𝑢3. The authors studied also the homoclinic solutions for (1.1) when 𝐶=1 and 𝑓(𝑥,𝑢)=𝑎(𝑥)𝑢|𝑢|𝜎, where 𝑎(𝑥) is a positive periodic function and 𝜎 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:𝑢(4)(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢(𝑡),𝑡(0,1),𝑢(0)=𝑢(1)=𝑢(0)=𝑢(1)=0,(1.3) where 𝑓[0,1]×[0,+)×(,0][0,+) 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:𝑢(6)(𝑡)=𝑓𝑡,𝑢(𝑡),𝑢(𝑡),𝑢(4)𝑢(𝑡),𝑡(0,1),(0)=𝑢(1)=𝑢(0)=𝑢(1)=𝑢(4)(0)=𝑢(4)(1)=0,(1.4) under the following assumptions on the nonlinear term.(𝐴1)𝑓[0,1]×[0,)×(,0]×[0,)[0,) is continuous and there exist functions 𝑎,𝑏,𝑐,𝑑,𝑚,𝑛𝐶([0,1,0,)) with 𝑎(𝑡)+𝑏(𝑡)+𝑐(𝑡)>0 and 𝑑(𝑡)+𝑚(𝑡)+𝑛(𝑡)>0 on [0,1] such that 𝑓||||||||(𝑡,𝑢,𝑝,𝑞)=𝑎(𝑡)𝑢𝑏(𝑡)𝑝+𝑐(𝑡)𝑞+𝑜𝑢,𝑝,𝑞,as𝑢,𝑝,𝑞0,(1.5) uniformly for 𝑡[0,1], and 𝑓||||||||(𝑡,𝑢,𝑝,𝑞)=𝑑(𝑡)𝑢𝑚(𝑡)𝑝+𝑛(𝑡)𝑞+𝑜𝑢,𝑝,𝑞,as(𝑢,𝑝,𝑞),(1.6) uniformly for 𝑡[0,1]. Here |(𝑢,𝑝,𝑞)|=𝑢2+𝑝2+𝑞2.(𝐴2)𝑓(𝑡,𝑢,𝑝,𝑞)>0 for 𝑡[0,1] and (𝑢,𝑝,𝑞)([0,)×(,0]×[0,)){(0,0,0)}.(𝐴3) There exist constants 𝑎0,𝑏0,𝑐0[0,) satisfying 𝑎20+𝑏20+𝑐20>0 and 𝑓(𝑡,𝑢,𝑝,𝑞)𝑎0𝑢𝑏0𝑝+𝑐0𝑞+𝑜(|(𝑢,𝑝,𝑞)|) for (𝑡,𝑢,𝑝,𝑞)[0,1]×[0,)×(,0]×[0,).

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(𝐴4)𝐴,𝐵,𝐶𝐶([0,1],[0,)) with 𝐴(𝑡)>0 or 𝐵(𝑡)>0 or 𝐶(𝑡)>0 with 𝑡[0,1].

Definition 2.1. We say 𝜇 is a generalized eigenvalue of the linear problem 𝑢(6)=𝜇𝐴(𝑡)𝑢𝐵(𝑡)𝑢+𝐶(𝑡)𝑢(4)𝑢,𝑡(0,1),(0)=𝑢(1)=𝑢(0)=𝑢(1)=𝑢(4)(0)=𝑢(4)(1)=0,(2.1) if (2.1) has nontrivial solutions.

Let 𝑒(𝑡)=sin𝜋𝑡,𝑡[0,1] and 𝑌=𝐶[0,1] be the Banach space equipped with the norm 𝑢=max0𝑡1|𝑢(𝑡)|. 𝑋 is defined as𝑋=𝑢𝐶4[]0,1𝑢(0)=𝑢(1)=𝑢(0)=𝑢(1)=𝑢(4)(0)=𝑢(4)[](1)=0,𝜀0,1,s.t.𝜀𝑒(𝑡)𝑢(4)[].(𝑡)𝜀𝑒(𝑡),𝑡0,1(2.2)

For any 𝑢𝑋, we have𝑢(𝑡)=1010𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝑢(4)(𝑠)𝑑𝑠𝑑𝜏,(2.3) where𝐻(𝑡,𝑠)=𝑡(1𝑠),0𝑡𝑠1,𝑠(1𝑡),0𝑠𝑡1.(2.4) Based on (2.3) and (1/𝜋4)𝑒(𝑡)=1010𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝑒(𝑠)𝑑𝑠𝑑𝜏, we come to𝜀𝜋4𝑒𝜀(𝑡)𝑢(𝑡)𝜋4𝑒[].(𝑡),𝑡0,1(2.5) Since 𝜀/𝜋4<𝜀, the norm of 𝑢𝑋 can be defined as𝑢𝑋=inf𝜀𝜀𝑒(𝑡)𝑢(4)[].(𝑡)𝜀𝑒(𝑡),𝑡0,1(2.6) It is not difficult to verify that (𝑋,𝑋) is a Banach space. Let𝑃=𝑢𝑋𝑢(4)(𝑡)0,𝑢[].(𝑡)0,𝑢(𝑡)0,𝑡0,1(2.7) Then the cone 𝑃 is normal and has a nonempty interior 𝑃.

Lemma 2.2. For 𝑢𝑋, then 𝑢𝑢𝑢𝑢𝑢(4)𝑢𝑋.

Proof. (1) By 𝑢(0)=𝑢(1), there is a 𝜉(0,1) such that 𝑢(𝜉)=0, and so 𝑢(𝑡)=𝜉𝑡𝑢(𝑠)𝑑𝑠,𝑡[0,𝜉]. Hence |𝑢(𝑡)||𝜉𝑡|𝑢(𝑠)|𝑑𝑠||10|𝑢(𝑠)|𝑑𝑠|𝑢. By 𝑢(0)=𝑢(1), there is a 𝜂(0,1), which makes 𝑢(𝜂)=0, thereby we come to 𝑢(𝑡)=𝜂𝑡𝑢(4)(𝑠)𝑑𝑠,𝑡[0,𝜂]. And we have |𝑢(𝑡)||𝜂𝑡|𝑢(4)(𝑠)|𝑑𝑠||10|𝑢(4)(𝑠)|𝑑𝑠|𝑢(4).
(2) Because of 𝑢(0)=0, we come to |𝑢(𝑡)||𝑡0|𝑢(𝑠)|𝑑𝑠|𝑢. Correspondingly, 𝑢𝑢 and we can obtain 𝑢𝑢 for the same reason.
(3) By the definition of 𝑋, we know |𝑢(4)(𝑡)|𝜀𝑒(𝑡)𝜀, that is, |𝑢(4)|𝜀. Correspondingly, we come to |𝑢(4)||𝑢|𝑋.
Based on the combination of (1), (2) and (3), the conclusion can be reached and the lemma is thus proved.

For any 𝑢𝑋, define a linear operator 𝑇𝑋𝑌 by 𝑇𝑢(𝑡)=101010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)𝐴(𝑠)𝑢(𝑠)𝐵(𝑠)𝑢(𝑠)+𝐶(𝑠)𝑢(4)(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥.(2.8)

Theorem 2.3. Assume that (𝐴4) holds, let 𝑟(𝑡) be the spectral radius of 𝑇, then Problem (2.1) has an algebraically simple eigenvalue, 𝜇1(𝐴,𝐵,𝐶)=(𝑟(𝑡))1, with a positive eigenfunction 𝜑()𝑃. 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 𝑢(𝑡)=𝜇𝑇𝑢(𝑡).
We define 𝑇𝑋𝑋.
In fact, for 𝑢𝑋, we have (𝑇𝑢)(4)(𝑡)=10𝐻(𝑡,𝑠)𝐴(𝑠)𝑢(𝑠)𝐵(𝑠)𝑢(𝑠)+𝐶(𝑠)𝑢(4)(𝑠)𝑑𝑠.(2.9) Combining this with the fact [𝐴(𝑠)𝑢(𝑠)𝐵(𝑠)𝑢(𝑠)+𝐶(𝑠)𝑢(4)(𝑠)]𝑌, it can be concluded that 𝑐0𝜔(𝑡)(𝑇𝑢)(4)(𝑡)𝑐0[],𝜔(𝑡),𝑡0,1(2.10) where 𝑐0=𝐴𝑢+𝐵𝑢+𝐶𝑢(4),𝜔(𝑡)=1010𝑡𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝑑𝑠𝑑𝜏=𝑡24(𝑡1)2.𝑡1(2.11) Thus, there exist corresponding constants 𝜌1,𝜌2, which make 𝜌1𝑒(𝑡)𝜔(𝑡)𝜌2[].𝑒(𝑡),𝑡0,1(2.12) Consequently, we obtain 𝑇𝑢𝑋, thus 𝑇(𝑋)𝑋. The assertion is proved.
If 𝑢𝑃, then [𝐴(𝑠)𝑢(𝑠)𝐵(𝑠)𝑢(𝑠)+𝐶(𝑠)𝑢(4)(𝑠)]0,𝑠[0,1], and correspondingly, (𝑇𝑢)(4)(𝑡)=10𝐻(𝑡,𝑠)𝐴(𝑠)𝑢(𝑠)𝐵(𝑠)𝑢(𝑠)+𝐶(𝑠)𝑢(4)(𝑠)𝑑𝑠0,(𝑇𝑢)(𝑡)=10𝐻(𝑡,𝜏)(𝑇𝑢)(4)(𝜏)𝑑𝜏0,(𝑇𝑢)(𝑡)=1010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)(𝑇𝑢)(4)(𝜏)𝑑𝜏𝑑𝑥0.(2.13) So, 𝑇𝑢𝑃, and correspondingly 𝑇(𝑃)𝑃.
Because 𝑇(𝑋)𝐶6[0,1]𝑋, and 𝐶6[0,1]𝑋 is compactly embedded in 𝑋, thus we obtain that 𝑇𝑋𝑋 is completely continuous.
Next, we will prove that 𝑇𝑋𝑋 is strongly positive.
(1) For any 𝑢𝑃{0}, if 𝐴(𝑡)>0 on [0,1], then there exists a constant 𝑘>0 such that 𝑇𝑢(𝑡)𝑘101010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)𝑢(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥=𝑇1𝑢(𝑡).(2.14) It is easy to verify that there exists 𝑟1>0, such that 𝑇1𝑢(𝑡)𝑟1𝑒(𝑡) on [0,1]. Thus 𝑇𝑢(𝑡)𝑟1𝑒(𝑡).
For any 𝑢𝑃{0}, if 𝐵(𝑡)>0 on [0,1], then there exists a constant 𝑘1>0 such that 𝑇𝑢(𝑡)101010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)𝐵(𝑠)𝑢(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥𝑘1101010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)𝑢(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥=𝑘11010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝑢(𝑠)𝑑𝜏𝑑𝑥=𝑇2𝑢(𝑡).(2.15) Then there exists 𝑟2>0, which makes 𝑇2𝑢(𝑡)𝑟2𝑒(𝑡) on [0,1]. Thus, 𝑇𝑢(𝑡)𝑟2𝑒(𝑡).
For any 𝑢𝑃{0}, if 𝐶(𝑡)>0 on [0,1], similarly, we can verify that there exists 𝑟3>0, which makes 𝑇𝑢(𝑡)𝑟3𝑒(𝑡) on [0,1].
Hence we obtain 𝑇𝑢(𝑡)𝑟𝑒(𝑡), for all 𝑡[0,1], in which 𝑟=min{𝑟1,r2,𝑟3}.
(2) Thanks to the definition of 𝑇, we come to (𝑇𝑢)(4)(𝑡)=10𝐻(𝑡,𝑠)𝐴(𝑠)𝑢(𝑠)𝐵(𝑠)𝑢(𝑠)+𝐶(𝑠)𝑢(4)(𝑠)𝑑𝑠.(2.16) For any 𝑢𝑃{0}, if 𝐴(𝑡)>0 on [0,1], there exist 𝑘>0 and 𝑟4>0 such that (𝑇𝑢)(4)(𝑡)10𝐻(𝑡,𝑠)𝐴(𝑠)𝑢(𝑠)𝑑𝑠𝑘10𝐻(𝑡,𝑠)𝑢(𝑠)𝑑𝑠𝑟4𝑒(𝑡).(2.17) For any 𝑢𝑃{0}, if 𝐵(𝑡)>0 on [0,1], there exist 𝑘1>0 and 𝑟5>0 such that (𝑇𝑢)(4)(𝑡)10𝐻(𝑡,𝑠)𝐵(𝑠)𝑢(𝑠)𝑑𝑠𝑘110𝐻(𝑡,𝑠)𝑢(𝑠)𝑑𝑠=𝑘1𝑢(𝑠)𝑟5𝑒(𝑡).(2.18) To 𝑢𝑃{0}, if 𝐶(𝑡)>0 on [0,1], there exist 𝑘2>0 and 𝑟6>0 such that (𝑇𝑢)(4)(𝑡)10𝐻(𝑡,𝑠)𝐶(𝑠)𝑢(4)(𝑠)𝑑𝑠𝑘210𝐻(𝑡,𝑠)𝑢(4)(𝑠)𝑑𝑠=𝑘2𝑢(𝑠)𝑟6𝑒(𝑡).(2.19) Hence we obtain (𝑇𝑢)(4)(𝑡)𝑟𝑒(𝑡), for all 𝑡[0,1], where 𝑟=min{𝑟4,𝑟5,𝑟6}.
(3) It is easy to come to (𝑇𝑢)(𝑡)=1010𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝐴(𝑠)𝑢(𝑠)𝐵(𝑠)𝑢(𝑠)+𝐶(𝑠)𝑢(4)(𝑠)𝑑𝑠𝑑𝜏,(2.20) for all 𝑢𝑃{0}, if 𝐴(𝑡)>0 on [0,1], there exist constants 𝑘>0 and 𝑟7>0 such that (𝑇𝑢)(𝑡)1010𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝐴(𝑠)𝑢(𝑠)𝑑𝑠𝑑𝜏𝑘1010𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝑢(𝑠)𝑑𝑠𝑑𝜏𝑟7𝑒(𝑡),(2.21) for all 𝑢𝑃{0}, if 𝐵(𝑡)>0 on [0,1], there exist constants 𝑘1>0 and 𝑟8>0 such that (𝑇𝑢)(𝑡)1010𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝐵(𝑠)𝑢(𝑠)𝑑𝑠𝑑𝜏𝑘11010𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝑢(𝑠)𝑑𝑠𝑑𝜏=𝑘110𝐻(𝑡,𝜏)𝑢(𝜏)𝑑𝜏𝑟8𝑒(𝑡),(2.22) for all 𝑢𝑃{0}, if 𝐶(𝑡)>0 on [0,1], there exist constants 𝑘2>0 and 𝑟9>0 such that (𝑇𝑢)(𝑡)1010𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝐶(𝑠)𝑢(4)(𝑠)𝑑𝑠𝑑𝜏𝑘21010𝐻(𝑡,𝜏)𝐻(𝜏,𝑠)𝑢(4)(𝑠)𝑑𝑠𝑑𝜏=𝑘2𝑢(𝑡)𝑟9𝑒(𝑡).(2.23) Hence we obtain (𝑇𝑢)(𝑡)𝑟𝑒(𝑡), for all 𝑡[0,1], where 𝑟=min{𝑟7,𝑟8,𝑟9}.
By (1), (2), and (3), we have 𝑇𝑢𝑃.
According to Krein-Rutman theorem, we know that 𝑇 has a single algebraic eigenvalue 𝑟(𝑇)>0 which corresponds to the eigenvector 𝜑()𝑃. Furthermore, there is no other eigenvalues with corresponding positive eigenfunctions. Correspondingly, 𝜇1(𝐴,𝐵,𝐶)=(𝑟(𝑡))1 is an algebraic single eigenvalue of Problem (2.1) with a corresponding positive eigenvector 𝜑()𝑃, 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 (𝐴1), (𝐴2), and (𝐴3) hold. Assume that either 𝜇1(𝑑,𝑚,𝑛)<1<𝜇1(𝑎,𝑏,𝑐)or𝜇1(𝑎,𝑏,𝑐)<1<𝜇1(𝑑,𝑚,𝑛).(3.1) Then (1.4) has at least one positive solution.

Proof. Define that 𝐿𝐷(𝐿)𝑌 by 𝐿(𝑢)=𝑢(6), 𝑢𝐷(𝐿), where 𝐷(𝐿)=𝑢𝐶6[]0,1𝑢(0)=𝑢(1)=𝑢(0)=𝑢(1)=𝑢(4)(0)=𝑢(4)(1)=0.(3.2) It is easy to verify that 𝐿1𝑌𝑋 is compact.
Let 𝜁,𝜉𝐶([0,1]×[0,)×(,0]×[0,)), and satisfy 𝑓𝑓(𝑡,𝑢,𝑝,𝑞)=𝑎(𝑡)𝑢𝑏(𝑡)𝑝+𝑐(𝑡)𝑞+𝜁(𝑡,𝑢,𝑝,𝑞),(𝑡,𝑢,𝑝,𝑞)=𝑑(𝑡)𝑢𝑚(𝑡)𝑝+𝑛(𝑡)𝑞+𝜉(𝑡,𝑢,𝑝,𝑞).(3.3) Obviously, by the condition (𝐴1), we have lim||||(𝑢,𝑝,𝑞)0𝜁(𝑡,𝑢,𝑝,𝑞)||||[],(𝑢,𝑝,𝑞)=0,uniformlyfor𝑡0,1lim|(𝑢,𝑝,𝑞)|𝜉(𝑡,𝑢,𝑝,𝑞)||(||[].𝑢,𝑝,𝑞)=0,uniformlyfor𝑡0,1(3.4) Let ̃𝜉||𝜉||||||[](r)=max(𝑡,𝑢,𝑝,𝑞)0𝑢,𝑝,𝑞𝑟,𝑡0,1,(3.5) It is easy to see the fact that ̃𝜉(𝑟) is monotone, not decreasing and lim𝑟̃𝜉(𝑟)𝑟=0.(3.6)
Let us consider 𝐿𝑢=𝜆𝑎(𝑡)𝑢𝑏(𝑡)𝑢+𝑐(𝑡)𝑢(4)+𝜆𝜁𝑡,𝑢,𝑢,𝑢(4),𝜆>0,(3.7) as a bifurcation problem from the trivial solution 𝑢0.
It is easy to verify that (3.7) is equivalent to equation 𝑢(𝑡)=𝜆101010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)𝐴(𝑠)𝑢(𝑠)𝐵(𝑠)𝑢(𝑠)+𝐶(𝑠)𝑢(4)(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥+𝜆101010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)𝜁𝑡,𝑢,𝑢,𝑢(4)𝑑𝑠𝑑𝜏𝑑𝑥=𝑅(𝜆,𝑢).(3.8) From the proof of Theorem 2.3, we know that 𝑇𝑋𝑋 is strong positive and compact: 𝑇𝑢(𝑡)=101010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)𝑎(𝑠)𝑢(𝑠)𝑏(𝑠)𝑢(𝑠)+𝑐(𝑠)𝑢(4)(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥.(3.9) Define 𝐹[0,)×𝑋𝑋 by 𝐹(𝜆,𝑢)=𝜆101010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)𝜁𝑡,𝑢,𝑢,𝑢(4)𝑑𝑠𝑑𝜏𝑑𝑥,(3.10) then by (3.4) and Lemma 2.2, we know that when 𝑢𝑋0, 𝐹(𝜆,𝑢)𝑋=𝑜𝑢𝑋0partlyconsistentto𝜆.(3.11) Based on Theorem  2 in literature [11], we come to the following conclusion.
There exists an unbounded connected subset Γ for the following set: 𝐷𝑝𝑇=[𝜇(𝜆,𝑢)0,)×𝑃𝑢=𝜆𝑇𝑢+𝐹(𝜆,𝑢),𝑢int𝑃1(𝑎,𝑏,𝑐),0(3.12) such that (𝜇1(𝑎,𝑏,𝑐),0)Γ.
Next, we will verify the result of this theorem.
Obviously, any solution of (3.7), such as (1,𝑢), is the solution of problem (1.4). If we want to verify Γ passing through hyperplane {1}×𝑋, we only need to verify that Γ connects (𝜇1(𝑎,𝑏,𝑐),0) and (𝜇1(𝑑,𝑚,𝑛),).
Let (𝜂𝑛,𝑦𝑛)Γ and satisfy 𝜂𝑛+𝑦𝑛𝑋,𝑛.(3.13) Since (0,0) is the only solution of (3.7) when 𝜆=0 and Γ({0}×𝑋)=, we have 𝜂𝑛>0 for all 𝑛𝑁.
Case 1 (𝜇1(𝑑,𝑚,𝑛)<1<𝜇1(𝑎,𝑏,𝑐)). If we want to verify that (1.4) have at least one positive solution, we only need to verify that Γ connects (𝜇1(𝑎,𝑏,𝑐),0) and (𝜇1(𝑑,𝑚,𝑛),), that is, to verify 𝜇1(𝑑,𝑚,𝑛),𝜇1(𝑎,𝑏,𝑐){𝜆𝑅(𝜆,𝑢)Γ}.(3.14) The proof can be divided into the following two steps.Step 1. If we can verify that there exists a constant 𝑀>0 such that 𝜂𝑛[0,𝑀], where 𝑛𝑁, then this connects (𝜇1(𝑎,𝑏,𝑐),0) and (𝜇1(𝑑,𝑚,𝑛),).
By (3.13), when 𝑛, 𝑦𝑛𝑋, and we divide the two sides of equation 𝐿𝑦𝑛(𝑡)=𝜂𝑛𝑑(𝑡)𝑦𝑛𝑚(𝑡)𝑦𝑛+𝑛(𝑡)𝑦𝑛(4)+𝜂𝑛𝜉𝑡,𝑦𝑛(𝑡),𝑦𝑛(𝑡),𝑦𝑛(4)(𝑡),(3.15) with 𝑦𝑛𝑋 at the same time. Let 𝑦𝑛=𝑦𝑛/𝑦𝑛𝑋, then 𝑦𝑛 is bounded in 𝑋, and we have already known that 𝐿1 is compact and the bounded set is mapped as bicompact set, so there exists convergent subsequence in {𝑦𝑛(𝑡)} we might as well still mark it as {𝑦𝑛(𝑡)}; and satisfy 𝑦𝑛𝑦,𝑦𝑋,𝑦𝑋=1.(3.16) Furthermore, because |||𝜉𝑡,𝑦𝑛(𝑡),𝑦𝑛(𝑡),𝑦𝑛(4)|||(𝑡)𝑦𝑛𝑋̃𝜉𝑦𝑛𝑋𝑦𝑛𝑋,(3.17) according to (3.6) and Lemma 2.2, we obtain lim𝑛|||𝜉𝑡,𝑦𝑛(𝑡),𝑦𝑛(𝑡),𝑦𝑛(4)|||(𝑡)𝑦𝑛𝑋=0.(3.18) Hence there is 𝑦(𝑡)=101010𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)𝜂𝑑(𝑠)𝑢(𝑠)𝑚(𝑠)𝑢(𝑠)+𝑛(𝑠)𝑢(4)(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥,(3.19) where 𝜂=lim𝑛𝜂𝑛. Hence 𝐿𝑦(𝑡)=𝜂(𝑑(𝑡)𝑦𝑚(𝑡)𝑦+𝑛(𝑡)𝑦(4)). Combined with Theorem 2.3, it is easy to obtain 𝜂=𝜇1(𝑑,𝑚,𝑛).(3.20) To sum up, Γ connects (𝜇1(𝑎,𝑏,𝑐),0) with (𝜇1(𝑑,𝑚,𝑛),).
Step 2. To verify the fact that arbitrary 𝑛𝑁, there exists 𝑚>0 such that 𝜂𝑛[0,𝑀].
Thanks to the Lemma  2.1 in [8], we only need to verify that nonlinear operator 𝑅(𝜆,𝑢) has linear function 𝑉, and there exists (𝜂,𝑦)(0,)×𝑃 such that 𝑦𝑋=1 and 𝜂𝑉𝑦𝑦. It follows from (𝐴3) that there exist 𝑎0,𝑏0,𝑐0[0,) such that 𝑎20+𝑏20+𝑐20>0, and 𝑓(𝑡,𝑢,𝑝,𝑞)𝑎0𝑢+𝑏0𝑝+𝑐0[]×[]×[𝑞,(𝑡,𝑢,𝑝,𝑞)0,10,)×(,00,).(3.21) To 𝑢𝑋, let 𝑉𝑢(𝑡)=101010𝑎𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)0𝑢(𝑠)𝑏0𝑢(𝑠)+𝑐0𝑢(4)(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥.(3.22) Then 𝑉 is the linear function of 𝑅(𝜆,𝑢).
Again, as 𝑉𝑒(𝑡)𝜋2=101010𝑎𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)0𝑒(𝑠)𝜋2𝑏0𝑒(𝑠)𝜋2+𝑐0𝑒(4)(𝑠)𝜋2=𝑑𝑠𝑑𝜏𝑑𝑥101010𝑎𝐻(𝑡,𝑥)𝐻(𝑥,𝜏)𝐻(𝜏,𝑠)0𝜋2𝑏0+𝑐0𝜋2=1𝑒(𝑠)𝑑𝑠𝑑𝜏𝑑𝑥𝜋6𝑎0𝜋2𝑏0+𝑐0𝜋2𝑒(𝑡),(3.23) that is, 𝑎0𝜋6+𝑏0𝜋4+𝑐0𝜋21𝑉𝑒(𝑡)𝜋2=𝑒(𝑡)𝜋2,(3.24) by the Lemma  2.1 in [8], we obtain ||𝜂𝑛||𝑎0𝜋6+𝑏0𝜋4+𝑐0𝜋21.(3.25) The conclusion is thus proved.Case 2 (𝜇1(𝑎,𝑏,𝑐)<1<𝜇1(𝑑,𝑚,𝑛)). If (𝜂𝑛,𝑦𝑛)Γ, which satisfies lim𝑛𝜂𝑛+𝑦𝑛𝑋=, and lim𝑛𝑦𝑛𝑋=, then we obtain 𝜇1(𝑎,𝑏,𝑐),𝜇1(𝑑,𝑚,𝑛){𝜆𝑅(𝜆,𝑢)Γ},(3.26) hence we have ({1}×𝑋)Γ.
Similar to Case 1, the verification of Case 2 can also be divided into two steps with the conclusion that Γ connects (𝜇1(𝑎,𝑏,𝑐),0) and (𝜇1(𝑑,𝑚,𝑛),). And we come to the conclusion that Γ passes through hyperplane {1}×𝑋 on 𝑅×𝑋, hence (1.4) have at least one positive solution.

Acknowledgments

The author are very grateful to the anonymous referees for their valuable suggestions and to be sponsored by the Tutorial Scientific Research Program Foundation of Education Department of Gansu Province China (1110-05).

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