JAMJournal of Applied Mathematics1687-00421110-757XHindawi Publishing Corporation16089110.1155/2012/160891160891Research ArticleMultiple Positive Solutions of Singular Nonlinear Sturm-Liouville Problems with Carathéodory Perturbed TermHanYuefeng1ZhangXinguang2LiuLishan3, 4WuYonghong4PickeringAndrew1College of International Economics and TradeJilin University of Finance and EconomicsJilin, Changchun 130117Chinactu.cc.jl.cn2School of Mathematical and Informational SciencesYantai UniversityShandong, Yantai 264005Chinaytu.edu.cn3School of Mathematical SciencesQufu Normal UniversityShandong, Qufu 273165Chinaqfnu.edu.cn4Department of Mathematics and StatisticsCurtin University of TechnologyPerthWA 6845 Australiacurtin.edu.au20120122012201213072011031120112012Copyright © 2012 Yuefeng Han 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.

By employing a well-known fixed point theorem, we establish the existence of multiple positive solutions for the following fourth-order singular differential equation Lu=p(t)f(t,u(t),u(t))-g(t,u(t),u(t)),0<t<1,α1u(0)-β1u(0)=0,γ1u(1)+δ1u(1)=0,α2u(0)-β2u(0)=0,γ2u(1)+δ2u(1)=0, with αi,βi,γi,δi0 and βiγi+αiγi+αiδi>0,    i=1,2, where L denotes the linear operator Lu:=(ru)-qu,rC1([0,1],(0,+)), and qC([0,1],[0,+)). This equation is viewed as a perturbation of the fourth-order Sturm-Liouville problem, where the perturbed term g:(0,1)×[0,+)×(-,+)(-,+) only satisfies the global Carathéodory conditions, which implies that the perturbed effect of g on f is quite large so that the nonlinearity can tend to negative infinity at some singular points.

1. Introduction

In this paper, we consider the existence of multiple positive solutions for the following fourth-order singular Sturm-Liouville boundary value problem involving a perturbed termLu=p(t)f(t,u(t),u′′(t))-g(t,u(t),u′′(t)),0<t<1,α1u(0)-β1u(0)=0,γ1u(1)+δ1u(1)=0,α2u′′(0)-β2u′′′(0)=0,γ2u′′(1)+δ2u′′′(1)=0, where αi,βi,γi,δi0 and βiγi+αiγi+αiδi>0,  i=1,2, and L denotes the linear operator Lu:=(ru′′′)-qu′′,rC1([0,1],(0,+)) and qC([0,1],[0,+)) and qC([0,1],[0,+)). The perturbed term, g:(0,1)×[0,+)×(-,+)[0,+), satisfies global Carathéodory’s conditions.

Equation (1.1) arises from many branches of applied mathematics and physics; for details, see . It mainly describes the deformation of an elastic beam for g(t,u,u′′)0; for example, under the Lidstone boundary condition,u(0)=u(1)=u′′(0)=u′′(1)=0, problem (1.1) is used to model such phenomena as the deflection of an elastic beam simply supported at the endpoints; see [1, 3, 5, 711]. Also, if the boundary condition of (1.1) is a Focal boundary condition, then it describes the deflection of an elastic beam having both end-points fixed, or having one end simply supported and the other end clamped with sliding clamps. In addition, the derivative u′′ in f is the bending moment term which represents the bending effect, see [1, 3, 5, 711, 13, 14, 16]. A brief discussion of the physical interpretation under some boundary conditions associated with the linear beam equation can be found in Zill and Cullen .

Recently, for the case where the nonlinearity f does not contain the bending moment term u′′, Ma and Wang  studied the existence of positive solutions for (1.1) subject to boundary conditions u(0)=u(1)=u′′(0)=u′′(1)=0 and u(0)=u(1)=u′′(0)=u′′′(1)=0 if f is superlinear or sublinear. In the case where f contains the bending moment term u′′ and under the particular boundary conditions, the authors of papers [9, 12] studied the existence of positive solutions for (1.1) when f satisfies the following growth condition:|f(t,x,y)-(αx-βy)|a|x|+b|y|+c, where α,β  a,b,c>0,a,b is enough small. But most of the above works were done on base of the assumptions that the nonlinearity is nonnegative and has no any singularity. In recent years, one found that the fourth-order changing-sign nonlinear problems also occur to the classical model for the elastic beam fixed at both ends, especially in the medium span or large span bridge constructions, this implies that it is necessary and quite natural to study fourth-order changing-sign boundary value problems.

In this paper, we focus on the particularly difficult and interesting situation, when (1.1) is singularly perturbed, so that the nonlinearity is allowed to change sign, even may tend to negative infinity. This problem has essential difference from those unperturbed problems of . We quote in the sequel some papers from the relevant bibliography devoted to this subject. In , Loud considered the existence of T-periodic solutions for a first-order perturbed system of ordinary differential equations by employing the so-called bifurcation functionf0(θ)=0Tz0(τ),ϕ(τ-θ,x0(τ),0)dτ. Moreover, the author of  also considered the case when θ0 is not a simple zero of f0, and the existence of T-periodic solutions of the above problem is associated with the existence of the roots of a certain quadratic equation. Recently, by using the exponential dichotomies and contraction mapping principle, Xia et al.  established some sufficient conditions of the existence and uniqueness of almost periodic solution for a forced perturbed system with piecewise constant argument. The other works, such as Khanmamedov , Wu and Gan , Makarenkov and Nistri , Liu and Yang , Clavero et al. , and Cui and Geng , are rich sources for application of perturbed problems.

Our main tool used for the analysis here is known as Guo-Krasnoselskii’s fixed point theorem, for the convenience of the reader, we now state it as follows.

Lemma 1.1 (see, [<xref ref-type="bibr" rid="B26">26</xref>]).

Let E be a real Banach space, PE a cone. Assume Ω1,Ω2 are two bounded open subsets of E with θΩ1,Ω̅1Ω2, and let T:P(Ω̅2Ω1)P be a completely continuous operator such that either

Txx,xPΩ1, and Txx,xPΩ2, or

Txx,xPΩ1, and Txx,xPΩ2.

Then, T has a fixed point in P(Ω¯2Ω1).

2. Preliminaries and Lemmas

The following definition introduces global Carathéodory’s conditions imposed on a map.

Definition 2.1.

A map g:(t,x,y)g(t,x,y) is said to satisfy global Crathéodory’s conditions if the following conditions hold:

for each (x,y)×, the mapping tg(t,x,y) is Lebesgue measurable;

for a.e. t[0,1], the mapping (x,y)g(t,x,y) is continuous on ×;

there exists a ρL1[0,1] such that, for a.e. t[0,1] and (x,y)×, we have |g(t,x,y)|ρ(t).

The following lemmas play an important role in proving our main results.

Lemma 2.2 (see, [<xref ref-type="bibr" rid="B27">27</xref>]).

Let ψ2 and ϕ2 be the solutions of the linear problems -(r(t)ϕ2(t))+q(t)ϕ2(t)=0,0<t<1,ϕ2(0)=β2,ϕ2(0)=α2,-(r(t)ψ2(t))+q(t)ψ2(t)=0,0<t<1,ψ2(0)=δ2,ψ2(0)=-γ2, respectively. Then,

ϕ2 is strictly increasing on [0,1] and ϕ2(t)>0 on (0,1];

ψ2 is strictly decreasing on [0,1] and ψ2(t)>0 on [0,1).

Setw2=-r(t)(ϕ2(t)ψ2(t)-ψ2(t)ϕ2(t)), by Liouville’s formula, one can easily show w2= constant >0.

As , we define Green’s function for the BVP:-(r(t)u(t))+q(t)u(t)=0,0<t<1,α2u(0)-β2u(0)=0,γ2u(1)+δ2u(1)=0, byG2(t,s)=1w2{ϕ2(t)ψ2(s),0ts1,ϕ2(s)ψ2(t),  0st1, then we have the following lemma.

Lemma 2.3.

For any (t,s)[0,1]×[0,1],  i=1,2, we have θ2G2(s,s)G2(t,t)G2(t,s)G2(s,s),(or  G2(t,t)), where θ2=w2ϕ2(1)ψ2(0).

Proof.

It follows from the monotonicity of ϕ2(t) and ψ2(t) that the right-hand side of (2.6) holds. For the left hand side, by the monotonicity of ϕ2(t) and ψ2(t), we have G2(t,s)=1w2{ϕ2(t)ψ2(s),0ts1,ϕ2(s)ψ2(t),0st1,1w2{ϕ2(t)ψ2(s)ϕ2(s)ψ2(t)ϕ2(1)ψ2(0),0ts1,ϕ2(s)ψ2(t)ϕ2(t)ψ2(s)ϕ2(1)ψ2(0),0st1,=w2ϕ2(1)ψ2(0)G2(t,t)G2(s,s),=θ2G2(t,t)G2(s,s). The proof is completed.

Also, it is well known the Green function for the boundary value problem-u′′=0,0<t<1,α1u(0)-β1u(0)=0,γ1u(1)+δ1u(1)=0, isG1(t,s)=1w1{(β1+α1s)(γ1+δ1(1-t)),  0st1,(β1+α1t)(γ1+δ1(1-s)),  0ts1, where w1=β1γ1+α1γ1+α1δ1. Lete(t)=1w1(β1+α1t)(γ1+δ1(1-t)), clearly,e(t)e(s)G1(t,s)e(s). Now, we define an integral operator S:C[0,1][0,1] bySv(t)=01G1(t,τ)v(τ)dτ, and, then, by (2.9), we have(Sv)′′(t)=-v(t),0<t<1,α1(Sv)(0)-β1(Sv)(0)=0,γ1(Sv)(1)+δ1(Sv)(1)=0.

In order to obtain existence of positive solutions to problem (1.1), we will consider the existence of positive solutions to the following modified problem-(rv)(t)+qv(t)=p(t)f(t,Sv(t),-v(t))-g(t,Sv(t),-v(t)),0<t<1,α2v(0)-β2v(0)=0,γ2v(1)+δ2v(1)=0.

Lemma 2.4.

Let u(t)=Sv(t),v(t)C[0,1]. Then, we can transform (1.1) into (2.15). Moreover, if vC([0,1],[0,+) is a solution of problem (2.15), then the function u(t)=Sv(t) is a positive solution of problem (1.1).

Proof.

It follows from (2.9) that u′′(t)=-v(t), put u′′(t)=-v(t) and u(t)=Sv(t) into (1.1), we can transform (1.1) into (2.15).

Conversely, if vC([0,1],[0,+)) is a solution of (2.15), let u(t)=Sv(t), we have u′′(t)=-v(t), thus u=Sv is a solution of (1.1). The proof of Lemma 2.4 is completed.

In the rest of the paper, we always suppose that the following assumptions hold.

p:(0,1)[0,+) is continuous and satisfies 0<01G2(s,s)p(s)ds<+.

f:[0,1]×[0,+)×(-,+)[0,+) is continuous.

g:[0,1]×[0,+)×(-,+)[0,+) satisfies global Crathéodory’s condition and 01ρ(s)ds>0.

Remark 2.5.

It follows from (B1), (B3) and from the monotonicity of ϕ2(t),ψ2(t) that there exists (a,b)(0,1) such that 0<abG2(s,s)p(s)dsabG2(s,s)[p(s)+ρ(s)]ds01G2(s,s)[p(s)+ρ(s)]ds01G2(s,s)p(s)ds+ϕ2(1)ψ2(0)w201ρ(s)ds<+. So for convenience, in the rest of this paper, we define serval constants as follows: K=01G2(s,s)[p(s)+ρ(s)]ds,  μ2=ϕ2(a)ψ2(b)ϕ2(1)ψ2(0),l=μ2abG2(s,s)p(s)ds,μ1=(β1+α1a)(γ1+δ1(1-b))θ2w101G1(s,s)G2(s,s)ds.

Lemma 2.6.

Assume (B3) is satisfied. Then, the boundary value problem -(r(t)y)(t)+q(t)y(t)=ρ(t),0<t<1,α2y(0)-β2y(0)=0,γ2y(1)+δ2y(1)=0, has a unique solution y(t)=01G2(t,s)ρ(s)ds, which satisfies y(t)G2(t,t)01ρ(s)ds.

Proof.

First, y(t)=01G2(t,s)ρ(s)ds solves the BVP (2.20), and it is the unique solution of the BVP (2.20), since -(r(t)y)(t)+q(t)y(t)=0 with boundary conditions α2y(0)-β2y(0)=0,γ2y(1)+δ2y(1)=0 only has a trivial solution. Finally, it follows from (2.6) and (B3) that (2.22) holds.

Define a modified function [·]* for any zC[0,1] by[z(t)]*={z(t),z(t)0,0,  z(t)<0. We consider the following approximating problem-(r(t)x)(t)+q(t)x(t)=p(t)f(t,S[x(t)-y(t)]*,-[x(t)-y(t)]*)-g(t,S[x(t)-y(t)]*,-[x(t)-y(t)]*)+ρ(t),0<t<1,α2x(0)-β2x(0)=0,γ2x(1)+δ2x(1)=0.

Lemma 2.7.

If x(t)y(t) for any t[0,1] is a positive solution of the BVP (2.25), then S(x-y) is a positive solution of the singular perturbed differential equation (1.1).

Proof.

In fact, if x is a positive solution of the BVP (2.25) such that x(t)y(t) for any t[0,1], then, from (2.25) and the definition of [z(t)]*, we have -(r(t)x)(t)+q(t)x(t)=p(t)f(t,S(x(t)-y(t)),-(x(t)-y(t)))-g(t,S(x(t)-y(t)),-(x(t)-y(t)))+ρ(t),0<t<1,α2x(0)-β2x(0)=0,γ2x(1)+δ2x(1)=0. Let v=x-y, then -(r(t)v)(t)+q(t)v(t)=-(rx)(t)+qx(t)+(ry)(t)-qy(t), which implies that -(r(t)x)(t)+q(t)x(t)=-(r(t)v)(t)+q(t)v(t)-(r(t)y)(t)+q(t)y(t)=-(r(t)v)(t)+q(t)v(t)+ρ(t),α2x(0)-β2x(0)=0,γ2x(1)+δ2x(1)=0. Thus, (2.26) becomes (2.15), that is, x-y is a positive solution of the differential equation (2.15). By Lemma 2.4, u=S(x-y) is a positive solution of the singular perturbed differential equation (1.1). This completes the proof of Lemma 2.7.

Thus, the BVP (2.25) is equivalent to the integral equationx(t)=01G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)  -g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]ds. Hence, we will look for fixed points x(t)y(t),t[0,1], for the mapping T defined on E:=C([0,1],[0,+)) by(Tx)(t)=01G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]ds.

The basic space used in this paper is E=C([0,1];), where is a real number set. Obviously, the space E is a Banach space if it is endowed with the norm as follows:u=maxt[0,1]|u(t)|, for any uE. LetP={xE:x(t)θ2G2(t,t)x}, where θ2 is defined by (2.7), then P is a cone of E.

Lemma 2.8.

Assume that (B1)–(B3) hold. Then, T:PP is well defined. Furthermore, T:PP is a completely continuous operator.

Proof.

For any fixed xP, there exists a constant L>0 such that xL. And then, [x(s)-y(s)]*x(s)xL,|S[x(s)-y(s)]*|L01G1(t,s)ds(β1+α1)(γ1+δ1)w1L.

On the other hand, since g satisfies global Carathéodory’s condition, we have g(·,u(·),v(·))L1(0,1). Accordingly, Tx in (2.29) is continuous on [0,1], and, by (2.32), (Tx)(t)=01G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]ds01G2(s,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+2ρ(s)]dsN01G2(s,s)[p(s)+ρ(s)]ds<+, where N=max(t,u,v)[0,1]×[0,((β1+α1)(γ1+δ1)/w1)L]×[-L,0]f(t,u,v)+2. This implies that the operator T:PE is well defined.

Next, for any xP, by (2.6), we have Tx=max0t101G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]ds01G2(s,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]ds.

On the other hand, from (2.6), we also have (Tx)(t)=01G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]dsθ2G2(t,t)01G2(s,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]ds. So (Tx)(t)θ2G(t,t)Tx,t[0,1], which yields that T(P)P.

At the end, according to the Ascoli-Arzela Theorem, using standard arguments, one can show T:PP is a completely continuous operator.

3. Main ResultsTheorem 3.1.

Suppose (B1)–(B3) hold. In addition, assume that the following conditions are satisfied.

There exists a constant r>max(2K,01ρ(s)dsθ2) such that for any (t,u,v)[0,1]×[0,((β1+α1)(γ1+δ1)/w1)r]×[-r,0],  f(t,u,v)(r/K)-2, where θ2 and K are defined by (2.7) and (2.19), respectively.

There exists a constant R>2r such that, for any (t,u,v)[0,1]×[(1/2)μ1R,((β1+α1)(γ1+δ1)/w1)R]×[-R,-(1/2)μ2R], f(t,u,v)Rl, where μ1,μ2,l are defined by (2.19).

lim|u|+|v|+maxt[0,1]f(t,u,v)|u|+|v|=0.

Then, the singular perturbed differential equation (1.1) has at least two positive solutions u1,u2, and there exist two positive constants n1,n2 such that u1(t)n1e(t),u2(t)n2e(t), for any t[0,1].

Proof.

Let Ω1={xP:x<r}. Then, for any xΩ1,s[0,1], we have [x(s)-y(s)]*x(s)xr,|S[x(s)-y(s)]*|r01G1(t,s)ds(β1+α1)(γ1+δ1)w1r.

It follows from (S1) that Tx=maxt[0,1](Tx)(t)=max0t101G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]ds01G2(s,s)[(rK-2)p(s)+2ρ(s)]dsrK01G2(s,s)[p(s)+ρ(s)]ds=r=x. Therefore, Txx,xPΩ1.

On the other hand, let Ω2={xP:x<R} and Ω2={xP:x=R}. Then, for any xΩ2,t[0,1], noticing R>2r and (2.22), we have x(t)-y(t)x(t)-G2(t,t)01ρ(s)dsx(t)-x(t)θ2R01ρ(s)dsx(t)-rRx(t)12x(t)12θ2G2(t,t)R. So by (3.7), for any xΩ2,t[a,b], we have 12μ2R=ϕ2(a)ψ2(b)2w2θ2Rx(t)-y(t)R,12μ1R=(β1+α1a)(γ1+δ1(1-b))θ2R2w101e(s)G2(s,s)ds12θ2R01G1(t,s)G2(s,s)dsS(x(t)-y(t))(β1+α1)(γ1+δ1)w1R.

It follows from (S2), (3.8), and (2.6) that, for any xΩ2,t[a,b], Tx01G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]dsθ2G2(t,t)01G2(s,s)p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)dsθ2G2(t,t)abG2(s,s)p(s)f(s,S[x(s)-y(s)],x(s)-y(s))dsθ2ϕ2(a)ψ2(b)w2abG2(s,s)p(s)dsRl=μ2abG2(s,s)p(s)dsRl=R=x. So we have Txx,xPΩ2.

Next, let us choose ɛ>0 such that ɛ01G2(s,s)p(s)ds<1. Then, for the above ɛ, by (S3), there exists N>R>0 such that, for any t[0,1] and for any |u|+|v|N, f(t,u,v)ɛ(|u|+|v|). Let σ=max(t,u,v)[0,1]×[0,((β1+α1)(γ1+δ1)/w1)N]×[-N,0]f(t,u,v)+2, take R*=σ01G2(s,s)[p(s)+ρ(s)]ds+201G2(s,s)ρ(s)ds1-ɛ(1+((β1+α1)(γ1+δ1)/w1))01G2(s,s)p(s)ds+N, then R*>N>R.

Now let Ω3={xP:x<R*} and Ω3={xP:x=R*}. Then, for any xPΩ3, we have Tx=max0t101G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]ds01G2(s,s)[p(s)f(s,S[x(s)-y(s)]*,[x(s)-y(s)]*)+2ρ(s)]ds(max(t,u,v)[0,1]×[0,((β1+α1)(γ1+δ1)/w1)N]×[-N,0]f(t,u,v)+2)01G2(s,s)[p(s)+ρ(s)]ds+01G2(s,s)[p(s)ɛ(S[x(s)-y(s)]*+[x(s)-y(s)]*)+2ρ(s)]dsσ01G2(s,s)[p(s)+ρ(s)]ds+01G2(s,s)[p(s)ɛ(1+(β1+α1)(γ1+δ1)w1)x+2ρ(s)p(s)ɛ(1+(β1+α1)(γ1+δ1)w1)x  ]ds=σ01G2(s,s)[p(s)+ρ(s)]ds+201G2(s,s)ρ(s)ds+ɛ(1+(β1+α1)(γ1+δ1)w1)×01G2(s,s)p(s)dsx<R*=x, which implies that Txx,xPΩ3. By Lemma 1.1, T has two fixed points x1,x2 such that rx1Rx2.

It follows from r>01ρ(s)dsθ2 that x1(t)-y(t)x1(t)-G2(t,t)01ρ(s)dsx1(t)-x1(t)θ2r01ρ(s)ds=(1-01ρ(s)dsθ2r)x1(t)θ2r(1-01ρ(s)dsθ2r)G2(t,t)=m1G2(t,t)>0,t(0,1).

As for (3.18), we also find a positive constant m2 such that x2(t)-y(t)m2G2(t,t)>0,t(0,1).

Let ui(t)=S(xi-y)(t),(i=1,2), then ui(t)>0,t(0,1)(i=1,2),ui(t)=S(xi-y)(t)mi01G1(t,s)G2(s,s)dsmi01e(s)G2(s,s)dse(t)=nie(t). By Lemma 2.7, we know that the singular perturbed differential equation (1.1) has at least two positive solutions u1,u2 satisfying u1(t)n1e(t),u1(t)n2e(t),t[0,1], for some positive constants n1,n2. The proof of Theorem 3.1 is completed.

Theorem 3.2.

Suppose (B1)–(B3) hold. In addition, assume that the following conditions are satisfied.

There exists a constant r>201ρ(s)dsθ2 such that, for any (t,u,v)[0,1]×[(1/2)μ1r,((β1+α1)(γ1+δ1)/w1)r]×[-r,-(1/2)μ2r], f(t,u,v)rl, where θ2 and μ1,μ2,l are defined by (2.7) and (2.19), respectively.

There exists a constant R>max{r,((r/l)+2)K} such that, for any (t,u,v)[0,1]×[0,((β1+α1)(γ1+δ1)/w1)R]×[-R,0], f(t,u,v)RK-2, where K,l are defined by (2.19) and r is defined by (S4).

(S6)lim|u|+|v|mint[a,b]f(t,u,v)|u|+|v|=+. Then, the singular perturbed differential equation (1.1) has at least two positive solutions u1,u2, and there exist two positive constants n1,n2 such that u1(t)n1e(t),u2(t)n2e(t), for any t[0,1].

Proof.

Firstly, let Ω1={xP:x<r}. Then, for any xΩ1,t[0,1], by (2.22), we have x(t)-y(t)x(t)-G2(t,t)01ρ(s)dsx(t)-x(t)θ2r01ρ(s)ds12x(t)12θ2G2(t,t)r. So, by (3.26), for any xΩ1,t[a,b], we have 12μ2r=ϕ2(a)ψ2(b)2w2θ2rx(t)-y(t)r,12μ1r=(β1+α1a)(γ1+δ1(1-b))θ2r2w101e(s)G2(s,s)ds12θ2r01G1(t,s)G2(s,s)dsS(x(t)-y(t))(β1+α1)(γ1+δ1)w1r.

It follows from (S4), (3.27), and (2.6) that, for any xΩ2,t[a,b], Tx01G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]dsθ2G2(t,t)01G2(s,s)p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)dsθ2G2(t,t)abG2(s,s)p(s)f(s,S[x(s)-y(s)],x(s)-y(s))dsθ2ϕ2(a)ψ2(b)w2abG2(s,s)p(s)dsrl=μ2abG2(s,s)p(s)dsrl=R=x. Therefore, we have Txx,xPΩ1.

Next, by (S5), we have R>r and RK-2>rl>0.

Let Ω2={xP:x<R}. Then, for any xΩ2,s[0,1], we have [x(s)-y(s)]*x(s)xR,|S[x(s)-y(s)]*|R01G1(t,s)ds(β1+α1)(γ1+δ1)w1R.

It follows from (S5) that Tx=maxt[0,1](Tx)(t)=max0t101G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]ds01G2(s,s)[(RK-2)p(s)+2ρ(s)]dsRK01G2(s,s)[p(s)+ρ(s)]ds=R=x. Therefore, Txx,xPΩ2.

On the other hand, choose a large enough real number M>0 such that 12μ22MabG2(s,s)p(s)ds1.

From (S6), there exists N>R such that, for any t[a,b], f(t,u,v)M(|u|+|v|),|u|+|v|N. Take R*>max{2Nμ2,R}, then R*>R>r. Let Ω3={xP:x<R*}, for any xPΩ3 and for any t[a,b], we have x(t)-y(t)x(t)-G2(t,t)01ρ(s)dsx(t)-01ρ(s)dsθ2R*x(t)12x(t)θ2G2(t,t)2R*12μ2R*N>0. So, for any xPΩ3,t[a,b], by (3.35), (3.37), we have Tx01G2(t,s)[p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)-g(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)+ρ(s)]dsθ2G2(t,t)01G2(s,s)p(s)f(s,S[x(s)-y(s)]*,-[x(s)-y(s)]*)dsθ2G2(t,t)abG2(s,s)p(s)f(s,S[x(s)-y(s)],x(s)-y(s))dsθ2ϕ2(a)ψ2(b)w2abG2(s,s)p(s)M[|S(x(s)-y(s))|+|(x(s)-y(s))|]dsθ2ϕ2(a)ψ2(b)w2abG2(s,s)p(s)M(|x(s)-y(s)|)ds=μ2MabG2(s,s)p(s)ds×12μ2R*=12μ22MabG2(s,s)p(s)ds×R*R*=x. Thus, Txx,xPΩ3.

By Lemma 1.1, T has two fixed points x1,x2 such that rx1Rx2.

Noticing that r>201ρ(s)dsθ2, we have x1(t)-y(t)x1(t)-G2(t,t)01ρ(s)dsx1(t)-x1(t)θ2r01ρ(s)ds=(1-01ρ(s)dsθ2r)x1(t)12x1(t)12θ2rG2(t,t)=m1G2(t,t)>0,t(0,1). As for (3.41), we also can find a positive constant m2 such that x2(t)-y(t)m2G2(t,t)>0,t(0,1). Let ui(t)=S(xi-y)(t),(i=1,2), then ui(t)>0,t(0,1)  (i=1,2),ui(t)=S(xi-y)(t)mi01G1(t,s)G2(s,s)dsmi01e(s)G2(s,s)dse(t)=nie(t). By Lemma 2.7, we know that the singular perturbed differential equation (1.1) has at least two positive solutions u1,u2 satisfying u1(t)n1e(t),u1(t)n2e(t),t[0,1], for some positive constants n1,n2. The proof of Theorem 3.2 is completed.

An example Consider the following singular perturbed boundary value problem-(etu′′′)(t)+2etu′′(t)=1t(1-t)f(t,u(t),u′′(t))-sin(u(t))+arctan(u(t))(1+(π/2))t,u(0)=0,u(1)=0,u(0)-2u(0)=0,u(1)+u(1)=0, wheref(t,x,y)={tx+y2100,  (t,x,y)[0,1]×[0,10]×[-10,0],41.88x+4.075y,  (t,x,y)[0,1]×[10,13.93]×[-100,-10],200(1+t2)ex/13.93+siny,(t,x,y)[0,1]×[13.93,100]×[-100,-422.89],23.2469(x+|y|)1/2,(t,x,y)[0,1]×[100,+)×(-,-422.89]. Then, the BVP (3.45) has at least two positive solutions u1(t) and u2(t) such thatu1(t)2.3924t(1-t),t[0,1],u2(t)24.833t(1-t),t[0,1].

Proof.

In fact, let r(t)et,q(t)2et,p(t)=1t(1-t),ρ(t)=1t,g(t,x,y)=sinx+arctany(1+(π/2))t, then |g(t,x,y)|1t.

The corresponding Green’s functions can be written by G1(t,s)={s(1-t),0st1,t(1-s),0ts1,G2(t,s)=13e{(53es+13e-2s)(13et+23e-2t),0st1,(53et+13e-2t)(13es+23e-2s),0ts1, and, for (t,s)[0,1]×[0,1], we have t(1-t)s(1-s)G1(t,s)s(1-s) and 15e2+e-1G2(s,s)G2(t,t)G2(t,s)G2(s,s)(or  G2(t,t)).

Now, take [1/4,3/4][0,1], then we haveK=01G2(s,s)[p(s)+ρ(s)]ds=01((5/3)es+(1/3)e-2s)((1/3)es+(2/3)e-2s)3e[1s(1-s)+1s]ds0.6206,μ2=ϕ2(a)ψ2(b)ϕ2(1)ψ2(0)=ϕ2(1/4)ψ2(3/4)ϕ2(1)ψ2(0)=((5/3)e1/4+(1/3)e-1/2)((1/3)e3/4+(2/3)e-3/2)(5/3)e+(1/3)e-20.4374,l=μ2abG2(s,s)p(s)ds=0.43741/43/4((5/3)es+(1/3)e-2s)((1/3)es+(2/3)e-2s)3es(1-s)ds0.2786,μ1=(β1+α1a)(γ1+δ1(1-b))θ2w101G1(s,s)G2(s,s)ds=116(5e2+e-1)01s(1-s)((5/3)es+(1/3)e-2s)((1/3)es+(2/3)e-2s)3eds8.4578,θ2=w2ϕ2(1)ψ2(0)=p(0)[ψ2(0)ϕ2(0)-ψ2(0)ϕ2(0)]ϕ2(1)ψ2(0)4.938. On the other hand, since01ρ(s)ds=011sds=2, then (B1)–(B3) are satisfied.

Choose r=10, then10=r>max{2K,01ρ(s)dsθ2}={1.2412,0.4050}, and, for any (t,x,y)[0,1]×[0,10]×[-10,0],f(t,x,y)11rK-214.1134. So the condition (S1) is satisfied.

On the other hand, we take R=100, then R>2r=20, and, for any (t,x,y)[0,1]×[(1/2)μ1R,((β1+α1)(γ1+δ1)/w1)R]×[-R,-(1/2)μ2R]=[0,1]×[13.93,100]×[-100,-422.89], we have f(t,x,y)542.65Rl358.93, this implies that the condition (S2) holds. Next, we have lim|x|+|y|+maxt[0,1]f(t,x,y)|x|+|y|=lim|x|+|y|+23.2469(|x|+|y|)-1/2=0. Thus, (S3) also holds. By Theorem 3.1, the BVP (3.45) has at least two positive solutions u1(t) and u2(t). Sincen1=13e(θ2r-01ρ(s)ds)01s(1-s)(53es+13e-2s)(13es+23e-2s)ds2.3924,n2=13e(θ2R-01ρ(s)ds)01s(1-s)(53es+13e-2s)(13es+23e-2s)ds24.833, thus u1(t)2.3924t(1-t),t[0,1],u2(t)24.833t(1-t),t[0,1].

Acknowledgment

The authors were supported financially by the National Natural Science Foundation of China (11071141) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017).

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