By establishing a maximal principle and constructing upper and lower solutions, the existence of positive solutions for the eigenvalue problem of a class of fractional differential equations is discussed. Some sufficient conditions for the existence of positive solutions are established.

1. Introduction

In this paper, we discuss the existence of positive solutions for the following eigenvalue problem of a class fractional differential equation with derivatives-Dtαx(t)=λf(t,x(t),Dtβx(t)),t∈(0,1),Dtβx(0)=0,Dtγx(1)=∑j=1p-2ajDtγx(ξj),
where λ is a parameter, 1<α≤2, α-β>1, 0<β≤γ<1, 0<ξ1<ξ2<⋯<ξp-2<1, aj∈[0,+∞) with c=∑j=1p-2ajξjα-γ-1<1, and Dt is the standard Riemann-Liouville derivative. f:(0,1)×(0,+∞)×(0,+∞)→[0,+∞) is continuous, and f(t,u,v) may be singular at u=0,v=0, and t=0,1.

As fractional order derivatives and integrals have been widely used in mathematics, analytical chemistry, neuron modeling, and biological sciences [1–6], fractional differential equations have attracted great research interest in recent years [7–17]. Recently, ur Rehman and Khan [8] investigated the fractional order multipoint boundary value problem:Dtαy(t)=f(t,y(t),Dtβy(t)),t∈(0,1),y(0)=0,Dtβy(1)-∑i=1m-2ζiDtβy(ξi)=y0,
where 1<α≤2, 0<β<1, 0<ξi<1, ζi∈[0,+∞) with ∑i=1m-2ζiξiα-β-1<1. The Schauder fixed point theorem and the contraction mapping principle are used to establish the existence and uniqueness of nontrivial solutions for the BVP (1.2) provided that the nonlinear function f:[0,1]×ℝ×ℝ is continuous and satisfies certain growth conditions. But up to now, multipoint boundary value problems for fractional differential equations like the BVP (1.1) have seldom been considered when f(t,u,v) has singularity at t=0 and (or) 1 and also at u=0,v=0. We will discuss the problem in this paper.

The rest of the paper is organized as follows. In Section 2, we give some definitions and several lemmas. Suitable upper and lower solutions of the modified problems for the BVP (1.1) and some sufficient conditions for the existence of positive solutions are established in Section 3.

2. Preliminaries and Lemmas

For the convenience of the reader, we present here some definitions about fractional calculus.

Definition 2.1 (See [<xref ref-type="bibr" rid="B1">1</xref>, <xref ref-type="bibr" rid="B6">6</xref>]).

Let α>0 with α∈ℝ. Suppose that x:[a,∞)→ℝ. Then the αth Riemann-Liouville fractional integral is defined by
Iαx(t)=1Γ(α)∫at(t-s)α-1x(s)ds
whenever the right-hand side is defined. Similarly, for α∈ℝ with α>0, we define the αth Riemann-Liouville fractional derivative by
Dαx(t)=1Γ(n-α)(ddt)(n)∫at(t-s)n-α-1x(s)ds,where n∈ℕ is the unique positive integer satisfying n-1≤α<n and t>a.

Remark 2.2.

If x,y:(0,+∞)→ℝ with order α>0, then
Dtα(x(t)+y(t))=Dtαx(t)+Dtαy(t).

Lemma 2.3 (See [<xref ref-type="bibr" rid="B6">6</xref>]).

One has the following.

(1) If x∈L1(0,1),ν>σ>0, then IνIσx(t)=Iν+σx(t),DtσIνx(t)=Iν-σx(t),DtσIσx(t)=x(t).

(2) If ν>0,σ>0, then Dtνtσ-1=Γ(σ)Γ(σ-ν)tσ-ν-1.

Lemma 2.4 (See [<xref ref-type="bibr" rid="B6">6</xref>]).

Let α>0. Assume that x∈C(0,1)∩L1(0,1). ThenIαDtαx(t)=x(t)+c1tα-1+c2tα-2+⋯+cntα-n,
where ci∈ℝ(i=1,2,…,n), and n is the smallest integer greater than or equal to α.

and for t,s∈[0,1], we haveki(t,s)≤(1-s)α-γ-1Γ(α-β),i=1,2.

Lemma 2.5.

Let h∈C(0,1); If 1<α-β≤2, then the unique solution of the linear problem
-Dtα-βy(t)=h(t),t∈(0,1),y(0)=0,Dtγ-βy(1)=∑j=1p-2ajDtγ-βy(ξj)
is given by
y(t)=∫01K(t,s)h(s)ds,
where
K(t,s)=k1(t,s)+tα-β-11-∑j=1p-2ajξjα-γ-1∑j=1p-2ajk2(ξj,s)
is the Green function of the boundary value problem (2.9).

Proof.

Applying Lemma 2.4, we reduce (2.9) to an equivalent equation:
y(t)=-Iα-βh(t)+c1tα-β-1+c2tα-β-2,c1,c2∈R.
From (2.12) and noting that y(0)=0, we have c2=0. Consequently the general solution of (2.9) is
y(t)=-Iα-βh(t)+c1tα-β-1.
Using (2.13) and Lemma 2.3, we have
Dtγ-βy(t)=-Dtγ-βIα-βh(t)+c1Dtγ-βtα-β-1=-Iα-γh(t)+c1Γ(α-β)Γ(α-γ)tα-γ-1=-∫0t(t-s)α-γ-1Γ(α-γ)h(s)ds+c1Γ(α-β)Γ(α-γ)tα-γ-1.Thus,
Dtγ-βy(1)=-∫01(1-s)α-γ-1Γ(α-γ)h(s)ds+c1Γ(α-β)Γ(α-γ),
and for j=1,2,…,p-2,
Dtγ-βw(ξj)=-∫0ξj(ξj-s)α-γ-1Γ(α-γ)h(s)ds+c1Γ(α-β)Γ(α-γ)ξjα-γ-1.
Using Dtγ-βy(1)=∑j=1p-2ajDtγ-βy(ξj), (2.15), and (2.16), we obtain
c1=∫01(1-s)α-γ-1h(s)ds-∑j=1p-2aj∫0ξj(ξj-s)α-γ-1h(s)dsΓ(α-β)(1-∑j=1p-2ajξjα-γ-1).
So the unique solution of the problem (2.9) is
y(t)=-∫0t(t-s)α-β-1Γ(α-β)h(s)ds+tα-β-11-∑j=1p-2ajξjα-γ-1×{∫01(1-s)α-γ-1Γ(α-β)h(s)ds-∑j=1p-2aj∫0ξj(ξj-s)α-γ-1Γ(α-β)h(s)ds}=-∫0t(t-s)α-β-1Γ(α-β)h(s)ds+∫01(1-s)α-γ-1tα-β-1Γ(α-β)h(s)ds+tα-β-11-∑j=1p-2ajξjα-γ-1∑j=1p-2aj∫01(1-s)α-γ-1ξjα-γ-1Γ(α-β)h(s)ds-tα-β-11-∑j=1p-2ajξjα-γ-1∑j=1p-2aj∫0ξj(ξj-s)α-γ-1Γ(α-β)h(s)ds=∫01(k1(t,s)+tα-β-11-∑j=1p-2ajξjα-γ-1∑j=1p-2ajk2(ξj,s))h(s)ds=∫01K(t,s)h(s)ds.
The proof is completed.

From (2.11), we obtainK(t,s)≥tα-β-11-∑j=1p-2ajξjα-γ-1∑j=1p-2ajk2(ξj,s)=tα-β-1M(s).From (2.8), we have
K(t,s)=k1(t,s)+tα-β-11-∑j=1p-2ajξjα-γ-1∑j=1p-2ajk2(ξj,s)≤(1-s)α-γ-1Γ(α-β)+∑j=1p-2aj1-∑j=1p-2ajξjα-γ-1(1-s)α-γ-1Γ(α-β)≤(1+∑j=1p-2aj1-∑j=1p-2ajξjα-γ-1)(1-s)α-γ-1Γ(α-β). The proof is completed.

Consider the modified problem of the BVP (1.1):-Dtα-βy(t)=λf(t,Iβy(t),y(t)),y(0)=0,Dtγ-βy(1)=∑j=1p-2ajDtγ-βy(ξj).

Lemma 2.7.

Let x(t)=Iβy(t)andy(t)∈C[0,1]; then problem (1.1) is turned into (2.22). Moreover, if y∈C([0,1],[0,+∞)) is a solution of problem (2.22), then the function x(t)=Iβy(t) is a positive solution of the problem (1.1).

Proof.

Substituting x(t)=Iβy(t) into (1.1) and using Definition 2.1 and Lemmas 2.3 and 2.4, we obtain
Dtαx(t)=dndtnIn-αx(t)=dndtnIn-αIβy(t)=dndtnIn-α+βy(t)=Dtα-βy(t),Dtβx(t)=DtβIβy(t)=y(t).
Consequently, Dtβx(0)=y(0)=0. It follows from Dtγx(t)=dn/dtnIn-γx(t)=(dn/dtn)In-γIβy(t)=(dn/dtn)In-γ+βy(t)=Dtγ-βy(t) that Dtγ-βy(1)=∑j=1p-2ajDtγ-βy(ξj). Using x(t)=Iβy(t), y∈C[0,1], we transform (1.1) into (2.22).

Now, let y∈C([0,1],[0,+∞)) be a solution for problem (2.22). Using Lemma 2.3, (2.22), and (2.23), one has-Dtαx(t)=-dndtnIn-αx(t)=-dndtnIn-αIβy(t)=-dndtnIn-α+βy(t)=-Dtα-βy(t)=λf(t,Iβy(t),y(t))=λf(t,x(t),Dtβx(t)),0<t<1.
Noting
Dtβx(t)=DtβIβy(t)=y(t),Dtγx(t)=Dtγ-βy(t),
we have
Dtβx(0)=0,Dtγx(1)=∑j=1p-2ajDtγx(ξj).It follows from the monotonicity and property of Iβ that
Iβy∈C([0,1],[0,+∞)).
Consequently, x(t)=Iβy(t) is a positive solution of the problem (1.1).

Definition 2.8.

A continuous function ψ(t) is called a lower solution of the BVP (2.22), if it satisfies
-Dtα-βψ(t)≤λf(t,Iβψ(t),ψ(t)),ψ(0)≥0,Dtγ-βψ(1)≥∑j=1p-2ajDtγ-βψ(ξj).

Definition 2.9.

A continuous function ϕ(t) is called an upper solution of the BVP (2.22), if it satisfies
-Dtα-βϕ(t)≥λf(t,Iβϕ(t),ϕ(t)),ϕ(0)≤0,Dtγ-βϕ(1)≤∑j=1p-2ajDtγ-βϕ(ξj).

By Lemmas 2.5 and 2.6, we have the maximal principle.

Lemma 2.10.

If 1<α-β≤2 and y∈C([0,1],R) satisfies
y(0)=0,Dtγ-βy(1)=∑j=1p-2ajDtγ-βy(ξj),and -Dtα-βy(t)≥0 for any t∈(0,1), then y(t)≥0, for t∈[0,1].

To end this section, we present here two assumptions to be used throughout the rest of the paper.

f∈C((0,1)×(0,∞)×(0,∞),[0,+∞)) is decreasing in u and v, and for any (u,v)∈(0,∞)×(0,∞),

limσ→+∞σf(t,σu,σv)=+∞

uniformly on t∈(0,1).

For any μ,ν>0, f(t,μ,ν)≢0, and

∫01(1-s)α-γ-1f(s,μL(s),μG(s))ds<+∞.3. Main Results

The main result is summarized in the following theorem.

Theorem 3.1.

Provided that (B1) and (B2) hold, then there is a constant λ*>0 such that for any λ∈(λ*,+∞), the problem (1.1) has at least one positive solution x(t), which satisfies x(t)≥ℒ(t), t∈[0,1].

Proof.

Let E=C[0,1]; we denote a set P and an operator Tλ in E as follows:
P={y∈E:there exists positive numberlysuch thaty(t)≥lyG(t),t∈[0,1]},(Tλy)(t)=λ∫01K(t,s)f(s,Iβy(s),y(s))ds,for anyy∈P.
Clearly, P is a nonempty set since 𝒢(t)∈P. We claim that Tλ is well defined and Tλ(P)⊂P.

In fact, for any ρ∈P, by the definition of P, there exists one positive number lρ such that ρ(t)≥lρ𝒢(t) for any t∈[0,1]. It follows from Lemma 2.6 and (B2) that(Tλρ)(t)=λ∫01K(t,s)f(s,Iβρ(s),ρ(s))ds≤λM∫01(1-s)α-γ-1f(s,Iβρ(s),ρ(s))ds≤λM∫01(1-s)α-γ-1f(s,lρL(s),lρG(s))ds<+∞.

Setting B=maxt∈[0,1]ρ(t)>0, from (B2), we have f(t,B/Γ(β+1),B)≢0. By the continuity of f(t,u,v) on (0,1)×(0,∞)×(0,∞), we have ∫01𝔐(s)f(s,B/Γ(β+1),B)ds>0. On the other hand,IβB=1Γ(β)∫0t(t-s)β-1Bds=BtββΓ(β)≤BΓ(β+1),M(s)=∑j=1p-2ajk2(ξj,s)1-∑j=1p-2ajξjα-γ-1≤∑j=1p-2aj(1-s)α-γ-1Γ(α-β)(1-∑j=1p-2ajξjα-γ-1).From (3.3), one has
0<∫01M(s)f(s,BΓ(β+1),B)ds≤∫01M(s)f(s,IβB,B)ds≤∫01M(s)f(s,Iβρ(s),ρ(s))ds≤∑j=1p-2ajΓ(α-β)(1-∑j=1p-2ajξjα-γ-1)∫01(1-s)α-γ-1f(s,Iβρ(s),ρ(s))ds<+∞.It follows from Lemma 2.6 and (3.3) that
(Tλρ)(t)≥λG(t)∫01M(s)f(s,Iβρ(s),ρ(s))ds=lρ′G(t),
where
lρ′=λ∫01M(s)f(s,Iβρ(s),ρ(s))ds.Using (3.3) and (3.6), we know that Tλ is well defined and Tλ(P)⊂P.

Next we will focus on the upper and lower solutions of problem (2.22). From (B1) and (3.2), we know that the operator Tλ is decreasing in y. Using∫01K(t,s)f(s,L(s),G(s))ds≥G(t)∫01M(s)f(s,L(s),G(s))ds,∀t∈[0,1],and letting
λ1=1∫01M(s)f(s,L(s),G(s))ds,we have
λ1∫01K(t,s)f(s,L(s),G(s))ds≥G(t),∀t∈[0,1].

On the other hand, letting b(t)=∫01K(t,s)f(s,ℒ(s),𝒢(s))ds, since f(t,u,v) is decreasing with respect to u and v, for any λ>λ1, we have∫01K(t,s)f(s,λIβb(s),λb(s))ds≤∫01K(t,s)f(s,λ1Iβb(s),λ1b(s))ds≤∫01K(t,s)f(s,L(s),G(s))ds≤M∫01(1-s)α-γ-1f(s,L(s),G(s))ds<+∞.From (3.2), (3.3), and (B1), for all (u,v)∈(0,∞)×(0,∞), we have
limμ→+∞μf(t,μu,μv)=+∞uniformly on t∈(0,1). Thus there exists large enough λ*>λ1>0, such that, for any t∈(0,1),
λ*f(s,λ*L(s),λ*G(s))≥1∫01M(s)ds.From Lemma 2.6, one has
λ*∫01K(t,s)f(s,λ*L(s),λ*G(s))ds≥∫01K(t,s)ds∫01M(s)ds≥∫01G(t)M(s)ds∫01M(s)ds=G(t),∀t∈[0,1].Letting
ϕ(t)=λ*∫01K(t,s)f(s,L(s),G(s))ds=λ*b(t),ψ(t)=λ*∫01K(t,s)f(s,λ*Iβb(s),λ*b(s))ds,and using Lemmas 2.3 and 2.7, we obtain
ϕ(t)=λ*∫01K(t,s)f(s,L(s),G(s))ds≥G(t),t∈[0,1],ϕ(0)=0,Dtγ-βϕ(1)=∑j=1p-2ajDtγ-βϕ(ξj),ψ(t)=λ*∫01K(t,s)f(s,λ*Iβb(s),λ*b(s))ds≥G(t),t∈[0,1],ψ(0)=0,Dtγ-βψ(1)=∑j=1p-2ajDtγ-βψ(ξj).
Obviously, ϕ(t),ψ(t)∈P. By (3.16), we have
G(t)≤ψ(t)=(Tλ*ϕ)(t),G(t)≤ϕ(t),∀t∈[0,1],
which implies that
ψ(t)=(Tλ*ϕ)(t)=λ*∫01K(t,s)f(s,Iβϕ(s),ϕ(s))ds≤λ*∫01K(t,s)f(s,L(s),G(s))ds=ϕ(t),∀t∈[0,1].Consequently, it follows from (3.17)-(3.18) that
Dtψ(t)+λ*f(t,Iβψ(t),ψ(t))=Dt(Tλ*ϕ)(t)+λ*f(t,Iβ(Tλ*ϕ)(t),(Tλ*ϕ)(t))≥Dt(Tλ*ϕ)(t)+λ*f(t,Iβϕ(t),ϕ(t))=-λ*f(t,Iβϕ(t),ϕ(t))+λ*f(t,Iβϕ(t),ϕ(t))=0,Dtϕ(t)+λ*f(t,Iβϕ(t),ϕ(t))=-λ*f(t,L(t),G(t))+λ*f(t,Iβϕ(t),ϕ(t))≤-λ*f(t,L(t),G(t))+λ*f(t,L(t),G(t))=0.From (3.16) and (3.18)–(3.20), we know that ψ(t)andϕ(t) are upper and lower solutions of the problem (2.22), and ψ(t),ϕ(t)∈P.

Define the function F and the operator Aλ* in E byF(t,y)={f(t,Iβψ(t),ψ(t)),y<ψ(t),f(t,Iβy(t),y(t)),ψ(t)≤y≤ϕ(t),f(t,Iβϕ(t),ϕ(t)),y>ϕ(t),(Aλ*y)(t)=λ*∫01K(t,s)F(s,y(s))ds,∀y∈E.
It follows from (B1) and (3.21) that F:(0,1)×[0,+∞)→[0,+∞) is continuous. Consider the following boundary value problem:
-Dtα-βy(t)=λ*F(t,y),t∈[0,1],y(0)=0,Dtγ-βy(1)=∑j=1p-2ajDtγ-βy(ξj).Obviously, a fixed point of the operator Aλ* is a solution of the BVP (3.22). For all y∈E, it follows from Lemma 2.6, (3.21), and ψ(t)≥𝒢(t) that
(Aλ*y)(t)≤λ*M∫01(1-s)α-γ-1F(s,y(s))ds≤λ*M∫01(1-s)α-γ-1f(s,Iβψ(s),ψ(s))ds≤λ*M∫01(1-s)α-γ-1f(s,L(s),G(s))ds<+∞.So Aλ* is bounded. From the continuity of F(t,y) and K(t,s), it is obviously that Aλ*:E→E is continuous.

From the uniform continuity of K(t,s) and the Lebesgue dominated convergence theorem, we easily get that Aλ*(Ω) is equicontinuous. Thus from the Arzela-Ascoli theorem, Aλ*:E→E is completely continuous. The Schauder fixed point theorem implies that Aλ* has at least one fixed point w such that w=Aλ*w.

Now we proveψ(t)≤w(t)≤ϕ(t),t∈[0,1].
Let z(t)=ϕ(t)-w(t),t∈[0,1]. Since ϕ(t) is the upper solution of problem (2.22) and w is a fixed point of Aλ*, we have
w(0)=0,Dtγ-βw(1)=∑j=1p-2ajDtγ-βw(ξj).

From (3.17), (3.18), and the definition of F, we obtainf(t,Iβϕ(t),ϕ(t))≤F(t,y(t))≤f(t,Iβψ(t),ψ(t)),∀y∈E,f(t,L(t),G(t))≥f(t,Iβψ(t),ψ(t)),∀t∈[0,1].So
f(t,Iβϕ(t),ϕ(t))≤F(t,y(t))≤f(t,L(t),G(t)),∀y∈E.From (3.18) and (3.20), one has
Dtα-βz(t)=Dtα-βϕ(t)-Dtα-βw(t)=-λ*f(t,L(t),G(t))+λ*F(t,w(t))≤0,∀t∈[0,1].By (3.27), (3.28), and Lemma 2.10, we get z(t)≥0 which implies that w(t)≤ϕ(t) on [0,1]. In the same way, we have w(t)≥ψ(t) on [0,1]. Thus we obtain
ψ(t)≤w(t)≤ϕ(t),t∈[0,1].
Consequently, F(t,w(t))=f(t,Iβw(t),w(t)),t∈[0,1]. Then w(t) is a positive solution of the problem (2.22). It thus follows from Lemma 2.7 that x(t)=Iβw(t) is a positive solution of the problem (1.1).

Finally, by (3.29), we havew(t)≥ψ(t)≥G(t).Thus,
x(t)=Iβw(t)=1Γ(β)∫0t(t-s)β-1w(s)ds≥1Γ(β)∫0t(t-s)β-1G(s)ds=L(t).

Corollary 3.2.

Suppose that condition (B1)holds, and that for any μ,ν>0, f(t,μ,ν)≢0, and
∫01f(s,μL(s),μG(s))ds<+∞.Then there exists a constant λ*>0 such that for any λ∈(λ*,+∞), the problem (1.1) has at least one positive solution x(t), which satisfies x(t)≥ℒ(t), t∈[0,1].

We consider some special cases in which f(t,u,v) has no singularity at u,v=0 or t=0,1.

We give the following assumption.

(B*1)f∈C((0,1)×[0,∞)×[0,∞),(0,+∞)) is decreasing in u,v.

Then, f(t,u,v) is nonsingular at u=v=0 and for all u,v≥0, f(t,u,v)>0,t∈(0,1), which implies that f(t,0,0)>0,t∈(0,1). Thuslimμ→+∞μf(t,0,0)=+∞,uniformly fort∈(0,1)

naturally holds; we then have the following corollary.

Corollary 3.3.

If (B*1) holds and

(B*2)∫01(1-s)α-γ-1f(s,0,0)ds<+∞,
then there exists a constant λ*>0 such that for any λ∈(λ*,+∞), the problem (1.1) has at least one positive solution x(t), which satisfies x(t)≥ℒ(t), t∈[0,1].Proof.

In the proof of Theorem 3.1, we replace the set P by
P1={y∈E:y(t)≥0,t∈[0,1]}and the inequalities (3.18)–(3.20) by
0≤ψ(t)=Tλ0,0≤ϕ(t)=Tλψ(t)≤Tλ0=ψ(t).Since Tλ0,Tλψ(t)∈P, we have
Dtα-βTλ0+f(t,IβTλ,Tλ0)=-f(t,0,0)+f(t,IβTλ0,Tλ0)≤0,Dtα-βTλψ(t)+f(t,IβTλψ(t),Tλψ(t))=-f(t,Iβψ(t),ψ(t))+f(t,IβTλψ(t),Tλψ(t))≥0,t∈[0,1].
The rest of the proof is similar to that of Theorem 3.1.

If f(t,u,v) is nonsingular at u=0,v=0 and t=0,1, we have the conclusion.

Corollary 3.4.

If f(t,u,v):[0,1]×[0,∞)×[0,∞)→(0,+∞) is continuous and decreasing in uandv, the problem (1.1) has at least one positive solution x(t), which satisfies x(t)≥ℒ(t), t∈[0,1].

Example 3.5.

Consider the existence of positive solutions for the following eigenvalue problem of fractional differential equation:
-Dt3/2x(t)=λet(1-t)1/8(x-1/2(t)+(Dt1/8x(t))-1/8),Dt1/8x(0)=0,Dt3/8x(1)=2Dt3/8x(12)-Dt3/8x(34).

Let
f(t,u,v)=1et(1-t)1/8(u-1/2+v-1/8),(t,u,v)∈(0,1)×(0,+∞)×(0,+∞).
Then f∈C((0,1)×(0,+∞)×(0,+∞),(0,+∞)) is decreasing in uandv, and for any (u,v)∈(0,∞)×(0,∞),
limσ→+∞σf(t,σu,σv)=limσ→+∞σ1/2u-1/2+σ7/8v-1/8et(1-t)1/8=+∞,uniformly on t∈(0,1). Thus (B1) holds.

On the other hand, for any μ,ν>0 and t∈(0,1),f(t,μ,ν)=1et(1-t)1/8(μ-1/2+ν-1/8)≢0,L(t)=∫0t(t-s)-7/8s3/8Γ(1/8)ds=Γ(11/8)Γ(3/2)t1/2.

thus we have∫01(1-s)α-γ-1f(s,μL(s),μG(s))ds=∫01(1-s)1/8[1es(1-s)1/8(μ-1/2L-1/2(s)+μ-1/8G-1/8(s))1es(1-s)1/8]≤∫01[μ-1/2(Γ(11/8)Γ(3/2)t1/2)-1/2+μ-1/8s-3/64]ds=∫01[μ-1/2(Γ(11/8)Γ(3/2)t-1/4)+μ-1/8s-3/64]ds<+∞,

which implies that (B2) holds. From Theorem 3.1, there is a constant λ*>0 such that for any λ∈(λ*,+∞) the problem (3.38) has at least one positive solution x(t) andx(t)≥L(t)=Γ(11/8)Γ(3/2)t1/2≈1.003t1/2,t∈[0,1].

Acknowledgments

This work is supported financially by the National Natural Science Foundation of China (11071141, 11126231) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017).

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