AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation31465610.1155/2009/314656314656Research ArticleThe Existence of Positive Solution to Three-Point Singular Boundary Value Problem of Fractional Differential EquationTianYuansheng1ChenAnping1,22ZhouYong1Department of MathematicsXiangnan UniversityChenzhou 423000Chinaxnu.edu.cn2School of Mathematics and Computational ScienceXiangtan UniversityXiangtan 411105Chinaxtu.edu.cn200902072009200913052009230620092009Copyright © 2009This 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.

We investigate the existence of positive solution to nonlinear fractional differential equation three-point singular boundary value problem: Dqu(t)+f(t,u(t))=0, 0<t<1, u(0)=0, u(1)=αD(q1)/2u(t)|t=ξ, where 1<q2 is a real number, ξ(0,1/2], α(0,+) and αΓ(q)ξ(q1)/2<Γ((q+1)/2),Dq is the standard Riemann-Liouville fractional derivative, and fC((0,1]×[0,+),[0,+)),limt+0f(t,)=+ (i.e., f is singular at t=0). By using the fixed-point index theory, the existence result of positive solutions is obtained.

1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and the applications of such constructions in various sciences such as physics, mechanics, chemistry, and engineering. For details, see  and the reference therein. However, up to our knowledge, most of those papers have studied the existence and multiplicity of solution (or positive solution) to the initial value problem of nonlinear fractional differential equations; see [1, 4, 10, 11].

Recently, there are a few paper considering the Dirichlet-type boundary value problem for nonlinear ordinary differential equations of fractional order; see . Delbosco  has investigated the nonlinear Dirichlet-type problem

uq-1Dqu(t)=f(u(t)),0<t<1,1<q<2,u(0)=u(1)=0. He has proved that if f(u) is Lipschitizan function, then the problem has at least one solution u(t) in a certain subspace of C[0,1] in which fractional derivative has a Holder property. When f(t,u) is continuous on [0,1]×[0,+), by the use of some fixed-point theorem on cones, Bai and Lü  and Zhang  have given the existence of positive solutions to the equation

Dqu(t)+f(t,u(t))=0,0<t<1, with boundary condition u(0)=u(1)=0,u(0)+u'(0)=u(1)+u'(1), respectively.

This paper is to study the existence of positive solution for the three-point singular boundary value problem of nonlinear fractional differential equation Dqu(t)+f(t,u(t))=0,0<t<1,u(0)=0,u(1)=αD(q-1)/2u(t)|t=ξ. By using the fixed-point index theory, where 1<q2 is a real number, ξ(0,1/2], α(0,+) satisfy that αΓ(q)ξ(q-1)/2<Γ((q+1)/2), Dq is the standard Riemann-Liouville fractional derivative, and the function f satisfies the following condition:

fC((0,1]×[0,+),[0,+)), limt+0f(t,·)=+, there exists a constant b:0<b<1 such that tbf(t,u(t)) is continuous function on [0,1]×[0,+).

The organization of this paper is as follows. In Section 2, we present some necessary definitions and Preliminary results that will be used to prove our main results. The proof of our main result is given in Section 3. In Section 4, we will give an example to ensure our main result.

2. Preliminaries

The material in this section is basic in some sense. For the reader’s convenience, we present some necessary definitions from fractional calculus theory and preliminary results.

Definition 2.1.

The fractional integral of order q>0 of a function x:(0,+)R is given by Iqx(t)=1Γ(q)0t(t-s)q-1x(s)ds, provided that the right side is pointwise defined on (0,).

Definition 2.2.

The fractional derivative of order q>0 of a continuous function x:(0,+)R is given by Dqx(t)=1Γ(n-q)(ddt)n0tx(s)(t-s)q-n+1ds, where n=[q]+1, provided that the right side is pointwise defined on (0,).

Lemma 2.3 (see [<xref ref-type="bibr" rid="B11">7</xref>]).

(1) If xL(0,1), ρ>σ>0, then DσIρx(t)=Iρ-σx(t).

If ρ>0, λ>0, then Dρtλ-1=(Γ(λ)/Γ(λ-ρ))tλ-ρ-1.

Lemma 2.4 (see [<xref ref-type="bibr" rid="B3">12</xref>]).

Assume that xC(0,1)L(0,1) with a fractional derivative of order q>0 that belongs to C(0,1)L(0,1). Then IqDqx(t)=x(t)+A1tq-1+A2tq-2++ANtq-N,AiR, i=1,2,,N, where N is the smallest integer greater than or equal to q.

Lemma 2.5.

If yC(0,1)L(0,1) and 1<q2, ξ(0,1), α satisfy that αΓ(q)ξ(q-1)/2Γ((q+1)/2), then the problems Dqu(t)+y(t)=0,0<t<1,u(0)=0,u(1)=αD(q-1)/2u(t)|t=ξ have the unique solution u(t)=-0t(t-s)q-1Γ(q)y(s)ds+tq-1Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{01(1-s)q-1Γ(q)y(s)ds-α0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}.

Proof.

By applying Lemma 2.4, we may reduce (2.4) to an equivalent integral equation u(t)=-Iqy(t)+A1tq-1+A2tq-2 for some A1,A2R. Consequently the general solution of (2.4) is u(t)=-0t(t-s)q-1y(s)Γ(q)ds+A1tq-1+A2tq-2. Note that u(0)=0, we have A2=0 and u(1)=-01(1-s)q-1y(s)Γ(q)ds+A1. On the other hand, by (2.6) and Lemma 2.3, we have D(q-1)/2u(t)=-D(q-1)/2Iqy(t)+A1D(q-1)/2tq-1=-I(q+1)/2y(t)+A1Γ(q)Γ((q+1)/2)t(q-1)/2. Therefore D(q-1)/2u(t)|t=ξ=-0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds+A1Γ(q)Γ((q+1)/2)ξ(q-1)/2. By u(1)=αDpu(t)|t=ξ, combine with (2.8) and (2.10), we obtain A1=Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{01(1-s)q-1Γ(q)y(s)ds-α0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}. So, the unique solution of problem (2.4) is u(t)=-0t(t-s)q-1Γ(q)y(s)ds+tq-1Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{01(1-s)q-1Γ(q)y(s)ds-α0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}. The proof is completed.

Lemma 2.6.

If yC((0,1),[0,+))L(0,1) and 1<q2, ξ(0,1/2], α(0,+) satisfy that αΓ(q)ξ(q-1)/2<Γ((q+1)/2), then the unique solution of the problem (2.4) u(t)=-0t(t-s)q-1Γ(q)y(s)ds+tq-1Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{01(1-s)q-1Γ(q)y(s)ds-α0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}=01G(t,s)y(s)ds+αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2×{0ξ((1-s)q-1ξ(q-1)/2-(ξ-s)(q-1)/2)y(s)ds+ξ1(1-s)q-1ξ(q-1)/2y(s)ds}. is nonnegative on [0,1], where G(t,s)={[t(1-s)]q-1-(t-s)q-1Γ(q),0st1,[t(1-s)]q-1Γ(q),0ts1.

Proof.

By Lemma 2.5, the unique solution of problem (2.4) is u(t)=-0t(t-s)q-1Γ(q)y(s)ds+tq-1Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{01(1-s)q-1Γ(q)y(s)ds-α0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}=-0t(t-s)q-1Γ(q)y(s)ds+tq-1Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/201(1-s)q-1Γ(q)y(s)ds-αtq-1Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds=-0t(t-s)q-1Γ(q)y(s)ds+(1+αΓ(q)ξ(q-1)/2Γ((q+1)/2)-αΓ(q)ξ(q-1)/2)tq-101(1-s)q-1Γ(q)y(s)ds-αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2y(s)ds=-0t(t-s)q-1Γ(q)y(s)ds+01(1-s)q-1tq-1Γ(q)y(s)ds+αtq-1ξ(q-1)/2Γ((q+1)/2)-αΓ(q)ξ(q-1)/201(1-s)q-1y(s)ds-αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2y(s)ds=0t[t(1-s)]q-1-(t-s)q-1Γ(q)y(s)ds+t1[t(1-s)]q-1Γ(q)y(s)ds+αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{01(1-s)q-1ξ(q-1)/2y(s)ds-0ξ(ξ-s)(q-1)/2y(s)ds}=01G(t,s)y(s)ds+αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2×{0ξ((1-s)q-1ξ(q-1)/2-(ξ-s)(q-1)/2)y(s)ds+ξ1(1-s)q-1ξ(q-1)/2y(s)ds}.

Observing the expression of G(t,s), it is clear that G(t,s)>0 for s,t(0,1). On the other hand, by ξ(0,1/2], we have (1-s)2(1-sξ),s[0,ξ]. Hence (1-s)q-1(1-sξ)(q-1)/2. It implies that (1-s)q-1ξ(q-1)/2(ξ-s)(q-1)/2. Therefore u(t) is nonnegative on [0,1].

Lemma 2.7.

Let 1<q2, ξ(0,1/2], α(0,+) satisfy that αΓ(q)ξ(q-1)/2<Γ((q+1)/2) and y:(0,1][0,+) is continuous, and limt+0y(t)=+. Suppose that there exists a constant b:0<b<1 such that tby(t) is continuous function on [0,1]. Then the unique solution of (2.4) u(t)=-0t(t-s)q-1Γ(q)y(s)ds+tq-1Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{01(1-s)q-1Γ(q)y(s)ds-α0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}. is continuous on [0,1].

Proof.

Since tby(t) is continuous in [0,1], thus there exists a constant L>0 such that |tby(t)|L for all t[0,1] and u(t)=-1Γ(q)0t(t-s)q-1s-b·sby(s)ds+tq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)01(1-s)q-1s-b·sby(s)ds-αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2s-b·sby(s)ds. For any t0[0,1], we will prove u(t)u(t0) (tt0, t[0,1]). For the convenience, the proof is divided into three cases.Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M114"><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>0</mml:mn></mml:mrow></mml:msub><mml:mo>=</mml:mo><mml:mn>0</mml:mn></mml:math></inline-formula>).

It is easy to know that u(0)=0. For all t(0,1], we have |u(t)-u(t0)||1Γ(q)0t(t-s)q-1s-b·sby(s)ds|+|tq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)01(1-s)q-1s-b·sby(s)ds|+|αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2s-b·sby(s)ds|LΓ(q)0t(t-s)q-1s-bds+Ltq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)01(1-s)q-1s-bds+αLtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2s-bds=Ltq-bΓ(q)B(1-b,q)+Ltq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)B(1-b,q)+αLξ(1-b+(q-1)/2)tq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2B(1-b,(q+1)/2)=LΓ(1-b)Γ(1-b+q)tq-b+LΓ(1-b)Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+q)tq-1+αLΓ(1-b)Γ((q+1)/2)ξ(1-b+(q-1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+(q+1)/2)tq-10(t0), where B denotes beta function.

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M119"><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>0</mml:mn></mml:mrow></mml:msub><mml:mo>∈</mml:mo><mml:mrow><mml:mo stretchy="false">(</mml:mo><mml:mrow><mml:mn>0,1</mml:mn></mml:mrow><mml:mo stretchy="false">)</mml:mo></mml:mrow></mml:math></inline-formula>, for all <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M120"><mml:mi>t</mml:mi><mml:mo>∈</mml:mo><mml:mo stretchy="false">(</mml:mo><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>0</mml:mn></mml:mrow></mml:msub><mml:mo>,</mml:mo><mml:mn>1</mml:mn><mml:mo stretchy="false">]</mml:mo></mml:math></inline-formula>).

We have |u(t)-u(t0)|  =|-1Γ(q)0t(t-s)q-1y(s)ds+1Γ(q)0t0(t0-s)q-1y(s)ds+(tq-1-t0q-1)Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{01(1-s)q-1Γ(q)y(s)ds-α0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}||1Γ(q)0t0((t-s)q-1-(t0-s)q-1)s-b·sby(s)ds+1Γ(q)t0t(t-s)q-1s-b·sby(s)ds|+|(tq-1-t0q-1)Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)01(1-s)q-1s-b·sby(s)ds|+|α(tq-1-t0q-1)Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2s-b·sby(s)ds|LΓ(q)0t0((t-s)q-1-(t0-s)q-1)s-bds+LΓ(q)t0t(t-s)q-1s-bds+(tq-1-t0q-1)Γ((q+1)/2)L[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)01(1-s)q-1s-bds+α(tq-1-t0q-1)LΓ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2s-bds=LΓ(q)(tq-b-t0q-b)B(1-b,q)+(tq-1-t0q-1)Γ((q+1)/2)L[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)B(1-b,q)+α(tq-1-t0q-1)Lξ(1-b+(q-1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2B(1-b,q+12)=LΓ(1-b)Γ(1-b+q)(tq-b-t0q-b)+LΓ(1-b)Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+q)(tq-1-t0q-1)+αLΓ(1-b)Γ((q+1)/2)ξ(1-b+(q-1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+(q+1)/2)(tq-1-t0q-1)0(tt0).

Case 3 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M122"><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>0</mml:mn></mml:mrow></mml:msub><mml:mo>∈</mml:mo><mml:mrow><mml:mo stretchy="false">(</mml:mo><mml:mrow><mml:mn>0,1</mml:mn></mml:mrow><mml:mo stretchy="false">]</mml:mo></mml:mrow></mml:math></inline-formula>, for all <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M123"><mml:mi>t</mml:mi><mml:mo>∈</mml:mo><mml:mo stretchy="false">[</mml:mo><mml:mn>0</mml:mn><mml:mo>,</mml:mo><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>0</mml:mn></mml:mrow></mml:msub><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>).

The proof is similar to the step 2. Here, we omit it.

The main tool of this paper is the following well-known fixed-point index theorem (see ).

Lemma 2.8.

Let E be a Banach space, PE is a cone. For r>0, define Ωr={uPu<r}. Assume that T:Ω¯rP is a completely continuous such that Tuu for uΩr={uPu=r}.

If Tuu for uΩr, then i(T,Ωr,P)=0.

If Tuu for uΩr, then i(T,Ωr,P)=1.

3. Main Results

For the convenience we introduce the following notations: C1=Γ(1-b)Γ(1-b+q),C2=Γ(1-b)Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2(1Γ(1-b+q)+αξ(1-b+(q-1)/2)Γ(1-b+(q+1)/2)),C3=αΓ(1-b)ξ(q-1)/2Γ((q+1)/2)-αΓ(q)ξ(q-1)/2(Γ(q)Γ(1-b+q)-Γ((q+1)/2)ξ1-bΓ(1-b+(q+1)/2)).

Remark 3.1.

If 1<q2, 0<b<1, 0<ξ1/2, α>0, and Γ((q+1)/2)>αΓ(q)ξ(q-1)/2, then C3=α[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]{0ξ((1-s)q-1ξ(q-1)/2-(ξ-s)(q-1)/2)·s-bds+ξ1(1-s)q-1ξ(q-1)/2·s-bds}=αΓ(1-b)ξ(q-1)/2Γ((q+1)/2)-αΓ(q)ξ(q-1)/2(Γ(q)Γ(1-b+q)-Γ((q+1)/2)ξ1-bΓ(1-b+(q+1)/2))>0.

Let E=C[0,1] be a Banach spaces with the maximum norm u=max0t1|u(t)|. Define the cone PE by P={uEu(t)0,0t1}.

The positive solution which we consider in this paper is the form u(0)=0, u(t)>0, 0<t1, uE.

Define an operator T:PE by Tu(t)=-0t(t-s)q-1Γ(q)f(s,u(s))ds+tq-1Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{01(1-s)q-1Γ(q)f(s,u(s))ds-α0ξ(ξ-s)(q-1)/2Γ((q+1)/2)f(s,u(s))ds}.

Lemma 3.2.

Assume that condition (H1) holds. Then the operator T:PP is completely continuous.

Proof.

If condition (H1) holds, by Lemmas 2.6 and 2.7, we have T(P)P. Let u0P and u0=a0; if uP and u-u0<1, then u<1+a0:=a. By the continuous of tbf(t,x), we know that tbf(t,x) is uniformly continuous on [0,1]×[0,a]. Thus, for all ε>0, there exists a δ>0 (δ<1) such that |tbf(t,x1)-tbf(t,x2)|<εC1+C2 for all t[0,1] and x1,x2[0,a] with |x1-x2|<δ. Obviously, if u-u0δ, then u0(t),u(t)[0,a] and |u(t)-u0(t)|<δ for each t[0,1]. Hence, we have |tbf(t,u(t))-tbf(t,u0(t))|<εC1+C2 for all t[0,1],uP with u-u0<δ. It follows from (3.6) that Tu-Tu0=max0t1|Tu(t)-Tu0(t)|max0t1{1Γ(q)0t(t-s)q-1s-b|sbf(s,u(s))-sbf(s,u0(s))|ds+tq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)×01(1-s)q-1s-b|sbf(s,u(s))-sbf(s,u0(s))|ds+αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2×0ξ(ξ-s)(q-1)/2s-b|sbf(s,u(s))-sbf(s,u0(s))|ds}<max0t1{εΓ(q)(C1+C2)0t(t-s)q-1s-bds+εtq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2](C1+C2)Γ(q)01(1-s)q-1s-bds+αεtq-1[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2](C1+C2)0ξ(ξ-s)(q-1)/2s-bds}=max0t1{εΓ(q)(C1+C2)tq-bB(1-b,q)+εtq-1Γ((q+1)/2)B(1-b,q)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2](C1+C2)Γ(q)+αεtq-1ξ(1-b+(q-1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2](C1+C2)B(1-b,q+12)}{Γ(1-b)Γ(1-b+q)+Γ((q+1)/2)Γ(1-b)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+q)+αΓ(1-b)Γ((q+1)/2)ξ(1-b+(q-1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+(q+1)/2)}ε(C1+C2)=ε. By the arbitrariness of u0, we have that T:PP is continuous.

Let ΩP be bounded; that is, there exists a positive constant M such that Ω{uPuM}. Since tbf(t,u(t)) is continuous on [0,1]×[0,+), we let N=max(t,u)[0,1]×[0,M]tbf(t,u(t))+1. For all uΩ, we have |Tu(t)||1Γ(q)0t(t-s)q-1s-b·sbf(t,u(t))ds|+|tq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)01(1-s)q-1s-b·sbf(s,u(s))ds|+|αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2s-b·sbf(s,u(s))ds|NΓ(q)0t(t-s)q-1s-bds+Ntq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)01(1-s)q-1s-bds+αNtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2s-bds=NΓ(q)tq-bB(1-b,q)+Ntq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)B(1-b,q)+αNξ(1-b+(q-1)/2)tq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2B(1-b,q+12)NΓ(1-b)Γ(1-b+q)+NΓ((q+1)/2)Γ(1-b)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+q)+αNΓ(1-b)Γ((q+1)/2)ξ(1-b+(q-1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+(q+1)/2)=(C1+C2)N. Hence T(Ω) is bounded.

On the other hand, given ε>0, set δ=min{12(ε4NC1)1/(q-b),12(ε2NC2)1/(q-1),ε4NC1}. For each uΩ, we will prove that if t1,t2[0,1] and 0<t2-t1<δ, then |Tu(t2)-Tu(t1)|<ε. In fact, similar to the proof of Lemma 2.7, we have |Tu(t2)-Tu(t1)|NΓ(1-b)Γ(1-b+q)(t2q-b-t1q-b)+NΓ((q+1)/2)Γ(1-b)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+q)(t2q-1-t1q-1)+αNΓ(1-b)Γ((q+1)/2)ξ(1-b+(q-1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+(q+1)/2)(t2q-1-t1q-1)=NC1(t2q-b-t1q-b)+NC2(t2q-1-t1q-1). In the following, the proof is divided into three cases. Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M199"><mml:mi>δ</mml:mi><mml:mo>≤</mml:mo><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo><</mml:mo><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msub><mml:mo><</mml:mo><mml:mn>1</mml:mn></mml:math></inline-formula>, <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M200"><mml:mi>q</mml:mi><mml:mo>-</mml:mo><mml:mi>b</mml:mi><mml:mo>-</mml:mo><mml:mn>1</mml:mn><mml:mo>≤</mml:mo><mml:mn>0</mml:mn></mml:math></inline-formula>).

|Tu(t2)-Tu(t1)|NC1(t2q-b-t1q-b)+NC2(t2q-1-t1q-1)NC1(q-b)δq-b-1(t2-t1)+NC2(q-1)δq-2(t2-t1)<2NC1δq-b+NC2δq-12NC1·ε4NC12q-b+NC2·ε2NC22q-1=ε2×2q-b+ε2×2q-1<ε.

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M202"><mml:mi>δ</mml:mi><mml:mo>≤</mml:mo><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo><</mml:mo><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msub><mml:mo><</mml:mo><mml:mn>1</mml:mn></mml:math></inline-formula>, <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M203"><mml:mi>q</mml:mi><mml:mo>-</mml:mo><mml:mi>b</mml:mi><mml:mo>-</mml:mo><mml:mn>1</mml:mn><mml:mo>></mml:mo><mml:mn>0</mml:mn></mml:math></inline-formula>).

|Tu(t2)-Tu(t1)|NC1(t2q-b-t1q-b)+NC2(t2q-1-t1q-1)NC1(q-b)(t2-t1)+NC2(q-1)δq-2(t2-t1)<2NC1δ+NC2δq-12NC1·ε4NC1+NC2·ε2NC22q-1=ε2+ε2×2q-1<ε.

Case 3 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M205"><mml:mn>0</mml:mn><mml:mo>≤</mml:mo><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow></mml:msub><mml:mo><</mml:mo><mml:mi>δ</mml:mi></mml:math></inline-formula>, <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M206"><mml:msub><mml:mrow><mml:mi>t</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msub><mml:mo><</mml:mo><mml:mn>2</mml:mn><mml:mi>δ</mml:mi></mml:math></inline-formula>).

|Tu(t2)-Tu(t1)|NC1(t2q-b-t1q-b)+NC2(t2q-1-t1q-1)NC1t2q-b+NC2t2q-1NC1(2δ)q-b+NC2(2δ)q-1NC1·ε4NC1+NC2·ε2NC2=ε4+ε2<ε.

Therefore, T(Ω) is equicontinuous. The Arzela-Ascoli Theorem implies that T(Ω) is compact. Thus, the operator T:PP is completely continuous.

We obtain the following existence results of the positive solution for problem (1.4).

Theorem 3.3.

If condition (H1) holds and assume further that there exist two positive constants R>r>0 such that

tbf(t,u)>r/C3, for (t,u)[0,1]×[0,r];

tbf(t,u)<R/(C1+C2), for (t,u)[0,1]×[0,R],

then problem (1.4) has at least one positive solution u such that r<u<R.

Proof.

Problem (1.4) has a solution u=u(t) if and only if u is a solution of the operator equation u=Tu. In order to apply Lemma 2.8, we separate the proof into the following two steps.Step 1.

Let Ωr:={uPu<r}. For any uΩr, we have u=r and 0u(t)r for all t[0,1]. Observing the expression of G(t,s) (see (2.14), it is clear that G(1,s)=0. By assumption (H2), we have Tu(1)=01G(1,s)f(s,u(s))ds+αΓ((q+1)/2)-αΓ(q)ξ(q-1)/2{0ξ((1-s)q-1ξ(q-1)/2-(ξ-s)(q-1)/2)f(s,u(s))ds+ξ1(1-s)q-1ξ(q-1)/2f(s,u(s))ds}=αΓ((q+1)/2)-αΓ(q)ξ(q-1)/2{0ξ((1-s)q-1ξ(q-1)/2-(ξ-s)(q-1)/2)t-b·tbf(t,u(t))ds+ξ1(1-s)q-1ξ(q-1)/2·t-b·tbf(t,u(t))ds}>αr[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]C3×{0ξ((1-s)q-1ξ(q-1)/2-(ξ-s)(q-1)/2)·s-bds+ξ1(1-s)q-1ξ(q-1)/2·s-bds}=αr[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]C3{ξ(q-1)/201(1-s)q-1s-bds-0ξ(ξ-s)(q-1)/2s-bds}=αr[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]C3{ξ(q-1)/2B(1-b,q)-ξ1-b+(q-1)/2B(1-b,q+12)}=αΓ(1-b)ξ(q-1)/2Γ((q+1)/2)-αΓ(q)ξ(q-1)/2(Γ(q)Γ(1-b+q)-Γ((q+1)/2)ξ1-bΓ(1-b+(q+1)/2))·rC3=r. So Tuu,uΩr. By Lemma 2.8, we have i(T,Ωr,P)=0.

Step 2.

Let ΩR:={uPu<R}. For any uΩR, we have u=R and 0u(t)R for all t[0,1]. By assumption (H3), for t[0,1], we get |Tu(t)||1Γ(q)0t(t-s)q-1s-b·sbf(t,u(t))ds|+|tq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)01(1-s)q-1s-b·sbf(s,u(s))ds|+|αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/20ξ(ξ-s)(q-1)/2s-b·sbf(s,u(s))ds|<RΓ(q)(C1+C2)0t(t-s)q-1s-bds+Rtq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)(C1+C2)01(1-s)q-1s-bds+αRtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2(C1+C2)0ξ(ξ-s)(q-1)/2s-bds=RΓ(q)(C1+C2)tq-bB(1-b,q)+Rtq-1Γ((q+1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(q)(C1+C2)B(1-b,q)+αNξ(1-b+(q-1)/2)tq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2(C1+C2)B(1-b,q+12){Γ(1-b)Γ(1-b+q)+Γ((q+1)/2)Γ(1-b)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+q)+αΓ(1-b)Γ((q+1)/2)ξ(1-b+(q-1)/2)[Γ((q+1)/2)-αΓ(q)ξ(q-1)/2]Γ(1-b+(q+1)/2)}·RC1+C2=R. Therefore Tuu,uΩR. By Lemma 2.8, we have i(T,ΩR,P)=1. Combine with (3.18) and (3.21), we have i(T,ΩRΩ¯r,P)=i(T,ΩR,P)-i(T,Ωr,P)=1-0=1. Therefore, T has a fixed point uΩRΩ¯r. Then problem (1.4) has at least one positive solution u such that r<u<R.

4. Example

Let q=3/2, α=ξ=1/2. We consider the following boundary value problem: D3/2u(t)+f(t,u(t))=0,0<t<1,u(0)=0,u(1)=12D1/4u(t)|t=1/2, where f(t,u)=14t-1/2(118u2+t4+1). Let b=1/2. By simple computation, we have C11.7724,C23.9824,C30.2635. Choosing R=6, r=1/16, we have t1/2f(t,u)=14(118u2+t4+1)1<RC1+C21.0426,(t,u)[0,1]×[0,6],t1/2f(t,u)=14(118u2+t4+1)14>rC314.216,(t,u)[0,1]×[0,116]. By Theorem 3.3, problem (4.1) has at least one solution u such that 1/16<u<6.

Acknowledgments

This work was supported by the Scientific Research Foundation of Hunan Provincial Education Department (05A057, 08C826) was also supported by the Aid Program for Science and Technology Innovative Research Team in Higher Educational Institutions of Hunan Province, and the Construct Program of the Key Discipline in Hunan Province.

BabakhaniA.Daftardar-GejjiV.Existence of positive solutions of nonlinear fractional differential equationsJournal of Mathematical Analysis and Applications20032782434442MR197401710.1016/S0022-247X(02)00716-3ZBL1027.34003BaiC.-Z.FangJ.-X.The existence of a positive solution for a singular coupled system of nonlinear fractional differential equationsApplied Mathematics and Computation20041503611621MR203966210.1016/S0096-3003(03)00294-7ZBL1061.34001El-SayedA. M. A.Nonlinear functional-differential equations of arbitrary ordersNonlinear Analysis: Theory, Methods & Applications1998332181186MR1621105ZBL0934.34055Daftardar-GejjiV.BabakhaniA.Analysis of a system of fractional differential equationsJournal of Mathematical Analysis and Applications20042932511522MR205389410.1016/j.jmaa.2004.01.013ZBL1058.34002KilbasA. A.TrujilloJ. J.Differential equations of fractional order: methods, results and problems—IApplicable Analysis2001781-2153192MR188795910.1080/00036810108840931ZBL1031.34002KilbasA. A.TrujilloJ. J.Differential equations of fractional order: methods, results and problems—IIApplicable Analysis2002812435493MR192846210.1080/0003681021000022032ZBL1033.34007PodlubnyI.Fractional Differential Equations1999198San Diego, Calif, USAAcademic Pressxxiv+340Mathematics in Science and EngineeringMR1658022YuC.GaoG.Existence of fractional differential equationsJournal of Mathematical Analysis and Applications200531012629MR216067010.1016/j.jmaa.2004.12.015ZBL1088.34501ZhangS.The existence of a positive solution for a nonlinear fractional differential equationJournal of Mathematical Analysis and Applications20002522804812MR180018010.1006/jmaa.2000.7123ZBL0972.34004ZhangS.Existence of positive solution for some class of nonlinear fractional differential equationsJournal of Mathematical Analysis and Applications20032781136148MR196347010.1016/S0022-247X(02)00583-8ZBL1026.34008DelboscoD.RodinoL.Existence and uniqueness for a nonlinear fractional differential equationJournal of Mathematical Analysis and Applications19962042609625MR142146710.1006/jmaa.1996.0456ZBL0881.34005BaiZ.H.Positive solutions for boundary value problem of nonlinear fractional differential equationJournal of Mathematical Analysis and Applications20053112495505MR216841310.1016/j.jmaa.2005.02.052ZBL1079.34048DelboscoD.Fractional calculus and function spacesJournal of Fractional Calculus199464553MR1301228ZBL0829.46018ZhangS.Positive solutions for boundary-value problems of nonlinear fractional differential equationsElectronic Journal of Differential Equations200636112MR2213580ZBL1096.34016DeimlingK.Nonlinear Functional Analysis1985Berlin, GermanySpringerxiv+450MR787404