# The Existence of Positive Solution to  Three-Point Singular Boundary Value Problem of  Fractional Differential Equation

Authors: Yuansheng Tian; Anping Chen
Journal: Abstract and Applied Analysis
 (2009)

---

## Abstract

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(q−1)/2u(t)|t=ξ, where 1<q≤2 is a real number, ξ∈(0,1/2], α∈(0,+∞) and αΓ(q)ξ(q−1)/2<Γ((q+1)/2),Dq is the standard Riemann-Liouville fractional derivative, and f∈C((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.


---

## Body

## 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 [1–10] 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 [12–14]. Delbosco [13] has investigated the nonlinear Dirichlet-type problem(1.1)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 [12] and Zhang [14] have given the existence of positive solutions to the equation(1.2)Dqu(t)+f(t,u(t))=0,0<t<1,
with boundary condition
(1.3)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(1.4)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<q≤2 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:(H1)
f∈C((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 Section2, 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 orderq>0 of a function x:(0,+∞)→R is given by
(2.1)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 orderq>0 of a continuous function x:(0,+∞)→R is given by
(2.2)Dqx(t)=1Γ(n-q)(ddt)n∫0tx(s)(t-s)q-n+1ds,
where n=[q]+1, provided that the right side is pointwise defined on (0,∞).Lemma 2.3 (see [7]).
(1) If x∈L(0,1), ρ>σ>0, then DσIρx(t)=Iρ-σx(t).(2)
Ifρ>0, λ>0, then Dρtλ-1=(Γ(λ)/Γ(λ-ρ))tλ-ρ-1.Lemma 2.4 (see [12]).
Assume thatx∈C(0,1)∩L(0,1) with a fractional derivative of order q>0 that belongs to C(0,1)∩L(0,1). Then
(2.3)IqDqx(t)=x(t)+A1tq-1+A2tq-2+⋯+ANtq-N,Ai∈R, i=1,2,…,N, where N is the smallest integer greater than or equal to q.Lemma 2.5.
Ify∈C(0,1)∩L(0,1) and 1<q≤2, ξ∈(0,1), α∈ℝ satisfy that αΓ(q)ξ(q-1)/2≠Γ((q+1)/2), then the problems
(2.4)Dqu(t)+y(t)=0,0<t<1,u(0)=0,u(1)=αD(q-1)/2u(t)|t=ξ
have the unique solution
(2.5)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-α∫0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}.Proof.
By applying Lemma2.4, we may reduce (2.4) to an equivalent integral equation
(2.6)u(t)=-Iqy(t)+A1tq-1+A2tq-2
for some A1,A2∈R. Consequently the general solution of (2.4) is
(2.7)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
(2.8)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
(2.9)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
(2.10)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
(2.11)A1=Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2{∫01(1-s)q-1Γ(q)y(s)ds-α∫0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}.
So, the unique solution of problem (2.4) is
(2.12)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-α∫0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}.
The proof is completed.Lemma 2.6.
Ify∈C((0,1),[0,+∞))∩L(0,1) and 1<q≤2, 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)
(2.13)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-α∫0ξ(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
(2.14)G(t,s)={[t(1-s)]q-1-(t-s)q-1Γ(q),0≤s≤t≤1,[t(1-s)]q-1Γ(q),0≤t≤s≤1.Proof.
By Lemma2.5, the unique solution of problem (2.4) is
(2.15)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-α∫0ξ(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)/2∫01(1-s)q-1Γ(q)y(s)ds-αtq-1Γ((q+1)/2)Γ((q+1)/2)-αΓ(q)ξ(q-1)/2∫0ξ(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-1∫01(1-s)q-1Γ(q)y(s)ds-αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2∫0ξ(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)/2∫01(1-s)q-1y(s)ds-αtq-1Γ((q+1)/2)-αΓ(q)ξ(q-1)/2∫0ξ(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 ofG(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
(2.16)(1-s)2≥(1-sξ),∀s∈[0,ξ].
Hence
(2.17)(1-s)q-1≥(1-sξ)(q-1)/2.
It implies that
(2.18)(1-s)q-1ξ(q-1)/2≥(ξ-s)(q-1)/2.
Therefore u(t) is nonnegative on [0,1].Lemma 2.7.
Let1<q≤2, 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)
(2.19)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-α∫0ξ(ξ-s)(q-1)/2Γ((q+1)/2)y(s)ds}.
is continuous on [0,1].Proof.
Sincetby(t) is continuous in [0,1], thus there exists a constant L>0 such that |tby(t)|≤L for all t∈[0,1] and
(2.20)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)/2∫0ξ(ξ-s)(q-1)/2s-b·sby(s)ds.
For any t0∈[0,1], we will prove u(t)→u(t0) (t→t0, t∈[0,1]). For the convenience, the proof is divided into three cases.Case 1 (t0=0).
It is easy to know thatu(0)=0. For all t∈(0,1], we have
(2.21)|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)/2∫0ξ(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)/2∫0ξ(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-1→0(t→0),
where B denotes beta function.Case 2 (t0∈(0,1), for all t∈(t0,1]).
We have(2.22)|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-α∫0ξ(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)/2∫0ξ(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)/2∫0ξ(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(t→t0).Case 3 (t0∈(0,1], for all t∈[0,t0)).
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 [15]).Lemma 2.8.
LetE be a Banach space, P⊂E is a cone. For r>0, define Ωr={u∈P∣∥u∥<r}. Assume that T:Ω¯r→P is a completely continuous such that Tu≠u for u∈∂Ωr={u∈P∣∥u∥=r}.(1)
If∥Tu∥≥∥u∥ for u∈∂Ωr, then i(T,Ωr,P)=0.(2)
If∥Tu∥≤∥u∥ for u∈∂Ωr, then i(T,Ωr,P)=1.


## 3. Main Results

For the convenience we introduce the following notations:(3.1)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.
If1<q≤2, 0<b<1, 0<ξ≤1/2, α>0, and Γ((q+1)/2)>αΓ(q)ξ(q-1)/2, then
(3.2)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.
LetE=C[0,1] be a Banach spaces with the maximum norm ∥u∥=max0≤t≤1|u(t)|. Define the cone P⊂E by
(3.3)P={u∈E∣u(t)≥0,0≤t≤1}.
The positive solution which we consider in this paper is the formu(0)=0, u(t)>0, 0<t≤1, u∈E.
Define an operatorT:P→E by
(3.4)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-α∫0ξ(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:P→P is completely continuous.Proof.
If condition(H1) holds, by Lemmas 2.6 and 2.7, we have T(P)⊂P. Let u0∈P and ∥u0∥=a0; if u∈P 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
(3.5)|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-u0∥≤δ, then u0(t),u(t)∈[0,a] and |u(t)-u0(t)|<δ for each t∈[0,1]. Hence, we have
(3.6)|tbf(t,u(t))-tbf(t,u0(t))|<εC1+C2
for all t∈[0,1],u∈P with ∥u-u0∥<δ. It follows from (3.6) that
(3.7)∥Tu-Tu0∥=max0≤t≤1|Tu(t)-Tu0(t)|≤max0≤t≤1{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}<max0≤t≤1{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}=max0≤t≤1{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:P→P is continuous.
LetΩ⊂P be bounded; that is, there exists a positive constant M such that Ω∈{u∈P∣∥u∥≤M}. Since tbf(t,u(t)) is continuous on [0,1]×[0,+∞), we let
(3.8)N=max(t,u)∈[0,1]×[0,M]tbf(t,u(t))+1.
For all u∈Ω, we have
(3.9)|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)/2∫0ξ(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)/2∫0ξ(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, givenε>0, set
(3.10)δ=min{12(ε4NC1)1/(q-b),12(ε2NC2)1/(q-1),ε4NC1}.
For each u∈Ω, we will prove that if t1,t2∈[0,1] and 0<t2-t1<δ, then
(3.11)|Tu(t2)-Tu(t1)|<ε.
In fact, similar to the proof of Lemma 2.7, we have
(3.12)|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 (δ≤t1<t2<1, q-b-1≤0).
(3.13)|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-1≤2NC1·ε4NC12q-b+NC2·ε2NC22q-1=ε2×2q-b+ε2×2q-1<ε.Case 2 (δ≤t1<t2<1, q-b-1>0).
(3.14)|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-1≤2NC1·ε4NC1+NC2·ε2NC22q-1=ε2+ε2×2q-1<ε.Case 3 (0≤t1<δ, t2<2δ).
(3.15)|Tu(t2)-Tu(t1)|≤NC1(t2q-b-t1q-b)+NC2(t2q-1-t1q-1)≤NC1t2q-b+NC2t2q-1≤NC1(2δ)q-b+NC2(2δ)q-1≤NC1·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:P→P 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(H2)
tbf(t,u)>r/C3, for (t,u)∈[0,1]×[0,r];(H3)
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.
LetΩr:={u∈P∣∥u∥<r}. For any u∈∂Ωr, we have ∥u∥=r and 0≤u(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
(3.16)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)/2∫01(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
(3.17)∥Tu∥≥∥u∥,∀u∈∂Ωr.
By Lemma 2.8, we have
(3.18)i(T,Ωr,P)=0.Step 2.
LetΩR:={u∈P∣∥u∥<R}. For any u∈∂ΩR, we have ∥u∥=R and 0≤u(t)≤R for all t∈[0,1]. By assumption (H3), for t∈[0,1], we get
(3.19)|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)/2∫0ξ(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
(3.20)∥Tu∥≤∥u∥,∀u∈∂ΩR.
By Lemma 2.8, we have
(3.21)i(T,ΩR,P)=1.
Combine with (3.18) and (3.21), we have
(3.22)i(T,ΩR∖Ω¯r,P)=i(T,ΩR,P)-i(T,Ωr,P)=1-0=1.
Therefore, T has a fixed point u∈ΩR∖Ω¯r. Then problem (1.4) has at least one positive solution u such that r<∥u∥<R.


## 4. Example

Letq=3/2, α=ξ=1/2. We consider the following boundary value problem:
(4.1)D3/2u(t)+f(t,u(t))=0,0<t<1,u(0)=0,u(1)=12D1/4u(t)|t=1/2,
where
(4.2)f(t,u)=14t-1/2(118u2+t4+1).
Let b=1/2. By simple computation, we have
(4.3)C1≈1.7724,C2≈3.9824,C3≈0.2635.
Choosing R=6, r=1/16, we have
(4.4)t1/2f(t,u)=14(118u2+t4+1)≤1<RC1+C2≈1.0426,(t,u)∈[0,1]×[0,6],t1/2f(t,u)=14(118u2+t4+1)≥14>rC3≈14.216,(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.



---

*Source: 314656-2009-07-15.xml*