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

Abstract

We investigate the existence of positive solution to nonlinear fractional differential equation three-point singular boundary value problem: 𝐷^𝑞𝑢(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),𝐷^𝑞 is the standard Riemann-Liouville fractional derivative, and 𝑓∈𝐶((0,1]×[0,+∞),[0,+∞)),lim_𝑡→+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 [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

𝑢^𝑞−1𝐷^𝑞𝑢(𝑡)=𝑓(𝑢(𝑡)),0<𝑡<1,1<𝑞<2,𝑢(0)=𝑢(1)=0. (1.1) 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

𝐷^𝑞𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1, (1.2) with boundary condition 𝑢(0)=𝑢(1)=0,𝑢(0)+𝑢^′(0)=𝑢(1)+𝑢^′(1), (1.3) respectively.

This paper is to study the existence of positive solution for the three-point singular boundary value problem of nonlinear fractional differential equation 𝐷^𝑞𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑢(0)=0,𝑢(1)=𝛼𝐷^(𝑞−1)/2𝑢(𝑡)|_𝑡=𝜉. (1.4) 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), 𝐷^𝑞 is the standard Riemann-Liouville fractional derivative, and the function 𝑓 satisfies the following condition:

(𝐻₁)𝑓∈𝐶((0,1]×[0,+∞),[0,+∞)), lim_𝑡→+0𝑓(𝑡,⋅)=+∞, there exists a constant 𝑏∶0<𝑏<1 such that 𝑡^𝑏𝑓(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 𝐼^𝑞1𝑥(𝑡)=Γ(𝑞)^𝑡₀(𝑡−𝑠)^𝑞−1𝑥(𝑠)𝑑𝑠, (2.1) 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 𝐷^𝑞1𝑥(𝑡)=(𝑑Γ(𝑛−𝑞))𝑑𝑡^𝑛𝑡₀𝑥(𝑠)(𝑡−𝑠)^{𝑞−𝑛+1}𝑑𝑠, (2.2) 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 𝐷^𝜎𝐼^𝜌𝑥(𝑡)=𝐼^𝜌−𝜎𝑥(𝑡).(2)If 𝜌>0, 𝜆>0, then 𝐷^𝜌𝑡^𝜆−1=(Γ(𝜆)/Γ(𝜆−𝜌))𝑡^{𝜆−𝜌−1}.

Lemma 2.4 (see [12]). Assume that 𝑥∈𝐶(0,1)∩𝐿(0,1) with a fractional derivative of order 𝑞>0 that belongs to 𝐶(0,1)∩𝐿(0,1). Then 𝐼^𝑞𝐷^𝑞𝑥(𝑡)=𝑥(𝑡)+𝐴₁𝑡^𝑞−1+𝐴₂𝑡^𝑞−2+⋯+𝐴_𝑁𝑡^𝑞−𝑁, (2.3)𝐴_𝑖∈𝑅, 𝑖=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝑞.

Lemma 2.5. If 𝑦∈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 𝐷^𝑞𝑢(𝑡)+𝑦(𝑡)=0,0<𝑡<1,𝑢(0)=0,𝑢(1)=𝛼𝐷^(𝑞−1)/2𝑢(𝑡)|_𝑡=𝜉 (2.4) have the unique solution 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝑞−1+𝑡Γ(𝑞)𝑦(𝑠)𝑑𝑠_𝑞−1Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{¹₀(1−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠−𝛼^𝜉₀(𝜉−𝑠)^(𝑞−1)/2Γ((𝑞+1)/2)𝑦(𝑠)𝑑𝑠}. (2.5)

Proof. By applying Lemma 2.4, we may reduce (2.4) to an equivalent integral equation 𝑢(𝑡)=−𝐼^𝑞𝑦(𝑡)+𝐴₁𝑡^𝑞−1+𝐴₂𝑡^𝑞−2 (2.6) for some 𝐴₁,𝐴₂∈𝑅. Consequently the general solution of (2.4) is 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝑞−1𝑦(𝑠)Γ(𝑞)𝑑𝑠+𝐴₁𝑡^𝑞−1+𝐴₂𝑡^𝑞−2. (2.7) Note that 𝑢(0)=0, we have 𝐴₂=0 and 𝑢(1)=−¹₀(1−𝑠)^𝑞−1𝑦(𝑠)Γ(𝑞)𝑑𝑠+𝐴₁. (2.8) On the other hand, by (2.6) and Lemma 2.3, we have 𝐷^(𝑞−1)/2𝑢(𝑡)=−𝐷^(𝑞−1)/2𝐼^𝑞𝑦(𝑡)+𝐴₁𝐷^(𝑞−1)/2𝑡^𝑞−1=−𝐼^(𝑞+1)/2𝑦(𝑡)+𝐴₁Γ(𝑞)𝑡Γ((𝑞+1)/2)^(𝑞−1)/2. (2.9) Therefore 𝐷^(𝑞−1)/2𝑢(𝑡)|_𝑡=𝜉=−^𝜉₀(𝜉−𝑠)^(𝑞−1)/2Γ((𝑞+1)/2)𝑦(𝑠)𝑑𝑠+𝐴₁Γ(𝑞)𝜉Γ((𝑞+1)/2)^(𝑞−1)/2. (2.10) By 𝑢(1)=𝛼𝐷^𝑝𝑢(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{¹₀(1−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠−𝛼^𝜉₀(𝜉−𝑠)^(𝑞−1)/2Γ((𝑞+1)/2)𝑦(𝑠)𝑑𝑠}. (2.11) So, the unique solution of problem (2.4) is 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝑞−1+𝑡Γ(𝑞)𝑦(𝑠)𝑑𝑠_𝑞−1Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{¹₀(1−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠−𝛼^𝜉₀(𝜉−𝑠)^(𝑞−1)/2Γ((𝑞+1)/2)𝑦(𝑠)𝑑𝑠}. (2.12) The proof is completed.

Lemma 2.6. If 𝑦∈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) 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝑞−1+𝑡Γ(𝑞)𝑦(𝑠)𝑑𝑠_𝑞−1Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{¹₀(1−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠−𝛼^𝜉₀(𝜉−𝑠)^(𝑞−1)/2=Γ((𝑞+1)/2)𝑦(𝑠)𝑑𝑠}₁₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠+𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2×{^𝜉₀((1−𝑠)^𝑞−1𝜉^(𝑞−1)/2−(𝜉−𝑠)^(𝑞−1)/2)𝑦(𝑠)𝑑𝑠+¹_𝜉(1−𝑠)^𝑞−1𝜉^(𝑞−1)/2𝑦(𝑠)𝑑𝑠}. (2.13) is nonnegative on [0,1], where 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. (2.14)

Proof. By Lemma 2.5, the unique solution of problem (2.4) is 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝑞−1+𝑡Γ(𝑞)𝑦(𝑠)𝑑𝑠_𝑞−1Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{¹₀(1−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠−𝛼^𝜉₀(𝜉−𝑠)^(𝑞−1)/2Γ((𝑞+1)/2)𝑦(𝑠)𝑑𝑠}=−^𝑡₀(𝑡−𝑠)^𝑞−1+𝑡Γ(𝑞)𝑦(𝑠)𝑑𝑠_𝑞−1Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/210}(1−𝑠)^𝑞−1−Γ(𝑞)𝑦(𝑠)𝑑𝑠𝛼𝑡^𝑞−1Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2Γ((𝑞+1)/2)𝑦(𝑠)𝑑𝑠=−^𝑡₀(𝑡−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠+(1+𝛼Γ(𝑞)𝜉^(𝑞−1)/2Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2)𝑡^𝑞−11₀(1−𝑠)^𝑞−1−Γ(𝑞)𝑦(𝑠)𝑑𝑠𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2𝑦(𝑠)𝑑𝑠=−^𝑡₀(𝑡−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠+¹₀(1−𝑠)^𝑞−1𝑡^𝑞−1+Γ(𝑞)𝑦(𝑠)𝑑𝑠𝛼𝑡^𝑞−1𝜉^(𝑞−1)/2Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/210}(1−𝑠)^𝑞−1−𝑦(𝑠)𝑑𝑠𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2=𝑦(𝑠)𝑑𝑠_𝑡0[𝑡(1−𝑠)]^𝑞−1−(𝑡−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠+¹_𝑡[𝑡(1−𝑠)]^𝑞−1+Γ(𝑞)𝑦(𝑠)𝑑𝑠𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{¹₀(1−𝑠)^𝑞−1𝜉^(𝑞−1)/2𝑦(𝑠)𝑑𝑠−^𝜉₀(𝜉−𝑠)^(𝑞−1)/2=𝑦(𝑠)𝑑𝑠}₁₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠+𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2×{^𝜉₀((1−𝑠)^𝑞−1𝜉^(𝑞−1)/2−(𝜉−𝑠)^(𝑞−1)/2)𝑦(𝑠)𝑑𝑠+¹_𝜉(1−𝑠)^𝑞−1𝜉^(𝑞−1)/2𝑦(𝑠)𝑑𝑠}. (2.15)(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)
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)²𝑠≥(1−𝜉),∀𝑠∈[0,𝜉]. (2.16) Hence (1−𝑠)^𝑞−1𝑠≥(1−𝜉)^(𝑞−1)/2. (2.17) It implies that (1−𝑠)^𝑞−1𝜉^(𝑞−1)/2≥(𝜉−𝑠)^(𝑞−1)/2. (2.18) Therefore 𝑢(𝑡) is nonnegative on [0,1].

Lemma 2.7. Let 1<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 lim_𝑡→+0𝑦(𝑡)=+∞. Suppose that there exists a constant 𝑏∶0<𝑏<1 such that 𝑡^𝑏𝑦(𝑡) is continuous function on [0,1]. Then the unique solution of (2.4) 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝑞−1+𝑡Γ(𝑞)𝑦(𝑠)𝑑𝑠_𝑞−1Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{¹₀(1−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠−𝛼^𝜉₀(𝜉−𝑠)^(𝑞−1)/2Γ((𝑞+1)/2)𝑦(𝑠)𝑑𝑠}. (2.19) is continuous on [0,1].

Proof. Since 𝑡^𝑏𝑦(𝑡) is continuous in [0,1], thus there exists a constant 𝐿>0 such that |𝑡^𝑏𝑦(𝑡)|≤𝐿 for all 𝑡∈[0,1] and 1𝑢(𝑡)=−Γ(𝑞)^𝑡₀(𝑡−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏+𝑡𝑦(𝑠)𝑑𝑠_𝑞−1Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏−𝑦(𝑠)𝑑𝑠𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏⋅𝑠^𝑏𝑦(𝑠)𝑑𝑠. (2.20) 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 that 𝑢(0)=0. For all 𝑡∈(0,1], we have |𝑢(𝑡)−𝑢(𝑡₀1)|≤|Γ(𝑞)^𝑡₀(𝑡−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏𝑡𝑦(𝑠)𝑑𝑠|+|^𝑞−1Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏𝑦(𝑠)𝑑𝑠|+|𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏⋅𝑠^𝑏≤𝐿𝑦(𝑠)𝑑𝑠|Γ(𝑞)^𝑡₀(𝑡−𝑠)^𝑞−1𝑠^−𝑏+𝑑𝑠𝐿𝑡^𝑞−1Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏+𝑑𝑠𝛼𝐿𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏=𝑑𝑠𝐿𝑡^𝑞−𝑏Γ(𝑞)𝐵(1−𝑏,𝑞)+𝐿𝑡^𝑞−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)}𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2=𝐵(1−𝑏,(𝑞+1)/2)𝐿Γ(1−𝑏)𝑡Γ(1−𝑏+𝑞)^𝑞−𝑏+𝐿Γ(1−𝑏)Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2𝑡]Γ(1−𝑏+𝑞)^𝑞−1+𝛼𝐿Γ(1−𝑏)Γ((𝑞+1)/2)𝜉^{(1−𝑏+(𝑞−1)/2)}[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2𝑡]Γ(1−𝑏+(𝑞+1)/2)^𝑞−1→0(𝑡→0), (2.21)(15)(16)(17)(18)(19)(20)(21)(22)(23) where 𝐵 denotes beta function.Case 2 (𝑡₀∈(0,1), for all 𝑡∈(𝑡₀,1]). We have |𝑢(𝑡)−𝑢(𝑡₀1)|=|−Γ(𝑞)^𝑡₀(𝑡−𝑠)^𝑞−11𝑦(𝑠)𝑑𝑠+Γ(𝑞)^𝑡0₀(𝑡₀−𝑠)^𝑞−1+𝑦(𝑠)𝑑𝑠(𝑡^𝑞−1−𝑡₀^𝑞−1)Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{¹₀(1−𝑠)^𝑞−1Γ(𝑞)𝑦(𝑠)𝑑𝑠−𝛼^𝜉₀(𝜉−𝑠)^(𝑞−1)/21Γ((𝑞+1)/2)𝑦(𝑠)𝑑𝑠}|≤|Γ(𝑞)^𝑡0₀((𝑡−𝑠)^𝑞−1−(𝑡₀−𝑠)^𝑞−1)𝑠^−𝑏⋅𝑠^𝑏1𝑦(𝑠)𝑑𝑠+Γ(𝑞)^𝑡_𝑡0(𝑡−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏𝑦(𝑠)𝑑𝑠|+|(𝑡^𝑞−1−𝑡₀^𝑞−1)Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏𝑦(𝑠)𝑑𝑠|+|𝛼(𝑡^𝑞−1−𝑡₀^𝑞−1)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏⋅𝑠^𝑏≤𝐿𝑦(𝑠)𝑑𝑠|Γ(𝑞)^𝑡0₀((𝑡−𝑠)^𝑞−1−(𝑡₀−𝑠)^𝑞−1)𝑠^−𝑏𝐿𝑑𝑠+Γ(𝑞)^𝑡_𝑡0(𝑡−𝑠)^𝑞−1𝑠^−𝑏+𝑑𝑠(𝑡^𝑞−1−𝑡₀^𝑞−1)Γ((𝑞+1)/2)𝐿[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏+𝑑𝑠𝛼(𝑡^𝑞−1−𝑡₀^𝑞−1)𝐿Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏=𝐿𝑑𝑠Γ(𝑞)(𝑡^𝑞−𝑏−𝑡₀^𝑞−𝑏)𝐵(1−𝑏,𝑞)+(𝑡^𝑞−1−𝑡₀^𝑞−1)Γ((𝑞+1)/2)𝐿[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2+]Γ(𝑞)𝐵(1−𝑏,𝑞)𝛼(𝑡^𝑞−1−𝑡₀^𝑞−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)(𝑡^𝑞−𝑏−𝑡₀^𝑞−𝑏)+𝐿Γ(1−𝑏)Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(1−𝑏+𝑞)(𝑡^𝑞−1−𝑡₀^𝑞−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)(𝑡^𝑞−1−𝑡₀^𝑞−1)→0(𝑡→𝑡₀). (2.22)(24)(25)(26)(27)(28)(29)(30)(31)(32)(33)(34)(35)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. Let 𝐸 be a Banach space, 𝑃⊂𝐸 is a cone. For 𝑟>0, define Ω_𝑟={𝑢∈𝑃∣‖𝑢‖<𝑟}. Assume that 𝑇∶Ω_𝑟→𝑃 is a completely continuous such that 𝑇𝑢≠𝑢 for 𝑢∈𝜕Ω_𝑟={𝑢∈𝑃∣‖𝑢‖=𝑟}.(1)If ‖𝑇𝑢‖≥‖𝑢‖ for 𝑢∈𝜕Ω_𝑟, then 𝑖(𝑇,Ω_𝑟,𝑃)=0.(2)If ‖𝑇𝑢‖≤‖𝑢‖ for 𝑢∈𝜕Ω_𝑟, then 𝑖(𝑇,Ω_𝑟,𝑃)=1.

3. Main Results

For the convenience we introduce the following notations: 𝐶₁=Γ(1−𝑏)𝐶Γ(1−𝑏+𝑞)₂,=Γ(1−𝑏)Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2(1+Γ(1−𝑏+𝑞)𝛼𝜉^{(1−𝑏+(𝑞−1)/2)}𝐶Γ(1−𝑏+(𝑞+1)/2)),₃=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)𝜉^1−𝑏Γ(1−𝑏+(𝑞+1)/2)). (3.1)(36)(37)

Remark 3.1. If 1<𝑞≤2, 0<𝑏<1, 0<𝜉≤1/2, 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]{^𝜉₀((1−𝑠)^𝑞−1𝜉^(𝑞−1)/2−(𝜉−𝑠)^(𝑞−1)/2)⋅𝑠^−𝑏+𝑑𝑠_1𝜉(1−𝑠)^𝑞−1𝜉^(𝑞−1)/2⋅𝑠^−𝑏=𝑑𝑠}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)𝜉^1−𝑏Γ(1−𝑏+(𝑞+1)/2))>0. (3.2)
Let 𝐸=𝐶[0,1] be a Banach spaces with the maximum norm ‖𝑢‖=max_0≤𝑡≤1|𝑢(𝑡)|. Define the cone 𝑃⊂𝐸 by 𝑃={𝑢∈𝐸∣𝑢(𝑡)≥0,0≤𝑡≤1}. (3.3)
The positive solution which we consider in this paper is the form 𝑢(0)=0, 𝑢(𝑡)>0, 0<𝑡≤1, 𝑢∈𝐸.
Define an operator 𝑇∶𝑃→𝐸 by 𝑇𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝑞−1+𝑡Γ(𝑞)𝑓(𝑠,𝑢(𝑠))𝑑𝑠_𝑞−1Γ((𝑞+1)/2)Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{¹₀(1−𝑠)^𝑞−1Γ(𝑞)𝑓(𝑠,𝑢(𝑠))𝑑𝑠−𝛼^𝜉₀(𝜉−𝑠)^(𝑞−1)/2Γ((𝑞+1)/2)𝑓(𝑠,𝑢(𝑠))𝑑𝑠}. (3.4)

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 𝑡^𝑏𝑓(𝑡,𝑥), we know that 𝑡^𝑏𝑓(𝑡,𝑥) is uniformly continuous on [0,1]×[0,𝑎]. Thus, for all 𝜀>0, there exists a 𝛿>0 (𝛿<1) such that |𝑡^𝑏𝑓(𝑡,𝑥₁)−𝑡^𝑏𝑓(𝑡,𝑥₂𝜀)|<𝐶₁+𝐶₂ (3.5) 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 |𝑡^𝑏𝑓(𝑡,𝑢(𝑡))−𝑡^𝑏𝑓(𝑡,𝑢₀𝜀(𝑡))|<𝐶₁+𝐶₂ (3.6) for all 𝑡∈[0,1],𝑢∈𝑃 with ‖𝑢−𝑢₀‖<𝛿. It follows from (3.6) that ‖𝑇𝑢−𝑇𝑢₀‖=max_0≤𝑡≤1|𝑇𝑢(𝑡)−𝑇𝑢₀(𝑡)|≤max_0≤𝑡≤1{1Γ(𝑞)^𝑡₀(𝑡−𝑠)^𝑞−1𝑠^−𝑏|𝑠^𝑏𝑓(𝑠,𝑢(𝑠))−𝑠^𝑏𝑓(𝑠,𝑢₀+𝑡(𝑠))|𝑑𝑠_𝑞−1Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)×¹₀(1−𝑠)^𝑞−1𝑠^−𝑏|𝑠^𝑏𝑓(𝑠,𝑢(𝑠))−𝑠^𝑏𝑓(𝑠,𝑢₀+(𝑠))|𝑑𝑠𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2×^𝜉₀(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏|𝑠^𝑏𝑓(𝑠,𝑢(𝑠))−𝑠^𝑏𝑓(𝑠,𝑢₀(𝑠))|𝑑𝑠}<max_0≤𝑡≤1{𝜀Γ(𝑞)(𝐶₁+𝐶₂)^𝑡₀(𝑡−𝑠)^𝑞−1𝑠^−𝑏+𝑑𝑠𝜀𝑡^𝑞−1Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2](𝐶₁+𝐶₂)Γ(𝑞)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏+𝑑𝑠𝛼𝜀𝑡^𝑞−1[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2](𝐶₁+𝐶₂)^𝜉0(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏𝑑𝑠}=max_0≤𝑡≤1{𝜀Γ(𝑞)(𝐶₁+𝐶₂)𝑡^𝑞−𝑏𝐵(1−𝑏,𝑞)+𝜀𝑡^𝑞−1Γ((𝑞+1)/2)𝐵(1−𝑏,𝑞)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2](𝐶₁+𝐶₂+)Γ(𝑞)𝛼𝜀𝑡^𝑞−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}𝜀]Γ(1−𝑏+(𝑞+1)/2)(𝐶₁+𝐶₂)=𝜀. (3.7) By the arbitrariness of 𝑢₀, we have that 𝑇∶𝑃→𝑃 is continuous.
Let Ω⊂𝑃 be bounded; that is, there exists a positive constant 𝑀 such that Ω∈{𝑢∈𝑃∣‖𝑢‖≤𝑀}. Since 𝑡^𝑏𝑓(𝑡,𝑢(𝑡)) is continuous on [0,1]×[0,+∞), we let 𝑁=max_{(𝑡,𝑢)∈[0,1]×[0,𝑀]}𝑡^𝑏𝑓(𝑡,𝑢(𝑡))+1. (3.8) For all 𝑢∈Ω, we have 1|𝑇𝑢(𝑡)|≤|Γ(𝑞)^𝑡₀(𝑡−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏𝑡𝑓(𝑡,𝑢(𝑡))𝑑𝑠|+|^𝑞−1Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏𝑓(𝑠,𝑢(𝑠))𝑑𝑠|+|𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏⋅𝑠^𝑏≤𝑁𝑓(𝑠,𝑢(𝑠))𝑑𝑠|Γ(𝑞)^𝑡₀(𝑡−𝑠)^𝑞−1𝑠^−𝑏𝑑𝑠+𝑁𝑡^𝑞−1Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏+𝑑𝑠𝛼𝑁𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏=𝑁𝑑𝑠𝑡Γ(𝑞)^𝑞−𝑏𝐵(1−𝑏,𝑞)+𝑁𝑡^𝑞−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)}𝑡^𝑞−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. (3.9) Hence 𝑇(Ω) is bounded.
On the other hand, given 𝜀>0, set 1𝛿=min{2(𝜀4𝑁𝐶₁)^{1/(𝑞−𝑏)},12(𝜀2𝑁𝐶₂)^1/(𝑞−1),𝜀4𝑁𝐶₁}. (3.10) For each 𝑢∈Ω, we will prove that if 𝑡₁,𝑡₂∈[0,1] and 0<𝑡₂−𝑡₁<𝛿, then |𝑇𝑢(𝑡₂)−𝑇𝑢(𝑡₁)|<𝜀. (3.11) In fact, similar to the proof of Lemma 2.7, we have |𝑇𝑢(𝑡₂)−𝑇𝑢(𝑡₁)|≤𝑁Γ(1−𝑏)Γ(1−𝑏+𝑞)(𝑡₂^𝑞−𝑏−𝑡₁^𝑞−𝑏)+𝑁Γ((𝑞+1)/2)Γ(1−𝑏)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(1−𝑏+𝑞)(𝑡₂^𝑞−1−𝑡₁^𝑞−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)(𝑡₂^𝑞−1−𝑡₁^𝑞−1)=𝑁𝐶₁(𝑡₂^𝑞−𝑏−𝑡₁^𝑞−𝑏)+𝑁𝐶₂(𝑡₂^𝑞−1−𝑡₁^𝑞−1). (3.12) In the following, the proof is divided into three cases. Case 1 (𝛿≤𝑡₁<𝑡₂<1, 𝑞−𝑏−1≤0). |𝑇𝑢(𝑡₂)−𝑇𝑢(𝑡₁)|≤𝑁𝐶₁(𝑡₂^𝑞−𝑏−𝑡₁^𝑞−𝑏)+𝑁𝐶₂(𝑡₂^𝑞−1−𝑡₁^𝑞−1)≤𝑁𝐶₁(𝑞−𝑏)𝛿^{𝑞−𝑏−1}(𝑡₂−𝑡₁)+𝑁𝐶₂(𝑞−1)𝛿^𝑞−2(𝑡₂−𝑡₁)<2𝑁𝐶₁𝛿^𝑞−𝑏+𝑁𝐶₂𝛿^𝑞−1≤2𝑁𝐶₁⋅𝜀4𝑁𝐶₁2^𝑞−𝑏+𝑁𝐶₂⋅𝜀2𝑁𝐶₂2^𝑞−1=𝜀2×2^𝑞−𝑏+𝜀2×2^𝑞−1<𝜀. (3.13)Case 2 (𝛿≤𝑡₁<𝑡₂<1, 𝑞−𝑏−1>0). |𝑇𝑢(𝑡₂)−𝑇𝑢(𝑡₁)|≤𝑁𝐶₁(𝑡₂^𝑞−𝑏−𝑡₁^𝑞−𝑏)+𝑁𝐶₂(𝑡₂^𝑞−1−𝑡₁^𝑞−1)≤𝑁𝐶₁(𝑞−𝑏)(𝑡₂−𝑡₁)+𝑁𝐶₂(𝑞−1)𝛿^𝑞−2(𝑡₂−𝑡₁)<2𝑁𝐶₁𝛿+𝑁𝐶₂𝛿^𝑞−1≤2𝑁𝐶₁⋅𝜀4𝑁𝐶₁+𝑁𝐶₂⋅𝜀2𝑁𝐶₂2^𝑞−1=𝜀2+𝜀2×2^𝑞−1<𝜀. (3.14)Case 3 (0≤𝑡₁<𝛿, 𝑡₂<2𝛿). |𝑇𝑢(𝑡₂)−𝑇𝑢(𝑡₁)|≤𝑁𝐶₁(𝑡₂^𝑞−𝑏−𝑡₁^𝑞−𝑏)+𝑁𝐶₂(𝑡₂^𝑞−1−𝑡₁^𝑞−1)≤𝑁𝐶₁𝑡₂^𝑞−𝑏+𝑁𝐶₂𝑡₂^𝑞−1≤𝑁𝐶₁(2𝛿)^𝑞−𝑏+𝑁𝐶₂(2𝛿)^𝑞−1≤𝑁𝐶₁⋅𝜀4𝑁𝐶₁+𝑁𝐶₂⋅𝜀2𝑁𝐶₂=𝜀4+𝜀2<𝜀. (3.15)
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)𝑡^𝑏𝑓(𝑡,𝑢)>𝑟/𝐶₃, for (𝑡,𝑢)∈[0,1]×[0,𝑟];(𝐻₃)𝑡^𝑏𝑓(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 Ω_𝑟∶={𝑢∈𝑃∣‖𝑢‖<𝑟}. For any 𝑢∈𝜕Ω_𝑟, 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 𝑇𝑢(1)=¹₀+𝛼𝐺(1,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{^𝜉₀((1−𝑠)^𝑞−1𝜉^(𝑞−1)/2−(𝜉−𝑠)^(𝑞−1)/2)𝑓(𝑠,𝑢(𝑠))𝑑𝑠+¹_𝜉(1−𝑠)^𝑞−1𝜉^(𝑞−1)/2=𝛼𝑓(𝑠,𝑢(𝑠))𝑑𝑠}Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2{^𝜉₀((1−𝑠)^𝑞−1𝜉^(𝑞−1)/2−(𝜉−𝑠)^(𝑞−1)/2)𝑡^−𝑏⋅𝑡^𝑏+𝑓(𝑡,𝑢(𝑡))𝑑𝑠_1𝜉(1−𝑠)^𝑞−1𝜉^(𝑞−1)/2⋅𝑡^−𝑏⋅𝑡^𝑏>𝑓(𝑡,𝑢(𝑡))𝑑𝑠}𝛼𝑟[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]𝐶₃×{^𝜉₀((1−𝑠)^𝑞−1𝜉^(𝑞−1)/2−(𝜉−𝑠)^(𝑞−1)/2)⋅𝑠^−𝑏𝑑𝑠+¹_𝜉(1−𝑠)^𝑞−1𝜉^(𝑞−1)/2⋅𝑠^−𝑏=𝑑𝑠}𝛼𝑟[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]𝐶₃{𝜉^{(𝑞−1)/21}₀(1−𝑠)^𝑞−1𝑠^−𝑏𝑑𝑠−^𝜉₀(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏=𝑑𝑠}𝛼𝑟[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]𝐶₃{𝜉^(𝑞−1)/2𝐵(1−𝑏,𝑞)−𝜉^{1−𝑏+(𝑞−1)/2}𝐵(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)𝜉^1−𝑏𝑟Γ(1−𝑏+(𝑞+1)/2))⋅𝐶₃=𝑟. (3.16)(38)(39)(40)(41)(42)(43)(44)(45)(46) So ‖𝑇𝑢‖≥‖𝑢‖,∀𝑢∈𝜕Ω_𝑟. (3.17) By Lemma 2.8, we have 𝑖(𝑇,Ω_𝑟,𝑃)=0. (3.18)Step 2. Let Ω_𝑅∶={𝑢∈𝑃∣‖𝑢‖<𝑅}. For any 𝑢∈𝜕Ω_𝑅, we have ‖𝑢‖=𝑅 and 0≤𝑢(𝑡)≤𝑅 for all 𝑡∈[0,1]. By assumption (𝐻₃), for 𝑡∈[0,1], we get 1|𝑇𝑢(𝑡)|≤|Γ(𝑞)^𝑡₀(𝑡−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏𝑡𝑓(𝑡,𝑢(𝑡))𝑑𝑠|+|^𝑞−1Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏⋅𝑠^𝑏𝑓(𝑠,𝑢(𝑠))𝑑𝑠|+|𝛼𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^{(𝑞−1)/2𝜉0}(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏⋅𝑠^𝑏<𝑅𝑓(𝑠,𝑢(𝑠))𝑑𝑠|Γ(𝑞)(𝐶₁+𝐶₂)^𝑡₀(𝑡−𝑠)^𝑞−1𝑠^−𝑏+𝑑𝑠𝑅𝑡^𝑞−1Γ((𝑞+1)/2)[Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2]Γ(𝑞)(𝐶₁+𝐶₂)¹⁰(1−𝑠)^𝑞−1𝑠^−𝑏+𝑑𝑠𝛼𝑅𝑡^𝑞−1Γ((𝑞+1)/2)−𝛼Γ(𝑞)𝜉^(𝑞−1)/2(𝐶₁+𝐶₂)^𝜉0(𝜉−𝑠)^(𝑞−1)/2𝑠^−𝑏=𝑅𝑑𝑠Γ(𝑞)(𝐶₁+𝐶₂)𝑡^𝑞−𝑏+𝐵(1−𝑏,𝑞)𝑅𝑡^𝑞−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)}𝑡^𝑞−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)/2𝑅]Γ(1−𝑏+(𝑞+1)/2)}⋅𝐶₁+𝐶₂=𝑅. (3.19)(47)(48)(49)(50)(51)(52)(53)(54)(55)(56) Therefore ‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝜕Ω_𝑅. (3.20) By Lemma 2.8, we have 𝑖(𝑇,Ω_𝑅,𝑃)=1. (3.21) Combine with (3.18) and (3.21), we have 𝑖(𝑇,Ω_𝑅⧵Ω_𝑟,𝑃)=𝑖(𝑇,Ω_𝑅,𝑃)−𝑖(𝑇,Ω_𝑟,𝑃)=1−0=1. (3.22) Therefore, 𝑇 has a fixed point 𝑢∈Ω_𝑅⧵Ω_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: 𝐷^3/21𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑢(0)=0,𝑢(1)=2𝐷^1/4𝑢(𝑡)|_𝑡=1/2, (4.1) where 1𝑓(𝑡,𝑢)=4𝑡^−1/2(1𝑢18²+𝑡⁴+1). (4.2) Let 𝑏=1/2. By simple computation, we have 𝐶₁≈1.7724,𝐶₂≈3.9824,𝐶₃≈0.2635. (4.3) Choosing 𝑅=6, 𝑟=1/16, we have 𝑡^1/21𝑓(𝑡,𝑢)=4(1𝑢18²+𝑡⁴𝑅+1)≤1<𝐶₁+𝐶₂𝑡≈1.0426,(𝑡,𝑢)∈[0,1]×[0,6],_1/21𝑓(𝑡,𝑢)=4(1𝑢18²+𝑡⁴1+1)≥4>𝑟𝐶₃≈114.216,(𝑡,𝑢)∈[0,1]×[0,16]. (4.4) 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.