Existence of Positive Solutions for a Kind of Fractional Boundary Value Problems

Abstract

We are concerned with the following nonlinear three-point fractional boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)+𝜆𝑎(𝑡)𝑓(𝑡,𝑢(𝑡))=0, 0<𝑡<1, 𝑢(0)=0, and 𝑢(1)=𝛽𝑢(𝜂), where 1<𝛼≤2, 0<𝛽<1, 0<𝜂<1, 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative, 𝑎(𝑡)>0 is continuous for 0≤𝑡≤1, and 𝑓≥0 is continuous on [0,1]×[0,∞). By using Krasnoesel'skii's fixed-point theorem and the corresponding Green function, we obtain some results for the existence of positive solutions. At the end of this paper, we give an example to illustrate our main results.

1. Introduction

Fractional differential equations have been of great interest recently. With the development of nonlinear science, the researchers found that nonlinear fractional differential equations could describe something's changing rules more accurately. Therefore, it is significant to study nonlinear fractional differential equations to solve the nonlinear problems. Recently, many researchers paid attention to existence and multiplicity of solution of the boundary value problem for fractional differential equations with different boundary conditions, such as [1–10]. Bai and Lü [1] investigated the existence and multiplicity of positive solutions for nonlinear fractional boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)(1)+𝑓𝑡,𝑢))=0,0<𝑡<1,=𝑢=0, (1) where 1<𝛼≤2 is a real number, 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative, and 𝑓∶[0,1]×[0,∞)→[0,∞) is continuous. By means of some fixed-point theorems on cone, some existence and multiplicity results of positive solutions are obtained. Agarwal et al. [2] investigated the existence of positive solutions for the singular fractional boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)+𝑓𝑡,𝑢(𝑡),𝐷^𝜇𝑢(𝑡𝑢(0)(1))=0,0<𝑡<1,=𝑢=0, (2) where 1<𝛼≤2,𝜇>0 are real numbers, 𝑓 is positive, and 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative. By means of a fixed point theorem on cone, the existence of positive solutions is obtained. Delbosco and Rodino [3] investigated the nonlinear Dirichlet-type problem: 𝑥^𝛼−1𝐷^𝛼₀₊𝑦(𝑥)𝑦(𝑥𝑦(0)(1)=𝑓),1<𝛼<2,0≤𝑥≤1,=𝑦=0. (3) They proved that if 𝑓(𝑦) is a Lipschitzian function, then the problem has at least one solution 𝑦(𝑥) in a certain subspace of 𝐶[0,1]. In this paper, we study the following three-point fractional boundary value problem. Consider 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑓((𝑡𝑢(0)𝜂(1),+𝜆𝑎𝑡,𝑢))=0,0<𝑡<1,=0,𝛽𝑢=𝑢 (4) where 1<𝛼≤2,0<𝛽<1,0<𝜂<1,𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative, 𝑎(𝑡)>0 is continuous, and 𝑓≥0 is continuous on [0,1]×[0,∞). By using Krasnoesel'skii's fixed-point theorem, we get the existence of at least one positive solution.

2. Background Materials and Preliminaries

For convenience of the readers, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature [1–5].

Definition 1. The fractional integral of order 𝛼>0 of a function 𝑦∶(0,∞)→𝑅 is given by 𝐼^𝛼₀₊𝑦(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠, (5) provided that the right side is pointwise defined on (0,∞).

Definition 2. The fractional derivative of order 𝛼>0 of a function 𝑦∶(0,∞)→𝑅 is given by 𝐷^𝛼₀₊𝑦(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀𝑦(𝑠)(𝑡−𝑠)^{𝛼−𝑛+1}𝑑𝑠,𝑛=[𝛼]+1, (6) where 𝑛=[𝛼]+1, provided that the right side is pointwise defined on (0,∞).

Lemma 3. Let 𝛼>0, if we assume 𝑢∈𝐶(0,1)∩𝐿(0,1), then the fractional differential equation 𝐷^𝛼₀₊𝑢(𝑡)=0 has 𝑢(𝑡)=𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2+⋯𝐶_𝑁𝑡^𝛼−𝑁,𝐶_𝑖∈𝑅,𝑖=1,2,…,𝑁 (7) as unique solutions.

Lemma 4. Assume that 𝑢∈𝐶(0,1)∩𝐿(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐶(0,1)∩𝐿(0,1). Then 𝐼^𝛼₀₊𝐷^𝛼₀₊𝑢(𝑡)=𝑢(𝑡)+𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2+⋯𝐶_𝑁𝑡^𝛼−𝑁 (8) for some 𝐶_𝑖∈𝑅,𝑖=1,2,…,𝑁.

Definition 5. Let 𝐸 be a real Banach space. A nonempty closed convex set 𝐾⊂𝐸 is called cone of 𝐸 if it satisfies the following conditions: (1)𝑥∈𝐾,𝜎>0  implies  𝜎𝑥∈𝐾;(2)𝑥∈𝐾,−𝑥∈𝐾  implies  𝑥=0.

Lemma 6. Let 𝐸 be a Banach space and let 𝐾⊂𝐸 be a cone in 𝐸. Assume that Ω₁ and Ω₂ are open subsets of 𝐸 with 0∈Ω₁,Ω₁⊂Ω₂. Let 𝑇∶𝐾∩(Ω₂⧵Ω₁)→𝐾 be a completely continuous operator. In addition, suppose that either (H1)‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝐾∩𝜕Ω₁ and ‖𝑇𝑢‖≥‖𝑢‖,∀𝑢∈𝐾∩𝜕Ω₂ or(H2)‖𝑇𝑢‖≥‖𝑢‖,∀𝑢∈𝐾∩𝜕Ω₁ and ‖𝑇𝑢‖≤‖𝑢‖,∀𝑢∈𝐾∩𝜕Ω₂
holds. Then 𝑇 has a fixed point in 𝐾∩(Ω₂⧵Ω₁).

In the following, we present the Green function of fractional differential equation boundary value problem.

Lemma 7 (see [4]). Let 𝑦∈𝐶[0,1], then the boundary value problem 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑢(0)𝜂(1)+𝑦=0,0<𝑡<1,=0,𝛽𝑢=𝑢 (9) has a unique solution 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (10) where 𝐺(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−11−𝛽𝜂^𝛼−1×(1−𝛽𝜂^𝛼−1)Γ(𝛼)⁻¹,0≤𝑠≤min𝑡,𝜂≤1;[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−11−𝛽𝜂^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),0<𝜂≤𝑠≤𝑡≤1;[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),0≤𝑡≤𝑠≤𝜂<1;[𝑡(1−𝑠)]^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),0≤max𝑡,𝜂≤𝑠≤1. (11)

Lemma 8. The function 𝐺(𝑡,𝑠) defined by (11) satisfies 𝐺(𝑡,𝑠)>0  for  𝑡,𝑠∈(0,1).

Proof. (1) For 0≤𝑠≤min{𝑡,𝜂}≤1, 𝐺()=𝑡,𝑠[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−11−𝛽𝜂^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼). (12) Let 𝐹(𝑡,𝑠,𝜂)=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝛽𝜂^𝛼−1). There is 𝜕𝐹𝑡,𝑠,𝜂𝜕𝜂=−𝛽𝑡^𝛼−1(𝛼−1)(𝜂−𝑠)^𝛼−2+(𝛼−1)(𝑡−𝑠)^𝛼−1𝛽𝜂^𝛼−2=𝛽(𝛼−1)(𝑡−𝑠)^𝛼−1𝜂^𝛼−2−𝑡^𝛼−1(𝜂−𝑠)^𝛼−2≤𝛽(𝛼−1)𝑡^𝛼−1𝜂^𝛼−2−𝑡^𝛼−1(𝜂−𝑠)^𝛼−2=𝛽(𝛼−1)𝑡^𝛼−1𝜂^𝛼−2−(𝜂−𝑠)^𝛼−2. (13) With 0<𝛽<1,1<𝛼≤2, there is 𝜕𝐹/𝜕𝜂<0. Hence, function 𝐹(𝑡,𝑠,𝜂) is monotonically decreasing in 𝜂. With 0<𝜂<1, 𝐹(𝑡,𝑠,𝜂)>𝐹(𝑡,𝑠,𝜂)_min=𝐹(𝑡,𝑠,1). Let 𝜂=1, there is 𝐹(𝑡,𝑠,1)=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(1−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1=1−𝛽1−𝛽[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1=1−𝛽(𝑡−𝑡𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1. (14) With 0<𝑠≤𝑡<1,1<𝛼≤2, we have (𝑡−𝑡𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1≥0. With 1−𝛽>0, there is 𝐹(𝑡,𝑠,1)≥0, then 𝐹(𝑡,𝑠,𝜂)>𝐹(𝑡,𝑠,1)≥0, there is 𝐺(𝑡,𝑠)>0.
(2) For 0<𝜂≤𝑠≤𝑡<1, 𝐺(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−11−𝛽𝜂^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼). (15) Let 𝐹(𝑡,𝑠,𝜂)=[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝛽𝜂^𝛼−1); it is obvious that function 𝐹(𝑡,𝑠,𝜂) is monotonically increasing in 𝜂, when 𝜂=0, 𝐹(𝑡,𝑠,𝜂)_min=𝐹(𝑡,𝑠,0)=[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1=(𝑡−𝑡𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1. (16) With 0≤𝑠≤𝑡<1,0<𝛼−1≤1, there is (𝑡−𝑡𝑠)^𝛼−1≥(𝑡−𝑠)^𝛼−1; we have 𝐹(𝑡,𝑠,𝜂)>𝐹(𝑡,𝑠,0)≥0, then 𝐺(𝑡,𝑠)>0.
(3) For 0<𝑡≤𝑠≤𝜂<1, 𝐺(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼). (17) Let 𝐹(𝑡,𝑠,𝜂)=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−1, then 𝐹(𝑡,𝑠,𝜂) is monotonically decreasing in 𝜂, 𝐹(𝑡,𝑠,𝜂)_min=𝐹(𝑡,𝑠,1)=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(1−𝑠)^𝛼−1=1−𝛽[𝑡(1−𝑠)]^𝛼−1. (18) With 0<𝛽<1,0<𝑡≤𝑠<1, there is 𝐹(𝑡,𝑠,𝜂)>0, then 𝐹(𝑡,𝑠,𝜂)>𝐹(𝑡,𝑠,1)≥0, hence 𝐺(𝑡,𝑠)>0.
(4) For 0≤max{𝑡,𝜂}≤𝑠≤1, 𝐺(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼)>0. (19) Therefore, 𝐺(𝑡,𝑠)>0 for 𝑡,𝑠∈(0,1). The proof is complete.

Lemma 9. 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) for 𝑡,𝑠∈[0,1].

Proof. To get 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠), firstly, we prove the following inequality: 𝑡^𝑎−𝑠^𝑎(𝑡−𝑠)^𝑎≤1−𝑠^𝑎(1−𝑠)^𝑎, (20) where 𝑎=𝛼−1, 0≤𝑠<𝑡≤1. Let 𝑓(𝑡)=(𝑡^𝑎−𝑠^𝑎)/(𝑡−𝑠)^𝑎, then 𝑓^′(𝑡)=𝑎𝑡^𝑎−1(𝑡−𝑠)^𝑎−𝑎𝑡^𝑎−𝑠^𝑎(𝑡−𝑠)^𝑎−1(𝑡−𝑠)^2𝑎=𝑎(𝑡−𝑠)^𝑎−1(𝑡−𝑠)^2𝑎⋅𝑡^𝑎−1(𝑡−𝑠)−𝑡^𝑎−𝑠^𝑎=𝑎(𝑡−𝑠)^𝑎−1(𝑡−𝑠)^2𝑎⋅𝑠𝑠^𝑎−1−𝑡^𝑎−1. (21) With 1<𝛼≤2, there is 𝑎−1≤0; with 𝑠<𝑡, there is 𝑠^𝑎−1>𝑡^𝑎−1; then 𝑓^′(𝑡)>0. It means that 𝑓(𝑡) is monotonically increasing in 𝑡, then 𝑓(𝑡)≤𝑓(1); we have 𝑡^𝑎−𝑠^𝑎(𝑡−𝑠)^𝑎≤1−𝑠^𝑎(1−𝑠)^𝑎. (22) Here we prove 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠).
(1) For 0≤𝑠≤min{𝑡,𝜂}≤1, 𝐺()=𝑡,𝑠[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−11−𝛽𝜂^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),𝐺(𝑠,𝑠)=[𝑠(1−𝑠)]^𝛼−1−𝛽𝑠^𝛼−1(𝜂−𝑠)^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼). (23) With (1−𝛽𝜂^𝛼−1)Γ(𝛼)>0,  𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) is equivalent to 𝑡^𝑎−𝑠^𝑎(𝑡−𝑠)^𝑎(1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎≤1−𝛽𝜂^𝑎 (24) or 𝑡^𝑎−𝑠^𝑎(1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎≤1−𝛽𝜂^𝑎(𝑡−𝑠)^𝑎. (25)(a)For 𝑠=𝑡,(𝑡^𝑎−𝑠^𝑎)[(1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎]=(1−𝛽𝜂^𝑎)(𝑡−𝑠)^𝑎=0;(b)for 𝑠=𝑡=𝜂=1, (𝑡^𝑎−𝑠^𝑎)[(1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎]=(1−𝛽𝜂^𝑎)(𝑡−𝑠)^𝑎=0;(c)for 𝑠=𝜂≤𝑡, 𝑡^𝑎−𝑠^𝑎(1−𝑠)^𝑎−𝛽𝜂−𝑠^𝑎=𝑡^𝑎−𝜂^𝑎1−𝜂^𝑎,1−𝛽𝜂^𝑎(𝑡−𝑠)^𝑎=1−𝛽𝜂^𝑎𝑡−𝜂^𝑎, (26) if 𝑡=𝜂, (𝑡^𝑎−𝑠^𝑎)[(1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎]=(1−𝛽𝜂^𝑎)(𝑡−𝑠)^𝑎=0; if 𝑡>𝜂, from (22), we have (𝑡^𝑎−𝜂^𝑎)/(𝑡−𝜂)^𝑎≤(1−𝜂^𝑎)/(1−𝜂)^𝑎≤(1−𝛽𝜂^𝑎)/(1−𝜂)^𝑎;(d)for 0≤𝑠<𝑡≤1, 0≤𝑠<𝜂≤1, if (1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎<0; it is obvious that (𝑡^𝑎−𝑠^𝑎)[(1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎]≤(1−𝛽𝜂^𝑎)(𝑡−𝑠)^𝑎. If (1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎>0, by (22), there is (𝜂^𝑎−𝑠^𝑎)/(𝜂−𝑠)^𝑎≤(1−𝑠^𝑎)/(1−𝑠)^𝑎; then (𝜂−𝑠)^𝑎≥1−𝑠^𝑎(1−𝑠)^𝑎𝜂^𝑎−𝑠^𝑎. (27)
Consider 𝑡^𝑎−𝑠^𝑎(𝑡−𝑠)^𝑎(1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎≤1−𝑠^𝑎(1−𝑠)^𝑎(1−𝑠)^𝑎−𝛽(𝜂−𝑠)^𝑎=1−𝑠^𝑎−𝛽1−𝑠^𝑎(1−𝑠)^𝑎(𝜂−𝑠)^𝑎≤1−𝑠^𝑎−𝛽1−𝑠^𝑎(1−𝑠)^𝑎⋅(1−𝑠)^𝑎1−𝑠^𝑎⋅𝜂^𝑎−𝑠^𝑎=1−𝛽𝜂^𝑎−1−𝛽𝑠^𝑎≤1−𝛽𝜂^𝑎. (28) It means that 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) holds.
(2) For 0<𝜂≤𝑠≤𝑡≤1, 𝐺(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−11−𝛽𝜂^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),𝐺(𝑠,𝑠)=[𝑠(1−𝑠)]^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼). (29) With (1−𝛽𝜂^𝛼−1)Γ(𝛼)>0,   𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) is equivalent to [𝑡(1−𝑠)]^𝑎−(𝑡−𝑠)^𝑎1−𝛽𝜂^𝑎≤[𝑠(1−𝑠)]^𝑎 (30) or 𝑡^𝑎−𝑠^𝑎(1−𝑠)^𝑎≤(𝑡−𝑠)^𝑎1−𝛽𝜂^𝑎. (31) From (22), we have ((𝑡^𝑎−𝑠^𝑎)/(𝑡−𝑠)^𝑎)(1−𝑠)^𝑎≤1−𝑠^𝑎; with 𝜂≤𝑠,0<𝛽<1, we have 1−𝑠^𝑎≤1−𝜂^𝑎≤1−𝛽𝜂^𝑎; then ((𝑡^𝑎−𝑠^𝑎)/(𝑡−𝑠)^𝑎)(1−𝑠)^𝑎≤1−𝛽𝜂^𝑎; with 𝑠≤𝑡, there is 𝑡^𝑎−𝑠^𝑎(1−𝑠)^𝑎≤(𝑡−𝑠)^𝑎1−𝛽𝜂^𝑎. (32) It means that 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) holds.
(3) For 0≤𝑡≤𝑠≤𝜂<1, 𝐺(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),𝐺(𝑠,𝑠)=[𝑠(1−𝑠)]^𝛼−1−𝛽𝑠^𝛼−1(𝜂−𝑠)^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼). (33) With (1−𝛽𝜂^𝛼−1)Γ(𝛼)>0, 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) is equivalent to [𝑡(1−𝑠)]^𝑎−𝛽𝑡^𝑎(𝜂−𝑠)^𝑎≤[𝑠(1−𝑠)]^𝑎−𝛽𝑠^𝑎𝜂−𝑠^𝑎 (34) or 𝑡^𝑎−𝑠^𝑎(1−𝑠)^𝑎≤𝛽𝑡^𝑎−𝑠^𝑎(𝜂−𝑠)^𝑎. (35) With 0<𝛽<1, 𝑠≤𝜂, there is 𝛽(𝜂−𝑠)^𝑎≤(𝜂−𝑠)^𝑎<(1−𝑠)^𝑎; with 𝑡≤𝑠, there is 𝑡^𝑎−𝑠^𝑎(1−𝑠)^𝑎≤𝛽𝑡^𝑎−𝑠^𝑎(𝜂−𝑠)^𝑎. (36) It means that 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) holds.
(4) For 0<max{𝑡,𝜂}≤𝑠≤1, 𝐺(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),𝐺(𝑠,𝑠)=[𝑠(1−𝑠)]^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼). (37) It is obvious that 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠). The proof is complete.

Lemma 10. 𝐺(𝑡,𝑠)≥𝑞(𝑘)𝐺(𝑠,𝑠) for 𝑡,𝑠∈[1/𝑘,1−(1/𝑘)], where 𝑘>2 is an integer and 𝑞(𝑘)())=min1−𝛽(1−1/𝑘^2(𝛼−1)−(1−(2/𝑘))^𝛼−1(1/4)^𝛼−1,1−𝛽(1/𝑘)^2(𝛼−1)(1/4)^𝛼−1,(𝑘−1)^𝛼−1/𝑘^2(𝛼−1)(1/4)^𝛼−1. (38)

Proof. From (11), we have 𝐺(𝑠,𝑠)=[𝑠(1−𝑠)]^𝛼−1−𝛽𝑠^𝛼−1(𝜂−𝑠)^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),𝑠≤𝜂,[𝑠(1−𝑠)]^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),𝑠>𝜂. (39) It is obvious that [𝑠(1−𝑠)]^𝛼−1/(1−𝛽𝜂^𝛼−1)Γ(𝛼)≥([𝑠(1−𝑠)]^𝛼−1−𝛽𝑠^𝛼−1(𝜂−𝑠)^𝛼−1)/(1−𝛽𝜂^𝛼−1)Γ(𝛼), for 𝑡,𝑠∈[1/𝑘,1−(1/𝑘)], max𝐺(𝑠,𝑠)=[𝑠(1−𝑠)]^𝛼−1(1−𝛽𝜂^𝛼−1)Γ(𝛼)_𝑠=1/2=(1/4)^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼). (40)
(1) For 1/𝑘≤𝑠≤min{𝑡,𝜂}≤1−(1/𝑘), let 𝐹𝑡,𝑠,𝜂=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−11−𝛽𝜂^𝛼−1. (41) By Lemma 8, for 𝜂=1, 𝐹(𝑡,𝑠)=𝐹(𝑡,𝑠,𝜂)_min=𝐹(𝑡,𝑠,1)=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(1−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1=1−𝛽1−𝛽[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1=1−𝛽(𝑡−𝑡𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1,𝜕𝐹=𝜕𝑠1−𝛽(𝛼−1)(𝑡−𝑠)^𝛼−2−𝑡^𝛼−1(1−𝑠)^𝛼−2. (42) With 1<𝛼≤2,0≤𝑠≤𝑡≤1, then (𝑡−𝑠)^𝛼−2−𝑡^𝛼−1(1−𝑠)^𝛼−2>0; with 0<𝛽<1,𝜕𝐹/𝜕𝑠>0. It means that 𝐹(𝑡,𝑠) is monotonically increasing in 𝑠. Similarly, 𝜕𝐹=(𝜕𝑡1−𝛽𝛼−1)(𝑡−𝑡𝑠)^𝛼−2(1−𝑠)−(𝑡−𝑠)^𝛼−2<0. (43) It means that 𝐹(𝑡,𝑠) is monotonically decreasing in 𝑡. Therefore 𝐹(𝑡,𝑠)≥min𝐹(𝑡,𝑠)=𝐹(𝑡,𝑠)_{𝑡=1−(1/𝑘),𝑠=1/𝑘}=11−𝛽1−𝑘^2(𝛼−1)−21−𝑘^𝛼−1,()=()−min𝐺𝑡,𝑠1−𝛽1/𝑘1/𝑘^2𝛼−1−(1−(2/𝑘))^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),min𝐺(𝑡,𝑠)()=())max𝐺𝑠,𝑠1−𝛽(1−1/𝑘^2(𝛼−1)−(1−(2/𝑘))^𝛼−1(1/4)^𝛼−1. (44)
(2) For 1/𝑘<𝜂≤𝑠≤𝑡≤1−(1/𝑘), let 𝐹𝑡,𝑠,𝜂=[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−11−𝛽𝜂^𝛼−1. (45) By Lemma 8, for 𝜂=0, 𝐹(𝑡,𝑠)=𝐹(𝑡,𝑠,𝜂)_min=𝐹(𝑡,𝑠,0)=[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1=(𝑡−𝑡𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1,𝜕𝐹=(𝜕𝑠𝛼−1)(𝑡−𝑠)^𝛼−2−𝑡^𝛼−1(1−𝑠)^𝛼−2>0. (46) It means that 𝐹(𝑡,𝑠) is monotonically increasing in 𝑠𝜕𝐹=(𝜕𝑡𝛼−1)(1−𝑠)^𝛼−1𝑡^𝛼−2−(𝑡−𝑠)^𝛼−2<0. (47) It means that 𝐹(𝑡,𝑠) is monotonically decreasing in 𝑡𝐹(𝑡,𝑠)≥min𝐹(𝑡,𝑠)=𝐹(𝑡,𝑠)_{𝑡=1−(1/𝑘),𝑠=(1/𝑘)}=11−𝑘^2(𝛼−1)−21−𝑘^𝛼−1,()=())min𝐺𝑡,𝑠1−𝛽(1−1/𝑘^2(𝛼−1)−(1−(2/𝑘))^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),min𝐺(𝑡,𝑠)()=())max𝐺𝑠,𝑠1−𝛽(1−1/𝑘^2(𝛼−1)−(1−(2/𝑘))^𝛼−1(1/4)^𝛼−1. (48)
(3) For 1/𝑘≤𝑡≤𝑠≤𝜂<1−(1/𝑘), let 𝐹𝑡,𝑠,𝜂=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(𝜂−𝑠)^𝛼−1. (49) By Lemma 8, for 𝜂=1, 𝐹(𝑡,𝑠)=𝐹(𝑡,𝑠,𝜂)_min=𝐹(𝑡,𝑠,1)=[𝑡(1−𝑠)]^𝛼−1−𝛽𝑡^𝛼−1(1−𝑠)^𝛼−1=1−𝛽[𝑡(1−𝑠)]^𝛼−1,𝜕𝐹=()𝑡𝜕𝑠𝛼−1^𝛼−1𝛽−1(1−𝑠)^𝛼−2<0; (50) then 𝐹(𝑡,𝑠) is monotonically decreasing in 𝑠. Consider 𝜕𝐹=()𝑡𝜕𝑡𝛼−11−𝛽^𝛼−2(1−𝑠)^𝛼−1>0; (51) then 𝐹(𝑡,𝑠) is monotonically increasing in 𝑡. Therefore, 𝐹(𝑡,𝑠)≥min𝐹(𝑡,𝑠)=𝐹(𝑡,𝑠)_{𝑡=1/𝑘,𝑠=1−(1/𝑘)}=1−𝛽𝑘^2(𝛼−1),min𝐺(𝑡,𝑠)=1−𝛽1/𝑘^2(𝛼−1)1−𝛽𝜂^𝛼−1Γ(𝛼),min𝐺(𝑡,𝑠)()=max𝐺𝑠,𝑠1−𝛽(1/𝑘)^2(𝛼−1)(1/4)^𝛼−1. (52)
(4) For 1/𝑘≤max{𝑡,𝜂}≤𝑠≤1−(1/𝑘), let 𝐹(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1. (53) It is obvious that 𝐹(𝑡,𝑠) is monotonically increasing in 𝑡 and monotonically decreasing in 𝑠; then 𝐹(𝑡,𝑠)≥min𝐹(𝑡,𝑠)=𝐹(𝑡,𝑠)|_{𝑡=1/𝑘,𝑠=1−(1/𝑘)}=1𝑘−1𝑘^2𝛼−1; (54) we have ()=()−min𝐺𝑡,𝑠1/𝑘1/𝑘^2𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼); (55) there is min𝐺(𝑡,𝑠)()=max𝐺𝑠,𝑠(𝑘−1)^𝛼−1/𝑘^2(𝛼−1)(1/4)^𝛼−1. (56) Let 𝑞(𝑘)())=min1−𝛽(1−1/𝑘^2(𝛼−1)−(1−(2/𝑘))^𝛼−1(1/4)^𝛼−1,1−𝛽(1/𝑘)^2(𝛼−1)(1/4)^𝛼−1,(𝑘−1)^𝛼−1/𝑘^2(𝛼−1)(1/4)^𝛼−1. (57) Therefore, 𝐺(𝑡,𝑠)/𝐺(𝑠,𝑠)≥min𝐺(𝑡,𝑠)/max𝐺(𝑠,𝑠)=𝑞(𝑘)≠0.
The proof is complete.

We consider the Banach space 𝑋=𝐶[0,1] equipped with standard norm ‖𝑢‖=max_0≤𝑡≤1|𝑢(𝑡)|,𝑢∈𝑋. We define a cone 𝑃 by 𝑃=𝑢∈𝑋∶𝑢(𝑡)≥0,𝑡∈[0,1],min_{𝑡∈[1/𝑘,1−1/𝑘]}𝑢(𝑡)≥𝑞(𝑘)‖𝑢‖. (58) Define an integral operator 𝑇∶𝑃→𝑋 by (𝑇𝑢)(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,0≤𝑡≤1,𝑢∈𝑃. (59)

Lemma 11. It holds the following. (1)𝑇∶𝑃→𝑃 is completely continuous.(2)𝑢(𝑡) is a positive solution of the fractional boundary value problem (4) if and only if 𝑢(𝑡) is a fixed point of the operator 𝑇 in cone 𝑃.

Proof. For 𝑢∈𝑃,𝑡∈[1/𝑘,1−(1/𝑘)], by Lemma 10, 𝐺(𝑡,𝑠)≥𝑞(𝑘)𝐺(𝑠,𝑠); for 𝑡,𝑠∈[1/𝑘,1−(1/𝑘)], 𝑇𝑢(𝑡)=𝜆¹₀𝐺()𝑎(𝑠)𝑓((𝑠(𝑘)𝑡,𝑠𝑠,𝑢))𝑑𝑠≥𝜆𝑞¹₀𝐺()𝑎(𝑠)𝑓((𝑠(𝑘)𝑠,𝑠𝑠,𝑢))𝑑𝑠=𝜆𝑞max_{1/𝑘≤𝑡≤1−1/𝑘}¹₀𝐺()𝑎(𝑠)𝑓((𝑠(𝑘(𝑡.𝑡,𝑠𝑠,𝑢))𝑑𝑠=𝑞)‖𝑇𝑢)‖ (60) It is obvious that min_{1/𝑘≤𝑡≤1−1/𝑘}(𝑇𝑢)(𝑡)≥𝑞(𝑘)‖𝑇𝑢‖,𝑇𝑃⊆𝑃. Thus 𝑇𝑃⊂𝑃. In addition, standard arguments show that 𝑇 is completely continuous.
(2) It is obvious that 𝑢(𝑡) is the positive solution of BVP (4) if and only if 𝑢(𝑡)=𝜆¹₀𝐺(𝑡,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,0≤𝑡≤1. (61) It can be proved by the definition of integral operator 𝑇.

3. Main Results

We denote some important constants as follows: 𝐴=¹₀𝐺(𝑠,𝑠)𝑎(𝑠)𝑞(𝑘)𝑑𝑠,𝐵=¹₀𝐺()𝑎(𝑠)𝐹𝑠,𝑠𝑑𝑠,₀=lim_𝑢→0+supmax_0≤𝑡≤1𝑓(𝑡,𝑢(𝑡))𝑢(𝑡),𝑓₀=lim_𝑢→+∞infmin_0≤𝑡≤1𝑓(𝑡,𝑢(𝑡))𝑢(𝑡),𝐹_∞=lim_𝑢→0+supmax_0≤𝑡≤1𝑓(𝑡,𝑢(𝑡))𝑢(𝑡),𝑓_∞=lim_𝑢→+∞infmin_0≤𝑡≤1𝑓(𝑡,𝑢(𝑡))𝑢(𝑡). (62) Here we assume that 1/𝐴𝑓_∞=0 if 𝑓_∞→∞, 1/𝐵𝐹₀=∞ if 𝐹₀→0, 1/𝐴𝑓₀=0 if 𝑓₀→∞, and 1/𝐵𝐹_∞=∞ if 𝐹_∞→0.

Theorem 12. Suppose that 𝐴𝑓_∞>𝐵𝐹₀, then for each 𝜆∈(1/𝐴𝑓_∞,1/𝐵𝐹₀), BVP (4) has at least one positive solution.

Proof. We choose 𝜀>0 sufficiently small such that (𝐹₀+𝜀)𝜆𝐵≤1. By the definition of 𝐹₀, we can see that there exists 𝑙₁>0, such that 𝑓(𝑡,𝑢(𝑡))≤(𝐹₀+𝜀)𝑢(𝑡) for 0≤𝑢≤𝑙₁. For 𝑢∈𝑃 with ‖𝑢‖=𝑙₁, we have ‖𝑇𝑢(𝑡)‖=𝜆¹₀𝐺(𝑠,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠≤𝜆¹₀𝐺(𝑠,𝑠)𝑎(𝑠)𝐹₀𝑢(𝑠)𝐹+𝜀𝑑𝑠≤𝜆₀+𝜀‖𝑢‖¹₀𝐺()𝑎(𝑠)𝐹𝑠,𝑠𝑑𝑠=𝜆𝐵₀𝑢‖≤‖𝑢‖.+𝜀‖ (63) Then we have ‖𝑇𝑢‖≤‖𝑢‖. Thus if we let Ω₁={𝑢∈𝑋∶‖𝑢‖<𝑙₁}, then ‖𝑇𝑢‖≤‖𝑢‖  for  𝑢∈𝑃∩𝜕Ω₁. We choose 𝛿>0 and 𝑐∈(0,1/4) such that 𝜆(𝑓_∞−𝛿)𝐴≥1. There exists 𝑙₂>𝑙₁>0, such that 𝑓(𝑡,𝑢)≥(𝑓_∞−𝛿)𝑢 for 𝑢>𝑙₂. Therefore, for each 𝑢∈𝑃 with ‖𝑢‖=𝑙₂, we have ‖𝑇𝑢(𝑡)‖=𝜆¹₀𝐺(𝑠,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠≥𝜆¹_𝑐𝐺(𝑠,𝑠)𝑎(𝑠)𝑓_∞𝑢(𝑠)𝑓−𝛿𝑑𝑠≥𝜆_∞−𝛿‖𝑢‖¹_𝑐𝐺()𝑎(𝑠)𝑞(𝑘)𝑓𝑠,𝑠𝑑𝑠≥𝜆_∞‖𝑢‖≥‖𝑢‖.−𝛿𝐴⋅ (64) Thus if we let Ω₂={𝑢∈𝐸∶‖𝑢‖<𝑙₂}, then Ω₁⊂Ω₂ and ‖𝑇𝑢‖≥‖𝑢‖ for 𝑢∈𝑃∩𝜕Ω₂. Condition (H1) of Krasnoesel'skii's fixed point theorem is satisfied. So there exists a fixed point of 𝑇 in 𝑃. This completes the proof.

Theorem 13. Suppose that 𝐴𝑓₀>𝐵𝐹_∞, then for each 𝜆∈(1/𝐴𝑓₀,1/𝐵𝐹_∞), BVP (4) has at least one positive solution.

Proof. Choose 𝜀>0 sufficiently small such that (𝑓₀−𝜀)𝜆𝐴≥1. From the definition of 𝑓₀, we see that there exists 𝑙₁>0, such that 𝑓(𝑡,𝑢)≥(𝑓₀−𝜀)𝑢 for 0<𝑢≤𝑙₁. If 𝑢∈𝑃 with ‖𝑢‖=𝑙₁, we have ‖𝑇𝑢(𝑡)‖=𝜆¹₀𝐺()𝑎(𝑠)𝑓((𝑠𝑓𝑠,𝑠𝑠,𝑢))𝑑𝑠≥𝜆₀‖𝑢‖𝐴≥‖𝑢‖.−𝜀 (65) Let Ω₁={𝑢∈𝑋∶‖𝑢‖<𝑙₁}; then we have ‖𝑇𝑢‖≥‖𝑢‖ for 𝑢∈𝑃∩𝜕Ω₁. Choose 𝛿>0,𝑐∈(0,1/4); then we have 𝜆(𝐹_∞+𝛿)𝐴≤1. There exists 𝑙₂>𝑙₁>0, such that 𝑓(𝑡,𝑢)≤(𝐹_∞+𝛿)𝑢 for 𝑢>𝑙₂. For each 𝑢∈𝑃with‖𝑢‖=𝑙₂, we have ‖𝑇𝑢(𝑡)‖=𝜆¹₀𝐺(𝑠,𝑠)𝑎(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠≤𝜆¹_𝑐𝐺(𝑠,𝑠)𝑎(𝑠)𝐹_∞𝑢(𝑠)𝐹+𝛿𝑑𝑠≤𝜆_∞+𝛿‖𝑢‖¹_𝑐𝐺()𝑎(𝑠)𝑞(𝑘)𝐹𝑠,𝑠𝑑𝑠≤𝜆_∞‖𝑢‖≤‖𝑢‖.+𝛿𝐴⋅ (66) Let Ω₂={𝑢∈𝐸∶‖𝑢‖<𝑙₂}; then Ω₁⊂Ω₂, and we have ‖𝑇𝑢‖≤‖𝑢‖ for 𝑢∈𝑃∩𝜕Ω₂. Condition (H2) of Krasnoesel'skii's fixed-point theorem is satisfied. So there exists a fixed point of 𝑇 in 𝑃. This completes the proof.

4. An Example

Example 1. Consider the following three-point fractional boundary value problem: 𝐷^3/2₀₊𝑢(𝑡)((𝑡𝑢(0)(1)=1+𝜆𝑓𝑡,𝑢))=0,0<𝑡<1,=0,𝑢2𝑢13, (67) where 𝑓(𝑡,𝑢(𝑡))=((𝑡+1)(2014𝑢³+𝑢)/(𝑢²+1))+𝑢(sin𝑢+1),𝑎(𝑡)=1.
By calculations, 𝐺(𝑠,𝑠)=[𝑠(1−𝑠)]^𝛼−1−𝛽𝑠^𝛼−1(𝜂−𝑠)^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),0≤𝑠=𝑡≤𝜂≤1,[𝑠(1−𝑠)]^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼),0≤𝜂≤𝑡=𝑠≤1,₁₀𝐺(𝑠,𝑠)𝑑𝑠=^𝜂₀[𝑠(1−𝑠)]^𝛼−1−𝛽𝑠^𝛼−1𝜂−𝑠^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼)+𝑑𝑠_1𝜂[𝑠(1−𝑠)]^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼)=𝑑𝑠₁₀[𝑠(1−𝑠)]^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼)−𝑑𝑠_𝜂0𝛽𝑠^𝛼−1𝜂−𝑠^𝛼−11−𝛽𝜂^𝛼−1Γ(𝛼)=1𝑑𝑠1−(1/2)(1/3)^1/2Γ()×3/2₁₀[𝑠(1−𝑠)]^1/2−𝑑𝑠_01/312𝑠^1/2(𝜂−𝑠)^1/2𝑑𝑠=0.58832. (68) Let 𝑘=10, then 𝑞(𝑘)())min1−𝛽(1−1/𝑘^2(𝛼−1)−(1−(2/𝑘))^𝛼−1(1/4)^𝛼−1,1−𝛽(1/𝑘)^2(𝛼−1)(1/4)^𝛼−1,(𝑘−1)^𝛼−1/𝑘^2(𝛼−1)(1/4)^𝛼−1.𝑓=0.00557,_∞=lim_𝑢→+∞infmin_0≤𝑡≤1(𝑡+1)2014𝑢²+1𝑢²𝐹+1+sin𝑢+1=2014,₀=lim_𝑢→0+supmax_0≤𝑡≤1(𝑡+1)2014𝑢²+1𝑢²+1+sin𝑢+1=4. (69) Consider 𝐴=¹₀𝐺(𝑠,𝑠)𝑎(𝑠)𝑞(𝑘)𝑑𝑠=0.00328,𝐵=¹₀𝐺(𝑠,𝑠)𝑎(𝑠)𝑑𝑠=0.58832. (70)(71) The condition 𝐴𝑓_∞>𝐵𝐹₀ is obtained. From Theorem 12, we see that if 𝜆∈(0.15138,0.42494), the problem (67) has a positive solution.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are grateful to the reviewers for the valuable suggestions. This paper was supported by the Fundamental Research Funds for the Central Universities (no. 2652012141) and Beijing Higher Education Young Elite Teacher Project.