The Existence of Positive Solutions for Boundary Value Problem of the Fractional Sturm-Liouville Functional Differential Equation

Abstract

We study boundary value problems for the following nonlinear fractional Sturm-Liouville functional differential equations involving the Caputo fractional derivative: ^𝐶𝐷^𝛽(𝑝(𝑡)^𝐶𝐷^𝛼𝑢(𝑡)) + 𝑓(𝑡,𝑢(𝑡−𝜏),𝑢(𝑡+𝜃))=0, 𝑡∈(0,1), ^𝐶𝐷^𝛼𝑢(0)=^𝐶𝐷^𝛼𝑢(1)=(^𝐶𝐷^𝛼𝑢(0))^′′=0, 𝑎𝑢(𝑡)−𝑏𝑢^′(𝑡)=𝜂(𝑡), 𝑡∈[−𝜏,0], 𝑐𝑢(𝑡)+𝑑𝑢^′(𝑡)=𝜉(𝑡), 𝑡∈[1,1+𝜃], where ^𝐶𝐷^𝛼, ^𝐶𝐷^𝛽 denote the Caputo fractional derivatives, 𝑓 is a nonnegative continuous functional defined on 𝐶([−𝜏,1+𝜃],ℝ), 1<𝛼≤2, 2<𝛽≤3, 0<𝜏, 𝜃<1/4 are suitably small, 𝑎,𝑏,𝑐,𝑑>0, and 𝜂∈𝐶([−𝜏,0],[0,∞)), 𝜉∈𝐶([1,1+𝜃],[0,∞)). By means of the Guo-Krasnoselskii fixed point theorem and the fixed point index theorem, some positive solutions are obtained, respectively. As an application, an example is presented to illustrate our main results.

1. Introduction

Fractional calculus is a branch of mathematics, it is an emerging field in the area of the applied mathematics that deals with derivatives and integrals of arbitrary orders as well as with their applications. The origins can be traced back to the end of the seventeenth century. During the history of fractional calculus, it was reported that the pure mathematical formulations of the investigated problems started to be addressed with more applications in various fields. With the help of fractional calculus, we can describe natural phenomena and mathematical models more accurately. Therefore, fractional differential equations have received much attention, and the theory and its applications have been greatly developed; see [1–6].

Recently, there have been many papers focused on boundary value problems of fractional ordinary differential equations [7–19] and initial value problems of fractional functional differential equations [10, 20–27]. But the results dealing with the boundary value problems of fractional functional differential equations with delay are relatively scarce [28–32]. It is well known that in practical problems, the behavior of systems not only depends on the status just at the present but also on the status in the past [33]. Thus, in many cases, we must consider fractional functional differential equations with delay in order to solve practical problems. Consequently, our aim in this paper is to study the existence of solutions for boundary value problems of fractional functional differential equations.

In 2005, by means of the fixed point index theorem, Bai and Ma [34] established some criteria for the existence of solutions for the boundary value problem expressed by second order differential equations with delay: −𝑝(𝑡)𝑦^′(𝑡)^′(𝑡)𝑦(𝑡)()((),(𝑡)+𝑞=𝑓𝑡,𝑦𝑡−𝜏,𝑦𝑡+𝑎),𝑡∈0,1𝛼𝑦−𝛽𝑦^′(𝑡)(𝑡)[],(𝑡)=𝜂,𝑡∈−𝜏,0𝛾𝑦+𝛿𝑦^′(𝑡)=𝜉(𝑡),𝑡∈1,1+𝜑, (1) where 0<𝜏, 𝜑<1/4 are suitably small and 𝑝∈𝐶¹([0,1],(0,∞)), 𝑞∈𝐶([0,1],[0,∞)), 𝑞≤0.

In 2011, Li et al. [21] investigated the existence of positive solutions for the nonlinear Caputo fractional functional differential equation: 𝐷^𝛼₀₊𝑢(𝑡)+𝑎(𝑡)𝑓𝑢_𝑡=0,0<𝑡<1,1<𝛼≤2, (2) where 𝐷^𝛼₀₊ is the Caputo fractional order derivative, subject to the following boundary conditions: −𝑎𝑢(𝑡)+𝑏𝑢^′(𝑡)(𝑡)[],(𝑡)=𝜉,𝑡∈−𝜏,0𝑐𝑢+𝑑𝑢^′(𝑡)=𝜂(𝑡),𝑡∈1,1+𝛽. (3) They obtained the existence results of positive solutions by using some fixed point theorems.

In 2012, Zhao et al. [22] studied the existence of positive solutions for the nonlinear Caputo fractional functional differential equation: 𝐷^𝑞₀₊𝑦(𝑡)+𝑟(𝑡)𝑓𝑦_𝑡()(],𝑦=0,∀𝑡∈0,1,𝑞∈𝑛−1,𝑛_(𝑖)(0)=0,0≤𝑖≤𝑛−3,𝛼𝑦^(𝑛−2)(𝑡)−𝛽𝑦^(𝑛−1)(𝑡)=𝜂(𝑡),𝑡∈[−𝜏,0],𝛾𝑦^(𝑛−2)(𝑡)+𝛿𝑦^(𝑛−1)(𝑡)=𝜉(𝑡),𝑡∈[1,1+𝑎]. (4) By constructing a special cone and using the Guo-Krasnoselskii fixed point theorem, they obtained the existing results.

Motivated by the works above, in this paper, we study the existence of positive solutions of boundary value problems for nonlinear fractional functional differential equation: ^𝐶𝐷^𝛽𝑝(𝑡)^𝐶𝐷^𝛼𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡−𝜏),𝑢(𝑡+𝜃))=0,𝑡∈(0,1_𝐶),𝐷^𝛼𝑢(0)=^𝐶𝐷^𝛼𝑢(1)=^𝐶𝐷^𝛼𝑢(0)^′′(𝑡)=0,𝑎𝑢−𝑏𝑢^′(𝑡)(𝑡)[],(𝑡)=𝜂,𝑡∈−𝜏,0𝑐𝑢+𝑑𝑢^′(𝑡)=𝜉(𝑡),𝑡∈[1,1+𝜃], (5) where ^𝐶𝐷^𝛼,^𝐶𝐷^𝛽 denote the Caputo fractional derivatives, 𝑓 is a nonnegative continuous functional defined on 𝐶([−𝜏,1+𝜃],ℝ), 1<𝛼≤2, 2<𝛽≤3, 0<𝜏, 𝜃<1/4 are suitably small, 𝑎,𝑏,𝑐,𝑑>0 with 𝜌∶=𝑎𝑐+𝑎𝑑+𝑏𝑐>0 and 𝜂∈𝐶([−𝜏,0],[0,∞)), 𝜉∈𝐶([1,1+𝜃],[0,∞)) with 𝜂(0)=𝜉(1)=0, and 𝑝 is a positive measurable continuous function defined on [0,1] and satisfies the following condition: 0<_𝐸𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠≤¹₀𝐺(𝑠,𝑠)𝑝(𝑠)𝑑𝑠<+∞, (6) where 𝐸={𝑡∈[0,1]∣0≤𝑡+𝑎≤1,−𝜏≤𝑎≤𝜃}, 𝑝(0)=𝑝(1)=𝑙, and 𝑙 is a constant.

When 𝛼=𝛽=1, problem (5) is reduced to the problem of second order differential equations with delay and has been studied by Bai and Ma [34]. To the best of our knowledge, no one has studied the existence of positive solutions for boundary value problem (5). Key tools in finding our main results are fixed point index theorem and the Guo-Krasnoselskii fixed point theorem, and our main results of this paper are to extend and supplement some results in [21, 22, 34].

The paper is organized as follows. In Section 2, we will introduce some definitions and lemmas to prove our main results. In Section 3, we investigate the existence of positive solution for boundary value problems (5) by the Guo-Krasnoselskii fixed point theorem and the fixed point index theorem. As an application, an example is presented to illustrate our main results.

2. Preliminaries

In the following section, we introduce the definitions and lemmas which are used throughout the paper.

Definition 1 (see [4]). The fractional integral of order 𝛼 (𝛼>0) of a function 𝑓∶(𝑡₀,+∞)→ℝ is given by 𝐼^𝛼𝑓(𝑡)=1Γ(𝛼)^𝑡_𝑡0𝑓(𝑠)(𝑡−𝑠)^1−𝛼𝑑𝑠,𝑡>𝑡₀, (7) where Γ(⋅) is the gamma function, provided that the right side is pointwise defined on (𝑡₀,+∞).

Definition 2 (see [4]). The Caputo fractional derivative of order 𝛼 (𝑛−1<𝛼<𝑛) of a function 𝑓∶(𝑡₀,+∞)→ℝ is given by ^𝐶𝐷^𝛼𝑓(𝑡)=1Γ(𝑛−𝛼)^𝑡_𝑡0𝑓^(𝑛)(𝑠)(𝑡−𝑠)^{𝛼+1−𝑛}𝑑𝑠,𝑡>𝑡₀, (8) where Γ(⋅) is the gamma function, provided that the right side is pointwise defined on (𝑡₀,+∞).
Obviously, the Caputo derivative for every constant function is equal to zero.

From the definition of the Caputo derivative, we can acquire the following statements.

Lemma 3 (see [5]). Let 𝑓(𝑡)∈𝐿¹[𝑡₀,∞). Then, ^𝐶𝐷^𝛼𝐼^𝛼𝑓(𝑡)=𝑓(𝑡),𝑡>𝑡₀,0<𝛼<1. (9)

Lemma 4 (see [5]). Let 𝛼>0. Then, 𝐼^𝛼𝐶𝐷^𝛼𝑓(𝑡)=𝑓(𝑡)−𝑐₁−𝑐₂𝑡−⋯−𝑐_𝑛𝑡^𝑛−1, (10) for some 𝑐_𝑖∈ℝ, 𝑖=1,2,…,𝑛, where 𝑛 is the smallest integer greater than or equal to 𝛼.

Assume that 𝑥₀ is the solution of (5) with 𝑓≡0; then it can be expressed as 𝑥₀(𝑡)=𝑥₀()[][]𝑥−𝜏;𝑡𝑡∈−𝜏,00𝑡∈0,1₀(𝜃;𝑡)𝑡∈[1,1+𝜃], (11) where 𝑥₀()=1−𝜏;𝑡𝑏𝑒^{𝑎𝑡/𝑏0}_𝑡𝑒^{−𝑎𝑠/𝑏}𝜂(𝑠)𝑥𝑑𝑠,₀()=1𝜃;𝑡𝑑𝑒^{−𝑐𝑡/𝑑𝑡}₁𝑒^{𝑐𝑠/𝑑}𝜉(𝑠)𝑑𝑠. (12)

Next, we introduce the Green function of boundary value problems for fractional functional differential equations.

Lemma 5. Let 1<𝛼≤2 and 𝑓∶[0,1]×(0,∞)×(0,∞)→ℝ be continuous. Then, the boundary value problem for fractional functional differential equation ^𝐶𝐷^𝛼𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡−𝜏),𝑢(𝑡+𝜃))=0,𝑡∈(0,1),𝑎𝑢(𝑡)−𝑏𝑢^′(𝑡)(𝑡)[],(𝑡)=𝜂,𝑡∈−𝜏,0𝑐𝑢+𝑑𝑢^′(𝑡)=𝜉(𝑡),𝑡∈[1,1+𝜃] (13)(14) has a unique solution: 𝑢(𝑡)=𝑢(−𝜏;𝑡)𝑡∈[−𝜏,0₁₀]𝐺()𝑓(()([]𝑢()[],𝑡,𝑠𝑠,𝑢𝑠−𝜏,𝑢𝑠+𝜃))𝑑𝑠𝑡∈0,1𝜃;𝑡𝑡∈1,1+𝜃 (15) where 𝑢(−𝜏;𝑡)=𝑒^{(𝑎/𝑏)𝑡}1𝑏⁰_𝑡𝑒^{−(𝑎/𝑏)𝑠}𝜂(𝑠)(0),𝑢()𝑑𝑠+𝑢𝜃;𝑡=𝑒^{−(𝑐/𝑑)𝑡}1𝑑^𝑡₁𝑒^{(𝑐/𝑑)𝑠}𝜉(𝑠)𝑑𝑠+𝑢(1),𝐺()=1𝑡,𝑠Γ(𝛼)𝑔₁()𝑔𝑡,𝑠0≤𝑠≤𝑡≤1₂(𝑡,𝑠)0≤𝑡≤𝑠≤1,𝑔₁(𝑡,𝑠)=−(𝑡−𝑠)^𝛼−1+𝑏+𝑎𝑡×𝑐()𝑏𝑐+𝑎𝑐+𝑎𝑑1−𝑠^𝛼−1+𝑑(𝛼−1)(1−𝑠)^𝛼−2,𝑔₂(𝑡,𝑠)=𝑏+𝑎𝑡𝑐()𝑏𝑐+𝑎𝑐+𝑎𝑑1−𝑠^𝛼−1+𝑑(𝛼−1)(1−𝑠)^𝛼−2. (16)(17)(18)

Proof. It is easy to know that 𝑢(𝑡)=𝑢(−𝜏;𝑡) for 𝑡∈[−𝜏,0] and 𝑢(𝑡)=𝑢(𝜃;𝑡) for 𝑡∈[1,1+𝜃]. From (13), we know that 𝐼^𝛼𝐶𝐷^𝛼𝑢(𝑡)=−𝐼^𝛼𝑓(𝑡,𝑢(𝑡−𝜏),𝑢(𝑡+𝜃)),1<𝛼≤2. (19) From Lemma 4, we have 𝑢(𝑡)1=−Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑓(𝑠,𝑢(𝑠−𝜏),𝑢(𝑠+𝜃))𝑑𝑠+𝑐₁+𝑐₂𝑡. (20) Then, we get that 𝑢^′(𝑡)1=−Γ(𝛼−1)^𝑡₀(𝑡−𝑠)^𝛼−2𝑓(𝑠,𝑢(𝑠−𝜏),𝑢(𝑠+𝜃))𝑑𝑠+𝑐₂. (21) According to (14) and conditions 𝜂(0)=𝜉(1)=0, we imply that 𝑎𝑢(0)−𝑏𝑢^′(0)=0,𝑐𝑢(1)+𝑑𝑢^′(1)=0. (22) Then, we obtain 𝑐₂=𝑎×𝑐𝑎𝑐+𝑏𝑐+𝑎𝑑Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝑓(()(+𝑑𝑠,𝑢𝑠−𝜏,𝑢𝑠+𝜃))𝑑𝑠Γ()×𝛼−1₁₀(1−𝑠)^𝛼−2𝑓(()(,𝑐𝑠,𝑢𝑠−𝜏,𝑢𝑠+𝜃))𝑑𝑠₁=𝑏×𝑐𝑎𝑐+𝑏𝑐+𝑎𝑑Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝑓(()(+𝑑𝑠,𝑢𝑠−𝜏,𝑢𝑠+𝜃))𝑑𝑠Γ()×𝛼−1₁₀(1−𝑠)^𝛼−2𝑓(𝑠,𝑢(𝑠−𝜏),𝑢(𝑠+𝜃))𝑑𝑠. (23) Therefore, 𝑢(𝑡)1=−Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑓(()(+𝑠,𝑢𝑠−𝜏,𝑢𝑠+𝜃))𝑑𝑠𝑎𝑡+𝑏×𝑐𝑎𝑐+𝑏𝑐+𝑎𝑑Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝑓(()(+𝑑𝑠,𝑢𝑠−𝜏,𝑢𝑠+𝜃))𝑑𝑠Γ()×𝛼−1₁₀(1−𝑠)^𝛼−2𝑓(𝑠,𝑢(𝑠−𝜏),𝑢(𝑠+𝜃))𝑑𝑠=¹₀𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠−𝜏),𝑢(𝑠+𝜃))𝑑𝑠, (24) where 𝐺(𝑡,𝑠) is defined by (17). The proof is completed.

Lemma 6. Let 1<𝛼≤2, 2<𝛽≤3 and 𝑓∶[0,1]×(0,∞)×(0,∞)→ℝ be continuous. Then, the boundary value problem for fractional functional differential equation ^𝐶𝐷^𝛽𝑝(𝑡)^𝐶𝐷^𝛼𝑢(𝑡)=𝑓(𝑡,𝑢(𝑡−𝜏),𝑢(𝑡+𝜃)),𝑡∈(0,1_𝐶),𝐷^𝛼𝑢(0)=^𝐶𝐷^𝛼𝑢(1)=^𝐶𝐷^𝛼𝑢(0)^′′(𝑡)=0,𝛼𝑢−𝑏𝑢^′(𝑡)(𝑡)[],(𝑡)=𝜂,𝑡∈−𝜏,0𝑐𝑢+𝑑𝑢^′(𝑡)=𝜉(𝑡),𝑡∈[1,1+𝜃] (25) has a unique solution: 𝑢(𝑡)=𝑢(−𝜏;𝑡)𝑡∈[−𝜏,0₁₀]𝐺()×1𝑡,𝑠𝑝(𝑠)¹₀𝐻()((),𝑢([]𝑢()[],𝑠,𝑣×𝑓𝑣,𝑢𝑣−𝜏𝑣+𝜃))𝑑𝑣𝑑𝑠𝑡∈0,1𝜃;𝑡𝑡∈1,1+𝜃 (26) where 𝐺(𝑡,𝑠) is defined by (17), and 𝐻()=1𝑠,𝑣Γ𝛽𝑠(1−𝑣)^𝛽−1−(𝑠−𝑣)^𝛽−1𝑠()0≤𝑣≤𝑠≤11−𝑣^𝛽−10≤𝑠≤𝑣≤1. (27)

Proof. From (25), we know that 𝐼^𝛽𝐶𝐷^𝛽𝑝(𝑡)^𝐶𝐷^𝛼𝑢(𝑡)=𝐼^𝛽𝑓(𝑡,𝑢(𝑡−𝜏),𝑢(𝑡+𝜃)). (28) From Lemma 4, we have ^𝐶𝐷^𝛼𝑢(𝑡)=1𝑝(𝑡)Γ𝛽×^𝑡₀(𝑡−𝑣)^𝛽−1𝑓(()(+𝑐𝑣,𝑢𝑣−𝜏,𝑢𝑣+𝜃))𝑑𝑣₁+𝑐₂𝑡+𝑐₃𝑡²𝑝(𝑡). (29) According to the conditions ^𝐶𝐷^𝛼𝑢(0)=^𝐶𝐷^𝛼𝑢(1)=(^𝐶𝐷^𝛼𝑢(0))^′′=0 and 𝑝(0)=𝑝(1)=𝑙, we derive that 𝑐₁=0,𝑐₃𝑐=0,₂1=−Γ𝛽¹₀(1−𝑣)^𝛽−1𝑓(𝑣,𝑢(𝑣−𝜏),𝑢(𝑣+𝜃))𝑑𝑣. (30) Therefore, 𝑝(𝑡)^𝐶𝐷^𝛼𝑢(𝑡)=1Γ𝛽^𝑡₀(𝑡−𝑣)^𝛽−1𝑓(()(−𝑡𝑣,𝑢𝑣−𝜏,𝑢𝑣+𝜃))𝑑𝑣Γ𝛽¹₀(1−𝑣)^𝛽−1𝑓(()(=1𝑣,𝑢𝑣−𝜏,𝑢𝑣+𝜃))𝑑𝑣Γ𝛽^𝑡₀(𝑡−𝑣)^𝛽−1𝑓(()(−𝑡𝑣,𝑢𝑣−𝜏,𝑢𝑣+𝜃))𝑑𝑣Γ𝛽^𝑡₀(1−𝑣)^𝛽−1𝑓(()(−𝑡𝑣,𝑢𝑣−𝜏,𝑢𝑣+𝜃))𝑑𝑣Γ𝛽¹_𝑡(1−𝑣)^𝛽−1𝑓(𝑣,𝑢(𝑣−𝜏),𝑢(𝑣+𝜃))𝑑𝑣=−¹₀𝐻(𝑡,𝑣)𝑓(𝑣,𝑢(𝑣−𝜏),𝑢(𝑣+𝜃))𝑑𝑣. (31) Thus, 𝑝(𝑡)^𝐶𝐷^𝛼𝑢(𝑡)+¹₀𝐻(𝑡,𝑣)𝑓(𝑣,𝑢(𝑣−𝜏),𝑢(𝑣+𝜃))𝑑𝑣=0. (32) Then boundary value problem (25) is equivalent to the following problem: ^𝐶𝐷^𝛼𝑢(𝑡)+1𝑝(𝑡)×¹₀𝐻()𝑓(()([],(𝑡)𝑡,𝑣𝑣,𝑢𝑣−𝜏,𝑢𝑣+𝜃))𝑑𝑣=0,𝑡∈0,1𝑎𝑢−𝑏𝑢^′(𝑡)(𝑡)[],(𝑡)=𝜂,𝑡∈−𝜏,0𝑐𝑢+𝑑𝑢^′(𝑡)=𝜉(𝑡),𝑡∈[1,1+𝜃]. (33) Lemma 5 implies that boundary value problem (33) has a unique solution: 𝑢(𝑡)=𝑢(−𝜏;𝑡)𝑡∈[−𝜏,0₁₀]𝐺()×1𝑡,𝑠𝑝(𝑠)¹₀𝐻()(()([]𝑢()[],𝑠,𝑣×𝑓𝑣,𝑢𝑣−𝜏,𝑢𝑣+𝜃))𝑑𝑣𝑑𝑠𝑡∈0,1𝜃;𝑡𝑡∈1,1+𝜃 (34) where 𝐺(𝑡,𝑠) and 𝐻(𝑠,𝑣) are defined as (17) and (27), respectively. The proof is completed.

The following properties of the Green function play important roles in this paper.

Lemma 7 (see [21]). The function 𝐺(𝑡,𝑠) defined by (17) satisfies the following conditions: (1)𝐺(𝑡,𝑠) is continuous on [0,1]×[0,1];(2) for 𝑏>((2−𝛼)/(𝛼−1))𝑎, we have 𝐺(𝑡,𝑠)>0 for 𝑡,𝑠∈(0,1); (3) 𝐺(𝑡,𝑠)≤𝐺(𝑠,𝑠) for 𝑡,𝑠∈(0,1); (4) there exist positive numbers 𝑧,𝑧^∗ such that 𝑧𝑌(𝑠)≤𝐺(𝑡,𝑠)≤𝑧^∗𝑌(𝑠), (35) where 𝑌(𝑠)=(𝑏+𝑎𝑠)((𝑐/Γ(𝛼))(1−𝑠)^𝛼−1+(𝑑/Γ(𝛼−1))(1−𝑠)^𝛼−2).

Remark 8. From the definition of 𝐺(𝑡,𝑠) and 𝑌(𝑠), we know that 𝐺(𝑠,𝑠)=𝑌(𝑠)/(𝑎𝑐+𝑎𝑑+𝑏𝑐). Furthermore, its easy to get from the proof of Lemma 3.3 in [21] that there exists a positive number 𝛾₁<1 such that 𝐺(𝑡,𝑠)≥𝛾₁𝐺(𝑠,𝑠)∀𝑡,𝑠∈(0,1), (36) where 𝛾₁=𝑧(𝑎𝑐+𝑎𝑑+𝑏𝑐).

Lemma 9 (see [18]). The function 𝐻(𝑠,𝑣) defined by (27) satisfies the following conditions: (1)𝐻(𝑠,𝑣)>0 for 𝑠,𝑣∈(0,1); (2)there exists a positive function 𝛾₂(𝑣)∈𝐶(0,1) for 𝑣∈(0,1) such that min_{1/4≤𝑠≤3/4}𝐻(𝑠,𝑣)≥𝛾₂(𝑣)𝑀(𝑣),max_0≤𝑠≤1𝐻(𝑠,𝑣)≤𝑀(𝑣), (37) where 𝑀(𝑣)=𝐻𝑠₀=,𝑣𝛽−2𝛽−1^2/(2−𝛽)(1−𝑣)^{(𝛽−1)2/(𝛽−2)}+𝑣(1−𝑣)^𝛽−1Γ𝛽,𝑠₀=𝛽−1^{−1/(𝛽−2)}(1−𝑣)^{(𝛽−1)/(𝛽−2)}+𝑣,𝑠₀∈[𝑣,1]. (38)

The following lemmas are fundamental in proof of our main results.

Lemma 10 (see [35]). Let 𝐸 be a Banach space, and let 𝐾⊂𝐸 be a cone. Assume that Ω₁, Ω₂ are open and bounded subsets of 𝐸 with 0∈Ω₁, Ω₁⊂Ω₂, and let 𝑇∶𝐾∩(Ω₂⧵Ω₁)→𝐾 be a completely continuous operator such that(i)‖𝑇𝑢‖≤‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₁, and ‖𝑇𝑢‖≥‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₂; (ii)‖𝑇𝑢‖≥‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₁, and ‖𝑇𝑢‖≤‖𝑢‖, 𝑢∈𝐾∩𝜕Ω₂.
Then, 𝑇 has a fixed point in 𝐾∩(Ω₂⧵Ω₁).

Lemma 11 (see [36]). Assume that 𝑋 is a Banach space and 𝐾⊂𝑋 is a cone. Let 𝐾_𝑟={𝑢∈𝐾;‖𝑢‖<𝑟}. Furthermore, assume that 𝑇∶𝐾→𝐾 is compact and 𝑇𝑢≠𝑢 for 𝑢∈𝜕𝐾_𝑟={𝑢∈𝐾;‖𝑢‖=𝑟}. Thus, one has the following conclusions: (i)if ‖𝑢‖≤‖𝑇𝑢‖ for 𝑢∈𝜕𝐾_𝑟, then 𝑖(𝑇,𝐾_𝑟,𝐾)=0; (ii)if ‖𝑢‖≥‖𝑇𝑢‖ for 𝑢∈𝜕𝐾_𝑟, then 𝑖(𝑇,𝐾_𝑟,𝐾)=1.

3. Main Results

In this section, we discuss the existence of positive solutions for boundary value problem (5).

Let 𝑋=𝐶[−𝜏,1+𝜃] be a Banach space with the maximum norm ‖𝑢‖_{[−𝜏,1+𝜃]}=max_{−𝜏≤𝑡≤1+𝜃}|𝑢(𝑡)| for 𝑢∈𝑋. Let 𝐾 be a cone in 𝑋 defined by 𝐾=𝑢∈𝑋;𝑢(𝑡)≥0for[];𝑢(𝑡)𝑡∈−𝜏,1+𝜃≥𝛾₁‖𝑢‖foreach𝑡∈[0,1]. (39)

Let 𝑥(𝑡)=𝑢(𝑡)−𝑥₀(𝑡). Noting that 𝑥(𝑡)≡𝑢(𝑡) for 𝑡∈[0,1], we have 𝑥(𝑡)=𝑒^{𝑎𝑡/𝑏}𝑥(0)𝑡∈[−𝜏,0₁₀]𝐺()×1𝑡,𝑠𝑝(𝑠)¹₀𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀()[]𝑒𝑣+𝜃𝑑𝑣𝑑𝑠𝑡∈0,1_{−𝑐𝑡/𝑑}𝑥(1)𝑡∈[1,1+𝜃], (40) where 𝑥+𝑥₀(𝑣−𝜏)∶=𝑥(𝑣−𝜏)+𝑥₀(𝑣−𝜏),𝑥+𝑥₀(𝑣+𝜃)∶=𝑥(𝑣+𝜃)+𝑥₀(𝑣+𝜃). (41) Define an operator 𝑇 on 𝐾 as follows: (𝑡)=𝑒𝑇𝑥)(^{𝑎𝑡/𝑏10}𝐺()×10,𝑠𝑝(𝑠)×¹₀𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀×([]𝑣+𝜃))𝑑𝑣𝑑𝑠𝑡∈−𝜏,0₁₀𝐺()×1𝑡,𝑠𝑝(𝑠)×¹₀𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀×([]𝑒𝑣+𝜃))𝑑𝑣𝑑𝑠𝑡∈0,1_{−𝑐𝑡/𝑑10}𝐺()×11,𝑠𝑝(𝑠)×¹₀𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀×([].𝑣+𝜃))𝑑𝑣𝑑𝑠𝑡∈1,1+𝜃 (42)

Lemma 12. The operator 𝑇∶𝐾→𝐾 is completely continuous.

Proof. For 𝑥∈𝐾, we have (𝑇𝑥)(𝑡)≥0, 𝑡∈[−𝜏,1+𝜃] from properties of 𝐺(𝑡,𝑠) and 𝐻(𝑠,𝑣). It follows from (55) that we have for 𝑡∈[−𝜏,0], (𝑇𝑥)(𝑡)≤(𝑇𝑥)(0), (43) and for 𝑡∈[1,1+𝜃], (𝑇𝑥)(𝑡)≤(𝑇𝑥)(1). (44) Hence, we get that ‖𝑇𝑥‖_{[−𝜏,1+𝜃]}=‖𝑇𝑥‖_[0,1]. (45) By Lemma 7 and Remark 8, we have ‖(𝑇𝑥)(𝑡)‖_[0,1]≤¹₀𝐺()×1𝑠,𝑠𝑝(𝑠)¹₀𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀(𝑣+𝜃)𝑑𝑣𝑑𝑠, (46) and (𝑇𝑥)(𝑡)≥¹₀𝛾₁𝐺()×1𝑠,𝑠𝑝(𝑠)¹₀𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀(𝑣+𝜃)𝑑𝑣𝑑𝑠≥𝛾₁‖(𝑇𝑥)(𝑡)‖_[0,1]=𝛾₁‖(𝑇𝑥)(𝑡)‖. (47) Thus, 𝑇(𝐾)⊆𝐾.
Let Ω be a bounded subset in 𝐾, that is, there exists a positive constant 𝜁>0 such that ‖𝑥‖≤𝜁, for all 𝑥∈Ω. Let 𝑄∶=max_{0≤𝑣≤1,0≤𝑥≤𝜁}|𝑓(𝑣,(𝑥+𝑥₀)(𝑣−𝜏),(𝑥+𝑥₀)(𝑣+𝜃))|+1. Then, for 𝑥∈Ω, in view of Lemma 9, we have (𝑇𝑥)(𝑡)=¹₀𝐺()×1𝑡,𝑠𝑝(𝑠)¹₀𝐻()(()(𝑠,𝑣×𝑓𝑣,𝑢𝑣−𝜏,𝑢𝑣+𝜃))𝑑𝑣𝑑𝑠≤𝑄¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠. (48) Hence, 𝑇(Ω) is bounded in 𝐾.
Now, we divide three cases to prove that 𝑇(Ω) is equicontinuous.
Case  1. If 𝑢∈Ω, 0<𝑡<1.
Then, (𝑇𝑥)^′(𝑡≤−𝑄)Γ()×𝛼−1₁₀(𝑡−𝑠)^𝛼−21𝑝(𝑠)¹₀𝑀(𝑣)+𝑑𝑣𝑑𝑠𝑎𝑄Γ(𝛼)×)(𝑏𝑐+𝑎𝑐+𝑎𝑑₁₀𝑐(1−𝑠)^𝛼−1+𝑑(𝛼−1)(1−𝑠)^𝛼−2×1𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠∶=𝑄₁. (49)
Case  2. If 𝑢∈Ω, −𝜏≤𝑡≤0.
Then, (𝑇𝑥)^′(𝑡=𝑎)𝑏𝑒^{𝑎𝑡/𝑏1}₀𝐺()×10,𝑠𝑝(𝑠)¹₀𝐻()((),(≤𝑠,𝑣×𝑓𝑣,𝑢𝑣−𝜏𝑣+𝜃))𝑑𝑣𝑑𝑠𝑎𝑄𝑏¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠∶=𝑄₂. (50)
Case  3. If 𝑢∈Ω, 1≤𝑡≤1+𝜃.
Then, (𝑇𝑥)^′(𝑡=−𝑐)𝑑𝑒^{−𝑐𝑡/𝑑}×¹₀𝐺()×11,𝑠𝑝(𝑠)¹₀𝐻()(()(≤𝑠,𝑣×𝑓𝑣,𝑢,𝑣−𝜏,𝑢𝑣+𝜃))𝑑𝑣𝑑𝑠𝑐𝑄𝑑¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠∶=𝑄₃. (51) Let 𝛿₁=𝜖₁/max{𝑄₁,𝑄₂,𝑄₃}. Then, for 𝑡₁,𝑡₂∈[−𝜏,1+𝜃], |𝑡₂−𝑡₁|<𝛿₁, we obtain (𝑇𝑥)𝑡₁−(𝑇𝑥)𝑡₂≤^𝑡2_𝑡1(𝑇𝑥)^′(𝑡𝑄)𝑑𝑡≤max₁,𝑄₂,𝑄₃𝑡₂−𝑡₁<𝜖₁. (52) Thus, 𝑇(Ω) is equicontinuous.
Next, we show that 𝑇∶𝐾→𝐾 is continuous. For any 𝑥_𝑛, 𝑥∈𝐾, 𝑛=1,2… with ‖𝑥_𝑛−𝑥‖_{[−𝜏,1+𝜃]}→0 as 𝑛→∞. Then, for 𝑡∈[0,1], we have 𝑇𝑥_𝑛(𝑡)−(𝑇𝑥)(𝑡)⩽sup_𝑣∈[0,1]𝑓𝑣,𝑥_𝑛+𝑥₀(𝑣−𝜏),𝑥_𝑛+𝑥₀(𝑣+𝜃)−𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀()×𝑣+𝜃₁₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠. (53) This implies that ‖𝑇𝑥_𝑛−𝑇𝑥‖_{[−𝜏,1+𝜃]}→0 as 𝑛→∞. Hence, 𝑇 is continuous. According to the Ascoli-Arzelà Theorem, 𝑇 is completely continuous. The proof is completed.

For convenience, we give some conditions, which will play roles in this paper for 𝑓(𝑡,𝜇,𝜈) as follows(𝐻₁)liminf_{𝜇+𝜈→0}⁺min_𝑡∈[0,1]𝑓(𝑡,𝜇,𝜈)/(𝜇+𝜈)=∞.(𝐻₂)liminf_{𝜇+𝜈→+∞}min_𝑡∈[0,1](𝑓(𝑡,𝜇,𝜈))/(𝜇+𝜈)=∞.(𝐻₃)limsup_{𝜇+𝜈→+∞}min_𝑡∈[0,1](𝑓(𝑡,𝜇,𝜈))/(𝜇+𝜈)=0.

Lemma 13. Let (𝐻₁) and (𝐻₂) hold. Then, there exist positive numbers 0<𝑟₀<𝑅₀<∞ such that 𝑖𝑇,𝐾_𝑟,𝐾=0𝑓𝑜𝑟0<𝑟≤𝑟₀,𝑖𝑇,𝐾_𝑅,𝐾=0𝑓𝑜𝑟𝑅₀≤𝑅. (54)

Proof. Choose 𝐿>0 such that 𝛾₁𝐿^3/4_1/4𝐺121,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)1𝑑𝑣𝑑𝑠>2. (55) By (𝐻₁), there is an 𝑟₀>0 such that 0<𝑟≤𝑟₀ implies that 𝑓,[].𝑣,𝜇,𝜈≥𝐿𝜇+𝜈𝜇,𝜈≥0,𝜇+𝜈≤2𝑟,𝑣∈0,1 (56) Thus, for 𝑥∈𝜕𝐾_𝑟, we have 𝑟=‖𝑥‖≥𝑥(𝑣−𝜏)≥𝛾₁‖𝑥‖≥𝛾₁1𝑟,𝑣∈4,34,𝑟=‖𝑥‖≥𝑥𝑣+𝜑≥𝛾₁‖𝑥‖≥𝛾₁1𝑟,𝑣∈4,34. (57) In view of (55) and (56), we get that (1𝑇𝑥)2≥^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀()≥𝑣+𝜃𝑑𝑣𝑑𝑠_3/41/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)×𝑓(𝑣,𝑥(𝑣−𝜏),𝑥(𝑣+𝜃))𝑑𝑣𝑑𝑠≥𝐿^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)×(𝑥(𝑣−𝜏)+𝑥(𝑣+𝜃))𝑑𝑣𝑑𝑠≥2𝛾₁𝑟𝐿^3/4_1/4𝐺121,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)‖𝑥‖.𝑑𝑣𝑑𝑠>𝑟= (58) Hence, ‖𝑇𝑥‖>‖𝑥‖. It is obvious that 𝑇𝑥≠𝑥 for 𝑥∈𝜕𝐾_𝑟. Therefore, by Lemma 11, we conclude that 𝑖(𝑇,𝐾_𝑟,𝐾)=0.
In the same way, for the same 𝐿>0 satisfying (55), (𝐻₂) implies that there is 𝑅₁>0 such that 𝑓𝑣,𝜇,𝜈≥𝐿𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈≥𝑅₁,𝑣∈[0,1]. (59) Choose 𝑅₀𝑅>max𝑟,¹2𝛾₁. (60) Thus, for 𝑥∈𝜕𝐾_𝑅, 𝑅≥𝑅₀, we have 𝑅=‖𝑥‖≥𝑥(𝑣−𝜏)≥𝛾₁‖𝑥‖≥𝛾₁𝑅≥𝛾₁𝑅₀>𝑅₁21,𝑣∈4,34,𝑅=‖𝑥‖≥𝑥(𝑣+𝜃)≥𝛾₁‖𝑥‖≥𝛾₁𝑅≥𝛾₁𝑅₀>𝑅₁21,𝑣∈4,34. (61) In view of (55) and (59), we get that (1𝑇𝑥)2≥^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀()≥𝑣+𝜃𝑑𝑣𝑑𝑠_3/41/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)×𝑓(𝑣,𝑥(𝑣−𝜏),𝑥(𝑣+𝜃))𝑑𝑣𝑑𝑠≥𝐿^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)×(𝑥(𝑣−𝜏)+𝑥(𝑣+𝜃))𝑑𝑣𝑑𝑠≥2𝛾₁𝐿𝑅^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)‖𝑥‖.𝑑𝑣𝑑𝑠>𝑅= (62) Therefore, ‖𝑇𝑥‖>‖𝑥‖. By Lemma 11, we conclude that 𝑖(𝑇,𝐾_𝑅,𝐾)=0. The proof is completed.

Lemma 14. Let (𝐻₃) hold. Then, there exists a 0<𝑅₀<∞ such that 𝑖𝑇,𝐾_𝑅,𝐾=1𝑓𝑜𝑟𝑅₀≤𝑅. (63)

Proof. By (𝐻₃), for any 0<𝜖₂<12¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠⁻¹, (64) there exists 𝑅^′>0 such that 𝑓𝑣,𝜇,𝜈≤𝜖₂𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈≥𝑅^′,𝑣∈[0,1]. (65) Setting 𝐶∶=max_𝑣∈[0,1]max_{0≤𝜇,𝜈,𝜇+𝜈≤𝑅′}𝑓𝑣,𝜇,𝜈−𝜖₂𝜇+𝜈+1, (66) we get that 𝑓𝑣,𝜇,𝜈≤𝜖₂𝜇+𝜈+𝐶,𝜇,𝜈≥0,𝑣∈[0,1]. (67) Choose 𝑅₀≥𝐶+2𝜖₂𝑥₀∫¹₀𝐺(𝑠,𝑠)1/𝑝(𝑠)∫¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠1−2𝜖₂∫¹₀𝐺(𝑠,𝑠)1/𝑝(𝑠)∫¹₀𝑀(𝑣).𝑑𝑣𝑑𝑠 (68) By (67), for 𝑥∈𝜕𝐾_𝑅, 𝑅≥𝑅₀, we have (𝑡)=𝑇𝑥)(₁₀𝐺()×1𝑡,𝑠𝑝(𝑠)¹₀𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀()≤𝑣+𝜃𝑑𝑣𝑑𝑠₁₀𝐺()×1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀()≤𝑣+𝜃𝑑𝑣𝑑𝑠₁₀𝐺()×1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)×𝜖₂𝑥+𝑥₀()+𝑣−𝜏𝑥+𝑥₀)≤(𝑣+𝜃+𝐶𝑑𝑣𝑑𝑠₁₀𝐺()×1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)2𝜖₂‖𝑥‖+𝑥₀+𝐶𝑑𝑣𝑑𝑠≤2𝜖₂𝑅¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)+𝑑𝑣𝑑𝑠𝐶+2𝜖₂𝑥₀×¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)‖𝑥‖.𝑑𝑣𝑑𝑠<𝑅= (69) Hence, ‖𝑇𝑥‖<‖𝑥‖. By Lemma 11, 𝑖(𝑇,𝐾_𝑅,𝐾)=1. The proof is completed.

Now, we prove the existence of solutions for boundary value problem (5) by using the Guo-Krasnoselskii fixed point theorem.

Theorem 15. Let 𝑏>(2−𝛼)/(𝛼−1), and let (𝐻₁) and (𝐻₃) hold. Then, boundary value problem (5) has at least a positive solution.

Proof. By (𝐻₃), there exists a 𝑍₁>0 such that 𝑓𝑣,𝜇,𝜈≤𝜀₁𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈≥𝑍₁,𝑣∈[0,1], (70) where 0<𝜀₁<12¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠⁻¹. (71)
Taking 𝐶∶=max_𝑣∈[0,1]max_{0≤𝜇,𝜈,𝜇+𝜈≤𝑍1}𝑓𝑣,𝜇,𝜈−𝜀₁𝜇+𝜈+1, (72) we get that 𝑓𝑣,𝜇,𝜈≤𝜀₁𝜇+𝜈+𝐶,𝜇,𝜈≥0,𝑣∈[0,1]. (73) Assume that 𝑍₁≥𝐶+2𝜀₁𝑥₀∫¹₀𝐺(𝑠,𝑠)1/𝑝(𝑠)∫¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠1−2𝜀₁∫¹₀𝐺(𝑠,𝑠)1/𝑝(𝑠)∫¹₀𝑀(𝑣).𝑑𝑣𝑑𝑠 (74) By (73), for any 𝑥∈𝐾 satisfied ‖𝑥‖=𝑍₁ and 𝑡∈[0,1], we have 𝑡=|(𝑇𝑥)()|₁₀𝐺()×1𝑡,𝑠𝑝(𝑠)¹₀𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀()≤𝑣+𝜃𝑑𝑣𝑑𝑠₁₀𝐺()×1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀(≤𝑣+𝜃)𝑑𝑣𝑑𝑠₁₀𝐺()×1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)×𝜀₁𝑥+𝑥₀()+𝑣−𝜏𝑥+𝑥₀()≤𝑣+𝜃+𝐶𝑑𝑣𝑑𝑠₁₀𝐺()×1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)2𝜀₁‖𝑥‖+𝑥₀+𝐶𝑑𝑣𝑑𝑠≤2𝜀₁𝑍₁¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)+𝑑𝑣𝑑𝑠𝐶+2𝜀₁𝑥₀¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠<𝑍₁=‖𝑥‖. (75) Now, if we let Ω₁=𝑥∈𝑋∶‖𝑥‖<𝑍₁, (76) then (76) shows that ‖𝑇𝑥‖_{[−𝜏,1+𝜃]}≤‖𝑥‖_{[−𝜏,1+𝜃]}for𝑥∈𝐾∩𝜕Ω₁. (77)
On the other hand, by (𝐻₁), there is an 𝑟₀>0 such that 0<𝑍₂≤𝑟₀ implies 𝑓𝑣,𝜇,𝜈≥𝐿𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈≤2𝑍₂,𝑣∈[0,1], (78) where 𝐿>0 satisfies 𝛾₁𝐿^3/4_1/4𝐺121,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)1𝑑𝑣𝑑𝑠≥2. (79) Thus for 𝑥∈𝐾 and ‖𝑥‖=𝑍₂, 𝑡∈[0,1], we have 𝑍₂=‖𝑥‖≥𝑥(𝑣−𝜏)≥𝛾₁‖𝑥‖≥𝛾₁𝑍₂1,𝑣∈4,34,𝑍₂=‖𝑥‖≥𝑥𝑣+𝜑≥𝛾₁‖𝑥‖≥𝛾₁𝑍₂1,𝑣∈4,34. (80) In view of (78) and (79), we get that 1(𝑇𝑥)2≥^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀(𝑣+𝜃)𝑑𝑣𝑑𝑠≥^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)×𝑓(𝑣,𝑥(𝑣−𝜏),𝑥(𝑣+𝜃))𝑑𝑣𝑑𝑠≥𝐿^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)×(𝑥(𝑣−𝜏)+𝑥(𝑣+𝜃))𝑑𝑣𝑑𝑠≥2𝛾₁𝑍₂𝐿×^3/4_1/4𝐺121,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)𝑑𝑣𝑑𝑠≥𝑍₂=‖𝑥‖. (81) Now, if we let Ω₂=𝑥∈𝑋∶‖𝑥‖<𝑍₂, (82) then (81) shows that ‖𝑇𝑥‖_{[−𝜏,1+𝜃]}≥‖𝑥‖_{[−𝜏,1+𝜃]}for𝑥∈𝐾∩𝜕Ω₂. (83)
Thus, by the first part of Lemma 10, 𝑇 has a fixed point 𝑥∈𝐾∩(Ω₂⧵Ω₁) with 𝑍₁≤‖𝑥‖_{[−𝜏,1+𝜃]}≤𝑍₂, and accordingly, 𝑥 is a positive solution of problem (5). The proof is completed.

Theorem 16. Let 𝑏>(2−𝛼)/(𝛼−1), and let (𝐻₂) and (𝐻₃) hold. Then, boundary value problem (5) has at least a positive solution.

Proof. By (𝐻₂) for any 𝐿>0 satisfying (126), there is a 𝑅₀>0 such that 𝑓𝑣,𝜇,𝜈≥𝐿𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈≥𝑅₀,𝑣∈[0,1]. (84) Choose 𝑍₃>max𝑍₁,𝑅₀2𝛾₁. (85) Thus for 𝑥∈𝐾, ‖𝑥‖=𝑍₃ and 𝑡∈[0,1], we have 𝑍₃=‖𝑥‖≥𝑥(𝑣−𝜏)≥𝛾₁‖𝑥‖≥𝛾₁𝑍₃>𝑅₀21,𝑣∈4,34,𝑍₃=‖𝑥‖≥𝑥(𝑣+𝜃)≥𝛾₁‖𝑥‖≥𝛾₁𝑍₃>𝑅₀21,𝑣∈4,34. (86) In view of (55) and (59), we get that 1(𝑇𝑥)2≥^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀()≥𝑣+𝜃𝑑𝑣𝑑𝑠_3/41/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)((),𝑥(×𝑓𝑣,𝑥𝑣−𝜏𝑣+𝜃))𝑑𝑣𝑑𝑠≥𝐿^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)×(𝑥(𝑣−𝜏)+𝑥(𝑣+𝜃))𝑑𝑣𝑑𝑠≥2𝛾₁𝑍₃𝐿^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)𝑑𝑣𝑑𝑠≥𝑍₃=‖𝑥‖. (87)
Now, if we let Ω₁=𝑥∈𝑋∶‖𝑥‖<𝑍₃, (88) then (144) shows that ‖𝑇𝑥‖_{[−𝜏,1+𝜃]}≥‖𝑥‖_{[−𝜏,1+𝜃]}for𝑥∈𝐾∩𝜕Ω₁. (89)
On the other hand, from Theorem 15, by (𝐻₃), there exists a 𝑍₁>0 such that 𝑓𝑣,𝜇,𝜈≤𝜀₂𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈≥𝑍₁,𝑣∈[0,1], (90) where 0<𝜀₂<12¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠⁻¹. (91) Now, if we let Ω₂=𝑥∈𝑋∶‖𝑥‖<𝑍₁, (92) then (119) shows that ‖𝑇𝑥‖_{[−𝜏,1+𝜃]}≤‖𝑥‖_{[−𝜏,1+𝜃]}for𝑥∈𝐾∩𝜕Ω₂. (93)
Thus, by the second part of Lemma 10, 𝑇 has a fixed point 𝑥∈𝐾∩(Ω₂⧵Ω₁) with 𝑍₃≤‖𝑥‖_{[−𝜏,1+𝜃]}≤𝑍₁, and accordingly, 𝑥 is a positive solution of problem (5). The proof is completed.

Next, we study the existence of solutions for boundary value problem (5) by the fixed point index theorem. For this purpose, we first give the following lemma.

Lemma 17. Let 𝐿₀∶=(𝜆∫¹₀𝐺(𝑠,𝑠)((1/𝑝(𝑠))∫¹₀𝑀(𝑣)𝑑𝑣)𝑑𝑠)⁻¹. Assume that lim_𝑢→∞max_[]𝑡∈0,1𝑓(𝑡,𝑢)𝑢=𝐿∈0,𝐿₀. (94) If there exists a function 𝑔∶[0,1]→ℝ such that 𝑓()(𝑡)1𝑡,𝑢≥𝑔𝑓𝑜𝑟𝑡∈4,34, (95) Then boundary value problem for the following fractional functional differential equation: ^𝐶𝐷^𝛽𝑝(𝑡)^𝐶𝐷^𝛼𝑢(𝑡)+𝜆𝑓(𝑡,𝑢)=0,𝑡∈(0,1_𝐶),𝐷^𝛼𝑢(0)=^𝐶𝐷^𝛼𝑢(1)=^𝐶𝐷^𝛼𝑢(0)^′′(𝑡)=0,𝑎𝑢−𝑏𝑢^′(𝑡)(𝑡)[],(𝑡)=𝜂,𝑡∈−𝜏,0𝑐𝑢+𝑑𝑢^′(𝑡)=𝜉(𝑡),𝑡∈[1,1+𝜃] (96) has at least a positive solution for 𝜆>0.

Proof. Let 𝜆>0. We define the operator 𝑇_𝜆 on 𝐾 by 𝑇_𝜆𝑢(𝑡)=𝜆¹₀𝐺()1𝑡,𝑠𝑝(𝑠)¹₀𝐻(𝑠,𝑣)𝑓(𝑣,𝑢)𝑑𝑣𝑑𝑠. (97) In view of Lemma 12, it can be verified that 𝑇_𝜆 is completely continuous. Then, we prove that 𝑇_𝜆 has a fixed point in 𝐾.
Set 𝜖₀=𝐿₀−𝐿. From the assumption, for any given 𝜖₀>0, there exists 𝑢₀>0 such that 𝑓(𝑡,𝑢)≤𝐿+𝜖₀𝑢∀𝑡∈[0,1],𝑢≥𝑢₀. (98) For 𝑢∈𝐾 and ‖𝑢‖=𝑟₁∶=max{𝜆,𝛾₁⁻¹𝑢₀}, we have 𝑇_𝜆𝑢(𝑡)≤𝜆¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑓(𝑣,𝑢)𝑑𝑣𝑑𝑠≤𝜆𝐿₀𝑟₁¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠≤𝑟₁=‖𝑢‖. (99) Now, if we let Ω₁={𝑢∈𝐾∶‖𝑢‖<𝑟₁}, then it shows that 𝑇_𝜆𝑢(𝑡)≤‖𝑢‖,for𝑢∈𝐾∩𝜕Ω₁. (100)
Denote 𝑀₀∶=^3/4_1/4𝐺()1𝑠,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)𝑔(𝑣)𝑑𝑣𝑑𝑠. (101) For 𝑢∈𝐾 and ‖𝑢‖=𝜆𝛾₁𝑀₀, we have 𝑇_𝜆𝑢(𝑡)≥𝜆^3/4_1/4𝛾₁𝐺()×1𝑠,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)𝑔(𝑣)𝑑𝑣𝑑𝑠≥𝜆𝛾₁𝑀₀=‖𝑢‖. (102) Now, if we let Ω₂={𝑢∈𝐾∶‖𝑢‖<𝜆𝛾₁𝑀₀}, then it shows that 𝑇_𝜆𝑢(𝑡)≥‖𝑢‖,for𝑢∈𝐾∩𝜕Ω₂. (103)
Thus, by Lemma 10, 𝑇_𝜆 has a fixed point 𝑢∈𝐾∩(Ω₂⧵Ω₁) between 𝑟₁ and 𝜆𝛾₁𝑀₀. Accordingly, 𝑢 is a positive solution to (96). The proof is completed.

From Lemma 17, we assume that 𝜆₁>0 is an eigenvalue of −^𝐶𝐷^𝛽𝑝(𝑡)^𝐶𝐷^𝛼𝑢=𝜆₁𝑢 (104) subject to the following conditions: 𝑎𝑢(0)−𝑏𝑢^′(0)=0,𝑐𝑢(1)+𝑑𝑢^′(1)=0, (105) and the corresponding eigenfunction 𝜑₁(𝑡)>0 for 𝑡∈(0,1) such that 𝜑_1[0,1]=max_𝑡∈[0,1]𝜑₁(𝑡)=1. (106)

For convenience, we give some conditions as follows.(𝐻₄)liminf_{𝜇+𝜈→0}⁺min_𝑡∈[0,1]𝑓(𝑡,𝜇,𝜈)/(𝜇+𝜈)>(1/2)𝜆₁(1+𝑀₁), where 𝑀₁=(𝜏+𝜃)(1−𝜏−𝜃)𝜑^′_1[0,1]+2(𝜏+𝜃)∫₀^{1−𝜏−𝜃}𝛾₁𝜑₁(𝑡+𝜏)𝑑𝑡+∫¹_𝜏+𝜃𝛾₁𝜑₁().𝑡−𝜃𝑑𝑡 (107)(𝐻₅)liminf_{𝜇+𝜈→+∞}min_𝑡∈[0,1]𝑓(𝑡,𝜇,𝜈)/(𝜇+𝜈)>(1/2)𝜆₁(1+𝑀₂), where 𝑀₂=(𝜏(1−𝜏)+𝜃(1−𝜃))𝜑^′_1[0,1]+𝜏+𝜃∫₀^1−𝜏𝛾₁𝜑₁(𝑡+𝜏)𝑑𝑡+∫¹_𝜃𝛾₁𝜑₁().𝑡−𝜃𝑑𝑡 (108)(𝐻₆)limsup_{𝜇+𝜈→+∞}max_𝑡∈[0,1]𝑓(𝑡,𝜇,𝜈)/(𝜇+𝜈)>(1/2)𝜆₁(1−𝑚), where 𝑚=(𝜏(1−𝜏)+𝜃(1−𝜃))𝜑^′_1[0,1]−+𝜏+𝜃_11−𝜏𝛾₁𝜑₁(𝑡)𝑑𝑡−^𝜃₀𝛾₁𝜑₁(𝑡)×𝑑𝑡_01−𝜏𝛾₁𝜑₁()+𝑡+𝜏𝑑𝑡_1𝜃𝛾₁𝜑₁(𝑡−𝜃)𝑑𝑡+𝜏+𝜃⁻¹<1. (109)(𝐻₇) There is a ℎ₁>0 such that 0≤𝜇≤ℎ₁+‖𝑥₀‖_[−𝜏,0], 0≤𝜈≤ℎ₁+‖𝑥₀‖_[1,1+𝜃], and 𝑡∈[0,1] imply that 𝑓𝑡,𝜇,𝜈<𝑘ℎ₁, (110) where 𝑘=(∫¹₀𝐺(𝑠,𝑠)((1/𝑝(𝑠))∫¹₀𝑀(𝑣)𝑑𝑣)𝑑𝑠)⁻¹.(𝐻₈) There is a ℎ₂>0 such that 𝛾₁ℎ₂≤𝜇≤ℎ₂, 𝛾₁ℎ₂≤𝜈≤ℎ₂, and 𝑡∈[0,1] imply that 𝑓𝑡,𝜇,𝜈>𝜗ℎ₂, (111) where 𝜗=(∫^3/4_1/4𝐺((1/2),𝑠)((1/𝑝(𝑠))∫^3/4_1/4𝛾₂(𝑣)⋅𝑀(𝑣)𝑑𝑣)𝑑𝑠)⁻¹.

Lemma 18. Let (𝐻₇) hold. Then, 𝑖(𝑇,𝐾_ℎ1,𝐾)=1.

Proof. Let 𝑥∈𝜕𝐾_ℎ1. Then, 0≤𝑥+𝑥₀(𝑣−𝜏)≤‖𝑥‖+𝑥_{0[−𝜏,1−𝜏]}=ℎ₁+𝑥_0[−𝜏,0],𝑣∈[0,1],0≤𝑥+𝑥₀(𝑣+𝜃)≤‖𝑥‖+𝑥_{0[𝜃,1+𝜃]}=ℎ₁+𝑥_0[1,1+𝜃],𝑣∈[0,1]. (112) Thus, for 𝑣∈[0,1], we have by (𝐻₇) that (𝑡)≤𝑇𝑥)(₁₀𝐺()×1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀(𝑣+𝜃)𝑑𝑣𝑑𝑠<𝑘ℎ₁¹₀𝐺()1𝑠,𝑠𝑝(𝑠)¹₀𝑀(𝑣)𝑑𝑣𝑑𝑠=ℎ₁=‖𝑥‖. (113) Hence, ‖𝑇𝑥‖<‖𝑥‖. By Lemma 11, we get 𝑖(𝑇,𝐾_ℎ1,𝐾)=1. The proof is complete.

Lemma 19. Let (𝐻₈) hold. Then, 𝑖(𝑇,𝐾_ℎ2,𝐾)=0.

Proof. Let 𝑥∈𝜕𝐾_ℎ2. Then, ℎ₂=‖𝑥‖≥𝑥(𝑣−𝜏)≥𝛾₁‖𝑥‖≥𝛾₁ℎ₂1,𝑣∈4,34,ℎ₂=‖𝑥‖≥𝑥(𝑣+𝜃)≥𝛾₁‖𝑥‖≥𝛾₁ℎ₂1,𝑣∈4,34. (114) Thus, for 𝑣∈[0,1], we have by (𝐻₈) that (𝑇𝑥)(𝑡)≥^3/4_1/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝐻(𝑠,𝑣)×𝑓𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀()≥𝑣+𝜃𝑑𝑣𝑑𝑠_3/41/4𝐺12×1,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)×𝑓(𝑣,𝑥(𝑣−𝜏),𝑥(𝑣+𝜃))𝑑𝑣𝑑𝑠>𝜗ℎ₂^3/4_1/4𝐺121,𝑠𝑝(𝑠)^3/4_1/4𝛾₂(𝑣)𝑀(𝑣)𝑑𝑣𝑑𝑠=ℎ₂=‖𝑥‖. (115) Hence, ‖𝑇𝑥‖>‖𝑥‖. By Lemma 11, we get 𝑖(𝑇,𝐾_ℎ2,𝐾)=0. The proof is complete.

Theorem 20. Let (𝐻₄), (𝐻₅), and (𝐻₇) hold. Then, boundary value problem (5) has at least two positive solutions 𝑢₁ and 𝑢₂ such that 0<𝑢_1[0,1]<ℎ₁<𝑢_2[0,1]. (116)

Proof. According to Lemma 18, we get that 𝑖(𝑇,𝐾_ℎ1,𝐾)=1. Let 0<𝑚<1<𝑛 and 𝑓₁(𝑣,𝜇,𝜈)=(𝜇+𝜈)^𝑚+(𝜇+𝜈)^𝑛, for 𝜇,𝜈≥0. Then, 𝑓₁(𝑣,𝜇,𝜈) satisfy (𝐻₁) and (𝐻₂).
Define 𝑇₁∶𝐾→𝐾 by 𝑇₁𝑥(𝑡)=𝑒^{𝑎𝑡/𝑏}×¹₀𝐺()×10,𝑠𝑝(𝑠)×¹₀𝐻(𝑠,𝑣)×𝑓₁(𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀×([],𝑣+𝜃))𝑑𝑣𝑑𝑠,𝑡∈−𝜏,0₁₀𝐺()×1𝑡,𝑠𝑝(𝑠)×¹₀𝐻(𝑠,𝑣)×𝑓₁𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀×([],𝑒𝑣+𝜃))𝑑𝑣𝑑𝑠,𝑡∈0,1_{−𝑐𝑡/𝑑10}𝐺()×11,𝑠𝑝(𝑠)×¹₀𝐻(𝑠,𝑣)×𝑓₁𝑣,𝑥+𝑥₀(𝑣−𝜏),𝑥+𝑥₀×([].𝑣+𝜃))𝑑𝑣𝑑𝑠,𝑡∈1,1+𝜃 (117) It is easy to know that 𝑇₁ is a completely continuous operator.
Also, according to Lemma 13, we have that there exist 0<𝑟₀<𝑅₀<∞, such that 𝑖𝑇₁,𝐾_𝑟,𝐾=0,0<𝑟≤𝑟₀,𝑖𝑇₁,𝐾_𝑅,𝐾=0,𝑅≥𝑅₀. (118)(119)
Define 𝐴∶[0,1]×𝐾→𝐾 by 𝐴(𝑠,𝑥)=(1−𝑠)𝑇𝑥+𝑠𝑇₁𝑥. For any 𝑠₁,𝑠₂∈[0,1], 𝜔>0, we have 𝐴𝑠₁,𝑥−𝐴𝑠₂≤,𝑥1−𝑠₁𝑇𝑥+𝑠₁𝑇₁𝑥−1−𝑠₂𝑇𝑥−𝑠₂𝑇₁𝑥≤𝑠₁−𝑠₂𝑇₁𝑥+‖𝑇𝑥‖. (120) Note that ‖𝑇₁𝑥‖+‖𝑇𝑥‖ is uniformly bounded in 𝐾_𝜔. Thus, 𝐴(𝑠,𝑥) is continuous on 𝑥∈𝐾_𝜔 uniformly for 𝑠∈[0,1]. Then, we conclude that 𝐴 is a completely continuous operator on [0,1]×𝐾_𝜔.
By (𝐻₄) and definition of 𝑓₁, there exists a 𝜖₃>0 and 0<𝑟₁≤𝑟₀ such that 𝑓≥1𝑡,𝜇,𝜈2𝜆₁1+𝑀₁+𝜖₃𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈≤2𝑟₁[],𝑓,𝑡∈0,1₁≥1𝑡,𝜇,𝜈2𝜆₁1+𝑀₁+𝜖₃𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈≤2𝑟₁,𝑡∈[0,1]. (121)
Next, we prove that 𝐴(𝑠,𝑥)≠𝑥 for all 𝑠∈[0,1] and 𝑥∈𝜕𝐾_𝑟1.
In fact, if there exist 𝑠₁∈[0,1] and 𝑥₁∈𝜕𝐾_𝑟1 such that 𝐴(𝑠₁,𝑥₁)=𝑥₁, then 𝑥₁(𝑡) satisfies the following: −^𝐶𝐷^𝛽𝑝(𝑡)^𝐶𝐷^𝛼𝑥₁(𝑡)=1−𝑠₁𝑓𝑡,𝑥₁+𝑥₀(𝑡−𝜏),𝑥₁+𝑥₀(𝑡+𝜃)+𝑠₁𝑓₁𝑡,𝑥₁+𝑥₀(𝑡−𝜏),𝑥₁+𝑥₀(𝑡+𝜃) (122) and the following boundary conditions: 𝑎𝑥₁(𝑡)−𝑏𝑥^′₁(𝑡)=0,𝑡∈[−𝜏,0],𝑐𝑥₁(𝑡)+𝑑𝑥^′₁(𝑡)=0,𝑡∈[1,1+𝜃]. (123) Multiplying both sides of (122) by 𝜑₁(𝑡), then integrating from 0 to 1 and using Green’s formula, we obtain 𝜆₁¹₀𝑥₁(𝑡)𝜑₁(𝑡)=𝑑𝑡₁₀−^𝐶𝐷^𝛽𝑝(𝑡)^𝐶𝐷^𝛼𝑥₁(𝑡)𝜑₁(𝑡)=𝑑𝑡1−𝑠₁×¹₀𝑓𝑡,𝑥₁+𝑥₀(𝑡−𝜏),𝑥₁+𝑥₀(𝑡+𝜃)𝜑₁(𝑡)𝑑𝑡+𝑠₁¹₀𝑓₁𝑡,𝑥₁+𝑥₀(𝑡−𝜏),𝑥₁+𝑥₀(𝑡+𝜃)𝜑₁(𝑡)𝑑𝑡. (124) Thus, we know that 𝜆₁¹₀𝑥₁(𝑡)𝜑₁(𝑡)≥𝑑𝑡1−𝑠₁×_𝜏^1−𝜃𝑓𝑡,𝑥₁+𝑥₀(𝑡−𝜏),𝑥₁+𝑥₀(𝑡+𝜃)×𝜑₁(𝑡)𝑑𝑡+𝑠_1𝜏^1−𝜃𝑓₁𝑡,𝑥₁+𝑥₀(𝑡−𝜏),𝑥₁+𝑥₀(𝑡+𝜃)×𝜑₁(𝑡)=𝑑𝑡1−𝑠_1𝜏^1−𝜃𝑓𝑡,𝑥₁(𝑡−𝜏),𝑥₁(𝑡+𝜃)𝜑₁(𝑡)𝑑𝑡+𝑠_1𝜏^1−𝜃𝑓₁𝑡,𝑥₁(𝑡−𝜏),𝑥₁(𝑡+𝜃)𝜑₁(𝑡)≥1𝑑𝑡21−𝑠₁𝜆₁1+𝑀₁+𝜖₃×_𝜏^1−𝜃𝑥₁(𝑡−𝜏)+𝑥₁(𝑡+𝜃)𝜑₁(𝑡)+1𝑑𝑡2𝑠₁𝜆₁1+𝑀₁+𝜖₃×_𝜏^1−𝜃𝑥₁(𝑡−𝜏)+𝑥₁(𝑡+𝜃)𝜑₁(𝑡)=1𝑑𝑡2𝜆₁1+𝑀₁+𝜖₃×_𝜏^1−𝜃𝑥₁(𝑡−𝜏)𝜑₁(𝑡)+𝑑𝑡_𝜏1−𝜃𝑥₁(𝑡+𝜃)𝜑₁(𝑡)=1𝑑𝑡2𝜆₁1+𝑀₁+𝜖₃×₀^{1−𝜃−𝜏}𝑥₁(𝑡)𝜑₁()+𝑡+𝜏𝑑𝑡_1𝜏+𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡. (125) Then, we have 𝜆₁𝑀₁+𝜖₃×₀^{1−𝜃−𝜏}𝑥₁(𝑡)𝜑₁()+𝑡+𝜏𝑑𝑡_1𝜏+𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡≤2𝜆₁¹₀𝑥₁(𝑡)𝜑₁(𝑡)𝑑𝑡−𝜆₁₀^{1−𝜃−𝜏}𝑥₁(𝑡)𝜑₁()+𝑡+𝜏𝑑𝑡_1𝜏+𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡=𝜆₁₀^{1−𝜃−𝜏}𝑥₁(𝑡)𝜑₁(𝑡)−𝜑₁(𝑡+𝜏)𝑑𝑡+𝜆₁¹_{1−𝜏−𝜃}𝑥₁(𝑡)𝜑₁(𝑡)𝑑𝑡+𝜆₁¹_𝜏+𝜃𝑥₁(𝑡)𝜑₁(𝑡)−𝜑₁(𝑡−𝜃)𝑑𝑡+𝜆₁₀^𝜏+𝜃𝑥₁(𝑡)𝜑₁(𝑡)𝑑𝑡≤𝜆₁𝜏(1−𝜃−𝜏)𝜑^′_1[0,1]𝑥₁+𝜆₁(𝜃+𝜏)𝜑_1[0,1]𝑥₁+𝜆₁𝜃(1−𝜃−𝜏)𝜑^′_1[0,1]𝑥₁+𝜆₁(𝜃+𝜏)𝜑_1[0,1]𝑥₁=𝜆₁(𝜃+𝜏)(1−𝜃−𝜏)𝜑^′_1[0,1]+2(𝜃+𝜏)𝑥₁. (126) By the definition of 𝐾, we have ₀^{1−𝜃−𝜏}𝑥₁(𝑡)𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜏+𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡≥𝛾₁𝑥₁₀^{1−𝜃−𝜏}𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜃+𝜏𝜑₁(𝑡−𝜃)𝑑𝑡. (127) Combining with (126), we get that 𝜆₁𝑀₁+𝜖₃𝛾₁₀^{1−𝜃−𝜏}𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜃+𝜏𝜑₁(𝑡−𝜃)𝑑𝑡≤𝜆₁(𝜃+𝜏)(1−𝜃−𝜏)𝜑^′_1[0,1]+2(𝜃+𝜏), (128) which is a contradiction for definition of 𝑀₁ in (𝐻₄). Thus, 𝐴(𝑠,𝑥)≠𝑥 for 𝑥∈𝜕𝐾_𝑟 and 𝑠∈[0,1]. In view of (118) and homotopy invariance of the fixed point index, we obtain 𝑖𝑇,𝐾_𝑟1,𝐾=𝑖𝐻(0,⋅),𝐾_𝑟1𝐻(),𝐾=𝑖1,⋅,𝐾_𝑟1,𝐾=𝑖𝑇₁,𝐾_𝑟1,𝐾=0. (129) On the other hand, by (𝐻₅) and definition of 𝑓₁, there are 𝜖₄>0 and 𝑅₁>ℎ₁ such that 𝑓≥1𝑡,𝜇,𝜈2𝜆₁1+𝑀₂+𝜖₄𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈>𝑅^′[],𝑓,𝑡∈0,1₁≥1𝑡,𝜇,𝜈2𝜆₁1+𝑀₂+𝜖₄𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈>𝑅^′,𝑡∈[0,1]. (130) Setting 𝐶∶=max_𝑡∈[0,1]max_{0≤𝜇,𝜈,𝜇+𝜈≤𝑅′}𝑓𝑡,𝜇,𝜈−𝜆₁1+𝑀₂+𝜖₄𝜇+𝜈+max_𝑡∈[0,1]max_{0≤𝜇,𝜈,𝜇+𝜈≤𝑅′}𝑓₁−𝜆𝑡,𝜇,𝜈₁1+𝑀₂+𝜖₄𝜇+𝜈+1, (131) and it follows that 𝑓≥1𝑡,𝜇,𝜈2𝜆₁1+𝑀₂+𝜖₄[],𝑓𝜇+𝜈+𝐶,𝜇,𝜈≥0,𝑡∈0,1₁≥1𝑡,𝜇,𝜈2𝜆₁1+𝑀₂+𝜖₄[].𝜇+𝜈+𝐶,𝜇,𝜈≥0,𝑡∈0,1 (132) Then, we prove that there exists a 𝑅₁≥𝑅^′ such that 𝐴(𝑠,𝑥)≠𝑥 for all 𝑠∈[0,1] and 𝑥∈𝐾, ‖𝑥‖≥𝑅₁. In fact, if there exist 𝑠₁∈[0,1] and 𝑥₁∈𝐾 such that 𝐴(𝑠₁,𝑥₁)=𝑥₁, we get that 𝜆₁¹₀𝑥₁(𝑡)𝜑₁(𝑡)≥1𝑑𝑡21−𝑠₁𝜆₁1+𝑀₂+𝜖₄×¹₀𝑥₁+𝑥₀(𝑡−𝜏)+𝑥₁+𝑥₀(𝑡+𝜃)×𝜑₁(𝑡)𝑑𝑡−1−𝑠₁𝐶¹₀𝜑₁(𝑡)+1𝑑𝑡2𝑠₁𝜆₁1+𝑀₂+𝜖₄×¹₀𝑥₁+𝑥₀(𝑡−𝜏)+𝑥₁+𝑥₀(𝑡+𝜃)×𝜑₁(𝑡)𝑑𝑡−𝑠₁𝐶¹₀𝜑₁(𝑡)≥1𝑑𝑡2𝜆₁1+𝑀₂+𝜖₄×¹₀𝑥₁(𝑡−𝜏)+𝑥₁(𝑡+𝜃)𝜑₁(𝑡)𝑑𝑡−𝐶¹₀𝜑₁(𝑡)≥1𝑑𝑡2𝜆₁1+𝑀₂+𝜖₄×¹_𝜏𝑥₁(𝑡−𝜏)𝜑₁(𝑡)𝑑𝑡+₀^1−𝜃𝑥₁(𝑡+𝜃)𝜑₁(𝑡)𝑑𝑡+𝐶¹₀𝜑₁(𝑡)=1𝑑𝑡2𝜆₁1+𝑀₂+𝜖₄×₀^1−𝜏𝑥₁(𝑡)𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡+𝐶¹₀𝜑₁(𝑡)𝑑𝑡. (133) Then, we have 𝜆₁𝑀₂+𝜖₄×₀^1−𝜏𝑥₁(𝑡)𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡≤𝜆₁₀^1−𝜏𝑥₁(𝑡)𝜑₁(𝑡)−𝜑₁(𝑡+𝜏)𝑑𝑡+𝜆₁¹_1−𝜏𝑥₁(𝑡)𝜑₁(𝑡)𝑑𝑡+𝜆₁¹_𝜃𝑥₁(𝑡)𝜑₁(𝑡)−𝜑₁(𝑡−𝜃)𝑑𝑡+𝜆₁^𝜃₀𝑥₁(𝑡)𝜑₁(𝑡)𝑑𝑡+2𝐶¹₀𝜑₁(𝑡)𝑑𝑡≤𝜆₁𝜏(1−𝜏)𝜑^′_1[0,1]𝑥₁+𝜆₁𝜏𝜑_1[0,1]𝑥₁+𝜆₁𝜃(1−𝜃)𝜑^′_1[0,1]𝑥₁+𝜆₁𝜃𝜑_1[0,1]𝑥₁+2𝐶𝜑_1[0,1]=𝜆₁𝜏(1−𝜏)+𝜃(1−𝜃)𝜑^′_1[0,1]×𝑥+𝜃+𝜏₁+2𝐶. (134) By the definition of 𝐾, we get that ₀^1−𝜏𝑥₁(𝑡)𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡≥𝛾₁𝑥₁₀^1−𝜏𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜃𝜑₁(𝑡−𝜃)𝑑𝑡. (135) Combining with (134), we conclude that 𝜆₁𝑀₂+𝜖₄𝛾₁𝑥₁₀^1−𝜏𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜃𝜑₁(𝑡−𝜃)𝑑𝑡≤𝜆₁(𝜏(1−𝜏)+𝜃(1−𝜃))𝜑^′_1[0,1]+𝜃+𝜏𝑥₁+2𝐶, (136) that is, 𝑥₁≤2𝐶𝜖₄𝛾₁∫₀^1−𝜏𝜑₁(𝑡+𝜏)𝑑𝑡+∫¹_𝜃𝜑₁()=𝑡−𝜃𝑑𝑡𝑅. (137) Let 𝑅₁=1+max{𝑅^′,𝑅}. Then, we obtain 𝐴(𝑠,𝑥)≠𝑥 for 𝑥∈𝜕𝐾_𝑟 and 𝑠∈[0,1]. In view of (118) and homotopy invariance of the fixed point index, we obtain 𝑖𝑇,𝐾_𝑅1,𝐾=𝑖𝐴(0,⋅),𝐾_𝑅1,𝐾=𝑖𝐴(1,⋅),𝐾_𝑅1𝑇,𝐾=𝑖₁,𝐾_𝑅1,𝐾=0. (138) By using additivity, we get that 𝑖𝑇,𝐾_𝑅1⧵𝐾_ℎ1,𝐾=−1,𝑖𝑇,𝐾_ℎ1⧵𝐾_𝑟1,𝐾=1. (139) Since 𝑢_𝑖[0,1]=𝑥_𝑖+𝑥_0[0,1]=𝑥_𝑖[0,1]=𝑇𝑥_𝑖[0,1]=𝑇𝑥_𝑖=𝑥_𝑖, (140) it follows that 𝑢₁(𝑡) and 𝑢₂(𝑡) satisfy 0<𝑢_1[0,1]=𝑥₁<ℎ₁<𝑥₂=𝑢_2[0,1]. (141) The proof is completed.

Theorem 21. Let (𝐻₆), (𝐻₇), and (𝐻₈) hold. Assume that 0<ℎ₁<ℎ₂; then boundary value problem (5) has at least two positive solutions 𝑢₁ and 𝑢₂ such that ℎ₁<𝑢_1[0,1]<ℎ₂<𝑢_2[0,1]. (142)

Proof. According to Lemmas 18 and 19, we obtain 𝑖𝑇,𝐾_ℎ1,𝐾=1,𝑖𝑇,𝐾_ℎ2,𝐾=0. (143)(144) By (𝐻₆) and definition of 𝑓₁, there are 𝜖₅>0 and 𝑅₂>ℎ₂ such that 𝑓≤1𝑡,𝜇,𝜈2𝜆₁1−𝑀₃−𝜖₅𝜇+𝜈,𝜇,𝜈≥0,𝜇+𝜈>𝑅^′,𝑡∈[0,1]. (145) Let 𝐶∶=max_𝑡∈[0,1]max_{0≤𝜇,𝜈,𝜇+𝜈≤𝑅′}𝑓−𝜆𝑡,𝜇,𝜈₁1−𝑀₃−𝜖₅𝜇+𝜈+1, (146) and it follows that 𝑓≤1𝑡,𝜇,𝜈2𝜆₁1−𝑀₃−𝜖₅[].𝜇+𝜈+𝐶,𝜇,𝜈≥0,𝑡∈0,1 (147)
Define 𝐴∶[0,1]×𝐾→𝐾 by 𝐴(𝑠,𝑥)=𝑠𝑇𝑥. For any 𝑠₁,𝑠₂∈[0,1], 𝜔>0, we have 𝐴𝑠₁,𝑥−𝐴𝑠₂,𝑥≤𝑠₁−𝑠₂‖𝑇𝑥‖. (148) Thus, 𝐴(𝑠,𝑥) is continuous on 𝑥∈𝐾_𝜔 uniformly for 𝑠∈[0,1]. Then, we conclude that 𝐴 is also a completely continuous operator on [0,1]×𝐾_𝜔.
Next, we prove that there exists a 𝑅₁≥𝑅^′ such that 𝐴(𝑠,𝑥)≠𝑥 for all 𝑠∈[0,1] and 𝑥∈𝐾, ‖𝑥‖≥𝑅₁.
In fact, if there exist 𝑠₂∈[0,1] and 𝑥₂∈𝐾 such that 𝐴(𝑠₂,𝑥₂)=𝑥₂, then 𝑥₂(𝑡) satisfies the following: −^𝐶𝐷^𝛽𝑝(𝑡)^𝐶𝐷^𝛼𝑥₂(𝑡)=𝑠₂𝑓₁𝑡,𝑥₂+𝑥₀(),𝑥𝑡−𝜏₂+𝑥₀(𝑡+𝜃),𝑡∈(0,1). (149) Multiplying both sides of (149) by 𝜑₁(𝑡) then integrating from 0 to 1 and using Green’s formula, we get that 𝜆₁¹₀𝑥₂(𝑡)𝜑₁(𝑡)=𝑑𝑡₁₀−^𝐶𝐷^𝛽𝑝(𝑡)^𝐶𝐷^𝛼𝑥₂(𝑡)𝜑₁(𝑡)𝑑𝑡=𝑠₂¹₀𝑓₁𝑡,𝑥₂+𝑥₀(𝑡−𝜏),𝑥₂+𝑥₀(𝑡+𝜃)𝜑₁(𝑡)𝑑𝑡. (150) Thus, we get that 𝜆₁¹₀𝑥₁(𝑡)𝜑₁(𝑡)𝑑𝑡=𝑠₁¹₀𝑓𝑡,𝑥₁+𝑥₀(),𝑥𝑡−𝜏₁+𝑥₀(𝑡+𝜃)𝜑₁(𝑡)≤1𝑑𝑡2𝜆₁1−𝑀₃−𝜖₅×¹₀𝑥₁+𝑥₀)+𝑥(𝑡−𝜏₁+𝑥₀(𝑡+𝜃)𝜑₁(𝑡)𝑑𝑡+𝐶¹₀𝜑₁(𝑡)=1𝑑𝑡2𝜆₁1−𝑀₃−𝜖₅×^𝜏₀𝑥₁+𝑥₀(𝑡−𝜏)𝜑₁(𝑡)+𝑑𝑡_1𝜏𝑥₁+𝑥₀(𝑡−𝜏)𝜑₁(𝑡)+𝑑𝑡_01−𝜃𝑥₁+𝑥₀(𝑡+𝜃)𝜑₁(𝑡)+𝑑𝑡_11−𝜃𝑥₁+𝑥₀(𝑡+𝜃)𝜑₁(𝑡)𝑑𝑡+𝐶¹₀𝜑₁(𝑡)=1𝑑𝑡2𝜆₁1−𝑀₃−𝜖₅×^𝜏₀𝑥₁+𝑥₀(𝑡−𝜏)𝜑₁(𝑡)+𝑑𝑡_01−𝜏𝑥₁(𝑡)𝜑₁()+𝑡+𝜏𝑑𝑡_1𝜃𝑥₁(𝑡)𝜑₁()+𝑡−𝜃𝑑𝑡_11−𝜃𝑥₁+𝑥₀(𝑡+𝜃)𝜑₁(𝑡)𝑑𝑡+𝐶¹₀𝜑₁(𝑡)≤1𝑑𝑡2𝜆₁1−𝑀₃−𝜖₅×𝜏𝑥₁+𝑥₀𝜑_1[0,1]+₀^1−𝜏𝑥₁(𝑡)𝜑₁()+𝑡+𝜏𝑑𝑡_1𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡+𝜃𝑥₁+𝑥₀𝜑_1[0,1]+𝐶𝜑_1[0,1]≤12𝜆₁1−𝑀₃−𝜖₅×(𝜏+𝜃)𝑥₁+𝑥₀+₀^1−𝜏𝑥₁(𝑡)𝜑₁()+𝑡+𝜏𝑑𝑡_1𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡+𝜑_1[0,1]+𝐶. (151) Then, we have 𝜆₁𝑀₃+𝜖₅×₀^1−𝜏𝑥₁(𝑡)𝜑₁()+𝑡+𝜏𝑑𝑡_1𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡+(𝜏+𝜃)𝑥₁+𝑥₀≤𝜆₁₀^1−𝜏𝑥₁(𝑡)𝜑₁(𝑡+𝜏)−𝜑₁(𝑡)𝑑𝑡+𝜆₁¹_𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)−𝜑₁(𝑡)𝑑𝑡+𝜆₁(𝜃+𝜏)𝑥₁+𝑥₀−𝜆₁¹_1−𝜏𝑥₁(𝑡)𝜑₁(𝑡)𝑑𝑡−𝜆₁^𝜃₀𝑥₁(𝑡)𝜑₁(𝑡)𝑑𝑡+2𝐶≤𝜆₁𝜏(1−𝜏)𝜑^′_1[0,1]𝑥₁+𝜆₁𝜃(1−𝜃)𝜑^′_1[0,1]𝑥₁−𝜆₁𝛾₁¹_1−𝜏𝜑₁(𝑡)𝑑𝑡𝑥₁+𝜆₁(𝜃+𝜏)𝑥₁+𝑥₀−𝜆₁𝛾₁^𝜃₀𝜑₁(𝑡)𝑑𝑡𝑥₁+2𝐶. (152) By the definition of 𝐾, we get that ₀^1−𝜏𝑥₁(𝑡)𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜃𝑥₁(𝑡)𝜑₁(𝑡−𝜃)𝑑𝑡≥𝛾₁𝑥₁₀^1−𝜏𝜑₁(𝑡+𝜏)𝑑𝑡+¹_𝜃𝜑₁(𝑡−𝜃)𝑑𝑡. (153) This combined with (126) gives, 𝜆₁𝑀₃+𝜖₅×𝛾₁₀^1−𝜏𝜑₁()+𝑡+𝜏𝑑𝑡_1𝜃𝜑₁(𝑡−𝜃)𝑑𝑡+𝜏+𝜃𝑥₁≤𝜆₁𝜏(1−𝜏)𝜑^′_1[0,1]+𝜃(1−𝜃)𝜑^′_1[0,1]−+𝜃+𝜏_11−𝜏𝜑₁(𝑡)−𝑑𝑡_𝜃1𝜑₁(𝑡)𝑑𝑡𝑥₁+2𝐶, (154) that is, 𝑥₁≤2𝐶+𝜆₁(𝜃+𝜏)𝑥₀𝜖₅𝛾₁∫₀^1−𝜏𝜑₁(𝑡+𝜏)𝑑𝑡+∫¹_𝜃𝜑₁()=𝑡−𝜃𝑑𝑡+𝜏+𝜃𝑅. (155) Let 𝑅₁=1+max{𝑅^′,𝑅}. Then, we obtain 𝐴(𝑠,𝑥)≠𝑥 for 𝑥∈𝜕𝐾_𝑟 and 𝑠∈[0,1]. In view of (143), (144), and homotopy invariance of the fixed point index, we obtain 𝑖𝑇,𝐾_𝑅1,𝐾=𝑖𝐴(1,⋅),𝐾_𝑅1,𝐾=𝑖𝐴(0,⋅),𝐾_𝑅1𝑇,𝐾=𝑖₁,𝐾_𝑅1,𝐾=1. (156) By using additivity, we get that 𝑖𝑇,𝐾_𝑅1⧵𝐾_ℎ2,𝐾=1,𝑖𝑇,𝐾_ℎ2⧵𝐾_ℎ1,𝐾=−1. (157) Thus 𝑇 has fixed points 𝑥₁ and 𝑥₂ in 𝐾_𝑅1⧵𝐾_ℎ2 and 𝐾_ℎ2⧵𝐾_ℎ1, which means that 𝑢₁(𝑡)=𝑥₁(𝑡)+𝑥₀(𝑡) and 𝑢₂(𝑡)=𝑥₂(𝑡)+𝑥₀(𝑡) are positive solutions of boundary value problem (5) and ℎ₁<𝑢_1[0,1]=𝑥₁<ℎ₂<𝑥₂=𝑢_2[0,1]. (158) The proof is completed.

Corollary 22. Let (𝐻₄), (𝐻₇), and (𝐻₈) hold. Assume that 0<ℎ₁<ℎ₂. Then, boundary value problem (5) has at least two positive solutions 𝑢₁ and 𝑢₂ such that 0<𝑢_1[0,1]<ℎ₁<𝑢_2[0,1]. (159)

Corollary 23. Let (𝐻₅), (𝐻₇) and (𝐻₈) hold. Assume 0<ℎ₂<ℎ₁. Then boundary value problem (5) has at least two positive solutions 𝑢₁ and 𝑢₂ such that ℎ₂<𝑢_1[0,1]<ℎ₁<𝑢_2[0,1]. (160)

4. Example

In this section, we will present an example to illustrate our main results.

Consider the following fractional functional differential equations with delay: ^𝐶𝐷^5/21𝑡−2^2𝐶𝐷^3/2𝑢(𝑡)+(𝑡+1)²+𝑢1𝑡−51+𝑢𝑡+5^1/3=0,𝑡∈[0,1_𝐶],𝐷^3/2𝑢(0)=^𝐶𝐷^3/2𝑢(1)=^𝐶𝐷^3/2𝑢(0)^′′𝑢(𝑡)=0,−2𝑢^′(𝑡)−1=sin𝑡,𝑡∈3(𝑡),03𝑢+5𝑢^′(𝑡)4=cos𝑡,𝑡∈1,3. (161)

By a simple computation, we can get that 𝑏>((2−𝛼)/(𝛼−1))𝑎 and min_[]𝑡∈0,1𝑓𝑡,𝜇,𝜈𝜇+𝜈=min_𝑡∈[0,1](𝑡+1)²+(𝑢+𝑣)^1/3=1𝑢+𝑣+1𝑢+𝑣(𝑢+𝑣)^2/3⟶∞, (162) as 𝑢+𝑣→0⁺.

Similarly, we can obtain min_𝑡∈[0,1]𝑓(𝑡,𝜇,𝜈)/(𝜇+𝜈)→0 as 𝑢+𝑣→+∞.

Then, conditions (𝐻₁) and (𝐻₃) are satisfied. Then, by Theorem 15, boundary value problem (161) has at least a positive solution.

Acknowledgments

The authors sincerely thank the reviewers for their valuable suggestions and useful comments that have led to the present improved version of the original paper. This research is supported by the Natural Science Foundation of China (11071143), Natural Science Outstanding Youth Foundation of Shandong Province (JQ201119), Shandong Provincial Natural Science Foundation (ZR2012AM009, ZR2011AL007), Natural Science Foundation of Educational Department of Shandong Province (J11LA01).