Positive Solutions of Eigenvalue Problems for a Class of Fractional Differential Equations with Derivatives

Abstract

By establishing a maximal principle and constructing upper and lower solutions, the existence of positive solutions for the eigenvalue problem of a class of fractional differential equations is discussed. Some sufficient conditions for the existence of positive solutions are established.

1. Introduction

In this paper, we discuss the existence of positive solutions for the following eigenvalue problem of a class fractional differential equation with derivatives−𝓓𝓓_𝐭𝛼𝑥(𝑡)=𝜆𝑓𝑡,𝑥(𝑡),𝓓_𝐭𝛽𝑥(𝑡),𝑡∈(0,1),_𝐭𝛽𝑥(0)=0,𝓓_𝐭𝛾𝑥(1)=^𝑝−2_𝑗=1𝑎_𝑗𝓓_𝐭𝛾𝑥𝜉_𝑗,(1.1) where 𝜆 is a parameter, 1<𝛼≤2, 𝛼−𝛽>1, 0<𝛽≤𝛾<1, 0<𝜉₁<𝜉₂<⋯<𝜉_𝑝−2<1, 𝑎_𝑗∈[0,+∞) with 𝑐=∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1}<1, and 𝓓_𝐭 is the standard Riemann-Liouville derivative. 𝑓∶(0,1)×(0,+∞)×(0,+∞)→[0,+∞) is continuous, and 𝑓(𝑡,𝑢,𝑣) may be singular at 𝑢=0,𝑣=0, and 𝑡=0,1.

As fractional order derivatives and integrals have been widely used in mathematics, analytical chemistry, neuron modeling, and biological sciences [1–6], fractional differential equations have attracted great research interest in recent years [7–17]. Recently, ur Rehman and Khan [8] investigated the fractional order multipoint boundary value problem:𝓓𝑦(_𝐭𝛼0𝑦)(=𝑡)0=,𝑓𝓓𝑡,_𝐭𝛽𝑦(𝑡),𝓓_𝐭𝛽𝑦(𝑡),𝑡∈(0,1),𝑦(1)−^𝑚−2_𝑖=1𝜁_𝑖𝓓_𝐭𝛽𝑦𝜉_𝑖=𝑦₀,(1.2) where 1<𝛼≤2,  0<𝛽<1, 0<𝜉_𝑖<1,  𝜁_𝑖∈[0,+∞) with ∑^{𝑚−2𝑖=1}𝜁_𝑖𝜉^{𝑖𝛼−𝛽−1}<1. The Schauder fixed point theorem and the contraction mapping principle are used to establish the existence and uniqueness of nontrivial solutions for the BVP (1.2) provided that the nonlinear function 𝑓∶[0,1]×ℝ×ℝ is continuous and satisfies certain growth conditions. But up to now, multipoint boundary value problems for fractional differential equations like the BVP (1.1) have seldom been considered when 𝑓(𝑡,𝑢,𝑣) has singularity at 𝑡=0 and (or) 1 and also at 𝑢=0,𝑣=0. We will discuss the problem in this paper.

The rest of the paper is organized as follows. In Section 2, we give some definitions and several lemmas. Suitable upper and lower solutions of the modified problems for the BVP (1.1) and some sufficient conditions for the existence of positive solutions are established in Section 3.

2. Preliminaries and Lemmas

For the convenience of the reader, we present here some definitions about fractional calculus.

Definition 2.1 (See [1, 6]). Let 𝛼>0 with 𝛼∈ℝ. Suppose that 𝑥∶[𝑎,∞)→ℝ. Then the 𝛼th Riemann-Liouville fractional integral is defined by 𝐼^𝛼𝑥(𝑡)=1Γ(𝛼)^𝑡𝑎(𝑡−𝑠)^𝛼−1𝑥(𝑠)𝑑𝑠(2.1) whenever the right-hand side is defined. Similarly, for 𝛼∈ℝ with 𝛼>0, we define the 𝛼th Riemann-Liouville fractional derivative by 𝓓^𝛼𝑥(𝑡)=1Γ(𝑛−𝛼)𝑑𝑑𝑡^(𝑛)^𝑡𝑎(𝑡−𝑠)^{𝑛−𝛼−1}𝑥(𝑠)𝑑𝑠,(2.2)where 𝑛∈ℕ is the unique positive integer satisfying 𝑛−1≤𝛼<𝑛 and 𝑡>𝑎.

Remark 2.2. If 𝑥,𝑦∶(0,+∞)→ℝ with order 𝛼>0, then 𝓓_𝐭𝛼(𝑥(𝑡)+𝑦(𝑡))=𝓓_𝐭𝛼𝑥(𝑡)+𝓓_𝐭𝛼𝑦(𝑡).(2.3)

Lemma 2.3 (See [6]). One has the following.

(1) If 𝑥∈𝐿¹(0,1),𝜈>𝜎>0, then 𝐼^𝜈𝐼^𝜎𝑥(𝑡)=𝐼^𝜈+𝜎𝑥(𝑡),𝓓_𝐭𝜎𝐼^𝜈𝑥(𝑡)=𝐼^𝜈−𝜎𝑥(𝑡),𝓓_𝐭𝜎𝐼^𝜎𝑥(𝑡)=𝑥(𝑡).(2.4)

(2) If 𝜈>0,𝜎>0, then 𝓓_𝐭𝜈𝑡^𝜎−1=Γ(𝜎)Γ(𝜎−𝜈)𝑡^{𝜎−𝜈−1}.(2.5)

Lemma 2.4 (See [6]). Let 𝛼>0. Assume that 𝑥∈𝐶(0,1)∩𝐿¹(0,1). Then𝐼^𝛼𝓓_𝐭𝛼𝑥(𝑡)=𝑥(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2+⋯+𝑐_𝑛𝑡^𝛼−𝑛,(2.6) where 𝑐_𝑖∈ℝ(𝑖=1,2,…,𝑛), and 𝑛 is the smallest integer greater than or equal to 𝛼.

Let𝑘₁(𝑡,𝑠)=⎧⎪⎨⎪⎩𝑡^{𝛼−𝛽−1}(1−𝑠)^{𝛼−𝛾−1}−(𝑡−𝑠)^{𝛼−𝛽−1}𝑡Γ(𝛼−𝛽),0≤𝑠≤𝑡≤1,_{𝛼−𝛽−1}(1−𝑠)^{𝛼−𝛾−1}𝑘Γ(𝛼−𝛽),0≤𝑡≤𝑠≤1,₂(𝑡,𝑠)=⎧⎪⎨⎪⎩(𝑡(1−𝑠))^{𝛼−𝛾−1}−(𝑡−𝑠)^{𝛼−𝛾−1}(𝑡(1−𝑠))Γ(𝛼−_{𝛼−𝛾−1}𝛽),0≤𝑠≤𝑡≤1,Γ(𝛼−𝛽),0≤𝑡≤𝑠≤1,(2.7)

and for 𝑡,𝑠∈[0,1], we have𝑘_𝑖(𝑡,𝑠)≤(1−𝑠)^{𝛼−𝛾−1}Γ(𝛼−𝛽),𝑖=1,2.(2.8)

Lemma 2.5. Let ℎ∈𝐶(0,1); If 1<𝛼−𝛽≤2, then the unique solution of the linear problem 𝑦(0)−=𝓓0,_{𝐭𝛼−𝛽}𝓓𝑦(𝑡)=_{𝐭𝛾−𝛽}ℎ(𝑡),𝑡∈(0,1),𝑦(1)=^𝑝−2_𝑗=1𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝑦𝜉_𝑗(2.9) is given by 𝑦(𝑡)=¹⁰𝐾(𝑡,𝑠)ℎ(𝑠)𝑑𝑠,(2.10) where 𝐾(𝑡,𝑠)=𝑘₁(𝑡,𝑠)+𝑡^{𝛼−𝛽−1}1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1𝑝−2}_𝑗=1𝑎_𝑗𝑘₂𝜉_𝑗,𝑠(2.11) is the Green function of the boundary value problem (2.9).

Proof. Applying Lemma 2.4, we reduce (2.9) to an equivalent equation: 𝑦(𝑡)=−𝐼^𝛼−𝛽ℎ(𝑡)+𝑐₁𝑡^{𝛼−𝛽−1}+𝑐₂𝑡^{𝛼−𝛽−2},𝑐₁,𝑐₂∈ℝ.(2.12) From (2.12) and noting that 𝑦(0)=0, we have 𝑐₂=0. Consequently the general solution of (2.9) is 𝑦(𝑡)=−𝐼^𝛼−𝛽ℎ(𝑡)+𝑐₁𝑡^{𝛼−𝛽−1}.(2.13) Using (2.13) and Lemma 2.3, we have 𝓓_{𝐭𝛾−𝛽}𝑦(𝑡)=−𝓓_{𝐭𝛾−𝛽}𝐼^𝛼−𝛽ℎ(𝑡)+𝑐₁𝓓_{𝐭𝛾−𝛽}𝑡^{𝛼−𝛽−1}=−𝐼^𝛼−𝛾ℎ(𝑡)+𝑐₁Γ(𝛼−𝛽)Γ(𝛼−𝛾)𝑡^{𝛼−𝛾−1}=−^𝑡0(𝑡−𝑠)^{𝛼−𝛾−1}Γ(𝛼−𝛾)ℎ(𝑠)𝑑𝑠+𝑐¹Γ(𝛼−𝛽)Γ(𝛼−𝛾)𝑡^{𝛼−𝛾−1}.(2.14)Thus, 𝓓_{𝐭𝛾−𝛽}𝑦(1)=−¹⁰(1−𝑠)^{𝛼−𝛾−1}Γ(𝛼−𝛾)ℎ(𝑠)𝑑𝑠+𝑐¹Γ(𝛼−𝛽)Γ(𝛼−𝛾),(2.15) and for 𝑗=1,2,…,𝑝−2, 𝓓_{𝐭𝛾−𝛽}𝑤𝜉_𝑗=−^𝜉𝑗₀𝜉_𝑗−𝑠^{𝛼−𝛾−1}Γ(𝛼−𝛾)ℎ(𝑠)𝑑𝑠+𝑐¹Γ(𝛼−𝛽)Γ(𝛼−𝛾)𝜉^{𝑗𝛼−𝛾−1}.(2.16) Using 𝓓_{𝐭𝛾−𝛽}𝑦(1)=∑^{𝑝−2𝑗=1}𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝑦(𝜉_𝑗), (2.15), and (2.16), we obtain 𝑐₁=∫¹⁰(1−𝑠)^{𝛼−𝛾−1}ℎ(𝑠)𝑑𝑠−∑^{𝑝−2𝑗=1}𝑎_𝑗∫^𝜉𝑗₀𝜉_𝑗−𝑠^{𝛼−𝛾−1}ℎ(𝑠)𝑑𝑠Γ(𝛼−𝛽)1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1}.(2.17) So the unique solution of the problem (2.9) is 𝑦(𝑡)=−^𝑡0(𝑡−𝑠)^{𝛼−𝛽−1}Γ(𝛼−𝛽)ℎ(𝑠)𝑑𝑠+1−∑𝑡^{𝑝−2𝑗=1}𝑎^{𝛼−𝛽−1}_𝑗𝜉^{𝑗𝛼−𝛾−1}×¹⁰(1−𝑠)^{𝛼−𝛾−1}Γ(𝛼−𝛽)ℎ(𝑠)𝑑𝑠−^𝑝−2_𝑗=1𝑎_𝑗^𝜉𝑗₀𝜉_𝑗−𝑠^{𝛼−𝛾−1}=−_𝑡0Γ(𝛼−𝛽)ℎ(𝑠)𝑑𝑠(𝑡−𝑠)^{𝛼−𝛽−1}Γ(𝛼−𝛽)ℎ(𝑠)𝑑𝑠+¹⁰(1−𝑠)^{𝛼−𝛾−1}𝑡^{𝛼−𝛽−1}+𝑡Γ(𝛼−𝛽)ℎ(𝑠)𝑑𝑠_{𝛼−𝛽−1}1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1𝑝−2}_𝑗=1𝑎_𝑗¹⁰(1−𝑠)^{𝛼−𝛾−1}𝜉^{𝑗𝛼−𝛾−1}−𝑡Γ(𝛼−𝛽)ℎ(𝑠)𝑑𝑠_{𝛼−𝛽−1}1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1𝑝−2}_𝑗=1𝑎_𝑗^𝜉𝑗₀𝜉_𝑗−𝑠^{𝛼−𝛾−1}=Γ(𝛼−𝛽)ℎ(𝑠)𝑑𝑠₁₀𝑘¹(𝑡,𝑠)+𝑡^{𝛼−𝛽−1}1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1𝑝−2}_𝑗=1=𝑎_𝑗𝑘₂𝜉_𝑗,𝑠ℎ(𝑠)𝑑𝑠₁₀𝐾(𝑡,𝑠)ℎ(𝑠)𝑑𝑠.(2.18) The proof is completed.

Lemma 2.6. The function 𝐾(𝑡,𝑠) has the following properties.

(1)𝐾(𝑡,𝑠)>0,for𝑡,𝑠∈(0,1)(2)𝑡^{𝛼−𝛽−1}𝔐(𝑠)≤𝐾(𝑡,𝑠)≤𝑀(1−𝑠)^{𝛼−𝛾−1},for𝑡,𝑠∈[0,1],

where𝔐(𝑠)=∑^{𝑝−2𝑗=1}𝑎_𝑗𝑘₂𝜉_𝑗,𝑠1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1},𝑀=1+∑^{𝑝−2𝑗=1}𝑎_𝑗1−𝜉^{𝑗𝛼−𝛾−1}Γ(𝛼−𝛽)1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1}.(2.19)

Proof. It is obvious that (1) holds.
From (2.11), we obtain𝐾(𝑡,𝑠)≥𝑡^{𝛼−𝛽−1}1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1𝑝−2}_𝑗=1𝑎_𝑗𝑘₂𝜉_𝑗,𝑠=𝑡^{𝛼−𝛽−1}𝔐(𝑠).(2.20)From (2.8), we have 𝐾(𝑡,𝑠)=𝑘₁(𝑡,𝑠)+𝑡^{𝛼−𝛽−1}1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1𝑝−2}_𝑗=1≤(1−𝑠)_{𝛼−𝛾−1}𝑎_𝑗𝑘₂𝜉_𝑗,𝑠Γ(𝛼−𝛽)+∑^{𝑝−2𝑗=1}𝑎_𝑗1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1}(1−𝑠)^{𝛼−𝛾−1}≤1+∑_{𝑝−2𝑗=1}Γ(𝛼−𝛽)𝑎_𝑗1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1}(1Γ−(𝛼𝑠)−𝛽)^{𝛼−𝛾−1}.(2.21) The proof is completed.

Consider the modified problem of the BVP (1.1):𝑦(0)−=𝓓0,_{𝐭𝛼−𝛽}𝓓𝑦(𝑡)_{𝐭𝛾−𝛽}=𝜆𝑓𝑡,𝐼^𝛽𝑦(𝑡),𝑦(𝑡),𝑦(1)=^𝑝−2_𝑗=1𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝑦𝜉_𝑗.(2.22)

Lemma 2.7. Let 𝑥(𝑡)=𝐼^𝛽𝑦(𝑡)and𝑦(𝑡)∈𝐶[0,1]; then problem (1.1) is turned into (2.22). Moreover, if 𝑦∈𝐶([0,1],[0,+∞)) is a solution of problem (2.22), then the function 𝑥(𝑡)=𝐼^𝛽𝑦(𝑡) is a positive solution of the problem (1.1).

Proof. Substituting 𝑥(𝑡)=𝐼^𝛽𝑦(𝑡) into (1.1) and using Definition 2.1 and Lemmas 2.3 and 2.4, we obtain 𝓓_𝐭𝛼𝑥(𝑡)=𝑑^𝑛𝑑𝑡^𝑛𝐼^𝑛−𝛼𝑥(𝑡)=𝑑^𝑛𝑑𝑡^𝑛=𝑑𝐼^𝑛−𝛼𝐼^𝛽𝑦(𝑡)_𝑛𝑑𝑡^𝑛𝓓𝐼^{𝑛−𝛼+𝛽}𝑦(𝑡)=𝓓_{𝐭𝛼−𝛽}𝑦(𝑡),_𝐭𝛽𝑥(𝑡)=𝓓_𝐭𝛽𝐼^𝛽𝑦(𝑡)=𝑦(𝑡).(2.23) Consequently, 𝓓_𝐭𝛽𝑥(0)=𝑦(0)=0. It follows from 𝓓_𝐭𝛾𝑥(𝑡)=𝑑^𝑛/𝑑t^𝑛𝐼^𝑛−𝛾𝑥(𝑡)=(𝑑^𝑛/𝑑𝑡^𝑛)𝐼^𝑛−𝛾𝐼^𝛽𝑦(𝑡)=(𝑑^𝑛/𝑑𝑡^𝑛)𝐼^{𝑛−𝛾+𝛽}𝑦(𝑡)=𝓓_{𝐭𝛾−𝛽}𝑦(𝑡) that 𝓓_{𝐭𝛾−𝛽}𝑦(1)=∑^{𝑝−2𝑗=1}𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝑦(𝜉_𝑗). Using 𝑥(𝑡)=𝐼^𝛽𝑦(𝑡), 𝑦∈𝐶[0,1], we transform (1.1) into (2.22).
Now, let 𝑦∈𝐶([0,1],[0,+∞)) be a solution for problem (2.22). Using Lemma 2.3, (2.22), and (2.23), one has−𝓓_𝐭𝛼𝑥(𝑡)=−𝑑^𝑛𝑑𝑡^𝑛𝐼^𝑛−𝛼𝑥(𝑡)=−𝑑^𝑛𝑑𝑡^𝑛𝐼^𝑛−𝛼𝐼^𝛽𝑦(𝑡)=−𝑑^𝑛𝑑𝑡^𝑛=𝜆𝑓𝑡,𝐼_𝛽𝐼^{𝑛−𝛼+𝛽}𝑦(𝑡)=−𝓓_{𝐭𝛼−𝛽}𝑦(𝑡)𝑦(𝑡),𝑦(𝑡)=𝜆𝑓𝑡,𝑥(𝑡),𝓓_𝐭𝛽𝑥(𝑡),0<𝑡<1.(2.24) Noting 𝓓_𝐭𝛽𝑥(𝑡)=𝓓_𝐭𝛽𝐼^𝛽𝑦(𝑡)=𝑦(𝑡),𝓓_𝐭𝛾𝑥(𝑡)=𝓓_{𝐭𝛾−𝛽}𝑦(𝑡),(2.25) we have 𝓓_𝐭𝛽𝑥(0)=0,𝓓_𝐭𝛾𝑥(1)=^𝑝−2_𝑗=1𝑎_𝑗𝓓_𝐭𝛾𝑥𝜉_𝑗.(2.26)It follows from the monotonicity and property of 𝐼^𝛽 that 𝐼^𝛽𝑦∈𝐶([0,1],[0,+∞)).(2.27) Consequently, 𝑥(𝑡)=𝐼^𝛽𝑦(𝑡) is a positive solution of the problem (1.1).

Definition 2.8. A continuous function 𝜓(𝑡) is called a lower solution of the BVP (2.22), if it satisfies 𝜓(0−)≥𝓓0,_{𝐭𝛼−𝛽}𝜓𝓓(𝑡)_{𝐭𝛾−𝛽}≤𝜆𝑓𝑡,𝐼^𝛽𝜓(𝑡),𝜓(𝑡),𝜓(1)≥^𝑝−2_𝑗=1𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝜓𝜉_𝑗.(2.28)

Definition 2.9. A continuous function  𝜙(𝑡)  is called an upper solution of the BVP (2.22), if it satisfies 𝜙(0)−≤𝓓0,_{𝐭𝛼−𝛽}𝓓𝜙(𝑡)_{𝐭𝛾−𝛽}≥𝜆𝑓𝑡,𝐼^𝛽𝜙(𝑡),𝜙(𝑡),𝜙(1)≤^𝑝−2_𝑗=1𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝜙𝜉_𝑗.(2.29)

By Lemmas 2.5 and 2.6, we have the maximal principle.

Lemma 2.10. If 1<𝛼−𝛽≤2 and 𝑦∈𝐶([0,1],𝑅) satisfies 𝑦(0)=0,𝓓_{𝐭𝛾−𝛽}𝑦(1)=^𝑝−2_𝑗=1𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝑦𝜉_𝑗,(2.30)and −𝓓_{𝐭𝛼−𝛽}𝑦(𝑡)≥0 for any 𝑡∈(0,1), then 𝑦(𝑡)≥0, for 𝑡∈[0,1].

Set𝒢(𝑡)=𝑡^{𝛼−𝛽−1},ℒ(𝑡)=𝐼^𝛽𝒢(𝑡)=1Γ(𝛽)^𝑡0(𝑡−𝑠)^𝛽−1𝑠^{𝛼−𝛽−1}𝑑𝑠=Γ(𝛼−𝛽)Γ(𝛼)𝑡^𝛼−1.(2.31)

To end this section, we present here two assumptions to be used throughout the rest of the paper.(B1)𝑓∈𝐶((0,1)×(0,∞)×(0,∞),[0,+∞)) is decreasing in 𝑢 and 𝑣, and for any (𝑢,𝑣)∈(0,∞)×(0,∞),lim_𝜎→+∞𝜎𝑓(𝑡,𝜎𝑢,𝜎𝑣)=+∞(2.32)

uniformly on 𝑡∈(0,1).(B2) For any 𝜇,𝜈>0, 𝑓(𝑡,𝜇,𝜈)≢0, and¹⁰(1−𝑠)^{𝛼−𝛾−1}𝑓(𝑠,𝜇ℒ(𝑠),𝜇𝒢(𝑠))𝑑𝑠<+∞.(2.33)

3. Main Results

The main result is summarized in the following theorem.

Theorem 3.1. Provided that (B1)   and (B2) hold, then there is a constant 𝜆^∗>0 such that for any 𝜆∈(𝜆^∗,+∞), the problem (1.1) has at least one positive solution 𝑥(𝑡), which satisfies 𝑥(𝑡)≥ℒ(𝑡), 𝑡∈[0,1].

Proof. Let 𝐸=𝐶[0,1]; we denote a set 𝑃 and an operator 𝑇_𝜆 in 𝐸 as follows: 𝑃=𝑦∈𝐸𝑇∶thereexistspositivenumber𝑙_𝑦suchthat𝑦(𝑡)≥𝑙_𝑦𝒢(𝑡),𝑡∈[0,1],(3.1)_𝜆𝑦(𝑡)=𝜆¹⁰𝐾(𝑡,𝑠)𝑓𝑠,𝐼^𝛽𝑦(𝑠),𝑦(𝑠)𝑑𝑠,forany𝑦∈𝑃.(3.2) Clearly, 𝑃 is a nonempty set since 𝒢(𝑡)∈𝑃. We claim that 𝑇_𝜆 is well defined and 𝑇_𝜆(𝑃)⊂𝑃.
In fact, for any 𝜌∈𝑃, by the definition of 𝑃, there exists one positive number 𝑙_𝜌 such that 𝜌(𝑡)≥𝑙_𝜌𝒢(𝑡) for any 𝑡∈[0,1]. It follows from Lemma 2.6 and (B2) that𝑇_𝜆𝜌(𝑡)=≤𝜆𝜆𝑀¹⁰𝐾₁₀(𝑡,𝑠)𝑓𝑠,𝐼^𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠≤𝜆𝑀¹⁰(1−𝑠)^{𝛼−𝛾−1}𝑓𝑠,𝐼^𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠(1−𝑠)^{𝛼−𝛾−1}𝑓𝑠,𝑙_𝜌ℒ(𝑠),𝑙_𝜌𝒢(𝑠)𝑑𝑠<+∞.(3.3)
Setting 𝐵=max_𝑡∈[0,1]𝜌(𝑡)>0, from (B2), we have 𝑓(𝑡,𝐵/Γ(𝛽+1),𝐵)≢0. By the continuity of 𝑓(𝑡,𝑢,𝑣) on (0,1)×(0,∞)×(0,∞), we have ∫¹⁰𝔐(𝑠)𝑓(𝑠,𝐵/Γ(𝛽+1),𝐵)𝑑𝑠>0. On the other hand,𝐼^𝛽𝐵=1Γ(𝛽)^𝑡0(𝑡−𝑠)^𝛽−1𝐵𝑑𝑠=𝐵𝑡^𝛽𝔐(𝑠)=∑_{𝑝−2𝑗=1}𝛽Γ(𝛽)≤Γ(𝛽𝐵+1),𝑎_𝑗𝑘₂𝜉_𝑗,𝑠1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1}≤∑^{𝑝−2𝑗=1}𝑎_𝑗(1−𝑠)^{𝛼−𝛾−1}Γ(𝛼−𝛽)1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1}.(3.4)From (3.3), one has 0<¹⁰𝔐(𝑠)𝑓𝑠,𝐵Γ(𝛽+1),𝐵𝑑𝑠≤¹⁰≤∑𝔐(𝑠)𝑓𝑠,𝐼^𝛽𝐵,𝐵𝑑𝑠≤¹⁰𝔐(𝑠)𝑓𝑠,𝐼^𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠_{𝑝−2𝑗=1}𝑎_𝑗Γ(𝛼−𝛽)1−∑^{𝑝−2𝑗=1}𝑎_𝑗𝜉^{𝑗𝛼−𝛾−1}¹⁰(1−𝑠)^{𝛼−𝛾−1}𝑓𝑠,𝐼^𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠<+∞.(3.5)It follows from Lemma 2.6 and (3.3) that 𝑇_𝜆𝜌(𝑡)≥𝜆𝒢(𝑡)¹⁰𝔐(𝑠)𝑓𝑠,𝐼^𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠=𝑙^𝜌𝒢(𝑡),(3.6) where 𝑙^𝜌=𝜆¹⁰𝔐(𝑠)𝑓𝑠,𝐼^𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠.(3.7)Using (3.3) and (3.6), we know that 𝑇_𝜆 is well defined and 𝑇_𝜆(𝑃)⊂𝑃.
Next we will focus on the upper and lower solutions of problem (2.22). From (B1) and (3.2), we know that the operator 𝑇_𝜆 is decreasing in 𝑦. Using¹⁰𝐾(𝑡,𝑠)𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠≥𝒢(𝑡)¹⁰𝔐(𝑠)𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠,∀𝑡∈[0,1],(3.8)and letting 𝜆₁=1∫¹⁰𝔐(𝑠)𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠,(3.9)we have 𝜆₁¹⁰𝐾(𝑡,𝑠)𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠≥𝒢(𝑡),∀𝑡∈[0,1].(3.10)
On the other hand, letting 𝑏(𝑡)=∫¹⁰𝐾(𝑡,𝑠)𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠, since 𝑓(𝑡,𝑢,𝑣) is decreasing with respect to 𝑢 and 𝑣, for any 𝜆>𝜆₁, we have¹⁰≤𝐾(𝑡,𝑠)𝑓𝑠,𝜆𝐼^𝛽𝑏(𝑠),𝜆𝑏(𝑠)𝑑𝑠≤¹⁰𝐾(𝑡,𝑠)𝑓𝑠,𝜆₁𝐼^𝛽𝑏(𝑠),𝜆₁𝑏(𝑠)𝑑𝑠₁₀<+∞𝐾.(𝑡,𝑠)𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠≤𝑀¹⁰(1−𝑠)^{𝛼−𝛾−1}𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠(3.11)From (3.2), (3.3), and (B1), for all (𝑢,𝑣)∈(0,∞)×(0,∞), we have lim_𝜇→+∞𝜇𝑓(𝑡,𝜇𝑢,𝜇𝑣)=+∞(3.12)uniformly on 𝑡∈(0,1). Thus there exists large enough 𝜆^∗>𝜆₁>0, such that, for any 𝑡∈(0,1), 𝜆^∗𝑓𝑠,𝜆^∗ℒ(𝑠),𝜆^∗𝒢(𝑠)≥1∫¹⁰𝔐(𝑠)𝑑𝑠.(3.13)From Lemma 2.6, one has 𝜆^∗¹⁰𝐾(𝑡,𝑠)𝑓𝑠,𝜆^∗ℒ(𝑠),𝜆^∗𝒢(𝑠)𝑑𝑠≥∫¹⁰𝐾(𝑡,𝑠)𝑑𝑠≥∫∫¹⁰𝔐(𝑠)𝑑𝑠₁₀𝒢(𝑡)𝔐(𝑠)𝑑𝑠∫¹⁰𝔐(𝑠)𝑑𝑠=𝒢(𝑡),∀𝑡∈[0,1].(3.14)Letting 𝜙(𝑡)=𝜆^∗¹⁰𝜓(𝑡)=𝜆_∗𝐾(𝑡,𝑠)𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠=𝜆^∗𝑏(𝑡),¹⁰𝐾(𝑡,𝑠)𝑓𝑠,𝜆^∗𝐼^𝛽𝑏(𝑠),𝜆^∗𝑏(𝑠)𝑑𝑠,(3.15)and using Lemmas 2.3 and 2.7, we obtain 𝜙(𝑡)=𝜆𝜙^∗∫(¹⁰0)𝐾=(𝑡0,,𝑠)𝑓𝓓(𝑠,ℒ_{𝐭𝛾−𝛽}(𝑠),𝒢(𝑠))𝑑𝑠≥𝒢(𝑡),𝑡∈[0,1],𝜓(𝑡)=𝜆_∗𝜙(1)=∑^𝑝−2_𝑗=1𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝜙𝜉_𝑗,∫¹⁰𝜓𝐾(0()𝑡,=𝑠)0𝑓,𝑠𝓓,𝜆^∗𝐼^𝛽_{𝐭𝛾−𝛽}𝑏(𝑠),𝜆^∗𝑏(𝑠)𝑑𝑠≥𝒢(𝑡),𝑡∈[0,1],𝜓(1)=∑^𝑝−2_𝑗=1𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝜓𝜉_𝑗.(3.16) Obviously, 𝜙(𝑡),𝜓(𝑡)∈𝑃. By (3.16), we have 𝒢(𝑡)≤𝜓(𝑡)=𝑇_𝜆^∗𝜙(𝑡),𝒢(𝑡)≤𝜙(𝑡),∀𝑡∈[0,1],(3.17) which implies that 𝜓(𝑡)=𝑇_𝜆^∗𝜙(𝑡)=𝜆^∗¹⁰≤𝜆_∗𝐾(𝑡,𝑠)𝑓𝑠,𝐼^𝛽𝜙(𝑠),𝜙(𝑠)𝑑𝑠¹⁰𝐾(𝑡,𝑠)𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠=𝜙(𝑡),∀𝑡∈[0,1].(3.18)Consequently, it follows from (3.17)-(3.18) that 𝓓_𝐭𝜓(𝑡)+𝜆^∗𝑓𝑡,𝐼^𝛽𝜓(𝑡),𝜓(𝑡)≥=𝓓𝓓_𝐭𝐭𝑇_𝜆^∗𝜙(𝑡)+𝜆^∗𝑓𝑡,𝐼^𝛽𝑇_𝜆^∗𝜙(𝑡),𝑇_𝜆^∗𝜙(𝑡)=−𝜆_∗𝑇_𝜆^∗𝜙(𝑡)+𝜆^∗𝑓𝑡,𝐼^𝛽𝜙(𝑡),𝜙(𝑡)𝓓𝑓𝑡,𝐼^𝛽𝜙(𝑡),𝜙(𝑡)+𝜆^∗𝑓𝑡,𝐼^𝛽𝜙(𝑡),𝜙(𝑡)=0,(3.19)_𝐭𝜙(𝑡)+𝜆^∗𝑓𝑡,𝐼^𝛽𝜙(𝑡),𝜙(𝑡)≤=−−𝜆𝜆^∗𝑓_∗(𝑡,ℒ(𝑡),𝒢(𝑡))+𝜆^∗𝑓𝑡,𝐼^𝛽𝜙(𝑡),𝜙(𝑡)𝑓(𝑡,ℒ(𝑡),𝒢(𝑡))+𝜆^∗𝑓(𝑡,ℒ(𝑡),𝒢(𝑡))=0.(3.20)From (3.16) and (3.18)–(3.20), we know that 𝜓(𝑡)and𝜙(𝑡) are upper and lower solutions of the problem (2.22), and 𝜓(𝑡),𝜙(𝑡)∈𝑃.
Define the function 𝐹 and the operator 𝐴_𝜆^∗ in 𝐸 by𝐹(𝑡,𝑦)=⎧⎪⎨⎪⎩𝑓𝑡,𝐼^𝛽𝑓𝑡,𝐼_𝛽𝜓(𝑡),𝜓(𝑡),𝑦<𝜓(𝑡),𝑓𝑡,𝐼_𝛽𝑦(𝑡),𝑦(𝑡),𝜓(𝑡)≤𝑦≤𝜙(𝑡),𝐴𝜙(𝑡),𝜙(𝑡),𝑦>𝜙(𝑡),_𝜆∗𝑦(𝑡)=𝜆^∗¹⁰𝐾(𝑡,𝑠)𝐹(𝑠,𝑦(𝑠))𝑑𝑠,∀𝑦∈𝐸.(3.21) It follows from (B1) and (3.21) that 𝐹∶(0,1)×[0,+∞)→[0,+∞) is continuous. Consider the following boundary value problem: 𝑦(0−)𝓓=0,_{𝐭𝛼−𝛽}𝑦𝓓(𝑡)=_{𝐭𝛾−𝛽}𝜆^∗𝐹(𝑡,𝑦),𝑡∈[0,1],𝑦(1)=^𝑝−2_𝑗=1𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝑦𝜉_𝑗.(3.22)Obviously, a fixed point of the operator 𝐴_𝜆^∗ is a solution of the BVP (3.22). For all 𝑦∈𝐸, it follows from Lemma 2.6, (3.21), and 𝜓(𝑡)≥𝒢(𝑡) that 𝐴_𝜆^∗𝑦(𝑡)≤≤𝜆𝜆^∗_∗𝑀¹⁰(1−𝑠)^{𝛼−𝛾−1}𝐹(𝑠,𝑦(𝑠))𝑑𝑠≤𝜆_∗𝑀¹⁰(1−𝑠)^{𝛼−𝛾−1}𝑓𝑠,𝐼^𝛽𝜓(𝑠),𝜓(𝑠)𝑑𝑠<+∞𝑀.¹⁰(1−𝑠)^{𝛼−𝛾−1}𝑓(𝑠,ℒ(𝑠),𝒢(𝑠))𝑑𝑠(3.23)So 𝐴_𝜆^∗ is bounded. From the continuity of 𝐹(𝑡,𝑦) and 𝐾(𝑡,𝑠), it is obviously that 𝐴_𝜆^∗∶𝐸→𝐸 is continuous.
From the uniform continuity of 𝐾(𝑡,𝑠) and the Lebesgue dominated convergence theorem, we easily get that 𝐴_𝜆^∗(Ω) is equicontinuous. Thus from the Arzela-Ascoli theorem, 𝐴_𝜆^∗∶𝐸→𝐸 is completely continuous. The Schauder fixed point theorem implies that 𝐴_𝜆^∗ has at least one fixed point 𝑤 such that 𝑤=𝐴_𝜆^∗𝑤.
Now we prove𝜓(𝑡)≤𝑤(𝑡)≤𝜙(𝑡),𝑡∈[0,1].(3.24) Let 𝑧(𝑡)=𝜙(𝑡)−𝑤(𝑡),𝑡∈[0,1]. Since 𝜙(𝑡) is the upper solution of problem (2.22) and 𝑤 is a fixed point of 𝐴_𝜆^∗, we have 𝑤(0)=0,𝓓_{𝐭𝛾−𝛽}𝑤(1)=^𝑝−2_𝑗=1𝑎_𝑗𝓓_{𝐭𝛾−𝛽}𝑤𝜉_𝑗.(3.25)
From (3.17), (3.18), and the definition of 𝐹, we obtain𝑓𝑡,𝐼𝑓^𝛽𝜙(𝑡,(ℒ𝑡),(𝜙𝑡)(,𝑡𝒢)(≤𝑡))𝐹≥(𝑡𝑓,𝑦(𝑡𝑡,)𝐼)_𝛽≤𝑓𝑡,𝐼^𝛽𝜓(𝑡),𝜓(𝑡),∀𝑦∈𝐸,𝜓(𝑡),𝜓(𝑡),∀𝑡∈[0,1].(3.26)So 𝑓𝑡,𝐼^𝛽𝜙(𝑡),𝜙(𝑡)≤𝐹(𝑡,𝑦(𝑡))≤𝑓(𝑡,ℒ(𝑡),𝒢(𝑡)),∀𝑦∈𝐸.(3.27)From (3.18) and (3.20), one has 𝓓_{𝐭𝛼−𝛽}𝑧(𝑡)==−𝓓𝜆_{𝐭𝛼−𝛽∗}𝜙(𝑡)−𝓓_{𝐭𝛼−𝛽}𝑤(𝑡)≤0,𝑓∀(𝑡𝑡,∈ℒ[(0𝑡,)1,]𝒢.(𝑡))+𝜆^∗𝐹(𝑡,𝑤(𝑡))(3.28)By (3.27), (3.28), and Lemma 2.10, we get 𝑧(𝑡)≥0 which implies that 𝑤(𝑡)≤𝜙(𝑡) on [0,1]. In the same way, we have 𝑤(𝑡)≥𝜓(𝑡) on [0,1]. Thus we obtain 𝜓(𝑡)≤𝑤(𝑡)≤𝜙(𝑡),𝑡∈[0,1].(3.29) Consequently, 𝐹(𝑡,𝑤(𝑡))=𝑓(𝑡,𝐼^𝛽𝑤(𝑡),𝑤(𝑡)),𝑡∈[0,1]. Then 𝑤(𝑡) is a positive solution of the problem (2.22). It thus follows from Lemma 2.7 that 𝑥(𝑡)=𝐼^𝛽𝑤(𝑡) is a positive solution of the problem (1.1).
Finally, by (3.29), we have𝑤(𝑡)≥𝜓(𝑡)≥𝒢(𝑡).(3.30)Thus, 𝑥(𝑡)=𝐼^𝛽𝑤(𝑡)=1Γ(𝛽)^𝑡0(𝑡−𝑠)^𝛽−1𝑤(𝑠)𝑑𝑠≥1Γ(𝛽)^𝑡0(𝑡−𝑠)^𝛽−1𝒢(𝑠)𝑑𝑠=ℒ(𝑡).(3.31)

Corollary 3.2. Suppose that condition (B1)holds, and that for any 𝜇,𝜈>0, 𝑓(𝑡,𝜇,𝜈)≢0, and ¹⁰𝑓(𝑠,𝜇ℒ(𝑠),𝜇𝒢(𝑠))𝑑𝑠<+∞.(3.32)Then there exists a constant 𝜆^∗>0 such that for any 𝜆∈(𝜆^∗,+∞), the problem (1.1) has at least one positive solution 𝑥(𝑡), which satisfies 𝑥(𝑡)≥ℒ(𝑡), 𝑡∈[0,1].

We consider some special cases in which 𝑓(𝑡,𝑢,𝑣) has no singularity at 𝑢,𝑣=0 or 𝑡=0,1.

We give the following assumption.

(B^∗1)𝑓∈𝐶((0,1)×[0,∞)×[0,∞),(0,+∞)) is decreasing in 𝑢,𝑣.

Then, 𝑓(𝑡,𝑢,𝑣) is nonsingular at 𝑢=𝑣=0 and for all 𝑢,𝑣≥0, 𝑓(𝑡,𝑢,𝑣)>0,𝑡∈(0,1), which implies that 𝑓(𝑡,0,0)>0,𝑡∈(0,1). Thuslim_𝜇→+∞𝜇𝑓(𝑡,0,0)=+∞,uniformlyfor𝑡∈(0,1)(3.33)

naturally holds; we then have the following corollary.

Corollary 3.3. If (B^∗1) holds and
(B^∗2)¹⁰(1−𝑠)^{𝛼−𝛾−1}𝑓(𝑠,0,0)𝑑𝑠<+∞,(3.34) then there exists a constant 𝜆^∗>0 such that for any 𝜆∈(𝜆^∗,+∞), the problem (1.1) has at least one positive solution 𝑥(𝑡), which satisfies 𝑥(𝑡)≥ℒ(𝑡), 𝑡∈[0,1].Proof. In the proof of Theorem 3.1, we replace the set 𝑃 by 𝑃₁={𝑦∈𝐸∶𝑦(𝑡)≥0,𝑡∈[0,1]}(3.35)and the inequalities (3.18)–(3.20) by 0≤𝜓(𝑡)=𝑇_𝜆0,0≤𝜙(𝑡)=𝑇_𝜆𝜓(𝑡)≤𝑇_𝜆0=𝜓(𝑡).(3.36)Since 𝑇_𝜆0,𝑇_𝜆𝜓(𝑡)∈𝑃, we have 𝓓𝓓_{𝐭𝛼−𝛽}𝑇_𝜆0+𝑓𝑡,𝐼^𝛽𝑇_𝜆,𝑇_𝜆0=−𝑓(𝑡,0,0)+𝑓𝑡,𝐼^𝛽𝑇_𝜆0,𝑇_𝜆0≤0,_{𝐭𝛼−𝛽}𝑇_𝜆𝜓(𝑡)+𝑓𝑡,𝐼^𝛽𝑇_𝜆𝜓(𝑡),𝑇_𝜆𝜓(𝑡)=−𝑓𝑡,𝐼^𝛽𝜓(𝑡),𝜓(𝑡)+𝑓𝑡,𝐼^𝛽𝑇_𝜆𝜓(𝑡),,𝑇𝑡_𝜆∈𝜓([𝑡0),1≥].0(3.37) The rest of the proof is similar to that of Theorem 3.1.

If 𝑓(𝑡,𝑢,𝑣) is nonsingular at 𝑢=0,𝑣=0 and 𝑡=0,1, we have the conclusion.

Corollary 3.4. If 𝑓(𝑡,𝑢,𝑣)∶[0,1]×[0,∞)×[0,∞)→(0,+∞) is continuous and decreasing in 𝑢and𝑣, the problem (1.1) has at least one positive solution 𝑥(𝑡), which satisfies 𝑥(𝑡)≥ℒ(𝑡), 𝑡∈[0,1].

Example 3.5. Consider the existence of positive solutions for the following eigenvalue problem of fractional differential equation: −𝓓_𝐭3/2𝑥(𝑡)=𝜆𝑒^𝑡(1−𝑡)^1/8𝑥^−1/2𝓓_𝐭1/8(𝑡)+𝓓_𝐭1/8𝑥(𝑡)^−1/8,𝑥(0)=0,𝓓_𝐭3/8𝑥(1)=2𝓓_𝐭3/8𝑥12−𝓓^𝐭3/8𝑥34.(3.38)

Let 𝑓(𝑡,𝑢,𝑣)=1𝑒^𝑡(1−𝑡)^1/8𝑢^−1/2+𝑣^−1/8,(𝑡,𝑢,𝑣)∈(0,1)×(0,+∞)×(0,+∞).(3.39) Then 𝑓∈𝐶((0,1)×(0,+∞)×(0,+∞),(0,+∞)) is decreasing in 𝑢and𝑣, and for any (𝑢,𝑣)∈(0,∞)×(0,∞), lim_𝜎→+∞𝜎𝑓(𝑡,𝜎𝑢,𝜎𝑣)=lim_𝜎→+∞𝜎^1/2𝑢^−1/2+𝜎^7/8𝑣^−1/8𝑒^𝑡(1−𝑡)^1/8=+∞,(3.40)uniformly on 𝑡∈(0,1). Thus (B1) holds.

On the other hand, for any 𝜇,𝜈>0 and 𝑡∈(0,1),𝑓(𝑡,𝜇,𝜈)=1𝑒^𝑡(1−𝑡)^1/8ℒ(𝑡)=_𝑡0𝜇^−1/2+𝜈^−1/8≢0,(𝑡−𝑠)^−7/8𝑠^3/8Γ(1/8)𝑑𝑠=ΓΓ((131//28))𝑡^1/2.(3.41)

thus we have¹⁰(1−𝑠)^{𝛼−𝛾−1}𝑓(𝑠,𝜇ℒ(𝑠),𝜇𝒢(𝑠))𝑑𝑠=¹⁰(1−𝑠)^1/8𝑒^𝑠(1−1𝑠)^1/8𝜇^−1/2+𝜇_−1/8ℒ^−1/2(𝑠)≤𝒢^−1/8(𝑠)₁₀𝜇^−1/2ΓΓ((131//28))𝑡^1/2^−1/2+𝜇^−1/8𝑠^−3/64=𝑑𝑠₁₀𝜇^−1/2ΓΓ((131//28))𝑡^−1/4+𝜇^−1/8𝑠^−3/64𝑑𝑠<+∞,(3.42)

which implies that (B2) holds. From Theorem 3.1, there is a constant 𝜆^∗>0 such that for any 𝜆∈(𝜆^∗,+∞) the problem (3.38) has at least one positive solution 𝑥(𝑡) and𝑥(𝑡)≥ℒ(𝑡)=Γ(11/8)Γ(3/2)𝑡^1/2≈1.003𝑡^1/2,𝑡∈[0,1].(3.43)

Acknowledgments

This work is supported financially by the National Natural Science Foundation of China (11071141, 11126231) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017).