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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 512127, 16 pages
http://dx.doi.org/10.1155/2012/512127
Research Article

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

1School of Mathematical and Informational Sciences, Yantai University, Yantai, Shandong 264005, China
2School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, China
3Department of Mathematics, Faculty of Science, Mahidol University, Bangkok 10400, Thailand
4Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia

Received 2 January 2012; Accepted 15 March 2012

Academic Editor: Shaoyong Lai

Copyright © 2012 Xinguang Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 derivatives𝓓𝐭𝛼𝑥(𝑡)=𝜆𝑓𝑡,𝑥(𝑡),𝓓𝐭𝛽𝓓𝑥(𝑡),𝑡(0,1),𝐭𝛽𝑥(0)=0,𝓓𝐭𝛾𝑥(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝑥𝜉𝑗,(1.1) where 𝜆 is a parameter, 1<𝛼2, 𝛼𝛽>1, 0<𝛽𝛾<1, 0<𝜉1<𝜉2<<𝜉𝑝2<1, 𝑎𝑗[0,+) with 𝑐=𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1<1, and 𝓓𝐭 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 [16], fractional differential equations have attracted great research interest in recent years [717]. Recently, ur Rehman and Khan [8] investigated the fractional order multipoint boundary value problem:𝓓𝐭𝛼𝑦(𝑡)=𝑓𝑡,𝑦(𝑡),𝓓𝐭𝛽𝑦(𝑡),𝑡(0,1),𝑦(0)=0,𝓓𝐭𝛽𝑦(1)𝑚2𝑖=1𝜁𝑖𝓓𝐭𝛽𝑦𝜉𝑖=𝑦0,(1.2) where 1<𝛼2,  0<𝛽<1, 0<𝜉𝑖<1,  𝜁𝑖[0,+) with 𝑚2𝑖=1𝜁𝑖𝜉𝑖𝛼𝛽1<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]×× 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. Suppose that 𝑥[𝑎,). Then the 𝛼th Riemann-Liouville fractional integral is defined by 𝐼𝛼1𝑥(𝑡)=Γ(𝛼)𝑡𝑎(𝑡𝑠)𝛼1𝑥(𝑠)𝑑𝑠(2.1) whenever the right-hand side is defined. Similarly, for 𝛼 with 𝛼>0, we define the 𝛼th Riemann-Liouville fractional derivative by 𝓓𝛼1𝑥(𝑡)=𝑑Γ(𝑛𝛼)𝑑𝑡(𝑛)𝑡𝑎(𝑡𝑠)𝑛𝛼1𝑥(𝑠)𝑑𝑠,(2.2)where 𝑛 is the unique positive integer satisfying 𝑛1𝛼<𝑛 and 𝑡>𝑎.

Remark 2.2. If 𝑥,𝑦(0,+) with order 𝛼>0, then 𝓓𝐭𝛼(𝑥(𝑡)+𝑦(𝑡))=𝓓𝐭𝛼𝑥(𝑡)+𝓓𝐭𝛼𝑦(𝑡).(2.3)

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

(1) If 𝑥𝐿1(0,1),𝜈>𝜎>0, then 𝐼𝜈𝐼𝜎𝑥(𝑡)=𝐼𝜈+𝜎𝑥(𝑡),𝓓𝐭𝜎𝐼𝜈𝑥(𝑡)=𝐼𝜈𝜎𝑥(𝑡),𝓓𝐭𝜎𝐼𝜎𝑥(𝑡)=𝑥(𝑡).(2.4)

(2) If 𝜈>0,𝜎>0, then 𝓓𝐭𝜈𝑡𝜎1=Γ(𝜎)𝑡Γ(𝜎𝜈)𝜎𝜈1.(2.5)

Lemma 2.4 (See [6]). Let 𝛼>0. Assume that 𝑥𝐶(0,1)𝐿1(0,1). Then𝐼𝛼𝓓𝐭𝛼𝑥(𝑡)=𝑥(𝑡)+𝑐1𝑡𝛼1+𝑐2𝑡𝛼2++𝑐𝑛𝑡𝛼𝑛,(2.6) where 𝑐𝑖(𝑖=1,2,,𝑛), and 𝑛 is the smallest integer greater than or equal to 𝛼.

Let𝑘1𝑡(𝑡,𝑠)=𝛼𝛽1(1𝑠)𝛼𝛾1(𝑡𝑠)𝛼𝛽1Γ𝑡(𝛼𝛽),0𝑠𝑡1,𝛼𝛽1(1𝑠)𝛼𝛾1𝑘Γ(𝛼𝛽),0𝑡𝑠1,2(𝑡,𝑠)=(𝑡(1𝑠))𝛼𝛾1(𝑡𝑠)𝛼𝛾1(Γ(𝛼𝛽),0𝑠𝑡1,𝑡(1𝑠))𝛼𝛾1Γ(𝛼𝛽),0𝑡𝑠1,(2.7)

and for 𝑡,𝑠[0,1], we have𝑘𝑖(𝑡,𝑠)(1𝑠)𝛼𝛾1Γ(𝛼𝛽),𝑖=1,2.(2.8)

Lemma 2.5. Let 𝐶(0,1); If 1<𝛼𝛽2, then the unique solution of the linear problem 𝓓𝐭𝛼𝛽𝑦(𝑡)=(𝑡),𝑡(0,1),𝑦(0)=0,𝓓𝐭𝛾𝛽𝑦(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝑦𝜉𝑗(2.9) is given by 𝑦(𝑡)=10𝐾(𝑡,𝑠)(𝑠)𝑑𝑠,(2.10) where 𝐾(𝑡,𝑠)=𝑘1𝑡(𝑡,𝑠)+𝛼𝛽11𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1𝑝2𝑗=1𝑎𝑗𝑘2𝜉𝑗,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)=𝐼𝛼𝛽(𝑡)+𝑐1𝑡𝛼𝛽1+𝑐2𝑡𝛼𝛽2,𝑐1,𝑐2.(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)=𝐼𝛼𝛽(𝑡)+𝑐1𝑡𝛼𝛽1.(2.13) Using (2.13) and Lemma 2.3, we have 𝓓𝐭𝛾𝛽𝑦(𝑡)=𝓓𝐭𝛾𝛽𝐼𝛼𝛽(𝑡)+𝑐1𝓓𝐭𝛾𝛽𝑡𝛼𝛽1=𝐼𝛼𝛾(𝑡)+𝑐1Γ(𝛼𝛽)𝑡Γ(𝛼𝛾)𝛼𝛾1=𝑡0(𝑡𝑠)𝛼𝛾1Γ(𝛼𝛾)(𝑠)𝑑𝑠+𝑐1Γ(𝛼𝛽)𝑡Γ(𝛼𝛾)𝛼𝛾1.(2.14)Thus, 𝓓𝐭𝛾𝛽𝑦(1)=10(1𝑠)𝛼𝛾1Γ(𝛼𝛾)(𝑠)𝑑𝑠+𝑐1Γ(𝛼𝛽)Γ(𝛼𝛾),(2.15) and for 𝑗=1,2,,𝑝2, 𝓓𝐭𝛾𝛽𝑤𝜉𝑗=𝜉𝑗0𝜉𝑗𝑠𝛼𝛾1Γ(𝛼𝛾)(𝑠)𝑑𝑠+𝑐1Γ(𝛼𝛽)𝜉Γ(𝛼𝛾)𝑗𝛼𝛾1.(2.16) Using 𝓓𝐭𝛾𝛽𝑦(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝑦(𝜉𝑗), (2.15), and (2.16), we obtain 𝑐1=10(1𝑠)𝛼𝛾1(𝑠)𝑑𝑠𝑝2𝑗=1𝑎𝑗𝜉𝑗0𝜉𝑗𝑠𝛼𝛾1(𝑠)𝑑𝑠Γ(𝛼𝛽)1𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1.(2.17) So the unique solution of the problem (2.9) is 𝑦(𝑡)=𝑡0(𝑡𝑠)𝛼𝛽1Γ𝑡(𝛼𝛽)(𝑠)𝑑𝑠+𝛼𝛽11𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1×10(1𝑠)𝛼𝛾1Γ(𝛼𝛽)(𝑠)𝑑𝑠𝑝2𝑗=1𝑎𝑗𝜉𝑗0𝜉𝑗𝑠𝛼𝛾1Γ(𝛼𝛽)(𝑠)𝑑𝑠=𝑡0(𝑡𝑠)𝛼𝛽1Γ(𝛼𝛽)(𝑠)𝑑𝑠+10(1𝑠)𝛼𝛾1𝑡𝛼𝛽1+𝑡Γ(𝛼𝛽)(𝑠)𝑑𝑠𝛼𝛽11𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1𝑝2𝑗=1𝑎𝑗10(1𝑠)𝛼𝛾1𝜉𝑗𝛼𝛾1𝑡Γ(𝛼𝛽)(𝑠)𝑑𝑠𝛼𝛽11𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1𝑝2𝑗=1𝑎𝑗𝜉𝑗0𝜉𝑗𝑠𝛼𝛾1=Γ(𝛼𝛽)(𝑠)𝑑𝑠10𝑘1𝑡(𝑡,𝑠)+𝛼𝛽11𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1𝑝2𝑗=1𝑎𝑗𝑘2𝜉𝑗=,𝑠(𝑠)𝑑𝑠10𝐾(𝑡,𝑠)(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)𝑡𝛼𝛽1𝔐(𝑠)𝐾(𝑡,𝑠)𝑀(1𝑠)𝛼𝛾1,for𝑡,𝑠[0,1],

where𝔐(𝑠)=𝑝2𝑗=1𝑎𝑗𝑘2𝜉𝑗,𝑠1𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1,𝑀=1+𝑝2𝑗=1𝑎𝑗1𝜉𝑗𝛼𝛾1Γ(𝛼𝛽)1𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1.(2.19)

Proof. It is obvious that (1) holds.
From (2.11), we obtain𝑡𝐾(𝑡,𝑠)𝛼𝛽11𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1𝑝2𝑗=1𝑎𝑗𝑘2𝜉𝑗,𝑠=𝑡𝛼𝛽1𝔐(𝑠).(2.20)From (2.8), we have 𝐾(𝑡,𝑠)=𝑘1𝑡(𝑡,𝑠)+𝛼𝛽11𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1𝑝2𝑗=1𝑎𝑗𝑘2𝜉𝑗,𝑠(1𝑠)𝛼𝛾1+Γ(𝛼𝛽)𝑝2𝑗=1𝑎𝑗1𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1(1𝑠)𝛼𝛾1Γ(𝛼𝛽)1+𝑝2𝑗=1𝑎𝑗1𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1(1𝑠)𝛼𝛾1.Γ(𝛼𝛽)(2.21) The proof is completed.

Consider the modified problem of the BVP (1.1):𝓓𝐭𝛼𝛽𝑦(𝑡)=𝜆𝑓𝑡,𝐼𝛽,𝑦(𝑡),𝑦(𝑡)𝑦(0)=0,𝓓𝐭𝛾𝛽𝑦(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝑦𝜉𝑗.(2.22)

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

Proof. Substituting 𝑥(𝑡)=𝐼𝛽𝑦(𝑡) into (1.1) and using Definition 2.1 and Lemmas 2.3 and 2.4, we obtain 𝓓𝐭𝛼𝑑𝑥(𝑡)=𝑛𝑑𝑡𝑛𝐼𝑛𝛼𝑑𝑥(𝑡)=𝑛𝑑𝑡𝑛𝐼𝑛𝛼𝐼𝛽=𝑑𝑦(𝑡)𝑛𝑑𝑡𝑛𝐼𝑛𝛼+𝛽𝑦(𝑡)=𝓓𝐭𝛼𝛽𝓓𝑦(𝑡),𝐭𝛽𝑥(𝑡)=𝓓𝐭𝛽𝐼𝛽𝑦(𝑡)=𝑦(𝑡).(2.23) Consequently, 𝓓𝐭𝛽𝑥(0)=𝑦(0)=0. It follows from 𝓓𝐭𝛾𝑥(𝑡)=𝑑𝑛/𝑑t𝑛𝐼𝑛𝛾𝑥(𝑡)=(𝑑𝑛/𝑑𝑡𝑛)𝐼𝑛𝛾𝐼𝛽𝑦(𝑡)=(𝑑𝑛/𝑑𝑡𝑛)𝐼𝑛𝛾+𝛽𝑦(𝑡)=𝓓𝐭𝛾𝛽𝑦(𝑡) that 𝓓𝐭𝛾𝛽𝑦(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝑦(𝜉𝑗). Using 𝑥(𝑡)=𝐼𝛽𝑦(𝑡), 𝑦𝐶[0,1], we transform (1.1) into (2.22).
Now, let 𝑦𝐶([0,1],[0,+)) be a solution for problem (2.22). Using Lemma 2.3, (2.22), and (2.23), one has𝓓𝐭𝛼𝑑𝑥(𝑡)=𝑛𝑑𝑡𝑛𝐼𝑛𝛼𝑑𝑥(𝑡)=𝑛𝑑𝑡𝑛𝐼𝑛𝛼𝐼𝛽𝑑𝑦(𝑡)=𝑛𝑑𝑡𝑛𝐼𝑛𝛼+𝛽𝑦(𝑡)=𝓓𝐭𝛼𝛽𝑦(𝑡)=𝜆𝑓𝑡,𝐼𝛽𝑦(𝑡),𝑦(𝑡)=𝜆𝑓𝑡,𝑥(𝑡),𝓓𝐭𝛽𝑥(𝑡),0<𝑡<1.(2.24) Noting 𝓓𝐭𝛽𝑥(𝑡)=𝓓𝐭𝛽𝐼𝛽𝑦(𝑡)=𝑦(𝑡),𝓓𝐭𝛾𝑥(𝑡)=𝓓𝐭𝛾𝛽𝑦(𝑡),(2.25) we have 𝓓𝐭𝛽𝑥(0)=0,𝓓𝐭𝛾𝑥(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝑥𝜉𝑗.(2.26)It follows from the monotonicity and property of 𝐼𝛽 that 𝐼𝛽[],[𝑦𝐶(0,10,+)).(2.27) Consequently, 𝑥(𝑡)=𝐼𝛽𝑦(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 𝓓𝐭𝛼𝛽𝜓(𝑡)𝜆𝑓𝑡,𝐼𝛽,𝜓(𝑡),𝜓(𝑡)𝜓(0)0,𝓓𝐭𝛾𝛽𝜓(1)𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝜓𝜉𝑗.(2.28)

Definition 2.9. A continuous function  𝜙(𝑡)  is called an upper solution of the BVP (2.22), if it satisfies 𝓓𝐭𝛼𝛽𝜙(𝑡)𝜆𝑓𝑡,𝐼𝛽,𝜙(𝑡),𝜙(𝑡)𝜙(0)0,𝓓𝐭𝛾𝛽𝜙(1)𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝜙𝜉𝑗.(2.29)

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

Lemma 2.10. If 1<𝛼𝛽2 and 𝑦𝐶([0,1],𝑅) satisfies 𝑦(0)=0,𝓓𝐭𝛾𝛽𝑦(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝑦𝜉𝑗,(2.30)and 𝓓𝐭𝛼𝛽𝑦(𝑡)0 for any 𝑡(0,1), then 𝑦(𝑡)0, for 𝑡[0,1].

Set𝒢(𝑡)=𝑡𝛼𝛽1,(𝑡)=𝐼𝛽1𝒢(𝑡)=Γ(𝛽)𝑡0(𝑡𝑠)𝛽1𝑠𝛼𝛽1𝑑𝑠=Γ(𝛼𝛽)𝑡Γ(𝛼)𝛼1.(2.31)

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

uniformly on 𝑡(0,1).(B2) For any 𝜇,𝜈>0, 𝑓(𝑡,𝜇,𝜈)0, and10(1𝑠)𝛼𝛾1𝑓(𝑠,𝜇(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)(𝑡), 𝑡[0,1].

Proof. Let 𝐸=𝐶[0,1]; we denote a set 𝑃 and an operator 𝑇𝜆 in 𝐸 as follows: 𝑃=𝑦𝐸thereexistspositivenumber𝑙𝑦suchthat𝑦(𝑡)𝑙𝑦[],𝑇𝒢(𝑡),𝑡0,1(3.1)𝜆𝑦(𝑡)=𝜆10𝐾(𝑡,𝑠)𝑓𝑠,𝐼𝛽𝑦(𝑠),𝑦(𝑠)𝑑𝑠,forany𝑦𝑃.(3.2) Clearly, 𝑃 is a nonempty set since 𝒢(𝑡)𝑃. We claim that 𝑇𝜆 is well defined and 𝑇𝜆(𝑃)𝑃.
In fact, for any 𝜌𝑃, by the definition of 𝑃, there exists one positive number 𝑙𝜌 such that 𝜌(𝑡)𝑙𝜌𝒢(𝑡) for any 𝑡[0,1]. It follows from Lemma 2.6 and (B2) that𝑇𝜆𝜌(𝑡)=𝜆10𝐾(𝑡,𝑠)𝑓𝑠,𝐼𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠𝜆𝑀10(1𝑠)𝛼𝛾1𝑓𝑠,𝐼𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠𝜆𝑀10(1𝑠)𝛼𝛾1𝑓𝑠,𝑙𝜌(𝑠),𝑙𝜌𝒢(𝑠)𝑑𝑠<+.(3.3)
Setting 𝐵=max𝑡[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,𝐼𝛽1𝐵=Γ(𝛽)𝑡0(𝑡𝑠)𝛽1𝐵𝑑𝑠=𝐵𝑡𝛽𝐵𝛽Γ(𝛽),Γ(𝛽+1)𝔐(𝑠)=𝑝2𝑗=1𝑎𝑗𝑘2𝜉𝑗,𝑠1𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1𝑝2𝑗=1𝑎𝑗(1𝑠)𝛼𝛾1Γ(𝛼𝛽)1𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾1.(3.4)From (3.3), one has 0<10𝐵𝔐(𝑠)𝑓𝑠,Γ(𝛽+1),𝐵𝑑𝑠10𝔐(𝑠)𝑓𝑠,𝐼𝛽𝐵,𝐵𝑑𝑠10𝔐(𝑠)𝑓𝑠,𝐼𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠𝑝2𝑗=1𝑎𝑗Γ(𝛼𝛽)1𝑝2𝑗=1𝑎𝑗𝜉𝑗𝛼𝛾110(1𝑠)𝛼𝛾1𝑓𝑠,𝐼𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠<+.(3.5)It follows from Lemma 2.6 and (3.3) that 𝑇𝜆𝜌(𝑡)𝜆𝒢(𝑡)10𝔐(𝑠)𝑓𝑠,𝐼𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠=𝑙𝜌𝒢(𝑡),(3.6) where 𝑙𝜌=𝜆10𝔐(𝑠)𝑓𝑠,𝐼𝛽𝜌(𝑠),𝜌(𝑠)𝑑𝑠.(3.7)Using (3.3) and (3.6), we know that 𝑇𝜆 is well defined and 𝑇𝜆(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 𝑇𝜆 is decreasing in 𝑦. Using10𝐾(𝑡,𝑠)𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠𝒢(𝑡)10[]𝔐(𝑠)𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠,𝑡0,1,(3.8)and letting 𝜆1=110𝔐(𝑠)𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠,(3.9)we have 𝜆110[]𝐾(𝑡,𝑠)𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠𝒢(𝑡),𝑡0,1.(3.10)
On the other hand, letting 𝑏(𝑡)=10𝐾(𝑡,𝑠)𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠, since 𝑓(𝑡,𝑢,𝑣) is decreasing with respect to 𝑢 and 𝑣, for any 𝜆>𝜆1, we have10𝐾(𝑡,𝑠)𝑓𝑠,𝜆𝐼𝛽𝑏(𝑠),𝜆𝑏(𝑠)𝑑𝑠10𝐾(𝑡,𝑠)𝑓𝑠,𝜆1𝐼𝛽𝑏(𝑠),𝜆1𝑏(𝑠)𝑑𝑠10𝐾(𝑡,𝑠)𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠𝑀10(1𝑠)𝛼𝛾1𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠<+.(3.11)From (3.2), (3.3), and (B1), for all (𝑢,𝑣)(0,)×(0,), we have lim𝜇+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), 𝜆𝑓𝑠,𝜆(𝑠),𝜆1𝒢(𝑠)10𝔐(𝑠)𝑑𝑠.(3.13)From Lemma 2.6, one has 𝜆10𝐾(𝑡,𝑠)𝑓𝑠,𝜆(𝑠),𝜆𝒢(𝑠)𝑑𝑠10𝐾(𝑡,𝑠)𝑑𝑠10𝔐(𝑠)𝑑𝑠10𝒢(𝑡)𝔐(𝑠)𝑑𝑠10𝔐[].(𝑠)𝑑𝑠=𝒢(𝑡),𝑡0,1(3.14)Letting 𝜙(𝑡)=𝜆10𝐾(𝑡,𝑠)𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠=𝜆𝜓𝑏(𝑡),(𝑡)=𝜆10𝐾(𝑡,𝑠)𝑓𝑠,𝜆𝐼𝛽𝑏(𝑠),𝜆𝑏(𝑠)𝑑𝑠,(3.15)and using Lemmas 2.3 and 2.7, we obtain 𝜙(𝑡)=𝜆10[],𝜙𝐾(𝑡,𝑠)𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠𝒢(𝑡),𝑡0,1(0)=0,𝓓𝐭𝛾𝛽𝜙(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝜙𝜉𝑗,𝜓(𝑡)=𝜆10𝐾(𝑡,𝑠)𝑓𝑠,𝜆𝐼𝛽𝑏(𝑠),𝜆[],𝑏(𝑠)𝑑𝑠𝒢(𝑡),𝑡0,1𝜓(0)=0,𝓓𝐭𝛾𝛽𝜓(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝜓𝜉𝑗.(3.16) Obviously, 𝜙(𝑡),𝜓(𝑡)𝑃. By (3.16), we have 𝒢𝑇(𝑡)𝜓(𝑡)=𝜆𝜙[](𝑡),𝒢(𝑡)𝜙(𝑡),𝑡0,1,(3.17) which implies that 𝑇𝜓(𝑡)=𝜆𝜙(𝑡)=𝜆10𝐾(𝑡,𝑠)𝑓𝑠,𝐼𝛽𝜙(𝑠),𝜙(𝑠)𝑑𝑠𝜆10𝐾[].(𝑡,𝑠)𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠=𝜙(𝑡),𝑡0,1(3.18)Consequently, it follows from (3.17)-(3.18) that 𝓓𝐭𝜓(𝑡)+𝜆𝑓𝑡,𝐼𝛽𝜓(𝑡),𝜓(𝑡)=𝓓𝐭𝑇𝜆𝜙(𝑡)+𝜆𝑓𝑡,𝐼𝛽𝑇𝜆𝜙𝑇(𝑡),𝜆𝜙(𝑡)𝓓𝐭𝑇𝜆𝜙(𝑡)+𝜆𝑓𝑡,𝐼𝛽𝜙(𝑡),𝜙(𝑡)=𝜆𝑓𝑡,𝐼𝛽𝜙(𝑡),𝜙(𝑡)+𝜆𝑓𝑡,𝐼𝛽𝓓𝜙(𝑡),𝜙(𝑡)=0,(3.19)𝐭𝜙(𝑡)+𝜆𝑓𝑡,𝐼𝛽𝜙(𝑡),𝜙(𝑡)=𝜆𝑓(𝑡,(𝑡),𝒢(𝑡))+𝜆𝑓𝑡,𝐼𝛽𝜙(𝑡),𝜙(𝑡)𝜆𝑓(𝑡,(𝑡),𝒢(𝑡))+𝜆𝑓(𝑡,(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 𝐴𝜆 in 𝐸 by𝑓𝐹(𝑡,𝑦)=𝑡,𝐼𝛽𝑓𝜓(𝑡),𝜓(𝑡),𝑦<𝜓(𝑡),𝑡,𝐼𝛽𝑓𝑦(𝑡),𝑦(𝑡),𝜓(𝑡)𝑦𝜙(𝑡),𝑡,𝐼𝛽𝐴𝜙(𝑡),𝜙(𝑡),𝑦>𝜙(𝑡),𝜆𝑦(𝑡)=𝜆10𝐾(𝑡,𝑠)𝐹(𝑠,𝑦(𝑠))𝑑𝑠,𝑦𝐸.(3.21) It follows from (B1) and (3.21) that 𝐹(0,1)×[0,+)[0,+) is continuous. Consider the following boundary value problem: 𝓓𝐭𝛼𝛽𝑦(𝑡)=𝜆[],𝐹(𝑡,𝑦),𝑡0,1𝑦(0)=0,𝓓𝐭𝛾𝛽𝑦(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝑦𝜉𝑗.(3.22)Obviously, a fixed point of the operator 𝐴𝜆 is a solution of the BVP (3.22). For all 𝑦𝐸, it follows from Lemma 2.6, (3.21), and 𝜓(𝑡)𝒢(𝑡) that 𝐴𝜆𝑦(𝑡)𝜆𝑀10(1𝑠)𝛼𝛾1𝐹(𝑠,𝑦(𝑠))𝑑𝑠𝜆𝑀10(1𝑠)𝛼𝛾1𝑓𝑠,𝐼𝛽𝜓(𝑠),𝜓(𝑠)𝑑𝑠𝜆𝑀10(1𝑠)𝛼𝛾1𝑓(𝑠,(𝑠),𝒢(𝑠))𝑑𝑠<+.(3.23)So 𝐴𝜆 is bounded. From the continuity of 𝐹(𝑡,𝑦) and 𝐾(𝑡,𝑠), it is obviously that 𝐴𝜆𝐸𝐸 is continuous.
From the uniform continuity of 𝐾(𝑡,𝑠) and the Lebesgue dominated convergence theorem, we easily get that 𝐴𝜆(Ω) is equicontinuous. Thus from the Arzela-Ascoli theorem, 𝐴𝜆𝐸𝐸 is completely continuous. The Schauder fixed point theorem implies that 𝐴𝜆 has at least one fixed point 𝑤 such that 𝑤=𝐴𝜆𝑤.
Now we prove[].𝜓(𝑡)𝑤(𝑡)𝜙(𝑡),𝑡0,1(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 𝐴𝜆, we have 𝑤(0)=0,𝓓𝐭𝛾𝛽𝑤(1)=𝑝2𝑗=1𝑎𝑗𝓓𝐭𝛾𝛽𝑤𝜉𝑗.(3.25)
From (3.17), (3.18), and the definition of 𝐹, we obtain𝑓𝑡,𝐼𝛽𝜙(𝑡),𝜙(𝑡)𝐹(𝑡,𝑦(𝑡))𝑓𝑡,𝐼𝛽𝑓𝜓(𝑡),𝜓(𝑡),𝑦𝐸,(𝑡,(𝑡),𝒢(𝑡))𝑓𝑡,𝐼𝛽𝜓[].(𝑡),𝜓(𝑡),𝑡0,1(3.26)So 𝑓𝑡,𝐼𝛽𝜙(𝑡),𝜙(𝑡)𝐹(𝑡,𝑦(𝑡))𝑓(𝑡,(𝑡),𝒢(𝑡)),𝑦𝐸.(3.27)From (3.18) and (3.20), one has 𝓓𝐭𝛼𝛽𝑧(𝑡)=𝓓𝐭𝛼𝛽𝜙(𝑡)𝓓𝐭𝛼𝛽𝑤(𝑡)=𝜆𝑓(𝑡,(𝑡),𝒢(𝑡))+𝜆[].𝐹(𝑡,𝑤(𝑡))0,𝑡0,1(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),𝑡0,1(3.29) Consequently, 𝐹(𝑡,𝑤(𝑡))=𝑓(𝑡,𝐼𝛽𝑤(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)=𝐼𝛽𝑤(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)=𝐼𝛽1𝑤(𝑡)=Γ(𝛽)𝑡0(𝑡𝑠)𝛽11𝑤(𝑠)𝑑𝑠Γ(𝛽)𝑡0(𝑡𝑠)𝛽1𝒢(𝑠)𝑑𝑠=(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,𝜇(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)(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.

(B1)𝑓𝐶((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,𝑣0, 𝑓(𝑡,𝑢,𝑣)>0,𝑡(0,1), which implies that 𝑓(𝑡,0,0)>0,𝑡(0,1). Thuslim𝜇+𝜇𝑓(𝑡,0,0)=+,uniformlyfor𝑡(0,1)(3.33)

naturally holds; we then have the following corollary.

Corollary 3.3. If (B1) holds and
(B2)10(1𝑠)𝛼𝛾1𝑓(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)(𝑡), 𝑡[0,1].Proof. In the proof of Theorem 3.1, we replace the set 𝑃 by 𝑃1[]}={𝑦𝐸𝑦(𝑡)0,𝑡0,1(3.35)and the inequalities (3.18)–(3.20) by 0𝜓(𝑡)=𝑇𝜆0,0𝜙(𝑡)=𝑇𝜆𝜓(𝑡)𝑇𝜆0=𝜓(𝑡).(3.36)Since 𝑇𝜆0,𝑇𝜆𝜓(𝑡)𝑃, we have 𝓓𝐭𝛼𝛽𝑇𝜆0+𝑓𝑡,𝐼𝛽𝑇𝜆,𝑇𝜆0=𝑓(𝑡,0,0)+𝑓𝑡,𝐼𝛽𝑇𝜆0,𝑇𝜆0𝓓0,𝐭𝛼𝛽𝑇𝜆𝜓(𝑡)+𝑓𝑡,𝐼𝛽𝑇𝜆𝜓(𝑡),𝑇𝜆𝜓(𝑡)=𝑓𝑡,𝐼𝛽𝜓(𝑡),𝜓(𝑡)+𝑓𝑡,𝐼𝛽𝑇𝜆𝜓(𝑡),𝑇𝜆[].𝜓(𝑡)0,𝑡0,1(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)(𝑡), 𝑡[0,1].

Example 3.5. Consider the existence of positive solutions for the following eigenvalue problem of fractional differential equation: 𝓓𝐭3/2𝜆𝑥(𝑡)=𝑒𝑡(1𝑡)1/8𝑥1/2𝓓(𝑡)+𝐭1/8𝑥(𝑡)1/8,𝓓𝐭1/8𝑥(0)=0,𝓓𝐭3/8𝑥(1)=2𝓓𝐭3/8𝑥12𝓓𝐭3/8𝑥34.(3.38)

Let 1𝑓(𝑡,𝑢,𝑣)=𝑒𝑡(1𝑡)1/8𝑢1/2+𝑣1/8,(𝑡,𝑢,𝑣)(0,1)×(0,+)×(0,+).(3.39) Then 𝑓𝐶((0,1)×(0,+)×(0,+),(0,+)) is decreasing in 𝑢and𝑣, and for any (𝑢,𝑣)(0,)×(0,), lim𝜎+𝜎𝑓(𝑡,𝜎𝑢,𝜎𝑣)=lim𝜎+𝜎1/2𝑢1/2+𝜎7/8𝑣1/8𝑒𝑡(1𝑡)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),1𝑓(𝑡,𝜇,𝜈)=𝑒𝑡(1𝑡)1/8𝜇1/2+𝜈1/80,(𝑡)=𝑡0(𝑡𝑠)7/8𝑠3/8Γ(1/8)𝑑𝑠=Γ(11/8)𝑡Γ(3/2)1/2.(3.41)

thus we have10(1𝑠)𝛼𝛾1𝑓(𝑠,𝜇(𝑠),𝜇𝒢(𝑠))𝑑𝑠=10(1𝑠)1/81𝑒𝑠(1𝑠)1/8𝜇1/21/2(𝑠)+𝜇1/8𝒢1/8(𝑠)10𝜇1/2Γ(11/8)𝑡Γ(3/2)1/21/2+𝜇1/8𝑠3/64=𝑑𝑠10𝜇1/2Γ(11/8)𝑡Γ(3/2)1/4+𝜇1/8𝑠3/64𝑑𝑠<+,(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)(𝑡)=Γ(11/8)𝑡Γ(3/2)1/21.003𝑡1/2[].,𝑡0,1(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).

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