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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 723507, 17 pages
http://dx.doi.org/10.1155/2012/723507
Research Article

Existence of Positive Solution for Semipositone Fractional Differential Equations Involving Riemann-Stieltjes Integral Conditions

1Water Transportation Planning & Logistics Engineering Institute, College of Harbor, Coastal and Offshore Engineering, Hohai University, Jiangsu, Nanjing 210098, China
2National Research Center for Resettlement, Hohai University, Jiangsu, Nanjing 210098, China

Received 13 May 2012; Accepted 11 July 2012

Academic Editor: Yong Hong Wu

Copyright © 2012 Wei Wang and Li Huang. 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

The existence of at least one positive solution is established for a class of semipositone fractional differential equations with Riemann-Stieltjes integral boundary condition. The technical approach is mainly based on the fixed-point theory in a cone.

1. Introduction

In this paper, we discuss the existence of positive solutions for the following singular semipositone fractional differential equation with nonlocal condition: 𝒟𝐭𝛼𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),𝒟𝐭𝛽𝒟𝑥(𝑡)+𝑞(𝑡),𝑡(0,1),𝐭𝛽𝑥(0)=𝒟𝐭𝛽+1𝑥(0)=0,𝒟𝐭𝛽𝑥(1)=10𝒟𝐭𝛽𝑥(𝑠)𝑑𝐴(𝑠),(1.1) where 2<𝛼3, 0<𝛽<1, and 𝛼𝛽>2, 𝒟𝐭 is the standard Riemann-Liouville derivative. 10𝒟𝐭𝛽𝑥(𝑠)𝑑𝐴(𝑠) denotes the Riemann-Stieltjes integral, and 𝐴 is a function of bounded variation. 𝑓(0,1)×[0,+)×(,0][0,+) is continuous, 𝑞(0,1)(,+) is Lebesgue integrable.

Differential equations of fractional order have been recently proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in physics, engineering (like traffic, transportation, logistics, etc.), mechanics, chemistry, and so forth, (see [15]). There has been a significant development in the study of fractional differential equations in recent years, see the monographs of Kilbas et al. [6], Lakshmikantham et al. [7], Podlubny [4], Samko et al. [8], the survey by Agarwal et al. [9] and some recent results [1014].

On the other hand, the nonlocal condition given by a Riemann-Stieltjes integral is due to Webb and Infante in [1517] and gives a unified approach to many BVPs in [15, 16]. Motivated by [1517], Hao et al. [18] studied the existence of positive solutions for 𝑛th-order singular nonlocal boundary value problem: 𝑥(𝑛)𝑥(𝑡)+𝑎(𝑡)𝑓(𝑡,𝑥(𝑡))=0,0<𝑡<1,(𝑘)(0)=0,0𝑘𝑛2,𝑥(1)=10𝑥(𝑠)𝑑𝐴(𝑠),(1.2) where 𝑎 can be singular at 𝑡=0,1, 𝑓 also can be singular at 𝑥=0, but there is no singularity at 𝑡=0,1. The existence of positive solutions of the BVP (1.2) is obtained by means of the fixed point index theory in cones.

More recently, Zhang [19] considered the following problem whose nonlinear term and boundary condition contain integer order derivatives of unknown functions 𝒟𝛼𝑥(𝑡)+𝑞(𝑡)𝑓𝑥,𝑥,,𝑥(𝑛2)𝑥=0,0<𝑡<1,𝑛1<𝛼𝑛,(0)=𝑥(0)==𝑥(𝑛2)(0)=𝑥(𝑛2)(1)=0,(1.3) where 𝒟𝛼 is the standard Riemann-Liouville fractional derivative of order 𝛼, 𝑞 may be singular at 𝑡=0 and 𝑓 may be singular at 𝑥=0, 𝑥=0,,𝑥(𝑛2)=0. By using fixed point theorem of the mixed monotone operator, the unique existence result of positive solution to problem (1.3) was established. And then, Goodrich [20] was concerned with a partial extension of the problem (1.3) by extending boundary conditions 𝒟𝛼𝑥𝑥(𝑡)=𝑓(𝑡,𝑥(𝑡)),0<𝑡<1,𝑛1<𝛼𝑛,(𝑖)(0)=0,0𝑖𝑛2,𝒟𝛼𝑥(1)=0,1𝛼𝑛2,(1.4) and the author derived the Green’s function for the problem (1.4) and showed that it satisfies certain properties, then by using cone theoretic techniques, a general existence theorem for (1.4) was obtained when 𝑓(𝑡,𝑥) satisfies some growth conditions.

Recently, Rehman and Khan [21] investigated the multipoint boundary value problems for fractional differential equations of the form: 𝒟𝐭𝛼𝑦(𝑡)=𝑓𝑡,𝑦(𝑡),𝒟𝐭𝛽𝑦(𝑡),𝑡(0,1),𝑦(0)=0,𝒟𝐭𝛽𝑦(1)𝑚2𝑖=1𝜁𝑖𝒟𝐭𝛽𝑦𝜉𝑖=𝑦0,(1.5) where 1<𝛼2, 0<𝛽<1, 0<𝜉𝑖<1, 𝜁𝑖[0,+) with 𝑚2𝑖=1𝜁𝑖𝜉𝑖𝛼𝛽1<1. By using the Schauder fixed point theorem and the contraction mapping principle, the authors established the existence and uniqueness of nontrivial solutions for the BVP (1.5) provided that the nonlinear function 𝑓[0,1]×× is continuous and satisfies certain growth conditions. Since 10𝒟𝐭𝛽𝑥(𝑠)𝑑𝐴(𝑠) covers the multipoint BVP and integral BVP as special case, the fractional differential equations with the Riemann-Stieltjes integral condition also were extensively studied by many authors, see [22, 23]. In [23], Zhang and Han considered the existence of positive solution of the following singular fractional differential equation: 𝒟𝐭𝛼𝑥𝑥(𝑡)+𝑓(𝑡,𝑥(𝑡))=0,0<𝑡<1,𝑛1<𝛼𝑛,(𝑘)(0)=0,0𝑘𝑛2,𝑥(1)=10𝑥(𝑠)𝑑𝐴(𝑠),(1.6) where 𝛼2 and 𝑑𝐴(𝑠) can be a signed measure. Some growth conditions were adopted to guarantee that (1.6) has an unique positive solution, moreover, the authors also gave the iterative sequence of the solution, an error estimation, and the convergence rate of the positive solution.

Fractional differential equations like (1.1), with nonlinearities which are allowed to change sign and boundary conditions which contain nonlocal condition given by a Riemann-Stieltjes integral with a signed measure, are rarely studied. This type problems are referred to as semipositone problems in the literature, which arise naturally in chemical reactor theory [24]. In the recent work [25], by constructing a modified function, Zhang and Liu studied the existence of positive solution of a class of semipositone singular second-order Dirichlet boundary value problem, and when 𝑓 is superlinear, a sufficient condition for the existence of positive solution is obtained under the simple assumptions.

Motivated by the above work, in this paper, we establish the existence of positive solutions for the semipositone fractional differential equations (1.1) when 𝑓 is superlinear and involves fractional derivatives of unknown functions.

2. Preliminaries and Lemmas

Definition 2.1 (see [4, 5]). The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑥(0,+) is given by 𝐼𝛼1𝑥(𝑡)=Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑥(𝑠)𝑑𝑠,(2.1) provided that the right-hand side is pointwise defined on (0,+).

Definition 2.2 (see [4, 5]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a function 𝑥(0,+) is given by 𝒟𝐭𝛼1𝑥(𝑡)=𝑑Γ(𝑛𝛼)𝑑𝑡𝑛𝑡0(𝑡𝑠)𝑛𝛼1𝑥(𝑠)𝑑𝑠,(2.2) where 𝑛=[𝛼]+1, [𝛼] denotes the integer part of number 𝛼, provided that the right-hand side is pointwise defined on (0,+).

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

Proposition 2.4 (see [4, 5]). (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)

Proposition 2.5 (see [4, 5]). Let 𝛼>0, and 𝑓(𝑥) is integrable, then 𝐼𝛼𝒟𝐭𝛼𝑥(𝑡)=𝑓(𝑥)+𝑐1𝑥𝛼1+𝑐2𝑥𝛼2++𝑐𝑛𝑥𝛼𝑛,(2.6) where 𝑐𝑖(𝑖=1,2,,𝑛), 𝑛 is the smallest integer greater than or equal to 𝛼.
Let 𝑥(𝑡)=𝐼𝛽𝑦(𝑡),𝑦(𝑡)𝐶[0,1], by standard discuss, one easily reduces the BVP (1.1) to the following modified problems, 𝒟𝐭𝛼𝛽𝑦(𝑡)=𝑓𝑡,𝐼𝛽𝑦(𝑡),𝑦(𝑡)+𝑞(𝑡),𝑦(0)=𝑦(0)=0,𝑦(1)=10𝑦(𝑠)𝑑𝐴(𝑠),(2.7) and the BVP (2.7) is equivalent to the BVP (1.1).

Lemma 2.6 (see [26]). Given 𝑦𝐿1(0,1), then the problem, 𝒟𝐭𝛼𝛽𝑥(𝑡)+𝑦(𝑡)=0,0<𝑡<1,𝑥(0)=𝑥(0)=0,𝑥(1)=0,(2.8) has the unique solution 𝑥(𝑡)=10𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠,(2.9) where 𝐺(𝑡,𝑠) is the Green function of the BVP (2.8) and is given by 1𝐺(𝑡,𝑠)=[]Γ(𝛼𝛽)𝑡(1𝑠)𝛼𝛽1[],0𝑡𝑠1,𝑡(1𝑠)𝛼𝛽1(𝑡𝑠)𝛼𝛽1,0𝑠𝑡1.(2.10)

Lemma 2.7 (see [26]). For any 𝑡,𝑠[0,1], 𝐺(𝑡,𝑠) satisfies: 𝑡𝛼𝛽1(1𝑡)𝑠(1𝑠)𝛼𝛽1Γ(𝛼𝛽)𝐺(𝑡,𝑠)𝑠(1𝑠)𝛼𝛽1Γ𝑡(𝛼𝛽1),or𝛼𝛽1(1𝑡)Γ.(𝛼𝛽1)(2.11)

By Lemma 2.6, the unique solution of the problem, 𝒟𝐭𝛼𝑥(𝑡)=0,0<𝑡<1,𝑥(0)=𝑥(0)=0,𝑥(1)=1,(2.12) is 𝑡𝛼𝛽1. Let 𝒞=10𝑡𝛼𝛽1𝑑𝐴(𝑡),=10𝑡𝛼𝛽1(1𝑡)𝑑𝐴(𝑡)(2.13) and define 𝒢𝐴(𝑠)=10𝐺(𝑡,𝑠)𝑑𝐴(𝑡),(2.14) as in [25], one can get that the Green function for the nonlocal BVP (2.7) is given by 𝑡𝐻(𝑡,𝑠)=𝛼𝛽1𝒢1𝒞𝐴(𝑠)+𝐺(𝑡,𝑠).(2.15) Throughout paper one always assumes the following holds.

(𝐇𝟎)𝐴 is a increasing function of bounded variation such that 𝒢𝐴(𝑠)0 for 𝑠[0,1] and 0𝒞<1, where 𝒞 is defined by (2.13).

Define 𝑞+(𝑡)=max{𝑞(𝑡),0},𝑞(𝑡)=max{𝑞(𝑡),0}.(2.16) One has the following Lemma.

Lemma 2.8. Let 1<𝛼𝛽2 and (𝐇𝟎) hold, then the unique solution 𝑤(𝑡) of the linear problem, 𝒟𝐭𝛼𝛽𝑤(𝑡)=𝑞𝑤(𝑡),𝑡(0,1),(0)=𝑤(0)=0,𝑤(1)=10𝑤(𝑠)𝑑𝐴(𝑠),(2.17) satisfies 𝑤(𝑡)𝜂𝛾(𝑡),(2.18) where 1𝜂=𝒞Γ(𝛼𝛽1)1+10𝑞(𝑠)𝑑𝑠,𝛾(𝑡)=𝑡𝛼𝛽1(1𝑡)+𝑡1𝒞𝛼𝛽1.(2.19)

Proof. By (2.11) and that 𝐴(𝑡) is a increasing function of bounded variation, we have 𝒢𝐴(𝑠)=10𝐺(𝑡,𝑠)𝑑𝐴(𝑡)10𝑡𝛼𝛽1𝒞Γ(𝛼𝛽1)𝑑𝐴(𝑡)=.Γ(𝛼𝛽1)(2.20) Consequently, 𝑤(𝑡)=10𝐻(𝑡,𝑠)𝑞(𝑡)𝑑𝑠=10𝑡𝛼𝛽1𝒢1𝒞𝐴𝑞(𝑠)+𝐺(𝑡,𝑠)𝑡(𝑠)𝑑𝑠𝛼𝛽1(1𝑡)Γ(𝛼𝛽1)10𝑞(𝑠)𝑑𝑠+𝒞𝑡𝛼𝛽1Γ(𝛼𝛽1)(1𝒞)10𝑞1(𝑠)𝑑𝑠𝒞Γ(𝛼𝛽1)1+𝑡𝛼𝛽1(1𝑡)+𝑡1𝒞𝛼𝛽110𝑞(𝑠)𝑑𝑠=𝜂𝛾(𝑡).(2.21)

Remark 2.9. (1) 𝛾(𝑡) satisfies (1𝒞)𝛾(𝑡)2(1𝒞)𝛾(𝑡)𝛼𝛽11,(2.22)
(2) 𝒢𝐻(𝑡,𝑠)𝐴(𝑠)+1𝒞𝑠(1𝑠)𝛼𝛽1Γ(𝛼𝛽1)=Φ(𝑠).(2.23)
In fact, since 1<𝛼𝛽1<2, the left side of (1) clearly holds. For right side of (1), from 𝒞, one gets 𝛾(𝑡)1+𝒞1𝒞1+=11𝒞,1𝒞(2.24) thus we have (1𝒞)𝛾(𝑡)𝛼𝛽11.(2.25) (2) is obvious from (2.11).
Now define a function [] for any 𝑧𝐶[0,1] by []𝑧(𝑡)=𝑧(𝑡),𝑧(𝑡)0,0,𝑧(𝑡)<0(2.26) and consider the following approximate problem of the BVP (2.7): 𝒟𝐭𝛼𝛽𝑣(𝑡)=𝑓𝑡,𝐼𝛽[]𝑣(𝑡)𝑤(𝑡)[],𝑣(𝑡)𝑤(𝑡)+𝑞+(𝑡),𝑣(0)=𝑣(0)=0,𝑣(1)=10𝑣(𝑠)𝑑𝐴(𝑠).(2.27)

Lemma 2.10. Suppose 𝑣 is a positive solution of the problem (2.27) and satisfies 𝑣(𝑡)𝑤(𝑡),𝑡[0,1], then 𝑣𝑤 is a positive solution of the problem (2.7), consequently, 𝐼𝛽[𝑣(𝑡)𝑤(𝑡)] also is a positive solution of the BVP (1.1).

Proof. In fact, if 𝑣 is a positive solution of the BVP (2.27) such that 𝑣(𝑡)𝑤(𝑡) for any 𝑡[0,1], then, from (2.27) and the definition of [𝑧(𝑡)], we have 𝒟𝐭𝛼𝛽𝑣(𝑡)=𝑓𝑡,𝐼𝛽(𝑣(𝑡)𝑤(𝑡)),(𝑣(𝑡)𝑤(𝑡))+𝑞+(𝑡),𝑣(0)=𝑣(0)=0,𝑣(1)=10𝑣(𝑠)𝑑𝐴(𝑠).(2.28)
Let 𝑦=𝑣𝑤, then we have 𝑤(0)=𝑤(0)=0,𝑤(1)=10𝑤(𝑠)𝑑𝐴(𝑠),(2.29) and 𝒟𝐭𝛼𝛽𝑦(𝑡)=𝒟𝐭𝛼𝛽𝑣(𝑡)𝒟𝐭𝛼𝛽𝑤(𝑡), which implies that 𝒟𝐭𝛼𝛽𝑣(𝑡)=𝒟𝐭𝛼𝛽𝑦(𝑡)+𝑞(𝑡).(2.30) Since 𝑞(𝑡)=𝑞+(𝑡)𝑞(𝑡), then (2.27) is transformed to (2.7), that is, 𝑣𝑤 is a positive solution of the BVP (2.27). By (2.7), 𝐼𝛽[𝑣(𝑡)𝑤(𝑡)] is a positive solution of the BVP (1.1).

It is well known that the BVP (2.27) is equivalent to the fixed points of the mapping 𝑇 given by (𝑇𝑣)(𝑡)=10𝑓𝐻(𝑡,𝑠)𝑠,𝐼𝛽[]𝑣(𝑠)𝑤(𝑠)[],𝑣(𝑠)𝑤(𝑠)+𝑞+(𝑠)𝑑𝑠.(2.31)

The basic space used in this paper is 𝐸=𝐶([0,1];), where is a real number set. Obviously, the space 𝐸 is a Banach space if it is endowed with the norm as follows: 𝑣=max[]𝑡0,1||||,𝑣(𝑡)(2.32) for any 𝑢𝐸. Let 𝑃=𝑣𝐸𝑣(𝑡)(1𝒞)𝛾(𝑡)4𝑣,(2.33) then 𝑃 is a cone of 𝐸.

For the convenience in presentation, we now present some assumptions to be used in the rest of the paper.

(𝐇𝟏) For any fixed 𝑡(0,1) and for any 𝑐(0,1], there exist constants 𝜇𝑖>1,𝑖=1,2,3,4, such that, for all (𝑡,𝑢,𝑣)(0,1)×[0,+)×(,+0], 𝑐𝜇1𝑓(𝑡,𝑢,𝑣)𝑓(𝑡,𝑐𝑢,𝑣)𝑐𝜇3𝑐𝑓(𝑡,𝑢,𝑣),𝜇2𝑓(𝑡,𝑢,𝑣)𝑓(𝑡,𝑢,𝑐𝑣)𝑐𝜇4𝑓(𝑡,𝑢,𝑣).(2.34)

(𝐇𝟐)10𝑞(𝑠)𝑑𝑠>0, for any 𝑠(0,1),𝑓(𝑠,1,1)>0, and 10Φ(𝑠)𝑓(𝑠,1,1)+𝑞+𝑟(𝑠)𝑑𝑠2(1+𝑟)𝜇1+𝜇2,(2.35) where 𝑟=8𝜂,1𝒞(2.36) and 𝒞,𝜂,Φ are defined by (2.13), (2.19), and (2.23), respectively.

Remark 2.11. If 𝑐>1, the reversed inequality of (2.34) holds, that is, for all (𝑡,𝑢,𝑣)(0,1)×[0,+)×(,+0], one has 𝑐𝜇3𝑓(𝑡,𝑢,𝑣)𝑓(𝑡,𝑐𝑢,𝑣)𝑐𝜇1𝑓(𝑡,𝑢,𝑣),𝑐𝜇4𝑓(𝑡,𝑢,𝑣)𝑓(𝑡,𝑢,𝑐𝑣)𝑐𝜇2𝑓(𝑡,𝑢,𝑣).(2.37)

Lemma 2.12. Suppose (𝐇𝟏)-(𝐇𝟐) holds, for any fixed 𝑡(0,1), 𝑓(𝑡,𝑢,𝑣) is nondecreasing in 𝑢 on [0,+) and nonincreasing in 𝑣 on (,0]; and for any [𝑎,𝑏](0,1), lim(𝑢,𝑣)(+,)min[]𝑡𝑎,𝑏𝑓(𝑡,𝑢,𝑣)|𝑢||𝑣|=+.(2.38)

Proof. Let 0𝑢1𝑢2. If 𝑢2=0, obviously 𝑓(𝑡,𝑢1,𝑣)𝑓(𝑡,𝑢2,𝑣) holds. If 𝑢20, let 𝑐0=𝑢1/𝑢2, then 0<𝑐01. It follows from (𝐇𝟏) that 𝑓𝑡,𝑢1,𝑣=𝑓𝑡,𝑐0𝑢2,𝑣𝑐𝜇30𝑓𝑡,𝑢2,𝑣𝑓𝑡,𝑢2.,𝑣(2.39) Thus 𝑓(𝑡,𝑢,𝑣) is nondecreasing in 𝑢 on [0,).
On the other hand, for any 𝑣1𝑣20, if 𝑣1=0, obviously 𝑓(𝑡,𝑢,𝑣1)𝑓(𝑡,𝑢,𝑣2) holds. If 𝑣10, let 𝑐1=𝑣2/𝑣1, then 0<𝑐11. By (𝐇𝟏), we have 𝑓𝑡,𝑢,𝑣2=𝑓𝑡,𝑢,𝑐1𝑣1𝑐𝜇41𝑓𝑡,𝑢,𝑣1𝑓𝑡,𝑢,𝑣1.(2.40) Thus 𝑓(𝑡,𝑢,𝑣) is nonincreasing in 𝑣 on (,0].
Now choose 𝑢>1 and 𝑣<1, then by Remark 2.11, we have 𝑓(𝑡,𝑢,𝑣)=𝑓(𝑡,|𝑢|,|𝑣|)|𝑢|𝜇3𝑓(𝑡,1,|𝑣|)|𝑢|𝜇3|𝑣|𝜇4𝑓(𝑡,1,1).(2.41) Thus for any [𝑎,𝑏](0,1), and any 𝑡[𝑎,𝑏], we have min[]𝑡𝑎,𝑏𝑓(𝑡,𝑢,𝑣)|𝑢||𝑣||𝑢|𝜇31|𝑣|𝜇41min[]𝑡𝑎,𝑏𝑓(𝑡,1,1)>0.(2.42) Therefore lim(𝑢,𝑣)(+,)min[]𝑡𝑎,𝑏𝑓(𝑡,𝑢,𝑣)|𝑢||𝑣|=+.(2.43)

Lemma 2.13. Assume that (𝐇𝟎)(𝐇𝟐) holds. Then 𝑇𝑃𝑃 is well defined. Furthermore, 𝑇𝑃𝑃 is a completely continuous operator.

Proof. For any fixed 𝑣𝑃, there exists a constant 𝐿>0 such that 𝑣𝐿. And then, []0𝑣(𝑠)𝑤(𝑠)𝑣(𝑠)𝑣𝐿,0𝐼𝛽[]𝑣(𝑠)𝑤(𝑠)=𝑡0(𝑡𝑠)𝛽1[]𝑣(𝑠)𝑤(𝑠)𝐿Γ(𝛽)𝑑𝑠.Γ(𝛽)(2.44) By (2.44) and (𝐇𝟏)-(𝐇𝟐), we have (𝑇𝑣)(𝑡)=10𝑓𝐻(𝑡,𝑠)𝑠,𝐼𝛽[]𝑣(𝑠)𝑤(𝑠)[],𝑣(𝑠)𝑤(𝑠)+𝑞+(𝑠)𝑑𝑠10𝐻𝑓𝐿(𝑡,𝑠)𝑠,Γ(𝛽)+1,(𝐿+1)+𝑞+𝐿(𝑠)𝑑𝑠Γ(𝛽)+1𝜇1(𝐿+1)𝜇210Φ(𝑠)𝑓(𝑠,1,1)+𝑞+(𝑠)𝑑𝑠<+,(2.45) which implies that the operator 𝑇𝑃𝐸 is well defined.
Next let 𝑓(𝑠)=𝑓(𝑠,𝐼𝛽[𝑣(𝑠)𝑤(𝑠)],[𝑣(𝑠)𝑤(𝑠)])+𝑞+(𝑠), for any 𝑣𝑃, by (2.11), we have 𝑇𝑣10𝑠(1𝑠)𝛼𝛽1𝑓Γ(𝛼𝛽1)𝑡(𝑠)𝑑𝑠+𝛼𝛽11𝒞10𝒢𝐴(𝑠)𝑓1(𝑠)𝑑𝑠1𝒞10𝑠(1𝑠)𝛼𝛽1𝑓Γ(𝛼𝛽1)𝑡(𝑠)𝑑𝑠+𝛼𝛽11𝒞10𝒢𝐴(𝑠)𝑓(𝑠)𝑑𝑠.(2.46) On the other hand, by (2.11), (2.22), and (2.46), we also have (𝑇𝑢)(𝑡)𝑡𝛼𝛽1(1𝑡)10𝑠(1𝑠)𝛼𝛽1𝑓Γ(𝛼𝛽)𝑡(𝑠)𝑑𝑠+𝛼𝛽11𝒞10𝒢𝐴(𝑠)𝑓1(𝑠)𝑑𝑠2𝑡𝛼𝛽1(1𝑡)10𝑠(1𝑠)𝛼𝛽1𝑓Γ(𝛼𝛽)𝑡(𝑠)𝑑𝑠+𝛼𝛽11𝒞10𝒢𝐴(𝑠)𝑓=1(𝑠)𝑑𝑠2𝑡𝛼𝛽1(1𝑡)10𝑠(1𝑠)𝛼𝛽1𝑓Γ(𝛼𝛽)𝑡(𝑠)𝑑𝑠+𝛼𝛽11𝒞10𝒢𝐴(𝑠)𝑓+𝑡(𝑠)𝑑𝑠𝛼𝛽12(1𝒞)10𝒢𝐴(𝑠)𝑓1(𝑠)𝑑𝑠2𝑡𝛼𝛽1(1𝑡)10𝑠(1𝑠)𝛼𝛽1𝑓Γ(𝛼𝛽)(𝑠)𝑑𝑠+𝑡𝛼𝛽11𝒞10𝑠(1𝑠)𝛼𝛽1𝑓Γ(𝛼𝛽)(+𝑡𝑠)𝑑𝑠𝛼𝛽12(1𝒞)10𝒢𝐴(𝑠)𝑓=(𝑠)𝑑𝑠(1𝒞)𝛾(𝑡)2×1(𝛼𝛽1)1𝒞10𝑠(1𝑠)𝛼𝛽1Γ𝑓(𝛼𝛽1)𝑡(𝑠)𝑑𝑠+𝛼𝛽12(1𝒞)10𝒢𝐴(𝑠)𝑓(𝑠)𝑑𝑠(1𝒞)𝛾(𝑡)12(𝛼𝛽1)1𝒞10𝑠(1𝑠)𝛼𝛽1𝑓Γ(𝛼𝛽1)𝑡(𝑠)𝑑𝑠+𝛼𝛽11𝒞10𝒢𝐴(𝑠)𝑓(𝑠)𝑑𝑠(1𝒞)𝛾(𝑡)411𝒞10𝑠(1𝑠)𝛼𝛽1𝑓Γ(𝛼𝛽1)𝑡(𝑠)𝑑𝑠+𝛼𝛽11𝒞10𝒢𝐴(𝑠)𝑓.(𝑠)𝑑𝑠(2.47) So we have (𝑇𝑣)(𝑡)(1𝒞)𝛾(𝑡)4[].𝑇𝑣,𝑡0,1(2.48) which yields that 𝑇(𝑃)𝑃.
At the end, using standard arguments, according to the Ascoli-Arzela Theorem, one can show that 𝑇𝑃𝑃 is continuous. Thus 𝑇𝑃𝑃 is a completely continuous operator.

Lemma 2.14 (see [27]). Let 𝐸 be a real Banach space, 𝑃𝐸 be a cone. Assume Ω1,Ω2 are two bounded open subsets of 𝐸 with 𝜃Ω1, Ω1Ω2, and let 𝑇𝑃(Ω2Ω1)𝑃 be a completely continuous operator such that either(1)𝑇𝑥𝑥, 𝑥𝑃𝜕Ω1, and 𝑇𝑥𝑥, 𝑥𝑃𝜕Ω2 or (2)𝑇𝑥𝑥, 𝑥𝑃𝜕Ω1, and 𝑇𝑥𝑥, 𝑥𝑃𝜕Ω2.
Then 𝑇 has a fixed point in 𝑃(Ω2Ω1).

3. Main Results

Theorem 3.1. Suppose (𝐇𝟎)(𝐇𝟐) hold. Then the BVP (1.1) has at least one positive solutions 𝑥(𝑡), and 𝑥(𝑡) satisfies 𝑥(𝑡)𝜂Γ(𝛽)Γ(𝛼𝛽)Γ(𝛼)1+𝑡1𝒞𝛼1𝛼𝛽𝛼𝑡𝛼[],𝑡0,1.(3.1)

Proof. Let 𝑟=8𝜂,1𝒞(3.2) and Ω1={𝑣𝑃𝑣<𝑟}. Then, for any 𝑣𝜕Ω1,𝑠[0,1], notice that 0<𝛽<1, we have []𝑣(𝑠)𝑤(𝑠)𝐼𝑣(𝑠)𝑣𝑟,𝛽[]𝑣(𝑠)𝑤(𝑠)=1Γ(𝛽)𝑠0(𝑠𝑡)𝛽1[]𝑣(𝑡)𝑤(𝑡)𝑟𝑑𝑡Γ(𝛽)𝑟.(3.3) By (𝐇𝟏) and (𝐇𝟐), one has 𝑇𝑣=max0𝑡110𝑓𝐻(𝑡,𝑠)𝑠,𝐼𝛽[]𝑣(𝑠)𝑤(𝑠)[],𝑣(𝑠)𝑤(𝑠)+𝑞+(𝑠)𝑑𝑠10Φ𝑓𝑟(𝑠)𝑠,Γ(𝛽),𝑟+𝑞+(𝑠)𝑑𝑠10Φ(𝑠)𝑓(𝑠,𝑟+1,(𝑟+1))+𝑞+(𝑠)𝑑𝑠2(𝑟+1)𝜇1+𝜇210Φ(𝑠)𝑓(𝑠,1,1)+𝑞+(𝑠)𝑑𝑠𝑟=𝑣.(3.4) Therefore, 𝑇𝑣𝑣,𝑣𝑃𝜕Ω1.(3.5) On the other hand, choose a real number 𝑀>0 such that 𝑎𝛼𝛽(1𝑏)𝛼𝛽×Γ(𝛼𝛽)𝑀(1𝒞)2𝑎𝛽𝑎64Γ(𝛽+1)𝛼𝛽1(1𝑏)+𝑎1𝒞𝛼𝛽121.(3.6)
From (2.38), there exists 𝑁>𝑟 such that, for any 𝑡[𝑎,𝑏], 𝑓(𝑡,𝑢,𝑣)𝑀|𝑢||𝑣|,|𝑢|,|𝑣|𝑁.(3.7) Take 𝑅>max1,8𝑁𝑎(1𝒞)𝛼𝛽1(1𝑏)+𝑎1𝒞𝛼𝛽1,8𝑁Γ(𝛽+1)(1𝒞)𝑎𝛽𝑎𝛼𝛽1(1𝑏)+𝑎1𝒞𝛼𝛽1,,𝑟(3.8) then 𝑅>𝑟. Let Ω2={𝑣𝑃𝑣<𝑅}, for any 𝑣𝑃𝜕Ω2 and for any 𝑡[𝑎,𝑏], we have 4𝑣(𝑡)𝑤(𝑡)𝑣(𝑡)𝜂𝛾(𝑡)𝑣(𝑡)1(1𝒞)𝑅𝑣(𝑡)2𝑣(𝑡)(1𝒞)𝛾(𝑡)8(𝑅1𝒞)8𝑎𝛼𝛽1(1𝑏)+𝑎1𝒞𝛼𝛽1𝐼𝑅𝑁>0,𝛽[]𝑣(𝑡)𝑤(𝑡)(1𝒞)8𝑎𝛼𝛽1(1𝑏)+𝑎1𝒞𝛼𝛽1𝑅𝑡0(𝑡𝑠)𝛽1Γ(𝛽)𝑑𝑠(1𝒞)𝑎𝛽𝑎8Γ(𝛽+1)𝛼𝛽1(1𝑏)+𝑎1𝒞𝛼𝛽1𝑅𝑁>0.(3.9) So for any 𝑣𝑃𝜕Ω2,𝑡[𝑎,𝑏], by (3.7)-(3.9), we have 𝑇𝑣10𝑓𝐻(𝑡,𝑠)𝑠,𝐼𝛽[]𝑣(𝑠)𝑤(𝑠)[],𝑣(𝑠)𝑤(𝑠)+𝑞+(𝑠)𝑑𝑠10𝐺(𝑡,𝑠)𝑓𝑠,𝐼𝛽[𝑣](𝑠)𝑤(𝑠)[𝑣],(𝑠)𝑤(𝑠)𝑡𝑑𝑠𝑠𝛼𝛽1(1𝑡)Γ(𝛼𝛽)𝑏𝑎𝑠(1𝑠)𝛼𝛽1𝑓𝑠,𝐼𝛽[𝑣](𝑠)𝑤(𝑠)[𝑣],(𝑠)𝑤(𝑠)𝑎𝑑𝑠𝛼𝛽1(1𝑏)Γ(𝛼𝛽)𝑏𝑎𝑠(1𝑠)𝛼𝛽1𝑀||𝐼𝛽[]||||||𝑎𝑣(𝑠)𝑤(𝑠)𝑣(𝑠)𝑤(𝑠)𝑑𝑠𝛼𝛽(1𝑏)𝛼𝛽×𝑀Γ(𝛼𝛽)(1𝒞)2𝑎𝛽𝑎64Γ(𝛽+1)𝛼𝛽1(1𝑏)+𝑎1𝒞𝛼𝛽12𝑅2𝑅2𝑅=𝑥.(3.10) Thus 𝑇𝑣𝑣,𝑣𝑃𝜕Ω2.(3.11) By Lemma 2.14, 𝑇 has at least one fixed points 𝑣 such that 𝑟||𝑣||𝑅.
In the end, 𝑣(𝑡)𝑤(𝑡)(1𝒞)𝛾(𝑡)4𝑣𝜂𝛾(𝑡)(1𝒞)𝛾(𝑡)4𝑟𝜂𝛾(𝑡)𝜂𝛾(𝑡)>0,𝑡(0,1).(3.12) Equation (3.12) implies that 𝑣(𝑡)>𝑤(𝑡),𝑡(0,1), and 𝑥(𝑡)=𝐼𝛽(𝑣(𝑡)𝑤(𝑡))𝜂𝐼𝛽𝛾(𝑡)=𝜂𝑡0(𝑡𝑠)𝛽1=𝛾(𝑠)𝑑𝑠𝜂Γ(𝛽)Γ(𝛼𝛽)Γ(𝛼)1+𝑡1𝒞𝛼1𝛼𝛽𝛼𝑡𝛼>0,𝑡(0,1).(3.13) So by Lemma 2.10, the BVP (1.1) has at least one positive solution 𝑥, and 𝑥 satisfies (3.13).

Example 3.2. Consider the following semipositone boundary value problem with fractional order 𝛼=21/8: 𝒟21/8𝑥𝑥(𝑡)=2𝒟(𝑡)+1/8𝑥(𝑡)31012×41(1𝑡)𝑡𝒟,0<𝑡<1,1/8𝑥(0)=𝒟9/8𝑥(0)=0,𝒟1/8𝑥(1)=10𝒟1/8𝑥(𝑠)𝑑𝐴(𝑠).(3.14) Let 1𝐴(𝑡)=0,𝑡0,4,11,𝑡4,34,32,𝑡4,,1(3.15) then the BVP (3.14) becomes the 4-Point BVP with coefficients 𝒟21/8𝑥𝑥(𝑡)=2𝒟(𝑡)+1/8𝑥(𝑡)31012×41(1𝑡)𝑡𝒟,0<𝑡<1,1/8𝑥(0)=𝒟9/8𝑥(0)=0,𝒟1/8𝑥(1)=𝒟1/8𝑥14+𝒟1/8𝑥34.(3.16) Then the BVP (3.14) has at least one positive solution 𝑥(𝑡), and 𝑥(𝑡) satisfies 𝑥(𝑡)62.42891.1357𝑡5/210𝑡2121/8[].,𝑡0,1(3.17)

Proof. Obviously, 𝛼=21/8,𝛽=1/8, and 0𝒞=10𝑡3/21𝑑𝐴(𝑡)=43/2+343/20.7745<1,=10𝑡3/23(1𝑡)𝑑𝐴(𝑡)=4143/2+14343/20.2561.(3.18) On the other hand, we have 𝐺𝐺(𝑡,𝑠)=1[](𝑡,𝑠)=𝑡(1𝑠)3/2Γ𝐺(5/2),0𝑡𝑠1,2[𝑡](𝑡,𝑠)=(1𝑠)3/2(𝑡𝑠)3/2𝒢Γ(5/2),0𝑠𝑡1,𝐴𝐺(𝑠)=214,𝑠+2𝐺2341,𝑠,0𝑠<4,𝐺114,𝑠+2𝐺234,1,𝑠43𝑠<4,𝐺114,𝑠+2𝐺134,3,𝑠4𝑠1.(3.19) Thus 𝒢𝐴(𝑠)0, 0𝒞<1 and 𝐴(𝑠) is increasing. So (𝐇𝟎) holds.
Take 𝑢𝑓(𝑡,𝑢,𝑣)=2𝑣31012×4([]11𝑡),(𝑡,𝑢,𝑣)(0,1)×0,+)×(,0,𝑞(𝑡)=𝑡,(3.20) then 𝑓𝑡,𝑥(𝑡),𝒟𝐭1/8𝑥=𝑥(𝑡)2𝒟(𝑡)+1/8𝑥(𝑡)31012×4(1𝑡),𝑞+(𝑡)=0,𝑞1(𝑡)=𝑡.(3.21) For any 0<𝑐1, we also have 𝑐3𝑓(𝑡,𝑢,𝑣)𝑓(𝑡,𝑐𝑢,𝑣)𝑐3/2𝑐𝑓(𝑡,𝑢,𝑣),2𝑓(𝑡,𝑢,𝑣)𝑓(𝑡,𝑐𝑢,𝑣)𝑐4𝑓(𝑡,𝑢,𝑣).(3.22) Then (𝐇𝟏) holds.
In the end, we notice 𝒢Φ(𝑠)=𝐴(𝑠)+1𝒞𝑠(1𝑠)𝛼𝛽1Γ𝒞(𝛼𝛽1)Γ+1(𝛼𝛽1)(1𝒞)Γ=1(𝛼𝛽1)Γ(𝛼𝛽1)(1𝒞)5.0041,(3.23) then 10Φ(𝑠)𝑓(𝑠,1,1)+𝑞+(𝑠)𝑑𝑠5.00411021012×4(1𝑡)𝑑𝑠1.33443×1011.(3.24) Moreover, 1𝜂=𝒞Γ(𝛼𝛽1)1+10𝑞(𝑠)𝑑𝑠9.0817,𝑟=8𝜂𝑟1𝒞=322.1889,2(1+𝑟)𝜇1+𝜇2=322.18892(1+322.1889)54.5688×1011,(3.25) which implies that (𝐇𝟐) holds.
According to Theorem 3.1, the BVP (3.14) has at least one positive solution 𝑥(𝑡), and 𝑥(𝑡) satisfies 𝑥(𝑡)62.42891.1357𝑡5/210𝑡2121/8[].,𝑡0,1(3.26)

Acknowledgments

The paper is Funded by the Humanities and Social Sciences Foundation of Ministry of Education (no. 09YJC630056), the National Natural Science Foundation of China (no. 51009060, no. 50909042), the Fundamental Research Funds for the Central Universities (no. 2009B13414), and the Priority Academic Program Development of Jiangsu Higher Education Institutions(Coastal Development Conservancy).

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