Abstract

By applying a fixed point theorem for mappings that are decreasing with respect to a cone, this paper investigates the existence of positive solutions for the nonlinear fractional boundary value problem: 𝐷𝛼0+𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0, 0<𝑡<1, 𝑢(0)=𝑢(0)=𝑢(1)=0, where 2<𝛼<3, 𝐷𝛼0+ is the Riemann-Liouville fractional derivative.

1. Introduction

Many papers and books on fractional calculus differential equation have appeared recently. Most of them are devoted to the solvability of the linear initial fractional equation in terms of a special function [14]. Recently, there has been significant development in the existence of solutions and positive solutions to boundary value problems for fractional differential equations by the use of techniques of nonlinear analysis (fixed point theorems, Leray-Schauder theory, etc.), see [5, 6] and the references therein.

In this paper, we consider the following boundary value problems of the nonlinear fractional differential equation 𝐷𝛼0+𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1,2<𝛼<3,𝑢(0)=𝑢(0)=𝑢(1)=0,(1.1) where 𝐷𝛼0+ is the standard Riemann-Liouville fractional derivative and 𝑓(𝑡,𝑥) is singular at 𝑥=0. Our assumptions throughout are(H1)𝑓(𝑡,𝑥)(0,1)×(0,)[0,) is continuous,(H2)𝑓(𝑡,𝑥) is decreasing in 𝑥, for each fixed 𝑡,(H3)lim𝑥0+𝑓(𝑡,𝑥)= and lim𝑥𝑓(𝑡,𝑥)=0, uniformly on compact subsets of (0,1), and(H4)0<10𝑓(𝑡,𝑞𝜃(𝑡))𝑑𝑡< for all 𝜃>0 and 𝑞𝜃 as defined in (3.1).

The seminal paper by Gatica et al. [7] in 1989 has had a profound impact on the study of singular boundary value problems for ordinary differential equations (ODEs). They studied singularities of the type in (H1)–(H4) for second order Sturm-Louiville problems, and their key result hinged on an application of a particular fixed point theorem for operators which are decreasing with respect to a cone. Various authors have used these techniques to study singular problems of various types. For example, Henderson and Yin [8] as well as Eloe and Henderson [9, 10] have studied right focal, focal, conjugate, and multipoint singular boundary value problems for ODEs. However, as far as we know, no paper is concerned with boundary value problem for fractional differential equation by using this theorem. As a result, the goal of this paper is to fill the gap in this area.

Motivated by the above-mentioned papers and [11], the purpose of this paper is to establish the existence of solutions for the boundary value problem (1.1) by the use of a fixed point theorem used in [7, 11]. The paper has been organized as follows. In Section 2, we give basic definitions and provide some properties of the corresponding Green's function which are needed later. We also state the fixed point theorem from [7] for mappings that are decreasing with respect to a cone. In Section 3, we formulate two lemmas which establish a priori upper and lower bounds on solutions of (1.1). We then state and prove our main existence theorem.

For fractional differential equation and applications, we refer the reader to [13]. Concerning boundary value problems (1.1) with ordinary derivative (not fractional one), we refer the reader to [12, 13].

2. Some Preliminaries and a Fixed Point Theorem

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions and properties can be found in the literature.

Definition 2.1 (see [3]). The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑓(0,)𝑅 is given by 𝐼𝛼0+1𝑓(𝑡)=Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)𝑑𝑠,(2.1) provided that the right-hand side is pointwise defined on (0,).

Definition 2.2 (see [3]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a continuous function 𝑓(0,)𝑅 is given by 𝐷𝛼0+1𝑓(𝑡)=𝑑Γ(𝑛𝛼)𝑑𝑡𝑛𝑡0𝑓(𝑠)(𝑡𝑠)𝛼𝑛+1𝑑𝑠,(2.2) where 𝑛1𝛼<𝑛, provided that the right-hand side is pointwise defined on (0,).

Definition 2.3. By a solution of the boundary value problem (1.1) we understand a function 𝑢𝐶[0,1] such that 𝐷𝛼0+𝑢 is continuous on (0, 1) and 𝑢 satisfies (1.1).

Lemma 2.4 (see [3]). Assume that 𝑢𝐶(0,1)𝐿(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐶(0,1)𝐿(0,1). Then 𝐼𝛼0+𝐷𝛼0+𝑢(𝑡)=𝑢(𝑡)+𝑐1𝑡𝛼1+𝑐2𝑡𝛼2++𝑐𝑁𝑡𝛼𝑁(2.3) for some 𝑐𝑖𝑅,𝑖=1,,𝑁, 𝑁=[𝛼].

Lemma 2.5. Given 𝑓𝐶[0,1], and 2<𝛼<3, the unique solution of 𝐷𝛼0+𝑢(𝑡)+𝑓(𝑡)=0,0<𝑡<1,𝑢(0)=𝑢(0)=𝑢(1)=0(2.4) is 𝑢(𝑡)=10𝐺(𝑡,𝑠)𝑓(𝑠)𝑑𝑠,(2.5) where 𝑡𝐺(𝑡,𝑠)=𝛼1(1𝑠)𝛼2(𝑡𝑠)𝛼1Γ𝑡(𝛼),0𝑠𝑡1,𝛼1(1𝑠)𝛼2Γ(𝛼),0𝑡𝑠1.(2.6)

Proof. We may apply Lemma 2.4 to reduce (2.4) to an equivalent integral equation 𝑢(𝑡)=𝐼𝛼0+𝑓(𝑡)+𝑐1𝑡𝛼1+𝑐2𝑡𝛼2+𝑐3𝑡𝛼3(2.7) for some 𝑐𝑖𝑅,𝑖=1,2,3. From 𝑢(0)=𝑢(0)=𝑢(1)=0, one has 𝑐1=10(1𝑠)𝛼2Γ(𝛼)𝑓(𝑠)𝑑𝑠,𝑐2=𝑐3=0.(2.8) Therefore, the unique solution of problem (2.4) is 𝑢(𝑡)=10𝑡𝛼1(1𝑠)𝛼21Γ(𝛼)𝑓(𝑠)𝑑𝑠Γ(𝛼)𝑡0(𝑡𝑠)𝛼1=𝑓(𝑠)𝑑𝑠𝑡0𝑡𝛼1(1𝑠)𝛼2(𝑡𝑠)𝛼1𝑓Γ(𝛼)(𝑠)𝑑𝑠+1𝑡𝑡𝛼1(1𝑠)𝛼2𝑓=Γ(𝛼)(𝑠)𝑑𝑠10𝐺(𝑡,𝑠)𝑓(𝑠)𝑑𝑠.(2.9)

Lemma 2.6. The function 𝐺(𝑡,𝑠) defined by (2.6) satisfies the following conditions: (i)𝐺(𝑡,𝑠)>0, 0<𝑡,𝑠<1,(ii)𝑞(𝑡)𝐺(1,𝑠)𝐺(𝑡,𝑠)𝐺(1,𝑠)=𝑠(1𝑠)𝛼2/(Γ(𝛼)) for 0𝑡,𝑠1, where 𝑞(𝑡)=𝑡𝛼1.

Proof. Observing the expression of 𝐺(𝑡,𝑠), it is clear that 𝐺(𝑡,𝑠)>0 for 0<𝑡,𝑠<1. For given 𝑠(0,1), 𝐺(𝑡,𝑠) is increasing with respect to 𝑡. Consequently, 𝐺(𝑡,𝑠)𝐺(1,𝑠) for 0𝑡,𝑠1.
If 𝑠𝑡, we have 𝐺(𝑡,𝑠)=𝑡(𝑡𝑡𝑠)𝛼2(𝑡𝑠)(𝑡𝑠)𝛼2Γ(𝛼)𝑡(𝑡𝑡𝑠)𝛼2(𝑡𝑠)(𝑡𝑡𝑠)𝛼2=Γ(𝛼)𝑠𝑡𝛼2(1𝑠)𝛼2Γ(𝛼)𝑠𝑡𝛼1(1𝑠)𝛼2Γ(𝛼)=𝑞(𝑡)𝐺(1,𝑠).(2.10) If 𝑡𝑠, we have 𝑡𝐺(𝑡,𝑠)=𝛼1(1𝑠)𝛼2Γ(𝛼)𝑠𝑡𝛼1(1𝑠)𝛼2Γ(𝛼)=𝑞(𝑡)𝐺(1,𝑠).(2.11)

Let 𝐸 be a Banach space, 𝑃𝐸 be a cone in 𝐸. Every cone 𝑃 in 𝐸 defines a partial ordering in 𝐸 given by 𝑥𝑦 if and only if 𝑦𝑥𝑃. If 𝑥𝑦 and 𝑥𝑦, we write 𝑥<𝑦. A cone 𝑃 is said to be normal if there exists a constant 𝑁>0 such that 𝜃𝑥𝑦 implies 𝑥𝑁𝑦. If 𝑃 is normal, then every order interval [𝑥,𝑦]={𝑧𝐸𝑥𝑧𝑦} is bounded. For the concepts and properties about the cone theory we refer to [14, 15].

Next we state the fixed point theorem due to Gatica et al. [7] which is instrumental in proving our existence results.

Theorem 2.7 (Gatica-Oliker-Waltman fixed point theorem). Let 𝐸 be a Banach space, 𝑃𝐸 be a normal cone, and 𝐷𝑃 be such that if 𝑥,𝑦𝐷 with 𝑥𝑦, then [𝑥,𝑦]𝐷. Let 𝑇𝐷𝑃 be a continuous, decreasing mapping which is compact on any closed order interval contained in 𝐷, and suppose there exists an 𝑥0𝐷 such that 𝑇2𝑥0 is defined (where 𝑇2𝑥0=𝑇(𝑇𝑥0)) and 𝑇𝑥0, 𝑇2𝑥0 are order comparable to 𝑥0. Then 𝑇 has a fixed point in 𝐷 provided that either: (i)𝑇𝑥0𝑥0 and 𝑇2𝑥0𝑥0;(ii)𝑥0𝑇𝑥0 and 𝑥0𝑇2𝑥0; or(iii)The complete sequence of iterates {𝑇𝑛𝑥0}𝑛=0 is defined and there exists 𝑦0𝐷 such that 𝑇𝑦0𝐷 with 𝑦0𝑇𝑛𝑥0 for all 𝑛N

3. Main Results

In this section, we apply Theorem 2.7 to a sequence of operators that are decreasing with respect to a cone. These obtained fixed points provide a sequence of iterates which converges to a solution of (1.1).

Let the Banach space 𝐸=𝐶[0,1] with the maximum norm 𝑢=max𝑡[0,1]|𝑢(𝑡)|, and let 𝑃={𝑢𝐸𝑢(𝑡)0,𝑡[0,1]}. 𝑃 is a norm cone in 𝐸. For 𝜃>0, let 𝑞𝜃(𝑡)=𝜃𝑞(𝑡),(3.1) where 𝑞(𝑡) is given in Lemma 2.6. Define 𝐷𝑃 by 𝐷=𝑢𝑃𝜃(𝑢)>0suchthat𝑢(𝑡)𝑞𝜃[](𝑡),𝑡0,1,(3.2) and the integral operator 𝑇𝐷𝑃 by (𝑇𝑢)(𝑡)=10𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠,(3.3) where 𝐺(𝑡,𝑠) is given in (2.6). It suffices to define 𝐷 as above, since the singularity in 𝑓 precludes us from defining 𝑇 on all of 𝑃. Furthermore, it can easily be verified that 𝑇 is well defined. In fact, note that for 𝑢𝐷 there exists 𝜃(𝑢)>0 such that 𝑢(𝑡)𝑞𝜃(𝑡) for all 𝑡[0,1]. Since 𝑓(𝑡,𝑥) decreases with respect to 𝑥, we see 𝑓(𝑡,𝑢(𝑡))𝑓(𝑡,𝑞𝜃(𝑡)) for 𝑡[0,1]. Thus, 010𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠10𝑓𝑠,𝑞𝜃(𝑠)𝑑𝑠<.(3.4) Similarly, 𝑇 is decreasing with respect to 𝐷.

Lemma 3.1. 𝑢𝐷 is a solution of (1.1) if and only if 𝑇𝑢=𝑢.

Proof. One direction of the lemma is obviously true. To see the other direction, let 𝑢𝐷. Then (𝑇𝑢)(𝑡)=10𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠, and 𝑇𝑢 satisfies (1.1). Moreover, by Lemma 2.6, we have (𝑇𝑢)(𝑡)=10𝐺(𝑡,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑞(𝑡)10𝐺[].(1,𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠=𝑞(𝑡)𝑇𝑢,𝑡0,1(3.5) Thus, there exists some 𝜃(𝑇𝑢) such that (𝑇𝑢)(𝑡)𝑞𝜃(𝑡), which implies that 𝑇𝑢𝐷. That is, 𝑇𝐷𝐷.
We now present two lemmas that are required in order to apply Theorem 2.7. The first establishes a priori upper bound on solutions, while the second establishes a priori lower bound on solutions.

Lemma 3.2. If 𝑓 satisfies (H1)–(H4), then there exists an 𝑆>0 such that 𝑢𝑆 for any solution 𝑢𝐷 of (1.1).

Proof. For the sake of contradiction, suppose that the conclusion is false. Then there exists a sequence {𝑢𝑛}𝑛=1 of solutions to (1.1) such that 𝑢𝑛𝑢𝑛+1 with lim𝑛𝑢𝑛=. Note that for any solution 𝑢𝑛𝐷 of (1.1), by (3.5), we have 𝑢𝑛(𝑡)=𝑇𝑢𝑛𝑢(𝑡)𝑞(𝑡)𝑛[],𝑡0,1,𝑛1.(3.6) Then, assumptions (H2) and (H4) yield, for 0𝑡1 and all 𝑛1, 𝑢𝑛(𝑡)=𝑇𝑢𝑛(𝑡)=10𝐺(𝑡,𝑠)𝑓𝑠,𝑢𝑛1(𝑠)𝑑𝑠Γ(𝛼)10𝑠(1𝑠)𝛼2𝑓𝑠,𝑞𝑢1(𝑠)𝑑𝑠=𝑁,(3.7) for some 0<𝑁<+. In particular, 𝑢𝑛𝑁, for all 𝑛1, which contradicts lim𝑛𝑢𝑛=.

Lemma 3.3. If 𝑓 satisfies (H1)–(H4), then there exists an 𝑅>0 such that 𝑢𝑅 for any solution 𝑢𝐷 of (1.1).

Proof. For the sake of contradiction, suppose 𝑢𝑛(𝑡)0 uniformly on [0,1] as 𝑛. Let 𝑀=inf{𝐺(𝑡,𝑠)(𝑡,𝑠)[1/4,3/4]×[1/4,3/4]}>0. From (H3), we see that lim𝑥0+𝑓(𝑡,𝑥)= uniformly on compact subsets of (0,1). Hence, there exists some 𝛿>0 such that for 𝑡[1/4,3/4] and 0<𝑥<𝛿, we have 𝑓(𝑡,𝑥)2/𝑀. On the other hand, there exists an 𝑛0𝑁 such that 𝑛𝑛0 implies 0<𝑢𝑛(𝑡)<𝛿/2, for 𝑡(0,1). So, for 𝑡[1/4,3/4] and 𝑛𝑛0, 𝑢𝑛(𝑡)=𝑇𝑢𝑛(𝑡)=10𝐺(𝑡,𝑠)𝑓𝑠,𝑢𝑛(𝑠)𝑑𝑠3/41/4𝐺(𝑡,𝑠)𝑓𝑠,𝑢𝑛(𝑠)𝑑𝑠𝑀3/41/4𝑓𝛿𝑠,2𝑑𝑠𝑀3/41/42𝑀𝑑s=1.(3.8) But this contradicts the assumption that 𝑢𝑛0 uniformly on [0,1] as 𝑛. Hence, there exists an 𝑅>0 such that 𝑅𝑢.

We now present the main result of the paper.

Theorem 3.4. If 𝑓 satisfies (H1)–(H4), then (1.1) has at least one positive solution.

Proof. For each 𝑛1, defined 𝑣𝑛[0,1][0,+) by 𝑣𝑛(𝑡)=10𝐺(𝑡,𝑠)𝑓(𝑠,𝑛)𝑑𝑠.(3.9) By conditions (H1)–(H4), for 𝑛1, 0<𝑣𝑛+1(𝑡)𝑣𝑛],(𝑡),on(0,1(3.10)lim𝑛𝑣𝑛[].(𝑡)=0uniformlyon0,1(3.11) Now define a sequence of functions 𝑓𝑛(0,1)×[0,+), 𝑛1, by 𝑓𝑛(𝑡,𝑥)=𝑓𝑡,max𝑥,𝑣𝑛(𝑡).(3.12) Then, for each 𝑛1, 𝑓𝑛 is continuous and satisfies (H2). Furthermore, for 𝑛1, 𝑓𝑛𝑓(𝑡,𝑥)𝑓(𝑡,𝑥)on(0,1)×(0,+),𝑛(𝑡,𝑥)𝑓𝑡,𝑣𝑛(𝑡)on(0,1)×(0,+).(3.13) Note that 𝑓𝑛 has effectively “removed the singularity” in 𝑓(𝑡,𝑥) at 𝑥=0, then we define a sequence of operators 𝑇𝑛𝑃𝑃, 𝑛1, by 𝑇𝑛𝑢(𝑡)=10𝐺(𝑡,𝑠)𝑓𝑛(𝑠,𝑢(𝑠))𝑑𝑠,𝑢𝑃.(3.14) From standard arguments involving the Arzela-Ascoli Theorem, we know that each 𝑇𝑛 is in fact a compact mapping on 𝑃. Furthermore, 𝑇𝑛(0)0 and 𝑇2𝑛(0)0. By Theorem 2.7, for each 𝑛1, there exists 𝑢𝑛𝑃 such that 𝑇𝑛𝑢𝑛(𝑥)=𝑢𝑛(𝑡) for 𝑡[0,1]. Hence, for each 𝑛1, 𝑢𝑛 satisfies the boundary conditions of the problem. In addition, for each 𝑢𝑛, 𝑇𝑛𝑢𝑛(𝑡)=10𝐺(𝑡,𝑠)𝑓𝑛𝑠,𝑢𝑛(𝑠)𝑑𝑠=10𝐺(𝑡,𝑠)𝑓𝑛𝑢𝑠,max𝑛(𝑠),𝑣𝑛(𝑠)𝑑𝑠10𝐺(𝑡,𝑠)𝑓𝑛𝑠,𝑣𝑛(𝑠)𝑑𝑠𝑇𝑣𝑛(𝑡),(3.15) which implies 𝑢𝑛𝑇(𝑡)=𝑛𝑢𝑛(𝑡)𝑇𝑣𝑛[](𝑡),𝑡0,1,𝑛N.(3.16) Arguing as in Lemma 3.2 and using (3.11), it is fairly straightforward to show that there exists an 𝑆>0 such that 𝑢𝑛𝑆 for all 𝑛𝑁. Similarly, we can follow the argument of Lemma 3.3 and (3.5) to show that there exists an 𝑅>0 such that 𝑢𝑛[](𝑡)𝑞(𝑡)𝑅,on0,1,for𝑛1.(3.17) Since 𝑇𝐷𝐷 is a compact mapping, there is a subsequence of {𝑇𝑢𝑛} which converges to some 𝑢𝐷. We relabel the subsequence as the original sequence so that 𝑇𝑢𝑛𝑢 as 𝑛.
To conclude the proof of this theorem, we need to show that lim𝑛𝑇𝑢𝑛𝑢𝑛=0.(3.18) To that end, fixed 𝜃=𝑅, and let 𝜀>0 be give. By the integrability condition (H4), there exists 0<𝛿<1 such that 𝛿0𝑠(1𝑠)𝛼2𝑓𝑠,𝑞𝜃(𝑠)𝑑𝑠<Γ(𝛼)2𝜀.(3.19) Further, by (3.11), there exists an 𝑛0 such that, for 𝑛𝑛0, 𝑣𝑛(𝑡)𝑞𝜃[],(𝑡)on𝛿,1(3.20) so that 𝑣𝑛(𝑡)𝑞𝜃(𝑡)𝑢𝑛[].(𝑡)on𝛿,1(3.21) Thus, for 𝑠[𝛿,1] and 𝑛𝑛0, 𝑓𝑛𝑠,𝑢𝑛𝑢(𝑠)=𝑓𝑠,max𝑛(𝑠),𝑣𝑛(𝑠)=𝑓𝑠,𝑢𝑛(𝑠),(3.22) and for 𝑡[0,1], 𝑇𝑢𝑛(𝑡)𝑢𝑛(𝑡)=𝑇𝑢𝑛(𝑡)𝑇𝑛𝑢𝑛=(𝑡)10𝑓𝐺(𝑡,𝑠)𝑠,𝑢𝑛(𝑠)𝑓𝑛𝑠,𝑢𝑛(𝑠)𝑑𝑠.(3.23) Thus, for 𝑡[0,1], ||𝑇𝑢𝑛(𝑡)𝑢𝑛||1(𝑡)Γ(𝛼)𝛿0𝑠(1𝑠)𝛼2𝑓𝑠,𝑢𝑛(𝑠)𝑑𝑠+𝛿0𝑠(1𝑠)𝛼2𝑓𝑢𝑠,max𝑛(𝑠),𝑣𝑛1(𝑠)𝑑𝑠Γ(𝛼)𝛿0𝑠(1𝑠)𝛼2𝑓𝑠,𝑢𝑛(𝑠)𝑑𝑠+𝛿0𝑠(1𝑠)𝛼2𝑓𝑠,𝑢𝑛2(𝑠)𝑑𝑠Γ(𝛼)𝛿0𝑠(1𝑠)𝛼2𝑓𝑠,𝑞𝜃(𝑠)𝑑𝑠<𝜀.(3.24) Since 𝑡[0,1] was arbitrary, we conclude that 𝑇𝑢𝑛𝑢𝑛𝜀 for all 𝑛𝑛0. Hence, 𝑢[𝑞𝑅,𝑆] and for 𝑡[0,1]𝑇𝑢(𝑡)=𝑇lim𝑛𝑇𝑢𝑛(𝑡)=𝑇lim𝑛𝑢𝑛(𝑡)=lim𝑛𝑇𝑢𝑛=𝑢(𝑡).(3.25)

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. The Project Supported by the National Science Foundation of China (10971179) and Research Award Fund for Outstanding Young Scientists of Shandong Province (BS2010SF023).