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Abstract and Applied Analysis
Volume 2011 (2011), Article ID 390543, 16 pages
http://dx.doi.org/10.1155/2011/390543
Research Article

Positive Solutions to Boundary Value Problems of Nonlinear Fractional Differential Equations

1School of Science, University of Jinan, Jinan, Shandong 250022, China
2Department of Mathematics and Statistics, Missouri University of Science and Technology Rolla, MO 65409-0020, USA

Received 23 September 2010; Revised 5 November 2010; Accepted 6 December 2010

Academic Editor: Josef Diblík

Copyright © 2011 Yige Zhao 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

We study the existence of positive solutions for the boundary value problem of nonlinear fractional differential equations 𝐷𝛼0+𝑢(𝑡)+𝜆𝑓(𝑢(𝑡))=0, 0<𝑡<1, 𝑢(0)=𝑢(1)=𝑢(0)=0, where 2<𝛼3 is a real number, 𝐷𝛼0+ is the Riemann-Liouville fractional derivative, 𝜆 is a positive parameter, and 𝑓(0,+)(0,+) is continuous. By the properties of the Green function and Guo-Krasnosel'skii fixed point theorem on cones, the eigenvalue intervals of the nonlinear fractional differential equation boundary value problem are considered, some sufficient conditions for the nonexistence and existence of at least one or two positive solutions for the boundary value problem are established. As an application, some examples are presented to illustrate the main results.

1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications; see [14]. It should be noted that most of papers and books on fractional calculus are devoted to the solvability of linear initial fractional differential equations on terms of special functions.

Recently, there are some papers dealing with the existence of solutions (or positive solutions) of nonlinear initial fractional differential equations by the use of techniques of nonlinear analysis (fixed-point theorems, Leray-Schauder theory, Adomian decomposition method, etc.); see [511]. In fact, there has the same requirements for boundary conditions. However, there exist some papers considered the boundary value problems of fractional differential equations; see [1219].

Yu and Jiang [19] examined the existence of positive solutions for the following problem: 𝐷𝛼0+𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑢(0)=𝑢(1)=𝑢(0)=0,(1.1) where 2<𝛼3 is a real number, 𝑓𝐶([0,1]×[0,+),(0,+)), and 𝐷𝛼0+ is the Riemann-Liouville fractional differentiation. By using the properties of the Green function, they obtained some existence criteria for one or two positive solutions for singular and nonsingular boundary value problems by means of the Krasnosel'skii fixed point theorem and a mixed monotone method.

To the best of our knowledge, there is very little known about the existence of positive solutions for the following problem:𝐷𝛼0+𝑢(𝑡)+𝜆𝑓(𝑢(𝑡))=0,0<𝑡<1,𝑢(0)=𝑢(1)=𝑢(0)=0,(1.2) where 2<𝛼3 is a real number, 𝐷𝛼0+ is the Riemann-Liouville fractional derivative, 𝜆 is a positive parameter and 𝑓(0,+)(0,+) is continuous.

On one hand, the boundary value problem in [19] is the particular case of problem (1.2) as the case of 𝜆=1. On the other hand, as Yu and Jiang discussed in [19], we also give some existence results by the fixed point theorem on a cone in this paper. Moreover, the purpose of this paper is to derive a 𝜆-interval such that, for any 𝜆 lying in this interval, the problem (1.2) has existence and multiplicity on positive solutions.

In this paper, by analogy with boundary value problems for differential equations of integer order, we firstly give the corresponding Green function named by fractional Green's function and some properties of the Green function. Consequently, the problem (1.2) is reduced to an equivalent Fredholm integral equation. Finally, by the properties of the Green function and Guo-Krasnosel'skii fixed point theorem on cones, the eigenvalue intervals of the nonlinear fractional differential equation boundary value problem are considered, some sufficient conditions for the nonexistence and existence of at least one or two positive solutions for the boundary value problem are established. As an application, some examples are presented to illustrate the main results.

2. Preliminaries

For the convenience of the reader, we give some background materials from fractional calculus theory to facilitate analysis of problem (1.2). These materials can be found in the recent literature; see [1921].

Definition 2.1 (see [20]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a continuous function 𝑓(0,+) is given by 𝐷𝛼0+1𝑓(𝑡)=𝑑Γ(𝑛𝛼)𝑑𝑡(𝑛)𝑡0𝑓(𝑠)(𝑡𝑠)𝛼𝑛+1𝑑𝑠,(2.1) where 𝑛=[𝛼]+1, [𝛼] denotes the integer part of number α, provided that the right side is pointwise defined on (0,+).

Definition 2.2 (see [20]). The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑓(0,+) is given by 𝐼𝛼0+1𝑓(𝑡)=Γ(𝛼)𝑡0(𝑡𝑠)𝛼1𝑓(𝑠)𝑑𝑠,(2.2) provided that the right side is pointwise defined on (0,+).

From the definition of the Riemann-Liouville derivative, we can obtain the following statement.

Lemma 2.3 (see [20]). Let 𝛼>0. If we assume 𝑢𝐶(0,1)𝐿(0,1), then the fractional differential equation 𝐷𝛼0+𝑢(𝑡)=0(2.3) has 𝑢(𝑡)=𝑐1𝑡𝛼1+𝑐2𝑡𝛼2++𝑐𝑁𝑡𝛼𝑁, 𝑐𝑖, 𝑖=1,2,,𝑁, as unique solutions, where 𝑁 is the smallest integer greater than or equal to α.

Lemma 2.4 (see [20]). 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,(2.4) for some 𝑐𝑖, 𝑖=1,2,,𝑁, where 𝑁 is the smallest integer greater than or equal to α.

In the following, we present the Green function of fractional differential equation boundary value problem.

Lemma 2.5 (see [19]). Let 𝐶[0,1] and 2<𝛼3. The unique solution of problem 𝐷𝛼0+𝑢(𝑡)+(𝑡)=0,0<𝑡<1,𝑢(0)=𝑢(1)=𝑢(0)=0(2.5) is 𝑢(𝑡)=10𝐺(𝑡,𝑠)(𝑠)𝑑𝑠,(2.6) where 𝑡𝐺(𝑡,𝑠)=𝛼1(1𝑠)𝛼1(𝑡𝑠)𝛼1Γ𝑡(𝛼),0𝑠𝑡1,𝛼1(1𝑠)𝛼1Γ(𝛼),0𝑡𝑠1.(2.7) Here 𝐺(𝑡,𝑠)is called the Green function of boundary value problem (2.5).

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

Lemma 2.6 (see [19]). The function 𝐺(𝑡,𝑠) defined by (2.7) satisfies the following conditions: (1)𝐺(𝑡,𝑠)=𝐺(1𝑠,1𝑡), for 𝑡,𝑠(0,1); (2)𝑡𝛼1(1𝑡)𝑠(1𝑠)𝛼1Γ(𝛼)𝐺(𝑡,𝑠)(𝛼1)𝑠(1𝑠)𝛼1, for 𝑡,𝑠(0,1); (3)𝐺(𝑡,𝑠)>0, for 𝑡,𝑠(0,1); (4)𝑡𝛼1(1𝑡)𝑠(1𝑠)𝛼1Γ(𝛼)𝐺(𝑡,𝑠)(𝛼1)(1𝑡)𝑡𝛼1, for 𝑡,𝑠(0,1).

The following lemma is fundamental in the proofs of our main results.

Lemma 2.7 (see [21]). Let 𝑋 be a Banach space, and let 𝑃𝑋 be a cone in 𝑋. Assume Ω1,Ω2 are open subsets of 𝑋 with 0Ω1Ω1Ω2, and let 𝑆𝑃𝑃 be a completely continuous operator such that, either (𝐴1)𝑆𝑤𝑤, 𝑤𝑃𝜕Ω1, 𝑆𝑤𝑤, 𝑤𝑃𝜕Ω2 or (𝐴2)𝑆𝑤𝑤, 𝑤𝑃𝜕Ω1, 𝑆𝑤𝑤, 𝑤𝑃𝜕Ω2. Then 𝑆 has a fixed point in 𝑃(Ω2Ω1).

For convenience, we set 𝑞(𝑡)=𝑡𝛼1(1𝑡),𝑘(𝑠)=𝑠(1𝑠)𝛼1; then 𝑞(𝑡)𝑘(𝑠)Γ(𝛼)𝐺(𝑡,𝑠)(𝛼1)𝑘(𝑠).(2.8)

3. Main Results

In this section, we establish the existence of positive solutions for boundary value problem (1.2).

Let Banach space 𝐸=𝐶[0,1] be endowed with the norm 𝑢=max0𝑡1|𝑢(𝑡)|. Define the cone 𝑃𝐸 by 𝑃=𝑢𝐸𝑢(𝑡)𝑞(𝑡)[]𝛼1𝑢,𝑡0,1.(3.1)

Suppose that 𝑢 is a solution of boundary value problem (1.2). Then 𝑢(𝑡)=𝜆10[]𝐺(𝑡,𝑠)𝑓(𝑢(𝑠))𝑑𝑠,𝑡0,1.(3.2)

We define an operator 𝐴𝜆𝑃𝐸 as follows: 𝐴𝜆𝑢(𝑡)=𝜆10[]𝐺(𝑡,𝑠)𝑓(𝑢(𝑠))𝑑𝑠,𝑡0,1.(3.3)

By Lemma 2.6, we have 𝐴𝜆𝑢𝜆Γ(𝛼)10𝐴(𝛼1)𝑘(𝑠)𝑓(𝑢(𝑠))𝑑𝑠,𝜆𝑢𝜆(𝑡)Γ(𝛼)10𝑞(𝑡)𝑘(𝑠)𝑓(𝑢(𝑠))𝑑𝑠𝑞(𝑡)𝐴𝛼1𝜆𝑢.(3.4) Thus, 𝐴𝜆(𝑃)𝑃.

Then we have the following lemma.

Lemma 3.1. 𝐴𝜆𝑃𝑃 is completely continuous.

Proof. The operator 𝐴𝜆𝑃𝑃 is continuous in view of continuity of 𝐺(𝑡,𝑠) and 𝑓(𝑢(𝑡)). By means of the Arzela-Ascoli theorem, 𝐴𝜆𝑃𝑃 is completely continuous.
For convenience, we denote𝐹0=lim𝑢0+sup𝑓(𝑢)𝑢,𝐹=lim𝑢+sup𝑓(𝑢)𝑢,𝑓0=lim𝑢0+inf𝑓(𝑢)𝑢,𝑓=lim𝑢+inf𝑓(𝑢)𝑢,𝐶1=1Γ(𝛼)10𝐶(𝛼1)𝑘(𝑠)𝑑𝑠,2=1Γ(𝛼)101𝐶(𝛼1)𝑞(𝑠)𝑘(𝑠)𝑑𝑠,3=1Γ(𝛼)101(𝛼1)𝑘(𝑠)𝑑𝑠.(3.5)

Theorem 3.2. If there exists 𝑙(0,1) such that 𝑞(𝑙)𝑓𝐶2>𝐹0𝐶1 holds, then for each 𝜆𝑞(𝑙)𝑓𝐶21,𝐹0𝐶11,(3.6) the boundary value problem (1.2) has at least one positive solution. Here we impose (𝑞(𝑙)𝑓𝐶2)1=0 if 𝑓=+ and (𝐹0𝐶1)1=+ if 𝐹0=0.

Proof. Let 𝜆 satisfy (3.6) and 𝜀>0 be such that 𝑓𝑞(𝑙)𝐶𝜀21𝐹𝜆0𝐶+𝜀11.(3.7) By the definition of 𝐹0, we see that there exists 𝑟1>0 such that 𝑓𝐹(𝑢)0+𝜀𝑢,for0<𝑢𝑟1.(3.8) So if 𝑢𝑃 with 𝑢=𝑟1, then by (3.7) and (3.8), we have 𝐴𝜆𝑢𝜆Γ(𝛼)10𝜆(𝛼1)𝑘(𝑠)𝑓(𝑢(𝑠))𝑑𝑠Γ(𝛼)10𝐹(𝛼1)𝑘(𝑠)0𝑟+𝜀1𝐹𝑑𝑠=𝜆0𝑟+𝜀1𝐶1𝑟1=𝑢.(3.9) Hence, if we choose Ω1={𝑢𝐸𝑢<𝑟1}, then 𝐴𝜆𝑢𝑢,for𝑢𝑃𝜕Ω1.(3.10)
Let 𝑟3>0 be such that𝑓𝑓(𝑢)𝜀𝑢,for𝑢𝑟3.(3.11) If 𝑢𝑃 with 𝑢=𝑟2=max{2𝑟1,𝑟3}, then by (3.7) and (3.11), we have 𝐴𝜆𝑢𝐴𝜆𝑢(𝑙)=𝜆10𝜆𝐺(𝑙,𝑠)𝑓(𝑢(𝑠))𝑑𝑠Γ(𝛼)10𝜆𝑞(𝑙)𝑘(𝑠)𝑓(𝑢(𝑠))𝑑𝑠Γ(𝛼)10𝑓𝑞(𝑙)𝑘(𝑠)𝜆𝜀𝑢(𝑠)𝑑𝑠Γ(𝛼)10𝑞(𝑙)𝑓𝛼1𝑞(𝑠)𝑘(𝑠)𝜀𝑢𝑑𝑠=𝜆𝑞(𝑙)𝐶2𝑓𝜀𝑢𝑢.(3.12) Thus, if we set Ω2={𝑢𝐸𝑢<𝑟2}, then 𝐴𝜆𝑢𝑢,for𝑢𝑃𝜕Ω2.(3.13) Now, from (3.10), (3.13), and Lemma 2.7, we guarantee that 𝐴𝜆 has a fixed-point 𝑢𝑃(Ω2Ω1) with 𝑟1𝑢𝑟2, and clearly 𝑢 is a positive solution of (1.2). The proof is complete.

Theorem 3.3. If there exists 𝑙(0,1) such that 𝑞(𝑙)𝐶2𝑓0>𝐹𝐶1 holds, then for each 𝜆𝑞(𝑙)𝑓0𝐶21,𝐹𝐶11,(3.14) the boundary value problem (1.2) has at least one positive solution. Here we impose (𝑞(𝑙)𝑓0𝐶2)1=0 if 𝑓0=+ and (𝐹𝐶1)1=+ if 𝐹=0.

Proof. Let 𝜆 satisfy (3.14) and 𝜀>0 be such that 𝑓𝑞(𝑙)0𝐶𝜀21𝐹𝜆𝐶+𝜀11.(3.15) From the definition of 𝑓0, we see that there exists 𝑟1>0 such that 𝑓𝑓(𝑢)0𝜀𝑢,for0<𝑢𝑟1.(3.16) Further, if 𝑢𝑃 with 𝑢=𝑟1, then similar to the second part of Theorem 3.2, we can obtain that 𝐴𝜆𝑢𝑢. Thus, if we choose Ω1={𝑢𝐸𝑢<𝑟1}, then 𝐴𝜆𝑢𝑢,for𝑢𝑃𝜕Ω2.(3.17)
Next, we may choose 𝑅1>0 such that𝑓𝐹(𝑢)+𝜀𝑢,for𝑢𝑅1.(3.18) We consider two cases.Case 1. Suppose 𝑓 is bounded. Then there exists some 𝑀>0, such that 𝑓(𝑢)𝑀,for𝑢(0,+).(3.19) We define 𝑟3=max{2𝑟1,𝜆𝑀𝐶1}, and 𝑢𝑃 with 𝑢=𝑟3, then 𝐴𝜆𝑢𝜆Γ(𝛼)10(𝛼1)𝑘(𝑠)𝑓(𝑢(𝑠))𝑑𝑠𝜆𝑀Γ(𝛼)10(𝛼1)𝑘(𝑠)𝑑𝑠𝜆𝑀𝐶1𝑟3𝑢.(3.20) Hence, 𝐴𝜆𝑢𝑢,for𝑢𝑃𝑟3=𝑢𝑃𝑢𝑟3.(3.21)Case 2. Suppose 𝑓 is unbounded. Then there exists some 𝑟4>max{2𝑟1,𝑅1}, such that 𝑓𝑟(𝑢)𝑓4,for0<𝑢𝑟4.(3.22) Let 𝑢𝑃 with 𝑢=𝑟4. Then by (3.15) and (3.18), we have 𝐴𝜆𝑢𝜆Γ(𝛼)10𝜆(𝛼1)𝑘(𝑠)𝑓(𝑢(𝑠))𝑑𝑠Γ(𝛼)10𝐹(𝛼1)𝑘(𝑠)+𝜀𝑢𝑑𝑠𝜆𝐶1𝐹+𝜀𝑢𝑢.(3.23) Thus, (3.21) is also true.In both Cases 1 and 2, if we set Ω2={𝑢𝐸𝑢<𝑟2=max{𝑟3,𝑟4}}, then 𝐴𝜆𝑢𝑢,for𝑢𝑃𝜕Ω2.(3.24) Now that we obtain (3.17) and (3.24), it follows from Lemma 2.7 that 𝐴𝜆 has a fixed-point 𝑢𝑃(Ω2Ω1) with 𝑟1𝑢𝑟2. It is clear 𝑢 is a positive solution of (1.2). The proof is complete.

Theorem 3.4. Suppose there exist 𝑙(0,1), 𝑟2>𝑟1>0 such that 𝑞(𝑙)>(𝛼1)𝑟1/𝑟2, and 𝑓 satisfy min(𝑞(𝑙)/(𝛼1))𝑟1𝑢𝑟1𝑟𝑓(𝑢)1𝜆(𝛼1)𝑞(𝑙)𝐶3,max0𝑢𝑟2𝑟𝑓(𝑢)2𝜆𝐶1.(3.25) Then the boundary value problem (1.2) has a positive solution 𝑢𝑃 with 𝑟1𝑢𝑟2.

Proof. Choose Ω1={𝑢𝐸𝑢<𝑟1}; then for 𝑢𝑃𝜕Ω1, we have 𝐴𝜆𝑢𝐴𝜆𝑢(𝑙)=𝜆10𝜆𝐺(𝑙,𝑠)𝑓(𝑢(𝑠))𝑑𝑠Γ(𝛼)10𝜆𝑞(𝑙)𝑘(𝑠)𝑓(𝑢(𝑠))𝑑𝑠Γ(𝛼)10𝑞(𝑙)𝑘(𝑠)min(𝑞(𝑙)/(𝛼1))𝑟1𝑢𝑟1𝑓(𝑢(𝑠))𝑑𝑠𝜆(𝛼1)𝑞(𝑙)𝐶3𝑟1𝜆(𝛼1)𝑞(𝑙)𝐶3=𝑟1=𝑢.(3.26)
On the other hand, choose Ω2={𝑢𝐸𝑢<𝑟2}, then for 𝑢𝑃𝜕Ω2, we have 𝐴𝜆𝑢𝜆Γ(𝛼)10𝜆(𝛼1)𝑘(𝑠)𝑓(𝑢(𝑠))𝑑𝑠Γ(𝛼)10(𝛼1)𝑘(𝑠)max0𝑢𝑟2𝑓(𝑢(𝑠))𝑑𝑠𝜆𝐶1𝑟2𝜆𝐶1=𝑟2=𝑢.(3.27) Thus, by Lemma 2.7, the boundary value problem (1.2) has a positive solution 𝑢𝑃 with 𝑟1𝑢𝑟2. The proof is complete.

For the reminder of the paper, we will need the following condition.(𝐻)(min𝑢[(𝑞(𝑙)/(𝛼1))𝑟,𝑟]𝑓(𝑢))/𝑟>0, where 𝑙(0,1).

Denote𝜆1=sup𝑟>0𝑟𝐶1max0𝑢𝑟,𝜆𝑓(𝑢)(3.28)2=inf𝑟>0𝑟𝐶3min(𝑞(𝑙)/(𝛼1))𝑟𝑢𝑟.𝑓(𝑢)(3.29)

In view of the continuity of 𝑓(𝑢) and (𝐻), we have 0<𝜆1+ and 0𝜆2<+.

Theorem 3.5. Assume (𝐻) holds. If 𝑓0=+ and 𝑓=+, then the boundary value problem (1.2) has at least two positive solutions for each 𝜆(0,𝜆1).

Proof. Define 𝑎𝑟(𝑟)=𝐶1max0𝑢𝑟.𝑓(𝑢)(3.30) By the continuity of 𝑓(𝑢), 𝑓0=+ and 𝑓=+, we have that 𝑎(𝑟)(0,+)(0,+) is continuous and lim𝑟0𝑎(𝑟)=lim𝑟+𝑎(𝑟)=0.(3.31) By (3.28), there exists 𝑟0(0,+), such that 𝑎𝑟0=sup𝑟>0𝑎(𝑟)=𝜆1;(3.32) then for 𝜆(0,𝜆1), there exist constants 𝑐1,𝑐2(0<𝑐1<𝑟0<𝑐2<+) with 𝑎𝑐1𝑐=𝑎2=𝜆.(3.33) Thus, 𝑐𝑓(𝑢)1𝜆𝐶1,for𝑢0,𝑐1,𝑐(3.34)𝑓(𝑢)2𝜆𝐶1,for𝑢0,𝑐2.(3.35)
On the other hand, applying the conditions 𝑓0=+ and 𝑓=+, there exist constants 𝑑1, 𝑑2(0<𝑑1<𝑐1<𝑟0<𝑐2<𝑑2<+) with𝑓(𝑢)𝑢1𝑞2(𝑙)𝜆𝐶3,for𝑢0,𝑑1𝑞(𝑙)𝑑𝛼12.,+(3.36) Then min(𝑞(𝑙)/(𝛼1))𝑑1𝑢𝑑1𝑑𝑓(𝑢)1𝜆(𝛼1)𝑞(𝑙)𝐶3,(3.37)min(𝑞(𝑙)/(𝛼1))𝑑2𝑢𝑑2𝑑𝑓(𝑢)2𝜆(𝛼1)𝑞(𝑙)𝐶3.(3.38) By (3.34) and (3.37), (3.35) and (3.38), combining with Theorem 3.4 and Lemma 2.7, we can complete the proof.

Corollary 3.6. Assume (𝐻) holds. If 𝑓0=+ or 𝑓=+, then the boundary value problem (1.2) has at least one positive solution for each 𝜆(0,𝜆1).

Theorem 3.7. Assume (𝐻) holds. If 𝑓0=0 and 𝑓=0, then for each 𝜆(𝜆2,+), the boundary value problem (1.2) has at least two positive solutions.

Proof. Define 𝑏𝑟(𝑟)=𝐶3min(𝑞(𝑙)/(𝛼1))𝑟𝑢𝑟.𝑓(𝑢)(3.39) By the continuity of 𝑓(𝑢), 𝑓0=0 and 𝑓=0, we easily see that 𝑏(𝑟)(0,+)(0,+) is continuous and lim𝑟0𝑏(𝑟)=lim𝑟+𝑏(𝑟)=+.(3.40) By (3.29), there exists 𝑟0(0,+), such that 𝑏𝑟0=inf𝑟>0𝑏(𝑟)=𝜆2.(3.41) For 𝜆(𝜆2,+), there exist constants 𝑑1, 𝑑2(0<𝑑1<𝑟0<𝑑2<+) with 𝑏𝑑1𝑑=𝑏2=𝜆.(3.42) Therefore, 𝑑𝑓(𝑢)1𝜆(𝛼1)𝑞(𝑙)𝐶3,for𝑢𝑞(𝑙)𝑑𝛼11,𝑑1,𝑑𝑓(𝑢)2𝜆(𝛼1)𝑞(𝑙)𝐶3,for𝑢𝑞(𝑙)𝑑𝛼12,𝑑2.(3.43)
 On the other hand, using 𝑓0=0, we know that there exists a constant 𝑐1(0<𝑐1<𝑑1) with𝑓(𝑢)𝑢1𝜆𝐶1,for𝑢0,𝑐1,(3.44)max0𝑢𝑐1𝑐𝑓(𝑢)1𝜆𝐶1.(3.45) In view of 𝑓=0, there exists a constant 𝑐2(𝑑2,+) such that 𝑓(𝑢)𝑢1𝜆𝐶1𝑐,for𝑢2.,+(3.46)
Let 𝑀=max0𝑢𝑐2𝑓(𝑢),𝑐2𝜆𝐶1𝑀.(3.47) It is easily seen that max0𝑢𝑐2𝑐𝑓(𝑢)2𝜆𝐶1.(3.48) By (3.45) and (3.48), combining with Theorem 3.4 and Lemma 2.7, the proof is complete.

Corollary 3.8. Assume (𝐻) holds. If 𝑓0=0 or 𝑓=0, then for each 𝜆(𝜆2,+), the boundary value problem (1.2) has at least one positive solution.

By the above theorems, we can obtain the following results.

Corollary 3.9. Assume (𝐻) holds. If 𝑓0=+, 𝑓=𝑑, or 𝑓=+, 𝑓0=𝑑, then for any 𝜆(0,(𝑑𝐶1)1), the boundary value problem (1.2) has at least one positive solution.

Corollary 3.10. Assume (𝐻) holds. If 𝑓0=0,𝑓=𝑑, or if 𝑓=0, 𝑓0=𝑑, then for any 𝜆((𝑞(𝑙)𝑑𝐶2)1,+), the boundary value problem (1.2) has at least one positive solution.

Remark 3.11. For the integer derivative case 𝛼=3, Theorems 3.23.7 also hold; we can find the corresponding existence results in [22].

4. Nonexistence

In this section, we give some sufficient conditions for the nonexistence of positive solution to the problem (1.2).

Theorem 4.1. Assume (𝐻) holds. If 𝐹0<+ and 𝐹<, then there exists a 𝜆0>0 such that for all 0<𝜆<𝜆0, the boundary value problem (1.2) has no positive solution.

Proof. Since 𝐹0<+ and 𝐹<+, there exist positive numbers 𝑚1,𝑚2,𝑟1, and 𝑟2, such that 𝑟1<𝑟2 and 𝑓(𝑢)𝑚1𝑢,for𝑢0,𝑟1,𝑓(𝑢)𝑚2𝑟𝑢,for𝑢2.,+(4.1) Let 𝑚=max{𝑚1,𝑚2,max𝑟1𝑢𝑟2{𝑓(𝑢)/𝑢}}. Then we have [𝑓(𝑢)𝑚𝑢,for𝑢0,+).(4.2) Assume 𝑣(𝑡) is a positive solution of (1.2). We will show that this leads to a contradiction for 0<𝜆<𝜆0=(𝑚𝐶1)1. Since 𝐴𝜆𝑣(𝑡)=𝑣(𝑡) for 𝑡[0,1], 𝐴𝑣=𝜆𝑣𝜆Γ(𝛼)10(𝛼1)𝑘(𝑠)𝑓(𝑣(𝑠))𝑑𝑠𝑚𝜆Γ(𝛼)𝑣10(𝛼1)𝑘(𝑠)𝑑𝑠<𝑣,(4.3) which is a contradiction. Therefore, (1.2) has no positive solution. The proof is complete.

Theorem 4.2. Assume (𝐻) holds. If 𝑓0>0 and 𝑓>0, then there exists a 𝜆0>0 such that for all 𝜆>𝜆0, the boundary value problem (1.2) has no positive solution.

Proof. By 𝑓0>0 and 𝑓>0, we know that there exist positive numbers 𝑛1,𝑛2,𝑟1, and 𝑟2, such that 𝑟1<𝑟2 and 𝑓(𝑢)𝑛1𝑢,for𝑢0,𝑟1,𝑓(𝑢)𝑛2𝑟𝑢,for𝑢2.,+(4.4) Let 𝑛=min{𝑛1,𝑛2,min𝑟1𝑢𝑟2{𝑓(𝑢)/𝑢}}>0. Then we get [𝑓(𝑢)𝑛𝑢,for𝑢0,+).(4.5) Assume 𝑣(𝑡) is a positive solution of (1.2). We will show that this leads to a contradiction for 𝜆>𝜆0=(𝑞(𝑙)𝑛𝐶2)1. Since 𝐴𝜆𝑣(𝑡)=𝑣(𝑡) for 𝑡[0,1], 𝐴𝑣=𝜆𝑣𝜆Γ(𝛼)10𝑞(𝑙)𝑘(𝑠)𝑓(𝑣(𝑠))𝑑𝑠>𝑣,(4.6) which is a contradiction. Thus, (1.2) has no positive solution. The proof is complete.

5. Examples

In this section, we will present some examples to illustrate the main results.

Example 5.1. Consider the boundary value problem 𝐷05/2+𝑢(𝑡)+𝜆𝑢𝑎=0,0<𝑡<1,𝑎>1,𝑢(0)=𝑢(1)=𝑢(0)=0.(5.1)
Since 𝛼=5/2, we have𝐶1=1Γ(𝛼)101(𝛼1)𝑘(𝑠)𝑑𝑠=Γ(5/2)1032𝑠(1𝑠)3/2𝐶𝑑𝑠=0.1290,2=1Γ(𝛼)101𝑞1(𝛼1)(𝑠)𝑘(𝑠)𝑑𝑠=Γ(5/2)1023𝑠5/2(1𝑠)5/2𝑑𝑠=0.0077.(5.2) Let 𝑓(𝑢)=𝑢𝑎,𝑎>1. Then we have 𝐹0=0, 𝑓=+. Choose 𝑙=1/2. Then 𝑞(1/2)=2/8=0.1768. So 𝑞(𝑙)𝐶2𝑓>𝐹0𝐶1 holds. Thus, by Theorem 3.2, the boundary value problem (5.1) has a positive solution for each 𝜆(0,+).

Example 5.2. Discuss the boundary value problem 𝐷05/2+𝑢(𝑡)+𝜆𝑢𝑏=0,0<𝑡<1,0<𝑏<1,𝑢(0)=𝑢(1)=𝑢(0)=0.(5.3)
Since 𝛼=5/2, we have 𝐶1=0.1290 and 𝐶2=0.0077. Let 𝑓(𝑢)=𝑢𝑏,0<𝑏<1. Then we have 𝐹=0,𝑓0=+. Choose 𝑙=1/2. Then 𝑞(1/2)=2/8=0.1768. So 𝑞(𝑙)𝐶2𝑓0>𝐹𝐶1 holds. Thus, by Theorem 3.3, the boundary value problem (5.3) has a positive solution for each 𝜆(0,+).

Example 5.3. Consider the boundary value problem 𝐷05/2+𝑢(𝑡)+𝜆200𝑢2+𝑢(2+sin𝑢)𝑢+1=0,0<𝑡<1,𝑎>1,𝑢(0)=𝑢(1)=𝑢(0)=0.(5.4)
Since 𝛼=5/2, we have 𝐶1=0.129 and 𝐶2=0.0077. Let 𝑓(𝑢)=(200𝑢2+𝑢)(2+sin𝑢)/(𝑢+1). Then we have 𝐹0=𝑓0=2,𝐹=600, 𝑓=200, and 2𝑢<𝑓(𝑢)<600𝑢.
(i)Choose 𝑙=1/2. Then 𝑞(1/2)=2/8=0.1768. So 𝑞(𝑙)𝐶2𝑓>𝐹0𝐶1 holds. Thus, by Theorem 3.2, the boundary value problem (5.4) has a positive solution for each 𝜆(3.6937,3.8759).(ii)By Theorem 4.1, the boundary value problem (5.4) has no positive solution for all 𝜆(0,0.0129).(iii)By Theorem 4.2, the boundary value problem (5.4) has no positive solution for all 𝜆(369.369,+).

Example 5.4. Consider the boundary value problem 𝐷05/2+𝑢𝑢(𝑡)+𝜆2+𝑢(2+sin𝑢)150𝑢+1=0,0<𝑡<1,𝑎>1,𝑢(0)=𝑢(1)=𝑢(0)=0.(5.5)
Since 𝛼=5/2, we have 𝐶1=0.129 and 𝐶2=0.0077. Let 𝑓(𝑢)=(𝑢2+𝑢)(2+sin𝑢)/(150𝑢+1). Then we have 𝐹0=𝑓0=2, 𝐹=1/50, 𝑓=1/150, and 𝑢/150<𝑓(𝑢)<2𝑢.
(i)Choose 𝑙=1/2. Then 𝑞(1/2)=2/8=0.1768. So 𝑞(𝑙)𝐶2𝑓0>𝐹𝐶1 holds. Thus, by Theorem 3.3, the boundary value problem (5.5) has a positive solution for each 𝜆(369.369,387.5968).(ii)By Theorem 4.1, the boundary value problem (5.5) has no positive solution for all 𝜆(0,3.8759).(iii)By Theorem 4.2, the boundary value problem (5.5) has no positive solution for all 𝜆(110810.6911,+).

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 manuscript. This research is supported by the Natural Science Foundation of China (11071143, 11026112, 60904024), the Natural Science Foundation of Shandong (Y2008A28, ZR2009AL003), University of Jinan Research Funds for Doctors (XBS0843) and University of Jinan Innovation Funds for Graduate Students (YCX09014).

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