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Abstract and Applied Analysis
VolumeΒ 2011Β (2011), Article IDΒ 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: JosefΒ DiblΓ­k

Copyright Β© 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+𝑒(𝑑)+πœ†π‘“(𝑒(𝑑))=0, 0<𝑑<1, 𝑒(0)=𝑒(1)=𝑒′(0)=0, where 2<𝛼≀3 is a real number, 𝐷𝛼0+ is the Riemann-Liouville fractional derivative, πœ† is a positive parameter, and π‘“βˆΆ(0,+∞)β†’(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 [1–4]. 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 [5–11]. 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 [12–19].

Yu and Jiang [19] examined the existence of positive solutions for the following problem: 𝐷𝛼0+𝑒(𝑑)+𝑓(𝑑,𝑒(𝑑))=0,0<𝑑<1,𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=0,(1.1) where 2<𝛼≀3 is a real number, π‘“βˆˆπΆ([0,1]Γ—[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+𝑒(𝑑)+πœ†π‘“(𝑒(𝑑))=0,0<𝑑<1,𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=0,(1.2) where 2<𝛼≀3 is a real number, 𝐷𝛼0+ is the Riemann-Liouville fractional derivative, πœ† is a positive parameter and π‘“βˆΆ(0,+∞)β†’(0,+∞) is continuous.

On one hand, the boundary value problem in [19] is the particular case of problem (1.2) as the case of πœ†=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 πœ†-interval such that, for any πœ† 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 [19–21].

Definition 2.1 (see [20]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a continuous function π‘“βˆΆ(0,+∞)→ℝ is given by 𝐷𝛼0+1𝑓(𝑑)=𝑑Γ(π‘›βˆ’π›Ό)𝑑𝑑(𝑛)ξ€œπ‘‘0𝑓(𝑠)(π‘‘βˆ’π‘ )π›Όβˆ’π‘›+1𝑑𝑠,(2.1) where 𝑛=[𝛼]+1, [𝛼] denotes the integer part of number Ξ±, 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 π‘“βˆΆ(0,+∞)→ℝ is given by 𝐼𝛼0+1𝑓(𝑑)=ξ€œΞ“(𝛼)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1𝑓(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 π‘’βˆˆπΆ(0,1)∩𝐿(0,1), then the fractional differential equation 𝐷𝛼0+𝑒(𝑑)=0(2.3) has 𝑒(𝑑)=𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2+β‹―+π‘π‘π‘‘π›Όβˆ’π‘, π‘π‘–βˆˆβ„, 𝑖=1,2,…,𝑁, as unique solutions, where 𝑁 is the smallest integer greater than or equal to Ξ±.

Lemma 2.4 (see [20]). Assume that π‘’βˆˆπΆ(0,1)∩𝐿(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐢(0,1)∩𝐿(0,1). Then 𝐼𝛼0+𝐷𝛼0+𝑒(𝑑)=𝑒(𝑑)+𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2+β‹―+π‘π‘π‘‘π›Όβˆ’π‘,(2.4) for some π‘π‘–βˆˆβ„, 𝑖=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to Ξ±.

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

Lemma 2.5 (see [19]). Let β„ŽβˆˆπΆ[0,1] and 2<𝛼≀3. The unique solution of problem 𝐷𝛼0+𝑒(𝑑)+β„Ž(𝑑)=0,0<𝑑<1,𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=0(2.5) is ξ€œπ‘’(𝑑)=10𝐺(𝑑,𝑠)β„Ž(𝑠)𝑑𝑠,(2.6) where ⎧βŽͺβŽͺ⎨βŽͺβŽͺβŽ©π‘‘πΊ(𝑑,𝑠)=π›Όβˆ’1(1βˆ’π‘ )π›Όβˆ’1βˆ’(π‘‘βˆ’π‘ )π›Όβˆ’1Γ𝑑(𝛼),0≀𝑠≀𝑑≀1,π›Όβˆ’1(1βˆ’π‘ )π›Όβˆ’1Ξ“(𝛼),0≀𝑑≀𝑠≀1.(2.7) Here 𝐺(𝑑,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(𝑑,𝑠) defined by (2.7) satisfies the following conditions: (1)𝐺(𝑑,𝑠)=𝐺(1βˆ’π‘ ,1βˆ’π‘‘), for 𝑑,π‘ βˆˆ(0,1); (2)π‘‘π›Όβˆ’1(1βˆ’π‘‘)𝑠(1βˆ’π‘ )π›Όβˆ’1≀Γ(𝛼)𝐺(𝑑,𝑠)≀(π›Όβˆ’1)𝑠(1βˆ’π‘ )π›Όβˆ’1, for 𝑑,π‘ βˆˆ(0,1); (3)𝐺(𝑑,𝑠)>0, for 𝑑,π‘ βˆˆ(0,1); (4)π‘‘π›Όβˆ’1(1βˆ’π‘‘)𝑠(1βˆ’π‘ )π›Όβˆ’1≀Γ(𝛼)𝐺(𝑑,𝑠)≀(π›Όβˆ’1)(1βˆ’π‘‘)π‘‘π›Όβˆ’1, for 𝑑,π‘ βˆˆ(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 π‘ƒβŠ‚π‘‹ be a cone in 𝑋. Assume Ξ©1,Ξ©2 are open subsets of 𝑋 with 0∈Ω1βŠ‚Ξ©1βŠ‚Ξ©2, and let π‘†βˆΆπ‘ƒβ†’π‘ƒ be a completely continuous operator such that, either (𝐴1)‖𝑆𝑀‖≀‖𝑀‖, π‘€βˆˆπ‘ƒβˆ©πœ•Ξ©1, ‖𝑆𝑀‖β‰₯‖𝑀‖, π‘€βˆˆπ‘ƒβˆ©πœ•Ξ©2 or (𝐴2)‖𝑆𝑀‖β‰₯‖𝑀‖, π‘€βˆˆπ‘ƒβˆ©πœ•Ξ©1, ‖𝑆𝑀‖≀‖𝑀‖, π‘€βˆˆπ‘ƒβˆ©πœ•Ξ©2. Then 𝑆 has a fixed point in π‘ƒβˆ©(Ξ©2⧡Ω1).

For convenience, we set π‘ž(𝑑)=π‘‘π›Όβˆ’1(1βˆ’π‘‘),π‘˜(𝑠)=𝑠(1βˆ’π‘ )π›Όβˆ’1; then π‘ž(𝑑)π‘˜(𝑠)≀Γ(𝛼)𝐺(𝑑,𝑠)≀(π›Όβˆ’1)π‘˜(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=𝐢[0,1] be endowed with the norm ‖𝑒‖=max0≀𝑑≀1|𝑒(𝑑)|. Define the cone π‘ƒβŠ‚πΈ by 𝑃=π‘’βˆˆπΈβˆΆπ‘’(𝑑)β‰₯π‘ž(𝑑)[]ξ‚Όπ›Όβˆ’1‖𝑒‖,π‘‘βˆˆ0,1.(3.1)

Suppose that 𝑒 is a solution of boundary value problem (1.2). Then ξ€œπ‘’(𝑑)=πœ†10[]𝐺(𝑑,𝑠)𝑓(𝑒(𝑠))𝑑𝑠,π‘‘βˆˆ0,1.(3.2)

We define an operator π΄πœ†βˆΆπ‘ƒβ†’πΈ as follows: ξ€·π΄πœ†π‘’ξ€Έξ€œ(𝑑)=πœ†10[]𝐺(𝑑,𝑠)𝑓(𝑒(𝑠))𝑑𝑠,π‘‘βˆˆ0,1.(3.3)

By Lemma 2.6, we have β€–β€–π΄πœ†π‘’β€–β€–β‰€πœ†ξ€œΞ“(𝛼)10𝐴(π›Όβˆ’1)π‘˜(𝑠)𝑓(𝑒(𝑠))𝑑𝑠,πœ†π‘’ξ€Έπœ†(𝑑)β‰₯ξ€œΞ“(𝛼)10π‘žβ‰₯(𝑑)π‘˜(𝑠)𝑓(𝑒(𝑠))π‘‘π‘ π‘ž(𝑑)β€–β€–π΄π›Όβˆ’1πœ†π‘’β€–β€–.(3.4) Thus, π΄πœ†(𝑃)βŠ‚π‘ƒ.

Then we have the following lemma.

Lemma 3.1. π΄πœ†βˆΆπ‘ƒβ†’π‘ƒ is completely continuous.

Proof. The operator π΄πœ†βˆΆπ‘ƒβ†’π‘ƒ is continuous in view of continuity of 𝐺(𝑑,𝑠) and 𝑓(𝑒(𝑑)). By means of the Arzela-Ascoli theorem, π΄πœ†βˆΆπ‘ƒβ†’π‘ƒ is completely continuous.
For convenience, we denote𝐹0=lim𝑒→0+sup𝑓(𝑒)𝑒,𝐹∞=lim𝑒→+∞sup𝑓(𝑒)𝑒,𝑓0=lim𝑒→0+inf𝑓(𝑒)𝑒,π‘“βˆž=lim𝑒→+∞inf𝑓(𝑒)𝑒,𝐢1=1ξ€œΞ“(𝛼)10𝐢(π›Όβˆ’1)π‘˜(𝑠)𝑑𝑠,2=1ξ€œΞ“(𝛼)101𝐢(π›Όβˆ’1)π‘ž(𝑠)π‘˜(𝑠)𝑑𝑠,3=1ξ€œΞ“(𝛼)101(π›Όβˆ’1)π‘˜(𝑠)𝑑𝑠.(3.5)

Theorem 3.2. If there exists π‘™βˆˆ(0,1) such that π‘ž(𝑙)π‘“βˆžπΆ2>𝐹0𝐢1 holds, then for each ξ‚€ξ€·πœ†βˆˆπ‘ž(𝑙)π‘“βˆžπΆ2ξ€Έβˆ’1,𝐹0𝐢1ξ€Έβˆ’1,(3.6) the boundary value problem (1.2) has at least one positive solution. Here we impose (π‘ž(𝑙)π‘“βˆžπΆ2)βˆ’1=0 if π‘“βˆž=+∞ and (𝐹0𝐢1)βˆ’1=+∞ if 𝐹0=0.

Proof. Let πœ† satisfy (3.6) and πœ€>0 be such that ξ€·ξ€·π‘“π‘ž(𝑙)βˆžξ€ΈπΆβˆ’πœ€2ξ€Έβˆ’1πΉβ‰€πœ†β‰€ξ€·ξ€·0𝐢+πœ€1ξ€Έβˆ’1.(3.7) By the definition of 𝐹0, we see that there exists π‘Ÿ1>0 such that 𝑓𝐹(𝑒)≀0ξ€Έ+πœ€π‘’,for0<π‘’β‰€π‘Ÿ1.(3.8) So if π‘’βˆˆπ‘ƒ with ‖𝑒‖=π‘Ÿ1, then by (3.7) and (3.8), we have β€–β€–π΄πœ†π‘’β€–β€–β‰€πœ†ξ€œΞ“(𝛼)10β‰€πœ†(π›Όβˆ’1)π‘˜(𝑠)𝑓(𝑒(𝑠))π‘‘π‘ ξ€œΞ“(𝛼)10𝐹(π›Όβˆ’1)π‘˜(𝑠)0ξ€Έπ‘Ÿ+πœ€1𝐹𝑑𝑠=πœ†0ξ€Έπ‘Ÿ+πœ€1𝐢1β‰€π‘Ÿ1=‖𝑒‖.(3.9) Hence, if we choose Ξ©1={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<π‘Ÿ1}, then β€–β€–π΄πœ†π‘’β€–β€–β‰€β€–π‘’β€–,forπ‘’βˆˆπ‘ƒβˆ©πœ•Ξ©1.(3.10)
Let π‘Ÿ3>0 be such that𝑓𝑓(𝑒)β‰₯βˆžξ€Έβˆ’πœ€π‘’,for𝑒β‰₯π‘Ÿ3.(3.11) If π‘’βˆˆπ‘ƒ with ‖𝑒‖=π‘Ÿ2=max{2π‘Ÿ1,π‘Ÿ3}, then by (3.7) and (3.11), we have β€–β€–π΄πœ†π‘’β€–β€–β‰₯π΄πœ†ξ€œπ‘’(𝑙)=πœ†10β‰₯πœ†πΊ(𝑙,𝑠)𝑓(𝑒(𝑠))π‘‘π‘ ξ€œΞ“(𝛼)10β‰₯πœ†π‘ž(𝑙)π‘˜(𝑠)𝑓(𝑒(𝑠))π‘‘π‘ ξ€œΞ“(𝛼)10ξ€·π‘“π‘ž(𝑙)π‘˜(𝑠)βˆžξ€Έβ‰₯πœ†βˆ’πœ€π‘’(𝑠)π‘‘π‘ ξ€œΞ“(𝛼)10π‘ž(𝑙)ξ€·π‘“π›Όβˆ’1π‘ž(𝑠)π‘˜(𝑠)βˆžξ€Έβˆ’πœ€β€–π‘’β€–π‘‘π‘ =πœ†π‘ž(𝑙)𝐢2ξ€·π‘“βˆžξ€Έβ€–βˆ’πœ€π‘’β€–β‰₯‖𝑒‖.(3.12) Thus, if we set Ξ©2={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<π‘Ÿ2}, then β€–β€–π΄πœ†π‘’β€–β€–β‰₯‖𝑒‖,forπ‘’βˆˆπ‘ƒβˆ©πœ•Ξ©2.(3.13) Now, from (3.10), (3.13), and Lemma 2.7, we guarantee that π΄πœ† has a fixed-point π‘’βˆˆπ‘ƒβˆ©(Ξ©2⧡Ω1) with π‘Ÿ1β‰€β€–π‘’β€–β‰€π‘Ÿ2, and clearly 𝑒 is a positive solution of (1.2). The proof is complete.

Theorem 3.3. If there exists π‘™βˆˆ(0,1) such that π‘ž(𝑙)𝐢2𝑓0>𝐹∞𝐢1 holds, then for each ξ‚€ξ€·πœ†βˆˆπ‘ž(𝑙)𝑓0𝐢2ξ€Έβˆ’1,ξ€·πΉβˆžπΆ1ξ€Έβˆ’1,(3.14) the boundary value problem (1.2) has at least one positive solution. Here we impose (π‘ž(𝑙)𝑓0𝐢2)βˆ’1=0 if 𝑓0=+∞ and (𝐹∞𝐢1)βˆ’1=+∞ if 𝐹∞=0.

Proof. Let πœ† satisfy (3.14) and πœ€>0 be such that ξ€·ξ€·π‘“π‘ž(𝑙)0ξ€ΈπΆβˆ’πœ€2ξ€Έβˆ’1πΉβ‰€πœ†β‰€ξ€·ξ€·βˆžξ€ΈπΆ+πœ€1ξ€Έβˆ’1.(3.15) From the definition of 𝑓0, we see that there exists π‘Ÿ1>0 such that 𝑓𝑓(𝑒)β‰₯0ξ€Έβˆ’πœ€π‘’,for0<π‘’β‰€π‘Ÿ1.(3.16) Further, if π‘’βˆˆπ‘ƒ with ‖𝑒‖=π‘Ÿ1, then similar to the second part of Theorem 3.2, we can obtain that β€–π΄πœ†π‘’β€–β‰₯‖𝑒‖. Thus, if we choose Ξ©1={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<π‘Ÿ1}, then β€–β€–π΄πœ†π‘’β€–β€–β‰₯‖𝑒‖,forπ‘’βˆˆπ‘ƒβˆ©πœ•Ξ©2.(3.17)
Next, we may choose 𝑅1>0 such that𝑓𝐹(𝑒)β‰€βˆžξ€Έ+πœ€π‘’,for𝑒β‰₯𝑅1.(3.18) We consider two cases.Case 1. Suppose 𝑓 is bounded. Then there exists some 𝑀>0, such that 𝑓(𝑒)≀𝑀,forπ‘’βˆˆ(0,+∞).(3.19) We define π‘Ÿ3=max{2π‘Ÿ1,πœ†π‘€πΆ1}, and π‘’βˆˆπ‘ƒ with ‖𝑒‖=π‘Ÿ3, then β€–β€–π΄πœ†π‘’β€–β€–β‰€πœ†ξ€œΞ“(𝛼)10≀(π›Όβˆ’1)π‘˜(𝑠)𝑓(𝑒(𝑠))π‘‘π‘ πœ†π‘€ξ€œΞ“(𝛼)10(π›Όβˆ’1)π‘˜(𝑠)π‘‘π‘ β‰€πœ†π‘€πΆ1β‰€π‘Ÿ3≀‖𝑒‖.(3.20) Hence, β€–β€–π΄πœ†π‘’β€–β€–β‰€β€–π‘’β€–,forπ‘’βˆˆπ‘ƒπ‘Ÿ3=ξ€½π‘’βˆˆπ‘ƒβˆΆβ€–π‘’β€–β‰€π‘Ÿ3ξ€Ύ.(3.21)Case 2. Suppose 𝑓 is unbounded. Then there exists some π‘Ÿ4>max{2π‘Ÿ1,𝑅1}, such that π‘“ξ€·π‘Ÿ(𝑒)≀𝑓4ξ€Έ,for0<π‘’β‰€π‘Ÿ4.(3.22) Let π‘’βˆˆπ‘ƒ with ‖𝑒‖=π‘Ÿ4. Then by (3.15) and (3.18), we have β€–β€–π΄πœ†π‘’β€–β€–β‰€πœ†ξ€œΞ“(𝛼)10β‰€πœ†(π›Όβˆ’1)π‘˜(𝑠)𝑓(𝑒(𝑠))π‘‘π‘ ξ€œΞ“(𝛼)10𝐹(π›Όβˆ’1)π‘˜(𝑠)βˆžξ€Έ+πœ€β€–π‘’β€–π‘‘π‘ β‰€πœ†πΆ1ξ€·πΉβˆžξ€Έ+πœ€β€–π‘’β€–β‰€β€–π‘’β€–.(3.23) Thus, (3.21) is also true.In both Cases 1 and 2, if we set Ξ©2={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<π‘Ÿ2=max{π‘Ÿ3,π‘Ÿ4}}, then β€–β€–π΄πœ†π‘’β€–β€–β‰€β€–π‘’β€–,forπ‘’βˆˆπ‘ƒβˆ©πœ•Ξ©2.(3.24) Now that we obtain (3.17) and (3.24), it follows from Lemma 2.7 that π΄πœ† has a fixed-point π‘’βˆˆπ‘ƒβˆ©(Ξ©2⧡Ω1) with π‘Ÿ1β‰€β€–π‘’β€–β‰€π‘Ÿ2. It is clear 𝑒 is a positive solution of (1.2). The proof is complete.

Theorem 3.4. Suppose there exist π‘™βˆˆ(0,1), π‘Ÿ2>π‘Ÿ1>0 such that π‘ž(𝑙)>(π›Όβˆ’1)π‘Ÿ1/π‘Ÿ2, and 𝑓 satisfy min(π‘ž(𝑙)/(π›Όβˆ’1))π‘Ÿ1β‰€π‘’β‰€π‘Ÿ1π‘Ÿπ‘“(𝑒)β‰₯1πœ†(π›Όβˆ’1)π‘ž(𝑙)𝐢3,max0β‰€π‘’β‰€π‘Ÿ2π‘Ÿπ‘“(𝑒)≀2πœ†πΆ1.(3.25) Then the boundary value problem (1.2) has a positive solution π‘’βˆˆπ‘ƒ with π‘Ÿ1β‰€β€–π‘’β€–β‰€π‘Ÿ2.

Proof. Choose Ξ©1={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<π‘Ÿ1}; then for π‘’βˆˆπ‘ƒβˆ©πœ•Ξ©1, we have β€–β€–π΄πœ†π‘’β€–β€–β‰₯π΄πœ†ξ€œπ‘’(𝑙)=πœ†10β‰₯πœ†πΊ(𝑙,𝑠)𝑓(𝑒(𝑠))π‘‘π‘ ξ€œΞ“(𝛼)10β‰₯πœ†π‘ž(𝑙)π‘˜(𝑠)𝑓(𝑒(𝑠))π‘‘π‘ ξ€œΞ“(𝛼)10π‘ž(𝑙)π‘˜(𝑠)min(π‘ž(𝑙)/(π›Όβˆ’1))π‘Ÿ1β‰€π‘’β‰€π‘Ÿ1𝑓(𝑒(𝑠))𝑑𝑠β‰₯πœ†(π›Όβˆ’1)π‘ž(𝑙)𝐢3π‘Ÿ1πœ†(π›Όβˆ’1)π‘ž(𝑙)𝐢3=π‘Ÿ1=‖𝑒‖.(3.26)
On the other hand, choose Ξ©2={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<π‘Ÿ2}, then for π‘’βˆˆπ‘ƒβˆ©πœ•Ξ©2, we have β€–β€–π΄πœ†π‘’β€–β€–β‰€πœ†ξ€œΞ“(𝛼)10β‰€πœ†(π›Όβˆ’1)π‘˜(𝑠)𝑓(𝑒(𝑠))π‘‘π‘ ξ€œΞ“(𝛼)10(π›Όβˆ’1)π‘˜(𝑠)max0β‰€π‘’β‰€π‘Ÿ2𝑓(𝑒(𝑠))π‘‘π‘ β‰€πœ†πΆ1π‘Ÿ2πœ†πΆ1=π‘Ÿ2=‖𝑒‖.(3.27) Thus, by Lemma 2.7, the boundary value problem (1.2) has a positive solution π‘’βˆˆπ‘ƒ with π‘Ÿ1β‰€β€–π‘’β€–β‰€π‘Ÿ2. The proof is complete.

For the reminder of the paper, we will need the following condition.(𝐻)(minπ‘’βˆˆ[(π‘ž(𝑙)/(π›Όβˆ’1))π‘Ÿ,π‘Ÿ]𝑓(𝑒))/π‘Ÿ>0, where π‘™βˆˆ(0,1).

Denoteπœ†1=supπ‘Ÿ>0π‘ŸπΆ1max0β‰€π‘’β‰€π‘Ÿ,πœ†π‘“(𝑒)(3.28)2=infπ‘Ÿ>0π‘ŸπΆ3min(π‘ž(𝑙)/(π›Όβˆ’1))π‘Ÿβ‰€π‘’β‰€π‘Ÿ.𝑓(𝑒)(3.29)

In view of the continuity of 𝑓(𝑒) and (𝐻), we have 0<πœ†1≀+∞ and 0β‰€πœ†2<+∞.

Theorem 3.5. Assume (𝐻) holds. If 𝑓0=+∞ and π‘“βˆž=+∞, then the boundary value problem (1.2) has at least two positive solutions for each πœ†βˆˆ(0,πœ†1).

Proof. Define π‘Žπ‘Ÿ(π‘Ÿ)=𝐢1max0β‰€π‘’β‰€π‘Ÿ.𝑓(𝑒)(3.30) By the continuity of 𝑓(𝑒), 𝑓0=+∞ and π‘“βˆž=+∞, we have that π‘Ž(π‘Ÿ)∢(0,+∞)β†’(0,+∞) is continuous and limπ‘Ÿβ†’0π‘Ž(π‘Ÿ)=limπ‘Ÿβ†’+βˆžπ‘Ž(π‘Ÿ)=0.(3.31) By (3.28), there exists π‘Ÿ0∈(0,+∞), such that π‘Žξ€·π‘Ÿ0ξ€Έ=supπ‘Ÿ>0π‘Ž(π‘Ÿ)=πœ†1;(3.32) then for πœ†βˆˆ(0,πœ†1), there exist constants 𝑐1,𝑐2(0<𝑐1<π‘Ÿ0<𝑐2<+∞) with π‘Žξ€·π‘1𝑐=π‘Ž2ξ€Έ=πœ†.(3.33) Thus, 𝑐𝑓(𝑒)≀1πœ†πΆ1ξ€Ί,forπ‘’βˆˆ0,𝑐1ξ€»,𝑐(3.34)𝑓(𝑒)≀2πœ†πΆ1ξ€Ί,forπ‘’βˆˆ0,𝑐2ξ€».(3.35)
On the other hand, applying the conditions 𝑓0=+∞ and π‘“βˆž=+∞, there exist constants 𝑑1, 𝑑2(0<𝑑1<𝑐1<π‘Ÿ0<𝑐2<𝑑2<+∞) with𝑓(𝑒)𝑒β‰₯1π‘ž2(𝑙)πœ†πΆ3ξ€·,forπ‘’βˆˆ0,𝑑1ξ€Έβˆͺξ‚΅π‘ž(𝑙)π‘‘π›Όβˆ’12ξ‚Ά.,+∞(3.36) Then min(π‘ž(𝑙)/(π›Όβˆ’1))𝑑1≀𝑒≀𝑑1𝑑𝑓(𝑒)β‰₯1πœ†(π›Όβˆ’1)π‘ž(𝑙)𝐢3,(3.37)min(π‘ž(𝑙)/(π›Όβˆ’1))𝑑2≀𝑒≀𝑑2𝑑𝑓(𝑒)β‰₯2πœ†(π›Όβˆ’1)π‘ž(𝑙)𝐢3.(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 π‘“βˆž=+∞, then the boundary value problem (1.2) has at least one positive solution for each πœ†βˆˆ(0,πœ†1).

Theorem 3.7. Assume (𝐻) holds. If 𝑓0=0 and π‘“βˆž=0, then for each πœ†βˆˆ(πœ†2,+∞), the boundary value problem (1.2) has at least two positive solutions.

Proof. Define π‘π‘Ÿ(π‘Ÿ)=𝐢3min(π‘ž(𝑙)/(π›Όβˆ’1))π‘Ÿβ‰€π‘’β‰€π‘Ÿ.𝑓(𝑒)(3.39) By the continuity of 𝑓(𝑒), 𝑓0=0 and π‘“βˆž=0, we easily see that 𝑏(π‘Ÿ)∢(0,+∞)β†’(0,+∞) is continuous and limπ‘Ÿβ†’0𝑏(π‘Ÿ)=limπ‘Ÿβ†’+βˆžπ‘(π‘Ÿ)=+∞.(3.40) By (3.29), there exists π‘Ÿ0∈(0,+∞), such that π‘ξ€·π‘Ÿ0ξ€Έ=infπ‘Ÿ>0𝑏(π‘Ÿ)=πœ†2.(3.41) For πœ†βˆˆ(πœ†2,+∞), there exist constants 𝑑1, 𝑑2(0<𝑑1<π‘Ÿ0<𝑑2<+∞) with 𝑏𝑑1𝑑=𝑏2ξ€Έ=πœ†.(3.42) Therefore, 𝑑𝑓(𝑒)β‰₯1πœ†(π›Όβˆ’1)π‘ž(𝑙)𝐢3ξ‚Έ,forπ‘’βˆˆπ‘ž(𝑙)π‘‘π›Όβˆ’11,𝑑1ξ‚Ή,𝑑𝑓(𝑒)β‰₯2πœ†(π›Όβˆ’1)π‘ž(𝑙)𝐢3ξ‚Έ,forπ‘’βˆˆπ‘ž(𝑙)π‘‘π›Όβˆ’12,𝑑2ξ‚Ή.(3.43)
 On the other hand, using 𝑓0=0, we know that there exists a constant 𝑐1(0<𝑐1<𝑑1) with𝑓(𝑒)𝑒≀1πœ†πΆ1ξ€·,forπ‘’βˆˆ0,𝑐1ξ€Έ,(3.44)max0≀𝑒≀𝑐1𝑐𝑓(𝑒)≀1πœ†πΆ1.(3.45) In view of π‘“βˆž=0, there exists a constant 𝑐2∈(𝑑2,+∞) such that 𝑓(𝑒)𝑒≀1πœ†πΆ1𝑐,forπ‘’βˆˆ2ξ€Έ.,+∞(3.46)
Let 𝑀=max0≀𝑒≀𝑐2𝑓(𝑒),𝑐2β‰₯πœ†πΆ1𝑀.(3.47) It is easily seen that max0≀𝑒≀𝑐2𝑐𝑓(𝑒)≀2πœ†πΆ1.(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 π‘“βˆž=0, then for each πœ†βˆˆ(πœ†2,+∞), 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=+∞, π‘“βˆž=𝑑, or π‘“βˆž=+∞, 𝑓0=𝑑, then for any πœ†βˆˆ(0,(𝑑𝐢1)βˆ’1), the boundary value problem (1.2) has at least one positive solution.

Corollary 3.10. Assume (𝐻) holds. If 𝑓0=0,π‘“βˆž=𝑑, or if π‘“βˆž=0, 𝑓0=𝑑, then for any πœ†βˆˆ((π‘ž(𝑙)𝑑𝐢2)βˆ’1,+∞), the boundary value problem (1.2) has at least one positive solution.

Remark 3.11. For the integer derivative case 𝛼=3, Theorems 3.2–3.7 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 πœ†0>0 such that for all 0<πœ†<πœ†0, the boundary value problem (1.2) has no positive solution.

Proof. Since 𝐹0<+∞ and 𝐹∞<+∞, there exist positive numbers π‘š1,π‘š2,π‘Ÿ1, and π‘Ÿ2, such that π‘Ÿ1<π‘Ÿ2 and 𝑓(𝑒)β‰€π‘š1𝑒,forπ‘’βˆˆ0,π‘Ÿ1ξ€»,𝑓(𝑒)β‰€π‘š2ξ€Ίπ‘Ÿπ‘’,forπ‘’βˆˆ2ξ€Έ.,+∞(4.1) Let π‘š=max{π‘š1,π‘š2,maxπ‘Ÿ1β‰€π‘’β‰€π‘Ÿ2{𝑓(𝑒)/𝑒}}. Then we have [𝑓(𝑒)β‰€π‘šπ‘’,forπ‘’βˆˆ0,+∞).(4.2) Assume 𝑣(𝑑) is a positive solution of (1.2). We will show that this leads to a contradiction for 0<πœ†<πœ†0∢=(π‘šπΆ1)βˆ’1. Since π΄πœ†π‘£(𝑑)=𝑣(𝑑) for π‘‘βˆˆ[0,1], ‖‖𝐴‖𝑣‖=πœ†π‘£β€–β€–β‰€πœ†ξ€œΞ“(𝛼)10(π›Όβˆ’1)π‘˜(𝑠)𝑓(𝑣(𝑠))π‘‘π‘ β‰€π‘šπœ†ξ€œΞ“(𝛼)‖𝑣‖10(π›Όβˆ’1)π‘˜(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 π‘“βˆž>0, then there exists a πœ†0>0 such that for all πœ†>πœ†0, the boundary value problem (1.2) has no positive solution.

Proof. By 𝑓0>0 and π‘“βˆž>0, we know that there exist positive numbers 𝑛1,𝑛2,π‘Ÿ1, and π‘Ÿ2, such that π‘Ÿ1<π‘Ÿ2 and 𝑓(𝑒)β‰₯𝑛1𝑒,forπ‘’βˆˆ0,π‘Ÿ1ξ€»,𝑓(𝑒)β‰₯𝑛2ξ€Ίπ‘Ÿπ‘’,forπ‘’βˆˆ2ξ€Έ.,+∞(4.4) Let 𝑛=min{𝑛1,𝑛2,minπ‘Ÿ1β‰€π‘’β‰€π‘Ÿ2{𝑓(𝑒)/𝑒}}>0. Then we get [𝑓(𝑒)β‰₯𝑛𝑒,forπ‘’βˆˆ0,+∞).(4.5) Assume 𝑣(𝑑) is a positive solution of (1.2). We will show that this leads to a contradiction for πœ†>πœ†0∢=(π‘ž(𝑙)𝑛𝐢2)βˆ’1. Since π΄πœ†π‘£(𝑑)=𝑣(𝑑) for π‘‘βˆˆ[0,1], ‖‖𝐴‖𝑣‖=πœ†π‘£β€–β€–β‰₯πœ†ξ€œΞ“(𝛼)10π‘ž(𝑙)π‘˜(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+𝑒(𝑑)+πœ†π‘’π‘Ž=0,0<𝑑<1,π‘Ž>1,𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=0.(5.1)
Since 𝛼=5/2, we have𝐢1=1ξ€œΞ“(𝛼)101(π›Όβˆ’1)π‘˜(𝑠)𝑑𝑠=ξ€œΞ“(5/2)1032𝑠(1βˆ’π‘ )3/2𝐢𝑑𝑠=0.1290,2=1ξ€œΞ“(𝛼)101π‘ž1(π›Όβˆ’1)(𝑠)π‘˜(𝑠)𝑑𝑠=ξ€œΞ“(5/2)1023𝑠5/2(1βˆ’π‘ )5/2𝑑𝑠=0.0077.(5.2) Let 𝑓(𝑒)=π‘’π‘Ž,π‘Ž>1. Then we have 𝐹0=0, π‘“βˆž=+∞. Choose 𝑙=1/2. Then βˆšπ‘ž(1/2)=2/8=0.1768. So π‘ž(𝑙)𝐢2π‘“βˆž>𝐹0𝐢1 holds. Thus, by Theorem 3.2, the boundary value problem (5.1) has a positive solution for each πœ†βˆˆ(0,+∞).

Example 5.2. Discuss the boundary value problem 𝐷05/2+𝑒(𝑑)+πœ†π‘’π‘=0,0<𝑑<1,0<𝑏<1,𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=0.(5.3)
Since 𝛼=5/2, we have 𝐢1=0.1290 and 𝐢2=0.0077. Let 𝑓(𝑒)=𝑒𝑏,0<𝑏<1. Then we have 𝐹∞=0,𝑓0=+∞. Choose 𝑙=1/2. Then βˆšπ‘ž(1/2)=2/8=0.1768. So π‘ž(𝑙)𝐢2𝑓0>𝐹∞𝐢1 holds. Thus, by Theorem 3.3, the boundary value problem (5.3) has a positive solution for each πœ†βˆˆ(0,+∞).

Example 5.3. Consider the boundary value problem 𝐷05/2+𝑒(𝑑)+πœ†200𝑒2ξ€Έ+𝑒(2+sin𝑒)𝑒+1=0,0<𝑑<1,π‘Ž>1,𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=0.(5.4)
Since 𝛼=5/2, we have 𝐢1=0.129 and 𝐢2=0.0077. Let 𝑓(𝑒)=(200𝑒2+𝑒)(2+sin𝑒)/(𝑒+1). Then we have 𝐹0=𝑓0=2,𝐹∞=600, π‘“βˆž=200, and 2𝑒<𝑓(𝑒)<600𝑒.
(i)Choose 𝑙=1/2. Then βˆšπ‘ž(1/2)=2/8=0.1768. So π‘ž(𝑙)𝐢2π‘“βˆž>𝐹0𝐢1 holds. Thus, by Theorem 3.2, the boundary value problem (5.4) has a positive solution for each πœ†βˆˆ(3.6937,3.8759).(ii)By Theorem 4.1, the boundary value problem (5.4) has no positive solution for all πœ†βˆˆ(0,0.0129).(iii)By Theorem 4.2, the boundary value problem (5.4) has no positive solution for all πœ†βˆˆ(369.369,+∞).

Example 5.4. Consider the boundary value problem 𝐷05/2+𝑒𝑒(𝑑)+πœ†2ξ€Έ+𝑒(2+sin𝑒)150𝑒+1=0,0<𝑑<1,π‘Ž>1,𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=0.(5.5)
Since 𝛼=5/2, we have 𝐢1=0.129 and 𝐢2=0.0077. Let 𝑓(𝑒)=(𝑒2+𝑒)(2+sin𝑒)/(150𝑒+1). Then we have 𝐹0=𝑓0=2, 𝐹∞=1/50, π‘“βˆž=1/150, and 𝑒/150<𝑓(𝑒)<2𝑒.
(i)Choose 𝑙=1/2. Then βˆšπ‘ž(1/2)=2/8=0.1768. So π‘ž(𝑙)𝐢2𝑓0>𝐹∞𝐢1 holds. Thus, by Theorem 3.3, the boundary value problem (5.5) has a positive solution for each πœ†βˆˆ(369.369,387.5968).(ii)By Theorem 4.1, the boundary value problem (5.5) has no positive solution for all πœ†βˆˆ(0,3.8759).(iii)By Theorem 4.2, the boundary value problem (5.5) has no positive solution for all πœ†βˆˆ(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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