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Abstract and Applied Analysis
VolumeΒ 2011Β (2011), Article IDΒ 390543, 16 pages
doi:10.1155/2011/390543
Research Article

Positive Solutions to Boundary Value Problems of Nonlinear Fractional Differential Equations

Yige Zhao,1Β Shurong Sun,1,2Β Zhenlai Han,1Β and Qiuping Li1

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 Diblík

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 𝐷 𝛼 0 + 𝑒 ( 𝑑 ) + πœ† 𝑓 ( 𝑒 ( 𝑑 ) ) = 0 , 0 < 𝑑 < 1 , 𝑒 ( 0 ) = 𝑒 ( 1 ) = 𝑒 β€² ( 0 ) = 0 , where 2 < 𝛼 ≀ 3 is a real number, 𝐷 𝛼 0 + is the Riemann-Liouville fractional derivative, πœ† is a positive parameter, and 𝑓 ∢ ( 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: 𝐷 𝛼 0 + 𝑒 ( 𝑑 ) + 𝑓 ( 𝑑 , 𝑒 ( 𝑑 ) ) = 0 , 0 < 𝑑 < 1 , 𝑒 ( 0 ) = 𝑒 ( 1 ) = 𝑒 ξ…ž ( 0 ) = 0 , ( 1 . 1 ) where 2 < 𝛼 ≀ 3 is a real number, 𝑓 ∈ 𝐢 ( [ 0 , 1 ] Γ— [ 0 , + ∞ ) , ( 0 , + ∞ ) ) , and 𝐷 𝛼 0 + 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: 𝐷 𝛼 0 + 𝑒 ( 𝑑 ) + πœ† 𝑓 ( 𝑒 ( 𝑑 ) ) = 0 , 0 < 𝑑 < 1 , 𝑒 ( 0 ) = 𝑒 ( 1 ) = 𝑒 ξ…ž ( 0 ) = 0 , ( 1 . 2 ) where 2 < 𝛼 ≀ 3 is a real number, 𝐷 𝛼 0 + is the Riemann-Liouville fractional derivative, πœ† is a positive parameter and 𝑓 ∢ ( 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 𝑓 ∢ ( 0 , + ∞ ) β†’ ℝ is given by 𝐷 𝛼 0 + 1 𝑓 ( 𝑑 ) = ξ‚€ 𝑑 Ξ“ ( 𝑛 βˆ’ 𝛼 )  𝑑 𝑑 ( 𝑛 ) ξ€œ 𝑑 0 𝑓 ( 𝑠 ) ( 𝑑 βˆ’ 𝑠 ) 𝛼 βˆ’ 𝑛 + 1 𝑑 𝑠 , ( 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 , + ∞ ) β†’ ℝ is given by 𝐼 𝛼 0 + 1 𝑓 ( 𝑑 ) = ξ€œ Ξ“ ( 𝛼 ) 𝑑 0 ( 𝑑 βˆ’ 𝑠 ) 𝛼 βˆ’ 1 𝑓 ( 𝑠 ) 𝑑 𝑠 , ( 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 𝑒 ∈ 𝐢 ( 0 , 1 ) ∩ 𝐿 ( 0 , 1 ) , then the fractional differential equation 𝐷 𝛼 0 + 𝑒 ( 𝑑 ) = 0 ( 2 . 3 ) has 𝑒 ( 𝑑 ) = 𝑐 1 𝑑 𝛼 βˆ’ 1 + 𝑐 2 𝑑 𝛼 βˆ’ 2 + β‹― + 𝑐 𝑁 𝑑 𝛼 βˆ’ 𝑁 , 𝑐 𝑖 ∈ ℝ , 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 𝑒 ∈ 𝐢 ( 0 , 1 ) ∩ 𝐿 ( 0 , 1 ) with a fractional derivative of order 𝛼 > 0 that belongs to 𝐢 ( 0 , 1 ) ∩ 𝐿 ( 0 , 1 ) . Then 𝐼 𝛼 0 + 𝐷 𝛼 0 + 𝑒 ( 𝑑 ) = 𝑒 ( 𝑑 ) + 𝑐 1 𝑑 𝛼 βˆ’ 1 + 𝑐 2 𝑑 𝛼 βˆ’ 2 + β‹― + 𝑐 𝑁 𝑑 𝛼 βˆ’ 𝑁 , ( 2 . 4 ) for some 𝑐 𝑖 ∈ ℝ , 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 β„Ž ∈ 𝐢 [ 0 , 1 ] and 2 < 𝛼 ≀ 3 . The unique solution of problem 𝐷 𝛼 0 + 𝑒 ( 𝑑 ) + β„Ž ( 𝑑 ) = 0 , 0 < 𝑑 < 1 , 𝑒 ( 0 ) = 𝑒 ( 1 ) = 𝑒 ξ…ž ( 0 ) = 0 ( 2 . 5 ) is ξ€œ 𝑒 ( 𝑑 ) = 1 0 𝐺 ( 𝑑 , 𝑠 ) β„Ž ( 𝑠 ) 𝑑 𝑠 , ( 2 . 6 ) where ⎧ βŽͺ βŽͺ ⎨ βŽͺ βŽͺ ⎩ 𝑑 𝐺 ( 𝑑 , 𝑠 ) = 𝛼 βˆ’ 1 ( 1 βˆ’ 𝑠 ) 𝛼 βˆ’ 1 βˆ’ ( 𝑑 βˆ’ 𝑠 ) 𝛼 βˆ’ 1 Ξ“ 𝑑 ( 𝛼 ) , 0 ≀ 𝑠 ≀ 𝑑 ≀ 1 , 𝛼 βˆ’ 1 ( 1 βˆ’ 𝑠 ) 𝛼 βˆ’ 1 Ξ“ ( 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 ( 1 βˆ’ 𝑠 , 1 βˆ’ 𝑑 ) , for 𝑑 , 𝑠 ∈ ( 0 , 1 ) ; (2) 𝑑 𝛼 βˆ’ 1 ( 1 βˆ’ 𝑑 ) 𝑠 ( 1 βˆ’ 𝑠 ) 𝛼 βˆ’ 1 ≀ Ξ“ ( 𝛼 ) 𝐺 ( 𝑑 , 𝑠 ) ≀ ( 𝛼 βˆ’ 1 ) 𝑠 ( 1 βˆ’ 𝑠 ) 𝛼 βˆ’ 1 , for 𝑑 , 𝑠 ∈ ( 0 , 1 ) ; (3) 𝐺 ( 𝑑 , 𝑠 ) > 0 , for 𝑑 , 𝑠 ∈ ( 0 , 1 ) ; (4) 𝑑 𝛼 βˆ’ 1 ( 1 βˆ’ 𝑑 ) 𝑠 ( 1 βˆ’ 𝑠 ) 𝛼 βˆ’ 1 ≀ Ξ“ ( 𝛼 ) 𝐺 ( 𝑑 , 𝑠 ) ≀ ( 𝛼 βˆ’ 1 ) ( 1 βˆ’ 𝑑 ) 𝑑 𝛼 βˆ’ 1 , for 𝑑 , 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 𝑃 βŠ‚ 𝑋 be a cone in 𝑋 . Assume Ξ© 1 , Ξ© 2 are open subsets of 𝑋 with 0 ∈ Ξ© 1 βŠ‚ Ξ© 1 βŠ‚ Ξ© 2 , and let 𝑆 ∢ 𝑃 β†’ 𝑃 be a completely continuous operator such that, either ( 𝐴 1) β€– 𝑆 𝑀 β€– ≀ β€– 𝑀 β€– , 𝑀 ∈ 𝑃 ∩ πœ• Ξ© 1 , β€– 𝑆 𝑀 β€– β‰₯ β€– 𝑀 β€– , 𝑀 ∈ 𝑃 ∩ πœ• Ξ© 2 or ( 𝐴 2) β€– 𝑆 𝑀 β€– β‰₯ β€– 𝑀 β€– , 𝑀 ∈ 𝑃 ∩ πœ• Ξ© 1 , β€– 𝑆 𝑀 β€– ≀ β€– 𝑀 β€– , 𝑀 ∈ 𝑃 ∩ πœ• Ξ© 2 . Then 𝑆 has a fixed point in 𝑃 ∩ ( Ξ© 2 ⧡ Ξ© 1 ) .

For convenience, we set π‘ž ( 𝑑 ) = 𝑑 𝛼 βˆ’ 1 ( 1 βˆ’ 𝑑 ) , π‘˜ ( 𝑠 ) = 𝑠 ( 1 βˆ’ 𝑠 ) 𝛼 βˆ’ 1 ; then π‘ž ( 𝑑 ) π‘˜ ( 𝑠 ) ≀ Ξ“ ( 𝛼 ) 𝐺 ( 𝑑 , 𝑠 ) ≀ ( 𝛼 βˆ’ 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 β€– 𝑒 β€– = m a x 0 ≀ 𝑑 ≀ 1 | 𝑒 ( 𝑑 ) | . Define the cone 𝑃 βŠ‚ 𝐸 by ξ‚» 𝑃 = 𝑒 ∈ 𝐸 ∢ 𝑒 ( 𝑑 ) β‰₯ π‘ž ( 𝑑 ) [ ] ξ‚Ό 𝛼 βˆ’ 1 β€– 𝑒 β€– , 𝑑 ∈ 0 , 1 . ( 3 . 1 )

Suppose that 𝑒 is a solution of boundary value problem (1.2). Then ξ€œ 𝑒 ( 𝑑 ) = πœ† 1 0 [ ] 𝐺 ( 𝑑 , 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 , 𝑑 ∈ 0 , 1 . ( 3 . 2 )

We define an operator 𝐴 πœ† ∢ 𝑃 β†’ 𝐸 as follows: ξ€· 𝐴 πœ† 𝑒 ξ€Έ ξ€œ ( 𝑑 ) = πœ† 1 0 [ ] 𝐺 ( 𝑑 , 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 , 𝑑 ∈ 0 , 1 . ( 3 . 3 )

By Lemma 2.6, we have β€– β€– 𝐴 πœ† 𝑒 β€– β€– ≀ πœ† ξ€œ Ξ“ ( 𝛼 ) 1 0 ξ€· 𝐴 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 , πœ† 𝑒 ξ€Έ πœ† ( 𝑑 ) β‰₯ ξ€œ Ξ“ ( 𝛼 ) 1 0 π‘ž β‰₯ ( 𝑑 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 π‘ž ( 𝑑 ) β€– β€– 𝐴 𝛼 βˆ’ 1 πœ† 𝑒 β€– β€– . ( 3 . 4 ) Thus, 𝐴 πœ† ( 𝑃 ) βŠ‚ 𝑃 .

Then we have the following lemma.

Lemma 3.1. 𝐴 πœ† ∢ 𝑃 β†’ 𝑃 is completely continuous.

Proof. The operator 𝐴 πœ† ∢ 𝑃 β†’ 𝑃 is continuous in view of continuity of 𝐺 ( 𝑑 , 𝑠 ) and 𝑓 ( 𝑒 ( 𝑑 ) ) . By means of the Arzela-Ascoli theorem, 𝐴 πœ† ∢ 𝑃 β†’ 𝑃 is completely continuous.
For convenience, we denote 𝐹 0 = l i m 𝑒 β†’ 0 + s u p 𝑓 ( 𝑒 ) 𝑒 , 𝐹 ∞ = l i m 𝑒 β†’ + ∞ s u p 𝑓 ( 𝑒 ) 𝑒 , 𝑓 0 = l i m 𝑒 β†’ 0 + i n f 𝑓 ( 𝑒 ) 𝑒 , 𝑓 ∞ = l i m 𝑒 β†’ + ∞ i n f 𝑓 ( 𝑒 ) 𝑒 , 𝐢 1 = 1 ξ€œ Ξ“ ( 𝛼 ) 1 0 𝐢 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑑 𝑠 , 2 = 1 ξ€œ Ξ“ ( 𝛼 ) 1 0 1 𝐢 ( 𝛼 βˆ’ 1 ) π‘ž ( 𝑠 ) π‘˜ ( 𝑠 ) 𝑑 𝑠 , 3 = 1 ξ€œ Ξ“ ( 𝛼 ) 1 0 1 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑑 𝑠 . ( 3 . 5 )

Theorem 3.2. If there exists 𝑙 ∈ ( 0 , 1 ) such that π‘ž ( 𝑙 ) 𝑓 ∞ 𝐢 2 > 𝐹 0 𝐢 1 holds, then for each ξ‚€ ξ€· πœ† ∈ π‘ž ( 𝑙 ) 𝑓 ∞ 𝐢 2 ξ€Έ βˆ’ 1 , ξ€· 𝐹 0 𝐢 1 ξ€Έ βˆ’ 1  , ( 3 . 6 ) the boundary value problem (1.2) has at least one positive solution. Here we impose ( π‘ž ( 𝑙 ) 𝑓 ∞ 𝐢 2 ) βˆ’ 1 = 0 if 𝑓 ∞ = + ∞ and ( 𝐹 0 𝐢 1 ) βˆ’ 1 = + ∞ if 𝐹 0 = 0 .

Proof. Let πœ† satisfy (3.6) and πœ€ > 0 be such that ξ€· ξ€· 𝑓 π‘ž ( 𝑙 ) ∞ ξ€Έ 𝐢 βˆ’ πœ€ 2 ξ€Έ βˆ’ 1 𝐹 ≀ πœ† ≀ ξ€· ξ€· 0 ξ€Έ 𝐢 + πœ€ 1 ξ€Έ βˆ’ 1 . ( 3 . 7 ) By the definition of 𝐹 0 , we see that there exists π‘Ÿ 1 > 0 such that 𝑓 ξ€· 𝐹 ( 𝑒 ) ≀ 0 ξ€Έ + πœ€ 𝑒 , f o r 0 < 𝑒 ≀ π‘Ÿ 1 . ( 3 . 8 ) So if 𝑒 ∈ 𝑃 with β€– 𝑒 β€– = π‘Ÿ 1 , then by (3.7) and (3.8), we have β€– β€– 𝐴 πœ† 𝑒 β€– β€– ≀ πœ† ξ€œ Ξ“ ( 𝛼 ) 1 0 ≀ πœ† ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 ξ€œ Ξ“ ( 𝛼 ) 1 0 ξ€· 𝐹 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 0 ξ€Έ π‘Ÿ + πœ€ 1 ξ€· 𝐹 𝑑 𝑠 = πœ† 0 ξ€Έ π‘Ÿ + πœ€ 1 𝐢 1 ≀ π‘Ÿ 1 = β€– 𝑒 β€– . ( 3 . 9 ) Hence, if we choose Ξ© 1 = { 𝑒 ∈ 𝐸 ∢ β€– 𝑒 β€– < π‘Ÿ 1 } , then β€– β€– 𝐴 πœ† 𝑒 β€– β€– ≀ β€– 𝑒 β€– , f o r 𝑒 ∈ 𝑃 ∩ πœ• Ξ© 1 . ( 3 . 1 0 )
Let π‘Ÿ 3 > 0 be such that 𝑓 ξ€· 𝑓 ( 𝑒 ) β‰₯ ∞ ξ€Έ βˆ’ πœ€ 𝑒 , f o r 𝑒 β‰₯ π‘Ÿ 3 . ( 3 . 1 1 ) If 𝑒 ∈ 𝑃 with β€– 𝑒 β€– = π‘Ÿ 2 = m a x { 2 π‘Ÿ 1 , π‘Ÿ 3 } , then by (3.7) and (3.11), we have β€– β€– 𝐴 πœ† 𝑒 β€– β€– β‰₯ 𝐴 πœ† ξ€œ 𝑒 ( 𝑙 ) = πœ† 1 0 β‰₯ πœ† 𝐺 ( 𝑙 , 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 ξ€œ Ξ“ ( 𝛼 ) 1 0 β‰₯ πœ† π‘ž ( 𝑙 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 ξ€œ Ξ“ ( 𝛼 ) 1 0 ξ€· 𝑓 π‘ž ( 𝑙 ) π‘˜ ( 𝑠 ) ∞ ξ€Έ β‰₯ πœ† βˆ’ πœ€ 𝑒 ( 𝑠 ) 𝑑 𝑠 ξ€œ Ξ“ ( 𝛼 ) 1 0 π‘ž ( 𝑙 ) ξ€· 𝑓 𝛼 βˆ’ 1 π‘ž ( 𝑠 ) π‘˜ ( 𝑠 ) ∞ ξ€Έ βˆ’ πœ€ β€– 𝑒 β€– 𝑑 𝑠 = πœ† π‘ž ( 𝑙 ) 𝐢 2 ξ€· 𝑓 ∞ ξ€Έ β€– βˆ’ πœ€ 𝑒 β€– β‰₯ β€– 𝑒 β€– . ( 3 . 1 2 ) Thus, if we set Ξ© 2 = { 𝑒 ∈ 𝐸 ∢ β€– 𝑒 β€– < π‘Ÿ 2 } , then β€– β€– 𝐴 πœ† 𝑒 β€– β€– β‰₯ β€– 𝑒 β€– , f o r 𝑒 ∈ 𝑃 ∩ πœ• Ξ© 2 . ( 3 . 1 3 ) Now, from (3.10), (3.13), and Lemma 2.7, we guarantee that 𝐴 πœ† has a fixed-point 𝑒 ∈ 𝑃 ∩ ( Ξ© 2 ⧡ Ξ© 1 ) with π‘Ÿ 1 ≀ β€– 𝑒 β€– ≀ π‘Ÿ 2 , and clearly 𝑒 is a positive solution of (1.2). The proof is complete.

Theorem 3.3. If there exists 𝑙 ∈ ( 0 , 1 ) such that π‘ž ( 𝑙 ) 𝐢 2 𝑓 0 > 𝐹 ∞ 𝐢 1 holds, then for each ξ‚€ ξ€· πœ† ∈ π‘ž ( 𝑙 ) 𝑓 0 𝐢 2 ξ€Έ βˆ’ 1 , ξ€· 𝐹 ∞ 𝐢 1 ξ€Έ βˆ’ 1  , ( 3 . 1 4 ) the boundary value problem (1.2) has at least one positive solution. Here we impose ( π‘ž ( 𝑙 ) 𝑓 0 𝐢 2 ) βˆ’ 1 = 0 if 𝑓 0 = + ∞ and ( 𝐹 ∞ 𝐢 1 ) βˆ’ 1 = + ∞ if 𝐹 ∞ = 0 .

Proof. Let πœ† satisfy (3.14) and πœ€ > 0 be such that ξ€· ξ€· 𝑓 π‘ž ( 𝑙 ) 0 ξ€Έ 𝐢 βˆ’ πœ€ 2 ξ€Έ βˆ’ 1 𝐹 ≀ πœ† ≀ ξ€· ξ€· ∞ ξ€Έ 𝐢 + πœ€ 1 ξ€Έ βˆ’ 1 . ( 3 . 1 5 ) From the definition of 𝑓 0 , we see that there exists π‘Ÿ 1 > 0 such that 𝑓 ξ€· 𝑓 ( 𝑒 ) β‰₯ 0 ξ€Έ βˆ’ πœ€ 𝑒 , f o r 0 < 𝑒 ≀ π‘Ÿ 1 . ( 3 . 1 6 ) Further, if 𝑒 ∈ 𝑃 with β€– 𝑒 β€– = π‘Ÿ 1 , then similar to the second part of Theorem 3.2, we can obtain that β€– 𝐴 πœ† 𝑒 β€– β‰₯ β€– 𝑒 β€– . Thus, if we choose Ξ© 1 = { 𝑒 ∈ 𝐸 ∢ β€– 𝑒 β€– < π‘Ÿ 1 } , then β€– β€– 𝐴 πœ† 𝑒 β€– β€– β‰₯ β€– 𝑒 β€– , f o r 𝑒 ∈ 𝑃 ∩ πœ• Ξ© 2 . ( 3 . 1 7 )
Next, we may choose 𝑅 1 > 0 such that 𝑓 ξ€· 𝐹 ( 𝑒 ) ≀ ∞ ξ€Έ + πœ€ 𝑒 , f o r 𝑒 β‰₯ 𝑅 1 . ( 3 . 1 8 ) We consider two cases.Case 1. Suppose 𝑓 is bounded. Then there exists some 𝑀 > 0 , such that 𝑓 ( 𝑒 ) ≀ 𝑀 , f o r 𝑒 ∈ ( 0 , + ∞ ) . ( 3 . 1 9 ) We define π‘Ÿ 3 = m a x { 2 π‘Ÿ 1 , πœ† 𝑀 𝐢 1 } , and 𝑒 ∈ 𝑃 with β€– 𝑒 β€– = π‘Ÿ 3 , then β€– β€– 𝐴 πœ† 𝑒 β€– β€– ≀ πœ† ξ€œ Ξ“ ( 𝛼 ) 1 0 ≀ ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 πœ† 𝑀 ξ€œ Ξ“ ( 𝛼 ) 1 0 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑑 𝑠 ≀ πœ† 𝑀 𝐢 1 ≀ π‘Ÿ 3 ≀ β€– 𝑒 β€– . ( 3 . 2 0 ) Hence, β€– β€– 𝐴 πœ† 𝑒 β€– β€– ≀ β€– 𝑒 β€– , f o r 𝑒 ∈ 𝑃 π‘Ÿ 3 = ξ€½ 𝑒 ∈ 𝑃 ∢ β€– 𝑒 β€– ≀ π‘Ÿ 3 ξ€Ύ . ( 3 . 2 1 ) Case 2. Suppose 𝑓 is unbounded. Then there exists some π‘Ÿ 4 > m a x { 2 π‘Ÿ 1 , 𝑅 1 } , such that 𝑓 ξ€· π‘Ÿ ( 𝑒 ) ≀ 𝑓 4 ξ€Έ , f o r 0 < 𝑒 ≀ π‘Ÿ 4 . ( 3 . 2 2 ) Let 𝑒 ∈ 𝑃 with β€– 𝑒 β€– = π‘Ÿ 4 . Then by (3.15) and (3.18), we have β€– β€– 𝐴 πœ† 𝑒 β€– β€– ≀ πœ† ξ€œ Ξ“ ( 𝛼 ) 1 0 ≀ πœ† ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 ξ€œ Ξ“ ( 𝛼 ) 1 0 ξ€· 𝐹 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) ∞ ξ€Έ + πœ€ β€– 𝑒 β€– 𝑑 𝑠 ≀ πœ† 𝐢 1 ξ€· 𝐹 ∞ ξ€Έ + πœ€ β€– 𝑒 β€– ≀ β€– 𝑒 β€– . ( 3 . 2 3 ) Thus, (3.21) is also true.In both Cases 1 and 2, if we set Ξ© 2 = { 𝑒 ∈ 𝐸 ∢ β€– 𝑒 β€– < π‘Ÿ 2 = m a x { π‘Ÿ 3 , π‘Ÿ 4 } } , then β€– β€– 𝐴 πœ† 𝑒 β€– β€– ≀ β€– 𝑒 β€– , f o r 𝑒 ∈ 𝑃 ∩ πœ• Ξ© 2 . ( 3 . 2 4 ) Now that we obtain (3.17) and (3.24), it follows from Lemma 2.7 that 𝐴 πœ† has a fixed-point 𝑒 ∈ 𝑃 ∩ ( Ξ© 2 ⧡ Ξ© 1 ) with π‘Ÿ 1 ≀ β€– 𝑒 β€– ≀ π‘Ÿ 2 . It is clear 𝑒 is a positive solution of (1.2). The proof is complete.

Theorem 3.4. Suppose there exist 𝑙 ∈ ( 0 , 1 ) , π‘Ÿ 2 > π‘Ÿ 1 > 0 such that π‘ž ( 𝑙 ) > ( 𝛼 βˆ’ 1 ) π‘Ÿ 1 / π‘Ÿ 2 , and 𝑓 satisfy m i n ( π‘ž ( 𝑙 ) / ( 𝛼 βˆ’ 1 ) ) π‘Ÿ 1 ≀ 𝑒 ≀ π‘Ÿ 1 π‘Ÿ 𝑓 ( 𝑒 ) β‰₯ 1 πœ† ( 𝛼 βˆ’ 1 ) π‘ž ( 𝑙 ) 𝐢 3 , m a x 0 ≀ 𝑒 ≀ π‘Ÿ 2 π‘Ÿ 𝑓 ( 𝑒 ) ≀ 2 πœ† 𝐢 1 . ( 3 . 2 5 ) Then the boundary value problem (1.2) has a positive solution 𝑒 ∈ 𝑃 with π‘Ÿ 1 ≀ β€– 𝑒 β€– ≀ π‘Ÿ 2 .

Proof. Choose Ξ© 1 = { 𝑒 ∈ 𝐸 ∢ β€– 𝑒 β€– < π‘Ÿ 1 } ; then for 𝑒 ∈ 𝑃 ∩ πœ• Ξ© 1 , we have β€– β€– 𝐴 πœ† 𝑒 β€– β€– β‰₯ 𝐴 πœ† ξ€œ 𝑒 ( 𝑙 ) = πœ† 1 0 β‰₯ πœ† 𝐺 ( 𝑙 , 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 ξ€œ Ξ“ ( 𝛼 ) 1 0 β‰₯ πœ† π‘ž ( 𝑙 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 ξ€œ Ξ“ ( 𝛼 ) 1 0 π‘ž ( 𝑙 ) π‘˜ ( 𝑠 ) m i n ( π‘ž ( 𝑙 ) / ( 𝛼 βˆ’ 1 ) ) π‘Ÿ 1 ≀ 𝑒 ≀ π‘Ÿ 1 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 β‰₯ πœ† ( 𝛼 βˆ’ 1 ) π‘ž ( 𝑙 ) 𝐢 3 π‘Ÿ 1 πœ† ( 𝛼 βˆ’ 1 ) π‘ž ( 𝑙 ) 𝐢 3 = π‘Ÿ 1 = β€– 𝑒 β€– . ( 3 . 2 6 )
On the other hand, choose Ξ© 2 = { 𝑒 ∈ 𝐸 ∢ β€– 𝑒 β€– < π‘Ÿ 2 } , then for 𝑒 ∈ 𝑃 ∩ πœ• Ξ© 2 , we have β€– β€– 𝐴 πœ† 𝑒 β€– β€– ≀ πœ† ξ€œ Ξ“ ( 𝛼 ) 1 0 ≀ πœ† ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 ξ€œ Ξ“ ( 𝛼 ) 1 0 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) m a x 0 ≀ 𝑒 ≀ π‘Ÿ 2 𝑓 ( 𝑒 ( 𝑠 ) ) 𝑑 𝑠 ≀ πœ† 𝐢 1 π‘Ÿ 2 πœ† 𝐢 1 = π‘Ÿ 2 = β€– 𝑒 β€– . ( 3 . 2 7 ) Thus, by Lemma 2.7, the boundary value problem (1.2) has a positive solution 𝑒 ∈ 𝑃 with π‘Ÿ 1 ≀ β€– 𝑒 β€– ≀ π‘Ÿ 2 . The proof is complete.

For the reminder of the paper, we will need the following condition. ( 𝐻 ) ( m i n 𝑒 ∈ [ ( π‘ž ( 𝑙 ) / ( 𝛼 βˆ’ 1 ) ) π‘Ÿ , π‘Ÿ ] 𝑓 ( 𝑒 ) ) / π‘Ÿ > 0 , where 𝑙 ∈ ( 0 , 1 ) .

Denote πœ† 1 = s u p π‘Ÿ > 0 π‘Ÿ 𝐢 1 m a x 0 ≀ 𝑒 ≀ π‘Ÿ , πœ† 𝑓 ( 𝑒 ) ( 3 . 2 8 ) 2 = i n f π‘Ÿ > 0 π‘Ÿ 𝐢 3 m i n ( π‘ž ( 𝑙 ) / ( 𝛼 βˆ’ 1 ) ) π‘Ÿ ≀ 𝑒 ≀ π‘Ÿ . 𝑓 ( 𝑒 ) ( 3 . 2 9 )

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

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

Proof. Define π‘Ž π‘Ÿ ( π‘Ÿ ) = 𝐢 1 m a x 0 ≀ 𝑒 ≀ π‘Ÿ . 𝑓 ( 𝑒 ) ( 3 . 3 0 ) By the continuity of 𝑓 ( 𝑒 ) , 𝑓 0 = + ∞ and 𝑓 ∞ = + ∞ , we have that π‘Ž ( π‘Ÿ ) ∢ ( 0 , + ∞ ) β†’ ( 0 , + ∞ ) is continuous and l i m π‘Ÿ β†’ 0 π‘Ž ( π‘Ÿ ) = l i m π‘Ÿ β†’ + ∞ π‘Ž ( π‘Ÿ ) = 0 . ( 3 . 3 1 ) By (3.28), there exists π‘Ÿ 0 ∈ ( 0 , + ∞ ) , such that π‘Ž ξ€· π‘Ÿ 0 ξ€Έ = s u p π‘Ÿ > 0 π‘Ž ( π‘Ÿ ) = πœ† 1 ; ( 3 . 3 2 ) then for πœ† ∈ ( 0 , πœ† 1 ) , there exist constants 𝑐 1 , 𝑐 2 ( 0 < 𝑐 1 < π‘Ÿ 0 < 𝑐 2 < + ∞ ) with π‘Ž ξ€· 𝑐 1 ξ€Έ ξ€· 𝑐 = π‘Ž 2 ξ€Έ = πœ† . ( 3 . 3 3 ) Thus, 𝑐 𝑓 ( 𝑒 ) ≀ 1 πœ† 𝐢 1 ξ€Ί , f o r 𝑒 ∈ 0 , 𝑐 1 ξ€» , 𝑐 ( 3 . 3 4 ) 𝑓 ( 𝑒 ) ≀ 2 πœ† 𝐢 1 ξ€Ί , f o r 𝑒 ∈ 0 , 𝑐 2 ξ€» . ( 3 . 3 5 )
On the other hand, applying the conditions 𝑓 0 = + ∞ and 𝑓 ∞ = + ∞ , there exist constants 𝑑 1 , 𝑑 2 ( 0 < 𝑑 1 < 𝑐 1 < π‘Ÿ 0 < 𝑐 2 < 𝑑 2 < + ∞ ) with 𝑓 ( 𝑒 ) 𝑒 β‰₯ 1 π‘ž 2 ( 𝑙 ) πœ† 𝐢 3 ξ€· , f o r 𝑒 ∈ 0 , 𝑑 1 ξ€Έ βˆͺ ξ‚΅ π‘ž ( 𝑙 ) 𝑑 𝛼 βˆ’ 1 2 ξ‚Ά . , + ∞ ( 3 . 3 6 ) Then m i n ( π‘ž ( 𝑙 ) / ( 𝛼 βˆ’ 1 ) ) 𝑑 1 ≀ 𝑒 ≀ 𝑑 1 𝑑 𝑓 ( 𝑒 ) β‰₯ 1 πœ† ( 𝛼 βˆ’ 1 ) π‘ž ( 𝑙 ) 𝐢 3 , ( 3 . 3 7 ) m i n ( π‘ž ( 𝑙 ) / ( 𝛼 βˆ’ 1 ) ) 𝑑 2 ≀ 𝑒 ≀ 𝑑 2 𝑑 𝑓 ( 𝑒 ) β‰₯ 2 πœ† ( 𝛼 βˆ’ 1 ) π‘ž ( 𝑙 ) 𝐢 3 . ( 3 . 3 8 ) 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 𝑓 0 = + ∞ or 𝑓 ∞ = + ∞ , then the boundary value problem (1.2) has at least one positive solution for each πœ† ∈ ( 0 , πœ† 1 ) .

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

Proof. Define 𝑏 π‘Ÿ ( π‘Ÿ ) = 𝐢 3 m i n ( π‘ž ( 𝑙 ) / ( 𝛼 βˆ’ 1 ) ) π‘Ÿ ≀ 𝑒 ≀ π‘Ÿ . 𝑓 ( 𝑒 ) ( 3 . 3 9 ) By the continuity of 𝑓 ( 𝑒 ) , 𝑓 0 = 0 and 𝑓 ∞ = 0 , we easily see that 𝑏 ( π‘Ÿ ) ∢ ( 0 , + ∞ ) β†’ ( 0 , + ∞ ) is continuous and l i m π‘Ÿ β†’ 0 𝑏 ( π‘Ÿ ) = l i m π‘Ÿ β†’ + ∞ 𝑏 ( π‘Ÿ ) = + ∞ . ( 3 . 4 0 ) By (3.29), there exists π‘Ÿ 0 ∈ ( 0 , + ∞ ) , such that 𝑏 ξ€· π‘Ÿ 0 ξ€Έ = i n f π‘Ÿ > 0 𝑏 ( π‘Ÿ ) = πœ† 2 . ( 3 . 4 1 ) For πœ† ∈ ( πœ† 2 , + ∞ ) , there exist constants 𝑑 1 , 𝑑 2 ( 0 < 𝑑 1 < π‘Ÿ 0 < 𝑑 2 < + ∞ ) with 𝑏 ξ€· 𝑑 1 ξ€Έ ξ€· 𝑑 = 𝑏 2 ξ€Έ = πœ† . ( 3 . 4 2 ) Therefore, 𝑑 𝑓 ( 𝑒 ) β‰₯ 1 πœ† ( 𝛼 βˆ’ 1 ) π‘ž ( 𝑙 ) 𝐢 3 ξ‚Έ , f o r 𝑒 ∈ π‘ž ( 𝑙 ) 𝑑 𝛼 βˆ’ 1 1 , 𝑑 1 ξ‚Ή , 𝑑 𝑓 ( 𝑒 ) β‰₯ 2 πœ† ( 𝛼 βˆ’ 1 ) π‘ž ( 𝑙 ) 𝐢 3 ξ‚Έ , f o r 𝑒 ∈ π‘ž ( 𝑙 ) 𝑑 𝛼 βˆ’ 1 2 , 𝑑 2 ξ‚Ή . ( 3 . 4 3 )
 On the other hand, using 𝑓 0 = 0 , we know that there exists a constant 𝑐 1 ( 0 < 𝑐 1 < 𝑑 1 ) with 𝑓 ( 𝑒 ) 𝑒 ≀ 1 πœ† 𝐢 1 ξ€· , f o r 𝑒 ∈ 0 , 𝑐 1 ξ€Έ , ( 3 . 4 4 ) m a x 0 ≀ 𝑒 ≀ 𝑐 1 𝑐 𝑓 ( 𝑒 ) ≀ 1 πœ† 𝐢 1 . ( 3 . 4 5 ) In view of 𝑓 ∞ = 0 , there exists a constant 𝑐 2 ∈ ( 𝑑 2 , + ∞ ) such that 𝑓 ( 𝑒 ) 𝑒 ≀ 1 πœ† 𝐢 1 ξ€· 𝑐 , f o r 𝑒 ∈ 2 ξ€Έ . , + ∞ ( 3 . 4 6 )
Let 𝑀 = m a x 0 ≀ 𝑒 ≀ 𝑐 2 𝑓 ( 𝑒 ) , 𝑐 2 β‰₯ πœ† 𝐢 1 𝑀 . ( 3 . 4 7 ) It is easily seen that m a x 0 ≀ 𝑒 ≀ 𝑐 2 𝑐 𝑓 ( 𝑒 ) ≀ 2 πœ† 𝐢 1 . ( 3 . 4 8 ) 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 𝑓 0 = 0 or 𝑓 ∞ = 0 , then for each πœ† ∈ ( 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 𝑓 0 = + ∞ , 𝑓 ∞ = 𝑑 , or 𝑓 ∞ = + ∞ , 𝑓 0 = 𝑑 , then for any πœ† ∈ ( 0 , ( 𝑑 𝐢 1 ) βˆ’ 1 ) , the boundary value problem (1.2) has at least one positive solution.

Corollary 3.10. Assume ( 𝐻 ) holds. If 𝑓 0 = 0 , 𝑓 ∞ = 𝑑 , or if 𝑓 ∞ = 0 , 𝑓 0 = 𝑑 , then for any πœ† ∈ ( ( π‘ž ( 𝑙 ) 𝑑 𝐢 2 ) 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 𝐹 0 < + ∞ 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 𝐹 0 < + ∞ and 𝐹 ∞ < + ∞ , there exist positive numbers π‘š 1 , π‘š 2 , π‘Ÿ 1 , and π‘Ÿ 2 , such that π‘Ÿ 1 < π‘Ÿ 2 and 𝑓 ( 𝑒 ) ≀ π‘š 1 ξ€Ί 𝑒 , f o r 𝑒 ∈ 0 , π‘Ÿ 1 ξ€» , 𝑓 ( 𝑒 ) ≀ π‘š 2 ξ€Ί π‘Ÿ 𝑒 , f o r 𝑒 ∈ 2 ξ€Έ . , + ∞ ( 4 . 1 ) Let π‘š = m a x { π‘š 1 , π‘š 2 , m a x π‘Ÿ 1 ≀ 𝑒 ≀ π‘Ÿ 2 { 𝑓 ( 𝑒 ) / 𝑒 } } . Then we have [ 𝑓 ( 𝑒 ) ≀ π‘š 𝑒 , f o r 𝑒 ∈ 0 , + ∞ ) . ( 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‘š 𝐢 1 ) βˆ’ 1 . Since 𝐴 πœ† 𝑣 ( 𝑑 ) = 𝑣 ( 𝑑 ) for 𝑑 ∈ [ 0 , 1 ] , β€– β€– 𝐴 β€– 𝑣 β€– = πœ† 𝑣 β€– β€– ≀ πœ† ξ€œ Ξ“ ( 𝛼 ) 1 0 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑣 ( 𝑠 ) ) 𝑑 𝑠 ≀ π‘š πœ† ξ€œ Ξ“ ( 𝛼 ) β€– 𝑣 β€– 1 0 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑑 𝑠 < β€– 𝑣 β€– , ( 4 . 3 ) which is a contradiction. Therefore, (1.2) has no positive solution. The proof is complete.

Theorem 4.2. Assume ( 𝐻 ) holds. If 𝑓 0 > 0 and 𝑓 ∞ > 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 𝑓 0 > 0 and 𝑓 ∞ > 0 , we know that there exist positive numbers 𝑛 1 , 𝑛 2 , π‘Ÿ 1 , and π‘Ÿ 2 , such that π‘Ÿ 1 < π‘Ÿ 2 and 𝑓 ( 𝑒 ) β‰₯ 𝑛 1 ξ€Ί 𝑒 , f o r 𝑒 ∈ 0 , π‘Ÿ 1 ξ€» , 𝑓 ( 𝑒 ) β‰₯ 𝑛 2 ξ€Ί π‘Ÿ 𝑒 , f o r 𝑒 ∈ 2 ξ€Έ . , + ∞ ( 4 . 4 ) Let 𝑛 = m i n { 𝑛 1 , 𝑛 2 , m i n π‘Ÿ 1 ≀ 𝑒 ≀ π‘Ÿ 2 { 𝑓 ( 𝑒 ) / 𝑒 } } > 0 . Then we get [ 𝑓 ( 𝑒 ) β‰₯ 𝑛 𝑒 , f o r 𝑒 ∈ 0 , + ∞ ) . ( 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 ) 𝑛 𝐢 2 ) βˆ’ 1 . Since 𝐴 πœ† 𝑣 ( 𝑑 ) = 𝑣 ( 𝑑 ) for 𝑑 ∈ [ 0 , 1 ] , β€– β€– 𝐴 β€– 𝑣 β€– = πœ† 𝑣 β€– β€– β‰₯ πœ† ξ€œ Ξ“ ( 𝛼 ) 1 0 π‘ž ( 𝑙 ) π‘˜ ( 𝑠 ) 𝑓 ( 𝑣 ( 𝑠 ) ) 𝑑 𝑠 > β€– 𝑣 β€– , ( 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 𝐷 0 5 / 2 + 𝑒 ( 𝑑 ) + πœ† 𝑒 π‘Ž = 0 , 0 < 𝑑 < 1 , π‘Ž > 1 , 𝑒 ( 0 ) = 𝑒 ( 1 ) = 𝑒 ξ…ž ( 0 ) = 0 . ( 5 . 1 )
Since 𝛼 = 5 / 2 , we have 𝐢 1 = 1 ξ€œ Ξ“ ( 𝛼 ) 1 0 1 ( 𝛼 βˆ’ 1 ) π‘˜ ( 𝑠 ) 𝑑 𝑠 = ξ€œ Ξ“ ( 5 / 2 ) 1 0 3 2 𝑠 ( 1 βˆ’ 𝑠 ) 3 / 2 𝐢 𝑑 𝑠 = 0 . 1 2 9 0 , 2 = 1 ξ€œ Ξ“ ( 𝛼 ) 1 0 1 π‘ž 1 ( 𝛼 βˆ’ 1 ) ( 𝑠 ) π‘˜ ( 𝑠 ) 𝑑 𝑠 = ξ€œ Ξ“ ( 5 / 2 ) 1 0 2 3 𝑠 5 / 2 ( 1 βˆ’ 𝑠 ) 5 / 2 𝑑 𝑠 = 0 . 0 0 7 7 . ( 5 . 2 ) Let 𝑓 ( 𝑒 ) = 𝑒 π‘Ž , π‘Ž > 1 . Then we have 𝐹 0 = 0 , 𝑓 ∞ = + ∞ . Choose 𝑙 = 1 / 2 . Then √ π‘ž ( 1 / 2 ) = 2 / 8 = 0 . 1 7 6 8 . So π‘ž ( 𝑙 ) 𝐢 2 𝑓 ∞ > 𝐹 0 𝐢 1 holds. Thus, by Theorem 3.2, the boundary value problem (5.1) has a positive solution for each πœ† ∈ ( 0 , + ∞ ) .

Example 5.2. Discuss the boundary value problem 𝐷 0 5 / 2 + 𝑒 ( 𝑑 ) + πœ† 𝑒 𝑏 = 0 , 0 < 𝑑 < 1 , 0 < 𝑏 < 1 , 𝑒 ( 0 ) = 𝑒 ( 1 ) = 𝑒 ξ…ž ( 0 ) = 0 . ( 5 . 3 )
Since 𝛼 = 5 / 2 , we have 𝐢 1 = 0 . 1 2 9 0 and 𝐢 2 = 0 . 0 0 7 7 . Let 𝑓 ( 𝑒 ) = 𝑒 𝑏 , 0 < 𝑏 < 1 . Then we have 𝐹 ∞ = 0 , 𝑓 0 = + ∞ . Choose 𝑙 = 1 / 2 . Then √ π‘ž ( 1 / 2 ) = 2 / 8 = 0 . 1 7 6 8 . So π‘ž ( 𝑙 ) 𝐢 2 𝑓 0 > 𝐹 ∞ 𝐢 1 holds. Thus, by Theorem 3.3, the boundary value problem (5.3) has a positive solution for each πœ† ∈ ( 0 , + ∞ ) .

Example 5.3. Consider the boundary value problem 𝐷 0 5 / 2 + 𝑒 ξ€· ( 𝑑 ) + πœ† 2 0 0 𝑒 2 ξ€Έ + 𝑒 ( 2 + s i n 𝑒 ) 𝑒 + 1 = 0 , 0 < 𝑑 < 1 , π‘Ž > 1 , 𝑒 ( 0 ) = 𝑒 ( 1 ) = 𝑒 ξ…ž ( 0 ) = 0 . ( 5 . 4 )
Since 𝛼 = 5 / 2 , we have 𝐢 1 = 0 . 1 2 9 and 𝐢 2 = 0 . 0 0 7 7 . Let 𝑓 ( 𝑒 ) = ( 2 0 0 𝑒 2 + 𝑒 ) ( 2 + s i n 𝑒 ) / ( 𝑒 + 1 ) . Then we have 𝐹 0 = 𝑓 0 = 2 , 𝐹 ∞ = 6 0 0 , 𝑓 ∞ = 2 0 0 , and 2 𝑒 < 𝑓 ( 𝑒 ) < 6 0 0 𝑒 .
(i)Choose 𝑙 = 1 / 2 . Then √ π‘ž ( 1 / 2 ) = 2 / 8 = 0 . 1 7 6 8 . So π‘ž ( 𝑙 ) 𝐢 2 𝑓 ∞ > 𝐹 0 𝐢 1 holds. Thus, by Theorem 3.2, the boundary value problem (5.4) has a positive solution for each πœ† ∈ ( 3 . 6 9 3 7 , 3 . 8 7 5 9 ) .(ii)By Theorem 4.1, the boundary value problem (5.4) has no positive solution for all πœ† ∈ ( 0 , 0 . 0 1 2 9 ) .(iii)By Theorem 4.2, the boundary value problem (5.4) has no positive solution for all πœ† ∈ ( 3 6 9 . 3 6 9 , + ∞ ) .

Example 5.4. Consider the boundary value problem 𝐷 0 5 / 2 + 𝑒 ξ€· 𝑒 ( 𝑑 ) + πœ† 2 ξ€Έ + 𝑒 ( 2 + s i n 𝑒 ) 1 5 0 𝑒 + 1 = 0 , 0 < 𝑑 < 1 , π‘Ž > 1 , 𝑒 ( 0 ) = 𝑒 ( 1 ) = 𝑒 ξ…ž ( 0 ) = 0 . ( 5 . 5 )
Since 𝛼 = 5 / 2 , we have 𝐢 1 = 0 . 1 2 9 and 𝐢 2 = 0 . 0 0 7 7 . Let 𝑓 ( 𝑒 ) = ( 𝑒 2 + 𝑒 ) ( 2 + s i n 𝑒 ) / ( 1 5 0 𝑒 + 1 ) . Then we have 𝐹 0 = 𝑓 0 = 2 , 𝐹 ∞ = 1 / 5 0 , 𝑓 ∞ = 1 / 1 5 0 , and 𝑒 / 1 5 0 < 𝑓 ( 𝑒 ) < 2 𝑒 .
(i)Choose 𝑙 = 1 / 2 . Then √ π‘ž ( 1 / 2 ) = 2 / 8 = 0 . 1 7 6 8 . So π‘ž ( 𝑙 ) 𝐢 2 𝑓 0 > 𝐹 ∞ 𝐢 1 holds. Thus, by Theorem 3.3, the boundary value problem (5.5) has a positive solution for each πœ† ∈ ( 3 6 9 . 3 6 9 , 3 8 7 . 5 9 6 8 ) .(ii)By Theorem 4.1, the boundary value problem (5.5) has no positive solution for all πœ† ∈ ( 0 , 3 . 8 7 5 9 ) .(iii)By Theorem 4.2, the boundary value problem (5.5) has no positive solution for all πœ† ∈ ( 1 1 0 8 1 0 . 6 9 1 1 , + ∞ ) .

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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