Abstract and Applied Analysis

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Nonlinear Functional Analysis of Boundary Value Problems: Novel Theory, Methods, and Applications

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Volume 2012 |Article ID 512127 | https://doi.org/10.1155/2012/512127

Xinguang Zhang, Lishan Liu, Benchawan Wiwatanapataphee, Yonghong Wu, "Positive Solutions of Eigenvalue Problems for a Class of Fractional Differential Equations with Derivatives", Abstract and Applied Analysis, vol. 2012, Article ID 512127, 16 pages, 2012. https://doi.org/10.1155/2012/512127

Positive Solutions of Eigenvalue Problems for a Class of Fractional Differential Equations with Derivatives

Academic Editor: Shaoyong Lai
Received02 Jan 2012
Accepted15 Mar 2012
Published20 May 2012

Abstract

By establishing a maximal principle and constructing upper and lower solutions, the existence of positive solutions for the eigenvalue problem of a class of fractional differential equations is discussed. Some sufficient conditions for the existence of positive solutions are established.

1. Introduction

In this paper, we discuss the existence of positive solutions for the following eigenvalue problem of a class fractional differential equation with derivativesβˆ’π““π­π›Όξ€·π‘₯(𝑑)=πœ†π‘“π‘‘,π‘₯(𝑑),𝓓𝐭𝛽𝓓π‘₯(𝑑),π‘‘βˆˆ(0,1),𝐭𝛽π‘₯(0)=0,𝓓𝐭𝛾π‘₯(1)=π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύπ‘₯ξ€·πœ‰π‘—ξ€Έ,(1.1) where πœ† is a parameter, 1<𝛼≀2, π›Όβˆ’π›½>1, 0<𝛽≀𝛾<1, 0<πœ‰1<πœ‰2<β‹―<πœ‰π‘βˆ’2<1, π‘Žπ‘—βˆˆ[0,+∞) with βˆ‘π‘=π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1<1, and 𝓓𝐭 is the standard Riemann-Liouville derivative. π‘“βˆΆ(0,1)Γ—(0,+∞)Γ—(0,+∞)β†’[0,+∞) is continuous, and 𝑓(𝑑,𝑒,𝑣) may be singular at 𝑒=0,𝑣=0, and 𝑑=0,1.

As fractional order derivatives and integrals have been widely used in mathematics, analytical chemistry, neuron modeling, and biological sciences [1–6], fractional differential equations have attracted great research interest in recent years [7–17]. Recently, ur Rehman and Khan [8] investigated the fractional order multipoint boundary value problem:𝓓𝐭𝛼𝑦(𝑑)=𝑓𝑑,𝑦(𝑑),𝓓𝐭𝛽𝑦(𝑑),π‘‘βˆˆ(0,1),𝑦(0)=0,𝓓𝐭𝛽𝑦(1)βˆ’π‘šβˆ’2𝑖=1πœπ‘–π““π­π›½π‘¦ξ€·πœ‰π‘–ξ€Έ=𝑦0,(1.2) where 1<𝛼≀2,  0<𝛽<1, 0<πœ‰π‘–<1, β€‰πœπ‘–βˆˆ[0,+∞) with βˆ‘π‘šβˆ’2𝑖=1πœπ‘–πœ‰π‘–π›Όβˆ’π›½βˆ’1<1. The Schauder fixed point theorem and the contraction mapping principle are used to establish the existence and uniqueness of nontrivial solutions for the BVP (1.2) provided that the nonlinear function π‘“βˆΆ[0,1]×ℝ×ℝ is continuous and satisfies certain growth conditions. But up to now, multipoint boundary value problems for fractional differential equations like the BVP (1.1) have seldom been considered when 𝑓(𝑑,𝑒,𝑣) has singularity at 𝑑=0 and (or) 1 and also at 𝑒=0,𝑣=0. We will discuss the problem in this paper.

The rest of the paper is organized as follows. In Section 2, we give some definitions and several lemmas. Suitable upper and lower solutions of the modified problems for the BVP (1.1) and some sufficient conditions for the existence of positive solutions are established in Section 3.

2. Preliminaries and Lemmas

For the convenience of the reader, we present here some definitions about fractional calculus.

Definition 2.1 (See [1, 6]). Let 𝛼>0 with π›Όβˆˆβ„. Suppose that π‘₯∢[π‘Ž,∞)→ℝ. Then the 𝛼th Riemann-Liouville fractional integral is defined by 𝐼𝛼1π‘₯(𝑑)=ξ€œΞ“(𝛼)π‘‘π‘Ž(π‘‘βˆ’π‘ )π›Όβˆ’1π‘₯(𝑠)𝑑𝑠(2.1) whenever the right-hand side is defined. Similarly, for π›Όβˆˆβ„ with 𝛼>0, we define the 𝛼th Riemann-Liouville fractional derivative by 𝓓𝛼1π‘₯(𝑑)=𝑑Γ(π‘›βˆ’π›Ό)𝑑𝑑(𝑛)ξ€œπ‘‘π‘Ž(π‘‘βˆ’π‘ )π‘›βˆ’π›Όβˆ’1π‘₯(𝑠)𝑑𝑠,(2.2)where π‘›βˆˆβ„• is the unique positive integer satisfying π‘›βˆ’1≀𝛼<𝑛 and 𝑑>π‘Ž.

Remark 2.2. If π‘₯,π‘¦βˆΆ(0,+∞)→ℝ with order 𝛼>0, then 𝓓𝐭𝛼(π‘₯(𝑑)+𝑦(𝑑))=𝓓𝐭𝛼π‘₯(𝑑)+𝓓𝐭𝛼𝑦(𝑑).(2.3)

Lemma 2.3 (See [6]). One has the following.

(1) If π‘₯∈𝐿1(0,1),𝜈>𝜎>0, then 𝐼𝜈𝐼𝜎π‘₯(𝑑)=𝐼𝜈+𝜎π‘₯(𝑑),π““π­πœŽπΌπœˆπ‘₯(𝑑)=πΌπœˆβˆ’πœŽπ‘₯(𝑑),π““π­πœŽπΌπœŽπ‘₯(𝑑)=π‘₯(𝑑).(2.4)

(2) If 𝜈>0,𝜎>0, then π““π­πœˆπ‘‘πœŽβˆ’1=Ξ“(𝜎)𝑑Γ(πœŽβˆ’πœˆ)πœŽβˆ’πœˆβˆ’1.(2.5)

Lemma 2.4 (See [6]). Let 𝛼>0. Assume that π‘₯∈𝐢(0,1)∩𝐿1(0,1). Then𝐼𝛼𝓓𝐭𝛼π‘₯(𝑑)=π‘₯(𝑑)+𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2+β‹―+π‘π‘›π‘‘π›Όβˆ’π‘›,(2.6) where π‘π‘–βˆˆβ„(𝑖=1,2,…,𝑛), and 𝑛 is the smallest integer greater than or equal to 𝛼.

Letπ‘˜1⎧βŽͺ⎨βŽͺβŽ©π‘‘(𝑑,𝑠)=π›Όβˆ’π›½βˆ’1(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1βˆ’(π‘‘βˆ’π‘ )π›Όβˆ’π›½βˆ’1Γ𝑑(π›Όβˆ’π›½),0≀𝑠≀𝑑≀1,π›Όβˆ’π›½βˆ’1(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1π‘˜Ξ“(π›Όβˆ’π›½),0≀𝑑≀𝑠≀1,2⎧βŽͺ⎨βŽͺ⎩(𝑑,𝑠)=(𝑑(1βˆ’π‘ ))π›Όβˆ’π›Ύβˆ’1βˆ’(π‘‘βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1(Ξ“(π›Όβˆ’π›½),0≀𝑠≀𝑑≀1,𝑑(1βˆ’π‘ ))π›Όβˆ’π›Ύβˆ’1Ξ“(π›Όβˆ’π›½),0≀𝑑≀𝑠≀1,(2.7)

and for 𝑑,π‘ βˆˆ[0,1], we haveπ‘˜π‘–(𝑑,𝑠)≀(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1Ξ“(π›Όβˆ’π›½),𝑖=1,2.(2.8)

Lemma 2.5. Let β„ŽβˆˆπΆ(0,1); If 1<π›Όβˆ’π›½β‰€2, then the unique solution of the linear problem βˆ’π““π­π›Όβˆ’π›½π‘¦(𝑑)=β„Ž(𝑑),π‘‘βˆˆ(0,1),𝑦(0)=0,π““π­π›Ύβˆ’π›½π‘¦(1)=π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύβˆ’π›½π‘¦ξ€·πœ‰π‘—ξ€Έ(2.9) is given by ξ€œπ‘¦(𝑑)=10𝐾(𝑑,𝑠)β„Ž(𝑠)𝑑𝑠,(2.10) where 𝐾(𝑑,𝑠)=π‘˜1𝑑(𝑑,𝑠)+π›Όβˆ’π›½βˆ’1βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1π‘βˆ’2𝑗=1π‘Žπ‘—π‘˜2ξ€·πœ‰π‘—ξ€Έ,𝑠(2.11) is the Green function of the boundary value problem (2.9).

Proof. Applying Lemma 2.4, we reduce (2.9) to an equivalent equation: 𝑦(𝑑)=βˆ’πΌπ›Όβˆ’π›½β„Ž(𝑑)+𝑐1π‘‘π›Όβˆ’π›½βˆ’1+𝑐2π‘‘π›Όβˆ’π›½βˆ’2,𝑐1,𝑐2βˆˆβ„.(2.12) From (2.12) and noting that 𝑦(0)=0, we have 𝑐2=0. Consequently the general solution of (2.9) is 𝑦(𝑑)=βˆ’πΌπ›Όβˆ’π›½β„Ž(𝑑)+𝑐1π‘‘π›Όβˆ’π›½βˆ’1.(2.13) Using (2.13) and Lemma 2.3, we have π““π­π›Ύβˆ’π›½π‘¦(𝑑)=βˆ’π““π­π›Ύβˆ’π›½πΌπ›Όβˆ’π›½β„Ž(𝑑)+𝑐1π““π­π›Ύβˆ’π›½π‘‘π›Όβˆ’π›½βˆ’1=βˆ’πΌπ›Όβˆ’π›Ύβ„Ž(𝑑)+𝑐1Ξ“(π›Όβˆ’π›½)𝑑Γ(π›Όβˆ’π›Ύ)π›Όβˆ’π›Ύβˆ’1ξ€œ=βˆ’π‘‘0(π‘‘βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1Ξ“(π›Όβˆ’π›Ύ)β„Ž(𝑠)𝑑𝑠+𝑐1Ξ“(π›Όβˆ’π›½)𝑑Γ(π›Όβˆ’π›Ύ)π›Όβˆ’π›Ύβˆ’1.(2.14)Thus, π““π­π›Ύβˆ’π›½ξ€œπ‘¦(1)=βˆ’10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1Ξ“(π›Όβˆ’π›Ύ)β„Ž(𝑠)𝑑𝑠+𝑐1Ξ“(π›Όβˆ’π›½)Ξ“(π›Όβˆ’π›Ύ),(2.15) and for 𝑗=1,2,…,π‘βˆ’2, π““π­π›Ύβˆ’π›½π‘€ξ€·πœ‰π‘—ξ€Έξ€œ=βˆ’πœ‰π‘—0ξ€·πœ‰π‘—ξ€Έβˆ’π‘ π›Όβˆ’π›Ύβˆ’1β„ŽΞ“(π›Όβˆ’π›Ύ)(𝑠)𝑑𝑠+𝑐1Ξ“(π›Όβˆ’π›½)πœ‰Ξ“(π›Όβˆ’π›Ύ)π‘—π›Όβˆ’π›Ύβˆ’1.(2.16) Using π““π­π›Ύβˆ’π›½βˆ‘π‘¦(1)=π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύβˆ’π›½π‘¦(πœ‰π‘—), (2.15), and (2.16), we obtain 𝑐1=∫10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1βˆ‘β„Ž(𝑠)π‘‘π‘ βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—βˆ«πœ‰π‘—0ξ€·πœ‰π‘—ξ€Έβˆ’π‘ π›Όβˆ’π›Ύβˆ’1β„Ž(𝑠)π‘‘π‘ ξ‚€βˆ‘Ξ“(π›Όβˆ’π›½)1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1.(2.17) So the unique solution of the problem (2.9) is ξ€œπ‘¦(𝑑)=βˆ’π‘‘0(π‘‘βˆ’π‘ )π›Όβˆ’π›½βˆ’1Γ𝑑(π›Όβˆ’π›½)β„Ž(𝑠)𝑑𝑠+π›Όβˆ’π›½βˆ’1βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1Γ—ξƒ―ξ€œ10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1Ξ“(π›Όβˆ’π›½)β„Ž(𝑠)π‘‘π‘ βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—ξ€œπœ‰π‘—0ξ€·πœ‰π‘—ξ€Έβˆ’π‘ π›Όβˆ’π›Ύβˆ’1ξƒ°ξ€œΞ“(π›Όβˆ’π›½)β„Ž(𝑠)𝑑𝑠=βˆ’π‘‘0(π‘‘βˆ’π‘ )π›Όβˆ’π›½βˆ’1ξ€œΞ“(π›Όβˆ’π›½)β„Ž(𝑠)𝑑𝑠+10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1π‘‘π›Όβˆ’π›½βˆ’1+𝑑Γ(π›Όβˆ’π›½)β„Ž(𝑠)π‘‘π‘ π›Όβˆ’π›½βˆ’1βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1π‘βˆ’2𝑗=1π‘Žπ‘—ξ€œ10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1πœ‰π‘—π›Όβˆ’π›Ύβˆ’1βˆ’π‘‘Ξ“(π›Όβˆ’π›½)β„Ž(𝑠)π‘‘π‘ π›Όβˆ’π›½βˆ’1βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1π‘βˆ’2𝑗=1π‘Žπ‘—ξ€œπœ‰π‘—0ξ€·πœ‰π‘—ξ€Έβˆ’π‘ π›Όβˆ’π›Ύβˆ’1β„Ž=ξ€œΞ“(π›Όβˆ’π›½)(𝑠)𝑑𝑠10ξƒ©π‘˜1𝑑(𝑑,𝑠)+π›Όβˆ’π›½βˆ’1βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1π‘βˆ’2𝑗=1π‘Žπ‘—π‘˜2ξ€·πœ‰π‘—ξ€Έξƒͺ=ξ€œ,π‘ β„Ž(𝑠)𝑑𝑠10𝐾(𝑑,𝑠)β„Ž(𝑠)𝑑𝑠.(2.18) The proof is completed.

Lemma 2.6. The function 𝐾(𝑑,𝑠) has the following properties.

(1)𝐾(𝑑,𝑠)>0,for𝑑,π‘ βˆˆ(0,1)(2)π‘‘π›Όβˆ’π›½βˆ’1𝔐(𝑠)≀𝐾(𝑑,𝑠)≀𝑀(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1,for𝑑,π‘ βˆˆ[0,1],

whereβˆ‘π”(𝑠)=π‘βˆ’2𝑗=1π‘Žπ‘—π‘˜2ξ€·πœ‰π‘—ξ€Έ,π‘ βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1βˆ‘,𝑀=1+π‘βˆ’2𝑗=1π‘Žπ‘—ξ‚€1βˆ’πœ‰π‘—π›Όβˆ’π›Ύβˆ’1ξ‚ξ‚€βˆ‘Ξ“(π›Όβˆ’π›½)1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1.(2.19)

Proof. It is obvious that (1) holds.
From (2.11), we obtain𝑑𝐾(𝑑,𝑠)β‰₯π›Όβˆ’π›½βˆ’1βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1π‘βˆ’2𝑗=1π‘Žπ‘—π‘˜2ξ€·πœ‰π‘—ξ€Έ,𝑠=π‘‘π›Όβˆ’π›½βˆ’1𝔐(𝑠).(2.20)From (2.8), we have 𝐾(𝑑,𝑠)=π‘˜1𝑑(𝑑,𝑠)+π›Όβˆ’π›½βˆ’1βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1π‘βˆ’2𝑗=1π‘Žπ‘—π‘˜2ξ€·πœ‰π‘—ξ€Έβ‰€,𝑠(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1+βˆ‘Ξ“(π›Όβˆ’π›½)π‘βˆ’2𝑗=1π‘Žπ‘—βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1β‰€ξƒ©βˆ‘Ξ“(π›Όβˆ’π›½)1+π‘βˆ’2𝑗=1π‘Žπ‘—βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1ξƒͺ(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1.Ξ“(π›Όβˆ’π›½)(2.21) The proof is completed.

Consider the modified problem of the BVP (1.1):βˆ’π““π­π›Όβˆ’π›½ξ€·π‘¦(𝑑)=πœ†π‘“π‘‘,𝐼𝛽,𝑦(𝑑),𝑦(𝑑)𝑦(0)=0,π““π­π›Ύβˆ’π›½π‘¦(1)=π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύβˆ’π›½π‘¦ξ€·πœ‰π‘—ξ€Έ.(2.22)

Lemma 2.7. Let π‘₯(𝑑)=𝐼𝛽𝑦(𝑑)and𝑦(𝑑)∈𝐢[0,1]; then problem (1.1) is turned into (2.22). Moreover, if π‘¦βˆˆπΆ([0,1],[0,+∞)) is a solution of problem (2.22), then the function π‘₯(𝑑)=𝐼𝛽𝑦(𝑑) is a positive solution of the problem (1.1).

Proof. Substituting π‘₯(𝑑)=𝐼𝛽𝑦(𝑑) into (1.1) and using Definition 2.1 and Lemmas 2.3 and 2.4, we obtain 𝓓𝐭𝛼𝑑π‘₯(𝑑)=π‘›π‘‘π‘‘π‘›πΌπ‘›βˆ’π›Όπ‘‘π‘₯(𝑑)=π‘›π‘‘π‘‘π‘›πΌπ‘›βˆ’π›ΌπΌπ›½=𝑑𝑦(𝑑)π‘›π‘‘π‘‘π‘›πΌπ‘›βˆ’π›Ό+𝛽𝑦(𝑑)=π““π­π›Όβˆ’π›½π““π‘¦(𝑑),𝐭𝛽π‘₯(𝑑)=𝓓𝐭𝛽𝐼𝛽𝑦(𝑑)=𝑦(𝑑).(2.23) Consequently, 𝓓𝐭𝛽π‘₯(0)=𝑦(0)=0. It follows from 𝓓𝐭𝛾π‘₯(𝑑)=𝑑𝑛/𝑑tπ‘›πΌπ‘›βˆ’π›Ύπ‘₯(𝑑)=(𝑑𝑛/𝑑𝑑𝑛)πΌπ‘›βˆ’π›ΎπΌπ›½π‘¦(𝑑)=(𝑑𝑛/𝑑𝑑𝑛)πΌπ‘›βˆ’π›Ύ+𝛽𝑦(𝑑)=π““π­π›Ύβˆ’π›½π‘¦(𝑑) that π““π­π›Ύβˆ’π›½βˆ‘π‘¦(1)=π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύβˆ’π›½π‘¦(πœ‰π‘—). Using π‘₯(𝑑)=𝐼𝛽𝑦(𝑑), π‘¦βˆˆπΆ[0,1], we transform (1.1) into (2.22).
Now, let π‘¦βˆˆπΆ([0,1],[0,+∞)) be a solution for problem (2.22). Using Lemma 2.3, (2.22), and (2.23), one hasβˆ’π““π­π›Όπ‘‘π‘₯(𝑑)=βˆ’π‘›π‘‘π‘‘π‘›πΌπ‘›βˆ’π›Όπ‘‘π‘₯(𝑑)=βˆ’π‘›π‘‘π‘‘π‘›πΌπ‘›βˆ’π›ΌπΌπ›½π‘‘π‘¦(𝑑)=βˆ’π‘›π‘‘π‘‘π‘›πΌπ‘›βˆ’π›Ό+𝛽𝑦(𝑑)=βˆ’π““π­π›Όβˆ’π›½ξ€·π‘¦(𝑑)=πœ†π‘“π‘‘,𝐼𝛽𝑦(𝑑),𝑦(𝑑)=πœ†π‘“π‘‘,π‘₯(𝑑),𝓓𝐭𝛽π‘₯(𝑑),0<𝑑<1.(2.24) Noting 𝓓𝐭𝛽π‘₯(𝑑)=𝓓𝐭𝛽𝐼𝛽𝑦(𝑑)=𝑦(𝑑),𝓓𝐭𝛾π‘₯(𝑑)=π““π­π›Ύβˆ’π›½π‘¦(𝑑),(2.25) we have 𝓓𝐭𝛽π‘₯(0)=0,𝓓𝐭𝛾π‘₯(1)=π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύπ‘₯ξ€·πœ‰π‘—ξ€Έ.(2.26)It follows from the monotonicity and property of 𝐼𝛽 that 𝐼𝛽[],[π‘¦βˆˆπΆ(0,10,+∞)).(2.27) Consequently, π‘₯(𝑑)=𝐼𝛽𝑦(𝑑) is a positive solution of the problem (1.1).

Definition 2.8. A continuous function πœ“(𝑑) is called a lower solution of the BVP (2.22), if it satisfies βˆ’π““π­π›Όβˆ’π›½ξ€·πœ“(𝑑)β‰€πœ†π‘“π‘‘,𝐼𝛽,πœ“(𝑑),πœ“(𝑑)πœ“(0)β‰₯0,π““π­π›Ύβˆ’π›½πœ“(1)β‰₯π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύβˆ’π›½πœ“ξ€·πœ‰π‘—ξ€Έ.(2.28)

Definition 2.9. A continuous functionβ€‰β€‰πœ™(𝑑)  is called an upper solution of the BVP (2.22), if it satisfies βˆ’π““π­π›Όβˆ’π›½ξ€·πœ™(𝑑)β‰₯πœ†π‘“π‘‘,𝐼𝛽,πœ™(𝑑),πœ™(𝑑)πœ™(0)≀0,π““π­π›Ύβˆ’π›½πœ™(1)β‰€π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύβˆ’π›½πœ™ξ€·πœ‰π‘—ξ€Έ.(2.29)

By Lemmas 2.5 and 2.6, we have the maximal principle.

Lemma 2.10. If 1<π›Όβˆ’π›½β‰€2 and π‘¦βˆˆπΆ([0,1],𝑅) satisfies 𝑦(0)=0,π““π­π›Ύβˆ’π›½π‘¦(1)=π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύβˆ’π›½π‘¦ξ€·πœ‰π‘—ξ€Έ,(2.30)and βˆ’π““π­π›Όβˆ’π›½π‘¦(𝑑)β‰₯0 for any π‘‘βˆˆ(0,1), then 𝑦(𝑑)β‰₯0, for π‘‘βˆˆ[0,1].

Set𝒒(𝑑)=π‘‘π›Όβˆ’π›½βˆ’1,β„’(𝑑)=𝐼𝛽1𝒒(𝑑)=ξ€œΞ“(𝛽)𝑑0(π‘‘βˆ’π‘ )π›½βˆ’1π‘ π›Όβˆ’π›½βˆ’1𝑑𝑠=Ξ“(π›Όβˆ’π›½)𝑑Γ(𝛼)π›Όβˆ’1.(2.31)

To end this section, we present here two assumptions to be used throughout the rest of the paper.(B1)π‘“βˆˆπΆ((0,1)Γ—(0,∞)Γ—(0,∞),[0,+∞)) is decreasing in 𝑒 and 𝑣, and for any (𝑒,𝑣)∈(0,∞)Γ—(0,∞),limπœŽβ†’+βˆžπœŽπ‘“(𝑑,πœŽπ‘’,πœŽπ‘£)=+∞(2.32)

uniformly on π‘‘βˆˆ(0,1).(B2) For any πœ‡,𝜈>0, 𝑓(𝑑,πœ‡,𝜈)β‰’0, andξ€œ10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1𝑓(𝑠,πœ‡β„’(𝑠),πœ‡π’’(𝑠))𝑑𝑠<+∞.(2.33)

3. Main Results

The main result is summarized in the following theorem.

Theorem 3.1. Provided that (B1)   and (B2) hold, then there is a constant πœ†βˆ—>0 such that for any πœ†βˆˆ(πœ†βˆ—,+∞), the problem (1.1) has at least one positive solution π‘₯(𝑑), which satisfies π‘₯(𝑑)β‰₯β„’(𝑑), π‘‘βˆˆ[0,1].

Proof. Let 𝐸=𝐢[0,1]; we denote a set 𝑃 and an operator π‘‡πœ† in 𝐸 as follows: 𝑃=π‘¦βˆˆπΈβˆΆthereexistspositivenumber𝑙𝑦suchthat𝑦(𝑑)β‰₯𝑙𝑦[]ξ€Ύ,𝑇𝒒(𝑑),π‘‘βˆˆ0,1(3.1)πœ†π‘¦ξ€Έξ€œ(𝑑)=πœ†10𝐾(𝑑,𝑠)𝑓𝑠,𝐼𝛽𝑦(𝑠),𝑦(𝑠)𝑑𝑠,foranyπ‘¦βˆˆπ‘ƒ.(3.2) Clearly, 𝑃 is a nonempty set since 𝒒(𝑑)βˆˆπ‘ƒ. We claim that π‘‡πœ† is well defined and π‘‡πœ†(𝑃)βŠ‚π‘ƒ.
In fact, for any πœŒβˆˆπ‘ƒ, by the definition of 𝑃, there exists one positive number π‘™πœŒ such that 𝜌(𝑑)β‰₯π‘™πœŒπ’’(𝑑) for any π‘‘βˆˆ[0,1]. It follows from Lemma 2.6 and (B2) thatξ€·π‘‡πœ†πœŒξ€Έξ€œ(𝑑)=πœ†10𝐾(𝑑,𝑠)𝑓𝑠,πΌπ›½ξ€Έξ€œπœŒ(𝑠),𝜌(𝑠)π‘‘π‘ β‰€πœ†π‘€10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1𝑓𝑠,πΌπ›½ξ€Έξ€œπœŒ(𝑠),𝜌(𝑠)π‘‘π‘ β‰€πœ†π‘€10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1𝑓𝑠,π‘™πœŒβ„’(𝑠),π‘™πœŒπ’’ξ€Έ(𝑠)𝑑𝑠<+∞.(3.3)
Setting 𝐡=maxπ‘‘βˆˆ[0,1]𝜌(𝑑)>0, from (B2), we have 𝑓(𝑑,𝐡/Ξ“(𝛽+1),𝐡)β‰’0. By the continuity of 𝑓(𝑑,𝑒,𝑣) on (0,1)Γ—(0,∞)Γ—(0,∞), we have ∫10𝔐(𝑠)𝑓(𝑠,𝐡/Ξ“(𝛽+1),𝐡)𝑑𝑠>0. On the other hand,𝐼𝛽1𝐡=ξ€œΞ“(𝛽)𝑑0(π‘‘βˆ’π‘ )π›½βˆ’1𝐡𝑑𝑠=𝐡𝑑𝛽≀𝐡𝛽Γ(𝛽),βˆ‘Ξ“(𝛽+1)𝔐(𝑠)=π‘βˆ’2𝑗=1π‘Žπ‘—π‘˜2ξ€·πœ‰π‘—ξ€Έ,π‘ βˆ‘1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1β‰€βˆ‘π‘βˆ’2𝑗=1π‘Žπ‘—(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1ξ‚€βˆ‘Ξ“(π›Όβˆ’π›½)1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1.(3.4)From (3.3), one has ξ€œ0<10𝐡𝔐(𝑠)𝑓𝑠,ξ‚Άξ€œΞ“(𝛽+1),𝐡𝑑𝑠≀10𝔐(𝑠)𝑓𝑠,πΌπ›½ξ€Έξ€œπ΅,𝐡𝑑𝑠≀10𝔐(𝑠)𝑓𝑠,πΌπ›½ξ€Έβ‰€βˆ‘πœŒ(𝑠),𝜌(𝑠)π‘‘π‘ π‘βˆ’2𝑗=1π‘Žπ‘—ξ‚€βˆ‘Ξ“(π›Όβˆ’π›½)1βˆ’π‘βˆ’2𝑗=1π‘Žπ‘—πœ‰π‘—π›Όβˆ’π›Ύβˆ’1ξ‚ξ€œ10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1𝑓𝑠,πΌπ›½ξ€ΈπœŒ(𝑠),𝜌(𝑠)𝑑𝑠<+∞.(3.5)It follows from Lemma 2.6 and (3.3) that ξ€·π‘‡πœ†πœŒξ€Έξ€œ(𝑑)β‰₯πœ†π’’(𝑑)10𝔐(𝑠)𝑓𝑠,πΌπ›½ξ€ΈπœŒ(𝑠),𝜌(𝑠)𝑑𝑠=π‘™ξ…žπœŒπ’’(𝑑),(3.6) where π‘™ξ…žπœŒξ€œ=πœ†10𝔐(𝑠)𝑓𝑠,πΌπ›½ξ€ΈπœŒ(𝑠),𝜌(𝑠)𝑑𝑠.(3.7)Using (3.3) and (3.6), we know that π‘‡πœ† is well defined and π‘‡πœ†(𝑃)βŠ‚π‘ƒ.
Next we will focus on the upper and lower solutions of problem (2.22). From (B1) and (3.2), we know that the operator π‘‡πœ† is decreasing in 𝑦. Usingξ€œ10ξ€œπΎ(𝑑,𝑠)𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠β‰₯𝒒(𝑑)10[]𝔐(𝑠)𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠,βˆ€π‘‘βˆˆ0,1,(3.8)and letting πœ†1=1∫10𝔐(𝑠)𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠,(3.9)we have πœ†1ξ€œ10[]𝐾(𝑑,𝑠)𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠β‰₯𝒒(𝑑),βˆ€π‘‘βˆˆ0,1.(3.10)
On the other hand, letting βˆ«π‘(𝑑)=10𝐾(𝑑,𝑠)𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠, since 𝑓(𝑑,𝑒,𝑣) is decreasing with respect to 𝑒 and 𝑣, for any πœ†>πœ†1, we haveξ€œ10𝐾(𝑑,𝑠)𝑓𝑠,πœ†πΌπ›½ξ€Έξ€œπ‘(𝑠),πœ†π‘(𝑠)𝑑𝑠≀10𝐾(𝑑,𝑠)𝑓𝑠,πœ†1𝐼𝛽𝑏(𝑠),πœ†1ξ€Έβ‰€ξ€œπ‘(𝑠)𝑑𝑠10πΎξ€œ(𝑑,𝑠)𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠≀𝑀10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠<+∞.(3.11)From (3.2), (3.3), and (B1), for all (𝑒,𝑣)∈(0,∞)Γ—(0,∞), we have limπœ‡β†’+βˆžπœ‡π‘“(𝑑,πœ‡π‘’,πœ‡π‘£)=+∞(3.12)uniformly on π‘‘βˆˆ(0,1). Thus there exists large enough πœ†βˆ—>πœ†1>0, such that, for any π‘‘βˆˆ(0,1), πœ†βˆ—π‘“ξ€·π‘ ,πœ†βˆ—β„’(𝑠),πœ†βˆ—ξ€Έβ‰₯1𝒒(𝑠)∫10𝔐(𝑠)𝑑𝑠.(3.13)From Lemma 2.6, one has πœ†βˆ—ξ€œ10𝐾(𝑑,𝑠)𝑓𝑠,πœ†βˆ—β„’(𝑠),πœ†βˆ—ξ€Έβˆ«π’’(𝑠)𝑑𝑠β‰₯10𝐾(𝑑,𝑠)π‘‘π‘ βˆ«10𝔐β‰₯∫(𝑠)𝑑𝑠10𝒒(𝑑)𝔐(𝑠)π‘‘π‘ βˆ«10𝔐[].(𝑠)𝑑𝑠=𝒒(𝑑),βˆ€π‘‘βˆˆ0,1(3.14)Letting πœ™(𝑑)=πœ†βˆ—ξ€œ10𝐾(𝑑,𝑠)𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠=πœ†βˆ—πœ“π‘(𝑑),(𝑑)=πœ†βˆ—ξ€œ10𝐾(𝑑,𝑠)𝑓𝑠,πœ†βˆ—πΌπ›½π‘(𝑠),πœ†βˆ—π‘ξ€Έ(𝑠)𝑑𝑠,(3.15)and using Lemmas 2.3 and 2.7, we obtain πœ™(𝑑)=πœ†βˆ—βˆ«10[],πœ™πΎ(𝑑,𝑠)𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠β‰₯𝒒(𝑑),π‘‘βˆˆ0,1(0)=0,π““π­π›Ύβˆ’π›½πœ™(1)=π‘βˆ’2βˆ‘π‘—=1π‘Žπ‘—π““π­π›Ύβˆ’π›½πœ™ξ€·πœ‰π‘—ξ€Έ,πœ“(𝑑)=πœ†βˆ—βˆ«10𝐾(𝑑,𝑠)𝑓𝑠,πœ†βˆ—πΌπ›½π‘(𝑠),πœ†βˆ—ξ€Έ[],𝑏(𝑠)𝑑𝑠β‰₯𝒒(𝑑),π‘‘βˆˆ0,1πœ“(0)=0,π““π­π›Ύβˆ’π›½πœ“(1)=π‘βˆ’2βˆ‘π‘—=1π‘Žπ‘—π““π­π›Ύβˆ’π›½πœ“ξ€·πœ‰π‘—ξ€Έ.(3.16) Obviously, πœ™(𝑑),πœ“(𝑑)βˆˆπ‘ƒ. By (3.16), we have 𝒒𝑇(𝑑)β‰€πœ“(𝑑)=πœ†βˆ—πœ™ξ€Έ[](𝑑),𝒒(𝑑)β‰€πœ™(𝑑),βˆ€π‘‘βˆˆ0,1,(3.17) which implies that ξ€·π‘‡πœ“(𝑑)=πœ†βˆ—πœ™ξ€Έ(𝑑)=πœ†βˆ—ξ€œ10𝐾(𝑑,𝑠)𝑓𝑠,πΌπ›½ξ€Έπœ™(𝑠),πœ™(𝑠)π‘‘π‘ β‰€πœ†βˆ—ξ€œ10𝐾[].(𝑑,𝑠)𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠=πœ™(𝑑),βˆ€π‘‘βˆˆ0,1(3.18)Consequently, it follows from (3.17)-(3.18) that π““π­πœ“(𝑑)+πœ†βˆ—π‘“ξ€·π‘‘,πΌπ›½ξ€Έπœ“(𝑑),πœ“(𝑑)=π““π­ξ€·π‘‡πœ†βˆ—πœ™ξ€Έ(𝑑)+πœ†βˆ—π‘“ξ€·π‘‘,πΌπ›½ξ€·π‘‡πœ†βˆ—πœ™ξ€Έξ€·π‘‡(𝑑),πœ†βˆ—πœ™ξ€Έξ€Έ(𝑑)β‰₯π““π­ξ€·π‘‡πœ†βˆ—πœ™ξ€Έ(𝑑)+πœ†βˆ—π‘“ξ€·π‘‘,πΌπ›½πœ™ξ€Έ(𝑑),πœ™(𝑑)=βˆ’πœ†βˆ—π‘“ξ€·π‘‘,πΌπ›½ξ€Έπœ™(𝑑),πœ™(𝑑)+πœ†βˆ—π‘“ξ€·π‘‘,πΌπ›½ξ€Έπ““πœ™(𝑑),πœ™(𝑑)=0,(3.19)π­πœ™(𝑑)+πœ†βˆ—π‘“ξ€·π‘‘,πΌπ›½ξ€Έπœ™(𝑑),πœ™(𝑑)=βˆ’πœ†βˆ—π‘“(𝑑,β„’(𝑑),𝒒(𝑑))+πœ†βˆ—π‘“ξ€·π‘‘,πΌπ›½ξ€Έπœ™(𝑑),πœ™(𝑑)β‰€βˆ’πœ†βˆ—π‘“(𝑑,β„’(𝑑),𝒒(𝑑))+πœ†βˆ—π‘“(𝑑,β„’(𝑑),𝒒(𝑑))=0.(3.20)From (3.16) and (3.18)–(3.20), we know that πœ“(𝑑)andπœ™(𝑑) are upper and lower solutions of the problem (2.22), and πœ“(𝑑),πœ™(𝑑)βˆˆπ‘ƒ.
Define the function 𝐹 and the operator π΄πœ†βˆ— in 𝐸 by⎧βŽͺ⎨βŽͺβŽ©π‘“ξ€·πΉ(𝑑,𝑦)=𝑑,πΌπ›½ξ€Έπ‘“ξ€·πœ“(𝑑),πœ“(𝑑),𝑦<πœ“(𝑑),𝑑,𝐼𝛽𝑓𝑦(𝑑),𝑦(𝑑),πœ“(𝑑)β‰€π‘¦β‰€πœ™(𝑑),𝑑,πΌπ›½ξ€Έξ€·π΄πœ™(𝑑),πœ™(𝑑),𝑦>πœ™(𝑑),πœ†βˆ—π‘¦ξ€Έ(𝑑)=πœ†βˆ—ξ€œ10𝐾(𝑑,𝑠)𝐹(𝑠,𝑦(𝑠))𝑑𝑠,βˆ€π‘¦βˆˆπΈ.(3.21) It follows from (B1) and (3.21) that 𝐹∢(0,1)Γ—[0,+∞)β†’[0,+∞) is continuous. Consider the following boundary value problem: βˆ’π““π­π›Όβˆ’π›½π‘¦(𝑑)=πœ†βˆ—[],𝐹(𝑑,𝑦),π‘‘βˆˆ0,1𝑦(0)=0,π““π­π›Ύβˆ’π›½π‘¦(1)=π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύβˆ’π›½π‘¦ξ€·πœ‰π‘—ξ€Έ.(3.22)Obviously, a fixed point of the operator π΄πœ†βˆ— is a solution of the BVP (3.22). For all π‘¦βˆˆπΈ, it follows from Lemma 2.6, (3.21), and πœ“(𝑑)β‰₯𝒒(𝑑) that ξ€·π΄πœ†βˆ—π‘¦ξ€Έ(𝑑)β‰€πœ†βˆ—π‘€ξ€œ10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1𝐹(𝑠,𝑦(𝑠))π‘‘π‘ β‰€πœ†βˆ—π‘€ξ€œ10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1𝑓𝑠,πΌπ›½πœ“ξ€Έ(𝑠),πœ“(𝑠)π‘‘π‘ β‰€πœ†βˆ—π‘€ξ€œ10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1𝑓(𝑠,β„’(𝑠),𝒒(𝑠))𝑑𝑠<+∞.(3.23)So π΄πœ†βˆ— is bounded. From the continuity of 𝐹(𝑑,𝑦) and 𝐾(𝑑,𝑠), it is obviously that π΄πœ†βˆ—βˆΆπΈβ†’πΈ is continuous.
From the uniform continuity of 𝐾(𝑑,𝑠) and the Lebesgue dominated convergence theorem, we easily get that π΄πœ†βˆ—(Ξ©) is equicontinuous. Thus from the Arzela-Ascoli theorem, π΄πœ†βˆ—βˆΆπΈβ†’πΈ is completely continuous. The Schauder fixed point theorem implies that π΄πœ†βˆ— has at least one fixed point 𝑀 such that 𝑀=π΄πœ†βˆ—π‘€.
Now we prove[].πœ“(𝑑)≀𝑀(𝑑)β‰€πœ™(𝑑),π‘‘βˆˆ0,1(3.24) Let 𝑧(𝑑)=πœ™(𝑑)βˆ’π‘€(𝑑),π‘‘βˆˆ[0,1]. Since πœ™(𝑑) is the upper solution of problem (2.22) and 𝑀 is a fixed point of π΄πœ†βˆ—, we have 𝑀(0)=0,π““π­π›Ύβˆ’π›½π‘€(1)=π‘βˆ’2𝑗=1π‘Žπ‘—π““π­π›Ύβˆ’π›½π‘€ξ€·πœ‰π‘—ξ€Έ.(3.25)
From (3.17), (3.18), and the definition of 𝐹, we obtain𝑓𝑑,πΌπ›½ξ€Έξ€·πœ™(𝑑),πœ™(𝑑)≀𝐹(𝑑,𝑦(𝑑))≀𝑓𝑑,πΌπ›½ξ€Έπ‘“ξ€·πœ“(𝑑),πœ“(𝑑),βˆ€π‘¦βˆˆπΈ,(𝑑,β„’(𝑑),𝒒(𝑑))β‰₯𝑓𝑑,πΌπ›½πœ“ξ€Έ[].(𝑑),πœ“(𝑑),βˆ€π‘‘βˆˆ0,1(3.26)So 𝑓𝑑,πΌπ›½ξ€Έπœ™(𝑑),πœ™(𝑑)≀𝐹(𝑑,𝑦(𝑑))≀𝑓(𝑑,β„’(𝑑),𝒒(𝑑)),βˆ€π‘¦βˆˆπΈ.(3.27)From (3.18) and (3.20), one has π““π­π›Όβˆ’π›½π‘§(𝑑)=π““π­π›Όβˆ’π›½πœ™(𝑑)βˆ’π““π­π›Όβˆ’π›½π‘€(𝑑)=βˆ’πœ†βˆ—π‘“(𝑑,β„’(𝑑),𝒒(𝑑))+πœ†βˆ—[].𝐹(𝑑,𝑀(𝑑))≀0,βˆ€π‘‘βˆˆ0,1(3.28)By (3.27), (3.28), and Lemma 2.10, we get 𝑧(𝑑)β‰₯0 which implies that 𝑀(𝑑)β‰€πœ™(𝑑) on [0,1]. In the same way, we have 𝑀(𝑑)β‰₯πœ“(𝑑) on [0,1]. Thus we obtain [].πœ“(𝑑)≀𝑀(𝑑)β‰€πœ™(𝑑),π‘‘βˆˆ0,1(3.29) Consequently, 𝐹(𝑑,𝑀(𝑑))=𝑓(𝑑,𝐼𝛽𝑀(𝑑),𝑀(𝑑)),π‘‘βˆˆ[0,1]. Then 𝑀(𝑑) is a positive solution of the problem (2.22). It thus follows from Lemma 2.7 that π‘₯(𝑑)=𝐼𝛽𝑀(𝑑) is a positive solution of the problem (1.1).
Finally, by (3.29), we have𝑀(𝑑)β‰₯πœ“(𝑑)β‰₯𝒒(𝑑).(3.30)Thus, π‘₯(𝑑)=𝐼𝛽1𝑀(𝑑)=ξ€œΞ“(𝛽)𝑑0(π‘‘βˆ’π‘ )π›½βˆ’11𝑀(𝑠)𝑑𝑠β‰₯ξ€œΞ“(𝛽)𝑑0(π‘‘βˆ’π‘ )π›½βˆ’1𝒒(𝑠)𝑑𝑠=β„’(𝑑).(3.31)

Corollary 3.2. Suppose that condition (B1)holds, and that for any πœ‡,𝜈>0, 𝑓(𝑑,πœ‡,𝜈)β‰’0, and ξ€œ10𝑓(𝑠,πœ‡β„’(𝑠),πœ‡π’’(𝑠))𝑑𝑠<+∞.(3.32)Then there exists a constant πœ†βˆ—>0 such that for any πœ†βˆˆ(πœ†βˆ—,+∞), the problem (1.1) has at least one positive solution π‘₯(𝑑), which satisfies π‘₯(𝑑)β‰₯β„’(𝑑), π‘‘βˆˆ[0,1].

We consider some special cases in which 𝑓(𝑑,𝑒,𝑣) has no singularity at 𝑒,𝑣=0 or 𝑑=0,1.

We give the following assumption.

(Bβˆ—1)π‘“βˆˆπΆ((0,1)Γ—[0,∞)Γ—[0,∞),(0,+∞)) is decreasing in 𝑒,𝑣.

Then, 𝑓(𝑑,𝑒,𝑣) is nonsingular at 𝑒=𝑣=0 and for all 𝑒,𝑣β‰₯0, 𝑓(𝑑,𝑒,𝑣)>0,π‘‘βˆˆ(0,1), which implies that 𝑓(𝑑,0,0)>0,π‘‘βˆˆ(0,1). Thuslimπœ‡β†’+βˆžπœ‡π‘“(𝑑,0,0)=+∞,uniformlyforπ‘‘βˆˆ(0,1)(3.33)

naturally holds; we then have the following corollary.

Corollary 3.3. If (Bβˆ—1) holds and
(Bβˆ—2)ξ€œ10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1𝑓(𝑠,0,0)𝑑𝑠<+∞,(3.34) then there exists a constant πœ†βˆ—>0 such that for any πœ†βˆˆ(πœ†βˆ—,+∞), the problem (1.1) has at least one positive solution π‘₯(𝑑), which satisfies π‘₯(𝑑)β‰₯β„’(𝑑), π‘‘βˆˆ[0,1].Proof. In the proof of Theorem 3.1, we replace the set 𝑃 by 𝑃1[]}={π‘¦βˆˆπΈβˆΆπ‘¦(𝑑)β‰₯0,π‘‘βˆˆ0,1(3.35)and the inequalities (3.18)–(3.20) by 0β‰€πœ“(𝑑)=π‘‡πœ†0,0β‰€πœ™(𝑑)=π‘‡πœ†πœ“(𝑑)β‰€π‘‡πœ†0=πœ“(𝑑).(3.36)Since π‘‡πœ†0,π‘‡πœ†πœ“(𝑑)βˆˆπ‘ƒ, we have π““π­π›Όβˆ’π›½π‘‡πœ†ξ€·0+𝑓𝑑,πΌπ›½π‘‡πœ†,π‘‡πœ†0ξ€Έξ€·=βˆ’π‘“(𝑑,0,0)+𝑓𝑑,πΌπ›½π‘‡πœ†0,π‘‡πœ†0𝓓≀0,π­π›Όβˆ’π›½π‘‡πœ†ξ€·πœ“(𝑑)+𝑓𝑑,πΌπ›½π‘‡πœ†πœ“(𝑑),π‘‡πœ†ξ€Έξ€·πœ“(𝑑)=βˆ’π‘“π‘‘,πΌπ›½ξ€Έξ€·πœ“(𝑑),πœ“(𝑑)+𝑓𝑑,πΌπ›½π‘‡πœ†πœ“(𝑑),π‘‡πœ†ξ€Έ[].πœ“(𝑑)β‰₯0,π‘‘βˆˆ0,1(3.37) The rest of the proof is similar to that of Theorem 3.1.

If 𝑓(𝑑,𝑒,𝑣) is nonsingular at 𝑒=0,𝑣=0 and 𝑑=0,1, we have the conclusion.

Corollary 3.4. If 𝑓(𝑑,𝑒,𝑣)∢[0,1]Γ—[0,∞)Γ—[0,∞)β†’(0,+∞) is continuous and decreasing in 𝑒and𝑣, the problem (1.1) has at least one positive solution π‘₯(𝑑), which satisfies π‘₯(𝑑)β‰₯β„’(𝑑), π‘‘βˆˆ[0,1].

Example 3.5. Consider the existence of positive solutions for the following eigenvalue problem of fractional differential equation: βˆ’π““π­3/2πœ†π‘₯(𝑑)=𝑒𝑑(1βˆ’π‘‘)1/8ξ‚€π‘₯βˆ’1/2𝓓(𝑑)+𝐭1/8ξ€Έπ‘₯(𝑑)βˆ’1/8,𝓓𝐭1/8π‘₯(0)=0,𝓓𝐭3/8π‘₯(1)=2𝓓𝐭3/8π‘₯ξ‚€12ξ‚βˆ’π““π­3/8π‘₯ξ‚€34.(3.38)

Let 1𝑓(𝑑,𝑒,𝑣)=𝑒𝑑(1βˆ’π‘‘)1/8ξ€·π‘’βˆ’1/2+π‘£βˆ’1/8ξ€Έ,(𝑑,𝑒,𝑣)∈(0,1)Γ—(0,+∞)Γ—(0,+∞).(3.39) Then π‘“βˆˆπΆ((0,1)Γ—(0,+∞)Γ—(0,+∞),(0,+∞)) is decreasing in 𝑒and𝑣, and for any (𝑒,𝑣)∈(0,∞)Γ—(0,∞), limπœŽβ†’+βˆžπœŽπ‘“(𝑑,πœŽπ‘’,πœŽπ‘£)=limπœŽβ†’+∞𝜎1/2π‘’βˆ’1/2+𝜎7/8π‘£βˆ’1/8𝑒𝑑(1βˆ’π‘‘)1/8=+∞,(3.40)uniformly on π‘‘βˆˆ(0,1). Thus (B1) holds.

On the other hand, for any πœ‡,𝜈>0 and π‘‘βˆˆ(0,1),1𝑓(𝑑,πœ‡,𝜈)=𝑒𝑑(1βˆ’π‘‘)1/8ξ€·πœ‡βˆ’1/2+πœˆβˆ’1/8ξ€Έβ„’ξ€œβ‰’0,(𝑑)=𝑑0(π‘‘βˆ’π‘ )βˆ’7/8𝑠3/8Ξ“(1/8)𝑑𝑠=Ξ“(11/8)𝑑Γ(3/2)1/2.(3.41)

thus we haveξ€œ10(1βˆ’π‘ )π›Όβˆ’π›Ύβˆ’1ξ€œπ‘“(𝑠,πœ‡β„’(𝑠),πœ‡π’’(𝑠))𝑑𝑠=10(1βˆ’π‘ )1/8ξ‚Έ1𝑒𝑠(1βˆ’π‘ )1/8ξ€·πœ‡βˆ’1/2β„’βˆ’1/2(𝑠)+πœ‡βˆ’1/8π’’βˆ’1/8ξ€Έξ‚Ήβ‰€ξ€œ(𝑠)10ξƒ¬πœ‡βˆ’1/2ξ‚΅Ξ“(11/8)𝑑Γ(3/2)1/2ξ‚Άβˆ’1/2+πœ‡βˆ’1/8π‘ βˆ’3/64ξƒ­=ξ€œπ‘‘π‘ 10ξ‚Έπœ‡βˆ’1/2ξ‚΅Ξ“(11/8)𝑑Γ(3/2)βˆ’1/4ξ‚Ά+πœ‡βˆ’1/8π‘ βˆ’3/64𝑑𝑠<+∞,(3.42)

which implies that (B2) holds. From Theorem 3.1, there is a constant πœ†βˆ—>0 such that for any πœ†βˆˆ(πœ†βˆ—,+∞) the problem (3.38) has at least one positive solution π‘₯(𝑑) andπ‘₯(𝑑)β‰₯β„’(𝑑)=Ξ“(11/8)𝑑Γ(3/2)1/2β‰ˆ1.003𝑑1/2[].,π‘‘βˆˆ0,1(3.43)

Acknowledgments

This work is supported financially by the National Natural Science Foundation of China (11071141, 11126231) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017).

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