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Mathematical Problems in Engineering
VolumeΒ 2009, Article IDΒ 535209, 17 pages
http://dx.doi.org/10.1155/2009/535209
Research Article

Singular Positone and Semipositone Boundary Value Problems of Nonlinear Fractional Differential Equations

1School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, Jilin, China
2School of Mathematics and Computer, Harbin University, Harbin 150086, Heilongjiang, China

Received 8 January 2009; Accepted 15 April 2009

Academic Editor: VictoriaΒ Vampa

Copyright Β© 2009 Chengjun Yuan 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 present some new existence results for singular positone and semipositone nonlinear fractional boundary value problem D𝛼0+𝑒(𝑑)=πœ‡π‘Ž(𝑑)𝑓(𝑑,𝑒(𝑑)),  0<𝑑<1, 𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=π‘’ξ…ž(1)=0, where πœ‡>0, π‘Ž, and 𝑓 are continuous, π›Όβˆˆ(3,4] is a real number, and D𝛼0+ is Riemann-Liouville fractional derivative. Throughout our nonlinearity may be singular in its dependent variable. Two examples are also given to illustrate the main results.

1. Introduction

Fractional calculus has played a significant role in engineering, science, economy, and other fields. Many papers and books on fractional calculus, and fractional differential equations have appeared recently, (see [1–9]). It should be noted that most of papers and books on fractional calculus, are devoted to the solvability of initial fractional differential equations (see [3, 4]). Here, we consider positive solutions of nonlinear fractional differential equation conjugate boundary value problem involving Riemann-Liouville derivative: 𝐃𝛼0+𝑒𝑒(𝑑)=πœ‡π‘Ž(𝑑)𝑓(𝑑,𝑒(𝑑)),0<𝑑<1,(0)=𝑒(1)=π‘’ξ…ž(0)=π‘’ξ…ž(1)=0,(1.1) where πœ‡>0, π‘Ž, and 𝑓 are continuous. π›Όβˆˆ(3,4] is a real number, and 𝐃𝛼0+ is the Riemann-Liouville fractional derivative.

It is well known that in mechanics the boundary value problem (1.1) where 𝛼=4 describes the deflection of an elastic beam rigidly fixed at both ends. The integer order boundary value problem (1.2) has been studied extensively. For details, see for instance, the papers [10–13] and the references therein. In [10, 12], Yao considered π‘’ξ…žξ…žξ…žξ…ž(𝑑)=πœ†π‘“(𝑑,𝑒(𝑑)),0<𝑑<1,𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=π‘’ξ…ž(1)=0,(1.2) and using a Krasnosel'skii fixed-point theorem, derived a πœ†-interval such that, for any πœ† lying in this interval, the beam equation has existence and multiplicity on positive solution. In this paper, we will consider a more general situation, namely, the boundary value problem (1.1). To the best of our knowledge, there have been few papers which deal with the boundary value problem (1.1) for nonlinear fractional differential equation.

In this paper, in analogy with boundary value problems for differential equations of integer order, we firstly derive the corresponding Green's function named the fractional Green' function. Consequently problem (1.1) is reduced to an equivalent Fredholm integral equation of the second kind. Finally, using Krasnosel'skii's fixed-point theorems, the existence of positive solutions are obtained.

2. Preliminaries

For completeness, in this section, we will demonstrate and study the definitions and some fundamental facts of Riemann-Liouville derivatives of fractional order which can be found in [5].

Definition 2.1 (see [5, Definition 2.1]). The integral 𝐼𝛼0+1𝑓(π‘₯)=ξ€œΞ“(𝛼)π‘₯0𝑓(𝑑)(π‘₯βˆ’π‘‘)1βˆ’π›Όπ‘‘π‘‘,π‘₯>0,(2.1) where 𝛼>0, is called the Riemann-Liouville fractional integral of order 𝛼.

Definition 2.2 (see [5, page 36-37]). For a function 𝑓(π‘₯) given in the interval [0,∞), the expression 𝐃𝛼0+1𝑓(π‘₯)=𝑑Γ(π‘›βˆ’π›Ό)𝑑π‘₯π‘›ξ€œx0𝑓(𝑑)(π‘₯βˆ’π‘‘)π›Όβˆ’π‘›+1𝑑𝑑,(2.2) where 𝑛=[𝛼]+1,[𝛼] denotes the integer part of number 𝛼, is called the Riemann-Liouville fractional derivative of order 𝛼.
From the definition of Riemann-Liouville derivative, for πœ‡>βˆ’1, we have 𝐃𝛼0+π‘₯πœ‡=Ξ“(1+πœ‡)π‘₯Ξ“(1+πœ‡βˆ’π›Ό)πœ‡βˆ’π›Ό,(2.3) giving in particular 𝐃𝛼0+π‘₯π›Όβˆ’π‘š=0,π‘š=1,2,3,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

Lemma 2.3. Let 𝛼>0, then the differential equation 𝐃𝛼0+𝑒(𝑑)=0(2.4) has solutions 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 𝛼.
As 𝐃𝛼0+𝐼𝛼0+𝑒=𝑒, from Lemma 2.3, we deduce the following statement.

Lemma 2.4. Let 𝛼>0, then 𝐼𝛼0+𝐃𝛼0+𝑒(𝑑)=𝑒(𝑑)+𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2+β‹―+π‘π‘›π‘‘π›Όβˆ’π‘›,(2.5) for some π‘π‘–βˆˆβ„, 𝑖=1,2,…,𝑛, 𝑛 is the smallest integer greater than or equal to 𝛼.
The following Krasnosel'skii's fixed-point theorem will play a major role in our next analysis.

Theorem 2.5 (see [6]). Let 𝑋 be a Banach space, and let π‘ƒβŠ‚π‘‹ be a cone in 𝑋. Assume that 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 (1)‖𝑆𝑀‖≀‖𝑀‖, π‘€βˆˆπ‘ƒβˆ©πœ•Ξ©1, ‖𝑆𝑀‖β‰₯‖𝑀‖, π‘€βˆˆπ‘ƒβˆ©πœ•Ξ©2, or(2)‖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©2βˆ– Ξ©1).

3. Green's Function and Its Properties

In this section, we derive the corresponding Green's function for boundary-value problem (1.1), and obtain some properties of Green's function.

Lemma 3.1. Let β„Ž(𝑑)∈𝐢[0,1] be a given function, then the boundary-value problem, 𝐃𝛼0+𝑒𝑒(𝑑)=β„Ž(𝑑),0<𝑑<1,3<𝛼≀4,(0)=𝑒(1)=π‘’ξ…ž(0)=π‘’ξ…ž(1)=0,(3.1) has a unique solution ξ€œπ‘’(𝑑)=10𝐺(𝑑,𝑠)β„Ž(𝑠)𝑑𝑠,(3.2) where 1𝐺(𝑑,𝑠)=𝑑Γ(𝛼)π›Όβˆ’2(1βˆ’π‘ )π›Όβˆ’2[](π‘ βˆ’π‘‘)+(π›Όβˆ’2)(1βˆ’π‘‘)𝑠+(π‘‘βˆ’π‘ )π›Όβˆ’1𝑑,0≀𝑠≀𝑑≀1,π›Όβˆ’2(1βˆ’π‘ )π›Όβˆ’2[](π‘ βˆ’π‘‘)+(π›Όβˆ’2)(1βˆ’π‘‘)𝑠,0≀𝑑≀𝑠≀1.(3.3) Here 𝐺(𝑑,𝑠) is called Green's function of boundary-value problem (3.1).

Proof. By means of the Lemma 2.4, we can reduce (3.1) to an equivalent integral equation 𝑒(𝑑)=𝑐1π‘‘π›Όβˆ’1+𝑐2π‘‘π›Όβˆ’2+𝑐3π‘‘π›Όβˆ’3+𝑐4π‘‘π›Όβˆ’4+ξ€œπ‘‘0(π‘‘βˆ’π‘ )π›Όβˆ’1Ξ“(𝛼)β„Ž(𝑠)𝑑𝑠.(3.4) From 𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=π‘’ξ…ž(1)=0, we have 𝑐3=𝑐4=0 and 𝑐1=ξ€œ10(1βˆ’π‘ )π›Όβˆ’2(2π‘ βˆ’π›Όπ‘ βˆ’1)𝑐Γ(𝛼)β„Ž(𝑠)𝑑𝑠,2=ξ€œ10(π›Όβˆ’1)(1βˆ’π‘ )π›Όβˆ’2π‘ β„ŽΞ“(𝛼)(𝑠)𝑑𝑠.(3.5) Therefore, the unique solution of (3.1) is ξ€œπ‘’(𝑑)=10π‘‘π›Όβˆ’1(1βˆ’π‘ )π›Όβˆ’2(2π‘ βˆ’π›Όπ‘ βˆ’1)ξ€œΞ“(𝛼)β„Ž(𝑠)𝑑𝑠+10π‘‘π›Όβˆ’2(π›Όβˆ’1)(1βˆ’π‘ )π›Όβˆ’2π‘ ξ€œΞ“(𝛼)β„Ž(𝑠)𝑑𝑠+𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1=1Ξ“(𝛼)β„Ž(𝑠)π‘‘π‘ ξ€œΞ“(𝛼)𝑑0ξ€Ίπ‘‘π›Όβˆ’1(1βˆ’π‘ )π›Όβˆ’2(2π‘ βˆ’π›Όπ‘ βˆ’1)+π‘‘π›Όβˆ’2(π›Όβˆ’1)(1βˆ’π‘ )π›Όβˆ’2𝑠+(π‘‘βˆ’π‘ )π›Όβˆ’1ξ€»+ξ€œβ„Ž(𝑠)𝑑𝑠1π‘‘ξ€Ίπ‘‘π›Όβˆ’1(1βˆ’π‘ )π›Όβˆ’2(2π‘ βˆ’π›Όπ‘ βˆ’1)+π‘‘π›Όβˆ’2(π›Όβˆ’1)(1βˆ’π‘ )π›Όβˆ’2𝑠=ξ€œβ„Ž(𝑠)𝑑𝑠10𝐺(𝑑,𝑠)β„Ž(𝑠)𝑑𝑠.(3.6) The proof is finished.

Lemma 3.2. The function 𝐺(𝑑,𝑠) defined by (3.3) has the following properties: (1)𝐺(𝑑,𝑠)=𝐺(1βˆ’π‘ ,1βˆ’π‘‘),π‘“π‘œπ‘Ÿπ‘‘,π‘ βˆˆ[0,1];(2)π‘‘π›Όβˆ’2(1βˆ’π‘‘)2π‘ž(𝑠)≀𝐺(𝑑,𝑠)≀(π›Όβˆ’1)π‘ž(𝑠) and 𝐺(𝑑,𝑠)≀((π›Όβˆ’1)(π›Όβˆ’2)/Ξ“(𝛼))π‘‘π›Όβˆ’2(1βˆ’π‘‘)2 for 𝑑,π‘ βˆˆ[0,1],where π‘ž(𝑠)=((π›Όβˆ’2)/Ξ“(𝛼))𝑠2(1βˆ’π‘ )π›Όβˆ’2.

Proof. Observing the expression of 𝐺(𝑑,𝑠), it is clear that 𝐺(𝑑,𝑠)=𝐺(1βˆ’π‘ ,1βˆ’π‘‘) for 𝑑,π‘ βˆˆ[0,1]. In the following, we consider Ξ“(𝛼)𝐺(𝑑,𝑠).
For 0≀𝑠≀𝑑≀1, we have
Ξ“(𝛼)𝐺(𝑑,𝑠)=(π‘‘βˆ’π‘ )π›Όβˆ’1βˆ’(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2(π‘‘βˆ’π‘ )+(π›Όβˆ’2)(1βˆ’π‘‘)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2𝑠=(π‘‘βˆ’π‘ )(π‘‘βˆ’π‘ )π›Όβˆ’2βˆ’(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2ξ€»+(π›Όβˆ’2)(1βˆ’π‘‘)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2π‘ ξ€œ=βˆ’(π‘‘βˆ’π‘ )(π›Όβˆ’2)π‘‘βˆ’π‘‘π‘ π‘‘βˆ’π‘ π‘₯π›Όβˆ’3𝑑π‘₯+(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2(1βˆ’π‘‘)𝑠β‰₯βˆ’(π‘‘βˆ’π‘ )(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’3[]+(π‘‘βˆ’π‘‘π‘ )βˆ’(π‘‘βˆ’π‘ )(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2(1βˆ’π‘‘)𝑠=βˆ’(π‘‘βˆ’π‘ )(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’3(1βˆ’π‘‘)𝑠+(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2=(1βˆ’π‘‘)𝑠(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’3[βˆ’]β‰₯(1βˆ’π‘‘)𝑠(π‘‘βˆ’π‘ )+(π‘‘βˆ’π‘‘π‘ )(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2(1βˆ’π‘‘)2𝑠2=(π›Όβˆ’2)π‘‘π›Όβˆ’2(1βˆ’π‘‘)2𝑠2(1βˆ’π‘ )π›Όβˆ’2,(3.7) and ξ€œΞ“(𝛼)𝐺(𝑑,𝑠)=βˆ’(π‘‘βˆ’π‘ )(π›Όβˆ’2)π‘‘βˆ’π‘‘π‘ π‘‘βˆ’π‘ π‘₯π›Όβˆ’3𝑑π‘₯+(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2(1βˆ’π‘‘)π‘ β‰€βˆ’(π‘‘βˆ’π‘ )(π›Όβˆ’2)(π‘‘βˆ’π‘ )π›Όβˆ’3[](π‘‘βˆ’π‘‘π‘ )βˆ’(π‘‘βˆ’π‘ )+(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2(1βˆ’π‘‘)𝑠=βˆ’(π‘‘βˆ’π‘ )(π›Όβˆ’2)(π‘‘βˆ’π‘ )π›Όβˆ’3(1βˆ’π‘‘)𝑠+(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2ξ€Ί(1βˆ’π‘‘)𝑠=(π›Όβˆ’2)(1βˆ’π‘‘)𝑠(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2βˆ’(π‘‘βˆ’π‘ )π›Όβˆ’2≀(π›Όβˆ’2)2(1βˆ’π‘‘)2𝑠2(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’3≀(π›Όβˆ’1)(π›Όβˆ’2)π‘‘π›Όβˆ’3(1βˆ’π‘‘)2𝑠2(1βˆ’π‘ )π›Όβˆ’3≀(π›Όβˆ’1)(π›Όβˆ’2)𝑠2(1βˆ’π‘ )π›Όβˆ’2.(3.8)
For 0≀𝑑≀𝑠≀1, since 𝛼>3, we have
Ξ“(𝛼)𝐺(𝑑,𝑠)=(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2[](π‘ βˆ’π‘‘)+(π›Όβˆ’2)(1βˆ’π‘‘)𝑠β‰₯(π›Όβˆ’2)(π‘‘βˆ’π‘‘π‘ )π›Όβˆ’2(1βˆ’π‘‘)𝑠=(π›Όβˆ’2)π‘‘π›Όβˆ’2(1βˆ’π‘ )π›Όβˆ’2(1βˆ’π‘‘)𝑠β‰₯(π›Όβˆ’2)π‘‘π›Όβˆ’2(1βˆ’π‘‘)2𝑠2(1βˆ’π‘ )π›Όβˆ’2,Ξ“(𝛼)𝐺(𝑑,𝑠)=π‘‘π›Όβˆ’2(1βˆ’π‘ )π›Όβˆ’2[](π‘ βˆ’π‘‘)+(π›Όβˆ’2)(1βˆ’π‘‘)π‘ β‰€π‘‘π›Όβˆ’2(1βˆ’π‘ )π›Όβˆ’2[]𝑠+(π›Όβˆ’2)𝑠≀(π›Όβˆ’1)𝑑(1βˆ’π‘ )π›Όβˆ’2𝑠≀(π›Όβˆ’1)𝑠2(1βˆ’π‘ )π›Όβˆ’2≀(π›Όβˆ’2)(π›Όβˆ’1)𝑠2(1βˆ’π‘ )π›Όβˆ’2.(3.9) Thus π‘‘π›Όβˆ’2(1βˆ’π‘‘)2π‘ž(𝑠)≀𝐺(𝑑,𝑠)≀(π›Όβˆ’1)π‘ž(𝑠), for 𝑑,π‘ βˆˆ[0,1]. Combining 𝐺(𝑑,𝑠)=𝐺(1βˆ’π‘ ,1βˆ’π‘‘), we have 𝐺(𝑑,𝑠)≀(π›Όβˆ’1)π‘ž(1βˆ’π‘‘)=(π›Όβˆ’1)(π›Όβˆ’2)𝑑Γ(𝛼)π›Όβˆ’2(1βˆ’π‘‘)2[],for𝑑,π‘ βˆˆ0,1.(3.10) This completes the proof.

We note that 𝑒(𝑑) is a solution of (1.1) if and only if

ξ€œπ‘’(𝑑)=πœ‡10𝐺(𝑑,𝑠)π‘Ž(𝑠)𝑓(𝑒(𝑠))𝑑𝑠,0≀𝑑≀1.(3.11)

For our constructions, we will consider the Banach space 𝐸=𝐢[0,1] equipped with standard norm ‖𝑒‖=max0≀𝑑≀1‖𝑒(𝑑)β€–,π‘’βˆˆπ‘‹. We define a cone 𝐾 by 𝑑𝐾=π‘’βˆˆπ‘‹βˆ£π‘’(𝑑)β‰₯π›Όβˆ’2(1βˆ’π‘‘)2[]]ξ‚Όπ›Όβˆ’1‖𝑒‖,π‘‘βˆˆ0,1,π›Όβˆˆ(3,4.(3.12) Define an integral operator π΄βˆΆπΎβ†’π‘‹ by ξ€œπ΄π‘’(𝑑)=πœ‡10𝐺(𝑑,𝑠)π‘Ž(𝑠)𝑓(𝑒(𝑠))𝑑𝑠,0≀𝑑≀1,π‘’βˆˆπΎ.(3.13) Notice from (3.13) and Lemma 3.2 that, for π‘’βˆˆπΎ, 𝐴𝑒(𝑑)β‰₯0 on [0,1] and ξ€œπ΄π‘’(𝑑)=πœ‡10ξ€œπΊ(𝑑,𝑠)π‘Ž(𝑠)𝑓(𝑠,𝑒(𝑠))π‘‘π‘ β‰€πœ‡10(π›Όβˆ’1)π‘ž(𝑠)π‘Ž(𝑠)𝑓(𝑠,𝑒(𝑠))𝑑𝑠,(3.14) then βˆ«β€–π΄π‘’β€–β‰€10(π›Όβˆ’1)π‘ž(𝑠)π‘Ž(𝑠)𝑓(𝑠,𝑒(𝑠))𝑑𝑠 .

On the other hand, we have ξ€œπ΄π‘’(𝑑)=πœ‡10ξ€œπΊ(𝑑,𝑠)π‘Ž(𝑠)𝑓(𝑠,𝑒(𝑠))𝑑𝑠β‰₯πœ‡10π‘‘π›Όβˆ’2(1βˆ’π‘‘)2π‘žβ‰₯𝑑(𝑠)π‘Ž(𝑠)𝑓(𝑠,𝑒(𝑠))π‘‘π‘ π›Όβˆ’2(1βˆ’π‘‘)2πœ‡ξ€œπ›Όβˆ’110β‰₯𝑑(π›Όβˆ’1)π‘ž(𝑠)π‘Ž(𝑠)𝑓(𝑠,𝑒(𝑠))π‘‘π‘ π›Όβˆ’2(1βˆ’π‘‘)2π›Όβˆ’1‖𝐴𝑒‖.(3.15) Thus, 𝐴(𝐾)βŠ‚πΎ. In addition, standard arguments show that 𝐴 is completely continuous.

4. Singular Positone Problems

In this section we present some new result for the singular problem 𝐃𝛼0+𝑒𝑒(𝑑)=πœ‡π‘Ž(𝑑)𝑓(𝑑,𝑒(𝑑)),0<𝑑<1,3<𝛼≀4,(0)=𝑒(1)=π‘’ξ…ž(0)=π‘’ξ…ž(1)=0,(4.1) where πœ‡>0 and nonlinearity 𝑓 may be singular at 𝑒=0.

Using Theorem 2.5 we establish the following main result.

Theorem 4.1. Suppose that the following conditions are satisfied. π‘ŽβˆˆπΆ(0,1)∩𝐿1[][]⎧βŽͺβŽͺ⎨βŽͺβŽͺ⎩[][β„Ž0,1π‘€π‘–π‘‘β„Žπ‘Ž>0π‘œπ‘›(0,1)(4.2)π‘“βˆΆ0,1Γ—(0,∞)⟢(0,∞)π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ ,(4.3)𝑓(𝑑,𝑒)≀𝑔(𝑒)+β„Ž(𝑒)π‘œπ‘›0,1Γ—(0,∞)π‘€π‘–π‘‘β„Žπ‘”>0π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ π‘Žπ‘›π‘‘π‘›π‘œπ‘›π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘”π‘œπ‘›(0,∞),β„Žβ‰₯0π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ π‘œπ‘›0,∞)π‘Žπ‘›π‘‘π‘”π‘›π‘œπ‘›π‘‘π‘’π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘”π‘œπ‘›(0,∞),(4.4)βˆƒπΎ0π‘€π‘–π‘‘β„Žπ‘”(π‘₯𝑦)≀𝐾0π‘Žπ‘”(π‘₯)𝑔(𝑦)βˆ€π‘₯>0,𝑦>0,(4.5)0ξ€œ=πœ‡(π›Όβˆ’1)10ξ€·π‘ π‘ž(𝑠)π‘Ž(𝑠)π‘”π›Όβˆ’2(1βˆ’π‘ )2ξ€Έπ‘Ÿπ‘‘π‘ <∞,(4.6)βˆƒπ‘Ÿ>0π‘€π‘–π‘‘β„Žπ‘”(π‘Ÿ)+β„Ž(π‘Ÿ)>𝐾20π‘Ž0𝑔1ξ‚βŽ§βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺ⎩1π›Όβˆ’1,(4.7)π‘‡β„Žπ‘’π‘Ÿπ‘’π‘’π‘₯𝑖𝑠𝑑𝑠0<πœƒ<2(π‘β„Žπ‘œπ‘œπ‘ π‘’π‘Žπ‘›π‘‘π‘“π‘–π‘₯𝑖𝑑)π‘Žπ‘›π‘‘π‘Žπ‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ ,π‘›π‘œπ‘›π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘”π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›π‘”1∢(0,∞)⟢(0,∞),π‘Žπ‘›π‘‘π‘Žπ‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›β„Ž1∢[β„Ž0,∞)⟢(0,∞)π‘€π‘–π‘‘β„Ž1𝑔1π‘›π‘œπ‘›π‘‘π‘’π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘”π‘œπ‘›(0,∞)π‘Žπ‘›π‘‘π‘€π‘–π‘‘β„Žπ‘“(𝑑,𝑒)β‰₯𝑔1(𝑒)+β„Ž1[]⎧βŽͺ⎨βŽͺ⎩(𝑒)π‘“π‘œπ‘Ÿ(𝑑,𝑒)βˆˆπœƒ,1βˆ’πœƒΓ—(0,∞),(4.8)βˆƒ0<𝑅1<π‘Ÿ<𝑅2π‘…π‘€π‘–π‘‘β„Ž(𝑖=1,2),𝑖𝑔1(πœƒπ›Ό/π›Όβˆ’1)𝑅𝑖𝑔1𝑅𝑖𝑔1ξ€·(πœƒπ›Ό/π›Όβˆ’1)𝑅𝑖+𝑔1ξ€·π‘…π‘–ξ€Έβ„Ž1ξ€·(πœƒπ›Ό/π›Όβˆ’1)π‘…π‘–ξ€Έξ€œ<πœ‡πœƒ1βˆ’πœƒπΊ(𝜎,𝑠)π‘Ž(𝑠)𝑑𝑠,(4.9) here 𝐺(𝑑,𝑠) is Green's function and ξ€œπœƒ1βˆ’πœƒπΊ(𝜎,𝑠)π‘Ž(𝑠)𝑑𝑠=supπ‘‘βˆˆ[0,1]ξ€œπœƒ1βˆ’πœƒπΊ(𝑑,𝑠)π‘Ž(𝑠)𝑑𝑠.(4.10) Then (4.1) has two nonnegative solutions 𝑒𝑖 with 𝑅1<‖𝑒1β€–<π‘Ÿ<‖𝑒2β€–<𝑅2 and 𝑒𝑖(𝑑)>0 for π‘‘βˆˆ(0,1),𝑖=1,2.

Proof. First we will show that there exists a solution 𝑒2 to (4.1) with 𝑒2(𝑑)>0 for π‘‘βˆˆ(0,1) and π‘Ÿ<‖𝑒2β€–<𝑅2. Let Ξ©1={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<π‘Ÿ},Ξ©2=ξ€½π‘’βˆˆπΈβˆΆβ€–π‘’β€–<𝑅2ξ€Ύ.(4.11) We now show ‖𝐴𝑒‖<‖𝑒‖forπΎβˆ©πœ•Ξ©1.(4.12) To see this, let π‘’βˆˆπΎβˆ©πœ•Ξ©1. Then ‖𝑒‖=||𝑒||[0,1]=π‘Ÿ and 𝑒(𝑑)β‰₯(π‘‘π›Όβˆ’2(1βˆ’π‘‘)2/(π›Όβˆ’1))π‘Ÿ for π‘‘βˆˆ[0,1]. So for π‘‘βˆˆ[0,1], we have ξ€œ(𝐴𝑒)(𝑑)=πœ‡10ξ€œπΊ(𝑑,𝑠)π‘Ž(𝑠)𝑓(𝑠,𝑒(𝑠))π‘‘π‘ β‰€πœ‡10[𝑔]ξ€œ(π›Όβˆ’1)π‘ž(𝑠)π‘Ž(𝑠)(𝑒(s))+β„Ž(𝑒(s))𝑑𝑠=πœ‡10ξ‚»(π›Όβˆ’1)π‘ž(𝑠)π‘Ž(𝑠)𝑔(𝑒(𝑠))1+β„Ž(𝑒(𝑠))ξ‚Όξ€œπ‘”(𝑒(𝑠))π‘‘π‘ β‰€πœ‡10𝑠(π›Όβˆ’1)π‘ž(𝑠)π‘Ž(𝑠)π‘”π›Όβˆ’2(1βˆ’π‘ )2π‘Ÿπ›Όβˆ’1ξ‚Άξ‚»1+β„Ž(r)𝑔(π‘Ÿ)𝑑𝑠≀𝐾0π‘”ξ‚€π‘Ÿξ‚ξ‚»π›Όβˆ’11+β„Ž(π‘Ÿ)ξ‚Όπœ‡ξ€œπ‘”(π‘Ÿ)10𝑠(π›Όβˆ’1)π‘ž(𝑠)π‘Ž(𝑠)π‘”π›Όβˆ’2(1βˆ’π‘ )2𝑑𝑠=π‘Ž0𝐾20𝑔1[].π›Όβˆ’1𝑔(π‘Ÿ)+β„Ž(π‘Ÿ)(4.13) This together with (4.7) yields ‖𝐴𝑒‖=‖𝐴𝑒‖[0,1]<π‘Ÿ=‖𝑒‖,(4.14) so (4.12) is satisfied.
Next we show
‖𝐴𝑒‖>‖𝑒‖forπΎβˆ©πœ•Ξ©2.(4.15) To see this, let π‘’βˆˆπΎβˆ©πœ•Ξ©2 so ‖𝑒‖=‖𝑒‖[0,1]=𝑅2, and let 𝑒(𝑑)β‰₯(π‘‘π›Όβˆ’2(1βˆ’π‘‘)2/(π›Όβˆ’1))𝑅2 for π‘‘βˆˆ[0,1].
We have
ξ€œ(𝐴𝑒)(𝜎)=πœ‡10ξ€œπΊ(𝜎,𝑠)π‘Ž(𝑠)𝑓(𝑠,𝑒(𝑠))𝑑𝑠β‰₯πœ‡πœƒ1βˆ’πœƒπΊξ€Ίπ‘”(𝜎,𝑠)π‘Ž(𝑠)1(𝑒(s))+β„Ž1ξ€»ξ€œ(𝑒(s))𝑑𝑠=πœ‡πœƒ1βˆ’πœƒπΊ(𝜎,𝑠)π‘Ž(𝑠)𝑔1ξ‚»β„Ž(𝑒(𝑠))1+1(𝑒(𝑠))𝑔1ξ‚Ό(𝑒(𝑠))𝑑𝑠β‰₯𝑔1𝑅2ξ€Έπœ‡ξ€œπœƒ1βˆ’πœƒξƒ―β„ŽπΊ(𝜎,𝑠)π‘Ž(𝑠)1+1π‘ ξ€·ξ€·π›Όβˆ’2(1βˆ’π‘ )2𝑅/(π›Όβˆ’1)2𝑔1π‘ ξ€·ξ€·π›Όβˆ’2(1βˆ’π‘ )2𝑅/(π›Όβˆ’1)2𝑑𝑠β‰₯𝑔1𝑅2ξ€Έπœ‡ξ€œπœƒ1βˆ’πœƒξƒ―β„ŽπΊ(𝜎,𝑠)π‘Ž(𝑠)1+1ξ€·(πœƒπ›Ό/(π›Όβˆ’1))𝑅2𝑔1ξ€·(πœƒπ›Ό/(π›Όβˆ’1))𝑅2𝑑𝑠.(4.16) This together with (4.9) yields (𝐴𝑒)(𝜎)>𝑅2=‖𝑒‖.(4.17) Thus ||𝐴𝑒||>||𝑒||, so (4.15) is held.
Now Theorem 2.5 implies that 𝐴 has a fixed-point 𝑒2∈𝐾∩(Ξ©2⧡Ω1), that is, π‘Ÿβ‰€β€–π‘’2β€–=‖𝑒2β€–[0,1]≀𝑅 and 𝑒2(𝑑)β‰₯π‘ž(𝑑)π‘Ÿ for π‘‘βˆˆ[0,1]. It follows from (4.12) and (4.15) that ‖𝑒2β€–β‰ π‘Ÿ,‖𝑒2‖≠𝑅2, so we have π‘Ÿ<||𝑒2||<𝑅2.
Similarly, if we put
Ξ©1=ξ€½π‘’βˆˆπΈβˆΆβ€–π‘’β€–<𝑅1ξ€Ύ,Ξ©2={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<π‘Ÿ},(4.18) we can show that there exists a solution 𝑒1 to (4.1) with 𝑒1(𝑑)>0 for π‘‘βˆˆ(0,1) and 𝑅1<||𝑒1||<π‘Ÿ.
This completes the proof of Theorem 4.1.

Similar to the proof of Theorem 4.1, we have the following result.

Theorem 4.2. Suppose that (4.2)–(4.8) hold. In addition suppose ⎧βŽͺ⎨βŽͺβŽ©βˆƒ0<𝑅1𝑅<π‘Ÿπ‘€π‘–π‘‘β„Ž1𝑔1(πœƒπ›Ό/(π›Όβˆ’1))𝑅1𝑔1𝑅1𝑔1(πœƒπ›Ό/(π›Όβˆ’1))𝑅1+𝑔1𝑅1ξ€Έβ„Ž1(πœƒπ›Ό/(π›Όβˆ’1))𝑅1ξ€œ<πœ‡πœƒ1βˆ’πœƒπΊ(𝜎,𝑠)π‘ž(𝑠)𝑑𝑠.(4.19) Then (4.1) has a nonnegative solution 𝑒1 with 𝑅1<‖𝑒1β€–<π‘Ÿ and 𝑒1(𝑑)>0 for π‘‘βˆˆ(0,1).

Remark 4.3. If in (4.19) we have 𝑅1>π‘Ÿ, then (4.1) a nonnegative solution 𝑒2 with π‘Ÿ<‖𝑒2β€–<𝑅2 and 𝑒2(𝑑)>0 for π‘‘βˆˆ(0,1).
It is easy to use Theorem 4.2 and Remark 4.3 to write theorems which guarantee the existence of more than two solutions to (4.1). We state one such result.

Theorem 4.4. Suppose that (4.2)–(4.6) and (4.8) hold. Assume that βˆƒπ‘šβˆˆ{1,2,…} and constants 𝑅𝑖,π‘Ÿπ‘–(𝑖=1,…,π‘š), with π‘Ÿ1>𝑏0, and 0<𝑅1<π‘Ÿ1<𝑅2<π‘Ÿ2<β‹―<π‘…π‘š<π‘Ÿπ‘š.(4.20) In addition suppose for each 𝑖=1,…,π‘š that π‘Ÿπ‘–π‘”ξ€·π‘Ÿπ‘–ξ€Έξ€·π‘Ÿ+β„Žπ‘–ξ€Έ>𝐾20π‘Ž0𝑔1,π›Όβˆ’1(4.21) and 𝑅𝑖𝑔1(πœƒπ›Ό/(π›Όβˆ’1))𝑅𝑖𝑔1𝑅𝑖𝑔1(πœƒπ›Ό/(π›Όβˆ’1))𝑅𝑖+𝑔1ξ€·π‘…π‘–ξ€Έβ„Ž1(πœƒπ›Ό/(π›Όβˆ’1))π‘…π‘–ξ€œ<πœ‡πœƒ1βˆ’πœƒπΊ(𝜎,𝑠)π‘Ž(𝑠)𝑑𝑠(4.22) hold. Then (4.1) has nonnegative solutions 𝑦1,…,π‘¦π‘š with 𝑦𝑖(𝑑)>0 for π‘‘βˆˆ(0,1).

Example 4.5. Consider the boundary value problem 𝐃𝛼0+𝑒𝑒(𝑑)=πœŽβˆ’π‘Ž(𝑑)+𝑒𝑏(𝑑),π‘‘βˆˆ(0,1),3<𝛼≀4,𝑒(0)=𝑒(1)=π‘’ξ…ž(0)=π‘’ξ…ž(1)=0,0<π‘Ž<1<𝑏,(4.23) where 𝜎∈(0,𝜎0) is such that 𝜎0≀12π‘Ž1,(4.24) here π‘Ž1=(π›Όβˆ’1)(π›Όβˆ’2)ξ€œΞ“(𝛼)10π‘ π›Όβˆ’2(1βˆ’π‘ )2ξ€·π‘ π›Όβˆ’2(1βˆ’π‘ )2ξ€Έβˆ’π›Όξ€œπ‘‘π‘ =10𝑠(π›Όβˆ’2)(1βˆ’π‘Ž)(1βˆ’π‘ )2(1βˆ’π‘Ž)𝑑𝑠<∞.(4.25) Then (4.23) has two solutions 𝑒1,𝑒2 with 𝑒1(𝑑)>0,𝑒2(𝑑)>0 for π‘‘βˆˆ(0,1),𝑖=1,2.
To see this we will apply Theorem 4.1 with (here 0<𝑅1<1<𝑅2 will be chosen below)
𝑔(𝑒)=𝑔1(𝑒)=π‘’βˆ’π‘Ž,β„Ž(𝑒)=β„Ž1(𝑒)=𝑒𝑏,π‘Ž(𝑑)=𝜎,𝐾01=1,πœƒ=4.(4.26) Clearly (4.2)–(4.6) and (4.8) hold, and π‘Ž0=(πœŽπ‘Ž1/(π›Όβˆ’1)π‘Ž). Now (4.7) holds with π‘Ÿ=1 since π‘Ÿπ‘”=1(π‘Ÿ)+β„Ž(π‘Ÿ)2β‰₯π‘Ž1𝜎0>𝐾20π‘Ž0(π›Όβˆ’1)π‘Ž=𝐾20π‘Ž0𝑔1ξ‚π›Όβˆ’1.(4.27) Finally notice (4.9) is satisfied for 𝑅1 small and 𝑅2 large since 𝑅𝑖𝑔1ξ€·π‘…π‘–ξ€·β„Žξ€Έξ€½1+1ξ€Έξ€·(πœƒπ›Ό/π›Όβˆ’1)𝑅𝑖/ξ€·g1ξ€Έξ€·(πœƒπ›Ό/π›Όβˆ’1)𝑅𝑖=𝑅𝑖1+π‘Ž1+(π›Όβˆ’1)βˆ’(π‘Ž+𝑏)πœƒπ›Ό(π‘Ž+𝑏)π‘…π‘–π‘Ž+π‘βŸΆ0,(4.28) as 𝑅1β†’0,𝑅2β†’βˆž, since 𝑏>1. Thus all the conditions of Theorem 4.1 are satisfied so existence is guaranteed.

5. Singular Semipositone Problems

In this section we present a new result for the singular semipositone problem: 𝐃𝛼0+𝑒𝑒(𝑑)=πœ‡π‘Ž(𝑑)𝑓(𝑑,𝑒(𝑑)),0<𝑑<1,3<𝛼≀4,(0)=𝑒(1)=π‘’ξ…ž(0)=π‘’ξ…ž(1)=0,(5.1) where πœ‡>0 and nonlinearity 𝑓 may be singular at 𝑒=0.

Before we prove our main result, we first state a result.

Lemma 5.1. Suppose π‘ŽβˆˆπΏ1[0,1] with π‘Ž>0 on (0,1). Then the boundary value problem, 𝐃𝛼0+𝑒𝑒(𝑑)=π‘Ž(𝑑)𝑒(𝑑),0<𝑑<1,3<𝛼≀4,(0)=𝑒(1)=π‘’ξ…ž(0)=π‘’ξ…ž(1)=0,(5.2) has a solution 𝑀 with 𝑑𝑀(𝑑)β‰€π›Όβˆ’2(1βˆ’π‘‘)2πΆπ›Όβˆ’10[]forπ‘‘βˆˆ0,1,(5.3) here 𝐢0=(π›Όβˆ’1)2(π›Όβˆ’2)ξ€œΞ“(𝛼)10π‘Ž(𝑠)𝑒(𝑠)𝑑𝑠.(5.4) In fact, from Lemma 3.1, (5.2) has solution ξ€œπ‘€(𝑑)=10𝐺(𝑑,𝑠)π‘Ž(𝑠)𝑒(𝑠)𝑑𝑠.(5.5) According to Lemma 3.2, we have ξ€œπ‘€(𝑑)≀10(π›Όβˆ’1)(π›Όβˆ’2)π‘‘π›Όβˆ’2(1βˆ’π‘‘)2𝑑Γ(𝛼)π‘Ž(𝑠)𝑒(𝑠)𝑑𝑠=π›Όβˆ’2(1βˆ’π‘‘)2πΆπ›Όβˆ’10.(5.6) The above Lemma together with Theorem 2.5 establish our main result.

Theorem 5.2. Suppose that the following conditions are satisfied. π‘žβˆˆπΆ(0,1)∩𝐿1[]0,1π‘€π‘–π‘‘β„Žπ‘ž>π‘œπ‘›(0,1).(5.7)ξƒ―[]π‘“βˆΆ0,1Γ—(0,∞)βŸΆπ‘π‘–π‘ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ π‘Žπ‘›π‘‘π‘‘β„Žπ‘’π‘Ÿπ‘’π‘’π‘₯π‘–π‘ π‘‘π‘ π‘Žπ‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›π‘’βˆˆπΆ((0,1),(0,+∞))π‘€π‘–π‘‘β„Žπ‘“(𝑑,𝑒)+𝑒(𝑑)β‰₯π‘“π‘œπ‘Ÿ(𝑑,𝑒)∈(0,1)Γ—(0,∞),(5.8)⎧βŽͺβŽͺ⎨βŽͺβŽͺβŽ©π‘“βˆ—[][β„Ž(𝑑,𝑒)=𝑓(𝑑,𝑒)+𝑒(𝑑)≀𝑔(𝑒)+β„Ž(𝑒)π‘œπ‘›0,1Γ—(0,∞)π‘€π‘–π‘‘β„Žπ‘”>0π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ π‘Žπ‘›π‘‘π‘›π‘œπ‘›π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘”π‘œπ‘›(0,∞),β„Žβ‰₯0π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ π‘œπ‘›0,∞)π‘Žπ‘›π‘‘π‘”π‘›π‘œπ‘›π‘‘π‘’π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘”π‘œπ‘›(0,∞),(5.9)βˆƒπΎ0π‘€π‘–π‘‘β„Žπ‘”(π‘₯𝑦)≀𝐾0𝑔(π‘₯)𝑔(𝑦)βˆ€π‘₯>0,𝑦>0,(5.10)π‘Ž0=∫10𝑠𝐺(𝜎,𝑠)π‘ž(𝑠)π‘”π›Όβˆ’2(1βˆ’π‘ )2𝑑𝑠<∞,(5.11)βˆƒπ‘Ÿ>πœ‡πΆ0π‘Ÿπ‘€π‘–π‘‘β„Žπ‘”ξ€·ξ€·π‘Ÿβˆ’πœ‡πΆ0ξ€Έξ€Έ/(π›Όβˆ’1){1+(β„Ž(π‘Ÿ)/𝑔(π‘Ÿ))}β‰₯πœ‡πΎ0π‘Ž0,(5.12)⎧βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺ⎩1π‘‡β„Žπ‘’π‘Ÿπ‘’π‘’π‘₯𝑖𝑠𝑑𝑠0<πœƒ<2(π‘β„Žπ‘œπ‘œπ‘ π‘’π‘Žπ‘›π‘‘π‘“π‘–π‘₯𝑖𝑑)π‘Žπ‘›π‘‘π‘Žπ‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ ,π‘›π‘œπ‘›π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘”π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›π‘”1∢(0,∞)⟢(0,∞),π‘Žπ‘›π‘‘π‘Žπ‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘ π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘›β„Ž1∢[β„Ž0,∞)⟢(0,∞)π‘€π‘–π‘‘β„Ž1𝑔1π‘›π‘œπ‘›π‘‘π‘’π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘”π‘œπ‘›(0,∞)π‘Žπ‘›π‘‘π‘€π‘–π‘‘β„Žπ‘“(𝑑,𝑒)+𝑒(𝑑)β‰₯𝑔1(𝑒)+β„Ž1[](𝑒)π‘“π‘œπ‘Ÿ(𝑑,𝑒)βˆˆπœƒ,1βˆ’πœƒΓ—(0,∞),(5.13)π‘Žπ‘›π‘‘βˆƒπ‘…>π‘Ÿπ‘€π‘–π‘‘β„Žπ‘…π‘”1((πœ–πœƒπ›Ό/(π›Όβˆ’1))𝑅)𝑔1(𝑅)𝑔1((πœ–πœƒπ›Ό/(π›Όβˆ’1))𝑅)+𝑔1(𝑅)β„Ž1((πœ–πœƒπ›Όξ€œ/(π›Όβˆ’1))𝑅)β‰€πœ‡πœƒ1βˆ’πœƒπΊ(𝜎,𝑠)π‘ž(𝑠)𝑑𝑠,(5.14) here πœ–>0 is any constant (choose and fix it) so that 1βˆ’(πœ‡πΆ0/𝑅)β‰₯πœ– (note πœ– exists since 𝑅>π‘Ÿ>πœ‡πΆ0 in fact we can have πœ–=1βˆ’(πœ‡πΆ0/π‘Ÿ) ) and 𝐺(𝑑,𝑠) is Green's function and ξ€œπœƒ1βˆ’πœƒπΊ(𝜎,𝑠)π‘Ž(𝑠)𝑑𝑠=supπ‘‘βˆˆ[0,1]ξ€œπœƒ1βˆ’πœƒπΊ(𝑑,𝑠)π‘Ž(𝑠)𝑑𝑠.(5.15) Then (5.1) has a solution π‘¦βˆˆπΆ[0,1]∩𝐢2(0,1) with 𝑦(𝑑)>0 for π‘‘βˆˆ(0,1).

Proof. To show that (5.1) has a nonnegative solution we will look at the boundary value problem 𝐃𝛼0+𝑦(𝑑)=πœ‡π‘ž(𝑑)π‘“βˆ—(𝑑,𝑦(𝑑)βˆ’πœ™(𝑑)),0<𝑑<1,3<𝛼≀4,𝑦(0)=𝑦(1)=π‘¦ξ…ž(0)=π‘¦ξ…ž(1)=0,(5.16) where πœ™(𝑑)=πœ‡π‘€π‘€(𝑑),0≀𝑑≀1,(5.17) (𝑀 is as in Lemma 5.1).
We will show, using Theorem 2.5, that there exists a solution 𝑦1 to (5.16) with 𝑦1(𝑑)>πœ™(𝑑) for π‘‘βˆˆ(0,1). If this is true then 𝑒(𝑑)=𝑦1(𝑑)βˆ’πœ™(𝑑),0≀𝑑≀1 is a nonnegative solution (positive on (0,1)) of (5.1), since
𝐃𝛼0+𝑒(𝑑)=𝐃𝛼0+𝑦1(𝑑)βˆ’πƒπ›Ό0+πœ™(𝑑)=βˆ’πœ‡π‘ž(𝑑)π‘“βˆ—ξ€·π‘‘,𝑦1𝑓(𝑑)βˆ’πœ™(𝑑)+πœ‡π‘ž(𝑑)𝑒(𝑑)=βˆ’πœ‡π‘ž(𝑑)𝑑,𝑦1(𝑑)βˆ’πœ™(𝑑)+𝑒(𝑑)+πœ‡π‘ž(𝑑)𝑒(𝑑)=βˆ’πœ‡π‘ž(𝑑)𝑓𝑑,𝑦1ξ€Έ(𝑑)βˆ’πœ™(𝑑)=βˆ’πœ‡π‘ž(𝑑)𝑓(𝑑,𝑒(𝑑)),0<𝑑<1.(5.18) As a result, we will concentrate our study on (5.16). Let 𝐸,𝐾 as in Section 2, and let Ξ©1={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<π‘Ÿ},Ξ©2={π‘’βˆˆπΈβˆΆβ€–π‘’β€–<𝑅}.(5.19) Next let 𝐴∢𝐾∩(Ξ©2⧡Ω1)→𝐸  be defined by ξ€œ(𝐴𝑦)(𝑑)=πœ‡10𝐺(𝑑,𝑠)π‘ž(𝑠)π‘“βˆ—(𝑠,𝑦(𝑠)βˆ’πœ™(𝑠))𝑑𝑠,0≀𝑑≀1.(5.20) In addition, standard argument shows that 𝐴(𝑃)βŠ‚π‘ƒ and 𝐴 is completely continuous.
We now show
‖𝐴𝑦‖≀‖𝑦‖forπΎβˆ©πœ•Ξ©1.(5.21) To see this, let π‘¦βˆˆπΎβˆ©πœ•Ξ©1. Then ‖𝑦‖=‖𝑦‖[0,1]=π‘Ÿ and 𝑦(𝑑)β‰₯(π‘‘π›Όβˆ’2(1βˆ’π‘‘)2/(π›Όβˆ’1))π‘Ÿ for π‘‘βˆˆ[0,1]. Now for π‘‘βˆˆ(0,1), the Lemma 5.1 implies 𝑑𝑦(𝑑)βˆ’πœ™(𝑑)β‰₯π›Όβˆ’2(1βˆ’π‘‘)2π‘‘π›Όβˆ’1π‘Ÿβˆ’πœ‡π›Όβˆ’2(1βˆ’π‘‘)2πΆπ›Όβˆ’10β‰₯π‘‘π›Όβˆ’2(1βˆ’π‘‘)2ξ€·π›Όβˆ’1π‘Ÿβˆ’πœ‡πΆ0ξ€Έ>0,(5.22) so for π‘‘βˆˆ[0,1] we have ξ€œ(𝐴𝑦)(𝑑)=πœ‡10𝐺(𝑑,𝑠)π‘Ž(𝑠)π‘“βˆ—ξ€œ(𝑠,𝑦(𝑠)βˆ’πœ™(𝑠))π‘‘π‘ β‰€πœ‡10π‘ž[𝑔]ξ€œ(𝑠)π‘Ž(𝑠)(𝑦(𝑠)βˆ’πœ™(𝑠))+β„Ž(𝑦(𝑠)βˆ’πœ™(𝑠))𝑑𝑠=πœ‡10π‘žξ‚»(𝑠)π‘Ž(𝑠)𝑔(𝑦(𝑠)βˆ’πœ™(𝑠))1+β„Ž(𝑦(𝑠)βˆ’πœ™(𝑠))ξ‚Όξ€œπ‘”(𝑦(𝑠)βˆ’πœ™(𝑠))π‘‘π‘ β‰€πœ‡10ξ‚΅π‘ π‘ž(𝑠)π‘Ž(𝑠)π‘”π›Όβˆ’2(1βˆ’π‘ )2ξ€·π›Όβˆ’1π‘Ÿβˆ’πœ‡πΆ0ξ€Έξ‚Άξ‚»1+β„Ž(π‘Ÿ)𝑔(π‘Ÿ)π‘‘π‘ β‰€πœ‡πΎ0π‘”ξ‚΅π‘Ÿβˆ’πœ‡πΆ0π›Όβˆ’1ξ‚Άξ‚»1+β„Ž(π‘Ÿ)ξ‚Όξ€œπ‘”(π‘Ÿ)10ξ€·π‘ π‘ž(𝑠)π‘Ž(𝑠)π‘”π›Όβˆ’2(1βˆ’π‘ )2𝑑𝑠=πœ‡πΎ0π‘Ž0π‘”ξ‚΅π‘Ÿβˆ’πœ‡πΆ0π›Όβˆ’1ξ‚Άξ‚»1+β„Ž{π‘Ÿ}ξ‚Ό.𝑔{π‘Ÿ}(5.23) This together with (5.12) yields ‖𝐴𝑦‖=‖𝐴𝑦‖[0,1]β‰€π‘Ÿ=‖𝑦‖,(5.24) so (5.21) is satisfied.
Next we show
‖𝐴𝑦‖β‰₯‖𝑦‖forπΎβˆ©πœ•Ξ©2.(5.25) To see this let π‘¦βˆˆπΎβˆ©πœ•Ξ©2 so ‖𝑦‖=‖𝑦‖[0,1]=𝑅 and 𝑦(𝑑)β‰₯(π‘‘π›Όβˆ’2(1βˆ’π‘‘)2/(π›Όβˆ’1))𝑅 for π‘‘βˆˆ[0,1]. Also for π‘‘βˆˆ[0,1] we have 𝑑𝑦(𝑑)βˆ’πœ™(𝑑)=𝑦(𝑑)βˆ’πœ‡π‘€(𝑑)β‰₯π›Όβˆ’2(1βˆ’π‘‘)2π›Όβˆ’1π‘…βˆ’πœ‡πΆ0π‘‘π›Όβˆ’2(1βˆ’π‘‘)2β‰₯π‘‘π›Όβˆ’1π›Όβˆ’2(1βˆ’π‘‘)2π‘…ξ‚΅π›Όβˆ’11βˆ’πœ‡πΆ0𝑅β‰₯πœ–π‘‘π›Όβˆ’2(1βˆ’π‘‘)2π›Όβˆ’1𝑅.(5.26) As a result 𝑦(𝑑)βˆ’πœ™(𝑑)β‰₯πœ–πœƒπ›Ό[]π›Όβˆ’1𝑅forπ‘‘βˆˆπœƒ,1βˆ’πœƒ.(5.27) We have ξ€œ(𝐴𝑦)(𝜎)=πœ‡10𝐺(𝜎,𝑠)π‘ž(𝑠)π‘“βˆ—ξ€œ(𝑠,𝑦(𝑠)βˆ’πœ™(𝑠))𝑑𝑠β‰₯πœ‡01βˆ’πœƒπΊξ€Ίπ‘”(𝜎,𝑠)π‘ž(𝑠)1(𝑦(𝑠)βˆ’πœ™(𝑠))+β„Ž1ξ€»ξ€œ(𝑦(𝑠)βˆ’πœ™(𝑠))𝑑𝑠=πœ‡01βˆ’πœƒπΊ(𝜎,𝑠)π‘ž(𝑠)𝑔1ξ‚»β„Ž(𝑦(𝑠)βˆ’πœ™(𝑠))1+1(𝑦(𝑠)βˆ’πœ™(𝑠))𝑔1ξ‚Ό(𝑦(𝑠)βˆ’πœ™(𝑠))𝑑𝑠β‰₯πœ‡π‘”1ξ€œ(𝑅)01βˆ’πœƒξ‚»β„ŽπΊ(𝜎,𝑠)π‘ž(𝑠)1+1((πœ–πœƒπ›Ό/(π›Όβˆ’1))𝑅)𝑔1((πœ–πœƒπ›Όξ‚Ό/(π›Όβˆ’1))𝑅)𝑑𝑠.(5.28) This together with (5.14) yields (𝐴𝑦)(𝜎)β‰₯𝑅=‖𝑦‖.(5.29) Thus ‖𝐴𝑦‖β‰₯‖𝑦‖, so (5.25) is held.
Now Theorem 2.5 implies that 𝐴 has a fixed-point π‘¦βˆˆπΎβˆ©(Ξ©2⧡Ω1), that is, π‘Ÿβ‰€β€–π‘¦β€–=‖𝑦‖[0.1]≀𝑅 and 𝑦(𝑑)β‰₯π‘‘π›Όβˆ’2(1βˆ’π‘‘)2π‘Ÿ for π‘‘βˆˆ[0,1]. Thus 𝑦(𝑑) is a solution of (5.16) with 𝑦(𝑑)>πœ™(𝑑) for π‘‘βˆˆ(0,1). Thus (5.1) has a positive solution 𝑒(𝑑)=𝑦(𝑑)βˆ’πœ™(𝑑)> for π‘‘βˆˆ(0,1).

Example 5.3. Consider the boundary value problem 𝐃𝛼0+𝑦𝑦(𝑑)=πœ‡βˆ’π‘Ž(𝑑)+𝑦𝑏(𝑑)βˆ’1,0<𝑑<1,3<𝛼≀4,𝑦(0)=𝑦(1)=π‘¦ξ…ž(0)=π‘¦ξ…ž(1)=0,0<π‘Ž<1<𝑏,(5.30) where πœ‡βˆˆ(0,πœ‡0) is such that (π›Όβˆ’1)2(π›Όβˆ’2)πœ‡0ξ€·2Ξ“(𝛼)+(π›Όβˆ’1)2πœ‡0π‘Ž0ξ€Έ1/𝛼≀1,(5.31) here π‘Ž0=ξ€œ10ξ€·π‘ π‘ž(𝑠)π‘Ž(𝑠)π‘”π›Όβˆ’2(1βˆ’π‘ )2ξ€Έξ€œπ‘‘π‘ =10𝑠1βˆ’π‘Ž(π›Όβˆ’1)(1βˆ’π‘ )(π›Όβˆ’1)𝑑𝑠<∞.(5.32)
Then (5.30) has a solution 𝑦 with 𝑦(𝑑)>0 for π‘‘βˆˆ(0,1).
To see this we will apply Theorem 5.2 with (here 𝑅>1 will be chosen later, in fact here we choose 𝑅>1 so that πœ–=1/2 works, i.e., we choose 𝑅 so that 1βˆ’(πœ‡/(2Ξ“(𝛼)𝑅))β‰₯1/2),
𝑔(𝑦)=𝑔1(𝑦)=π‘¦βˆ’π‘Ž,β„Ž(𝑦)=β„Ž1(𝑦)=𝑦𝑏,π‘Ž(𝑑)=1,𝑒(𝑑)=π‘‘βˆ’1/2,a𝐾01=1,πœ–=21,πœƒ=4,𝐢0=(π›Όβˆ’1)2(π›Όβˆ’2)ξ€œΞ“(𝛼)10π‘ βˆ’1/2𝑑𝑠=(π›Όβˆ’2)2(π›Όβˆ’2).2Ξ“(𝛼)(5.33) Clearly (5.7)–(5.11) and (5.13) hold. Now (5.12) holds with π‘Ÿ=1 since πœ‡πΎ0π‘Ž0=πœ‡π‘Ž0<πœ‡0π‘Ž0≀12ξ‚΅1βˆ’πœ‡0𝐢0ξ‚Άπ›Όβˆ’1π‘Žβ‰€12ξ‚΅1βˆ’πœ‡π‘0ξ‚Άπ›Όβˆ’1π‘Ž=π‘Ÿ{1+(β„Ž(π‘Ÿ)/𝑔(π‘Ÿ))}π‘”ξ€·ξ€·π‘Ÿβˆ’πœ‡π‘€πΆ0ξ€Έξ€Έ,/(π›Όβˆ’1)(5.34) from (5.31). Finally notice (5.14) is satisfied for 𝑅 large since 𝑅𝑔1(πœ–πœƒπ›Ό/(π›Όβˆ’1))𝑅𝑔1(𝑅)𝑔1((πœ–πœƒπ›Ό/(π›Όβˆ’1))𝑅)+𝑔1(𝑅)β„Ž1((πœ–πœƒπ›Ό=𝑅/(π›Όβˆ’1))𝑅)1+π‘Ž1+(π›Όβˆ’1)βˆ’(π‘Ž+𝑏)ξ€Ίπœ–πœƒπ›Όξ€»π‘Ž+π‘π‘…π‘Ž+π‘βŸΆ0,(5.35) as π‘…β†’βˆž, since 𝑏>1. Thus all the conditions of Theorem 5.2 are satisfied so existence is guaranteed.

Acknowledgments

This paper is supported by Key Subject of Chinese Ministry of Education (no. 109051) and Scientific Research Fund of Heilongjiang Provincial Education Department (no. 11544032).

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