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Abstract and Applied Analysis
Volume 2011 (2011), Article ID 986575, 20 pages
http://dx.doi.org/10.1155/2011/986575
Research Article

Existence of Positive Solutions for m-Point Boundary Value Problem for Nonlinear Fractional Differential Equation

1Department of Mathematics, Faculty of Arts and Sciences, College of Education, P.O. Box 3771, Qassim-Unizah 51911, Saudi Arabia
2Department of Mathematics, Sciences Faculty for Girls, King Abdulaziz University, P.O. Box 30903, Jeddah 21487, Saudi Arabia

Received 23 January 2011; Revised 2 April 2011; Accepted 17 April 2011

Academic Editor: Paul Eloe

Copyright © 2011 Moustafa El-Shahed and Wafa M. Shammakh. 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 investigate an m-point boundary value problem for nonlinear fractional differential equations. The associated Green function for the boundary value problem is given at first, and some useful properties of the Green function are obtained. By using the fixed point theorems of cone expansion and compression of norm type and Leggett-Williams fixed point theorem, the existence of multiple positive solutions is obtained.

1. Introduction

In recent years, the existence of positive solutions multipoint boundary value problems of fractional order differential equations has been studied by many authors using various methods (see [17]).

The study of multipoint boundary value problems for linear second-order ordinary differential equations was initiated by II'in and Moiseev [8, 9].

Since then, nonlinear multipoint boundary value problems have been studied by several authors (see [1014]). Recently, in [15], the authors have studied the existence of at least one positive solution for the following 𝑛th-order three-point boundary value problem:𝑢(𝑛)[],(𝑡)+(𝑡)𝑓(𝑡,𝑢(𝑡))=0,𝑡𝑎,𝑏𝑢(𝑎)=𝛼𝑢(𝜂),𝑢(𝑎)=𝑢(𝑎)==𝑢(𝑛2)(𝑎)=0,𝑢(𝑏)=𝛽𝑢(𝜂),(1.1) where 𝑎<𝜂<𝑏,0𝛼<1,𝑓𝐶([𝑎,𝑏]×[0,),[0,)) and 𝐶([𝑎,𝑏]×[0,)) may be singular at 𝑡=𝑎 and 𝑡=𝑏.

Goodrich [16] considered the BVP for thehigher-dimensional fractional differential equation as follows:𝐷𝜈0+𝑦𝑦(𝑡)=𝑓(𝑡,𝑦(𝑡)),0<𝑡<1,𝑛1<𝜈𝑛,(𝑖)(𝐷0)=0,0𝑖𝑛2,𝛼0+𝑦(𝑡)𝑡=1=0,1𝛼𝑛2,(1.2) and a Harnack-like inequality associated with the Green function related to the above problem is obtained improving the results in [17].

Motivated by the aforementioned results and techniques in coping with those boundary value problems of fractional differential equations, we then turn to investigate the existence and multiplicity of positive solutions for the following BVP:𝐶𝐷𝛼𝑎+𝑢𝑢(𝑡)+𝑓(𝑡,𝑢(𝑡))=0,𝑎𝑡𝑏,𝑛1𝛼<𝑛,𝑛>2,(1.3)(𝑎)=𝑚2𝑖=1𝛽𝑖𝑢𝜂𝑖,𝑢(𝑎)=𝑢(𝑎)==𝑢(𝑛1)(𝑎)=0,𝑢(𝑏)=𝑚2𝑖=1𝛾𝑖𝑢𝜂𝑖,(1.4) where 𝑎<𝜂1<𝜂2<<𝜂𝑚2<𝑏,𝑚2𝑖=1𝛽𝑖<1,𝑚2𝑖=1𝛾𝑖<1 and 𝐶𝐷𝛼𝑎+ are the Caputo fractional derivative.

In this paper, we study the existence of at least one positive solution, existence of two positive solutions associated with the BVP (1.3)-(1.4) by applying the fixed point theorems of cone expansion and compression of norm type, and the existence of at least three positive solutions for BVP (1.3)-(1.4) by using Leggett-Williams fixed point theorem.

The rest of the paper is organized as follows. In Section 2, we introduce some basic definitions and preliminaries used later. In Section 3, the existence of multipoint boundary value problem (1.3)-(1.4) will be discussed.

2. Preliminaries

In this section, we introduce definitions and preliminary facts which are used throughout this paper.

Definition 2.1 (see [18]). For a function 𝑦(𝑎,)𝑅, the Caputo derivative of fractional order 𝛼>0 is defined as 𝐶𝐷𝛼𝑎+1𝑦(𝑡)=Γ(𝑛𝛼)𝑡𝑎(𝑡𝑠)𝑛𝛼1𝑦(𝑛)(𝑠)𝑑𝑠,𝑛1<𝛼𝑛.(2.1)

Definition 2.2 (see [18]). The standard Riemann-Liouville fractional derivative of order 𝛼>0 of a continuous function 𝑦(𝑎,)𝑅 is given by 𝐷𝛼𝑎+1𝑦(𝑡)=𝑑Γ(𝑛𝛼)𝑑𝑡𝑛𝑡𝑎(𝑡𝑠)𝑛𝛼1𝑦(𝑠)𝑑𝑠,(2.2) where 𝑛=[𝛼]+1, provided that the integral on the right-hand side converges.

Definition 2.3 (see [18]). The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑦(𝑎,)𝑅 is given by 𝐼𝛼𝑎+1𝑦(𝑡)=Γ(𝛼)𝑡𝑎(𝑡𝑠)𝛼1𝑦(𝑠)𝑑𝑠(2.3) provided that the integral on the right-hand side converges.

Definition 2.4 (see [19]). Let 𝐸 be a real Banach space. A nonempty closed convex set 𝐾𝐸 is called cone of 𝐸 if it satisfies the following conditions:(1)𝑥𝐾,𝜎0implies𝜎𝑥𝐾; (2)𝑥𝐾,𝑥𝐾implies𝑥=0.

Definition 2.5. An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.

Theorem 2.6 (see [20]). Let 𝐸 be a Banach space and 𝐾𝐸 is a cone in 𝐸. Assume that Ω1 and Ω2 are open subsets of 𝐸 with 0Ω1 and Ω1Ω2. Let 𝑇𝐾(Ω2Ω1)𝐾 be completely continuous operator. In addition, suppose either (i)𝑇𝑢𝑢,forall𝑢𝐾𝜕Ω1,and𝑇𝑢𝑢,forall𝑢𝐾𝜕Ω2or(ii)𝑇𝑢𝑢,forall𝑢𝐾𝜕Ω2,and𝑇𝑢𝑢,forall𝑢𝐾𝜕Ω1holds. Then, 𝑇 has a fixed point in 𝐾(Ω2Ω1).

Lemma 2.7. For 𝛼>0, the general solution of the fractional differential equation 𝐶𝐷𝛼𝑎+𝑢(𝑡)=0 is given by 𝑢(𝑡)=𝑐0+𝑐1(𝑡𝑎)+𝑐2(𝑡𝑎)2++𝑐𝑛1(𝑡𝑎)𝑛1,(2.4) where 𝑐𝑖𝑅,𝑖=0,1,2,,𝑛1.

Remark 2.8 (see [18]). In view of Lemma 2.7, it follows that 𝐼𝛼𝑎+𝐶𝐷𝛼𝑎+𝑢(𝑡)=𝑢(𝑡)+𝑐0+𝑐1(𝑡𝑎)+𝑐2(𝑡𝑎)2++𝑐𝑛1(𝑡𝑎)𝑛1,(2.5) for some 𝑐𝑖𝑅,𝑖=0,1,2,,𝑛1.

Definition 2.9. The map 𝜃 is said to be a nonnegative continuous concave functional on a cone 𝑃 of a real Banach space 𝐸 provided that 𝜃𝑃[0,) is continuous and 𝜃(𝜆𝑥+(1𝜆)𝑦)𝜆𝜃(𝑥)+(1𝜆)𝜃(𝑦),(2.6) for all 𝑥,𝑦𝑃,0𝜆1.

Lemma 2.10 (see [21]). Let 𝑃 be a cone in a real Banach space 𝐸,𝑃𝑐={𝑥𝑃𝑥<𝑐},𝜃 is a nonnegative continuous concave functional on 𝑃 such that 𝜃(𝑥)𝑥, for all 𝑥𝑃𝑐, and 𝑃(𝜃,𝑏,𝑑)={𝑥𝑃𝑏𝜃(𝑥),𝑥𝑑}. Suppose that 𝑇𝑃𝑐𝑃𝑐 is completely continuous and there exist positive constants 0<𝑎<𝑏<𝑑𝑐 such that (𝐶1){𝑥𝑃(𝜃,𝑏,𝑑)𝜃(𝑥)>𝑏}𝜙and𝜃(𝑥)>𝑏for𝑥𝑃(𝜃,𝑏,𝑑), (𝐶2)𝑇𝑥<𝑎for𝑥𝑃𝑎, (𝐶3)𝜃(𝑇𝑥)>𝑏for𝑥𝑃(𝜃,𝑏,𝑑)with𝑇𝑥>𝑑, then 𝑇 has at least three fixed points 𝑥1,𝑥2, and 𝑥3 with 𝑥1𝑥<𝑎,𝑏<𝜃2𝑥,𝑎<3𝑥with𝜃3<𝑏.(2.7)

Lemma 2.11. For a given 𝑦(𝑡)𝐶[𝑎,𝑏] and 𝑛1𝛼<𝑛, the unique solution of the boundary value problem 𝐶𝐷𝛼𝑎+𝑢𝑢(𝑡)+𝑦(𝑡)=0,𝑎𝑡𝑏,𝑛1𝛼<𝑛,𝑛>2,𝑛𝑁,(2.8)(𝑎)=𝑚2𝑖=1𝛽𝑖𝑢𝜂𝑖,𝑢(𝑎)=𝑢(𝑎)==𝑢(𝑛1)(𝑎)=0,𝑢(𝑏)=𝑚2𝑖=1𝛾𝑖𝑢𝜂𝑖,(2.9) is given by 𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠+𝑏𝑎𝐻𝑡,𝑠;𝜂1,,𝜂𝑚2𝑦(𝑠)𝑑𝑠,(2.10) where 1𝐺(𝑡,𝑠)=Γ(𝛼)(𝑏𝑠)𝛼1(𝑡𝑠)𝛼1,𝑎𝑠𝑡𝑏,(𝑏𝑠)𝛼1𝐻,𝑎𝑡𝑠𝑏,𝑡,𝑠;𝜂1,,𝜂𝑚2=𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝜂𝑖𝑠𝛼1𝛿2+Γ(𝛼)𝜇(𝑡)𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑠𝛼2𝛿1𝛿2Γ(𝛼1),𝑎𝑠𝜂𝑖,𝑖=1,2,,𝑚2,𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝛿2Γ(𝛼),𝜂𝑖𝛿𝑠𝑏,𝑖=1,2,,𝑚2,1=1𝑚2𝑖=1𝛽𝑖,𝛿2=1𝑚2𝑖=1𝛾𝑖,𝜇(𝑡)=𝑏𝑚2𝑖=1𝛾𝑖𝜂𝑖𝛿2𝑡.(2.11)

Proof. Using Remark 2.8, for arbitrary constants 𝑐𝑖𝑅,𝑖=0,1,2,,𝑛1, we have 𝑢(𝑡)=1Γ(𝛼)𝑡𝑎(𝑡𝑠)𝛼1𝑦(𝑠)𝑑𝑠+𝑐0+𝑐1(𝑡𝑎)+𝑐2(𝑡𝑎)2++𝑐𝑛1(𝑡𝑎)𝑛1=𝐼𝛼𝑎+𝑦(𝑡)+𝑐0+𝑐1(𝑡𝑎)+𝑐2(𝑡𝑎)2++𝑐𝑛1(𝑡𝑎)𝑛1.(2.12) In view of the relations 𝐶𝐷𝛼𝑎+𝐼𝛼𝑎+𝑢(𝑡)=𝑢(𝑡) and 𝐼𝛼𝑎+𝐼𝛽𝑎+𝑢(𝑡)=𝐼𝑎𝛼+𝛽+𝑢(𝑡) for 𝛼,𝛽>0, we obtain 𝑢(𝑡)=𝐼𝑎𝛼1+𝑦(𝑡)+𝑐1+2𝑐2(𝑡𝑎)++(𝑛1)𝑐𝑛1(𝑡𝑎)𝑛2,𝑢(𝑡)=𝐼𝑎𝛼2+𝑦(𝑡)+2𝑐2++(𝑛1)(𝑛2)𝑐𝑛1(𝑡𝑎)𝑛3,𝑢(𝑛1)(𝑡)=𝐼𝑎𝛼𝑛+1+𝑦(𝑡)+(𝑛1)!𝑐𝑛1.(2.13) Applying the boundary conditions (2.9), we find that 𝑐2=𝑐3==𝑐𝑛1=0,1𝑚2𝑖=1𝛽𝑖𝑐1=𝑚2𝑖=1𝛽𝑖𝐼𝑎𝛼1+𝑦𝜂𝑖,(2.14) then 𝑐1=𝑚2𝑖=1𝛽𝑖𝐼𝑎𝛼1+𝑦(𝜂𝑖)/𝛿1, and 𝑐0=𝐼𝛼𝑎+𝑦(𝑏)1𝑚2𝑖=1𝛾𝑖𝑚2𝑖=1𝛾𝑖𝐼𝛼𝑎+𝑦𝜂𝑖1𝑚2𝑖=1𝛾𝑖+(𝑏𝑎)𝑚2𝑖=1𝛾𝑖𝜂𝑖𝑎𝑚2𝑖=1𝛽𝑖𝐼𝑎𝛼1+𝑦𝜂𝑖1𝑚2𝑖=1𝛽𝑖1𝑚2𝑖=1𝛾𝑖=𝐼𝛼𝑎+𝑦(𝑏)𝛿2𝑚2𝑖=1𝛾𝑖𝐼𝛼𝑎+𝑦𝜂𝑖𝛿2+(𝑏𝑎)𝑚2𝑖=1𝛾𝑖𝜂𝑖𝑎𝑚2𝑖=1𝛽𝑖𝐼𝑎𝛼1+𝑦𝜂𝑖𝛿1𝛿2.(2.15) Substituting the values of the constants 𝑐𝑖,𝑖=0,1,2,,𝑛1, in (2.12), we obtain 𝑢(𝑡)=𝐼𝛼𝑎+𝐼𝑦(𝑡)+𝛼𝑎+𝑦(𝑏)𝛿2𝑚2𝑖=1𝛾𝑖𝐼𝛼𝑎+𝑦𝜂𝑖𝛿2+(𝑏𝑎)𝑚2𝑖=1𝛾𝑖𝜂𝑖𝑎𝑚2𝑖=1𝛽𝑖𝐼𝑎𝛼1+𝑦𝜂𝑖𝛿1𝛿2𝑚2𝑖=1𝛽𝑖𝐼𝑎𝛼1+𝑦𝜂𝑖𝛿1(𝑡𝑎)=𝐼𝛼𝑎+𝐼𝑦(𝑡)+𝛼𝑎+𝑦(𝑏)𝛿2𝑚2𝑖=1𝛾𝑖𝐼𝛼𝑎+𝑦𝜂𝑖𝛿2+𝜇(𝑡)𝛿1𝛿2𝑚2𝑖=1𝛽𝑖𝐼𝑎𝛼1+𝑦𝜂𝑖=𝐼𝛼𝑎+𝑦(𝑡)+𝐼𝛼𝑎+𝑦(𝑏)+1𝛿2𝛿2𝐼𝛼𝑎+𝑦(𝑏)𝑚2𝑖=1𝛾𝑖𝐼𝛼𝑎+𝑦𝜂𝑖𝛿2+𝜇(𝑡)𝛿1𝛿2𝑚2𝑖=1𝛽𝑖𝐼𝑎𝛼1+𝑦𝜂𝑖=𝑡𝑎(𝑡𝑠)𝛼1Γ(𝛼)𝑦(𝑠)𝑑𝑠+𝑏𝑎(𝑏𝑠)𝛼1Γ(𝛼)𝑦(𝑠)𝑑𝑠+𝑚2𝑖=1𝛾𝑖𝛿2𝑏𝑎(𝑏𝑠)𝛼1Γ(𝛼)𝑦(𝑠)𝑑𝑠𝑚2𝑖=1𝛾𝑖𝛿2𝜂𝑖𝑎𝜂𝑖𝑠𝛼1Γ(𝛼)𝑦(𝑠)𝑑𝑠+𝜇(𝑡)𝛿1𝛿2𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑎𝜂𝑖𝑠𝛼2Γ(𝛼1)𝑦(𝑠)𝑑𝑠=𝑡𝑎(𝑡𝑠)𝛼1𝑦Γ(𝛼)(𝑠)𝑑𝑠+𝑏𝑎(𝑏𝑠)𝛼1𝑦+Γ(𝛼)(𝑠)𝑑𝑠𝑚2𝑖=1𝛾𝑖𝛿2𝜂𝑖𝑎(𝑏𝑠)𝛼1𝜂𝑖𝑠𝛼1Γ+(𝛼)𝑦(𝑠)𝑑𝑠𝑚2𝑖=1𝛾𝑖𝛿2𝑏𝜂𝑖(𝑏𝑠)𝛼1Γ(𝛼)𝑦(𝑠)𝑑𝑠+𝜇(𝑡)𝛿1𝛿2𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑎𝜂𝑖𝑠𝛼2Γ(𝛼1)𝑦(𝑠)𝑑𝑠.(2.16)

Lemma 2.12. 𝜇(𝑡)=(𝑏𝑚2𝑖=1𝛾𝑖𝜂𝑖)𝛿2𝑡0, for 𝑡,𝜂𝑖[𝑎,𝑏],𝑖=1,2,,𝑚2.

Proof. Wehave𝜇=𝑏𝑚2𝑖=1𝛾𝑖𝜂𝑖𝛿2𝑡𝑏𝑚2𝑖=1𝛾𝑖𝜂𝑖1𝑚2𝑖=1𝛾𝑖𝑏𝑚2𝑖=1𝛾𝑖𝑏𝜂𝑖>0.(2.17)

Lemma 2.13. The functions 𝐺(𝑡,𝑠),𝐻(𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2) defined by (2.11) satisfy (i)𝐺(𝑡,𝑠)0,𝐻(𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2)0,forall𝑡,𝑠[𝑎,𝑏], (ii)min𝜏1𝑡𝜏2𝐺(𝑡,𝑠)𝜏0max𝑎𝑡𝑏𝐺(𝑡,𝑠)=𝜏0𝐺(𝑠,𝑠), for all 𝑡,𝑠(𝑎,𝑏),𝑎<𝜏1<𝜏2<𝑏,𝜏0=min𝜏1𝑡𝜏2𝜑(𝑡)=(𝑏𝜏2)/(𝑏𝑎),(iii)𝑁2𝑞(𝑠)𝐻(𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2)𝑁1𝑞(𝑠), where𝑞(𝑠)=(𝑏𝑠)𝛼2𝛿1𝛿2Γ(𝛼),𝑁1𝛿=(𝛼1)1(𝑏𝑎)𝑚2𝑖=1𝛾𝑖+𝑏𝑚2𝑖=1𝛽𝑖,𝑁2𝛿=min1𝑚2𝑖=1𝛾𝑖(𝛼1),𝛿1𝑚2𝑖=1𝛾𝑖𝑏𝜂𝑖,(2.18)(iv)min𝜏1𝑡𝜏2𝐻(𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2)𝜏max𝑎𝑡𝑏𝐻(𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2),𝑠(𝑎,𝑏), where 𝜏=𝑏𝑚2𝑖=1𝛾𝑖𝜂𝑖𝛿2𝜏2𝑏𝑚2𝑖=1𝛾𝑖𝜂𝑖𝛿2𝑎<1,𝑎<𝜏1<𝜏2<𝑏.(2.19)

Proof. It is clear that (i) holds. So, we prove that (ii) is true.
(ii) For 𝛼>1, in view of the expression for 𝐺(𝑡,𝑠), it follows that 𝐺(𝑡,𝑠)𝐺(𝑠,𝑠) for all 𝑠,𝑡[𝑎,𝑏], where 𝐺(𝑠,𝑠)=(𝑏𝑠)𝛼1/Γ(𝛼).
If 𝑎𝑠𝑡𝑏, we have 𝐺(𝑡,𝑠)=𝐺(𝑠,𝑠)(𝑏𝑠)𝛼1(𝑡𝑠)𝛼1(𝑏𝑠)𝛼1=(𝑏𝑠)𝛼2(𝑏𝑠)(𝑡𝑠)𝛼2(𝑡𝑠)(𝑏𝑠)𝛼1(𝑏𝑠)𝛼2[](𝑏𝑠)(𝑡𝑠)(𝑏𝑠)𝛼1=(𝑏𝑡)(𝑏𝑎)=𝜑(𝑡).(2.20) If 𝑎𝑡𝑠𝑏, then we have 𝐺(𝑡,𝑠)𝐺(𝑠,𝑠)=1(𝑏𝑡)(𝑏𝑎)=𝜑(𝑡).(2.21) Thus max𝑎𝑡𝑏𝐺(𝑡,𝑠)=𝐺(𝑠,𝑠),𝜑(𝑡)𝐺(𝑠,𝑠)𝐺(𝑡,𝑠)𝐺(𝑠,𝑠),𝑡,𝑠(𝑎,𝑏).(2.22) Therefore, min𝜏1𝑡𝜏2𝐺(𝑡,𝑠)𝜏0max𝑎𝑡𝑏𝐺(𝑡,𝑠)=𝜏0𝐺(𝑠,𝑠),𝑡,𝑠(𝑎,𝑏),𝑎<𝜏1<𝜏2<𝑏.(2.23)(iii) If 𝑎𝑠𝜂𝑖,𝑖=1,2,,𝑚2, then 𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2=𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝜂𝑖𝑠𝛼1𝛿2+Γ(𝛼)𝜇(𝑡)𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑠𝛼2𝛿1𝛿2Γ(𝛼1)𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝛿2Γ+(𝛼)(𝛼1)𝑏𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑠𝛼2𝛿1𝛿2Γ=1(𝛼)𝛿1𝛿2𝛿Γ(𝛼)1𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1+(𝛼1)𝑏𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑠𝛼2(𝛼1)(𝑏𝑠)𝛼2𝛿1𝛿2𝛿Γ(𝛼)1𝑚2𝑖=1𝛾𝑖(𝑏𝑠)+𝑏𝑚2𝑖=1𝛽𝑖(𝛼1)(𝑏𝑠)𝛼2𝛿1𝛿2𝛿Γ(𝛼)1(𝑏𝑎)𝑚2𝑖=1𝛾𝑖+𝑏𝑚2𝑖=1𝛽𝑖=𝑁1𝐻𝑞(𝑠),𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2=𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝜂𝑖𝑠𝛼1𝛿2+Γ(𝛼)𝜇(𝑡)𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑠𝛼2𝛿1𝛿2Γ(𝛼1)𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝜂𝑖𝑠𝛼1𝛿2=Γ(𝛼)𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼2𝜂(𝑏𝑠)𝑖𝑠𝛼2𝜂𝑖𝑠𝛿2=Γ(𝛼)(𝑏𝑠)𝛼2𝛿2Γ(𝛼)𝑚2𝑖=1𝛾𝑖𝑏𝜂𝑖𝑁2𝑞(𝑠).(2.24) If 𝜂𝑖𝑠𝑏,𝑖=1,2,,𝑚2, then we have 𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2=𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝛿2Γ(𝛼)=𝛿1𝑚2𝑖=1𝛾𝑖(𝑏𝑠)(𝑏𝑠)𝛼2𝛿1𝛿2Γ(𝛼)𝛿1𝑚2𝑖=1𝛾𝑖(𝑏𝑎)𝑞(𝑠)<𝑁1𝐻𝑞(𝑠),𝑡,𝑠;𝜂1,𝜂2,...,𝜂𝑚2=𝑚2𝑖=1𝛾𝑖(𝛼1)(𝑏𝑠)𝛼1(𝛼1)𝛿2Γ(𝛼)𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼2(𝛼1)𝛿2𝛿Γ(𝛼)1𝑚2𝑖=1𝛾𝑖(𝛼1)𝑞(𝑠)𝑁2𝑞(𝑠).(2.25)
(iv) Since 𝜕𝐻(𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2)/𝜕𝑡=(𝑚2𝑖=1𝛽𝑖(𝜂𝑖𝑠)𝛼2/𝛿1Γ(𝛼1))0, then 𝐻(𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2) is nonincreasing in 𝑡, so max𝑎𝑡𝑏𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2=𝐻𝑎,𝑠;𝜂1,𝜂2,,𝜂𝑚2=𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝜂𝑖𝑠𝛼1𝛿2+Γ(𝛼)𝑏𝑚2𝑖=1𝛾𝑖𝜂𝑖𝛿2𝑎𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑠𝛼2𝛿1𝛿2,Γ(𝛼1)min𝜏1𝑡𝜏2𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2=𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝜂𝑖𝑠𝛼1𝛿2+Γ(𝛼)𝑏𝑚2𝑖=1𝛾𝑖𝜂𝑖𝛿2𝜏2𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑠𝛼2𝛿1𝛿2=Γ(𝛼1)𝑚2𝑖=1𝛾𝑖(𝑏𝑠)𝛼1𝜂𝑖𝑠𝛼1𝛿2Γ(𝛼)+𝜏𝑏𝑚2𝑖=1𝛾𝑖𝜂𝑖𝛿2𝑎𝑚2𝑖=1𝛽𝑖𝜂𝑖𝑠𝛼2𝛿1𝛿2Γ(𝛼1)>𝜏max𝑎𝑡𝑏𝐻𝑡;𝜂1,𝜂2,,𝜂𝑚2.,𝑠(2.26)

3. Main Results

Let us denote by 𝐸=𝐶[𝑎,𝑏] the Banach space of all continuous real functions on [𝑎,𝑏] endowed with the norm 𝑢=max𝑎𝑡𝑏|𝑢(𝑡)| and 𝑃 the cone 𝑃=𝑢𝐸𝑢0,min𝜏1𝑡𝜏2[]𝑢(𝑡)𝜏𝑢,𝑡𝑎,𝑏,(3.1) where 𝜏=min{𝜏0,𝜏}, since 𝜏0, 𝜏 are constants do not depend on 𝑡.

Let the nonnegative continuous concave functional 𝜃 on the cone 𝑃 be defined by 𝜃(𝑢)=min𝜏1𝑡𝜏2𝑢(𝑡).

Set 𝑇𝑃𝐸 by𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠,𝑎𝑡𝑏,(3.2) where 𝐺(𝑡,𝑠),𝐻(𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2) are defined as in Lemma 2.11.

From (3.2) and Lemma 2.13, we havemin𝜏1𝑡𝜏2(𝑇𝑢(𝑡))𝑏𝑎𝜏0𝐺(𝑠,𝑠)+𝜏max𝑎𝑡𝑏𝐻𝑡,𝑠;𝜂1,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝜏𝑇𝑢.(3.3) Hence, we have 𝑇(𝑃)𝑃.

By standard argument, one can prove that 𝑇𝑃𝑃 is a completely continuous operator.

The Existence of One Positive Solution
We introduce the following definitions: 𝑓(𝑢)=sup[]𝑡𝑎,𝑏𝑓(𝑡,𝑢),𝑓(𝑢)=inf[]𝑡𝑎,𝑏𝑓𝑓(𝑡,𝑢),0=limsup𝑢0+𝑓(𝑢)𝑢,𝑓0=liminf𝑢0+𝑓(𝑢)𝑢,𝑓=limsup𝑢𝑓(𝑢)𝑢,𝑓=liminf𝑢𝑓(𝑢)𝑢,𝑀=𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑞(𝑠)𝑑𝑠1,𝑁=𝜏2𝜏1𝜏𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑑𝑠1.(3.4)
Theorem 3.1. Let 𝑓(𝑡,𝑢) be continuous on [𝑎,𝑏]×[0,)[0,). If there exist two positive constants 𝑟2>𝑟1>0 such that (𝐻1)𝑓(𝑡,𝑢)𝑀𝑟2,for(𝑡,𝑢)[𝑎,𝑏]×[0,𝑟2], (𝐻2)𝑓(𝑡,𝑢)𝑁𝑟1,for(𝑡,𝑢)[𝑎,𝑏]×[0,𝑟1], then the BVP (1.3)-(1.4) has at least a positive solution.Proof. We know that the operator 𝑇𝑃𝑃 defined by (3.2) is completely continuous.
(a) Let Ω2={𝑢𝐸𝑢<𝑟2}. For any 𝑢𝑃𝜕Ω2, we have 𝑢=𝑟2 which implies that 0𝑢(𝑡)𝑟2 for every 𝑡[𝑎,𝑏]: 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑀𝑟2𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑞(𝑠)𝑑𝑠𝑟2=𝑢,(3.5) which implies that 𝑇𝑢𝑢,𝑢𝑃𝜕Ω2.(3.6)
(b) Let Ω1={𝑢𝐸𝑢<𝑟1}. For any 𝑡[𝜏1,𝜏2],𝑢𝑃𝜕Ω1. We have 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑏𝑎𝜑(𝑡)𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑁𝑟1𝜏2𝜏1𝜏𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑑𝑠=𝑟1=𝑢,(3.7) which implies that 𝑇𝑢𝑢,𝑢𝑃𝜕Ω1.(3.8) In view of Theorem 2.6, 𝑇 has a fixed point 𝑢0𝑃(Ω2Ω1) which is a solution of the BVP (1.3)-(1.4).

The Existence of Two Positive Solutions
Theorem 3.2. Assume that all assumptions of Theorem 3.1, hold. Moreover, one assumes that 𝑓(𝑡,𝑢) also satisfies (𝐻3)𝑓=. Then, the BVP (1.3)-(1.4) has at least two positive solutions.Proof. At first, it follows from condition (𝐻1) that 𝑇𝑢𝑢,𝑢𝑃𝜕Ω2.(3.9) Further, it follows from condition (𝐻2) that 𝑇𝑢𝑢,𝑢𝑃𝜕Ω1.(3.10) Finally, since 𝑓=, there exists 𝜓>(𝜏2𝜏2𝜏1[𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)]𝑑𝑠)1 and 𝑟3>𝑟2 such that []𝑓(𝑡,𝑢)𝜓𝑢(𝑡),𝑡𝑎,𝑏,𝑢𝑟3.(3.11) Let 𝑟=max{2𝑟2,𝜏1𝑟3} and set Ω3={𝑢𝐸𝑢<𝑟}, then 𝑢𝑃𝜕Ω3impliesmin𝜏1𝑡𝜏2𝑢(𝑡)𝜏𝑢𝜏𝑟𝑟3, 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑏𝑎𝜑(𝑡)𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝜓𝑏𝑎𝜑𝐺(𝑡)(𝑠,𝑠)+𝑁2𝑞𝑢(𝑠)(𝑠)𝑑𝑠𝑟𝜓𝜏2𝜏1𝜏2𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑑𝑠>𝑟=𝑢.(3.12) Therefore, we have 𝑇𝑢𝑢,𝑢𝑃𝜕Ω3.(3.13) Thus, from (3.6), (3.8), (3.13), and Theorem 2.6, 𝑇 has a fixed point 𝑢1, in 𝑃(Ω2Ω1) and a fixed point 𝑢2, in 𝑃(Ω3Ω2). Both are positive solutions of BVP (1.3)-(1.4) and satisfy 𝑢0<1<𝑟2<𝑢2.(3.14)Theorem 3.3. Assume that 𝑓(𝑡,𝑢) be continuous on [𝑎,𝑏]×[0,)[0,). If the following assumptions hold: (𝐻1)𝑓0>𝜓, (𝐻2)𝑓>𝜓, (𝐻3) there exists a constant 𝜌>0 such that 𝑓(𝑡,𝑢)𝜌𝑀,(𝑡,𝑢)[𝑎,𝑏]×[0,𝜌],then the BVP (1.3)-(1.4) has at least two positive solutions 𝑢1 and 𝑢2 such that 𝑢0<1𝑢<𝜌<2.(3.15)Proof. At first, it follows from condition (𝐻1) that we may choose 𝜌1(0,𝜌) such that 𝑓(𝑡,𝑢)>𝜓𝑢,0<𝑢𝜌1,(3.16) where 𝜓 is defined as in Theorem 3.2. Set Ω1={𝑢𝐸𝑢<𝜌1}, and 𝑢𝑃𝜕Ω1; from (3.2) and Lemma 2.13, for 𝑎𝑡𝑏, we have 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑏𝑎𝜑(𝑡)𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝜓𝑏𝑎𝜑𝐺(𝑡)(𝑠,𝑠)+𝑁2𝑞𝑢(𝑠)(𝑠)𝑑𝑠𝜌1𝜓𝜏2𝜏1𝜏2𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑑𝑠>𝜌1=𝑢.(3.17) Therefore, we have 𝑇𝑢𝑢,𝑢𝑃𝜕Ω1.(3.18) Further, it follows from condition (𝐻2) that there exists 𝜌2>𝜌 such that 𝑓(𝑡,𝑢)>𝜓𝑢(𝑡),𝑢𝜌2.(3.19) Let 𝜌=max{2𝜌,𝜏1𝜌2}, set Ω2={𝑢𝐸𝑢<𝜌}, then 𝑢𝑃𝜕Ω2impliesmin𝜏1𝑡𝜏2𝑢(𝑡)𝜏𝑢𝜏𝜌𝜌2, 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑏𝑎𝜑(𝑡)𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝜓𝑏𝑎𝜑𝐺(𝑡)(𝑠,𝑠)+𝑁2𝑞𝑢(𝑠)(𝑠)𝑑𝑠𝜌𝜓𝜏2𝜏1𝜏2𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑑𝑠>𝜌=𝑢.(3.20) Therefore, we have 𝑇𝑢𝑢,𝑢𝑃𝜕Ω2.(3.21) Finally, let Ω3={𝑢𝐸𝑢<𝜌} and 𝑢𝑃𝜕Ω3. By condition (𝐻3), we have 𝑇𝑢(𝑡)𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑞(𝑠)𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑀𝜌𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑞(𝑠)𝑑𝑠=𝜌=𝑢,(3.22) which implies 𝑇𝑢𝑢,𝑢𝑃𝜕Ω3.(3.23) Thus, from (3.18), (3.21), (3.23), and Theorem 2.6, 𝑇 has a fixed point 𝑢1 in 𝑃(Ω3Ω1) and a fixed point 𝑢2, in 𝑃(Ω2Ω3). Both are positive solutions of BVP (1.3)-(1.4) and satisfy 𝑢0<1𝑢<𝜌<2.(3.24)Theorem 3.4. Assume that 𝑓(𝑡,𝑢) be continuous on [𝑎,𝑏]×[0,)[0,). If the following assumptions hold: (𝐻1)𝑓0=, (𝐻2)𝑓=, (𝐻3)there exists a constant 𝜌>0 such that 𝑓(𝑡,𝑢)𝜌𝑀,(𝑡,𝑢)[𝑎,𝑏]×[0,𝜌],then the BVP (1.3)-(1.4) has at least two positive solutions 𝑢1 and 𝑢2 such that 𝑢0<1<𝜌<𝑢2.(3.25) The proof of Theorem 3.4 is very similar to that of Theorem 3.3 and therefore is omitted.Theorem 3.5. Assume that 𝑓(𝑡,𝑢) be continuous on [𝑎,𝑏]×[0,)[0,). If the following assumptions hold: (𝐻1)𝑓0<𝑀, (𝐻2)𝑓<𝑀, (𝐻3)there exists a constant 𝑙>0 such that 𝑓(𝑡,𝑢)𝑁𝑙,(𝑡,𝑢)[𝑎,𝑏]×[𝜏𝑙,𝑙],then the BVP (1.3)-(1.4) has at least two positive solutions 𝑢1 and 𝑢2 such that 𝑢0<1𝑢<𝑙<2.(3.26)Proof. It follows from condition (𝐻1) that we may choose 𝜌3(0,𝑙) such that 𝑓(𝑡,𝑢)<𝑀𝑢,0<𝑢𝜌3.(3.27) Set Ω4={𝑢𝐸𝑢<𝜌3}, and 𝑢𝑃𝜕Ω4; from (3.2) and Lemma 2.13, for 𝑎𝑡𝑏, we have 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠<𝑀𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑞(𝑠)𝑑𝑠𝑢=𝑀𝑀1𝑢=𝑢.(3.28) Therefore, we have 𝑇𝑢𝑢,𝑢𝑃𝜕Ω4.(3.29) It follows from condition (𝐻2) that there exists 𝜌4>𝑙 such that 𝑓(𝑡,𝑢)<𝑀𝑢,𝑢𝜌4,(3.30) and we consider two cases.Case 1. Suppose that 𝑓 is unbounded, there exists 𝑙>𝜌4 such that 𝑓(𝑡,𝑢)𝑓(𝑡,𝑙) for 0<𝑢𝑙.Then, for 𝑢𝑃 and 𝑢=𝑙, we have 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑓𝑞(𝑠)𝑠,𝑙𝑑𝑠<𝑀𝑙𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑞(𝑠)𝑑𝑠=𝑙=𝑢.(3.31)Case 2. If 𝑓 is bounded, that is, 𝑓(𝑡,𝑢)𝑘 for all 𝑢[0,), taking 𝑙max{2𝑙,𝑘𝑀1}, for 𝑢𝑃 and 𝑢=𝑙, then we have 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑘𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑞(𝑠)𝑑𝑠=𝑘𝑀1𝑙=𝑢.(3.32) Hence, in either case, we always may set Ω5={𝑢𝐸𝑢<𝑙} such that 𝑇𝑢𝑢,𝑢𝑃𝜕Ω5.(3.33) Finally, set Ω6={𝑢𝐸𝑢<𝑙}, then 𝑢𝑃𝜕Ω6 and min𝜏1𝑡𝜏2𝑢(𝑡)𝜏𝑢=𝜏𝑙,(3.34) and by condition (𝐻3) and (3.2), we have 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,𝜂2,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑁𝑙𝑏𝑎𝜑(𝑡)𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑑𝑠𝑁𝑙𝜏2𝜏1𝜏𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑑𝑠=𝑙=𝑢.(3.35) Hence, we have 𝑇𝑢𝑢,𝑢𝑃𝜕Ω6.(3.36) Thus, from (3.29), (3.33), (3.36) and Theorem 2.6, 𝑇 has a fixed point 𝑢1 in 𝑃(Ω6Ω4) and a fixed point 𝑢2 in 𝑃(Ω5Ω6).Both are positive solutions of BVP (1.3)-(1.4) and satisfy 𝑢0<1𝑢<𝑙<2.(3.37)Theorem 3.6. Assume that 𝑓(𝑡,𝑢) be continuous on [𝑎,𝑏]×[0,)[0,). If the following assumptions hold:(𝐻1)𝑓0=0, (𝐻2)𝑓=0,(𝐻3)there exists a constant 𝜌>0 such that 𝑓(𝑡,𝑢)𝑁𝜌,(𝑡,𝑢)[𝑎,𝑏]×[𝜏𝜌,𝜌],then the BVP (1.3)-(1.4) has at least two positive solutions 𝑢1 and 𝑢2 such that 𝑢0<1<𝜌<𝑢2.(3.38) The proof of Theorem 3.6 is very similar to that of Theorem 3.5 and therefore omitted.

The Existence of Three Positive Solutions
Theorem 3.7. Let 𝑓(𝑡,𝑢) be continuous on [𝑎,𝑏]×[0,)[0,). If there exist constants 0<𝑎1<𝑎2𝑎3 such that the following assumptions (i)𝑓(𝑡,𝑢)<𝑀𝑎1,(𝑡,𝑢)[𝑎,𝑏]×[0,𝑎1], (ii)𝑓(𝑡,𝑢)𝑀𝑎3,(𝑡,𝑢)[𝑎,𝑏]×[0,𝑎3], (iii)𝑓(𝑡,𝑢)𝑁𝑎2,(𝑡,𝑢)[𝜏1,𝜏2]×[𝑎2,𝑎2/𝜏], hold, then BVP (1.3)-(1.4) has at least three positive solutions 𝑢1,𝑢2, and 𝑢3 with 𝑢1<𝑎1,𝑎2𝑢<𝜃2<𝑢2𝑎3,𝑎1<𝑢3𝑢,𝜃3<𝑎2.(3.39)Proof. We will show that all conditions of Lemma 2.10, are satisfied.
First, if 𝑢𝑃𝑎3, then 𝑢𝑎3. So, 0𝑢(𝑡)𝑎3,𝑡[𝑎,𝑏].
By condition (ii), we have 𝑇𝑢(𝑡)=𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑞(𝑠)𝑀𝑎3𝑑𝑠=𝑀𝑎3𝑏𝑎𝐺(𝑠,𝑠)+𝑁1𝑞(𝑠)𝑑𝑠=𝑎3,(3.40) which implies that 𝑇𝑢𝑎3,𝑢𝑃𝑎3. Hence 𝑇𝑃𝑎3𝑃𝑎3.
Next, by using the analogous argument, it follows from condition (i) that if 𝑢𝑃𝑎1, then 𝑇𝑢<𝑎1.
Choose 𝑢(𝑡)=(𝑎2+𝑎2/𝜏)/2,𝑡[𝑎,𝑏], it is easy to see that 𝑢(𝑡)=(𝑎2+𝑎2/𝜏)/2𝑃(𝜃,𝑎2,𝑎3),𝜃(𝑢)=(𝑎2+𝑎2/𝜏)/2>𝑎2.
Therefore, {𝑢𝑃(𝜃,𝑎2,𝑎2/𝜏)𝜃(𝑢)>𝑎2}𝜙. On the other hand, if 𝑢𝑃(𝜃,𝑎2,𝑎2/𝜏), then 𝑎2𝑢(𝑡)𝑎2/𝜏,𝑡[𝜏1,𝜏2]. By condition (iii), we have 𝑓(𝑡,𝑢(𝑡))𝑁𝑎2.
Hence, 𝜃(𝑇𝑢(𝑡))=min𝜏1𝑡𝜏2𝑇𝑢(𝑡)=min𝜏1𝑡𝜏2𝑏𝑎𝐺(𝑡,𝑠)+𝐻𝑡,𝑠;𝜂1,,𝜂𝑚2𝑓(𝑠,𝑢(𝑠))𝑑𝑠𝑁𝑎2𝑏𝑎𝜏𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑑𝑠>𝑁𝑎2𝜏2𝜏1𝜏𝐺(𝑠,𝑠)+𝑁2𝑞(𝑠)𝑑𝑠=𝑎2,(3.41) which implies that 𝜃(𝑇𝑢)>𝑎2, for 𝑢𝑃(𝜃,𝑎2,𝑎2/𝜏).
Finally, if 𝑢𝑃(𝜃,𝑎2,𝑎3) and 𝑇𝑢>𝑎2/𝜏, then 𝜃(𝑢)=min𝜏1𝑡𝜏2𝑇𝑢(𝑡)𝜏𝑇𝑢>𝑎2.(3.42) Thus, all the conditions of the Leggett-Williams fixed point theorem are satisfied by taking 𝑑=𝑎2/𝜏. Hence, the BVPs have at least three solutions in 𝑃, that is, three positive solutions 𝑢𝑖(𝑖=1,2,3) such that 𝑢1<𝑎1,𝑎2𝑢<𝜃2<𝑢2𝑎3,𝑎1<𝑢3𝑢,𝜃3<𝑎2.(3.43)
Example 3.8. Consider the problem 𝐷0(4.2)+1𝑢(𝑡)+3(1+𝑢𝑒𝑢𝑢)=0,𝑡(0,1),1(0)=4𝑢12,𝑢(0)=𝑢(0)=𝑢3(0)=0,𝑢(1)=4𝑢12,(3.44) where 𝛼=4.2, 𝑎=0, 𝑏=1, 𝛽=0.25, 𝛾=0.75, 𝜂=0.5, 𝜏1=0.25, 𝜏2=0.75,𝑁1=2.6,𝑁2=0.1758, 𝑁=168.9596, and 𝑀=1.6968,𝑓(𝑡,𝑢)=(1/3)(1+𝑢𝑒𝑢).
Since 𝑓(𝑡,𝑢)=(1/3)(1+𝑢𝑒𝑢) is a monotone increasing function on [0,), we take 𝑟1=0.001,𝑟2=0.8. We can get 𝑓(𝑡,𝑢)𝑓(0.8)=0.9268<𝑀𝑟2,𝑓(𝑡,𝑢)𝑓(0)=0.3333>𝑁𝑟1.(3.45) So, conditions (𝐻1) and (𝐻2) hold. By Theorem 3.1, the BVP (3.44) has at least one positive solution.

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