Positive Solutions for Nonlinear Fractional Differential Equations with Boundary Conditions Involving Riemann-Stieltjes Integrals

Abstract

We consider the existence of positive solutions for a class of nonlinear integral boundary value problems for fractional differential equations. By using some fixed point theorems, the existence and multiplicity results of positive solutions are obtained. The results obtained in this paper improve and generalize some well-known results.

1. Introduction

This paper is concerned with the existence of positive solutions to the following boundary value problem (BVP) for fractional differential equation: 𝐷^𝛼₀₊𝑢(𝑡)+𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),…,𝑢^(𝑛−2)(𝑡)𝑢(0)=0,0<𝑡<1,=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)𝑢=0,_(𝑛−2)(1)=𝛽𝑢^(𝑛−2), (1.1) where 𝛽[𝑣]=∫¹₀𝑣(𝑡)𝑑𝐴(𝑡) is a linear functional on 𝐶[0,1] given by a Riemann-Stieltjes integral with 𝐴 representing a suitable function of bounded variation, 𝐷^𝛼₀₊ is the Riemann-Liouville fractional derivative of order 𝑛−1<𝛼≤𝑛, 𝑛≥2, 𝑓∶[0,1]×ℝ^𝑛−1→ℝ⁺ satisfies the Carathéodory type conditions, ℝ=(−∞,+∞) and ℝ⁺=[0,+∞).

Fractional differential equations arise in the modeling and control of many real-world systems and processes particularly in the fields of physics, chemistry, aerodynamics, electrodynamics of complex media, and polymer rheology. Fractional differential equations also serve as an excellent tool for the description of hereditary properties of various materials and processes. Hence, intensive research has been carried out worldwide to study the existence of solutions of nonlinear fractional differential equations (see [1–25]). For example, by means of a mixed monotone method, Zhang [11] studied a unique positive solution for the singular boundary value problem 𝐷^𝛼₀₊𝑢(𝑡)+𝑞(𝑡)𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡),…,𝑢^(𝑛−2)(𝑡)𝑢(0)=0,0<𝑡<1,=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)=𝑢^(𝑛−2)(1)=0, (1.2) where 𝛼∈(𝑛−1,𝑛], 𝑛≥2, 𝐷^𝛼₀₊ is the standard Riemann-Liouville derivative, 𝑓=𝑔+ℎ is nonlinear, and 𝑔 and ℎ have different monotone properties.

Recently, nonlocal boundary value problems for fractional differential equations were investigated intensively [13–23]. In [14], Bai concerned the existence and uniqueness of a positive solution for the following nonlocal problem: 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝑢(0)𝜂(1),+𝑓𝑡,𝑢))=0,0<𝑡<1,=0,𝛽𝑢=𝑢 (1.3) where 1<𝛼≤2, 0<𝛽𝜂^𝛼−1<1, 0<𝜂<1, 𝐷^𝛼₀₊ is the standard Riemann-Liouville differentiation. The function 𝑓 is continuous on [0,1]×ℝ⁺.

In [20], El-Shahed and Nieto investigated the existence of nontrivial solutions for the following nonlinear 𝑚-point boundary value problem of fractional type: _𝑅𝐷^𝛼₀₊𝑢(𝑡)((𝑡[](]𝑢(0)+𝑓𝑡,𝑢))=0,𝑡∈0,1,𝛼∈𝑛−1,𝑛,𝑛∈ℕ,=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)=0,𝑢(1)=^𝑚−2_𝑖=1𝑎_𝑖𝑢𝜂_𝑖, (1.4) where 𝑛≥2, 𝑎_𝑖>0(𝑖=1,2,…,𝑚−2), 0<𝜂₁<𝜂₂<⋯<𝜂_𝑚−2<1, 𝑓∈𝐶([0,1]×ℝ,ℝ). Also the authors considered the analogous problem using the Caputo fractional derivative: _𝐶𝐷^𝛼₀₊𝑢(𝑡)((𝑡[](]𝑢(0)+𝑓𝑡,𝑢))=0,𝑡∈0,1,𝛼∈𝑛−1,𝑛,𝑛∈ℕ,=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)=0,𝑢(1)=^𝑚−2_𝑖=1𝑎_𝑖𝑢𝜂_𝑖. (1.5) Under certain growth conditions on the nonlinearity, several sufficient conditions for the existence of nontrivial solution are obtained by using the Leray-Schauder nonlinear alternative.

Inspired by the work of the above papers, the aim of this paper is to establish the existence and multiplicity of positive solutions of the BVP (1.1). We discuss the boundary value problem with the Riemann-Stieltjes integral boundary conditions, that is, the BVP (1.1), which includes fractional order two-point, three-point, multipoint, and nonlocal boundary value problems as special cases. Moreover, the 𝛽[⋅] in (1.1) is a linear function on 𝐶[0,1] denoting the Riemann-Stieltjes integral; the 𝐴 in the Riemann-Stieltjes integral is of bounded variation, namely, 𝑑𝐴 can be a signed measure. By using the Krasnosel’skii fixed point theorem, the Leray-Schauder nonlinear alternative and the Leggett-Williams fixed point theorem, some existence and multiplicity results of positive solutions are obtained.

The rest of this paper is organized as follows. In Section 2, we present some lemmas that are used to prove our main results. In Section 3, the existence and multiplicity of positive solutions of the BVP (1.1) are established by using some fixed point theorems. In Section 4, we give four examples to demonstrate the application of our theoretical results.

2. Basic Definitions and Preliminaries

We begin this section with some preliminaries of fractional calculus. Let 𝛼>0 and 𝑛=[𝛼]+1, where [𝛼] is the largest integer smaller than or equal to 𝛼. For a function 𝑓∶(0,+∞)→𝑅, we define the fractional integral of order 𝛼 of 𝑓 as 𝐼^𝛼₀₊𝑓(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑓(𝑠)𝑑𝑠, (2.1) provided the integral exists. The fractional derivative of order 𝛼>0 of a continuous function 𝑓 is defined by 𝐷^𝛼₀₊𝑓(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑓(𝑠)𝑑𝑠, (2.2) provided the right-hand side is pointwise defined on (0,+∞). We recall the following properties [26, 27] which are useful for the sequel. For 𝛼>0, 𝛽>0, we have 𝐼^𝛼₀₊𝐼^𝛽₀₊𝑓(𝑡)=𝐼^𝛼+𝛽₀₊𝑓(𝑡),𝐷^𝛼₀₊𝐼^𝛼₀₊𝑓(𝑡)=𝑓(𝑡). (2.3) As an example, we can choose a function 𝑓 such that 𝑓,𝐷^𝛼₀₊𝑓∈𝐶(0,+∞)∩𝐿¹_loc(0,+∞).

For 𝛼>0, the general solution of the fractional differential equation 𝐷^𝛼₀₊𝑢(𝑡)=0 with 𝑢∈𝐶(0,1)∩𝐿(0,1) is given by 𝑢(𝑡)=𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2⋅⋅⋅+𝑐_𝑛𝑡^𝛼−𝑛, (2.4) where 𝑐_𝑖∈𝑅(𝑖=1,2,…,𝑛). Hence for 𝑢∈𝐶(0,1)∩𝐿(0,1), we have 𝐼^𝛼₀₊𝐷^𝛼₀₊𝑢(𝑡)=𝑢(𝑡)+𝑐₁𝑡^𝛼−1+𝑐₂𝑡^𝛼−2⋅⋅⋅+𝑐_𝑛𝑡^𝛼−𝑛. (2.5) Set 𝐺₀()=1𝑡,𝑠Γ()[𝑡(𝛼−𝑛+21−𝑠)]^{𝛼−𝑛+1}−(𝑡−𝑠)^{𝛼−𝑛+1}[𝑡(,0≤𝑠≤𝑡≤1,1−𝑠)]^{𝛼−𝑛+1},0≤𝑡≤𝑠≤1. (2.6)

Lemma 2.1 (see [11]). Let 𝑦∈𝐶_𝑟[0,1](𝐶_𝑟[0,1]={𝑦∈𝐶[0,1],𝑡^𝑟𝑦∈𝐶[0,1],0≤𝑟<1}). Then the boundary value problem, 𝐷^{𝛼−𝑛+2}₀₊𝑣(𝑡)(𝑡)𝑣(0)(1)=𝑦,0<𝑡<1,𝑛−1<𝛼≤𝑛,𝑛≥2,=0,𝑣=0, (2.7) has a unique solution 𝑣(𝑡)=¹₀𝐺₀(𝑡,𝑠)𝑦(𝑠)𝑑𝑠. (2.8)

Lemma 2.2 (see [11]). The function 𝐺₀(𝑡,𝑠) defined by (2.6) satisfies the following properties: (i)𝐺₀(𝑡,𝑠)≥0, 𝐺₀(𝑡,𝑠)≤𝐺₀(𝑠,𝑠) for all 𝑡,𝑠∈[0,1];(ii)there exist a positive function 𝜌∈𝐶(0,1) and 0<𝜉<𝜂<1 such that min_{𝑡∈[𝜉,𝜂]}𝐺₀(𝑡,𝑠)≥𝜌(𝑠)𝐺₀(𝑠,𝑠),𝑠∈(0,1), (2.9)where 𝜌(𝑠)=𝜂(1−𝑠)^{𝛼−𝑛+1}−𝜂−𝑠^{𝛼−𝑛+1}[𝑠(1−𝑠)]^{𝛼−𝑛+1}(],𝜉,𝑠∈0,𝑟𝑠^{𝛼−𝑛+1},𝑠∈[𝑟,1). (2.10)

By (2.4), the unique solution of the problem 𝐷^{𝛼−𝑛+2}₀₊𝑣(𝑡)𝑣(0)(1)[𝑣]=0,0<𝑡<1,𝑛−1<𝛼≤𝑛,𝑛≥2,=0,𝑣=𝛽 (2.11) is 𝛾(𝑡)=𝑡^{𝛼−𝑛+1}, with 𝛽[𝑣] replaced by 1. As in [21], the Green’s function for boundary value problem (2.11) is given by 𝐺()=𝛾(𝑡)𝑡,𝑠𝛾𝒢(𝑠)1−𝛽+𝐺₀(𝑡,𝑠), (2.12) where 𝒢(𝑠)∶=∫¹₀𝐺₀(𝑡,𝑠)𝑑𝐴(𝑡).

Lemma 2.3. Let 0≤𝛽[𝛾]<1 and 𝒢(𝑠)≥0 for 𝑠∈[0,1], the Green function 𝐺(𝑡,𝑠) defined by (2.12) has the following properties: (i)𝐺(𝑡,𝑠)≥0,𝐺(𝑡,𝑠)≤(1+𝛽[1]/(1−𝛽[𝛾]))𝐺₀(𝑠,𝑠) for all 𝑡,𝑠∈[0,1];(ii)min_{𝑡∈[𝜉,𝜂]}𝐺(𝑡,𝑠)≥min_{𝑡∈[𝜉,𝜂]}𝐺₀(𝑡,𝑠)≥𝜌(𝑠)𝐺₀(𝑠,𝑠), 𝑠∈(0,1).

Proof. By Lemma 2.2, it is easy to prove this lemma, so we omit it.

Let 𝑋=𝐶[0,1]. It follows that (𝑋,‖⋅‖) is a Banach space, where ‖⋅‖ is defined by the supernorm ‖𝑥‖=sup_𝑡∈[0,1]|𝑥(𝑡)|. 𝑃={𝑥∈𝑋∶𝑥(𝑡)≥0,𝑡∈[0,1]}. Clearly 𝑃 is a cone of 𝑋. Now, in the following, we give the assumptions to be used throughout the rest of this paper.(H₁)𝐴 is a function of bounded variation, 𝒢(𝑠)≥0 for 𝑠∈[0,1] and 0≤𝛽[𝛾]<1.(H₂)𝑓∶[0,1]×ℝ^𝑛−1→ℝ⁺ satisfies the following conditions of Carathéodory type:(i)𝑓(⋅,𝑥) is Lebesgue measurable for each fixed 𝑥∈ℝ^𝑛−1;(ii)𝑓(𝑡,⋅) is continuous for a.e. 𝑡∈[0,1].

In order to overcome the difficulty due to the dependance of 𝑓 on derivatives, we consider the following modified problem: 𝐷^{𝛼−𝑛+2}₀₊𝑣(𝑡)+𝑓𝑡,𝐼^𝑛−2₀₊𝑣(𝑡),𝐼^𝑛−3₀₊𝑣(𝑡),…,𝐼¹₀₊𝑣(𝑡)(𝑡𝑣(0)(1)[𝑣],,𝑣)=0,0<𝑡<1,=0,𝑣=𝛽 (2.13) where 𝑛−1<𝛼≤𝑛, 𝑛≥2.

Lemma 2.4. The nonlocal fractional order boundary value problem (1.1) has a positive solution if and only if the nonlinear fractional integrodifferential equation (2.13) has a positive solution.

Proof. If 𝑢 is a positive solution of the fractional order boundary value problem (1.1), let 𝑣(𝑡)=𝐷^𝑛−2₀₊𝑢(𝑡). Then from the boundary value conditions of (1.1) and the definition of the Riemann-Liouville fractional integral and derivative, we have𝑣(𝑡)=𝐷^𝑛−2₀₊𝑢(𝑡)=𝑢^(𝑛−2)(𝑡),𝐼¹₀₊𝑣(𝑡)=𝐼¹₀₊𝑢^(𝑛−2)(𝑡)=1Γ(1)^𝑡₀𝑢^(𝑛−2)(𝑠)𝑑𝑠=𝑢^(𝑛−3)(𝑡),𝐼²₀₊𝑣(𝑡)=𝐼²₀₊𝑢^(𝑛−2)(𝑡)=1Γ(2)^𝑡₀(𝑡−𝑠)𝑢^(𝑛−2)(𝑠)𝑑𝑠=𝑢^(𝑛−4)(𝑡),⋮𝐼^𝑛−2₀₊𝑣(𝑡)=𝐼^𝑛−2₀₊𝑢^(𝑛−2)(𝑡)=1Γ(𝑛−2)^𝑡₀(𝑡−𝑠)^𝑛−3𝑢^(𝑛−2)(𝑠)(𝑡),𝐷𝑑𝑠=𝑢_{𝛼−𝑛+20+}𝑣(𝑡)=1Γ()𝑑2𝑛−𝛼−2𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{2𝑛−𝛼−3}𝑢^(𝑛−2)(𝑠)=1𝑑𝑠Γ()𝑑2𝑛−𝛼−3𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{2𝑛−𝛼−4}𝑢^(𝑛−3)(𝑠)=1𝑑𝑠=⋯Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑢(𝑠)𝑑𝑠=𝐷^𝛼₀₊𝑢(𝑡), (2.14)(2.15) which imply that 𝑣(0)=𝑢^(𝑛−2)(0)=0, 𝑣(1)=𝑢^(𝑛−2)(1)=𝛽[𝑢^(𝑛−2)]=𝛽[𝑣]. Thus 𝑣(𝑡) is a positive solution of the nonlinear fractional integrodifferential equation (2.13).
On the other hand, if 𝑣 is a positive solution of the nonlinear fractional integrodifferential equation (2.13), let 𝑢(𝑡)=𝐼^𝑛−2₀₊𝑣(𝑡), then by (2.3) and the definition of the Riemann-Liouville fractional derivative, we have 𝑢^′(𝑡)=𝐷¹₀₊𝑢(𝑡)=𝐷¹₀₊𝐼^𝑛−2₀₊𝑣(𝑡)=𝐷¹₀₊𝐼¹₀₊𝐼^𝑛−3₀₊𝑣(𝑡)=𝐼^𝑛−3₀₊𝑣(𝑡),𝑢^′′(𝑡)=𝐷²₀₊𝑢(𝑡)=𝐷²₀₊𝐼^𝑛−2₀₊𝑣(𝑡)=𝐷²₀₊𝐼²₀₊𝐼^𝑛−4₀₊𝑣(𝑡)=𝐼^𝑛−4₀₊𝑣(𝑡),⋮𝑢^(𝑛−3)(𝑡)=𝐷^𝑛−3₀₊𝑢(𝑡)=𝐷^𝑛−3₀₊𝐼^𝑛−2₀₊𝑣(𝑡)=𝐷^𝑛−3₀₊𝐼^𝑛−3₀₊𝐼¹₀₊𝑣(𝑡)=𝐼¹₀₊𝑣(𝑡),𝑢^(𝑛−2)(𝑡)=𝐷^𝑛−2₀₊𝑢(𝑡)=𝐷^𝑛−2₀₊𝐼^𝑛−2₀₊𝑣(𝑡)(𝑡),𝐷=𝑣_𝛼0+𝑢(𝑡)=𝑑^𝑛𝑑𝑡^𝑛𝐼^𝑛−𝛼₀₊𝑢(𝑡)=𝑑^𝑛𝑑𝑡^𝑛𝐼^𝑛−𝛼₀₊𝐼^𝑛−2₀₊𝑣(𝑡)=𝑑^𝑛𝑑𝑡^𝑛𝐼^{2𝑛−𝛼−2}₀₊𝑣(𝑡)=𝐷^{𝛼−𝑛+2}₀₊𝑣(𝑡)=−𝑓𝑡,𝐼^𝑛−2₀₊𝑣(𝑡),𝐼^𝑛−3₀₊𝑣(𝑡),…,𝐼¹₀₊𝑣(𝑡)(𝑡)(𝑡),𝑣=−𝑓𝑡,𝑢,𝑢^′(𝑡),…,𝑢^(𝑛−3)(𝑡),𝑢^(𝑛−2)(𝑡),0<𝑡<1, (2.16) which imply that 𝑢(0)=𝑢′(0)=⋯=𝑢^(𝑛−3)(0)=0,   𝑢^(𝑛−2)(0)=𝑣(0)=0,   𝑢^(𝑛−2)(1)=𝑣(1)=𝛽[𝑣]=𝛽[𝑢^(𝑛−2)]. Moreover, it follows from the monotonicity and property of 𝐼^𝑛−2₀₊ that 𝐼^𝑛−2₀₊𝑣(𝑡)∈𝐶([0,1],ℝ⁺). Consequently, 𝑢(𝑡)=𝐼^𝑛−2₀₊𝑣(𝑡) is a positive solution of the fractional order boundary value problem (1.1).

By Lemma 2.4, we will concentrate our study on (2.13). We here define an operator 𝑇∶𝑃→𝑃 by 𝑇𝑣(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)𝑑𝑠,𝑡∈[0,1]. (2.17) Clearly, 𝑣 is a fixed point of 𝑇 in 𝑃, and so 𝑣 is a positive solution of BVP (2.13).

In order to prove our main results, we need the following lemmas.

Lemma 2.5 (see [28]). Let 𝑋 be a real Banach space, Ω be a bounded open subset of 𝑋, where 𝜃∈Ω, 𝑇∶Ω→𝑋 is a completely continuous operator. Then, either there exist 𝑥∈𝜕Ω, 𝜇∈(0,1) such that 𝜇𝑇(𝑥)=𝑥, or there exists a fixed point 𝑥^∗∈Ω.

Lemma 2.6 (see [29]). Let 𝑋 be a real Banach space, 𝑃 be a cone in 𝑋. Assume that Ω₁ and Ω₂ are two bounded open sets of 𝑋 with 𝜃∈Ω₁ and Ω₁⊂Ω₂. Let 𝑇∶𝑃∩(Ω₂⧵Ω₁)→𝑃 be a completely continuous operator such that either(i)‖𝑇𝑥‖≤‖𝑥‖, 𝑥∈𝑃∩𝜕Ω₁ and ‖𝑇𝑥‖≥‖𝑥‖, 𝑥∈𝑃∩𝜕Ω₂, or(ii)‖𝑇𝑥‖≥‖𝑥‖, 𝑥∈𝑃∩𝜕Ω₁ and ‖𝑇𝑥‖≤‖𝑥‖, 𝑥∈𝑃∩𝜕Ω₂.Then T has a fixed point in 𝑃∩(Ω₂⧵Ω₁).

Lemma 2.7 (see [30, 31]). Let 𝑃 be a cone in a real Banach space 𝑋, 𝑃_𝑐={𝑥∈𝑃∶‖𝑥‖<𝑐},  𝜑 be a nonnegative continuous concave functional on 𝑃 such that 𝜑(𝑥)≤‖𝑥‖ for all 𝑥∈𝑃_𝑐, and 𝑃(𝜑,𝑏,𝑑)={𝑥∈𝑃∶𝑏≤𝜑(𝑥),‖𝑥‖≤𝑑}. Suppose that 𝑇∶𝑃_𝑐→𝑃_𝑐 is completely continuous and there exist positive constants 0<𝑎<𝑏<𝑑≤𝑐 such that(C₁){𝑥∈𝑃(𝜑,𝑏,𝑑)∶𝜑(𝑥)>𝑏}≠𝜙 and 𝜑(𝑇𝑥)>𝑏 for 𝑥∈𝑃(𝜑,𝑏,𝑑),(C₂)‖𝑇𝑥‖<𝑎 for 𝑥∈𝑃_𝑎,(C₃)𝜑(𝑇𝑥)>𝑏 for 𝑥∈𝑃(𝜑,𝑏,𝑐) with ‖𝑇𝑥‖>𝑑.Then T has at least three fixed points 𝑥₁,𝑥₂, and 𝑥₃ satisfying 𝑥₁<𝑎,𝑏<𝜑𝑥₂,𝑎<𝑥₃with𝜑𝑥₃<𝑏. (2.18)

Remark 2.8. If 𝑑=𝑐, then condition (C₁) of Lemma 2.7 implies condition (C₃) of Lemma 2.7.
For notational convenience, we introduce the following constants: 𝐿₁=^𝜂_𝜉𝜌(𝑠)𝐺₀(𝑠,𝑠)𝑑𝑠,𝐿₂=𝛽[1]1+1−𝛽𝛾¹₀𝐺₀(𝑠,𝑠)𝑑𝑠, (2.19) and a nonnegative continuous concave functional 𝜑 on the cone 𝑃 defined by 𝜑(𝑣)=min_{𝜉≤𝑡≤𝜂}|𝑣(𝑡)|. (2.20)

3. Main Results

In this section, we present and prove our main results.

Theorem 3.1. Assume that (H₁) and (H₂) hold and there exist nonnegative functions ℎ₁,ℎ₂,…,ℎ_𝑛−1∈𝐿[0,1] such that 𝑓𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1−𝑓𝑡,𝑦₁,𝑦₂,…,𝑦_𝑛−1≤^𝑛−1_𝑖=1ℎ_𝑖(𝑡)𝑥_𝑖−𝑦_𝑖, (3.1) for  almost  every 𝑡∈[0,1] and  all (𝑥₁,𝑥₂,…,𝑥_𝑛−1), (𝑦₁,𝑦₂,…,𝑦_𝑛−1)∈𝑅^𝑛−1.
If 0<¹₀𝐺₀(𝑠,𝑠)^𝑛−1_𝑖=1ℎ_𝑖(𝑠)𝛽[1]𝑑𝑠<1+1−𝛽𝛾⁻¹, (3.2) then BVP (1.1) has a unique positive solution.

Proof. We will show that 𝑇 is a contraction mapping. For any 𝑣₁,𝑣₂∈𝑃 and 1≤𝑖≤𝑛−2, by the definition of fractional integral, we obtain 𝐼^𝑖₀₊𝑣₁(𝑡)−𝐼^𝑖₀₊𝑣₂(𝑡=1)(𝑖−1)!^𝑡₀(𝑡−𝑠)^𝑖−1𝑣₁(𝑠)−𝑣₂(𝑠)≤1𝑑𝑠(𝑖−1)!^𝑡₀(𝑡−𝑠)^𝑖−1𝑣₁(𝑠)−𝑣₂(𝑠≤1)𝑑𝑠(𝑖−1)!^𝑡₀(𝑡−𝑠)^𝑖−1𝑑𝑠𝑣₁−𝑣₂=1𝑡𝑖!^𝑖𝑣₁−𝑣₂≤𝑣₁−𝑣₂. (3.3) So, for any 𝑣₁,𝑣₂∈𝑃, by (3.3) and Lemma 2.3, we have 𝑇𝑣₁(𝑡)−𝑇𝑣₂(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)−𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠)(𝑠≤,𝑣)𝑑𝑠₁₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)−𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)𝑑𝑠≤¹₀𝛽[1]1+𝛾𝐺1−𝛽⁰(𝑠,𝑠)^𝑛−1_𝑖=1ℎ_𝑖(𝑠)𝐼^{𝑛−1−𝑖}₀₊𝑣₁(𝑠)−𝐼^{𝑛−1−𝑖}₀₊𝑣₂(𝑠≤𝛽[1])𝑑𝑠1+1−𝛽𝛾¹₀𝐺₀(𝑠,𝑠)^𝑛−1_𝑖=1ℎ_𝑖(𝑠)𝑑𝑠𝑣₁−𝑣₂. (3.4) This implies that 𝑇𝑣₁−𝑇𝑣₂≤𝜅𝑣₁−𝑣₂, (3.5) where 𝜅=(1+𝛽[1]/(1−𝛽[𝛾]))∫¹₀𝐺₀(𝑠,𝑠)∑^𝑛−1_𝑖=1ℎ_𝑖(𝑠)𝑑𝑠∈(0,1). By the Banach contraction mapping principle, we deduce that 𝑇 has a unique fixed point 𝑣^∗. Thus, by Lemma 2.4, 𝑢^∗(𝑡)=𝐼^𝑛−2₀₊𝑣^∗(𝑡) is a unique positive solution of BVP (1.1).

Lemma 3.2. Assume that (H₁) and (H₂) hold and the following conditions are satisfied.(H₃) There exist nonnegative real-valued functions 𝑞,𝑝₁,𝑝₂,…,𝑝_𝑛−1∈𝐿[0,1] such that 𝑓𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1≤𝑞(𝑡)+^𝑛−1_𝑖=1𝑝_𝑖(𝑡)𝑥_𝑖, (3.6)for  almost  every 𝑡∈[0,1] and  all (𝑥₁,𝑥₂,…,𝑥_𝑛−1)∈𝑅^𝑛−1.
Then 𝑇∶𝑃→𝑃 is a completely continuous operator.

Proof. For any 𝑣∈𝑃, as 𝐺(𝑡,𝑠)≥0 for all 𝑡,𝑠∈[0,1], we have 𝑇𝑣(𝑡)≥0, so 𝑇(𝑃)⊂𝑃. Let 𝐷⊂𝑃 be any bounded set. Then there exists a constant 𝐿>0 such that ‖𝑣‖≤𝐿 for any 𝑣∈𝐷. Moreover for any 𝑣∈𝐷, 𝑠∈[0,1], 𝑣(𝑠)≤‖𝑣‖≤𝐿. Proceeding as for (3.3), we obtain 𝐼^{𝑛−1−𝑖}₀₊𝑣(𝑠)=𝐼^{𝑛−1−𝑖}₀₊𝑣(𝑠)≤‖𝑣‖≤𝐿,𝑖=1,2,…,𝑛−1. (3.7) Thus, |𝑇𝑣(𝑡)|=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠)(𝑠)≤,𝑣𝑑𝑠₁₀𝛽[1]1+𝛾𝐺1−𝛽⁰(𝑠,𝑠)𝑞(𝑠)+^𝑛−1_𝑖=1𝑝_𝑖(𝑠)𝐼^{𝑛−1−𝑖}₀₊𝑣(𝑠≤𝛽[1])𝑑𝑠1+1−𝛽𝛾¹₀𝐺₀(𝑠,𝑠)𝑞(𝑠)+^𝑛−1_𝑖=1𝑝_𝑖(𝑠𝑣≤𝛽[1])‖‖𝑑𝑠1+𝛾()1−𝛽𝐿+1¹₀𝐺₀(𝑠,𝑠)𝑞(𝑠)+^𝑛−1_𝑖=1𝑝_𝑖(𝑠)𝑑𝑠<+∞. (3.8) Therefore, 𝑇(𝐷) is uniformly bounded.
Now we show that 𝑇(𝐷) is equicontinuous on [0,1]. Since 𝐺(𝑡,𝑠) is continuous on [0,1]×[0,1], 𝐺(𝑡,𝑠) is uniformly continuous on [0,1]×[0,1]. Hence, for any 𝜀>0, there exists a constant 𝛿₀>0 such that for any 𝑠∈[0,1], 𝑡,𝑡′∈[0,1], when |𝑡−𝑡′|<𝛿₀, it holds 𝐺(𝑡,𝑠)−𝐺𝑡^′,𝑠<1+(𝐿+1)¹₀𝑞(𝑠)+^𝑛−1_𝑖=1𝑝_𝑖(𝑠)𝑑𝑠⁻¹𝜀. (3.9) Consequently, for any 𝑡,𝑡′∈[0,1] and |𝑡−𝑡′|<𝛿₀, we have 𝑇𝑣(𝑡)−𝑇𝑣𝑡^′≤¹₀𝐺(𝑡,𝑠)−𝐺𝑡^′,𝑠𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠)(𝑠)≤,𝑣𝑑𝑠₁₀𝐺(𝑡,𝑠)−𝐺𝑡^′,𝑠𝑞(𝑠)+^𝑛−1_𝑖=1𝑝_𝑖(𝑠𝑣≤())‖‖𝑑𝑠𝐿+1¹₀𝐺(𝑡,𝑠)−𝐺𝑡^′,𝑠𝑞(𝑠)+^𝑛−1_𝑖=1𝑝_𝑖(𝑠)𝑑𝑠<𝜀. (3.10) This implies that 𝑇(𝐷) is equicontinuous. Thus according to the Ascoli-Arzela Theorem, 𝑇(𝐷) is a relatively compact set.
In the end, we show that 𝑇∶𝑃→𝑃 is continuous. Assume that 𝑣_𝑚,𝑣₀∈𝑃(𝑚=1,2,…), 𝑣_𝑚→𝑣₀(𝑚→+∞), then 𝑣_𝑚(𝑡)−𝑣₀(𝑡)≤𝑣_𝑚−𝑣₀→0, (3.11) and ‖𝑣_𝑚‖≤𝐿 (𝑚=0,1,2,…), where 𝐿 is a positive constant. Keeping in mind that 𝑓 satisfies Carathéodory conditions on [0,1]×ℝ^𝑛−1, we have lim_𝑚→+∞𝑓𝑡,𝐼^𝑛−2₀₊𝑣_𝑚(𝑡),𝐼^𝑛−3₀₊𝑣_𝑚(𝑡),…,𝐼¹₀₊𝑣_𝑚(𝑡),𝑣_𝑚(𝑡)=𝑓𝑡,𝐼^𝑛−2₀₊𝑣₀(𝑡),𝐼^𝑛−3₀₊𝑣₀(𝑡),…,𝐼¹₀₊𝑣₀(𝑡),𝑣₀(𝑡),fora.e.𝑡∈[0,1]. (3.12) Proceeding as for (3.3), for 𝑚∈ℕ we obtain 𝐼^{𝑛−1−𝑖}₀₊𝑣_𝑚(𝑠)=𝐼^{𝑛−1−𝑖}₀₊𝑣_𝑚(𝑠)≤𝑣_𝑚≤𝐿𝑖=1,2,…,𝑛−1. (3.13) This together with (3.6), 0≤𝑓𝑡,𝐼^𝑛−2₀₊𝑣_𝑚(𝑡),𝐼^𝑛−3₀₊𝑣_𝑚(𝑡),…,𝐼¹₀₊𝑣_𝑚(𝑡),𝑣_𝑚(𝑡)≤𝑞(𝑡)+𝐿^𝑛−1_𝑖=1𝑝_𝑖(𝑡). (3.14) The Lebesgue dominated convergence theorem gives lim_𝑚→+∞¹₀𝑓𝑠,𝐼^𝑛−2₀₊𝑣_𝑚(𝑠),𝐼^𝑛−3₀₊𝑣_𝑚(𝑠),…,𝐼¹₀₊𝑣_𝑚(𝑠),𝑣_𝑚(𝑠)−𝑓𝑠,𝐼^𝑛−2₀₊𝑣₀(𝑠),𝐼^𝑛−3₀₊𝑣₀(𝑠),…,𝐼¹₀₊𝑣₀(𝑠),𝑣₀(𝑠)𝑑𝑠=0. (3.15) Now we deduce from (3.15), Lemma 2.3𝑇𝑣_𝑚(𝑡)−𝑇𝑣₀(𝑡=₁₀)𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣_𝑚(𝑠),𝐼^𝑛−3₀₊𝑣_𝑚(𝑠),…,𝐼¹₀₊𝑣_𝑚(𝑠),𝑣_𝑚(𝑠)−𝑓𝑠,𝐼^𝑛−2₀₊𝑣₀(𝑠),𝐼^𝑛−3₀₊𝑣₀(𝑠),…,𝐼¹₀₊𝑣₀(𝑠),𝑣₀(𝑠)≤𝛽[1]𝑑𝑠1+1−𝛽𝛾¹₀𝐺₀(𝑠,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣_𝑚(𝑠),𝐼^𝑛−3₀₊𝑣_𝑚(𝑠),…,𝐼¹₀₊𝑣_𝑚(𝑠),𝑣_𝑚(𝑠)−𝑓𝑠,𝐼^𝑛−2₀₊𝑣₀(𝑠),𝐼^𝑛−3₀₊𝑣₀(𝑠),…,𝐼¹₀₊𝑣₀(𝑠),𝑣₀(𝑠)≤𝛽[1]𝑑𝑠1+1−𝛽𝛾max_𝑠∈[0,1]𝐺₀(𝑠,𝑠)¹₀𝑓𝑠,𝐼^𝑛−2₀₊𝑣_𝑚(𝑠),𝐼^𝑛−3₀₊𝑣_𝑚(𝑠),…,𝐼¹₀₊𝑣_𝑚(𝑠),𝑣_𝑚(𝑠)−𝑓𝑠,𝐼^𝑛−2₀₊𝑣₀(𝑠),𝐼^𝑛−3₀₊𝑣₀(𝑠),…,𝐼¹₀₊𝑣₀(𝑠),𝑣₀(𝑠)𝑑𝑠 (3.16) that ‖𝑇𝑣_𝑚−𝑇𝑣₀‖→0, as 𝑚→+∞. So 𝑇∶𝑃→𝑃 is continuous. Therefore 𝑇∶𝑃→𝑃 is completely continuous.

Remark 3.3. If 𝑓∶[0,1]×ℝ^𝑛−1 is continuous, by similar argument as above, we can show that 𝑇 is completely continuous.

Theorem 3.4. Assume that (H₁)–(H₃) hold. If ¹₀𝐺₀(𝑠,𝑠)^𝑛−1_𝑖=1𝑝_𝑖(𝑠)𝑑𝑠<1+𝛽[1]1−𝛽[𝛾]⁻¹, (3.17) then BVP (1.1) has at least one positive solution.

Proof. Let {‖𝑣‖}[1]/𝛾∫Ω=𝑣∈𝑃∶<𝑟,where𝑟=1+𝛽1−𝛽¹₀𝐺₀(𝑠,𝑠)𝑞(𝑠)𝑑𝑠1−1+𝛽[1]/1−𝛽𝛾∫¹₀𝐺₀(𝑠,𝑠)∑^𝑛−1_𝑖=1𝑝_𝑖(𝑠)𝑑𝑠, (3.18) we have Ω⊂𝑃. From Lemma 3.2, we know that 𝑇∶Ω→𝑃 is completely continuous. If there exists 𝑣∈𝜕Ω,𝜇∈(0,1) such that 𝑣=𝜇𝑇𝑣, (3.19) then by (H₃) and (3.19), we have 𝑣(𝑡)=𝜇𝑇𝑣(𝑡)=𝜇¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)𝑑𝑠≤𝜇¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠)(𝑠𝛽[1],𝑣)𝑑𝑠≤𝜇1+1−𝛽𝛾¹₀𝐺₀(𝑠,𝑠)𝑞(𝑠)𝑑𝑠+¹₀𝐺₀(𝑠,𝑠)^𝑛−1_𝑖=1𝑝_𝑖(𝑠)𝑑𝑠‖𝑣‖, (3.20) which implies that ‖𝑣‖𝛽[1]≤𝜇1+1−𝛽𝛾¹₀𝐺₀(𝑠,𝑠)𝑞(𝑠)𝑑𝑠+𝑟¹₀𝐺₀(𝑠,𝑠)^𝑛−1_𝑖=1𝑝_𝑖(𝑠)<𝛽[1]𝑑𝑠1+1−𝛽𝛾¹₀𝐺₀(𝑠,𝑠)𝑞(𝑠)𝑑𝑠+𝑟¹₀𝐺₀(𝑠,𝑠)^𝑛−1_𝑖=1𝑝_𝑖(𝑠)𝑑𝑠=𝑟. (3.21) This means that 𝑣∉𝜕Ω. By Lemma 2.5, 𝑇 has a fixed point 𝑣∈Ω. By Lemma 2.4, BVP (1.1) has at least one positive solution 𝑢(𝑡)=𝐼^𝑛−2₀₊𝑣(𝑡).

Theorem 3.5. Assume that (H₁)–(H₃) hold. If there exist two positive constants 𝑟₁<𝑟₂ such that (i)𝑓(𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)≤𝐿₂⁻¹𝑟₂, for (𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)∈[0,1]×[0,𝑟₂]×⋯×[0,𝑟₂],(ii)𝑓(𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)≥𝐿₁⁻¹𝑟₁, for (𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)∈[0,1]×[0,𝑟₁]×⋯×[0,𝑟₁],where 𝐿₁, 𝐿₂ are defined by (2.19), then BVP (1.1) has at least one positive solution.

Proof. Let Ω₂={𝑣∈𝑃∶‖𝑣‖<𝑟₂}. For any 𝑣∈𝜕Ω₂, we have ‖𝑣‖=𝑟₂ and 0≤𝑣(𝑡)≤𝑟₂ for every 𝑡∈[0,1]. Similar to (3.7), for 0≤𝑣(𝑠)≤𝑟₂, we have 0≤𝐼^{𝑛−1−𝑖}₀₊𝑣(𝑠)=𝐼^{𝑛−1−𝑖}₀₊𝑣(𝑠)≤‖𝑣‖≤𝑟₂,𝑖=1,2,…,𝑛−1. (3.22) It follows from condition (i) that 𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)≤𝐿₂⁻¹𝑟₂,for(𝑠,𝑣)∈[0,1]×0,𝑟₂. (3.23) Thus, for any 𝑣∈𝜕Ω₂, by (3.23) and Lemma 2.3, we have |𝑇𝑣(𝑡)|=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠)(𝑠)≤,𝑣𝑑𝑠₁₀𝐺(𝑡,𝑠)𝐿₂⁻¹𝑟₂𝑑𝑠≤1+𝛽[1]/1−𝛽𝛾𝐿₂⁻¹𝑟₂¹₀𝐺₀(𝑠,𝑠)𝑑𝑠=𝑟₂=‖𝑣‖,𝑡∈[0,1], (3.24) which means that ‖𝑇𝑣‖≤‖𝑣‖,𝑣∈𝜕Ω₂. (3.25)
On the other hand, let Ω₁={𝑣∈𝑃∶‖𝑣‖<𝑟₁}. For any 𝑣∈𝜕Ω₁, we have ‖𝑣‖=𝑟₁ and 0≤𝑣(𝑡)≤𝑟₁ for every 𝑡∈[0,1]. Similar to (3.23), from condition (ii), we can get 𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)≥𝐿₁⁻¹𝑟₁,for(𝑠,𝑣)∈[0,1]×0,𝑟₁. (3.26) Hence for any 𝑡∈[𝜉,𝜂], 𝑣∈𝜕Ω₁, by (3.26) and Lemma 2.3 we have |𝑇𝑣(𝑡)|=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠)(𝑠)≥,𝑣𝑑𝑠₁₀𝐺(𝑡,𝑠)𝐿₁⁻¹𝑟₁𝑑𝑠≥𝐿₁⁻¹𝑟₁^𝜂_𝜉𝐺(𝑡,𝑠)𝑑𝑠≥𝐿₁⁻¹𝑟₁^𝜂_𝜉𝜌(𝑠)𝐺₀(𝑠,𝑠)𝑑𝑠=𝑟₁=‖𝑣‖. (3.27) Thus we get ‖𝑇𝑣‖≥‖𝑣‖,𝑣∈𝜕Ω₁. (3.28) By (3.25), (3.28), and Lemma 2.6, 𝑇 has a fixed point 𝑣∈Ω₂⧵Ω₁ such that 𝑟₁≤‖𝑣‖≤𝑟₂. By Lemma 2.4, BVP (1.1) has at least one positive solution 𝑢(𝑡)=𝐼^𝑛−2₀₊𝑣(𝑡).

Theorem 3.6. Assume that (H₁)–(H₃) hold. If there exist constants 0<𝑎<𝑏<𝑐 such that (I)𝑓(𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)<𝐿₂⁻¹𝑎, for (𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)∈[0,1]×[0,𝑎]×⋯×[0,𝑎],(II)𝑓(𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)≤𝐿₂⁻¹𝑐, for (𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)∈[0,1]×[0,𝑐]×⋯×[0,𝑐],(III)𝑓(𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)≥𝐿₁⁻¹𝑏, for (𝑡,𝑥₁,𝑥₂,…,𝑥_𝑛−1)∈[𝜉,𝜂]×[(𝑏/(𝑛−2)!)𝜉^𝑛−2,𝑐]×[(𝑏/(𝑛−3)!)𝜉^𝑛−3,𝑐]×⋯×[𝑏𝜉,𝑐]×[𝑏,𝑐],where 𝐿₁, 𝐿₂ are defined by (2.19), then BVP (1.1) has at least three positive solutions 𝑢₁, 𝑢₂, and 𝑢₃ satisfying 𝐷^𝑛−2₀₊𝑢₁<𝑎,𝑏<𝜑𝐷^𝑛−2₀₊𝑢₂<𝐷^𝑛−2₀₊𝑢₂𝐷≤𝑐,𝑎<^𝑛−2₀₊𝑢₃,𝜑𝐷^𝑛−2₀₊𝑢₃<𝑏. (3.29)

Proof. We will show that all conditions of Lemma 2.7 are satisfied.
First, if 𝑣∈𝑃_𝑐, then ‖𝑣‖≤𝑐. So we have 0≤𝑣(𝑡)≤𝑐, 𝑡∈[0,1]. Similar to (3.23), it follows from condition (II) that 𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)≤𝐿₂⁻¹𝑐,for(𝑠,𝑣)∈[0,1]×[0,𝑐]. (3.30) Thus, for any 𝑣∈𝑃_𝑐, by (3.30), we have |𝑇𝑣(𝑡)|=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠)(𝑠)≤,𝑣𝑑𝑠₁₀𝐺(𝑡,𝑠)𝐿₂⁻¹𝛽[1]𝑐𝑑𝑠≤1+𝛾𝐿1−𝛽²⁻¹𝑐¹₀𝐺₀(𝑠,𝑠)𝑑𝑠=𝑐, (3.31) which means that ‖𝑇𝑣‖≤𝑐, 𝑣∈𝑃_𝑐. Therefore, 𝑇∶𝑃_𝑐→𝑃_𝑐. By Lemma 3.2, we know that 𝑇∶𝑃_𝑐→𝑃_𝑐 is completely continuous.
Next, similar to (3.30) and (3.31), it follows from condition (I) that if 𝑣∈𝑃_𝑎 then ‖𝑇𝑣‖<𝑎. So the condition (𝐶₂) of Lemma 2.7 holds.
Now, we take 𝑣(𝑡)=(𝑏+𝑐)/2, 𝑡∈[0,1]. It is easy to see that 𝑣(𝑡)=(𝑏+𝑐)/2∈𝑃(𝜑,𝑏,𝑐), and so 𝜑(𝑣)=min_{𝜉≤𝑡≤𝜂}|𝑣(𝑡)|=𝑏+𝑐2>𝑏, (3.32) where 𝜑(𝑣) is defined by (2.20). This proves that {𝑣∈𝑃(𝜑,𝑏,𝑐)∶𝜑(𝑣)>𝑏}≠𝜙.
On the other hand, if 𝑣∈𝑃(𝜑,𝑏,𝑐), then 𝑏≤𝑣(𝑡)≤𝑐, 𝑡∈[𝜉,𝜂]. By the definition of fractional integral, for any 𝑡∈[𝜉,𝜂], 1≤𝑖≤𝑛−2, we obtain 𝑏𝜉𝑖!^𝑖≤𝑏(𝑖−1)!^𝑡₀(𝑡−𝑠)^𝑖−1𝑑𝑠≤𝐼^𝑖₀₊𝑣(𝑡)=1(𝑖−1)!^𝑡₀(𝑡−𝑠)^𝑖−1𝑣(𝑠)≤𝑐𝑑𝑠(𝑖−1)!^𝑡₀(𝑡−𝑠)^𝑖−1≤𝑐𝑑𝑠𝜂𝑖!^𝑖≤𝑐. (3.33) It follows from (3.33) and condition (III) that 𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)≥𝐿₁⁻¹𝑏,for𝑠∈𝜉,𝜂,𝑣∈𝑃𝜑,𝑏,𝑐. (3.34) Hence we have 𝜑(𝑇𝑣)=min_{𝜉≤𝑡≤𝜂}|𝑇𝑣(𝑡)|=min_{𝜉≤𝑡≤𝜂}¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)𝑑𝑠≥𝐿₁⁻¹𝑏min_{𝜉≤𝑡≤𝜂}¹₀𝐺(𝑡,𝑠)𝑑𝑠>𝐿₁⁻¹𝑏^𝜂_𝜉𝜌(𝑠)𝐺₀(𝑠,𝑠)𝑑𝑠=𝑏, (3.35) which implies that 𝜑(𝑇𝑣)>𝑏, for 𝑣∈𝑃(𝜑,𝑏,𝑐). This shows that condition (C₁) of Lemma 2.7 is also satisfied.
By Lemma 2.7 and Remark 2.8, BVP (2.13) has at least three positive solutions 𝑣₁,𝑣₂, and 𝑣₃ such that ‖𝑣₁‖<𝑎, 𝑏<𝜑(𝑣₂)<‖𝑣₂‖≤𝑐, and 𝑎<‖𝑣₃‖, 𝜑(𝑣₃)<𝑏. By Lemma 2.4, BVP (1.1) has at least three positive solutions 𝑢_𝑖(𝑡)=𝐼^𝑛−2₀₊𝑣_𝑖(𝑡), (𝑖=1,2,3). By (2.3), we have 𝐷^𝑛−2₀₊𝑢_𝑖(𝑡)=𝐷^𝑛−2₀₊𝐼^𝑛−2₀₊𝑣_𝑖(𝑡)=𝑣_𝑖, 𝑖=1,2,3. So 𝑢₁,𝑢₂,𝑢₃ are three positive solutions of BVP (1.1) satisfying 𝐷^𝑛−2₀₊𝑢₁<𝑎,𝑏<𝜑𝐷^𝑛−2₀₊𝑢₂<𝐷^𝑛−2₀₊𝑢₂𝐷≤𝑐,𝑎<^𝑛−2₀₊𝑢₃,𝜑𝐷^𝑛−2₀₊𝑢₃<𝑏. (3.36) The proof of Theorem 3.6 is completed.

4. Examples

Example 4.1. Consider the following problem: 𝐷^7/2₀₊𝑢(𝑡)+(1−𝑡)³𝑒^𝑡𝑢(1+𝑒^𝑡)+1)(1+𝑢2𝑡²sin²1𝑢′+4𝑡𝑢^′′𝑢(0)=0,0<𝑡<1,=𝑢^′(0)=𝑢^′′(0)=0,𝑢^′′(1)=𝛽𝑢^′′. (4.1) Let 𝛽[𝑢^′′]=(1/2)𝑢^′′(1/2). Then 𝐺₀()=1𝑡,𝑠Γ()()]3/2[𝑡1−𝑠^1/2−(𝑡−𝑠)^1/2[𝑡(,0≤𝑠≤𝑡≤1,1−𝑠)]^1/2𝒢(𝑠)=1,0≤𝑡≤𝑠≤1.2𝐺₀12,𝑠≥0,𝛽[1]=¹₀(𝑡)=1𝑑𝐴2,𝛽𝛾=¹₀𝑡^1/2(𝑡)=√𝑑𝐴24<1. (4.2)(4.3) Let 𝑓=()𝑡,𝑥,𝑦,𝑧1−𝑡³𝑒^𝑡𝑥(1+𝑒^𝑡)+1)(1+𝑥2𝑡²sin²1𝑦+4ℎ𝑡𝑧,₁(𝑡)=(1−𝑡)³𝑒^𝑡1+𝑒^𝑡,ℎ₂(𝑡)=12𝑡²,ℎ₃(𝑡)=14𝑡. (4.4) Then 𝑓 is a nonnegative continuous function on [0,1]×(ℝ⁺)³ and, for any (𝑡,𝑥₁,𝑦₁,𝑧₁) and (𝑡,𝑥₂,𝑦₂,𝑧₂)∈[0,1]×(ℝ⁺)³, satisfies 𝑓𝑡,𝑥₁,𝑦₁,𝑧₁−𝑓𝑡,𝑥₂,𝑦₂,𝑧₂≤ℎ₁(𝑡)𝑥₁−𝑥₂+ℎ₂(𝑡)𝑦₁−𝑦₂+ℎ₃(𝑡)𝑧₁−𝑧₂. (4.5) So we have ¹₀𝐺₀(𝑠,𝑠)³_𝑖=1ℎ_𝑖(𝑠)1𝑑𝑠≤Γ(3/2)¹₀(𝑠(1−𝑠))^1/2(1−𝑠)³+𝑠²=𝐵()()()+𝑠𝑑𝑠3/2,9/2+𝐵7/2,3/2+𝐵5/2,3/2Γ()=3/233√128<𝛽[1]𝜋≈0.45696081+1−𝛽𝛾⁻¹=4−√26−√2≈0.5638901, (4.6) where 𝐵(⋅,⋅) denotes a Beta function. So all conditions of Theorem 3.1 are satisfied. Thus, by Theorem 3.1, BVP (4.1) has at least one positive solution.

Example 4.2. Consider the following problem: 𝐷^5/2₀₊𝑢(𝑡)+12𝑡−𝑡²()+1ln1+𝑢2𝑡²𝑢′+𝑡³𝑢(0)+sin𝑡=0,0<𝑡<1,=𝑢^′(0)=0,𝑢^′(1)=𝛽𝑢^′, (4.7) where 𝛽[𝑢′]=∫¹₀𝑢′(𝑠)𝑑𝐴(𝑠) with 𝐴(𝑠)=10,𝑠∈0,4,12,𝑠∈4,9,9161,𝑠∈.16,1 (4.8) Set 𝑓=1𝑡,𝑥,𝑦2𝑡−𝑡²()+1ln1+𝑥2𝑡²𝑦+𝑡³+sin𝑡,𝑝₁(𝑡)=12𝑡−𝑡²,𝑝₂(𝑡)=12𝑡²,𝑞(𝑡)=𝑡³+1. (4.9) Then 𝑓∶[0,1]×[0,+∞)×[0,+∞)→[0,+∞) is continuous and 𝑓𝑡,𝑥,𝑦≤𝑝₁(𝑡)𝑥+𝑝₂(𝑡)𝑦+𝑞(𝑡). (4.10) As in [21], 𝛽[𝛾]=∫¹₀𝛾(𝑡)𝑑𝐴(𝑡)=21/4+(−1)×9/16=1/4<1, 𝒢(𝑠)=∫¹₀𝐺₀(𝑡,𝑠)𝑑𝐴(𝑡)≥0, ¹₀𝐺₀(𝑠,𝑠)𝑝₁(𝑠)+𝑝₂(𝑠)1𝑑𝑠≤Γ(3/2)¹₀(𝑠(1−𝑠))^1/2𝑠−𝑠²+𝑠²=𝐵()()𝑑𝑠5/2,5/2+𝐵7/2,3/2Γ()=13/28√<𝛽[1]𝜋≈0.221556731+1−𝛽𝛾⁻¹=37≈0.42857143, (4.11) where 𝐵(⋅,⋅) denotes a Beta function and 𝐺₀(𝑡,𝑠) is defined by (4.2). So all conditions of Theorem 3.4 are satisfied. Thus, by Theorem 3.4, BVP (4.7) has at least one positive solution.

Example 4.3. Consider the following problem: 𝐷^5/2₀₊𝑢(𝑡)+𝑡²5ln(1+𝑢)+𝑡𝑒^𝑡𝑢′10+10𝑒^𝑡+sin𝑡+1202𝑢(0)=0,0<𝑡<1,=𝑢^′(0)=0,𝑢^′(1)=𝛽𝑢^′, (4.12) where 𝛽[𝑢′]=∫¹₀𝑢′(𝑠)𝑑𝐴(𝑠) with 𝐴(𝑠) as given by (4.8). Set 𝑓=𝑡𝑡,𝑥,𝑦²5ln(1+𝑥)+𝑡𝑒^𝑡𝑦10+10𝑒^𝑡+sin𝑡+1202,𝑝₁(𝑡)=𝑡²5,𝑝₂(𝑡)=𝑡𝑒^𝑡10+10𝑒^𝑡,𝑞(𝑡)=sin𝑡+1202. (4.13) Then 𝑓∶[0,1]×[0,+∞)×[0,+∞)→[0,+∞) is continuous and 𝑓𝑡,𝑥,𝑦≤𝑝₁(𝑡)𝑥+𝑝₂(𝑡)𝑦+𝑞(𝑡). (4.14) By Example 4.2, 𝛽[𝛾]=∫¹₀𝛾(𝑡)𝑑𝐴(𝑡)=1/4<1, 𝒢(𝑠)=∫¹₀𝐺₀(𝑡,𝑠)𝑑𝐴(𝑡)≥0, where 𝐺₀(𝑡,𝑠) is defined by (4.2). As in [1, 3], we also take 𝜉=1/4,𝜂=3/4, then 𝐿₁⁻¹=^3/4_1/4𝜌(𝑠)𝐺₀(𝑠,𝑠)𝑑𝑠⁻¹𝐿≈13.6649,₂₋₁=𝛽[1]1+1−𝛽𝛾⁻¹¹₀𝐺₀(𝑠,𝑠)𝑑𝑠⁻¹=127√𝜋≈0.967182. (4.15) Choosing 𝑟₁=1/30,𝑟₂=1, we have 𝑓𝑡,𝑥,𝑦≤0.85≤𝐿₂⁻¹𝑟₂∈[]×[]×[],𝑓,for𝑡,𝑥,𝑦0,10,10,1𝑡,𝑥,𝑦≥0.5≥𝐿₁⁻¹𝑟₁∈[]×1,for𝑡,𝑥,𝑦0,10,×1300,.30 (4.16) So all conditions of Theorem 3.5 are satisfied. Thus, by Theorem 3.5, BVP (4.12) has at least one positive solution.

Example 4.4. Consider the following problem: 𝐷^5/2₀₊𝑢(𝑡)+𝑓𝑡,𝑢(𝑡),𝑢^′(𝑡)𝑢(0)=0,0<𝑡<1,=𝑢^′(0)=0,𝑢^′(1)=𝛽𝑢^′, (4.17) where 𝛽[𝑢′]=∫¹₀𝑢′(𝑠)𝑑𝐴(𝑠) with 𝐴(𝑠) as given by (4.8). Set 𝑓=𝑡𝑡,𝑥,𝑦²()100ln1+𝑥+15𝑦²+𝑡,∈[]×[]×[],𝑡1000𝑡,𝑥,𝑦0,10,10,1₂()+100ln1+𝑥292+12𝑡𝑦+,∈[]×ℝ1000𝑡,𝑥,𝑦0,1⁺²⧵[0,1]²,𝑝₁(𝑡)=𝑡²100,𝑝₂(𝑡)(𝑡)=𝑡=15,𝑞+1000292. (4.18) Then 𝑓𝑡,𝑥,𝑦≤𝑝₁(𝑡)𝑥+𝑝₂(𝑡)𝑦+𝑞(𝑡). (4.19) By Example 4.3, 𝛽[𝛾]=∫¹₀𝛾(𝑡)𝑑𝐴(𝑡)=1/4<1, 𝒢(𝑠)=∫¹₀𝐺₀(𝑡,𝑠)𝑑𝐴(𝑡)≥0, 𝐿₁⁻¹=^3/4_1/4𝜌(𝑠)𝐺₀(𝑠,𝑠)𝑑𝑠⁻¹𝐿≈13.6649,₂₋₁=𝛽[1]1+1−𝛽𝛾⁻¹¹₀𝐺₀(𝑠,𝑠)𝑑𝑠⁻¹=127√𝜋≈0.967182. (4.20) Choosing 𝑎=1/20, 𝑏=1, 𝑐=100, we have 𝑓𝑡,𝑥,𝑦≤0.044845<𝐿₂⁻¹∈[]×1𝑎≈0.048359,for𝑡,𝑥,𝑦0,10,×1200,,𝑓20𝑡,𝑥,𝑦≥14.5025≥𝐿₁⁻¹∈1𝑏≈13.6649,for𝑡,𝑥,𝑦4,34×14×[],𝑓,1001,100𝑡,𝑥,𝑦≤65.501≤𝐿₂⁻¹𝑐≈96.7182,for𝑡,𝑥,𝑦∈[0,1]×[0,100]×[0,100]. (4.21) So all conditions of Theorem 3.6 are satisfied. Thus, by Theorem 3.6, BVP (4.17) has at least three positive solutions 𝑢₁,𝑢₂, and 𝑢₃, satisfying max_0≤𝑡≤1𝑢^′₁(𝑡<1)20,1<min_{1/4≤𝑡≤3/4}𝑢^′₂(𝑡)<max_0≤𝑡≤1𝑢^′₂(𝑡1)≤100,20≤max_0≤𝑡≤1𝑢^′₃(𝑡)≤100withmin_{1/4≤𝑡≤3/4}𝑢^′₃(𝑡)<1. (4.22)

Acknowledgments

The first and second authors were supported financially by the National Natural Science Foundation of China (11071141, 11126231) and the Natural Science Foundation of Shandong Province of China (ZR2011AQ008, ZR2010AM017). The third author was supported financially by the Australia Research Council through an ARC Discovery Project grant.