Upper and Lower Solutions Method for a Class of Singular Fractional Boundary Value Problems with p-Laplacian Operator

Abstract

The upper and lower solutions method is used to study the 𝑝-Laplacian fractional boundary value problem 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝑢(𝑡)))=𝑓(𝑡,𝑢(𝑡)), 0<𝑡<1, 𝑢(0)=0, 𝑢(1)=𝑎𝑢(𝜉), 𝐷^𝛼₀₊𝑢(0)=0, and 𝐷^𝛼₀₊𝑢(1)=𝑏𝐷^𝛼₀₊𝑢(𝜂), where 1<𝛼,𝛾⩽2,0⩽𝑎,𝑏⩽1,0<𝜉,𝜂<1. Some new results on the existence of at least one positive solution are obtained. It is valuable to point out that the nonlinearity 𝑓 can be singular at 𝑡=0,1 or 𝑢=0.

1. Introduction

It is well know that the upper and lower solutions method is a powerful tool for proving the existence results for boundary value problem. It has been used to deal with many multipoint boundary value problem of integer ordinary differential equations (see, e.g., [1–3] and references therein).

Recently, boundary value problems of nonlinear fractional differential equations have aroused considerable attention. Many people pay attention to the existence and multiplicity of solutions or positive solutions for boundary value problems of nonlinear fractional differential equations by means of some fixed point theorems, such as the Krasnosel'skii fixed-point theorem, the Leggett-Williams fixed-point theorem, and the Schauder fixed-point theorem (see [4–8]). To the best of our knowledge, the upper and lower solutions method is seldom considered in the literatures, and there are few papers devoted to investigate 𝑝-Laplacian fractional boundary value problems.

In this paper, we deal with the following 𝑝-Laplacian fractional boundary value problem: 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝑢(𝑡)))=𝑓(𝑡,𝑢(𝑡)),0<𝑡<1,𝑢(0)=0,𝑢(1)=𝑎𝑢(𝜉),𝐷^𝛼₀₊𝑢(0)=0,𝐷^𝛼₀₊𝑢(1)=𝑏𝐷^𝛼₀₊𝑢(𝜂),(1.1) where 1<𝛼,𝛾⩽2,0⩽𝑎,𝑏⩽1,0<𝜉,𝜂<1, and 𝐷^𝛼 is the standard Riemann-Liouville fractional differential operator of order 𝛼, 𝜙_𝑝(𝑠)=|𝑠|^𝑝−2𝑠, 𝑝>1, (𝜙_𝑝)⁻¹=𝜙_𝑞,(1/𝑝)+(1/𝑞)=1. By using upper and lower solutions method, the existence results of at least one positive solution for the above fractional boundary value problem are established, and an example is given to show the effectiveness of our results. It is valuable to point out that the nonlinearity 𝑓 can be singular at 𝑡=0,1 or 𝑢=0.

The remaining part of the paper is organized as follows. In Section 2, we will present some definitions and lemmas. In Section 3, some results are given. In Section 4, we present an example to demonstrate our results.

2. Basic Definitions and Preliminaries

In this section, we present some necessary definitions and lemmas.

Definition 2.1 (see [9]). The integral 𝐼^𝛼₀₊1𝑦(𝑡)=Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠,(2.1) where 𝛼>0, is called the Riemann-Liouville fractional integral of order 𝛼.

Definition 2.2 (see [10]). For a function 𝑦(𝑡) given in the interval [0,∞), the expression 𝐷^𝛼₀₊1𝑦(𝑡)=(𝑑Γ(𝑛−𝛼))𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑦(𝑠)𝑑𝑠,(2.2) where 𝑛=[𝛼]+1, and [𝛼] denotes the integer part of number 𝛼, is called the Riemann-Liouville fractional derivative of order 𝛼.

Remark 2.3. From the definition of the Riemann-Liouville fractional derivative, we quote for 𝜇>−1, then 𝐷^𝛼₀₊𝑡^𝜇=Γ(1+𝜇)𝑡Γ(1+𝜇−𝛼)^𝜇−𝛼.(2.3)
In particular, 𝐷^𝛼₀₊𝑡^𝛼−𝑚=0,𝑚=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

Lemma 2.4 (see [4]). Assume that 𝑢∈𝐶(0,1)∩𝐿(0,1) with a fractional derivative of order 𝛼>0 that belongs to 𝐶(0,1)∩𝐿(0,1). Then 𝐼^𝛼₀₊𝐷^𝛼₀₊𝑢(𝑡)=𝑢(𝑡)+𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2+⋯+𝐶_𝑁𝑡^𝛼−𝑁,(2.4) for some 𝐶_𝑖∈𝑅,𝑖=1,2,…𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼.

Lemma 2.5 (see [6]). Let 𝑦∈𝐶[0,1] and 1<𝛼⩽2, the unique solution of 𝐷^𝛼₀₊𝑢(𝑡)+𝑦(𝑡)=0,0<𝑡<1,𝑢(0)=0,𝑢(1)=𝑎𝑢(𝜉),(2.5) is 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠,(2.6) where 𝐺(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−𝑎𝑡^𝛼−1(𝜉−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)(1−𝑎𝜉^𝛼−1)Γ(𝛼),0⩽𝑠⩽𝑡⩽1,𝑠⩽𝜉,[𝑡(1−s)]^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)(1−𝑎𝜉^𝛼−1)Γ(𝛼),0<𝜉⩽𝑠⩽𝑡⩽1,[𝑡(1−𝑠)]^𝛼−1−𝑎𝑡^𝛼−1(𝜉−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)Γ(𝛼),0⩽𝑡⩽𝑠⩽𝜉⩽1,[𝑡(1−𝑠)]^𝛼−1(1−𝑎𝜉^𝛼−1)Γ(𝛼),0⩽𝑡⩽𝑠⩽1,𝜉⩽𝑠.(2.7)

Lemma 2.6. Let 𝑦∈𝐶[0,1], 1<𝛼,𝛾⩽2, 0<𝜉,𝜂<1, 0⩽𝑎,and𝑏⩽1. The fractional boundary value problem 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝑢(𝑡)))=𝑦(𝑡),0<𝑡<1,𝑢(0)=0,𝑢(1)=𝑎𝑢(𝜉),𝐷^𝛼₀₊𝑢(0)=0,𝐷^𝛼₀₊𝑢(1)=𝑏𝐷^𝛼₀₊𝑢(𝜂),(2.8) has a unique solution 𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑦(𝜏)𝑑𝜏)𝑑𝑠,(2.9) where 𝐻(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛾−1−𝑏₁𝑡^𝛾−1(𝜂−𝑠)^𝛾−1−(𝑡−𝑠)^𝛾−1(1−𝑏₁𝜂^𝛾−1)(1−𝑏₁𝜂^𝛾−1)Γ(𝛾),0⩽𝑠⩽𝑡⩽1,𝑠⩽𝜂,[𝑡(1−𝑠)]^𝛾−1−(𝑡−𝑠)^𝛾−1(1−𝑏₁𝜂^𝛾−1)(1−𝑏₁𝜂^𝛾−1)Γ(𝛾),0<𝜂⩽𝑠⩽𝑡⩽1,[𝑡(1−𝑠)]^𝛾−1−𝑏₁𝑡^𝛾−1(𝜂−𝑠)^𝛾−1(1−𝑏₁𝜂^𝛾−1)Γ(𝛾),0⩽𝑡⩽𝑠⩽𝜂⩽1,[𝑡(1−𝑠)]^𝛾−1(1−𝑏₁𝜂^𝛾−1)Γ(𝛾),0⩽𝑡⩽𝑠⩽1,𝜂⩽𝑠,(2.10)𝑏₁=𝑏^𝑝−1 and 𝐺(𝑡,𝑠) is defined by (2.7).

Proof. At first, by Lemma 2.4, the (2.8) is equivalent to the integral equation 𝜙_𝑝(𝐷^𝛼₀₊𝑢(𝑡))=𝐼^𝛾₀₊𝑦(𝑡)+C₁𝑡^𝛾−1+𝐶₂𝑡^𝛾−2,𝐶₁,𝐶₁∈𝑅,(2.11) that is, 𝜙_𝑝(𝐷^𝛼₀₊𝑢(𝑡))=^𝑡₀(𝑡−𝜏)^𝛾−1Γ(𝛾)𝑦(𝜏)𝑑𝜏+𝐶₁𝑡^𝛾−1+𝐶₂𝑡^𝛾−2.(2.12) By the boundary conditions 𝐷^𝛼₀₊𝑢(0)=0,𝐷^𝛼₀₊𝑢(1)=𝑏𝐷^𝛼₀₊𝑢(𝜂), we have 𝐶₂=0,𝐶₁=−¹₀(1−𝜏)^𝛾−1Γ(𝛾)(1−𝑏₁𝜂^𝛾−1)𝑦(𝜏)𝑑𝜏+^𝜂₀𝑏₁(𝜂−𝜏)^𝛾−1Γ(𝛾)(1−𝑏₁𝜂^𝛾−1)𝑦(𝜏)𝑑𝜏.(2.13) Therefore, the solution 𝑢(𝑡) of boundary value problem (2.8) satisfies 𝜙_𝑝(𝐷^𝛼₀₊𝑢(𝑡))=^𝑡₀(𝑡−𝜏)^𝛾−1Γ(𝛾)𝑦(𝜏)𝑑𝜏−¹₀𝑡^𝛾−1(1−𝜏)^𝛾−1Γ(𝛾)(1−𝑏₁𝜂^𝛾−1)+𝑦(𝜏)𝑑𝜏_𝜂0𝑏₁𝑡^𝛾−1(𝜂−𝜏)^𝛾−1Γ(𝛾)(1−𝑏₁𝜂^𝛾−1)𝑦(𝜏)𝑑𝜏=−¹₀𝐻(𝑡,𝜏)𝑦(𝜏)𝑑𝜏.(2.14) Consequently, 𝐷^𝛼₀₊𝑢(𝑡)+𝜙_𝑞(∫¹₀𝐻(𝑡,𝜏)𝑦(𝜏)𝑑𝜏)=0. So, fractional boundary value problem (2.8) is equivalent to the following problem: 𝐷^𝛼₀₊𝑢(𝑡)+𝜙_𝑞(¹₀𝐻(𝑡,𝜏)𝑦(𝜏)𝑑𝜏)=0,0<𝑡<1,𝑢(0)=0,𝑢(1)=𝑎𝑢(𝜉).(2.15) Lemma 2.5 implies that fractional boundary value problem (2.8) has a unique solution 𝑢(𝑡)=¹₀𝐺(t,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑦(𝜏)𝑑𝜏)𝑑𝑠.(2.16) The proof is completed.

Lemma 2.7. Let 0⩽𝑎,𝑏⩽1, 0<𝜉,𝜂<1, 1<𝛼,and𝛾⩽2. The functions 𝐺(𝑡,𝑠) and 𝐻(𝑡,𝑠) defined by (2.7) and (2.10), respectively, are continuous on [0,1]×[0,1] and satisfy(1)𝐺(𝑡,𝑠)⩾0,𝐻(𝑡,𝑠)⩾0,for𝑡,𝑠∈[0,1];(2)𝐺(𝑡,𝑠)⩽𝐺(𝑠,𝑠),𝐻(𝑡,𝑠)⩽𝐻(𝑠,𝑠)for𝑡,𝑠∈[0,1];(3)𝐺(𝑡,𝑠)/𝐺(1,𝑠)⩾𝑡^𝛼−1,for𝑡,𝑠∈(0,1).

Proof. The proofs of part (1) and part (2) can be obtained in [6]. Here, we only prove part (3).
If 0<𝑠⩽𝑡<1,𝑠⩽𝜉, then 𝐺(𝑡,𝑠)=𝐺(1,𝑠)[𝑡(1−𝑠)]^𝛼−1−𝑎𝑡^𝛼−1(𝜉−s)^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)𝑎𝜉^𝛼−1(1−𝑠)^𝛼−1−𝑎(𝜉−𝑠)^𝛼−1=𝑡^𝛼−1⋅(1−𝑠)^𝛼−1−𝑎(𝜉−𝑠)^𝛼−1−(1−(𝑠/𝑡))^𝛼−1(1−𝑎𝜉^𝛼−1)𝑎𝜉^𝛼−1(1−𝑠)^𝛼−1−𝑎(𝜉−𝑠)^𝛼−1⩾𝑡^𝛼−1.(2.17)
If 0<𝜉⩽𝑠⩽𝑡<1, then 𝐺(𝑡,𝑠)=𝐺(1,𝑠)[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)𝑎𝜉^𝛼−1(1−𝑠)^𝛼−1=𝑡^𝛼−1⋅(1−𝑠)^𝛼−1−(1−(𝑠/𝑡))^𝛼−1(1−𝑎𝜉^𝛼−1)𝑎𝜉^𝛼−1(1−𝑠)^𝛼−1⩾𝑡^𝛼−1.(2.18)
If 0<𝑡⩽𝑠⩽𝜉<1, then 𝐺(𝑡,𝑠)=𝐺(1,𝑠)[𝑡(1−𝑠)]^𝛼−1−𝑎𝑡^𝛼−1(𝜉−𝑠)^𝛼−1𝑎𝜉^𝛼−1(1−𝑠)^𝛼−1−𝑎(𝜉−𝑠)^𝛼−1=𝑡^𝛼−1⋅(1−𝑠)^𝛼−1−𝑎(𝜉−𝑠)^𝛼−1𝑎𝜉^𝛼−1(1−𝑠)^𝛼−1−𝑎(𝜉−𝑠)^𝛼−1⩾𝑡^𝛼−1.(2.19)
If 0<𝑡⩽𝑠<1,𝜉⩽𝑠, then 𝐺(𝑡,𝑠)=𝐺(1,𝑠)[𝑡(1−𝑠)]^𝛼−1𝑎𝜉^𝛼−1(1−𝑠)^𝛼−1=𝑡^𝛼−1⋅(1−𝑠)^𝛼−1𝑎𝜉^𝛼−1(1−𝑠)^𝛼−1⩾𝑡^𝛼−1.(2.20) The above inequalities imply that part (3) of Lemma 2.7 holds.

From Lemmas 2.5 and 2.7, it is easy to obtain the following lemma.

Lemma 2.8. Let 0⩽𝑎⩽1, 0<𝜉<1, 1<𝛼⩽2. If 𝑦(𝑡)∈𝐶[0,1], and 𝑦(𝑡)⩾0, then fractional boundary value problem (2.5) has a unique solution 𝑢(𝑡)⩾0,𝑡∈[0,1].
Let 𝐸={𝑢∶𝑢,𝜙_𝑝(𝐷^𝛼₀₊𝑢)∈𝐶²[0,1]}. Now one introduces the following definitions about the upper and lower solutions of fractional boundary value problem (1.1).

Definition 2.9. A function 𝛼(𝑡) is called a lower solution of fractional boundary value problem (1.1), if 𝛼(𝑡)∈𝐸 and 𝛼(𝑡) satisfies 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝐷𝛼(𝑡)))⩽𝑓(𝑡,𝛼(𝑡)),0<𝑡<1,1<𝛼,𝛾⩽2,𝛼(0)⩽0,𝛼(1)⩽𝑎𝛼(𝜉),_𝛼0+𝛼(0)⩾0,𝐷^𝛼₀₊𝛼(1)⩾𝑏𝐷^𝛼₀₊𝛼(𝜂).(2.21)

Definition 2.10. A function 𝛽(𝑡) is called an upper solution of fractional boundary value problem (1.1), if 𝛽(𝑡)∈𝐸 and 𝛽(𝑡) satisfies 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝐷𝛽(𝑡)))⩾𝑓(𝑡,𝛽(𝑡)),0<𝑡<1,1<𝛼,𝛾⩽2,𝛽(0)⩾0,𝛽(1)⩾𝑎𝛽(𝜉),_𝛼0+𝛽(0)⩽0,𝐷^𝛼₀₊𝛽(1)⩽𝑏𝐷^𝛼₀₊𝛽(𝜂).(2.22)

3. Main Result

In this section, our objective is to give and prove our main results.

For the sake of simplicity, we assume that(𝐻₁)𝑓(𝑡,𝑢)∈𝐶[(0,1)×(0,+∞),[0,+∞)] and 𝑓(𝑡,𝑢) is nonincreasing relative to 𝑢;(𝐻₂)For any constant 𝜇>0, 0<∫¹₀𝐻(𝑠,𝑠)𝑓(𝑠,𝜇𝑠^𝛼−1)𝑑𝑠<+∞;(𝐻₃)There exist a continuous function 𝑚(𝑡) and a positive number 𝜆 such that 𝑚(𝑡)⩾𝜆𝑡^𝛼−1,𝑡∈[0,1], and¹₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑓(𝜏,𝑚(𝜏))𝑑𝜏)𝑑𝑠=𝑛(𝑡)⩾𝑚(𝑡),₁₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑓(𝜏,𝑛(𝜏))𝑑𝜏)𝑑𝑠⩾𝑚(𝑡).(3.1)

We define 𝑃=∶{𝑢∈𝐶[0,1]∶thereexistsanumber𝜆_𝑢>0𝑠𝑢𝑐ℎ𝑡ℎ𝑎𝑡𝑢(𝑡)⩾𝜆_𝑢𝑡^𝛼−1,𝑡∈[0,1]}. And define an operator 𝑇 by 𝑇𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏)𝑑𝑠,∀𝑢∈𝑃.(3.2)

Theorem 3.1. Suppose that conditions (𝐻₁)∼(𝐻₃) are satisfied, then the boundary value problem (1.1) has at least one positive solution 𝜌(𝑡), which satisfies 𝜌(𝑡)⩾𝜇𝑡^𝛼−1 for some positive number 𝜇.

Proof. We will divide our proof into four steps.
Step 1. We prove that 𝑇(𝑃)⊆𝑃.
Firstly, from Lemma 2.7 and conditions (𝐻₁)∼(𝐻₂), for any 𝑢∈𝑃, there exists 𝜆_𝑢>0, such that ¹₀𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏⩽¹₀𝐻(𝜏,𝜏)𝑓(𝜏,𝜆_𝑢𝜏^𝛼−1)𝑑𝜏<+∞.(3.3) Thus, 𝑇𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀⩽𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏)𝑑𝑠₁₀𝐺(𝑠,𝑠)𝑑𝑠⋅𝜙_𝑞(¹₀𝐻(𝜏,𝜏)𝑓(𝜏,𝜆_𝑢𝜏^𝛼−1)𝑑𝜏)<+∞.(3.4)
Secondly, it follows from Lemma 2.7 that 𝑇𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝐺(1,𝑠)𝐺(1,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏)𝑑𝑠⩾𝑡^𝛼−11₀𝐺(1,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑓(𝜏,𝑢(𝜏))𝑑𝜏)𝑑𝑠=𝜆_𝑇𝑢𝑡^𝛼−1,∀𝑡∈[0,1].(3.5) Consequently, 𝑇 is well defined and 𝑇(𝑃)⊆𝑃.
In the meanwhile, by direct computations, we can obtain 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝐷(𝑇𝑢)(𝑡)))=𝑓(𝑡,𝑢(𝑡)),0<𝑡<1,1<𝛼,𝛾⩽2,(3.6)(𝑇𝑢)(0)=0,(𝑇𝑢)(1)=𝑎(𝑇𝑢)(𝜉),_𝛼0+(𝑇𝑢)(0)=0,𝐷^𝛼₀₊(𝑇𝑢)(1)=𝑏𝐷^𝛼₀₊(𝑇𝑢)(𝜂).(3.7)Step 2. We will prove that the functions 𝛼(𝑡)=𝑇𝑛(𝑡), 𝛽(𝑡)=𝑇𝑚(𝑡) are lower and upper solutions of fractional boundary value problem (1.1), respectively.
From (𝐻₁), we know that 𝑇 is nonincreasing relative to 𝑢. Combining (𝐻₃), we have 𝑚(𝑡)⩽𝑛(𝑡)=𝑇𝑚(𝑡),𝑇𝑛(𝑡)⩽𝑛(𝑡)=𝑇𝑚(𝑡),𝑡∈[0,1].(3.8) Therefore, 𝛼(𝑡)⩽𝛽(𝑡).
By Step 1, we know 𝛼(𝑡),𝛽(𝑡)∈𝑃. And by (3.6)∼(3.8), we obtain 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊(𝛼)(𝑡)))−𝑓(𝑡,𝛼(𝑡))⩽𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊(𝑇𝑛)(𝑡)))−𝑓(𝑡,𝑛(𝑡))=0,𝛼(0)=0,𝛼(1)=𝑎𝛼(𝜉),𝐷^𝛼₀₊𝛼(0)=0,𝐷^𝛼₀₊𝛼(1)=𝑏𝐷^𝛼₀₊𝐷𝛼(𝜂),_𝛾0+(𝜙_𝑝(𝐷^𝛼₀₊(𝛽)(𝑡)))−𝑓(𝑡,𝛽(𝑡))⩾𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊(𝑇𝑚)(𝑡)))−𝑓(𝑡,𝑚(𝑡))=0,𝛽(0)=0,𝛽(1)=𝑎𝛽(𝜉),𝐷^𝛼₀₊𝛽(0)=0,𝐷^𝛼₀₊𝛽(1)=𝑏𝐷^𝛼₀₊𝛽(𝜂),(3.9) that is, 𝛼(𝑡) and 𝛽(𝑡) are lower and upper solutions of fractional boundary value problem (1.1), respectively.Step 3. We will show that the fractional boundary value problem 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝑢(𝑡)))=𝑔(𝑡,𝑢(𝑡)),0<𝑡<1,1<𝛼,𝛾⩽2,𝑢(0)=0,𝑢(1)=𝑎𝑢(𝜉),𝐷^𝛼₀₊𝑢(0)=0,𝐷^𝛼₀₊𝑢(1)=𝑏𝐷^𝛼₀₊𝑢(𝜂),(3.10) has a positive solution, where 𝑔(𝑡,𝑢(𝑡))=𝑓(𝑡,𝛼(𝑡))if𝑢(𝑡)<𝛼(𝑡),𝑓(𝑡,𝑢(𝑡))if𝛼(𝑡)⩽𝑢(𝑡)⩽𝛽(𝑡),𝑓(𝑡,𝛽(𝑡))if𝑢(𝑡)>𝛽(𝑡).(3.11) Thus, we consider the operator 𝐴∶𝐶[0,1]→𝐶[0,1] defined as follows: 𝐴𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑔(𝜏,𝑢(𝜏))𝑑𝜏)𝑑𝑠,(3.12) where 𝐺(𝑡,𝑠) is defined as (2.7), 𝐻(𝑡,𝑠) is defined as (2.10). It is clear that 𝐴𝑢⩾0,forall𝑢∈𝑃, and a fixed point of the operator 𝐴 is a solution of the fractional boundary value problem (3.10).
Since 𝛼(𝑡)∈𝑃, there exists a positive number 𝜆_𝛼 such that 𝛼(𝑡)⩾𝜆_𝛼𝑡^𝛼−1,𝑡∈[0,1]. It follows from (𝐻₂) that ¹₀𝐻(𝜏,𝜏)𝑔(𝜏,𝛼(𝜏))𝑑𝜏⩽¹₀⩽𝐻(𝜏,𝜏)𝑓(𝜏,𝛼(𝜏))𝑑𝜏₁₀𝐻(𝜏,𝜏)𝑓(𝜏,𝜆_𝛼𝜏^𝛼−1)𝑑𝜏<+∞.(3.13) Consequently, for all 𝑢(𝑡)∈𝐶[0,1], we have 𝐴𝑢(𝑡)=¹₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀⩽𝐻(𝑠,𝜏)𝑔(𝜏,𝑢(𝜏))𝑑𝜏)𝑑𝑠₁₀𝐺(𝑠,𝑠)𝜙_𝑞(¹₀⩽𝐻(𝜏,𝜏)𝑔(𝜏,𝑢(𝜏))𝑑𝜏)𝑑𝑠₁₀𝐺(𝑠,𝑠)𝑑𝑠⋅𝜙_𝑞(¹₀𝐻(𝜏,𝜏)𝑓(𝜏,𝛼(𝜏))𝑑𝜏)<+∞,(3.14) which implies that the operator 𝐴 is uniformly bounded.
On the other hand, since 𝐺(𝑡,𝑠) is continuous on [0,1]×[0,1], it is uniformly continuous on [0,1]×[0,1]. So, for fixed 𝑠∈[0,1] and for any 𝜀>0, there exists a constant 𝛿>0, such that any 𝑡₁,𝑡₂∈[0,1] and |𝑡₁−𝑡₂|<𝛿,|𝐺(𝑡₁,𝑠)−𝐺(𝑡₂𝜀,𝑠)|<𝜙(∫¹₀(𝜏,𝜏)𝑓(𝜏,𝜆_𝛼𝜏^𝛼−1)𝑑𝜏).(3.15) Then, for all 𝑢(𝑡)∈𝐶[0,1], |A𝑢(𝑡₂)−𝐴𝑢(𝑡₁)|⩽¹₀|𝐺(𝑡₂,𝑠)−𝐺(𝑡₁,𝑠)|𝜙_𝑞(¹₀⩽𝐻(𝜏,𝜏)𝑔(𝜏,𝑢(𝜏))𝑑𝜏)𝑑𝑠₁₀|𝐺(𝑡₂,𝑠)−𝐺(𝑡₁,𝑠)|𝜙_𝑞(¹₀⩽𝐻(𝜏,𝜏)𝑓(𝜏,𝛼(𝜏))𝑑𝜏)𝑑𝑠₁₀|𝐺(𝑡₂,𝑠)−𝐺(𝑡₁,𝑠)|𝑑𝑠⋅𝜙_𝑞(¹₀𝐻(𝜏,𝜏)𝑓(𝜏,𝜆_𝛼𝜏^𝛼−1)𝑑𝜏)<𝜀,(3.16) that is to say, 𝐴 is equicontinuous. Thus, from the Arzela-Ascoli Theorem, we know that 𝐴 is a compact operator. By the Schauder's fixed-point theorem, the operator 𝐴 has a fixed point; that is, the fractional boundary value problem (3.10) has a positive solution.Step 4. We will prove that fractional boundary value problem (1.1) has at least one positive solution.
Suppose that 𝜌(𝑡) is a solution of (3.10), we only need to prove that 𝛼(𝑡)⩽𝜌(𝑡)⩽𝛽(𝑡),𝑡∈[0,1].

Let 𝜌(𝑡) be a solution of (3.10). We have 𝜌(0)=0,𝜌(1)=𝑎𝜌(𝜉),𝐷^𝛼₀₊𝜌(0)=0,𝐷^𝛼₀₊𝜌(1)=𝑏𝐷^𝛼₀₊𝜌(𝜂).(3.17) In addition, the function 𝑓(𝑡,𝑢) is nonincreasing in 𝑢, we know that 𝑓(𝑡,𝛽(𝑡))⩽𝑔(𝑡,𝜌(𝑡))⩽𝑓(𝑡,𝛼(𝑡)),𝑡∈[0,1].(3.18) It follows from (3.8) and (𝐻₃) that 𝑓(𝑡,𝑛(𝑡))⩽𝑔(𝑡,𝜌(𝑡))≤𝑓(𝑡,𝑚(𝑡)),𝑡∈[0,1].(3.19) By (3.6) and 𝑚(𝑡)∈𝑃, we obtain 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝛽(𝑡)))=𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊(𝑇𝑚)(𝑡)))=𝑓(𝑡,𝑚(𝑡)),𝑡∈[0,1].(3.20) Together with (3.7), (3.17)–(3.20), we obtain 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝛽(𝑡)))−𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝐷𝜌(𝑡)))=𝑓(𝑡,𝑚(𝑡))−𝑔(𝑡,𝜌(𝑡))⩾0,𝑡∈[0,1],(𝛽−𝜌)(0)=0,(𝛽−𝜌)(1)=𝑎(𝛽−𝜌)(𝜉),_𝛼0+(𝛽−𝜌)(0)=0,𝐷^𝛼₀₊(𝛽−𝜌)(1)=𝑏𝐷^𝛼₀₊(𝛽−𝜌)(𝜂).(3.21) Let 𝑧(𝑡)=𝜙_𝑝(𝐷^𝛼₀₊𝛽(𝑡))−𝜙_𝑝(𝐷^𝛼₀₊𝜌(𝑡)), we obtain 𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝛽(𝑡)))−𝐷^𝛾₀₊(𝜙_𝑝(𝐷^𝛼₀₊𝜌(𝑡)))⩾𝑓(𝑡,𝛽(𝑡))−𝑔(𝑡,𝜌(𝑡))⩾0,𝑡∈[0,1](3.22) and 𝑧(0)=0,𝑧(1)−𝜙_𝑝(𝑏)𝑧(𝜂)=0.

By Lemma 2.8, we know 𝑧(𝑡)⩽0,𝑡∈[0,1], which implies that 𝜙_𝑝(𝐷^𝛼₀₊𝛽(𝑡))⩽𝜙_𝑝(𝐷^𝛼₀₊𝜌(𝑡)),𝑡∈[0,1].(3.23) Since 𝜙_𝑝 is monotone increasing, we have 𝐷^𝛼₀₊𝛽(𝑡)⩽𝐷^𝛼₀₊𝜌(𝑡), that is, 𝐷^𝛼₀₊(𝛽−𝜌)(𝑡)⩽0. By Lemma 2.8 and (3.21), we have (𝛽−𝜌)(𝑡)⩾0. Therefore, 𝛽(𝑡)⩾𝜌(𝑡),𝑡∈[0,1].

In the similar way, we can prove that 𝛼(𝑡)⩽𝜌(𝑡),𝑡∈[0,1]. Consequently, 𝜌(𝑡) is a positive solution of fractional boundary value problem (1.1). This completes the proof.

Theorem 3.2. If 𝑓(𝑡,𝑢)∈𝐶([0,1]×[0,+∞),[0,+∞)) is nonincreasing relative to 𝑢 and 𝑓(𝑡,𝑢) does not vanish identically for any 𝑢>0, then the fractional boundary value problem (1.1) has at least one positive solution 𝜌(𝑡), which satisfies 𝜌(𝑡)⩾𝜇𝑡^𝛼−1 for some positive number 𝜇.

The proof is similar to Theorem 3.1, we omit it here.

4. Example

Example 4.1. As an example, we consider the fractional boundary value problem 𝐷^4/3₀₊(𝜙_𝑝(𝐷^3/2₀₊𝑢(𝑡)))=𝑡²+1√𝑢1,0<𝑡<1,𝑢(0)=0,𝑢(1)=31𝑢(2),𝐷^3/2₀₊𝑢(0)=0,𝐷^3/2₀₊1𝑢(1)=2𝐷^3/2₀₊2𝑢(3).(4.1)

Proof. It is clear that (𝐻₁) holds. For any 𝜆>0, ¹₀𝐻(𝜏,𝜏)𝑓(𝜏,𝜆𝜏^((3/2)−1))𝑑𝜏=¹₀𝐻(𝜏,𝜏)(𝜏²+1√𝜆𝜏^1/4)𝑑𝜏<+∞,(4.2) which implies that (𝐻₂) holds.
For 0<𝑟<1, 𝑓(𝑡,𝑢)=𝑡²+(1/√𝑢)⩽𝑡²+(1/√𝑟𝑢)⩽𝑟^−1/2𝑓(𝑡,𝑢). Let 𝑎(𝑡)=𝑡^1/2, by (3.5), we have 𝑏(𝑡)=∶𝑇𝑎(𝑡)=¹₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑓(𝜏,𝑎(𝜏))𝑑𝜏)𝑑𝑠∈𝑃,(4.3) and 𝑇𝑏(𝑡)=𝑇²𝑎(𝑡)∈𝑃, that is, there exist positive numbers 𝜆₁, 𝜆₂, such that 𝑇𝑎(𝑡)⩾𝜆₁𝑎(𝑡), 𝑇²𝑎(𝑡)⩾𝜆₂𝑎(𝑡).
Choose a positive number 𝜆₀⩽{1,𝜆₁,𝜆₂^4/3} and combining the monotonicity of 𝑇, we have 𝑇(𝜆₀𝑎(𝑡))⩾𝑇𝑎(𝑡)⩾𝜆₁𝑎(𝑡)⩾𝜆₀𝑇𝑎(𝑡),₂(𝜆₀𝑎(𝑡))⩾𝜆₀^1/4𝑇²𝑎(𝑡)⩾𝜆₀^1/4𝜆₂𝑎(𝑡)⩾𝜆₀𝑎(𝑡).(4.4) Taking 𝑚(𝑡)=𝜆₀𝑡^1/2, then, 𝑛(𝑡)=𝑇𝑚(𝑡)=¹₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑓(𝜏,𝜆₀𝜏^1/2)𝑑𝜏)𝑑𝑠⩾𝜆₀𝑡^1/2=𝑚(𝑡),𝑇𝑛(𝑡)=𝑇²𝑚(𝑡)=¹₀𝐺(𝑡,𝑠)𝜙_𝑞(¹₀𝐻(𝑠,𝜏)𝑓(𝜏,𝑇(𝜆₀𝜏^1/2))𝑑𝜏)𝑑𝑠⩾𝜆₀𝑡^1/2=𝑚(𝑡),(4.5) that is to say, the condition (𝐻₃) holds. Theorem 3.1 implies that the fractional boundary value problem (4.1) has at least one positive solution.

Acknowledgments

This work was jointly supported by Science Foundation of Hunan Provincial under Grants 2009JT3042, 2010GK3008 and 10JJ6007, the Construct Program of the Key Discipline in Hunan Province, Aid Program for Science and Technology Innovative Research Team in Higher Educational Institutions of Hunan Province.