Uniqueness of Positive Solutions for a Perturbed Fractional Differential Equation

Abstract

We are concerned with the existence and uniqueness of positive solutions for the following nonlinear perturbed fractional two-point boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)+𝑓(𝑡,𝑢,𝑢^′,…,𝑢^(𝑛−2))+𝑔(𝑡)=0, 0<𝑡<1, 𝑛−1<𝛼≤𝑛, 𝑛≥2, 𝑢(0)=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)=𝑢^(𝑛−2)(1)=0, where 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative. Our analysis relies on a fixed-point theorem of generalized concave operators. An example is given to illustrate the main result.

1. Introduction

In this paper, we are interested in the existence and uniqueness of positive solutions for the following nonlinear perturbed fractional two-point boundary value problem: 𝐷^𝛼₀₊𝑢(𝑡)+𝑓𝑡,𝑢,𝑢^′,…,𝑢^(𝑛−2)(𝑡)𝑢(0)+𝑔=0,0<𝑡<1,𝑛−1<𝛼≤𝑛,𝑛≥2,=𝑢^′(0)=⋯=𝑢^(𝑛−2)(0)=𝑢^(𝑛−2)(1)=0, (1.1) where 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative and 𝑔∶[0,1]→[0,+∞) is continuous.

Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, economics, engineering, and biological sciences; see [1–15], for example. In recent years, the study of positive solutions for fractional differential equation boundary value problems has attracted considerable attention, and fruits from research into it emerge continuously. For a small sample of such work, we refer the reader to [16–26] and the references therein. On the other hand, the uniqueness of positive solutions for nonlinear fractional differential equation boundary value problems has been studied by some authors; see [19–22, 25], for example.

When 𝑔(𝑡)≡0, Yang and Chen [22] investigated the existence and uniqueness of positive solutions for the problem (1.1) by means of a fixed-point theorem for 𝑢₀ concave operators. They present the following result.

Theorem 1.1 (see [22]). Assume that(𝐻₁)𝑓∈𝐶([0,1]×[0,+∞)×𝑅^𝑛−2→[0,+∞)),𝑓(𝑡,𝑦₁,𝑦₂,…,𝑦_𝑛−1) is increasing for 𝑦_𝑖≥0, 𝑖=1,2,…,𝑛−1, and 𝑓 is not identically vanishing;(𝐻₂) for any 𝑡∈[0,1], 𝑦_𝑖≥0, 𝑖=1,2,…,𝑛−1, there exist constants 𝑚₁, 𝑚₂, 𝑚₁≤0≤𝑚₂<1 such that 𝑐^𝑚2𝑓𝑡,𝑦₁,𝑦₂,…,𝑦_𝑛−1≤𝑓𝑡,𝑐𝑦₁,𝑐𝑦₂,…,𝑐𝑦_𝑛−1≤𝑐^𝑚1𝑓𝑡,𝑦₁,𝑦₂,…,𝑦_𝑛−1,∀0<𝑐≤1; (1.2)(𝐻₃)𝑚₂<(2𝑚₁+1)/3. Then the problem (1.1) with 𝑔(𝑡)≡0 has a unique positive solution.

In this paper, we will remove the condition (𝐻₃) and improve on (𝐻₂). And we will use a fixed-point theorem of generalized concave operators to show the existence and uniqueness of positive solutions for the problem (1.1). Our main result is summarized in the following theorem.

Theorem 1.2. Assume that (𝐻₁) holds and(𝐻₄) for any 𝑐∈(0,1) and 𝑦_𝑖≥0, 𝑖=1,2,…,𝑛−1, there exists a number 𝜑(𝑐)∈(𝑐,1) such that 𝑓𝑡,𝑐𝑦₁,𝑐𝑦₂,…,𝑐𝑦_𝑛−1≥𝜑(𝑐)𝑓𝑡,𝑦₁,𝑦₂,…,𝑦_𝑛−1. (1.3)Then the problem (1.1) has a unique positive solution 𝑢^∗=𝐼^𝑛−2₀₊𝑣^∗∈𝐶([0,1],[0,+∞)) which satisfies 𝜆(𝑣^∗)𝑡^{𝛼−𝑛+1}(1−𝑡)≤𝑣^∗(𝑡)≤𝜇(𝑣^∗)𝑡^{𝛼−𝑛+1}(1−𝑡), 𝑡∈[0,1], where 𝜇(𝑣^∗)≥𝜆(𝑣^∗)>0.

Remark 1.3. Some examples of 𝜑(𝑡) which satisfy the condition (𝐻₄) are (1)𝜑(𝑡)=𝑡^𝑟,forall𝑡∈(0,1), where 𝑟∈(0,1),(2)𝜑(𝑡)=𝑡(1+𝜂(𝑡)) with 0<𝜂(𝑡)≤1/𝑡−1, for all 𝑡∈(0,1).

Remark 1.4. It is easy to see that the condition (𝐻₄) is weaker than the condition (𝐻₂). Moreover, we do not need the condition (𝐻₃) in our main result.

2. Preliminaries and Previous Results

For the convenience of the reader, we present here some definitions, lemmas, and basic results that will be used in the proof of our theorem.

Definition 2.1 (see [4, Definition 2.1]). The integral 𝐼^𝛼₀₊𝑓(𝑥)=1Γ(𝛼)^𝑥₀𝑓(𝑡)(𝑥−𝑡)^1−𝛼𝑑𝑡,𝑥>0, (2.1) is called the Riemann-Liouville fractional integral of order 𝛼, where 𝛼>0 and Γ(𝛼) denotes the gamma function.

Definition 2.2 (see [4, page 36-37]). For a function 𝑓(𝑥) given in the interval [0,∞), the expression 𝐷^𝛼₀₊𝑓(𝑥)=1Γ()𝑑𝑛−𝛼𝑑𝑥^𝑛𝑥₀𝑓(𝑡)(𝑥−𝑡)^{𝛼−𝑛+1}𝑑𝑡, (2.2) where 𝑛=[𝛼]+1, [𝛼] denotes the integer part of number 𝛼, is called the Riemann-Liouville fractional derivative of order 𝛼.
By using the same method in [21], the problem (1.1) can be transformed into the following boundary value problem: 𝐷^{𝛼−𝑛+2}₀₊𝑣(𝑡)+𝑓𝑡,𝐼^𝑛−2₀₊𝑣(𝑡),𝐼^𝑛−3₀₊𝑣(𝑡),…,𝐼¹₀₊𝑣(𝑡)(𝑡(𝑡)𝑣(0)(1),𝑣)+𝑔=0,0<𝑡<1,=𝑣=0. (2.3) Moreover, from Lemma 2.5 and Lemma 2.7 in [21], we can easily obtain the following result.

Lemma 2.3. If 𝑣∈𝐶([0,1],[0,+∞)) is a positive solution of the problem (2.3), then 𝑢(𝑡)=𝐼^𝑛−2₀₊𝑣(𝑡) is a positive solution of the problem (1.1). On the other hand, if 𝑣∈𝐶([0,1],[0,+∞)) is a positive solution of the problem (2.3), then the solution is 𝑣(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)+𝑔(𝑠)𝑑𝑠,𝑡∈[0,1], (2.4) where 𝐺()=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. (2.5) Here 𝐺(𝑡,𝑠) is called the Green function of the problem (2.3). Evidently, 𝐺(𝑡,𝑠)≥0 for 𝑡,𝑠∈[0,1].

The following property of the Green function plays important roles in this paper.

Lemma 2.4 (see [22]). The Green function 𝐺(𝑡,𝑠) in Lemma 2.3 has the following property: 𝛼−𝑛+1Γ()𝑠()𝛼−𝑛+21−𝑠^{𝛼−𝑛+1}𝑡^{𝛼−𝑛+1}()()≤11−𝑡≤𝐺𝑡,𝑠Γ()()𝛼−𝑛+21−𝑠^𝛼−𝑛𝑡^{𝛼−𝑛+1}(1−𝑡)for𝑡,𝑠∈[0,1]. (2.6)

In the sequel, we present some basic concepts in ordered Banach spaces for completeness and a fixed-point theorem which we will be used later. For convenience of readers, we suggest that one refers to [27, 28] for details.

Suppose that (𝐸,‖⋅‖) is a real Banach space which is partially ordered by a cone 𝑃⊂𝐸, that is, 𝑥≤𝑦 if and only if 𝑦−𝑥∈𝑃. If 𝑥≤𝑦 and 𝑥≠𝑦, then we denote 𝑥<𝑦 or 𝑦>𝑥. By 𝜃 we denote the zero element of 𝐸. Recall that a nonempty closed convex set 𝑃⊂𝐸 is a cone if it satisfies (i) 𝑥∈𝑃, 𝜆≥0⇒𝜆𝑥∈𝑃; (ii) 𝑥∈𝑃, −𝑥∈𝑃⇒𝑥=𝜃.

𝑃 is called normal if there exists a constant 𝑀>0 such that, for all 𝑥,𝑦∈𝐸, 𝜃≤𝑥≤𝑦 implies ‖𝑥‖≤𝑀‖𝑦‖; in this case 𝑀 is called the normality constant of 𝑃. If 𝑥₁,𝑥₂∈𝐸, the set [𝑥₁,𝑥₂]={𝑥∈𝐸∣𝑥₁≤𝑥≤𝑥₂} is called the order interval between 𝑥₁ and 𝑥₂.

For all 𝑥,𝑦∈𝐸, the notation 𝑥∼𝑦 means that there exist 𝜆>0 and 𝜇>0 such that 𝜆𝑥≤𝑦≤𝜇𝑥. Clearly, ∼ is an equivalence relation. Given ℎ>𝜃 (i.e., ℎ≥𝜃 and ℎ≠𝜃), we denote by 𝑃_ℎ the set 𝑃_ℎ={𝑥∈𝐸∣𝑥∼ℎ}. It is easy to see that 𝑃_ℎ⊂𝑃 is convex and 𝜆𝑃_ℎ=𝑃_ℎ for all 𝜆>0.

In a recent paper [28], Zhai et al. considered the following operator equation: 𝑥=𝐴𝑥+𝑥₀. (2.7) They established the existence and uniqueness of positive solutions for the above equation, and they present the following interesting result.

Theorem 2.5 (see Theorem 2.1 in [28]). Let ℎ>𝜃 and 𝑃 be a normal cone. Assume that(𝐷₁)𝐴∶𝑃→𝑃 is increasing;(𝐷₂)𝑥₀∈𝑃 satisfies 𝐴ℎ+𝑥₀∈𝑃_ℎ;(𝐷₃) for any 𝑥∈𝑃 and 𝑡∈(0,1), there exists 𝜑(𝑡)∈(𝑡,1) such that 𝐴(𝑡𝑥)≥𝜑(𝑡)𝐴𝑥.Then the operator equation (2.7) has a unique solution in 𝑃_ℎ.

Remark 2.6. An operator 𝐴 is said to be generalized concave if 𝐴 satisfies condition (𝐷₃).

3. Proof of Theorem 1.2

In this section, we apply Theorem 2.5 to study the problem (1.1), and we obtain a new result on the existence and uniqueness of positive solutions. The method used here is new to the literature and so is the existence and uniqueness result to the fractional differential equations.

In our considerations we will work in the Banach space 𝐶[0,1]={𝑥∶[0,1]→𝐑iscontinuous} with the standard norm ‖𝑥‖=sup{|𝑥(𝑡)|∶𝑡∈[0,1]}. Notice that this space can be equipped with a partial order given by 𝑥,𝑦∈𝐶[0,1],𝑥≤𝑦⟺𝑥(𝑡)≤𝑦(𝑡)for𝑡∈[0,1]. (3.1) Set 𝑃={𝑥∈𝐶[0,1]∣𝑥(𝑡)≥0,𝑡∈[0,1]}, the standard cone. It is clear that 𝑃 is a normal cone in 𝐶[0,1] and the normality constant is 1.

Proof of Theorem 1.2. Let ℎ(𝑡)=𝑡^{𝛼−𝑛+1}(1−𝑡), 𝑡∈[0,1]. Then 𝑃_ℎ=𝑥∈𝑃∣andthereexist𝜆(𝑥),𝜇(𝑥)>0suchthat𝜆(𝑥)ℎ≤𝑥≤𝜇(𝑥)ℎ. (3.2) For any 𝑣∈𝑃, we define 𝐴𝑣(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)𝑑𝑠,𝑥₀(𝑡)=¹₀𝐺(𝑡,𝑠)𝑔(𝑠)𝑑𝑠, (3.3) where 𝐺(𝑡,𝑠) is given as in Lemma 2.3. Noting that 𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠),𝑣(𝑠)≥0 and 𝐺(𝑡,𝑠)≥0, it follows from (𝐻₁) that 𝐴∶𝑃→𝑃. In the sequel we check that 𝐴 and 𝑥₀ satisfy all assumptions of Theorem 2.5.
Firstly, we prove that 𝐴∶𝑃→𝑃 is an increasing operator. In fact, for 𝑣_𝑖∈𝑃, 𝑖=1,2 with 𝑣₁≤𝑣₂, we know that 𝑣₁(𝑡)≤𝑣₂(𝑡), 𝑡∈[0,1], by the monotonicity of Riemann-Liouville fractional integral 𝐼^𝛿₀₊,𝛿>0 and (𝐻₁), 𝐴𝑣₁(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣₁(𝑠),𝐼^𝑛−3₀₊𝑣₁(𝑠),…,𝐼¹₀₊𝑣₁(𝑠),𝑣₁(𝑠)≤𝑑𝑠₁₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝑣₂(𝑠),𝐼^𝑛−3₀₊𝑣₂(𝑠),…,𝐼¹₀₊𝑣₂(𝑠),𝑣₂(𝑠)𝑑𝑠=𝐴𝑣₂(𝑡). (3.4) That is 𝐴𝑣₁≤𝐴𝑣₂. Hence, the condition (𝐷₁) in Theorem 2.5 is satisfied.
Next we show that the condition (𝐷₃) holds. From (𝐻₄), for any 𝛾∈(0,1) and 𝑣∈𝑃, we obtain 𝐴𝛾𝑣(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊𝛾𝑣(𝑠),𝐼^𝑛−3₀₊𝛾𝑣(𝑠),…,𝐼¹₀₊(𝑠)(𝑠)=𝛾𝑣,𝛾𝑣𝑑𝑠₁₀𝐺(𝑡,𝑠)𝑓𝑠,𝛾𝐼^𝑛−2₀₊𝑣(𝑠),𝛾𝐼^𝑛−3₀₊𝑣(𝑠),…,𝛾𝐼¹₀₊𝑣(𝑠)(𝑠≥,𝛾𝑣)𝑑𝑠₁₀𝐺(𝑡,𝑠)𝜑𝛾𝑓𝑠,𝐼^𝑛−2₀₊𝑣(𝑠),𝐼^𝑛−3₀₊𝑣(𝑠),…,𝐼¹₀₊𝑣(𝑠)(𝑠)𝛾(𝑡)[].,𝑣𝑑𝑠=𝜑𝐴𝑣,𝑡∈0,1 (3.5) That is 𝐴(𝛾𝑣)≥𝜑(𝛾)𝐴𝑣, for all 𝑣∈𝑃, 𝛾∈(0,1). So the condition (𝐷₃) in Theorem 2.5 is satisfied. Now we show that the condition (𝐷₂) is also satisfied. On one hand, it follows from (𝐻₁) and Lemma 2.4 that 𝐴ℎ(𝑡)=¹₀𝐺(𝑡,𝑠)𝑓𝑠,𝐼^𝑛−2₀₊ℎ(𝑠),𝐼^𝑛−3₀₊ℎ(𝑠),…,𝐼¹₀₊ℎ(𝑠)(𝑠)(𝑡),ℎ𝑑𝑠≥ℎ¹₀𝛼−𝑛+1Γ()𝑠()𝛼−𝑛+21−𝑠^{𝛼−𝑛+1}𝑓𝑠,𝐼^𝑛−2₀₊ℎ(𝑠),𝐼^𝑛−3₀₊ℎ(𝑠),…,𝐼¹₀₊ℎ(𝑠)(𝑠=,ℎ)𝑑𝑠𝛼−𝑛+1Γ()ℎ(𝑡)×𝛼−𝑛+2₁₀𝑠(1−𝑠)^{𝛼−𝑛+1}𝑓𝑠,𝐼^𝑛−2₀₊ℎ(𝑠),𝐼^𝑛−3₀₊ℎ(𝑠),…,𝐼¹₀₊ℎ(𝑠),ℎ(𝑠)𝑑𝑠,𝑡∈[0,1]. (3.6) On the other hand, also from (𝐻₁) and Lemma 2.4, we obtain (𝑡)≤1𝐴ℎΓ()ℎ(𝑡)𝛼−𝑛+2¹₀(1−𝑠)^𝛼−𝑛𝑓𝑠,𝐼^𝑛−2₀₊ℎ(𝑠),𝐼^𝑛−3₀₊ℎ(𝑠),…,𝐼¹₀₊ℎ(𝑠),ℎ(𝑠)𝑑𝑠,𝑡∈[0,1]. (3.7) Let 𝜆(ℎ)=𝛼−𝑛+1Γ(𝛼−𝑛+2)¹₀𝑠(1−𝑠)^{𝛼−𝑛+1}𝑓𝑠,𝐼^𝑛−2₀₊ℎ(𝑠),𝐼^𝑛−3₀₊ℎ(𝑠),…,𝐼¹₀₊ℎ(𝑠)(𝑠)𝜇(ℎ)=1,ℎ𝑑𝑠,Γ(𝛼−𝑛+2)¹₀(1−𝑠)^𝛼−𝑛𝑓𝑠,𝐼^𝑛−2₀₊ℎ(𝑠),𝐼^𝑛−3₀₊ℎ(𝑠),…,𝐼¹₀₊ℎ(𝑠),ℎ(𝑠)𝑑𝑠. (3.8) Since 𝑓 is continuous and 𝑓≢0, we can get 0<𝜆(ℎ)≤𝜇(ℎ). Consequently, 𝜆(ℎ)ℎ(𝑡)≤𝐴ℎ(𝑡)≤𝜇(ℎ)ℎ(𝑡). (3.9) Next we consider 𝑥₀. If 𝑔(𝑡)≡0, then 𝑥₀(𝑡)≡0; if 𝑔(𝑡)≢0, let 𝑙=max_𝑡∈[0,1]𝑔(𝑡), then 𝑙>0. It is easy to prove that 0≤𝑥₀(𝑡)≤𝑙Γ()ℎ(𝑡)𝛼−𝑛+2¹₀(1−𝑠)^𝛼−𝑛𝑙𝑑𝑠=()Γ()ℎ(𝑡).𝛼−𝑛+1𝛼−𝑛+2 (3.10) Hence, 0≤𝑥₀≤𝑙(𝛼−𝑛+1)Γ(𝛼−𝑛+2)ℎ. (3.11) Further, 𝜆(ℎ)ℎ≤𝑥₀𝜇(ℎ)+𝑙+𝐴ℎ≤(𝛼−𝑛+1)Γ(𝛼−𝑛+2)ℎ. (3.12) Hence 𝑥₀+𝐴ℎ∈𝑃_ℎ. Finally, using Theorem 2.5, 𝑣=𝐴𝑣+𝑥₀ has a unique solution 𝑣^∗ in 𝑃_ℎ. That is, 𝑣^∗ is a unique positive solution of the problem (2.3) in 𝑃_ℎ. So there are 𝜇(𝑣^∗), 𝜆(𝑣^∗) with 𝜇(𝑣^∗)≥𝜆(𝑣^∗)>0 such that 𝜆(𝑣^∗)𝑡^{𝛼−𝑛+1}(1−𝑡)≤𝑣^∗(𝑡)≤𝜇(𝑣^∗)𝑡^{𝛼−𝑛+1}(1−𝑡), 𝑡∈[0,1]. From Lemma 2.3, 𝑢^∗=𝐼^𝑛−2₀₊𝑣^∗ is the solution of the problem (1.1). Evidently, 𝑢^∗∈𝐶([0,1],[0,+∞)) is a unique positive solution of the problem (1.1).

Remark 3.1. Let 𝑓≡𝐶>0. Then the conditions (𝐻₁), (𝐻₄) are satisfied and the problem (2.3) has a unique solution 𝑣(𝑡)=∫¹₀𝐺(𝑡,𝑠)[𝐶+𝑔(𝑡)]𝑑𝑠, 𝑡∈[0,1]. From Lemma 2.4, the unique solution 𝑣 is a positive solution and satisfies 𝑣∈𝑃_ℎ=𝑃_{𝑡𝛼−𝑛+1(1−𝑡)}. So 𝑢=𝐼^𝑛−2₀₊𝑣 is a unique positive solution of the problem (1.1).
To illustrate how our main result can be used in practice we present an example.

Example 3.2. Consider the following problem: 𝐷^5/2₀₊𝑢(𝑡)+𝑎(𝑡)𝑢^1/3(𝑡)+𝑢^′(𝑡)^1/4(𝑡)𝑢(0)+𝑔=0,0<𝑡<1,=𝑢^′(0)=𝑢^′(1)=0, (3.13) where 𝑎,𝑔∶[0,1]→[0,+∞) are continuous with 𝑎≢0.
In this example, we have 𝛼=5/2. Let 𝑓(𝑡,𝑥,𝑦)=𝑎(𝑡)[𝑥^1/3+𝑦^1/4]. Evidently, 𝑓(𝑡,𝑥,𝑦) is increasing in 𝑥 for 𝑡∈[0,1], 𝑦≥0, and increasing in 𝑦 for 𝑡∈[0,1], 𝑥≥0. Moreover, 𝑓≢0. Set 𝜑(𝛾)=𝛾^5/12,𝛾∈(0,1). Then 𝑓𝑡,𝛾𝑥,𝛾𝑦=𝑎(𝑡)𝛾^1/3𝑥^1/3+𝛾^1/4𝑦^1/4≥𝜑𝛾𝑓𝑡,𝑥,𝑦,𝑡∈[0,1],𝑥,𝑦≥0. (3.14) Hence, all the conditions of Theorem 1.2 are satisfied. An application of Theorem 1.2 implies that the problem (3.13) has a unique positive solution.

Acknowledgments

This research was supported by the Youth Science Foundation of China (11201272) and the Science Foundation of Business College of Shanxi University (2012050).