A Note on Fractional Equations of Volterra Type with Nonlocal Boundary Condition

Abstract

We deal with nonlocal boundary value problems of fractional equations of Volterra type involving Riemann-Liouville derivative. Firstly, by defining a weighted norm and using the Banach fixed point theorem, we show the existence and uniqueness of solutions. Then, we obtain the existence of extremal solutions by use of the monotone iterative technique. Finally, an example illustrates the results.

1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, and so forth. There has been a significant theoretical development in fractional differential equations in recent years (see [1–18]). Monotone iterative technique is a useful tool for analyzing fractional differential equations.

In [3], Jankowski considered the existence of the solutions of the following problem: 𝐷^𝑞𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),^𝑡₀𝑘()𝑥(𝑠)(],𝑥(0)𝑡,𝑠𝑑𝑠,0<𝑞<1,𝑡∈0,𝑇=𝑟, (1) where 𝑓∈𝐶([0,𝑇]×𝑅²,𝑅), 𝑥(0)=𝑡^1−𝑞𝑥(𝑡)|_𝑡=0 by using the Banach fixed point theorem and monotone iterative technique.

Motivated by [3], in this paper we investigate the following nonlocal boundary value problem: 𝐷^𝛼𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),^𝑡₀𝑘()𝑥(𝑠)(𝑡)(],𝑥(0)(𝑥),𝑡,𝑠𝑑𝑠≡𝐹𝑥,0<𝛼<1,𝑡∈0,𝑇=𝑔 (2) where 𝑓∈𝐶([0,𝑇]×𝑅²,𝑅), 𝑔∶𝐶_1−𝛼([0,𝑇])→𝑅 is a continuous functional, 𝐽=[0,𝑇],𝑥(0)=𝑡^1−𝛼𝑥(𝑡)|_𝑡=0, and 𝑘(𝑡,𝑠)∈𝐶(Δ,𝑅); here Δ={(𝑡,𝑠)∈𝐽×𝐽∶0≤𝑠≤𝑡≤𝑇}.

Firstly, the nonlocal condition can be more useful than the standard initial condition to describe many physical and chemical phenomena. In contrast to the case for initial value problems, not much attention has been paid to the nonlocal fractional boundary value problems. Some recent results on the existence and uniqueness of nonlocal fractional boundary value problems can be found in [1, 2, 12, 14, 18]. However, discussion on nonlocal boundary value problems of fractional equations of Volterra type involving Riemann-Liouville derivative is rare. Secondly, in [3], in order to discuss the existence and uniqueness of problem (1), Jankowski divided 𝑞∈(0,1) into two situations to discuss; one is 0<𝑞≤1/2 with an additional condition and the other is 1/2<𝑞<1. In this paper, we unify the two situations without using the additional condition. Thirdly, for the study of differential equation, monotone iterative technique is a useful tool (see [9, 10, 16, 17]). We know that it is important to build a comparison result when we use the monotone iterative technique. We transform the differential equation into integral equation and use the integral equation to build the comparison result which is different from [3]. It makes the calculation easier and is suitable for the more complicated forms of equations.

The paper is organized as follows. In Section 2, we present some useful definitions and fundamental facts of fractional calculus theory. In Section 3, by applying Banach fixed point theorem, we prove the existence and uniqueness of solution for problem (2). In Section 4, by the utility of the monotone iterative technique, we prove that (2) has extremal solutions. At last, we give an example to illustrate our main results.

2. Preliminaries

Let 𝐶_1−𝛼(𝐽,𝑅)={𝑥∈𝐶((0,𝑇],𝑅)∶𝑡^1−𝛼𝑥(𝑡)∈𝐶(𝐽,𝑅)} with the norm ‖𝑥‖_𝐶1−𝛼=max_𝑡∈𝐽|𝑡^1−𝛼𝑒^−𝜆𝑡𝑥(𝑡)|, where 𝜆 is a fixed positive constant which will be fixed in Section 3. Obviously, the space 𝐶_1−𝛼(𝐽,𝑅) is a Banach space. Now, let us recall the following definitions from fractional calculus. For more details, one can see [5, 11].

Definition 1. For 𝛼>0, the integral 𝐼^𝛼𝑓(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑓(𝑠)𝑑𝑠 (3) is called the Riemann-Liouville fractional integral of order 𝛼.

Definition 2. The Riemann-Liouville derivative of order 𝛼(𝑛−1<𝛼≤𝑛) can be written as 𝐷^𝛼𝑓(𝑡)=𝑑𝑑𝑡^𝑛𝐼^𝑛−𝛼𝑓(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑓(𝑠)𝑑𝑠,𝑡>0. (4)

Lemma 3 (see [5]). Let 𝑛−1<𝛼≤𝑛. If 𝑓(𝑡)∈𝐿(0,𝑇) and 𝐷^𝛼−𝑛₀₊𝑓(𝑡)∈𝐴𝐶^𝑛[0,𝑇], then one has the following equality: 𝐼^𝛼𝐷^𝛼𝑓(𝑡)=𝑓(𝑡)−^𝑛_𝑖=1𝐷^𝛼−𝑖𝑓(𝑡)_𝑡=0𝑡^𝛼−𝑖Γ().𝛼−𝑖+1 (5)

3. Existence and Uniqueness of Solutions

In what follows, to discuss the existence and uniqueness of solutions of nonlocal boundary value problems for fractional equations of Volterra type involving Riemann-Liouville derivative, we suppose the following.(H1) There exist nonnegative constants 𝐿₁, 𝐿₂, and 𝑊 such that |𝑘(𝑡,𝑠)|≤𝑊, for all (𝑡,𝑠)∈Δ, and 𝑓𝑡,𝑣₁,𝑣₂−𝑓𝑡,𝑢₁,𝑢₂≤𝐿₁𝑣₁−𝑢₁+𝐿₂𝑣₂−𝑢₂,∀𝑡∈𝐽,∀𝑣₁,𝑣₂,𝑢₁,𝑢₂∈𝑅. (6)(H2) There exists a nonnegative constant 𝐿₃∈(0,1) such that 𝑔𝑢₁−𝑔𝑢₂≤𝐿₃𝑢₁−𝑢_{2𝐶1−𝛼},∀𝑡∈𝐽,∀𝑢₁,𝑢₂∈𝐶_1−𝛼(𝐽). (7)

Lemma 4. Let (H1) hold. 𝑥∈𝐶_1−𝛼(𝐽) and 𝑥 is a solution of the following problem: 𝐷^𝛼𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),^𝑡₀𝑘()𝑥(𝑠)(𝑡),𝑥(0)(𝑥),𝑡,𝑠𝑑𝑠≡𝐹𝑥=𝑔 (8) if and only if 𝑥(𝑡) is a solution of the following integral equation: 𝑥(𝑡)=𝑔(𝑥)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝐹𝑥(𝑠)𝑑𝑠. (9)

Proof. Assume that 𝑥(𝑡) satisfies (8). From the first equation of (8) and Lemma 3, we have 𝑥(𝑡)=𝐼^1−𝛼₀₊𝑥(𝑡)_𝑡=0𝑡^𝛼−1Γ(𝛼)+𝐼^𝛼₀₊(𝑡)(𝑥)𝑡𝐹𝑥=𝑔^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝐹𝑥(𝑠)𝑑𝑠. (10) Conversely, assume that 𝑥(𝑡) satisfies (9). Applying the operator 𝐷^𝛼 to both sides of (9), we have 𝐷^𝛼𝑥(𝑡)=𝐹𝑥(𝑡). (11) In addition, by calculation, we can conclude 𝑥(0)=𝑡^1−𝛼𝑥(𝑡)|_𝑡=0=𝑔(𝑥). The proof is completed.

Theorem 5. Let (H1), (H2) hold, 𝑓∈𝐶(𝐽×𝑅²,𝑅),and𝑘∈𝐶(Δ,𝑅). Then problem (2) has a unique solution.

Proof. Define the operator 𝑁∶𝐶_1−𝛼(𝐽)→𝐶_1−𝛼(𝐽) by 𝑁𝑥(𝑡)=𝑔(𝑥)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝐹𝑥(𝑠)𝑑𝑠. (12) It is easy to check that the operator 𝑁 is well defined on 𝐶_1−𝛼(𝐽). Next we show that 𝑁 is a contradiction operator on 𝐶_1−𝛼(𝐽). For convenience, let 1𝜌≡𝑞^1/𝑞1−𝐿₃×𝐿₁Γ(𝛼)𝑇^{𝑝𝛼−𝑝+1}Γ𝑝𝛼−𝑝+1²Γ2𝑝𝛼−2𝑝+2^1/𝑝+𝐿₂𝑊(𝛼)𝑇𝛼Γ^𝑝𝛼+1Γ𝑝𝛼−𝑝+1Γ𝑝𝛼+1Γ2𝑝𝛼−𝑝+2^1/𝑝, (13) and choose 11<𝑝<,11−𝛼𝑝+1𝑞=1,𝜆>𝜌^𝑞, (14) where 𝜆 is a positive constant defined in the norm of the space 𝐶_1−𝛼(𝐽).
Then, for any 𝑥,𝑦∈𝐶_1−𝛼(𝐽), we have from (H1), (H2), and the Hölder inequality (𝑁𝑥)(𝑡)−(𝑁𝑦)(𝑡)_𝐶1−𝛼=max_{𝑡∈[0,𝑇]}𝑡^1−𝛼𝑒^−𝜆𝑡(𝑁𝑥)(𝑡)−𝑁𝑦(𝑡)≤max_{[]𝑡∈0,𝑇}𝑒^−𝜆𝑡𝑔(𝑥)−𝑔𝑦+max_{[]𝑡∈0,𝑇}1Γ(𝛼)𝑡^1−𝛼𝑒^{−𝜆𝑡𝑡}₀(𝑡−𝑠)^𝛼−1𝐹𝑥(𝑠)−𝐹𝑦(𝑠)𝑑𝑠≤𝐿₃𝑥−𝑦_𝐶1−𝛼+max_{[]𝑡∈0,𝑇}𝐿₁Γ(𝛼)𝑡^1−𝛼𝑒^{−𝜆𝑡𝑡}₀(𝑡−𝑠)^𝛼−1𝑥(𝑠)−𝑦(𝑠)𝑑𝑠+max_{[]𝑡∈0,𝑇}𝐿₂𝑊Γ(𝛼)𝑡^1−𝛼𝑒^−𝜆𝑡×^𝑡₀(𝑡−𝑠)^𝛼−1𝑠₀𝑥(𝜏)−𝑦(𝜏)𝑑𝜏𝑑𝑠≤𝐿₃𝑥−𝑦_𝐶1−𝛼+max_{[]𝑡∈0,𝑇}𝐿₁Γ(𝛼)𝑡^1−𝛼𝑒^{−𝜆𝑡𝑡}₀(𝑡−𝑠)^𝛼−1𝑠^𝛼−1𝑒^𝜆𝑠𝑑𝑠𝑥−𝑦_𝐶1−𝛼+max_{[]𝑡∈0,𝑇}𝐿₂𝑊(𝛼)𝑡𝛼Γ^1−𝛼𝑒^{−𝜆𝑡𝑡}₀(𝑡−𝑠)^𝛼−1𝑠^𝛼𝑒^𝜆𝑠𝑑𝑠𝑥−𝑦_𝐶1−𝛼≤𝐿₃𝑥−𝑦_𝐶1−𝛼+max_{[]𝑡∈0,𝑇}𝐿₁Γ(𝛼)𝑡^1−𝛼𝑒^{−𝜆𝑡𝑡}₀(𝑡−𝑠)^𝛼−1𝑠^𝛼−1𝑝𝑑𝑠^1/𝑝×^𝑡₀𝑒^𝜆𝑠𝑞𝑑𝑠^1/𝑞𝑥−𝑦_𝐶1−𝛼+max_{[]𝑡∈0,𝑇}𝐿₂𝑊(𝛼)𝑡𝛼Γ^1−𝛼𝑒^{−𝜆𝑡𝑡}₀(𝑡−𝑠)^𝛼−1𝑠^𝛼𝑝𝑑𝑠^1/𝑝×^𝑡₀𝑒^𝜆𝑠𝑞𝑑𝑠^1/𝑞𝑥−𝑦_𝐶1−𝛼≤𝐿₃𝑥−𝑦_𝐶1−𝛼+max_{[]𝑡∈0,𝑇}𝐿₁Γ(𝛼)𝑒^−𝜆𝑡𝑡^{𝑝𝛼−𝑝+11}₀1−𝜂^{𝑝𝛼−𝑝}𝜂^{𝑝𝛼−𝑝}𝑑𝜂^1/𝑝×𝑒^𝜆𝑡𝜆𝑞^1/𝑞𝑥−𝑦_𝐶1−𝛼+max_{[]𝑡∈0,𝑇}𝐿₂𝑊(𝛼)𝑒𝛼Γ^−𝜆𝑡𝑡^𝑝𝛼+11₀1−𝜂^{𝑝𝛼−𝑝}𝜂^𝑝𝛼𝑑𝜂^1/𝑝×𝑒^𝜆𝑡𝜆𝑞^1/𝑞𝑥−𝑦_𝐶1−𝛼≤𝐿₃𝑥−𝑦_𝐶1−𝛼+𝐿₁Γ(𝛼)×𝑇^{𝑝𝛼−𝑝+1}Γ𝑝𝛼−𝑝+1²Γ2𝑝𝛼−2𝑝+2^1/𝑝1𝜆𝑞^1/𝑞𝑥−𝑦_𝐶1−𝛼+𝐿₂𝑊(𝛼)𝑇𝛼Γ^𝑝𝛼+1Γ𝑝𝛼−𝑝+1Γ𝑝𝛼+1Γ2𝑝𝛼−𝑝+2^1/𝑝×1𝜆𝑞^1/𝑞𝑥−𝑦_𝐶1−𝛼≤𝐿₃+𝐿₁Γ(𝛼)𝑇^{𝑝𝛼−𝑝+1}Γ𝑝𝛼−𝑝+1²Γ2𝑝𝛼−2𝑝+2^1/𝑝1𝜆𝑞^1/𝑞+𝐿₂𝑊(𝛼)𝑇𝛼Γ^𝑝𝛼+1Γ𝑝𝛼−𝑝+1Γ𝑝𝛼+1Γ2𝑝𝛼−𝑝+2^1/𝑝×1𝜆𝑞^1/𝑞𝑥−𝑦_𝐶1−𝛼. (∗) According to 𝜆>𝜌^𝑞 and the Banach fixed point theorem, the problem (2) has a unique solution. The proof is completed.

Remark 6. Theorem 5 is an essential improvement of [3, Theorem 1].

4. The Monotone Iterative Technique for Problem (2)

In this section, the monotone iterative technique is presented and constructed for problem (2). This method leads to a simple and yet efficient linear iterative algorithm. It yields two sequences of iterations that converge monotonically from above and below, respectively, to a solution of the problem.

Let 𝑀,𝑁∈𝐶(𝐽). We may assume |𝑀(𝑡)|≤𝑀₁, |𝑁(𝑡)|≤𝑁₁, for all 𝑡∈𝐽, 𝜎∈𝐶_1−𝛼(𝐽). Then, according to Lemma 4 and Theorem 5, the following linear problem 𝐷^𝛼𝑥(𝑡)−𝑀(𝑡)𝑥(𝑡)−𝑁(𝑡)^𝑡₀𝑘()𝑥(𝑠)(𝑡),(]𝑥(0)𝑡,𝑠𝑑𝑠=𝜎𝑡∈0,𝑇,0<𝛼<1,=𝑡^1−𝛼𝑥(𝑡)_𝑡=0=𝑔(𝑥) (15) has a unique solution which satisfies 𝑥(𝑡)=𝑔(𝑥)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑀(𝑠)𝑥(𝑠)+𝑁(𝑠)^𝑠₀𝑘(𝑠,𝜏)𝑥(𝜏)𝑑𝜏+𝜎(𝑠)𝑑𝑠. (16)

Lemma 7. Let 0<𝛼<1, 𝑀,𝑁∈𝐶(𝐽), |𝑀(𝑡)|≤𝑀₁, |𝑁(𝑡)|≤𝑁₁. Suppose that 𝑀₁𝑇^𝛼Γ(𝛼)Γ()+𝑁2𝛼¹𝑊𝑇^𝛼+1Γ(𝛼)Γ(2𝛼+1)<1 (17) and 𝑝∈𝐶_1−𝛼(𝐽) satisfies the problem 𝑝(𝑡)≤𝑝(0)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑀(𝑠)𝑝(𝑠)+𝑁(𝑠)^𝑠₀𝑘()𝑝(𝜏)𝑝(0)𝑠,𝜏𝑑𝜏𝑑𝑠.≤0, (18) Then 𝑝(𝑡)≤0 for all 𝑡∈(0,𝑇].

Proof. Suppose that the inequality 𝑝(𝑡)≤0, for all 𝑡∈(0,𝑇], is not true. Therefore, there exists at least a 𝑡_∗∈(0,𝑇] such that 𝑒^{−𝜆𝑡∗}𝑡_∗^1−𝛼𝑝(𝑡_∗)>0. Without loss of generality, we assume 𝑒^{−𝜆𝑡∗}𝑡_∗^1−𝛼𝑝(𝑡_∗)=max{𝑒^−𝜆𝑡𝑡^1−𝛼𝑝(𝑡)∶𝑡∈(0,𝑇]}=𝜌₁>0.
We obtain that 𝑒^−𝜆𝑡𝑡^1−𝛼𝑝(𝑡)≤𝑒^−𝜆𝑡𝑝(0)+𝑒^−𝜆𝑡𝑡^1−𝛼Γ(𝛼)×^𝑡₀(𝑡−𝑠)^𝛼−1×𝑀(𝑠)𝑝(𝑠)+𝑁(𝑠)^𝑠₀𝑘()𝑝(𝜏)≤𝑒𝑠,𝜏𝑑𝜏𝑑𝑠_−𝜆𝑡𝑡^1−𝛼Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑀(𝑠)𝑝(𝑠)+𝑁(𝑠)^𝑠₀𝑘()𝑝(𝜏)≤𝑒𝑠,𝜏𝑑𝜏𝑑𝑠_−𝜆𝑡𝑀₁𝑡^1−𝛼Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑠^𝛼−1𝑒^𝜆𝑠𝑒^−𝜆𝑠𝑠^1−𝛼𝑝(𝑠+𝑒)𝑑𝑠_−𝜆𝑡𝑁₁𝑊𝑡^1−𝛼Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×^𝑠₀𝜏^𝛼−1𝑒^𝜆𝜏𝑒^−𝜆𝜏𝜏^1−𝛼𝑝(𝜏)𝑑𝜏𝑑𝑠. (19) Let 𝑡=𝑡_∗; we have 𝑒^{−𝜆𝑡∗}𝑡_∗^1−𝛼𝑝𝑡_∗≤𝑀₁𝑒^{−𝜆𝑡∗}𝑡_∗^1−𝛼Γ(𝛼)×^𝑡∗₀𝑡_∗−𝑠^𝛼−1𝑠^𝛼−1𝑒^𝜆𝑠𝑒^−𝜆𝑠𝑠^1−𝛼𝑝(𝑠+𝑁)𝑑𝑠₁𝑊𝑒^{−𝜆𝑡∗}𝑡_∗^1−𝛼Γ(𝛼)×^𝑡∗₀𝑡_∗−𝑠^𝛼−1×^𝑠₀𝜏^𝛼−1𝑒^𝜆𝜏𝑒^−𝜆𝜏𝜏^1−𝛼𝑝(𝜏𝜌)𝑑𝜏𝑑𝑠₁≤𝑀₁𝑡_∗^1−𝛼Γ(𝛼)^𝑡∗₀𝑡_∗−𝑠^𝛼−1𝑠^𝛼−1𝑑𝑠𝜌₁+𝑁₁𝑊𝑡_∗^1−𝛼𝛼Γ(𝛼)^𝑡∗₀𝑡_∗−𝑠^𝛼−1𝑠^𝛼𝑑𝑠𝜌₁,𝜌₁≤𝑀₁𝑇^𝛼Γ(𝛼)Γ()+𝑁2𝛼¹𝑊𝑇^𝛼+1Γ(𝛼)Γ()𝜌2𝛼+1¹. (20) So 𝑀₁𝑇^𝛼Γ(𝛼)Γ()+𝑁2𝛼¹𝑊𝑇^𝛼+1Γ(𝛼)Γ(2𝛼+1)≥1. (21) This is a contradiction. Hence 𝑝(𝑡)≤0 for all 𝑡∈(0,𝑇]. The proof is completed.

Definition 8. We say that 𝑥₀∈𝐶_1−𝛼(𝐽) is called a lower solution of problem (2) if 𝑥₀(𝑡)≤𝑥₀(0)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝐹𝑥₀(𝑠)(],𝑥𝑑𝑠,𝑡∈0,𝑇₀(0)≤𝑔𝑥₀. (22) We say that 𝑦₀∈𝐶_1−𝛼(𝐽) is called an upper solution of problem (2) if 𝑦₀(𝑡)≥𝑦₀(0)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝐹𝑦₀(𝑠)(],𝑦𝑑𝑠,𝑡∈0,𝑇₀(0)≥𝑔𝑦₀. (23)
In the following discussion, we need the following assumptions.(H3) Assume that 𝑔∶𝐶_1−𝛼(𝐽)→𝑅 is a nondecreasing continuous function, 𝑓(𝑡,𝛽₁,𝛽₂)∈𝐶_1−𝛼(𝐽) for all 𝑡∈𝐽, 𝑥₀≤𝛽₁≤𝑦₀, ∫^𝑡₀𝑘(𝑡,𝑠)𝑥₀(𝑠)𝑑𝑠≤𝛽₂≤∫^𝑡₀𝑘(𝑡,𝑠)𝑦₀(𝑠)𝑑𝑠. 𝑥₀ and 𝑦₀ are lower and upper solutions of problem (2), respectively, and 𝑥₀≤𝑦₀.(H4) Consider 𝑓𝑡,𝑣₁,𝑣₂−𝑓𝑡,𝑢₁,𝑢₂≥𝑀(𝑡)𝑣₁−𝑢₁+𝑁(𝑡)𝑣₂−𝑢₂, (24) where 𝑥₀≤𝑢₁≤𝑣₁≤𝑦₀, ∫^𝑡₀𝑘(𝑡,𝑠)𝑥₀(𝑠)𝑑𝑠≤𝑢₂≤𝑣₂≤∫^𝑡₀𝑘(𝑡,𝑠)𝑦₀(𝑠)𝑑𝑠. 𝑀,𝑁∈𝐶(𝐽).
Let [𝑥₀,𝑦₀]={𝑧∈𝐶_1−𝛼(𝐽)∶𝑥₀(𝑡)≤𝑧(𝑡)≤𝑦₀(𝑡),𝑥₀(0)≤𝑧(0)≤𝑦₀(0)}.

Theorem 9. Let inequality (17), (H2)–(H4) hold. Then there exist monotone sequences {𝑥_𝑛},{𝑦_𝑛} ⊂ [𝑥₀,𝑦₀] which converge uniformly to the extremal solutions of (2) in [𝑥₀,𝑦₀], respectively.

Proof. This proof consists of the following three steps.
Step 1. Construct the sequences {𝑥_𝑛},{𝑦_𝑛}.
For any 𝜂∈[𝑥₀,𝑦₀], we consider the following linear problem: 𝐷^𝛼𝑥(𝑡)(𝑡)𝑥(𝑡)(𝑡)−𝑀−𝑁^𝑡₀𝑘()𝑥(𝑠)(𝑡),𝑡,𝑠𝑑𝑠=𝑓𝑡,𝜂^𝑡₀𝑘()𝜂(𝑠)(𝑡)𝜂(𝑡)(𝑡)𝑡,𝑠𝑑𝑠−𝑀−𝑁^𝑡₀𝑘()𝜂(𝑠)(],𝑥(0)𝜂.𝑡,𝑠𝑑𝑠,𝑡∈0,𝑇=𝑔 (25) By Theorem 5, (25) has a unique solution which satisfies 𝑥(𝑡)=𝑥(0)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑓𝑠,𝜂(𝑠),^𝑠₀𝑘()𝜂(𝜏)(𝑠)𝜂(𝑠)(𝑠)(𝑠)𝑠,𝜏𝑑𝜏−𝑀−𝑥−𝑁^𝑠₀𝑘()×𝜂(𝜏)(𝜏)𝑥(0)𝜂.𝑠,𝜏−𝑥𝑑𝜏𝑑𝑠,=𝑔 (26) Define an operator 𝐴∶[𝑥₀,𝑦₀]→[𝑥₀,𝑦₀] by 𝑥=𝐴𝜂. It is easy to check that the operator 𝐴 is well defined on [𝑥₀,𝑦₀]. Let 𝜂₁,𝜂₂∈[𝑥₀,𝑦₀] with 𝜂₁≤𝜂₂.
Setting 𝑝(𝑡)=𝑧₁(𝑡)−𝑧₂(𝑡), 𝑧₁(𝑡)=𝐴𝜂₁(𝑡), and 𝑧₂(𝑡)=𝐴𝜂₂(𝑡), by (26), we obtain 𝑝(𝑡)=𝑝(0)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑓𝑠,𝜂₁(𝑠),^𝑠₀𝑘(𝑠,𝜏)𝜂₁(𝜏)𝑑𝜏−𝑓𝑠,𝜂₂(𝑠),^𝑠₀𝑘(𝑠,𝜏)𝜂₂(𝜏)(𝑠)𝜂𝑑𝜏−𝑀₁(𝑠)−𝑧₁(𝑠)+𝑀(𝑠)𝜂₂(𝑠)−𝑧₂(𝑠)−𝑁(𝑠)^𝑠₀𝑘(𝑠,𝜏)𝜂₁(𝜏)−𝑧₁(𝜏)(𝑠)𝑑𝜏+𝑁^𝑠₀𝑘(𝑠,𝜏)𝜂₂(𝜏)−𝑧₂(𝜏≤𝑝(0)𝑡)𝑑𝜏𝑑𝑠_𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑀(𝑠)𝑝(𝑠)+𝑁(𝑠)^𝑠₀𝑘(𝑠,𝜏)𝑝(𝜏)𝑑𝜏𝑑𝑠. (27) Besides, 𝑝(0)=𝑧₁(0)−𝑧₂(0)=𝑔𝜂₁−𝑔𝜂₂≤0. (28) By Lemma 7, we know 𝑝(𝑡)≤0, 𝑡∈(0,𝑇]. It means that 𝐴 is nondecreasing. Obviously, we can easily get that 𝐴 is a continuous map. Let 𝑥_𝑛=𝐴𝑥_𝑛−1, 𝑦_𝑛=𝐴𝑦_𝑛−1, 𝑛=1,2,….
Step 2. The sequences {𝑡^1−𝛼𝑥_𝑛}, {𝑡^1−𝛼𝑦_𝑛} converge uniformly to 𝑡^1−𝛼𝑥^∗, 𝑡^1−𝛼𝑦^∗, respectively.
In fact, 𝑥_𝑛, 𝑦_𝑛 satisfy the following relation: 𝑥₀≤𝑥₁≤⋯≤𝑥_𝑛≤⋯≤𝑦_𝑛≤⋯≤𝑦₁≤𝑦₀. (29) Setting 𝑝(𝑡)=𝑥₀(𝑡)−𝑥₁(𝑡) and 𝑥₀(𝑡) is a lower solution of problem (2): 𝑝(𝑡)≤𝑥₀(0)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑓𝑠,𝑥₀(𝑠),^𝑠₀𝑘(𝑠,𝜏)𝑥₀(𝜏)−𝑥𝑑𝜏𝑑𝑠₁(0)𝑡^𝛼−1−1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑓𝑠,𝑥₀(𝑠),^𝑠₀𝑘(𝑠,𝜏)𝑥₀(𝜏)(𝑠)𝑥𝑑𝜏−𝑀₀(𝑠)−𝑥₁(𝑠)−𝑁(𝑠)^𝑠₀𝑘()×𝑥𝑠,𝜏₀(𝜏)−𝑥₁(𝜏)=𝑝(0)𝑡𝑑𝜏𝑑𝑠_𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑀(𝑠)𝑝(𝑠)+𝑁(𝑠)^𝑠₀𝑘(𝑠,𝜏)𝑝(𝜏)𝑑𝜏𝑑𝑠. (30) Besides, 𝑝(0)=𝑥₀(0)−𝑥₁(0)≤𝑔𝑥₀−𝑔𝑥₀=0. (31) By Lemma 7, we can obtain that 𝑥₀≤𝑥₁ for all 𝑡∈(0,𝑇]. Similarly, we can show that 𝑦₁≤𝑦₀ for all 𝑡∈(0,𝑇]. Applying the operator 𝐴 to both sides of 𝑥₀≤𝑥₁, 𝑦₁≤𝑦₀, and 𝑥₀≤𝑦₀, we can easily get (29). Obviously, the sequences {𝑡^1−𝛼𝑥_𝑛}, {𝑡^1−𝛼𝑦_𝑛} are uniformly bounded and equicontinuous. Then by using the Ascoli-Arzela criterion, we can conclude that the sequences {𝑡^1−𝛼𝑥_𝑛}, {𝑡^1−𝛼𝑦_𝑛} converge uniformly on (0,𝑇] with lim_𝑛→∞𝑡^1−𝛼𝑥_𝑛=𝑡^1−𝛼𝑥^∗, lim_𝑛→∞𝑡^1−𝛼𝑦_𝑛=𝑡^1−𝛼𝑦^∗ uniformly on (0,𝑇].
Step 3. 𝑥^∗, 𝑦^∗ are extremal solutions of (1).
𝑥^∗, 𝑦^∗ are solutions of (1) on [𝑥₀,𝑦₀], because of the continuity of operator 𝐴. Let 𝑧∈[𝑥₀,𝑦₀] be any solution of (1). That is, 𝑧(𝑡)=𝑧(0)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1(𝑠)𝑧(0)(𝑧).𝐹𝑧𝑑𝑠,=𝑔 (32) Suppose that there exists a positive integer 𝑛 such that 𝑥_𝑛(𝑡)≤𝑧(𝑡)≤𝑦_𝑛(𝑡) on (0,𝑇]. Let 𝑝(𝑡)=𝑥_𝑛+1(𝑡)−𝑧(𝑡); we have 𝑝(𝑡)=𝑥_𝑛+1(0)𝑡^𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑓𝑠,𝑥_𝑛(𝑠),^𝑠₀𝑘(𝑠,𝜏)𝑥_𝑛(𝜏)(𝑠)𝑥𝑑𝜏−𝑀_𝑛(𝑠)−𝑥_𝑛+1(𝑠)−𝑁(𝑠)^𝑠₀𝑘()×𝑥𝑠,𝜏_𝑛(𝜏)−𝑥_𝑛+1(𝜏)−𝑧(0)𝑡𝑑𝜏𝑑𝑠_𝛼−1−1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑓𝑠,𝑧(𝑠),^𝑠₀𝑘()𝑧(𝜏)≤𝑝(0)𝑡𝑠,𝜏𝑑𝜏𝑑𝑠_𝛼−1+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×𝑀(𝑠)𝑝(𝑠)+𝑁(𝑠)^𝑠₀𝑘()𝑝(𝜏)𝑝(0)=𝑥𝑠,𝜏𝑑𝜏𝑑𝑠,_𝑛+1(0)−𝑧(0)=𝑔𝑥_𝑛−𝑔(𝑧)≤0. (33) By Lemma 7, we know that 𝑝(𝑡)≤0 on (0,𝑇], which implies 𝑥_𝑛+1(𝑡)≤𝑧(𝑡) on (0,𝑇]. Similarly, we obtain that 𝑧(𝑡)≤𝑦_𝑛+1(𝑡) on (0,𝑇]. Since 𝑥₀(𝑡)≤𝑧(𝑡)≤𝑦₀(𝑡) on (0,𝑇], by induction we get that 𝑥_𝑛(𝑡)≤𝑧(𝑡)≤𝑦_𝑛(𝑡) on (0,𝑇] for every 𝑛. Therefore, 𝑥^∗(𝑡)≤𝑧(𝑡)≤𝑦^∗(𝑡) on (0,𝑇] by taking 𝑛→∞. Thus, we completed this proof.

5. An Example

Example 1. Consider the following problem: 𝐷^1/2𝑥(𝑡)1=𝑡+𝑥(𝑡)+16030^𝑡₀(𝑠)(],𝑥(0)(𝑥)=𝜂𝑡𝑠𝑥𝑑𝑠,𝑡∈0,1=𝑔𝑥𝜂12,0<𝜂<1. (34) Obviously, 𝑇=1, 𝛼=1/2, 𝑘(𝑡,𝑠)=𝑡𝑠, and 𝑓(𝑡,𝑣₁,𝑣₂)=𝑡+(1/60)𝑣₁+(1/30)𝑣₂.
Let 𝑤=1, 𝐿₁=1/60, 𝐿₂=1/30, and 𝐿₃=1/12.
It is easy to check that |𝑘(𝑓𝑡,𝑠)|≤1,𝑡,𝑣₁,𝑣₂−𝑓𝑡,𝑢₁,𝑢₂≤1𝑣60¹−𝑢₁+1𝑣30²−𝑢₂,𝑔𝑥₁−𝑔𝑥₂≤1𝑥12¹−𝑥_{2𝐶1−𝛼}. (35) So, (H1) and (H2) are satisfied. By the choice of 𝑝=3/2, 𝑞=3, we can get that 𝜆>𝜌³ and 𝜌≡(4/(11×3^1/3)){(1/20Γ(1/2))[Γ(1/4)²/Γ(1/2)]^2/3+(1/15Γ(1/2))[Γ(1/4)Γ(7/4)]^2/3}. According to Theorem 5, the problem (34) has a unique solution.
Consider the same equation as (34), taking 𝑥₀(𝑡)=0, 𝑦₀(𝑡)=𝑡^−1/2+6, and then we have 𝑦₀(0)=1.
Moreover, 𝑦₀(𝑡)=𝑡^−1/2+6≥𝑡^−1/2+1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1×1𝑠+𝑠60^−1/2+1+630^𝑠₀𝑠𝜏𝜏^−1/2𝑦+6𝑑𝜏𝑑𝑠,₀(0)𝜂=1≥^1/2+𝜂122,0<𝜂<1. (36) On the other hand, it is easy to check that (H3) holds. And let 𝑀(𝑡)=1/(𝑡−1), 𝑁(𝑡)=cos𝑡/30, and then we have 𝑓𝑡,𝑣₁,𝑣₂−𝑓𝑡,𝑢₁,𝑢₂≥1𝑣𝑡−1¹−𝑢₁+cos𝑡𝑣30²−𝑢₂, (37) where 𝑥₀≤𝑢₁≤𝑣₁≤𝑦₀, ∫^𝑡₀𝑘(𝑡,𝑠)𝑥₀(𝑠)𝑑𝑠≤𝑢₂≤𝑣₂≤∫^𝑡₀𝑘(𝑡,𝑠)𝑦₀(𝑠)𝑑𝑠 . So (H4) is satisfied. Obviously, 𝑀₁=1/30, 𝑁₁=1, and then we can get 𝑀₁𝑇^𝛼Γ(𝛼)Γ()+𝑁2𝛼¹𝑊𝑇^𝛼+1Γ(𝛼)Γ()=2𝛼+131𝜋^1/260<1. (38) Inequality (17) holds. All conditions of Theorem 9 are satisfied, so problem (34) has extremal solutions.

Acknowledgments

The project is supported by NNSF of China Grant nos. 11271087, 61263006 and Guangxi Scientific Experimental (China-ASEAN Research) Centre no. 20120116.