Abstract

We consider a new class of boundary value problems of nonlinear fractional differential equations with fractional separated boundary conditions. A connection between classical separated and fractional separated boundary conditions is developed. Some new existence and uniqueness results are obtained for this class of problems by using standard fixed point theorems. Some illustrative examples are also discussed.

1. Introduction

In this paper, we investigate the existence of solutions for a fractional boundary value problem with fractional separated boundary conditions given by where denotes the Caputo fractional derivative of order , is a given continuous function, and are real constants, with .

Fractional calculus has recently gained much momentum as extensive and significant progress on theoretical and practical aspects of the subject can easily be witnessed in the literature. As a matter of fact, the tools of fractional calculus have been effectively applied in the modelling of many physical and engineering problems. The recent development in the theory and methods for fractional calculus indicates its popularity. For some recent work on fractional boundary value problems, See [116] and the references therein.

2. Linear Problem

Let us recall some basic definitions of fractional calculus [1, 3].

Definition 2.1. For a continuous function , the Caputo derivative of fractional order is defined as where denotes the integer part of the real number .

Definition 2.2. The Riemann-Liouville fractional integral of order is defined as provided the integral exists.
To define the solution of the boundary value problem (1.1) we need the following lemma, which deals with linear variant of the problem (1.1).

Lemma 2.3. For a given , the unique solution of the problem is given by where

Proof. It is well known [3] that the solution of fractional differential equation in (2.3) can be written as Using ( is a constant), , , (2.6) gives From the boundary condition , we have By the boundary condition , we get which, on inserting the value of , gives Using (2.5) in the above equation, we obtain Substituting the values of and in (2.6), we get (2.4).

Remark 2.4. In the limit , it has been observed that the solution (2.4) of problem (2.3) is not reduced to the solution of the resulting problem given by The solution of (2.12) is where . However, we notice that the solution (2.4) of problem (2.3) does not depend on the parameter (appearing in the boundary conditions of (2.3)). Thus we conclude that the parameter is of arbitrary nature. Furthermore, it has been found that the solutions (2.4) and (2.13) coincide by taking in (2.13). Hence, for a particular choice of in problems (2.3) and (2.12), the two problems have the same solution.

3. Main Results

Let denotes the Banach space of all continuous functions from endowed with the norm defined by .

In view of Lemma 2.3, we define an operator by Observe that the problem (1.1) has solutions if and only if the operator equation has fixed points.

Now we are in a position to present our main results. The methods used to prove the existence results are standard; however, their exposition in the framework of problem (1.1) is new.

Theorem 3.1. Suppose that is a continuous function and satisfies the following assumption:(A1), for all , . Then the boundary value problem (1.1) has a unique solution provided

Proof. Setting and choosing , where we show that , where . For , we have Now, for and for each , we obtain As , therefore is a contraction. Thus, the conclusion of the theorem followed by the contraction mapping principle (Banach fixed point theorem).

Example 3.2. Consider the following fractional boundary value problem Here, , , , , , , , is arbitrary, and . As , therefore, is satisfied with . Further, , and Thus, by the conclusion of Theorem 3.1, the boundary value problem (3.6) has a unique solution on .
Now, we prove the existence of solutions of (1.1) by applying Krasnoselskii’s fixed point theorem [17].

Theorem 3.3 (Krasnoselskii’s fixed point theorem). Let be a closed, bounded, convex, and nonempty subset of a Banach space . Let be the operators such that (i) whenever ; (ii) is compact and continuous; (iii) is a contraction mapping. Then there exists such that .

Theorem 3.4. Let be a jointly continuous function satisfying the assumption . In addition one assumes that (A2), for all , and . Then the problem (1.1) has at least one solution on if

Proof. Letting , we choose a real number satisfying the inequality and consider . We define the operators and on as For , we find that Thus, . It follows from the assumption (A1) together with (3.8) that is a contraction mapping. Continuity of implies that the operator is continuous. Also, is uniformly bounded on as
Now we prove the compactness of the operator .
In view of (A1), we define , and consequently we have which is independent of . Thus, is equicontinuous. Hence, by the Arzelá-Ascoli Theorem, is compact on . Thus, all the assumptions of Theorem 3.3 are satisfied. So the conclusion of Theorem 3.3 implies that the boundary value problem (1.1) has at least one solution on .

Our next existence result is based on Leray-Schauder nonlinear alternative [18].

Lemma 3.5 (nonlinear alternative for single-valued maps). Let be a Banach space, a closed, convex subset of an open subset of and . Suppose that is a continuous, compact (i.e., is a relatively compact subset of ) map. Then either (i) has a fixed point in , or(ii)there is a (the boundary of in ) and with .

Theorem 3.6. Let be a jointly continuous function. Assume that: (A3) there exist a function and a nondecreasing function such that , for all ;(A4) there exists a constant such that Then the boundary value problem (1.1) has at least one solution on .

Proof. Consider the operator defined by (3.1). We show that maps bounded sets into bounded sets in . For a positive number , let be a bounded set in . Then Thus Next we show that maps bounded sets into equicontinuous sets of . Let with and , where is a bounded set of . Then we obtain Obviously the right hand side of the above inequality tends to zero independently of as . As satisfies the above assumptions, therefore it follows by the Arzelá-Ascoli theorem that is completely continuous.
Let be a solution. Then, for and using the computations in proving that is bounded, we have
Consequently, we have
In view of (A4), there exists such that . Let us set Note that the operator is continuous and completely continuous. From the choice of , there is no such that for some . Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 3.5), we deduce that has a fixed point which is a solution of the problem (1.1). This completes the proof.

In the special case when and we have the following corollary.

Corollary 3.7. Let be a continuous function. Assume that there exist constants , where and such that for all , . Then the boundary value problem (1.1) has at least one solution.

Example 3.8. Consider the following boundary value problem: Here, , , , , , , , is arbitrary, and Clearly and Thus, all the conditions of Corollary 3.7 are satisfied, and consequently the problem (3.21) has at least one solution.

Acknowledgment

The research of Bashir Ahmad was partially supported by Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia.