Abstract

This paper investigates a new class of antiperiodic boundary value problems of higher order fractional differential equations. Some existence and uniqueness results are obtained by applying some standard fixed point principles. Some examples are given to illustrate the results.

1. Introduction

Boundary value problems of fractional differential equations involving a variety of boundary conditions have recently been investigated by several researchers. It has been mainly due to the occurrence of fractional differential equations in a number of disciplines such as physics, chemistry, biology, economics, control theory, signal and image processing, biophysics, blood flow phenomena, aerodynamics, andfitting of experimental data. For details and examples, see [1–5]. The recent development of the subject can be found, for example, in papers [6–16].

The mathematical modeling of a variety of physical processes gives rise to a class of antiperiodic boundary value problems. This class of problems has recently received considerable attention; for instance, see [17–24] and the references therein. In [22], the authors studied a Caputo-type antiperiodic fractional boundary value problem of the form

In this paper, we investigate a new class of antiperiodic fractional boundary value problems given by where denotes the Caputo fractional derivative of order and is a given continuous function. Some new existence and uniqueness results are obtained for problem (2) by using standard fixed point theorems.

2. Preliminaries

Let us recall some basic definitions [1–3].

Definition 1. The Riemann-Liouville fractional integral of order for a continuous function is defined as provided the integral exists.

Definition 2. For times absolutely continuous function , the Caputo derivative of fractional order is defined as where denotes the integer part of the real number .

Notice that the Caputo derivative of a constant is zero.

Lemma 3. For any , the unique solution of the linear fractional boundary value problem is where is Green’s function (depending on and ) given by

Proof. We know that the general solution of equation , can be written as [3] for some constants , , and . Using the facts ( is a constant), we get Applying the boundary conditions for the problem (5), we find that Substituting the values of , , and in (8), we get the solution (6). This completes the proof.

Remark 4. For , the solution of the antiperiodic problem is given by [18] where is
If we let in (7), we obtain We note that the solutions given by (14) and (15) are different. As a matter of fact, (15) contains an additional term: . Therefore the fractional boundary conditions introduced in (2) give rise to a new class of problems.

Remark 5. When the phenomenon of antiperiodicity occurs at an intermediate point , the parametric-type antiperiodic fractional boundary value problem takes the form whose solution is where is given by (7). Notice that when .

3. Existence Results

Let denotes a Banach space of all continuous functions defined on into endowed with the usual supremum norm.

In relation to (2), we define an operator as

Observe that the problem (2) has a solution if and only if the operator has a fixed point.

For the sequel, we need the following known fixed point theorems.

Theorem 6 (see [25]). Let be a Banach space. Assume that is a completely continuous operator and the set is bounded. Then has a fixed point in .

Theorem 7 (see [25]). Let be a Banach space. Assume that is an open bounded subset of with and let be a completely continuous operator such that Then has a fixed point in .

Now we are in a position to present the main results of the paper.

Theorem 8. Assume that there exists a positive constant such that for , . Then the problem (2) has at least one solution.

Proof. First, we show that the operator is completely continuous. Clearly continuity of the operator follows from the continuity of . Let be bounded. Then, for all together with the assumption , we get which implies that .
Furthermore, Hence, for , we have
This implies that is equicontinuous on , by the Arzela-Ascoli theorem, the operator is completely continuous.
Next, we consider the set and show that the set is bounded. Let , then . For any , we have Thus, for any . So, the set is bounded. Thus, by the conclusion of Theorem 6, the operator has at least one fixed point, which implies that (2) has at least one solution.

Theorem 9. Let there exists a positive constant such that with , where is a positive constant satisfying Then the problem (2) has at least one solution.