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`Discrete Dynamics in Nature and SocietyVolume 2015, Article ID 973783, 7 pageshttp://dx.doi.org/10.1155/2015/973783`
Research Article

## On Antiperiodic Nonlocal Three-Point Boundary Value Problems for Nonlinear Fractional Differential Equations

1Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
2Departamento de Análisis Matemático, Facultad de Matemáticas, Universidad de Santiago de Compostela, 15782 Santiago de Compostela, Spain

Received 4 November 2014; Accepted 3 April 2015

#### Abstract

We introduce boundary value conditions involving antiperiodic and nonlocal three-point boundary conditions. We solve a nonlinear fractional differential equation supplemented with those conditions. We obtain some existence results for the given problem by applying some standard tools of fixed point theory. These results are well illustrated with the aid of examples.

#### 1. Introduction

In recent years, several kinds of boundary value problems of nonlinear fractional differential equations, supplemented with a variety of boundary conditions (including Dirichlet, Neumann, mixed, periodic, antiperiodic, multipoint, integral type, and nonlocal), have been investigated by several researchers. This investigation includes a wide collection of results ranging from theoretical to analytic and numerical methods. For details and examples, see [19] and the references therein. This surge has been mainly due to the extensive applications of fractional operators in basic and technical sciences and engineering. One can easily witness from literature (special issues and books) on the topic that the tools of fractional calculus have helped in improving the mathematical modeling of several phenomena of practical nature; for instance, see [1017].

In this paper, we study a new class of problems of fractional differential equations supplemented with antiperiodic and three-point nonlocal boundary conditions. Precisely, we consider the following fractional problem:where is the usual Caputo fractional derivative, is a continuous function, are positive real constants, and is a nonnegative constant. Further, it is assumed that .

We emphasize that the second boundary condition in (1) can be interpreted as the sum of scalar multiples of the values of the derivative of the unknown function at and is proportional to the value of the unknown function at an arbitrary value . In case of ,  , problem (1) reduces to the one with antiperiodic boundary conditions. Thus the proposed problem generalizes antiperiodic problems to semiantiperiodic three-point nonlocal problems.

#### 2. Auxiliary Lemma and Notations

In order to define the solutions for the given problem, we consider the following lemma.

Lemma 1. Let . Then the following linear semiantiperiodic three-point fractional boundary value problem has a unique solution given by

Proof. It is well known that the solution of equation ,   can be written aswhere are unknown arbitrary constants. Since , (4) givesUsing the boundary conditions in (4), we get Solving system of (5) and (6) for and , it is found that Substituting these values of and in (4) completes solution (3).
Let denote the Banach space of all continuous functions from into endowed with the usual supremum norm defined by ,  .
In view of Lemma 1, we consider a fixed point problem equivalent to the nonlinear antisymmetric problem (1) as follows: where the operator is defined as Next we set

#### 3. Main Results

In this section, we present our main results. The first result relies on classical Banach’s contraction mapping principle.

Theorem 2. Let be a continuous function satisfying the Lipschitz condition; that is, it exists such that .
Then problem (1) has a unique solution if , where is given by (10).

Proof. In the first step, it will be shown that , where with , and . For , using , we get which implies that , where we have used (10). Now, for , and for each , we obtain Since by the given assumption, the operator is a contraction. Thus, by Banach’s contraction mapping principle, there exists a unique solution for problem (1). This completes the proof.

The next existence result is based on the following Schaefer’s fixed point theorem [18], Th. 4.3.2.

Theorem 3. Let be a Banach space. Assume that is completely continuous operator and the set is bounded. Then has a fixed point in .

Theorem 4. Assume that there exists a positive constant such that for . Then problem (1) has at least one solution.

Proof. We first show that the operator is completely continuous. Obviously continuity of the operator follows from continuity of . Let be a bounded set. By the assumption , for , we havewhich implies that . Further, we find thatHence, for , we haveThis implies that is equicontinuous on . Thus, by the Arzelà-Ascoli theorem, the operator is completely continuous.
Next, we consider the set and show that the set is bounded. Let ,  and then . For any , we have and for any . So the set is bounded. Thus, the conclusion of Theorem 3 applies and the operator has at least one fixed point. This, in turn, implies that problem (1) has at least one solution on .

Now we show the existence of solutions for problem (1) by means of Leray-Schauder degree theory.

Theorem 5. Suppose that there exist constants and such that for all . Then problem (1) has at least one solution.

Proof. Define a ball with radius by where will be fixed later. Then, it is enough to show that the operator (given by (10)) is such that Now we set Then, by Arzelà-Ascoli theorem, is completely continuous. If condition (20) holds true, then the following Leray-Schauder degrees are well defined and, by the homotopy invariance of topological degree, we have that where denotes the unit operator. By the nonzero property of Leray-Schauder degree, for at least one . In order to justify condition (20), it is assumed that for some and for all so that which, on taking norm over the interval , yields where is given by (10). In consequence we have Letting , (20) holds. This completes the proof.

The next result is based on Krasnoselskii’s fixed point theorem [18], Th. 4.4.1.

Theorem 6. Let be a continuous function satisfying and , and .Then problem (1) has at least one solution on if

Proof. For , let us define a closed set (ball) and introduce the operators and on as For , it is easy to show that , where is given by (10). This implies that .
In view of condition (26), the operator is a contraction. Continuity of the operator follows from that of . Also, implies that is uniformly bounded on . Furthermore, with and , we have independent of as . This shows that is relatively compact on . Hence, we infer by the Arzelà-Ascoli theorem that is compact on . Thus all the conditions of Krasnoselskii’s fixed point theorem hold true. Hence, problem (1) has at least one solution on . This completes the proof.

Finally, we make use of Leray-Schauder nonlinear alternative to show the existence of solutions for problem (1).

Lemma 7 (nonlinear alternative for single valued maps [19]). Let be a closed, convex subset of a Banach space and be an open subset of with . Suppose that is continuous, compact (i.e., is a relatively compact subset of ) map. Then either (i) has a fixed point in or(ii)there is (the boundary of   in ) and with .

Theorem 8. Let be a continuous function. Assume that there exist a function and a nondecreasing function such that ;there exists a constant such that where is given by (10).
Then problem (1) has at least one solution on .

Proof. As a first step, we show that the operator defined by (10) maps bounded sets into bounded sets in . For a positive number , let be a bounded set in . Then, for together with and , we obtain (as before)Next, it will be shown that maps bounded sets into equicontinuous sets of . Let and . Then where Clearly, the right-hand side of the above inequality tends to zero independent of as . Thus, by the Arzelà-Ascoli theorem, the operator is completely continuous.
Let be a solution for the given problem. Then, for , as before, we obtain In view of , there exists such that . Let us choose .
Notice that the operator is continuous and completely continuous. From the choice of , there is no such that for some . Consequently, the conclusion of Lemma 7 applies and hence the operator has a fixed point which is a solution of problem (1). This completes the proof.

##### 3.1. Examples

Consider a three-point boundary value problem of nonlinear fractional differential equations given by Here ,  , and . With the given values, it is found that(a)In (34), let us choose a continuous function

Clearly ,  , and , where and is given by (36). Thus problem (34)-(35) with given by (37) has a unique solution on by Theorem 2.(b)Taking

in (34), one can notice that . Thus, with ,  , the assumptions of Theorem 5 are satisfied and consequently its conclusion applies to problem (34)-(35) with given by (38).(c)Choosing

in (34), we find that , where Using condition (), that is, , we get . Clearly all the conditions of Theorem 8 are satisfied. Hence there exists a solution of problem (34)-(35) with given by (39).

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

Jorge Losada acknowledges financial support by Xunta de Galicia under grant Plan I2C ED481A-2015/272.

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