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`Abstract and Applied AnalysisVolume 2013 (2013), Article ID 658617, 8 pageshttp://dx.doi.org/10.1155/2013/658617`
Research Article

## On the Solvability of Caputo -Fractional Boundary Value Problem Involving -Laplacian Operator

Eastern Mediterranean University, Gazimagusa, Mersin 10, Turkey

Received 2 April 2013; Accepted 10 June 2013

Copyright © 2013 Hüseyin Aktuğlu and Mehmet Ali Özarslan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We consider the model of a Caputo -fractional boundary value problem involving -Laplacian operator. By using the Banach contraction mapping principle, we prove that, under some conditions, the suggested model of the Caputo -fractional boundary value problem involving -Laplacian operator has a unique solution for both cases of and . It is interesting that in both cases solvability conditions obtained here depend on , , and the order of the Caputo -fractional differential equation. Finally, we illustrate our results with some examples.

#### 1. Introduction

In this section we will give some basic definitions and results that will be needed in the sequel. For more details about the theory of -calculus, fractional calculus, and -fractional calculus, we refer readers to [110].

Let be a fixed real number. Then for any , The -binomial function is defined for all as where is not a positive integer. It is easy to see that The -analog of Euler's gamma function is denoted by and defined as The following theorem will be used to compare values of , the usual gamma function, with values of for a fixed .

Theorem 1 (see [11]). For , one has

It is known that for , The nabla -derivative of the function is defined by The nabla -integral of is defined by Jackson in [12] and Thomae in [13] showed that the -beta function, which is defined by has the following -integral representation: The fundamental theorem of -calculus states that and if is continuous at , then Moreover, where the derivative is applied with respect to .

The nabla -fractional derivative of with respect to and for all is given by Moreover, the -fractional integral of order is defined by The -order Caputo -fractional derivative of a function is defined by where and denotes the greatest integer less than or equal to .

The following lemma enables us to transfer Caputo -fractional differential equations into an equivalent -fractional integral equation.

Lemma 2 (see [3]). Assume that and is defined on a suitable domain. Then and if , then

On the other hand the operator , where is called the -Laplacian operator. It is easy to see that , where . The following properties of -Laplacian operator will be used in the rest of the paper. (P1) if ,  , and , then ; (P2) if and then, .

#### 2. A Model of Caputo -Fractional Boundary Value Problem Involving -Laplacian Operator

In this paper, our main aim is to prove the existence and uniqueness of the solution for the following Caputo -fractional boundary value problem involving the -Laplacian operator: where ,  , and .

Note that, the boundary value problem given in (19) is antiperiodic for .

In the following lemma we obtain a -integral equation which is equivalent to the Caputo -fractional boundary value problem given in (19).

Lemma 3. Assume that ,, and . Then are equivalent to the following -integral equation: where ,  , and .

Proof. Using (20) and the fact that, we have or equivalently, Applying -fractional integral operator to both sides and using Lemma 2, we get Using ,   for in (24), we obtain According to (13) and (14), we have Taking in both sides of (25) and (26), we get Solving equations obtained by the given boundary value conditions and , it follows that Substituting (28) into (25) gives (21) which completes the proof.

#### 3. Solvability of the Caputo -Fractional Boundary Value Problem

This section is devoted to the solvability of the Caputo -fractional boundary value problem given in (19). In the first part we shall prove the existence and uniqueness of the solution, and then we shall illustrate our main results with some examples.

Recall that is a Banach space with the norm . Now consider , with Then is a continuous and compact operator.

Theorem 4. Suppose that ,   is fixed, and the following conditions hold:
and with such that Then the boundary value problem (19) has a unique solution.

Proof. Inequality given in (31) implies that On the other hand using and (32), we have
Similarly, Finally using (34) in (35), we get Since we have In other words, where .
By condition (30), we get , which implies that is a contraction. As a consequence of the Banach contraction mapping theorem and Lemma 3, the boundary value problem given in (19) has a unique solution.

Theorem 5. Suppose that ,    , and the following conditions hold for a fixed ,
, , and with such that Then the boundary value problem (19) has a unique solution.

Remark 6. When , Theorems 4 and 5 reduce to Theorems 3.1 and 3.2 of [14].

Theorem 7. Suppose that , and the following conditions hold for a fixed . There exists a nonnegative function with such that and there exists a constant with Then the boundary value problem (19) has a unique solution.

Proof. By (42), we can get that for all . By the definition of , we have Using (P2) and (45) gives
Therefore, Since we have Using -Beta function and the fact that , we get where By condition (43), we get which implies that is a contraction; therefore boundary value problem given in (19) has a unique solution.

Next, we give some examples to illustrate our results.

Example 8. Consider the following Boundary value problem where Then , and take ,  , and . Using Theorem 1 and the fact that for any fixed , we have Moreover, it can be easily seen that Finally, Therefore as a consequence of Theorem 4, boundary value problem given in (53) has a unique solution.

Example 9. Consider the following boundary value problem: Then Therefore boundary value problem given in (58) reduces to the boundary value problem given in (53), and it has a unique solution.

Example 10. Now consider the following antiperiodic boundary value problem: where Then , and take . Using Theorem 1 and taking , we get that On the other hand, Therefore by Theorem 7, the antiperiodic boundary value problem given in (60) has a unique solution.

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