- Research Article
- Open Access

# Nonlocal Boundary Value Problem for Impulsive Differential Equations of Fractional Order

- Liu Yang
^{1, 2}Email author and - Haibo Chen
^{1}

**2011**:404917

https://doi.org/10.1155/2011/404917

© Liu Yang and Haibo Chen. 2011

**Received:**18 September 2010**Accepted:**4 January 2011**Published:**17 January 2011

## Abstract

We study a nonlocal boundary value problem of impulsive fractional differential equations. By means of a fixed point theorem due to O'Regan, we establish sufficient conditions for the existence of at least one solution of the problem. For the illustration of the main result, an example is given.

## Keywords

- Positive Constant
- Fractional Order
- Fractional Derivative
- Fractional Calculus
- Fractional Differential Equation

## 1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in various fields, such as physics, mechanics, aerodynamics, chemistry, and engineering and biological sciences, involves derivatives of fractional order. Fractional differential equations also provide an excellent tool for the description of memory and hereditary properties of many materials and processes. In consequence, fractional differential equations have emerged as a significant development in recent years, see [1–3].

where denotes the Caputo's fractional derivative.

Ahmad and Sivasundaram [14] studied a class of four-point nonlocal boundary value problem of nonlinear integrodifferential equations of fractional order by applying some fixed point theorems.

where , is a continuous function, and are continuous functions, with , , , , , , and are two continuous functions. We will define in Section 2.

To the best of our knowledge, this is the first time in the literatures that a nonlocal boundary value problem of impulsive differential equations of fractional order is considered. In addition, the nonlinear term involves . Evidently, problem (1.5) not only includes boundary value problems mentioned above but also extends them to a much wider case. Our main tools are the fixed point theorem of O'Regan. Some recent results in the literatures are generalized and significantly improved (see Remark 3.6)

The organization of this paper is as follows. In Section 2, we will give some lemmas which are essential to prove our main results. In Section 3, main results are given, and an example is presented to illustrate our main results.

## 2. Preliminaries

At first, we present here the necessary definitions for fractional calculus theory. These definitions and properties can be found in recent literature.

where the right side is pointwise defined on .

where , denotes the integer part of the number , and the right side is pointwise defined on .

where , , .

where , , .

Second, we define

, and exist with .

. Let ; it is a Banach space with the norm , where .

Like the definition 2.1 in [16], we give the following definition.

Definition 2.5.

A function with its Caputo derivative of order existing on is a solution of (1.5) if it satisfies (1.5).

To deal with problem (1.5), we first consider the associated linear problem and give its solution.

Lemma 2.6.

Proof.

Substituting the value of into (2.10) and (2.13), we obtain (2.7).

Now, we introduce the fixed point theorem which was established by O'Regan in [22]. This theorem will be applied to prove our main results in the next section.

Denote by an open set in a closed, convex set of a Banach space . Assume that . Also assume that is bounded and that is given by , in which is continuous and completely continuous and is a nonlinear contraction (i.e., there exists a nonnegative nondecreasing function satisfying for , such that for all ), then either

has a fixed point , or

there exists a point and with , where represent the closure and boundary of , respectively.

## 3. Main Results

Now, we make the following hypotheses.

is continuous. There exists a nonnegative function with on a subinterval of . Also there exists a nondecreasing function such that for any .

where .

Now, we state our main results.

Theorem 3.1.

Assume that , , and are satisfied; moreover, , where , then the problem (1.5) has at least one solution.

Proof.

The proof will be given in several steps.

Step 1.

The operator is completely continuous.

Taking into account the uniform continuity of the function on , we get that is equicontinuity on . By the Lemma 5.4.1 in [23], we have as relatively compact. Due to the continuity of , , , it is clear that is continuous. Hence, we complete the proof of Step 1.

Step 2.

is bounded.

Combining with the property that is bounded (Step 1), we have bounded on . Hence, we can assume that , is a constant.

Step 3.

is a nonlinear contraction.

Let , , . By , we obtain , and , . Since , we have , that is, is a nonlinear contraction .

Step 4.

in Lemma 2.7 does not occur.

Therefore, . However, it contradicts with (3.8).

Hence, by using Steps 1–4, Lemmas 2.6 and 2.7, has at least one fixed point , which is the solution of problem (1.5).

Next, we will give some corollaries.

Corollary 3.2.

Assume that , , and are satisfied; moreover, , where ; then the problem (1.5) has at least one solution.

Assume that,

()(sublinear growth), is continuous. There exists a nonnegative function with on a subinterval of . Also there exists a constant , such that for any .

Corollary 3.3.

Assume that , , and are satisfied, then the problem (1.5) has at least one solution.

Assume that

is continuous. There exists a nonnegative function with on a subinterval of . Also there exists a constant such that for any ,

Corollary 3.4.

Assume that , , and are satisfied, then the problem (1.4) has at least one solution.

Assume that

is continuous. There exists a nonnegative function with on a subinterval of . for any .

Corollary 3.5.

Assume that , , and are satisfied, then the problem (1.4) has at least one solution.

Remark 3.6.

Compared with Theorem 3.2 in [16], Corollary 3.5 does not need conditions , and . Moreover, we only need .

Example 3.7.

where . Here, , , . Let and , then we can see that holds. Choosing , , we can easily obtain that holds. Let , then we have that also holds. Moreover, , . Hence, we get for any given . Therefore, By Theorem 3.1, the above problem (3.13) has at least one solution for .

## Authors’ Affiliations

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