#### Abstract

In this paper, we study a class of boundary value problem (BVP) with multiple point boundary conditions of impulsive p-Laplacian operator fractional differential equations. We establish the sufficient conditions for the existence of solutions in Banach spaces. Our analysis relies on the Kuratowski noncompactness measure and the Sadovskii fixed point theorem. An example is given to demonstrate the main results.

#### 1. Introduction

With the development of the theory of fractional order calculus, fractional differential equations are getting more and more extensively used (see[115]). For instance, the impulsive fractional differential equations are widely used in various scientific fields, such as the problem of dynamics of populations subject to abrupt changes, harvesting, diseases, and so on. Lakshmikantham et al. [16], Bainov and Simeonov [17], and Benchohra et al. [18] have done in-depth studies on this issue. Moreover, the p-Laplacian operator differential equation was first proposed by Leibenson [19] in order to study the problem of turbulent flow in a porous medium. He converted this problem into the existence of solution of the following differential equation: is the p-Laplacian operator, is the inverse function of with , and satisfy . In recent years, many results about the solutions of the p-Laplacian operator fractional differential equation BVP have been obtained (see [2026]). The research of the solutions of the BVP with p-Laplacian operator and with impulsive has been attracting increasing interest.

Zhao and Gong [27] study the solution of the following impulsive fractional differential equations with generalized periodic boundary conditions:By means of the Schauder and Guo- Krasnosel’skii fixed point theorem, they get the existence of single and multiple positive solutions of the above BVP. By the technique of the Guo-Krasnosel’skii fixed point theorem and the Leggett-Williams theorem, Wang et al. [28] obtained the results of the following BVP.The results about the BVP with multiple point boundary conditions of impulsive p-Laplacian operator fractional differential equations are few, especially in Banach space.

In this paper, we study the following BVP with multiple point boundary conditions of impulsive p-Laplacian operator fractional differential equations in Banach space : where is the zero element of , and are the fractional derivatives of order , , , with . , , and are continuous. The impulsive point set satisfies ; we denote , . We denote by and the right and left limits of at the point , i.e., and . We establish the existence of solution to BVP (4). by the technique of the Kuratowski noncompactness measure and the Sadovskii fixed point theorem. The main innovations of this paper are as follows. Firstly, we study the impulsive fractional differential equations with the p-Laplacian Operator. Secondly, we study the BVP in Banach space. Thirdly, the nonlinear term contains the derivatives .

The paper is organized as follows. In Section 2, we recall some definitions and lemmas. In Section 3, the main results of this paper are discussed. Finally, one example is given in Section 4.

#### 2. Preliminaries

First, we recall some definitions and preliminary.

Definition 1 (see [29]). Let , .
The fractional integral of order of is given by the fractional derivative of order of is given by where , denotes the integer part of and the right side integral is pointwise defined on .

Lemma 2 (see[30, 31]). Let be integrable, , . Then
(1) where , , .
(2)

Definition 3 (see [32]). Let be a bounded set in a real Banach space ; the Kuratowski noncompactness measure of is given by where denote the diameters of .

Remark 4. From the definition, it is obvious that .

Lemma 5 (see [33]). If is bounded and equicontinuous, then is continuous on and , , where for each .

Definition 6 (see [32]). Let and be real Banach spaces, , and be a continuous and bounded operator; is called a k-set contraction operator if there exists a constant , for any bounded set in , such that .

Remark 7 (see [34]). is called a strict set contraction operator if . It is clear that a strict set contraction operator is a condensing operator.
In the following, we define the basic space of this paper. Denote and it is easy to see that is Banach space with the norm The basic space used in this paper is . The Kuratowski noncompactness measures in , , and DC(I) are denoted by , , and , respectively.

The following Sadovskii fixed point theorem is needed for the proof of our main results.

Lemma 8 ((Sadovskii)(see [32])). Let be a bounded, closed, and convex subset of the Banach space . If the operator is condensing, then has a fixed point in .

#### 3. Existence of Solutions

Before proceeding further, let us give some denotations as follows:

For simplicity of presentation, we list some conditions.

are nonnegative functions and satisfy

For any , , is uniformly continuous on .

For any ,, , and , .

For all and all bounded subsets , there exist such that with .

Lemma 9. Given , the following BVP, has a unique solution satisfying the following.

Proof. From (15) and Lemma 2, we knowBecause of , we can obtain that . According to the definition of the p-Laplacian operator it follows thatThereforeIf , thenFor and , we haveWhen , we can also obtainand taking the boundary condition of (15) into consideration, some tedious manipulation yields the following.Substituting (24) and (25) into (19), we can get (16), which implies that the solution of BVP (15) is given by (16).

From Lemma 9, we can establish the following conclusion.

Lemma 10. If is satisfied, then BVP (4) has a unique solution satisfying

Proof. The proof is almost identical to Lemma 9, with the major change being the substitution of for .

Remark 11. From Lemma 10, we can get the conclusion that the solutions to the BVP (4) are equivalent to the fixed point of the following operator: Taking derivative to both sides of (27), we have

Lemma 12. If , , and are satisfied, then operator is continuous and bounded.

Proof. ​​
Step 1. For any , we prove that is well defined and . From condition , we haveTogether with the definition of operator , we haveAs above, from (28), a tedious calculation givesFrom (30) and (31), we have that is well defined and for any .
Step 2. We prove that is bounded. For any , from (29)−(31), we getSo is bounded operator.
Step 3. It is time to prove that is continuous. Let , and for any , , there exists , when , satisfying that . So is a bounded subset of ; let such that ; taking limit, we obtain . From , we know that, for any and , there exists ; when , we haveTaking , for any , , and , according to (27), (28), we haveThus So is continuous.

Lemma 13. If are satisfied, is a bounded subset of . Then and are equicontinuous on .

Proof. In fact, from the boundedness of , that is, for any , there exists such that . Suppose that with ; by the mean value theorem, we have the following.