#### Abstract

The existence of mild solutions for second-order impulsive semilinear neutral functional differential equations with nonlocal conditions in Banach spaces is investigated. The results are obtained by using fractional power of operators and Sadovskii's fixed point theorem.

#### 1. Introduction

The study of impulsive functional differential equations is linked to their utility in simulating processes and phenomena subject to short-time perturbations during their evolution. The perturbations are performed discretely and their duration is negligible in comparison with the total duration of the processes. That is why the perturbations are considered to take place “instantaneously” in the form of impulses. The theory of impulsive differential and functional differential equations has been extensively developed; see the monographs of Bainov and Simeonov [1], Lakshmikantham et al. [2], and Samoilenko and Perestyuk [3], where numerous properties of their solutions are studied, and detailed bibliographies are given.

This paper is devoted to extending existing results to second-order differential equations. To be precise, in [4], the authors used Sadovsii's fixed point theorem for a condensing map to establish existence results for first-order impulsive semilinear neutral functional differential inclusions with nonlocal conditions. Here, we obtain existence results for second-order semilinear impulsive differential equations with nonlocal conditions of the form

where is the infinitesimal generator of a strongly continuous cosine family of bounded linear operators in . Also, and are given functions to be specified later.

Other results on second order functional differential equations with and without impulsive effect can be founded in the monographs [5–8].

This paper is organized as follows. In Section 2, we recall briefly some basic definitions and lemmas. The existence theorem for (1.1) and its proof are arranged in Section 3. Our approaches are based on Sadovskii's fixed point theorem, and the theory of strongly continuous cosine families.

#### 2. Preliminaries

*Definition 2.1 (see [9]). *A one-parameter family of bounded linear operators in the Banach space is called a strongly continuous cosine family if and only if(i) for all (ii)(iii) is strongly continuous in on for each fixed We define the associated sine family , by

We make the following assumption on :

is the infinitesimal generator of a strongly continuous cosine family of bounded linear operators from into itself.The infinitesimal generator of a strongly continuous cosine family is the operator defined by

where

We define

Lemma 2.2 (see [9]). *If be a strongly continuous cosine family in then*(i)*there exist constants and so that for all and
*(ii)*if then and *

It is proved in [10] that for the fractional powers exist as close linear operator in , for and for

We assume in addition the following assumption:

for maps onto and is , so that is a Banach space when endowed with the form We denote this Banach space byDenote We define the following classes of functions:

and there exist with :??

and there exist with where and represent the restriction of and to , respectively, and

Obviously, is a Banach space with the norm and is also a Banach space with the norm

*Definition 2.3. *A function is said to be a mild solution of (1.1) if(i);(ii)(iii)(iv)the restriction of to the interval is continuous and the following integral equation is verified:

For (1.1), we assume that the following hypotheses are satisfied: for some

()there exists a constant such that is a continuous function, and satisfies the Lipschitz condition, that is, there exists a constant such thatfor any Moreover, there exists a constant such that the inequality

holds for any

()the function satisfies the following conditions:(i)for each the function is continuous, and for each , the function is strongly measurable,(ii)for each positive number , there is a positive function such thatwhere

() is continuous and satisfies that(i)there exist positive constants and such that (ii) is a completely continuous map;() are all bounded, that is, there exist constants such that for () is completely continuous.#### 3. Main Result

Theorem 3.1. *Let If the hypotheses are satisfied, then (1.1) has a mild solution provided that
**
where
*

*Proof. *Consider the space with morm We should now show that the operator defined by
has a fixed point. This fixed point is then a solution of (2.6).

For each positive number let Then for each is clearly a bounded close convex set in B. We claim that there exists a positive integer such that If it is not true, then for each positive integer there is a function but that is, for some where denotes is dependent on However, on the other hand, we have

Dividing on both sides by and taking the lower limits as we get This is a contradiction with the formula (3.2). Hence for some positive integer

Next we will show that the operator has a fixed point on which implies that (1.1) has a mild solution. For this purpose, we decompose as where the operators are defined on respectively, by

for and we will verify that is a contraction while is a completely continuous operator.

To prove that is a contraction, we take arbitrarily. Then for each and by condition we have that

Thus Therefore, by assumption (see (3.1)), we see that is a contraction.

To prove that is completely continuous, firstly we prove that is continuous on Let then by (i), we have Since by the dominated convergence theorem, we have

Thus, is continuous.

Next, we prove that is a family of equicontinuous functions. Let Then for each we have

and similarly
The right-hand sides are independent of and tend to zero as since are uniformly continuous for and the compactness of for implies the continuity in the uniform operator topology.

The compactness of follows from that of and Lemma 2.2.

This shows that maps into a family of equicontinuous functions.

It remains to prove that is relatively compact in

Obviously, by condition (ii), is relatively compact in Let be fixed and For we define

Since are compact operators, the set is relatively compact in for every Moreover, for every we have
Therefore, there are relatively compact sets arbitrarily close to the set Hence, the set is relatively compact in

Thus, by Arzela-Ascoli theorem, is a completely continuous operator. Those arguments enable us to conclude that is a condensing map on and by the fixed point theorem of Sadovskii, there exists a fixed point for on Therefore, the nonlocal Cauchy problem with impulsive effect (1.1) has a mild solution. The proof is completed.

#### Acknowledgments

The work of the M. Li is supported by NNSF of China (no. 10971139). The work of C. Kou is supported by NNSF of China (nos. 10701023 and 10971221). The authors would like to thank the anonymous referee for his/her remarks about the evaluation of the original version of the manuscript.