#### Abstract

This paper is concerned with the existence of mild and strong solutions on the interval for some neutral partial differential equations with nonlocal conditions. The linear part of the equations is assumed to generate a compact analytic semigroup of bounded linear operators, whereas the nonlinear part satisfies the Carathëodory condition and is bounded by some suitable functions. We first employ the Schauder fixed-point theorem to prove the existence of solution on the interval for that is small enough, and, then, by letting and using a diagonal argument, we have the existence results on the interval . This approach allows one to drop the compactness assumption on a nonlocal condition, which generalizes recent conclusions on this topic. The obtained results will be applied to a class of functional partial differential equations with nonlocal conditions.

#### 1. Introduction

The purpose of this paper is to study the existence of mild and strong solutions for the following neutral evolution problem with nonlocal initial conditions: in a Banach space, whereandgenerates an analytic compact semigroupon. The functions,, andwill be specified later. The Cauchy problem with the nonlocal conditionwas first considered by Byszewski , and since it reflects physical phenomena more precisely than the classical initial conditiondoes, this issue has gained enormous attention in the past several years. For more detailed information about the importance of nonlocal initial conditions in applications, we refer to the works of Byszewski , Byszewski and Lakshmikantham , and to many other authors  and the references therein.

Equation (1) has been studied by many authors under various assumptions on the linear part, the nonlinear terms,, and the nonlocal conditionsee, for example, . A basic approach to this problem is to define the solution operatorby and to use various fixed-point theorems, including Schauder fixed-point theorem, Banach contraction principle, Leray-Schauder alternative, and Sadovskii fixed-point theorem, to show thathas a fixed point, which is the mild solution of (1). When using fixed-point theorems, it is necessary that the semigroupgenerated by the linear part of (1) be compact; that is,is a compact operator, for all, so that the norm continuity of, for, becomes a key point in the study of the existence of mild solutions. Thus, because of the absence of compactness of the solution operator at, most of the papers on the relevant topics (e.g., [810, 14]) assume complete continuity on the nonlocal term. However, it is too restrictive in terms of applications.

Recently, Liang et al.  observed the nonlocal Cauchy problem [1, 3, 4, 6] that the nonlocal conditionis completely determined on, for some; that is, suchignores the fact that; for instance, in [4, 6], the functionis given by where’s are given constants, and in this case, we have measurements atrather than just at. Thus, by assuming that there is asuch that the authors utilize fixed-point theorem twice to deduce the existence results. More recently, Liu and Yuan  gave existence results using Schauder fixed-point theorem and a limiting process under the following hypothesis.

There is asuch thatand, for all,, withandwith the nonlinear termbeing bounded by an integrable function.

Motivated by the works in [15, 16], we drop the compactness assumption on the nonlocal conditionand discuss the existence of solutions for (1). The obtained results generalize recent conclusions on this topic.

The present work is organized as follows. Section 1 is devoted the introduction of the problem we studied. Section 2, we explain some known notations and results we will use. The basic hypotheses on (1) are also given in this section. In Section 3, we study the existence of mild solutions to (1) and in Section 4, we investigate some conditions for (1) to come up with strong solutions. In Section 5, an example is given to illustrate the existence results.

#### 2. Preliminaries

Throughout this paper,will be a fixed real number,will be a Banach space with norm, andis the infinitesimal generator of a compact analytic semigroup of uniformly bounded linear operatorssuch that. Then, there exists a constantsuch that, forand it is possible to define the fractional power, for, as a closed linear operator on its domainwith inverse(see ). The followings are the basic properties of.

Theorem 1 (see , pages 69–75). The following assertions hold:(i)is a Banach space with the norm, for.(ii), for each.(iii)for eachand.(iv)For every,is bounded on, and there existandsuch that (v)is a bounded linear operator inwith.(vi)If, then.

Letbe the Banach spaceendowed with the norm. Then, we denote bythe operator norm of, that is,, and letbe the Banach spaceendowed with the supnorm given by and, for any, set. Moreover, letbe the Banach spaceendowed with the supnorm given by The following hypotheses are the basic assumptions of this paper.

(H1) There existandsuch that the functionsatisfies for alland.

(H2) The functionsatisfies the following conditions.(i)For each, the functionis continuous, and, for each, the functionis strongly measurable.(ii)For eachand, there exists a positive functionsuch that and there is asuch that

(H3) The functionis continuous, and there exist constantssuch that for.

#### 3. Mild Solutions

Definition 2. A continuous functionis called a mild solution of (1) onif, for each, the functionis integrable on, and the following equation is satisfied: for all.
To see the existence of mild solution of nonlocal problem (1), we will, in view of (12), locate the fixed point of a mappingdefined onby For this, we first observe the following result, where for all, we let.

Lemma 3. Assume that hypotheses (H1)–(H3) are satisfied, and, in addition, there holds the following inequality:
Then,, for some.

Proof. Suppose, on the contrary, that, for each, there existandsuch that. Then, we have Dividing the two sides byand taking the lower limit as, we have which is a contradiction. This completes the proof.

By Lemma 3, we see that the mappingdefined by (13) mapsinto itself. We will show thathas a fixed point in. To see this, note first thatis continuous by the continuity of,and. We decomposeas, where We show thatis a contraction inandis a compact operator in.

Lemma 4. Assume that hypotheses (H1)–(H4) are satisfied. If, thenis a contraction in.

Proof. Observe that, forand, we have the assumption (H1) as follows: Hence, wherewhich is, by (H4), less than 1. Thus,is a contraction.

Lemma 5. Assume that hypotheses (H1)–(H4) are satisfied, and, in addition, the following is given.
There exists asuch that,and, for any, withand.
Then the problem (1) has at least one mild solution infor some.

Proof. Letbe given by (H5), and let For any, letbe defined by Now, we defineonby Then, by Lemma 3, we see that. Consideras the sum, where andis defined onby With a similar argument as in the proof of Lemma 4, one sees thatis a contraction on.
For the compactness of, note first thatis continuous by the continuity ofand. Now, to show that the setis relatively compact in, we will prove that, for each, the two sets are relatively compact inand that are equicontinuous families of functions on. In fact, it follows from (H3) and the compactness of, forthat, for each, Moreover, since for each, and, the set is relatively compact, then, in view of we see by (H2) that there are relative compact sets arbitrarily close to, and, hence,is also relatively compact in. Now, by the norm continuity of, for, we see that independently of, and, hence,is an equicontinuous family of functions on. Finally, letbe arbitrarily small, and we see that which is, by the norm continuity of, for, arbitrarily small and independent ofas. Therefore,is an equicontinuous family of functions onand so is. It follows from Arzela-Ascoli’s theorem thatis relatively compact on. Thus, the mappingdefined by (24) is compact.
By the fixed-point theorem of Sadovskiĭ , this shows thathas a fixed point in; that is, there is asuch that Now, define a functiononby Then,on, and. Consequently, (H5) guarantees that That is,is a mild solution of (1).

For the main results in this section, we introduce a family of nonlocal neutral problems as follows. Firstly, we define, for each, an operatoronby for all. It is clear thatis bounded onand, and, hence,. Now, for each, we defineby by andby Consider the following nonlocal neutral problem: In view of (35)–(38), the following result is an immediate corollary of Lemmas 4 and 5.

Lemma 6. Suppose that (H1)–(H4) are satisfied. Then, for any, the problem has at least one mild solution in.

Theorem 7. Suppose that, hypotheses (H1)–(H4) are satisfied. Then, problem (1) has at least one mild solution infor some.

Proof. Choose a decreasing sequenceso that, and, then, by Lemma 6, we see that for each, there is ansuch that
Now, for each, we defineonby Then, (39) implies thathas a fixed point inwhich is a mild solution for the nonlocal Cauchy problem. Decomposeas, where With the same argument as in the proof of Lemma 4, we have thatis a contraction. Furthermore, since the sequencelies in, then a similar argument as in the proof of Lemma 5 (see (27)–(31)) shows that, for each, the sets are both relatively compact inand that the sequence of functions is equicontinuous on. Hence, it follows from Ascoli-Arzela theorem that Now, letbe a decreasing sequence such that, and letbe a subsequence of. Then, a similar argument as in the proof of Lemma 5 insures thatis an equicontinuous sequence of functions on. Thus, Ascoli-Arzela theorem guarantees that the sequence Thus, by (44) and (45), we see thatis relatively compact in, and, hence, we can select a subsequence ofdenoted by, which is a Cauchy sequence in. By a similar process, we can select a subsequence ofdenoted by, which is a Cauchy sequence in. Repeat the above argument, and use a diagonal argument to obtain a subsequence ofdenoted by. Then, for every,is a Cauchy sequence in, and thus, we can define the functionby It is clear thatis strongly measurable,, and It therefore follows from Lebesgue’s dominated convergence theorem that there is a subsequenceofsuch that This shows that the sequenceis relatively compact on, and, hence, by the continuity of, it follows that By (44) and (49), we see the relative compactness ofon. Thus, there is a subsequence ofdenoted byand a functionsuch that It is clear that. Since then (50) and the uniform continuity ofimply that. By taking limits in (39), we see thatis a mild solution of (1) and this completes the proof.

We will consider the case more generally; that is, the nonlocal conditionis defined onrather than.

Theorem 8. Suppose that, hypotheses (H1) and (H2) are satisfied, and, in addition, there hold the following hypotheses.
The functionis continuous, and inequality (11) also holds.
, where If inequality (H4) holds, then, problem (1) has at least one mild solution in, for some.

Proof. Letandbe the sequences defined as in the proof of Theorem 7. With the same arguments as in the proof of Lemma 4, we see that, for some. Moreover, it follows from the same arguments as in the proof of Theorem 7 that (44), and (45) also hold, and, for every subsequenceof, there exist a subsequenceand a functionsuch thatis continuous onand, for every, Letbe given. It follows from (H7) and (53) that there is asuch that and that, for every, there is ansuch thatimplies that Choosethat is large enough so that, and defineby Thus, (H7), (54), and (55) insure that And, hence, by the continuity ofand the compactness of, for, (49) is also valid in this case. Therefore, a similar argument as in the last paragraph of the proof of Theorem 7 shows the existence of a mild solution for (1).

#### 4. Strong Solutions

Definition 9. A mild solutionis called a strong solution ifis continuously differentiable onwithand satisfies (1).
In the following, we establish a result of a strong solution for (1).

Theorem 10. Letbe a reflexive Banach space. Suppose that there hold the following hypotheses.
The functionis a continuous function and there existssuch that for alland.
is Lipschitz continuous; that is, there exists a constantsuch that for all.
The functionis continuous,, for all, and for some.
There holds the following inequality: Ifand inequality (H4) also holds with;is replaced by; and, respectively, then (1) has a strong solution on.

Proof. Letbe the operator defined by (13). By (H8), (H9) and (H10), one can use a similar argument as in the proof of Lemma 3 to deduce that there is asuch that. For this, consider the set for someandthat are large enough. It is clear thatis nonempty, convex, and closed. We will prove thathas a fixed point on. Obviously, from the proofs of Lemmas 4 and 5 and Theorem 7, it is sufficient to show that, for any, We first fix an elementand observe that, for any, whereand. Now, Thus, from (H8), (H9), and (H10), it follows that whereis a constant independent of. Since (H11) implies that then whenever Therefore,has a fixed pointwhich is a mild solution of (1). By the above calculation, we see that, for this, all of the functions are Lipschitz continuous, respectively. Sinceis Lipschitz continuous onand the spaceis reflexive, then a result of  asserts thatis a.e. differentiable onand. A similar argument shows that,,, andalso have this property. Furthermore, with a standard argument as in  (Theorem), we have So the following holds, for almost all: This shows thatis also a strong solution to the nonlocal Cauchy problem (1), and the proof is completed.

The following result is an immediate corollary of Theorems 8 and 10.

Corollary 11. Suppose that the hypotheses (H7)–(H9), and (H11) are satisfied, and in addition, there holds the following hypotheses.
The functionis continuous,, for all, and inequality (60) sustains.
Ifand inequality (H4) also holds with; andis replaced by;, respectively, then (1) has a strong solution on.

#### 5. An Example

In the last section, our existence results will be applied to solve the following system: whereandequipped withnorm.

The operatordefined by Then,generates a compact, analytic semigroupof uniformly bounded linear operators. It is well known that, and, thus, the fractional powers ofare well-defined where the eigenvalues ofareand the corresponding normalized eigenvectors are,. Moreover, with, and the operatoris given by with domain.

We need the following assumptions to solve (73) with our results.

(A1) The functionsatisfies the following conditions.(a)is welldefined and measurable with (b), for each.

(A2) The functionsatisfies the following conditions.(a)For each, the functionis measurable.(b)For each, the functionis continuous.(c)There is ansuch that

(A3) The functionsandsatisfy the following conditions, respectively:(a),(b)and there is ansuch that Letbe the Banach spaceequipped with supnorm, letbe defined by