Abstract

In this paper, we consider a kind of neutral measure evolution equations with nonlocal conditions. By using semigroup theory and fixed point theorem, we can obtain sufficient conditions for the controllability results of such equations. Finally, an example is given to verify the reliability of the results.

1. Introduction

In the past decades, the theory of impulsive differential equations has been fully developed. An impulsive differential equation is used to simulate the evolution process of a system under perturbation during continuous evolution [13]. However, this type of system allows only a limited number of discontinuities within a limited range. As a result, it cannot simulate some complex phenomena, such as Zeno’s behavior. However, the dynamic system with discontinuous trajectory is modeled by a measure differential equation or measure-driven equation [410]. Measure differential equations (MDEs) were studied in the early days [1118]. In 1971, Das first studied measure differential equations. For a complete introduction of measure differential systems, we can refer to [12, 13].

On the other hand, the complete controllability of several nonlinear dynamic systems, such as stochastic systems of fractional order and dynamic systems of impulsive differential equations, has been extensively studied. Recently, some authors had discussed existence, stability, and nonlocal controllability of the measure evolution equation [9, 15, 1922]. In the past few years, the existence and controllability of fractional abstract functional differential development systems with nonlocal conditions have been fully studied [2, 2331]. However, the controllability problem of neutral measure evolution equations with nonlocal conditions is seldom studied. Therefore, whether the appropriate solution of the system exists for any given control and whether the system is approximately controllable are studied.

In the paper, we will study the following neutral measure evolution equations with nonlocal conditions:where , is a specified function. The state variable takes values in Banach space X with the norm . generates a uniformly bounded analytic semigroup in Banach space . is a nondecreasing left-continuous function. The control function takes values in another Banach space , where is a control set. The set and represent the regulated functions space on and , respectively. The rest of this paper is organized as follows. In Section 2, some notations and preparation are given about Kurzweil–Henstock–Stieltjes integrals and regulated functions. In Section 3, we obtain the existence results for measure evolution system (1) by using Schauder’s fixed point theorem. In Section 4, based on Krasnoselskii’s fixed point theorem, we establish a controllability result for mild solutions of system (1). An example is given to prove validity of the results we obtained in Section 5.

2. Preliminaries

In this section, we will review some concepts and main results regarding Kurzweil–Henstock–Stieltjes integrals and regulated functions.

Consider a function . A tagged partition of the interval with division points and tags is called -fine if .

Definition 1. (See [22]). Let . A function is said to be Kurzweil-Henstock-Stieltjes integrable with respect to (w.r.t.) on (shortly, K–H–Stieltjes integrable) if there exists a vector such that for every , there is a gauge on withfor every -fine partition of . The K–H–Stieltjes integrability is preserved on all subintervals of . The function is called the K–H–Stieltjes primitive of w.r.t. on . Let be a space of all -ordered K–H–Stieltjes integral regulated functions from to with respect to , with the norm defined by

Definition 2. (See [7]). Let be a Banach space with a norm and be a closed interval of the real line. A function is called regulated on if the limitsexist and are finite. The space of all regulated functions is denoted by . It is well known that the set of discontinuities of a regulated function is at most countable and that the space is a Banach space endowed with the norm . Let , for any element , and we define the norm .
Let be another separable reflexive Banach space where control function takes values. Let be bounded, and admissible control set .

Definition 3. (See [7]). A set is called equiregulated if for every and , there is a such that(i)If and , then (ii)If and , then

Lemma 1. (See [7]). Consider the functions and such that is regulated and exists. Then, the function is regulated and satisfieswhere and , where and denote the left limit and the right limit of the function at , respectively.

Lemma 2. (See [7]). Let be a Banach space. Assume that is equiregulated, and for every , the set is relatively compact in . Then, the set is relatively compact in .

Lemma 3. (See [7]). Let be a Banach space and be a bounded, closed, and convex set. If the operator is completely continuous, then has a fixed point on .

Lemma 4. (See [7]). Let be a sequence of functions from to . If converges pointwisely to as and the sequence is equiregulated, then converges uniformly to .
For , can be defined as a closed linear invertible operator with its domain being dense in . We denote by the Banach space endowed with norm which is equivalent to the graph norm of . We have for , and the embedding is continuous [32].

Lemma 5. (See [32]). The following properties hold:(1)If , then and the embedding is compact whenever the resolvent operator of is compact(2)For every , there exists such that

Lemma 6. (See [31]). Let be a closed convex nonempty subset of a Banach space . We suppose that and map into such that(i)(ii) is continuous and is contained in a compact set(iii) is a contraction with constant Then, there is a with .

3. Existence Results

In this section, we will study the existence of the mild solution of system (1). We first give the definition of solutions for measure system (1). Using the methods same as in [33], we can obtain the following definition.

Definition 4. The function is called a mild solution of system (1) on if it satisfies the following measure integral equation:We introduce the following assumptions:(H1): let , and is a compact operator for every (H2): for each , the function is continuous and there exists a constant , and such that , and for any , the function is strongly measurable and satisfies the Lipschitz condition,and the inequality(H3): the function satisfies the following:(i) is measurable for all , and is continuous for a.e.(ii)There is a function and a nondecreasing continuous function such thatfor all , almost all , and (H4): is continuous and compact, and there exist positive constants and such that(H5): is a linear and bounded operator, so there is a positive constant such that

Theorem 1. If assumptions (H1)–(H5) hold, then problem (1) has a mild solution provided that

Proof. We define the operator byLet be a constant and , where is a bounded, closed, and convex set, .Step I: we prove that there exists a constant such that . Assuming that this conclusion is not true, for each , there will exist such that .According to (H1)–(H5),In the formula (13), , combining (13) with the fact , we can obtainDividing both sides of (14) by and taking limit as , we havewhich contradicts.Step II: is equiregulated.whereThe combination of the compactness of semigroup and its strong continuity shows the continuity of in the sense of uniform operator topology and are bounded, and applying dominated convergence theorem, , as , independently on particular choices of . Combining and with (H2), .as . Moreover, let , according to Lemma 1, and is a regulated function; then,as . Also, we can follow the similar procedure to show as for each . Therefore, is equiregulated on .Step III: is a continuous operator. Let be a convergent sequence in with as . According to hypothesis (H2)–(H5) and the boundedness of , we have, for each ,as , and then, by the dominated convergence theorem, we getIn addition, the analysis same as in Step II demonstrates that is equiregulated on . This property and the abovementioned verification together with Lemma 4 imply that converges uniformly to as , namely,Therefore, is a continuous operator on .Step IV: for every , the set is relatively compact in .The case where is trivial, since . Let be fixed, and let be a given real number satisfying . For every , we defineSince is compact, the set is relatively compact in for every , . On the other hand, for every in view of assumption (H2)-(H3), we haveBy the left continuity of and Lemma 1, as , . Therefore there are relatively compact sets arbitrarily close to the set . Hence, for each , is a relatively compact set in . Step II and Step IV together with Lemma 2 imply that the set is relatively compact in . Hence, is a completely continuous operator on . By Schauder’s fixed point theorem (Lemma 3), has a fixed point in , which is a mild solution of measure control system (1). The proof is completed.

4. Controllability Result

In this section, to investigate the controllability of system (1), we first give the definition of controllability.

Definition 5. The system (1) is said to be controllable on the interval if for every initial function and , there exists a control such that the mild solution of (1) satisfies .
We assume the following conditions:(H6): for all , we assume that function satisfiesand for all , function satisfies(H7): the linear operator , defined by , has an invertible operator taking values in , and there exists a positive constant such that (H8): where (H9):

Theorem 2. If hypotheses (H1)–(H9) are satisfied, system (1) is controllable on .

Proof. Using assumption (H7) for an arbitrary function , we define the controlIn what follows, it suffices to show that when using this control, the operator defined byhas a fixed point from which it follows that this fixed point is a mild solution of system (1). Clearly, , from which we conclude that the system is controllable. Let , where is taken as for while for , it is defined as .
We define the operators and byObviously, the operator has a fixed point if only if operator has a fixed point. Thus, we shall employ Krasnoselskii’s fixed point theorem.Step I: we have to show , i.e., any implies that . It is easy from hypotheses (H1–H9) and Lemma 1 to see thatand thus, condition (i) in Lemma 6 is verified.Step II: next, we need to show that operator is continuous.Therefore, given , there exists such that implies , which proves that operator is continuous.Step III: we shall show that maps into an equicontinuous family. For , we havewhere , . , as . Hence, is equicontinuous. Since the case or is very simple, the proof of the equicontinuities for the two cases is omitted.Subsequentially, we shall show that is precompact. Let be fixed and be a real number satisfying . For , we define is a compact operator, and is relatively compact in . For every . Moreover, for every , we have is precompact in . is uniformly bounded. By the Arzela–Ascoli theorem, it is concluded from the uniform boundedness, equicontinuity, and precompactness of the set that is compact.Step IV: is equiregulated on .whereAccording to Step II of Theroem 1, we can know that is equiregulated on .Step V: we show that operator is a contraction with constant , and we haveTherefore, all the conditions of Krasnoselskii’s fixed point theorem are satisfied, and thus, operator has a fixed point in . From this, it follows that operator has a fixed point and, hence, system (1) is controllable on . This completes the proof.

5. Example

Consider the following equation:

Let , we define by and are absolutely continuous, and . Then, generates a strongly continuous semigroup . Furthermore, has a discrete spectrum, and the eigenvalues are , with corresponding normalized eigenvectors . We also use the following properties:(i)For each . In particular, is a uniformly stable semigroup and .(ii)For each . In particular, .(iii)The operator is given by , on the space .Obviously, Lemma 5 and (H1) are satisfied. Let , for . Then, function is continuous. Let , where is a constant. We defineLet , and we can see that .Therefore, (H3) is satisfied., and . Supposing that and letting be a convergence sequence on ,as . Thus, is a continuous operator. Using the same method, we can obtain that, for any , as . Moreover, . Therefore, (H4) is satisfied.(iv)The function is measurable and .(v)The function is measurable, and ; let

From (iv) it is clear that is a bounded linear operator on . Furthermore, , and .

In fact, from the definition of and (v), it follows thatwhere is defined by .

From (v) we know that is a bounded linear operator with . Hence, we can write , which implies that (H2) holds. Hence, according to Theorem 1, system (38) has a mild solution provided that the condition of Theorem 1 holds. Moreover, we can impose suitable hypotheses to verify the assumptions stated in Theorem 2. Thus, system (38) is controllable on .

6. Conclusions

In this paper, the issue on approximate controllability of neutral measure evolution equations has been addressed, which can model a large class of hybrid systems without any restriction on their Zeno behavior. Firstly, by adopting Schauder’s fixed point theorem, the existence results of mild solutions for this type of measure control system corresponding to some control function are obtained. Then, the approximate controllability results are provided. Finally, we also use an example to illustrate the main result. Furthermore, we will investigate measure functional evolution equations of Sobolev type in the next work.

Data Availability

There are no underlying data in the results.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors read and approved the final manuscript.

Acknowledgments

This work was supported by the Outstanding Young Science and technology Training program of Xinjiang (2019Q022), National Natural Science Foundation of China (11961069 and 11861068), Natural Science Foundation of Xinjiang (2019D01A71 and 2018D01A27), Scientific Research Programs of Colleges in Xinjiang (XJEDU2018Y033), and Xinjiang Normal Uniersity Graduate Research Innovation Project (XSY202002002).