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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 850529, 20 pages
http://dx.doi.org/10.1155/2012/850529
Research Article

Control Systems Described by a Class of Fractional Semilinear Evolution Equations and Their Relaxation Property

1School of Mathematical Science and Computing Technology, Central South University, Changsha, Hunan 410075, China
2Department of Mathematics, Shaoxing University, Shaoxing, Zhejiang 312000, China

Received 29 May 2012; Accepted 16 December 2012

Academic Editor: Irena Lasiecka

Copyright © 2012 Xiaoyou Liu and Xi Fu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We consider a control system described by a class of fractional semilinear evolution equations in a separable reflexive Banach space. The constraint on the control is a multivalued map with nonconvex values which is lower semicontinuous with respect to the state variable. Along with the original system we also consider the system in which the constraint on the control is the upper semicontinuous convex-valued regularization of the original constraint. We obtain the existence results for the control systems and the relaxation property between the solution sets of these systems.

1. Introduction

Let and . We consider the following control system described by a class of fractional semilinear evolution equations of the form: with the mixed nonconvex constraint on the control Here is the Caputo fractional derivative of order , is a finite real number, is the infinitesimal generator of a strongly continuous semigroup in a separable reflexive Banach space , ( is the space of continuous linear operators from into ), is a nonlinear function, and is a multivalued mapping with closed values that is not necessarily convex. The space is a separable, reflexive Banach space modeling the control space.

We denote by the space of all continuous functions from into with the supremum norm given by for . Let be the open unit ball centered at zero. Consider the multivalued map here stands for the closed convex hull of a set. The map (1.4) is usually called the convex upper semicontinuous regularization of .

Along with the constraint (1.2) on the control we also consider the constraint on the control. Note that usually we have .

Definition 1.1. A solution of the control system (1.1), (1.2) is defined to be a pair consisting of a trajectory and a control satisfying (1.1) and the inclusion (1.2) a.e.

A solution of the control system (1.1), (1.5) is defined similarly. We denote by , (, ) the sets of all solutions, all admissible trajectories of the control system (1.1) and (1.2) (the control system (1.1) and (1.5)).

Relaxation property [1] has important ramifications in control theory. There are many papers dealing with the verification of the relaxation property for various classes of control systems. For example, we refer to [25] for nonlinear evolution inclusions or equations, [6, 7] for control problems of subdifferential type and the references therein. In this paper, we investigate this property for control systems described by fractional semilinear evolution equations. We will prove that is a compact set in and where the bar stands for the closure in .

Fractional calculus has recently gained much attentions due to its numerous applications in science and engineering. Examples can be found in various disciplines such as mechanics, electrophysics, signal and image processing, thermodynamics, biophysics, aerodynamics, and economics, (see [812] for instance). For some recent results on fractional differential equations, we can refer to [1317]. As for the study of fractional semilinear differential equations, we can refer to Zhou and Jiao [18, 19], Wang and Zhou [20] for the existence results. The issue of approximate controllability was considered by Kumar and Sukavanam [21], Sakthivel et al. [22]. Wang and Zhou in [23] were concerned with the optimal control settings.

2. Preliminaries and Assumptions

Let be a closed interval of the real line with the Lebesgue measure and the -algebra of measurable sets. The norm of the space (or ) will be denoted by (or ). For any Banach space the symbol stands for equipped with the weak topology. The same notation will be used for subsets of . In all other cases we assume that and its subsets are equipped with the strong (normed) topology.

We first recall the following known definitions from fractional differential theory. For more details, please see [11, 12].

Definition 2.1. The fractional integral of order with the lower limit zero for a function is defined as provided the right hand side is point-wise defined on , where is the gamma function.

Definition 2.2. The Riemann-Liouville derivative of order with the lower limit zero for a function is defined as

Definition 2.3. The Caputo derivative of order with the lower limit zero for a function is defined as

If is an abstract function with values in , then integrals which appear in Definitions 2.1 and 2.2 are taken in Bochner's sense.

We now proceed to some basic definitions and results from multivalued analysis. For more details on multivalued analysis, see the books [24, 25].

We use the following notations: is the set of all nonempty closed subsets of , is the set of all nonempty, closed and bounded subsets of , and is the set of all nonempty, closed, and convex subsets of .

On , we have a metric known as the “Hausdorff metric” and defined by where is the distance from a point to a set . We say a multivalued map is -continuous if it is continuous in the Hausdorff metric .

We say that a multivalued map is measurable if for every closed set . If , then measurability of means that , where is the -algebra of subsets in generated by the sets , , , and is the -algebra of the Borel sets in .

Suppose are two Hausdorff topological spaces and . We say that is lower semicontinuous in the sense of Vietoris (l.s.c. for short) at a point if for any open set , , there is a neighborhood of such that for all . is said to be upper semicontinuous in the sense of Vietoris (u.s.c. for short) at a point if for any open set , , there is a neighborhood of such that for all . For the properties of l.s.c and u.s.c, see the book [24].

Besides the standard norm on (here is a separable, reflexive Banach space), , we also consider the so called weak norm: The space furnished with this norm will be denoted by . The following result establishes a relation between convergence in and convergence in .

Lemma 2.4 (see [5]). If a sequence is bounded and converges to in , then it converges to in .

We assume the following assumptions on the data of our problems in the whole paper.

: The operator generates a strongly continuous semigroup , in , and there exists a constant such that . For any , is compact.

Remark 2.5. Let us take and define the operator by with the domain : are absolutely continuous, , and . Then , , where , , . Clearly generates a compact semigroup in and it is given by , . In such a case, it is easy to see that H(A) holds [22].

: The operator is such that

: The function of Carathéodory type satisfies: there exists a constant such that for a.e. and all , , where and .

: The multivalued map is such that:

: For any , there exists a function such that for a.e. and for any , , and , there is a such that We note that the condition similar to H(M) was also assumed in [6, 7].

From the Definitions 2.1 and 2.2 and the results obtained in [18, 19], Definition 1.1 can be rewritten in the following form.

Definition 2.6. A function is a (mild) solution of the system (1.1), (1.2) if and there exists such that a.e. on and
A similar definition can be introduced for the system (1.1) and (1.5). Here and is a probability density function defined on [26], that is It is not difficult to verify that

Lemma 2.7 (see [18, 19]). Let H(A) hold, then the operators and have the following properties.(1) For any fixed , and are linear and bounded operators, that is, for any , (2) and are strongly continuous.(3) For every , and are compact operators.

Lemma 2.8 (see [27, Theorem 3.1]). Let be continuous and nonnegative on . If where , is a non-negative, monotonic increasing continuous function on and is a positive constant, then where is the Mittag-Leffler function defined for all by

3. Auxiliary Results

In this section, we will give some auxiliary results needed in the proof of the main results. We begin with the a priori estimation of the trajectory of the control systems.

Lemma 3.1. For any admissible trajectory of the control system (1.1) and (1.5), that is, , there is a constant L such that

Proof. Let any , from Definition 2.6, we know that there exists a with a.e. and Then by Lemma 2.7, we get From H(h) and Hölder inequality, we have Similarly, by H(g)(3) and H(U)(3), Combining (3.4), (3.5) with (3.3), we obtain From the above inequality, using the well-known singular version of the Gronwall inequality (see Lemma 2.8), we can deduce that there exists a constant such that .
Let . Let be the -radial retraction, that is, This map is Lipschitz continuous. We define . Obviously, satisfies H(U)(1) and H(U)(2). Moreover, by the properties of , we have for a.e. , all and all the estimates Hence, Lemma 3.1 is still valid with substituted by . Therefore, we assume without any loss of generality that for a.e. , and all Similarly, we can assume that for a.e. and all Let It follows from assumption H(g) that for any , the function is measurable in and continuous in almost everywhere. Hence, for any measurable functions and , the function is scalarly measurable [28]. The separability of the space implies that the function is measurable. Therefore, according to H(g) and H(h), for any and , the function is an element of the space . Hence we can consider the operator defined by the rule

Lemma 3.2. The operator is sequentially continuous as an operator from into .

Proof. Suppose that in and in . Take an arbitrary and any . H(g) and the equality imply that is a scalarly measurable function from to . Hence it is measurable. Consider a subsequence , , of the sequence , , converging to a.e. in . By H(g), H(h), and (3.10), we have Using the preceding four formulae and Lebesgue's theorem of dominated convergence, we obtain Then it follows from (3.16) that Since and is arbitrary, by (3.17) and (3.18), we deduce that It follows from (3.9), (3.10), and (3.12) that is a subset of which is a metrizable compact set in . If the sequence , , does not converge to in , then it has a subsequence , , such that none of its subsequences converges to in . Applying the above arguments to this very subsequence , , we obtain a contradiction. The lemma is proved.

Lemma 3.3. For a.e. , the multivalued map defined by (1.4) from to is u.s.c. with convex closed values.

Proof. From the definition of , it is clear that is closed convex valued. Since is increasing (in the sense of inclusion), and letting we obtain Let and be an open set in such that . By (3.21), we can find an such that For an arbitrary , we can find a such that . Therefore we obtain Then it is clear that , for all . This means that is u.s.c.

Let be a dense countable subset of the ball . We put

Lemma 3.4. For a.e. , let be defined by (3.20), then we have where the closure is taken in .

Proof. We recall that for any subset . Hence it is sufficient to prove that for a.e. , That the left hand side of (3.26) is contained in its right hand side is obvious. Let , then for some . Now let , . Since a.e. , is l.s.c. at and , there is a sequence , converging to (Proposition 1.2.6 [24]). Due to , we have . Since is arbitrary, we can get . Therefore (3.26) holds. The lemma is proved.

Now we consider the following auxiliary problem: It is clear that for every , (3.27) has a unique (mild) solution which is given by

The following lemma describes a property of the solution map which is crucial in our investigation.

Lemma 3.5. The solution map is continuous from to .

Proof. Consider the operator defined by We know is linear. From simple calculation, one has that is, the operator is continuous from into , hence is also continuous from into .
Let any and suppose that for any , ( is a constant). Next we will show that is completely continuous.(a) From (3.30), we know that is uniformly bounded for any and .(b) is equicontinuous on . Let and any , we get By using analogous arguments as in Lemma 3.1, we find For , , it is easy to see that . For and be enough small, we have Combining the estimations for , and , and letting and in , we obtain that is equicontinuous. For more details, please see [19].(c) The set is relatively compact in . Clearly, is compact, and hence, it is only necessary to consider . For each , , and be arbitrary, we define where From the compactness of (), we obtain that the set is relatively compact in for any and . Moreover, we have In virtue of (2.10), the last term of the preceding inequality tends to zero as and . Therefore, there exist relatively compact sets arbitrarily close to the set , . Hence the set , is also relatively compact in .
Since is a convex compact metrizable subset of , it suffices to prove the sequential continuity of the map . Now let such that
By the property of the operator , we have in . Since is bounded, there is a subsequence of the sequence such that in for some . From the facts that we obtain that and in .
From the definitions of the operators and , we have that . Then due to the arguments above, we have in . This completes the proof of the lemma.

4. Existence Results for the Control Systems

In the present section, we are interested in the existence results for the control systems (1.1), (1.2) and (1.1), (1.5).

Let , from Lemma 3.5, we have that is a compact subset of . It follows from formulae (3.9), (3.10), and (3.12) that . Let be defined by

Theorem 4.1. The set is nonempty and the set is a compact subset of the space .

Proof. By the hypothesis H(U)(1), we have that for any measurable function , the map is measurable and has closed values. Therefore it has measurable selectors [29]. So the operator is well defined and its values are closed decomposable subsets of . We claim that is l.s.c. Let , and let be a sequence converging to . It follows from Lemma 3.2 in [30] that there exists a sequence such that Since the map is l.s.c., by the Proposition 1.2.26 in [24], the function is u.s.c. for a.e. . It follows from (4.2) that for a.e. This together with (3.9) implies that in . Therefore the map is l.s.c. By Proposition 2.2 in [31], there is a continuous function such that Consider the map defined by . Due to Lemma 3.5 and the continuity of , the map is continuous from into . Then by Lemma 3.2, we deduce that the map is continuous from into . It follows from (3.9), (3.10), (3.12), and (3.13) that for every . Therefore, the map is continuous from into . Since is a convex metrizable compact set in , Schauder's fixed point theorem implies that this map has a fixed point , that is, . Let and , then we have and . That means which implies that is a solution of the control system (1.1) and (1.2). Hence is nonempty.
It is easy to see that . Since is compact in and is metrizable convex compact in , we have that is relatively compact in . Hence to complete the proof of this theorem, it is sufficient to prove that is sequentially closed in .
Let be a sequence converging to in . Denote According to Lemma 3.2, in . Since and , , Lemma 3.5 implies that Hence, to prove that , we only need to verify that a.e. .
Since in , by Mazur's theorem we have From Lemma 3.3, we have that for a.e. , the map is u.s.c., then by Proposition 1.2.61 in [24], the map is -upper semicontinuous. Therefore from assertion (b) of Proposition 1.2.86 in [24], the map has property . Hence we have In virtue of (4.8) and (4.9), and for a.e. , , , we obtain that a.e. . This means that is compact in . The proof is complete.

5. Main Results

In this section, we will prove the relaxation result. But first, we give a lemma which is important in the proof of our relaxation theorem.

Lemma 5.1. For any pair , there exists a sequence of simple functions and a sequence , , such that

Proof. Let . From Lemma 3.4, we have that for a.e. , The map is measurable (see Propositions 2.3 and 2.6 in [29]) and, by (3.9), is integrally bounded. Therefore, from (5.4) and Theorem 2.2 in [32], we have that there exists an such that We know that the map is measurable and its value are closed, then following Theorem 5.6 in [29] (also Proposition 2.2.3 in [24]), there exists a sequence of measurable selectors , , such that Therefore we have From (5.5) and (5.7), according to Lemma 1.3 in [33] (also see Proposition 2.3.6 [24]), there is a finite measurable partition of such that where is the characteristic function of the set . Now let Formula (5.9) implies that is a simple function, and (5.1), (5.2) hold. By Lemma 2.4, (5.5) and (5.8), we obtain that (5.3) holds. The lemma is proved.

Now we are ready to present our main result.

Theorem 5.2. The set is compact in and the following relation holds where the bar stands for the closure in .

Proof. Let and , be given as in Lemma 5.1. Put , for fixed , we consider the function defined by It is clear that the function is measurable and the function is continuous (in view of H(g) and the fact that if is a measurable function, is a measurable real-valued function). According to the Theorem 2.4 in [34], there exists a sequence of nested (in the sense of inclusion) closed sets , , such that the map is l.s.c. on and is continuous on . Let the multivalued map be defined by For every the graph of the map is an open subset of . Let the map be defined by Hypothesis H(M) together with (5.12) implies that is nonempty for a.e. and all . Since the map is l.s.c. on , , and the graph of the map is an open subset of , , then according to Proposition 1.2.47 in [24], we obtain that the map is l.s.c. on , . Hence the map is l.s.c. on , . Therefore, for every continuous function the map is measurable and the map is l.s.c. on for a.e. .
It is clear that . Consider the system (1.1) with the constraint on the control. The arguments used in the proof of the Theorem 4.1 enable us to obtain the existence of a solution and Now by and , , we have