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Abstract and Applied Analysis
Volume 2010, Article ID 307409, 16 pages
http://dx.doi.org/10.1155/2010/307409
Research Article

On the Controllability of a Differential Equation with Delayed and Advanced Arguments

Departamento de Matemáticas Puras y Aplicadas, Universdad Simón Bolívar, Caracas 1080-A, Venezuela

Received 4 January 2010; Revised 26 March 2010; Accepted 20 April 2010

Academic Editor: Roman Šimon Hilscher

Copyright © 2010 Raúl Manzanilla et al. 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

A semigroup theory for a differential equation with delayed and advanced arguments is developed, with a detailed description of the infinitesimal generator. This in turn allows to study the exact controllability of the equation, by rewriting it as a classical Cauchy problem.

1. Introduction

In this paper, we will study the exact controllability of a functional differential equation with both delayed and advanced arguments. Such equations are often referred to in the literature as mixed-type functional differential equations (MTFDE) or forward-backward equations. The study of this type of equations is less developed compared with other classes of functional equations. Interest in MTDFEs is motivated by problems in optimal control [1] and applications, for example, in economic dynamics [2] and travelling waves in a spatial lattice [3]. See also [4]. In all these references, the reader can find interesting examples and applications.

In order to achieve our goal, first, we rewrite the equation as a classical Cauchy problem in a certain Banach space. Then we introduce the associated semigroup and its infinitesimal generator and prove some important properties of these operators (including some spectral properties). This will allow us to characterize the exact controllability, by applying a result of Bárcenas and Diestel (see [5]).

2. Preliminary Results

In [6], the following differential-difference equation is considered where and is differentiable in .

Equation (2.1) may be written as or equivalently From this we have that in order to find the solution on the interval , it is necessary to know its value on the interval , with being a positive integer. In particular, to determine the solution on the interval , it is necessary to know it on the interval .

Accordingly, is defined for as where the function belongs to the space . The solution of the initial value problem (2.1), (2.4) is constructed via an iterative process using the step derivation method.

It is where and are constants not all necessarily different from zero.

This solution may be extended to the left by rewriting (2.1) as which allows to yield an expression for analogous to (2.5).

In order to assure the existence, differentiability, and uniqueness of the solution , it is demanded that must satisfy the relationship for

Further, if a differentiable solution exists, then it belongs to the space (see [6, Theorems and ]).

Let be the following nonempty, closed subspace of the topological space : The space is endowed with the topology induced by the following countable system of seminorms: The convergence in this topology means the uniform convergence of the function and each of its derivatives of any order. We denote A sequence converges to if and only if tends to , as tends to infinity, for each .

For each , the operator is defined on the solutions of (2.1) as follows: The following result originally appears on [7].

Theorem 2.1. The family defines a strongly continuous semigroup on .

Proof. That and for each are straightforward from the definition of . Since , the domain of is . On the other hand, the function belongs to for each , because if , then for each fixed . Additionally, for each , , for , or equivalently Hence .
In order to prove that is continuous for each fixed , we will prove that there exists for each and some constant such that In view that and using formula (2.5) with , one obtains It only remains to prove that , as , for each .
In fact, let . We have that as for each , as for each
Assuming that and taking into account the uniform convergence of in the closed interval , it follows that as .

Some basic definitions and concepts on controllability are recalled below. Let and be Banach spaces. We consider the inhomogeneous differential linear system: where is the infinitesimal generator operator of a strongly continuous semigroup ; is a bounded linear operator and is a strongly measurable essentially bounded function.

Let be a nonempty separable weakly compact convex subset of .

We recall (see [8]) that is controllable with respect to if there exist and a control such that in (2.18).

The controllability map on for some is the linear map defined by Now, one says that (2.18) is exactly controllable on if every point in can be reached from the origin at time , that is, if , which is equivalent to In other words, The set is called the set of admissible controls of (2.18), while the set is called the set of accessible points of (2.18). Therefore, the system (2.18) is controllable if , for each .

We will make use of the following theorem, which will be applied to problem (2.1),(2.4).

Theorem 2.2 (Bárcenas and Diestel [5]). Let X and U be Banach spaces. Let be a bounded linear operator and the infinitesimal generator of a -semigroup on whose dual semigroup is strongly continuous on . Suppose is nonempty separable weakly compact convex subset of containing . Then, for each , if and only if for each ,

The Bárcenas-Diestel Theorem is an important and recent achievement on exact controllability. Throughout the literature on optimal control in Banach Spaces, hypotheses like “separable and reflexive” are frequently encountered. Using techniques from Banach space theory and the theory of vector measures, the authors show how to remove the hypothesis of reflexivity (thus giving considerably greater generality to the resulting conclusions) and translate the question of accessibility of controls to a problem in semigroups of operators, namely, given a -semigroup of operators on a Banach space , under what conditions is the dual semigroup strongly continuous on ?

It should be noted that, for each fixed and each , a bounded linear functional is defined by means of . The maximum in Theorem 2.2 exists as a consequence of a now classical result of James [9], stating that a weakly closed subset of a Banach space is weakly compact if and only if each continuous linear functional on attains a maximum on .

On the other hand, the following result is proven in [10]: for each , the mapping of to that takes to is continuous (see also [5]), and so the integral in Theorem 2.2 exists in the common Riemann sense.

3. The Cauchy Problem

We will formulate the problem (2.1),(2.4) in the form (2.18). One should observe that we are working on a topological space which is not a Banach space. Let us consider the space , endowed with the product topology, where is defined in (2.8), and let be the closed subspace of all pairs in such that . On we define the following map: for each and , where is the solution (2.5) of the initial value problem (2.1),(2.4).

Now, one has the following result.

Theorem 3.1. The operator satisfies(i) for all ;(ii) is a -semigroup in .

Proof. (i) The linearity of follows from the linearity of its components. In order to see the continuity for each , let us bear in mind that we are working with the product topology. The continuity and convergence for this topology are coordinatewise. We say that is close or tends to if is close to and is close to , for each , , or, equivalently, if for each is close to , where is defined as above. In the case of the linear map , the second coordinate is the continuous semigroup (2.11), and so, if and , we have, for each If is close to , then, in particular, is close to , for each , and thus, by the previous estimate, is close to . Being linear, this is enough to prove that it is continuous.
(ii) Now we will check the semigroup properties. Obviously, . To prove , one defines the function , where is the solution of (2.1),(2.4). Therefore, satisfies By the definition of , one has On the other hand, According to Theorem 2.1, we have that ; since is continuous, then . Consequently, and is a -semigroup in .

Lemma 3.2. Let be the infinitesimal generator associated to the semigroup . For sufficiently large, the resolvent is given by where
(i) ;(ii) with , for .Further, satisfies the following relation:(iii) .

Proof. (i) From [8, Lemma ], for ( is the growth bound of the semigroup), one has One defines for It is observed that is a solution of the differential equation The variation of constants formula for (3.10) shows that is equal to (i).
(iii) Since, as it was defined above, one has after an integration by parts, which is (iii)
(ii) On the other hand, Therefore, which is (ii) for sufficiently large.

Theorem 3.3. Let be the semigroup defined by (3.1). Then its infinitesimal generator is given by where the domain of is

Proof. Consider the operator defined by with We will prove that . Suppose is a sufficiently large number such that Lemma 3.2 is true. In this case, for , one can write by differentiating (i) from Lemma 3.2.
Now, let us see that To do this, one takes and defines We will show that is injective.
Suppose that there exists such that which implies As is different from zero, then and so ; thus is injective. Consequently, which implies that In other words, for every , one has that it is equivalent to Then , as we wanted to see.

For each , let be the space defined in (2.8) provided with the topology of , as a closed subset of . is a Banach space. Therefore, , for all , is also a Banach space with the norm Now, let be the closed subspace of of all pairs such that .

The semigroup , defined in the same form as , is now a semigroup defined on a Banach space for all .

Lemma 3.4. If denote the infinitesimal generator of , then for all , where is the infinitesimal generator of .

Proof.

Theorem 3.5. Let be the infinitesimal generator of .
The spectrum of is discrete and it is defined by where for all , and the multiplicity of each eigenvalue is finite. For every , there exist only a finite number of eigenvalues in . If , then , where satisfies , is an eigenvector of with eigenvalue . On the other hand, if is an eigenvector of with eigenvalue , then with .

Proof. According to the previous Lemma 3.2, for sufficiently large, one has where and are as (i) and (ii), respectively in Lemma 3.2. Denote by the extension of (3.31) to , that is, A simple calculation shows that if satisfies , then is a bonded linear operator from to . Furthermore, for these we have and is injective. Therefore, , the resolvent operator of .
Since , then .
On the other hand, if , then there exists a such that .
The following element of , belongs to and Then .
Let be an element in with .
For this , one has , and from [8, Corollary ], one concludes that is invertible in , where is defined as before. Thus and . Since is an entire function, it has finitely many zeros in the compact set (see in [8, Theorem ]) and we have showed that in the rest of there are none. Therefore, there are only a finite number of eigenvalues in .
Let be an eigenvector of with eigenvalue . From the definition of , one obtains for which gives .
Since , one has . Using the first equation of the definition of , which shows that . The other implication is obvious.
Finally, we will show that the multiplicity of each eigenvalue is finite. From Lemma 3.2, one has where is given in (ii) of Lemma 3.2. We deduce from this expression that the resolvent operator, as an operator from to , is the sum of an operator with finite range and an integral operator. The first operator is compact (see [8, Lemma ]), and so is the second one, as we will see. Therefore, is compact as an operator of , for all . From [8, Theorem and Lemma ], one obtains that the eigenvalue's multiplicity is finite for and .
In order to prove the compactness of the second operator, let us observe that in [11, Example , page 277], it is seen that the operator is compact in . Let us suppose that and let be the closed unit ball in ; then is the unit ball of and it is valid that in the topology of .
Therefore, it is enough to prove that is compact in . Let be the closed unit ball of ; then which implies . If is the closure of in , then, as , is compact in .
Now, let be a covering consisting of open balls in of . Each is contained in an open ball of . Therefore, is an open covering of in , hence there is a finite open covering . At last, is a covering of and so is compact in .
Since , is compact in .

4. Controllability

Let be a Banach space. One will consider the following linear system: where is the infinitesimal generator of the semigroup , is a bonded linear operator, is a strongly measurable, essentially bounded function.

In this section, we will study the controllability of the system (4.1). The mild solution of (4.1) is given by Further, we will suppose that is a separable weakly compact convex subset of .

For , the set is called the set of admissible controls of (4.1), while the set is called the set of accessible points of (4.1).

The system (4.1) is controllable if , for each .

Let be as (4.1). Consider for each , the system where .

Note that, in this case, is also a bounded operator. In fact, since for each , , we can write , where and are bounded linear operators. If is a convergent sequence to , then which converges to zero as tends to .

Let us consider and as before.

The system (4.5) is controllable if , for each , where

Lemma 4.1. .

Proof. If , there exist such that . Since and , for all , then . If , then there exists, for every , , such that . In particular, for , and as and , then , that is, .

Before stating the main result on controllability, we need to prove the following lemma.

Lemma 4.2. The map , defined by , is compact.

Proof. To prove this lemma, we will follow the next five steps.
(1) The Kondrasov's Theorem [12, 13] gives that the following canonical injection is compact where with being the derivative of order in the sense of distributions. is endowed with the norm This implies that is a Banach space.
(2) Using ( ), for , . Thus, for , if , then . Therefore, .
(3) By ( ) and the definition of , it follows that Hence, for every bounded sequence in with the topology of , it can be found a convergent subsequence that converges to in ; but is continuous and . Thus converges to , and so this subsequence converges to in .
(4) To prove that is compact, it must be proven that for every bounded sequence in , a subsequence can be found such that the sequence converges in . As , then the previous fact is equivalent to obtain a convergent subsequence from in .
(5) Using ( ) and the conclusion of ( ), it is obtained that for all , is compact as a bounded linear operator from to .

Theorem 4.3. for each (i.e., the system (4.1) is controllable) if and only if for every and each .

Proof. First, we will prove that , for all , is compact, for all ; in fact, The first term has finite rank, therefore, it is compact. According to Lemma 4.2, the second term is compact in . Consequently, is compact, as an operator in . In particular, it is compact as an operator in . Hence, the adjoint is strongly continuous in (see, e.g., [14]). On the other hand, in view of Lemma 4.1, if and only if , for all and this is true, after Theorem of Bárcenas-Diestel [5], if and only if for all and each .

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