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

Authors: Raúl Manzanilla; Luis Gerardo Mármol; Carmen J. Vanegas
Journal: Abstract and Applied Analysis
 (2010)

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## 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.


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## Body

## 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(2.1)x′(t)=x(t-1)+x(t+1),
where t≥0 andx:[-1,+∞)→ℂn is differentiable in [0,+∞).Equation (2.1) may be written as (2.2)x(t+1)=x′(t)-x(t-1)
or equivalently (2.3)x(t)=x′(t-1)-x(t-2).
From this we have that in order to find the solution x(t) on the interval [m,m+1], it is necessary to know its value on the interval [m-2,m], with m being a positive integer. In particular, to determine the solution on the interval [1,2], it is necessary to know it on the interval [-1,1].Accordingly,x(t) is defined for t∈[-1,1] as(2.4)x(t)=φ(t)={φ1(t),t∈[-1,0],φ2(t),t∈[0,1],
where the function φ belongs to the space C∞([-1,1],ℂn). The solution of the initial value problem (2.1), (2.4) is constructed via an iterative process using the step derivation method.It is(2.5)x(t)=∑i=0mc1iφ(i)(t-m)+∑i=0mc2iφ(i)(t-m-1),
where c1i and c2i are constants not all necessarily different from zero.This solution may be extended to the left by rewriting (2.1) as (2.6)x(t-1)=x′(t)-x(t+1)
which allows to yield an expression for x(t) analogous to (2.5).In order to assure the existence, differentiability, and uniqueness of the solutionx(t), it is demanded that x(t) must satisfy the relationship (2.7)φ(n)(0)=φ(n-1)(-1)+φ(n-1)(1)
for n=1,2,3,….Further, if a differentiable solutionx(t) exists, then it belongs to the space C∞([-1,+∞],ℂn) (see [6, Theorems 3.1 and 3.2]).LetM be the following nonempty, closed subspace of the topological space C∞([-1,1],ℂn):(2.8)M={φ∈C∞([-1,1],ℂn):φ(n)(0)=φ(n-1)(-1)+φ(n-1)(1),n=1,2,3,…}.
The space C∞([-1,1],ℂn) is endowed with the topology induced by the following countable system of seminorms: (2.9)Pk(f)=maxt∈[-1,1]∥f(k)(x)∥ℂn,fork=0,1,2,….
The convergence in this topology means the uniform convergence of the function and each of its derivatives of any order. We denote(2.10)∥f∥k=∑i=0kPi(f),for eachk=0,1,2,….
A sequence {fn}n=1∞ converges to f if and only if ∥fn-f∥k tends to 0, as n tends to infinity, for each k.For eacht≥0, the operator T(t) is defined on the solutions x(t) of (2.1) as follows:(2.11)T(t)x(θ)=x(t+θ),θ∈[-1,1].
The following result originally appears on [7].Theorem 2.1.
The family{T(t)}t≥0 defines a strongly continuous semigroup on L(M).Proof.
ThatT(0)=I and T(t+s)=T(t)T(s) for each t,s≥0 are straightforward from the definition of T(t).Since x(θ)|θ∈[-1,1]=φ(θ)∈M,the domain of T(t) is M.On the other hand, the function
(2.12)yt(θ):=T(t)x(θ)=x(t+θ)
belongs to M for each t≥0, because if x(t)∈C∞([-1,1],ℂn), then yt(θ)∈C∞([-1,1],ℂn) for each fixed t≥0. Additionally, for each t≥0, yt(n)(0)=yt(n-1)(-1)+yt(n-1)(1), for n=1,2,3,…, or equivalently
(2.13)x(n)(t)=x(n-1)(t-1)+x(n-1)(t+1),n=1,2,….
Hence yt(θ)∈M.
In order to prove thatT(τ) is continuous for each fixed τ≥0, we will prove that there exists nk for each k and some constant c≥0 such that
(2.14)∥T(τ)x(θ)∥k≤c∑i=0nk∥x(θ)∥i.
In view that
(2.15)Pk(T(τ)x(θ))=maxθ∈[-1,1]|x(k)(τ+θ)|≤maxt∈[[τ]]-1,[τ]+2]|x(k)(t)|≤maxt∈[τ]-1,[τ]]|x(k)(t)|+maxt∈[[τ]],[τ]+1]|x(k)(t)|+maxt∈[[τ]]+1,[τ]+2]|x(k)(t)|
and using formula (2.5) with m=[τ], one obtains
(2.16)maxθ∈[-1,1]|x(k)(τ+θ)|≤∑i=0m-1c1imaxt∈[m-1,m]|φ(i+k)(t-(m-1))|+∑i=0m-1c2imaxt∈[m-1,m]|φ(i+k)(t-(m-1)-1)|+∑i=0mc3imaxt∈[m,m+1]|φ(i+k)(t-m)|+∑i=0mc4imaxt∈[m,m+1]|φ(i+k)(t-m-1)|+∑i=0m+1c5imaxt∈[m+1,m+2]|φ(i+k)(t-(m+1))|+∑i=0m+1c6imaxt∈[m+1,m+2]|φ(i+k)(t-(m+1)-1)|≤∑i=0m-1(c1i+c2i)∑i=0m-1maxθ∈[-1,1]|φ(i+k)(θ)|+∑i=0m(c3i+c4i)∑i=0mmaxθ∈[-1,1]|φ(i+k)(θ)|+∑i=0m+1(c5i+c6i)∑i=0m+1maxθ∈[-1,1]|φ(i+k)(θ)|≤c1∥φ∥m+k-1+c2∥φ∥m+k+c3∥φ∥m+k+1≤(c1+c2+c3)(∥φ∥m+k-1+∥φ∥m+k+∥φ∥m+k+1)=c(∥φ∥m+k-1+∥φ∥m+k+∥φ∥m+k+1).
It only remains to prove that T(t)x→T(t0)x, as t→to, for each x∈M.
In fact, lett0≥0. We have that limt→t0T(t)x=T(t0)x⇔limτ→0T(t0+τ)x=T(t0)x⇔∥T(t0+τ)x(θ)-T(t0)x(θ)∥k→0 as τ→0 for each k=0,1,…, ⇔maxθ∈[-1,1]|x(k)(t0+τ+θ)-x(k)(t0+θ)|→0 as τ→0 for each k=0,1,….
Assuming that0≤[τ]≤1 and taking into account the uniform convergence of x(k)(t) in the closed interval [a,b]:=[t0-2,t0+2](t,t'∈[a,b]), it follows that
(2.17)maxθ∈[-1,1]|x(k)(t0+τ+θ)-x(k)(t0+θ)|≤max|t-t'|≤[τ]|x(k)(t)-x(k)(t′)|→0
as τ→0.Some basic definitions and concepts on controllability are recalled below. LetU and X be Banach spaces. We consider the inhomogeneous differential linear system:(2.18)ẋ(t)=Ax(t)+Bu(t),x(0)=x0∈D(A),
where A:X→X is the infinitesimal generator operator of a strongly continuous semigroup (St)t≥0; B:U→U is a bounded linear operator and u:[0,+∞)→U is a strongly measurable essentially bounded function.LetΩ be a nonempty separable weakly compact convex subset of U.We recall (see [8]) that x0∈X is controllable with respect to x1 if there exist t_≥0 and a control u∈L∞([0,t_];U) such that x(t_)=x1 in (2.18).The controllability map on[0,t_] for some t_≥0 is the linear map (2.19)Bt_:L∞([0,t_];U)→X
defined by (2.20)Bt_u=∫0t_S(t_-s)Bu(s)ds.
Now, one says that (2.18) is exactly controllable on [0,t_] if every point in X can be reached from the origin at time t_, that is, if ran(Bt_)=X, which is equivalent to(2.21)x1=∫0t_S(t_-s)Bu(s)ds,0=-x1+∫0t_S(t_-s)Bu(s)ds,or0=S(t_)x0+∫0t_S(t_-s)Bu(s)ds.
In other words,(2.22)ran(Bt_)=X⇔0∈{S(t_)x0+∫0t_S(t_-s)Bu(s)ds:uis a control inL∞([0,t_];U)}.
The set Ωt_={u∈L∞([0,t];U):u∈Ωa.e.} is called the set of admissible controls of (2.18), while the set At_(x0)={S(t_)x0+∫0t_S(t_-s)Bu(s)ds:u∈Ωt_} is called the set of accessible points of (2.18). Therefore, the system (2.18) is controllable if 0∈At_(x0), for each t_>0.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. LetB:U→X be a bounded linear operator and A:X→X the infinitesimal generator of a C0-semigroup (St)t≥0 on X whose dual semigroup is strongly continuous on (0,+∞). Suppose Ω is nonempty separable weakly compact convex subset of U containing 0. Then, for each t_>0, 0∈At_(x0) if and only if for each x*∈X*,
(2.23)x*(S(t_)x0)+∫0t_maxv∈Ωx*(S(t)Bv)dt≥0.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 ac0-semigroup (S(t))t≥0 of operators on a Banach space X, under what conditions is the dual semigroup strongly continuous on (0,∞)?It should be noted that, for each fixedt≥0 and each x*∈X*, a bounded linear functional u*∈U* is defined by means of u*(v):=x*(S(t)Bv). The maximum in Theorem 2.2 exists as a consequence of a now classical result of James [9], stating that a weakly closed subset C of a Banach space Z is weakly compact if and only if each continuous linear functional on Z attains a maximum on C.On the other hand, the following result is proven in [10]: for each x*∈X*, the mapping of [0,τ] to [0,∞) that takes t∈[0,τ] to maxv∈Ωx*(S(t)Bv) 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 ℂn×M, endowed with the product topology, where M is defined in (2.8), and let N be the closed subspace of all pairs (r,f) in ℂn×M such that f(0)=r. On N we define the following map:(3.1)S(t)(φ(0)φ(·))=(x(t)x(t+θ))=(∑i=0mc1iφ(i)(t-m)+∑i=0mc2iφ(i)(t-m-1)x(t+θ))
for each t≥0 and θ∈[-1,1], where x(t) is the solution (2.5) of the initial value problem (2.1),(2.4).Now, one has the following result.Theorem 3.1.
The operatorS satisfies(i)
S(t)∈L(N) for all t≥0;(ii)
S(t) is a C0-semigroup in N.Proof.
(i) The linearity ofS(t) follows from the linearity of its components. In order to see the continuity for each t≥0,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 (f(0),f) is close or tends to (g(0),g) if ∥f(0)-g(0)∥ℂn is close to 0 and ∥f-g∥k is close to 0,for each k,(k=0,1,2,…),or, equivalently, if for each k(3.2)∥f(0)-g(0)∥ℂn+∥f-g∥k
is close to 0, where ∥·∥k is defined as above. In the case of the linear map S(t), the second coordinate is the continuous semigroup (2.11), and so, if c1=max1≤i≤m|c1i| and c2=max1≤i≤m|c2i|, we have, for each k(3.3)∥x(t)∥ℂn+∥x(t+θ)∥k≤∑i=0mc1∥φ(i)(t-m)∥ℂn+∑i=0mc2∥φ(i)(t-m-1)∥ℂn+c∑i=0nk∥φ(·)∥i≤(c1+c2)∥φ(·)∥m+c∑i=0nk∥φ(·)∥i.
If (φ(0),φ(·)) is close to (0,0), then, in particular, ∥φ(·)∥k is close to 0, for each k, and thus, by the previous estimate, S(t)(φ(0)φ(·)) is close to (0,0). Being S(t) linear, this is enough to prove that it is continuous.
(ii) Now we will check the semigroup properties. Obviously,S(0)=I. To prove S(t+s)=S(t)S(s), one defines the function h(t)=x(t+s), where x(·) is the solution of (2.1),(2.4). Therefore, h(t) satisfies
(3.4)ḣ(t)=h(t-1)+h(t+1),0≤t,h(0)=x(s),h(θ)=φ(s+θ),θ∈[-1,1].
By the definition of S, one has
(3.5)(h(t)h(t+·))=S(t)(x(s)x(s+·))=S(t)S(s)(φ(s)φ(·)).
On the other hand,
(3.6)S(t+s)(φ(0)φ(·))=(x(t+s)x(t+s+·))=(h(t)h(t+·))=S(t)S(s)(φ(0)φ(·)).
According to Theorem 2.1, we have that limt→0+x(t+s)=x(s); since φ is continuous, then limt→0+x(t)=φ(0). Consequently, limt→0+S(t)=I and S(t) is a C0-semigroup in N.Lemma 3.2.
LetA be the infinitesimal generator associated to the semigroup S(t). For α∈ℝ sufficiently large, the resolvent is given by
(3.7)(αI-A)-1(φ(0)φ(·))=(g(0)g(·)),
where
(i)
g(θ)=eαθg(0)-∫0θeα(θ-s)φ(s)ds,θ∈[-1,1];(ii)
g(0)=(Δ(α))-1[φ(0)+∫-10e-α(θ+1)φ(θ)dθ-∫01e-α(θ-1)φ(θ)dθ] withΔ(λ)=(λ-eλ-e-λ)I, forλ∈C.Further, g satisfies the following relation:(iii)
αg(0)=φ(0)+g(-1)+g(1).Proof.
(i) From [8, Lemma 2.1.11],for α≥ω0(ω0 is the growth bound of the semigroup), one has
(3.8)(αI-A)-1(φ(0)φ(·))=∫0+∞e-αtS(t)(φ(0)φ(·))dt=∫0+∞e-αt(x(t)x(t+·))dt.
One defines for θ∈[-1,1](3.9)g(θ)=∫0+∞e-αtx(t+θ)dt=∫θ+∞e-α(s-θ)x(s)ds.
It is observed that g(θ) is a solution of the differential equation
(3.10)dgdθ(θ)=αg(θ)-x(θ).
The variation of constants formula for (3.10) shows that g(θ) is equal to (i).
(iii) Since, as it was defined above,(3.11)g(θ)=∫0+∞e-αtx(t+θ)dt,θ∈[-1,1],
one has after an integration by parts,
(3.12)αg(0)=α∫0+∞e-αtx(t)dt=φ(0)+∫0+∞e-αt(x(t-1)+x(t+1))dt=φ(0)+g(-1)+g(1)
which is (iii)
(ii) On the other hand,(3.13)αg(0)=φ(0)+∫0-1e-αtx(t+1)dt+∫-1+∞e-αtx(t+1)dt+∫01e-αtx(t-1)dt+∫1+∞e-αtx(t-1)dt=φ(0)+∫-1+∞e-αtx(t+1)dt+∫1+∞e-αtx(t-1)dt+∫01e-αtφ(t-1)dt-∫-10e-αtφ(t+1)dt=φ(0)+eαg(0)+e-αg(0)+∫01e-αtφ(t-1)dt-∫-10e-αtφ(t+1)dt.
Therefore,
(3.14)[(α-(eα+e-α))I]g(0)=φ(0)+∫-10e-α(θ+1)φ(θ)dθ-∫01e-α(θ-1)φ(θ)dθ
which is (ii) for α sufficiently large.Theorem 3.3.
LetS(t) be the semigroup defined by (3.1). Then its infinitesimal generator is given by
(3.15)A(φ(0)=rφ(·))=(φ(-1)+φ(1)dφdθ(·)),
where the domain of A is
(3.16)D(A)={(rφ(·))∈N:φis absolutely continuous, anddφdθ∈M}.Proof.
Consider the operatorÃ defined by
(3.17)Ã(rφ(·))=(φ(-1)+φ(1)dφdθ(·))
with
(3.18)D(Ã)={(rφ(·))∈N:φis  absolutely  continuous, anddφdθ∈M}.
We will prove that A=Ã. Suppose α is a sufficiently large number such that Lemma 3.2 is true. In this case, for (rφ(·))∈N, one can write
(3.19)(αI-Ã)(αI-A)-1(rφ(·))=(αI-Ã)(g(0)g(·))=(αg(0)-g(1)-g(-1)αg(·)-dgdθ(·))=(rφ(·))
by differentiating (i) from Lemma 3.2.
Now, let us see that(3.20)(αI-A)(αI-Ã)-1(rφ(·))=(rφ(·)).
To do this, one takes (rφ(·))∈D(A) and defines
(3.21)(r1φ1(·)):=(αI-A)-1(αI-Ã)(rφ(·)).
We will show that (αI-Ã) is injective.
Suppose that there exists(r0φ0(·))∈D(A) such that
(3.22)(00)=(αI-Ã)(r0φ0(·))=(αI-Ã)(φ0(0)φ0(·))=(αφ0(0)-φ0(-1)-φ0(1)αφ0(·)-dφ0dθ(·))
which implies
(3.23)αφ0(0)-φ0(-1)-φ0(1)=αφ0(0)-φ0(0)eα-φ0(0)e-α=(α-eα-e-α)φ0(0)=0,φ0(θ)=φ0(0)eαθ.
As α-eα-e-α is different from zero, then φ0(0)=r0=0 and so φ0(·)=φ0(0)eα(·)=0; thus (αI-Ã) is injective. Consequently,
(3.24)(αI-Ã)(r1φ1(·))=(αI-Ã)(αI-A)-1[(αI-Ã)(rφ(·))]=(αI-Ã)(rφ(·))
which implies that
(3.25)(r1φ1(·))=(rφ(·)).
In other words, for every (rφ(·))∈D(A), one has
(3.26)(αI-A)-1(αI-Ã)(rφ(·))=(rφ(·)),
that it is equivalent to
(3.27)(αI-Ã)(rφ(·))=(αI-A)(rφ(·)).
Then Ã=A, as we wanted to see.For eachm∈ℕ, let Mm be the space M defined in (2.8) provided with the topology of Cm([-1,1],ℂn), as a closed subset of Cm([-1,1],ℂn). Mm is a Banach space. Therefore, ℂn×Mm, for all m∈ℕ, is also a Banach space with the norm (3.28)∥(rf)∥ℂn×Mm=∥r∥ℂn+∥f∥Cm([-1,1],ℂn).
Now, let Nm be the closed subspace of ℂn×Mm of all pairs (r,f) such that r=f(0).The semigroupSm(t):Nm→Nm, defined in the same form as S(t), is now a semigroup defined on a Banach space for all m∈ℕ.Lemma 3.4.
IfAm denote the infinitesimal generator of Sm(t), then Am=A for all m∈ℕ, where A is the infinitesimal generator of S(t).Proof.
(3.29)limt→01t∥Sm(t)(φ(0)φ(·))-(φ(0)φ(·))-tA(φ(0)φ(·))∥ℂn×Mm=limt→01t∥(x(t)x(t+θ))-(φ(0)φ(·))-(t(φ(-1)+φ(1))tdφdθ(·))∥ℂn×Mm=limt→0[1t∥x(t)-φ(0)-t(φ(-1)+φ(1))∥ℂn+1t∥x(t+θ)-φ(·)-tdφdθ(·)∥Mm]≤limt→0[1t∥x(t)-φ(0)-t(φ(-1)+φ(1))∥ℂn+1t∑k=0mPk(x(t+θ)-φ(·)-tdφdθ(·))]=0.Theorem 3.5.
LetA be the infinitesimal generator of S(t).
The spectrum ofA is discrete and it is defined by
(3.30)σ(A)={λ∈ℂ:det(Δλ)=0},
where Δλ=(λ-eλ-e-λ)I for all λ∈ℂ, and the multiplicity of each eigenvalue is finite. For every δ∈ℝ, there exist only a finite number of eigenvalues in Cδ+={s∈ℂ:Re(s)>δ}. If λ∈σ(A), then (reλ·r), where r≠0 satisfies Δλr=0, is an eigenvector of A with eigenvalue λ. On the other hand, if Φ is an eigenvector of A with eigenvalue λ, then Φ=(reλ·r) with Δλr=0.Proof.
According to the previous Lemma3.2,for α∈ℝ sufficiently large, one has
(3.31)(αI-A)-1(rφ(·))=(g(0)g(·)),
where g(θ) and g(0) are as (i) and (ii), respectively in Lemma 3.2. Denote by Qλ the extension of (3.31) to ℂ, that is,
(3.32)Qλ(rφ(·)):=(g(0)g(·)).
A simple calculation shows that if λ∈ℂ satisfies det(Δλ)≠0, then Qλ is a bonded linear operator from N to N. Furthermore, for these λ we have (λI-A)Qλ=I and (λI-A) is injective. Therefore, Qλ=(λI-A)-1=ϱ(A), the resolvent operator of A.
Since{λ∈ℂ:det(Δλ)≠0}⊆ϱ(A), then σ(A)⊂{λ∈ℂ:det(Δλ)=0}.
On the other hand, ifdet(Δλ)=0, then there exists a ξ∈ℂn such that (λI-e-λ-eλ)(ξ)=0.
The following element ofN, z0=(ξeλ·ξ) belongs to D(A) and
(3.33)(λI-A)z0=(λξ-e-λξ-eλξλeλθξ-ddθeλθξ)=(00).
Then {λ∈ℂ:det(Δλ)=0}⊂σ(A).
Letλ be an element in Cδ+ with eδ+e-δ<|λ|.
For thisλ, one has |eλ+e-λ|≤eδ+e-δ<|λ|, and from [8, Corollary A.4.10], one concludes that λI-eλ-e-λ is invertible in Nm, where Nm is defined as before. Thus det(Δλ)≠0 and λ∈ρ(A). Since det(λI-eλ-e-λ) is an entire function, it has finitely many zeros in the compact set Cδ+¯∩{λ∈ℂ:|λ|≤eδ+e-δ} (see in [8, Theorem A.1.4.6b]) and we have showed that in the rest of Cδ+ there are none. Therefore, there are only a finite number of eigenvalues in Cδ+.
LetΦ=(rφ(·)) be an eigenvector of A with eigenvalue λ. From the definition of A, one obtains (dφ/dθ)(θ)=λφ(θ) for θ∈[-1,1] which gives φ(θ)=eλθφ(0).
SinceΦ∈D(A), one has φ(0)=r. Using the first equation of the definition of A, eλ+e-λr=λr which shows that Δλr=0. The other implication is obvious.
Finally, we will show that the multiplicity of each eigenvalue is finite. From Lemma3.2, one has
(3.34)(αI-A)-1(φ(0)φ(·))=(g(0)eαg(0))+(0-∫0·eα(·-s)φ(s)ds),
where g(0) is given in (ii) of Lemma 3.2. We deduce from this expression that the resolvent operator, as an operator from Nm to Nm, is the sum of an operator with finite range and an integral operator. The first operator is compact (see [8, Lemma A.3.22a]), and so is the second one, as we will see. Therefore, (αI-A)-1 is compact as an operator of L(Nm), for allm∈ℕ. From [8, Theorem A.4.18 and Lemma A.4.19], one obtains that the eigenvalue's multiplicity is finite for (αI-A) and A.
In order to prove the compactness of the second operator, let us observe that in [11, Example 1, page 277], it is seen that the operator
(3.35)Tφ(a)=∫0aeα(a-s)φ(s)ds
is compact in C[0,a]. Let us suppose that a≤1 and let ℬ be the closed unit ball in Cm[-1,1]; then ℬ∩M is the unit ball of Mm and it is valid that T(ℬ∩M)¯⊆T(ℬ)¯ in the topology of Cm[-1,1].
Therefore, it is enough to prove thatT(ℬ)¯ is compact in Cm[-1,1]. Let ℬ' be the closed unit ball of C[0,a]; then ℬ⊆ℬ' which implies T(ℬ)⊆T(ℬ′). If 𝒲 is the closure of T(ℬ) in C[0,a], then, as T(ℬ)⊆T(ℬ′), 𝒲 is compact in C[0,a].
Now, let{Vα}α∈J be a covering consisting of open balls in Cm[-1,1] of 𝒲. Each Vα is contained in an open ball ℬα of C[0,a]. Therefore, {ℬα}α∈J is an open covering of 𝒲 in C[0,a], hence there is a finite open covering {ℬαi}i=1k. At last, {Vαi:ℬαi⊆Vαi}i=1k is a covering of 𝒲 and so 𝒲 is compact in Cm[-1,1].
SinceT(ℬ)¯Cm[-1,1]⊆T(ℬ)¯C[0,a]=𝒲, T(ℬ)¯ is compact in Cm[-1,1].


## 4. Controllability

LetU be a Banach space. One will consider the following linear system:(4.1)x′(t)=Ax(t)+Bu(t),x(0)=x0,x0∈D(A),x(θ)=φ(θ),θ∈[-1,1],
where A:D(A)⊆N→N is the infinitesimal generator of the semigroup (S(t))t≥0, B:U→N is a bonded linear operator, u:[0,∞)→U 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 (4.2)x(t)=S(t)x0+∫0tS(t-s)Bu(s)ds.
Further, we will suppose that Ω is a separable weakly compact convex subset of U.Forτ>0, the set
(4.3)Ω̃τ={u∈LU∞[0,τ]:u∈Ωa.e.}
is called the set of admissible controls of (4.1), while the set
(4.4)Aτ(x0)={S(τ)x0+∫0τS(τ-s)Bu(s)ds:u∈Ω̃τ}
is called the set of accessible points of (4.1).The system (4.1) is controllable if 0∈Aτ(x0), for each τ>0.Letx0,φ,U,u,B be as (4.1). Consider for each m∈ℕ, the system(4.5)x′(t)=Amx(t)+Bu(t),x(0)=x0,x0∈D(A),x(θ)=φ(θ),θ∈[-1,1],
where B:U→Nm.Note that, in this case,B is also a bounded operator. In fact, since for each u∈U, Bu∈Nm, we can write Bu=(B1u,B2u), where B1:U→ℂn and B2:U→Mm are bounded linear operators. If {uj}j=0∞⊆U is a convergent sequence to u, then (4.6)∥Buj-Bu∥ℂn×Mm=∥(B1ujB2uj)-(B1uB2u)∥ℂn×Mm=∥(B1uj-B1uB2uj-B2u)∥ℂn×Mm=∥B1uj-B1u∥ℂn+∥B2uj-B2u∥Mm≤∥B1uj-B1u∥ℂn+Pm(B2uj-B2u)
which converges to zero as j tends to +∞.Let us considerΩ and Ω̃τ as before.The system (4.5) is controllable if 0∈Aτm(x0), for each τ>0, where
(4.7)Aτm(x0)={Sm(τ)x0+∫0τSm(τ-s)Bu(s)ds:u∈Ω̃τ}.Lemma 4.1.
Aτx0=⋂m=1∞Aτmx0.Proof.
Ifz∈Aτx0,there exist u∈Ω̃τ such that z=S(τ)x0+∫0τS(τ-s)Bu(s)ds. Since S(τ)x0=Sm(τ)x0 and S(τ-s)Bu(s)=Sm(τ-s)Bu(s), for allm∈ℕ, then z∈⋂m=1∞Aτmx0. If z∈⋂m=1∞Aτmx0, then there exists, for every m∈ℕ, um∈Ω̃τ, such that z=Sm(τ)x0+∫0τSm(τ-s)Bum(s)ds. In particular, for m=1, z=S1(τ)x0+∫0τS1(τ-s)Bu1(s)ds and as S(τ)x0=S1(τ)x0 and S(τ-s)Bu1(s)=S1(τ-s)Bu1(s), then z=S1(τ)x0+∫0τS1(τ-s)Bu1(s)ds, that is, z∈Aτx0.Before stating the main result on controllability, we need to prove the following lemma.Lemma 4.2.
The mapTt:Mm→Mm, defined by Tt(u)(θ)=u(t+θ), 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
(4.8)Hm(-1,1)↪cC0([-1,1])ifm∈ℕ,
where
(4.9)Hm(-1,1)={f∈L2(-1,1):f(k)∈L2(-1,1),k=1,…,m}
with f(k) being the derivative of order k in the sense of distributions. Hm(-1,1) is endowed with the norm
(4.10)∥f∥m2=∑k=0m∫-11|f(k)(t)|2dt.
This implies that Hm[-1,1] is a Banach space.
(2) Using (1), for m=1, H1[-1,1]↪cC0([-1,1]). Thus, for m≥1, if f∈Hm+1(-1,1), then f(m)∈H1(-1,1). Therefore, Hm+1(-1,1)↪cCm([-1,1]).
(3) By (2) and the definition of Mm+1, it follows that
(4.11)Mm+1⊆Cm+1([-1,1])⊆Hm+1(-1,1)↪cCm([-1,1]).
Hence, for every bounded sequence {fk}k∈ℕ in Mm+1 with the topology of Cm([-1,1]), it can be found a convergent subsequence {fkl}l∈ℕ⊆{fk}k∈ℕ that converges to f in Cm([-1,1]); but f(m) is continuous and f(m+1)∈L2(-1,1). Thus {fkl(m+1)}l∈ℕ converges to f(m+1)a.e, and so this subsequence converges to f in Mm+1.
(4) To prove thatTt is compact, it must be proven that for every bounded sequence {ui}i∈ℕ in Mm, a subsequence {uij}j∈ℕ can be found such that the sequence {Tt(uij)}j∈ℕ converges in Mm. As Tt(ui)(θ)=ui(t+θ), then the previous fact is equivalent to obtain a convergent subsequence {uij}j∈ℕ from {ui}i∈ℕ in Mm.
(5) Using (3) and the conclusion of (4), it is obtained that for all t∈ℝ, Tt is compact as a bounded linear operator from Mm to Mm.Theorem 4.3.
0∈Aτx0 for each τ>0 (i.e., the system (4.1) is controllable) if and only if
(4.12)x*(Sm(τ)x0)+∫0τmaxu∈Ωx*(Sm(t)Bu)dt≥0
for every m∈ℕ and each x*∈(Nm)*.Proof.
First, we will prove thatSm(t), for all t≥0, is compact, for all m∈ℕ; in fact,
(4.13)Sm(t)(rφ(·))=(x(t)x(t+θ))=(x(t)0)+(0x(t+θ)).
The first term has finite rank, therefore, it is compact. According to Lemma 4.2, the second term is compact in Mm. Consequently, Sm(t) is compact, as an operator in L(ℂn×Mm). In particular, it is compact as an operator in L(Nm). Hence, the adjoint (Sm(t))* is strongly continuous in (0,+∞) (see, e.g., [14]). On the other hand, in view of Lemma 4.1, 0∈Aτx0 if and only if 0∈Aτmx0, for allm∈ℕ and this is true, after Theorem 1.1 of Bárcenas-Diestel [5], if and only if x*(Sm(τ)x0)+∫0τmaxu∈Ωx*(Sm(t)Bu)dt≥0for allm∈ℕ and each x*∈(Nm)*.



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*Source: 307409-2010-07-07.xml*