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Controllability of a string submitted to unilateral constraint
Farid Ammar-Khodja
a,1, Sorin Micu
b,2, Arnaud Münch
c,∗,3aLaboratoire de Mathématiques de Besançon, UMR CNRS 6623, Université de Franche-Comté, 16 route de Gray, 25030 Besançon cedex, France bFacultatea de Matematica si Informatica, Universitatea din Craiova, 200585, Romania
cLaboratoire de Mathématiques de Clermont-Ferrand, Université Blaise Pascal, UMR CNRS 6620, Campus des Cézeaux, 63177 Aubière, France
Received 14 May 2009; received in revised form 17 December 2009; accepted 17 January 2010 Available online 13 February 2010
Abstract
This article studies the controllability property of a homogeneous linear string of length one, submitted to a time dependent obstacle (described by the function{ψ(t)}0tT) located below the extremityx=1. The Dirichlet control acts on the other extremity x=0. The string is modelled by the wave equationy−yxx=0 in (t, x)∈(0, T )×(0,1), while the obstacle is represented by the Signorini’s conditionsy(t,1)ψ(t),yx(t,1)0,yx(t,1)(y(t,1)−ψ(t))=0 in(0, T ). The characteristic method and a fixed point argument allow to reduce the problem to the analysis of the solutions atx=1. We prove that, for any T >2 and initial data(y0, y1)∈H1(0,1)×L2(0,1)withψ (0)y0(1), the system is null controllable with controls inH1(0, T ).
Two distinct approaches are used. We first introduce a penalized system iny, transforming the Signorini’s condition into the simpler oney,x(t,1)=−1[y(t,1)−ψ(t)]−,being a small positive parameter. We construct explicitly a family of controls of the penalized problem, uniformly bounded with respect toinH1(0, T ). This enables us to pass to the limit and to obtain a control for the initial equation. A more direct approach, based on differential inequalities theory, leads to a similar positive conclusion.
Numerical experiments complete the study.
©2010 Elsevier Masson SAS. All rights reserved.
Résumé
Cet article étudie les propriétés de contrôlabilité d’une corde homogène de longeur un, soumise à un obstacle dépendant du temps (décrit par la fonction{ψ(t)}0tT) à l’extrémitéx=1. Le contrôle Dirichlet agit à l’extrémitéx=0. La corde est modélisée par l’équation des ondesy−yxx=0 dans(t, x)∈(0, T )×(0,1), tandis que l’obstacle est représenté par les conditions de Signorini y(t,1)ψ(t),yx(t,1)0,yx(t,1)(y(t,1)−ψ(t))=0 sur(0, T ). La méthode des caractéristiques et un argument de point fixe permettent de réduire le problème à l’analyse des solutions enx=1. Nous prouvons que, pour toutT >2 et donnée initiale (y0, y1)∈H1(0,1)×L2(0,1)avecψ (0)y0(1), le système est contrôlable à zéro avec des contrôles dansH1(0, T ). Deux approches sont utilisées. On introduit tout d’abord un système pénalisé eny, transformant les conditions de Signorini en l’égalité y,x(t,1)=−1[y(t,1)−ψ(t)]−,étant un paramètre positif. On construit explicitement une famille de contrôle du problème pénalisé uniformément bornée par rapport àdansH1(0, T ). Cela nous permet de passer à la limite et d’obtenir un contrôle pour
* Corresponding author.
E-mail addresses:farid.ammar-khodja@univ-fcomte.fr (F. Ammar-Khodja), sd_micu@yahoo.com (S. Micu), arnaud.munch@math.univ-bpclermont.fr (A. Münch).
1 Partially supported by grant ANR-07-JC-183284 (Agence nationale de la recherche, France).
2 Partially supported by grant MTM2008-03541 funded by MICINN (Spain) and grant 420/2008 of CNCSIS (Romania).
3 Partially supported by grant ANR-07-JC-183284 (France) and grant 08720/PI/08 from Fundación Séneca (Spain).
0294-1449/$ – see front matter ©2010 Elsevier Masson SAS. All rights reserved.
doi:10.1016/j.anihpc.2010.02.003
le système initial. Une approche plus directe, relevant de la théorie des inéquations différentielles, conduit à un résultat positif similaire. Quelques applications numériques complètent l’étude.
©2010 Elsevier Masson SAS. All rights reserved.
MSC:35L85; 65M12; 74H45
Keywords:Nonlinear boundary controllability; Unilateral constraint; Penalization; Fixed point method
1. Introduction
LetT >0 andQT =(0, T )×(0,1). We consider the following system
⎧⎪
⎪⎨
⎪⎪
⎩
y−yxx=0, (t, x)∈QT,
y(t,0)=u(t ), t∈(0, T ),
y(t,1)ψ (t ), yx(t,1)0,
y(t,1)−ψ (t )
yx(t,1)=0, t∈(0, T ),
y(0, x)=y0(x), y(0, x)=y1(x), x∈(0,1),
(1.1)
where the symbol denotes the derivative with respect to the variablet. System (1.1) models the vibration of a homogeneous and linear string of length one in the time interval(0, T )submitted to an initial excitation(y0, y1)at timet=0. On the left extremityx=0 acts a control functionu(t ), whereas on the right extremityx=1, the string is limited by a lower, time dependent, obstacle so thaty(1, t )ψ (t )for allt >0. The functionψ (t )represents the position of the obstacle at each momentt∈ [0, T]. When the rod touches the obstacle, its reaction can be only upward, so thatyx(1, t )0 on the set{t: y(1, t )=ψ (t )}. When the rod does not touch the obstacle, the right end is free so thatyx(1, t )=0 on the set{t: y(1, t ) > ψ (t )}. These usual conditions, which permit to describe the presence of the obstacle, are calledunilateral Signorini conditions(see for instance [4]).
Various papers have been devoted to the existence and uniqueness of a solution of the boundary obstacle problem for the wave equation. Among them, we mention [5,6] (whose idea is used in our present work) and [10,11].
We investigate in this work the null boundary controllability of the nonlinear system (1.1) stated as follows: for anyT fixed large enough and any(y0, y1)in a given space, does there exist a Dirichlet controlu∈H1(0, T )which drives the corresponding solution of (1.1) to rest, i.e.
y(T )=y(T )=0 in(0,1)? (1.2)
More precisely, we prove the following theorem.
Theorem 1.1.LetT >2andψ∈H1(0, T )with the property that there existsT˜ ∈(2,min{3, T})such thatψ (t )0 for anyt∈ [ ˜T , T]. For any(y0, y1)∈H1(0,1)×L2(0,1)with
ψ (0)y0(1) (1.3)
there exists a control function u∈H1(0, T ) such that (1.1)admits a unique solution y ∈C([0, T], H1(0,1))∩ C1([0, T], L2(0,1))satisfyingy(T )=y(T )=0in(0,1).
To our knowledge, the exact controllability, when a unilateral constraint is involved, has not been studied so far. In the different context of stabilization, we mention the contribution [8] where the authors prove the exponential decay of the energy associated with the solution of a damped wave equation submitted to a boundary obstacle.
We approach this nonlinear controllability problem in a constructive way by using the characteristic method. This allows to compute the solution φ of the linear wave equation submitted to the initial condition (φ0, φ1)and the nonhomogeneous boundary conditions φ(t,0)=u(t ),φ(t,1)=f (t ), for any u, f ∈L2(0, T ). Consequently, the Dirichlet-to-Neumann mapAdefined byA(φ0, φ0, u, f )=φx(·,1)may also be computed explicitly (see Section 2).
Now, the controllability conditionsφ(T )=φ(T )=0 determineuas a function off in the interval(T −2, T ). On the other hand, at the right extremity x=1 of the string, the Signorini conditions are equivalent to the following inequations in(0, T )
⎧⎨
⎩
f−ψ0, t∈(0, T ),
A
φ0, φ1, u, f
0, t∈(0, T ),
(f−ψ )A
φ0, φ1, u, f
=0, t∈(0, T ).
(1.4) Hence, the problem is reduced to finduin[0, T −2]such thatf is a solution of (1.4). This may be regarded as a fixed point argument since we start with a nonhomogeneous problem and return to the nonlinear one by imposing conditions on(u, f ). A controluof (1.1) is found as a solution of these restrictions.
In Section 4, by using general results for differential inequalities, we describe a class of such controlsu∈H1(0, T ), assuming thatT is strictly greater than 2.
In Section 3, we obtain alternatively the controllability result by using a penalty method, classical in contact me- chanics, which consists in relaxing the Signorini inequations by the equationy,x(·,1)=−1[y(·,1)−ψ]−in(0, T ) where >0 denotes the penalized parameter. From a mechanical point of view−1may be interpreted as the stiffness of the obstacle which is not assumed to be perfectly rigid anymore. Therefore, a penetration of the obstacle is allowed atx=1. In the uncontrolled case, i.e. wheny(t,0)=0, the convergence ofy asgoes to zero toward a solution of (1.1) is discussed, for instance, in [5,11]. We remark that the penalty method described above is fundamental both for the theoretical study and the numerical approximation of solutions of any contact problem.
In Section 3, following the previous fixed point argument, we construct explicitly a class of couples(u, f), solutions of
A
φ0, φ1, u, f
=−1[f−ψ]−, t∈(0, T ), (1.5)
which are uniformly bounded with respect to(Sections 3.2 and 3.3). This property allows to pass to the limit and to obtain a control for (1.1) in Section 3.4. Relation (1.5) represents a fixed point type condition for the couple(u, f).
Note that we were able to estimate the dependence of the controluondue to the almost explicit formulas we have for it. With a more general approach, using for instance Schauder’s theorem like in [13] (see also [2]), it would be more difficult to obtain such precise estimates on.
Section 5 presents some numerical applications in agreement with the theoretical part while Section 6 concludes with some related extensions and open problems.
2. The control Dirichlet-to-Neumann map of a linear system LetT >0,QT =(0, T )×(0,1)and consider the following system:
⎧⎪
⎪⎨
⎪⎪
⎩
φ−φxx=0, (t, x)∈QT,
φ(t,0)=u(t ), t∈(0, T ),
φ(t,1)=f (t ), t∈(0, T ),
φ(0, x)=φ0(x), φ(0, x)=φ1(x), x∈(0,1),
(2.6)
whereu∈L2(0, T )is a control function andf ∈L2(0, T )is given. The following result is well known (see, for instance, [7]).
Proposition 2.1.
(1) If ((φ0, φ1), (u, f ))∈(L2(0,1)×H−1(0,1))×L2(0, T )2 there exists a unique solution φof (2.6)such that φ∈C([0, T], L2(0,1))∩C1([0, T], H−1(0,1))and a positive constantCsuch that
φ(t), φ(t )
L2(0,1)×H−1(0,1)Cφ0, φ1
L2(0,1)×H−1(0,1)+(u, f )
L2(0,T )2
.
(2) If((φ0, φ1), (u, f ))∈H1(0,1)×L2(0,1)×H1(0, T )2verifies the compatibility conditions
u(0)=φ0(0), f (0)=φ0(1), (2.7)
there exists a unique solution of (2.6)φ∈C([0, T], H1(0,1))∩C1([0, T], L2(0,1))and a positive constantC such that
φ(t), φ(t )
H1(0,1)×L2(0,1)Cφ0, φ1
H1(0,1)×L2(0,1)+(u, f )
H1(0,T )2
.
We define the space H= φ0, φ1
, (u, f )
∈H1(0,1)×L2(0,1)×H1(0, T )2, u(0)=φ0(0), f (0)=φ0(1) .
Given(φ0, φ1, f ), our aim is to find a family of explicit controlsufor which the solutionφof (2.6) satisfiesφ(T )= φ(T )=0 in(0,1). Setting
p=φ−φx, q=φ+φx, (2.8)
it follows that (2.6) is equivalent to
⎧⎪
⎪⎨
⎪⎪
⎩
p+px=q−qx=0, (t, x)∈QT, (p+q)(·,0)=2u, t∈(0, T ), (p+q)(·,1)=2f, t∈(0, T ), p0=φ1−φx0, q0=φ1+φx0, x∈(0,1).
(2.9)
If ((p0, q0), (u, f ))∈L2(0,1)2×H1(0, T )2system (2.9) admits a unique generalized solution(p, q)∈C([0, T], L2(0,1)2)(see for instance [9, Theorem 3.1, p. 650]). In view of (2.8), this solution corresponds to a solutionφof (2.6) satisfying
φ∈C
[0, T], H1(0,1)
∩C1
[0, T], L2(0,1) associated with the data((φ0, φ1), (u, f ))∈H.
Proposition 2.2. LetT ∈(2,3)and assume that((φ0, φ1), (u, f ))∈H. Then the solution(p, q) of (2.9)satisfies (p, q)(T )=0in(0,1)if and only if((φ0, φ1), (u, f ))satisfies
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎩
u(t )=f(t+1)+1
2q0(t ) ifT −2< t <1, u(t )=f(t+1)+f(t−1)−1
2p0(2−t ) if1< t < T −1, u(t )=f(t−1)−1
2p0(2−t ) ifT −1< t <2, u(t )+u(t−2)=f(t−1)+1
2q0(t−2) if2< t < T .
(2.10)
Proof. Solving system (2.9) using the characteristics method gives the expressions:
p(t, x)=
⎧⎪
⎪⎨
⎪⎪
⎩
p0(x−t ) if 0< t < x <1,
2u(t−x)−q0(t−x) if 0< x < t <1+x,
2u(t−x)−2f(t−x−1)+p0(2−t+x) if 1+x < t <2+x, 2u(t−x)−2f(t−x−1)+2u(t−x−2)−q0(t−x−2) if 2+x < t < T
(2.11)
and
q(t, x)=
⎧⎪
⎪⎨
⎪⎪
⎩
q0(x+t ) if 0< t <1−x,
2f(t+x−1)−p0(2−t−x) if 1−x < t <2−x,
2f(x+t−1)−2u(t+x−2)+q0(t+x−2) if 2−x < t <3−x, 2f(t+x−1)−2u(t+x−2)+2f(t+x−3)−p0(4−t−x) if 3−x < t < T .
(2.12)
It follows that p(T , x)=
2u(T −x)−2f(T−x−1)+p0(x−T +2) if 0< T−2< x <1, 2u(T −x)−2f(T−x−1)+2u(T −x−2)−q0(T−x−2) if 0< x < T −2<1 and
q(T , x)=
−2u(x+T −2)+2f(x+T −1)+q0(x+T−2) if 0< x <3−T ,
−2u(T+x−2)+2f(T+x−1)+2f(T +x−3)−p0(4−T −x) if 0<3−T < x <1.
Consequently,(p, q)(T )=0 in(0,1)if and only ifusatisfies
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎩
u(T−x)=f(T−x−1)−1
2p0(x−T +2) ifT −2< x <1, u(T−x)= −u(T −x−2)+f(T−x−1)+1
2q0(T −x−2) if 0< x < T −2, u(x+T −2)=f(x+T −1)+1
2q0(x+T−2) if 0< x <3−T , u(x+T −2)=f(T+x−1)+f(T+x−3)−1
2p0(4−T−x) if 3−T < x <1, which is equivalent to (2.10). 2
Remark 2.1.We point out that in Proposition 2.2, the values of the control functionsuare not prescribed on(0, T−2).
Consequently, there exists an infinite number of such control functions. The fact thatuis “free” in(0, T −2)plays a crucial role in the sequel.
Let us define the space Hc= φ0, φ1, u, f
∈H(2.10) is verified andu(T )=f (T )=0
. (2.13)
Corollary 2.1.LetT ∈(2,3)and(φ0, φ1, u, f )∈H. Then the solutionφof(2.6)satisfiesφ(T )=φ(T )=0in(0,1) if and only if(φ0, φ1, u, f )∈Hc.
Proof. From Proposition 2.2 and (2.8), it follows that (2.10) is a necessary and sufficient condition for φ(T )= φx(T )=0 on (0,1). Thus, in order to get φ(T )=0 on (0,1), it is necessary and sufficient to choose u(T )= f (T )=0. 2
Definition 2.1.TheDirichlet-to-Neumann mapassociated with the system (2.6) is the applicationA:H→L2(0, T ) defined by
A
φ0, φ1, u, f
=φx(.,1),
whereφis the solution of (2.6) associated with(φ0, φ1, u, f ).
TheControl Dirichlet-to-Neumann mapis the application Ac=A|Hc.
The following lemma gives a characterization of these two maps.
Lemma 2.1.LetT ∈(2,3)and(φ0, φ1, u, f )∈H. Then
A
φ0, φ1, u, f (t )=
⎧⎨
⎩
f(t )−p0(1−t ), 0< t <1, f(t )−2u(t−1)+q0(t−1), 1< t <2, f(t )+2f(t−2)−2u(t−1)−p0(3−t ), 2< t < T .
(2.14)
As a consequence of (2.10), if(φ0, φ1, u, f )∈Hc, then
Ac
φ0, φ1, u, f (t )=
⎧⎨
⎩
f(t )−p0(1−t ), 0< t <1, f(t )−2u(t−1)+q0(t−1), 1< t < T −1,
−f(t ), T −1< t < T ,
(2.15)
wherep0=φ1−φ0xandq0=φ1+φx0.
Proof. Taking into account the expressions ofp=φ−φxandq=φ+φxderived in (2.11) and (2.12), we get:
φx(t,1)=q(t,1)−p(t,1)
2 ,
which leads to (2.14). The expression (2.15) is obtained using (2.10). 2
Remark 2.2.Note that the expression ofAc(φ0, φ1, u, f )in (2.15) involves only the part ofudefined on(0, T −2), i.e. the “free” part ofu.
Remark 2.3.Clearly,Ac∈L(Hc;L2(0, T )). If moreover u, f, φ0, φ1
∈H2(0, T )×H2(0, T )×H2(0,1)×H1(0,1) with the compatibility conditions
u(0)= −φx0(0), f(T −1)=u(T −2)+1 2
φx0(T−2)+φ1(T−2) ,
thenAc(φ0, φ1, u, f )∈H1(0, T ). From (2.15) other regularity results can be easily derived.
3. A penalty method
We are going to study (1.1) by introducing the penalized problem:
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
y−y,xx=0, (t, x)∈QT,
y(t,0)=u(t ), t∈(0, T ),
y,x(t,1)=−1
y(t,1)−ψ (t )−
, t∈(0, T ), y(0, x)=y0(x), y(0, x)=y1(x), x∈(0,1),
(3.16)
where[·]−denotes the negative part so that[y(t,1)−ψ (t )]−= −min{0, y(t,1)−ψ (t )}and >0.
In the sequel we denote
p0=y1−yx0, q0=y1+yx0. (3.17)
Let us first give a definition of the weak solutions of (3.16).
Definition 3.1.For any(y0, y1)∈H1(0,1)×L2(0,1)andu∈H1(0, T )withu(0)=y0(0), aweak solution of (3.16)is a function
y∈C
[0, T], H1(0,1)
∩C1
[0, T], L2(0,1) with the property that there existsf∈H1(0, T )such that (1) f(0)=y0(1).
(2) yverifies (2.6) with nonhomogeneous terms(u, f)and initial data(y0, y1).
(3) A(y0, y1, u, f)=−1[f−ψ]−.
Note that relation (3) from Definition 3.1 represents a fixed point type condition ensuring that y,x(t,1)=−1
f(t )−ψ (t )−
=−1
y(t,1)−ψ (t )− ,
i.e. exactly the boundary condition onx=1 from (3.16). We have the following result of existence and uniqueness of solutions.
Proposition 3.1.LetT ∈(2,3). For any(y0, y1)∈H1(0,1)×L2(0,1)andu∈H1(0, T )withu(0)=y0(0), there exists a unique weak solution of (3.16).
Proof. From Proposition 2.1, for any (y0, y1)∈H1(0,1)×L2(0,1),u∈H1(0, T )withu(0)=y0(0)andf∈ H1(0, T )withf(0)=y0(1)there exists a unique solutiony∈C([0, T], H1(0,1))∩C1([0, T], L2(0,1))of (2.6).
Hence, the problem is reduced to show the existence and uniqueness off∈H1(0, T )withf(0)=y0(1)and such thatA(y0, y1, u, f)=−1[f−ψ]−. By taking into account (2.14) this is equivalent to prove that the following differential equation has a unique solution
⎧⎪
⎪⎨
⎪⎪
⎩ f(t )=
⎧⎨
⎩
−1[f(t )−ψ (t )]−+p0(1−t ), 0< t <1, −1[f(t )−ψ (t )]−+2u(t−1)−q0(t−1), 1< t <2, −1[f(t )−ψ (t )]−−2f(t−2)+2u(t−1)+p0(3−t ), 2< t < T , f(0)=y0(1).
(3.18)
In the interval(0,2)the right-hand side of Eq. (3.18) is Lipschitz with respect tofand the existence and unique- ness of solutions of (3.18) are a consequence of the classical results for the ordinary differential equations. In the interval(2, T )the functionf(t−2)is completely known, it belongs toL2(2, T )and may be seen as a nonhomoge- neous term. The existence and uniqueness of solution in this interval follow as before. 2
3.1. Controllability of the penalized problem
In this section we pass to study the controllability properties of (3.16).
Definition 3.2.Problem (3.16) isnull controllable in timeT if, for any(y0, y1)∈H1(0,1)×L2(0,1), there exists a controlu∈H1(0, T )withu(0)=y0(0)such that the corresponding weak solution of (3.16) verifiesy(T )= y(T )=0.
The following characterization of the controllability property of (3.16) is a direct consequence of Corollary 2.1.
Proposition 3.2.LetT ∈(2,3)andψ∈H1(0, T )withψ (T )0. Problem(3.16)is null controllable in timeT if and only if, for any(y0, y1)∈H1(0,1)×L2(0,1)withy0(1)ψ (0), there exist a controlu∈H1(0, T )and a function f∈H1(0, T )such that
(1) (y0, y1, u, f)∈Hc.
(2) yis the solution of (2.6)with nonhomogeneous terms(u, f)and initial data(y0, y1).
(3) Ac(y0, y1, u, f)=−1[f−ψ]−.
Proof. Indeed, if conditions (1)–(3) are fulfilled there exists a unique weak solution y of (3.16) with initial data (y0, y1)and controlu. Sinceyverifies (2.6), it follows from Corollary 2.1 that(y0, y1, u, f)∈Hcimpliesy(T )= y(T )=0 in(0,1).
Reciprocally, if (3.16) is null controllable, from the existence of a weak solution of (3.16) we deduce the existence off∈H1(0, T )such that(y0, y1, u, f)∈Hand conditions (2)–(3) are verified. Sincey(T )=y(T )=0 in(0,1) andyverifies (2.6), from Corollary 2.1, we deduce that(y0, y1, u, f)∈Hc. 2
Using (2.15), the controllability of (3.16) becomes equivalent to the following nonlinear control problem: given T ∈(2,3),
⎧⎪
⎨
⎪⎩
findf∈H1(0, T )withf(T )=0andu∈H1(0, T−2)withu(0)=y0(0)such that f(t )=G
t, f(t ), u(t )
, t∈(0, T ), f(0)=y0(1),
(3.19)
where
G
t, f(t ), u(t )
=
⎧⎪
⎨
⎪⎩
1
[f(t )−ψ (t )]−+p0(1−t ), t∈(0,1),
1
[f(t )−ψ (t )]−+2u(t−1)−q0(t−1), t∈(1, T −1),
−1[f(t )−ψ (t )]−, t∈(T −1, T ).
(3.20)
The study of the null controllability property of (3.16) will be carried out in two steps:
(1) In a first step, we will prove that, for any (y0, y1)∈H1(0,1)×L2(0,1)andψ∈H1(0, T ), satisfying (1.3), there exists a family of controlsu∈H1(0, T )such that the solution of (3.16) satisfiesy(T )=y(T )=0 on (0,1). At this level, we again make use of the Dirichlet-to-Neumann map and the characterization (3.19) of the controllability property.
(2) In a second step, we provide some estimates onu andy which will allow to pass to the limit inin order to obtain a solution of (1.1) satisfyingy(T )=y(T )=0 on(0,1).
3.2. Existence of solutions of the controlled penalized problem
Lemma 3.1.For any >0, problem(3.19)admits an infinite number of solutions(f, u).
Proof. On[0,1], letl∈H1(0,1)be the unique solution of
f(t )=−1
f(t )−ψ (t )−
+p0(1−t ), t∈(0,1),
f(0)=y0(1) (3.21)
and setf1,=l(1). Similarly, on(T−1, T ), letrbe the unique solution of the backward problem f(t )= −−1
f(t )−ψ (t )−
, t∈(T −1, T ),
f(T )=0 (3.22)
and setfT−1,=r(T −1).
We then consider the following nonlinear control problem:
⎧⎪
⎨
⎪⎩
findu∈H1(0, T −2)such thatu(0)=y0(0)andf∈H1(1, T−1)such that f(t )=−1
f(t )−ψ (t )−
+2u(t−1)−q0(t−1), t∈(1, T −1), f(1)=f1,, f(T −1)=fT−1,
(3.23)
which consists to find a controlusteering the solution of the differential equation f(t )=1
f(t )−ψ (t )−
+2u(t−1)−q0(t−1), t∈(1, T −1),
from the initial dataf1,to the final datafT−1,. We proceed as follows: we first consider the linear control problem
⎧⎨
⎩
findv∈H1(0, T −2)andθ∈H1(1, T −1)such that θ(t )=2v(t−1), t∈(1, T −1),
θ(1)=f1,, θ(T −1)=fT−1,.
(3.24)
Clearly, any(v, θ)∈H1(0, T −2)×H1(0, T −1)satisfying
⎧⎪
⎪⎨
⎪⎪
⎩
v(T−2)=1
2(fT−1,−f1,), v(0)=0,
θ(t )=2v(t−1)+f1,
(3.25)
is a solution of (3.24). Now let us choose
⎧⎪
⎪⎨
⎪⎪
⎩
θ(t )=2v(t−1)+f1,, 2u(t−1)=2v(t−1)−1
2v(t−1)+f1,−ψ (t )−
+q0(t−1), t∈(1, T −1), u(0)=y0(0),
(3.26)
wherevis any function fromH1(0, T−2)such thatv(T−2)=12(fT−1,−f1,)andv(0)=0.
It is straightforward that the couple(θ, u)defined by formulas (3.26) satisfies (3.19). Thus, a family of solutions (f, u)to problem (3.20) is constructed if we take
f=
⎧⎨
⎩
l, [0,1], θ, [1, T −1], r, [T −1, T],
(3.27) wherel,r and(θ, u)are given by (3.21), (3.22) and (3.26) respectively. 2
3.3. Boundedness of solutions of the controlled penalized problem
Now, we prove that the sequence(u, f)may be chosen uniformly bounded with respect toinH1(0, T ). From now on,Cdenotes a strictly positive constant that may vary from line to line but is independent on.
Lemma 3.2.There exists a constantC=C(y0, y1, ψ ) >0such that for any >0the functionf given by(3.27) satisfies the estimate:
f(t )2
C, t∈ [0,1] ∪ [T −1, T]. (3.28) If moreovery0(1)−ψ (0)0, then:
1
f(t )−ψ (t )−2
C, t∈ [0,1] ∪ [T −1, T], (3.29) t
0
f2
(s) dsC, t∈ [0,1], (3.30)
T t
f2
(s) dsC, t∈ [T −1, T]. (3.31)
Proof. We seth=f−ψso that problem (3.21) can be written as
⎧⎨
⎩
h(t )=1
h−(t )+p0(1−t )−ψ(t ), t∈(0,1), h(0)=y0(1)−ψ (0).
(3.32) Multiplying this equation byhand integrating over(0, t )fort <1, we get
h2(t )=h2(0)−2 t 0
h−(s)2
ds+2 t 0
p0(1−s)−ψ(s) h(s) ds
h2(0)+p02
L2(0,1)+ψ2
L2(0,1)
+ t
0
h2(s) ds.
From Gronwall’s lemma, we deduce that h2(t )C
y0(1)−ψ (0)2
+p02
L2(0,1)+ψ2
L2(0,1)
. Thus
f2(t )C
y0(1)−ψ (0)2
+p02
L2(0,1)+ ψ2H1(0,1)
, t∈(0,1). (3.33)
Similarly, problem (3.22) can be written as
h(t )= −1
h−(t )−ψ(t ), t∈(T−1, T ), h(T )= −ψ (T ).
Multiplying as previously this equation byhand integrating over(t, T )fort∈(T −1, T ), the same arguments lead to the estimate
h2(t )C
ψ2(T )+ψ2
L2(T−1,T )
. This implies
f2(t )C
ψ2(T )+ ψ2H1(T−1,T )
, t∈(T −1, T ). (3.34)
Estimates (3.33) and (3.34) prove the first part of the lemma.
We now multiply the equation ofhin (3.32) byhand integrating over(0, t )witht∈(0,1), we get t
0
h(s)2
ds+1
h−(t )2
= 1 2
h−(0)2
+ t 0
p0(1−s)−ψ(s)
h(s) ds.
If we assume thath(0)=y0(1)−ψ (0)0, thenh−(0)=0 and Cauchy–Schwarz inequality imply t
0
h(s)2
ds+1
h−(t )2
t 0
p0(1−s)−ψ(s)2
ds, t∈(0,1).
From this last inequality, it follows that 1
2 t
0
f(s)2
ds+1
f(t )−ψ (t )−2
t
0
ψ(s)2
ds+ t 0
p0(1−s)−ψ(s)2
ds, t∈(0,1).
With the same argument, we get on(T−1, T ):
T t
f(s)2
ds+1
f(t )−ψ (t )−2
Cψ2H1(T−1,T ), t∈(T−1, T ).
This ends the proof. 2
Remark 3.1.From (3.30) and (3.21) (resp. (3.31) and (3.22)) it can be deduced that 1
2 t 0
f(s)−ψ (s)−2
dsC, t∈ [0,1] (resp.
1 2
T t
f(s)−ψ (s)−2
dsC, t∈ [T −1, T]).
Note that, so far the only conditions imposed tofin[1, T −1]aref∈H1(1, T −1),f(1)=f1, andf(T− 1)=fT−1,. The next step is to prove thatf may be chosen such that the estimates (3.28)–(3.31) hold true on (1, T−1)too.
Lemma 3.3.There exists a sequence(f)>0⊂H1(1, T −1)such thatf(1)=f1,,f(T −1)=fT−1,and fH1(1,T−1)C, 1
2
T−1 1
f(t )−ψ (t )−2
dtC, uH1(0,T−2)C, whereuis the solution of (3.26).
Proof. Fix >0 sufficiently small. The idea behind the following construction offis to takef(t )ψ (t )in order to have[f(t )−ψ (t )]−=0 for allt∈(1, T−1). However, this is not possible iff1,< ψ (1)orfT−1,< ψ (T−1).
Taking into account (3.29), we can still keep bounded the integral12
T−1
1 ([f(s)−ψ (s)]−)2dsiff(t ) < ψ (t )in a neighborhood of lengthof 1 andT −1. Thus, iff1,< ψ (1)we have to construct a functionf in[1,1+]such that
f(1)=f1,, f(1+)=ψ (1+), 1 2
1+
1
f(s)−ψ (s)−2
dsC.
Analogously, iffT−1,< ψ (T −1)we have to find a functionf in[T −1−, T −1]such that
f(T −1)=fT−1,, f(T −1−)=ψ (T−1−), 1 2
T−1
T−1−
f(s)−ψ (s)−2
dsC.
In both cases,f−ψwill be an interpolation polynomial of degree one int. More precisely:
• Ifψ (1)−f1,<0 andψ (T−1)−fT−1,<0, we choose any functionf∈H1(1, T−1)satisfyingf(1)=f1, andf(T −1)=fT−1,. For instance:
f(t )=max
fT−1,−f1,
T −2 (t−1)+f1,, ψ (t )
, t∈(1, T−1).
• Ifψ (1)−f1,0 andψ (T−1)−fT−1,0,
f(t )=
⎧⎨
⎩
ψ (t )+(ψ (1)−f1,)(1(t−1)−1), t∈ [1,1+],
ψ (t ), t∈ [1+, T−1−],
ψ (t )+(ψ (T −1)−fT−1,)(−1+T−1−t), t∈ [T−1−, T −1].
(3.35)
• Ifψ (1)−f1,0 andψ (T−1)−fT−1,<0,
f(t )=
⎧⎪
⎪⎨
⎪⎪
⎩
ψ (t )+(ψ (1)−f1,)(1(t−1)−1), t∈ [1,1+],
ψ (t ), t∈ [1+,T2],
max{fT−1,T−ψ(T2)
2−1 (t−T2)+ψ (T2), ψ (t )}, t∈ [T2, T −1].
• Ifψ (1)−f1,<0 andψ (T−1)−fT−1,0,
f(t )=
⎧⎪
⎪⎨
⎪⎪
⎩
max{ψ(T2T)−f1,
2−1 (t−1)+f1,, ψ (t )}, t∈ [1,T2],
ψ (t ), t∈ [T2, T−1−],
ψ (t )+(ψ (T −1)−fT−1,)(−1+T−1−t), t∈ [T−1−, T −1].
By constructionf∈H1(1, T −1)and satisfiesf(1)=f1,,f(T−1)=fT−1,in all cases. Moreover, note that from (3.26) one has
2u(t−1)=f(t )−1
f(t )−ψ (t )−
+q0(t−1), t∈(1, T −1), u(0)=y0(0).
Consequently, the uniform boundedness ofu follows from the uniform estimates with respect to for f and
1
[f−ψ]−.
For the first case, note thatfψon(1, T −1)and that from (3.28),|fT−1,|and|f1,|are uniformly bounded with respect to . It is straightforward that this implies uniform bounds with respect to for fH1(1,T−1) and T−1
1 ([f(t )−ψ (t )]−)2dt=0.
