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A NNALES DE LA FACULTÉ DES SCIENCES DE T OULOUSE

M ARCELO M OREIRA C AVALCANTI Exact controllability of the wave equation with Neumann boundary condition and time- dependent coefficients

Annales de la faculté des sciences de Toulouse 6

e

série, tome 8, n

o

1 (1999), p. 53-89

<http://www.numdam.org/item?id=AFST_1999_6_8_1_53_0>

© Université Paul Sabatier, 1999, tous droits réservés.

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(2)

- 53-

Exact Controllability of the Wave Equation

with Neumann Boundary Condition and

Time-Dependent Coefficients(*)

MARCELO MOREIRA

CAVALCANTI(1)

Annales de la Faculte des Sciences de Toulouse Vol. VIII, 1, 1999

On considere la contrôlabilité exacte frontiere de l’équation

lorsque le controle est de type Neumann et 03A9 est un ouvert borne connexe

de I~~‘. On utilise la methode HUM

(Hilbert

Uniqueness

Method)

de J.-L.

Lions.

MOTS-CLÉS : controlabilite exacte, Neumann, coefficients dependant du temps.

ABSTRACT. - In this paper we study the exact boundary controllabil- ity for the equation

when the control action is of Neumann type and 03A9 is a bounded domain in 1L8’~. The result is obtained by applying HUM

(Hilbert

Uniqueness

Method)

due to J.-L. Lions

KEY-WORDS : Exact controllability, Neumann, time-dependent coefh-

cients.

AMS Classification : 93B05-35B37

(*) Recu le 20 février 1997, accepte le 10 novembre 1998

(~) Departamento de Matematica, Universidade Estadual de Maringa, Br-87020-900

Maringa-PR (Brazil)

Partially supported by CNPq/Brazil

(3)

1. Introduction

Let 52 be a bounded domain in with

C2 bondary r,

and let

Q

be

the finite

cylinder

52

x ] 0 ,

T

[

with lateral

boundary

E = r

x ] 0 ,

T

[.

We

consider the

following system

with

inhomogenous boundary

conditions

where

The

problem

of the exact

controllability for

the

system (1.1)

is formulated

as follows.

PROBLEM 1.1.- Given T > 0

large enough,

it is

possible, for

each

pair of

initial data

defined

in a suitable space, to

find

a control v such

that solution y =

y(x, t) of (1.1) satisfies

the condition

Let us note that when

a(t)

=

~3(t)

=

a(x)

=

1,

Problem 1.1 was studied

by

J .-L. Lions

[12] by using

HUM and also

by

I. Lasiecka and R.

Triggiani [10] by using

the ontoness

approach. Many

other authors studied the exact

controllability

of distributed

systems

with

time-dependent

or

x-dependent

coefhcients.

Among them,

we can cite J.

Lagnese [8]

who

firstly

worked

in

boundary controllability

of distributed systems with time

dependent

coefhcients and C.

Bardos,

G. Lebeau and J. Rauch

[2]

whose work treats to

the

geometric optics approach

in the case of space

dependent

coefhcients. In this

direction,

we can also

cite,

V. Komornik

[11]

who presents an

elementary

and constructive method to obtain the

optimal

estimates needed in HUM

(Hilbert Uniqueness Method)

for the exact

controllability

of some linear

evolution systems, R. Fuentes

[7],

L. A. Medeiros

[14],

M. Milla Miranda

[15],

M. Milla Miranda and L. A. Medeiros

[16],

J. A. Soriano

[17],

among others.

(4)

In this work we prove that

system (1.1)

is

exactly

controllable

by making

use of

HUM, c.f.

J.-L. Lions

[12].

For this

end,

we

employ

the

multiplier technique

to obtain the inverse

inequality.

When the coefhcients

depend

on

time,

after

making appropriated hypothesis

on

them,

the inverse

inequality

still remains

true;

but since standard arguments are not

applicable,

the

regularity

of backward

problem requires

a new

proof

which is the main task of this work

(Theorem 5.1 ) .

In

fact,

the

goal

of this work is to show that HUM can be

applied

to

the case of

time-dependent coefficients

with Neumann

boundary

condition.

In order to

simplify

the

computations,

we consider the

simple

operators defined

by (1.2). Howover,

with

appropriated changes,

we can extend our

results to those ones

given by

with

a(x, t)

>

fp

> 0 in

S2

x

(0, oo). Now,

when we consider the matricial operators

the usual

arguments

cannot be

applied

even if i

= j

and

t)

=

ai ( x )

(Remark 1,

Sect.

4).

Our paper is divided on sixth

chapters.

In Section

2,

we

give

notations

and state the

principal

result. In Section

3,

we consider the

homogeneous problem

and in Section 4 we establish the inverse

inequality.

In Section

5,

we

study

the backward

problem

and in the last section

(Sect. 6)

we

apply

HUM.

2. Notations and Main Result

Let

xO

E the unit exterior normal vector at x E

r, m(x) =

x E and

)

(5)

In what follows the

symbol

" ~ " denotes the inner

product

in II8’~. Let us

define

Let us introduce some notations that will be used

throughout

this work.

We are

going

to denote

( ~ , ~ ) and [ . the inner-product

and the norm of

L2(S2) respectively.

The norm in will be denoted

by [[

.

II.

Let A be the

operator

defined

by

the

triple L2(S2), a(u, v)}

where

We recall that the

Spectral

Theorem for

self-adjoint operators guatantees

the existence of a

complete

orthonormal

system (wv )

of

given by

the

eigenfunctions

of A. If are the

corresponding eigenvalues

of

A,

then

Av -

+00 as v -~ +00.

Besides,

Considering

in

D(A)

the norm

given by

the

graph,

that

is,

it turns that is a

complete

system in

D(A).

. In

fact,

it is known that is also a

complete

system in

Now,

since A is

positive, given

6 > 0 one has

(6)

In we consider the natural

topology given by

the norm

We observe that such

operators

are also

self-adjoint,

that

is,

D(Al/2)

= and

D(AO)

=

L2(S2).

We note that

A(t)

=

Here,

we are

using

the same

symbol

for both

operators

to

simplify

the notation.

We make the

following hypotheses :

(H2)

if n >

1,

(H3)

if n =

1,

Now we are in

position

to state our main result. Consider the

system,

We have the

following

result.

(7)

THEOREM 2.1.-

Suppose

that

assumptions

are

satisfied.

Then there exists a time

To

> 0 such that

for

T >

To

and initial data

{yo, yl}

E

L2(S2)

x

~H1(S2)~~,

there exists a control

such that the ultra-weak solution

(the

solution

of (2.IJ

is

defined by

the

transposition method,

see

(1 ‘~~~

y =

y(x, t) of (2.1~ satisfies

3. The

Homogeneous

Problem

In this section we

present

a standard result and a new one about the solutions of the

following homogeneous system

We have the

following

results.

THEOREM 3.1.-

Suppose

that

assumption (Hl)

holds.

Then, given

k E

{0,1, 2}

and

problem (~‘l.1)

possesses a

unique

solution B :

Q

--~ R such

that,

Moreover,

the linear map

is continuous.

(8)

Theorem 3.1 can be

proved

in a standard way

by applying

the Faedo-

Galerkin method and

using

the

spectral

considerations

given

in Section 2.

Next we consider the

homogeneous problem

which will be used in the

study

of the

regularity

of the solution of

(2.1).

.

THEOREM 3.2. - Given

f

E

D ~0, T D(A)),

the

unique

solution

of problem (~‘3.2J satisfies for

every t E

~ 0 , T

where C =

C(T).

Proof . -

Since

B~

=

Bl

= 0 and

f ~ E D ~0, T ; D(A)~ from

Theorem 3.1

the above

problem

has a

unique

solution B such that

Besides,

this solution satisfies the

identity

From

(3.3)

we

get

AB E

C°([0, T]; D(Al/2))

and therefore

(9)

This

togheter

with

assumption (HI) implies

that

Integrating

this

equality

and

noting

that

f (0)

= 0 we have

Replacing by f’ -

in the last

integral

we obtain

Now

integrating by

part and

noting

that

f (0)

=

0,

(10)

Replacing (3.6)

in

(3.5)

we have

From

(3.4)

and

(3.7)

it follows that

Denning o-~~ - &#x26;-~/~/

= ~. and

replacing by a-~/2~

+ in the

above

expression

we obtain

(11)

From

hypotheses (Hl), (H2)

and

(3.8)

there exists a constant C > 0

independent

of

f

and 8 such that

Applying

Gronwall’s

inequality

twice

(first

we consider the Gronwall in-

equality (1/2)g2(t) fo m(s)g(s)

ds where

g(t)

=

I

and

m(t)

=

2C(g(t) + IAf(t)I)

and after the usual

one),

we

get

In a similar way we also infer that

Using

the definition on ~p we obtain the desired

inequalities.

0

4. The Inverse

Inequality

In this section we construct a

special To

time

depending

on n, on

the functions

a(t), ,Q(t), a(t)

and also on the

geometry

of Q.

Taking

into account the

regularity of 0393, we

can define on F a unit exterior normal vector field

v(x)

on class

C1.

In the same way we can define a

family

of

(n - 1) tangent

vector field

{ T1 (x),

... ,

(x) ~

of class

C1

such that

the

family { v(x), T1 (x),

...

(x) ~

defines an orthonormal basis for each x E I‘ .

regular function,

we have

where

(12)

Defining

we obtain from

(4.1)

and

(4.2)

We observe that when

~03C6/~03BDA

= 0 on r then

~03C6/~03BD

= 0 since

Then, defining

y~ = ~p, ... we obtain from

(4.3)

and

consequently

Remark 1.2014 At this

point

we observe that when A is a matricial operator that

is,

when it is

given by

then we have

and therefore if = 0 we do not have

necessarely

that = 0 and

consequently

we cannot use the

identity

even if i

= j

and

t)

= As this

identity plays

an essential role to prove the inverse

inequality,

this case

requires

another treatment which

will not be considered in this work.

(13)

If 03C6

E

H2(O)

we can define in a natural way a continuous linear operator

such that

In

addition,

we can also define a continuous linear operator

where

Fo

is a

nonempty

open subset of r

(sometimes

the whole

r)

such that

Thus,

from

(4.7), (4.9)

and

by density arguments

it results that

Considering

the above

equality

we are able to define the

tangential gradient

Dropping

the index "2" in

(4.8)

to

simplify

the

notation,

we define the

adjoint operator

and from

(4.8)

and

(4.11)

we obtain the continuous linear operator

where "o" denotes

composition.

(14)

Hence,

for all

~, ~

E

H1 (T‘o),

In

particular

THEOREM 4.1.- Let B be the weak solution

(it

means that the initial data E

H1(52)

x

L2(S2)) of

the Problem

(~5.1). Then, if f

=

0,

where

and

Proof.-

We consider first that E

D(A)

x

Then,

from

Theorem

3.1,

there exists a

unique

solution e in the class

Multiplying (3.1), by 8’(t)

we obtain

Integrating

this

expression

from 0 to t and then

integrating by

parts we

get

(15)

Taking

into account

(4.14)

we can rewrite the above

expression

as follows

On the other

hand, differentiating E(t)

we have

So

where

The above

inequality gives,

Now, considering

it follows from

(4.15)

that

Finally, considering

we obtain the desired result

by using density arguments.

D

(16)

THEOREM 4.2.- Let q be a vector

field

such that q E Then each weak solution

~ of

Problem

(~.IJ satifies

Proof.

- First we prove the

identity

for the

strong (it

means that the

initial data E

D(A)

x

Hl (S2))

solutions of

(3.1)

and then the result follows

by

a

density

arguments.

So,

let us suppose that

By multiplying

the first

equation

of

(3.1) by

and

integrating

over

Q,

Integrating by

parts the left side of

equality (4.16)

we

get

On the other

hand,

since

(17)

we have from

(4.17)

that

We also have

Thus, combining (4.19)

and

(4.18)

we obtain

Now, evaluating

the

right

side of

(4.16)

we have from Green

identity

Combining (4.16), (4.20), (4.21)

and

(4.5)

we obtain the desired

identity.

0

(18)

The

To

time which Theorem 2.1 is defined

by

and

uniquely depends

on n,

a(t), ,~(t), a(t)

and on the

geometry

of S2.

THEOREM 4.3.-

Suppose

that

hypotheses (HI), (H2)

and

(H3)

hold and

that T >

Tp.

. Then

for

each weak solution

~ of (,~.1)

with

f

= 0 there exists

C > 0 such that:

(i~ if

n > 1 then

(ii) if

n = 1 then

Proof.

-

By using

the

identity given

in Theorem 4.2 with

q(x)

=

m(x)

=

x - we

get

after some calculations

(19)

On the other

hand,

Multiplying

the first

equation

of

(3.1) by §

and

integrating

on

Q

we have

Replacing (4.24)

in

(4.23)

it follows that

Now, substituting (4.25)

in

(4.22)

we obtain

(20)

Since =

max{ Ilm(x)11 ;

x E

52},

from

hypothesis (HI)

we have

From (4.26)

and

(4.27)

we

get

and from

hypothesis (H2)

and Theorem 4.1 we obtain

Next,

we estimate the

expression

From

hypothesis (HI)

and Theorem

4.1,

we

get,

and from

(4.29)

we obtain

(21)

From the above

inequality

we

get

which

together (4.28) implies

that

This

gives (i).

To prove

(ii),

we consider the

identity

(22)

Then,

it follows from

(4.22)

and

(4.31)

that

From

(H3)

we have that 0 ^y 1 and

therefore,

Then, by making

use of the same

arguments

of

(4.27)

and

(4.28),

from

(4.32)

we obtain

Defining

from

hypothesis (H2)

and

using

similar

arguments

to the case n >

1,

we obtain

(ii). D

(23)

THEOREM 4.4

(Inverse Inequality).

-

Suppose

that

(HI)-(H~)

hold and

let T >

Tp.

. Then

for

each

strong

solution

~ of (~.IJ

with

f

= 0 there exists

C > 0 such that:

(i) if

n > 1

Proof.

- We are

going

to prove the case

(i)

since

(it)

is

analougous.

Dropping

the terms which

give negative

contributions in Theorem 4.3 one

has

On the other

hand,

there exists a constant

C2

> 0 such that

Indeed, since 4;

is a

regular

solution of

(3.1),

then

and therefore

(24)

Defining

we have

and from

(4.35)

it follows that

h, h’

E

L2(0, T)

and hence h E

C°([0, T]).

Let

to E ~ [0 , T] be

a minimizer of h. Thus

and

consequently

But,

since to is a

minimizer,

we have

and then

Thus,

from

(4.36)

and

(4.37)

we obtain

V t E

[0, T],

which proves

(4.34). Combining (4.33)

and

(4.34)

we

get

the

desired result. 0

5. The Backward Problem

Let T >

To

where

To

is defined in the

previous section,

and consider the

following homogeneous problem:

(25)

According

to the inverse

inequality given

in Theorem

4.4,

the

expression

defines a norm in

D(A)

x

H1(S2).

So,

we are able to define the Hilbert space

equipped

with the

topology

where

~{~~,~~}~~,E~

is any

Cauchy

sequence in

~D(A)

x

defined

by

the

equivalence

relation

For every

{~~, ~~}

E

D(A)

x we have

Now,

since

D(A)

x

H1(S2)

in dense in

F,

we have

where the inclusion are continuous and

dense.

We can observe that

by

the construction of

F,

which means that if

{~°, ~1}

E F then

(26)

The

proof

of the above

regularities

are

given

in the

appendix.

We then consider the backward

problem

where §

is the

unique

solution of Problem

(5.1)

with initial data

~~~, ~1}

E

F.

We observe that the

operator 8/8t

is well defined on

E(x~) taking

into

account

(5.6)

and

considering

its

meaning:

V w E

H1 (0, T; L2 ~r(x~)~)

It is

important

to note this

operator

is not taken in the distributional sense.

On the other hand from

(5.7)

we obtain

Let w

be the solution of

(5.8)

defined

by

the

transposition method,

which

will be

precised

later. Let

~~~, ~1 ~

E

F, f

E

L1 (0, T ; H1 {S~))

and let

8 :

: f~ -~

R be the

unique

solution of

(27)

Multiplying (5.11) by 03C8

and

integrating by parts,

we obtain

formally

Replacing by

its value

given

in

(5.8)

we

get

from

(4.12)

and

(5.9)

Observing

this

expression

we define the functionnal

Thus,

from

(5.12)

and

(5.13)

we obtain

formally

that

Considering

Theorem 3.1 and the construction of

F,

we have that the functional

given by (5.13)

is

continuous,

that

is,

Indeed,

firt of all we note that the solution 0 of

(5.11 )

verifies 8

= ~ +~

,

where

81

and

82

are,

respectively,

the solutions of the

following problems:

(28)

Besides,

from

(5.13)

we can write for all

~~~, ~1}

E

D(A)

x and

i=1,2

and therefore from

(5.2)

and

(5.16)

we obtain

From

(5.17)

and Theorem 3.1 we have

By density arguments

we conclude that

inequality (5.18)

is valid for all

{~o ~1} {eo B1}

E F which proves

(5.15).

It follows that there exists a

unique triple pl, ~}

such that

and verifies

DEFINITION.- The

unique function ~

that

satisfies (5.19)

in called

solution

by transposition of

Problem

(5.8).

Now we state the main result of this

section,

which is a consequence of Theorem 3.2.

(29)

THEOREM 5.1.- The

unique

solution

by transposition ~ of

the Problem

(5.8)

has the

following regularity:

In a

addition,

the linear map

is continuous.

Proof

. - For every

f

E

D ~0, T D(A))

we have

where 8

and §

are,

respectively,

solutions of

By

the definition of F and from Theorem 3.1 it follows that

Indeed,

it is sufficient to prove

(5.22)

when the initial data

~~~, ~1}

E

D(A)

x because

by density arguments

we conclude the same when

f~°~m~}

E F.

(30)

We have

by

Schwarz

inequality

and Theorem 3.1

which concludes

(5.22).

On the other

hand,

from Theorem 3.2 we

get

which is the crucial

point

for control

problems involving time-dependent

coefficients.

In

fact,

before we prove

(5.23)

we observe that in the

right

side of

equation (5.20)

we have

/~ while

in the

right

side of

(5.23)

we have

f.

.

Besides,

we note

that when the coefficients do not

depend

on time

(see

for

example

the most

simple

case for the wave

equation),

it is not difficult to obtain the above

inequality by using

Theorem 3.1 and the

following

standard

argument.

If w is a solution to

problem

with

f

E

P (0, T; ; D(A)~ then

B = w~ is the solution of

But in our case, where we have

time-dependent coefficients,

this

arguments

fails

completely

and we need to solve it in other way. From Theorem 3.2 and

observing

that

8(t)

=

fo 8’(s)

ds after some calculations we obtain

(31)

which

implies

In addition

and therefore

From

(5.24)

and

(5.25)

we

get (5.23). Combining (5.22)

and

(5.23)

we

obtain

which is sufficient to prove the desired

regularity,

that

is,

In

fact,

let us define

Since

P(0, T ; ; D(A)~

is dense in

L1 (0, T; ; D(A)~

we can consider the

unique

linear continuous extension

S

of

S, given by

and, consequently,

if follows that

Now, given f with p

E

D(A)

and 8 E

D(0, T), according

to

(5.13),

(5.19), (5.27)

and

considering

the fact

that 80 = 81

= 0 we

obtain,

(32)

So, by (5.28)

it follows that

which

implies

that

Therefore S

= ’1/;/

in D~

(0, T

;

~D(A)~ ~)

and

(5.26)

is then

proved.

One observes that if in

(5.19)

we consider

f

= + B = cp~

with p

E

D(A3~2), ~

E

D(0, T)

and

~~ _ ~1 - 0,

we have

Since

a(t) >

ao >

0,

if follows that

Then,

from

(5.19), (5.26)

and

(5.29)

we obtain

(see

for

example,

J.-L. Lions and E.

Magenes [13,

Vol.

1,

Lemma

8.1])

which

makes

~(0)

and

~~(0) meaningful.

Using

the

regularity

of

~, considering f

= p

+

and B = cpr~

where p

E

D(A3~2), ~

E

CZ ~0, T) (first

we

get,

for

instance,

=

~T-t~ 2t

and

secondly

we can consider =

(T - t~2),

we obtain from

(5.19)

with

~o - ~i - 0,

Finally, considering f =

0 in

(5.19)

we conclude that

which ends the

proof.

D

(33)

6. HUM and Exact

Controllability

Let us define the linear

operator

A : F --~

F’ given by

which is continuous in view of Theorem 5.1.

Considering f

=

0, and 81

= in

(5.13)

and

(5.14),

we have

that

is,

This

implies immediately

that A is

injective

and

self-adjoint.

Then A

is an

isomorphism

from F to

F’. Therefore, given

E F~ we

obtain

{a(0)yl

E

F’

and

consequently

there exists a

unique

{~o , ~1}

E F such that

From

(6.1)

and

(6.2)

we have

Now we are

going

to finish the

proof

of Theorem 2.1. Since

L2(0)

x

~H1 (SZ)) ~, taking

into account the chain

we obtain E F’ and therefore in this case we deduce

(6.3).

Defining

in

(2.1)

the controls

(34)

from

(6.3),

the

uniqueness

of solutions to Problems

(2.1)

and

we

finally

conclude that

Thus Theorem 2.1 is

proved. Q

7.

Appendix

Since

D(A)

x

H1(S2)

is dense in

F,

there exist

{~~, ~~}

E

D(A)

x

such that

and

therefore, considering

the inverse

inequality,

According

to Theorem

(3.1),

for each v E N there exists

~"

E

C° ([ 0, , T ) D(A)) f1 Cl ~~ 0 , T] ;

which is the solution of

(3.1)

with

initial data

{~°,~~}

E

D(A)

x and

f

E

Thus,

from the

linearity

of

(3.1)

we have ,

wich

implies

that the

unique solution § : Q -.

R of

(3.1)

satisfies

(35)

On the other

hand,

from

(5.2)

we

obtain,

From the convergence in

(7.1)

we conclude that the

right

side of

(7.4)

converges to zero

when ~

and v goes to

infinity.

So

(~"), (~;,)

and

are,

respectively, Cauchy

sequences in

L2~E(x~)~

and

L2~E*(x~)~,

which proves

(5.6).

To prove

(5.7)

we need the

following

result.

LEMMA. - d R >

0,

3 C > 0 such that

satisfying II ~~~, II * >

R.

Proof . -

Consider E such that

II {~~, ~1 } II * >

R.

So

~~~,

is different from

{0, 0} and, consequently,

it is sufficient to prove that: b’ R >

0,

3 C >

0,

such that

with

Let us suppose it does not

happen,

that

is,

there exists

Ro

> 0 such that

dC > 0 3

{~C,~c}.E D(A) x

with =

1,

R and

1/C.

In the

particular

case when C =

1/R0 it

follows that

which is a contradiction.

So, (7.5)

is

proved

and

consequently

the lemma.

Let us consider

firstly ~~~,

E

D(A)

x and

suppose §

is the

strong

solution of

(5.1). Then §

E

C~ ( ~ 0 , T]; D (A) ) nC1 ([0 , T];

and

therefore,

(36)

Thus,

from Theorem 3.1 we obtain

Consider,

now,

{03C60, 03C61}

E F

and 03C6

the weak solution of

(5.1).

If

~~o ~1} - {p 0}

then

~

= 0 and the

regularity

in

(5.7)

follows

immediately.

Let us

consider {~~, ~1 ~

different from

{0, 0}.

Since

D(A)

x

Hl (S2)

is dense in F there exists

f ~~, ~v}

c

D(A)

x such that

Defining

there exists

{~~, ~~,} subsequence

of

{~~, ~v}

such that

Therefore

Thus,

from

(7.8)

and the above lemma 3 C = C

()) (§°, 03C61} )) F) ~

0 such

that " - ’> , /mii r 10

Let

{~~,}

be the sequence of strong solutions of

(5.1)

with

initial

data

{~~ ~ ~~}. Then,

from

(7.6)

and

(7.9)

there exists

Cl

=

Cl (II{~~,

0 such that

But,

from

(7.7)

we obtain

So,

from

(7.10)

and

(7.11)

we

conclude

that

(37)

Then,

there exists a

subsequence

that we will denote

by

the same notation

{~~,}

such that

On the other

hand,

from

(7.3)

we have

and from

(7.12)

and

(7.13)

we

have §

= x which proves

(5.7).

Acknowledgements

The author is very

grateful

to Professors Manuel Milla Miranda and Adauto Medeiros for their valuable

support

and

guidance

in the

develop-

ment of this reseach and to the referee for his constructives comments which resulted in this revised version of the paper. He should also thank God for

strenght

received.

References

[1]

BARDOS

(C.)

and CHENG

(C.) .

2014 Control and stabilization for the wave equation, part III: domain with moving boundary, SIAM J. Control and Optimization 19

(1981),

pp. 123-138.

[2]

BARDOS

(C.),

LEBEAU (G.) and RAUCH

(J.) .2014

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[3]

CIORANESCU (D.), DONATO

(P.)

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[5] FUENTES APOLAYA

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(J.) .

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[9]

LAGNESE

(J.)

and LIONS

(J.-P.) .

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of Thin Plates, R.M.A. Collection 6, Masson, Paris

(1988).

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LASIECKA

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KOMORNIK

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LIONS

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MILLA MIRANDA

(M.) .

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