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
esérie, tome 8, n
o1 (1999), p. 53-89
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- 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, n° 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
UniquenessMethod)
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
UniquenessMethod)
due to J.-L. LionsKEY-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
1. Introduction
Let 52 be a bounded domain in with
C2 bondary r,
and letQ
bethe finite
cylinder
52x ] 0 ,
T[
with lateralboundary
E = rx ] 0 ,
T[.
Weconsider the
following system
withinhomogenous boundary
conditionswhere
The
problem
of the exactcontrollability for
thesystem (1.1)
is formulatedas follows.
PROBLEM 1.1.- Given T > 0
large enough,
it ispossible, for
eachpair of
initial datadefined
in a suitable space, tofind
a control v suchthat solution y =
y(x, t) of (1.1) satisfies
the conditionLet us note that when
a(t)
=~3(t)
=a(x)
=1,
Problem 1.1 was studiedby
J .-L. Lions[12] by using
HUM and alsoby
I. Lasiecka and R.Triggiani [10] by using
the ontonessapproach. Many
other authors studied the exactcontrollability
of distributedsystems
withtime-dependent
orx-dependent
coefhcients.
Among them,
we can cite J.Lagnese [8]
whofirstly
workedin
boundary controllability
of distributed systems with timedependent
coefhcients and C.
Bardos,
G. Lebeau and J. Rauch[2]
whose work treats tothe
geometric optics approach
in the case of spacedependent
coefhcients. In thisdirection,
we can alsocite,
V. Komornik[11]
who presents anelementary
and constructive method to obtain the
optimal
estimates needed in HUM(Hilbert Uniqueness Method)
for the exactcontrollability
of some linearevolution 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.In this work we prove that
system (1.1)
isexactly
controllableby making
use of
HUM, c.f.
J.-L. Lions[12].
For thisend,
weemploy
themultiplier technique
to obtain the inverseinequality.
When the coefhcientsdepend
ontime,
aftermaking appropriated hypothesis
onthem,
the inverseinequality
still remains
true;
but since standard arguments are notapplicable,
theregularity
of backwardproblem requires
a newproof
which is the main task of this work(Theorem 5.1 ) .
In
fact,
thegoal
of this work is to show that HUM can beapplied
tothe case of
time-dependent coefficients
with Neumannboundary
condition.In order to
simplify
thecomputations,
we consider thesimple
operators definedby (1.2). Howover,
withappropriated changes,
we can extend ourresults to those ones
given by
with
a(x, t)
>fp
> 0 inS2
x(0, oo). Now,
when we consider the matricial operatorsthe usual
arguments
cannot beapplied
even if i= j
andt)
=ai ( x )
(Remark 1,
Sect.4).
’Our paper is divided on sixth
chapters.
In Section2,
wegive
notationsand state the
principal
result. In Section3,
we consider thehomogeneous problem
and in Section 4 we establish the inverseinequality.
In Section5,
we
study
the backwardproblem
and in the last section(Sect. 6)
weapply
HUM.
2. Notations and Main Result
Let
xO
E the unit exterior normal vector at x Er, m(x) =
x E and
)
In what follows the
symbol
" ~ " denotes the innerproduct
in II8’~. Let usdefine
Let us introduce some notations that will be used
throughout
this work.We are
going
to denote( ~ , ~ ) and [ . the inner-product
and the norm ofL2(S2) respectively.
The norm in will be denotedby [[
.II.
Let A be the
operator
definedby
thetriple L2(S2), a(u, v)}
where
We recall that the
Spectral
Theorem forself-adjoint operators guatantees
the existence of acomplete
orthonormalsystem (wv )
ofgiven by
theeigenfunctions
of A. If are thecorresponding eigenvalues
ofA,
thenAv -
+00 as v -~ +00.Besides,
Considering
inD(A)
the normgiven by
thegraph,
thatis,
it turns that is a
complete
system inD(A).
. Infact,
it is known that is also acomplete
system inNow,
since A ispositive, given
6 > 0 one has
In we consider the natural
topology given by
the normWe observe that such
operators
are alsoself-adjoint,
thatis,
D(Al/2)
= andD(AO)
=L2(S2).
We note thatA(t)
=Here,
we are
using
the samesymbol
for bothoperators
tosimplify
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 thesystem,
We have the
following
result.THEOREM 2.1.-
Suppose
thatassumptions
aresatisfied.
Then there exists a time
To
> 0 such thatfor
T >To
and initial data{yo, yl}
EL2(S2)
x~H1(S2)~~,
there exists a controlsuch that the ultra-weak solution
(the
solutionof (2.IJ
isdefined by
thetransposition method,
see(1 ‘~~~
y =y(x, t) of (2.1~ satisfies
3. The
Homogeneous
ProblemIn this section we
present
a standard result and a new one about the solutions of thefollowing homogeneous system
We have the
following
results.THEOREM 3.1.-
Suppose
thatassumption (Hl)
holds.Then, given
k E
{0,1, 2}
andproblem (~‘l.1)
possesses aunique
solution B :Q
--~ R suchthat,
Moreover,
the linear mapis continuous.
Theorem 3.1 can be
proved
in a standard wayby applying
the Faedo-Galerkin method and
using
thespectral
considerationsgiven
in Section 2.Next we consider the
homogeneous problem
which will be used in the
study
of theregularity
of the solution of(2.1).
.THEOREM 3.2. - Given
f
ED ~0, T D(A)),
theunique
solutionof problem (~‘3.2J satisfies for
every t E~ 0 , T
where C =
C(T).
Proof . -
SinceB~
=Bl
= 0 andf ~ E D ~0, T ; D(A)~ from
Theorem 3.1the above
problem
has aunique
solution B such thatBesides,
this solution satisfies theidentity
From
(3.3)
weget
AB EC°([0, T]; D(Al/2))
and thereforeThis
togheter
withassumption (HI) implies
thatIntegrating
thisequality
andnoting
thatf (0)
= 0 we haveReplacing by f’ -
in the lastintegral
we obtainNow
integrating by
part andnoting
thatf (0)
=0,
Replacing (3.6)
in(3.5)
we haveFrom
(3.4)
and(3.7)
it follows thatDenning o-~~ - &-~/~/
= ~. andreplacing by a-~/2~
+ in theabove
expression
we obtainFrom
hypotheses (Hl), (H2)
and(3.8)
there exists a constant C > 0independent
off
and 8 such thatApplying
Gronwall’sinequality
twice(first
we consider the Gronwall in-equality (1/2)g2(t) fo m(s)g(s)
ds whereg(t)
=I
andm(t)
=2C(g(t) + IAf(t)I)
and after the usualone),
weget
In a similar way we also infer that
Using
the definition on ~p we obtain the desiredinequalities.
04. The Inverse
Inequality
In this section we construct a
special To
timedepending
on n, onthe functions
a(t), ,Q(t), a(t)
and also on thegeometry
of Q.Taking
into account theregularity of 0393, we
can define on F a unit exterior normal vector fieldv(x)
on classC1.
In the same way we can define afamily
of
(n - 1) tangent
vector field{ T1 (x),
... ,(x) ~
of classC1
such thatthe
family { v(x), T1 (x),
...(x) ~
defines an orthonormal basis for each x E I‘ .regular function,
we havewhere
Defining
we obtain from
(4.1)
and(4.2)
We observe that when
~03C6/~03BDA
= 0 on r then~03C6/~03BD
= 0 sinceThen, 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 thatis,
when it isgiven by
then we have
and therefore if = 0 we do not have
necessarely
that = 0 andconsequently
we cannot use theidentity
even if i
= j
andt)
= As thisidentity plays
an essential role to prove the inverseinequality,
this caserequires
another treatment whichwill not be considered in this work.
If 03C6
EH2(O)
we can define in a natural way a continuous linear operatorsuch that
In
addition,
we can also define a continuous linear operatorwhere
Fo
is anonempty
open subset of r(sometimes
the wholer)
such thatThus,
from(4.7), (4.9)
andby density arguments
it results thatConsidering
the aboveequality
we are able to define thetangential gradient
Dropping
the index "2" in(4.8)
tosimplify
thenotation,
we define theadjoint operator
and from
(4.8)
and(4.11)
we obtain the continuous linear operatorwhere "o" denotes
composition.
Hence,
for all~, ~
EH1 (T‘o),
In
particular
THEOREM 4.1.- Let B be the weak solution
(it
means that the initial data EH1(52)
xL2(S2)) of
the Problem(~5.1). Then, if f
=0,
where
and
Proof.-
We consider first that ED(A)
xThen,
fromTheorem
3.1,
there exists aunique
solution e in the classMultiplying (3.1), by 8’(t)
we obtainIntegrating
thisexpression
from 0 to t and thenintegrating by
parts weget
Taking
into account(4.14)
we can rewrite the aboveexpression
as followsOn the other
hand, differentiating E(t)
we haveSo
where
The above
inequality gives,
Now, considering
it follows from
(4.15)
thatFinally, considering
we obtain the desired result
by using density arguments.
DTHEOREM 4.2.- Let q be a vector
field
such that q E Then each weak solution~ of
Problem(~.IJ satifies
Proof.
- First we prove theidentity
for thestrong (it
means that theinitial data E
D(A)
xHl (S2))
solutions of(3.1)
and then the result followsby
adensity
arguments.So,
let us suppose thatBy multiplying
the firstequation
of(3.1) by
andintegrating
over
Q,
Integrating by
parts the left side ofequality (4.16)
weget
On the other
hand,
sincewe have from
(4.17)
thatWe also have
Thus, combining (4.19)
and(4.18)
we obtainNow, evaluating
theright
side of(4.16)
we have from Greenidentity
Combining (4.16), (4.20), (4.21)
and(4.5)
we obtain the desiredidentity.
0The
To
time which Theorem 2.1 is definedby
and
uniquely depends
on n,a(t), ,~(t), a(t)
and on thegeometry
of S2.
THEOREM 4.3.-
Suppose
thathypotheses (HI), (H2)
and(H3)
hold andthat T >
Tp.
. Thenfor
each weak solution~ of (,~.1)
withf
= 0 there existsC > 0 such that:
(i~ if
n > 1 then(ii) if
n = 1 thenProof.
-By using
theidentity given
in Theorem 4.2 withq(x)
=m(x)
=x - we
get
after some calculationsOn the other
hand,
Multiplying
the firstequation
of(3.1) by §
andintegrating
onQ
we haveReplacing (4.24)
in(4.23)
it follows thatNow, substituting (4.25)
in(4.22)
we obtainSince =
max{ Ilm(x)11 ;
x E52},
fromhypothesis (HI)
we haveFrom (4.26)
and(4.27)
weget
and from
hypothesis (H2)
and Theorem 4.1 we obtainNext,
we estimate theexpression
From
hypothesis (HI)
and Theorem4.1,
weget,
and from
(4.29)
we obtainFrom the above
inequality
weget
which
together (4.28) implies
thatThis
gives (i).
To prove
(ii),
we consider theidentity
Then,
it follows from(4.22)
and(4.31)
thatFrom
(H3)
we have that 0 ^y 1 andtherefore,
Then, by making
use of the samearguments
of(4.27)
and(4.28),
from(4.32)
we obtain
Defining
from
hypothesis (H2)
andusing
similararguments
to the case n >1,
we obtain(ii). D
THEOREM 4.4
(Inverse Inequality).
-Suppose
that(HI)-(H~)
hold andlet T >
Tp.
. Thenfor
eachstrong
solution~ of (~.IJ
withf
= 0 there existsC > 0 such that:
(i) if
n > 1Proof.
- We aregoing
to prove the case(i)
since(it)
isanalougous.
Dropping
the terms whichgive negative
contributions in Theorem 4.3 onehas
On the other
hand,
there exists a constantC2
> 0 such thatIndeed, since 4;
is aregular
solution of(3.1),
thenand therefore
Defining
we have
and from
(4.35)
it follows thath, h’
EL2(0, T)
and hence h EC°([0, T]).
Let
to E ~ [0 , T] be
a minimizer of h. Thusand
consequently
But,
since to is aminimizer,
we haveand then
Thus,
from(4.36)
and(4.37)
we obtainV t E
[0, T],
which proves(4.34). Combining (4.33)
and(4.34)
weget
thedesired result. 0
5. The Backward Problem
Let T >
To
whereTo
is defined in theprevious section,
and consider thefollowing homogeneous problem:
According
to the inverseinequality given
in Theorem4.4,
theexpression
defines a norm in
D(A)
xH1(S2).
So,
we are able to define the Hilbert spaceequipped
with thetopology
where
~{~~,~~}~~,E~
is anyCauchy
sequence in~D(A)
xdefined
by
theequivalence
relationFor every
{~~, ~~}
ED(A)
x we haveNow,
sinceD(A)
xH1(S2)
in dense inF,
we havewhere the inclusion are continuous and
dense.
We can observe that
by
the construction ofF,
which means that if
{~°, ~1}
E F thenThe
proof
of the aboveregularities
aregiven
in theappendix.
We then consider the backward
problem
where §
is theunique
solution of Problem(5.1)
with initial data~~~, ~1}
EF.
We observe that the
operator 8/8t
is well defined onE(x~) taking
intoaccount
(5.6)
andconsidering
itsmeaning:
V w EH1 (0, T; L2 ~r(x~)~)
It is
important
to note thisoperator
is not taken in the distributional sense.On the other hand from
(5.7)
we obtainLet w
be the solution of(5.8)
definedby
thetransposition method,
whichwill be
precised
later. Let~~~, ~1 ~
EF, f
EL1 (0, T ; H1 {S~))
and let8 :
: f~ -~
R be theunique
solution ofMultiplying (5.11) by 03C8
andintegrating by parts,
we obtainformally
Replacing by
its valuegiven
in(5.8)
weget
from(4.12)
and(5.9)
Observing
thisexpression
we define the functionnalThus,
from(5.12)
and(5.13)
we obtainformally
thatConsidering
Theorem 3.1 and the construction ofF,
we have that the functionalgiven by (5.13)
iscontinuous,
thatis,
Indeed,
firt of all we note that the solution 0 of(5.11 )
verifies 8= ~ +~
,where
81
and82
are,respectively,
the solutions of thefollowing problems:
Besides,
from(5.13)
we can write for all~~~, ~1}
ED(A)
x andi=1,2
and therefore from
(5.2)
and(5.16)
we obtainFrom
(5.17)
and Theorem 3.1 we haveBy density arguments
we conclude thatinequality (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 thatand verifies
DEFINITION.- The
unique function ~
thatsatisfies (5.19)
in calledsolution
by transposition of
Problem(5.8).
Now we state the main result of this
section,
which is a consequence of Theorem 3.2.THEOREM 5.1.- The
unique
solutionby transposition ~ of
the Problem(5.8)
has thefollowing regularity:
In a
addition,
the linear mapis continuous.
Proof
. - For everyf
ED ~0, T D(A))
we havewhere 8
and §
are,respectively,
solutions ofBy
the definition of F and from Theorem 3.1 it follows thatIndeed,
it is sufficient to prove(5.22)
when the initial data~~~, ~1}
ED(A)
x becauseby density arguments
we conclude the same whenf~°~m~}
E F.We have
by
Schwarzinequality
and Theorem 3.1which concludes
(5.22).
On the other
hand,
from Theorem 3.2 weget
which is the crucial
point
for controlproblems involving time-dependent
coefficients.
In
fact,
before we prove(5.23)
we observe that in theright
side ofequation (5.20)
we have/~ while
in theright
side of(5.23)
we havef.
.Besides,
we notethat when the coefficients do not
depend
on time(see
forexample
the mostsimple
case for the waveequation),
it is not difficult to obtain the aboveinequality by using
Theorem 3.1 and thefollowing
standardargument.
If w is a solution to
problem
with
f
EP (0, T; ; D(A)~ then
B = w~ is the solution ofBut in our case, where we have
time-dependent coefficients,
thisarguments
failscompletely
and we need to solve it in other way. From Theorem 3.2 andobserving
that8(t)
=fo 8’(s)
ds after some calculations we obtainwhich
implies
In addition
and therefore
From
(5.24)
and(5.25)
weget (5.23). Combining (5.22)
and(5.23)
weobtain
which is sufficient to prove the desired
regularity,
thatis,
In
fact,
let us defineSince
P(0, T ; ; D(A)~
is dense inL1 (0, T; ; D(A)~
we can consider theunique
linear continuous extension
S
ofS, given by
and, consequently,
if follows thatNow, given f with p
ED(A)
and 8 ED(0, T), according
to(5.13),
(5.19), (5.27)
andconsidering
the factthat 80 = 81
= 0 weobtain,
So, by (5.28)
it follows thatwhich
implies
thatTherefore S
= ’1/;/
in D~(0, T ; ~D(A)~ ~)
and (5.26)
is thenproved.
One observes that if in
(5.19)
we considerf
= + B = cp~with p
ED(A3~2), ~
ED(0, T)
and~~ _ ~1 - 0,
we haveSince
a(t) >
ao >0,
if follows thatThen,
from(5.19), (5.26)
and(5.29)
we obtain(see
forexample,
J.-L. Lions and E.Magenes [13,
Vol.1,
Lemma8.1])
whichmakes
~(0)
and~~(0) meaningful.
Using
theregularity
of~, considering f
= p+
and B = cpr~where p
ED(A3~2), ~
ECZ ~0, T) (first
weget,
forinstance,
=~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 thatwhich ends the
proof.
D6. 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 havethat
is,
This
implies immediately
that A isinjective
andself-adjoint.
Then Ais an
isomorphism
from F toF’. Therefore, given
E F~ weobtain
{a(0)yl
EF’
andconsequently
there exists aunique
{~o , ~1}
E F such thatFrom
(6.1)
and(6.2)
we haveNow we are
going
to finish theproof
of Theorem 2.1. SinceL2(0)
x~H1 (SZ)) ~, taking
into account the chainwe obtain E F’ and therefore in this case we deduce
(6.3).
Defining
in(2.1)
the controlsfrom
(6.3),
theuniqueness
of solutions to Problems(2.1)
andwe
finally
conclude thatThus Theorem 2.1 is
proved. Q
7.
Appendix
Since
D(A)
xH1(S2)
is dense inF,
there exist{~~, ~~}
ED(A)
xsuch that
and
therefore, considering
the inverseinequality,
According
to Theorem(3.1),
for each v E N there exists~"
EC° ([ 0, , T ) D(A)) f1 Cl ~~ 0 , T] ;
which is the solution of(3.1)
withinitial data
{~°,~~}
ED(A)
x andf
EThus,
from the
linearity
of(3.1)
we have ,wich
implies
that theunique solution § : Q -.
R of(3.1)
satisfiesOn the other
hand,
from(5.2)
weobtain,
From the convergence in
(7.1)
we conclude that theright
side of(7.4)
converges to zero
when ~
and v goes toinfinity.
So(~"), (~;,)
andare,
respectively, Cauchy
sequences inL2~E(x~)~
andL2~E*(x~)~,
which proves(5.6).
To prove
(5.7)
we need thefollowing
result.LEMMA. - d R >
0,
3 C > 0 such thatsatisfying II ~~~, II * >
R.Proof . -
Consider E such thatII {~~, ~1 } II * >
R.So
~~~,
is different from{0, 0} and, consequently,
it is sufficient to prove that: b’ R >0,
3 C >0,
such thatwith
Let us suppose it does not
happen,
thatis,
there existsRo
> 0 such thatdC > 0 3
{~C,~c}.E D(A) x
with =1,
R and
1/C.
In the
particular
case when C =1/R0 it
follows thatwhich is a contradiction.
So, (7.5)
isproved
andconsequently
the lemma.Let us consider
firstly ~~~,
ED(A)
x andsuppose §
is thestrong
solution of(5.1). Then §
EC~ ( ~ 0 , T]; D (A) ) nC1 ([0 , T];
and
therefore,
Thus,
from Theorem 3.1 we obtainConsider,
now,{03C60, 03C61}
E Fand 03C6
the weak solution of(5.1).
If~~o ~1} - {p 0}
then~
= 0 and theregularity
in(5.7)
followsimmediately.
Let us
consider {~~, ~1 ~
different from{0, 0}.
SinceD(A)
xHl (S2)
is dense in F there exists
f ~~, ~v}
cD(A)
x such thatDefining
there exists
{~~, ~~,} subsequence
of{~~, ~v}
such thatTherefore
Thus,
from(7.8)
and the above lemma 3 C = C()) (§°, 03C61} )) F) ~
0 suchthat " - ’> , /mii r 10
Let
{~~,}
be the sequence of strong solutions of(5.1)
withinitial
data{~~ ~ ~~}. Then,
from(7.6)
and(7.9)
there existsCl
=Cl (II{~~,
0 such that
But,
from(7.7)
we obtainSo,
from(7.10)
and(7.11)
weconclude
thatThen,
there exists asubsequence
that we will denoteby
the same notation{~~,}
such thatOn the other
hand,
from(7.3)
we haveand from
(7.12)
and(7.13)
wehave §
= x which proves(5.7).
Acknowledgements
The author is very
grateful
to Professors Manuel Milla Miranda and Adauto Medeiros for their valuablesupport
andguidance
in thedevelop-
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
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