A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
S ERGE N ICAISE
About the Lamé system in a polygonal or a polyhedral domain and a coupled problem between the Lamé system and the plate equation. II : exact controllability
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 20, n
o2 (1993), p. 163-191
<http://www.numdam.org/item?id=ASNSP_1993_4_20_2_163_0>
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http://www.numdam.org/
and a
Coupled
ProblemBetween the Lamé System
and
the PlateEquation
II: Exact
Controllability
SERGE NICAISE
7. - Introduction
This paper is the second part of a
research,
whose purpose is tostudy
theregularity
of the solutions of someproblems
related to the linearelasticity theory
and the exact
controllability
of the associateddynamical problems.
Part I[22]
concerns the
regularity,
while PartII,
the exactcontrollability.
Forconvenience,
we have numbered the
paragraphs continuously: Paragraphs
1 to 6 form the first part, whileParaghraphs
7 to 12 forms this second part.We use the notations and definitions of Part I without comment. But
contrary
to PartI,
we consider here real Hilbert spaces; this means that all the functions we use are real-valued.Nevertheless,
all the resultsgiven
in Part Iremain true in the real
setting.
In order to avoid some
repetitions
andconfusions,
we divide this paper into two parts A andB, corresponding
to theproblem
studiedrespectively
inParagraphs
2 to 4 andParagraphs
5 to 6 of Part I.In Part
A,
westudy
the linearelasticity
system in apolygonal
domain ofthe
plane
or apolyhedral
domain of the space.Namely (see Paragraph
8 formore
details),
if Q denotes thisdomain,
we divide itsboundary
into twoparts rD U rN.
We consider thefollowing dynamical
linearelasticity
system(using
the classical notations of linear
elasticity theory):
Pervenuto alla Redazione il 20 Marzo 1991.
The
problem
of exactcontrollability
for(7.1 )-(7.4)
states as follows:given
T > 0
large enough,
for every initialdata üo, il
1 in suitable Hilbert spaces, is itpossible
to find controls v and wdriving
oursystem (7.1)-(7.4)
to rest at timeT, i.e.,
such that the solution £ of(7.1)-(7.4)
satisfiesUsing
his HilbertUniqueness
Method(HUM),
J.-L. Lions in[15]
answeredto this
question
when is smooth and when evenrN
orrD
is empty. In thissecond case, he also assumed that Q is star
shaped
with respect to apoint Here,
we remove all theseassumptions
andfollowing
Grisvard’s technics of[10] (especially Paragraph 5),
weadapt
HUM to oursetting.
As in[10],
we
impose
aregularity assumption,
which means that the weak solution of thestationary
Lamesystem belongs
to(H3~2+~(SZ))3,
for some 6 > 0. Let us notice that we gave in Part Igeometrical hypotheses
on SZ, whichimply
that thisregularity assumption
is fulfilled.Finally,
the fact that Q is notnecessarily
starshaped
with respect to apoint
leads to some difficulties onrN,
thepart
of theboundary
where weimpose
Neumannboundary
conditions.In Part
B,
we consider the exactcontrollability
of acoupled problem
between the linear
elasticity
system in the unit cube ofJR.3
with a crack and theplate equation
on this crack. Our motivation comes from aquestion
raisedin
Paragraph
6 of[3] concerning
the exactcontrollability
of theproblem they
obtained in
[3].
We answerpartially
to thisquestion
because wemodify
someboundary
conditions of theproblem
obtainedby [3] (we explain
at the end ofPart I the reasons of this
modification).
In view of theanalogy
between thisproblem
and theproblem
studied in[21],
we follow the method of[21]
toadapt
HUM to this new
problem.
As in[21],
the mainproblem
lies on the fact that the3D-part
of the weak solution of thestationary problem
has never theregularity H31116,
for some E > 0.Fortunately,
since it hasonly edge singularities along
the bottom of the
crack,
agood
choice of themultiplier
allows us to use HUM.Since we use
HUM,
thegeneral
method ofproof
of the exactcontrollability
of our
dynamical problems
is similar to those of[15], [10], [21].
This meansthat some results stated here seem to be identical with the
previous
ones; butof course
they
aredifferent, especially
theirproofs. Moreover,
some technicalproblems
are solved in a different way(for instance,
theidentity
withmultiplier).
A. The
elasticity system
in dimension 2 or 38. - Estimate of the energy
In all this Part
A,
we shall use the notations of thebeginning
of PartI,
i.e.,
fromParagraphs
1 to 4. So Q denotes a bounded open connected subset of JR1&, n E{2,3}.
But here we suppose that s2 isonly
on one side of itsboundary (so
Q has noslit!).
Forconvenience,
we also suppose that the set D is notempty.
We now introduce the operator A associated with the linear
elasticity system
We set H =
(L2(Q))n
and we recall thatSince Kom’s
inequality
holds onLipschitz domains (see [6]),
we know that the bilinear form au is V-coercive. Therefore it induces anisomorphism
,~ from Vinto VI defined
by
Since V is dense and
compactly
imbedded into H and the form an issymmetric,
it also induces a
positive selfadjoint operator
A from H intoH,
with acompact inverse,
definedby
- ~ I- - .... -6 ....,Since
H,
V and the form ao fulfil thehypotheses
of Remark 4.4 of[21],
Theorems 4.1 to 4.3 of[21]
may beapplied
to A. Inparticular,
wehave existence and
uniqueness
of thefollowing
waveequation: Given -
E V,ol
e H andf
EL1(O, T; H),
there exists aunique
solution0
ofAs
classically,
the estimate of the energy of the solutiony~
of(8.4)
withf = 6
will be obtainedby proving
anappropriate identity
withmultiplier.
Thisis our next
goal.
THEOREM 8.1. Let us assume that
Then ü E
DA fulfils (8.1)-(8.3)
withf =
Au.PROOF. To get
(8.1 ),
it suffices to notice that for all§Y
E(D (K2))n,
we haveNow, applying
Theorem6.6,
we see thatThis
implies
that J fulfils(8.3)..
Let us recall that in Theorems 2.3 and
4.5,
wegive geometrical
conditionson Q to ensure that
(8.5)
holds. We looked for these conditionsbecause,
as we shall seelater, (8.5)
is sufficient toapply
the HilbertUniqueness
Method ofJ.-L. Lions
[15]
to thissetting.
So in all this Part
A,
we shall assume, withoutrepeating it,
that theinclusion
(8.5)
holds.LEMMA 8.2.
If a
EDA
andm(x)
= x - xo, with some xo E R~. then we havewhere we use the notations
and 1’k
aui
denotes the outward normal derivativeof
ui on theface rk.
8v de
PROOF. Let us assume for a moment that g
belongs
to(H 2 (Q))n; applying
two times the Green
identity,
we obtainSince
H2(S2)
is dense inH 1/2,,(Q),
for some 6 > 0, we see that J EDA
fulfils(8.7)
because all the terms in(8.7)
have ameaning.
To prove that(8.7)
isequivalent
to(8.6),
we have to transformappropriately
theboundary
terms of(8.7).
For wedirectly
obtain the result since J fulfils(8.3).
On the otherhand,
theboundary
condition(8.2) implies
thatUsing
thisidentity (8.8)
in the definitions ofT(k)ü,
m ’ and weeasily
show thatUsing
these two identities in(8.7),
we arrive to(8.6).
Let us now fix xo E r. For
m(x)
= x - xo, we setWe are now
ready
to establish the main result of thisparagraph.
THEOREM 8.3. Let
~p
E0([0, T], DA)
n01([0, T], V) n 02([0, T], H)
be a solutionof (8.4)
withf
=6.
Then there exists a minimal timeTo
> 0 and a constant C > 0 such thatwhere
Eo
denotes the energyof 0
at time t = 0,namely
and
setting for
all k EI,
wedefine
PROOF.
By integration by
parts with respect to the variable t in(0, T)
andusing
the Greenidentity
inQ,
we show thatwhere we have set
Q
= S2 x(0, T).
Integrating (8.6)
over(0, T) (applied
to0(t))
andsubtracting
the obtainedidentity
to(8.11),
we getUsing
theidentity (4,24)
of[21],
which holds in the abstractsetting
of Remark4.4 of
[2J],
we show thatIn order to
conclude,
we need thefollowing inequality (8.14),
which is a directconsequence of
(8.8):
Inserting (8.13)
into(8.12), using
Schwarz’sinequality,
the coerciveness of theform au and
(8.14),
we obtain(8.9). ·
Let us now fix T >
To
such that theinequality (8.9)
holds. Then theapplication
is a norm
stronger
than the norm inducedby
V x H. As in[15], [10], [21],
wedefine F as the closure of
DA
x V for this new norm(obviously,
Fdepends
on xo and
T)
and we have thealgebraic
andtopological
inclusions:Arguing
as in Theorem 5.6 of[10],
Theorem 8.3 leads to the(by taking
into account the
hypothesis (8.5)).
PROPOSITION 8.4. Let e F and
f c L 1 (0, T; V),
then theunique solution 0
of(8.4)
fulfilsMoreover,
there exists a constant C > 0(independent
of100, ol I
andf )
suchthat
9. - Weak solutions of the
dynamical elasticity system
We transpose
Proposition
8.4 to get theTHEOREM 9.1. For all
do
E H,ill
EV’, ifk)
andv¡7)
Ei, j
E{ 1, ... , n },
k E N -, there existunique
Ü EL°° (o,
T;V’),
E
F’,
which are solutionsof
for
all wherey~
is theunique
solutionof
In order to
give
aninterpretation
to theequation (9.1 ),
we show that onthe
faces
where weimpose homogeneous
Neumannboundary conditions,
the ourward normal derivative is a linear combination of thetangential
derivatives.LEMMA 9.2. For all
kEN,
there existcoefficients
E R, anddfjlm
E R, i,j,
1 E{I,
...,n}
and m E11, ... ,
n -11
such that ufulfilling (8.3)
satisfies
where 1 ~ L m
forms
J an orthonormal basisof
I~n and19ut
denotes thetangential
derivativeof
in the directionof
PROOF.
By change
ofvariables,
thereobviously
exist coefficients E R such thatUsing
the definition ofT(k) 9
and(9.5),
we prove the existence of a vector whose components are linear combinations of(9u -,
,r,(k) forevery j
cm e
I n - 1},
such that(9"
where is the vector
Therefore,
theboundary
conditions(8.3)
isequivalent
toTaking
the innerproduct
of(9.7)
with¡jk)
and m E{ 1, ... , n - 1},
we seethat
(9.7)
isequivalent
toThis proves
(9.3). Moreover, (9.5)
and(9.3) obviously imply (9.4).
We can now say that a solution i1 of
(9.1 )
fulfilsformally (9.8)
to(9.12)
hereafter:
As
classically,
the solution 11 of(9.1 )
will be called a weak solution of theproblem (9.8)-(9.11 ).
Analogously
to[21],
we now show that the solution ü of Theorem 3.1 ismore
regular
and fulfils(9.12).
This is based on thefollowing
trace result.THEOREM 9.3. Let E k E D+ U N+ and E
0(£k),
i,j
E{ 1, ... , n },
k E V-. Then there exists v E( D (o, T, D (S2)))n fulfilling (9.10)-(9.11 ).
Moreover,for
all t E[0,
T ], v isequal
to zero in aneighbourhood of
the verticesof
SZand,
in dimension 3, also in aneighbourhood of
theedges ofQ.
PROOF. We
actually
solve thefollowing
moregeneral
trace result: Givenv-’’~k~,
E( D (~k ))n,
for all k Ejr,
find v E(D(0, T, D(?i)))n fulfilling
having
the samenullity property.
Butusing
theargumentations
of Lemma9.2,
we notice that
(9.13)
isequivalent
towhere
Z -,(k)
E(D(Y-k))’,
for all k E 7(z~k~
is a linear combination of and of thetangential
derivatives ofLet us fix l~ E 1.
By
an eventual rotation and atranslation,
we may suppose thatrk
is included in thehypersurface
zn = 0 and that the intersection between Q and thehalf-space
>0 }
is nonempty.Moreover,
thecompactness
of the support of and-Z,(k)
inYk imply
that there exists an open subsetVk
ofr~ satisfying Vk
Crk (where
we considerr~
as an open set ofJRn-1
I andwe take the closure of
Vk
in such thatSince Q is
nondegenerate,
we can find ?7k > 0 such thatLet us take a cut-off function
lbk
Esatisfying
We set
Owing
to theprevious remarks,
weeasily
check thatf1k)
fulfils(9.14)
onEk,
isequal
to zero in aneighbourhood
of the facesEj, j
E~B{~},
and has the desirednullity property.
We conclude
by taking
--THEOREM 9.4. Let u E
L°° (0, T; V’), 1,0 1
E F’ be the solutionsof (9.1 )
with data
go
E V,ul
E H, E(D(Ek))’,
k E D+ U N+ andv~~ ~
ED(yk), i,
j
E{ 1, ... , n },
k E N-. Then Ü EC( [0, T], H 1 (12))
n01([0, T], H)
andfulfils (9.8)-(9.10)
and(9.12).
PROOF. We
proceed
as in Theorem 5.3 of[21].
Let v E(D(0, T, Ð(Q))n
be the function built in Theorem 9.3satisfying (9.10)-(9. 11).
We setBy
Lemma 1.3.4 of[15],
there exists aunique solution 1$ E 0([0, T], V )
nC’ ([0, T], H)
nH2(0, T; V’)
ofWe now show that is the
unique
solution of(9.1 )
whenBy
Theorem 4.2 of[21],
it suffices to check(9.1) for §Y
EC([0, T], DA)
nC~([0,r],V)nC~([0,r]~). By integration by parts
over]0, T[,
weget
But the
following
Greenidentity
has ameaning
since~6(t)
EDA belongs
toH2
far from the vertices
and,
in dimension3,
also from theedges,
while isprecisely equal
to zero near the vertices and theedges
in dimension 3.Inserting
thisidentity
into theright-hand
side of(9.16), using (9.15)
andtaking
into account the
boundary
conditions fulfilledby ~6
andv,
theright-hand
sideof
(9.16)
isequal
toBy integration by parts
on Yk, for all 1~ E .Jl~+(this
is allowed sinceE
(D(Ek))n, vZ~ ~
E and0([0, T], DA))
andusing
theidentity (9.4),
we deduce that the
previous expression
isequal
to theright-hand
side of(9.1 ).
So £ fulfils
(9.1).
Now, 9
fulfils(9.8)
because(9.15) implies
thatin this distributional sense. This
completes
theproof
of Theorem 9.4..Before
going
on, let us notice that the solution u of Theorem 9.4 does notfulfil
(9.11 )
in a strong sensebecause y
is notenough regular
to fulfil(8.3).
Nevertheless, using
the arguments of the end ofParagraph
5 of[21]
replacing Proposition
4.10 of[21] by Proposition 8.4,
we conclude theTHEOREM 9.5. Under the
assumption of
Theorem9.1,
let 9, be the solutionsof (9.1 ),
then i1 EC( [0, T], V’)
nC 1 ( [0, T],
andsatisfies (9.12).
10. -
Application
of HUMIt is now easy to use HUM in order to conclude the exact
controllability
of the
elasticity
system(see [15], [10]
andParagraph
6 of[21]).
THEOREM 10.1. For all
go
E H,Ü1
EV’,
there exist E(L2 (1k))n, for
all k E Ð+ U
)1+,
andvi~ ~
EL2CLk), i, j E {I,
...,n 1,
k E .JV - such that the weak solution it EC([0, T], V’)
nC1([0, T], DA) of
theelasticity
system(9.8)-(9.11) (in
the senseof (9.1 )) satisfies
PROOF.
By Proposition 8.4,
for01 1
E F, there exists aunique
solutionSp
of(8.4)
withf = 0,
which fulfils(8.15)
to(8.17).
We
F’,
the solutions offor
all g E L 1 (o, T ; V ),
E F,where g
is theunique
solution ofThis is
possible owing
to Theorem 9.1by inverting
the order of time and because(8.16) implies
that-IkUij(i5) belongs
toL2(1k),
for all k E ).1-.On the other
hand,
Theorem 9.5 shows thatAs
classically,
we now define the operatorThe next lemma shows that A is an
isomorphism
and we conclude as inParagraph
6 of[21].
PROOF.
Applying
theidentity (10.2)
withg = 0
andusing
the definition ofo, ij (0),
we see thatThis
firstly
proves that A = A*.Secondly, taking (po,
and hence~,
we deduce(10.4)
in view of the definition(8.10)
of the norm in F..B. The
coupled problem
11. - An
identity
withmultiplier
In all this Part
B,
we use the notations ofParagraphs
5 and 6 of Part I of this paper[22].
We want to establish the exactcontrollability
of thedynamical problem
associated with theboundary
valueproblem (5.1)-(5.7).
As weexplain
in the
Introduction,
we use the same method as inParagraphs
4 to 6 of[21]
or in Part A.
Therefore,
weonly give
the great lines and the differences with theprevious
results.We gave in
Paragraph
5 the variational formulation ofproblem (5.1)-(5.7).
Since the next lemma proves that V is dense in H, we can associate with the form a
(defined by (5.8))
aselfadjoint
operator A from H to H with a compact inverse(see Paragraph 8).
So Theorems 4.1 to 4.3 of[21]
still hold for A.Moreover, Theorem 5.3 shows that for all
U
=(ü, ~)
EDA,
we haveLEMMA 11.1. V is dense in H.
PROOF.
Let 0
=(ii, 0
in H and let us fix E > 0. Thedensity
of inL2(W) gives
the existenceof q
ED(w)
such thatWe now introduce cut-off functions p5 E
depending
on the real parameter 6~]0,1[
[(b
will be determinedlater)
such thatand supp ~ We set
It is immediate that
In view of
( 11.1 ), choosing
6 such thatwe deduce that
Finally, owing
to thedensity
ofP(Q)
in thereexists 15
E(D(S2))3
suchthat
The conclusion follows
by setting
Indeed,
weeasily
check thatV:= (v, 1/) belongs
toV,
while theinequalitites
(11.1)
to(11.3) imply
thatThe second step consists in
establishing
anidentity
withmultiplier
as inLemma 8.2.
Unfortunately,
ifU
=(ü, ~)
EDA,
then Theorem 6.2 shows that u admits theexpansion (6.7)
andtherefore,
u has never theregularity
forsome s
> 3 .
Inspite
of this lack ofregularity,
as in Theorem 4.5 of[21], by
a2 p g Y y
good
choice of thepoint
xo, we can prove thisidentity. Contrary
to Theorem4.5 of
[21],
where we cannotseparate
Q and r to prove theidentity
withmultiplier,
here the bestregularity of ~
allows us toseparate S2
and w. Let usstart with the
3D-part,
which is very technical. Wepostpone
theproof
of the2D-part
to Theorem 11.7.THEOREM 11.2. Let us assume that
m(x)
= x - xo with XO =(0, 0, xo3).
Thenfor
allU
=(ü, ~) E DA,
we have:PROOF. For 6 > 0, let us denote
(we
recallthat zi
=(Oi - ~/2,
~pZ -x/2),
for i = 1 or -1, see
Paragraph 6)
Owing
to Theorem6.2,
we know that u admits theexpansion (6.7);
so itbelongs
to(H3/2+(Qd))3,
for some E > 0 and for 6sufficiently
small. Therefore u fulfils theidentity (8.7)
inU6;
that isby taking
into account theboundary
conditions fulfilledby J
and the fact thatWe now pass to the limit as 6 goes to zero.
Since 0
EDA,
the left-hand side of(11.5)
and the first term of theright-hand
side of(11.5)
tendrespectively
to the left-hand side of
(11.4)
and the first term of theright-hand
side of(11.4).
Moreover,
as in Lemma8.2,
theboundary
terms on r1 in(11.5)
are transformed into theboundary
term on r1 in(11.4).
So it remains theboundary
terms onr8
anda D8.
Let us denote themrespectively by 1,6
and128.
Using
the definition of T, weeasily
check thatBut
Proposition
6.3 asserts that1+(122(Ü) - 1-(122(Ü) E
for all pE] l, 2[;
while Theorem 6.6 and the Sobolev
imbedding
theoremimply
thatV2£
ELq(r),
for all q > 1.Therefore,
Hblder’sinequality
andLebesgue’s
boundedconvergence theorem allow us to conclude that
In view of the definition of
128,
there exists a constant M > 0 such thatBut it is
easily
seen thatand
using
Theorem 11.3hereafter,
we conclude thatSo, (11.6)
and(11.7)
prove theidentity (11.4).
THEOREM 11.3. Under the
assumption of
Theorem11.2, if U
=(Ü, ç)
EDA
there existpositive
constants K and60
such thatfor
all 6E]0,80[:
PROOF. Let us denote
by us,
thesingular part
of thedecomposition (6.7)
of 9. On one
hand,
Lemma 11.5 shows thatÜs
satisfies(11.8)
and(11.9).
Onthe other
hand,
Lemmas 6.3 and 11.4imply
thatur
fulfilsSince
( 11.10)
is stronger than( 11.8)
and( 11.9),
we concludeby
addition..LEMMA 11.4. Let u E
H 1 (S2),
then there exists a constant K > 0 such thatfor
all 6 > 0sufficiently small,
we haveIn
particular,
ufulfils
PROOF. It is similar to the
proof
of Lemma 4.7 of[21].
The difference is that wedecompose a D«5
into twoparts: a D5
=rs
where we setUsing
the coordinates(ri,
r2i,02i)
onCs
and thearguments
of Lemma 4.7 of[21]
onC§
where r isreplaced by
we can prove( 11.11 )..
Obviously,
we cannotapply
Lemmas 6.3 and 11.4 to thesingular
part~d,
of u.Nevertheless, using
itsexplicit expansion
and thespherical coordinates,
we prove
directly
that~6,
fulfils(11.8)
and(11.9).
LEMMA 11.5. Under the
assumption of
Theorem11.2, if ti
=(ic, ç) E DA,
then the
singular
partÜs of
9fulfils ( 11.8)
and( 11.9).
PROOF. We
only
prove(11.8)
and(11.9)
for apart
of thesingular part 118, namely
forThe
proof
is similar for the otherparts.
We remark that the matrix
A
1 is the transition matrix from thegradient
in cartesian coordinates to the
gradient
inspherical
coordinates(r 1, 91, ~p 1 ),
thismeans that for a scalar function u, we have
where we set
Since
Al
isuniformly bounded,
in order to prove(11.8),
it suffices toreplace V3u by
Let usfirstly compute
thegradient
ofJCi(ci) (defined
in Theorem
6.2). Using
the notations introduced in Lemma 6.5 and the Leibnizrule,
we see thatwhere for r21 ro, for some ro > 0, we have
(see
Remark 16.7 of[4]):
We shall now compute
As in Lemma
11.4,
wesplit
up thisintegral
into twoparts corresponding
tothe
boundary
ofCs
andC~
1 i.e.where we set
Using
the coordinates(ri ,
r2i,02i)
onC,,
we see thatT1
is characterizedby
for some rb > 1
(notice
that rb - 1, as 6 -0). Moreover,
weeasily
check thatthe surface measure du on
I1
isgiven by
Recalling
thatusing ( 11.14), (11.15)
and( 11.16),
we deduce that there existso
> 0 andKi
> 0 such that for all 6G]0, 6o [:
On one
hand, inequality (6.16)
shows that there existsK2
> 0 such that for 6On the other
hand, Proposition
11.6 hereafter(applied
with A= 0)
states thatSo we conclude that
Let us show the same result for Since the cut-off function ?yi 1 in
( 11.13)
isequal
to zero in aneighbourhood
ofS-i
I(see
Theorem6.2),
we cansay that there exists
K3
> 0 such thatUsing again ( 11.15),
we obtainBut we remark that for 6 small
enough,
there existpositive
constantsK5
andK6
such that for x Gr¡l
1 withr-,(x) e12321,we have
Passing
from the coordinates(r- 1,
r2, _ 1,62,-1)
to the coordinates(r 1,
r21,621),
wesee that on we have
But it is
always possible
to choosed00
> 0 such that for dE]0,800[,
the intervalr-i 1
e]l/2,3/2[
is sent into an interval included into]1/4,2[.
Sousing
thischange
of variable(11.23)
into(11.21)
andusing (11.22),
we getUsing again ( 11.18)
and( 11.19), by taking
into account(11.22),
we arrive toJoining together (11.20)
and(11.24),
we haveproved
thatThis is more than we need to prove
(11.8)
but we shall use it to show(11.9).
We argue as above for the derivatives with
respect
to81
and WI, but herewe
only
have boundedness.Indeed,
for the derivative with respect to81,
forinstance,
it appears a term boundedby
Since
au 1/ a01
=(-1 /2)r2~l~z cos(BZ, /2),
theprevious expression
is boundedby
Owing
toinequality (6.16),
it isonly
bounded.We have
just proved
that fulfils(11.8).
Now(11.14)
shows thatFor we
easily
check thatTherefore
(11.25), (11.8)
for andLebesgue’s
bounded convergence theoremimply ( 11.9). ·
Let us now
give
the result we need in( 11.19), concerning
theregularization
built with a function of mean zero.
PROPOSITION 11.6.
Let p
and setFor all Then
for
all 1PROOF. It is a direct consequence of the classical
regularization
theoremconcerning nonnegative
functions(see
for instance Lemma 2.18 of[1]) by splitting
up p into itspositive
andnegative
parts.We now
proceed
to theproof
of theidentity
withmultiplier
in the do- main w.THEOREM 11.7. Under the
assumption of
Theorem 11.2, allU
=(ü, ç)
EDA
fulfils:
or
equivalently
where v denotes the outward normal unit vector on the
boundary
9wof
w.PROOF. Theorem 6.6 proves
that d
EH7/2+e(w),
for some e > 0. SoPropositions
2 and 3 of[19] applied
withm(x)
= x - xo showthat ~
fulfils(11.28),
where the left-hand side is seen as aduality
bracket. But we show in Theorem 6.6 thatA2 ~
ELP(w),
for some pE] 1, 2[
andowing
to the Sobolevimbedding theorem, V2~ belongs
to(Lq(W))2,
for all q > 1;therefore,
we canreplace
thisduality
bracketby
a classicalintegral.
Finally, (11.28)
isequivalent
to(11.29)
becauseAdding
the identities(11.4)
and(11.29),
we obtain the so-calledidentity
with
multiplier;
which isanalogous
to theidentity (4.5)
obtained in[21].
COROLLARY 11.8. Under the
assumption of
Theorem11.2, 0
=(u, ç)
EDA satisfies
As in
Paragraph 8,
we are now able togive
the estimate of the energy.THEOREM 11.9. Let
0
=(,a, ç)
EC([0, TI, DA)n01([0, TI, V) n C2([o, TI, H)
be a solutionof
If
xo =(0, 0, X03), for
some X03 E then there exists a minimal timeTo
> 0 and a constant C > 0 such thatwhere,
asclassically, Eo
denotes the energyof Ü
at t = 0, and wedefine
As
previously,
we setThe
boundary of
w issplit
up asfollows:
where
6k
is an open linear segment. We also denotePROOF. The
proof
is almost identical with those ofProposition
4.8 of[21]
or Theorem 8.3above, i.e., by integrating (11.30) applied
toU(t)
over(0, T)
andadding
it with anidentity
similar to(8.11 ).
Theonly
difference is thefollowing
one: we need to show that there exists apositive
constant01
anda real number
C2
such thatUsing
the law of conservation of energy and the definition of a, thisinequality
holds if we have
- - - - -
The best choice is
01
=1/4, C2
=3/4.
The remainder of the
proof
is left to the reader. ·Fixing
once and for all XO =(o, 0,
and To > 0 such that( 11.32) holds,
we can define the
space F
as inParagraph
8. AndProposition
8.4 isreplaced by
thePROPOSITION 11.10. E F and
F E L 1 (o, T ; Y ),
then theunique
solution
CJ
=(ü, ~)
E0([0, T], V)
nC 1 ([0, T], H) of
fulfils
Moreover, their norms
depend continuously
on the sumof
the normof Ul }
in F with the norm
of F in L 1 (0,
T;V).
PROOF. There is no difference with the
proof
of Theorem 5.6 of[10]
since we use the
following lucky
feature: if U =(ü, ~)
EDA,
then 9 admits thedecomposition (6.7),
but thesingular part
has asupport
included intoD5~,
for some
60
> 0.Therefore,
the trace of 9(respectively
of itsderivatives)
onFi
Ur2
isequal
to the trace of itsregular part ur (respectively
of the derivativesof
it,).
12. - Exact
controllability
At this step, there is no
difficulty
to followParagraphs
9 and 10 and to obtain the exactcontrollability
of ourcoupled problem.
Theonly point
wehave to
explain
is a trace result. Togive it,
we shall suppose that62
and64
correspond
to the part of theboundary
of w such that x3 = -1 and 1respectively
and we set
THEOREM 12.1. Let
iflk)
E(0(£jk))~,
k E0+,
E(Ð(L2k))3, k
ek+,
E
i, j
E{I, 2, 3},
k E N - andv(k)
e Ed~, fulfilling if
k = 2 or 4,
v(k)
e n Then there exist w e(P(0,r,P(~)))~
and1./;
ED (O, T, D(w)) satisfying ( 12.1 ) to (12.6)
below:for
all IMoreover, ’IE
and V)
areequal
to zero in aneighbourhood of
the bottomof
thecrack
of
Q.PROOF. It is
proved locally
as Theorem 9.3. Let usonly
build twofunctions w
and V) fulfilling (12.1), (12.2),
the exactboundary
conditions onr 11 : = {x
Er 1 : X3 = -1 }
and (J 2, andhomogeneous boundary
conditions on the other faces. We claim(and
let check to thereader)
that these functions arewith the convention that = 0 if 1
ft
D’ andv~2~
= 0 ifd+, and q
is anappropriate
cut-off function.ig and V)
have thenullity property
of the theorem since andv~2~
areequal
to zero in aneighbourhood
of the bottom of thecrack.
Moreover,
theassumption v (2)
nD (Q 2_ ) implies
that 11 isequal
to zero in a
neighbourhood
of the face£2i :
={x
EL2:
x 1 =I}.
Using
the symmetrybetween U2
and Q 4 and the arguments of Theorem9.3,
we obtain the result..We can now
give
the weak formulation of the waveequation ( 11.33).
THEOREM 12.2. For all
Yo
E H,VI
EV’,
E(L2(L1k))3,
k E~+;
E
(L 2(12k ))3 , k
EV 1; ’1
EL2(L2k), i, j
E{ 1, 2, 3},
k E .N ; andv(k)
EL2«(J k),
k Ed+;
there existunique V
EL"(0, T, V’), E F’,
whichare solutions
of
for
allof
where is the
unique
solutionMoreover,
PROOF. Direct consequence of
Proposition
11.10 and Theorem12.1, arguing
as inParagraph
5 of[21].
Let us remark that thenullity property
in Theorem 12.1 for the function w is made in order to be able toapply
theGreen
identity (9.17)
to thepair (w(t), ü(t)),
whenti
=(ü, ç)
EC([0, T], DA)
(see
Theorem9.4)..
Formally,
the solutioniT
of(12.7)
isequal
to and satisfiesthe transmission and
boundary
conditions(12.1)
to(12.6),
the initial conditionsand the final condition
(12.9).
Finally, applying
the HilbertUniqueness
Method of J.-L. Lions[15],
wecan conclude the exact
controllability
of ourcoupled problem:
THEOREM 12.3. For all
Vo
E H,VI
EV’,
there exist Ek E
D+; v~2’~~
EL2(L2k), k
EjJ+;
EL2(L2k), ~1,2, 3~~ ~ E
M- andv(k)
EL2 (Uk),
k Ed+,
such that the solutionY of (12.7) fulfils
In Theorem
12.3,
we establish the exactcontrollability
for controls with support in k ED+,
L2k, k E N+ UN -,
andU k, k
E d~. From the mechanicalpoint
ofview,
it isinteresting
to notice that we get controlshaving
their supportsonly
concentrated on the externalboundary
of Q,i.e.,
no control on the bottom of the crackFo.
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