R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
M. C HIPOT
G. V ERGARA C AFFARELLI
Some results in viscoelasticity theory via a simple perturbation argument
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Viscoelasticity Theory
Via
aSimple Perturbation Argument.
M. CHIPOT - G. VERGARA CAFFARELLI
(*)
1. Introduction.
The
goal
of this note is toexplain
aperturbation argument
con-tained in
[4]
and to show how itprovides
existence anduniqueness
for the
theory
ofviscoelasticity
« without initial conditions >>.Let Q be a
smooth,
bounded open set ofRn, (n ~ 1 ).
The equa- tions of motion of a linear viscoelasticbody
aregiven by:
together,
forinstance,
with theboundary
condition:Here n =
3,
but thetheory applies
for any n. u is thedisplace-
ment of the
body
andf
=(~1, f2,
...,In)
theapplied
external force(*) Indirizzo
degli
AA.: M.Chipot:
Universite de Metz,D6partement
deMath6matiques,
Ile duSaulcy,
57045 Metz-Cedex(France);
G.Vergara
Caf-farelli : Universith di Trento,
Dipartimento
diMatematica,
38050 Povo, Trento(Italy).
which we suppose to be
given.
Forsimplicity,
we choose tostudy
the motion up to time t =
0,
of course, our results wouldapply
up to any time T.This kind of
problem
has been attackedby
numerouspeople (see
for instance
[1], [4], [6], [10], [12], [15], [18]). Usually, however,
thedisplacement u
issupposed
to be known up to some fixed timeto
0.So,
in this case, if we transfer the termto include it into
f,
theproblem
reads:Of course
together
with(1.3), (1.2 ), u
isprescribed
at timeto.
One has then to solve an initial value
problem (see [6]).
This
approach
has somephysical
limitation andconsidering
thefading
memory of the kernel this may lead toprefer
the formula- tion(1.1), (1.2).
In[10], [12]
Fichera studied thequasi-static
versionof this
problem,
i.e. the casewhere e equal
0 in(1.1).
We would liketo
investigate
here thedynamical
case.First,
we will transform theproblem, rewriting
it in an abstractsetting.
The mainingredients
to do so will beexplained
in section 2 where we will solve theproblem
in the case = 0. In section 3 wewill
explain
ourperturbation argument. Finally,
in section 4 we willapply
our results to(1.1), (1.2).
Note that our solution u, as well as our external force
f ,
will beassumed to have an
exponential decay
at - oo. This willguarantee
existence and
uniqueness.
Thenecessity
of such adecay
has beenestablished in
[4]. Indeed, considering
a unidimensional version of(1.1), (1.2),
one can show thatuniqueness
mayfail,
for instance forf
=0,
if thedecay
of the solution is notsupposed
to bestrong enough.
We refer the reader to
[4]
for details.If V is a Banach space, we denote
by WIt co ((- 00, 0); V),
. ( (-
co,0); Y’)
the set of functions defined on(-
oo,0)
with valuesin
V,
one or two times differentiable in the distributional sense with derivatives in.L°° ( ( - oo, 0 ) ; Y) .
All theequations
that we will writeinvolving
such functions are, of course,supposed
to hold in the dis-tributional sense and almost
everywhere.
2. An abstract existence and
uniqueness
result.Let V and .H two Hilbert spaces, their dual spaces.
We denote
by 11 «
the norm onV, by ( , )
the scalarproduct
on Hand
by ( I
itsnorm. ( , )
denotes theduality
bracket between V’and Y.
Moreover,
we suppose thatV
being
dense in H and the canonicalembedding
from V into Hbeing completely
continuous. Inparticular,
for some constant c one hasA
typical
situation is for instance whenand is a
smooth,
bounded open set ofR~, (n > 1). (We
refer thereader to
[13]
for definitions andproperties
of Sobolevspaces).
Let
a(u, v)
be abilinear, symmetric, continuous,
coercive formon
V,
i.e. such that:Clearly, a
defines aunique operator
A from V into V’through
theformula:
If we denote
by D,~
the setthen, A
can be viewed as an unboundedoperator
fromDA
into Hand one
has,
when H is identified with H’ :Moreover,
it is well known thatDA
is a Hilbert space when endowed with the normIn
fact,
we will find more convenient to use onDA
the normlaul.
These two norms are
equivalent
onDA . Indeed, by (2.2), (2.4), (2.6)
one has
Thus:
from which it results that
for some constant C.
DEFINITION. If Tr is a Banach space, normed
by 11,
, for v > 0we denote
by Loov((- oo, 0 ) ; V)
the set ofbounded,
measurable func- tions v from(- oo, 0)
into V such that for somepositive
constant C:Moreover,
we define onL~ ( (.-- oo, 0 ) ; V)
the normIt is easy to see that
L~ ( (- oo, 0) ; TT),
whenequiped
with this norm, is a Banach space.LEMMA 1. Assume
~>0.
Foroo, 0 ) ; R),
there existsa
unique
0090) ; R)
such that(The
firstequation
holds in the distributional sense and almost every-where).
Moreover one has:PROOF. Set
First,
due to ourdecay assumption
onf ,
thisintegral
makes sense.Indeed:
and this last function is
integrable
on(-
oo,0). Moreover, by (2.9), (2.10):
and thus
c>o, 0); R). Differentiating twice,
onegets:
(A simple justification
of this differentiation could be obtainedby expressing
in(2.9)
in terms of sin and
cos). Then, by
the sameargument
thanabove,
these
integrals
make sense and one has(2.8). So, (2.7), (2.8)
hold.Let us prove now that a is the
unique
solution of(2.7).
Forthis,
it is
enough
to show that thehomogeneous equation
has
only
0 as a solution.Note that it results from the first
equation
of(2.11)
that a is ofclass C2.
So,
if wemultiply
thisequation by
at weget:
Hence:
But now, due to
(2.11),
a, «11 - 0 when t - - oo.Thus,
at -~ const.But this constant must be 0 if one wants to have « - 0 when t 2013~ 2013 oo. This
implies
a =0,
and theproof
iscomplete.
LEMMA 2. Assume that
and
f , j
be solution of
one has the esti-
((2.12)
holds in the distributional sense, i.e. ina. e. in t) -
PROOF. Due to our
assumptions
one hasTaking
the scalarproduct
of(2.12)
and weget:
(Recall
that issymmetric). Integrating
between - oo and t andusing
the fact thatwe obtain:
Hence
by (2.4):
The results follows.
We are now able to prove:
THEOREM 1. Assume that
Then,
there exists aunique u
solution of:Moreover,
y one has:and
PROOF.
a) Let Vk
EDA
the basis of normalized orthonormaleigen-
functions of
A,
i.e. thevk’s
are such that:It is well known that the
vk’s
form a basis for H and that theI,ls
are
positive. (Recall
that we assumed the canonicalembedding
fromV into H to be
completely continuous.)
Taking
the scalarproduct
of(2.14) and vk
one hasclearly:
Hence,
if u, oo,0 ) ; H),
x/, =(u, Vk)
is theunique
solutionin 00,
0); R)
of(See
Lemma1.) Uniqueness
follows then from therepresentation
formula
b ) Conversely,
denoteby
ale theunique
solution inL~ ( (- oo, 0 ) ; R)
to
(2.1,8),
and setBy
Lemma 1 one hasclearly:
Hence
by
Lemma 2:From which it results
easily
that:Due to
(2.2)
and theequation
of(2.19)
we deducewhere C is some constant
independent
of N.Thus,
there exists asubsequence,
still denotedby
UN’ such that:Now it is easy to see that if
then for any v E H
(This
results from theequality
and the fact
oo, 0); ~~~.
So, taking
the scalarproduct
of(2.19)
and v E H one has in5)’(-
Taking
the limit in N weget
inÐ’ (-
oo,0)
But,
now,by
the formulawe have also in
D’(-
oo,0)
Moreover,
y forvEDA
we have:and thus
Since
D~
is dense in g this leads tofor any v E H.
Combining (2.21)-(2.23)
we obtainfor any v E g. From which is follows
easily
that u satisfies :(See
for instance[18], Chap. III,
Lemma1.1). Now, passing
to thelimit in
(2.20) implies clearly (2.15), (2.16).
3.
Existence
anduniqueness :
aperturbation argument.
One would like to consider here the
problem:
where for
any s > 0,
is a continuousoperator
fromD,~
intoH,
the above
integral being
understood in the usual sense.More
precisely
let us prove:THEOREM 2. Assume that
Moreover,
assume that s - is a elmapping
from(0, + oo)
into
H)
whereH)
denotes the space of continuous linear maps fromDA
into H. Then ifwhere
11 11
denotes the norm ofoperators
of£(D A’ H),
there exists aunique
solution to theproblem (3.1).
PROOF. Consider Thanks to our assump-
tions it is easy to check that
Moreover:
(G’
denotes the derivative of G as amapping
from R intoC(DA, H),
the fact that g, gt
belong
to.L~ ( (~--
oo,0) ; .H)
is clear from estimates similar to those that we will make below for h andht). Hence,
itresults from Theorem 1 that there exists a
unique n
such that:If we can prove that the map v --~ u has a
unique
fixedpoint
inZ~ ( (-
C>O0) ; J9~)y
we will be done. Forthis,
remark that ifE
LC:( -
oo,0); D,~)
and if UI, ’U2 denote thecorresponding
solutionsof
(3.3)
one has:Since
one has in H:
and
similarly
Hence
and thus
by (2.16)
weget
thatThus if
(3.2)
holds the map v 2013~ is a contraction from-~((2013
oo,0 ); DA)
into itself and the result follows.
REMARK 1. Provided we make suitable
assumptions
it is clear that it would bepossible
to handle a nonlinear termG(t - s, u(s))
insteadof
G(t - s) u(s)
in theintegral (3.1).
4.
Applications.
Let u be the solution to
(1.1), (1.2).
The framework of Theorem 2 is recovered if one sets:We assume here that the are smooth and
satisfy:
Moreover,
we suppose(with
the summationconvention):
where [$[
denotes the Euclidean norm of the itstranspose,
and a a
positive
constant. Note that(4.1) guarantees
thesymmetry
of a.
The
density
of thebody, ~O (x),
satisfies:It is well known from the
regularity theory
ofelliptic systems, (see
[16], [17])
thatNow,
if we assume that the are smooth one has:Thus,
from(4.4),
it is clear that if we set:we define that way a smooth map from
D.,4
into H and(1.1), (1.2)
can be
though
as(3.1). Moreover,
one hasclearly
for some constant Cand
I
denotes the norm in .g :(The ’
denotes the derivative ins).
Sol
if we set:and if we assume
where
O(e, a)
is some constantdepending
on e, a and on the normchosen on
DA (see
the formulae above for110(s)1I 11
and110’(8)11)
it isclear that Theorem 2
applies
and so weget
existence anduniqueness
of a solution to
(1.1), (1.2). (In fact,
due to ~o, one would need aslight
variant of what we
proved, replacing
uiiby
ouii in the L2 frameworkor more
generally Luii
where L is a linearoperator
in the H-frame-work. This doesn not create any
special difficulty.
Forinstance,
inthe case
where o
is a constant it isenough
to divide(1.1) by e
beforeto solve the
problem
and then argue on instead ofA).
REMARK 2.
Apparently
the condition(4.5 )
seems difficult to match.In
fact,
if G issmooth,
then(4.5)
holds f or vlarge enough (by
theLebesgue
convergencetheorem).
Assume that
f
is smooth and satisfies for some 0then,
for vlarge enough, (4.5)
holds(see
Remark2)
andf
satisfiesthe
assumptions
of Theorem 2.By
theuniqueness
of u we must haveand
(1.1 ), (1.2)
isequivalent
to(1.3), (1.2) together
with the initialcondition
Thus,
in this case, existence anduniqueness
of a solution to the initialboundary
valueproblem (1,.3 ), (1.2), (4.6)
is obtained as aparticular
case of our results.
(Recall
that we assume here thatf
issmooth).
Conversely,
iff
is extended to 0 for and if its extension is smooththe solution u to
(1.3), (1.2), (4.6)
extendedby
0 fort ~ to
satisfies(1.1), (1.2).
REMARK 3. In the above
analysis
it could bepossible
to writethe solution of
(3.1) directly
in terms of thesemi-group
associatedwith this
hyperbolic equation (and
obtain a formula like in(2.9))
this, however,
would introduce some technical refinements that wehave tried to avoid here.
Also,
it is not clear that thisapproach
would be much shorter nor more
general.
Recall also(see
Remark1)
that our method enable us to treat some mild nonlinear cases.
Acknowledgements.
This paper was written when the first authorwas
visiting
theUniversity
of Trento. We thank this institution for itssupport.
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Manoscritto pervenuto in redazione il 30 ottobre 1989.