R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
Y OUCEF A MIRAT D IDIER B RESCH J ÉRÔME L EMOINE
J ACQUES S IMON
Existence of semi-periodic solutions of steady Navier-Stokes equations in a half space with an exponential decay at infinity
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Navier-Stokes Equations in
aHalf Space
with
anExponential Decay at Infinity.
YOUCEF AMIRAT - DIDIER BRESCH
JÉRÔME
LEMOINE -JACQUES SIMON (*)
ABSTRACT - We prove the existence of a solution of
steady
Navier-Stokes equa- tions in ahalf-space
which isperiodic
in the directionsparallel
to the boun-dary
and whichdecays exponentially
fast with respect to the distance to theboundary.
In fact, thedecay
holds as soon as the energy over aperiod
isfinite.
1. Introduction.
Our
objective
is to prove the existence of a solution( u , ~ )
of thesteady
Navier-Stokesequations
in the half space
RE
which isperiodic
withrespect
to x, and x2, and whichdecay exponentially
fast as well as all its derivatives as X3 ~ 00.This
study
has been motivatedby
theanalysis
of theasymptotic
be-haviour of
hydrodynamic drag
of aplate
coveredby periodic asperities,
when their size goes to 0. We refer to
[3], [4]
in the case of Stokes flow.In a first
part,
we will prove the existence of a solution whose(*) Indirizzo
degli
AA.: Laboratoire deMath6matiques Appliqu6es,
CNRSUMR 6620, Blaise Pascal
University
(Clermont-Ferrand 2), 63177 Aubi6re cedex,France.
energy is finite
by
a contradiction method due to J.Leray [10],
see also C. Amick
[2],
W. Borchers and K. Pileckas[5].
In a second
part,
we will prove theexponential decay
of every such solutionby
atechnique
similar to the one in K. A.Ames,
L. E.Payne [1],
G. P. Galdi[6],
C. 0.Horgan,
L. T. Wheeler[8],
0. A.Ladyshenskaya,
V.A. Solonnikov
[9]
for Navier-Stokesequations
in a semi-infinite channel with Dirichlet condition on the lateralboundary.
There is an additionaldifficulty here,
which is that we can nolonger
use the Poincardinequali- ty
to bound theintegral f|u|2
over any crosssection
of thechannel,
I t
since u does no
longer
cancel on theboundary
when the Dirichlet condi- tion isreplaced by periodicity.
We will overcome thisdifficulty by
esta-blishing (Lemma 5)
a convenient estimate of the mean value of uover 2 t,
which allows us to use the
Poincare-Wirtinger inequality
instead of the Poinear6 one.2. Functional spaces and notations.
In all the paper
long,
for vector valued functionswe denote
the vector and scalar valued functions
respectively
definedby
We denote
and,
For any subset of
W,
we denote6D(Q)
the space of eoo functions withcompact support
in S~ and(J)’ (Q)
the space of distributions on Q.For 1 - r - ~o, we denote
r(S~) _ Iv
w E(Lr(S~))3 ~,
H1(S~) = W1~ 2(S~)
We define the space of functions which are
periodic
withrespect
to x, and x2 withperiods L1
andl2 by
equipped respectively
with the norm of andWe define similar spaces and
by replacing
in theabove definitions
RE e R :
0 X3~},
and we defineWper
1/r,as the trace spaces of
Finally,
we denote3. Main results.
All this paper
long,
we assume that theprescribed velocity
y on theboundary 2:
isperiodic
and that its flux across E cancels. Moreprecisely
Our first result
gives
the existence of aperiodic
solution in the whole space with a finiteenstrophy
over the«period»
O.THEOREM 1. There exists a soLution
of (1) satisfying
Our second main result
gives
theexponential decay
of such a solutionand of all its derivatives.
THEOREM 2. Let
(u, p) satisfy (1), (2)
and(3).
there existERg and P.
E R such thatfor
alt x such thatx3 ~
1 andfor
allI a I
~0 ,
zvhere c and a arepositive
numbers
depending only on l,
y , v , a and on anyarbitrary
bound Eof
.1 This means
that, given E, (4)
is satisfied for all(u, p)
such that~
It follows from
(4)
that u, p and all their derivatives areuniformly
continuous in O
t for any t
> 0. The uniformcontinuity
up to the boun-dary E,
that is the case t =0, requires,
inaddition,
y to be smooth.Moreover, by integration (4) gives,
for all m ~0,
1 ~ andt > 0,
with a norm bounded for all where
In
addition,
andsince the
enstrophy
is finite(for fluids,
the energy is is named theenstrophy).
- -
Finally,
let us remark that(u., p.)
isobviously unique
for agiven (u, p).
Itsatisfies, c, f. (32),
4. Proof of Theorem 1.
We will use the
following
twolemmas,
on the existence of a solenoidal extension of y and on Eulerequation,
which areproved
at the end of this section.LEMMA 3. There exists such that
LEMMA 4.
(i)
Let b EVm satisfy, for
all v EV~,
Then there exists such that
Moreover q is constant on Z and
(ii)
Let b E(Hper,
loc(e»3 satisfy V.
b =0, b 11:
= 0and, for
all m andfor
all v EVm ,
Then there exists such that
Moreover q is constant on E. 0
FIRST STEP: WEAK SOLUTION. First of
all,
let usremark, setting w
== u - h where h is defined
by
Lemma3,
that theequation (1)
isequivalent
to
We will find a weak solution w of these
equations,
which means that itsatisfies
(6),
and, denoting ~
the extensionby
0 for for allv E Vm ,
for allm,
At
first, given
aninteger
number m ~1,
we will prove the existence of anapproached
solution wm EVm
defined on the bounded setll m
suchthat,
for all
ve Vm,
Then,
we will see that remains bounded.Finally,
we willget
asolution w of
(6)-(8) by passing
to the limit on the solutions wm . Existenceof
a weak solution wmof (9).
To proveit,
we argue as in J.Leray [10] (see
also C. Amick[2],
W. Borchers and K. Pileckas[5]).
Thespace
V m being equipped
with the innerproduct
a
compact operator Nm
fromVm
into itself and a functionFm
EVm
are de-fined by
Then, by
the Rieszrepresentation theorem,
the variationalequation (9)
is
equivalent
to thefollowing equation
inVm :
To prove the existence of a solution wm of
(10),
we use theLeray-
Schauder
principle,
that is we have to prove that the set of allpossible
solutions of the
equation
is bounded in
V~ .
In order to concludeby contradiction,
let us suppose that there exists a sequence 1 in[ o , v -1 ] converging
toe[0,
and such thatDenoting
weget,
since for all s, the existence of asubsequence,
still denoted( b s ), converging weakly
inVm
to some b.Let us check that b satisfies the two
contradictory equations
The
equation (11)
for Agives,
for allMoreover,
for allv, w and z
inVm,
since V.w = 0 =0,
and then
Thus, choosing
weget
Therefore, letting
s - m andusing
in(L6(Am»3,
we obtain(12).
On the other
hand, dividing (14) by
andpassing
to the limitwith
respect to s,
we obtainBy
Lemma4,
we deduce that thereexists q
EW~r ~2 (~l m )
such that( b , q)
is solution of the Euler
equation (5). Therefore, multiplying (5) by h and integrating
overAt’
weget
Since q is constant on Z and on and we have
Reminding
that for all and we havewhere
8latA m = a~ m ~(~ U ~ m )
is the lateralboundary.
Thus theright-
hand side of
(17) vanishes,
which proves(13).
Thus,
since(12)
and(13)
arecontradictory,
the set of allpossible
solu-tions
w ~
of(11)
is bounded inVm
and weobtain, by Leray-Schauder prin- ciple,
the existence of a solutionWm e Vm
of(9).
Uniform
boundfor
Weproceed,
aspreviously, by
a con-tradiction
argument.
Let us suppose that there exists m such thatDefining
b ’ =llw,,, 11 -’w,,,,
we have = 1 for all m -> 1. There-fore, denoting
the extension ofby
0 for~3 ~
m, we deduce the exi- stence of asubsequence
denotedagain
such thatLet us check that b satisfies the two
contradictory equations
and
Since Wm
satisfies,
for allv e V m,
choosing
weobtain, using (16)
and(15),
and
therefore,
Passing
to the limit in thisequation,
since h has acompact support,
weobtain
(18).
Now, dividing (20) by extending
it on O andpassing
to thelimit with
respect
to rrz, we obtainUsing
Lemma 4 andarguing
aspreviously,
we deduce(19). Thus,
is
uniformly
bounded withrespect
to m.Existence
of
a solutions wof (6)-(8).
Letwm
be the extension of wmby
0 for m. Its
gradient
remains bounded in(L 2 ( O ) )9,
then there exist and asubsequence
still denoted(wm)
such thatas m - oo. The
equation V.wm
= 0gives V. w
=0,
and theperiodicity
ofium gives
the one of w, and thus w E(H I,, loc ( 0 ) )3·
Theequation (9)
is sa-tisfied for any v e
V~ provided
thatmo
m, then weget (8) by passing
tothe
limit,
moremaining
fixed.SECOND STEP: STRONG SOLUTION. Existence
of
p.Setting
u = w +h,
andchoosing v E ( ~( O ) )3
such theequation (8) yields
Then, by
de RhamTheorem,
see for instance[12],
thereexists p
E6D’(i9)
such thatSince u E
(Hper, loc(H))3,
thisequation gives Vp
e(H-1(At))3
for every fi-nite t,
whichimplies (see
forexample [13]
or[14])
that since~l
t is
aLipschitz
bounded domain.Extension
of
p.By
definition ofHper,
the function u is defined in the whole setRE ,
but p isonly
defined in the set 0. Itsgradient
has aperio-
dic extension due to
equation (1),
but this does not ensure the existence of aperiodic
extensionof p
itself.By (8), u satisfies,
for all v EV~
and forall m ~
1,
The factions and v
being periodic, is perio-
dic. Its
integral
over theperiod
0being
nullby (21),
so does itsintegral
over any translation of
0,
and inparticular
overChoosing
now such that V. v =0,
theequation gives
the existence of such thatin e’. In O n
O ’,
thegradient
of p -p ’ cancels, therefore p - p’
is con-stant.
By
addition of this constant top ’,
we obtain an extensionof p
which satisfies the
equation
in e U O ’ .By repeated extensions,
we ob-tain an which satisfies
equation (1)
in the whole setPeriodicity of p.
We denote Ph the translated function of pby II
in thedirection Since u is
periodic, (21) gives
nowp)
=0,
hence PZ1 -- p is
constant,
say c1.Moreover,
theregularity
results for Navier-Stokesequation (see [7]
or[15]
forinstance)
show that uand p
are smooth. Letus
multiply
theequation (1) by
a function v such thatwhere 0 r t oo.
Integrating by parts
over8,
weget
where n is the outer normal. The
integral
over 0 cancels due to the varia- tionalequation (21).
In theintegral
over3e,
theintegral
over the lowerboundary E
vanishes since vcancels,
and theintegral
ofau I an. v
overtwo
opposite
lateral boundaries cancels sinceau I an. v
isantiperiodic.
Fi-nally,
theequation
reduces towhere
Now,
we choose whereThen,
this reduces toThis proves that the constant value of is 0 and therefore
that p
isperiodic
withrespect
to xl .Similarly,
it isperiodic
withrespect
to X2, and theproof
of Theorem 1 iscomplete.
PROOF OF LEMMA 3. Let be such that 77 = y =
= 0 on E1. Since
is aLipschitz domain,
there exists(see
for instance[6]
Theorem 3.1 p.127) v E (Ho (~l 1 ) )3
such thatLet us denote
vper
theperiodic
extension of v over the horizontal direc- tions. the desiredproperties
are satisfiedby
choosing
PROOF OF LEMMA 4. Part
(i).
Let b eVm satisfy
By
de Rham Theorem(see
forexample [12]),
there exists such thatThis
equation gives
whichimplies (see
forexample [13]),
sinceAs
is aLipschitz
boundeddomain,
that q EWl, 3/2 (A m).
Using
the samearguments
than in theproof
of Theorem1,
we deducethe existence of a
periodic
extension which satisfies(22)
in the whole
0 ~3 ~}.
Let us now prove
that q
is constant on X andon Zm (the
constants can bedifferents).
Thisproperty
has been observedby
J.Leray [10]. Here,
we
adapt
theproof given by
Ch. J. Amick[2]
in the two dimensional case.We
have,
for all 0 E m,Since b = 0
on Z,
we havethen, using Hardy’s inequality (see
for instance Lemma 5.1 p. 94 in[11]),
we
get
Therefore,
since x3 E, we obtainBy
thePoinear6-Wirtinger inequality
the mean value4(t) over Z
satis-fies,
for almostall t,
and therefore
By (22)
and(23),
theright
hand side goes to 0 as E - 0 and therefore the tracesatisfies I q - Q( 0 )
= 0 which provesthat q
is constant on E.I
Similarly, q
is constanton
since b vanishes on it.Part
(ii). Now,
b is defined in the whole set and it vanishes on Xthen, using
the samemethod,
weget
aperiodic
function q in which is constant on Z..5. Proof of Theorem 2.
We will prove
(Lemmas
5 and6)
that and thereforeu(t)
go to a limit Then we will prove(Lemma 7)
anidentity
whichimplies (Lem-
ma
8)
theexponential decay of Finally,
thepointwise decay
wille
be deduced from
regularity properties
for Stokesequations given
atLemma 9.
In all the
sequel, (u, p), being
a solution of Navier-Stokesequations,
is smooth and therefore any of its derivatives is
pointwise
defined andintegrable
over every bounded domain or surface.For t2 ~ tl,
we denote that isWe will use the mean values over I and over the cross section t which are defined
by
The
Poinear6-Wirtinger inequality gives,
for anyr ~ 6,
At
first,
let us that 11 goes to a limitLEMMA 5. Let
(u, p) satisfy (1), (2)
and(3). Then, for all t,
and there exists a
unique
uooeRg
suchthat, for
all t >0,
PROOF.
Proof of (26).
Since u isperiodic
withrespect
to Xl and x2, for allpositive t
we haveTherefore, V . u = 0 implies
which may be written as
Then,
which proves
(26).
Proof of (27).
Since V - u =0,
the nonlinear term in(1)
may be writtenas
Then, integrating
the first andsecond
components
ofequation (1)
andusing
theperiodicity
of uand p
with
respect
to x, and X2, weobtain,
for i =1 or2,
Thus,
there exist real numbers Cl and c2 suchthat,
for all t >0,
Thanks to
(26), thus, by (25),
and therefore
Integrating
from 0 to t andusing
weget,
for all t >0,
which
implies
ci = 0.Now, according
to(28),
we may writeUsing (29),
weobtain,
for all t >0,
Therefore Tij
possesses a limit which satisfiesSince
u3 = 0,
thisgive (27).
LEMMA 6. Let
(u, p) satisfy (1), (2)
and(3). Theny , for
all t >0,
PROOF. The
inequality (27) implies
Therefore, by
thePoinear6-Wirtinger inequality (24),
which proves
(30).
On the otherhand, (27)
for t = 0gives
from which we obtain
(31).
From now, we will denote c various
positive
numbersdepending
atmost
on l, v,
y and E. Let us prove thefollowing identity.
PROOF. We remark that
(u, p)
is a solution ofMultiplying by u - integrating over ~
andusing
the identitiesfor all smooth vectors fields and w such =
0,
weget
Since
( u , ~ )
isperiodic
in the directions x, and x2 and =0,
it followsthat
Then, integrating
from T toi’,
weget,
for 0 ~ Ti’,
where
Due to the
hypothesis (3),
it follows from(35)
thatF( t )
has a limitF ~
as whichsatisfies,
for all i >0,
To
get
thelemma,
it remains to prove thatF ~
= 0. Forthis,
let us inte-grate
the both sides of(36)
from t to t + 1. Weobtain,
for t ~0,
Let us prove now that each term of the
right-hand
side of(37)
goes to 0 as t -~ ~ . Atfirst, by assumption (3),
thusFor the second
term, using (30),
we boundFor the third
term,
we remark thatUg
= 0 since~3=0, thus, using
the
Poincaré-Wirtinger inequality (24)
and(30),
we boundTo estimate the last term in
(37),
we will use thePoincaré- Wirtinger inequality
Using again Û3
= 0 and(24),
we boundLet us now
give
an estimate forbeing equipped
with the normfrom the
equation
we deduce thatBy
the Sobolev and H61derinequalities,
we boundThus, using (41), (42)
and(31),
weget
From
(3),
therefore(38), (39), (40)
and(43)
im-ply
that each term in theright-hand
side of(37)
goes to 0 as t --~ ~ . Con-sequently F ~
= 0 and(33)
holds.Before
going further,
let usimprove (40)
for a later use, as follows:Indeed
therefore
since is constant on and
Thus,
Integrating (27)
we boundthus, by (32), Then, using (31),
weget
Therefore, (44)
followsby
thePoinear6-Wirtinger inequality (24)
for r == 2 and 4.
LEMMA 8. Let
(u, p) satisfy (1), (2)
and(3). for
all t >0,
where c and a 1 are
positive
numbersdepending only
onl ,
y , v and E. oPROOF.
Integrating
the both sides of theidentity (33)
weget,
fort ~ 0 and
Using
thePoinear6-Wirtinger inequality on 2t
t(25)
and(27),
weobtain
Using respectively (43)
and(44)
on andsumming
over n from 0to
m -1,
weget
Then
(46) gives,
as m --~ ~ ,Setting
this
gives,
since for all t > 0:This
inequality
andy(t)
---> 0 asimply, by
a lemma of[8],
see[6]
Lemma 2.2 p.
315,
that is
(45).
Let us now focus on a
regularity property
ofperiodic
solutions ofStokes
equations.
LEMMA 9.
a solution
of
and let 0 d ~ 1 and s ~ d.
Then, for
all m ~ 0 and 1 q oo ,From now, the numbers c may
depend
on m, in additionto l , v ,
y and E.PROOF. We will deduce this
property
from a similar one for Dirichletboundary condition,
see[6], by
localization. We first note that it suffices to prove(48)
for s = 1 since thegeneral
result followsby
the translationGiven a
positive integer k,
we denoteWe consider a localization function such that in with supple
Vk
and we denotevk = ~ k v, ~ k = ~ k ~ .
Thenwhere
Since
(vk, ~ k)
satisifies the Dirichletboundary
conditions for any smooth domainQ,
such thatsupp yk c Q c Vk,
the Theorem IV.6.1 of[6]
gives
The definitions
of fk and gk imply
Since the statement of the lemma is
unchanged
if anarbitrary
constantis added to
0,
we can assumeand therefore the
equation (47) gives
Since
f
and v areperiodic
withrespect
to x, and xz,Finally, using (49) gives
By
arepeated
use, thisequation gives
which proves
(48)
for s = 1 sinceand,
m +1 /b,
We are now in a
position
to prove oursecond,
andlast,
mainresult.
PROOF OF THEOREM 2. Bound on u in
(H 2 (ll t + 1 ) )3,
Atfirst,
let us check thatThe
inequality (48)
for m =0, q
=3/2
and 6 =1/2 gives
By
H61derinequality,
for all t ~0,
which is bounded
by (3)
and(31),
thus theright-hand
side of(51)
is bounded for all t ~1 /2. Therefore, by
Sobolevtheorem,
and thus
I Using
now(48)
for m =0,
q = 2and 6 = 1/2,
we
get (50).
Bound on u in
(H m (~l t + 1 ) )3
Due to theequation (34),
Lemma 9gives
For m ~
2, H’n (~l t ± a + a )
is amultiplicative algebra,
thusSince
I
is boundedby (32),
weget,
for 0 ~ s ~1,
By m repeated
use of thisinequality
with 6=1 /m,
we deduce from(50)
thatExponential decacy of u -
~ooNow,
theprevious inequality yields,
form > 2,
This
inequality
also holds for m = 0 and 1since, by
Sobolevtheorem,
c which
gives
a convenient bound for the nonlinear term in theright-hand
side of(52). Then, by m
+ 1repeated
use with now 3 ==1 /( m + 1),
weget
By (30)
theright-hand
side is boundedby
andby (45)
it isbounded
by c exp ( - cr 1 t/2 ). Thus, by Sobolev theorem,
Exponential decay By
Lemma9,
the pressure term be added to the left-hand side of(52),
and thereforeof
(53). Thus,
This
inequality
for m=0implies
the existenceof poo E R
such thatwhich ends the
proof
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