Numerical analysis of a penalization method for the three-dimensional motion of a rigid body in an incompressible viscous fluid

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by  Claire Bost

Numerical analysis of a penalization method for the
three-dimensional motion of a rigid body in an
9 incompressible viscous fluid
0
0
2
C. Bost, G.-H. Cottet and E. Maitre

n
Universit´e de Grenoble and CNRS, Laboratoire Jean Kuntzmann
a
J BP 53, 38041 Grenoble Cedex 9, France

4
1 January 14, 2009

]
P
A Abstract
. We present and analyze a penalization method wich extends the the method of [2] to
h
the case of a rigid body moving freely in an incompressible fluid. The fluid-solid system is
t
a viewedasasinglevariabledensityflowwithaninterfacecapturedbyalevelsetmethod. The
m solid velocity is computed by averaging at avery time the flow velocity in the solid phase.
[ This velocity is used to penalize the flow velocity at the fluid-solid interface and to move
the interface. Numerical illustrations are provided to illustrate our convergence result. A
1
discussion of ourresult in thelight of existing existence results is also given.
v
8
5 1 Introduction
9
1
. In this paper we are concerned with the numerical analysis of a penalization method for the two-
1
way interaction of a rigid body with an incompressible fluid in three dimensions. The traditional
0
numerical approach to deal with fluid-structure problems is the so-called ALE (for Arbitrary
9
0 Lagrangian Eulerian) method where fluids (resp solids) are described in an Eulerian (resp La-
: grangian) framework. Fluids are computed on a moving mesh fitting the solids and stress and
v
velocity continuity are used to derive the appropriate boundary conditions on the fluid/solid in-
i
X terface. A convergence proof of a finite-element method based on this approach can be found in
r [18].
a
Alternate methods can be devised where the whole fluid-solid system is seen as a multiphase
flow and the fluid/solid interface is captured implicitly rather than explicitly. Likewise, the inter-
face continuity conditions are recovered in an implicit fashion. The rigid motion inside the solid
phase can be enforcedthrough a Lagrangemultiplier [11]. The method we consider here is of this
type but the rigid motion is approximately satisfied in the solid through penalization.
Penalization methods have already been considered in the past for this problem. In [6] the
authors considered a single solid ball and worked inside the frame moving with its center. Then
they penalized the mean velocity of the (virtual) fluid inside this ball. Their method is restricted
tooneball. In[19,15]thepenalizationisappliedtothedeformationtensorinsidethebody. In[19]
this method is used to prove the existence of solutions for the fluid-solid interaction variational
problem in two dimensions. In [15] it is used together with a two-dimensional finite element
method in a variational framework. Here the penalization is applied to the flow velocity itself.
The method thus extends the one devised and analyzed in [2] in the case of a rigid solid with
prescribed motion.
In our method the determination of the body velocity is part of the problem. This velocity,
insteadoftheflowvelocity,isusedtomovethesolidphase. Thishasacrucialpracticalimportance,
in particular for problems with large displacements and strong shear, since it ensures that the
solid remains rigid at the discrete level, although the rigidity constraint in the flow field is only
1
approximately satisfied. A vorticity formulation of the method and its validation on a number of
2D and 3D reference cases are given in [7]. An outline of the paper is as follows. In section 2 we
recallthe weak formulationof the problem and we describe the penalization method. Section 3 is
devoted to the convergenceproof. In section 4 we provide some numericalillustrations. Section 5
is devoted to some concluding remarks. The proofs of some technical results used in section 3 are
given in the appendix.
2 Weak formulation and penalized problem
Let Ω be an open bounded domain of R3, filled with a viscous incompressible and homogeneous
fluid of density ρ > 0 and viscosity µ > 0. Inside this domain, we consider the motion of an
f
immersed homogeneous rigid solid of density ρ > 0 during a time interval [0,T], T > 0, chosen
s
sothatthe solidnevercomes incontactwith∂Ω. Fort [0,T],we denoteby Ω (t)andΩ (t) the
f s
∈
non-emptyfluidandsolidopenconnecteddomains,withΩ (t) Ω (t)=Ω andΩ (t) Ω (t)= .
s f f s
∪ ∩ ∅
The center of mass of the solid is denoted by x (t), its mass and inertia tensor by M and J(t).
G
Without loss of generality we assume that M =1. Then
x (t)= ρ xdx, J(t)= ρ (r2I r r)dx
G s s
− ⊗
ZΩs(t) ZΩs(t)
where r(x,t)=x x (t). The system is subject to a body density force g (usually gravity).
G
−
2.1 Weak formulation
The basic formulation of this fluid-solid coupling is the following : given initial conditions,
x′ (0)=v0, ω (0)=ω0, u=u0, Ω (0)=Ω0 (1)
G g u u s s
supplemented with
x (0)= xdx, X (x,0)=x, (2)
G s
ZΩ0s
find t Ω (t) and (x,t) (u(x,t),p(x,t)) solution for t>0 of
s
→ →
ρ (u +(u )u) 2µdiv(D(u))+ p=ρ g on Ω (t), (3)
f t f f
·∇ − ∇
divu=0 on Ω (t), (4)
f
u=0 on ∂Ω, (5)
u=x′ +ω r on ∂Ω (t), (6)
G u× s
x′′(t)=g+ (Σn)ds, (7)
G
Z∂Ωs(t)
J(t)ω′(t)= ω (t) (J(t)ω (t))+ ρ (r g)dx+ r (Σn)ds, (8)
u − u × u s × ×
ZΩs(t) Z∂Ωs(t)
Ω (t)=X (t,Ω0), (9)
s s s
∂X
s =x′ (t)+ω (t) r(X (t),t), (10)
∂t G u × s
where n denotes the unit outward normal on ∂Ω (t), and Σ is the fluid stress tensor.
s
In this formulation the last two equations describe the rigid motion of Ω (t). In order to give a
s
weak formulation of this problem, let us introduce some function spaces. From now on, u will
denote the velocity field on the whole computational domain Ω. We define
= u H1(Ω),divu=0 , = u L2(Ω), divu=0, u n=0 on ∂Ω
V { ∈ 0 } H { ∈ · }
2
and, with the notations of [19] extended to the three dimensional case,
(t)= u , D(u)=0 in Ω (t)
s
K { ∈V }
= u , (V ,ω ) R3 R, u=V +ω r in Ω (t) .
u u u u s
{ ∈V ∃ ∈ × × }
Next we define a density on the whole domain by setting ρ = ρ χ + ρ χ , where χ
s Ωs(t) f Ωf(t) A
denotes the characteristic function of set A, which takes value 1 inside A and 0 outside. Let
us note Q = Ω ]0,T[. Then the weak formulation is the following [14]: given initial conditions
×
H0 =χ ,ρ0 =ρ H0+ρ (1 H0)andu=u0 (0), find (x,t) (ρ(x,t),u(x,t),H(x,t)) such
Ω0 s f
s − ∈K →
that
u L∞(0,T, ) L2(0,T, ), H,ρ (0,T;Lq(Ω)) q 1,
∈ H ∩ V ∈C ∀ ≥
u(t) (t) for a.e. t ]0,T[, with Ω (t)= x Ω, H(x,t)=1 ,
 ∈K ∈ s { ∈ }
∀∀ψξ ∈∈ZHCΩ11[((ρQQu))·,∩∂ψtLξ(T2+()0(=,ρT(u0;,K·∇(t))u),−2µD(u)):D(ξ)+ρg·ξ] dx= ddtZΩρu·ξdx, (11)
T ∂ψ
H +Hu ψdxdt+ H0ψ(0)dx=0,
Note that weZZc00oTuZZldΩΩρeq∂∂u∂ψtitv+aleρnutl·y∇·h∇ψavdexddte+finZeΩdZΩρρ0ψ=(0ρ)Hdx+=ρ0.(1 H), as ρ0 is piecewise constant,
s f
−
and transported by the same velocity field than H.
2.2 Penalized problem
For η >0, we consider the following penalized problem:
given (ρ (0)=ρ0,u (0)=u0,H (0)=χ ), to find (ρ ,u ,p ,H ), with
η η η η Ω0s η η η η
ρ ,H L∞(]0,T[ Ω), u L∞(0,T; ) L2(0,T; ), p L2(Q)
η η η η
∈ × ∈ H ∩ V ∈
solution on Q of
∂u 1
η
ρ +(u )u 2µdiv(D(u ))+ p + ρ H (u u )=ρ g (12)
η η η η η η η η η,s η
∂t ·∇ − ∇ η −
(cid:18) (cid:19)
divu =0 (13)
η
1
u = ρ u H dx+ J−1 ρ (r u )H dx r (14)
η,s M η η η η η η× η η × η
η ZΩ (cid:18) ZΩ (cid:19)
ρ +u . ρ =0 (15)
ηt η ∇ η
H +u . H =0 (16)
ηt η,s ∇ η
We set Ωη(t) = x Ω, H (x,t) = 1 . In equation (14) we divided the first term by M =
s { ∈ η } η
ρ H dx, which is not constant in time in general. On contrary we have Ωη(t) = H dx=
Ω η η | s | Ω η
Ω0 since u is divergencefree and H vanishes on∂Ω (we assumedno contactof the solid with
|R s| η,s η R
∂Ω). The inertia tensor is defined as
J = ρ H (r2I r r )dx= ρ (r2I r r )dx.
η ZΩ η η η − η⊗ η ZΩηs(t) η η − η ⊗ η
with r =x x =x ρ H xdx.
Fora η R3 −0 G,aηTJ a−= Ω η ρη r a2dx min(ρ ,ρ ) r a2dx(seeestimate(18)).
∈ \{ } η R Ωηs(t) η| η× | ≥ s f Ωηs(t)| η× |
This last quantity being strictly positive for an open nonempty integration set, J is nonsingular
R R η
(we recall that Ωη(t) = Ω0 >0).
| s | | s|
3
Before stating our convergence result, a few remarks are in order.
Firstone maywonderabout the well-posednessof the aboveproblem. Howeverit willdirectly
result from the a priori estimates and convergence arguments given in the following that this
problemdoeshaveatleasta weaksolution. Indeed, these argumentscouldeasilybe usedto show
the convergence of the solutions to a linearized version - or finite-dimension approximation - of
(12)-(16). Next we can observe that in this model we penalize the difference between u and the
η
projectionof u onto velocity fields rigid in the solid domain, namely u (see lemma 3.1 below).
η η,s
The density is transported with the original velocity field so that estimates on the Navier-Stokes
equations are easier to obtain. The characteristic function is transported by the rigid velocity
so that the shape of Ωη(t) remains undeformed (this is exactly the Eulerian counterpart of (9-
s
10)). As observed in [7] this has a practical importance (in particular it means that the rigid
solid can be recoveredexactly through simple algebra from its initial shape). As far as numerical
analysisis concerned,it alsoprovides”forfree” regularityproperties onthe computed rigidbody,
as soon as the initial body is smooth. The price to pay is that the level sets of ρ and H do not
η η
coincide, i.e. in generalwe do not haveρ H =ρ as in the non penalized formulation. Note also
η η s
that in principle we should prescribe a boundary value for H on ∂Ω when u is inward. Since
η η,s
our analysis is restricted to times when the solid body does not approach the boundary of the
computational box, we can take this boundary value to be zero, which amounts to solve (16) on
Rn and take its restriction to Ω.
Inthefollowingsectionswewillprovetheconvergenceofatleastasubsequenceof(ρ ,u ,p ,H )
η η η η
to the weak solution defined above. Next section starts with some a priori estimates which will
provide weak convergence of subsequences. In section 3.3 we will have to use more sophisticated
tools adapted from [19] to get some strong convergence in u which will allow us to pass to the
η
limit in nonlinear terms of (P ). More precisely we prove the following result.
η
Theorem 2.1. Under the regularity assumptions of section 2, let (ρ ,u ,p ,H ) a solution of
η η η η
(P ). Then there exists a subsequence of (ρ ,u ,H ) and functions (ρ,u,H) such that
η η η η
ρ ρ, H H strongly in C(0,T;Lq(Ω)) for all q 1,
η η
→ → ≥
u u strongly in L2(Q) and weakly in L2(0,T;H1(Ω)) L∞(0,T;L2(Ω))
η → 0 ∩
and such that (ρ,u,H), is a solution of (11).
Before proceeding to the proof, let us point out a few remarks. For a sake of simplicity in the
notations we have stated our penalization method and theorem for a single rigid body. It will be
apparent from the proof below that it readily extends to the case of severalbodies. Furthermore,
the time to which the convergence result is restricted, is essentially the time for which contact of
the rigidbody do not touchthe boundary of Ω (in the case of severalbodies it wouldbe the time
on which we can ensure that contact between bodies do not happen). As a result if we consider
periodic boundary conditions and a single body convergenceholds for all times.
3 Proof of theorem 2.1
Thefollowinglemmastatesthatu ,asdefinedin(P ),istheprojectionofu ontovelocityfields
η,s η η
which are rigid on Ωη(t).
s
Lemma 3.1. Let ξ be a rigid velocity field, i.e. such that ξ(x)=V +ω r(x) for some constant
ξ ξ
vectors V R3 and ω R3. Then if u is defined by (14) there holds×
ξ ξ η,s
∈ ∈
ρ H (u u ) ξdx=0. (17)
η η η η,s
− ·
ZΩ
Moreover, the result holds if ξ is a time dependent velocity field rigid in Ωη(t) at time t.
s
4
Proof. Let the mean translation and angular velocities be defined as
1
V = ρ H u dx ω =J−1 ρ H (r u )dx
u M η η η u η η η η× η
η ZΩ ZΩ
then
ρ H (u u ) ξdx = ρ H [u (V +ω r )] [V +ω r ] dx
η η η η,s η η η u u η ξ ξ η
− · − × · ×
ZΩ ZΩ
= V ρ H u dx+ω ρ H (r u )dx V V ρ H dx
ξ η η η ξ η η η η u ξ η η
· · × − ·
ZΩ ZΩ ZΩ
V ω ρ H r dx V ω ρ H r dx
u ξ η η η ξ u η η η
− · × − · ×
(cid:18) ZΩ (cid:19) (cid:18) ZΩ (cid:19)
ρ H (ω r ) (ω r )dx
η η u η ξ η
− × · ×
ZΩ
= V (M V )+ω (J ω ) V (M V )
ξ η u ξ η u u η ξ
· · − ·
V ω ρ H r dx V ω ρ H r dx
u ξ η η η ξ u η η η
− · × − · ×
(cid:18) ZΩ (cid:19) (cid:18) ZΩ (cid:19)
ρ H (ω r ) (ω r )dx.
η η u η ξ η
− × · ×
ZΩ
As (ω r ) (ω r )=(ω ω )r2 (r ω )(r ω ), we have ρ H (ω r ) (ω r )dx=
u× η · ξ× η ξ· u η− η· ξ η· u η η u× η · ξ× η
ZΩ
ω (J ω ).
ξ η u
·
Finally, by definition of r , ρ H r dx=0, and we get
η η η η
ZΩ
ρ H (u u ) ξdx = ω (J ω ) ω (J ω )=0.
η η η η,s ξ η u ξ η u
− · · − ·
ZΩ
3.1 Estimates for transport and Navier-Stokes equations
In all the sequel, C denotes a positive constant. At this stage, we consider a given time interval
[0,T]. The value to which T must be restricted will be given later in this section.
Standard estimates for transport equations (15) and (16) show that ρ and H are bounded
η η
in L∞(0,T;L∞(Ω)). More precisely, for all time t [0,T],
∈
ρ :=min(ρ ,ρ ) ρ (x,t) max(ρ ,ρ ) H (x,t) 0,1 a.e. x Ω. (18)
min s f η s f η
≤ ≤ ∈{ } ∈
Thus, up to extracting a subsequence, we can assume that
ρ ⇀ρ in L∞(0,T,L∞(Ω)) weak*, (19)
η
and
H ⇀H in L∞(0,T,L∞(Ω)) weak*, (20)
η
where H and ρ satisfy the bounds (18). We set Ω (t) = x Ω, H(x,t) = 1 . Concerning
s
{ ∈ }
Navier-Stokes equations, multiplying (12) by u and integrating on Ω, we get:
η
∂u
η
ρ +(u )u u dx 2µ div(D(u )) u dx+ u p dx
η η η η η η η η
∂t ·∇ · − · ·∇
ZΩ (cid:18) (cid:19) ZΩ ZΩ
1
+ ρ H (u u ) u dx= ρ g u dx.
η η η η,s η η η
η − · ·
ZΩ ZΩ
5
Classically we have from incompressibility and homogeneous boundary conditions on u ,
η
∂u 1 ∂(ρ u 2)
η η η
ρ +(u )u u dx= | | dx. (21)
η η η η
∂t ·∇ · 2 ∂t
ZΩ (cid:18) (cid:19) ZΩ
From Lemma 3.1 we get
ρ H (u u ) u dx= ρ H (u u )2dx,
η η η η,s η η η η η,s
− · −
ZΩ ZΩ
and since u is divergence free and vanishes on ∂Ω,
η
u p dx=0.
η η
·∇
ZΩ
Collecting terms we get, since from (18) H =H ,
η η
21ddtk√ρηuηk2L2(Ω)+µkD(uη)k2L2(Ω)+η1k√pρηHη(uη−uη,s)k2L2(Ω) ≤k√ρηuηkL2(Ω)kgkL∞(Q)kρηkL212(Q)
which upon time integration on [0,T] gives
2
k√ρη(t)uη(t)k2L2(Ω) + 2µkD(uη)k2L2(Q)+ ηk√ρηHη(uη−uη,s)k2L2(Q)
T
≤ k√ρη0uη0k2L2(Ω)+C k√ρη(s)uη(s)kL2(Ω)ds.
Z0
Applying Gronwall Lemma, Poincar´e inequality and bounds from (18) gives the following esti-
mates :
u bounded in L2(0,T,H1(Ω)), (22)
η 0
√ρηuη and uη bounded in L∞(0,T,L2(Ω)), (23)
1 1
√ρ H (u u ) and H (u u ) bounded in L2(0,T,L2(Ω)). (24)
√η η η η− η,s √η η η− η,s
Thus we can extract subsequences from ρ , u and H , still denoted by ρ , u and H , such that
η η η η η η
u ⇀u in L2(0,T,H1(Ω)) weak, (25)
η 0
√ρηuη ⇀χ and uη ⇀u in L∞(0,T,L2(Ω)) weak*, (26)
√ρηHηuη−√ρηHηuη,s →0 and Hηuη−Hηuη,s →0 in L2(0,T,L2(Ω)) strong. (27)
The identification of χ with √ρu results from strong convergence results proved by Lions and
DiPerna on transport equations. [9] theorem II.4, (25) and incompressibility imply
ρ ρ in C(0,T,Lq(Ω)) strong q [1,+ [ (28)
η
→ ∀ ∈ ∞
with ρ solution of
ρ +u ρ=0 on Ω ]0,T[,
t
·∇ ×
(ρ=ρ0 on Ω 0 .
×{ }
From this strong convergence we can pass to the limit in the product √ρηuη : given v
∈
Lq(0,T;Lr(Ω)) with q >2 and r > 6, we write
5
T T T
(√ρηuη−√ρu)vdxdt= (uη−u)√ρvdxdt+ (√ρη−√ρ)uηvdxdt.
Z0 ZΩ Z0 ZΩ Z0 ZΩ
FromtheinjectionofH1 intoL6 indimensionlessorequalto3thefirstintegralconvergestoward
0. Forthesecondintegralweusethestrongconvergence(following(28))of√ρη inLs forassuch
that u v is in Ls′ where s′ is the conjugate exponent of s. Thus we have
η
√ρηuη ⇀√ρu in Lq(0,T,Lr(Ω)) weak, for all q <2,r<6. (29)
6
3.2 Setting T and passing to the limit in the rigid velocity
This rigid velocity is defined by
u (x,t)=u (t)+ω (t) r (x,t),
η,s η,G η η
×
with
1
u (t)= ρ u H dx and ω (t)=J−1 ρ (r u )H dx.
η,G M η η η η η η η × η η
η ZΩ ZΩ
FirstwenotethatM = ρ H dxisboundedfrombelowindependentlyofηsinceρ isbounded
η Ω η η η
from below and H dx= Ω0 >0 does not depend on η or t. From the bounds on ρ , H and
Ω η R | s| η η
u it is straightforwardto show that
η
R
u (t) bounded in L∞(0,T).
η,G
Likewise, from the definition of J we observe that for a R3 0
η
∈ \{ }
aTJ a min(ρ ,ρ ) r a2dx>0.
η s f η
≥ | × |
ZΩs(t)
Moreover,the initialsolidis regularandtransportedby a rigidvelocity. We thus know thatthere
isaballofradiusR>0centeredonthecenterofgravityx includedintoΩη(t). Thentheabove
Gη s
estimates implies
aTJ a min(ρ ,ρ ) r a2dx=min(ρ ,ρ ) x a2dx=C(R)a2
η s f η s f
≥ | × | | × | | |
ZB(xGη,R) ZB(0,R)
with C(R)= 2R5π >0. Taking a=J−21b (J is symmetric) we get for all b R3 0 ,
15 η η ∈ \{ }
bTJ−1b= J−12b2 1 b2,
η | η | ≤ C(R)| |
which proves that each coefficient of J−1 is bounded independently of η and t. From the bounds
η
on u , H and ρ this implies that
η η η
ω (t) is bounded in L∞(0,T).
η
In particular this implies that the solid velocity u is bounded in L∞ by some constant M
η,s
independent of η and time. We can now define the maximum time for which the convergence
result will be proved. If we denote by d the initial distance between solid and the boundary ∂Ω
0
then choosing for instance T = d /2M ensures that the body will not touch the boundary for
0
t [0,T]. In all the sequel we will assume this value of T.
∈
From the above estimates we can ensure that there exists u (t) and ω(t) in L∞(0,T) such
G
that, up to the extraction of subsequences,
u ⇀u :=u +ω r in L∞(0,T,L∞(Ω)) weak*.
η,s s G
×
Now we point out that taking the gradient of the rigid velocity field u gives
η,s
0 ω3 ω2
− η η
u = ω3 0 ω1
∇ η,s  η − η
ω2 ω1 0
− η η
 
so that the u (and all subsequent space derivatives) is also bounded in L∞(0,T;L∞(Ω)). In
η,s
∇
particular
u ⇀u in L2(0,T,W1,∞(Ω)) weak*. (30)
η,s s
7
We now wish to prove that u (or equivalently u and ω) has a similar structure as u ,
s G η,s
that is, to pass to the limit in the expression defining u . Using (30), the already mentioned
η,s
compactness results of [9], applied to the transport equation on H now gives
η
H H a.e. and in C(0,T,Lp(Ω)) strong p [1,+ [ (31)
η
→ ∀ ∈ ∞
with H verifying
H +u . H =0 on Rn (0,T)
t s
∇ ×
(H =H0 on Rn 0 .
×{ }
Note that this Cauchy problem has been set in Rn because u does not vanish on ∂Ω, but H
s
vanishes outside Ω. However we can prove by passing to the limit in (27) that Hu= Hu . Thus
s
div(uH)=div(u H) and H verifies a transport equation with velocity field u on Ω (note that no
s
boundary conditions are needed on ∂Ω since u is zero on the boundary).
This convergence gives us the strong convergence of r in (0,T,Lp(Ω)), p 1, and from
η
C ∀ ≥
(28),(31) and (25) we can easily pass to the limit in the expression of u and ω to get
η,G η
ρuHdx
−1
u (t)= ZΩ and ω(t)= ρ(r2I r r)Hdx ρ(r u)Hdx.
G
− ⊗ ×
ρHdx (cid:18)ZΩ (cid:19) ZΩ
ZΩ
3.3 Strong convergence of u
η
The remaining part of the proof is more technical since we aim to prove the strong convergence
of at least a subsequence of u in order to be able to pass to the limit in the inertial term of
η
Navier-Stokesequations. Classically,this is obtainedthanks to a Fourier transformin time which
provides an estimate on some fractional time derivative of u which brings compactness [16, 22].
η
Here these technics can not be used since the solid is moving. We instead rely on tools developed
in [19].
Thereafter we will use, for σ >0 and r [0,1], the following notations
∈
Ω (t)= x Ω,d(x,Ω (t))<σ ,
s,σ s
• { ∈ }
0 = v L2(Ω), divv =0, v n=0 on ∂Ω ,
• V { ∈ · }
r = v Hr(Ω), divv =0, v =0 on ∂Ω (for r >0),
• V { ∈ }
r(t)= v(t) r,D(v(t))=0 in ′(Ω (t)) (which is a closed subset of r),
• Kσ { ∈V D s,σ } V
Pr(t) the orthogonal projection of r on r(t).
• σ V Kσ
To prove the strong convergence of a subsequence of u in L2(Q) we write
η
T 1 T T
u u2dxdt ρ(u2 u2) dxdt+ 2ρu (u u ) dxdt .
Z0 ZΩ| η− | ≤ ρmin Z0 ZΩ| η− | Z0 ZΩ| · − η | !
From (26) the second integral on the right side converges to 0, thus
T 1 T T
u u2dxdt ρ u2 ρu2 dxdt+ (ρ ρ)u2 dxdt .
Z0 ZΩ| η− | ≤ ρmin Z0 ZΩ| η η− | Z0 ZΩ| η − η| !
8
Moreover,by (22) and (28) the second integral on the right hand side converges to 0, and
T 1 T
u u2dxdt ρ u Pr(u ) ρu Pr(u) dxdt
Z0 ZΩ| η− | ≤ ρmin Z0 ZΩ| η η· σ η − · σ |
T T
+ ρ u (u Pr(u )) dxdt+ ρu (Pr(u) u) dxdt+l
Z0 ZΩ| η η· η− σ η | Z0 ZΩ| · σ − | η!
1
( ρ u Pr(u ) ρu Pr(u)
≤ ρ k η η· σ η − · σ kL1(Q)
min
+ kρηkL∞(Q)kuηkL2(Q)kPσr(uη)−uηkL2(Q)
+ kρkL∞(Q)kukL2(Q)kPσr(u)−ukL2(Q)+lη
where l 0 when η 0. Finally, as ρ is bounded in L∞(0,T,L∞(Ω)) and using (22), we get
η η
→ →
T 1
u u2dxdt ( ρ u Pr(u ) ρu Pr(u)
| η− | ≤ ρ k η η· σ η − · σ kL1(Q)
Z0 ZΩ min
+ C Pr(u ) u
k σ η − ηkL2(Q)
+ C Pr(u) u +l (32)
k σ − kL2(Q) η
This decomposition shows that the sought convergence essentially amounts to prove that (up to
the extraction of subsequences)
lim Pr(u) u =0 (33)
σ→0k σ − kL2(Q)
lim lim Pr(u ) u =0 (34)
σ→0η→0k σ η − ηkL2(Q)
lim lim ρ u Pr(u ) ρu Pr(u) =0 (35)
σ→0η→0k η η· σ η − · σ kL1(Q)
To prove (33-35) we will make use of some Lemma that we state now.
Lemma 3.2. Let (f ) be a sequence of functions bounded in Lp(0,T) for some p > 2 and con-
n
verging to 0 almost everywhere on [0,T]. Then (f ) converges strongly to 0 in L2(0,T).
n
Proof. Let ε > 0. From Egorov theorem ([5], p.75), the almost everywhere convergence implies
that there exists A [0,T] such that
ε
⊂
[0,T] A <ε
ε
| \ |
(fn 0 uniformly on Aε
→
which means
N N, n N, t A , f (t)2 <ε.
ε n
∃ ∈ ∀ ≥ ∀ ∈ | |
Therefore
(f (t))2dt εA εT.
n ε
≤ | |≤
ZAε
As from the bound in Lp norm,
(f (t))2dt 1qdt (f (t))pdt εqC
n n
≤ ≤
Z[0,T]\Aε Z[0,T]\Aε Z[0,T]
with 1 + 1 = 1, we get
q p 2
T
(f (t))2dt εT +εqC
n
≤
Z0
which proves the L2 convergence.
9
Lemma3.3. Letu(t) H1(Ω (t))suchthatu (t)=g(t)and(w(t),p(t)) H1(Ω Ω (t))
∈ s,σ |∂Ωs,σ(t) ∈ \ s,σ ×
L2(Ω Ω (t)) solution of the Stokes problem
s,σ
\
∆w(t)+ p(t)=0 on Ω Ω (t),
s,σ
− ∇ \
divw(t)=0 on Ω Ω (t),
 \ s,σ
w(t)=g(t) on ∂Ωs,σ(t),
w(t)=0 on ∂Ω.
Then there exists σ >0 andC >0, such that for all σ <σ , we have the following estimate:
0 0
w(t) 2 C u(t) u(t)
k kL2(Ω\Ωs,σ(t)) ≤ k kL2(Ωs,σ(t))k∇ kL2(Ωs,σ(t))
The proof of this result is postponed to the appendix.
Lemma 3.4. The limits H, u and u defined in (31), (25) and (30) verify
s
Hu=Hu . (36)
s
Proof. First we claim that
H u ⇀Hu in Lq(0,T,Lr(Ω)) weak, with q <2 and r <6 (37)
η η
Indeed, let us introduce v Lq(0,T,Lr(Ω)) with q >2 and r> 6. We have
∈ 5
T T T
(H u Hu) vdxdt= H(u u) vdxdt+ (H H)u vdxdt
η η η η η
− · − · − ·
Z0 ZΩ Z0 ZΩ Z0 ZΩ
From the injection of H1 into L6 in dimension less or equal to 3 and (25) we get
u ⇀u in L2(0,T,L6(Ω)) weak,
η
and, since H is bounded in L∞(Q),
T
lim H(u u) vdxdt=0.
η
η→0Z0 ZΩ − ·
Moreover from (31) we easily get
T
lim (H H)u vdxdt=0.
η η
η→0Z0 ZΩ − ·
We next show that
H u ⇀Hu in Lp(0,T,Lp(Ω)) weak, p [1,+ [. (38)
η η,s s
∀ ∈ ∞
Let v Lp(0,T,Lp(Ω)) with p>1. We write
∈
T T T
(H u Hu ) vdxdt= H(u u ) vdxdt+ (H H)u vdxdt.
η η,s s η,s s η η,s
− · − · − ·
Z0 ZΩ Z0 ZΩ Z0 ZΩ
As H is bounded in L∞(Q), with (30) we get:
T
lim H(u u ) vdxdt=0
η,s s
η→0Z0 ZΩ − ·
In addition, from (31) we get
T
lim (H H)u vdxdt=0.
η η,s
η→0Z0 ZΩ − ·
10