On an unsteady fluid–beam interaction problem
C´eline Grandmont∗ September 7, 2004
Abstract
We consider a two–dimensional viscous incompressible fluid gov-erned by the Navier–Stokes equations, interacting with an elastic beam located on one part of the fluid boundary. We do not neglect the de-formation of the fluid domain which consequently depends on the dis-placement of the structure. The purpose of this work is to study the solutions of this unsteady fluid–structure interaction problem, as the coefficient modeling the viscoelasticity (resp. the rotatory inertia) of the beam tends to zero. As a consequence, we obtain the existence of at least one weak solution for the limit problem (Navier–Stokes equa-tion coupled with Euler–Bernoulli equaequa-tion) as long as the beam does not touch the bottom of the fluid cavity.
Keywords: Fluid–Structure interaction, time dependent domain, existence of weak solutions.
1
Introduction.
Many physical phenomena deal with a fluid interacting with a moving or deformable structure. The model we focus on can be viewed as a first at-tempt to described the evolution of the blood flow past arteries (see [24]). We consider a viscous incompressible two–dimensional fluid described by the Navier–Stokes equations interacting with a one–dimensional elastic beam. One already knows that if a viscous term or the rotatory inertia are taken into account, there exists at least one weak solution to this problem (see [8]). In [8], existence of at least one weak solution for 3D Navier–Stokes coupled with a “viscous” plate in flexion is proved, and the same proof applies in
∗
the present case (i. e. for the “viscous” beam). From the mechanical point of view adding a viscous term is a way to introduce dissipation in the beam model and from the mathematical point of view, this is a way to regularize the structure velocity. Here the dissipation coming from the fluid enables us to pass to the limit in the coupled system as the additional viscous beam coefficient tends to zero, and thus obtain the existence of weak solutions of the limit problem (Navier–Stokes equations coupled with the classical Euler– Bernoulli equation). This limit system can also be obtained as the limit of the Navier–Stokes equation coupled with the Rayleigh–Love equation (with the rotatory inertia) with regularized initial beam velocity. This is, to our knowledge, the first theoretical result in this direction. In all the previous studies the structure velocity is quite regular because of the model or be-cause of the presence of a regularization operator in the equations. Most of the existing results are concerned with rigid body motions ([10], [13], [14], [16], [18], [20], [21], [26], [27], [29], [30]) or with the motion of a structure described by a finite number of modal functions ([15]) or a structure with additional “viscous” terms ([3], [8], [11]). Note that the present result ap-plies only to a beam in interaction with a 2D fluid. For the time being it is not clear how to obtain the same kind of result in the case of a plate interacting with a 3D fluid or a 1D membrane coupled with a 2D fluid.
In the first section we give the equations of the fluid–structure prob-lem for which we derive a priori energy estimates. Next, definitions and properties of the energy spaces are detailed and the weak formulation of the problem is given. In particular, we build suitable extensions of the fluid test functions and liftings of the structure test functions. In the second section, we state our main results, the third section being devoted to the obtention of compactness properties that enable us to pass to the limit in the equations as the “viscous” beam coefficent tends to zero (Section 4). In the appendix we prove a technical lemma (Lemma 3), for which we derive a regularity result of the solution of the steady–state Stokes equation in a “nonsmooth” domain (i. e. for which one cannot apply the standard regularity results).
1.1 Presentation of the problem
We assume that the fluid fills a two–dimensional cavity and interacts with a one–dimensional elastic structure, located on a part of the fluid domain boundary, the other part being rigid. At reference state, the elastic part
of the fluid boundary is located on (0, L) × {R}. We denote by ηε(t, x) the
vertical displacement with respect to the rest configuration. We assume that the beam is clamped so that ηε(t, 0) = ηε(t, L) = ∂xηε(t, 0) = ∂xηε(t, L) = 0.
The subscript ε underlines the dependence of the solution with respect to the parameter ε ≥ 0 which measures the “viscosity” of the beam (or the rotatory inertia). The reference domain for the system is denoted Ω and we will consider Ω = (−δ, L + δ) × (0, R), where δ > 0. At initial state the fluid occupies the domain Ωη0:
Ωη0 =(x, y) ∈ R
2, x ∈ (−δ, L + δ), 0 < y < R + η 0(x) ,
where η0 is a given vertical initial displacement of the elastic part and η0
is the extension by 0 on (−δ, 0) ∪ (1, 1 + δ) of η0. We consider δ > 0 for
technical reasons (see the Appendix). The rigid part of ∂Ωη0 is denoted by
Γ0. The domain occupied by the fluid at time t is denoted by Ωηε(t):
Ωηε(t) =(x, y) ∈ R
2/ x ∈ (−δ, L + δ), 0 < y < R + η
ε(t, x) , (1)
where ηε has been extended by 0 for x < 0, x > 1, this extension being
denoted by ηε. The fluid equations of conservation of mass and momentum write ∂tuε+ (uε· ∇)uε− ν∆uε+ ∇pε= f in Ωηε(t), div uε= 0 in Ωηε(t), uε(t, ·) = 0 on Γ0, uε(0, ·) = u0 in Ωη0, (2)
where uεdenotes the fluid velocity, and pεthe pressure field. The bulk force
f is given together with the initial velocity u0.
The equations describing the evolution of ηε are:
∂ttηε+ ∂xxxxηε+ ε∂xxxx∂tηε= g + (Tfε)2 in (0, L), ηε(t, 0) = ηε(t, L) = ∂xηε(t, 0) = ∂xηε(t, L) = 0, ηε(0, ·) = η0, ∂tηε(0, ·) = η1, (3)
where g denotes the exterior forces applied to the beam and Tε
f the surface
force exerted by the fluid on the structure. The parameter ε ( ε ≥ 0) measures the viscoelasticity of the beam. We could have consider another model taking into account the inertia of rotation of the beam. The additional
term is then −ε∂xx∂ttηε.
Since the fluid is viscous, it sticks to the structure and thus the velocities coincide (in a sense to be defined) at the interface:
uε(t, x, R + ηε(t, x)) = (0, ∂tηε(t, x))T , (t, x) ∈ (0, T ) × (0, L). (4)
This condition, together with the incompressibility of the fluid leads to Z L
0
∂tηε = 0. (5)
Condition (5) states that the global volume of the cavity is preserved. Note here that the pressure pε is not defined up to a constant but is uniquely
defined. Its average value ensures the global volume conservation of the fluid cavity. This average is in fact the Lagrange multiplier associated to the compatibility condition (5).
The surface force Tfε exerted by the fluid on the beam can be defined by: Z L 0 Tfε· ˆv = Z ∂Ωηε(t)\Γ0 (−2νD(uε) · nεt + pεnεt) · v, ∀v, (6)
where D(uε) = (∇uε+ (∇uε)T)/2 is the strain tensor, ˆv(t, x) = v(t, x, 1 +
ηε(t, x)), for (t, x) ∈ (0, T ) × (0, L) and nεt denotes the outer unit normal at
Γηε(t) = ∂Ωηε(t) \ Γ0: nεt = 1 p1 + (∂xηε)2 (−∂xηε, 1). As noted in [8], 2(D(uε) · nεt)2 = (∇uε· nεt)2.
This simplification comes from the fact that the displacement at the fluid– structure interface is only transverse and from the incompressibility of the fluid. Thus ∀v, such that v1(t, x, x + ηε(t, x)) = 0, (t, x) ∈ (0, T ) × (0, L)
Z L 0 Tfε· ˆv = Z Γηε(t) (−ν∇uε· nεt+ pεnεt)2 v2, (7)
In all what follows and without loss of generality we take L = R = 1.
1.2 A priori estimates
In this subsection we recall the a priori estimates satisfied by any solution, assuming that it is smooth enough. We multiply the Navier–Stokes equa-tions by uεand integrate over Ωηε(t), and we multiply the beam equation by
∂tηε and integrate over (0, 1) and add these two contributions. After
inte-grations by parts and taking into account the coupling conditions (equality of the velocities (4), and the definition of Tfε (7)) we obtain the following energy equality: 1 2 d dt Z Ωηε(t) |uε|2+ ν Z Ωηε(t) |∇uε|2 +1 2 d dt Z 1 0 (∂tηε)2+ 1 2 d dt Z 1 0 (∂xxηε)2+ ε Z 1 0 (∂xx∂tηε)2 = Z Ωηε(t) f · uε+ Z 1 0 g ∂tηε. (8)
Hence, using Cauchy–Schwarz, Young inequalities and Gronwall’s lemma: 1 2kuεk 2 L2(Ωηε(t))(t) + 2ν Z t 0 k∇uεk2L2(Ωηε(s))(s)ds +1 2k∂tηεk 2 L2(0,1)(t) + 1 2k∂xxηεk 2 L2(0,1)(t) + ε Z t 0 k∂xx∂tηεk2L2(0,1)(s)ds ≤ et 1 2ku0k 2 L2(Ω η0)+ 1 2kη1k 2 L2(0,L)+ 1 2k∂xxη0k 2 L2(0,1) +1 2 Z t 0 exp(t − s) kf k2L2(Ω η(s))(s) + kgk 2 L2(0,1)(s) ds (9)
Thus assuming that f ∈ L2(0, T ; L2(R2)), g ∈ L2(0, T ; L2(0, 1)), u0∈ L2(Ωη0),
η0 ∈ H02(0, 1), and η1 ∈ L2(0, 1), uε∈ L∞(0, T ; L2(Ωηε(t))), ∇uε∈ L 2(0, T ; L2(Ω ηε(t))), and ηε∈ W1,∞(0, T ; L2(0, 1)) ∩ L∞(0, T ; H02(0, 1)), moreover if ε > 0, ∂tηε∈ L2(0, T ; H02(0, 1)).
Consequently the spaces Lp(0, T ; L2(Ωγ(t))), L2(0, T ; H1(Ωγ(t))) need
to be defined, for γ belonging to W1,∞(0, T ; L2(0, 1)) ∩ L∞(0, T ; H02(0, 1)). We have also to give a sense to the equality of the velocities. Thus we are going to give some technical lemmas, definitions and properties.
1.3 Definitions– Preliminary properties
Let T > 0 and γ belong to W1,∞(0, T ; L2(0, 1)) ∩ L∞(0, T ; H02(0, 1)) such that for some positive M and α, M ≥ 1 + γ(t, x) ≥ α > 0 for all (t, x) ∈
[0, T ] × [0, 1]. The following continuous injection holds:
W1,∞(0, T ; L2(0, 1)) ∩ L∞(0, T ; H2(0, 1)) ,→ C0,1−θ([0, T ]; H2θ(0, 1)), for all 0 < θ < 1. In particular,
W1,∞(0, T ; L2(0, 1)) ∩ L∞(0, T ; H2(0, 1)) ,→ C0,β([0, T ]; C1([0, 1])) for all 0 < β < 1/4 and
W1,∞(0, T ; L2(0, 1)) ∩ L∞(0, T ; H02(0, 1)) ,→ C0,λ([0, T ]; C0([0, 1])) for all 0 < λ < 3/4.
The proof of the first injection relies on standard hilbertian interpolation inequalities (see [23]). The two others are deduced from the first one and from Sobolev injections in dimension one (see [5]).
Consequently, at each time t the function γ(t, ·) is Lipschitz continuous (and thus the solution ηε(t, ·) in energy space will be Lipschitz continuous,
for all ε ≥ 0). Note that this was not the case in [8] where the displacement was not C1 with respect to the space variables. In particular we can define all the functional spaces as in [8]. For the sake of completeness we recall the needed definitions. Let us define Ωγ(t) by
Ωγ(t) =(x, y) ∈ R2, x ∈ (−δ, 1 + δ), 0 < y < 1 + γ(t, x) ,
where γ is the extension by 0 on (−δ, 0) ∪ (1, 1 + δ) of γ. Note that γ belongs to W1,∞(0, T ; L2(−δ, 1 + δ)) ∩ L∞(0, T ; H02(−δ, 1 + δ)) and thus Ωγ(t) is an
open Lipschitz subset of R2, that is included in BM = (−δ, 1 + δ) × (0, M )
for all t ∈ (0, T ). Let bΩγ be the open domain of R3 defined by
b Ωγ =
[
t∈(0,T )
{t} × Ωγ(t).
We set dBM = (0, T ) × BM. One can define in a standard way the spaces
Lp(Ω
γ(t)), H1(Ωγ(t)), H01(Ωγ(t)), for every t, and Lp(bΩγ), H1(bΩγ), Lp( dBM),
H1( dBM). . . The space H0,Γ1 0(Ωγ(t)) will denote the subspace of H1(Ωγ(t))
of functions of zero trace on Γ0. We next define:
Lp(0, T ; Lq(Ωγ(t))) = ( v, v = v in bΩγ 0 in dBM\ bΩγ ∈ Lp(0, T ; Lq(B M)) ) ,
Lp(0, T ; H1(Ωγ(t))) =v ∈ Lp(0, T ; L2(Ωγ(t))), ∇v ∈ Lp(0, T ; L2(Ωγ(t))) . Remark that L2(0, T ; H1(Ωγ(t))) = n v ∈ L2(bΩγ), ∇v ∈ L2(bΩγ) o . We also define Lp(0, T ; H01(Ωγ(t))) =D(bΩγ) Lp(0,T ;H1(Ω γ(t))) , 1 ≤ p < +∞. Moreover we denote Vγp =v ∈ Lp(0, T ; H1(Ωγ(t))), div v = 0, v = 0 on (0, T ) × Γ0 , and Vp =v ∈ Lp(0, T ; H1(BM)), div v = 0, v = 0 on (0, T ) × (∂BM \ ((−δ, 1 + δ) × {M })) .
We next define a lifting of the deformations at interface: (t, x, y) ∈ (0, T ) × Ω, χγ(t, x, y) = x y(1 + γ(t, x)) .
Du to the regularity of γ, χγ belongs in particular to C0,β([0, T ]; C1(Ω)) ∩
L∞(0, T ; H2(Ω)), for all 14 > β > 0. Nevertheless we remark that χγ is C∞
with respect to y. Moreover since min(t,x)∈[0,T ]×[0,1]1 + γ(t, x) > 0, we have
i) ∇χγ is invertible in C0([0, T ]; C0(Ω)).
ii) χγ(t) is one to one on Ω, ∀t ∈ [0, T ].
iii) χγ(t) is a C1–diffeomorphism that maps Ω on Ωγ(t), ∀t ∈ [0, T ].
We define v ◦ χγ by (v ◦ χγ)(t, x, y) = v(t, x, y(1 + γ(t, x))). We easily have:
Lemma 1 Assuming that min
(t,x)∈[0,T ]×[0,1]1 + γ(t, x) > 0, the mapping v 7→
v ◦ χγ is an isomorphism from Lp(0, T ; H1(Ωγ(t))) on Lp(0, T ; H1(Ω)).
Remark 1 Thanks to the regularity of γ we could have defined the spaces Lp(0, T ; Hs(Ω
γ(t))), ∀p ≥ 1, ∀s ∈ [0, 1] by saying that v ∈ Lp(0, T ; Hs(Ωγ(t)))
if and only if v ◦ χγ ∈ Lp(0, T ; Hs(Ω)). If we choose not to this is because
Next we give a weak sense to the equality of the velocities. In [8], for v ∈ Lp(0, T ; H1(Ωγ(t))), we were only able to give a sense to the trace of
v ◦ χγ over (0, 1) × {1} in Lp(0, T ; L2(0, 1)). But in our context since the
deformed beam is Lipschitz continuous in space for all t, we have
Lemma 2 For all v ∈ Lp(0, T ; H1(Ωγ(t))), p ≥ 1, the trace (v ◦ χγ)|(0,1)×{1}
belongs to Lp(0, T ; H12(0, 1)) and
kv ◦ χγk
Lp(0,T ;H12(0,1))≤ CkvkLp(0,T ;H1(Ωγ(t))).
Moreover if v = 0 on (0, T )×Γ0then (v◦χγ)|(0,1)×{1}belongs to Lp(0, T ; H
1 2 00(0, 1)) and kv ◦ χγk Lp(0,T ;H12 00(0,1)) ≤ CkvkLp(0,T ;H1(Ωγ(t))).
We refer to [23] for the definition of the space H
1 2
00. Now we define different
type of lifting or extension operators. Extension Operator
Let us consider v ∈ Lp(0, T ; Lq(Ωγ(t))) and a vector function (0, b)T with b
belonging to Lp(0, T ; Lq(0, 1)). We denote by b the extension of b by 0 on (−δ, 0) ∪ (1, 1 + δ) and we define the function v by:
v = v in bΩγ (0, b)T in dBM \ bΩγ (10)
It is clear that v belongs to Lp(0, T ; Lq(BM)). Moreover if v ∈ Lp(0, T ; H1(Ωγ(t)))
and b ∈ Lp(0, T ; H01(0, 1)) with (v ◦ χγ)|(−δ,1+δ)×{1}= (0, b)T then v belongs
to Lp(0, T ; H1(BM)) and the following estimate holds:
kvkLp(0,T ;H1(B
M))≤ CM kvkLp(0,T ;H1(Ωγ(t))))+ kbkLp(0,T ;H1(0,1))
(11) Assuming moreover that div (v) = 0 (remark that implies that R01b = 0) then div (v) = 0.
Lifting Operator
We will define two types of lifting operator. • First, let us consider b ∈ Lp(0, T ; H1
0(0, 1)) such that R1 0 b = 0. We define R1(b) by R1(b) = (0, b)T in bΩγ\ bCα R((0,αyb)T) in bCα, (12)
where Cα = (−δ, 1 + δ) × (0, α) and R is a lifting operator from H
1 2(∂Cα)
onto H1(Cα) such that div (R(v)) = 0. The function R1(b) belongs to Vγp
and
kR1(b)kLp(0,T ;H1(B
M))≤ CαkbkLp(0,T ;H1(0,1)). (13)
• Next we consider b ∈ Lp(0, T ; H12
00(0, 1)) and such that
R1
0 b = 0, we would
like to build a fonction v in Lp(0, T ; H0,Γ1 0(Ωγ(t))) such that v◦χγ|(0,1)×{1}=
(0, b)T and div (v) = 0 in bΩγ. We consider the following Stokes problem:
−∆v + ∇q = 0 in Ωγ(t), div (v) = 0 in Ωγ(t), v(t, x, 1 + γ(t, x)) = (0, b(t, x))T on (0, 1), v = 0 on Γ0. (14)
Remark that the two boundary conditions can be reformulated by saying that v ◦ χγ = (0, yb)T, on ∂Ω and that yb belongs to Lp(0, T ; H
1 2(∂Ω))
since b ∈ Lp(0, T ; H
1 2
00(0, 1)). This problem has a unique solution (v, q)
in H1(Ωγ(t)) × L20(Ωγ(t)) for a. e. t where L20(Ωγ(t)) is the subspace of
L2(Ωγ(t)) of functions of zero average (see [17], [33]). The regularity with
respect to time can be obtained by looking at the regularity with respect to the time of (w, µ) = (v ◦ χγ, q ◦ χγ). We have
Lemma 3 The solution (v, q) of (14) belongs to Lp(0, T ; H1(Ωγ(t)))×Lp(0, T ; L20(Ωγ(t)))
and satisfies the following estimates: kvkLp(0,T ;H1(Ω γ(t)))+ kpkLp(0,T ;L20(Ωγ(t))) ≤ C(γ)kbk Lp(0,T ;H 1 2 00(0,1)) , (15) and kvkLp(0,T ;L2(Ω γ(t)))≤ C(γ)kbkLp(0,T ;L2(0,1)), (16)
where C(γ) depends only on M , α and on the norms of γ in W1,∞(0, T ; L2(0, 1)) and L∞(0, T ; H02(0, 1)).
This lemma is proved in the appendix. To obtain estimate (16) we use a transposition method that relies on a result of regularity of the solution of the Stokes problem in Ωγ(t) which is not smooth enough to apply standard
regularity results. We denote by Rγ the linear operator that associates to b
1.4 Weak formulation Let η0 ∈ H02(0, 1), (u0, η1) ∈ L2(Ωη0) × L 2(0, 1) such that min [0,1](1 + η0) > 0, div u0= 0 in Ωη0, u0· n = 0 on Γ0, (u0◦ χη0) · n0 = (0, η1) T · n 0 on (0, 1), Z 1 0 η1(x) = 0, (17)
where n0 denotes a normal to the initial position of the beam. We shall
say that (uε, ηε) is a weak solution of the considered model on (0, T ) if it
satisfies the following problem that will be denoted (Pε):
– uε∈ Vη2ε∩L ∞(0, T ; L2(Ω ηε(t))), ηε ∈ W 1,∞(0, T ; L2(0, 1))∩L∞(0, T ; H2 0(0, 1)), – For ε > 0, ∂tηε∈ L2(0, T ; H02((0, 1))), – uε(t, x, 1 + ηε(t, x)) = (0, 0, ∂tηε(t, x))T, for a. e. (t, x) ∈ (0, T ) × (0, 1), – for all (φ, b) ∈ (Vη2ε∩H1(bΩ ηε))×(L 2(0, T ; H2 0((0, 1)))×H1(0, T ; L2(0, 1))), such that φ(t, x, 1 + ηε(t, x)) = (0, 0, b(t, x))T, a. e. (t, x) ∈ (0, T ) × (0, 1), we have for a. e. t Z Ωηε(t) uε(t) · φ(t) − Z t 0 Z Ωηε(s) uε· ∂tφ + ν Z t 0 Z Ωηε(s) ∇uε : ∇φ + Z t 0 Z Ωηε(s) (uε· ∇)uε· φ − Z t 0 Z 1 0 (∂tηε)2b + Z 1 0 ∂tηε(t) b(t) − Z t 0 Z 1 0 ∂tηε∂tb + Z t 0 Z 1 0 ∂xxηε∂xxb + ε Z t 0 Z 1 0 ∂xx∂tηε∂xxb = Z t 0 Z Ωηε(t) f · φ + Z t 0 Z 1 0 g b + Z Ωη0 u0· φ(0) + Z 1 0 η1b(0) (18)
Remark 2 The test functions depends on the solution and thus, for ε > 0, on ε.
Remark 3 The trace of φ ∈ H1(bΩηε) at time t = 0 is well defined and
(C1Ωbηε
is dense in H1(bΩηε) since the domain is a continuous subgraph,
see for instance [1], Thm 2, p. 54) that Z Ωη0 |φ(0)|2+ Z Ωη(T ) |φ(T )|2 = 2 Z T 0 Z Ωηε(t) φ · ∂tφ + Z T 0 Z 1 0 |φ ◦ χγ|2∂tγ.
The right hand side of this equality makes sense for any φ ∈ H1(bΩηε) since
∂tγ ∈ L∞(0, T ; L2(0, 1)) and φ ◦ χγ ∈ L2(0, T ; H
1
2(0, 1)), which is
conti-nously embedded in L2(0, T ; L4(0, 1)).
2
Main Result
We make the following hypotheses on the data (bulk forces and initial data): f ∈ L2loc((0, +∞) × R2), g ∈ L2loc((0, +∞) × (0, 1)),
u0∈ L2(Ωη0), η1∈ L
2(0, 1), η
0 ∈ H02(0, 1),
(19) and we assume moreover that conditions (17) are satisfied.
First we recall that, for ε > 0, there exists at least one weak solution provided that the beam does not touch the bottom of the fluid cavity, in other words as long as minx∈[0,1](1 + ηε(t, x)) > 0.
Theorem 1 Let ε be strictly positive. Under assumptions (17), (19), and if minx∈[0,1]1 + η0(x) > 0, there exists Tε∗ ∈ (0, +∞] and a weak solution
(uε, ηε) of (Pε) on [0, T ], T < Tε∗. This solution satisfies energy estimates
for all T < Tε∗: kuεkL∞(0,T ;L2(Ω ηε(t)))+ k∇uεkL2(0,T ;L2(Ωηε(t))) +k∂tηεkL∞(0,T ;L2(0,1))+k∂xxηεkL∞(0,T ;L2(0,1))+ √ εk∂xx∂tηεkL2(0,T ;L2(0,1)) ≤ C(T, ku0kL2(Ω η0), kf kL2((0,T )×R2), kgkL2((0,T )×(0,1)), kη0kH02((0,1)), kη1kL2(0,1)),
where C > 0 is nondecreasing with respect to its arguments. Moreover we have the following alternative
- either Tε∗ = +∞, - or lim t→T∗ ε min [0,1](1 + ηε) = 0.
Remark 4 Since uε is bounded in L2(0, T ; H0,Γ1 0(Ωηε(t))) independently of
ε, the trace uε◦ χηε|(−δ,1+δ)×{1} is bounded in L2(0, T ; H
1
2(−δ, 1 + δ))
inde-pendently of ε and is equal to zero on (−δ, 0) ∪ (1, 1 + δ). Thus, thanks to the equality of the velocities, ∂tηεis bounded independently of ε in L2(0, T ; H
1 2
In all that follows, the characteristic function of bΩηε will be denoted by ρε
and ρεz will denote the function equal to z in bΩηε and zero elsewhere. The
main results of the present article are
Proposition 1 The sequence (Tε∗)ε>0is bounded from below away from zero
and the following convergences (up to subsequences) hold as ε goes to zero: ηε → η strongly in C0([0, T ]; C1[0, 1]),
ηε * η weakly in L2(0, T ; H02(0, 1)),
∂tηε → ∂tη strongly in L2(0, T ; L2(0, 1)),
ρεuε → ρu strongly in L2(0, T ; L2(BM)),
uε → u strongly in L2(0, T ; L2(BM)),
ρε∇uε * ρ∇u weakly in L2(0, T ; L2(BM)),
(20)
where T > 0 is a lower bound of Tε∗ independent of ε. Moreover u ◦ χη|(0,1)×{1} = (0, ∂tη)T.
This enables us to pass to the limit in the equation and thus obtain the Theorem 2 Under assumptions (17), (19), and if minx∈[0,1]1 + η0(x) > 0,
there exists T∗ ∈ (0, +∞] and a weak solution (u, η) of (P0) on [0, T ], T <
T∗. This solution satisfies energy estimates for all T < T∗: kukL∞(0,T ;L2(Ω η(t)))+ k∇ukL2(0,T ;L2(Ωη(t))) + k∂tηkL∞(0,T ;L2(0,1))+ kηkL∞(0,T ;H2 0(0,1)) ≤ C(T, ku0kL2(Ω η0), kf kL2((0,T )×R2), kgkL2((0,T )×(0,1)), kη0kH02(0,1), kη1kL2(0,1)), (21) where C > 0 is nondecreasing with respect to its arguments. The following alternatives are satisfied
- either T∗ = +∞, - or lim
t→T∗min[0,1](1 + η) = 0.
3
Compactness–Proof of Proposition 1
First we prove that Proposition 1 holds true. We have to verify that Tε∗ is bounded from below independently of ε, and obtain compactness properties
on (uε, ∂tηε) in order to prove the desired strong convergences that will
enable us to pass to the limit in (Pε) as ε goes to zero.
For ε > 0 the solution ηεis bounded independently of ε in L∞(0, T ; H02(0, 1))∩
W1,∞(0, T ; L2(0, 1)) for all T < T∗
ε. Thus ηε is bounded indepedently of ε
in C0,λ([0, T ]; C0([0, 1])), 0 < λ < 34. Consequently
1 + ηε(t, x) ≥ (1 + η0(x)) − Ctλ, ∀(t, x) ∈ [0, 1] × [0, Tε∗)
where C depends only on the data of the problem. Thus Tε∗is bounded from below by a time independent of ε. Let T be such that
∀ε > 0, min
(t,x)∈[0,T ]×[0,1](1 + ηε(t, x)) ≥ α > 0,
where α is chosen such that minx∈[0,1](1 + η0(x)) ≥ 2α > 0.
In what follows we prove strong convergence properties for the fluid and the structure velocities. The solution we build verifies estimate (9). In particular if M is chosen large enough 1 + ηε(t, x) ≤ M , ∀(t, x) ∈ [0, T ] ×
[0, 1]. The energy estimates are not sufficient to obtain the desired strong convergences. We are going to use the following lemma that characterizes the compact sets of Lp(0, T ; X) where X is a Banach space (see [28]). Lemma 4 Let X be a Banach space and F ,→ Lp(0, T ; X) with 1 ≤ p < ∞.
Then F is a relativelly compact set of Lp(0, T ; X) if and only if • nRt2
t1 f (t)dt, f ∈ F
o
is relativelly compact in X, ∀0 < t1 < t2 < T
• kf (t+h)−f (t)kLp(0,T ;X)−→ 0 as h goes to zero, uniformly with respect
to f in F . We set uε= uε in bΩηε ∂tηε in bBM \ bΩηε,
where ∂tηεdenotes the extension by zero of ∂tηεon (−δ, 0)∪(1, 1+δ). Thanks
to the energy estimates uεis bounded independently of ε in L∞(0, T ; L2(BM))
and L2(0, T ; H12(BM)).
Given any h > 0, we denote g−(t, ·) = g(t − h, ·) and g+(t, ·) = g(t + h, ·). We state the following lemma:
Lemma 5 Let T > 0 such that min[0,T ]×[0,1](1 + ηε) ≥ α > 0. We have
Z T 0 Z BM ρε|uε− u−ε|2+ Z T 0 Z 1 0 (∂tηε− ∂tη−ε)2 ≤ β (22) and Z T 0 Z BM |ρεuε− ρ−εu−ε|2 ≤ β, (23)
with ηε extended by ηε0 for t < 0 and uε and ∂tηε extended by 0 for t < 0,
and where ρε denotes the characteristic function of bΩηε.
Proof:
We first show that (22) implies (23). Indeed: |ρεuε− ρ−εu
−
ε|2 ≤ C ρε|uε− uε−|2+ |ρε− ρ−ε||u − ε|2 .
The estimate of the first contribution comes from (22). For the second contribution we have Z T 0 Z BM |ρε− ρ−ε||uε|2 ≤ Z T 0 kρε− ρ−εkL2(B M)kuεk 2 L4(B M)
Remember that ∂tηε is bounded in L∞(0, T ; L2(0, 1)) independently of ε,
Z BM |ρε− ρ−ε|2 = Z 1 0 |ηε− (ηε)−| = Z 1 0 Z t t−h ∂tηε(s)ds ≤ Z 1 0 Z t t−h |∂tηε(s)|ds ≤ Ch.
Moreover, since the extended velocity uεis bounded in L2(0, T ; H
1
2(BM))
in-dependently of ε and thanks to Sobolev injections uεis bounded in L2(0, T ; L4(BM)).
Consequently Z T 0 Z BM (ρε− ρ−ε)|uε|2 ≤ C√h. (24)
This shows (22) implies (23).
To prove (22) we are going to make a suitable choice for the test functions in the weak formulation satisfied by uε and ∂tηε. The same type of test
functions are used, for instance, in [25], where existence of weak solutions for the Navier–Stokes equations in a given time dependent domain is proven and in [8] where the same model we are looking at is studied but in dimension three and only for ε > 0. We have that uε and ∂tηε verify:
Z Ωηε(T ) uε(T ) · φ(T ) − Z T 0 Z Ωηε(t) uε· ∂tφ + ν Z T 0 Z Ωηε(t) ∇uε : ∇φ + Z T 0 Z Ωηε(t) (uε· ∇)uε· φ − Z T 0 Z 1 0 (∂tηε)2b + Z 1 0 ∂tηε(T ) b(T ) − Z T 0 Z 1 0 ∂tηε∂tb + Z T 0 Z 1 0 ∂xxηε∂xxb + ε Z T 0 Z 1 0 ∂xx∂tηε∂xxb = Z T 0 Z Ωηε(t) f · φ + Z T 0 Z 1 0 g b + Z Ωη0 u0· φ(0) + Z 1 0 η1b(0), for all (φ, b) ∈ (Vη2ε∩ H1(bΩηε)) × (L 2(0, T ; H2 0((0, 1))) ∩ H1(0, T ; L2(0, 1))), s. t. φ(t, x, 1 + ηε(t, x)) = (0, 0, b(t, x))T, for a. e. (t, x) ∈ (0, T ) × (0, 1) (25) We are going to study separately the low frequencies and the high fre-quencies of ∂tηε and take advantage of the fact that ∂tηε is bounded in
L2(0, T ; H
1 2
00(0, 1)) independently of ε. This implies that we can control the
space high frequencies of ∂tηε in L2(0, T ; L2(0, 1)).
We define a basis of H02(0, 1) ∩ L20(0, 1) by taking eigenfunctions (ξi)i∈N
defined by: Z 1 0 ∂xxξi∂xxb = λi Z 1 0 ξib, ∀b ∈ H02(0, 1) s.t. Z 1 0 b = 0, ξi ∈ H02(0, 1), Z 1 0 ξi= 0,
with (λi)i∈N the sequence of increasing eigenvalues : λi> 0, λi→ +∞. We
choose (ξi)i∈N orthonormal in L2(0, 1). We denote by dN0 the L2–projection
on the finite dimensional space span(ξi)0≤i≤N0 of any function d and by d
hf
the difference d − dN0. Thanks to the choice of the ξ
i the L2–projection on
the finite dimensional space span(ξi)0≤i≤N0 is stable in the L
2–norm as well
as in the H02–norm. We will use the following property that is obtained by Hilbertian interpolation: ∀d ∈ H 1 2 00(0, 1), kd hfk L2(0,1)≤ λ −1 8 N0kdkH12 00(0,1) (26) For σ > 1 we define vσ by vσ(x, y, z) = (σv1(x, y, σz), σv2(x, y, σz), v3(x, y, σz)).
We set b = Z t t−h ∂tηεN0(s)ds, and φ = Z t t−h uε− Rηε(∂tη hf ε ) σ(s)ds.
The integer N0 will be fixed later on and will be large enough. For well
chosen σ, they are admissible tests functions. Indeed, φ is divergence free thanks to the definitions of the lifting operator Rηε, the extension operator
v 7→ v and the operator v 7→ vσ. Moreover φ belongs to H1(0, T ; H1(BM)).
The function b belongs to H1(0, T ; H02(0, 1)). Both of them are bounded in the previous spaces independently of ε but not of N0. In what follows C
will denote a strictly positive constant that depends only on the data and not on ε and N0, and CN0 will also denote a strictly positive constant that
depends on the data and not on ε but may depend on N0. Moreover since
kηεkL∞(0,T ;H2 0(0,1))∩W1,∞(0,T ;L2(0,1))≤ C and L∞(0, T ; H2(0, 1))∩W1,∞(0, T ; L2(0, 1)) ,→ C0,λ([0, T ]; C0([0, 1])), 0 < λ < 3 4, we have kηε− ηε−kC0([0,T ]×[0,1])≤ Chλ. Thus if σ is such as σ ≥ 1 +2C α h λ, we have φ(t, x, 1 + ηε(t, x)) = 0, Z t t−h ∂tηεN0(s, x, y)ds T , on (0, 1).
In the sequel we choose σ = 1 +2C α h
λ, 0 < λ < 3
of test functions we made, (25) writes: − Z T 0 Z Ωηε(t) uε· ((vε)σ− (vε)−σ) + Z T 0 Z Ωηε(t) (uε· ∇)uε· Z t t−h (vε)σ + ν Z T 0 Z Ωηε(t) ∇uε: ∇ Z t t−h (vε)σ + Z Ωηε(T ) uε(T ) · Z T T −h (vε)σ − Z T 0 Z 1 0 ∂tηε∂t(ηεN0− (ηεN0)−) − Z T 0 Z 1 0 (∂tηε)2(ηεN0 − (ηNε0)−) + Z T 0 Z 1 0 ∂xxηε∂xx(ηεN0 − (ηNε0)−) + ε Z T 0 Z 1 0 ∂xx∂tηε∂xx(ηεN0− (ηεN0) − ) + Z 1 0 ∂tηε(T )(ηεN0(T ) − ηNε0(T − h)) = Z T 0 Z Ωηε(t) f · Z t t−h (vε)σ + Z T 0 Z 1 0 g(ηN0 ε − (ηεN0)−), (27) with vε= uε− Rηε(∂tη hf
ε ). The first term can be written:
− Z T 0 Z Ωηε(t) uε· ((vε)σ− (vε)−σ) = − Z T 0 Z Ωηε(t) uε· (uε− u−ε) − Z T 0 Z Ωηε(t) uε·(Rηε(∂tη hf ε )−Rηε(∂tη hf ε ) − )− Z T 0 Z Ωηε(t) uε·(((vε)σ−vε)−((vε)−σ−(vε)−)). We set I1= Z T 0 Z Ωηε(t) uε· (uε− u−ε). I1 = − 1 2 Z T 0 Z Ωηε(t) |uε|2+1 2 Z T 0 Z Ωηε(t) |u−ε|2−1 2 Z T 0 Z Ωηε(t) |uε− u−ε|2 = −1 2 Z T 0 Z Ωηε(t) |uε|2+1 2 Z T −h 0 Z Ωηε(t+h) |uε|2−1 2 Z T 0 Z Ωηε(t) |uε− u−ε|2 = 1 2 Z T −h 0 Z BM (ρ+ε−ρε)|uε|2− 1 2 Z T T −h Z Ωηε(t) |uε|2−1 2 Z T 0 Z Ωηε(t) |uε−u−ε|2. The same argument we use to prove (24) leads to
Z T 0 Z BM |ρ+ε − ρε||uε|2 ≤ C √ h. For the second term −R0T RΩ
ηε(t)uε· (Rηε(∂tη hf ε ) − Rηε(∂tη hf ε ) − ) we have, taking into account the energy estimates,
− Z T 0 Z Ωηε(t) uε· (Rηε(∂tη hf ε ) − Rηε(∂tη hf ε ) − ) ≤ CkuεkL2(0,T ;L2(Ω ηε(t)))kRηε(∂tη hf ε )kL2(0,T ;L2(B M)) ≤ C(kRηε(∂tηεhf)kL2(0,T ;L2(Ω ηε(t)))+ k∂tη hf ε kL2(0,T ;L2(0,1))).
Thanks to lemma 3, we obtain − Z T 0 Z Ωηε(t) uε· (Rηε(∂tη hf ε ) − Rηε(∂tη hf ε ) − ) ≤ Ck∂tηεhfkL2(0,T ;L2(0,1)).
The definition of ∂tηεhf and the fact that ∂tηεis bounded in L2(0, T ; H
1 2
00(0, 1))
independently of ε, imply, remembering (26), − Z T 0 Z Ωηε(t) uε·(Rηε(∂tη hf ε )−Rηε(∂tη hf ε ) − ) ≤ Ck∂tηhfε kL2(0,T ;L2(0,1))≤ Cλ −18 N0. We have also : Z T 0 Z Ωηε(t) uε· (((vε)σ− vε) − ((vε)−σ − v−ε)) ≤ 2kuεkL2(0,T ;L2(Ω ηε(t)))k(vε)σ− vε)kL2(0,T ;L2(Ωηε(t))) ≤ (σ − 1)kvεkL2(0,T ;H1(Ω ηε(t))) ≤ CN0(σ − 1) ≤ CN0h λ This yield I1 ≤ CN0h λ+ Cλ−18 N0 − 1 2 Z T 0 Z Ωηε(t) |uε− u−ε|2.
Now we take care of the convection term: I2 = Z T 0 Z Ωηε(t) (uε· ∇)uε· Z t t−h (vε)σ We have |I2| ≤ 1 2 Z T 0 kuεkL4(Ω ηε(t))k∇uεkL2(Ωηε(t)) Z t t−h k(vε)σkL4(Ω ηε(t)) ≤ √ h 2 Z T 0 kuεkL4(Ω ηε(t))k∇uεkL2(Ωηε(t)) Z t t−h k(vε)σk2L4(Ω ηε(t)) 12 ≤ C√h.
The next term to consider is I3= ν
Z T 0 Z Ωηε(t) ∇uε : ∇ Z t t−h (vε)σ . |I3| ≤ ν Z T 0 k∇uεkL2(Ω ηε(t)) Z t t−h k∇(vε)σkL2(Ω ηε(t)) ≤ ν√h Z T 0 k∇uεkL2(Ω ηε(t)) Z t t−h k∇(vε)σk2L2(Ω ηε(t)) 12 ≤ CN0√h.
The term I4= Z Ωηε(T ) uε(T ) · Z T T −h (vε)σ
can be estimated as follows:
|I4| ≤ kuε(T )kL2(Ωηε(T )) Z T T −h k(vε)σkL2(Ωηε(T )) ≤ √hkuε(T )kL2(Ω ηε(T )) Z T T −h k(vε)σk2L2(Ω ηε(T )) 1 2 ≤ C√h. We set I5= − Z T 0 Z 1 0 ∂tηε∂t(ηεN0 − (ηNε0)−).
We have thanks to the definition of ηN0
ε : I5 = − 1 2 Z T 0 Z 1 0 (∂tηεN0)2+ 1 2 Z T 0 Z 1 0 ((∂tηεN0)−)2− 1 2 Z T 0 Z 1 0 (∂tηNε0 − (∂tηεN0)−)2 = −1 2 Z T T −h Z 1 0 (∂tηεN0)2− 1 2 Z T 0 Z 1 0 (∂tηNε0 − (∂tηεN0) − )2 ≤ −1 2 Z T 0 Z 1 0 (∂tηεN0 − (∂tηεN0) − )2. For the next term we have:
|I6| = 1 2 Z T 0 Z 1 0 (∂tηεN0)2(ηεN0 − (ηεN0)−) ≤ 1 2 Z T 0 k∂tηεN0k2L4(0,1)kηεN0− (ηεN0)−kL2(0,1), but we have kηN0 ε − (ηεN0) −k L2(0,1)≤ Z t t−h k∂tηεN0kL2(0,1)≤ Ch.
Consequently taking into account the continous imbedding of H12(0, 1) in
L4(0, 1):
|I6| ≤ Ch.
The next term to consider is I7=
Z T 0 Z 1 0 ∂xxηεN0∂xx(ηNε0 − (ηεN0) − ). It can be estimated as follows |I7| ≤ Z T 0 k∂xxηN0 ε kL2(0,1) Z t t−h k∂xx∂tηεN0kL2(0,1) ≤ √h Z T 0 k∂xxηεN0kL2(0,1) Z t t−h k∂xx∂tηNε0k2L2(0,1) 12 ≤ CN0√h.
The additional viscous term gives for ε ≤ 1: |I8| = ε Z T 0 Z 1 0 ∂xx∂tηεN0∂xx(ηεN0− (ηεN0) − ) ≤ Z T 0 k∂xx∂tηεN0kL2(0,1) Z t t−h k∂xx∂tηεN0kL2(0,1) ≤ √h Z T 0 k∂xx∂tηεN0kL2(0,1) Z t t−h k∂xx∂tηNε0k2L2(0,1) 12 ≤ CN0√h. We set I9= Z 1 0 ∂tηεN0(T )(ηε(T ) − ηNε0(T − h)). |I9| = k∂tηε(T )kL2(0,1) Z T T −h k∂tηN0 ε kL2(0,1) ≤ Ch. Next I10 = Z T 0 Z Ωηε(t) f · Z t t−h (vε)σ and I11 = Z T 0 Z 1 0 g(ηN0 ε − (ηNε0)−)
can be estimated respectively by C√h and Ch. These estimates yield
Z T 0 Z Ωηε(t) |uε− u−ε|2+ Z T 0 Z 1 0 (∂tηεN0− (∂tηεN0) − )2 ≤ CN0 √ h + Cλ− 1 8 N0. Moreover Z T 0 Z 1 0 (∂tηεhf − (∂tηεhf) − )2 ≤ Cλ− 1 4 N0.
Then for all β > 0 if N0 is chosen large enough we have
Z T 0 Z Ωηε(t) |uε− u−ε|2+ Z T 0 Z 1 0 (∂tηε− (∂tηε)−)2 ≤ CN0 √ h +β 2, ∀ε > 0 Thus there exits h0 such that ∀h ≤ h0
Z T 0 Z Ωηε(t) |uε− u−ε|2+ Z T 0 Z 1 0 (∂tηε− (∂tηε)−)2 ≤ β, ∀ε > 0
This proves Lemma 5.
Thanks to Lemma 4, we obtain that ∂tηεis compact in L2(0, T ; L2(0, 1))
and that uε is compact in L2(0, T ; L2(BM)).
We now want to verify the convergences announced in Proposition 1 and to verify that the equality between the structure velocity and the fluid velocity at the interface holds in the limit.
Let T > 0 such that infεmin[0,T ]×[0,1](1 + ηε) ≥ α > 0. Let us denote
by (η, ˜u) the limit of a subsequence of (ηε, uε). We will denote any
subse-quence of (ηε, uε) by (ηε, uε). Thanks to the energy estimates and to the
compactness properties that have just been derived, we have the following convergences as ε goes to zero:
ηε → η uniformly in C0([0, T ]; C1[0, 1]), ηε * η weakly in L2(0, T ; H02(0, 1)), ∂tηε → ∂tη strongly in L2(0, T ; L2(0, 1)), uε → u˜ strongly in L2(0, T ; L2(BM)), ρεuε → ρ˜u strongly in L2(0, T ; L2(BM)). (28)
Moreover ρε∇uε tends to some z weakly in L2(0, T ; L2(BM)) as ε goes to
zero. It is easy to verify that, since ηε→ η in C0([0, T ] × [0, 1]), that z = 0
in dBM \ bΩη and z|Ωbη = ∇(˜u|Ωbη). Thus
ρε∇uε * ρ∇(˜u|Ωbη) in L
2(0, T ; L2(B
M)), (29)
Next we take care of the equality
uε(t, x, 1 + ηε(t, x)) = (0, ∂tηε(t, x))T
on (0, T ) × (0, 1). The right hand side converges to (0, ∂tη)T strongly in
L2(0, T ; L2(0, 1)). For the left hand side, the function ρε(uε − Rηε(∂tηε))
belongs to L2(0, T ; H1(BM)) and is bounded in this space independently of ε.
Thus a subsequence of ρε(uε−Rηε(∂tηε)) converges weakly in L
2(0, T ; H1(B M))
to w0. It is clear, thanks to the convergences of ηε, that w0 is equal to
ρ(˜u − Rη(∂tη)). Hence for a. e. t ˜ u = 0 on Γ0, and ˜ u ◦ χη = (0, ∂tη)T on (0, 1) × {1}.
4
Passage to the limit – Proof of Theorem 2
Next we pass to the limit in the weak formulation: Z Ωηε(t) uε(t) · φε(t) − Z t 0 Z Ωηε(s) uε· ∂tφε+ ν Z t 0 Z Ωηε(s) ∇uε: ∇φε +1 2 Z t 0 Z Ωηε(s) (uε· ∇)uε· φε− 1 2 Z t 0 Z Ωηε(s) (uε· ∇)uε· φε −1 2 Z t 0 Z 1 0 ∂tηε∂tηεb + Z 1 0 ∂tηε(t) b(t) − Z t 0 Z 1 0 ∂tηε∂tb + Z t 0 Z 1 0 ∂xx∂tηε∂xxb + Z t 0 Z 1 0 ∂xxηε∂xxb = Z t 0 Z Ωηε(t) f ·φε+ Z t 0 Z 1 0 g b+ Z Ω(0) uε0·φε(0)+ Z 1 0 ηe1b(0),
for a. e. t and for all (φε, b) ∈ (Vη2ε ∩ H1(bΩ
ηε)) × (L
2(0, T ; H2
0((0, 1))) ×
H1(0, T ; L2(0, 1))) such that φε(t, x, 1 + ηε(t, x, y)) = (0, 0, b(t, x))T, (t, x) ∈
[0, T ] × ω.
The fluid test functions should depend on ε. However, it is sufficient to consider test functions that do not depend on ε and that are admissible for any ε small enough.
First we consider test functions of the form (φ0, 0), such that φ0 belongs to D(∪t∈[0,T ]{t} × Ωη(t)) and div φ0 = 0. These test functions satisfy the property that φ0(t, ·) ∈ D(Ωη(t)), ∀t. For ε small enough, since ηεconverges
uniformly to η, φ0∈ D(∪t∈[0,T ]{t} × Ωηε(t)).
The second pair of test functions we consider is (φ1, b) where b belongs to L2(0, T ; H02(0, 1)) ∩ H1(0, T ; L2(0, 1)) withR01b = 0 and for a. e. t
φ1 = (0, b)T in dBM \ bCα R(0,yαb)T in bCα.
We have easily that φ1 belongs to L2(0, T ; H1(BM)) and div (φ1) = 0.
Moreover since min[0,T ]×[0,1](1 + ηε) ≥ α, φ1(t, x, 1 + ηε(t, x)) = (0, b(t, x))T
on (0, T ) × (0, 1), ∀ε. Furthermore we can choose the linear operator R such that kR(0,yb α) Tk L2(C α)≤ CkbkL2(0,1). (30)
It can be done by solving a Stokes problem in Cα. Indeed this type of
inequality can be obtained thanks to a transposition argument and relies on a regularity result for the Stokes problem which is true here since Cα is a
R is chosen such that (30) holds, we deduce that, for a. e. t ∈ I ⊂⊂ (0, T ), and for h small enough
kR(0,yb α) T(t) − R(0,yb α) T(t + h)k L2(C α)≤ Ckb(t) − b(t + h)kL2(0,1).
Since ∂tb ∈ L2(0, T ; L2(0, 1)) this implies that ∂tR(0,ybα)T ∈ L2(0, T ; L2(Cα))
and that ∂tφ1∈ L2(0, T ; L2(BM)). Consequently (φ1, b) is a pair of
admis-sible test functions for all ε.
With both type of test functions, it is easy to pass to the limit in the weak formulation as ε goes to zero. We obtain the existence of one weak solution on (0, T ) of (18) that satisfies energy estimates.
Eventually, we show that we can extend the solution, as long as we have min[0,T ]×[0,1](1 + η) > 0. We do exactly as in [8], but for the sake of
com-pletness we reproduce the proof here. We build an increasing sequence of times (Tk)k≥1 as follows. First we choose a time T1 > 0 such that there
exists a weak solution up to T1, with m1 = min[0,T1]×[0,1](1 + η) > 0.
Pos-sibly changing slightly T1, we may moreover assume that η(T1) ∈ H02(0, 1),
∂tη(T1) ∈ L2(0, 1) and u(T1) ∈ L2(Ωη(T1)) (since this is true for almost each
time).
Now, let k ≥ 1 and assume that we have built a solution up to some time Tk, with mk= min[0,Tk]×[0,1](1 + η) > 0. Our construction allows us to
build an extension of our solution, on some time interval starting from Tk.
Thanks to the energy estimate (21) (see also (9)), we have for s ≥ Tk for
any 0 < λ < 34 1 + η(s) ≥ 1 + η(Tk) − (s − Tk)λC(Tk, s) ≥ mk− (s − Tk)λC(Tk, s) , (31) with C(Tk, s) = ˜C ku(Tk)kL2(Ω η0), kη(Tk)kH02(0,1), k∂tη(Tk)kL20(0,1), Z s Tk exp(s − u)(kf kL2(Ω
η(u))(u) + kgkL2(0,1)(u))du
,
where ˜C is positive and nondecreasing with respect to its arguments, and C(Tk, s) ≤ C(0, s). This a priori estimate shows that if we let τk =
min{1, (mk/2C(Tk, Tk+ 1))
1
λ}, we can build a solution starting from u(Tk)
and η(Tk), ∂tη(Tk) up to the time Tk+ τk (this corresponds to choosing
close to Tk+τk(in [Tk+τk/2, Tk+τk]), in order to have also η(Tk) ∈ H02(0, 1),
∂tη(Tk+1) ∈ L2(0, 1) and u(Tk+1) ∈ L2(Ωη(Tk+1)).
If the sequence (Tk)k≥1 is infinite we let T∗ = supkTk. If T < +∞, it
must be that m = min[0,T∗]×[0,1](1 + η) = 0. Otherwise, we have mk ≥ m
for all k, hence τk ≥ min{1, (m/2C(0, T∗))
1
λ} > 0. But Tk+1 − Tk ≥ τk/2
and goes to zero, a contradiction. This achieves the proof of the theorem.
5
Appendix
Here we give the sketch of the proof of lemma 3. Let γ ∈ W1,∞(0, T ; L2(0, 1))∩
L∞(0, T ; H02(0, 1)) satisfying M ≥ min[0,T ]×[0,1](1+γ) ≥ α > 0. We consider
the following Stokes problem: −∆v + ∇q = 0 in Ωγ(t), div (v) = 0 in Ωγ(t), v(t, x, 1 + γ(t, x)) = (0, b(t, x))T on (0, 1), v = 0 on Γ0, (32) where b is given in L2(0, T ; H 1 2 00(0, 1) such that R1
0 b = 0. This problem has
a unique solution (v, q) in H1(Ωγ(t)) × L20(Ωγ(t)) for a. e. t (see [17], [33]).
The regularity with respect to the time can be obtained by looking at the regularity with respect to the time of (w, µ) = (v ◦ χγ, q ◦ χγ), that satisfies
after a change of variables −div ((Aγ∇)w) + (Bγ∇)µ = 0 in Ω, div (BTγw) = 0 in Ω, w = (0, b)T on (0, 1) × {1}, w = 0 on Γ0, (33)
where Aγ and Bγ are matrices depending on γ:
Aγ =
1
1 + γ(cof∇χγ)
Tcof∇χ
γ, Bγ= cof∇χγ,
where cof A denotes the cofactor matrix of the matrix A. These matrices verify the following properties:
• Aγ∈ C0([0, T ] × Ω), A γ symmetric. • ∃λ > 0, s. t. Aγ≥ λI, ∀γ ∈ W1,∞(0, T ; L2(0, 1)) ∩ L∞(0, T ; H2 0(0, 1)), s. t. min[0,T ]×[0,1](1 + γ) ≥ α > 0,
• Bγ ∈ C0([0, T ] × Ω) invertible.
• Aγ, Bγ ∈ C0([0, T ]; Xs) with Xs = H1(−δ, 1 + δ; Hs(0, 1)) for any
s ≥ 0.
• div (Bγ) = 0. This identity is known as the Piola’s identity (see [9]). First we are going verify the inf–sup condition (see [6]), to derive an estimate on w in L2(0, T ; H1(Ω)) and an estimate of µ in L2(0, T ; L2(Ω)). That will give estimate (15) after a change of variables. Next we prove that
kwkL2(0,T ;L2(Ω))≤ C(γ)kbkL2(0,T ;L2(0,1)),
by a transposition argument. In all what follows C(γ) will denote a strictly positive constante that depends only on the norms of γ in W1,∞(0, T ; L2(0, T )) and L∞(0, T ; H2(0, 1)), on M and α. The transposition argument relies on
the regularity of the solution of −div ((Aγ∇) ˜w) + (Bγ∇)˜µ = h in Ω, div (BγTw) = 0˜ in Ω, ˜ w = 0 on ∂Ω, (34) where h is given in L2(0, T ; L2(Ω)). But first of all we give a technical lemma
Lemma 6 Let g be in H1(Ω), let f be in Xs with s ≥ 1. Then f g belongs
to H1(Ω) and kg∂if kL2(Ω) ≤ Ckgk 1 2 H1(Ω)kgk 1 2 L2(Ω)kf kXs,
where ∂i denotes the derivation with respect to x as well as y.
Proof of Lemma 6.
The first assertion relies on the facts that Xs is continously embbeded in
L∞(Ω), ∂xf belongs to L2(−δ, 1 + δ; L∞(0, 1)), ∂yf belongs to L∞(−δ, 1 +
δ; L2(0, 1)) and that H1(Ω) = H1(−δ, 1+δ; L2(0, 1))∩L2(−δ, 1+δ; H1(0, 1)). For the second assertion we have
kg∂xf kL2(Ω) ≤ kgkL∞(−δ,1+δ;L2(0,1))k∂xf kL2(−δ,1+δ;L∞(0,1)) ≤ C Z 1 0 (supessx|g(x, y)|)2dy 12 kf kXs.
Using a Gagliardo–Nirenberg inequality in dimension one: kvkL∞(Λ)≤ Ckvk 1 2 L2(Λ)kvk 1 2 H1(Λ),
where Λ is a bounded interval of R, leads to kg∂xf kL2(Ω) ≤ Ckgk 1 2 L2(Ω)kgk 1 2 L2(0,1;H1(−δ,1+δ))kf kXs, ≤ Ckgk 1 2 L2(Ω)kgk 1 2 H1(Ω)kf kXs. Similarly kg∂yf kL2(Ω) ≤ kgkL2(−δ;1+δ;L∞(0,1))k∂yf kL∞(−δ,1+δ;L2(0,1)), ≤ Ckgk 1 2 L2(Ω)kgk 1 2 L2(−δ,1+δ;H1(0,1))kf kXs, ≤ Ckgk 1 2 L2(Ω)kgk 1 2 H1(Ω)kf kXs.
This ends the proof of lemma 6.
• Estimate of (w, µ) in L2(0, T ; H1(Ω)) × L2(0, T ; L2(Ω)).
Let zb be in L2(0, T ; H1(Ω)) such that zb = 0 on (0, T ) × Γ0 and
zb = (0, b)T on (0, T ) × ((0, 1) × {1}). Such a zb exists. Indeed if we
con-sider r a linear continuous lifting operator from H12(∂Ω) onto H1(Ω), then
r (0, yb)T satisfies the desired properties and the following estimate: kr (0, yb)T kL2(0,T ;H1(Ω))≤ Ckbk
L2(0,T ;H12 00(0,1))
.
Setting w1= w − r (0, yb)T = w − zb, the pair (w1, µ) is solution of
−div ((Aγ∇)w1) + (Bγ∇)µ = div ((Aγ∇)zb) in Ω,
div (BTγw1) = −div (BγTzb) in Ω,
w = 0 on ∂Ω.
Thanks to lemma 6, BTγzb belongs to L2(0, T ; H01(Ω)), thus −div (BγTzb) ∈
L2(0, T ; L20(Ω)) and kdiv (BTγzb)kL2(0,T ;L2(Ω))≤ C(γ)kbk L2(0,T,H 1 2 00(0,1)) . (35)
Remark 5 We could have also remark that div (BTγzb) = Bγ : ∇zb du to
Piola’s identity. That implies directly that −div (BγTzb) ∈ L2(0, T ; L2(Ω))
Now we build z2 satisfying z2 ∈ L2(0, T ; H01(Ω)) such that div (BγTz2) = −div (BT γzb) and kz2kL2(0,T ;H1 0(Ω))≤ C(γ)kbkL2(0,T ;H12 00(0,1)) ,
which is equivalent to the inf–sup condition for our Stokes–like problem. One knows (see [6]) that there exists C > 0 such that for any χ in L20(Ω) there exists z in H01(Ω) that satisfies div (z) = χ and kzkH1(Ω) ≤ CkχkL2(Ω).
We set ˜z = Bγ−Tz. We have, thanks to lemma 6, that ˜z ∈ H01(Ω), for all t (in particular ˜z ∈ C0([0, T ]; H01(Ω))) and
k˜zkH1(Ω)≤ CkBγkXskzkH1(Ω), s ≥ 1,
which gives, using the fact that Bγ ∈ C0([0, T ]; Xs),
k˜zkH1(Ω)≤ C(γ)kχkL2(Ω).
Moreover if χ ∈ L2(0, T ; L2
0(Ω)) then ˜z ∈ L2(0, T ; H01(Ω)) and
k˜zkL2(0,T ;H1(Ω))≤ C(γ)kχkL2(0,T ;L2(Ω)).
Furthermore div (BTγ˜z) = χ. Thus we are able to associate to χ = −div (BγTzb)
a vector z2 that satisfies the desired properties thanks to (35). Setting
w2 = w1− z2, the pair (w2, µ) satisfies
−div ((Aγ∇)w2) + (Bγ∇)µ = div ((Aγ∇)zb) + div ((Aγ∇)z2) in Ω,
div (BTγw2) = 0 in Ω,
w = 0 on ∂Ω.
Once again thanks to lemma 6 the right hand side of the first equation be-longs to L2(0, T ; H−1(Ω)) and can be estimated in this space by C(γ)kbk
L2(0,T ;H12 00(0,1))
. Thus using the ellipticity of Aγand the inf–sup condition we obtain by
stan-dard arguments that kw2kH1(Ω)+ kµ − 1 |Ω| Z Ω µkL2(Ω)≤ C(γ)kbk H 1 2 00(0,1) , for a. e. t, and kw2kL2(0,T ;H1(Ω))+ kµ − 1 |Ω| Z Ω µkL2(0,T ;L2(Ω))≤ C(γ)kbk L2(0,T ;H 1 2 00(0,1)) .
But if µ0 denotes µ − |Ω|1 R Ωµ, then R Ωµ = |Ω| |Ωγ(t)| R Ωµ0 det∇χγ. Conse-quently kwkL2(0,T ;H1(Ω))+ kµkL2(0,T ;L2(Ω))≤ C(γ)kbk L2(0,T ;H12 00(0,1)) , which yields kvkL2(0,T ;H1(Ω γ(t)))+ kqkL2(0,T ;L2(Ωγ(t)))≤ C(γ)kbk L2(0,T ;H12 00(0,1)) . Thus the first part of lemma 3 is proven. We now take care of the second estimate of lemma 3.
• Proof of estimate (16). Let h be given in L2(0, T ; L2(Ω)).
We have for a. e. t Z Ω w · h = Z (0,1)×{1} ((Aγ∇) ˜w · n − ˜µBγ· n)2b,
where ( ˜w, ˜µ) ∈ L2(0, T ; H01(Ω)) × L2(0, T ; L20(Ω)) is the solution of (34) and n is equal to (0, 1)T. We already know that
k ˜wkH1(Ω)+ k˜µkL2(Ω) ≤ C(γ)khkL2(Ω), for a. e. t, (36)
and
k ˜wkL2(0,T ;H1(Ω))+ k˜µkL2(0,T ;L2(Ω)) ≤ C(γ)khkL2(0,T ;L2(Ω)). (37)
Let us admit for the time being that
k ˜wkL2(0,T ;H2(Ω))+ k˜µkL2(0,T ;H1(Ω))≤ C(γ)khkL2(0,T ;L2(Ω)), (38)
then, du to the regularity of Aγ and Bγ, we deduce that
k(Aγ∇) ˜w · n − ˜µBγnkL2(0,T ;L2((0,1)×{1})))≤ C(γ)khkL2(0,T ;L2(Ω))
and thus
kwkL2(0,T ;L2(Ω))≤ C(γ)kbkL2(0,T ;L2(0,1)).
That implies (16) after a change of variables.
Now we are going to prove that (38) holds true. Several papers regard the regularity of the weak solution of the Stokes problem in various class of domain. In [7] the case of a C2–domain is investigated, in [2] the results are obtained for W2,∞–domains. For polygonal domains one can refer to [12],
[19], [22]. . . For regularity results obtain by nonlinear interpolation see [32]. For an abstract version of a regularity theorem in the spirit of the translation
method of L. Nirenberg, we refer to a note of L. Tartar [31]. Note that this abstract result applies here away from the corners. Nevertheless we give the main points of the proof for our very special case: we adapt (away from the corners of the domain) the translation method to prove that, for a. e. t, ( ˜w, ˜µ) ∈ H2(Ω) × H1(Ω) and that
k ˜wkH2(Ω)+ k˜µkH1(Ω)≤ C(γ)khkL2(Ω), (39)
where C(γ) depends on the W1,∞(0, T ; L2(0, 1)) and L∞(0, T ; H02(0, 1)) norms of γ, on α and M . Consequently by integrating the previous inequality with respect to the time we obtain (38).
Remark 6 Using lemma 6, one can show that this implies that ( ˜w ◦χ−1γ , ˜µ◦ χ−1γ ) belongs to L2(0, T ; H2(Ωγ(t))) × L2(0, T ; H1(Ωγ(t))).
Remark 7 Note that the proof described below applies in RN for W2,q– domains with q > N with only a few changes [4].
The first step is the step of localization. Then we study each local problem. Let us consider k open sets (Ui)1≤i≤k such that Ω ⊂ ∪ki=1Ui, and
such that Ui, 1 ≤ i ≤ 4 are neighbourghoods of the corners of Ω. Let us
consider θi ∈ D(Ui), 0 ≤ θi ≤ 1, and P θi = 1 in Ω. We set (wi, µi) =
(θiw, θ˜ iµ), it is easy to verify that it satisfies:˜ −div ((Aγ∇)wi) + (B γ∇)µi = fi in Ω ∩ Ui, div (BT γwi) = gi in Ω ∩ Ui, wi= 0 on ∂(Ω ∩ Ui), (40)
with (using the Einstein’s convention on the repeated indice)
fli = θihl+ 2(Aγ)km∂kθi∂mw˜l+ (Aγ)kmw˜l∂kmθi+ ∂k(Aγ)km∂mθiw˜l,
and
gi = (BγTw) · ∇θ˜ i. Since ( ˜w, ˜µ) ∈ H1
0(Ω)×L20(Ω) for a. e. t, taking into account the regularities
of Aγ, Bγ and thanks to lemma 6, fi ∈ L2(Ω ∩ Ui) and gi ∈ H1(Ω ∩ Ui) for
a. e. t. Moreover since ( ˜w, ˜µ) ∈ L2(0, T ; H01(Ω)) × L2(0, T ; L20(Ω)), we have fi ∈ L2(0, T ; L2(Ω ∩ U i)) and gi ∈ L2(0, T ; H1(Ω ∩ Ui)). Furthermore (36), (37) yield kfik L2(Ω∩U i)+ kg ik H1(Ω∩U i) ≤ Ci(γ)khkL2(Ω), for a. e. t, (41)
and kfikL2(0,T ;L2(Ω∩U i))+ kg ik L2(0,T ;H1(Ω∩U i))≤ Ci(γ)khkL2(0,T ;L2(Ω)). (42)
For 1 ≤ i ≤ 4, if the neibourghoods Uiare well chosen, then the matrices
Aγ and Bγare equal to the identity matrix in those neibourghoods and thus
the local problems reduce to standard non homogeneous Stokes problems. In [12], [22] it is shown that under compatibility conditions at the corner for gi (that is to say gi ∈ H
1 2
00(Γij) where Γij denote the edges associated
to the corners) the solution of (40) belongs to H2(Ω ∩ Ui) × H1(Ω ∩ Ui)
together with estimates with respect to the data. Here gi is equal to zero in
a neibourghood of its corresponding corner thus the result holds true. Remark 8 In the case where δ is equal to zero then the last argument is not valid any more and Aγ, Bγ are not equal to the identity matrix in the
neighbourhood of the corners. Consequently one can not apply directly the regularity results for the Stokes problem in polygonal domains. This is the reason why we took δ > 0.
For i > 4 the first remark that can be made is the following: if we assume that (wi, µi) belongs to H2(Ω ∩ U
i) × H1(Ω ∩ Ui) then, using lemma 6,
−div ((Aγ∇)wi) + (Bγ∇)µi and div (BγTwi) = Bγ: ∇wi belong respectively
to L2(Ω ∩ Ui) and H1(Ω ∩ Ui), for a. e. t. Thus one can hope that the
desired regularity holds true. To prove it we are going to regularize with respect to the space variables the matrices Aγ, Bγ. We denote by Aεγ, Bγε
these regularized matrices. We assume that they satisfies: Aε
γ ≥ λ2I, Aε
symetric and that there are bounded independently of ε in C0([0, T ]; Xs) for
all s ≥ 0 and converge respectively toward Aγ and Bγ in C0([0, T ]; Xs). Let
us denote by (wi,ε, µi,ε) the solution of (40) with Aεγ, Bγε instead of Aγ, Bγ,
i. e. the solution of −div ((Aε γ∇)wi,ε) + (Bεγ∇)µi,ε= fi in Ω ∩ Ui, div (BγεTwi,ε) = gi in Ω ∩ Ui, wi,ε= 0 on ∂(Ω ∩ Ui). (43)
Note that the inf–sup condition associated to this mixed problem is satisfied uniformly with respect to ε. Thus the solution verifies
kwi,εkH1(Ω∩U i)+ kµ
i,εk L2(Ω∩U
and kwi,εkL2(0,T ;H1(Ω∩U i))+ kµ i,εk L2(0,T ;L2(Ω∩U i)) ≤ Ci(γ)khkL2(0,T ;L2(Ω)). (45)
Furthermore it belongs to H2(Ω ∩ Ui) × H1(Ω ∩ Ui), for a. e. t (see [2], [7])
and satisfies estimates in those spaces. Moreover thanks to the regularity with respect to the time of the right hand sides fi, gi and of Aεγ, Bγε the pair (wi,ε, µi,ε) ∈ L2(0, T ; H2(Ω∩Ui))×L2(0, T ; H1(Ω∩Ui)). But those estimates
may depend on ε. The next step is to derive estimates of (wi,ε, µi,ε) in
H2(Ω ∩ Ui) × H1(Ω ∩ Ui), for a. e. t that do not depend on ε. To do so
we follow the same lines as in [2] pp 192 and after, where the translation method is used. Consequently we do not detail all the calculations but give only the main points. We denote by Dh, h ∈ R2\ {0}, h//ex the quotient
defined by
Dh(v)(·) =
v(· + h) − v(·) |h| , where ex denotes the unit vector of the x–axis.
For |h| small enough, it is easy to show, using the properties of the differ-ents quotidiffer-ents, that (Dh(wi,ε), Dh(µi,ε)) is the solution of Stokes–like
prob-lem as (43) with right hands sides where there are among others products of components of Dh(Aεγ) and of wi,ε. Taking into account the propreties of
Aεγ, Bγε and lemma 6, we can obtain for a. e. t
k∇Dh(wi,ε)kL2(Ω∩U i)+kDh(µ i,ε)k L2(Ω∩U i) ≤ C kAεγkXskw i,εk12 H2(Ω∩U i)kw i,εk12 H1(Ω∩U i) + kBεγkXskµi,εk 1 2 H1(Ω∩U i)kµ i,εk12 L2(Ω∩U i)+ kf ik L2(Ω∩U i)+ kg ik H1(Ω∩U i) , for s ≥ 1, where C does not depend on ε. Thus since (wi,ε, µi,ε) is bounded independently of ε in H1(Ω ∩ Ui) × L2(Ω ∩ Ui), for a. e. t (see (44)), and
du to (41), we have for a. e. t k∇Dh(wi,ε)kL2(Ω∩U i)+ kDh(µ i,ε)k L2(Ω∩U i) ≤ Ci(γ)khk 1 2 L2(Ω) kwi,εk 1 2 H2(Ω∩U i)+ kµ i,εk12 H1(Ω∩U i) + Ci(γ)khkL2(Ω). (46)
Then we use the equation satisfied by (wi,ε, µi,ε) in order to take care of ∂yywi,ε and ∂yµi,ε. We have, once again using lemma 6 and (44), for a. e. t
k∂yy(wi,ε)kL2(Ω∩U i)+ k∂yµ i,εk L2(Ω∩U i) ≤ Ci(γ)khk 1 2 L2(Ω) kwi,εk 1 2 H2(Ω∩U i)+ kµ i,εk12 H1(Ω∩U i) + Ci(γ)khkL2(Ω). (47) Consequently from (46, 47) we obtain that (wi,ε, µi,ε) is bounded indepen-dently of ε in H2(Ω ∩ Ui) × H1(Ω ∩ Ui), for a. e. t and
kwi,εk
H2(Ω)+ kµi,εkH1(Ω)≤ C(γ)khkL2(Ω), for a. e. t,
Letting ε go to zero we deduce that
kwikH2(Ω)+ kµikH1(Ω) ≤ C(γ)khkL2(Ω), for a. e. t.
Thus is inequality being true for any i, (39) holds true. Next (38) follows by taking the L2–norm with respect to the time of (39).
This ends the proof of lemma 3.
Acknowledgment
I would like to thanks Professor L. Tartar for pointing out usefull references.
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