Local null controllability of a fluid-solid interaction problem in dimension 3

(1)

Local null controllability of a fluid-solid interaction problem in dimension 3

M. Boulakia

^∗

and S. Guerrero

^∗

Abstract

We are interested by the three-dimensional coupling between an incompressible fluid and a rigid body. The fluid is modeled by the Navier-Stokes equations, while the solid satisfies the Newton’s laws. In the main result of the paper we prove that, with the help of a distributed control, we can drive the fluid and structure velocities to zero and the solid to a reference position provided that the initial velocities are small enough and the initial position of the structure is close to the reference position. This is done without any condition on the geometry of the rigid body.

1 Introduction

1.1 Statement of problem

We consider a rigid structure immersed in a viscous incompressible fluid. At timet, we denote by Ω_S(t) the domain occupied by the structure. The structure and the fluid are contained in a fixed bounded domain Ω⊂R³. Let O ⊂⊂Ω be the control domain. We suppose that the boundaries of Ω_S(0) and Ω are smooth (C⁴for instance) and that

ΩS(0)⊂Ω\ O, d(∂(Ω\ O),ΩS(0))>δ0>0 (1) For anyt >0, we note ΩF(t) := Ω\ΩS(t) the region occupied by the fluid andO ⊂⊂ Oe an open set. The time evolution of the eulerian velocity uand the pressure pof the fluid is governed by the incompressible Navier-Stokes equations: ∀t >0,∀x∈ΩF(t)

( (u_t+ (u· ∇)u)(t, x)− ∇ ·σ(u, p)(t, x) =v(t, x)ζ(x),

∇ ·u(t, x) = 0. (2)

The stress tensor is given by

σ(u, p) := 2µ(u)−pId,

where(u) :=¹₂(∇u+∇u^t) and the viscosity coefficientµis supposed to be positive. The functionζ∈C_c²(O) satisfiesζ= 1 inOe andv is a control force which acts over the system throughO.

At timet, the motion of the rigid structure is given by the positionb(t)∈R³ of the center of mass and by a rotation (orthogonal) matrix Q(t)∈M3×3(R). The domain Ω_S(t) is given byχ_S(t,Ω_S(0)), whereχ_S denotes the flow associated to the motion of the structure:

χS(t, y) =b(t) +Q(t)Q⁻¹₀ (y−b0) ∀y∈ΩS(0).

Here,Q0 andb0 are respectively the initial rotation matrix and the initial position of the solid.

∗Universit´e Pierre et Marie Curie-Paris 6, UMR 7598 Laboratoire Jacques-Louis Lions, Paris, F-75005 France. E-mails:

[email protected], [email protected].

(2)

Let r : (0, T)→R³ be the angular velocity. Then, the rotation matrix is the solution of the following system:





 dQ

dt(t) = (r×Q)(t) t∈(0, T), Q(0) =Q₀.

(3) For the equations of the structure, we denote bym >0 the mass of the rigid structure andJ(t)∈M3×3(R) its tensor of inertia at timet. This tensor is given by

J(t)d·d˜= Z

ΩS(0)

d×Q(t)(y−b₀)

· d˜×Q(t)(y−b₀)

dy ∀d,d˜∈R³. (4) One can prove that

J(t)d·d>C_J|d|² for alld∈R³,

where CJ is a positive constant independent of t >0. The equations of the structure motion are given by the balance of linear and angular momentum. We have, for allt∈(0, T)









 m¨b=

Z

∂Ω_S(t)

σ(u, p)n dγ, Jr˙= J r

×r+ Z

∂ΩS(t)

(x−b)×(σ(u, p)n)dγ.

(5)

In these equations, n is the outward unit normal to ∂ΩS(t). On the boundary of the fluid, the eulerian velocity has to satisfy a no-slip boundary condition. Therefore, we have, for allt >0

( u(t, x) = 0,∀x∈∂Ω,

u(t, x) = ˙b(t) +r(t)×(x−b(t)), ∀x∈∂ΩS(t).

(6) The system is completed by the following initial conditions:

u(0,·) =u0 in ΩF(0), b(0) =b0,b(0) =˙ b1, r(0) =r0, (7) which satisfy

u0∈H¹(ΩF(0)),∇ ·u0= 0 in ΩF(0), u0= 0 on∂Ω, u0(x) =b1+r0×(x−b0), x∈∂ΩS(0). (8) Let us now recall some of the most relevant results in interaction problems between a rigid structure and an incompressible fluid.

A local result was proved in [12], while the existence of global weak solutions is proved in [5] and [6]

(with variable density) and [18] (2D, with variable density); in this last paper, the existence of a solution is proved even beyond collisions. Later, the existence and uniqueness of strong global solutions in 2D was proved in [19] as well as the local in time existence and uniqueness of strong solutions in 3D.

In this paper, we prove the local null controllability of system (2)-(7). The same result was proved in [4]

and in [16] in dimension 2 provided that ΩS(0) satisfies some geometric properties. For the Burgers equation with a moving particle in dimension 1, the local null controllability was proved in [7]. In the absence of a solid, the local exact controllability to the trajectories of the Navier-Stokes equations was proved in [15].

This result was later improved in [9].

We state now the main result of this paper:

Theorem 1 There existsδ >0 such that for any(u₀, b₀, b₁, r₀, Q₀)satisfying (8),u₀∈H²(Ω_F(0))and ku₀k_H2(Ω_F(0))+|b₀|+|b₁|+|r₀|+|Q₀−Id|< δ, (9) there exists a control v∈L²(0, T;H¹(Ω)) such that the solution of (2)-(7) satisfies

u(T,·) = 0 in Ω_F(T), b(T) = 0, b(T) = 0, r(T˙ ) = 0, Q(T) =Id.

(3)

The proof of this result is based on a fixed-point argument. For this matter, we first consider a linearized system for which we prove the existence of controls inL²(0, T;H¹(Ω)) which drive the velocities to zero and the position of the structure to the desired reference position (b(T), Q(T)) = (0, Id).

This null controllability result is established with the help of a Carleman inequality for the associated adjoint system. To prove this Carleman inequality, we use a different and more concise method than the one presented in [15] and [9] and used in [4] and [16]: we first consider the parabolic equation satisfied by the curl of the solution (where the pressure does not appear) and establish a Carleman inequality for this parabolic problem in terms of two boundary integrals concerning some traces of the velocity. These boundary terms are then estimated thanks to regularity results which are stated and proved in the Appendix at the end of the paper.

1.2 A problem linearized with respect to the fluid velocity

Let us introduce







(ˆb,ˆr)∈H²(0, T)×H¹(0, T) (ˆb,b,˙ˆ ˆr)_|t=0= (b0, b1, r0).

(10) This allows us to define the following domains:

ΩbS(t) := ˆb(t) +Q(t)Qb ⁻¹₀ (ΩS(0)−b0)

and ΩbF(t) := Ω\ΩbS(t), where Qb is the solution of (3) with r replaced by ˆr. We suppose that the solid domain stays far away from∂(Ω\ O):

∃δ₁>0 :d(bΩ_S(t), ∂(Ω\ O))>δ₁ ∀t∈[0, T]. (11) Let us now define several notations which we will use all along the paper. We introduce the following spaces of functions defined on moving domains: forr, p∈N,

L²(L²) :=

(

umeasurable : Z T

0

Z

Ωb_F(s)

|u|²dx ds <+∞

) ,

L²(H^p) :=

(

u∈L²(L²) : Z T

0

kuk²_H_p_(b_Ω

F(s))ds <+∞

) ,

H^r(H^p) :=







u∈L²(L²) : Z T

0 r

X

β=0

k∂_t^βuk²_H_p_(b_Ω

F(s))ds <+∞





 , with the natural associated norms coming from the definition. On the other hand, we define

C⁰(L²) :={usuch thatu(s, x) :=e u(s, x)1

ΩbF(s)∈C⁰([0, T];L²(Ω))}

and

C^r(H^p) :={u:∂^β_t∂_x^αu∈C⁰(L²), ∀06β 6r, ∀06|α|6p}

with the associated norms given by

kuk_C0(L²):= max

t∈(0,T)ku(t)k_L2(bΩF(t))= max

t∈(0,T)keu(t)k_L2(Ω)

and

kuk_Cr(H^p):=

r

X

β=0

max

t∈(0,T)k∂_t^βu(t)k_Hp(bΩ_F(t)).

(4)

Let us now consider a velocity ˆusatisfying





 ˆ

u∈Zˆ:=H¹(L⁶)∩L²(W^2,6), ∇ ·uˆ= 0 x∈ΩbF(t), ˆ

u(t, x) = (b(t) + ˆ˙ˆ r(t)×(x−ˆb(t)))1_∂

Ωb_S(t)(x) x∈∂ΩbF(t).

(12)

Let us also introduce the spaces

Yˆ_k:=L²(H^2+k)∩H^1+k/2(L²) fork∈[−2,2]. Observe that ˆY_k is continuously imbedded in H¹(H^k).

Now, we consider the following linear system around (ˆu,ˆb,ˆr): for allt∈(0, T)











ut(t, x) + (ˆu· ∇)u(t, x)− ∇ ·σ(u, p)(t, x) =v(t, x)ζ(x) x∈ΩbF(t),

∇ ·u(t, x) = 0 x∈ΩbF(t),

u(t, x) = 0 x∈∂Ω,

u(t, x) = ˙b(t) +r(t)×(x−ˆb(t)) x∈∂Ωb_S(t), m¨b(t) =

Z

∂ΩbS(t)

(σ(u, p)n)(t, x)dγ, ( ˆJr)(t) = (( ˆ˙ Jr)ˆ ×r)(t) +

Z

∂Ωb_S(t)

(x−ˆb(t))×(σ(u, p)n)(t, x)dγ, u_|t=0=u0 in ΩF(0), b(0) =b0, b(0) =˙ b1, r(0) =r0.,

(13)

where ˆJ is defined by (4) withQreplaced byQ. The rotation matrixb Qis then defined by (3).

As we will see in Section 3, we will be interested in driving the solution of (13) to zero by means of L²(0, T;H¹(Ω)) controls. In order to do this, we will first obtain L²((0, T)×Ω) controls supported in a smaller open setO2⊂⊂Oefor the following linear system:











u^∗_t(t, x) + (ˆu· ∇)u^∗(t, x)− ∇ ·σ(u^∗, p^∗)(t, x) =v^∗(t, x)1_O₂(x) x∈ΩbF(t),

∇ ·u^∗(t, x) = 0 x∈ΩbF(t),

u^∗(t, x) = 0 x∈∂Ω,

u^∗(t, x) = ˙b^∗(t) +r^∗(t)×(x−ˆb(t)) x∈∂ΩbS(t), mb¨^∗(t) =

Z

∂Ωb_S(t)

(σ(u^∗, p^∗)n)(t, x)dγ, ( ˆJr˙^∗)(t) = (( ˆJr)ˆ ×r^∗)(t) +

Z

∂Ωb_S(t)

(x−ˆb(t))×(σ(u^∗, p^∗)n)(t, x)dγ, u^∗_|t=0=u0 in ΩF(0), b^∗(0) =b0,b˙^∗(0) =b1, r^∗(0) =r0.

(14)

Notice that the control force is slightly different from the one in (13).

In order to prove the null controllability of this system, we will prove a Carleman inequality for its adjoint

(5)

system. Let us introduce this system:











−ϕt(t, x)−(ˆu· ∇)ϕ(t, x)− ∇ ·σ(ϕ, π)(t, x) = 0 x∈ΩbF(t),

∇ ·ϕ(t, x) = 0 x∈ΩbF(t),

ϕ(t, x) = 0 x∈∂Ω,

ϕ(t, x) = ˙a(t) +ω(t)×(x−ˆb(t)) x∈∂ΩbS(t), m¨a(t) =−

Z

∂Ωb_S(t)

(σ(ϕ, π)n)(t, x)dγ, d

dt( ˆJ ω)(t) = (( ˆJr)ˆ ×ω)(t)− Z

∂ΩbS(t)

(x−ˆb(t))×(σ(ϕ, π)n)(t, x)dγ, ϕ_|t=T =ϕT inΩbF(T), a(T) =a^T₀,a(T˙ ) =a^T₁, ω(T) =ωT.

(15)

In the sequel, we will suppose thatϕT ∈L²(bΩF(T)) anda^T₀, a^T₁, ωT ∈R³.

The paper is organized as follows: in Section 2, we state and prove the Carleman inequality satisfied by the adjoint system. In Section 3, we deduce from this inequality an observability inequality and a controllability result for the linearized system. At last, in Section 4, we prove the null controllability of the non-linear system using a fixed point theorem.

2 Carleman inequality for the adjoint system

Let us first introduce the weight functions which we will use in the proof. Let β∈C⁰(W^2,∞)∩C¹(W^1,∞) satisfy

β= 0 on ∂Ωb_F(t), β >0 in Ωb_F(t), |∇β|>c₀>0 inΩb_F(t)\ O₀,

∂β

∂n 6−c1<0 on∂Ω, ∂β

∂n >c2>0 on∂ΩbS(t),

where O₀ ⊂⊂ O₂ is an open set. The existence of a functionβ satisfying the previous properties is proved in [4]. Let nowλbe a positive parameter,M :=kβk_C0(L^∞)and

α(t, x) :=e^(2k+2)λM −eλ(2kM+β(t,x))

t^k(T−t)^k , ξ(t, x) =eλ(2kM+β(t,x))

t^k(T−t)^k , α^∗(t) := e^(2k+2)λM −e^2kλM

t^k(T−t)^k , ξ^∗(t) = e^2kλM t^k(T −t)^k.

(16)

Here,k>24 is a constant.

Then, we can prove the following Carleman inequality:

Proposition 2 Let (ˆu,ˆb,r)ˆ be such that (10), (11) and (12) are satisfied. Then, there exist two constants C1(depending onΩ,O,δ0 andkukˆ Zˆ,kˆbk_W^1,∞_(0,T),kˆrk_L^∞_(0,T₎) andC2>0(just depending onΩ,Oandδ0) such that for all ϕT ∈L²(bΩF(T))and all a^T₀, a^T₁, ωT ∈R³ we have

s⁴λ⁶ Z T

0

Z

ΩbF(t)

e^−2sαξ⁵|ϕ|²dx dt+s⁴λ⁵ Z T

0

e^−2sα^∗(ξ^∗)⁴(|a|˙ ²+|ω|²)dt 6C2s⁵λ⁶

Z T 0

Z

O2

e^−2sαξ⁵|ϕ|²dx dt,

(17)

for allλ>C₁ and alls>C₁(T^k+T^2k), where(ϕ, π, a, ω)is the solution to (15).

(6)

Proof: All along the proof, C (resp. C) will stand for a positive constant just depending on Ω,b O and δ0

(resp. on Ω,O,δ₀ andkˆukZˆ,kˆbkW^1,∞(0,T),kˆrkL^∞(0,T)).

A) Carleman estimate for the heat equation

Let us apply the curl operator to the equation satisfied by (ϕ, π):

−(∇ ×ϕ)_t−(ˆu· ∇)(∇ ×ϕ)−µ∆(∇ ×ϕ) =L(ˆu, ϕ) inΩb_F(t), (18) where the right-hand side satisfies

|L(ˆu, ϕ)|6C|∇ˆu||∇ϕ| in ΩbF(t).

Therefore,∇ ×ϕfulfills a system of three heat equations. For this kind of systems, Carleman inequalities are well-understood since [10]. Here, we are going to use an inequality which has been proved in [4] (see section 2.1 in that reference). More precisely, we use the first inequality in page 21 of [4] by observing that, for the second term of the third line of that inequality, we havee^−2sV^∗γ²=|∇wτ|²(τ is the tangential vector field) with the notations of [4]. Using this forψ:=e^−2sα∇ ×ϕ, we can deduce

s³λ⁴ Z T

0

Z

Ωb_F(t)

ξ³|ψ|²dx dt+sλ² Z T

0

Z

Ωb_F(t)

ξ|∇ψ|²dx dt+s³λ³ Z T

0

Z

∂Ωb_S(t)

(ξ^∗)³|ψ|²dγ dt

+sλ Z T

0

Z

∂ΩbS(t)

ξ^∗|∇ψn|²dγ dt6C s³λ⁴ Z T

0

Z

O0

0

Z

O0

ξ|∇ψ|²dx dt

+sλ Z T

0

Z

∂Ωb_S(t)

ξ^∗|∇ψτ|²dγ dt+sλ² Z T

0

Z

∂Ωb_S(t)

ξ^∗|∇ψn||ψ|dγ dt

+ Z T

0

Z

∂bΩ_S(t)

|∇ψn||ψt+ (ˆu· ∇)ψ|dγ dt+ Z T

0

Z

Ωb_F(t)

e^−2sα|∇ˆu|²|∇ϕ|²dx dt

! ,

(19)

for allλ>Cb and all s>C(T^k+T^2k). Let us use Young’s inequality for the fourth and fifth terms in the right-hand side of this inequality. This yields:

s³λ⁴ Z T

0

Z

Ωb_F(t)

0

Z

Ωb_F(t)

ξ|∇ψ|²dx dt+s³λ³ Z T

0

Z

∂Ωb_S(t)

(ξ^∗)³|ψ|²dγ dt

+sλ Z T

0

Z

∂ΩbS(t)

ξ^∗|∇ψn|²dγ dt6C s³λ⁴ Z T

0

Z

O0

0

Z

O0

ξ|∇ψ|²dx dt

+sλ Z T

0

Z

∂Ωb_S(t)

ξ^∗|∇ψτ|²dγ dt+s⁻¹λ⁻¹ Z T

0

Z

∂Ωb_S(t)

(ξ^∗)⁻¹|ψ_t+ (ˆu· ∇)ψ|²dγ dt

+ Z T

0

Z

bΩ_F(t)

! ,

(20)

for all λ>Cb and all s >C(T^k+T^2k). We can estimate the term (ˆu· ∇)ψ in the fourth integral of the right-hand side of the previous inequality thanks to (12) (since ˆZ⊂C⁰(L^∞)):

s⁻¹λ⁻¹ Z T

0

Z

∂bΩ_S(t)

(ξ^∗)⁻¹|(ˆu· ∇)ψ|²dγ dt6εsλ Z T

0

Z

∂Ωb_S(t)

ξ^∗|∇ψ|²dγ dt forε >0 small enough provided thatλ>Cb_εands>C_εT^2k.

(7)

We come back to ϕ now. Observe that the boundary term concerning ψt in (20) can be bounded as follows:

s⁻¹λ⁻¹ Z T

0

Z

∂ΩbS(t)

(ξ^∗)⁻¹|ψt|²dγ dt62 s⁻¹λ⁻¹ Z T

0

Z

∂bΩS(t)

e^−2sα^∗(ξ^∗)⁻¹|∇ ×ϕt|²dγ dt

+s⁻¹λ⁻¹ Z T

0

Z

∂ΩbS(t)

|∂t(e^−sα^∗)|²(ξ^∗)⁻¹|∇ ×ϕ|²dγ dt

! .

Since|α^∗_t|6C(Tb +T²)(ξ^∗)^1+1/k, the last term can be absorbed by the third integral in the left-hand side of (20) by takingλ>Cb ands>C(T^2k−1+T^2k). Thus, using also that ∇β·τ = 0 on∂ΩbS(t) for the third term in the right-hand side of (20), we can rewrite estimate (20) in the following way:

s³λ⁴ Z T

0

Z

Ωb_F(t)

e^−2sαξ³|∇ ×ϕ|²dx dt+sλ² Z T

0

Z

bΩ_F(t)

e^−2sαξ|∇(∇ ×ϕ)|²dx dt

6C s³λ⁴ Z T

0

Z

O0

0

Z

O0

+sλ Z T

0

Z

∂ΩbS(t)

e^−2sα^∗ξ^∗|∇(∇ ×ϕ)τ|²dγ dt+s⁻¹λ⁻¹ Z T

0

Z

∂ΩbS(t)

e^−2sα^∗(ξ^∗)⁻¹|∇ ×ϕ_t|²dγ dt

+ Z T

0

Z

bΩ_F(t)

! ,

(21)

forλ>Cb ands>C(T^k+T^2k).

Let us obtain estimates on the second term in the right-hand side of (21). Let O1 be an open set with O0⊂⊂ O1⊂⊂ O2 andθ0∈C_c²(O1) be a positive function satisfyingθ0(x) = 1 for allx∈ O0. We apply the curl operator to the first equation in (15):

−(∇ ×ϕ)t− ∇ ×[(ˆu· ∇)ϕ]−µ∆(∇ ×ϕ) = 0 inΩbF(t).

Then, if we setρ(t, x) :=sλ²θ₀(x)e^−2sα(t,x)ξ(t, x), we multiply this equation byρ∇ ×ϕand we integrate by parts inO0, we obtain:

−1 2

d dt

Z

O0

ρ|∇ ×ϕ|²dx+1 2 Z

O0

ρt|∇ ×ϕ|²dx+ Z

O0

ρ[(ˆu· ∇)(∇ ×ϕ)]·(∇ ×ϕ)dx +

Z

O0

[(∇ρ)×(∇ ×ϕ)]·[(ˆu· ∇)ϕ]dx+ Z

O0

ρ∆ϕ·[(ˆu· ∇)ϕ]dx+µ Z

O0

ρ|∇(∇ ×ϕ)|²dx

−µ 2 Z

O0

∆ρ|∇ ×ϕ|²dx= 0,

(22)

fort∈(0, T). Next, we integrate betweent= 0 andt=T, we use ˆu∈Zˆ⊂C⁰(L^∞) and

|ρt|+|∆ρ|6Cs³λ⁴ξ³e^−2sα fors>C(T^k+T^2k) andλ>C.

This leads to:

sλ² Z T

0

Z

O0

6Cb s³λ⁴ Z T

0

Z

O1

0

Z

O1

e^−2sαξ⁻¹|∇ϕ|²dx dt

! ,

(23)

forλ>Cbands>C(T^k+T^2k). Observe that the second term in the right-hand side of this estimate can be bounded by the last one in (21) by takings>CT^2k. Next, we estimate the local term on∇ ×ϕ. In order

(8)

to do this, letθ1∈C_c²(O2) be a positive function satisfying θ1(x) = 1 for allx∈ O1. Then, integrating by parts in

s³λ⁴ Z T

0

Z

O2

θ1e^−2sαξ³(∇ ×ϕ)·(∇ ×ϕ)dx dt+sλ² Z T

0

Z

O2

θ1e^−2sαξ(∇ϕ)·(∇ϕ)dx dt we can prove that

s³λ⁴ Z T

0

Z

O1

0

Z

O1

e^−2sαξ|∇ϕ|²dx dt6Cεs⁵λ⁶ Z T

0

Z

O2

e^−2sαξ⁵|ϕ|²dx dt

+ε sλ² Z T

0

Z

O₂

e^−2sαξ|∇(∇ ×ϕ)|²dx dt+s³λ⁴ Z T

0

Z

O₂

e^−2sαξ³|∇ ×ϕ|²dx dt

!

forλ>Cands>CT^2k.

Thus, combining this with (21) and (23), we obtain s³λ⁴

Z T 0

Z

ΩbF(t)

0

Z

ΩbF(t)

6C s⁵λ⁶ Z T

0

Z

O₂

e^−2sαξ⁵|ϕ|²dx dt+sλ Z T

0

Z

∂Ωb_S(t)

e^−2sα^∗ξ^∗|∇(∇ ×ϕ)τ|²dγ dt

+s⁻¹λ⁻¹ Z T

0

Z

∂ΩbS(t)

e^−2sα^∗(ξ^∗)⁻¹|∇ ×ϕt|²dγ dt+ Z T

0

Z

ΩbF(t)

! ,

(24)

forλ>Cb ands>C(T^k+T^2k).

B) Elliptic estimates

Since∇ ·ϕ= 0, observe thatϕsatisfies the following boundary-value problem:







∆ϕ=−∇ ×(∇ ×ϕ) :=f0 inΩbF(t), ϕ= ( ˙a+ω×(x−ˆb))1_∂

Ωb_S(t):=g0 on∂ΩbF(t).

•Applying classical elliptic estimates, we have

kϕk_H1(bΩF(t))6C(kfb 0k_H−1(bΩF(t))+kg0k_H1/2(∂ΩbF(t)))6C(k∇ ×b ϕk_L2(bΩF(t))+|a|˙ +|ω|).

which directly leads to s³λ⁴

Z T 0

Z

Ωb_F(t)

e^−2sα^∗(ξ^∗)³|∇ϕ|²dx dt

6Cb s³λ⁴ Z T

0

Z

bΩ_F(t)

e^−2sαξ³|∇ ×ϕ|²dx dt+s³λ⁴ Z T

0

e^−2sα^∗(ξ^∗)³(|a|˙ ²+|ω|²)dt

! (25)

•We apply now the classical elliptic Carleman estimate which can be proved as in [10]:

κ⁴λ⁶ Z

ΩbF(t)

exp{2κe^λβ}e^4λβ|ϕ|²dx+κ²λ⁴ Z

ΩbF(t)

exp{2κe^λβ}e^2λβ|∇ϕ|²dx+κ⁴λ⁵ Z

∂ΩbF(t)

e^2κ|ϕ|²dγ

6C κ⁴λ⁶ Z

O2

exp{2κe^λβ}e^4λβ|ϕ|²dx+κλ² Z

ΩbF(t)

exp{2κe^λβ}e^λβ|f₀|²dx+κ²λ³e^2κ Z

∂ΩbF(t)

|∂_κg₀|²dγ

! ,

(9)

for anyκ>Cband any λ>C. Combining this withb H²elliptic estimates, we deduce that κ⁴λ⁶

Z

Ωb_F(t)

exp{2κe^λβ}e^4λβ|ϕ|²dx+κ²λ⁴ Z

Ωb_F(t)

exp{2κe^λβ}e^2λβ|∇ϕ|²dx+κ⁴λ⁵ Z

∂bΩ_F(t)

e^2κ|ϕ|²dγ +λ²

Z

Ωb_F(t)

exp{2κe^λβ}|D²ϕ|²dx6C

κ⁴λ⁶ Z

O2

exp{2κe^λβ}e^4λβ|ϕ|²dx

+κλ² Z

Ωb_F(t)

exp{2κe^λβ}e^λβ|∆ϕ|²dx+κ²λ³e^2κ(|a|˙ ²+|ω|²)

! , for anyκ>Cband any λ>C, where we have used thatb

kϕk_H_3/2_(∂

bΩ_F(t))6C(|b a|˙ +|ω|).

We setκ:= se^2kλM

t^k(T−t)^k and we multiply the previous inequality by exp

−2se^(2k+2)λM t^k(T−t)^k

. This yields

λ² Z T

0

Z

ΩbF(t)

e^−2sα(s⁴λ⁴ξ⁴|ϕ|²+s²λ²ξ²|∇ϕ|²+|D²ϕ|²)dx dt+s⁴λ⁵ Z T

0

Z

∂ΩbF(t)

e^−2sα^∗(ξ^∗)⁴|ϕ|²dγ dt

6C s⁴λ⁶ Z T

0

Z

O2

e^−2sαξ⁴|ϕ|²dx dt+sλ² Z T

0

Z

Ωb_F(t)

e^−2sαξ|∆ϕ|²dx dt

+s²λ³ Z T

0

e^−2sα^∗(ξ^∗)²(|a|˙ ²+|ω|²)dγ dt

! ,

(26)

forλ>Cbands>C(Tb ^k+T^2k). Observe that the terms|∇ϕ|²and|D²ϕ|²in the left-hand side of (26) allow to absorb the last term in (24) takingλ>Cb and s>CT^2k and using ˆu∈C⁰(W^1,3). Combining this with (25) and (24), we obtain

s⁴λ⁶ Z T

0

Z

Ωb_F(t)

e^−2sαξ⁴|ϕ|²dx dt+s³λ⁴ Z T

0

Z

bΩ_F(t)

e^−2sα^∗(ξ^∗)³|∇ϕ|²dx dt

+s⁴λ⁵ Z T

0

Z

∂ΩbF(t)

e^−2sα^∗(ξ^∗)⁴|ϕ|²dγ dt6C s⁵λ⁶ Z T

0

Z

O2

+sλ Z T

0

Z

∂Ωb_S(t)

e^−2sα^∗ξ^∗|∇(∇ ×ϕ)τ|²dγ dt+s⁻¹λ⁻¹ Z T

0

Z

∂Ωb_S(t)

e^−2sα^∗(ξ^∗)⁻¹|∇ ×ϕt|²dγ dt

+s³λ⁴ Z T

0

e^−2sα^∗(ξ^∗)³(|a|˙ ²+|ω|²)dx dt

! ,

(27)

forλ>Cb ands>C(Tb ^k+T^2k).

We notice that

Z

∂Ωb_S(t)

|ϕ|²dγ>C(|b a|˙ ²+|ω|²).

The proof of this inequality is given in [3] (lemma 1, section 4.1). This allows to absorb the last term in the

(10)

right-hand side of (27) thanks tos>CTb ^2k. For the moment, we have:

s⁴λ⁶ Z T

0

Z

Ωb_F(t)

0

Z

Ωb_F(t)

e^−2sα^∗(ξ^∗)³|∇ϕ|²dx dt

+s⁴λ⁵ Z T

0

e^−2sα^∗(ξ^∗)⁴(|a|˙ ²+|ω|²)dγ dt6C s⁵λ⁶ Z T

0

Z

O2

+sλ Z T

0

Z

∂Ωb_S(t)

e^−2sα^∗ξ^∗|∇(∇ ×ϕ)τ|²dγ dt+s⁻¹λ⁻¹ Z T

0

Z

∂Ωb_S(t)

e^−2sα^∗(ξ^∗)⁻¹|∇ ×ϕt|²dγ dt

! ,

(28)

forλ>Cb ands>C(Tb ^k+T^2k).

The rest of the proof is dedicated to the estimate of the two boundary terms B1:=sλ

Z T 0

Z

∂ΩbS(t)

e^−2sα^∗ξ^∗|∇(∇ ×ϕ)τ|²dγ dt and

B2:=s⁻¹λ⁻¹ Z T

0

Z

∂bΩS(t)

e^−2sα^∗(ξ^∗)⁻¹|∇ ×ϕt|²dγ dt.

C) Estimate of B₁ Let us define on (0, T)

θ1:=s^1/2λ^1/2e^−sα^∗(ξ^∗)^1/2

and set (ϕ^∗, π^∗,a˙^∗, ω^∗) :=θ1(ϕ, π,a, ω) together with˙ a^∗(T) = 0. These functions satisfy











−ϕ^∗_t(t, x)−(ˆu· ∇)ϕ^∗)(t, x)− ∇ ·σ(ϕ^∗, π^∗)(t, x) =−θ˙₁ϕ x∈Ωb_F(t),

∇ ·ϕ^∗(t, x) = 0 x∈Ωb_F(t),

ϕ^∗(t, x) = 0 x∈∂Ω,

ϕ^∗(t, x) = ˙a^∗(t) +ω^∗(t)×(x−ˆb(t)) x∈∂ΩbS(t), ma¨^∗(t) =−

Z

∂Ωb_S(t)

(σ(ϕ^∗, π^∗)n)(t, x)dγ+mθ˙1a,˙ d

dt( ˆJ ω^∗)(t) = (( ˆJr)ˆ ×ω^∗)(t)− Z

∂Ωb_S(t)

(x−ˆb(t))×(σ(ϕ^∗, π^∗)n)(t, x)dγ+ ˆJθ˙₁ω, ϕ^∗_|t=T = 0 inΩbF(T), a^∗(T) = ˙a^∗(T) = 0, ω^∗(T) = 0.

(29)

Here, we apply Corollary 9 (stated in the Appendix) with k0 = 13/9 and we deduce the existence of a constantCb such that

kθ1ϕk_L2(H^23/9)+kθ1ϕk_H1(H^5/9)+kθ1ak˙ _H23/18(0,T)+kθ1ωk_H23/18(0,T)

6C(kb θ˙1ϕk_L2(H^5/9)+kθ˙1ϕk_H5/18(L²)+kθ˙1ak˙ _H5/18+kθ˙1ωk_H5/18).

(30)

Since 23/9>5/2,B16Ckθb 1ϕk²_L₂_(H_23/9₎, it suffices to estimate all four terms in the right-hand side of (30).

C.1) Estimate ofkθ˙1ϕk_L2(H^5/9)

After an interpolation argument, we have

kϕk_H5/9(bΩF(t)) 6Ckϕkb ^4/9

L²(bΩ_F(t))kϕk^5/9

H¹(bΩ_F(t)).

(11)

Multiplying this inequality by ˙θ1, we obtain

θ˙1kϕk_H5/9(bΩ_F(t)) 6Csb ^2/9(ξ^∗)^2/9−4/(9k)( ˙θ1)^4/9kϕk^4/9

L²(bΩ_F(t))s^−2/9(ξ^∗)−2/9+4/(9k)( ˙θ1)^5/9kϕk^5/9

H¹(bΩ_F(t)). Applying now Young’s inequality, we get

θ˙1kϕk_H5/9(bΩF(t))6(εs^1/2(ξ^∗)^1/2−1/kθ˙1kϕk_L2(bΩF(t))+Cbεs^−2/5(ξ^∗)−2/5+4/(5k)θ˙1kϕk_H1(bΩF(t))). (31) Observe that

|θ˙1|6s^3/2λ^1/2e^−sα^∗(ξ^∗)^3/2+1/k, withs>C(Tb ^k+T^2k).

Integrating in time inequality (31), we obtain kθ˙1ϕk²_L2(H^5/9)6εs⁴λ

Z T 0

Z

Ωb_F(t)

e^−2sα^∗(ξ^∗)⁴|ϕ|²dx dt+Cεs^11/5λ Z T

0

Z

Ωb_F(t)

e^−2sα^∗(ξ^∗)11/5+18/(5k)|∇ϕ|²dx dt.

These two terms can be absorbed by the left hand-side of the Carleman inequality (28) provided thatk>9, s>C(T^k+T^2k) andλ>1.

C.2) Estimate ofkθ˙1ϕk_H5/8(L²)

Observe that

kθ˙1ϕk²_L2(L²)6Csb ³λ Z T

0

Z

bΩF(t)

e^−2sα^∗(ξ^∗)^3+2/k|ϕ|²dx dt and

kθ˙1ϕk²_H1(L²)6Cb s³λ Z T

0

Z

ΩbF(t)

e^−2sα^∗(ξ^∗)^3+2/k|ϕt|²dx dt+s⁵λ Z T

0

Z

bΩF(t)

e^−2sα^∗(ξ^∗)^5+4/k|ϕ|²dx dt

! . By an interpolation argument due to [20], we get

kθ˙₁ϕk²_H5/18(L²)6Cλb Z T

0

Z

Ωb_F(t)

e^−2sα^∗(s³(ξ^∗)^3+2/k|ϕ|²)^13/18(s³(ξ^∗)^3+2/k|ϕt|²)^5/18dx dt

+ Z T

0

Z

Ωb_F(t)

e^−2sα^∗(s³(ξ^∗)^3+2/k|ϕ|²)^13/18(s⁵(ξ^∗)^5+4/k|ϕ|²)^5/18dx dt

!

=Cλb Z T

0

Z

ΩbF(t)

e^−2sα^∗(s⁴(ξ^∗)⁴|ϕ|²)^13/18(s^2/5(ξ^∗)2/5+36/(5k)|ϕt|²)^5/18dx dt

+ Z T

0

Z

ΩbF(t)

e^−2sα^∗(s⁴(ξ^∗)⁴|ϕ|²)^13/18(s^12/5(ξ^∗)12/5+46/(5k)|ϕ|²)^5/18dx dt

! . In these two integrals, we apply Young inequality with parameters 18/13 and 18/5 and we find

kθ˙1ϕk²_H5/18(L²)6Cλsb ⁴ Z T

0

Z

ΩbF(t)

e^−2sα^∗(ξ^∗)⁴|ϕ|²dx dt+Cλsb ^2/5 Z T

0

Z

bΩF(t)

e^−2sα^∗(ξ^∗)2/5+36/(5k)|ϕt|²dx dt +Cλsb ^12/5

Z T 0

Z

bΩF(t)

e^−2sα^∗(ξ^∗)12/5+46/(5k)|ϕ|²dx dt.

The first and third integrals can be absorbed by the left hand side of (28) takingλ>C,b s>C(T^k+T^2k) andk>23/2 while the second integral is absorbed by the second term in the left-hand side of (30) (squared) takingλ>C,b s>C(T^k+T^2k) andk>24.

(12)

C.3) Estimate ofkθ˙1ak˙ _H5/18

We have

kθ˙1ak˙ ²_L2(0,T)6Csb ³λ Z T

0

e^−2sα^∗(ξ^∗)^3+2/k|a|˙ ²dt and

kθ˙₁ak˙ ²_H1(0,T)6Cb s³λ Z T

0

e^−2sα^∗(ξ^∗)^3+2/k|¨a|²dt+s⁵λ Z T

0

e^−2sα^∗(ξ^∗)^5+4/k|a|˙ ²dt

! . Using again an interpolation argument due to [20], we get

kθ˙1ak˙ ²_H5/18(0,T)6Cλb Z T

0

e^−2sα^∗(s³(ξ^∗)^3+2/k|a|˙ ²)^13/18(s³(ξ^∗)^3+2/k|¨a|²)^5/18dt

+ Z T

0

e^−2sα^∗(s³(ξ^∗)^3+2/k|a|˙ ²)^13/18(s⁵(ξ^∗)^5+4/k|a|˙ ²)^5/18dt

! . We apply Young inequality with parameters 18/13 and 18/5 and we find

kθ˙1ak˙ ²_H5/18(0,T)6Cλsb ^32/9 Z T

0

e^−2sα^∗(ξ^∗)32/9+23/(9k)|a|˙ ²dt+Cλsb ^2/5 Z T

0

e^−2sα^∗(ξ^∗)2/5+36/(5k)|¨a|²dt +Cλsb ⁴

Z T 0

e^−2sα^∗(ξ^∗)⁴|a|˙ ²dt.

(32)

The first and third integrals can be absorbed by the left-hand side of (28) takingλ>C,b s>C(T^k+T^2k) and k >23/4 while the second integral can be absorbed with the third term in the left-hand side of (30) provided thatλ>C,b s>C(T^k+T^2k) andk>12.

C.4) Estimate ofkθ˙1ωk_H5/18

In order to estimate this term, we proceed exactly as in stepC.3).

To conclude paragraphC), we put together stepsC.1)-C.4)and this gives the following estimate onB₁: B16ε s⁴λ⁶

Z T 0

Z

ΩbF(t)

0

Z

ΩbF(t)

e^−2sα^∗(ξ^∗)³|∇ϕ|²dx dt

+s⁴λ⁵ Z T

0

e^−2sα^∗(ξ^∗)⁴(|a|˙ ²+|ω|²)dγ dt

! (33)

forλ>Cb_ε ands>Cb_ε(T^k+T^2k) fork>24.

D) Estimate of B2

Let us define on [0, T]

θ2:=s^−1/2λ^−1/2e^−sα^∗(ξ^∗)^−1/2.

Then,θ2(ϕ, π,a, ω) satisfy system (29) with˙ θ1replaced byθ2. We notice that

|B2|6Ckθ2ϕ_tk²_L2(H^14/9), since 14/9>3/2.

Let us apply Corollary 9 fork₀= 4/9. For our system, the compatibility condition (71) is satisfied since, thanks to the weight functionθ₂, all the initial conditions are equal to zero. This yields, in particular:

kθ₂ϕ_tk_L2(H^14/9)6C(kb θ˙₂ϕk_L2(H^14/9)∩H^7/9(L²)+kθ˙₂ak˙ _H7/9(0,T)+kθ˙₂ωk_H7/9(0,T)).