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HAL Id: hal-01743783

https://hal.archives-ouvertes.fr/hal-01743783

Preprint submitted on 26 Mar 2018

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FEEDBACK STABILIZATION OF A

TWO-DIMENSIONAL FLUID-STRUCTURE INTERACTION SYSTEM WITH MIXED

BOUNDARY CONDITIONS

Jean-Pierre Raymond, Michel Fournié, Ndiaye Moctar

To cite this version:

Jean-Pierre Raymond, Michel Fournié, Ndiaye Moctar. FEEDBACK STABILIZATION OF A TWO-

DIMENSIONAL FLUID-STRUCTURE INTERACTION SYSTEM WITH MIXED BOUNDARY

CONDITIONS. 2018. �hal-01743783�

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FEEDBACK STABILIZATION OF A TWO-DIMENSIONAL FLUID-STRUCTURE INTERACTION SYSTEM WITH MIXED BOUNDARY CONDITIONS

FOURNI ´ E MICHEL

, NDIAYE MOCTAR

, AND RAYMOND JEAN-PIERRE

Abstract. We study the stabilization of a fluid-structure interaction system around an unstable stationary solution.

The system consists of coupling the incompressible Navier-Stokes equations, in a two dimensional polygonal domain with mixed boundary conditions, and a damped Euler-Bernoulli beam equations located at the boundary of the fluid domain.

The control acts only in the beam equations. The feedback is determined by stabilizing the projection of the linearized model onto a finite dimensional invariant subspace. Here we have resolved two important challenges for applications in this field. One is the fact that we prove a stabilization result around a non zero stationary solution, which is new for such fluid-structure interaction systems. The other one is that the feedback laws that we determine do not depend on the Leray projector used to get rid of the algebraic constraints of partial differential equations. This is essential for numerical aspects.

Key words. Fluid-structure interaction, feedback control, stabilization, Navier-Stokes equations, beam equation.

AMS subject classifications. 93B52, 93C20, 93D15, 76D55, 76D05, 74F10.

1. Introduction. We are interested in stabilizing, around a non-zero stationary solution, a fluid- structure interaction system coupling beam equations and the Navier-Stokes equations. The initial con- figuration for the fluid domain is denoted by Ω, and is refered as the reference configuration. Its boundary Γ is split into different parts Γ = Γ s ∪ Γ i ∪ Γ e ∪ Γ n , where Γ s is a flat part of the boundary occupied by elastic structures satisfying Euler-Bernoulli damped beam equations. For the fluid, Neumann boundary conditions are prescribed on Γ n , Navier type boundary conditions are prescribed on Γ e , and Dirichlet boundary conditions are prescribed on Γ i . The assumptions that are essential in our analysis are listed in Section 2. A particular configuration satisfying these assumptions corresponds to the right hand side of a wind tunnel, see Figure 1.1, in which a fluid flows around a thick plate, and two beams are located in the upper and lower boundaries of the plate. In Figure 1.1, the domain Ω is polygonal but the results of the paper can be easily extended to other geometrical configurations, see Remark 2.1. The stabilization problem for the associated semi-discrete model is studied in [8].

We assume that the inflow boundary condition in the computational domain Ω is a perturbed Blasius type profile, which is used to determine the stationary solution around which we want to stabilize the fluid-structure system. The goal is to use a force term as control in the beam equations in order to stabilize the full fluid-structure system. Since the structure is deformed under the action of the fluid, the domain occupied by the fluid at time t depends on the displacement η(t) of the structure, see Figure 1.2.

The fluid domain at time t is denoted by Ω η(t) and the fluid-structure interface by Γ η(t) . We use the notations

Q η = [

t∈(0,∞)

{t} × Ω η(t)

, Σ η = [

t∈(0,∞)

{t} × Γ η(t)

, Q = (0, ∞) × Ω, Σ s = (0, ∞) × Γ s , Σ i = (0, ∞) × Γ i , Σ e = (0, ∞) × Γ e , Σ n = (0, ∞) × Γ n . The Eulerian-Lagrangian system describing the evolution of the fluid-structure system is

u t − div σ(u, p) + (u · ∇)u = 0, div u = 0 in Q η ,

u = η t n on Σ η , u = g s + g p on Σ i , u · n = 0 and ε(u)n · τ = 0 on Σ e , σ(u, p)n = 0 on Σ n , u(0) = u 0 on Ω,

η tt − β∆ s η − γ∆ s η t + α∆ 2 s η = −σ(u, p)| Γ

η(t)

n η(t) p

1 + η x 2 · n + f s + f on Σ s , η = 0 on (0, ∞) × ∂Γ s , η x = 0 on (0, ∞) × ∂Γ s ,

η(0) = 0 on Γ s , η t (0) = η 2 0 on Γ s ,

(1.1)

where u and p stand for the fluid velocity and pressure, σ(u, p) is the Cauchy stress tensor σ(u, p) = 2 ν ε(u) − pI, ε(u) = 1

2 (∇u + (∇u) T ),

Institut de Math´ ematiques de Toulouse, UMR 5219, Universit´ e Paul Sabatier Toulouse III & CNRS, 31062 Toulouse Cedex, France (michel.fournie@math.univ-toulouse.fr, moctar.ndiaye@math.univ-toulouse.fr, jean- pierre.raymond@math.univ-toulouse.fr)

1

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U

computational domain

Fig. 1.1: Configuration of the wind tunnel.

ν is the fluid viscosity, α > 0, β ≥ 0 and γ > 0 are parameters of the structure, f s is a stationary force precisely defined below after (1.2), and n η(t) (resp. n) is the unit normal to Γ η(t) (resp. Γ s ) exterior to Ω η(t) (resp. Ω). Here ∆ s = ∂ xx stands for the Laplace-Beltrami operator on Γ s . Clamped boundary conditions are imposed at the extremities of the beams, and η is a vector valued function representing the displacements of the upper and lower beams. The inflow boundary condition g s ∈ H 2 (Γ i ; R 2 ) is independent of time and g p is a time dependent perturbation of g s , taking into account the fact that the inflow boundary condition of the computational domain Ω is not precisely known. The precise assumptions on g s are stated in Section 2.2. The function f s is also assumed to be time independent, and the control function f is taken in the form

f (t, x, y) =

n

c

X

i=1

f i (t)w i (x, y),

where the functions w i ∈ L 2 (Γ s ) are chosen in (8.4), so that some stabilizability condition stated in (8.7) is satisfied.

Let (u s , p s ) be a solution of the stationary Navier-Stokes equations

−div σ(u s , p s ) + (u s · ∇)u s = 0, div u s = 0 in Ω,

u s = 0 on Γ s , u s = g s on Γ i , u s · n = 0 and ε(u s )n · τ = 0 on Γ e , σ(u s , p s )n = 0 on Γ n . (1.2) We choose f s = −p s | Γ

s

in (1.1) 5 . Thus, (u, p, η) = (u s , p s , 0) is a stationary solution of system (1.1).

We assume that it is an unstable stationary solution of that system. The goal of the paper is to find a control variable f = (f 1 , · · · , f n

c

), in feedback form, able to stabilize system (1.1) around the stationary solution (u s , p s , 0), with any prescribed exponential decay rate −ω < 0, provided that g p , u 0 − u s and η 2 0 are small enough in appropriate functional spaces.

We follow a classical approach consisting of finding a feedback control law, stabilizing a linearized model, that is next applied to the nonlinear system. Let us explain why implementing this approach is not obvious and leads to new difficulties. In (1.1) the main nonlinearities come from the fact that the Navier-Stokes system is written in a time-dependent geometrical domain depending on the displacement of elastic structures. Therefore, before linearizing the system, we have to rewrite it in the reference configuration (0, ∞)× Ω (see Section 3). For the stability analysis of our linearized fluid-structure system and for stabilization issues, we have to study the direct and adjoint eigenvalue problems. These problems are not standard because the algebraic constraints of the direct eigenvalue problem are

div v = A 3 η 1 in Ω, v = η 2 n on Γ s ,

(see system (5.1)), while the algebraic constraints of the adjoint eigenvalue problem are div φ = 0 in Ω, φ = ξ 2 n on Γ s ,

(see system (6.2)). It is a consequence of the fact that the system is linearized around a non zero stationary solution. Because the algebraic constraints are different, studying the spectrum and performing a Jordan decomposition of the linearized operator is not standard. In order to study the spectrum of the linearized operator and to establish its link with the direct and adjoint eigenvalue problems in PDE formulation, we have to rewrite them into equivalent forms with eigenfunctions belonging to the same state space. This can be done by using the so-called Leray projector to transform the direct and adjoint eigenvalue problems.

This approach leads to direct and adjoint eigenvalue problems in the form of operator equations. Let

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us emphasize that operator formulation is needed to justify the eigenvalue analysis, and that the PDE formulation will be needed in [8] for numerical simulations. For that, we have to establish the equivalence between these two formulations. This is done in Section 7.

To the best of our knowledge, the results of the present paper are the first ones dealing with the stabilization of a fluid-structure system involving mixed boundary conditions in the fluid equations. The analysis of such a system seems also to be new in the literature.

Feedback stabilization of strong solutions of such a system around the null solution, by controls acting only in the equations of the structure, is treated in [24]. In [1], the authors obtained stabilization results by means of Dirichlet boundary controls acting in the fluid equations. Several existence results of strong and weak solutions, and some uniqueness results, have been proved in [6, 10, 2, 16, 11].

The plan of the paper is as follows. The functional setting and the assumptions are given in Section 2. The main result is stated in Section 3. The analysis of Oseen operator and the characterization of the pressure in the Oseen system are done in Section 4. In Section 5, we characterize the infinitesi- mal generator (A, D(A)) of the semigroup corresponding to the linearized system. We prove that this semigroup is analytic with compact resolvent. The adjoint of (A, D(A)) is characterized in Section 6.

The equivalences between eigenvalue problems are established in Section 7. In Section 8, we prove that the bi-orthogonality condition satisfied by the families of eigenfunctions of the direct and adjoint partial differential equations implies another bi-orthogonality condition satisfied by the families of eigenfunctions of the direct and adjoint operator equations. This new relationship is used to define a feedback control law independent of the Leray projector, and which can be easily calculated (see [8]).

Γ i

Γ i

Γ n

Γ s

Γ s

Ω Γ e

Γ e

Γ i

Γ i

Γ n

Γ η(t)

Γ η(t)

η(t) Γ e

Γ e

Fig. 1.2: Reference configuration (left) and deformed configuration (right).

2. Functional setting.

2.1. Notations. The geometrical domain Ω and Γ s , see Figure 1.2, are defined by Ω = ([0, L] × [−`, `]) \ ([0, ` s ] × [−e, e]), Γ s = (0, ` s ) × {e} ∪ (0, ` s ) × {−e},

where L > 0 is the length of the computational domain, 2` is its height, ` s is the length of the thick plate, and 2 e is the thickness of the plate. Thus equations (1.1) 5−7 describe the displacement of the upper beam located at (0, ` s ) × {e} and the lower beam located at (0, ` s ) × {−e}. We also have

Γ i = Γ i,1 ∪ Γ i,2 ∪ Γ i,3 with Γ i,1 = {0} × (−`, −e), Γ i,2 = {0} × (e, `), Γ i,3 = {` s } × (−e, e), Γ e =

(0, L) × {−`}

(0, L) × {`}

, Γ i,e = Γ i ∪ Γ e , Γ n = {L} × (−`, `), Γ 0 = Γ s ∪ Γ i,e . We introduce the functional spaces

L 2 (Ω) = L 2 (Ω; R 2 ), H s (Ω) = H s (Ω; R 2 ), ∀ s > 0, H s Γ

0

(Ω) = {v ∈ H s (Ω) | v = 0 on Γ s ∪ Γ i , v · n = 0 on Γ e }, ∀ s > 1 2 , V 0 n,Γ

0

(Ω) = {v ∈ L 2 (Ω) | div v = 0 on Ω, v · n = 0 on Γ 0 },

V s n,Γ

0

(Ω) = H s (Ω) ∩ V 0 n,Γ

0

(Ω), ∀ s > 0, V s Γ

i,e

(Ω) = {v ∈ H s (Ω) | div v = 0 in Ω, v = 0 on Γ i , v · n = 0 on Γ e }, ∀ s > 1 2 . We also introduce the product spaces

H = L 2 (Ω) × H 0 2 (Γ s ) × L 2 (Γ s ) and Z = V 0 n,Γ

0

(Ω) × H 0 2 (Γ s ) × L 2 (Γ s ).

3

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The space V n,Γ 0

0

(Ω) is equipped with the L 2 -norm, the spaces H s Γ

0

(Ω), V s Γ

i,e

(Ω) and V s n,Γ

0

(Ω) are equipped with the H s -norm. The inner product in H 0 2s ) is chosen at the beginning of Section 5.2, and the inner product in H and Z is defined in (5.2).

The space of boundary displacements. We consider displacements of the beams located on Γ s belonging to the space denoted by H 0

32

s ), which is identified with (H 0

32

(0, ` s )) 2 = (H

32

(0, ` s )∩ H 0 1 (0, ` s )) 2 . By writing H 0

32

s ), we emphasize the fact that η is a vector function with one component defined on (0, ` s ) × {e}

and the other one defined on (0, ` s ) × {−e}.

The space of inflow boundary conditions. On the boundary Γ i = Γ i,1 ∪ Γ i,2 ∪ Γ i,3 , we consider inflow conditions g belonging to the space

H(Γ i ) = n

g = (g 1 , g 2 ) | g 2 = 0, g 1 | Γ

i,3

= 0, g 1 | Γ

i,1

∈ H 2 (−`, −e), g 1 | Γ

i,2

∈ H 2 (e, `), g 1 0 (−`) = g 0 1 (`) = 0, g 1 (−e) = g 1 (e) = 0 o

. The space H(Γ i ) is equipped with the norm (g 1 , g 2 ) 7−→ kg 1 k H

2

((−`,−e)∪(e,`)) .

We denote by C = (C j ) 1≤j≤8 the set of the corners of Ω. For all −1 < δ < 1, s ∈ N , and for n = 1 or n = 2, we introduce the norms

kvk 2 H

s

δ

(Ω; R

n

) = X

|α|≤s

Z

r |∂ α v| 2 ,

where r stands for the distance to C, α = (α 1 , α 2 ) ∈ N 2 denotes a two-index, |α| = α 1 + α 2 is its length,

∂ α denotes the corresponding partial differential operator. We denote by H δ s (Ω; R n ) the closure of C (Ω) in the norm k·k H

s

δ

(Ω; R

n

) . We set

L 2 δ (Ω) = H δ 0 (Ω; R 2 ), H s δ (Ω) = H δ s (Ω; R 2 ) for all s > 0, H δ = H 2 δ (Ω) × H δ 1 (Ω) × H 4 (Γ s ) ∩ H 0 2 (Γ s )

× H 0 2 (Γ s ).

We also introduce the following spaces of time dependent functions

H 2,1 (Q ) = L 2 (0, ∞; H 2 (Ω)) ∩ H 1 (0, ∞; L 2 (Ω)), H δ 2,1 (Q ) = L 2 (0, ∞; H 2 δ (Ω)) ∩ H 1 (0, ∞; L 2 (Ω)), H 0 2,1 s ) = L 2 (0, ∞; H 0 2 (Γ s )) ∩ H 1 (0, ∞; L 2 (Γ s )), H 0 4,2 s ) = H 0 2,1 s ) ∩ H 4,2 s ),

Y = L 2 (0, ∞; L 2 (Ω)) × H 2,1 (Q ) × L 2 (0, ∞; L 2 (Γ s )), X δ = H δ 2,1 (Q ) × L 2 (0, ∞; H δ 1 (Ω)) × H 0 4,2 s ) × H 0 2,1 s ).

2.2. Assumptions.

Assumption 1. We assume that g s ∈ H(Γ i ), and that the system (1.2) admits a solution (u s , p s ) ∈ H 2 δ

0

(Ω) × H δ 1

0

(Ω) for some 0 < δ 0 < 1 2 specified in (4.4).

Assumptions 2 and 3. These assumptions are stated in Section 8.2, see (8.5) and (8.7).

In Assumption 2, we state that the parts of spectrum of the Oseen operator and that of the elastic structure contained in the half plane {λ ∈ C | Reλ ≥ −ω} are disjoint. In Assumption 3, we state a unique continuation property sufficient to verify the Hautus stabilization test for the linearized system.

Remark 2.1. The main results of the paper, Theorems 3.2 and 3.3, are proved for the geometrical configuration described in Section 2.1 and under Assumptions 1, 2 and 3. As mentioned in the intro- duction, Theorems 3.2 and 3.3 remain valid for geometrical configurations for which the regularity result stated in Theorem 4.1 is true. (See e.g. [21] where other geometrical configuration are considered.)

3. System in the reference configuration and main results. We denote by E a contin- uous linear operator from H 0 4,2 s ) into H 0 4,2 ((0, ∞) × (0, L) × {e, −e}) such that Eη(t, x) = 0 for (t, x) ∈ (0, ∞) × ((` s + L)/2, L). For beam displacements η ∈ H 0 4,2 s ) small enough, we define a C 1 -diffeomorphism T η(t) , from Ω η(t) into Ω, by setting

T η(t) (x, y) = (x, z) =

x, (` − e)y − `Eη(t, x, s(y)e)

` − e − s(y)Eη(t, x, s(y)e)

, (3.1)

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where the sign function s is defined by s(y) = −1 if y < 0 and s(y) = 1 if y ≥ 0.

Remark 3.1. To simplify the notation, in all the transformations, the nonlinear terms and the operators involving Eη for some η ∈ H 0 4,2 s ), we shall simply write η in place of Eη. We shall do the same abuse of notation when the partial derivatives of Eη appear in some expression.

For a given η ∈ H 0 4,2 s ) such that the mapping T η(t) is a C 1 -diffeomorphism from Ω η(t) into Ω for all t > 0, we say that u belongs to H δ 2,1 (Q η ) (resp. p belongs to L 2 (0, ∞; H δ 1 (Ω η ))) if and only if

u ·, T η(·) −1

belongs to H δ 2,1 (Q ) (resp. p ·, T η(·) −1

belongs to L 2 (0, ∞; H δ 1 (Ω))).

The main result of the paper is the following theorem.

Theorem 3.2. Let ω be positive. We assume that Assumptions 1, 2 and 3 are satisfied. There exists r > 0 such that, for all u 0 ∈ H 1 (Ω), η 2 0 ∈ H 0 1s ), and e ωt g p ∈ H 0 1 (0, ∞; H(Γ i )) satisfying u 0 − u s ∈ V 1 Γ

i,e

(Ω), (u 0 − u s )| Γ

s

= η 2 0 n and ku 0 − u s k H

1

(Ω) + kη 2 0 k H

1

0

s

) + ke ωt g p k H

1

0

(0,∞;H(Γ

i

)) ≤ r,

we can find f = (f 1 , · · · , f n

c

) ∈ H 1 (0, ∞; R n

c

), in feedback form, for which the nonlinear system (1.1) admits a solution (u, p, η) in H δ 2,1

0

(Q η ) × L 2 (0, ∞; H δ 1

0

(Ω η(t) )) × H 4,2 s ), for δ 0 defined in (4.4), sat- isfying

u t, T η(t) −1

− u s , η(t), η t (t) H

34+ε0

2

(Ω)×H

3

s

)×H

1

s

) ≤ Ce −ωt , ∀t > 0, and ke −ω · η 1 k L

s

) < e,

where C > 0 depends on r and 0 < ε 0 < 1 2 .

In order the prove that theorem, we make the change of unknowns

b u(t, x, z) := e ωt [u(t, T η(t) −1 (x, z)) − u s (x, z)], p(t, x, z) := b e ωt [p(t, T η(t) −1 (x, z)) − p s (x, z)], b η 1 (t, x) := e ωt η(t, x), η b 2 (t, x) := e ωt η t (t, x), b f (t) = ( f b i (t)) 1≤i≤n

c

:= e ωt f(t),

b g p (t, x, z) := e ωt g p (t, x, z) and b u 0 = u 0 − u s .

(3.2)

The quadruplet ( b u, p, b η b 1 , η b 2 ) satisfies the system

b u t − div σ( b u, p) + (u b s · ∇) u b + ( u b · ∇)u s − A 1 b η 1 − A 2 η b 2 − ω u b = e −ωt F f [ b u, p, b η b 1 , η b 2 ] in Q , div b u = A 3 η b 1 + e −ωt div F div [ u, b η b 1 ] in Q , u b = b η 2 n on Σ s , u b = b g p on Σ i ,

b u · n = 0 on Σ e , ε( u)n b · τ = 0 on Σ e , σ( u, b p)n b = 0 on Σ n , b η 1,t − η b 2 − ω η b 1 = 0 on Σ s ,

b η 2,t − β∆ s η b 1 − γ∆ s η b 2 + α∆ 2 s η b 1 − A 4 b η 1 − ω b η 2 = γ s p b + e −ωt F s [ b u, b η 1 ] + f b on Σ s , b η 1 = 0 on (0, ∞) × ∂Γ s , η b 1,x = 0 on (0, ∞) × ∂Γ s ,

b u(0) = u b 0 in Ω, η b 1 (0) = 0 on Γ s , η b 2 (0) = η 0 2 on Γ s ,

(3.3)

where γ s is the trace operator on Γ s , the nonlinear terms F f and F s are given in Appendix A, F div [ b u, η b 1 ], and the linear differential operators A 1 , A 2 , A 3 and A 4 are defined by

F div [ b u, η b 1 ] = 1

` − e (s(z) η b 1 u b 1 e 1 + (` − s(z)z) b η 1,x u b 1 e 2 ) , e 1 = (1, 0) T , e 2 = (0, 1) T , A 1 η b 1 = s(z) b η 1

` − e (p s,z e 2 + u s,2 u s,z − 2νu s,zz + νu s,1,xx e 1 + νu s,1,xz e 2 ) + (` − s(z)z) η b 1,x

` − e (p s,z e 1 − 2νu s,xz + u s,1 u s,z − νu s,1,zz e 2 − νu s,1,xz e 1 ) + ν η b 1,x

` − e (u s,1,z e 2 − u s,1,x e 1 ) − ν (` − s(z)z) b η 1,xx

` − e (u s,z + u s,1,z e 1 ), A 2 η b 2 = (` − s(z)z) η b 2

` − e u s,z , A 3 η b 1 = 1

` − e (s(z) b η 1 u s,1,x + (` − s(z)z) b η 1,x u s,1,z ), A 4 η b 1 = ν

s(z) b η 1

` − e u s,2,z − η b 1,x u s,1,z − 2γ s (A 3 η b 1 )

.

(3.4)

5

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We notice that we have used the condition div u b = A 3 b η 1 +e −ωt div F div [ b u, η b 1 ], and the boundary condition u b 1 | Σ

s

= 0 to replace 2ν b u 2,z , which appears in equation (3.3) 5 , by

2νγ s A 3 b η 1 + e −ωt div F div [ b u, η b 1 ] .

The term 2νγ s (A 3 η b 1 ) appears in A 4 η b 1 , while the term 2νγ s (div F div [ u, b b η 1 ]) is involved in the nonlinear term F s of the equation.

In the nonlinear terms F f , F s and F div , the functions b η 1 and η b 2 , which are a priori defined over [0, ` s ], are extended by using the operator E, see Remark 3.1. We can easily check that, if b u · n = 0 and ε( b u)n · τ = 0 on Σ e , then F div = F div [ b u, η b 1 ] satisfies the following boundary conditions

F div = 0 on Σ s , F div = 0 on Σ i , F div · n = 0 on Σ e , ε(F div )n · τ = 0 on Σ e , ε(F div )n = 0 on Σ n .

Moreover F div | t=0 = 0 since η b 1 | t=0 = 0. Theorem 3.2 is a direct consequence of the following result.

Theorem 3.3. Let η max be in (0, e) and ω be positive. We assume that Assumptions 1, 2 and 3 are satisfied. There exists r > 0 such that, for all u b 0 ∈ V 1 Γ

i,e

(Ω), η 2 0 ∈ H 0 1s ), and b g p ∈ H 0 1 (0, ∞; H(Γ i )) satisfying u b 0 | Γ

s

= η 2 0 n and

k b u 0 k H

1

(Ω) + kη 0 2 k H

1

0

s

) + kˆ g p k H

1

0

(0,∞;H(Γ

i

)) ≤ r,

we can find a control b f ∈ H 1 (0, ∞; R n

c

), in feedback form, for which the nonlinear system (3.3) admits a solution ( u, b b p, b η 1 , η b 2 ) ∈ X δ

0

, where δ 0 is introduced in (4.4), obeying

ke −ωt η b 1 (t)k L

s

) ≤ η max and k( u(t), b b η(t), η b t (t))k

H

34+ε0

2

(Ω)×H

3

s

)×H

1

s

) ≤ C, ∀t > 0, where C > 0 depends on r and 0 < ε 0 < 1 2 .

Remark 3.4. The estimate ke −ωt b η 1 (t)k L

s

) ≤ η max guarantees that, for all t > 0, the change of variables T e

−ωt

b η(t) is a C 1 −diffeomorphism from Ω e

−ωt

b η(t) into Ω.

Remark 3.5. In Theorems 3.2 and 3.3, we assume that η 1 (0), the initial displacement of the beam, is zero, see (3.3). As in [24], the initial displacement can be considered to be non zero. In that case, the compatibility conditions are more complicated (see [24] for compatibility conditions stated when η 1 (0) 6= 0).

4. The Oseen system. The goal of this section is to study the non-homogeneous Oseen system λ 0 w − div σ(w, π) + (u s · ∇)w + (w · ∇)u s = F, div w = h in Ω,

w = ηn on Γ s , w = 0 on Γ i , w · n = 0 and ε(w)n · τ = 0 on Γ e , σ(w, π)n = 0 on Γ n , (4.1) where F ∈ L 2 (Ω), η ∈ H

3 2

0 (Γ s ), h ∈ H 1 (Ω). We choose λ 0 > 0 such that λ 0

Z

|v| 2 + 2ν Z

|ε(v)| 2 + Z

[(u s · ∇)v + (v · ∇)u s ] · v ≥ ν

2 kvk H

1

(Ω) , ∀ v ∈ H 1 (Ω). (4.2) We shall say that (w, π) ∈ H 1 (Ω) × L 2 (Ω) is a weak solution of (4.1) if and only if it satisfies the following mixed variational formulation

Find (w, π) ∈ H 1 (Ω) × L 2 (Ω) such that a(w, φ) − b(φ, π) =

Z

F · φ for all φ ∈ H 1 Γ

0

(Ω), b(w, ψ) = Z

h ψ for all ψ ∈ L 2 (Ω), w = η n on Γ s , w = 0 on Γ i , w · n = 0 on Γ e ,

(4.3)

where a(w, φ) =

Z

0 w φ + 2νε(w) : ε(φ)) dx + Z

((u s · ∇)w + (w · ∇)u s )φ dx, b(w, ψ) = Z

div(w)ψ dx.

6

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In the case when u s = 0, that is for the Stokes system, if (F, h, η) ∈ L 2 (Ω) × H 1 (Ω) × H 0

32

s ), far from the set C, that is in Ω \ ∪ j=1,8 B(C j , ε), the solution (w, π) to system (4.1) belongs to H 2 (Ω \

∪ j=1,8 B(C j , ε)) × H 1 (Ω \ ∪ j=1,8 B(C j , ε)) for all ε > 0 (B(C j , ε) is the ball centered at C j of radius ε).

In the neighborhood of C, we can study the regularity of the solution to system (4.1) in weighted Sobolev spaces. According to [20, Theorem 9.4.5], there exists δ 0 < 1/2 such that, if (F, h, η) ∈ L 2 (Ω) × H 1 (Ω) × H

3 2

0 (Γ s ) and u s = 0, the solution (w, π) to system (4.1) belongs to H 2 δ

0

(Ω) × H δ 1

0

(Ω). In particular δ 0 < 1/2 can be chosen so that

1

3 < δ 0 < 1

2 . (4.4)

The regularity result for the Oseen system is stated in the following theorem.

Theorem 4.1. For all (F, h, η) ∈ L 2 (Ω) × H 1 (Ω) × H 0

32

s ), system (4.1) admits a unique solution (w, π) ∈ H 2 δ

0

(Ω) × H δ 1

0

(Ω) satisfying the estimate kwk H

2

δ0

(Ω) + kπk H

1

δ0

(Ω) ≤ C kFk L

2

(Ω) + khk H

1

(Ω) + kηk H

3/2

s

)

.

Proof. As mentioned above, if u s = 0, the existence of a unique solution to system (4.1) belonging to H 2 δ

0

(Ω)×H δ 1

0

(Ω) is already proved in [20, Theorem 9.4.5]. However, we have to notice that the variational formulation (4.3) is different from that in [20, (9.1.7)-(9.1.9)]. It is different because equation (4.1) 1 is written in terms of −div σ(w, π), while in [20] equation (9.1.1) is written in terms of −ν ∆w + ∇p. But the results stated in [20, Theorem 9.4.5] can be applied here to the Oseen system stated in (4.1).

The adaptation from the Stokes system to the Oseen system can be performed as in [21, Proof of Theorem 2.17, Step 2].

Remark 4.2. If in system (4.1) the boundary condition w = 0 on Γ i is replaced by w = g p on Γ i , we can prove that, for (F, h, η, g p ) ∈ L 2 (Ω) × H 1 (Ω) × H

3 2

0 (Γ s ) × H(Γ i ), the solution to system (4.1) satisfies kwk H

2

δ0

(Ω) + kπk H

1

δ0

(Ω) ≤ C kFk L

2

(Ω) + khk H

1

(Ω) + kηk H

3/2

s

) + kg p k H(Γ

i

) .

Lemma 4.3. We set ε 0 = 1 2 − δ 0 . Then ε 0 > 0, H 2 δ

0

(Ω) ⊂ H

32

0

(Ω) and H δ 1

0

(Ω) ⊂ H

12

0

(Ω), with continuous embeddings.

Proof. It follows from [20, Lemma 6.2.1].

4.1. The Leray projector. From [21, Lemma 2.2], we know that L 2 (Ω) = V 0 n,Γ

0

(Ω) ⊕ ∇H Γ 1

n

(Ω), where H Γ 1

n

(Ω) = {p ∈ H 1 (Ω) | p| Γ

n

= 0}. We introduce the orthogonal projection P from L 2 (Ω) onto V 0 n,Γ

0

(Ω), called the Leray projector. For all v in L 2 (Ω), we have P v = v − ∇q 1 v − ∇q v 2 , where q v 1 and q v 2 are the solutions to the equations

q 1 v ∈ H 0 1 (Ω), ∆q 1 v = div v in Ω, q v 1 = 0 on ∂Ω, q 2 v ∈ H 1 (Ω), ∆q 2 v = 0 in Ω, ∂q v 2

∂n = (v − ∇q v 1 ) · n on Γ 0 , q 2 v = 0 on Γ n .

(4.5)

Proposition 4.4. If v belongs to H 1 (Ω), then P v and (I − P )v belong to H

12

0

(Ω). Moreover, we have (I − P )v = ∇q, where q is the solution to the equation

q ∈ H 1 (Ω), ∆q = div v in Ω, ∂q

∂n = v · n on Γ 0 , q = 0 on Γ n . (4.6) Proof. We have to prove that the solution q to equation (4.6) belongs to H

32

0

(Ω). Since v belongs to H 1 (Ω), we know that v · n| Γ

0

belongs to H

12

(Γ 0 ). Thus, far from the corners of the domain Ω, the solution q is of class H 2 . We have to analyze the regularity at the corners corresponding either to a

7

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junction between two Neumann boundary conditions or to a junction between a Neumann and Dirichlet boundary condition. At the two corners corresponding to a Neumann-Dirichlet junction, the angle is equal to π 2 , the condition on Γ n is q = 0 and the condition on Γ 0 is ∂n ∂q ∈ H

12

0 ). Using an odd symmetry with respect to Γ 0 , we have to deal with a pure Neumann boundary condition on a flat boundary with the extension of ∂n ∂q belonging to H

12

−ε for all ε > 0, on that flat boundary. Thus q is of class H 2−ε , for all ε > 0, close to a Neumann-Dirichlet junction.

We may use the results in [12] to analyze the regularity at the corners corresponding to a Neumann- Neumann junction. From [12, Corollary 2.4.4, Remark 2.4.6, and Theorem 2.3.7 (iii)], it follows that q ∈ H s (Ω) for all s < 1 + 2 3 . In particular q ∈ H

32

0

(Ω). This completes the proof.

4.2. Expression of the pressure. The goal of this section is to express the pressure π in equation (4.1) in terms of w, F , η and h. The method used in [24] consists in calculating the divergence of equation (4.1) 1 in order to get an elliptic equation for π. This method does not work here. Indeed, formally π is the solution to the equation

∆π = −λ 0 h − div ((u s · ∇)w + (w · ∇)u s ) + div F in Ω, π = 2νε(w)n · n on Γ n ,

∂π

∂n = 2νdiv ε(w) · n − λ 0 w · n − ((u s · ∇)w + (w · ∇)u s ) · n + F · n on Γ 0 . (4.7) Since div ε(w) ∈ / L 2 (Ω), the boundary condition (4.7) 2 is not well-posed. Thus, we cannot use the classical transposition method because the Green formula needed for that is not valid (see [12, Theorem 1.5.3]).

We are going to use a variant of the classical transposition method to define an equation for the pressure π. First, we consider the equation

∆χ = ζ in Ω, ∂χ

∂n = 0 on Γ 0 , χ = 0 on Γ n . (4.8)

Lemma 4.5. For all ζ ∈ L 2 (Ω), equation (4.8) admits a unique solution χ belonging to H δ 2 (Ω), for all 1 3 < δ < 1, and satisfying

kχk H

2

δ

(Ω) ≤ C δ kζk L

2

(Ω) , where C δ depends on δ. In particular, this estimate is valid for δ = δ 0 .

Proof. It follows from [20, Theorem 6.5.4], [20, Lemma 6.2.1] and [15, Chapter 2].

Let us notice that

H δ 1 (Ω) , → L 2 −δ (Ω) for all 0 < δ < 1

2 . (4.9)

Indeed, from [19, page 399], it follows that H δ 1 (D) , → H −δ ε (D) for a dihedral domain D, when 1−δ = ε+δ and ε > 0. Thus we have H δ 1 (Ω) , → H −δ ε (Ω) for all 0 < δ < 1 2 and ε = 1 − 2δ. That proves (4.9), since H −δ ε (Ω) , → L 2 −δ (Ω).

Using (4.9), we can define k ∈ L(L 2 (Ω), R ) by k(ζ) = λ 0

Z

Γ

s

ηχ − λ 0

Z

h χ − Z

F · ∇χ + 2νhε(w), ∇ 2 χi L

2

−δ0

(Ω),L

2δ

0

(Ω)

−2ν Z

Γ

0

ε(w)n · ∇χ + Z

h

(u s · ∇)w + (w · ∇)u s

i · ∇χ, for all ζ ∈ L 2 (Ω),

(4.10)

where χ ∈ H δ 2

0

(Ω) is the solution to equation (4.8).

Lemma 4.6. Let (w, F, η, h) belong to H 2 δ

0

(Ω) × L 2 (Ω) × L 2s ) × H 1 (Ω). The operator k belongs to L(L 2 (Ω), R ) and we have

|k(ζ)| ≤ C

kηk L

2

s

) + khk L

2

(Ω) + kFk L

2

(Ω) + kwk H

2 δ0

(Ω)

kζk L

2

(Ω) for all ζ ∈ L 2 (Ω).

8

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Proof. The lemma follows from the following estimates

Z

Γ

s

ηχ

≤ Ckηk L

2

s

) kχk L

2

s

) ≤ Ckηk L

2

s

) kζk L

2

(Ω) ,

Z

≤ Ckhk L

2

(Ω) kχk L

2

(Ω) ≤ Ckhk L

2

(Ω) kζk L

2

(Ω) ,

Z

F · ∇χ

≤ CkFk L

2

(Ω) k∇χk L

2

(Ω) ≤ CkFk L

2

(Ω) kζk L

2

(Ω) ,

Z

Γ

0

ε(w)n · ∇χ

≤ Ckε(w)k L

2

0

) k∇χk L

2

0

) ≤ Ckwk

H

32+ε0

(Ω) kζk L

2

(Ω) ,

Z

h

(u s · ∇)w + (w · ∇)u s

i · ∇χ

≤ Ckwk H

1

(Ω) k∇χk L

2

(Ω) ≤ Ckwk H

1

(Ω) kζk L

2

(Ω) ,

hε(w), ∇ 2 χi L

2

−δ0

(Ω),L

2δ

0

(Ω)

≤ Ckε(w)k| L

2

−δ0

(Ω) k∇ 2 χk L

2

δ0

(Ω) ≤ Ckwk H

1+δ0

(Ω) kζk H

2 δ0

(Ω) . The last inequality comes from [20, Lemma 6.2.1].

We introduce the following operators:

• N s ∈ L(L 2s ), H 1 (Ω)) is defined by N s η = q where q is the solution to the equation

∆q = 0 in Ω, ∂q

∂n = η on Γ s , ∂q

∂n = 0 on Γ 0 \ Γ s , q = 0 on Γ n . (4.11)

• N div ∈ L(L 2 (Ω), H 1 (Ω)) is defined by N div h = q where q is the solution to the equation

∆q = h in Ω, ∂q

∂n = 0 on Γ 0 , q = 0 on Γ n . (4.12)

• N p ∈ L(L 2 (Ω), H 1 (Ω)) is defined by N p F = q 1 F + q F 2 where q 1 F and q 2 F are the solutions to equations (4.5) 1 and (4.5) 2 respectively, with v = F .

• N v ∈ L(H 2 δ

0

(Ω), L 2 (Ω)) is defined by N v w = q where q is the solution to the variational problem Find q ∈ L 2 (Ω) such that

Z

qζ = 2νhε(w), ∇ 2 χi L

2

−δ0

(Ω),L

2δ

0

(Ω) − 2ν Z

Γ

0

ε(w)n · ∇χ+

Z

h

(u s · ∇)w + (w · ∇)u s i

· ∇χ, for all ζ ∈ L 2 (Ω), where χ ∈ H δ 2

0

(Ω) is the solution to equation (4.8).

Thanks to the Lax-Milgram Theorem, we can easily prove that the operators N s , N div , N p and N v

are well-defined.

Remark 4.7. We notice that ∇N p = I − P.

Theorem 4.8. If (w, π) ∈ H 2 δ

0

(Ω) × H δ 1

0

(Ω) is a solution to system (4.1), where δ 0 obeys (4.4), then π is the unique solution to the problem

Find π ∈ L 2 (Ω) such that Z

π ζ dx = k(ζ), ∀ ζ ∈ L 2 (Ω). (4.13) As a consequence, π is determined by

π = −λ 0 N s η − λ 0 N div h + N p F + N v w.

Proof. Let (w, π) ∈ H 2 δ

0

(Ω) × H δ 1

0

(Ω) be the solution to system (4.1) and let ζ belong to L 2 (Ω).

Multiplying the first equation of the system by ∇χ, where χ is the solution to equation (4.8), we obtain λ 0

Z

w · ∇χ − Z

div σ(w, π) · ∇χ + Z

[(u s · ∇)w + (w · ∇)u s ] · ∇χ − Z

F · ∇χ = 0. (4.14) By using integration by parts, we get

λ 0

Z

w · ∇χ = −λ 0

Z

div w · χ + Z

∂Ω

w · nχ = −λ 0

Z

h χ + λ 0

Z

Γ

s

ηχ, (4.15)

9

(11)

and Z

div σ(w, π) · ∇χ = −hσ(w, π), ∇ 2 χi L

2

−δ0

(Ω),L

2δ

0

(Ω) + Z

∂Ω

σ(w, π)n · ∇χ

= −2νhε(w), ∇ 2 χi L

2

−δ0

(Ω),L

2δ

0

(Ω) + Z

π · ζ + Z

Γ

0

σ(w, π)n · ∇χ.

(4.16)

Let us notice that (4.16) 1 can be first obtained for χ ∈ C (Ω) and next by density for χ ∈ H 2 δ

0

(Ω). From equations (4.14)-(4.16), it follows that π is a solution of equation (4.13). Thanks to Lemma 4.6 and to the Lax-Milgram Lemma in L 2 (Ω), we prove that (4.13) admits a unique solution.

To prove the last part of the theorem, it is sufficient to verify that Z

Γ

s

ηχ = Z

Γ

s

∂N s η

∂n χ = − Z

ζN s η, Z

hχ = Z

∆N div hχ = Z

ζN div h and Z

F · ∇χ = Z

ζN p F .

(4.17)

Lemma 4.9. Let (w, π) ∈ H 2 δ

0

(Ω) × H δ 1

0

(Ω) be the solution of system (4.1), where δ 0 obeys (4.4).

Then N v w belongs to H

12

0

(Ω).

Proof. We start from the identity

π = −λ 0 N s η − λ 0 N div h + N p F + N v w.

We know that N p F, N div h and N s η belong to H 1 (Ω) and that π belongs to H δ 1

0

(Ω) ⊂ H

12

0

(Ω) (see Lemma 4.3). Hence N v w belongs to H

12

0

(Ω).

4.3. The Oseen operator. We introduce the Oseen operator (A, D(A)) in V n,Γ 0

0

(Ω) defined by D(A) = n

v ∈ V n,Γ

32

0

0

(Ω) | ∃q ∈ H

12

0

(Ω) such that div σ(v, q) ∈ L 2 (Ω), ε(v)n · τ = 0 on Γ e , σ(v, q)n = 0 on Γ n

o and Av = P div σ(v, q) − P [(u s · ∇)v + (v · ∇)u s ].

(4.18)

Theorem 4.10. The operator (A, D(A)) is the infinitesimal generator of an analytic semigroup on V 0 n,Γ

0

(Ω) and its resolvent is compact.

Proof. The proof is similar to that of [21, Theorem 2.8].

Proposition 4.11. The adjoint of (A, D(A)) in V n,Γ 0

0

(Ω) is defined by D(A ) = n

v ∈ V n,Γ

32

0

0

(Ω) | ∃q ∈ H

12

0

(Ω) such that div σ(v, q) ∈ L 2 (Ω), ε(v)n · τ = 0 on Γ e , σ(v, q)n + u s · nv = 0 on Γ n

o and A v = P div σ(v, p) + P(u s · ∇)v − P (∇u s ) T v.

Proof. See [21, Theorem 2.11].

Finally, we introduce the lifting operators L ∈ L(H 0

32

s ) × H 1 (Ω), H 2 δ

0

(Ω)) and L p ∈ L(H 0

32

s ) × H 1 (Ω), H δ 1

0

(Ω)) defined by

L(η, h) = w and L p (η, h) = π, (4.19)

where (w, π) is the solution to system (4.1) for F = 0.

Theorem 4.12. A pair (w, π) ∈ H 2 δ

0

(Ω) × H δ 1

0

(Ω) is solution to system (4.1) if and only if (λ 0 I − A)P w + (A − λ 0 I)P L(η, h) = P F, (I − P )w = ∇N s η + ∇N div h,

π = −λ 0 N s η − λ 0 N div h + N p F + N v w. (4.20)

10

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Proof. Let (w, π) ∈ H 2 δ

0

(Ω) × H δ 1

0

(Ω) be a solution of system (4.1). We set w b = w − L(η, h) and b π = π − L p (η, h).

The couple ( w, b b π) is solution to system (4.1) with (η, h) = (0, 0). Thus, w b belongs to D(A) and λ 0 P w b − AP w b = P F. Hence (λ 0 I − A)P w + (A − λ 0 I)P L(η, h) = P F. The last two equations in (4.20) come from Proposition 4.4 and Theorem 4.8.

Now, we suppose that (w, π) ∈ H 2 δ

0

(Ω) × H δ 1

0

(Ω) is solution to system (4.20). Since (I − P )w =

∇N s η + ∇N div h = (I − P )L(η, h), we have

w b = w − L(η, h) ∈ D(A). (4.21)

Thus, there exists π 1 ∈ H

12

0

(Ω) such that

A w b = Pdiv σ( w, π b 1 ) − P (u s · ∇) w b − P ( w b · ∇)u s and σ( w, π b 1 )n = 0 on Γ n . (4.22) From the first equation of system (4.20), it follows that

P [λ 0 w b − div σ( w, π b 1 ) + (u s · ∇) w b + ( w b · ∇)u s − F ] = 0.

Thus, there exists π 2 ∈ H Γ 1

n

(Ω) such that

λ 0 w b − div σ( w, π b 1 + π 2 ) + (u s · ∇) w b + ( w b · ∇)u s = F. (4.23) With equations (4.21), (4.22) 2 and (4.23), we obtain the system

λ 0 w b − div σ( w, π b 1 + π 2 ) + (u s · ∇) w b + ( w b · ∇)u s = F, div w b = 0 in Ω,

w b = 0 on Γ s , w b = 0 on Γ i , w b · n = 0 and ε( w)n b · τ = 0 on Γ e , σ( w, π b 1 + π 2 )n = 0 on Γ n . We deduce that (w, π), with

π = π 1 + π 2 + L p (η, h) = −λ 0 N s η − λ 0 N div h + N p F + N v w, is the solution to system (4.1).

5. Reformulation of the linearized system. For all (F f , F s 1 , F s 2 ) belonging to H = L 2 (Ω) × H 0 2 (Γ s ) × L 2 (Γ s ), we consider the system

λv − div σ(v, q) + (u s · ∇)v + (v · ∇)u s − A 1 η 1 − A 2 η 2 = F f , div v = A 3 η 1 in Ω, v = η 2 n on Γ s , v = 0 on Γ i , v · n = 0 and ε(v)n · τ = 0 on Γ e , σ(v, q)n = 0 on Γ n , λη 1 − η 2 = F s 1 on Γ s ,

λη 2 − β∆ s η 1 − γ∆ s η 2 + α∆ 2 s η 1 − A 4 η 1 = γ s q + F s 2 on Γ s , η 1 = 0 on ∂Γ s , η 1,x = 0 on ∂Γ s .

(5.1)

5.1. Properties of the operators A 1 , A 2 , A 3 and A 4 .

Proposition 5.1. The differential operators A 1 , A 2 , A 3 and A 4 obey A 1 ∈ L(H 0 2s ), L 2 (Ω)), A 2 ∈ L(L 2s ), L 2 (Ω)),

A 3 ∈ L(H 0 2s ), H 1 (Ω)) and A 4 ∈ L(H 0 2s ), H ε

0

s )).

Proof. For A 1 , A 2 and A 3 , the proposition follows from Lemmas 5.2 and 5.3 below, and for A 4 , it follows from [13, Proposition B.1].

Lemma 5.2. There exists C > 0 such that, for all (η, w) belonging to L 2 (Γ s )× H

12

0

(Ω), ηw belongs to L 2 (Ω) and

kηwk L

2

(Ω) ≤ Ckηk L

2

s

) kwk

H

12+ε0

(Ω) .

11

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Proof. Thanks to the continuous embeddings H

12

0

((±e, ±`) ×(0, ` s )) , → L 2 (±e, ±`; L (0, ` s )), and using the continuity of the extension operator E (see Remark 3.1), it follows that

Z

|ηw| 2 ≤ C Z −e

−`

Z `

s

0

|η(x, −e)w(x, z)| 2 dx dz + C Z `

e

Z `

s

0

|η(x, e)w(x, z)| 2 dx dz

≤ Ckηk 2 L

2

s

) kwk 2

H

12+ε0

(Ω) .

Lemma 5.3. There exists C > 0 such that, for all (η, w) belonging to H 0 1 (Γ s ) × L 2 δ

0

(Ω), ηw belongs to L 2 (Ω) and

kηwk L

2

(Ω) ≤ Ckηk H

1

0

s

) kwk L

2 δ0

(Ω) .

Proof. Let us denote by (C j ± ) 1≤j≤2 , the four corners corresponding to the boundary of Γ s , and by (x ± j , z j ± ) the coordinates of C j . Here the exponent ± refers to the upper and lower beams. Since η ± belongs to H 0 1 (Γ s ), using an appropriate corner, we have |η ± (x)| =

R x

x

±j

η x (s)ds

≤ kη x k L

2

s

)

x − x ± j

1 2

. Hence

Z

η ± w

2 ≤ kη x k 2 L

2

s

)

Z

x − x ± j

(|x − x ± j | 2 + |z − z ± j | 2 )

0

(|x − x ± j | 2 + |z − z ± j | 2 ) δ

0

w(x, z)

2 dxdz

≤ kη x k 2 L

2

s

)

Z

(|x − x ± j | 2 + |z − z j ± | 2 ) 1−2δ

0

(|x − x ± j | 2 + |z − z j ± | 2 ) δ

0

w(x, z)

2 dxdz.

Since 0 < δ 0 < 1 2 , it follows that kηwk L

2

(Ω) ≤ Ckηk H

1

0

s

) kwk L

2 δ0

(Ω) .

We denote by A 1 ∈ L(L 2 (Ω), H −2 (Γ s )), A 2 ∈ L(L 2 (Ω), L 2 (Γ s )), A 3 ∈ L(L 2 (Ω), H −2 (Γ s )) and A 4 ∈ L(L 2 (Γ s ), H −2 (Γ s )) the adjoints of A 1 , A 2 , A 3 and A 4 respectively, where H −2 (Γ s ) is the dual of H 0 2s ).

5.2. Definition of the unbounded operator (A, D(A)). For the beam, we introduce the un- bounded operator (A α,β , D(A α,β )) in H 0 2 (Γ s ) defined by

D(A α,β ) = H 4 (Γ s ) ∩ H 0 2 (Γ s ) and A α,β = β∆ s − α∆ 2 s . We equip the space H 0 2 (Γ s ) with the norm

(η, ξ) H

2

0

s

) = ((−A α,β )

12

η, (−A α,β )

12

ξ) L

2

s

) = Z

Γ

s

(β ∇ s η · ∇ s ξ + α∆ s η∆ s ξ),

and the spaces H = L 2 (Ω) × H 0 2 (Γ s ) × L 2 (Γ s ) and Z = V 0 n,Γ

0

(Ω) × H 0 2 (Γ s ) × L 2 (Γ s ) with the inner product

((v, η 1 , η 2 ), (φ, ξ 1 , ξ 2 )) H = (v, φ) L

2

(Ω) + (η 1 , ξ 1 ) H

2

0

s

) + (η 2 , ξ 2 ) L

2

s

) . (5.2) As in [24, Lemma 3.2], we can show that I + γ s N s is an automorphism in L 2 (Γ s ), and we can introduce M s ∈ L(Z ) and M s −1 ∈ L(Z) two automorphisms in Z defined by

M s =

I 0 0

0 I 0

0 γ s N div A 3 I + γ s N s

 and M s −1 =

I 0 0

0 I 0

0 −(I + γ s N s ) −1 γ s N div A 3 (I + γ s N s ) −1

 . Remark 5.4. We notice that, in M s , γ s N s corresponds to the so-called added mass effect (see [5]).

Here, the added mass operator M s is not symmetric.

12

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We are going to define the infinitesimal generator (A, D(A)) in Z of the linearized system around (0, 0, 0, 0) associated with (3.3). For that we first introduce

D(A) := n

(P v, η 1 , η 2 ) ∈ V

1 2

0

n,Γ

0

(Ω) × H 4 (Γ s ) ∩ H 0 2 (Γ s )

× H 0 2 (Γ s ) | P v − P L(η 2 , A 3 η 1 ) ∈ D(A) o . Lemma 5.5. If (P v, η 1 , η 2 ) belongs to D(A), then N v (P v+∇N s η 2 +∇N div A 3 η 1 ) belongs to H

12

0

(Ω).

Proof. Let (P v, η 1 , η 2 ) belong to D(A). Then, there exists F ∈ V 0 n,Γ

0

(Ω) such that (λ 0 I − A)P v + (A − λ 0 I)P L(η 2 , A 3 η 1 ) = F.

We set v = P v + ∇N s η 2 + ∇N div A 3 η 1 and ρ = −λ 0 N s η 2 − λ 0 N div A 3 η 1 + N v v. Then, according to Theorem 4.12, we know that (v, ρ) is the unique solution of the equation

λ 0 v − div σ(v, ρ) + (u s · ∇)v + (v · ∇)u s = F, div v = A 3 η 1 in Ω,

v = η 2 n on Γ s , v = 0 on Γ i , v · n = 0 and ε(v)n · τ = 0 on Γ e , σ(v, ρ)n = 0 on Γ n . (5.3) According to Lemma 4.9, we conclude that N v (P v + ∇N s η 2 + ∇N div A 3 η 1 ) belongs to H

12

0

(Ω).

We are now in position to introduce the unbounded operator (A, D(A)) in Z where D(A) is defined above and

A = M s −1

A P A 1 + (λ 0 I − A)P L(0, A 3 · ) P A 2 + (λ 0 I − A)P L( · , 0)

0 0 I

0 A α,β + γ s N p A 1 + A 4 γ∆ + γ s N p A 2

 + M s −1 A p , (5.4) where (A p , D(A p )) is the unbounded operator in Z defined by D(A p ) = D(A) and

A p (P v, η 1 , η 2 ) T = (0, 0, γ s N v (P v + ∇N s η 2 + ∇N div A 3 η 1 )) T , ∀ (P v, η 1 , η 2 ) ∈ D(A). (5.5) 5.3. Resolvent of A in Z . In what follows, we are going to consider system (5.1) either when λ ∈ C and (F f , F s 1 , F s 2 ) are complex valued functions (in which case the solutions are also complex valued functions), or when λ ∈ R and (F f , F s 1 , F s 2 ) are real valued functions (in which case the solutions are also real valued functions). The context will clearly indicate in which case we shall be.

The goal is to prove that the unbounded operator (A, D(A)) has a compact resolvent.

Proposition 5.6. Let (F f , F s 1 , F s 2 ) belong to H and λ belong to C . A quadruplet (v, q, η 1 , η 2 ) ∈ H δ

0

is solution to system (5.1) if and only if

λ(P v, η 1 , η 2 ) T = A(P v, η 1 , η 2 ) T + M s −1 (P F f , F s 1 , F s 2 + γ s N p F f ) T , (I − P )v = ∇N s η 2 + ∇N div A 3 η 1 ,

q = −λN s η 2 − λN div A 3 η 1 + N p A 1 η 1 + N p A 2 η 2 + N v (P v + ∇N s η 2 + ∇N div A 3 η 1 ) + N p F f .

(5.6)

Proof. It follows from Theorem 4.12.

Proposition 5.7. Let (F f , F s 1 , F s 2 ) belong to Z. If λ ∈ R is large enough, then system (5.1) admits a unique solution (v, q, η 1 , η 2 ) ∈ H δ

0

satisfying the estimate

kvk H

2

δ0

(Ω) + kqk H

1

δ0

(Ω) + kη 1 k H

4

s

) + kη 2 k H

2

s

) ≤ Ck(F f , F s 1 , F s 2 )k Z .

Proof. Step 1. Lifting of the divergence condition. We are going to rewrite system (5.1) in the form of a system with divergence free condition. For that, we make the change of unknowns

v b = ( b v 1 , b v 2 ) = v − L (0, A 3 η 1 ) and p b = p − L p (0, A 3 η 1 ) , (5.7) where L and L p are introduced in (4.19). We have b v = η 2 n on Γ s , which gives

η 2 = b v 2 on Γ s and η 1 = 1 λ b v 2 + 1

λ F s 1 on Γ s .

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We introduce the operator L λ in H 0 2 (Γ s ) defined by

L λ = λ 2 − (β + γλ)∆ s + α∆ 2 s − A 4 − γ s L p (0, A 3 · ).

We choose λ large enough so that L λ satisfies a coercivity condition in H 0 2s ). From Proposition 5.1, it follows that L λ ∈ L(H 0 2s ), H −2s )), and with the Lax-Milgram Theorem, we can prove that L λ is an isomorphism from H 0 2s ) into H −2s ). The operator L λ is also an isomorphism from H 4s ) ∩ H 0 2s ) into L 2s ). The system satisfied by ( b v, p, η b 1 , η 2 ) can be rewritten in the form

λ b v − div σ( b v, p) + (u b s · ∇) b v + ( b v · ∇)u s − A 5 b v = G f in Ω, div v b = 0 in Ω, b v = λL −1 λ γ s pn b + G b n on Γ s , b v = 0 on Γ i , b v · n = 0 on Γ e , ε( b v)n · τ = 0 on Γ e , σ( b v, p)n b = 0 on Γ n , λη 1 − η 2 = F s 1 on Γ s , L λ η 1 = γ s p b + G s on Γ s ,

η 1 = 0 on ∂Γ s , η 1,x = 0 on ∂Γ s ,

(5.8)

where

A 5 b v = 1

λ A 1 γ s b v 2 + A 2 γ s v b 2 + λ 0 − λ

λ L(0, A 3 γ s b v 2 ) − 1

λ L p (0, A 3 γ s b v 2 ), G f = F + 1

λ A 1 F s 1 + λ 0 − λ

λ L(0, A 3 F s 1 ) − 1

λ L p (0, A 3 F s 1 ), G s = F s 2 + λF s 1 − γ∆F s 1 , G b = λL −1 λ G s − F s 1 .

From Theorem 4.1 and Proposition 5.1, it follows that there exists C > 0 such that kA 5 b vk L

2

(Ω) ≤ Ckγ s b v 2 k H

2

0

s

) for all v b 2 ∈ H 0 2 (Γ s ), and

kG f k L

2

(Ω) + kG s k L

2

s

) + kG b k H

2

0

s

) ≤ Ck(F f , F s 1 , F s 2 )k Z . Step 2. Existence of weak solutions for system (5.8). We introduce the space E = n

b v = ( b v 1 , b v 2 ) ∈ H 1 (Ω) | div v b = 0 in Ω, b v = 0 on Γ i , v · n = 0 on Γ e , b v 1 = 0 on Γ s , b v 2 | Γ

s

∈ H 0 2s ) o , equipped with the norm k b vk E =

k b vk H

1

(Ω) + kL λ

12

b v 2 | Γ

s

k L

2

s

)

12

. We set a λ ( b v, φ) = λ

Z

Ω b v · φ + 2ν Z

ε( b v) : ε(φ) + Z

((u s · ∇) v b + ( v b · ∇)u s ) · φ − Z

A 5 b v · φ + 1 λ

Z

Γ

s

L

1 2

λ b v 2 L

1 2

λ φ 2 , and ` λ (φ) =

Z

G f · φ + 1 λ

Z

Γ

s

L λ

12

G b L λ

12

φ 2 . If ( b v, p) b ∈ H 2 δ

0

(Ω) × H δ 1

0

(Ω) is a solution to system (5.8) 1−3 , then b v is solution to the variational problem Find b v ∈ E such that a λ ( b v, φ) = ` λ (φ), ∀φ ∈ E. (5.9) Thanks to the Lax-Milgram Theorem, problem (5.9) admits a unique solution. Moreover it can be shown that if b v ∈ E is the solution to (5.9), then there exists a pressure p b ∈ L 2 (Ω) such that the pair ( b v, p) is a b variational solution to the system

λ 0 b v − div σ( b v, b p) + (u s · ∇) b v + ( b v · ∇)u s = (λ 0 − λ) b v + G f − A 5 b v, div b v = 0 in Ω,

b v = v b 2 n on Γ s , v b = 0 on Γ i , b v · n = 0 and ε( b v)n · τ = 0 on Γ e , σ( b v, p)n b = 0 on Γ n . (5.10) By variational solution, we understand a solution to the mixed variational problem associated with (5.10), as introduced in (4.3). Since b v belongs to V 1 (Ω), (λ 0 − λ) v b + G f − A 5 b v belongs to L 2 (Ω) and b v 2 belongs to H

3 2

0 (Γ s ). Thus, from Theorem 4.1, it follows that ( v, b p) belongs to b H 2 δ

0

(Ω) × H δ 1

0

(Ω) and k b vk H

2

δ0

(Ω) + k pk b H

1

δ0

(Ω) ≤ C(k(λ 0 − λ) b v + G f − A 5 b vk L

2

(Ω) + k b v 2 k

H

32

0

s

) ) ≤ Ck(F f , F s 1 , F s 2 )k Z .

14

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