Modeling and Analysis of an Elastic Problem with Large Displacements and Small Strains

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Modeling and analysis of an elastic problem in large

displacements and small strains

C. Grandmont, Y. Maday and P. M´

etier

7th April 2004

Abstract

In this paper we present and analyse a dynamical geometrically nonlinear formulation that models the motion of a two–dimensional elastic structure in large displacements–small strains. In a first part (section 2) we derive the equations describing the motion of the body. In a second part (section 3), existence of a weak solution is proven using a Galerkin method. We also prove that the solution is unique.

1 Introduction

In this paper, we present and analyse a dynamical geometrically nonlinear formulation that models the motion of a two–dimensional structure in large displacements–small strains. The starting point of the formulation is to seperate the rigid body motion from purely elastic displacements. This is an idea that has been used in the past two decades in engineering sciences and is referred in literature as the corotational kinematic description (see [6]).

In a first part, we derive the 2D model (section 2). We give a criterion that enables us to associated to deformations that are “close” to rigid body motion a unique translation, rotation and elastic displacement. This criterion was considered by de Veubeke in [7]. Writing the Lagrange equations of the problem, two equivalent weak formulations are obtained. We next obtain a priori estimates for the solution. Unlike the linearized elasticity for which the rotations are not well described the model take into account the real rotations of the body and therefore is nonlinear. Nevertheless, this model is linear with respect to the elastic displacement, thus it can be seen as a model between the linearized elasticity and the standard system of elasticity for a Saint–Venant material for instance.

In the second part, we study the solvability of the problem. The proof of the existence of at least one local solution is based on a Galerkin method. We prove that the discrete problem has one solution bounded in the energy spaces. Because of the presence of nonlinear terms, additional bounds are needed in order to get compactness and obtain the continuous problem as the limit of the discrete one. These estimates are obtained by looking at the low and high frequencies of the acceleration separately. Then, we prove in two steps that the solution is unique. The first step consists in formal calculations and in a second part thoses calculations are justified using admissible test functions. Finally we prove the existence of a maximal solution.

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2 Modeling

2.1 Decomposition of the deformation

Let Ω be a lipchitz domain of R2_{. We suppose that Ω is the reference configuration of an elastic}

body. We denote by φ : Ω −→ R2the deformation of this body: to each point ξ ∈ Ω we associate the point x = φ(ξ) of the deformed configuration φ(Ω). The associated displacement is denoted by u = φ − Id

R2 We want to obtain a model in order to describe the motion of this structure in large displacements and small strains. The key idea is to separate the displacements due to elastic behavior from the one due to rigid body motions. We thus consider that the deformation can be decomposed in

1. a translation whose vector is τ : ξ 7−→ ξ + τ with τ ∈ R2. 2. a rotation : ξ 7−→ Rθ

−→

Gξ, of angle θ and centre G, where G denotes the centre of gravity of Ω and where Rθ= cos θ − sin θ sin θ cos θ , θ ∈ R. 3. an elastic deformation : Ω −→ R2 ξ 7−→ ξ + d(ξ). The deformation can be written as follows:

∀ξ ∈ Ω, φ(ξ) = τ + Rθ(

−→

Gξ + d(ξ)) (1)

The rotation can be viewed as a rotation in R3 of angle θ and axis −→ez, where −→ez denotes a unit

vector orthogonal to the motion’s plane R2. We first remark that for all φ in L2

(Ω), for all τ in R2

, for all θ in R, d is defined uniquely thanks to (1) by d(ξ) = RTθ(φ(ξ) − τ ) −

−→

Gξ and d belongs also to L2(Ω).

The first step is to be able to associate a unique triplet (τ , θ, d) to a given φ in L2(Ω). This will be possible for deformations φ “close” to rigid deformations in a sense that will be made precise latter on. The criterion we introduce is the minimization of the L2–norm of d. We thus consider the following mapping:

Kφ: R2× R −→ R

(τ , θ) 7−→ Kφ(τ , θ) = || RTθ(φ − τ ) −

−→ Gξ ||2_L2_(Ω)

and we search the minimizer (τ , θ) of this functional. Using the fact that Z Ω −→ Gξ = 0 we have that Kφ(τ , θ) = || φ − τ ||2− 2 Z Ω φ · Rθ −→ Gξ + ||−Gξ ||→ 2. We want to solve (P) min θ ∈ [0, 2π[ τ ∈ R2 Kφ(τ , θ) .

This problem has at least one solution since Kφ(τ , θ) = k1(τ ) + k2(θ), where k1 is a coercive

strictly convex mapping, and k2 is continuous and 2π-periodic (k1(τ ) = || φ − τ ||2, k2(θ) =

−2R

Ωφ · Rθ

−→

Gξ + ||−Gξ ||→ 2). The optimality conditions of order one and two are: ∂Kφ

∂τ (τ , θ) = 0 ,

∂2_K φ

∂τ2 (τ , θ) > 0 in the sense of symmetric matrix ,

∂Kφ

∂θ (τ , θ) = 0 ,

∂2Kφ

∂θ2 (τ , θ) ≥ 0 .

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They can be written in an equivalent way: τ = 1 |Ω| Z Ω φ , Z Ω φ ∧ Rθ −→ Gξ = 0 , Z Ω φ · Rθ −→ Gξ ≥ 0 . (3)

The first condition τ = 1 |Ω|

Z

Ω

φ determines uniquely τ . The two other conditions can be rewritten as follows: Z Ω φ ∧ Rθ −→ Gξ = Z Ω φ ∧−Gξ→ cos θ + Z Ω φ ·−Gξ→ sin θ = 0, Z Ω φ · Rθ −→ Gξ = − Z Ω φ ∧−Gξ→ sin θ + Z Ω φ ·−Gξ→ cos θ ≥ 0. (4)

Thus ifR_Ωφ ∧−Gξ 6= 0 or→ R_Ωφ ·−Gξ 6= 0, we can define a unique θ in [0, 2π[. To this unique solution→ (τ , θ), we associate d defined by d(ξ) = RT

θ(φ(ξ) − τ ) −

−→

Gξ, ∀ξ ∈ Ω. Next, taking into account the first condition of (3), and since G is the center of gravity of Ω, we remark that d has a zero average in Ω. Moreover the conditions (4) can be rewritten as follows:

Z Ω φ ∧ Rθ −→ Gξ = Z Ω d ∧−Gξ = 0→ and Z Ω φ · Rθ −→ Gξ = Z Ω (−Gξ + d) ·→ −Gξ ≥ 0.→ (5)

Remark that, taking into account the first condition of (4), if Z Ω φ ∧−Gξ 6= 0 or→ Z Ω φ ·−Gξ 6= 0, then→ the second condition of (4) (or the second condition of (5)) is in fact a strict inequality.

Thus we have established the proposition: Proposition 1 For all φ ∈ L2(Ω) such that:

Z Ω φ ∧−Gξ→ 2 + Z Ω φ ·−Gξ→ 2 6= 0 , (6)

there exists a unique triplet (τ , θ, d) ∈ R2× [0, 2π[×L2

(Ω) satisfying τ + Rθ(

−→

Gξ + d) = φ and such that d satisfies:

Z Ω d = 0 , Z Ω d ∧−Gξ = 0 and→ Z Ω (d +−Gξ) ·→ −Gξ > 0 .→ (7) In particular (τ , θ) is the unique solution of (P), and d = RT

θ(φ − τ ) −

−→ Gξ.

In all what follows, we will suppose that φ satisfies (6). This assumption is justified since we are interested in deformations that are close to rigid deformations. Indeed, in the rigid case, we have d ≡ 0 and consequently

Z Ω φ · Rθ −→ Gξ = Z Ω |−Gξ|→2> 0.

Now that we have obtained a suitable representation of the deformations, let us make precise the sets we will use. Let s be a positive real number. We set:

Xs= ( φ ∈ Hs(Ω); Z Ω φ ·−Gξ→ 2 + Z Ω φ ∧−Gξ→ 2 6= 0 ) , Es= d ∈ Hs(Ω); Z Ω d = 0, Z Ω d ∧−Gξ = 0→ , Ys= d ∈ Es; Z Ω (−Gξ + d) ·→ −Gξ > 0→ , Zs= R2× [0, 2π[×Ys.

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Here H0(Ω) denotes the space L2(Ω). One can note that E0 is a close subspace of L2(Ω) whose

vectors are orthogonal (for the scalar product of L2_{(Ω)) to the translations and infinitesimal}

rotations (ξ 7→ −→ez∧

−→

Gξ). We now introduce the mapping

H : R2× R × L2(Ω) −→ L2(Ω) (τ , θ, d) 7−→ φ = τ + Rθ(

−→ Gξ + d)

.

The mapping satisfies the following proposition:

Proposition 2 The mapping H : R2× R × L2(Ω) −→ L2(Ω) (τ , θ, d) 7−→ φ = τ + Rθ(

−→ Gξ + d)

is a C1 -diffeo-morphisme from Zsonto Xs. For all φ ∈ Xs, the differential DF (φ) of the inverse F of H is the

linear mapping from Hs(Ω) onto R2× R × Esdefined by DF (φ).v = τ (v), θ(v), d(v) where

τ (v) = 1 |Ω| Z Ω v, θ(v) = 1 R Ω( −→ Gξ + d) ·−Gξ→ Z Ω v · (−→ez∧ Rθ −→ Gξ), (8) and d(v) = RθT v − 1 |Ω| Z Ω v − 1 R Ω( −→ Gξ + d) ·−Gξ→ Z Ω v · (−→ez∧ Rθ −→ Gξ) − →_e z∧ Rθ( −→ Gξ + d) ! , (9) with (τ , θ, d) = F (φ).

Proof: We first prove the proposition for s = 0. Using the previous definitions, Proposition 1 says that for all φ in X0, there exists a unique triplet (τ , θ, d) ∈ Z0 such that: H(τ , θ, d) = φ.

Furthermore, H(Z0) ⊂ X0. Indeed for all (τ , θ, d) ∈ Z0, since:

Z Ω φ ∧ Rθ −→ Gξ = Z Ω d ∧−Gξ→ and Z Ω φ · Rθ −→ Gξ = Z Ω (−Gξ + d) ·→ −Gξ,→ we have that Z Ω φ ∧ Rθ −→ Gξ = 0 and Z Ω φ · Rθ −→ Gξ > 0 hold. But if R Ωφ · Rθ −→ Gξ (which is equal to −R Ωφ ∧ −→ Gξsin θ +R Ωφ · −→ Gξcos θ) is strictly positive then φ ∈ X0. Consequently, the mapping H is one to one from Z0onto X0. We are going

to show that this is a C1_{-diffomorphism from Z}

0onto X0 and study its differential together with

the differential of F = H−1.

The mapping H is of class C∞, and we have in particular:

∀(τ , θ, d) ∈ R2× R × E0, DH(τ , θ, d).(τ , θ, d) = τ + θ−→ez∧ Rθ(

−→

Gξ + d) + Rθd. (10)

We want to prove that ∀(τ , θ, d) ∈ Z0, ∀v ∈ L2(Ω), there exists a unique triplet (τ (v), θ(v), d(v)) ∈

R2× R × E0 such that DH(τ , θ, d).(τ (v), θ(v), d(v)) = v. Thanks to the definition of E0 and G

we have the orthogonal decomposition of L2_(Ω):

L2(Ω) = h−→ex, −→eyi ⊥

⊕D−→ez∧

−→ GξE⊕E⊥ 0.

this yields, for all θ ∈ R,

L2(Ω) = h−→ex, −→eyi ⊥

⊕D−→ez∧ Rθ

−→

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Considering d ∈ L2_{(Ω) such that}

Z

Ω

(−Gξ + d) ·→ −Gξ 6= 0, we deduce from (11) that→

L2(Ω) = h−→ex, −→eyi ⊕

D₋_→ ez∧ Rθ(

−→

Gξ + d)E⊕Rθd; d ∈ E0 . (12)

This implies that for each v ∈ L2

(Ω), there exists a unique triplet (τ (v), θ(v), d(v)) ∈ R2

× R × E0

such that:

v = τ (v) + θ(v)−→ez∧ Rθ(

−→

Gξ + d) + Rθd(v).

Moreover we easily obtain

τ (v) = 1 |Ω| Z Ω v, θ(v) = 1 R Ω( −→ Gξ + d) ·−Gξ→ Z Ω v · (−→ez∧ Rθ −→ Gξ), and thus d(v) = RTθ(v − τ (v) − θ(v)−→ez∧ Rθ( −→ Gξ + d)) = RTθ v − 1 |Ω| Z Ω v − 1 R Ω( −→ Gξ + d) ·−Gξ→ Z Ω v · (−→ez∧ Rθ −→ Gξ) − → ez∧ Rθ( −→ Gξ + d) ! . (13)

Consequently for all (τ , θ, d) ∈ Z0, DH(τ , θ, d) is an isomorphism from R2× R × E0 onto L2(Ω).

And we have:

DF (φ).v = τ (v), θ(v), d(v) , where φ ∈ X0and v ∈ L2(Ω).

For s > 0, we see easily that if φ ∈ Xs then (τ , θ, d) ∈ Zs (and reciprocally). Moreover we

deduce from (13) that if v ∈ Hs_{(Ω) then d(v) ∈ E}

s. Thus the mapping F is a C1-diffeomorphism

from Xsonto Zs, and the components of its differential at each φ ∈ Xs are given by (8)-(9).

2.2 Derivation of the model: weak formulation of the equations

In this subsection, we derive a dynamical model that describes the motion of an homogeneous, isotropic, elastic body in large displacements, small strains. We assume that the reference con-figuration Ω of the elastic media is a natural state. Under these hypotheses we can suppose that the material is a Saint Venant–Kirchhoff material. Its stored energy for a given displacement u is (cf for instance [2], Theorem 4.4.3 page 155):

ˇ

W (E(u)) = λ

2[Tr(E(u))]

2

+ µTr[E(u)]2

where λ and µ denote the Lam´e constants of the material and E is the Green–Saint Venant strain tensor:

E(u) =1 2

∇u + ∇uT+ ∇uT∇u.

An important remark is that, for a displacement u and the associated translation, angle of rotation and elastic displacement (τ , θ, d), we have E(u) = E(d), which traduces that the strain tensor depends only on the elastic displacement and do not depend on rigid body motion. The stored energy thus appears as an invariant with respect to the rigid body motions. Assuming that ∇d is small, the linearized versions of the strain tensor and of the stored energy function are:

|ε(d) =1 2 ∇d + (∇d)T, ˆ W(d) =λ 2[Tr(|ε(d))] 2 + µTr[|ε(d)]2. (14)

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We would like to describe the dynamical behavior of such a material under the hypothesis that d is “small” (in a sense to precise). Let T be a non negative real number. For any t ∈ [0, T ], the deformation φ(·, t) is decomposed at each time t according to (1):

∀ξ ∈ Ω, φ(ξ, t) = τ (t) + Rθ(t)(

−→

Gξ + d(ξ, t)). We write the Lagrange equations satisfied by u = φ − Id

R2 (one can refer to [1]or [4] for details on Lagrange equations). We denote by ˙u = ∂tu the body velocity, and we choose u (respectivelly

˙

u) as the generalized coordinate (resp. velocity). We introduce the external work function: T (u, ˙u) = Z Ω f · u + Z ∂Ω g · u, (15)

where f denotes the applied body force and g the applied surface force. For the elastic displace-ment d, we define the stored energy function ˆW by (14), and the strain energy associated to u by: W (u, ˙u) = W(d) = Z Ω ˆ W(d). Finally the kinetic energy is denoted by EC = ρS

Z

Ω

( ˙u)2/2 where ρS is the body density. The

lagrangian of the stucture is then L(u, ˙u) = EC+ T − W , and the Lagrange equations of the

system read as follows

∀T > 0, ∀v, v(·, t = 0) = v(·, t = T ) = ∂tv(·, t = 0) = ∂tv(·, t = T ) ≡ 0 , Z T 0 ∂L ∂ ˙u(u, ˙u).∂tv + ∂L ∂u(u, ˙u).v

= 0. (16)

For all u, ˙u, v and ˙v, we can see easily that ∂EC

∂u (u, ˙u).v = 0 ,

∂EC ∂ ˙u (u, ˙u). ˙v = ρS Z Ω ˙ u · ˙v , ∂T

∂u(u, ˙u).v = Z Ω f · v + Z ∂Ω g · v , ∂T ∂ ˙u(u, ˙u). ˙v = 0 , ∂W ∂ ˙u(u, ˙u). ˙v = 0 . Recalling that u = φ − Id

R2, using the fact that Duφ(u) = Id and Proposition 2 we obtain:

∀v, ∂W

∂u(u, ˙u).v = Z Ω DdW(d). [Dˆ φd(φ). (Duφ(u).v)] = Z Ω DdW(d).dˆ

where d is associated to v through (9). Moreover

DdW(d).b = λTr(|ε(d))Tr(|ε(b)) + 2µTr[|ε(d)|ε(b)],ˆ

where I2 is the identy matrix of R2. Using the notation A : B = Tr(BTA), ∀A, B ∈ M2(R) this

leads to,

∂W

∂u(u, ˙u).v = Z

Ω

σ(d) : |ε(d),

with σ(d) = λTr(|ε(d))I2+ 2µ|ε(d). Finally, for v sufficiently smooth such as v(·, 0) = v(·, T ) ≡ 0

and ∂tv(·, 0) = ∂tv(·, T ) ≡ 0, we have: Z T 0 ρS Z Ω ∂tu · ∂tv + Z Ω f · v + Z ∂Ω g · v − Z Ω σ(d) : |ε(d) = 0,

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with d associated to v through (9). From the previous equation we deduce the weak formulation of the equations satisfied by u and by the associated elastic displacement d:

 



∀v sufficiently smooth and d associated to v through (9), ρS Z Ω (∂ttu) · v + Z Ω σ(d) : |ε(d) = Z Ω f · v + Z ∂Ω

g · v in the sense of distributions D0(0, T ). (17) This system has to be completed by initial data: u(·, t = 0) = u0 and ∂tu(·, t = 0) = u1.

We have obtained equations that model the motion of a structure with large displacements and small strains, that are valid for elastic displacements d satisfying

Z Ω (−Gξ + d) ·→ −Gξ > 0 (or at→ least Z Ω

(−Gξ + d) ·→ −Gξ 6= 0). This model belongs to the class of the so called geometrically nonlinear→ models. The weak formulation we obtain is not surprising since we have a term of acceleration depending on the global displacement and a term of mechanical energy depending only on the elastic displacement. Note however that this formulation is not standard since the test functions depend on the solution.

Remark 1 From this weak formulation, a strong formulation can be deduced at least formally. Denoting by Pu: H1(Ω) −→ H1(Ω) the linear operator which to any v associates the vector d

defined by (9), the weak formulation (17), becomes: Z Ω f · v + Z ∂Ω g · v − ρS Z Ω (∂ttu) · v = Z Ω σ(d) : ∇(Pu(v)) = − Z Ω PuT[div(σ(d))] · v + Z ∂Ω h PuT(σ(d) · n) i · v Consequently for v smooth enough, this leads to the following equations

ρS∂ttu − PuT[div(σ(d))] = f in Ω × (0, T ) ,

PuT(σ(d) · n) = g on ∂Ω × (0, T ) .

(18) Those equations are nonlinear since the operator Pu depends on the solution.

2.3 Dynamic of the translation, rotation and elastic displacement

In this subsection we derive a set of three equations equivalent to (17), these equations describe the dynamic of the translation τ , the angle of rotation θ and the elastic displacement d. These weak formulations are obtained by taking appropriate test functions in (17).

For a given θ ∈ R and a given d ∈ Y1, we remark that, as what was done to obtain (12),

H1(Ω) is the sum of three subspaces: H1(Ω) = h−→ex, −→eyi ⊕

D₋_→ ez∧ Rθ(

−→

Gξ + d)E⊕Rθd; d ∈ E1 . (19)

Thus we will choose the test fonction v in (17), succesively belonging to < −→ex, −→ey>, < −→ez∧ (

−→ Gξ + d) > then equal to Rθd where d ∈ E1 :

• If the mass of the structure is denoted by m, m = ρS|Ω|, we obtain for v ∈< −→ex, −→ey>:

m¨τ = ρS Z Ω ∂ttu = Z Ω f + Z ∂Ω g (20)

since d has a zero average and u = τ + Rθ(

−→

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• Choosing −→ez∧ Rθ(

−→

Gξ + d) as a test function in (17), we obtain: ρS Z Ω Rθ( −→ Gξ + d) ∧ ∂ttu = Z Ω Rθ( −→ Gξ + d) ∧ f + Z ∂Ω Rθ( −→ Gξ + d) ∧ g. (21)

Here the exterior product has to be taken as an operator defined on R2. In all what follows the exterior product ∧ stands for both the exterior product of R3_{and the operator defined}

on R2. We next express u with respect to τ , θ and d and differentiate twice: ∂ttu − ¨τ = ¨Rθ(

−→

Gξ + d) + 2 ˙Rθ∂td + Rθ∂ttd. (22)

Thus this gives a first contribution: Z Ω Rθ( −→ Gξ + d) ∧ ¨Rθ( −→ Gξ + d) = Z Ω Rθ( −→ Gξ + d) ∧h ¨θ−→ez∧ Rθ( −→ Gξ + d) + ˙θ−→ez∧ ( ˙θ−→ez∧ Rθ( −→ Gξ + d))i = ¨θJ (t) ρS + ( ˙θ)2Rθ Z Ω (−Gξ + d) ∧ (−→ →ez∧ (−→ez∧ ( −→ Gξ + d))) = ¨θJ (t) ρS where J (t) = ρS Z Ω

(−Gξ + d)→ 2 denotes the inertia momemtum at time t, and where we have used the fact that ˙Rθ. = ˙θ−→ez∧ Rθ. and that

−→

Gξ + d and −→ez∧ (−→ez∧ (

−→

Gξ + d)) are colinear. The second contribution can be written

Z Ω Rθ( −→ Gξ + d) ∧ ˙Rθ(∂td) = ˙θRθ Z Ω (−Gξ + d) ∧ (−→ →ez∧ ∂td) = ˙θ Z Ω (−Gξ + d) · ∂→ td = θ˙ 2 d dt J(t) ρS .

Using (22) together with the two previous equalities and using the fact that τ is orthogonal to −→ez∧ Rθ( −→ Gξ + d) yields: ρS Z Ω Rθ( −→ Gξ + d) ∧ ∂ttu = d dt ˙θJ + ρS Z Ω (−Gξ + d) ∧ ∂→ ttd.

Moreover since d is orthogonal to the infinitesimal rotation ξ 7→ −→ez∧

−→ Gξ we have, ρS Z Ω Rθ( −→ Gξ + d) ∧ ∂ttu = d dt ˙θJ + ρS Z Ω d ∧ ∂ttd. Then (21) becomes: d dt ˙θJ + ρS Z Ω d ∧ ∂ttd = Z Ω Rθ( −→ Gξ + d) ∧ f + Z ∂Ω Rθ( −→ Gξ + d) ∧ g. (23)

• Finally to derive the equations satisfied by d, we choose v = Rθd as a test function in (17),

with d ∈ E1: ρS Z Ω ∂ttu · Rθd + Z Ω σ(d) : |ε(d) = Z Ω f · Rθd + Z ∂Ω g · Rθd. (24)

Once again using (22) and the fact that d is orthogonal to the translations: Z Ω ∂ttu · Rθd = Z Ω ¨_R_θ₍−_{Gξ + d) + 2 ˙}→ _R_θ_∂_t_{d + R}_θ_∂_tt_d · Rθd.

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Note we have also: Z Ω ¨ Rθ( −→ Gξ + d) · Rθd = Z Ω h ¨_θ−→_e z∧ ( −→ Gξ + d) + ˙θ−→ez∧ ( ˙θ−→ez∧ ( −→ Gξ + d))i· d = ¨θ−→ez· Z Ω d ∧ d − ( ˙θ)2 Z Ω (−Gξ + d) · d→ since d again is orthogonal to the infinitesimal rotation. We also have:

Z Ω ˙ Rθ∂td · Rθd = ˙θ Z Ω (−→ez∧ ∂td) · d = ˙θ Z Ω ∂td ∧ d . The previous equalities and (24) leads to:

∀d ∈ E1, ρS Z Ω ∂ttd · d + 2 ˙θ Z Ω ∂td ∧ d + ¨θ Z Ω d ∧ d − ( ˙θ)2 Z Ω (−Gξ + d) · d→ + Z Ω σ(d) : ∇d = Z Ω f · Rθd + Z ∂Ω g · Rθd. (25)

These equations (20)-(23)-(25) have to be completed by initial data:    τ (0) = τ0, θ(0) = θ0, d(·, t = 0) = d0, with Z Ω (−Gξ + d→ 0) · −→ Gξ > 0 , ˙ τ (0) = τ1, θ(0) = θ˙ 1, ∂td(·, t = 0) = d1, where: ( τ0+ Rθ0( −→ Gξ + d0) − −→ Oξ := u0, τ1+ Rθ0d1+ θ1 − → ez∧ Rθ0( −→ Gξ + d0) := u1.

Remark 2 The weak formulation (25) corresponds to the weak formulation of the equations that describe the motion of a linearized elastic body in rotation at velocity ˙θ, equations written in the moving frame where Coriolis and centrifugal terms appear.

Remark 3 The equations (20), (23), (25) can be also obtained by using the Lagrangian with q1=

τ , q2 = θ, q3 = d as generalized coordinates. Thus L = L(q1, q2, q3, ˙q1, ˙q2, ˙q3) and

d dt ∂L ∂ ˙qi − ∂L ∂qi

= 0 for i = 1, 2, 3 give (20), (23) and (25).

2.4 A priori estimates

In the present subsection we derive a priori estimates assuming that there exists a solution u of (17) smooth enough. In a first step we establish an energy equality then energy estimates for the global system and in a second step we find bounds on the different components (τ , θ, d) of the displacement u.

We take v = ∂tu = ˙τ + ˙θ−→ez ∧ Rθ(

−→

Gξ + d) + Rθ∂td as a test function in (17), which is

equivalent to multiply (20) by ˙τ , (23) by ˙θ, take ∂td as a test function in (25) and add these three

contributions. The first term gives the derivative with respect to time of the structure kinetic energy. Moreover, setting

((d, b)) = Z

Ω

(λTr(|ε(d))Tr(|ε(b)) + 2µ|ε(d) : |ε(b)) , and ||| b |||2 = ((b, b)) ,

and observing that ||| d |||2= 2W(d), we obtain the following energy equality: ρS 2 d dt Z Ω |∂tu|2 + d dt[W(d)] = Z Ω f · ∂tu + Z ∂Ω g · ∂tu. (26)

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Let us now establish a priori estimates. For t ∈ (0, T ), we write: Z ∂Ω g · ∂tu = d dt Z ∂Ω g · u − Z ∂Ω ∂tg · u,

and we denote by E(t) the structure energy: E(t) = 1 2ρS|| ∂tu(t) || 2 L2(Ω)+ 1 2||| d(t) ||| 2 . Then, integrating (26) over (0, t), we have:

E(t) = E(0) − Z ∂Ω g(0) · u(0) + Z t 0 Z Ω f(s) · ∂tu(s)ds + Z ∂Ω g(t) · u(t) − Z t 0 Z ∂Ω ∂tg(s) · u(s)ds

This leads to the following estimate:

E(t) ≤ a0+ 1 2ρS || f ||2 L2(0,T ;L2(Ω))+ Zt 0 E(s)ds + C|| g(t) ||_H−1/2_(∂Ω)|| u(t) ||_H1(Ω) + C Z t 0 || ∂tg(s) ||H−1/2_(∂Ω)|| u(s) ||_H1(Ω)ds, (27) with a0= E(0) + Z ∂Ω g(0) · u0

and where we used that for any b ∈ H1_(Ω):

|| b ||_H1/2(∂Ω)≤ C|| b ||H1(Ω).

Next we estimate || u(t) ||_H1(Ω)with respect to E(t). Remembering that u = τ +Rθ(

−→ Gξ +d)−−Oξ,→ it gives : || u(t) ||2_H1_(Ω)≤ 3 || τ (t) ||2_H1_(Ω)+ || Rθ −→ Gξ −−Oξ ||→ 2_H1_(Ω)+ || d(t) || 2 H1_(Ω) . (28) It is clear that: || Rθ −→ Gξ −−Oξ ||→ 2_H1_(Ω)≤ C

where the constant C depends on Ω, and that: || τ (t) ||2 H1(Ω) = || τ (t) || 2 L2(Ω)= 1 |Ω| Z Ω u 2 = 1 |Ω| Z t 0 Z Ω ∂tu(s)ds + τ (0)|Ω| 2 ≤ 2 kτ0k2L2(Ω)+ t Z t 0 || ∂tu(s) ||2_L2(Ω)ds ≤ 2 kτ0k2L2(Ω)+ T Z t 0 E(s)ds .

Furthermore since b 7→ ||| b ||| is a norm on E1that is equivalent to the norm || · ||_H1_(Ω) (cf [5]),

there exists a constant C such that:

∀b ∈ E1, || b ||2_H1_(Ω)≤ C||| b |||

2

. Then from (28) the following estimates holds:

|| u(t) ||2 H1(Ω) ≤ 3 2 kτ0k2L2_(Ω)+ T Z t 0 E(s)ds + C + C||| d(t) |||2 ≤ C(1 + Z t 0 E(s)ds + E(t)) (29)

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where C is a non negative constant depending on the data. From estimate (27) we then have:

E(t) ≤ a0+ 1 2ρS || f ||2 L2(0,T ;L2(Ω))+ Z t 0 E(s)ds +γ 2|| g(t) || 2 H−1/2_(∂Ω)+ 1 2γ|| u(t) || 2 H1(Ω) +C 2 || ∂tg(s) ||2_L2_{(0,T ;H}−1/2_(∂Ω))+ Z t 0 || u(s) ||2_H1_(Ω)ds ,

where γ is a postive constant. Choosing γ large enough and using (29) we obtain E(t) ≤ A1+ A2 Z t 0 E(s)ds + A3 Z t 0 Z s 0 E(r)dr ds,

where Ai are constants depending on the data (initial data, forces, Ω, T ). Remark that A2 > 0.

We then apply the Gronwall Lemma to the function t 7→ E(t) +A3 A2

Z t

0

E(s)ds, assuming that f ∈ L2(0, T ; L2(Ω)) and g ∈ H1(0, T ; H−1/2(∂Ω)).

This gives bounds on ∂tu in L∞(0, T ; L2(Ω)) and d in L∞(0, T ; H1(Ω)). From these bounds and

assuming that d ∈ C0([0, T ]; Y0) we obtain thanks to (8) and (9) applied to v = ∂tu that ˙τ ,

˙

θ, ∂td are bounded respectively in L∞(0, T ), L∞(0, T ) and L∞(0, T ; L2(Ω)). Consequently any

solution (τ , θ, d) of (20)-(23)-(25) is bounded, at least formally, in W1,∞(0, T ) × W1,∞(0, T ) × (W1,∞_{(0, T ; L}2_{(Ω)) ∩ L}∞

(0, T ; H1_(Ω))).

This is in those energy spaces we are going to work.

3 Main results

We now state existence and uniqueness results.

Theorem 1 Let τ0, τ1 be given in R2, θ0, θ1 be real numbers, d0 ∈ Y1, d1 ∈ L2(Ω), f ∈

L2loc(0, +∞; L2(Ω)) and g ∈ Hloc1 (0, +∞; H −1/2

(∂Ω)) ∩ L2loc(0, +∞; L1(∂Ω)). There exists T ∗

> 0 such that for all T < T∗there exists a unique triplet (τ , θ, d) satisfying

                                     τ ∈ H2(0, T ), θ ∈ H2(0, T ), d ∈ H2_{(0, T ; (H}1_(Ω))0 ) ∩ W1,∞_{(0, T ; L}2_{(Ω)) ∩ L}∞ (0, T ; H1_(Ω)), d ∈ C0([0, T ]; Y0) such that m¨τ = Z Ω f + Z ∂Ω g in L2(0, T ) d dt J ˙θ−→ez+ ρS Z Ω d ∧ ∂td = Z Ω Rθ( −→ Gξ + d) ∧ f + Z ∂Ω Rθ( −→ Gξ + d) ∧ g in D0(0, T ) ∀b ∈ E1, ρS Z Ω ∂ttd · b + 2 ˙θ Z Ω ∂td ∧ b + ¨θ Z Ω d ∧ b − ( ˙θ)2 Z Ω (−Gξ + d) · b→ + Z Ω σ(d) : ∇b = Z Ω f · Rθb + Z ∂Ω g · Rθb in D 0 (0, T ) (30) together with τ (0) = τ0, ˙τ (0) = τ1, θ(0) = θ0, ˙θ(0) = θ1, d(0) = d0 and ∂td(0) = d1.

Moreover the following alternative holds: — T∗= +∞ or — lim t−→T∗ Z Ω (−Gξ + d) ·→ −Gξ = 0.→

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This result is equivalent to:

Theorem 2 Let u0 ∈ H1(Ω) such that u0 +

−→

Oξ ∈ X1 and u1 ∈ L2(Ω). Assume that f ∈

L2

loc(0, +∞; L2(Ω)) and g ∈ Hloc1 (0, +∞; H −1/2

(∂Ω)) ∩ L2

loc(0, +∞; L1(∂Ω)). There exists T ∗

> 0 such that for all T < T∗there exists a unique

u ∈ H2(0, T ; (H1(Ω))0) ∩ W1,∞(0, T ; L2(Ω)) ∩ L∞(0, T ; H1(Ω)) and φ = u + Id_R2 ∈ C0([0, T ]; X0) solution of            ∀v ∈ H1(Ω), ρS d dt Z Ω ∂tu · v + Z Ω σ(d) : ∇b = Z Ω f · v + Z ∂Ω g · v in D0(0, T ) , with b associated to v though (9), d associated to u and defined by Proposition 2, u(·, t = 0) = u0, ∂tu(·, t = 0) = u1.

(31) Moreover the following alternative holds

— T∗= +∞ or — lim t−→T∗ Z Ω (−Gξ + d) ·→ −Gξ = 0.→

The proof of these theorems will be split into several steps. First we prove that there exists a local solution to (30), then that this solution is unique. The proof of the existence of at least one solution of (30) is based on a Galerkin method. We then extend this solution until Z

Ω

(−Gξ + d) ·→ −Gξ > 0 remains valid.→ To conclude we prove that Theorem 1 is equivalent to Theorem 2.

4 Proof of Theorem 1

4.1 Existence of a local solution

The first remark is that the existence and uniqueness of τ is straightforward. Thus we will look at the coupled system describing the dynamics of θ and d.

4.1.1 Modal approximation

In many applications, it is sufficient to consider that d belongs to a well chosen space of finite dimension, and thus to approximate our system by a discrete one. A natural basis is, in the case we deal with, the family of eigenfunctions associated with all non negative eigenvalues of the linearized elasticity operator Σ. This family of eigenfunctions is a basis of E1. Indeed let us

consider the operator of linearized elasticity Σ with homogeneous Neumann boundary conditions. This is a selfadjoint operator of L2(Ω) whose inverse is a bounded operator from E0onto E0. We

define the eigenfunctions ψi, i ∈ IN by:

−div(σ(ψi)) = λiψi in Ω ,

σ(ψi).n = 0 on ∂Ω .

(32) The family (ψi)_i∈_{IN is a basis of H}1(Ω) that we choose othonormal in L2(Ω). We denote by

ψ0,1=_|Ω|11/2 −→ ex ||−e→x||, ψ0,2= 1 |Ω|1/2 −→ ey ||−→ey||, and ψ0,3= − →_e z∧ −→

Gξ/||−Gξ ||→ _L2(Ω)the eigenvectors associated

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rotations. This family of eigenfunctions is a basis of Ker(Σ). Consequently E1 =< ψi >_i∈_IN∗,

and we approximate d by a linear combinaison of a finite number N ∈ IN∗of modal functions: dN=PN_j=1αj(t)ψj.

This will be used in what follows to prove the existence of a solution of our problem: the family of eigenfunctions (ψi)_i∈_IN∗ _{will be chosen as a Galerkin basis of E}₁_.

In this subsection we prove that the approximate problem is wellposed and that it converges towards the continuous one. We set dN(ξ, t) =

N

X

j=1

αj(t)ψj(ξ). The problem (23)-(25) is

approxi-amated by the following discrete problem: find (θN, dN) satisfying

             d dt JN(t) ˙θN + ρS Z Ω dN∧ ∂ttdN= Z Ω RθN( −→ Gξ + dN) ∧ f + Z ∂Ω RθN( −→ Gξ + dN) ∧ g 1 ≤ i ≤ N, ρS Z Ω ∂ttdN· ψi+ 2 ˙θN Z Ω ∂tdN∧ ψi+ ¨θN Z Ω dN∧ ψi− ( ˙θN)2 Z Ω (−Gξ + d→ N) · ψi + Z Ω σ(dN) : ∇ψi= Z Ω f · RθNψi+ Z ∂Ω g · RθNψi (33) together with initial conditions:

θN(0) = θ0, dN(·, t = 0) = dN 0,

˙

θN(0) = θ1, ∂tdN(·, t = 0) = dN 1,

where dN 0= Π0N(d0) (resp. dN 1= Π0N(d1)) with Π0N denoting the L2(Ω)–orthogonal projector

operator on EN =< ψi; 1 ≤ i ≤ N >. Since (ψi)_i∈_{IN is an hilbertian basis of (E}1, ((., .))) and of

(L2_(Ω),R

Ω. · .), we have as N goes to infinity:

dN 0−→ d0in H1(Ω) and dN 1−→ d1 in L2(Ω) . (34)

The discrete system can be written as

A(X) ¨X = h(t, X, ˙X) (35)

where XT = [θ, α1, . . . , αN] ∈ RN +1, and where, using the fact that (ψi)iis orthonormal in L2(Ω),

the mass matrix A(X) ∈ MN +1(R) is given by:

A(X) =      JN ρS Z Ω dN∧ ψi T 1≤i≤N ρS Z Ω dN∧ ψi 1≤i≤N (ρSδi,j)1≤i,j≤N      .

At this step of the proof the fact that the Galerkin basis is a familly of eigenvectors associated to the linear elasticity operator plays no role. It will in the next subsection. We prove now that the mass matrix is invertible ifR

Ω( −→ Gξ + dN) · −→ Gξ 6= 0. Indeed, for BT_{= (β, β} 1, . . . , βN) ∈ RN +1, and setting b =PN i=1βiψi, we have: BT· A(X)B = JNβ2+ 2ρSβ Z Ω dN∧ b + ρS Z Ω |b|2 = ρS Z Ω h β−→ez∧ RθN( −→ Gξ + dN) i2 + 2ρS Z Ω h β−→ez∧ RθN( −→ Gξ + dN) i · RθNb + ρS Z Ω |RθNb| 2 = ρS β− →_e z∧ RθN( −→ Gξ + dN) + RθNb 2 L2_(Ω)≥ 0. (36)

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Thus remembering that (12) holds if Z

Ω

(−Gξ + d→ N) ·

−→

Gξ 6= 0 we deduce that A(X) is a positive definite matrix.

We now apply the Carath´eodory Theorem (cf [3] page 43) in U × RN +1with:

U = ( (θ, α1, . . . , αN) ∈ R/2πZZ × RN; N X i=1 αiψi∈ Y0 ) .

Then the system of ODE (35) has a unique solution X = (θ, α1, . . . αN) ∈ H2(0, TN) N +1 , TN > 0. In particular, dN = N X i=1

αiψi∈ C0([0, TN]; Y0). Moreover the solution satisfies energy

estimates. Indeed multiplying (23) by θN then (25) by αifor 1 ≤ i ≤ N and summing up these

contributions, we obtain, as for the continuous problem, that vN = ˙θN−→ez ∧ RθN(

−→

Gξ + dN) +

RθN∂tdN is bounded in L

∞

(0, TN; L2(Ω)) independently of N and that:

|| dN||_L∞_(0,T

N;H1(Ω))≤ C (37)

with C depending only on the data and independent of N . Let δ be a real number such that R

Ω(

−→ Gξ + d0) ·

−→

Gξ > δ > 0. For N sufficiently large we haveR

Ω( −→ Gξ + dN0) · −→ Gξ > δ. If we set Yδ0= b ∈ L2(Ω), Z Ω (−Gξ + b) ·→ −Gξ ≥ δ→ ,

we deduce from the continuity in time of dN that there exists TNδ ≤ TN such that dN ∈

C0_{([0, T}δ

N]; Yδ0). Then expressing ˙θNand ∂tdN with respect to vN(see (8) and (9)) we have that

˙

θN and ∂tdNare bounded respectively in L∞(0, TNδ) and L ∞

(0, TNδ; L 2

(Ω)), and these bounds are independent of N but depend on δ. Consequently, there exists a time Tδ_{such that d}

Nis bounded

in C0([0, Tδ]; Y0δ) independently of N . Thus TNδ can be chosen independently of N .

Remark 4 Using (8) and (9), we obtain by (36) that for all t ∈ [0, Tδ]: BT· A(X)B = ρS β− → ez∧ RθN( −→ Gξ + dN) + RθNb 2 L2_(Ω)≥ α(δ)|B| 2 RN +1 (38)

where | · |_RN +1 is the euclidian norm in RN +1and where α(δ) is of order δ.

In all what follows we fix δ > 0 satisfyingR

Ω(

−→ Gξ + d0) ·

−→

Gξ > δ and we work on [0, Tδ] such that dN is bounded in C0([0, Tδ]; Yδ0) independently of N . For simplicity we set T = Tδ.

4.1.2 Additional estimate

The previous estimates enable us to pass to the limit in a weak sense in the discrete system. Nevertheless, we are not able to identify the limit function and nor prove that it corresponds to the solution of the continuous problem. For instance we know that ˙θN and ( ˙θN)2 converge in

L∞(0, T ) weak–star respectively to ˙θ and χ but we do not know whether χ = ( ˙θ)2. We are thus going to derive additional estimates on ¨θNand ∂ttdNthat will provide us compactness properties.

We prove that ¨θN is bounded in L∞(0, T ) and that ∂ttdNis bounded in L2(0, T ; (H1(Ω))0). The

analysis uses a duality argument and relies on the “special” Galerkin basis we have chosen ([10]). First we recall the properties of Π0

N the L2–projector on EN and of Π1N the projector on EN

with respect to the semi–norm ||| · |||. Thanks to the choice of the special basis, we have that ∀b ∈ H1

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Consequently, the restriction of Π0N to H 1

(Ω) is stable with respect to the semi–norm ||| · |||. Moreover since ||| · ||| is a norm on E1 equivalent to the H1–norm, there exists constants such

that:

∀b ∈ H1

(Ω), CkΠ0N(b)kH1_(Ω)≤ ||| Π0_N(b) ||| ≤ ||| b ||| ≤ C0kbk_H1_(Ω).

This implies that the restriction of Π0N to H 1

(Ω) is stable with respect to the H1–norm. At this stage an important remark is that if the estimates are carried out without paying attention, we obtain the desired result under a condition of smallness of the data (as annonced in [9]). We only explain quickly why (since the same kind of estimates will be detailed right after). Indeed for any b in H1(Ω) and with bN= Π0N(b) we have from the dynamic of dN (see (33)):

ρS Z Ω ∂ttdN· b = ρS Z Ω ∂ttdN· bN = − ρS 2 ˙θN Z Ω ∂tdN∧ bN+ ¨θN Z Ω dN∧ bN− ( ˙θN)2 Z Ω (−Gξ + d→ N) · bN − ((dN, bN)) + Z Ω f · RθNbN+ Z ∂Ω g · RθNbN. (39)

Using the energy estimates and the stability of the orthogonal projector Π0N in the L2–norm, as

well as in the H1–norm, we obtain by duality

|| ∂ttdN ||_L2(0,T ;(H1(Ω))0₎≤ C1+ C2|| ¨θN||_L2(0,T ), (40)

where Ci, i = 1, 2 depend on the data. The equation describing the dymanic of θN also couple

¨ θN and ∂ttdN: JN(t)¨θN = ρS Z Ω ∂ttdN∧ dN+ Z Ω RθN( −→ Gξ + dN) ∧ f + Z ∂Ω RθN( −→ Gξ + dN) ∧ g −2ρSθ˙N Z Ω (−Gξ + d→ N) · ∂tdN.

For any t ∈ [0, T ], dN(t) ∈ Yδ0, and thus:

0 <√ρSδ ≤ √ ρS Z Ω (−Gξ + d→ N) · −→ Gξ ≤√ρS|| −→ Gξ + dN ||_L2_(Ω)|| −→ Gξ ||_L2_(Ω)≤ p JN(t)|| −→ Gξ ||_L2_(Ω).

Using the energy estimates, this leads to

|| ¨θN ||L2(0,T )≤ C3+ C4|| ∂ttdN ||L2(0,T ;(H1(Ω))0₎ (41)

where Ci, i = 3, 4 depend on the data, on δ but are independent of N . The inequalities (40)

and (41), finally give

|| ∂ttdN||_L2_{(0,T ;(H}1_(Ω))0₎≤ C5+ C6|| ∂ttdN||_L2_{(0,T ;(H}1_(Ω))0₎.

with Ci, i = 5, 6 depending on the data, on δ but are independent of N . Thus if the data are

small enough, C6 can be chosen strictly less than 1 and we obtain the desired estimate. Since we

do not want these kinds of restrictions we are going to estimate separatly the high frequencies and the low frequencies of ∂ttdN. Let us consider N0∈ IN∗that will be chosen later on and such

that N ≥ N0. We separate dN into two parts:

dN= βN+ ωN where βN= N0−1 X i=1 αiψi and ωN= N X i=N0 αiψi. We set HfN0_{= hψ} i; i ∈ IN, i ≥ N0i.

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Estimate on the high frequencies

We introduce the orthogonal projection Π0,hf_N from L2_{(Ω) on E}N_{∩ Hf}N0_{, for which we have}

∀b ∈ L2(Ω), Z Ω ∂ttdN· Π0,hfN (b) = Z Ω ∂ttΠ0,hfN (dN) · b = Z Ω ∂ttωN· b . (42)

Taking bN = Π0,hf_N (b) as a test function in (39) with b ∈ H1(Ω) and using (42) the following

equality holds: ρS Z Ω ∂ttωN· b = −ρS ¨ θN Z Ω dN∧ bN+ 2 ˙θN Z Ω ∂tdN∧ bN− ( ˙θN)2 Z Ω (−Gξ + d→ N) · bN − ((dN, bN)) + Z Ω f · RθNbN+ Z ∂Ω g · RθNbN. Thus: Z Ω ∂ttωN· b ≤h|¨θN| || dN||_L2_(Ω)+ 2| ˙θN| || ∂tdN||_L2_(Ω)+ | ˙θN|2|| −→ Gξ + dN||_L2_(Ω) + 1 ρS || f ||_L2(Ω) || bN||L2(Ω)+ C ρS || g ||_H−1/2_(∂Ω)|| bN||H1(Ω)+ 1 ρS ||| dN||| ||| bN||| . (43)

Thanks to the choice of the special basis and to the equivalence of the norm ||| · ||| and || · ||_H1(Ω)

on E1 we have that the restriction of Π0,hfN to H 1

(Ω) is stable with respect to the H1–norm. Moreover, ∀b ∈ H1(Ω), kΠ0,hfN (b)kL2(Ω)≤ 1 pλN0 ||| Π0,hfN (b) ||| ≤ 1 pλN0 ||| b ||| ≤ C pλN0 kbkH1(Ω)

This leads to: Z Ω ∂ttωN· b ≤2| ˙θN| || ∂tdN||_L2_(Ω)+ |¨θN| || dN||_L2_(Ω)+ | ˙θN|2|| −→ Gξ + dN||_L2_(Ω) +1 ρS || f ||_L2_(Ω) C pλN0 || b ||_H1_(Ω)+ C(|| g ||_H−1/2_(∂Ω)+ || dN||_H1_(Ω))|| b ||_H1_(Ω).

Taking the L2–norm in time we get by duality: || ∂ttωN||L2(0,T ;(H1(Ω))0₎≤ C pλN0 2|| ˙θN ||L∞_{(0,T )}|| ∂tdN||L2(0,T ;L2(Ω)) + || ¨θN||L2(0,T )|| dN ||L∞_{(0,T ;L}2(Ω))+ || ˙θN|| 2 L∞_{(0,T )}|| −→ Gξ + dN ||L2(0,T ;L2(Ω)) + 1 ρS || f ||_L2(0,T ;L2(Ω)) + C|| g ||_L2(0,T ;H−1/2_(∂Ω))+ C|| dN ||L2(0,T ;H1(Ω)).

Now thanks to the energy estimates θNis bounded in W1,∞(0, T ), dNis bounded in L∞(0, T ; H1(Ω))∩

W1,∞(0, T ; L2(Ω)), those bounds beeing independent of N . That yield: || ∂ttωN ||_L2_{(0,T ;(H}1_(Ω))0₎≤ C +

C pλN0

|| ¨θN||_L2_{(0,T )} (44)

where C depends on the initial data, on the exterior forces and on δ. We remark that unlike inequality (40), in (44) we have a control on the constant in front of || ¨θN||L2(0,T )which can be

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Estimate on the low frequencies

We now look at the low frequency part βN of dN. We multiply the first equation of (33) by ¨θN,

and choose ∂ttβN as a test function in the weak formulation describing the dynamic of the elastic

displacement. Summing up those contributions it comes: JN(¨θN)2+ 2ρSθ¨N Z Ω dN∧ ∂ttβN+ ρS Z Ω |∂ttβN|2 = ¨θN Z Ω RθN( −→ Gξ + dN) ∧ f + Z ∂Ω RθN( −→ Gξ + dN) ∧ g − ρS Z Ω dN∧ ∂ttωN −2ρSθ˙N Z Ω (−Gξ + d→ N) · ∂tdN + Z Ω f · RθN∂ttβN+ Z ∂Ω g · RθN∂ttβN − Z Ω σ(dN) : ∇∂ttβN+ ρS( ˙θN)2 Z Ω (−Gξ + d→ N) · ∂ttβN− 2ρSθ˙N Z Ω ∂tdN∧ ∂ttβN (45) The left hand side of (45) can be written using (36):

JN(¨θN)2+2ρSθ¨N Z Ω dN∧∂ttβN+ρS Z Ω |∂ttβN|2= ρS ¨ θN−→ez∧ RθN( −→ Gξ + dN) + RθN∂ttβN 2 L2_(Ω). (46) Let us introduce v = θ−→ez∧RθN( −→ Gξ+dN)+RθNd ∈ L 2

(Ω), and remember that dN ∈ C0([0, T ]; Yδ0),

we have thanks to (8) and (9):

|θ| ≤1 δ|| −→ Gξ ||_L2_(Ω)|| v ||_L2_(Ω), || d ||L2(Ω)≤ || v − θ−→ez∧ RθN( −→ Gξ + dN) ||_L2_(Ω)≤ 1 +C δ || v ||L2(Ω),

because dN is bounded in L∞(0, T ; L2(Ω)). Thus:

|θ|2+ || d ||2_L2(Ω)≤ C|| v || 2 L2(Ω),

where C depends on the data and on δ−1.Thus for θ = ¨θN and d = ∂ttβN we have

|¨θN|2+ || ∂ttβN||2_L2_(Ω)≤ C ¨ θN−→ez∧ RθN( −→ Gξ + dN) + RθN∂ttβN 2 L2_(Ω). (47)

We now estimate the right hand side of (45) (taking into account (46) and (47) to obtain a lower bound for the left hand side). Consequently:

|¨θN|2+ || ∂ttβN||2_L2_(Ω)≤ C|¨θN| ρS|| dN||_H1_(Ω)|| ∂ttωN ||_(H1_(Ω))0 +||−Gξ + d→ N||H1(Ω) h || f ||_L2(Ω)+ C|| g ||H−1/2_(∂Ω)+ 2ρS| ˙θN||| ∂tdN||L2(Ω) i +h|| f ||_L2(Ω)+ ρS| ˙θN|2|| −→ Gξ + dN||L2(Ω)+ 2ρS| ˙θN||| ∂tdN ||L2(Ω) i || ∂ttβN||L2(Ω) +hC|| g ||_H−1/2_(∂Ω)+ C|| dN||_H1_(Ω) i || ∂ttβN||_H1_(Ω) , (48) where we used: Z Ω dN∧ ∂ttωN = Z Ω 0 −1 1 0 dN · ∂ttωN ≤ || dN||H1(Ω)|| ∂ttωN||(H1(Ω))0.

We next remark that the following inverse inequality holds || ∂ttβN ||H1(Ω)≤ C

p

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Thus by integrating (48) over (0, T ) and thanks to the H¨older inequality the following inequality holds: || ¨θN || 2 L2_{(0,T )}+ || ∂ttβN||2_L2_{(0,T ;L}2_(Ω))≤ C|| ¨θN||_L2_{(0,T )}|| dN||_L∞_{(0,T ;H}1_(Ω))|| ∂ttωN||_L2_{(0,T ;(H}1_(Ω))0₎ +C|| ¨θN||L2(0,T )|| −→ Gξ + dN||L∞_{(0,T ;H}1(Ω)) || f ||_L2(0,T ;L2(Ω))+ C|| g ||L2(0,T ;H−1/2_(∂Ω)) +2ρS|| ˙θN||_L∞_{(0,T )}|| ∂tdN||_L2_{(0,T ;L}2_(Ω)) + C|| f ||_L2_{(0,T ;L}2_(Ω)) +ρS|| ˙θN|| 2 L4_{(0,T )}|| −→ Gξ + dN||_L∞_{(0,T ;L}2_(Ω))+ 2ρS|| ∂tdN ||_L∞_{(0,T ;L}2_(Ω)) + h || g ||_L2_{(0,T ;H}−1/2_(∂Ω))+ || dN||_L2_{(0,T ;H}1_(Ω)) i CpλN0 || ∂ttβN ||_L2_{(0,T ;L}2_(Ω)),

Thanks to the energy estimates we deduce: || ¨θN|| 2 L2_{(0,T )}+ || ∂ttβN||2_L2_{(0,T ;L}2_(Ω))≤ CN0|| ∂ttβN||L2_{(0,T ;L}2_(Ω)) +|| ¨θN||_L2_{(0,T )} C|| ∂ttωN||_L2_{(0,T ;(H}1_(Ω))0₎+ C ,

where C depends on the data (initial data, exterior forces, δ) and CN0 depends on the data and

on λN0. Then using Young inequality:

|| ¨θN|| 2

L2_{(0,T )}+ || ∂ttβN||_L22_{(0,T ;L}2_(Ω))≤ CN0+ C|| ¨θN||L2_{(0,T )}|| ∂ttωN ||_L2_{(0,T ;(H}1_(Ω))0₎, (49)

where once again C depends on the data (initial data, exterior forces, δ) and CN0 depends on the

data and on λN0.

Final estimate

From (44) and (49) the following inequality holds: || ¨θN|| 2 L2(0,T )+ || ∂ttβN ||L22(0,T ;L2(Ω))≤ CN0+ C pλN0 || ¨θN|| 2 L2(0,T ),

Thus choosing N0sufficently large with respect to the data and δ we obtain bounds on || ¨θN ||_L2_{(0,T )}

and on || ∂ttβN||L2(0,T ;L2(Ω)). Back into (44) it shows that || ∂ttωN||L2(0,T ;(H1(Ω))0₎ is also

bounded. All those bounds are independent of N .

To summarize, we have estimated ¨θN in L2(0, T ) and ∂ttdN in L2(0, T ; (H1(Ω))0)

indepen-dently of N , for any T such that dN∈ C0([0, T ]; Y0δ).

4.1.3 Passage to the limit

In this subsection, we pass to the limit as N −→ +∞ in the discrete sytem (33). We have just proven that (θN)N and (dN)N are bounded independently of N respectively in H2(0, T ) and

H2_{(0, T ; (H}1_(Ω))0

) ∩ W1,∞_{(0, T ; L}2_{(Ω)) ∩ L}∞

(0, T ; H1_(Ω)).

• We rewrite the first equation of (33) as follows: d dt JNθ˙N+ ρS Z Ω dN∧ ∂tdN = Z Ω RθN( −→ Gξ + dN) ∧ f + Z ∂Ω RθN( −→ Gξ + dN) ∧ g. (50)

Remind now that dN is bounded in W1,∞(0, T ; L2(Ω)) ∩ L∞(0, T ; H1(Ω)), from the sequence of

imbeddings:

W1,∞(0, T ; L2(Ω)) ∩ L∞(0, T ; H1(Ω)) ⊂ H1((0, T ) × Ω),→Lc 2((0, T ) × Ω) , where the last injection is compact, we derive that there exists a subsequence1 _(d

η)η of (dN)N

that converges towards d strongly in L2(0, T ; L2(Ω)). Moreover ˙θN being bounded in H1(0, T ),

there exists a subsequence such that ˙θη−→ ˙θ in C0([0, T ]). Thus :

Jηθ˙η−→ J ˙θ in L2(0, T ).

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Moreover, ∂tdη* ∂td weakly in L2(0, T ; L2(Ω)), and dη−→ d strongly in L2(0, T ; L2(Ω)), thus: Z Ω dη∧ ∂tdη−→ Z Ω d ∧ ∂td weakly in L1(0, T ).

Next, θN being bounded in H1(0, T ) c

,→C0([0, T ]), we obtain that Rθη −→ Rθ uniformly on

[0, T ]. The following convergences: dη −→ d strongly in L2(0, T ; L2(Ω)), dη * d weakly in

L2_{(0, T ; H}1_{(Ω)) enable us to pass to the limit in the forcing terms:}

Z Ω Rθη( −→ Gξ + dη) ∧ f −→ Z Ω Rθ( −→ Gξ + d) ∧ f strongly in L2(0, T ) Z ∂Ω Rθη( −→ Gξ + dη) ∧ g * Z ∂Ω Rθ( −→ Gξ + d) ∧ g weakly in L2(0, T ) Next we pass to the limit in the equation (50) to obtain:

d dt J ˙θ + ρS Z Ω d ∧ ∂td = Z Ω Rθ( −→ Gξ + d) ∧ f + Z ∂Ω Rθ( −→ Gξ + d) ∧ g in D0(0, T ) (51) Due to the regularities of the solution the equality (51) takes place in L2_{(0, T ).}

• For 1 ≤ i ≤ N fixed, we pass to the limit in the remainder of the discrete system (33). Since dN is bounded in H2(0, T ; (H1(Ω))0) ∩ L∞(0, T ; H1(Ω)), the linear terms are easy to deal with:

Z Ω ∂ttdη· ψi * Z Ω ∂ttd · ψi Z Ω σ(dη) : ∇ψi * Z Ω σ(d) : ∇ψi weakly in L2(0, T ).

Next since ˙θη−→ ˙θ in C0([0, T ]) and ∂tdη* ∂td weakly in L2(0, T ; L2(Ω)) we have:

˙ θη Z Ω ∂tdη∧ ψi * θ˙ Z Ω ∂td ∧ ψi weakly in L2(0, T ).

Moreover ¨θN is bounded in L2(0, T ), then ¨θη * ¨θ weakly in L2(0, T ); then reminding that

dη−→ d strongly in L2(0, T ; L2(Ω)) (for instance) we obtain:

¨ θη Z Ω dη∧ ψi * θ¨ Z Ω d ∧ ψi weakly in L1(0, T ).

Similarly, since ˙θη−→ ˙θ in C0([0, T ]) and dη−→ d strongly in L2(0, T ; L2(Ω)) then:

( ˙θη)2 Z Ω (−Gξ + d→ η) · ψi −→ ( ˙θ)2 Z Ω (−Gξ + d) · ψ→ i strongly in L2(0, T ).

Finally as previously the forcing terms give: Z Ω Rθηψi· f −→ Z Ω Rθψi· f Z ∂Ω Rθηψi· g −→ Z ∂Ω Rθψi· g in C0([0, T ]).

Consequently for any given i, we have: ρS Z Ω ∂ttd · ψi+ 2 ˙θ Z Ω ∂td ∧ ψi+ ¨θ Z Ω d ∧ ψi− ( ˙θ)2 Z Ω (−Gξ + d) · ψ→ i + Z Ω σ(d) : ∇ψi= Z Ω f · Rθψi+ Z ∂Ω g · Rθψi in L1(0, T ). (52)

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Since E1 =< ψi >i≥1, the equation is satisfied for any function b ∈ E1. Moreover, due to the

regularities of the solution the equality (52) takes place actually in L2_{(0, T ). We next check that}

d ∈ C0_{([0, T ]; Y}δ

0). Indeed dN∈ C0([0, T ]; Y0δ), thus for all t ∈ [0, T ],

R ΩdN = 0, R ΩdN∧ −→ Gξ = 0 andR Ω( −→ Gξ + dN) · −→

Gξ ≥ δ > 0. Since dN−→ d strongly in C0([0, T ]; L2(Ω)) we have:

∀t ∈ [0, T ], Z Ω d = 0, Z Ω d ∧−Gξ = 0 and→ Z Ω (−Gξ + d) ·→ −Gξ ≥ δ > 0 ,→ which state that d ∈ C0([0, T ]; Y0δ).

We next verify that the initial conditions are satisfied. The angle θN is bounded in H2(0, T )

which is compactly embeded into C1([0, T ]), then θη(0) = θ0 −→ θ(0) and ˙θη(0) = ˙θ1 −→ ˙θ(0).

Similarly, the fact that dN is bounded in H2(0, T ; (H1(Ω))0) ∩ W1,∞(0, T ; L2(Ω)), implies that

(cf [11], theorem 3. 1 p. 23) dN ∈ C0([0, T ]; L2(Ω)) and ∂tdN ∈ C0([0, T ]; (H1(Ω))0), and that

dη(0) −→ d(0) in L2(Ω), ∂tdη(0) −→ ∂td(0) in (H1(Ω))0. But from (34), we have dη(0) =

dη0−→ d0 in H1(Ω) and ∂tdη(0) = dη1−→ d1 in L2(Ω). Consequently we obtain d(0) = d0 in

H1(Ω) and ∂td(0) = d1 in L2(Ω).

Finally since d ∈ H2_{(0, T ; (H}1_(Ω))0

) ∩ W1,∞_{(0, T ; L}2_{(Ω)), we deduce some additional}

regulari-ties. We have (see [11] lemme 8.2 p. 298) that d ∈ C0([0, T ]; H1(Ω)w) and ∂td ∈ C0([0, T ]; L2(Ω)w),

where H1_(Ω)

w (resp. L2(Ω)w.) denotes the space H1(Ω) (resp. L2(Ω)) endowed with its weak

topology.

Thus we have proven the existence of a local solution of (30) for any data. In the next section we prove that this solution is unique.

4.2 Uniqueness of the solution

We suppose that there exists two solutions (θi, di), i = 1, 2 of the system (30) (the uniqueness

of the translation is straightfoward). We denote by the subscript i the quantities associated with (θi, di), like the rotation Ri= Rθi. We recall the system satisfied by the solutions:

             d dt(Ji ˙ θi) + ρS Z Ω di∧ ∂ttdi= Z Ω Ri( −→ Gξ + di) ∧ f + Z ∂Ω Ri( −→ Gξ + di) ∧ g in L2(0, T ) ∀b ∈ E1, ρS Z Ω ∂ttdi· b + 2 ˙θi Z Ω ∂tdi∧ b + ¨θi Z Ω di∧ b − ( ˙θi)2 Z Ω (−Gξ + d→ i) · b + Z Ω σ(di) : ∇b = Z Ω f · Rib + Z ∂Ω g · Rib in L2(0, T ) where we denoted by Ji= ρSR_Ω( −→

Gξ + di)2 the inertia momentum associated with di. To prove

the uniqueness of the rotation angle and of the displacement d, we divide our proof in two steps. The first step is done in a formal way assuming that ∂td belongs to L∞(0, T ; E1) and consequently

is an admissible test function. This step consists in algebraic calculations so that we rewrite the system as an equality (see (58)) in terms of θ = θ2− θ1 and d = d2− d1. We then estimate the

terms on the right hand side of this equality. Finally we apply the Gronwall lemma to conclude (for the regularized case). The second step adressess the general case (where ∂td only belongs

to L∞(0, T ; L2(Ω))) thanks to a regularization argument.

Step 1.

On one hand, we subtract the equations concerning the angles θ2and θ1, and we multiply the

new equation by ˙θ to get: ˙ θd dt ˙θ2J2− ˙θ1J1 + ρSθ˙ Z Ω (d2∧ ∂ttd2− d1∧ ∂ttd1) = ˙θ Z Ω h R2( −→ Gξ + d2) − R1( −→ Gξ + d1) i ∧ f + ˙θ Z ∂Ω h R2( −→ Gξ + d2) − R1( −→ Gξ + d1) i ∧ g .

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Then, by making d and θ appear, we have: ˙ θd dt ˙θJ1+ ˙θ2(J2− J1) + ρSθ˙ Z Ω d1∧ ∂ttd + ρSθ˙ Z Ω d ∧ ∂ttd2= Z Ω h (R2− R1)( −→ Gξ + d2) + R1d i ∧ f + ˙θ Z ∂Ω h (R2− R1)( −→ Gξ + d2) + R1d i ∧ g . (53) On the other hand, we subtract the equations concerning the perturbations d2 and d1, and we

have for all b ∈ E1:

ρS Z Ω ∂ttd · b + 2 ˙θ Z Ω ∂td1∧ b + 2 ˙θ2 Z Ω ∂td ∧ b + ¨θ Z Ω d1∧ b + ¨θ2 Z Ω d ∧ b +T ( ˙θ1, d1, b) − T ( ˙θ2, d2, b) + Z Ω σ(d) : ∇b = Z Ω f · (R2− R1)b + Z ∂Ω g · (R2− R1)b , (54) where T is defined by T (θ, d, b) = (θ)2 Z Ω (−Gξ + d) · b.→

Let us assume in this step only that ∂td ∈ L∞(0, T ; E1). So, it is now licit to take b = ∂td in (54):

ρS Z Ω ∂ttd · ∂td + 2 ˙θ Z Ω ∂td1∧ ∂td + ¨θ Z Ω d1∧ ∂td + ¨θ2 Z Ω d ∧ ∂td +T ( ˙θ1, d1, ∂td) − T ( ˙θ2, d2, ∂td) + ((d, ∂td)) = Z Ω f · (R2− R1)∂td + Z ∂Ω g · (R2− R1)∂td . (55) We now set F = Ff + Fg, which contains the terms of the external forces (coming from (53) and (55)): Ff = ˙θ Z Ω h (R2− R1)( −→ Gξ + d2) + R1d i ∧ f + Z Ω f · (R2− R1)∂td Fg = ˙θ Z ∂Ω h (R2− R1)( −→ Gξ + d2) + R1d i ∧ g + Z ∂Ω g · (R2− R1)∂td .

Thus, by adding each side of (53) and (55), we obtain: ˙ θ d dt( ˙θJ1) + ρSθ˙ Z Ω d1∧ ∂ttd + ρS Z Ω ∂ttd · ∂td + ρSθ˙ Z Ω ∂td1∧ ∂td + ρSθ¨ Z Ω d1∧ ∂td + ˙θd dt ˙θ2(J2− J1) + ρSθ˙ Z Ω d ∧ ∂ttd2+ ρSθ˙ Z Ω ∂td1∧ ∂td + ρSθ¨2 Z Ω d ∧ ∂td + ρS(T ( ˙θ1, d1, ∂td) − T ( ˙θ2, d2, ∂td)) + 1 2 d dt((d, d)) = F . (56) Let us set y(t) = ρS ˙ θ−→ez∧ R1( −→ Gξ + d1) + R1∂td 2 L2(Ω)= | ˙θ| 2 J1+ 2ρSθ˙ Z Ω d1∧ ∂td + ρS Z Ω |∂td|2, (57)

and remark that: ˙ y(t) 2 = ¨θ ˙θJ1+ | ˙θ| 2J˙1 2 + ρSθ¨ Z Ω d1∧ ∂td + ρSθ˙ Z Ω ∂td1∧ ∂td + ρSθ˙ Z Ω d1∧ ∂ttd + ρS Z Ω ∂td · ∂ttd .

Consequently, the terms of the first line of (56) between the square brackets are equal to y(t)˙

2 +

( ˙θ)2J˙1

2. Thus we deduce that: ˙ y(t) 2 + 1 2 d dt((d, d)) = F − R , (58)

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where: R = ( ˙θ)2J˙1 2 + ˙θ d dt ˙θ2(J2− J1) + ρSθ˙ Z Ω d ∧ ∂ttd2 + ρSθ˙ Z Ω ∂td1∧ ∂td + ρSθ¨2 Z Ω d ∧ ∂td + ρS(T ( ˙θ1, d1, ∂td) − T ( ˙θ2, d2, ∂td))

Furthermore, we rewrite T ( ˙θ1, d1, ∂td) − T ( ˙θ2, d2, ∂td) using the identity:

a21b1− a22b2 = a1b1(a1− a2) + a1a2b1− a2b2(a2− a1) − a2b2a1 = −(a2− a1)(a1b1+ a2b2) − a1a2(b2− b1) . It gives T ( ˙θ1, d1, ∂td) − T ( ˙θ2, d2, ∂td) = ( ˙θ1)2 Z Ω (−Gξ + d→ 1) · ∂td − ( ˙θ2)2 Z Ω (−Gξ + d→ 2) · ∂td = − ˙θ Z Ω h ˙θ1( −→ Gξ + d1) + ˙θ2( −→ Gξ + d2) i · ∂td − ˙θ1θ˙2 Z Ω d · ∂td.

So R can be rewritten as follows: R = ( ˙θ)2J˙1 2 + ˙θ ¨θ2(J2− J1) + ˙θ ˙θ2( ˙J2− ˙J1) + ρS ˙ θ Z Ω d ∧ ∂ttd2+ ρSθ˙ Z Ω ∂td1∧ ∂td + ρSθ¨2 Z Ω d ∧ ∂td − ρSθ˙ Z Ω h ˙θ1( −→ Gξ + d1) + ˙θ2( −→ Gξ + d2) i · ∂td − ρSθ˙1θ˙2 Z Ω d · ∂td . (59)

Now that we have rewritten the energy equality satisfied by d and θ we estimate F and R. • Estimate of R.

We first begin with the terms that do not involve ¨θ2 nor ∂ttd2. We have:

( ˙θ)2_J_˙ 1 2 = ρS| ˙θ|2 Z Ω (−Gξ + d→ 1) · ∂td1 ≤ ρS|| −→ Gξ + d1||L∞_{(0,T ;L}2(Ω))|| ∂td1||L∞_{(0,T ;L}2(Ω))| ˙θ| 2_{≤ C| ˙} θ|2,

with a constant C independent of time because d1 ∈ W1,∞(0, T ; L2(Ω)). For the same reason, we

get: ˙ θ Z Ω ∂td1∧ ∂td ≤ | ˙θ||| ∂td1||_L2_(Ω)|| ∂td ||_L2_(Ω)≤ C| ˙θ||| ∂td ||_L2_(Ω).

Reminding that ˙θ2∈ L∞(0, T ) and ∂td2∈ L∞(0, T ; L2(Ω)), we get:

˙ θ ˙θ2( ˙J2− ˙J1) = 2ρS| ˙θ|| ˙θ2| Z Ω d · ∂td2+ Z Ω (−Gξ + d→ 1) · ∂td ≤ C| ˙θ||| d ||_L2_(Ω)+ || ∂td ||_L2_(Ω) . Furthermore, since d1, d2∈ L∞(0, T ; L2(Ω)) and ˙θ1, ˙θ2∈ L∞(0, T ), we have as well:

˙ θ Z Ω h ˙θ1( −→ Gξ + d1) + ˙θ2( −→ Gξ + d2) i · ∂td ≤ C| ˙θ||| ∂td ||_L2_(Ω), and: ˙ θ1θ˙2 Z Ω d · ∂td ≤ C|| d ||_L2_(Ω)|| ∂td ||_L2_(Ω).

We now treat the terms that require regularity on the second order derivatives with respect to time: ˙ θ ¨θ2(J2− J1) = ρS| ˙θ||¨θ2| Z Ω (2−Gξ + d→ 2+ d1) · d ≤ C1(t)| ˙θ||| d ||L2(Ω)

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with C1(·) ∈ L2(0, T ) because ¨θ2∈ L2(0, T ). For the same reason, it comes: ¨ θ2 Z Ω d ∧ ∂td ≤ |¨θ2||| d ||_L2_(Ω)|| ∂td ||_L2_(Ω)≤ C1(t)|| d ||_L2_(Ω)|| ∂td ||_L2_(Ω).

Next the last term of R gives: ˙ θ Z Ω d ∧ ∂ttd2 ≤ | ˙θ||| ∂ttd2||(E1)0|| d ||H1(Ω)≤ C2(t)| ˙θ||| d ||H1(Ω).

with C2(·) ∈ L2(0, T ) because ∂ttd2 ∈ L2(0, T ; (E1)0). For the term R, we finally find the following

estimate: R ≤ Ch| ˙θ|2+ | ˙θ||| ∂td ||_L2(Ω)+ || d ||L2(Ω) + || d ||_L2(Ω)|| ∂td ||L2(Ω) i +C1(t)|| d ||L2(Ω)(| ˙θ| + || ∂td ||L2(Ω)) + C2(t)| ˙θ||| d ||H1(Ω) ≤ C(t)h| ˙θ|2+ || ∂td ||2L2(Ω)+ || d || 2 H1(Ω) i

thanks to Young’s inequality ≤ C(t)h| ˙θ|2+ || ∂td ||2L2(Ω)+ C||| d ||| 2i , (60) with still C(·) ∈ L2_{(0, T ).} • Estimates of F .

The terms where the bulk forces appears can be estimated by: |Ff| ≤ ˙ θ Z Ω h (R2− R1)( −→ Gξ + d2) + R1d i ∧ f + Z Ω f · (R2− R1)∂td ≤ || f ||_L2(Ω) h C| ˙θ|(|R2− R1| + || d ||L2(Ω)) + |R2− R1||| ∂td ||L2(Ω)) i

because d2∈ L∞(0, T ; L2(Ω)). Taking now into account the fact that sines and cosines are both

lipschitz continuous, we have |R2− R1| ≤ C|θ2− θ1| and thus:

|Ff| ≤ C(t)| ˙θ||θ| + | ˙θ||| d ||_L2(Ω)+ |θ||| ∂td ||L2(Ω)

≤ C(t)|θ|2+ | ˙θ|2+ || d ||_L22_(Ω)+ || ∂td ||2_L2_(Ω)

with C(·) ∈ L2_{(0, T ) because f ∈ L}2_{(0, T ; L}2_(Ω)).

The first term in Fg can be easily upperbounded as previously by: ˙ θ Z ∂Ω h (R2− R1)( −→ Gξ + d2) + R1d i ∧ g ≤ | ˙θ||| g ||_H−1/2_(∂Ω) |R2− R1||| −→ Gξ + d2||_H1/2_(∂Ω)+ C|| d ||_H1/2_(∂Ω) ≤ C| ˙θ||θ2− θ1| + || d ||_H1_(Ω) ≤ C|θ|2 + | ˙θ|2+ ||| d |||2

because g ∈ L∞(0, T ; H−1/2(Ω)), d2 ∈ L∞(0, T ; L2(Ω)). We treat the second term in Fg

sepa-rately2 _by: Z ∂Ω g · (R2− R1)∂td = d dt Z ∂Ω g · (R2− R1)d − Z ∂Ω ∂tg · (R2− R1)d − Z ∂Ω g · ˙θ2−→ez∧ R2− ˙θ1−→ez∧ R1 d = d dt Z ∂Ω g · (R2− R1)d − Z ∂Ω ∂tg · (R2− R1)d − Z ∂Ω g · ˙θ−→ez∧ R2+ ˙θ1−→ez∧ (R2− R1) d . 2_Considering Z ∂Ω g · (R2− R1)∂td

≤ C|θ| || ∂td ||H1(Ω)(because ∂td ∈ E1) would bring a term of the form ||| ∂td |||

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On one hand we have: Z ∂Ω ∂tg · (R2− R1)d ≤ C|| ∂tg ||_H−1/2_(∂Ω)|θ||| d ||_H1_(Ω)≤ C(t)|θ|||| d ||| with C(·) ∈ L2_{(0, T ) because ∂}

tg ∈ L2(0, T ; H−1/2(∂Ω)). And on the other hand, we have:

Z ∂Ω g · ˙θ−→ez∧ R2+ ˙θ1−→ez∧ (R2− R1) d ≤ C|| g ||_H−1/2_(∂Ω) | ˙θ| + |θ| || d ||_H1_(Ω) ≤ C|θ|2+ | ˙θ|2+ ||| d |||2 because ˙θ1 ∈ L∞(0, T ).

Thus, for the external forces terms, we finally get: F ≤ C(t)|θ|2+ | ˙θ|2+ || ∂td ||2_L2_(Ω)+ ||| d ||| 2 + d dt Z ∂Ω g · (R2− R1)d (61) with C(·) ∈ L2(0, T ).

• Uniqueness of a regular solution.

Thanks to (58), (60) and (61) the following estimate holds: 1 2 d dt y(t) + ||| d ||| 2_{≤ C(t)}_|θ|2 + | ˙θ|2+ || ∂td ||2L2(Ω)+ ||| d ||| 2 + d dt Z ∂Ω g · (R2− R1)d . We now integrate the previous inequality in time on (0, t) ⊂ (0, T ). Using the fact that ˙θ(0) = 0, d(0) = ∂td(0) ≡ 0 and y(0) = 0, we get:

y(t) + ||| d |||2≤ Zt 0 C(s)|θ(s)|2 + | ˙θ(s)|2+ || ∂td(s) ||2L2(Ω)+ ||| d(s) ||| 2 ds + h(t) (62) with h(t) =R ∂Ωg(t) · (R2(t) − R1(t))d(t). Furthermore, θ(t) = Rt 0θ(s)ds. Thus (62) becomes:˙ y(t) + ||| d |||2≤ Z t 0 C(s) Z s 0 ˙ θ(u)du 2 + | ˙θ(s)|2+ || ∂td(s) ||2_L2_(Ω)+ ||| d(s) ||| 2 ! ds + h(t) ≤ C Z t 0 ( ˙θ(u))2du + Z t 0 C(s)| ˙θ(s)|2+ || ∂td(s) ||2_L2_(Ω)+ || d(s) || 2 E1 ds + h(t) . (63) For the function h, we have:

|h(t)| ≤ C|| g ||_H−1/2_(∂Ω)|θ|||| d ||| ≤ C Z t 0 ( ˙θ(u))2du 1₂ ||| d(t) |||, (64) with C depending on the data and more particularly on || g ||_L∞_{(0,T ;H}−1/2_(∂Ω)).

Since d1∈ C0([0, T ]; Y0δ), we have by (8), and (9) with d replaced by d1 that there exists C > 0

such that: ∀β ∈ R, ∀b ∈ E0, |β|2+ || b ||2_L2(Ω)≤ ρSC β− → ez∧ R1( −→ Gξ + d1) + R1b 2 L2(Ω) . Thus | ˙θ(t)|2+ || ∂td(t) ||2_L2_(Ω)≤ Cy(t).

The inequality (63) then becomes (taking into account (64) and using Young inequality): | ˙θ(t)|2+ || ∂td(t) ||2_L2_(Ω)+ || d(t) || 2 E₁ ≤ C Z t 0 ( ˙θ(u))2du + C Z t 0 C(s)| ˙θ(s)|2+ || ∂td(s) ||2L2(Ω)+ || d(s) || 2 E1 ds +1 2|| d(t) || 2 E1. (65)

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By setting: z(t) = | ˙θ(t)|2+ || ∂td(t) ||2_L2_(Ω)+ 1 2|| d(t) || 2 E1, we have: z(t) ≤ Z t 0 C(s)z(s)ds .

with C(·) ∈ L2(0, T ). By the Gronwall lemma, we obtain z(t) ≡ 0 on (0, T ) (because θ(0) = 0, ˙

θ(0) = 0, d(·, 0) ≡ 0 and ∂td(·, 0) ≡ 0), so that, in the case where ∂td ∈ L∞(0, T ; E1), we have

θ1≡ θ2 and d1≡ d2.

Step 2. Uniqueness of the solution.

We now adapt the previous proof to the case where ∂td is no more in L∞(0, T ; E1), but only

in L∞(0, T ; L2(Ω)), and we give the main differences compared to the regular case. To do so, we keep the previous notations and we set dN = Π0N(d) the orthogonal projection

3

from L2(Ω) onto EN _{of the difference d = d}

2− d1. Thus dN and especially ∂tdN are in L∞(0, T ; E1), and it

is licit to take b = ∂tdN as a test-function in (54). Doing so we obtain:

ρS Z Ω ∂ttd · ∂tdN+ 2 ˙θ Z Ω ∂td1∧ ∂tdN+ 2 ˙θ2 Z Ω ∂td ∧ ∂tdN+ ¨θ Z Ω d1∧ ∂tdN +¨θ2 Z Ω d ∧ ∂tdN+ T ( ˙θ1, d1, ∂tdN) − T ( ˙θ2, d2, ∂tdN) + ((d, ∂tdN)) = Z Ω f · (R2− R1)∂tdN+ Z ∂Ω g · (R2− R1)∂tdN. (66) As previously we set yN(t) = ρS ˙ θ−→ez∧ R1( −→ Gξ + d1) + R1∂tdN 2

L2_(Ω) for almost every t ∈ (0, T ).

We notice thatR

Ω∂ttd · ∂tdN =

R

Ω∂ttdN· ∂tdN and ((d, ∂tdN)) = ((dN, ∂tdN)) because of the

definitions of dN and EN. Then adding (53) and (66), we find, as in (58):

1 2 d dt yN+ ||| dN||| 2_{= F}N − RN, (67) with FN= F_{f + F}N _{g :}N FN f = ˙θ Z Ω h (R2− R1)( −→ Gξ + d2) + R1d i ∧ f + Z Ω f · (R2− R1)∂tdN FN g = ˙θ Z ∂Ω h (R2− R1)( −→ Gξ + d2) + R1d i ∧ g + Z ∂Ω g · (R2− R1)∂tdN,

and where RN is given by:

RN= ρSθ˙ Z Ω d1∧ (∂ttd − ∂ttdN) + ( ˙θ)2 ˙ J1 2 + ˙θ ¨θ2(J2− J1) + ˙θ ˙θ2( ˙J2− ˙J1) + ρS ˙ θ Z Ω d ∧ ∂ttd2 + ρSθ˙ Z Ω ∂td1∧ ∂tdN+ ρSθ¨2 Z Ω d ∧ ∂tdN+ 2ρSθ˙2 Z Ω ∂td ∧ ∂tdN − ρSθ˙ Z Ω h ˙θ1( −→ Gξ + d1) + ˙θ2( −→ Gξ + d2) i · ∂tdN− ρSθ˙1θ˙2 Z Ω d · ∂tdN.

Most of the terms in RN are the ones appearing in R with, in some cases, d replaced by dN.

There are two main differences beetween RN and R. The first additional term ρSθ˙

R

Ωd1∧ (∂ttd − 3_{There is no confusion possible between d}

1, d2and dNif we assume N ≥ 3 (which is reasonable since we want to

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∂ttdN) comes from (53) and from ˙yN/2 due to the definition of yN. The second extra term is

2ρSθ˙2

R

Ω∂td ∧ ∂tdN. We shall prove that these terms go to zero as N goes to infinity.

• Estimate of RN

We now recall the estimates of the different terms of RN_{. These upper bounds are obtained just}

as in the regular case : ( ˙θ)2J˙1 2 ≤ C| ˙θ|2, | ˙θ ¨θ2(J2− J1)| ≤ C(t)| ˙θ||| d ||L2(Ω), ˙ θ Z Ω ∂td1∧ ∂tdN ≤ C| ˙θ||| ∂tdN||_L2_(Ω), ¨ θ2 Z Ω d ∧ ∂tdN ≤ C(t)|| d ||_L2_(Ω)|| ∂tdN||_L2_(Ω), ˙ θ Z Ω h ˙θ1( −→ Gξ + d1) + ˙θ2( −→ Gξ + d2) i · ∂tdN ≤ C| ˙θ||| ∂tdN||_L2_(Ω), ˙ θ1θ˙2 Z Ω d · ∂tdN ≤ C|| d ||_L2(Ω)|| ∂tdN||L2(Ω).

We now treat the terms of RN _{that need sharper estimates:}

˙ θ Z Ω d1∧ (∂ttd − ∂ttdN) ≤ C| ˙θ||| ∂ttd − ∂ttdN||(H1(Ω))0, ˙ θ ˙θ2( ˙J2− ˙J1) = 2ρS| ˙θ|| ˙θ2| Z Ω d · ∂td2+ Z Ω (−Gξ + d→ 1) · (∂tdN− ∂tdN+ ∂td) ≤ C| ˙θ|(|| d ||_L2_(Ω)+ || ∂tdN||_L2_(Ω)+ || ∂td − ∂tdN||_L2_(Ω)) , ˙ θ Z Ω d ∧ ∂ttd2 ≤ C(t)| ˙θ|(||| d − dN||| + ||| dN |||) , ˙ θ2 Z Ω ∂td ∧ ∂tdN = ˙ θ2 Z Ω (∂td − ∂tdN) ∧ ∂tdN ≤ C|| ∂td − ∂tdN||L2(Ω)|| ∂tdN||L2(Ω), with C(·) ∈ L2(0, T ). • Estimates of FN_.

For the term F_{f we have directly:}N |FN

f | ≤ C(t)

| ˙θ||θ| + | ˙θ||| d ||_L2(Ω)+ |θ||| ∂tdN||L2(Ω)

with C(·) ∈ L2_{(0, T ). We next have to treat F}N

g . First we have ˙ θ Z ∂Ω [(R2− R1)( −→ Gξ + d2)] ∧ g ≤ C|θ|| ˙θ| , Z ∂Ω g · (R2− R1)∂tdN≤ d dt Z ∂Ω g · (R2− R1)dN + C(t)(| ˙θ| + |θ|)||| dN||| , and ˙ θ Z ∂Ω R1d ∧ g ≤ | ˙θ| Z ∂Ω R1(d − dN) ∧ g + Z ∂Ω R1dN∧ g ≤ C| ˙θ|(||| d − dN||| + ||| dN|||) .

Summing up all these estimates, we arrive at: 1 2 d dt yN+ ||| dN||| 2 ≤ C(t)|θ|2+ | ˙θ|2+ ||| dN|||2+ || ∂tdN ||2_L2(Ω) +|| d ||2_L2(Ω)+ || ∂td − ∂tdN||2L2(Ω)+ || ∂ttd − ∂ttdN||2(H1(Ω))0 +||| d − dN|||2 + d dt Z ∂Ω g · (R2− R1)dN .

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From the estimates of the terms of RN and in the terms of FN we obtain easily an estimate analogous to (65): | ˙θ(t)|2+ || ∂tdN(t) ||_L22_(Ω)+ || dN(t) ||2E₁ ≤ Zt 0 C(s)| ˙θ(s)|2+ || ∂tdN(s) ||2L2(Ω)+ || dN(s) ||2L∞_(E 1) ds + wN(t) where: wN(t) = C0 Zt 0 C(s)||| (d − dN)(s) |||2+ || (∂td − ∂tdN)(s) ||2_L2_(Ω) +|| (∂ttd − ∂ttdN)(s) ||2(H1(Ω))0 ds . We set: zN(t) = | ˙θ(t)|2+ || ∂tdN(t) ||2_L2_(Ω)+ || dN(t) ||2E₁, thus zN(t) ≤ Z t 0 C(s)z(s)ds + wN(t) .

Then, by the Gronwall lemma, we obtain: zN(t) ≤ wN(t) exp Z t 0 C(s)ds .

But, as d − dN goes to zero strongly in H2(0, T ; (H1(Ω))0) ∩ H1(0, T ; L2(Ω)) ∩ L2(0, T ; E1), we

have lim

N −→∞wN(t) = 0 for almost every t ∈ (0, T ). ThenN −→∞lim zN(t) = 0 implies θ ≡ 0 (because

θ(0) = 0) and: lim N −→∞|| dN|| 2 L∞_(0,t;E₁₎= lim N −→∞|| ∂tdN|| 2 L∞_(0,t;L2_(Ω))= 0 ,

which implies d ≡ 0. This ends the proof of the uniqueness of the rotation angle and of the elastic displacement.

4.2.1 Existence and uniqueness of a maximal solution

We have thus proven that there exists a time T > 0 such that on the interval (0, T ) the solution of (30) exists and is unique, under the condition d ∈ C0_{([0, T ]; Y}δ

0) where δ is given such that

0 < δ <R

Ω(

−→ Gξ + d0) ·

−→

Gξ. By a standard argument there exists a unique maximal solution on (0, T∗). Next we verify that

— T∗= +∞ or — lim t−→T∗ Z Ω (−Gξ + d) ·→ −Gξ = 0.→

Indeed thanks to the regularities of the solution and to the energy estimates satisfied by the solution, we can extend it untillR

Ω(

−→

Gξ + d) ·−Gξ > 0. That concludes the proof of Theorem 1.→

4.3 Proof of Theoreme 2

In this subsection we prove that Theorem 1 is equivalent to Theorem 2. Starting from the system (30) satisfied by the translation the rotation and the elastic displacement we build a global displacement and the system satisfied by it. We consider: