Debayan Maity
email: debayan.maity@u-bordeaux.fr

Jorge San

Takéo Takahashi
email: takeo.takahashi@inria.fr

Jorge San Martín

Marius Tucsnak
email: marius.tucsnak@u-bordeaux.fr

Analysis of a simplified model of rigid structure floating in a viscous fluid

Keywords: Viscous shallow water equations, floating structure, fluid-structure interaction, strong solutions, return to equilibrium. AMS subject classifications. 35Q35, 74F10. Contents

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1 Introduction

The motion of rigid bodies in a fluid is a widely studied subject in both engineering and mathematical literature. This is due to important applications (naval and aerospace engineering, biology, medicine, . . . ) and to the mathematical challenges, namely the existence of free boundaries, raised by the corresponding mathematical models. The study of the case in which the rigid bodies are completely immersed in the fluid goes back to Euler, Kelvin and Kirchhoff. Due to the important number of works (namely in the last two decades) devoted to this subject, the corresponding mathematical questions may be considered by now well understood, for various types of fluids (ideal, viscous incompressible, compressible, . . . ). We refer, for instance, to [START_REF]Particles in flows[END_REF] and references therein for a concise description of recent progress in this field. The case when the solids are floating, thus only partially immersed and interacting with the water waves, has been much less studied in the literature, essentially in the case of an ideal fluid. We refer to the classical work of John [START_REF] John | On the motion of floating bodies. I[END_REF][START_REF]On the motion of floating bodies. II. Simple harmonic motions[END_REF] or to the monograph Falnes [START_REF] Falnes | Ocean waves and oscillating systems: linear interactions including wave-energy extraction[END_REF] for the linearized theory and to the review paper of Lannes [START_REF] Lannes | On the dynamics of floating structures[END_REF] for the description of recent progress in the field. As far as we know, the case of a viscous fluid has not been tackled, at least from a mathematical view point. The aim of this work is to partially fill this gap by considering a PDE system modeling the coupled motion of a free surface viscous fluid and of a solid, constrained to move in the vertical direction only, which is floating on this free surface. The main contributions of this work are:

• Proposing, via a Hamiltonian formalism, a new mathematical model which takes into account the viscosity of the fluid, for the coupled of a free boundary shallow fluid and of a rigid body floating on the surface.

• Proving the existence and uniqueness of strong solutions (locally in time or globally in time for small data).

• Proposing a reduced model and performing numerical simulations in the linear case.

The content of the subsequent sections is outlined below. In Section 2 we derive the governing equations using a Hamiltonian formalism. In Section 3 we show that our model is energetically consistent and we state our main existence and uniqueness results. In Section 4 we write the governing equation in an equivalent form obtained by "eliminating" the unknown functions involving the floating solid. Section 5 is devoted to the proof of the local in time existence and uniqueness result, whereas in Section 6 we prove global in time existence and uniqueness of solutions for initial data which are close to an equilibrium state. Finally, in Section 7 we analyze the return to equilibrium problem and in Section 8, we present some numerical simulations for this problem.

Modelling

In this section, we derive, using a Hamiltonian formalism, the simplified model which is analyzed in the remaining part of this work. The fluid is described using the viscous Saint Venant equations, whereas the solid obeys Newton's second law. To couple the two models we use a Hamiltonian formalism, passing by the following steps:

• Introduce the total energy of the fluid-floating object system within the Saint-Venant approximation and define conservation of mass as a constraint, following Petit and Rouchon [START_REF] Petit | Dynamics and solutions to some control problems for water-tank systems[END_REF].

• Use of a Hamiltonian formalism to derive the governing equations, by combining the approach introduced in [START_REF] Petit | Dynamics and solutions to some control problems for water-tank systems[END_REF] for the inviscid case with the methodology used in Gay-Balmaz and Yoshimura [START_REF] Gay-Balmaz | A free energy lagrangian variational formulation of the Navier-Stokes-Fourier system[END_REF] in the case the Navier-Stokes-Fourier system, which allows us to include the viscous effects in the system.

To describe our model we need some notation, which is introduced below. Denote respectively by ρ(t, x), v(t, x) and h(t, x) the density, the velocity, and the height of the free surface of the fluid (see Figure 1). These quantities depend on the time t 0 and on the position x ∈ [0, ]. We also denote by H(t) the height of the cylinder. The mass M of the rigid body is a positive constant. We also assume that the cylinder moves only vertically and we set I := [a, b] := the projection of the cylinder on the flat bottom, E := (0, ) \ I,

with I ⊂ (0, ).

The function H, which depends only on time is supposed to satisfy 0 < H(t) < min(h(t, a -), h(t, b + )) (t 0), so that the cylinder is immersed into the fluid and does not touch the bottom.

We are now on a position to derive the governing equations. The first one is the equation of mass conservation:

∂h ∂t + ∂(hv) ∂x = 0 (t > 0, x ∈ [0, ]). (2.2)
For x ∈ I the height of fluid is given by the position of the cylinder, i.e.: h(t, x) = H(t) (t > 0, x ∈ I).

(2.3) Equations (2.2) and (2.3) appear below as constraints in our Hamiltonian formalism.

In order to derive the other equations, we introduce the kinetic energy K f and the potential energy U f of the fluid:

K f = 1 2 (0, ) ρhv 2 dx, U f = (0, ) 1 2 ρgh 2 + e(s(t, x)) dx,
where g is the gravity acceleration, and e(s) represents the internal energy density which is a function of the density of entropy s. We assume that the fluid is homogeneous and incompressible, so that that its density ρ is a constant.

We also introduce the kinetic energy K f and the potential energy U f of the solid:

K s = 1 2 M Ḣ2 U s = M gH.
We obtain below the governing equations by writing the stationarity for the total action of the system under the constraints (2.2) and (2.3). To this aim, it seems more convenient, in particular to treat the continuity restriction (2.2), to introduce the variable q defined by

q(t, x) = h(t, x)v(t, x) (t > 0, x ∈ [0, ]). (2.4)
The total action of the system is given by

A(h, v, H) = (K f + K s ) -(U f + U s ) = T 0 0 1 2 ρhv 2 - 1 2 ρgh 2 -e(s(t, x)) dx + 1 2 M Ḣ2 -M gH dt.
In order to include the restrictions (2.2)-(2.3), we introduce two Lagrange multipliers:

λ 1 (t, x) (t 0, x ∈ [0, ]), λ 2 (t, x) (t 0, x ∈ I).
With the above notation, the governing equations are obtained as stationarity conditions for the Lagrangian defined by

L(h, q, H, λ 1 , λ 2 ) = T 0 0 1 2 ρ q 2 h - 1 2 ρgh 2 + ρλ 1 ∂h ∂t + ∂q ∂x -e(s(t, x)) dx + I λ 2 H -h dx + 1 2 M Ḣ2 -M gH dt. (2.5)
To achieve this aim, we consider virtual displacements of the trajectory, given by the independent variations δh(x, t), δH(t) and δq. Since q is a velocity type quantity, whereas h and H are displacements, we introduce the function ϕ and its variation δϕ defined by

∂ϕ ∂t = q and ∂(δϕ) ∂t = δq. (2.6)
The governing equations are obtained by imposing that δL = 0 for any virtual displacement δh(t, x), δH(t) and δϕ(t, x) such that

δh(0, x) = δh(T, x) = δH(0) = δH(T ) = δϕ(0, x) = δϕ(T, x) = 0, (2.7) δϕ(t, 0) = δϕ(t, ) = 0. (2.8)
Moreover, to take in consideration friction forces inside the fluid, we follow the approach in [START_REF] Gay-Balmaz | A free energy lagrangian variational formulation of the Navier-Stokes-Fourier system[END_REF] by assuming that the entropy trajectory s and its variation δs satisfy the variational constraint

∂e ∂s δs = µ h ∂q ∂x ∂(δϕ) ∂x (2.9)
where the constant µ > 0 is the viscosity of the fluid, together with the phenomenological constraint

∂e ∂s ṡ = µ h ∂q ∂x 2 .
Note that the above expression of this friction term leads to an nonstandard form of the viscous Saint-Venant equations, see Remark 2.1 below.

Using (2.5) we get

δL = T 0 0 ρ q h (δq) - 1 2 ρ q 2 h 2 (δh) -ρgh(δh) + ρλ 1 ∂(δh) ∂t + ∂(δq) ∂x - ∂e ∂s (δs) dx + I λ 2 δH -δh dx + M Ḣ(δ Ḣ) -M g(δH) dt.
The above formula and (2.6) yield:

δL = T 0 0 ρ q h ∂(δϕ) ∂t - 1 2 ρ q 2 h 2 (δh) -ρgh(δh) + ρλ 1 ∂(δh) ∂t + ∂ ∂x ∂(δϕ) ∂t - ∂e ∂s (δs) dx + I λ 2 (δH ) -(δh) dx + M Ḣ(δ Ḣ) -M g(δH) dt. (2.10)
Considering (2.7) and integrating by parts in time the formula (2.10), we find

δA = T 0 0 -ρ ∂ ∂t q h δϕ -ρ ∂λ 1 ∂t ∂ ∂x δϕ - ∂e ∂s δs dx + 0 - 1 2 ρ q 2 h 2 -ρgh -ρ ∂λ 1 ∂t δh dx - I λ 2 δh dx + I λ 2 dx -M Ḧ -M g δH dt. (2.11)
Using the model (2.9) in (2.11) we get

δA = T 0 0 -ρ ∂ ∂t q h (δϕ) -ρ ∂λ 1 ∂t ∂ ∂x (δϕ) -µ 1 h ∂q ∂x ∂(δϕ) ∂x dx + 0 - 1 2 ρ q 2 h 2 -ρgh -ρ ∂λ 1 ∂t (δh) dx - b a λ 2 (δh) dx + b a λ 2 dx -M Ḧ -M g (δH) dt. (2.12)
Now we integrate by parts in space but we need to take care about the fact that some quantities can present discontinuities at the interfaces solid-liquid. We thus introduce the notation

[f ] a := f (a + ) -f (a -), [f ] b := f (b + ) -f (b -) (2.13)
and we use the following formula for the integration by parts with discontinuities:

0 uv dx = - 0 u v dx -[uv] a -[uv] b . (2.14)
We thus obtain

δA = T 0 0 -ρ ∂ ∂t q h + ρ ∂ ∂x ∂λ 1 ∂t + ∂ ∂x µ h ∂q ∂x (δϕ) dx + ρ ∂λ 1 ∂t + µ h ∂q ∂x a (δϕ)(t, a) + ρ ∂λ 1 ∂t + µ h ∂q ∂x b (δϕ)(t, b) + 0 - 1 2 ρ q 2 h 2 -ρgh -ρ ∂λ 1 ∂t (δh) dx - b a λ 2 (δh) dx + b a λ 2 dx -M Ḧ -M g (δH) dt. (2.15)
Since δA = 0 for any virtual displacement, we get the following system of equations

∂ ∂t q h - ∂ ∂x ∂λ 1 ∂t = µ ρ ∂ ∂x 1 h ∂q ∂x (t > 0, x ∈ [0, ]), (2.16
)

ρ ∂λ 1 ∂t + µ h ∂q ∂x a = 0 (t > 0), (2.17)
ρ ∂λ 1 ∂t + µ h ∂q ∂x b = 0 (t > 0), (2.18)
- ∂λ 1 ∂t = 1 2 q 2 h 2 + gh + λ 2 ρ (t > 0, x ∈ I), (2.19)
- ∂λ 1 ∂t = 1 2
q 2 h 2 + gh (t > 0, x ∈ E) (2.20) M Ḧ = b a λ 2 dx -M g (t > 0). ( 2
∂h ∂t + ∂q ∂x = 0 (t > 0, x ∈ I ∪ E), (2.22) ∂ ∂t q H + ∂ ∂x 1 2 q 2 H 2 + gH + p ρ = µ ρH ∂ 2 q ∂x 2 (t > 0, x ∈ I), (2.23)
∂ ∂t q h + ∂ ∂x 1 2 q 2 h 2 + gh = µ ρ ∂ ∂x 1 h ∂q ∂x (t > 0, x ∈ E), (2.24) 1 2 q 2 H 2 + gH + p ρ - µ Hρ ∂q ∂x (a + ) = 1 2 q 2 h 2 + gh - µ hρ ∂q ∂x (a -) (t > 0), (2.25) 1 2 q 2 H 2 + gH + p ρ - µ Hρ ∂q ∂x (b -) = 1 2 q 2 h 2 + gh - µ hρ ∂q ∂x (b + ) (t > 0), (2.26) M Ḧ(t) = -M g + b a p(t, x) dx (t > 0), (2.27)
where we have also rewritten (2.2) and used (2.3) in the solid part. In the solid part we can simplify once again this system, using the fact that as h(t, x) = H(t) does not depend on x, then equation (2.22) implies that q is a first degree polynomial.

Then our model for the rigid rectangle solid, constrained to move in the vertical direction, which is floating at the surface of a viscous fluid is

∂h ∂t + ∂q ∂x = 0 (t > 0, x ∈ I ∪ E), (2.28) ∂ ∂t q H + ∂ ∂x 1 2 q 2 H 2 + gH + p ρ = 0 (t > 0, x ∈ I), (2.29)
∂ ∂t q h + ∂ ∂x 1 2 q 2 h 2 + gh = µ ρ ∂ ∂x 1 h ∂q ∂x (t > 0, x ∈ E), (2.30) 1 2
q 2 H 2 + gH + p ρ - µ Hρ ∂q ∂x (t, a + ) = 1 2
q 2 h 2 + gh - µ hρ ∂q ∂x (t, a -) (t > 0), (2.31) 1 2
q 2 H 2 + gH + p ρ - µ Hρ ∂q ∂x (t, b -) = 1 2 q 2 h 2 + gh - µ hρ ∂q ∂x (t, b + ) (t > 0), (2.32) M Ḧ(t) = -M g + b a p(t, x) dx (t > 0). (2.33)
Remark 2.1. As mentioned above, our modeling approach yields a nonstandard viscous term in the Saint-Venant system, more precisely in the right hand side of (2.31). Indeed, different forms of this term are generally used in most of the literature on viscous Saint-Venant equations. We mention in this direction Kloeden [START_REF] Kloeden | Global existence of classical solutions in the dissipative shallow water equations[END_REF], Sundbye [START_REF] Sundbye | Global existence for the Dirichlet problem for the viscous shallow water equations[END_REF], Gerbeau and Perthame [START_REF] Gerbeau | Derivation of viscous Saint-Venant system for laminar shallow water; numerical validation[END_REF] or the review paper Bresch [START_REF] Bresch | Shallow-water equations and related topics[END_REF], which consider, once written in dimension one and with our notation, the term

µ hρ ∂ ∂x h ∂ ∂x q h
in the right hand side of (2.31). However, other viscous terms have been considered in the literature, see, for instance, [START_REF] Bresch | Shallow-water equations and related topics[END_REF], Bernardi and Pironneau [START_REF] Bernardi | On the shallow water equations at low Reynolds number[END_REF], Orenga [START_REF] Orenga | Un théorème d'existence de solutions d'un problème de shallow water[END_REF] or Rodríguez and Taboada-Vázquez [START_REF] Rodríguez | From Navier-Stokes equations to shallow waters with viscosity by asymptotic analysis[END_REF]. Comparing the viscosity term we introduce with the more commonly one used used, for instance, in [START_REF] Gerbeau | Derivation of viscous Saint-Venant system for laminar shallow water; numerical validation[END_REF], shows that their difference is given by

µ ρ ∂ ∂x 1 h ∂q ∂x - µ hρ ∂ ∂x h ∂ ∂x q h = µq ρh 2 ∂ 2 h ∂x 2 - µq ρh 3 ∂h ∂x 2 .
Passing to dimensionless variables and denoting by ε the size of the solution, the term in the right hand side of the above formula is of higher order in ε than the other terms in the momentum equation. Therefore, within the global precision of the shallow water approximation, the two viscosity terms are equivalent. Our choice of the viscous term in the right hand side of (2.31) seems an appropriate one in the case of viscous shallow water model of a fluid interacting with a floating solid for at least two reasons:

• It is energetically consistent, see Proposition 3.1 below.

• In our case, the smoothing effect coming from the viscous term applies to the flux q instead of the velocity field v as in [START_REF] Kloeden | Global existence of classical solutions in the dissipative shallow water equations[END_REF] and [START_REF] Sundbye | Global existence for the Dirichlet problem for the viscous shallow water equations[END_REF] (note that in our case, unlike in the case of non viscous Saint Venant equations, q and v have different regularity properties). An advantage of our approach is that, since we use the variables h and q we do not need any information on the sign of the velocity v at the solid-fluid interface.

Energy estimate and main result

By extending the pressure by 0 in the fluid part, by taking ρ = 1 and by adding boundary and initial conditions, we can rewrite system (2.28)-(2.33) as follows:

∂h ∂t + ∂q ∂x = 0 (t > 0, x ∈ I ∪ E), (3.1)
∂ ∂t q h + ∂ ∂x q 2 2h 2 + gh + p -µ ∂ ∂x 1 h ∂q ∂x = 0 (t > 0, x ∈ I ∪ E), (3.2)
q(t, 0) = q(t, ) = 0 (t > 0), (3.3)
p(t, x) = 0 (t 0, x ∈ E), (3.4)
h(t, x) = H(t) (t 0, x ∈ I), (3.5)
[q(t,

•)] a = [q(t, •)] b = 0 (t > 0), (3.6)
p(t, •) + q 2 (t, a) 2h 2 (t, •) + gh(t, •) - µ h ∂q ∂x (t, •) a = 0 (t 0), (3.7)
p(t, •) + q 2 (t, b) 2h 2 (t, •) + gh(t, •) - µ h ∂q ∂x (t, •) b = 0 (t 0), (3.8)
M Ḧ(t) = -M g + b a p(t, x) dx (t > 0), (3.9)
h(0, x) = h 0 (x) (x ∈ E), (3.10)
q(0, x) = q 0 (x) (x ∈ [0, ]), (3.11)
H(0) = H 0 . (3.12)
Note that from (3.1) and (3.5) it follows that ∂q ∂x (t, x) = -Ḣ(t) (t 0, x ∈ I).

(3.13)

In particular, the initial velocity of the solid is given by

Ḣ(0) = - ∂q 0 ∂x .
Let us note that the spatial derivatives in the above equations and in the whole paper correspond to the spatial derivatives of the restrictions for each subdomain. In particular, we have the classical formula for any function f , smooth on E ∪ I:

0 df dx dx = f ( ) -f (0) -[f ] b -[f ] a . (3.14)
We first show that the system (3.1)-(3.12) satisfies an energy estimate:

Proposition 3.1. Let us write

E(t) = 1 2 0 q 2 (t, x) h(t, x) + gh 2 (t, x) dx + 1 2 M Ḣ2 (t) + M gH(t) (t 0) (3.15)
and let us consider a smooth solution (h, q, H, p) of system (3.1)-(3.12). Then we have the following relation:

Ė(t) = -µ 0 1 h(t, x) ∂q ∂x (t, x) 2 dx (t 0). (3.16)
Proof. From (3.15), it follows that

Ė(t) = 0 1 2 q h 2 ∂h ∂t + q ∂ ∂t q h + gh ∂h ∂t dx + M Ḣ(t) Ḧ(t) + M g Ḣ(t).
By combining the above formula with (3.1), (3.2) and (3.4) it follows that

Ė(t) = - 0 q 2 2h 2 + gh ∂q ∂x + q ∂ ∂x q 2 2h 2 + gh -qµ ∂ ∂x 1 h ∂q ∂x dx - b a q(t, x) ∂p ∂x (t, x) dx + M Ḣ(t) Ḧ(t) + M g Ḣ(t) (t 0). (3.17) Since q 2 2h 2 + gh ∂q ∂x + q ∂ ∂x q 2 2h 2 + gh = ∂ ∂x q q 2 2h 2 + gh ,
we can combine (3.3), (3.6) and (3.17) to obtain

Ė(t) = q(t, a) q 2 (t, a) 2h 2 (t, •) + gh(t, •) a + q(t, b) q 2 (t, b) 2h 2 (t, •) + gh(t, •) b + 0 qµ ∂ ∂x 1 h ∂q ∂x dx - b a q(t, x) ∂p ∂x (t, x) dx + M Ḣ(t) Ḧ(t) + M g Ḣ(t) (t 0).
Integrating by parts in the two integrals in the above formula and using (3.7) and (3.8), together with (3.6), it follows that

Ė(t) = -µ 0 1 h ∂q ∂x 2 dx + b a p(t, x) ∂q ∂x (t, x) dx + M Ḣ(t) Ḧ(t) + M g Ḣ(t) (t 0).
Finally, inserting (3.13) and (3.9) in the last formula we obtain the desired energy estimate (3.16).

One can check that the stationary solutions of (3.1)-(3.9) can be parametrized by H > 0 by setting

h := H + M b -a , p := M g b -a (3.18)
and in that case

h S (x) = h x ∈ E H x ∈ I , q S (x) = 0 and p S (x) = 0 x ∈ E p x ∈ I . (3.19)
In the statements below, we also need the following notation: we write P k (I), k 0 the set of polynomial functions of degree k. Let us state our main results:

Theorem 3.2. Let us assume that H 0 , h 0 , q 0 ∈ R × H 1 (E) × H 1 (0, ) with H 0 > 0, h 0 > 0 in E, H 0 < min(h 0 (a -), h 0 (b + )) (3.20) q 0 (0) = q 0 ( ) = 0. (3.21)
Let K > 0 be such that

|H 0 | + h 0 H 1 (E) + q 0 H 1 (0, ) K, 1 K h 0 (x) K for x ∈ E. (3.22)
Then, there exists T > 0, depending only on K such that the system (3.1)-(3.12) admits a unique strong solution

H ∈ H 2 (0, T ), h ∈ H 1 (0, T ; H 1 (E)) ∩ C 1 ([0, T ]; L 2 (E)), (3.23) q ∈ C 0 ([0, T ]; H 1 (0, )) (3.24) q |E ∈ H 1 (0, T ; L 2 (E)) ∩ C 0 ([0, T ]; H 1 (E)) ∩ L 2 (0, T ; H 2 (E)), (3.25)
q |I ∈ H 1 (0, T ; P 1 (I)), (3.26)

p |I ∈ L 2 (0, T ; P 2 (I)). (3.27)
Moreover, there exists a constant K T > 0 such that

1 K T h 0 (t, x), H(t) K T , for all t ∈ (0, T ), x ∈ E, H(t) < min(h(t, a -), h(t, b + ))
for all t ∈ (0, T ).

Theorem 3.3. For any H > 0, the system (3.1)-(3.12) is locally well-posed around the stationary state defined by (3.18)- (3.19). More precisely, there exists η 0 > 0 such that, for all η ∈ (0, η 0 ) there exists two constants ε > 0 and C > 0, such that, for any

H 0 , h 0 , q 0 ∈ R × H 1 (E) × H 1 (0, ) with E h 0 (x) dx + H 0 (b -a) = M |E| |I| + H , q 0 (0) = q 0 ( ) = 0, (3.28)
|H 0 -H| + h 0 -h H 1 (E) + q 0 H 1 (0, ) ε, (3.29)
there exists a unique solution of (3.1)-(3.12) with

H ∈ H + H 2 (0, ∞), h ∈ h + H 1 (0, ∞; H 1 (E)) ∩ C 1 b ([0, ∞); L 2 (E)), (3.30) q ∈ C b ([0, ∞); H 1 (0, )) (3.31) q |E ∈ H 1 (0, ∞; L 2 (E)) ∩ C b ([0, ∞); H 1 (E)) ∩ L 2 (0, ∞; H 2 (E)), (3.32)
q |I ∈ H 1 (0, ∞; P 1 (I)) (3.33) p |I ∈ M g |I| + L 2 (0, ∞; P 2 (I)) (3.34) E h(t, x) dx + H(t)(b -a) = M |E| |I| + H (t 0), (3.35) h(t, x) h 2 , H(t) H 2 , for all t ∈ (0, ∞), x ∈ E, (3.36)
H(t) < min(h(t, a -), h(t, b + )) for all t ∈ (0, ∞). (3.37)
satisfying the estimate

e η(•) H -H H 2 (0,∞) + e η(•) h -h H 1 (0,∞;H 1 (E)) + e η(•) q L ∞ (0,∞;H 1 (0, )) + e η(•) q| E H 1 (0,∞;L 2 (E))∩L 2 (0,∞;H 2 (E)) Cε.
4 Equations (3.1)-(3.12) as a parabolic system on E

In this section we eliminate pressure from (3.1)-(3.12) and obtain an equivalent system involving the restrictions of h and q to E, together with the traces of q at x = a, b and H. More precisely, let us set q a (t) := q(t, a -) = q(t, a + ), q b (t

) := q(t, b -) = q(t, b + ). (4.1)
From (3.13), we deduce

Ḣ = - q b -q a b -a in I, (4.2)
and

q(t, x) = q a (t) x -b a -b + q b (t) x -a b -a , ∂q ∂x (t, x) = q b (t) -q a (t) b -a (t 0, x ∈ I). (4.3)
Combining (3.2), (3.5) and (3.13), we get

∂p ∂x = -H ∂ ∂t q H 2 = - q H + 2 q Ḣ H 2 (t 0, x ∈ I). (4.4)
We recall that if

f ∈ P 2 ([a, b]), then we have the following standard formulas b a f (x)dx = f (a)(b -a) + f (a) (b -a) 2 3 + f (b) (b -a) 2 6 , (4.5) b a f (x)dx = f (b)(b -a) -f (a) (b -a) 2 6 -f (b) (b -a) 2 3 . (4.6)
In particular, since p is a second degree polynomial with respect to x, we deduce b a p(t, x) dx

= p(t, a)(b -a) + - qa (t) H(t) + 2 q a (t) Ḣ(t) H(t) 2 (b -a) 2 3 + - qb (t) H(t) + 2 q b (t) Ḣ(t) H(t) 2 (b -a) 2 6 (4.7) and b a p(t, x) dx = p(t, b)(b -a) -- qa (t) H(t) + 2 q a (t) Ḣ(t) H(t) 2 (b -a) 2 6 -- qb (t) H(t) + 2 q b (t) Ḣ(t) H 2 (b -a) 2 3 . (4.8)
Combining (4.7), (4.8), (4.2) and (3.9) yields

M + (b -a) 3 3H qa -M - (b -a) 3 6H qb = -M g(b -a) + p(•, a)(b -a) 2 + 2q 2 a -q a q b -q 2 b (b -a) 2 3H 2 (t > 0), (4.9)
and

-M - (b -a) 3 6H qa + M + (b -a) 3 3H qb = M g(b -a) -p(•, b)(b -a) 2 -2q 2 b -q a q b -q 2 a (b -a) 2 3H 2 (t > 0). (4.10)
Inverting the above linear system, we get

qa qb = S(H)   -M g(b -a) + p(•, a)(b -a) 2 + 2q 2 a -q a q b -q 2 b (b-a) 2 3H 2 M g(b -a) -p(•, b)(b -a) 2 -2q 2 b -q a q b -q 2 a (b-a) 2 3H 2   , (4.11)
where for all H > 0, S(H) is the following symmetric positive matrix:

S(H) := H (b -a) 3 M + (b-a) 3 12H M + (b-a) 3 3H M -(b-a) 3 6H M -(b-a) 3 6H M + (b-a) 3 3H . ( 4
qa qb = R H, q(•, a), ∂ ∂x q(•, a -), h(•, a -), q(•, b), ∂ ∂x q(•, b + ), h(•, b + ) , (4.13)
where

R H, q(•, a), ∂ ∂x q(•, a -), h(•, a -), q(•, b), ∂ ∂x q(•, b + ), h(•, b + ) = S(H)   -M g(b -a) + p(•, a)(b -a) 2 + 2q 2 a -q a q b -q 2 b (b-a) 2 3H 2 M g(b -a) -p(•, b)(b -a) 2 -2q 2 b -q a q b -q 2 a (b-a) 2 3H 2   , (4.14)
and p(•, a) and p(•, b) are given by:

p(t, a) = q 2 (t, a) 2h 2 (t, a -) + gh(t, a -) - µ h ∂q ∂x (t, a -) - q 2 (t, a) 2H 2 (t) -gH(t) + µ q b (t) -q a (t) H(b -a) (4.15) p(t, b) = q 2 (t, b) 2h 2 (t, b + ) + gh(t, b + ) - µ h ∂q ∂x (t, b + ) - q 2 (t, b) 2H 2 (t) -gH(t) + µ q b (t) -q a (t) H(b -a) . (4.16)
Finally, the system (3.1)-(3.12) writes in the equivalent form Ḣ = -q b -q a b -a (t 0), (4.17)

∂h ∂t (t, x) + ∂q ∂x (t, x) = 0 (t > 0, x ∈ E), (4.18
)

∂q ∂t + ∂ ∂x q 2 h + gh 2 2 = h ∂ ∂x µ h ∂q ∂x (t > 0, x ∈ E), (4.19)
q(t, 0) = q(t, ) = 0 (t > 0), (4.20)

q(t, a) = q a (t), q(t, b) = q b (t) (t > 0), (4.21) qa qb = R H, q(•, a), ∂ ∂x q(•, a -), h(•, a -), q(•, b), ∂ ∂x q(•, b + ), h(•, b + ) , (4.22)
with R defined by (4.14), (4.15) and (4.16).

Local in time existence and uniqueness

In this section we prove Theorem 3.2. In view of the reduction performed in Section 4, our main result in Theorem 3.2 can be restated as:

Theorem 5.1. Assume that H 0 , h 0 , q 0 , q a,0 , q b,0 ∈ R × H 1 (E) × H 1 (E) × R 2 ,
with

H 0 > 0, h 0 > 0 in E, H 0 < min(h 0 (a -), h 0 (b + )), (5.1)
q 0 (0) = q 0 ( ) = 0, q 0 (a) = q a,0 , q 0 (b) = q b,0 .

(5.2)

Let K > 0 be such that

|H 0 | + h 0 H 1 (E) + q 0 H 1 (E) + |q 0,a | + |q 0,b | K, 1 K h 0 (x) K for x ∈ E. (5.3)
Then, there exists T > 0, depending only on K, such that the system (4.17)-(4.22) admits a unique strong solution

H ∈ H 2 (0, T ), h ∈ H 1 (0, T ; H 1 (E)) ∩ C 1 ([0, T ); L 2 (E)), (5.4
)

q ∈ H 1 (0, T ; L 2 (E)) ∩ C([0, T ); H 1 (E)) ∩ L 2 (0, T ; H 2 (E)), (5.5
)

q a ∈ H 1 (0, T ), q b ∈ H 1 (0, T ) (5.6)
Moreover, there exists a constant K T > 0 such that

1 K T h(t, x), H(t) K T (t ∈ (0, T ), x ∈ E), H(t) < min(h(t, a -), h(t, b + )) (t ∈ (0, T )).
The proof of the above theorem relies on classical estimates on linear parabolic problems, combined with the use of the Banach fixed point theorem. The strategy we adopt here is based on the fact that the system (4.17)-(4.22) can be rewritten as

Ḣ = - q b -q a b -a (t > 0), (5.7
)

∂h ∂t + ∂q ∂x = 0 (t > 0, x ∈ E), (5.8)
∂q ∂t -µ ∂ 2 q ∂x 2 = F 1 (h, q) (t > 0, x ∈ E), (5.9)
q(t, 0) = q(t, ) = 0, (t > 0), (5.10)

q(t, a) = q a (t), q(t, b) = q b (t) (t > 0), (5.11)
qa qb = F 2 (H, q, q a , q b ) (t > 0), (5.12) h(0, x) = h 0 (x), q(0, x) = q 0 (x) (x ∈ E), (5.13)
H(0) = H 0 , q a (0) = q a,0 , q b (0) = q b,0 , (5.14)
where

F 1 (h, q) = - 2q h ∂q ∂x + q 2 h 2 ∂h ∂x + gh ∂h ∂x - µ h ∂h ∂x ∂q ∂x , (5.15)
and

F 2 (H, q, q a , q b ) = R H, q(•, a), ∂ ∂x q(•, a -), h(•, a -), q(•, b), ∂ ∂x q(•, b + ), h(•, b + ) , (5.16)
with R defined by (4.14), (4.15) and (4.16). Next, by replacing F 1 and F 2 by given source term f 1 and f 2 we obtain the following linear system :

Ḣ = - q b -q a b -a (t > 0), (5.17) ∂h ∂t + ∂q ∂x = 0 (t > 0, x ∈ E), (5.18) ∂q ∂t -µ ∂ 2 q ∂x 2 = f 1 (t > 0, x ∈ E), (5.19)
q(t, 0) = q(t, ) = 0, (t > 0), (5.20) q(t, a) = q a (t), q(t, b) = q b (t) (t > 0), (5.21)

qa qb = f 2 (t > 0), (5.22) h(0, x) = h 0 (x), q(0, x) = q 0 (x) (x ∈ E), (5.23)
H(0) = H 0 , q a (0) = q a,0 , q b (0) = q b,0 .

(5.24)

We have the following regularity result for the system (5.17)-(5.24)

Theorem 5.2. Let us assume that H 0 , h 0 , q 0 , q a,0 , q b,0 ∈ R×H 1 (E)×H 1 (E)×R 2 and satisfies the compatibility condition (5.2). Then, for all 0 < T 1, f 1 ∈ L 2 (0, T ; L 2 (E)) and f 2 ∈ L 2 (0, T ; R 2 ) the system (5.17)-( 5.24) admits a unique solution

H ∈ H 2 (0, T ), h ∈ H 1 (0, T ; H 1 (E)) ∩ C 1 ([0, T ); L 2 (E)), q ∈ H 1 (0, T ; L 2 (E)) ∩ C([0, T ); H 1 (E)) ∩ L 2 (0, T ; H 2 (E)), q a ∈ H 1 (0, T ), q b ∈ H 1 (0, T ).
Moreover, there exists a constant C independent of T such that

H H 2 (0,T ) + h H 1 (0,T ;H 1 (E)) + h L ∞ (0,T ;H 1 (E)) + q L 2 (0,T ;H 2 (E)) + q H 1 (0,T ;L 2 (E)) + q L ∞ (0,T ;H 1 (E)) + (q a , q b ) H 1 (0,T ;R 2 ) C |H 0 | + h 0 H 1 (E) + q 0 H 1 (E) + |q a,0 | + |q b,0 | + (f 1 , f 2 ) L 2 (0,T ;L 2 (E)×R 2 )
, (5.25) for all 0 < T 1.

Proof. Let us remark that the linear system (5.17)-(5.24) can be solved in "cascades": Eq. (5.22) can be solved independently and admits a unique solution (q a , q b ) ∈ H 1 (0, T ; R 2 ). Next we solve Eq. (5.17) and we obtain H ∈ H 2 (0, T ). Using the standard regularity results for parabolic equations with non-homogeneous boundary conditions, we also have that

q ∈ L 2 (0, T ; H 2 (E)) ∩ H 1 (0, T ; L 2 (E)) ∩ L ∞ (0, T ; H 1 (E)).
Finally, using the regularity of q, we obtain the desired regularity of h from Eq. (5.18). To obtain estimates with continuity constant independent of time T we can proceed as in [START_REF] Haak | Mathematical analysis of the motion of a rigid body in a compressible Navier-Stokes-Fourier fluid[END_REF]Theorem 5.3]. Now we estimate the nonlinear terms defined in (5.15)-(5.16).

Proposition 5.3. Let us assume that H 0 , h 0 , q 0 , q a,0 , q b,0

∈ R × H 1 (E) × H 1 (E) × R 2 such that (5.1)-(5.
3) holds. There exist T ∈ (0, 1), δ > 0 and C = C(M, T ) > 0 such that for T ∈ (0, T ] and for (f

1 , f 2 ) ∈ L 2 (0, T ; L 2 (E) × R 2 ) satisfying (f 1 , f 2 ) L 2 (0,T ;L 2 (E)×R 2 )
1,

the solution (H, h, q, q a , q b ) of (5.17)-(5.24) verifies

F 1 L 2 (0,T ;L 2 (E)) + F 2 L 2 (0,T ;R 2 ) CT δ .
Proof. We choose T < 1 and T ∈ (0, T ]. The constant appearing in this proof depends only on K and independent of T. First of all, from (5.25) we first obtain

H H 2 (0,T ) + h H 1 (0,T ;H 1 (E)) + h L ∞ (0,T ;H 1 (E)) + q L 2 (0,T ;H 2 (E)) + q H 1 (0,T ;L 2 (E))
+ q L ∞ (0,T ;H 1 (E)) + (q a , q b ) H 1 (0,T ;R 2 ) C. (5.26) Applying Hölder's inequality together with the above estimate we deduce that

H -H 0 L ∞ (0,T ) C √ T , h -h 0 L ∞ (0,T ;H 1 (E)) C √ T .
In particular, there exists T such that for all T ∈ (0, T ] the following holds:

H(t) H 0 2 , h(t, x) 1 2K , t ∈ (0, T ), x ∈ E, (5.27) h L ∞ ((0,T )×E) + H L ∞ (0,T ) C, (5.28)
h L 2 (0,T ;H 1 (E)) + H L 2 (0,T ) C √ T , (5.29) 1 h L ∞ (0,T ;H 1 (E)) + 1 h L ∞ ((0,T )×E) + 1 H L ∞ (0,T ) C.
(5.30)

In a similar manner, we also obtain

q a L ∞ (0,T ) + q b L ∞ (0,T ) C, q a L 2 (0,T ) + q b L 2 (0,T ) C √ T . (5.31)
On the other hand, using an interpolation inequality one has following estimate (see, for instance estimate, (6.13) of [START_REF] Haak | Mathematical analysis of the motion of a rigid body in a compressible Navier-Stokes-Fourier fluid[END_REF])

q L 2 (0,T ;H 1+s (E)) CT (1-s)/4
, s ∈ (0, 1).

(5.32)

Using the above estimate, (5.26) and the Sobolev embedding we get

q L ∞ ((0,T )×E) C, ∂q ∂x L 2 (0,T ;L ∞ (E))
CT (1-s)/4 , s ∈ (1/2, 1) (5.33)

We are now in a position to estimate the non-linear terms defined in (5.15)-(5.16). Estimate of F 1 :

F 1 L 2 (0,T ;L 2 (E))
CT δ , for some δ > 0.

(5.34)

•Estimate of the first term of F 1 : Using (5.26), (5.27) and (5.33), we have

- 2q h ∂q ∂x L 2 (0,T ;L 2 (E)) C q L 2 (0,T ;H 1 (E)) C √ T q L ∞ (0,T ;H 1 (E)) C √ T .
•Estimate of the second and third term of F 1 : Using (5.26), (5.27) and (5.33) we have

q 2 h 2 ∂h ∂x + gh ∂h ∂x L 2 (0,T ;L 2 (E)) C h L 2 (0,T ;H 1 (E)) C √ T .
•Estimate of the last term of F 1 : Using (5.26), (5.27) and (5.33) we obtain

µ h ∂h ∂x ∂q ∂x , L 2 (0,T ;L 2 (E)) C ∂h ∂x L ∞ (0,T ;L 2 (E)) ∂q ∂x L 2 (0,T ;L ∞ (E)) CT (1-s)/4 ,
for some s ∈ (1/2, 1). Thus we have proved (5.34). Estimate of F 2 : F 2 L 2 (0,T ;R 2 ) CT δ , for some δ > 0.

(5.35)

•Estimate of S(H) defined in (4.12) : Using (5.27) -(5.29), we have the following estimates (5.37)

S(H) L ∞ (0,T ;R 4 ) C, S (
We provide only the estimate of third term of p(t, a). Estimates of the other terms are similar. Using (5.30) and (5.32) we have

µ h ∂q ∂x (•, a -) L 2 (0,T ) C 1 h ∂q ∂x L 2 (0,T ;H s (0,a)) C 1 h L ∞ (0,T ;H 1 (0,a)) ∂q ∂x L 2 (0,T ;H s (0,a)) CT (1-s)/4 ,
for some s ∈ (1/2, 1). Thus we have (5.37). Finally, combining (5.36), (5.37) and (5.31), we obtain (5.35).

We are now in a position to prove one of Theorem 5.1.

Proof of Theorem 5.1. Let T be the constant in Proposition 5.3. For T ∈ (0, T ], we consider the ball

B T := (f 1 , f 2 ) ∈ L 2 (0, T ; L 2 (E) × R 2 ) ; (f 1 , f 2 ) L 2 (0,T ;L 2 (E)×R 2 ) 1 (5.38)
and the map Ξ :

(f 1 , f 2 ) ∈ B T → (F 1 (h, q), F 2 (H, q, q a , q b )), (5.39)
where (H, h, q, q a , q b ) is the solution to the system (5.17)-( 5.24) and where F 1 and F 2 are given by (5.15)- (5.16). By Proposition 5.3, we have that Ξ(B T ) ⊂ B T for T small enough. With similar calculation as Proposition 5.3, we can also show that for small T , Ξ |B T is a strict contraction. This completes the proof of Theorem 5.1.

Global in time existence and uniqueness

In this section, we linearize the system (4.17)-(4.22) around a stationary state and we study the corresponding linear system. Note that, in order to prove global existence for small data of our original problem (4.17)-(4.22) we need this linearization to be exponentially stable, so that we cannot use the linearized problem (5.17)-(5.24), introduced for the local existence result.

Linearization around a stationary state

We consider the stationary state given by (3.19) and we define

H := H -H, h := h -h S = h -h in E H in I .
From (4.15)-(4.16) and (3.18), we deduce

p(t, a) = q 2 (t, a) 2 1 ( h + h) 2 (t, a -) - 1 ( H + H) 2 (t) + g( h(t, a -) -H(t)) + g M b -a - µ h ∂q ∂x (t, a -) + µ q b (t) -q a (t) H(b -a) + µ h h ( h + h) ∂q ∂x (t, a -) -µ q b (t) -q a (t) b -a H ( H + H)H (6.1) p(t, b) = q 2 (t, b) 2 1 ( h + h) 2 (t, b + ) - 1 ( H + H) 2 (t) + g( h(t, b + ) -H(t)) + g M b -a - µ h ∂q ∂x (t, b + ) + µ q b (t) -q a (t) H(b -a) + µ h h ( h + h) ∂q ∂x (t, b + ) -µ q b (t) -q a (t) b -a H ( H + H)H . (6.2) Thus -M g(b -a) + p(•, a)(b -a) 2 + 2q 2 a -q a q b -q 2 b (b -a) 2 3H 2 = (b -a) 2 g( h(•, a -) -H) - µ h ∂q ∂x (•, a -) + µ(q b -q a ) H(b -a) + q 2 (•, a) 2 1 ( h + h) 2 (•, a -) - 1 ( H + H) 2 + µ h h ( h + h) ∂q ∂x (t, a -) -µ q b -q a b -a H ( H + H)H + 2q 2 a -q a q b -q 2 b 3( H + H) 2 (6.3) and M g(b -a) -p(•, b)(b -a) 2 -2q 2 a -q a q b -q 2 b (b -a) 2 3H 2 = (b -a) 2 -g( h(•, b + ) -H) + µ h ∂q ∂x (•, a -) - µ(q b -q a ) H(b -a) - q 2 (•, b) 2 1 ( h + h) 2 (•, b + ) - 1 ( H + H) 2 + µ h h ( h + h) ∂q ∂x (t, b + ) + µ q b -q a b -a H ( H + H)H - 2q 2 b -q a q b -q 2 a 3( H + H) 2 (6.4)
Then ( H, h, q, q a , q b ) satisfies

˙ H = - q b -q a b -a (t 0), (6.5)
∂ h ∂t (t, x) + ∂q ∂x (t, x) = 0 (t > 0, x ∈ E), (6.6
)

∂q ∂t + gh ∂ h ∂x -µ ∂ 2 q ∂x 2 = F 1 ( h, q) (t > 0, x ∈ E), (6.7)
q(t, 0) = q(t, ) = 0 (t > 0), (6.8)

q(t, a) = q a (t), q(t, b) = q b (t) (t > 0), (6.9) qa qb = (b -a) 2 S(H)    g( h(•, a -) -H) - µ h ∂q ∂x (•, a -) + µ H q(•, b) -q(•, a) b -a -g( h(•, b + ) -H) + µ h ∂q ∂x (•, b + ) - µ H q(•, b) -q(•, a) b -a   
+ F 2 ( H, h, q, q b , q a ) (6.10)

where

F 1 ( h, q) := - ∂ ∂x q 2 h + h + g h 2 2 - µ (h + h) 2 ∂ h ∂x ∂q ∂x , (6.11)
F 2 ( H, h, q, q b , q a ) :

= (b -a) 2 S( H + H) -S(H)    g( h(•, a -) -H) - µ h ∂q ∂x (•, a -) + µ H q b -q a b -a -g( h(•, b + ) -H) + µ h ∂q ∂x (•, a -) - µ H q b -q a b -a   
+ (b -a) 2 S( H + H) F 3 ( H, h, q, q b , q a ) F 4 ( H, h, q, q b , q a ) , (6.12)

F 3 ( H, h, q, q b , q a ) := q 2 a 2 1 ( h + h) 2 (•, a -) - 1 ( H + H) 2 + µ h h ( h + h) ∂q ∂x (•, a -) -µ q b -q a b -a H ( H + H)H + 2q 2 a -q a q b -q 2 b 3( H + H) 2 (6.13) F 4 ( H, h, q, q b , q a ) := - q 2 b 2 1 ( h + h) 2 (•, b + ) - 1 ( H + H) 2 - µ h h ( h + h) ∂q ∂x (•, b + ) + µ q b -q a b -a H ( H + H)H - 2q 2 b -q a q b -q 2 a 3( H + H) 2 (6.14)
Using (3.10), (3.11) and (3.12), we add the following initial conditions to system (6.5)-(6.14):

h(0, x) = h 0 (x) (x ∈ E), (6.15)
q(0, x) = q 0 (x) (x ∈ E), (6.16)
H(0) = H 0 , q a (0) = q a,0 q b (0) = q b,0 , (

where H 0 = H 0 -H, h 0 := h 0 -h, q a,0 := q 0 (a), q b,0 := q 0 (b).

In order to study the above equations, we consider the following linear system

˙ H = - q b -q a b -a (t 0), (6.18)
∂ h ∂t (t, x) + ∂q ∂x (t, x) = 0 (t > 0, x ∈ E), (6.19
)

∂q ∂t + gh ∂ h ∂x -µ ∂ ∂x ∂q ∂x = f 1 (t > 0, x ∈ E), (6.20)
q(t, 0) = q(t, ) = 0 (t > 0), (6.21)

q(t, a) = q a (t), q(t, b) = q b (t) (t > 0), (6.22) qa qb = S 0 (H)    µ H q(•, b) -q(•, a) b -a + g h(•, a -) -H - µ h ∂q ∂x (•, a -) - µ H q(•, b) -q(•, a) b -a -g h(•, b + ) -H + µ h ∂q ∂x (•, b + )    + f 2 (6.23)
where

S 0 (H) = H (b -a) M + (b-a) 3 12H M + (b-a) 3 3H M -(b-a) 3 6H M -(b-a) 3 6H M + (b-a) 3 3H , (6.24)
and where f 1 and f 2 are now given.

Study of the Linear Problem

In this section we use the methodology developed in [START_REF] Maity | A maximal regularity approach to the analysis of some particulate flows[END_REF] in order to prove that the linear problem introduced above can be described by an analytic C 0 semigroup in an appropriate Hilbert space and then we show that this semigroup is exponentially stable. To this end we begin by defining the following spaces:

X = [H, h, q] ∈ C × H 1 (E) × L 2 (E) | E h 0 (x) dx + H 0 (b -a) = 0}, U = C 2 , (6.25) Z = [H, h, q] ∈ C × H 1 (E) × H 2 (E) | E h 0 (x) dx + H 0 (b -a) = 0, q 0 (0) = q 0 ( ) = 0 . (6.26)
We also introduce the following operators:

L : D(L) → X, L   H h q   =     -q(b)-q(a) b-a -dq dx -gh dh dx + µ d 2 q dx 2     , D(L) = Z, (6.27) G : D(G) → U, G   H h q   = q(a) q(b) , D(G) = Z, (6.28)
and

A := L| Ker(G) . (6.29)
The above definition yields

D(A) = C × H 1 (E) × (H 2 ∩ H 1 0 )(E) and A   H h q   =     0 -dq dx -gh dh dx + µ d 2 q dx 2     .
(6.30) Lemma 6.1. The operator A : D(A) → X generates an analytic semigroup.

Proof. It clearly suffices to prove that the operator A defined by

D( A) = H 1 (E) × (H 2 ∩ H 1 0 )(E), A h q = -dq dx -gh dh dx + µ d 2 q dx 2 h q ∈ D( A) ,
generates an analytic semigroup in X = H 1 (E) × L 2 (E). To this aim, noticing that the operator defined by

P h q = 0 -gh dh dx is in L( X)
, it suffices to prove that the operator A 1 defined by

D(A 1 ) = D( A), A 1 h q = -dq dx µ d 2 q dx 2 h q ∈ D( A) ,
generates an analytic semigroup in X. Denoting (6.31) this is equivalent to the existence of ε ∈ (0, π/2) and γ 0 such that

Σ ε,γ = {λ ∈ C \ {0} | |argλ| π -ε, |λ| > γ},
Σ ε,γ ⊂ ρ(A 1 )
and the set λ(λI

-A 1 ) -1 | λ ∈ Σ ε,γ
is bounded in L( X). To do this, for f g ∈ X we notes that the equation

(λI -A 1 ) h λ q λ = f g , writes λh λ + dq λ dx = f, (6.32) λq λ -µ d 2 q λ dx 2 = g, (6.33) q(0) = q(a) = q(b) = q( ) = 0. (6.34)
Classical results imply the existence of γ, C 0 and ε ∈ (0, π/2) such that for every λ ∈ Σ ε,γ the unique solution of (6.33) and ( 6.34) satisfies

λq λ L 2 (E) + q λ H 2 (E) C g L 2 (E) (g ∈ L 2 (E)). ( 6

.35)

Inserting the above estimate in (6.32) we obtain that

λh λ H 1 (E) f H 1 (E) + C g L 2 (E) (f ∈ H 1 (E), g ∈ L 2 (E)).
The last estimates and (6.35) imply that the family of operators

λ(λI -A 1 ) -1 (λ ∈ Σ ε,γ )
is bounded in L( X), which ends the proof.

Using the notation

z =   H h q   , u = q a q b , ( 6
.36) the linear system (6.18)-( 6.24) writes

ż = Lz +   0 0 f 1   , Gz = u, (6.37) u = Cz + f 2 , (6.38)
where L and G have been defined in (6.27) and (6.28), respectively, and

Cz = S 0 (H)   µ H q(b)-q(a) b-a + g h(a -) -H -µ h dq dx (a -) -µ H q(b)-q(a) b-a -g h(b + ) -H + µ h dq dx (b + )   . (6.39)
We introduce the operator A by:

D(A) = z u ∈ Z × U; Gz = u (6.40)
and

A z u = Lz Cz (6.41)
Then the linear system (6.18)-( 6.24) can be written as

d dt z u = A z u + Φ 1 Φ 2 (6.42)
where

Φ 1 =   0 0 f 1   , Φ 2 = f 2 .
Theorem 6.2. The operator A is the infinitesimal generator of an analytic semigroup on X × U.

Proof. From a trace theorem, one can check the following property on C:

∀δ > 0, ∃k(δ) Cz C 2 δ z Z + k(δ) z X . (6.43)
Then by [START_REF] Maity | A maximal regularity approach to the analysis of some particulate flows[END_REF]Theorem 1.25] and Lemma 6.1 we deduce that A : D(A) → X × U generates an analytic semigroup.

We now show the following property on the resolvent set ρ(A) of A: Proposition 6.3. The resolvent set of A contains the closed right half plane of C:

C + := {λ ∈ C ; Re(λ) 0} ⊂ ρ(A).
Proof. We first show that 0 ∈ ρ(A), i.e. that A is boundedly invertible. Assume F ∈ X × U and let us solve the equation

A z 0 u 0 = F z 0 u 0 ∈ D(A) . (6.44) Denoting z 0 u 0 =       H 0 h 0 q 0 q a,0 q b,0       and F =       f 1 f 2 f 3 f 4 f 5      
, the above system writes

- q 0 (b) -q 0 (a) b -a = f 1 , (6.45) - dq 0 dx (x) = f 2 (x) (x ∈ E), (6.46) -gh dh 0 dx + µ d 2 q 0 dx 2 = f 3 (x) (x ∈ E), (6.47) S 0 (H)   µ H q0(b)-q0(a) b-a + g h 0 (a -) -H 0 -µ h dq0 dx (a -) -µ H q0(b)-q0(a) b-a -g h 0 (b + ) -H 0 + µ h dq0 dx (b + )   = f 4 f 5 (6.48)
q 0 (0) = q 0 ( ) = 0, (6.49) q 0 (a) = q a,0 , q 0 (b) = q b,0 . (6.50)

We first note that from (6.46) and ( 6.49) it follows that (6.25)), equation ( 6.45) is also satisfied. Moreover, from (6.50) it follows that

q 0 (x) = - x 0 f 2 (ξ) dξ for x ∈ [0, a] x f 2 (ξ) dξ for x ∈ [b, ]. (6.51) Remark that since   f 1 f 2 f 3   ∈ X (see
q a,0 = - a 0 f 2 (ξ) dξ, q b,0 = b f 2 (ξ) dξ. (6.52)
Multiplying both sides of (6.48) by S 0 (H) -1 and using (6.45), (6.46), we obtain

  g h 0 (a -) -H 0 g h 0 (b + ) -H 0   =   µ H f 1 -µ h f 2 (a -) + M + (b-a) 3 3H f4 (b-a) 2 -M -(b-a) 3 6H f5 (b-a) 2 µ H f 1 -µ h f 2 (b + ) + M -(b-a) 3 6H f4 (b-a) 2 -M + (b-a) 3 3H f5 (b-a) 2   . (6.53)
On the other hand, by combining (6.46) and ( 6.47) it follows that

d dx (h 0 -H 0 )(x) = - f 3 (x) gh - µ gh df 2 dx (x) (x ∈ E), (6.54)
so that, according to (6.53), we have

h 0 (x) -H 0 = µ gH f 1 + M + (b -a) 3 3H f 4 g(b -a) 2 -M - (b -a) 3 6H f 5 g(b -a) 2 + 1 gh a x f 3 (ξ) dξ - µ gh f 2 (x) (x ∈ [0, a]
), (6.55)

h 0 (x) -H 0 = µ gH f 1 + M + (b -a) 3 3H f 4 g(b -a) 2 -M - (b -a) 3 6H f 5 g(b -a) 2 - 1 gh x b f 3 (ξ) dξ - µ gh f 2 (x) (x ∈ [b, ]). (6.56)
The above formulas, combined with the the fact that E (h 0 (x) -H 0 ) dx + H 0 = 0, give an explicit formula for H 0 in terms of f k , with k ∈ {1, . . . , 5}, together with the existence of some positive constant K 0 such that

|H 0 | K 0 F X×U (F ∈ X × U) .
The above estimate, together with (6.55) and (6.56) imply the existence of K 1 > 0 such that

h 0 L 2 (E) K 1 F X×U (F ∈ X × U) .
The last two estimates and (6.54), (6.51), (6.52) imply that there exists K 2 > 0 such that

z 0 u 0 X×U K 2 F X×U (F ∈ X × U) .
This shows that 0 ∈ ρ(A).

Assume now λ ∈ C + \ {0} and F = [f 1 , f 2 , f 3 , f 4 , f 5 ] , the equation (λI -A) z 0 u 0 = F, writes λH 0 + q 0 (b) -q 0 (a) b -a = f 1 (6.57) λh 0 (x) + dq 0 dx (x) = f 2 (x) (6.58) λq 0 (x) + gh dh 0 dx -µ d 2 q 0 dx 2 = f 3 (x) (6.59)
q 0 (0) = q 0 ( ) = 0, (

q 0 (a) = q a,0 , q 0 (b) = q b,0 , (6.61)

λ q a,0 q b,0 -S 0 (H)   µ H q0(b)-q0(a) b-a + g h 0 (a -) -H 0 -µ h dq0 dx (a -) -µ H q0(b)-q0(a) b-a -g h 0 (b + ) -H 0 + µ h dq0 dx (b + )   = f 4 f 5 . (6.62)
From (6.62) and ( 6.24), it follows that

µ H q 0 (b) -q 0 (a) b -a + g h 0 (a -) -H 0 - µ h dq 0 dx (a -) = M + (b -a) 3 3H (λq a,0 -f 4 ) (b -a) 2 + M - (b -a) 3 6H (f 5 -λq b,0 ) (b -a) 2 (6.63)
- µ H q 0 (b) -q 0 (a) b -a -g h 0 (b + ) -H 0 + µ h dq 0 dx (b + ) = M - (b -a) 3 6H (f 4 -λq a,0 ) (b -a) 2 + M + (b -a) 3 3H (λq b,0 -f 5 ) (b -a) 2 . (6.64)
We next transform (6.57)-(6.64) into a boundary value problem for q 0 by eliminating h 0 , H 0 , q a,0 , q b,0 from the above mentioned equations. Firstly, from (6.58) and (6.59) we deduce

λ h q 0 - µ h + g λ d 2 q 0 dx 2 = f 3 h - g λ df 2 dx =: φ 1 ∈ L 2 (E). (6.65)
Next, using (6.57), (6.58) and (6.61) in (6.63)-(6.64) it follows that

µ h + g λ dq 0 dx (a -) = µ H + g λ + λM b -a q 0 (b) -q 0 (a) b -a - λ(b -a) 6H (2q 0 (a) + q 0 (b)) + φ 2 , (6.66) µ h + g λ dq 0 dx (b + ) = µ H + g λ + λM b -a q 0 (b) -q 0 (a) b -a + λ(b -a) 6H (q 0 (a) + 2q 0 (b)) + φ 3 , (6.67)
with

φ 2 := g f 2 (a -) λ - f 1 λ + M + (b -a) 3 3H f 4 (b -a) 2 -M - (b -a) 3 6H f 5 (b -a) 2 , φ 3 := g f 2 (b + ) λ - f 1 λ + M - (b -a) 3 6H f 4 (b -a) 2 -M + (b -a) 3 3H f 5 (b -a) 2 .
At this stage we remark that the fact that C + \ {0} ⊂ ρ(A) readily follows as soon as we have shown that for every λ ∈ C + \ {0} ⊂ ρ(A) and F ∈ X × U, the system formed by (6.65) and (6.66)-(6.67), together with (6.60) admits a unique solution q 0 with

q 0 H 2 (E) K(λ) F X×U (F ∈ X × U).
We then recover h 0 , H 0 , q a,0 , q b,0 by using (6.57), (6.58) and (6.61). Now, to solve (6.65), (6.60) and (6.66)-( 6.67), it is sufficient to prove the existence and uniqueness of a weak solution. The H 2 regularity can then be obtained in a standard way.

Let us set

V := ψ ∈ H 1 (E) | ψ(0) = ψ( ) = 0 .
Then we can check (6.65), (6.60) and (6.66)-(6.67) are equivalent to

B λ (q, ψ) = E φ 1 ψ dx + φ 2 ψ(a) -φ 3 ψ(b) (ψ ∈ V ), (6.68)
where

B λ (q, ψ) := E λ h qψ + µ h + g λ dq 0 dx dψ dx dx - g λ + µ H + λM b -a q 0 (b) -q 0 (a) b -a - λ(b -a) 6H [2q 0 (a) + q 0 (b)] ψ(a) + g λ + µ H + λM b -a q 0 (b) -q 0 (a) b -a + λ(b -a) 6H [q 0 (a) + 2q 0 (b)] ψ(b) (q, ψ ∈ V ).
We notice that B λ writes more conveniently as

B λ (q, ψ) = E λ h qψ + µ h + g λ dq 0 dx dψ dx dx + g λ + µ H + λM b -a (q 0 (b) -q 0 (a))(ψ(b) -ψ(a) b -a + λ(b -a) 6H
q 0 (a)ψ(a) + q 0 (b)ψ(b) + (q 0 (b) + q 0 (a))(ψ(b) + ψ(a)) (q, ψ ∈ V ). (6.69)

From the above formula and the Poincaré inequality, it follows that for any λ ∈ C + \ {0} there exist positive constants C = C(λ), α = α(λ) such that

|B λ (q, ψ) C q V ψ V , Re B λ (q, q) α q 2 V (q, ψ ∈ V ),
thus B λ is a bounded and coercive form on V . Moreover, the right hand side of (6.68) clearly defines a bounded linear functional on V . Thus the conclusion follows by the complex version of the Lax-Milgram Lemma (see, for instance, Arendt et al. [START_REF] Arendt | Form methods for evolution equations, and applications[END_REF]Lemma 5.4]).

Proposition 6.4. The operator A generates an exponentially stable semigroup (e tA ) t 0 on X × U.

In other words, there exist constant C > 0 and η 0 > 0 such that e tA H 0 , h 0 , q 0 , q a,0 , q b,0 X×U Ce -η0t H 0 , h 0 , q 0 , q a,0 , q b,0 X×U . (6.70)

Proof. By Lemma 6.1 we have the existence of C > 0 such that for any λ ∈ Σ ε,γ with ε ∈ (0, π/2),

(λI -A) -1 L(X×U ) C.
Next, by Proposition 6.3 and noting the fact that C + \ Σ ε,γ is a compact set, we obtain

(λI -A) -1 L(X×U ) C, for any λ ∈ C + ∪ Σ ε,γ . This yields that {λ ∈ C | Reλ -η} ⊂ ρ(A).
Finally applying Proposition 2.9 [2, p. 120] we deduce the exponential stability of the semigroup generated by the operator A.

We conclude from Theorem 6.2 and Proposition 6.4 the following result: Proposition 6.5. Let η ∈ [0, η 0 ) where η 0 is the constant introduced in Proposition 6.4. Then for any H 0 , h 0 , q 0 , q a,0 , q b,0

∈ R × H 1 (E) × H 1 (E) × R 2
with E h 0 (x) dx + H 0 (b -a) = 0, q 0 (0) = q 0 ( ) = 0, q 0 (a) = q a,0 , q 0 (b) = q b,0 , (6.71)

and for any e η(•) f 1 ∈ L 2 (0, ∞; L 2 (E)), e η(•) f 2 ∈ L 2 (0, ∞; R 2 ) the system (6.18)-(6.23), with (6.15)-(6.17) admits a unique solution

e η(•) H ∈ H 1 (0, ∞), e η(•) h ∈ H 1 (0, ∞; H 1 (E)), (6.72) e η(•) q ∈ H 1 (0, ∞; L 2 (E)) ∩ C b ([0, ∞); H 1 (E)) ∩ L 2 (0, ∞; H 2 (E)), (6.73) e η(•) q a ∈ H 1 (0, ∞), e η(•) q b ∈ H 1 (0, ∞), (6.74)
E h(t, x) dx + H(t)(b -a) = 0 (t 0). (6.75)

Moreover, there exists a positive constant C depending on η such that

e η(•) H H 1 (0,∞) + e η(•) h H 1 (0,∞;H 1 (E)) + e η(•) q H 1 (0,∞;L 2 (E))∩C b ([0,∞);H 1 (E))∩L 2 (0,∞;H 2 (E)) + e η(•) (q a , q b ) H 1 (0,∞;R 2 ) C | H 0 | + h 0 H 1 (E) + q 0 H 1 (E) + e η(•) (f 1 , f 2 ) L 2 (0,∞;L 2 (E)×R 2 )
. (6.76)

Fixed point

Let us fix η ∈ (0, η 0 ), where η 0 is the constant introduced in Proposition 6.4. We consider for all ε > 0 the ball

B ε := (f 1 , f 2 ) ∈ L 2 (0, ∞; L 2 (E) × R 2 ) ; e η(•) (f 1 , f 2 ) L 2 (0,∞;L 2 (E)×R 2 ) ε (6.77)
and the map Ξ : (f 1 , f 2 ) ∈ B ε → (F 1 ( h, q), F 2 ( H, h, q, q b , q a )) (6.78) where ( H, h, q, q b , q a ) is the solution of (6.18)-(6.23), with (6.15)-(6.17) associated with (f 1 , f 2 ), and where F 1 and F 2 are given by (6.11)-(6.14). We take

| H 0 | + h 0 H 1 (E) + q 0 H 1 (E) ε (6.79)
so that (6.76) yields

e η(•) H H 1 (0,∞) + e η(•) h H 1 (0,∞;H 1 (E)) + e η(•) q H 1 (0,∞;L 2 (E))∩C b ([0,∞);H 1 (E))∩L 2 (0,∞;H 2 (E))
+ e η(•) (q a , q b ) H 1 (0,∞;R 2 ) Cε. (6.80)

Taking ε small enough, we deduce from the above estimate that

H - 1 2 H, h - 1 2 h ((t, x) ∈ (0, ∞) × E). (6.81)
From the above estimate and from the the Sobolev embeddings

H 1 (0, ∞) ⊂ L ∞ (0, ∞) and H 1 (E) ⊂ L ∞ (E), we deduce 1 h + h L ∞ ((0,∞)×E) + 1 h + h L ∞ (0,∞;H 1 (E)) C, e η(•) q L ∞ ((0,∞)×E)
Cε, e η(•) h

L ∞ ((0,∞)×E)
Cε, (6.82)

for some constant C indepedent of ε. For simplicity, we assume now that ε 1. (6.83)

We deduce from (6.11), (6.80) and the above estimate that

e η(•) F 1 ( h, q) L 2 (0,∞;L 2 (E)) Cε 2 . (6.84)
From (6.80) and (6.82), we have

e η(•) h ( h + h) ∂q ∂x L 2 (0,∞;H 1 (E))
Cε 2 , (6.85)

and thus e η(•) F 3 ( H, h, q, q b , q a ) F 4 ( H, h, q, q b , q a )

L 2 (0,∞;R 2 ) Cε 2 . (6.86)
We also deduce from (6.80) that

e η(•)    g( h(•, a -) -H) - µ h ∂q ∂x (•, a -) + µ H q b -q a b -a -g( h(•, b + ) -H) + µ h ∂q ∂x (•, a -) - µ H q b -q a b -a    L 2 (0,∞;R 2 )
Cε.

From (4.12), (6.81) and (6.80), we have

S( H + H) L ∞ (0,∞;R 4 ) C, S( H + H) -S(H) L ∞ (0,∞;R 4 ) Cε. (6.87)
We obtain from (6.12) and from the above estimates e η(•) F 2 ( H, h, q, q b , q a ) L 2 (0,∞;R 2 ) Cε 2 . (6.88)

This shows that for small ε, Ξ defined by (6.78) satisfies Ξ(B ε ) ⊂ B ε . With similar calculations, we can also obtain that for small ε, Ξ |Bε is a strict contraction. This gives the existence and uniqueness of a solution for system (4.17)-(4.22). We have now to come back to system (3.1)-(3.12). We see that (4.17 This concludes the proof of Theorem 3.3.

Return to the Equilibrium

The return to the equilibrium problem is a particular configuration of the floating structure problem, which consists in starting from a configuration where the solid is not at its equilibrium state and with water at rest. The motion of the solid is often described in the literature by a linear integro-differential equation, known as the Cummins equation, which has been obtained in [START_REF] Cummins | The impulse response function and ship motions[END_REF], in the case of an inviscid fluid filling the whole space and assuming small waves amplitude. A similar equation taking in consideration some nonlinear effects has been derived in [START_REF] Lannes | On the dynamics of floating structures[END_REF]. The classical linear Cummins equation, in the case of vertical displacements of a floating structure reads:

(M + m ∞ ) Ḧ(t) = αH + K * Ḣ, (7.1)
where H denotes the displacement of the structure from the equilibrium position, M denotes the mass of the structure, m ∞ is the added mass at infinite frequency, α is the hydrostatic stiffness and K is the radiation force impulse response function (see, for instance, [START_REF] Cummins | The impulse response function and ship motions[END_REF]). In this section, our aim is to derive a similar equation taking into consideration the viscosity of the fluid and possibly the presence of an exterior boundary for the fluid. To this aim, we use a linearized version of the system (3.1)-(3.12) as departure point.

Let h and H be the equilibrium height for the fluid and the solid respectively. Then we have

h = H + M b -a ,
where M is the mass of the solid (see (3.18)). For simplicity, let us assume that

h = 1, M = 1, g = 1, p = 1 b -a .
The system (3.1)-(3.12), linearized around the trajectory (H, h, q, p) = (h, H, 0, p), reads as

∂h ∂t + ∂q ∂x = 0, x ∈ E, (7.2)
∂q ∂t + ∂h ∂x -µ ∂ 2 q ∂x 2 = 0, x ∈ E, (7.3)
h(t, a -) -µ ∂q ∂x (t, a -) = p(t, a + ) + H(t) -µ ∂q ∂x (t, a + ) (7.4) h(t, b + ) -µ ∂q ∂x (t, b + ) = p(t, b -) + H(t) -µ ∂q ∂x (t, b -) (7.5) ∂H ∂t + ∂q ∂x = 0, x ∈ I, (7.6
)

∂q ∂t + ∂p ∂x = 0, x ∈ I, (7.7)
Ḧ(t) = b a p(t, x) dx. (7.8)
As we are interested in the return to the equilibrium problem we consider the above system with the following initial data

h(0, x) = h 0 , q(0, x) = 0, H(0) = H 0 , (7.9)
where h 0 is a constant, h 0 = 0, H 0 = 0 and

h 0 |E| + H 0 (b -a) = 0. (7.10)
The last condition means that the volume of fluid is the same in the initial configuration and at equilibrium and it is necessary only in the case when the fluid is bounded.

Remark 7.1. The system (7.2)-(7.8) is equivalent to the system (6.18)-(6.24). Therefore by Proposition 6.5, the system (7.2)-(7.9) admits a unique solution

H ∈ H 2 (0, ∞), h ∈ H 1 (0, ∞; H 1 (E)) ∩ C 1 b ([0, ∞); L 2 (E)), q ∈ C b ([0, ∞); H 1 (0, )) q |E ∈ H 1 (0, ∞; L 2 (E)) ∩ C b ([0, ∞); H 1 (E)) ∩ L 2 (0, ∞; H 2 (E)), q |I ∈ H 1 (0, ∞; P 1 (I)), p |I ∈ L 2 (0, ∞; P 2 (I)).
Our aim is to show that, in the above configuration, we can eliminate h, q and p from the above system to obtain an integro-differential equation of Cummins type for H. To this aim, we first differentiate (7.7) with respect to x and using (7.4) -(7.6) we have

             ∂ 2 p ∂x 2 = Ḧ(t), x ∈ I, p(t, a + ) = p a (t) := h(t, a -) -µ ∂q ∂x (t, a -) -H(t) -µ Ḣ(t), p(t, b -) = p b (t) := h(t, b + ) -µ ∂q ∂x (t, b + ) -H(t) -µ Ḣ(t). (7.11)
We decompose p as p = p 1 + p 2 , (7.12)

where p 1 solves Substituting the values of p 1 and p 2 in (7.8), we obtain the following equation

∂ 2 p 1 ∂x 2 = Ḧ(t), p 1 (t, a) = p 1 (t, b) = 0, (7.13
1 + 1 12 (b -a) 3 Ḧ(t) = p a (t)(b -a) + (p b (t) -p a (t)) b -a 2 = b -a 2 (p a (t) + p b (t)) .
Using the expression of p a and p b from (7.11) we can rewrite the above equation as

1 + 1 12 (b -a) 3 Ḧ(t) = b -a 2 h(t, a -) -µ ∂q ∂x (t, a -) + h(t, b + ) -µ ∂q ∂x (t, b + ) -(b -a) H(t) + µ Ḣ(t) . (7.17)
To simplify we assume symmetry of the fluid-structure configuration, i.e., we suppose that the solid is symmetric around the axis x = x 0 where x 0 = 1 2 (b + a) and that = a + b. Thus we have h(t, a -) = h(t, b + ) q(t, b + ) = -q(t, a -). (7.18) Then by taking the boundary trace of all the terms in (7.2), we obtain

∂q ∂x (t, a -) = ∂q ∂x (t, b + ).
Therefore, (7.17) can be written as

1 + 1 12 (b -a) 3 Ḧ(t) = (b -a) h(t, a -) -µ ∂q ∂x (t, a -)) -(b -a) H(t) + µ Ḣ(t) . (7.19)
We now need to express h(t, a -) -µ ∂q ∂x (t, a -) in terms of H and Ḣ. For x ∈ I we have q(t, b) -q(t, a) = -(b -a) Ḣ(t). (7.20) Using (7.18), we obtain

q(t, a) = b -a 2 Ḣ(t), q(t, b) = - b -a 2 Ḣ(t). (7.21) From (7.2)-(7.
3), q satisfy the following equation for

x ∈ [0, a] ∂ 2 q ∂t 2 - ∂ 2 q ∂x 2 -µ ∂ 3 q ∂t∂x 2 = 0, q(t, 0) = 0, q(t, a) = b -a 2 Ḣ(t), q(0, x) = ∂q ∂t (0, x) = 0. (7.22)
For f ∈ L 1 [0, ∞) let f be the Laplace transform of f . Applying this transform to both sides of (7.22) we obtain

s 2 q -(1 + sµ) ∂ 2 q ∂x 2 = 0, q(s, 0) = 0, q(s, a) = b -a 2 Ḣ, Re(s) > 0. (7.23) Therefore q(s, x) = A(s)e -s √ 1+sµ x + B(s)e s √ 1+sµ x , (7.24)
where z → √ z is the principal branch of the square root function.

Using the boundary conditions (7.23) it follows that

A(s) + B(s) = 0, A(s)e -sa √ 1+sµ + B(s)e sa √ 1+sµ = b -a 2 Ḣ.
Therefore, (7.24) yields

q(s, x) = b -a 2 e -sx √ 1+sµ -e sx √ 1+sµ e -sa √ 1+sµ -e sa √ 1+sµ Ḣ(s) (x ∈ [0, a], Re s > 0). ( 7

.25)

From the above formula it follows that

∂ q ∂x (s, a) = - b -a 2 f a (s) Ḣ(s) (Re s > 0), (7.26)
where

f a (s) = s √ 1 + sµ e -sa √ 1+sµ + e sa √ 1+sµ e -sa √ 1+sµ -e sa √ 1+sµ
(Re s > 0).

On the other hand, taking Laplace transform of (7.2) we have The function F growths like √ Re s when Re s → ∞, thus F is a distribution. Consequently, the second term in the right hand side of the first equation in (7.30) involves fractional derivatives of H, as explained with more details below. More precisely, in the remaining part of this section we consider the case in which the fluid domain is unbounded and we derive a generalized Cummins equation, taking in consideration the viscosity.

h(s, a) = h 0 s - 1 s ∂ q ∂x (s, a). Therefore, h(s, a) -µ ∂ q ∂x (s, a) = h 0 s - b -a 2 F (s) Ḣ(s) with F (s) = - 1 s + µ f a (s). ( 7
Let E = (-∞, a) ∪ (b, ∞) and take the initial data (for the linearized problem (7.2)-(7.9)) h 0 = 0, q 0 = 0, H 0 = 0.

In this case, for x ∈ (-∞, a) we have the following equation satisfied by q(s, x) (instead of (7.23)):

s 2 q -(1 + sµ)
∂ 2 q ∂x 2 = 0, q(s, x) → 0 as x → -∞, q(s,

Numerical Simulations

In this section, we discuss some numerical simulations for the system nonlinear (4.17)-(4.22) and we compare with the results obtained via linearization. To this aim, consider some subdivisions of the intervals [0, a] and [b, ], and we use them to discretize the equations with respect to the space variable. Since the procedure is similar for the two intervals, we only present the method for the interval [0, a]. To simplify, we also assume that the subdivision (x k ) of [0, a] is uniform:

x k = k n a, k ∈ {0, . . . , n}, δx = a n .

We also consider the subdivision ( x k ):

x k = k -1/2 n a, k ∈ {1, . . . , n}.

Our aim is compute approximations of h and q respectively on ( x k ) and on (x k ):

h k ≈ h(•, x k ) q k ≈ q(•, x k ).
In particular, q 0 = 0 and q n ≈ q a where q a is one of the unknowns of (4.17)-(4.22). With the above notation, it is natural to approximate the derivative of q by the following formula:

∂q ∂x k = q k -q k-
where

F k = q 2 k h k + g h 2 k 2 .
In the above formulas, we see that we need to compute q k and h k . For the height of fluid, we consider the formula

h k = h k+1 + h k 2 , k ∈ {1, . . . , n -1},
and for q we use an upwind approximation:

q k = q k+1 + q k 2 or q k + q k-1 2
, k ∈ {1, . . . , n -1}, according to the sign of q k+1 +q k 2 .

Note that in (8.1), we don't need to compute the derivative of q k for k = 0 (q 0 = 0) or for k = n since q n corresponds to q a and its derivative is obtained through the equation (4.22).

Using the above numerical scheme, we can consider the return to the equilibrium problem, in its linear and nonlinear version. We place ourselves in the configuration described in Section 7. In particular, we take the initial conditions (7.9), with different values of h 0 , H 0 and µ (see Figure 2 and Figure 3).

3

  Energy estimate and main result 8

  Figure 1: Configuration

•

  H) L 2 (0,T ;R 4 ) Estimate p(t, a) and p(t, b) defined in (4.15) -(4.16): We claim that p(•, a) L 2 (0,T ) + p(•, b) L 2 (0,T ) CT δ , for some δ > 0.

  ) yields (3.1) in I and there exists a unique p ∈ M g |I| + L 2 (0, ∞; P 2 (I)) satisfying (4.4) and b a p(t, x) dx = -M Ḧ(t) + M g = M qb -qa b -a + M g. Using (4.5) and (4.6), we deduce that p(t, a) and p(t, b) satisfies (4.9) and (4.10). We deduce from this and from (4.22) that (3.7) and (3.8) hold.

-a 2 F 2 F 7 . 2 .

 2272 .27) Let F (t) denotes the inverse Laplace transform of F (s). By taking the inverse Laplace transform in the above identity we obtainh(t, a -) -µ ∂q ∂x (t, a -) = h 0 -b a) 3 Ḧ = (b -a)h 0 -(b -a) 2 * Ḣ -(b -a)H -µ(b -a) Ḣ. (7.29)Therefore, we have proved the following propositionProposition Assume that = a + b, i.e.that the solid is symmetric around the axis x = x 0 , x 0 = 1 2 (a + b). Then, for the return to the equilibrium problem (7.2)-(7.10), the position of the solid is completely determined by the integro-differential equation a)3 Ḧ = -H 0 |I| 2 |E| -(b -a) 2 2 F * Ḣ -(b -a)H -µ(b -a) Ḣ, H(0) = H 0 , Ḣ(0) = 0, (7.30) and with F such that F (s) = -1 + sµ e

  the above expressions in(7.19), the equation for H reads as1 + 1 12 (b -a) 3 Ḧ = -(b -a) 2 F * Ḣ -(b -a)H -µ(b -a) Ḣ,(7.36)

Figure 2 :Figure 3 :

 23 Figure 2: H 0 = 9, µ = 5, 0.5, 50 34

* Takéo Takahashi acknowledges the support of the Agence Nationale de la Recherche, project IFSMACS, ANR-15-CE40-0010. † Debayan Maity and Marius Tucsnak acknowledge the support of the Agence Nationale de la Recherche -Deutsche Forschungsgemeinschaft (ANR -DFG), project INFIDHEM, ID ANR-16-CE92-0028.

1

where F (s) = 1 + µs.

(7.37) Equation (7.36) can be further transformed in an equation involving a fractional derivative. Indeed, (7.37) can be rewritten

where F 1 (s) = √ µs, (7.39)

Using tables of Laplace transforms we see that

On the other hand, (7.39) implies that F 1 is (up to a multiplicative constant) the fractional derivative of order 1 2 of the Dirac mass at t = 0, that is

Combining (7.36)-( 7.42) we obtain the Cummins type equation

which is the same as Eq. (57) of [START_REF] Lannes | On the dynamics of floating structures[END_REF].

We notice in both experiments that the convergence to the equilibrium is depending a lot of the viscosity. When the viscosity is large, both linear and nonlinear problem converges to the equilibrium very quickly. For instance, if µ = 50 we see that at t = 100 the solutions are close to the equilibrium. On the contrary, when the viscosity is small (µ = 5 or 0.5) trajectories do not converge to the equilibrium as fast as the previous one. Moreover, these simulations seem to support the conjecture that the return to equilibrium takes place without imposing initial data close to equilibrium.

We also see that the differences between the nonlinear problem and the linear problem are stronger for small viscosities, and we remark that, at least for these numerical tests, the nonlinear system goes faster than the linear system to the equilibrium.