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FEEDBACK STABILIZATION OF A FLUID–STRUCTURE MODEL

JEAN-PIERRE RAYMOND

Abstract. We study a system coupling the incompressible Navier-Stokes equations in a 2D rectangular type domain with a damped Euler-Bernoulli beam equation, where the beam is a part of the upper boundary of the domain occupied by the fluid. Due to the deformation of the beam the fluid domain depends on time. We prove that this system is exponentially stabilizable, locally about the null solution, with any prescribed decay rate, by a feedback control corresponding to a force term in the beam equation. The feedback is determined, via a Riccati equation, by solving an infinite time horizon control problem for the linearized model. A crucial step in this analysis consists in showing that this linearized system can be rewritten thanks to an analytic semigroup of which the infinitesimal generator has a compact resolvent.

Key words. Fluid-structure interaction, feedback control, stabilization, Navier-Stokes equations, Beam equation AMS subject classifications. 93B52, 93C20, 93D15, 35Q30, 76D55, 76D05, 74F10

1. Setting of the problem. Let Ω be the rectangular domain (0, L) × (0, 1) ⊂ R

2

, with boundary Γ. Let us set Γ

s

= (0, L) × {1}, the upper part of the boundary of Ω, and Γ

0

= Γ \ Γ

s

. For a given function η from Γ

s

× (0, ∞) into (−1, ∞), we denote by Ω

η(t)

and Γ

s,η(t)

the sets

η(t)

= n

(x, y) | x ∈ (0, L), 0 < y < 1 + η(x, t) o , Γ

s,η(t)

= n

(x, y) | x ∈ (0, L), y = 1 + η(x, t) o . For 0 < T < ∞ or T = ∞, we also use the notation

Σ

0T

= Γ

0

× (0, T ), Σ

T

= Γ × (0, T ), Q

T

= Ω × (0, T ), Q e

T

= S

t∈(0,T)

η(t)

× {t}, Σ

sT

= Γ

s

× (0, T ), Σ e

sT

= S

t∈(0,T)

Γ

s,η(t)

× {t}.

We consider the following fluid-structure model coupling the Navier-Stokes equations with a damped Euler-Bernoulli beam equation:

u

t

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

,

u = η

t

~ e

2

on Σ e

s

, u = 0 on Σ

0

, u(0) = u

0

in Ω

η(0)

= Ω

η0 1

, η

tt

− βη

xx

− δη

txx

+ αη

xxxx

= ρ

1

p + H (u, η) + f on Σ

s

, η = 0 and η

x

= 0 on

0, L × (0, ∞), η(0) = η

10

and η

t

(0) = η

02

in Γ

s

,

(1.1)

with

H (u, η) = −ρ

2

ν (∇u + ∇u

T

)(−η

x

~ e

1

+ ~ e

2

) · ~ e

2

,

σ(u, p) = ν (∇u + ∇u

T

) − p I, ~ e

1

= (1, 0), ~ e

2

= (0, 1).

In this setting ν > 0 is the fluid viscosity, α > 0, β ≥ 0, and δ > 0 are the adimensional rigidity, stretching, and friction coefficients of the beam, ρ

1

and ρ

2

are positive constants related to the density of the fluid and the density of the structure (see [4]), f is a control function. Our objective is to determine f in feedback form, able to stabilizes the system (1.1) (in an appropriate space) with a prescribed exponential decay rate −ω < 0, locally about (0, 0, 0, 0). Existence of a local strong solution for system (1.1) with f = 0 has been proved in [4] (with periodic boundary conditions on the lateral boundary of Ω), under smallness conditions on the data, while existence of Hopf solutions for a slightly different model is proved

Universit´e de Toulouse, UPS, Institut de Math´ematiques, 31062 Toulouse Cedex 9, France,

& CNRS, Institut de Math´ematiques, UMR 5219, 31062 Toulouse Cedex 9, France, email: raymond@math.univ-toulouse.fr 1

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in [13] (see also [8] and [12] for other models for the beam equation and for existence results in the three dimensional case). To the author knowledge nothing is known about control and stabilization of such a system. To study the control system (1.1), as in [4], we make a change of variable in order to rewrite system (1.1) in the cylindrical domain Ω × (0, ∞) and we denote by (ˆ u, p) the image of (u, p) by this ˆ transformation. Since we are looking for solutions satisfying a prescribed exponential decay rate −ω, we rewrite the system as a first order system by setting η = η

1

and η

t

= η

2

and we study the control system satisfied by (˜ u, p, ˜ η ˜

1

, η ˜

2

) = e

ωt

(ˆ u, p, η ˆ

1

, η

2

). We linearize the system satisfied by (˜ u, p, ˜ η ˜

1

, η ˜

2

) about (0, 0, 0, 0) and we determine a feedback control, able to stabilize the linearized system satisfied by (v, p, η

1

, η

2

), by solving an infinite time horizon control problem. Next we prove that this linear feedback law, applied in the nonlinear system satisfied by (˜ u, p, ˜ η ˜

1

, η ˜

2

), is able to stabilize the nonlinear system provided that the intial condition is small enough in an appropriate norm.

The analysis that we do for the linearized system is completely new for this type of fluid–structure system. Indeed we show that the linearized system satisfied by (v, p, η

1

, η

2

) is equivalent to a system of the form

d dt

 P v

η

1

η

2

= A

ω

 Pv

η

1

η

2

 + B

 0 ¯ 0 f

 ,

 P v(0)

η

1

(0) η

2

(0)

=

 Pv

0

η

10

η

20

 ,

(I − P )v(t) = (I − P )D(η

2

(t) ~ e

2

χ

Γs

),

(1.2)

where P is the so-called Leray projector and D is the Dirichlet operator associated with the stationary Stokes equation (P and D are defined in section 3, while A

ω

and B are defined in section 4). This type of decomposition of velocity fields, into Pv and (I − P)v, has already been introduced for the Navier- Stokes equations with nonhomogeneous boundary conditions in [22]. Finding again this decomposition for system (1.2) is not totally obvious because the pressure, which is eliminated in the Navier-Stokes equations thanks to the projector P , also appears in the beam equation. Rewriting the system satisfied by (v, p, η

1

, η

2

) in the form (1.2) is crucial to prove the stabilizability of this system. Indeed, we show that the operator (A

ω

, D(A

ω

)) is the infinitesimal generator of an analytic semigroup on the space H = V

n0

(Ω) × (H

02

s

) ∩ L

20

s

)) × L

20

s

) and has a compact resolvent in this space (for the precise definitions of these spaces we refer to section 3). We show that the stabilizability of system (1.2) reduces to proving an approximate controllability result for a projected system. Such an approximate controllability result can be deduced from [18] in the case of a rectangular domain (see also [19, 20] for supplementary approximate controllability results).

The plan of the paper is as follows. Section 2 is devoted to rewriting system (1.1) in a fixed domain and to the obtention of a linearized system. We study the semigroup of the linearized system and properties of its infinitesimal generator in section 3. Existence and regularity results for the linearized system are stated in section 4. We study the stabilizability of the linearized system in section 5. Three feedback control laws for the linearized system (1.2) are introduced in section 6. The first one is a feedback law for system (1.2) written as a system of partial differential equations, involving the pressure (see system (2.6)). The second one is a feedback law obtained by the classical approach introduced in [2] or in [15].

In that case the pressure is eliminated since it does not appear in (1.2). The corresponding feedback law is defined via the solution to a Riccati equation of the form

Π e ∈ L(H), Π = e Π e

≥ 0, ΠA e

ω

+ A

ω

Π e − ΠBB e

Π + e C

C = 0.

(See equation (6.2) for the definition of C.) Since A

ω

, which is determined in section 3.5, cannot be interpreted only in terms of partial operators (contrarily to A

ω

), we introduce a third feedback law obtained by solving a Riccati equation of the form

Π b ∈ L( H), b Π = b Π b

≥ 0, ΠA b

ω

+ A

]ω

Π b − ΠBB b

]

Π + b I = 0,

where H b is the space H equipped with another inner product (see section 3.5), A

]ω

∈ L( H) is the adjoint b of A

ω

∈ L( H) and b B

]

∈ L( H, L b

20

s

)) is the adjoint of B ∈ L(L

20

s

), H). The main interest of this b approach is that A

]ω

can be interpreted in terms of partial differential operators (which can be helpful for

2

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numerical calculations). Moreover, we are able to establish the precise relationship between the feedback operators obtained by the first approach and the third one.

The optimal control problems corresponding to the first approach are studied in details in sections 7 and 8.1. In these sections, all the calculations are made in a very simple way via integrations by parts.

Therefore they can be easily checked and do not need a sophisticated functional analysis framework.

However the feedback law corresponding to the first approach is expressed in terms of an operator Π which is not, at that stage, characterized by a Riccati equation. This is why the third approach is helpful even if in that case the representation of the state and adjoint systems via A

ω

and A

]ω

cannot be avoided.

To deal with the nonlinear closed loop system, we first study the nonhomogeneous linearized closed loop system in section 9. The main results of the paper are stated in section 10 (Theorems 10.2 and 10.3). Some Lipschitz properties of the nonlinear terms in the nonlinear system are established in section 11. These properties are next used in section 12 in the proof of the main results.

Let us finally give some references which are connected to the present work. The control of a channel flow with periodic boundary conditions have been studied in [5, 31, 32, 33]. We think that the results in those papers may be very useful to study the control of a channel flow coupled with a beam equation, with periodic boundary conditions at the lateral boundary {0} × [0, L] ∪ {L} × [0, L]. This will be investigated in a future work. Let us also mention some controllability results obtained for systems coupling the Navier-Stokes equations with finite dimensional solid-structure models [6, 21, 26] (and see also [25] for a simplified model). These controllability resuls are mainly based on results first obtained for the Navier- Stokes equations in [3]. In those models the controls act in the fluid equation and not in the structure equation as in (1.1). Thus the problems are quite different. The feedback stabilization of the Navier- Stokes equations in the three dimensional case is studied in [24]. It can be a starting point to study the stabilization of systems similar to (1.1) in the 3D case.

2. The linearized system. The solutions to system (1.1) obey 0 =

Z

η(t)

div u(t) = Z

Γs,η(t)

u(t) · n(t) = Z

Γs

η

t

(t) = Z

L

0

η

t

(x, t)dt, since the unit normal to Γ

s,η(t)

outward Ω

η(t)

is

n(t) = −η

x

(t)

p 1 + η

x2

(t) , 1 p 1 + η

2x

(t)

!

T

.

Thus we must choose η

20

in the space

L

20

s

) = n

η ∈ L

2

s

) | Z

Γs

η = 0 o .

If η

01

also belongs to L

20

s

), then we have Z

Γs

η(t) = 0 and Z

Γs

η

t

(t) = 0 for all t ≥ 0.

Everywhere throughout the paper we shall choose η

10

and η

02

with zero mean value over Γ

s

. If we denote by M

s

the orthogonal projection in L

2

s

) onto L

20

s

), the equation satisfied by η in system (1.1) must be written in the form

η

tt

− βη

xx

− δη

txx

+ αM

s

xxxx

) = M

s

1

p + H (u, η) + f ) on Σ

s

. Observe that due to the boundary conditions

η = 0 and η

x

= 0 on

0, L × (0, ∞), we have (for solutions regular enough and when η

01

and η

20

belong to L

20

s

))

Z

Γs

η

tt

= 0, Z

Γs

η

xx

= 0, and Z

Γs

η

txx

= 0,

3

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but we do not necessarily have

Z

Γs

η

xxxx

= 0.

This is why, in the equation satisfied by η, we have to write M

s

xxxx

) in place of η

xxxx

. But for simplicity we shall skip the writing of M

s

in the different equations, except if we want to stress on the role of the operator M

s

(which is for example the case when we shall define the operator (A

ω

, D(A

ω

))).

We consider system (1.1) for initial conditions u

0

such that div u

0

= u

01,x

+ u

02,y

= 0 and obeying the compatibility condition

u

0

= 0 on Γ

0

, u

0

(x, 1 + η(x, 0)) = u

0

(x, 1 + η

10

(x)) = η

02

(x)~ e

2

for x ∈ (0, L). (2.1) As in [4], for a given function η : (0, L) × (0, T ) 7→ R satisfying η > −1, we consider the changes of variables

T

η

: (x, y, t) 7−→ (x, z, t) =

x, y

1 + η(x, t) , t

and T

η(t)

: (x, y) 7−→ (x, z) =

x, y

1 + η(x, t)

. (2.2) The mapping T

η0

1

is defined in a similar way. The mapping T

η(t)

transforms Ω

η(t)

into Ω = (0, L) × (0, 1).

Setting

ˆ

u(x, z, t) = u(x, y, t), p(x, z, t) = ˆ p(x, y, t), the nonlinear system (1.1) is rewritten in the form

ˆ

u

t

+ (ˆ u · ∇)ˆ u − ν∆ˆ u − ∇ˆ p = ˆ F (ˆ u, p, η), ˆ div ˆ u = ˆ G(ˆ u, η) in Q

, u ˆ = η

t

~ e

2

on Σ

s

, u ˆ = 0 on Σ

0

, u(0) = ˆ ˆ u

0

in Ω,

η

tt

− βη

xx

− δη

txx

+ αη

xxxx

= ρ

1

p ˆ + ˆ H (ˆ u, η) + f on Σ

s

, η = 0 and η

x

= 0 on

0, L × (0, ∞), η(0) = η

10

and η

t

(0) = η

20

in Γ

s

,

(2.3)

where ˆ u

0

(x, z) = u

0

(x, y) = u

0

(x, z(1 + η(x, 0))) = u

0

(x, z(1 + η

01

(x))) = u

0

◦ T

η−10 1

(x, z), F(ˆ ˆ u, p, η) ˆ

= −η u ˆ

t

+

t

+ νz

η2 x

1+η

− η

xx

ˆ u

z

−2zη

x

u ˆ

xz

+ η u ˆ

xx

+

z2ηx2−η 1+η

ˆ u

zz

+z(η

x

p ˆ

z

− η p ˆ

x

)~ e

1

− (1 + η)ˆ u

1

u ˆ

x

+ (zη

x

u ˆ

1

− u ˆ

2

)ˆ u

z

,

G(ˆ ˆ u, η) = −η u ˆ

1,x

+ zη

x

u ˆ

1,z

= div ( ˆ w) with ˆ w = −η u ˆ

1

~ e

1

+ zη

x

u ˆ

1

~ e

2

, and

H ˆ (ˆ u, η) = ρ

2

ν

η

x

1+η

u ˆ

1,z

+ η

x

u ˆ

2,x

2+η1+η2x

u ˆ

2,z

= −2ρ

2

ν u ˆ

2,z

+ ρ

2

ν

η

x

1+η

u ˆ

1,z

+ η

x

u ˆ

2,x

η2x1+η−2η

u ˆ

2,z

. Due to (2.1), we can see that

div (ˆ u

0

− w(0)) = 0 in Ω, ˆ u ˆ

0

− w(0) = 0 on Γ ˆ

0

, u ˆ

0

− w(0) = ˆ η

20

~ e

2

on Γ

s

. (2.4) For −ω < 0, we make the following change of variables:

u ˜ = e

ωt

u, ˆ p ˜ = e

ωt

p, ˆ η ˜

1

= e

ωt

η, η ˜

2

= e

ωt

η

t

.

4

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The system (2.3) is transformed into

u ˜

t

+ e

−ωt

(˜ u · ∇)˜ u − ν∆˜ u − ∇ p ˜ − ω˜ u = e

−ωt

F ˜ (˜ u, p, ˜ η ˜

1

, η ˜

2

), div ˜ u = e

−ωt

G(˜ ˜ η

1

, u) ˜ in Q

,

˜

u = ˜ η

2

~ e

2

on Σ

s

, u ˜ = 0 on Σ

0

, u(0) = ˆ ˜ u

0

in Ω,

˜

η

1,t

= ˜ η

2

+ ω η ˜

1

on Σ

s

,

˜

η

2,t

− ω η ˜

2

− β η ˜

1,xx

− δ η ˜

2,xx

+ α˜ η

1,xxxx

= ρ

1

p ˜ − 2νρ

2

u ˜

2,z

+ e

−ωt

H(˜ ˜ u, η ˜

1

) + ˜ f on Σ

s

,

˜

η

1

= 0 and η ˜

1,x

= 0 on

0, L × (0, ∞),

˜

η

1

(0) = η

01

and η ˜

2

(0) = η

20

in Γ

s

,

(2.5)

with

f ˜ = e

ωt

f,

F ˜ (˜ u, p, ˜ η ˜

1

, η ˜

2

) = −˜ η

1

(˜ u

t

− ω u) + ˜

z η ˜

2

+ νz

η˜2 1,x

eωt+ ˜η1

− η ˜

1,xx

˜ u

z

−2z η ˜

1,x

u ˜

xz

+ ˜ η

1

u ˜

xx

+

z2η˜21,x−e−ωtη˜1 eωt+ ˜η1

˜ u

zz

+z(˜ η

1,x

p ˜

z

− η ˜

1

p ˜

x

)~ e

1

− (1 + e

−ωt

η ˜

1

)˜ u

1

u ˜

x

+ (ze

−ωt

η ˜

1,x

u ˜

1

− u ˜

2

)˜ u

z

, G(˜ ˜ η

1

, u) = ˜ −˜ η

1

u ˜

1,x

+ z η ˜

1,x

u ˜

1,z

= div (−˜ η

1

u ˜

1

~ e

1

+ z˜ η

1,x

u ˜

1

~ e

2

) , H ˜ (˜ u, η ˜

1

) = ν

e−ωt˜η1,x

eωt+ ˜η1

u ˜

1,z

+ ˜ η

1,x

u ˜

2,x

η˜

2 1,x

eωt+ ˜η1

u ˜

2,z

+

2eeωt−ωt+ ˜ηη˜11

u ˜

2,z

.

If we linearize (2.5) about (0, 0, 0, 0), we obtain the system v

t

− div σ(v, p) − ωv = 0,

div v = 0 in Q

,

v = η

2

~ e

2

on Σ

s

, v = 0 on Σ

0

, v(0) = v

0

in Ω, η

1,t

= η

2

+ ωη

1

on Σ

s

,

η

2,t

− ωη

2

− βη

1,xx

− δη

2,xx

+ αM

s

η

1,xxxx

= M

s

1

p − 2νv

2,z

+ f ) on Σ

s

, η

1

= 0 and η

1,x

= 0 on

0, L × (0, ∞), η

1

(0) = η

01

and η

2

(0) = η

20

in Γ

s

.

(2.6)

Observe that

v

1,x

+ v

2,z

= 0 implies v

2,z

|

Γs

= 0,

if for example v belongs to L

2

(0, ∞; H

2

(Ω)). This is why the term −2νv

2,z

will be dropped out from the equation satisfied by η

2

. Let us notice that ˜ u

2,z

cannot be dropped out in system (2.5).

3. Definition of an analytic semigroup.

3.1. Transformation of system (2.6). Let us recall that L

2

(Ω) = L

2

(Ω; R

2

) admits the following orthogonal decomposition

L

2

(Ω) = V

0n

(Ω) ⊕ grad H

1

(Ω), with

V

0n

(Ω) = n

y ∈ L

2

(Ω) | div y = 0, y · n = 0 on Γ o ,

5

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and let us denote by P : L

2

(Ω) 7−→ V

0n

(Ω) the so-called Leray or Helmholtz projector. We also introduce the notations

V

0

(Ω) = n

y ∈ L

2

(Ω) | div y = 0 o

, H

10

(Ω) = H

01

(Ω; R

2

), H

2

(Ω) = H

2

(Ω; R

2

), V

2

(Ω) = H

2

(Ω) ∩ V

0

(Ω), V

10

(Ω) = H

10

(Ω) ∩ V

0n

(Ω), V

−1

(Ω) = (V

10

(Ω))

0

, L

20

(Ω) = n

p ∈ L

2

(Ω) | R

p = 0 o

, H

σ

(Ω) = H

σ

(Ω) ∩ L

20

(Ω), V

nσ

(Ω) = H

σ

(Ω) ∩ V

0n

(Ω) for σ ≥ 0, for σ < 0, H

σ

(Ω) = (H

−σ

(Ω))

0

, (H

−σ

(Ω))

0

is the dual of H

−σ

(Ω) with L

20

(Ω) as pivot space, L

20

s

) = n

η ∈ L

2

s

) | R

Γs

η = 0 o

, L

20

(Γ) = n

π ∈ L

2

(Γ) | R

Γ

π = 0 o , H

σ

s

) = H

σ

s

) ∩ L

20

s

) and H

σ

(Γ) = H

σ

(Γ) ∩ L

20

(Γ) for σ ≥ 0,

for σ < 0, H

σ

(Γ) = (H

−σ

(Γ))

0

where (H

−σ

(Γ))

0

is the dual of H

−σ

(Γ) with L

20

(Γ) as pivot space, for σ < 0, H

σ

s

) = (H

−σ

s

))

0

, (H

−σ

s

))

0

is the dual of H

−σ

s

) with L

20

s

) as pivot space.

We denote by A

0

= νP ∆ the Stokes operator in V

0n

(Ω) with domain D(A

0

) = V

2

(Ω) ∩ V

10

(Ω).

It is well known that, by the extrapolation method, the Stokes operator can be extended as an unbounded operator in (V

2

(Ω) ∩ V

01

(Ω))

0

with domain V

0n

(Ω). This extension will be still denoted by A

0

, and we shall see that it does not lead to confusion. The operator P may also be extended to a bounded operator from H

−1

(Ω) (the dual of H

10

(Ω) with L

2

(Ω) as pivot space) to V

−1

(Ω) (the dual of V

10

(Ω) with V

0n

(Ω) as pivot space) by the formula

hP u, Φi

V−1(Ω),V10(Ω)

= hu, Φi

H−1(Ω),H10(Ω)

for all Φ ∈ V

10

(Ω).

In that case P is a projector in H

−1

(Ω) but no longer an orthogonal projector.

We only need to consider system (2.6) in the case when ω = 0. Following [22], it is convenient to rewrite the equation satisfied by v in system (2.6) (for ω = 0) into two equations, one satisfied by P v and the other one by (I − P)v. More precisely we have

P v

0

= A

0

P v + (−A

0

)P D(η

2

~ e

2

χ

Γs

), v(0) = v

0

in Ω, (I − P )v(t) = (I − P)D(η

2

(t)~ e

2

χ

Γs

).

In this setting A

0

is the Stokes operator in (V

2

(Ω) ∩ V

10

(Ω))

0

with domain V

0n

(Ω), χ

Γs

denotes the characteristic function of Γ

s

, and D is defined by Dg = w, where (w, q) is the solution to the Dirichlet problem

−ν∆w + ∇q = 0 and div w = 0 in Ω, w = g on Γ.

We shall also set

D

s

η

2

= D(η

2

~ e

2

χ

Γs

).

This rewriting is a way to eliminate the pressure in the equation satisfied by v. However, since the pressure p also appears in the equation satisfied by η

2

, we have to express p in terms of P v and (I −P )v.

For that we can notice that (I − P)v is the gradient of the function q ∈ H

1

(Ω) solution to the Neumann problem

∆q(t) = 0 in Ω, ∂q(t)

∂n = η

2

(t) on Γ

s

, ∂q(t)

∂n = 0 on Γ

0

. (3.1)

We denote by N

s

∈ L(L

20

s

), H

3/2

(Ω)) the operator defined by N

s

η

2

(t) = q(t). In [22] it is shown that the pressure p appearing in the first equation in (2.6) satisfies

p = π − q

t

,

6

(7)

where q

t

is the time derivative of q and π(t) is the solution of the other Neumann problem

∆π(t) = 0 in Ω, ∂π(t)

∂n = ν∆Pv(t) · n on Γ. (3.2)

Let us notice that ∆P v(t) · n is well defined in H

−1/2

(Γ) if ∆P v(t) belongs to L

2

(Ω). Indeed in that case ∆P v(t) ∈ L

2

(Ω) and div (∆Pv(t)) = 0 ∈ L

2

(Ω). Moreover h∆P v(t) · n, 1i

H−1/2(Γ),H1/2(Γ)

(see [28, Chapter 1, Theorem 1.2]). Therefore if the solution to system (2.6) is such that P v ∈ L

2

(0, ∞; V

2

(Ω)), the solution π to the above Neumann problem belongs to L

2

(0, ∞; H

1

(Ω)). We de- note by N

0

∈ L(H

−1/2

(Γ), H

1

(Ω)) the operator defined by N

0

(ν∆Pv(t) · n) = π(t), when ∆P v(t) · n ∈ H

−1/2

(Γ). We denote by γ

s

the modified trace operator on Γ

s

defined by

γ

s

p = M

s

(p|

Γs

) = p|

Γs

− 1

s

| Z

Γs

p for all p ∈ H

σ

(Ω) with σ > 1/2.

Thus we have

M

s

(p(t)|

Γs

) = M

s

(π(t) − q

t

(t))|

Γs

= ν γ

s

N

0

∆P v(t) · n − γ

s

N

s

η

2,t

(t).

We can now rewrite the equation satisfied by η

2

in (2.6) in the form

(I + ρ

1

γ

s

N

s

2,t

− ωη

2

− βη

1,xx

− δη

2,xx

+ αM

s

η

1,xxxx

= ρ

1

νγ

s

N

0

∆P v(t) · n + M

s

f on Σ

s

. Lemma 3.1. The operator I + ρ

1

γ

s

N

s

is an automorphism in L

20

s

).

Proof. The operator γ

s

N

s

, considered as an operator belonging to L(L

20

s

)), is symmetric, positive, and compact. Indeed if q = N

s

η and ˜ q = N

s

η, we have ˜

0 = Z

∆q q ˜ = Z

Γs

η γ

s

N

s

η ˜ − Z

Γs

γ

s

N

s

η η, ˜ for all η, η ˜ ∈ L

20

s

). Thus γ

s

N

s

is symmetric. Moreover

0 = Z

∆q q = − Z

|∇q|

2

+ Z

Γs

η γ

s

N

s

η,

from which we deduce that γ

s

N

s

is nonnegative. If 0 =

Z

Γs

η γ

s

N

s

η = Z

|∇q|

2

,

we have q = C = 0 and

∂q∂n

= η = 0, which proves that γ

s

N

s

is positive. Since γ

s

N

s

∈ L(L

20

s

), H

1

(Γ)), it is clear that γ

s

N

s

is a compact operator in L

20

s

). Thus I + ρ

1

γ

s

N

s

is symmetric and positive and it is an automorphism in L

20

s

).

In order to write the system satisfied by (Pv, η

1

, η

2

) as an evolution equation, we introduce the unbounded operator (A

α,β

, D(A

α,β

)) in L

20

s

) defined by

D(A

α,β

) = H

4

s

) ∩ H

02

s

) ∩ L

20

s

), A

α,β

η = βη

xx

− αM

s

η

xxxx

.

Let us notice that (A

α,β

, D(A

α,β

)) is a selfadjoint operator in L

20

s

). Since A

α,β

is an isomorphism from D(A

α,β

) to L

20

s

), it can be extended as an isomorphism from L

20

s

) to (D(A

α,β

))

0

(the dual of D(A

α,β

) with L

20

s

) as pivot space), and from H

02

s

) ∩ L

20

s

) into (H

02

s

) ∩ L

20

s

))

0

. The space

H = V

0n

(Ω) × (H

02

s

) ∩ L

20

s

)) × L

20

s

) will be equipped with the inner product

(v, η

1

, η

2

), (w, ζ

1

, ζ

2

)

H

= ρ

1

(v, w)

Vn0(Ω)

+ (η

1

, ζ

1

)

H2

0s)

+ (η

2

, ζ

2

)

L2

0s)

,

7

(8)

with

1

, ζ

1

)

H2

0s)

= Z

Γs

(−A

α,β

)

1/2

η

1

(−A

α,β

)

1/2

ζ

1

= Z

Γs

(βη

1,x

ζ

1,x

+ αη

1,xx

ζ

1,xx

) dx.

We define the unbounded operator (A, D(A)) in H by D(A) =

n

(P v, η

1

, η

2

) ∈ V

2n

(Ω) × (H

4

∩ H

02

∩ L

20

)(Γ

s

) × (H

02

∩ L

20

)(Γ

s

) | P v − P D

s

η

2

∈ V

2

(Ω) ∩ V

10

(Ω) o , and

A =

I 0 0

0 I 0

0 0 (I + ρ

1

γ

s

N

s

)

−1

A

0

0 (−A

0

)P D

s

0 0 I

ρ

1

ν γ

s

N

0

(∆(·) · n) A

α,β

δ∆

s

 ,

where ∆

s

=

∂x22

s

. We define the unbounded operator (A

s

, D(A

s

)) in H

s

= (H

02

s

) ∩ L

20

s

)) × L

20

s

) by A

s

=

0 I

A

α,β

δ∆

s

!

, D(A

s

) = (H

4

s

) ∩ H

02

s

) ∩ L

20

s

)) × (H

02

s

) ∩ L

20

s

)).

It can be easily shown that A

s

is an isomorphism from D(A

s

) into H

s

. Now, it is clear that, for ω = 0, we can rewrite system (2.6) in the form

d dt

 P v

η

1

η

2

= A

 P v

η

1

η

2

 ,

 Pv(0)

η

1

(0) η

2

(0)

=

 Pv

0

η

01

η

02

 ,

(I − P )v(t) = (I − P)D(η

2

(t) ~ e

2

χ

Γs

).

The rewriting of system (2.6) when ω 6= 0 is done in (4.1).

Proposition 3.2. The norm

(Pv, η

1

, η

2

) 7−→ k(Pv, η

1

, η

2

)k

H

+ kA

0

P v + (−A

0

)P D

s

η

2

k

V0

n(Ω)

+ kA

s

1

, η

2

)k

Hs

is a norm on D(A) equivalent to the norm (Pv, η

1

, η

2

) 7−→ kPvk

V2

n(Ω)

+ kη

1

k

H4s)

+ kη

2

k

H2 0s)

.

Proof. For λ > 0, λI − A

s

is an isomorphism from D(A

s

) to H

s

(see e.g. section 3.4). Thus (η

1

, η

2

) 7→ k(η

1

, η

2

)k

Hs

+kA

s

1

, η

2

)k

Hs

is a norm equivalent to (η

1

, η

2

) 7→ kη

1

k

H4s)

+kη

2

k

H2

0s)

. Since (−A

0

) is an isomorphism from V

2

(Ω) ∩ V

10

(Ω) to V

0n

(Ω), there exist positive constants C

1

and C

2

such that

C

1

kP v − P D

s

η

2

k

V2

n(Ω)

≤ kA

0

Pv + (−A

0

)P D

s

η

2

k

V0

n(Ω)

≤ C

2

kPv − P D

s

η

2

k

V2 n(Ω)

. Moreover D

s

∈ L(H

03/2

s

), V

2

(Ω)) (see Lemma 3.10) and A

s

∈ L(D(A

s

), H

s

), therefore we have

k(P v, η

1

, η

2

)k

H

+ kA

0

P v + (−A

0

)P D

s

η

2

k

V0n(Ω)

+ kA

s

1

, η

2

)k

Hs

≤ C(kP vk

V2

n(Ω)

+ kη

1

k

H4s)

+ kη

2

k

H2 0s)

).

To prove the reverse inequality we write kPvk

V2

n(Ω)

+ kη

1

k

H4s)

+ kη

2

k

H2 0s)

C1

1

kA

0

(P v − P D

s

η

2

)k

Vn0(Ω)

+ kP D

s

η

2

k

V2n(Ω)

+ kη

1

k

H4s)

+ kη

2

k

H2 0s)

C1

1

kA

0

(P v − P D

s

η

2

)k

Vn0(Ω)

+ kη

1

k

H4s)

+ Ckη

2

k

H2

0s)

.

8

(9)

The proof is complete.

Theorem 3.3. The operator (A, D(A)) is the infinitesimal generator of an analytic semigroup on H, and the resolvent of A is compact.

To prove this theorem, we rewrite A in the form A = A

1

+ B

0

, with

A

1

=

A

0

0 (−A

0

)P D

s

0 0 I

0 A

α,β

δ∆

s

and

B

0

=

0 0 0

0 0 0

ρ

1

ν(I + ρ

1

γ

s

N

s

)

−1

γ

s

N

0

(∆(·) · n)) K

s

A

α,β

δK

s

s

 ,

with K

s

= (I + ρ

1

γ

s

N

s

)

−1

− I.

Theorem 3.4. The operator (A

1

, D(A

1

)) is the infinitesimal generator of a strongly continuous semigroup on H.

Proof. Step 1. We first show that the unbounded operator ( A e

1

, D( A e

1

)) in V

−1

(Ω) × H

s

, defined by D( A e

1

) =

n

(P v, η

1

, η

2

) ∈ V

1n

(Ω) × (H

4

∩ H

02

∩ L

20

)(Γ

s

) × (H

02

∩ L

20

)(Γ

s

) | P v − P D

s

η

2

∈ V

10

(Ω) o and A e

1

=

A

0

0 (−A

0

)P D

s

0 0 I

0 A

α,β

δ∆

s

 ,

is the infinitesimal generator of a strongly continuous semigroup on V

−1

(Ω) × H

s

. We endow V

−1

(Ω) with the norm

v 7−→

(−A

0

)

−1

v, v

V10(Ω),V−1(Ω)

1/2

, and H

s

with the norm k · k

H2

0s)×L20s)

. For λ > 0, we have

( A e

1

− λI )(P v, η

1

, η

2

), (P v, η

1

, η

2

)

V−1(Ω)×Hs

= −kP vk

2V0

n(Ω)

+ (P D

s

η

2

, P v)

V0

n(Ω)

− λkP vk

2V−1(Ω)

− λk(η

1

, η

2

)k

2H

s

− δkη

2

k

L2 0s)

.

Thus, for λ > 0 big enough, ( A e

1

− λI, D( A e

1

)) is dissipative in V

−1

(Ω) × H

s

. It can also be shown that it is maximal. Thus, for λ > 0 big enough, ( A e

1

− λI, D( A e

1

)) is the infinitesimal generator of a semigroup of contractions on V

−1

(Ω) × H

s

, and ( A e

1

, D( A e

1

)) is the infinitesimal generator of a strongly continuous semigroup on V

−1

(Ω) × H

s

.

Step 2. Let us consider the evolution equation d

dt

 P v

η

1

η

2

= A e

1

 Pv

η

1

η

2

 ,

 P v(0)

η

1

(0) η

2

(0)

=

 P v

0

η

10

η

20

. (3.3)

Let us recall that (A

s

, D(A

s

)) is the infinitesimal generator of an analytic semigroup on H

s

(see e.g.

[9, 29]). Let us notice that the solution (Pv, η

1

, η

2

) to equation (3.3) can be solved by first determining (η

1

, η

2

) and next P v. Thus, if (Pv

0

, η

10

, η

02

) ∈ V

−1

(Ω) × H

s

, the solution (Pv, η

1

, η

2

) to equation (3.3) is such that η

1

∈ H

3,3/2

sT

) and η

2

∈ H

1,1/2

sT

) for all T > 0 (see e.g. [1, Chapter 3, Corollary 2.1]).

9

(10)

From [22, Theorem 2.7] it follows that if (Pv

0

, η

10

, η

02

) ∈ H then Pv ∈ H

1,1/2

(Q

T

)∩ C([0, T ]; V

n0

(Ω)), and (Pv, η

1

, η

2

) ∈ C([0, T ]; H). Therefore the restriction of the semigroup (e

tAe1

)

t∈R+

to H is a semigroup on H. It is easy to verify that its domain is D(A

1

) = D(A).

We are going to prove the two following theorems.

Theorem 3.5. The operator (A

1

, D(A

1

)), with D(A

1

) = D(A), is the infinitesimal generator of an analytic semigroup on H = V

0n

(Ω) × (H

02

s

) ∩ L

20

s

)) × L

20

s

).

Theorem 3.6. The operator (B

0

, D(A

1

)) is A

1

-bounded with relative bound zero.

The first claim in Theorem 3.3 clearly follows from Theorems 3.5 and 3.6 (see [14, Chapter 9, Corollary 2.5]). The second claim is proved in section 3.4.

3.2. Proof of Theorem 3.5. Now we are going to estimate the resolvent of A

1

. We have (λI − A

1

)

−1

= (λI − A

0

)

−1

0 (λI − A

0

)

−1

(−A

0

)P D

s

(λI − A

s

)

−1

0 (λI − A

s

)

−1

! .

Since (λI − A

0

)

−1

(−A

0

)P D

s

= −λ(λI − A

0

)

−1

P D

s

+ P D

s

, we obtain (λI − A

1

)

−1

=

(λI − A

0

)

−1

0 −λ(λI − A

0

)

−1

P D

s

+ P D

s

(λI − A

s

)

−1

0 (λI − A

s

)

−1

! .

From [9] (see also [29, section 2.2]), we know that there exist a ∈ R and π/2 < θ

0

< π such that k(λI − A

s

)

−1

k

L(Hs)

≤ C

s

|λ − a| for all λ ∈ S

a,θ0

, (3.4) where

S

a,θ0

= n

λ ∈ C | λ 6= a, |arg(λ − a)| < θ

0

o .

For the Stokes resolvent we have

k(λI − A

0

)

−1

f k

Vn0(Ω)

≤ C

0

|λ| kΘk

V0n(Ω)

for all λ ∈ S

0,θ1

, (3.5) with π/2 < θ

1

< π. We can choose θ

0

= θ

1

and a > 0. Thus if (f , Θ) ∈ V

0n

(Ω) × H

s

, we have

(λI − A

1

)

−1

f Θ

!

=

(λI − A

0

)

−1

f − λ(λI − A

0

)

−1

P D

s

(λI − A

s

)

−1

Θ

2

+ P D (λI − A

s

)

−1

Θ

2

(λI − A

s

)

−1

Θ

! .

From (3.4) and (3.5), it follows that

k(λI − A

s

)

−1

Θk

Hs

|λ−a|Cs

kΘk

Hs

, kP D (λI − A

s

)

−1

Θ

2

k

V0

n(Ω)

C|λ−a|P DCs

kΘk

Hs

, kλ(λI − A

0

)

−1

P D (λI − A

s

)

−1

Θ

2

k

V0

n(Ω)

C0|λ−a|CP DCs

kΘk

Hs

for all λ ∈ S

a,θ0

. By combining the previous estimates we obtain

(λI − A

1

)

−1

f Θ

!

V0

n(Ω)×Hs

≤ C

0

|λ| kf k

V0n(Ω)

+ C

0

C

P D

C

s

|λ − a| kΘk

Hs

+ C

P D

C

s

|λ − a| kΘk

Hs

+ C

s

|λ − a| kΘk

Hs

, for all λ ∈ S

a,θ0

, which proves the analyticity of the semigroup generated by A

1

.

10

(11)

3.3. Proof of Theorem 3.6. We set

B

1

=

0 0 0

0 0 0

ρ

1

ν (I + ρ

1

γ

s

N )

−1

γ

s

N

0

(∆(·) · n)) 0 0

, B

2

=

0 0 0

0 0 0

0 K

s

A

α,β

0

 ,

and

B

3

=

0 0 0

0 0 0

0 0 δK

s

s

 .

Lemma 3.7. The operator (B

1

, D(A

1

)) is A

1

-bounded with relative bound zero.

Proof. Let us prove that, for all ε > 0, there exists C

ε

> 0 such that kγ

s

N

0

(∆v · n)k

L2

0s)

≤ εkvk

V2n(Ω)

+ C

ε

kvk

V0n(Ω)

, (3.6) for all v ∈ V

2n

(Ω). To prove (3.6), we argue by contradiction. We assume that there exists a sequence (v

k

)

k

⊂ V

2n

(Ω) such that

s

N

0

(∆v

k

· n)k

L2

0s)

= 1, kv

k

k

V0

n(Ω)

−→ 0 and kv

k

k

V2

n(Ω)

≤ M,

for some M > 0. Therefore, without loss of generality, we can assume that there exists v ∈ V

2n

(Ω) such that

v

k

* 0 in V

2n

(Ω), ∆v

k

· n * 0 in H

−1/2

(Γ) and ∆v

k

· n −→ 0 in H

−1/2−ε

(Γ), for all 0 < ε ≤ 1/2. From [7, Lemma A.5], we know that γ

s

N

0

is bounded from H

−1

s

) to L

20

s

). Thus

γ

s

N

0

(∆v

k

· n) −→ 0 in L

20

s

), which is in contradiction with

s

N

0

(∆v

k

· n)k

L2

0s)

= 1.

Thus (3.6) is proved. The lemma is a direct consequence of (3.6), of Lemma 3.1 and Proposition 3.2.

Lemma 3.8. There exists 0 < θ

2

< 1 such that B

2

is bounded from D((−A

1

)

θ2

) into H.

Proof. Let (φ

k

)

k≥1

be an orthonormal basis in L

20

s

) constituted of eigenvectors of the operator ρ

1

γ

s

N and let λ

k

> 0 be the eigenvalue associated with φ

k

. We have

(I + ρ

1

γ

s

N

s

)f =

X

k=1

(1 + λ

k

)f

k

φ

k

.

Thus

(I + ρ

1

γ

s

N

s

)

−1

f =

X

k=1

f

k

1 + λ

k

φ

k

, and

K

s

f = (I − (I + ρ

1

γ

s

N

s

)

−1

)f =

X

k=1

λ

k

1 + λ

k

f

k

φ

k

.

Since the operator A

α,β

is an isomorphism from H

4

s

) ∩ H

02

s

) ∩ L

20

s

) into L

20

s

) and from L

20

s

) into (H

4

s

) ∩ H

02

s

) ∩ L

20

s

))

0

, by interpolation it is also continuous from H

4−ε

s

)∩ H

02

s

) ∩ L

20

s

) into H

−ε

s

) for all 0 ≤ ε ≤ 1.

11

(12)

Denoting by (A

α,β

f )

k

the coefficient of A

α,β

f in the basis (φ

k

)

k≥1

, we have kK

s

A

α,β

f k

2L2

0s)

= P

∞ k=1

λ2k

(1+λk)2

(A

α,β

f )

2k

≤ P

k=1

λ

2k

(A

α,β

f )

2k

= kρ

1

γ

s

N

s

A

α,β

f k

2L2 0s)

≤ C

ε

kA

α,β

f k

2H−εs)

≤ C

ε

kf k

2H4−εs)

,

for all f ∈ H

4−ε

s

) ∩ H

02

s

) ∩ L

20

s

) and all 0 ≤ ε < 1/2. Indeed γ

s

N

s

is continuous from H

−ε

s

) into L

20

s

) if 0 ≤ ε < 1/2 (see e.g. [7, Lemma A.5]). Since

H

4−ε

s

) ∩ H

02

s

) ∩ L

20

s

)

⊃ D((−A

1

)

(4−ε0)/4

), for all 0 ≤ ε

0

< ε < 1/2, the proof is complete.

Lemma 3.9. There exists 0 < θ

3

< 1 such that B

3

is bounded from D((−A

1

)

θ3

) into H.

Proof. The proof is very similar to that of the previous Lemma and is left to the reader.

Theorem 3.6 is a direct consequence of Lemmas 3.7, 3.8 and 3.9.

3.4. Resolvent of A. In this section we want to show that the resolvent of A is compact. For that we study the stationary problem

λv − div σ(v, p) = f and div v = 0 in Ω, v = η

2

~ e

2

on Γ

s

, v = 0 on Γ

0

,

λη

1

− η

2

= g in Γ

s

,

λη

2

− βη

1,xx

− δη

2,xx

+ αM

s

η

1,xxxx

= M

s

1

p + h) in Γ

s

, η

1

= 0 and η

1,x

= 0 on

0, L ,

(3.7)

where f ∈ V

n0

(Ω), g ∈ H

02

s

) ∩ L

20

s

), h ∈ L

20

s

), λ ∈ R and λ > 0. This system is equivalent to λv − div σ(v, p) = f and div v = 0 in Ω,

v = (λη

1

− g)~ e

2

on Γ

s

, v = 0 on Γ

0

, λη

1

− η

2

= g in Γ

s

,

λ

2

η

1

− βη

1,xx

− δλη

1,xx

+ αM

s

η

1,xxxx

= M

s

1

p + h + λg − δλg

xx

) in Γ

s

, η

1

= 0 and η

1,x

= 0 on

0, L .

(3.8)

We denote by L the unbounded operator in L

20

s

) with domain H

4

s

) ∩ H

02

s

) ∩ L

20

s

) defined by Lη = λ

2

η − βη

xx

− δλη

xx

+ αM

s

η

xxxx

.

The operator L is also an isomorphism from H

4

s

) ∩ H

02

s

) ∩ L

20

s

) into L

20

s

) and from H

02

s

) ∩ L

20

s

) into (H

02

s

) ∩ L

20

s

))

0

. Thus, we can rewrite the system (3.8) in the form

λv − div σ(v, p) = f and div v = 0 in Ω,

v = (λL

−1

M

s

1

γ

s

p + h + λg − δλg

xx

) − g)~ e

2

on Γ

s

, v = 0 on Γ

0

, λη

1

− η

2

= g in Γ

s

,

λ

2

η

1

− βη

1,xx

− δλη

1,xx

+ αM

s

η

1,xxxx

= ρ

1

γ

s

p + h + λg − δλg

xx

in Γ

s

, η

1

= 0 and η

1,x

= 0 on

0, L .

(3.9)

We consider the system

λv − div σ(v, p) = f and div v = 0 in Ω,

v = λρ

1

L

−1

s

p)~ e

2

+ f~ e

2

on Γ

s

, v = 0 on Γ

0

, (3.10)

12

(13)

where f ∈ H

02

s

) ∩ L

20

s

) stands for λL

−1

M

s

(h + λg − δλg

xx

) − g. We set E = n

w ∈ V

1

(Ω) | v = 0 on Γ

0

, v

1

= 0 on Γ

s

, v

2

|

Γs

∈ H

02

s

) ∩ L

20

s

) o . The space E, equipped with the norm

kvk

E

=

kvk

2V1(Ω)

+ kL

1/2

v

2

|

Γs

k

2L2 0s)

1/2

,

is a Hilbert space because L

1/2

is an isomorphism from H

02

s

) ∩ L

20

s

) onto L

20

s

).

Multiplying the first equation in (3.10) by w ∈ E, after integration we obtain Z

(λv · w + ν ∇v : ∇w) + Z

Γs

p w

2

= Z

f w.

Using

λρ

1

γ

s

p = Lv

2

− Lf in (H

02

s

) ∩ L

20

s

))

0

, we obtain

Z

(λv · w + ν ∇v : ∇w) + 1 λρ

1

Z

Γs

L

1/2

v

2

L

1/2

w

2

= Z

f w + 1 λρ

1

Z

Γs

L

1/2

f L

1/2

w

2

.

Next, we set

a(v, w) = Z

(λv · w + ∇v : ∇w) + 1 λρ

1

Z

Γs

L

1/2

v

2

L

1/2

w

2

and

`(w) = Z

f w + 1 λρ

1

Z

Γs

L

1/2

f L

1/2

w

2

.

Thus system (3.10) is equivalent to

a(v, w) = `(w) for all w ∈ E,

λρ

1

γ

s

p = Lv

2

− Lf in (H

02

s

) ∩ L

20

s

))

0

. (3.11) With the Lax-Milgram theorem, we can prove that the variational problem

Find v ∈ E such that a(v, w) = `(w) for all w ∈ E, (3.12) has a unique solution. Indeed, for all H

02

s

) ∩ L

20

s

), we have

Z

Γs

L

1/2

η L

1/2

η = Z

Γs

λ

2

|η|

2

+ β |η

x

|

2

+ α|η

xx

|

2

≥ ρkηk

2H2 0s)

, for some ρ > 0.

The solution v ∈ E to the above variational problem obeys

kvk

E

≤ C(kf k

Vn0(Ω)

+ kL

1/2

f k

L2s)

).

Since f = λL

−1

M

s

(h + λg − δλg

xx

) − g, we have kvk

E

≤ C(kf k

Vn0(Ω)

+ kL

−1/2

hk

L2

0s)

+ kL

1/2

gk

L2

0s)

) ≤ C(kf k

Vn0(Ω)

+ khk

L2

0s)

+ kgk

H2

0s)

).

Therefore

kv

2

|

Γs

k

H2

0s)

≤ C(kf k

V0n(Ω)

+ khk

L2

0s)

+ kgk

H2

0s)

).

13

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