DSpace at VNU: BOUNDARY STABILIZATION OF THE NAVIER-STOKES EQUATIONS IN THE CASE OF MIXED BOUNDARY CONDITIONS

We study the boundary feedback stabilization, around an unstable stationary solu-tion, of a two dimensional fluid flow described by the Navier–Stokes equations with mixed boundary conditi

BOUNDARY STABILIZATION OF THE NAVIER–STOKES

PHUONG ANH NGUYEN† AND JEAN-PIERRE RAYMOND‡

Abstract We study the boundary feedback stabilization, around an unstable stationary

solu-tion, of a two dimensional fluid flow described by the Navier–Stokes equations with mixed boundary conditions The control is a localized Dirichlet boundary control A feedback control law is deter- mined by stabilizing the linearized Navier–Stokes equations around the unstable stationary solution.

We prove that the linear feedback law locally stabilizes the Navier–Stokes system Thus, we extend results previously known in the case when the boundary of the geometrical domain is regular and when only Dirichlet boundary conditions are present to the case with mixed boundary conditions.

Key words incompressible Navier–Stokes equations, feedback control, mixed boundary

condi-tions

AMS subject classifications 93B52, 93C20, 93D15, 35Q30, 76D55, 76D05, 74D07 DOI 10.1137/13091364X

1 Introduction We consider the Navier–Stokes equations with mixed

bound-ary conditions, of Dirichlet and Neumann types, in two dimensional domains We areinterested in stabilizing such fluid flows, in a neighborhood of an unstable stationarysolution, by a Dirichlet boundary control This type of problem has already beenconsidered in [3, 4, 41, 43, 45] in the case when only Dirichlet boundary conditionsare involved in the model

The presence of mixed boundary conditions is the source of several difficulties deed, in the case of mixed boundary conditions the solutions to the stationary Stokesequations [36], or the solutions to the stationary Navier–Stokes equations are less reg-ular than in the case when only Dirichlet conditions are present; see, e.g., [37] Theexistence of solutions to the stationary Navier–Stokes equations is guaranteed onlyfor small data [39, 21] As far as we know, the existence of solutions to the instation-ary Navier–Stokes equations with mixed boundary conditions, and nonhomogeneousDirichlet conditions whose normal component is not zero, is not yet studied in theliterature We refer to [26] for the case when the normal component of the Dirich-let condition is zero and where there is no junction between Neumann and Dirichletboundary conditions

In-From the point of view of control theory, the null controllability of the linearizedNavier–Stokes equations is nowadays well understood when only Dirichlet conditionsare present and when the boundary of the domain occupied by the fluid is regular; see,e.g., [17] As far as we know, there is no similar result in the case of mixed boundaryconditions Therefore, the extension of results obtained in [41] to the case of mixedboundary conditions is not trivial

∗Received by the editors March 20, 2013; accepted for publication (in revised form) July 23, 2015;

published electronically September 15, 2015.

http://www.siam.org/journals/sicon/53-5/91364.html

†International University, Vietnam National University - Ho Chi Minh City, Quarter 6, Linh

Trung Ward, Thu Duc District, Ho Chi Minh City, Vietnam (npanh@hcmiu.edu.vn) This author was partially supported by VIASM.

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

CNRS, Institut de Math´ ematiques, UMR 5219, 31062 Toulouse Cedex, France (jean-pierre.raymond@ math.univ-toulouse.fr) This author was partially supported by the ANR-project CISIFS 09-BLAN- 0213-03 and by the project ECOSEA from FNRAE.

3006

We are interested in fluid flows modeled by the Navier–Stokes equations in atwo dimensional bounded domain Ω, with a boundary Γ = Γd ∪ Γ n, when Dirichletboundary conditions are applied on Γd= Γo ∪Γ i ∪Γ c, while a homogeneous Neumannboundary condition is prescribed on Γn The subset Γc, relatively open in Γd, corre-sponds to the control zone An inflow boundary condition is prescribed on the subset

Γi, while homogeneous boundary conditions are imposed on Γo Precise assumptions

on Ω and Γ are stated in section 2.1 We assume that (w s , q s)∈ H1(Ω;R2)× L2(Ω)

is a stationary solution to the equation

(1.1)

(w s · ∇)w s − div σ(w s , q s ) = f s , div w s= 0 in Ω,

w s= 0 on Γc ∪ Γ o , w s = u s on Γi , σ(w s , q s )n = g s on Γn ,

where σ(w s , q s ) = ν( ∇w s+ (∇w s)T)− q s I is the usual Cauchy stress tensor, ν > 0

is the viscosity of the fluid, the inflow velocity u s is applied on Γi , f s and g s arestationary data

A particular configuration which satisfies the assumptions stated in section 2.1corresponds to a fluid flow around a circular cylinder in a rectangular channel, with

a Neumann boundary condition at the end of the channel, and a parabolic profile u s

in the inflow boundary condition; see [1, 2] and Figure 1 Associated with equation(1.1), we consider the controlled Navier–Stokes system

n = Γn × (0, ∞), w(0) = w s + z0 on Ω.

Here, M is a truncation function, precisely described later on (see assumption (H4)

in section 2.1), used to localize the action of the control u on a part of Γ c , and z0is a

perturbation whose presence will destabilize the stationary solution w s As noticed in

[42] in the case of Dirichlet boundary condition, the initial condition w(0) = w s + z0

for all φ ∈ {z ∈ L2(Ω;R2) | div z = 0, z · n = 0 on Γ d } This is equivalent to

Πw(0) = Π(w s + z0), where Π is the so-called Leray projector introduced in section

2.2 Since z0is chosen such that Πz0= z0, we shall write Πw(0) = Πw s + z0.

Our goal is to find a feedback control law K ∈ L(L2(Ω;R2), L2(Γ

e ωt (w(t) − w s) H ε(Ω;R 2 )−→ 0 as t tends to infinity

for a given decay rate −ω < 0, and for z0 small enough in H ε(Ω;R2), and such

that div z0 = 0 in Ω and z0· n = 0 on Γ d Similar results have been obtained inthe case when only Dirichlet boundary conditions are prescribed on Γ; see [41, 43].The case of Neumann boundary conditions is also studied in [5] The results in [5]cannot be adapted to deal with the case of mixed boundary conditions with junctionpoints between the Dirichlet and Neumann boundary conditions In all these previousresults, the feedback control law is determined by looking for a feedback stabilizing

the linearized Navier–Stokes equations around the stationary solution (w s , q s), which

is next applied to the nonlinear system Here, we follow the same approach

Setting z = w − w s and p = q − q s , the nonlinear system satisfied by (z, p) is

o because M is supported in Γ c ; see assumption (H4).

The linearized system is

in section 4 We shall see that the local stabilization result is stated for initial data in

H ε(Ω;R2)∩ V0

n,Γ d (Ω) for any 0 < ε < 1/2 (V0

n,Γ d(Ω) is the subspace of divergence free

functions in L2(Ω;R2) whose normal trace is zero on Γ

d; see section 2.2) Contrary

to the case of the two dimensional Navier–Stokes equations with Dirichlet boundary

conditions, we cannot hope to have such a result with ε = 0, that is, with initial data

in V0

n,Γ d(Ω); see Remark 4.3 Thus the fixed point method used in [3, 4, 41, 43, 45] in

the space L2(0, ∞; H1(Ω;R2))∩H 1/2 (0, ∞; L2(Ω;R2)), when only Dirichlet boundary

conditions are involved, is not appropriate here

By choosing initial data in H ε(Ω;R2)∩ V0

n,Γ d (Ω) with 0 < ε < 1/2, we are able to

prove that the solutions to the closed loop linearized Navier–Stokes equations belong

n,Γ d(Ω)), while the contribution of the feedback control of finite

di-mension provides the part which belongs to H1(0, ∞; H 3/2+ε0(Ω;R2)) This result

is established in Theorem 3.5 for the nonhomogeneous closed loop linearized Navier–Stokes system, and it is an essential step in the proof of our local feedback stabilizationresult Let us finally stress that, in the proof of Theorem 3.5, we strongly use thefact that the feedback is of finite dimension, and it depends only on the unstablecomponent of the state variable, and that the unstable component of the state vari-able belongs to a finite dimensional space of regular functions (functions belonging to

H 3/2+ε0(Ω;R2)).

Let us also mention that additional references on the stabilization of the Navier–Stokes equations may be found in [1, 2, 7, 8, 9, 11, 18, 19, 27]

2 The linearized Navier–Stokes equations.

2.1 Assumptions We denote byJ = {J1, · · · , J N J } the set of corner vertices

corresponding to a junction between either two Dirichlet boundary conditions or aDirichlet and a Neumann boundary condition We have to make some assumptions

on the junctions between Γn and Γd

(H1) Ω is a bounded domain in R2 (in particular Ω is connected), and Γ\ J

is a submanifold of class C2. For any J

k ∈ J , there exists r k > 0 such that {x ∈ R2| dist (x, J k)≤ r k } ∩ Γ is the union of two segments.

(H2) Γn is either a segment, or a union of a finite number of disjoint segments,

or, more generally, a union of a finite number of regular connected components of Γ.There is no junction between two segments with Neumann boundary conditions, andthe junction between a segment with a Neumann boundary condition and a segment

of Γd is a right angle

(H3) Γd is a union of a finite number of regular connected components of Γ.The angles of junctions between two segments with Dirichlet boundary conditions are

strictly less than π.

(H4) Γc is a connected component of Γ of class C3. We also assume that

dist(Γc , J ) > 0 We assume that the function M is a nonnegative function,

de-fined on Γd, with support in Γc , taking values in [0, 1], of class C2, and positive in a

nonempty relatively open subset Γ+

(H5) We assume that the solution (w s , q s ) to (1.1) belongs to H 3/2+ε o(Ω;R2)×

H 1/2+ε o (Ω), where 0 < ε o < 1/2 is the exponent appearing in Theorem 2.5 We

also assume that (w s , q s)|Ωδ,J, with Ωδ,J = {x ∈ Ω | dist(x, J ) > δ}, belongs to

H2(Ω

δ,J ) for all δ > 0.

We make no explicit assumption on Γiand Γoexcept that Γi ∪Γ ois relatively open

in Γd The only assumption we need is actually the regularity condition (w s , q s) ∈

n;R2), satisfying some compatibility

con-ditions, (1.1) admits a solution (w s , q s ) in H 3/2+ε o(Ω;R2)× H 1/2+ε o(Ω) We claimthat there exist configurations for which such solutions exist For that, we refer tothe appendix

The condition ‘Γc is of class C3’ is used to prove that the solution (φ, ψ) to (3.3) is

of class H3for φ and H2for ψ in a neighborhood of Γ+

c (see (H4) for the assumptions

on Γ+

In order to study the linearized Navier–Stokes equations with nonhomogeneous

boundary conditions w = M u not satisfying the condition

Γc M u · n dx = 0, we have

to construct particular solutions, and for that we have to define tubular domains in avery precise way

Definition 2.1 An infinite tube of width δ > 0 is a closed connected subset in

R2 limited by two simple curves C ∞

1 and C2∞ of class C3, distant from δ, that is such

that the distance from any point of C ∞

1 to C2∞ is δ Such a tube will be denoted by

2 contain exactly one point Such a tube

will be denoted by T (δ; C1, C2; S1, S2), where C1 ⊂ C ∞

1 and C2 ⊂ C ∞

2 are such that

∂ T (δ; C1, C2; S1, S2), the boundary of the finite tube, is equal to C1∪ C2∪ S1∪ S2.

Each point x ∈ T (δ; C1, C2; S1, S2) belongs to a unique segment S x ⊂

T (δ; C1, C2; S1, S2) orthogonal to C1 and to C2, and a unique curve C x ⊂

T (δ; C1, C2; S1, S2) parallel to C1 and C2. Thus any point can be localized by

its arc length coordinate on C x , denoted by (x), and its transverse coordinate

ρ(x) ∈ [−δ/2, δ/2] in S x, with the convention that the transverse coordinate of the

points belonging to C1 is −δ/2, while the transverse coordinate of points belonging

to C2 is δ/2 We also make the convention that the arc length coordinate of points

in S1 is zero and that the arc length coordinate of a point x ∈ S2is the length of the

curve C x To each x ∈ T (δ; C1, C2; S1, S2), we associate τ (x), the unitary tangent vector to C x at point x oriented in the sense of increasing arc length coordinates.

We can now state an assumption needed to construct solutions to the linearizedNavier–Stokes equations with nonhomogeneous boundary conditions (see Theorem2.16)

(H6) There exist a finite tubeT (δ; C1, C2; S1, S2) of width δ > 0, and a

nonneg-ative function η belonging to C ∞

c (R), with values in [0, 1], with compact support in

(−δ/2, δ/2), and obeying η(0) = 1, such that

(2.1)

(i) S2⊂ Γ n , S1⊂ R2\ Ω;

(ii) (T (δ; C1, C2; S1, S2)\ S2)∩ Γ = T (δ; C1, C2; S1, S2)∩ Γ c := Γc,T;

(iii) For all x ∈ T (δ; C1, C2; S1, S2), we assume that C x ∩ Γ c,T is

reduced to a point, denoted by γ c (x), such that C x ∩ Ω = {x ∈ C x | (x)

the unitary normal to Γc exterior to Ω

Let us notice that condition (2.2) is not restrictive Indeed, if there exists a tube

T (δ; C1, C2; S1, S2) satisfying (2.1), we can always modify T (δ; C1, C2; S1, S2) in a

neighborhood of S1 in such a way that (2.2) is satisfied Figure 1 is an example of a

domain with a tube satisfying (H6).

2.2 Some function spaces In order to write (1.5) as a controlled system, we

have to introduce the Leray projector associated with the boundary conditions of ourproblem We shall define the Oseen operator in section 2.4

o

2 1

1

Fig 1 Domain with a tube satisfying ( H6).

In the case of mixed Dirichlet/Neumann boundary conditions, we introduce thespace

n,Γ d (Ω) will be denoted by Π, and called

the Leray projector for the above decomposition.

Proof This type of result may be deduced from results in [26] in the case when

Γd and Γn have no junction point In the case of the present paper, let us give a shortproof of that result

Let us notice that, due to the divergence formula, V0

for all z ∈ L2(Ω;R2), and the proof is complete.

To define the Oseen operator, we introduce the space

V1

Γd(Ω) ={z ∈ H1(Ω;R2)∩ V0

n,Γ d(Ω)| z = 0 on Γ d }.

Remark 2.3 Due to assumption (H2), we can extend Ω to a bounded Lipschitz

domain Ωe ⊂ R2 in such a way that Γ

n ⊂ Ω e and Γd ⊂ ∂Ω e Thus, we have

If we identify V0

n,Γ d (Ω) with its dual, and if V −1

Γd (Ω) denotes the dual of V1

n(Ωe) for the

L2-topology (see, e.g., [23]).

We also introduce the intermediate spaces

2.3 The Stokes equation with homogeneous boundary conditions

Be-fore defining the Oseen operator in the next section, we first consider the equation

The following theorem is deduced from [38, Theorem 9.1.5]

Theorem 2.4 Let us assume that f ∈ L2(Ω;R2) Equation (2.3) admits a

unique solution (z, p) ∈ V1

Γd(Ω)× L2(Ω) and

z V1

Γd(Ω)+p L2 (Ω)≤ Cf L2 (Ω;R 2 ).

According to [25, page 174] (see also [39]), since there is no reentrant corner at

a junction between two Dirichlet boundary conditions, the restriction to Ωδ,J d,n of

the solution (z, p) to (2.3) belongs to H2(Ω

Thus, to study the regularity of solutions to (2.3), we introduce weighted Sobolev

spaces as in [39, 38, 36] For β > 0, we introduce the norms

where r j stands for the distance to the junction point J j ∈ J d,n , k = (k1, k2) ∈

N2 denotes a two-index, |k| = k1+ k2 is its length, ∂ k denotes the corresponding

partial differential operator, and z = (z1, z2) We denote by W 2,2

β (Ω)) According to [39, 38, 36], in order to determine

the exponent β of weighted Sobolev spaces in which the solution to (2.3) will belong when f ∈ L2(Ω;R2), we have to consider the complex roots λ to the equation

(2.6) λ2sin2(π/2) − cos2(λπ/2) = λ2− cos2(λπ/2) = 0.

Let us notice that, if 0≤ Re λ < 1, (2.6) is equivalent to

cos(λθ)(λ2sin2(θ) − cos2(λθ)) = 0

when θ = π/2, which is the equation that we have to consider in the case of a junction point between a Dirichlet and a Neumann condition with angle θ (see, e.g., [36, page

761]) We have the following regularity result

Theorem 2.5 Let us assume that f ∈ L2(Ω;R2) The unique solution (z, p) ∈

β (Ω;R2)× W β 1,2 (Ω) follows from [38, Theorem 9.4.5] The solution λ c to (2.6)1

with the smallest positive real part is such that Re λ c ≈ 0.59 and Re λ c > 0.58; see

[10, page 71] Thus, the solution (z, p) to (2.3) belongs to W 2,2

β (Ω;R2)× W β 1,2(Ω), if

1− β ≤ 0.58 In particular, we can choose β = 0.42 According to [10, Proposition

A.1], z belongs to H 3/2+0.08 The proof is complete.

Remark 2.6 Analogous results to those in Theorem 2.5 are stated in three dimensions in the case of polyhedral cones; see [36] and [37, Theorem 4.2] Similar results in two dimensions are also stated in [36, Lemma 2.9], but for higher regularity conditions, and in [35, Lemma 2.6] with the same regularity as in Theorem 2.5 for

second order elliptic systems

With Theorem 2.4, we know that, for a given f , the pressure p corresponding to (2.3) is uniquely defined To find an equation for p, in terms of z, we first notice that (z, p) = (z, p0) + (0, p1), where (z, p0) is the solution to equation

z = 0 on Γd , σ(z, p0)n = 0 on Γn ,

and p1∈ H1

Γn(Ω) is characterized by ∇p1= (I − Π)f.

Since z belongs to H 3/2+ε0(Ω;R2) and p0 belongs to H 1/2+ε0(Ω), p0|Γ and

ν( ∇z +(∇z) T )n ·n|Γbelong to H ε0(Γ) Now, since Πf belongs to V n,Γ0 d(Ω), by taking

the divergence of the first equation in (2.7), we obtain Δp0 = 0 and div(−νΔz +

∇p0) = 0 in Ω Thus the trace (∇p0 − νΔz) · n|Γ = Πf · n|Γ is well defined in

H −1/2 (Γ) Using the fact that p

d) is the trace space, defined in [25, Page 26],

as a subspace of functions belonging to H 3/2(Γ

d) and satisfying some compatibilityconditions at each junction points ofJ d,n Therefore the equation

But we do not give the precise definition of solutions to (2.8) in the sense of

transpo-sition, because we do not need to define p0precisely, and we do not want to introduce

d) (thetrace space introduced in [25], already mentioned above)

2.4 The Oseen operator We first define the Stokes operator (A0, D(A0))corresponding to the boundary conditions in system (1.5) We set

where ε0 is the exponent appearing in Theorem 2.5.

The condition div σ(z, p) ∈ L2(Ω;R2) allows us to define σ(z, p)n in H −1/2(Γ;R2).

Thus σ(z, p)n |Γn = 0 is meaningful in H −1/2(Γ

n;R2), where H −1/2(Γ

n;R2) is the

dual of the space H 1/2

00 (Γn;R2) defined in [33] Since we know that (z, p) belongs to

H 3/2+ε0(Ω;R2)× H 1/2+ε0(Ω), σ(z, p)n |Γn is also defined as a function belonging to

H ε0(Γn;R2).

Remark 2.7 If f ∈ L2(Ω;R2) and if (z, p) is the solution to (2.3), then we easily

see that z belongs to D(A0) and (f, p) is a pair of functions for which −div σ(z, p) =

f ∈ L2(Ω;R2) and σ(z, p)n = 0 on Γ

Conversely, if z ∈ D(A0) and if p ∈ H 1/2+ε0(Ω) are such that −div σ(z, p) =

f ∈ L2(Ω;R2) and σ(z, p)n = 0 on Γ

n , then we can verify that (z, p) is the solution

to (2.3) We can write p in the form p = p0+ p1, where (z, p0) is the solution

to (2.7) and p1 ∈ H1

Γn(Ω) is defined by ∇p1 = (I − Π)f And we obviously have

A0z = Π div σ(z, p) = div σ(z, p0)

We can directly see that p is uniquely defined by z and f ∈ L2(Ω;R2) for which

divσ(z, p) = f Indeed, if p ∈ H 1/2+ε0(Ω) and q ∈ H 1/2+ε0(Ω) obey divσ(z, p) = divσ(z, q) = f and σ(z, p)n = 0 = σ(z, q)n on Γ n, we have∇p − ∇q = 0 in Ω and

(p − q)n = 0 on Γ n Thus p = q.

The Oseen operator (A, D(A)) is defined by D(A) = D(A0) and Az = A0z − Π ((w s · ∇)z + (z · ∇)w s )

Theorem 2.8 The operator (A, D(A)) is the infinitesimal generator of an lytic semigroup on V0

ana-n,Γ d (Ω) Its resolvent is compact.

Proof The proof follows from the following inequality

(λ0I − A)z, zL2 (Ω;R 2 )≥ ν

2z2

V Γd1(Ω) ∀z ∈ D(A),

with λ0 > 0 large enough (see the proof in [41] in the case of Dirichlet boundary

conditions We rewrite it in the case of mixed boundary conditions for the convenience

of the reader) For all z ∈ D(A), we have

We notice that the boundary terms are zero because z = 0 on Γ d and σ(z, p)n = 0 on

Γn Using the classical inequality (in two dimensions)

Γd(Ω) Thus, from Theorem 2.12 in [12, Chapter 1], it

follows that A is the infinitesimal generator of an analytic semigroup on V0

Let us prove that its resolvent is compact We first verify that, for λ0 > 0 for

which (2.9) is true, the solution to the equation (λ0I − A)z = f belongs to V1

Γd(Ω) if

f ∈ V0

n,Γ d(Ω) This follows from the Lax–Milgram lemma Finally, we use the fact that

the imbedding from V1

Γd (Ω) into V0

n,Γ d(Ω) is compact The proof is complete

Remark 2.9 Due to assumption (H3), the restriction of functions in D(A) to

where f ∈ L2(Ω;R2) We shall say that (φ, ψ) ∈ V1

where (·, ·) L2(Ω;R 2 )is the inner product in L2(Ω;R2), a( ·, ·) and b(·, ·) are, respectively,

defined in (2.10) and (2.4) The analogue of Theorems 2.4 and 2.5 is stated below.Theorem 2.10 Let us assume that f ∈ L2(Ω;R2) Equation (2.11) admits a

We can now characterize the adjoint of (A, D(A)).

Theorem 2.11 The adjoint operator of (A, D(A)) is defined by

Proof The result is standard Let us recall the main steps of the proof Let us

define (A  , D(A  )) as an unbounded operator in V0

for all z ∈ D(A) and all φ ∈ D(A ) Thus D(A ) ⊂ D(A ∗ ) and A ∗ φ = A  φ for all

φ ∈ D(A  ) Next, it is enough to show that ((e tA)∗)

t≥0 = (e tA )t≥0 to conclude

This last result can be checked by making an integration by parts using the equation

satisfied by z(t) = e tA z0 and the one satisfied by φ(t) = e (T −t)A 

Γd (Ω) This extension will be still

denoted by Π Moreover, we have

Γd (Ω), and let (f n)n be a sequence in V0

Γd (Ω) The proof is complete

Theorem 2.13 Let λ0> 0 be as in (2.9) We have

D((λ0I − A) 1/2 ) = V1

Γd (Ω), D((λ0I − A ∗)1/2 ) = V1

Γd (Ω), and(2.14)

(νI − A0)1/2 is an isomorphism from VΓ1d (Ω) into V n,Γ0 d(Ω).

We can extend νI − A0 as an unbounded operator in (D(A0)) with domain

V0

n,Γ d(Ω) We will make no distinction of notation between the unbounded operator

and the bounded operator from its domain into the space in which the unbounded

operator is defined Using this abuse of notation, we know that νI − A0 is an morphism fromD(A0) into V0

iso-n,Γ d (Ω) and from V0

n,Γ d(Ω) into (D(A0)) It is also an

isomorphism from V1

Γd (Ω) into V −1

Γd (Ω) This follows from the Lax–Milgram theorem

Thus (νI −A0)1/2 is an isomorphism from V1

Step 2 Proof of (2.14) We repeat for λ0I − A what we have written for νI − A0,except that we need the identity proved before, namely, 

un-that λ0I − A is an isomorphism from D(A) into V0

n,Γ d (Ω) and from V0

n,Γ d(Ω) into(D(A ∗)) It is also an isomorphism from V1

Γd (Ω) into V −1

Γd (Ω) This follows from

(2.9) and the Lax–Milgram theorem Thus, V1

Γd (Ω) is the domain of λ0I − A in

V −1

Γd (Ω) We notice that

(−A0)−1 (λ0I − A)z, zV0

n,Γd(Ω)

≤ C(−A0)−1 z  1/2 L2 (Ω;R 2 )w s  H 3/2(Ω;R 2 )z 3/2 V0

n,Γd(Ω)

and (−A0)−1 z  L2 (Ω;R 2 ) ≤ C(−A0)−1/2 z  L2 (Ω;R 2 ) From these estimates, as in the

proof of Theorem 2.8, it follows that we can choose λ0> 0 large enough so that

((λ0I − A)z, z) V −1

Γd(Ω)=

(−A0)−1 (λ

0I − A)z, zV0

n,Γd(Ω)≥ 0.

Thus, λ0I − A, considered as an unbounded operator in V −1

and the proof of (2.15)1 is complete

Step 4 Proof of (2.15)2 To prove (2.15)2, let us denote by z f the solution tothe equation

(λ0I − A)z f = f.

Let ε ∈ (0, 1/2) be given fixed We know that λ0I − A is an isomorphism from

(D((λ0I − A ∗)1/2−ε/2)) into D((λ0I − A) 1/2+ε/2) Thus we have

2.5 The linearized Navier–Stokes equations with nonhomogeneous

boundary conditions In the case when u ∈ L2(0, T ; L2(Γ

c;R2)), solutions to

sys-tem (1.5), over the interval (0, T ) in place of (0, ∞), may be defined by transposition.

Definition 2.14 A function z ∈ L2(0, T ; L2(Ω;R2)) is a solution to system

(1.5) in the sense of transposition, over the interval (0, T ), if and only if div z = 0 in

As in the case of Dirichlet boundary conditions (see [42]), we can prove the lowing theorem

fol-Theorem 2.15 Assume that z0∈ V0

n,Γ d (Ω) and u ∈ L2(0, T ; L2(Γ

c;R2)) Then

the system (1.5) admits a unique solution in the sense of Definition 2.14 Moreover,

we have the following estimate

z L2(0,T ;L2 (Ω;R 2 )) ≤ C(z0 V0

n,Γd(Ω)+u L2(0,T ;L2 (Γc;R 2 ))).

Proof The proof can be done as in [42, Theorem 2.2] See also [44] It follows from

the fact that f → (σ(φ, ψ)n, φ(0)) is linear and continuous from L2(0, T ; L2(Ω;R2))

into L2(0, T ; L2(Γ

In order to rewrite the Oseen equations (1.5) as a controlled system, we look for

the solution (z, p) to system (1.5) in the form (z, p) = (y, q) + (w, π), where (w, π) is

a lifting of the boundary condition z = M u on Γ c We define the lifting operator D

by setting DM u = w, where

(2.20) λ0w − div σ(w, π) + (w s · ∇)w + (w · ∇)w s = 0, div w = 0 in Ω,

w = M u on Γd , σ(w, π)n = 0 on Γn ,

and λ0 is chosen so that (2.9) is satisfied When u is a time dependent function over

the time interval (0, ∞), the solution (w, π) to (2.20) also depends on t, and (2.20) is

solved for all t ∈ (0, ∞).

Theorem 2.16 Assume that M u ∈ H 1/2(Γ

d;R2) The equation

(2.21)

div v = 0 in Ω,

v = M u on Γ d , (∇v + (∇v) T )n = 0 on Γn , admits a solution satisfying

(2.22) v H1 (Ω;R 2 )≤ CMu H 1/2(Γd;R 2 ).

In addition, if M u ∈ H 3/2(Γ

d;R2), then we have

(2.23) v H2 (Ω;R 2 )≤ CMu H 3/2(Γd;R 2 ).

Proof Due to assumption (H6), there exist a tubeT (δ; C1, C2; S1, S2), of width

δ > 0, and a nonnegative function η ∈ C ∞

c (R), with compact support in (−δ/2, δ/2), satisfying η(0) = 1, obeying the condition (2.2).

We look for a solution v to (2.21) in the form v = v0+ α1v1, where v0is a solutionto

length coordinates (see assumption (H6) for the precise definition of ρ(x), (x), and

τ (x)), and α1 is chosen to have

Step 1 Choice for v1 We first choose two points x0∈ C1and x1∈ C1 such that

(x0) < (x1) and the closed tube T (δ; C1, C2; S x0, S x1)) limited by S x0 and S x1 is

included in Ω Let us recall that S x0 (respectively, S x1) is the unique segment of length

δ, containing x0(respectively, x1), included inT (δ; C1, C2; S1, S2) and orthogonal to

C1 and C2 Let θ ∈ C2(T (δ; C1, C2; S1, S2)) be a nonnegative function, equal to 1 in

T (δ; C1, C2; S1, S x0) and equal to 0 inT (δ; C1, C2; S x1, S2).

We set

v1(x) = θ(x) η(ρ(x)) τ (x) in T (δ; C1, C2; S1, S2) and v1= 0 elsewhere Since C1and C2are of class C3, and since η is with compact support in ( −δ/2, δ/2), it

is clear that v1belongs to C2(Ω) Due to the definition of θ, θ(x) = 1 for all x ∈ Γ c,T,thus (2.25)1 is satisfied Since η is with compact support in ( −δ/2, δ/2), v1 = 0 on

Γd \ Γ c,T Since θ is equal to 0 in T (δ; C1, C2; S x1, S2)), we have (∇v1+ (∇v1)T )n = 0

on Γn

Step 2 Existence of v0 Notice that 

Γc v0 · n dx = −α1 

Ωdiv v1dx Since

dist(Γc , J ) > 0, as in [22, Chapter III, proof of Theorem 3.2], we can construct a

vector field v0∈ H1(Ω;R2) with compact support inT (δ; C1, C2; S1, S x2)⊂ Ω ∪ Γ c,T,

with x2∈ C1 such that (x2) < L(C1) (L(C1) denoting the length of C1), satisfying

v0= M u − α1v1 on Γc and v0= 0 on Γd \ Γ c, and such that

where ε0∈ (0, 1/2) is the exponent introduced in Theorem 2.5.

Proof Step 1 We first consider the case when w s ≡ 0 We use the solution v to

(2.21), constructed in the proof of Theorem 2.16 In particular, we shall use the fact

that v = v0+ α1v1 and that v0 is with compact support in T (δ; C1, C2; S1, S x2) ⊂

Ω∪ Γ c,T We look for the solution w to (2.20) with w s ≡ 0 in the form w = v + ζ

with v = v0+ α1v1 The equation satisfied by ζ is

λ0ζ − div σ(ζ, π) = −λ0v + νdiv ( ∇v0+ (∇v0)T ) + να1div (∇v1+ (∇v1)T ), div ζ = 0 in Ω, ζ = 0 on Γd , σ(ζ, π)n = 0 on Γn

When M u ∈ H 1/2(Γ

1 belongs to H2(Ω;R2), v

0 belongs to H Γ\Γ1 c(Ω;R2)

because v0 is with compact support in a neighborhood of Γc, and therefore we may

identify νdiv ( ∇v0+ (∇v0)T ) with an element in H −1

Γd(Ω;R2) With the Lax–Milgram

theorem, we can show that ζ belongs to H1

Γd(Ω;R2).

When M u ∈ H 3/2(Γ

d;R2), v belongs to H2(Ω;R2) and νdiv ( ∇v+(∇v) T) belongs

to L2(Ω;R2) From Theorem 2.5, it follows that ζ belongs to H 3/2+ε0(Ω;R2), and the

proof is complete in the case when w s ≡ 0.

Step 2 We now consider the case when w s

to (2.20) in the form (w, π) = (w0, π0) + (w1, π1), where (w0, π0) is the solution to

(2.20) corresponding to w s ≡ 0 and (w1, π1) is the solution to

λ0w1− div σ(w1, π1) + (w s · ∇)w1+ (w1· ∇)w s=−(w s · ∇)w0− (w0· ∇)w s ,

div w1= 0 in Ω, w1= 0 on Γd , σ(w1, π1)n = 0 on Γn

We treat the case when M u ∈ H 3/2(Γ

d;R2) The case when M u ∈ H 1/2(Γ

d;R2) can

be treated similarly Since w s and w0 belong to H 3/2+ε0(Ω;R2), then

(w s · ∇)w0+ (w0· ∇)w s belongs to H 1/2+ε0(Ω;R2) We deduce that (w

1, π1)

be-longs to H 3/2+ε0(Ω;R2)× H 1/2+ε0(Ω) The corresponding estimate for (w, π) can be

easily deduced The proof of the theorem is complete

In order to write system (1.5) satisfied by (z, p) as a controlled system, we follow the approach used in [42] We assume that u ∈ H1(0, T ; H 1/2(Γ

c;R2)), and we extend

it by zero to Σd \ Σ c As already mentioned at the beginning of section 2.3, we look

for (z, p) in the form (z, p) = (w, π) + (y, q), where (w, π) is the solution to (2.20) The equation satisfied by (y, q) is

conditions We rewrite it in the case of mixed boundary conditions for the convenience

of the reader)...

(1.5) in the sense of transposition, over the interval (0, T ), if and only if div z = in< /i>

As in the case of Dirichlet boundary conditions (see [42]), we can prove the lowing theorem... while the transverse coordinate of points belonging

to C2 is δ/2 We also make the convention that the arc length coordinate of points

in S1