Stabilization of the 2D incompressible Euler system in an infinite strip

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Ann. I. H. Poincaré – AN 30 (2013) 737–762

www.elsevier.com/locate/anihpc

Stabilization of the 2D incompressible Euler system in an infinite strip

Hayk Nersisyan

CNRS UMR 8088, Département de Mathématiques, Université de Cergy-Pontoise, Site de Saint-Martin, 2 avenue Adolphe Chauvin, F95302 Cergy-Pontoise Cedex, France

Received 19 September 2011; received in revised form 9 May 2012; accepted 18 December 2012 Available online 8 January 2013

Abstract

The paper is devoted to the study of a stabilization problem for the 2D incompressible Euler system in an infinite strip with boundary controls. We show that for any stationary solution(c,0)of the Euler system there is a control which is supported in a given bounded part of the boundary of the strip and stabilizes the system to(c,0).

1. Introduction

We consider the incompressible two-dimensional Euler system

˙

u+ u,∇u+ ∇p=0, divu=0, (1.1)

whereu=(u1, u2)andpare unknown velocity field and pressure of the fluid, and u,∇v=

3 i=1

u_i(t, x) ∂

∂xi

v.

The space variablex=(x₁, x₂)belongs to the stripDdefined by D:=

(x1, x2): x1∈R, x2∈(−1,1)

. (1.2)

Let us take two open intervals(a, b), (a+d, b+d)⊂Rand denote

Γ0=(a, b)× {1} ∪(a+d, b+d)× {−1}. (1.3)

The aim of this paper is the study of stabilization of (1.1) with boundary controls supported byΓ0. System (1.1) is completed with the boundary and initial conditions

u·n=0 onΓ \Γ₀, (1.4)

u(x,0)=u₀(x), (1.5)

whereΓ :=∂Dandnis the outward unit normal vector onΓ. In particular, (1.4) is equivalent tou₂=0 onΓ \Γ₀.

E-mail address:[email protected].

http://dx.doi.org/10.1016/j.anihpc.2012.12.002

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For any integers0 we denote byH^s(D)the space of vector functionsu=(u₁, u₂)whose components belong to the Sobolev space of ordersand by · s,Dthe corresponding norm. If there is no confusion, we drop the indexD.

In the cases=0, we write · := · 0.For any integers >0 we defineH^s(D)as the space of distributionsuinD with∇u∈H^s⁻¹(D). We equipH^s(D)with the semi-norm

u_H^s(D):= ∇us−1.

We denote byH˙^s(D)the quotient spaceH^s(D)/R. The following theorem is our main result.

Main result.Lets4be an integer. Then for any constantc∈Rand initial functionu0∈H^s(D)that decays fast at infinity and satisfies the relations

divu0=0, u0·n=0 onΓ \Γ0

there exists a solution(u, p)∈C(R₊, C(D)∩ ˙H^s(D))×C(R₊,H˙^s(D))of(1.1), (1.4)and(1.5)such that

tlim→∞u(·, t )−(c,0)

L^∞+ ∇us−1+ ∇ps−1

=0.

For the exact statement see Theorem3.1. In this formulation the control is not given explicitly, but we can assume that control acts on the system as a boundary condition onΓ₀.

Before turning to the ideas of the proof, let us describe in a few words some previous results on the controllability of Euler and Navier–Stokes systems. Coron [7] introduced the return method to study a stabilization problem for ODE’s, then using this method he proved in[8]the exact boundary controllability of 2D incompressible Euler system in a bounded domain. Glass [14]generalized this result for 3D Euler system. Chapouly [6] using return method proved the global null controllability of the Navier–Stokes system in rectangle. Recently, Glass and Rosier[15]proved the controllability of the motion of a rigid body, which is surrounded by an incompressible fluid. The asymptotic stabilization of 2D Euler equation by stationary feedback boundary controls is studied by Coron[10]and Glass[16].

Controllability of Euler and Navier–Stokes systems with distributed controls is studied in[2,13,19,20]; see also the book[11]for further references.

Notice that the above papers concern the problem of controllability of the fluid in a bounded domain. In this paper, we develop Coron’s return method to get the controllability of velocity of 2D Euler system in an unbounded strip.

This method consists in reducing the controllability of nonlinear system to the linear one. To this end, one constructs a particular solution(u, p)of Euler system and a sequence of balls{Bi}coveringD, such that

(P) Any ballB_idriven by the flow ofuleavesDthroughΓ₀at some time.

Then the linearized system arounduis controllable. In our case, since the domainDis unbounded, the number of balls B_i is infinite, thus we cannot construct a bounded functionu, whose flow moves all balls outsideDin a finite time.

However, we can find a particular solutionusuch that property(P )holds in infinite time. This proves the stabilization of linearized system in infinite time.

To show that controllability of linearized system implies that of the nonlinear system, we need to prove that(P ) also holds for anyu˜sufficiently close tou. This is obvious in the case of bounded domain. In our case, to prove this, we need some additional properties foru. In particular, we need to construct a solutionu, which decays at infinity faster than 1/x²₁. As our particular solutionuis a combination of the Green functions of the Laplacian with Neumann boundary condition, we need to prove that Green functions decay at infinity. This property is a consequence of elliptic regularity and some explicit formulas for solutions of the Laplace equation in a strip.

The paper is organized as follows. In Section 2, we give preliminaries on Poisson and Euler equations in an unbounded strip. The main results of the paper are presented in Section3. In Section4, we construct the particular solutionu. In Appendix A, we prove an auxiliary result used in Section2.

Notation.LetJ_T := [0, T ). The space of continuous functionsu:J_T →X is denoted byC(J_T, X). For any integer s0 ors= ∞, we denote

C_b^s(D)=

u∈C^s(D): uL^∞(D)<∞ .

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We setH˙^∞(D):=_∞

s=0H˙^s(D). Define S(D):=

u∈L²(D): |x₁|^α∂^βu(x₁, x₂)∈L²(D)for anyα∈R₊, β∈Z²₊ . For a vector fieldu=(u₁, u₂)we set

curlu=∂₁u₂−∂₂u₁.

The interior of a setKis denoted by int(K). LetB(x₀, r)be the closed ball inR²of radiusrcentred atx₀. We denote byCa universal constant whose value may change from line to line.

2. Preliminaries

In this section, we present some auxiliary results on Poisson and Euler equations in an unbounded strip. The methods used in their proofs are well known and in many cases we confine ourselves to a brief description of the main ideas.

2.1. Poisson equations in an unbounded strip

First, let us describe the spacesH˙^s(D).

Proposition 2.1.For any integers1we have (i) The spaceH˙^s(D)is complete.

(ii) H^s(D)= {u∈H_loc^s (D): ∇u∈H^s⁻¹}.

(iii) Ifs3, then for anyu∈H^s(D)there is a constantCdepending onusuch that u(x₁, x₂) C|x₁| +C

holds for allx∈D.

Proof. Let {un} ⊂ ˙H^s(D)be a Cauchy sequence. Then there is v∈H^s⁻¹(D)such that∇un→v inH^s⁻¹(D)as n→ ∞, and for anyϕ∈C^∞₀ (D)such that divϕ=0, we have

0= lim

n→∞(∇u_n, ϕ)_L2=(v, ϕ)_L2.

Hence,v= ∇z, wherez∈ ˙H^s(D). This proves thatH˙^s(D)is complete. Now let us prove assertion (ii). Clearly the space in the right-hand side is contained in H˙^s(D). Let us take a functionu∈ ˙H^s(D), a compact setK⊂D and let us show thatu∈H^s(K). Take two functionsχ , χ₁∈C₀^∞(D)and a compact setK₁⊂Dwith int(K1)⊃Ksuch thatχ=1 inK1andχ1=1 inK˜1:=suppχ. Then there existsr∈Nsuch thatχ1u∈H⁻^r(D). This implies that u∈H⁻^r(K˜1), hence

(χ u)=2∇χ∇u+χ u+uχ∈H^min(⁻^r^;^s⁻²⁾(K˜1).

The elliptic regularity implies χ u∈H^min(⁻^r⁺²^;^s)(D), thus u∈H^min(⁻^r⁺²^;^s)(K1). Repeating this argument for a compact setK2⊂K1with int(K2)⊃Kwe can show thatu∈H^min(⁻^r⁺⁴^;^s)(K2). Iterating this, we getu∈H^s(K).

This completes the proof of assertion (ii).

It is easy to see that (ii) implies (iii). Indeed, from (ii) we get u(x₁, x₂)=

x1

0

∂₁u(y, x₂)dy+u(0, x2).

The Sobolev inequality yields (iii). 2

Now we summarize some facts about Poisson equation. Let us take a non-negative functionγ ∈C₀^∞(R)such that suppγ= [a, b]andγ=0 in(a, b)and define

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Fig. 1. DomainD.˜

D˜ :=

(x1, x2): x1∈R, x2∈

−1−γ (x1−d),1+γ (x1)

(2.1) (see Fig. 1).

Let us takeD=DorD= ˜Dand consider the Dirichlet problem for the Poisson equation:

u=f inD, (2.2)

u=0 onΓ, (2.3)

whereΓ=∂Dandf∈L²(D). We say thatu∈H₀¹(D)is a solution of (2.2), (2.3) if

D

∇u∇θdx= −

D

f θdx

for anyθ∈H₀¹(D). We have the following result for the well-posedness of this problem.

Proposition 2.2. For any integer s0 and for any f ∈H^s(D)problem (2.2), (2.3) has a unique solution u∈ H^s⁺²(D). Moreover,

us+2Cfs, (2.4)

whereCdepends only ons.

Proof. The existence of the solutionu∈H₀¹(D)is a consequence of the Riesz representation theorem. Clearly, we have

∇u²Cfu. (2.5)

The Poincaré inequality applied tou(x1,·)gives uC∂₂u.

Combining this with (2.5), we obtain

u1Cf. (2.6)

To show the regularity of the solution and estimate (2.4), we need the following lemma.

Lemma 2.3.For any integers1we have H^s

D

= z∈L²

D

: curlz∈H^s⁻¹ D

, divz∈H^s⁻¹ D

, z·n∈H^s⁻^1/2 Γ

,

wherenis the outward unit normal vector onΓ. Moreover, any functionz∈H^s(D)satisfies the inequality zsC

z + curlzs−1+ divzs−1+ z·ns−1/2

, whereCdepends only ons.

The proof of this lemma is given in Appendix A. Let us denotez= ∇^⊥u:=(∂2u,−∂1u). Then curlz= −u=

−f, divz=0. Notice that (2.3) implies that z·n=0. It follows from Lemma2.3 and inequality (2.6) that z∈ H^s⁺¹(D)andzs+1Cfs. Thus, we obtainu∈H^s⁺²(D)and (2.4). 2

Let us takeg∈H¹(D)and consider the Neumann problem for the Poisson equation:

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u=divg inD, (2.7)

∂u

∂n=g·n onΓ. (2.8)

We say thatu∈ ˙H¹(D)is a solution of (2.7), (2.8) if for anyθ∈H¹(D)we have

D

∇u∇θdx=

D

g∇θdx.

Proposition 2.4.For any integers1 andg∈H^s(D)problem (2.7), (2.8)has a unique solutionu∈ ˙H^s⁺¹(D).

Moreover,

u_H_˙^s+1Cgs. (2.9)

Proof. The Riesz representation theorem implies the existence of the solution u∈ ˙H¹(D). Lemma2.3applied to z:= ∇ugives (2.9). 2

Now we consider the problem

Ga=∂1δa inD,˜ (2.10)

∂G_a

∂n =0 on∂D,˜ (2.11)

whereδ_ais the Dirac delta function concentrated ata=(a₁, a₂)∈ ˜D.

Proposition 2.5.Problem(2.10), (2.11)has a solutionG_a∈C^∞(D˜\ {a}). Moreover, the following assertions hold:

(i) For any open neighbourhoodQofa and for any integers1, the solutionG_a is uniquely determined by the additional condition that it belongs toH˙^s(D˜\Q).

(ii) For anyx∈ ˜D\ {a}

∇Ga(x)= − 1 2π

|x−a|²−2(x1−a₁)²

|x−a|⁴ ,−2(x1−a₁)(x₂−a₂)

|x−a|⁴

+ψa(x), (2.12)

whereψ_a∈H^∞(D).˜

(iii) Leta∈ ˜D\D, thenG_a∈ ˙H^∞(D)and for any integers1i, j2we have

∂_i∂_jG_a(x₁, x₂)∈S(D). (2.13)

(iv) For any fixedx∈ ˜Dthe functionG_a(x)is analytic ina∈ ˜D\ {x}.

Proof. The existence of a solutionG_a∈C^∞(D˜ \ {a})will be established when proving assertion (ii). To prove the uniqueness of the solution, we assume that there are two solutionsG_1,a andG_2,a. ForG˜ =G_1,a−G_2,a we have

G˜ =0 inD,˜

∂G˜

∂n =0 on∂D.˜

Letχ∈C₀^∞(D)˜ withχ=1 inQ. Then (χG)˜ =h,

whereh∈C₀^∞(D). The elliptic regularity for a bounded domain implies that˜ χG˜ ∈H^∞(D). Since˜ G˜ ∈ ˙H^s(D˜ \Q), we getG˜ ∈ ˙H^s(D). It follows from Proposition˜ 2.4thatG˜ =0.

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To prove (ii), we seek the solution in the form

G_a=∂₁(F_aχ )+u_a, (2.14)

whereF_a(x)= −_2π¹ ln|x−a|is the fundamental solution of the Laplace operator in R²,χ∈C₀^∞(D),˜ χ is 1 in a neighbourhood ofa. Thenu_amust be the solution of the problem

u_a= −∂₁(2∇F_a· ∇χ+F_aχ ):=∂₁f inD,˜

∂u_a

∂n =0 on∂D.˜

Sincef∈C₀^∞(D), applying Proposition˜ 2.4forg=(f,0), we conclude that this problem has a solutionu_a∈H^∞(D).˜ Property (2.12) follows from the construction ofG_a.

Now let us show (2.13). We have thatG_asatisfies the following problem inD:

G_a=0 inD, (2.15)

∂Ga

∂n =ϕ onΓ, (2.16)

whereϕ∈C^∞(Γ )and suppϕ⊂Γ₀. To show that the second derivatives of the solution belong toS(D), let us apply the Fourier transform inx1to (2.15), (2.16). We obtain

d²

dx₂²Gˆ_a−ξ²Gˆ_a=0 inD, dGˆ_a

dx₂(ξ,−1)= ˆϕ1(ξ ), dGˆ_a

dx₂(ξ,1)= ˆϕ₂(ξ ),

whereGˆ_a,ϕˆ₁andϕˆ₂ are Fourier transforms ofG_a,ϕ(·,−1)andϕ(·,1), respectively. The solution of this ODE is given by

Gˆ_a(ξ, x₂)= ϕˆ2− ˆϕ1

2ξsinh(ξ )cosh(ξ x2)+ ϕˆ2+ ˆϕ1

2ξcosh(ξ )sinh(ξ x2).

Sinceϕ₁andϕ₂are compactly supported, we have F(∂_i∂_jG_a)∈S(D), 1i, j2,

whence it follows that∂_i∂_jG_a∈S(D). This completes the proof of (iii).

LetΩ be any domain such thatΩ⊂ ˜DandΩ∩(D˜ \D)= ∅. Then for any fixedx∈Ω the functionG_a(x)is analytic ina∈Ω\ {x}. Indeed, letχin (2.14) be 1 inΩ. Then the analyticity ofG_a(x)is consequence of the facts thatF_ais analytic inaandu_ais a linear operator inF_a. SinceG_ais the unique solution of (2.10), (2.11), we have the analyticity ofG_a(x)inD˜ \ {x}. 2

2.2. Euler equations in an unbounded strip We consider the incompressible Euler system:

˙

u+ u,∇u+ ∇p=0, divu=0 inD, (2.17)

u·n=0 onΓ, (2.18)

u(x,0)=u₀(x). (2.19)

It is well known that ifD is a bounded domain or ifD=R², then problem (2.17)–(2.19) is well posed in various function spaces (e.g., see[17,18,22]).

In this subsection, we study the well-posedness of Euler system inDdefined by (1.2).

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Definition 2.6.For any integers3 we say that(u, p)is a solution of Euler system if(u, p)∈C(J_T, H^s(D))× C(J_T,H˙^s⁺¹(D))and (2.17) is satisfied in the sense of distributions.

Let us show that the Euler system is equivalent to the problem

˙

w+ u,∇w=0, w(x,0)=curlu₀(x), (2.20)

curlu=w, divu=0, u·n|Γ =0. (2.21)

Clearly, if(u, p)is a solution of the Euler system, then (2.20), (2.21) hold. Now let us show that to any solution (u, w)∈C

J_T, H^s(D)

∩C¹

J_T, H^s⁻¹(D)

×C

J_T, H^s⁻¹(D)

of (2.20), (2.21) there corresponds a unique solution (u, p)∈C(J_T, H^s(D))×C(J_T,H˙^s⁺¹(D))of (2.17)–(2.19).

From (2.20) and (2.21) it follows that curl

˙

u+ u,∇u

=0.

Hence, there existsp∈C(JT,H˙^s(D))such that−∇p= ˙u+ u,∇u. It is easy to see that

−div∇p=div

u,∇u

= 2 i,j=1

∂_iu_j∂_ju_i∈H^s⁻¹, curl∇p=0,

−∂p

∂n =

u,∇u

·n= u,∇(u· ˜n)− 2 i,j=1

u_ju_i∂_jn˜_i

= − 2 i,j=1

u_ju_i∂_jn˜_i∈H^s⁻^1/2,

wheren˜is a regular extension ofn. Thus, it follows from Lemma2.3that∇p∈C(J_T, H^s(D)), whence we conclude thatp∈C(J_T,H˙^s⁺¹(D)).

We have the following result on the local well-posedness of Euler system. The ideas used in the proof of existence of a solution play an important role in the study of stabilization problem (see Section3). Therefore we present a rather complete proof, even though we do not really need this result.

Theorem 2.7.Lets4. For anyu₀∈H^s(D)satisfying the conditions divu₀=0,

u₀·n=0 onΓ,

there is T_∗ = T_∗(u₀s) such that system (2.17)–(2.19) has a unique solution (u, p) ∈ C(J_T_∗, H^s(D)) × C(J_T_∗,H˙^s⁺¹(D)).

Proof. Uniqueness.To prove the uniqueness, we argue as in the case of bounded domain. We assume that there are two solutionsu1andu2. Then forv=u1−u2, we have

˙

v+ u₁,∇v+ v,∇u₂+ ∇p=0,

divv=0, v·n|Γ =0, v(x,0)=0. (2.22)

Multiplying (2.22) byvand integrating overD, we get

∂_tv(·, t )²−

D

u₁,∇v·vdx+Cv(·, t )²−

D

∇p·vdx, (2.23)

whereC >0 is a constant depending only onu₂. Sinceu₁·n=0, the first term on the right-hand side of (2.23) is zero. Let us show that the last term is also zero. Let us denote

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Ω_(R):=

x∈D: |x₁|< R , and letχ∈C^∞(D)be such that

χ (x)=

0, ifx /∈Ω₍₂₎, 1, ifx∈Ω₍₁₎. Clearly, we have

Rlim→∞

D

χ x

R

∇p(x)·v(x)dx=

D

∇p(x)·v(x)dx.

On the other hand, integrating by parts, we obtain

D

χ x

R

∇p(x)·v(x)dx= −

Ω(2R)\Ω_(R)

∇χ x

R p(x)

R ·v(x)dx.

Sincep∈ ˙H^s⁺¹, from assertion (iii) of Proposition2.1we have sup

x∈Ω(2R)

p(x) R

< C,

whereCdoes not depend onR. Thus, dominated convergence theorem yields

D

∇p(x)·v(x)dx=0.

Applying the Gronwall inequality to (2.23), we obtainv=0.

Existence.To prove the existence of the solution, we shall need the following result.

Lemma 2.8.Letu˜∈C(R+, H^s),u˜·n|Γ×R+=0,f ∈C(R+, H^s)andw0∈H^s,s3. Then the problem

∂_tw+ ˜u,∇w=f, (2.24)

w(x,0)=w0, (2.25)

has a unique solutionw∈C(R₊, H^s), which satisfies the inequality w(·, t )

sw0s+ t 0

f (·, τ )

s+Cw(·, τ )

s∇ ˜u(·, τ )

s−1

dτ. (2.26)

Proof. Let us denote byφ^g:D˜×R+→ ˜Dthe flow associated tog, i.e., the solution of the problem

∂φ^g

∂t =g φ^g, t

, φ^g(x,0)=x.

Since (2.24), (2.25) is an inhomogeneous transport equation, its solution is given by w

φ^u^˜(x, t), t

=w0(x)+ t 0

f

φ^u^˜(x, τ ), τ dτ.

Let us derive formally inequality (2.26). Taking the ∂^α := _∂x^∂^αα, |α|s derivative of (2.24) and multiplying the resulting equation by∂^αw, we get

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1 2

d

dt∂^αw²=

D

∂^αf ∂^αwdx−

D

∂^α(u˜· ∇)w·∂^αwdx

D

(u˜· ∇)∂^αw·∂^αwdx

+ fsws+C∇ ˜us−1w²s.

Integrating by parts, one verifies that the first integral in the right-hand side vanishes. Integrating in time, we obtain (2.26). 2

Lemma 2.9.Letw∈H^s, s0. Then the problem

curlz=w, (2.27)

divz=0, (2.28)

z·n|Γ =0 (2.29)

has a unique solutionz∈H^s⁺¹. Moreover, there isC >0depending only onssuch that

zs+1Cws. (2.30)

Proof. Let us consider the following Dirichlet problem for the Poisson equation:

v=w inD, v=0 onΓ.

By Proposition2.2,v∈H^s⁺²andvs+2Cws. Then forz= −∇^⊥vproperties (2.27)–(2.30) are satisfied. 2 We now return to the proof of the theorem. The proof is based on some ideas from[3]and[5].

Step 1.Let

E:H^k(D)→H^k R²

, 0ks+1

be an extension operator. Letρ∈S(R²)be the function such that ˆ

ρ(ξ )=

exp

−₁_−|^|^ξ^|_ξ²_|2

, |ξ|<1,

0, |ξ|1.

DefineJ_m:H^s(D)→H^s⁺¹(D)by J_m(v):=

m²ρ(mx)∗E(v)

D. (2.31)

Foru0∈H^s(D)we defineu^m₀ :=Jm(u0). Then u^m₀ →u₀ inH^s(D), u^m₀

sCu₀s, u^m₀

s+1mCu₀s, (2.32)

u^m₀ −u^k₀

s=o(1) and u^m₀ −u^k₀

1=o 1

m^s⁻¹

asm→ ∞, (2.33)

where (2.33) holds uniformly ink > m. Using Lemmas2.8and2.9, we define the sequencesu^m∈C(R+, H^s⁺¹)and w^m∈C(R+, H^s)by

u⁰=u₀,

∂_tw^m⁺¹+ u^m,∇

w^m⁺¹=0, w^m⁺¹(0)=curlu^m₀⁺¹, curlu^m⁺¹=w^m⁺¹, divu^m⁺¹=0, u^m⁺¹·n|Γ =0.

Our strategy is to show that sequenceu^mis convergent and the limit is the solution of Euler system. From (2.26) we derive

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w^m(·, t )

icurlu^m₀

i+C1

t 0

w^m(·, τ )

iu^m⁻¹(·, τ )

idτ (2.34)

fori=s−1, s.

Step 2.In this step, we show that there exists a timeT_∗=T_∗(u₀s)such that for anyt∈J_T_∗ w^m(·, t )

s−1Cu^m₀

s, w^m(·, t )

sCu^m₀

s+1mCu₀s. (2.35)

By induction, let us prove fori=s−1, sthe inequality w^m(·, t )

iy_m(t), (2.36)

whereCdoes not depend onmandy_m(t)is the solution of

˙

ym=C1y²_m, ym(0)=curlu^m₀

i. (2.37)

Clearly (2.36) holds form=0 for a sufficiently largeC. Assume that it holds also form−1 and let us prove it form.

From the construction ofρˆ we haveu^m₀⁻¹iu^m₀i, hencey_m₋₁y_m. Thus, from (2.34), (2.37) and induction hypothesis, we have

w^m(·, t )

i−y_mC₁ t 0

w^m(·, τ )

iu^m⁻¹(·, τ )

i−y_m² dτ

C1

t 0

ymw^m(·, τ )

i−ym

dτ.

Inequality (2.36) follows from the Gronwall inequality. It is easy to see that (2.36) yields (2.35).

Step 3. Now let us show thatw^m converges in C(J_T_∗, H^s⁻¹). In view of Lemma 2.9, sequenceu^m converges in C(J_T_∗, H^s)and the limituis the solution of Euler problem.

Notice that form < kwe have

∂_t

w^m−w^k +

u^k⁻¹,∇

w^m−w^k

=

u^k⁻¹−u^m⁻¹,∇

w^m. (2.38)

DenoteK^m,k(t):= w^m(·, t )−w^k(·, t )s−1. Lemma2.8implies K^m,k(t)u^m₀ −u^k₀

s+C t

0

K^m,k(τ )u^k⁻¹(·, τ )

s−1

+u^m⁻¹(·, τ )−u^k⁻¹(·, τ )

s−1w^m(·, τ )

s

dτ. (2.39)

On the other hand, w^m

sCm, u^m⁻¹−u^k⁻¹

s−1u^m⁻¹−u^k⁻¹^s⁻¹¹

1 u^m⁻¹−u^k⁻¹^s^s⁻−²1

s . (2.40)

Assume for a moment that

U^m,k:=w^m⁻¹−w^k⁻¹o 1

m^s⁻¹

. (2.41)

Substituting (2.40) into (2.39) and using (2.33) and (2.41), we obtain K^m,k(t)o(1)+C

t 0

K^m,k(τ )u^k⁻¹(·, τ )

s−1

dτ.

Using the Gronwall inequality, we obtain the convergence ofw^minC(J_T_∗, H^s⁻¹(D)).

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Step 4.To complete the proof of the theorem, it remains to show (2.41). Taking the scalar product of (2.38) with w^m−w^kinL², we get

U^m,k(t)Cu^m₀ −u^k₀

1+C t 0

U^m⁻^1,k⁻¹(t₁)dt1.

Iterating this inequality, one deduces U^m⁺^p,k⁺^p(t)Cu^m⁺^p

0 −u^k₀⁺^p

1+C t 0

U^m⁺^p⁻^1,k⁺^p⁻¹(t₁)dt1

Cu^m⁺^p

0 −u^k₀⁺^p

1+C t 0

Cu^m⁺^p⁻¹

0 −u^k₀⁺^p⁻¹

1+C

t1

0

U^m⁺^p⁻^2,k⁺^p⁻²(t₂)

dt2dt1

Cu^m₀⁺^p−u^k₀⁺^p

1+C t 0

Cu^m₀⁺^p⁻¹−u^k₀⁺^p⁻¹

1+ · · ·

+C

t_p₋₁

0

Cu^m₀⁺¹−u^k₀⁺¹

1+C

tp

0

U^m,k(tp)

dtp· · ·dt2dt1. Hence, for anyt∈J_T_∗we obtain

U^m⁺^p,k⁺^p p j=1

C^p⁻^j⁺¹T_∗^p⁻^j

(p−j )! u^m₀⁺^j−u^k₀⁺^j

1+C^p⁺¹T_∗^p p! max

t∈[0,T_∗]U^m,k. (2.42)

Since ∞ j=1

C^j⁺¹T_∗^j j! <∞,

inequalities (2.33) and (2.42) imply (2.41). 2 Remark 2.10.We have the following assertions:

• Adapting the Beale–Kato–Majda criterion (see [4]) for an unbounded strip, one can prove that the solution of (2.17)–(2.19) is global in time. However, we shall not need this result.

• Let us take any non-zero functiong∈H^s⁻^1/2(Γ ). If the homogeneous boundary condition (2.18) is replaced by u·n|Γ =g, then, the result of Theorem2.7holds if we add the boundary condition

curlu=φ whereu·n <0.

See[23]for the case of a bounded domain.

3. Main result

LetDandΓ₀be defined by (1.2) and (1.3). Consider the Euler system:

˙

u+ u,∇u+ ∇p=0 inD×(0,∞), (3.1)

divu=0, (3.2)

u·n=0 onΓ \Γ₀×R₊, (3.3)

u(x,0)=u₀(x). (3.4)

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For any integerswe denote X^s(D)=C

R₊, C_b(D)∩ ˙H^s(D) ,

andx₁ :=(1+x₁²)^1/2. The following theorem is our main result.

Theorem 3.1.For any constantsα, β >0,c∈Rand integers4, for any initial datau₀∈H^s(D)such that

divu₀=0, (3.5)

u₀·n=0 onΓ \Γ₀, (3.6)

exp

αx1²⁺^β

curlu0(x1, x2)

s−1<∞ (3.7)

there is a solution(u, p)∈X^s(D)×C(R+,H˙^s(D))of(3.1)–(3.4)with

tlim→∞u(·, t )−(c,0)

L^∞(D)+∇u(·, t )

s−1+∇p(·, t )

s−1

=0. (3.8)

As explained in the Introduction, in this formulation the control is not given explicitly, but we can assume that control acts on the system as a boundary condition onΓ0. So we show that there exists controlη∈C^∞(R+, H^s⁻^1/2(Γ0)) such that there is a solution of our system withu·n|Γ₀=ηverifying (3.8). Moreover, we show thatη(x, t )=η₁(x, t)+ μ(t )u₀(x)·n(x), whereη₁∈C^∞(Γ₀×R+)does not depend onu₀, limt→∞η₁(·, t ) =0 andμ∈C₀^∞([0,∞))is a non-negative function such thatμ(0)=1.As we mentioned in Remark2.10, we are not able to show that this solution is unique.

Using a standard scaling argument for Euler system, we can reduce this theorem to a small neighbourhood of the origin.

Theorem 3.2.There existsε >0such that for anyu0∈H^s(D)andc∈Rverifying(3.5)–(3.7)and u0s< ε, |c|< ε

there is a solution(u, p)∈X^s(D)×C(R+,H˙^s(D))of(3.1)–(3.4)satisfying(3.8).

Proof of Theorem3.1. Letε >0 be the constant in Theorem3.2. Take anyu₀∈H^s(D)andc∈Rverifying (3.5)–

(3.7). LetM >0 be such that u0

M

s

< ε, c

M < ε.

By Theorem3.2, there exists a solution(u_M, p_M)of (3.1)–(3.3) with initial conditionu_M(0)=^u_M⁰, such that

tlim→∞

u_M(·, t )− c

M,0 L^∞(D)

+∇u_M(·, t )

s−1+∇p_M(·, t )

s−1

=0.

Then(u, p)=(Mu_M(x, Mt ), M²p_M(x, Mt))is a solution of our system withu(0)=u₀and it satisfies (3.8). 2 Proof of Theorem3.2. The proof of this theorem is based on generalization of the Coron return method to the case of an unbounded strip. It consists in construction of a particular solution(u, p)of (3.1)–(3.3) such that the solution of linearized system around(u, p)verifies property (3.8). Then, in the small neighbourhood ofu, we construct a solution uof Euler system satisfying (3.8).

Step 1.In this step, we construct a particular solution(u, p)of (3.1)–(3.3) such that any point of stripD, driven by the flow ofu, leavesDat some time. LetDˆ ⊂R²be the strip

Dˆ :=

(x₁, x₂): x₁∈R, x₂∈(−2,2) .

Let us admit the proposition below, which is proved in Section4.1.

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Proposition 3.3.There are scalar functionsθⁱ∈C¹(Dˆ×R₊)with∇θⁱ∈X^s(D), open ballsˆ Bⁱ, a sequenceτ_i⊂R₊, constantsL,λand an integerN∈Nsuch that the following properties are true.

1. Covering. For any integerk0, we have [k, k+1] × [−1,1] ⊂

N j=1

B^2kN⁺^j, (3.9)

[−k−1,−k] × [−1,1] ⊂ N j=1

B^(2k⁺^1)N⁺^j. (3.10)

In particular, the union of ballsBⁱ coversDand any square[k, k+1] × [−1,1]is covered byNballs.

2. Support.

suppθⁱ⊂ ˆD×(0, τi). (3.11)

3. Vector field. The time dependent vector field∇θⁱis divergence-free inDand tangent toΓ \Γ₀and∂D:ˆ

θⁱ=0 inD× [0, τi], (3.12)

∂θⁱ

∂n =0 on(Γ \Γ₀)∪∂Dˆ× [0, τi]. (3.13) 4. Time decay. For anyi1we have

∇θⁱ(·, t )

X^s(D)ˆ 1

i for anyt∈ [0, τi], (3.14)

τ_iLi. (3.15)

5. Flow. For anyi1andc∈Rwith|c|< λthe flow associated with∇θⁱ+(c,0)is such that φ^∇^θⁱ⁺^(c,0)

Bⁱ, τ_i

⊂ ˆD\D. (3.16)

Moreover, there are two closed ballsB˜1,B˜2⊂ ˆD\Dsuch that ∞

i=1

φ^∇^θⁱ⁺^(c,0) Bⁱ, τ_i

⊂ ˜B₁∪ ˜B₂. (3.17)

Let us sett₀=0, ti=2

i j=1

τj, ti+1/2=t_i+t_i₊₁

2 , i1. (3.18)

We defineθin the following way:

θ (x, t )=θⁱ(x, t−t_i₋₁) fort∈ [t_i₋₁, t_i₋_1/2], (3.19) θ (x, t )= −θⁱ(x, ti−t ) fort∈ [ti−1/2, ti]. (3.20) Notice that from the construction oft_i we havet_i−t_i₋_1/2=τ_i. Thus (3.11) shows thatθ∈C¹(Dˆ ×R+)and∇θ∈ X^s(D). We defineˆ

u:= ∇θ+(c,0), p:= −∂_tθ−|∇θ|²

2 −c∂₁θ.

Then(u, p)is a solution of (3.1)–(3.3). Indeed, by construction,(u, p) satisfies (3.1). Properties (3.12) and (3.13) imply (3.2) and (3.3), respectively. Moreover, it follows from (3.14), (3.16) that for anyi∈N, we have