of Periodic Flows in a Two Dimensional Channel

Normal Feedback Stabilization

of Periodic Flows in a Two Dimensional Channel

Ionut ¸ Munteanu

Communicated by Viorel Barbu

AbstractWe consider a two-dimensional incompressible channel flow with periodic condition along one axis. We stabilize the linearized system by a boundary feedback controller with vertical velocity

observation, which acts on the normal component of the velocity only. The stability is achieved without any a priori condition on the viscosity coefficient, that is on Reynolds number.

KeywordsNavier-Stokes equations·Fourier decomposition·analyticC₀-semigroups·feedback controller AMS Subject Classification93B52, 93C20, 93D15, 35Q30, 76D55,76D05

1 Introduction

In this article, we treat the problem of boundary control of a viscous and incompressible fluid flow in a two-dimensional channel. Our objective is to globally stabilize the parabolic equilibrium profile in channel flow. The mathematical theory of stabilization of equilibrium solutions to Newtonian fluid flows has been developed, as a principal tool, to attenuate or eliminate the turbulence. The first result, that directly addresses flow stabilization, is the Coron’s [1] result on stabilization of Euler’s equations.

The same problem, we study here, has been investigated also in the papers [2] and [3]. We mention that this work is an improvement of the paper [2], in the next sense: in [2], in order to achieve stability, the normal velocity is controlled on the both walls of the channel. Here, we want to achieve exponential stability by controlling the normal component of the velocity field on one wall only. Even if, in both papers, the results rely on the Fourier decomposition of the flow and stabilization at the level of coefficients, the technique used in the present paper, to design the feedback controller, is totally different from the one in [2]. Hence, the feedback controller, which assures the exponential stability, has a totally different structure than the one in [2]. Moreover, the feedback is easily manageable from computational point of view, as we shall see in what follows.

The author gratefully acknowledges the support of the project POSDRU/88/1.5/S/47646, cofinantiated by the European Social Found, the Operational Sectorial Programme Development of Human Resources 2007-2013 and by CNCSIS project PN II IDEI ID 70/2008

”O. Mayer” Institute of Mathematics of the Romanian Academy and ”Al.I. Cuza” University, Department of Mathematics 700506 Ia¸si, Romania, e-mail: [email protected]

The analysis in [3] is motivated by Lyapunov stabilization results in tangential velocity actuation.

There are two main issues that appear in the work [3]: the first one is that the boundary controller acts on the tangential component of the velocity field. It is explained there that tangential actuation is technologically feasible because of the work on synthetic jets of Glezer [4]. In [4] it is shown that a teamed up pair of synthetic jets can achieve an angle of 85^◦from the normal direction, with the same momentum as wall normal actuation. The tangential velocity actuation is generated using arrays of rotating disks.

In the present work, the goal is to use only wall normal actuation, in order to obtain stability. The second problem is that the result in [3] is guaranteed only for sufficiently low values of the Reynolds number.

However, in simulations, they demonstrate that the control law has a stabilizing effect far beyond the value required in the theorem (five or more orders of magnitude). In our work, the stability is achieved without any a priori condition on the Reynolds number.

The feedback, designed in the present work, is not limited to two dimensional channel flows. It applies equally well to three dimensional channels, for stabilization. Since this is nontrivial, stabilization of three- dimensional channel flows, using the same technique as in the present paper, represents the topic of a future work.

The paper is organized as follows. In Section 2, we present the problem. In Section 3 and Section 4, we introduce the Fourier functional settings and discuss some properties of the operators which appear.

In Section 5, we state the main result of this work. The proof of the main theorem is presented in Section 9 and collects all the results from Section 6, Section 7 and Section 8. Finally, we conclude with some remarks in Section 10.

2 Preliminaries

In this paper, we consider a 2-D incompressible channel flow evolving in a semi-infinite rectangle, defined by (x, y)∈]− ∞,∞[×[0,1], with the walls located aty= 0 andy= 1.

The dynamic of the flow is governed by the dimensionless and incompressible 2-D Navier-Stokes equations:

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

u_t−ν∆u+uu_x+vu_y=p_x, t≥0, x∈R, y∈]0,1[, vt−ν∆v+uvx+vvy =py, t≥0, x∈R, y∈]0,1[, u_x+v_y= 0, t≥0, x∈R, y∈]0,1[,

u(t, x,0) =u(t, x,1) = 0, v(t, x,0) = 0, v(t, x,1) = Ψ(t, x),

u(t, x+ 2π, y) =u(t, x, y), v(t, x+ 2π, y) =v(t, x, y), p(t, x+ 2π, y) =p(t, x, y),

∀t≥0,∀x∈R,∀y∈]0,1[,

(1)

with initial data

u(0, x, y) =uo(x, y), v(0, x, y) =vo(x, y), x∈R, y∈]0,1[.

Here (u(t, x, y), v(t, x, y)) is the velocity field and p(t, x, y) is the unknown pressure (for more details about the Navier-Stokes equations see, for example, [5] and [6]). Both the velocity field and the pressure are assumed to be 2π−periodic in the first spatial coordinatex. Of course, this is not true in reality, but, it is often assumed, for mathematical convenience, because it does not alter the essential features of the behavior of the flow (for more explanations see, for example, [7]).

In order to compute the equilibrium state, we put the boundary control to be zero, i.e., v= 0, at y= 1, ∀x∈R,

and consider a steady-state flow with zero wall-normal velocity, i.e., (U(x, y),0). From the divergence free, we first get thatU_x≡0, this means thatU =U(y). Substituting (U,0) into equation (1), we obtain

−νU⁰⁰(y) =p^e_x(x, y), p^e_y(x, y)≡0, (2) wherep^e is the equilibrium pressure corresponding to the steady-state solution (U,0) (by⁰ we mean the derivative with respect to the second spatial coordinatey, i.e., _∂y^∂ ). From the relation above, we deduce thatp^e≡p^e(x) andp^e_x=−νU⁰⁰=const. Hence, U⁰⁰⁰ ≡0 . This yields

U(y) =C(y²−y), y∈]0,1[, (3)

whereC∈R−. In the following, we takeC=−_2ν^a, wherea∈R+.

This is a parabolic laminar flow profile. It is well known that the stability property, of the stationary flow (U,0), varies with the Reynolds number ¹_ν (forν small the flow is unstable, forν large enough the flow is stable).

Our aim here is the stabilization of this flow profile by a boundary controller on the wally= 1, that is v(t, x,1) = Ψ(t, x), t≥0, x∈R.

There is no action, however, iny= 0, for streamwise or inside the channel. Therefore, only the normal velocityvis controlled on the wall y= 1.

By the substitutions (u, v)→(u, v)−(U,0), p→p−p^e, we are readily led, via (1) and (2), to the study of null stabilization of the system

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



ut−ν∆u+uxU+vU⁰+uux+vuy =px, x∈R, y∈]0,1[, v_t−ν∆v+v_xU+uv_x+vv_y =p_y, x∈R, y∈]0,1[, ux+vy= 0,

u(t, x,0) =u(t, x,1) = 0, v(t, x,0) = 0, v(t, x,1) = Ψ(t, x),

u(t, x+ 2π, y) =u(t, x, y), v(t, x+ 2π, y) =v(t, x, y), p(t, x+ 2π, y) =p(t, x, y),

∀t≥0,∀x∈R,∀y∈]0,1[,

(4)

with initial data

u(0, x, y) =u⁰(x, y) :=uo(x, y)−U(y), v(0, x, y) =v⁰(x, y) :=vo(x, y), x∈R, y∈]0,1[.

The linearization of (4) is the following system:

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ut−ν∆u+uxU+vU⁰=px, vt−ν∆v+vxU =py, ux+vy= 0,

u(t, x,0) =u(t, x,1) = 0, v(t, x,0) = 0, v(t, x,1) = Ψ(t, x),

u(t, x+ 2π, y) =u(t, x, y), v(t, x+ 2π, y) =v(t, x, y), p(t, x+ 2π, y) =p(t, x, y),

∀t≥0,∀x∈R,∀y∈(0,1),

(5)

with initial datau⁰, v⁰.

The main result here (see Theorem 5.1 below) is that the exponential stability of (5) can be achieved by a normal boundary, finite-dimensional feedback controller, of the form

Ψ(t, x) =−ν X

0<|k|≤S

(L⁻²_k RkLkvk(t))⁰⁰⁰(1)e^ikx,

where vk(t, y) = R2π

0 v(t, x, y)e^−ikxdx,0 < |k| ≤ S, more precisely: for every initial data (u⁰, v⁰), the corresponding solution of the closed-loop system (5) satisfies

k(u(t), v(t))k²≤Ce^−αtk(u⁰, v⁰)k², t≥0,

for someC, α >0. Here,k · k denotes the norm of (L²( ]0,2π[×]0,1[ ))² andS is given by relation (17) below. The expression of the stabilizable feedback controller involves linear operatorsRk:X → X, which are selfadjoint and satisfy Riccati algebraic equations inH

L⁻¹_k Rkz0k, L⁻¹_k Akz0k

+1

2ν²|(L⁻²_k Rkz0k)⁰⁰⁰(1)|²=1

2kL⁻¹_k z0kk²,∀z0k ∈H,

for all 0<|k| ≤S. Ak =FkL⁻¹_k , whereFk is given by (12) and Lk is given by (11). <·,·>andk · k stand for the scalar product and the norm inH, respectively. H being the complexified space ofL²(0,1).

X is the dual space H²(0,1)∩H₀¹(0,10

.

In a few words, our approach is the following one: we reduce first, via Fourier functional setting, the linear system (5) to an infinite parabolic system, and prove the boundary stabilization of it (see Theorem 5.2 and notations (50) below). Next, using a standard technique-minimization of a cost functional, we prove the feedback exponential stability of the infinite parabolic system (see Theorem 8.1 below). Finally, we plug that feedback controller, determined in the previous step, into the linear system (5), and prove the feedback asymptotical stability of it (see Theorem 5.1 below).

The boundary stabilization of Navier-Stokes equations, with deterministic tangential feedback con- trollers, was studied, for example, in [8],[9]. However, there are few results on boundary stabilization with normal controllers, and most refer to periodic flows in 2-D channels [10],[3],[11]. Most of these stabiliza- tion results are established for large Reynolds number with exception of [2], [12], where the stabilizable feedback controller is constructed without any a priori condition onν, which is also the case of this paper.

The stochastic approach, for the normal stabilization of periodic flows in a channel, is developed in the work [13] (for more details about stabilization of Navier-Stokes flows, one can check, for example, the book [14]).

3 The Fourier Functional Setting

LetL²(Q), Q=]0,2π[×]0,1[, be the space of all functionsu∈L²_loc(R×]0,1[), which are 2π−periodic in x. These functions are characterized by their Fourier series

u(x, y) =X

k∈Z

uk(y)e^ikx, uk= ¯u−k, ∀k∈Z, such that

X

k∈Z

Z 1 0

|uk(y)|²dy <∞.

The norm inL²(Q) is defined as

|u|L²(Q):= 2πX

k∈Z

|uk|²_L2(0,1)

!¹₂ .

Since is no danger of confusion, we shall denote byk · kthe norm in bothL²(Q) andL²(0,1). We set (L²(Q))²:=

(u, v) :u, v ∈L²(Q) .

We denote also byk · kthe norm in (L²(Q))², defined as

k(u, v)k:= kuk²+kvk²¹₂ .

We return to system (5) and rewrite it in terms of Fourier coefficients (u_k)_k∈Z, (v_k)_k∈Z, (p_k)_k∈Z, (ψk)_k∈Z. We have

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(uk)t−νu⁰⁰_k+ (νk²+ikU)uk+U⁰vk=ikpk, a.e. in ]0,1[, (vk)t−νv⁰⁰_k+ (νk²+ikU)vk=p⁰_k, a.e. in ]0,1[,

ikuk+v_k⁰ = 0, a.e. in ]0,1[,

uk(0) =uk(1) = 0, vk(0) = 0, vk(1) =ψk,

(6)

with initial datau⁰_k, v⁰_k.Here u:=X

k∈Z

uk(t, y)e^ikx, v:=X

k∈Z

vk(t, y)e^ikxandp:=X

k∈Z

pk(t, y)e^ikx, Ψ :=X

k∈Z

ψk(t)e^ikx.

In what follows, we shall considerH to be the complexified space ofL²(0,1). We denote also byk · k the norm inH and by<·,·>the scalar product. We shall denote byH^m(0,1),m= 1,2, ..., the standard Sobolev spaces on ]0,1[ and

H₀¹(0,1) :=

v∈H¹(0,1) :v(0) =v(1) = 0 , H₀²(0,1) :=

v∈H²(0,1)∩H₀¹(0,1) :v⁰(0) =v⁰(1) = 0 . First, let us treat the case whenk= 0. In this case, system (6) has the next form

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(u0)t−νu⁰⁰₀+v0U⁰ = 0, a.e. in ]0,1[, (v₀)_t−νv⁰⁰₀ =p⁰₀, a.e. in ]0,1[, v⁰₀= 0, a.e. in ]0,1[,

u₀(0) =u₀(1) = 0, v₀(0) = 0, v₀(1) =ψ₀.

(7)

Taking into account the divergence free condition,v⁰₀= 0, ∀t≥0, and the boundary condition, v₀(0) = 0, ∀t≥0, we deduce thatv0≡0. Which implies also thatψ0≡0. Thus,u0 satisfies the next equation

( (u₀)_t−νu⁰⁰₀ = 0, y∈]0,1[,

u0(0) =u0(1) = 0. (8)

Multiplying (scalarly) equation (8) by u0, and using the Poincar´e inequality, we get that, for some constantC₀>0, we have

1 2

d

dtku0(t)k²+C0νku0k²≤0, t≥0.

This implies that

ku0(t)k²≤e^−2C0νtku⁰₀k², ∀t≥0. (9) This means that, for the Fourier modesu0, v0, the exponential stability holds true without any controller.

That is why, from now on, we consider only the case whenk6= 0.

Reducingpk from the first two equations from (6), we obtain

ik(vk)t−ikνv⁰⁰_k+ik²(νk+iU)vk−(u⁰_k)t+νu⁰⁰⁰_k −k(νk+iU)u⁰_k−ikU⁰uk−U⁰v⁰_k−U⁰⁰vk= 0.

Taking into account thatuk=−_ik¹v_k⁰, we have

ik(v_k)_t−ikνv_k⁰⁰+ik²(νk+iU)v_k+ 1

ik(v_k⁰⁰)_t− ν

ikv_k^iv+1

i(νk+iU)v_k⁰⁰−U⁰⁰v_k = 0, t≥0, y∈]0,1[.

Finally,

( (v⁰⁰_k−k²vk)t−νv^iv_k + (2νk²+ikU)v⁰⁰_k−k(νk³+ik²U+iU⁰⁰)vk= 0, t≥0, y∈]0,1[,

v⁰_k(0) =v_k⁰(1) = 0, vk(0) = 0, vk(1) =ψk(t). (10) For each 06=k∈Z, we denote byLk :D(Lk)⊂H →H andFk :D(Fk)⊂H→H the operators

L_kv:=−v⁰⁰+k²v,D(L_k) =H²(0,1)∩H₀¹(0,1), (11) Fkv:=νv^iv−(2νk²+ikU)v⁰⁰+k(νk³+ik²U +iU⁰⁰)v, D(Fk) =H⁴(0,1)∩H₀²(0,1). (12) With these notations, equation (10) becomes

( (L_kv_k)_t+F_kv_k = 0, t≥0, y∈]0,1[,

v_k⁰(0) =v⁰_k(1) = 0, vk(0) = 0, vk(1) =ψk(t). (13) Consider the solutionwk=wk(t, y) to the equation

( θkwk+Fkwk = 0, y∈]0,1[, t≥0,

w⁰_k(0) =w⁰_k(1) = 0, w_k= 0, w_k(1) =ψ_k. (14) (Forθ_k positive and sufficiently large, there exists a unique solutionw_k for equation (14)). Then, sub- tracting (13) and (14), we obtain that

(L_kv_k)_t+F_k(v_k−w_k)−θ_kw_k = 0, t≥0.

Equivalently,

(Lk(vk−wk))t+Fk(vk−wk) =θkwk−(Lk(wk))t, (15) v_k−w_k ∈ D(Fk).

In order to represent equation (15) as an abstract boundary control system, we consider the operators A_k:D(Ak)⊂H →H, defined by

Ak:=FkL⁻¹_k , D(Ak) =

v∈H :L⁻¹_k v∈ D(Fk) . (16) We have

Lemma 3.1 The operator −Ak generates a C0−analytic semigroup on H, and for each λ ∈ ρ(−Ak), (λI+A_k)⁻¹ is compact. Moreover, one has

σ(−Ak)⊂ {λ∈C:<λ≤0},∀|k|> S, where

S := 1

√ν

1 + a

√2ν ¹₂

. (17)

Hereσ(−Ak)is the spectrum of−Ak andρ(−Ak)is the resolvent set of −Ak.

Proof. The proof can be found in [13, Lemma 1]. 2 Remark 3.1 For |k| > S, S given by (17), let us consider the system (10) without any boundary controller, that is

( (v⁰⁰_k−k²v_k)_t−νv^iv_k + (2νk²+ikU)v⁰⁰_k−k(νk³+ik²U+iU⁰⁰)v_k= 0, t≥0, y∈]0,1[,

v⁰_k(0) =v_k⁰(1) = 0, vk(0) = 0, vk(1) = 0. (18)

Multiplying (scalarly) equation (18) byv_k and taking the real part of the result, we get that 1

2 d

dt kv_k⁰k²+k²kvkk²

+νkv_k⁰⁰k²+ 2νk²kv_k⁰k²+νk⁴kvkk²=−<

ik

Z 1 0

U⁰v⁰_kv¯_kdy

. Hence,

1 2

d

dt kv_k⁰k²+k²kvkk²

+ 2νk²kv⁰_kk²+νk⁴kvkk²≤k

Z 1 0

U⁰v_k⁰v¯_kdy

≤ 3

2νk²kv⁰_kk²+ 2 3ν

Z 1 0

|U⁰|²|¯vk|²dy

≤ 3

2νk²kv⁰_kk²+ a²

6ν³kvkk²≤ 3

2νk²kv⁰_kk²+1

2νk⁴kvkk², since|k|> S=^√1_ν

1 + ^√a

2ν

¹₂

.The above estimates imply that d

dt kv⁰_kk²+k²kvkk²

+νk²(kv⁰_kk²+k²kvkk²)≤0, t≥0.

This yields

kv⁰_k(t)k²+k²kv_k(t)k²≤e^−νk2t k(v_k⁰)⁰k²+k²kv⁰_kk²

, t≥0. (19)

Taking into account thatuk=−_ik¹v_k⁰, relation (19) implies that kuk(t)k²+kvk(t)k²≤e^−νS2t ku⁰_kk²+kv⁰_kk²

, ∀t≥0,|k|> S. (20) Relation (20) says that, for |k| > S, the exponential asymptotical stability holds true for the system (10) (equivalently (13)) without any boundary controller. Hence, we have to control the system (13) (equivalently, (10)) for 0<|k| ≤S only.

Now, coming back to system (15), we set

yk(t) :=Lk(vk(t)−wk(t)), and write it as

y_k(t) =e^−Akty_k(0) + Z t

0

e^−Ak(t−s)(θ_kw_k(s)−(L_k(w_k(s))_s)ds,

=e^−Aktyk(0)−Lk(wk(t)) +e^−AktLk(wk(0))+ (21)

+ Z t

0

e^−Ak(t−s)(θk(wk(s)) + ( ˜Fk(wk(s)))ds, where ˜Fk :H →(D(F_k^∗))⁰ is the extension ofFk toH, defined by

( ˜Fkv, ψ) :=

Z 1 0

v(y)F_k^∗ψ(y)dy, ∀ψ∈ D(Fk) =D(F_k^∗). (22) The space (D(F_k^∗))⁰ is the completion of the space H in the norm k|fk|:= k(λI +F_k)⁻¹fk, for some λ∈Rsufficiently large. HereF_k^∗ is the dual operator of Fk.

Define similarly ˜Ak, the extension of Ak to H. Likewise −Ak, the operator −A˜k generates a C₀−analytic semigroup on H. In the same way is defined the extension of L_k to an operator from H to (H₀¹(0,1)∩H²(0,1))⁰, again denotedLk.

Then, the above equation (21) can be rewritten as d

dt(Lkvk(t)) + ˜Ak(Lk(vk)(t)) = (θk+ ˜Fk)(wk(t)), t≥0. (23) For eachψ∈C, we denote byDkψ:=w∈H⁴(0,1) the solution to the equation

( θkw+Fkw= 0, ∀y∈]0,1[,

w(0) =w(1) = 0, w⁰(0) = 0, w⁰(1) =ψ. (24) (The operator Dk is called the Dirichlet map associated with θk+Fk) It is easy to see that the dual ((θ_k+ ˜F_k)D_k)^∗ is given by

((θk+ ˜Fk)Dk)^∗φ=νφ⁰⁰⁰(1), (25) for allφ∈H⁴(0,1), φ⁰(0) =φ⁰(1) = 0, φ(0) = 0.With this notation, equation (23) can be rewritten as

d

dt(Lkvk(t)) + ˜Ak(Lk(vk)(t)) = (θk+ ˜Fk)(Dkψk(t)), t >0 (26) and initial datav_k⁰.

Remark 3.2 Note that the systems (26), (13) and (10) are all equivalent.

Remark 3.3 Equation (26) is understood in the following weak sense:

d

dtLkvk(t), φ

+hLkvk,A^∗_kφi=hψk(t),((θk+Fk)Dk)^∗φi,∀φ∈ D(A^∗_k).

4 Further Results Concerning the Linear Operators A

j=1the eigenvalues of−Ak(λ^k_jis repeated according to its multiplicity m^k_j). Hence, ¯λ^k_j∞

j=1are the eigenvalues of the dual operator−A^∗_k, of−A_k. Denote by

φ^k_j, j= 1,2, ...

and

φ^k∗_j , j= 1,2, ... the corresponding eigenfunctions for−Ak and−A^∗_k, respectively.

Remark 4.1 Concerning the eigenvalues and eigenfunctions of the extended operator−A˜k, it is known that they coincide with the eigenvalues and eigenfunctions of−A_k, respectively. The same holds true for the dual operators−A˜^∗_k.

As mentioned before (see Remark 3.1), we have to stabilize system (26) (equivalently (13)) for 0<

|k| ≤S only. So, let us consider k∈ Zsuch that 0 <|k| ≤ S. We know, from Lemma 3.1, that −A_k generates aC0−analytic semigroup and has a compact resolvent inH, so, it has only a finite numberNk

of eigenvaluesλ^k_j with<λ^k_j ≥0, the unstable eigenvalues. Denote byM_k :=m^k₁+m^k₂+...+m^k_N

k (m^k_j is the multiplicity ofλ^k_j, j= 1,2, ..., Nk).

The next lemma is the key result of this paper. It shows that we can choose the eigenfunctionsφ^k∗_j , j= 1,2, ..., Nk, of the dual operator−A^∗_k, corresponding to the unstable eigenvalues ¯λ^k_j, j= 1,2, ..., Nk, such that (φ^k∗_j )⁰⁰⁰(1)>0.

Lemma 4.1 Let λ¯^k_j, for some 0 < |k| ≤ S and j ∈ {1, ..., Nk}, be an unstable eigenvalue of −A^∗_k. Then, we can choose the dual eigenfunctionφ^k∗_j , corresponding to the unstable eigenvalueλ¯^k_j, such that (φ^k∗_j )⁰⁰⁰(1)>0.

Proof. We shall prove that we can choose the dual eigenfunctionsφ^k∗_j such that (φ^k∗_j )⁰⁰⁰(1)6= 0. Then, replacing eventuallyφ^k∗_j by−φ^k∗_j , we obtain that we have (φ^k∗_j )⁰⁰⁰(1)>0, as claimed. The proof consists of three steps.

Step 1. For a functionf : [0,1]→C, let us denote by ˇf : [0,1]→Cthe function fˇ(y) :=f(1−y), ∀y∈[0,1].

We say that the functionf : [0,1]→Cis symmetric iff f(y) = ˇf(y),∀y ∈[0,1], and anti-symmetric ifff(y) =−fˇ(y),∀y∈[0,1].

In this step, we show that we can choose a basis, of the dual eigenfunction space, consisting of symmetric or anti-symmetric functions.

Let us denote by ¯λ:= ¯λ^k_j, the unstable eigenvalue. Ifφ^∗ :=φ^k∗_j is an eigenfunction corresponding to

¯λ, thenφ^∗ satisfies the next boundary value problem

( ν(φ^∗)^iv−(2νk²−ikU+ ¯λ)(φ^∗)⁰⁰+ 2ikU⁰(φ^∗)⁰+ (k²¯λ+νk⁴−ik³U)φ^∗= 0, a.e. in ]0,1[,

φ^∗(0) =φ^∗(1) = 0,(φ^∗)⁰(0) = (φ^∗)⁰(1) = 0. (27)

Let us observe that, ifφ^∗ is solution to (27), then ˇφ^∗ is also a solution to (27), because of the symmetric form of the equation.

Let us denote by H, the fourth-dimensional linear space of the solutions to the fourth-order linear homogeneous differential equation

ν(φ^∗)^iv−(2νk²−ikU+ ¯λ)(φ^∗)⁰⁰+ 2ikU⁰(φ^∗)⁰+ (k²λ¯+νk⁴−ik³U)φ^∗= 0, a.e. in ]0,1[. (28) Then, the eigenfunction space can be written as the linear spaceE, defined below

E:={φ∈ H:φ(0) =φ(1) = 0,(φ)⁰(0) = (φ)⁰(1) = 0}.

It is easy to see that the dimension ofE is≤2. We claim that we can find a basis, for this linear space, consisting of symmetric functions or anti-symmetric functions. Indeed, let us assume that there isφ∈ E which is niether symmetric nor anti-symmetric. Then, the next two functionsφ₁:=φ+ ˇφandφ₂:=φ−φˇ are both in E. Moreover, φ1 6= 0, φ2 6= 0 and the system {φ1, φ2} is linear independent, becauseφ1 is symmetric andφ₂ is anti-symmetric. This, together with the fact that the dimension ofE is≤2, proves our claim.

Hence, we can assume that the corresponding eigenfunctionφ^∗ is symmetric or anti-symmetric. As- sume, for example, thatφ^∗ is symmetric. The other case can be treated similarly.

We want to show that we have (φ^∗)⁰⁰⁰(1)6= 0. Let us assume, by contradiction, that we have (φ^∗)⁰⁰⁰(1) = 0. From the symmetry we get also that (φ^∗)⁰⁰⁰(0) = 0.

Let us denote byLk :D(Lk)⊂H →H the next linear operator

Lkφ:=νφ^iv−(2νk²+ikU+λ)φ⁰⁰+k(νk³+ik²U+iU⁰⁰+kλ)φ, ∀φ∈ D(Lk).

It is easy to see thatφ^∗ satisfiesL^∗_kφ^∗= 0, whereL^∗_k is the dual operator ofL_k. We have 0 =

Z 1 0

L^∗_kφ^∗φdy= Z 1

0

φ^∗L¯kφdy+ν((φ^∗)⁰⁰(1)φ⁰(1)−(φ^∗)⁰⁰(0)φ⁰(0)),∀φ∈ D(Lk), (29) taking into account the boundary conditions forφ^∗, i.e.,

φ^∗(0) =φ^∗(1) = 0,(φ^∗)⁰(0) = (φ^∗)⁰(1) = 0,(φ^∗)⁰⁰⁰(0) = (φ^∗)⁰⁰⁰(1) = 0.

From (29) we get that

Z 1 0

|φ^∗|²dy= 0, (30)

ifφsatisfies

( L_kφ=φ^∗,

φ⁰(0) =φ⁰(1) = 0. (31)

Relation (30) implies thatφ^∗≡0, which is in contradiction with the fact thatφ^∗ is an eigenfunction.

This implies that the assumption (φ^∗)⁰⁰⁰(1) = 0 is false, which leads to our wanted result. So, in order to complete the proof, it remains to show that there exists a solutionφto the equation (31).

Step 2. We claim that there exists a function φ₁such that ( Lkφ1= 0, y∈]0,1[,

φ⁰₁(0)−φ⁰₁(1)6= 0. (32) The proof of this claim will be given in the last step of the proof. In this step, we shall prove that, under the above claim, there exists a solution to the equation (31) and so, we obtain the wanted result.

Let us construct the next functionφ2:=φ1+ ˇφ1.As before, the equationLkφ1= 0 is symmetric, and this implies that ifφ1 is a solution then ˇφ1 is also a solution to the equation. Hence, we have







Lkφ₂= 0,

φ⁰₂(0) =φ⁰₁(0)−φ⁰₁(1)6= 0, φ₂ is symmetric.

(33)

Letφ3 be a solution to the equation Lkφ3 =φ^∗, such thatφ3 is symmetric. There exists a symmetric solution to the above equation becauseφ^∗ is symmetric. Indeed, let φ4 be a solution for the equation Lkφ4= ¹₂φ^∗. If we takeφ3:=φ4+ ˇφ4, we haveLkφ3= ¹₂φ^∗+¹₂φ^∗=φ^∗ andφ3 is symmetric, as wanted.

Now, we defineφ5:=−^φ_φ⁰³0⁽⁰⁾

2(0)φ2+φ3.We have thatφ5is welldefined, and moreover,φ5satisfies







Lkφ5=φ^∗, φ₅is symmetric,

φ⁰₅(0) = 0 and because of the symmetry φ⁰₅(1) = 0.

(34)

So, we can takeφ:=φ₅.

Step 3. In the last step we show that there exists a function φ1 such that ( Lkφ1= 0, y∈]0,1[,

φ⁰₁(0)−φ⁰₁(1)6= 0. (35) We assume, by contradiction, that this is not true. Hence, for every solutionψ to the equationLkψ= 0 we haveψ⁰(0)−ψ⁰(1) = 0.

Let us denote byH1 the linear space of the solutions to the equation Lkψ= 0, and byE1 the linear subspace ofH₁, defined by

E1:={ψ∈ H1:ψ⁰⁰⁰(0)−ψ⁰⁰⁰(1) = 0}. Let anyψ∈ E₁. Define Ψ :=ψ+ ˇψ. We have

Ψ⁰(0) =ψ⁰(0)−ψ⁰(1) = 0,Ψ⁰(1) =ψ⁰(1)−ψ⁰(0) = 0, sinceψ∈ H1. Also,

Ψ⁰⁰⁰(0) =ψ⁰⁰⁰(0)−ψ⁰⁰⁰(1) = 0,Ψ⁰⁰⁰(1) =ψ⁰⁰⁰(1)−ψ⁰⁰⁰(0) = 0,

sinceψ∈ E1.

Let us denote by Φ := Ψ⁰⁰−k²Ψ. The equationL_kΨ = 0 can be rewritten in the next form

νΦ⁰⁰−(νk²+ikU+λ)Φ +ikU⁰⁰Ψ = 0, (36) and

Ψ⁰⁰−k²Ψ = Φ. (37)

Observe that, since Ψ⁰(0) = Ψ⁰(1) = 0 and Ψ⁰⁰⁰(0) = Ψ⁰⁰⁰(1) = 0, we have Φ⁰(0) = Φ⁰(1) = 0.

If we multiply (scalarly) equation (36) by Φ and equation (37) by Ψ, we get

−ν Z 1

0

|Φ⁰|²dy−(νk²+λ) Z 1

0

|Φ|²dy−ik Z 1

0

U|Φ|²dy+ikU⁰⁰ Z 1

0

Ψ ¯Φdy= 0, (38) and

− Z 1

0

|Ψ⁰|²dy−k² Z 1

0

|Ψ|²dy= Z 1

0

Φ ¯Ψdy. (39)

From (39) we see thatR1

0 Ψ ¯Φdy is a real number. Using this and taking the real part of (38), we obtain that

−ν Z 1

0

|Φ⁰|²dy−(νk²+<λ) Z 1

0

|Φ|²dy= 0.

Sinceλis an unstable eigenvalue, we have that<λ >0. So, the relation above yields Φ≡0.

It is easy to see that Φ≡0 implies Ψ≡0, which implies thatψ=−ψ. Hence,ˇ

ψ∈ E1⇒ψ=−ψ.ˇ (40)

Let us consider the subspaces ofH₁, defined below S:=

ψ∈ H1:ψ= ˇψ , AS :=

ψ∈ H₁:ψ=−ψˇ ,

which are the symmetric subspace ofH₁and the anti-symmetric subspace ofH₁, respectively. It is easy to see thatS =

ψ∈ H1:ψ⁰(¹₂) =ψ⁰⁰⁰(¹₂) = 0 and AS =

ψ∈ H1:ψ(¹₂) =ψ⁰⁰(¹₂) = 0 .This implies that dim_CS= dim_CAS = 2.

Relation (40) implies thatE1⊂ AS.Let us define the next subspace ofH1

F₁:={ψ∈ H₁:ψ⁰⁰⁰(0) = 0}.

We have dim_CF1= 3. Since dim_CS= 2, dim_CH1= 4 andF1,S ⊂ H1, we have that F₁∩ S 6={0}.

Hence, there exists 06=ψ∈ F1∩ S. ψ∈ F1 implies thatψ⁰⁰⁰(0) = 0, and, from the symmetry (because ψ∈ S),ψ⁰⁰⁰(1) = 0. These yield thatψ⁰⁰⁰(0)−ψ⁰⁰⁰(1) = 0 andψ∈ H₁. We conclude that ψ∈ E₁⊂ AS. Finally, we haveψ∈ S ∩ AS, which impliesψ≡0, which is absurd. Hence, the assumption made is not true. This means that there exists a functionφ₁such that

( Lkφ1=φ^∗, y∈]0,1[,

φ⁰₁(0)−φ⁰₁(1)6= 0. (41)

As we mentioned earlier, this completes the proof. 2

We finally need one more result concerning the Dirichlet mapsDk,0<|k| ≤S.

Proposition 4.1 For all0<|k| ≤S,Dk :C→H is a linear continuous operator.

Proof. Let (ψ_n) be a complex sequence tending to zero when n goes to infinity. Remember that Dkψn∈H⁴:=qn is the solution to the equation

( θkqn+Fkqn= 0, a.e. in ]0,1[,

q_n⁰(0) =q_n⁰(1) = 0, qn(0) = 0, qn(1) =ψn. (42) Or, by (12),

( θkqn+νq^iv_n −(2νk²+ikU)q_n⁰⁰+k(νk³+ik²U+iU⁰⁰)qn= 0, a.e. in ]0,1[,

q⁰_n(0) =q⁰_n(1) = 0, q_n(0) = 0, q_n(1) =ψ_n. (43) Multiplying (scalarly) equation (43) byq_n, we get that

θkkqnk²+q_n⁰⁰⁰(1)ψn+νkq⁰⁰_nk²+ Z 1

0

(2νk²+ikU(y))|q_n⁰(y)|²dy+ik Z 1

0

U⁰(y)q⁰_n(y)¯qn(y)dy+ (44)

+νk⁴kq_nk²+i Z 1

0

(k³U(y) +kU⁰⁰(y))|q_n(y)|²dy

= 0.

Taking the real part of equation (44), we obtain that

θkkqnk²+<(q⁰⁰⁰_n(1)ψn) +νkq⁰⁰_nk²+ 2νk²kq_n⁰k²+νk⁴kqnk²= (45)

=−<

ik

Z 1 0

U⁰(y)q_n⁰(y)¯qn(y)dy

. We have

−<

ik

Z 1 0

U⁰(y)q⁰_n(y)¯qn(y)dy

≤

ik Z 1

0

U⁰(y)q_n⁰(y)¯qn(y)dy

≤ (46)

≤ |k|

Z 1 0

|U⁰(y)q_n⁰(y)qn(y)|dy.

Taking in the inequality|αβ| ≤ ¹₂

ν|k||α|²+_ν|k|¹ |β|²

,α:=q_n⁰(y) andβ:=U⁰(y)q_n(y), yields

|k|

Z 1 0

|U⁰(y)q⁰_n(y)qn(y)|dy≤1 2νk²

Z 1 0

|q_n⁰(y)|²dy+ 1 2ν

Z 1 0

|U⁰(y)|²|qn(y)|²dy (47)

≤1

2νk²kq_n⁰k²+ a²

8ν³kq_nk²≤2νk²kq_n⁰k²+ a² 8ν³kq_nk². Hence, by (45), (46) and (47), we obtain

θ_k− a² 8ν³

kq_nk²+<(q⁰⁰⁰_n(1)ψ_n)≤0. (48) Lettingnto go to infinity in relation (48), yields

0≤

θ_k− a² 8ν³

limkqnk²≤0, sinceψ_n→0 andθ_k is sufficiently large such thatθ_k−_8ν^a2₃ ≥0.

Hence limkqnk= 0. Thus, we obtained that: ifψn→0 inCthenqn=Dkψn →0 inH. SinceDk is

linear, we have thatD_k is continuous fromCtoH. 2

5 Main Results

The main result of this paper is the next theorem.

Theorem 5.1 There exists a finite-dimensional feedback controllerΨ, of the form Ψ(t, x) =−ν X

0<|k|≤S

(L⁻²_k RkLkvk(t))⁰⁰⁰(1)e^ikx, (49) where

vk(t, y) = Z 2π

0

v(t, x, y)e^−ikxdx,0<|k| ≤S,

such that, once inserted into equation(5), the corresponding solution of the closed-loop system(5)satisfies k(u(t), v(t))k²≤Ce^−αtk(u⁰, v⁰)k², t≥0,

for someC, α >0. Where,R_k:X → X are linear selfadjoint operators such that (i) Rk :H →H,

(ii) Rk satisfy Riccati algebraic equations inH L⁻¹_k Rkz0k, L⁻¹_k Akz0k

+1

2ν²|(L⁻²_k Rkz0k)⁰⁰⁰(1)|²= 1

2kL⁻¹_k z0kk²,∀z0k ∈H,

for all 0<|k| ≤S,S given by (17). Ak =FkL⁻¹_k , where Fk is given by (12) andLk is given by (11).

X = (H²(0,1)∩H₀¹(0,1))⁰.

In order to prove Theorem 5.1, we first control the system (26) (equivalently (13)), for all 0<|k| ≤S, such that to achieve the exponential asymptotical stability of it, that is Theorem 5.2 below (see notations (50)).

For simplicity, we are going to omit the symbol ˜ (since is no danger of confusion), also we are going to redefine some symbols, i.e.,

z_k:=L_kv_k,B_k := (θ_k+ ˜F_k)D_k. (50) With these notations, equation (26) becomes

d

dtzk(t, y) +Akzk(t, y) =Bkψk(t), t >0, (51) zk(0, y) =z0k(y),

wherez0k(y) :=Lkv_k⁰(y).

First, we shall prove that

Theorem 5.2 For all 0 <|k| ≤S, there exists a controller ψk such that the corresponding solution zk

to the system(51)and the controller have the exponential decay

d dtψk(t)

+|ψk(t)| ≤Cα₁e^−α1tkz0kk,kzk(t)k ≤Cα₀e^−α0tkz0kk, ∀t≥0, whereα0 is given by (53)below

0< α₀<<λ_Nk+1, andα1>0 is given by Lemma6.1below.

From the previous section, we have that the eigenvalues of−Akareλ^k_j, j= 1,2, ..., and the eigenfunctions φ^k_j, j = 1,2, .... The eigenfunctions of the dual operator−A^∗_k, of −A_k, are φ^∗k_j , j = 1,2, .... Moreover, the operator−Ak has a finite number of unstable eigenvalues (λ^k_j)^N_j=1^k.

We denote byZ_N^u

k :=span

φ^k_j ^N_j=1^k , andZ_N^s

k :=span φ^k_j ^∞_j=N

k+1. Denote byP_Nk the projection and its adjoint P_N^∗

k, defined by PN_k:=− 1

2πi Z

Γ

(λI+Ak)⁻¹dλ; P_N^∗_k :=− 1 2πi

Z

¯Γ

(λI+A^∗_k)⁻¹dλ,

where Γ (its conjugate ¯Γ, respectively) separates the unstable spectrum from the stable one of −Ak

(−A^∗_k, respectively). We set

−A^u_Nk:=PN_k(−Ak), −A^s_Nk:= (I−PN_k)(−Ak), (52) for the restrictions of−Ak toZ_N^u

k andZ_N^s

k, respectively. This projections commute with−Ak. We then have that the spectra of−Ak onZ_N^u

kandZ_N^s

k coincide with λ^k_j ^Nk

j=1 and λ^k_j ^∞

j=Nk+1, respectively (for more details about the projectionP_N, see [15]).

Moreover, since−Akgenerates aC0−analytic semigroup onH, its restriction−A^s_NktoZ_N^s

kgenerates likewise a C₀−analytic semigroup on Z_N^s

k. This implies that−A^s_N

k satisfies the spectrum determined growth condition onZ_N^s

k, and so, we have

ke^−AsNktk_L(H,H)≤C_α0e^−α0t,∀t≥0, (53) for someα0<|<λN_k+1|.

The system (51) can accordingly be decomposed as

zk =zN_k+ζN_k, zN_k:=PN_kzk, ζN_k := (I−PN_k)zk, where applyingP_Nk and (I−P_Nk) on (51), we obtain

onZ_N^u_k: d

dtzN_k+A^u_NkzN_k =PN_k(Bkψk), zN_k(0) =PN_kz0k, (54) onZ_N^s

k: d

dtζN_k+A^s_N

kζN_k = (I−PN_k)(Bkψk), ζN_k(0) = (I−PN_k)z0k, (55) respectively.

6 Stabilization of the Unstable z

-System (54) on Z

k

.

Lemma 6.1 For all0<|k| ≤S, there existα1, Cα₁ >0 and a controllerψk such that, once inserted in (54), yields the estimate

kzN_k(t)k ≤Cα₁e^−α1tkz0kk, t≥0, herezN_k is the solution to(54), for the corresponding controllerψk.

Moreover, the controllerψk can be chosen of classC¹ and such that

|d

dtψk(t)|+|ψk(t)| ≤Cα₁e^−α1tkz0kk, t≥0.

Proof. Let us decomposezN_k as

zN_k(t, y) =

Mk

X

j=1

z_j^k(t)φ^k_j(y),

wherez^k_j(t)∈C,∀t≥0, j= 1, ..., Mk. We introduce that zN_k in equation (54), and obtain

M_k

X

j=1

d dtz_j^k(t)

φ^k_j+z_j^k(t)A^u_Nkφ^k_j

=PN_k(Bkψk).

We multiply (scalarly) the above equation withφ^k∗_l , l= 1, ..., M_k, and get that d

dtz_l^k+

Mk

X

j=1

A^u_N

kφ^k_j, φ^k∗_l z_j^k=

P_NkB_kψ_k, φ^k∗_l

, l= 1, ..., M_k. (56)

Let us observe that we can assumeP_N^∗

k(φ^k∗_l ) =φ^k∗_l , l= 1, ..., Mk. We have PN_kBkψk, φ^k∗_l

=

Bkψk, P_N^∗_kφ^k∗_l

=ψkB^∗_kφ^k∗_l = by (25) =ψkν(φ^k∗_l )⁰⁰⁰(1), l= 1, ..., Mk. Denote by

b^k_l :=ν(φ^k∗_l )⁰⁰⁰(1), l= 1, ..., M_k. (57) We set

Zk:=









 z₁^k z₂^k ...

z^k_M

k









 , Bk :=









 b^k₁ b^k₂ ...

b^k_M

k











, (58)

and

Λk :=















 A^u_N

kφ^k₁, φ^k∗₁ A^u_N

kφ^k₂, φ^k∗₁

...

A^u_N

kφ^k_M

k, φ^k∗₁ A^u_N

kφ^k₁, φ^k∗₂ A^u_N

kφ^k₂, φ^k∗₂

...

A^u_N

kφ^k_M

k, φ^k∗₂

... ... ... ...

A^u_N

kφ^k₁, φ^k∗_M

k

... ...

A^u_N

kφ^k_M

k, φ^k∗_M

k

















. (59)

With these notations, equation (56) becomes d

dtZk+ Λ_kZk=B_kψ_k, t≥0. (60)

(Note thatBk is a column matrix, and by Bkψk we mean the product between a scalar and a column matrix. This is because, in our case,ψ_k =ψ_k(t) only)

We claim that, for µ > 0 large enough, the controller ψk := −µB_k^TZk exponentially stabilizes the system (60). HereB_k^T is the transpose of the matrixBk, i.e.,

B^T_k =

b^k₁ b^k₂ ... b^k_Mk

. (61)

It is easy to see that, by Lemma 4.1, we haveb^k_l >0, l= 1, ..., Mk, which implies that, forµ >0 large enough, we have

Λk+µBkB_k^T Z,Z

R^Mk ≥αkZk²

R^Mk, ∀Z ∈R^Mk, for someα >0. Hence, − Λ_k+µB_kB_k^T

is a Hurwitz matrix, which gives the exponential stability, as claimed.