of Periodic Flows in a Two Dimensional Channel

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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 85from 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:









ut−ν∆u+uux+vuy=px, t≥0, x∈R, y∈]0,1[, vt−ν∆v+uvx+vvy =py, t≥0, x∈R, y∈]0,1[, ux+vy= 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),



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 thatUx≡0, this means thatU =U(y). Substituting (U,0) into equation (1), we obtain

−νU00(y) =pex(x, y), pey(x, y)≡0, (2) wherepe 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 thatpe≡pe(x) andpex=−νU00=const. Hence, U000 ≡0 . This yields

U(y) =C(y2−y), y∈]0,1[, (3)

whereC∈R. In the following, we takeC=−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 1ν (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−pe, we are readily led, via (1) and (2), to the study of null stabilization of the system









ut−ν∆u+uxU+vU0+uux+vuy =px, x∈R, y∈]0,1[, vt−ν∆v+vxU+uvx+vvy =py, 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),



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:









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),



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


(L−2k RkLkvk(t))000(1)eikx,


where vk(t, y) = R

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−1k Rkz0k, L−1k Akz0k


2|(L−2k Rkz0k)000(1)|2=1

2kL−1k z0kk2,∀z0k ∈H,

for all 0<|k| ≤S. Ak =FkL−1k , 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∈L2loc(R×]0,1[), which are 2π−periodic in x. These functions are characterized by their Fourier series

u(x, y) =X


uk(y)eikx, uk= ¯u−k, ∀k∈Z, such that



Z 1 0

|uk(y)|2dy <∞.

The norm inL2(Q) is defined as

|u|L2(Q):= 2πX



!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 (uk)k∈Z, (vk)k∈Z, (pk)k∈Z, (ψk)k∈Z. We have





(uk)t−νu00k+ (νk2+ikU)uk+U0vk=ikpk, a.e. in ]0,1[, (vk)t−νv00k+ (νk2+ikU)vk=p0k, a.e. in ]0,1[,

ikuk+vk0 = 0, a.e. in ]0,1[,

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


with initial datau0k, v0k.Here u:=X


uk(t, y)eikx, v:=X


vk(t, y)eikxandp:=X


pk(t, y)eikx, Ψ :=X



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 byHm(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





(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.


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


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−ikνv00k+ik2(νk+iU)vk−(u0k)t+νu000k −k(νk+iU)u0k−ikU0uk−U0v0k−U00vk= 0.

Taking into account thatuk=−ik1vk0, we have


ik(vk)t−ikνvk00+ik2(νk+iU)vk+ 1

ik(vk00)t− ν


i(νk+iU)vk00−U00vk = 0, t≥0, y∈]0,1[.


( (v00k−k2vk)t−νvivk + (2νk2+ikU)v00k−k(νk3+ik2U+iU00)vk= 0, t≥0, y∈]0,1[,

v0k(0) =vk0(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

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

( (Lkvk)t+Fkvk = 0, t≥0, y∈]0,1[,

vk0(0) =v0k(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,

w0k(0) =w0k(1) = 0, wk= 0, wk(1) =ψk. (14) (Forθk positive and sufficiently large, there exists a unique solutionwk for equation (14)). Then, sub- tracting (13) and (14), we obtain that

(Lkvk)t+Fk(vk−wk)−θkwk = 0, t≥0.


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

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

Ak:=FkL−1k , D(Ak) =

v∈H :L−1k 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+Ak)−1 is compact. Moreover, one has

σ(−Ak)⊂ {λ∈C:<λ≤0},∀|k|> S, where

S := 1


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

( (v00k−k2vk)t−νvivk + (2νk2+ikU)v00k−k(νk3+ik2U+iU00)vk= 0, t≥0, y∈]0,1[,

v0k(0) =vk0(1) = 0, vk(0) = 0, vk(1) = 0. (18)

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

2 d

dt kvk0k2+k2kvkk2

+νkvk00k2+ 2νk2kvk0k2+νk4kvkk2=−<


Z 1 0


. Hence,

1 2


dt kvk0k2+k2kvkk2

+ 2νk2kv0kk2+νk4kvkk2≤k

Z 1 0


≤ 3

2νk2kv0kk2+ 2 3ν

Z 1 0


≤ 3

2νk2kv0kk2+ a2

3kvkk2≤ 3


2νk4kvkk2, since|k|> S=1ν

1 + a


.The above estimates imply that d

dt kv0kk2+k2kvkk2

+νk2(kv0kk2+k2kvkk2)≤0, t≥0.

This yields

kv0k(t)k2+k2kvk(t)k2≤e−νk2t k(vk0)0k2+k2kv0kk2

, t≥0. (19)

Taking into account thatuk=−ik1vk0, relation (19) implies that kuk(t)k2+kvk(t)k2≤e−νS2t ku0kk2+kv0kk2

, ∀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

yk(t) =e−Aktyk(0) + Z t



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

+ Z t


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

( ˜Fkv, ψ) :=

Z 1 0

v(y)Fkψ(y)dy, ∀ψ∈ D(Fk) =D(Fk). (22) The space (D(Fk))0 is the completion of the space H in the norm k|fk|:= k(λI +Fk)−1fk, for some λ∈Rsufficiently large. HereFk 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 Lk 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 ((θk+ ˜Fk)Dk) 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


dt(Lkvk(t)) + ˜Ak(Lk(vk)(t)) = (θk+ ˜Fk)(Dkψk(t)), t >0 (26) and initial datavk0.

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:


dtLkvk(t), φ

+hLkvk,Akφi=hψk(t),((θk+Fk)Dk)φi,∀φ∈ D(Ak).

4 Further Results Concerning the Linear Operators A

k For every 06=k∈Z, denote by λkj

j=1the eigenvalues of−Akkjis repeated according to its multiplicity mkj). Hence, ¯λkj

j=1are the eigenvalues of the dual operator−Ak, of−Ak. Denote by

φkj, j= 1,2, ...


φk∗j , j= 1,2, ... the corresponding eigenfunctions for−Ak and−Ak, 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−Ak, 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 −Ak generates aC0−analytic semigroup and has a compact resolvent inH, so, it has only a finite numberNk

of eigenvaluesλkj with<λkj ≥0, the unstable eigenvalues. Denote byMk :=mk1+mk2+...+mkN

k (mkj is the multiplicity ofλkj, j= 1,2, ..., Nk).

The next lemma is the key result of this paper. It shows that we can choose the eigenfunctionsφk∗j , j= 1,2, ..., Nk, of the dual operator−Ak, corresponding to the unstable eigenvalues ¯λkj, j= 1,2, ..., Nk, such that (φk∗j )000(1)>0.

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


Proof. We shall prove that we can choose the dual eigenfunctionsφk∗j such that (φk∗j )000(1)6= 0. Then, replacing eventuallyφk∗j by−φk∗j , we obtain that we have (φk∗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:= ¯λkj, the unstable eigenvalue. Ifφ :=φk∗j is an eigenfunction corresponding to

¯λ, thenφ satisfies the next boundary value problem

( ν(φ)iv−(2νk2−ikU+ ¯λ)(φ)00+ 2ikU0)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)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:=νφiv−(2νk2+ikU+λ)φ00+k(νk3+ik2U+iU00+kλ)φ, ∀φ∈ D(Lk).

It is easy to see thatφ satisfiesLkφ= 0, whereLk is the dual operator ofLk. We have 0 =

Z 1 0

Lkφφdy= Z 1


φkφdy+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

|2dy= 0, (30)


( Lkφ=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.


Letφ3 be a solution to the equation Lkφ3, 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 andφ3 is symmetric, as wanted.

Now, we defineφ5:=−φφ030(0)

2(0)φ23.We have thatφ5is welldefined, and moreover,φ5satisfies



Lkφ5, φ5is symmetric,

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


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∈ H1000(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 equationLkΨ = 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|2dy−(νk2+λ) Z 1


|Φ|2dy−ik Z 1


U|Φ|2dy+ikU00 Z 1


Ψ ¯Φdy= 0, (38) and

− Z 1


0|2dy−k2 Z 1


|Ψ|2dy= Z 1


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|2dy−(νk2+<λ) Z 1


|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∈ H10(12) =ψ000(12) = 0 and AS =

ψ∈ H1:ψ(12) =ψ00(12) = 0 .This implies that dimCS= dimCAS = 2.

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

F1:={ψ∈ H1000(0) = 0}.

We have dimCF1= 3. Since dimCS= 2, dimCH1= 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, 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 (ψn) 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[,

qn0(0) =qn0(1) = 0, qn(0) = 0, qn(1) =ψn. (42) Or, by (12),

( θkqn+νqivn −(2νk2+ikU)qn00+k(νk3+ik2U+iU00)qn= 0, a.e. in ]0,1[,

q0n(0) =q0n(1) = 0, qn(0) = 0, qn(1) =ψn. (43) Multiplying (scalarly) equation (43) byqn, we get that

θkkqnk2+qn000(1)ψn+νkq00nk2+ Z 1


(2νk2+ikU(y))|qn0(y)|2dy+ik Z 1


U0(y)q0n(y)¯qn(y)dy+ (44)

+νk4kqnk2+i Z 1


(k3U(y) +kU00(y))|qn(y)|2dy

= 0.

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

θkkqnk2+<(q000n(1)ψn) +νkq00nk2+ 2νk2kqn0k2+νk4kqnk2= (45)



Z 1 0


. We have



Z 1 0


ik Z 1



≤ (46)

≤ |k|

Z 1 0


Taking in the inequality|αβ| ≤ 12

ν|k||α|2+ν|k|1 |β|2

,α:=qn0(y) andβ:=U0(y)qn(y), yields


Z 1 0

|U0(y)q0n(y)qn(y)|dy≤1 2νk2

Z 1 0

|qn0(y)|2dy+ 1 2ν

Z 1 0

|U0(y)|2|qn(y)|2dy (47)


2νk2kqn0k2+ a2

3kqnk2≤2νk2kqn0k2+ a23kqnk2. Hence, by (45), (46) and (47), we obtain

θk− a23

kqnk2+<(q000n(1)ψn)≤0. (48) Lettingnto go to infinity in relation (48), yields


θk− a23

limkqnk2≤0, sinceψn→0 andθk is sufficiently large such thatθka23 ≥0.

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

linear, we have thatDk 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


(L−2k RkLkvk(t))000(1)eikx, (49) where

vk(t, y) = Z


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,Rk:X → X are linear selfadjoint operators such that (i) Rk :H →H,

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


2|(L−2k Rkz0k)000(1)|2= 1

2kL−1k z0kk2,∀z0k ∈H,

for all 0<|k| ≤S,S given by (17). Ak =FkL−1k , 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.,

zk:=Lkvk,Bk := (θk+ ˜Fk)Dk. (50) With these notations, equation (26) becomes


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

wherez0k(y) :=Lkvk0(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<<λNk+1, andα1>0 is given by Lemma6.1below.


From the previous section, we have that the eigenvalues of−Akareλkj, j= 1,2, ..., and the eigenfunctions φkj, j = 1,2, .... The eigenfunctions of the dual operator−Ak, of −Ak, are φ∗kj , j = 1,2, .... Moreover, the operator−Ak has a finite number of unstable eigenvalues (λkj)Nj=1k.

We denote byZNu

k :=span

φkj Nj=1k , andZNs

k :=span φkj j=N

k+1. Denote byPNk the projection and its adjoint PN

k, defined by PNk:=− 1

2πi Z


(λI+Ak)−1dλ; PNk :=− 1 2πi




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

(−Ak, respectively). We set

−AuNk:=PNk(−Ak), −AsNk:= (I−PNk)(−Ak), (52) for the restrictions of−Ak toZNu

k andZNs

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


k coincide with λkj Nk

j=1 and λkj

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

Moreover, since−Akgenerates aC0−analytic semigroup onH, its restriction−AsNktoZNs

kgenerates likewise a C0−analytic semigroup on ZNs

k. This implies that−AsN

k satisfies the spectrum determined growth condition onZNs

k, and so, we have

ke−AsNktkL(H,H)≤Cα0e−α0t,∀t≥0, (53) for someα0<|<λNk+1|.

The system (51) can accordingly be decomposed as

zk =zNkNk, zNk:=PNkzk, ζNk := (I−PNk)zk, where applyingPNk and (I−PNk) on (51), we obtain

onZNuk: d

dtzNk+AuNkzNk =PNk(Bkψk), zNk(0) =PNkz0k, (54) onZNs

k: d


kζNk = (I−PNk)(Bkψk), ζNk(0) = (I−PNk)z0k, (55) respectively.

6 Stabilization of the Unstable z


-System (54) on Z




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

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

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


dtψk(t)|+|ψk(t)| ≤Cα1e−α1tkz0kk, t≥0.


Proof. Let us decomposezNk as

zNk(t, y) =





wherezkj(t)∈C,∀t≥0, j= 1, ..., Mk. We introduce that zNk in equation (54), and obtain




d dtzjk(t)



We multiply (scalarly) the above equation withφk∗l , l= 1, ..., Mk, and get that d






kφkj, φk∗l zjk=

PNkBkψk, φk∗l

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

Let us observe that we can assumePN

kk∗l ) =φk∗l , l= 1, ..., Mk. We have PNkBkψk, φk∗l


Bkψk, PNkφk∗l

kBkφk∗l = by (25) =ψkν(φk∗l )000(1), l= 1, ..., Mk. Denote by

bkl :=ν(φk∗l )000(1), l= 1, ..., Mk. (57) We set


 z1k z2k ...



 , Bk :=

 bk1 bk2 ...



, (58)


Λk :=

 AuN

kφk1, φk∗1 AuN

kφk2, φk∗1




k, φk∗1 AuN

kφk1, φk∗2 AuN

kφk2, φk∗2




k, φk∗2

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


kφk1, φk∗M


... ...



k, φk∗M


. (59)

With these notations, equation (56) becomes d

dtZk+ ΛkZk=Bkψk, 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,ψkk(t) only)

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

BTk =

bk1 bk2 ... bkMk

. (61)

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

Λk+µBkBkT Z,Z

RMk ≥αkZk2

RMk, ∀Z ∈RMk, for someα >0. Hence, − Λk+µBkBkT

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




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