NORMAL FEEDBACK STABILIZATION OF -

NORMAL FEEDBACK STABILIZATION OF PERIODIC FLOWS

IN A 2-D CHANNEL

IONUT MUNTEANU

Abstract. We consider a 2-D incompressible channel flow with periodic con-dition along one axis. We stabilize the linearized system by a boundary feed-

back controller with vertical velocity observation which acts on the normal

component of the velocity only. The stability is achived without any a prioricondition on the viscosity coefficient ν, that is on Reynolds number.

1. Introduction

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

The dynamic of the flow is governed by the dimensionless incompressible 2-DNavier-Stokes equations:

ut − ν∆u+ uux + vuy = px, x ∈ R, y ∈ (0, 1),vt − ν∆v + 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),∀t ≥ 0,∀x ∈ R,∀y ∈ (0, 1),

(1.1) ow1

with initial data

u(0, x, y) = u0(x, y), v(0, x, y) = v0(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 pres-sure. We assume that the velocity field and the pressure are 2π−periodic in the firstspatial coordinate x. Of course this is not true in reality but it is often assumedfor mathematical convenince because it does not alter the essential features of thebehaviour of the flow (for more explanations see for example [8]).

In order to compute the equilibrium state, we put the boundary control to bezero, 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 that Ux ≡ 0, this means that U = U(y).Substituting U into equation (1.1) we obtain

−νU ′′(y) = pex(x, y), pey(x, y) ≡ 0, (1.2) ow2

Key words and phrases. Navier-Stokes equation, analytic C0-semigroup, feedback controller.

1

2 IONUT MUNTEANU

where pe is the equilibrium pressure corresponding to the steady-state solution(U, 0) (by ′ we mean the derivative with respect to the second spatial coordinatey). From the relation above we deduce that pe ≡ pe(x) and pex = −νU ′′ = const.Hence U ′′′ ≡ 0 . This yields

U(y) = C(y2 − y), y ∈ (0, 1), (1.3) UU

where C ∈ R−. In the following, we take C = − a2ν , where a ∈ R+.

This is a parabolic laminar flow profile. It is well known that the stabilityproperty of the stationary flow (U, 0) varies with the Reynolds number 1

ν (for νsmall the flow is unstable, for ν big enough the flow is stable).

Our aim here is the stabilization of this flow profile by a boundary controller onthe wall y = 1, that is:

v(t, x, 1) = Ψ(t, x), t ≥ 0, x ∈ R.There is no action, however, in y = 0, for streamwise or inside the channel. There-fore, only the normal velocity v is controlled on the wall y = 1.

By the substitutions (u, v) → (u, v) − (U, 0), p → p − pe, we are readly led via(1.1),(1.2) to the study of null stabilization of the system

ut − ν∆u+ uxU + vU ′ + 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),∀t ≥ 0,∀x ∈ R,∀y ∈ (0, 1),

(1.4) f1

with initial data

u(0, x, y) = u0(x, y) = u0(x, y)−U(y), v(0, x, y) = v0(x, y) = v0(x, y), x ∈ R, y ∈ (0, 1).

The linearization of (1.4) is the following system:

ut − ν∆u+ uxU + vU ′ = 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),∀x ∈ R,

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),

(1.5) f2

with initial data u0, v0.The main result here (see Theorem 3.1 below) is that the exponential stability

of (1.5) can be achived by a normal boundary finite-dimensional feedback con-troller of the form Ψ(t, x) = −ν

∑|k|≤S(L−2

k RkLkvk(t))′′′(1)eikx, where vk(t, y) =∫ 2π

0v(t, x, y)e−ikxdx, |k| ≤ S, more precisely: for every initial data (u0, v0) the

corresponding solution of the closed-loop system (1.5) satisfies

||(u(t), v(t))||2 ≤ Ce−αt||(u0, v0)||2, t ≥ 0,

for some C,α > 0. Here || · || denotes the norm of (L2((0, 2π) × (0, 1)))2 and S isgiven by relation (2.9), below.

NORMAL FEEDBACK STABILIZATION OF PERIODIC FLOWS IN A 2-D CHANNEL 3

The expresion of the stabilizable feedback controller involves linear operatorsRk : H → H, which are self-adjoint and satisfy Riccati algebraic equations in H

< L−1k Rkz0, L

−1k Akz0 > +

1

2ν2|(L−2

k Rkz0)′′′(1)|2 =1

2||L−1

k z0||2,∀z0 ∈ H,

for all |k| ≤ S, S given by (2.9). Ak = FkL−1k , where Fk is given by (2.4) and Lk

is given by (2.3). H = L2(0, 1).In a few words, our approach is the following one: we reduce first, via Fourier

functional setting, the linear system (1.5) to an infinite parabolic system and provethe boundary stabilization of it (see Theorem 3.2 and notations (3.2) below). Next,using a standard tehnique-minimization of a cost functional- we prove the feedbackexponential stability of the infinite parabolic system (see Theorem 3.3 below). Fi-nally, we plug that feedback controller, determined in the previous step, into thelinear system (1.5), and prove the feedback asymptotical stability of it (see Theorem3.1 below).

The boundary stabilization of Navier-Stokes equations with deterministic tan-gential feedback controllers was studied for example in [5],[10],[12]. However thereare few results on boundary stabilization with normal controllers and most referto periodic flows in 2-D channels [1],[2],[15]. Most of these stabilization results areestablished for large Reynolds number with exception of [4], [14], where the stabi-lizable feedback controller is constructed without any a priori condition on ν, whichis also the case of this paper.

2. The Fourier functional setting

Let L2(Q), Q = (0, 2π)× (0, 1) be the space of all functions u ∈ L2loc(R× (0, 1))

which are 2π−periodic in x. These functions are characterized by their Fourierseries

u(x, y) =∑k∈Z

uk(y)eikx, uk = u−k, u0 = 0,

∑k∈Z

∫ 1

0

|uk(y)|2dy <∞.

The norm in L2(Q) is defined as

|u|L2(Q) =

(∑k∈Z|uk|2L2(0,1)

) 12

.

Since is no danger of confusion we shall denote by || · || the norm in both L2(Q)and L2(0, 1).

We set

(L2(Q))2 ={

(u, v) : u, v ∈ L2(Q)}.

We denote also by || · ||, the norm in (L2(Q))2, defined as

||(u, v)|| =(||u||2 + ||v||2

) 12 .

Since the modes u0 = 0 for all u ∈ L2(Q), from now on we only consider k ∈ Z,with k 6= 0.

We now return to system (1.5) and rewrite it in terms of Fourier coefficients(uk)k∈Z, (vk)k∈Z, (pk)k∈Z, (ψk)k∈Z. We have

4 IONUT MUNTEANU

(uk)t − νu′′k + (νk2 + ikU)uk + U ′vk = ikpk, a.e. in (0, 1),

(vk)t − νv′′k + (νk2 + ikU)vk = p′k, a.e. in (0, 1),ikuk + v′k = 0, a.e. in (0, 1),

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

(2.1) f3

with initial data u0k, v

0k.

Here u =∑k∈Z uk(t, y)eikx, v =

∑k∈Z vk(t, y)eikx and p =

∑k∈Z pk(t, y)eikx,

Ψ =∑k∈Z ψk(t)eikx.

Reducing pk from the first two equations from (2.1), we obtain

ik(vk)t−ikνv′′k+ik2(νk+iU)vk−(u′k)t+νu′′′k −k(νk+iU)u′k−ikU ′uk−U ′v′k−U ′′vk = 0

Taking into account that uk = − 1ikv′k, we have

ik(vk)t−ikνv′′k+ik2(νk+iU)vk+1

ik(v′′k )t−

ν

ikvivk +

1

i(νk+iU)v′′k−U ′′vk = 0, t ≥ 0, y ∈ (0, 1).

Finally, (v′′k − k2vk)t − νvivk + (2νk2 + ikU)v′′k − k(νk3 + ik2U + iU ′′)vk = 0,t ≥ 0, y ∈ (0, 1),

v′k(0) = v′k(1) = 0, vk(0) = 0, vk(1) = ψk(t).(2.2) f4

In what follows we shall consider H to be the complexified space of L2(0, 1). Wedenote also by || · ||, the norm in H and by < ·, · >, the scalar product. We shalldenote by Hm(0, 1), m = 1, 2, .. the standard Sobolev spaces on (0, 1) and

H10 (0, 1) =

{v ∈ H1(0, 1) : v(0) = v(1) = 0

},

H20 (0, 1) =

{v ∈ H2(0, 1) ∩H1

0 (0, 1) : v′(0) = v′(1) = 0},

H30 (0, 1) =

{v ∈ H3(0, 1) ∩H2

0 : v′′(0) = v′′(1) = 0}.

For each 0 6= k ∈ Z, we denote by Lk : D(Lk) ⊂ H → H and Fk : D(Fk) ⊂ H →H the operators

Lkv = −v′′ + k2v,D(Lk) = H2(0, 1) ∩H10 (0, 1), (2.3) Lk

Fkv = νviv − (2νk2 + ikU)v′′ + k(νk3 + ik2U + iU ′′)v, (2.4) Fk

D(Fk) = H4(0, 1) ∩H20 (0, 1).

With these notations, equation (2.2) becomes{(Lkvk)t + Fkvk = 0, t ≥ 0, y ∈ (0, 1),

v′k(0) = v′k(1) = 0, vk(0) = 0, vk(1) = ψk(t).(2.5) v1

Consider the solution wk = wk(t, y) of the equation{θkwk + Fkwk = 0, y ∈ (0, 1), t ≥ 0,

w′k(0) = w′k(1) = 0, wk = 0, wk(1) = ψk.(2.6) f5

(For θk positive and sufficiently large, there exists a unique solution wk forequation (2.6)). Then, subtracting (2.5) and (2.6), we obtain

(Lkvk)t + Fk(vk − wk)− θkwk = 0, t ≥ 0.

Equivalently,


(Lk(vk − wk))t + Fk(vk − wk) = θkwk − (Lk(wk))t, (2.7) f7

vk − wk ∈ D(Fk).

In order to represent equation (2.7) as an abstract boundary control system, weconsider the operators Ak : D(Ak) ⊂ H → H defined by

Ak = FkL−1k , (2.8) Ak

D(Ak) ={v ∈ H : L−1

k v ∈ D(Fk)}.

We have

l1 Lemma 2.1. The operator −Ak generates a C0− analytic semigroup on H and foreach λ ∈ ρ(−Ak) (the resolvent set of −Ak), (λI + Ak)−1 is compact. Moreover,one has

σ(−Ak) ⊂ {λ ∈ C : <λ ≤ 0} ,∀|k| > S,

where

S =1√ν

(1 +

a√2ν

) 12

. (2.9) L

Here σ(−Ak) is the spectrum of −Ak.

Proof. The proof can be found in [7, Lemma 1]. �

b9 Remark 2.1. For |k| > S, S given by (2.9), let us consider the system (2.2)without any boundary controller, that is (v′′k − k2vk)t − νvivk + (2νk2 + ikU)v′′k − k(νk3 + ik2U + iU ′′)vk = 0,

t ≥ 0, y ∈ (0, 1),v′k(0) = v′k(1) = 0, vk(0) = 0, vk(1) = 0.

(2.10) alf4

Multiplying (scalary) equation (2.10) by vk and taking the real part of the result,we get that

1

2

d

dt

(||v′k||2 + k2||vk||2

)+ν||v′′k ||2+2νk2||v′k||2+νk4||vk||2 = −<

(ik

∫ 1

0

U ′v′kvkdy

).

Hence

1

2

d

dt

(||v′k||2 + k2||vk||2

)+ 2νk2||v′k||2 + νk4||vk||2 ≤ k

∣∣∣∣∫ 1

0

U ′v′kvkdy

∣∣∣∣≤ 3

2νk2||v′k||2 +

2

3ν

∫ 1

0

|U ′|2|vk|2dy

≤ 3

2νk2||v′k||2 +

a2

6ν3||vk||2 ≤

3

2νk2||v′k||2 +

1

2νk4||vk||2,

since |k| > S = 1√ν

(1 + a√

2ν

) 12

.

The above estimates imply that

d

dt

(||v′k||2 + k2||vk||2

)+ νk2(||v′k||2 + k2||vk||2) ≤ 0, t ≥ 0.

This yields

||v′k(t)||2 + k2||vk(t)||2 ≤ e−νk2t(||(v0

k)′||2 + k2||v0k||2), t ≥ 0. (2.11) az1

6 IONUT MUNTEANU

Taking into account that uk = − 1ikv′k, relation (2.11) implies that

||uk(t)||2 + ||vk(t)||2 ≤ e−νS2t(||u0

k||2 + ||v0k||2), t ≥ 0, |k| > S. (2.12) az2

Due to relation (2.12) we have that for |k| > S, the exponential asymptoticalstability holds true for the system (2.2) (equivalently (2.5)) without any boundarycontroller. Hence, we have to control the system (2.5) (equivalently (2.2)) for|k| ≤ S, only.

Now, coming back to system (2.7), we set

yk(t) = Lk(vk(t)− wk(t))

and write it as

yk(t) = e−Aktyk(0) +

∫ t

0

e−Ak(t−s)(θkwk(s)− (Lk(wk(s))s)ds

= e−Aktyk(0)− Lk(wk(t)) + e−AktLk(wk(0))+ (2.13) f400

+

∫ t

0

e−Ak(t−s)(θk(wk(s)) + (Fk(wk(s)))ds,

where Fk : H → (D(F ∗k ))′ is the extension of Fk to H defined by

(Fkv, ψ) =

∫ 1

0

v(y)F ∗kψ(y)dy, (2.14) ss

∀ψ ∈ D(Fk) = D(F ∗k ).The space (D(F ∗k ))′ is the completion of the space H in the norm |||f ||| =

||(λI + Fk)−1f ||, for some λ ∈ R sufficiently large. Where F ∗k is the dual operatorof Fk.

Define similarly Ak, the extension of Ak to H. Likewise −Ak, the operator −Ak

generates a C0−analytic semigroup on H. In the same way is defined the extensionof Lk to an operator from H to (H1

0 (0, 1) ∩H2(0, 1))′, again denoted Lk.Then, the above equation (2.13) can be rewritten as

d

dt(Lkvk(t)) + Ak(Lk(vk)(t)) = (θk + Fk)(wk(t)), t ≥ 0. (2.15) f8

For each ψ ∈ C, we denote by Dkψ = w ∈ H4(0, 1) the solution to the equation{θkw + Fkw = 0,∀y ∈ (0, 1),

w(0) = w(1) = 0, w′(0) = 0, w′(1) = ψ.(2.16) tt

(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 = νφ′′′(1), (2.17) f9

for all φ ∈ H4(0, 1), φ′(0) = φ′(1) = 0, φ(0) = 0.With this notation, equation (2.15) can be rewritten as

d

dt(Lkvk(t)) + Ak(Lk(vk)(t)) = (θk + Fk)(Dkψk(t)), t > 0, (2.18) f11

and initial data v0k.

Remark 2.2. Note that the systems (2.18), (2.5) and (2.2) are all equivalent.


Remark 2.3. Equation (2.18) is understood in the following weak sense:

<d

dtLkvk(t), φ > + < Lkvk,A

∗kφ >=< ψk(t), ((θk + Fk)Dk)∗φ >,∀φ ∈ D(A∗k).

2.1. Further results concerning the linear operators Ak. For every 0 6= k ∈Z, denote by

(λkj)∞j=1

, the eigenvalues of −Ak (λkj is repeated according to its mul-

tiplicity mkj ). Hence

(λkj)∞j=1

are the eigenvalues of the dual operator −A∗k, of −Ak.

Denote by{φkj , j = 1, 2, ...

}and

{φ∗kj , j = 1, 2, ...

}the corresponding eigenfunctions

for −Ak and −A∗k, respectively.

M3 Remark 2.4. Concerning the eigenvalues and eigenfunctions of the extended op-erator −Ak, it is known that they coincide with the eigenvalues and eigenfunctions

of −Ak. The same holds true for the dual operators −A∗k.

As mentioned before (see Remark 2.1), we have to stabilize system (2.18) (equiv-alently (2.5)) for |k| ≤ S only. So, let us consider k ∈ Z such that |k| ≤ S. FromLemma 2.1 we know that −Ak generates a C0−analytic semigroup and has a com-pact resolvent in H, so it has only a finite number Mk of eigenvalues λkj with

<λkj ≥ 0, the unstable eigenvalues. Denote by Nk = mk1 + mk

2 + ... + mkMk

(mkj is

the multiplicity of λkj , j = 1, 2, ...,Mk).

The algebraic multiplicity of λkj is the dimension of the range of the projectionoperator

Pj = − 1

2πi

∫Γj

(λI + Ak)−1dλ

where Γj is a smooth closed curve encircling λkj . The geometric multiplicity of λkjis the dimension of the eigenfunction space corresponding to λkj . In general, thealgebraic multiplicity is greater than the geometric multiplicity, but when are equal,the eigenvalue is called semi-simple. Let us consider the assumption

(A1) All the eigenvalues λkj , with j = 1, ...,Mk, are semi-simple.When proving the stabilization of the system (2.18) via the Kalman’s Criterion

we shall assume first that hypothesis (A1) holds true in order to simplify the argu-ment. But, at the end of the day, we shall show that the same results of stabilityhold true without assuming (A1).

Next, we will see that the geometric multiplicity of each eigenvalue of the oper-ators −Ak is 1. And so, if an eigenvalue is semi-simple then it is simple. This isimportant when we will compute the Kalman matrix of controlabillity, and this isthe reason why we need only one controller on the wall y = 1, and not on the bothwalls.

p1 Proposition 2.1. Let λ be an eigenvalue of the linear operator −Ak. If λ is semi-simple then it is simple.

Proof. First we show the next lemmas.

u1 Lemma 2.2. Let λ ∈ C and φ∗ be a solution to the equation

λφ∗ + A∗kφ∗ = 0 in (0, 1), (2.19) u2

φ∗(0) = φ∗(1) = 0, (φ∗)′(0) = (φ∗)′(1) = 0.

If (φ∗)′′′(1) = 0, then φ∗ ≡ 0.

8 IONUT MUNTEANU

Proof. Equation (2.19) reduces, in fact, to the next boundary value problem (sinceAk = FkL

−1k , Fk given by (2.4) and Lk given by (2.3)):

ν(φ∗)iv(y)− (2νk2 − ikU(y) + λ)(φ∗)′′(y) (2.20) u3

+2ikU ′(y)(φ∗)′(y) + (k2λ+ νk4 − ik3U(y))φ∗(y) = 0, y ∈ (0, 1),

φ∗(0) = φ∗(1) = 0, (φ∗)′(0) = (φ∗)′(1) = 0.

Let φ be any solution of the equation

λφ+ Akφ = 0, (2.21) u4

φ(0) = φ′(0) = 0.

Multiplying (scalary) equation (2.20) by φ, yields

ν[(φ∗)′′′(1)φ(1)− (φ∗)′′′(0)φ(0)]− ν[(φ∗)′′(1)φ′(1)− (φ∗)′′(0)φ′(0)] (2.22) u5

+ν[(φ∗)′(1)φ′′(1)− (φ∗)′(0)φ′′(0)]− ν[φ∗(1)φ′′′(1)− φ∗(0)φ′′′(0)]

+(φ∗)′(1)(αφ)(1)− (φ∗)′(0)(αφ)(0)− φ∗(1)(αφ)′(1) + φ∗(0)(αφ)′(0)

+φ∗(1)(βφ)(1)− φ∗(0)(βφ)(0)+ < φ∗, λφ+ Akφ >= 0,

where α(y) = −(2νk2 − ikU(y) + λ) and β(y) = 2ikU ′(y).Taking into account the boundary conditions of the equations (2.20) and (2.21)

and the hypothesis (φ∗)′′′(1) = 0, and the fact that φ satisfies equation (2.21), wehave that equation (2.22) implies

(φ∗)′′(1)φ′(1) = 0. (2.23) k2

It is easy to see that there exists a solution φ to equation (2.21) such that φ′(1) 6= 0.Indeed, since (2.21) is a fourth order equation, let us consider a fundamental systemof solutions of it {φ1, φ2, φ3, φ4}. Obviously, we can not have φ′i(1) = 0 for alli = 1, ..., 4. So, let us assume that φ′1(1) 6= 0. Moreover, we can assume that

det

φ2(0) φ3(0) φ4(0)φ′2(0) φ′3(0) φ′4(0)φ′2(1) φ′3(1) φ′4(1)

6= 0 (2.24) k1

Now consider the function φ defined by

φ = φ1 + C1φ2 + C2φ3 + C3φ4,

where C1, C2, C3 satisfy the system C1φ2(0) + C2φ3(0) + C3φ4(0) = −φ1(0),C1φ

′2(0) + C2φ

′3(0) + C3φ

′4(0) = −φ′1(0),

C1φ′2(1) + C2φ

′3(1) + C3φ

′4(1) = 0.

(there exists a unique solution of the above system because of (2.24))Now it is easy to see that the above φ satisfies the equation (2.21) and φ′(1) 6= 0.Hence, due to (2.23), we have that (φ∗)′′(1) = 0. Thus φ∗ is an analytic solution

to the fourth order equation (2.20) with

φ∗(1) = (φ∗)′(1) = (φ∗)′′(1) = (φ∗)′′′(1) = 0.

This implies φ∗ ≡ 0.�

Y1 Lemma 2.3. The geometric multiplicity of each eigenvalue λ of −Ak is 1.


Proof. Consider two eigenfunctions φ∗1 and φ∗2 of the dual operator −A∗k, corre-sponding to the eigenvalue λ. We have

λφ∗j + A∗kφ∗j = 0 in (0, 1), (2.25) u7

φ∗j (0) = φ∗j (1) = 0, (φ∗j )′(0) = (φ∗j )

′(1) = 0, j = 1, 2.

By Lemma 2.2, since φ∗j 6= 0, j = 1, 2 we have (φ∗j )′′′(1) 6= 0, j = 1, 2. Hence

µ =(φ∗1)′′′(1)(φ∗2)′′′(1) 6= 0 is well-defined.

Consider the function w(y) = φ∗1(y)− µφ∗2(y). Then, we have

λw + A∗w = 0,

w(0) = w(1) = 0, w′(0) = w′(1) = 0,

and w′′′(1) = 0. Hence, by Lemma 2.2, we obtain w ≡ 0, i.e., φ∗1 = µφ∗2. This meansthat the dimension of the eigenfunction space is 1, i.e., the geometric multiplicityof the eigenvalue λ is 1. And this implies that the geometric multiplicity of λ is 1,also. �

The proof of Proposition 2.1 follows immediately from the definition of a semi-simple eigenvalue and Lemma 2.3. �

rem Remark 2.5. Assuming (A1), we have by Proposition 2.1, that the multiplicitiesmkj , j = 1, 2, ...,Mk are all equal to 1 and Mk = Nk. Moreover, it follows by the

assumption (A1) that the system{φ∗kj}Nk

j=1can be chosen in such a way that

< φkl , φ∗kj >= δlj , l, j = 1, ..., Nk.

We can conclude now with the next remark.

M11 Remark 2.6. Assuming (A1) we have due to Remark 2.4 and Remark 2.5 that

for all |k| ≤ S, operator −Ak has a finite number of unstable eigenvalues (λkj )Nkj=1,

which are simple. Moreover, the corresponding eigenfunctions (φkj )Nkj=1, (φ

∗kj )Nk

j=1 of

−Ak and −A∗k, respectively, can be chosen in such a way that

< φkl , φ∗kj >= δlj , l, j = 1, ..., Nk.

lp1 Lemma 2.4. The algebraic multiplicity of each unstable eigenvalue λkj , j = 1, ..., Nkof −Ak is ≤ 2.

Proof. Let f ∈ H30 (0, 1) and X(λ, y) = (λI + A∗k)−1f. By the definition of Ak we

see that X is the solution to the equation

νXiv−(λ+2νk2−ikU)X ′′+2ikU ′X ′+(λk2−ik3U+νk4)X = Lkf on (0, 1) (2.26) lp2

X(0) = X(1) = 0, X ′(0) = X ′(1) = 0.

By the variation of constant formula we may therefore represent the solution Xto (2.26) as (assuming λ is not eigenvalue)

X(λ, y) =

4∑s=1

Xs(λ, y)

(Cs +

∫ y

0

Ws(ξ, λ)Lkf(ξ)dξ

)where {X1, X2, X3, X4} is a fundamental system of solutions for homogeneous sys-tem (2.26), Ws are analytic functions in λ and Cs are constants.

We may choose {X1, X2, X3, X4} in such a way that

|det(X1, X2, X3, X4)| ≥ ρ > 0,∀λ ∈ C,

10 IONUT MUNTEANU

andX1(λ, 0) = 0, X2(λ, 0) = 0, X ′1(λ, 0) = 0, X ′2(λ, 0) = 0,

det(Xs(λ, 0), X ′s(λ, 0))4s=3 6= 0,∀λ ∈ C,

andX3(λ, 1) = 0, X4(λ, 1) = 0, X ′3(λ, 1) = 0, X ′4(λ, 1) = 0,

det(Xs(λ, 1), X ′s(λ, 1))2s=1 6= 0,∀λ ∈ C.

Then, by (2.26), we see that

X(λ, 0) =

4∑s=3

Xs(λ, 0)Cs = 0, X ′(λ, 0) =

4∑s=3

X ′s(λ, 0)Cs = 0,

because f ∈ H30 (0, 1).

Since det(Xs(λ, 0), X ′s(λ, 0))4s=3 6= 0,∀λ ∈ C, we get that Cs = 0, s = 3, 4.

Also by (2.26), we get that

X(λ, 1) =

2∑s=1

Xs(λ, 1)

(Cs +

∫ 1

0


)= 0,

X ′(λ, 1) =

2∑s=1

X ′s(λ, 1)

(Cs +

∫ 1

0


)= 0,

because f ∈ H30 (0, 1). Since det(Xs(λ, 1), X ′s(λ, 1))1

s=1 6= 0,∀λ ∈ C, we obtain that

Cs = −∫ 1

0

Ws(ξ, λ)Lkf(ξ)dξ, s = 1, 2.

Then

P ∗j (f) = − 1

2πi

∫Γj

(λI+A∗k)−1fdλ = − 1

2πi

∫Γj

(2∑s=1

Xs(λ, y)

∫ 1

y


)dλ

= −∫ 1

y

Lkf(ξ)dξ1

2πi

2∑s=1

∫Γj

Xs(λ, y)Ws(ξ, λ)dλ

= −2∑s=1

ηsΦs(y)

where ηs ∈ C and Φs ∈ H10 (0, 1) are given functions. This implies that rank P ∗j ≤ 2.

Hence, the algebraic multiplicity of λkj and λkj is ≤ 2 as claimed. �

We finally need one more result concerning the Dirichlet maps Dk, |k| ≤ S.

remarca1 Proposition 2.2. For all |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,∀y ∈ (0, 1), (2.27) ww1

q′n(0) = q′n(1) = 0, qn(0) = 0, qn(1) = ψn.

Or, equivalently, by (2.4)

θkqn + νqivn − (2νk2 + ikU)q′′n + k(νk3 + ik2U + iU ′′)qn = 0, y ∈ (0, 1), (2.28) w1

q′n(0) = q′n(1) = 0, qn(0) = 0, qn(1) = ψn.


Multiplying (scalary) equation (2.28) by qn, we get that

θk||qn||2+q′′′n (1)ψn+ν||q′′n||2+

∫ 1

0

(2νk2+ikU(y))|q′n(y)|2dy+ik

∫ 1

0

U ′(y)q′n(y)qn(y)dy+

(2.29) w2

+νk4||qn||2 + i

(∫ 1

0

(k3U(y) + kU ′′(y))|qn(y)|2dy)

= 0.

Taking the real part of equation (2.29), we obtain that

θk||qn||2 + < (q′′′n (1)ψn) + ν||q′′n||2 + 2νk2||q′n||2 + νk4||qn||2 = (2.30) w3

= −<(ik

∫ 1

0

U ′(y)q′n(y)qn(y)dy

).

We have

−<(ik

∫ 1

0


)≤∣∣∣∣ik ∫ 1

0


∣∣∣∣ ≤ (2.31) W2

≤ |k|∫ 1

0

|U ′(y)q′n(y)qn(y)| dy.

Taking in the wellknown inequality |αβ| ≤ 12

(ν|k||α|2 + 1

ν|k| |β|2)

, α = q′n(y)

and β = U ′(y)qn(y), yields

|k|∫ 1

0

|U ′(y)q′n(y)qn(y)| dy ≤ 1

2νk2

∫ 1

0

|q′n(y)|2dy +1

2ν

∫ 1

0

|U ′(y)|2|qn(y)|2dy

(2.32) W1

≤ 1

2νk2||q′n||2 +

a2

8ν3||qn||2 ≤ 2νk2||q′n||2 +

a2

8ν3||qn||2.

Hence, by (2.30), (2.31) and (2.32), we obtain

(θk −a2

8ν3)||qn||2 + < (q′′′n (1)ψn) ≤ 0. (2.33) w5

Letting n to go to infinity in relation (2.33), yields

0 ≤ (θk −a2

8ν3) lim ||qn||2 ≤ 0,

since ψn → 0, and θk is sufficiently large such that θk − a2

8ν3 ≥ 0.Hence lim ||qn|| = 0. Thus, we obtain that: if ψn → 0 in C then qn = Dkψn → 0

in H. Since Dk is linear, we have that Dk is continuous.�

3. Feedback stabilization of the linearized system (1.5)

The main result of the paper is the next theorem.

T1 Theorem 3.1. There exists a finite-dimensional feedback controller Ψ of the form

Ψ(t, x) = −ν∑|k|≤S

(L−2k RkLkvk(t))′′′(1)eikx, (3.1)

where

vk(t, y) =

∫ 2π

0

v(t, x, y)e−ikxdx, |k| ≤ S,

12 IONUT MUNTEANU

such that once inserted into equation (1.5), the corresponding solution of the closed-loop system (1.5) satisfies

||(u(t), v(t))||2 ≤ Ce−αt||(u0, v0)||2, t ≥ 0,

for some C,α > 0. Here Rk : H → H are linear self-adjoint operators which satisfyRiccati algebraic equations in H

< L−1k Rkz0, L

−1k Akz0 > +

1

2ν2|(L−2

k Rkz0)′′′(1)|2 =1

2||L−1

k z0||2,∀z0 ∈ H,

for all |k| ≤ S, S given by (2.9). Ak = FkL−1k , where Fk is given by (2.4) and Lk

is given by (2.3). H = L2(0, 1)

In order to prove Theorem 3.1, we first control the system (2.18) (equivalently(2.5)), for every |k| ≤ S, such that to have the exponential asymptotical stabilityof the system, that is Theorem 3.2 below (see notations (3.2)).

Let us fix k ∈ Z such that 0 < |k| ≤ S, S given by relation (2.9).For simplicity we are going to omit the index k and the symbol ˜ (since is no

danger of confusion), also we are going to renote some symbols, i.e.,∥∥∥∥∥∥∥∥−Ak = A,Lkvk = z,

(θk + Fk)Dk = B,ψk = ψ.

(3.2) REX

With these notations, equation (2.18) becomes

d

dtz(t, y)−Az(t, y) = Bψ(t), t > 0, (3.3) f112

z(0, y) = z0(y),

where z0(y) = Lkv0k(y).

We shall prove that

l11 Theorem 3.2. There exists a controller ψ such that the corresponding solution zto the system (3.3) and the controller have the exponential decay

| ddtψ(t)|+ |ψ(t)| ≤ Cα1

e−α1t||z0||, ||z(t)|| ≤ Cα0e−α0t||z0||,∀t ≥ 0,

where α0 is given by (3.5) below

0 < α0 < <λN+1,

and α1 > 0 is given by Lemma 3.1 below.

By the notations (3.2) and the previous section we have that the eigenvaluesof A are λj = λkj , j = 1, 2, ..., and the eigenfunctions φj = φkj , j = 1, 2, ... The

eigenfunctions of the dual operator A∗, of A, are φ∗j = φ∗kj , j = 1, 2, .... Moreover,

operator A has a finite number of unstable eigenvalues (λj)Nj=1.

We denote by ZuN = span {φj}Nj=1, and ZsN = span {φj}∞j=N+1.

Denote by PN the projection and its adjoint P ∗N , defined by

PN = − 1

2πi

∫Γ

(λI −A)−1dλ;P ∗N = − 1

2πi

∫Γ

(λI −A∗)−1dλ,


where Γ (respectively, its conjugate Γ) separates the unstable spectrum from thestable one of A (respectively, A∗). We set

AuN = PNA;As

N = (I − PN )A (3.4) f12

for the restrictions of A to ZuN and ZsN , respectively. This projections commute

with A. We then have that the spectra of A on ZuN and ZsN coincide with {λj}Nj=1

and {λj}∞j=N+1, respectively. (For more details about the projection PN see [11]).

Moreover, since A generates a C0−analytic semigroup on H, then its restrictionAsN to ZsN generates likewise a C0−analytic semigroup on ZsN . This implies that

AsN satisfies the spectrum determined growth condition on ZsN , and so we have

||eAsN t||L(H,H) ≤ Cα0

e−α0t,∀t ≥ 0, (3.5) f13

for some α0 < |<λN+1|.The system (3.3) can accordingly be decomposed as

z = zN + ζN , zN = PNz, ζN = (I − PN )z,

where applying PN and (I − PN ) on (3.3), we obtain

on ZuN :d

dtzN −Au

NzN = PN (Bψ), zN (0) = PNz0, (3.6) f100

on ZsN :d

dtζN −As

NζN = (I − PN )(Bψ), ζN (0) = (I − PN )z0, (3.7) f101

respectively.

3.1. Stabilization of the unstable zN -system (3.6) on ZuN .

l12 Lemma 3.1. There exist α1 > 0 and a controller ψ such that once inserted in(3.6), yields the estimate

||zN (t)|| ≤ Cα1e−α1t||z0||, t ≥ 0,

here zN is the solution to (3.6), for the corresponding controller ψ.Moreover, the controller ψ can be chosen of class C1 and such that

| ddtψ(t)|+ |ψ(t)| ≤ Cα1

e−α1t||z0||, t ≥ 0.

Proof. We shall assume first that (A1) holds true. In virtue of Remark 2.4 andRemark 2.6 we have that the unstable eigenvalues (λj)

Nj=1 of A are simple and the

corresponding eigenfunctions (φj)Nj=1, (φ

∗j )Nj=1 can be chosen in such a way that

< φl, φ∗j >= δlj , l, j = 1, ..., N.

Let us decompose zN as

zN (t, y) =

N∑j=1

zj(t)φj(y).

We introduce that zN in equation (3.6), and obtain

N∑j=1

{(d

dtzj(t))φj − zjAφj

}= PN (Bψ).

We multiply (scalary) the above equation with φ∗l , and obtain

d

dt(zl)− λlzl = (PNBψ, φ∗l ), l = 1, .., N. (3.8) f20

14 IONUT MUNTEANU

Let us observe that we can assume P ∗N (φ∗l ) = φ∗l , l = 1, ..., N . We have

(PNBψ, φ∗l ) = (Bψ, P ∗Nφ∗l ) = ψB∗φ∗l =

= by (2.17) = ψν(φ∗l )′′′(1), l = 1, ..., N.

Denote by

bl = ν(φ∗l )′′′(1), l = 1, ..., N. (3.9) Y8

We set

z =

z1

z2

...zN

, (3.10) f16

B =

b1b2...bN

, (3.11) RE1

and

Λ =

λ1 0 .. 00 λ2 .. 0... ... .. ..0 .. 0 λN

(3.12) RE2

With these notations, equation (3.8) becomes

d

dtz− Λz = Bψ (3.13) f21

Remark 3.1. Note that B is a column matrix, and by Bψ we mean the productbetween a scalar and a column matrix. This is because, in our case ψ = ψ(t), only.

So, the question is whether the pair {Λ, B} is controllable. To see this we shalltest the Kalman’s criterion of controllability. The Kalman controllability matrix is

[B,ΛB,Λ2B, ...,ΛN−1B] =

b1 λ1b1 ... λN−1

1 b1b2 λ2b2 ... λN−1

2 b2...

.... . .

...

bN λNbN ... λN−1N bN

.

The Kalman matrix is of size N ×N . We shall prove that it is of full rank.It is easy to see that the rank of the Kalman matrix [Λ, B] is full if and only if

bl 6= 0, l = 1, ..., N , since the Kalman matrix is of Vandermonde type. By Lemma2.2, we have (φ∗l )

′′′(1) 6= 0, for all l = 1, ..., N , since φ∗l 6= 0, l = 1, ..., N . Hence,bl 6= 0, l = 1, ..., N (see (3.9)), so the Kalman’s controllability condition is satisfied.

Next we assume that assumption (A1) does not hold anymore.The general case of eigenvalues.The general case of eigenvalues (i.e., the case when the eigenvalues are not all

semi-simple) can be treated in the same manner. The only difference appears whenwe compute the Kalman matrix of controllability.

By Lemma 2.3 we know that the geometric multiplicity of each unstable eigen-value is 1, by Lemma 2.4 we know that the algebraic multiplicity of each unstableeigenvalue is ≤ 2. These combined, yields: for each λj we may choose a linear


independent system of generalized eigenfunctions {φji}4j=1 ,mj

i=1 ,mj− the algebraic

multiplicity of λj , such that the matrix Λ corresponding to AuN in this basis has

the Jordan canonical form; i.e.,

Λ =

J1 0 00 J2 0

0. . . JM

.

Where Jl = λl or Jl =

(λl 10 λl

), l = 1, ...,M .

Note that Jl cannot be of the form Jl =

(λl 00 λl

), since the geometric multi-

plicity of λl is 1. If Jl was of that form, then, when we compute the Kalman matrix,we obtain two lines which are proportional, hence the rank of the Kalman matrixis not full anymore. In that case, in order to have a full rank matrix, one has tointroduce another controller (i.e., a controller on the wall y = 0). But, here, it isnot the case. Hence, just one normal controller on the wall y = 1 is still enough.

For simplicity we shall assume that J1 =

(λ1 10 λ1

)and Jl = λl, l = 2, ...,M .

Other possible situations are treated similarly.

In this case, the matrix Λ has the form Λ =

λ1 1 ... 00 λ1 ... 00 0 λ2 ...0... ... ... ...0 0 ... λM

.

For this case, the Kalman matrix [Λ, B] is

[B,ΛB,Λ2B, ...,ΛN−1B] =

b1 λ1b1 + b2 ... λN−1

1 b1 + (N − 1)λN−21 b2

b2 λ1b2 ... λN−11 b2

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

bN λMbN ... λN−1M bN

.

Remember that N is the sum of the algebraic multiplicities of the unstableeigenvalues, in our case N = M + 1.

Multiplying the first line of [Λ, B] by b2, the second by b1 and taking the differenceof them, yields

det([Λ, B]) =

∣∣∣∣∣∣∣∣∣∣0 b22 ... (N − 1)λN−2

1 b221 λ1 ... λN−1

1

b3 λ2b3 ... λN−12 b3

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

bN λMbN ... λN−1M bN

∣∣∣∣∣∣∣∣∣∣= (3.14) Y2

= b22b3...bN

∣∣∣∣∣∣∣∣0 1 ... (N − 1)λN−2

1

1 λ1 ... λN−11

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

1 λM ... λN−1M

∣∣∣∣∣∣∣∣ .

Denote by J(x1, ..., xN ) =

∣∣∣∣∣∣∣∣1 x1 ... xN−1

1

1 x2 ... xN−12

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

1 xN ... xN−1N

∣∣∣∣∣∣∣∣. J is a Vandermonde type

determinat, and it is known that

16 IONUT MUNTEANU

J =∏

1≤i<j≤N

(xj − xi). (3.15) Y4

With this notation, by (3.14), we have that

det([Λ, B]) = b22b3...bN∂

∂x1J(λ1, λ1, λ2, ..., λM ) (3.16) Y6

Using (3.15), we get that

∂

∂x1J(x1, x2, ..., xN ) = −

∏1<i<j≤N

(xj − xi)

· N∑j=2

∏Nl=2 (xl − x1)

xj − x1

. (3.17) Y5

Thus, (3.16) and (3.17) combined, yields

det([Λ, B]) = −b22b3...bN

∏1<i<j≤M

(λj − λi)

· M∏j=3

(λl − λ1)

(3.18) Y7

Since bl 6= 0, l = 1, ..., N and the unstable eigenvalues are mutually distinct, weconclude that det([Λ, B]) 6= 0, i.e. rank of [Λ, B] is full in this case, also.

Since the Kalman’s criterion of controllability is satisfied we know from the finite-dimensional theory, that there exists a vector q in CN , such that the spectrum ofthe matrix [Λ +Bq] may be arbitrarily preassigned, in particular to lie on the lefthalf-plane {λ : <λ < −<λN+1}, as desired. Thus, returning form CN to ZuN , thereexists a vector Ξ in ZuN , such that Ξ =< zN ,Ψ >, whereby the closed loop systemcorresponding to (3.6) is given by

d

dtzN −Au

NzN =< zN ,Ξ >, (3.19) kk1

rewritten asd

dtzN = A

uzN , zN = eA

utzN (0), (3.20) kk2

where Au

= AuN + Ξ.

In conclusion, there exist α1 > 0 and Ξ ∈ ZuN such that the solution to (3.6)corresponding to the controller ψ =< zN ,Ξ >, satisfies the estimate

|ψ(t)|+ ||zN (t)|| = ||qzN (t)||+ ||zN (t)|| ≤

≤ (|q|+ 1)||eAutzN (0)|| ≤ Cα1

e−α1t||zN (0)||, t ≥ 0.

Moreover, since the solution zN is of class C1, the controller ψ is of class C1 andsatisfies the exponential decay claimed. �

3.2. Stabilization of system (3.3). In this section we prove the Theorem 3.2 byusing Lemma 3.1.

Proof of Theorem 3.2By Lemma 3.1, there exists a controller ψ which exponentially stabilizes the

finite dimensional zN−system (3.6) on ZuN . We next recall the infinite-dimensionalζN−system (3.7) on ZsN : ζ ′N −As

NζN = (I − PN )(Bψ), ζN (0) = (I − PN )z0. Weestimate in H the corresponding variation of parameter formula

ζN (t) = eAsN tζN (0) +

∫ t

0

eAsN (t−τ)(I − PN )(Bψ)dτ (3.21) m1

Remember that B = (θk + Fk)Dk. Hence, relation (3.21) becomes


ζN (t) = eAsN tζN (0) +

∫ t

0

eAsN (t−τ)(I − PN )((θk + Fk)Dk)ψ)dτ = (3.22) m2

= eAsN tζN (0) +

∫ t

0

eAsN (t−τ)(I − PN )θkDkψdτ +

∫ t

0

eAsN (t−τ)(I − PN )FkDkψdτ.

By Proposition 2.2 we have that Dk : C → H is a continuous linear operator.Hence, in virtue of Lemma 3.1 and relation (3.5), we have∣∣∣∣∫ t

0

eAsN (t−τ)(I − PN )θkDkψdτ

∣∣∣∣ ≤ ∫ t

0

C||eAsN (t−τ)|||ψ|dτ ≤ (3.23) m3

≤ C∫ t

0

e−α0(t−τ)e−α1τ ||z0||dτ ≤ Ce−α0t||z0||.

For the third term of the RHS of (3.22), since AsN = −(I − PN )Ak and Ak =

FkL−1k , we have that∫ t

0

eAsN (t−τ)(I − PN )FkDkψdτ = −

∫ t

0

eAsN (t−τ)As

NLkDkψdτ = (3.24) m4

= −∫ t

0

d

dτ

(eA

sN (t−τ)

)LkDkψdτ = eA

sN tLkDkψ(0)−LkDkψ(t)+

∫ t

0

eAsN (t−τ)LkDk

d

dτψ(τ)dτ.

Relation (3.24), togheter with (3.5) and Lemma 3.1, implies∣∣∣∣∫ t

0

eAsN (t−τ)(I − PN )FkDkψdτ

∣∣∣∣ ≤ Ce−α0t||z0||+Ce−α1t||z0||+C∫ t

0

e−α0(t−τ)e−α1τ ||z0||dτ ≤ Ce−α0t||z0||.

(3.25) m5

Finally, relations (3.22), (3.23), (3.25) and (3.5) combined, yields

||ζN (t)|| ≤ Ce−α0t||z0||,∀t ≥ 0. (3.26) pe

Since z = zN + ζN , we have, in virtue of Lemma 3.1 and (3.26), that

||z(t)|| ≤ Cα0e−α0t||z0||,∀t ≥ 0.

W5 Remark 3.2. By Theorem 3.2 and notations (3.2), we have that there exists acontroller ψk, such that once inserted into equation (3.3), the corresponding solutionof (3.3) satisfies the estimate

||Lkvk(t)|| ≤ Cα0e−α0t||Lkv0

k||,∀t ≥ 0. (3.27) W4

Taking the scalar product between Lkv and v, for some v ∈ H, we get that

||v′||2 + k2||v||2 =< Lkv, v > .

Hence, by the Schwarz inequality, we have

||v′||2 + k2||v||2 ≤ |(Lkv, v)| ≤ ||Lkv||||v||. (3.28) az3

This implies that||v|| ≤ ||Lkv||. (3.29) lp7

Taking Cα0 := Cα0 ||Lk||L(H,H), relations (3.27) and (3.29) imply that

||vk(t)|| ≤ Cα0e−α0t||v0

k||, t ≥ 0. (3.30) lp8

Returning to the notations (3.2), relation (3.30) can be rewritten as

||L−1k z(t)|| ≤ Cα0e

−α0t||L−1k z0||,∀t ≥ 0. (3.31) lp18

18 IONUT MUNTEANU

3.3. Feedback stabilization of the system (3.3). The next theorem gives afeedback representation of the controller in Remark 3.2 using a standard tehnique:minimization of a cost-functional.

pop1 Theorem 3.3. For all |k| ≤ S, there exists a feedback controller of the form ψk =−ν(L−2

k Rkz)′′′(1), such that, once inserted into equation (3.3), the corresponding

solution of the closed-loop system (3.3) satisfies

||L−1k z(t)|| ≤ Ce−γt||L−1

k z0||,∀t ≥ 0.

Here Rk : H → H is a linear self-adjoint operator which satisfies the next Riccatialgebraic equation in H

− < L−1k Rkz0, L

−1k Az0 > +

1

2ν2|(L−2

k Rkz0)′′′(1)|2 =1

2||L−1

k z0||2,∀z0 ∈ H.

Proof. Let us first observe that the operator A is the extension to H of the operatorAk. Hence, it is corectly to consider the system (3.3) in the space H and look fora feedback representation of the controller in Remark 3.2 via the linear quadraticcontrol problem

φ(z0) = min1

2

∫ ∞0

(||L−1k z(t)||2 + |ψk(t)|2)dt, (3.32) e1

subject to ψk ∈ L2(0,∞;H) and

d

dtz(t)−Az(t) = Bψk, z(0) = z0. (3.33) e2

Let us first show that the optimization problem is well-posed on the state spaceH, i.e., φ(z0) <∞,∀z0 ∈ H.

We must show that with z0 ∈ H arbitrary, there exists some control ψk ∈L2(0,∞;H) such that the corresponding solution z of (3.33) satisfy z ∈ L2(0,∞;H).Indeed, Remark 3.2 provide the wanted result (see relation (3.31)). From the ex-ponential stability we deduce also that there exists some constant a2 such that

φ(z0) ≤ a2||L−1k z0||2,∀z0 ∈ H. (3.34) e3

It is easy to see that the map φ(z)→ z ∈ H is continuous, thus ||z|| ≤ cφ(z). Thiswith relation (3.34), show that there exist constants a1 and a2 such that:

a1||z0||2 ≤ φ(z0) ≤ a2||z0||2,∀z0 ∈ H. (3.35) e4

Thus, by (3.35), there is a linear non-negative selfadjoint operator Rk : H → Hassociated with the linear symmetric form φ(·), such that

φ(z0) =1

2< Rkz0, z0 >,∀z0 ∈ H. (3.36) e5

Rk ∈ L(H,H).By the dynamic programming principle, for each 0 < t < T , the optimal solution

(ψ∗k, z∗) to (3.32), (3.33) is also the solution to the optimization problem

min

{1

2

∫ T

t

(||L−1k z(s)||2 + |ψk(s)|2)ds+ φ(z(T )), subject to (3.32), z(t) = z∗(t)

},

(3.37) e6

z∗(t) ∈ H as initial condition, where z∗(T ) ∈ H as well.By the maximum principle, we obtain that


ψ∗k(t) = B∗qT = νq′′′T (1), a.e. t ∈ (0, T ),

L−2k Rkz

∗(t) = −qT (t),∀t ∈ [0, T ], (3.38) e7

ψ∗k(t) = B∗(L−2k Rkz

∗(t)),∀t ≥ 0,

where qT is the solution to the dual equation:

d

dtqT + A∗qT = L−2

k z,∀t ∈ (0, T ),

qT (T ) = −L−2k Rkz

∗(T ). (3.39)

Finally, we show that Rk is a solution to a Riccati type equation. To this end,we first notice that, again by the dynamic programming principle and (3.36), wehave

1

2< Rz∗(t), z∗(t) >= φ(z∗(t)) =

1

2

∫ ∞t

(||L−1k z∗(s)||2 + |ψ∗k(s)|2)ds,∀t ≥ 0.

(3.40) e8

Differentiating (3.40) in t and using self-adjointness of Rk on H and equation(3.33), yields:

−(L−1k Az∗(t), L−1

k Rkz∗(t)) +

1

2ν2|(L−2

k Rkz∗(t))′′′(1)|2 =

1

2||L−1

k z∗(t)||2, t ≥ 0.

(3.41)Which implies, by setting t = 0, that Rk satisfies the Riccati equation

(L−1k Az0, L

−1k Rkz0) +

1

2ν2|(L−2

k Rkz0)′′′(1)|2 =1

2||L−1

k z0||2,∀z0 ∈ H. (3.42) e11

Let us observe that since z0 ∈ H and Rk ∈ L(H,H) we have that Rkz0 ∈ H.Hence, L−2Rkz0 ∈ H4(0, 1). Thus it makes sense to consider the third derivativeof L−2

k Rkz0 in the above Riccati algebraic equation.In the end, using the classical Datko’s Theorem, we finally obtain the exponential

decay claimed. �

pop3 Remark 3.3. Applying Theorem 3.3 for all |k| ≤ S (see notations (3.2)), we obtainthat:

for all |k| ≤ S, there exist constants Cγk , γk > 0 and a feedback controller

ψk of the form ψk(t) = −ν(L−2k RkL

−1k vk(t))′′′(1), such that, once inserted into the

equation (2.18), the corresponding solution of the closed-loop system (2.18) satisfiesthe exponential decay

||vk(t)||2 ≤ Cγke−γkt||v0k||2,∀t ≥ 0. (3.43) pop2

There exists a constant Ck > 0 such that ||v|| ≤ Ck||L−1k v||, for v ∈ D(L−1

k ). Hence,by (3.43), we have that (by taking Cγk := CkCγk)

||Lkvk(t)||2 ≤ Cγke−γkt||v0k||2, t ≥ 0. (3.44) az6

Then, using the Schwarz inequality, one can obtain as in Remark 3.2 that

||v′k||2 ≤ ||Lkvk||||vk||. (3.45) az7

It follows by (3.43),(3.44) and (3.45) that

||v′k(t)||2 ≤ Cγke−γkt||v0k||2, t ≥ 0. (3.46) az8

Now, taking into account that uk = − 1ikv′k, relation (3.46) implies that

||uk(t)||2 ≤ Cγke−γkt||v0k||2, t ≥ 0, |k| ≤ S. (3.47) az9

20 IONUT MUNTEANU

Finally, relations (3.43) and (3.47) imply that(||uk(t)||2 + ||vk(t)||2

)≤ 2Cγke

−γkt(||u0

k||2 + ||v0k||2), t ≥ 0, |k| ≤ S. (3.48) az10

3.4. Feedback stabilization of the linear system (1.5). Next, we will proveTheorem 3.1 by collecting all the results obtained in the previous sections.

Proof of Theorem 3.1Let us define the feedback controller (see Remark 3.3)

Ψ(t, x) = −ν∑|k|≤S

(L−2k RkLkvk(t))′′′(1)eikx, (3.49) pop4

where

vk(t, y) =

∫ 2π

0

v(t, x, y)e−ikxdx, |k| ≤ S.

Here Rk is given by Theorem 3.3.If we plug that controller into system (1.5), we have, by Remark 3.3, that the

Fourier modes of u and v, for all |k| ≤ S, satisfy(||uk(t)||2 + ||vk(t)||2

)≤ 2Cγke

−γkt(||u0

k||2 + ||v0k||2),∀t ≥ 0, (3.50) W8

for some Cγk , γk > 0. (see (3.48))By Remark 2.1, we have that the Fourier modes of u and v, for all |k| > S,

satisfy (||uk(t)||2 + ||vk(t)||2

)≤ e−νS

2t(||u0

k||2 + ||v0k||2),∀t ≥ 0 (3.51) W7

(see (2.12))Thus, by (3.50) and (3.51), yields∑

k

(||uk(t)||2 + ||vk(t)||2

)≤ Ce−αt

∑k

(||u0

k||2 + ||v0k||2), (3.52) W10

where C = max {2Cγk : |k| ≤ S, 1} and α = min{γk : |k| ≤ S, νS2

}.

By the definition of the norm in (L2(Q))2 and relation (3.52) we finally get that

||(u(t), v(t))||2 ≤ Ce−αt||(u0, v0)||2, t ≥ 0. (3.53) W12

i.e. the exponential decay claimed.

Remark 3.4. Theorem 3.1 is a global stabilizable result for the linearization equa-tion (1.5) with actuation on the wall y = 1 and the stabilizing effect is independentof how large is the Reynolds number. It should be emphasized that the Riccatiequation (3.42), which provides the feedback controller (3.49), is easily manageablefrom computational point of view since it is associated with a parabolic boundarycontrol system on (0, 1), whose structure is identic for different wave numbers k.

Acknowledgement. This work is financial supported by project POSDRU/88/1.5/S/47646,cofinantiated by the European Social Found, the Operational Sectorial ProgrammeDevelopment of Human Resources 2007-2013 and by CNCSIS project PN II IDEIID 70/2008.


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Ionut Munteanu

”O. Mayer” Institute of Mathematics of the Romanian Academy, ”Al. I. Cuza” Univer-

sity, Department of Mathematics700506 Iasi, Romania

E-mail address: [email protected]

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