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Control and Stabilization of the Benjamin-Ono Equation on a Periodic Domain

Felipe Linares, Lionel Rosier

To cite this version:

Felipe Linares, Lionel Rosier. Control and Stabilization of the Benjamin-Ono Equation on a Periodic Domain. Transactions of the American Mathematical Society, American Mathematical Society, 2015, 367 (7), pp.4595-4626. �hal-00734445�

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ON A PERIODIC DOMAIN

FELIPE LINARES AND LIONEL ROSIER

Abstract. It was proved by Linares and Ortega in [24] that the linearized Benjamin-Ono equation posed on a periodic domain Twith a distributed control supported on an arbitrary subdomain is exactly controllable and exponentially stabilizable. The aim of this paper is to extend those results to thefull Benjamin-Ono equation. A feedback law in the form of a localized damping is incorporated in the equation. A smoothing effect established with the aid of a propagation of regularity property is used to prove the semi-global stabilization inL2(T) of weak solutions obtained by the method of vanishing viscosity. The local well-posedness and the local exponential stability inHs(T) are also established for s >1/2 by using the contraction mapping theorem. Finally, the local exact controllability is derived in Hs(T) for s >1/2 by combining the above feedback law with some open loop control.

1. Introduction The Benjamin-Ono (BO) equation can we written as

ut+Huxx+uux = 0,

whereu=u(x, t) denotes a real-valued function of the variablesx∈Randt∈R, andHdenotes the Hilbert transform defined as

d

Hu(ξ) =−isgn(ξ) ˆu(ξ).

This integro-differential equation models the propagation of internal waves in stratified fluids of great depth (see [4, 33]) and turns out to be important in other physical situations as well (see [9, 18, 26]). Among noticeable properties of this equation we can mention that: (i) it defines a Hamiltonian system; (ii) it admits infinitely many conserved quantities (see [6]); (iii) it can be solved by an analogue of the inverse scattering method (see [2]); (iv) it admits (multi)soliton solutions (see [6]).

In this paper, we consider the BO equation posed on the periodic domain T=R/(2πZ):

ut+Huxx+uux = 0, x∈T, t∈R, (1.1) where the Hilbert transformH is defined now by

(Hdu)k=−isgn(k)ˆuk.

1991 Mathematics Subject Classification. 93B05, 93D15, 35Q53.

Key words and phrases. Benjamin-Ono equation; Periodic domain; Unique continuation property; Propagation of regularity; Exact controllability; Stabilization.

1

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The two first conserved quantities are

I1(t) = Z

T

u(x, t)dx and

I2(t) = Z

T

u2(x, t)dx.

From the historical origins [4, 33] of the BO equation, involving the behavior of stratified fluids, it is natural to think I1 and I2 as expressing conservation of volume (or mass) and energy, respectively.

The Cauchy problem for the equation (1.1) in the real line has been intensively studied for many years ([45, 17, 1, 32, 31, 20, 19, 46, 5, 16, 29, 13, 14]). In the periodic case, there have been several recent developments. (See for instance [28, 30, 29] and the references therein.) The best known result so far [28, 29] is that the Cauchy problem is well-posed in the space

H0s(T) ={u∈Hs(T); ˆu0 := 1 2π

Z

T

u(x)dx= 0}

for s≥0. Moreover, the corresponding solution map (u0 → u) is real analytic from the space H00(T) to the spaceC([0, T], H00(T)).

In this paper we will study the equation (1.1) from a control point of view with a forcing term f =f(x, t) added to the equation as a control input:

ut+Huxx+uux=f(x, t), x∈T, t∈R, (1.2) where f is assumed to be supported in a given open set ω ⊂ T. The following exact control problem and stabilization problem are fundamental in control theory.

Exact Control Problem: Given an initial state u0 and a terminal state u1 in a certain space, can one find an appropriate control inputf so that the equation (1.2) admits a solution u which satisfies u(·,0) =u0 and u(·, T) =u1?

Stabilization Problem: Can one find a feedback law f =Ku so that the resulting closed- loop system

ut+Huxx+uux=Ku, x∈T, t∈R+ is asymptotically stable as t→+∞?

Those questions were first investigated by Russell and Zhang in [44] for the Korteweg-de Vries equation, which serves as a model for propagation of surface waves along a channel:

ut+uxxx+uux =f, x∈T, t∈R. (1.3) In their work, in order to keep themassI1(t) conserved, the control input is chosen to be of the form

f(x, t) = (Gh)(x, t) :=a(x)

h(x, t)− Z

T

a(y)h(y, t)dy

where h is considered as a new control input, anda(x) is a given nonnegative smooth function such that {x∈T; a(x)>0}=ω and

2π[a] = Z

T

a(x)dx= 1.

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For the chosen a, it is easy to see that d

dt Z

T

u(x, t)dx= Z

T

f(x, t)dx= 0 ∀t∈R for any solution u=u(x, t) of the system

ut+uxxx+uux =Gh, x∈T, t∈R. (1.4) Thus the mass of the system is indeed conserved.

The control of dispersive nonlinear waves equations on a periodic domain has been extensively studied in the last decade: see e.g. [44, 40, 23] for the Korteweg-de Vries equation, [27] for the Boussinesq system, [42] for the BBM equation, and [11, 38, 21, 41, 22] for the nonlinear Schr¨odinger equation. By contrast, the control theory of the BO equation is at its early stage.

The following results are due to Linares and Ortega [24].

Theorem A.[24] Let s≥0 and T >0 be given. Then for any u0, u1 ∈Hs(T) with [u0] = [u1] one can find a control input h∈L2(0, T, Hs(T)) such that the solution of the system

ut+Huxx =Gh, u(x,0) =u0(x) (1.5)

satisfies u(x, T) =u1(x).

In order to stabilize (1.5), Linares and Ortega employed a simple control law h(x, t) =−Gu(x, t).

The resulting closed-loop system reads

ut+Huxx =−GGu.

Theorem B. [24] Let s≥0 be given. Then there exist some constants C >0 and λ > 0 such that for any u0∈Hs(T), the solution of

ut+Huxx=−GGu, u(x,0) =u0(x) satisfies

ku(·, t)−[u0]kHs(T) ≤Ce−λtku0−[u0]kHs(T) ∀t≥0.

The extension of those results to the full BO equation (1.4) turns out to be a very hard task.

Indeed, it is by now well known that the contraction principle cannot be used to establish the local well-posedness of BO in H0s(T) for s ≥0. The method of proof in [28, 29] used strongly Tao’s gauge transform, and it is not clear whether this approach can be followed when an additional control term is present in the equation.

For the sake of simplicity, we shall assume from now on that [u0] = 0, so that u(t) has a zero mean value for all times.

To stabilize the BO equation, we consider the following feedback law f =−G(D(Gu))

where Duck=|k|uˆk. Scaling in (1.3) byu gives (at least formally) 1

2ku(T)k2L2(T)+ Z T

0 kD12(Gu)k2L2(T)dt= 1

2ku0k2L2(T). (1.6)

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This suggests that the energy is dissipated over time. On the other hand, (1.6) reveals a smoothing effect, at least in the region {a > 0}. Using a propagation of regularity property in the same vein as in [11, 21, 22, 23], we shall prove that the smoothing effect holds everywhere, i.e.

kukL2(0,T;H12(T))≤C(T,ku0k). (1.7) Using this smoothing effect and the classical compactness/uniqueness argument, we shall first prove that the corresponding closed-loop equation is semi-globally exponentially stable.

Theorem 1.1. Let R >0 be given. Then there exist some constants C =C(R) and λ=λ(R) such that for any u0 ∈ H00(T) with ku0k ≤ R, the weak solutions in the sense of vanishing viscosity of

ut+Huxx+uux=−GDGu, u(x,0) =u0(x) (1.8) satisfy

ku(t)k ≤Ce−λtku0k ∀t≥0.

A weak solution of (1.8) in the sense of vanishing viscosity is a distributional solution of (1.8) u∈Cw(R+, H00(T))∩L2loc(R+, H

1 2

0(T)) that may be obtained as a weak limit in a certain space of solutions of the BO equation with viscosity

ut+ (H−ε)uxx+uux =−GDGu, u(x,0) =u0(x) (1.9) asε→0+(see below Definition 2.11 for a precise definition). The issue of theuniquenessof the weak solutions in the sense of vanishing viscosity seems challenging.

Using again the smoothing effect (1.7), one can extend (at least locally) the exponential stability from H00(T) to H0s(T) fors >1/2.

Theorem 1.2. Let s ∈ (12,2]. Then there exists ρ > 0 such that for any u0 ∈ H0s(T) with ku0kHs(T) < ρ, there exists for all T > 0 a unique solution u(t) of (1.8) in the class C([0, T], H0s(T))∩L2(0, T, Hs+

1 2

0 (T)). Furthermore, there exist some constants C >0 andλ >0 such that

ku(t)ks ≤Ce−λtku0ks ∀t≥0.

Finally, incorporating the same feedback lawf =−G(D(Gu)) in the control input to obtain a smoothing effect, one can derive an exact controllability result for the full equation as well.

Theorem 1.3. Let s ∈ (12,2] and T >0 be given. Then there exists δ > 0 such that for any u0, u1 ∈H0s(T) satisfying

ku0kHs(T) ≤δ, ku1kHs(T)≤δ

one can find a control input h∈L2(0, T, Hs−12(T))such that the system (1.4) admits a solution u∈C([0, T], H0s(T))∩L2(0, T, Hs+

1 2

0 (T)) satisfying

u(x,0) =u0(x), u(x, T) =u1(x).

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Note that it would be desirable to have a control input h in the class L2(0, T, Hs(T)), but this will require to adapt the analysis in [28, 29]. Note also that a global controllability result inH00(T) would follow from Theorems 1.1 and 1.3 if Theorem 1.3 were also true fors= 0.

The paper is organized as follows. Section 2 is concerned with the local well-posedness and the stability properties of (1.8). We first prove the global well-posedness of (1.9) in the energy space H00(T), by using classical energy estimates (Theorem 2.1). Next, we establish several technical properties, namely a commutator estimate (Lemma 2.5), a propagation of regularity property (Propositions 2.7 and 2.16), and a unique continuation property (Proposition 2.8) that are used to derive the exponential stability of (1.9) with a decay rate independent of ε (Theorem 2.10).

This leads to the proofs of Theorems 1.1 and 1.2. Finally, the control properties of (1.4) are investigated in Section 3.

2. Stabilization of BO with a localized damping 2.1. Semi-global exponential stabilization in L2(T).

Pick any function

a∈C(T,R+) with Z

T

a(x)dx= 1 (2.10)

decomposed as a(x) =P

k∈Zkeikx.

We are interested in the stability properties of the BO equation with localized damping ut+Huxx+ (u2

2 )x=−G(D(Gu)), u(0) =u0, (2.11) where

d

Huk =−isgn(k)ˆuk, Ddsuk=|k|sk, (Gu)(x) =a(x)(u(x)− Z

T

a(y)u(y)dy). (2.12) We shall assume that u0 ∈H00(T), where for anys∈R,

H0s(T) ={u=X

k∈Z

ˆ

ukeikx ∈Hs(T); ˆu0 = 0}. Let (u, v) = R

Tu(x)v(x)dx denote the usual scalar product in L2(T) with kuk = kukL2(T) as associated norm, and for any s ∈ R, let (u, v)s = ((1−∂x2)s2u,(1−∂x2)s2v) denote the scalar product inHs(T) with corresponding normkuks = (u, u)

1

s2. Lethxi:= (1 +|x|2)12 for anyx∈R. Note that fors <0 andu∈Hs(T), Guhas to be understood as

Gu=a u− hu, aiHs(T),Hs(T)

.

Assuming that u0 ∈H00(T), we obtain (formally) by scaling in (2.11) byu that 1

2ku(T)k2+ Z T

0 kD12(Gu)k2dt= 1

2ku0k2. (2.13)

This suggests that the energy is dissipated over time. On the other hand, (2.13) reveals a smoothing effect, at least in the region {a > 0}. Using a propagation of regularity property in

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the same vein as in [11, 21, 22, 23], we shall prove that the smoothing effect holds everywhere, i.e.

u∈L2(0, T;H12(T)). (2.14)

Of course, a rigorous derivation of (2.13) requires enough regularity foru, e.g.

u∈L2(0, T, H1(T))∩C([0, T], H00(T)). (2.15) As there is a gap between (2.14) and (2.15), we are let to put some artificial viscosity in (2.11) (parabolic regularization method) to derive in a rigorous way the energy identity for the ε−BO equation

ut+Huxx+uux=εuxx−G(D(Gu)), u(0) =u0. (2.16) We shall prove the global well-posedness (GWP) of (2.16) inH00, together with the semi-global exponential stability in H00 with a decay rateuniforminε >0. Lettingε→0, this will give the semi-global exponential stability in H00 of the weak solutions u∈Cw([0,+∞), H00(T)) of (2.11) obtained as limits of the (strong) solutions of (2.16). The (difficult) issue of the uniqueness of a weak solution to (2.11) will not be addressed here.

We first establish the GWP of (2.16).

Theorem 2.1. Let ε > 0 and u0 ∈H00(T). Then for anyT >0 there exists a unique solution u∈C([0, T], H00(T))∩L2(0, T;H1(T)) of (2.16). Moreover

u∈C((0, T], H2(T))∩C1((0, T], H1(T)), (2.17) and for any t≥0

1

2ku(t)k2+ε Z t

0 kux(τ)k2dτ+ Z t

0 kD12(Gu)(τ)k2dτ = 1

2ku0k2. (2.18) Proof: The proof of Theorem 2.1 is divided into five parts. Note that the weak smoothing effect (2.14) will be established later, as it is not needed here.

Step 1. Linear Theory We consider the linear system

ut+ (H−ε)uxx+G(D(Gu)) = 0, u(0) =u0.

Let Au = (H−ε)uxx with domain D(A) = H02(T) ⊂ H00(T), and Bu = G(D(Gu)). Clearly G ∈ L(Hr(T), H0r(T)) for all r ∈ R, hence B ∈ L(H01(T), H00(T)). Let θ0 ∈ (arctanε−1, π/2).

Then, for θ0 <|argλ| ≤π, we have k(A−λ)−1k ≤sup

k6=0|(ε+isgnk)k2−λ|−1 ≤ C

|λ|·

It follows thatA is a sectorial operator (see [15, Definition 1.3.1]) inH00(T). Note thatσ(A) = {(ε+isgnk)k2; k∈Z}. Therefore, Re σ(A)≥ε and A−α is meaningful for all α >0. Since for all s >0

kAs2uk2Hs(T)≤CX

k6=0

|ε+isgnk|−s|uˆk|2 ≤Ckuk2L2(T)

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we infer that BA12 ∈L(H00(T)). It follows from [15, Corollary 1.4.5] that the operator A:=

A+B is also sectorial, so that −A generates an analytic semigroup S(t)

t≥0 = (e−tA)t≥0 on H00(T) according to [15, Theorem 1.3.4]. Note that, by [15, Theorem 1.4.8],D((A+B+λ)α) = D(Aα) =H0(T) for all α≥0 and λ >0 large enough, hence

S(t)H0s(T)⊂H0s(T), ∀t >0, ∀s≥0.

Let us derive estimates for the solutions of the Cauchy problem

ut+Au=f, u(0) =u0. (2.19)

For anyT >0 and anys∈N, let

Ys,T =C([0, T];H0s(T))∩L2(0, T;H0s+1(T)) (2.20) be endowed with the norm

kukYs,T =kukL(0,T;Hs(T))+kukL2(0,T;Hs+1(T)). (2.21) Lemma 2.2. We have for some constant C0 =C0(ε, s, T)

kukYs,T ≤C0 ku0ks+kfkL1(0,T,Hs(T))

,

u denoting the mild solution of (2.19) associated with (u0, f)∈H0s(T)×L1(0, T, H0s(T)).

Proof of Lemma 2.2. It is well known from classical semigroup theory that kukL(0,T,Hs(T))≤C ku0ks+kfkL1(0,T,Hs(T))

·

Next we estimatekukL2(0,T,Hs+1(T)). We first assume u0 ∈H0s+2(T) and f ∈C([0, T];H0s+2(T)), so that u ∈ C([0, T];H0s+2(T))∩C1([0, T];H0s(T)). Taking the scalar product of each term of (2.19) byu inHs(T) results in

1

2ku(t)k2s+ε Z t

0 kuxk2sdτ+ Z t

0

(G(D(Gu)), u)sdτ = 1

2ku0k2s+ Z t

0

(f, u)sdτ. (2.22) The identity (2.22) is also true for u0 ∈ H0s(T) and f ∈ L1(0, T, H0s(T)), by density. The following claim is needed.

Claim 1. For anys∈R, there exists a constant C=C(s)>0 such that

− G(D(Gu)), u

s≤Ckuk2s− kD12(Gu)k2s ∀u∈H0s+1(T).

Proof of Claim 1. We have G(D(Gu)), u

s = (1−∂x2)s2G(D(Gu)),(1−∂2x)s2u

= [(1−∂x2)s2, G]D(Gu),(1−∂x2)2su) +(G(1−∂2x)s2D(Gu),(1−∂x2)s2u)

=: I1+I2. Since a∈C(T), we easily obtain that

k[(1−∂x2)2s, G]uk ≤Ckuks−1. It follows that

|I1| ≤Ckuk2s.

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On the other hand

I2 = (1−∂x2)s2D(Gu), G(1−∂x2)s2u

= k(1−∂x2)s2D12(Gu)k2+ ((1−∂x2)s2(Gu), D[G,(1−∂x2)2s]u), hence

−I2≤Ckuk2s− kD12(Gu)k2s· The claim is proved.

Combining Claim 1 with (2.22), we obtain that for t=T 1

2ku(T)k2s+ε Z T

0 kux(τ)k2sdτ+ Z T

0 kD12(Gu)k2s

≤ 1

2ku0k2s+Ckuk2L2(0,T,Hs(T))+1

2kuk2L(0,T,Hs(T))+1

2kfk2L1(0,T,Hs(T))

≤C ku0k2s+kfk2L1(0,T,Hs(T))

·

The proof of Lemma 2.2 is achieved.

Remark 2.3. We observe that whenu0≡0 in (2.19) then k

Z t

0

S(t−τ)f(τ)dτkYs,T ≤C(ǫ, s, T)kfkL1(0,T,Hs(T)), (2.23) and when f ≡0 in (2.19)

kS(t)u0kYs,T ≤C(ǫ, s, T)ku0kHs(T). (2.24) Step 2. Local Well-posedness in H0s(T), s≥0

We prove the following

Proposition 2.4. Let s ≥ 0. For any u0 ∈ H0s(T), there exists some T > 0 such that the problem (2.16) admits a unique solution u∈Ys,T.

Proof. Write (2.16) in its integral form u(t) =S(t)u0

Z t

0

S(t−τ)(uux)(τ)dτ

where the spatial variable is suppressed throughout. For given u0 ∈H0s(T), letr >0 andT >0 be constants to be determined. Define a map Γ on the closed ball

B =

v∈Ys,T; kvkYs,T ≤r of Ys,T by

Γ(v)(t) =S(t)u0− Z t

0

S(t−τ)(vvx)(τ)dτ.

We aim to prove that Γ contracts inB forT small enough andr conveniently chosen. To that end, we shall prove the following estimates

kΓ(v)kYs,T ≤ C0ku0ks+C1T14kvk2Ys,T, ∀v∈B, (2.25) kΓ(v1)−Γ(v2)kYs,T ≤ C1T14(kv1kYs,T +kv2kYs,T)kv1−v2kYs,T ∀v1, v2 ∈B. (2.26)

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From Lemma 2.2 and Remark 2.3, it is adduced that kΓ(v1)−Γ(v2)kYs,T ≤ Ckv1vx1−v2v2xkL1(0,T,Hs(T))

≤ C Z T

0 kv1−v2kLkv1+v2ks+1+kv1+v2kLkv1−v2ks+1

≤ CT14kv1−v2kYs,T kv1kYs,T +kv2kYs,T

where we used the fact that

Z T

0 kvk2Ldt≤C Z T

0 kvk1kvkdt≤C√

TkvkL(0,T,L2(T))kvkL2(0,T,H1(T)).

This yields (2.26). (2.25) follows from Lemma 2.2, Remark 2.3 and (2.26). Choosing r >0 and

T >0 so that

r = 2C0ku0ks,

2rC1T1412, (2.27)

we obtain that

kΓ(v1)kYs,T ≤r, kΓ(v1)−Γ(v2)kYs,T ≤ 1

2kv1−v2kYs,T

for any v1, v2 ∈B. Thus, with this choice ofr andT, Γ is a contraction inB. Its fixed-point is the unique solution of (2.16) in B.

Step 3. Global Well-Posedness in H00(T).

Assume thatu0∈H00(T). We first establish (2.18) for 0≤t≤T. Sinceu∈Y0,t, we have that Z t

0 kuuxk2−1dτ ≤ C Z t

0 ku2k2

≤ C Z t

0 kuk3kuxkdτ

≤ C√

tkuk4Y0,t.

Thus each term in (2.16) belongs to L2(0, t, H−1(T)). Scaling in (2.16) byu yields Z t

0 hut+ (H−ε)uxx+uux+G(D(Gu)), uiH−1(T),H1(T)dτ = 0.

We have that for a.e. τ ∈(0, t)

h(H−ε)uxx, uiH−1(T),H1(T)=−((H−ε)ux, ux) =εkuxk2, huux, uiH−1(T),H1(T) = (uux, u) = 0,

hG(D(Gu)), uiH−1(T),H1(T) = (G(D(Gu)), u) =kD12(Gu)k2.

(2.18) follows at once, and we infer thatku(t)k ≤ ku0k. Using the standard extension argument, one sees that u is defined on R+ withu ∈Y0,T for all T >0. Furthermore, with the constants C0 and C1 given in Step 2 for s= 0 andT = (8C0C1ku0k)−4, we obtain

ku(nT +·)kY0,T ≤2C0ku(nT)k ≤2C0ku0k.

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Step 4. Global Well-Posedness in H02(T).

Pick any u0 ∈H02(T). By Proposition 2.4 and Step 3, (2.16) admits a unique solutionu ∈Y0,T for each T >0, which belongs to Y2,T0 for someT0 >0. We just need to show that T0 may be taken as large as desired. Letv=ut. Ifu∈Y2,T, thenv∈Y0,T and it satisfies

vt+ (H−ε)vxx+ (uv)x =−G(D(Gv)), v(0) =v0 (2.28) where

v0 :=− {(H−ε)u0,xx+u0u0,x+G(D(Gu0))} ∈H00(T).

We may write (2.28) in its integral form v(t) =S(t)v0

Z t

0

S(t−s)(uv)x(s)ds.

Let Γ(w)(t) = S(t)v0−Rt

0S(t−s)(uw)x(s)ds for w ∈ Y0,T. Computations similar to those in Step 2 lead to

kΓwkY0,T ≤ C0kv0k+C1T14kukY0,TkwkY0,T, kΓ(w1)−Γ(w2)kY0,T ≤ C1T14kukY0,Tkw1−w2kY0,T

where the constantsC0 andC1 depend only on εforT <1. Therefore Γ contracts inB={w∈ Y0,θ; kwkY0,θ ≤r:= 2C0kv0k}, provided that

C1θ14kukY0,θ ≤ 1 2·

Its fixed point gives the unique solution of the integral equation inB. Pickθ fulfilling θ <min{(8C0C1ku0k)−4,1}.

Then, from Step 2, we have that

ku(nθ+·)kY0,θ ≤2C0ku0k

for all n∈Nand thatw may be extended to [nθ,(n+ 1)θ] inductively by using the contraction mapping theorem (replacingv0byw(θ),w(2θ), etc.). Therefore,wis defined onR+and it holds kw(nθ+·)kY0,θ ≤2C0kw(nθ)k ≤(2C0)n+1kv0k. (2.29) By uniqueness of the solution of the integral equation, we have that v(t) = w(t) as long as 0 < t < T and v ∈ Y0,T. (2.29) shows that kv(t)k =kw(t)k is uniformly bounded on compact sets of R+, namely

kvkY0,T ≤C(T,ku0k)kv0k. The same is true for ku(t)k2, by (2.16). Indeed, since

kuuxk ≤ kukL(T)kuxk ≤ kuk54kuxxk34 ≤Cδkuk5+δkuxxk, we infer from (2.16) that

k(H−ε)uxx(t)k ≤C(T,ku0k)ku0k2+C(kuk+kuk5) +δkuxxk hence

ku(t)k2 ≤C(T,ku0k)ku0k2.

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Using the standard extension argument, one sees that u(t)∈H02(T) for all t≥0 withu∈Y2,T for all T >0.

Step 5. Smoothing effect from H00(T) to H02(T).

Pick anyu0∈H00(T). Then the solutionuto (2.16) belongs toY0,1. Therefore, for a.e. t0∈(0,1), u(t0) ∈ H01(T). The solution of (2.16) in Y1,T issued from u(t0) at t = 0 must coincide with u(t0+t) in [0, T], by uniqueness of the solution of (2.16) in Y0,T. In particular, u(t1)∈H02(T) for a.e. t1 > t0. Again by uniqueness we conclude thatu∈C([t1,+∞), H02(T)) for a.e. t1 >0, so that

u∈C((0,+∞), H02(T))∩C1((0,+∞), H00(T)).

The proof of Theorem 2.1 is complete.

The following commutator lemma, used several times in the proof of the property of propa- gation of regularity, is a periodic version of a result from [10].

Lemma 2.5. Let N⊂Zbe a set such that for some constant C >0

hni+hki ≤Chn−ki, ∀n6∈N, ∀k∈N. (2.30) Let P be the projector on the closure of Span{eikx; k∈N} in L2(T), namely

P(X

k∈Z

ˆ

ukeikx) =X

k∈N

ˆ ukeikx.

Let a∈C(T) and let p∈N, q ∈N. Then there exists some constant C =C(a, p, q)>0 such that for all v∈L2(T)

k∂xp[a, P]∂qxvk ≤Ckvk. (2.31) Remark 2.6. Note that condition (2.30) is fulfilled in the following cases: (i) N = N; (ii) N is a finite set, or the complement of a finite set in Z. It follows that (2.31) is true with P =H= (−i)(PN−PN). Note, however, that condition (2.30) and (2.31) are not true when N= 1 + 2Z (pick e.g. a(x) =eix).

Proof of Lemma 2.5. Let N, a, p and q be as in the statement of the lemma, and pick any v∈C(T). Decompose aand v in using Fourier series

v(x) =X

n∈Z

ˆ

vneinx, a=X

n∈Z

ˆ aneinx,

and denote by 1N the characteristic function of N, defined by 1N(n) = 1 if n ∈ N, and 0 otherwise. Then

[a, P]v = a(P v)−P(av)

= a(X

n

1N(n)ˆvneinx)−P(X

n

(X

k

ˆ

an−kk)einx)

= X

n

X

k

ˆ

an−kk(1N(k)−1N(n))

! einx.

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Taking derivatives, one obtains

xp[a, P]∂xqv=X

n

X

k

ˆ

an−k(ik)qk(1N(k)−1N(n))

!

(in)peinx =: Σ1−Σ2

where Σ1 (resp. Σ2) is the sum over the (n, k) with n 6∈N and k∈ N (resp. with n ∈N and k 6∈N). Let us estimate Σ1 only, the estimate for Σ2 being similar. Since a∈C(T), for any s∈Nthere exists some constant Cs>0 such that

|ˆal| ≤Cshli−s ∀l∈Z. (2.32) Then, for s >sup{2p+ 1,2q+ 1},

1k2L2(T) = kX

n6∈N

(X

k∈N

ˆ

an−kˆvk(ik)q(in)p)einxk2L2(T)

= CX

n6∈N

X

k∈N

ˆ

an−kk(ik)q

2

|n|2p

≤ Ckvk2X

n6∈N

X

k∈N

hn−ki−2s|n|2p|k|2q

≤ Ckvk2X

n6∈N

X

k∈N

(hni+hki)−2s|n|2p|k|2q

≤ Ckvk2

where we used the Cauchy-Schwarz inequality, (2.32) and (2.30). SinceC(T) is dense inL2(T),

the proof is complete.

The propagation of regularity property we need is as follows.

Proposition 2.7. Let a ∈ C(T,R+), ε > 0, α ∈ R, T >0, and R > 0 be given. Pick any v0 ∈H00(T) with kv0k ≤R and let v∈C([0, T];H00(T))∩L2(0, T, H1(T))∩C((0, T], H2(T)) be such that

vt+ (H−ε)vxx+αvvx=−G(D(G v)), x∈T, t∈(0, T) (2.33)

v(0) =v0. (2.34)

Then there exists some constant C =C(T)>0 (independent of ε, α and R) such that Z T

0 kD12vk2dt≤C(R24R6). (2.35) Proof of Proposition 2.7. Pick any t0 ∈(0, T). Let (f, g)L2

t,x := RT t0

R

Tf(x, t)g(x, t)dxdt denote the scalar product in L2(t0, T, L2(T)). C will denote a constant which may vary from line to line, and which may depend on T, but not on t0, ε, α and R. Setting Lv := vt+Hvxx, f :=εvxx−G(D(Gv)) andg:=−αvvx, we have that

Lv =f+g.

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Pick any ϕ∈ C(T), and set Av =ϕ(x)v. Noticing that L is formally skew-adjoint, we have that

([L, A]v, v)L2

t,x = (L(ϕv)−ϕ(Lv), v)L2

t,x

= (ϕv, Lv)L2

t,x+ [(ϕv, v)]Tt0 −(Lv, ϕv)L2

t,x

so that

|([L, A]v, v)L2

t,x| ≤2|(f+g, ϕv)L2

t,x|+ 2kϕkL(T)R2. We first notice that

|(f, ϕv)L2

t,x| ≤ |(vx, ε(ϕv)x)L2

t,x|+|(D(Gv), G(ϕv))L2

t,x|

≤ Cε Z T

0

Z

T

(|v|2+|vx|2)dt+C Z T

0 kD12(Gv)k2dt +

Z T

0 {|(D(Gv),[G, ϕ]v)|+|(D12(Gv),[D12, ϕ](Gv))|}dτ

≤ CR2

where we used (2.18) and classical commutator estimates. (Note that Theorem 2.1 is still true when α= 1 is replaced by any value α∈R.) On the other hand

|(g, ϕv)L2

t,x|=|(αvvx, ϕv)L2

t,x|= |α|

3 |(v3, ϕx)L2

t,x|. From Sobolev embedding and the fact that the L2−norm is nonincreasing

kvkL3 ≤ kvk12kvk

1 2

L6 ≤CR12kvk

1 2 1 2· Therefore,

|(g, ϕv)L2

t,x| ≤ C|α| Z T

t0

kvk3L3dt

≤ C|α|R32T14 Z T

t0

kvk21

2

dt 34

≤ Cδ−3α4R6T+δ Z T

t0

kD12vk2dt where δ >0 will be chosen later on. On the other hand

[L, A]v = [H∂x2, ϕ]v

= H (∂x2ϕ)v+ 2(∂xϕ)(∂xv) +ϕ∂x2v

−ϕH∂x2v

= [H, ϕ]∂x2v+H (∂x2ϕ)v

+ 2[H, ∂xϕ]∂xv+ 2(∂xϕ)H∂xv. (2.36)

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It follows from Lemma 2.5 and Remark 2.6 that

|([H, ϕ]∂x2v, v)L2

t,x|+| H((∂x2ϕ)v), v

L2t,x|+| [H, ∂xϕ]∂xv, v

L2t,x|

≤Ckvk2L2(0,T;L2(T))

≤CR2. (2.37)

Therefore

|(∂xϕH∂xv, v)L2

t,x| ≤C(R2−3α4R6) +δ Z T

t0

kD12vk2dt.

Let b∈C0(ω), whereω={x∈T; a(x)>0}. Then b=a˜b with ˜b∈C0(ω) and Z T

t0

kD12(bv)k2dt ≤ 2 Z T

t0

k[D12,˜b](av)k2+k˜bD12(av)k2 dt

≤ C Z T

t0

(kvk2+kD12(av)k2)dt

≤ C Z T

0 kvk2+kD12(Gv)k2+kD12ak2| Z

T

a(y)v(y, t)dy|2 dt

≤ CR2. (2.38)

Pick anyx0∈T. Thenb2(x)−b2(x−x0) =∂xϕfor some ϕ∈C(T). Noticing thatH∂x =D, we have that

|(b2(x)H∂xv, v)L2

t,x| = |(bDv, bv)L2

t,x|

≤ |([b, D]v, bv)L2

t,x|+|(D(bv), bv)L2

t,x|

≤ Ckvk2L2(0,T;L2(T))+ Z T

t0

kD12(bv)k2dt

≤ CR2 by (2.38). It follows that

|(b2(x−x0)Dv, v)L2

t,x| ≤C(R2−3α4R6) +δ Z T

t0

kD12vk2dt.

Using a partition of unity and choosing δ >0 small enough, we infer that

|(Dv, v)L2

t,x| ≤C(R24R6) +1 2

Z T

t0

kD12vk2dt.

This gives

Z T

t0

kD12vk2dt≤C(R24R6),

where C=C(T). Letting t0→ 0 yields the result.

A unique continuation property is also required.

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Proposition 2.8. Let α∈R, ε≥0, c∈L2(0, T), and u∈L2(0, T;H00(T))be such that

ut+ (H−ε)uxx+αuux= 0 in T×(0, T), (2.39) u(x, t) =c(t) for a.e. (x, t)∈(a, b)×(0, T) (2.40) for some numbers T >0 and 0≤a < b≤2π. Then u(x, t) = 0 for a.e. (x, t)∈T×(0, T).

Proof. From (2.40), we obtain that uxx(x, t) = (uux)(x, t) = 0 for a.e. (x, t) ∈ (a, b)×(0, T).

Thus, by using (2.39),

Huxx =−ut=−ct in (a, b)×(0, T).

Therefore, for almost every t∈(0, T), it holds

uxxx(·, t)∈H−3(T), (2.41)

uxxx(·, t) = 0 in (a, b), (2.42) Huxxx(·, t) = 0 in (a, b). (2.43) Pick a time tas above, and set v=uxxx(·, t). Decompose v as

v(x) =X

k∈Z

ˆ vkeikx,

the convergence of the Fourier series being inH−3(T). Then in (a, b) 0 =iv−Hv= 2iX

k>0

ˆ vkeikx.

Since v is real-valued, we also have that ˆv−k = ˆvk for all k. The following lemma for Fourier series is needed.

Lemma 2.9. Let s∈Rand let v(x) =P

k≥0keikx be such that v∈Hs(T) and v= 0 in (a, b).

Then v≡0.

Proof of Lemma 2.9. It is clearly sufficient to prove the property fors=−p, wherep∈N. Let us proceed by induction on p. Assume first that p= 0. Then

X

k≥0

|ˆvk|2 <∞. (2.44)

Introduce the set U ={z∈C; |z|<1} and the Hardy space (see e.g. [43]) H2(U) ={f :U →C; f is holomorphic inU and lim sup

r→1

Z π

−π|f(re)|2dθ <∞}

Let f(z) =P

k≥0ˆvkzk. Then, by [43, Thm 17.10] and (2.44), we have that f ∈H2(U). On the other hand, by [43, Thm 17.10 and Thm 17.18], it holds that

f(e) := lim

r→1f(re) exists for a.e. θ∈(0,2π); (2.45) f(e) =X

k≥0

ˆ

vkeikθ =v(θ) inL2(T); (2.46) Iff 6≡0, then f(e)6= 0 for a.e. θ∈(0,2π). (2.47)

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Since

f(e) =v(θ) = 0 for a.e. θ∈(a, b),

it follows from (2.47) that f ≡0. Therefore ˆvk = 0 for all k ≥0, hence v ≡0. This gives the result for p= 0. Assume now that the result has been proved for s=−p for somep ∈N, and pick any v ∈H−p−1(T), decomposed as v(x) =P

k≥0ˆvkeikx, and such that v ≡0 in (a, b). Let w(x) =P

k>0 ˆ vk−1

ik eikx. Thenw∈H−p(T) and wx=X

k>0

ˆ

vk−1eikx =eixv, sowx= 0 on (a, b) and we have, for some constant C∈C,

w(x) =C on (a, b). (2.48)

Introducing the function ˜w(x) =w(x)−C, we infer from (2.48) and the induction hypothesis that ˜w≡0 onT, which yields v≡0 onT. This completes the proof of Lemma 2.9.

With Lemma 2.9 we infer that for a.e. t ∈ (0, T), uxxx(., t) = 0 in T, hence with (2.40) u(x, t) =c(t) a.e. inT×(0, T). From (2.39) we infer thatct = 0, which, combined with the fact that u∈L2(0, T;H00(T)), gives thatu(x, t) = 0 a.e. inT×(0, T). The proof of Proposition 2.8

is complete.

We are now in a position to state a stabilization result for theε-BO equation. We stress that the decay rate does not depend on ε.

Theorem 2.10. Let R > 0. There exist some numbers λ > 0 and C > 0 such that for any ε∈(0,1]and any u0∈H00(T) with ku0k ≤R, the solution u of (2.16) satisfies

ku(t)k ≤Ce−λtku0k ∀t≥0.

Proof. Note that ku(t)k is nonincreasing by (2.18), so that the exponential decay is ensured if ku((n+ 1)T)k ≤ κku(nT)k for some κ < 1. To prove the theorem, it is thus sufficient (with (2.18)) to establish the following observability inequality: for any T > 0 and any R > 0 there exists some constantC(T, R)>0 such that for anyε∈(0,1] and anyu0 ∈H00(T) withku0k ≤R, it holds

ku0k2 ≤C

ε Z T

0 kux(t)k2dt+ Z T

0 kD12(Gu)k2dt

, (2.49)

where u denotes the solution of (2.16). Fix anyT >0 and any R > 0, and assume that (2.49) fails. Then there exist a sequence (un0) inH00(T) and a sequence (εn) in (0,1] such that for each nwe have kun0k ≤R, and

kun0k2 > n

εn Z T

0 kunx(t)k2dt+ Z T

0 kD12(G un)k2dt

.

Let αn=kun0k ∈(0, R]. Extracting a sequence if needed, we may assume thatαn→α ∈[0, R]

and εn→ε∈[0,1]. Letvn=unn. Thenvn solves

vtn+ (H−εn)vnxxnvnvnx =−G(D(Gvn)), vn(0) =v0n (2.50)

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withv0n∈H00(T) andkvn0k= 1. Again, we have that 1

2kvn(t)k2n Z t

0 kvxnk2dτ + Z t

0 kD12(Gvn)k2dτ = 1

2kv0nk2 ∀t >0, (2.51) 1 =kvn0k2 > n

εn

Z T

0 kvnx(t)k2dt+ Z T

0 kD12(Gvn)k2dt

. (2.52)

We infer from Proposition 2.7 that Z T

0 kD12vnk2dt≤C. (2.53)

This yields

kG(D(Gvn))kL2(0,T;H12(T))+k(H−ε)vxxn kL2(0,T;H32(T))≤C.

On the other hand, for any δ >0

kvnvxnkH32δ(T) ≤Ck(vn)2kH12δ(T) ≤Ck(vn)2kL1(T)≤Ckvnk2≤C thus

nvnvnxkL2(0,T;H32−δ(T))≤C.

It follows that (vnt) is bounded inL2(0, T;H32−δ(T)). Combined with (2.53) and Aubin-Lions’

lemma, this gives that for a subsequence still denoted by (vn), we have vn→v inL2(0, T;Hα(T)) ∀α < 1

2, vn→v inL2(0, T;H12(T)) weak, vn→v inL(0, T;L2(T)) weak∗ for some functionv∈L2(0, T;H

1 2

0 (T))∩L(0, T;L2(T)). In particular, (vn)2 →v2 inL1(T×(0, T)).

Letting n→ ∞ in (2.52), we obtain that Z T

0 kD12(Gv)k2dt= 0,

hence Gv= 0 a.e. onT×(0, T). Recall that ω={x∈T; a(x)>0}. Then v(x, t) =

Z

T

a(y)v(y, t)dy =:c(t) for a.e. (x, t)∈ω×(0, T).

Note thatc∈L(0, T). Taking the limit in (2.50) gives

vt+ (H−ε)vxx+αvvx= 0, inT×(0, T),

v(x, t) =c(t) for a.e. (x, t)∈ω×(0, T).

It follows from Proposition 2.8 thatv ≡0. Thus, extracting a subsequence still denoted by (vn), we have that vn(·, t)→0 in L2(T) for a.e. t∈(0, T). Using (2.51)-(2.52), we infer that vn0 →0 inL2(T). This contradicts the fact that kvn0k= 1 for all n.

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