stabilization of water wavest. alazard (ens paris-saclay) stabilization versailles 2017 18 / 26...

Stabilization of water waves

Thomas Alazard

CNRS & École normale supérieure Paris-Saclay

Journée “Contrôle et EDP”, LMV-USVQ12 octobre 2017

T. Alazard (ENS Paris-Saclay) Stabilization Versailles 2017 1 / 26

Question: Generation and absorption of water waves in wave tanks

Control theory for the water-wave equations

There are many control results for equations describing water waves :• Benjamin-Ono, KdV, Saint-Venant;

see works by Cerpa, Crépeau, Coron, Dubois, Glass, Guerrero, Laurent,Linares, Ortega, Petit, Rosier, Rouchon, Russell, Zhang....

Here we consider the dynamics of an incompressible, irrotational liquid flow• moving under the force of gravitation and surface tension,• in a time-dependent domain with a free boundary.

Key features :• the equation is quasi-linear,• the domain has a free boundary and the problem is non local.

Non perturbative problem.


The equations

The fluid domain Ω has a free surface. At time t ≥ 0 ,

Ω(t) = (x, y) ∈ [0, L]× R : −h < y < η(t, x) ,

where η is an unknown.

y

x0 L

0

−h

Ω(t)

For simplicity: we consider only 2D water waves.

Incompressible liquid in a domain Ω with a free surface. At time t ≥ 0 ,

Ω(t) = (x, y) ∈ [0, L]× R : −h < y < η(t, x) .

The equations are

∂tv + v · ∇v +∇(P + gy) = 0 in Ω

div v = 0 in Ω

v · n = 0 on the bottom and walls

∂tη =√

1 + η2x v · n on the free surface

P − Pext = κ∂x

(ηx√

1 + η2x

)on the free surface,

g gravity, P pressure, Pext external pressure, κ surface tension.

We assume that κ = 1 .

Reduction to the case

Ω(t) = (x, y) ∈ T× R : −h < y < η(t, x) .

Periodization


Moreover we assume that curlx,y v = 0 so that v = ∇x,yφ where

∆x,yφ = 0 in Ω(t), φy = 0 on y = −h.

Following Zakharov, set

ψ(t, x) = φ(t, x, η(t, x)),

and introduce the Dirichlet to Neumann operator G(η) defined by

G(η)ψ =√

1 + |∇xη|2 ∂nφy=η

=(φy − ηxφx)

y=η

.

The problem is then given by two nonlinear equations on the free surface:∂tη = G(η)ψ,

∂tψ + gη − κH(η) +1

2(ψx)2 − 1

2

(ηxψx +G(η)ψ

)21 + η2

x

= 0.

This system is quasi-linear and non local.


Moreover we assume that curlx,y v = 0 so that v = ∇x,yφ where

∆x,yφ = 0 in Ω(t), φy = 0 on y = −h.

Following Zakharov, set

ψ(t, x) = φ(t, x, η(t, x)),

and introduce the Dirichlet to Neumann operator G(η) defined by

G(η)ψ =√

1 + |∇xη|2 ∂nφy=η

=(φy − ηxφx)

y=η

.

The problem is then given by two nonlinear equations on the free surface:∂tη = G(η)ψ,

∂tψ + gη − κH(η) +1

2(ψx)2 − 1

2

(ηxψx +G(η)ψ

)21 + η2

x

= 0.

This system is quasi-linear and non local.T. Alazard (ENS Paris-Saclay) Stabilization Versailles 2017 7 / 26

Classical problem

Pionneering works:

Zakharov, Nalimov, Yoshihara, Craig, Wu, Beyer–Günther

To quote only results concerning gravity-capillary waves :

Ambrose–Masmoudi, Shatah–Zeng, Coutand–Shkoller,Córdoba–Córdoba–Gancedo, Masmoudi–Rousset, Lannes, Pusateri,Christianson–Hur–Staffilani, A.– Burq–Zuily, Germain–Masmoudi–Shatah,Castro–Córdoba–Fefferman–Gancedo–Gómez Serrano,Fefferman–Ionescu–Lie, Coutand –Shkoller, Ifrim–Tataru, Ionescu–Pusateri,de Poyferré–Nguyen, Deng–Ionescu–Pausader–Pusateri, Berti-Delort...


Generation of water waves

Pext

Ω(0)

y

x

Ω(T )

y

xTheorem (A, Baldi and Han-Kwan). Given- a time T > 0 ,- an initial state (ηin, ψin) and a final state (ηfinal, vfinal) , small enoughin Hs+1/2(T)×Hs(T) with s large enough,- a domain ω = (a, b) ,there is Pext(t, x) supported in [0, T ]× ω such that the unique solutionwith initial data (η, ψ) = (ηin, ψin) satisfies (η, ψ)|t=T = (ηfinal, ψfinal) .



Pext

Ω(0)

y

x

Ω(T )

y

x

Theorem (A, Baldi and Han-Kwan). Given- a time T > 0 ,- an initial state (ηin, ψin) and a final state (ηfinal, vfinal) , small enoughin Hs+1/2(T)×Hs(T) with s large enough,- a domain ω = (a, b) ,there is Pext(t, x) supported in [0, T ]× ω such that the unique solutionwith initial data (η, ψ) = (ηin, ψin) satisfies (η, ψ)|t=T = (ηfinal, ψfinal) .



Pext

Ω(0)

y

x

Ω(T )

y

xTheorem (A, Baldi and Han-Kwan). Given- a time T > 0 ,- an initial state (ηin, ψin) and a final state (ηfinal, vfinal) , small enoughin Hs+1/2(T)×Hs(T) with s large enough,- a domain ω = (a, b) ,there is Pext(t, x) supported in [0, T ]× ω such that the unique solutionwith initial data (η, ψ) = (ηin, ψin) satisfies (η, ψ)|t=T = (ηfinal, ψfinal) .


After wave generation, wave absorption is the most important mechanismin a wave tank.

Many problems require to study the propagation in unbounded domains(open sea). On the other hand, the numerical or experimental analysis ofthe water-wave equations requires to work in a bounded domain. Toovercome this problem, a classical method consists in damping outgoingwaves in an absorbing zone


The most popular experimental wave absorbers are passive absorbers. Theyconsist of a beach with a mild slope. The principle is that, when arriving tothe artificial beach, the steepening of the forward face of waves and theirsubsequent overturning dissipates energy.T. Alazard (ENS Paris-Saclay) Stabilization Versailles 2017 10 / 26

StabilizationClassical numerical absorbers in numerical wave tanks: sponge boundarylayer surrounding the computational boundary where one absorbs (ordamps) outgoing waves. Related to the stabilization problem.

Pext

Ω(t)

y

xL− δ L0

0

−h

Figure: Absorption of water waves in the neighborhood of x = L by means of anexternal counteracting pressure produced by blowing above the free surface.

Denote by H(t) the energy of the fluid at time t :

H(t) =g

2

∫ L

0η2 dx+ κ

∫ L

0

(√1 + η2

x − 1)

dx︸︷︷︸potential energy

+1

2

∫∫Ω(t)|∇x,yφ|2 dy dx︸︷︷︸

kinetic energy

.

Goal : find Pext such that(i) one has supp ∂xPext(t, ·) ⊂ [L− δ, L] ;(ii) H decays exponentially to 0 .

But WW might blow up in finite time.... Here exponential decay means thefollowing: find Pext such that

H(T ) ≤ C

TH(0).

H(T ) ≤ C

TH(0)︸︷︷︸

damping for T > C

⇒ H(nT ) ≤(C

T

)nH(0)︸︷︷︸

exponential decay

.


First question: how to force the energy to decay ?

Hamiltonian damping. Zakharov observed that the equations have thehamiltonian form

∂η

∂t=δHδψ

,∂ψ

∂t= −δH

δη− Pext.

ThendHdt

=

∫ [δHδη

∂η

∂t+δHδψ

∂ψ

∂t

]dx = −

∫∂η

∂tPext dx.

If Pext = χ∂tη with χ ≥ 0 and F arbitrary, thendHdt≤ 0.

This choice is widespread : Cao–Beck–Schultz, Clément, Grilli, Bonnefoy,Ducrozet, Baker–Meiron–Orszag,...Other choices exist : Clamond–Fructus–Grue–Kristiansen.


Numerical simulation for the linearized equation, with Emmanuel Dormy.


TheoremAssume that

Pext(t, x) = χ(x)∂tη (with χ(x) well chosen).

i) There exist two positive constants δ, C , depending explicitely ong, κ, L , such that, if

‖η‖W 2,∞ ≤ δ,

and if the solution exists on the time interval [0, T ] , then

H(T ) ≤ C

TH(0).

ii) There exists a constant c∗ such that, if

‖η0‖H7/2 + ‖ψ0‖H3 ≤ ε,

then the solution exists and is O(ε) on a time interval of size c∗/ε .

Choices for the cut-off function

x

χ1(x)

L−L

1

x

χ2(x)

ε

L−L


Cauchy problem

Linearized equation (neglect gravity):

u = ψ − i|Dx|12 η

satisfies the dispersive equation

∂tu+ i |Dx|32 u = Pext.

Similar diagonalization of the nonlinear equations, based on

• study in Eulerian coordinates (Zakharov, Craig-Sulem, Lannes)• complete paralinearization of the equations (A-Métivier)• symmetrization (A-Burq-Zuily)• normal forms (A-Delort; A-Baldi)

(Oversymplifying) One can rewrite the WW system as:

∂u

∂t+ V (u)∂xu+ i |Dx|

34(c(u) |Dx|

34 u)

= Pext

where V, c are real-valued functions.T. Alazard (ENS Paris-Saclay) Stabilization Versailles 2017 17 / 26

Cauchy problem

One can think of the damped WW system as:

∂u

∂t+ V (u)∂xu+ i |Dx|

32 u+ χ(x)u = 0.

To prove uniform estimates in time O(ε−1) is equivalent to prove uniformestimates in time O(1) for

∂u

∂t+ V (u)∂xu+

i

ε|Dx|

32 u+

χ(x)

εu = 0.

We do not have uniform H1 estimate for

∂v

∂t+χ(x)

εv = 0.

As Mesognon-Gireau, we exploit an ellipticity property and proceed in twosteps:• we prove L2 -estimates for Lku with L = i |Dx|3/2 + χ(x) ,• we deduce Sobolev estimates using the equation.


Damping (decreasing energy) is easy but stabilization (exp decay to 0 ) ismore difficult (since the equation is quasi-linear and nonlocal).

Our goal is to find Pext such that (i) suppPext(t, ·) ⊂ [L− δ, L] and

(ii) H is decreasing and (iii)

∫ T

0H(t) dt ≤ CH(0).

Then

H(T ) ≤ 1

T

∫ T

0H(t) dt ≤ C

TH(0)︸︷︷︸

damping for T > C

.

First problem: compute ∫ T

0H(t) dt.


Lemma (An exact identity)Consider m ∈ C∞([0, L]) such that m(0) = m(L) = 0 . Set

ζ = ∂x(mη)− 1

4η +

1−mx

2η, ρ = (m− x)ηx +

(5

4+mx

2

)η.

Then, for smooth enough solutions of the gravity water-wave equations, definedfor t ∈ [0, T ] ,

1

2

∫ T

0

H(t) dt+Q =

∫∫Pext ζ dxdt−

∫ζψ dx

T0

+

∫∫ (1−mx

2ψ + (x−m)ψx

)∂tη dx dt

+

∫∫∫ρx φx φy dy dx dt,

where

Q =

∫ T

0

∫ L

0

(h

2+ρ

2

)φ2x(t, x,−h) dx dt+

L

2

∫ T

0

∫ η(t,L)

−hφ2y(t, L, y) dy dt.

If η and ∂xη are small enough then Q ≥ 0 and one can absorb∫∫∫ρx φx φy dy dx dt in the energy.

Then

1

4

∫ T

0H(t) dt ≤

∫∫Pext ζ dx dt−

∫ζψ dx

T0

+

∫∫ (1−mx

2ψ + (x−m)∂xψ

)G(η)ψ dx dt.

Similar inequality with surface tension (but more difficult).

Proof of the identity.First tool: multiplier method seen as an approximate conservation law.

Recall that there is invariance by spatial translation :

d

dsH(ηs, ψs)

s=0

= 0 with

ηs(t, x) = η(t, x+ s)

ψs(t, x) = ψ(t, x+ s).

Here we compute

d

dsH(ηs, ψs)

s=0

with

ηs(t, x) = η(t, x+ sm(x))

ψs(t, x) = ψ(t, x+ sm(x)).

We have

d

dsH(ηs, ψs)

s=0

=

∫ T

0

∫ [(∂tη)(m∂xψ)− (∂tψ)(m∂xη)

]dx dt.

We compute this integral in two different ways : integration by parts andusing the equations.

Second tool: a Pohozaev identity.

Next, we split∫(G(η)ψ)m∂xψ dx =

∫(G(η)ψ)x∂xψ dx+

∫(G(η)ψ)(m− x)∂xψ dx.

We have a Pohozaev identity for the Dirichlet to Neumann operator:∫(G(η)ψ)(x∂xψ) dx = Σ +

∫(η − xηx)

(φ2x − φ2

y + 2φx φyηx)

y=ηdx

where Σ = Σ(t) is a positive term given by

Σ(t) =h

2

∫φ2x(t, x,−h) dx+

L

2

∫ η(t,L)

−hφ2y(t, L, y) dy.

Third tool: an identity.

All the nonlinear terms are of the form:∫ρ(φ2x − φ2

y + 2φx φyηx)

y=ηdx.

The energy controls∫∫|∇x,yφ|2 dy dx , not

∫|∇x,yφ|2

y=η

dx .

We use the identity:∫ρ(φ2x − φ2

y + 2φx φyηx)

y=ηdx = −

∫∫ρx φx φy dy dx

+1

2

∫ρφ2

x|y=−h dx,

which relies on

∂y(ρφ2

y − ρφ2x

)+ 2∂x

(ρφxφy

)= 2ρxφxφy.


Recall that

i)1

4

∫ T

0H(t) dt+Q ≤

∫∫Pext ζ dx dt−

∫ζψ dx

T0

+

∫∫ (1−mx

2ψ + (x−m)∂xψ

)G(η)ψ dx dt,

Q ≥∫ T

0

∫ L

0

(h

2+ρ

2

)φ2x(t, x,−h) dx dt,

ii) Pext(t, x) = χ(x)∂tη,

iii)d

dtH(t) = −

∫ L

0Pext∂tη dx.

Since ∂tη = G(η)ψ , one deduces the inequality∫ T

0

∫ L

0Pext(t, x)2 dx dt ≤ ‖χ‖L∞

∫ T

0

∫ L

0χ(∂tη)2 dx dt

≤ −‖χ‖L∞

∫ T

0

d

dtH(t) dt ≤ ‖χ‖L∞ H(0).


Estimate the contribution of∫∫(x−m)ψxG(η)ψ dx dt.

One has|x−m(x)| ≤ Lχ(x) = L(1−mx(x)),

and ∫ T

0

∫ L

−Lχ(G(η)ψ)2 dx dt ≤ ‖χ‖L∞ H(0).

There exists a constant A , depending only on ‖ηx‖L∞x, such that∫ L

−Lχψ2

x dx ≤ A∫ L

−Lχ(G(η)ψ)2 dx+A

∫ L

−Lφ2x|y=−h dx

− 2A

∫ L

−L

∫ η(t,x)

−hχxφxφy dy dx.

(1)

We obtain∫ T

0 H(t) dt ≤ CH(0) .The special choice for χ is used only to control the mean value of ψ .

Thank you!T. Alazard (ENS Paris-Saclay) Stabilization Versailles 2017 26 / 26

stabilization of water wavest. alazard (ens paris-saclay) stabilization versailles 2017 18 / 26...

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