Control and stabilization of the incompressible Euler equation with free surface
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00:00
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Transkript: English(automatisch erzeugt)
00:16
very much for the introduction, thanks also very much for the invitation, I'm very happy to speak here. I apologize for, because many of you have already seen a
00:25
part of this talk about observability and control, the new point maybe is about stabilization of water wave. So the problem I am considering in the following is the generation and absorption of water wave, and here we are going to consider the problem in a rectangular tank, which means that we
00:44
are considering a bounded container with a fluid inside, and the container has vertical walls, which are vertical, the container has a bottom which is flat, and also you have right angle everywhere, ok, you have right angle everywhere. So there are two pictures, one with a 3D fluid and one with a 2D fluid.
01:12
For the sake of simplicity, we are considering only 2D waves, ok, we are going to consider a 2D wave, but in principle, the result should apply also for a three-dimensional fluid.
01:28
I am going to present results in collaboration with Pietro Baldi and Daniel Ancua, about the controllability of 2D gravity capillary wave, and also results about observability and stabilization, which apply for either 2D or 3D wave, with or without surface tension.
01:51
So the problem of controllability and stabilization of water wave has been studied a lot, for equation describing water wave in sub-asymptotic regime, like the Benjamino equation, or the KdV equation, or the Sambeno equation. And there are many
02:07
many people that have worked on this topic, and to name a few I can quote Jean-Michel Coron, Camille Laurent, Felipe Linares, James Mortega, and also Petit, Rosier and Roussel. Here the difference with this work is that we are considering the full model, that is we are considering the dynamics of an incompressible liquid
02:28
in a free domain omega, with a free surface. This is a key point, we are considering a free domain which is time dependent. And to fix notation, we assume that at time t, at a given time t, the free domain is of the following form.
02:46
This is the set of points x and y in 0, L times R, such that y is bigger than minus h and smaller than eta of t and x. And the fact that the domain has a free surface means that eta is an unknown of the problem.
03:07
We are going to consider a free domain, which depends on time omega of t in a bounded container, and the free surface is the curve with the equation y equals eta of t x.
03:24
So the physical parameters here are the depth, which is h, and the size of the domain, which is L. This is the free domain. Inside the free domain, the free motion is determined by its Eulerian velocity, which is vector v from omega to R2,
03:48
and the equations by which the motion is to be determined are well known for two centuries. These are the incompressible Euler equation dt v plus v gradient v plus the gradient of p plus gy is equal to 0.
04:06
Capital P is the free pressure, g is the acceleration of gravity, so this is the usual Euler equation. This is the incompressibility constraint, divergence of v is equal to 0 inside the free domain.
04:23
So inside the free domain, the equations are well known, and we have also boundary conditions on the walls, on the bottom, and on the free surface. These boundary conditions are the following. First, we assume that v dot n is equal to 0 both on the bottom and on the walls.
04:46
The usual solid wall boundary condition. Then the problem is determined by two boundary conditions on the free surface. Firstly, we assume that on the free surface, that is here, we no longer have v dot n equal to 0,
05:05
we have instead dt of eta is equal to the square root of 1 plus the derivative of eta, eta x squared times v dot n. And secondly, this is a kinematic equation that describes the motion of the free surface.
05:22
And the second equation is the balance of forces across the free surface. This is an equation for the difference between the pressure inside the fluid and the external pressure, the pressure in the air. We assume that the jump of pressure at the free surface is proportional to the curvature.
05:45
That is, p minus px is equal to minus kappa times dx of eta x divided by square root of 1 plus eta x squared. So the physical parameters are g and kappa, the surface tension coefficient, p is the pressure, and px is the external pressure.
06:04
In this talk, we consider the case with surface tension, which means that we assume that kappa is equal to 1. And another assumption that we make is that v is irrotational, the curl of v is zero.
06:35
The important notations are eta is the free surface elevation, v is the Eulerian fluid velocity,
06:48
we assume that curl of v is equal to zero. And hence, we can write v as a gradient of some function phi, which is the velocity potential. This is a scalar function.
07:05
Okay, now I am going to describe the problem of controllability. The question is very easy to explain. What happens with boundary conditions at the point of triple junction between the solid and the fluid?
07:23
This is not a problem here. Because as I will explain later, we can extend this problem to a problem defined on the wall line. So it is hitting it vertically, for example? Yes. So there is always a right angle? There is always a right angle. The right angle condition is propagated by the equation.
07:45
But I have to say glad has a point, which is... I don't understand. Because glad has a point, because every interface has a free energy. An interface between water and wall does, water and air does, air and wall does, and is generally a meniscus which applies pressure, which applies a force to...
08:04
So saying that it is normal, is assuming that there is no wall? Yes, I am making this assumption. I am making this assumption and this assumption is propagated in time. So the problem is the following.
08:20
You have a container with a fluid at rest inside, okay? And you want, by some means, to generate a wave. Starting from this fluid, you want to generate a wave. That is, you want to generate a free surface, you want to generate a fluid motion inside, and you want to be able to do that in some time capital T.
08:44
To generate a wave, you have of course many ways to act on a fluid, on a container. Here we are going to consider what is possibly the simplest way, according to the mathematical analysis, which is to blow above the free surface. Of course, if you blow above the free surface, you are going to generate a wave.
09:03
And the question is, is it possible to blow above a localized or confined portion of the free surface and generate a wave? More precisely, the question is the following. The mathematical question is the following. Given some time T, that is given in advance,
09:22
given a finite state, eta final, v final, in some space X, given an interval AB, the domain omega, which is equal to the interval AB, find the pressure, the value of the external pressure at the free surface, which is a function of T and X, which is supported in zero T times omega,
09:43
such that the unique solution of the water wave equation with initial data zero, is such that the solution coincides at time T, capital T, with the finite state that is given in advance. This is a controllability question.
10:02
And then there is an observation that in this case, one can see that the assumption that the fluid is irrotational is meaningful, because you start from something that is irrotational, and you have a conservative force, so the flow will be irrotational at any time.
10:21
So this is the question. Now, the first thing to do is to reduce ourselves to the case where there is no lateral walls, and this is done in a standard way. We reduce ourselves to the case where omega of T is in fact the state of point X and Y in R times R, so that Y is larger than minus H and smaller than eta.
10:42
And to do that, we use a classical periodization and symmetrization process. We first symmetrize the fluid domain, and then we extend it by periodicity. By soldering, we are reduced with the fluid, which is period. This raises, of course, a lot of questions about regularity,
11:02
because when you symmetrize, you could introduce singularities. There are some subtle questions about regularity and the analysis of the Cauchy problem at a low regularity level. I am not going to discuss this question. This was studied with Nicolas Bjork and Claude Sully for the case without surface tension,
11:21
and more recently by Thibaut Despoifers for the case with surface tension. So now I can state the first result, which is about controllability, and it is a local result, controllability of two-dimensional gravity capillary waterway. For that, I just need to introduce one more notation.
11:43
The observation is the following, as you have already seen in the talk. Since divergence of v is equal to zero, phi is an harmonic function. However, you know that the Neumann derivative is zero on the wall and the bottom
12:04
because of the solid wall boundary condition. So phi is fully determined by its traits on the free surface. And this is why it is convenient to introduce a function psi,
12:21
which is the evaluation of phi on the free surface. The nice point is that this is a function of t and x, whereas this function is defined on a domain with free surface. Whereas this function is a function of time and x, which is in the terms.
12:41
The other notation is that h mu is the usual subspace of order mu, and the index n will refer to the fact that the function with mean value is zero. So a statement of the main result with Pietro Baldi and Daniel Antoine is the following. Given any time t, it can be arbitrarily small, and given any non-empty interval, including in 0, L,
13:04
there exists an index L, S, large enough, and a number m0, small enough, such that for any initial data and any final data in this space, Hs plus one half m times Hs, such that the norm initially is smaller than m0,
13:24
and the norm at the final point is smaller than m0, you can find an external pressure which is continuous in time with value in Hs, supported in zero t times omega,
13:41
such that the Cauchy problem with data at time 0, given by the initial data, has a unique solution which is continuous in time with value in Hs plus one half times Hs, which coincides at time t with the final state that is given in advance. Okay, so this is a local controllability result.
14:02
Controllability means that we are able to steer this state to this one in a finite time. Local means that we are working in a ball around zero. It's not a global result, it's a local result. We are working in a neighborhood of zero.
14:22
And one important point is that the result is true in any time t. For controllability, it is harder to work in small times than in large time, of course, because you want to reach a state in a small time, it is harder than to reach a state in a longer time. What are your estimates on p?
14:40
Hm? What are the estimates? Exactly. And zero depends on t. When t is very small, m0. But how does px blow up as t goes to zero? I don't understand. What is the estimate on p, the pressure? Ah, just p here. Yes. Everything blows up when t goes to zero.
15:05
This can be computed by the proof. This is an optimization. For linear problem, it's exponential. Okay. So not better. Yes, but only should be worse. Does s actually depend on omega? I imagine if you rely on...
15:22
Yes, the constant depends on omega. Little s. Does little s depend on omega? No. No, no, s is universal. In our proof, it is something like 15, but... Does p have a sign? p have a sign? No. That could be negative.
15:41
You have to suck. To not just blow. Yes, we push and remove. Yes, that's a good question to try to find a pressure which is always positive.
16:01
Okay. So I am going to say only a few words about the proof. The proof uses in a crucial way that we know a lot of things about the study of the Cauchy problem. We know a lot of things because there have been many, many studies which have been already described in previous talks, so I just skip this part.
16:26
The key point is that this is a dispersive equation and we are using in a crucial way the fact that the equation is dispersive. This can be most easily seen at the level of the linearized equation. If you neglect gravity to simplify the notation,
16:42
if you neglect the nonlinear term to consider a linearized equation, and if instead of working with psi and eta, which are real value functions, you decide to work with this complex value function u, which is psi minus i dx to the one half eta, then you find that u satisfies this Schrodinger type equation,
17:03
dt u plus i dx to the three half u is equal to the external pressure. So to control this equation is extremely easy because you know explicitly the solution using the Fourier transform.
17:22
So the starting point to study the controllability of the nonlinear equation is to use a similar diagonalization of the nonlinear equation. And this can be done. And this is based in the study in our proof. This is based on the study in Eulerian coordinates, which goes back to the work by Zakharov, Krek-Suleman and Lam,
17:41
and the fact that you can completely parallelize the equation, and the fact that once you have completely parallelized the equation, you can symmetrize them, and you can furthermore consider various normal forms. In the end, we can manipulate the equation quite easily. And if you allow me to oversimplify,
18:00
the result is that you can rewrite the water wave system as the following equation. dt u plus v of u dx u plus i dx to the three over four acting on c of u times dx to the three over four u is equal to the external pressure, where v and c are real value functions.
18:21
So this is an equation similar to this one. So, as I said, the linearized system at the origin has constant coefficients and can be controlled by mean of Fourier analysis or many other methods. This is not enough for our problem since the problem is quasi-linear.
18:40
Since the problem is quasi-linear, it's not enough to know what happened for the linearized system at the origin. You cannot apply an implicit function theorem working only at the origin. You have to apply a scheme that is based on the analysis of the linearized equation in the neighborhood of the origin. So necessarily, you have to use a scheme,
19:02
and this scheme has to be based on the analysis of an equation with variable coefficients. And indeed, we seek the external pressure as a limit of solution to approximate control problems with variable coefficients. We use a quasi-linear scheme. So in some sense, the main point is to study a linear equation with variable coefficients which are fixed.
19:23
Let us denote it. You consider a reference state u bar, and you consider the operator p, which is dt plus v dx plus i, this term, where v and c are the previous functions evaluated at u bar. And we want to control this equation.
19:44
To do so, we are going to transform the operator p. In several steps. The first step is to flatten the coefficient of the leading order term. And to do so, we use a change of variable of that form. We use a change of variable in x,
20:03
replacing the function h, let's say, by h of tx and plus kappa of tx. And it is convenient to multiply here by a pre-factor, which is such that the L2 norm in x is preserved.
20:20
By so doing, you can find that p is conjugated to this operator, q, which is dt plus some coefficient w dx plus i dx to the 3r plus r, where r is of order 0. You can further assume by some simple transformation that the mean value of w is 0.
20:43
It is simple to conjugate this operator to this one, but it is not entirely trivial because the equation is non-local. You have a change of variable. And also because you see here, you have the cancellation of the sub-principal term of order one-half. You go from an equation of order three-half to an equation of order three-half.
21:03
We should have a sub-principal term of order one-half that in fact disappears simply because we choose in an appropriate way a pre-factor. So this is the first step. The second step in the analysis is to conjugate this equation to the same
21:23
without the term w dx. To do so, we seek an operator A such that the commutator of A with this term i dx to the three-half plus w dx A is a zero-order operator.
21:44
We want to eliminate this term by means of a commutator. If you seek A as a pseudo-differential operator, you can do that. This is something we did with Pretro-Baldi to study the stunning wave problem, the small devices problem.
22:01
And we found an operator of this form. A is a pseudo-differential operator whose symbol is of the form some amplitude q of tx and xi times an exponential of i beta of tx xi to the one-half where the function beta in the phase is the sum of a function depending only on time
22:26
and some periodic primitive of w. It is possible to find a periodic primitive because w has mean value zero. Then with this operator, you can conjugate this evolution equation to this one
22:44
with constant coefficient plus a reminder term of order zero. Maybe one has to take care because here A is a pseudo-differential operator but it is a pseudo-differential operator in this exotic class
23:00
A zero rho rho with rho is equal to one-half and this reflects the fact that the equation is quasi-linear. Because if you consider instead a semi-linear equation, here you put a coefficient two or you consider the Binge-Amino equation, you find a similar conjugation, for example, with a pseudo-differential operator
23:20
in the standard Amanda class as one zero. So this is how we proceed the scheme of the proof. There are so many things to do, of course, but just to explain to you that we reduce ourselves to a constant coefficient case.
23:42
So now we know the phase of oscillation in time of the solution and these phases are the ones given by integrating this evolution equation and taking into account the fact that in the conjugation process, we have an exponential factor that modifies the phase.
24:06
So in the end, we are reduced to the following inequality from harmonic analysis, which is called Ingham inequality. What is an Ingham inequality?
24:21
This is a Planchet type inequality which applies for pseudo-periodic function. Why? Planchet is for periodic function. Ingham is for pseudo-periodic function. And here, the variable x is seen as a parameter. Okay, and we drop it. What is important is the dependence in time.
24:41
So the inequality that we are using is the following one. For some given real value function beta, which is a smooth function C3, you introduce these phases of oscillation. Mu n of t, which is the sine of n, n is an integer, times n to the three half times t plus beta of t times n to the one half.
25:06
These are exactly the phase of oscillation of this term, n to the three half. And here, you introduce a phase beta times n to the one half. And then we can prove the following result.
25:20
For any time t, strictly positive, there are constants C of t and delta of t, such that if beta is small enough in l infinity norm, smaller than delta, then for any l2 sequence wl, the integral in time of the l2 norm of this function
25:43
controls the l2 norm of w. This is like Blanchard inequality except that this function is not periodic. Okay, it's not periodic.
26:02
Does C grow like t? No, you can compute explicitly everything. When t is larger than some constant, depending on the... If you consider only the high frequency case, you have a very good control of the constant.
26:24
If you consider a sequence where n is larger than some number, you have a very good control of the constant. When you consider t very small, you have exponential factor also. There is a low frequency problem here.
26:42
So this is income type inequality. When you don't have beta, this inequality goes back to work of ingram when t is large enough and can balance Lemrod and Arrow for any time t. Here we are forced to consider the case where beta is non-zero.
27:04
The fact this introduces a sub-principle problem, because here you have difference of order one, but it is not a perturbative term, simply because exponential half of i n to the one half t minus one is not small.
27:25
This is not a perturbative problem, but we can follow the proof given by Arrow in the case beta equals zero and adjust it to consider this case. Once you have this ingram type inequality where x is seen as a parameter, you apply it for every x and you integrate in x
27:43
and you obtain an observability inequality. And I am going to state this observability inequality. So you consider a domain omega, which is a, b, including in t, and a given time t, positive. You assume that v solves this evolution equation without source term
28:04
and an initial data, v naught. You assume that your coefficient v and the difference between c and one is small enough. Then you have this observability inequality. The integral in time of the integral over omega of v of t x squared dx dt
28:25
is bigger than the energy, which is the integral over the torus of v of zero x squared dx. This is an observability inequality. By looking at the solution only above omega,
28:44
you can have information about the energy everywhere, the energy of the solution. This is an observability inequality. And this is indeed a key point. To conclude from observability to controllability is based on a usual argument that is called the Hilbert uniqueness method,
29:02
which is a duality argument going back to the work of Jacques Williams. The idea is very simple. It is based on Riesz Representation Theorem. This inequality tells you that you have a cohesive bilinear form. So you can apply Riesz Representation Theorem
29:21
that tells you that something exists. And the thing that exists is related to the fact that your control exists. And since you have a scalar product which is localized, the things that exist will also be localized. So you have the existence of a control that is localized.
29:41
This is only for the linearized equation at some state, a different state, in L2, then you have to prove a similar result in Sobolev spaces and you have to prove some stability estimates to prove that the scheme is converging. This is quite technical and I don't want to enter into the details of this part. What I would like to do is to explain something
30:03
instead about stabilization, which will be also in the end related to another way of thinking, observability. What I want to discuss now is unless you have questions on this part. So isn't the right way to think that basically the observability is this inequality?
30:23
So that you can say for any system, there's nothing to do with the... It's basically any system that's the inequality you're looking to prove. Yes, for wave equation, for Schrodinger equation, for all equation observability inequalities and inequality of that type. You look at the solution, it's a quantitative unique continuation result. If the solution vanishes above 0T times omega,
30:43
then the solution is zero everywhere. So controllability has to do with wave generation, which is of course very important in a wave time, but there is also in a wave time another thing which is very important, which is wave absorption.
31:01
The reason is very simple. The reason is that many problems in water wave theory require to study the behavior of wave propagating in unbounded domain, like the wave you can see on the open sea. And then for obvious reasons, the numerical or experimental analysis of the water wave equation requires to work in a bounded domain.
31:22
Of course. So you have to do something to simulate working in bounded domain. Numerically, there are two approaches that can be used. Either you can truncate the domain by introducing an artificial boundary and you seek for artificial boundary conditions.
31:40
This goes back to work by Inquist and Meider. Or you can also try to dump outgoing wave in an absorbing zone surrounding the computational boundary. You add a layer surrounding the computational domain where you dump outgoing wave. This is for the numerical analysis.
32:01
I am going to present something which has to do with the experimental study of water wave. I will show you a movie. This movie was done at the École Centrale de Nantes. I visited Felicien Beuford and Guillaume du Crozet, who kindly showed me a wave tank
32:21
and kindly allowed me to show this movie. So in this movie you are going to see a wave tank. This is a swimming pool. A very giant swimming pool, 50 meters long. Here you have something like 25 meters. The depth of the basin is 5 meters.
32:41
And you have two sides of the wave tank. On one side you generate wave. On the left hand side you generate wave. And on the right hand side you are going to generate wave like the swell on the open sea. And here you are going to dump the wave. Because otherwise the waves are going from the left to the right.
33:01
And then they bounce back in the basin. And this is chaos. So here you generate, here you dump. So this is a picture of the swimming pool.
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This is a picture of the swimming pool. This is the left side of the tank where this paddle will be used to generate the wave. And on the right side of the tank there is an observing device which is extremely simple. This is a passive device which you introduce an artificial beach.
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Like a kind of beach you have on the coast. Waves are coming to this beach and are going to blow up of course like on the beach. And blow up dissipates energy. I am going to show you a picture, a schematic picture of the beach.
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Here it is a, ok, the plan. Here it is a picture of the beach. From above it is here, ok. It does not touch the bottom of the wave tank. Here you can see wave generated in the wave tank.
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They travel from the left to the right. These are island on linear wave. They propagate. And ultimately, ok, they dissipate all their energy when they reach the beach.
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You see that nothing is going back in the swimming pool. This is for the experimental study of water wave. Of course, this is not something you are going to do
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to stabilize mathematically or numerically the water wave equation because we don't understand blow up, of course. If you want to stabilize the equation mathematically or numerically, there is something much simpler to do, which is to dissipate energy in a surrounding layer
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by using again the idea of blowing above the free surface. Here there is a picture, the wave generating in a wave tank as above, periodic wave traveling to the right. And then we want to find a way to blow above the free surface such that we dissipate the energy.
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We want to absorb water wave in the neighborhood of x equals L by means of an external counteracting pressure. The problem is the following. Let us denote by h the energy of the fluid at time t, which is the sum of two terms, the potential energy and the kinetic energy.
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Potential energy is the L2 norm of eta plus the integral on the free domain of the square root of 1 plus eta x square. And the kinetic energy is one half of the L2 norm of the velocity. And the goal is to find a feedback law for the pressure Px
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such that Px will be included in, compactly supported in L minus delta L. Px will be located here. The energy is decreasing and the integral in time of the energy will be controlled by a constant time h0.
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And what I have forgotten to write is that we want a feedback law, such as we want that Px is defined as a function of eta psi. Okay, it's a passive wave. It is a function of eta psi by contrast with the control problem.
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So why do you, we want to make the energy decrease of course because we want to dump energy. This condition, what is the reason of this condition? Simply because if this is true and if h is decreasing, h of t is smaller than the mean of h
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that is smaller than 1 divided by t of the integral of h of t which is, because of this, smaller than capital C divided by capital T, h of 0. Which means that you have a dumping whenever capital T is bigger than C. Of course, h of t will be smaller than some fraction time h of 0.
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Which means that in fact you have exponential decay simply because you can iterate this h at time n times capital T will be smaller than capital C divided by t to the n times the initial energy. So whenever you have this property
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and whenever the fluid exists on long enough time interval you have exponential decay of the energy. Which is stabilization. So now the question is to find a feedback law that allows to obtain this property. So the first question is to find a feedback law
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that allows to dissipate energy and this can be done straightforwardly using the Hamiltonian formulation. We are going to find an Hamiltonian dumping. Based on the fact that the equations are Hamiltonian they can be written under this form dT of eta is equal to the derivative of the energy with respect to psi
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dT of psi is equal to minus the derivative of the energy with respect to eta minus the external pressure. Then the derivative of the energy can be written in this form of course you use the cancellation coming from the fact that the equation is Hamiltonian and it remain only this part
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that is minus the integral of dT eta times the external pressure. So you want that this is negative to dissipate energy so simply you choose for example px as a cut of function chi multiplying dT of eta then of course this term will be positive
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and dT of eta will be negative. And this explains that this feedback law this feedback law is widespread in the numerical analysis of water wave
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whenever you want to dump energy to dump outgoing wave in a surrounding layer. And the question is to understand why this feedback law which is localized allows to prove a global decay of the energy.
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I am going to show you a numerical experiment done with Emmanuel Dormi it is extremely easy because it is an experiment for linearized equation just to test it. It is extremely easy for linearized equation because you can solve the equation using Fourier transform and you are going to see that it works very well.
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Here we are generating wave by blowing here above the free surface on the left and we generate a periodic wave train that is travelling to the right and on the right we are using this damping and it works very well. We are able to produce a travelling wave propagating from the left to the right and never bouncing back
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or at least not so much reflection. Ok, so it works very well, at least numerically. The next statement says that it works also this theorem explains in some sense the previous stabilization result. Statement is the following
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you assume that as above, the external pressure is equal to chi of x times dt eta minus some time dependent function just to ensure that px has mean value zero where chi is some well chosen coefficient. Then you can prove two things Firstly, there exists two positive constants
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delta and c depending only on the physical parameters which are g, gravity, kappa, surface tension coefficient h, the depth and l, the size of the container such that if eta and its first derivative in l infinity norm are smaller than delta
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and if the solution exists on the time interval 0c then you have h of t smaller than h of 0
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divided by 2 for any t larger than c, capital C because here there exists c and for any t larger than c So you have dk of the energy but this raises of course the question of is it possible to have a solution which exists on the time interval 0c
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and is it possible to iterate this argument to obtain exponential dk and this question is solved at least partially by the second statement There exists a constant C star such that if the initial data are small enough in this space smaller than epsilon
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then the solution exists and is of size epsilon on a time interval of size C star divided by epsilon So you can prove the existence of the solution and arbitrary long time interval So I am going to say just a few words about the proof
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Firstly, some words about the studio of the Cauchy problem Excuse me Firstly, a word about the fact that I claim that chi has to be well chosen at least for my method How I choose chi is the following I start from a function kappa
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then I define a function m then I define the cutoff function chi I start from a cutoff function kappa supported in minus l, l even and equal to 1 on a large time interval Then I define a function m which is a multiplier which is equal to x kappa of x
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Then chi of x is equal to 1 minus the derivative of m This produces a cutoff function chi which is not as usual as usual you will define something like that This is the kind of function chi we use
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Now a word about the Cauchy problem As above, we are working in a container We symmetrize and we periodize So we are reduced to study the Cauchy problem for the weather wave equation with periodic data And with the previous analogy
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you can think of the damp water wave system as this equation dt u plus v of u dx u plus i dx to the 3 half u plus chi of x dx u is equal to 0 If you want to prove that the solution with size epsilon exists on a time interval of size 1 over epsilon you need to prove as usual
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uniform estimates in epsilon for this equation You divide this linear term by epsilon Here there is no epsilon here because I am assuming that v depends linearly on u So you see that the damping term does not help
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for the study of the Cauchy problem because in fact it adds a term with variable coefficients A singular term So you have to take care It's not difficult but you have to take care to this point It's not straightforward to prove that the size
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the life span is 1 over epsilon It's not difficult but it is not straightforward So I am not entering into the detail because it's a classical analysis What is much more interesting from my point of view is to study the stabilization
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As I said, damping to prove that the energy is decreasing is extremely easy because this is based only on the Hamiltonian formulation of the equation What is more interesting is to prove that the energy converges to 0 It is more difficult because the equation is quasi-linear and non-local At first when I studied this problem
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my motivation was to do this proof was to remove on purpose para-differential calculus and micro-local analysis This was on purpose because I wanted to discuss with engineers, people working in physics I wanted on purpose to remove micro-local analysis and I wanted to use instead global analysis, instead of working with
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functions of x and xi, working only with integrals functions of integrals of some functions f and dx And by so doing I found the proof that I would not be able to find using para-differential calculus OK, so
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this is maybe the important thing to remember is that the proof is based only on global quantities We seek exact identities using global quantities and using six principles The first is the variational principle for water wave or for Euler equation The second is to use the multipliers method We use also a principle
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of equipartition of energy positive type identities This is for the Dirichlet-Tunayman operator We use as far as possible conservation laws and we use the Hamiltonian formulation These are used as principles And I am going to explain only this, the multipliers method
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How it works for the water wave equation Firstly, what is the multiplier method in general? In the simplest case Excuse me, in the simplest case It is used to compute this quantity. You want to compute the integral in time of the energy Let us consider, for example, the 1D wave
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equation with Dirichlet-Bandari condition 1D wave equation, Dirichlet-Bandari condition You have this nice identity that can be obtained by multiplying the equation by x dxu You multiply by x dxu and you integrate by parts in time and space In time and space. And what do you obtain?
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You obtain this. The integral in time of dxu evaluated at x equal 1 is equal to 2 times this quantity at time t minus this quantity at time 0 plus the integral in time of the energy
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For example, for the wave equation, it is very easy to obtain an expression for the integral in time of the energy, which is this term which depends only on the boundary behavior of wave, something localized It is localized on x equal 1 and a term that can be neglected. Why can you
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neglect this term? The intuition is very simple This is an integral in time integral over 0t and this does not depend in time This is the difference at two different values When t is very large, this quantity will be very large, while this one will have a fixed size You can drop this term. You can compare the integral in time of the energy with something
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which is localized This is the key point The key identity for a wetter wave which is quite surprising, is that you can obtain also an exact identity where all the terms have very simple expressions using the multiplier method
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Instead of working here with a weight which is x I work with a weight which is a general infinity function vanishing at both ends m of L equals 0 You have two coefficients zeta and rho, depending in a very simple way on m and eta
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and such that with this weight, if you neglect surface tension to have a simpler statement you have this relation The integral in time of the energy plus some term q is equal to something depending on the pressure, a boundary integral in time an observation term
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and a non-linear term I will comment on this term and I'll stop quickly q is a positive term so you can neglect it It is positive, you can neglect because it has a good sign This is an observation term Why? Because y minus
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m of x and x minus m are both equal to 0 and 0 L minus delta Their weights are equal to 0 in this direction by the choice of the weight L This is an observation term It depends only on the solution near x equals L
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and what is interesting is that all the terms here are quadratic which means that the identity is the same for the linearized equation The only term coming from the non-linearity is this one The integral over the free domain of the derivative of rho times phi x times phi y
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and this non-linear term can be very easily compared to the energy because the energy is controlling phi x squared plus phi x squared So the only cubic non-linear term is this one and it can be controlled by the energy easily, provided that rho of x is in L infinity norm
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smaller than one half which is a very reasonable assumption on the smallest condition of eta and I am stopping here Thank you very much for your attention Are there any questions?
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Quite a few slides ago you did a change of variables in psi I think to make C of u constant Is there any analogue of that transformation outside of one dimension? The transformation we did are specific to one dimension
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Maybe not this one but this one is very specific to dimension one In dimension two you have to use micro-cal analysis in a smarter way Can we go back to the preceding one? In two dimensions you also have to use micro analysis here? You can flatten the coefficient here
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also in dimension two To flatten the coefficient is something that you can do For example, for this problem with surface tension this was done by for the study of Stryker's estimate for water wave Here what is really specific to dimension one is the find of this
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In dimension two you cannot do this you have to use micro-cal analysis in a smarter way Is it true that in two dimensions the difficulty
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is that the difficulty is very low In dimension two you have to do something else but there are a lot of micro-cal analysis tools that you can use The difficulty is to prove low frequency estimates Even for in-game there is this difficulty of proving low frequency estimates but there are some tricks
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that you can use in dimension one In dimension two you have to do something else Any questions? Does that change the variable in dimension one? Is that related to Riemann mapping or Schrodinger equation? Does it change the variable?
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To understand the change of variable you can do exactly the same thing for Schrodinger equation You consider a Schrodinger equation with a coefficient inside and if you want to remove to flatten the coefficient you have to make a change of variable of that form This will generate a term of order one which will be a convective derivative which is with an imaginary coefficient
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which is quite weird but it is armless and the fact that it is armless can be understood by things that it can be removed by using an appropriate pre-factor and the pre-factor is chosen in such a way that the change of variable, the change of a norm preserves the L2 norm which is not the case if you make only a change of variable
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So if you make only a change of variable you no longer preserve the L2 norm so you no longer keep the structure of the equation Ok, more questions? You know I just have a quick question So at the end when you explain the multiplier method Is this how you know how to select the chi?
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Is that what tells you how to select the chi? Yes, this is the kind of chi comes from the fact that I am using a multiplier method to prove the stabilization result No, no, no, in fact this is also ok The choice of the chi is more subtle It is in what I have not explained Later, how to use this inequality to prove stabilization is non-trivial
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And at some point you need much more identities And for one of these identities I am forced to choose chi in this way Thank you very much