 OK, thank you, and thank you for this invitation here. So in it I want to talk about the dynamics of floating structures. So as Nicolas told me, it was an ambitious title, but the ambition will stop at the title. Essentially, what I will do is just to formulate the problem. So the problem is it was motivated by a project with the Beckham in Bilbao, the back center of applied math, and Tectalia, which is a company working on wave energy. So the problem is that they want to make electricity by using the motion of some floating structure, and the motion is created by the incoming waves. And especially, they are working on the mooring system. You see these cables here that are holding this structure. And they want to know what will happen when they will have big waves, and if they can handle and hold the structures. So of course, the key word here is that we will have some nonlinear effects, and this is what they wanted to understand. So the problem that I could also do it for both, any kind of floating structure, is that I want to find the motion of the wave. So this is a classical water wave problem if I didn't have the floating structure here. The only new thing is that I have a floating body here. So what does it mean? It means that I have a rigid body, and a part of which, which I will call the wetted region. So this is why I have a w here, is in contact with the water, and this part of the bottom of the boat, I will say that it's a parametrize function zeta w. So w is for wetted region. And otherwise, this is a standard notation, the same as in Thomas' talks. I will just put it here to remember them on the slide. So here I have zeta w. Otherwise, the surface of the water is always parametrized by a function zeta of t and x. x is a horizontal dimension, and the total depth of the fluid will be h. So if I have the rest at z is equal to zero, the bottom will be at minus h0. I could handle without any problem some non-flat bottoms, but I will do it for simplicity with a flat bottom. Okay, so we want to understand this double free boundary problem. It's a double free boundary problem because we have one free boundary here, which is a standard water wave problem. And the second one is that you have to know the dynamics of this boundary problem, which is what is the geometry of the wetted region and how it evolves. So for this talk, I will use this notation for interior and exterior region. Okay, so I didn't update my file. So here, if I take the projection of the wetted region on the horizontal plane, I find some two-dimensional domain, which I will call i of t, like i, like interior, and the complement of this region on the x variable, so in rd, will be called e of t, which is the exterior domain. So the restriction of any function defined on the horizontal variables, the restriction on the exterior domain will be denoted by f e, and the restriction on the interior domain will be denoted by f i, okay? And so what are the equations? So the equation, of course, are the basic equation that we have seen for the standard water waves equation. So you have, first, in the fluid domain. So this will be Euler equation, so acceleration of the fluid particle here, dtu plus u grad u, is equal to the pressure forces, so minus one over rho, so rho is the constant density of the fluid, gradient of p, and you have here the force due to gravity. The fluid will be assumed to be incompressible and also irrotational. And then I need some bottom and surface boundary condition, so the bottom boundary condition is a standard one that the fluid is impermeable, so the velocity is tangential at the bottom, and at the surface, I have the standard kinematic condition at the surface, which relates the time derivative of zeta to the normal component of the velocity, so zeta is the same, zeta is all this, even on the interior region, I call it zeta. And so this kinematic condition, which states that any fluid particle which is at the surface of the fluid stays at the surface of the fluid, this is valid everywhere. And the other condition is exactly like in Thomas' talk, I have a pressure at the surface, the pressure is equal to the atmospheric pressure which will be assumed to be constant, I don't have surface tension effects, this is different with Thomas' talk, so the pressure will be constant in the exterior region here and here, but I will have exactly as in Thomas' talk, I will have non-constant pressure in some region of the domain, and for me this region will be the interior region here. So in this interior region, I don't know what is the pressure, this is different with Thomas' talk, is that the pressure here, for me it's not a given function, this is an unknown of the problem. So my boundary condition for the pressure is just that P in the exterior region, the surface pressure in the exterior region is constant. I don't have any condition in the interior region for the pressure. So of course you have to complement this with equations in the interior region and coupling condition at the boundary between the two regions. So in the interior region, of course what you will have is that you have a constraint, the constraint is that the parametrization of the surface of the fluid coincides with the bottom of the boat. So with my notation, it means that the interior part of zeta, so the restriction of zeta to the interior region, coincide with the parametrization of the bottom of the boat. Okay, so this is a constraint in the interior region. And now, if you want to give the coupling condition at the border gamma t, which is the border between the interior and the exterior region, you need to introduce some variables. So it's not maybe it's not standard variables in the water waves. The coupling condition will be given on the discharge. So this discharge is q, which is the vertical integral of the horizontal velocity. Okay? So the physical condition is that when you, the discharge should be a continuous function. And when you cross this gamma t, so when you come from the exterior to the interior region, you assume that the discharge is constant, which means that the exterior value of the discharge is equal to the interior value of the discharge on the boundary. And then, you also have condition for the surface elevation and the pressure elevation. So here for this talk, I will assume that the borders of my rigid body are not vertical. Of course, if I had vertical, so I can generalize this condition, I would just do it in a case which is not vertical, because in the case of a vertical wall, of course, you have a discontinuity in the surface elevation. In my case, I don't have a discontinuity. It's, I assume that it is continuous, like in this picture here, I have that the exterior value of zeta is equal to the interior value of zeta. You don't have discontinuity here of the surface elevation. And also, the pressure, the interior pressure here is equal at the corner to the atmospheric, to the exterior pressure, which is atmospheric pressure. Okay, so I have three continuity conditions. David, does that take into account the inertia of the mass? Yeah, right. At this moment, yes, everything is taken into account in this. The question of all this, and then what I'm going to show, is as I said, I will have very few theorems, but just to formulate the effects that are acting on the boat and then. So this is a problem which is not new and which was formulated by Fritz John in the two papers in 49 and 50. And she did it in a very simplified way. So for him, he had this rigid body here. And what did he do? He simplified the equations. He said, okay, I have a potential flow. So this is also true here because we have incompressibility and irrotationality. And he assumed that he had some linear flow models. So he neglected all the nonlinearities. So the kinematic condition became this one. And the Bernoulli equation on the surface became this one. So you have basically a linear wave equation, a non-local wave equation for the surface elevation. So this is linear equation. And then he also neglected the variation of the weighted zone. He said, okay, the weighted zone is constant with time. So if he removed one of the boundary problems, free boundary problems, which is determining the boundary of this boundary and this boundary. And then what he did and which is not completely correct is that he used some boundary condition and the coupling here and here, which are not completely correct because he worked with this potential variable, which are not very adapted to the problem. As we saw, good condition, coupling condition is on the discharge, which is a non-local quantity if you interpret it in terms of velocity potential. And also he assumed that he had some time harmonic motion. So he could replace the dt zeta by some high lambda of zeta, which transformed the problem into a spectral problem. And so you have been hundreds of papers on this spectral problem. And then how do you recover the motion of the body? Well, it's just that you use the linear Bernoulli equation in the field, which gives you the pressure in terms of the velocity potential and the surface elevation here, the hydrostatic pressure. So once you have P, then you apply Newton's law for the rigid body. So M times the second derivative of the position of the center of mass is given by its weight here and the pressure forces exerted by the fluid. So it's integral on this boundary of the interior pressure, the pressure here that is completed like this, times the normal vector. And you also have an equation for the angular velocity if you want. I just put it for the center of mass, but it's more general. Okay, so this is a very simplified problem. And actually this is the only thing that there is and which is used by engineers. Now by the software, for instance, WAMIT is used a lot. These are these linear models. And of course you cannot use this software to understand nonlinear effect, in particular the nonlinear effect that I mentioned you about the mooring systems. So what people also do is big CFD computation with 3D Navier-Stokes equation, 3-phatheic because you have air, solid and water. So it's very complicated, very heavy. And you cannot use this to simulate, for instance, arrays of wave energy convectors when you have several of them. So this is out of reach in terms of numerical costs. On the other hand, you have recent works on the motion of totally immersed solid. Okay, so they are in a domain which is fixed. So the domain can be RD plus one or Omega, like in this picture. And in this case, the situation is much simpler because you have your body which is here. You have as before Newton's law in the body. So with the pressure forces. And you have earlier equation, but you don't have the free surface, of course. And this is coupled throughout this condition, continuity of the normal component of the velocity at the border of the solid, which is that the normal velocity of the fluid is equal to the normal velocity of the solid here. Okay, so you have some papers studying the local well-positives for this. And more recently, when you are going, if you want to get some more, some finer information on the motion, you have to use an added mass effect, which is something that goes back to, I think, D'Alembert on the linear case and when this was immersed in RD plus one, so you don't have boundaries. And essentially, it is that you can prove that at least part of the pressure force here, you can put it under this form, which is a negative number times a second derivative of the position of the center of mass plus other terms. And so you see that this term, you can therefore put it on the left-hand side and instead of having M G dot dot, you have M plus MA G dot dot. So this is like, it acts like an added mass in Newton laws. So this is very important, also very important for numerical purpose. Then of course, when you are working in the domain with the boundary, this added mass effect depends strongly on the position of the object. So this is quite complicated. And of course, if you want to handle this in the free surface case, you expect to have some real difficulty because the domain is moving and is a free boundary. And moreover, in order to compute this added mass effect, you are obliged to solve at each time a three-dimensional elliptic equation in the free domain. So this is, if you have in mind read application, this is numerically very expensive. And also, in this case, you don't have a difficulty which is of course the interior-exterior coupling. So the dynamics of this free boundary problem here. And also just to say that this added mass effect is very important in many free interaction problems and especially for your numerical stability of the codes. Okay, so actually what I will do here, the outline of this talk is that I want to propose another approach because this approach, as I said, is if we could generalize it to the free surface case, then we will have some very heavy things to do, in particular to compute the added mass effect to solve every time three-dimensional elliptic equation. So I want to use a new set of variables. And the first step will be to formulate the water-web equation in a new set of variables, which is zeta and vbar, vbar which is the horizontal variable. Average velocity, which is just q divided by h. So q is h vbar, if you want to. So that you see that the boundary condition on the discharge is very adapted to this formulation because you can state it as a continuity on vbar. You don't have any normal effect on this. Then I will include the soline and show how the coupling occurs. And which is the novelty here and this is why, essentially, we will gain one dimension in the computation of this coupling between the solide and the fluid is that I will show that the interior pressure, this is essentially the main node of the problem, which governs the motion of the body, I will formulate the problem as a compressible and compressible model and I will show that the pressure, interior pressure can be seen as a Lagrange multiplier that you can compute with a two-dimensional instead of a three-dimensional problem. Then I will analyze this interior pressure, show the added mass effect, which is up here in the overview because the added mass effect is, essentially, it says that when the fluid, solide moves in a fluid, it, of course, has to accelerate itself, but it has to accelerate also the fluid which is around him. In the case of a floating structure, it's, essentially, you don't know which part of the fluid it has to accelerate, so it's a bit more complicated. And then I will show that this strategy to see the interior pressure as a Lagrange multiplier, you can apply it on a reduced model, like non-universal water equation that Thomas mentioned, and you can also use it on discretized model and you can find some very efficient numerical codes. Okay, so let me formulate the equation in this set of variables, zeta and viva. So the first one, the equation on zeta, so this is a reformulation of the kinematic equation, so this is very easy. You take the incompressibility condition, so v is the horizontal component of the velocity, w is the vertical one, so this is divergence-free. You integrate from the bottom to the top and you find this equation, which is standard conservation of mass equation, dT zeta plus the divergence of the discharge is equal to zero. This is exact. This doesn't use irrotationality assumption. This is very robust and exact. Okay, so what we need now is the equation on viva. Actually, I will give an equation on q, which is h, viva. So if you want to do this, the first thing that you are going to do is that you first recover the pressure from the vertical component of the Euler equation. So the vertical component of the Euler equation, this is the equation on v, so dT v plus two grad vw, excuse me, plus gravity plus one over all dZp is equal to zero and you integrate this between some height z and the surface. You know that at the surface, the pressure is zero, I mean it's the atmospheric pressure, and so you will find p minus at the surface, so atmospheric pressure minus p at height z is equal to this integral. Okay, so this gives you an expression for the pressure. So you see that the first term here is the so-called hydrostatic pressure. This is a pressure due to the gravity forces and this nonlinear pressure is called the non-hydrostatic pressure. So I will just call it like this to simplify. And then you take the horizontal component of Euler equation, dT v plus two grad v plus one over u grad p is equal to zero. You plug this p here and you integrate from the bottom to the top. So you see that this integration would give an equation on dT hv bar. And this is what you obtain, dT hv bar plus some term. So this is a vertical integration of v times of v plus this, which is due to the hydrostatic component of the pressure and the integral of the gradient of the non-hydrostatic component of the pressure. This, to simplify notation, I will call it h times the non-hydrostatic acceleration. Just notation. Okay, so I have a set of equation. Set of equation, this is an exact set of equation. I have no approximation. The question is, is it a closed set of equation? Which means that, can I express all these terms here as function of zeta, excuse me, nv bar? Okay, so you have to prove that if I give you zeta and v bar, you are able to reconstruct essentially the velocity field in the fluid domain. So this is the case. So you define first this space to be the set of admissible velocity field. So you look for velocity field, which are in L2 of the domain, which are divergence free and curl free. And that has homogeneous normal components at the bottom. So the proposition is the following one. So you take zeta, which is w1 infinity. So you cannot assume more on the surface parameterization. You see that you have an angle here at the contact line. So the regularity you have to work with is very low. It's lip sheets. You cannot have more than this. So the first is that you take a velocity field. You take the average of its horizontal component. And so this you can prove that it works as a trace, essentially. Because you see that you, if you satisfy this and this, essentially you is in h1. Its trace will be in h1 half of the boundary. And the average is also in h1 half. Okay, so it works as a trace. And then, of course, you need the reconstruction mapping. So you start with a vector, which would be an average vector, and you want to reconstruct the velocity field. So the way you do it is that you take the gradient for velocity potential, which you find by solving this, this has to be harmonic in the free domain. And its value at the surface should be the inverse of the Dirichlet Neumann operator that Thomas mentioned applied to this quantity. So just the Dirichlet Neumann operator, we know that it is essentially a projection from homogenous sobri-spaces of order 1 half into the subspace of h minus 1 half, which are the derivative of function in h1 half. Okay, so you see that you have the regularity that you need, the space you need. So this is well defined in a homogenous sobri-lef for the 1 half, okay? And of course, this reconstruction mapping is the right inverse to the average mapping, okay? So this proved that this set of equation is closed and you can work, you can see this as an equation on zeta and V bar. And okay, so now the question is what happens if I put a floating body? So the only difference with the analysis I have made is that I have this red term, which is exactly as in Thomas' talk, I have the pressure applied at the surface, except that the pressure at the surface is given by the atmospheric pressure in the exterior region, but on the interior region, this is a non-none function that you have to determine. So... You said it's a right inverse, so you know something about whether... Left inverse, left inverse is you have to define the range correctly. Because you cannot... But the range is somehow dense or something like that? No, it's not dense, but you can define it, yes. It's... You cannot do it in an easy way, but you... What you need is just the right inverse property. But if you want, so then on each range, it also has a left inverse. But it's not subjective on each one half. Okay, and so of course, the way you are going to determine this interior pressure is by using the constraint. And the constraint is that on the interior region, zeta surface parametrization should be equal to the parametrization of the bottom of the boat. So therefore, if you look at this, the first equation, you know that the divergence of hg bar is minus dt of this w, okay? So you see that in the interior region, this pressure is a Lagrange multiplier associated to this constraint. So you write the equations, and this is in the interior region, it is an incompressible problem. So what should be... I mean, I should insist is that... So without any... So the waterway problem will start with the Euler, which is incompressible. But then when you write the equation in hv bar variable, the structure of the equation is compressible. You see that you have typical compressible Euler equation here, okay? So in zeta v bar variable, the problem is compressible. And then in the interior region, the problem becomes incompressible, okay? But incompressible with, how to say, constraint, the divergence constraint, which is this one. It's not the divergence fritz, the divergence given by this. And the equation is this one. So it's just a two-dimensional equation. And so you can compute explicitly, actually I mean, you have find a simple equation for the interior pressure. So you just define afs, which is the acceleration of the pressure phase that you would obtain, which means it's dT squared of zeta in the case without any floating object. You see that if you didn't have any floating object, then you could compute dT squared of zeta by taking time derivative of this, which should be equal to the divergence of this. And this, so this is equal to the divergence of all this if you don't have any floating object. So this is this quantity. And so the reason why you have an interior pressure is because in the case of a floating structure, the second time derivative of the boat parameterization is different from this quantity. So this creates an interior pressure that you can solve by a very simple two-dimensional equation in the interior domain. And here I use the boundary condition on the pressure. The pressure should be continuous, so I put it here. So this gives me the equation for the interior pressure. And what you can prove is that you can use this to remove the constraint. Because having the constraint in the equation is you have something you have to check to handle every time. This is a bit heavy, but you can remove it if you just define a pressure like this. So if you have some solution, so a solution, what does it mean? It means zeta. It's in its V bar, and it means also to know this curve here, gamma t. So if you have a solution of the equation satisfying this set of equation with the surface pressure given in the exterior region by the atmospheric pressure and the interior region by this simple elliptic problem, then for all time, of course, if it is satisfied initially, for all time, the parametrization, the surface of the water will correspond to the surface of the boat in the interior region. So you don't need to have this structure. It is automatically satisfied. So you can forget about it in the analysis. And once again, this interior pressure, you find it interpreting it as a counter-multiply like this. You just have a two-dimensional elliptic problem on this region instead of a three-dimensional if you have generalized the approach by friction. Yeah, I'm going to at this point you have to determine. Actually, this set of equation, OK. I'm not able to show that this for the full equation is well-posed because, of course, you need to have several open problems in terms of regularity. For instance, you have some surface which is elliptic. You have many open problems, but then on the reduced model, in some cases, I am able to prove well-poseness which means that for this equation, well-poseness means that you have a smooth curve here and the regular function zeta and regular function v in the interior and in the exterior with, of course, an angle here that propagates. And this is the kind of well-poseness theorem that you have. OK, so now you have to couple this with the dynamic of the solid. So if you want to look at this, you look at the equation for the interior pressure, minus divergence h grad p, given by this. And I want to show that there is some adid mass effect. So adid mass effect is that part of the force here is some kind of negative number applied to the second derivative of the position of the center of mass. So from the continuity of the normal velocity here, you say that dt zeta in the interior region is the normal velocity of the solid and the normal velocity of the solid. So the velocity of the solid at the border is given by the velocity of the center of mass. So I need to. So the center of mass g is here. So it has some velocity Ug. And this is the vector Rg. And omega is the angular velocity. So this is standard solid mechanics. There's nothing to do. And in particular, you can decompose the pressure into three terms. So this is the pressure equation. So one component of it will be the contribution for this term. The contribution for this term is the contribution that you would have if the solid were not moving if it was just fixed. So in this case, in particular, you have the arcane median force and everything is in P1. So just forget about it. Then you see that what I need is a second time derivative. And if I want to see the adid mass effect, what I need is a first derivative of U and omega. So when I take the derivative of this, so the derivative can hit this one and this one. So this will be in P2. So this will be the candidate to create the adid mass effect. And the time derivative can also hit R and N here. So this creates some geometrical terms. Essentially, this is the second fundamental form at the bottom of the boat. And then I have to couple this with Newton's law. So Newton's law is an equation for the velocity of the center of mass and an equation for its angular velocity. So for the center of mass, you have the weight and you have the force exerted by the fluid on the surface, which is integral of P times the normal vector. And for the angular velocity, you have the torque, which is applied, which is given by this equation. So this is standard. OK, since you're in 3D, the inertia matrix here is time-dependent, but I don't insist on this. So what I want to insist is on the creation of the adid mass effect. So I will just assume that omega is equal to 0 just to simplify. I think that you have a sphere, for instance. And so let's see how the adid mass effect occurs. So as I said, P could be decomposed. And the second component was given by solving this elliptic equation, minus divergence h grad P2, was given by this term. And the corresponding force is the integral of P2 times n. So how can you put this as something like negative operator applied to the time derivative of the velocity of the center of mass? So this is a very simple computation. So you have to define elementary potential. So this is inspired by the Kirchhoff potential. The way you do it, so n has three components. So you define three velocity potential, elementary potential in the interior region. Just by solving, the jth elementary potential is solved, this elliptic equation applied to the jth component of the normal vector with zero boundary condition. So you have three of them. And then the computation is just this one. So the force is P applied to n. But you know that n, by definition of the elementary potential, is the sum for j from 1 to 3 of minus divergence of h grad P of ij times the vector ej. So this is the definition of the elementary potential. Then this operator is self-adjoint. You put it on the P. So now it's the advantage of h grad P applied to phi. So this is a vector of phi 1, phi 2, phi 3. But then you use the equation on pi 2, which tells you that this is u dot applied to n. So just replace. And then so you just rewrite it to put the dot u on the right-hand side. So you write it like this. And then you say that n, you say again that it is using the definition of the elementary potential, this is minus divergence of h grad phi. So you write it like this. And then you integrate by part 1. And you find that f2 is minus m, so which is a 3 by 3 matrix, times dot u, where m is this quantity given in terms of the elementary potential. And you see that this is a gram matrix, so this is a positive matrix. And so in the end, you can rewrite the equation under this form. So m plus positive matrix times dot u is equal to the rest of the forces. And in general, if you have also the angular velocity, you have a 6 by 6 mass inertia matrix, which is positive. And so then if you want to understand how it happens, you can look at dimension 1. In dimension 1, you can make explicit computations. So in dimension 1, the interior region is just an interval x minus x plus. And the added mass matrix, you can just compute it. And it is the square of the oscillating part of the function defined in the interior region. So how do you define an oscillating function? You remove its average. And the average way to define it, and it's the way, in some sense, to determine the quantity of free that you have to accelerate, is just the integral from x minus to x plus of f over h divided by the integral of 1 over h. Because this is kind of average. So it's explicit, and you can really do things very easily. So as you said in your question, you need to understand the evolution of the interior region. So in the case of dimension 1, this is an interval. So I will do it for the case of the interval. So everything is in the equation. But if I want to make it more explicit, I will show you how to do it. So of course, we can do it for the two-dimensional general case. But I will just do it for this one for simplicity. So everything is encoded, actually, in this continuity equation. So as you remember, I said that I had surface elevation on the water depth. It's the same. It should be continuous at this point here. Which means that h exterior of t applied to x plus minus is equal to h interior of t applied to x plus minus. You time differentiate this. So let's emphasize dt h e plus x dot dx of h. And the same in the interior part. And then you say that in the exterior part, you know that you solved the equation. So dt h, you will replace it by minus dx of h v bar. So this was a kinematic equation. So here, you don't need to do it. dt of h interior is a function which is given in terms of the velocity of the center of mass. It's angular velocity, everything. So this is a non-function in some sense. And so from this, you recover the equation for x dot. So x dot is given by this function. So you have a contact line equation. And maybe it's important to comment it because it's singular contact lines, quite singular. For instance, you could compare it to have an idea with a vacuum boundary condition. So for vacuum boundary condition, you have been some work for the isentropic Euler equation, especially by Nader and Zhang and Kutang Scholar, where you want to understand the formation of vacuum for isentropic Euler equation, which is very similar to this problem. So actually, the vacuum boundary condition is a particular case which could correspond to say that in the interior region, you have h is equal to 0. So the depth vanishes in the interior region. And then what happens in this case? So in this case, if I look at the vacuum boundary condition, so x dot is given by this. So if I have vacuum boundary condition, h i is equal to 0. So this vanishes. This vanishes. OK, so you have this. And for this one, you expand the derivative. So it's h dxv, but h is equal to 0, so it vanishes. So you just have v dxh. And the dxh exterior simplifies. And you find that dot x is equal to v at the contact point. So this means that you have a kinematic condition for the dynamic of the boundary. And this means that if you want to solve this boundary problem, then you will essentially have a diffomorphism that will have the regularity of the velocity. Now for this contact line equation, you lose one derivative more. So it's much more singular, because you have to handle this derivative. So this is a generalization of this one, but which is not a kinematic boundary condition. OK, and so I said that I mentioned this work about vacuum in the isentropic Euler equation. So isentropic Euler equation or shallwater equation are the same? So let's now try to see if we can apply this strategy for simplified model if you are interested in doing this for applications. So how can you simplify the models? So these are the full equations. And the shallwater equation lies on several assumptions that you can prove. In the case without floating objects, you can prove them rigorously. The first one is that this would share the Reynolds tensor, because it's in some sense the integral of the integration as an average that would replace a statistical average in turbulence. Then you say that this is just the first term, which is the square of the average. So this is the basic stuff. So you make this approximation, which is a very robust one. And the second one, so you replace this in the equation. So it's much simpler. And the second approximation that you make is that you neglect these non-hydrostatic terms. You say that they are equal to 0. So if you want to include a dispersive effect, work with Boussinesq equation, or thing like that, you cannot neglect them. You have to take them into account. So you have a more complicated model. But in the Savano equation, you don't have this term. So you have a very simple system. And so the equation like this, you have the pressure term here. And of course, if you use the pressure term that I gave before, the equation for the interior pressure that I gave before, it will no longer be a Lagrange multiplier for this system of equation. Of course, you have to adapt the equation for the interior pressure by doing the same approximation. So in this term, free surface acceleration, you just make the same approximation. So you have now a good Lagrange multiplier. And you can do exactly the same kind of analysis with simpler terms for this system. And for those who know, there is a remark here. It's that you could rewrite this system as a congested flow model, which means that you could write it as a Savano equation with a pressure term here. And with a saturating term here, which says that i should be w, so under the boat. So when h is inferior to hw, so when you are in the exterior region, you have no constraint. So it means that p bar is equal to 0. But then when you saturate the constraint, when h is equal to the depth under the boat, then the pressure is not 0. And it gives you this. So people derive many models like this, essentially with g is equal to 0 for pressure less earlier equation, if you want. So you have this equation rising in traffic flow, in granular flow, in hydrodynamics in pipes. And the idea is that you are saturated some constant. So these are called congested flow. So you can put it under this form. It's a bit different because of the g. And you could do this also for the full equation, of course. OK. And then you could extend this to dispersive models by adding more terms in the equation, more complicated terms, so I won't do it. But you can do the same strategy. And then you can also do the same strategy at the discrete level. So I will not insist. But just saying that if you, for instance, the Savano equation in dimension 1, you can put them in conservative form. So you can use any finite volume formulation. So you have some discretization for the flux here. You can choose any discretization. What is important, that you choose your flow model. So you have first order, second order, approximation for the flux is what you want. But then you compute the discretization of the source term in a consistent way as a discrete Fourier multiplier to your numerical scheme. So you can do this. And if you do this, as I did for the continuous case, you can remove the constraint that h is equal to hw under the boat. It will propagate at machine precision. So you have just to solve an equation on the computational domain without handling the coupling zone. And so this is, in terms of numerical efficiency, is the same as for the standard flow model without objects. Almost the same. So it goes very fast. And just to show you that I'm not lying, I just made some computations. So in this case, I have a vertical, just vertical bottom. And the idea is that I allowed my object to move only vertically. Just assume that you can contract like this. I have all the computations if you want. So here you see that I just plotted the free surface elevation. And you see that I'm solving it on this interval. I put a discrete pressure term. And you see that this is a surface elevation. And because of this pressure term, you see that it takes exactly the form of my object here. So this is I solving zeta on my numerical model for all the domain. I'm not saying artificially that this should be equal to the boat. And I don't put it as a constraint. It's just automatic because I have chosen the discretization as a discrete Lagrange multiplies. But you haven't given inertia to the body in this case. Yeah, wait, wait, wait, wait. This is the case of fixed body. And so this is the same. Now I draw the body to the side. You can understand what happens. So the wave comes, you see. And then there's nothing here. And part of the flow goes below and creates a wave on the right. This is what happens. So if you had a wall, of course, it would be reflected. So part of it is reflected here. And part of it goes under. And you see that in orange, orange is the discharge. So for a fixed body, the discharge should be constant. dT zeta plus dx of q is equal to 0. So if dT zeta is equal to 0, dx of q is 0. Of course, it varies with time. And it is continuous. This is one of our boundary conditions. Then if now this is a forced motion. So I put the object in forced vertical motion. And so in this case, we create the wave. And you see that in this case, now the discharge under the boat is not constant anymore. Now it's linear. But it's still continuous. And it creates the wave. So it's vertical motion. And of course, the most interesting case is to have this floating case. So in this case, I take an object like this. And I take its volume density. And from its volume density at its shape, I am able to compute the place where I should have the equilibrium. So it's above the equilibrium. I just drop it. And if you drop it, so I'm sorry. So you see it goes down. And it will go and it will stabilize until it's equilibrium point. So for this one, I have the inertia. How do you get such nice picture? I mean, this is in terms of computational, this is real-time computation. If you compare to that. Did you say this is shallow water or the fully? Yeah, no. This is shallow water, of course. I could do it for Boussinesque. Of course, for the fully nonlinear equation, I mean, you could try in 1D with point vortex method, maybe. And so this is not the one. And the last one is the one we are concerned about for the application I was mentioning for the resistance of the mooring system. Is that then you send a wave? And it will start to float because of the wave. And so now we have some very simple model that we can use. So we can put several of them and see what happens. And it's very easy, numerically, very efficient. So we can use it for a real application. And of course, what remains to do is now is the math. And it's really for another talk. So thank you very much. About the bubbles. I thought at one point there is no vorticity. But there is vorticity in that. No, this is irrotational. Putting vorticity, I mean, it's more complicated, but it's not out of reach. So is there some mechanism for generating vorticity from the bubbles? From this, if I generate vorticity, no, I don't think. Yeah, it's true because here you have singularity. You have boundary effects. I don't think that you create vorticity. No, you don't create vorticity. You would create vorticity if you had wave breaking in addition. Then you dissipate in addition, because this is, I didn't say it, but you have some conservation law for energy. And so I think it's OK for the vorticity. I didn't check, but. You've restricted the motion of this to be vertical. But in the bigger picture, you also had the omega and the angular. Yes. So for the woman. So OK, the basic. And here also, I treated the case that I did not mention in the equation, which is vertical walls. So for this, I have discontinuous zeta, discontinuous pressure. I have two. Then in this case, I contributed later the numerics. In this case, this point would move. So I will have this here. I don't have this familiar problem because this is constant. But I presented this one because for this one, I can do some numerical analysis very cleanly and prove it. For this one, it works numerically, but the numerical analysis is OK. But I can see this moving. And then, yes. And then you can couple with the motion like this. And also, the next step or so is to create things here. I assume that it's a graph. Then I will need to allow. For this one, if you start moving like this. So I have two regions. Maybe a sphere. But for this one, it's OK. I understand right that basically, you'll be able to recover the contact line afterwards after you solve the. For this one, you can handle the contact line. The problem is in the. So you get it from the formulation. Yes. Without having to propagate the contact line. Yes, that's right.