On the dynamics of floating structures
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00:00
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Transkript: Englisch(automatisch erzeugt)
00:15
OK, thank you. And thank you for this invitation here. So, indeed, I want to talk on the dynamics of floating
00:20
structures. So, as Nicola 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 BECCAB in Bilbao, Bass Center of Applied Math, and Tectalia, which is a company working on wave
00:40
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.
01:01
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, I could also do it for both, any kind of floating structure, is that they want to find the motion of the wave.
01:23
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.
01:42
So this is why I have a w here. It is in contact with the water. And this part of the bottom of the boat, I will say that it's a parameterized function, zeta w. So w is for wetted region. And otherwise, these are standard notation, the same as in Thomas Stokes. I will just put it here to remember them on the slide.
02:05
So here, I have zeta w. Otherwise, the surface of the water is always parameterized by a function zeta of t and x. x is a horizontal dimension. And the total depth of the fluid will be h.
02:23
So if I have the rest at z is equal to 0, 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. So we want to understand this double free boundary problem.
02:41
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, how it evolves. So for this talk, I will use this notation for interior and exterior region.
03:04
OK, 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.
03:24
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
03:40
will be denoted by Fd, and the restriction on the interior domain will be denoted by Fi. And so what are the equations? So the equation, of course, the basic equation that we have seen for the standard water-wave equation. So you have first in the free domain. So this will be earlier equation,
04:00
so acceleration of the free particle. Here, dtu plus u grad u is equal to the pressure forces, so minus 1 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.
04:21
And then I need some bottom and surface boundary conditions. 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.
04:41
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 the free 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
05:00
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. See the difference 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,
05:23
I will have a non-constant pressure in some region of the domain. And for me, this region will be in the interior region here. So in this interior region, I don't know what is the pressure. This is a difference with Thomas Talk, is that the pressure here, for me, it's not a given function.
05:40
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 of 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
06:01
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,
06:21
it means that the interior part of zeta, so the restriction of zeta to the interior region, coincides with the parametrization of the bottom of the boat. 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
06:41
and the exterior region, you need to introduce some variables. So it's not, maybe it's not standard variables in the waterways. The coupling condition will be given on the discharge. So the discharge is Q, which is the vertical integral
07:04
of the horizontal velocity. So the physical condition is that when you, the discharge should be a continuous function. And when you cross this gamma t,
07:20
so when you come from the exterior to the interior region, you assume that the discharge is constant, which means that the Q, 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.
07:41
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 the case which is not vertical, because in the case of a vertical wall, of course, you have a discontinuity in the surface elevation.
08:00
In my case, I don't have a discontinuity. 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. So I have three continuity conditions.
08:22
David, does that take into account the inertia of the mass of the body? Yes, everything is taken into account in this. The question of all this, and then what I'm going to show, is that as I said, I would have very few theorems, but just to formulate the effects that are acting on the boat.
08:41
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, she had this rigid body here. And what did he do? She simplified the equations.
09:00
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 model. So he neglected all the nonlinearities. So the kinematic condition became this one and the Bernoulli equation at the surface became this one. So you have basically a linear wave equation,
09:22
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. She said, okay, the weighted zone is constant with time. So if he removed one of the boundary problem, free boundary problems, which is determining the boundary of this boundary
09:40
and this boundary. And then what he did, and which is not completely correct, is that he used some boundary condition at 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,
10:03
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 she 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.
10:23
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 then 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.
10:41
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,
11:00
the pressure here that is computed like this, times the normal vector. And you also have an equation for the angular velocity, if you want. I just put this for the center of mass, but it's more detailed. Okay, so this is a very simplified problem. And actually, this is the only thing that there is,
11:24
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 I mentioned you about the mooring systems.
11:42
So what people also do is big CFD computation with 3D Navier-Stokes equation, three-phatting 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,
12:00
and you have several of them. So this is height 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 the domain which is fixed. So the domain can be RD plus one, or omega,
12:20
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 Euler equation, but you don't have the free surface, of course. And this is coupled throughout this condition,
12:41
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 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 finer information on the motion,
13:00
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
13:24
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 m a g dot dot. So this is like, it acts like an added mass in Newton's laws.
13:42
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,
14:00
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 a real application,
14:23
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 fluid interaction problems,
14:42
and especially for numerical stabilities 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,
15:02
then we would 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-wave equation in a new set of variables, which is zeta and v-bar,
15:21
v-bar which is the horizontal, average velocity, which is just q divided by h. So q is h v-bar, if you want. So you see that the boundary condition on the discharge
15:44
is very adapted to this formulation because you can state it as a continuity on v-bar, you don't have any non-local effect on this. Then I will include the solen and show how the coupling occurs. And which is the novelty here, and this is why essentially we would gain one dimension
16:02
in the computation of this coupling between the solid and the fluid, is that I will show that the interior pressure, which is essentially the main anode of the problem, which governs the motion of the body, I will formulate the problem as a compressible model.
16:21
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 top view, because the added mass effect is essentially it says
16:40
that when the solid moves in the 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, in such a way 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,
17:04
you can apply it on a reduced model, like non-linear shallow 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 v bar.
17:22
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.
17:42
This is 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 the equation on v bar. Actually, I will give an equation on q, which is h v bar. So if you want to do this,
18:00
the first thing that you are going to do is that you first recover the pressure from the vertical component of the earlier equation. So the vertical component of the earlier equation, this is the equation on v, so dt v plus u grad v, w, excuse me, plus gravity plus one over rho dzp is equal to zero, and you integrate this between some height z and the surface.
18:21
You know that at the surface, the pressure is zero. I mean, this is 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.
18:42
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 the earlier equation, dt v plus u grad v plus one over u grad p is equal to zero. You plug this p here,
19:00
and you integrate from the bottom to the top. So you see that this integration would give an equation on dt h v bar, and this is what you obtain, dt h v bar plus some term, so you see the 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.
19:20
This, to simplify notation, I would 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 this term here
19:42
as function of zeta, excuse me, and v bar? Okay, so you have to prove that if I give you zeta v bar, you are able to reconstruct essentially the velocity field in the free domain. So this is the case. So you define first this space to be the set of admissible velocity field.
20:02
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 parametrization.
20:24
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 Lipschitz. You cannot have more than this. So the first is that you take a velocity field. You take the average of its horizontal components. And so this, you can prove that it works as a trace, essentially,
20:41
because you see that u, if u is this and this, essentially u 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,
21:01
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. It has to be harmonic in the field domain. And its value at the surface should be the inverse of the Dirichlet number operator that Thomas mentioned applied to this quantity. So just the Dirichlet number operator,
21:22
we know that it is essentially a projection from homogeneous sublet spaces of order one half into the subspace of h minus one half, which are the derivative of function in h1 half. Okay, so you see that you have the regularity
21:41
that you need, the space needed. So this is well defined in homogeneous subwoofer for the one half. And of course, this reconstruction mapping is a 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.
22:03
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
22:20
in the exterior region. But on the interior region, this is an unknown function that you have to determine. So... You said it's a right inverse. So you know something about whether... Left inverse, you have to define the range correctly. Because you cannot... The range is somehow dense or something?
22:42
No, it's not dense, but you can define it. You cannot do it in an easy way, but what you need is just the right inverse property. But if you want, so then it's on its range, it's also have a left inverse.
23:00
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 parameterization, should be equal to the parameterization of the bottom of the boat. So therefore, if you look at this, the first equation,
23:21
you know that the divergence of h g bar is minus dt of zeta 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, this is an incompressible problem.
23:40
So what should be, maybe I should insist, is that, so without any, so the waterway problem, we start with Euler, which is incompressible. But then when you write the equation in h v bar variable, the structure of the equation is compressible. You see that you have typical compressible
24:00
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, divergence constraint,
24:22
which is this one, it's not divergence fritz, 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 you have to find a simple equation for the interior pressure. So you just define A f s, which is the acceleration of the fresh surface, that you would obtain, which means it's dt square of zeta,
24:44
in the case without any floating object. You see that if you didn't have any floating object, then you could compute dt square of zeta by taking time derivative of this, which should be equal to divergence of this. And this gives, so this is equal to divergence of all this if you don't have any floating object. So this is this quantity.
25:01
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.
25:20
And here, I use a boundary condition on the pressure. The pressure should be continuous, so I put it here, okay? 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 the pressure like this.
25:42
So if you have some solution, so a solution, what does it mean? It means zeta, it's in this 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,
26:03
then for all time, of course, if it is satisfied initially, for all time, the surface of the water will correspond to the surface of the boat in the interior region, okay? So you don't need to have this structure. It is automatically satisfied,
26:21
so you can forget about it in the analysis. And once again, this interior pressure, you find it, interpreting it as a Lagrange multiplier like this, you just have a two-dimensional elliptic problem on this region instead of a three-dimensional if you had generalized the approach by Fritz Schoen.
26:42
Yeah, I'm going to do it. At this point, you have to determine. Actually, this is all, this set of, okay. I'm not able to show that this, for the full equation, is well-posed because, of course, you need to, there are several open problems in terms of regularity. For instance, you have some surface which is deep sheets. You have many open problem, but then on the reduced model, in some cases,
27:01
I am able to prove well-posedness theorem, which means that for this equation, well-posedness 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-posedness theorem that you have.
27:23
Okay, 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 pressures, minus divergence h grad p, divided by this. And I want to show that there is some added mass effect. So added mass effect is that part of the force here
27:43
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
28:02
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 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.
28:22
So this is that 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 a fix.
28:42
So in this case, in particular, you have the Archimedean 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 added mass effect, what I need is a first derivative of u and omega. So when I take the derivative of this, the derivative can hit this one and this one.
29:00
So this will be in p2. So this will be the candidate to create the added 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.
29:21
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
29:41
is given by this expression. 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 equation of the added mass effect. So I will just assume that omega is equal to 0 just to simplify. You can just think that you have a sphere, for instance. And so let us see how the added mass effect occurs.
30:02
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
30:22
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
30:42
just by solving the jth elementary potential is solve this elliptic equation applied to jth component of the normal vector with zero boundary condition. So you have three of them. And then the computation is just this one. It's very easy. So the force is p applied to n.
31:02
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 phi of phi j times the vector eg. So this is the definition of the elementary potential. Then this operator is self-adjoint. You put it on the p.
31:22
So now it's the difference of edge grad p applied to phi. So phi is a vector of phi 1, phi 2, phi 3. But then you use the equation on phi 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.
31:40
And then you say that n, you say again that this is using the definition of the elementary potential. This is minus divergence of edge grad phi. You write it like this. And then you integrate by part 1. And you find that f2 is minus m, which is a 3 by 3 matrix, times dot u, where m is this quantity given
32:01
in terms of the elementary potential. And you see that this is a grand matrix. So this is a positive matrix. And so in the end, you can rewrite the equation under this form. So m plus a positive matrix times dot u is equal to the rest of the forces. And in general, if you have also the angular velocity,
32:21
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 computation. 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
32:41
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. It's just the integral from x minus x plus of f over h divided by the integral of 1 over h. This is kind of a friend.
33:01
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,
33:21
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. The same should be continuous at this point here.
33:40
And this point. Which means that h interior of t applied to x plus minus is equal to h interior of t applied to x plus minus. You time differentiate this. So I'm left hand side is dt h e plus x dot dx of h. And the same in the interior part.
34:00
And then you say that in the exterior part, you know that you solved the equation. So dt h, you replace it by minus dx of h v bar. So this was a kinematic equation. 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.
34:22
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 line. It's quite singular. For instance, you could compare it to have an idea with vacuum boundary condition.
34:42
So for vacuum boundary condition, you have been some work for the isentropic Euler equation, especially by Nader and Zhang and Goutange-Koller, 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
35:03
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? In this case, if I look at the vacuum boundary condition, x dot is given by this.
35:21
So if I have vacuum boundary condition, hi is equal to 0. So this vanishes. This vanishes. 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 interior simplifies.
35:43
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 lift, then to solve this boundary problem, then you will essentially have a deformorphism
36:02
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. We have to handle this derivative. So this is a generalization of this one, which is not a kinematic boundary condition.
36:21
OK. And so I mentioned this work about vacuum in the isentropic Euler equation. So isentropic Euler equation or shallow water equation are the same. So let's now try to see if we can apply this strategy for simplifying model if you are interested in doing this for applications. So how can you simplify the models?
36:43
So these are the full equations. And the shallow water 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, which is a kind of tensor, because it's in some sense the integral of the,
37:04
if you think about the integral 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. OK. So this is the basic stuff.
37:21
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 zero. So if you want to include the dispersive effect, work with the Boussinesq equation,
37:41
nothing like that, you cannot think like that. You have to take them into account, so you have a more complicated model. But in the 7R 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,
38:01
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, Frisch's fast acceleration, you just make the same approximation. OK. So you have now a good Lagrange multiplier,
38:22
and you can do exactly the same kind of analysis with simpler terms for this system. And for those who know, there's a remark here, is that you could write this system as a congested flow model, which means that you could write it as a 7R equation with a pressure term here, and with a saturating term here,
38:43
which says that it's not, I'm sorry, it's not I, it should be W, so under the boat. So when H is inferior to H W, so when you are in the exterior region, you have no constraint. So it means that P bar is equal to zero. But then when you saturate the constraint,
39:02
when H is equal to the depth under the boat, then the pressure is not zero, and it gives you this. So people derived many models like this, essentially with G is equal to zero, for pressurize earlier equation if you want. So you have this equation rising in traffic flow, in granular flow, in hydrodynamics in pipes.
39:20
And the idea is that you are saturated some constant, so these are called congested flow. So you can put it under this form, so it's a bit different because of the G. And you could do this also for the full equation, of course. And then you could extend this to dispersive models by adding more terms in the equation, more complicated terms.
39:41
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, for instance, the 7R equation in dimension one, you can put them in conservative form. You can use any finite volume formulation. So you have some discretization for the flux here,
40:00
you can choose any discretization. What is important is that you choose your flow model. So you have first order, second, fourth 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,
40:21
you can remove the constraints that h is equal to hw on the board. 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
40:40
without object, almost the same. So it goes very, very fast. And just to show you that I'm not lying, I just made some computation. So in this case, I have a vertical, just vertical bottom. And the idea is that I put, I allow my object to move only vertically. Just assume that you can control it like this.
41:01
I have all the computation if you want. So here, you see that you just, 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 the surface elevation. And because of this pressure term, you see that it takes exactly the form of my object here. So this is,
41:21
I'm 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 multiplier. But you haven't given inertia to the body
41:40
in this case. Yeah, wait, wait, wait, wait. This is a case of fixed body. And so this is the same. And now I draw the body. Just that you can understand what happens. So the wave comes, you see, and then there is nothing here. And part of the flow goes below and creates a wave on the right. This is what happens. So it's not like you are at the wall. Of course, it will reflect it.
42:01
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 zero. So if dt zeta is equal to zero, dx of q is zero. Of course, it varies with time and it is continuous. This is one of our boundary conditions.
42:24
Then now this is a force motion. So I put the object in force 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.
42:42
Now it's linear. But it's still continuous and it creates the wave. So it's vertical motion. And of course, the way, the most interesting case is to have this floating case. In this case, I take an object like this and I take its volumetric density and from its volumetric density and its shape,
43:02
I am able to compute the place where I should have the equilibrium. So it's above the equilibrium. I just drop it and drop it. I'm sorry. So you see, it goes down
43:20
and it will stabilize until its equilibrium point. OK, so for this one, I have the inertia, I have the... 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?
43:43
Yeah, no, this is shallow water, of course. I could do it for Boussinesq Of course, for the fully nonlinear equation, I mean, there is not even... You could try in 1D with point vortex method, maybe.
44:02
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.
44:20
And so, OK, 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 to read for another talk. So thank you very much.
44:54
I thought at one point there is no vorticity, but there is vorticity in the... No, this is irrational.
45:01
Putting vorticity, I mean, more complicated, but it's not out of reach. So is there some mechanism for generating vorticity? From this, if I generate vorticity... No, I don't think... Yeah, it's true, because here you have singularity,
45:20
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 energy, because this is... I didn't say it, but you have some conservation law
45:40
for energy. And so I think it's okay for the vorticity. I didn't check. You've restricted the motion of this. But in the bigger picture, you also had the omega and the angular. Yes.
46:04
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. Then in this case, I contributed the numerics. In this case, this point would move. So I would have this...
46:20
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 improve it. For this one, it works numerically, but the numerical analysis is okay. But I can see this moving. And then, yes, and then you can couple with the motion like this.
46:44
And also, the next step also 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, okay, so I have two regions. But for this one...
47:01
Do you understand that basically you'll be able to recover the contact line afterwards, after you solve the... For this one, yeah, you can handle the contact line. The problem in the... So you get it from the formulation
47:21
without having to propagate the contact line. Yes, that's right.