 very much for the invitation. Can you see the text? Okay, very good. Can't make it bigger. So, right. Okay, so I will talk about the motion of a liquid body which could be a small water drop or something big like a galaxy. But it's a boundary. It's a bounded domain. And the idea is that it's a without surface tension or gravity for simplicity. So we from the beginning time we have some water drop here and each point have the velocity and the water drop is going to move but also change shape. So the boundary condition here is that the boundary moves with velocity vector field at the boundary and that the pressure is zero outside. And this defines a boundary, free boundary problem. So loosely speaking the velocity tells the boundary where to move but the boundary is the zero set of the pressure and the pressure determines the acceleration. So as we will see the regularity, I mean I mean the people that know about this problem, but regularity of the boundary is very important here. That's the most important thing. Okay, so we have Euler's compressible equations. We have a density and this is Euler's equation and then the continuity equation. And I use the notation here that you sum over repeated indices. Okay, so the incompressible case which people have talked about mostly so far is when the density is constant. And in the compressible case you assume that the pressure is some increasing function of the density. And then the two cases, there is the case of a liquid which just means that the pressure is, well the pressure is zero at the boundary because it's zero outside, but the density is a positive constant on the boundary. And the gas is the case when the density is zero at the boundary. And as I said the boundary condition is that the velocity vector field is tangential to the boundary and that the pressure is zero on the boundary. An initial condition is that you give the domain and you give the velocity and density. Now for the compressible case you're going to have lots of compatibility condition. If you want the solution in HR you need our compatibility condition. This is because we will see it becomes a wave equation with a zero boundary condition and such a wave equation needs compatibility conditions to be solvable in solvable spaces. Okay, so this is just a bit of history. From local existence was first known for analytic data. And then the question was if there was instability in a Ray-Li-Teller instability in solvable of norms. And then you need a physical condition to assume that there is no instability because otherwise there is a counter example of David Abin. But also C.U.W. showed us that in general in the irritation in the incompressible irritation on case this physical condition is always satisfied. And but in general you have to assume it and you assume it at time zero and you hope that well by continuity will be true for some small time so at least you have local existence. But one possibility of it breaking down is that this becomes zero. Okay, so then in incompressible case I did some work and other people did work also for local existence. And for the compressible case of a liquid I proved local existence assuming this physical condition and then for a gas I proved them a priori bounds and then also a Coat and small improve local existence. Now there's a much longer story now which I can't even fit the rest of the talk. Many people have talked about different ways of proving local existence now in particular in the incompressible irritation on case. It seems like there are as many ways of proving it as there are people working on it now. So I can't mention all and also of course there is a case of global existence now that many people in the audience have worked on that I apologize for not mentioning so now. Okay so this is a little bit about the intuition here. So in the irrotational incompressible case then if you take the divergence of the equation the top equation you get using that the divergence of the velocity serial it becomes an elliptic equation for the pressure. So Laplace P should be zero and P is zero at the boundary. Now by strong maximum principle the normal derivative of P is negative so the physical condition is satisfied and the equation was for what wave problem if it would be ill post when the wave turns over and this is what C of O told us that it's not. Now just to show you how why the boundary is important here. Well the whole problem is the boundary. If in this case if it wasn't for the boundary you could ask the invert this equation and one derivative of P would be like with the velocity or two derivative P would be like one derivative of the velocity. So one derivative of P would be like the velocity so the right hand side would just be like the velocity. So this just becomes an ordinary differential equation in function spaces if it wasn't for the boundary. So this shows that the whole problem is with the regularity of the boundary. Okay so you introduce to try to deal with this it's easier at least for me to introduce Lagrangian coordinates in which the boundary becomes fixed. So you just follow the flow lines of the velocity vector field. They add that in T and it straightens out in these variables T and Y. And everything looks nice. The material derivative that occurred in Euler's equations just becomes DDT partial derivative with respect to T in the Lagrangian coordinates. However this what before was a partial derivative with respect to X has more complicated expressions. So you're trying to somehow stay on both sides to try to use the simple expression for DDX on this side and simple expression for DDT on this side. Okay so Euler's equation then becomes just that the material derivative of the velocity is the gradient of the pressure and this is the continuity equation. And then there is a relation between the acobia and determinant of the change of variable and the density. They're in proportion proportional like this apart from the initial condition. Okay so for this problem you can see that you have an energy conservation. Now Daniel told us that we should look at the energies on the boundary but I'm doing the opposite. I'm looking at the energies in the interior. And for the compressible case and also when the curl is non-varnishing it's coming in to look at the energies in the interior. So you can see that there is an exact energy conservation. Q is formed from the pressure is a function of the density. So you have to take this anti-derivative and perform Q. And you use that the domain in the y-variables becomes fixed. And the change of variable satisfies that the T of real times the change of variable is zero. This just follows actually from... Okay when you take the time derivative of the determinant it becomes the trace of the derivative. So it becomes just the time derivative of the determinant just becomes the divergence of the velocity. This is a well-known formula for taking the derivative of the determinant. Okay so when you take the derivative of energy it's convenient to change them to the fixed coordinates. And you get that the time derivative commute in the fixed coordinates. So for the first time here you use Euler's equation. The Tv is equal to minus dIp. And for the second you use that... Well you use this formula for Q. And then you use that the continuity equation that the T rho is equal to divergence of P. Well you use it in the next step actually. Here you integrate by parts now. So you have a gradient of the pressure. You integrate by parts. On the one hand it can fall on the velocity. And then there's a relation then between the divergence of the velocity and the T of the density. So that's cancelled. On the other hand you get the boundary term. And the boundary condition that the pressure is here on the boundary can be understood as a physical condition that energy should be conserved. Because this is a physical energy, right? So in order for your energy to be conserved you have to have that the pressure is here at the boundary. Okay. So then in the incompressible case of incompressible liquid together with crystal lulu we prove them energy bounds. And the interesting thing is that when you go up to higher order energies you have to add a boundary term. And this is theta here. It's the second fundamental form of the boundary. And you get an energy estimate like that which gives a local energy bound. So the energy depends on this physical condition but it also depends on some like first order infinity bounds for the first order derivative of the velocity and of the second fundamental form. And then with Cotan's Scholar we looked at the compressible gas. But then you have to have energies that vanishes completely in the interior. And you have this row which vanishes at the boundary, the density. So the norms actually most of them vanish here at the boundary. And we prove the similar energy bound. But what I want to talk about today is some energy bound for the compressible liquid with chendul lulu. And so what we prove is that higher order energy bounds. I had proven existence before using Nash-Möse but it was sort of not at all explicit what the energy bounds were. It was existing in some big sub-level space. So they look like this. So you have the velocity as like before those terms were here in the in compressible case but you have a new term here. In particular you have a bit higher order one more time derivative extra of the density. You have a wave equation for the density in addition that you get an estimate for. In the previous case this was of course constant. So you get an energy bound but moreover our energy bounds are uniform. So we prove that the incompressible limit exists. So this is done in the following way. Instead of thinking of the pressure as a function of the density you could think of the density as a function of a pressure. And then the incompressible limit is when this function is converges to a constant function. And then we show that the solution to the corresponding problem converges to a solution of the case when divergence is zero. If initial data satisfied the divergence of the velocity initially is zero. And also there has to be some compatibility conditions satisfied. And then the question is I mean what we wanted to do was of course prove long-time existing for a free boundary for compressible Euler. And the hope is that now if we can prove that the incompressible limit exists it should be close to the incompressible case for which we know that we have long-time existence. But in freedom mention anyway there's a blow up Fritz John. Tom Sider is using the methods of Fritz John prove them a blow up for all small lay for most there's a condition of some integral being positive. So you can't hope to get global existence like in the incompressible case. So a question. So you expect rho and phi to have the same regularity right? Rho and p. Well this has more derivatives one more time derivative actually. In space they have the same regularity. But they have one more time. Yeah that's what we can prove. So now to prove this it's natural to reformulate the equation. So it looks like Euler's incompressible equation and you could do that and you can absorb this rho in p by introducing h prime of rho to be phi prime of rho. So this is this called the enthalpy. So it just looks like Euler's equation now. In incompressible equation the only difference is that well you have this equation the t E of h is divergence of b. And this is the t of the coordinate of the velocity where E is now some given smooth function such that the E is positive on the boundary where h is 0 and it increases with h. And also we have those conditions have to be satisfied. So we now want to pause to limit when E goes to becomes a constant E prime goes to 0. So we have to have and our estimate holds under these conditions which allows us to pause to the limit but it doesn't allow us to deal with the case of a gas where E is 0 at the boundary to begin with. And it shouldn't really because these norms shouldn't work for that. You should have interior norm for that. Okay so you assume a physical condition. Now what you get here if you take one more if you take divergence of the first equation and take the time derivative of the second you get the wave equation for enthalpy which is similar. So you just see that if you commute with the t, commute divergence here you get divergence of b. So the t of divergence of b which is the t squared of E of h and then there is the commentator coming from this which is this term. So this is similar then to the wave equation for the pressure for the incompressible case. Only difference is this term here that in the limit we hope will go to 0. Okay so the energies for Euler's equation look like this. The main thing is the energy for Euler's equation itself that looks like this. I will soon describe what this is. There is also two other terms. There's the energy for the wave equation down here that you also have to have control over and there's the energy for curl which has a better equation so it's easier to control. Now you may ask why do I need curl when I have all these things up here? Well it turns out this this Q here is something which is a positive definite quadratic form but it doesn't contain all components at the boundary. It only contains tangential components at the boundary. So this norm here is the metric in all components minus the normal components. So just a tangential it's a projection onto this at the boundary this is projection onto the tangential component of this tensor projection of this derivative partial s to the boundary and this is necessary for the energy estimate to hold to do this. At least if you do it like this otherwise the energy estimate wouldn't hold. And that's why you need an exercise made for curl to help you as well. Okay, so the energy we bound to prove is something like this. Time derivative of the energy is bounded by function times the energy. So it's linear in the highest order but it depends also on the previous energy. But at least inductively you can get bounced and for all energy as long as all those quantities are bounded. In particular the normal derivative has to be positive, has to be bounced for the second fundamental form and first of derivatives of velocity and some derivatives of the enthalpy also. Now these derivatives you can then get back afterwards because you have the energy doesn't contain bounds for all norm but you can then afterwards bound all norms with the energy using elliptic estimates. So then you can by several of lemma bound all these quantities as well. So you get local existence. So again can you say one word about this W, what is W, how to do? Yeah, this is the energy for the wave equation. And you prove that separately and it's actually easier to prove because this highest order energy you can see directly that you don't have problem with boundary condition because when you integrate. So H is like rho. H is like rho, yeah. So instead of having H to be a function of rho. I now think of rho as a function of H. Because I want to make it look more like incompressible, Euler's equations. And this energy actually, so you have a wave equation, the T squared E of H minus Laplacian of H equal to some right hand side F. And well you can commute with time derivatives, but if you don't commute with time derivative, well this is essentially E prime H times the T squared H. And you just multiply with the T of H on both sides. And in this order, because well the first time in integrating space, right? So the first time it just becomes d dt, d dt of integral E prime H times dt H squared over two. And the other term you integrate by parts, but here the T H is here at the boundary. So you don't have any problem with boundary conditions. So you get gradient H squared dx. This is just well equal to integral F dt H and then you use Cauchy spots as usual. So the energy estimate is actually simple to obtain for just time derivatives. But then you have to transfer it to space derivatives and that's why you can get, you can get one more time derivative for the energy. Okay, let's see. But what, and the estimate for curl is almost trivial because we know that when you take curl, the gradient of the enthalpy and the right disappears. So this is, there's no problem bounding this. It's very easy to bound it. So the main problem is this term and all the terms here comes with exactly the right constant for it to cancel. If you want, you could think about actually, well, so in the incompressible case, this middle term will disappear. So if you like to think of a simplified case, you can look at that case. And also you don't need time derivatives in the incompressible case. You can just put s equal to r here. So if you want to think about the simple case, you can think about that when we go over the proof, which I want it to do now, okay. So this is going to be a little bit sketchy, I probably should write up a little bit more. So we had, yeah, so we had this energy and what I've done here is just let the time derivative fall on each of these terms, right? Neglecting all sorts of lower, all the terms when they fall on the coefficients in q and the measure because it can be controlled. So here you have these three terms and to make it simple, you can skip this line and you can think of k as being zero if you want. Okay, so now you want to use Euler's equation, but you also want to use that the commutator between dt and partial has this form. And the fact of the matter is that this partial is not, it's actually, the way to think about it is that this is actually dx, dy, right? So partial i is the y a, the xi times dt, y a. So it actually contains the coordinates. And so it's a main term actually when the derivative, the commutator has a main term. Okay, let's see. So in the first term here, it's actually not the main term, the commutator. This here says that the dt commutes and you use the equation, the t of the velocity is minus the gradient of h. That's the first term, you replace it by that. And then the next term is what we use on the boundary. The t, okay, again we think of k is zero. And here the main term is actually the commutator, so over here. So it becomes this, right? The t of, when it commutes with one of partial derivatives, it becomes dv. And then all the other derivative, the main term is when all other derivative falls on the v as well. So the main term in the commutator is this. Plus when the dt has commuted. And then this is for the interior term. So we can skip the interior term for now. The first term here, we want to do something similar to what we did before, namely integrate by parts and use the divergence in zero. So now when you take the time derivative, this becomes the gradient of h. And then you, I skipped a few steps here, but it becomes the gradient of h. And then you integrate by parts and the gradient of h falls on the velocity instead. Just like it did before, for the energy. And so here you get divergence of v, which is the same as the t of h. So it corresponds to this term. It cancels this term. That was the same as before. But in addition, you get the boundary term here when you integrate by parts. And so you have, I should write this out. So what you had was the top term we're looking at. So q of, and let's say r of k, 0, 2, d, i, v times dt, dr, v, j, dx. That's the term. And what you did was that you commuted. And then it becomes q of dr, v, dt. You used this equation here. No, this equation, yes. So this just becomes dr, dt, vj. But the dt, vj was the dj of h. So that's the first step. And now you integrate by parts of the domain. And you get two terms. One is minus when dj falls here, or there's a delta ia also. So that's q of di, dr, vi, dr, h, dx, plus a boundary term. These are all interior terms. And the boundary term is the normal component of v. So it becomes niq of partial r, vi, times partial r, h, ds. That should be it, yes. Let's see if I have that. Yeah, this should be this term. OK, so this is a boundary term. Now I have another boundary term up here. And how does that look like? Well, I have to replace it with this. So that boundary term will look like the integral of q of partial r, h, dt, partial r, h, mu, rho, ds. And here I use the commentator. So I replace it with this one. So it becomes minus dr, v, j, times dj, h. OK. So now the idea is that those two terms should cancel each other. What a little that I may not have gotten the signs right. So this term should cancel with this term. And the reason for this is that this dj, h is normal at the boundary, because h vanishes at the boundary. So this goes in the normal direction times the normal derivative of h. So this is the normal component of this tensor, which is the same tensor I have here, the normal component. So this was important that those cancel each other. And that's how the energy was constructed, in fact, so that they would cancel each other. There's one more thing that is important. And that's why we have this q here. There is a term which is low order, but it requires some work. There's an additional term at the boundary that looks like q of partial r, h, times partial r dt, h dx. Yes. Now, we do have, I said before that we have an additional interior estimate for dt, h. But we would have problem estimating this actually, the norm of this at the boundary. So this is where this projection comes in. And this is what I will end talking about now. So the point is that this q makes it so that at the boundary, it's nothing but the norm of the tangential component or projection of this tensor to the boundary. Now, how do you project, what does that mean? Well, so I should write that up too. So what is the projection of a tensor to the boundary? Well, so this should be delta i, j. Yeah, it's that you take away the normal component. So a tangential derivative, for instance, applied to q, what does that mean? It means that you take pi i, j, nabla, j applied to q. So it differentiates and it's a projection of this derivative to the boundary, right? And then you know that then there's a formula. You want to calculate nabla i, nabla j applied to q and then project i, i prime, pi i, j, j prime. And what you want to do is you want to sneak this projection in here. If you can sneak the projection in here, it becomes, yes, the tangential derivatives. But you make a mistake, which is exactly nabla i, nabla i pi j, j prime, applied to nabla j, applied to q. And the mistake you make is exactly this when the derivative falls on the normal. And this is the second fundamental form. So this basic calculation, what it becomes is this. If you have a second derivative and you project both indices, it's the same as first projecting plus something involving the second fundamental form and the normal derivative. So when you look at this term here, it looks bad, but it's actually, because the th is here at the boundary, it's actually just like order r minus one in the th at the boundary. So you have, and more of, so that's one thing. Moreover, in the energy, so that's for estimating the error terms, but moreover in the energy, you had this term in the original energy. And this is now equal to, well, the highest order is equal to this. This is the highest order term actually, this is a lower order, because the boundary is the most important thing. So that's actually equal to the norm of the tangential derivatives, r minus two of the second fundamental form squared at the boundary. So the energy gives us estimate for the second fundamental form in this way. And that allows us then to control the error terms. They're not dangerous, right? Because they're going to have exactly the same second fundamental form in front of it. So, yeah, basically this is the technique. The problem was that it doesn't work to prove existence in this norm, so you have to do it in that, because you're basically using Eulerian coordinates. These derivatives here are ds is the Eulerian derivatives. And you sort of have to fix yourself in the Lagrangian frame to prove existence. But the energy estimates works out nicely this way in all cases. Could you motivate us a little bit? Why do you construct your energy in this way? You're involving two? Why? Yeah, why did we know? So this started with Dmitry and I did it, yes. Where did it come from? Well, it came just from the knowledge of this formula here, right? Because it came actually from that you wanted this pressure term to cancel, right? This term, you had to get it to cancel, because this was higher order. And there was no way of doing it if we couldn't somehow get this to cancel. And it sort of came from the knowledge of this formula. And the understanding, I guess, that the regularity of the boundary is the important thing. So nothing like you cannot express q by something else? No, I can't do that. I don't know any interpretation. The first two terms has an interpretation. These terms come from the energy, right? Yes, the generalization of the energy, the first two terms. The physical energy. Because if R is zero, it reduces to exactly the physical energy, the first two terms. And you just plug in the derivatives? And then you take the derivative and when you say that in order for it to cancel, you have to add this boundary term. Because before when you integrated by part, this term didn't happen because some other pressure was zero at the boundary. But now because you have derivatives here, it's no longer zero. So you realize you have to add a boundary term. I guess that's the way it works. So it's more or less like a gas? Yeah, it's a gas. So without the first two and then you realize you have to have the same term? Work. And then the next step is to add in this queue because you realize that this term has to cancel. But it would have been nice if there was a physical interpretation. I think maybe Shathau, some other people, some students of David are being tried to interpret it as curvature on this Arnold space of eodesics where you think of Euler's equation as an action of some minimizing some action. And I think they could interpret this term maybe as some sort of curvature in this infinite dimensional space. I think there might be some sort of interpretation like that. Does this have anything to do with the energy you and me use in the geometry? This is the metering, the one with the metering. But without the middle term, yeah. Okay, so the difference between this and the one with the metering is you add an extra middle term. Add a middle term and also time derivatives because we only have space. And also you have to add the equation for them. You have to add a double, okay. You have to add, well, it was previous but you had to add a wave equation also. The estimate for the wave equation and you had to add the time derivative up to the highest order actually for this to work and you had to add in the compressible case a middle term also. It wasn't too easy actually to get it to work. You actually have to get everything to work together. It was quite delicate actually. But I guess it was that you realized we can have one or more of the enthalpy. You can allow yourself one more time derivative. That's what made it work. But it was much more complicated. And also these estimates are uniform now and so you can pass to the incompressible limit which should be interesting to study what happens with long-time existence in the compressible case. Slightly compressible. Yeah. So maybe you mentioned something about that. So what happens when you look at the gas rather than the fluid? When you look at the gas, then you have completely interior energies. The energies we had one can probably do a little bit better about these ones. So there's no boundary term in that case. And instead you have these functions that gross like vanishes at the boundary. So there are certain powers of the distance to the boundary essentially. And these are the norms you have to use in that case. So you are saying that you also have a paper with gas. Yeah. So then you don't have a boundary term first of all because all these energies has to... Well, the energy we used... I suspect you can do a little bit more similar to the energies I used today but the energy we used was this. It's the same in the sense that they involved a lot of time derivative. Somehow it's convenient to take time derivative up to highest order when you study this compressible case. That's one thing that they have in common. And you certainly need those interior integrals, not just boundary integrals. So I'm more about the compressible incompressible limit. So what can you say about that? Well, so what we say is that we first reformulate it as that we think of the density being a function of the enthalpy or the pressure. And the density is, I guess, rho zero at the boundary. I guess it has your domain. So it's rho zero here and then I guess positive increasing in the interior. And the incompressible limit would be something like rho zero, I guess plus rho over k or something like that. Something so that this converges to just rho zero. So k goes to infinity here. So k is like one over the number. Okay, if that's what it's called. Right, so then what you can show is that you have to, for the compressible case, you have to have initial data compatible with that the limit exists. So you have to have the divergence of the initial data. Vs here has to be zero, for instance. So that has to happen. And you also have to have, prepare your initial data properly. And then you can show that the limit of the incompressible solution, the limit of the compressible solution becomes a solution to the incompressible equation. So for your estimate, you need the condition of the pressure at some point, right? Yes, the normal derivative of the pressure. Yes. So, but that will survive, right? That condition is this one. I'm not sure that's your question. Because this will require a time of existence that depends on k. Well, we have a uniform time of existence independent of k. And this is just by continuity. This is true times zero. And this will be true for some small term time assuming that the norms don't behave badly, right? As long as they are bounded. This at time t, right? It's equal to this at time zero. Yeah, but you need... Plus you need a bound for the time derivative, right? A local bound for the time derivative. But this we have. We have, I mean, we have uniform bounds for all sorts of things. Paper will come out soon, so you can check.