 I would like to thank the organizers for the invitation and for this very nice conference. I enjoyed all the talks so far. So what I'm going to talk about is water wave. So basically, the question is, how to understand the wave in the middle of the ocean. So you have some wave form. Sometimes you want to understand how does it evolve, what kind of typical behavior it is. So for the water in the middle of the ocean, we use this basic assumption that the air density is 0. That means we neglect the wind. And the water density is 1. And the water region is, in general, below the air region. But of course, you can overturn like this. And we assume that the water is inviscid incompressible irrotational. So inviscid incompressible are reasonable assumptions for the water in the middle of the ocean, in this large scale. Irrotational is assumption that we do have rotational waves, rotational fluids. But here we assume it's irrotational because we know that if the fluid is initially irrotational, it remains irrotational for all time. We neglect surface tension because the surface tension is responsible for the ripples. So we only care about the large wave motion. So we just neglect surface tension. And there's a gravity. So this is a basic equation. And we have seen already last week in the lecture of Professor Tataru. So this is the first equation is Euler equation. The left-hand side is acceleration v is velocity. And the right-hand side is force term. So here we assume the density of the fluid is 1. So this is really force equal to mass density times acceleration. And divergent 0, coke 0. So incompressible irrotational. And because we assume the surface tension 0, and the air density is 0. So the only pressure force is from the air is 0 on the interface. So the pressure is 0 everywhere on the interface. So now for this type of problem, there's an important sign condition that's called Taylor sign condition, which was discovered by G.I. Taylor around 1949. So basically it says that the pressure has to be increasing in the inward normal direction. Because if it's decreasing, then there's a force rip open the fluids. And then the bubble will form. Then, of course, this formulation will not hold anymore. Because here we are assuming that the domain initially is, I mean, you have new boundaries form inside the domain. So this problem is no longer well set. So this is an important assumption. So of course, water waves is a very common phenomenon. And the study of it can trade back long, long time ago. So I just give you, on the top line, some great names, Stokes, Levi-Chervita, and G.I. Taylor. And I just give you a brief review of some branches of water wave. So here a lot of work exists in the existence of solitary waves and periodic waves, starting from maybe even earlier than Levi-Chervita. Of course, we have Stokes earlier, but we have Levi-Chervita, Friedrich Hayes, and these people. And actually still more work is produced. For example, the very last person, Miles Wheeler, is now in Quran. It's, I think, two years out of PhD. And there's also a lot of numerical works. So this list is by no means complete. So about these are some important papers. So what's interesting is this problem, apparently, in the 50s, 60s, and 70s there. So these work, either numerical or, I mean, existence of solitary wave. Let me mention that actually it's still a widely open subject on the stability of solitary wave. So we have very little known about the stability. We do have this Benjamin Fair instability for the Stokes wave. That's modulation instability. And there is a dynamical system approach given by the proof by Tom Bridges and Alexander Mielke. And for solitary wave, I'm only aware of the work of Pago and the Sun on the spectral stability of small solitary waves. And Zuling has instability of large solitary waves. And of course, there are numerical works to demonstrate the stability and instability. But for time-dependent problem, in the 50s, 60s, and 70s, very little could be done. So basically, most work were in analytic class. And we have the work of Kano Nishida and of Sannikov. So the reason that the work was done only in analytic class is for the time-dependent problem is that if you really want to work on the solvive class, you have to understand what is the nature of the quasi-linearization of the equation. So whether this equation is hyperbolic type or elliptic type. Because in analytic class, it doesn't really matter. But if you want to build energy that in solvive type energy to close, you have to know what type of quasi-linearization is this. OK, so the first work in this direction was given by Nadymov for infinite depths to the water wave. And there he assumed interface flat initial velocity small. So this paper was written in Russian. Now I put an important name here, even though he didn't write a paper on this. But Nishida was a very good great Japanese mathematician. And he was visiting Paris in the late 70s. So he saw the paper of Nadymov. And he could read Russian. So he translated it into English and gave it to his student, Yoshihara. So Yoshihara, if you read the introduction of Yoshihara, he basically says that I followed the approach of Nadymov. And what he did is he could prove for finite depths to these small data. Well, post this for a short time. OK, and then we have seen the work of Walter Craig in 1985, where the main contribution is he get KDV asymptotics, deriving regularity from the water wave equation. OK, so now there is another important work is the Tom Bill, Tom Hoh and the Rowan grab. So in this work, the importance of tons of Taylor sign condition becomes apparent. So they basically say that if I just take a solution of the what, suppose we have a solution water wave equation. And if we linearize about this given solution, then the problem is linearly well opposed if this condition holds. So this shows importance of this assumption. So therefore, and we look at this and we just think whether this could be true. And we actually proved it. In the first proof was given in 97. And then I gave another proof that in 97, the papers on 2D and then 99 was 3D. So the first thing we proved is that the Taylor sign condition always holds for incompressible irrotational water wave of infant depths. As long as the interface is non-self-intercepting. So actually, in the 97 paper, I gave a precise formula, which is still very important for the work I'm going to present today. And in 97, 99, and in the 3D case, I used maximum principle. It's a more vague type of result. OK, so now the key, the important new tool we introduced in this paper was to use Riemann mapping. As explained by Professor Tataru last week, the Riemann map, so there is a Dirichlet Neumann operator in this problem, which is fully nonlinear. It's difficult to understand. In fact, we just by looking at the work of you, Shihara and Nalimov, it's quite apparent, like the derivation of quasi-linearization is messy and then only work with small data. And we realized the Riemann mapping just flattened the domain and can simplify considerably the problem. And we can understand the problem much better. So we used the Riemann mapping valuable and we derived. So the second important part of the paper is to derive the quasi-linear equation in the Riemann mapping valuable. And we realized that this equation actually is of hyperbolic type. And so here I just want to emphasize that, well, in 2D, it looks like, OK, you use Riemann mapping. So what about 3D? So in 3D, there is no Riemann mapping. So the basic idea is you just try to understand as much as possible using as much tools as possible available. So for example, we use Riemann mapping and we derive the quasi-linear equation. And then we find that actually Riemann mapping is not needed. So in 3D, we actually just find that we can actually just use Lagrangian coordinates. And actually the same proof also work for 2D. So use Lagrangian coordinates also can prove local well postness in solvable class. But this is all for S sufficiently large. OK, so let me just also mention that actually Riemann mapping, as mentioned by Professor Tataru, is really not a new tool. So it's commonly used to study potential flow, 2D potential flow. And for example, it was used in the paper of Zakharov, Dychenko, Vashiliyev, where an adhesion norm operator actually is absolute value of k. So to do the numerics, it becomes much easier, right? So you just calculate the Fourier modes as evolution by absolute value of k, and then it's much easier. And here in the work of Asanikov, he studied, he apparently the equation becomes simpler in his setting, but he still studied in the analytic class. So he didn't derive the quasi-linear equation there. So in both these previous works, quasi-linear equation was not derived. OK, so the local well postness seems to have been extended to many more complex settings. For example, you can include non-zero surface tension and finite depths. And you can also include non-zero vorticity. So let me just mention that. So far, finite depths always, the height, OK, so the bottom of the fluid and the interface do not interact. So this is the fluid, and this is bottom. So the minimum height, so the height, has a strict lower bar. So this is the bottom. So in this case, but in all these work, the Taylor-San condition do not hold automatically, and you have to assume it. So in the crystal lembrad, the energy estimate was obtained by fluids with non-zero vorticity. And for example, Ambrose-Masmoudi gave the approved for non-zero surface tension. And the W-Lion was finite depths. But importantly, it's in Euler coordinates. And we have more work, OK, so in the general setting. And more recently, we also see compressible fluids with free boundary, right? So yesterday, we have just seen the proof of Hans and his student, Chen Yunru. OK, so by now, there are also quite a lot of work on global behavior for small, smooth, and localized data. So here, in order to go to global, it's very important to have localized data. So in the 2D case, in this work, we were able to show that if the size of the data is epsilon, then the solution and the data is sufficiently smooth and localized, then you can solve the equation for the time e to the 1 over epsilon, OK? So the main idea in this work is that we find a nonlinear change of coordinates and a nonlinear change of unknown. So as explained by Professor Tataru last week, OK? So I mean, if you perform the Shata type of normal form change, and there was a loss of derivative, OK? So it depends on in which level you're looking at. So if you look at the velocity potential and interface of the loss of derivative, but if you look at the velocity and acceleration level of the equation, then there's a small divisor at origin. So anyway, the Fourier symbol of the transformation is not bounded. So you have to deal with this. So the way we dealt with there is to find a nonlinear change of unknown plus a nonlinear change of coordinates. So this new unknown in this new coordinate system actually satisfy a equation that contains no quadratic nonlinearity. So the quadratic nonlinearity can completely remove. That's why we can, and using the dispersion relation in 2D, we can prove almost global. And for this global 3D paper, we use a similar idea. So this idea, so as I said earlier, always this is our strategy, is we try to understand the 2D case as well as enough, and then we hope that we can extend it to 3D. OK, so Jermaine Massoudi-Shata gave a different approach independently for the global existence, and they also get scattering for the 3D case. So the method they use is the so-called, they call it spacetime resonance method. And now the 2D work has been extended independently by Ionescu, Pesateri, and Arasadolo, and to global, and where they also showed, both showed modified scaling. But all of them are small, smooth, and localized data. And last week, we have just seen another proof by Daniel Tataru using modified energy. And so now there, so I don't think this is, I can just make a complete list and just mention a few. So this is, Shresen Wang is a student of Ionescu and wrote a paper on 3D water wave with global and small, smooth localized data and with flat bottom. And we also seen the lecture of Daniel last week and for flat bottom in the 2D case. So for cubic lifespan. So essentially I would, so even though this work is not global, but I would group them together because the essential idea is the same that the quadratic non-linearity can be removed one way or the other. So now I mentioned this very important new work. Then Ionescu, Pasatari, and Posada, okay, they look different, is it very? Okay, it's okay. Yeah, I see, the order is correct. Else before S, right? Okay, so this is for 3D gravity capillary water wave. So apparently with both gravity and capillary, the resonance set is no longer just one point, it's really a circle. But they can actually still handle it and got global existence for localized, smooth data, small, smooth data. So here I also mentioned our recent work with Lydia Berry, Shuangmiao, and Sorab Shoshani where we studied self-gradient rotating fluids. Okay, so here the main difference is the gravity is no longer constant. The gravity is a non-linear one, which is a self-gravitating one. And still we can find a change of unknown and a change of coordinate system so that the quadratic non-linearity is just removed. There's no quadratic non-linearity in this new unknown, for this new unknown and in this new coordinate system. Okay, so there are also efforts in other directions, such as try to prove well-posedness in low-regularity class. So the reason is, of course, if you assume less on the data and on the solution, maybe you are able to get the scaling level and then you could prove global existence. Okay, so what was achieved by Aladair Burke-Zuri is that they're able to prove local well-posedness in sub-level spaces and the regularity equivalent to requiring the interface is C3 half, plus a little bit, and then they use strict house estimate to lower the regularity, just C3 half minus a little bit. So more or less it's C3 half. And last week we have seen the presentation of Daniel in the 2D case, so in a similar class, I would say. Somewhere in between, as explained by Daniel. Okay, so let me also mention there was also work of the called about Feyferman group on so-called splash singularity. So splash singularity basically says it is possible to have a non-self-intercepting interface it becomes self-intercepting at certain later times. So there exists something like this and becomes at later time become self-intercepting. So the basic idea is you just start with something like this and you find a velocity field to put open and then you use a time reversibility to prove there is one to go from here to there. Okay, so the 3D is given by Corten's Scholar. So since I'm mentioning this, so maybe I'm not going to mention later. So in fact, so remember I mentioned that we derived the quasi-linear equation for the water wave equation in the Riemann mapping variable. In fact, if you look at the quasi-linear equation, this is the quasi-linear equation about the velocity and acceleration. So you don't see the interface anymore. So in fact, this quasi-linear equation applies to self-intercepting interface as well. As long as you have a conformal mapping takes a lower half plane to a fluid. For example, this is a Riemann surface. If you can just regard it as a Riemann surface. Suppose you have a conformal mapping to take it to there and the quasi-linearized equation holds. Of course, here you are cheating in this case, right? Because in the real fluid, when the fluid intersects like this situation, the pressure is not zero here. But of course, you pretend your pressure is just continuous to be zero, then that's not physical. But theoretically, the quasi-linear equation actually do not rule out self-intercepting wave. But in order to show that, when you solve the self-similar solution, you actually don't know whether your interface is self-intercepting or not because you are solving equation about velocity and acceleration. But when you try to recover the solution of the water wave equation, you have to cut off at the time when it becomes self-intercepting. Because after that, it's no longer physical. Okay. So the real contribution of the Phefermann and the Kodoba group is to find the velocity field is really, they show there is a velocity field to pull it open. But actually, they approach, they didn't use my quasi-linear equation. Okay, so let me just, they actually just did the whole thing from scratch. Okay, so this is the picture we really want to understand today. So if you look at this, this is not, this really seems to be quite singular. So of course, it's like, it's probably not even C1. It looks like there's a kink on it. So how do we understand this picture? So if we look at this, for the first approach we want to say is, I mean, where do we start? So we try to say, maybe this thing is self-similar at the kink. At the kink, it could be self-similar. So let's see whether we can, I mean, because this is a very important method in PDE just to find self-similar solution. So in other, so the basic question is, what type of singular behavior we find in water wave and where does it come from? What are some basic structure? So the first thing we want to find is, are there any self-similar solutions? Okay, so now in order to find self-similar solution, we have to understand what kind of scaling is important here. So if we look at this wave, we think that, well, if you really look at the wave, you see this thing happens, the kink happens is due to the motion. So this means that the convection should be important. This should be driven by convection or the velocity. So this means we need to use a hyperbolic scale. So spatial valuable is scale like time valuable. So in other words, you try to plug in, you answer us look like t to some alpha f of z over t, something like this. And if you try to, if you use this answer, you have to ignite gravity, because the gravity scales differently. Time is x over t square. So it's like gravity is scale different from, so we neglect gravity and surface tension. Surface tension also has a different scaling, but still we need to assume Taylor-San condition hold, because we know Taylor-San condition is a stability condition. So in order to see the wave, this has to be true, otherwise we won't see the wave. Okay, so what we realized, first thing is by plugging the answers is that, well, if there is such a wave, the wave have to look like this, concave up on one side, and it's possible to turn a degree and then concave up again. The concave up, concavity is due to the requirement negative dpdn is non-nactive. So this is really like that. If you write down it, it's very clear that, you know that concavity is secondary to positive, okay? And another thing is that if such thing exists, this angle has to be less than pi over two and this wide angle at infinity is bigger than pi over two. Okay, so this is in order for the wave to not roll up infinitely times, okay? Okay, so the question is, well, is this anything real, like anything real like this, concave up on both sides, right? It's kind of because I'm not so sure, actually I was not so sure at that time because I don't really see waves that much in Ann Arbor. Okay, we have a river, we don't have ocean. Okay, so anyway, what we could find is something concave up on both sides. So now I'm going to show you some photo I took. This is in San Diego and to convince you what we found actually is not far away from reality, it seems. So this is, I was standing here and looking at the wave underneath, okay? So you see the waves coming in and then reflect out, okay? So the wave reflect out like this and then some wave train forms. The next thing is they click up, right? Come on, crest up and break, okay? So if we look at the crest, look like it's concave up on both sides, okay? So it's, so because we saw this photo, actually not we saw this first. Because I saw this, I went to actually find the solution because I knew the answers for a while but I was not convinced enough. This is anything meaningful to find such a thing, okay? So, okay, so now, and we also see this, this is a photo I took from internet, okay? So concave up on both sides. And now once you see, and once you realize things are concave up, actually you see it almost everywhere, okay? So this is on the river here. In Paris I was standing on the bridge and then this boat going forward and then the wave behind it, okay? So this, so you can basically say, think that the way the fluid is pushed out and outside the velocity is zero. So there's a relative motion of the velocity and then the things look like concave up, okay? So it looks like we could be happy now, okay? And this is concave up also. So, but of course we, it doesn't really mean a lot these photos because in real situation there are many factors in it. And in the equation we are thinking that we assume there's no wind, no surface tension. I mean it's quite idealized. So still this is a question to be answered what how relevant is a self-similar solution? We really want to understand the relevance of this, okay? Okay, so I'm going to change a topic for now. Just forget about self-similar. So we are going to go back to self-similar later. So for the moment, so for the moment, so I just want to change topic. So remember that, so far what has been studied is there's no interaction of the fixed boundary and the fluid. And also all the situation required as really a strict lower bound of the negative dpdn has a positive lower bound, okay? Okay, so very common of course we really want to understand how does the interface interact with fixed boundary, okay? So even if of course in the real situation if it's a sand it's very difficult to model. But even if we assume this is a smooth rigid boundary it's not easy to understand, okay? Okay, so how does the interface interact with a fixed boundary, okay? So of course on a rigid smooth boundary that's what you would have to assume v dot n equal to zero. That means the free particle only slide along the interface. Okay, so this is a difficult situation to understand. But okay, so but there's one particular situation it seems that, and this is equation that you want to write down, okay? So the same equation plus this boundary condition, okay? So there is a particular simple situation that we can understand. So this is like just imagine you have a cup, okay? If you have a cup and you have just a vertical boundary, okay? You have a vertical boundary and then you shake the fluid in the cup, right? So you shake your cup and then you see how the fluid interacts with it. There's no fluid here. Very little fluid, don't worry. So how does this interact, right? So why this is easy because you know so if you have fluid going up like this so you think about this fluid going up and like this but roughly you can also think, you know this is if you just symmetrize, do a symmetric reflection, it's equivalent to this problem, right? So you reflect the fluid, okay? And you can just remove the solid boundary, right? So these two problems as a purely equivalent. So if you have something symmetric and the free particle has to go only up and down along the interface, right? So this satisfies automatically the boundary condition. Okay, so I thought this is a quite easy problem. So I gave it to my student, right, Kenzie? Because I thought, and I waited, actually I thought about it and I thought, oh, that's a good problem for students. So I waited for a student to arrive and then gave it to him. So why don't you study this, okay? So there are two situations here, okay? So one situation is you make an angle like this. Another situation is you make an angle, just a 90 degree angle, right? So if you reflect by 90 degree and you remove and you can see that the problem is reduced to something like smooths, okay? So in fact, this situation has been studied by Alazard books really in their paper. So in fact, they also, they really just did the reflection and reduced in the 90 degrees case and basically they reduced to a smooth case, okay? So the whole domain. So boundary or no boundary, it's by reflection that you remove the boundary and they studied in the C3 half setting, okay? So I told Rafe that, well, we have to avoid the 90 degree situation, right? Because it's difficult to tell people why you're new, okay? Anything you get is anything new. So we want to understand whether it's possible to form an angle which is not 90, okay? So that's the main question. So whether this can be not 90, okay? So the question is, is it possible for the interface to interact with a wall with a non-90 degree, okay? Okay, so to put the problem easier, I just told Rafe that, okay, well, why don't you just, you know, assume there's only one angle? So you have a cup, okay? You have two sides, you can reflect on both sides. Let's assume that on one side is just a 90 degree angle and on the other side is a not 90. So just try to understand what type of angle can this be, let's assume this 90. So you reflect, so this is a graph, reflect on this side and then remove the singularity here, right? There's no singular, so there's only one angle you have to understand. So it turns out that, so the question is whether this angle can be other than 90 and we want to understand in the frame, answer the question in terms of getting a proteasimate and local existence, okay? So it turns out just by trying to understand whether it's possible to have one non-90 angle and we can actually treat very general situation, okay? So what we can answer is quite general, quite general. The answer is yes, it can be not 90, okay? So in fact, we can treat arbitrary situations right. It's not just one angle here, it can be many angles inside the free domain, okay? So this angle can be many, okay? And we also don't have to treat a fixed, two fixed boundary here, so you can have, so either you have a fixed boundary and this, you can have as many as angle as you want, okay? Or you can just treat a better way like this, okay? So if there is angle, okay? This angle is, if there is angle, this angle with this vertical boundary has to be mu, has this to be less than 90 degree, okay? If there is, it does form angle which is not 90, then it has to be less than 90. In other words, it is not possible to have angle like this, okay? So this angle is not possible, okay? Such a situation is not possible and of course we can also have angles in the interior, right? So for the interior angle, this angle has to be no more than 180, okay? So in other words, we don't see water wave like this, okay? Like this, okay? So of course that's quite obvious, right? Every kid knows if you ask them to draw, you will never draw a graph like this, okay? Okay, so everybody knows this and actually this fact is purely just determined by the water wave equation, okay? So I will explain why, okay? So in fact, this is really just by purely coming from the quasi-linear equation we derived in the 97, you can understand these simple facts, okay? Okay, so now what we did, what we have, so the main result is the following. So we are able to construct another framework. This framework is more general than just sublif class, okay? So include all the sublif class, I mean this four S big enough, less than three, okay? So four less than two maybe, okay? And we, but this framework can also include interface with angle crest, okay? So now this interface, so if and we obtain a purely estimate and we obtain the local existence in this region, which include sublif and include angle crest, okay? And what I want to really say is that this, by now we can say that by understanding, this is a more fitting framework to study water wave equation. So in other words, previously we had studied, we used the sublif class, you know, when we did it, we always feel there was something not completely fit, okay? So here it seems that this is quite fit, okay? So I have a feeling that this we got, we finally got things right now, okay? But of course things can still be better, okay? Okay, so the purely estimate was done jointly with Ralph Kinsey, my student, okay? So in other words, okay, just explanation, okay? So such a picture is possible, okay? So now, so this, so how do we understand this, okay? So I guess in this community, nobody questions. So in fact, I gave this lecture in many other community and they say, well, I mean, what you do is really just, not right because, you know, in order for wave like this to form, you need wind, okay? Chinese, there was Chinese phrase says, no wind, no wave. Okay? So how come you have this, right? But the basic fact about wave is that, you know, we know that a storm can form in very far place, okay? And once it's formed, then the wave is formulated, formed, wave forms, but this wave can travel for long distance. So the wave in front of you could be coming from far away. It doesn't have to come from the wind right above it, okay? So this could still be a situation that under a vacuum, okay? This is not caused by the wind right blowing above it, can be wind blow like far, far away, okay? And this is a way of travel. So in other words, once it forms, it stays like this and we're not, okay? So this is possible. Okay, so the main difficulty to study wave with singularity angle crest is that actually, so we can still prove negative DPDN positive, non-negative. So in fact, this fact has already been proved in 97, okay? So if you look at the formula, you do have lower bound, greater than equal to zero, okay? But the negative DPDN equal to zero at the wall where there is a non-90 degree, okay? So if you form angle like this, negative DPDN equals to zero and the negative DPDN equal to zero in all the angle crest. So negative DPDN equals to zero where there's a singularity angle crest, okay? So the main difficulty is if you allow angle crest in your wave, then negative DPDN becomes zero in these singularities, okay? Okay, so why this is a case, okay? So let's just derive briefly. Actually, I'm going to derive briefly the quasi-linear equation because it can be done pretty easily, okay? So this is the interface in Lagrangian coordinates. So we write down the Lagrangian coordinates. So taking derivative to t, you get velocity. To derivative to t, you get acceleration. And we normalize gravity equal to one, so minus i is a gravity. And the pressure zero equal to zero on interface means that the gradient P is pointing in the normal direction of the interface. So we write down negative gradient P is because z alpha, z alpha is a tangential vector, okay? z alpha and i z alpha is the normal vector, okay? So a is just a real value quantity and precisely is given by minus the Jacobian times DPDN. So has the same sign as negative DPDN. Okay, so then we can write down the Euler equation, right? So the Euler equation is acceleration equal to gravity on the right-hand side. So this is acceleration minus gravity equal to gradient P. Okay, so this is negative gradient P, okay? And now divergence co-equal to zero means that the velocity field bar is a holomorphic function, okay? Being holomorphic, you can write it in terms of the Hilbert transform, okay? So this is a direct consequence of the Cauchy integral formula. And here I will show you the difficulty of not using remap mapping, okay? So you can see that. Oh, I'm going to show you the difficulties here, okay? But so this Hilbert transform actually is written like this, okay? So difficulty is you have the unknown in the denominator. Just too nonlinear to understand this, okay? So that's why we use remap mapping in the first paper, okay? Just to understand this. But this quasi-linear derivation actually was the one version of 99. So this is after we understand it in the 2D case in 97 using remap mapping, okay? So here this derivation actually is in Lagrangian coordinates. So briefly speaking is just by taking one time, we realize, okay? So of course, I mean, this is a normal procedure of taking quasi-linearization is just, you just take derivatives to the equation, right? You get, but the question is which equation you take derivative to? Do you take time derivative or space derivative or which one, okay? So it turns out that we only need to take one time derivative of the first equation, okay? And then, so this had become two term, right? So one of the terms is higher order and one's lower order. So we just move one of the terms to the left. So the left-hand side, the higher order term, the right-hand side, the lower term. And the reason is this quantity a is real value and you use the projection i minus h to project and you get a lot of commutative. I don't go into detail. So the left-hand side shows clearly the Taylor sign, whereas it is, okay? So u is z t bar. So if z t bar is holomorphic, then i d alpha of a holomorphic function, i d alpha of a holomorphic function actually is really the normal derivative. Dirichlet Neumann operator acting on norm derivative. Okay, so this is the Dirichlet Neumann operator. So this is how it goes about. So the left-hand side is the main operator of the water wave equations like this. And we know the Dirichlet Neumann operator is a positive operator, right? And this a is the net-to-dipedian times this. So this clearly shows that if net-to-dipedian has a strictly lower bound as this is hyperbolic type, right? If net-to-dipedian's negative, it's elliptic type and it's ill-posed. So because we have shown net-to-dipedian has a strictly lower bound in the smooth case, so the problem is hyperbolic type and we were able to prove the energy estimate. So now the question is the situation we are facing is net-to-dipedian actually equal to zero sometimes. So if it's equal to zero, then we are in trouble, right? So the lower order term on the right may not be lower order anymore, okay? So the type of the equation could change, okay? So that's the difficulty, okay? So okay, so I already explained that we really need to use Riemann mapping. So this was actually still the derivation of 97, so we actually use the Riemann mapping. The reason of using Riemann mapping is here, okay? So we just rewrite everything in terms of Riemann mapping variable, okay? So it takes the Riemann mapping from the fluid domain to the lower half plane, okay? And now we just rewrite everything in terms of Riemann mapping variables of compulsive with H inverse acceleration compulsive. So capital, the sine capital sine means it's in the Riemann mapping variable, okay? So everything is the same, right? So this quantity capital A is just, we have this Jacobian that's come from Chandler, right? So the first equation basically is of the same type except in the, you just do the Chandler, okay, here. What's really make a difference is the second question. So the velocity field, okay? Now velocity field, so the velocity field after composed with the Riemann mapping becomes a holomorphic function in the lower half plane which has a boundary value as CT bar, okay? So to characterize a holomorphic function in the lower half of plane then we only need to use this flat Hilbert transform which is much more simpler and it's linear, okay? So just using this, so let me just keep the notation that, okay, so capital Z is phi inverse and Z bar is this, okay? So now we two, this was still the early work in order to show the Taylor-San condition pose we actually just multiply the equation to both sides by Z alpha bar and we get A1 and we derive the formula for A1 and we show that A1 actually have a lower bound. So this is a precise formula and this formula just implies Nectar DPDN is non-nectar, right? Because we can write down Nectar DPDN is A1 divided by this is greater than equal to zero, okay? So A1, this quantity A1 has a lower bound which is one, okay, that's very important. So, okay, so if the interface is smooth then this is a Jacobian of the Riemann mapping, right? The Jacobian of Riemann mapping have upper and lower bound so obviously it has a strictly positive lower bound and that was the case dealt with in 97. So now you have, this guy is actually has singularity so that means this bound may not hold, okay? I think I, did I skip some transparency? Anyway, that's okay, I thought I have a simple argument. Okay, anyway, I don't know, that's okay, actually I don't have time now, right? So, okay, so this is a quasi-linear equation. I just give you an idea how it looks, okay? So let me just, so since, I mean, you have seen the lecture of Daniel last week and they look quite different from his. Actually, this is, the difference, main difference is really the notation. So if you, I mean, this equation, very, his equation or this equation, I mean, overlap in a large extent, okay? Same structure. Same equations. But the same very, but it has a very different meaning in your notation as in our notation. Okay, so, yeah, so this B, particularly this B is the same B, okay? So, so, so, I mean, that's not surprising, right? It's just really a matter of how to use this equation. I mean, we use, we have, I mean, because we're dealing with the same equation and of course we have, and we use the same variable. If we get different equations then, I mean, it's scary for the young people, okay? If equations different, that would be pretty bad. Okay, so this is a very important equation as a consequence of this derivation, okay? So, ZTT plus I, Z alpha bar is I, A1, so we just use it as like you just divide over, okay? So this is the same equation as this one, okay? So that's the same equation, okay? So A1 has a lower bound greater than equal to one. So in other words, one over Z alpha is similar to ZTT minus I, because A1 is a low order term. Ah, this is a transparency. So, so here I give you an argument which is very early on given by my student, Rafe-Kinsey, more formal student, okay? So, so he said very quickly that negative dp, dn equals to zero, we are in trouble, okay? So why negative dp, dn equals to zero? So this equation, the, the, all your question, if you write down in terms of components, you can rearrange like this. And tangent of mu, okay? So this nu is this angle, okay? This angle, this angle nu. So of course this is your tangent vector x alpha, y alpha, right, tangent nu is given by this. So now we know the fluid particle only move up and down, okay? So xt equals to zero, okay? So of course xtt is also equal to zero along this line, okay? So this means the denominator is always zero. Now in order for mu to be not equal to 90, then ytt plus one has to be zero, right? Because otherwise, nu is pi over two. Okay, so this means that grade in p has to be zero because now xtt plus i is zero. So grade in p has to be zero. So grade in p has to be zero, that means n dot grade in p equals zero. So in other words, a zero at the corner. Remember a is, a is like one over c of times n dot grade in p, okay? So this is like n dot grade in p equals zero, okay? Okay, so this is a very simple argument. Now why, so I give you also a derivation why this angle has to be less than pi over two. So the reason it is less than pi over two is a following. So you just look at the Riemann mapping, okay? So it's very simple. You look at the Riemann mapping. Can I borrow a few minutes? I will only need five minutes, five minutes. Too much? I started late. I started late. Okay. So if you look at the Riemann mapping, okay? So this is like, you know, this is, so this is your angle mu, right? This is phi, okay? So you take Riemann mapping from here to here. So in order to take the angle mu to pi over two, you need at this point a power law to open up, okay? So this is a power law, it says that, okay? It's given by this relation. So if you take derivative, the Riemann mapping, z alpha prime is like alpha prime R minus one. So if this angle is bigger than pi over two, then this R has to be bigger than one, that means z alpha has to be zero at the corner. And z alpha has to be zero at the corner, means dt has to be infinity, right? If you look at this formula, because A1 is greater than equal to one, okay? So dt is infinity. So this means y dt has to be infinity at the corner, because x dt, we already know is zero, right? Because x dt is zero around this line, okay? Okay, so, but remember that was the formula for the tangent of the angle. So this means that tangent of mu has to be infinity, right? Because the denominator of zero at the top is infinity. So at the end, if you start with anything bigger than pi over two, it has to be pi over two, okay? So you cannot have anything bigger than pi over two. So this is purely determined by this relation, which is the Euler equation, okay? Okay, so let me just show you how we construct energy function, okay? So we use this weighted derivative, okay? So still we use this so-called quasi-linear equation, even though in this case it's not very clear it's still hyperbolic type, and even, okay? Because A could be zero somewhere. But we still use it, and we use a weighted derivative, okay? And now, of course, there is an obvious energy coming from the left-hand side. I mean, this is elementary, but this doesn't work because this becomes infinity if you want to deal with anything singular. Okay, so what ends up is we use these two weighted energy, and you take one derivative to apply to this one, two derivative apply to this one. So it's difficult to see from here, so let me just explain what are we doing in this energy. So this energy is equivalent to this one, okay? So this is a quantity. So basic quantity is the acceleration on the surface taking derivatives in L2, okay? So now if you want to take further derivatives, you have to put appropriate weights, okay? So what is capital D alpha? A capital D alpha is one over the alpha, the alpha, okay? So now two derivatives, one over the alpha, okay? And the T alpha. So now this guy is basically in L2. Now if you want one more derivative to be in L2 because if it's singular, you take derivative become more singular. But remember that in the angle crest, precisely one over the alpha is zero, right? So if there's a singularity, one over the weight is zero, okay? So this means in order for the derivative to be in L2, you have to put two weights to weight it down. So two zero to weight it down, and this is stealing L2. So overall, so overall you get this energy, okay? So this is weighted energy, and I just want to emphasize everything here has a derivative. So in other words, we don't really require the surface itself is in L2, okay? So it's like, you know, the behavior at infinity can be wired, okay? The velocity at infinity could be wired. Only the derivative is in L2, okay? Everything has a derivative. So things can be quite big at infinity here, okay? Okay, so I don't see the talk about difficulty. So why is this energy works, okay? So anyway, this energy works, but what makes us believe this works because you know the real derivation is long, right? You have to really have a confidence, otherwise you won't be able to continue. So the real reason is really I just go back, okay? The self-similar solution, I plug in the self-simulation, I find this finite. So I say it must work because it took us a while to find the right energy, okay? What's the right weight, the self-similar? And I also want to emphasize that when I say this energy allows angle crest, this angle, interior angle in fact, cannot be all the way about less than 180. In the energy, this angle has to be less than 90 degrees, same as the self-similar solution, okay? So in other words, the stokes wave of maximum height do not have finite energy because the stokes wave's interior angle's 120 degree, okay? So what we proved is first we have a poly-esimate and then we have the local existence in this framework. So assuming the data satisfy this new energy's finite, then there is a time, depend only on this new energy so that you can solve for this finite time and this energy remains finite. So let me just remark that actually, so this was observed first by Ralph Kinsey using a heuristic argument and Cetan gave my current student here, he's here, Cetan, gave a rigorous proof as they showed that if initially you have an interface like this piecewise smooth and contain some angle, if initially you have something like this, then the angle actually do not change for the time period the solution exists, okay? Okay, so in order to prove the existence, we use approximation arguments using the previous result, okay? But this is a crucial, this is a crucial criteria we need because when we use the previous results, all the existing time depend on the solverif norm, okay? So we need a time do not depend on solverif norm but only depend on this new energy. So this theorem says that for any given smooth data, okay, you can solve for finite time, okay? And now if we let t start to be the maximum existing time, then either it's infinity, then we are all happy or if it's finite, then the interface, this energy has to blow up, okay? So when the singularity happens at finite time, then this energy, so this give us a blow up criteria in other words, a criteria to find some singularities, okay? So I, thank you very much. One comment and one question. So one comment on the difference of notations, just to clarify things. So the meaning of the tangeribbative and the notations that you are using and the tangeribbative and the notations that we're using are different. In the notations that you're using, the tangeribbative is the Lagrange tangeribbative. That's true. So it does not commute with the alpha derivative. That's true. So the notations that we're using, the tangeribbative and the alpha derivative are in the same frame, so they commute. My question, so. No, actually we use the same notation because why? Because actually, because I go too fast. Okay. Went too fast. Let me just answer. You see, this is a quasi-linear equation. So this is really the Lagrange and, this is Lagrange material derivative. So you do, you also have dt plus bd alpha, right? Yeah. So you have two first order equations and we have one second order equation. So in other words, it's really that, so I checked because I go back after your lecture. I checked that your w is essentially z of minus one and your r is ct bar. Okay. So these are the two quantities, your evolution. And if you like, you can say that we are using, so this is Daniel Tartaru and Mihalyer Ivry. And what we are using is, you can say, because it's second order zt and ztt, okay, ztt, right? So this capital ztt is dt plus bd alpha, zt bar, okay? So why I'm saying that these two, to a large extent, is equivalent. So again, we go back to this magic formula here. One over z alpha is in terms of regularities like ztt, okay? So you see, there, no, of course I mean, we cannot do different things, right? Because that's one of us, I said it would be wrong. And I saw some questions, so the questions are following. So in these solutions that you obtained, do you obtain solutions at all possible angles, less than 180 degrees? No, no, no. That's a very good question. So in here, the energy only allowed here is less than 90 degrees, okay, it has. So this is consistent with self-seminar solution. Yes, but this question, I think, can you obtain any of this angle process in a solution? Yes? For self-seminar, yes, yeah. No, no, no, for that level of solution, can you have solutions which evolve at any possible angle less than 180 degree in this approach? No, I say that the energy only allowed this angle to be less than 90. And so the statement, okay, let me show you the statement. So somehow you can take initial data that has many of these angles if they are less than 90, then you can solve for small time. That's true, no, no. Of course, so in fact, let me clarify. So of course, if you have data smooth, you can solve. No problem, that fits in the previous work. So what's new is you can also take data like this, but not arbitrary. So you need this finite. So in order for this finite, it allows interface with angle less than 90. So you allow this, not every interface with less than 90 satisfy this finite, but the opposite, okay. So their interface with less than 90 degree satisfy this energy to be in finite and then you can solve for finite time. And as shown by C. Tom, that actually, this angle do not change during the time the solution exists. Maybe, no, okay, I have one question, in fact. You are passing time. But my question, just yes or no. Okay, okay, okay, okay. I'm counting seconds. You say your self-similar solution was in the energy space. No, no, no, no, no, no. Okay, so first, okay. Oh, oh, oh, oh, oh, yes, yes, yes. Oh, good question. No, I wish, which energy, yes, the answer, yes, yes. That's a good question, okay. No, no, if you plug in the self-similar solution here, the energy is finite. Yes, but not this energy. This energy, Hamiltonian. Hamiltonian, no. No, everything has. Because we, in fact, here, I emphasize that we do not require Z alpha in L infinity. So this is equivalent to, Nectadepedian has a lower bound, positive. And we also do not require one over Z alpha in L infinity. So we only require one point to be in bounded. Yeah, yeah, question, yeah? No, no, but private question, yes, I think. Okay, okay, I think it's,