 Okay, so as you see, I cheated a little bit and I just wrote some things that were on the board last time, so nothing new in there. You don't need to copy those, but I will use them for reference during this last part. So the thing that I want to start discussing right now is the problem of obtaining lifespan estimates for small data. So that's a problem, lifespan. So in a nutshell, the question that you want to ask here is you want to assume that your initial data is small, and the question that you ask is what is the lifespan of the solutions, t, which perhaps will depend on epsilon, or at least the best estimate that you can give for the lifespan of the solutions. Now here when I'm talking about the lifespan of the solutions, one doesn't really track the time up to which the solutions actually blow up, but what you try to track is the time up to which the solution still retains this initial size of epsilon. You don't go into regimes where the solutions actually grow. So at least we haven't tried to do that, and we don't know how to do that for these problems. And I'll begin with a few very simple theoristics. So let me take two very simple model problems. So suppose I start with the linear equation, and now I put in here some quadratic perturbation, and let's say I know how to do energy estimates for this problem. So if my energy is like some norm of u squared, then the energy estimates for this problem would look like the time derivative of the energy will be smaller than the norm of u times the energy. And you look at this inequality, imagine you can do some grown wall. And so if you're looking for the time up to which your solution stays of size epsilon, assuming that your initial data u0, let's say, is of size epsilon, then you will see that you get no contribution from the grown wall term up to a time. So your grown wall estimate will say that e of t is smaller than e of 0, e to the power t. And here you'll put norm of u, which is epsilon. And so if this guy is epsilon square, then you get that this guy is also epsilon square provided that your time is small enough. And small enough for this time means that time needs to be smaller than 1 over epsilon. And so 1 over epsilon, you might call it quadratic lifespan. It's the standard lifespan that's caused by a generic nonlinearity, which is quadratic. So let me write this down. And then the second model, I will do the same. But now instead of taking a quadratic nonlinearity, I'll take a cubic nonlinearity. And if you have something like this, then your energy estimate will look like d dt of e is smaller than norm of u square, because now you have three us in here times the energy. You apply grown wall, you conclude that e of t is smaller than e of 0 times e to the power time t times this coefficient, which I want it to be of size epsilon square. And then I can propagate an epsilon square bound from here to an epsilon square bound here, provided this term does not matter. In other words, provided that t is smaller than 1 over epsilon square. And so we're going to refer to this as a cubic lifespan. And of course, you can imagine continuing on and on. If you take a quadratic nonlinearity and so on, your lifespan bound improves. And so given this kind of considerations, the question that I want to ask is what kind of lifespan bounds can we get for this four model problems that I presented to you. And of course, the moment you have energy estimates for each of these problems, you're going to immediately get quadratic lifespan bounds. Now if you look at the nonlinearities that occur in these equations, you'll see there are lots of quadratic nonlinearities. So you might imagine that lifespan bounds, that quadratic lifespan bounds would be optimal. However, an observation of Chateau, more than about, let's see, 30 years ago, tells us that there are actually many interesting scenarios where you can move from quadratic lifespan to cubic lifespan using what is called a normal form method. And what Chateau's idea was the following. Let's say I have my original equation and then I can try to make a change of variable to switch from my initial variable u to a variable v, which is u plus a quadratic correction. Let me call this by b of u and u. And suppose you can do this in such a way so that when you write down the equation for v, you will get something like dt minus lv is something cubic. So you want to choose this quadratic perturbation so that in dt minus l, let me put a bracket in here, so that all the quadratic terms cancel and you're left with only cubic estimates. And then, of course, in this new equation for v, you might say I should get cubic lifespan. So this would give you a nice way to move from quadratic lifespan to cubic lifespan, right? But there is a catch here. And the catch is that when you write this equation for v, really the cubic term here would be cubic term in u, but you'd like it to be a cubic term in v. In other words, you want to have a good way of moving between u and v. So for this to work, you need the map between u and v to be invertible. So need invertibility u to v. And since you have smallness, and this is a quadratic form, this invertibility simply means that this bilinear map has to be bounded. And in Schattach's original example, which was some form of applying Gordon equation, this was indeed happening. And he was getting a bounded expression in here, and he was able to apply this normal form method very nicely. So the question is now whether there's any chance of applying this normal form method to any of these four problems. And the first thing you have to realize when you think of this is that you cannot produce a normal form for any problem. Your problem has to have some very specific structure in order for this to work. And the first key concept that you need to take into account when you do such an analysis is that of resonant interactions. And what is a resonant interaction? A resonant interaction is when you look at your original equation and you put two different waves in here. And the question is if the interaction of the two waves will generate a third wave, which will be evolved by the linear flow. And in Fourier analysis terms, let's say you have one wave which is localized at frequency tau 1 psi 1. Let's say tau 1 is a time frequency, psi 1 is a space frequency. You compose this, so this is your one wave, let's call it U1. A second wave will be localized at frequency tau 2 and psi 2. All right, and this is your function U2. And then when you let these two functions interact, okay, 1 U2, you're going to get possibly a third wave. This will leave that frequency, let's say tau 1 plus tau 2 tau 3 psi 1 plus psi 2. And you ask whether this frequency in here also corresponds to linear waves. In other words, whether this frequency is in the characteristic set of our operator. So suppose your L has a symbol L of psi. This means that tau 1 psi 1 corresponds to a wave, which means that tau 1 is L of psi 1. tau 2 will be L of psi 2. These are just two pairs of frequencies on the characteristic set. And then the output frequency has to have the same property. So tau 1 plus tau 2 is equal to L of psi 1 plus psi 2. So this is what we call a resonant interaction. And so what happens when you try to compute a normal form is that if you have no such resonant interactions, you are going to get some sort of smooth normal form. Whereas if you have some resonant interactions at those resonant interactions, your normal form is going to have some singularity. And when I say singularity for the normal form, the way I'm thinking about this bilinear operator B, I'm thinking of it in Fourier terms. So it's a generalization of a multiplication, if you want. So it's like the way you write the product of two functions in the Fourier space as a convolution, and to that you add a weight. And this is what we call a bilinear translation invariant operator. And such operators will be described by their symbols. And so this operator B in particular will be described by its symbol B of let's say psi 1, psi 2, or psi 1 and psi 2 would be the entry frequencies of the first component and of the second component. And so you'd like this to be a nice object, but the moment you encounter such resonant interactions, this symbol is going to have a singularity. Simply because the algebraic computation of evaluating B will involve a denominator which looks like 1 over L of C1 plus L of C2 minus L of C1 plus C2. So the first thing to understand when you try to compute the normal form is whether you have such resonant interactions. If you don't have them, you're in better shape. If you have them, the question is whether you might be able to deal with them in some other way. And what does it mean to deal with them in some other way? That might possibly mean that maybe this symbol, sorry, this nonlinearity here that I simply took to be quadratic, maybe it also has some structure. And maybe the structure of that nonlinearity will be in such a way that it will vanish exactly when this resonant interaction occurs. So maybe you will have such resonant interactions from a geometric standpoint, but the amplitude of this resonant interaction is equal to zero or goes to zero as you approach the resonance. And so we want to look at all of these problems and try to understand whether you have such resonant interactions or not. This is somehow a key point of the analysis. And so I want to show that to you to show to you what happens in some pictures. One thing that I already mentioned before is these four problems have different dispersion relations. One complication that occurs with respect to this very simple model that I have considered in here is here I have considered an equation which is of order one in time. Here what we have is a system and the system is like an equation of order two in time. So the dispersion relation will be more complicated. So for each psi we're going to have two relevant tals that correspond to maybe the two roots of a characteristic polynomial. So let's see how these dispersion relations look like. And actually maybe I'm going to use this board to draw the dispersion relations because I want to keep them. Also in problem one in model one up there and that's the one described by the equations that I wrote in here. The dispersion relation will be tau square plus psi is equal to zero. And here psi is constrained to be negative. Actually psi is constrained to be negative in all of these examples because we're looking only at holomorphic functions as I discussed last talk. And so if I try to do a picture of this what you'll get is a parabola like this. And now try to visualize what happens when you try to add two points of this parabola and see if the sum of the corresponding vectors ends up again on the parabola. Okay. This is a computation that you can do algebraically but it's also a computation that you can visualize in here because all that you have to visualize is suppose I take this parabola and I shift it to another point of this parabola. So let's say I want to shift it here. And now you ask whether this parabola and this parabola intersect and they obviously cannot intersect and that has to do with the convexity of this region under the parabola. So let me use a dotted line in here because this is sort of a secondary construction. All right. So in this case the only way you're going to get resonant interactions is if one of your interaction frequencies is at zero, right? Because if I shift the parabola at zero then it's still the parabola and then it overlaps on top of itself. So I'm going to have resonant interactions when one of the frequencies is zero. And here you have to be careful when I say that one of the frequencies is zero. You really have three scenarios if you want. You can have psi one is equal to zero or you could have psi two is equal to zero or you could have psi one plus psi two being equal to zero. So either one of the input frequencies is zero or the output frequency is zero. So psi one is zero, psi two is zero or psi one plus psi two is equal to zero. So this is good news and bad news. It's good news in that we don't have too many resonant interactions is bad news in that we have some resonant interactions and we have to deal with those. But since you have so few resonant interactions you might now legitimately ask the question whether the quadratic nonlinearity, the symbol of the quadratic nonlinearity vanishes in this region. And that would give you some nice cancellation and perhaps the potential to construct a normal form. The second model in there, the dispersion relation is tau square plus psi cube is equal to zero and again psi is negative. So if I try to depict this, you're going to get two curves with a power three half, right? And then by drawing essentially a picture similar to this, you will see that again the same thing happens. You can only have resonant interactions if one of the frequencies is zero. Okay, model three, this is where I was telling you that some symmetry gets lost. The principal symbol here is tau square minus I think it's something like c tau plus psi square is equal to zero. This is still a parabola that's too big. But now this parabola intersects the x-axis tau psi obviously. This parabola intersects the what did I do wrong here? An oxy in here. Right, so this will intersect the x-axis at zero and c. So your parabola would look something like this. Okay, and so you have these two branches corresponding to the two roots of the characteristic equation and they are essentially obtained by looking only at the portion of this full parabola that is inside, that is to the left of the y-axis. Okay, and again you have the same conclusion that you only have resonant interactions when one of the frequency is zero. And finally the fourth model is okay that would be tau square minus psi hyperbolic tangent of psi is equal to zero. And so what happens in this case is you get a picture which looks a little bit like the picture in case one, but where the behavior at zero is linear. So you have square root behavior at infinity like a parabola, but at zero you have a linear behavior. All right, here you have a linear an oblique asymptote. And you see the same phenomena that the only case you're going to have resonant frequencies is if one of the frequencies is equal to zero. So, to summarize the outcome of this discussion, if we try to construct a normal form for each of these problems, we're going to face at least one issue that we can see in this picture that you might have singularities at frequency zero. And then there is a second difficulty that you're going to face, which is not apparent in these pictures. And that has to do with the fact that our equation is quasi-linear rather than semi-linear. So this bad nonlinear by linear interactions will occur at the highest order in terms of the derivatives. So this second issue would be that you might have singularities, which means unboundedness at high frequencies. So you worry about zero frequencies, you worry about high frequencies. And if you think about these two issues, issue one might be resolved if your equation has special structure. The second issue cannot be resolved if your equation has special structure because it has to do with the fact that the equation is quasi-linear and no matter how you'll change coordinates in here is never going to become a semi-linear equation. And so you see in here that Schattach's normal form method cannot be applied directly. And just to exemplify my discussion, in case one over here, I will explicitly write down the normal form for you because that's the case when the normal form is quite simple. So my function w is replaced by w tilde, which is w plus w. Let's call this guy w2, which is the correction, where w2, my quadratic correction, which plays the role of b over there, is equal to two times the projection of the real part of w times w alpha. And for q, we have q tilde is equal to q plus q2, where q2 is defined in a very similar form. So it's two times the projection of real part of w times q alpha. So in this first equation, we are very fortunate to obtain a rather obvious, rather simple, easy to write normal form, where the corrections are by linear, quite explicit. They are not exactly multiplication operators. They involve this projection, but other than this projection, they're very simple. And now let's see whether we have run into these two issues, number one and number two. And at frequency zero, you see no singularity in here. And the fact that you see no singularity will tell us that our equation, this particular equation here, has enough structure so that the resonant interaction at frequency zero dies out. There's an actual no structure in there. On the other hand, if you look at the issue number two, you will see that the correction, the quadratic correction actually involves w alpha and q alpha, involves derivatives. So every time you correct, you end up with one extra derivative in the equation. And so now if I were to try to write down the equation for w tilde and q tilde, I will get an equation which, where the non-linearity somehow is cubic. So I'll have dt of w tilde minus or plus d alpha of q tilde is equal to maybe cubic plus higher. And the equation for q tilde, dt of q tilde minus i w tilde, i g w tilde is equal to again cubic plus higher. But this cubic and higher terms will contain in particular expressions like w alpha alpha, q alpha alpha. So they'll contain higher order terms. So it's not possible to invert, first of all, this maps in any meaningful way. And it's not possible to try to work directly with these equations for the normal form variables. I'm going to call w tilde and q tilde normal form variables. Ideally, you'd like to work directly with them, but you cannot because their equation for them has too many unbounded terms on the right. So then the question is, what can one do in terms of applying Schattach's normal form method to our problem? All right, let's see how this works. And so I will begin my discussion simply by stating a theorem. And I will state the theorem in a rather vague way without putting sobolev spaces because that would take too long. My theorem says the following that in these equations 1 through 4, so in all of these four models, we have cubic lifespan bounds. So if you start with initial data of size epsilon and the initial data of size epsilon will be measured in the same sobolev spaces that I put on the board last time, then our solution will exist up to a time of epsilon to the power minus 2, so to a cubic time. And so what I'll try to give you some idea about is how we go about proving such theorems. And so the main idea, if you want, is a method that we called modified energy. This is an idea that we came up with actually before we started looking at this problem. So we were first working on a simpler model, something called the Berger-Hilbert equation, which also appears in fluid as an approximation to fluid models. And what this method is in a nutshell, it says the following that even though you cannot transform the equation using a normal form transformation using Schattach's method, what you can do instead is you can modify your energy. So instead of modifying the equation, you modify the energy. And in all of these four problems, there is a suitable modification of the energy, which satisfies good energy estimates. And let me explain what this means. So this is modify the energy functional rather the equation. All right, so to get to the heart of this, let me introduce, as terminology, the following notion of a normal form energy. And so a normal form energy, suppose I want to compute to write it down. Let's say w differentiated k times and r differentiated k minus one times. These are the diagonal variables that we had before. Should be equal to. And what you see on top is that w tilde and q tilde, the normal form variables, satisfy a good equation in the sense that the terms on the right are cubic and higher. So here I'll put the linear energy of w tilde, let's say differentiated k times and q tilde differentiated k times. All right, so I just take the linear energy, but the linear energy associated to the normal form variables. And what the estimate that you will have for this linear energy will be the following d dt of say normal form energy of these variables will be smaller than. And we know, because of the form of the equation for w tilde and q tilde, that the terms in here will be terms which are quartic and higher. All right, but the problem that we face in here we face in here is that the quartic and higher terms in this normal form energy have too many derivatives in them. So you cannot close, you cannot do a gran wall in here. Let's say if this guy has ordered k, then maybe this guy will have ordered k plus 2. You cannot close. On the other hand, we have the quadratic energy estimates that I showed you before which say that we have some quadratic energy of the same variables with the same index k. And the derivative of this is smaller than with implicit constant depending on our parameter A that I showed you before and with a constant depending on B. And here I'll put the same energy and so on. So this is an inequality where you can do gran wall, but you only have a linear contribution in there. So this is your standard energy estimate. And so what you would like to have instead is to get these two inequalities to become friends. You would like to have the best of both worlds. You'd like to have an estimate that is correct in terms of the derivatives, but where your errors are quartic and higher. So in here I will say maybe that what we want is another energy function. Let me call it E3 of the same variables so that d dt of E3 to be smaller than. And now I want to put here something with a quadratic coefficient. And my quadratic coefficient because of scaling considerations has to be AB E3 of k and so on. So I'd like to have some energy functional which shares the property of the second inequality in that we have the same energy on the left and on the right so you can apply a gran wall estimate. At the same time we would like to have something that shares the property of the first inequality in here in that the errors that we have in here are quartic and higher. So if you count the order of this you have two in here and order two in here. So this is something of order quartic. So the question is how do we get to put together these two objects to get this energy. And this is in a nutshell the idea of this method that we apply to each of these four problems. And what I want to show you are some of the issues that you face when you try to implement this idea. Let me try to put this a little bit lower. Alright so since you want this expression in here to have only quartic and higher orders and you want the normal form energy to have quartic and higher order it's clear that the way you want to choose this modified energy is you want to choose this modified energy so that it agrees very well with the normal form energy. So what one thing we need is to know that this modified energy E3 is equal to the normal form energy plus quartic terms. Because then the error that you make when you compute the derivative of this versus computing the derivative of this will be quartic and higher. Alright so there are three steps to achieving this and I'm going to summarize them in here. So step one what you want to do here is you want to start with the normal form energy and the normal form energy is a functional of W and Q. And this normal form energy if you look at the way it is expressed it will have quadratic terms, cubic terms and quartic terms. The quartic terms in this normal form energy we do not care about them because they will give no contribution to this relation. So I take this and I'm going to discard the quartic terms. Another good reason to discard the quartic terms is that these terms will always be unbounded. There's no way you can make them bounded so you'd better not have anything to do with them. The second step, okay there might be a little bit more than three steps, is the following and this is specific to this problem. So I was telling you that or as originally one writes the equations in terms of this physical variables if you want the initial position parameterized in holomorphic coordinates and the holomorphic velocity potential again parameterized in holomorphic coordinates. The natural thing is to work with the diagonal variables W, alpha and r. So what you'd like to do is you'd like to re-express this normal form energy in terms of this diagonal variable. So this is very specific to this problem and so I can write this in a nutshell as a following. So I want to start with W and Q and I want to replace this by some let's say still normal form energy but maybe I'm going to put here an equal sign W, alpha and r and when I do this I allow myself to do again to make quartic errors, okay? So I'm just adapting my normal form energy to to my diagonal variables. The next step, I'm looking at this diagonal if you want normal form energy and I try to separate it into a leading order term and lower order term. So if you want a high frequency part and the low frequency part say E normal form maybe high plus E normal form low where high simply means that you have the leading order derivatives in this expression. All right and the last step is the following. So the lower order terms will be harmless from the perspective of losing derivatives in the estimate. So when you differentiate the lower order terms you're not going to lose derivatives. So now you still face the issues of losing derivatives in the leading order part, okay? And you don't want to lose those derivatives and so then what you do is you compare this high frequency part of the normal form energy with our quadratic if you want energy. So E2 of let's say Wk r k minus 1, okay? And what you would like to have is essentially an equality between this plus quartic. So what you want to do is you want to correct in some sense this high frequency normal form in a way that's inspired by the quasi-linear energy. So you're adapting this high frequency part to the quasi-linear structure of the equation. And so this is a summary of the steps that one has to undertake but without much in terms of details and I will take a few minutes to show you how this actually works out in the simplest case where I was able to write down the normal form for you. So let's see what I want to erase and what I want to keep. Maybe I can erase the theorem actually because I want to keep the board above. And in doing this I want to there's one or two features that I want to point out to you which we never get to point out in any regular length talk in this analysis. So what would be this normal form energy Wk r k minus 1? Well given what I have been telling you what what we would do in here is we would take the linear energy of the normal form variables. So this would be norm of W plus W to differentiated n minus 1 times sorry differentiated k times. So it's norm of this in L2 square plus the norm of and I'll put the bracket in here dk of q plus q2 in h dot one half square. Very simple expression. Now if you think for a moment this expression the W2 contains expressions that look like W alpha. When I square it I'm going to get in here expressions that look like W k plus 1 times W k plus 1 right. So I'm going to get k plus 1 derivatives as opposed to my energy which is supposed to have k derivatives. This is obviously bad this will come with the bar actually and not only this but they cannot cancel with anything. But fortunately this comes from the interaction of this term with itself so that would be a quartic interaction. You can discard this. Another way you can think about this procedure is that when you talk about normal forms you actually can define the normal form in more than one way. All that's important to have is the the the good quadratic part of the normal form and you cannot q b can hire terms as many as you want and produce extra cancellations. But the easiest way to think about this you just discard the self-interaction of this guy with itself. And so what I would be left in here with is and I'm not going to discuss this term because what happens with this term also happens with this term. So what I will have in here is I'll have norm of W differentiated k times in L2 which is sort of a good guy plus two times and I'll have the inner product of W differentiated of order k with W2 and now I'll write down what W2 is so I'll get a factor of 4 in here and here I'll have I can drop the projection because W is holomorphic and so this is the inner product of W with differentiation k times of the real part of W times W alpha. Okay and this is an L2 inner product right and now let's count the derivatives in this relation. Here I have k derivatives but here I have k plus 1 derivatives so I'm going to say oops I'm in trouble right still too many derivatives. However what happens is that your enemy is when this k derivatives go all on W alpha right and so you end up with something that looks like W k and here times inner product with the real part of W times W bar k there will be an integral W bar k plus 1 right and the real part of this integral if you want. Okay and now what's critical in here is that you have a real coefficient and so you can integrate by parts right you can integrate by parts and you can move the one of one of these derivatives to W and when you do this you really kill two birds with one stone on one hand you eliminate what would have been otherwise an unbounded term on the other hand you eliminate the W that is alone and you replace it with a W alpha thereby moving toward using the diagonal variables W alpha and r so what is behind this cancellation in here this integration by parts well what's behind it is that when you look at this normal form correction there is some anti-symmetry to it it has some structure it's not just any quadratic form it has some anti-symmetry in other words the operator if you want q i'll write this for q the operator q goes into two projection of real part of W times q q alpha this operator the principal part of this operator is anti-symmetric right and the fact that it's anti-symmetric is what gives you this cancellation so to deal with a high frequency issue not only you need to have this normal form but you also need some anti-symmetry in there and my my intuition about this is the fact that you get this anti-symmetry is essentially directly related to the fact that you had good estimates for the original equation so it's it's it's essentially something that that should come for free and indeed it it comes for free in all of these models so this is one issue that one one quick computation that I wanted to show you show to you oh maybe I want to ask a question like one step before like when you had those loss of derivatives you said you I mean it's okay because it's a quadratic term can you I mean maybe you explained it but why I mean like yeah you have loss of derivatives so you have to be also worried about it oh maybe you mentioned it but I missed it I have loss of derivatives in here right but I'm using this just at the algebraic level I'm not doing any estimates for it this is to my mind just an algebraic expression that I'm going to use it to eventually produce my good energy so I will never try to do any estimates with this expression it's just an intermediate step in constructing the right energy so you can think of this if you want as a trilinear symbol doing doing this computing this energy is essentially computing the linear energy plus a trilinear correction to it which you can express as a symbol you don't care about boundedness or unbound this at that stage the only place when you care about boundedness is at the final step so you only want to prove estimates for your final energy once you know you have the right object then you prove estimates okay but but this one I will never try to estimate okay so so this is one issue that arises and maybe I the the next thing that I wanted to to show to you is where the quasi linear correction comes in in this analysis so let me erase this and let me I'm going to jump into the middle of one computation and you'll have to believe me that this is actually what what happens so when I'm looking at this normal form energy written in terms of the diagonal variables form of w alpha and r the leading term in this expression would be the norm of w or if you want the integral of w differentiated k times in l2 well squared plus the imaginary part of r differentiated k minus one times times i times i r differentiated k times d alpha and this would be simply the l2 norm of w differentiated k times and this would be the h one half norm of r differentiated k minus one times all right and and I should say this comes with the factor of g now if you remember from last talk what was the expression of the if you want quasi linear energy for the linearized equation it looked very much like this but with one modification namely that instead of g we had the non-linear factor g plus a which was the normal derivative of the pressure right now a when you add it to this term a is a quadratic expression you add it to another quadratic expression so this would be a quadratic term so this term would be absolutely irrelevant to this computation up to this stage all right but at this stage if I want to have a good bound for the derivative of this expression I have to throw in the quasi linear structure and in this case this simply means that I'm going to add in the factor of the the a term in here and so if I do this this is exactly the last step in here where I'm going to replace the high frequency maybe I'm going to put a high in here just to clarify things with simply by by adding an a into g so I might say oh but this this is a fairly simple thing it was even this was even simpler for the burger silver model that we looked at earlier however if you look at the model two up there where you have surface tension then this computation gets quite elaborate and instead of simply changing one coefficient for a single energy you have to somehow compute three layers of this energy and correct this three layers successively so the analysis gets gets very complicated and I don't want to get into that but it's still possible to do it so this this analysis here is simple for the models one three and four but rather involved for the model two and the reason for this is that the model two has a more complicated quasi linear structure and so more corrections of this nature are needed but but this is the idea and so the computations that I did here for you I did them for this case one where you have a very simple and explicit normal form okay and now I want to quickly show you on the on the laptop how the normal form looks like look like for models two three and four and you can guess why I'm not writing this on the board or you will see that in a moment so okay all right so this is how the normal form looks like for capillary waves and what I'm writing up there are the two corrections w two and q two the corrections to w and q and you see that this time these objects do not realize as explicit operators but instead they realize as bilinear translation invariant operators which can be described via their symbols and if you look at the symbols in here maybe I should say one thing that you see three variables three four variables psi eta and eventually zeta you'll see it on the next slide and those choice of signs of these variables is so that the denominators in here do not vanish so here if you keep psi and eta both negative which is the case the denominator does not vanish in this terminology in here the the indices h and a stand for holomorphic and anti holomorphic interaction so here you have essentially something that looks like the product of w with w and the anti holomorphic part is something that looks like the product of w with w bar okay so you get some some messy symbols in here but these are nice smooth symbols and you have to work with them if you want to figure out this normal form energy these are the other four symbols so seven guys to compute it took us a little bit of time to make sure that we get the correct numbers but believe me that we do get the correct numbers this denominator that you see in there is exactly the denominator that arises when you compute the resonant the resonance relation so that l of psi plus psi one plus l of psi two minus l of psi one plus psi two that's reflected by the denominator that's where the denominator will vanish but fortunately in this case it vanishes only when all three frequencies are zero and what this tells you is that there's still some low frequency null condition you don't get a bad singularity at frequency zero even though by the by the dispersion relation you might expect to get one all right the situation changes if i move to problem three and here you have the normal form this is better and it's worse it's better in the sense that it's more explicit it's just algebraic and we don't know why you get an algebraic expression for it and again it took us quite a bit to to to do this computation but i promise you that this is correct so in principle you'd expect an infinite expansion in terms of powers of c but this this doesn't really happen and this expansion stops at the power c fourth and then the worst term that you get in here are inverse derivatives so related to my initial discussion what you see in here the one important thing to observe in here is that you have inverse derivatives in other words you have a singularity at frequency zero so if you had some null condition in here which i'm claiming you still do and you'll see in a moment why then this null condition is weaker than at my first than for my first two two models because of the presence of this inverse derivatives now if i go back to to this board and you see the computations that i did in here to try to argue for you that you have some cancellation at high frequencies when you compute normal form energies here you want to see that you also have cancellations at low frequencies okay so you want to to see that when you compute this normal form energies this inverse derivatives vanish and i want to give you one heuristic answer for that why why why that happens and that happens because of of symmetrizations and again let me discuss just the w term so i'm looking at w maybe i'll transform this into w plus some bilinear form of w and w right and then in the normal form i will see the inner product of w with some expression that looks like a bilinear expression in w right very simple and of course if this were both if neither of these were conjugated then when i construct this form b to uniquely determine it i make sure that it's symmetric right but now when i take the inner product of this with w there's an extra symmetrization that happens okay so before i had symmetry with respect to two variables now i symmetrize again and i'll symmetrize with respect to the three variables and this is the key computation because when you do this extra symmetrization those inverse derivatives miraculously cancel and to my mind this is exactly the point that you have some extra null structure in here that is not strong enough to make this inverse derivatives disappear from the normal form but it's enough to make the inverse derivatives disappear from the normal form energy okay so it's it's it's one cancellation which the way we do it is we directly compute it and it's actually not very hard to compute that this cancellation happens because you don't have to work with all of those terms you only need to look at those which have inverse derivatives in them all right and so this is what happens here and now the the fourth model that i wanted to tell you about where you have the finite bottom again you have an expansion of the quadratic corrections in terms of bilinear forms and for each of these bilinear forms we want to compute their symbols okay and here the symbols look a little bit different in the sense that you're going to see a lot of hyperbolic functions so this j is the one that appears in the dispersion relation in here the omega that you see in here is the one that measures three wave resonances and so you will see this omega in all of these denominators and so we had seven expressions these three are the simple ones this this is the rest of it okay now it's not immediately obvious but if you look at this omega and compute what it is you will see that omega vanishes of order two when c is zero or and that i zero and zeta is zero so here psi by symmetry you choose psi and zeta and zeta to be to add up to zero so we only care about the symbols on this hyperplane and on this hyperplane this vanishes of order two when psi is zero and eta is zero and when zeta is zero which means that when you compute all of these symbols you're still going to get something that looks like inverse derivatives at frequency zero so you still don't have a full normal form okay and despite the fact that you don't have a full normal form that that you have singularities in the normal form when you write down the normal form energy and do all the symmetrizations you have cancellations and these cancellations will tell you that actually no inverse derivatives survive in the normal form energy just to make to draw another analogy with perhaps a much easier problem if you look at at this last case sorry if you look at this last case when you have finite bottom and you have this dispersion relation like this if you look at waves which are very close to frequency zero so if you look at this portion of this dispersion relation but just at one branch not at both branches this branch is going to be something linear plus something cubic so it's going to look like something psi plus xi cube psi is nothing it's just a transport operator but xi cube is the symbol for the k dv that that correspond to the k dv equation so waves which are on a single branch here and close to frequency zero will be well approximated by the k dv equation if you try to take the k dv equation and compute a normal form for the k dv equation you're going to get inverse derivatives because the normal form for the k dv equation would mean inverting the mu ram app associated to the k dv equation you're going to get inverse derivatives in there nevertheless when you compute energies for k dv you're going to get lots and lots of conserved energies so our analysis in here is a reflection of the same fact that you see very easily in the context of the k dv equation all right so i'm done with the slides all right and what i want to do for the last 10 minutes is i want to briefly touch on the topic that i was supposed to i was planning to do for this third lecture and discuss the following related question namely what happens if your initial data instead of being small is small and localized and so the idea here is that if you have your initial data with some small spatial localization various frequencies of this solution you expect them to evolve essentially linearly according to the dispersion relations that you see on these two boards and different waves will go into different directions different frequencies will move into different directions depending on on the dispersion relation and so you might hope to be able to get more information for these solutions than for the solutions with general small data because here this spreading means that different frequencies will not interact very much after a certain time and so let me write down one so one theorem here uh and this theorem that's the following that if so for the models one and two if initial data is small and localized then the solution is global and furthermore its energies so the energies of w and q will grow at a very slow rate so t to the power small constant epsilon epsilon uh is the size of the initial data and you have the point wise decay measured for suitable norms of w and q maybe let me write this for a and b so a and b our control parameters are smaller than one over square root nt and if you're wondering where this one over square root of t comes from this one over square root of t comes from the dispersive decay of of waves and from the fact that all of these curves that you see in here have non-vanishing curvature so you have one dimensional dispersive decay and that's exactly one over root t and I'll put an epsilon in here this theorem has a long history for the equation one was first proved not for a global time but almost globally by cj a couple of years ago then the first global result was independently obtained by Allahzard and the lord and by UNESCO and Pusatari and then together with Mihaila in this paper that I'm talking about we gave an alternate and we think simpler proof of the same result using the setting that we have with holomorphic coordinates for this problem number two we we proved the global result again about two years ago and then there was also some there were some improvements to that due to UNESCO and Pusatari now what is maybe interesting about this result I don't have a lot of time to discuss it but one thing that we know for sure is that that result does not apply to the other two models the models three and four and this is for a reason that I mentioned before that those two models have small solitons so if you start with this pattern in here you'll see for a while spreading and then solitons will emerge from your profile might emerge or maybe might not emerge depending on what your initial profile is but what we do know is that solitons are arbitrarily small so regardless of how small you choose your initial data solitons are still possible in this problem the other observation that I wanted to to make in here has to do with how you prove this theorem and I'll show you a very simple scheme of this proof with one sort of missing bit and one this this scheme is based on the following common feature of equations one and two namely that very simple fact that these two equations have scaling right so the fact that you have scaling which is a continuous symmetry of the equation that means that you can look at the derivative of the solution with respect to the scaling parameter okay so you look at sw and sq for some meanings of the scaling derivative in here but this is simply derivative of the solution when you rescale it and you differentiate with respect to scaling parameter and what this pair will do because of the fact that that the equation has scaling this will solve the linearized equation okay and if this solves the linearized equation this means that you don't have to do any extra work to get estimates for this guy right so you can do energy estimates you can do the estimates for this cubic energy if you want of s and here I'll put the diagonalization of sw and sq because this energy is with respect to diagonalized variables in the linearized equation okay and this derivative will be smaller than our a times b times the same energy okay and so since a and b grow like one over square root of decay like one epsilon over square root of t and epsilon over square root of t this means that you have for free energy estimates for this equation for as long as you for for this scaling derivatives for as long as you control a's and b's and if you do a ground wall in here you're almost done and you're almost done up to a very large time you're done up to a very large time and this time is the almost global time of CJ's result so in some sense CJ's result is based on using the dispersive decays in decay in here and so you might ask well where do I get my dispersive decay for a and b where I get this one over root t decay certainly I'm not going to get it by studying the linear decay of ways with constant coefficients it has to come in non-linearly and there are two ingredients to that so the first ingredient is what we call the Kleinermann sub-olivine qualities adapted to this problem and this will tell you that a and b are smaller than one over root t times norm of w and q maybe in w and r if you want w alpha and r in some hk plus plus this object in here okay so norm of the scaling derivatives so this is very similar to Kleinermann's vector field method but the problem is that when you try to apply this method globally you only get t to the power epsilon growth in here so using this inequality combined with the energy estimates will only tell you that a and b are smaller than epsilon over root t times e to the power times t to the power constant times epsilon square and this is where the almost global result comes from because you ask how small should be the t be chosen so that this term is irrelevant and that's the almost global result right and so why can't you get a global result from here and I'll make a long story very short that's because when you look at solutions to this equation you do not get linear scattering at infinity instead you get something that's called modified scattering and modified scattering the modified scattering that you get in these equations is identical to the modified scattering that you get in a very familiar model which is the cubic nls equation in one space dimension so if you take the cubic nls equation in one space dimension so i ut minus laplace u is u module so value of u square then for this equation for small dates small localized data are going to have global solutions but these global solutions are not going to scatter linearly they're going to scatter in a modified fashion at infinity and so together with Michela we also came up with a new idea to to treat this modified scattering problem we're not the first ones to to address this problem but i think the way we're addressing it is more is much more efficient than in previous works maybe which go which is which are of several varieties so but i maybe i don't have time to mention a bunch of names now but what i want you to say is that together with Michela we wrote a short paper where we explain this modified scattering analysis for the cubic nls equation which is much easier to swallow than the same analysis for the water wave equation so if you would like to understand that i would invite you to read that simpler paper before you try your hand at understanding what happens to for for the water wave models and sorry for keeping you a little bit extra but i think it's i'm okay to stop now okay just a very quick comment on your last this applied to for modified kdv this was applied by my my former phd student benjamin harrop griffin yeah yeah there are some different issues that arise when you when you look at mkdv that's my question so modified energy message is applied elsewhere how do you know this is a work well no well as far as the modified energy method so this is a different question right but that that we applied we first used it in that context of the burgers hillbark model that i mentioned earlier and then for each for this four models my my intuition is that that might also help streamline some of the more classical results concerning global world poisonous but we did not have time to look into that so that's yeah that's my question is of course if you work on it you know very well it's expected say i just randomly give you a problem and then you run into a similar issue of higher loss activity and then the modified energy message so so so my my my my intuition is that provided you have good quasi linear estimates which are maybe not cubic but quadratic this modified energy method is bound to work in terms of this high frequency issue that i have described it's you asked me for a guess i gave you my guess it's not a theorem no it's a philosophical thing it's a theorem in these four cases plus the one that we have considered before do you need to work in high regularities of spaces for doing this kind of analysis well for doing i did not speak about some of regularity so for doing the cubic lifespan estimates we need to work exactly at the same level of regularities in the local well-posedness result nothing more but to do this kind of analysis for the global result you need a little bit of extra regularity in order to be able to properly use this climb amount so below type estimates and if i remember correctly mihal might correct me i think we have six derivatives for gravity waves and maybe 10 derivatives for capillary waves we did not really try to optimize that number but we did try to get some control over it so that's maybe an open question to try to improve on that in the absence of the assumption that your data is localized do you still get some kind of a time integrity of a and your capital and capital b quantities the the other source for time integrability of a and b quantities is over short times and that comes possibly from three cards estimates three cards estimates might allow you to lower the regularity of the initial data and still retain the integrability of b the integrability of a is not an issue a is lower than b a you expect it to be bounded a is is tied to the critical norm a you once a gets unbounded every all hell breaks loose you don't want that but b that's that's the right question whether b is integrable and b might get better integrabilities properties because of strict arts inequalities but you can pursue this with the sum of the a la czar the works really people who have worked extensively on this