 First, I would like to send an organizer for the invitation. It's a real pleasure to be here, to be part of this conference. As the title say, I will talk about the question related with the KDV, and first, I will say a few words about the introduction. Then I will talk about the K-Wall poseness and uniqueness, and all these things will be related. And after that, I will talk about propagation of regularities of this equation. Then, as many people know, is that the KDV with power one here was first introduced by Corte V. De Vries, or Boussiness, in more than 100 years ago, and later became very famous because it was the first model that was proved to be completely integrable, and that was a resort of Garnet Green, Cruz Calamura. It's a fact that this method applies also for K-Q2, as the previous speaker said here. Then there is another equation that carries more or less the same story, is, okay, I have to say that even if when you are K-Q1 and 2, you have infinitely many conservation law, if your solution is real, and if you are for general power K, you have at least three conservation law. Okay, everything that I say applies to the Benjaminon equation that is this one, one derivative in T, two derivatives in X composed with the Hilbertron 4, and you have exactly this nonlinearity than the KDV. The Hilbertron 4 is defined accordingly to this rule. And this equation was introduced first in propagation of nonlinear waves and then was proved to be a completely integrable later on. And if you put the power here K, you have that for K-Q1, you have infinite conservation law, but for K larger than one, then you have at least three. That means they're running parallel, in some sense. But at least three or at most? Well, you have three that you have these three that are helpful. And you will see that you don't have anything more than that for K-Q1 to 2, you will see that you have a problem. Okay, then we're going to talk about well postness for the KDV and let's say something about well postness. If you can prove well postness using contraction principle, then you get automatically that the map data solution is as smooth if the nonlinearity is as smooth. That is by free. If you use contraction principle. Right? If you don't use contraction principle and you include the notion of well postness, then, for example, in the quasi-linear case, then it's very hard to prove that. Right? I mean, sometimes you have to give up to prove everything in the best space and we will return to that at the end of the talk. You will see that there are some examples that you give up, but still you have existing uniqueness and you have. Okay, what is known is that if S is larger than 3.5, this is a local well post by Bonnet Smith and Cato. And interestingly enough, this method that Bonnet Smith did is the one that you use for quasi-linear problems. Even though this is not a quasi-linear problem, you don't use it here anymore. It's the one that you use when you want to prove persistent and continuous dependent. And the one by Cato is the one that you use in RN for quasi-linear hyperbolic symmetric system. However, for the KDV, things got better and everything here is related with the contraction principle. You can prove that up to minus 3.5, I'm not going to say they're written there. Then you can go here. You can go up to minus 3.5 with contraction. And this is the best that you can do if you want contraction. You cannot have a map that says data solution is smooth below here. And this is a result with Carlos and Luis that we proved. More than that, if you go to minus 1, this is a result of Molyneux that you have no hope whatsoever to have continuous dependent. Doesn't matter what method you use. The map data solution is not defined even in the distribution sense for target. Then from here down, you cannot have contraction. From here down, you have no hope to have anything. Okay, that is related with there. However, if you go for power 2, you have that one quarter. And let me write these numbers. K general i, KDV, if K is 1, 2, and 3, then the S is minus 3 quarter. Here will be one quarter. And in the power 3, you have that the result, at least locally, is minus 1, 6. And this is the scaling. By the way, the scaling here is minus 3 half. Then you have an example that you don't reach the scaling. Okay, in the power 3, you have this situation that you can prove local onto minus 1, 6, and then you have a gap between local and global. And if you reach the homogeneity in this power or any power, you have the extra information is that for a small data, you have global solution. And here that can take complex value because there is no conservation law that you need. Okay, what happens if K is larger than 2, then you reach the homogeneity. And this is something that we proved a long time ago with Carlos and Luis. What happens if K is equal to 4, then this is a Resolo. On Ivan and Fran, in 2002, and more recently, they give a more detailed explanation of the blow-up in that case. And two weeks ago, we saw that if you don't live with integers, you extend that to real, then you can go beyond the power 5 and still have a blow-up. And this is a Resolo. Okay, what is known for the Benjaminon equation? Okay, for the Benjaminon equation, you have for power 1, you have this result that it can be 0 or 1 quarter depend how strong you want your uniqueness. And in fact, if you assume the result 1 quarter, it was proved that you have, this is more in it, and proved that you have unconditional uniqueness. Then 1 quarter is 2, okay? If you go from power 2, then Kenny and Takaoka proved that you have 1 quarter, I'm sorry, 1 half, and very recent, like a week ago, Ivan and the dear proved that you have blow-up. For this, you can see that you don't, this is a question Sergio that you asked. Then if you have power 2 here, then you can preserve until half of the derivatives. If you are small, but no, if you are large, you can have blow-up. And if you have more power, then you reach the homogeneity when you pass 4, and you have the Resol 1 first. Then for the Benjaminon, no? So all these results were a bit confused. So all these results now are concerned with global existence. Local. This is... No, no, no, you have blow-up here. For k equal to 2, you have blow-up. No, you have local existence. And this is open for, what you say is open for larger power. Okay, and then for the k generalized, you know the Resol R0 here, one half and one third. And then you can see a strange thing is that power 1 and 3 are better than 2. Okay, that will bother if you are in linear theory, but I mean since we are in a linear equation, we have to accept, right? And both equations have the same situation. And these are sharp. I mean it's not that you're going to, that is a matter of, it's a fact. What's the reason? Okay, the reason is if you take the symbol, right? And you are very close to the symbol. If you take convolution twice, it's better than you take convolution three times because you get off of the symbol. Okay, I don't know if that is a good explanation. Did that make sense? Okay, that is a matter of how the symbol. And then if you take convolution four times, it's better. And then what? Three is better than four times is two, good, bad, good again. Okay, okay, there is another issue in all this, is that none of these Resol can be proved by contraction. Then you have to, and this is a, the following Resol is due to Molinesz's unset code is if we write this equation, let me write it here because we are going to live with that. And then alpha equal one is KDV and alpha equal one half Benjamin Uno. And then you ask if the power is one, that means that you are quadratic, when can you apply contraction principle? Well, you need to, the dispersion has to be better or equal to the KDV. And that is Resol saying contraction only work if this, if you are in power one, this is more than two or two. However, if you put more powers, contraction work in the moment that you are better, if you put more power, then contraction work is you are strictly more than one half. And can you reach one half? Well, if you are a small data, yes. If you are a large data, contraction doesn't, is unknown. Okay, then we will use this local well postings in the, in the sense we are going to talk about decay. And this is something that has been mentioned several times. If you are in the KDV, you send the chart class in the chart class. And in fact, this is for the generalized KDV. And if you want, you can reach the chart class using this space. This is the regularity. This is the decay. And you map this space into itself only if you have this condition. If you don't have that condition, this is not true. You need the regularity to surpass the decay to, for the flow solution to do that. Otherwise it's false. Okay. And you will see that for the Benjamin or you don't have anything like this. You don't send chart class to chart class. It's a fact. Okay. Okay, this is in the case of symmetric. And if you are in the non-symmetric K, you have the following result of Cato that say, suppose that you're in, you have enough derivatives at, at a time that Cato wrote that you need H2 derivatives, right? And you have exponential decay to the right. Then you preserve the exponential decay for the positive. You become infinity automatically. That means that it's some kind of parabolic. Why you have this parabolicity? Because if you consider this operator in this space, this operator become this one. And this one have the heat equation inside. Well, I don't, I wouldn't call it parabolic. I would say that singularities run fast. They don't get suppressed. Okay. We will get there. And that statement is not true. If you don't quantify. Okay. Well, I'm not allowed to talk very much. No, come on Walter. We, we, it's okay. You can, we know each other for so long. Okay. You don't, well, I mean, you will see that you can have a singularity that peer and peer forever. Okay. We will get there. Okay. We, I will say that, they say we will call it parabolic. And start because Walter doesn't like it. Okay. We will, okay. Now we are going to talk about uniqueness result. And I choose two kinds of uniqueness result is if you have two solutions and you have some condition, you are going to prove that these two solutions are equal. Or you have something much weaker is if you have one function and something happened, this function is zero. Okay. The first condition is much better because you have any two solutions. For the second, you are assuming that the second solution is zero. And you will see these two different. I choose this result. I didn't know that Jean-Claude will be here, but okay. This is a Jean-Claude result. Say the following. If you have two general solutions for the KDB somewhere here, and they agree in an open set, they have to be equal everywhere. And these two solutions are very general. Okay. Something like that I don't know if this is true for the Benjamino. Okay. Okay. The second uniqueness is something that only applies to the KDB. And is the following. That's a result of Tarama that says suppose that you are in the situation that Cato have, that you have exponential decay to the positive side, even with one half, I mean not as strong as Cato. Then the solution become analytic for T positive and in X. That is a extremely strong result. Why you only apply this to one solution is because you use the inverse category. A result like this, as far as I know, is unknown if you are not in the KDB. For power, two, three, four, and so forth. Okay. That means that this is also some kind of parabolicity. You don't like it? Okay. Is if you have some analyticity after one. Okay. Then what kind of result can you prove in this direction? And you want result that apply to two solutions. Then if you go like this, we have the following result that we prove with Luis E. Carlos and Luis B. Right. Is the following. Suppose that you have two general solutions with some property of the KDB with some power and they have enough regularity of decay for sure this is not sharp. And they satisfy that at time zero they are close. That means that they are in L2 with this weight. And at time one they are in L2 with this weight. Then they are equal. Okay. You are not assuming anytime that they are equal. Right. You are only assuming that they stay together to close. Okay. And then you are going to say, well, this is a easy business to do. You take an equation. I mean, why this three half? Okay. Let's first. Why the three half appear? This is the decay of the 80 functions. Right. Then you say, okay, very simple. I take any dispersive equation. I ask, if you decay more than the fundamental solution at two times, then you have to be equal. Okay. Translate yourself to the Schrodinger equation and you say that you don't have decay. Then this statement is false. Right. This is a, there is some parabolicity here. Okay. There is an issue is that the fundamental solution decay. Right. In general, I mean for the Schrodinger equation, the fundamental solution doesn't decay. Then you have a completely different business is, if you ask too much, for example, for the Schrodinger equation, suppose that you ask the same theorem. We have a theorem that we prove that we're not going to give the same, but for example, for the Schrodinger equation, if you ask that they have compact support at time zero, the thing that you conclude is that at time different than zero, they don't decay exponential. It's quite the opposite. If you ask too much, immediately you have to have a four-liter. And the question is that you have some kind of uncertainty principle that is the thing that they have to be applying. Okay. Then the issue is, is this sharp? Well, depending on what I mean you have asked me for this constant. I may not say that it's sharp, but this is the justification of the three-half. This is what we proved no long ago with Pedri Zaza from the University of Medellina, Colombia, and Linares at IMPA Brazil is the following. Suppose that you asked to have this decay initially. Then what happened? You know that you are not going to preserve this decay. What happened is what you expect, is that this decay start to deteriorate when time goes on and deteriorate at this level. And you can see that when t is equal to 1, this constant is just a number. It's not big. OK. Then in that sense, you have the right scaling. Sorry. You have the right scaling, the one that you expect, because you have 3 half here. This is 1 half because everything should come in powers of 3 and so forth. Then the question is, OK, now I have to give Walter, this is not a parabolic result. OK. Why? This is not a completely dispersive result either, because it's not the kind of result that you expect for the Schrodinger equation. But on the other hand, it's not also what is a parabolic result. What is a parabolic result? This is an advertisement. This is a parabolic result. Suppose that you have the heat equation. Suppose that you have a potential that is in L infinity. Suppose that you have a solution of that, that is what you expect. Suppose that you think this function is at any time, at any time, one time, the condition is at one time, bounded, then you are zero. And this is sharp. You can check easily. And there is no side condition, I'm sorry, of the potential. There is no potential that can be as big as you wish. Then in some sense, the problem with that you need two times for the KDV is because you are reversible, right? And in the heat equation, you one time, and L2 will be enough. OK. OK. Then after all this, you're going to say, OK, what can you say for the Benjamin-Hono? OK, for the Benjamin-Hono, you have the following situation. It's suppose that you consider this space that you have S regularity and R decay in L2. And suppose that you intersect this with function that have average zero. And remember that the Benjamin-Hono preserve that. If that makes sense, you can integrate the data that is in L1, you will be able to say that that is preserved. Then what is known for the Benjamin-Hono is something that we prove with Hermann-Fonseca that is also from Colombia, right? Is the following, as you expect, is this. OK, this phi is not sharp. If your decay is less than phi half, the flow solution is preserved. You preserve that decay. You start with phi is not sharp, but this is E. You have phi half. If you add two times, you reach phi half, then it's because you have mean value zero. If you have mean value zero and you reach 7 half, it's OK. Every synom to 7 half, you will preserve it. Now, if you reach 7 half three times, then you are zero. Then we pass from two times, Kdb one time, heat equation, and three times Benjamin-Hono. And then you say, well, maybe this is wrong. No, it's a fact that if you start with this data that have decay more than 7 half, and this momentum is different than zero, you will have the same decay only one more time exactly at this time. There is some magic cancellation between the nonlinear term and the linear term that give you that. And never again, never before. Of course, and then all this is sharp. Nothing of this is known if you take the different of two solutions. Nothing of these, well, some of that can be extended for all the powers. For power, more than two. OK, now we are going to talk about the last subject, this propagation of regularity. OK, then we are going to work with energy estimate. And when you work with energy estimate for the Kdb, I have to work Kdb is this. In any case, suppose that you want to work with energy estimate, and you don't want to see this. This is an skew symmetric operator, right? Then you need to control this to keep going. That this is the quantity that you need. I mean, if you are lazy, and this is what we need because we are going to use this equation. OK, and then we are going to work with these kind of solutions is how much in order to have this, to have this integrability for the Kdb, you need three-quarter of derivatives. And sometimes you need three-quarter plus. Why is it because the stricter estimate that we prove gives you that you gain one-quarter of derivatives? OK, you gain derivatives in the Kdb, you don't gain derivatives in the Turing. You have to do with the curvature. OK, we have this solution. OK, now the Turing is the following. And I want to remark something before saying this. In this local Turing, this time can be arbitrarily large if K is equal to 1, 2, 3. And then you have the other remark is the following. If your data has S derivatives but doesn't have more than S prime, then you will always obey this law. OK, you will state in Hs as good as you are. And if you are not good, you cannot get there, right? Because you solved it by one, completely trivial. Then we have the following result is the following. And this is also with Pedriz Asa and Felipe Linares. And suppose that you have a data for which you can apply your local assistant Turing or global. Suppose that this data for some x to infinity happened to have more derivatives. Then the solution has the property that for forwarding time, all the regularity move to minus infinity. Then this is something that what they say is that there is a statement that say the singularity of the solution of the KDB move with infinite speed to minus infinity, right? We are going to see that this is false. If you don't quantify what do you mean for regularity, this is false. And the quantification is this one. That should be the quantification of that. Moreover, even that you have only m derivatives in half the line whenever, you gain n plus 1 because you have the local smoothing and so forth. Then this is a statement about that the regularity move with infinite speed to the left. At the end of the board, OK? We will see that this is not true in general. You have to have this condition. And of course, you are not saying that the solution, this is false. Here you are better, then you will be better at any later time in this semi-line. Of course, the solution here depends on everything. You don't have enough propagation speed or anything like that. OK, there are some questions. The first question is something that Kenji Nakanishi asked me. Is the following? OK, you cannot be in L2. You have this situation. If you don't start in H1, you will never be in H1. OK, how far are you from being in H1? Well, if you have the function to be some power at minus infinity, then you will be in L2 with this weight. OK, if I don't put this weight here, this is false, right? I have to put that weight. I mean, if you put one here, you have one, then it will be OK. And this is inside the proof. You need G to be an integer, not close to 0, right? What? No, no, sorry. The T is in the bottom. OK, then instead of giving a colorari, let me give you a picture of what you can say about the solution. Suppose let me make it in H1 to make everything integer. Suppose that you have data that is in H1. Suppose that you have an x0. And here, u sub 0 belong to Hm x sub 0 infinity, n very large. Suppose that you have a corner here, then u sub 0 is not in H2 here, OK? What can you say about the solution? OK, you can say the following. For T negative, let's suppose that we have global assistance. For T negative, you have that you cannot apply the previous theorem. You cannot apply the previous theorem. Then you have that the second derivatives of the solution that never belongs to L2 doesn't matter how bad you start for any b. In the past, you never went here H2, because if you were H2, then you would be H2 there, right? Then since decay implied regularity, and this is a result of Krusko-Flaminkis, you have this also, right? Then you have this situation that immediately you know something about what happened here. Then you apply the theorem, and you see that you, before you apply the theorem, you see that everything that happened here for the Kdb happened here. Then you have the same situation here for T positive, but for X in this side. But you have integrability if you help the function. Then you have that this is finite, and you will never see this corner again in the future. You may have seen it in the past, but maybe not in the future. You have all the information here. You don't have this, but you have this. OK, this is a corollary. And then you have a question. Then if you go to the theorem, OK, this is a theorem with Pedro and Felipe. You are assuming that you have integer derivatives. Well, can you remove the fact that you have integer derivatives? Suppose this condition involved that you are in the fractional solar space. Then the answer, do you have the same result? Then the answer is yes, and this is a node that will appear in one or two weeks with Felipe, Carlos, and Luis. OK, is this related with the Kdb only? No, you can also prove that for the Benjaminon equation. The proof is more complicated. You have to do it the iteration twice because you advance in half, right? Then you can do it for the Benjaminono. It's the same result, a little bit more complicated, but that's OK. Then if you can do it for the Benjaminono and for the Kdb, the question is, is this related with integrability? OK, then you go here and you say, for this equation, the same equation that we have here, if alpha is equal to 1, you have Kdb. If alpha is equal to 1 half, you have Benjaminono. When this propagation of regularity is known, only for these two parameters. I'm not claiming that it's not true for the other, but I only know how to prove for that. Then the question is, is this related with integrability? Well, the answer should be no. And let me give you some examples. And this is an example, consider that you have a quasi-linear Kdb of this type, and then you have to put the hypothesis here on this coefficient and in this coefficient. They are sinfinity bounded and so forth. And this is the important hypothesis, is that you have some strictly positive dispersion. Why? This result is not true if you have compact tone. I mean, the result of the propagation of regularity is not true. I mean, if you have compact tone, for example, you will have something. Then here, you are not C2, but here you are extremely nice. Then if you have a propagation of regularity, they should something happen. Then you have to kill this possibility. And these are these hypotheses and these hypotheses. And then under these two hypotheses, then you have a result of Kreka-Peller-Strauss say the following. If you have seven derivatives, then this problem, this problem for the quasi-linear, have a unique solution that belong here and hold Cato result. Well, if you want to believe that seven is sharp, I mean, you go to the proof and it's very close to be sharp, right? I mean, you use this argument. OK, you need a little bit of an improvement, but if you follow exactly the argument, you will prove this. Then you are not only L infinity. You are continuous here. And then you need some continuous dependent because you want to pass to the limit, right? And here it is a typical thing that you are losing delta. Why you are losing delta? Because you don't want to apply the bonus mid argument because it's very involved and that's not part of the issue. But this is a typical, if you don't want to lose that delta, you have to use that. And then you get the result. And this is with Derek Smith and Linares. Last year, we proved, if you start with seven derivatives, but in the half-lane, you have more derivatives, then it's exactly what the KDV have. You have the propagation and you have the integrability and you gain one derivative. Then it's not a matter that you have constant coefficient and you are semi-linear, by no means. Then do you preserve other regularity? And this will answer the question that if you make the statement that for solution of the KDV, the singularity move with infinite speed to the right is false. You have to quantify. This is a result of bonus mid, bonus solve, I'm sorry, is the following. I don't seem that they stated like this, but if you follow their proof, you get that. You can start with an extremely nice function that is in H1. Therefore, if the time is not an integer, you will be C1. Of course, you will be forever in H1, but you will be C1. And when the time is an integer, you will have a corner at the origin. The corner appear and disappear, appear and disappear, and so forth. Therefore, you are not going to tell me that the singularity propagate with infinite speed. They do? OK. It just comes from the thing. Well, OK. OK. OK, suppose that you are picking, you say, well, OK, then maybe this infinity is not good. What about HP? This space. Well, the theorem is exactly the same. You can have data that have a lot of regularity with p more than 2 and j derivatives, and you don't stay there, not even in the half length of something like this. OK. And then finally, we have the question. And I think I will finish ahead of time. It's OK? OK. OK, the question is, do you have? There are some models that people have proposed to imitate the KDV, right? And it has been mentioned here. Someone mentioned here the BBM equation. And then you have the situation for the BBM equation is, then it was proved that this local well posed by Nicholas and Jerry in L2. Then what happened with the bonus mid equation with propagation of regularity? This is what happened. If you have a singularity here, the singularity will be there all the time. If you are Ck plus theta in this interval, you will be Ck plus theta here forever. Nothing move. Nothing. Then you don't have something like this, right? There is no influence anywhere on that, OK? And this is the result. And then you can say, well, what happened if you go to the Kamasa Hall, right? And then, of course, if you take the Kamasa Hall, that is an integrable system. And if this K factor here is not there, then you have something called PCON solution that are not C1. And they are traveling. Then it's a deep result of Bresson and Constantine prove uniqueness. No, they prove existence. And Bresson and Chen and Sun prove uniqueness of this kind of solution. See, you are starting with something that is in H1, and you watch. I mean, this is the result. And then you are not proving that this is continuous. You are not having any continuous dependent of the data. And you have this. I mean, you are losing a lot related with the whirlpool. However, if you restrict yourself to live here, everything that happened with the BVN happened here if you start with Lipschitzon, OK? And this is a result with Sideris. And we are not writing this yet. And I think that I will stop here. Thank you. It's OK, Walter. Somebody should have that. So the principal result in the middle of your several slides is that HM, so we'll have regularity on one side of the axis. I guess it's on the right. It probably gets to the left. But can you replace that with decay? Yes, and that is a result of Krusko-Framinghi. That is Krusko-Framinghi. And for derivatives? If you have for decay, you have a decay. Yes, in one side, and only for T positive. Only for T positive, decay in the other side, T negative. This is Krusko-Framinghi. I don't understand these last results. So the first one is globally in time. Then you say, you belongs to? OK, the issue is this. If you start with something that is lipchis for the Kamasa Hall, then you can prove that it's still lipchis up to some time. And after that time, it may not be lipchis anymore. It will be a holder. But still, you can extend it. Yes, I mean, the singularity will be that the H1 north cannot blow up. This one, you say along the flow, I don't understand. It's regular as the data in minus T e along the flow. Yeah, you can. OK. You have the flow is, this is your flow. And that is well-defined. Along that, you are as good at initial data up to some time. And after that, OK. Because the flow never appears. The flow could have been as though the flow could be the flow map. Yes, the flow map and the flow map. OK, other questions? Yes, on that, for blow-up, so is there a proof of blow-up solution of this kind? Well mentioned, so it's a possibility, but is there a proof of such a possibility at times? It blows up in the C1 north, but you can continue this solution. Yeah, but you mean it's abstract. Is there initial data? Yes, yes. People have identified some data, some condition that this happened. That's a constant, you have a paper of? Even Molinez have some condition. Can I start? Yes. OK, all of them here. But it's done. It's done, yes, yes, yes. But the singularity is that you will be holder. This is, still you will be holder. Do you know which has flow, which will go here? What's that? Which exponent? 1 1⁄2, because you are in H1 forever. 1 1⁄2. Again? Yes, yes, yes.