On special regularity properties of a class of dispersive equations
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Ponce, Gustavo
This talk is concerned with special properties of solutions to the IVP for the k-generalized KdV equation \pa_tu+\pa_x^3u+u^k \pa_xu=0 on the line. We shall discuss results concerning well posedness, unique continuation and propagaion of regularity, and compare them with similar results for other nonlinear dispersive models (in particular Benjamin Ono equation). This is joint work with P. Isaza and F. Linares
 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. At the title, I will talk about equation related with the KDV. First, I will say a few words about introduction. Then I will talk about the K-well poseness and uniqueness, and all these things will be related. 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 Boussinesse, 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 result of Garnet Green-Kruz-Karamura. It's a fact that this method applied also for K-quart to two, as the previous speaker said here. Then there is another equation that carries more or less the same story is, even if you are K-quart to one and two, 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 applied to the Benjamin-Ohno equation that is this one. One derivative in T, two derivatives in X, composed with the Hilbert-Tron-4, you have exactly the same nonlinearity than the KDV. The Hilbert-Tron-4 is defined accordingly to this rule. This equation was introduced first in propagation of nonlinear wave, and then was proved to be a complete integrable later on. If you put the power here K, you have that for K-quart to one, 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. At least three or at most? Well, you have three that you have these three that are helpful. You will see that you don't have anything more than that for K-quart to two. You will see that you have a problem. 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 a smooth if the nonlinearity is smooth. That is by free if you use contraction principle. If you don't use contraction principle and you include the notion of well postness, then for example in the Quasilline case, then it's very hard to prove that. 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. What is known is that if S is larger than 3.5, this is a local well post by Bonnet-Smith and Cato. Interesting enough, this method that Bonnet-Smith did is the one that you use for Quasilline problems. Even though this is not a Quasilline problem, you don't use it here anymore, it's the one that you use when you want to prove persistent and continuous dependent. The one by Cato is the one that you use in the Rn for Quasilline hyperbolic symmetric system. However, for the Kdb, things got better and everything here is related with the contraction principle. You can prove that up to minus 3 quarter, I'm not going to say they are written there. Then you can go here, you can go up to minus 3 quarter with contraction. This is the best that you can do if you want contraction. You cannot have a map that says data solution is set mood below here. 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 Mollinx 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. 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, Kdb, if K is 1, 2, and 3, then the S is minus 3 quarter, here will be one quarter. In the power 3, you have 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. 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. If you reach the homogeneity in this power or any power, the extra information is that for a small data, you have global solution. Here, that can take complex value because there is no conservation law that you need. What happened if K is larger than 2, then you reach the homogeneity, and this is something that we prove long time ago with Carlos and Luis. What happened if K is equal to 4, then this is a Resolute. On Ivan and Fran in 2002, more recently they give a more detailed explanation of the blow-up in that case. 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 blow-up. This is a result of Young. What is known for the Benjaminon equation? For the Benjaminon equation, for power 1, you have this result that it can be 0 or 1 quarter, depends how strong you want your uniqueness. In fact, if you assume the result 1 quarter, it was proved that you have, this is more in it and a lot, proved that you have unconditional uniqueness. Then 1 quarter is 2. 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 this is a question, Sergio, that you asked. Then if you have power 2 here, then you press R until half of the ribates. If you are small but no, if you are large, you can have blow-up. If you have more power, then you reach the homogeneity when you pass 4 and you have the result 1 first. Then for the Benjaminon. So all these results were a bit confused. So all these results now are concerned with global existence. Local. This is- Local well-positiveness. But local well-positiveness, I thought you can for all k, you can go all the way to- No, no, no. You have low up here. For k equal to 2, you have low up. So you have positive. No, you have local existence. Systems and you have low up. Yeah, okay. So it's not global. That's what I mean. Yeah. This is open for, what you say is open for larger power. Okay. And then for the k generalized, you know the results are zero 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 linear, 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're very close to the symbol. 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. Does that make sense? Okay. That is a matter of how the symbol. And then you take convolution four times, it's better. Then the word- Three times and four times is two. Good, bad, good again. Okay. Okay. There is another issue in all this, is that none of these results can be proved by contraction. Then you have to, and this is a, the following result is due to Molinets-Solz and Z-Co, is if we write this equation, let me write it here, because we are going to leave with that. And then alpha equal to one is Kdv, and alpha equal to one-half, Benjamin O'Noll. 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 a, these results say the contraction only works if this is your power one, this is more than two or two. Okay. However, if you put more powers, contraction works in the moment that you are better, you put more power, then contraction works if 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. Okay. Okay. Then we will use this local well postness 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 truark class in the truark class. And in fact, this is for the generalized Kdv. And if you want, you can reach the truark 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. 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 venjamino, no, you don't have anything like this. You don't send truark class to truark class. It's a fact. 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 have enough derivatives at the time that Cato wrote that you need H2 derivatives. And you have exponential decay to the right. Then you preserve the exponential decay for the positive. You become infinity automatically. That means that this is some kind of parabolic. Why you have this parabolicity? Because if you consider this operator in this space, this operator becomes this one. And this one has the heat equation inside. Well, I don't know. I wouldn't call it parabolic. I would say that singularities run fast. They don't get suppressed. Okay. We will get there. That statement is not true if you don't quantify. Okay? Well, I'm not allowed to talk very much, so I won't talk. No, come on, Walter. It's okay. You can't. We know each other for so long. Okay. Well, I mean, you will see that you can have a singularity that Pia and Pia forever. Okay? We will get there. Okay, I will say that we will call it parabolic and start because Walter doesn't like it. Okay? Okay, now we are going to talk about uniqueness result. And I choose two kinds of uniqueness result. It's if you have two solutions and you have some conditions, you are going to prove that these two solutions are equal. Or you have something much weaker. That 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 are 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 is this true for the Benjamin Olo. Okay? Okay? Okay? The second uniqueness is something that only apply to the KDB. And it's 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. And the solution becomes 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 2, 3, 4, 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 all. Okay? Okay. Then what kind of result can you prove in this direction? And you want results 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. You 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. Right? And they satisfy that at time zero they are close. I mean 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 any time 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 these three half. Okay, let's first. Why these three half appear? This is the decay of the 80 functions. Right? Then you say, okay. Very simple. 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 are not going to give the term, 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. You ask too much immediately you have to have a four-liter. The question is that you have some kind of uncertainty principle that is the thing that they do, that they have to be applied. Okay. Then the issue is, is this sharp? Well, depending on what I mean, you ask 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 Pedro Sasa from the University of Medellina, Colombia and Linares at IMPA Brazil is the following. Suppose that you ask to have this decay initially. Then what happened? You know that you are not going to preserve this decay. What happens is what you expect is that this decay starts to deteriorate when time goes on and deteriorate at this level and you can see that when t is equal to one this constant is just a number. It's not big. Okay. And in that sense, you have the right scaling. You have the right scaling, the one that you expect because you have three-half here. This is one-half because everything should come in powers of three. Then the question is, okay, now I have to give Walter, this is not a parabolic result. Okay? Why? This is not that it completely disperses the 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? Okay, 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, the potential can be as big as you wish. Okay? Then in some sense, the problem with, you need two times for the KDB is because you are reversible. Right? And in the heat equation, you one time and L2 will be enough. Okay? Okay. Then after all this, you're going to say, okay, what can you say for the Benjaminono? Okay, for the Benjaminono, you have the following situation is 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 Benjaminono 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 Benjaminono is something that we prove with Hermann von Seckert that is also from Colombia. Right? Is the following, as you expect, is this. Okay, this phi is not sharp. If you are decaying less than phi half, the flow solution is preserved. You preserve that decay. You start with phi is not sharp, but this is E. If you have phi half, if you are two times you reach phi half, then it's because you have mean value zero. If you have mean value zero and you reach seven half, it's okay. Every single to seven half, you will preserve it. Now if you reach seven half three times, then you are zero. Then we pass from two times Kdb one time, heat equation and three times Benjaminono. 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 seven half and this momentum is different than zero, you will have the same decay only one more time at exactly this time. There is some magic cancellation between the nonlinear term and the linear term that gives you that. Never again, never before. Of course, and then all this is sharp. Nothing of this is known if you take the difference of two solutions. Nothing of this, well, some of that can be extended for other powers, for power more than two. Okay. Now we are going to talk about the last subject is propagation of regularity. Okay, okay. 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. 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. Okay. And then we are going to work with these kind of solutions is how much in order to have this to have this integrity for the Kdb. You need three quarter of derivatives and sometimes you need three quarter plus. Why is because this tricker estimate that we prove give you that you gain one quarter of derivatives. Okay, you gain derivatives in the Kdb, you don't gain derivatives in the charting. You have to do with the curvature. Okay, we have this solution. Okay, now the theorem is the following and I want to remark something before saying this. In this local theorem, this time can be arbitrarily large if K is equal to one to three. And then you have the other remark is the following. If your data has derivatives but doesn't have more than s prime, then you will always obey this law. Okay, you will state in HS as good as you are and if you are not good, you cannot get there, right? Because you solve it by one. Completely trivial. Then we have the following result is the following and this is also with Pedri Sascha and Felipe Linare. And suppose that you have a data for which you can apply your local assistant theorem on global. Suppose that this data for some x to infinity happen to have more derivatives. Then the solution has the property that for forwarding time, all the regularity move to minus infinity. Okay? And 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 infinity 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, you have, even that you have only m derivatives in half the line, whenever, you gain m plus 1 because you have the local smoothing. Okay? Then this is a statement about that the regularity move with infinity speed to the left. Okay? 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 than you will be better at any later time in this semi-line. Of course, the solution here depends on everything, right? You don't have to find a propagation speed or anything like that. Okay, there are some questions. The first question is something that Kenji Nakanishi asked me is the following. Okay, you cannot be in L2. You have this situation. If you don't start in H1, you will never be in H1. Okay, 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. Okay, 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 okay. And this is inside the proof. You need G to be a integer, not close to zero, right? What? No, no, sorry. That is in the bottom. Yeah. Okay. 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 zero belongs to HM X sub zero infinity, M very large. Suppose that you have a corner here, then U sub zero is not in H2 here. Okay. What can you say about the solution? Okay. 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 never belongs to L2. It doesn't matter how bad you start for any B. In the past, you never went here H2, because you were H2, then you will be H2 there. Then since decay implies regularity, and this is a Resolus Krusko-Flaminkis, you have this also. 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. 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. This is a corollary. Then you have a question. Then if you go to the theorem, this is the 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. Is this related with the KDB only? No. You can also prove that for the Benjamin-Ohno equation. The proof is more complicated. You have to do it the iteration twice because you advance in half. Then you can do it for the Benjamin-Ohno. It's the same result, a little bit more complicated, but that's okay. Then if you can do it for the Benjamin-Ohno and for the KDB, the question is, is this related with integrability? Okay. 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, you have Benjamin-Ohno, 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? The answer should be no. Let me give you some examples. This is an example, consider that you have a quasi-linear KDB of this type. Then you have to put the hypothesis here and this coefficient and in this coefficient, okay, they are infinity bounded and so forth. This is the important hypothesis, is that you have some strict deposit this person. 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, right? Then here you are not C2, but here you are extremely nice. Then if you have a propagation of regularity, that should something happen. Then you have to kill this possibility and these are these hypothesis and these hypothesis. Then under these two hypothesis, then you have a result of Cree-Capellar-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 cataract results. If you want to believe that seven is sharp, I mean, you go to the proof and it's very close to be sharp. I mean, you use this argument. 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. Here 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. This is a typical, if you don't want to lose that delta, you have to use that. Then you get the result. This is with Derek Smith and Linarez. Last year we proved if you start with seven derivatives, but in the half-lane, you have more derivatives than 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. Right, Nomin? Then do you preserve other regularity? 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, right? Is the following. I don't think 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. When the time is an integer, you will have a corner at the origin. The corner appear, disappear, appear, disappear, and so forth. You are not going to tell me that the singularity propagate with infinite speed. They do? Okay. It just comes from there. Okay. Well, okay. Okay. Okay. Okay. Suppose that you are picking, you say, well, okay, then maybe this infinite 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 two and J derivatives and you don't stay there, not even in the half length of something like this. Okay. And then finally, we have the question and I think I will finish ahead of time. Yeah? It's okay? Okay. The question is, do you have, there are some models that people have proposed to imitate the KDV or right? Oh, then I mean, have we mentioned here, someone mentioned here the BBM equation and then you have the, 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, there is no influence anywhere on that. Okay. 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, this is, in this, if you take the Kamasa Hall, that is an integral system and if this K factor here is not there, then you have something called P-con solution that are not C1 and they are traveling. And it's a deep result of Brezan and Constantine through uniqueness, no, they prove existence and Brezan and Chen and Sun prove uniqueness of this kind of solution. See, you are starting with something that is in H1 as you watch. I mean, this is the result. And then you are not proving that this is continuous. You are not having any continuous dependence of the data and you have this. I mean, you are losing a lot related with the workforce. However, if you restrict yourself to live here, everything that happened with the BBN happened here if you start with libGIS-on. Okay. And this is a result with CIDERIS and we are not writing this yet. And I think that I will stop here. Thank you. It's okay. What? Somebody should have a question. So the principle result in the middle of your several slides is that HM, so we will 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-Framing. That is Krusko-Framing. If you have for decay, you have decay. Yes. In one side and only for T positive. Only for T positive. Decay in the other side. This is Krusko-Framing. I don't understand this last result. So the first one is globally in time. Then you say you belong to... Okay. The issue is this. What... If you start with something that is lipchis for the Kamasa Hall, right, 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 nor cannot blossom. Okay. This one, you say along the flow, I don't understand. It's regular as data in minus T of E along the flow. Yeah. You can... You can extend it. Okay. So you have the flow is... Okay, let me see. This is your flow. Oh, okay. Okay. And that is well defined. So on that, you are as good at initial data up to some time. And after that, okay. Because the flow never appears. The flow could be the flow. Okay. Other questions? Yes. On that, for blow-up. So is there a proof of blow-up solution of this kind? Well, you mentioned. So it's a possibility, but is there a proof of such a possibility at the end? It blows up in the C1 nor, but you can continue this solution. Yes, but you mean it's abstract. Is there initial data? Yes. People have identified some data, some condition that this happened. Even Molinet have some condition. Can I start? Yes. Okay. 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. Okay. Do you know which one is for, which one is for? What's that? One half because you are in H1 forever. One half. Okay. Okay. Thank you. Any more questions or comments? So thank you again. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you.
 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.  At the title, I will talk about equation related with the KDV.  First, I will say a few words about introduction.  Then I will talk about the K-well poseness and uniqueness,  and all these things will be related.  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 Boussinesse, 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 result of Garnet Green-Kruz-Karamura.  It's a fact that this method applied also for K-quart to two, as the previous speaker said here.  Then there is another equation that carries more or less the same story is,  even if you are K-quart to one and two, 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 applied to the Benjamin-Ohno equation that is this one.  One derivative in T, two derivatives in X,  composed with the Hilbert-Tron-4,  you have exactly the same nonlinearity than the KDV.  The Hilbert-Tron-4 is defined accordingly to this rule.  This equation was introduced first in propagation of nonlinear wave,  and then was proved to be a complete integrable later on.  If you put the power here K,  you have that for K-quart to one,  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.  At least three or at most?  Well, you have three that you have these three that are helpful.  You will see that you don't have anything more than that for K-quart to two.  You will see that you have a problem.  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  a smooth if the nonlinearity is smooth.  That is by free if you use contraction principle.  If you don't use contraction principle and you include the notion of well postness,  then for example in the Quasilline case,  then it's very hard to prove that.  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.  What is known is that if S is larger than 3.5,  this is a local well post by Bonnet-Smith and Cato.  Interesting enough, this method that Bonnet-Smith did is the one that you use for Quasilline problems.  Even though this is not a Quasilline problem,  you don't use it here anymore,  it's the one that you use when you want to prove persistent and continuous dependent.  The one by Cato is the one that you use in the Rn for Quasilline hyperbolic symmetric system.  However, for the Kdb, things got better and everything here is related with the contraction principle.  You can prove that up to minus 3 quarter,  I'm not going to say they are written there.  Then you can go here, you can go up to minus 3 quarter with contraction.  This is the best that you can do if you want contraction.  You cannot have a map that says data solution is set mood below here.  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 Mollinx 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.  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, Kdb, if K is 1, 2, and 3,  then the S is minus 3 quarter,  here will be one quarter.  In the power 3,  you have 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.  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.  If you reach the homogeneity in this power or any power,  the extra information is that for a small data,  you have global solution.  Here, that can take complex value because there is no conservation law that you need.  What happened if K is larger than 2,  then you reach the homogeneity,  and this is something that we prove long time ago with Carlos and Luis.  What happened if K is equal to 4,  then this is a Resolute.  On Ivan and Fran in 2002,  more recently they give a more detailed explanation of the blow-up in that case.  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 blow-up.  This is a result of Young. What is known for the Benjaminon equation?  For the Benjaminon equation,  for power 1, you have this result that it can be 0 or 1 quarter,  depends how strong you want your uniqueness.  In fact, if you assume the result 1 quarter,  it was proved that you have,  this is more in it and a lot,  proved that you have unconditional uniqueness.  Then 1 quarter is 2.  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 this is a question,  Sergio, that you asked.  Then if you have power 2 here,  then you press R until half of the ribates.  If you are small but no,  if you are large, you can have blow-up.  If you have more power,  then you reach the homogeneity when you pass 4 and you have the result 1 first.  Then for the Benjaminon.  So all these results were a bit confused.  So all these results now are concerned with global existence.  Local. This is-  Local well-positiveness.  But local well-positiveness,  I thought you can for all k,  you can go all the way to-  No, no, no. You have low up here.  For k equal to 2, you have low up.  So you have positive.  No, you have local existence.  Systems and you have low up.  Yeah, okay. So it's not global.  That's what I mean.  Yeah. This is open for,  what you say is open for larger power.  Okay. And then for the k generalized,  you know the results are zero 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 linear,  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're very close to the symbol.  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.  Does that make sense?  Okay. That is a matter of how the symbol.  And then you take convolution four times, it's better.  Then the word-  Three times and four times is two.  Good, bad, good again.  Okay. Okay.  There is another issue in all this,  is that none of these results can be proved by contraction.  Then you have to, and this is a,  the following result is due to Molinets-Solz and Z-Co,  is if we write this equation, let me write it here,  because we are going to leave with that.  And then alpha equal to one is Kdv,  and alpha equal to one-half, Benjamin O'Noll.  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 a, these results say the contraction only works  if this is your power one, this is more than two or two.  Okay.  However, if you put more powers, contraction works  in the moment that you are better,  you put more power, then contraction works  if 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.  Okay.  Okay.  Then we will use this local well postness 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 truark class  in the truark class.  And in fact, this is for the generalized Kdv.  And if you want, you can reach the truark 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.  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 venjamino,  no, you don't have anything like this.  You don't send truark class to truark class.  It's a fact.  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 have enough derivatives at the time  that Cato wrote that you need H2 derivatives.  And you have exponential decay to the right.  Then you preserve the exponential decay for the positive.  You become infinity automatically.  That means that this is some kind of parabolic.  Why you have this parabolicity?  Because if you consider this operator in this space,  this operator becomes this one.  And this one has the heat equation inside.  Well, I don't know.  I wouldn't call it parabolic.  I would say that singularities run fast.  They don't get suppressed.  Okay.  We will get there.  That statement is not true if you don't quantify.  Okay?  Well, I'm not allowed to talk very much, so I won't talk.  No, come on, Walter.  It's okay.  You can't.  We know each other for so long.  Okay.  Well, I mean, you will see that you can have a singularity  that Pia and Pia forever.  Okay?  We will get there.  Okay, I will say that we will call it parabolic and start  because Walter doesn't like it.  Okay?  Okay, now we are going to talk about uniqueness result.  And I choose two kinds of uniqueness result.  It's if you have two solutions and you have some conditions,  you are going to prove that these two solutions are equal.  Or you have something much weaker.  That 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 are 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 is this true for the Benjamin Olo.  Okay?  Okay?  Okay?  The second uniqueness is something that only apply to the KDB.  And it's 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.  And the solution becomes 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 2, 3, 4, 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 all.  Okay?  Okay.  Then what kind of result can you prove in this direction?  And you want results 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.  You 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.  Right?  And they satisfy that at time zero they are close.  I mean 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 any time 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 these three half.  Okay, let's first.  Why these three half appear?  This is the decay of the 80 functions.  Right?  Then you say, okay.  Very simple.  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 are not going to give the term, 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.  You ask too much immediately you have to have a four-liter.  The question is that you have some kind of uncertainty principle that is the thing that  they do, that they have to be applied.  Okay.  Then the issue is, is this sharp?  Well, depending on what I mean, you ask 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 Pedro Sasa from the University of Medellina, Colombia  and Linares at IMPA Brazil is the following.  Suppose that you ask to have this decay initially.  Then what happened?  You know that you are not going to preserve this decay.  What happens is what you expect is that this decay starts to deteriorate when time goes  on and deteriorate at this level and you can see that when t is equal to one this constant  is just a number.  It's not big.  Okay.  And in that sense, you have the right scaling.  You have the right scaling, the one that you expect because you have three-half here.  This is one-half because everything should come in powers of three.  Then the question is, okay, now I have to give Walter, this is not a parabolic result.  Okay?  Why?  This is not that it completely disperses the 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?  Okay, 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, the potential can be as big as you wish.  Okay?  Then in some sense, the problem with, you need two times for the KDB is because you are reversible.  Right?  And in the heat equation, you one time and L2 will be enough.  Okay?  Okay.  Then after all this, you're going to say, okay, what can you say for the Benjaminono?  Okay, for the Benjaminono, you have the following situation is 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 Benjaminono 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 Benjaminono is something that we prove with Hermann von Seckert  that is also from Colombia.  Right?  Is the following, as you expect, is this.  Okay, this phi is not sharp.  If you are decaying less than phi half, the flow solution is preserved.  You preserve that decay.  You start with phi is not sharp, but this is E.  If you have phi half, if you are two times you reach phi half, then it's because you  have mean value zero.  If you have mean value zero and you reach seven half, it's okay.  Every single to seven half, you will preserve it.  Now if you reach seven half three times, then you are zero.  Then we pass from two times Kdb one time, heat equation and three times Benjaminono.  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 seven half  and this momentum is different than zero, you will have the same decay only one more  time at exactly this time.  There is some magic cancellation between the nonlinear term and the linear term that gives  you that.  Never again, never before.  Of course, and then all this is sharp.  Nothing of this is known if you take the difference of two solutions.  Nothing of this, well, some of that can be extended for other powers, for power more  than two.  Okay.  Now we are going to talk about the last subject is propagation of regularity.  Okay, okay.  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.  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.  Okay.  And then we are going to work with these kind of solutions is how much in order to have  this to have this integrity for the Kdb.  You need three quarter of derivatives and sometimes you need three quarter plus.  Why is because this tricker estimate that we prove give you that you gain one quarter  of derivatives.  Okay, you gain derivatives in the Kdb, you don't gain derivatives in the charting.  You have to do with the curvature.  Okay, we have this solution.  Okay, now the theorem is the following and I want to remark something before saying this.  In this local theorem, this time can be arbitrarily large if K is equal to one to three.  And then you have the other remark is the following.  If your data has derivatives but doesn't have more than s prime, then you will always obey  this law.  Okay, you will state in HS as good as you are and if you are not good, you cannot get  there, right?  Because you solve it by one.  Completely trivial.  Then we have the following result is the following and this is also with Pedri Sascha  and Felipe Linare.  And suppose that you have a data for which you can apply your local assistant theorem  on global.  Suppose that this data for some x to infinity happen to have more derivatives.  Then the solution has the property that for forwarding time, all the regularity move to  minus infinity.  Okay?  And 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 infinity 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, you have, even that you have only m derivatives in half the line, whenever, you  gain m plus 1 because you have the local smoothing.  Okay?  Then this is a statement about that the regularity move with infinity speed to the left.  Okay?  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 than you will be better at any later time in this semi-line.  Of course, the solution here depends on everything, right?  You don't have to find a propagation speed or anything like that.  Okay, there are some questions.  The first question is something that Kenji Nakanishi asked me is the following.  Okay, you cannot be in L2.  You have this situation.  If you don't start in H1, you will never be in H1.  Okay, 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.  Okay, 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 okay.  And this is inside the proof.  You need G to be a integer, not close to zero, right?  What?  No, no, sorry.  That is in the bottom.  Yeah.  Okay.  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 zero belongs to HM X sub zero infinity, M  very large.  Suppose that you have a corner here, then U sub zero is not in H2 here.  Okay.  What can you say about the solution?  Okay.  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 never belongs to L2.  It doesn't matter how bad you start for any B. In the past, you never went here H2, because  you were H2, then you will be H2 there.  Then since decay implies regularity, and this is a Resolus Krusko-Flaminkis, you have this  also.  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.  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.  This is a corollary.  Then you have a question.  Then if you go to the theorem, this is the 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.  Is this related with the KDB only?  No.  You can also prove that for the Benjamin-Ohno equation.  The proof is more complicated.  You have to do it the iteration twice because you advance in half.  Then you can do it for the Benjamin-Ohno.  It's the same result, a little bit more complicated, but that's okay.  Then if you can do it for the Benjamin-Ohno and for the KDB, the question is, is this  related with integrability?  Okay.  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, you have Benjamin-Ohno, 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?  The answer should be no.  Let me give you some examples.  This is an example, consider that you have a quasi-linear KDB of this type.  Then you have to put the hypothesis here and this coefficient and in this coefficient,  okay, they are infinity bounded and so forth.  This is the important hypothesis, is that you have some strict deposit this person.  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, right?  Then here you are not C2, but here you are extremely nice.  Then if you have a propagation of regularity, that should something happen.  Then you have to kill this possibility and these are these hypothesis and these hypothesis.  Then under these two hypothesis, then you have a result of Cree-Capellar-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 cataract results.  If you want to believe that seven is sharp, I mean, you go to the proof and it's very  close to be sharp.  I mean, you use this argument.  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.  Here 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.  This is a typical, if you don't want to lose that delta, you have to use that.  Then you get the result.  This is with Derek Smith and Linarez.  Last year we proved if you start with seven derivatives, but in the half-lane, you have  more derivatives than 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.  Right, Nomin?  Then do you preserve other regularity?  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, right?  Is the following.  I don't think 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.  When the time is an integer, you will have a corner at the origin.  The corner appear, disappear, appear, disappear, and so forth.  You are not going to tell me that the singularity propagate with infinite speed.  They do?  Okay.  It just comes from there.  Okay.  Well, okay.  Okay.  Okay.  Okay.  Suppose that you are picking, you say, well, okay, then maybe this infinite 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 two and J derivatives and  you don't stay there, not even in the half length of something like this.  Okay.  And then finally, we have the question and I think I will finish ahead of time.  Yeah?  It's okay?  Okay.  The question is, do you have, there are some models that people have proposed to imitate  the KDV or right?  Oh, then I mean, have we mentioned here, someone mentioned here the BBM equation and then you  have the, 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, there is no influence anywhere on that.  Okay.  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, this is, in this, if you take the Kamasa Hall, that is an integral  system and if this K factor here is not there, then you have something called P-con solution  that are not C1 and they are traveling.  And it's a deep result of Brezan and Constantine through uniqueness, no, they prove existence  and Brezan and Chen and Sun prove uniqueness of this kind of solution.  See, you are starting with something that is in H1 as you watch.  I mean, this is the result.  And then you are not proving that this is continuous.  You are not having any continuous dependence of the data and you have this.  I mean, you are losing a lot related with the workforce.  However, if you restrict yourself to live here, everything that happened with the BBN  happened here if you start with libGIS-on.  Okay.  And this is a result with CIDERIS and we are not writing this yet.  And I think that I will stop here.  Thank you.  It's okay.  What?  Somebody should have a question.  So the principle result in the middle of your several slides is that HM, so we will 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-Framing.  That is Krusko-Framing.  If you have for decay, you have decay.  Yes.  In one side and only for T positive.  Only for T positive.  Decay in the other side.  This is Krusko-Framing.  I don't understand this last result.  So the first one is globally in time.  Then you say you belong to...  Okay.  The issue is this.  What...  If you start with something that is lipchis for the Kamasa Hall, right, 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 nor cannot blossom.  Okay.  This one, you say along the flow, I don't understand.  It's regular as data in minus T of E along the flow.  Yeah.  You can...  You can extend it.  Okay.  So you have the flow is...  Okay, let me see.  This is your flow.  Oh, okay.  Okay.  And that is well defined.  So on that, you are as good at initial data up to some time.  And after that, okay.  Because the flow never appears.  The flow could be the flow.  Okay.  Other questions?  Yes.  On that, for blow-up.  So is there a proof of blow-up solution of this kind?  Well, you mentioned.  So it's a possibility, but is there a proof of such a possibility at the end?  It blows up in the C1 nor, but you can continue this solution.  Yes, but you mean it's abstract.  Is there initial data?  Yes.  People have identified some data, some condition that this happened.  Even Molinet have some condition.  Can I start?  Yes.  Okay.  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.  Okay.  Do you know which one is for, which one is for?  What's that?  One half because you are in H1 forever.  One half.  Okay.  Okay.  Thank you.  Any more questions or comments?  So thank you again.  Thank you.  Thank you.  Thank you.  Thank you.  Thank you.  Thank you.  Thank you.  Thank you.
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