 So thank you, Tomá, and thank you all of the organizers for doing this. And it is a real, real pleasure to be here and to celebrate Frank's change of color in the hair, right, around the years. So as many of you, I have also made the exercise to remember when we met for the first time. It was at the institute, really, of Artan's study. It was, well, I guess, time flies many years ago. And I happened to go there because Carlos was... We visited in Jamburgen, and Jamburgen had invited also Frank. I guess he wanted to learn about blow-up, and then why not invite Frank for that. Anyway, so it happened that he was there, and it was kind of a very nice, nice period. We made two little kids. He was, of course, a very, very nice and familiar. Wilhelm was there also, by the way. And so it was kind of a reduced group. It was not the kind of program that you have so many people around that at some point there are so many things happening that it's hard to speak to anybody. But we had the three of us. Wilhelm was around. I guess he would also appear, or was visiting, I don't remember. And so we talked a lot. And the memory I have of myself of that time is that definitely it was what I would call a linear guy. And Frank, of course, was a nonlinear guy. But anyway, we were managed to talk. I don't know in what language, because our English is kind of peculiar in our conversations. I always had the feeling that when Rebeca, his wife, listened to us and said, come on, what are these people saying? But anyway, so yeah, we talk a lot. And I guess we are still talking, right, Frank? And there's many years after, and so far so good. So anyway, I hope I move a little bit, not too much from this statement of being a linear guy. And also, I have the opportunity. I mean, this was really a time of celebration, so you can use big words or strong words. So I, well, I was brave enough. I mean, if I have to give a talk on Frank's birthday, I have a... I should speak out of this if I can, or at least I should be allowed to use these words. So it is not clear what is the meaning of a blow-up here, but it's part of the talk. And of course, probably you will wonder how is it possible to put all these things together, blow-up, and 1D cubic and less, but this is part of the talk. And also, and just to finish this slide, it is not just by myself, but part of the talk is also a survey of things we have been doing along the years. And in all this business, Valeria Vanica, who is in the audience also, is playing a fundamental role. Okay, so let's move forward. This I have kind of a summary, and the starting of the summary is the 1D cubic and less. It could be with a plus or with a minus. But interestingly, they are not the small solutions. But in any case, we can do it for either plus or minus, and they are not the rough solutions. So the main part of the talk, if you want the main new thing, is this theorem with Valeria Renato Lucca and Nicolaus Betkoff, that we uploaded in the archive like one month ago. In fact, this was just after my visit to Paris. I have become kind of a religious person, and one of my first commandments is to visit Paris at least one month a year since I met this guy and tried to keep this as much as possible. And I really want to use this opportunity not just to thank Frank, but also the community around Frank and in Paris, because for me it has been really before and after. All these things I'm going to talk about, it couldn't have been possible if I wouldn't have been invited by different people, different departments. In this city, in particular, one of the departments was this one, Sergei, and invited by Frank. Anyway, so this is, as I said, we put it in the archive, not so long ago. And this is the statement. So imagine I'm telling you the data time one, saying that if I multiply the data by dc to the minus i squared, ix squared over four, then what I get is kind of a nice four times p periodic function. And I assume h stands for the solve of space, and zero plus means that I assume just epsilon derivatives in L2. Then the solution blows up at t equals zero. Of course, this one here and this one quarter are related, as you will see in the proof. This is one possible way of describing the result. Another possible way could be let's assume that now I'm giving you the information in the Fourier side. So what I say is that the Fourier transform of the, say the condition of time one multiplied now by e to the ic squared. And again, this plus here, this one here and this one are related. Now, because you see you are using the Fourier transform, so there is this changes of two pi for pi, depending also on the definition. If you put a one half there, then your life could be easier. So the result is that if I do that, then I am claiming that this is a two pi periodic function. And again, I am assuming epsilon derivatives in L2, then the solution blows up at t equals zero. And this is, say, the first part of the talk. And the second part of the talk will be mainly things that we already knew how to prove, and this was with Valeria, that nevertheless the u can be continued for negative time as a geometric solution. And of course you have a lot of quotations in here. And the talk is really to try to explain what do I mean by these words, here, here, and here. And the reason why I'm going from one to zero is because we have already done the other way around, which is to start at zero with Valeria and go up to one. And in a sense, once you know that you can do it, it is easier from an analytical point of view at least what matters regarding the cubic NLS. But really the difficulty there is the geometric part. But when you construct the solution from one to zero, then the geometric part is essentially the same. So really the new thing is that you can go from one to zero. So this is the summary. And as I said, I guess I am allowed to use beautiful words. And if you learn something from these blow-up people, is that you have to look for, say, universal, right, profiles, or like story tones or things like that, that can tell you how things are growing up or whatsoever. So I wonder if these two examples are universal from any kind of point of view. The first one, I'm pretty sure of it. The second one, it is not so clear, although I think it is. So the first one is very easy. You give me a parameter a, which is a real number. And this is a function of x and t. And what I came is this is a solution of cubic NLS, of the equation I wrote in the previous slide. And now if I try to understand it in terms of the statement of the theorem, what happens is that I pass this guy here, and of course the log of one is zero. So this is essentially nothing but a constant, which is your free parameter. And then the conclusion is that first, you could think that you are dealing with some kind of self-similar solution, and it's almost true. There is a correction of the phase, which is due to the rotation in variance of the equation that Daniel insisted a lot in his talk. But it is not self-similar, but it's almost self-similar. And then, of course, the other conclusion is that this blows up in which sense that, well, you cannot define the solution at time zero. Why? Because when you pass this correction of the phase to the left, then you obtain a limit. So because this has a limit and this does not have a limit, then the solution cannot have a limit. So as simple as that. So if you want, this is my first notion that something is going on at time zero. This was an observation that Carlos, can you, Gustavo Ponte and myself did a long time ago, and it was more than related to the scalings. It was related to another group of symmetries of cubic NLS, which is very important, which is the so-called Galilean variance, of Galilean symmetry, which will appear in a minute. So this is the very first simple example. But I want to convince you that this example matters. And then I have to recall some very old and relevant results by Toroza Wabalt, the scattering of 1D cubic NLS. That is all we know that it was really very true because it was an open question for quite some time and the reason was that it was not the usual scattering. It has to be long range. And you can reformulate it as follows. You can start with a profile as good as you want and then you build up this guy, which is very similar to my first example, except that you have this amplitude and it depends on your profile left. And then what Othawa did was to construct solutions of this problem such that you have an error and the error goes to zero in appropriate topology when t goes to infinity. But of course another of the symmetries of cubic NLS is the scaling. So you can now play a little bit with the scaling. I say instead of starting with just with f, let's start with f lambda of xi direction. I'm sorry, direction. The reason why I'm using psi here, which is x over t is because really this is really more on the free level and the classical linear Schrodinger equation. This guy is really the Fourier transform of the initial conditions. So that's the reason why I'm calling this L. So now try to do the same in L. In lambda now you have a family of solutions. And at least formally this family of solutions converges really to epsilon A and A is nothing but f of zero. That really I prefer to understand it as the integral of the anti Fourier transform of f. And really this is the scattering behavior of a nice solution at time zero. So it doesn't matter what is the functional setting you start with, that if you want to construct things that are, say, well behaved with respect to scaling, you leave the class. EA has no decay at infinity. This is also very clear, for example, in modified KB, which is like a sister equation of cubic errors. So from this point of view I claim this DCI is a relevant solution. And it is non perturbative as Daniel insisted the other day because really it is a solution of a non-linear OD, which is this one. And it will be pretty clear in a minute why this is the right hand side. As you see there is no Laplacian here. Laplacian has been erased. But on the other hand you have a 1 over T that comes from somewhere. And then you have to deal with this guy. This is a non-linear guy. The only thing that modifies is the phase, not the amplitude, but it is there. All right, so now the second example is much, in a sense, is much more weird. Now I call it D, and D is standing for Dirac. From the Dirac comb, not for... Okay, so this Cm is just a constant. M eventually will have a meaning, a geometrical meaning, which is the number of sides of a regular polygon. But you can imagine it's just a constant. And then what I have here is nothing but just a Fourier series, which of course is very much related to the free evolution. And now what I have to tell you is that this will be kind of as before, it's a solution of the constant coefficient, the linear Schrodinger equation. But I have to modify it in order to fit with the non-linearity. So now what is the change of phase? It is you have, as before, the logarithmic correction with the amplitude to the square. And now you have a completely new y. And you have this R of M that will be clear in a minute what is the meaning of that. Observe that you are summing in the non-resonant set, which is typically called in this setting, which M is different from zero, so you're ready to integrate there in time and to obtain a synthetic expansion of this phase in terms of negative powers of t. So now that this is a solution in a sense, of course this is very weird because there is no notion of computing the absolute value squared of this guy. There is no way of speaking of that, so you have to make meaning of what you mean by dissolving the cubic NLS. So it's just the existence of a solution. This is something that we did, Valerio and I, with Bravin, and it was one year ago, I guess, or two years ago. So then the next computation I need you to pay attention to is simple, but it is important, which is the Poisson-Sovetian formula, right? So Poisson-Sovetian formula tells you that you can instead of if you put delta, this is the rad comb, that's the reason of the D. So what is the Fourier transform of this guy? It's nothing but this with t equals 0. So now to add in these frequencies, the only thing you have to do is to look at free evolution. So if you go from right to left, I guess you agree that this is true. But I'm sorry, but the next three lines are kind of important. So now you can say what is the solution? We know what is the solution. This is the free evolution, this is the delta, so this is the fundamental solution just translated in J. So you write it and this is it, okay? And now you make the expansion of the square and this is what you obtain, right? So you obtain something which is very reminiscent of the first example, right? It's nothing by the fundamental solution with one delta and you essentially obtain something very similar to that but now, well, it is at this location, x over 2t and the frequency is now 1 over 40, right? This is the version of the conformal transformation at the free level. So anyway, so the conclusion is that if I look at time one and don't pay too much attention at 4 pi, 2 pi and whatever, because quite likely they are not correct, let's assume that 2 pi is 1. So essentially what I have on the left is that it's at a given time, so I assume this is 1 and I am re-scaling and I am saying what is the solution. So time one is just this guy that of course is periodic but it is quite rough, right? So it is a distribution, that's what it is. So what we claim is that then what I wrote in the first line is a solution. So I guess that I need to tell you what is this R of m and by this eventually it will be the cardinality. So this is the cardinality of a set that measures the resonance set that because you are in 1D cubic analysis is extremely simple but that doesn't mean that the arithmetic function that is behind that is not important. So let's try to play a small modification of the previous game. The previous game was I put deltas in the integers, right? I look at the free evolution and now I put instead of the coefficient 1, I put a general coefficient a j that there is no reason why it's going to be constant so it has to depend on time. And I know from the very first example that I have to make a modification of the phase because otherwise I will not be able to find the source, right? So that this answer is so simple, it's nice and gives you some game which is a very simple idea is due to quita that I forgot to mention in the slide. So anyway, so now we reproduce what we are doing on the conjecture and this is what I am now more or less telling you how you go from time 0 to time 1, right? So you say let's assume that a j are given by, capital a j are given by some, let us call it Fourier coefficients and I look for a solution in a perturbative way where rj of t goes to 0. So and I know that this should be the answer or this should be called you want the right gauge. Anyway, so then because of precisely previous computation I have done in the, I mean the computation I have done in the previous slide, it turns out that if u is your unknown, it's much better to look at what is called the conformal transformation of the u that because essentially this is the lens transformation that Nicola was using before but when you don't have the harmonic oscillator so it's something that happens at the free level. And now what is the equation? u is the solution of cubic NLS and you wonder what is the equation for this new unknown v and this was precisely in the previous talk and it is the same as before but now it is non-autonomous you have the 1 over t that appear in the first study I wrote in the second slide of the talk. So in a few words, I want to find solutions of this type and the way of constructing it is instead of working with the u I work with the conformal transformation I do the computations and I end up with another cubic NLS which is essentially the same except that it is non-autonomous I have the 1 over t that of course is non-integral and that's the reason why these logs appear and also I am changing time so what was t between 0 and 1 now it's becoming t between 1 and infinity but imagine that tau is in between 1 and 2 you say come on then the v is going to be periodic and then you are looking for solutions in the v so quite likely you are in business and this is what happens so now you call Bourguin and say what he did and what he did was essentially this he looks for a solution of this problem in this way and renormalized with respect to the phase you expect to be the right one and you end up in an ODE system for the coefficients vj of tau and what is the relation between this vj of tau and the aj is very simple it's just this one okay so yes the equation depends on j this equation? yes in this aj is good because this aj is given yes but the sum is over j no it does not depend on j because when you do all this together what you end up is something which is periodic right so this is the way I am writing it is this the question? this is the sum ah the sum, the sum I'm sorry the sum is in j yeah the sum is in j this is nothing but a Fourier series but of course this is what Bourguin did you put your Fourier coefficient depending on time and try to solve for this new Fourier coefficient I think the question was the third line this one no no no ah sorry this is I'm sorry this is wrong yes yeah so this is what is what you have to put here okay yeah it is a fun, I'm sorry they are completely right I guess too many j's so instead of aj of t instead of aj you have to put some a of d a aj of t and you really this part of the business what is what you have to put here so it will be clear in a minute what I mean by this I guess it was too much copy paste so well anyway so this is the audio system you finish with and of course so it is a cubic nls you plug your Fourier series and you end up with an audio system for each Fourier coefficient and this is the j, okay I'm sorry so now that you fix the you want to compute what is the dynamics I mean the audio of the j Fourier coefficient then this is the guy you have to put here and it's because of the logarithmic correction of the face I wrote okay is it clear now good so well and then you of course I have clarified the 1 over t and this is the reason of the log correction of the face because you really want to kill this guy at zero otherwise you have a problem of course this is not a new idea in fact it's a new idea as far as the first time I saw it was with again trying to solve this kind of problem not in the 1d case but in higher so then if because you assume that bj at time 0 time 1 in this time infinity in this case is going to be a j square so this goes to 0 at time infinity with the power rate that I will allow you to integrate in time alright so this is it so now finally what is the and of course these are the the frequencies that appears because you have the log of t and you have the you have the bj here and you have the three other guys and then this comes with this particular expression and now what is the RM the RM is that remember that I started with j1, j2, j3 so this M is the difference the corresponding difference of the squares so you factorize and this is the resonance set which is has a very simple expression in the 1d case but it's better to write it this way you have given an integer that is written as in fact it's an even integer and now for all the pairs that this thing happens then you have to sum in this RM so really in the case in the example 2 the bj's are a constant right the bj's are a constant and then what you count is just the cardinality of this guy which is the number essentially the number of divisors of M which is an arithmetic simple function but that really the behavior the behavior of M is complex so then this is the second example so the second example is in the pre if I put all the bj's to be a constant then I can reduce it to a finite system that the only thing I have to do is to do the corresponding integral in time of 1 over t of this series that's the worst the reason of this so the theorem we prove is first goes from the statement of the theorem what says is that we go from 1 to 0 so I don't fix a priori the fj's right I don't know which disguise are and secondly that well I assume not all the Fourier coefficients are identically neither they are identically there is only one which is the first example 1 0 neither it is the second example which are all the Fourier coefficients are the same equal to a constant right so I want to do something in between and what I assume and the theorem states that I have a theorem as long as these a j's are in L2 epsilon a little L2 in this case because these a j's are the Fourier coefficients so h epsilon means that they are in L2 against an epsilon weight so this is the statement of the theorem so in this case of course the value at 1 it is precisely what I was saying it is just the Fourier series that I renormalize with respect to the phase because it's much easier to give the dynamics of the Fourier coefficient when it is written in this way alright so now let me start to convince you that this there is some kind of blow up here this is what I will try to explain in this slide because my functional setting is kind of peculiar let's put it this way so I have the free evolution and then of course what we know is that if I use Fourier transform then if I renormalize by the phase in time 1 i to the IC2 then right this guy is not changing time because all the modes are frozen by this phase so if I renormalize the phase nothing happens and in particular if I start with something which is periodic so it's something like this then it will remain periodic and in fact all the Fourier coefficients are the same and this is what we are starting with I start telling you that at time 1 this is periodic so what I expect is to work with not with u hat but with the corresponding renormalization of the phase and the first question is why is this going to be periodic I mean there is no reason that the periodicity of this guy will be stable because of the non-linearity well it turns out that it is and essentially this is one of the few examples if you try to do this in modified kdb you definitely are dead this does not happen and there is a very simple way of convincing yourself you write the equation for the omega that now will be an integral differential equation because you are in the Fourier side but you do you write your guys in terms of the Fourier transform and again you have the difference of the squares that factorize as before I did it before with the Fourier coefficients but now I do it with with psi and it is exactly the same factorization but definitely this is invariant under translations so if I change c1, c2 and c3 by the same amount this does not change at all so if I start with something which is periodic it is going to be periodic and this is really the solutions the framework we construct I could tell you what is the similar statement in the physical space but I guess I am getting out of time so better if I rush a little bit because otherwise I guess I will not tell many things so this is the theorem the theorem is that if I start with an initial condition at time 1 that is after the multiplication by the right phase is periodic then there exists a solution of the 1dq we can analyze that grows up infinitely in the following sense if I start with this frame this frame is broken at time 0 it remains in the continuous I mean the omega is still periodic it remains in the hs for the torus but this does not happen in t equals 0 it is not a question of size really the size of the Fourier coefficients are the supremum remains the same is a question of the phases so what we say is that the phase at time 0 has this logarithmic behavior where now of course the difficulty is to find what is this number what is the right choice you have to put in order to have a limit and that is the reason why this way back from 1 to 0 is more complicated and this is it this is my notion of law I have a structure and of course the L2 norm is preserved because this is not a question that I don't have a conservation law I do have a conservation law which is the L2 norm in the period so that means that the sum of the aj square is something that remains constant is a question that the phase blows up at every aj and at the phase space is exactly the same except that the phase will go to the delta and x-j but it will not have a limit unless you put this logarithmic correction right it's like I started with the modified direct comb I put not the same weights in all the deltas but these aj's and unless I make this modification of the phase in each j I will not have the limit so this is the precise statement ok so hence the theorem this is what I said before this theorem in a sense interpolates between the two examples you could wonder what happens if instead of the aj's being in L2 with epsilon weight it's in Llp quite likely there are things that you can do in this direction but at least with the smallness if you want to avoid the smallness you will have to work and really then the problem becomes just to have problems in the geometric side and of course the interesting thing is what happens when p equals plus infinity probably that is too much but there is a scenario within this setting that maybe could be affordable but I will not speak about this so now this is of course a notation that at some point was familiar to me after talking to Frank and you want to obtain some some blow up result then you have to keep track of the compact symmetries or the non-compact symmetries so which are the non-compact symmetries here are the translations in space and and time then you also have the validating variance but of course the translations in time are used to determine what is the blow up time this is where the translation in time is used otherwise also including the scaling I am in a framework which is invariant by these symmetries of course by translations because the question itself is translation invariant but also by the Galilean invariance because I am discovering my space precisely in Fourier side and I am saying it is a 2-periodic and also by scaling because I am starting with delta functions if you want at time 0 scaling invariance so this is one of the bonus of the result of course I have to fix some setting which is this persistent and also notice that my condition in physical space is L2 log I mean I am not assuming any kind of decay at infinity although it is true that I am fixing the structure right so it is an L2 log result alright so about the previous results of course as you can imagine this is a very old problem I started with Casenave and I am not writing the years because what likely I would be wrong so it is by Casenave and Casenave is with 2 S then the neighbor and Velo and of course Susumi for L2 again for L2 in the periodic case then about this there was a first result by Carlos Kenny Gustavo once and myself and this was for the focusing case and for the non-focusing and father was extended by Chris Colleander and Tao and there are many other important things in terms of the growth of solid norms and there is a huge literature on this and as far as I know these are the names then with respect to well postness and S here stands for the solid class and it is the the non-focusing solid class this is a hard problem and there was the first results by Koh and Tataru using the complete integrability then by Kilipe, Visan and Sang and then by Harrop, Kilipe and Visan that go all the way up to minus 1 half which is the critical exponent so all these results are sub-critical but of course the class is I mean it is huge because all the class of course it is not Galilean invariant but in this regime Galilean invariance penalizes you so it's something that you really have to deal with so and then but then there is another possible setting that is much closer to the Fourier analysis point of view which is to assume that that the Fourier transform lives in some LP that global in space and of course this is translation invariant which by is infinity so if you want the previous result can be said as a sub-space of this L infinity hat which is the way we typically call this space of where you can say that you have a global world-post mystery where this is local and then this was something that started with Anna Vargas and then Alex Gurnock really put it in this framework and the final result which is this one where that also includes the periodic case at the critical level is essentially the very much in the spirit of what I have told you so far is work with Valeria and Brabine and more recently and this is that he will defend in a few weeks a student of Valeria which is Gouvernin now about the proof I have essentially given you all the ingredients except that now I have to guess what is the the the question that Tomas was asking what is this guy and of course the lambda m is the same the RM is the same and now what is the first observation the first observation is that first you try to understand what is the dynamic of the action what can you say about the amplitude and it turns out that this is really nice so in fact you can integrate and this is of course the proof you can integrate this from one to infinity remember that tau changes the direction of time so you are going from one to infinity so this is your dynamical system and then what you can say is that that the integral of the right hand side is well defined at infinity but of course it is not absolutely convergent so it is an oscillator to integrate and I think this happens also at the usual QNLS and my impression is that we haven't paid too much attention to this fact because you see it is precisely in these oscillations in how so because this is integrable next slide what you can say is that you have a limit so at least you control the modulus of the guy you are looking for this limit then you do the previous trick and you subtract this part and then you take the leading order which is the ODE for each J and then you renormalize and you see that this guy has a limit what I am saying is that what I think is very interesting in the dynamics is in which sense this limit is what are the fluctuations and these fluctuations see this arithmetic function and this arithmetic function has some say stochastic properties which are not the usual ones they are much closer to properties like the one that are called so this was the selection this is what I was trying to say but for example you pick up this example and you try to analyze how are these fluctuations really you have this typical intermittent phenomena that I guess people say in weak turbulence like alright and here I mean you can measure things I am not saying just waving this is not just hand waving this was what I am trying to say so which are the relating systems the relating systems are the geometric flow of curves in 3D which is the so called binormal flow and Liya stands for localized induction approximation and then the two examples would be that I start with a curve in 3D that has a corner at times at x equals 0 so this is the corner so you have two vectors, you need vectors that are not the same and you have this line on the left, on the right and this line on the left this is one of the systems the second example which is when all the coefficients are the same if you make the right choice of the CM I was mentioning before and this is a delicate issue and I will extract a regular polygon so you are looking at the evolution of this geometric equation for starting with a regular polygon and I will show you a video in a minute and that you can expect to have a solution for this guys is something really not is something that I have never expected but fortunately enough Valeria call our attention we here is the Laos our attention of paper is as much where they do I mean they have theorems but they do in particular this numerical simulation of the evolution of a curve which is given by a square of a triangle numerically and then you start to believe that you can do math with that anyway so there is another related system which is now in the physical literature is related to what is called the Heisenberg change or the Landau-Liszi equation which is the one you obtain looking at the derivative with respect to x of the curve the tangent vector and this is it you differentiate formally here you immediately obtain this case so now we are dealing with guys these leaves in 3D these leaves the target of this is the unit sphere so I am speaking about systems and the connection is a very old and beautiful idea by Hasim Otto that if you take the evolution of the tangent vector and you use for example you can just use the frenet frame or the parallel frame and you call alpha and beta your curvatures you say your geometric quantities then you build up a complex function that amazingly solves this this cubic NLS but with the freedom of gauge you have the freedom always to use this capital A of t which is the one that allows me to in a sense forget about this this loss of information of the face so this is what I am trying to say that the geometric structures implies that the singularities of you are human are harmless and the solution can be continued as a geometric object but of course life is not that easy in fact there are part of the geometric object that does not survive which is precisely the frame of the normal vector and this you have to raise a blow up but again here you use model how are the beautiful words modulation here and then you will be able to do your business why because you identify the asymptotic profile which is given by a similar solution and now I need to say a few words about the self-similar solution because this is the one of the key objects so the self-similar solution with respect to scaling will be written in this way and what it is kind of interesting something that honestly I don't know why but this is what happens is that the simplest way of understanding this nonlinear profile is by looking at the Fourier transform and you have to do a nonlinear change of variable in the descent of this e to the ITC square I am modifying all the time so when you do and this is because eventually you are going to deal with the derivative of the tangent but this is the key guy, I mean you compute the Fourier transform and you do a nonlinear change of variable and this is your good and known and then you identify what is the the singularity is this OD which is a singular equation very simple so you have really linearized the problem of course you have also linearized the problem through the Frenet frame but this is this is the guy and this is something in particular and now your object which is this is the important object which is the derivative of the tangent vector the Fourier transform is nothing but this say the solution of this equation evaluated here in particular what happens and this is not easy this is where the difficulty comes that the information you have is one part I mean because eventually in order to continue the solution geometrically you have to find the information you lose in the phase from somewhere so this information has to be in the in the OD and it happens that at a frequency zero what you know essentially is they say the the angle of your solution the two lines at infinity but the renormalization of the normal vector I mean the modulation is an information that you have at infinity so you have to solve the matching problem which is you have same information at zero and I have to tell you precisely what happens at plus infinity and as you know this happens with very few equations in part of the literature this is like a hint that you are dealing with a completely integrable system that in fact it is the case the cubic analysis is a completely integrable system but the conclusion of all this is that the Fourier transform for some reason is playing a role so this is the dynamic I was saying so well this is this was done by Gerard and Schmetz and it was nothing too complicated they use a scheme of of bad key which was a very old scheme and it's in fact just finite differences and the only thing I wanted to insist is that amazingly you go up and up with the same guy this is there is no reason why this will happen but why I think that the similar solution is the universal thing is because you see when you close to the singularity in each corner what you see you have to believe me is the similar solution and the similar solution has the usual two scales of the Schrodinger equation which is the one given by the phase which is similar but then you have the other linear scale whichever the oscillations start to appear and of course because you are in a periodic setting all these waves start to interact and create very complicated things as the ones that you see here this is something that the Laothan and I put in the picture which is to illuminate what is the trajectory of a point and you see this guy which is pretty complicated and has is very famous is related to the to the so-called Riemann's no differentiable function so now I guess I am three minutes okay so to finish that really there are things that are happening which is so this is kind of a mixture of linear and nonlinear you are measuring in some kind of equilibrium with most of the dynamics is something that you recognize at the linear level this is a complicated dynamics that survives even with nonlinear interaction but there are facts that are not completely linear and there is some cascade of energy or some transfer of energy you don't want to use strong word which is more or less as follows first of all the natural quantity t is for the energy is this and you can write it in the conservation law so this is a natural energy but I was trying to convince you that you can understand the solution also in Fourier transform side and it turns out that you can identify this energy associated to this guy in the Fourier transform side as an scattering energy ok so this is really the energy of the system you see it in the Fourier transform side when you look far away at a given time in frequency but on the other hand even though you have the L2 conservation law there is some growth of is L2 in a very precise sense so I don't have the time to explain but anyway so what happens is really the Fourier modes even if I start with a very nice as I said the function v can be very nice the periodic function so it can be h infinity so it is not a question that it is not bounding at time 1 however when you look at the Fourier transform and you go to t going to 0 the Fourier transform blows up and this is some kind of say transfer of energy which is complementary to the well known of the IT and Honey-Pausadeur Esbetkov and Bezilia so let me finish just with a simple idea of the proof because the proof is mainly written in the Fourier and with this I finish so it is written in the Fourier side and of course here I am going to not be imprecise but at least to try to convince you where the new resonance comes from because it is a mixture of a non-local guy with a local guy so I start with the equation for the Frenet equation this is tx is the curvature but it is better to write it in this way using the complex notation and n is the complex vector that tells me how the coordinates of the normal vector are changing with time and of course this is symplectic you have to construct an orthonormal frame so then remember that you is essentially this guy times a nice function because it is periodic but it is nice, it can be as regular in fact I have started with just two Fourier modes negative one and one and it is very simple now we know how to do it in a quite straightforward way to say that t goes to two given vectors which are really frozen so then I can so then I have to imagine that you want to understand how we send what you are saying is that the derivative behaves as essentially at infinity because the problem is to make this integral finite at infinity is a given constant times a very well known guy so you are going to integrate it by parts and this is the role of the different scales in the Schrodinger equation that when you integrate by parts even though you have here the parabolic scaling when you integrate by parts you start with you end up with the hyperbolic scale and this is your basic integration by parts so you are going to take good powers of t here now t is in between 0 and 1 so I am in physical space but you have to pay with the distance of xa to one of your Fourier modes, imagine r is going to be the negative one so then you are going to plug this into here so this is what I am saying and it is going to be essentially this guy times what happens close to 0 this is part of the hand waving but at least it is pretty clear that after integration by parts this is what you are going to obtain and it is not so simple to say why I can now freeze the time in 0 but now if I take this in the upper equation and I wrote what is the u then immediately I see that I have created new modes which comes from the differences of the phases so the new modes are just the differences so if I only have two modes then I will only have a new mode which is the difference but it comes with a very precise frequency which is 1 over t so the Fourier transform this value x over 2t will be as big as this and of course x minus r now this is a key point that because you have more than one Fourier mode you lose the symmetry from left to right and then even though this was true for say in absolute value even then so that if you interact you are going to have the principal value because you have two corners then really you have this guy in this region and when you compute the coefficient which is precisely this will cancel this phase and you have to integrate 1 over x which gives you a log so this is the picture and with this I finished so I have the two corners then because of the self-similar solution I know that nothing happens in the self-similar region square root of t but as soon as I am far there I start to see all these oscillations you saw in the picture and they start to interact each other and then I scale t you start to have new modes that do not decay too fast and eventually give you a logarithmic connection and so let me just go to the last one ok it's clear what is there right so that's all so thank you Luisa do you have some questions or remarks so you started with regular polygon but you are here regular polygon well we have no idea even with a regular polygon the only thing we have is the candidate to the solution so example 2 is we say this is a solution in a very very weak way and then what we know when I put it in the computer and I look at the polygons that happens at the rational times fit with the evolution but we don't even know how to construct the curve for irrational times so this is we are working on that but this is because all these Fourier series I mean it's much harder than to deal with the Riemann's function Riemann's function at the end is absolutely integrable and all this Talbot effect a few experts here on this Talbot effect this it took quite a long time to understand Riemann's function and so now it's understood but the guy we have to deal with for constructing the curve for irrational times it has one less derivative so Riemann's function is 1 over j square and the one you have to deal with is 1 over j so this is much more delicate and of course you have to solve for a net frame for that so this is very much in the rough very rough path things but here you have a very good description in terms of Fourier series of what it is your curvature and torsion so then your question is now what happens if I have say something like this right of course you can do numerical experiments and then you see that essentially you can superpose this is the reason why I think that the regular polygons are kind of universal because it's like you have two different periods one is this and the other one is this one so you will generate like one for one period the another solution for the other period and then you have to look at the interaction like if you had two solitudes right so this quite likely can be doable but not yet other questions no one comment the comment is about the solution EA when you look at it in self-similar variable it's a penalty solution what do you mean by self-similar variables x by x over square root of t well it is not periodic it's e to the i x square this is the point that say the kind of strange behavior that's happening is that you are maybe if you look at the time you will have your p i s x square and this solution x in s it's periodic this is the solution I don't know I will say no no because you have time also we will explain later because I don't think I am understanding you and a second question sorry for this stupid question the parameter is in higher dimension because it seems that I would say I guess that maybe Daniel is agreeing with this the difficulty here is one dimension if you look at of course what happens in higher dimensions I am not saying that the higher dimensions are easier I mean the Schrodinger Mathems these people did in 2D was really to the force but I am speaking about something which is critical and also to understand this long range scattering and as far as I know this is a basic question that we don't understand yet we don't have a good threshold theorem of when you have scattering when you don't have for example all these solutions are huge so they are not as small and in a sense all the behavior is one of the long range scattering and this is because I am using the conformal transformation that kills the solitons of course but for me this is not clarified at all so definitely higher d is very interesting but I mean I have enough job here many things to do here ok, no more questions thank you again