A continuous family of energies for the Korteweg-de Vries equation...
Koch, Herbert
A continuous family of energies for the Korteweg - de Vries equation and the cubic nonlinear Schrödinger equation The KdV equation and the cubic NLS equation are known to have an infinite number of conserved energies. I will explain the construction of a continuous family of conserved energies interpolating the classical energies
 Well, thank you. I'm grateful for being here and I want to thank the organizers for this nice and interesting conference. I'm going to talk about concept energies and this is joint work, concept energies for NLS, MKDV and KDV and this is joint work with Daniel Tataru. And before I state exactly what we do, I want to give the equations. So, there is the NLS equation. It's I one-dimensional IDTU plus UXX is equal to plus or minus U squared U. So, plus is the minus case is focusing, then there are solitons and the plus case is defocusing and there are no solitons. Then there is the M modified KDV equation, which is UT plus UXX and I realize that the first thing I didn't want to put attention to plus minus 6, what do you want to write it? U squared, write it like this, 2 U squared UX is equal to 0. Now, I want to add in a plus or minus and then it's the focusing or the defocusing one. I want to write it with the absolute value here in order to allow complex solutions and it's a complex MKDV, otherwise it's the real one. And then there's a KDV equation, which is UT plus UXX minus 6 U UX is equal to 0. And all these equations are pretty connected. There are integrable and the integrability shows in the existence of a lux pair and I want to write it in the following fashion as a system. Psi x, Psi is a that's the x derivative of a map to C2 is equal to, I know I only write it for the defocusing case, for the plus case, I see U bar minus plus I see C. So this is the lux equation. C is a complex parameter and the amazing fact is that if you complemented with an equation for the time derivative is equal to I, now I have to look it up because I can't remember these things. Plus U squared, value U squared to I see U plus UX minus to I see U bar minus U bar X to C squared plus U squared C. So these are two first order equations and the interesting, the very interesting structure is that if we want to solve them simultaneously for given data at one point, general, then this is only possible if a compatibility condition is satisfied and the compatibility condition is exactly the defocusing NLS here on this side. So and the connection to MK2V is the following. So for MK2V there's the same structure with the same operator here, the same equation here and a different equation on this side. So there's a different time equation for the time derivative and then there's the same structure. This is solvable for given data at a point let's say 0, 0, X equal to 0, T equal to 0 for general data, if and only if then U satisfies the MK2V equation. And for the focusing case there are some sign changes which I don't want to go into. So the only case, the only sign change here is a minor sign and here it is more complex but this equation doesn't play a big role for what I'm going to tell. Now what is the effect of that? Well suppose we take C, let's say we say imaginary part of C, larger equal to 0. Then we might look for your solutions. I mean it's a 2 by 2 system so there is a two dimensional space of solutions and I want to normalize them on the left. So if I want to have that C behaves like e to the minus i Cx, 0 on the left, so the case to the left for X to minus infinity then on the right, let's suppose that U is compactly supported, then on the right it decomposes in a linear combination of two, there's similar fundamental system on the right, then on the right we get the decomposition that C behaves like t to the minus 1 of C e to the minus i Cx and well here we have to be a bit more careful as X tends to plus infinity if C is the real number. So we get a linear combination of the two solutions which you would have of a fundamental system which would have a few which would be equal to 0 and I use this to define t to the minus 1 which works well even if the imaginary part of C is positive and there's a reflection coefficient, it's called the transmission coefficient and the reflection coefficient, so t is the transmission coefficient. Now the fact that the equation, the analysis equation is the compatibility condition for solving these equations allows us to solve this equation at plus infinity and check what happens with this plus infinity and minus infinity and check what happens with the function C. So at plus or minus infinity we don't see the effect of U, so what we see here is a constant coefficient differential operator, it gives the time derivative of C in terms of C here and no U, so we can check this and if we check this on the left then there is some time evolution, if we do it on the right there is some time evolution and what we see there is that t of C is independent of time, if we have a solution to the analysis equation. So whatever, if you look for quantities which are conserved for NLS or MKTV or KTV, we simply have to look out the transmission coefficient, have to see what we can define it and this is a conserved quantity at each point, we may integrate it, we may do whatever we want with it and this gives a conserved quantity. So now this is some sort old, fairly old, goes back to well in this context to Aplovitz for AKNS, this is Zegor, Newell and K-Kalp, sorry, so they did it for the non-inheriting equation and then there is an expansion of t of C which is equal to minus 1 over 2i times a sum, this is formal, we should try to make this, this is a formal sum and then there are the energies EjC to the minus J minus 1, so this is at C tending to i infinity or depending on what you assume on the initial data and how it is, it is the Laurent series at infinity and then E1 is equal to the integral u squared dx and E2 is equal to the integral ux, sorry, u2 is equal to i times u u bar x momentum and E3, I did something stupid, I want to have this one, 0, 1 and 2 equal to the integral ux squared plus 2 times u to the power 4, no, it is u to the power 4 dx, which one is it? It has a 2 here, so and then you can continue and it is classical to expand these things and to get recursive formulas and what I want to do is I want to, we did in the John Dewey Pistanio, we sort of interpolated between these quantities in order to get a continuous family of conserved quantities, well that is not exactly what we did, so more precisely we interpolated between the even ones and not even that, we interpolated between linear combinations of the even ones. So here is what we get, well the things which we define, so let us first look at the focusing case for n or less in MKDV, then ES can be defined as an integral over, well at least for Schwartz functions, for Schwartz potentials as an integral over the real line, over the logarithm, it is always the logarithm which occurs of the transmission coefficient, evaluated at the real axis, so for that we need some decay on some Schwartz functions and we would have to study the decay which we do not do, but let us suppose that the logarithm of t is a Schwartz function, so say a bit more about when these things are defined later, so the one way to define it is to integrate over the real axis and to put a weight which is something like the Fourier weight for sub-Lav spaces, c to the 2s, so this is for s larger than minus a half and then another way of expressing that is to move the contour of integration to the line from i to i times infinity and that gives the second formula which is equal to 4 sin pi s integral from 1 to infinity, so now we are writing this in a real fashion, tau squared minus 1 to the power s times this logarithm evaluated at i tau over 2, I see I forgot factor 1 over 2, it should be evaluated at psi over 2 on the left hand side, so we evaluate it at i tau over 2 and then we have to do some corrections because the weight c to the s goes at infinity polynomially, so we have to correct that in order to be able to do a contour integration, so we subtract the, oh I messed up the, I wanted to call these guys h, so h1, h2, yeah so we correct, we have to correct the behavior at infinity a little bit, thank you, oh sorry, yeah, so that's the only place where I, wrote it, okay so, so we have to correct the behavior at infinity in order to get, to be able to work with a contour integration, to move the contour and if we do it in this fashion then this doesn't depend on how many terms we correct if we have the regularity and so we get this formula here, what you see here is that if s is an integer then the sine pi s vanishes and here you get the linear combination of the classical conserved quantities, so in that sense we interpolate between the classical conserved quantities, well then we can do the same thing for the focusing, NLS if you do the same thing for the focusing NLS then we have to adjust for eigenvalues of the AKNS operator, so the AKNS operator has eigenvalues in the upper half plane and we have to adjust for them and there is a function there's a way of doing that, so we have to add a sum over the eigenvalues with a multiplicity mj over some, with some function psi and nothing changes on the right hand side. Now for KTV that's the same structure, first if s is equal is larger than minus 1 then this is a generalization of the classical formula of Fadiev, there is a different function psi, we have to adjust for the eigenvalues, so the eigenvalues in this case are minus, no kappa k squared and on the line from i to infinity there's a similar expression as for the NLS equation and again we have to adjust for the classical energies. Something special happens at s equal to minus 1, so the left hand side makes sense for minus 1 and for the right hand side there is this part of the integral which has the effect of converging to a delta function combined with the sine pi s, so the outcome is then 8 times the transmission coefficient evaluated at half minus this integral u and here one has to do some normalization which I don't want to take much about, talk much about. So these are conserved quantities for NLS and MKTV, they go back to AKNS or to Fadiev, this relation between the different contour integrals and these are the objects we want to work with. Now defining these objects for Schwarz functions is trivial because then the scattering transform becomes something nice, the logarithm of t is a Schwarz function on the real axis and then it's an exercise in complex analysis maybe not an entirely easy one to get these formulas. So as I said before since t is independent of time whenever you have a solution whatever you do with t is going to give you a conserved quantity. So the question is how to relate them to things we work with in analysis, how to relate them to sub-lawed spaces. And I want to give the results first and then explain a bit about the proof. Thank you. Yeah I tried it before but it's always a bit different when I do it on the blackboard. Is it okay? I guess you still can have an impression of the formulas even if you don't see all of them. So the theorem is Daniel Tataru. So what we do is we look at the right hand side. We define the quantities by the right hand side and then it turns out that it makes all the sense with the left hand side. So the first thing is if u is in Hs then both sides are well defined. Both sides are well defined. But basically you want to look at the more on the thing on the right hand side. And so the map from Hs to Es is continuous in u and also in S whenever u is sufficiently regular. And it is analytic if i over 2 is not an eigenvalue. Well you can't expect it to be an eigenvalue if this so this is in the focusing case because then this function xi is not analytic at 1. So this has to be excluded. But otherwise it is analytic and it's jointly analytic in S and u in the appropriate sense. So whenever u is in Hs then it's analytic in S for S and then Es is analytic in u and S when S is less than sigma. So I don't want to write that but in the appropriate sense it is as all the properties one would want to have it. And the nice thing is we can play with both sides in order to get whatever is more convenient. Second, well how does it compare? So this says that it's analytic so whenever u is in Hs then this is defined but the question is does this control the Hs now? Well so we can look at Es of u minus u in Hs where it is basically the same as the left hand side but for the Fourier transform evaluated I didn't correct all the xi over 2sz. So this is on the Fourier side similar to the integral on the left hand side and this is less or equal to a constant times u in Hs squared and now I want to use a strange norm here and I want to explain it later on with u in L2, D2, D2 is less than delta for some delta. So this tells us that the quadratic part of I mean it's analytic so we can do an analytic expansion in u at 0 and the quadratic part of that is exactly the Hs norm. So for small if this quantity is small I'm going to say more about that in a minute if this quantity is small then this quantity and the Hs norm are pretty close together. And also what happens in that case is this function xi of the eigenvalue kappa j is larger or equal to a constant times kappa j to the power 2, sorry the real part of kappa j, Japanese brackets to the power 2s times the imaginary part of kappa j. So all, if this is small then all quantities on the left hand side of the formula are non-negative. So there's a similar statement for kdv, so this is NLS bigger than minus 1 half here. And there's a similar theorem for kdv where there is the same sort of statement for S larger than minus 1, that's the same statement. And if S equal to minus 1 then it's the same statement but for H minus 1 we have to take out the potentials so that minus 1 over 4 is an eigenvalue of the Schrodinger operator. Which makes sense because then we want to evaluate the singular function, this logarithm at 1 half and that we can, the logarithm at 1 minus 1 at 0 and that we can't. Now some corollaries, the first corollary, corollary if you use this on scaling then we get that if u0 is in Hs, I don't specify which problem but if you are in this range then the soup of u of t in Hs is less than infinity and it could be made more precise for the solution for NLS MKdv whatsoever. The second corollary is, well if you look at solitons, solitons correspond to eigenvalues at pre-sales and anything else what you want to define in terms of the spectrum then they are all stable in Hs for all S in that range. So this is a question of scaling, you might scale so if you look at the formulas, now we look at the left hand side, then in the left hand side if the smallest condition is satisfied then every term is positive and if you look at the second line then the soliton is the one where the logarithm of t vanishes on the real axis and there is one contribution corresponding to the eigenvalue so this is the lowest value ES can take whenever you have the soliton so this ES is a Lappano functional and it's minimized at the soliton and similarly at the pre-thor which is also characterized in terms of the spectrum. So there is one thing on the blackboard which I didn't explain and I don't want to go into it, at least not too much but I want to give some information, I have this du2. So what this du2 is, it's something like H minus a half, H plus a half if you don't get derivative. So H a half is a space with half a derivative and it doesn't have any good structure because there is no embedding into continuous functions, L1 doesn't embed into H a half so this space doesn't have the homogeneous space. This space is not good for most things we do in analysis, this du2 is a replacement of it and it has all the good properties one would wish to have or many of the good properties one would wish to have from a space with half a derivative and it's still on the same scale, it's on the same scale also in the sense that if you look at the best of space then B1 half, 1 1 embeds into u2, embeds into 1 half, 1 infinity so it's a very close space to this H a half. And du2, these are the formal derivatives, they are like things with H minus a half, same sort of embedding and then there come similar spaces for the same properties v2 and dv2 and there is a good duality between these things so what do you have here, you have embeddings into bounded functions, you have limits at infinity for these functions, you have embeddings of L1 into du2 into dv2 and so on. So all the things which don't work for H a half or H minus a half, many of those things work for u2 and v2. So in particular what is relevant here, so if you look at this L2, du2 is defined by the norm so we take a function u and then we multiply it by a smooth cutoff function x minus j, so this is something which lives with the bump of size 1, we take this thing into u2 and this is pieces and then we take an L2 summation over that and similarly this Lp. So we take small pieces, we sum them in L2 and Lp and then this L2, if you look at this L2 du2 norm then this is less or equal to a constant times u in Hs for all s larger than minus a half. So whenever you could possibly hope to get a good bound like that then this quantity is going to be bounded and it's smaller than, it's controlled in terms of all sub-left spaces so we could replace it by any Hs here, H minus a half plus epsilon and then the same statement is true. So can I ask you a question though? But if you take a u in S between like minus 1 and minus 1 half, that could be infinity so you have a bunch of things but you don't have line 2, is that correct? And I take u in Hs, purely in Hs but S less than 1 half, minus less than minus 1 half, then you have the first part of the theorem, you don't have the second part of the theorem then it is. No, so the assumption S larger than minus 1 half is for both parts. So nothing of that makes sense, I'm not quite sure about some sides of the equation but nothing makes sense if you. And what's the analog for the KDV part of the u space? So I should have given that, so for the KDV part you have to replace, simply replace L to du2 by H minus 1 half and then minus 1 to take H minus 1. Okay, well what I present here is connected to a lot of literature and I can't go into all of that. I mean the inverse scattering part for this AKNS system is due to the up to its carb, KNNN and Ziko, Fadeev and many others, I guess there's no point in repeating this the history inverse scattering and I'm sure that there are other people in the audience who can do that much better than I could do it. For the a priori estimate, I should say there is a sort of a history which is different, there are priori estimates by Christkul Jandantau for the non-linear Schrodinger equation, I did some priori estimates, not uniform in time but control over all times somehow with Daniel Tataru up to H minus 4th for the non-linear Schrodinger equation. There is a lot of work on the KTV equation, I did some work on the KTV equation in H minus 1 with Tristan Backmaster, some uniform in time, a priori estimate in H minus 1 which used a bit of inverse scattering but not much of the structure. There is the work which was then simultaneously by Kilipe, Wichain and Zhang, and Rohan gave a talk on it last week on the KTV equation for S, I think the talk was for S between minus 1 and S and 0 but I guess also some extension beyond 0 probably 1. Yeah, what I think I said was 1. And you worked with the, so you worked with this term here and I have to collect a title. Yeah, that is important. So you worked with this term and this gives, this allows to show an equivalence of this single term here with the Hs norm, the integrated versions of that which lead probably to something similar than the formula on the right hand side with a single correction at j equal to 0 which is the L2 norm so it would probably give the same formula as this one in that range. Okay, so I want to have to give them up again. Now how do we prove it? Well, the strategy is to bound the, to control the integral part on the right hand side. This is the most essential part. Once we have control of the integral part on the right hand side, it also gives the control by limits, by certain limits of the whole right hand side and then if you look at the default crossing part, this allows to, I mean first we look at the transmission coefficient not in the whole half plane but in the, so we start with looking at the, we have some smallness conditions. We control, we try to control the transmission coefficient T of C away from the real axis and the smallness condition ensures that we can rescale things so that whenever we get it somewhere, we also get it let's say below one half, over two and then at least morally we have a harmonic function, the logarithm of T is a harmonic function which in the defaults in case has a real part less than zero. So then we are able to interpret this as such a harmonic function on the half space and then the real part of that is going to be a measure and that allows to define the left hand side. So if you use then that in the defaulting part that T of C is less than equal to one then the logarithm has the real part of the logarithm has a sign and then we can take the trace at the real part and that gives an interpretation of the formulas on the left hand side. For the focusing case one also has to use then the backload transform but I don't want to go into that. What I want to do is I want to explain how we can control the integrals. Well so we go back to the AKNS system. So the AKNS system was C1 prime is equal to minus IcC1 plus uc2, C2 prime is equal to IcC2 plus u bar C1. And what we want is we want a solution so that we want C behaves like e to the minus Icx0 at x equal to minus infinity. So then we can try to solve this equation recursively. So we want to get a power expansion in terms of u. So in the first step we integrate this equation and then we get C21 is equal to the integral from minus infinity to x times e to the Icx, maybe I could take that up with a y, y minus x u bar of x e to the minus Icx dx. And then we take, we look at C, so this stands for the iteration C12 and we get something similar. And then- So you have a prime, I mean you have one and prime. So and then we get the first, we get t is equal to 1 plus the limit as x tends to infinity to the 1, 1 of x plus, let me do this iteration and we call this guy t2 of C e to the Icx of this guy. We have to multiply by this and then we get t4 and these terms have a pretty nice structure. So if you do the math then this t2j of C is equal to an integral and here we have x1 less than y1, so we iterate this thing up to y2j. Here we get e to the 2ic the sum over the yj by case minus the sum over the x case times u of y1, u of yj, u bar of x1, u bar of xj, dy dx. So we do the iteration and then you get a pretty nice representation. The imaginary part of C is positive, so this is negative and this thing is larger than this one and we have the y's on the right hand side so that leads to exponential decay here whenever we are in the other half plane. Okay, so you do that, then you try to control these things and I guess I'm running a bit out of time so I have to decide what I want to focus on. So then the first type of estimate is that this t2j of C is bounded by, I guess I don't want to go into that, a constant times the u e to the minus i real part of Cx u in L2 and I put the imaginary part here, Cu du2 to the power j. So this is where the properties of the spaces du2 come in nicely. You have to take to put the oscillatory part into u and then they give very naturally this estimate here where this imaginary part of C corresponds to the localization in space and then you have two intervals of size imaginary part of C to the minus 1. This is exactly the scale on which this exponential decays, so there is this concentration. I wanted to say a bit about that but I guess I'm going to skip that. So this is the first estimate and it allows to get convergence provided this quantity is small and this is basically where the condition in the theorem comes in. So far so good. Then we don't want to get estimates in terms of this thing, we want to get estimates in terms of Hs norms. So how do we go from here to Hs norms? Well we can do embeddings, we can use embeddings and then we get some variable. We can use embeddings here and look at the scaling parameter and get powers of the imaginary part and this is going to allow to get some of the integrals. But it only goes up to a certain point and then we have to look at the logarithm of this thing. I know I want to go to the logarithm. Actually the way we did it, I think I did make another sign error. Anyhow, so we want to take the logarithm of t to the minus one. So this is t to the minus one is the sum of integrals which I write in this fashion. So this is the trivial integral, this here is the first integral which we have here, this is the second integral. So I wrote it here in red, I hope that it's visible. So this codifies the information that x1 is less than y1, less than x2, less than y2 and then we do the integral. And then we have this with three things, with four things. Then if you take the logarithm, you get a lot of combinatorics. You can split the integration domains into the sets where we order the variables and this is messy. This is really messy. So I did it for the first part, I got the second part, explain what this integral stands for in a minute. I think I computed by hand these things here up to here and then I was convinced that the structure somehow persists but at MSRI I was stuck with that, we were stuck with that and didn't know what to do. But it was important for us to get this structure, but explain what this structure is. So this goes, so for every arc which is going up, I put an x, for every arc which is going down I put a y and then I get here an ordering that x1 is less than x2, less than x3, less than y1. This is going up, that's x4, less than x4, less than y3, less than y4. The indices don't matter but this codifies this integral here. The nice thing here is that we always have the last y on the right of every x. So this here has a much better decay on the domain of integration as the original integral because the decays exponentially whenever two of the variables are far apart. So whereas this information here leads to decay of order basically at most imaginary part of c to the minus j because this is the size of the corresponding simplex, here we get a decay like imaginary part of c to the minus 2j plus 1 because the last y is on the right. So the last, you can't split these things into, no matter what we do, this decays exponentially whenever we put things apart. Here we could, in these integrals we could have the situation that y1 is close to x1 and y2 is close to x2 but x2 is far away from y1 and then it wouldn't get decay. I pay for that with the combinatorics. So at MSI there was the fortune situation that in this program there was people from probability and I showed this question to Martin Heier and Martin Heier told me that I should look at Hopf-Algebras and that it would be related to a famous formula in Hopf-Algebras. So my colleagues in Bonn are still, when I told them about Hopf-Algebras they were in the colleagues in Algebras they were excited. Finally a use. So a thing can be used. Well but then I tried to get it into context with the Milner-Mur theorem and that didn't work. So the basic thing is it's not a big effort to show that this is a Hopf-Algebras because it's very close to a standard Hopf-Algebras which I didn't know but that can be, that's not so difficult, the Schaffler-Algebras and the Milner-Mur theorem would give the conclusion if that would be a co-commutative Hopf-Algebras but it's a commutative one. So it sounds almost the same but it is not but then still if one tries to do things by food basically then all integrals are connected. Then one gets the structures. So where are we? Well we have this estimate here on the t's which transfers to the expansion of the logarithm of t. So we get the estimates we need whenever j is large compared to s. So if you look at the high order polynomial then things are easy but since we get that we have to look at the lower polynomial case by case and if you look at the conserved quantities then it's clear that something has to happen because if s is less than 1, less than 0 then there is no conserved quantity which hurts. At s equal to 0 you get the L2 norm so you have to handle the L2 norm. This is easy. This is easy because with the L2 norm if you check t2 then this is equal to 0. This is equal to u in hs squared so that requires simple calculation so the two term is okay. Now at the level of the next term at the t4 term that corresponds to the next at the level of the t4 term you expect the L4 norm to the power 4 to show up at the level s equal to 1 so this has to be taken care of. Now with this connected structure you get up to this point and the idea to go beyond that is pretty simple. This is doing integration by parts in the integration with respect to y1. If you do an integration by parts you get a derivative which falls on this. So if you do an integration by parts with respect to this exponential so you gain a factor 1 over c and the derivative falls either on this then you get the same formula with the derivative or you get to the boundary terms and then you have an integral of one dimension less and then you continue to do that and if you end up with no one-dimensional integral in the end you get the conserved quantities and otherwise you do similar estimates as I explained before. So I guess I went a bit over time. I'm sorry for that. Thank you for the attention. So this is the structure of Freyholm term. I have no idea. Freyholm term. No. Probably it has something to do with Freyholm term. I have no idea. I guess you know that better than I do. I mean in KGV, I mean that way, so I knew that the matrix case was for the scalar case. This is exactly the operation in passing from the series of T. Yeah. The thing along the series is sort of restricting the technical diagrams just taking the term there. So the real problem I wanted to make was, okay, so again the matrix sitting out of the detail. But in the scalar setting it's actually a general phenomenon that the transmission coefficient is the Freyholm term in it because the integral kernel is semi-separable. It means the function of the lesser times the function of the greater. And then what is my point? If you ask a problem that's something, then you might recognize it because that's exactly what makes processes on the line Markovian. Well, right. I mean why is the Einstein-Waldberg process Markovian? It's because the kernel is semi-separable. Yeah. So I'm not familiar with these things. But what I think is true is that there's a lot of combinatorics behind which can be expressed in various ways and it should be connected. I mean one thing with the Hopf-Algerprass, the reason why Martin Heierer is interested in the Hopf-Algerprass is I think different. It's closer to what I presented here and not the Freyholm determinant. But the thing which is striking is there's this theorem. It's not difficult to get. And it's not difficult to usage in order to compute these things to any order. But people I asked were sort of stuck with explaining what the coefficients or this expansion should look like or the coefficients should look like. So this is sort of big miracle. It gives connected integrals which is what we need but we don't have any access to the size of the coefficients. And maybe the Freyholm determinant is allowed to get that more explicitly. There's a lot of combinatorics behind which I don't understand. I have to go open. Stanley's enumerative combinatorics exactly has a theorem which says if you have a generating function generated with the coefficients and the number of objects of this type, then the logarithm of your generating function is exactly the generating function of the number of connected objects of this type. OK, that sounds, yeah. So I would be interested in a reference. I'd like to get that here. I was happy to get that but it's clearly related to many things. Well, yeah. Oh, I just comment on this estimate. I think C is independent of J. And if so, it's articulating that you're really creating a transmission coefficient is a holomorphic function of u and u bar. Yes. And these are the Taylor coefficients. Yes. It's exactly. So sort of the McKean-Truberwitz picture of the world which is objects are holomorphic with regard to functions. Comments or questions? So if not, thank you again. Thank you.
 Well, thank you. I'm grateful for being here and I want to thank the organizers for this  nice and interesting conference. I'm going to talk about concept energies and this is  joint work, concept energies for NLS, MKDV and KDV and this is joint work with Daniel  Tataru. And before I state exactly what we do, I want to give the equations. So, there  is the NLS equation. It's I one-dimensional IDTU plus UXX is equal to plus or minus U  squared U. So, plus is the minus case is focusing, then there are solitons and the plus case  is defocusing and there are no solitons. Then there is the M modified KDV equation, which  is UT plus UXX and I realize that the first thing I didn't want to put attention to plus  minus 6, what do you want to write it? U squared, write it like this, 2 U squared UX is equal  to 0. Now, I want to add in a plus or minus and then it's the focusing or the defocusing  one. I want to write it with the absolute value here in order to allow complex solutions  and it's a complex MKDV, otherwise it's the real one. And then there's a KDV equation,  which is UT plus UXX minus 6 U UX is equal to 0. And all these equations are pretty connected.  There are integrable and the integrability shows in the existence of a lux pair and I  want to write it in the following fashion as a system. Psi x, Psi is a that's the x  derivative of a map to C2 is equal to, I know I only write it for the defocusing case, for  the plus case, I see U bar minus plus I see C. So this is the lux equation. C is a complex  parameter and the amazing fact is that if you complemented with an equation for the  time derivative is equal to I, now I have to look it up because I can't remember these  things. Plus U squared, value U squared to I see U plus UX minus to I see U bar minus  U bar X to C squared plus U squared C. So these are two first order equations and the  interesting, the very interesting structure is that if we want to solve them simultaneously  for given data at one point, general, then this is only possible if a compatibility condition  is satisfied and the compatibility condition is exactly the defocusing NLS here on this  side. So and the connection to MK2V is the following. So for MK2V there's the same structure  with the same operator here, the same equation here and a different equation on this side.  So there's a different time equation for the time derivative and then there's the same  structure. This is solvable for given data at a point let's say 0, 0, X equal to 0, T  equal to 0 for general data, if and only if then U satisfies the MK2V equation. And for  the focusing case there are some sign changes which I don't want to go into. So the only  case, the only sign change here is a minor sign and here it is more complex but this  equation doesn't play a big role for what I'm going to tell.  Now what is the effect of that? Well suppose we take C, let's say we say imaginary part  of C, larger equal to 0. Then we might look for your solutions. I mean it's a 2 by 2 system  so there is a two dimensional space of solutions and I want to normalize them on the left.  So if I want to have that C behaves like e to the minus i Cx, 0 on the left, so the case  to the left for X to minus infinity then on the right, let's suppose that U is compactly  supported, then on the right it decomposes in a linear combination of two, there's similar  fundamental system on the right, then on the right we get the decomposition that C behaves  like t to the minus 1 of C e to the minus i Cx and well here we have to be a bit more  careful as X tends to plus infinity if C is the real number. So we get a linear combination  of the two solutions which you would have of a fundamental system which would have a  few which would be equal to 0 and I use this to define t to the minus 1 which works well  even if the imaginary part of C is positive and there's a reflection coefficient, it's  called the transmission coefficient and the reflection coefficient, so t is the transmission  coefficient. Now the fact that the equation, the analysis  equation is the compatibility condition for solving these equations allows us to solve  this equation at plus infinity and check what happens with this plus infinity and minus  infinity and check what happens with the function C. So at plus or minus infinity we don't see  the effect of U, so what we see here is a constant coefficient differential operator,  it gives the time derivative of C in terms of C here and no U, so we can check this and  if we check this on the left then there is some time evolution, if we do it on the right  there is some time evolution and what we see there is that t of C is independent of time,  if we have a solution to the analysis equation. So whatever, if you look for quantities which  are conserved for NLS or MKTV or KTV, we simply have to look out the transmission coefficient,  have to see what we can define it and this is a conserved quantity at each point, we  may integrate it, we may do whatever we want with it and this gives a conserved quantity.  So now this is some sort old, fairly old, goes back to well in this context to Aplovitz  for AKNS, this is Zegor, Newell and K-Kalp, sorry, so they did it for the non-inheriting  equation and then there is an expansion of t of C which is equal to minus 1 over 2i  times a sum, this is formal, we should try to make this, this is a formal sum and then  there are the energies EjC to the minus J minus 1, so this is at C tending to i infinity  or depending on what you assume on the initial data and how it is, it is the Laurent series  at infinity and then E1 is equal to the integral u squared dx and E2 is equal to the integral  ux, sorry, u2 is equal to i times u u bar x momentum and E3, I did something stupid,  I want to have this one, 0, 1 and 2 equal to the integral ux squared plus 2 times u to  the power 4, no, it is u to the power 4 dx, which one is it?  It has a 2 here, so and then you can continue and it is classical to expand these things  and to get recursive formulas and what I want to do is I want to, we did in the John Dewey  Pistanio, we sort of interpolated between these quantities in order to get a continuous  family of conserved quantities, well that is not exactly what we did, so more precisely  we interpolated between the even ones and not even that, we interpolated between linear  combinations of the even ones.  So here is what we get, well the things which we define, so let us first look at the focusing  case for n or less in MKDV, then ES can be defined as an integral over, well at least  for Schwartz functions, for Schwartz potentials as an integral over the real line, over the  logarithm, it is always the logarithm which occurs of the transmission coefficient, evaluated  at the real axis, so for that we need some decay on some Schwartz functions and we would  have to study the decay which we do not do, but let us suppose that the logarithm of t  is a Schwartz function, so say a bit more about when these things are defined later,  so the one way to define it is to integrate over the real axis and to put a weight which  is something like the Fourier weight for sub-Lav spaces, c to the 2s, so this is for s larger  than minus a half and then another way of expressing that is to move the contour of integration  to the line from i to i times infinity and that gives the second formula which is equal  to 4 sin pi s integral from 1 to infinity, so now we are writing this in a real fashion,  tau squared minus 1 to the power s times this logarithm evaluated at i tau over 2, I see  I forgot factor 1 over 2, it should be evaluated at psi over 2 on the left hand side, so we  evaluate it at i tau over 2 and then we have to do some corrections because the weight  c to the s goes at infinity polynomially, so we have to correct that in order to be  able to do a contour integration, so we subtract the, oh I messed up the, I wanted to call  these guys h, so h1, h2, yeah so we correct, we have to correct the behavior at infinity  a little bit, thank you, oh sorry, yeah, so that's the only place where I,  wrote it, okay so, so we have to correct the behavior at infinity in order to get, to be  able to work with a contour integration, to move the contour and if we do it in this fashion  then this doesn't depend on how many terms we correct if we have the regularity and  so we get this formula here, what you see here is that if s is an integer then the sine  pi s vanishes and here you get the linear combination of the classical conserved quantities,  so in that sense we interpolate between the classical conserved quantities, well then  we can do the same thing for the focusing, NLS if you do the same thing for the focusing  NLS then we have to adjust for eigenvalues of the AKNS operator, so the AKNS operator  has eigenvalues in the upper half plane and we have to adjust for them and there is a  function there's a way of doing that, so we have to add a sum over the eigenvalues with  a multiplicity mj over some, with some function psi and nothing changes on the right hand  side. Now for KTV that's the same structure, first  if s is equal is larger than minus 1 then this is a generalization of the classical  formula of Fadiev, there is a different function psi, we have to adjust for the eigenvalues,  so the eigenvalues in this case are minus, no kappa k squared and on the line from i to  infinity there's a similar expression as for the NLS equation and again we have to adjust  for the classical energies. Something special happens at s equal to minus  1, so the left hand side makes sense for minus 1 and for the right hand side there is this  part of the integral which has the effect of converging to a delta function combined  with the sine pi s, so the outcome is then 8 times the transmission coefficient evaluated  at half minus this integral u and here one has to do some normalization which I don't  want to take much about, talk much about. So these are conserved quantities for NLS  and MKTV, they go back to AKNS or to Fadiev, this relation between the different contour  integrals and these are the objects we want to work with.  Now defining these objects for Schwarz functions is trivial because then the scattering transform  becomes something nice, the logarithm of t is a Schwarz function on the real axis and  then it's an exercise in complex analysis maybe not an entirely easy one to get these  formulas. So as I said before since t is independent  of time whenever you have a solution whatever you do with t is going to give you a conserved  quantity. So the question is how to relate them to things we work with in analysis, how  to relate them to sub-lawed spaces. And I want to give the results first and then explain  a bit about the proof. Thank you.  Yeah I tried it before but it's always a bit different when I do it on the blackboard.  Is it okay? I guess you still can have an impression of the formulas even if you don't  see all of them. So the theorem is Daniel Tataru. So what we do is we look at the right  hand side. We define the quantities by the right hand side and then it turns out that  it makes all the sense with the left hand side. So the first thing is if u is in Hs  then both sides are well defined. Both sides are well defined. But basically you want to  look at the more on the thing on the right hand side. And so the map from Hs to Es is  continuous in u and also in S whenever u is sufficiently regular. And it is analytic  if i over 2 is not an eigenvalue. Well you can't expect it to be an eigenvalue if this  so this is in the focusing case because then this function xi is not analytic at 1. So this  has to be excluded. But otherwise it is analytic and it's jointly analytic in S and u in the  appropriate sense. So whenever u is in Hs then it's analytic in S for S and then Es  is analytic in u and S when S is less than sigma. So I don't want to write that but in  the appropriate sense it is as all the properties one would want to have it. And the nice thing  is we can play with both sides in order to get whatever is more convenient. Second, well  how does it compare? So this says that it's analytic so whenever u is in Hs then this  is defined but the question is does this control the Hs now? Well so we can look at Es of u  minus u in Hs where it is basically the same as the left hand side but for the Fourier  transform evaluated I didn't correct all the xi over 2sz. So this is on the Fourier side  similar to the integral on the left hand side and this is less or equal to a constant times  u in Hs squared and now I want to use a strange norm here and I want to explain it later on  with u in L2, D2, D2 is less than delta for some delta. So this tells us that the quadratic  part of I mean it's analytic so we can do an analytic expansion in u at 0 and the quadratic  part of that is exactly the Hs norm. So for small if this quantity is small I'm going  to say more about that in a minute if this quantity is small then this quantity and the  Hs norm are pretty close together. And also what happens in that case is this function  xi of the eigenvalue kappa j is larger or equal to a constant times kappa j to the power  2, sorry the real part of kappa j, Japanese brackets to the power 2s times the imaginary  part of kappa j. So all, if this is small then all quantities on the left hand side  of the formula are non-negative. So there's a similar statement for kdv, so this is NLS  bigger than minus 1 half here. And there's a similar theorem for kdv where there is  the same sort of statement for S larger than minus 1, that's the same statement. And if  S equal to minus 1 then it's the same statement but for H minus 1 we have to take out the  potentials so that minus 1 over 4 is an eigenvalue of the Schrodinger operator. Which makes sense  because then we want to evaluate the singular function, this logarithm at 1 half and that  we can, the logarithm at 1 minus 1 at 0 and that we can't.  Now some corollaries, the first corollary, corollary if you use this on scaling then  we get that if u0 is in Hs, I don't specify which problem but if you are in this range  then the soup of u of t in Hs is less than infinity and it could be made more precise  for the solution for NLS MKdv whatsoever. The second corollary is, well if you look at  solitons, solitons correspond to eigenvalues at pre-sales and anything else what you want  to define in terms of the spectrum then they are all stable in Hs for all S in that range.  So this is a question of scaling, you might scale so if you look at the formulas, now  we look at the left hand side, then in the left hand side if the smallest condition is  satisfied then every term is positive and if you look at the second line then the soliton  is the one where the logarithm of t vanishes on the real axis and there is one contribution  corresponding to the eigenvalue so this is the lowest value ES can take whenever you  have the soliton so this ES is a Lappano functional and it's minimized at the soliton  and similarly at the pre-thor which is also characterized in terms of the spectrum.  So there is one thing on the blackboard which I didn't explain and I don't want to go into  it, at least not too much but I want to give some information, I have this du2.  So what this du2 is, it's something like H minus a half, H plus a half if you don't  get derivative.  So H a half is a space with half a derivative and it doesn't have any good structure because  there is no embedding into continuous functions, L1 doesn't embed into H a half so this space  doesn't have the homogeneous space.  This space is not good for most things we do in analysis, this du2 is a replacement of  it and it has all the good properties one would wish to have or many of the good properties  one would wish to have from a space with half a derivative and it's still on the same scale,  it's on the same scale also in the sense that if you look at the best of space then  B1 half, 1 1 embeds into u2, embeds into 1 half, 1 infinity so it's a very close space  to this H a half.  And du2, these are the formal derivatives, they are like things with H minus a half, same  sort of embedding and then there come similar spaces for the same properties v2 and dv2  and there is a good duality between these things so what do you have here, you have  embeddings into bounded functions, you have limits at infinity for these functions, you  have embeddings of L1 into du2 into dv2 and so on.  So all the things which don't work for H a half or H minus a half, many of those things  work for u2 and v2.  So in particular what is relevant here, so if you look at this L2, du2 is defined by  the norm so we take a function u and then we multiply it by a smooth cutoff function  x minus j, so this is something which lives with the bump of size 1, we take this thing  into u2 and this is pieces and then we take an L2 summation over that and similarly this  Lp.  So we take small pieces, we sum them in L2 and Lp and then this L2, if you look at this  L2 du2 norm then this is less or equal to a constant times u in Hs for all s larger than  minus a half.  So whenever you could possibly hope to get a good bound like that then this quantity  is going to be bounded and it's smaller than, it's controlled in terms of all sub-left  spaces so we could replace it by any Hs here, H minus a half plus epsilon and then the same  statement is true.  So can I ask you a question though?  But if you take a u in S between like minus 1 and minus 1 half, that could be infinity  so you have a bunch of things but you don't have line 2, is that correct?  And I take u in Hs, purely in Hs but S less than 1 half, minus less than minus 1 half,  then you have the first part of the theorem, you don't have the second part of the theorem  then it is.  No, so the assumption S larger than minus 1 half is for both parts.  So nothing of that makes sense, I'm not quite sure about some sides of the equation but  nothing makes sense if you.  And what's the analog for the KDV part of the u space?  So I should have given that, so for the KDV part you have to replace, simply replace L  to du2 by H minus 1 half and then minus 1 to take H minus 1.  Okay, well what I present here is connected to a lot of literature and I can't go into  all of that.  I mean the inverse scattering part for this AKNS system is due to the up to its carb,  KNNN and Ziko, Fadeev and many others, I guess there's no point in repeating this  the history inverse scattering and I'm sure that there are other people in the audience  who can do that much better than I could do it.  For the a priori estimate, I should say there is a sort of a history which is different,  there are priori estimates by Christkul Jandantau for the non-linear Schrodinger equation, I  did some priori estimates, not uniform in time but control over all times somehow with  Daniel Tataru up to H minus 4th for the non-linear Schrodinger equation.  There is a lot of work on the KTV equation, I did some work on the KTV equation in H minus  1 with Tristan Backmaster, some uniform in time, a priori estimate in H minus 1 which  used a bit of inverse scattering but not much of the structure.  There is the work which was then simultaneously by Kilipe, Wichain and Zhang, and Rohan gave  a talk on it last week on the KTV equation for S, I think the talk was for S between  minus 1 and S and 0 but I guess also some extension beyond 0 probably 1.  Yeah, what I think I said was 1.  And you worked with the, so you worked with this term here and I have to collect a title.  Yeah, that is important.  So you worked with this term and this gives, this allows to show an equivalence of this  single term here with the Hs norm, the integrated versions of that which lead probably to something  similar than the formula on the right hand side with a single correction at j equal to  0 which is the L2 norm so it would probably give the same formula as this one in that  range.  Okay, so I want to have to give them up again.  Now how do we prove it?  Well, the strategy is to bound the, to control the integral part on the right hand side.  This is the most essential part.  Once we have control of the integral part on the right hand side, it also gives the  control by limits, by certain limits of the whole right hand side and then if you look  at the default crossing part, this allows to, I mean first we look at the transmission  coefficient not in the whole half plane but in the, so we start with looking at the, we  have some smallness conditions.  We control, we try to control the transmission coefficient T of C away from the real axis  and the smallness condition ensures that we can rescale things so that whenever we get  it somewhere, we also get it let's say below one half, over two and then at least morally  we have a harmonic function, the logarithm of T is a harmonic function which in the defaults  in case has a real part less than zero.  So then we are able to interpret this as such a harmonic function on the half space and  then the real part of that is going to be a measure and that allows to define the left  hand side.  So if you use then that in the defaulting part that T of C is less than equal to one  then the logarithm has the real part of the logarithm has a sign and then we can take  the trace at the real part and that gives an interpretation of the formulas on the left  hand side.  For the focusing case one also has to use then the backload transform but I don't want  to go into that.  What I want to do is I want to explain how we can control the integrals.  Well so we go back to the AKNS system.  So the AKNS system was C1 prime is equal to minus IcC1 plus uc2, C2 prime is equal  to IcC2 plus u bar C1.  And what we want is we want a solution so that we want C behaves like e to the minus  Icx0 at x equal to minus infinity.  So then we can try to solve this equation recursively.  So we want to get a power expansion in terms of u.  So in the first step we integrate this equation and then we get C21 is equal to the integral  from minus infinity to x times e to the Icx, maybe I could take that up with a y, y minus  x u bar of x e to the minus Icx dx.  And then we take, we look at C, so this stands for the iteration C12 and we get something  similar.  And then-  So you have a prime, I mean you have one and prime.  So and then we get the first, we get t is equal to 1 plus the limit as x tends to infinity  to the 1, 1 of x plus, let me do this iteration and we call this guy t2 of C e to the Icx  of this guy.  We have to multiply by this and then we get t4 and these terms have a pretty nice structure.  So if you do the math then this t2j of C is equal to an integral and here we have x1 less  than y1, so we iterate this thing up to y2j.  Here we get e to the 2ic the sum over the yj by case minus the sum over the x case times  u of y1, u of yj, u bar of x1, u bar of xj, dy dx.  So we do the iteration and then you get a pretty nice representation.  The imaginary part of C is positive, so this is negative and this thing is larger than  this one and we have the y's on the right hand side so that leads to exponential decay  here whenever we are in the other half plane.  Okay, so you do that, then you try to control these things and I guess I'm running a bit  out of time so I have to decide what I want to focus on.  So then the first type of estimate is that this t2j of C is bounded by, I guess I don't  want to go into that, a constant times the u e to the minus i real part of Cx u in L2  and I put the imaginary part here, Cu du2 to the power j.  So this is where the properties of the spaces du2 come in nicely.  You have to take to put the oscillatory part into u and then they give very naturally this  estimate here where this imaginary part of C corresponds to the localization in space  and then you have two intervals of size imaginary part of C to the minus 1.  This is exactly the scale on which this exponential decays, so there is this concentration.  I wanted to say a bit about that but I guess I'm going to skip that.  So this is the first estimate and it allows to get convergence provided this quantity  is small and this is basically where the condition in the theorem comes in.  So far so good.  Then we don't want to get estimates in terms of this thing, we want to get estimates in  terms of Hs norms.  So how do we go from here to Hs norms?  Well we can do embeddings, we can use embeddings and then we get some variable.  We can use embeddings here and look at the scaling parameter and get powers of the imaginary  part and this is going to allow to get some of the integrals.  But it only goes up to a certain point and then we have to look at the logarithm of  this thing.  I know I want to go to the logarithm.  Actually the way we did it, I think I did make another sign error.  Anyhow, so we want to take the logarithm of t to the minus one.  So this is t to the minus one is the sum of integrals which I write in this fashion.  So this is the trivial integral, this here is the first integral which we have here,  this is the second integral.  So I wrote it here in red, I hope that it's visible.  So this codifies the information that x1 is less than y1, less than x2, less than y2  and then we do the integral.  And then we have this with three things, with four things.  Then if you take the logarithm, you get a lot of combinatorics.  You can split the integration domains into the sets where we order the variables and  this is messy.  This is really messy.  So I did it for the first part, I got the second part, explain what this integral stands  for in a minute.  I think I computed by hand these things here up to here and then I was convinced that the  structure somehow persists but at MSRI I was stuck with that, we were stuck with that and  didn't know what to do.  But it was important for us to get this structure, but explain what this structure is.  So this goes, so for every arc which is going up, I put an x, for every arc which is going  down I put a y and then I get here an ordering that x1 is less than x2, less than x3, less  than y1.  This is going up, that's x4, less than x4, less than y3, less than y4.  The indices don't matter but this codifies this integral here.  The nice thing here is that we always have the last y on the right of every x.  So this here has a much better decay on the domain of integration as the original integral  because the decays exponentially whenever two of the variables are far apart.  So whereas this information here leads to decay of order basically at most imaginary  part of c to the minus j because this is the size of the corresponding simplex, here we  get a decay like imaginary part of c to the minus 2j plus 1 because the last y is on the  right.  So the last, you can't split these things into, no matter what we do, this decays exponentially  whenever we put things apart.  Here we could, in these integrals we could have the situation that y1 is close to x1  and y2 is close to x2 but x2 is far away from y1 and then it wouldn't get decay.  I pay for that with the combinatorics.  So at MSI there was the fortune situation that in this program there was people from  probability and I showed this question to Martin Heier and Martin Heier told me that  I should look at Hopf-Algebras and that it would be related to a famous formula in Hopf-Algebras.  So my colleagues in Bonn are still, when I told them about Hopf-Algebras they were  in the colleagues in Algebras they were excited.  Finally a use.  So a thing can be used.  Well but then I tried to get it into context with the Milner-Mur theorem and that didn't  work.  So the basic thing is it's not a big effort to show that this is a Hopf-Algebras because  it's very close to a standard Hopf-Algebras which I didn't know but that can be, that's  not so difficult, the Schaffler-Algebras and the Milner-Mur theorem would give the conclusion  if that would be a co-commutative Hopf-Algebras but it's a commutative one.  So it sounds almost the same but it is not but then still if one tries to do things by  food basically then all integrals are connected.  Then one gets the structures.  So where are we?  Well we have this estimate here on the t's which transfers to the expansion of the logarithm  of t.  So we get the estimates we need whenever j is large compared to s.  So if you look at the high order polynomial then things are easy but since we get that  we have to look at the lower polynomial case by case and if you look at the conserved quantities  then it's clear that something has to happen because if s is less than 1, less than 0 then  there is no conserved quantity which hurts.  At s equal to 0 you get the L2 norm so you have to handle the L2 norm.  This is easy.  This is easy because with the L2 norm if you check t2 then this is equal to 0.  This is equal to u in hs squared so that requires simple calculation so the two term is okay.  Now at the level of the next term at the t4 term that corresponds to the next at the  level of the t4 term you expect the L4 norm to the power 4 to show up at the level s equal  to 1 so this has to be taken care of.  Now with this connected structure you get up to this point and the idea to go beyond  that is pretty simple.  This is doing integration by parts in the integration with respect to y1.  If you do an integration by parts you get a derivative which falls on this.  So if you do an integration by parts with respect to this exponential so you gain a factor  1 over c and the derivative falls either on this then you get the same formula with the  derivative or you get to the boundary terms and then you have an integral of one dimension  less and then you continue to do that and if you end up with no one-dimensional integral  in the end you get the conserved quantities and otherwise you do similar estimates as  I explained before.  So I guess I went a bit over time.  I'm sorry for that.  Thank you for the attention.  So this is the structure of Freyholm term.  I have no idea.  Freyholm term.  No.  Probably it has something to do with Freyholm term.  I have no idea.  I guess you know that better than I do.  I mean in KGV, I mean that way, so I knew that the matrix case was for the scalar case.  This is exactly the operation in passing from the series of T.  Yeah.  The thing along the series is sort of restricting the technical diagrams just taking the term  there.  So the real problem I wanted to make was, okay, so again the matrix sitting out of the  detail.  But in the scalar setting it's actually a general phenomenon that the transmission coefficient  is the Freyholm term in it because the integral kernel is semi-separable.  It means the function of the lesser times the function of the greater.  And then what is my point?  If you ask a problem that's something, then you might recognize it because that's exactly  what makes processes on the line Markovian.  Well, right.  I mean why is the Einstein-Waldberg process Markovian?  It's because the kernel is semi-separable.  Yeah.  So I'm not familiar with these things.  But what I think is true is that there's a lot of combinatorics behind which can be  expressed in various ways and it should be connected.  I mean one thing with the Hopf-Algerprass, the reason why Martin Heierer is interested  in the Hopf-Algerprass is I think different.  It's closer to what I presented here and not the Freyholm determinant.  But the thing which is striking is there's this theorem.  It's not difficult to get.  And it's not difficult to usage in order to compute these things to any order.  But people I asked were sort of stuck with explaining what the coefficients or this expansion  should look like or the coefficients should look like.  So this is sort of big miracle.  It gives connected integrals which is what we need but we don't have any access to the  size of the coefficients.  And maybe the Freyholm determinant is allowed to get that more explicitly.  There's a lot of combinatorics behind which I don't understand.  I have to go open.  Stanley's enumerative combinatorics exactly has a theorem which says if you have a generating  function generated with the coefficients and the number of objects of this type, then the  logarithm of your generating function is exactly the generating function of the number of connected  objects of this type.  OK, that sounds, yeah.  So I would be interested in a reference.  I'd like to get that here.  I was happy to get that but it's clearly related to many things.  Well, yeah.  Oh, I just comment on this estimate.  I think C is independent of J. And if so, it's articulating that you're really creating  a transmission coefficient is a holomorphic function of u and u bar.  Yes.  And these are the Taylor coefficients.  Yes.  It's exactly.  So sort of the McKean-Truberwitz picture of the world which is objects are holomorphic  with regard to functions.  Comments or questions?  So if not, thank you again.  Thank you.
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Well, thank you. I'm grateful for being here and I want to thank the organizers for this nice and interesting conference. I'm going to talk about concept energies and this is joint work, concept energies for NLS, MKDV and KDV and this is joint work with Daniel Tataru. And before I state exactly what we do, I want to give the equations. So, there is the NLS equation. It's I one-dimensional IDTU plus UXX is equal to plus or minus U squared U. So, plus is the minus case is focusing, then there are solitons and the plus case is defocusing and there are no solitons. Then there is the M modified KDV equation, which is UT plus UXX and I realize that the first thing I didn't want to put attention to plus minus 6, what do you want to write it? U squared, write it like this, 2 U squared UX is equal to 0. Now, I want to add in a plus or minus and then it's the focusing or the defocusing one. I want to write it with the absolute value here in order to allow complex solutions and it's a complex MKDV, otherwise it's the real one. And then there's a KDV equation, which is UT plus UXX minus 6 U UX is equal to 0. And all these equations are pretty connected. There are integrable and the integrability shows in the existence of a lux pair and I want to write it in the following fashion as a system. Psi x, Psi is a that's the x derivative of a map to C2 is equal to, I know I only write it for the defocusing case, for the plus case, I see U bar minus plus I see C. So this is the lux equation. C is a complex parameter and the amazing fact is that if you complemented with an equation for the time derivative is equal to I, now I have to look it up because I can't remember these things. Plus U squared, value U squared to I see U plus UX minus to I see U bar minus U bar X to C squared plus U squared C. So these are two first order equations and the interesting, the very interesting structure is that if we want to solve them simultaneously for given data at one point, general, then this is only possible if a compatibility condition is satisfied and the compatibility condition is exactly the defocusing NLS here on this side. So and the connection to MK2V is the following. So for MK2V there's the same structure with the same operator here, the same equation here and a different equation on this side. So there's a different time equation for the time derivative and then there's the same structure. This is solvable for given data at a point let's say 0, 0, X equal to 0, T equal to 0 for general data, if and only if then U satisfies the MK2V equation. And for the focusing case there are some sign changes which I don't want to go into. So the only case, the only sign change here is a minor sign and here it is more complex but this equation doesn't play a big role for what I'm going to tell. Now what is the effect of that? Well suppose we take C, let's say we say imaginary part of C, larger equal to 0. Then we might look for your solutions. I mean it's a 2 by 2 system so there is a two dimensional space of solutions and I want to normalize them on the left. So if I want to have that C behaves like e to the minus i Cx, 0 on the left, so the case to the left for X to minus infinity then on the right, let's suppose that U is compactly supported, then on the right it decomposes in a linear combination of two, there's similar fundamental system on the right, then on the right we get the decomposition that C behaves like t to the minus 1 of C e to the minus i Cx and well here we have to be a bit more careful as X tends to plus infinity if C is the real number. So we get a linear combination of the two solutions which you would have of a fundamental system which would have a few which would be equal to 0 and I use this to define t to the minus 1 which works well even if the imaginary part of C is positive and there's a reflection coefficient, it's called the transmission coefficient and the reflection coefficient, so t is the transmission coefficient. Now the fact that the equation, the analysis equation is the compatibility condition for solving these equations allows us to solve this equation at plus infinity and check what happens with this plus infinity and minus infinity and check what happens with the function C. So at plus or minus infinity we don't see the effect of U, so what we see here is a constant coefficient differential operator, it gives the time derivative of C in terms of C here and no U, so we can check this and if we check this on the left then there is some time evolution, if we do it on the right there is some time evolution and what we see there is that t of C is independent of time, if we have a solution to the analysis equation. So whatever, if you look for quantities which are conserved for NLS or MKTV or KTV, we simply have to look out the transmission coefficient, have to see what we can define it and this is a conserved quantity at each point, we may integrate it, we may do whatever we want with it and this gives a conserved quantity. So now this is some sort old, fairly old, goes back to well in this context to Aplovitz for AKNS, this is Zegor, Newell and K-Kalp, sorry, so they did it for the non-inheriting equation and then there is an expansion of t of C which is equal to minus 1 over 2i times a sum, this is formal, we should try to make this, this is a formal sum and then there are the energies EjC to the minus J minus 1, so this is at C tending to i infinity or depending on what you assume on the initial data and how it is, it is the Laurent series at infinity and then E1 is equal to the integral u squared dx and E2 is equal to the integral ux, sorry, u2 is equal to i times u u bar x momentum and E3, I did something stupid, I want to have this one, 0, 1 and 2 equal to the integral ux squared plus 2 times u to the power 4, no, it is u to the power 4 dx, which one is it? It has a 2 here, so and then you can continue and it is classical to expand these things and to get recursive formulas and what I want to do is I want to, we did in the John Dewey Pistanio, we sort of interpolated between these quantities in order to get a continuous family of conserved quantities, well that is not exactly what we did, so more precisely we interpolated between the even ones and not even that, we interpolated between linear combinations of the even ones. So here is what we get, well the things which we define, so let us first look at the focusing case for n or less in MKDV, then ES can be defined as an integral over, well at least for Schwartz functions, for Schwartz potentials as an integral over the real line, over the logarithm, it is always the logarithm which occurs of the transmission coefficient, evaluated at the real axis, so for that we need some decay on some Schwartz functions and we would have to study the decay which we do not do, but let us suppose that the logarithm of t is a Schwartz function, so say a bit more about when these things are defined later, so the one way to define it is to integrate over the real axis and to put a weight which is something like the Fourier weight for sub-Lav spaces, c to the 2s, so this is for s larger than minus a half and then another way of expressing that is to move the contour of integration to the line from i to i times infinity and that gives the second formula which is equal to 4 sin pi s integral from 1 to infinity, so now we are writing this in a real fashion, tau squared minus 1 to the power s times this logarithm evaluated at i tau over 2, I see I forgot factor 1 over 2, it should be evaluated at psi over 2 on the left hand side, so we evaluate it at i tau over 2 and then we have to do some corrections because the weight c to the s goes at infinity polynomially, so we have to correct that in order to be able to do a contour integration, so we subtract the, oh I messed up the, I wanted to call these guys h, so h1, h2, yeah so we correct, we have to correct the behavior at infinity a little bit, thank you, oh sorry, yeah, so that's the only place where I, wrote it, okay so, so we have to correct the behavior at infinity in order to get, to be able to work with a contour integration, to move the contour and if we do it in this fashion then this doesn't depend on how many terms we correct if we have the regularity and so we get this formula here, what you see here is that if s is an integer then the sine pi s vanishes and here you get the linear combination of the classical conserved quantities, so in that sense we interpolate between the classical conserved quantities, well then we can do the same thing for the focusing, NLS if you do the same thing for the focusing NLS then we have to adjust for eigenvalues of the AKNS operator, so the AKNS operator has eigenvalues in the upper half plane and we have to adjust for them and there is a function there's a way of doing that, so we have to add a sum over the eigenvalues with a multiplicity mj over some, with some function psi and nothing changes on the right hand side. Now for KTV that's the same structure, first if s is equal is larger than minus 1 then this is a generalization of the classical formula of Fadiev, there is a different function psi, we have to adjust for the eigenvalues, so the eigenvalues in this case are minus, no kappa k squared and on the line from i to infinity there's a similar expression as for the NLS equation and again we have to adjust for the classical energies. Something special happens at s equal to minus 1, so the left hand side makes sense for minus 1 and for the right hand side there is this part of the integral which has the effect of converging to a delta function combined with the sine pi s, so the outcome is then 8 times the transmission coefficient evaluated at half minus this integral u and here one has to do some normalization which I don't want to take much about, talk much about. So these are conserved quantities for NLS and MKTV, they go back to AKNS or to Fadiev, this relation between the different contour integrals and these are the objects we want to work with. Now defining these objects for Schwarz functions is trivial because then the scattering transform becomes something nice, the logarithm of t is a Schwarz function on the real axis and then it's an exercise in complex analysis maybe not an entirely easy one to get these formulas. So as I said before since t is independent of time whenever you have a solution whatever you do with t is going to give you a conserved quantity. So the question is how to relate them to things we work with in analysis, how to relate them to sub-lawed spaces. And I want to give the results first and then explain a bit about the proof. Thank you. Yeah I tried it before but it's always a bit different when I do it on the blackboard. Is it okay? I guess you still can have an impression of the formulas even if you don't see all of them. So the theorem is Daniel Tataru. So what we do is we look at the right hand side. We define the quantities by the right hand side and then it turns out that it makes all the sense with the left hand side. So the first thing is if u is in Hs then both sides are well defined. Both sides are well defined. But basically you want to look at the more on the thing on the right hand side. And so the map from Hs to Es is continuous in u and also in S whenever u is sufficiently regular. And it is analytic if i over 2 is not an eigenvalue. Well you can't expect it to be an eigenvalue if this so this is in the focusing case because then this function xi is not analytic at 1. So this has to be excluded. But otherwise it is analytic and it's jointly analytic in S and u in the appropriate sense. So whenever u is in Hs then it's analytic in S for S and then Es is analytic in u and S when S is less than sigma. So I don't want to write that but in the appropriate sense it is as all the properties one would want to have it. And the nice thing is we can play with both sides in order to get whatever is more convenient. Second, well how does it compare? So this says that it's analytic so whenever u is in Hs then this is defined but the question is does this control the Hs now? Well so we can look at Es of u minus u in Hs where it is basically the same as the left hand side but for the Fourier transform evaluated I didn't correct all the xi over 2sz. So this is on the Fourier side similar to the integral on the left hand side and this is less or equal to a constant times u in Hs squared and now I want to use a strange norm here and I want to explain it later on with u in L2, D2, D2 is less than delta for some delta. So this tells us that the quadratic part of I mean it's analytic so we can do an analytic expansion in u at 0 and the quadratic part of that is exactly the Hs norm. So for small if this quantity is small I'm going to say more about that in a minute if this quantity is small then this quantity and the Hs norm are pretty close together. And also what happens in that case is this function xi of the eigenvalue kappa j is larger or equal to a constant times kappa j to the power 2, sorry the real part of kappa j, Japanese brackets to the power 2s times the imaginary part of kappa j. So all, if this is small then all quantities on the left hand side of the formula are non-negative. So there's a similar statement for kdv, so this is NLS bigger than minus 1 half here. And there's a similar theorem for kdv where there is the same sort of statement for S larger than minus 1, that's the same statement. And if S equal to minus 1 then it's the same statement but for H minus 1 we have to take out the potentials so that minus 1 over 4 is an eigenvalue of the Schrodinger operator. Which makes sense because then we want to evaluate the singular function, this logarithm at 1 half and that we can, the logarithm at 1 minus 1 at 0 and that we can't. Now some corollaries, the first corollary, corollary if you use this on scaling then we get that if u0 is in Hs, I don't specify which problem but if you are in this range then the soup of u of t in Hs is less than infinity and it could be made more precise for the solution for NLS MKdv whatsoever. The second corollary is, well if you look at solitons, solitons correspond to eigenvalues at pre-sales and anything else what you want to define in terms of the spectrum then they are all stable in Hs for all S in that range. So this is a question of scaling, you might scale so if you look at the formulas, now we look at the left hand side, then in the left hand side if the smallest condition is satisfied then every term is positive and if you look at the second line then the soliton is the one where the logarithm of t vanishes on the real axis and there is one contribution corresponding to the eigenvalue so this is the lowest value ES can take whenever you have the soliton so this ES is a Lappano functional and it's minimized at the soliton and similarly at the pre-thor which is also characterized in terms of the spectrum. So there is one thing on the blackboard which I didn't explain and I don't want to go into it, at least not too much but I want to give some information, I have this du2. So what this du2 is, it's something like H minus a half, H plus a half if you don't get derivative. So H a half is a space with half a derivative and it doesn't have any good structure because there is no embedding into continuous functions, L1 doesn't embed into H a half so this space doesn't have the homogeneous space. This space is not good for most things we do in analysis, this du2 is a replacement of it and it has all the good properties one would wish to have or many of the good properties one would wish to have from a space with half a derivative and it's still on the same scale, it's on the same scale also in the sense that if you look at the best of space then B1 half, 1 1 embeds into u2, embeds into 1 half, 1 infinity so it's a very close space to this H a half. And du2, these are the formal derivatives, they are like things with H minus a half, same sort of embedding and then there come similar spaces for the same properties v2 and dv2 and there is a good duality between these things so what do you have here, you have embeddings into bounded functions, you have limits at infinity for these functions, you have embeddings of L1 into du2 into dv2 and so on. So all the things which don't work for H a half or H minus a half, many of those things work for u2 and v2. So in particular what is relevant here, so if you look at this L2, du2 is defined by the norm so we take a function u and then we multiply it by a smooth cutoff function x minus j, so this is something which lives with the bump of size 1, we take this thing into u2 and this is pieces and then we take an L2 summation over that and similarly this Lp. So we take small pieces, we sum them in L2 and Lp and then this L2, if you look at this L2 du2 norm then this is less or equal to a constant times u in Hs for all s larger than minus a half. So whenever you could possibly hope to get a good bound like that then this quantity is going to be bounded and it's smaller than, it's controlled in terms of all sub-left spaces so we could replace it by any Hs here, H minus a half plus epsilon and then the same statement is true. So can I ask you a question though? But if you take a u in S between like minus 1 and minus 1 half, that could be infinity so you have a bunch of things but you don't have line 2, is that correct? And I take u in Hs, purely in Hs but S less than 1 half, minus less than minus 1 half, then you have the first part of the theorem, you don't have the second part of the theorem then it is. No, so the assumption S larger than minus 1 half is for both parts. So nothing of that makes sense, I'm not quite sure about some sides of the equation but nothing makes sense if you. And what's the analog for the KDV part of the u space? So I should have given that, so for the KDV part you have to replace, simply replace L to du2 by H minus 1 half and then minus 1 to take H minus 1. Okay, well what I present here is connected to a lot of literature and I can't go into all of that. I mean the inverse scattering part for this AKNS system is due to the up to its carb, KNNN and Ziko, Fadeev and many others, I guess there's no point in repeating this the history inverse scattering and I'm sure that there are other people in the audience who can do that much better than I could do it. For the a priori estimate, I should say there is a sort of a history which is different, there are priori estimates by Christkul Jandantau for the non-linear Schrodinger equation, I did some priori estimates, not uniform in time but control over all times somehow with Daniel Tataru up to H minus 4th for the non-linear Schrodinger equation. There is a lot of work on the KTV equation, I did some work on the KTV equation in H minus 1 with Tristan Backmaster, some uniform in time, a priori estimate in H minus 1 which used a bit of inverse scattering but not much of the structure. There is the work which was then simultaneously by Kilipe, Wichain and Zhang, and Rohan gave a talk on it last week on the KTV equation for S, I think the talk was for S between minus 1 and S and 0 but I guess also some extension beyond 0 probably 1. Yeah, what I think I said was 1. And you worked with the, so you worked with this term here and I have to collect a title. Yeah, that is important. So you worked with this term and this gives, this allows to show an equivalence of this single term here with the Hs norm, the integrated versions of that which lead probably to something similar than the formula on the right hand side with a single correction at j equal to 0 which is the L2 norm so it would probably give the same formula as this one in that range. Okay, so I want to have to give them up again. Now how do we prove it? Well, the strategy is to bound the, to control the integral part on the right hand side. This is the most essential part. Once we have control of the integral part on the right hand side, it also gives the control by limits, by certain limits of the whole right hand side and then if you look at the default crossing part, this allows to, I mean first we look at the transmission coefficient not in the whole half plane but in the, so we start with looking at the, we have some smallness conditions. We control, we try to control the transmission coefficient T of C away from the real axis and the smallness condition ensures that we can rescale things so that whenever we get it somewhere, we also get it let's say below one half, over two and then at least morally we have a harmonic function, the logarithm of T is a harmonic function which in the defaults in case has a real part less than zero. So then we are able to interpret this as such a harmonic function on the half space and then the real part of that is going to be a measure and that allows to define the left hand side. So if you use then that in the defaulting part that T of C is less than equal to one then the logarithm has the real part of the logarithm has a sign and then we can take the trace at the real part and that gives an interpretation of the formulas on the left hand side. For the focusing case one also has to use then the backload transform but I don't want to go into that. What I want to do is I want to explain how we can control the integrals. Well so we go back to the AKNS system. So the AKNS system was C1 prime is equal to minus IcC1 plus uc2, C2 prime is equal to IcC2 plus u bar C1. And what we want is we want a solution so that we want C behaves like e to the minus Icx0 at x equal to minus infinity. So then we can try to solve this equation recursively. So we want to get a power expansion in terms of u. So in the first step we integrate this equation and then we get C21 is equal to the integral from minus infinity to x times e to the Icx, maybe I could take that up with a y, y minus x u bar of x e to the minus Icx dx. And then we take, we look at C, so this stands for the iteration C12 and we get something similar. And then- So you have a prime, I mean you have one and prime. So and then we get the first, we get t is equal to 1 plus the limit as x tends to infinity to the 1, 1 of x plus, let me do this iteration and we call this guy t2 of C e to the Icx of this guy. We have to multiply by this and then we get t4 and these terms have a pretty nice structure. So if you do the math then this t2j of C is equal to an integral and here we have x1 less than y1, so we iterate this thing up to y2j. Here we get e to the 2ic the sum over the yj by case minus the sum over the x case times u of y1, u of yj, u bar of x1, u bar of xj, dy dx. So we do the iteration and then you get a pretty nice representation. The imaginary part of C is positive, so this is negative and this thing is larger than this one and we have the y's on the right hand side so that leads to exponential decay here whenever we are in the other half plane. Okay, so you do that, then you try to control these things and I guess I'm running a bit out of time so I have to decide what I want to focus on. So then the first type of estimate is that this t2j of C is bounded by, I guess I don't want to go into that, a constant times the u e to the minus i real part of Cx u in L2 and I put the imaginary part here, Cu du2 to the power j. So this is where the properties of the spaces du2 come in nicely. You have to take to put the oscillatory part into u and then they give very naturally this estimate here where this imaginary part of C corresponds to the localization in space and then you have two intervals of size imaginary part of C to the minus 1. This is exactly the scale on which this exponential decays, so there is this concentration. I wanted to say a bit about that but I guess I'm going to skip that. So this is the first estimate and it allows to get convergence provided this quantity is small and this is basically where the condition in the theorem comes in. So far so good. Then we don't want to get estimates in terms of this thing, we want to get estimates in terms of Hs norms. So how do we go from here to Hs norms? Well we can do embeddings, we can use embeddings and then we get some variable. We can use embeddings here and look at the scaling parameter and get powers of the imaginary part and this is going to allow to get some of the integrals. But it only goes up to a certain point and then we have to look at the logarithm of this thing. I know I want to go to the logarithm. Actually the way we did it, I think I did make another sign error. Anyhow, so we want to take the logarithm of t to the minus one. So this is t to the minus one is the sum of integrals which I write in this fashion. So this is the trivial integral, this here is the first integral which we have here, this is the second integral. So I wrote it here in red, I hope that it's visible. So this codifies the information that x1 is less than y1, less than x2, less than y2 and then we do the integral. And then we have this with three things, with four things. Then if you take the logarithm, you get a lot of combinatorics. You can split the integration domains into the sets where we order the variables and this is messy. This is really messy. So I did it for the first part, I got the second part, explain what this integral stands for in a minute. I think I computed by hand these things here up to here and then I was convinced that the structure somehow persists but at MSRI I was stuck with that, we were stuck with that and didn't know what to do. But it was important for us to get this structure, but explain what this structure is. So this goes, so for every arc which is going up, I put an x, for every arc which is going down I put a y and then I get here an ordering that x1 is less than x2, less than x3, less than y1. This is going up, that's x4, less than x4, less than y3, less than y4. The indices don't matter but this codifies this integral here. The nice thing here is that we always have the last y on the right of every x. So this here has a much better decay on the domain of integration as the original integral because the decays exponentially whenever two of the variables are far apart. So whereas this information here leads to decay of order basically at most imaginary part of c to the minus j because this is the size of the corresponding simplex, here we get a decay like imaginary part of c to the minus 2j plus 1 because the last y is on the right. So the last, you can't split these things into, no matter what we do, this decays exponentially whenever we put things apart. Here we could, in these integrals we could have the situation that y1 is close to x1 and y2 is close to x2 but x2 is far away from y1 and then it wouldn't get decay. I pay for that with the combinatorics. So at MSI there was the fortune situation that in this program there was people from probability and I showed this question to Martin Heier and Martin Heier told me that I should look at Hopf-Algebras and that it would be related to a famous formula in Hopf-Algebras. So my colleagues in Bonn are still, when I told them about Hopf-Algebras they were in the colleagues in Algebras they were excited. Finally a use. So a thing can be used. Well but then I tried to get it into context with the Milner-Mur theorem and that didn't work. So the basic thing is it's not a big effort to show that this is a Hopf-Algebras because it's very close to a standard Hopf-Algebras which I didn't know but that can be, that's not so difficult, the Schaffler-Algebras and the Milner-Mur theorem would give the conclusion if that would be a co-commutative Hopf-Algebras but it's a commutative one. So it sounds almost the same but it is not but then still if one tries to do things by food basically then all integrals are connected. Then one gets the structures. So where are we? Well we have this estimate here on the t's which transfers to the expansion of the logarithm of t. So we get the estimates we need whenever j is large compared to s. So if you look at the high order polynomial then things are easy but since we get that we have to look at the lower polynomial case by case and if you look at the conserved quantities then it's clear that something has to happen because if s is less than 1, less than 0 then there is no conserved quantity which hurts. At s equal to 0 you get the L2 norm so you have to handle the L2 norm. This is easy. This is easy because with the L2 norm if you check t2 then this is equal to 0. This is equal to u in hs squared so that requires simple calculation so the two term is okay. Now at the level of the next term at the t4 term that corresponds to the next at the level of the t4 term you expect the L4 norm to the power 4 to show up at the level s equal to 1 so this has to be taken care of. Now with this connected structure you get up to this point and the idea to go beyond that is pretty simple. This is doing integration by parts in the integration with respect to y1. If you do an integration by parts you get a derivative which falls on this. So if you do an integration by parts with respect to this exponential so you gain a factor 1 over c and the derivative falls either on this then you get the same formula with the derivative or you get to the boundary terms and then you have an integral of one dimension less and then you continue to do that and if you end up with no one-dimensional integral in the end you get the conserved quantities and otherwise you do similar estimates as I explained before. So I guess I went a bit over time. I'm sorry for that. Thank you for the attention. So this is the structure of Freyholm term. I have no idea. Freyholm term. No. Probably it has something to do with Freyholm term. I have no idea. I guess you know that better than I do. I mean in KGV, I mean that way, so I knew that the matrix case was for the scalar case. This is exactly the operation in passing from the series of T. Yeah. The thing along the series is sort of restricting the technical diagrams just taking the term there. So the real problem I wanted to make was, okay, so again the matrix sitting out of the detail. But in the scalar setting it's actually a general phenomenon that the transmission coefficient is the Freyholm term in it because the integral kernel is semi-separable. It means the function of the lesser times the function of the greater. And then what is my point? If you ask a problem that's something, then you might recognize it because that's exactly what makes processes on the line Markovian. Well, right. I mean why is the Einstein-Waldberg process Markovian? It's because the kernel is semi-separable. Yeah. So I'm not familiar with these things. But what I think is true is that there's a lot of combinatorics behind which can be expressed in various ways and it should be connected. I mean one thing with the Hopf-Algerprass, the reason why Martin Heierer is interested in the Hopf-Algerprass is I think different. It's closer to what I presented here and not the Freyholm determinant. But the thing which is striking is there's this theorem. It's not difficult to get. And it's not difficult to usage in order to compute these things to any order. But people I asked were sort of stuck with explaining what the coefficients or this expansion should look like or the coefficients should look like. So this is sort of big miracle. It gives connected integrals which is what we need but we don't have any access to the size of the coefficients. And maybe the Freyholm determinant is allowed to get that more explicitly. There's a lot of combinatorics behind which I don't understand. I have to go open. Stanley's enumerative combinatorics exactly has a theorem which says if you have a generating function generated with the coefficients and the number of objects of this type, then the logarithm of your generating function is exactly the generating function of the number of connected objects of this type. OK, that sounds, yeah. So I would be interested in a reference. I'd like to get that here. I was happy to get that but it's clearly related to many things. Well, yeah. Oh, I just comment on this estimate. I think C is independent of J. And if so, it's articulating that you're really creating a transmission coefficient is a holomorphic function of u and u bar. Yes. And these are the Taylor coefficients. Yes. It's exactly. So sort of the McKean-Truberwitz picture of the world which is objects are holomorphic with regard to functions. Comments or questions? So if not, thank you again. Thank you.
connected. There are integrable and the complemented with an equation for the time order to get a continuous family of conserved the NLS equation and again we have to adjust for the classical energies. Something special happens non-linear Schrodinger equation, I did some the generating function of the number of