 Veliko bilo. Počutke, prvno, s tešem organizacijem. Čelo se, da se počutim. Prvo, da je nekako vse vse vse vredno, ali za vsega bila, bilo očaj. Neko, zelo prijezdi. Od palju minutnih vsega. Počutim počutim počutim. Tukaj, da sem počutila 3D, ali baham, da sem zelo počutim, da bi se izvajila tektiv. V nekaj težiši je zelo. To je očaj skupniko z Fabrizio Pranšonem So, what's the plan? Well, I just start very, very simple situation about the Cochitheorie. So I want to say very quickly how to solve the Cochitheorie locally and global in time, and once we have a global solution, you want to understand the properties for large times. In particular we are interested in the issues of the growth of the solvable unknowns to deduce some upper bound, the polynomial bound that Jullio was mentioning before, in v svoj svoj svoj zelo. Zelo smo nekaj stručen, nekaj je nekaj ator, nekaj nekaj periodist, je tudi tudi 3D in 2D kompaktive, nekaj nekaj nekaj in nekaj nekaj nekaj klasične strategije v Burgeni, da se počuče tudi z polinoma glasboj, kaj je način vzelo modificovati energij. Vzelo smo, da v 2D More general results, in in 3D I think that you get some new result that as far as you can see you cannot get by using the bourguent technique. At least we were not able to do that. And then if there is time, but maybe not, I will say how this technique can give even better result in one day. Again about this issue of the sides of the growth. Now, what is the problem? It is the same, my Giuliolla, a few minutes ago. And here we assume, in general, for simplicity I assume a pure power non-linearity, but since now I tell you that this technique that we use allows you to deal also with more general potential. On the contrary it is not completely clear to me that the bourguent technology can work also for rougher non-linear potential. But first of all, please let's work on the pure power case. We are in this setting, so MD will be a d-dimensional compact manifold. Laplacian G is the Laplacian Beltrami operator. S is the regularity of the initial data. And at least you hope that for an initial data in Hs you propagate the regularity Hs for every times. And you want to understand these growth of Hs. But let's go step by step. And the special case that we will be interested in is the following. So 2D, essentially M2 is a general compact manifold. And P is any integer non-linearity. So no restriction to the cubic case. And in 3D we can deal with the cubic case. And we can say something in the sub-cubic case. But without any assumption on the topology of the manifold. So this is the usual. We learned what is the focusing between this morning and the day. And so as you can see, I assume here the sine plus, which is the same, is the opposite of the other one. So we get a positive energy. So what is sure is that the H1 non, if the solution exists, cannot blow up, not even at infinity. So no problem about the growth of H1. And let me spend a few words about the local and global Cauchy theory. Because if you want to start to analyze the long-time behavior, let's start by the local, by the global existence of one solution. So what is the status of the art, as far as I know? So in 2D, give me any 2D compact manifold. Give me any non-linearity, no restriction, no necessarily cubic. Then the Cauchy problem is locally well-posed in H1. So once you are locally well-posed in H1, for free, you globalize just by the conservation. What is the other result that you can prove? Well, in 3D, give me any 3D compact manifold. Give me any initial data in H1 plus epsilon. Then for cubic, then you can solve the cubic NLS in this regularity. And you see already in 3D the globalization is not free. Because you can solve locally in time in H1 plus epsilon. So this is essentially what I said before. So in 2D the globalization is free because you are at the level of the energy. In 3D the globalization argument is much more involved. It is due to Bjerke Gerard and Zvetnikov. OK, but I will come back to this approach later on. But just let me say a few words. So what is already this first point here? The second point here is not free. You need to work, especially on generic compact manifold, because the free result is the following. If you are in 2D, what is really easy and trivial is to solve locally in time in H1 plus epsilon. So already the globalization in 2D, if you just rely on the sobolef embedding, is not free. And in 3D, just by using this sobolef embedding, you have to work in H3 over 2 plus epsilon. In the previous slide they said H1 plus epsilon, H1. Why you can do that? Well, because instead of just using sobolef embedding, we have to use the Strikard's estimate. So you have to put in the game the dispersion in a more serious way, not just by L2 estimate plus sobolef embedding. And I also want to mention one result that was not mentioned today at least, which is the, as far as you know, the first global existence result for NLS in 2D, which is the result by Brésil Saint-Galouet, who said the following, they are able to globalize in 2D for cubic NLS any solution with initial date in H1 plus epsilon. That you cannot do that just by conservation law, but you have to add this extra sobolef embedding. You see, this is a sobolef embedding. In 2D you want to control L infinity in principle with H1, but you cannot. So you need to add some high frequency contribution, but this high frequency contribution can be hidden in the log. And this is something that helps to globalize. And you globalize, and moreover, you get also a bound, which is double exponential. So H1 plus epsilon cannot blow up in finite time, and moreover, you have the double exponential bound. How you can improve the situation? Where to improve the situation, you have to play with the dispersion. So you start to look, as Juliola said before, at some smoothing effect on the free waves. So you look at free waves, you take an initial data phi, you take the free evolution, and you start to look at some regularity, which is not necessarily L infinity, L2, but you start to play with Lpq, with the pq that move under this condition, which is a typical for Strikard's estimate. What you see is that, in general, if you are on RD, you can control this quantity by L2. If you are on a generic compact manifold of dimension D, you have to lose derivatives, and the loss is of the order 1 over p, where p is the regularity in time. Of course, in this situation, that Juliola was mentioned before, where m2 is T2, you can do much better. But this result is true in general, for every compact manifold. So this is the main difference between this thing and the one on T2. And this result, as far as I know, was proved in the same time, I don't know the details of this thing, but between Bjorki, Gerard and Svetkov, in the same time by Stafilani and Tatar. So we want to take advantage of this to solve our Cauchy problem at least local in time at low regularity. More precisely, in h1 in 2D, h1 plus epsilon in 3D. And, OK, let me go further. And let me explain very quickly why this Strikart's estimate helps to solve at least in 2D with initial data in h1, not anymore h1 plus epsilon. Well, you see, the game is very easy. The slide before implies this Strikart's estimate. You see, I lose 1 over r derivative if I take lr regularity, and this number here exactly the one for which r and this number here is Strikart's admissible. This estimate is proved, and thanks to a sobolef embedding, I go in l infinity. How many derivatives I need, I need 1 minus 1 over r derivative, and this 1 minus r derivative gives me h1, and thanks to this estimate, you see, I can control l infinity just by h1, that I cannot do for the time independent sobolef embedding, and that, thanks to dispersion, along three waves. If you combine this estimate with this free, essentially estimate on the dual meltdown, you get global existence in h1 and global solution because of the conservation. This is the 2D situation, so the globalization comes for free. Now, which are the questions we are interested in? Well, as I told you, we have an initial data in h1 and you have a global solution in h1. You can also prove without big efforts that the hs regularities preserved along with the evolution. So if you start from hs, which is very big, for every time you remain hs, but what is not so clear is whether or not this hs norm remains bounded or not. If there are solutions that whose hs norm grows or not, and if it grows, how fast you can grow. The second, even more basic question is there at least one solution that grows when t goes to plus infinity at least for other orders. The answer to this question, at least on 2D compact manifold is still open, but there are some results in other situations, like cylindrical symmetry on the manifold and so on. We have several fundamental paper on this topic, starting from Burgen, here it is a list of names, not necessarily working with Schrodinger equation, but the same question you can ask for generic dynamical systems. Here it is a list of names and I don't want to stay too much on this slide, but let me come now to the first remark already done by Giuliolla. If you start to ask yourself the growth of sobolev norms in R2 it is impossible simply because the nonlinear solution scatter. Since they scatter they behave for large times like free waves and for free waves Hs is bounded, is constant so at least you cannot run. So M2 is not so interesting so in some sense what is more interesting is the compact case. And another remark is that ok, once you have a reasonable cosci teori in H1, which is the case thanks to the estimate by Giuliolla and Tataru in the Birkjard and Zvetkov, then for free you get unexponential growth on the Hs norm. Why? Simply because in general just by tame estimate you can show that the Hs norm after an increment tau that depends just on the size of initial data you can control these Hs with a constant time this. You iterate this bound and for free you get exponential. Which is better than the double exponential in Galway for example. And moreover this is Yes. Not to growth bound. At the most the growth can be exponential. An upper bound for the growth. But can we do something better or not? Well there is the already mentioned work by Bougain who worked in T2 cubic nonlinearity and was able to show that in fact you have polynomial growth. And for instance I don't say that this is a big issue but for instance if you look at the same question for higher order nonlinearity I don't know any written results in the literature telling you that you have polynomial growth. But the technique by Bougain in this context should work and what is the basic idea of Bougain the basic idea of Bougain is to prove that this is a nonlinearity. It is based on some smoothing estimate in excess B spaces but the key estimate is this one. You see? Hs minus at an increment tau which depends just on the size of phi minus the initial data t with one constant but the power here has to be below 2. If you are below 2 then just by an elementary argument to introduce the quantity alpha n of time tau n then you write down this estimate implies this thing here this inequality here and just by simple iteration you forget this polynomial bound. N to 1 over epsilon means polynomial bound. But how Bougain could prove this kind of bound here? It is a fundamental that the constant c here is equal to 1. If you get another constant the best you can do is exponential but here you can get a big constant less exponents. This is the key. The idea of Bougain is very simple. We have to measure the difference in Hs at two different time. T1 and T2. What to do? We start with the fundamental theorem of calculus. This is the entity of this. You compute this time delivery but this is real part of this quantity. And now what it does? It replace ut by the equation. When you replace ut by the equation ut bar that just by simple self-adjointness argument is imaginary, so it disappears and you end up just with nonlinear part replaced in ut bar. Imaginary part. And here for simplicity they call 2s1 equal s. Well s is the regularity of the sobolev known that you want to understand. And now it is the key point. If you work a bit you see that this Laplacian 2s1 is Laplacian s1 to the square so you can put s1 derivative on the nonlinearity. Now when Laplacian s1 falls here you get zero because you have imaginary parts. So the most complicated thing is to deal when the Laplacian falls on this term, more specifically not in the bar and this is the most dangerous term that you have to control. But at the end of the story you start to control Hs you still get in this quantity Laplacian square. So it is not so clear how you can go down to 2 minus epsilon on the growth. But the magic thing is that here you have the bar. And this is what the bourgain spaces like a lot. The bar may give some miracle. So you have to play with these XSB spaces you have to spend some regularity in time given by B to gain something in space and you can go down with the power of Laplacian s1. What is the definition of X? So X, I'm not really an expert of this, but I can try. So you have to write this. One possibility is to take the solution the this can be adapted to general dispersive equation. So you act with the free dynamics on your function U and then you take the usual Hs hb in t hs in x. But acting with the free evolution. And the so you have B is in some sense the regularity in time, s is the regularity in x and thanks to this definition you can play a lot to push some regularity from B to s. This is the miracle that allows us in some sense to get a smaller power than 2 in x. And this is the game of bourguine. Now what we can say what is our result? Well the result that we can prove is the following. Ok, what is the the assumption? We assume that we can control l4 for free waves with s0 epsilon loss s0 can be chosen to be equal to epsilon if m2 is t2 but in the worst case we can prove in these two papers here that if s0 is 1 fourth then it is always satisfied. So it is never empty this assumption at the level of free waves. And our theorem is the following. Consider nls on 2d with any non-linearity not necessarily the cubic one no extra assumption on m2 then you have this polynomial loss. You can also this s0 one responsible of the loss of derivative in the free stricards estimate. And this s is the regularity that you want to measure. And this we can prove without seeing any difference between cubic, higher order non-linearity and so on and so forth and can be adapted as I told you not necessarily to the pure power case. It is just a question in some sense of integration by path as we shall see. It is a different strategy compared with the one by bourguine. And you should also say that the previous result at least in the cubic case was obtained by zonk and they also have to say that we never used the XSB spaces at least here. Probably by combining this technique with the XSB we can still improve. This is something that we are thinking about. But for the moment let me go really with the simplest thing that you can do. And what is the key point? The key point is that we don't use we introduce some modified energies. I should say that the user of modified energies is not something in you in the literature it has been used at least in those papers but even in other papers. But what we did is to combine first of all every problem as its own modified energy. Second thing we combined this with dispersion which is something not to don't be found in some sense. And how to proceed? We proceed like this. We introduced this energy which looks a bit strange I agree with you but I will do some comments in a few minutes. The leading term is the H2 norm square but we subtract some quantities which are there with proper coefficient. Because when we compute the entity of this modified energy, see what happens. We have the first time just one derivative so no second derivative on the contrary, Burgen had Laplacian to the square. We have just one derivative one derivative. Here we have one derivative and here we have dTU but dTU if you replace the equation gives you Laplacian U but to the power 1 not to the power 2 as Burgen did before but to the equation caused by this time derivative you gain one power of Laplacian and here it is where you gain the polynomial bounds. And just to give yes so, what to do? Well, notice that this energy is constructed in such a way that you have some algebraic cancellation where you compute time derivative but essentially the leading term to the square. Because in the second term you have the Laplacian U which appears in a linear way and here it does not even appear. So essentially at e2 if you control e2 essentially you are controlling h2. And so if you integrate this relation here what you get? You get that e2 but essentially h2 to the square minus h2 to the square right hand side here. And the most delicate term to be controlled at the following one you have this term and here you just use elder nothing more than that. You put this dtu in l2 you put u in l infinity because you are 1d you are 2d and in 2d essentially you can control l infinity by h1 plus epsilon and you just need to put to grad u square in l2 and this is w14 to the square. So at the end of the story what you have to control is essentially this l4 in time w14 in x and this you can do just by the dispersion you do that and you conclude the proof that does not depend on the structure of the nonlinearity, does not depend on the fact that you have a pure power and so on and so forth. This is the 2d that we did for an exercise to understand how to deal with the 3d case and just to give a very rough idea of where this energy comes from where you see the beginning is exactly the computation of bourguine bourguine said, ok let's compute dndt of h2 square so you have to compute dndt of this plus a nu l2 square this first identity is the same as bourguine also here you replace by the equation and you end up with this the point is that bourguine said ok I am enough like this so is the following this time dt we put outside you see and if you put dndt outside that can be absorbed in the energy of course you have to move dt also on the nonlinearity and you keep doing this game and at the end of the story you construct step by step the modified energy this is how this modification appears but after this cancellation the bad term by bourguine is disappeared and the remaining term involves just Laplacian nu to power 1 no more Laplacian nu to power square which was the dangerous term and of course we don't exploit too much the algebraic structure of the reminder for the reminder we just put absolute value and we put other but probably playing more with the reminder we can also improve what we do and another important thing is that this computation does not depends on if you are in the flat case or not even if apparently there are hidden some integration by parts just because this fundamental identity that is of course true in the flat case is true without any modification on every compact reminder does not depends on the curvature so the tool that you need is essentially hidden here this is how you deal with the piece of the energy but let me go further so let me come now to what we can do in the 3D case so in the 2D case every 2D compact manifold forever 2D compact manifold ever in no linearity we have growth polynomial growth what we want to do in 3D where in 3D on a genetic compact manifold first of all let's ask ourselves for which no linearity we have a global dynamic first of all and the unique case that I know in sub cubic no linearity so if you give me something which is a cubic plus epsilon and you don't assume nothing on the geometry of the manifold I don't even know if there is a global solution but in this situation you have a global solution and the point is how to guarantee that you can globalize the solution because you see assume that we start in 3D we are in 3D we take initial data H2 the local existence result comes for free just by sobolef embedding because 2 is above 3 over 2 question, can we globalize or not? which you cannot do that for free and as far as you know the first answer to these questions is by Bjerke Geradensecki and they give you very quickly their strategy in this way you can see where you gain if you use the modified energies the strategy is very simple put down the integral equation satisfied by U and you apply Minkowski inequality so the L infinity H2 norm of the solution is controlled by H2 of the initial data just by the Duhamel formula plus L1 H2 of the non-linear term and the non-linear term is cubic now let's go to the second step H2 of this cubic term can be controlled by H2 at power guard and of course you have to put this term in L infinity and you get L2 in time L infinity in x to the power 2 question how to control L2 L infinity because you cannot rely anymore on a sobolef embedding if you rely on sobolef embedding you lose other power of H2 and you cannot do that so here it comes the idea of Bjerke Geradensecki to use the dispersion so the Strikert's estimate was mentioned before and the target is the following the target is to show that L2 L infinity to the square can be controlled by this log you see if I can put here the log then I imagine here a log I am exactly in the position to apply a kind of gronval lemma that gives for free also a double exponential bound now how to control for the non-linear solution L2 L infinity you have to use the so called Strikert's estimate this version just a few comments about the Strikert's that you need to use well in general when you write down Strikert's estimate what you have to do well you have a non-linear solution in general you need also a forcing term because you have to work with the dual mill operator with initial condition equal phi and in general what you expect you expect to control some norm of v by using what by using some hs norm of the initial data which is what you expect if you don't have forcing term plus another norm that involves just f this is not the case of this family of Strikert's estimate for two reasons the first reason is that you have the project of pn and this is something that is not a serious problem because the project commutes with the equation so if you localize the forcing term and the initial data you have no problem but the key point is that here you don't take some norms of the initial datum phi you have to exploit the solution himself for f you have a norm but for the here you don't have just phi you have the evolution of course if f is equal to zero it's the same thing because h1 half is preserved along the evolution this is essentially the h1 half norm of the initial data but if f is not zero you don't control anymore in the standard way by Strikert's this is the key remark and why you have to do that because if you play like in a non clever way you lose derivative on the forcing term and it kills you so in this way you don't lose derivative on the forcing term but the price that you pay is that here you have the whole solution not just the initial data and this is technically this comes from what well it comes from the fact that when you write down the Strikert's estimate on a compact manifold when the frequency grows the times in which you have the Strikert's they come with semi classical times so what to do is to start for semi classical times and then you sum up if you sum up and then you apply the so called Kriste Kiselyev argument you lose two derivatives and you are killed in this way by this cutoff in time you are safe because you don't lose nothing here which is fundamental ok and now how do they proceed they say ok now imagine that in our case of course f is youtube because we want to apply this estimate no linear solution and what you do well you sum up, you take those estimatives since you want to control 12 infinity and there you control 126 you have to lose one half derivative you lose one half derivative one half derivative here becomes one derivative here and there you lose one half derivative and then they sum up because this is an estimate true for localized solution but you have to sum up on the frequencies so you have a sum of this term here plus a sum of those term here over capital N and you just use Minkowski on the left hand side this is where you lose a bit because use Minkowski and when you sum up you have to lose something but let me come a bit later on this point and essentially what is the estimate we arrive at this estimate here and now let's check that this is not a problem this is not a problem why well because you see in W1 half 6 over 5 I want to remove the sum but to remove the sum is free if I can lose derivative I can put one derivative here and then remove the sum and when I remove the sum you see what I can do I put a gradient in L2 and U squared in L3 and I get h1 and 6 squared which are conserved by the Hamiltonian so the unique term which is dangerous to sum up is the term coming from here when you sum up the L2 in time h1 in x of the projected pieces of the solution and how do they solve the problem well you split this sum in high frequency and low frequency part where n0 is a frequency that you will select later on you see this sum you split below n0 above n0 when you are above n0 you can gain 1 over nier if you lose h2 this is for free and when you work at low frequency you just do koši švarts you take a square root of log of n0 which is the amount of the dyadic number below n0 and then you sum up the square you continue this estimate and at the end you see where the log comes from just because you have to select in appropriate way the frequency n0 but what is the problem is that you have to sum up without the square if you could sum up with the square then you lose nothing and this is what we can do with the modified energies we avoid the usage of L infinity and the mean koši you can use just square of quantity so you can use all the power of the harmonic analysis with the squared function and now the conclusion at least for them is easy because now I have controlled L2 in time L2 in time L infinity by this log you replace there but since you have this one times the log you have double explanation now what we want to prove for 3D in which sense we can improve we can improve in the following sense we have two kind of results the first one is for cubic the same nonlinearity treated by Bjerge, Gerard and Svetikov and what we say is that in the cubic case global existence is already proved by themselves but we will prove the global existence with a better bound we just take explanation which is what you expect in some sense but you cannot prove with their strategy and the other thing that we can prove is that if you take something slightly below cubic so in the sub cubic case then you get also polynomial but why in this statement I put HM with M generic where I put H2 because if P is not enough regular then it is not so clear that higher order sublif norm higher regularity is preserved but of course in our situation we can regularize a bit of nonlinearity and still the proof is true and here you can really take HM with every M but since I wanted to focus just in the pure power case then the unique thing that we can do is the to work with H2 but still you get a polynomial growth on a genetic compact 3D manifold which as far as you know was not proved elsewhere and I don't see how the Burgen technology can work because here it is already a problem in H1 now this is something and just let me tell you why by using the modified energy we can go from double exponential so the modified energy that we use are exactly the same I have introduced before in 2D and you can use also in higher dimensions if you want so this is the E2 that should play the role of H2 but it produces in the computation of the derivative such a cancellation that cancel the worst term this is the key idea on the contrary but Patrick with his collaborator they start by putting absolute value 2ML formula absolute value and Minkowski here you don't use this you start by computing d dt of E2 and if you compute this you get this term which is not complicated the more complicated is this one because you have dTU that if you replace the equation is a Laplacianu but still to the power 1 here you have Laplacianu to the power 2 so you have a lot of room to try to play the ground volume and how to control this term in 3D well it's really very simple because simply you see you put dTU in L2 this is essentially H2 then you put this grad U square you put in L6 and you get T2W16 to the square and you remain with this extra term U that you put in L6 this is just another inequality this L infinity L6 is under control just by because you can control H1 so you end up with this quantity ok so the quantity that before for them was L2 L infinity now becomes L2W16 ok and the W16 you can deal without summing up the square not by taking Winkowski and you don't lose nothing when you sum up in some sense why let me give some details so the starting point is the same stricards estimate as before still localized at the frequency pN but before we started with the L we took one half derivative if here we take one full derivative ok you take this and what to do now I don't sum up this quantity sum up the square if you sum up the square on the N I sum up also the square here and I sum up also the square here now you see the nice thing is that here you can use planche RL to sum up by using high frequency low frequency this time that before was our enemy and now is our friend because it is for free L2 in time H3 over 2 square but here I have to sum up the squares but it is not a problem because you see when I want to sum up this W16 project I take the solution I project on the with the L2P I take a sequence and sum up the square this is L6 in time L6 in x L2 in time discrete L2 in N but since 2 is smaller than 6 I can reverse the sum ability I can put smaller 2 in time and this is nothing else that L6 of the squared function but L6 of the squared function by harmonic analysis is equivalent to L6 and so you conclude that you can really end up with this estimate the same as before by removing the project that I could do if I work with an infinity this is the the key point and once you have this now to control L2 W16 is essentially easy because you see I have L2 H3 over 2 and yeah this F becomes the non-linearity and here I just use Gallardo Nuremberg inequality I put H2 to the one half H1 to the one half this is conserved this is H2 to the one half when I take the square I take H1 but I add another H2 before so you take H2 square controlled with H2 square which is a sharp case to apply exactly grombal getting the exponential growth no need to lose more on the exponential and why in the subcubic you gain polynomial growth well it is simple because here I can work with 3 over 2 minus epsilon and if I apply the Gallardo Nuremberg inequality I take on H2 one half minus epsilon so I control H2 square with H2 to 2 minus epsilon which is what you need for the polynomial so this is nothing else than that so simple idea is to combine modify the energy with the correct stricards estimate to be careful when you sum up the square and the nice thing is that you have the correct monotonicity between the exponent that allows you to interchange L6 and the smaller 2 and then you apply just harmonic analysis result on the squared functions and you get this exponential value just one word I said everything about H2 how to deal with H1 million for example the point is that if you start to compute the modified energy associated with the power k of the Laplacian it starts to be a bit nasty the situation but what is the idea well instead of considering power of Laplacian consider power of the time derivative what is the advantage is that the time derivative they are straight so they commute very nicely when you apply Leibnitz rule and so on and so forth no need to take care of the Laplacian to to commute to the Laplacian with other terms and essentially once you control this then you have controlled this hk to sum of 2 so the correct modified energy at higher level order is this one you have to compute d d t which is not really nice it is the result but essentially you can isolate 2 or 3 bad terms and other terms can be controlled without dispersion essentially at this point you just apply older process I don't exclude that by combining now with the Bulgarian spaces you can get even better things and this is the situation I just I want to say that the unique point where you use differential geometry is this identity but you don't even use differential geometry because there is no curvature here it is just an algebraic identity that you can compute chart by chart and that's it and I just wanted to say one last slide to say how this technique we think can be used to improve further in 1d ok if you look at the result we have proved for instance in 2d of course if you are on the 1d tors I can consider every solution in the 1d tors as a solution in the 2d tors by a trivial extension so the same bound we get in t2 should work for t but it want to be a bit better if you work to our result in t2 what we did well we get here the growth t what I say is that at least for h2 we are able to prove that you get square root of t and we have some computation showing that this is true in hm for general m but you can gain order you can gain something on t and what is the new tools in 1d that we don't know how to use in 2d well as usual introduce the thing that the key point is to introduce this quantity here the capital N the j and the capital T that satisfy this couple of transport equation ok why this because all the terms that you see when you write down the derivative of the modified energy can be written in terms of j t and n and to have this nice solution this nice relation absolute for instance how does it help you see what you get when you compute the second derivative you get this term and this term for instance one trivial relation between this term here and this term here is this one you see I rewrite I like this I put dTn then you have n dx u2 that can be written in terms of j and then like this like j2 plus dx n2 then this dTn by the equation is minus dx j so je minus dx j j2 and the terms coming from dx n2 is exactly 2 so you have a non trivial relation between the first term and the second term if I find another relation you have done and to find another relation is to exploit now here I don't use the t I use the t because the t involves this dx u2 I get another identity starting from the t another system you have a system satisfied by 1 and 2 that gives you something nice that you can control because you see here the nice thing is that at the end you have a relation telling you that this difference can be controlled by at the integral of a term that involves four derivative shared on every term and once you have four derivative shared on four terms you need to get the square root of t function of course to do that for h1 million is not I mean we have some ideas but this is the way how this modified energy can be exploited in the 1D to get even better balance and I think that I thank you very much and I stop here in the case of 3D compact manifolds well of course without any other assumption for the moment there is three cuts inequalities we have but there are examples like of course teorai but also spheres or gel soul manifolds for which in fact it is possible to solve up to the quintic and in that case it is certainly possible using the bogyan spaces to get polynomial here but using your using my technique I didn't try but for sure once you have for instance if you look at the work by Svetkov, Er and Atar 1 t3 they can reach the quintic and the pausa there in your next could deal the large data case by using these UV spaces I think that you can get polynomial bounds by using my technique I didn't try but probably I didn't try yes, yes, probably just by an elder and the would think here is that that you need is not in h1 it's in h2 and all the tools of low regularity are needed just for the globalization so probably all the technicality needed when you ask for h1 are not needed it's unclear for me the technique of bogyan if it works to get the exponential bound in a generic 3D compact manifold that's what I'm saying probably but it's not completely clear unless you don't have a vector structure that allows you to use the bogyan spaces like t3 spheres I agree with you the modified energy do you have any intuition of what does it have the physical meaning for me it has just algebraic meaning just I don't see really but you see the construction is that you have to do a step more compare it with bogyan just this dndt you don't replace by the equation but you extract another term of the energy and you gain one power of the Laplacian for free without any effort no, no, absolutely no, this is the point we don't use bogyan resting all the refuel all the plus tricards so just the linear tricards