 Ох, извините. Спасибо. Про нелиньер-шреддингер эквизиумы, это один из самых известных примеров, для которых теория уплывается, и один из самых простых, я думаю, так. Итак, я буду использовать both slides and the blackboard, и сейчас я буду использовать некоторые нотения. Я буду их поставить на blackboard. Давайте возьмем dimension t более чем или equal to 2, и consider the torus of the dimension d, with period L, so this is rd factor L times zd, and here L will be a large parameter, and further it will go to infinity. Let me note that of course the volume of this torus is L to the power d. So let me consider a function, complex function, on this torus, and I will denote by such norm, just the L2 norm of this function, normalized over the volume of the torus. So L to the power minus d, integral over tdL, absolute value of the function squared dx. So this is just L to norm, normalized per volume. Also I will write my functions on the torus, I will develop them in the Fourier series, and I will choose convenient normalization, namely I will have factor L to the power minus d over 2, and then just the Fourier series, so I take a sum over S from zdL. Yeah, what is zdL? zdL, this is just lattice zd divided over L. So when L grows, my lattice becomes more and more frequent. Here I have space of the size L minus 1. So I take a sum over this space over zdL, then the Fourier coefficients vS, e to the power 2 pi imaginary unit S times X. So why this normalization is convenient, it's easy to see, sorry, here, here, this is squared norm, of course. Why this normalization is convenient, it's easy to see by applying the partial valentinity. Then I can write the L to norm in the following form. Here I will have due to this factor L to the power minus d, and then the sum over all S from my lattice zdL, this step L to the power minus 1, absolute value of the Fourier coefficients squared. So you see, when L will go to infinity, this lattice becomes more and more frequent, and it will be convenient to approximate sums of these forms by integrals over Rd. When we will redefine appropriately the function, the Fourier coefficients vS to Rd from the lattice. So that's why this normalization is convenient. Okay, so now let me explain what is written here. This is just nonlinear Schrodinger equation with cubic nonlinearity, where nonlinearity is slightly modified, so I have subtract this normalized L to norm, and also I have a small parameter nu, which will go to zero, then we will send it to zero. And our equation is on our torus, which sides of which will grow. So this modification of nonlinearity is rather innocent, it's quite frequently used by people working on Hamiltonian PDEs, because it keeps the main futures of the equation, but reduces some noncritical technicalities. And also let me know that if you consider such function u' which equals to e to the power ET L to norm of u squared times u, where u is the solution of my Schrodinger equation, then this function will satisfy just usual Schrodinger equation without this modification. So it's indeed rather innocent modification. Our equation is Hamiltonian equation. It preserves as usual Schrodinger the L to norm. Let me write it down. So I will erase this, but instead I will write that the L to norm is preserved. And the objective of the wave turbulence is to study solutions of this equation under the limit, which below I will call lim, when nu goes to zero, so the nonlinearity goes to zero, and the period L goes to infinity. So this is called lim. Nu goes to zero, and L goes to infinity. The meaning of this limit is not quite clear in the sense that in general this limit is not commutative. So let me mention some existing results. The limit when nu goes to zero, but the size of the torus L is fixed, it's treated in a number of applications. This one of Juan Guan, Sergey Kuxin and Alberto Mayoki. But this is far from the wave turbulence, because L is fixed. Due to our acknowledge, the only mathematical works addressing this limit when nu goes to zero, L goes to infinity is these two works. One of 2016, another one is just a reason. It was just published several days before we put on archive our work. In the first one, L goes to infinity quite slow, so it goes to infinity much slower than nonlinearity, goes to zero, in the sense that L goes to infinity much slower than nu to the power minus 1. And the results of this work are far from the prediction of the wave turbulence, and this is more kind of averaging. The results of this work, they are more in spirit of the wave turbulence, because they hold only for relatively short time interval, and this is also more kind of averaging. So, still, as I have mentioned, there are plenty of physical works that contain some different, but still consist an approach to treatment of this limit, and none of them were regularly justified, despite the interest in mathematical community. So, in these works, physicists usually speak about pumping the energy to low modes and dissipating it in high modes, but in their level of rigor, these words do not have trace in the equation, and to make this rigorous, they suggested to consider the analysis equation dumped by viscosity and driven by random force. So, random force make pumping the energy to low modes, and dumping the viscosity takes away energy from high modes. So, and that's what we are going to do. To explain this, it's convenient to write our equation in the slow time. So, I set tau equal to nu t, and then my equation takes such form. So, in front of the nonlinearity, we do not have small parameter, but we have such a fast rotating term. But from now on, by dot, I denote the derivative with respect to this slow time, and I will work only in this slow time. So, now we consider the following equation. So, we have the same analysis equation written in the slow time. I have additional parameter rho. Here I will comment in a few moments what is this. On the right-hand side, we put viscosity. If r star equals to 1, this is just viscosity. If r star is larger than this, this is hyperviscosity. And eta dot, this is a random noise. So, I will explain what is this random noise, but before let me comment on the scaling of parameters, you see, if I rewrite this equation back in the usual time, not in the slow, but in the usual time, I will get again small parameter nu here, and I will get also nu here. So, it means that our perturbation, this perturbation is weak. But it isn't clear at all why it should be of the same size that nonlinearity. That's why we put here some scaling parameter rho, which is positive, and we will scale it with nu in such a way that the limit we are looking for exists. And let me mention that, in fact, rho will be not small, but large parameter. It means that nonlinearity is much stronger that our perturbation of the system. So, about rho star we will assume only that it is just positive. What is about the noise? The noise is given by its Fourier series. You see, here I have the same scaling parameter L to the power minus D, like here, like here. Then the Fourier coefficients are bs times beta s, bs are just some real numbers, but they obtained as a restriction of Schwarz function to my lattice. So, in particular, these coefficients are fast decaying. So, it means that I pump the energy mostly in low modes, but in high modes the pumping is quite weak. And these are standard complex independent binary processes. That is each beta s obtained as some beta s1 plus imaginary unit beta s2, where beta s1 and beta s2 are standard real independent binary processes. So, yes. We can write the balance of energy of this equation, and then we will find that the averaged energy per volume, what is this average energy per volume? For example, this quantity expectation of the L2 norm, I recall that my L2 norm is normalized over the volume, is a further one uniformly in all my parameters. So, it means that the scaling, which we have chosen, is correct. About this equation, yes, I should mention that if rho star is sufficiently large, it is well posed, and also it's known that this equation is mixing. So, there is unique stationary measure, and there is convergence to this stationary measure. And we are interested in the limit, when nu goes to zero and L goes to infinity in this equation. Yeah, I'll call it dDnLs, because it's a dump-driven nLs equation. And we specify this limit by the following two possibilities. So, the first possibility, I'll write it down, the second I will not write on the blackboard. L is more than or equal to nu to the power minus 2 minus delta. Delta is positive. So, you see that L goes to infinity sufficiently fast. We just find this precise meaning of this relation, I mean that the precise power just during our computations. Or the first possibility, just first L goes to infinity, first the size of the torus goes to infinity, then the non-linearity goes to zero. So, we'll write our equation in the Fourier space. So, I write the corresponding equation to the coefficients of the Fourier series. Let me write it down on the blackboard. It has the form p s dot minus nu to the power minus 1 absolute value of s squared v s. So, this is the input of this faster rotating Laplacian. Then we have input of viscosity plus gamma s v s. So, this is the viscosity. And gamma s here is 1 plus squared absolute value of s to the power r star. Then, on the right-hand side, we have non-linearity e rho L to the power minus d. Some over s1 s2 s3 from that dL v s1 v s2 v s3 bar, and then the noise plus bs bts dot. So, you see that due to the definition of the noise in the Fourier series, we obtain that it's just diagonal. Here, sorry, I forgot this delta prime 1 2 3 s. And what is this delta prime? This is 1, if s1 plus s2 equals to s3 plus s. This is 0, otherwise. If it would be just like this, then it would be natural to denote this just by delta 1 2 3 s. This is just common notation. But we have additional restriction that the set s1 s2 does not coincide with the set s3 s. And this restriction comes from the fact that we modified our non-linearity. So, somehow, we subtract from the non-linearity integrable term, which pose some technical problems. So, and we will work only in the Fourier representation, so I will forget about the initial equation, I will look only at this equation. What are we interested in? The central object in the wave turbulence is a so-called energy spectrum, which is in red in the blackboard. Energy spectrum. Energy spectrum, this is denoted by ns of tau. This is the expectation of the squared absolute value of the amplitude of the Fourier coefficient, number s. And the objective of the wave turbulence is to study the behavior of this energy spectrum under the limit nu goes to 0, l goes to infinity. So, you see, people do not interested in the behavior of the solution itself because most probably just, I mean, in the equation we cannot take this limit. But what are people interested in, in such an observable? So, let me make a small comment. Why is it so? And what does it mean? Look, we know that our equation, if we switch off the noise and the viscosity, we know that our equation preserves l to norm. So, particularly it preserves the sum of these squared amplitudes because of the partial identity. But, of course, it doesn't mean that each of these squared is conserved, just their sum is conserved. Then we turn on the damping and the noise, and we are doing the following just to make a picture for a moment. Let me assume that everything is dependent only on absolute value of s, not on s itself. And then we have damping and, sorry, we have injection of energy by the noise. And it mostly acts on low modes, injection of energy. We have damping, which mostly acts on high modes. Actually, why? Because, as I already told, bs, they are fast decaying. So, we have stochastic perturbation mostly, we have injection of energy mostly in low modes. But on the contrary, gamma s, which is our viscosity, is large for high s. So, in particular, then it's natural to expect that we have some flow of energy from high modes to low modes, which is the function ns, in which we are interested in. So, when the question is how this function behaves when non-linearity goes to zero and the size of the system goes to infinity. This question, morally, is related to what about Jean-Pierre Hoekman who was talking, because he had a chain of oscillators, he had high energy in first oscillator, then he had damping in the last oscillator, and he was looking how the energy propagates through the chain. So, morally, somehow these questions are related, but in our case, but, of course, the problems are very different. And he was interested in much more quantitative questions, because the goal of the wave turbulence, one of the main goals is to show that under these limits the energy spectrum satisfies so-called wave kinetic equation. Wave kinetic equation is an equation of such form. This is closed equation for the energy spectrum. So, ns dot equals to some operator, which is called wave kinetic integral. This is an integral operator of n. We will write it down, plus some term, which comes from the viscosity, plus some term, which comes from the noise. And here, already, s belongs to Rd, not to that dL, because we have taken the limit L to infinity. So, you see, this equation is a closed equation for the energy spectrum. In particular, we do not see any trace of arguments of Vs. I mean that each Vs is a complex number, so it has absolute value and argument. And here we have trace... Why don't I talk? We have trace only of absolute values. So, somehow, under this limit, all the angles should average out. And this is quite usual type of questions in statistical mechanics, to find, under some limits, to find the closed equation on some macroscopic observables, like the energy spectrum. Okay. So, now what are we doing? For the moment, there is no model for which this convergence would be justified of the energy spectrum to the solution of the kinetic equation. And what we are doing here? So, we take the initial conditions, such that at some negative moment of time our system is at rest. This is just technical condition, it's simply to work with, because anyway it's quite technical and this assumption is quite natural. If T equals to infinity, it means that at minus infinity we are at rest, that now we are in stationary region. Then we write our solution of our equation as a formal series in rho. I recall that rho is this parameter in front of non-linearity. I have it here on the blackboard. And then we take only quadratic part, only quadratic truncation of the series. So, we take terms till the second one, and call this object quasi-solution. Let me write it down. So, we have formal series vs equals to vs0 plus vs1 plus rho vs1 plus rho squared vs2 plus etc. And we throw away the rest of this formal series the tail of formal series and denote this object p large s and call it quasi-solution. And we will study instead of the real energy spectrum, I mean instead of the energy spectrum of the solution itself, we will study the energy spectrum of this quasi-solution. And we will denote it by ns large. So, actually this is exactly what physicists do because they always try that they study the solution, but in reality they make such type of formal decomposition and they throw away exactly terms starting from the fourth one in this series. So, our first goal is to try just to make rigorously what physicists do in a not rigorous way. The second goal will be to study the higher order terms of this formal decomposition. Of course, we hope to prove that this quasi-solution is somehow close to our solution. But for the moment we cannot do this. I will comment on this later. So, now, as I already told, we study the energy spectrum and our goal is to show that under our limits L to infinity nu to zero to show that approximately this energy spectrum satisfies the wave kinetic equation. So, since here I have the truncation till the order rho squared in the energy spectrum I will have truncation till the order power 4 and for the further need let me write it down on the blackboard so ns0 plus rho ns1 plus rho squared ns2 plus rho to the power 3 ns3 plus rho to the power 4 ns4 and now our goal before just proceeding to study the limit and we should choose scaling of this parameter rho in such a way that non-trivial limit exists here. Yes, thank you for this question. Just we stop at minimal order which will give non-trivial result. You will see in a minute why is it so. I will explain this in the item of my talk. So, now I can say the following thing. So, we can go till higher orders as far as you want but simultaneously we should take d larger and larger the dimension of the torus. For dimension of the torus more than or equal to 2 we can take only till this order. Yes, I take square of Yes, yes, yes, I miss some terms here. I mean if you take the whole expansion if you take the whole expansion then you compute formally the square of this series then in this and in this term you will have some additional terms but in fact they change absolutely nothing. So, we can add them there and nothing will change. So, that's okay. Good. So, my next goal is to choose a scaling for rho. Scaling for rho. To this end I need to estimate the terms ns1, 2, 3 and 4. ns0 this is just the absolute value of vs0 squared but what is vs0? vs0 is the solution of the linear equation just Gaussian process and this is of order 1, of course. So, this guy is of order 1. Then simple computation shows that ns1 just equals to 0 and this is the reason why we cannot take only these two guys. So, this will be just trivial. Linear equation. Concerning ns2 we have theorem. Theorem ns2 behaves as new. So, we can just find ns2 explicitly because here we can find all the terms explicitly. I mean, higher term is more involved with computation but for ns2 it is still reasonable and then it turns out that's possible to find asymptotics of ns2 under this limit and asymptotically it behaves as new. Symptotically it behaves as new. And ns3 and ns4 they are bounded by new squared. So, it means that the only reasonable scaling is rho should behave as new to the power minus one half rho equals to square root of epsilon new to the power minus one half and here epsilon this is just a small but fixed number. So, it will not go to zero simultaneously with new or something like that. This is just fixed. So, rho is a large parameter because new goes to zero. Why is it so? Indeed, if we take such scaling then this term will have size just epsilon just epsilon this term will have size just epsilon squared and this term in fact it also can be bounded by epsilon squared because you'll have here epsilon 3 over half and you'll have some additional power of new and new is small so you can bound it by epsilon squared. So, in this quasi-energy spectrum the main non-trivial input gives 2nd term NS2 and this term they'll be smaller they'll be just given error. In fact, our goal is to show somehow that this term is governed by the wave kinetic equation. So, in particular this energy spectrum represents in such form NS equals to NS0 plus epsilon times some NS1 which depends on epsilon but here's a further one so this NS this one is given by these 3 terms multiplied by new to the power minus 1 so and the guy of order 1 this will be this one. Okay so what is the time now already 40 minutes past so let me say just a couple of words about the proof of the theorem I have written here some formulas but I will not enter into details the first issue is to prove that NS2 asymptotically behaves as new to this end we just write what is NS2 this is expectation of the absolute squared of VS1 and some another term formulas for NS3 and NS4 are more complicated I do not write them down we can just inductively compute V0 we know that this is a Gaussian process linear equation VS1 and then we can compute VS2 here for example just throw down formula for almost VS1 but I called it AS1 because this is VS1 which is slightly modified just I didn't tell you that in fact we do not work in this V variables we work in the so called interaction representation which is much more convenient for this subject but since I do not enter into details of the proofs I just use it only one time so I will briefly explain what is this and then I will forget about this interaction representation so interaction representation these are variables AS of tau which are obtained by rotation from the variable and the point here is that when you pass to these variables you just kill this large term in your equation but instead you get some additional fast rotating term in this equation sorry in the nonlinearity okay so here I wrote down this A1 then to compute for example this term we should have something like product something like square absolute value of this guy and then we should take the expectation so you see here you have summation over ZdL then you have L to some negative power and in the result after taking the expectation you will get some of such form some sums of S1 in this particular case so sum over S1S2 and L to the power minus 2d and you approximate it by by integrals and this is the only place in fact where we use that L is sufficiently large we approximate all sums by integrals and this is the key for the first developments but this is the only place where we use that L is large and then we represent these terms after approximating the sums by the integrals as such singular integrals as such singular integrals look here you have some nice fun for example some Schwartz function in the denominator but in the denominator you have some non-degenerate quadratic form plus new squared which is separated from zero and it turns out that it's possible to find asymptotes of such integrals this is not trivial some nice geometry is inside of this proof and it turns out that these integrals they are of the size new and in fact the asymptotes can be written down just explicitly and it is given by integration of the manifold where this quadratic form equals to zero of course, because otherwise you see that if this quadratic form is not zero then the denominator is of order one and the integral is of order new squared but the answer is that it is in fact of order new that is because the integration of the manifold where this quadratic form degenerates now this was about the proof of this theorem and now we know the scaling for rho and we can study the energy spectrum of quasi-solution NS and we'll prove that that it converges that it's approximately governed by the wave kinetic equation and to formulate the result I should define what is the wave kinetic equation precisely to the extent I need to define the wave kinetic integral so let X be a real function on RD then the wave kinetic integral is the following integral so what do we have here I recall that gamma is a trace of dissipation here it is here we have like a sum of four integrals in each we have degree 3 and this comes from the fact that nonlinearity has degree 3 and we have some denominator and the integration is performed over the surface where such equation equals to zero and here there is S3 here there is no S3 should be expressed through S1 S2 and S in such a way so to understand where from this expression comes just look at this delta prime that's exactly what is written here and equality to zero of this of this expression that's exactly what I have just told about the fact that asymptotics of integrals we studied here is given by integrating over the set where this quadratic form vanishes because in fact when we deduce this kinetic equation this relies also on analysis of such integrals and the quadratic form is exactly this one and I call it we call it the resonant surface why just because when you pass from this A this V variable to this interaction representation you will get here first rotating term and the first rotation will be given so you will have here term e to the power I into the power minus 1 and then let me call it omega123S and omega123S this is exactly what is written here so we keep in the wave kinetic equation resonant frequencies so wave kinetic integral possesses some nice properties I will not speak about them just as an example if x is a Schwarz function then k of x also is a Schwarz function sorry so what is the wave kinetic equation this equation of such form here I write m not n I will denote the solutions by m m dot equals to epsilon this is epsilon times the kinetic integral minus the trace of viscosity plus the trace of the noise and initial conditions are the same as for our system Theorem wave kinetic equation is well posed I do not speak in which sense to save time and its solution takes the following form so we have solution of linear equation plus epsilon plus error of order of order epsilon so let me write it down here ms of tau equal to m0 s of tau plus epsilon ms1 of tau epsilon of course ms0 of tau if you compute it you will just find that it equals to ns0 which is the same that in the energy spectrum and the main result the first main result is the following to compare our energy spectrum of our quasi solution we should extend it to a function on rd and only on the dl it can be done naturally and it becomes a Schwarz function of rd and then theorem that if we fix any r if epsilon is sufficiently small then supremum here we have some growing with s if r is positive factor and then the difference of solution of the wave kinetic equation minus the energy spectrum is less than is bounded by epsilon squared is bounded by epsilon squared and look, but we have that the non-trivial term of the energy spectrum is of the size epsilon and non-trivial term I mean the term which feels on linearity of the wave kinetic equation is also of the size epsilon but the difference epsilon squared those size epsilon squared so it means that indeed the solution of the wave kinetic equation approximate the energy spectrum of the quasi solution and this holds for any time provided that new of course new is sufficiently small and we remember that L should be large that's a good question we do not know the problem that is we do not know of course the next thing we would like to do is to prove that the quasi solution is close to the just to the solution and then to prove similar result for the solution yes physicists below in a minute I will show some reasons why it can be not true but physicists they always consider quasi solution anything we do not know anything about solutions we speak only about quasi solutions and about higher order terms of formal series and it will come now so this is the theorem for the quasi solutions well and about the proof of the theorem I will say almost nothing just the only thing that I want to tell that this theorem, this previous theorem is indeed in the heart of the proof of our main result because in order to get the main result we just analyze the increments the increments in time of the standard spectrum and compare them with increments in time of the wave kinetic equation and then when you compute in the increments in time of the energy spectrum you should compute the of such forms that I explained before so now let us go back to the formal decomposition which is not truncated so before I considered truncation up to order 2 and now I consider the whole formal decomposition and let us consider the corresponding energy spectrum which I also compute formally from this decomposition and we would like to study the higher order terms of in this series of course we would like to prove the series conversion that this approximates well the solution that the series is asymptotic but let me show that what we can prove so here NS0 is yeah, sorry before I wrote in quite imprecise way I just forgot completely the dependence of s everything which we considered these are fast decaying functions and below by s sharp I will denote various various Schwarz function so what's important here is that they are fast decaying so then as before NS0 is some Schwarz function so NS0 is the same that for quasi solution of course 1 is still 0 NS2 is also the same NS3 and NS4 how it was how it was noted they are slightly different from NS3 and NS4 there but they still satisfy the same estimate by the way about this estimate I didn't tell anything before about this but we do not know to compute asymptotics for them only for NS2 for these guys we can get only this estimate for example for NS4 if you want to do this by hands you have to estimate like about 20 integrals each integral this takes about 2 pages so it's something something terrible and you use the kind of stationary phase method there but this estimate we did use from following result which I will mention in a minute so assuming that we know this it's natural to to suppose that it's natural to to assume that NSK is bounded by new to the power K over 2 look indeed if this is if this is so then when I consider a term number K of my formal series for the energy spectrum rho of course should be the same as before so it's epsilon to the power K over 2 new to the power minus K over 2 and if this estimate is satisfied then we have here also new to the power K over 2 and this means that that terms of our formal series they decay the power of epsilon so the formal series this is a good formal series which is uniform with respect to new but if this estimate is not satisfied if this estimate is not satisfied that means that just the terms of our formal series they explode when new goes to zero and if they explode it's hard to believe that just the series cannot be asymptotic and it's hard to believe that the solution the solution which should be given by some of the series satisfies the prediction of the wave turbulence that is a similar theorem so we indeed need to prove this estimate if in the prediction of the wave turbulence and moreover if this estimate is satisfied it's your question then we can consider any tracation of this series and it will satisfy our theorem because the higher the terms will give input of just smaller order in epsilon trying to prove this estimate we get the following theorem which is our second main result you see we have this fast decay factor in S then we have even slightly better than new to the power K over 2 which is in accordance with this estimate for 3 here because we have this 3 then here you have 2 like here not 3 over half like here but we have maximum with new to the power D it means that we can consider higher order tracation of this series only if D is sufficiently large and D should be larger and larger if you want to consider high and high order tracation of this series here in this estimate we should assume L grows efficiently fast namely this thing is less than the right hand side of this inequality but for us if L goes to infinity fast it's okay it doesn't pose problems so the question is if this estimate is optimal maybe just we didn't manage to maybe it's a problem of our techniques that we didn't manage to prove this estimate so trying to understand if this estimate is optimal or not we prove the following, we get the following result in fact when getting estimate for this NSK the case term of the formal decomposition for the energy spectrum we represent Aegis NSK as a finite sum which is parameterized by certain diagrams which following standard notion standard terminology we call fame diagrams so this is some finite sum and this sum is obtained I mean it isn't something artificial it's very natural way to parameterize to parameterize this sum maybe if I have time I will explain what are these fame diagrams in couple of words some natural representation as a sum of finite number of terms and then each integral this is just given each integral is of f this is some fast oscillating integral and when proving the estimate for this NSK from the previous theorem this one we just prove it for every integral just prove it for every integral but then for K large that is K more than 2D in this case here maximum is attained on nu to the power D so this is a bad case this is a bad case we find for each K some subsets of the sets of all fame diagrams such that the integrals from these subsets they are indeed of the size nu to the power D so they are large this estimate cannot be improved and these integrals they seem to be of really different nature so there is no reason to have some consideration inside we just wanted to check that there is no considerations of course we found considerations so we then we understood that this is a general situation if you take a sum of this set of like big fame diagrams then you obtain that it is bounded by nu to the power K over 2 exactly what we need but in fact we get even even better estimate so this does not contradict to our desired estimate which I recall NSK less or equal than nu to the power K over 2 but it doesn't give positive answer because this set of diagrams is very special we just can treat it by hands for other diagrams we know we believe that also there are large diagrams maybe slightly smaller but still large which do not satisfy the desired estimate but also there are some constellations but we do not know how far do they go I mean so you see we do not know if this this estimate which you want to prove is true or not and it seems due to this theorem that this question is at least non-trivial and these constellations there are some stupid constellations like you have integral which plus plus integral is minus which gives zero they are not pairwise and also these are constellations only in a few several terms of the decomposition of the sum is nu so this isn't just zero so that's what what we know so our next goal of course is to this quasi-solution approximates the real solution and to prove our first theorem for the solution otherwise to find a model for which it is so because the techniques we develop we believe that it can be applied to different models as well let me say a couple of words about the proof of this solution I think that I have like 3 minutes more because we started 3 minutes later so the first thing you should do you should define Feynman diagrams and come in to A what are the Feynman diagrams it's just quite a simple thing when you compute the terms of this series first you find Vs0 this is just Gaussian process which is the solution of the linear equation then you find Vs1 and when you are computing Vs1 you have equation here Vs1 and here you will have 3 times Vs0 because you collect terms which are multiplied by root of the power 1 and so on and so forth and then for example for Vs2 in this term you will have several possibilities here you have Vs1 Vs1 Vs0 another one here you have Vs2 Vs0 Vs0 so it's convenient to represent these different possibilities in the following way Vs2 for example you have several trees one is Vs1 Vs1 Vs0 another one is Vs2 Vs0 Vs0 here you have bar and so on and so forth but when you compute your energy spectrum you should multiply Vs by Vs bar and you take like a product of these trees and then you you should couple these trees via their leaves because their leaves are Gaussian processes and when you take the expectation like you apply weak formula and then you should couple these trees via their leaves these can be defined in a convenient way which we believe is quite is quite non-standard in such a way I mean we add some additional vertex in these trees which corresponds to this delta which doesn't exist in the tree and it turns out that this is a very convenient language and just looking attentively at these trees you can find coordinates in which the integrals ESF take just explicit and nice form because in general if you work in initial coordinates you have some terrible summations which are subject to number of linear relations due to these deltas but looking at the diagrams you can find the coordinates in which they take such form and the point is that just this fast oscillating this fast frequency are in L which is in fact time and quadratic in Z Z this is the SI where we changed coordinates and it turns out that these integrals can be analyzed by a kind of like method which is inspired by the stationary phase stationary phase method the difficulty is that the phase function here degenerates and this is bad for the stationary phase method but nevertheless we get just for the integrals for these integrals in some general settings this estimate so thank you I think that's all I wanted to tell