 So my talk will be about joint work with Massimiliano Berti, so containing almost global solutions for the periodic gravity-capilarity equation. So we consider a fluid, say water, which lies between a flat bottom, given by this line, and a free surface, which is a graph of some function y equals eta of t and of x. So this function eta will be periodic, and we assume that the fluid is incompressible and irrotational, so that in the water the velocity of the fluid will be given by the gradient of some potential phi, and phi will solve the Laplace equation, Laplacian phi equals 0. And we assume a boundary condition at the bottom, which is that the normal derivative of phi should vanish, and so if we are given the value of this potential on the free surface, we may, solving this equation, deduce the value of phi and solve the velocity everywhere. So the motion of the fluid is determined by the function eta and by the restriction psi of capital phi to the free surface. And so if you write the early equations inside the water, and if you write a convenient boundary condition at the free surface, you get an evolution equation for eta and psi, which is the so-called Craig-Sulem-Zakar formulation of the gravity-capilarity wave equations that I described now. So the first equation in that system is dt eta equals g of eta psi, where g of eta is the Dirichlet-Nelmann operator, which is defined the following way. So given a function psi, you solve the Dirichlet problem with boundary value psi, and so you find a potential capital phi, and then you compute a normal derivative of that potential, dy phi minus dx eta dx phi, at the free surface y equals eta of tn of x. And this defines g of eta psi, which is just the usual Dirichlet-Nelmann operator. So the second equation says that dt psi is given by some complicated expression, say minus g eta minus one-half of dx psi to the square plus g of eta acting on psi plus x psi dx eta to the square divided by twice one plus dx eta to the square plus kappa h of eta, where h of eta is some explicit function that's the Dirichlet derivative of eta prime over the square root of one plus eta prime squared, eta prime denoting just dx eta. And so we have two parameters in that equation, namely g, which is the gravity, and kappa, which is the surface tension. So we shall consider these equations when x belongs to t1, that is, we consider these equations with functions which are periodic in x, and we want to solve that system for long times. So let me give some non-results. So I shall not read all the names which are on this slide. I will just say that concerning local existence theory for the preceding system with smooth enough data, the most basic work was due to CJU was proved that when x belongs to r, or r2 instead of t1, when kappa, the surface tension is zero, and when the fluid has infinite depths, that local solutions exist over some time interval. So then there have been quite a lot of other works treating different other cases. The only one I shall mention explicitly is the one of Reuser, which obtained local existence in the periodic framework for finite depths and for positive kappa, that is, the problem we're interested in. And then there have been quite a lot of works devoted to long-time existence when you consider small and decaying data. So in that case, you may exploit the dispersive properties of the equation so that the solutions of the linearized equation will decay when time goes to infinity. And this allows you to prove that even more over the data of small and smooth, then you may get a solution defined over a long-time interval and, even in some cases, a global solution. So I shall not read all these names or contributors. I will just say that there are nevertheless also some results which have been obtained when you consider a periodic, not necessarily periodic, but data which do not decay at infinity, by Ephraim and Tataru and also by Alasarburg and Zuley. And, well, it has been shown in different cases. Let's say, for instance, for a periodic Cauchy data, you have a so-called cubic lifespan. That is, if you take Cauchy data of size epsilon, you get solutions defined over an interval of time of length epsilon to the minus 2, while local existence theory provides a solution defined over an interval of time of length epsilon to the minus 1. And so what we want to do is to try, under a convenient assumption, to do better than getting this cubic lifespan. Well, you should ask this question to Daniel and Mihail Ephraim. I guess that this is because quadratic, this corresponds to quadratic nonlinearity. And then, when you have local existence theory, you get 1 over epsilon. If you have a cubic nonlinearity, you get 1 over epsilon to the square. But, of course, here, the nonlinearity is quadratic and, nevertheless, you get this cubic lifespan. So that's why the origin, I guess, of the terminology. So, let me give the assumptions we shall be working with. Let me introduce some operator, just this matrix, S, which is a matrix 1, 0, 0, minus 1. And let me say that a solution, eta psi of our capillarity-gravity system, is reversible if and only if, when we compute the value of the solution at time minus t, it is equal to the matrix S acting on the solution computed at time t. So, of course, it follows immediately from this equality that if this property holds, then the second component psi has to vanish at time t equals 0. And conversely, if you start from a Cauchy data with the second component vanishes at time 0, then your solution will be reversible that it will satisfy this equality. And this just follows from the fact that your equation, maybe written as, say, eta dot psi dot, equals some function f of eta psi, where f has the property that making at this matrix S at the left on f of eta psi, you get something which is equal to minus f of s of eta psi. And so you deduce immediately from that property that under that assumption, the solution is reversible. And so to state the main theorem, I will use this property of reversibility and I will use also a couple of notation that I introduced now. So we shall look for the first component, eta of our solution in a Sobolev space, h s plus 1 over 4 of t1. So periodic functions which are Sobolev on each period. We shall assume moreover that these functions are even and we shall assume that the average of eta is 0. So if these properties are satisfied at the initial time, then they hold for any positive time as a consequence of the structure of the equation. So concerning the second component of eta psi, psi we shall take it in a Sobolev space, h s minus 1 over 4, of even functions on the torus on t1 and actually I'm taking an homogeneous Sobolev space that is a Sobolev space modular constants. So we do that because the system is actually well defined on functions modular constants. So we may consider the signal equation of the system, the one on psi as projected on functions modular constants and it's natural to do so because actually only the gradient of psi if you want to add the physical meaning, not psi itself. And now that I have introduced these notations and definitions, I may state the main theorem. So the theorem says that you may find the zero measure subset, capital N, of 0 plus infinity to the square and for any couple, g kappa of gravity and surface tension taken outside this subset of zero measure for any given integer capital N there is some index of gravity as zero, large enough. Such that if you take any s larger than as zero well you may find some positive constants c epsilon zero such that for any epsilon small than epsilon zero for any function eta zero in the space that I introduced above with zero average and even with norm in that space small than epsilon if you solve your capital gravity system with cos data at equal zero given by eta zero and zero then you get a unique solution continuous on some interval with values in the natural subspace where the length of the interval of existence is bounded from below by c epsilon to the minus n. So in other words the theorem says that you may get almost global solutions that these solutions define on an interval of length epsilon to the minus n for any given a priori n if you take cos data which are small enough which are smooth enough which are reversible in that sense in the sense that the second component should be zero at equal zero and if moreover you have taken your parameters outside some exceptional subset of zero measure so before giving some elements of proof let me make a remark which is that actually you may reduce to the case when g equals one just using the homogeneity of the equation and then the parameter that you take outside a subset of zero measure is just the parameter kappa and let me say also that this theorem provides almost global solutions for the Cauchy problem under the assumptions that I indicated we do not know if one can obtain global solutions for the Cauchy problem or e to the one over epsilon sorry or even e to the one over epsilon we do not know that but there are non-trivial examples of global solutions that is Cauchy time-periodic solutions which have been constructed by Bertie and Montalto and which will be the subject of the talk of Massimiliano tomorrow the initial data is special because we want to have these reversible solutions so this corresponds to functions which satisfy the properties that I wrote before so we need the psi times t equals okay no, we need these assumptions and I shall explain why later well, you know I'm not the kind of person who's making conjectures on the air so I will not no, psi no, psi physically is the restriction of the potential to the restriction of the potential to the free surface so if it's 0 then there's 0 initial velocities you raise the surface and then you let it go but with no okay so now we shall in a first step try to forget the explicit expression of the equation and reduce ourselves to some parallel differential formulation first of all let me fix some notations so if a is a symbol so say a function of x and psi such that d psi beta a is bounded from above by psi power m minus beta for a given m and any beta when associated to this a an operator or a parallel operator in the following way one takes chi some cutoff function with more enough support equal to 1 close to 0 and one defines from the function a a new function a chi in the following way one takes the free transform of a relatively to the variable this gives a hat of x hat psi one multiplies by the cutoff chi of x hat of x i and takes inverse free transform and then in that way you define the Bonivet quantization of the function a as the operator associating to some test function u a new function given by this integral 1 over 2 pi integral e to the ix minus a chi of x over 2 u of y d y x i in that way one gets an operator which is bounded from hs to hs minus m for ns if you assume that your symbol a is periodic in x your operator will send a periodic function of hs of t1 to functions of hs minus m of t1 and then Then using the celebrated bonus parallelization formula, you may rewrite the equation we started from as a parallel differential equation. That is, you may write the initial equation and that's the following form. You have capital DT, where capital DT will be 1 over i d over DT, minus the operator associated to a matrix of symbols, capital A. The symbols depend on eta psi because we treat an ordinary equation and all that operator acting on the unknown eta psi should be equal to some smoothing operator, that is an operator that will gain rho derivatives for rho given a large number that we may choose if we work in spaces of smoothing functions acting on eta psi. So usually when you do that from a nice quasi-linear republic equation, you end up with a system for which you can get easily some energy estimates and then eventually a local existence theory. But in the case of the water wave equation, it is well known that when you do that you get into trouble because the matrix of symbols A that you get has eigenvalues whose imaginary port goes to infinity when psi goes to infinity. In other words, you have instabilities which prevent you from getting energy estimates without derivative losses from such a formulation when you apply this polarization method to a water wave equation. So this problem has been solved years ago, first of all by Cidre Wu who designed some Lagrangian formulation of the equations that a law wants to overcome the difficulty. Can I say something about that, that Nanymov invented that? Yeah, but it was for small data. It's the same formulation? Yeah, okay. It's the same calculation. So then there has been another approach that has been used relying on the so-called good and known of Alignac by Alazard Metivier, Alazard Bechvili, Lan. And this is the method which I'll adopt. And then, more recently, Anterefrim and Tataru proposed a third approach using some holomorphic coordinates. And so we shall use, as I said, this good and known approach which consists in saying that these problems that I was mentioning comes from the fact that we didn't use the good and known to express a problem, and that instead of writing the equation on eta psi, as I did, one should introduce instead omega, which is psi minus, well, some operator acting on eta, where B, the symbol of this operator, is just a function, an explicit function, g of eta psi plus d psi d x eta divided by 1 plus d x eta to the square. And it turns out that if you write the system in these unknowns, then things are much better. So more precisely, if we start from eta psi, the solution of our equation defined on some time interval, let me introduce a complex unknown the following way, called lambda kappa of d, the operator d, the appropriate tangent of d divided by 1 plus kappa d square to the 1 over 4, this is the operator of order minus 1 over 4. And defined from the good and known omega and from eta, a complex valid function, u, which is just lambda kappa of d omega plus i lambda kappa of d minus 1 eta. Next, introduce capital U, the vector u, u bar, and write your equation under a parallelized form on capital U. So when you do that, you obtain the following. Well, capital DT minus the operator with a new matrix of symbols, capital A of u, d x psi acting on u, equals some smoothing operator, our view, acting on u. And now what is important is the information we have on the structure of that matrix A. So A is a matrix of symbols of order three-halves. So the main contribution in terms of the order is given by some Fourier multiplier, m kappa of psi, where m kappa of psi is given explicitly by psi, the appropriate tangent of psi, power one-half, times 1 plus kappa psi square to the one-half. So this is a symbol of order three-halves, which is multiplied by 1 plus zeta of u tx, zeta being some real valued function, multiplied by this matrix, 1, 0, 0, minus 1. And you have also another contribution of order three-halves, m kappa zeta, multiplied by a matrix, 0, minus 1, 1, 0. Then you have also contributions of positive but lower order, namely lambda one-half of u times some matrix, and lambda one of u times the identity matrix. These two symbols of order one-half and one-respectively, so they are of positive order, but it turns out that their imaginary port is of order j minus one, so of negative order, well, or at least non-positive order for lambda one. And then you have contributions of non-positive order given by symbols lambda minus one-half, over the minus one-half, lambda zero, over the zero. And what is important here is that when you compute the eigenvalues of that new matrix, you get that the imaginary port of these eigenvalues is bounded when psi goes to infinity. Which was the property which was not true in the initial formulation. And this property comes from the fact that using this good unknown, you make appear a symbol lambda one-half here, which has an imaginary part of negative order. While you are doing that starting from the initial unknown psi and eta instead of omega and eta, you would have got a similar expression, but the imaginary port of this symbol would have been of order still one-half. And this is what would be responsible of the fact that the eigenvalues of the matrix would have had an imaginary port also of order one-half that may go to infinity when psi goes to infinity. And this is the point that creates instability in the initial formulation. So now that we have this form of the equation, let me also introduce three properties satisfied by the matrix of symbols A that will play an essential role. So recall that I define reversibility using some linear operator S. And now I will call S the matrix minus zero one one zero. That's just a translation of the same operator, but on the new basis we are using in the complex formulation of our equation. And the matrix of symbols that you get for the expression of your system satisfies three properties. So the first one says that if you compute these matrix of symbols A at u, t, x minus psi and conjugate it, you get minus S A of u, t, x, psi, S. So this property just means that when you consider the associated operator, make it at on v and take the conjugate, what you obtain is minus S is the operator with symbol A acting on S acting on v bar. And this property is a property that just reflects the fact that we are working with a system on capital U, which is u bar, whereas the second equation of the system is obtained from the first one by conjugation. So this is just a property coming from the fact that we started from an equation from a system with real valued functions. So second property satisfied by the matrix A is that A of u, t minus x minus psi is equal to A of u, t, x, psi. And this means that the associated operator in particular preserves even functions, which is a property we need since we shall be working with even functions. And finally, the third property is the so-called reversibility, which says that if you compute the matrix A at SU, t, x, psi, it is equal to minus S A of u, t, x, psi, S. So the meaning of this property is as follows. If you consider the right-hand side of our system, which was op A of u acting on u, and if you apply to it S, this is equal to minus the operator with symbol A of SU acting on SU. And this just reflects the fact that the system is reversible. The property that I used before when I wrote that S of some function f of vtab psi was equal to minus f of S of vtab psi. So that's the third property we shall be using to obtain the result of long-time existence. And now let me give some ideas on the proof. So the first step is a sequence of reductions that have been used in other contexts by Alzar Baldi, by Betty Montalto. So I will not give any detail about these reductions. I will just say that making a diagonalization of the symbol of your operator. And then making some change of variable, or actually some paracomposition, and making conjugations by convenient free integrals operators. You may reduce the systems that I wrote before to a system which has constant coefficients. So in other words, you may introduce some new and known capital V. That may be expressed explicitly from U. That will be of the same size of U in HS when both terms are small enough. And such that these new and known will solve a new equation that may be written as follows. So you have capital DT minus m cap of dx. The free multiplier that I introduced before, which is dx, a public tangent of dx to the one-half, 1 plus cap of square dx square to the one-half, times some function 1 plus zeta underline of U and of T, which is only a function of T. So this operator is a constant coefficient operator at fixed time. So this multiplies the diagonal matrix 1, 0, 0 minus 1. And then you have another operator, capital H of U, T, dx, where capital H is also a diagonal matrix of constant coefficient, two different operators, which are over the 1. And what is important is that the imaginary part of the symbol H is over the 0, while H itself is over the 1. And finally, this operator H satisfies the properties, the three properties that I introduced before, the reality, parity preservation, and reversibility conditions. And in the right-hand side of the equation, you have our view acting on V, where R is again some smoothing operator that satisfies also these three properties. So in other words, we have reduced ourselves to a new equation, where in the left-hand side you have just constant coefficient operators. And so it's very easy now to get energy estimates on that equation. Well, actually, when you make an energy estimate on such an equation, the real part of H gives a self-adjoint operator, which disappears in energy estimates. And so what you are left with is just a contribution coming from the imaginary part of H, which is an operator over the 0, so an operator which is bounded on V2. And moreover, since we have constant coefficients, we may commute as many derivatives as we want, and so we can write as well L2 or sub-Ref energy estimates. So in other words, if you write on the preceding equation a very basic energy estimate, you get that the sub-Ref norm of the solution at time t is bounded by the sub-Ref norm at time 0, plus constant, the integral from 0 to t of the operator imaginary part of H, of u tau tx acting on V at time tau in Hs d tau. So if we knew in addition that this symbol imaginary part of H not only is over the 0, but that it vanishes like norm of u in Hs to the n when u goes to 0, then the proof will be finished. Because in this case, we could write the preceding inequality, saying that the sub-Ref norm of V at time t is bounded by the sub-Ref norm of V at time 0, plus an integral from 0 to t of the norm of u at time tau in Hs power n, norm of V at time tau in Hs d tau. And since we work with cosidata of size epsilon, and since we aim at propagating the fact that the solution stays of such a size over some time interval, morally in this integral, this norm of u to the n is of size epsilon to the n, so that some elementary bootstrap argument will show you that if you consider this inequality only four times smaller than c divided by epsilon to the n for small positive c, then you will be able to prove an a priori bound, saying that the left-hand side is more than k times epsilon, where k is some fixed constant. And so this a priori bound will tell you that the solution may be extended up to such a time. But of course, to do so, we would need to have this property that the imaginary part of the vanishes, like a large power view, when u goes to 0. And this is not true, in H, the vanishes like u power 1 when u goes to 0. So we have to do a first step in order to arrive to such a property. And so this step will be a normal forms method. So to explain this, let me call b of capital U tx i some diagonal matrix of symbols over 0 to be determined. And let me look for again a new variable, a new known v tilde, given by the exponential of b of u tx acting on v. So in other words, I'm starting from the equation that was on the preceding slide, namely dt minus m cap of dx times 1 plus zeta underline 1 0 0 minus 1 plus h of u t dx acting on v equals some smoothing operator acting on v. And what I'm doing is just that I intertwine this operator by this exponential. So since I have constant coefficients operators, this exponential of b of u tx commutes with all these operators. And so when I'm making the conjugation of this equation by this exponential, the only new contribution that will appear will come from the conjugation of dt with xb, that is I will get another term dtb of u tx. So that the new unknown v tilde satisfies an equation which is capital dt minus dtb of u tx minus m cap of dx 1 plus zeta underline 1 0 0 minus 1 plus h of u tx v tilde equals r of u v tilde. And now what we have to do since I recall that what we want to do is to get rid of the contributions to the imaginary part of h that are vanishing when u goes to 0 at the order lower than some u to the n for a given n. So in other words what we want to do is to choose capital B such that this term will cancel the contributions to coming from mh that do not vanish at load order when u goes to 0. So why do you worry only about canceling the contribution from h and not r? Oh no of course I have also to do the same thing for r later, of course of course, but here I'm you know the method is pretty much the same in both cases and so I explain it on h. So I have to choose B so that dtb minus i imaginary part of h of u tx sorry should vanish like u to the n when u goes to 0. So to construct B I'm looking for it as a sum of expressions Bp of u u u which are homogeneous of degree p, p going from 1 up to n minus 1 the last degree that I want to cancel out each Bp being so p linear map that restricts to the diagonal u equals u u1 equals u2 etc. So now let me compute dtb. When I compute dtb the t derivative will act successively on each argument of Bp and I will replace the corresponding dtu using the equation which was that dtu is sum for a multiplier m cap of dx times sum matrix k which was 1 0 0 minus 1 acting on capital U plus some non-linear terms which will generate contributions of higher degree of homogeneity. So if I write what I get at the level of the contributions homogenous of degree p so I will get Bp in which I have replaced one of the arguments by the linear part of the equation this j corresponding to the place of this argument going from 1 to p. The non-linearities will give terms of higher degree of homogeneity that I may forget at that level and I want to choose Bp in order to get rid of the contribution to mh homogenous of degree p that I may write as sum in the report of hp of u u dx i and now this is essentially the equation to which I reduce myself to finish the proof. So I wrote again that equation at the top of this slide and to solve it let me decompose each argument capital U as the sum of pi and the sum in n of pi n plus capital U plus pi n minus capital U where what I denoted by pi n is just the spectral projector associated to the nth eigen mode of the Laplacian on the circle and where since capital U is a two vector the plus and the minus denote projection on the first or the second component on that vector. So if I replace each argument U by such a decomposition and if I look at what happens when I make act this operator on for instance pi n plus capital U one sees immediately that the result of this operation acting on such a localized function is given by multiplication of the function by the symbol m kappa computed at n. So in other words when we write the preceding equation replacing the arguments by pi n 1 plus U, pi n l plus U, pi n l plus 1 minus U up to pi n p minus U when you make act this operator on one of the first l terms you just get multiplication by m kappa of nj with a plus sign and of course this multiplication by a function may move out of VP by denierity. In the same way for the last terms when you make act such an operator on the last terms you get multiplication by some m kappa of nj with a minus sign. So that finally when you compute this left hand side where you replace each argument by a spectrally localized one what you obtain is just BP computed at these arguments multiplied by some function dl which is just given by this sum m kappa of nj from 1 to l minus the same sum from l plus 1 to p. So our problem is to determine BP to in order that the equation be satisfied and so we have just to divide by dl and so we have to know that dl is not zero. So clearly there is a case when dl will be zero whatever you do which is a case when you have the same number of terms in both terms and when you have a two by two cancellations between one term in the first term and one term in the second one this may happen. In this case dl will be identically zero and so I have to distinguish two cases the first one being the case when such a scenario does not happen. So let me assume that I am not in the preceding case that is in the case when p is even l is p over 2 and the set n1 and l coincides with nl plus 1np since in this case dl the function dl vanishes identically. So the lemma says that if you are not in this case then if the parameter kappa surface tension is taken outside a convenient subset of zero measure then you may ensure a small device is estimate saying that not only dl does not vanish but moreover dl is bounded from below by the constant times n1 plus it represents p power minus some integer n0. So in other words when we are not in the forbidden case we shall be able to divide by dl that is we shall be able to to solve the equation we were interested in dl times bl equals i in the report of hl since dl does not vanish. So of course when we make such a division we lose some large power over n1 plus np which means that we lose some derivatives on the coefficients of our equation but this does not matter because these the coefficients of the equation are low frequencies that is we may afford to lose a large number of derivatives on these coefficients if of course we work with smooth enough functions. Okay so this solves the equation we wanted to study in the case which except in the case p even a equals p over 2 and n1l l equals nl plus 1l. As we have seen in that case this dl will vanish identically whatever the value of the parameter kappa and so to solve the equation in that last case we use the three properties that I was mentioning reality parity preservation and reversibility because combining these three properties one may check that the right hand side of the equation in hl well mhp in this case computed at p1 plus u up to pp minus u with n1 and p satisfying this equality vanishes identically so that in the remaining case the equation to be solved was just 0 equals 0 and we know how to solve such an equation and to consequently we have been able to solve in all cases this equation and we have seen before that solving this equation was a low end one to eliminate the contributions to mhl which have a low degree of homogeneity and that doing that we are able to get by energy inequality our long-time existence result over an interval of time of length epsilon to minus n and being the level at which we stop this process of elimination so this concludes so the proof of the theorem and this concludes also my talk which is fortunate because my time is essentially over are there some questions that's a very interesting cancellation in this in this reversible and parity setting so in the general case that is allowing cosigned insides that I mean because those those somehow from my Hamiltonian point of view those terms are very important and those are ones which which really are important to the dynamics and so I yeah and so I imagine that yeah they they would cause the action frequency uh uh non degeneracy so I'm just uh wondering what happens in the general case so you know in the general case you know what what one is able to do in other settings and what one would you would like to be able to do here would be to use the fact that the equation is Hamiltonian and then the terms that you cannot eliminate eliminate the term for which these dial vanishes correspond to terms depending only on the actions so these terms and so these terms they cannot grow they have so they have known do not grow because they depend just on the actions and so these are terms that you would need to eliminate in a Birkhoff normal form using the Hamiltonian character of the equation and so why didn't we do use this approach well this is because of the good unknown that I introduced before when you pass from the initial unknown to the good unknown the reversibility condition is trivially preserved but the point is that the passage from the good unknown to the new and from the old unknown to the new unknown uh is not something that is Hamiltonian or at least we do not know to do that in an Hamiltonian way if you we could say design some Hamiltonian way of doing that step then I'm pretty confident that we could write the rule proof in the Hamiltonian framework and so get rid of these conditions of parity and and reversibility but that doesn't work in this case in any case right that would not work in this case you have to go to another yeah yeah another point of view maybe I have a short question concerning this property the one you spoke so uh uh usually the papers I know the parameter is in the zero order term of the equation of the symbol so here your kappa is in the main part of the symbol right yeah so it's something which I never saw it it makes really a difference in the analysis or actually well you don't mention much about this immigration dl how you prove it also no so this is this follows from essentially from some old result with jeremie which was applying to uh um say to some general functions subalytic functions satisfying some conditions so the fact that the so the fact that it depends this m kappa which was a written here so the fact that this uh you know the fact that the kappa here is in front of a derivative of i order does not matter at all in the proof of this this lower bound you what what what what is important is the fact that when you consider this function as a function of xi the dependence on kappa is essentially a non-trivial dependence and this is what allows you to to to obtain this small device also estimate at least to to say that when you evaluate this or actually what I call dl only at integers then moving a little bit kappa you may arrange so that these uh lower bounds are satisfied so the fact that kappa is here or here does not does not matter at all actually you could you know factor out kappa and get a parameter here and you wouldn't change anything in terms of the method of proof can you make a comparison with the result of of the tarot and which is a norm of epsilon so well so the basic the basic point you know the basic stocking point is the same that you want to get rid of terms of lower degree in the nonlinearity so to get their cubic result you have to get rid of the quadratic term of the nonlinearity oh no no in the in that case there is no parameter but this is because this is for the yeah but the point is that they don't have parameter because it's at the level of the quadratic nonlinearity for which you will have essentially no resonance no and they don't need disparity conditions also they don't need disparity conditions also they don't need disparity condition but disparity condition actually we need it only you know to to deal with the terms the bad terms the ones for which you your you cannot make nonzero your function dl and so you needed only to get rid of terms of odd degree in your nonlinearity you needed to get rid of a cubic term you do not need it you do not need it to get rid of a quadratic term or a quartic term okay so up to up to those cubic terms you would be the proof would be the same yeah yeah except that it's not exactly the same framework because yeah so you would not you would you be able to recover that result in other words without the generic condition well i i suppose we could but we we didn't try to to write it down we we we want you know this general thing