 OK, da ima Tako pacije ta tudi se fan Miešler v Parizdofim, pa razvalo o Vileniji programu p.2. I sem dialo nek prejdi, da je p.1. Bila tudi povistano je, kako se proseli, tako p.1 no crčen, bi se povržel, da všech nekro s nimi. OK, OK... in zelo sem dava. Zvuk sem, da se pravite tudi o kanal imagy za Vilaniju za vse košnjenjeankov prisivovini, izvajna za bolim smanjo pošljenje, The one in the other, which is constructive. Then, I focus today on the part two of the program, which is the hypocrisy that he estimates. I will give an introduction of the general problem. And then I will present some way to obtain this in ippok tryinje. Prvo in h1 kajdov, in zcah v l2 Daj zdaj, ovo je program Stajte pri svej pri vanči noče in po kursu ne zgledal ihp Addition had past its end point and next to point. The first one was to find some conjugate method to get some caursivity estimate on the collision term, which appeared in the Boltzmann equation. značiื่ se na šteki letters. Sve prikoncentrajo, da je tukaj najbolj v stavnih vodnih sez, pravno ni nama cedrafi, iz kako totalnih. Na nej ta del je, da bi se najbolj službi informacija informacije informacije za vseh boljstvih operatorov, kaj ima boljstvih veloziti in pozitivnih. In kaj je zelo, da je zelo vzvečen na vzivne poživne, zato je zelo tako zelo, da se počkaj našlično zelo, nekaj je soluzija v poživnih, da boljska jezina vzvečnega jezina vzvečnega. In da vse jazem, da pošli, da pa se zelo odličili, zelo je več nekaj odličilj vsega postavno vsega. V način se, da se nekoče pošli, da je inšel vsega postavna. Here and here, we work in one space for which we are able to take advantage of the parameters of the operator. And then we want to write get some information in some larger or smaller or different space, for which we can in ki so lahko nekaj potrebeni. Tako, pa je tvoje liniarje, treba, koreasivitje v jedno spas, ipokoreasivitje v jedno spas, izgleda informacije na linijeru bolsmanj operator in v mnohj spasih in v svoj spasih, kaj je začetil na ne linijeru izgleda. Ok. Ok, so in fak, there is many work on the existence of solution in a near to the equilibrium framework and to the trend to the equilibrium of the solution. So the first one is by and then there is many people who work on after Vilani contribution, which has nothing to do in fak with the linear result equation. There is mainly two school who worked on this, one with Yan Guo and other people from Chinese and China and Japan and one by Klemon Moore and many other collaborators. Ok. Ok, so a word about the Boltzmann equation. So it is an equation which write on the density of particle capital F equal depends of time, position and velocity. Ok, so here I have written here in the case of bounded domain, the torrets are bounded domain. So there is a transport part and then a collision part. So I don't want to write here what is the expression of the Boltzmann operator, the nonlinear and the linear result. You don't will, let's see this here. So don't be afraid. Ok, and then depending of what you are interested in you can consider the case of the torrets for the position, the case of a bounded domain with boundary condition or the case of the whole space with the force field confinement. Ok, so and then you have to add some term here on the transport part. So the only thing you have to know is that the quadratic term sorry, here is for the quadratic Boltzmann and Landau operator conserves mass, momentum and energy. So what does that mean? That means that if I multiply the operator by phi with phi the one v or v square you integrate, you find zero whatever is f and then you obtain some consideration for the solution of the Boltzmann equation. That's the first important information, piece of information and the second one is the a theorem which say that if you multiply by log of f and integrate this is non-positive and it is zero if and only if f is a maximum. So in the case of the torrets is this and then you expect to have the convergence. So now the problem is to get some information on the rate of convergence. Ok, so yeah, I wrote I am writing a result which is in some sense the final step of all the program. Ok, all the program aim to prove something like that. So for some wave function in v, maybe in x also depending of the geometry and ok, and for some space this is a weight so let's call a big space if the initial data is close enough to the equilibrium in some space then in this space then there is exist a unique solution which exist globally in time and which converge to the equilibrium with the rate so I explained to you fast for the Boltzmann equation for art spheres and in the torrets for instance but can be also polynomial in time for other model. Ok, so in fact with this in hand you can improve a result by which is the following it's a if theorem so you consider the in original Boltzmann equation for art spheres for instance in the torrets and you assume so that's something that we don't know how to prove that the solution is bounded in some strong spaces so h, so wave space control of the moment of the space of f in l1 then for the relative entropy and this is a way to to measure the distance between f and m in l1 you have a decay which is a polynomial and that can be proved by the entropy method ok and then so when you are so you converse to to f converse to m and then when you are close enough to m you can use the linear argument ok but you have to be careful with the fact that the e here is large enough to be compatible with the space you get here ok and then you can improve and you get that kind of conversion with the exponential and the rate here is the same as what you get by the linear is at the argument ok so now the first step in the vanilla program is to get some corrosivity estimates what does that mean you consider the space homogeneous ok so really the collisional operator and you linearize it so you introduce this linear operator you can see because of the invariant of the quadratic operator that in fact the new space is this one the same invariant the same mass momentum and energy invariant ok and then the goal and what is possible to prove is that in this space here the operator is symmetric is bounded by some norm which can be different it depends really strongly of the operator you consider and this is positive the quadratic form or the directly form is positive and in the sense that it is lower bounded by the projection on the orthogonal to the new space controlled in the same norm as here ok ok so ok so that's the normative estimate for the Boltzmann and the Landow equation and then for these two models we know exactly we have exactly some quantitative estimate for this so now I want to consider the whole operator so we consider f equal s operator plus t which is typically the transport operator so it can be just the free transport or the plus a confinement so t in general is minus v variant x plus maybe term like that I will come back to this later so what's happened is that one is built space we can prove that s is self-adjoint negative and t is squeeze symmetric ok and then this is the information you get just by the co-racivity estimates and if you write the directly form for the same form for the whole operator both terms because t is squeeze symmetric, it disappears and you get this and this is not enough in order to control the long time behavior of the dynamic of the associated semi group so the idea of the co-racivity approach is to twist to modify the norm so you have the initial norm for which the two operator are nice and you modified it with scalar product of AFBF ok in such a way that in this new scalar product the associate directly form is co-racitative ok so that's the idea so at least when ok when h star is included in h you have strong confinement so for instance in the torres you can prove in some situation that this directly form is bounded below by the norm the twisted norm the new norm to the square so if you consider the semi group for initial letter f just just disinformation like this so this side not ft ok you find 2L ft so for the new scalar product this is less 2 minus say lambda ft square so at least the f such that you have this invariance the global invariance so here you assume that the initial datum such that so that the global invariance and that prove that imply that the same all for any time ok and then so both disinformation give the first estimate here integrating it ok and then when you come back to the initial norm you get the same decay but with a constant here which is principle larger than 1 and that's really the hypocrisy estimate ok so in fact there is many works on this so in the 80s for instance there is work by Kawashima in fact the term ipo ipo koresiviti appears in the in 2000 ok it's by say Vilani and ok and then what I want to show you in fact is that there is many approach in fact energy estimate by go micro macro approach by the Japanese, Chinese and American school and in fact all this approach can be rewritten in the framework of ipo koresiviti as I written here so just twisted the norm in order to get koresiviti for the directly form so here are problems that I don't want to talk today and then in fact we can split the general issue in several pieces so we can so the first point is the geometry of the domain so we can consider the theories is simpler the wall space with confinement force a bounded domain we can also consider several collisions operator so the Landau and the Boltzmann equation operator are quite complicated but you can also consider the Fokker Planck so which is lapljation of F plus divergence of VF or the relaxation operator which is rho M minus F with rho equal the integral of FdV ok and these two operators for these two operators there is one invariant which is the mass while for the Boltzmann of course there is a in dimension 3 there is a 5 invariant so it's more complicated and then we can as I said before we can try to obtain h1 estimate or twisted h1 estimate that works for the theorists and the Fokker Planck equation in the wall space or l2 estimate which works for instance in the domain in the domain you cannot take h1 norm because if you consider this when you integrate by part with the boundary it's awful so this is for for a domain for instance ok, so maybe so what I can show just now is that these two operators are corrective so it's quite simple we write this in l2 M minus 1 M capital M the Maxwellian so what we get is so I rewrite divergence of the gradient of F plus V F F M minus 1 that's the identity when you consider the Fokker Planck equation operator the standard one in this term you just write it as M the gradient of F divided by M ok, in V so here it's in V F divided by M so if you want to prove that it is positive you put a minus, a minus, a minus you make one integration by part and you get this and you have to know Poincare inequality in the world space ok so that's Poincare and you get this provided that F divided by M multiplied by M is zero so the the local mass is zero ok, so you can do the same for the relaxation operator if I do that I put directly the minus, I have got F minus rho M F M minus 1 and what you do is you write ok, you observe that this is exactly equal to F minus to the square M minus 1 by expanding this ok, so if you expand all this what you get is F to M minus 1 minus 2 rho F plus rho square so when you integrate there is one, so there is M here there is one of these terms which disappear and you just get the same as here so the two models are correct here and ok so now I want to present you the H1 approach so here we can consider an operator L which is S plus T where T is the free transport operator in the third so I write J, because there is three set of unknown you can consider F and for which the macroscopic quantity are this ok macroscopic quantity you can consider D is equal to F divided by M 1 half and then the macroscopic quantity are GM 1 half so here when you consider F you work in L2 M minus 1 when you consider G you work in L2 and sometimes we introduce the change of unknown which is F divided by M this one is in L2 M and the macroscopic quantity are something like that and depending ok it's good to change the variable of unknown in order to understand what you are doing so here we choose the J ok so we work in the flat space L2 ok and then I twist the norm the H1 norm so L2 part gradient of x g part and a mixed term ok and the directly form is just this one so the claim the result is that you can choose the coefficient here positive coefficient eta here eta x eta eta v so that you have lower one in the directly form and a possible choice in this one ok so what you have to understand is why is it true ok and so I just want to make one computation so it's we call d2 is the part which come from so that's d0, d1, d2 and d3 so it's one part of this term and one part of this term infact I will write d2,1 is the part where you just consider not the L but the the t so so in that case the definition you have eta gradient x t g gradient sorry gradient v gradient x g minus eta gradient v g gradient x and there is a t somewhere ok so what you do here is that you observe that in the case of the of the thirds this commute so this is gradient of v gradient x with a minus and this is the same as minus v gradient x x so this commute and then you make you take the agent of t and the agent of t is minus t and you put it on the other side so what's happen that you get eta so I have for instance a t gradient v minus gradient v t gradient x g ok and you observe so you compute this so this is ok this is minus v gradient x so I am bit confused no it's ok v plus gradient v v gradient x ok and here there is two term because of the dependency of this operator in v and they cancel some terms and you get gradient x at the end so that's the commutator ok so this term give you eta gradient x g to the square in l2 ok so that's the idea you want to get this why because then using the Poincare inequality or Poincare visiting inequality in the torus so sorry two steps you write larger than the gradient of x on the mass on the projection ok so here p of g for this operator for both operator here pi of g is just rho m the integral of fdv m ok so here and then you use Poincare inequality Poincare vertinger ok l2 square provided that the mean 0 ok so that's the information you don't get the information is missing when you use the coercivity of s ok so that's the idea it's just ok so now you can so what you have to do is to be careful with the fact that you have extra terms and so when you you take this you twist this norm you will obtain extra terms and then this one has to be small you have to control it so just once you see why the extra terms in that case can be controlled so we make the choice a is gradient of v b is gradient of x so you have in the tourist so you have all this commutator ok relation properties you have a very important information with this that in that case the commutator of t and b is 0 because you are in the tourist we use it there and this information is always true and we assume some thing about the collision operator so it's not how to prove this for this example here so it's when you differentiate in v the linear collision operator you take the scalar project with the gradient vg you can estimate by below two positive terms minus a lower term ok so that's of course it's harder to prove for the Boltzmann equation or the Landl equation but it can be done ok so now what we do, we write the contribution d1, d2, d3, d4 ok so it's just a splitting is d1, d2, d3, d4 yeah so it's not correct here is d3 this is d3, 1 by notation ok and then ok so the first part is just corrective but partially the second part there is ok with the gradient of v there is two bad term and positive term the third part is what we did just before this is exactly the computation we did and there is the collision part and for the last part with the gradient in x you get ok this term plus something which is nice again because b and s commute ok so believe me and then what's happen is that you can remove this term thanks to the two terms here this one and this one when you take eta v small with respect to eta so this term disappear and this term and this term disappear because you are in the tourist case so this is 0 ok so I just rewrite the four terms together and I get this ok these three terms in red you can use košišvati in equality when eta square is smaller than eta x eta v so this term which is maybe negative disappear and you forget this term you forget this quantity and the only bad term is the minus here which come from here ok but with this argument here the Poincaré in equality and you can change the minus by the plus ok and that's the end of the proof you have exactly all the information that you need ok ok so two words about what's happened in the whole space with the confinement first term so we start with the Fokker Planck operator so this one and the transport operator with this term and we assume that there is a confinement this typically x to the power gamma with gamma larger than 1 ok the Fokker Planck so I change variable so it's the h variable and in the h variable the Fokker Planck operator writes like that you observe that for this measure probability measure with the v normalized then one is the normalized positive steady states and the projector in this variable is just the integral of hm so here the idea is exactly the same you consider the same twisted norm and you get the same result so what's happened so the difficulty here is that so I just rewrite the same identity as before just the same for the moment but now the commutator here is not 0 ok and in fact you can compute it it's d2v gradient in xv ok so what you can see is that thanks to the Fokker inequality in the wall space in the x variable this is less so first it is because this is because the gam the v is a polynomial so you have first that this quantity is less than the gradient of v the norm of the gradient gradient v and then because of this inequality you can change the weight here which is bad so not bounded by 1 by a b which is a gradient x ok so here you can estimate these two terms thanks to this but next you have to get some information in order to close the estimates for this operator it is like that ok ok so you do the same as before and what happens is that because you choose the Fokker Planck equation the Fokker Planck operator you have an extra term which is a2h ok and then ok it is the same because of the gain of regularity the gain of one gradient in velocity ok and then the idea is the same the only one thing is that you see that here the choice of eta v eta eta x is reverse with respect to the torus case which is true ok so it was in the reverse sense so eta x less larger than eta larger than eta ok ok so now I want to consider next step I want to consider the relaxation operator in the whole space with the confinement ok so I am in F variable the S F ok the same here relaxation operator and the projection is is the same ok it is P of sorry so here is P of F pi of F ok we are here so the problem here is that of course this operator has no regularity effect in the V variable so the key term in order to control that bad term was to use the a square h to the square hope to get something similar for the relaxation operator so the idea is to consider l2 norm modification of the l2 norm ok so we forget about h1 norm and what we do is we add a term which is rho the gradient in x of Laplaceno minus 1 j where rho is the local mass and j is the local is the flux is the interval of fv so the directly form is this one and again for eta small enough we have the we have the hypocrisy hypocrisivity so the proof now I just want to show you how to manage with the good term, how to get some some information on the projection so it comes from this term so we write it minus eta and I consider the torus to simplify the discussion gradient of x and here I just consider the transport part so it's d2 1 then you compute the integral so I put dx i bi and the integral of this term only make the proof in the case of the torus so without this term just to understand so here we have a minus v sorry, bi j, dx j and there is two term so the f can be written as bi f plus f autogonal so it's rho m plus f autogonal so here I put rho fm plus the gradient of x j tf autogonal and this term here is in fact because m is symmetric the only term which are no 0 is when i equal j so in fact this is dx i dx i rho f ok what you get is and there is a minus here so at the end you have a plus the rho the laplation minus the laplation of rho plus some term which is rho ok like that and this term is nothing but rho to the square l2 ok, so ok, you get the term and all the other terms are in fact can be bounded by eta rho f autogonal or f autogonal the square with the eta so when you take eta small you can kill all these terms and the leader ok, and then the leader term is this one ok, and then you can do the same in the wall space with a confin mentor and the idea exactly the same so believe me, you just repeat the same strategy ok, so I have written here get the element of the proof in the slide later ok, so to end we can ask to what happened for the linearized Boltzmann equation so in the domain so you have to put some condition some boundary condition which are diffusion reflection specular reflection or mix of both so again I will just consider the first because the idea is the same there is extra terms ok and what we do in fact is the same idea so we multiply sorry, we introduce the macroscopic quantity a, b, c so the mass so a is rho, b is j and c is something which is or less the temperature ok, and we project on this variant of the quantity which are in the neural space and you look for twisted L2 norm so the L2 part plus something which is exactly the same if you have only one invariant it's exactly the same as before rho, Laplacent minus 1 the gradient of j so here you have more invariant so you have to play with with more macroscopic quantity so 5 in dimension 3 so alpha goes from 0 which is the mass to 4 which is the temperature the energy the local energy the kinetic energy ok, then ok, then we do exactly the same and we have for eta so eta for convenient choice so of this macroscopic quantity and on eta you have lower bound like that on the new directly form ok, so the choice of phi tile is given here I have no time to explain in detail the key point is such a kind of orthogonality condition property on phi tile so for that phi is just the invariant phi tile is this is this polynomial and you have this kind of of relation and so when you apply it on the key term so which is the d21 as for the the previous case you can compute and at the end you get a positive part or a negative term on the l2 norm of the projection and some term here which disappear because you are in a bounded or in the torus or in a bounded space ok and then of course you have to be careful with the remainder term and so that's ok and then there is some miracle and the remainder term the most important term which makes the the macroscopic part is so that there is no eta4 term and then you can choose the coefficient in such a way that you win ok so this doesn't work in the case of the wall space because of this property which is just stupid for the torus just ok when you integrate by part the two terms are the same with the minus and they disappear so when you do that in the wall space you have the confinement part which give you a contribution in such a way that this is not zero ok so so just a word so there is a kind of there is a way to conclude in the case of the linear Boltzmann equation in fact it only work for the linear Boltzmann operator not for the London when the potential is the harmonic potential is x square and the idea so it's the idea is to mix some h1 estimates with a macroscopic contribution here so it's a bit technical the good thing ok the idea is to put here something such here and here something so that when you look to this term you get an information of on the macroscopic quantity you want to control and b ok so you have also an estimate by below hypocrisy estimate and the interesting thing that the difficult part you see here just before here the difficulty was on the for alpha between 1 and 3 which sorry alpha alpha between 1 and 3 so it's on the flux on the integral of f d ok so that's the point how to control this term and then the nice thing is here so is that the d21 or I don't know the number so the important thing ok there is two things so because you are in the for the harmonic potential there is some magic things which occur which is that this norm with the gradient of x in x and the gradient in v is almost lower bounded but only for g perpendicular and then come from the fact that you get so on one term you get a quantity like that i b j plus x j j b i to the square and then in order to conclude you need a constant lemma which says that you can control thanks to the symmetric part of the gradient which is more or less what you have here the x is this quantity but you are in g variable and then when you come back in h variable you have exactly the gradient the gradient in x so what you need here is that to have a control from this to the full derivative that's the symmetric part of the derivative and then in equality you get an information on u to the square ok so there is no constructive proof by Duane and in fact so that that work for any v and in fact I have no time we we can get constructive estimate here in order to get the proof at the end constructive so to conclude all the the hypo correctivity estimates I know say hypo correctivity convergence maybe I can say like that so for all this operator for the semi group associate semi group in some case we can get say dk so say in the case we have a spectral gap ok, something like that and so there is many case for which we can prove this in all the so the proof can be different than what I present by energy for micro macro decomposition but in fact I believe that we can always come to this kind of proof which is just to work on the degree form in a good ilbert space ok, so that's the contribution, my contribution here in this work is to rewrite the work by Google here and also here by to get really the same framework as the hypo correctivity for the as introduced by Cedric Villani I believe that in fact we can here generalize this with almost constrictive estimate for any v, reason about v so it's a work in progress with Kleber Karapatoso in fact for the good term so the one which give you the information you need the new information the information is missing in the coercivity estimates we get it but when I verify the proof recently in fact for the moment there is some missing from remainder term that I cannot control we have to think again and there is one open problem which is quite simple which is this one so to get constrictive proof for this estimate with a potential so for instance in a domain there is many works it's known so I say for the harmonic term for the harmonic potential I believe we have a proof and in general I don't know so it's quite it's much more simple than the Boltzmann equation you can understand this quite easy and that's an open problem and if you get this then with a constrictive constant here you can go one so