 O tem skupanju ljudi. Ljubiče. O tem, da se je kozalo, da se je kozalo in da se je obštavek. Ljudi, da se je nekozalo. Zato se dva nekaj je boj, kot je ljudi na trg. Kozalo, da se je nekozalo. Tudi, da se najdeš, kako je vsem nekozalo. Tudi, da seje instructor, kako se je zelo spravil in vse vso komponular, kako, v svoj različenju, različenju, kaj je vse vse vse vse vse vse vse kaj je vse vse vse vse vse kaj je vse we were first looking for an entropy, then for a strict entropy, and finally when it's possible to a link between the entropy and the entropy dissipation. So let's do this now for the Lando equation and it's really the heart of the whole series of lecture. So we look for a link between entropy and entropy dissipation for Lando operator. So let me write maybe the main proposition there and it's maybe number 7, I'm not really sure. Let's consider the set E of functions which are functions of V which are non-negative and such that the following normalization is satisfied that is F as an integral equal to 1, so this is defined on Rn. We suppose that the momentum F v dv is equal to 0 and that the energy the kinetic energy which is F v square over 2 is equal to N over 2. So this is just a normalization which enables to write down a theorem which is a little more easy to read than in general case. So taking the F in this set, it's possible to show that the entropy dissipation associated to the Lando equation when the cross-section psi is the cross-section z to the square and we will see how this appear naturally in the entropy dissipation. So this quantity here is always bigger than some constant Cf times the relative feature information associated to F which is this quantity here so this is actually the relative feature information associated to here the logarithmic gradient of the Gaussian normalized. So here one has to understand that the Gaussian the Maxwellian we are looking for in this proposition is related to these constraints here so this is the standard centered normalized Gaussian. So we want to show this and here Cf is a constant which is bigger than an absolute constant which is strictly bigger than zero whenever h of f which is the integral of f log f on Rn is dominated by a given constant h bar so if you prefer Cf is bigger than a constant which depends on h bar which is let's say an upper an upper bound on the possible entropy h of f So this is really what we are looking for that is a lower bound on the dissipation associated to land operator in terms of something that we hope to be able to link with the entropy which is defined here so anticipating a little on the sequel if you remember what we did for the Fokker-Plank equation we already showed thanks to the logarithmic of f in equality of growth that this quantity is in fact bigger than this quantity minus the same quantity taken at point f infinity where f infinity is the centered normalized Maxwellian. So if we are able to show this then we will have the exponential decay to voards equilibrium for the solution of the Landau equation with this cross-section here OK So this corresponds to what we want to show and let me now show you a proof which is not really complete but in which I will really try to point out the part which for my point of view is the most important so let's start from the definition of the Landau entropy dissipation I think it's written here so as you can see if in this formula in the formula that you have at the bottom of the slide you take the function psi to be exactly z gives z to the square you end up with the following so let's do it in dimension n on the slide it's written in dimension 2 but it's the same in any dimension so you have f of v f of w times now this is just v minus w to the square and the operator pi here is just remember this was also in the slides of Irene this is a projector which is defined like this with a notation which I hope is understandable and which is just the projector onto the orthogonal set to the vector v minus w and so this matrix is applied as let's say a quadratic form to vector which is here so if you like this notation let's say this is the matrix taken on the vector and then this is a scalar product with the vector itself now if you remember maybe the first thing to do is to sorry is to use the multiplication here in order to get rid of the part which is here and you see at this level what is the interest of really considering this specific cross-section which is the only one which gives you this simplification here and we will see that it plays an important role in the sequel and this is now let's use let's say the standard equality coming from linear algebra that when you have two vectors you can decompose them I mean you look at the norm the euclidean norm of x to the square you multiply it by y to the square then you can write that this is the square of the scalar product plus the square of the cross product ok so now if you write it in this way you see that this quantity here is exactly the cross product of v minus w with gradient f over f of v minus gradient f over f of w up to up to a constant anyway there is one half here so let me not compute the constant and let's write it with coordinates so this is the sum of our all possible i and j and here I will get so v minus w i djf over f minus djf over f but taken at point v and point w minus the same with indices in the in the following way that is first j and then i so this is to the square and this is dv dw ok so this is just standard algebraic manipulations between the and if you remember this is exactly the quantity which I called qijf of vw so this is exactly the constant times the sum on all possible i and j actually i different from j if not this is equal to zero so this can be rewritten in this way dw and now at this point it is a very good idea to sorry come back to the formula which is written here that is varies a way of transforming qijf in this quantity here because in the relative feature information which is written here so let's just do it let's use the formula which is written here I can write dif over f of v as both this determinant divided by the determinant which is at the bottom which does not depend on v it is a given it is something which depends on f so let's call it df, d like determinant and let's say to the minus one because it is under the other determinant and the determinant above I will not write it again in detail or maybe yes so you have here let's write it after all so we have 1 wiwj we have qij we have qij times wi plus vj minus wj and qij wj minus bi minus wi like this wi, wi2 wi, wj we can write things in this way and then I will try to start from the formula which is here and try to bound it from above but by what I have and what I have is exactly this so I take let's say the absolute value of this quantity and now I bound from above if you look at this determinant you can see that it's made of terms in which you have you pick one of those things in the first column you pick one of those terms in the last column and then you pick one of those terms in the second column and then you do sums of this so the number of sum is not that high it's a 3 by 3 determinant so let it be put in the constant and then you have an integral which corresponds to the first column so something like which will be bounded anyway by this something which will be bounded by the same but with a square so like this the third column and finally something which comes out from the second column and which will be bounded by something like q i j in absolute value times let's say 1 plus w at worst and then one has to be a little careful because of those terms but let's say so we have this plus terms which are at worst like I forgot about the f of w here there is maybe something like f of w times 1 plus v plus w sorry it's a little like this ok and in fact in the determinant many terms of terms like this but now we assumed that f has a mass and an energy which is bounded by absolute constant so this term and this term they are just constant and in this term I can remove everything except the 1 plus v so all of this can be bounded by something like this so this is a constant this is a constant and I end up with just q i j so let me write it in a completely precise way this is something which depends on v and w and I have this 1 plus w I have f of w to take into account and all of this is plus 1 plus v like this I hope it can be understood ok and now what is the link between this quantity and this one as you can see first this is one component of this but then there is a v here which is missing in what I wrote down and actually I will not prove anything about this quantity because it is a little too technical to be shown in such a short time so I will look just at this quantity for the rest and then you see that in order to get what I need I have to take the square to multiply by f and to integrate respect to what I have here so I will take this quantity here I will take the square I will multiply by f and I will integrate in v and then what I hope is that this can be bounded by this quantity here let's try to do it all of this but I will keep the last line the integral of gradient f over f to the square times f this is the quantity here in which I just eliminated the v so this quantity is bounded by so let's call the variable v according to the inequality here this is bounded by df to the minus 1 times the integral over v of the square of the integral of q ij f of v w times 1 plus w times f of w dw like this plus 1 plus v all of this is integrated is taken to the square and is integrated in v against f but I hope that at this level it's clear what has to be done in order to get something like this you see here that you have the square outside of the integral so you want to get it inside but the only way to do that is to do košiš fart so you have to do košiš fart choosing in a good way what you want to put in the square and what you want to put outside so let's do it let's do it quietly first I will say that the square of the sum is less than 2 times the square of this one plus the square of this one and the first term which comes out from the 1 plus v which is just the integral of f 1 plus v square the 2 has been absorbed in the constant and the second term let's absorb the 2 in the constant also is an integral so this is an integral over rn so this is an integral over rn of an integral in which the square of this quantity so like this so I'm glad with that because it's what appears here then I have to put f of v and f of w so I take the square root of f here I put it to the square and I get f here so like this and then in the second one the square has already been used like this and here I used half of the f so I used the square root of the f so I have to put also the f at this level I hope I did not write something wrong at this level let me say that yes you have still to multiply by f of v and integrate in v here so it's like this now this one this is just mass plus energy so it's a constant so let's say one if you put it in the df minus one and then here you also have a constant which is basically the same at this level so you can regroup the integral in v and w and you end up exactly with this term here so this is for a given so this is one point which is maybe interesting so actually I did not do it with the gradient I did it with one of the components and then I used just one j which is taken arbitrary among the other components so now if I want to get it for the whole gradient then for each of the component I have to pick another one and of course all of this is adding up here in the end it's just a finite number and I can use here the whole sum on the two i and j ok so did I really solve the problem at this level let's have a look it remains to show a certain number of things this is exactly up to a constant d of f ok so this is what we computed at this level so what we get is f to the minus 1 plus lambda with the cross section z gives the square of f and what I would like to have is that what I would like is to remove the one here it does not appear in the equation here and instead of gradient f over f I would like to have this one so you can maybe believe me at this level that if you do it carefully putting plus v here this is exactly what you need to remove the one here but it's not really direct I mean I could not just add it in just a few minutes get it in this way ok so I will not show more than this for this inequality then the last part is to control this df to the minus 1 so the point is how do I relate df to the minus 1 which is this determinant that you have at the bottom here and a quantity which has to be strictly positive when f is known to have an entropy which is less than something so this I will not develop in detail but this amounts to say exactly the same thing as I said this morning that is this determinant here it's 0 exactly when 1 wi and wj are linearly dependent according to the measure f of w dw which means that this determinant is 0 when f is concentrated on a 0 measure set which is a hyperplane in R3 and so if you suppose that f has an entropy which is bounded a priori then thanks to the entropy structure of the Boltzmann equation or I mean here it's not necessary let's say the entropy is something which is controlled in L log L and so it controls the way in which f can be concentrated on 0 measure sets so I think it's enough to be convinced that this is working so when all of this is done and especially when once seriously puts the v here and eliminates the one you really get this entropy-entropy dissipation estimate that I was looking for for now for lectures and one ends up with the theorem which is written here and which actually already appeared in a paper that Cédric Villeni and I wrote together at the beginning of the 2000 that is you suppose that f is basically in this set E and you can show that provided that you take the right the right for the computation but not the right for physics cross section here you can show that the Lando entropy dissipation is bigger than this quantity which is exactly the quantity which I showed in this on the blackboard and then one has to remember that thanks to the sobolef the logarithmic sobolef in equality of growth it's possible to go from here to the relative entropy respect to the Maxwellian which here is the centered normalized Gaussian so at the end of the day we have really proven that the entropy dissipation is bigger than the entropy which actually depends only on an upper bound of the entropy and so the consequence of this is that if you now you look at a solution of the Lando equation which is reasonable in the sense that it satisfies rigorously the entropy structure that is there is an entropy there is a dissipation of entropy and the derivative of the entropy entropy dissipation then provided that the initial datum lies in the set which is here you can show that f converges towards its equilibrium the center of normalized Maxwellian exponentially fast with constants here which can be in principle computed the first one is just let's say the entropy minus the entropy at time t at time zero minus the entropy of the equilibrium up to a constant which is related to the Shechar-Hullback inequality but I will not say much more here and the constant which is here is actually related to the initial entropy because of the constant here which depends on the initial entropy but it's possible let's say in principle explicitly compute those constants if I give you an initial datum in principle you could be you could give me the constant which appear here so that was a theorem which was proven in this paper of 2000 actually I wanted just to say one word about the fact that thanks to a very clever link between Boltzmann entropy dissipation especially for this cross-section Toskani and Villani and then Villani alone a little later were able to prove the same result basically for the Boltzmann equation in a special case which is sometimes called super heart spheres and so this is also a brick in this extra result so those results usually are called Cherchinianis conjecture in the literature because they were first closed by Cherchiniani in the 80s so everything is perfect when you take the cross-section z give z square which is sometimes called Maxwell molecules but unfortunately this cross-section has nothing to do with physics either in the Landau equation in which you should take the Coulomb one or in the Boltzmann equation in which super heart sphere is something which is invented by mathematicians but which corresponds to no gas actually so all of this has to be in some sense extended to cases which have to do with physics and so that was the work that I tried to perform more recently and actually I was able to do it only in the case of the Landau equation and in the case of the Boltzmann equation my hope was that it was possible to use this link obtained by Cherchiniani and Villani some time ago to get better results for the Cherchiniani conjecture for the Boltzmann equation but up to now I have not been able to obtain it so I will present something only on the Landau equation let me begin maybe by this inequality here so for this last part of the lecture I will go faster I will show only slides so I hope you will forgive me the let's say the extension to the case of the Landau equation with the true Coulomb interaction which is represented here by v minus v minus w to the minus 1 takes this form here which is not as beautiful as what was obtained for this specific cross section so you can have a look to the differences this is exactly the same as here so there is no difference at this point but inside the relative feature information you have now this extra weight which is very bad when v is large because when v is large this will be small and so this is not as good as having a constant here and so in some sense in the case of the Landau interaction you have problems with large velocities which is something which has been known for a very long time but large velocities are problematic for the Landau equation with Coulomb interaction so that's the first thing but then you also have in the computation actually it's in the computation which corresponds to what I wrote there when you are doing the Cauchy-Schwarz inequality and things like that in this case you have to use a moment of f which is of much higher order than the second moment which is here and for which we know that there is propagation and actually conservation for the Landau equation so you have here a quantity which is not naturally conserved by the evolution of the Landau equation with Coulomb interaction so you have two difficulties if you wish and one which is related to this extra moment that you need in order to get an inequality but what I would like to emphasize is that this result which I obtained one year before so it was in this paper here so it was two years before this paper appeared so this inequality is obtained rather quickly from what we saw together in the two last lectures by modifying the weights the coefficients etc which was not really apparent in the proof that we did together with Cedric Villani in 2000 which was based mainly basically the same ingredients but in which we did not really identify this inversion formula between q and gradient f over f so the new I would say improvement here is really this inversion formula which in some sense makes it quite easy to do new proofs if you give me something which looks like the Landau equation and so now I think that we have released something which is really robust but of course if you use it in cases in which there are troubles you will find the trouble somewhere in the inequality anyway so that is what is obtained at the level of the Coulomb potential my feeling is that this is optimal so I am not completely sure but my feeling is that this is optimal that is you cannot remove the weight and you cannot remove the moment I am not completely sure but I would be very surprised if this were not optimal in this setting how can we use this in order to produce a large-time behavior result for the Landau equation in the most interesting case which is the Coulomb case so this was obtained in a paper that I wrote together with Kleber-Karapatoso and Lingling He last year and in fact which was recently improved I will explain then how by Kleber-Karapatoso and Stefan Mischler very recently which says basically the following now you look at the Landau equation with the Coulomb potential which is represented here so the real physical Landau equation and you take an initial datum which has an entropy but which also has a lot of moments so here this means that you have a stretch exponential function which is integrable against the initial datum so it is a lot of extra assumptions respect to the theorem that could be produced thanks to this inequality here then there exists a global weak solution of the Landau equation so this was known from the previous paper and it's possible to at this level to eliminate the notion of H solutions which was invented by Cedric Villani in the late 90s thanks to something which I will maybe take some time to explain at the very end of the lecture so anyway there is a good weak solution let's say for the equation and it's possible to show the following large time behavior estimate that F converges towards the centered normalized Gaussian in L1 with a rate which is not exponential which is still not far from being exponential in the sense that you have here exponential to the minus t to some power so the power here is 1 over 7 and there is a small logarithmic correction that you can forget moreover the constant here is something that can be completely computed if you know well the initial datum like for example if you know L1 norm in a good weighted space and the rate can be improved if you suppose some more things on the initial datum this is a contribution of Carapatozo and Micheler in this so be more precise this rate is not optimal and so in their work they were able to obtain optimal rate which I think is exponential minus t to the 2 third if I am not mistaken but anyways the optimal rate is not exponential convergence it's a different power and this has been known since at least Kaflisch in the 80s because when you linearize you see that you see that there is no spectral gap that you have some difficulty with weight and you can somehow increase the rate that you want to get so the way to go from this result to this result consists actually in using this result up to a certain time t in which you enter a small neighborhood of m in a norm which is like this one let's say L1 and then to do a spectral analysis so around m but this spectral analysis is done in a space which is not L1 it's done in a space which is much smaller which is basically L2 with a weight which is Gaussian which is a natural space let's say for the linearized equation and then you have to play a game which is a rather technical one in order to see that the spectrums the spectra are not when you extend your space from L2 with a Gaussian weight to L1 and this is something which was I think first proposed by Clemence Moore I think there was a Stéphane Michelaire and also was the third one should have noted it maybe I did at the end yes Maria Pia, Maria Pia Guldani so it's actually this paper here so this was first proposed in 2013 and actually in this case one has to use a quite refined version of this first result here because it's you don't get an exponential decay at the end you obtain a decay which is like straight exponential so this is what I wanted to show for the Coulomb case let me say that among the ingredients varies so this entropy entropy dissipation estimate but I would like to show also the two or three other ingredients which enable to get the straight exponential decay so the next ingredient is a Bacry-M reversion of the Sobolev logarithmic inequality and this is something which is related to the weight that appears in the fissure information here you had the standard fissure information relative fissure information and so the Sobolev the log Sobolev inequality you have to use is a standard one which is the growth log Sobolev inequality here because of the weight you need to use this refined version which is due to Bacry and Emery so that is the second ingredient which has to be used which was not used in the non-physical case then there is a theory of propagation of moments in this specific case of the Landau equation and to be more precise one needs to propagate which are stretched exponential weighted moments to be precise so one has to show that those moments do not grow too fast like they must grow less than exponentially and in fact it's possible to show that they grow at most like 1 plus t so that is a second ingredient which has to be used in the proof and the last ingredient I would like to show to you there are two more ingredients so let me show the very last if I wrote it no so let me show this one when you do the the computations you see that at some point you have to control not only the L1 moment the stretch exponential L1 moment but actually a moment of F log F that is something which appeared already in the old papers by Toskani and Villani from the late 90s that such moments are important for that kind of computations and here it happened that we had to control those moments so the idea was to try to interpolate cleverly between moments in L1 which are here of the same kind of quantities and extra regularity on F and so given by the formula interpolation formula here and this extra regularity is actually something which is itself computed from the same estimate here and this is the last thing I would like to present to you in this lecture that is this entropy-entropy dissipation estimate does not only tell you how to converge fast toward the equilibrium it also tells you something on the regularity of the solutions of the equation let me show to you very briefly how it works one corollary of this which is very simple to get is that the entropy dissipation associated to the equation is bigger than the same quantities times I hope I will not be times the integral in L3 of F mistaken we should have this I hope I did not make mistake with the exponents but I think it's okay this one is like gradient F to the square over F so it's linear in terms of F and it's also the case here so as you can see when you know that the entropy dissipation is bounded you will know that F is naturally in L3 and so the reason for this is a direct let's say sobolev inequality because this is a fissure information sobolev proof but this is really just two lines when you have that gradient F to the square over F is bounded so this is the quantity here if you forget about the V you can write this as a constant times the gradient of the square root of F to the square and then you can do a sobolev inequality in from H1 to L6 because you are in dimension 3 so this is only for dimension 3 so this will be bigger than a constant times square root of F in norm L6 to the square and this is nothing but what I have written here so it's quite natural that solutions of the Landau equation will have an entropy dissipation which is bounded once you integrate in time and so they will naturally live in L1 in time with value in L3 in V so then when you deal with the equation itself you have at your disposal this L3 norm but here I did everything in the setting of the bad the good for mass but the bad for physics cross section and so no weights appear unfortunately in the Coulomb case there are weights and so there is this minus 3 here which is a weight related to the L3 norm which is a negative weight and it's only this that you can control when you look at the entropy dissipation with the Coulomb potential so you have some loss also at this level so as you can see one also needs a certain number of interpolations and here this one is actually not very hard to get you have to interpolate between weights and regularity in order to do things so actually the last ingredient which is not written in my slides is a sort of treatment of the differential inequalities that you get at the end in which it's not really a direct gronwale argument as we saw in the first lecture it's a variant which is much more complicated but which is once you know what you want to get that you can really handle and at the end because of this you do not get the exponential decay but this stretch exponential decay is presented ok, let me maybe finish by saying a few words about recent progresses on the same subject so actually the group which has been working a lot on this is the group which is based in Dauphin and Cambridge I would say here Maria Pia, I think she's Washington now but if I'm not mistaken and Kleber at that time was also in Dauphin Danila, of course, she's in Dauphin and Isabel Tristani was also in Dauphin she's now in ANS film ok, anyway they really all worked systematically to get the best possible decay for spatially homogeneous or spatially inhomogeneous kinetic equations with collision operator so either Boltzmann or Landau and you can see that first was treated the case which is maybe the most traditional one that is Boltzmann with angular cutoff and hard potentials I don't want to discuss too much what this means but it is a traditional Boltzmann equation Irene presented moderate soft potentials and let's say not moderate soft potentials I would say so this was obtained a little later in the case of the Landau equation and also the case of the spatially inhomogeneous equation with data close to equilibrium was treated recently by Erotono and Tristani and maybe the most recent one was two ones where appeared really recently anyway so is this extension of what I presented now by Karapatov so in Michel and all of these results are based on this spectral gaps in large spaces which I did not present at all in the lectures which I think are very interesting tool in order to get optimal results of convergence towards the equilibrium and which marry very well with this entropy let's say business since usually the best way to get optimal rate consists in first using an entropy method which tells you that in an amount of time that you can control you will get close to the equilibrium, you will get in a small neighborhood of the equilibrium and then you finish by using linearized theory and this enlarged spaces theory so I think maybe I will stop now I think it's a good time maybe to stop with the lecture