Partial Regularity in Time for the Landau Equation with Coulomb Interaction
Golse, François
Whether there is global regularity or finite time blow-up forthe space homogeneous Landau equation with Coulomb potential is a longstanding open problem in the mathematical analysis of kinetic models. Thistalk shows that the Hausdorff dimension of the set of singular times of theglobal weak solutions obtained by Villanis procedure is at most 1/2.
 Thank you very much. So thanks to the organizers, thanks to Xi and Mihai for organizing this beautiful conference. I'm very sorry that I cannot be in Marseille, but anyway, so it's already very nice to have a conference like this when you're locked down in Paris, even though locked down in a week since. All right, so what I want to report on today is a recent result with Maya Pierre-Gualdany, Cyrille Lambert and Alexi Vasseur on partial regularity for the Landau equation with collo-interaction. And so this is going to appear in the analysis of Ecole Normale Supérieure. All right, so here is the space homogeneous Landau equation with unknown f, the distribution function, that's the function of time and velocity. So the space variable is removed. So f, t and v is non-negative function and the Landau equation, which is probably familiar to most people working in kinetic equations here. So recalled for convenience is d by dt of f equals to the divergence in v of the integral of the R3 of A of v minus w grad v minus grad w of f of tv, f of tw, dw. So you integrate for w running through R3 and this equation is posed on the whole of R3. So here A of z is proportional to the Hessian of the norm of z, the Euclidean norm of z. So in other words, if you compute what it is, so if you put the coefficient to be 1 over 8 pi, so it's 1 over 8 pi the norm of z times pi of z, where pi of z is the orthogonal projection on the orthogonal of the line span by z. So you can equivalently recast this equation in non-conservative form in which case it reads d by dt of f of t and v equals to some diffusion matrix which you obtain by taking the convolution in v of the matrix Aij of z with f, so you take the convolution in v and this multiplies dvi dvj of f of tv and there is a term which is precisely f of t and v squared. The fact that here you have a purely local term is characteristic of the Coulomb interaction. So it's because when you take the second derivative of this, you get the direct measure at z equals 0. So here you have a purely local term which is f squared and of course this is a dangerous term. So if you have here a constant diffusion matrix, this is a semi-linear heat equation, we know this blow up in finite time. But here if you know of course this will promote blow up, but if blow up builds up, I mean if f increases, of course this will make the diffusion matrix increase as well and you can hope that the diffusion matrix will offset the effect of the f squared terms here. So in other words you can hope for a balance between the f squared term here and the diffusion term there. But nevertheless there is an open question as to whether there is global existence of classical solutions or finite time blow up for the Cauchy problem for the Landau equation set on R3. So however in the late 1990s Cédric Villanille came up with a good notion of weak global solutions defined therefore for all time. He called them H solutions because the Boltzmann H theorem plays a big role in the existence of the solution itself. So an H solution in the terminology introduced by Cédric is a continuous function with values and distributions. So it's non-negative function, already it's a measure. And it's also L1 in time with values in the weighted space L1 minus 1 over 3. So already L1 minus 1 over 3 refers to this family of weighted space. What I call LP minus k, well LPk sorry, LPk, so function g is an LPk if g modulus to the p is integrable with the weight 1 plus v squared to the k over 2. And you have this formula for the norm. So here it means that you are integrable with the weight proportional to 1 over 1 plus norm of v. So it's a function in this functional space which satisfies conservation of mass, momentum and energy with respect to the initial data and which satisfies the H inequality, the fact that the H function at time t is less than or equal to the H function initially. And the equation is satisfied in the following sense. The weak formulation of the Landau equation is the integral of f in of v times phi of 0 and v dv. That's the integral from 0 to t of the integral in v of f times dt phi is equal to this term that involves the entropy dissipation integrand that you find in the dissipation formula for the H function here. So here you have the integral from 0 to t, the integral in v and w of the difference of capital phi at v and at w, where capital phi is the grad of little phi, the test function. In the product with the projection on v minus w of the integral capital F grad v minus grad w of capital F, where capital F is the square root of f of v f of w divided by v minus w and there is a phi that is floating around. So this is the weak formulation of the, this is the weak formulation which Villani proposed to define global solutions to the Landau equation. Good, nevertheless for the purpose of proving partial regularity, we need more information on the solutions but fortunately with the same procedure by which Villani constructed global H solution, the same procedure, the same approximating scheme will lead, will result, will produce what we have called suitable solution by analogy with the paper by Kafferli Konean-Wer for partial regularity on Navier-Sos. So what is a suitable solution of the Landau equation? So it's a solution which satisfies the following truncated entropy inequality. If I look at the relative entropy at the level kappa, which I define as h plus of g and kappa to be the integral of kappa times little h plus of g over kappa dv, where h plus is the function generating the h function. So it's z log z minus z minus one truncated, so truncated for z above one, right? So if you want this is z log z minus z minus one times the indicator that z is larger than one. Okay, all right, so now h plus of f times t2 with respect to kappa plus some dissipation term which involves some constant c prime e integrated between t1 and t2, the integral from t1 to t2 of the LQ norm of the grad v of f to the one over q where f is larger than kappa. So you take the L2 norm in time of this thing to the square dt should be less than or equal to the truncated entropy at time t1. And there is an additional term which is proportional to kappa times the integral from t1 to t2 and the integral of all v of f minus kappa plus. Okay, so if you take the approximation scheme proposed by Villani to construct h solutions and you pass through the limit and you pay a little more attention to the properties of the sequence, then you arrive at this inequality which is satisfied for all time in on the half line except possibly a negligible set of times, okay, which is satisfied for certain values of q which I'm going to explain in a minute and with a certain dissipation constant which I'm going to comment on in a minute also. All right, so okay, now partial regularity in time has to do with the fact that you want to say that the set of times where the solution possibly becomes singular, there's no longer a classical solution, whatever you want to call it, is going to be small in some sense. So here we call a regular time of f, a suitable solution on some interval of the half line, it's a time tau in that interval, such that f is in L infinity on the time interval tau minus epsilon tau for some epsilon and for all these. So here we localize only in time but this is global in velocity. So this is the set of, this is singular times, so the set of singular times, in other words, the times which are not regular times in the interval i with a note by s of fi, okay. So now if you take suitable solution to another equation on some time interval cross r3 for all positive t, I'm sorry, I mean on the half line cross r3 with some initial data f in, so the initial data has a decay to any order in V, right, so it has fast decay integrating in V, it has bounded initial entropy, right. And in that case, you can prove that the house of dimension of the set of singular times for any suitable solution which has this f initial as initial data is at most one half, okay. All right, so let me say a few words, let me explain how the proof is organized. Okay, first, first we need, we need the global existence of such suitable solutions. So as I said, the suitable solutions that are constructed by using the same approximation scheme as was used by Cedric Villani to construct H solutions. And if you do that, here's what you get, if you take any f in which is in L1, non-negative on r3, and which is integral ball against the weight of normal V to the k, which has finite entropy initially, then such an f initial will launch a suitable solution with that initial data. With a Q, so if you remember in the definition of suitable solution, there is a, there is a Lebesgue exponent in the dissipation here. So with some Q that is related to the decay in V of the initial data, so specifically the Q which you observe in the dissipation term is related to the decay by this formula here. So Q is equal to 2k over k plus 3. And the constant C prime E, that depends, that's the dissipation constant in the suitable solution inequality, depends on capital T, depends on Q, and depends functionally on f initial. Okay. And okay, there is always some negligible set of times which you don't care about. Okay. Good. So this is the existence theory for such solutions, which as I said are constructed as the, by the same approximation scheme as was used by Villani for H solutions. Now if you look at the, if you look at the entropy production term that comes from the Boltzmann H theorem, this is a non-local term, but there is a rather recent and very important observation, very important theorem by Deville-Etts, which allowed replacing this non-local dissipation integrand for the Boltzmann H theorem applied to the Lander equation by purely local terms. So here's the, here's the Deville-Etts theorem. So if you take a function f, which is integrable and has finite energy, right, so L12 means you can integrate with respect to 1 plus d squared dv, right, and which has a finite entropy. So f log f is in L1. Then if you look on the right-hand side here, this quantity is exactly the entropy dissipation integrand which appears in the Lander equation when you do the Boltzmann H theorem. Well the Deville-Etts result is that up to some constant, which you can compute explicitly. This is bounded below by the L2 norm of the gradient of square root of f with a weight in v, right. So this is a Fisher information if you want with the weight of the order of v to the minus 3, okay. And again, so the constant cd that appears here depends explicitly on the mass, momentum, energy, and entropy of f. And therefore if you apply it, you know, this is an inequality which is true for all f, okay, you can apply it of course at each time on a solution to the Lander equation. The corollary of this theorem in the paper by Deville-Etts, which appeared I think five years ago, is that you have propagation of moments for a Lander. So if you assume that f initial is in L1k, so it has finite k finite moments with k larger than 2, for 2 which is obvious, this is the conservation of energy. Then and if in addition f in as a finite entropy, then if f is an h solution with this f in as initial data, then f is going to be L infinity over finite time with values in L1k. So in other words, you propagate k moments in finite time, okay. So maybe if you let t go to infinity, this is not going to be bounded in that space, but for finite time you propagate k moments, okay. Good. So with this, let me look at the truncated h theorem which is used in the definition of these suitable solutions to Lander. So you want to compute d by dt of the truncated entropy truncated at the level kappa. Of course, if you don't truncate, this is a standard h theorem. If you truncate at the level kappa, there is one piece that is dissipative, which I denote by d1, which involves essentially the same dissipation integrant as would appear for the standard h theorem in the case of the Lander equation, except that here instead of having grad v of f over f minus grad w of f over f to the square, I have the grad v localized where f is larger than kappa, and here I have the grad w localized where f is larger than kappa to the square integrating in all variables. All right, so that's the dissipation term, and this is the nice part of the equation, but of course, this truncation gives rise to other terms which are not nice and which we put on the right hand side. Nevertheless, these terms, however not nice, can be computed explicitly. If you compute them, you find that you have something of the form minus the integral of a times the grad of vf, where f is larger than kappa, times the grad in w of f, where f is less than kappa. You can integrate by parts in v and w, and you let the derivatives bear on the collision kernel. If you do that, that will be the integral of minus the divergence in v, the divergence in w of the tensor, a of v minus w. In fact, if you compute it, this is equal to the Dirac measure at v equals w, and here you have f of tv minus kappa plus times kappa minus f of tw minus kappa minus. In the approximation process, this is not going to be identically the Dirac measure, but this is at least going to be non-negative. This extra term here is less than or equal to kappa, which comes from here, times the integral of f minus kappa plus. Now we find that if you want, this is a depleted non-linearity. We call that depleted non-linearity because if you return to the non-conservative form of the Landau equation, which is written here, if you multiply both sides of the equation by log of f, and if you integrate, here you would observe something that grows like f squared log of f, which would be very bad. But nevertheless, if you look at, if you do this computation here, then the non-linear term that comes from this Dirac measure that appears when you take the double derivative of a, it doesn't grow like f squared log of f. It grows like f, it's kappa, the integral of f minus kappa plus, which is tamed as a non-linearity compared to the entropy. So this is good news because this is going to help us controlling the equation. Unfortunately, not completely. That's why we have partial regularity and not full regularity. Good. So now, as I said, the sketch should prove a proposition one, the existence proof. Well, I mean, you truncate, you replace a collision kernel of Landau with a truncated variance. In other words, you truncate one of the z at a level n. Right? You check that it satisfies this inequality here. You use the Deville-Let's theorem to bounce the dissipation plus the integral of f minus kappa plus below by this fissure information with the weight proportional to the v to the minus three. And to remove the weight, you use the Deville-Let corollary, so in other words, the propagation of moments, to remove this decaying weight at the expense of lowering the power that appears here. So by further inequality, you control this expression here below by the gradient with respect to v of f to the one of a q, where f is larger than kappa. And here, q belongs to the interval one, two. So q is always less than two. And you have this further inequality here. And you enter that in the truncated entropy inequality here. And you get the announced inequality in the definition of the suitable solution. Good. All right. So now we have suitable solutions. How is that going to help us proving partial regularity? Well, this is done by the DeGeorgi method. I mean, actually, the first part of the DeGeorgi method, which consists in gaining L infinity out of the energy or L2 bound. So here is how it works. So you take let f be a suitable solution to the Landau equation with some coarsivity constant c prime e positive. And here I'm going to assume that q is less than two, but at least 6 fifths. You need something slightly larger than one. And specifically, you need 6 fifths. All right. So that's technical. It doesn't matter. What is important is that q should be free to approach to as much as possible. What is interesting is when q is very near two. Okay. Well, so for such an f, then there exists an eta naught, which depends on q and on that constant c prime e positive, such that if the entropy of f truncated at the level of one half integrated in time between, say, one eighth and one is less than eta naught, then f is going to be bounded by two, right, is going to be bounded by two on a smaller time interval, say on one half one cross R3. So for time between one half and one and for all v. Almost everywhere. Good. So how does that work? Well, I mean, this is a standard, this is a standard, the Georgie procedure for power equation. So you pick a sequence of times which grow from one fourth to one half. You pick a sequence of levels which grow from one to two. And then you replace, you replace the entropy by this nonlinearity here, which if you want is comparable to the entropy. So you look at fk plus of T and v to be mu of f to the one of a q minus kappa sub k to the one of a q positive part. And out of this, you construct the quantity ak, which is the supremum of the integral of fk plus to the power q dv for T between tk and one and the integral from tk to one of the grad v of fk plus in lq norm in velocity, you raise that to the square and you integrate in t. Okay, so this is the quantity ak, which on which you apply the Georgie procedure. So you control this by using the truncated entropy inequality, which is characteristic of suitable solutions. And you can show that this ak is at a size and inequality of the form, ak plus one controlled by some constants. There is something that grows exponentially fast, some lambda to the k. What is important is that you have ak to the beta and here beta is larger than one. So provided that a naught is small enough, then ak tends to zero as k tends to plus infinity. So you control this a naught by the truncated entropy and you conclude by Fatou Lemma that since ak tends to zero as k tends to plus infinity, this tells you exactly that f is going to be less than or equal to two on that interval here corresponding to the level at which you truncate for k going to infinity here and the interval that you obtain for k going to infinity there. Okay, very good. So that's the standard Georgie arguments, but in the language of the lambda equation. All right, so now this is not enough for partial regularity and we're going to now apply another argument but at the same nature, but now we have truncated and zoomed on the values of f, on the set of values of f. This is absolutely characteristic of the Georgie method to look at the level sets of f. Now we're going to do some scaling on the function on the solution to the lambda equation. And so here's that first, suppose that you have f as suitable solution to the lambda equation on some interval 0, 1 say, and with an exponent q, again close enough to 2 by I mean here, here q has to be just between four thirds and two four thirds is higher than six fifths. But okay, so all right, so but still less than two. With this q define gamma to be five q minus six over two q minus two. What I'm saying is that there exists parameter eta one, which depends on q and on the dissipation constants in the truncated entropy inequality. And there exists a delta one, which is zero and one such that if the lim soup of the integral of the Lq norm to the square of the grad V of f to the one over q, where f is larger than epsilon to the minus gamma, which you integrate between one minus epsilon to the gamma and one, right, which you multiply by epsilon to the gamma minus three. So if you assume that this lim soup is less than eta one, then f is going to be an infinity on the strip one minus delta one one cross R three. Now, if you have this proposition proposition three and vitally type covering arguments on the half line, it's classical to deduce from that, that if you look at the house of measure three minus gamma over gamma dimensional house of measure of the set of singular times, this is going to be finite. Okay, three minus gamma over gamma. This is equal to q over five q minus six if you do the arithmetic here. And the main theorem will follow by translation. Okay, because here I just I have localized in a subinterval not a half line and letting q going to two, right. So I remind you that q cannot be equal to two because I've used this, the devilette propagation of moments to remove the weights in the in the Fisher information, the vanishing weight at infinity. So I've used a little bit of q. So q cannot be exactly equal to two. But nevertheless, you can let q going to two converge to q to two. And for q equals two, you see immediately that this is equal to one half again. So at the end of the day, I prove that the house of measure at any dimension less than one half of the singular set is finite. So that gives me that gives me the the house of dimension as an answer. All right, so how does one prove this? Well, first you do a scaling, right, like, like so. So you introduce fn of t and v to be epsilon n to the gamma f of one plus epsilon to the gamma t minus one and epsilon v epsilon and two to the minus n. So this scaling here will give you a suitable solution to the lambda equation. This preserves the lambda equation. What's important to give you a suitable solution to the lambda equation with the same creativity constant. That's the thing that is absolutely crucial in this business. You don't you don't change. You don't touch the the constant C prime E and okay, so then you define capital Fn in terms of little fn was the same nonlinearity mu that sort of replaces the the F log F by the mean of R and R square. Right. Okay, so that's not very important. Anyway, so this assumption to the fact that this limb soup is less than eta one tells you that there exists a capital N large enough such that this quantity here is going to be less than more than eta one say eight eta one. Okay, good. So then with this, you use the order inequality and the sub-elephant equality as in the proof of proposition two. You isolate the term grad V of fn plus one in L two in time Q and V, which is a folder eta one. And you show that if you look at xm, there's going to be the soup of integral of fn plus m to the Q for T between one half and one here. You don't shrink the interval. Interval is a waste. One half one. I mean in terms of capital F. And you show that this quantity here satisfies a recursion inequality of this form, right? Xm plus one less than rho times the max of one and xm to the alpha plus the max of one and xm minus one to the alpha. You start from initial conditions x naught and x one less than or equal to capital M. So capital M is going to be a large enough power of two. So the M is say the first N for which this is going to be less than eight eta one. Here alpha is going to be Q over three and rho is some constant depending on Q eta one to the Q over two. Okay. Whatever. Here Q is less than two. So the alpha is less than one. So this is the opposite of the traditional, the Georgie argument. So when you iterate this inequality here, there's an easy induction that tells you that at some point all the xm are going to be below one. Okay. All right. So x2m is going to be less than or equal to the max of two rho to the one minus alpha to the M divided by one minus alpha M to the alpha to the M. Alpha is less than or equal to one. So that converges to zero. So at the end of the day, you find that for some M naught, this is going to be small compared to one. And now with this M naught, you scale back to the original variables and you find that F capital N plus M naught plus three, if I remember, satisfy the assumption in proposition two. In other words, satisfies the fact that the truncated entropy at the level one-half on the time interval one-eighths one is less than eta naught. And then you find the N infinity bound which you need, which you want. Okay. Well, that's it. All right. So final remarks. Well, I mean, this is a partial regularity result in the same style as the Le Réthier theorem for Navier-Stokes. Right? I mean, the Le Réthier paper for Navier-Stokes, you already find the fact, it's not written like this in this paper, but you already find the fact that a half-storey dimension of singular times for Le Réthier solution on Navier-Stokes is the most one-half. Well, in fact, the Devillein, if you look at the Devillein theorem, it puts along that equation the same class as 3D Navier-Stokes in terms of the back exponents, right? For this weight, one plus V to the minus 3, which could be important or not. But I don't know. Anyway, so if you look at Navier-Stokes, Le Réthier retails you that the solution is L infinity in time with values in L2 and X, with grad U, which is L2 in time and X. So here, if you look at Langdorff, square root of F is L infinity in time with values in L2 and V. So that's the conservation of us. And the Devillein theorem tells you that the grad V of the square root F is L2 in time and velocity except for this weight. All right. But if you, yeah, apart from the weight, this is very much in the same number. These are very much the same exponents. So it's natural to ask whether you have partial regularity in T and V in the same style as the Kaffareli and Nihon-Bert. Why not? I don't know. But more generally, you could ask yourself whether you have conditional regularity in the style of the papers by Serrin and all subsequent papers where you assume some local Le Bag exponents above a certain critical threshold. And out of this, you deduce regularity. So I mean, if you take P equals infinity, so the Serrin criterion, if you copied for square root of F would be exactly this. So I don't know if this is true or not. This is only true for P equals infinity and K large number 5. So this is a paper by Sylveste in 2017. So it's also a result by Guadagnan and Gilan in 2016. And there's recently a result in that direction by my student, Emmanuel Ben-Carrot. Okay. Thank you very much. I think Renjun has some questions in the chat. I think he says he's referring to the stability in the nonlinear case. Just did you have any comments on it? Renjun, do I am asking on the chat? I was in it a question for Jean. Yeah. Not for... It was a question for Jan. I thought... It's a very strange problem. Sorry. It's a long one. So... Someone else is typing some questions. Yeah. Samir is typing. Samir is typing. Yeah, I don't know. I didn't see... I thought that one was new. I didn't just realize now. No, well... Well, maybe there's some stability for... Yeah. Yeah. Samir wrote there, yes? Yeah. So... Well, unfortunately not because... I mean, with this technique... Hi, Samir. So with this technique... You know, this technique, somehow you truncate the velocity. You get rid of the weight, okay? If you... Oh my God, I did the wrong thing. If you return... So where is that? So that's, I think, here. You see, at this step, you remove... You replace the weight in V as appears in the... In the devilette inequality. You trade that for the bag... The bag exponents slightly less than two. For some gradient of F to the one of a Q. Okay, Q is slightly less than two. So here... I mean, if you change the... If you change the potential, what will change is this weight, typically. Right? But this weight, you throw it away by this procedure of the propagation of moments. So I doubt that with this argument... I mean, I doubt that this argument, you will see the difference between the potential that will be less singular than Q. Of course, that will be a very nice question. We know that if the exponent is larger than minus two, I think no problem at all. You propagate regular IT and all that. Exactly. I think there's the answer by the way that... But it's not the same argument. This is a totally different argument. Right? So unfortunately, these partial regularity techniques will not see the difference between Coulomb and possibly nicer potentials. Right? For instance, we don't know what happens between minus two and minus three. I mean, probably this is the same as Coulomb period.
 Thank you very much.  So thanks to the organizers, thanks to Xi and Mihai for organizing this beautiful conference.  I'm very sorry that I cannot be in Marseille, but anyway, so it's already very nice to have  a conference like this when you're locked down in Paris, even though locked down in a week  since.  All right, so what I want to report on today is a recent result with Maya Pierre-Gualdany,  Cyrille Lambert and Alexi Vasseur on partial regularity for the Landau equation with collo-interaction.  And so this is going to appear in the analysis of Ecole Normale Supérieure.  All right, so here is the space homogeneous Landau equation with unknown f, the distribution  function, that's the function of time and velocity.  So the space variable is removed.  So f, t and v is non-negative function and the Landau equation, which is probably familiar  to most people working in kinetic equations here.  So recalled for convenience is d by dt of f equals to the divergence in v of the integral  of the R3 of A of v minus w grad v minus grad w of f of tv, f of tw, dw.  So you integrate for w running through R3 and this equation is posed on the whole of  R3.  So here A of z is proportional to the Hessian of the norm of z, the Euclidean norm of z.  So in other words, if you compute what it is, so if you put the coefficient to be 1 over  8 pi, so it's 1 over 8 pi the norm of z times pi of z, where pi of z is the orthogonal projection  on the orthogonal of the line span by z.  So you can equivalently recast this equation in non-conservative form in which case it reads  d by dt of f of t and v equals to some diffusion matrix which you obtain by taking the convolution  in v of the matrix Aij of z with f, so you take the convolution in v and this multiplies  dvi dvj of f of tv and there is a term which is precisely f of t and v squared.  The fact that here you have a purely local term is characteristic of the Coulomb interaction.  So it's because when you take the second derivative of this, you get the direct measure at z equals  0.  So here you have a purely local term which is f squared and of course this is a dangerous  term.  So if you have here a constant diffusion matrix, this is a semi-linear heat equation, we know  this blow up in finite time.  But here if you know of course this will promote blow up, but if blow up builds up, I mean if  f increases, of course this will make the diffusion matrix increase as well and you can hope that  the diffusion matrix will offset the effect of the f squared terms here.  So in other words you can hope for a balance between the f squared term here and the diffusion  term there.  But nevertheless there is an open question as to whether there is global existence of  classical solutions or finite time blow up for the Cauchy problem for the Landau equation  set on R3.  So however in the late 1990s Cédric Villanille came up with a good notion of weak global solutions  defined therefore for all time.  He called them H solutions because the Boltzmann H theorem plays a big role in the existence  of the solution itself.  So an H solution in the terminology introduced by Cédric is a continuous function with values  and distributions.  So it's non-negative function, already it's a measure.  And it's also L1 in time with values in the weighted space L1 minus 1 over 3.  So already L1 minus 1 over 3 refers to this family of weighted space.  What I call LP minus k, well LPk sorry, LPk, so function g is an LPk if g modulus to the  p is integrable with the weight 1 plus v squared to the k over 2.  And you have this formula for the norm.  So here it means that you are integrable with the weight proportional to 1 over 1 plus norm  of v.  So it's a function in this functional space which satisfies conservation of mass, momentum  and energy with respect to the initial data and which satisfies the H inequality, the fact  that the H function at time t is less than or equal to the H function initially.  And the equation is satisfied in the following sense.  The weak formulation of the Landau equation is the integral of f in of v times phi of  0 and v dv.  That's the integral from 0 to t of the integral in v of f times dt phi is equal to this term  that involves the entropy dissipation integrand that you find in the dissipation formula for  the H function here.  So here you have the integral from 0 to t, the integral in v and w of the difference  of capital phi at v and at w, where capital phi is the grad of little phi, the test function.  In the product with the projection on v minus w of the integral capital F grad v minus grad  w of capital F, where capital F is the square root of f of v f of w divided by v minus w  and there is a phi that is floating around.  So this is the weak formulation of the, this is the weak formulation which Villani proposed  to define global solutions to the Landau equation.  Good, nevertheless for the purpose of proving partial regularity, we need more information  on the solutions but fortunately with the same procedure by which Villani constructed  global H solution, the same procedure, the same approximating scheme will lead, will  result, will produce what we have called suitable solution by analogy with the paper by Kafferli  Konean-Wer for partial regularity on Navier-Sos.  So what is a suitable solution of the Landau equation?  So it's a solution which satisfies the following truncated entropy inequality.  If I look at the relative entropy at the level kappa, which I define as h plus of g and kappa  to be the integral of kappa times little h plus of g over kappa dv, where h plus is the  function generating the h function.  So it's z log z minus z minus one truncated, so truncated for z above one, right?  So if you want this is z log z minus z minus one times the indicator that z is larger than  one.  Okay, all right, so now h plus of f times t2 with respect to kappa plus some dissipation  term which involves some constant c prime e integrated between t1 and t2, the integral  from t1 to t2 of the LQ norm of the grad v of f to the one over q where f is larger than  kappa.  So you take the L2 norm in time of this thing to the square dt should be less than or equal  to the truncated entropy at time t1.  And there is an additional term which is proportional to kappa times the integral from t1 to t2  and the integral of all v of f minus kappa plus.  Okay, so if you take the approximation scheme proposed by Villani to construct h solutions  and you pass through the limit and you pay a little more attention to the properties  of the sequence, then you arrive at this inequality which is satisfied for all time  in on the half line except possibly a negligible set of times, okay, which is satisfied for  certain values of q which I'm going to explain in a minute and with a certain dissipation  constant which I'm going to comment on in a minute also.  All right, so okay, now partial regularity in time has to do with the fact that you want  to say that the set of times where the solution possibly becomes singular, there's no longer  a classical solution, whatever you want to call it, is going to be small in some sense.  So here we call a regular time of f, a suitable solution on some interval of the half line,  it's a time tau in that interval, such that f is in L infinity on the time interval tau  minus epsilon tau for some epsilon and for all these.  So here we localize only in time but this is global in velocity.  So this is the set of, this is singular times, so the set of singular times, in other words,  the times which are not regular times in the interval i with a note by s of fi, okay.  So now if you take suitable solution to another equation on some time interval cross r3 for  all positive t, I'm sorry, I mean on the half line cross r3 with some initial data f in,  so the initial data has a decay to any order in V, right, so it has fast decay integrating  in V, it has bounded initial entropy, right.  And in that case, you can prove that the house of dimension of the set of singular times  for any suitable solution which has this f initial as initial data is at most one half,  okay.  All right, so let me say a few words, let me explain how the proof is organized.  Okay, first, first we need, we need the global existence of such suitable solutions.  So as I said, the suitable solutions that are constructed by using the same approximation  scheme as was used by Cedric Villani to construct H solutions.  And if you do that, here's what you get, if you take any f in which is in L1, non-negative  on r3, and which is integral ball against the weight of normal V to the k, which has  finite entropy initially, then such an f initial will launch a suitable solution with  that initial data.  With a Q, so if you remember in the definition of suitable solution, there is a, there is  a Lebesgue exponent in the dissipation here.  So with some Q that is related to the decay in V of the initial data, so specifically  the Q which you observe in the dissipation term is related to the decay by this formula  here.  So Q is equal to 2k over k plus 3.  And the constant C prime E, that depends, that's the dissipation constant in the suitable  solution inequality, depends on capital T, depends on Q, and depends functionally on  f initial.  Okay.  And okay, there is always some negligible set of times which you don't care about.  Okay.  Good.  So this is the existence theory for such solutions, which as I said are constructed  as the, by the same approximation scheme as was used by Villani for H solutions.  Now if you look at the, if you look at the entropy production term that comes from the  Boltzmann H theorem, this is a non-local term, but there is a rather recent and very important  observation, very important theorem by Deville-Etts, which allowed replacing this non-local  dissipation integrand for the Boltzmann H theorem applied to the Lander equation by  purely local terms.  So here's the, here's the Deville-Etts theorem.  So if you take a function f, which is integrable and has finite energy, right, so L12 means  you can integrate with respect to 1 plus d squared dv, right, and which has a finite  entropy.  So f log f is in L1.  Then if you look on the right-hand side here, this quantity is exactly the entropy dissipation  integrand which appears in the Lander equation when you do the Boltzmann H theorem.  Well the Deville-Etts result is that up to some constant, which you can compute explicitly.  This is bounded below by the L2 norm of the gradient of square root of f with a weight  in v, right.  So this is a Fisher information if you want with the weight of the order of v to the minus  3, okay.  And again, so the constant cd that appears here depends explicitly on the mass, momentum,  energy, and entropy of f.  And therefore if you apply it, you know, this is an inequality which is true for all f,  okay, you can apply it of course at each time on a solution to the Lander equation.  The corollary of this theorem in the paper by Deville-Etts, which appeared I think five  years ago, is that you have propagation of moments for a Lander.  So if you assume that f initial is in L1k, so it has finite k finite moments with k larger  than 2, for 2 which is obvious, this is the conservation of energy.  Then and if in addition f in as a finite entropy, then if f is an h solution with this f in  as initial data, then f is going to be L infinity over finite time with values in L1k.  So in other words, you propagate k moments in finite time, okay.  So maybe if you let t go to infinity, this is not going to be bounded in that space,  but for finite time you propagate k moments, okay.  Good.  So with this, let me look at the truncated h theorem which is used in the definition  of these suitable solutions to Lander.  So you want to compute d by dt of the truncated entropy truncated at the level kappa.  Of course, if you don't truncate, this is a standard h theorem.  If you truncate at the level kappa, there is one piece that is dissipative, which I  denote by d1, which involves essentially the same dissipation integrant as would appear  for the standard h theorem in the case of the Lander equation, except that here instead  of having grad v of f over f minus grad w of f over f to the square, I have the grad  v localized where f is larger than kappa, and here I have the grad w localized where f is  larger than kappa to the square integrating in all variables.  All right, so that's the dissipation term, and this is the nice part of the equation,  but of course, this truncation gives rise to other terms which are not nice and which  we put on the right hand side.  Nevertheless, these terms, however not nice, can be computed explicitly.  If you compute them, you find that you have something of the form minus the integral of  a times the grad of vf, where f is larger than kappa, times the grad in w of f, where  f is less than kappa.  You can integrate by parts in v and w, and you let the derivatives bear on the collision  kernel.  If you do that, that will be the integral of minus the divergence in v, the divergence  in w of the tensor, a of v minus w.  In fact, if you compute it, this is equal to the Dirac measure at v equals w, and here  you have f of tv minus kappa plus times kappa minus f of tw minus kappa minus.  In the approximation process, this is not going to be identically the Dirac measure,  but this is at least going to be non-negative.  This extra term here is less than or equal to kappa, which comes from here, times the  integral of f minus kappa plus.  Now we find that if you want, this is a depleted non-linearity.  We call that depleted non-linearity because if you return to the non-conservative form  of the Landau equation, which is written here, if you multiply both sides of the equation  by log of f, and if you integrate, here you would observe something that grows like f  squared log of f, which would be very bad.  But nevertheless, if you look at, if you do this computation here, then the non-linear  term that comes from this Dirac measure that appears when you take the double derivative  of a, it doesn't grow like f squared log of f.  It grows like f, it's kappa, the integral of f minus kappa plus, which is tamed as a  non-linearity compared to the entropy.  So this is good news because this is going to help us controlling the equation.  Unfortunately, not completely.  That's why we have partial regularity and not full regularity.  Good.  So now, as I said, the sketch should prove a proposition one, the existence proof.  Well, I mean, you truncate, you replace a collision kernel of Landau with a truncated  variance.  In other words, you truncate one of the z at a level n.  Right?  You check that it satisfies this inequality here.  You use the Deville-Let's theorem to bounce the dissipation plus the integral of f minus  kappa plus below by this fissure information with the weight proportional to the v to the  minus three.  And to remove the weight, you use the Deville-Let corollary, so in other words, the propagation  of moments, to remove this decaying weight at the expense of lowering the power that  appears here.  So by further inequality, you control this expression here below by the gradient with  respect to v of f to the one of a q, where f is larger than kappa.  And here, q belongs to the interval one, two.  So q is always less than two.  And you have this further inequality here.  And you enter that in the truncated entropy inequality here.  And you get the announced inequality in the definition of the suitable solution.  Good.  All right.  So now we have suitable solutions.  How is that going to help us proving partial regularity?  Well, this is done by the DeGeorgi method.  I mean, actually, the first part of the DeGeorgi method, which consists in gaining L infinity  out of the energy or L2 bound.  So here is how it works.  So you take let f be a suitable solution to the Landau equation with some coarsivity constant  c prime e positive.  And here I'm going to assume that q is less than two, but at least 6 fifths.  You need something slightly larger than one.  And specifically, you need 6 fifths.  All right.  So that's technical.  It doesn't matter.  What is important is that q should be free to approach to as much as possible.  What is interesting is when q is very near two.  Okay.  Well, so for such an f, then there exists an eta naught, which depends on q and on that  constant c prime e positive, such that if the entropy of f truncated at the level of  one half integrated in time between, say, one eighth and one is less than eta naught,  then f is going to be bounded by two, right, is going to be bounded by two on a smaller  time interval, say on one half one cross R3.  So for time between one half and one and for all v.  Almost everywhere.  Good.  So how does that work?  Well, I mean, this is a standard, this is a standard, the Georgie procedure for power  equation. So you pick a sequence of times which grow from one fourth to one half.  You pick a sequence of levels which grow from one to two.  And then you replace, you replace the entropy by this nonlinearity here, which if you want  is comparable to the entropy.  So you look at fk plus of T and v to be mu of f to the one of a q minus kappa sub k to  the one of a q positive part.  And out of this, you construct the quantity ak, which is the supremum of the integral  of fk plus to the power q dv for T between tk and one and the integral from tk to one  of the grad v of fk plus in lq norm in velocity, you raise that to the square and you integrate  in t.  Okay, so this is the quantity ak, which on which you apply the Georgie procedure.  So you control this by using the truncated entropy inequality, which is characteristic  of suitable solutions.  And you can show that this ak is at a size and inequality of the form, ak plus one controlled  by some constants.  There is something that grows exponentially fast, some lambda to the k.  What is important is that you have ak to the beta and here beta is larger than one.  So provided that a naught is small enough, then ak tends to zero as k tends to plus infinity.  So you control this a naught by the truncated entropy and you conclude by Fatou Lemma that  since ak tends to zero as k tends to plus infinity, this tells you exactly that f is  going to be less than or equal to two on that interval here corresponding to the level at  which you truncate for k going to infinity here and the interval that you obtain for  k going to infinity there.  Okay, very good.  So that's the standard Georgie arguments, but in the language of the lambda equation.  All right, so now this is not enough for partial regularity and we're going to now apply another  argument but at the same nature, but now we have truncated and zoomed on the values of  f, on the set of values of f.  This is absolutely characteristic of the Georgie method to look at the level sets of f.  Now we're going to do some scaling on the function on the solution to the lambda equation.  And so here's that first, suppose that you have f as suitable solution to the lambda  equation on some interval 0, 1 say, and with an exponent q, again close enough to 2 by  I mean here, here q has to be just between four thirds and two four thirds is higher than  six fifths.  But okay, so all right, so but still less than two.  With this q define gamma to be five q minus six over two q minus two.  What I'm saying is that there exists parameter eta one, which depends on q and on the dissipation  constants in the truncated entropy inequality.  And there exists a delta one, which is zero and one such that if the lim soup of the integral  of the Lq norm to the square of the grad V of f to the one over q, where f is larger  than epsilon to the minus gamma, which you integrate between one minus epsilon to the  gamma and one, right, which you multiply by epsilon to the gamma minus three.  So if you assume that this lim soup is less than eta one, then f is going to be an infinity  on the strip one minus delta one one cross R three.  Now, if you have this proposition proposition three and vitally type covering arguments  on the half line, it's classical to deduce from that, that if you look at the house  of measure three minus gamma over gamma dimensional house of measure of the set of singular times,  this is going to be finite.  Okay, three minus gamma over gamma.  This is equal to q over five q minus six if you do the arithmetic here.  And the main theorem will follow by translation.  Okay, because here I just I have localized in a subinterval not a half line and letting  q going to two, right.  So I remind you that q cannot be equal to two because I've used this, the devilette  propagation of moments to remove the weights in the in the Fisher information, the vanishing  weight at infinity.  So I've used a little bit of q.  So q cannot be exactly equal to two.  But nevertheless, you can let q going to two converge to q to two.  And for q equals two, you see immediately that this is equal to one half again.  So at the end of the day, I prove that the house of measure at any dimension less than  one half of the singular set is finite.  So that gives me that gives me the the house of dimension as an answer.  All right, so how does one prove this?  Well, first you do a scaling, right, like, like so.  So you introduce fn of t and v to be epsilon n to the gamma f of one plus epsilon to the  gamma t minus one and epsilon v epsilon and two to the minus n.  So this scaling here will give you a suitable solution to the lambda equation.  This preserves the lambda equation.  What's important to give you a suitable solution to the lambda equation with the same creativity  constant.  That's the thing that is absolutely crucial in this business.  You don't you don't change.  You don't touch the the constant C prime E and okay, so then you define capital Fn in  terms of little fn was the same nonlinearity mu that sort of replaces the the F log F by  the mean of R and R square.  Right.  Okay, so that's not very important.  Anyway, so this assumption to the fact that this limb soup is less than eta one tells  you that there exists a capital N large enough such that this quantity here is going to be  less than more than eta one say eight eta one.  Okay, good.  So then with this, you use the order inequality and the sub-elephant equality as in the proof  of proposition two.  You isolate the term grad V of fn plus one in L two in time Q and V, which is a folder  eta one.  And you show that if you look at xm, there's going to be the soup of integral of fn plus  m to the Q for T between one half and one here.  You don't shrink the interval.  Interval is a waste.  One half one.  I mean in terms of capital F.  And you show that this quantity here satisfies a recursion inequality of this form, right?  Xm plus one less than rho times the max of one and xm to the alpha plus the max of one  and xm minus one to the alpha.  You start from initial conditions x naught and x one less than or equal to capital M.  So capital M is going to be a large enough power of two.  So the M is say the first N for which this is going to be less than eight eta one.  Here alpha is going to be Q over three and rho is some constant depending on Q eta one  to the Q over two.  Okay.  Whatever.  Here Q is less than two.  So the alpha is less than one.  So this is the opposite of the traditional, the Georgie argument.  So when you iterate this inequality here, there's an easy induction that tells you that  at some point all the xm are going to be below one.  Okay.  All right.  So x2m is going to be less than or equal to the max of two rho to the one minus alpha  to the M divided by one minus alpha M to the alpha to the M. Alpha is less than or  equal to one.  So that converges to zero.  So at the end of the day, you find that for some M naught, this is going to be small compared  to one.  And now with this M naught, you scale back to the original variables and you find that  F capital N plus M naught plus three, if I remember, satisfy the assumption in proposition  two.  In other words, satisfies the fact that the truncated entropy at the level one-half on  the time interval one-eighths one is less than eta naught.  And then you find the N infinity bound which you need, which you want.  Okay.  Well, that's it.  All right.  So final remarks.  Well, I mean, this is a partial regularity result in the same style as the Le Réthier  theorem for Navier-Stokes.  Right?  I mean, the Le Réthier paper for Navier-Stokes, you already find the fact, it's not written  like this in this paper, but you already find the fact that a half-storey dimension of singular  times for Le Réthier solution on Navier-Stokes is the most one-half.  Well, in fact, the Devillein, if you look at the Devillein theorem, it puts along that  equation the same class as 3D Navier-Stokes in terms of the back exponents, right?  For this weight, one plus V to the minus 3, which could be important or not.  But I don't know.  Anyway, so if you look at Navier-Stokes, Le Réthier retails you that the solution is  L infinity in time with values in L2 and X, with grad U, which is L2 in time and X.  So here, if you look at Langdorff, square root of F is L infinity in time with values  in L2 and V.  So that's the conservation of us.  And the Devillein theorem tells you that the grad V of the square root F is L2 in time  and velocity except for this weight.  All right.  But if you, yeah, apart from the weight, this is very much in the same number.  These are very much the same exponents.  So it's natural to ask whether you have partial regularity in T and V in the same style as  the Kaffareli and Nihon-Bert.  Why not?  I don't know.  But more generally, you could ask yourself whether you have conditional regularity in  the style of the papers by Serrin and all subsequent papers where you assume some local  Le Bag exponents above a certain critical threshold.  And out of this, you deduce regularity.  So I mean, if you take P equals infinity, so the Serrin criterion, if you copied for  square root of F would be exactly this.  So I don't know if this is true or not.  This is only true for P equals infinity and K large number 5.  So this is a paper by Sylveste in 2017.  So it's also a result by Guadagnan and Gilan in 2016.  And there's recently a result in that direction by my student, Emmanuel Ben-Carrot.  Okay.  Thank you very much.  I think Renjun has some questions in the chat.  I think he says he's referring to the stability in the nonlinear case.  Just did you have any comments on it?  Renjun, do I am asking on the chat?  I was in it a question for Jean.  Yeah.  Not for...  It was a question for Jan.  I thought...  It's a very strange problem.  Sorry.  It's a long one.  So...  Someone else is typing some questions.  Yeah.  Samir is typing.  Samir is typing.  Yeah, I don't know.  I didn't see...  I thought that one was new.  I didn't just realize now.  No, well...  Well, maybe there's some stability for...  Yeah.  Yeah.  Samir wrote there, yes?  Yeah.  So...  Well, unfortunately not because...  I mean, with this technique...  Hi, Samir.  So with this technique...  You know, this technique, somehow you truncate the velocity.  You get rid of the weight, okay?  If you...  Oh my God, I did the wrong thing.  If you return...  So where is that?  So that's, I think, here.  You see, at this step, you remove...  You replace the weight in V as appears in the...  In the devilette inequality.  You trade that for the bag...  The bag exponents slightly less than two.  For some gradient of F to the one of a Q.  Okay, Q is slightly less than two.  So here...  I mean, if you change the...  If you change the potential, what will change is this weight, typically.  Right?  But this weight, you throw it away by this procedure of the propagation of moments.  So I doubt that with this argument...  I mean, I doubt that this argument, you will see the difference between the potential that will be less singular than Q.  Of course, that will be a very nice question.  We know that if the exponent is larger than minus two, I think no problem at all.  You propagate regular IT and all that.  Exactly.  I think there's the answer by the way that...  But it's not the same argument.  This is a totally different argument.  Right?  So unfortunately, these partial regularity techniques will not see the difference between Coulomb and possibly nicer potentials.  Right?  For instance, we don't know what happens between minus two and minus three.  I mean, probably this is the same as Coulomb period.
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Thank you very much. So thanks to the organizers, thanks to Xi and Mihai for organizing this beautiful conference. I'm very sorry that I cannot be in Marseille, but anyway, so it's already very nice to have a conference like this when you're locked down in Paris, even though locked down in a week since. All right, so what I want to report on today is a recent result with Maya Pierre-Gualdany, Cyrille Lambert and Alexi Vasseur on partial regularity for the Landau equation with collo-interaction. And so this is going to appear in the analysis of Ecole Normale Supérieure. All right, so here is the space homogeneous Landau equation with unknown f, the distribution function, that's the function of time and velocity. So the space variable is removed. So f, t and v is non-negative function and the Landau equation, which is probably familiar to most people working in kinetic equations here. So recalled for convenience is d by dt of f equals to the divergence in v of the integral of the R3 of A of v minus w grad v minus grad w of f of tv, f of tw, dw. So you integrate for w running through R3 and this equation is posed on the whole of R3. So here A of z is proportional to the Hessian of the norm of z, the Euclidean norm of z. So in other words, if you compute what it is, so if you put the coefficient to be 1 over 8 pi, so it's 1 over 8 pi the norm of z times pi of z, where pi of z is the orthogonal projection on the orthogonal of the line span by z. So you can equivalently recast this equation in non-conservative form in which case it reads d by dt of f of t and v equals to some diffusion matrix which you obtain by taking the convolution in v of the matrix Aij of z with f, so you take the convolution in v and this multiplies dvi dvj of f of tv and there is a term which is precisely f of t and v squared. The fact that here you have a purely local term is characteristic of the Coulomb interaction. So it's because when you take the second derivative of this, you get the direct measure at z equals 0. So here you have a purely local term which is f squared and of course this is a dangerous term. So if you have here a constant diffusion matrix, this is a semi-linear heat equation, we know this blow up in finite time. But here if you know of course this will promote blow up, but if blow up builds up, I mean if f increases, of course this will make the diffusion matrix increase as well and you can hope that the diffusion matrix will offset the effect of the f squared terms here. So in other words you can hope for a balance between the f squared term here and the diffusion term there. But nevertheless there is an open question as to whether there is global existence of classical solutions or finite time blow up for the Cauchy problem for the Landau equation set on R3. So however in the late 1990s Cédric Villanille came up with a good notion of weak global solutions defined therefore for all time. He called them H solutions because the Boltzmann H theorem plays a big role in the existence of the solution itself. So an H solution in the terminology introduced by Cédric is a continuous function with values and distributions. So it's non-negative function, already it's a measure. And it's also L1 in time with values in the weighted space L1 minus 1 over 3. So already L1 minus 1 over 3 refers to this family of weighted space. What I call LP minus k, well LPk sorry, LPk, so function g is an LPk if g modulus to the p is integrable with the weight 1 plus v squared to the k over 2. And you have this formula for the norm. So here it means that you are integrable with the weight proportional to 1 over 1 plus norm of v. So it's a function in this functional space which satisfies conservation of mass, momentum and energy with respect to the initial data and which satisfies the H inequality, the fact that the H function at time t is less than or equal to the H function initially. And the equation is satisfied in the following sense. The weak formulation of the Landau equation is the integral of f in of v times phi of 0 and v dv. That's the integral from 0 to t of the integral in v of f times dt phi is equal to this term that involves the entropy dissipation integrand that you find in the dissipation formula for the H function here. So here you have the integral from 0 to t, the integral in v and w of the difference of capital phi at v and at w, where capital phi is the grad of little phi, the test function. In the product with the projection on v minus w of the integral capital F grad v minus grad w of capital F, where capital F is the square root of f of v f of w divided by v minus w and there is a phi that is floating around. So this is the weak formulation of the, this is the weak formulation which Villani proposed to define global solutions to the Landau equation. Good, nevertheless for the purpose of proving partial regularity, we need more information on the solutions but fortunately with the same procedure by which Villani constructed global H solution, the same procedure, the same approximating scheme will lead, will result, will produce what we have called suitable solution by analogy with the paper by Kafferli Konean-Wer for partial regularity on Navier-Sos. So what is a suitable solution of the Landau equation? So it's a solution which satisfies the following truncated entropy inequality. If I look at the relative entropy at the level kappa, which I define as h plus of g and kappa to be the integral of kappa times little h plus of g over kappa dv, where h plus is the function generating the h function. So it's z log z minus z minus one truncated, so truncated for z above one, right? So if you want this is z log z minus z minus one times the indicator that z is larger than one. Okay, all right, so now h plus of f times t2 with respect to kappa plus some dissipation term which involves some constant c prime e integrated between t1 and t2, the integral from t1 to t2 of the LQ norm of the grad v of f to the one over q where f is larger than kappa. So you take the L2 norm in time of this thing to the square dt should be less than or equal to the truncated entropy at time t1. And there is an additional term which is proportional to kappa times the integral from t1 to t2 and the integral of all v of f minus kappa plus. Okay, so if you take the approximation scheme proposed by Villani to construct h solutions and you pass through the limit and you pay a little more attention to the properties of the sequence, then you arrive at this inequality which is satisfied for all time in on the half line except possibly a negligible set of times, okay, which is satisfied for certain values of q which I'm going to explain in a minute and with a certain dissipation constant which I'm going to comment on in a minute also. All right, so okay, now partial regularity in time has to do with the fact that you want to say that the set of times where the solution possibly becomes singular, there's no longer a classical solution, whatever you want to call it, is going to be small in some sense. So here we call a regular time of f, a suitable solution on some interval of the half line, it's a time tau in that interval, such that f is in L infinity on the time interval tau minus epsilon tau for some epsilon and for all these. So here we localize only in time but this is global in velocity. So this is the set of, this is singular times, so the set of singular times, in other words, the times which are not regular times in the interval i with a note by s of fi, okay. So now if you take suitable solution to another equation on some time interval cross r3 for all positive t, I'm sorry, I mean on the half line cross r3 with some initial data f in, so the initial data has a decay to any order in V, right, so it has fast decay integrating in V, it has bounded initial entropy, right. And in that case, you can prove that the house of dimension of the set of singular times for any suitable solution which has this f initial as initial data is at most one half, okay. All right, so let me say a few words, let me explain how the proof is organized. Okay, first, first we need, we need the global existence of such suitable solutions. So as I said, the suitable solutions that are constructed by using the same approximation scheme as was used by Cedric Villani to construct H solutions. And if you do that, here's what you get, if you take any f in which is in L1, non-negative on r3, and which is integral ball against the weight of normal V to the k, which has finite entropy initially, then such an f initial will launch a suitable solution with that initial data. With a Q, so if you remember in the definition of suitable solution, there is a, there is a Lebesgue exponent in the dissipation here. So with some Q that is related to the decay in V of the initial data, so specifically the Q which you observe in the dissipation term is related to the decay by this formula here. So Q is equal to 2k over k plus 3. And the constant C prime E, that depends, that's the dissipation constant in the suitable solution inequality, depends on capital T, depends on Q, and depends functionally on f initial. Okay. And okay, there is always some negligible set of times which you don't care about. Okay. Good. So this is the existence theory for such solutions, which as I said are constructed as the, by the same approximation scheme as was used by Villani for H solutions. Now if you look at the, if you look at the entropy production term that comes from the Boltzmann H theorem, this is a non-local term, but there is a rather recent and very important observation, very important theorem by Deville-Etts, which allowed replacing this non-local dissipation integrand for the Boltzmann H theorem applied to the Lander equation by purely local terms. So here's the, here's the Deville-Etts theorem. So if you take a function f, which is integrable and has finite energy, right, so L12 means you can integrate with respect to 1 plus d squared dv, right, and which has a finite entropy. So f log f is in L1. Then if you look on the right-hand side here, this quantity is exactly the entropy dissipation integrand which appears in the Lander equation when you do the Boltzmann H theorem. Well the Deville-Etts result is that up to some constant, which you can compute explicitly. This is bounded below by the L2 norm of the gradient of square root of f with a weight in v, right. So this is a Fisher information if you want with the weight of the order of v to the minus 3, okay. And again, so the constant cd that appears here depends explicitly on the mass, momentum, energy, and entropy of f. And therefore if you apply it, you know, this is an inequality which is true for all f, okay, you can apply it of course at each time on a solution to the Lander equation. The corollary of this theorem in the paper by Deville-Etts, which appeared I think five years ago, is that you have propagation of moments for a Lander. So if you assume that f initial is in L1k, so it has finite k finite moments with k larger than 2, for 2 which is obvious, this is the conservation of energy. Then and if in addition f in as a finite entropy, then if f is an h solution with this f in as initial data, then f is going to be L infinity over finite time with values in L1k. So in other words, you propagate k moments in finite time, okay. So maybe if you let t go to infinity, this is not going to be bounded in that space, but for finite time you propagate k moments, okay. Good. So with this, let me look at the truncated h theorem which is used in the definition of these suitable solutions to Lander. So you want to compute d by dt of the truncated entropy truncated at the level kappa. Of course, if you don't truncate, this is a standard h theorem. If you truncate at the level kappa, there is one piece that is dissipative, which I denote by d1, which involves essentially the same dissipation integrant as would appear for the standard h theorem in the case of the Lander equation, except that here instead of having grad v of f over f minus grad w of f over f to the square, I have the grad v localized where f is larger than kappa, and here I have the grad w localized where f is larger than kappa to the square integrating in all variables. All right, so that's the dissipation term, and this is the nice part of the equation, but of course, this truncation gives rise to other terms which are not nice and which we put on the right hand side. Nevertheless, these terms, however not nice, can be computed explicitly. If you compute them, you find that you have something of the form minus the integral of a times the grad of vf, where f is larger than kappa, times the grad in w of f, where f is less than kappa. You can integrate by parts in v and w, and you let the derivatives bear on the collision kernel. If you do that, that will be the integral of minus the divergence in v, the divergence in w of the tensor, a of v minus w. In fact, if you compute it, this is equal to the Dirac measure at v equals w, and here you have f of tv minus kappa plus times kappa minus f of tw minus kappa minus. In the approximation process, this is not going to be identically the Dirac measure, but this is at least going to be non-negative. This extra term here is less than or equal to kappa, which comes from here, times the integral of f minus kappa plus. Now we find that if you want, this is a depleted non-linearity. We call that depleted non-linearity because if you return to the non-conservative form of the Landau equation, which is written here, if you multiply both sides of the equation by log of f, and if you integrate, here you would observe something that grows like f squared log of f, which would be very bad. But nevertheless, if you look at, if you do this computation here, then the non-linear term that comes from this Dirac measure that appears when you take the double derivative of a, it doesn't grow like f squared log of f. It grows like f, it's kappa, the integral of f minus kappa plus, which is tamed as a non-linearity compared to the entropy. So this is good news because this is going to help us controlling the equation. Unfortunately, not completely. That's why we have partial regularity and not full regularity. Good. So now, as I said, the sketch should prove a proposition one, the existence proof. Well, I mean, you truncate, you replace a collision kernel of Landau with a truncated variance. In other words, you truncate one of the z at a level n. Right? You check that it satisfies this inequality here. You use the Deville-Let's theorem to bounce the dissipation plus the integral of f minus kappa plus below by this fissure information with the weight proportional to the v to the minus three. And to remove the weight, you use the Deville-Let corollary, so in other words, the propagation of moments, to remove this decaying weight at the expense of lowering the power that appears here. So by further inequality, you control this expression here below by the gradient with respect to v of f to the one of a q, where f is larger than kappa. And here, q belongs to the interval one, two. So q is always less than two. And you have this further inequality here. And you enter that in the truncated entropy inequality here. And you get the announced inequality in the definition of the suitable solution. Good. All right. So now we have suitable solutions. How is that going to help us proving partial regularity? Well, this is done by the DeGeorgi method. I mean, actually, the first part of the DeGeorgi method, which consists in gaining L infinity out of the energy or L2 bound. So here is how it works. So you take let f be a suitable solution to the Landau equation with some coarsivity constant c prime e positive. And here I'm going to assume that q is less than two, but at least 6 fifths. You need something slightly larger than one. And specifically, you need 6 fifths. All right. So that's technical. It doesn't matter. What is important is that q should be free to approach to as much as possible. What is interesting is when q is very near two. Okay. Well, so for such an f, then there exists an eta naught, which depends on q and on that constant c prime e positive, such that if the entropy of f truncated at the level of one half integrated in time between, say, one eighth and one is less than eta naught, then f is going to be bounded by two, right, is going to be bounded by two on a smaller time interval, say on one half one cross R3. So for time between one half and one and for all v. Almost everywhere. Good. So how does that work? Well, I mean, this is a standard, this is a standard, the Georgie procedure for power equation. So you pick a sequence of times which grow from one fourth to one half. You pick a sequence of levels which grow from one to two. And then you replace, you replace the entropy by this nonlinearity here, which if you want is comparable to the entropy. So you look at fk plus of T and v to be mu of f to the one of a q minus kappa sub k to the one of a q positive part. And out of this, you construct the quantity ak, which is the supremum of the integral of fk plus to the power q dv for T between tk and one and the integral from tk to one of the grad v of fk plus in lq norm in velocity, you raise that to the square and you integrate in t. Okay, so this is the quantity ak, which on which you apply the Georgie procedure. So you control this by using the truncated entropy inequality, which is characteristic of suitable solutions. And you can show that this ak is at a size and inequality of the form, ak plus one controlled by some constants. There is something that grows exponentially fast, some lambda to the k. What is important is that you have ak to the beta and here beta is larger than one. So provided that a naught is small enough, then ak tends to zero as k tends to plus infinity. So you control this a naught by the truncated entropy and you conclude by Fatou Lemma that since ak tends to zero as k tends to plus infinity, this tells you exactly that f is going to be less than or equal to two on that interval here corresponding to the level at which you truncate for k going to infinity here and the interval that you obtain for k going to infinity there. Okay, very good. So that's the standard Georgie arguments, but in the language of the lambda equation. All right, so now this is not enough for partial regularity and we're going to now apply another argument but at the same nature, but now we have truncated and zoomed on the values of f, on the set of values of f. This is absolutely characteristic of the Georgie method to look at the level sets of f. Now we're going to do some scaling on the function on the solution to the lambda equation. And so here's that first, suppose that you have f as suitable solution to the lambda equation on some interval 0, 1 say, and with an exponent q, again close enough to 2 by I mean here, here q has to be just between four thirds and two four thirds is higher than six fifths. But okay, so all right, so but still less than two. With this q define gamma to be five q minus six over two q minus two. What I'm saying is that there exists parameter eta one, which depends on q and on the dissipation constants in the truncated entropy inequality. And there exists a delta one, which is zero and one such that if the lim soup of the integral of the Lq norm to the square of the grad V of f to the one over q, where f is larger than epsilon to the minus gamma, which you integrate between one minus epsilon to the gamma and one, right, which you multiply by epsilon to the gamma minus three. So if you assume that this lim soup is less than eta one, then f is going to be an infinity on the strip one minus delta one one cross R three. Now, if you have this proposition proposition three and vitally type covering arguments on the half line, it's classical to deduce from that, that if you look at the house of measure three minus gamma over gamma dimensional house of measure of the set of singular times, this is going to be finite. Okay, three minus gamma over gamma. This is equal to q over five q minus six if you do the arithmetic here. And the main theorem will follow by translation. Okay, because here I just I have localized in a subinterval not a half line and letting q going to two, right. So I remind you that q cannot be equal to two because I've used this, the devilette propagation of moments to remove the weights in the in the Fisher information, the vanishing weight at infinity. So I've used a little bit of q. So q cannot be exactly equal to two. But nevertheless, you can let q going to two converge to q to two. And for q equals two, you see immediately that this is equal to one half again. So at the end of the day, I prove that the house of measure at any dimension less than one half of the singular set is finite. So that gives me that gives me the the house of dimension as an answer. All right, so how does one prove this? Well, first you do a scaling, right, like, like so. So you introduce fn of t and v to be epsilon n to the gamma f of one plus epsilon to the gamma t minus one and epsilon v epsilon and two to the minus n. So this scaling here will give you a suitable solution to the lambda equation. This preserves the lambda equation. What's important to give you a suitable solution to the lambda equation with the same creativity constant. That's the thing that is absolutely crucial in this business. You don't you don't change. You don't touch the the constant C prime E and okay, so then you define capital Fn in terms of little fn was the same nonlinearity mu that sort of replaces the the F log F by the mean of R and R square. Right. Okay, so that's not very important. Anyway, so this assumption to the fact that this limb soup is less than eta one tells you that there exists a capital N large enough such that this quantity here is going to be less than more than eta one say eight eta one. Okay, good. So then with this, you use the order inequality and the sub-elephant equality as in the proof of proposition two. You isolate the term grad V of fn plus one in L two in time Q and V, which is a folder eta one. And you show that if you look at xm, there's going to be the soup of integral of fn plus m to the Q for T between one half and one here. You don't shrink the interval. Interval is a waste. One half one. I mean in terms of capital F. And you show that this quantity here satisfies a recursion inequality of this form, right? Xm plus one less than rho times the max of one and xm to the alpha plus the max of one and xm minus one to the alpha. You start from initial conditions x naught and x one less than or equal to capital M. So capital M is going to be a large enough power of two. So the M is say the first N for which this is going to be less than eight eta one. Here alpha is going to be Q over three and rho is some constant depending on Q eta one to the Q over two. Okay. Whatever. Here Q is less than two. So the alpha is less than one. So this is the opposite of the traditional, the Georgie argument. So when you iterate this inequality here, there's an easy induction that tells you that at some point all the xm are going to be below one. Okay. All right. So x2m is going to be less than or equal to the max of two rho to the one minus alpha to the M divided by one minus alpha M to the alpha to the M. Alpha is less than or equal to one. So that converges to zero. So at the end of the day, you find that for some M naught, this is going to be small compared to one. And now with this M naught, you scale back to the original variables and you find that F capital N plus M naught plus three, if I remember, satisfy the assumption in proposition two. In other words, satisfies the fact that the truncated entropy at the level one-half on the time interval one-eighths one is less than eta naught. And then you find the N infinity bound which you need, which you want. Okay. Well, that's it. All right. So final remarks. Well, I mean, this is a partial regularity result in the same style as the Le Réthier theorem for Navier-Stokes. Right? I mean, the Le Réthier paper for Navier-Stokes, you already find the fact, it's not written like this in this paper, but you already find the fact that a half-storey dimension of singular times for Le Réthier solution on Navier-Stokes is the most one-half. Well, in fact, the Devillein, if you look at the Devillein theorem, it puts along that equation the same class as 3D Navier-Stokes in terms of the back exponents, right? For this weight, one plus V to the minus 3, which could be important or not. But I don't know. Anyway, so if you look at Navier-Stokes, Le Réthier retails you that the solution is L infinity in time with values in L2 and X, with grad U, which is L2 in time and X. So here, if you look at Langdorff, square root of F is L infinity in time with values in L2 and V. So that's the conservation of us. And the Devillein theorem tells you that the grad V of the square root F is L2 in time and velocity except for this weight. All right. But if you, yeah, apart from the weight, this is very much in the same number. These are very much the same exponents. So it's natural to ask whether you have partial regularity in T and V in the same style as the Kaffareli and Nihon-Bert. Why not? I don't know. But more generally, you could ask yourself whether you have conditional regularity in the style of the papers by Serrin and all subsequent papers where you assume some local Le Bag exponents above a certain critical threshold. And out of this, you deduce regularity. So I mean, if you take P equals infinity, so the Serrin criterion, if you copied for square root of F would be exactly this. So I don't know if this is true or not. This is only true for P equals infinity and K large number 5. So this is a paper by Sylveste in 2017. So it's also a result by Guadagnan and Gilan in 2016. And there's recently a result in that direction by my student, Emmanuel Ben-Carrot. Okay. Thank you very much. I think Renjun has some questions in the chat. I think he says he's referring to the stability in the nonlinear case. Just did you have any comments on it? Renjun, do I am asking on the chat? I was in it a question for Jean. Yeah. Not for... It was a question for Jan. I thought... It's a very strange problem. Sorry. It's a long one. So... Someone else is typing some questions. Yeah. Samir is typing. Samir is typing. Yeah, I don't know. I didn't see... I thought that one was new. I didn't just realize now. No, well... Well, maybe there's some stability for... Yeah. Yeah. Samir wrote there, yes? Yeah. So... Well, unfortunately not because... I mean, with this technique... Hi, Samir. So with this technique... You know, this technique, somehow you truncate the velocity. You get rid of the weight, okay? If you... Oh my God, I did the wrong thing. If you return... So where is that? So that's, I think, here. You see, at this step, you remove... You replace the weight in V as appears in the... In the devilette inequality. You trade that for the bag... The bag exponents slightly less than two. For some gradient of F to the one of a Q. Okay, Q is slightly less than two. So here... I mean, if you change the... If you change the potential, what will change is this weight, typically. Right? But this weight, you throw it away by this procedure of the propagation of moments. So I doubt that with this argument... I mean, I doubt that this argument, you will see the difference between the potential that will be less singular than Q. Of course, that will be a very nice question. We know that if the exponent is larger than minus two, I think no problem at all. You propagate regular IT and all that. Exactly. I think there's the answer by the way that... But it's not the same argument. This is a totally different argument. Right? So unfortunately, these partial regularity techniques will not see the difference between Coulomb and possibly nicer potentials. Right? For instance, we don't know what happens between minus two and minus three. I mean, probably this is the same as Coulomb period.
homogeneous Landau equation with unknown f, the as to whether there is global existence of classical solutions or finite time blow up for that the house of dimension of the set of has this f initial as initial data is at most one mean, this is a partial regularity result in the dimension of singular times for Le Réthier