 Okay, so what I want to do is show you how to prove global regularity in 2D given everything that we've done so far. And then I want to talk about continuation criteria in terms of norms in 3D and show you what we know about that and how to prove it in some cases that are easier to explain. Okay, so what's left to prove after everything that we've discussed is that k in L infinity is finite. So in 2D, so what we really need to prove the continuation criteria that I talked about last time was that the integral, the time integral of k along characteristics needs to remain finite for all fixed times. There should be a constant such that it's finite. That constant can definitely depend on the time. And then you have a solution that can be continued slightly beyond that time. It turns out in 2D and also in 2.5D you can just prove that k is in L infinity. And this is a fact that you can prove if you work hard. In 3D you can't prove this. You have to actually really use the integral of k along characteristics. We'll talk about that later. So we need two estimates initially. Okay, let's say this is one quantity that we need to bound. Somehow, we want to show that this is completely bounded. The other quantity is as follows. Okay, so we want to estimate these two integrals. And here kg can be thought of as... So you can take these decompositions that we talked about in the last lectures for the fields and split them up in a convenient way and prove point-wise estimates for the kernels and get a complicated expression. And I'm claiming that initially what we want to do is estimate completely the two integrals like this. So is it clear that that kg is better than e to e? Those components they have to look like that. Okay, so if you remember from the first lecture, these components are controlled by something that I called the good conservation law. And so that's what you do. You basically get rid of kg by using the good conservation law and you can... The part is good and now you're trying to control the rest. Yeah, okay, so this is just the first step. Right, this part is good so we can get rid of it and throw it away. And then there are other things we need to control. So this is just the first estimate. So the main difficulty here is that this term is nonlinear. It involves kg times f and it has singularities. It has two order growth in p from this singularity and then that's killed by one order growth in p. So you have a first order growth here. Up there you have a... also you have a first order growth. It's linear but you also have this extra t minus s hanging around. And the main problem is that you want to control this by constant. You don't want to control it by higher order norms. So to control these singularities you split. So I claim you can prove a point-wise thing. Okay, so I proved this p0 squared estimate yesterday. You can also prove a couple of other estimates and these are basically the three regimes that you worry about when you're close to the boundary of the cone. That's this one. So remember that c is x minus y over t minus s. And this one is just the other case where you can be big when you worry about the size of that angle. So you split these integrals in terms of the size of either p0 squared or the size of 1 minus c. So you can even split this in terms of the size of p0. And then you use the good conservation law so you can control these components by a conservation law and throw that away. And then you can control f after doing, after the splitting by the conservation laws. And you can prove that these two integrals are finite by this idea. Okay, and then you have an estimate. So this, once that's done, so once, so this is, okay, so being able to control these two things is one of the things that you can do in 2D that is harder to do in 3D and that's part of the problem. But once those things are bounded, then you can prove point-wise estimates for k. So k is the fields and it's expressed in terms of the ST decomposition. And then once you bound these quantities, you can show that this is less than or equal to box inverse Epsilon B integral over R2. Okay, so this is what you can prove. You can, you can work a little bit and, oh, I should say one other thing. So always, okay, so for everything I'm going to say today, box inverse f represents the solution to box u equals f with zero initial data. So it's the solution to the wave equation with zero initial data. It's just an integral. So you just use the integral formulas. They're hiding in these expressions up there. I wrote it on, I put it on the slide yesterday. So once these two quantities are bounded, you can work a little bit and get a bound like this where this k zero parentheses is a integral of the initial data that I won't write down. It can be bounded by your assumptions on the initial data. And then you get box inverse Epsilon U of B times the integral of R2 of f over P. And then you get Epsilon, you can get an arbitrary Epsilon to the minus one-tenth times the integral of P zero f dp to the two-fifths. And then you can also get an Epsilon to the three-tenths. What? No, I'm missing a box here. That's for sure. This is not constant. It depends on x and t. Okay, so you can prove an estimate like that. And so now it's also, it's inside the two-fifths is outside. Okay, so you can prove an estimate like this. And now you're ready to use a Stryckart's estimates. And using Stryckart's estimates, you can get... So you're thinking of this as a function of t and x. Yes. And you are right to bound it. I mean, you are using those formula, right? We have already used those. So you have a very complicated expression for k in terms of... For E and B using the ST decomposition, it's expressed in terms of integrals of f and... You are thinking of k as E and E and to solve them, you have to infer on something that you are bounding by these things. Yes. So how, I mean, and the epsilon, how does the epsilon come in? You just put it in and you can... How does it come in? What do you mean? Do you mean, how am I going to use it later? Or do you mean, how do I get it in the first place? You cut in terms of epsilon and you calculate integrals and you get that these quantities are finite and what's left is an epsilon. It's complicated. It's over 100 pages of paper and there's a lot of interpolation and a lot of cutting integrals into several different small parts and there's a lot of little epsilons and deltas everywhere and then you, but basically here you just, you cut the integrals and you do splitings like this and you calculate the, you calculate exact integrals of singularities after using holders inequality with carefully chosen exponents and you get this, something like this times an integral which is an epsilon. Okay, but if I explain that carefully it would be my whole lecture. So we want lq is control, okay. So use, so now we can use a strict card estimates. In fact, use the improvement of the strict cards due to, there's an s there, what? Yes, that's coming, but not yet. So, okay, so I decided to give you the precise estimate in one special case in 3D, not in 2D. I can give it to you also in 2D, but it's going to be ugly. Okay, can you make the difference between this one and the standard? Okay, so there's a, for, what kind of estimate do you do? It's strict cards, it's just, no, but everybody knows strict cards. Oh, so you want me to state this, strict cards? Yeah, let me do that. So, okay, yeah, let me grab something else. So I get this right. Okay, so, okay, so basically there are well-known family of estimates called strict card estimates and foski improve the range of exponents that you can use for the problem that we're considering that I wrote down over there. So for the forcing term with zero initial data, you can get a larger range of exponents. Yeah, this is better if I write over here. So then you consider you, you can write down the solution formula, and then you can say that, okay, so this is the strict card estimate in, well, it's actually still quite general, but if u is the solution to the wave equation with the forcing capital F with zero initial data, then it's written in with this solution formula, and then you can prove an estimate that reduces the Soblov space, sorry, the LPLQ space. So you have a ULQ1 in time, LR1 in space is less than or equal to F LQ2 prime, LR2 prime in space. And then, and these are the assumptions are, yes, there is awful. So, okay, now you see why I was trying to sweep this under the rug. So this is your strict arts estimate and these and you can prove it in any range, satisfying this equality for Q1, R1 and Q2 prime, R2 prime, and these two inequalities and Q1 and Q2 and R1 and R2 in this range, where R2 prime and Q2 prime are the holder conjugates of R2 and Q2. Oh, yeah, yeah, I copied the wrong one. Um, sorry. So this is two and this is two and this is, this is, oh, I can allow one here and there's another condition. So, okay, so you can choose a good collection of exponents. I, I couldn't think of a good way to actually motivate the exponents you choose in 2D and I do motivate it and when I do 3D, so I just left that part out. But, okay, so I need to speed up. So then you can apply strict arts estimates and you obtain something like K is in, you obtain a bound for K and now QTLRX. Okay, so now we want to use this bound to, so, so without trying to go into indices, but just to understand what you are doing. Yes. So here you took this, this estimate. Yes. And you want to apply to, each time we have this box minus one, you want to apply some estimate like that. Yes. And then, so normally you get K here less or equal than K or, I mean like the same bound and then you want to close it. This is the, this is the next thing I'm going to talk about. So, so you are not assuming this. You are going to show us this, right? We aren't assuming anything. This is 2D and we can prove global existence for large data. Or once you have this estimate, then you can get some control on K in these spaces and then we're going to use it and you'll understand how I'm going to use it momentarily. That's my question. How do you get this control? K in LQLR? From here. Using those. Using Stryckart's estimates. Okay. So you, you just slap an LQLR norm everywhere and a bounded above using the triangle inequality. And then choose the right and then. Yes. Yes. This is my plan to talk about this stuff. Okay. So bound moments. So the next thing to do is bound moments. So, so if you multiply the Vlasov equation by P0 to the n, then you get df dt plus p hat dot grad xf plus k tilde dot grad pf equals 0. k tilde, k tilde is the Lorentz force. So you get this equation. You integrate it. You integrate by parts and you get, okay. So you can do this. So you want to bound the moment. So you take the Vlasov equation, you multiply by P0 to the n, you integrate in x and p and you can get an inequality like that after integrating by parts in x gives you 0 and in p that reduces the power of P0 to P0 to the n minus 1. And then the game is to control this. And this is the same one. Yeah, it works. Okay. So then Holder gives you that the right hand side is less than or equal to P0 to the n minus 1 f l n plus 2 over n plus 2 minus 1 x l 1 p k in l n plus 2 x. Okay. So by Holder you can do that. Let me not explain it too much. More than that. And then you have interpolation. You have the following interpolation inequality. Actually, let me say this. Okay. So you have this interpolation inequality that I may want to refer to again later. So I wrote it over here and I want to erase it. So this is actually not so it's such a big deal, but it is something that we couldn't find anywhere. So we proved it by the P0 to the sg in this m plus d over s plus d l 1 is bounded above by a constant that depends on the infinity norm of g and P0 to the mg in l 1 to the power s plus d over m plus d where now d is the P dimension and you can do this in any space or P dimension and m has to be bigger than s. So the idea here is to just write the P integral of P0 to the s and split into a large part and a small part and then optimize in your splitting. Okay. So then by interpolation this is less than or equal to that coordinate. This is let's call that alpha tilde. I don't know. And then you divide and you get d e t of P0 to the nf of 1 is less than or equal to 1 plus k. Let me just write the final thing. Okay. So you divide and you integrate in time after using this interpolation. I didn't want to calculate this alpha on the fly. So I just called it alpha tilde. So you get a constant that comes from this interpolation inequality. You divide to the other side and then you integrate and you get that the nth moment of f is bounded by a constant that depends on the initial data plus the l m plus 1 to the power m plus 1 norm of the fields. And now you see the answer. Oh, right. I forgot to write that one. Okay. So you integrate in time and you get an l 1 norm in time and l m plus 2 norm in space. And then you so now you see that you can control the high moments of f by the high l p norms of k. And this is what motivates which strict arts estimate you try to use. So you want some high l r control of k. And you don't really care about the q. You can sort of use it as a parameter. It's not such a big deal. You do want one there but you can since we're doing this for arbitrary large times and we're doing it for large data, you can sort of throw away a lot of powers of t and not really worry about it. Yes. So so now the game is to control this and you control it by strict arts norm. Yes, that's exactly right. Okay, so I'm going to make a big jump. Okay, so I'm going to make a big jump. But what can you do if you try to explain this craziness? Then it takes another four lectures. So you use strict arts moment interpolation like this. So this is the tool you use. This kind of thing is a tool you use and you use strict arts. You use the conservation laws and you do a lot of optimization and you choose a good epsilon. That's a inverse power of the nth moment of f. And so you're basically bound in k up there using the point-wise bound for k that I wrote down. And then you're using strict arts and then you're doing a lot of holder inequality and a lot of optimization and moment interpolation and the conservation laws and choosing a good epsilon. And at the end of the day you get four n greater than 13. Okay, this is what you get. So the nth norm of f in L infinity by the moment bounds, when you do all this stuff to the right hand side k term, you get that it's bounded by data times f p0 to the n L infinity l1 to the power alpha times f p0 to the n L infinity t l1 x to the power of a prime and f p0 and lq prime l1 x to the power p. So here q prime is in one infinity and it's strictly less than infinity and that's important. So what we want to do is prove that f is bounded, prove that the high moment of f is bounded. So let me briefly switch to slides and then we can come back here. I start with slides one slide really. Okay so um actually so what can you do? You can uh it turns out that alpha prime is bigger than alpha at least you can assume that for n larger than 13 say and you can also suppose without loss of generality that the p0 to the nth moment of f in infinity is bigger than one because if it's smaller than one than you're done then the whole problem is solved and then you can divide through by f p0 to the n alpha prime in this kind of an inequality and you obtain this kind of an inequality. So f p0 to the n infinity l1 is bounded above by a constant times one plus f p0 to the n one minus alpha prime lq l1 and this so this automatically gives you that f p0 to the n is bounded for large data. If you can get a sub-linear bound it's important that this one minus alpha prime is less than one then it's a homework problem to prove that f p0 to the n is bounded. Furthermore I claim that we got good bounds only one exponentially growing bounds in the theorem so let me sort of tell you quickly how to do that. So if you have this ground wall type inequality it can be proved via a continuity argument so let g0 be positive and a non-decreasing function and support sorry gt and then gt satisfies the inequality is less than or equal to m of t one plus g in lp in time for some p finite then there is a and for some non-decreasing positive function m of t then g of t is less than or equal to m of t times e to the m of t to the power p. You can prove this by a simple continuity argument where you assume the bound is true and you use the inequality to actually get a better bound and that automatically implies that the bound is true. Okay so from this lemma and from the previous inequality we know that f the nth moment of any nth moment of f for n greater or equal to 13 is bounded. Assuming that's true initially and then you go back to the moment bounds that you had before so you go back to the original bound for k this one that i wrote down on the board and you can once you know that the high moments of f are bounded you can use the same this is kind of like this almost circular argument you use this estimate to prove that the high moments are f are bounded then you come back to this estimate and you use that the high moments are bounded to prove that k is impacting ill infinity um let me not sort of explain in detail how you do that but you choose a good l you choose a good l p l q norm like l one in time l two minus epsilon epsilon in space and you you use the stricard's estimates and then you get that the l infinity norm of k is bounded um so in 2d and in 2 and a half d you can actually prove that the l infinity norm of k is bounded you can't prove it and through we don't know how to prove it in 3d so this is this is 2d um so now we have a continuation in criterion 3d involving the boundedness of norms so uh this is what i want this is the quantity that we want to study m theta q where m theta q is p zero to the theta f l infinity time l q in space l one in momentum and we want this to be finite and we would like to say that if m theta q is bounded then the solution can be continued beyond t for certain theta q um so you can if you're familiar with navier stokes you can compare this to a continuation criteria for navier stokes um they have the so-called uh prody lati hanskai aserian conditions that are motivated by scaling where if the navier stokes equation is bounded in l p l q then um your solution is finite there's a long history of studying that and there's a scaling for navier stokes that predicts what uh the reasonable ranges of this of the navier stokes quantity are let me not go into that more than that except to say that uh for uh of lasso maxwell i'm not aware of a reasonable heuristic that's really convincing in the same way the scaling is convincing for what is the uh full range that you can ever hope to uh realistically prove as continuation criteria for this um because if you rescale of lasso maxwell you get lasso boson and it's a different equation and it's known to be globally regular um and so but the known estimates for the conservation laws give as i've shown uh before that p zero f and l infinity l one is finite and this for m theta q is theta q is one one and additionally it is known that uh yeah so you can you also have the l infinity um conservation law and uh then you can use interpolation like what you have here and choose your snm uh carefully and you get that the l infinity and time l four thirds norm in the ball is uh bounded by constant and the constant is doesn't depend on r so you can take our infinity and use the whole space and this is q r equals zero four thirds uh sorry this is theta q equals zero four thirds and uh that's what we know for what is known to be bounded and you can interpolate between these two bounds and get a range of q and theta i won't really worry about that too much um but we are generally still very far away from these bounds um so uh oh the title of this slide is misleading so you want m theta q to be bounded in the first estimate that uh of this form is that i'm aware of is from glassy Strauss and they prove that uh p zero f in l infinity uh x with uh theta equal one is would be a continuation criteria this has the physical significance of being the kinetic energy density and uh so the title of the slide is accidentally misleading um because this continuation criteria actually holds even when you don't have comeback support um you just need uh the solution to point wise decay polynomially so this is an l infinity statement about the decay of f um as p goes to infinity okay so that's the first criteria now you have a whole collection of criteria um so the first one that i knew about is uh due to Pallard who's uh in Orsay and uh he said that if theta is bigger than four over q and q is between six and infinity uh then uh you have a continuation criteria again you can always interpolate with the uh like this with the l infinity bound that's known to get an l one bound um and to get that the l one bound implies so there's what you can prove and then there's what you can say to compare with uh the known thing easily so uh Pallard's Pallard's uh estimate is uh they like to compare to the known thing is q equal one and theta greater than 19 so what you what you want is q equal uh one and theta equal to one so you're 18 away there and then uh these this group from Victoria Canada in 2010 showed that the endpoint case is actually also correct so theta equals zero q equals uh infinity um is actually correct um and uh so then uh Pallard very recently in 2014 uh showed that um on the other side you can find a very uh interesting and nice cancellation in the uh the std composition and knock these uh estimates down to theta equals zero and q equals six so again you have theta equals zero q equals six compared to theta equals zero q equals four thirds is known so you're still far away but it's much better than uh say theta equals zero and uh q equal infinity um so okay and then this is what so all of those previous results are in the range of a compact support so uh we finished this paper in 2014 and uh with non-compact support um you can get theta greater than two over q and q between two and infinity and uh by interpolation uh this would give you q equal one theta equals five so this knocks you down a lot um and then uh there was another paper uh after that by Kunze from uh who's in Germany and he comes uh he's actually as far as I know he's written two papers unblessive Maxwell one was his thesis in the uh in the eighties or the nineties and then he wrote this paper in 2015 uh and uh he but he comes from this old uh school of people uh like Gerhard Rhein and uh Fafel Moser and Horst and Bat from this chain of advisors and students out of Germany and uh he uh proved that he actually beat us in the compact support case so we were doing non-compact support and he went back to he he went back to compact support and he said for q between one and infinity you have this theta if you look at his paper this isn't exactly what he stated but he stated it in a weird way that's hard to compare to what we did so I changed it and for q equal one this requires theta greater than three so he knocked it down to three um so he had some new ideas that were very nice and I'll tell you about later and but he also used some of our ideas like the strict arts estimates and uh he and he used compact support so compact support really saves you um an order of uh one order of p which makes the estimates uh one order of p better um and his proof really uh you can't we tried and you can't really modify his proof to the non-compact support case directly um and but Neil Patel who's a graduate student working with me is finishing soon uh very recently uh was able to get the currently best continuation criteria in this direction so with without compact support he can get q equals one and theta greater than three and so this improves the q equal one theta greater than five and with compact support he can get q equal one and theta greater than two plus three over five so it doesn't look um that much bigger than uh theta greater than three but it's uh for me it's significant because it goes beyond what you can get from the strict arts estimates and i'm still optimistic that you can go further than this with the uh technique so we really thought that uh uh kunze had gotten the best uh continuation criteria that you can get using these methods with strict arts estimates um but it turns out if you go back to strict arts original paper he used a sharp averaging operator estimate for the sphere um and if i remember if i i think the history is uh that he claimed this estimate was true and he didn't actually prove it and later uh but it was correct and later other people wrote down the sharp proof of the sharp averaging operator estimate for the sphere and in case you can use that to uh go beyond what you get with strict arts estimates uh so this is uh what you can do i was going to switch to his blackboard now but uh how much time do i have what do you want to know uh when normally you are 15 minutes okay okay so okay so um we have a lot on the board that can be reused um not that this okay so okay so in 3d uh the estimate that you get uh to about the moments is n plus 3 n plus 3 and uh you have to assume the bound uh the integral from 0 to t ds k of s x of s is less than or equal to c of t um and so uh we start by uh assuming this bound and we we want to uh bound a uh then we have we have this moment uh type inequality and we want to use uh figure out how to control k so basically the strategy is similar to the 2d strategy except one you want to assume a bound on m theta q let's just do m okay i'm going to use this notation a lot so um m1 plus 2 means a number bigger than 1 um so we want to assume this is bounded and use it to show that this is bounded and we want to do that via moment bounds um so oh so ideally what you want is um we want a good strict arts estimate uh to bound k um but what is it okay so uh this is the point why the estimate that you can get in 3d it's worse than what you get in 2d and that's uh part of the problem and you want to uh plug this into the moment bound up there and you want to use a good strict arts estimate um but the ideal strict arts estimate is problematic um okay this is the ideal uh estimate that we want to use um unfortunately it's false there are there are no encounter examples you can't do this um but you can um you can do this okay i'm going to use the notation l2 plus l infinity minus l2 plus so plus means you're bigger than the number there and a minus means you're smaller than the number there infinity minus means you're finite and smaller than infinity um so the ideal strict arts estimate is false in 3d uh so but so that makes your life much more complicated um and um let me give you a flavor of how to estimate um one term so but so you instead of using the ideal strict arts estimate you use a uh almost as good strict arts estimate and then you get you have to estimate everything okay so then uh okay so i should say uh yeah this is i said it okay okay so this is what you get um and this is um so you need to uh so eventually since you'll probably ask uh we want to uh we have this bound one and we have uh the moment bound and so we want to bound k from above by something then we want to bound the moments and then we want to use the moments to bound uh the k and then we want to do all that conditional on the m one plus two quantity being bounded and then we want to use all that together to show that the interval of k along the characteristics is finite um so so let's go um so let's bound the first term so uh by a holder let me just call this one okay uh so what did i do i almost took a k in an infinity and uh to pay the price for that i uh made this two a little bit larger um one in both cases yes okay it doesn't matter um yeah that's probably a typo but uh i'll just put infinity here we ignore the um so in a lot of these types of arguments you don't really have to pay so close attention do the exponent time of course you want to be correct but you can uh at the end of the day what you want is um um oh wait i don't want to put infinity there though i want to put it here okay uh yeah so then um okay yeah so i have this uh moment bound up there p p to the n f is less than one plus k to the lm press three um then i have this bound for k here uh point wise bound and then i use uh the strict arts estimate that i wrote there in k so in the bound for k and then i um get this and then i'm going to estimate these terms are above in terms of the conservation laws and the uh moment bound and other things i'm going to do interpolation and stuff like that and then um you use that to prove that the high moments are bounded and then you prove that uh k is bounded and then uh you have that the m plus two implies that this continuation criteria is true and this continuation criteria in terms of the integral along a k is uh bounded and we proved that last time uh that the continuation criteria is what what is the assumption so here that's and that's the assumption the assumption which is the one above right yeah the integral from zero to tds of k of s along the characteristics is bounded is finite this is the assumption that so that if you assume that you can continue the solution and now we're saying if you assume that it implies the the assumption on m implies the assumption on the integral of k along characteristics so that implies that m is a continuation criteria okay um yeah so basically this condition of the characteristic so first like in the case of compact support yes it's true i mean like having compact support and that keeps compact support yes right so this is somehow a way of generalizing that yes this is a this is a generalization of the compact support assumption that it is continues to be true in the non-compactly supported case absolutely okay um because of time constraints um i would like to uh switch back to slides okay so we have this bound for k and um then you plug in the uh uh the strict estimate that i mentioned and then you get uh in two terms that you need to estimate one of them you can estimate uh using holder and the conservation law so you get that this first term is bounded because it's in l2 so you can get this exactly in l2 with an exponent that's less than one and then you can get this which is with something that's finite but large with an exponent that's close to zero and then you can get uh this leftover part by a holder basically um oh and then using and then this leftover part that i mentioned to you uh just a second ago can be bounded using the uh interpolation inequality by oh we are in the wrong place the same stuff uh sorry so uh yeah similar story but uh that was the q a because infinity case so uh similar sorry we need to bound these two terms and uh using and you do it using holder's inequality so first you and you get k is l1 l infinity and this is uh that should be an l infinity there l2 something that's bigger than something that's bigger than two and then for k uh l1 l infinity you bound it by the l2 conservation law and something uh that's like l infinity a little less and then since f is in l infinity uh you can also increase this a little bit to get that a p0 fdp l1 l2 you increase the power a little bit and you reduce the l2 a little bit and you have a l infinity that's hide of f that's hiding in the constant and then you can bound this by m theta two which is the quantity that we're assuming is finite and you get this less than one so collecting all of this you get that this term which was the first term we wanted to estimate is bounded above by constant times an exponent that's less than one two plus l infinity minus there's two plus we don't really care about the l infinity minus is important and the sublinear bound one minus is important um so then you can bring the k to the other side um okay let me be quick here so then for the other term you can do a bunch of interpolation and bound it by m theta two to some power that you don't care about times the p0 to the power n to the power alpha m plus 3 l1 l1 um and basically you do that by using a holder and this interpolation again so we're using this interpolation a lot um and then um what you get is this inequality 10 here so you get k l2 plus lm plus 3 is less than or equal to p0 to the nf to the power alpha over m plus 3 l1 l1 so here we choose in all the previous things we choose infinity minus to be m plus 3 where m plus 3 is a high moment and then uh the important thing here is that this alpha is strictly less than one so then uh this we plug in the standard moment bounds that I stated before p0 to the nf l1 l1 is bounded by k lm plus 3 and then you plug in uh so you have the l1 here but you can trivially go up to l2 plus paying with it by some power of capital t that you don't care about and then you get that p0 to the n in l1 is bounded by p0 to the n in l1 times the power which is smaller than one and that is a sub linear bound so it tells you that p0 to the n is less than one and it tells you to that k l2 plus lm plus 3 is also less than one uh less than a big constant which you can approximate to be one uh this is unfortunately this not good enough for k so our goal is to show that m theta 2 finite implies k uh the integral of k along the characteristics is finite and in the uh 2d and 2 and a half d case you can just directly prove that k is in l infinity in the 3d case which implies the conninguration criterion 3d you don't know that we don't know how to prove that um but what you can do is use a change of variable uh due to uh pelard uh so pelard didn't prove this estimate but he he proved related estimates uh that uh and he did a change of variable which is what the basically what we use um that so actually in 3d you really have to use prove that the integral of k along the characteristics is bounded you can't just prove that it's uh k is bounded without caring about the characteristics this is really a statement along the characteristics so you say that the integral of k along the characteristics is bounded by the l4 norm in x and the l1 norm in p of k f to the p0 and uh then you also have another term f to the p0 and l4 l1 um and then uh because you have high moment bounds for f you can prove that the quantities above are bounded by uh basic um holder and things like that um okay so i tell you um pelard's uh inequality so you you uh have integrals like this and then you end up with this map pi which is the characteristics time that's prime minus sw omega where omega is on the sphere and this is a uh this turns out to be a c1 diffeomorphism and it has a nice tucobian that you can calculate the tucobian is singular but under uh certain conditions on the powers p its integral is finite and uh so that's why you do you just have integrals like this and you uh use uh this tucobian uh so the what's really going on here is that we're converting in 3d the integral on uh the boundary of the cone to the integral on a solid region so then you can uh really use the lp norms in the whole space um otherwise it's hard to do that um and this is the only thing you can do and in 2d you can do this uh almost arbitrarily because the way the inverse of the box is uh almost arbitrary and then i'm and then uh in 3d you have to be more careful and you get restrictions on the lp norms that come up that come out of when this uh jacobian integral will be finite when you do something like a holder okay um so the last thing i want to tell you about is neil patel's improvement this would be quick so uh kunze introduced this quantity so he changed uh the kind of things that you study and he said well you can study sigma minus one and you can just isolate this and prove estimates for this quantity and uh streamline everything and then uh patel further also used a similar quantity where you have a power one half here instead of one it calls the capital sigma minus one and then you can uh with some work streamline the field estimates and you get uh you split them into ks one ks two and kt where kt is less than uh w two sigma minus one ks two using the good conservation law is less than w two sigma squared minus one to the one half and ks one is less than box inverse absolute value of k capital five minus one um and w two is uh this integral of the average on the sphere so and this is uh this is an averaging operator um and this so you and then uh you have this strict arts estimate sharp estimate known to be sharp estimate for the averaging operator on the sphere and you can generalize that to uh an estimate for the w two operator directly and this streamlines a lot of stuff and it holds for a range of exponents that i won't punish you by explaining and then uh there's a there's a new argument to prove that uh if the three plus epsilon's moment is bounded then the five plus epsilon's moment is bounded and from uh my favorite was jonathan luke uh the five plus moment is is bounded is a continuation criteria therefore the three plus moment three plus epsilon moment being bounded is also a continuation criteria and that's how you uh reduce um okay so thank you very much um and uh do you expect global systems for general data or some kind of singularity? oh so um that's a very good question and um i like to answer that question by saying uh i don't know i mean any i saying that anything that i will give you an answer but let me give you a slightly longer answer um the uh in general you don't know and uh anything any uh comment that i make is just speculation that having been said uh for this equation when gamma equals plus one at least uh there's uh no physical argument that i'm aware of to uh produce singularities and uh you don't if you look in the physics no one uh gives an argument to produce singularities there's a lot of things like stability instability there's a huge i didn't talk at all but there's a huge number of steady states um that you can uh talk about and uh the the big problem is that the uh when you do things this way the uh the estimates that you can prove with the techniques that we have so far at least and that we've been able to figure out uh give you a gran wall that doesn't say that anything is bounded um so that doesn't help but i give you you have to reduce the powers somehow uh if you want me to speculate yeah i think this is globally regular for the gamma equal plus one place in uh 3d yeah i think that uh it's probably going to be regular okay minus it could be for minus uh is not as uh obvious i mean so uh for the lasso poisson case uh either plus or minus you have uh global regularity except in the relativistic case which cannot be derived from this model then when in the minus one case uh you have uh singularities and i mentioned the papers that study that on monday um okay thank you uh other questions this is more kind of a global question yes um so you said that these um max of blossom equations describe plasmas yes like heavily charged particles so um is there some kind of a connection with the landau equation which also describes um yes oh that's that's a very good question so is there a connection between vlasa maxwell and uh the landau equation from plasma physics i've actually studied the landau equation quite a bit um in my life and it is a uh very hard uh to view non-local diffusion equation where the uh diffusion coefficient is non-local and singular and uh that is uh one of the few examples in the especially homogenous bolzmann equation theory where you still don't know global regularity for larger data and it's a really important open problem in my opinion um okay so uh actually if you go to landau lift sheets um there is a very very non-rigorous um derivation of the landau equation from uh lasso of poisson that uses uh you just uses uh all sorts of uh approximations that aren't justified and in addition uses uh contour uh integral techniques that are only valid for analytic functions um even though you're working with arbitrary unknown functions that you don't necessarily understand but and and also uh in the the derivation of the landau equation is a little bit um hard to justify because it involves a logarithmic infinity that uh landau said oh it's a logarithm who cares um but uh the physical constant in front of the diffusion coefficient in the landau equation is actually a log that goes to infinity um that you just cut off um and there are also um difficulties in the assumptions that you use physically to you make uh you make an assumption on the particle on the two particle interactions when you even heuristically derive the landau equation from an underlying model of dynamics of many particle dynamics uh that uh you also have to think about carefully so um given all the caveats yes there are connections between these two equations um another connection is that uh you can as i have done and other people have done you can study uh lasso poisson landau and lasso maxwell landau which is the lasso equation instead of zero on the right hand side you have the landau collision operator on the right hand side the landau collision operator is thought to be a higher-order effect than the transport effects from the uh plus of systems so we uh we have a