 Today I'm going to talk for a while about the relativistic glass-of-maxile system in 2D and It will start on the with the slides and if there's time we'll Do some proofs on the chalkboard at the end Okay, so what is 2D relativistic glass-of-maxile Well, let's start again by reminding you what is 3D relativistic glass-of-maxile You saw this slide Yesterday hasn't changed, but I should remind you Before we do a dimensional reduction So you have the transport equation for the density function f and you have Maxwell's equations for the electric and magnetic terms e and b and They have constraints in terms of the divergence of e and b and there's The charge row and the current J This is relativistic glass-of-maxile Let you look at it for a second. Oh Two people have asked me to if they could see my slides I'm really happy to share them They're on the computer here. I can also put them on my web page or something It might be kind of hard to write everything down On the other hand, it's hard to Say say everything without putting in on slides, so There's a you have to make a choice and this is the one I made but I don't so What I'm thinking is don't try to write everything down and just you can have my slides if you want Okay, so To study 2D glass-of-maxile we make the following on sorts you considered you Think of everything as being 2D, so you drop the x3 and the p3 term and then you have And you make it further reduction than that you assume that e is of the form e1 e2 0 Where e is only a function of tx1 x2 and b is of the form 0 0 b Where we Abuse notation by using b inside the vector and outside the vector briefly And b is also a function of t x1 x2 and then f is a function of t x1 x2 p1 and p2 And then you obtain what we call 2D lots of Maxwell, which is The same the same equation except you see you get a much simpler thing here It's just e1 plus p2b and e2 minus p1b gradient pf and Maxwell's equations simplify a lot. You only have one derivative on the right-hand side for each term the Divergence of b term disappears and you have divergence of e equals a row And you have the charge in the current again Except now they're integrated over our 2 instead of our 3 and everywhere you plug in x is x1 x2 0 and p is p1 p2 0 And then this is the equation. I should point out to be honest that this is not Full 2D relativistic loss of Maxwell if you don't make this on sauce at the top then you get More singular terms if you just assume everything doesn't depend upon The third variable then you get more singular terms and for that system As far as I know zero is really known because the more singular terms are actually fairly singular but This is what we're doing and this reduction is propagated by the system So we can honestly do 2 and a half D. So that's a better This is what you do in 2D And now you also have the 2 and a half D relativistic loss of Maxwell in this case You just assume that x is of the form x1 x2 0 and p has the full three variables then f e and b depend upon T x1 x2 p1 p2 and p3 respectively And you have all the components And you plug this into 3d plus a Maxwell and you can the 2.5d Plus a Maxwell system I'll write it down on the next slide. It's important to point out This is 2d in this is 2d in space, but 3d in the momentum And there's an additional conservation law here. So in 2 and a half D Classy and Schaefer in 97 observed that the conservation of the projection Momentum in the p3 direction Propagated by the conservation laws is conserved that means that v3 plus a3 Txv is equal to a constant in time What is a3? Well, it says already on the slide a3 is a gauge. So you can assume In Maxwell's equations and other systems you can assume that you have an underlying gauge and That would be of the form B is equal to the curl of a1 a2 a3 I won't go into the complexities of what it means to assume there exists a gauge But you can do it and that's what it is and then You can show that a3 By the equations that a3 satisfies you can show that a3 is bounded in time and That allows you to show that the variation of the v3 characteristic in times if you soup over st and r where you have s In and t are the time variables that you're changing and then r is the fixed time variable from the characteristic These this variation is bounded And this is an important conservation law and it allows you to prove existence in two and a half to you without this There's the proof the proof really needs this to Prove existence and uniqueness Okay, so this is the system This is lasso max in two and a half T. You have partial T p1 at Grading X part p2 plus p2 hat Partial derivative of X and then you have the Lorentz force dotted with Gradient p of f and you have the Divergence free condition the divergence conditions the divergence of e is equal to roll and the divergence of b is equal to zero and Then you can do the same reduction for Maxwell's equations and you get Hey wave equations for satisfy by e and b in a The important thing to notice here is that the while the wave equations are entirely 2d so You have the advantage of working in 2d here which helps and again you have the charge in the current. They're the same as in 3d Okay, so this is the point that I want to make is that For box u equals f and 2d where the box is just the wave operator that we wrote on the previous slide without the box notation Then you can notice that the solution to the linear inhomogeneous wave equation in our tube is zero initial data is Given explicitly by this formula so you have So the important point here for us is that you're integrating over an entire Interior of a solid cone. You're not just integrating on the boundary and that allows you to Precisely use the conservation laws whereas in 3d similar solution to the wave equation was your initial data when you invert the box is the integration over a over the Boundary of the cone. So you're you're integrating over a lower dimensional region than what you get in the in most of the conservation laws And that can cause a problem when you try to control things So That helps so notice some axles equations are still essentially 2d and Then you can integrate over a solid region and then you can bound more things by the conservation laws Which makes the 2d case easier to solve Okay I'm saying that integrating over the boundary of a cone in 3d and makes it harder to control Stuff being vague about what I mean by stuff by the conservation laws You can still do it in some cases or in particularly using the good using good conservation law But in 2d you're integrating over a solid region. This is a 2d region and so you can When you do LP estimates and you have these when you do have these STD compositions Then you can bound more stuff by conservation laws and then you have less Stuff to worry about how to bound Yes, this is not a statement about decay, so that's right So you can also think about some sometimes when you do a dispersive problems you talk about in 3d you have more decay than in 2d and that helps you when you're trying to close estimates and improve global existence problems this is a different phenomenon the phenomena is that when you do these STD compositions and you have Expressions for EMB that are on the next slide in terms of integrals and then you have the conservation laws that I talked about last time then in in 2d you don't have a Huygens principle and you have an In the the inverse of the box is a integral over a solid region And that gives you better control in terms of the conservation laws That enables you to prove global existence. So, yeah, we would like to have Decay I mean what were the kind of things we do in this business for without making a size assumption on the data Give you bounds that are very very far away from decay and I'll talk about them later, too Okay, good, okay, so so this Glassy Strauss 1986 paper introduced this STD composition that I talked about last time and it gave us a And then there were subsequent decompositions that I also talked about last time and it gave us a way to Express the fields in terms of integrals that are not derivatives that gave us a way to do estimates and then in 1988 glassy Schaefer wrote a conditional paper proving Stuff about a conditional existence and uniqueness for 2d and 2 and a half d Lassa Maxwell and They but they were unable to prove that and they use the 3d STD composition and they with this approach They were unable to prove Unconditional global existence and uniqueness and it took about 10 years to realize For them to come out with a better decomposition. So the 3d STD composition is not really the right thing to do in 2d Because of this phenomenon that I talked about on the previous slide that you're actually when you invert the box you're integrating over a solid region and so you They had this alternates decomposition. So the s is the same. It's just the transport operator in 2d And then the t is not the same you have this See which is y minus x over t minus tau. This is the unit vector on the boundary of the 2d cone It's not a unit vector on the interior and then you renormalize for a certain regime and you get partial x minus c partial t times this weight function in terms of c and With this decomposition integration by parts can be performed on the interior of the backward wave cone And 2d if you go back to the 3d decomposition here You're basically this is it. I won't read it to you again, but here the integration by parts is on the Lateral surface of a characteristic cone. So it's not on the full interior. It's just done lateral surfaces And the 2d STD composition is better cancellation properties with respect to the box And so you get lower order singularities in your final expressions that you get so again you What you're doing here is you're inverting you have the wave equations for max for max those equations you invert the box and then you get these derivatives on the right-hand side you express them in terms of these ST variables and you integrate by parts and then you can decompose the fields E and B so you can express E as E i 0 E i s and E i t and you can similarly express B Which I won't write down carefully where the E 0 and the B 0 terms are functions of the initial data and E i t has this form So E i t is this is it's the t term It's a term that you get from the applying the divergence theorem to the t operator and it's f over the The wave kernel except you get this extra t minus s term here, and then you get this this complicated polynomial type expression One minus p hat squared c i p i one plus p hat dot c i want to explain these expressions last time But instead I'll explain them today. You get a similar about worse Expression for yes, so you have this e si thing, which is a this complicated polynomial and then you dot that with the Lorentz fourth times f and you divide by the wave Kernel and 2d so this is The ET term is linear enough but more singular and the s term is not linear, but it has structure So Yeah, so I wanted to just explain quickly Or not so quickly Chuck is over here What you get with these expressions. So how do you estimate these things? So recall? C is Y minus x over t minus s and Is a unit vector is a unit vector on the boundary of the cone you're integrating over but It's smaller than that on the interior and if you take C equal to Minus p over p which you can do on the boundary then One plus p hat c which is these singularities that you get in the denominator denominator up there is of the form one minus absolute value of p over P zero and you can calculate this exactly You complete the square And You get that one minus absolute value p over p zero is p zero minus absolute value p over p zero Which is p zero squared minus absolute value of p squared over p zero times p plus absolute value p and this is one over p zero p zero plus absolute value p and this is Greater than or equal to one over p zero square So this gives you the the bad news The bad news is that I claim this last time, but I didn't prove it so there's a quick proof So the bad news is that one plus p hat dot c to the minus one is like one over is like p zero squared So you see that in all of these terms you have a one plus p hat dot c squared and That gives you bad growth. So the worst growth is p zero to the fourth and The other terms are have less growth. You have like p zero cubed and p zero growth in these different terms and that's Really bad because if you want to bound The fields you don't want to lose momentum growth. So this causes big problems But it's not a death sentence Because the numerators also have Cancellation that's hiding and I want to explain this so So you have cancellation in particular you have The easiest one is one minus p hat square This is actually very similar to what we just did it's P zero squared minus p squared over p zero squared, which is One over p zero squared. So there's a so the worst term up there has a P zero to the fourth coming out of the denominator, but then it has a one over p zero squared Helping you out in the numerator and it has even more than that. So I can Explain this for you right here So if you take c i plus P i hat I want to convince you that this term has hidden cancellation that Helps you reduce the power of these singularities in the denominator Say square it and you bound it above by c plus p hat Square the whole thing not just one term and you expand this Next so you can just bound it above by two times One plus p hat c Okay, so you expand that you get c dot c plus p hat dot p hat plus two p hat dot c and the first two terms that I just Spoke are bounded by one and so you get this bound and so The absolute value of c i plus p hat i is Less than or equal to one plus p hat c to the one half So you can automatically quickly reduce the worst singularity in the ET term down to from One plus p hat dot c squared down to one plus P hat dot c to the one half So you you kill a lot just by paying close attention to the cancellation Yeah, so But it may not look like there's he has much cancellation in the ES term It turns out that there is and it's so this Delta ij minus p Hat i p hat j term term you can do something Sort of like this not exactly, but you can get a similar type of thing if you're careful And then what about this term the cj minus c hat plus p hat Pj can you get cancellation out of that? Yeah, you can so um Let me just quickly show you so You can add and subtract so you're right Cj plus p hat j minus P hat j One plus p hat c You you add and subtract P hat j and then you pull the p h a out of the last term and then This in absolute value is less than by the previous calculation One plus p hat c to the one half so that's that's the best that's The the first term is the one half the second term is one But you just bound it above by the one half because it's easier because it's smaller Okay, so it turns out that there's even more going on in this ES term Then just these cancellations there you have to split up into the A term that contains good components from the good conservation law and The rest and you can do that But this is all I wanted to say about this right now Okay, so now in 2d or in two and a half to you have this Series of works of glassy Schaefer on global existence for large data, so you assume that the conservation law the L2 norms in EMB and the P0 f Is bounded initially? Oh, that should say f zero there. Excuse me That's the initial left. It's not the f for all time and so so here you see what I've been saying If it just briefly let me return to the question from before so if in 2d you're integrating over a whole Solid region and then you have these L2 conservation laws for EMB This is stated as the initial data, but you can continue to bound for all time when you're integrating over a solid region you can Bound from above the fields by this conservation law when you're integrating over a Lower dimensional boundary of a cone like you are in 3d, then you can't bound from above the Terms by this L2 conservation law, so that's the thing that you're losing in 3d versus 2d Okay, so you have these so let me return to the existence theorem So you have that you assume that the L infinity norm is bounded You assume that the conservation laws are bounded you assume that you have a compact support So the soup of P such that the initial data is non-zero is finite and then you work in C in C0 and C1 and C2 spaces so and they really do work in C1 and C2 spaces Not so below spaces, so you assume that f0 is C1 and easier and B0 and C2 you assume all the constraints are satisfied you assume you're positive and in Two and a half to you assume some initial additional things that I'm not mentioning Then there exists a unique global and time C1 solution to the 2d or two and a half T relativistic plus a Maxwell system and furthermore the fields grow So the fields grow double exponentially in time. So this was the question from before about you should have decay for the wave equation We don't have we don't know the K we know From a glassy Schaefer. We know double exponential growth So that's that's pretty strong powerful growth in time of the alphanorms of EMB Okay So let me explain for you in one slide just in a couple minutes again. I just want to give you a picture I don't want you to understand I'm not trying to explain this to you in too much detail But you this is a brief outline of glassy Schaefer's proof So to find the momentum support by P of T is one plus the soup over the absolute value of P of all the Times where f is not zero in all of backward time and in all of space So this is the largest that the support ever was going backwards in time from where you are now to zero and Then you decompose the field the 2d field with the 2d ST decomposition And then as I you do L infinity estimates So you bound the fields in the distribution function using a lot of careful and hard L infinity estimates and you prove That The support is bounded above by The integral so p of t is bound above by constant plus a constant That depends on time plus the integral from zero to t of p of s log of p of s And this is where you get the double exponential growth of the field. So If you study Groton wall, then you see that Something being bounded above by log log sorry by itself times log of itself not log log is Is double exponential growth and if you want less than double exponential growth, then you need a better Groton wall Okay, so there's an additional conservation law I told you about it before and that's also important in the Two and a half t proof you have to use that carefully And so this is what you get for 2d less of Maxwell in the Relativistic case and two and a half t it's worth pointing out that for existence and uniqueness If you go to the non-relativistic case Then the situation is Is worse you have to go to lower dimensions so for non-relativistic plus the max well the characteristics can explode because the velocities are Not bound the There's no p hat. There's just v and v can go to infinity. Of course you assume that the momentum support is finite But initially and you hope to prove that For all time, but it's not as easy in one and a half t and The best results in the large are in one and a half d And uh, there's a very some paper by glassy pankovitch schaefer in this direction Okay, so uh, again, I was since I want to give details. Let me sort of briefly Explain Something more here Um So how do you get um So you have the characteristics Every suit Okay, so these are the characteristics uh the x t s and d v d s and you can integrate the v uh characteristic in time And you get v of s t x p Is less than or equal to v zero t x p Plus the integral from zero to s d tau of Okay, so I tried to write that briefly. Um But so I wanted to answer the question on this board, uh How the heck do you bound the support? Um, and you bound the support using the, uh characteristics, um So these are the characteristics And you integrate the v characteristic in time and you slap absolute values on everything And then you get this v characteristic is bounded above by the initial characteristic Plus the absolute value of the uh fields Evaluated at t At tau and the characteristic x and now if you're bounded initially Then um And if you can prove bounds for the fields Then you can Turn this into you can take the soup over this V of s characteristic and get a bound for p So this will this v will turn into p and then your your hard work is to bound e and b And you do that with l infinity estimates and the st decomposition and stuff like this Okay, um, yeah So let's continue Okay, so one of our goals was to prove Existence and uniqueness in 2d without a compact support And so this is the first theorem that i'm telling you about Um, so we switch from ck spaces to sobola spaces And you consider initial data f zero in the sobola space h3 Which is an un-negative and obeys a bunch of bounds on the next slide. Don't worry about it yet Um, and uh, you're in your fields e and b are also in h3 So we don't really need to assume high regularity for the fields um And you satisfy the constraints that you don't assume compactly supported initial data Then there exists a unique global in time c1 or h3 solution to vasemex well in 2d or in 2.5 g this is uh The joint work with luke and you can get much better growth of the fields it's still far away from what we want But exponential growth is much lower than double exponential growth And this is even sort of relevant to physical considerations When you think about deriving the equation, um, but So this is what we can do in 2 e and 2.5 t and it's also instructive What we do in 3d um So here are the assumptions on the initial data you have f0 Is in l infinity And uh, you have some high moment bound For example, you assume that the l1 norms in x and p of f with an nth moment are finite initially Where I say n is greater than 13 and you assume It weighted l2 bound for At least three derivatives. Um We actually do this in a scale of spaces so you can go up to as many derivatives as you want That's pretty standard, but I'm just dating here for three And then you have these funny Variation assumptions. I want to explain Where they come from and why you use them. So, um Let me for every r's or furthermore Uh You have the l infinity norm of x times the integral In dp of the soup of f0 of the variation Of f0 near p And w less than equal to r that's bounded by a constant that depends on r for every single r um, so And you have a similar assumption for the gradient In x and p of f0 and you have a similar assumption For the gradient in x and p of the l infinity norms. Um So why do we do this? This is uh, this is the next thing I want to explain What so these assumptions are odd, um, they're unusual at least So I want to explain why we do this before I do that Let me just say well, you can make simpler assumptions if you want. So, uh, the previous odd assumptions are Are implied if you assume that f0 Is uh sufficient decaying sufficiently rapidly for example Minus 16 minus epsilon for some small epsilon And if the gradient is point wise decaying sufficiently rapidly For example like p0 minus 6 log to minus 2 Uh, you could easily state these assumptions and you and they'd be less weird But uh, they're not they're also significantly less general and they're not what you use Um, and so I want to go back here and explain How you use these assumptions um Okay, so uh, you can go back here do the characteristics Um, and uh So if you have compact support you can Do this inequality Um, if you don't have compact support you can You can't do this inequality anymore because this Characteristic at zero is going to be uh infinite But you can still try to uh, assume or prove that this integral is finite um So you can uh You can't integrate the characteristics anymore, but you can still use that intuition and try to uh Say well, what do you get if you assume that this integral is finite and what you get? um In reverse order is uh the following so, um So you integrate So you integrate from zero to t And then you get that uh x zero t x p minus x Is also going to go to r over 2 say and uh v zero of t x p minus p Is less than or equal to r prime over 2 there's some Okay, so if you assume that this integral is bounded then you can integrate the x And the v characteristics and you can get that um These kinds of bounds so you have some constant which bounds the variation of the characteristics so uh In 2d you can prove this bound in 2 and a half t you can prove this bound in 3d this bound is a conditional assumption And then when you integrate from zero to t you can prove that the variation Of the characteristics is controlled Given if you have a control of this bound um, and then uh You can use the characteristics Again, so also So since the density satisfies the transport equation, uh, we talked about this last time You can express the solution in terms of the initial data and the characteristics And uh, what you want to do Is prove good bounds for uh, integrals involving f so these assumptions give you exactly what you need um, so from From 11 Okay, so it turns out that when you do these um Um St the compositions that I showed you for the field you want to get estimates of various quantities involving f And integrals of f one of those quantities you have to believe me is something like this the integral over dp of p zero cube f of t xp And then you try to bound that in a space like l infinity in t over a finite interval and l infinity in x And uh, so what do you do? Well, you plug in the expression that I put up here And you use assumption 11 and you and you use that these these uh Quantities don't vary too much conditional on this assumption. So coming back to this assumption You can prove that this a quantity like this Is bounded By one And that uh is very uh helpful And uh in a similar way you can prove uh from from 13 that uh Okay, so in a similar I haven't told you what b of t is yet. I'm about to but uh from an assumption like 13 you can Show that uh the gradient of x and p which is another quantity that comes up if you take the l infinity norm over that And you take a finite time interval you take x and you take the p variable Then just using this expression and you then you differentiate it you end up differentiating the characteristics And you use that the variation The variations are bounded And you use assumption 13 Then you end up seeing that uh The quantity like this is bounded above by uh The maximum value of the first derivative of the backward characteristics. So uh b of t Is one plus uh the stoop gradient um p of x and v. Okay, so here b is the uh Just for simplicity we write as one plus the maximum over the backward characteristics So it's the stoop of the gradient in x and p of x and v Where you're the x's and v's are the backward characteristics So, uh, yeah, so I wanted to tell you what these assumptions are and I wanted to tell you why you use them. So that's the story So that's how you use them and uh Okay, so the uh similar theorem for 2 and a half t relative to the black sum Maxwell without compact support Uses the uh Additional conservation law in 2 and a half d Although since we're not assuming compact support you have to uh where we use an assumption like this that uh The fifth plus some small delta moment um in the p3 direction Is bounded so this is This is actually a statement that you can prove because it's time dependent But so you make an assumption similar to the ones I just explained on the board and then you uh Can prove that it's propagated using the characteristics. Um, and using the conservation law Okay, so for now, that's all I oh, I guess not. Um, so I'll give you a quick brief outline of the 2d and the 2 and a half t theorem um, so you bound the moments the nth Uh, the nth powers of f integrated instead of the supreme of the momentum momentum support Many previous works rely heavily on L infinity estimates and uh But that's L infinity estimates are a problem So while the moments of f can be controlled by the L infinity norms of e and b these L infinity norms of e and b Cannot be controlled by the moment. So you have to do something else Um, and this is where you start to use lp estimates and uh Strict arts estimates. Um, and in fact in 2d You don't just use strict arts estimates. You use the improvement Of the standard strict arts due to foci foci foci foci, excuse me, and the tiger is a Tiger the Elaborated on foci um And so these foci, uh strict arts estimates, uh, in particular give you a better range of exponents in the case when you're working with uh, Uh, a forced wave equation with zero initial data and so and that's exactly what the kind of situation that we're reducing to Okay, so strict arts estimates capture the smoothing effect of the wave equation and give you better sort of long time asymptotics And that's why we get better estimates in this situation And uh, I'll explain these techniques in more detail when we use 3d case. So let's just jump right to that because So again, uh, I won't read this slide carefully, but let me just remind you that you have the glassy Strauss theorem that says that if uh the uh Support of the distribution function remains bounded Then it doesn't grow to infinity Then you can continue the solution indefinitely um, and the question is can you have a continuation criterion in 3d That does not involve the momentum support or does not involve compactly supported functions And making that assumption initially it seems like a strong assumption to make considering that you're working with a relativistic model that's supposed to That be a better model for large velocities for large momentum, but uh, okay, so uh, Again, we come back to this kind of thing that I wrote on the board over here. So if you let p of t be the momentum support That I explained to you previously that and you have the characteristics again for x and v And if p zero is finite then the continuation criteria that the Momentum support remains finite is equivalent to this condition that the fields remain bounded Um, which is also what we wrote over here Um, on the other hand that you can assume that the fields The the l1 integral of the fields along the characteristics Remain bounded even without assuming that the momentum support is a compact um, I there's a hidden uh Tx and p in these slides that I'm not mentioning and this is what we do. So 17 that this condition makes sense even when the momentum support is not compact And uh, so that's what you can do okay, so, um Now state of theorem, uh, so given uh, initial data would satisfy the uh, constraints on the divergences and given The initial data is in a weighted uh, h so below space hd where d is like a rhythm four. Um Then you can prove the local existence theorem And uh, here the weight is just p zero to the three over two log of one plus p. It's used for uh Technical reasons in the local existence theorem. We need some decay um And then if you assume this condition on the l1 integral of the fields of long characteristics Then if the bound 17 holds up to time t, uh, then you can continue your local solution Further past t. So this is a continuation criterion Um, that's the theorem. So you can continue your solution beyond t and the solution stays in the sub-level of spaces hd So you can you can continue the solution up to time t plus epsilon for any small epsilon and if you have a bound like this for all t Then uh, you have a global solution because you have a solution that exists for any large amount of time Okay, so, uh, there are similar assumptions in uh, 3d. So I won't spend as much time on them Um, it's a little bit different. So you assume the infinity norm is bounded. You assume some moment Some uh, collection of moments are bounded. I didn't go back and calculate it. Um, but it's It's not that big. It's like 15 or something. Um And you assume that some the sobolev norms and l2 based sobolev norms are finite here. We We use d greater than or equal to 4. We haven't done any like low regularity type work. We're just doing the basic Sobolev local existence theorem And you have related, uh, crazy assumptions again, I definitely don't want you to write down this slide Um, but, um, I just want to show it to you that I explained it more carefully in the 2d case Because it's easier to explain in that case But these are the sort of brutal assumptions that you use. There are the 2d analogs Of the 3d assumptions. You have certain norms and you assume that the variation Of these functions initially is not too big Um, you assume it's controllable. It doesn't blow up. So for every r, there's a different constant cr Such that these things are finite where you soup over The variation away from a point and then that allows you to, uh, control the characteristics Okay, so here is a quick outline of the proof. Um You have a quick outline of the local part of the proof. So you prove local existence via standard energy estimates in hk the local proof Immediately implies that the solution main remains regular in 3d If this quantity is bounded so you can remain regular beyond capital t If this integral is bounded up to t. So you have the lmp norm of, uh The fields the first derivative of the field down the second derivative of the fields You have for some reason we have to use l3 and then we have the l infinity l2 norm of the weighted first derivatives Of the distribution function f. So this is an automatic continuation criterion that comes out of the local existence theorem Um, and then you assume that 17 remains bounded where 17 is again the, uh l1 norm of the integral of the fields of long characteristics and then, um You show that this integral implies this continue this the bound, um of e and b up to uh time t integrated along the characteristics implies, um The continuation criterion that comes directly out of the local existence theorem. That's 26. That's the integral So this one look this one is significantly more general and easier to use. So you basically reduce to it um Okay, so yes Was there no one said anything? Okay. Um So, uh Now I was planning to switch to the blackboard, but I don't know if I'll have time to really say everything Um Let's try Yeah, okay, so we can start in the middle I erased it Okay, so we let b of t be uh The soup over everything of the first derivative of the backward characteristics And you let f of t be So we're basically estimating the characteristics So you have the suit that b of t is the soup over the backward characteristics and f of t Is the soup over the forward characteristics um And now you have the characteristics up here. So it's uh It's basically automatic That you have an estimate like Okay, so uh the forward characteristics can be bounded by this is the ds integral And this is ill infinity x so the forward characteristics, uh If you just if you integrate both of them and you take a derivative um and uh It's just this straightforward calculation that you can bound the forward characteristics by uh a constant times the integral of the forward characteristics times the ill infinity norm of the field I use the middle one first, right? Okay, and uh The goal is to bound the field so Then St decomposition bound for Okay, so I claim that if you combine the st decomposition The assumed bound for e and b and ground wall You get a horrible bound for k Okay, so uh, I claim that um, you have this estimate and You'll just have to believe me or read the paper Um, so you can bound you can use st decomposition. You can use the assume bound for the So remember we're trying to bound K itself the first derivative of k and the second derivative of k I'm only going to explain to you how to bound the first derivative of k Because then we're going to run out of time and um I'll explain how to do How to bound k itself tomorrow And uh, so you get this bound so you get the p 3 f moment And you get the p 3 gradient x p f moment And then you get this delta parameter that comes out of estimating a singular part of the integrals And uh, so how do we so how do we estimate these terms? This is a homework problem Except I already told you the answer um Okay, so you use the assumptions um, so these crazy assumptions that I told you about are useful and uh, so by the assumptions then p 0 3 f and infinity p infinity x l 1 the key is less than or equal to 1. Um I need new chalk. Okay, so by the assumption by this Assumption and by the assumptions on the uh initial data these crazy variation assumptions that I asked that I uh Showed you and by the uh Method that I showed you to use those crazy assumptions to estimate things like this you estimate that and um also Okay, so again by the assumptions you can show that uh p cubed gradient x p and uh gradient x p by itself is bound above by uh The characteristics um by the backward characteristics And then uh Also By the very nice matrix calculation that I like That is not so original um You can bound the backward characteristics by a power of the forward characteristics And you can bound the forward characteristics by a power of the backward characteristics using uh You just take the determinant of one and use the properties of characteristics and uh Use uh kramer's rule and you get That you can bound one by the other and then um Okay, so you plug all this in and uh We we go back to the top Um, so we have f of t is bounded by integral f of s gradient uh k of s And we have gradient of k is bounded by all this stuff And we have p zero cubed l one is bounded by one And we have p zero cubed gradient x l one is bounded by b um and uh We have an extra bound for gradient x p f that I just stated Then uh choose Delta of t equal to t Over b of t Now you know why I wrote b of t as one plus the soup Because I want delta to be between zero and t And so you choose this delta and then you get uh f of t Is less than or equal to one Plus the integral from zero to t on f of s log b of s ts Okay, so now if the log wasn't there you'd be dead. Um, you wouldn't be able to do anything But since the log so now we're going to bound b of t by f of t and pull out the constant Okay, so uh, this is uh what you do um So what we've shown here is that f is bounded the super the characteristics is bounded and then um b is also bounded So we see that uh f of t is uniformly bounded and then p of t is also uniformly bounded Double exponentially growing but still uniformly bounded So the forward characteristics and the backward characteristics are both uniformly bounded and then um This is bounded And this is bounded And that's bounded so everything is bounded. So therefore gradient x k as infinity t is bounded So we're just sort of Moving around in a circle But not exactly doing Doing slightly non-circular logic. So we prove some things are bounded and then Go back to the previous inequality and use it again to prove that the thing is actually bounded instead of just ground wall um Okay, so k so gradient k is bounded uh because of all these other bounds and then also uh w3 gradient x p f And um say l 1 t l infinity x l 2 p Is less than or equal to um Now you see why I did the l infinity bound for gradient x p f Okay, so you can do an interpolation. This is the this w 3 gradient x p l 1 l infinity x l 2 p Is the thing that we want to bound to have a continuation criteria But we haven't shown that that's bounded all we've shown is that uh p 0 cubed gradient x p in that norm is bounded And the gradient x p and f in that norm is bounded previously. You didn't know why I put this norm here It was just an odd thing, but now you know why It's because uh, you can interpolate for some theta between these two norms and see that this quantity is bounded So then uh, that's how it works. Um Okay, so uh, you can do a similar Type of argument for the second derivative of k and um Then you have um, you almost have the continuation criteria. The only thing missing is this um Yeah, I'm having to prove this bound that the The interval on characteristic is is finite in 3d. That's a Assumption that you don't know that we don't know how to prove that if you could prove that you solve the big problem um in 2d and 2 and a half d. I haven't explained to you how to prove it and uh It's uh, it involves bounding the moments and And uh, it involves using stricard estimates and I will uh explain it to you on tomorrow Okay, thank you So in your local existence theorem, you Softed in uh, with the weights in the momentum term and this is um P0 to a three-halves log of 1 plus P0 Okay, so I'm sure it's kind of wondering like do you expect this to be optimal or Uh, no, um I don't expect it to be optimal. I uh excellent We made no effort other than uh using the smallest thing that worked in order to uh Use that weight um the the idea is that you do uh Holders inequality in certain places and when so you basically you take the equation You hit it with a bunch of derivatives and you multiply it by the solution With the same number derivatives and you integrate and then you get something that resembles an l2 norm Minus a bunch of other stuff and you have to control the other stuff and To control the other stuff you use uh so below of embedding theorems Use holders inequality and use all the various other tools that uh people normally do to prove local existence and uh At one point you need to control an integral In a holder's inequality to get from an l1 norm to an l2 norm and uh that step you P0 to the three-halves uh log of 1 plus P0 is Just barely bigger than P0 to the minus 3 and it's enough to control and a three-dimensional integral And that's why we use that weight Okay, thanks. So tomorrow I'll talk about the uh Bound for k and I'll talk about further uh the state of the art that as I know it for continuation criteria For uh 3d Lassa Maxwell and uh most of it will be on the board. There'll be some more proof Okay, thank you