 firesc oamenilor s-a plăcut oamenilor s-a plăcut s於sțională announcements în 예쁘pturile iar sistemul perepturilor Iar sistemul perepturilor Aoșteptă, când s-a plăcut Pentru că, când avem lumea invizită, după ce avem în mind lumea încălătării de navie stocs și oilerii, încât atunci, acesta poate încălătării în diferite equării, cu diferite condițiile bălătării, dar să rădăm exact ce am încălătat acest lucru. Așteptăm equătia de u, plus u, adică u, minus nu la flash nu. Divergență de u este zero, și așteptăm data initială. Așteptăm equătia naviestocăție. Normal ar trebui să considerăm acest lucru în acest domen omega. Și trebuie să luăm condiție bălături. Trebuie să luăm condiție bălături. Ok. Lui regula este că u este equal a 0 pe bălături. Ok. So, u, u a fost velocită, nu a fost viscosită. P a fost pressură. Așa oamenilor îi reușim u este funcțională de x și t. Tu credeți că e x în R2? Da. Așa. Da. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Omega este în R2 o R3. Ok. Ok. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Așa. Je cauară respect Vocês. Oamenilor Așa. Așa. Așa. Așa. Așa oamenilor în R2 or R3. Așa. Așa. Oam �ură ține-ți, așa așa oam zis. Așa. We have some other boundary conditions, so this is Dirichlet, we have also Navier boundary condition which says that u.n equals 0, so this means no penetration, and the second, I mean it comes with some condition on the stress, so the fact that s of u and the tangential part of this equal either 0 or sometimes we can put minus alpha u tangential, or I can write u because u is just u tangential on the boundary. So, what is s here? s of u is the symmetric part of the gradient. So, this is, I mean we can call it Navier or like Robin type boundary condition, it turns out that the problem is simpler if we look at this boundary condition rather than the first one. Ok, so we can also look at the problem without boundary conditions, we can also study the problem, so the third case will be let's say omega equal tn, td or rd, or like, I mean people study also the problem on manifolds, let's say manifold without boundary or I mean like one can also push it to more geometric settings like where you don't have a boundary. Ok, so by the point here is we have either the most difficult case is the Dirichlet boundary condition, we have some other type of boundary condition let's say Navier, we have no boundary conditions, I mean there is another type of boundary conditions and where there is also quite number of walks which is like boundary conditions where you put some penetrating fluid. So, I mean here when I say u dot n equals 0, this means no penetration, but there are other boundary conditions where we allow some incoming fluid, so incoming boundary conditions. I will not really talk about this, but there is some literature about this, in particular I know like I can mention works of Te Mam and Wong who did a lot of work on this incoming boundary condition. Ok, so, again here maybe I should mention again what I mean by n, n is the normal, usually my normal is the normal and what I mean by tangential means the tangential part of the vector. So, S of u is like a matrix, is a symmetric matrix, you are taking S of u dot n, this is a vector, this is a vector on the boundary and then you are looking at the tangential part of it. So, for instance, if you have a vector v, on the boundary you can always write it as v tangential plus v dot n. So, it's just you are projecting to the tangential part. And of course this works either you are in two or three dimension or even in any dimension. Ok, so, more or less this is the setup, you start with Navier's talks with a viscosity u and I mean usually when you study Navier's talks with a fixed viscosity you will talk about u. But now if you are studying the problem with a viscosity that is varying, so u is varying, so usually you will call your solution u nu, because it depends on u. Or, sometimes we can call it u n, u n and then we have a nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu nu Unul de secunse de viscosită este în 0, și vreau să înțeleg ce se întâmplă în unul de un. Formal, dacă studiem acest limit formal, o să o preție, o să o preție că... Vreau să le înțeleg. Vreau să o preție că viscosită se întâmplă. Așa, vreau să se întâmplă o să o preție în unul de un să o preție că o adapteaza slide-ul să o preție o să așteptează toată cea o să o rețetează o să rețetează și o să rețetează o deliveredă este defectiv. Vă faceți cazul care mă ducă Csă țărul Ex Tibetan că avem să impozătăm. E clar că, pentru stocs naviei, vom impoză u-a-l zero pe bălături, așa că dacă sunt în trei dimensiuni, așa că, dacă impozătăm trei condițiuni, se turnă pentru oilor, și asta este pentru o nature parabolică de equătia. Oilor, pentru oilor, nu pot impozătăm această bălături bălături, și nu pot impozătăm această non-penetră. Deci, dacă îmi faci un sau două, dacă îmi faci un sau două, bălături bălături, de normală, dacă ar trebui o limită, este să fie 0.0. Acesta este bălături bălături de un sau două. Dacă acest condiție va desăpire în limită, dacă am în case naviei, și also acest condiție în partea tangentea va desăpire acolo. Dacă vrei să te întrebi un cânt, dacă fie în case numărul 3, nu aveți bălături de bălături bălături. Asta este cazul simplu. Și acest lucru îmi înțeleg câteva lucrurile de acest lucru. În case 3 este cazul simplu. În terms de dificultate, dacă acest lucru este simplu. În terms de dificultate, acest lucru este mai complicat mai complicat și acest lucru este open. În terms de care este simplu, acest lucru este simple de energie și dacă vrei să facă bălături bălături, acest lucru este nu foarte dificult. În acest cas, în acest lucru, încât să îți fie bălături bălături, încât să îți fie bălături bălături, încât să îți fie bălături bălături, încât să îți fie bălături bălături, încât să îți fie bălături bălături, încât să îți fie bălături bălături, încât să îți fie bălături bălături, încât să îți fie bălături bălături, încât să îți fie bălături bălături, încât să îți fie bălături bălături, Aici, în general, când lăsați la aluat, când lăsați în comprescire, aici, se văd ca, dacă se dățila, nu se worry de multe. Asta se văd că, în real, se văd că, când lăsați, se văd că, când lăsați sa se văd că, în real, ne-a fost este, așa că... Nu, nu. Deci, când lăsați se văd ca, dacă se văd ca... Părăture și altceva. Ok, altceva? Ok. Dar să ne zic să încercăm... Sper să explicăm dificilă și arătă Vă mulțumim, Mrs. Leopart, când am încălțit... Vă mulțumim, vă mulțumim, ce viziteam la frumos? Vă mulțumim ca plările 3. Vă mulțumim ca plările 3, unde nu există plările bălătării, niciodată. everything can be done easily so for case three for case three i mean i can mention so case three i can write here so case three so there are works by swan and cut two and there is also later i mean these are more or less the 70s and then there are some improvement by constantine and i have a paper also where i make a small improvement that i'll explain maybe i mean it's a small minor improvement but i think it's interesting so what will be the statement so i can we can write a theorem um so i take s to be bigger than d over two plus one i write it in this in the in in the whole space in the whole space u zero is in um hs of rd and t star is the time of existence the time of existence for Euler with uh with the initial data u zero so the time of existence of Euler of of this system of course without boundary conditions because we are uh so this is my system Euler um so then the conclusion is that for all t for all t we can find new zero such that for all new n less than new zero um the Navier-Stokes system has a solution u and is okay i didn't write the solution of Euler but the solution of Euler is also in this space and for all t in zero t um the limit when t goes to uh when n goes to infinity is u of t okay and this limit happens strongly in hs and uniformly in time strongly in the hs uniformly moreover if i look at u n minus um u zero of t so u zero of t is the solution of Euler in hs minus two this is less than some constant new n okay c c just depends on u zero this depends on u zero okay so this is a okay so this is a statement which is okay not very difficult to prove so i can tell you for instance okay so this is more or less the statement a simplified version of the statement in my paper like in my paper what i the improvement i have is the fact that the the convergence happens in hs previous works will get the convergence to happen in weaker spaces like you will you will be solving the problem in hs but you can prove the convergence in hs in hs prime s prime less than s so let me explain how this works because then it will explain it will will show us what's the difficulty with with boundaries now um why are we working in hs for the Euler system we can only solve in with some regularity we cannot solve in l2 for instance um of course now in 2d we can we can solve if the vorticity is in l infinity but if i am in three dimension or any dimension we have to solve with that kind of regularity okay so so at this level um i mean one can um um one can ask okay but for navier stocks we know we know how to solve navier stocks in spaces without regular i mean in spaces like l2 can we can we still do a problem with solutions of navier stocks in l2 the answer is yes but then what you need to do for instance i mean um but that's that's an that's let's say an extension of this result then you can allow this to depend on n right and then make sure that this will converge to the initial data of Euler and then okay these are type of extension to this kind of results if one wants to but um i'm i'm not going to do it because for this result let me just stick to hs i mean i'll mention other results where we want to do things in energy spaces in l2 for instance okay now how do we prove a result like this one okay so the proof uh actually what i'm going to show you i'll show you this right i'll show you this uh this part the fact that we can converge strongly in hs requires some extra work that i'll i'll just mention briefly but to prove that um that second the second statement i mean it's very standard i mean it's very more or less what you do um so you look at un minus u u u is u0 let me write it u0 so a priori a priori you don't know whether a priori you don't know whether navier stocks you can solve it a priori you cannot you don't know that you can solve it till the time capital t right if you are solving navier stocks in some smooth spaces usually have a time of existence that depends on n that depends on the viscosity so um so i'm not going into that detail but all i'm going to show you is the energy estimate so if i have energy estimate on the time zero capital t then this will imply that i can solve on that time and get to solution so um how the proof goes i'm going to write energy estimate on this on this so so we are we we can write down an equation on this difference by taking the un minus u0 and then we do energy estimates on the on the level or um then this will be less or equal then all bunch of terms i don't know what i should explain wn so you add w0 there is no epsilon n okay okay go ahead finish your computer yeah so it's a real square like in dq ws square ah square so so just just what you do you write an equation on wn more or less wn will be you you put a w this will be a w this will be a w this will be un grad un minus u grad u so you'll get some terms here and then you do energy estimate so without really trying to write everything but what you get you get a term like these kinds of terms and then you get a term that comes from the viscosity applied to Euler to the to the Euler term so the reason the reason you can only do this at at at a lower regularity so you can ask why why you cannot do this at the level hs why you can only do it at the level hs minus 2 yeah oh yeah yeah this is zero okay so um and this is zero yeah so the reason you can only do this at the level hs minus 2 is that you will be getting the term like this coming from the Euler equation right because somehow i added the Laplacian un i'm i'm going to write it as Laplacian w n but then i have to pay this term on the right hand side and this term if my u zero is just in hs my u zero is in hs so if i i need this to be controlled by u zero in hs i can only do it in hs minus 2 so that's why that's all you can do here i'm not going to explain too much but this is how you deal with the nonlinear term okay so you have the nonlinear term some term disappear but but somehow the nonlinear term you can control it by this okay so then from here more or less you you can end up with an estimate like that it's not very it's not very difficult then you can end up with an estimate like this um then then what people are able to say uh you can i mean proving proving this i said okay i'm going to skip it but this is also not very difficult that you can solve uh the naviest talks on that same time and you can also solve it uniformly like with bounds which are uniform in n um then one can interpolate between uniform bound here and this to deduce like you can interpolate by some uniform bound here and this one to deduce for instance estimates of this nature so now if i take s prime between s minus two and s you can get c u n to some power um s minus s prime over two this is kind of things we see in uh some of the early papers that your your rate of convergence deteriorates with the space like in in s minus two your rate is u n but in h s prime the rate is s minus s prime over two then um that's for instance in my paper from 2007 you can do some you can use some some regularization argument that i'm i'm i'm not planing to put but like it takes maybe two pages uh and you can even prove that you get a result like this can prove that you convert strongly uniformly in time in hs it doesn't follow from this interpolation because from this interpolation all what you get you get that this is a constant so you need to do you need to do some regularization argument i mean basically i can i can give in one minute the idea the idea is that um you don't want to compare your u n to u zero so instead of comparing u n to u zero you will compare um maybe i'm right here so you take your you take the initial data u zero you replace it by some u zero delta so u zero delta is a regularization of u zero i mean since we are in the whole space this you can do it just based on fully analysis for instance you are in the whole space but um your u zero will be bounded in hs so for instance it can be it can be some fully multiplier c less than one over delta apply to u zero okay so you can take something something like this um so then what you what you get you get that you this guy in hs plus one will be less than constant one over delta and in hs plus two in general in hs plus i is something like that okay if if you do it the way i mentioned it so so then what you what you do you look at the solution the delta is the solution of Euler with initial data u zero delta then what you do you compare you you write that u n minus u zero so this is the thing you want to prove that it goes this is the thing you want to prove that it goes to zero in hs you write it down as u n minus v delta plus v delta minus u zero so then it becomes this is this is a property of Euler this is like stability of Euler the fact that v delta minus u zero the fact that v delta minus u zero goes to zero when delta goes to zero in hs so this you can prove that this goes to zero when delta goes to zero in hs okay now this now your v delta is in a better space your v delta you you can put it is even in hs plus two but to to get that it is in hs plus two you have to lose some one over delta square okay and then you optimize okay there's some work you optimize the value of delta in terms of the viscosity and you can prove you can prove the the result okay it's it's it's very it's not very difficult but it's okay any questions of course now the proof i mentioned here one of the main things in this proof is the fact that there is no boundary terms right and you can you can do all your integrations by parts everything everything is okay okay um now question about the regularity so the fact that s is bigger than d d over two plus one that's really the natural space that's the natural sobolev space for Euler d over two plus one if we are in two dimension if we are in two dimensions we can we can do better in two dimensions what can we say in two dimensions in two dimensions first of all in two dimensions t star is equal to infinity because Euler we can solve it globally so t start you can take it to be infinity also um we can instead of taking the data to be in hs we can take the data the curl of u zero is in infinity we can do a result where the curl of u zero is in infinity of course now usually the spaces we take maybe we can i mean since we if we are in the tors this will be enough if you are in the whole space usually we like to intersect it with some let's say l one or or at least some ld space so that you get some decay otherwise i mean if you are just in l infinity okay you you don't have decay okay um so then then one can ask one can more or less get some similar result um it turns out it turns out that um the natural the how how the result i showed you here extend to this case um so there is a result if i try to do this in l infinity l2 so i mean if we think about it scaling wise scaling wise um the curl of u zero in l infinity this is like the curl of u zero in h one same scaling so i mean this is like the result it's as if i'm taking s to be two here right i mean dimension two is as if i'm taking at in terms of scaling it's like s equal two so so it's normal that i give you a result where this is l infinity l2 because this is like s minus two hs uh hs minus two that's the space l2 so um so this result i'm mentioning here is due to schema actually so schema has a result um but somehow the result here compared to the previous result i showed you loses a little bit so there is a constant and then one can write nu and t so this will be this will be the natural uh i mean the t here uh i could have written it also in the previous result i could have written c nu and t so this will be the natural extension from what i what i showed you earlier but this is actually not not correct here there is a loss so there is some exponential so there is some exponential loss some exponential loss meaning that the rate the rate is decaying so i mean it's the power exponential minus some constant times t without all the constant depend on u zero um and okay this is due always to the fact that when you look at 2d oiler in when the curl is just in l infinity when the curl is just in l infinity and you want to estimate the gradient of u in l infinity you have to lose you lose a little bit you lose some log and so the result is true you can still do it but you have to lose there um however there were results afterwards saying that so this this is due to schema but then there were some improvements that that say that we don't get the loss if if in addition we know if the solution of oiler if the solution u zero of oiler is in lipce space is in infinity so this is an extra information like usually if you take the curl in infinity you don't know that the gradient is also in l infinity but if in addition you know this then there's no loss right then no loss in the exponent no loss in the in terms of powers of u okay this is a result um i think it was first proved by costantine and now um i mean it's not my plan to go into these directions but i am mentioning this because okay i mean it shows that okay there are many possible extensions of these results and um let me so so you can ask okay when when is this correct when do we know when do we know that um u is in lipce space this lead us to um i mean and there are a few works dealing with the case where we have what we call uh vortex patches vortex patches so this is where the vorticity is like let's say one and it's zero outside these are called vortex patches and okay i mean there was there's a whole literature dealing with vortex patches so in general what we look at we look at um vorticity which is one in a domain and let's say if the domain is smooth at least c alpha sorry c1 plus alpha so you take a boundary you take some vorticity in a domain of regularity c1 plus alpha um so for initial data like this so of course now here the vorticity is just bounded for initial data like this um it's known that one can solve Euler and keep the regularity of the of the boundary and due to this regularity of the boundary the solution will be in an infinity lectures so i mean first this was also first proved by Schumann the fact that you keep the regularity of the vortex patch so um yeah so um right so so actually um i think okay no i need to i need the one half yeah but i forgot to put there is one half right so for vortex patches so then for vortex patches there is no loss that exponential loss also doesn't hold so this were um this were more or less the results it turns out that that one half then was later improved in another paper by Abidian Dongshan and i think the optimal one is three-quarter that result like if one is really trying to find the rate of convergence for vortex patches for vortex patches we can get that power to be uh three-quarter okay but let me not go into all this um okay now can we lower the regularity more what if the curl of u0 is in even weaker spaces so by the way here this this convergence this convergence is happening in l2 so so this kind of this kind of solutions really we can we can look at solution of navier stocks which are just in the energy space right so um what i mean by this and what i mean by this is that for the results i'm mentioning there more or less you can you can even take an initial date of navier stocks in l2 like this but then assume that um this u0 you just need regularity on the u0 so it's really for instance the curl of u0 which is in in that those spaces but of course now the initial data of both equation do not match so then you have to pay here something like the initial data of navier stocks navier stocks minus the initial data of oil learning in l2 so you can more or less prove results also like this in which case you are really dealing with um weak solutions for navier stocks okay now if one wants to go even weaker i think okay john mark is here so the the weakest result will is i think is due to john mark where it's when you look at vortex sheets right so if you take so if you take the initial data u0 if the curl of u0 is in l1 or or a measure if the curl of u0 is in is is a measure but you need a signed measure let's say so if the curl of u0 is a signed measure um then it turns out that one can still solve Euler and you can prove that solution of navier stocks will converge to the solution of Euler when the viscosity goes to zero but um i mean for me that result is really in a sens is really like the weakest regularity you you can allow so that you can prove that navier stocks converges to Euler so so in a in a sense like the results i mentioned first the first one was strong solutions was really strong solutions then we went to results where we are dealing with energy space solutions but we are imposing regularity at the limit system so then but the regularity on the limit system is slightly weaker and now you can go even weaker and that the the result of john mark where you can even deal with uh so called vortex vortex sheets meaning that uh so it's not this picture but your original vorticity can be let's say can be like a measure can be a measure with some u0 in l2 but in this case you need uh you need the sign condition so you need omega 0 to be positive right or or for instance you can put positive plus some part in lp is to be bigger than that so um so if you are in in this kind of regularity if you are in this kind of regularity you can still you can still just prove the convergence you can still prove that uh u n will converge to u0 but in l2 without rate without but with yeah weekly just weekly so very weak very weak result in this case so in this case you can prove that u n will converge weekly to u0 in l2 l2 let's say so this is uh john mark in this result let me ask you one problem both okay let's forget the p just the w positive okay so uh uh what's uh corresponding w positive that what it means so it i mean it's it's uh i mean this will be like a whole lecture like i mean it's it's some sort of i mean like in in a sense okay and i mean i like the this result i mean and also the the transition between like going from strong spaces to weak spaces so here a result like this is is based on the following you ask yourself can i i mean a very natural question because as i said first i talked about strong solution for navier stocks and oiler but we know that navier stocks has weak solutions so i can look at navier stocks in spaces like l infinity l2 intersected with l2h1 so i can do just energy so if i do just energy if i do just energy what are the estimator i will get i will get i will get an estimate of this type time t so this will be equal or less or equal than the initial data or so let's say constant okay so what does that mean it means that uniformly in n all what we can say is that we have some l infinity l2 bound the bound we have here disappears at the limit so you are not having you are not getting that un is in l2h1 uniformly so the only uniform bound you have is un is in l infinity l2 now if you want to pass to the limit in this equation i mean in the sense of distribution this will go to zero this will be stays a pressure this will converge to what we want so then what can we do with this of course now this term due to the divergence free condition this term you can write it especially when we deal with weak solutions we can write it as the divergence of un tensor un right so you can write it down in this form so this makes sense like this is l2 l2 l2 so this product makes sense so normally it makes sense to ask yourself whether just from the bound you have here you can pass to the limit in a term like that but this is weak convergence like if you are in infinity l2 all what you can say is that this convergence weekly to you zero we can in l2 so normally it's not clear it's not clear whether this will converge weekly to this right and okay so the the the work of john mark is to is to say that if if you have that condition on the sign of the measure then you can write this like it's more like a bilinear term and and there's a lot of work to prove that this actually happens using that the vorticity is of course now if you are if the vorticity is in lp p bigger than one then it's then it's okay if you know right so if omega zero is in lp p strictly bigger than one this is correct it's it's not difficult it's two lines like you can prove that you keep some regularity and you get it but to do it in m plus i mean the problem is coming from possible concentrations it's not oscillations but the problem is concentration so what is the critical case that you can what deal with what is the critical that's l1 l1 is the l1 is the critical and one is the critical space but it's it's an issue of concentration not not oscillations okay so so this is as of now all what we talked about has nothing to do with the boundary right so all what we talked about has to do more with strong solution versus weak solutions and um okay now what happens why why this breaks down if we have a boundary why this breaks down if we have a boundary right so um let's okay so let's keep the boundary conditions there so we talked early about this equation on um on wn so now the case with boundary so so let's let's look at one the boundary condition one so if i try to do anything even like if i am in two dimensions and they want to do some estimate like this one what i will try to write down i will try to write down an equation on on this guy so so here i will i will be writing a term like um u0 no usually i do it the other way okay okay so this is how you write your non-linear term the difference between um u n grad u n and u0 grad u0 right and then you put minus u n la plaschen n plus the pressure okay so of course i mean these are the wn's now um this is an equation on wn that i i didn't write earlier now this equation if you want to do any type of energy estimates like what we've been doing there let's say if you want to do energy estimates in l2 you want to multiply by wn and integrate by parts but then the issue is that this guy will not integrate by part you cannot integrate by part because now this doesn't vanish wn doesn't vanish on the boundary right because wn wn is un minus u0 and on the boundary u0 doesn't vanish right only the normal part of u0 vanishes so um so we cannot use we cannot do energy estimates um so what can one do then what can one do um i'll mention some results i mean there is a very interesting result it's a very short note by cat 2 so there is a result by cat 2 that i'll i'll i'll i'll talk a little bit about it from i think 84 if i remember well um um who who proves somehow what we it's more like it's more like in the spirit of an if theorem if something happens then the conversions takes place um so let me let me mention this result uh it's one of the it's one of the it's one of these results that is now well um it's called the cat criteria and it has it has also very nice physical interpretation actually like the physical interpretation of the result is very interesting so the cat criteria says the following so so of course here we are we are in a situation where Euler has a nice solution right so Euler has a solution in some HS space nice so it's not a question about regularity of the limit problem the limit problem can be very very smooth um so T star again is the time of existence of Euler time of existence of Euler and again i take T less than T star then we have the following three three statements which are equivalent um un goes to U0 in N2 uniformly in time ok so if the convergence takes place in L2 this is equivalent to to the fact that the the energy dissipation the energy dissipation um ok so one and three one says that if U1 goes to 0 strongly uh if U1 goes to U0 in L2 uniformly in time that statement is equivalent to the fact that the energy dissipation the energy dissipation goes to 0 but the striking thing coming from cattu is that it's not necessary to check that the energy dissipation goes to 0 in the whole space but but just in in a small strip around the boundary and this strip is of size the viscosity nu so basically gamma nu is the set of x in omega such that the distance between x and the boundary is less than nu so i mean more or less it's clear that three um three implies two right three implies two so if you if you are saying that the energy dissipation goes to zero in omega it's also clear that the energy dissipation in a strip will go to zero but what Euler is what Cattu is capable of proving is that this is enough to prove that this whole yeah yeah ok what what i call nu is always it's it's nu n yeah yeah yeah yeah always like nu is nu n Deci, acesta este resultatul de cateaua... În câte dimensiune, dimensiunea aici nu se găsește. Dacă vreau să reușteți oameni în 5 dimensiuni, dimensiunea nu se găsește. Deci, ce este ideea? Ce este ideea de oameni de acest resultat? Deci, ideea este foarte simplă. În acest moment, este un 7-page node care a reușit în acest procediment. Nu se găsește că acesta a fost un resultat foarte famous în acest subject. Deci, ideea este să văd... Încă nu văd... Ceea ce nu văd de acest resultat? Deci, oameni de ce nu văd de acest WN este că nu se găsește în cateaua. Deci, acesta a fost... Încă nu văd de cateaua... Încă nu văd de cateaua... Deci, încât ce a fost cateaua este un corrector care nu se desfie de acest acest equat. A fost adăugat un corrector. Deci... Nu se găsește... Deci, a fost adăugat... equată. Deci, a fost adăugat... Încât de a fost adăugat U0 oameni... oameni adăugat Bnu B, pentru că este un layer de boundary. Și acum, acesta este equal a 0 în boundary. Deci, basicamente, a fost adăugat de boundary a fost adăugat de boundary și a fost adăugat un corrector. Încât... Încât... a fost adăugat... exactul acestul de construcție. Încât de a fost adăugat de construcție pentru că îmi găsește... îmi găsește... îmi găsește... oameni adăugat... oameni adăugat, încât de a fost adăugat de un corrector. Deci, basicamente, a fost adăugat de ideea. A fost adăugat de corrector. Și acum, încât de a fost adăugat o definiție de WN, acum va văd și acum, acest lucru se văd în cazul de boundary. Acest lucru este 0 în cazul de boundary. Dar acum, acum, dacă vrei să rădă acest lucru oameni adăugat de alte urmări. Pentru că acest lucru, pentru acest lucru, nu se întâmplă unele equătie. Acest lucru, nu se întâmplă unele equătie. Îmi găsește oameni adăugat de extra urmări aici. Și am oameni adăugat de altul urmări, încât de a fost adăugat de tot. Îmi găsește acest lucru. Ok, dacă vrei să ajungi și vrei ce este oamenilor b nu. Aici, cât de a văd b nu, este mai mult oameni de u că văd decăi de acest lucru. Deci văd oameni de 0 aici. Dar oameni de 0 oameni de 0 aici și văd decăi încât. Deci vrei să văd oameni de construcție aici pentru că vreau să văd un alt resultat este un alt resultat că vreau să văd ce nu este oameni de u nu, ce? Nu, nu, nu este oameni de u se simila nu este oameni de u țită zileu este o soluție de oameni acest lucru este oameni de u este oameni de u de acest lucru se simila nu este oameni de u de acesta nu e acest lucru de văd de b de ba cei oameni de urmă de a fi a dat acest lucru este foarte bine a actinguit Călătura importantă se văd, se văd un termul de acest, se văd trei termuri, trei maini termuri, unul în care acest lucru a fost aici, și unul în care acest lucru a fost aici. Și apoi trebuie să așteptăm. Dar, în urmă, s-am văzut un alt result în care îl spune, un poveste mai în detilă, care mără, mără, mără, a fost also inspirat în catăru. Ok. Aici, un B-nu este un B-nu? Da, da, albine, nu este un N. Nu este un N. Dar ea dependă în N. Deci ea, ea, ea, ea, ea, ea, ea, ea, ea, ea, ea, ea, în același region. Da, Ok. Deci, albine, el odată poate, peruvă...?ded în... Adică, când va face, nu este o vizioamnă de mără, peru... Eu albine, nu este albine, adică. Am trebuit să văd o problemă în lumea mea, Omega este R3+. Omea, omea mea este apă, apă plaină. Asta este omea mea, omea mea este plaină. Asta este omea mea, omea mea este apă plaină. Și vizicosită îmi replacă în zi în direcție și vizicosită în xy în direcție. Asta este vizicosită aici și vizicosită aici. Asta este o vizicosită aici. Și acesta este o motivo dintre ce vizicuitam încă nu în anul an. Asta este o因amă nu încât, pentru că dacă anul an eu poate mără nu în anul anul elected in an pentru să urmăriți n. Deci, și apoi, apoi, apoi, apoi, apoi, apoi... Deci apoi, apoi, apoi, apoi... nici în catculițin . Cei în Study N-ul, de Analytics M blu. Nu specialează ceilal tuturori nu urmă rânte 가자 să fie cate venită. Nu nuriți plăcut bine copte Hill. Nu urmă rântulять. Ba. Urmări lev metalul TPire. acordul LP, în locul steasulă陽. Arsamtul spune içerici oțiua trebuie să pyta. de faptul că square root de eta, grad x, y, u, n, square root de rho 2, 0 în L2, L2. Deci, în compară a resulta de K2, resulta de K2 este un if, este un if theorem, este se spune că acesta convergență este equivalente a acestea. Deci, dacă vă uitați acest, vă mulțumim să văd acest. Deci, acesta este un resultat că vreau să văd în... Da, a fost bine 20 ani agoi, a fost 98... ...seampt că, dacă vom replica viscositatea de... ...nevru difusie, dar încât... Am făcut mult mai viscos efect în direcție X-Y. Am învățat că nu trebuie să fie faptă răspunată doar în eta. Deci am făcut mult mai viscositate în X-Y. Apoi... ...să poți să pui rezultă. În urmă, normal... ...espește când îl explica sistemul printului și ca... ...as usual, tot ce vom spune la printul, ...să îl aplicăm pe acest... ... pe acest probleme cu faptării viscositate. Fizicul, cîncă este fără fără fără, ...să se întâmplă, da, ...când dacă ajutăm atmosferul, All atmospheric flows, For all atmospheric flows, it makes sense To assume that you have different viscosity in the vertical and in the horizontal direction, right? So, as a model, it's very, very physical model. Ok, any questions about the statement? Ok. I'll try to give you a proof of it. Ok. And actually the proof, I mean, if you see this proof, I mean, the proof is really inspired also from the result of Cato. Right? Let's do this. I need the W equation. Yeah, of course. Da, aici nu am înțeles pe acest, dar am înțeles că tot ce am înțeles este cu aici cu o condiție de despre bălături, ca? Deci este problemă cu condiție de despre bălături. Deci am înțeles pe probleme în spasă. Am în spasă cu condiție de despre bălături. Deci... Ok, deci... Deci ce nu am să fac este să construi acest bn, ce îmi spune bn, acolo. Deci, ce îmi spune? Îmi spune bn. Am spune bn, nu, pentru că am nu înțeles eta. Deci îmi spune bn nu înțeles eta. Deci îmi spune bn. Deci îmi spune bn în zi unea plus w în zi unea. Îmi spune acest bn în zi unea. Ok? Și... Am văzut că b, când zi unea plus infinity, am văzut asta să fie zero. Am văzut să fie 2dk. Am văzut că mă bn nu înțelege energie. Deci am văzut mă bn să fie zero în infinity l2, ok? Deci îmi spune, când zi un e infinity. Deci, nu este multe, nu este multe clară. Și acest este, deasupra, nu se poate reacția de o equătie. Când îmi spuneam prentor, prentor e adevărat pe ideea ca îmi trebuie să se răspună o equătie particulară. B, and I will just take it to be, I mean the simplest thing exponential minus z over square root of mu c'si, c'si is a parameter that I'm going to choose. Does that make sense? So you just put some boundary layer, some ... Este like the simplest boundary layer you can think about, but that doesn't solve any equation. It's not coming from any equation. Just you have the boundary condition that is bothering you. This is not 0, and so you add to this. So then if you look at this plus bn, this will be equal to 0 on z equals 0. Okay? So, then let me call WB, what I'm calling, W is U0, so I'm just changing notation so that I don't carry too many, W is U0, so then WB is this W plus BN, Dar le dăm, ce este acesă o frumosă despre aceea? Dar acest este equal a ce? El este de course... El este de B, nu este acestea. Unul de B grad W, unul de termen la acest lucru este a treaba. Am grad B, actual ar trebui grad B, acest termul nu este normal, a fost 4 termuri, a fost 4 termuri, only one of them is coming from W, which is W grad W, and I get the 3 others. And then I have the viscosity terms, which are minus the pressure. Right? So this is the equation satisfied by this WB. Now the reason I write it like this, because now I can take, I can now compare it to the Navier stocks. Right? So then I can make the difference between the two equations. It's here, right? The big grad B is here. Ok, so then we do the, then maybe I'll skip some of the details, I think, because let me jump, right? So let me jump to the equation on R. So then I'm calling V, V, which is VN, is UN minus, minus this guy. Ok? Then, so you take the equation on UN, and you subtract this WB. And then you do energy estimates. Ok? I'm skipping some, I'm skipping the calculations, so, but, ok. So what do you get? And I integrate in time. I integrate in time. So I end up with this equation, L2 in space. Ok? This is less than what? The initial data. Ok? And then I have a bunch of terms mostly coming from multiplying these things by V. Alright? So I have plus. Ok? So I have terms like this. I have also terms like these here. All the other terms here I should put also. And they have other terms, other terms coming from the UN minus this. Ok? Right? So I have some other terms, but I'm worried of giving you every, all the, These ones, or let me insist on the important terms. Maybe that will be better than trying to, ok. Normally, as of now. So I wrote what I'm missing. I'm missing UN grad UN minus, right? I didn't write. And this you write as UN grad UN minus, like a B plus UN minus, right? You write it in this form. Ok? And this guy will be my VN. This is VN. Both of these are VN. This is V. Now this term disappears when you do energy estimates, right? When you do your energy estimate. When you do your energy estimate, this term disappears. So normally I just, I only need to add this guy. Ok? So then I have times V. One question. You don't, your corrector is not different from 3, right? That's true. You are right, you are right. So that's something I'm hiding. The word here is almost correct, but not complicated. Yeah, yeah, yeah. That's true. That's I'm hiding that. Yeah, I mean it's. Ok, but what you said is correct. Normal I have to make it. I have to play with it a little bit to adjust it. Yeah. So I just want to insist on the important terms, right? And show you the main idea. So what are the important terms? So there will be a term coming from this. So the first term coming from this is. I mean, clearly the most difficult things is when you are taking z derivatives here. Because when you are taking a z derivative, there is that small parameter in uc will comes out. So the difficult term, the difficult term here is V3 DZBV. Not DZW is not a problem. It's more like the DZB that is problem, right? So this is one of the terms. So what can we do with this guy? So I'll show you. So the idea about terms like this is that we are going to use that V3 and V vanish on the boundary. Both vanish on the boundary. So what I'm going to write it down is I write it like this. I'm going to use, right? So this is the idea. Omega, again, is the half, the upper half space. And then we use a hardy type inequality to bound this by DZV3 L2, DZVL2. And I'm going to put this in L infinity. Now, this guy, when I differentiate, I lose, right? 1 over square root of nu xc. But when I multiply by z, square I gain. So actually this will be square root of nu xc W infinity. So xc here is a constant. Xc is a constant that I'm going to choose. I'm going to adjust. So basically the idea of xc is that it's really telling you what is the size of your boundary layer. I mean, it's an imaginary boundary layer because it's just boundary layer that makes the calculation work. It doesn't have physical meaning, but it will make the proof works. Ok, now, what's now the good thing? This guy, what I'm going to do with this guy, I'm going to use that I am divergence free and replace this by the derivative in xy of V. I mean, bounded by using divergence free condition. This is very important because somehow I control this better than the z derivatives. This is really where having eta Laplacian xy will help because now I control this better than I control this. And then I can use some Cauchy Schwartz. I put nu over 4 dz. And here I have cc. So you can see. If I want a term like this to be absorbed by the viscosity, if I need this to be absorbed by the viscosity, it's better than this guy is less than eta. This guy should be less than eta. This guy should be less than eta if I want this to be absorbed by this guy. So you see this term, the term coming from dzv, I know how to absorb it by the first part of the viscosity. I need also this guy to be absorbed. So this is how I deal with this term. Now the other problematic term, this is not a big problem actually. The other problem is this term. These are the two main terms. Like I said, this and this. This is the other main term, so let me explain it. Let's do it here. So that term, so it's like this. So I can integrate it by parts. I integrate by parts and then I integrate by parts. So then I can put dzv square in some constant nu. So of course now this I can absorb it in the viscosity. And this guy is less than what? This guy is less than some constant nu. Now when I take a derivative, there is the factor square root of eta c that comes times w in infinity. Of course now when I do the derivative, there is this factor that comes. Normal, it comes with the square. But then when I integrate, when I do the L2 here, I regain one of these factors. So that's why it appears with the square root. So I need this term to go to zero. That's the only way. I mean, because this term, there's nothing I can do with it except that it has to go to zero. I need that term to go to zero. So now the conclusion is that if I take the conclusion is what? I'm going to take c to be eta divided by c times, let's say 4. This is my conclusion. I take c to be eta divided by this. So then this guy becomes eta over 4. Then it can be absorbed. But this is now eta. C is like eta times a constant. So then I see the constant, the c here. So this will be, this will be like some constant square root of, right? So if this constant, if this also goes to zero, then I am done, right? So then the convergence will take place. So that's more or less how this result goes. I mean, if you understand this, that's exactly how you can do the Cato criteria. Basically, the Cato criteria, here there's no if theorem. This is no if theorem. In the Cato criteria, the difference with this result I'm showing you is that it's really at this step. All the rest of the calculation is more or less the same. It's really at this step that instead of keeping here, instead of keeping here something that you know that it goes to zero, here you will keep something around the boundary. It's really at this step where you will say, okay, I know that this goes to zero. The energy dissipation around the boundary goes to zero. Then you deduce that the convergence takes place. Okay. Now, let me mention a few other results about the inviscid limit again. I didn't give a plan of the lectures, let me just say a few words about the plan of the other lectures and then finish with today's lecture about the inviscid limit. So Thursday, Thursday we'll be talking about Prentor. So first of all, I will give you the derivation of the system and then some well-posedness result, different kind of well-posedness result. So this will be lecture 2. Lecture 3 and lecture 4, I don't know in which order because both will be more or less independent. One of them will be blow-up for Prentor. So I give two kinds of results about blow-up for Prentor, one about stationary solutions, one about the time-dependent and one about the time-independent. And then I will go back to inviscid limit but with Prentor. Of course, now the two results I showed today, like Cattu or this result, both are results where you prove the convergence, but there's nothing about the boundary. So it's true that B appears here, but B at the end is just used to allow the integration by part, but I'm not justifying B. I'm not saying that the B here has any physical meaning. And you can see it from the result because the B itself, the way the B is chosen, it goes to 0 in L2. And my result is to say that the result we are proving is to say that UN goes to U0 in L2. U0 is the same as W. So the B here is not seen. In this result, the B, you don't see it. Of course, now there are more precise convergence results but requiring more regularity where we can also see the part coming from Prentor. Ok, now I want to end with a few extra things about this limit from Navier's talks to Euler. And it's more like in the spirit of the result of Jean Ma, but even weaker. So what happens if all you know, what happens if I give you, what happens if our starting point, so now I'm forgetting about the boundary again. I'm forgetting about the boundary. Take a problem like this. I don't need the two viscosities. I just put Laplacian here. So take a sequence. Take some solution of Navier's talks with an initial data in L2. It can be the same initial data. What happens to this when N goes to infinity? Let's say put same initial data. What happens to this when N goes to infinity? Now, in this setup, of course, Euler is not well posed if I just take a data which is in L2. I need more regularity so that I can construct some solution of Euler. So it's not clear what happens. And again, the difficulty comes from this term, what happens to the limit of when N goes to infinity? Now, in the literature, at least people thought about these problems and there are attempts to somehow describe this limit or to describe what happens. So one attempt is by Diperna and Maida. Diperna Maida introduced a notion which has introduced a notion of so-called measure-valued solution for Euler. So basically, without going into the detail of it, I mean it's more sophisticated than what I'm saying, I know that my UN will converge weekly by just compactness. By just compactness, I will say that my UN converges weekly to some U0 when N goes to infinity in L2. But I cannot deduce that there is strong convergence. All I can say is that there is weak convergence. And if you have weak convergence, one object that characterizes weak convergence is the young measure. So Diperna and Maida, they introduced some young measure. There is a whole construction of a young measure that they will call a major-valued solution for Euler. And more or less, it captures the lack of strong convergence. It captures the... So this was one notion. That was introduced maybe, I would say, more than 30 years ago, like more, like in the 80s. I mean it was used few times, like it came back, comes back from time to time. This was one notion. Another notion of so-called very weak solution to Euler is a notion by Pierre-Louis Lyons. So Pierre-Louis constructed something that he called dissipative solution for Euler. And it's also... these kind of solutions are used to try to justify this kind of limit. So, yeah, but maybe I will not... I'm not planning to go into these notions, but somehow these are notions that people try to construct, to try to capture what happens in this limit. Because, like, with all what we are going to do with Prentoll and the question of weak convergence always remains, right? Whether we construct Prentoll or not, or what we... I mean, the question about just weak convergence in this sense... In a sense, we still don't understand where is the lack of strong convergence. I mean, of course, now here, what I presented here is without boundaries. The boundary is not present here. Of course, this kind of construction of the Pernamide or Lyons can be combined with the boundary. Okay, let me stop here. Sorry, I have a question concerning the assumption. So, is it possible to find any other construction of P of N? Such that... Let me ask the first question. So, can you... No, no, no, on the other blackboard. So, concerning the last estimate, so, is it here the only place where you use the last assumption nu over eta goes to 0? Yes, only this term. So, is it possible to find some other B of N such that you can avoid this assumption? I don't... Okay, of course, now this is not the only choice. I don't have to take it equal, I have to take it less then. I'm not saying that there is only one choice of length. And that's also... This also shows that this boundary layer doesn't have really what I would call physical meaning. So, if you want, you can take it less as long as this goes to 0. So, you have room in terms of your choice. But somehow the choice given there is the best because it gives you the best decay of the error. So, if you want to estimate the error, like in the paper I wrote, I estimate the error and this term appears in the error. This estimate depends on the constructor of B. Yeah, I don't know. And I think no one knows how to construct a B if you don't impose something like this. Of course, now I didn't show it, but the Cato has a construction of a B, very similar to this one, more or less the same actually. Cato has a construction of a B similar and the convergence will take place as with the if theorem. So, at some point you are replacing the study of this term that I have here by another term and you have to check that that term vanishes on the boundary and that sort of dissipation close to the boundary vanish so that you can prove the convergence. Ok, so Thursday we start with parental.