 So first of all I want to thank the organizers for inviting me. It's it's really a pleasure to be here and I really enjoyed being here for a few days now and it's Okay, so I'm going to talk about a joint work with Jacob Bedrosian, Pierre Germain So Jacob is in Maryland, Pierre Germain is also at NYU and this is about hydrodynamic stability So the It's long. I mean it is related I think to the talk of Carlos because it's also about Understanding something about long-time behavior, but here in fluid equation so Hydrodynamic stability at high Reynolds number, okay, so maybe this is a slightly different I mean it is exactly the same title, but I Wanted to insist on the high Reynolds number and I'll explain I'll explain that So what is our starting point here the starting point is the Navier-Stokes equation So let me write down the Navier-Stokes equation I'm going more or less to assume that people are familiar with this. So I'm not going to recall Navier-Stokes and so on so If you are not familiar with that Okay, try to read a little bit about it or maybe ask me questions. So This is the equation We'll be looking at So V is the velocity P is the pressure and We are incompressible. So the divergence of V is zero and we take this with an initial data Okay, so this is the equation. This is the equation. We are interested in So immediately now I'm going to decide What is the domain the type of domain I'm going to look at? Okay So again nu is what I call the Reynolds number minus one. So this is the viscosity Which will be Bigger or equal than zero when nu is equal to zero we recover the Euler equation Right nu equals zero is in the inviscid case that corresponds to Euler equation So here V is a function of t and x and it is in Rd d will be two or three in our case. So we are either in dimension two or in dimension three Okay, so I'm going to use the following notation in terms of how I denote the three component of the velocity So my V will be either Will be it will be like either V1 V2 So it's not a square, but it's It's more the three components or Depending on which case we are it will be either this or this So this is this if we are in two dimension and this will be if we are in three dimensions. Okay, so Usually whenever we look at the Navier-Stokes equation or Euler will be also talking about boundary conditions Here I'm going to take really like the simplest example Where there's no boundaries. So my domain will be in dimension two if I am in D equal to My domain so maybe I should say that x is in Here as I will say x is in some domain omega My domain omega will be the torus times R And if I am in dimension three my domain will be the torus times R times the torus Okay, I'll explain why I'm taking this So this means that x is in the torus y is in R x y z Okay So then in that case I don't have to I Don't have to specify boundary conditions Except the fact that I will require functions to be periodic in x and periodic in z. Okay Okay, so now the the question we are interested in here is To try to understand The asymptotic stability of the simplest possible Non-trivial stationary solution. Okay, so the question is what is I mean, of course V equals 0 is a stationary solution But okay, that's what we can understand the stability of that and there are that will Correspond to like small data Global existence or things like that, but then we want to go one step further and consider Non-trivial stationary solution so now For the Navier-Stokes equation the easiest the simplest not the simplest non-trivial stationary solution is the quet flow So we are interested in the stability of Quet flow Okay, so what is the quet flow? So let me write it down first in 2d So if we are in 2d the quet flow correspond to the case where V So V normally now is a function of x and y And with my notations here V will be y 0 Okay so if you take if you take your velocity to be y 0 and You plug that in into the equation This gives you a stationary solution. Okay The Laplacian the Laplacian just vanishes and V grad V V grad V also vanishes there. Okay, so everything actually All the terms vanish, right? Okay, so same thing in dimension 3 in dimension 3 the the in dimension 3 the quet flow will be Will be y 0 0 Yeah, so it's period so so it is So why is in the whole space? Why is in the whole space but x and z are periodic? Okay, so so so the flow the way you have to think about the flow that I can draw it in two dimensions So if this is my x This is my y. So if I am in two dimensions the way you have to think about your flow is It is a flaw like that. Okay. This is your flaw, right? So maybe I can draw it like It only depends. It only depends on why right? It only depends on why but it is About the first component, right? I mean, that's also a reason why it is divergence free because when you compute DX of y is zero. So so it is divergence free Okay, so this is what we are interested in is and to understand the stability of of this Stationary solution I mean more precisely. I mean it more precisely we are more interested in Long-time behavior. We are more interested in asymptotic stability. So I want to perturb this and prove that When t goes to infinity, I converge back to my to my stationary solution. Okay Okay, so this is this is more or less the the question that we want to address. So Let me just say one thing before If new is equal to zero So mainly and namely if we are in the Euler case, right? If I am in the Euler case Actually this this stationary solution Comes with the whole family of other stationary solutions which are called shear flows Okay, so I mean in the case of Navier-Stokes. This is actually like an Isolated station is I mean isolated. I mean you can multiply it by a constant add a small add a small constant I'm not considering that but the only type of stationary solutions for For the Navier-Stokes case, I mean it has to be it has to be an affine function here So that the Laplacian vanishes, but if you are in the Euler case, I mean all shear flows Shea flows are stationary solutions So basically anything of the type anything of the type Phi of y zero Where phi is any function of any function you you can think about let's say has enough smoothness so that you can define things These these will be also stationary solutions. Okay, but if you wanted to be also a stationary solution for Navier-Stokes Then you want phi of y to be an affine function Okay, so I mean there there is a small This is important difference between the two cases because It tells you what what you can expect when t goes to infinity, right? It turns out like in in the case in the case of Euler if you take this Stationary solution and you perturbed I mean you can perturbed a little bit by something of this type and then you stay there So in the case of Euler you should not necessarily expect that necessary you go back to What you started with right because you have whole family of stationary solutions, which are just close to your To your quit flow. Whereas in the case of Navier-Stokes if you start from this you perturb it Of course now if when you perturbed and you keep the behavior at infinity the same Then I cannot really perturbed by multiplying by by constant close to one or adding small constant because I'm fixing the behavior Infinity so then in that case I should expect maybe that I will go back to this Okay, so that's that's more or less at least the the goal of what? What I want to talk about in this Three three courses Okay Should I keep yeah, maybe I can write here. Oh, yeah, let's use the third one that okay now Okay, so So basically what I would like to do is to take V and write it down as my Quit flow plus a small perturbation. So let me now use notation in three dimensions So I'll be going sometimes working with two dimensions sometimes working with three dimensions But basically let me Do notations in three dimensions. So then I'm going to write my V as y zero zero Plus you So you is the small perturbation. So think of you as small perturbation small perturbation Okay, now it's a small exercise Small exercise to rewrite the equation satisfied by you, right? So it's a small exercise to rewrite the the and basically I mean the whole exercise is just The non-linear term you have V grad V V grad V now becomes four terms One of them disappears which is when you do the quit flow you put it twice that disappears and then you get Some other terms. So then let me write it down this this What we get in that case Why the ex you So you is a vector, right? I mean you have to keep in mind that you is a vector I'm not putting necessarily the the arrow on top of it Okay, so it turns out that there is another term here that comes minus u two zero zero That's okay Okay, so so I mean this is just the term that comes I Mean there is a term like u two zero zero, but I wrote it on the other side. Okay? Yeah You are right. Okay. Thank you. Um, okay, so Yeah, I mean I keep sometimes switching from where I write things. So so right. That's so That's a good. Okay. So this is the equation This is the equation and I mean like you have to think a little bit about where this comes from I mean just comes from In in in your term V grad V Right, so you have a term V grad V and In this term if I put here you You grad Y zero zero Right If you look at this term this gives you a u two on the second component, right? So this is equal exactly To you two zero zero, right? Okay? So again, I mean this is your arm Yeah, yeah, yeah, yeah, I insisted on that V1 V2 those are indices like It's not a power and I I Did it in purpose to write to insist on that there. Okay. So um the So this is the equation now. We are looking at right so the question the question is the following Um So I want to understand I take this equation I take a u which is an initial data for you which is small and Some you zero or you initial and the question is Can I prove that if? You zero in some norm is small Then then I have global existence here, then I have a global solution Then we have a global solution That remains small or remains under control and goes back to zero We have a global solution you and you of T Goes to zero and T goes to infinity Okay, so that's more or less the type of question here. I was not precise in Writing down the equation because what will be important here is the norms In which norm I am small Okay, so there are two question. I mean there are at least two important question is Which norm and what do I mean by small? What do I mean by small? And it turns out okay, it turns out that if you fix the viscosity if you fix the viscosity The the question is not difficult like the the if you if you have a fixed viscosity then It's true that the statement is correct and that's was known from a long time ago And they think I mean it was known even like in the 19th century the end of 19th century This is related to questions that people were interested in basically The end of 19th century like people like Reynolds really these are questions that they they they were interested in and there were even experiments about it and During that period there was something called Somerfeld paradox and the somerfeld paradox is exactly related to the question We want to consider here. Basically, it's the fact that if you linearize If you linearize the equation and I'll be talking about the linearized problem If you linearize the the your your equation, then you expect To have linear stability But on the other hand people doing experiment were not really able to see the quite flow as as something that they can consider to be stable, right and And and the reason which I think is related to the result I'll be presenting here is the fact that the the smallness Here will depend on the viscosity, right? So the smallest depends on the viscosity. So if Okay so so the the question is How to relate How to relate the norm here to the smallness here, okay? so so now in the physics literature like It's I think it's more recent notation. It's more recent terminology There is a If I if I try to rewrite the the the question it's it's more or less the following so given a norm Given a norm that I'm going to call an initial norm Mathematically, this will be either sobolef or jeuvre So for us it will be sobolef or jeuvre I'll I'll give you an exercise. I think since we have many students. I'll give you an exercise about jeuvre No, so it will be the sobolef or jeuvre in the physics literature This will well what they will be calling rough data and smooth I Think in there in the physics. I mean like in the numerics. They do. I think they consider analytic But it turns out the mathematical result works also with jeuvre. I'll Say a few things about that so Then the question will be the following so what is What is the smallest gamma bigger equal to zero such that if your initial data in your norm in your initial norm is Smaller than new to the gamma So here much smaller means I think I maybe I should sit right in town Smaller than some fixed constant but this constant can be small so what is the smallest gamma such that if You start like that Then your U of tf like your final this will be like the initial this will be the final stays less than one for all time and It goes to zero. Okay, so more or less. This is how how you should understand the the the problem We want to address and gamma will have a name gamma will be called stability threshold Right, so gamma will be the stability threshold and I mean this is In recent years has been Um Studied a lot in the physics literature Maybe I call it Of course, of course the reason this is called the stability threshold. I mean like if you try to find the optimal gamma What you should also expect? We also expect that We can find initial data we can find initial data such that u0 initial is Larger Than some New to the gamma and And you have instability and There exists a time t There exists a time t that depends of course on the viscosity such that You at time t in this final norm Is larger than some fixed number delta. Okay becomes of order one in finite time Okay, so basically this is what one should that One should try to prove basically, I mean even though like we are not able to prove all this But this is how one should try to Characterize the optimal gamma like which is like the optimal size of your initial perturbation so you have a Size of initial perturbation if you start smaller than that size then you get Stability and you get also asymptotic stability and you can find Some candidates which are slightly larger or which are exactly at the boundary for which You don't have stability. Okay Okay, so that's more or less the How how we should ask the question So it turns out it turns out for some technical reasons, I mean like I mean, it's not technical, but I think it's important I mean one has to think a little bit about it that this type of result We will be able to prove such kind of result, but with another condition With the fact that this has to be Smaller than some other With some gamma prime No gamma prime less than Okay, so we will be able to give results of this type But I will not allow my initial perturbation to be Very large Right, it has to be still under control and the reason is that it's it's like it's I will The way we have to understand this result. It's really You have to follow your solution till it reaches the size one So then I have to start really small so that I can follow it Okay, so so let me write down a few results and I think for those I think this is like a Table that Pierre wrote when he gave his talk so I'm Going to redo the same thing. So what what do we what do we know about this problem as as of now? So I will draw a table to tell you What what we know Okay, so if we are if we are in dimension two If we are jeuvre So in dimension two if we are in jeuvre we know that gamma is equal to two. Oh, sorry gamma is equal to zero And we can we can take gamma to be zero In the case of so ball F It's we know that gamma is less or equal than one half. So what I mean by less or equal than one half It means that we don't have We don't have the second part like we don't have we don't know whether this is the optimal gamma or not In dimension three in jeuvre a gamma is equal to one and in so ball F Gamma is less or equal than three halves Okay, okay, so this is more or less the the table I Mean there are different works about these Right, so yeah, let me not now Give names, but I'll I'll I'll I'll get to giving names and so on No No, no, so so basically basically like in the so ball F case in the so ball F case We only have this type of result We don't have this one and there is there is a reason behind it There is a reason behind it is that like this type of results this type of results in the jeuvre case is that you are constructing Precisely a way how the solution is going to grow and you are controlling precisely how how this growth is happening, right? so So then I mean you are starting large you are starting large enough basically because you are starting like you'll be starting Like at this size or at this size so you can really follow it in the so ball F case You have to be even much smaller because of the nonlinear term the nonlinear term in this case is worse Right, so then you you may have nonlinear effects that you don't control so then as of now all what we can prove is that You can prove the stability part, but whether that is optimal or not you we don't know So so basically basically it means that It means that in the jeuvre case you can decide you can compute precisely what is the Instability route how how you go from being small till being size one you can follow it completely in the jeuvre case In the so ball F nonlinearity become more important and maybe there are other nonlinear effect that Makes the picture different So that's that's the reason basically Okay Other questions or Yes Where does the expectation come from that there should be such a scaling with the viscosity at all? That's a good I Mean that okay, I mean that's what you observe so but I think I Would say like the physics literature people realized this some time ago and they They started thinking about that and Yes, so I mean I would say like in the end of the 19th century People were not thinking in this way. I guess and that's how the paradox started like there was a real paradox that I mean One can read people can read about it like you can type somerfield paradox and it will come exactly about this question Like I mean you have a flow at the linearized level at the linearized level you expect it to be stable But it's not something you can observe easily in experiments And the reason the reason is that the perturbation you can allow are very very tiny compared to the viscosity and Okay, so I Mean I think I think it's not it's not that difficult to To to guess that that's really the right way of I mean now now now maybe It's it's not difficult like if you start playing with with the equation. It's not difficult to To guess that that's really the right way of asking the question But I wouldn't I I I don't have Yeah, one one thing one can do is if you look at the linear the linearized problem and you try to Study how the how the linear how the linearized I mean I mean the linearized problem has has a dissipative effect Right and the dissipative effect is based on the viscosity, right? So you expect that if you have a larger viscosity, you have more dissipation So then you can allow larger perturbations, right? I think it makes sense, right other questions So here I'm not very precise yet on Deciding norms in which norm I'm I'm looking It turns out it turns out that the choice of norm here Will depend actually on the coordinate system I use so If if I don't do anything with the coordinate if I keep the same x y If I keep the same x y here I will only allow things like L2 or maybe h1 No, but but but actually I'm not going to I'm not going to be working in my original Coordinate system. I will make a change of coordinate and then this norm will be actually The more or less the same jewelry or the more or more less the same so but less nor So so it's L2 in the original in original or here jewelry or Sobo left in After change of variable Okay This is easy to understand I'll explain it after change of variable I mean, it's it's really the next thing I'm going to talk about which is the Necessity of using a change of variable in this problem Number So here what I mean by sober left it means sober left with s large enough I think in the papers more or less There's I mean, we didn't really try To optimize what is the sober left you need but it's more less like around four or five derivatives I think it's it's true that it's an interesting question to ask Okay What happened in energy space for instance? I don't know right I Think that will be also a question to to ask but here sober left means let's say HS s Larger than five let's say Okay, I mean we have to spend some derivatives. I mean the whole proof uses Regularity, I mean one one of the ideas about the proof is that use regularity for decay I mean there will always be this sort of theme and so we need we have to lose a little bit of derivatives in in the statements Okay, other questions Yeah, okay, so there's an exercise I'm going to give you now So Jovrey is I think he's French or Italian it French, okay No, I yeah, what is what is Jovrey? Yeah, so I think he's I mean the paper is 1980 yeah, so so the So so Jovrey Jovrey regularity is something which lies between see infinity and Analytic right so so somehow somehow I mean if you think about regularity if you think about analytic has been very regular You have see infinity here. Here you have analytic You want to put things in between right so putting things in between meaning that okay You have to quantify in a way this see infinity regularity And here you get some Jovrey Okay, and the exercise I want to give you I mean the reason behind this exercise even though I think I'm not going to use it is that in in the last few years. There were many few papers about especially in fluid using the Jovrey regularity and some of them use the Fourier side of the Jovrey regularity and others use the physical side and Like it's not an easy exercise. I mean like okay. I mean, it's not it's not completely trivial to see that these two sides are The same right that the space is the same but I mean since Since I think it's a nice exercise and it may be useful. So let me give you this exercise so I think I think the original definition of Jovrey which the original definition of Jovrey is based on Saying that okay F So let me just write it for one variable X So I'm one variable X which is in on the in the torus. What do we mean by F is? Jovrey of class One over s. Okay, so this is another thing which is not Why I write it one over s because I want s to be between My s will be between zero and one So one over s is bigger than one Okay, one over s is bigger than one one over s equal one This is analytic and One over s Yeah, I mean one over s large So when one over s gets larger The space gets weaker. Okay, so that's why I prefer to use s I Use I prefer to use s than one over s But the definition is say that f is Jovrey of class one over s if the following holds If f is c infinity So you can take any number of derivatives and If you take for all x in the torus Let's say we are in the torus if I take J derivatives it will grow like some Derivative grow in the following way J factorial To the power One over s and There is a tau to the power J Okay, and there exists I should say there exists C zero and tau So F is Jovrey of class one over s if F is c infinity and There exists C zero tau Positive such that For all x in your torus if you take J derivatives your J derivative can grow in this Manor, okay, so so you can observe that If if one over s is equal to one that's analytic If one over s is large you are allowing derivative to grow More right so the larger one over s if one over s gets larger and larger your space is Closer and closer from just being c infinity, okay, so it's a way of quantifying It's a way of quantifying how How derivative grow Okay, so this is a physical side definition this is a physical side definition of of the Jovrey Regularity Okay, so now the exercise is to prove that To prove that this is actually equivalent to the following fact now As I said here, I am in the torus so my f I can look at the Fourier transform and Of course since here. I am in the torus my Fourier transform will go from Z to C okay, and This is actually equivalent to the fact that there exists constants now C zero and C so the C zero here and this are not necessarily the same Exponential minus C Okay to the power s Okay, so so so basically the exercise is to prove that If f satisfies this with some C zero and tau Then it satisfies this with some C zero and C okay so it's a way of Quantifying how The Fourier transform decays Okay It's not difficult to see that The constant C being big is like tau being small like C is big Is like Tau small Okay So right so here the way I wrote it here I mean the way the definition of the space is is a c infinity Is a an element? I mean like if you think about it this way if you think about it this way This is like an L infinity This is an L infinity type space right you are thinking about it as an L infinity type space in the way I'm you I'm going to use it. I'll be using an L to Type space right so so I can I mean it's a good I mean your question is perfect because that's really what Motivates then the choice of the norms Yeah before let me just give me one second before I go into the non When s is equal to one when s is equal to one this is exactly the analytic case and See see will be just the radius of analyticity, right? So when s is equal to one So we are in the analytic case and see will be like the radius of analyticity Yeah, my hand writing is getting worse Okay So I leave this exercise for you maybe next Tomorrow, maybe I'll start by Sketching very fast how you can prove this Or maybe someone if someone wants to do it in five minutes tomorrow that would be good Right so yeah, so as I said, this is like this is an L infinity based norm I mean the way I'm going to state things which will be more like with L2 based norms and Um, basically the type of norms. I'll be talking about when I say they are Gevray when I say they are Gevray. I mean I Mean the following things will always I will take s to be actually one half and that's I Think we think it is the optimal thing even though I Will say like in the 2d and the 3d even though the s is the same S is one half, but it's for completely different reasons. I mean there's no clear reason why it's the same and okay, I mean that was like But I mean it's it's I think it's more like a coincidence than than a deep reason And the norm will be Okay, so let me write it in 2d I'm going to consider d equal to because at least it will start fix I will start fixing notations So now I have a function of x y Remember that x is in the torus y is in the whole space. So f hat Will be a function of k Eta k is in z Eta will be in R. I mean if I put z also Let me put z so that I Can fix notation for the whole Okay, so that's the Fourier and as I said, I'll be using more the Fourier side of it So it's more like this sort of decay, but I don't like using L infinity based Spaces I want to use L to base space because I'll be using energy estimates things like that So an L to base space here will be so So normally I have an s And I have a radius right s s will be fixed in the whole proof So basically I don't need to really specify it so but it will depend on lambda lambda is related to the weight you see here so The norm is you just sum over Lk in z You integrate for eta in R and then you take f f hat Square and Now I put a weight Exponential to lambda unfortunately. I'm not having enough space here K eta L square d eta Okay, that's more or less the L to base norm For lambda bigger than zero. I mean think of lambda as being the C here Maybe slightly smaller than the C Right, so if you have a C here if you take a lambda slightly smaller than the C then that that will be That will be an L to base norm An L to base norm for your jewelry space Sorry, no, this is not square. This is s Okay, that's s s s is the exponent I have here Okay, the reason the reason I mean I Should be writing s here but like in the whole in the whole Lectures the s will be like fixed. I said s equal one half, but it's like strictly bigger than one half one half is like the Threshold, but we cannot reach us but s will be fixed lambda will change with time like I Will start with the lambda and then I can allow lambda to become weaker So I Can allow C to change So these are spaces with two. I mean like it's really one has to think of them as Saying that s is fixed is like you think that your sobola of norm is fixed But you can allow your norm to grow and basically instead of allowing My norm here to grow. I will be keeping norm under control. I will allow lambda To shrink to be smaller and smaller Right, so in a statement like that the reason I'm having an initial and the final The final is really allowing the lambda to getting smaller right But getting smaller with time in in a controlled way and so on Okay, so Any other questions about the norms? Yes, so You are making this assumption that you're working in the periodic case with respect to those two variables Okay, okay, so so So to remind yeah, so basically there are two things why why why hold first why do the whole space in y? So if if I don't do the whole space in why I Cannot do periodic in why because why is not a periodic function? But the more natural question is to try to study Something between Like let's say y equals zero and y equal one let's say and you have your flow like that and like experiments will be doing this, right? So if you do that the issue Mathematically there are Tremendous issues I mean then you have other problems due to boundary layers, right? So this is this has to do with boundary layers and the problem The question becomes completely different and there are there's a whole literature about this So that's not the goal of what I'm going to talk about here but Like even doing the limit between Navier stocks and Euler in in the case is an open problem To understand what happens here So that's that's about why we take white in the whole space I mean there is also I think there's a physical reason even I Mean we want now to insist about it The actually the 2d model especially the 2d model the 2d model having seeing like the fact with that we have 2d Euler 2d Euler is more related to problems coming even like from plasma physics. So in plasma physics There are set of variables for which your plasma will behave exactly like 2d Euler And they think it comes more in cases where more or less you should not be having boundaries even though I'm not maybe very Familiar with that, but I think the the model itself to the Euler in this set of in this geometry I think is more motivated from plasma physics than from fluids Okay, so that's the first that's about why we take y in the whole space. So now yx in the periodic I mean, there are issues if you if you take x in the whole space you get issues with low frequencies So there are low frequency issues So so that's why we prefer to take it in in in a periodic setting and Also, it shows better the the idea of mixing that I'm going to talk about. I mean like the whole the whole Proof is based on mixing and you understand mixing better in the periodic setup I mean, it's not completely impossible that these results still work in the whole space in x I mean, I think there should be some other version of that, but um as of now we I mean we have okay, so actually I Mean there is there is a paper There is a paper that we posted with Jacob and Muho, so there's a paper with Jacob and Muho about the land out dumping in the whole space. I Mean, I'm not going to really I mean since you ask the question Okay, since you ask the question. I mean there there is I mean I was not planning to talk about it right now, but maybe at the end but there is I mean this problem is There is a brother to this problem, which is the London dumping, right? London dumping for the Vlas of Poisson for the Vlas of case for Vlas of Poisson, I mean this was I mean related to the to the work of Muho and Villani about London dumping and and in this case I mean we have like X and V X Usually is considered in the torus V in the whole space This is like the work of Muho Villani is what they work in this case Exactly the reason X is in the torus is the same reason my X is in the torus and Then we have a paper now with with Muho and Jacob doing X in the in RD and the whole difficulties is Studying low frequencies is issues with low frequencies and so on Okay, so Other questions I still have I still have how Then 10 minutes 15 minutes. Okay. Okay. Good. Oh Okay, good good Yeah, I think I went much slower than what I wanted to do, but I think that's okay so Right, so I think what I'm going to do now is maybe give you like a small plan of what I intend to do even though maybe I will not be able to Do the whole plan but anyway, so So the proofing results of this so as of now I'm not going to give statements more or less the statements are I Think you can I mean if you if you take any one of those spaces the way I wrote them there The way I gave you the exercise I mean you use a norm like that more or less you see what type of statements I want to prove but the reason I'm not stating the statement right now There are also change of variables that I have to do the right way and so on but so that's why I'll first start by Trying to give you a Linear dynamics first, right? So so so the Okay, so I'll start by 2d. So what what do I mean by linear dynamic? Okay, so if you have this problem first of all you want to understand the linear problem and then from understanding the linear problem You can try to do something about the nonlinear problem the nonlinear problem. So In 2d In 2d actually what we what what we do we look at the vorticity and it turns out Linear dynamics so first 2d So we we look at the vorticity. So again the vorticity For those who are not very familiar in the following quantity And then it turns out that now if you write your equation on the vorticity the u equation So you remember I had the u equation that I just erased the u equation Maybe it's you know, I think I always oh no, it's here. Okay. Good. Okay So if I take the curl of this equation, right? If I take this equation and take the curl of it In 2d in 2d what I get as of now let me write it for the nonlinear problem But then immediately I'll be looking at the at the linear problem What you get you get a following equation, okay? And this is an easy exercise. I'm not going to do it If I am in the I mean, this is the nonlinear term the nonlinear term is here This is the nonlinear term. So if I if I just want to look at the linear If I just want to look at the linear problem. These are the linear terms. Okay, so those are the linear problems So actually in the 2d case This equation is very interesting because I can recover you from omega So they so all I have to do is just there is a scalar There is a scalar variable that they have to study and they can recover you from omega By the B1 Savard law. So the fact that you is the grad of Laplacian minus 1 of omega Okay, grad burp means minus dy dx Laplacian minus 1 of omega I mean, this is at the nonlinear level at the linear level this disappears and they just get a Transport diffusion equation for omega this happens only in 2d in 3d things gets more complicated But it turns out that in 2d, this is a very interesting quantity and it simplifies thing tremendously and I think most of the differences between the two cases is due to the fact that in 2d we have a scalar It's like a scalar problem basically in 3d is there is more Okay, so now Okay, I mean you have you have this this equation you have this equation about the vorticity what what is the natural thing to do What is the natural thing to do? So even even if if I am in the Navier even if I am in the oil in the oiler case This will not be there and they will just get this term, right? So the natural thing to do if you are if you look at if you see an equation like this the natural thing to do is to make a change of variable right to get rid of that transport term Because like this transport term Gives you growth of norms, but for a wrong reason maybe like it's just because things are Rotating a lot turning around I mean, it's an easy exercise to check that if you take just this transport equation and you consider High sobolev norms they will be growing Because of this y here So so the natural change of variable so Let me now forget about this term. I'm just going to study the linear problem So basically I'm going to study. I Just want to explain few things about the following equation at the linear level, right? So again here. So now we are in 2d So remember my x is in in the torus my y is in the whole space And I will be thinking sometimes with the viscosity sometimes without the viscosity to understand different effects Some inviscid effect and some viscous effect So first of all the coordinate the change of coordinate consists. I mean, I would like to get rid of this So the change of variable is just this one Okay, so this is the natural change of variable because then If I if I make this change of variable and consider capital U to be and my omega Okay Yeah, I think okay Let me keep the nonlinear term so that at least you can see the change of variable at the nonlinear level And then I will I will just so if I do the change of if I take this equation if I take this equation And you'll see how the nonlinear term transforms and I make this change of variable I make this change of variable then the equation I get Is the following thing so I Get an equation on this capital omega to be up on this Phi Okay, let me explain now. What are the the notations? so grad U Becomes grad L of capital U Laplacian U Becomes Laplacian L of capital U and grad L is the following dx dy minus T dx Okay So it's a it's an easy change of variable. I mean, this is type of things you will do you explain to do that in calculus how How the river to fin why so if I take the river to fin why? The river to fin small y is like the river to fin capital y minus T Times the river to fin capital X. Okay. Okay makes sense and then Laplacian L is just this square okay, and Phi Phi is like the stream function in the new coordinate system So Laplacian L Phi is omega Okay, so that's for the nonlinear term To understand the nonlinear term if I were in the linear case, this will not be there. We'll just get dt omega equal to this Okay, okay, so this is a small exercise again like you can For I mean you can make this change of variable. I think it's a good exercise of change of variable Take your omega to be the solution of this with you given by this formula make the following change of variable and prove that Big omega will be solving the following equation with the following things, okay The gradient here means gradient in capital X In capital X capital Y Just a small remark at the nonlinear level. There is here a hidden important cancellation hidden important cancellation, which I Will I mean we use in in in this work we use sometimes the Terminology like null conditions to To mimic the same type of things that are observed in dispersive equations like to mention like derivatives that behave in in a good way and here there is one Aspect of that at the nonlinear level even though I want to explain it a little bit very fast You see there. I have gradients with L sometimes. I mean the maybe the notation is somehow confusing But when you see a derivative hitting small u this means small x small y right These are small x small y If you see a derivative hitting capital U, it means large x large y But then these derivatives the natural way they transform they transform into these derivatives with With this sort of growth But then an interesting observation is that if you take the The following term so the following term has there's a derivative that comes from here and the derivative that comes from here Right, so so basically there is There's a term of this type psi is the stream function. So psi. I will laplacian psi is omega here So normally this when I transform it if I transform it It should become right If I use exactly that rule Okay, but then I mean it's not difficult to see that actually there is a constellation here and this becomes this At the nonlinear level, this is very important that this happens like that I mean the the This is a nice nonlinear Constellation, I mean it's just it's just the following fact. I mean if you think about it this way I should be doing okay. So this is this one and The other one is here Okay, so you see that This times this and this times this they they cancelled so somehow this guy And this guy they cancel right so that's why from the L I can get rid of it right so this was any this was an important observation because I mean It's it's one of many actually other Important constellation coming from the specific nonlinear term. So basically that what this means that if you took if instead of taking The right Navier stocks if you come and you choose a different nonlinearity Then a constellation like this will will not be there and then I don't know how how how there is how we can manage, right? Okay, so No, yeah, I mean whenever you tell me to stop I stop I mean because I mean I am in the middle of a small calculations But I think no, I think no, no, no, no, no, no, I think I am good I think that's where because then I will start talking no no because then I'm going to start talking about enhanced dissipation So I think I think this is yeah, I think okay, so this is another good exercise for those who want to do exercises I gave two exercises, but Change your variable and the exercise about your very norms Other questions, yeah For the pressure Okay, so I think okay, so this is a good question in terms of okay What happened to the pressure turns out in this problem? in This right so so maybe the way I answer the question in in many problems about Navier stocks and Euler the pressure You completely ignore it because you think of it as a Lagrange multiplier and in many Here You'll see like the pressure will be important in few places Because like the three components, let's say I'm talking about let's say about the 3d case in the in the 2d case The pressure disappears because I'd be working with the vorticity So I will not really be talking too much about pressure, but in the 3d result the three components they have different Interpretations and they play different roads and so I will not be working really with vorticity and Then the pressure will be important So I will be talking about linear part of the pressure non-linear part of the pressure So it there will be there will be a visible pressure that I have to I have to talk about But of course at the end of the day the pressure will be following the The velocity and it will it will be Slave to the velocity But it will have an important role in 3d But if you assume more of the energy then it gets better to carry on in the next three And I'm just asking whether the moment will grow as your regularity As long as it's large So okay, so Here like if you take Sobolev if you take S much larger Like let's say if you take if you make as larger and larger the gamma will not change So it will not be so so either you are here or here like they will as of now Okay, I don't know about Like let's say intermediate thing between these two for instance So either either you are in Sobolev and Then this is your gamma Let's say if you are in 2d or this is your gamma if you are in 3d Or you go to the jeuvre that I'm giving there like s bigger than one half And then this is the gamma So even if I improve the Sobolev regularity, I will not change the gamma If you decrease it to the energy, I don't know that's another issue How do we define Laplace inverse? How do we define Laplace inverse? Okay, so Laplace inverse here you are let's say here you are you are in the in the torus times r Right So you are in the domain like that so you can you can define it on the Fourier side You can define it on the Fourier side. Okay, maybe I didn't There are statements about zero average and so on that you have what No, no, no, no, I mean like you just need like the whole average to be Okay, so here you are thought you are asking question basically about low frequencies how we deal with low frequencies, right so We need the small statement about We need some weighted norm on omega in y One weight actually we need like one weight so that omega decays well when y goes to infinity and Then there is an issue about the average like you can always assume that the average the whole average is zero That's that's all so the only issue is when k is equal to zero when k is different from zero k k meaning the The Fourier variable associated to this capital X But that's yeah, that's one of the assumption you can always make Yeah, yeah, I think I think there are yeah, yeah, but I think there are possibilities of extending to many other I know like for the 2d There's the beta I mean you can include the beta effect and they think Yeah, my yeah, I mean there are works also about including the beta effect You know like this term coming from the Coriolis force. I Think that's an an easy adaptation