 That's it. izgledaj z nekaj, da smo vlasti v maju. Zdaj... Zdaj sem... Zdaj sem izgledaj, izgledaj, izgledaj, prantul system. V svoju dana, v svoju dana sem izgledaj prantul system in prantul system. Zdaj... Stacionalne prantorne system je prantorne, ali nekaj nekaj prantorne. Tako to je. Tako to je izgleda na u. In potem smo vse in kompressivnosti kondition, da imajo vse vse vse vse vse vse vse vse. potem včasno zelo je otekljamo vzal u boju tako že bomo vzal u boju, če je tako professionela vzal, napisak nekaj a polo na prostvenstven cans do, OK, primetno so bo vzal tko, More or less, we are interested in solving this in the domain x bigger than zero, y bigger than zero. So basically, the picture should have in mind, so this is x, you take some x equal zero here, you impose an initial condition, which is this u zero of y, and then this u zero of y is positive, right? u zero of y would be positive, so then this guy is positive, and then the equation itself, you will think about it as some evolution equation in x, okay? So x plays the role of time, right? So then you want to solve this till some time or some x x star, right? So this is where you are going to solve, right? So the idea is really we are thinking about this as some evolution equation in x, okay? So this is the stationary prank tool. So today the plan is to discuss three types of results. I'm not going to go into details of all these things, but there will be three types of result that I'll be mentioning, and more or less these three types, I will be discussing more or less exactly similar three types of result for the for the full system with the u t term on Wednesday. So the first result will be a result about blow-up for the blow-up result. I mean the term blow-up maybe is not the best term used, but it's more separation. That's the term that we use usually, the separation of boundary layer. So then there is another type of result which is long time asymptotics or large x asymptotics, but of course you'll be seeing that large x behavior. And then another type of result that, okay, I will not go too much into details about it, but validity of the prank tool system. Okay, so this is related to what we discussed last time. Validity of prank tool. Validity of prank tool means that if I take a solution of Navier's talks, it can be well approximated by prank tool. Right? So, and these are, I mean I'll be mentioning two results here. Okay, so these are more or less the plan for today. For Wednesday we'll be talking about a result about separation, but for the evolution problem and also some results about validity for the evolution problem also. Okay, any questions? No. Okay, so for one and two, I mean in one case we will only be solving till some value x star, in the other case we will be solving globally. What's the difference? The difference will always come from the pressure, whether the pressure is helping you or is acting against you. Right? So, more or less here this equation, you can solve it as long as you is positive. As long as you are positive, as long as you are positive, this looks like some evolution equation in x. Right? So your enemy is when you start become, tries to become negative, that's your enemy. So, so here that you are forcing you by this guy. Right? So if this guy is bigger equal than zero, maybe you expect that you can solve this globally. Whereas if this term is negative, that's where it's acting against you. So, that's what we call favorable or adverse pressure condition. Okay, so let me first recall a result from Olenik, a result where she proves local existence for the system. I think I may have mentioned this maybe around 64. So, I mean, I will try to state more or less how the result goes. Yeah, maybe just one remark, I didn't insist on this again. So, this guy is the Euler flow. Right? So, normally the Prandtl system that we are normally, I mean, we will be having Prandtl here, and here we have Euler. Right? So, the limit y goes to infinity, when y goes to infinity here, that corresponds to Euler on the boundary. Right? So, basically this guy, if you are in the Euler flow, this corresponds to the boundary value of Euler. And, of course, now it's not difficult to see that there is a compatibility between this guy and this guy, because when y goes to infinity, you expect this guy to go to zero, and you will be having only these two terms. So, normally what you, what we expect to have, what we should be having is that u Euler. Right? So, this equation, this is the x derivative of, this is the x derivative. This is, and basically this equation is the trace of the Euler equation on the boundary. So, if you take the Euler equation up and you take the trace on the boundary, that's what you get. Of course, I mean, for what we are doing today, we don't care about this. I mean, we are just looking at Prandtl. But, I mean, if you are trying to understand the coupling between the two, that's where it takes place. The other thing here in this model is that the Prandtl system does not affect the Euler flow. Right? We are only seeing, we are only seeing the effect of the Euler flow on the Prandtl. I mean, I was thinking about adding a fourth point, but I don't think I will have time. But let me just put it, but I will not talk about it. But there are, related to this question, I mean, there are slightly more involved model than Prandtl, that somehow try, I mean, like people saw that the Prandtl system is not always the good thing. So, they wanted to find models where what happens here can affect what happens here. Right? So, they wanted to see some feedback of the Prandtl on the Euler. So, then there are models, there are a few, a lot of physics papers dealing with a few models. The mathematics about it are still, still, I mean, there are starting, there are a few results now, that I can mention by Tali Var, David Gerard Vary, and then there is also some recent work of Samir Iyer and Vikol about trying to understand what is called triple deck models, where there are three layers here, or interactive boundary layer models. I mean, all these are models where you try to see a feedback from the Prandtl on the Euler. Ok, I will not be talking about them here. Ok, now, if I go back now to the result I want to mention about from Olenik, so you take some u0, which is, let's say, c2α bounded, some c2α function, which is bounded, u0 of 0 is 0, and u0 prime of 0 is positive, the limit of u0 is, and u0 of y, ok. So, basically, I mean, you have some profile, you have a profile of this nature, like it, your profile is more or less something like that, ok. This is y, this is u0 of y, ok, so it's positive. Ok, now, in this assumption, in this assumption, this is not necessarily monotone, so it can be, it can even do this, right, in this assumption it can do this, for instance, for here, this is also valid, ok. So, the other assumption, so this is first assumption on u0, we have another assumption that, let's say, this is c1, there is a compatibility condition, there is a compatibility condition that states that is a big O of y2, when y goes to 0, which is more or less, you can understand that, because we, oh, maybe I didn't write it down, right, I didn't write down the fact that u at y equal 0 equal v at y equal 0 equal 0, right, so we have also this boundary condition. Actually, from that boundary condition, normally, what you have, you have that u behaves like some constant times y at 0, and v behaves like constant times y2, right, v is like y2 at the boundary, but then the transport term, this plus this, the transport term behaves like y2 at the boundary, so then, since the transport term behaves like y2, this plus this should behave like y2, so that's your compatibility condition, right, so the compatibility condition is coming from the fact that you expect these things to happen at 0, okay? Okay, so then, these are the assumption of Olinik, and then the conclusion is that, if we have these, then there exists x star positive, sej that the Prandtl system has a local solution, which is, let's say, c1 in 0 x star, open, I mean, it can blow up at x star, times r plus, okay? So, so basically is that you can solve, you can solve from x equal 0 till x equal x star, and at x star something bad can happen, or some separation happens, okay? Now, the solution constructed by Olinik satisfies some extra, so it satisfies some regularity, so, I mean, I mean, at least like you can prove that, two derivatives are under control, then so on. So, let's say u is bounded, continuous, derivatives in y, derivatives in y actually are bounded and continuous till including, I mean, till x star, but derivatives in x, dxu, v, or dxu and v are locally bounded, are locally bounded, which is more or less understood. I mean, the reason is that, when you approach the separation, I mean, the separation will be the point x star where u, where u will be, can't touch 0 in a sense, but then you lose control on this kind of derivatives. So, so this is the, some of the property that we have, okay, now this is regularity, the other property is some sort of non-degeneracy, which says that u of x, y is positive for, okay, and so that's the, and then the other important remark is that if the pressure is favorable, if the pressure is favorable, meaning that if, if the pressure is, if the pressure is negative, dp, if this term is negative, then x star equal infinity, okay, if the pressure is not acting against u, then you can prove, you can do it globally. Now, what happens if we have separation? What is separation here? Now, if x star, if x star is finite, so meaning if the pressure is acting against u, so there are two possible scenarios, there are two possible scenarios. The first one is that the y derivative at zero of, the y derivative at zero at the point x star is zero. Remember, like when we started, we started like this, right, right, so we, our starting, our starting slope is strictly positive, or there exists y star positive, such that u of x star, y star equals zero, okay. Let me try to explain these two things. Usually, like here, okay, here I'm putting y like this and u like that. Sometimes when we draw these profiles, we like drawing them slightly the other way around, like I'll be drawing them like, like you see them in some of the pictures. So, so, so the profile we started with, the profile we start with is let's say profile like that. So, again, here I'm more or less changing the, this is my u, u is this, right. Now, now I change, this is u, this is u of y. So, y is like this and this is my u of y. Okay, so I just change the axis. This is how usually we draw them and I'll explain why. I mean the idea is that you think about this as being x and you are trying to see how your u of y evolves with x. Okay, so that's why we like, many times you like drawing it this way. So, this will be your initial data. Okay, so this will be your initial profile, for instance. So, then what can happen? Either this happens. So, this happens means u, y equal zero, u, y equal zero means that your profile becomes something like that, right. Because remember this is my y, this is my u, u of y. At the point x star. So, this will be the point x equal zero. Right, so my profile that was like this, due to the pressure term, the unfavorable pressure term, it is pushing. So, then you get this. So, this will correspond to one. That's the first, this is the first scenario. The second scenario, the second scenario, the second scenario is that something like this happens. It touches here. So, the second scenario can only happen if your profile is not monotony in y, right? Your profile, so this is y, this is u of y. The second scenario can only happen if you are not monotony in y. Is this clear? Of course, now, this is also, this is theorem, this goes back to Olenik. This is, this is, goes back to Olenik, yeah. Yeah, so this is a theorem, like what can, the two cases where you lose, where you cannot solve globally is either this one or this one. Of course, now, if this happens, and this is what we usually expect, there are a lot of question about how can you extend the solution afterwards. Physically, what happens here, this is the point of separation, that we call separation of the boundary layer. And then what you expect afterwards is that you get a flow, which looks like that, right? So, we have a flow that goes backward. Now, in terms of well posedness of that sort of transport equation, this is bad, I mean, like, so one has to understand this really differently. Like, you cannot just think about, just thinking about access being time will not be working very well, I mean, you have to think about that. And that's, as of now, it's not very clear how one can attack these problems mathematically. Okay, now, there's one important case, there's one important case where we can actually rule out this. If our origin of, if our starting profile u0 is monotone, monotone means it has to be increasing, right? If it is monotone, so it's like the picture I draw here. Then only one can happen, then two cannot happen. The reason that two cannot happen is that this monotonicity you keep it. You can prove that if u0 of y is monotone, then u of x, ux of y as a function of y stays monotone, right? So, if you start, if you start with a profile like this, which is monotone, you keep it monotone. So, if you keep it monotone, you can never have this. You can only get this. Proving this is not completely trivial. If you want to go through the whole proof, it uses some sophisticated version of maximum principle. It's one level, it's more than just maximum principle. You use something called comparison principle, which is like maximum principle, but it allows you to compare. OK, any questions? OK, now the first result I want to mention, as I said, is to describe the separation, right? So, an important question is to try to understand what happens here exactly, right? How does this separation happen, OK? So, that's the first result I want to describe. So, let me write down this theorem. So, this is a theorem that... So, it's with Anlorda Livar. OK, so, I mean, there are a lot of assumptions for this theorem. I'm not going to give all the assumptions, but maybe some of the important ones. But, more or less, the conclusion of the theorem is that it gives us... It tells us exactly how this guy... How this guy goes to zero as a function of x. So, if I define... So, let me start by a short version. If I define lambda of x to be Uy of x0, U of x0, OK? So, that's the derivative in y, right? Then, we know that lambda of x will go to zero when x goes to x star. That's really the definition of x star being the separation point. But we can say more. I mean, we can say that lambda of x behaves like some constant square root of x star minus x. And we can be more precise about this. But this is really the short version of the result. OK? OK, so, what are the assumptions? Yeah, so, I'll be giving the assumptions. So, this is what we call, at least what we think, to be the stable separation process. OK? And this confirms some of the formal asymptotics done more or less in the 40s, 50s, and 60s. So, at least some of the Goldstein has some formal asymptotics, Stewart, so on. They have some formal asymptotics. There is also an interesting, Landau himself also has some very interesting heuristic about why it should be like that. OK. OK, now, what is the, what are the assumptions? So, let's say we need some regularity. I think seven derivatives are OK. U0 increasing. We start with some lambda 0, which is the U0 prime, at 0, which is positive. OK. We have to take this to be already small. So, our lambda 0 will be small. So, the second type of assumptions are assumptions that are the second types. These are assumptions that will be used in maximum principal arguments. Sorry, so you are saying that lambda 0 is small compared to what? OK. All right, so it's good that your question is good. It will be small compared to 1, and the reason is that I'm going to take the pressure term to be 1, right? So, here in my result, this is equal to minus 1, OK? So, I already, I already, I already, I already, I'm already choosing a size here, which is 1. So, then lambda 0 will be small compared to that, OK? So, then U0 double prime minus 1 be infimum between y square and 1. OK. So, and then we have also an assumption on the fourth derivative. The fourth derivative is of U0 is bounded from above and below. OK. I mean, as of now, it's not clear where these things are used. I mean, all these assumptions are not completely necessary, and you can relax them if you want. Modulo, making the proof more complicated, right? But, but what is behind, behind it at the end, and the end of the day, is that what we really need, which is now 0.3, right here, now 0.3 is that our U0, we can write it as some approximate U0 plus V0. So, the U0 approximate is some sort of approximate profile that we can construct by hand, and then we try to close some energy estimate on this V0, right? So, that, that will be the idea. Now, our U0 app will be what? Lambda zero Y, OK, so that's the derivative, plus Y squared over 2. The Y squared over 2, that's, that's related to the Y squared over 2, that's related to the fact that the pressure is minus one. The pressure term is minus one, so it's related exactly to this condition, right? That condition tells you that this term, this term has to be, this term has to be Y squared over 2, plus, OK, also there's no Y3 term, also because of that. Y to the 3 term, there's no Y to the 3 term because of that. So, so, so you see, the, this guy, this guy is like the compatibility condition at the first order, like you just take the equation and that's the first compatibility condition. The next term that you are seeing here, they are coming from higher order compatibility condition that your solution should satisfy. So, I'm not going to write everything, but we need some other terms. But you mean at zero. Yeah, yeah, this is at zero. Yeah, at zero. OK, so I'm not, I'm not writing everything, but we need this till maybe the order Y to the 10, for instance. We need to know the terms. I mean, you can compute them, right? You can compute them just from, from the equation. OK. And then our V0, what we, what we can say about V0, after we remove all those terms, V0 is less than some constant, plus some, another constant, Y8. Yeah, maybe I don't want to go into some unnecessary details, but let me here write down everything, so that at least you get the sense about how things are scaling. So, right, so maybe this doesn't make a lot of sense, because I didn't write all the terms here. I mean, there are some other terms, and you have to push this till, like, Y to the 11, more or less. You need some sort of good compatibility condition at the origin. So, but then after you remove those terms, you want your rest, the error here, to be of this size. OK. I mean, basically, you need it to vanish till the order Y to the 7, right? So, the other terms, they come just from scaling. Whenever you make the scaling, that's the natural thing to put. OK. One important, I mean, the important thing in all this, even though, I mean, this is really the one of the technical parts that I will not be discussing too much, but if I put remarks whenever, if one looks carefully, is that V0 is much smaller than the approximate solution, and this happens if lambda 0 is small. Under that condition, oh, yeah, I should mention that this we need this for Y small, and more precisely, we need it in this region. I'm just writing down what is needed in the proof, but at this level, you just think about it as Taylor expansion at 0, right? So, you just do the Taylor expansion at 0, use the equation, and then you can, if your solution has enough regularity, then all these terms are completely necessary. So, the second remark, which is related to what I mentioned there, that the monotonicity of U0, the monotonicity of U0 is propagated, again, I insist, like, we are prescribing some sort of Taylor expansion till the order 7, and this is needed to derive our estimates, because we'll be taking higher derivatives, so that's why we need a Taylor expansion at a higher level. So then, the conclusion is what? The conclusion is that there exists some c0, positive, there exists epsilon0 positive, such that if our lambda0 is less than epsilon0, then there exists x star, phi star, phi star, phi nit. Actually, more precisely, x star is some v go of lambda0 square. So, it's really your x star here is not that big, such that lambda of x, lambda of x being the number that they define there, lambda of x will behave like some constant square root of x star minus x, as x goes to x star. So, it's more or less, it gives us exactly what is the rate of, what is the rate of separation. OK, now c0, c0 is the number that appears here, there. The c0 is the same in the equivalent, so it's the same, because it's the same. Yes. OK, so this is the statement. This is the statement. I'll just give you ideas about the proof without going into all details. Of course, I mean, this is really just the statement about the lambda of x. I mean, the proof has in it the fact that v is much, I mean, is normally what you are proving is that v is much smaller than the u app. OK, now how are we going to prove this result? So, we have this lambda of x, lambda of x is the, I mean, that's what we are interested in. We are trying to, we are trying to find how lambda of x goes to zero, right? I mean, more precisely, we are trying to see what is the stable, what is the stable behavior of lambda of x. I mean, we are not, we are not necessarily saying that all separation should happen with that lambda of x, and as of now, we don't know. I mean, maybe there may be other ways, but we are trying to identify the stable behavior. OK, so let's start by asking a question. What are the, what are the, what are the scalings that keep the equation invariant? OK, so what are the invariant scaling of the equation? So, the fact that I put a pressure, the fact that I put a pressure reduces a little bit the allowed scalings. Right, so if I don't put a pressure, if I say I have terms like this, if I don't put a pressure, there are more scalings in these equations. And actually, I will be discussing one later on, which is related to the so-called Blasius profile. For the Blasius profile, we scale x like y square. For the Blasius profile. But here we are putting a pressure, so this is minus one. So, here the, if you solve the equation, if you solve p, then if I look at u lambda, then I need to put here lambda square. OK, lambda is not necessarily the one there, but afterwards it will be more or less. Right, so if you solve this equation and you do this scaling, then this will also solve the equation, the same one, right? It's not difficult to see. With minus one? Yeah, to keep it minus one. I mean, if you don't put this, if you put, no, if you don't put this, then you have more scalings. You can, putting this reduces the scaling. OK, so this is the natural scaling that one would like to do. And so then, so I think, so all this is inspired, of course, from works on singularity formation for Schrodinger, like Frank and Pierre. So, inspired by those works, like what we introduce, the following change of coordinate, we look at u of s y to be lambda minus two. OK, where s ds over dx. Yeah, maybe in my notation, I should have. OK, so maybe it's confusing the lambda here and there, but that's more or less. I mean, you are extracting the profile. You are extracting the profile from. OK, now this change of so this is now the object we are interested in. This is the object we are interested in. Again, the way we are scaling, the way we are scaling things, the reason we are putting lambda, the reason we are putting four, the reason we are putting two here, I think these are the things that one should ask, why we are putting it like that, is really motivated by the fact that the equation is invariant under this scale. OK, so now the nice thing that happens is that this profile u of s y satisfies now a nice equation. Nice equation. So capital u solves the following. So it's just change of coordinate. So OK, I have to write down what is b. So b is equal to minus 2 lambda x lambda 3 which is also minus minus 2 lambda s over lambda. OK, so that's really the equation in the new coordinate system. OK, so basically all what we did is just to explain how that comes. It's not difficult. OK, so we started from the x y coordinate. We make the change of coordinate to s capital y where ds over dx is 1 over lambda 4 and capital y is small y over lambda. Right? So we just make that change of coordinate. So then just you rewrite the equation and that's what you... Oh, yeah. And also we made the u of s y exactly that guy. 2 u of... The other thing is that in this formulation I'm not using v. I mean you may ask why what's the meaning of this guy, for instance. But this is just v. But written in the new coordinate system. Right? So basically like whenever you take derivatives you have... these are like the transport term. The reason you get these extra two terms that's coming from when the derivatives is hitting the scaling that when you are differentiating the lambda. Right? Right? So when you are differentiating the lambda you are getting these two extra terms. And this is coming from the u y y in the new coordinate system it becomes this guy. Is that clear? You want me to go... You can spend like 5-10 minutes to do this but maybe you can skip that. Okay. So the other thing the other thing is that the limit when y goes to infinity of it becomes some u infinity of s. So the u infinity of s is corresponding... it corresponds to this guy. To the u Euler. As I mentioned the u Euler satisfies normally the u Euler satisfies u Euler prime u Euler prime okay. So and you can you can see what this means means that you can take your u Euler to be the square root of some x zero minus x. Maybe there is a tool. Maybe there is a tool. Okay. You can always write it down something like that because the u Euler square over 2 prime equal minus 1. So you can solve it exactly. The derivative is in x. This is minus 1 so you can it should be of that form. Now when you change this u Euler into the s variable it becomes something like that. But then this guy now will solve I'm sorry but so if you solve that so u Euler if you take this okay. So the x star is before x zero Yeah the x star will be before x zero. x zero is can be large. x zero is he has to be a theorem here okay so u zero is defined I mean okay I mean here I don't need to I don't need to tell you that x zero is coming from here I can tell you from the beginning I tell you okay this is my profile I'm choosing some u Euler of this nature and then you make all those assumption but all those assumption will imply that x zero has to be slightly large cannot be too small exactly but it doesn't have to be very large like x zero is yeah okay so this equation this equation that happens at infinity for that system translates here to something in the s variable that has this extra b okay I mean when you do the change of coordinate you get this extra term okay it's a good exercise I mean I really encourage you to do all these small calculations so that you see how the change of coordinate is happening okay so the last thing about this system that another important fact about it is the fact that I mean and that was the reason behind all this change of coordinate the change of coordinate will also imply that I write it down here because I think this is an important thing is that why at zero what should this be when we do the change of coordinate so we started lambda lambda was u lambda x zero okay and we made this change of coordinate right so now if I look at capital U lambda right capital U lambda capital U y when I take the y derivative here there is a lambda that comes out right so this will be what lambda minus 1 U U y but U y U y was lambda this is lambda so this is 1 okay so the scaling was done so that U lambda is identically 1 make sense now the next term the y square over 2 term by the way the y square over 2 term is something that stays all the time this expansion at zero in kondition at zero that derivative like the second derivative at zero has to stay always like that and of course I mean that will like this equation will satisfy it also because of the scaling I mean that the scaling really keeps the the y square term you can see it like if you take the second derivative here the this so now now you can just look at this you can forget about the original system you can forget completely about the original system I mean the the condition we have on the on the initial data you can translate it into the new on capital U like at the end we'll be doing estimate in this in this coordinate system so you can completely forget about the original system the only thing that you have to keep in mind is also lambda also lambda you can forget about it because lambda doesn't doesn't really appear here so the fact that B is this guy this I will only need this I will only need to go back but as of now I can forget about this guy so then all I'm having I need to solve this equation I need to solve this equation but globally in S maybe I didn't insist on this but when when X goes to X star we expect S to go to infinity and then the question is ok can we can we solve this and what is the natural behavior of B what I mean how should we choose B of course now when you get to this point I don't know whether like saying that B is like 1 over S is ok we will be proving that B is 1 over S is the stable behavior but I don't know whether there is a clear precise intuition why why it has to be 1 over S it's not clear right of course it's not always the case yeah but I mean at least for us it was motivated by and also the use of the letter B and you can explain why so we will prove that B of S behaving like 1 over S is some sort of stable behavior right so again I want to insist on how now we are going to try to solve the original problem first I forget about the original problem I only look at this equation I can forget about this one I look at this equation with this prescribed behavior with this behavior at infinity with this compatibility condition at infinity and with the original profile of course now the my original solution U0 I can translate it and now my goal is to find the goal is to find so the goal is to find some and B I want to solve globally I want to find U of Sy and some B that solve that equation so honestly I don't know if you get to this point and you don't have a precise intuition about what should be B if you don't say okay I want B to be one of S I don't know how to proceed I find that at some point there is something you can get you need some intuition from somewhere to say okay let me let me assume that B to be like that you do some calculations and then you close it at the end of course there will be a place where no, maybe I am slightly wrong so while constructing the approximate solution while constructing the approximate solution there is a place where you want BS plus B2 to be more or less 0 but of course at the beginning to start your expansion you should have in mind that B should be small otherwise you will not even start so okay now let me just give you without going into the whole detail the steps of the proof okay so I will just give the steps of the proof so step one is to construct an approximate solution right so we want to construct an approximate solution in what it turns out that there are two ways of thinking about the problem and more or less they give you both work the first way is to say okay B is small let me expand in B you can try to expand in B you can look at you can say okay my starting point is I will call it I think I am calling it u0 or u1 let me be consistent so u1 u1 will be y plus y2 over 2 okay I mean you can check it is not difficult to check if that if you put that into the equation if you put that into the equation it solves the equation you put B equal 0 you put B equal 0 you plug that in so it is something that is independent of S so the first two terms they cancel I am taking B equal 0 and then this solves the equation with B equal 0 solves P with B equal 0 and in the language of the Schrodinger equation this will correspond to a soliton for the ground state for the problem of course it doesn't solve it doesn't satisfy the good boundary conditions because it's unbounded I mean whereas we expect this guy when y goes to infinity to go to something finite of course I mean this guy can grow with S because of the rescaling okay so you have u1 and then and then you can keep solving you can keep expanding in B so I will write it down as plus Bt1 plus B2t2 so on I am expanding in B for instance I mean if the Bt1 term the Bt1 term should be and if you think a little bit about it you need dyy of t1 dyy of t1 this has to cancel the term coming from exactly this one I just show you the first guy I will not go into the okay so the Bt1 term the Bt1 term the Bt1 term here the Bt1 term here will cancel the term coming from BU1 these two terms as of now I don't need to know as of now I don't need to know what is the law of B how B behaves as a function of S so this you can calculate and then you can calculate t1 which is some minus a4 so this gives us t1 equal minus a4 y4 a4 is 1 over 48 you just compute you can see all these guys here leading order you have a term like this will be like some constant times y2 so then what happens is that and this is where where there is the choice of B will come let me just explain here and then maybe we will so now if I look at if I call u2 equal u1 plus Bt1 and then I plug u2 in the equation I plug u2 in the equation so as I will write down equation applied to u2 so equation I mean by that equation of u means this guy plus 1 so normally what I am trying to do is to write equation of u equal 0 so if I write down if I write down the equation applied to u2 what is the error that I find I will find so there are some calculations that one has to do so basically u2 is an approximate solution it's not a solution because when you put equation of u2 you don't find 0 but you find something ok you find something and you find a term which is growing like y to the 5 term y to the 6 term like y to the 8 ok now comes the place where I want to tell you what is Bs how I should choose B right and again I mean this is also a little bit inspired by works of Frank and Pierre so you look here and you see the term which has y to the 8 has B times B2 so this is B3 so this coefficient is smaller than this one or this one right this coefficient is smaller than this one ok so you can convince yourself that you can forget this one as this level so you have this one and this one now this guy grows like y to the 5, this guy grows like y to the 6 and it's better to make this 0 because then this growth will be weaker so that's how you choose your B you choose it so that this is like 0 right so you say ok I want Bs plus B2 to be 0 at least whenever when you are constructing your approximate solution like afterwards it will be it will not be exactly 0 but it will be lower order terms but in the construction of the approximate solution in the construction of the approximate solution we can just replace B by so then I'm going to choose Bs plus B2 to be 0 which means like B equal 1 over s right so whenever I'm going to construct my approximate solution each time I see Bs each time I see B I can replace it by 1 over s ok so then you can keep constructing your approximate solution to any order right and so then I can say that I'm going to construct t so u3 will be my u2 plus B t2 and then I can write down the equation on Bt2 so basically the main term that you have to cancel is this guy and that's why your t2 your t2 is some sort of minus a7 sorry so there is B2 here ok and a7 is also explicit a7 is 1 over a4 ok ok so so then what happens let me just show this I was not really planning to go into this detail but I just want to show you at the end how the law closes but when you do it this way you are looking at your solution as what I'm calling the approximate solution plus some error and the whole thing is to close some estimate on this guy now since we expanded till this level so since we removed this guy this guy it will start with terms like y to the 7 which is like consistent with this kind of with this guy so so basically our v3 sy we can see what are the first terms in the expansion you can compute them of course now when I write down the precise thing I put back vS and actually the control of v3 will allow me to control this guy this is how the control of this guy will be coming it will come from controlling v3 over y to the 7 plus other terms let me not write them down but the term like y to the 8 and so on but this is more or less at the end of the day this is how the how the law on vS plus v2 will be derived it will be derived from getting some good estimate on v3 ok ok, maybe I should almost finish with this let I have a question about the formal approach if he decided to cancel the other guy in YouTube this one? I don't know, I see that you have odd power of power of y and do you think that this will make up here something like even power of y? no, no, no actually they are here there are also some power 8 so it's not really so the next term has a power 8 and this coefficient is explicit so it's not question about even or odd but for me it's more like a question about stability if you try to if you try to cancel this guy I think my thinking that if you cancel this guy you get a better behavior at zero but it's bad at infinity so in particula if you cancel this guy then this term here instead of being 7 it will be 8 so you are losing at 2 places actually you are losing in the fact that you have a worse behavior at infinity first but the other thing which is bad is that you need to control v3 over y to the 8 so that you control this I mean you have a different coefficient of course I mean you get that coefficient so it will I mean I don't think it works and maybe that corresponds to some unstable behaviors but I have no idea in the case which I know it doesn't work in some case for example there is no solution of such it's a prototype where the approximate solution has it's more the fact that you have an approximate solution doesn't mean automatically that you have a solution which looks like there is no so what are the other steps of the proof what are the other steps of the proof so the real approximate solution here we talked about the approximate solution but for y small actually why not very small like why here can be what we discussed is why like s to some power alpha it can go to ok at some point we have to choose alpha I mean it's there is room in choosing alpha more or less we are choosing at some point we are choosing alpha equal to over 7 I mean it's related also to the powers we had there but there is freedom so what we want to do is to take a cutoff right and to plug in u3 actually it turns out even though at this level normally u3 is enough because when we go to u3 we have v3 and v3 controls everything but I mean it's usual thing we push it even to u4 ok so this is the approximate solution for y smaller than as to the alpha but then what happens afterwards afterwards I have to I want to reach this guy I want to reach this guy now my b is 1 over s my b is 1 over s so b is 1 over s so basically this guy u infinity should be like s plus 1 right so u infinity is like s plus 1 ok ok you can check that will more or less do the drop so it's like s let's say it's like s like the difference between not important because all the norms we are going to use they they are weighted norms so what happens far away is not that important ok I mean you cannot really make it y square but it's like constant that infinity is fine because of the weights we are going to use so then what we do more or less here actually what I'm going to do is subtracting y square over 2 ok for some small reason it's just a choice but plus 1 over b some function theta because the y square will be coming from here so the theta behaves like y square over 2 for y small and it behaves like 1 for y large ok so basically you just take that profile that we constructed we only keep it when y is less than s to the alpha then we want our solution to behave like there is a region where it will behave more or less like y square over 2 and then it will behave like 1 over b which is s so in a sense there are more or less like 3 regions so there is this region s alpha there is a region which is more or less s to the 1 half and then there is a region at infinity so here it's really the approximate solution here it's more or less y2 over 2 and here it's like 1 over s sorry s so that's how your approximate solution looks like ok so this is the approximate solution then then the next part of the proof is to perform some good energy estimate on the rest which is v right so then we write our solution as u u up plus v and then we need some estimate on v now there are 2 types of estimate so we will do some maximum principle type estimate and the second type are energy energy type with weight ok so and it's very important that I need whatever norm I'm going to use on v so I need a norm on v but I need this norm to control the trace of v divided by y7 so I need this to control v so v is what I call v v3 so I need this to control I need a norm that controls that and that one thing and also I need to make sure that this whenever I'm estimating this I want this to be better than 1 over s2 plus some eta so those are the two things the requirements is that I need a norm to be strong enough so that it controls this guy then it controls bs plus b2 and I need it to be smaller than this because if I am smaller than this then I can deduce that my b is like 1 over s plus something smaller ok so that's at the end the whole game so so any questions about this ok so can you be more precise about n ok, so n n can be for instance an example of n is that if you can take something like this if you take something like that that will at least it will do this of course now in our choice we take something more complicated with weights and so on right but that's really the point that whenever you take trace whenever you apply trace theorems you can control it at the end we are using something adapted to the linearized operator right ok maybe yeah, I can say a few words about that nice I don't think it's technical I think there is a good idea there so maybe I can explain it so you see like the way the operator here is given the way the operator is given it's not really an evolution in s you have us here but you have also us here so this is not very nice if you want to do energy estimates and you are slightly you don't know what to use so it turns out there is an idea you have to put these two together and ok I mean these difficulties also slightly related to what we see whenever we do Rayleigh operators when one looks at linearization of Euler we get sometimes things like that or also linearization of Prentor so so an idea that we introduced is to introduce operators of this nature so this L so if I do this if I introduce this then actually my equation my equation will look like this my equation can look like LU US minus BU2 so just introduce this this notation so somehow you combine these two terms together then what you do is that you apply LU minus 1 to everything and you can check that LU minus 1 makes sense and you can apply it and then it turns out that LU minus 1 applied to this you can compute it, it's simple LU minus 1 applied to this is simple you can compute and then what you end up with at the end and this is one of the version of the equation that we use so let me write it down so this is a I mean there are some interesting cancellations that happened that minus LU so this is your new equation I mean here we use the fact that LU minus 1 applied to U2 give you this LU minus 1 applied to this give you this I mean there are identity that you so then the good thing is that it only appears here now so this is again this sort of things I think Frank is more seems looks like type of things that also you you see like this sort of transport and the fact that we put the fact that you differentiate more you get more dumping also right so now you differentiate more you get more dumping so then you differentiate till you get this kind of dumping that would be the idea of the energy estimates now we will not be using this kind of things because this kind of things do not compute very well but it's more it's more like applying this kind of operator many times in this kind of operator related to this guy and you apply it so that's any questions about this so I think we spent most of the time talking about one result so let me mention I want to finish by talking about few other papers by other people but that I found interesting in that are recent papers about results related to printer and it turns out like the interesting things that some of the difficulties that we saw here like for instance this kind of this kind of object they are also hidden in all those works like at some point you have to deal with the fact that your equation if you want to think about it as an evolution you need to do something to this U.S. because the U.S. appears in two terms so ok, so I'm going to mention two one result that I found interesting so let me mention this result by some year I heard about long time behavior of it's about long time behavior of printer but when the pressure is 0 ok, so of course now the separation that we have here was really driven by the pressure it's really the pressure that is pushing you to become negative in a sense so if you don't put a pressure if you look at printer with zero pressure so if you look at this equation ok same initial data everything is same so in this case the result of Olenik the result of Olenik gives you global existence ok, so Olenik gives you global existence so so then so Prentol had a student Prentol in 1904 wrote this equation, had a student his name was Blasius and maybe few years after Prentol, maybe in 1908 or he looked at self-similar solution of this so as I said when you put zero somehow you have much more scalings right and one scaling which is maybe more natural is like some sort of parabolic scaling where you scale y like square root of x you can scale y like square root of x and if you do that so if you introduce eta equal y over square root of x and you can add plus x zero if you want and that's how it appears afterwards but if you look at this guy and you try to find solution of this equation then you have profiles which are these Blasius profiles and which are like in application turns out to be really very important like all physical, I mean many of the profiles that we see as solution for Prentol I mean when you don't have a pressure are coming from this because that's really the long time the large x behavior of this Prentol with zero pressure so then you can look at solutions it's not difficult to see that you can look at a solution that behaves like this so you can find the solution that that behaves like that I mean because this will satisfy the incompressibility this satisfies the incompressibility and then you plug it in and you derive an equation on F so there is an equation on F you get that F should satisfy this so it becomes just an ODE and then you fix I mean you need this to be zero because that's the initial condition and then you also fix F prime at infinity so you can fix F prime at infinity and the fact that also F of eta over eta goes to 1 at infinity so you fix all this you fix all this then you can solve that ODE you get an F you scale it like this and that gives you a solution so these are solutions of of course here we are fixing some behavior at infinity right so again, since you don't have a pressure since you don't have a pressure it means that it means that u Euler has to be a constant because remember like the u Euler and the pressure are are related so this is the case where u at infinity goes to 1 so this is the case that u of y goes to 1 when y goes to infinity ok so Blazius constructed this guy this solution, 1908 now what this guy Samir did like recently is to prove that solutions of this equation they relax to they converge to the Blazius profile so this is a long time behavior type of result so I should mention that his result was already proved actually but what I like about his result is more like the method used and which has some potential for other things so so as I said so Olenik gives global existence Blazius this is 1908 constructed this sort of particular solution now there is a result from 66 by Seren who proved using maximum principle using maximum principle techniques but not only maximum principle but also you need to go to the so called von Meiss coordinate we mention this last time von Meiss change of coordinate so you make this change of coordinate and then use maximum principle technique to prove that the u if you take a u here you start with an initial data which is not the Blazius profile you can prove that u will relax to some Blazius profile so the so you get the fact that you relaxes to Blazius so there is some sort of dumping some sort of dumping that pushes you to look like Blazius as x goes to infinity so it's more like a relaxation result so I mean I found this interesting because even that proof it's not completely okay so what what this guy did is to re-study that but more using energy estimates finding the right energies and so on which is I think very important in this business to be able to do things by energy estimates now this is one type of result that are recent that I found interesting so there is another two other results that I want to mention but maybe don't have too much time but the two other results are one by by go and same guy and the other one is by David Gerard Veret and and Yesunuri Mayakawa more or less around the same time more or less about the same time which is about the validity of the parental equation validity validity of the stationary parental both are from so somehow the parental system we derived it based on some asymptotics nothing I mean the result of conversions to parental are very few and like in the evolution case some of these results use analyticity and ok, Wednesday I will mention another result that I have with Gerard Veret and Mayakawa where we have to use we need jevre regularity but it turns out in the stationary case it can be done in sobolev so that's the interesting thing in sobolev so they can prove the validity of the parental in sobolev regularity so it turns out like the setup they have are different for Gerard for David and Yesunuri they look at the problem in the periodic setting in X so you look at something periodic in X you take some solution of Navier stocks with periodic data so there is no boundary condition to be imposed here and here because it's periodic and they put also some assumption about the profile for parental which is independent of X so they are taking some solution which is independent of X of parental and they can prove that solution of Navier stocks will converge to the solution of parental I mean the important thing for the days that you have a way of deriving parental rigorously from Navier stocks so that's the setup here the setup here actually is more interesting like more general because they are working in in here like between they have a flow that goes in so they have to impose some boundary condition but the type of boundary condition they have to impose are quite complicated like you cannot I mean it's similar to the type of approximate thing that you impose so they have an approximate solution that they compute to very high order and they need to be compatible with the boundary condition ok so they impose that they have to choose L small same here they need there is a condition about smallness in L and same thing they can prove that solution of Navier stocks will converge to the solution of parental I mean both proofs are really long this one has like 100 pages both proofs are really not very trivial one of the interesting fact about this even though this kind of things seem different from what I was talking about here but let's say an object like this guy also has to be used in both proofs like you need to understand how you can extract UX from that transport to structure ok let me stop here