 Thank you all for coming for the third lecture and as promised I will stop talking about 3D Euler today and start talking about the 3D Navier Stokes. So of course some of the at the level of the technology of the proofs will be related to the previous talk and I hope Tristan in the other three lectures will tell you more about the fancy results for Euler. Okay so we have Navier Stokes incompressible. In the first lecture we've discussed quite a bit about what is known and in particular LeRae constructed weak solutions for any initial data in L2 divergence 3 and I will write LeRae weak solution. Of course he did not call them like that and of course the uniqueness of LeRae solutions is open and it's one of the beautiful questions in PDEs. We do now quite a bit about conditional results which imply uniqueness such as any if any scaling invariant norm is a priori finite then the solution is smooth and unique right. So this type of results go back to Foyash and Prodi, Kisilev and Adizhinskaya, Serin, a lot of people. Instead today I want to give the proof of the following theorem which is joined to work with Tristan and it states the following that there exists a tiny number beta such that the following holds for any kinetic energy profile E smooth C1 is fine. There exists a weak solution U of 3D Navier Stokes with the following properties. One I'll tell you how smooth U is and the answer is not very smooth. U is continuous in time with values in H beta, beta is that beta so it's a bit better than L2. At this regularity level you may say well if the velocity is so bad can you even make sense of gradients or of the vorticity or things like this and the answer is yes. The vorticity is a bit better than L1 so this is the vorticity it's a bit better than L1 and the disturbing part of this theorem is that the kinetic energy of U is exactly that function that I've called E of t. Okay so this is the theorem I will prove today and notice very importantly that I have called this a weak solution and not an array weak solution. Okay so there's several consequences of this theorem. I want to state one more theorem before I'm going to discuss let me actually discuss straight up the consequences and then I'll discuss a couple more things. So what are some consequences? Should I define weak solution again? I've done it in the first lecture maybe I should define it again. Okay so we call U which is weakly continuous in time with values in L2 is called a weak solution or distributional of 3D Navier Stokes if three things hold for almost every time U is weakly incompressible so the divergence of U is zero in the sense of distributions and the equations are also satisfied in the sense of distributions. So the integral over R integral over the torus so I'm considering throughout my talk the equations post on the torus with zero boundary with periodic boundary conditions and the functions have zero mean because the mean we can always just remove to base the heat equation. So U dot dt phi plus U grad phi plus U Laplacian phi dx dt is zero for all phi compactly supported and if you want to study the Cauchy problem you should add one more term thank you thank you because otherwise we need to write a pressure term. Okay so you notice that there's two U's there so to give meaning you just need L2 space time we've added the continuity in time if you want to discuss the Cauchy problem and this notion of solutions of course the first paper I've seen this defined explicitly like this is a paper of Serrin from 62 but maybe Pierre Gilles can tell me if this was maybe defined earlier I don't know but this is the first paper I've seen it written explicitly like this. Okay okay okay so yeah what is the difference with Lorentz week? Yes I will erase the word consequences and discuss the notion of solutions. So before I tell you the difference with Lorentz let me mention an old result of Fabes Jones in Riviera from 72 which essentially says weak is mild. It says if you have U which is a weak solution in the sense of this definition if and only if and by the way I should impose some regularity for this any LP LQ more than two works but let's write it like this if and only if U of t is e to the t nu Laplacian initial data plus integral zero to t the heat semi-group the ray projector divergence of U tensor U. So if you treat Navier-Stokes as a heat equation with force then of course ODE says you should do Duhamel and then you have the heat propagator acting on f convolved with f and this is what's written here just a mild formulation of Navier-Stokes because f is the Ray projector applied to diverges of U tensor U and this is called the Ossine or mild solution so this is really the same as that what it is not the same as is the Ray because of two things one is important one is I don't think so important the Ray solutions as was mentioned earlier also have the energy dissipation rate finite right so that means U should be square integrable in time with a gradient in L2 right because the Ray wanted to make sense of the energy inequality that the energy at time t plus the energy dissipation rate is less than the energy at time zero so he needed or he got it for free in my definition I've never mentioned this space although in terms of scaling it has exactly the same scale so remember that Navier-Stokes is invariant under this rescaling U sub lambda of xt is lambda u of lambda x lambda square t this is the parabolic scaling this is having to do with the fact that there's a nonlinear term with a single derivative in it if U is a solution so is U lambda and then for any space time space you can ask when is this behaving like a power of lambda and this space and that space behave the same but of course one encodes different kind of information and the other way in which these solutions are not Leray and I will emphasize this one because I think this is the important one is that they do not necessarily obey no energy inequality so mild solutions don't have embedded in them the energy inequality because if you write this form down good luck proving out of there the energy inequality without actually knowing the solution was kind of good already and this is I think really the fundamental difference between just a mild solution or weak solution however you want to call it and the Leray solution and by the way we discussed this in the first lecture but the Ray's solutions actually obey not just this but also this the energy inequality not just time 0 and t but almost every s bigger than 0 and all t bigger than s so it's not just that the Leray solutions have this energy inequality they have this one that you can look at you can start at any time and look a bit later almost every time not just start at 0 these two things are very different from each other double star is contains a lot more information than star double star is believe it or not much more powerful okay so what is this theorem that I've stayed on the first board how is this fitting in in the in the program so the conjecture let me write now some consequences so if I take my energy e in advance to be decreasing I'm able or we are able to construct infinitely many dissipative weak solutions you can also improve you can prove non uniqueness if e let's say increases a little bit then non uniqueness in what way well I can start with any data in L2 prove that serum with a prescribed energy the Leray solution is another solution whose energy goes down they're not the same or equivalently if you start with zero data the only the Ray's solution starting with zero zero this doesn't have to be like this and third there's no backwards uniqueness this is an open problem again posed by serin so in this paper he asked if those solutions can reach zero in finite time so you start somewhere non-zero can you dissipate everything in finite time and the answer is well in that regularity class yes and if you end up at zero you don't necessarily have to come from zero so there's no backwards uniqueness for mild solutions what other consequences did I want to mention I wanted to mention the bound for the vorticity I wanted to say that the first bullet in the theorem implies the second bullet automatically and the reason is when you take a curl by the way I wrote curl because div is a priori bounded if you know different curl you know everything if you just apply curl here you can read a very beautiful exposition in Pierre Gilles book this is a calderon segment operator on space time with the parabolic distance so in particular you'll get that the vorticity has the same space time regularity as u squared if u is a bit better than l2 then u squared is a bit better than l1 and that's what stated there okay so the vorticity estimate is for free now I want to mention before I get to the proof a couple open problems and the first one I will state is that non-uniqueness for mild solutions in the indirect class so this open problem is not the same open problem as that one okay because mild solutions they don't know what the energy inequality is they don't have it so what I'm hopeful about is that you can prove that the l'array class by itself is not sufficient to recover any kind of meaningful uniqueness and more somehow a stronger version of this theorem would be non uniqueness for mild solutions in c0lp for p less than 3 or c0hs for s less than a half again this would be showing that scaling dominates the regularity problem and the uniqueness problem at least in the class of mild solutions because we know since Fujita and Kato that for h1 half data you have local uniqueness since the result of Kato for l3 data well it's a bit for mild solutions it's not due to Kato right so you can read that in his book also alright what are other by the way this one after we've proven our theorem some people got interested in it and Terry Tao on his blog proved that this is true if you send the dimension to infinity so if instead of three space dimensions you send the spatial dimension to infinity he proved that using our method you can get this result but of course you want to prove this in three dimensions lastly and this is maybe a bit these two I actually strongly believe are doable with modifications of this technique the C is a bit of a stretch same as a but also with this is a bit of a stretch I am willing to bet that using these techniques you cannot get double star star I'm not so sure about so these are some open problems of increasing level of difficulty and of course the uniqueness of Lyray solutions is yet another level and I should mention that in terms of in terms of this that there's a beautiful conjecture by Gi and Sverak I think it's 14 which states that not only should Lyray solutions not be unique but in fact as soon as you depart for initial data from L3 and you just go to week L3 and what's a function in week L3 in three dimensions think 1 over x that's not an incompressible vector field but that's the behavior both at zero at an infinity that you should have non uniqueness and their theorem states actually the following consider a self-similar solution a forward self-similar solution to 3d Navier-Stokes which behaves like this then linearized Navier-Stokes around it you get a linear operator if the spectrum of that linear operator is unstable then theorem non uniqueness follows now proving that the spectrum of that operator is unstable is very hard and to date it's not known there is a beautiful numerical simulation by Julien Guillaume who I think is in Paris and Sverak so this is a numerical simulation of the spectrum they compute on the computer the spectrum and they get that it's you do have unstable spectrum so this is a completely different strategy of establishing non uniqueness of Lyray and it's very promising so somehow this is coming from the top by just going a bit below L3 and Tristan and I we went from the bottom by going above L2 and of course they are just different techniques okay I did want to say and I should say it that in the same paper with Tristan we also considered the vanishing viscosity limit and the theorem we wrote in that same paper states that if you give me an Euler solution any kind just further continuous given a further continuous weak solution of 3d Euler in particular the ones that were constructed to resolve the on-sugger conjecture they can be attained as vanishing viscosity limits of Navier-Stokes solutions so there exists a sequence of viscosities that goes to zero and the sequence un of weak solutions of Navier Stokes in that sense with those viscosities such that let me call it U bar such that UN converges to U bar strongly in C0 L2 so in the energy norm in the energy class so all these solutions have finite kinetic energy what they don't have is a finite energy dissipation rate actually if you were to look at the energy it looks almost like a Kantor-Lebesgue function except the Kantor-Lebesgue function is constant most of the time this would be the dissipative hopefully Tristan will mention this result also that's a different result no no it cannot possibly be true because if you have a solution of Navier Stokes which is held there it's smooth if we could prove that then I would start with that theory yes yes so okay so for the rest of the talk I want to mention the proof and the point is that the proof of constructing these very non-physical solutions goes by actually incorporating something from physics into the construction and that's intermittency and Professor Raell knows more about intermittency than I will ever learn but the point is that if you build intermittency into convex integration so you then you can so you use physics to construct via these things a very non-physical object so the the the fundamental difference between what I've presented for Euler and what I will present now is that I will call this intermittent convex integration and this is really the technique I want to describe today for the remaining time I want to really just carry the message of this technique what's intermittent about what I'm gonna do is the following the solution that we have constructed for Euler was based on adding plane waves signs and cosines a sign and a cosine no matter the frequency wherever you sit on the torus it looks mostly the same right intermittency deals with deviations from homogeneity in turbulence so in particular it's trying to capture sporadic high intensity events so we want to build into this geometric construction building blocks which are not spatially homogeneous they instead have very concentrated regions where a lot of stuff is happening and then there are seas of calmness and then there's another very turbulent quote-unquote region and so on and in addition the same thing will happen in time so in in time they will look pretty wild as well they're not going to be just the same everywhere in time okay so to build to build this I need to recall the framework of just convex integration before I can again say what's intermittent about it and let me start by remembering the Euler the Navier Stokes Reynolds relaxed system and so on okay so the goal is to construct approximate solutions which solve the Navier Stokes Reynolds equation so you have dt of uq q is a natural number as before plus gradient of a pressure minus nu Laplacian so far I've written the momentum equation and now instead of equaling zero it's equal an error which is always a divergence of a traceless symmetric matrix and the divergence of uq is zero so our goal just like in Euler is to construct a sequence uq rq to this equation such that rq will go to zero and the increments will also go to zero the first condition states that when you as q goes to infinity the first condition simply states that when you send q to infinity this error term will die okay in the limit there will be no more error because you're going to send a norm of it to zero I didn't on purpose write what norm I will tell you what the second condition says that in this iteration scheme the increments are going to go very fast to zero so in particular you will get a convergence here is now in what norm in Euler these were C0 norms for Navier-Stokes you cannot possibly do C0 norms because again if you have a continuous weak solution of Navier-Stokes it's smooth so you have to go below and we know that we want at least finite kinetic energy because if we are below kinetic energy level then we cannot even mention the word weak solution so if u is an L2 type object what should R be remember that what happens here is that you're building a cascade from the nonlinear term to cancel the stress right that was the the message of the of the talk yesterday that you're using you're building by hand an energy cascade so R is going to be like u squared and if u is an L2 type function R must be an L1 type function and just like yesterday I'm not writing the time integrability they will always be continuous in time so these are always in space and in time it's always C0 but I'll never write it because it's always going to be C0 in time next we need to quantify this convergence just like last time so introduce a frequency parameter lambda q morally speaking please think of uq living at frequency lambda q and below okay what do I mean by living at that frequency not really in the sense of projections but more in a Gaussian sense okay the main tail of the Gaussian is in that ball and just like before I will take this to be super exponential a not to the q but to the b to the q where a is huge and b is huge so there's an incredible separation between these frequencies if you were to look at this a priori they would be very sparse a priori in particular I will use frequently that the next frequency is the old one to the power b and b is huge okay then you have amplitude this is how we measure the frequency of objects but in order to compute norms we need to compute the amplitude of objects and these parameters we have denoted in the last lecture by delta sub q and it's just an inverse power of the frequency parameter and bet I steinie in space this is a Fourier in X it's actually important you ask this because there will be a time frequency also but I'm not writing it on purpose yet down but there will be a time frequency which is huge it's almost like lambda q squared okay yeah okay so we've introduced a Fourier frequency parameter and amplitude parameter so now let's quantify these two statements okay so I want to quantify these two statements so inductively I will propagate three pieces of information one is that my sequence uq is uniformly bounded in L2 it's a bit below one always and you're leaving yourself a bit of room then my stress will go to zero in L1 like this amplitude parameter with a tiny constant there universal constant and then in order to close my scheme I should know something more about uq than its L2 norm something about its derivative and what I will write is that uq in C1 in space and time okay so if you think of an L2 object which is living at frequency lambda q how much should it cost to go to C1 well one derivative should cost the lambda q but then you should also go from L2 to L infinity in three dimension that costs a bit more three halves and one it's five halves so to give ourselves some room let's write lambda q to the four okay this is just to give ourselves some room or it's because of the time derivative because the time derivative is more like the Laplacian and now you have to estimate the Laplacian so you give yourself some room okay so throughout the scheme we will propagate so the goal is now this for every natural number q construct the solution of the system with these three properties so you don't need any information about the increment you do and in this process I did not say anything about this so I will add it on this board you want that the increment is in L2 by the way I should stress this L1 because this is gonna be very important for the intermittency part this is L1 and this is L2 this in L2 should be less than again the stress one should think of it as the square root quote-unquote of uq so if this is of size delta q plus one then the increment should be square root of that so it's not three properties but four properties that I want to propagate now let's quickly show that those properties imply the theorem oh and by the way this thing requires more okay I will not I will only construct a solution with increasing energy because otherwise I'm gonna have to state stuff about energy and I don't want to do that so we'll construct a solution which increases the energy and then the same initial data it's LeRae solution that will decrease energy so nonuniqueness follows so the first iterate u0 just like in Euler we just take a shear flow so this is a shear flow multiplied by t this of course does not solve Navier-Stokes if you stick this into Navier-Stokes the nonlinear term will die the pressure is not needed because it's incompressible but the dt will land here and the Laplacian will land there so this will not solve Navier-Stokes it's gonna solve Navier-Stokes with error and this error just like last time can be written down explicitly but now we have viscosity so like always these rqs have to be traceless symmetric the trace is zero and it's obviously symmetric and in order to start the procedure I need to make sure that these are true these three rq not not this one because there's no increment yet just these three at level zero for convenience you don't have to do this but so that I don't spend much time let's set that number to be one because otherwise I have to explain a little bit about that so take viscosity to equal lambda zero to the minus two just for simplicity of the presentation so that this is one okay what is the kinetic energy of you well it's a parabola goes like t squared times whatever sine squared integrates to on the torus so this is energy this is time so if the time is up less than half or something this is automatically true because t will be less than half so this is a quarter it's automatically true what's the l1 norm of r0 cosines have l1 norm of order one and the only parameter that you have here is one over lambda zero so if lambda zero is huge automatically this is going to be true by just taking lambda zero huge how about this one how much does a c1 norm cost well a time derivative costs nothing a space derivative costs lambda zero but we allowed ourselves to cost lambda zero to the four so it's all good so if you take lambda zero huge and of course I can class classified this then I am allowed to start the induction scheme once the induction steam is started it produces this infinite sequence uq and like in Euler exactly like in Euler this condition will guarantee that I converge to something finite so the limiting that exists you which is a limit as q goes to infinity of uq strongly in l2 okay that's just follows from uniform boundedness it follows from this estimate u minus u0 which is just nothing but the sum of increments so there's a telescoping sum u0 is this explicit function so let's estimate the distance from the limiting solution to the original starting point well that means I need to sum the increments so let's put an HS norm here then I'm going to have lesser equal the HS norm here by interpolation I have the l2 norm to the 1 minus s and the h1 norm to the s by assumption a gradient I know what it costs me even in l infinity the torus has finite measure so this will be less than a lambda q plus 1 to the 4s because this costs more than that this one is the increment I know exactly its size it's delta q plus 1 to the one half and I remember who delta is so one half kills that too so it's lambda q plus 1 to the minus beta times 1 minus s and if 4s minus beta times 1 minus s is negative it means that this is a geometrically summable series and I get here less than lambda 1 to this power what does this condition mean take s small right beta is something when you send s to zero this will be true it's it's not even that complicated what is it beta okay so this is why you have strong convergence in l2 because in fact you have strong convergence in HS and moreover I have I wanted to do this bound this less than this because it says that the limiting solution not just in l2 but in HS is very close this is a negative number right is in a small tube of width of that width around more original guy so the kinetic energy of the limiting solution can be at most this and at time one half it's at least this so you've increased the energy from here to here and that's of course a very non-physical thing to do okay so what remains now is to prove the inductive procedure first step like an Euler is to modify there's an inherent derivative loss problem just like in a typical Nash Moser scheme so you modify so you recall you call you L to be UQ modified in space and time with some modifier RL to be RQ modified in space and time with some modifier L is a length scale it's tiny I will tell you exactly what it is when you modify Navier-Stokes just like Euler in all the linear terms the modification just goes through so I have dt UL minus nu Laplace in UL plus the gradient of a pressure but in the nonlinear terms it doesn't go through so you have to commute modification past the product and the error is a commutator keep in mind the circle on top of the stress this means that it's not just a symmetric matrix but it's also traceless and this commutator up to a sign is UL tensor UL minus UQ tensor UQ modified and to estimate the difference between a function between a product of modified objects and the modification of the product right which is not zero we have a simple commutator estimate that the L1 norm is essentially less than the L1 norm of UQ times UL minus UQ in L infinity it's one estimate you can do it's not very hard I didn't propagate L1 bounds so I have L2 bounds instead I can do hurl there because my torus has finite so I can put a 2 there so then this is less than my L2 norm is less than one so let me not write one and this costs how much L times the gradient the gradient is exactly what I've propagated lambda q to the 4 because again this is less than L times the gradient in L infinity and the gradient in L infinity is exactly what I have bound so somewhere I will need to keep track of parameters so I guess this board is a good board maybe I'll try not to erase that board even for the second lecture I will choose L to satisfy two things one is that that thing is tiny how do I quantify tiny it's an inverse power of the next frequency alpha is yet another parameter it's a tiny parameter but it's of the next frequency that's very important and moreover during this modification I also need to be careful because accidentally I don't want to modify it's such a small scale that I've already gone beyond my next frequency so L inverse can't be too large so let's make it large but not too large to alpha the same alpha is a tiny number but keep in mind this number B is huge so because of this relation something which has the index Q is absolutely destroyed by anything which has the index Q plus one because I can make B huge okay so whenever you see something with the index Q or L for that matter it can be completely annihilated by Q plus one okay so then and I think I'll write one more line and then we take a break we have this equation but we want to get to this equation with Q replaced by Q plus one because that's the inductive procedure right so you declare W Q plus one to be this increment so I want to get here so I'm going to add to this guy a perturbation what is this perturbation obey DT W Q plus one and I'm going to arrange terms in a very funny way and I will only be able to explain this arrangement in the second lecture then I have here all the linear terms plus some kind of pressure term the pressure is irrelevant equals the new stress and there's somewhere are this guy where should I write the irrelevant thing this is irrelevant the commutator stress which is here due to our choice of L already obeys this bound with Q replaced by Q plus one so this is already much less in fact so this is already done I don't have to do anything with it so just like in Euler the game is design W Q plus one with this constraint so we need I need a half because I went from you Q to you L that's a half and now from you L to you Q plus one another half design W Q plus one such that given an error I will get another error which is smaller so in some sense ideally you would solve this is equal to zero and then you would be done turns out that you can do that for short time but not for a long time and the amount of time you can do that for is very tiny so instead we're going to do something very different and this very different thing will be called so it's we're going to construct as a superposition of what we call intermittent jets and I'll tell you what they are they're very beautiful explicit things in which there's a network of pipes in the torus and there's water flowing through these pipes at some incredible speed in a very precise geometric arrangement so I in the next hour after we take a short break I will convince you first of all what kind of object do I need and then I'm going to prove that the object exists and that object will be called an intermittent jet and I should add that this intermittent jets we wrote them not in the paper with Tristan but in another joint paper with Tristan and with Maria Colombo so in that paper with Maria Colombo where we also proved partial regularity in time this is where these intermittent jets are defined okay so let's take a quick break all right so we are left on that board and it's the same as in Euler so far except there is one more term and I've arranged the terms on on the line in a different way that's all so far so what we did next in Euler we wrote the linear algebra lemma so this time let's write the original lemma written by Nash in which which he used for the asymmetric embedding problem and it basically says that stresses can be spanned by primitive metrics so let me just write this down there exists a set lambda there exist many sets but I will need just one which is on the unit sphere with rational entries and for all elements in this set which I'm going to call Xi I have a smooth function gamma Xi defined in the ball of radius one half around the identity in the space of symmetric matrices you actually just need a positive matrix you don't need to be in the ball of radius one half you just need a positive matrix such that M any matrix in that set can be written as this sum so the only thing that's different from the lemma for Euler is that this is Xi tensor Xi as opposed to the identity matrix minus that okay this is a primitive matrix which is a vector tensor itself okay so that means that when we design the principal part of the perturbation just like in Euler we have some amplitudes which we're going to construct in terms of the all stress and then we have some vectors which we will want to point in the direction of Xi because we want from u tensor u to get Xi tensor Xi so instead of writing a general vector which points in a direction Xi let me write Xi times a scalar function so this is a scalar function and it's fast this is the slow part and this is okay so if this is true if this is my ansatz then of course this tensor product of the principal part of the correction should be a double sum but I will insist that these bump functions w Xi have this joint support is empty I'm just gonna require this for the sake of it they're gonna have this joint support so when I multiply actually get a single sum it makes bookkeeping a bit easier it's not fundamental so then I get Xi tensor Xi and the square of that bump function okay so remember that we want this quadratic form w q plus one tensor w q plus one to cancel RL and this is gonna be a slow object but here I have a fast object so in what's the slow part of it the slow part is the mean and in order for me to apply this lemma it's very convenient to set the mean to be one it's gonna be seven of them now if you have a bump function just let's think of this w as a bump function if I have a bump function and I square it I still get a bump function right I will have to make it periodic because I'm working on the torus after all so when you subtract from a bump function it's mean and again it's periodic what frequency does this object live at and I guess it depends right if you think of you know just sign sigma x this is a periodic thing and you square it and you subtract its mean you get cosine of 2 sigma x so it's related to sigma but this is if it has one mode what if it has many modes bump functions tend to have many modes not one mode so let me put as a requirement that w psi is periodic at some scale I will tell you what sigma is but what should I want from sigma this somehow is slow it has frequency related to L so I wanted the period of this to be higher than L actually if you make it much bigger than L to the 5 then I can put a constant it doesn't matter okay it's it's gonna be very fast in particular lambda q plus 1 will do if sigma is lambda q plus 1 that's gonna destroy any L but if it's lambda q plus 1 to the power 1 over 10 it's also gonna destroy any L because L has this alpha there and if I make alpha less than 1 over 20 you know so sigma will be something like lambda q plus 1 to the 1 over 10 I'm I'm ignoring a minus 5 sorry about that and of course you can't have this period to be larger than lambda q plus 1 because otherwise you've exceeded your next frequency the next property you would maybe want is that your perturbation is not faster than the next frequency available which is lambda q plus 1 so maybe what you want is that every single gradient hitting on this bump costs at most lambda q plus 1 times a constant the constant has to do with exactly how I normalize things so because I've normalized it in L2 I can put L2 norm here but you know what I mean every derivative should cost at most that much the other thing is that if I want this to be incompressible then when I take the divergence to leaving order it's gonna land on the fast object right and the divergence of Xi times a bump function it's Xi dot grad so if I want this to be leading order incompressible I will need that Xi dot grad is small so what does it mean small compared to what the divergence is much smaller than the curl that's what it means to be roughly incompressible okay so I will want that the divergence I'm gonna put this in quotations the divergence of Xi w Xi which is nothing but Xi dot grad is much smaller than the curl if this is true then I can add a small perturbation to make it incompressible if it's not true I cannot make it incompressible with a small perturbation I would need a large perturbation so keep in mind this is never incompressible the principal part you need to make it incompressible by adding a tiny thing and how tiny that object is it's gonna be the quotient now let's go back here this is now one so we would like to cancel RL right so that means we should define these amplitudes in terms of these functions gamma now what's different in Navier-Stokes and in Euler in Euler RL was naturally a C0 object so it means I know everywhere in space how big it is and I can naturally define a square root of it but here RL is naturally an L1 object it is actually C0 because I've modified it but the bounds which are good are L1 bounds so that means it's not the same everywhere so how I cut it starts to actually matter so it's useful to think of this as square root of RL in terms of estimates every single estimate I'm gonna prove is the same estimate for square root of RL except what exactly do I mean by that right RL is not positive I mean what what do I mean by this so okay so to define the amplitudes you introduce a function rho of xt which is trying to mimic the square root function so it's a constant related to that constant let's actually write it because then so that C is that C delta Q plus 1 chi of RL divided by C delta Q plus 1 where chi is a smoothened out function so what is this rho rho is now because I've avoided 0 right when you take derivatives of square root you're worried about 0 chi is never 0 okay so I've somehow removed the 0 also raw has LP norms when RL is much less than this by the way why did I divide by this because that's exactly what it's what is there so inside of the parentheses I have a unit size L1 function so this unit size at one function could either be small and that says chi spits out 1 and then the LP norm will be just delta Q plus 1 because it's a constant or it spits out that guy itself in which case these deltas cancel and I get the LP norm of RL so row is something which has the same LP norms as R but it doesn't have the issues near R is equal to 0 and then once I have this row I can really tell you who a who the a's are a psi of RL is nothing but row to the one half these gammas of identity minus RL divided by row why did I do this besides this beautiful property row has another beautiful property that when you divide RL by row you see RL divided by row essentially this is point wise less than a half I started with an L1 function I cooked up another L1 function which is that one and it's at least twice as big always that means that this object is going to be in the ball of radius one half around the identity so that means I'm allowed to plug in those matrices and this is the proper renormalization because now this term which is underlined there becomes this so a squared will mean row times gamma squared gamma squared will give me back the matrix so I could get row times the matrix so it's row times which matrix this one and voila you've cancelled RL modulo and I multiple of the identity which you're going to hide in the pressure anyway it's not zero row depends on x and t you're really changing the pressure in an untrivial way okay so that's settled now comes something a bit more interesting I want my perturbation to obey that estimate it's L2 norm should be of size delta q plus 1 to the one half the L2 norm of this is the L2 norm of this product and these I've normalized in L2 to be one so you're very tempted to put L2 L infinity what's the L infinity norm of A this is an order one object so the L infinity norm is the L infinity norm of row to the one half what is row in L infinity delta q plus one times RL in L infinity that I don't have access to I had access to RL in L1 so it seems that there's a fundamental issue either I've normalized in the wrong way but then you will discover I have other issues because then the number one which I have written here wouldn't be the number one and then I have a much bigger problem or I should not to herder and the answer is you should not to her there because whenever you multiply a slow times a fast object I guess this is a lesson for the younger people in the audience never apply herder it's a loss it's a huge loss so how much does the gradient cost on a so a first of all it has a size in L2 the size of a in L2 is like row in L1 right just because of this what is rowing L1 this plus this in L1 so it's both of size delta q plus one so this has a certain L2 norm and that quantifies its amplitude now how fast is it how much does a gradient on I xi costs how much does it cost every single gradient well you can hit the gradient here or there if it falls on RL it costs an L inverse how about if it falls on raw well if it falls on raw it also costs something related to L okay there's all sorts of square roots there's divisions unpleasant stuff it costs at most L to the minus 5 okay which is why this this is some algebra I'm not gonna do it because I'm gonna embarrass myself so you have a function which technically oscillates at frequencies below L to the minus 5 and you're multiplying it with a torus divided by sigma periodic function where sigma is much more than this theorem and this is a we call it the decoupling lemma you may call it the homogenization lemma this is actually less than the L2 norm of this times the L2 norm of that fake helder and the one comes from the fact that I've normalized my L2 norm and I forgot the one-half somewhere right so actually because of this decoupling lemma which says that slow times fast obeys strange helder which is we actually will be in business if whatever I'm adding namely the incompressibility corrector and whatever else I'm gonna add is smaller then I'm gonna be okay and again because of this they put compressibility corrector will be smaller what's next let's try to estimate the Laplacian because we've never really discussed anything about the Laplacian so let's try to look at this term and see what is its contribution to the divergence of Rq plus one so I need inverse divergence of Laplacian of the principle part of the perturbation inverse Laplacian sorry inverse divergence of the Laplacian is also known as the gradient right and in what norm do we estimate this in it's a stress all stresses are in L1 and this is a key thing this is naturally an L2 object but we're estimating it in L1 that should ring some kind of bell so this is of course roughly the same as the gradient in L1 and when the gradient hits on the product it costs more when it hits the fast guy and because of this property it costs exactly lambda q plus one so this is gonna be less than some over Xi lambda q plus one the L1 norm of this again I have my fake L fake herder inequality and the L1 norm of prime whatever that means all right so what is the L1 norm of a a is naturally an L2 object I don't have access to its L1 norm because I don't have access to the L1 half norm of R there is no such thing so I'm just gonna do herder this is a bump function which is normalized in L2 so when I try to estimate when you take a derivative it's still a bump when you try to estimate it the difference between the L1 norm and the L2 norm should be related to the size of the support herder says it should be the size the Lebesgue measure of the support to the power one half times one or something of order one and what do we want we want this to be less than that now you can unpack what this means and what this will mean is that W Xi is more than 2d thin meaning the Lebesgue measure of its support of a single bump is much less than lambda q plus 1 to the minus 2 y minus 2 because I'm gonna take the power one half here and I need to beat this guy but in addition I need to also get below there so I need to add the minus epsilon for some epsilon I think it's for beta b if I do the computation or beta b squared something like this no it's for beta b actually so this is the key point of all of this the building block has to be what do I call it more than 2d thin if you imagine something which is a tube that has a long axis and two small axes because of condition for these small axes can be not more than lambda q plus 1 inverse so the total measure of this long tube will be lambda q plus 1 to the minus 2 and not better if you think of a ball where the radius is related to lambda q plus 1 to the minus 1 then this is cubically small what you require here is more than 2d and the emphasis is on more so this will be true if I succeed in getting six because whatever epsilon I get I will make beta to be so tiny that I can go below that absolute okay so that looks very promising what else do we need oh by the way I'm not going to write this but we've estimated this in L1 you should believe me that these in L1 are much smaller because you didn't even pay a gradient right inverse divergence here is a gradient the gradient costs a lot multiplication by this costs nothing so this is incredibly much larger than that so if I succeed in doing this we have killed the entire line there so what's left well dt and the divergence hitting this at high frequency namely the divergence landing there what was the lesson learned from Euler the lesson learned from Euler was that if I want this high order oscillation error to not matter I should ensure by the way when the divergence hits it kills the mean anyway right so what I should want is that the divergence of psi to be roughly a pressure gradient that's the lesson we have from Euler and by equal I mean up to small error okay if you succeed with this then you know that there's not much there's no fundamental enemy left and then you're hopeful so now you ask the question does there exist an object with these seven properties so some kind of nice normalization property support properties some periodicity roughly incompressible this says the frequency solves stationary Euler and is more than 2d thin and the conjecture I have and I think Tristan agrees and this is a beautiful Liouville theorem waiting to be proven by somebody that this is the empty set I should add that I want this so with this psi right because I also need to span matrices right I really need these directions to span matrices so so let me write a family to emphasize that I need six of them and the reason that this doesn't exist are these two properties if I erase either one of them it's very easy to construct an object if I erase the last one I just take a bump function supported in a ball it's gonna obey all properties how about if I erase six but keep seven and that was known that if you raise key six but keep seven such an object was known and these are the Mikado flows of Sara Daneri and last was a key I was trying to find the year 17 so what are Mikado flows their shear flows but very specific shear flows if you take the box the original box you truncate it at length lambda q plus 1 inverse which is the only length that you should have in view of sorry it's the smallest length that you should have in view of property for and now you blow up one of these things and on this blown up thing there's a tube this is the direction psi and this is a cross section of the tube so it's a cutoff which is acting if this would be the easy axis will be only acting in the radial direction and if this was the easy axis it would be axis symmetric okay it has zero mean so so there's some flows like this there are some flows like this and why does this solve Euler well let's write down what that means the divergence of psi okay let's actually write what this is this is equal to psi psi dot grad these guys oscillate in the directions orthogonal to psi not in the direction of psi so this is zero now this is of course periodized so there's a that's why we insisted to have rational directions because we want to periodize them and not to become dense so this obeys all the properties except six because if you check how small is this object in this direction it's a it's a very long periodized tube if you were undo the torus and you view it on R it would be an actual tube it's very or it's order one in this direction and it's order lambda q plus one to the minus two in this direction so this is exactly two dimensionally small and by the way it's also incompressible this is an exact solution of Euler this is just a shear flow okay so this doesn't work because we need more than 2d so what's next and this is where the intermittent jets come in the natural thing to make it three dimensionally small would be to cut it in this direction to introduce some kind of smooth cutoff at some scale doesn't have to be like this one right this one is it just needs to be a bit more than 2d so minus something you you can cut it right but you introduce a fast cutoff all of a sudden these terms will matter a lot these terms are in no way pressures because if you take the curl of this expression you will get something huge they're not precious you can't hide them so there's something fundamentally wrong with cutting them you it just doesn't solve stationary Euler and the punchline is this guy is not our enemy it is our friend if instead of saying it's a pressure gradient what if you say it's the time derivative of somebody what does that mean if this blob of fluid would move in this tube it's a network of tubes which are periodized would move at some kind of linear speed you could make this a time derivative and that's what I want to convince you of and then we're going to add one more corrector which is going to be called a temporal corrector which is designed exactly to cancel this so the intermittent jets will basically have these networks of tubes in which there are these ellipsoidal fluid parcels which move through these tubes at some incredible speed and this is the temporal oscillation that I'm going to I mentioned earlier and now I'm going to have to introduce it so for the sake of clarity let us denote psi to be easy so that I don't have to tilt things so then what I'm going to do is I'm going to introduce two new parameters which have to do with the parallel length and the perpendicular length the perpendicular length will be related to this parameter r-perp and the parallel length will be a parameter r parallel and what I'm going to define them to be is as if these were order one so let me just write them because then so I introduce a cutoff of phi we've used already phi r-perp it's a cutoff in the directions perpendicular if you want to make it radial make it radial why did I normalize it like this because the L2 norm of phi r-perp is invariant it's the same as the L2 norm of phi which I'm going to choose one so phi is a bump function on the unit ball phi r-perp is a bump function on the ball of radius r-perp but with the same L2 norm so it's a spike then I have psi r parallel and I'm adding t there and why am I adding t this is psi is e3 so this is dz so this is the guy I'm gonna have to deal with so I'm gonna do actually you know what let me do this and I'll do the time dependence later let me just define a bump function right now it's 1d only so I need to put one half so I have these two bump functions at this point and now I will define w these bump functions that I have there have small support right so how do I periodize them I view that support as sitting within the cube of side lengths one and then I repeat okay so those are periodized to unit that's actually quite important but my property there was that these are periodic with some different thing so I need to artificially pump in periodicity so what I'm gonna do is I'm gonna say this is phi r-perp off not x1 and x2 but some sigma times x1 and the sigma is this one this is sigma times so this cuts in this direction times the cut in the other direction same periodicity of course and time and mu is this new parameter it has to do with how fast these blobs of fluid move through the networks of tubes and they move of course linearly it's t and they move in the direction of the tube which is in this case e3 I've described this I'm only missing one thing to have an honest definition I've defined for you what the tube corresponding to e3 is but there's many of them right so I need to put a rotation matrix there okay fine but I also want this but in three dimensions and last line sorrow observed this and this was used essentially by I said when he saw the on-sector conjecture in three dimensions you can put tubes such that they don't ever intersect so all I have to do is shift these tubes a little bit so that they don't exactly intersect and you can do this only in 3d in 2d any two tubes would intersect so I'm not write this what's the upshot well since I'm lifting the board it would be smart to write down what they are now I left the room there to be filled in later but I might as well fill it now how much volume do these tubes take by the way well they take square of that times square of that times this so they take up volume which is our perp squared times our parallel that's how much volume this so now we'll want to estimate this what I will have is our perp squared our parallel so whatever I define them to be this needs to be more than two and I think this is something that works it's not universal I forgot already because there's so many choices 6 on 7 and 4 on 7 this turns out to work the first one squared is 12 over 7 which is not quite 14 over 7 but when you add the other one is 16 over 7 16 over 7 is definitely more than 2 so this will be check so now still have maybe a bit more than 10 minutes but not much more so let me now convince you that these intermittent jets we call them jets because they move like this actually obey those seven properties and then I'm gonna try to convince you of the fact that this combination saves this term okay and then I think I'll stop so property one normalize to unit size well yes of course they're normalized you have Fubini integration in this direction times integration in this second one the supports don't touch again you just shift the tubes a little bit the supports don't touch third one this is periodic with that scale with those properties so let's check what is sigma so for us sigma is lambda q plus 1 r perp which with these choices is lambda q plus 1 to the 1 over 7 1 over 7 is definitely less than 1 and lambda q plus 1 to the 1 over 7 will definitely beat L to the minus 5 by making alpha tiny L to the minus 5 will be 10 alpha so if I make 10 alpha less than 1 over 7 I'm good next a gradient should not cost more than lambda q plus 1 okay so let's hit it with a gradient if the gradient is in x1 or x2 it lands here what do you pay so d1 and d2 acting on cost this times 1 over that how about d3 it's a bit less right because you still pay this cost some periodization but you're dividing by r parallel and this is much less than that because with this choice the quotient is 2 7 so it's 5 7 the fact that this is slower is very important and the next property will show you the divergence is e3 dot right it's d3 so the divergence will cost that much the curl will point in those two directions the curl will cost much more so while this object is not incompressible it's to leading order incompressible and that means we can add a small perturbation to make it incompressible I've already convinced you of the support and of course I'm cannot convince you of this but what I can convince you of plus dt of something because as I said for stationary you can't do it essentially what we'll building is not a stationary solution of Euler but a time-dependent solution of Euler by hand essentially to leading order so let's do the computation with this building block let's compute the divergence of e3 w e3 tensor e3 double ah let's just do it in general why am I even bothering so this is psi psi dot grad psi squared psi dot grad which in this case is e3 cannot land on phi right so this is psi phi sub psi squared so I'm not writing I'm just to emphasize that it's related to psi and then psi dot grad what we've called psi what is psi dot grad in this case will be d3 turns out to be equal to 1 over mu dt this guy didn't depend on time neither does this I've convinced you that this is the derivative of somebody and that somebody has this beautiful factor of 1 over mu in front if mu is huge this object is tiny without a time derivative right so we need to incorporate a temporal corrector I will erase the micado flows and define the temporal corrector all right so there's an incompressibility corrector and this is the temporal corrector and it's designed such that it's time derivative the time derivative of the temporal corrector will balance the time derivative coming from the principal part so I think it should be 1 over mu sum over psi it should be incompressible so let's write the very projector of psi it's exactly what I have there well which is exactly w psi squared okay let me so this is I just define this object how much l2 norm does it contain well it's 1 over mu times the l2 norm of this keep in mind these are squads so this is like the l4 norm to the one-half and this guy is spiky so the l4 norm is much larger than the l2 norm but you have the 1 over mu to help you and if you choose mu properly I can guarantee that this has l2 norm which is tiny because the loss you have from growing from l2 to l4 think of Bernstein is accommodated by choosing mu very large now dt of wq plus 1 plus plus what the term I have erased oh yes yes yes yes thank you that's very important thank you thank you absolutely so plus what so the guy we were afraid of is this one when the divergence hits the high frequency object this is what we were afraid of and what I will try to convince you now is that this is a pressure gradient the l'array projector of a function is the function minus the gradient of the inverse Laplacian of the divergence of the function so it's a the function minus a gradient so dt of this is a gradient I'm not even going to bother writing it minus dt of this and this guy by our computation is positive 1 over mu do they cancel well not entirely because the time derivative could get on a but that's gonna be slow and what's the only term I never talked about where should I do this I'll do it I'll erase the parameters you see when I wrote this I've somehow hid under the rug that this divergence could have landed there right so let me add it so also need to worry about so as I said that's one of the guys am I allowed to write projection on nonzero modes for a function minus its mean so okay that's one term that I never talked about and there's actually other terms coming from the fact that we've added this temporal corrector so for instance we never said what happens when dt falls on the principal part we said what it happens with dt on the temporal corrector but we never said what happens to it on the principal part we also never said what happens to the Laplacian of the temporal corrector everything else is noise these actually matter now of course these have to be hit with an inverse divergence all of these and so is this let me convince you that all of these are good how about this guy inverse derivative of the Laplacian of the temporal corrector this costs a gradient I need to estimate this in L one because it's a stress how much does the gradient cost at most lambda q plus one times the L one norm of the temporal corrector modulo the fact that the rape director is not bounded on L one so there's some log losses it's the L one norm of this which is the same as the L two norm of that but I have a one over mu so if I make my speed much larger than lambda q plus one this term will be obliterated so you should ask well can I choose mu to be anything I want and the answer is no this term will force you inverse divergence would gain you a frequency so it's gonna gain because this is high frequency right the temporal derivative costs something costs a mu and that there's still the L one norm of this which I can only bond with the L two norm so what am I missing please somebody I mean I know what but it's again this trick L one and I've written here the L two norm I need to get the support these things are thin support to the one half without it you're screwed okay so let's check does there exist a new which obeys both of them because it seemed without writing this that these are contradictory right so the answer is yes because this is more than lambda q inverse so this says that mu should be more than lambda q plus one but less than lambda q plus two plus one squared so mu if you take it more than lambda q plus one this is small and if you take it less than lambda q plus one squared this is small so you're good we took it nine over seven in the paper whatever how about this I left it last well they're both of them right whether you have well actually no no this is actually much smaller because it has a one over a mu in front the DT falls on the slow guy it costs an L but you gain a mu that just destroys it so I have left last the most dangerous term it doesn't look like it and this is the term for which intermittency does not come for free in Euler if you remember when the gradient hit the slow guy that actually allowed us to go to on stagger one-third directly this was the best term in Euler and the reason was that the minimal frequency in the square was lambda q plus one but now it's not lambda q plus one it's the sigma so when I invert the divergence I don't get lambda q plus one I get the minimal frequency of the square which is sigma and well okay it still works because of this sigma still beats so the sigma gain you have from the fact that this has frequencies only larger than sigma and you apply a negative one-order operator against the loss which is an L2-5 it's still good but if you try to improve the scheme and get more regularity that's the enemy okay so I'm gonna close now because I think I've given the idea but let me just say in closing that to construct these in all the schemes you construct very non-physical things but underneath the proof you're actually finding by hand rich families of either stationary solutions or time-dependent solutions of your PDEs so under the hood you're actually solving the PDE in a classical sense but it's just under the hood and I'll finish with them thank you very much