First of all, I would like to thank the organizers for the invitation here. I'm very honored to be at IHES and in this summer school. Maybe one word about the other affiliation, the name of the math department in Montpellier, which sounds a little bit like the name of the lab here. CNRS lab here, which is Laboratoire Alexander Grotendick. So we are Montpellier Institute, well, Alexander Grotendick, Montpellier Institute. Because, of course, Grotendick is famous for the work he has achieved here. And the institute here was actually essentially created for him. And he was initially a student at the University of Montpellier. His first year at the university was completed in Montpellier. And he finished his career in Montpellier as a full professor. He finished in the early 90s, I think. So, well, Montpellier and IHES are two important locations for Grotendick. And, well, he loved Isocahedra, the 20 facet polyhedra. And he loved to teach about that. That's what my colleagues in Montpellier taught me. Okay, now the talk. I will start with what is called here, Geometric Optics, what can be called also WKB expansion. WKB expansion usually are met for hyperbolic equations or hyperbolic systems. The idea, and that's how I will use them today, is to rule out some energy estimates. Essentially, we construct particular solutions and we test energy estimates. And these particular solutions can tell you, you know, these kinds of estimates cannot be true. So, as often in this framework, there will be a small parameter, which eventually goes to zero, so here it's named epsilon. So, and I will start with that. So, I forgot to say, this is essentially a review talk. So, I will present three kinds of results. Hopefully, there will be not too many technicalities. I will try to give a flavor of the method of the proof, but no rigorous proof, say, formal proofs only. So, what do I mean by WKB expansion in the framework of Schrodinger equations? So, every time there is a derivative in time or in space, there is a factor epsilon. And the initial data are under the form of amplitude, which does not depend on epsilon. And rapid oscillation e to the i phi not divided by epsilon. The same size, of course, complex value. The space variable belongs to begin with to Rd. And WKB expansion consists in seeking and approximate, at least formulation of the exact solution like the initial data. So, and how do we find a and phi? Well, we plug this into the equation. We order the powers of epsilon and cancel from the highest power, well, from the larger term, the lowest power, to as many terms as possible. So, the first term we get, so, is when we differentiate the phase only. So, every time we differentiate the phase, there is one over epsilon factor. But when we differentiate, we have an epsilon factor. So, it becomes of all the one. And this is the ICONOL equation, which is a Hamiltonian-Coby equation. So, if you consider the gradient of phi, it's a multi-deep Burgers equation. So, you have to expect finite time singularity. And the next term in the expansion is a transport equation for the amplitude. And the coefficients of the transport equation are given by the phase. Okay, for enough and some examples. If the initial phase is linear in X, then we have an explicit formula. So, it's linear in XT, the evolution. And there is some dispersion relation, as these physicists call it. And you see that the solution to this Hamiltonian-Coby equation is global in time. On the other hand, if you start with a quadratic phase, you still have an explicit formula that shows you that in finite time, when you can compute exactly at what time the phase becomes singular. Now, nonlinear case. So, in the talk, I will consider only de-focusing nonlinearities. So, there is a plus here. I will write the energy on another slide. So, there is no blow up in the sense of formation of singularity. And because it's nonlinear, the size in terms of epsilon here, of the initial data is an important parameter. But since I consider homogeneous nonlinearity, I can keep the same initial data as before. And then there is an extra, the coupling constant depends on epsilon. And I want to proceed as in the linear case. And this is sensible thanks to the gauge invariance. When you plug this into the nonlinearity, here you have modulus of A to the two sigma. And there is one oscillation only. So, you still have one oscillation carried by one exponential. So, it makes sense. So, let's do the same trick as before. And then, of course, the discussion will depend on the value of alpha. So, let me write the equation once and for all. And we seek c of t and x, well, approximately like, I will not be too precise about the meaning of this. And we plug it into the equation. And now, so the distinction, if alpha is positive, so the Iconol equation, the Hamiltonian acoby, doesn't see the nonlinearity yet. Well, supposedly. On the other hand, if alpha equals zero, so this will be of order one. And the left hand side with phi will also be of order one. So I will just keep this equation for the rest of the talk. That's the only thing that really matters to me. And I will rather write alpha at least one. Because as a matter of fact, I'm really cheating here. It's much more tricky than that. Okay, so with same initial data. Now the transport equation, so it's terms of order epsilon. So if alpha is larger than one, you still don't see the nonlinear term. You see it on the other hand, if alpha is equal to one and this becomes a nonlinear transport equation, which you can essentially solve. Well, I will not mention today, no. And if alpha is less than one, well, mysterious. But today, we don't pay too much attention to that. So there are two critical values, which I will use during the talk. Alpha equal one, which is the smallest value of alpha for which there is the nonlinear term in these two equations. And alpha equals zero, which is the worst case, because I wanted to consider non-negative alpha. Where everything becomes coupled, and especially here, which is hidden. So the only remark I want you to remember today is if alpha is at least one, this is the same icon only question as before. The linear one. And on the other hand, if alpha equals zero, the coupling here is so strong that even if you start with phi not equals zero. Because you have, so, at time zero, this is zero. But unless you start with the zero initial datum, this will not be zero. So you start from zero, but dt phi at time zero is given by that. So dt phi is not zero. So even if you have no oscillation at time zero, instantaneously rapid oscillation because it's phi divided by epsilon will appear. So that's all I want you to remember. Now back to NLS with no, not yet, small parameter, so defocusing. So now the two of the conserved quantities I will be interested in. So the L2 norm, the mass or power of number of particles depends on the physical context and the energy. So the energy is a sum of two positive terms. Okay, so now this equation enjoys two important invariances. So the scaling invariance. So if u is a solution then for any positive lambda, this is a solution to the same equation. And this scaling leaves the homogeneous sub left norm H s dot invariant. So forget about time here for s equal sc which is given by this formula. So maybe I can keep this formula here because not so many people use the same notation for the non-enlarity, so I keep the formula here. And the other invariant I want to mention is the Galilean invariant. So if u is a solution, then so is this quantity for every vector v of rd. And the norm which is preserved by this transform is the L2 in space norm. So and this, this is a clue that, okay, if sc is positive, you can expect some notion of well-posedness. So I don't want to give a precise definition, at least if s is larger than sc. So the notion of well-posedness I'm using is the same as Joachim in a previous talk. That is, I want local well-posedness not depending on the profile of the initial data, but on the size of the initial data only. So equality would be valid with a different notion of well-posedness. So and for sc, so sc non-negative means actually that the non-enlarity is L2 critical or L2 super critical. So the numerality means that. So any positive regularity larger than sc will give rise to a locally well-posed Cauchy problem. And if sc is negative, so if you are L2 subcritical, then to me I've shown that for every non-negative regularity, the problem is locally well-posed from HS to HS. And why do you have to assume that s is non-negative? It's because of the Galilean invariance. Okay, some result concerning the lack of well-posedness. So the first part of the talk, I will always assume that sc is positive, strictly positive. So that means once again that the non-enlarity I consider is L2 super critical, possibly energy subcritical, energy critical, energy supercritical. And I want to consider, I start from not enough regularity. So s, the HS regularity of the initial data, s is between 0 and sc. And the first result in this direction is due to the both of the supercritical, energy supercritical wave equations, so 3D wave equation, de-focusing non-linearity. And it's energy supercritical, that means that p is at least 7. So, well, I won't give the statement, but the result by Le Bo used too many properties, well, he proved too many properties in order to prove his result, and the proof was greatly simplified by Gui Metivier in a Bobacchi seminar. For non-linear Schrodinger equation, well, similar results. The statement will be probably on the next slide. It's due to Chris Collinder and Tau and Berg-Jarrad Vettkopf, and the argument I will resume on a formal level is the following. You start with concentrated initial data. So think of A0 as a compactly supported, smooth compactly supported initial data. Okay, now I have a small parameter H, which is going to zero. And the initial datum is concentrating at the origin at scale H. Now there is a renormalizing factor M, and if I want my initial data to be bounded in HM, this boils down to this inequality. And thanks to a scaling I will show later on. We can actually relate, so the non-linear Schrodinger equation without epsilon, to what happens in the geometrical optics regime is the supercritical case, meaning that here I consider alpha equals 0. So the method of the proof in these papers is the following. For a very short time, so the following, so alpha equals 0. There is an OD approximation, which is the following. It consists in neglecting the spatial derivatives here. So the approximate solution, which is considered is this one. So it seems to be a non-linear ODE, but if you have ever seen it, you know it's a linear ODE because the modulus is conserved. So this is the modulus of the initial datum. My initial datum after scaling is A0 independent of epsilon. So the solution is given explicitly by A0 by this formula. So what you see is that here you have oscillation like T over epsilon. So it gets rapidly oscillatory. So you can prove by energy estimates that the approximation of this function Pc by this phi is true for a very short time. I mean epsilon log epsilon, log epsilon to some power. So this is really short. But this is enough when T over epsilon is like log epsilon, which is unbounded. So you have become rapidly oscillatory. And then if you are rapidly oscillatory, HS norms become unbounded. So that's essentially the picture. And then I will focus on the phenomenon called norm inflation. And well, so the statement about finite loss of regularity, so that explains the first part of the title, is the following. So the cubic case sigma equal 1 was given with one proof. The defocusing higher, well, more than cubic was proven in a joint work with Thomas Alazar for defocusing non-linearity. And at the same time by Laurent Oman and his proof included allowed to consider focusing non-linearity, which is not possible in the first two papers. Actually, these three papers use three different techniques for the rigorous part I will not explain today. And there would be a fourth proof available now. OK, so the statement is following. So SC is positive, L2 supercritical, S between 0 and SC. We can find smooth, rapidly decaying, so Schwartz-class initial data, a sequence of initial data, whose HS norm goes to 0, a sequence of time going to 0, along which the nonlinear evolution blows up in HK. So the result by Chris Collender and Tao is for K equal S. So here we allow K to be strictly smaller than S. So this is why I call that loss of regularity. So because, OK, the denominator is larger than 1, so this is positive. So this is finite loss of regularity because, OK, K is bounded from below. And we'll see why you cannot expi- well, at least we know we have the L2 invariant. So we could not go below L2. So as a corollary, another way to state that is that the local Cauchy problem is not well posed from HS to HK. So that means even if you're allowed to pay some loss of regularity in your energy estimate, the loss of regularity has to be at least this one. So there will be no energy estimates for these kinds of norms. So, well, final word about the numerology is let's consider S sub in the sharp value so that HS is embedded into L2 sigma plus 2. L2 sigma plus 2 is the potential energy in the Hamiltonian. So let me write it here. So the energy, so it corresponds to that. So how do we control this guy when this is not possible with the first kinetic term that is when we are energy supercritical? So the value of S is given by that, so it's numerology. And what is this formula? This formula is exactly this with S equal S sub. So what is the previous statement with S equal S sub? So that means we are, in space dimension, at least 3. This means that we are energy supercritical. So we consider data which are small in HS sub. So in particular, the mass, which is the L2 norm, goes to 0, the energy. So S sub is larger than 1, so this term goes to 0. S sub is so that this term is controlled by the S sub norm. So both terms go to 0, so the energy goes to 0. We know it's conserved, so it will go to 0 for all time. On the other hand, the numerology, because of this formula, tells you that every norm above the energy will blow up instantaneously. And this is, of course, short because for K equal 1, we get the first term of the energy. So at least for energy supercritical equations, we know the numerology is sharp. For energy subcritical, I don't know. So and the argument is different from the argument by Lobo, but the statement is the same. And the spirit, the formal proof is the same. So I have a question. If I remember when I came in the proof of Lobo, he gets like one initial data for which he gets loss of figure left. Whereas here, you need like a sequence. So the only difference that we have is that you have finite speed of propagation for the wave. No, but well, can I ask you to ask me the question at the end? Because there is a longer answer. OK, so the formal proof is the following. So like Chris Collin.tau and Birkjar Advetkov, we consider concentrated initial data. So this is so that the hs norm is of order 1. But hs plus blows up as h goes to 0. And OK, if you want to introduce semi-classical parameter, you play with these factors. So you modify the parabolic scaling with this epsilon. So the goal is to get this with alpha equals 0. OK, so the formula that comes out is this one. And you see that epsilon and h go to 0 simultaneously exactly because you are considering a supercritical situation, s is smaller than sc. So we have this. And the initial data do not oscillate. So the relation between, well, given by scaling is that one. And the key phenomenon I wanted you to remember is this one, for a time tau of order 1. So that means we're not using this. We're using something which is much more sophisticated. And the proof, I will say a word about it in the remark, the proof is completely different. So c is epsilon oscillatory in the sense that the phase has become of order 1. So we have this epsilon oscillation here. So now, well, let's plug this into the formula. So that means for, we have this. So it's at t h square epsilon. So in h dot m will be of order, so this. So c is epsilon oscillatory. So that means when we differentiate m times, we have epsilon to the negative m. We plug the formula for epsilon, get h s minus m minus m sigma. And this power is negative exactly for m strictly larger than the formula that was in the statement. So this is where the loss of regularity comes from. Because for c, you start with something which is not oscillatory, but it becomes oscillatory, rapidly oscillatory, instantaneously. So it's at time tau of order 1, which is much larger than this epsilon log epsilon. And actually, to prove the OD approximation, you just use energy estimates in a high order suborder of spaces in short time, because eventually you invoke granular inequality. But if you want to go up to time of order 1, you need quasi-linear analysis in the focusing case or analytic setting, which also covered the focusing case, which means that for fixed epsilon, this is a, well, you solved the equation by a fixed point ideal moment. This is a perturbation of the free flow. So it's semi-linear analysis. So when introducing this semiclassical parameter epsilon with alpha equal 0, in a way, the problem has turned much more nonlinear. So to justify this one line, well, is the tough part, which I will not say more about today. OK, now, what if a side remark, essentially, what if the Laplacian is modified? So you consider a Fourier multiplier, P. The important assumption is that P is real valued. So the group, because P is real valued, the free group is unitary on every suborder of space based on L2. So essentially, if P is homogeneous, well, the scaling yields this critical index. And indeed, you can prove that there is norm inflation in the same sense as in a paper by Chris Collin de Hetaou. I'm not claiming loss of regularity. Just norm inflation. Heal poseness from Hs to Hs. And the proof is the same. So same proof. What do I mean by same? It's the approximation here. Essentially, what we do is that. So it's not P because, well, the notation are not consistent. But essentially, the proof is this one. So we approximate this by that. And OK, so there are two cases. So either P is homogeneous. So this will be really unbounded or P is bounded. And actually, the difference is what is the critical Sobolev regularity. So we know that there is, of course, local well poseness above HD over 2 because it's an algebra embedded into L infinity. And actually, when P is bounded, you can prove norm inflation, essentially by this ODE approximation, for any S strictly smaller than d over 2. And OK, so the side remark is that this first statement implies that there is no striccote, well, isotropic striccote estimate improving on Sobolev embedding. So the statement is following. So striccote and miscigal repairs are given by this algebraic relation. And Sobolev embedding tells you that OK, LQ is controlled by this Sobolev norm. But in Sobolev space, it's based on L2. The group is unitary. So we have actually equality from this line to this line. And OK, I'm considering finite time interval. So order inequality roughly gives that. So here is what is given by Sobolev estimate. So the corollary is the following. So suppose that for some finite time intervals, there is an admissible pair, a k, a real k, such that you have something like that. So remember yesterday in the talk by Nicolas, he recalled the result by Birchard-Vetkov that states that on compactly supported manifold, you can take k equal 1 over p, which is half what you get by Sobolev embedding. So there is an improvement. Here what I'm saying is that if p is bounded, the Fourier multiplier, then k, if this is true, k is larger, is at least 2 over p. So you cannot improve in such a statement. You cannot improve over Sobolev embedding. And well, indirect proof is the following. So you can solve the, suppose you have some gain. So it's really rewriting the proof by Birchard-Vetkov on the 2D manifold. So suppose you have something that improves over Sobolev embedding. That is, k is strictly smaller than 2 over p. That is, 1 over p, for instance, in the case of a compactly supported manifold. So that you have this. Then you can solve the non-linear Schrodinger equation, locally in time, with this numerology. OK. So it's well posed in HS, where S is here. Sigma is here. So the p is given by the striccard inequality. So we take sigma equal 1 to infer the corollary. So that means cubic defocusing non-linearity. And the proof, essentially, relies on a fixed point in this space. And this is where we use this striccard slide inequality. And the fact that, since we have this assumption, then we have the embedding, which is very convenient to treat non-linear terms. So it's just rewriting the proof by Birchard-Vetkov. And then, OK, suppose you had something like that, for some case, strictly smaller than 2 over p, then you could solve the problem below the hd over 2. But you know you can't. So there is no such inequality. So last part of the talk is lack of, so it will be infinite loss of regularity. So now I consider negative regularity. So we know that, so there was this threshold here. And there was the other threshold given by Galilean invariance, which is s equals 0. So now I'm below s equals 0. So there are already many results concerning the lack of local well-posedness. So in 1D, Chris Cagliander-Tau proved that there is no local well-posedness from hs to hk, provided s is negative. And this is true for any k. So no matter how near to and minus infinity k is, it's not locally well-posed. So this is already a hint of infinite loss of regularity. So there will be a more precise statement later. So it was refined by Molina in the cubic 1D case. And so I will present some generalizations in the next slide. And again, if I want to translate it into the terms of geometric optics, WKB expansion, here is the picture. We start with a semi-classical Schrodinger equation, but with an epsilon here. So this is alpha now. So from no one, I will consider only alpha equal 1. So I will just keep this. So the Iconal equation for a single phase WKB expansion is this one. So the idea is that you start from two rapid oscillation, one of which is not an oscillation at all. And actually, you can compute exactly the leading order approximation. So that means you can determine a0 and a1 of t. So this is exactly the solution to the Iconal equation with initial datum x1. I've shown the explicit formula. In that case, it's this one. And the main idea is that this is true up to a reminder of order epsilon in the Wiener algebra Fourier of L1. And why is Wiener algebra a convenient framework? Is that, OK, the name highly suggests this is an algebra. This is indeed an algebra thanks to Young's inequality, embedded into L infinity. And the nice thing is that the W norm of this guide does not see the plane oscillation. So OK, so as I said, we have a formula for a0, a1. They are coupled. And unlike the first part of the talk, which was transferred from low frequency to high frequency, the appearance of rapid oscillation. Here, when you consider negative subalive spaces, the phenomenon is transferred from high to low frequencies. So high frequency is this one because it's rapidly oscillatory. Low frequency is this one because there is no oscillation. And the idea is that you take alpha0 small, alpha1 large, and then alpha1 will feed alpha0, and a0 will be large. So that's the rough picture for this case. OK. And now, some more quantitative statements. So Chris Cologne and Tao have also proved lack of well-posedness with norm inflation in the same sense as before from HS to HS. So the target is the same as the initial space. In dimension, at least two. I will explain why two later on. The argument of Bezinaru and Tao, based on a Picard iterative scheme, makes it possible to prove rather easily the lack of well-posedness from HS to HK for every k. So like the initial case on the circle I mentioned before. So now, the result I won't mention too much today is the following. Suppose we are at least in dimension two. Sigma is a natural number. That means the non-linearity is smooth. And well, there is this upper bound, which is not very pleasant, but we don't know how to get rid of it. So if s is larger than this value, so we would expect rather s negative. But we have this value, which is at least better than negative d over 2 because d is at least 1. So this is smaller than, well, 1 over 2 sigma plus 1 is larger than 1. So and the statement is the same as before, in a way. So we have small initial data, which are smooth and rapidly decaying. So they go to 0. The sequence goes to 0. And a sequence of time going to 0, along which all, absolutely all, the sub-left norms blow up at the same time. So this is why I call this infinite loss of regularity. So you cannot expect any energy estimate involving a sub-left norm on the left-hand side and the initial data in that space. OK, so I will rather insist on the periodic case. Because there, the assumption is simply s negative. And the other, OK, so now, periodic case. So well, this is a joint work with Thomas Kepler and he's more expert than I am on the periodic case. So he noticed that the proof was valid in four-year-long space, which are generalized sub-left space. So I don't want to describe them too much. But I think that if p is 2, this is the HS sub-left space. And this is something slightly more general. So you can read the statement with p equal to. That's, I mean, that's already the main picture. So suppose d sigma is at least 2. So it turns out that that means you are L2 critical or L2 super critical. So the way it comes out in the proof is that either, so sigma is always an integer. So either sigma is 1 and more. And d is at least 2. Or in 1d, we can allow quantic non-linearity and higher. So this is just to rule out the 1d cubic, essentially. So the 1d cubic is the second case because it's less satisfactory. So pick any negative s. Then we can find a sequence of smooth initial data with, say, HS norm goes to 0. And a sequence of time going to 0, along which all the sub-left norm blow up at the same time. So it's any sub-left regularity all here. So the same as on Rd, except that here is a small satisfactory. And in the cubic 1d case, we have this abstraction that s has to be smaller than negative 2 third. And then the statement is exactly the same. So negative 2 third, we know for sure it's not sharp in order for the problem not to be locally well posed. For instance, by the result by 0, 0, 1d cubic, it proved that if s is at most negative 1 half, the same statement is true, but with R equal s. So it's norm inflation, not loss of regularity, but norm inflation. And also, there is a modification in the cubic case, 1d or higher, because it's been known since, at least, the work by Chris Collier and Tao that the 0 mode plays a crucial role in all instability issues. So there is this V-order equation that consists in removing the 0 Fourier mode from the nonlinearity. But actually, the proof does not see the difference when you remove that. So the statement remains valid. OK, so one word about multiphase. So weakly nonlinear geometric optics, weakly is because here we have an epsilon. And that means that the Iconol equation is the same as nonlinear in the linear Schrodinger equation. So weakly means that the only thing that may see the nonlinearity is not the phase, but the amplitude. So we start from possibly infinitely many linear phases, with this small parameter epsilon. And we look for an approximate solution of the same form. So the idea is, how can we choose phi j, and how do we choose a j? So phi j, well, they will solve the Iconol equation. And since we start with linear in x phases, we have an explicit formula, which is recalled here. So we know the phi j. So the goal is to determine the evolution of the amplitude hj. And eventually, the picture is the following, is that we will start with alpha 0 equals 0, so no 0 Fourier mode. And by nonlinear interactions of nonzero Fourier mode, the 0 Fourier mode will appear instantaneously. So there will be a transfer from high frequency to low frequency, the lowest possible frequency, which is 0 mode. OK, when you plug such an expansion into the equation, you have to look at the nonlinearity. So every time you see a psi here, you plug this. So you have many sums, 2 sigma plus 1 sums. And the phases you get, because of gauge invariant, the new phases you get is this one. And since each of these terms in linear in tx, the outcome is linear in tx. So it will be of this form. And what we will call resonance, and I think you will hear about that later on during this summer school, is if the outcome that you get by nonlinear interaction is also a solution to this icon only equation. And what physicists call dispersion relation is exactly that omega is length of j squared over 2. And the natural definition of resonance sets is, so this is what you get by combinations of such phases. So let's say that phi j will be kj.x minus kj squared over 2t. So you take linear combinations of this like above. So it's kj. So the linear combination in x will give you this relation. So we have j.x. And we want this to be a solution to the icon only equation, that is the resonance condition is satisfied, which is given by this other algebraic relation. So the resonance set is this one. And the claim is that only this resonance set matters. Because if you have the first relation, but not the second relation, because we work with integers, the non-stationary phase argument tells you that in the equation, when you plug this into the equation, it gives rise to a source term, which will be of order epsilon squared. Because here we have epsilon. And then energy estimates in Binner algebra, for instance, tells you that this gives rise to a reminder of order epsilon. So only this resonance set matters. In the cubic case, so in the cubic case we have only three terms there, the description of the resonance set is given in the paper by Collion de Kiel, Stafilani, Takaoka, and Tao that Giulia mentioned yesterday. And the resonance set is fully described by an algorithm based on the completion of rectangles. So you essentially use Pythagoras theorem to get this description. And for Quintic and higher non-linearities, it's more intricate. But we don't have to fully describe, because as I said before, it's enough to start from non-zero phases and create the zero phase. And that's all we want. We don't need a full description. So a remark, if you can generate a resonance in a cubic case, you can generate the same resonance by just copying the last phase for higher order non-linearities. What you can do in cubic, you can do for higher non-linearity. OK, so the key example. So which is an example of the rectangle completion is you start from these three phases. So this is why we have to assume these at least two. So the last d minus 2 coordinates, if any, can be taken to zero. They don't matter. So it's a 2D phenomenon. You start from these three phases. So x1 over epsilon, x1 plus x2 over epsilon, and x2 over epsilon. And then, of course, it's easy to check these relations. So zero is a resonance. So that means by the completion of rectangle, you start from three non-zero Fourier modes. And by non-linear interaction, you expect to create the zero Fourier mode. So transfer from high to low frequency by this example only. So the amplitude here are given in the periodic case by this relation. So you restrict to resonance set. If you didn't restrict, that would be just rewriting the non-linear Schrodinger equation in Fourier variables. So you restrict to this because that's only what matters. And so the word of caution is that it's not because you know that phases resonate that this term will actually appear. So in the 1D case, where cubic 1D case, so you have this completion of rectangles algorithm, so you have flat rectangles because it's 1D. Actually, this equation, you can see that it simplifies in a way that the modulus of each term is conserved. So if you start from zero, nowhere you will remain zero. So that's why the previous statement was bound to 2D and higher dimension. This was a multi-D phenomenon. And then, OK, now if you are at least two dimensional and consider the previous examples, well, essentially, you start from three oscillations and they will create zero mode. If the associated amplitude is the same for each of the three, you can see that the zero mode is indeed created by this equation. OK, no, if it's 1. So just the word. So once again, a non-stationary phase argument makes it possible to show that in L2 intersected with the vener algebra, this is indeed a good description of the exact solution. And well, maybe just to finish, well, it's probably too late. So the idea, well, I would just comment with hands. So you use a scaling. Scaling in a periodic case may be tricky. So that's why epsilon n is a carefully chosen sequence going to zero. Carefully means we want to remain periodic. We use scaling in order for c given by the scaling by this relation. We want c to solve this in order to use the previous approximation. So that means for u0, we consider this. We measure the hs norm of this guy. So we have this factor. And we take the hs norm. So every time, well, that means you take this power of epsilon n to the s. This negative power of epsilon n to the s. So the hs norm at time zero is of this order. So s is negative. This is this. I want this guy to go to zero. So this is one condition. And you see that any positive b, well, you can always find positive b so that this is true because s is non-zero. So there is creation of the zero mode by this algorithm, by this geometric optics approximation. So once again, you rescale time. So a time which was of order one here becomes a time going to zero in terms of u. And we can bound from below any, absolutely any sub left norm by the zero mode. So that's the only idea. And the zero mode has appeared. And in front, we have this factor. And if beta is positive, this, of course, goes to infinity. So that's it. And now it's just algebra. And we see that, essentially, because for us, when the zero mode is concerned, every sub left norm is equivalent. So that's the idea to conclude. So we know that the approximation also makes sense at the level of the solution u. So maybe one word about the quintic and higher non-inertity. So one example is enough. So Grébert and Thomas gave some examples where they wanted the first and third phase to be equal. So with Thomas, we constructed other examples based on Pythagorean triplets. So I was happy to pretend I was doing some number theory for once. And essentially, the approach is the same as before. You create, from this, you create the zero mode. So you start from five mode, and they create the zero mode. And the same picture as before. So we don't have to fully describe the non-inert thing. And the 1D cubic, so 1D cubic, as I said before, you cannot create the zero mode at leading order. So at leading order is important. So we look at the next term in the WKB expansion. The next term will be of order epsilon. And essentially, we construct it. And the outcome, everyone is tied, including the speaker. So I would say that the outcome is, if you start from non-zero mode here, the zero mode will not appear at this order, but it will appear here. So we have an explicit formula, which is nice. And then when you do the same numerology as before, if you want the initial data to be small in HS, the requirement is this one.  And this is why you get the negative 2 third critical exponent by this approach. And probably by this approach, you cannot improve the statement. But it's certainly not sharp. OK, thank you. Question before. Oh, I mean, my understanding here is in the terminal, why are you using the term loss of regularity? Because it means that you start with a fixed data in space, and then at the time, you are out of that space. But here, it's not exactly. It's not exactly. It tells you that if you expect energy estimates, they will involve loss of regularity. So it's not exactly constructive. It's just that you rule out some energy estimates. And in the L2 supercritical case, you know that you have a minimum amount of regularity to be able to be ready to lose. So with this lower bound on k given by this numerology, and in negative regularity, there's no way you can expect any subalif estimate. So that's the whole picture. Just to go a little further on with this question, but you expect a situation where you have to start to perform the equation, you have to do the blow up. OK, one of your parameters might not work out enough.   If you are energy supercritical, you can use scaling to blow up arbitrarily soon. But beside that, I don't have any answer. These types of analysis never lead to examples of type 2. Type 2 is the post-master, somehow your minds don't grow, but you still have the post-master. Not that I know, but it's a very interesting question. I've never thought of it this way.


How do you do? My name is Jean-Pierre Hormont. As you see, I'm indulging in a Frenchman's favorite pastime, sipping an aperitif in a sidewalk cafe. And where, where else but Paris? My beautiful city of Paris. I say my city because I was born here. But since I spent so much time in Hollywood and in New York, I've learned to know and love America. So much so that I like to call America and France both my contents. I've just finished a picture here. And I'm glad I delayed my return to Hollywood, for I have just seen something that, as a Frenchman and an ex-soldier who believes in a free world, a free way of life, has made me very proud. It is the story of Comzy, the United States Army Communications' own Europe. The supply force, which is the very live blood, the staying power of the American Defense Network in Europe. I've had the opportunity to witness the Comzy story. And in a moment, I would like to show it to you. Comzy began with a visit to its headquarters at Orléans, France. It was here, during my first talks with the commander of Comzy, Major General Henry R. Westphalinger, that I began to realize for the first time how vast is the job of equipping and supporting a modern army. The commander had arranged for me to see at first hand how Comzy does this massive job, so that I could report it to you with understanding. There was, as you will discover, much to see. When I was asked to tell the Comzy story to the American people, I had no idea what an intensely dramatic story it was. It is a story which begins with the needs of combat units stationed in France, Germany, and in Italy. A munition supplies the equipment to the right man at the right time. This, in simple terms, is the vital mission of Comzy. It is not a question of any one weapon or one type of ammunition. Comzy must supply every weapon system in the defense arsenal of the United States Army. For example, the M-60 machine gun, which fires the NATO caliber 7-62 cartridge, about 60 of them in the space of five seconds, or 700 to the minute. Or consider the small but deadly hand grenade, an essential for the fighting foot soldier. 0.2-inch mortar, using high explosive shells. And the versatile recoilless rifle, which makes mobile-light artillery out of a jeep. The guided anti-tank missile SS-11, directed by its operator with deadly pin-point accuracy. 105-millimeter armored howitzer. Mainstay of artillery firepower. Must be supplied along with yet another kind of ammunition. Larger still, the 280-millimeter cannon, firing its massive projectiles. And still there is more. Here is a tank, flamethrower, whose stream of jellied petroleum delivered in flame to an enemy is highly affected. Every of liquid flame by air is an extension of the flamethrower technique, and of the supply responsibility of Comzee. For the armored divisions, the new M-60 tank provides increased range, speed, firepower. The tanks themselves, of course, together with their ammunition, parts and fuel, are supplied by Comzee. The means of swift movement, delivery of the right kind and the right amount of firepower to the proper place at the proper moment. All of these essentials of success in combat depend upon availability of the equipment, supplies, ammunition, weapons, fuel and services which make that success possible. This is part, only part, of the supply mission of Comzee. The next stop on my itinerary was to visit the men. Recently arrived from the United States, these men were not left in doubt as to why they were in France, what their mission with Comzee would be. They were now part of the working team, and their key role in this free world defensive alliance was made clear by the officer who welcomed them. You have been assigned to an organization known as the Communication Zone, a vital part of the United States Army in Europe. We are here to fulfill our commitment as part of the NATO team. You're in France primarily to do one thing, stop the spread of communist aggression. If war should come, Comzee supply lines and American troops would be the mainstay, the bulwark of the NATO forces. Many battles have hinged upon the availability of materiel. The job of troops stationed here, your job is to see that the United States Army throughout Europe gets this materiel. It takes men, competent men to operate the trucks and the ships and the barges that move the millions of tons of military equipment through Comzee channels each year. Your work is cut out for you. Yes, your job, to move the weapons, ammunition, fuel, tanks, and missiles. This is the task of the United States Army Communication Zone Europe. Yes, France is the logical place for Comzee with good seaports such as here, in Bordeaux, La Palice, and San Jose. France also offers the advantage of excellent railway facilities for the overland transport of heavy equipment. Also good roads for truck transport, though they are maybe at times a rather tight squeeze in some picturesque village. And of course there are modern super highways, direct high-speed links to the key areas of Europe. And finally, a number of excellent airfields able to handle the largest of cargo aircraft. All this combines to make France the logical setting for the story of Comzee, an organization preoccupied with movement, logistics, the military science of putting the right supplies in the right place at the right time. The fighting man who needs a fresh clip of rifle ammunition has a very specific requirement. No substitute is accepted. Can the crew of an M60 tank replace a broken track block with one design for the M113 personnel carrier? So it goes. A million and one items of material. Each item a problem in itself, and each handled smoothly with disciplined efficiency by the United States Communications Zone, Europe. One of Comzee's most effective innovations in solving problems of supply is a new method of fast delivery called roll-on roll-off.  Compare the speed, efficiency and economy of this technique with the older, more time-consuming and therefore money-consuming method of unloading cargo at the dock side. This roll-on roll-off technique saves thousands of man-hours and tens of thousands of dollars while delivering vital military cargo faster than ever before. Another time-saving aspect of this technique is the relay action which keeps the material moving without delay toward its destination. At relay points, one tractor is exchanged for another quickly, smoothly. A system of quick connect fittings reduces the time required by several minutes. In moments, the loaded trailer is again on its way to the people who need it through clumsy depots in France, Germany and Italy. These relay points and depots form a vast and specialized network across free Europe. In France, there is Verdun, also Captieux, a central supply point for ammunition. Agrand, with its many acres of warehouses and service buildings. Bracken, a general depot where vehicles of every type are processed for delivery. Chineau and many others across France. Moving onward into Germany, there are depots at Miesel, Kaiserslautern, Pium Assentments and others. A thousand separate activities each designed to meet specific needs. To the south, in Italy, still more of Komzi's truck transports are on the move over routes which are often picturesque as well as purposeful. Camp Darby near Leghorn is one of the main supply centers in Italy. But of course, transport by truck is only one method used by Komzi in its supply mission. Through a vast network of pipelines built by Komzi, flow millions of gallons of gasoline, aviation gas, jet fuel, diesel fuel. Together they form Komzi's largest single item of supply. One that demands a fuel distribution system of tremendous complexity and truly massive capacity. Huge control centers to direct the pipeline flow. Tank farms to provide working reserves. Pumping stations, distribution points and automation to coordinate the routing of fuels to destinations all across free Europe. Komzi takes nothing for granted. Here, in its petroleum laboratory at Dongge, all petroleum products are pre-tested for quality and performance. This is only one of many quality control laboratories staffed and operated by Komzi. Petroleum distribution begins at Dongge in France, where tankers from the United States unload their liquid cargo. There are also emergency facilities for offshore unloading. From the central point at Dongge, the pipeline goes underground, then across France. And ultimately underneath the Rhine, going on to link up with NATO pipelines which lead into Holland, Belgium and West Germany. Komzi's pipeline network saves the United States government millions of dollars each year in commercial storage charges. And it moves petroleum products at one-tenth the cost of shipment by rail or water. So effective is this pipeline system that it is able to meet all the requirements of both the United States Army and Air Force, as well as supplying our Canadian and German allies in Europe. In coastal cities of Western Europe, I found yet another aspect of Komzi's supply mission. From busy ports such as Bremerhaven in Germany, Leghorn in Italy, and Rotterdam in Holland, comes a steady flow of heavy stuff, construction machinery, tanks, armored personnel carriers. The equipment is loaded onto river barges to be floated to destinations up the Rhine. This presents at times an interesting contrast, the modern weapons of today's army passing the ancient castles of another era. To handle the diversified fields required by the new complex army, Komzi has introduced highly specialized troops and units such as the United States Army Aerial Support Center in Normandy. The Aerial Support Center utilizes C-130 aircraft stationed at Everoo. In addition, the fields of Everoo, Drew, and St. Andre are at our disposal. Komzi therefore stands ready to support the combat troops via airdrop, anytime, anyplace, with everything that these troops might need. At St. Andre, we store and maintain personnel and cargo parachutes in support of the Army in Europe. As required, Komzi's parachutists jump with the loads to assist in the recovery of critical equipment. At Komzi's Aerial Support Centers, skilled personnel use the latest techniques in rigging and airdrop, techniques which have been developed and perfected many of them by Komzi personnel. Komzi parachute soldiers are trained to parachute nearly any item of combat equipment and to go with it to ensure its safe arrival in support of United States ground forces all over Europe. Working with the United States Air Force, this specialized element of Komzi makes certain that priority cargo gets to the men who need it and gets there in a hurry. Today, the demands of preparedness take many and highly specialized forms. For example, at key points throughout Europe, Komzi operates equipment repair installations whose versatility is remarkable. The skilled personnel of these Komzi plants can and do repair anything military from the largest tanks to wet springs. Some of these installations even show a respectable gold profit as well through repair work done for other NATO nations. Virtually nothing is discarded here, as Komzi maintains its unremitting effort to reduce defense costs overseas. At Land Gris in Germany, one discovers an unexpected aspect of the Komzi responsibility. Here, two specialized talents are called for, as Komzi soldiers take up the role of handler, teacher and leader of the four-footed soldiers of the K9 Corps. The dogs are carefully selected for qualities of health, intelligence and strength, since their work and the training which prepares them for it requires an abundance of all three. These sentry dogs, strange as it may seem, are also a considerable cost-saving item in the Komzi scheme of things. For this is an area where hundreds of border points and security installations must be guarded day and night. A single well-trained dog with his handler can do the sentry work of several soldiers, and these dogs are well-trained indeed. They learn to negotiate obstacles far more severe than any they are likely to encounter in their actual work. Komzi provides both the training and the dogs in pursuit of its basic objective, to meet every need, large or small, of the United States Army forces in Europe. The result is a large reserve of highly trained dogs, from which can be drawn the keen-eyed and sharp-eared sentries which have earned their title Four-Foot Bereda. Together with their handlers, they continue to stand watch at keen installations and along the remote frontiers which divide free Europe from the Iron Curtain countries. Also, my tour of Komzi, as well as considerable reading and study of the subject, I can truthfully state that Komzi operates a complex larger than any company in the world and does it with a speed and efficiency that is literally fingertip control. A precision, a depth of control which men may direct, but only the electronic memory banks of a computer brain can encompass. This is automatic data processing. The very core of Komzi's ability to respond instantly to requirements in the field. In seconds with automation, skilled operators can complete a bookkeeping job which not long ago took days to accomplish. And at night when the workload is lighter, the same equipment produces a document known as the status report. Two inches in thickness, it is a complete detailed evaluation of the entire Komzi's supply inventory. It is equipment such as this that makes possible Komzi's pinpoint control of a million and more items of defense. The ceaseless hum of automation, maintaining stock levels, controlling the flow of orders from the field, routing equipment by the fastest means available. With limited time, we've touched upon only a few of the functions, depots, storage sites, seaports and specialized installations of Komzi which form a strategic network of supply and service across Western Europe. These are the facilities which collectively make it possible for Komzi to supply everything necessary to the massive United States Army establishment. From the smallest weapon part or tool to heavy equipment, construction supplies, the mountains of material go to support and sustain the Allied effort in Europe. The tangible means of ensuring the strength and the survival of freedom. The operation of supplying war material to our armies, everything from tiny springs to giant cannon, is a logistical problem of the highest order, a problem well met by the people of Komzi. For it is known that the side that can first recoup from an atomic attack is the side that will win. Our side is ready. For today, helping to maintain the security of the world's free nations is the modern, forward-looking military creation known as the United States Army Communications Zone, Europe. Here you see the full strength of an army division, the Fourth Armour, assembled in a single formation. Thousands of men, tanks, self-propelled guns, trucks, an impressive array of mobile firepower. Yet, this is only one of several divisions included in the United States Army's establishment in Europe. And every weapon, every vehicle, every round of ammunition and gallon of gasoline, every button on every uniform throughout this massive military force depends upon the dedication and know-how of our male logistical specialists, the dedication and know-how of the men of Komzi.


Hi everyone, Joe for Jaspy's CaseBreaks.com coming at you with 2020-2021 Upper Deck Trilogy Hockey, one box random team number one So we're gonna do the break first and then at the end of the break we'll re-randomize everybody's names and then You'll be assigned some teams and SP authentic brand new release just dropped today But again as always we'll do the break first big. Thanks everyone here for making it happen. There are all 31 teams Let's roll it randomize it one and three four times one two three and Four no, I cannot pay fines with the break credit in the state of California Mike Tower. I Don't know if that that's I don't know if that's currency outside of Jaspy's case race comm one of three four times for the teams one two Three and a one different dice roll for the giveaway part as always Four times penguins down to the blue jackets All right, so here's the first half ish of the list right here Here's the second half ish of the list right here Once we get once the once the Kraken starts appearing in checklist and we'll get exactly Both halves of the list but for now odd number of teams so I think you can still go to a Live traffic school as long as you didn't get a second ticket in the past 18 months What happens if I get a second ticket? That's ever happened to me Not yet knock on wood All right, so no trade windows in these filler breaks Let's just dive right into it big. Thanks everyone here once again for getting in on it Appreciate it. Yeah, Jason did online traffic schools for being taken back in 2020 Jason I was asking everybody if any if anybody Like went to an actual traffic school like live where you have to spend like four hours and a Barely air-conditioned room and go through all that sort.  Oh Second ticket. Oh, they report to the NBN the insurance. Yeah, I don't want that Okay, I'm a car world champion along with him Maybe future world champion right there gave the lardy rookie super stage World champion Nate McKinnon Alexander Romanoff to 499 rookie premier level level 2 That's for Montreal. That's for Michael. We got Pierre Olivier Joseph, that's for Pittsburgh. That's gonna go to Edward and the penguins and behind Nick Robertson is oh a signature puck Now my hockey knowledge isn't very deep. Is this guy any good the defender from the coyotes Nicholas Jalmarson Please be good Please be a good player Adam cupperman with Arizona Card looks cool though. There's Martin cowl to 999 And we got Lucas Carlson rookie premier autographs 399. Yeah, Jalmarson. Is that guy any good Jack? Here's one for Chicago. I hope he's good too. This is gonna go to Ryan in the Blackhawks Some nice cards out of this filler box here and Stephen Lorenz to 999 All right, not too shabby Don't know him Adam says but looks like a cool card. Well, hopefully Hopefully he's good. All right Get everyone's names one through 31 new dice new list and There are the teams right there. So we got a couple teams in break one and then a few more teams in break two So it's gonna be the top nine Top nine after Top nine after five one and a four five times good luck one two Three four and one more fifth and final time Five right there. So from ten on down Very close Stephen P, but from ten through thirty one. Thank you very much for trying We'll get him next time though. I Think we do have some more pig your teams Available on jazby's case breaks comm without a filler yet So if you want to avoid a future filler breaks three and four are loaded up get your team straight up Congrats to happy times for Michael is happy Han mark Mike tower Steven P got in Steven Logan Han Jack and Adam They're happy because they're in One of the next two breaks if not both after five times. Oh, look at this Adam got his Rangers There you go, Adam So we'll put a little rooftop next to your name so you know you won that in the filler So Adam you got your New York Rangers in break one Jack with the Senators in break one Han with the Senators in break two Logan with the Blackhawks in break two Steven P. Canadiens in break two Mike tower with the Sabres in break one Mark with the Sabres in break two Han with the Sharks in Break two and Michael with the Jets in break two. We'll see you for breaks one and two back to back Separate videos jazby's case race calm. Bye. Bye


How important are the weekends when it comes to fitness and nutrition? Well, it's more important than you think. Watch this. Our next caller is Peyton from West Virginia. Peyton, what's happening? How can we help you? Good. Good. How are you guys doing? I wrote some stuff down because I forget easily, so just bear with me. But I've been listening for about a year or two, a really big fan. Obviously, I love you guys talking about nutrition and lifting. But I really like when you guys talk about your personal family lives. Listening to you guys talk about families and stuff just makes me want to be a dad and everything right now.  Some background on me. 23 years old, obviously. Six feet tall, currently 108 pounds. I'm assuming I'm around, probably around 10 body fat. I haven't officially got it measured. But I've been lifting for years now just because of sports from high school and everything. But around COVID, I really started taking the nutritional and supplemental side pretty seriously.  I was at about 185 to 190. I got down to 160 pounds. And I thought I was in really good shape. I was super skinny, super lean. But then looking back on now, I'm like, dude, I was like a straight skeleton. So start last fall, I started, I balked up, started actually taking creatine then, and then got up to, right around February, I got up to about 208 pounds. Wow. Wow. Yeah, I know. Pretty intense. I probably put on a little bit more fat than I wanted to. I was still kind of lean. I was pretty good still with my diet. But then I cut back down and I'm 180 pounds at the end of June. And on June 30th, I was 180 pounds. So I'm in the best shape of my life. Obviously, I don't think so. I body dysmorphia, still kicks in and everything. But I just wanted to talk about kind of like the mental side of fitness and nutrition. So when I first started balking up and I'm putting on size and muscle and everything, I was really pushing myself hard, training till failure, drop sets, super sets a lot. Just every single lip, I just wanted to feel that burn because I just loved absolutely feeling that. So I've been listening to you guys talk about how this probably isn't the most efficient pushing till failure on every single set. So I just wanted to see what, like the first step is to realizing that that's not the answer and then like, or if there's like a way to push till failure without like over training and everything. So that's kind of like the main question about the fitness part.  Yeah. Whatever link you sent us, I think those are the pictures when you bulked up. Would you play football? Do you play football in high school? I did not play football. I played into my freshman and then I played basketball, baseball, and then I actually ran cross country. So okay.  Yeah, I would say the body fat you think you put on that was too much is not too much. No, I think you're doing fine, bro.  Cause that is too much. That is too much intensity. How do you move away from that? Okay. Okay. So that's a challenge and that's a, that's a continual challenge as you stay on this path and on this journey. The best, the best metric to measure, not perfect, but the best one to measure is our objective metrics. The mirror is never objective. The mirror always is subjective. So you could try to make it as objective as you want, but it's a kind of a tough one. But, but objective measure will be like your strength. Follow maps, power lift, bro. Yeah. Maps, power lift, maps and a ball. That's power lift would be great for him because he's measuring that, right? You know what you're tracking that, the whole point of that is designed for you to get stronger and then going, going to bulk. Yeah. Feed yourself, go back on the bulk. Yeah. And so this picture that I actually sent you guys, that was probably when I was leaning out of my cut, I didn't show you pictures of what my full bulk was, but I was still kind of, I was that lean, but the right, those pictures I just sent you guys, that was full cut now, obviously, not before I was, when I was skinny slash lean. So, yeah, I think you look better filled out for sure.  Yeah. Yeah. No, you look, you look okay, bro. Yeah. I mean, okay, here's, here's the thing too, right? So, and you got great results and you were training to failure. Like, I mean, this is, you're, you're literally talking about what all of us did. I mean, we were the same thing in our early 20s, train the same way. And, and here's the thing, and you've probably heard us say this because you've been listening to the show for a long time is that you, you think that the more you put into it, the harder you go, the more the results are going to be because that is like how almost everything else in life works. It's not, it's not that way with physiology. It's not that way with nutrition and training. It's not. It's not the harder you go, the more you do, the more results, the better you're going to be. It's, there is a very, there's a sweet spot and that's unique to everybody. So, you know, even though we're giving you advice here, you know, it's going to be different from each person. You're going to kind of figure that out. A great, a great test of this is to be true to a whole program. I would tell you, all right, Peyton, just trust the process, follow one of these programs and never go to failure.  Do not always stop with two in the tank. Always allow yourself, it doesn't mean you're not trying to add weight to the bar. Every time you go back to the gym the next day or the next time you're hitting chest, it's like, okay, last week I was doing this. Let's try and increase by a couple pounds and see if I can do that. But get rid of the failure. Get rid of the, I need my buddy to come over and spot me for the last rep or two and train through a whole program that way and watch what happens. You'll be surprised. Okay. Yeah. All right. For sure. All right. Awesome. You're going to crush if you do that. Your body's going to respond. If you respond as well as you already have at your age with what you were doing, you're going to respond really, really well.  Yeah.  You don't need to do that. You really don't need to do that in order to see progress. Yeah. And then back, you know, and back to what Adam was saying about hard, redefine hard. Everybody thinks hard means grueling. Hard just means challenging. And oftentimes, oftentimes, especially with someone who has a hard work mentality, harder for that person means smarter. Because they just want to go, they just want to make it grueling. So it's not necessarily harder in the sense that it hurts more. It's hard in the sense that it's more challenging for you to do it in a smart way. So if you redefine it that way, it'll help you. Yeah, sometimes you need to throttle the intensity and you need to, you know, get after it. Sometimes you need to just focus and throttle more on the discipline of actually doing what's right for your body in order to get results. So, you know, that's just a totally different mentality. And this is, too, you're sort of the example of like, if somebody were to ask us in hindsight, like how we would go back to a 20 year old self, what we would tell ourselves, like you have that opportunity right now to shift and completely do things in a way that's beneficial. Yeah, earlier than we did, right? Yeah.  Yep, perfect. Yeah. All right. And then just one last question again, kind of with the mental side, but this is more with the nutrition. And it has to do with cheat days. So I'm very consistent in like my diet throughout the week, like in like 200 tunes, 50 grams of protein a day, enough fat. And then if I need to cut weight, calorie deficit, calorie surplus for bulk. But then when the weekend hits, it's like a completely different story. I'm still young, and I'm usually consuming alcohol. And then even when I go on vacation, that's whenever I'm like away from my diet of the whole time. I just want to know how much like alcohol, like a couple of beers on the weekend is affecting that or anything. You're overthinking it. Yeah. Payton, you're overthinking it, bro. Listen, when you're in your 40s, you're not going to look back and like, man, I shouldn't have, I should not have enjoyed myself so straight. Like don't go binge drinking.  Right.    If it becomes about the alcohol, well, now you're doing something wrong.  I'm going to tell you that right now. You're not going to get much out of it. I will add this though.  I'm drinking beer. I'm doing everything. Something that worked really well for me that might work for you, which is like, instead of telling myself I can't have the beer, I can't have the fast food or can't do these things, it's like, I'm going to at least hit my protein intake and get my training in.  And then after that, if I'm out with the boys at night and we're going out drinking and we're going to go have a restaurant, I'm going to have the chips and dip, like I'm going to enjoy myself. But those two, those two like rules for myself for the weekend of getting my lift in and then hitting my protein intake and then the rest, whatever, that will keep you on pace. Cause you'll be surprised how much two days if you kind of just have the extreme on off attitude, how that can really stall your progress because you go so extreme one or the other. So at least have a good foundation, but to Sal's point, you know, you're not, you're young. This is a time to enjoy life a little bit too. So you don't need to be, you know, anal about not having a beer with your buddies when you guys are having a good time. Hey, how do you paint? I'm 23. Hey, are you going to school? You're working. I just graduated. I just got my master's in software engineering. So I'm just, I just finished. Oh, you got, I know the case. Yeah. I could smell your mentality. You're very, very focused and disciplined. That's great. Yeah. That's great, man.   Yeah, that makes sense. Yeah. Okay. Yeah. I'm pretty good. Like, um, like you said on, I'm good on the weekends of, of, um, of like tracking protein and stuff like that. But again, the alcohol, but then the difference is when I go on vacation, it's like, it's all out. Like I'm going there to consume as much food as I can. And I know if you guys have talked about this, like don't eat so quickly and stuff like that. And so that is like, that is one thing that I need to work on. Actually, the last vacation would help me. I don't know. This is weird, but I started just eating like sunflower seeds because it just kept my hands moving and my mouth moving to like just not like eat so much food. And when I'm on vacation, so that was another thing. I love that. That's a nice hat. Yeah, but I'm going to tell you. Yeah. It was weird. It was so weird. I think I know what the problem is though. You're too regimented, too strict on the week. And so the weekend becomes a release. That's what I was saying. That's the point I was alluding to. I was the same way in my 20s. I was dialed Monday through Friday, Saturday and Sunday. I had the complete opposite attitude of, oh, but the belt comes on. This was also a big, uh, you know, plateau breaker for me. I was stuck in this rut for a long time. And even as a trainer, I couldn't quite figure out, why am I, why am I not able to get more shredded or look like this? And then this was, this was way before all these great tools that we have now, like the aura rings and we had, we didn't have that. So I could track once those tools came out. I saw what was happening. It was like, I was just totally over consuming bad calories, not hitting my protein intake, not training on the weekends.  I was still in good shape. I was a trainer, right? So to the average person, but I was trying to get to the next level. I couldn't get to the next level because I was so on off. And so that was why my advice to you is just, just don't go off the rails and have a couple of small rules. I love to eat in sunflower seeds. So you're not snacking on some other, maybe loosen up during the week a little bit. So it doesn't feel like you have to completely disconnect. Make sense.  That's a sign that the week may be a little too regimented or too, whatever, too stressed. So either mindset needs to change or some of the things that you're doing during the week need to change a little bit. So, yeah. Yeah, Peyton. Yeah, that's all I had, but thank you so much. Peyton, we're going to send you a mass power lift. Do you have an Instagram? Okay. Yeah, I do. Yeah. Are you dating anybody or are you single? I'm single. I'm single. We're going to help you out right now. What's your Instagram? So how's we get you laid? All right, it's pk underscore fitness underscore nutrition. Yeah, we got a six foot tall software engineer. He's a hot guy, everybody. He's got too much discipline. Ladies, we got you. Hey, you're welcome. Look out. Look out. You got it. Yeah. Thanks for calling in, man. All right. Yeah. Thank you, guys. Thank you. Thanks, buddy. Good kid. Yeah, good kid. At that age. Yeah. I'm telling him to loosen up. Most 23-year-olds have flooded with DMs.   I mean, he's built a lot like how I was built. And I do remember that was the... Bro, he was jacked at that picture. Yeah, he looked great when he was bulking, only putting calories in and stuff like that. But I mean, of course, you always want more, right? So as a kid, you're like, you know, I could get more, I could do more. And like, the weekends were... You've heard me talk about it. Like, winning the weekends became like a huge change. Totally. And to your point about allowing that free... So my idea was like, okay, when I shifted away from the pizza, beer, drinking all the crazy stuff on Saturday and said, okay, I'm going to win Saturday and Sunday. And then if I still want pizza, beer and those things, I'm just going to have it Monday through Friday. So I actually have more days that I could have it if I wanted it. What I found was when I'm working, when I'm in my routine, I craved it less. I didn't want it as much. And then if I felt like I was deprived because I was winning the weekend and I was being so good on the weekend, I would then allow myself on Tuesday night to go have it. It was great. And algae I came up with, it was like about being jet lagged over the weekend. And, you know, wasn't that yours? It's a good one. You're doing the handoff. Basically, I started trying to make up sleep by going to bed late on Friday, Monday comes around your jet lag. But, but yeah, I mean, it's, you know, that's a... By the way, forget the kid we just talked to. A lot of people are like that when they go on vacation or they look at like a break, because their life is so unmanageable in terms of regimen and stress. They go off the rails.  So it's like a mindset shift and maybe some of the, some of the structure needs a shift. He's going to crush it, dude.


Cool. Hi, my name is Corey Hinker. I am a programmer at MOTE, and this talk is about some of the work I've been doing with partitioning, and specifically with ranges, and my interest there was that there is a lot of partitioning solutions out there that cover time series data, something where there's a notion of a new partition and old partitions that could be aged off, and this particular, our particular need didn't really fit that. So that got me looking into more about how to, what we wanted to do was partition by client names. We have some text named client data, and it's very lumpy, and we couldn't just go by simple, you know, the A's go here, the B's go there. We had to get much more detailed than that, and that got me thinking about could we leverage ranges in native Postgres to do that. So, show of hands, how many people have used ranges in Postgres? Okay, some of this is going to be a little bit basic for you. What I'm going to be talking about is an introduction to ranges. I'm going to be occasionally griping about things that I wish were there and aren't, and then I'm going to get around to, once I've established all that, just a quick overview of what we did with range partitioning. I like ranges. They allow you to indicate in the data that two different values are connected, start, time, and time.  If those things are range, there is no doubt the intention of the data. Likewise, when you're writing statements that use ranges, you're allowed to use operators that more accurately convey your meaning. Do these two hotel stays overlap? Things like that. You can also index on ranges. It's just, it's not somebody last night said that Bryn was possible, but I hadn't heard that before and didn't check it before this presentation. And it also allows you to do exclusion constraints, which is useful for things like resource contention. No two people can be staying in the same hotel room at the same time. And the thing that interests me most, and why I made this a talk here, is that no other relational database has them, or at least they didn't the last time I looked. Ranges go by standard mathematical notation. And if you're familiar with that, then you're going to find that how ranges are expressed to be very natural. If it's a parentheses on the left side, that is an exclusive, meaning that it does not include the actual value indicated. So in this case, values greater than x. If it's a square bracket, then it that it does include, so it is values greater than or equal to x. Likewise on the right side, that is the upper bound. Exclusive at inclusive. If no value is given, that means there is no lower bound or no upper bound. So the set of exclusive comma exclusive means everything, every possible set for that data type. Likewise, the word empty means none of the values. There's several ways you can construct ranges. The first is casting from text, and in that case, you need to specify your bound. Inclusivity, exclusivity, the actual bound value, comma, the high value, and the inclusivity, exclusivity of that end. If you omit one, that means that that side is unbounded. It does not matter if you call it exclusive or inclusive, it will be considered infinite and take all values to, in this case, all values to the right of low, will be part of that range. There's also a constructor function, in which case you pass in the low and high parameters, which can be null. And then you give the two bracket bounds as a pair at the end. Now, if you omit that, the default is inclusive on the left side, exclusive on the right side. There is no polymorphic way to do this. That was a small gripe of mine. So you can't select to a range of a type with arbitrary low and high bounds and an exclusivity constraint. What types exist? You've got your basic integers, numerics, time stamps, dates. They left off Boolean, I guess because the range isn't that big. And the one that really got me was there was no text range. Anybody care to guess why there would be no text range? Collation sequences. Okay, what letter is greater than another letter? It depends. You would essentially need one text range type per collation sequence on your system. That can vary. And as we've learned, unfortunately, the collation sequences on the operating system can change out from underneath you. So that could be a very big problem. If they wanted to create all the possible text range types, they would have to spend one oid per, and it didn't strike them as worth it. So you must create them on your own. In our case, the actual ordering didn't matter. We just wanted classification and getting our data into smaller buckets for easier querying because we were going to make heavy use of GIN indexes on text search. So the smaller you can make those tables, the faster it's going to be. So in this case, we just used a text range with the C collation, which is probably certainly the most stable. When you have range values, you can decompose them into their component attributes. In this example, I've created various stages of water. Please don't get all science class on me because I'm sure that the state changes involve ice and water being both at the same temperature. But for this purpose, you see that we've created ice as being defined as any temperature below 32 and not including. Water takes over at 32, goes up to but not including 212. Steam takes over at 212 and goes up as high as it can. You'll see that I used a different way to construct the ranges on each at each instance. The first one is just a standard text coercion. The second one is the explicit constructor and the third one there is the constructor omitting the final parameter, which then defaults again to exclusive on the right and inclusive on the left. Just for fun, I added in another state which has no temperatures. There's a series of functions that you can use to extract those attributes back out. They are lower, lower-inclusive, lower-infinite, upper-inclusive, upper-infinite and is empty. There's a few more but they're not specifically tied to extracting specific scalar values back out of a range. I don't remember whose blog it was that had this example to set your nulls to be displayed as that little pillow character. It was only when I was making these slides that I learned that the travel industry calls that the pillow character and it's also currency in a couple of Nordic countries. So it wasn't maybe the best choice but I find it really useful. So here you can see that I have decomposed all of the scalar values back out of the ranges and there really aren't any surprises there. The most interesting ones would be that you can see where it was unbounded. There will be a null value as is the case for the low value of ice and the high value of steam. In those cases it will also be infinite as you can see for the low value of ice and the high value of steam. In the case of heat death there is of course no values there so you're going to have to contend with the fun of null arithmetic. Ranges come in two forms. There is discrete and continuous. And the way you tell them apart is if that data type has a concept of a next value. If it does then you're dealing with a discrete data type. That's your integers, that's your dates. In continuous in the case of text you can always add an additional character on to tie break to otherwise similar strings. And you can always add a smaller number in the case of number range. You can see that there is no smallest number you can get where you will have two continuous values described differently that are actually equal. In the case of discrete types you will see that I have created a value from 1 to 10 for in 4 range. But the canonical description of that shifts it to inclusive on the left, exclusive on the right so it becomes 211. They are the same value internally. That is just how it chooses to store them. Some operators just don't work the way you expect them to. In this case less than, less than equal, greater than, greater than equal. The concepts don't really map well. So internally what it does is it tests on the lower bound or the lower bound scaler first and then uses the upper bound as a tie breaker. There's not a lot of real world cases where you would actually need something like that. And I thought about it for a while, like what would it be better for testing? If a median value, would you actually be testing like say the cardinality of the range, something like that? I couldn't come up with anything and it turns out they are needed internally for indexes anyway. Some other operators that are of value, this one is strictly to the left of which is to say that the normalized upper bound, again you have to remember to shift things to the canonical description. If the normalized upper bound is less than the normalized lower bound of B, it is strictly to the left. And as you can see one to three non-inclusive is strictly to the left of three to five because only one of them actually has three. However if you switch that exclusive bound to inclusive it then becomes false. Strictly to the right of is the exact opposite of course. And in this case I decided to do a little more illustration with fun you can have coursing text values into data types. Casting the string today into a date works. That's kind of fun. As does tomorrow and yesterday. It's not quite as good as the UNIX date command so you can't say two days ago or at least I wasn't able to get it to do that. So in this case the question is a range of today and up to tomorrow, strictly to the right of yesterday and up to today. The answer is true. Again if you change the bound from exclusive to inclusive it becomes false. Some operators I'm really not sure what the need for them was and this is an example does not extend to the right of. So basically is any element, if any element of A is greater than the greatest element of B then it does extend to the right of and that would be false. And as you can see here I've taken our common values and tried to show the similar tests does not extend to the left of. Yeah these I have not found a good use for these but they exist and I mentioned them because there's some operators I thought really ought to exist and don't. Adjacent. Now this one's actually useful. There is no overlap between two sets and there's no gap between them. The most interesting part of this is it doesn't matter which one is on the left or the right and you'll see that I actually put the lower set the lower range on the right in the first case. And in the other ones I showed an instance where there is a gap and then an instance where there is an overlap containment. These are the same as the is this point in a circle operators. I don't have a good name for them so the value that is on the greater than less than or pointy side must fit entirely within the range on the at side. The side that is testing for containment can be a scalar or can itself be another range. And you can see here I've shown that one is in the set of one to four inclusive. 20 up to 30 is in the set of one to 100 one inclusive. The date infinity is in the set of all possible dates. The everything range does contain the empty range. And the everything range for integers does not contain null. The answer is no welcome to the Boolean black hole overlaps pretty straightforward with one catch. It means do they have a common value. So as you can see 20 up to 30 overlapping with one to 100 that's true. The everything set overlapping with empty. Well that doesn't work. We just saw that well so I tried again is empty is the range empty in the everything range. Yes it is this empty overlap with the everything range. No it doesn't because there's no values in the empty range. They have no value in common therefore they don't overlap. You have the operators for plus union whatever you want to call it. And there's also the range merge function. This takes two ranges and combines them together. It only combines them together if there is not a gap. So in the second example you can see that I created or I attempted to create a range that would have had a gap and you get an error. New in nine five there is a function that will cobble those together into one large superset range. But if you're on nine for earlier you can go over to PG X N and install range type functions. Nice little extension written by yours truly. And you can have that capability on earlier versions. The intersection operator is what values do these two things have in common. In the case where the in four range they do not in these two do not overlap. They both get up to four but neither one includes it. You can see that their resulting range is empty. And in the case of one to four inclusive and four to one hundred also inclusive on the lower end exclusive on the upper end. Got to remember that default. They have the value of four in common. That is if you'll remember very strangely represented as four and up to five. The difference does what you would expect from a the except set difference function. The catch here is that you cannot create disjoint ranges. If you try to subtract out a range that is in the middle of the larger range you get an error. And that's something that I'll be addressing later. What I wanted was that ability to do a range split. If you have partitions and you want to pull out a specific value and make a partition just for that one value or for that very narrow range. You would like to have a partition on the left and a partition on the right. Maybe if they if unless you had picked a value that happened to be already on the edge of an existing range. So in this hypothetical case every time you see the hypothetical prompt that means that this was not actually run. It's just something that I dreamed up and thought would be nice. In this case I would like to see a function where if you tried to split the range one to one hundred inclusive. Take that range and it split out of it two to four you would get the three ranges one up to two two up to five and five to one hundred inclusive. Another missing operator couldn't there's not a lot of characters left so this is the best I could do. But it thought it would be nice to have something where you can test whether two different ranges share a bound. And in this case the side of the equal determines which bound you're checking on the left side for lower on the right side for upper. And they should be pretty simple for you guys to read over I don't think I have to walk through them quite as much right along. I would like it if the strictly to the left of strictly to the right of operators allowed you to compare individual elements against a range. This is very important if you are say writing an insert trigger that is going to be used for range partitioning. And you need to look at a specific value and figure out which section of the ranges you need to be looking at. As it is I end up doing containment on a middle range then trying left then trying right. It's not the most efficient but generally speaking in large bulk operations you're testing for which partition you're going to end up in first and then directly inserting into that. You can simulate this functionality by casting the scalar value to a singleton range in this case one one inclusive on both sides. And you can get that functionality it's just you can extend Postgres to make things elegant and right now this isn't elegant. So that's why I'm complaining to y'all about it. Another one I would like is essentially a string comparison comparing an element to a range where if that value falls within the range it returns zero. If it goes strictly to the left give me a negative one if it falls to the right give me a positive one basically mimicking string compare. And in this case you can see that the integer one would have fallen within the range one to ten inclusive. The integer one would fall to the left of one above one and up to four inclusive. Four would have fallen into one to four and four would have fallen to the right of something that was up to four but not inclusive. This I implemented as the element range comp function also in range type functions and there will be a URL for that at the end. I would like a test to see if a range happens to contain exactly one value. That would be useful for knowing if there was any point in trying to further subdivide a partition that was growing too large. If it only contains one particular value you're not going to find any luck there. And as you can see four to five not inclusive is a singleton because only the value four is in there but four to five inclusive is not because there is four and there is five. Another thing that will surprise you if you're writing a range partitioning extension is that you can make check constraints on values being contained in a range and that check constraint will work. It however will not work for constraint exclusion when you're doing a query. After some digging I've found out that the check clauses that are considered when doing constraint exclusion on selects and updates and things like that only look for equal to greater than less than and between. I think there may be a couple of others regardless containment is not there so I had to back manufacture the where clauses that would be used to create the check constraints. And it occurred to me that this would also be useful if I was some of the data that we're pulling in is actually derivative of data that is on a machine that doesn't have this extension in fact it's not even Postgres it's a redshift cluster. And they're not going to understand ranges anytime soon so I was going to have to be able to generate that sequel dynamically for making the remote call. So in this case these functions get lower bound condition expression upper bound condition expression and the full bounds expression came in handy. They all have they default to a variable of X but you can specify one as you see in the third example for a way to if you know what your column name is going to be on the other side. I doubt most people are partitioning by X. Okay so our use case is a type ahead search which is very much like what you see when you're doing Google searches as you add more characters typed it refines the search.  So what we needed was the ability to make very fast.