 Let me start the last lecture. Before, I mentioned that there's notes on these lectures and much more material on the website of IHES. Just to clarify, it's not on the website of the special trimester, but on the real first page of the website of IHES. And you will see my picture there, so you go to my picture. OK. So the last lecture will deal with the proof of this theorem. So I'm reminding you of this. We flashed it last time. This is the non-radial case. And there are two cases. The first case is a finite time. Then we have a finite set of singular points with a look at 1. Then there are J star traveling waves. And notice that in this case, J star is bigger than or equal to 1. So there is at least one traveling wave. OK. This is the finite time blow-up case. And a radiation term, v0, v1, a well-chose and sequence of times, tm, and scales that give us the scales at which the traveling waves live, which are, we can order them They're all much, much smaller than the self-similar rate. And they are much, much smaller than each other if you order them properly. OK. Now there are centers for the traveling waves, which we call CJM. And these centers are all strictly contained in the well-scaled ball. Around the singular point. OK. So there's a beta here, which is strictly less than 1. And this is a very important fact. And there's a connection between the sequence of times, the center of the traveling wave, and the direction of the traveling wave. And that's given by this formula. OK. And this comes out of the proof. It's a very important fact. So what you should think about by this statement is that all the traveling waves remain strictly inside the inverted call. OK. Because they're centered at a point that's strictly inside. And they're rescaled at a scale that goes to 0 much faster than the width of the call. In the definition of lj, isn't that, I mean, in the numerator, isn't that c and j minus x star? Well, yes. Yes, let me do the following. Fix x star equal to 0, belonging to s. OK. OK. Yeah, so we think of x star as 0. And then I don't have to say any more. But thank you. Yeah, so implicitly in everything I said I was thinking that x star is 0. OK. So we have this inverted cone. We have the traveling waves that are somehow concentrated inside. And they're concentrated strictly inside by an amount given by this number beta. OK. And then our solution is written as the radiation term, the sum, this finite sum of modulated traveling waves, and an error that goes to 0. And these traveling waves live at different scales, or their distances are apart, and so on, as given by these orthogonality conditions. OK, so that tells you that each wave doesn't see the other one. OK. So that's the statement in the finite time case. And in the infinite time case, so there's only one infinity, so we don't have to worry about the different singular points. This is only the point of infinity. And so the first point is that we can extract the scattering profile. And that's this linear solution that has this property. And I said a few words as to how you go about it. And then there's also a j star and j star traveling waves. Now here in the infinite time case, j star can be 0, because that would be the case where the solution scatters. So for instance, if the data is very small, then j star will be 0, because the solution will scatter. OK. And then we have scales, which are again smaller than the self-similar scale. We have centers. Now 0 is the right center here, because everything goes to infinity. And there's a connection between the direction of the traveling wave, the time sequence, and the center. And again, this traveling waves live strictly inside the light core by an amount beta. And then the formula is the same and with a similar orthogonality of the parameters. So this is the result that we're going to sketch its proof. OK. So I'm going to sketch the proof in the case 1 of the finite time blow up. The proof of the second case, the infinite time, once you've extracted the scattering profile is completely similar. So it's enough to go through the ideas for one of them, provided we know that there's already the scattering profile extracted, which we know. OK. So end of audio-visual presentation. So this proof has a number of steps. It starts by some basic estimate. From this estimate, we make a preliminary decomposition. And then with an error that will go to 0, but in not as strong a way as we wish, then we retweak the sequence of times and improve the properties of the error. We improve some more the properties of the error. And at that point, we get an error which tends to 0 in the dispersive sense. And then we show that somehow we can still produce a channel of energy to make it go to 0 in the energy sense, even though, in general, that is false. But the thing is, is that we prepare the error through these preliminary steps in such a way so that the counter examples disappear. And for this well-prepared sequence of errors, we can find the channel of energy. And that finishes the proof. But it's a proof that requires patience, because it has many, many steps. But I'm going to start with the first step, which is the starting point somehow. Oh, is that one I like? So the starting point is something that one can refer to as a Morowitz-type estimate in analogy with the Morowitz estimate that appears in the theory of wave maps. And the first proof of this estimate went by analogy with the theory of wave maps. I'm going to present a different proof that's not in those notes that uses the ideas that Frank Merle and I used in our paper on the ground state conjecture. There, one key step was the elimination of self-similar finite time blow-up for compact solutions. And we did that by introducing something called self-similar coordinates. And we will see that one can use self-similar coordinates here. And that's what I'm going to do now. So the claim, so let me assume that T plus is 1 and U, a solution of NLW. I have to watch the time carefully today. So it's a finite blow-up solution and 0 belongs to the singular set. Then for all T, belonging to, let's say, 1 half 1, we have the estimate that the integral from T to 1, the integral for x less than 1 minus T, dt U minus x over 1 minus T dot grad U minus 1 half over 1 minus T U squared. The x dt over 1 minus T is less than log of 1 over 1 minus T to the power 1 half, OK? So the way to think about this estimate or the way I think about this estimate is that it's a replacement in the non-radial case of the estimate on the Chisaro means of the T derivative that we saw coming in the radial proof, the first radial proof with De Caer and Merle, OK? So that's what this is. And of course, that one with just dt fails in the non-radial case, but this is a replacement. So the next thing I want to explain is why this is a good estimate. Why is this a good estimate? We know that this guy inside is bounded. This term, this is a U, is controlled by the gradient by the use of Hardish inequality. So that term inside remains bounded, OK? But then it's a bounded object and I integrate against dt over 1 minus T. If I just replace this by 1, I get log of 1 over 1 minus T, right? But the bound here is with a power 1 half, which is strictly smaller than 1. So that means as T goes to 1, the average goes to 0. That's the meaning of this bound, right? Because I'm improving on the constant 1, which gives me the logarithm to the first power. Here I get the logarithm to the half power, so that's better. So that meant that the thing had to vanish in average, OK? So now let's prove this. Now the constant 1 half comes by the scaling law, OK? This is a consequence of scaling. And if we were in dimension D, it would be D minus 2 over 2. This is what? So there's no coincidences here, OK? These are things that are the way they have to be. So maybe my notation is not so good, OK? So now it's a bit better. So the proof of this goes by steps, OK? So in the study of focusing on linear wave equation, the first problem that you encounter is that the density of energy doesn't have a definite sign. It can be positive, it can be negative. That means that the flux, which arises when you differentiate the energy outside of a cone as a function of time, also has not a definite sign. And the definite sign of the flux is something that is very helpful when studying nonlinear wave equations. For instance, in the study of wave maps, which is very similar to this, the density of energy is non-negative, and therefore the flux already has a sign. So you're in a better position. So what we're going to show now is that, in fact, we can make sense out of the flux, even though it may have a variable sign. So the first step is the control of the flux, OK? And for those of you who don't know what the flux is, I will write it down, OK? I'm sure that there's a fair number of you. So the point here is that there is the regular part of the solution v of t. So this is a regular solution so that the support of the difference u of t minus v of t is contained in the inverted cone. So on the boundary of the cone, u equals v. And v is a regular solution. So since v is a regular solution, I know that v is in L5t, L10x, past t equal 1. OK, that's the Strikert's estimate for regular solutions. And I also know, that's not meant to be a letter A, right? It's an inverted cone. So the other thing we know, of course, is that dt of v belongs to L infinity t, L2x, OK? That's in the energy space. So the dt derivative is in L2. So as a consequence of this, d dt of v to the 6 belongs to L1t, right? Because I take the derivative of v to the 6 as v to the 5 times d dt v. And I use those two estimates. I use them first in the x variable. I use Helder's inequality shorts. And then I take the L infinity out and keep the other one. So this is immediate, OK? But this, by an obvious trace inequality, shows that v6 belongs to L1. Let me call this cone gamma, OK? So if a function is, on a domain of a graph of a Lipschitz function, has a derivative in L1 in t, it restricts to an L1 function of the boundary. And the proof is fundamental theorem of calculus. But v on the boundary of the cone equals u. So that means that u is in L6. But this is precisely the negative term in the flux, the term that we don't like. But this controls it, OK? This is a very simple observation, OK? So let me now tell you what the flux is. I lost my flux, no. So I'm going to write the energy flux identity. I got to subtract the term here. The same expression at t1. Maybe I do this. This is the, and the energy flux is what's in here. Now in the linear wave equation, we don't have the term 1, 6 of u to the 6. But this is in the non-linear equation. But this guy is totally controlled. Because u to the 6 is exactly in L1 of the boundary of the cone. So that tells us that we can control this guy. And this guy is the real flux, OK? So I write that down. So that's a corollary. This is bounded for all t1, t2, OK? Because these two terms are bounded. And this whole term is bounded because it's integral. OK, I'm going to get one more bound. And then we're going to be ready to swing here. Because if you look at that formula, when x is on the boundary of the cone, x over 1 minus t is the same as x over length of x, right? So I'm already beginning to see things here. Question or no? Yes. In fact, you wrote it in the sense for global in the blow-up cases. No, it depends on the direction. I'm blowing up at t equal to 1. No, no, because t plus is 1. I'm doing at the blow-up case t plus equal to 1. It's from t0 to 1. Yes. This is correct. OK, so let's move on. So if you see, then there's another term on the right, which is u over 1 minus t. And I need to control that on the boundary also. So to control that on the boundary, I will use a hardy's inequality. So what I will do is I will call f of x to be u of x, comma, 1 minus absolute value of x. So which is exactly u restricted to the boundary of my cone. It's u restricted to the boundary of my cone. And now what happens is that the absolute value of the gradient in x of f of x squared is exactly this quantity. Maybe the factor of 2 might be in the wrong place, but it's OK. This is correct. OK, so we go on. So this is the f. And now we apply hardy's inequality to this f. Hardy's inequality applied to this f says that f squared over x over x squared is less than or equal to the constant times gradient of f of x squared. This is controlled. So we get this. And since x equals 1 minus t on the boundary of the cone, we control this term. So what this says is that I also have like this. Because 1 minus t is x on that boundary. So now we control that quantity on the boundary. Nothing 3D and higher. Yes, it's whole space. Yeah, I don't know. Yeah, OK. So the next step is we're going to use all this, introducing self-similar variables. So we introduce self-similar variables, and that puts us in a completely different universe somehow. We renormalize the problem. So for x less than 1 minus t, I'm going to call y to be x over 1 minus t and s to be the log of 1 over 1 minus t. So the s now goes from 0 to infinity. And the cone goes to y less than 1. And now I'm going to define a new function w of y and s. And this is exactly 1 minus t to the 1 half u of x and t. So I replace u of x and t by a renormalized one, and I call the variables like that. It's a definition. Now in the work on the ground state problem, our w's had zero boundary values on the sphere, y equal to 1. Here they don't, but that is replaced by these flux estimates and the hardy estimate. That allows us to control the boundary values that are not 0, OK? So the first thing we do is we write down the equation for w. For w has its own equation. And it's good to have a big blackboard here because it's a long equation. So that's the equation. And rho of y is an important function for us. 1 minus y squared to the minus 1 half. So let's look briefly at this equation. So it looks a little bit like a wave equation, right? It has two derivatives in S and two derivatives in y here, OK? But this equation is not elliptic, right? At y equals to 1, this equation degenerates in two ways. Pardon me? You compute the divergence of a big field minus a Pascal. What did I do there? Let me check my formula, you're right? OK. There's a y there, OK? So the equation, thank you. So the equation, first of all, degenerates when y equals to 1 because of the rho, right? There's a 1 minus rho. The rho is somewhere here. And you see that at y equals to 1, it becomes infinite. There's another problem that's, in a sense, worse, let's say. Because this one at least gets compensated, which is that you are subtracting a one-dimensional direction. So there's a rank 1 thing there. Now, if y is strictly less than 1, this can be absorbed by that. But at y equals to 1, there's no way to absorb it. Even if you remove the rho, it's not elliptic. You lose the direction y, OK? So this is one thing about this equation. The other thing is that there's another second order term, which mixes the y and the s, OK? So you could see it as a problem or you could see it as a help. In fact, it will help, OK? So let me just continue by some simple facts. And this is a calculation. And this is just a chain rule. Nothing more than that. So this expresses dds of w in terms of u. And how do you see this connects to what I'm trying to do? If you look on the board over there and you compare with this, you see that that's precisely what I have with a factor of 1 minus t to the 3 halves, which is the change of measure. Once I square it, it's 1 minus t cubed. And that's precisely the change of measure. So what this tells you is that the control that I want over there basically is a control of the dds of w. Not basically, it's precisely. And what is my information about the flux? My information about the flux is precisely the fact that integral from 0 to infinity integral over y1 of ds w squared d sigma ds is finite. That's exactly what my flux information tells me, together with the Hardy inequality, after change of variables. So what I need to do then is to prove this estimate is exactly the estimate on the top, after I change variables. Manipulating something which is infinity. No, because yes in the end, yes. But here I go from s1 to s2. And s2 is less than infinity. I don't have anything it's infinity. Of log of 1 over 1 minus t. 1 minus t. 1, yeah. No, I think you are blowing up at t equal to 1. Yes, that's where I am. So you cannot start from 1. That's the blow up point. And up to the blow up point I have the bound. You have the inverted complex of r. It's OK, but it doesn't really matter. No, no, no. It's a mistake, right? No, no, no. The friend part is right, because this would mean that the integral is your integral in t here. Yeah. How do you say it's converged? Yeah. You look at the rest. No, so I should write this as follows. But that's the only point, OK? This is the point, OK? So but this is what we want to prove. And this is precisely what I wrote over there. When we change variables, the logarithm gives me a 1 over 1 minus t. And then this becomes that. So now I'll spend five minutes discussing why that's the case. And the idea is now, let me see how I do this. So the idea is that the W equation has a natural energy associated. And this natural energy has some monotonicity properties. The only problem is that this energy cannot be defined because it's infinite. So we do a regularization of the natural energy, OK? So I introduce for an epsilon positive e epsilon of s. OK, so this is a quantity. When epsilon is 0, this would be the true energy. How do you, what do I mean by that? If you multiply the equation by ddsw times rho and you integrate by parts and neglect boundary terms, you would see that this is the energy once epsilon is 0. Now, for this, I need to tell you what rho epsilon is. So rho epsilon just adds an epsilon squared inside the weight so that it doesn't blow up at the boundary, OK? But, pardon me, yes. So now what we do is we compute the energy, the derivative of e epsilon, by multiplying our equation by ddsw rho epsilon instead of rho. And then there will be boundary terms. And I'll tell you just what you get. And this is a very compact formula. So this is something, this is something very good because e epsilon, you can think of, it's easy to see it's bounded by, let's say, 1 over epsilon. So when we integrate in s, we integrate the derivative, we use the fundamental theorem, we can make the integral convergent. And this is positive, so this is a very good thing because this is what I'm trying to control. Now, there are two more terms. This is the unfortunate part. The first term comes from the boundary integrals. And it's precisely 1 over epsilon, the integral of ddsw squared on the boundary, which is controlled. So this one is something else that's of the order of 1 over epsilon. And that would be just fine. I can integrate it and control it. Unfortunately, there's a third term, OK? So the third term, it's precisely that. So the formula is still very compact. Now, this term has some good features. One good feature is that I have epsilon squared in front, which is small, let's say. Another good feature is that I have ddsw to the power 1. And so you could think that by Cauchy-Schwarz, I could hide it into the first term. The bad news is that I don't know anything about the integrability in S of y, don't grad w. And all I can do, basically, is use its boundedness. Y is less than 1, so it's just the gradient. And the gradient by change of variables becomes the gradient of u, OK? So then you play. And now we're going to choose S1 and S2. And you choose this. You choose epsilon to be exactly that. And if you do Cauchy-Schwarz in the appropriate way, you obtain that S2. And that's the end of this argument, OK? And so the fact that we don't get the whole integral, which is what I had secretly hoped for in my subconscious when I wrote this, you only get something that's controlled in finite intervals, but you get a better control than you would, OK? OK. And I explained why this is a good estimate. So now let's move on and see what you get out of this. Now I'm going to go to a slightly different notation. It is easier in what follows to make the blow up point be 0 and the solution going forward from 0, OK? So I'll just write that down. So I'll draw the picture. So this is the picture now. And this is the blow up point. And the solution is defined here, OK? And then the estimate then translates. And 0 is again in S. And we have some neighborhood R star where we don't see anything, no other singular points, OK? And our estimate then translates. And because we are going down instead of going up, some of the signs have changed. And I thought that's what you thought I was mixing up, Frank. Yeah. So this is there. So this is the same as this. So now the idea now is to extract information from this inequality. So this inequality is an inequality about Cesaro means, OK, again, Cesaro means go to 0. So you can use a Tauberian argument to pass going to a subsequence from Cesaro means to real means, yes. Yes, this is most definitely there. Otherwise, this wouldn't be of much use, right? OK? So it's the averages that are getting small. OK, so now we use Tauberian argument. There exists mu n tending to 0. And I actually need to do it at two different sequences of times. But this is a minor point. And don't pay too much attention to the fractions. OK, I'll draw a picture of what they mean. So the picture is I got mu n. I got mu 2n. And I divide this into three parts. Here is T1n. And here is T2n. That's the picture of what those fractions mean. And what happens in this setting is that the actual averages and even more supremums of averages go to 0, distance to 0. So we pass from Cesaro means to averages. But instead of averages, we get maximal functions of averages. And so somehow here, you have to use the harder to do maximal function and the weak type 1, 1 inequality. So there's a little real variable argument here. OK? And now I'll tell you later why we need two sequences of times. Let's ignore that. At those two sequences of times, we can do now an analysis of the profiles in the profile decomposition using this estimate. OK? And so that gives us the first decomposition. So is there as well T in the denominator dx? No, here, no, because the top. OK? So now we get to the analysis of the profiles. I'm not going to be very lengthy in the discussion, but of course, in the actual proof, this is a lengthy step. I would say it's not a difficult step, but it's a lengthy step. So now we do a profile decomposition. I had the two sequences, but we ignore that. And so we get blocks, uj. These blocks have centers and scales. OK? So we're going to analyze cases. The first case, case one, is when, let's say, T, the T translation is 0. And the lambda n, the scaling factor, is self-similar. OK, this is one case. Now what happens in this case? Once you input this information into such a profile, what you get is that the corresponding profiles verifies a first order equation. And that's exactly this expression here when you study it. And you see that the vanishing of this, if you look at it at this scale, gives you that this thing is identically 0. So you get a transport equation. And what does this transport equation mean? This, of course, is a first order equation. I can integrate it, right? So this equation means that uj0 of xt is precisely T plus 1 to the minus 1 half times c of x over 1 plus T for some function Pc. You can see that any function of this form verifies this. And any function that verifies this is of this form. And this function has to verify the nonlinear wave equation, because that's what our blocks do. But then this is a self-similar solution to the nonlinear wave equation. And Merle and I had proved that there are no such things. So this cannot happen by the result of myself and Merle in 08. uj0 does not exist. So the second case is when, again, Tj0n is identically 0. But this time, lambda nj0 is much, much smaller than Tn, OK? So this is a true case that arises. And if you play with the profiles and that inequality, what you end up proving is that d dt of uj0 plus l0 plus lj0, gradient of uj0 is 0, where lj0 is the limit of cj0, cj0n over Tn. And cj0 is the center of the profile. That's what happens when this is concentrating much more than Tn from this guy you eliminate and x over T then becomes this limit. And so you end up having this transport equation. And what does this transport equation mean? It means that uj0 of xt is nothing more than some function f of x minus lj0 times T. That's what this transport equation means. So this is a traveling wave solution. We know what all the traveling wave solutions are. They are, first of all, this lj0 is strictly less than 1. And I have a solution to the elliptic equation and a Lorentz transformation that gives me this. And that gives me the profiles, the solitary waves in the decomposition. Now, there's a case 3 in the analysis, which is when Tj0n over lambda j0n tends to infinity. So those are the last remaining profiles. And these are the profiles that somehow come from infinity. And all we do is we'll put them in the error. Now, you use the fact that for a linear solution, the l6 norm goes to 0's time goes to infinity. So using that and this, you end up with a decomposition like you want, but with the error going to 0 in l6. Very good. And now we start improving. So the first step is improved by the fact that you use two different times like I did there. And then I tune the time and I get one in between where something better happens. So I'll say what that is. And here, you use a virial identity for this. What you find, this additional property. So how do you find this? You use the virial identity where you take a T derivative and you get that difference. And then you integrate that. And at time T1n and time T2n, we use the decomposition that we've already obtained with the l6 errors to prove better bounds that allow us to conclude this. So that's the next step. And now I'm going to find, I now find a third Tn between T1n and T2n. So from this, which is a Tauberian condition, I go to an actual convergence for a subsequence by a real variable argument. So I get T2, T1n, smaller than Tn, smaller than T2n. And I have, this is the old condition, but in addition, from this condition I obtain is less than or equal to 0. And here you have to do something because you can produce the first condition or you can produce the second condition but you need to produce them simultaneously. So there's a little argument for that. Again, it's a real variable argument. Now from these two conditions, I prove something about the error because of the following two cancellation conditions on solitary waves, OK? For any solitary wave and any Lorentz transformation, this is 0. And by the definition of a solitary wave, this is 0. Because these two properties hold for the solitary waves, I first do the decomposition at Tn by the first condition and I use those properties on my decomposition but the soliton parts get killed by these properties and so the error inherits these properties. So now I improve the properties of the error. So from being in L6, now I get a whole bunch of other properties of the error. And since time is running short, I'll just list them, OK? But now I no longer have to change the sequence at this point. So the sequence of errors now will verify. So this will be the tangential gradient to the sphere. I'll be more specific in a minute. So this, the region of integration is B lambda Tn, the union complement of B Tn for any lambda less than 1. And this is the tangential portion of the gradient. And then epsilon 1, and also goes to 0 in the same region. Yes? No, this is the whole gradient, but I'm not quite up to the edge of the ball, OK? So the domain is only for the third one? This domain is only for the third one, OK? So it was a parenthesis, yeah. And there's a third property, which is very important. All of these go to 0. And if I look at the linear solution corresponding to this data, this also tends to 0. So I narrow down the properties of my error. OK, and now comes the punchline. The punchline is that if you have a sequence of functions that verify these five properties, there's a channel of energy for the linear solution. So I'll write down that lemma. What's the property in the second line of these three terms? No, the line above, the line above. Yeah, all of them go to 0, OK? So let me put in the punchline now. I'm sorry if I'm getting, I have two minutes, right? I may take five. No, five. To subtract from our discussions. OK, so we have a sequence epsilon 0n, epsilon 1n, which is a bounded sequence in H1 cross L2. And it verifies these properties. So this verifies if the inf in n of the H1 cross L2 norm is positive, which is the bad situation we want to avoid, OK? Then all eta naught in 0, 1, I have channels of energy outside. So why is this? Well, there's a simple proof of this fact using virial identities. So you can do this by integration by parts. Now how will we choose eta naught? The eta naught will be chosen so that we skip all the solitons. We'll choose the eta naught bigger than this beta that confines all the solitons. So on the part outside, the solitons are not present. And we just have the regular part of the solution and the error. But the error verifies all these properties and so it sends a channel of energy. And our solution is almost like the thing without solitons because the solitons have small amount of energy outside. And that channel of energy creates a contradiction if this doesn't go to 0. And that's how the proof concludes. I understood that this is choosing this sequence of Tn is connected to this Tauberian argument. But when you have some Caesarean property, some Caesarean property, just think of just one sequence positive such that the Caesarean goes to 0, then usually you could say that it goes to 0 to a subsequence of density 1. Is there a way to say that this sequence Tn is... It's very fat, yes. It is very fat. This is true that it is very fat. The thing that we don't know is it may be fat but it may be very sparse too. Yeah, that's a problem. That's the problem. So I think that to pass from one sequence to a general sequence, there is a fundamental problem. And the fundamental problem can be broken up maybe in two parts. First, you want to have more control of the sequence of times in which you do this. And that involves perhaps a better study of that integral. I don't know. The other thing that is there in the problem is that you have to understand how multisolidons collide. Because that's what happens from one time to the next time. You don't know how they will interact. And the property of collision of multisolidons is a difficult one to study.