 So the last time we showed how to extract the scattering profile in the infinite-time case. And now today and next time we will sketch the proof of the decomposition in the finite-time blow-up case. So we have a solution that blows up at time 1, let's say, that remains bounded in H1 cross L2 up to time 1. And we want to now produce a decomposition into solitons. So that's the task. So in the notes, so there's a first step in which we prove the decomposition with a weaker error than what we'd like. And then we will go through several stages to improve the error. And in order to improve the error, sometimes we will have to slightly tune the sequence of times. And at some point we don't tune the sequence of times anymore, but then we show better and better properties of the error. Now the first step is inspired by work on wave maps. This goes back to work on grilakis on wave maps. And it's what we call a Morowitz estimate, or a Morowitz identity. The fact that there's a way to use ideas from wave maps for the energy-critical wave equation goes back to work that we did with a caught lorry and a slug. So this is the approach that's presented in the notes. But today I'm going to present a different approach, which I think is more concise conceptually. And it allows, I think, hopefully for more progress as we move along. So because of that, we'll use the blackboard for the beginning. And then in the notes is the other proof, so you will be able to read that. So in the finite blow-up case, we can always make it to be one by scaling. Not the problem. So we're going to approve the following claim. And we call this the crucial Morowitz estimate. Although in the new forming, which we're going to do it, you do not recognize this as coming from a Morowitz identity. And in fact, I think you can recover the usual Morowitz identities, even in the wave map case from this point of view. So the claim is that we assume T plus equals to one, and that the origin is a singular point. So it's a point where even locally I cannot continue the solution anymore. Okay, we introduced this notion earlier. So what it means is that it is not a regular point. And a regular point is a point where if you take integrals over small balls around that point, you get that to be uniformly small for all times up to one. Those are the regular points. And this is a singular point, so it's a point where there's concentration. That's what that means. And let me assume that it's spaced by one from all the other singular points. Since there are, finally, many, I can always pretend that. Okay? This is just so as not to have to write more. It's not important. It's the constant C such that four zero less than T up to one. No, maybe not. Let's say a half. So this a half is symbolic, and it depends on the fact that the other singular points are at distance further than one. Otherwise, you have to go sufficiently close to one. Okay? So we have an estimate, and the estimate is that the integral from T to one of the integral of X less than one minus T of the following expression. So we get a logarithmic estimate. But what's crucial here is that, well, first of all, this integral is finite. That's already very good. What is interesting here is that the power there is less than one. Okay? So the original proof had the power three quarters. We can get the power one half using this approach. Okay? Now, why is the power one half interesting? Instead, why is the power less than one interesting? It's because the finalness of this, suppose that this was just bounded, this inner integral. Then, of course, the integral up to one minus T would behave like the logarithm of one over one minus T to the power one. But somehow, we're only getting the power one half. So that means that in some sense, the inner integral is vanishing with the rate as you go to T equal to. Okay? And the power two is dimension dependent. Okay? So the number is different in dimension three. And this has to do with the scaling of the equation. That's exactly what it is. Now, in wave maps, there is no, this factor doesn't appear. And that's because there's no factor in front when you rescale wave maps. Okay? And so that's how this thing is working. Now, the proof of this that I'm going to show now is a very, from my point of view, is a very satisfying one. Because it uses the ideas that we already seen that come from the work of Merlin and myself where we showed that there cannot be a compact solution which has self-similar scaling. Okay? And if you recall the proof of that, which was given, I don't know, a month ago or something like that, it used the introduction of self-similar coordinates. Okay? And then some integration by parts. Now, the reason one could do that in the compact situation was that the boundary values on this inverted cone, let me put the inverted cone here, the boundary values of a compact solution are forced to be zero. And therefore, we had some boundary terms that disappeared because that was zero. Now, in this case, we can no longer assume that the thing is zero. So we have to somehow handle these boundary terms. And these boundary terms are huge. Okay? So we will see. So I'm going to prove the claim. So the first step is what we call the energy flux estimate. And we recall we have this solution that blows up at t equal to one. So this is t plus equal one. We recall that there is a v of t which is a regular solution at t equal to one. So this is not one that blows up, but it's one that moves continuously up to t plus equal to one. So with the property that the support of u of t minus v of t is contained in the inverted cone. Okay? So that outside the inverted cone, our solution looks like a regular solution. And we saw that this was a consequence of finite speed of propagation, basically. You consider the weak limit as you tend to one. And you now solve the nonlinear wave equation with that initial data at time one. And that's the v. Okay? So we have the object. Now, since v is a regular solution, we have two facts. Of course, the norm, let me just put it like that. So the v arrow and u arrow notation means the pair u, d, d, t, u and v, d, d, t, v. Okay? So it's just a shorthand. So this is a finite, but moreover, the spacetime norm, the Strikatz norm, is finite. Because this is a regular solution, so by the finite blow-up criterion, that norm is finite. Because it doesn't blow up. Okay? Is this okay? Okay, so then... In the last one, x is in the cone of everywhere? No, x is everywhere, because this is a regular solution. Okay? So this is what this local well-posedness theory gives you. Okay? So the next remark I'm going to do about the v is that as a consequence of these two properties, and this is a very remarkable fact, but it's trivial, right? From the two other properties and the live neutrals, right? v to the sixth, you take d, d, d, t, you get v to the fifth times d, d, t. You do the norm in x, you do Cauchy-Schwarz, you get dL10 times dL2, and then the other one is in L1 and the other one is in L8, so this matches. But now, this implies that v to the sixth belongs to L1 in t and x and the boundary of the cone, d sigma. That's a standard trace theorem, right? If you have the t derivative in L1, you don't need to even invoke a theorem, you just use the fundamental theorem. But on the boundary of the cone, v to the sixth equals u to the sixth by this support. Okay? So therefore, now we've proven that u to the sixth belongs to L1, and this is huge. This tells us a lot, because now we combine this with the flux. So I'll write down the flux. So, well, maybe I'll write this explicitly here. Okay? This is the same. I'm just writing this in coordinate. So now I'm going to write the energy flux. The energy flux is the following identity. So let's take t1 and t2 less than 1. So the energy flux is fundamental in the study of wave equations, and this is a well-known identity. This is the standard flux identity. And you prove this by calculating d dt of the density of the energy and then using the fundamental theorem and the divergence theorem. There's nothing more than that. Now one of the big difficulties in the focusing wave equation comes from the fact that this object in here is not non-negative. It can possibly change sign. And this is what causes a lot of the difficulties that we see. But this observation tells you that you can control the negative part, the key thing. We can control the negative part because it's exactly on the boundary of the cone where your blow-up solution equals a regular solution. So the corollary of this is that y. Why is this true? Well, we use this identity by the fact that the h1 cross l2 norm is bounded. This term is bounded. This term is bounded by the previous thing. The l6 term is bounded. So what I have is bounded. I'm just worried that there is a square here and not in the first line. There's a square where? I'll try the bracket here. It's because here this shouldn't be there. You're worried because I made a typo. Clearly this thing is non-negative because it's the half here. Or completing the square. Another way of writing this, which will sometimes be used, is that I'll introduce some notation. I hope you can read this by definition. This is the tangential part of the gradient. This is the definition of that one. And the reason why this follows from that is that if you just expand this square, you get what's on top. So it's telling you some control on the boundary of these objects. Now I will need one other term, which is a hardy term here. So I will add this to the corollary. This is also bounded. Now this needs a proof. You can't just... But you'll see that things work very neatly here because let me define f to be this function. This is a function only of x. I'm going to use hardy on this function of x. So let's calculate the gradient of this function of x. This holds just by the chain rule. So you see that exactly the gradient of this f is the flux. And since the flux is in L2, this gradient is in L2. And now I can use hardy. And on the boundary of the cone, 1 minus t equals x. So now we've got the three things that we want. So did I call this step one? I did. Well now I'm going to go to step one prime. Step one prime introduces self-similar coordinates, which we did already when we studied compact solutions in the grand state conjecture. So this will be the same. So suppose that x is less than 1 minus t. In between these two sets of lectures, I was giving another set of lectures where there was a blackboard with a hole and I lost the eraser. It went down the hole. The whole thing grounded to a hole. So inside the cone, y is x over 1 minus t. So that y is less than 1. And s is minus the log of 1 minus t. So our new universe now is the sy variables. And this is, as I mentioned in previous lectures, this is the analog of the parabolic self-similar variables that were introduced by Giga and Kohn. And that's what we use here. So now we define w of ys to be 1 minus t to the 1 half u of xt. This was just like we did earlier. And this 1 half power of 1 minus t is responsible for the coefficient 1 half in front of the u. That's where that comes from. We'll get to that. This is just information. It's not something that's supposed to be obvious. So now we introduce some weights. And this weight obviously blows up at y equals to 1 which corresponds to the boundary of the cone. And that's why when we were dealing with compact objects we needed to have things that vanished on the boundary of the cone. But now all we're going to do is regularize this weight. We'll work with this weight instead and then choose epsilon appropriately. So now we have s goes from 0 to infinity now by this change of variables. Where t goes between 0 and 1, s goes from 0 to infinity and y is less than 1 because we're looking inside the cone. And now I write the equation for w which we have already seen. The equation is the same as before. So this is the equation that w verifies. So it is a wave equation but it has a nonlinear wave equation that you cannot avoid but it has some interesting features. First the elliptic part degenerates as y goes to 1. So it degenerates the elliptic equation. And the second thing is that there is this extra term here. What I'm going to do is copy this properly. There's a gradient and a dds. So it's a second order derivative and that's a kind of parabolic type term. And that's why this becomes a little bit parabolic in some similar coordinates. Anyway, so this is the equation. And now I'm just going to tell you two more things. This we can compute in terms of u. So the s derivative of w with light. Oh, I'm sorry. Should I lower this? Oh, maybe we need light there. It's okay now? No, it should be okay now. Okay, so if we look at this expression here you see that it corresponds exactly to what I have inside my integral. I just have to divide by the power 1 minus d to the 3 halves. And so that's the point that this expression inside is the self-similar time derivative. Okay? You first move this for y, and then you take it in. I haven't done anything yet. I haven't done anything yet. This is just the formula. There's no epsilon in this formula. Right. And this equation has the rho. It doesn't have the rho epsilon. Okay? Which is singular at y equals to 1. And I'll tell you when the rho epsilon comes. And then I put the gradient in y. This one is much simpler, right? So from these two formulas we get that the h1 norm is bounded. Okay? Because that's just the change in measure. We know what ddy is. We know what ddy is. It just gives you the right factor. This just comes from the fact that each one of these objects, once you weight it appropriately, is in alto. And for you, you have to use the regular Hardy inequality. Now I tell you what the flux estimate gives me. Together with this estimate. The flux estimate, together with this estimate, gives me that the integral from 0 to infinity integral of y equals to 1 of dsw of ys squared the sigma in y ds is a sum constant. Why is that? Because, well, dds has this term in it. And this one, once you change the variables corresponds exactly to the flux term. Corresponds to the first term there. And this one corresponds to this term. So it's just a change of variable. So you see the boundary values may be bad, but there's something good about them. The s derivative can be integrated. And finally, my claim boils down to the following. And this is just the change of variables and the formula for dds. You just have to change variables in the y. And in ds. See the dds change of variables produces the factor of 1 over 1 minus 3 because s is the log. So this is all fits, right? It's all fitting perfectly. So now, how do I prove the estimate? And now, maybe can people see here? So now what we do is we introduce an energy. Step two is so I define e epsilon of s an energy which now depends on e epsilon. So that's my definition. So it's a definition. Now this is the natural if I, instead of having rho epsilon I had rho, this would be the natural energy associated to the to this equation. If you multiply by d wds and integrate by parts and assume that all the boundary terms you get exactly this you have to do ds w times rho. Multiply and integrate by parts which is what you always do for the wave equation. That's how you deduce the energy. Okay? If you do that you get the energy for epsilon and all the boundary terms disappear you get the energy for epsilon equal to 0. Okay? Now unfortunately the boundary terms don't disappear because nothing is 0. And second, we don't know that when epsilon is 0 this expression is finite. Because it's singular y equals to 1 and why should it be finite? That's the energy corresponding to the first line of the equation. No, do the second one. These two terms combine to today. Yes, it includes everything. So you have to be a little patient when calculating. But everything cancels out and it gives you that. Okay? But for us this this rho is too singular because we don't have 0 on the boundary. In the case when we were working with the compact objects they were 0 on the boundary and we could use the E0 energy. And that's what we did in our paper. So now okay, so we put the epsilon. So the interesting thing is the time derivative the s derivative of the epsilon. That's what you need to look at. And now the amazing thing is that there's a clean formula. So this is the very nice thing that happens here. Okay, so this is the formula. And I guess we were very surprised when you could actually calculate it. And it's very compact. So let's look a little. So how do you prove this first? You do the same as you would do trying to show that the energy for the wave equation is constant in time. You multiply the equation by W times rho epsilon. But Ws I mean the s derivative of W times rho epsilon and you integrate by parts. Yeah, square in epsilon in the second term. No, it's a power one. It's too strange because they're essentially the same for when epsilon is positive or negative. But then some of the signs will change. But it is a power one because if you look at if epsilon is the same for when epsilon is negative or positive, right? Yes. Oh, I see what you're saying. Let me think about it. Is it? It would be absolute value, epsilon. What you get is epsilon squared square root. No, it is epsilon. It is not. The reason that it is epsilon is what happens with this when you go to y equal to 1. It's exactly epsilon squared square root. The way you prove this is you multiply by d d s W times rho epsilon and you integrate by parts. And of course if you're going to do it at home take your time because you will most of the times you will get it wrong and then you will have I know by experience, okay? But this is the formula. Okay, so now let's understand this formula a little bit. Okay? What we want so first let's assume that W was zero on the boundary, okay? Like the old case. When W is zero on the boundary d d s W is also zero because it's a cylinder. So this term would disappear. This term you make epsilon equal to zero goes to zero. This term disappears. And in this term you get 1 plus epsilon squared so you get 1. Now when we integrate if the epsilon is if the energy is bounded you get the integral of this thing and the integral of this thing is exactly what you're trying to control. So this is the right kind of thing. We want to integrate and so the best thing to integrate is the derivative because by the fundamental theorem in our case what? In our case when we integrate in s this term is the one we already know from the flux. The integral of this term is convergent by the flux. So that's somehow what saves you here. You have the flux and therefore I think I wrote it here, yeah. The integral is convergent. What's the flux estimate? Of course I'm not going to make epsilon zero. If I make epsilon zero I die. Okay? So you say if you pretend to take epsilon zero we're left with the first term and then we're integrating. And if e is bounded what happens with y? Why it got integrated? There's no right is just an energy and then we integrate it and then by this magical identity you control this integral because this is I'm sorry because this is a positive quantity which is what you really want to control. So that's how this magic works. So now what do we do when epsilon is not when we don't have the boundary terms? So the first observation I'm going to use all the black boards and then move on back to the transparencies. The first observation is that the epsilon the energy is bounded by one over epsilon. Why? Because I'll just I'll write the formula there's only one remark because of this. In the definition this is rho epsilon rho epsilon so that and the rest is covered by unhardy. And this term when I integrate it is bounded so if I do an integral here the integral of this is going to be the e epsilon at the two endpoints which is one over epsilon this is a constant times a one over epsilon and then I have these two terms this is the one I like and this one is a horrible term but it has the redeeming feature that there's an epsilon squared in it. There's an epsilon squared and and it's mixed it has a gradient and the dds. So now we're going to just use Cauchy Schwarz. So we will integrate between s1 and s2 and use the fundamental theorem. Okay? Then one plus epsilon squared integral between s1 and s2 integral of y less than one ds w squared one plus epsilon squared minus y squared to the three halves will be less than or equal to c over epsilon that comes from this term from this term and this term and then plus so now I'm going to do Cauchy Schwarz keeping this weight okay? But the epsilon squared I will keep with this term which is the one I don't like and the other one I leave alone and then I put a small constant in front of this one that I absorb by the left that I'm left with is the following c over epsilon plus now here I will get an epsilon to the fourth when I square and in the denominator I have a one over epsilon cubed right? So I get an epsilon to the one over an epsilon then I have the length of the integral because I'm not using anything on that I'm just doing it in Y and that's it and maybe instead of one here I have one over eight after I hit that one term I just did Cauchy Schwarz I kept these two things together pardon me? It comes it comes from the Cauchy Schwarz Cauchy Schwarz, what do you need one S? What S? Why do you need one S? You lose a little bit it's just a symbol for something smaller than one I don't care what constant I have as far as a universal constant so all I'm trying to say here is that I lose a little bit in the constant here and I increase a little bit in the constant there What information do you have on the Y dot Y is less than one so I can throw out the Y and the gradient is in L2 so it's not an infinity problem so the Y doesn't hurt me what hurts me is that I can't integrate in S but now I'm gonna so once I have this bound the first thing I say I'm gonna throw away this thing after all it's bigger than one it's only bad it only grows if I bound it from below I can throw it away and now I just choose epsilon to make the two terms equal and that's the proof so is this okay? so you just have to have faith because after all the computations are not difficult but you have to think that it will work but the key the key point that we realized later was the fact that the dds of this W has this direct relationship with the Mora-Witz thing it has to do with the scaling of the equation alright? and this does work also for wave maps this kind of argument gives you control of the Mora-Witz type estimates in wave maps which are in the work of the servants and the taru and Tao they are very important and this gives another way of looking at that too okay all these computations have nothing to do with criticality you could have it for with criticality it's the scaling playing against the criticality otherwise you get other numbers then the thing that is a little bit the only thing that you have from is the W5 in the equation no but it's also what is the expression that appears here because this will always appear so what I will say though is that you are right to a point this can be used in subcritical cases too that in fact Zag and Merle used this in some subcritical cases now what it buys you is unclear anyway I think this is a new perspective on this kind of argument okay and if you look at the paper I will flash the slides now one thing we are going to do next the reason I used t plus equal to 1 here is because the calculations were done in my paper with Merle at t plus equal to 1 but going on it's more convenient to instead of using this picture to use this picture but this being the blow up it's just a little bit more convenient and the effect it has is that this minus here becomes a plus pointing in the opposite direction okay other than that it's all the same but I didn't want to redo all the calculations so I just borrowed that okay so this is what I'm saying here and I'm reminding you about the singular set and the weak limit but now the blow up point is t equals 0 okay so the picture is so the estimate now had the power three fourths we got it now with the power one half but they are equally good for what I'm going to do next but one half is better than three fourths because it's smaller and hopefully there's further improvements and the only thing different now is instead of 1 minus t you have t and that's the point of using that picture t is nicer than 1 minus t and the sign is the opposite so I'll go through the proof here which it wasn't complete but it's a very lengthy thing and you don't really see clearly where things are coming from I think this calculation is quite illuminating okay so now let's move on so now we have to improve things so that's the process is that okay you get something and the reason this is a very nice estimate is I'm saying this again here is because this thing grows slower now in average than this log and so it has to vanish at some rate on average okay and so one can think of this as a Tauberian argument okay so this is telling something about the Cesare Psalms and we want something about the object itself so that's where we need to pass to a subsequence right the classical Tauberian argument says that if you have Cesare's ability for a subsequence you have convergence and this is the reason why we now need to pass to a subsequence so it's actually a very classical thing and it's very easy to trace why okay so let's move on then so now we're going to do the actual Tauberian argument in this context and of course we're going to use a bit of modern technology to do that we're not going to follow the original things we're going to use the hardly little with maximal function for example that actually is comes in very handy here so now there's some real variable type of arguments the next lemma tells me that I can find the sequence remember the time is now going to zero instead of to one okay so now I can find a sequence of times and I'm going to get two different sequences of times and if you recall in the proof of the extraction of the scattering profile we had two times, two sequences of times one was Tn and the other one was Tn minus alpha over 10 because we needed to do some integration and we needed to control both endpoints in the integration and the reason for this is similar here and what we will do is first produce these objects such that the averages are actually going to zero and not only that but something enhanced where you can also integrate a little bit bigger in T still take the average and a little bit bigger in X still take the average and those goes to zero and we want to choose these guys so that they are actually separated you think of an interval and you're choosing you split it into three parts and one of them is one extreme part and the other one is the other extreme part so this is a very standard thing that you do in this kind of problem okay except the numerology I don't know don't pay too much attention about the fractions it's the it's the point of the thirds of the interval okay so that there's a separation between them and they are of the same size okay so we're going to produce a sequence, two sequences with these two properties okay why is it important that they are the same size because you need that there are differences of the size of each one the difference is the length of the interval you need the length of the interval to be okay so how do we do this okay so now we're going to use this more always estimate first so we pick a big capital J and we look at T1 equal to 4 to the minus J and T2 equals 2 to the minus J we will apply it in that case so then T2 over T1 log gives me J to the 3 fourths we could get J to the 1 half but we don't care at this point okay and then I'm splitting it into the intervals so this is just a splitting of the interval between 2 to the 4 to the minus J and 2 to the minus J into equal size intervals now since the sum is smaller than this by pigeonhole there's one that's smaller than the minus 1 quarter which is 1 minus 3 quarters because there's J of them okay so this is just the pigeonhole argument if all of them are bigger than this I have J of them I have bigger than J times J to the minus a quarter and that's J to the 3 quarters but it is smaller so one of them is big one of them is small if they are all big I get a contradiction okay so there's one little J like that and now I look at the sequence made of the 2 to the J's and 4 to the minus J's where these are the chosen mu J's and I choose a decreasing subsequence okay so this is a sequence that's going to zero and so the 3 four is because you are using the estimate with the log the 3 four that's exactly otherwise it's one after the other okay and this is precisely the way that you use such an estimate where instead of having if you had J here you couldn't do this there's no way to gain okay so this is what we do we get to the J to the minus of fourth and then we can make this a decreasing and we get that this will tend to zero so now we did the first part now I want to call G of t this function of t which is the thing that I control by the Morowitz estimate and which I just gave better control and now I know that this averages of G of t tend to zero that's how I chose the mu J's so since they tend to zero after passing to a subsequence and relaveling the indices I can assume that they are four to the minus J okay just a simple argument and now I look at the hardly little maximal function of G times chi of mu J up to 2 mu J okay so that's a function and now I'm going to use the weak type 1, 1 inequality so the measure of the set of t in mu J and 2 mu J where the maximal function is bigger than 2 to the minus J is less than or equal to 2 to the J times 4 to the minus J times mu J 1 norm so that's why I have 4 here and 2 here and I'm left with a 2 okay just a weak type 1, 1 inequality so what this tells me is that the set where it's big is getting very, very small so there are a lot of t's in which it is small and that's how I choose my t' there's a lot of room because I have a very small fraction of the total and so that's why we can choose them like that and now the G remember was integrated only up to t why can I go to c times t because between t and c times t, u equals v and v is a regular solution for the regular solution and the maximal function is exactly what you need to control these things this is exactly the maximal function it's just the definition of the maximal function so now we have our two sequences of times we will first use just each one of them and then we will put them together so the first step is to use each one of them and we will get a decomposition for one and a decomposition for the other then we will use those two decompositions to get some improved estimates and then we will find a new sequence of time you have to be doing things gradually you can't get everything at once so now we can give a preliminary decomposition so suppose we have a sequence tn such that this condition star holds then we can have a preliminary decomposition so we have a j0 scales centers strictly less than tn Lorentz parameters which are less than 1 traveling waves such that u is the regular part plus the modulated solitons plus an error but in what sense is this an error it's only an error as an error in the sense that it's L6 norm goes to 0 remember I want the h1 cross L2 norm going to 0 but first I'm going to just get the L6 norm going to 0 so this is the first step so I'll explain how you do this the L6 norm is only for epsilon 0 right epsilon 1 is not in N6 but epsilon 0 is okay can you prove this more with these stricter snores for free or not no we have to work very hard for the stricter snores we'll get that at the very end which is even less than an error so what we will do is we will see but this comes at the end at the step before the last one and then from the stricter snores you have to go to the actual energy okay but you will see this will see next time that's a lot of extra things are needed for that let's see let's go with this for now let's try to see this I mean this is the starting point we're going to use an inequality star remember this more of a type inequality but now we chose the interval as well so the first point is we pick a cut-off function phi which is 1 on B3 and support it on B4 and now I'm going to look at u0N u1N which is u times phi of x over TN okay so I truncate my u at time TN the TN is the sequence of times for which I have this enhance more of its estimate okay now I claim that it's enough to prove the decomposition for this sequence why is that? because the difference this thing equals u-v plus an error that goes to 0 why why is that? so let's start basically there are two cases suppose that we look at x bigger than TN okay if x is bigger than TN u of TN and v of TN are equal so this part is 0 what happens with this guy? if x is bigger than TN u is v so v is a regular solution and so for a regular solution since I'm truncating up to a small size T7 it goes to 0 so that's the part where x is bigger than TN how about the part where x is smaller than TN? when x is smaller than TN this guy is equal to u because of the choice of phi so all I'm left here is v but v is regular and it's in the region where x is smaller than TN so everything goes to 0 so that's how you prove this statement okay so that tells me that all I have to do is handle this term r0 is the amount where you don't meet any other singular point okay you can think of it as 1 if you want okay so that they don't interfere with each other okay so now we go on so I'm sorry but we have to use the profile decomposition at this point so what we do is we do a profile decomposition we decompose our sequence into these blocks and they are linear blocks plus an error which tends to 0 in the dispersive sense so far so good I'm going to divide the t's tj's I'll assume that they're either identically 0 or the limit goes to plus or minus infinity okay I always can do that and I have the of course the pseudo orthogonality conditions the first observation is that u0n u1n goes to 0 outside x bigger than TN is that because it's u cutoff more or less at TN and outside TN u is v so this is really v and since I have the cutoff on v the cutoff is up to 3TN then this is small so this is true so since the u0n and since the use of n01 are localized in x less than or equal to TN that gives me some control of the parameters in the profiling composition okay so this goes back to Bahoury and Gerard you just have to take my word for this so what you get for free from this Bahoury Gerard the accent got put in the wrong place it was not a French speaking typist okay so what you get is immediately from this from the localization property all the scaling parameters are controlled by constant times T7 and the translation parameters are controlled by T7 because away from T7 nothing is happening the thing is just zero so everything is controlled like that and the other thing that you get and this is kind of handy I mean it's not too terribly important but it is handy here is that this limit is non zero for at most one index J that can be only one where the lambda Jn equals T7 then is what localizes the sequence okay and we will call it there's only one for which this is not zero okay so there's only one for which the scaling is self-similar okay that's what this means and for that one if it existed we could change the profile again to make them equal by rescaling the profile and now by extraction this sequence is bounded right because of this property so the limit will always exist I'm not saying now that it's less than one yet it's bounded so I can assume by extraction that all these limits exist okay now we will divide the profiles into three cases the first one is suppose that there is a TJ0 that there is a J0 for which these two things are equal what we will do is in this case the associated non-linear profile is actually exactly self-similar and with compact support and because of the theorem that Merrill and I proved those things cannot exist so there can be no self-similar profile that's the first case I will explain how you prove this I'm just trying to show you the big structure of the proof the second case is still the TJN in 0 but the lambda J is much smaller than TN remember TJ the J is always bounded by a constant times T7 okay so they're either comparable or lambda J is much smaller and the last case is when TJN is not 0 but when it is not 0 it means that TJN over lambda JN goes to infinity because we said that we could change the TJNs to be either 0 or going to infinity okay we will see that the profiles from case 3 can be put into the error they automatically have small L6 norm okay I will explain that in a second so we will not need to take care of these profiles but in this case in case 2 we will show that the LJs are all less than 1 and that the profile is actually a traveling wave and once we have this the decomposition follows alright okay so once this is done the result follows for case 3 the point is that for any linear solution the L6 norm goes to in the energy space the L6 norm goes to 0 as T goes to infinity and if minus TJN over lambda JN goes to infinity that's where the profile the linear profile is evaluated so each one goes to 0 but we can combine them because we have a Pythagorean expansion for the L6 norm to the 6 right we and this is the proof that the L6 norm goes to 0 it uses finite speed of propagation this proof there's other proofs this proof just uses finite speed of propagation and the dispersive ester so because of these things the third case can be moved into the error term so we just have to understand the first two cases now the other point is that in the second case the case where we have the Q's there can only be finitely many and the number that you can have depends only on the H1 cross L2 norm because if you recall for all non-linear elliptic solutions the gradient has a lower bound the lower bound is the lower bound of W so if you had too many of them by the Pythagorean expansion you would violate the boundedness of the H1 cross L2 norm so that always gives you an upper bound on how many of the Q's you can have so now I'm going to try to explain why in the first case we have the self-similar case which is not possible and in the second case we get the solitary wave for case 1 so we're in case 1 so that means that lambda j0n equals T7 and Tj0n is 0 now in this case it's not hard to show that Cjn0 since they're going to 0 you can take them all equal to 0 by changing the profile and that the linear profile and hence the non-linear profile have compact support of size 1 so after using properties, the standard properties of the profile decomposition and the theorem you can see that the more or less estimate gets inherited by this non-linear profile and what we get is the T derivative, the x over T plus 1 derivative, the u over 2 dt and this now has to be identically 0 because you make the n parameter go and the scaling gives you this this is true not everywhere but in a sufficiently big region so now we get that this thing is identically 0 now this is a first-order equation so you can integrate it by characteristics and so now you can say what this means so we have this first-order equation and there's only one kind of solution this is a formula for the solution for some functions Psi and since uj0 was compactly supported Psi is compactly supported now we also know that uj0 is a solution of a non-linear wave equation so this since it equals it is a solution of a non-linear wave equation so this is a self-similar solution to the non-linear wave equation with compact support the theorem of Merlin and I shows that it is 0 so this profiles couldn't have appeared because they lead to a contradiction now let's look at the other profiles case 2 and let's say that the one we're looking at is the first one now t1n is identically 0 lambda1n is much much smaller than tn and l1 is this limit now because of the fact that the lambda1n is so much smaller than the tn this term does not contribute when you plug it in that will go to 0 by the scaling and then you can prove that x over t see where l1 goes goes to c1 over tn the profile is concentrating around c1 over tn so x over t becomes l1 so in the limit in this non-linear profile decomposition you get that this guy verifies that this is 0 so that means that now the first order equation is this first order equation so that means that it's a traveling wave and it's a traveling wave solution and we have proved with duke and merrill that traveling wave solutions are exactly solutions of the elliptic equations Lorentz transformed according to this direction so that already you have to use that theorem to show that l1 has to be less than 1 what we showed is that if you have a traveling wave the speed has to be strictly less than 1 and here is a solution of the elliptic equation Lorentz transformed so that's how we get that this guy is the traveling wave and so this now just gives the decomposition in this preliminary form but it gives it for the two sequences of times that we constructed because it gives it for any sequence of times for which this average goes to 0 this Morowitz type average and the reason we want this averages instead of for each t is that in this computation of these limits you want to take weak limits and so you need to use compactness and if you go to two variables l2 log is compact if you are contained in h1 in space time and that's the reason why you need to pass to the averages okay to be able to use that local weak convergence gives strong convergence locally but that okay that works in fact this is something that we had used with Duke Aére and Merle in our first paper in the radial case passing to two arrows instead of one now we are going to improve the bounds on the errors using virial identities okay we have to switch the sequence of times so how do you do this so the claim is now there is another sequence of times such that for this sequence of times we have a decomposition and the error in addition to having the l6 norm going to 0 has some further good properties the tangential part of the gradient goes to 0 remember that anyway that outside t sub n everything goes to 0 the gradient outside t sub n goes to 0 that's no news but for any slightly smaller ball it also goes to 0 so the energy is concentrating near the boundary of the ball the same is true for the t derivative and then there is a third or fourth property which is fundamental which is that your solution is becoming outgoing which is there is a relationship between the spatial derivatives and the time derivatives and this is precisely the expression in the Morowitz formula and that is going to 0 so that's the first thing that you do so I will show you now how to do this but I want to address your question about the Strickards' norm the Strickards' norm doesn't come at this time what happens is that once you have these properties then you can prove that the Strickards' norm goes to 0 but not before you have these properties but if you prefer the composition it comes with a smallness but there is the extra profiles that we are throwing in there remember that we threw in the profiles that were scattering profiles that's the point is we have to kill those and we kill those by using this ok the proof of that I don't know how much of that I will describe is in the same spirit as the extraction of the scattering profile ok and what allows you to make that succeed is the fact that you have this properties so you go back to that proof and you can extract what you are using and it's precisely that you have these properties ok so let's say those profiles are at infinity alright but let's not skip steps let's do this ok so that's what we are going to do today the last thing we are going to do today is this so to do this is where we need the two times so what we will do is prove a further estimate at these two times for which we have the decomposition and the only way we can prove it is by using that you already have the decomposition but fortunately it's enough to only have the L6 norm going to 0 ok so we recall that we have this that's the the kind of what we get from the Morowitz estimate using the maximal function and then we have these other two times for which we can put the maximal function here ok this is what we have and at each one of these two times therefore by the previous theorem we can do a solid decomposition where the error now goes to 0 and L6 ok so we have all of this all the orthogonality conditions and this goes to 0 and L6 we will use improved L2 estimates at those two times in which I have this decomposition so I take a small epsilon and I'm going to estimate the L2 norm of my solution in that ball of radius TIN on which I have the decomposition so in the decomposition I have the V I have the epsilon and I have the the solitons and the solitons I split the L2 norm into the part inside this union and the part outside this union ok and those are the balls centered at the center of the soliton with the radius epsilon T sub IN where the epsilon is the number that's given now remember that I have an a priori bound on how big the j's are I can just they were a fixed number that depends only on the L2 norms the supremum of the L2 norms ok so this part is just the decomposition and the triangle inequality now I'm going to use helder inequality to go from the L2 norm to the L6 norm so I go to the L6 norm and if you do the calculation because it's three dimensions it's L6 the ball volume of the ball is T to the cubed you get TYN and here TYN ok that's just a helder the passage from L2 to L6 now here I do helder in each one of these balls it's the union I I use each one of the balls and the balls are radius epsilon TIN so I get epsilon TIN here remember these guys are completely orthogonal so I can treat them as if the sum of the sum is the sum of the L6 norms ok so this is fine and then the last term outside I just used the fact that this is of radius TIN and I get that now I'm going to recognize what happens to each one of these guys this norm because v is regular goes to zero with n because I'm shrinking the support this one I know that goes to zero with n because that's what I proved in my decomposition that's the property that I have this one well this is just a uniform constant now because I know how many I have and this one remember we have good point wise bounds and the gradient point wise bounds now this one because I'm away from the center by an amount epsilon look at the L6 norm I gain power of epsilon ok and so I have either little o of TIN for this term and this term or epsilon TIN for this term and this term so the conclusion since this is true for each epsilon is that this ratio goes to zero because I had little o of TIN and epsilon TIN is arbitrary so what I've now shown is that this L2 norm is going to zero with a faster than TIN and I I mean this is something you want and you cannot prove it unless you have the decomposition already but fortunately you only need the L6 norm in the air ok now we're going to improve now we're going to use that we have two times and I'm going to do a virial type argument so to do the virial argument what I do is I multiply my equation by U and integrate over X less than T and T in this interval ok because I multiply the equation by something I get zero so I got this three things so I get all of these things and now I'm going to integrate by parts T and in X when I integrate by parts in X what do I get here is the no it is a U what I get is the flux times U that's from the divergence in X and then from the T terms I get UDTU UDTU and then I get what I didn't integrate by parts which is this solid integral ok so now I'm going to show that this three terms go to zero faster than a T2n over T1n this is where we want the T2n minus T1n is of the size of each one ok pardon me the first identity comes from intelligible parts no no because this is an identity when we take it and it falls here or it falls here and then I subtract and here the Laplacian falls here or then I have the gradient squared and then I add it and then I get U to the six because that's U to the five times U ok so I really literally just straight zero I didn't integrate and now I integrate alright the object of this is to try to show that this in average is small and so I do that I look first at this at the term at the surface integral which is this guy I use Cauchy Schwartz and then I get the flux to the power one half and I get the U squared to the power one half here I could have used the Harley term but somehow I use the the L6 term by using Helder here I get the length and a little low of one so I get a little low of t to the 2L minus t1 ok so that because of the control of the flux gives me that this goes to zero faster than the the average goes to zero for this term where do I get the gain I get the gain from the U squared but remember that this term appears only at the endpoints of the time integral and it's at the endpoints of the time interval where I already know the decomposition and so I can prove this L2S term so I get it so the conclusion is that the average of this quantity tends to zero and I also had this and after passing to a subsequence and numbering something that goes to zero can be 4 to the minus n and another thing that goes to zero can be 8 to the minus n and 8 is twice 4 ok alright so what do we do next this one you can do things initially you think because it's a coercive quantity it's bigger than or equal to zero thing this guy is not a coercive quantity so now we have to use another argument and the argument that we use to handle this non-coercive quantity is an argument that Howard Jia and I had in a previous paper on wave maps in the equivariant case ok so we pull back another argument here so are you ok to go for 5 more minutes yeah ok let's go for 5 more minutes see what happens if people want me to repeat next time I'll repeat some other things ok oops so I look at this guy and again I use the maximal function and I get that the maximal function where it's bigger than 2 to the minus n is more than 4 minus n times the length of the interval ok same argument as before from so now I'm going to see what to do with this with the other term this kind of a little bit bad term initially we know that this thing is bounded so that's at least we know that and what I'm saying is that this bound together with this average bound implies that the thing has to be that the set of points where this object is small is big is substantial ok and this is a real variable argument this is a standard kind of probabilistic argument but the thing that you have to do which is a little bit different than usual is that you don't know that this thing has a sign but you replace that by the fact that you know that it's bounded ok so you have to split the thing into the positive and the negative part and argue with those and I gave the argument here but let's assume that this that we believe this this is really a probabilistic argument a real variable argument so and since this here where it's big is small and the set of points where this is small is large we can combine and have a sequence of points where both things happen at the same time where both favorable events happen at the same time that the maximal function goes to 0 and this thing here maybe will not go to 0 the limit will be less than or equal to 0 because I have a sign here ok so from these two things ok so here I am telling you you show that the maximal function you can find the sequence at which the maximal function goes to 0 and the lim soup of this thing is negative because it will be smaller than 2 to the minus for every end so less than or equal to 0 the lim soup so now what happens now what happens is that at this sequence the end because I know this I can do the decomposition but just with an L6 error ok now I will use the fact that for a solitary wave this thing is essentially 0 so I kill all the solitary waves so that tells me that the error has to have this property but remember the error goes to 0 and L6 so the negative the L6 part is thrown out so here is the fact for any elliptic solution for any Lorentz parameter this quantity is 0 so this is a crucial identity that needs to be used here and how do you prove that? Well you calculate I mean we know that this is 0 from the elliptic equation and then you see how the Lorentz transform affects this and if you are patient and you do the calculation you get that this is always 0 and the probability of the parameters and the fact that the L6 norm goes to 0 and for the regular part there is no contribution because it is regular and this is on a smaller and smaller set so this goes to 0 then we get this this thing and the next thing is of course we use the Morawitz thing that tells us already that this goes to 0 because we know that this in L2 goes to 0 also divided by Tn whenever you know that the L6 norm goes to 0 in the error you get that this goes to 0 then you get rid of this thing then you know that this thing is going to 0 and now we see what effect that has on the solitons remember that for the solitons because they are traveling waves this is 0 you know that there is a connection between the L and the C's and the T therefore and you know that this thing is concentrating so X over Tn is really Lj plus a small error in the place where the soliton is concentrating so that tells me that when I oops when I do this calculation on the soliton I essentially get this plus a little error but this is 0 because it's a soliton so here we are using where the Lorentz parameter and the translation parameter are linked to get this so I think I am going to go one more step so what we conclude at the end of this for this error we have these two facts because on the solitons you get 0 and on the regular part you get 0 and you have it on the U so I will show you next time how from these two facts we can get all the conditions that I stated at the beginning so we will start with that next time and then we will see that from all of these conditions you can eliminate the profiles that scatter to to infinity in the error and so then you will get that the error also has the strictest norm going to 0 and then you have to come up with an argument to show that the energy norm goes to 0 because that's what we are looking for and here comes a true channel of energy argument remember that when we discussed the radial case I said that the channel of energy is not true in the non-radial case for the linear equation what happens is that for the linear equation once you have the additional properties that these errors have you can prove a channel of energy and I will show you how to do that next time and that's how you then show that the thing goes to 0 in energy but you need all of this preparation to be able to get to the channel of energy and it all matches I mean somehow it's a it was meant to be ok so we stop for now then thank you for your patience 3D is relevant somewhere no this works the same way for 3, 4, 5 and 6 now for higher than 6 there is a problem having to do with how much smoothness the non-linearity has and then it should still be true but it will be more technical and I don't think we want to be more technical than what we are already ok but it's only that point so in your no, I estimate you don't really need that factor you need anything that would give you a little over the channel yes, yes there is quite a bit of room there but I should say that the better estimate that you get there the more control you will have on the sequence of times that you can pick the more chance you have of passing to a general sequence so that is the real reason why we are desperately trying to get rid of that log even we would like to show eventually that this infinite integral is convergent and that may not exactly be true but there will be some version of that and then if that is true then you can choose a lacunary sequence of times and you can choose a lacunary sequence of times then you have much more of a chance to passing to a general sequence but many years may pass until that happens well, yeah, but it's been 10 years