 So maybe I do a little bit of a summary of what we've done so far. By the way, after today, or at least soon after, the notes will be posted on the IHES website. They will be the notes up to today. The other ones are not ready. Okay, so what we've done so far is we introduced the energy critical wave equation in the focusing case as a good model in which to try to study long-term behavioral solutions for large solutions. And so the first thing we did is we dealt with the so-called ground state conjecture. So the ground state is an elliptic solution which has minimal energy among elliptic solutions. And so it's a static solution of the nonlinear wave equation. And we proved the result that if the energy is smaller than the energy of the ground state, two things can happen. If the gradient is bigger than the gradient of the ground state, then you blow up in both time directions. If the gradient is smaller, then you exist globally and scatter in both time directions. And this was done through this concentration compactness rigidity theorem method. Okay, so that was what we did. And then now we're going to go to trying to understand how the dynamics behave above the ground state level of energy. And so for that we start in the radial case. And this is what we're going to do today and show that there is a complete soliton resolution in that situation for solutions that remain bounded in energy. And last time I introduced this tool, this channel of energy method, which is a way to measure dispersion and is the key tool in this proof. Okay, so maybe I recall. So first I start recalling a little calculation. This is just nothing more than an integration by parts. It says that the integral from A to infinity, now the R is inside the DDR. Oh, remember we're doing all of this in 3D. Okay, and I made some comments as to what happens in higher D. So there's this simple calculation that the DDR expression where the R is inside equals exactly the DDR where the R is outside minus this constant term, A H squared, reflecting the endpoint. And it's important that it's a minus here, so this is always smaller than this. And as a consequence, this channel of energy inequality tells us that for radial solution of the linear wave equation in 3D for T positive or T negative, when you look at the energy outside for R bigger than absolute value of T plus A, you get that that's controlled from below by this expression which corresponds to the time equals zero thing. Okay, and this is true for all T. So for all time you remain having this block of energy outside R bigger T plus A, either in the positive or in the negative direction. Okay, and the proof of this is a straightforward integration argument. I mean it's a completely elementary fact. Okay, so now we're going to see the key point in this radial soliton resolution is that this kind of channel of energy exists also in the nonlinear setting provided you're not the one soliton, which is doubly. And that's the characterization of doubly. Okay, and then using this you can do the job of proving the soliton resolution. So I introduced this notation that's convenient for us where we truncate your solution outside. So all of these arguments are done by going outside somehow because the inside is unknowable to us, but we can tinker with the outside. And that's very well adapted to find its p-thepropagation. Okay, so you tilde is U0 for R bigger than capital R, and then we fill it in as a constant in the middle. And U1 tilde we're just brutal, we make it zero in the middle. And the point is that the H1 cross L2 norm of this new function is exactly the outside part of the H1 cross L2 norm. And the first proposition is U is a global in-time radial solution of the nonlinear wave equation in 3D. Okay, and suppose that both of these limits for both for negative time and for positive time are zero for some R. Then the consequence of this is that you have two options for the data. It's either compactly supported or it equals W except for a compact set. So for R large, the data has to be either zero or W. And that's the only way you can have both of these limits to be zero for some R, okay? And of course, if U0 U1 is compactly supported, let's say in the ball of radius R, then this is true because it's just zero there by finite speed. And if it equals W outside the ball of radius R, this is also true because for W this is true, okay? So this is a way in which we can capture W, okay? So the first part is to prove this, okay? So the proof of this is sort of an elliptic proof in nature. So you argue as if you pretend that you're studying elliptic equations, okay? And this is in some sense has some philosophical similarity with the moving plane method of Giedesnie and Herbert, okay? But of course we're doing hyperbolic PDE but still, okay? Okay, so there are two lemmas. The first lemma says the following. Suppose that U is, oh, I wanted to make one little comment here. It's a trivial comment but it will help us understand this. Of course, if it's true for an R0, it's true for any R bigger than R0 because those are smaller quantities, okay? So just keep that in mind. It's not that I'm wedded to the same R. I can always make it larger if I need to, okay? So now let me take U as in the proposition and let me call V R times U of RT. You may recall that I mentioned earlier in the proof of this outer energy lower bounds that for a solution of the linear wave equation in 3D, which is radial, if I multiply it by R, I get a solution of the 1D wave equation, okay? And this is something that is in the back of our minds. So the statement, so I call that V, V0, or this should be V1 to be the corresponding Cauchy data. And my assumption is that the outer part of the H1 cross L2 norm from some number R0 on is small enough. Okay, this delta naught will be a sufficiently small number. And of course, if I have any data, I can always make this true by going far out enough, no matter what delta naught is, because it just goes to 0 by 5 netness. Okay. Then I can bound this outer H1 cross L2 norm in terms of V0 to the 10th divided by R0 to the 5. Okay, and this corresponds to the fact that the non-linearity is U to the 5th. Okay. And the second part is that the differences in V0 can be controlled by this. And this, in turn, can be controlled by that. So let me first point out that these bounds follow from this one. Okay. The fact that this follows from that is nothing more than the fundamental theorem in Cauchy-Schwarz. Okay. So this part follows from this by fundamental theorem of Cauchy-Schwarz. Okay. Now how do you pass from this to that? Remember, V0 over R, V0 over R squared, I'm sorry. Yeah, so I'm going to write V0 over R to the 5th over R squared as V0 times V0 to the 4th over R squared. Okay. I peel off one V0. And V0 to the 4th over R squared is V0 squared over R squared. And V0 squared over R is R times U0 R. And we know that R times U0 R is controlled by this kind of expression. And so therefore it will be delta naught. And so that's how you get the delta naught squared. Okay. So this is exactly the calculation I started out by showing you. I'll show it again so that you see it again. It's this calculation here. Okay. So in particular this thing is bounded by that. So those are the preliminaries but there's still something to show, right? Things haven't gone away. And this is quite an important estimate. So the idea is that knowing information from the solution of the nonlinear wave equation at infinity on both time directions gives you information at time zero. Okay. So this is what's happening. So how do we prove it? So this is for the passage from the top inequality to the second one. So to prove the first inequality we do the corresponding thing. We did this with the nonlinear solutions. We also do it with the linear solution. For the linear solution I have this outer energy inequality. That's precisely my corollary that I showed at the beginning. Either for t positive or for t negative I get this lower bound. Okay. And now I truncate at r zero and let u tilde l be the corresponding linear solution. Now here I use the local theory of the Cauchy problem. I know that if this delta zero is small enough the linear solution is close to the nonlinear solution. And that's the good thing about going on the exterior which because on the exterior it can make things small and therefore use the local theory of the Cauchy problem and then pass to the stage where I haven't done a perturbation by finite speed of propagation. Okay. So this difference is bounded by this to the fifth that comes from the local theory of the Cauchy problem and this to the fifth by definition is this. That's my definition of u0, u1, u0. And now u0 and v0 are connected. v0 is r times u0 and by the calculation I started out with this equals that. So these are all equalities. Okay. Now I know my... Okay. So the difference is controlled by this. So this one in particular is bounded by twice this plus that. Okay. That's the triangle inequality. Okay. So this difference is smaller than this. So each one of them has this kind of bound where this is the bound for the difference squared. Okay. I'm just using a... What am I using? a times b is less than twice a squared plus b squared. Okay. But by finite speed of propagation having truncated doesn't affect either the nonlinear or the linear solutions in this region. And now I use my lower bound on this linear thing given by the outer energy inequality, which is true for either positivity or negativity. I combine this and I had a one half and so before I had a two now I have a four. Okay. So I go to the initial data. And now I have this inequality for either t positive or for t negative. But both of those limits on this thing are zero. So I choose t positive or two negative according to which one is used in this inequality and for either one the limit is zero. So I pass to the limit. And therefore what I have is that this is bounded by this. Okay. Because this two guys don't have t's. So I pass t and the t disappears. Now I split this into this term and this term. If you think about it this is the term I really want to get. That was the stated term in the inequality. It was expressed in terms of v but when you translate v in terms of u you get that. And now this is the fifth power. And this is the one power. But this thing is small because it's bounded by delta naught. So the fifth power is bounded by a small constant times the one power and I can hide it. So this is the end of the proof of this bound. Okay. So let me go back to the statement so that you see that this is what we have proved. We know that this is small and then we obtain that this is small. I'm sure that the total energy is smaller than that of the... No, no, no. No, I avoid that by only working with the outside where I can make things small. I'm now working with large data. I don't care about the size. Okay. It doesn't affect this. Okay. Okay. The second statement is that v zero is like before. And now I can show that v zero has a limit, l. And I can tell you at what speed the limit is attained. So the proof goes in... There's a few steps involved. So just be patient. The first thing is eventually if I'm going to have a limit it's going to be bounded. Right? If I get a limit it will be bounded. But I can't do that in one stage. So I first are going to prove a weak growth condition at infinity. And the growth that I prove is by the power one-tenth. Of course by taking delta not small enough I can improve the one-tenth to any epsilon that I want. But one-tenth is enough. Okay? So why is this true? Well, this is just a triangle. I'm going to do some kind of iterative argument. And it's in this style that elliptic theory works. Right? Do iterations. Think of the Moser theory. Okay. Or the Georgie. So we go to 2n plus one. We want to compare with the previous stage which is 2 to the n. We use the triangle inequality. But now remember we have a good bound for differences. My good bound for differences gives me this. The one comes from this term. And the difference term has a c0 delta not squared. That was the last bound in my lemma one. Okay? But now all I have to do is choose delta not small enough such that this is bounded by 2 to the one-tenth. This is a number that's bigger than one. This is another one that's bigger than one. And I just choose this small enough such that this is true. Okay? And then I just iterate. And iterating this gives me that v0 to the 2n is bounded by 2 to the nr0 is bounded by 2 to the nr0 to the one-tenth. So I get the desired growth in this dyadic sequence. But I have my difference estimate that allows me to estimate things that are close to each other. So from the dyadic estimate I pass to the general hours. And that's how I prove this. Now this still doesn't give me the existence of the limit. But let me now improve to get the existence of the limit. Okay? So how do we prove this? Well, we know this difference estimate. But we know also that this is smaller than the power one-tenth. And one-tenth to the fifth is one-fifth. Right? It's a one-half, I'm sorry. So then you see that we get 2 minus five-tenths, which is 2 to the 3n over 2. And now that's a convergent series in n. And so I can just telescope things. This sum is finite so the limit exists. So I got my limit existing along a dyadic sequence. But I know that the oscillation on dyadic intervals doesn't change too much. So I get that my limit exists everywhere. Okay? But now that the limit exists I improve my first bound. Now it's bounded because the limit exists. So in my bound the v0 to the 5 here gets replaced by c. And now I do the tail sum and you see that v0 over r minus l is bounded by 1 over r squared. Okay? Just by telescoping some. So let's go back to... I'm going back too much. So we've proven this. You can see we're using very simple considerations here. It's not high-tech here. So now I will prove... Now I will prove proposition one. Proposition one says that if those two limits at t going to plus and minus infinity are zero, your only choice is that your data is compactly supported or that it is w outside. Okay? So that's what we have to prove. We will use both limits. So here we distinguish the limit is l. What happens depends on who l is. If l is zero then we have to have compact support. And if l is different from zero we will get a plus or a minus w. Alright? So let's prove that. So the first case is l equals zero. So we first start out with this difference bound that we had before. We choose delta naught so small that this is less than a fourth. Now we use the triangle inequality and we see that v zero at two n plus one is bigger than three fourths at two n. Okay? It's just a triangle inequality. And now I iterate. I put the four thirds here and I iterate. So v zero r is less than four thirds to the n v zero of two to the n r. Now I use that l is zero. Since l is zero when I take the difference with l I don't get any difference and I just get that this decays quadratically. Two n r squared which is four n r squared. And now I put that input in here and I get v zero r is three to the minus n r squared for all n for r larger than r zero. And I let n goes to infinity, go to infinity and therefore v zero is compactly supported. It's supported in r zero. I still have to catch v one. But for v one I have this bound. My first lemma bounded the whole thing in terms of v zero. So if v zero is zero the integral of v one squared has to be zero. So v one also has to be zero. Okay? So that's the case l equals zero. The case l is not zero it's basically the same. So we just have to show that there is a lambda and a sign such that this thing has compact support. Okay that's what this means. So now I recall my formula for w. Of course we know how what the asymptotics of this w is because it's completely explicit. So for large r if I scale it by lambda I get that this difference is bounded by c over r cubed for r large. That's just from the formula. Okay? And this is true for whatever lambda is chosen. So now I'll choose the lambda to match the l. Okay that's how I choose lambda. So that lambda is l squared over three. And how do I choose plus or minus w according to whether l is positive or negative. And now all my choices are made. Okay? So once I've chosen lambda and the sign I get this inequality for r large where the relationship between lambda and l is this one. And now to make things easier I rescale and maybe I replace u by minus u so that I can assume that this inequality holds for r large. And now I have to show that for r large u0 is w. That is the difference the case too fast then they have no choice but to be actually equal. So how do we do that? Okay. I let little h be u minus w and capital h to be r times h. Remember I always multiply by r and pass to that equation. I will show that if r0 is large and r0 is larger than this capital r0 this outer integral for the h is bounded by a small constant times h0 squared over r0. And this constant is chosen to be small enough that if this inequality is true if we do the previous argument that we used in lambda one this guy will have to be 0 for large r. So the whole thing is now to prove this inequality. And the idea is that little h satisfies not the nonlinear wave equation because it's a difference with w but the equation that the difference satisfies. So it's like a nonlinear wave equation corresponding to the linear equation linearized around w. So that's what we do. So I'll verify this equation which is the linearized equation around w. Now the next idea is that in this linearized equation you want to show the think of this still as a perturbation of the linear wave equation. Okay. But you can't quite do that unless the potential is small enough. Of course we have space time here. So the way to make it small enough is again to chop it off on a cone. In a cone of course this w is time independent so you have to be a bit watchful there. But if we go on a cone and the cone is truncated large enough is placed large enough to the side then the space time norms of w are restricted to this cone will be small enough in fact this is a perturbation of the linear wave equation. And finite speed still applies. We can truncate both the potential and the data and use finite speed of propagation to see that it's the same as if we haven't truncated the potential and the data. Because this is all completely local provided we are or outside of light cones. Okay. And so this is how we truncate the potential and we obtain a local theory of the Cauchy problem. We show that again the nonlinear solution is a small perturbation of the linear solution. We use the outer energy lower bounds for the linear equation and then we get to that. Okay. And so that's the end of this proposition. This is still a nonlinear equation. Of course. But it's a nonlinear equation with small data. Right? Because of the truncation. It's a nonlinear equation with a potential term and small data. But both the potential and the data are small. And this is independently whether you started from something large or not. Right? This is the kind of the beauty of the approach. Okay. So the first proposition tells us that you cannot have both limits going to zero unless you're either compactly supported or you're compactly supported per duration of W. Now I'm going to study each one of those cases. The case is left out by this first proposition. All right? So I assume that it is not a compact support per duration of W. So then I claim that I have this channel of energy. This channel of energy. This dispersive bound is a 40 positive or 40 negative. Okay? And how do I prove that? Well, if those two limits in the first proposition are zero then I'm compact support or compactly supported of W. So this is ruled out by assumption so I have to be compact support. And otherwise, one of those two limits is not zero. If one of those two limits is not zero an easy argument lets me get this fact. So all I need to do now is deal with the case when U0, U1 is compactly supported. So let U0, U1 be of compact support but of course it cannot be zero. If zero I can't do this. There's no lower bound possible for zero. Okay, so I think I'm going to draw a little picture to show you the idea. This row is the diameter of the support of U0, U1. Okay, that's the definition. Definition is a little bit strange because sets of measures zero cannot count and that's all. We're dealing in U1 is just an L2 function so a set of measures zero doesn't play a row. But really this is just the diameter of the support. Now since my function is of compact support but it is not zero this row is some positive number which is finite. That's all I'm, that's obvious. So how am I going to proceed? I'm going to proceed by looking at the very edge of the support. So I'm going to look in this part. Now in this part no matter how small I take it going in here there's always some energy in here because this row is the true diameter of the support. I can't make it any smaller. So I will always pick out some energy going a little bit inside. On the other hand by having gone a little bit inside the energy is certainly not zero but it's also small. And that's how I proceed. Now I am in a small data problem and I have my linear solution close to the nonlinear solution and then I use the channel of energy for the linear solution. So let's go to the proof. So that's the idea without the, did I write details? Okay, so that's the idea of the proof. Now one important point you remember that time zero had this minus a times h squared term that comes because of the fact that I did not have the h1 cross l2 norm but I put the ddr there because the solution is zero here that term can be absorbed by the other term. So for compactly supported data the term that appears is always bounded from below by a constant times the energy. And this is an important fact. Okay, alright. So finally the third proposition deals with the case where I have a compact support perturbation of W. Okay, that case is there. And so far I haven't dealt with that term. So let me explain this thing. So it is a compact support perturbation of W but it's not identically zero meaning I'm not exactly W. I'm only W far out. So that's the situation. So and I'm going to call rho the diameter of the support of this guy. As before. So let me first read the second conclusion before we read the first. The second conclusion tells me that if rho is large enough so if it is equal to W for very, very long piece then I still have the outer the lower bound in the channel. Okay. So if for a very long bit on W I have this channel energy bound. And what does the first bit of the thing say? Okay, suppose that it equals to W almost every, you know, starting from a very small piece. Then if I evolve in time either for T positive or for T negative the size of the support will grow linearly. And so if I wait long enough I'm in a situation where I can apply the second part. That's the statement. So now let me prove it. So the second part is proven by using this linearized equation about W truncated. And that's why you need to make the R0 large to make all the norms small and the data small. So the second part we've already basically seen how to prove. The first part that has now a twist. For the second part where we want the support to grow linearly we will linearize around W but without truncating W at all. So now we won't truncate W. How do we get the smallness? By only arguing it by only arguing for small time. That's how we will get smallness. So W is globally defined. Therefore for any finite time if I'm close enough to W then I exist up to time minus R0, R0 for any R0. Because of the perturbation there. That's provided that the data is close enough to W. So this is the perturbation equation for our non-linear equation. What I call the long-time perturbation. There. Now I make the R prime. What was R prime? I need to go back to see what R prime is. R prime is again something almost at the edge of that support. I take an R prime almost at the edge and it's for R bigger than R prime that I'm going to have the growth of the support for all positive time or for all negative time. So if R prime is close enough I can still make the norm small and the new data is just a truncated data. So now I solve the wave equation with this data and I've got this identity and this defined on time minus R0 and R0. By finite speed if R is larger than this then the H's don't live there anymore. They're zero. That's where the support is and so everything is W. So my solution will be W for R bigger than this plus T for all this T. And if you think about what this means it means that the diameter of this support is smaller than the diameter of this support plus the absolute value of T. So there's sublinear growth. And now I want to use the channel of energy argument to show that the growth is actually linear. It's not sublinear. So that's another way of understanding this outer energy inequalities. They force the support to grow linearly either for positive time or for negative time. And of course that's why they restrict it to a situation where the strong Huygens principle holds. Right, well in this case three times. I'm sorry. I got excited here. So now I have to show this. That's what I'm trying to show. So I only have to show bigger than or equal because the other inequalities already show. So I look at this. I have a good local theory and I have a small perturbation of the usual linear flow and so on. So I get the outer energy inequality for this GL because it holds for all linear solutions. And this is the extra term that I was saying that when you are in compact support situations you can absorb this by that because this is just small. You can make this be smaller than a quarter and so this is bigger than a quarter times that. And this is what you get out of compact support. So now I know that for all T positive or for all T negative this is true. Now this first we get it for G which is close to GL and then we can replace G by H tilde by finite speed. Okay? So I inserted this R naught that's in between there to give myself a bit of room and make that small. Now once I have that I get the lower bound that I want instead of with rho with rho naught but since this is true for any rho naught I get it for rho. So at least I get it for a very short time and now I have to iterate. Now what happens the one thing that you have to be careful about is that this lower bound is either for T positive or for T negative and it could be that you went a little bit and it was true for T negative but then the next time is going to be true for T positive and then you can't continue but that is ruled out by the support properties. If you chase out what the support properties of this imply this can't happen so it has to be consistent either for negative or for positive and that's why you can always advance. Okay? So that requires a little bit of thinking and a few drawings and I'll skip that. Okay? So the meaning of this proposition if you combine propositions two and three is that as long as we're patient we're willing to wait a little bit of time we will always have the lower bound unless we're plus or minus W. Right? Because there are three possibilities you are compact support but non-zero then you always have the lower bound. If you are neither compactly supported nor W plus a compactly supported thing you have the lower bound and if you are W plus a compact support object then you have it for either positive or negative time as soon as you wait long enough that the size of the support is large. So all possibilities are accounted for. All right? But long enough is finite because you cannot iterate. No, it is finite. It's a fixed number R0 that comes from these proofs. Okay? Okay. So the next thing I'm going to do so we're going to use this thing to give the soliton resolution in the radial case. So the way that's going to go is through profiles. So I'm going to translate this result in terms of profiles just to make it easy to see how it's used. Okay? There's no new math here. There's just notation. Okay? So I consider a radial non-zero profile that's an object of this kind and of course I don't have to consider the zero profiles because you just throw them away. They're zero. So I only consider non-zero profiles and if you recall one of these three options are always holding after some reduction. The time translation is either zero or tjn over lambda jn in absolute value goes to infinity. You can go with it to plus infinity or to minus infinity. Okay? And now I'm going to assume let me get this thing out. So assume that one of the following holds either it's not a compact support perturbation of w or minus w or if it is a compact support perturbation the diameter of the support is large enough. Okay? Then I have the outer energy inequality. I have the outer energy lower bound for all t positive or for all t negative. Okay? So that's the conclusion. Okay. Now if the tjn is zero, r0 this is nothing else than what we've already proved. Now if we are in the scattering situation you prove this by a simpler argument. So I'll skip that simple argument. So now we have our conclusions stated in terms of profiles. Remember that the j is just a notation here. It doesn't play any role. It's to recall that it will be used as an index in a profile decomposition. So now let's turn to this solid-term resolution in the radial case. So this is the theorem. Okay? So it takes a little while to digest. So let you be a radial solution of a nonlinear wave equation in 3D. And there's no size assumption. Just a radial solution. Then there's three options. The first option is that I have finite time blow up and it's type one. So that means that the norm has a limit and the limit is infinity. In that case I have nothing to say. There's some different theory to study. All right? So that's not a very exciting thing because I had said nothing. So the second stage is again T plus is finite but I have type two blow up. That is to say I assume that the norm remains bounded as I approach T plus. Then I have a solid-term resolution. I have a radiation term and I have a number capital J. I have sines plus or minus. I have scaling parameters lambda one up to lambda capital J. One much smaller than the next one, much smaller than the next one and so on. And they're all much smaller than the linear rate. And my solution writes as sum of modulated solitons plus the radiation term plus something that goes to zero as T goes to T plus. Okay? But one thing to be observed is that the union of A and B shows you that you cannot have mixed asymptotics. There cannot be a sequence for which you go to plus infinity and another sequence for which you remain bounded. So that's already a big corollary regardless of this decomposition. But of course you use the decomposition to prove that. And then there's the third case. The third case is that your solution exists globally. Okay? In this case we have this radiation term. We have J's. This capital J now could be zero. Here there always has to be at least one solitary wave if you have finite time blow up. But in the infinite case you may not have finite. You may not have any wave because if your solution scatters there won't be any W. Okay? And then we have sines, we have scaling parameters. They're all much smaller than T. And my solution writes as the sum of the solitary waves plus a small error. And an immediate colliery is that solutions that exist for all time are bounded. Because are you assuming that the energy is smaller than that of the... No assumption. Except that the soup of the H1 cross L2 norm is finite. In this case I don't even have to assume that it's a colliery of the result. So in the global case I make no assumption except that it's global. Any other questions before I go on? So here you will be heavily using the fact that you don't cover the standard stationary solution as than the translator and booster of the value. Translator and booster of the value. Yeah. Than the translated... No translation because it's radial. So just the scales of W. So in July we will go to the case where I have translates and Lorentz boosts of W and all other elliptic solutions. But you said there were no other... Of radial type. Or in the non-radial case... Non-radial. Yeah. So that's why the non-radial case is a big zoo of traveling case. Okay? That's why the radial and the non-radial case are very different. In the non-radial case you have to account for a huge number of... I mean there's a continuum of solutions of the elliptic problem and they're not classified. So I'm just saying now I'm doing propaganda for the next set of lectures. Okay, so maybe I sell tickets at the end. All right. But I will provide notes of the last lectures too. But at the end of the last lecture. Okay? Any other questions? Should we move on to see how one could prove this from what we've done? Okay. So in the radial case I'll concentrate on this case. And then in the non-radial case I will concentrate on this case just for, you know, to vary things a little bit. All right? So the first case then is a global in-time solution for T-positives. And the first step is to show that at least for one sequence of times it is bounded. Okay? And this is basically the argument that I showed you if we go back to this ground state conjecture. There's an argument in which we said that if the energy was smaller than W but the gradient was bigger then it blew up. Remember there was this differential inequality argument working with the L2 norm. Okay? If you use that argument carefully here you will see that the limit of the gradient squared is always at most three times the energy because otherwise it will blow up in finite time and so it won't be that T plus is infinite. So the point is that the fact that it exists for all time gives you a constraint on its size at least for a sequence and this is what this is true. Okay? So now what we will do is build up on the fact that there is a sequence of times in which it is bounded. Of course we have much more to do but that's one beginning and I'm not going to give this argument because it's basically almost the repetition of the one I've already done. The next step is to extract the free wave the radiation term in the infinite time case. And in order to do that you first extract it for a sequence of times on which you know that the thing is bounded and then you show that if you can extract it for a sequence of time you can extract it for all time. So this is the following step if you use a radial solution you have a linear solution such that provided your finite distance within the light cone the energy norm goes to zero. So the meaning is that at any distance from the boundary of the light cone any global solution behaves like a linear one. So all the nonlinear action is strictly contained in the light cone. Now this is a real theorem and I'm not going to prove it now because I will prove its version in the radial case in July. In the non-radial, yes, thank you. So in July I will give the non-radial proof and I don't want to give both proofs. Okay? So we will assume that. You are not worried by the scale the value that's paid? No, no, no, no. Because it always gets rescaled in these parameters, lambda, that are much faster than the time. So the tails all go away. They have small norm. I'm not saying that it is zero outside. I'm just saying that it's small outside in the limit, okay? So you might have the tails but they are small, right? So that is an important difference between the finite time the finite time blow-up and the infinite time. In the finite time blow-up we could take away everything outside the light cone. In the infinite time we can take out only most of it. Okay? Because there's this tail. But that doesn't bother us. Eventually we'll deal with that. All right? So the next, the next day the next step is to see how our dispersive estimates imply certain things about the profiles. And this is the channel of energy argument. This is an important argument. So I have a globally defined radial solution. Then I claim that there is no sequence of times tn with the following properties. This equals u sub n outside of rho sub n. And I have a profile decomposition that starts where I put all the w's first. I have the radiation term and then I have the other possible profiles and the error. And let's assume that we're in the non-linear in the globally defined scattering case. But moreover for one of them, the ones that are not w I have for all t positive or all t negative this channel of energy or if I don't have it in the profiles I have it on the remainders. That eventually the remainders have a channel of energy. And I claim that this isn't possible. Okay? And we call it the inductive lemma because it's proved by induction. Okay, so let me... So we will assume that this inductive lemma so there's no decomposition where either from here or from here you have a channel of energy. So you see that this kind of morally forces all of this to all have been w's because we know that we have a channel of energy where the only possibility is that those things are w. And not only that, but that this error has to go to 0. And that's why this dispersive property implies the soliton resolution. Okay, but we'll see that more formally later on. But this is what we're gearing to. Okay, so let me give the proof of this. So the first thing is that you can see that this limit, the rows are how far the u equals the un's and then you compare to the time interval to the time tn at which everything holds. They have a comparability. And then the second thing is that the scaling parameters whenever you have a w have to be smaller than the tn's. Not smaller than the tn's. Anyway, so these are preliminary considerations. So the next thing that we do is we look at the solution of the nonlinear wave equation which asymptotically equals the linear solution that we're given. So it's kind of the wave operator result. And we can assume that this is defined for all positive time. So the statement is proved by induction on the number of w's in this profile decomposition. So the first case is when there is no w in the profile decomposition. Okay? So in this case, let un be the solution. And so now, remember, we have a profile decomposition where all of these guys scatter and so the solutions exist for all time. And so our approximation theorem works perfectly here. And I can write the nonlinear solution in terms of the other nonlinear solutions, the nonlinear profiles. And so the nonlinear solution corresponding to this data can be expanded as a sum of this nonlinear thing corresponding to the radiation plus the nonlinear profiles plus the remainder plus the further remainder which is very small. So assume first A, A means the channel of energy that will hold either for this or for this but for positive time. And then I will consider the case of negative time. Okay? If it holds for negative times, then this guys or this have a channel. But by orthogonality in the profile decomposition, this is inherited by that. If one of these things have a channel, this remainder is very small, so this thing has to have a channel. And this is what we conclude. That this term outside has this lower bound. That's just by orthogonality and the assumption that one of these guys has a channel or there's no uniformity that this guys uniformly have a channel. Okay? So let's see what this says. But by finite speed, remember that this U n's are just the U truncated outside rho n. And so by finite speed this is what you conclude. Now, I change variables. I call this T tilde and so x is between bigger than T plus rho n minus T n. And this is positive. This is just from this channel of energy. But remember the V and the VL I constructed in such a way that this goes to 0 for any A. So I fix T n and for any A this goes to 0 and this is a contradiction. So I can't have that lower bound. Now I have to consider the other case. The case when the channel goes for negative time. If the channel goes for negative time I still use... Oh, I'm sorry. I still use this expansion but now I use it for T equal minus T n. If I use it for T minus T n I get this minus that has a lower bound on this region. But remember that by finite speed this equals that outside here. So what we get is this. But this also cannot hold because as T n tends to infinity since T n goes to infinity this integral is outside further and further and these are fixed functions. So that integral can't have a lower bound. So the end conclusion is that I never can have this channel of energy. And that's how this... So that's how we use the channel of energy in this proof by showing that there can be no profile decomposition where some of the profiles have this. So this was just the case J where there were no W's. Now you have to treat with a case when you have a W inductively. And so what you do is you take one of those W's and you just truncate it for R large. One of them. And then you play this very same game again and you reduce to the previous case. So that's how the induction proceeds. All right. So now we come to the main step. Yes. You are saying that none of the UJs or the W can give a positive contribution to that inequality? The W cannot give any positive contribution because it's zero. Okay? So the only possible contribution comes from the others. The ones that aren't double. But if one of them has a positive contribution you reach a contradiction. Okay? Okay, so now let's go to the main step in the proof of the decomposition. All right? So how do I do it? I start with a sequence of times on which I know it is bounded. And I know that there is one at least. Okay? That was the fact that the lymph was less than three times the energy. And the energy has to be positive. Okay? Then VL is the radiation term that we've already chosen. Then after extraction we have the decomposition for this sequence of time. So we will do this sequence of time for this decomposition for the sequence of times in which we know boundedness. And then there will be a separate step that says that if I have the decomposition for one sequence of time I will have boundedness on all sequences of times. And so on. Okay? So we will get to that. So the logic is a bit intricate. Anyway, we go on. So suppose that this is not true. Then I have a profile decomposition where some of these are W's, but some of them are not. Right? That's what it means that I don't have. Or they're all W's and this guy doesn't go to zero in energy. Okay? Those are the possibilities. So again the limits are minus or plus or they're zero. And for all lambda this is not that or they're all zero. The ones that are not W, but this thing doesn't go to zero. That's my other case. And I'm going to rule all of this out by using my inductive lemma and my dispersive characterization. Okay? Now this is a dense lemma but it condenses the whole proof. Okay? So we get some control on the parameters and then to carry out the proof we split into various case. In each case we use the lemma on profiles to reduce to the situation where x is bigger than rho n with the sum of rescale W's and globally defined profiles. That's the important thing. We can go outside and all our profiles are globally defined and scattering and they create energy channels in cones because they're not W and this is what our dispersive property holds, gives. So and this gives a contradiction by this inductive lemma. Now we can perform this directly along the given sequence of times unless the profile which is somehow concentrated further from zero is a compact support perturbation of W with small diameter. That's our bad case where we cannot say that we have the channel of energy. But then we just change the sequence of times by using our proposition that says that we can grow the support. And then we use the approximation theorem to pass on this sequence of times that we started to the new sequence of times. And in the new sequence of times we get the contradiction. So in either way we get a contradiction. And of course the details of such an argument are long. But this is the basic principle of how the proof works. So if you know this from my faith you can prove it. So this proves it for the first sequence of times in which it was bound. So how do we conclude? So these arguments are not immediate but they are more or less standard. But let me go through them. So the first step is that once I have the decomposition for one sequence of times for which this holds I first show two facts. First that the gradient converges to j times the gradient of the w squared. And second that the t derivatives go to zero. If I have this I immediately get the boundedness of the energy. Normal. Right? Because if this has a limit it's bounded. If this has a limit it's bounded. That's all that I have to say. So from this I can already go to other sequences of times. But it also fixes how many w's each time I get the decomposition for a sequence of times. I have. Because j is the ratio of this and that. And this depends only on you and this is a number. So j is fixed also. So the number of bubbles is fixed and the boundedness follows from this. So this seems to, this is a big improvement. So how do we prove this? So let's first show that this is true. So assume that this isn't true. Then because along t sub n we have the right limit because of this decomposition. Along t sub n clearly this equals j times w. Because the difference goes to zero and the norm of this is the sum of the norms by the orthogonality of the parameters. So I know what I want along the sequence t sub n for which I have the decomposition. So I assume that this doesn't hold for all t's. Then there's a sequence for which t and one for which this limit is this plus epsilon. For some epsilon which is non-zero but which I can choose. Why is that? This is the intermediate value theorem and the continuity of the flow. I can go in between and choose any intermediate value that I want between the limit not existing and the fixed limit that I have. So this is just the intermediate value theorem and continuity of the flow that allows me to do this. Now let's fix an epsilon to be determined in a little bit. This epsilon will not be going to zero. It will be some fixed thing. Okay? Because of this... Yes? Yes, but if it is infinite in between being finite and being infinite it takes all the values. Right? They can always do this. So that's the intermediate value theorem that allows for that. Okay. So this guy is bounded in L2 because of this. Because VL is bounded in L2. Therefore it is bounded in L6 by the embedding. Now remember that the energy is conserved and the energy is the gradient squared plus d dt squared minus u to the sixth squared and that's a constant. The gradient squared and the u to the sixth are bounded. The energy is constant so that means that the d dt is always also bounded. So this is bounded. So now I have in this sequence I can do the decomposition because now I am bounded. Right? So I do the decomposition and therefore this limit has to be some integer times grad w squared. That's what the decomposition gives me an integer and so on. And now I choose an epsilon so that this is never an integer times grad w squared. And that's a contradiction. Right? I have a countable number of possible values and I can always miss them. So that's the end of the proof. So that's why the limit had to exist and equal j times the gradient of w squared. And the proof that this limit is zero is the same because along our sequence t sub n the limit is zero because I get w comma zero and so a similar argument. That allows me. I'm choosing another sequence. I always have this place. Right? So now I've proven this. I know how many bubbles I have and I know that I have boundedness for all time. And now the rest is basically a continuity argument. Now what hampers the continuity argument at first is that your lambdas, the lambda j's that you can now choose for any sequence but they may not depend very nicely on the sequence. They may not depend continuously. And if they don't depend continuously I can't do a continuity argument. So the first thing to do is we replace the lambdas by another lambda tilde that I'm going to construct in a minute which is continuous. Okay? So I will tell you how the scaling parameters can be constructed. So that's the next step. So I'm going to redo the choice of the scaling. So I I define now this number b sub j for any little j is the contribution of j minus one bubbles plus the contribution of the first bubble restricted to the unit ball. Okay? So this is a number. Nice number. Alright? And now I define lambda j of t which will be my new scaling parameter the infimum of all the lambdas such that this is bigger than or equal to b j. And if you think about it this defines lambda j of t to be continuous. So I have a continuous function. So what does it have to do with everything else? So the next claim is that if theta n is any sequence going to infinity after extraction I can have a decomposition but where now my lambdas are the ones given by the previous definition. So originally I had some lambdas which I didn't control. Now I can choose this lambdas. Okay? So let's prove that. Okay? So this is recalibrating the lambdas somehow. We know that there's some sequence lambda j one n for which this is true. And now I will show that the lambda one j n's are comparable to lambda j n and therefore I can swap one for the other. If our zero is positive by the decomposition this equals that. This is because I have the decomposition with the lambda one j n's. Now I look at the case if our zero is less than one then this by the definition of this this has to be strictly smaller for large n. And if this were bigger than one it would have to be bigger for large n. So if you think about that carefully what this means is that this in the limit are the same as that. But since the limit of the ratio is one I can swap one by the other at the price of the error which still goes to zero. And this gives the claim that now for any sequence I can move the lambdas and this moving is continuously. So I'm building up now what's my next problem? My next problem is the signs. That's the last thing I have to work out because I can't have one combination of signs for one t and another for one t n and another combination for a different one. And this is always the same. But this is somehow easier or well I don't know somehow not hard because this takes only two values. So by continuity once I have one value I should always be able to keep the same set of values otherwise something drastic will happen and I'll get out of where I am. And that will be the next step. Okay? So that's another continuity argument. So this is the choice of signs and end of the proof. So I'm going to consider n tuples j tuples of signs alpha one up to alpha j. And for any delta I'll consider functions which are close to this rescaled w's with the signs given by alpha j's and such that the lambda j's have the orthogonality property measured by delta. And this I'll call a sub i delta. Okay? So what is the point? The point is is that if this delta is very small and I have one of these j tuples and a different j tuple and one is in one of these guys and the other one is in the other they're apart. Because what's the point? If you look at w plus a rescaled w plus or minus w minus a rescaled w they're a fair amount apart. They can't be cancellations. And that's why these things are close enough and that gets preserved under perturbations. And that's the reason for this to be true. And once you have this it tells you that if you chose your signs originally and you start flowing and you're always close to a continuous thing which is your original thing you can't change the signs. You're applying it to u minus vl. And so the choice of signs remains forever. Once you start being close enough and that's what gives it for all sequences of times this decomposition. So this is that and that concludes the proof. I'm a bit confused about the choice of the TN. You first have a sequence of TN which can be anything. It's the one on which it is bounded. Other than that it's anything. Then you extract until you get the laundry that you want to have. No, no, so let's backtrack. First you have a sequence TN and for some subsequence of it you get the decomposition. Then you prove that the norm is bounded using this because the gradient minus the radiation of the solution minus the radiation squared has a limit which is J times the gradient of W which is something you prove. Okay? And the d dt of the solution minus goes to zero. So now you have now you prove boundedness times. Whatever sequence. Now you take another sequence of times arbitrary. Our first theorem said that whenever you have a sequence of time for a subsequence you get the decomposition. So now you get the decomposition for some subsequence and then you prove once you've introduced these scaling factors that the scalings can be changed to be these ones that are continuous. That means you extract a subsequence so as to keep just the lambda J that you want. No, no, you don't keep the you have all the lambda J's you replace them by the ones you like. Because they are comparable in size. But you can do that without changing the TN? Yes, because the the ratio the ratio tends to 1. You prove that this ratio tends to 1. And that's why you can do it. And then you have the continuity on the lambdas. And then you get the fact that you can avoid changing the signs. And now you have that the limit is the same for whatever sequence of times is. For subsequence you have a subsequence for which you have the right answer but it's always the same answer. So the subsequence doesn't matter. With condors you know that the limit is unique and therefore it is strong. It's strong convergence. But it's convergence, yeah. But it's convergence. That's all that matters. Very good. Other questions? Maybe I'll make some comments. I'm allowed to make comments. Yeah. Okay, so the comment is to build up for the non-radial case. So in the non-radial case we have this whole huge collection of possible traveling waves. So whenever you have any solution of the elliptic equation and you do any Lorentz transformation of it you have a possible traveling wave. So to attempt to get a dispersive type or a channel type characterization for all those things seems completely impossible. So in non-radial case you have to have a different strategy. You cannot use this kind of argument. That's the first thing. The other thing is this outer energy inequalities that we've been using are false in the non-radial case. It's not that we don't know how to prove them, they're just false. So you need to change the way in which you obtain your channels of energy. And that's what will happen in July. It's the explanation of how you overcome those things. But you still keep some of the energy inequalities from the speed of propagation. Of course, you have the finite speed of propagation and this is huge. But another problem, do not depend only on the finite speed of propagation. Right. Anyway, so all I'm saying is that it is considerably complicated. But you have to understand the non-radial case. If you don't start by something, you can never hope to do this. To keep the idea in mind. Yes. And the hope. Okay, very good. Let's find the speaker. Thank you.