 No, I think at the technical level the lectures will be completely independent Okay, thank you Okay, but so I maybe just say a few words to recall what we did in the previous lectures so we considered the energy critical nonlinear wave equation in the focusing case and And the goal is to understand the Asymptotics the long-time asymptotics for large solutions Okay, that's the general goal then I I Fert oh, I'm sorry so I started by Describing the work with Frank Merrill in which we proved what we call the The ground state conjecture so that if the energy is smaller than the energy of the ground state Which we call w which is a radial solution of the elliptic equation and it's the minimal energy solution Among nonzero solutions Then there's a dichotomy is the h1 norm of the data is smaller Then the one of w you have global existence and scattering in both time directions And if the h1 norm is bigger you have a finite time blow-up in both time directions And the case of equality is ruled out by variational considerations So that was the first result that I Described and in doing so we introduced this Concentration compactness rigidity theorem method that allows one To reduce the thing to proving that certain solutions with a compactness property Do not exist Okay, and in that I want to point this out because it will be Interesting from the point of view of my lecture on Wednesday One crucial point was that there are no cells similar Blow-up solutions which are compact and The proof of that introduced a similar coordinates Okay, and we will see that revisited tomorrow on Wednesday Then after that we started with a general study of type 2 solutions and so type 2 blow-up solutions are solutions which Sees to exist in finite time, but have uniformly bounded norm in the energy space and as I was saying this is a phenomenon which is very characteristic of energy critical situations and the first examples were constructed by Krieger-Schlag and Tataru in the three-dimensional case and then Iléa and Raphael constructed solutions in the four-dimensional case and gender engine the five-dimensional case Of this kind of solution. So we're talking about the non-empty set of objects and then we introduced the soliton resolution conjecture which gives Precise description of the asymptotics of solutions which remain bounded until their final time in the energy space and description is that asymptotically a solution looks like a finite sum of rescaled traveling waves plus a Radiation term that's a solution of a linear equation plus a term that goes to zero to infinity as time goes to the final time And then the last thing that we did in the first sequence of lectures was to prove this result for the radial case Okay, and the proof in the radial case Boiled things down to a dynamical characterization of the only radial elliptic solution, which is the ground state and That was what we called a channel of energy argument. We showed that all Solutions of the non-linear equation that exists for both positive and negative time Create a channel of energy outside the light cone unless they are a rescaled okay, and From that we could prove this soliton resolution conjecture So that's the summary of what we did before Okay So now we continue so the subject of this week's lectures is the non-radial case Okay, I was going backwards. Okay So in the non-radial case Some new difficulties arise because now we We have to deal with all the traveling wave solutions to the equation and We've proven already and we saw that in the last Series that traveling wave solutions are precisely Lorentz transformations of elliptic solutions But when we're in the non-radial case, there are many elliptic solutions and there's no classification of them so in the in the radial case we have an explicit Radial solution there's a formula for it. It's one over one plus x squared over six the whole quantity square root So I mean we have very specific information Now the general solutions to the elliptic equation have many flavors as far as we know and they're not classified at all okay So we have a very large set of traveling waves the second thing is in in our dynamical characterization of W we used a certain Outer energy inequality for radial solutions of the three-dimensional wave equation Now that corresponding estimate fails in the non-radial case so that's So a different approach is needed so We believe that an approach that is purely based on dynamical properties of the traveling waves since this is such a and will the enlarge set of objects is unrealistic, so we're gonna proceed in a different way Okay So let me state the result that we're going to be describing So we take a solution that remains bounded in the in the energy space So let's assume that first that T plus is finite And let S be the set of singular points so a singular point is a point where there's energy concentration and We showed that the set of singular points is always non-empty and Finite in the case when the blow-up time is finite okay so We fix an element in the among the singular set and Then the result is that there's a number J star which is bigger than or equal to 1 a little radius R star this radius has the The role that it separates all the singular points so that you're staying away from the other singular points okay, and the time sequence T n which converges to T plus and This is a well-chosen time sequence It's not an arbitrary time sequence and here is a main difference between this result and the radial result in the radial result We were able to prove this for all sequences of times Okay So there's a well-chosen sequence of times so far Scales and the J n which go to zero faster than the self-similar rate positions CJ and which are Strictly inside the ball of radius T star minus T n around x star for some beta which is strictly smaller than one and Directions L J which are given as the limit of the vector CJ n over T plus minus T n and notice that Because of this the limit after passing to a further subsequence is always well defined And it's always strictly less than one and traveling waves Such that within this ball we have the decomposition that our solution equals This radiation term I fixed h1 cross L2 term plus the sum of modulated solitons So these are the traveling wave solutions plus an error that goes to zero and this Traveling waves Don't see each other because of the orthogonality property of the parameters Okay, so this is precisely the solid and resolution result Along a well-chosen sequence of times in the finite law case Okay, see the question. So your J star depends only on What the J star depends basically on this And that's a premium So presumably for the same J star, you could have another sequence of times For the same Is that this holds inside be okay, so it just separates the different singular points Okay, so that they don't see each other That's the purpose of this Okay, other questions Now comes the other possibility, which is that there's an infinite time of existence Then there's the radiation term which is a solution to the linear equation That extracts the most scattering you can have in the you So that's what we call the scattering profile Okay, and what it does is outside any finite distance from the Light cone the difference between you and this linear solution is zero Okay, then there's also numbers J star now here We are allowing J star to be zero and that would be the case when the solution actually scatters A time sequence which is well-chosen going to infinity Scales on the Jn which are much smaller than the cell similar rate Positions which are strictly inside the light cone Directions lj for the traveling waves and Traveling waves such that the solution equals the linear solution Plus the sum of modulated traveling waves plus an error that goes to zero Infinity and then this traveling waves do not see each other because of the orthogonality property Okay, and here there's no r star because there's only one infinity Okay, so this is precisely what one would hope to prove Of course one would hope to prove it for all sequence of times T n's but for now we have it for one sequence of times That's what shows that's an optimistic state for now Okay, okay, so the let me just go on Yeah, the land that is less than T n that means that some of these are traveling with No, no, no, no, no they're concentrated Yes, they concentrate faster than the light They are dividing I Mean they can open but not as far as the white and the light home that goes to infinity now Yeah, and then everything happens it's strictly inside this light Okay, that's what we're saying So I will if you remember what I did in the first part of the lectures I proved the the soliton resolution in the radial case for t plus equals infinity So in the non-radial case, I'll do the finite case just for Variation but the proofs are more or less the same once The scattering profile and this is a typo. This should be a v sub L Once we have the scattering profile the two proofs are more or less identical Okay So what we're gonna begin doing now is Show how to extract this scattering profile. That's will be our first step once we do that then we will Only work in the finite time case because the two things will be the same But the extraction of the scattering profile is a highly non-trivial thing at this at this level, okay So this is a work with the Kyron Merle at the beginning of the year and Maybe I should have said So this theorem is due to the Kyron Gia myself and Merle Okay So the J is how Gia so to So Who is this scattering profile? Well, the thing that you do is you apply the linear group at minus t to the solution you Then you take the weak limit of that and the solution corresponding to this weak limit That's what the scattering profile is and I don't know why in this page I'm calling it UL and in the previous one I called it VL, but They are the same. There's only one scattering profile. So let's go on Okay So the let me first summarize the proof Okay So they keep well, I mean just explain the idea is that we're gonna rule out any block So we know we have this idea of how to decompose a solution into blocks Which are non-linear profiles and about who these are profile decomposition But I'm not gonna go into Details of what they are More than that. So we we're gonna see that there is no non-linear block in Your solution which remains close to the light cone for very large time and This is what should happen if you believe that The only non-linear objects are these traveling waves because the traveling waves that are traveling at Speed strictly less than one So there can't be any non-linear object that travels at speed one and that's what this proof shows Okay How do we show it we show it by combining a family of formulas which are called virial identities which we saw in the previous lectures okay, so one tool that we use in in in this is actually a Part of the theory of the linear wave equation Which it was developed by Gerald Friedlander in the 80s 70s 60s and 50s. Yes, and somehow people working in The Melrose school of scattering theory Kept using this but everybody else more or less forgot About this and this turns out to be an important tool here and we rediscovered it independently but then we found out that Friedlander had already done this so we will use the notation ddr for the radial derivative and one over our grads of omega for the tangential derivative to the sphere, okay So these are more or less standard Okay, so I'll give you this the linear theorem This linear theorem allows to associate the profile at infinity to any solution of the linear wave equation Okay, it says the following suppose we have any solution to the linear wave equation with data in the energy space okay, then always the Tangential derivative goes to zero as T goes to infinity and The hardy term goes to zero as T goes to infinity the L6 term also goes to zero the L6 norm and There is a profile at infinity. There's a unique function g plus Which is an L2 of r cross Sn minus 1 such that The properly rescaled t derivative Equal at infinity equals this object and The properly rescaled r derivative at infinity equals this object Furthermore, there's an isometry the energy the linear energy of this solution Equals the L2 norm in r cross Sn minus 1 of this g plus and The map that maps the Cauchy data to the g plus is a bijective Isometry between the two spaces and of course there's a corresponding g minus as time goes to minus infinity and This g plus and g minus are called the radiation fields associated to VL So somehow there's a way to associate a profile at plus and minus infinity Which is an asymptotic state for any solution of a linear wave equation Okay, and as I said We Reproved this result not knowing that Fred Lander already had it But then we found out that he did so somehow one can use this This objects to patch up together solutions of the wave equation coming from infinity. Okay, so this is how we're gonna use this Oh, yeah, I know there's one more thing. I want to mention here There's a well-known property of solutions of the linear wave equation, which is what's called the equipartition of energy That the spatial energy and the time energy asymptotically are the same In fact, this shows a much stronger version of that because this has a minus and this has the plus and this goes to that and This goes to that Therefore the equipartition of energy follows immediately from this But it's a much this is a much stronger Statement than equipartition of energy. Oh By the way at the end of this series By Friday, let's say afternoon. This notes will be posted on the IHES Website The first part is already posted in the IHES website Okay, so oh by the way this was for all n bigger than or equal to 3 Everything else is also for all n bigger than or equal to 3, but I will only be working on R3 Okay, so Here's another typo and I don't know if you have noticed but there is a Oh There's a French notation and there's a US notation the French notation the T comes first And in the US notation the T comes last and I've been using the US notation And then here I lapsed into French notation. I've been polluted by French collaborators and But it should be that the T goes second here Okay, okay, so now it this the striccarts norm that we will be working with is the L5 in TL 10 in X norm Okay, but we will localize it to sets of this form by taking first This spatial integral and then the time integral and if we have an interval I I will call S of I to be SR 3 cross I And then If we have an F in L1 cross L2 You have a solution you and it verifies the striccarts estimate in terms of the L1 cross L2 It's measurable Let's say No, well, we will We will use it for domains that have the dependence property Okay But in in this definition, there's no need for that But it will always be in in case of a finite dependence property. Okay Okay. Yeah, so then there's a little claim here is that you have a solution like that Then in fact asymptotically it looks like a free solution without any F And that's the same proof as the proof of scattering in In in standard situations. Okay, and I will use this In this proof, okay, so now I will sketch the proof of this of the extraction of the scattering profile So first I take the I choose any sequence going to infinity and I choose a weak limit Okay, this I can always do because this thing is bounded And now I will call vl The linear solution corresponding to this weak limit And what I want to show Is that this limit holds Now notice that really I want to show this for a negative Because if I know it for some a I will know it for any bigger a But it's convenient for me to state the result as saying that it holds for all a real as you will see in a second So, uh, if a is negative that covers the outcome No, it's a moved up It covers the outside but moved up Okay, and if I know it for a negative I know it for a positive Because of continuity Okay, so now I'm going to assume that this does not hold and I'm going to reach a contribution So the first claim is that there is an optimal a bar For which the thing holds So there is an a bar such that if I'm above it my desired property Which is that this space time norm is fine it holds And that for all a bigger than a bar This behaves as if it were a linear solution. So this should be zero And that for all a bigger than a bar This desired limit holds And this a bar is optimal because this norm is infinite Okay, so I'm going to prove this And it's in in this statement that it's convenient that I'm working with all r instead of just with negative r because The sketch of proof of this I define a bar as you would normally do it You take the infimum of the good a's But you have to show that the set is non-empty Now the set is non-empty because if a is very large and positive This is certainly finite because if I chop up the data It will have small norm and then it will coincide outside a light cone with something that scatters So then this will be true And that's the reason for for doing this okay So this set is certainly non-empty and Clearly for any a bigger than a bar This thing is finite by definition Now this a bar could be real or could be minus infinity Right, I mean I probably I don't know that it isn't The fact that I'm assuming that the theorem does not hold Forces this to be real And that is how we will reach a contradiction Okay, because if this is minus infinity This would hold for all a and that's precisely what I'm trying to prove so okay, so Now let me show the second Hypothesis The second conclusion So let a be bigger than a bar and now I'm going to concoct A solution of an inhomogeneous problem. I will call it v And what it will have is u chopped off x bigger than t plus a u to the fifth Okay, so that this by assumption is in l1 l2 because a is bigger than a bar and this norm is finite So since this is an inhomogeneous term in l1 cross l2 We have a linear solution that has this property That was the first statement that I made that was just like scattering okay now By finite speed of propagation u has to equal v Outside the light cone because outside the light cone both u and v verify the same wave equation So then in that previous statement, I had a v. I can put a u in here In this region And since u equals v and for linear solutions, these terms go to zero the same is now true for you Now I will show that this vl of a Minus the desired vl goes to zero and this will finish the proof of two Okay, now how do I prove that I will just say it and look at the associated radiation fields The radiation field associated to v sub l and the one to v sub al They're both linear solutions. So they always have radiation fields And now what one can check by the definitions and a little bit of patience is that for any direction in s2 As long as I stay away from a the two radiation fields coincide And since the two radiation fields coincide And one is approaching its own radiation field and so is the other one the difference has to be small In exactly the region that I want so this proves this This first two properties now I I still have to prove Yeah, no, I have also three So now I have to show that this the strickers norm at a bar is not finite So if it were finite Then I could make it small by moving the t high Right because it's an it's an integral norm then Because this is a well a well defined object and it's completely linear at this point If this is small I can make it smaller. I can still give up a delta one over two So here I get delta one over four then I get a delta one over two then I get a delta one And then from this if delta one is very small I can conclude for the non-linear solution the same thing But this is a contradiction to the definition of a bar Because I have a smaller thing for which the thing is finite So that's how one concludes that the norm is infinite Now so now I've got a lemma that tells me what's happening with this a bar But this is a completely radial lemma In order to Conclude here I have to now chop up into the different directions So I'm going to define the set of regular directions and the set of singular directions Okay So let me first define what I mean by a regular direction So I have a regular direction if I can go a little bit below a bar Provided I stay in a sufficiently small angle Okay, so in this direction I can go a little bit below And that's what I will call a regular direction And a direction which is no This is the the usual angle and a direction which is not regular is singular Okay So the the main result About this setup is that the set of regular directions is non-empty but finite Okay So this is our first step and then we will obtain a contradiction by showing that it has to be empty That if this a bar Was finite the set of regular directions has to be empty. Okay, so that's the structure of the proof So the main tool improving the fact that the set of Regular directions is non-empty and finite is this statement There is a delta 2 such that if omega Satisfies that for some epsilon positive This linear object Has smaller than delta 2 norm And what's important here is that I'm allowing the region to go below a bar And I'm keeping the angle small According to this t And the square root of t comes on the curvature Okay So the statement is that if this is small enough, but not arbitrarily small, there's a fixed threshold Then the thing has to be a regular. Okay So that's the the the tool that we use to prove that The singular set is non-empty and finite So let me just say this claim is not Difficult to prove using the compactness of the sphere The the previous proposition and strict estimates with elementary geometric arguments and finite speed of propagation Because this will show suppose that There is no point that is singular then every point is regular and this satisfies this is satisfied Then I cover my sphere by finitely many points where this holds And then because of that I can go slightly below a bar So there has to be at least one singular point Why do I have to have finitely many because If I am bigger than this delta over two if I'm singular I have to be bigger than a fixed delta over two And if I have infinitely many points the deltas over two get too many Okay I they get multiplied by a large number for any largeness So this is the proposition and this proposition is not so difficult to prove but we will we will use it So now we come to the the heart of the matter Which is to show That this singular set is empty And this will give us a contradiction Okay so at this point You're pretty much stuck with having to use a profile decomposition And I've tried to avoid discussing very much about profile decompositions Because whenever I do that in lectures, I see all eyes glaze over Okay So anyway, but here we need a profile decomposition. I I recall what the profile decomposition is If I have a bounded sequence in h1 cross l2 after passing to a subsequence I can rewrite it as a sum of rescaled linear solutions fixed linear solutions plus an error And this error has small strictest norm as I take more and more profiles Okay Now these are linear solutions and associated to each linear solution. There's a scale A translation parameter and a time parameter And associated with this object and a linear solution I can find What we call the non-linear profile Which asymptotically behaves like the linear profile at the sequence of Times overscaled Okay and The parameters the scaling translation And time translation parameters verify orthogonality properties that allow us to keep the The profiles separate And one of the consequences of this is that there's orthogonality properties For the profiles in the h1 cross l2 more Okay So just keep these things in mind and So the first task is to find Properties of the fact Of the singular set in terms of the profile decomposition Okay So I will call I will have two profile properties profile property one and profile property two Okay, profile property one deals just with the radial situation. There's not a There's not a direction chosen here Okay, it tells me that if I have a profile decomposition for this solution For which the a bar Is real then there's no profile With the properties that the translation parameter x j n Minus the time parameter at which I'm taking the profile decomposition Stays further away than the a bar And the same when I incorporate also the time parameter and the scaling parameters Are all shrinking to zero and this bar here is a typo Now what I will post will have the typos because I'm just post the pdf file But I trust that there's so few that you will remember Or else curse me for it. Okay Okay So We will argue let let me sketch the proof which is very short We will argue by contradiction. Let's assume that the time translate thing parameter is always zero Remember that when we do a profile decomposition We can reduce to a situation where the t j n's are either zero Or t j n in absolute value over lambda j n tends to infinity Okay, those are the three cases Otherwise by changing the profiles you can always reduce to this case. Okay In fact in the last series of lectures I explained how there is a certain lack of uniqueness in the profiles that's Tied up to the possibility of making these changes, but that's the only lack of uniqueness in the profiles Okay Okay, remember that these profiles are obtained as weak limits And so the fact that this is the profile in the profile decomposition Means that this Weak limit gives me exactly the energy of the linear energy of the profile Okay And the definition of the scaled profile the modulated profile is this Because t j n is zero otherwise we would have minus t j n over lambda j n here Okay So now i'm going to do this calculation I do the gradient dot U j and I know what this goes to goes to this Which is positive because this is a non-zero profile now because This profile is one which is concentrated in this region Right i'm trying to prove that there is no profile different from zero So I assume that there is And i'm going to reach a contradiction Okay, because of this because of this property this integral is concentrated in that region Now in this region Because this is there's an epsilon here This is the property one. This is bigger than a plus two epsilon. There's an epsilon here outside a bar I can replace u by this v l. We already proved that So I do that and I still have a little low one But here now I can replace this region By the whole thing with v l again by this property here Is where it's concentrating So I get that Now I can use the the asymptotic properties of v l given by the radiation by the radiation field And that Will tell me that this integral actually goes to zero Because this is a linear integral And using the radiation field. I will get that this is zero So then this object on the one hand goes to this and on the other hand goes to zero. That's a contradiction The uj has to be zero Okay, so this is another way in which you use this radiation fields you have a Sequence of time tending to infinity and then you have an asymptotics Now then there's the case where limit of t j n over lambda j n is infinity and this is worked out here Here you use the non the radiation field associated to the uj l And then you get a contradiction in a similar way Okay Now we come to the profile property two Profile property two deals with a singular direction And profile property two tells you the following suppose I have There is a there is a fixed number delta three Such that if t n goes to infinity and omega is a singular point After extraction I have a profile decomposition such that One of the profiles which i'm calling u1 Is substantial and it's concentrating around in the direction of omega and the addition so this is the the spatial concentration And it's in the direction of omega in addition the scaling goes to zero and the time Translation the the time concentration is also going to zero Okay, and this is because otherwise If there wasn't any concentration of the of the stricter's norm In near this direction omega I would have a regular point So since it's a singular point there has to be something like this, okay Now the proof is you use the first proposition and some geometric arguments and you And you obtain that result Okay, so these are the preliminaries now we're going to get to the heart of the matter The heart of the matter is this proposition Suppose u and a bar are as before and epsilon is given Then I can find the sequence of times going to infinity and A number alpha positive such that I can make The x1 plus d dt plus d dx 2 plus d dx 3 Concentrated along the e1 direction Up to level alpha within epsilon. Okay, so i'm going to And the interesting thing is okay if I had just d dx 1 d dx 2 d dx 3 and d dt This would be much better But I can't Hope for that much, but if I can add the d dx 1 and the d d 2 I can get it Okay So the first thing that i'm going to do is show you how From this proposition I can show that there is no singular element There's no singular direction And what I will do is I will show that this Implies that e1 That's tied up to this d dx 1 Enter the direction e1 in here. I will show that e1 cannot be a singular direction But of course since e1 is an arbitrary direction after rotation that shows that the singular set is empty And that will be the contradiction Okay So in here this lemma prove it will be proved without using that e1 is a singular direction Just prove this for the direction e1 And then prove that this implies that e1 cannot be singular Yeah, but you can impose the choice of one because if you wrote it Yes, then I can if I prove this I can prove that none is singular Yeah, I can prove this for any other direction So I choose one but it could be proved for any other direction So for another direction you will have a different sequence? Possibly But I will show that I have this this sequence of time and this sequence of time gives me That e1 any direction on s2 you will get a sequence Right and using that sequence I can show that that direction cannot be singular And the sequence depends on the direction Yeah, okay So let's use this So that that is okay. That's part of the logic here Or the e-logic if you like anyway So let's show that this implies that e1 cannot be a singular direction Okay, and this is enough and then of course I have to prove the proposition And the proof of the proposition is the main part of the proof Okay, so up to now what we've been doing is Peeling layers so as to get to the heart and this is the heart So I'm now going to show how proposition 2 implies e1 does not belong to s So I I now have to do a little digression to do that Okay So I introduce A new norm on the energy space So if I take a fun a pair in the energy space, I I define the ease of one norm To be g plus d dx1 l2 plus d dx2 plus d dx3 Okay, let me first convince you that this is a norm You at first you could have Of course the triangle inequality is obvious, but what is not obvious at least Without thinking for a second is that if the norm of fg is zero fg has to be zero Okay, so let's show that Suppose that the norm of fg is zero then f Has two partials that are zero Right, and it's an h1 Then the third partial has to be zero Right, there can't be an h1 function that depends on only one variable Okay, so then d dx1 is zero, but then because of this being zero g zero So this is a norm Okay, so okay Now the the interesting thing Is that for the linear equation We have two Conservation laws right first we have conservation of the energy, but we also have conservation of the momentum And if we combine those two You can see that this Quantity is independent of t Because what you are adding is two d dx1 D dt and that's the momentum in the direction of e1 Okay So this is independent of t because of that Now we can plug this into the profile decomposition and we get a new Pythagorean expansion So We have a profile decomposition and we have this Pythagorean expansion Because in in proving this all you use is the orthogonality of the parameters and the invariance under the flow Okay, so this is this Pythagorean expansion for the e1 norm Okay And the next thing that we need Is the following claim Suppose somebody gives me a beta and an m Then there's an epsilon such that if I have something in the energy space We're bounded by m energy norm, but very small E1 norm Then the striccarts norm can be made as small as I like So somehow being small in this norm forces the striccarts norm to be small So how do I prove this? Well, I Prove it by contradiction. If not I have a sequence which is uniformly bounded whose norm tends to zero and whose striccarts norm tends to zero Right, this is what I need to prove. That's the same as this To show that this is true Let's take a profile decomposition for this Now because of the Pythagorean expansion All the profiles have to be zero because this norm is tending to zero Because each profile will have zero e1 norm and the e1 norm being zero Means that the thing is zero Therefore, I'm only left with a dispersive term in the Pythagorean in the profile decomposition And the dispersive term has striccarts norm tending to zero. So that's the end of the story So that proves this Is this clear? This statement follows Because if this goes to zero, I have no profiles and therefore I have this Okay So now let me go to the proof that e1 does not belong To s and now I need to recall for you profile property two The profile property two says that If I have a tn going to infinity and an omega in s Then for some subsequence This property holds this property holds this property holds and there's a delta three Which is substantial, okay And the beta comes from this claim That if I have a uniform bound and this is smaller than epsilon Then this has to be smaller than delta than beta So I have an m given by the supremum I'll choose beta to be delta three over two Where delta three is the thing that gives me the Substantiality of the first profile in profile property two By the proposition I can find the sequence t prime n and an alpha such that this limb soup is small This gives me the sequence from this sequence. I take a profile decomposition Verifying the properties of profile property two So now I'm going to show that the This implies The right smallness of the e1 norm So since this is an inner product associated to this norm And we have the Pythagorean expansion for this norm. We also have this weak limit For the profiles remember we we use this before For the h1 cross l2 norm. We now can use it for the e1 norm But we know by the profile property two That this that this profile is concentrated where this is less than alpha. So this part is small Now I look at this inner product And I split it into this region and this region in this region. This is small and in this region I get a bound of epsilon Therefore, I get this whole thing is bounded by epsilon plus something that goes to zero with n But this gives me the e1 e1 norm squared And that means that this thing has e1 norm less than epsilon But if the e1 norm is less than epsilon Given the way I chose my betas This means that the s norm of e1 is less than delta 3 over 2 because beta was delta 3 over 2 But the property 2 said that it had to be bigger than delta 3 And so that's the contradiction So this tells me that all I have to do is prove proposition 2 And that is the heart of the matter Okay And the next hour will be spent improving proposition 2 So is the logic at least clear? Of course some of the calculations you have to do at home But this is the outline okay, so proposition 2 is is the The important thing Given epsilon we can find the sequence t prime n and an alpha such that this limb soup is less than epsilon Okay, this is our task That was proposition 2 You can trust me that this was proposition 2 Okay, so let's prove this and this is the thing that is proved using virial identities This is the heart of the matter that shows you that things cannot be traveling at speed 1 Near the near the light core, okay Okay, so now we start a different Chapter in this So the first thing we're going to do is we're going to produce some kind of the defect measure Okay So let tau n be any sequence Of time standing to infinity I will call rho of x t To be u to the 6 plus the gradient. This is an e1 not an epsilon 1 squared and then the Hardy term So what what what's this rho representing? It's I'm scaling the u at the singular point And I'm doing it beyond a bar where things go wrong and then I have I put in the the Linear and the energy the linear energy, but I also throw in the l6 norm and the Hardy norm Okay For for good measure So the the statement is that after we're given a sequence After extraction, I have a non-negative Radon measure such that Rows of At time tn goes weakly to this non-negative measure mu tilde But the the support of the u tilde Has to be an x1 negative Okay, less than or equal to zero because and that's where the The fact that omega is a singular That a bar is there Makes it be that way Okay, okay the first Statement is obvious right because these are bounded Bounded measure has a subsequence the a bounded sequence of measures has a subsequence that converges weakly Nothing more to be said. So the important thing is the support. So in proposition three We showed that if I went an epsilon outside a bar The u converges to the vl If I go an epsilon tilde outside the a bar The u can be replaced by the vl And by using that the plane is outside the tangent sphere This holds The first statement is true and these two statements are also true because these two quantities went to zero So we just want to to show the the support But since the u equals the vl in in this region If I show if I choose that this is going to zero for any compacted supported thing This has to be supported in x1 less than or equal to epsilon for every epsilon And therefore in x1 less than or equal to zero Okay So i'm going to you show this And this is now a linear solution. So what am I going to use? Now the point is that the times are going to infinity. So what am I going to use the radiation field? okay, that's What i'm usually using So this is what I Have this will be the radiation field I first change variables And now I change this by the radiation field And and make a small error Now I change variables here And now because this has compact support if omega is different from e1 this goes to zero And therefore by dominated conversions the thing goes to zero Okay, I have two different directions and then one is going to infinity I exit the support okay Now we have our our defect measure mu tilde And I'm going to decompose it into a delta mass at the origin plus a remainder And the remainder has the property that it charges The origin zero Right, I had no way of knowing that this is an absolutely continuous measure I don't gonna I'm not gonna claim that Okay So I just take the the part corresponding to the origin and take the rest So the next lemma says the following Let tn tend to infinity and epsilon be given Then after extraction there exist two non-negative radon measures mu zero and mu one And two non-negative numbers Such that the mu zero and mu one don't charge the origin The row at tn converges weakly to mu zero plus c zero delta zero The row at tn minus alpha over ten Goes weakly to mu one plus c one delta zero The supports of the mu j's are in x one less than or equal to zero And here is the the punch line. There's also an alpha And mu zero in the ball of radius less than alpha is smaller than epsilon And mu one in the ball of radius seven alpha over ten is smaller than epsilon Okay Each one of these two things is obvious from the definitions Because mu one and mu zero charge zero to the origin So in a small ball the they charge a small amount The only point is that there's this uniformity That I can use a something proportional to the alpha In both cases, okay, and so what would you use to uniformize things? There's only one thing you can do and let's use finite speed of propagation Because here we're working with a wave equation and so by finite speed of propagation And small data theory You can see That you can make the same Provided there It's because We're only alpha over ten Apart and we're allowing alpha in axis Okay, so that's where the Wave equation plays a role finite speed of propagation now, uh, there's an important fact here Which is an extra piece of information that this tells you Which is if we define rho of x t t n To be the same expression that you had before Where before you had t n, but now you put t Then For all t between t n minus alpha over ten and t n this will be smaller than two epsilon Provided you're always charging these annuals Okay, so yes So it is why do you need the same alpha the same so very oh because otherwise my my my proof will not yeah I need them to be comparable Okay, it's because of the geometry of cones everything has to be Any any other question? Okay Some alpha over ten seven alpha over ten may presently that means that the sum of the two should be Should be something exactly yeah, yeah, these are not fixed Exactly Yeah, that's that's the that's the only point So this property rho star Shows you that uniformly In this range of t n's and in this range of axes things are small Okay So that's what we should take from this Okay Now we're going to review virial identities That we've seen in the first part of the class, but I mean Of the course, but we'll review them so I don't know why this got there Okay, so that's a typo So little r is always the tail These are the tails Phi is a cutoff function, which is one for x less than one quarter and zero for x bigger than one and phi sub r is phi scaled Okay Then we have this collection of identities Okay, these are our virial identities The first one tells us what x dot grad u times d dt u That's when we take a d dt We get this constant minus three halves Plus a half grad u square minus u to the sixth and then We have to take a a cutoff function because x grows And then there's an error and the error is bounded by the tail Okay The second one Oh Don't copy the second one. The plus should be a minus. That's a typo Okay, whenever the gradient square then the u to the sixth appear they have to appear with opposite signs because we're in the focusing case Okay, so this is a minus So then the other virial is where we have u d dt u And then we have the same type of thing, but with different numbers here It's one and here's minus one and here is minus three halves and here's one Then this virial is related. I don't know if it's a virial this identity Is related to the Invariance of the momentum And it says that if you take the energy density And you multiply it by x The derivative of that is the momentum Okay And we have to chop it off And then we get the error And we will also be using the two conservation laws Okay, the the derivative of the Energy zero and the derivative of the momentum is zero And of course we will always be using also truncated versions of them So if we put the face of r here, we get a little of capital r and so on And how do you prove these things you prove them by brute force, right? You multiply the equation for by an appropriate object And you integrate by parts. Okay, so the usual multiplier method So now I'm going to take a sequence t sub n tending to infinity I'll call phi sub alpha What before I was calling phi sub r, but now I have an alpha And it'll be convenient to define use of n of x to be this trans translated and scaled U So our definitions And our previous lemma Give us information about these un's This converges weakly to this this converges weakly to that And then there were the smaller than alpha things Okay And now I'm going to introduce four quantities A sub n is the d dt truncated by phi alpha B sub n is the gradient squared C sub n is the u to the six And d sub n is dx one times d dt Which is the first component of the moment And they're all truncated by this By this phi sub alpha and then I will be averaging these quantities for t between t n minus alpha and t sub n Okay And I will be proving things about the averages and from proving something about the averages is how I will manufacturer another sequence t prime That comes from the average, okay And then the the terms That are averaged have upper bars Okay Now I'm going to rephrase the virial identities and my information in terms of these quantities, okay the first one Well, the first virial was this We have the phi alpha We have And this is an o of epsilon This o of epsilon comes from that property star Okay, remember the star was the thing that was uniform for t between t n minus alpha over 10 and t n Okay, so that was the first virial where this is the Time derivative. This is the space derivative. This is the l6 norm And this one is the the second virial where I take the time derivative and then the coefficients are all one Or minus one the third one was the The derivative of the x1 times the energy density which gave me the momentum Here is x1 minus the momentum. Here is just in in the x1. So I get the E1 component of the momentum plus o of epsilon This is the Constancy of the momentum truncated And this is the constancy of the energy truncated Okay So that's what this This Things are and the capital of epsilon is uniform in t. All right So what I will do is uh Force algebraic relationships between these objects that Come From my assumptions, okay so Since we're taking an average And this is roughly constant Is the difference between its average is smaller than epsilon, which is the point was difference But then since I'm averaging I get an alpha And the same for the energy Okay, pardon me. This one is a plus. Yes. Thank you. Yeah, that's a plus Okay, so now I'm going to prove three Facts minus d n bar. So this tells me The first fact is that in the situation wherein the energy equals the momentum With the minus sign That's what the first one says the second one says that this This combination is almost zero And the the last one is that this combination is the momentum plus almost zero So let me show you the the proofs of this I will only show you a but you will see once I prove a how all the others go They're very similar Okay, so a relates the energy to the momentum and why how do I relate the energy to the momentum remember there's a A relationship that the derivative of x one times the energy Density equals the momentum. That's what I will use Okay, so I integrate that relationship between t n and t n minus alpha over 10 And then take the average so I get minus d n bar The average of the momentum And here I get minus 10 over alpha that's this is the x one times the energy density at time t n Minus at time t n minus alpha over 10 Okay, so this is the fundamental theorem And this 10 over alpha is the length of the integral Okay, so this is just the fundamental theorem Should I go back to the identity that I used to get this? I take the integral between t n and t n minus alpha over over 10 And then average and I get this and then I get this at the end points by the fundamental theorem Okay, so that's what I have so here What will happen what? I'll explain in words what happens this guy will be small because I have it at t n and This is a consequence of that star bound In here, I have it at t n minus alpha over 10. Then I have to change the space Scaling to match I have to make a space translation to match when I make a space translation I will get y one plus alpha over 10 And now the y one term will be small And then the alpha over 10 term is the one that will give me the energy Okay, so let me now Do the calculation The first term is this I have to understand this quantity I have the x one And this is exactly phi alpha over x d mu zero x in the limit now It's important that I have x one here Because x one vanishes at the origin And that's why I didn't get the delta mass of the origin in that convergence. Okay And this is precisely less than epsilon over alpha All right, that was one of the properties that this measure d mu zero had so that's why I get that For the other one I have to do this change of variables So what I have is this and I now I do that change of variables And of course x one has a y one part And then a part with no y one which is this one And this one is the one that gives me Precisely when I take the average the the average of the energy And this one now it's everything is centered the right way will give me epsilon alpha So I have this of course I have the energy, but I have it at the at the at the wrong time But the energy is roughly constant so I can change the time So I have it here So I I get this identity but I can change the energy and I get it here And this is precisely the first one of those identities Was what I called a So now I'm going to add a and c when I add a And c when I add a and c I get b n bar minus a m bar minus two-thirds c n bar equals capital o of epsilon Now I add b to it if you remember b has b n minus a n And I think it's plus c n a bar So the a n minus b n cancels The b n minus a n And what I end up is with the fact that c bar n is small is o of epsilon So then I plug this back to the To the first thing and this is small So I can put it with the o of epsilon and so I have this One half a n plus one half b n bar plus d n bar is o of epsilon But now let's remember what b n bar b n and a n had a n had the d dt squared b n has the gradient squared And they have the factor one half and d n bar is precisely dx one times d dt So now I complete that into the square And you see that that's precisely the e one norm The e one norm has d dx one plus d dt squared plus d dx two squared plus d dx three squared So I just come you know So I get this And that's what I want to show is small. That's my conclusion I mean this kind of expression being small But it's average in time So since the average in time is small, there has to be one sequence of times For which the thing is small and to here it was for every t n. So this is the first time This is how you choose the t prime and passing from the average to the particular Set of t prime ends But as we saw in the proof that that ruled out That it was a singular point Any sequence was enough It didn't matter that So any sequence killed that so Now all that you have to do Is see where the supports are and see that you have this and that creates your alpha Instead of alpha is alpha over 10 instead of epsilon c over epsilon But that concludes this proof as you see. I mean it's an elaborate proof You cannot just sit down and do it. Okay. Yeah I just say we have the option of continuing Or if you're tired, we can we can stop for now and restart next time. Uh, maybe some way Uh Okay, but It will have to change subject So maybe it's a good time to stop maybe more questions if there are more questions Yeah, this is a big chunk to have swallowed. Yes. People have to understand Secrets of time depend on the point when you obtain a result for all time, but only the Yeah Absolutely, no, no, I I have no Yes, I I have no desire to to go on but Okay, so let's let's go back First the statement of the proposition The proposition is a bar is what we had epsilon is given Then there is a sequence and an alpha such that this Is small. Okay, so I think maybe what would be good is to review that this implies that e1 Is not a singular point I think that that would be I saw a hand there I don't know if I misunderstood but before you said that if you had the same proposition that there exists a sequence of times But instead of having the dt you had it like some you said that that would be better Yeah, why? Or we could say it would immediately kill all the profiles Even if it's just you have along this subsequent along the sequence. Yes. Yeah Because I think that's the thing that people Don't understand Is as soon that you have it for a sequence when you get for all time, but only in one special direction Okay, okay And you are working on one yes, you have to work one direction at a time. Yes, all the sequence depends on the direction Okay, thank you. But the thing is that you're killing that direction. So you have a Some direction on your sites and you want to eliminate it And to eliminate it we find the sequence of time that has all these properties But then that implies that that couldn't have happened Okay, but let's go through the proof that this proposition implies that e1 is not a singular direction Okay Let's review So we introduce this norm and why do we introduce this norm? Well, let's go back to proposition two It appears here and why does it appear here as we saw in the proof of this? Uh, algebraic identities. This is the thing that appears naturally when you have the This virial identities and you see that The energy equals the momentum and so then you can combine them and when you combine them you get precisely this type of term It was the one half d dt squared plus one half gradient squared plus d dx one d dt is exactly this That would be zero on the function of t minus x1, right? Okay But of course those functions are not allowed because they can't be in the space Okay, so So we introduce this thing is a natural object For our proof Then the first observation is again you have to combine energy and momentum here And because they are both independent of time This quantity is independent of time. Okay So this is a constant thing that you're using The square Okay, and because this is time independent and we have the orthogonality properties of the parameters This is immediately implies a Pythagorean identity for the profile decomposition Okay, and the proof is the same as the one for the usual energy If you recall the proof from the usual energy, the only thing you need is that the energy is invariant Then you're going to relate Being small in this e1 norm to not having profiles To having that this trick at norm is small. Okay So this statement tells you that Provided you have a uniform bound If this norm is small Then this norm has to be small so to prove that What you're trying to show is that provided this is uniformly bounded if this goes to zero this goes to zero Okay So if this holds and this goes to zero then this has to go to zero. This is what we need to show To show that We just take So we have these two properties and we're going to show that these two properties means mean That there's no non zero profile in the profile decomposition Okay Why is that? Because there's the Pythagorean expansion So each profile squared is controlled in the e1 norm is controlled by this e1 norm And we know that if something has e1 norm that is zero it is zero That was the first thing that we showed that this this is a norm And that's so therefore all the profiles are zero from this two facts and the Pythagorean expansion Once you have that all the profiles are zero That means that only the remainder is in the profile decomposition the remainder has this property So that proves this claim So you want to have it for any t and you You use the fact that you can have it for any subsequent for any sequence of here Yes Because the limits are identified The limit is zero And having a discrete sequence is necessary to have a profile decomposition. Yes, of course, maybe there's some other But I mean that's For a subsequence. So now we're going to prove that e1 cannot be in s By using proposition 2 and this Property 2 of the profiles So proposition 2 tells me that I'm in this situation and The first profile is substantial With the substantiality delta 3 Should that that's written here. It's right here This is even a well thought out Transparency, okay So now I'm going to contradict this and this Using this property So I have this sequence t prime and I use that for a subsequence I have a profile decomposition and the profile decomposition has these properties Now I will use the the weak limit property that tells me that the inner product With u gives me The the norm of the profile and this is just a consequence of the Pythagorean expansion orthogonality of parameters, okay Now from our assumptions These two assumptions it tells me where the profile lives It lives in this region For any alpha because the other parts are no not where the You know, these profiles always are concentrated near the light cone once you undo the scaling And so this is where this lives You can use the radiation field or the concentration near the light cone either way Okay Now I look at this inner product and I split it into this part and the other part The part corresponding to this Gives me an error that goes to zero And the the rest I just use the Cauchy Schwartz for the inner product and I get these two things So this is bounded by that Now this converges to this quantity squared So the conclusion is that this is less than epsilon But my claim tells me that if this is less than epsilon The s norm is less than beta and beta was delta three over two So this is delta three over two But the profile property two told me that it had had to be bigger than delta over three So what I'm using is that if somebody gives me a sequence For a subsequence, I have a profile decomposition and I then just work with that and my profile property two Worked for any given sequence. Is this more clear? I will not say that this is not a tricky proof. It is a tricky proof So let me let me just say that what what I will do next time Is now I will abandon the infinite time because the finite time and the infinite time behave the same way once we have this part So this is the part that's the main difference between the finite time and the infinite time, but we've done it So then I'm going to go to the finite time case and prove the decomposition So that will be the last two lectures For a sequence of times, yeah And I'll say something about how to find this sequence of time So how close are you do you think to the continuous? Uh far Pardon me. Do you have some strategy you think or? We have several attempts, but I'm not willing to say more than that. I mean There's a I think it is a very difficult question to pass It's already difficult to prove it for a sequence of times But to pass to the general case you need to understand collisions of solitary waves And this is a very tricky thing. Okay, so that's what's Now in the radial case, of course, you have only one solitary wave and then This dispersive properties of Solutions allow you to understand that Thank you