 Yes, so what I will do now is Recapitulate what we did last time, okay, so that we can see the logic. Hopefully, okay So I will do this on the board and then we will go back. This is where we stopped Okay, but I will remind you where we wanted to prove and so on okay, so Yes, the notes The notes will be available on the web starting next week Yes, tonight Anyway, but there Well, but there's the one proof that will not be on the web because it's not typed so that I showed last time Okay, okay So let's the situation is we have a solution This is time and this is the blow-up point Okay, and the solution remains bounded as we approach the blow-up point and now we went to prove the decomposition into some of modulated solitons plus the Radiation term which is a just obtained by the weak limit and An error that goes to zero for a properly selected sequence, okay So this is our aim So the first step was the control of the flux so We have V a regular solution and by that I mean it's a solution that continues below zero Okay, and we know that we have such a thing such that they support Of u of t Minus v of t is contained Okay, so that we take the weak limit of the u We take the solution with that beta times zero going forward and the difference will be supported in the column By finite speed of propagation Okay And now using the fact that this v is a regular solution We can control the flux because the flux lives on the boundary of the cone and on the boundary of the cone the solution equals the regular solution by this property, okay, and we did this by combining strict estimates with the uniform boundedness of the energy Okay, so that was the first part the second part Was a Moral words estimate so basically we looked at points at times Zero to one less than t2 Small oh and we are in a situation of course that zero Belongs to the singular set so that the solution cannot be continued locally Beyond zero for negative time and we know that there's finally many of such singular points So we'll just concentrate near each one Okay, so this is a Moral words estimate says that the integral between t1 and t2 the integral over x less than t of d dt u Plus, I don't know if you will see Can you can you see what I'm writing So and if you remember we proved this by going to sell similar variables and Introducing a sell similar energy and computing Sell similar regularized energy and computing the derivative of that So the third step is the first decomposition So so why is this an interesting estimate? This is an interesting estimate because the power of the long is less than one Right because if we had the power of one, this would just be the boundedness of Determine side which is our hypothesis, but The fact that we get better than that tells us that in some sense in some average sense This is going to zero and we have this logarithmic rate that which it is going to zero the square root of the law Okay So the first decomposition follows from First a real variable argument Okay, so the real variable argument So I let's let me call this a Tauverian argument. I can find the sequence mu n Tending to zero such that But not only that I get an enhancement of it using a hard-to-ditude maximal function So I can find Maybe I draw a picture now here. Maybe I should use some color So that you can see but I don't have color really doesn't matter. Okay, so I split my interval in thirds and In the two far out points I can find two times t1n and t2n such that The supremum zero less than Tau less than t i n over, I don't know, 20 some number like that of the integral of 1 over tau tau minus t i n less than 20 something like that the integral over x less than T Or ct for i equals 1 and 2 so that's some kind of improved Going to zero thing that and the argument to prove this is a using the We type on one inequality for the maximal operator and so Okay, and the reason I can always put ct here is I can go to t By using the more of its identity and pass t u equals v which is a regular solution So that contribution is negligible So this is what the top For any see any see Yes, I mean we want to make it a 10 or something like that, okay For any so the correct logic is for any see This is true, but the sequence does not of times does not depend on the sea Okay, okay So let me see Can people read what I'm doing in my little corner there? Okay. That's the first. That's the tabir and part and be Is that in the integral in time is this what it's T minus T and T. I am this two times Yeah, but in the integral there in the in which integral is in the Yeah, yeah, right. So that's it's integral over T minus T and I Yes, I will tau and the soup is in tau tau is fixed so Oh I conveniently didn't call it. Okay, so this yeah, it's yeah, it is t Okay, I Just take the average and I think the supreme okay So this is what my tabirian argument gives that that's essentially a real variables argument okay, and Now I use my conclusion is to show by analyze Analysis of profiles this implies that if Let me call this condition if star I Have the decomposition with error Tending to zero in L6 And so how do I do that? Well, I analyze the different possibilities of the profiles and use the inequality I Mean use this facts and According to the scaling of the profile I get either a self-similar solution which is ruled out by the theorem of Merlin myself or If the Concentration is faster than some self-similar I can produce a traveling wave Okay, and that gives me the decomposition Except that the error only tends to zero in L6 And now I have to work to make the error better and for Will be Second Now the second decomposition Has two parts the first part is there is a corollary of Whenever B holds I can get a bound on L2 norms the L2 norm goes faster than This is in N little low in N. So whenever I have a decomposition With an L6 error going to zero I get this improvement And this is a consequence of the L6 term going to zero and Of the flux estimates and of the known estimates for the solitary waves Okay, so that's the first part Now the second part To do this question on that remember we use the decomposition Inserted absolute values use the triangle inequality L2 norms. We controlled either by Helder's inequality or by knowing this and then there were a term And that involved the solitons and those by the point-wise bounds that we know on the solitons one to zero Yes And no not yet Now this You see here that what the correct statement here is is whenever you have a sequence of time tn For which you have a sol a decomposition With L6 error going to zero The L2 norm has this property Okay, so it's not a property of two times but of each individual Okay, the second part Is the virial argument at this point we use a virial argument So basically what we do is we multiply the equation by you We write zero we integrate and Use the fundamental theorem and control terms the terms at the two end points are the Ti ends And those terms we control Using this kind of thing and then we have lateral terms that come from integration by parts on the lateral side And those we control by the control of the flux Okay, and what is the conclusion of that? the conclusion of that is that run over T2n minus T1n Integral between T1n and T2n integral of x less than t of grad u squared minus dTu squared minus u to the sixth Tends to zero. So we're almost there and now we use a Second real variable argument. So I go here See second real variable argument knowing this So let's say b plus the mu n This implies that I can find a time Tn in between somewhere in here Okay I will find the time Tn Somewhere in between and let me put the two things Together that I know at this Tn This tends to zero and now this quantity Notice that the thing I'm integrating could be negative Okay, this is not a coercive quantity, but what I get is that the limb soup in n of the integral over x less than Tn of grad u of Tn squared minus d dt u of Tn squared minus u to the sixth Tn dx dt is less than or equal to zero Okay, so these are the things that I do by this is my second real variable argument now d now at this sequence of time Tn's because I have this The first property I can do a decomposition without six-norm going to zero and I also have the L2 norm going to zero Because one implies the other Okay Then I'm going to use two properties of it of a the solitary work waves Question so what's the difference between the star and this one? I'm sorry, which was the star The star is that it's at a different point at the different time Yeah, and here is this Tn that's somewhere in between the two Tis Not necessarily somewhere between the two But it's the same statement, right? It's the same statement, but at a different point. Okay, because it's I do not prove that the same time No, because I'm going to have a second property Because they write that both of them have to be hold at the same time That that's the point. Okay, so you use those two times to reduce another third time where you have the same But you have another property, okay? So this is the logic and Now if Ql is a traveling wave, I know the following straight This is a consequence of the elliptic equation And the second one is just the fact that it's a traveling wave in the direction of L okay, and The li's that appear in my decomposition are the limits in n of the C i ends over the times Tn where this C i n is the Center where the the traveling waves are centered at Okay, now with this information I can Use this part to deduce that if this holds for you Once I have the soliton that I once I have the soliton decomposition it must hold also for the error Because in the soliton part I get zero Yes, that's what they are that that's the Proof Don't you have this this is the last thing the point-wise point-wise equal to zero GTql plus L upgrade and ql isn't that the point that's point-wise equal to zero, yes But I'm going to use it in this way Yes C and I for the Tn that you found Tn this one is Tn because I'm at Tn now Okay, and the I is a different thing here. Maybe I call it J Okay, I Forgot about the other two times now now that I produce my third time. I can forget about the other two So this tells me that the error has to verify that yes That line that you're pointing at do you mean to integrate in T also? No, this is point-wise in X That's what I That's what I gained my my my real variable arguments Oh, oh then I put a DT I I Misspoke so the the first thing is that now This property is inherited by the error, but the error Goes to zero in L6, so I'm going to be able to take out that part. Okay, that's the first thing the second thing Now is that this will be inherited also by the error because The L is essentially X over T and because this is zero all the terms coming from the solitary waves get killed and so Then I get to my Transparency so I get to this The reason that the L2 term goes away is again because of the L2 norm going to zero in here, so I get rid of the L2 of the u-term and So the offshoot of this is that I can conclude these two properties on the error This is exactly What's inherited from this and the other one is what's inherited from this the u I can throw away because of the L2 property and The other term that all the solitons can be taken away because of this and I'm just left with With the epsilon okay, and here since I have a supremum If I take tau going to zero I catch the limit okay, and so I have this and I have that Okay, and now these are the two inequalities that will allow me to show further properties of the error Okay, okay, so now let's go on from this two inequalities So is it clear how I got to those two inequalities? Okay, it's all this stuff that I wrote on the boards got me to those two inequalities. Okay, so that's been my What I've gained so far All right Okay, so we have these two things Okay, now here From this I can show that it follows that the difference of these two L2 norms goes to zero All right, that's it Reverse triangle inequality, okay, but the DDT Because of this is asymptotically bigger than that so I conclude That this thing which is always non-negative because I'm integrating over x less than Tn is Less than or equal to zero by this and that so therefore I Obtained this So this is just triangle inequality to get here and combining the two estimates Okay, now from this since the integrand is bigger than or equal to zero now if I go to a smaller Region pick any lambda on any region x over lambda Tn will be less than lambda And so the whole gradient will tend to zero Okay, so that tells me that if I shrink from the ball by any small amount the energy goes to zero So the energy at least the spatial energy is concentrating on the boundary now that implies That the tangential energy is going to zero. Why is that? Well, if I go to the integral for x less than lambda Tn This is clear, but instead of choosing a lambda I now choose a sequence lambda n tending to 1 and The x over length of x transforms into x over Tn along that sequence and Then I get from the two inequalities. I get this this comes From this and that it's just the simple thing Now, I also know that this goes to zero because of this and therefore When x is smaller than lambda Tn This thing goes to zero also the t integral the t integral so I get control of the t integral. Is that clear? And we this is what we have to do in the next step, but to do that we have to do a lot of work Okay To to be able to show that in the whole energy you're going to zero is a whole extra step Okay, so we have to kill the concentration on the boundary Okay, so let me review the statement. We've just proved the claim is that there is a new sequence T7 such that we have the decomposition with an error which besides being in L6 Has the tangential Part of the gradient going to zero The gradient outside always goes to zero in this decomposition because everything is localized When I go a little bit inside it goes to zero because I just proved that the same is true for the L2 norm of the DDT and This is the expression Which is in here and That goes to zero That was the other thing I had And I have now proved this That's what we've just proved. It's precisely this statement. Yeah, we've proved all of this What were the first assumptions for the for this claim? Like the assumption is what? the assumption is that We have the uniform boundedness of the H1 cross L2 norm as T goes to zero which is the blow up point then we find a sequence T7 and a Collection of solitary waves such that Once I remove the regular part the difference is the sum of the solitary waves modulated plus an error that's Has goes to zero in this sense Okay, and now what I want is that the error go to zero in energy norm But I have to work two more steps for this, okay Do you have any sort of like a control of how It's epsilon one thank you. That's a typo. It's the DDT. Yeah Yeah, no, no, no, it's a type of DDT Yeah, it's what we have in here. So what is missing at this point is taking the lambda to be one right But before we can do that We have to do one extra step the extra step is that we want to show that when we Apply the linear salute that this linear solution corresponding to this data Has triggered snob going to zero So we need to first before showing that it goes to zero in energy We need to show that it goes to zero in the dispersive sense and after that We have to have a new mechanism To show that I'm going to zero in energy and that will be a new channel of energy argument. So I will Go in the steps, okay So the next step and now I don't have to change my sequence of times anymore This properties are enough to show that from this Automatically follows that the linear solution Goes to zero in the dispersive sense. Okay So the proof of this is Very analogous To the proof of the extraction of the scattering profile that we did on Monday In fact, it's the same ideas that go into that So what we're gonna do is do a profile decomposition on the on this error and show that all the profiles are zero and That's what it means To go to zero in the dispersive sense Okay, but I will not repeat all that Because it's so similar to the one thing that we've seen and details will be in the in the notes Okay, okay, but what I want is there's a property of solutions to the linear wave equation Which I want to discuss briefly. Okay, so this is kind of an aside Okay, so this is a Property of solutions to the linear wave equation with data in H1 cross L2 Then it tells you basically that the energy concentrates around the boundary of the light Okay, so this everybody will agree with that now to prove this statement when you are in odd Dimensions three I remind you is odd You just use the strong Huygens principle You approximate your data by compact and supported data for the compact and supported data You see where the solution lives and then this follows automatically Now for even dimensions, this doesn't work okay now you could use the The radiation field that we introduced on Monday to prove this this is an immediate consequence of the radiation field The problem is that in our proof of the existence of this radiation field we use this property So you don't want to do that Now there's another proof of this fact that follows from standard estimates on the standard improved dispersive estimates on the linear wave equation that Are due to Dan corner and his collaborators the dispersive estimate for solutions on of the Wave equation says that if you have a nice initial data in 3d, let's say the gradient the case like 1 over t Okay, but the improvement of Dan corner and collaborators is that away from the Light cone there's an improved rate of decay And if you use that this follows immediately also But I don't want to use any of these proofs. I'm gonna use a different proof Why do I want to use a different proof because it's an introduction to what? the channel of energy argument is that we're gonna use In a few minutes, okay And as I'm mentioning this is only interested interesting in even dimensions But doesn't matter I mean we want to have the proof of these things in all dimensions So I should also mention that there's another approach in the radial case using a Fourier analysis That's a in my work with a cotton slide Anyway, so let's go To a proof of this so this is a proof with the care and Merrill So the first point is that by density you can assume that your data is in C0 infinity Okay, there's nothing to say Next I'm going to use a virial identity to prove this Okay, and Here is a virial identity that you use and Of course, so this is a virial identity for the linear wave equation. You prove it by integration by parts There's nothing more here Okay, and you can trust me that these are the right coefficients for for this to do The linear energy has the one half. Okay, all right Okay, so now I integrate this between zero and T All right, so This is a constant. So when I integrate I get the linear term And then I get the two boundary terms on this Okay, so I just use the fundamental theorem and that the Primitive of one is T. Okay Now comes another Important point in this and what follows For the linear wave equation, of course We have conservation of energy But there's also conservation of the L2 cross H-1 norm Just by integrating in space. I mean that's Okay, and that means Since C0 infinity is contained in H-1 and here it's important to have compact support Okay So that means that the L2 norm of the solution to the wave equation remains bounded as T goes to infinity when the when the Initial data has compact support So you would think that it grows like T, but no it doesn't it remains bounded And you can see this also directly from the formula for the solution using the cosine and the sine The H-1 means that you have some decay at infinity in space. Yes, and since I have a compact support Okay, so there's nothing okay so using this from this previous inequality This term is controlled because the L2 norm of use controlled DDT use controlled and This term is controlled and this term is controlled. Okay So what we get is The absolute value of this term is bigger than this All right So all the L2 terms are controlled by that remark and The others are controlled by the energy So that's a lower bound and now I'm going to proceed and give an upper bound and then I'll compare So suppose that M is the diameter of the support and Then by finite speed the support is contained in M plus T and Here this is valid in both even and odd dimensions. Okay, so I didn't cheat here This is the usual finite speed and Let me pick an R so I'm going to Split the integral in X Bigger than or equal to T minus R and X less than or equal to T minus R When X is bigger than or equal to T minus R. I bound X by T plus M And when X is smaller than T minus R. I bound it by T minus R Okay Now yes No, M is fixed and R is a given number. Okay now The two T terms combine and since I have the product and The H1 norm has to the squares But with the half I can combine them into this Okay Now the minus R term again, I do I do So let me go to the M term In the M term I add and subtract the part corresponding to X less than T minus R times M The part that I added Once I use this Cauchy-Schwarz again gives me this The part I subtract Combines with the other part to give me a minus R over 2 Okay, so I finally end up with this and I Put together both inequalities and I get this inequality and now I divide by R and When I take T going to infinity and R going to infinity this thing goes to 0 and This thing is 0 by finite speed by the finite speed that I can use so that proves my My result Okay So this is a way to show that all the energy concentrates on the boundary of the cone But I only use integration by parts. I only use variable identities for this Okay, and this is will be important for us Okay, so this is a kind of a side remark now. Let's keep all this these things are The thing that you need to rule the profiles that come from infinity and so The next claim which I said I wasn't gonna prove because it's similar to the extraction of the scattering profile that we saw Monday This is the next step So in this sequence of times we have the decomposition the L6 norm goes to 0 We have this equality this CJ's are all strictly contained When we normalize them in a ball of some radius beta this beta is a number that you should remember for later Okay, it's somehow where all the solitons are contained One No, just the the space normally because that's where you control the grade in the now But also you have these properties and in addition The dispersive norm goes to 0 so that's my statement up to now And I just want to make one comment once we have this And we couldn't do this before we automatically have that there has to be a soliton It couldn't be that there's no soliton Because if there's no soliton if the dispersive norm goes to 0 by this a Local Cauchy theory and the perturbation theory the the solution wouldn't blow up Would exist for all time so But you couldn't say that until you prove that the dispersive norm goes to 0 Okay, so it could have been that in my decomposition I didn't have any solitons. That would be really stupid, right? So no It's okay Right, okay That's the only Comment that I Pass the proof Okay, I'm sorry. I went too far So now is the last step of the proof and this is the fundamental extra ingredient We need to kill these guys We need to show that their energy norm goes to 0 and so far We know places where this energy norm goes to 0, but we don't know that it goes to 0 everywhere Right, it could be concentrating on the boundary of the sphere And now we're gonna use this channel of energy argument to prove that that's the case Okay, and remember that the channel of energy That we used in the radial case allowed us to prove this dispersive property of Solutions which are not the soliton which is what we use to prove the soliton resolution in the radial case And we know that those things that that out Outer energy lower bound fails for The non-radial case Okay, this fails is counter example and not only that but in even dimensions It's also fails in the radial case so Somehow we will produce a channel of energy that doesn't see these counter examples Okay, so what we will prove is a new Very simple estimate for the linear wave equation. Okay, but it's an estimate that will Be interesting only when the data when these things have some good properties. So let me just Let me just state dilemma first So I give myself a pair in H1 cross L2 which is supported in the ball radius R Okay, and I saw the linear wave with this data Then I'll call E0 the energy and E-1 the L2 cross H-1 energy and I'm allowed to use this guy because I have the complex support and This are both constant in time So the statement is that for any eta 0 between 0 and R and all T positive there's I Can have a lower bound for the amount of energy Outside X bigger than T plus 8 and not For any 8 and not between 0 and R Okay, and this is the bound as the first term here another term that depends on a eta 0 and The product of these two to the square root with a minus now a Priory this may be completely useless because what's on the right might be negative right and this is not interesting But we will see that we are in a good situation where that's not the case But you see this is the other the type of lower bound that those outer energy inequalities Were giving the advantage of this one is that it is true While those were not In the non-radial case so the proof of this is very similar to this Concentration on the boundary of the cone and that's why I gave that one first So that the beginning is the same up to here is the same I Again integrate. Oh, I already integrated here. I just put the terms on the other side Now I again split the integrals into T bigger X bigger than T plus 8 or not and X less than T plus 8 Okay, so perhaps it's not so interesting to go through the details of this proof But all it is is basically the same argument as before It has to do the accounting And you get the lemma so the important thing now is this statement And what will be really important for us will be the corollary It's a depending on how you split we use the support to get the T plus R and then we force the Internet Okay, so the important thing is this corollary the corollary tells me that if I have a bounded sequence Which is well prepared in The sense that the energy is concentrating on the sphere the L6 norm is going to zero the tangential energy is going to zero and The solution is outgoing in this sense the epsilon one plus DDR epsilon zero and Plus this tends to zero if I look at the solution with this data and This infimum of the H1 cross L2 norm is positive then for any eta between zero and one so Basically the support is morally one because outside of one everything is small Then we do have a true channel of energy. So if the data is well prepared then the Amount of energy outside T plus eta naught is controlled by the total energy and I can choose the eta naught to be anything smaller than the support so that will be the corollary from the Inequality we had before. Okay, so I'll explain how the corollary falls. So the first observation is that Okay, I'm gonna I Don't have it Exactly compact support. I only have it up to an error that goes to zero I will force now compact support and I do that by localization Fix some are slightly bigger than one that I will choose later on pick a cut-off function and Look at the difference The cut-off function is identically one in the ball of readers one and absolute the till this the difference goes to zero because in the part outside one there was no energy the energy went to zero and Inside when I take the difference I get zero. So this is obvious. Okay The next thing is that now I note That this sequence will go to zero weekly in H1 cross L2 Why does it go to zero weekly in H1 cross L2 goes to zero weekly just because its energy is concentrated in a thin annuals The energy has then it has to go weakly to zero because when I test against the function The contribution or anything annulus is negligible and outside I go to zero Okay Now since this goes to zero weekly It will go to zero in L2 cross H minus one by the relative compactness So this E minus one contribution will go to zero. So That's where you put X word T No, B1 is the ball of radius one The ball in X of radius one X. Yes. Thanks Okay, I haven't evolved. I'm just looking at the data Okay, okay, so this is true and this data now goes to zero in L2 cross H minus one and now I look at the The corresponding solution to the wave and now I just use the lemma. I'm I mean In condition to use the lemma because the lemma just said that it had to be supported in some ball of radius Some radius and the radius is one And I pick it is r bigger than one. So this is the Exactly the statement of the lemma that I just discussed Now this term will go to zero because the E minus one goes to zero So I get that but remember that this data is well prepared Okay, the data is well prepared means that the T derivative plus the DDR is almost zero Okay, and it also means that away from the boundary of the ball of radius one. There's no energy So I can replace this quantity first by the energy and then With a negative because X is pointing outside and Then I get one minus eight and oh You know Remember, this is practically DDR times DDT, but DDR is Practically DDT so this is practically DDT squared and DDT squared is one half DDT squared plus one half DDR squared And that isn't the only energy I have because the tangential gradient was negligible okay, so The combination of those two things gives me that and now I take r very close to one and I divide and Then I get this Which is what I want if the supremum is not zero I can take large enough and that this is true. So all this preparation and this initial data culminates with the fact that you now have a channel of energy for it and The basic idea of the energy channel method is that you can pass from Dispersively going to zero to going to zero in energy by using this method So that's the conclusion of the proof which I'll show now. Okay, so of course this has been completely linear right This is a linear lemma. Why can I use this for the nonlinear problem? And that's where the fact that the dispersive norm is going to zero comes in if the dispersive norm norm goes to zero the linear solution and the nonlinear solution are close to each other and that's just the Local theory of the Cauchy problem and its perturbation Okay And that's why it was important to prove that the dispersive norm is going to zero So that's the step that allows you to pass from linear to nonlinear Okay, so now is the conclusion of the proof. So we're near the end Okay, so suppose by contradiction that for some subsequence this doesn't go to zero So it's bounded from below now. I have to rescale by t sub n my previous inequalities. Okay, but that's a Standard and immediate and once I rescale What I have This is my channel of energy now instead of a to naught. I have a to naught times t sub n And I have bigger than a to naught and because this is uniformly bounded For Some subsequence. I have the mu naught and I don't know why I put eight. I could have put four. Okay, and This is gonna be true for any a to naught In zero one So now I'm going to tell you how I will choose the a to naught Remember that we had it in our in our decomposition We had a number beta Which indicated how? In how much inside the ball the solitons are Okay Right the beta was an upper bound on these numbers The a to naught will be chosen slightly bigger than this beta But still less than one And what's the effect of that the effect of that is that for x bigger than a to naught you don't see the solitons Because of their concentration, okay Yes, because that's a given by the energy When the energy of the soliton tends to infinity as l tends to 1 Okay, so that that is crucial is absolutely essential for this area But we this is okay. All right, so I now define W0 n plus comma w1 n to be the radiation term plus the error Yes, that is my solution where I forgot the solitons Okay, now since this a to naught is bigger than beta The difference between you and this guy outside is small Because the solitons are negligible in this region. So that's why we need to Have this outer energy information for any a to naught because we don't know where the solitons are going to be Okay, and now I look at the solution of the non-linear wave equation This things just have to do with staying away from the other singular points. So let's ignore them Now we know That the dispersive norm of this is going to zero So the Approximation theorem gives us an expansion for the solution with this data It tells us that it's the V the solution for that plus the epsilon the linear solution for these things because the Since the dispersive norm goes to zero the linear and the non-linear norm Solutions are close to each other and an error which becomes small Okay, and it becomes small not just in the dispersive sense, but even in the energy That's the approximation now because of this inequality initially and finite speed of propagation this inequality propagates in time and what we get is That u is close to w in a X bigger than t minus tn plus a to naught tn That's what finite speed of propagation does when t equals tn This is a to zero tn, which is this so and this delta is smaller than this are not It's just some small parameter here and doesn't play much of a role. It's just is separating the different singular points We have this decomposition for W n We have a channel property for epsilon that L and then W n Because of this Decomposition this guy is small. This is a regular solution So on small sets its contribution is small and this is the thing that has the channel So that means that W here. It's a channel and this is what I get now I can put t here by again by The concentration property of epsilon zero epsilon one there in the in the ball of radius t sub n and Finan speed of propagation so that grows like t the rest is negligible So I have that but in this region this guy is almost you So I can replace it by you paying a small price and the small price is that I went from One over sixteen to one over thirty two and this is valid for all t's in this range Okay Now I apply this at delta over two So I get this but notice that when T n goes to zero this region disappears and This is a regular because it's time delta over two is away from the singular time This is a regular function in the energy norm. So how could its integral over nothing be positive? This is what the channel of energy argument does So this is the contradiction to the fact that the epsilon ends Had a lower bound so the epsilon ends couldn't have had a lower bound So they go to zero and if you remember in the other channel of energy arguments we've seen There's something that happens for either positive time or negative time here We don't need that because we have outgoing data So the flow of time is always forward and this concludes it so this is the last step Well that I had a decomposition with an error having certain properties I have to use all the property I Have to use all the properties of the error that I've had up to now Yes, because the the otherwise I don't know Otherwise, I don't know this lower bound Right because in the general result the right-hand side could be negative Except that for this particularly well massaged a sequence I Have a good control in this way Okay, so One could conclude from this that this is a correct approach to this problem Because everything matches Just sort of in the same one Yeah, if you had you had a theorem which says you have the steps long Which on a sequence of time satisfies all these things so whatever Strength you have that result for you get the full result when you need a certain amount of results because You need all of the all of these things come in different places and showing that from So it's on like page 81. You had a list of properties, right? And so those list of properties are necessary to show that this thing has a lower bound First thing that there you don't need the dispersive property that you're going to zero But you need the dispersive property to pass from linear to nonlinear Okay, okay So I know that this is a Somewhat convoluted Maybe not the easiest easiest thing to absorb, but I hope at least I gave you some feeling for how this proof goes but before Asking for more questions. I just want to make a few comments Okay So we've proven this decomposition But you could ask first is Do you really need all these? Solitons in such a decomposition Okay, maybe you only need one This is a question that many people ask And it turns out that you need more than one and they as T goes to infinity This was proven in a paper of Martel and Merle In dimension 5 you have multi solitons appearing in the asymptotics as time goes to infinity the reason for dimension 5 is that this What's very important is the fatness of the tails of Of the solitons and the higher the dimension the less fat they are And that's why their argument works in dimension 5 Now in the finite time blow up. It still hasn't been proven as far as I know Yes, I guess here Hasn't been proven that you need that there Solutions with two bubbles that have finite time blow up, but I believe that this is Within the reach of the techniques developed by Jacek Gendres and his thesis that will be defended on Monday At the polytechnic Then there's the question of passing to other sequences of times instead of one and I already said something about that, but I'll repeat it. This is a Difficult problem it has to do with how multi solitons collide and It will take some time to understand Okay So this is all I want to say about that again then How specific is this to this problem? Okay, so this is something else to understand Of course, I mean for this problem all of this worked as if it was made for it, right? Well, I think that for other non-linear wave equations maybe geometric wave equations this approach should also apply so For wave maps for example for critical wave maps, I believe that this will work Hopefully eventually also for the wave version of young males at some point then What happens without finite speed of propagation? So in Schrodinger cases or correct the race cases or things of that type I think this is a really good direction for the future. I Don't want to say more than that but I think that they You know This shows that things like that can be done. So One should have the courage to to think about those problems. Okay, and Now let me see. I have one more slide Thank You Frank That's That's the final slide and I really appreciate the interest and the sustained attention over many lectures. So, thank you Questions or comments? You explained this before so just don't remember I said use that for clangor and you see that as an intermediate thing between Schrodinger and the Yeah, I think in some sense that that is the case because for small frequencies Well, and I have no results for clangor You have no voice. I have no results. So it's okay For clangor Other questions Or comments or Question about directions Very long-term well, I think I discussed a little bit of that but Any other question is welcome. What would survive in this in subcritical situations? Well, so far nothing Yeah, I mean we Yeah, that I of course that's an important question. It's a another you know important future direction They're the client Gordon You still have an approximate scaling for high frequencies, right, but But I don't know it's difficult to understand what replaces the energy channels in that case The outer energy in equalities are all false Yes, you need to have solitons, right? So if you just make subcritical No, that's why you need to have to you need to have two terms Okay, so you have to have something and I guess there are Is a natural example for that many of many examples many problems Yeah, many many problems But of course, that's good because that's more work to do You don't want to do something that finishes things off more it's better to start things off The speaker again for all the series of lectures