 Okay, so after the preliminary discussions of further properties of these critical elements, we see that the contradiction comes in this grand scheme from this theorem. The theorem says that if you have a solution below the energy of W, gradient is below that one of W, which has a momentum zero and has the compactness property where the x and the lambda are continuous, the lambda is positive, and if T plus is finite, we have this lower bound and this support property, and if T plus is infinite, we can assume that the lambda is bounded from below, x of zero is zero and lambda of zero is one. Okay, the conclusion is that u is zero. So this is a complicated description of the zero function. Okay, so we see that the critical element, we were able to manufacture a critical element which had all these properties, but we also were able to prove that the critical element had positive energy, so it couldn't be the zero function. So this critical element couldn't have existed, and since it couldn't have existed, the theorem had to hold. Okay, so that was the scheme of the proof. Any question up to now? The lower bound of lambda of T says it's a self-similar or it just says that it's bounded from below? No, it's just bounded from below. No, we're not saying that it is bounded above and below. Okay, so it's a one-sided bound. Of course, this bound from below implies such a bound, right? If it's going to go to infinity, it has to be big. No, but of course in the proof, we will see that in the case when T plus is plus infinity, we have to have a bound from both above and below. But that's part of the proof. It's not an assumption, okay? Any other question? Okay. So again, the proof or something like this cannot be done quickly. So we have to be patient here. So we do cases. So we will do first the case when T plus is plus infinity, and then we'll do the case when T plus is finite. So we start with the case T plus is plus infinity and we need a little bit of preparation. We pick a cut-off function, phi, which is one on the ball of radius one, and zero outside the ball of radius two, and we scale it by r. We call that phi sub r. And then there's another object that's needed which is x, but x has to be scaled. So we multiply by phi of x over r. So we have to truncate x because it grows too much. Okay, but x is needed. Okay, and now we will call the remainder r of capital R the integral for large x of all of these quantities. All quantities are all controlled by the energy norm, basically. We have the L6 norm, we're in R3, so that's controlled by the gradient in L2, and we have u squared over x squared, which because we're in dimension three, by hardness and equalities controlled by gradient u squared. Okay? So all of these quantities have the same homogeneity. And now, as I mentioned at some point, I think maybe the first day, in elliptic equations, there's this well-known identity of Pohojave that allows to prove that certain things have to be zero. And these are hyperbolic analogs of the Pohojave identity. Okay? That's the meaning of this. And they are more than one because we have space and time. Okay? So the first identity, this, you know, you should remember that c is basically x and phi is basically one, truncated. Okay? So the first identity is that the derivative in t of x grad u dt u is equal to this, minus three-halves plus one-half. And then there's a remainder because we've truncated and the remainder is controlled in this way. Now, if all, we can't just write x grad u dt u because that's a divergent integral. But if it were convergent, then the identity would be that this derivative equals that and there's no error. Okay? So this is just a way to make a divergent integral convergent. And then the second one is that u d dt u is grad du squared minus grad du squared plus u to the sixth plus o r of r. And the important thing to notice is that these two right-hand sides are roughly the same but with different coefficients. So these informations combine because the coefficients are different. Okay? We will see that later. And then the third one, in some form we've already seen, if you remember we had this proof that the critical elements have to have a momentum zero. And the reason was because if we performed the Lorentz transformation we could decrease the energy and then if the momentum was not zero. And that is basically what's reflected in this identity. That if you multiply the density of energy by x and you differentiate, you get the momentum and the negative of the momentum. Okay? All right. So I'll have to use these identities. I know that nobody can remember them but you'll trust me that I'm using them correctly. Okay? And once you have the nodes you can verify that your trust was not misplaced. Okay? So we start with the proof and now we're going to go back to our variational estimates. Okay? Our variational estimates play a role. I start out with a data which is non-zero but which verifies our size assumptions and then we know that the energy is positive that in the whole interval of existence since its t plus is infinite is for all t positive this is controlled by the energy and this is controlled below by the x gradient squared and therefore if I add ut squared I have that c of e. Okay? So basically for each t I have bounds from above and from below for various quantities. Okay? And the next thing is okay I have to use the compactness of this k otherwise I can't show that it is zero because there's plenty of non-zero solutions so the compactness of the k tells you that tails are uniformly small. Okay? This is what the compact set of functions in L2 does. It's tails are all uniformly small. So this translates because this scaling parameter is uniformly bounded from below into saying that given epsilon I can find an r naught such that all of these guys are smaller than epsilon times the energy. The energy is a convenient positive number to put there. Okay? So as long as I integrate outside the ball around x of t over lambda of t there's not much energy. I mean at this end outside the ball there's not much energy and for this is where we use the compactness and the fact that lambda is strictly bounded from below is a positive constant. Okay so these are the things we need to remember. So this is a very full slide. Let's try to digest it. There are two lemmas. The first lemma says that for epsilon this c is a constant it's an absolute constant. So if epsilon is small and r is very big you can find the t0 such that t0 is controlled by c times r and for 0 less than t less than t0 this is below r minus r naught plus epsilon and at t0 it is exactly equal. Okay? So how do we think about this? Remember that one of our normalizations is that x of 0 is 0 and lambda of 0 is 1. That was in our statement. So at t equals 0 this ratio certainly is less than this. Right? And now I flow in t. Remember now that x of t and lambda of t are continuous. That was one of my assumptions. So I flow until I hit r minus r0 epsilon the first time. And that's t0. And the statement of the lemma is that I hit the first time below c times r for some constant c. I hit that first time that the threshold of r minus r0 epsilon by c times r. This has to be proved but that's at least what the lemma says. Okay? Now let me just say that if we were dealing with the radial case suppose that we were in the radial case then x of t would be always 0. And then this already gives a contradiction. Okay? But we're not in the radial case so we have to do one extra thing. And the extra thing has to involve somehow the momentum. Of course the momentum of any radial function is 0. You should remember that. It doesn't depend on any equation. You take a radial function a pair of radial functions you multiply the gradient one times the other you get 0. Okay? Okay the second lemma says that for epsilon small r large then this t0 in fact has to be bounded from below by 1 over epsilon r over epsilon. Okay? So the first one gave us a bound from above the second one gives us a bound from below and the bound from below and the bound from above are not compatible for epsilon 0 and that's why we get a contradiction. That's why no such function can exist. Right? 1 over epsilon is not smaller than c. So it's all in these two lemmas. So of these two lemmas the first one is a little bit easier than the second one but neither one of them is that terribly hard provided you know how to combine these virial identities. So let's do the first one. We assume that it is not true and then I never up to c times r I never reach the value r minus r not of epsilon. Okay? And we will see that that's impossible. If my lemma is not true I never reach this value in the required integral. Okay? So now I look at zr which is the sum of the two quantities in the first two virial identities. And now the sum is what works well in 3D in higher D you have to put coefficients there. Okay? And this is because of how the different terms are weighted in the virial identities. Okay? So if we sum let's see what the derivative is. So it turns out that it combines to be this and then the error. Okay? So that's what it combines. Now what do we know about this part? Well what we know from our variational estimates is that this quantity is coercive so this is strictly positive. So this whole thing is a basically minus the energy. That's what we know from our variational estimates. But we have to deal with the error. Now let's understand what happens for x larger than r because this error gives us an integration over x bigger than r. Okay? If x is bigger than r and t is between 0 and c r this thing is bounded from below by r 0 epsilon by the triangle inequality, right? Because this is bigger than r minus r minus r 0 minus epsilon so you get r 0 epsilon. So this whole thing will be bigger than r 0 plus epsilon. So if x is bigger than r my region of integration is contained on where this is bigger than r 0 epsilon but the compactness gave me that the integral over that whole region of integration is small. So this error is bounded by this. So the error cannot cancel what we have on the main term. That's the conclusion of this. So the derivative is negative and of the size of c times z because this guy cannot reach to cancel what this gives because epsilon is small. Now I'm going to give a point wise bound on z a bound from above and then I'll combine the two things and I'll get the lemma. Okay the point wise bound from above is okay in here let's look at the large the first term the large thing is x right? x is bounded by r and the rest is just simply bounded by the energy because of course she shorts and the second term is the same because I multiply and divide by the length of x and u over the length of x by hardly is again bounded by the energy so I also get r times the energy for the second piece so all in all I get that z is bounded by r e okay now I integrate between 0 and t c r and I use the fundamental theorem okay so I get this from integrating time and this from the bounds at t equals 0 and t equals t0 so I get this is bounded by that but if I chose r if I chose c so large that c times c tilde is bigger than 2c1e I get the contradiction so I had to have reached c r for this large c so that's how this lemma is proved the other one is a bit more complicated and remember that this is the one that's crucial in the non-radial case because it's crucial in the non-radial case that momentum has to come and so we're going to look for 0 less than t0 to the function y sub r of t which is the one where when you take the derivative you get the momentum or minus the momentum okay that's why you use this function so let's look at what it says we know that for t less than t0 x of t over lambda of t is less than r minus r0 of epsilon so if x is bigger than r quantity is bigger than r0 of epsilon okay it's just the same as before now the momentum is 0 so the main term in the derivative of this thing is 0 if you recall the derivative of this thing is minus the momentum if the momentum is 0 then that's 0 and so we just get the error and the error because for x bigger than r this is big is big is controlled by r of r and r of r is controlled by epsilon e so therefore this whole thing is less than the length of the interval which is t0 times epsilon e so that's my bound on this difference and now I'm going to see that this bound gives me problems okay so the first thing I'm going to bound is yr of 0 so yr of 0 all I'm going to do is split the region of integration in x less than r0 and x bigger than r0 in the part of x less than r0 the size of r gets bounded by r0 and the rest is bounded by the energy so I get this piece and in the part where x is bigger than r size of r of x is bounded by x for x less than 2r and 0 otherwise so I replace the size of r by r and what I get is an integral outside r0 epsilon okay the bound of this I keep and this outside r0 of epsilon by the definition of little r is bound and the compactness is bounded by e epsilon r so this is my bound for yr of 0 so I get a bound from above now I'm going to get a bound from below for yr of t0 and there will be a balance over these things okay so for yr of t0 I split the region of integration in this region and it's component in this region necessarily x has to be bigger than than r because for x bigger than r this is bigger okay so I can replace psi r of x by x because that's what it is for x less than r so I have this now in this region the psi r is always bounded by r and by the compactness I get epsilon e so that's my bound for the tail and now I've got this part inside okay so this is what I'm dealing with now in the part inside I add and subtract minus x0 t0 of lambda t0 okay I add it and I subtract it remember that this is the thing that's exactly in size r minus r0 so I know exactly what the size of this from below is okay this now I'm just left with the density of energy but I don't quite get the energy because I'm only integrating on this region so I add the part outside to get the full energy so the bound is that this is bigger than this from the contribution of this and the whole integral of that minus the part that I add the region outside r0 of epsilon and so it's controlled by epsilon e and this guy is still r minus r0 epsilon and then this guy in here the absolute value of this is less than r0 by the way I chose the region of integration and the rest is the density of energy and so I get this and you know the negative signs don't matter because we know that the energy is positive anyway it's a bound from above and I'm bound from below by the negative of this yes are you sure that the energy density also no, no I'm not I'm not but I'm bounding from above because I have a minus sign and so then I get all the terms I got the 2 okay so my total of this from below is all of that and if you look at it if I make r large and epsilon small so this doesn't hurt because of the epsilon small small this doesn't hurt because I'm going to take the r large the epsilon small tells me that this term doesn't hurt me and the r large tells me that this doesn't hurt me and so I can bound all from below by er over 4 why everyone is asking that you have an integral x plus x t0 over lambda t0 times the energy density and you get to bound on that yes so I put I'm going to bound from above I'm going to get bound from it's absolute value from above and since I'm bounding from below then I get a negative sign so I put the absolute value inside the integral I'm not assuming that the energy density is positive to get the estimate no because the absolute value is outside now I put it inside because the integral the absolute value of the integral is less than or equal to the integral of the absolute value then the absolute value of this is bounded by r0 and the absolute value of each one of these terms is controlled by the energy and then I I get a negative okay so no I didn't cheat there but it's good to keep me honest okay so let's go on and here you see by taking r larger and epsilon small it's very easy to see that I can get a bound from by er over 4 and now I put all my bounds together and the point is that on the right I had the epsilon because the momentum was so the derivative had the bound by epsilon and therefore T0 is less than epsilon over er over 8 and that's the statement of the lemma okay so that takes care of the case r of T plus equals infinity and we regard that as the easy case here so now we're going to go to the case T plus is finite and we know about the support and we know about the lower bound and we know about the momentum okay but of course if T plus is finite I can rescale to make it one one rescale so it's easier to write things with T plus equal 1 so the first step is to prove that lambda not only is bigger than constant over 1 minus T but it's also smaller over 1 minus T so lambda is forced to be of the size of 1 over 1 minus T okay and this is a big step the fact that you can prove that and this is the last time we will use the virial identities once we get to that we've spent all our virial money we cannot do anything else so we have to find something else then we assume not and then you have a sequence tending to 1 such that this tends to infinity and we will see that this is absurd okay so it will be a contradiction so now I look at Z of T is the same object that I had before but I don't need to truncate now why don't I need to truncate because of the support I have compact support and so X is bounded so I can make these integrals so what's Z prime then it's a clean formula it's the same as before but now there's no remainder so by the variational estimates again Z prime now is bounded by minus C E for all T and the next thing I'm going to say is that the limit of Z of T as T tends to 1 is 0 why is that? well let's look at the first part X is smaller than 1 minus T so X this gives me a 1 minus T here and the rest is bounded by the energy so I don't have any so that tends to 0 the second part I multiply and divide by absolute value of X absolute value of X is still bounded by 1 minus T and then U over X in L2 by Hardy is controlled by the energy again so this is true and the derivative is bounded from below so by the fundamental theorem I get this lower bound then Z of T is bigger than constant times 1 minus T and I will see now that this and this are not compatible and that's where the contradiction will come alright I do that so we show that Z of Tn over 1 minus Tn tends to 0 and this is a contradiction because I've just shown that it had to be bounded below ok so now I'm going to exploit that the moment is 0 remember in the definition of Z I had X but for free I can put X plus Tn X of Tn over lambda Tn because the momentum is 0 ok and I have the division by 1 minus Tn I have the division of 1 minus Tn so now there's a few steps but they're not too bad ok so let's somebody give me an epsilon and now I'm going to look at these terms in the region where X plus X of Tn over lambda Tn is less than epsilon times 1 minus Tn so for the first part this is epsilon times 1 minus Tn so I get this bound for the next part I multiply and divide by X plus X of Tn over lambda of Tn and I can apply Hardy at any origin and that's why that term gives me the same bound ok for fixed Tn I can apply Hardy in X where with the origin wherever I choose it to be and I choose it to be at minus X over Tn over lambda Tn ok so then we have this bound so this part of the integral divided by 1 minus Tn gives me epsilon so it's what I mean the good situation now I have to see that the other part is not too large ok so in order to do that I first have to control X of Tn over lambda of Tn ok the center and I claim that it has to be less than twice 1 minus Tn really twice I mean 2 not c times 2 ok why is that if this is not true the intersection the intersection of these two balls is empty, that's triangle inequality and therefore this integral is 0 over here because this is supported in here and since this doesn't meet the support the integral is 0 ok now what happens on the other part the other part is this integral that's the complement in here I can just rescale things and change variables and I get this but remember this object is compact and I'm integrating outside something and my assumption is that lambda of Tn times 1 minus Tn goes to infinity so what happens to this integral it goes to 0 so all of all of the gradient squared integrates to something that goes to 0 this cannot be because then my solution would be 0 and I'm assuming that it has finite time blow up so it's not the 0 solution therefore X lambda of Tn over lambda of Tn is indeed smaller than 2 times 1 minus Tn ok so I got the size of X of Tn over lambda of Tn so now I'm going to this is the quantity I have to estimate and of course X is smaller than 1 minus Tn and X of Tn over lambda of Tn is less than 2 times 1 minus Tn so this whole quantity is less than 3 when I divide by 1 minus Tn and I have that and now this again goes to 0 by the fact that this goes to infinity and the compactness and the other term is handled similarly and then I get the contradiction that Z of Tn over 1 minus Tn tends to 0 while I saw that it was bounded from below so at the end lambda of Tn times 1 minus Tn tended to infinity could not hold so lambda of Tn is bounded by a constant over 1 minus Tn and lambda then is of the size of 1 over 1 minus Tn so at this stage of the game then we have found that lambda of T is of the size of 1 over 1 minus Tn and believe me at this point you run out of usage of the virial identities so you have to do something else so the first step is to show that because of that by some general principle the X of T can be gotten rid of and you can replace the lambda precisely by 1 minus T and this is a compact object okay and this is a very simple argument I don't think I want to go through it just a few lines not much to do so please believe me that now we can assume that X of T is 0 and lambda of T equals 1 over 1 minus Tn and now we are in the self-similar situation we have a family of solutions which have a solution which has the compactness property but the scaling parameter is 1 over 1 minus T and I have to show that this cannot happen and this is somehow a crucial point in the theory it's a in fact crucial all the way up to the soliton resolution and it's always a very important step in critical problems to be able to rule out self-similar objects but here we are not ruling a solution which is exactly self-similar but one which is self-similar up to compactness so now what do we do at this stage one may be stuck and the thing here is as I said in the first lecture there's a lot of elliptic theory now we go to a part which is really parabolic in this theory so we're going to think of the self-similar case with the parabolic situation and when you're dealing with parabolic situations and you want to understand the blow-up you introduce self-similar variables that's the method that was pioneered by Giga and Korn and it plays a role here okay so we go to that so somehow you change universe you introduce similar variables this was introduced by Giga and Korn in the parabolic case Marilyn and Zag used it in the wave equation case but in a subcritical regime in fact their work was in the power smaller than the conformally invariant power which is strictly smaller than the energy critical power and somehow the formalism works differently below or above this critical power anyway so what you do is you introduce a function w so now our new variables will be y and s y is x over 1 minus t and s is the log of 1 over 1 minus t and you introduce w as 1 minus t to the 1 half u of x t and in the y s variables it's like that and it's defined for s between 0 and infinity and the support is always in the ball of radius 1 because x over 1 minus t was less than 1 on the support okay and now for technical reasons we have to introduce a parameter delta and the technical reasons is so as not to divide by infinity okay or multiply by infinity you don't want to do that so we introduce this regularizing parameter delta which corresponds to looking at the solution shifted in time by delta okay so now y is x over 1 plus delta minus t and s is log of 1 over 1 plus delta minus t and w y as delta this to the 1 half times u and then we get that and now this w is not defined up to plus infinity it's up to defined up to log of 1 over delta now and when delta goes to 0 that gets bigger and bigger and the support of this thing if you do the calculation it's now always within 1 minus delta so you never get to delta equal to 1 and that allows us to do calculations near y equals 1 that would not be allowed otherwise so the next thing is what is the equation for this w both w and w delta now you calculate the equation and it turns out that it's a some kind of a non-linear wave equation that's not surprising there's a power wave that's not surprising you get some extra terms in the dds that look harmful at first in the end they help you and then the elliptic part is this where a row where did I write row oh row you see this coefficient row that generates at y equals 1 so this equation now is a degenerative elliptic equation so this is the price you pay now we get a degenerative elliptic equation instead of ordinary Laplacian and we get this other this one is a lower order term and this one is a lower order term so we don't think too much about them but we get an extra term of order 2 we have to see what does that do now a few comments about this elliptic equation if you if you analyze it you see that this can be written row 1 over row divergence and then the identity minus this rank 1 matrix y tends to y okay so if y is strictly smaller than 1 i minus that rank 1 matrix is an elliptic matrix right because I don't reach the 1 as y tends to 1 I'm losing one direction of ellipticity okay so this is an elliptic equation with smooth coefficients for y less than 1 but it's degenerate that y equals to 1 so what we're gonna do now is our universe is w and we're gonna try to compute what are the formulas in the w universe to be able to kill this w okay and that's how we're gonna kill ourselves similar solution that's the plan so first there's some easy easy remarks w is 0 at the boundary because u was 0 at the boundary remember u when x equals 1 minus t is 0 x over 1 minus t is y so when y equals 1 w is 0 and that's extremely important can you remind us why is u 0 because we had the support property in the inverted cone remember from the compactness you get and then we risqué well in the h1 sense so that's why it's only 0 in the h1 sense okay it's not the point wise sense it's in the sense of traces in some of the spaces so this is the correct technical and the gradient square is controlled because when you change variables in the y that's why you have the factor 1 minus t to the 1 half it keeps the home it keeps the scaling the w to the 6 the same thing and this one is a combination of things and if you check it it's still true does the c depend on delta? no uniformly in delta whenever there is a delta dependence I will show it in the pictures yeah now this is uniformly in delta and even true for delta equals 0 now there's another inequality that you get which is for free which is w squared is integrable against this high power weight and that's the hardy inequality where instead of being at the point it's at the circle so this is another hardy inequality so all of these bounds are uniform in delta now we introduce the right energy for this equation so it's this expression and now you see what the purpose of the delta is the delta means means that since y is supported in y less than 1 minus delta this expression converges because where this weight becomes degenerate my solution is 0 so this is an honest quantity I'm allowed to write this at least for delta positive how do I decide that this is the energy? well you multiply by the s derivative of w and you integrate by parts and this times the rho is the same way you deduce the energy for the wave equation now the thing is that this is not an energy that's conserved because of the extra terms in s derivatives but we don't worry this is better for us the derivative of the energy is a positive quantity so this is the miracle of this the derivative of the energy is this integral here so the energy is increasing by the way when the powers are smaller than the conformal power which is 3 in this case the energy is decreasing instead of increasing but anyway here it's increasing and again I have this now enormous power of 3 halves but the thing still converges because the w is 0 near y equal 1 and therefore it's s derivative it's a cylinder now is also 0 so I can write this so that's my first formula my second formula calculates so this is the derivative of the energy the second formula calculates and just to tell you how you get this you look at the equation and instead of multiplying by dsw times rho you multiply by w times rho and you integrate by parts and so this is our formula and of course these formulas may look awful at first but they grow over you get used to them and in fact they're very very nice and then the final thing is that I'm approaching the final time of existence which is log of 1 over delta and I look at my energy in the limit I get exactly the usual energy the usual energy of my solution and this is just a calculation there's nothing more than a calculation but see now we combine 1 and and 3 and you obtain immediately that this energy is bounded from above right because at the last time it is bounded so and it's increasing so before it's bounded and this is a very non-trivial bound believe me you don't see this with a naked eye independently of delta I'm sorry Carlos this is why does it go to d when s goes to log of 1 but I'm cautious here because I still have my weight so why yes so everything they compensate they compensate with the shrinking okay so it's a calculation that you do term by term it's surprising at first but remember there's the weight outside there's the 1 minus d to the 1 half in the w so you go back above you remember what you did in normalization and you computed that other question so now I'm going to tell you our first improvement our first improvement is this if it were very very naive you would get a bound of 1 over delta from the way the support is made because the support is in the ball of radius 1 minus delta so that's how far you can go in 1 minus y squared and what I'm saying here is that that's a bad bound there's an improvement and there's an improvement that's so strong that you get a log not just the power, better power okay how do you get the improvement so here really this is a truly parabolic thinking you're going to have to find the right test function to put into the equation to produce this and in the mozer theory of parabolic equations or Nash's theory logarithms play a role okay and so what you do is you do this so what this is I test my equation with w times log okay and then you get this very nice formula and you see that here in the part inside this bracket there's no weight there's only the log weight while here we have this weight so the log weight in here gives you the log of 1 over delta bound and so when we integrate the derivative goes away and we get the 1 over delta bound and now here when we integrate remember this is a bound for the integral from 0 to 1 it's not for each S it's for the integral by integrating you make things better and so this is the first term this is good and now we look at the others this one is good this one is good gives you the right bound and in this one you just use Cauchy Schwartz and then you're done so that's how you control this by a log and now you start cranking a machine where you're now going to improve the estimates now we got a log now we're going to get the square root of a log for one of the terms this one let's say okay so let me explain why this is true and that sits in this formula here let me put this term on the other side this term doesn't harm us because the energy is bounded from above remember that's the monotonicity of the energy and the fact that I knew what its limit so this term is okay this terms what do I do with them well the W term I have no trouble because of my hardy inequality remember W squared supports up to 1 minus y squared squared so this one is harmless and here well I have a dangerous term but I use Cauchy Schwartz and I use my previous bound and that gives me the square root of the log because of Cauchy Schwartz okay and then now I've got this terms my first one which somehow looks awful has the right sign so I can ignore it I just throw it away has the right sign and the other ones are fine by Cauchy Schwartz and what I just proved and the fact that on W I can put a lot okay so this is a very old structure again you're telling me that something if I take two derivatives of something right this is what you're saying right you're just going above so this is the very old structure of the self-similar equation so it's precisely that and you get information so but you have to know how to use it of course so the first thing you get is this and the second thing the second term follows from this one because if you remember the only negative term in the energy is the one with this and the energy is bounded from above and so I just integrate the energy okay the only term in the energy that's negative is the one that comes from this so I get that now I'm now going to go and improve this bound and what is interesting here is that the length of this is log 1 over delta but my bound is the power 1 half of log of 1 minus delta so I've been able to show that this is going to zero in some way from this bound so how do I do that well I just express this as the difference of the energy at this time and at this time the bound from above at this time I just throw away from below I use the minus log the previous step so that's all this is but now this is a very powerful lemma because of the corollary the corollary is that there's an s bar sub delta which is between 1 and log 1 over delta to the 3 fourths such that on something of length 1 eighth I get a bound by 1 over 1 eighth so this lemma follows from this one just by a pigeon hole argument I split the long interval into smaller intervals and count how many I can have and I have the computation down here I do it in this joint's intervals of this length the number is this 5 8 minus 1 eighth is 1 half and then I get that I'll tell you where I'm having I'm having to try to prove that w is independent of s I can do some manipulation to make this independent of s if I make the ds derivative zero that's how you make it and that's what I'm trying to here I'm going to try to make the dds zero and you see that since in this interval I get a very good bound when I take the average on the interval I get a very good bound because the interval is long therefore I'll be able to find the sequence in which this goes to zero there is some relation between 1 eighth and 3 fourths and presumably you could take any these numbers the important thing is that on a long interval I get the bound that goes to zero okay so now you do a little bit of we're going to eventually have to take delta going to zero for all of this to work and so I have to get a convergence and for that I use the compactness because I get all my things are compact and so when I take delta j is going to zero I can get compactness so I can find the sequence delta j going to zero such that this guy is converged to a W star and the W star is in fact independent of s but the next thing about it is that it okay it's independent of s but it cannot be zero because if it is zero it's a limiting point of one of the things in the compact set my original compact set and that means that U is very can be taken very small by modulation but then by the small data theory I cannot blow up in finite time okay so this W star cannot be zero so this W star of course will have to solve the same equation except that there are no s's there anymore because it's independent of s and so what I get is this degenerate elliptic equation for W star and I know that W star is not zero and I know that W star is zero on y equal to one now the crucial next point is that there's some extra bounds for my W star yeah this should be a star and it's that these two objects are finite so this limiting object is zero in the boundary but in the priori I have no idea why would this integral be finite and the point is that because of the way we chose it we took this W ys delta j's and we had this uniform bounds on them because of our formulas what Pierre called the Virial formula so we get these things that are uniformly bounded in j and therefore in the limit the W star has this boundedness but the W star doesn't depend on s so I can take away the s integral and just get the bound in y so capital s is capital s I'll show you what it is is the interval where this converges is an interval from the local existence theory okay so I got these things and now I conclude by using a unique continuation so this is where there's a connection between what I'm presenting well one of the connections and Adamar because Adamar was the first person to insist that one study unique continuation for equations without analytical coefficients and he did this in connection with the wave equation and the first person to actually be able to do something like what Adamar proposed was a Carleman in his work in the 30s and 40s Carleman's work that gives basically that this W star is 0 after after a lot of things so we have to show now that if we have a W star solving the degenerate elliptic equation with the two additional bounds it has to be 0 for y less than 1 minus 8 to 0 I have a linear operator with smooth coefficients with critical non-linearity by a well known argument due to Trudinger in elliptic theory we can show that W star is then bounded here once W star is bounded I can forget about the non-linearity and consider W to the 5 to be W to the 4th times W with and called W to the 4th V and say I have a bounded potential so now I have an elliptic equation or kind of an elliptic equation with a bounded potential and Carleman's theorem is that if it vanishes in an open set it has to be 0 but so if I can show that W star is 0 near y equals to 1 then that propagates inside so the whole action is near y equals to 1 so instead of giving you the actual proof I'll model the problem in slightly easier coordinates to read so I'm going to flatten the circle and so if I flatten the circle and I look at what my equation looks like it looks like this R now is the perpendicular direction and there I have a degenerate operator and there's some sign here that is wrong this is minus a half ok but ok so this is a typo this is minus a half in the tangential variables disease I don't have any degeneracy I have the real Laplacian times W star and then I have this W star to the 5 now I have our new estimates on this W star and these are our crucial estimates translated in this way and now there's a trick here which is that this degenerate equation can be desingularized by changing R to be a squared because it's the square root and if I do that and I call V W star of A squared Z then the A is that so our equation now becomes non-degenerate and I know that the thing is 0 but of course that doesn't allow me to say that the solution is 0 I need extra information I need the whole Cauchy data to be 0 but my extra information so these are bounds and my extra information gives me this that this thing is finite and the degeneracy of the equation forces the DDA desingularized equations to vanish on the boundary and now I have the whole Cauchy data vanishing and now I can use unique continuation or uniqueness in the Cauchy problem if you like to show that W star is 0 and that finishes the proof so it's a kind of a long journey but that finishes the proof of the theorem so let me conclude with this part of the course with a formulation that's slightly different of this alternative theorem so suppose that the energy is less than the energy of W if the whole H1 cross L2 norm is small then the solution exists forever and scatters and if it is big it blows up in both directions and the equality doesn't hold before our formulation was without the U1 but it turns out that the formulation with the U1 and without the U1 are equivalent and this is very odd and the first claim shows you why this is purely variational if the energy is less than the energy of W then the gradient is less than the gradient of W if and only if the other one happens and the same for above and this is playing with the variational estimates that we have it's because of this energy constraint so we leave this with that and now we're going to proceed if E of U0 U1 is less than E of W0 then and the gradient U0 under this hypothesis these two are equivalent but if you are given the energy E of U E of U0 and U1 you get an explicit bound for gradient of U0 squared less than gradient of W squared and presently if you add U1 squared the value will be different yes but it will still be small yes I'm not saying that they are equal no no but one is smaller if and only if the other one is smaller okay alright so now where we're heading next is to prove this soliton resolution in the radial case that's our next task and that will be what we will do until the end of Monday and then the second part of the course will prove the soliton resolution in the non-radial case for a sequence of times okay so far we are not radial no so far I avoided being radial but then to do this more difficult thing you first have to do it in the radial case so the first step is to study give a general study of solutions which are type 2 solutions so that means that they cease to exist in finite time but the norm remains bounded and as I mentioned in the first lecture there's many examples by now of such solutions and this is where the blow up happens by concentration so now I'm going to okay and I recall also that type 1 blow up means that the norm actually goes to infinity but a priori there could be something that's neither type 1 nor type 2 if it's bounded on one sequence and unbounded on another then it's neither type 1 nor type 2 and we will show that in the radial case that cannot happen there isn't such a thing okay there are no mixed asymptotics now the first thing I want to clarify is what does it mean to be a type 2 blow up solution and for that I introduced the notion of regular and singular points okay a point is called regular if or every epsilon there is an r that's independent of t such that this thing becomes small so this is the usual norm and I'm adding the norm for safety so what it means is that there's no concentration at the blow up time near this point at this point and we call x0 not regular if x0 is not regular we call it singular and now we call s the set of singular points so now let's first talk a little bit about these notions the theorem says this is a theorem with de Caire and Merrill that there's always at least one singular point if I'm type 2 blow up so that's what happened there was at least one singular point but that there's only finally many such and the finally many depends on this energy bound how many other say energy norm bound okay moreover you always have a weak limit as you approach the final time yeah this is a t tends to t plus and if you stay away from each one of these finally many singular points actually approaching strongly so this is the result and I will explain a little bit about the proof of this in a few minutes now then there's going to be a definition so of course you t for any sequence tn converges weakly to some limit after subsequence because of the boundedness assumption the point of this first statement is that the limit is independent of the sequence okay but of course if you believe that there's only finitely many points and that outside those points the limit is strong that gives you the uniqueness of the weak limit okay that's just functional analysis so the really important points are that there's only finitely many of these points and that away from the singular points this limit is strong so now let u be as in the theorem v be the solution which at the final time has this data v0 v1 so we call this v the regular part of u at t plus and the difference between u and v the singular part now the thing is that it's very easy to see that because of this strong limit away from each singular point the support of u minus the singular part is in this inverted cones centered around each of the singular points so let me maybe I draw a picture so the support has to be there because everywhere that I cut here from here on the two solutions are very very close so by finite speed of propagation there can't be anything except in this inverted cone in which they are different okay so now let me prove could it happen that you have two singular points which are reached at different times I'm worried that the backward cone of the right so not in the way I'm setting things up because I stop at the first time that there is a singular point so the later when you don't want to see I stop I mean that is possible if one develops instead the Cauchy horizon of the solution but in this way of looking at things we're stopping the first time there's a singular and then we just rest so that never happens because of that so the main point in the proof of our theorem is the following lemma there is a fixed number delta 1 such that if at this time t0 this is small okay the t plus is 1 in this and t plus belongs to 0 1 and I cut to this neighborhood then this has to have a limit and the reason for this basically is the small data theory I take the delta 1 to be constant in the small data theory and then I make a solution that equals phi times u of t0 dt of t0 at time t0 and then the small data theory gives me that that has a limit as you approach t equal to 1 because I had small data and finite speed of propagation will make it equal to this okay so this point is very is very simple and the next point is the corresponding point at infinity okay where I do it outside instead of near point but the corollary is the following if I have a singular point this thing has to remain bounded from below because if it were smaller I could use the previous lemma and the convergence in h1 cross l2 means that it will be a regular point now this already gives me that there's finitely many singular points because if there were infinitely many I can approach this becomes smaller and smaller as t goes to 1 and then I would get if I had k points I would get the bound of k times delta 1 for this h1 cross l2 norm but the h1 cross l2 norm is uniformly bounded so the number of points k has to be bounded okay so this already tells me that there's finitely many singular points and as you see here is the proof of this thing so now I'm going to start my preparation for the proof of the photon resolution thing in the radial case and so we do a little bit of gear shift and we're going to study some properties of linear radial solutions okay so for a solution of a linear wave equation the linear energy is 1 half of gradient x plus gradient t plus d dt squared density of the energy is that and recall this identity if I take the t derivative of the linear energy that's the same as the space divergence of this quantity okay and this is of course just differentiation and this is what gives you the fact that the energy is constant right because if we integrate this in x that's the derivative of the energy and the integral of the spatial divergence is 0 so the derivative of the energy integral is 0 so energy conservation follows for this so now I'm going to introduce for all non-negative numbers A what I call the outer energy which is that I integrate outside the the light cone displaced by A okay this is a terminology so the claim is that for all A bigger than or equal to 0 this is a decreasing function of t for t positive an increasing function of t for t negative I mean the way you picture it is if I integrate on a smaller set I should get something smaller of course this isn't really a justification for this claim because the function I'm integrating changes with each t but that's a mnemonic device to remember whether the increases or decreases okay so how do we prove that this is really true we integrate the derivative of the energy on x bigger than t and have to show that it's either bigger than or equal to 0 or less than or equal to 0 and we integrate it by this formula integration by parts so what we get is that the difference at time s0 and time t0 you can now do the integration by the divergence theorem in spacetime and what you get is this identity and this quantity here is called the flux okay and it will be important for us when we do the soliton resolution in the non-radial case but here it's important to us that it appears here and gives us that monotonicity so because of this monotonicity I always have the following limits that exist and this is a kind of interesting formula you can quantify by how much it increases or decreases in terms of the flux so now I'm going to prove this outer energy inequality that I had in the very first lecture it says that if v is a radial solution of a linear wave equation in 3d and a is a non-negative number then for all t positive or all t negative this is bigger than that roughly speaking it says that the outer energy either for t going to plus infinity or for t going to minus infinity does not tend to zero has a limit which is a fixed positive number which is this okay now this isn't quite true because I put the R inside instead of outside and we will I'll discuss what the difference is in a few minutes okay but let me prove this fact and as I mentioned the first time this is something really D'Alembert could have proved and the constant is one half so how do you do it suppose you take a radial solution of a linear wave in R3 then Rt to be R, V of Rt and you extend this to be oddly for t less than zero as it's well known it solves the ordinary wave equation in 1d this is a well known fact and you can check it there's nothing nothing but to differentiate okay and I will call f0 times V0 and f1 Rv1 which are the Cauchy data for this f now another important fact is that hard is inequality remember it says that V squared over R squared is integrable in 3d but the I mean the measure is R squared dR so what you get is that this thing is finite and because this thing is finite this f0 is actually in h1 because of course if we differentiate the V0 this is in h1 because V0 is in h1 but if we differentiate dR we get just V0 and V0 is in L2 by this fact okay so we have a data in h1 and in L2 for the wave equation in 1d now I look at the quantities z1 and z2 where I look at ddr plus ddt and ddr minus ddt and this parenthesis belongs in here anyway so of course the sum of the squares of this 2 gives me that so because the sum of the squares gives me this at least 2 at least one of them is bigger than 1 fourth all of that okay and this is what we'll choose whether this happens for t positive or for t negative so let's assume that it's i equal to 1 that has this being positive so that this one has this and then I will prove the theorem for t positive for t negative now the wave equation in 1d remember z1 has ddt plus ddr so ddt minus ddr of z1 is 0 okay that's so the wave equation factors as 2 transport equations in 1d that's all I'm using now because of that for tau smaller than r if I look at the tau derivative of this function I get 0 this function is constant so I have that this guy is constant along characteristics now I'm going to do it for t less than 0 and I start I say this is the same as that by this equality now I change variables and I get that and now I say that this is less than or equal to that and here I'm using this inequality and then this was bigger than the initial data that I have and I proved my inequality and all I used was that I have the 1d heat equation and therefore I have constant things on characteristics and I could decide whether it's for t positive or for t negative according to which of the two parts of the solution dominates so it's a decomposition into in going and outgoing waves and the in going waves are more powerful than the outgoing waves and that's all and that tells you what the time direction to use so as you can see I mean there's not much in this proof so now we will use the following calculation very often remember that I had this quantity here r h of r and I will develop it I use the product rule that I developed the square and then the other two terms combine to this and then this one of course I can integrate and I get that so now you see what's the difference between this and this precisely this so this term is always smaller than that one by this calculation so as a corollary on the left hand side we can replace what we had by the outer energy because that's bigger than this and we always have that now you could ask well why would you do something wasteful okay and the reason I do this is because I don't really care but for because for t going to infinity this guy will go to zero so for solutions of the wave equation so it will make no contribution why is that because we have this dispersive bounds suppose that v is in c0 infinity and solves the wave equation then for large time v of xt is bounded by 1 over t in 3d right n minus 1 over 2 is the decay rate 3 minus 1 over 2 is 1 okay and so that the extra term doesn't play a role now at time 0 though you cannot replace this by that and this is easy to see once you think of it because you choose your data v0 to be the Newtonian potential for r bigger than a and let's say constant to be 1 over a for 0 and a okay and v1 to be 0 so what is the solution of the wave equation with this data for r bigger than a plus t it's precisely 1 over r by uniqueness because the Newtonian potential solves the the plus equation away from the origin now what happens to this guy when you calculate on 1 over r the limit you get 0 the outer limit so on the left hand side you get 0 but of course this quantity is not 0 but if you replace that by that it is 0 because r times 1 over r is 1 and so the derivative is 0 so that's why we have to formulate it in this way which is the way you prove it I mean you can't prove something false but it wouldn't be a good idea now so you could ask what do you do in higher dimensions and there are corresponding inequalities in higher dimensions and to do that you can you think about what you're doing here about what you're doing here in the following way what you're doing is doing the orthogonal projection to the complement of this one-dimensional subspace of h1 cross l2 of r bigger than a given by 1 over r,0 so you take away the orthogonal projection to the bad subspace one-dimensional subspace and then that's what you have and that's the inequality and then once you see it this way in higher dimensions there are more terms to subtract and so in every odd dimension there's an inequality of this type where you subtract more and more terms depending on the dimension and it's always a finite dimensional space that you subtract but the dimension increases when the dimension of the space increases and that inequality is still true with constant one-half you are still using the one power r? no, no then you have to use 1 over n-2 but then other derivatives and so on have to come in okay now there's a few extra things that I want to do today the first is an extra property of the wave equation so this is a suppose you have a linear solution I'm not necessarily assuming that it's radial in this in this setting and I have parameters I rescale things and then I have a wave from the surface of the light cone the limit is 0 now to prove this in 3D first we can assume that the scaling parameters are all 1 by rescaling now you approximate your solution by compactly supported data which you can always do by density and then the next thing you do is you use the stronger Huygens principle and that tells me that this thing is supported here and so this integral is not just tending to 0 it is 0 and that's the end of this proof so you can do this in any odd dimension now how about this fact in even dimensions you cannot prove it like this but it is still true in even dimensions and we will see a proof of this maybe we'll see a proof so the meaning of this result is that morally for a large time the energy of your solution is concentrating on the boundary of the light cone that's what this is telling you because away from the boundary of the light cone which is this there's nothing we're going to notice the following thing which is a non-zero it's a dispersive property that non-zero solutions of the linear wave equation have and that's the following for all t bigger than 0 or less than or equal to 0 there is an r and an eta such that for all t bigger than or equal to 0 less than or equal to 0 there's always outside energy outside r this is always true for the wave equation and it's true for all infinite time so how do we prove that we have the tools to prove this it's very simple since it's non-zero the data is non-zero and this thing equals that because remember by integration by parts this went up at 0 minus v0 squared a it would be 0 times v0 squared so at 0 the two versions are the same so this quantity is non-zero but since it's non-zero we can give up a little bit on the integral and find it bounded from below once we have that this thing is bounded from below our corollary to our outer energy inequality proves that for either t positive or for t negative this is bounded by 8 and that's the proof okay so to do the soliton resolution in the radial case the key tool is to extend this to solutions of a non-linear wave radial solutions on the non-linear wave equation in 3d which are not the soliton w so this is our next task and this will last until the end for w this will be false because w if you think about it behaves like the Newtonian potential for large x it's 1 over 1 plus x squared to the 1.5 for large x this is 1 over x and why doesn't the Newtonian potential play a role in the linear case after all it's a linear object because you cannot close it at r equal to 0 to be a solution up to r equal to 0 and you can then find this r by where you stop being the Newtonian potential so there's some trickiness here that you have to be aware of okay so the first task is to show this and so in order to do this we will introduce the following notation this is a useful notation we're using it we're going to use it all the time suppose somebody gives me a u0 u1 in h1 cross l2 radial and a number r which is positive I'm going to call u0 tilde u1 tilde depending on r c sub r to be the solution where the u1 tilde I just chop off by 0 that's just in l2 so I'm allowed to do that and the other one I just fill in by u0 of r and the point of this truncation is this equality that the outer norm is exactly the norm of this truncated one alright and now we will use these things and the first proposition I'm not going to prove it but I just want to explain what it means is that if u is a global in time solution and such that for some r for both positive t and negative t this limit is 0 instead of being positive then there's not much that I can say about this the data is either compactly supported and of course if it's compactly supported such a thing will happen by taking r much larger than the support or it minus a scaled w is compactly supported and those are the two options so when this dispersive property doesn't hold in either side that's just something very specific about your solution and of course this is a non-trivial proposition but the key tool improving this non-trivial proposition is this outer energy bounds that we saw so we'll continue on Monday I think everybody is tired by the end of the week so we'll stop a little bit earlier as you didn't get two slides before about this concentration near the light comb I get some doubt that it is true no I'm teasing you so here we have lambda much bigger than think of lambda to be 1 and what goes to infinity is tn not 1 over tn so this should be tn over lambda n ok so that probably is the source of your doubt no the tn is going to infinity and lambda n is 1 ok so that was the typo other question ok so maybe one more question sorry don't so the result about weak limit which does not depend on sequence but this the proof heavily depends on finance and for other you have to find another proof but yes and this is an important use of finance