3/7 The energy critical wave equation
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00:00
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Transcript: English(auto-generated)
00:11
Okay, so after the preliminary discussions of further properties of these critical elements,
00:22
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, whose gradient is below that one of W, which has a momentum zero, and has the compactness property, where the
00:49
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
01:04
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
01:30
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
01:44
it couldn't have existed, the theorem had to hold. Okay, so that was the scheme of the proof. Any questions up to now? The lower bound of Flanders, he says it's similar, we just say that it's bounded.
02:06
No, this is bounded from, no. We're not saying that it is bounded above and below, so it's a one-sided bound. Of course, this bound from below implies such a bound, right?
02:27
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,
02:40
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 questions? Okay, so again, you know, the proof of something like this cannot be done quickly.
03:07
We have to be patient here. So we do cases. So we will do first the case when T plus is plus infinity, okay? And then we'll do the case when T plus is finite, okay?
03:31
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
03:43
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.
04:03
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.
04:23
Of all of these quantities, these 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 hardy's inequality, is controlled by gradient u squared, okay?
04:45
So all of these quantities have the same homogeneity, all right? And now, as I mentioned at some point, I think maybe the first day,
05:02
in elliptic equations, there's this well-known identity of Pohorzhaev that allows to prove that certain things have to be zero. And these are hyperbolic analogs of the Pohorzhaev identity.
05:23
That's the meaning of this. And there are more than one because we have space and time, okay? So the first identity, this, you know, you should remember that
05:44
c is basically x and phi is basically one, truncated, okay? So the first identity is that the derivative in T of x squared u dT u is equal to this, minus three-halves plus one-half.
06:03
And then there's a remainder because we've truncated, and the remainder is controlled in this way. Now, we can't just write x grad u dT u because that's a divergent integral.
06:22
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 dT u is grad T u squared
06:42
minus grad u 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.
07:01
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
07:20
have to have a momentum zero. And the reason was because if we perform the Lorentz transformation, we could decrease the energy 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
07:44
and you differentiate, you get the momentum and the negative of the momentum, okay? All right, so I'll have to use these identities.
08:04
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,
08:24
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,
08:42
that in the whole interval of existence, since it's t plus is infinite 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,
09:02
I have that c of e, okay? So basically, for each t, I have bounds from above and from below for various quantities, okay?
09:22
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.
09:44
Its 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
10:00
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.
10:24
Outside the ball, there's not much energy. And this is where we use the compactness and the fact that lambda is strictly bounded from below by its positive constant, all right? Okay, so these are the things we need to remember.
10:46
So this is a very full slide. Let's try to digest it. There are two lemmas. The first lemma says that for epsilon small, this c is an absolute constant.
11:05
So if epsilon is small and r is very big, I can find the t zero such that t zero is controlled by c times r, and for zero less than t less than t zero,
11:20
this is below r minus r naught plus epsilon, and at t zero it is exactly equal. So how do we think about this? Remember that one of our normalizations is that x of zero is zero and lambda of zero is one.
11:41
Okay? That was in our statement. So at t equals zero, 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 r zero of epsilon the first time.
12:07
And that's t zero. Okay? And the statement of the lemma is that I hit the first time below c times r for some constant c. Okay?
12:20
I hit that first time, that threshold of r minus r zero of epsilon by c times r. Okay? 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,
12:43
suppose that we were in the radial case, then x of t would be always zero. And then this already gives a contradiction. But we're not in the radial case, so we have to do one extra thing.
13:01
And the extra thing has to involve somehow the momentum. Of course, the momentum of any radial function is zero. You should remember that. It doesn't depend on any equation. You take a pair of radial functions, you multiply the gradient of one times the other, you get zero.
13:23
Okay? Okay, the second lemma says that for epsilon small r large, then this t zero, in fact, has to be bounded from below
13:42
by one over epsilon, r over epsilon. 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 zero, and that's why we get a contradiction.
14:02
That's why no such function can exist. One over epsilon is not smaller than c. So it's all in these two lemmas.
14:22
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.
14:41
We assume that it is not true, and then I never, up to c times r, I never reach the value r minus r naught of epsilon. Okay? And we will see that that's impossible. If my lemma is not true,
15:01
I never reach this value in the required interval. 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.
15:21
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.
15:40
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?
16:00
Well, what we know from our variational estimates is that this quantity is coercive. So this is strictly positive. So this whole thing is basically minus the energy.
16:21
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
16:42
over x bigger than r. Okay? If x is bigger than r, and t is between 0 and Cr, this thing is bounded from below by r0 epsilon by the triangle inequality, right? Because this is bigger than r minus r minus r0 minus epsilon,
17:04
so we get r0 epsilon. So this whole thing will be bigger than r0 plus epsilon. So if x is bigger than r, my region of integration is contained where this is bigger than r0 epsilon,
17:21
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.
17:41
That's the conclusion of this. So the derivative is negative of the size of c times z, because this guy cannot reach to cancel what this gives,
18:01
because epsilon is small. Now I'm going to give a pointwise bound on z, a bound from above. And then I'll combine the two things and I'll get the lemma. Okay, the pointwise bound from above is, okay, in here, let's look at the first term.
18:24
The large thing is x, right? x is bounded by r, and the rest is just simply bounded by the energy, of Cauchy-Schwarz. And the second term is the same,
18:44
because I multiply and divide by the length of x. And u over the length of x by Hardy is again bounded by the energy. And so I also get r times the energy for the second piece.
19:00
So all in all, I get that z is bounded by Re. Okay, now I integrate between 0 and Tcr, and I use the fundamental theorem, okay? So I get this from integrating zr prime,
19:22
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,
19:42
I get the contradiction. So I had to have reached Cr for this large C, all right?
20:01
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.
20:22
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?
20:41
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, this whole quantity is bigger than r0 of epsilon, okay?
21:04
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,
21:21
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,
21:42
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.
22:03
Okay. So the first thing I'm going to bound is yr of 0, okay? 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.
22:24
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,
22:44
and 0 otherwise, so I replace the size of r by r, and what I get is an integral outside r0 of epsilon, okay? So the bound of this I keep,
23:02
and this outside r0 of epsilon, by the definition of little r, 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.
23:20
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 its complement.
23:45
In this region, necessarily, x has to be bigger 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.
24:03
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?
24:26
So this is what I'm dealing with. Now in the part inside, I add and subtract minus x0 t0 of lambda t0, and that.
24:42
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,
25:00
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,
25:25
which is in 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.
25:43
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
26:00
because we know that the energy is positive. Anyway, it's a bound from above, and I'm bounding from below by the negative of this. Yes? Are you sure that the energy density is also you? No. No, I'm not. I'm not, but I'm bounding from above because I have a minus sign.
26:23
And so then I get all the terms. I got the two. Okay? So my total control of this from below is all of that, and if you look at it, if I make r large and epsilon small,
26:44
so this doesn't hurt because of the epsilon 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.
27:05
The way I'm asking is that you have an integral x plus xt0 over lambda t0 times the energy density, and you get a bound on that. Yes, so I'm going to bound from above. I'm going to get a bound from its absolute value from above,
27:21
and since I'm bounding from below, then I get a negative sign. So I put the absolute value inside the integral. Are you not assuming that the energy density is positive to the estimate? No, because the absolute value is outside. Now I put it inside because the absolute value of the integral
27:42
is less than or equal to the integral of the absolute value. Then the absolute value of this is bounded by r naught, and the absolute value of each one of these terms is controlled by the energy. I separate each term, yeah, okay. And then I get a negative.
28:01
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 large and epsilon small, it's very easy to see that I can get a bound by e r over 4. And now I put all my bounds together,
28:24
and the point is that on the right I had the epsilon because the momentum was zero, so the derivative had the bound by epsilon. And therefore t naught is less than epsilon over e r over 8.
28:43
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, okay?
29:01
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 rescaling.
29:21
So it's easier to write things with t plus equal one. So the first step is to prove that lambda not only is bigger than constant over 1 minus t, but it's also smaller than the constant over 1 minus t.
29:41
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.
30:03
We cannot do anything else. So we have to find something else then, okay? So we assume naught, and then you have a sequence tending to 1 such that this tends to infinity.
30:27
And we will see that this is absurd, okay? So it will be a contradiction. So now I look at z of t. It's the same object that I had before, but I don't need to truncate now.
30:45
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.
31:05
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 ce for all t. And the next thing I'm going to say is that the limit of z of t
31:24
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 this gives me a 1 minus t here. And the rest is bounded by the energy.
31:41
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.
32:01
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.
32:23
And I will see now that this and this are not compatible. And that's where the contradiction will come. I do that. So we show that z of tn over 1 minus tn tends to 0.
32:45
And this is a contradiction because I've just shown that it had to be bounded below. So now I'm going to exploit that the moment is 0. Remember, in the definition of z, I had x.
33:01
But for free, I can put x plus tn. x of tn over lambda tn because the momentum is 0. And I have the division by 1 minus tn. I have the division of 1 minus tn.
33:20
So now there's a few steps, but they're not too bad. So let 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.
33:50
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.
34:07
And I can apply Hardy at any origin. And that's why that term gives me the same bound. For fixed tn, I can apply Hardy in x with the origin wherever I choose it to be.
34:24
And I choose it to be at minus x over tn over lambda of tn. So then we have this bound. So this part of the integral divided by 1 minus tn gives me epsilon. So I'm in a good situation.
34:44
Now I have to see that the other part is not too large. So in order to do that, I first have to control x of tn over lambda of tn.
35:00
The center. And I claim that it has to be less than twice 1 minus tn. Really twice. I mean two. Not c times 2. Two. OK, why is that? If this is not true, the intersection of these two balls is empty.
35:22
That's a triangle inequality. And therefore this integral is zero over here because this is supported in here. And since this doesn't meet the support, the integral is zero. Now what happens on the other part?
35:45
The other part is this integral. That's the complement. In here, I can just rescale things and change variables. And I get this.
36:00
Now 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 zero.
36:20
So all of the gradient squared integrates to something that goes to zero. This cannot be because then my solution would be zero. And I'm assuming that it has a finite time blowup. So it's not the zero solution. Therefore, x of tn over lambda of tn is indeed smaller than 2 times 1 minus tn.
36:48
So I got the size of x of tn over lambda of tn. So now this is the quantity I have to estimate. And of course, x is smaller than 1 minus tn.
37:05
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 zero by the fact that this goes to infinity and the compactness.
37:27
And the other term is handled similarly. And then I get the contradiction that z of tn over 1 minus tn tends to zero. Well, I saw that it was bounded from below.
37:42
So at the end, the assumption that 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.
38:02
OK? So at this stage of the game then, we have found that lambda of t is of the size of 1 over 1 minus t. And believe me, at this point, you run out of usage of the virial identities. So you have to do something else.
38:23
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.
38:44
And this is a very simple argument. I don't think I want to go through it. It's just a few lines. It's not much to do. So please believe me that now we can assume that x of t is zero and lambda of t equals 1 over 1 minus tn.
39:05
So now we are in the self-similar situation. We have a family of solutions. We have a solution which has the compactness property, but the scaling parameter is 1 over 1 minus t.
39:20
And I have to show that this cannot happen. And this is somehow a crucial point in the theory. It's 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.
39:43
But here we're not ruling a solution which is exactly self-similar, but one which is self-similar up to compactness. So now what do we do?
40:02
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.
40:21
So we're going to think of the self-similar case as if we were dealing with a parabolic situation. And when you're dealing with parabolic situations and you want to understand the blowup, you introduce self-similar variables. That's the method that was pioneered by Giga and Kohn,
40:42
and it plays a role here. So we go to that. So somehow you change the universe. You introduce self-similar variables. This was introduced by Giga and Kohn in the parabolic case.
41:02
Merle and Zag used it in the wave equation case, but in a sub-critical regime. In fact, their work was in powers smaller than the conformally invariant power, which is strictly smaller than the energy critical power.
41:21
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,
41:42
and s is the log of 1 over 1 minus t. And you introduce w as 1 minus t to the one-half u of xt, and in the ys variables, it's like that.
42:01
And it's defined for s between 0 and infinity, and the support is always in the bowl of radius 1 because x over 1 minus t was less than 1 on the support.
42:20
And now for technical reasons, we have to introduce a parameter delta. And the technical reasons is so as not to divide by infinity 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.
42:48
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 s delta is this to the one-half times u.
43:03
And then we get that. And now this w is not defined up to plus infinity. It's 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,
43:24
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?
43:42
Both w and w delta, now you calculate the equation. And it turns out that it's some kind of a nonlinear wave equation. That's not surprising. There's a power 5. That's not surprising.
44:02
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,
44:27
where rho, where did I write rho? Oh, rho. Okay. You see this coefficient rho that generates at y equals 1.
44:43
So this equation now is a degenerate elliptic equation. So this is the price you pay. Now we get a degenerate elliptic equation instead of the ordinary Laplacian.
45:01
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 over the 2. And we have to see what does that do.
45:21
Now, a few comments about this elliptic equation. If you analyze it, you see that this can be written as rho, 1 over rho, divergence, and then the identity minus this rank 1 matrix, y tends to y.
45:46
So if y is strictly smaller than 1, I minus that rank 1 matrix is an elliptic matrix, because I don't reach the 1.
46:00
But as y tends to 1, I'm losing one direction of ellipticity. So this is an elliptic equation with smooth coefficients for y less than 1, but it's degenerate at y equals to 1.
46:27
So what we're going to do now is our universe is W. And we're going to try to compute what are the formulas in the W universe and see if we're going to be able to kill this W.
46:44
And that's how we're going to kill ourselves similar solutions. That's the plan. So first there are some easy remarks.
47:01
W is 0 at the boundary because U was 0 at the boundary. Remember, U when x equals 1 minus t is 0. So x over 1 minus t is y. So when y equals 1, W is 0. And that's extremely important.
47:22
Can you remind us, why is U 0? Because we had the support property in the inverted cone. Remember, from the compactness you get the… and then we rescale… Well, in the H1 sense.
47:41
So that's why it's only 0 in the H1 sense. Okay, it's not the pointwise sense. It's in the sense of traces in subsyllable spaces. So this is the correct technical.
48:01
And the gradient squared is controlled because when you change variables in the y, that's why you have the factor 1 minus t to the 1 half in front. 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, this is still true.
48:25
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 uniform in delta. And even true for delta equals 0.
48:46
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.
49:07
So this is another Hardy inequality. So all of these bounds are uniform in delta.
49:22
Now we introduce the right energy for this equation. So it's this expression.
49:42
And now you see what the purpose of the delta is. The delta 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.
50:08
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?
50:22
Well, you multiply by the S derivative of W and you integrate by parts. And this times the rho. And this is what you get. The same way you deduce the energy for the wave equation.
50:42
Now the thing is that this is not an energy that's conserved because of the extra terms in S derivatives. But, OK, we don't worry. This is better for us.
51:00
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,
51:24
which is 3 in this case, the energy is decreasing instead of increasing. But anyway, here it's increasing. And again, OK, I have this now enormous power of 3 halves, not just 1 half.
51:43
But the thing still converges because the W is 0 near y equal 1. And therefore its S derivative, it's a cylinder now, is also 0. So I can write this.
52:00
So that's my first formula. My second formula calculates... So this is the derivative of the energy. The second formula calculates the primitive of the energy. And just to tell you how you get this, you look at the equation and instead of multiplying by dsW times rho,
52:23
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,
52:43
but they grow over here. OK, 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.
53:01
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. It's nothing more than a calculation.
53:20
But you see, now we combine 1 and 3 and you obtain immediately that this energy is bounded from above. Right, because at the last time it is bounded, and it's increasing, so before it's bounded.
53:42
And this is a very non-trivial bound. Believe me, you don't see this with the naked eye. OK, independent with delta. So I'm sorry Carlos, this is confusing. Why does dt del omega s,
54:01
why does it go to e when s goes to log of 1 over delta? I'm confused here because I still have my weight, so why? Yes, so everything comps, they compensate. They compensate with the shrinking. OK, so it's a calculation that you do term by term.
54:25
Yeah, 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. That plays a role. So you go back above, you remember what you did in renormalization, and you computed that. Yeah, OK.
54:47
Other questions? So now I'm going to tell you our first improvement.
55:01
Our first improvement is this. If you 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.
55:23
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, a better power. OK? How do you get the improvement? Well, so here, really this is a truly parabolic thinking.
55:45
You're going to have to find the right test function to put into the equation to produce this. And in the Moser theory of parabolic equations,
56:02
or Nash's theory, logarithms play a role. And so what you do is you do this. OK, there's an identity. So what this is, is that I test my equation with w times log.
56:30
And then you get this very nice formula. And you see that here, in the part inside this bracket, there's no weight.
56:45
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,
57:02
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.
57:21
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-Schwarz. Then you're done.
57:43
OK, 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've got a log. Now we're going to get the square root of a log for one of the terms.
58:03
This one, let's see, OK? So let me explain why this is true. And that sits in this formula here.
58:22
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 its limit. So this term is OK. These terms, what do I do with them?
58:43
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,
59:00
but I use Cauchy-Schwarz and I use my previous bound. And that gives me the square root of the log because of Cauchy-Schwarz. And then now I've got these terms. My first one, which somehow looks awful,
59:23
has the right sign, so I can ignore it. I just throw it away. It has the right sign. And the other ones are fine by Cauchy-Schwarz and what I just proved. And the fact that on w I can put a lot.
59:46
I'm sorry, Cauchy, this is a virial structure again. You're telling me that something, if I take two derivatives of something, I get your guide? This is what you're saying, right? You're just going above. So this is the virial structure of the self-simulation.
01:00:01
Yeah, so it's precisely that. And you get information. But you have to know how to get it. Of course. So the first thing you get is this.
01:00:21
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.
01:00:43
OK, 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.
01:01:09
But my bound is the power one-half of log of one minus delta. So I've been able to show that this is going to zero in some way from this bound.
01:01:24
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. And then the bound from below I use the minus log in the previous step.
01:01:54
So that's all this is. But now this is a very powerful lemma because of the corollary.
01:02:05
The corollary is that there is an s bar sub delta, which is between 1 and log 1 over delta to the three-fourths, such that on something of length one-eighth I get a bound by 1 over one-eighth.
01:02:32
So this lemma follows from this one just by a pigeonhole argument. I split the long interval into smaller intervals and count how many I can have.
01:02:48
And then I get this. I have the computation down here. I do it in this joint's intervals of this length. The number is this. Five-eighth minus one-eighth is one-half. And then I get that.
01:03:08
I'll tell you where I'm heading. I'm heading to try to prove that w is independent of s. I can do some manipulation to make this independent of s.
01:03:22
If I make the ds derivative zero, that's how you make it independent of s. And that's what I'm trying to have here. I'm going to try to make the d ds zero. And you see that since in this interval I get a very good bound,
01:03:50
when I take the average and 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.
01:04:00
There is some relation between one-eighth and three-quarters, and presumably you could take any... Yeah, these numbers. The important thing is that on a long interval I get the bound that goes to zero.
01:04:21
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 going to zero, I can get compactness.
01:04:46
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.
01:05:01
But the next thing about it is that 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 can be taken very small by modulation.
01:05:24
But then by the small data theory, I cannot blow up in finite time. So this w star cannot be zero. So this w star of course will have to solve the same equation
01:05:40
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.
01:06:04
Now the crucial next point is that there are some extra bounds for my w star. Yeah, this should be a star. And it's that these two objects are finite.
01:06:23
So this limiting object is zero in the boundary. But at 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 y s delta j's
01:06:46
and we had these 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.
01:07:03
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. Capital S, I'll show you what it is,
01:07:21
is the integral where this converges. It's an integral from the local existence theory.
01:07:40
Okay, so I got these things. And now I conclude by using unique continuation. So this is where there's a connection between what I'm presenting, well, one of the connections, and Adam R, because Adam R was the first person to insist that one should study unique continuation
01:08:04
for equations without analytic coefficients. And he did this in connection with the wave equation. And the first person to actually be able to do something
01:08:27
like what Adam R proposed was Carleman in his work in the 30s and 40s. And it is Carleman's work that gives basically that this w star is zero, okay, after a lot of things.
01:08:51
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 zero.
01:09:04
For y less than one minus eta zero, I have a linear operator with smooth coefficients with critical nonlinearity. By a well-known argument due to Trudinger in elliptic theory,
01:09:22
we can show that w star is a bounded here, okay? Once w star is bounded, I can forget about the nonlinearity and consider w to the five to be w to the fourth times w and call w to the fourth v and say I have a bounded potential.
01:09:44
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 zero, okay?
01:10:04
So if I can show that w star is zero near y equals to one, then that propagates inside. So the whole action is near y equals to one. So instead of giving you the actual proof,
01:10:22
I'll model the problem in slightly easier coordinates to read. So I'm gonna flatten the circle. And so if I flatten the circle and I look at what my equation looks like, it looks like this, where r now is the perpendicular direction,
01:10:44
and there I have a degenerate operator and there's some sign here that is wrong. This is minus a half, okay? But, okay, so there's a typo, this is minus a half.
01:11:01
In the tangential variables, the z's, I don't have any degeneracy, I have the real Laplacian. Then I have c times w star and then I have this w star to the five. Now I have our new estimates on this w star and these are our crucial estimates translated in this way.
01:11:27
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.
01:11:41
And if I do that, and I call v w star of a squared z, then dA is that. So our equation now becomes non-degenerate and I know that the thing is zero, but of course that doesn't allow me to say that the solution is zero,
01:12:01
I need extra information, I need the whole Cauchy data to be zero. But my extra information, so these are bounds, and my extra information gives me this, that this thing is finite.
01:12:20
And the degeneracy of the equation forces the d dA in the desingularized equation 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,
01:12:40
to show that the w star is zero. And that finishes the proof. So it's kind of a long journey, but that finishes the proof of the theorem. Okay? So let me conclude with this part of the course
01:13:05
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,
01:13:24
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.
01:13:41
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.
01:14:03
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.
01:14:23
And now we're going to proceed. Yes? If E of u, u0, u1 is less than E of w0, then the gradient u0... No, then under this hypothesis, these two are equivalent.
01:14:44
But if you are given the energy 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,
01:15:01
then the bound will be different. Yes, but it would still be small. I'm not saying that they are equal. But one is smaller if and only if the other one is small. OK? All right. So now where we're heading next
01:15:21
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 is to prove the soliton resolution in the non-radial case for a sequence of times.
01:15:42
OK? So far, you were 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.
01:16:02
So the first step is to give a general study of solutions which are type 2 blow-up solutions. So that means that they cease to exist in finite time,
01:16:22
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.
01:16:45
So now I'm going to... OK, 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
01:17:03
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.
01:17:21
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 introduce the notion of regular and singular points. A point is called regular if for every epsilon
01:17:41
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 Hardy norm for safety. OK? So what it means is that there's no concentration
01:18:02
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.
01:18:29
So now let's first talk a little bit about these notions. The theorem says, this is a theorem with Duque, Heyer, and Merrill,
01:18:42
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 finitely many such.
01:19:00
And the finitely many depends on this energy bound. How many are they saying? Energy norm bound. OK. Moreover, you always have a weak limit as you approach the final time. Yeah, this is as t tends to t plus. And if you stay away from each one of these finitely many singular points,
01:19:26
you are actually approaching strongly. So this is the result. And I will explain a little bit about the proof of this in a few minutes.
01:19:45
Now then, there's going to be a definition. So, of course, u t for any sequence t n 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.
01:20:02
OK. But, of course, if you believe that there's only finitely many points and that outside those points the limit is strong, that gives to you the uniqueness of the weak limit. OK, that's just functional analysis.
01:20:24
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. OK. So, now let u be as in the theorem, v be the solution,
01:20:44
which at the final time has this data v zero v one. So we call this v the regular part of u at t plus and the difference between u and v the singular part.
01:21:02
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 these inverted cones centered around each of the singular points.
01:21:25
So maybe I draw a picture.
01:21:41
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.
01:22:08
So now let me prove… Could it happen that you have two singular points which are reached at different times? Or is it the backward cone of the…
01:22:22
Right, so no, 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 one you don't want to see. Right, I stop. I mean, that is possible if one develops instead the Cauchy horizon of the solution.
01:22:42
But in this way of looking at things, we're stopping the first time there's a singularity. And then we just rest. So that never happens because of that. OK? OK.
01:23:02
So the main point in the proof of our theorem is the following lemma. If there is a fixed number delta one such that if at this time t zero this is small,
01:23:22
OK, the t plus is one in this, and t plus belongs to zero one, and I cut to this neighborhood, then this has to have a limit.
01:23:40
And the reason for this basically is the small data theory. I take the delta one to be the constant in the small data theory, and then I make a solution that equals phi times u of t zero dt of u of t zero times t zero.
01:24:04
And then the small data theory gives me that that has a limit as you approach t equal to one because I had small data. And finite speed of propagation will make it equal to this. OK?
01:24:21
So this point is very simple. And then the next point is the corresponding point at infinity, OK, where I do it outside instead of near a point.
01:24:43
But a 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,
01:25:03
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,
01:25:21
this becomes smaller and smaller as t goes to one, and then I would get, if I had k singular points, I would get the bound of k times delta one 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.
01:25:45
OK? So this already tells me that there's finitely many singular points. And as you see here is the proof of this thing. OK. So now I'm going to start my preparation
01:26:06
for the proof of the soliton resolution thing in the radial case. And so we do a little bit of a gear shift, and we're going to study some properties of linear radial solutions.
01:26:24
OK? So for a solution of the linear wave equation, the linear energy is one half of gradient x plus gradient t plus d dt squared.
01:26:45
And so the 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.
01:27:05
And this is, of course, just the differentiation. And this is what gives you the fact that the energy is constant. Because if we integrate this in x, that's the derivative of the energy,
01:27:21
and the integral of a spatial divergence is zero. So the derivative of the energy integral is zero. So energy conservation follows with this.
01:27:43
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 light cone, displaced by a. OK?
01:28:03
This is a terminology. So the claim is that for all a bigger than or equal to zero, this is a decreasing function of t for t positive and an increasing function of t for t negative.
01:28:29
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
01:28:41
because the function I'm integrating changes with each t. But that's a mnemonic device to remember whether it increases or decreases. OK? So how do we prove that this is really true?
01:29:04
We integrate the derivative of the energy on x bigger than t plus a and have to show that it's either bigger than or equal to zero or less than or equal to zero. And we integrate it by this formula, integration by parts.
01:29:21
So what we get is that the difference at time s zero and time t zero, you can now do the integration by the divergence theorem in space-time. And what you get is this identity. And this quantity here is called the flux.
01:29:41
OK? 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,
01:30:01
I always have the following limits that exist. And this is a kind of an interesting formula. You can quantify by how much it increases or decreases
01:30:20
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,
01:30:43
then for all t positive or all t negative, this is bigger than that. So roughly speaking, it says that the outer energy, either for t going to plus infinity or for t going to minus infinity,
01:31:01
does not tend to zero. It has a limit which is a fixed positive number, which is this. OK? Now, this isn't quite true because I put the r inside instead of outside.
01:31:23
And I'll discuss what the difference is in a few minutes. But let me prove this fact. And as I mentioned the first time, this is something really D'Alembert could have proved.
01:31:41
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, if you look at f of r, t to the r, V of r, t, and you extend this to be oddly, oddly
01:32:01
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.
01:32:21
OK? And I will call f0 r times V0 and f1 r V1, which are the COSY data for this f. Now, another important fact is that Hardy's inequality,
01:32:43
remember it says that V squared over r squared is integrable in 3D, but the measure is r squared dr. So what you get is that this thing is finite. And because this thing is finite,
01:33:01
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 the r, we get just V0, and V0 is in L2 by this fact. OK? So we have data in H1 and in L2
01:33:23
for the wave equation in one dimension. Now I look at the quantities Z1 and Z2 where I look at d dr plus d dt, and d dr minus d dt, and this parenthesis belongs in here.
01:33:43
So, of course, the sum of the squares of these two gives me that.
01:34:01
So because the sum of the squares gives me this, at least, over 2, at least one of them is bigger than 1 fourth, all of that. OK? And this is what we'll choose, whether this happens for t positive or for t negative.
01:34:24
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 negative. Now, the wave equation in 1D,
01:34:43
remember Z1 has d dt plus d dr, so d dt minus d dr of Z1 is 0. So the wave equation factors as two transport equations in 1D. That's all I'm using.
01:35:01
And 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,
01:35:20
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.
01:35:41
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.
01:36:03
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 ingoing and outgoing waves, and you have to decide whether the ingoing waves are more powerful than the outgoing waves,
01:36:21
and that's all. That tells you what the time direction is. So as you can see, there's not much in this proof. Now we will use the following calculation very often.
01:36:44
Remember that I had this quantity here, rH of r, and I will develop it. I use the product rule. I get that. I develop the square, and then the other two terms combine to this.
01:37:01
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. It's precisely this. Okay? So this term is always smaller than that one
01:37:21
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,
01:37:42
well, why would you do something wasteful? Okay? And the reason I do this is because I don't really care, because for T going to infinity, this guy will go to zero. So for solutions of the wave equation, we'll make no contribution.
01:38:02
Why is that? Because we have these dispersive bounds. Suppose that v is in C zero infinity and solves the wave equation, then for a large time, v of x t is bounded by one over t in 3D. Right? N minus one over two is the decay rate.
01:38:22
Three minus one over two is one. Okay? And so that extra term doesn't play a role. Now, at time zero, though, you cannot replace this by that. And this is easy to see once you think of it,
01:38:43
because you choose your data v zero to be the Newtonian potential for r bigger than a, and let's say constant to be one over a for zero and a. Okay? And v one to be zero. So what is the solution of the wave equation
01:39:02
with this data for r bigger than a plus t? It's precisely one over r by uniqueness. Because the Newtonian potential solves the Laplace equation away from the origin.
01:39:22
Now, what happens to this guy? When you calculate on one over r the limit, you get zero, the outer limit. So on the left-hand side, you would get zero. But of course, this quantity is not zero.
01:39:41
But if you replace that by that, it is zero because r times one over r is one, and so the derivative is zero. Okay? 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.
01:40:02
Well, you could, 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.
01:40:21
And to do that, 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
01:40:43
of H1 cross L2 of r bigger than a given by one over r comma zero. 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.
01:41:02
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 there's always a finite dimensional space that you subtract,
01:41:22
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 over r? No, then you have to use one over n minus two.
01:41:40
But then other derivatives and so on have to come in. 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 we have a linear solution. I'm not necessarily assuming that it's radial in this setting.
01:42:05
And I have parameters, I rescale things, and then I have that away from the surface of the light cone, the limit is zero. Now to prove this in 3D,
01:42:24
first we can assume that the scaling parameters are all one 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 Strong-Huygens Principle.
01:42:46
And that tells me that this thing is supported here. And so this integral is not just tending to zero, it is zero. And that's the end of this proof. So you can do this in any odd dimension.
01:43:01
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, okay? Maybe we'll see a proof. So the meaning of this result is that morally, for a large time,
01:43:21
the energy of your solution is concentrating on the boundary of the light cone. That's what this is telling you. It's away from the boundary of the light cone, which is this, there's nothing.
01:43:41
Okay? So now I'm going to notice the following thing, which is a non-zero, is a dispersive property that non-zero solutions of the linear wave equation have.
01:44:06
And that's the following. For all t bigger than zero or less than or equal to zero, there is an r and an eta such that for all t bigger than or equal to zero
01:44:23
or less than or equal to zero, there's always outside energy, outside r. This is always true for the wave equation. And it's true for all infinite time.
01:44:42
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. Okay? And this thing equals that.
01:45:03
Because remember, by integration by parts, this went up at zero minus v zero squared. A, it would be zero times v zero squared. So at zero, the two versions are the same. So this quantity is non-zero.
01:45:22
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,
01:45:44
this is bounded by eta. And that's the proof. So to do the soliton resolution in the radial case, the key tool is to extend this to solutions of the nonlinear wave, radial solutions
01:46:02
of the nonlinear wave equation in 3D, which are not the soliton, w. Okay? So this is our next task. And this will last until the end.
01:46:23
For w, this will be false. Because w, if you think about it, behaves like the Newtonian potential for large x. Right? It's one over one plus x squared to the one half.
01:46:41
For large x, this is one over x. And why doesn't the Newtonian potential play a role in the linear case? After all, it's a linear object. So you cannot close it at r equal to zero to be a solution after r equal to zero. And you can then find this r
01:47:03
by where you stop being the Newtonian potential. Okay? So there's some trickiness here that you have to be aware of. Very good.
01:47:21
So the first task is to show this. So in order to do this, we will introduce the following notation. This is a useful notation. We're going to use it all the time. Suppose somebody gives me u0, u1 in H1 cross L2,
01:47:43
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 zero. That's just in L2, so I'm allowed to do that.
01:48:02
And the other one, I just fill in by u0 r. And the point of this truncation is this equality. That the outer norm is exactly the norm of this truncated one. All right?
01:48:22
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,
01:48:44
and such that for some r, for both the positive t and negative t, this limit is zero instead of being positive, then there's not much that I can say about this. Then the data is either compactly supported,
01:49:02
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.
01:49:21
So when this dispersive property doesn't hold in either side, it tells you something very specific about your solution. And of course this is a non-trivial proposition. But the key tool improving this non-trivial proposition
01:49:42
is this outer energy balance that we saw. So we'll continue on Monday. I think everybody's tired by the end of the week, so we'll stop a little bit earlier.
01:50:03
I didn't get two slides before about this concentration near the light cone. I get some doubt. That it is true? No, I'm teasing you, Jesse. So here we have lambda much bigger than...
01:50:21
Think of lambda to be one. And here what goes to infinity is tn, not one over tn. So this should be tn over lambda n.
01:50:40
So that probably is the source of your doubt. No, the tn is going to infinity, and lambda n is one. So that was the typo.
01:51:00
Other questions? Okay, so maybe one more question. So there was this result about weak limit, which does not depend on sequence. But this, the proof, heavily depends on finite speed. Of course, yes. And for other...
01:51:20
You have to find another proof. Yes. And this is an important use of the finite speed of propagation.
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