Okay, folks, so good afternoon. And our goal was to actually prove interior regularity for weak solutions off elliptic differential equations. Alright? So let me remind you how somebody situation u is bounded open subset of our rent. >> And we have the operator L given by l u is minus D plus B plus C, where AIJ is positive-definite and symmetric for all x and u. >> And so that's the operator. We are given that f is an L2, u and u, and each one is a weak solution. 4f L u equals f on you. Okay? >> Show, show that for any v which is compactly contained in U, we have hue is an H2 v and we have the estimate. >> So who was originally in each one, but now be, prove you as NH2, right? That's a great regularity in the interior. >> Only though it comes with an S2, it less than equal to c times plus times two. >> That was our goal, right? >> So I'm not going to repeat. This will actually also be, have I forgot to add here, so a B, C, or an L infinity, you, that's also true. So this is this has For the elliptic operator, all this stuff. So this is the new thing. So given f and l to u and u and h one u and u is a weak solution of that. We explain the definition of what that means. And we have to show that actually because of the ellipticity in this positive definiteness, you is actually an H2 v for any v compatibly contained in U. And it comes with an estimate. >> Okay. >> So we had discussed the statement of this and I had also shown you that this will follow from. >> So to prove this, he showed it was enough to prove the following estimate, right? >> We had to just show the following, that this difference, a difference quotient or the first order derivative of u on L2 or fee was bounded by a constant times, right? And I did not explain how you can replace this H one u by L2 you, but I said I will give that as a homework, right? But so let's call this for today's lecture. Let's call it one. Alright, so our goal was to prove this estimate with C dependent only on u, v and the coefficients of OK, we had to prove this. And this is for all small. >> And actually we just remember we just need to prove this for just a sequence going to 0. >> But that's why we had to do it for some sequence. So we prove it for just small enough H. That's what we're going to do. >> Okay, so we also showed that it was showed wildtype. >> Yeah, just said homework. >> It was enough to prove one. >> When the BIM C was 0 because they're really taught them they can just sort of be absorbed in this. >> Ok? >> So here is where we stand right now. Okay? >> And this is where PR, so given u is, sorry, you're given F is an l to u, and u is in each one you, okay? >> And it's a weak solution of value equals f. So what we have is ai j d u d v is equal to integral f times v for all v n h 01. Let's call it two. Ok? That's what we are given. Any idea is positive-definite. >> Of course, we have to show that dk h u squared everything, no L2 of v is less than c times this. >> Okay? That's what you have to show less clothes for h, small enough. Okay, so and I discussed the idea of the proof, OK, that we are going to have. Sorry, this is squared. So this is what I'm going to call my second-order term. Okay? And these things are first-order terms. This has like gradient of u and then it's quotient, difference quotient. So it's the second-order term, like a second-order derivative. And f, I'm thinking about data out, so I'll just included my first-order term, and that's the first order derivative of u. Alright? So we have to do this and remember my trick is going to be That if you get S square is less than C1 times second-order, first-order plus c2 times f squared, then epsilon trick will give us three. Okay? So this is what I'm going to do. I'm going to get S square, ok? And that is going to come from this, the S square from the left-hand side. >> Ok? >> So S square will come from there, but it's actually not going to be. So this will be greater than or equal to S square minus some terms which I can, which will be a product of S and N, f and f squared. So because it's greater than S square minus some terms which are this. I can bring the minus terms to the right, right? I mean, if I get, what I am trying to say is if I get S square minus c1, SF minus c2, f square less than equal to c one prime sf plus c2 prime f square. If I get that, then I can bring it all to this side. Alright? And I will get that. So that is what I am going to do. Alright, so the, this term here, the S squared, this is the crucial thing to get this S square, and that's going to come from here. >> Okay? >> And then this stuff is going to be from here. So the important, most important relation is that from this left hand side to get this, ok, that's the most critical thing is S square. So when I'm doing my calculations, I will point out to you where is the square, whereas the SF and where does the f square? >> Okay? >> So the S square from the left here should have a positive sign. It's required to, otherwise it's not going to work. And then negative signs can be with S, f and f squared on this side doesn't matter. Okay? So again, so I am not writing the haven't written the proof that I'm just trying to go over this again, so let me just copy the whole thing again, right? So you are given F is unable to u, u is in H one, u, u d v is equal to o. >> In each one, you have to show C1, L2. >> You can do everything with squares. >> All right. >> That's what you have to show. Okay? So how are we going to get from this here, this second-order term, right? This is just first-order terms. If I just take V to be, remember, I have to choose my v to be an 80 u, one u. Okay? So how am I going to get this sort of second-order term, Israeli first-order, second-order quotient. It's by choosing this v appropriately. Ok? So, so you are given V is compactly contained in U. >> Ok? >> So this is sorry for small enough. Okay? Now, given this v is contained in u, because this quotient difference here, even if I'm working on v, this caution difference uses you and a larger set, right? So it makes it a little inconvenient. So what we have to do is we have to deal with it. We have to introduce two more spaces. Set, say here is my u, here is my v. Then I'm going to introduce w1, which is bigger than W2, which has even bigger. >> Okay? >> So choose open sets W1, W2, such that v is compact cotangent. W1, which is compactly contained in W2, which has compactly content in New York. All right, so then the other issue is, remember I can only use v, which are in eight, which means, you know, its support has to be inside you, right? So I cannot choose vi to be just the difference of the difference quotient off of you because that will just get bigger. So I have to first cut off things. >> So choose chi. >> Belong to c is here to infinity R n such that chi is one Henri and support of chi is inside W. >> Okay? >> So let me redraw this picture again so that we see it. Here is you. So let's start with V. >> Okay? >> Then there is w1 surrounding it, W2 surrounding it, then there is eo. So chi is one on this. >> And its support is inside this. >> Okay? So when I take the quotient difference on w1, its support is going to lie in W2 and then its works. Okay? So you have to do this tricky with, okay? So I'm going to take, so h is something that's two times h less than the distance of w1 from. Because these things. So take the distance of w1 from the boundary of that Bluetooth v to be e to the power minus h chi-square of Yoav pitch. You can note this has then we belongs to, of course, each one of you. And notice that support, right, chi is cut-off on W1. So and then you take the difference quotient different. So you get slightly bigger, but you get support of v is in W2, so v belongs to h. Remember this? I'm going to use it later on. Okay? Alright, so below, okay, we're going to have indices. >> I'm just going to drop the K, okay? >> When I write, I will mean it. Otherwise there'll be too many indices going around. >> Okay? >> So I just want to don't want to confuse it. So I'll just use DH. Well, I really mean it. Alright, so let us remember here, because the calculation, we get a little messy. So I just wanted to see, I want to get second-order difference, which is the quotient difference quotient of grad u to be bounded by the L2 norm of this is u, not v, and the h one norm of u. Okay? Second-order, and I'll call this first-order alcohol, this first-order, alright? And what we have for going for us is just this relation. Okay. And I'm going to use this p. All right? So let's start with the, so we get, therefore we get d t. >> Remember I dropped the K D minus h of chi-square of d H of u. >> Ok? Now remember that these things are supported inside W2, right? So I can do the 0 extension and I'm going to do this integral r. And remember, because what I want to do is I want to integrate by parts. Okay? Now remember I want this. Cool. Let's go back here. I want a d, k, h, and grad u, right. So I have a DH here with a detail, but I have another DH here. I want to bring it to this side. Writes I wanted to be an integration by part. So that is best done on R n, right? Because there was, he was home. It's okay. So I just wanted to convert this integral, one integral r n. And I can do that because the support of both of these things is contained inside W2. So I could just 0 extension. So therefore I can write this as r in AIJ you okay? Second thing is the DJs and the D minus h, Let's just a quotient, right? So it's quotient and derivative just commute. So you can just rewrite this as t minus H d j of chi square d H of u. >> And this is also where I've used the fact that d j of t k minus Hs. >> Okay? >> That's, that's, there's nothing much there, it's just a writer and the expression, then you will see they are the same. Okay, so now let's work with the left hand side. Okay? So let's call this, I forget what number we are on three. Okay, so let's look at the left hand side, this side, which is supposed to give me the S squared, the second-order square term Okay. So first thing I do is I need to shift as t minus H to the right. I'm sorry I made a mistake. We have to choose has already got a minus. I want to choose with a minus. So this minus here, minus here, minus here. Okay? And the reason I chose minus has because when I integrate by parts, the minus becomes a plus. So left hand side is integral from that integration by parts formula for quotient differences is d i j d u times d j kai squared e h e. Okay? Alright, so now this is second order term. And this is like a second-order term except as an idea and as a chi-squared, but we'll handle that. So you'll see we're headed towards getting a second-order square. We're not quite there yet. Okay, I don't want the chi-square inside here, the DJ, and I don't want the AIJ inside here, but that is from the product rule formula for caution differences. So this has equal to shift this more so long expressions, okay? Product rule says, okay, you apply ds to this and it just results in a shift. >> So it's sorry, no, it's AIG h, which means the shaft th ok. So d h d plus then you put the DH Olympus and d times. >> Okay, so now this DJ I want to put inside. So one way is to take it all the way through. The second is to actually apply it to the chi squared. I'm just going to call it chi prime at some data where two, it doesn't matter which one, okay? It's not important what derivative it is as long as it's just some sort of data. Ok? Now remember, this just means it's a shift by H. Not even a caution with us to shift. Okay. So the left hand side has four terms. And let's just, before I do anything, let us see. Okay, so this is a second-order terms. This is a second-order, second-order, second-order, this is quotient of derivative question is something that some first-order term, okay? Because it's bounded by a number, and this is a first-order term. This is a first-order too. So if you notice the left hand side is second-order, second-order, second-order square. That's at Tom, I want everything else. This is first-order and second-order can be a negative. I can put a negative here and bring it to the right. No problem. First-order, first-order, no problems. >> Okay? >> And I just want to, why is it bounded? >> This is a minor thing, but I want you to think about it, right? >> Notice that this quotient has eight at the bottom, right? So you have to think about this. But anyway, I'll just say it's bounded. So this is equal to or greater than or equal to, I. Keep this term R n, then it's a h t. This chi squared is going to be important. Chi of DAH, DAH, DAH, DJ you. Okay, so AIJ off the second-order, second-order. So this is where the positive definiteness will play a role is greater than or equal to everything else. I can put a minus because everything else is difference of a first-order term, N times the second-order term. >> Okay? >> So everything else is the second-order, second-order multiplied. Everything else is first-order, second-order, first-order squared. So it can go to the right. So I can do it with a minus, minus times some constant of RN. Okay? It's first-order, second-order absolute chi squared. Well, I don't need the absolute value for that. >> Chi-square, DIO, second-order ph, and commute. >> So it doesn't matter. I can write this, THE DJ you and you can take the absolute value, okay? And then you get first starter and that there are all these constants. So plus absolute Chi, absolute chi prime. So this is second-order, second-order, first-order, second-order, first-order, first-order. Okay? So this has worked, squint work. >> And now H is positive definite, right? >> Aig is positive definite. And even if you shift, it doesn't matter to be positive definite. So this has greater than or equal to okay? So this is absolute value of chi of DH. That's what that vector, right? So it's the grad u minus some constant times this term here. >> I just have to, okay? >> So I'm going to use the Cauchy-Schwarz inequality. So, so I'm going to take one chi with this and one chi with that, right? So I get norm, well, it's summation. So I'll just put a grad and its overall Iran but with the chi because supporters the chi, this has an L2 of q times d d H of u. So there's a chi also there. Each of you that's from these two terms. And then these two terms, and there's this guy I'm going to take with that. >> Then you will see in a second why I need those types. >> Okay, so this is the S square terms that I needed. That was exactly what I needed. That is S square. Okay? This has first-order, this has exactly, this says, alright, this first-order. Okay, this is first-order, but it's not exactly the kind of first, not the norm of u. But remember that the difference quotient of you is estimated by the h one norm, right? And remember chi is in supported, so chi supported inside w1. Okay? So this is going to play a role later on. So let's call this number for left hand side. This was the most crucial estimate. Okay, I haven't really done the hard part already. Let's, let's go back here. Ok, we have this relation for every v in a two-to-one you. So I'm going to use this B here, which is Chi-square teak chi is okay, let's just go back here. What is chi? Chi is one here and supported inside W1. So when you take the difference of chi, chi times u, you will utmost get support in W2 if you choose your h small enough. Alright, that is the idea. So v is in each one you support of v is in W2, and so v is an eighth. Okay? So you plug it into this relation because u is a weak solution of that. And we get this on this relation just plugging in right and take it to RN because that support, then we started working with the left-hand side here. Okay? No, I don't want the DH here. I want basically a quadratic form. And you know, chi times D DID a2. So I need to move this to the left by integration by parts. I get that. Well, then you use the product rule. So there is a shaft times that shift times a derivative of this plus the derivative of AIJ times this. And again, apply the product rule here, the actual product rule, you get that. Ok, so we have second-order, second-order multiplied first-order, second-order is no problem. You put it on the right hand side with a minus and you put fn f first-order. So this was the crucial term this times that, and we get that. Let me get rid of all these. Okay, so keep the second-order. Second-order, that's the one which is going to give me x squared, okay? And everything else I can manage. This is first-order times. First-order, second-order God is not a problem. I've talked about that before. >> So this times this is this term, then first times first is this term. >> Alright? So these are easy to estimate. This is the crucial one, this one is the crucial one. And then from positive definiteness of the AIG, AIG has positive net translated still positive definite. I get theta times that vector norm of that vector. Okay? So these are the S square term, that's what I want. And then the rest I can easily manage height. This is apply the Holder's inequality, L2 norm, the grad. Okay? And this is the second order term. And I'm going to put one chi with this because that is exactly what I have here. Okay? So this is first order, this is a second-order, But it's no problem first time, second I know how to handle and then first-order term. And this th, every time I have a DH, because this domain is slightly larger, I need to put a chi with it, chi or a chi prime. So that's first-order, first-order. Okay? This is really the hard part. We've done the hard part of the estimate. So I'm calling this four. So this is the left hand side is greater than or equal to that. Now, what is the right hand side? That's the relatively easy part. Ok? This is red and it's f, f times the v, which was d minus h chi square th of you. Okay? So right hand side is going to be less than or equal to, so this is less than or equal to the f is supported in use. >> So this is L2, y2 times the normal. >> D minus h square d h you L2 of R n. And then two is less than or equal to. So this actually looks like a second-order derivative, but actually we will see that it's really related to a first-order derivative. Okay? Yeah, so this one is actually sort of misleading. >> It looks like it's a second-order derivative, but it's really going to be a first order derivative, right? >> I said it wrong, sorry. This is like a second order derivative, but it is multiplied with the first-time second-order. So it's going to be okay, right? First time, second, we know how to deal with we'll take a little bit of effort. >> So now, what is that? So okay, now chi is supported in w1, and if you shift it, okay, then the support actually gets to be w two, this whole thing supported in Bluetooth. >> Ok, you can check that. >> So this is, so this is the first-order term, this is second order term. >> Okay? And so this is less than equal to norm f L2. Ok? Now remember that we prove that the difference quotient of anything in H1, okay, is bounded by the gradient of that. So this is less than or equal to this from proposition, Proposition two. It's the gradient of, I'll do it in two steps. So this is d minus H, square H of U of L2 of W2. Okay? And remember W2 is compactly contained in, sorry. Okay, so W2 is compactly contained in you. So why Proposition two Okay. >> You it can be estimated by the gradient of the bigger set, okay? >> That's from Proposition two. Okay? Now this is almost a second-order term, exactly like the one we have on the left here, right? We have the keys outside. But that's not a problem. You can just do product rule here. And so this is equal to norm f L2, y2 times. Firstly, you do without the, you put the derivative inside chi-square gradient of t h of u. Plus then the second time we put the derivative, you get th, a few L2, okay? And this is less than or equal to sum. They should be constant everywhere. Okay? >> And this I can do, I can just take only one chi, which is exactly the way everything is set up already. >> And this is chi DHL Phew and chi supported in W one, so it's L2 of up to 100. So this is first-order. That's exactly my second-order. This is first-order. So we put that, therefore putting everything together, okay? From the left hand side, this was the only positive thing. So I keep this one the left and I bring everything on to the right. >> So theta times integral of chi of t squared is less than or equal to. >> So let me write all the terms which have a second-order enough first-order second-order term. Whether you write D, they commute, so it doesn't matter. C1 times all the pills which have this chi DH L2, y2 times this f L to you. Ok? Now this is from the right hand side. What is there from the left hand side, which has that term, this term, this the second-order term. And it's multiplied by grad u of L2, y2. So second order, first order, then we have this times that, which has leftover these two terms. So plus c2 times this first term. Everything has no manageable. There's nothing. >> We've already done all the hard work from here, these two terms. >> All right, so this is a second-order, second-order, this is fine. This is 5n squared, this is S, this is OK. And now I just need to replace this and this. >> So this is actually, let's, let's just do it one more step. >> Okay, so this is less than equal to c1 times. I didn't want to copy the whole thing all over again. Okay. So noting so let's call this the whole thing. So noting dA2 >> One is less than or equal to this is from propositional to this one. >> Chi th, you too are n less than equal to c3 times that's less than or equal to S3. So therefore, five gives us the one that we want to. >> Now, five gives us, do you just have to let it below here, times plus S4 times. >> So it's this other view. >> So c4 times 2k plus this. >> Okay, so now we can use the trick. So you'll get that trip, you will get something like chi DH of grad u to u square is less than or equal to the square. See five times. >> Then because of the support, since K is equal to one, we get norma L2, you square school grad. >> Alright? And that completes the proof. Okay? So I know it looks long, but it's you just need to remember what are the things you want with a positive sign? And what are the things you want with a negative sign or getting on the right hand side. And that's all there is. And as I said, there are just a few places where you really need okay? You have to keep track of what is second-order, what is first-order? If it's Second time Second, you need it. And if it's on the left side, you need with the positive sign. And that's what this is, right? And then everything else, which is first time, second time, second, or first, second times first doesn't matter. That's you can handle it later. So that is it. So that is from the left hand side, it's second squared. And then everything is first-time second right hand side. So this is actually, so this is not as trivial as I was making it out in the beginning. This is a second-order dump, right? So you will have to make it with the first times second. That's what you'll have to do and that's what we do here. So that's Cauchy-Schwarz first time, second, but it's not quite the Second we want, right? It's second. The form of the second term should be this, That's on the positive side. So the second-order term has to be exactly of this kind. We can't be anything, something different. So I cannot have two D minus. I need to have a gradient in there. So I use the fact that this has L to R N, so it's supported in W2 because of the support of chi and the shifts. And then I use proposition to get a gradient. Here. Now that's starting to look I look like the actual term that we have there, but I don't want a chi-square inside. The chi-squared has to be outside here. So, but then I can use the product rule. So I get this. And then if you apply Chi, You get a first-order term. And now we are okay. Alright, so both on the left and the right, we need a first. We have to sort of, we get a first at times, a second-order. So and then the rest is just putting it together. So just remember, think of is that a first-order, second-order? Is at first-order first-order or is it a second time? Second. You have to keep track of that and then the rest follows. Okay? So that's the proof. So here we show that if u is a weak solution, then in H1 and H2. Alright, so what I want to talk about is higher-order regularity. And I'm just going to state the result. The proof is actually fairly standard and it has some simple manipulation, so I will give that as a homework. So what does the higher-order regularity theorem say? So it says the following. So if we just copy it carefully, suppose, suppose u is bounded open subset, okay? And you have this operator L U, which is minus dw plus B plus C is an, is an electric operator. Okay? So all that is okay? Not remember that to prove that you knew was a solution in H one than you needed to be C1s. Okay? You needed to be able to differentiate this. Now, if you want to prove that you as more regular, which means NYU has more derivatives, then you need to be able to do. These need to be differentiable more. So there's an elliptic operator with, so you want to prove you were has m plus two derivatives. Then you need a IgA c m plus one. B i belongs to C, B, i and C. Both Okay. Suppose all that right? So if instead of F being an L2, f is an h m of u m derivatives and u is in H1, mu is a weak solution. Okay? Then you actually see if M was 0, then we showed you was h to see you again, if f is an H and then you get two more derivatives because L is a second-order elliptic operator. So whatever is the regularity of F, u has two more. For every v compactly contained in U, U is m plus two derivatives in the interior. And it comes with an estimate norm f h n plus norm of u. Okay? So that's the higher-order regularity that if f has an F, if F was an L2 was originally just engage one. Another weak solution than actually we showed you was H2 because L is elliptic. That's the main thing, right? If f has m derivatives, more AIFF is H M of u, then you will have two more derivatives. Then whatever f has, so it's m plus two and becomes with an estimate. >> This was always okay. >> And I should just says higher-order interior regularity. That's, it's actually done by induction. It's really just, you just basically apply the induction. They just trade where there is nothing much in it. Give it as a homework problem. You can read the book. It has a proof, very simple, and I will ask you to do a special case of it in the homework. Okay? >> So proof, excuse me, read the book. >> That's a homework problem. Okay? Alright, so next what we're going to do is actually how do you get regularity up to the boundary? So we had so far and then everything with a shift operator. Now shift operator there disadvantages that here as you, but if you want to get to the boundary, then you can't. Shift operator takes you outside. So chuffed operators, the quotient can only be done for things interior. How do you deal with the boundary? Actually, it turns out you don't have to deal with here is a shift operator at all. Because what you can do is you can straighten things, straightened things, and then just work in there and just use the PDE itself, and it turns out it's much easier. Okay, so that's what we're going to do next. We will do regularity up to the boundary. Okay? And it's done without the difference quotient becomes a difference quotient takes you outside. Okay? So we'll use a simple thing going to straighten things out. And it actually turns out to be somewhat simpler, actually. Okay. That's it.
Interior regularity II
From Rakesh Rakesh May 04, 2020
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