Okay, So I want to, what we talked about last time was the trace map. Alright, so let's see, just to, you know, there were all these things about approximating. So Sobolev spaces. That was really just a way so that I could prove theorems about Sobolev spaces. First theorem we proved was extension theorem. If you have HKU, you can extend it to a function h k are n number 1, s thing which I stated, which all well, which is, which I was stated, which was this theorem, which is the trace theorem. Right? If u is a bounded open subset of R n with a smooth boundary. Okay? Then there is when you'd loop. So the question is traced thing is, if you have a function in HKU, can you talk of its boundary values? Right? That's the question. Now, if you have a function in C infinity u bar, or even C U bar, right? Continuous, uniformly continuous function on you. If shown that it has a unique extension to u bar by taking limits while your personal boundaries. So unique continuous extension to u bar. In particular, you can take this value restriction on the boundary. Correct? Let me repeat that. If you have a uniformly continuous function on you, we've seen that was a homework problem. It has a unique extension. A continuous function on your bar. Meaning you can, there's a unique possible value on the boundary. Right? So if your function and uniformly continuous function, or what we call C U bar, it has a unique trace on the boundary. C infinity u bar is dense in HKU. Everything in C infinity u bar has a restriction on the boundary because it's in C U bar. Okay? So the question is can that restriction map be extended to HKU? Right? So that's what this theorem is saying. Then there is a bounded linear map T from HKU. You lose one derivative such that u is u bar with stricter to you. Okay? What does that mean? That there is, we know that if you take a function in C infinity u bar, meaning all the derivatives are uniformly continuous. You, in particular you is uniformly continuous on here, so it has a unique extension. So this makes sense. Write this d u equals u bar, makes sense for us, which is in C infinity u bar. Okay? Now, if I see infinity u bar is sitting here in HKU, it's a dense subset of that. So if I could prove that that map, which is defined on C infinity u bar is a bounded linear operator in the HKU norm and h k minus one, then it will have a unique extension. Correct? So this is already well-defined. All I needed to prove, to prove this theorem was that I needed to, just, all I needed was this. I'm just writing it. Okay, let's see. If I can prove this estimate. For all functions u and C infinity u bar, then automatically it will have a unique extension. Because this is dense. All these functions are dense in HKU. So from that theorem, so originally my trace map is defined from C infinity u bar h k minus delta u, right? I just send you the restriction of u bar on the boundary. That's my map. If you have a uniformly continuous function on you, it has a restriction and it's a smooth function. So it doesn't h k minus one delta u. If I could prove this estimate, the star estimate. Then it will have a unique extension as a map to each SKU to that. Correct. So that's why, okay. Alright. So that is what the tree is. Theorem says if we take essentially now you can forget all the estimate. All it says is if you give me a function in HKU, it has a trace, it has a value on the boundary, which is a lot get to h k minus one u. And this map is linear and bounded. Okay? That's what it's saying is not just a trace, but actually we have this estimate. Right. Now I said to you last time that by usual tricks which I'll talk about today a little bit, all you have to do is prove it for when you as the half-plane. You prove it when you as the half-plane. Okay? So that is, you prove that you have a map from h k r plus n to h k minus one. Now what is the boundary of that? Right? This is x n equal to zero. So it's an n minus one dimensional plane. So it's really just that, right? You're sending basically you too. U x n equal to zero. We take the trace, right? But this makes sense for functions which are uniformly continuous on the half-plane. Because then they have a limit as you approach the boundary. And it says it has a unique extension. If you have a function in e.g. one of the half-plane, it has a restriction on x n equal to zero, and that'll be a function and L2 of R n minus one. Okay? I said that if you can prove this, then from this you will follow the current for general view, okay, which I'll talk about in a bit. Okay? Then I showed you how this relation, how this is true, because you can prove that in one dimension is less than or equal to just by using that integral fundamental theorem of calculus. Okay, that was the proof. So I'm just recalling what we did. So the only part which was missing was how do you reduce this for a general bounded domain to the half-plane case? Right? In fact, there is actually a most serious question. What do you mean by this object? Right? Delta of u is the boundary of us regions with some surface. How do you define the struggle of spaces on the surface? I've talked a Sobolev spaces on open sets in R N. A surface is not an open set in r n. How do you even divine Sobolev spaces? Okay? So that is the question I'm going to address. And then after that. Okay? So e.g. right, Let's think in R3, I take the unit sphere. If your x is, you know, all the unit ball in R three. By the way. So this is in the notes, it's in the first chapter. Actually I've talked about some of this first chapter and later. Okay? So then delta u is the unit sphere, which I'm going to call S. Or would use S for something else. Slips, leave it like that. Okay, so I have or even a circle, just think of a circle. The circle is easier. How do I, what do I mean by firstly, what are functions on the circle? Right? So you can think of if you're doing Cellco, say n equals two case. If you're on the circle, if you have a function you want the circle, then just use a function of the angle. Is a periodic function. Repeats itself every two pi. So you can straighten it out. It sets the functional periodic function on the line. Okay? No problem defining serverless spaces there, right? You can talk of derivatives and everything. You just have to keep it periodic. It's trickier now when you do it in R3. Okay, because now you cannot talk of periodicity. So basically, there is a way to straighten out the circle into a plane. As long as you just assume periodicity, there is no way to straighten out, sorry, your circle to a line. There's no way to straighten out a sphere into a plane so easily. Right? Are you with me on this? Right. Okay. So let's do it. N equals three. So suppose u is a function on the unit sphere. Colon comma just now. Okay, Just like that, I am saving S for something else. So, okay. So you want to talk about Sobolev spaces, right? First thing is, say instead of talking HKU, HK of Gamma, Let's talk of K equal to zero case. Okay? How do you define L2 functions on the unit sphere? Define L2 functions. So it's right. So it's all the functions on the unit sphere such that integral on the unit sphere is finite. L2. Okay? How do we define this? How do you integrate on the unit sphere? Parametrization? Correct. So how do we integrate on any surface by using a parametrization? Okay? So question. How do you integrate, right? Okay. Correct. Parametrization. Okay. So how do we integrate on any surface or a general delta, you use a parametrization. Delta U. Okay? So let's say you have some bizarre surface. Okay? So this is your delta u. Okay? How do you integrate on that? Okay? Use F here. So she said exactly right, you use parametrization not, you'll not find one parametrization for the whole boundary. For the sphere. You're lucky, right? Spherical coordinates are there. There is one parametrization excepted drop one point. Okay? But most surfaces you will not be able to find one single parametrization. So what you will have to do is you'll have to use a partition of unity. You have to break them up into pieces. Right? Remember that partition of unity. So some chi sub I. And then, you know, you take you, you is Sigma Chi. Sorry. I'm just doing it very quickly here because the idea is more important than the details. Details are in the notes very carefully written. Then you know, you call these things U sub i. And u sub i's are going to be supported just in one batch. Right? And then on that patch, you have a parametrization. Okay, so breakup delta u into batches. What I call surface patches has some by which have a metro physicians use a partition of unity corresponding to these patches. Okay? Then on the boundaries, so u is, I'm just repeating what I wrote before. With the loci supported. On the patch has high. So if you want to integrate like use square ds over delta u. Okay? Well, just to v. It's the same thing as integrating you Cai on delta u. And now because it's eukaryotes supported on SIs, are you integrate? Does make sense what I'm saying here, right? So you break up you into sums. Each, each piece has supported on a surface patch. And then integrating that thing on this whole delta u is same thing as integrating it on this surface patch. So you really need to know how to integrate on a surface patch. And it's exactly as you said, the surface patch will have a parametrization. So I'm going to now give you she's correct. So once you are given the parametrization, how do you integrate on that? Right now you may remember the formula from R3, R4 to 43. You have a parametrization. In UV. You define the vector r, which is x psi y j z k. Then you take ru, cross RV, and then take magnitude, okay? And actually do the something similar in higher dimensions. It's there in the notes. But I'm going to choose a simpler parametrization. There's another way to parametrize it. Okay, and I'm going to write about that. So here's the definition. A surface patch S in R n, right here is my surface patch. So what is that? So I'm going to take C surface patches that okay. Let's, before I go, let us talk of an example. Right? What are the simplest kind of surface we have in our three simplest type, not kind, or type of surface. In R3, right? It's this. You have x, y, z, and you have some surface. Z equals f x, y. Right? This is the simplest type of surface we deal with an artery. So what you're doing and so you know, and it's sitting on top, some open set W. So you have x, y in an open set w. And then the surface here is surface patch S is equal to f x y, all the points on this. So basically S consists of all the points x, y, f x y. Where x, y is in some open set W. Write f of x. Y is some map where f is a map from w to r, which is the height function. That's a simpler surface patch in R3. You can do the same thing in R n minus one. You have n minus one plane at the bottom. And then the Zen is the z direction. So you need a function from R n minus one plane, which gives you the height. Okay? So it's generalization is the following. Surface. Patch. S in our Ren says off an open set, w and r n minus one. This is the base, right? This is your base. Then you need the height function. Smooth function from W2 are this gives you the height of each point of the base. And S has all the points. So the way you write as x prime F x prime, x prime in that video, right? It's just this unit. So x prime is x comma y and f x is the Z. So that's what a surface patches. Okay? Now, you will not always solve for the fifth x, y, z. You can solve for z in terms of x and y. You can solve for y in terms of x and z. So you can swap around, okay? So I'm not going to emphasize that, but it could be, you could solve for one of the variables. So e.g. if you're doing a sphere, right, Well Paul points near that. You can use this height. Or you know, for points near there, you can use this kind of height function. But if you're trying to do for points near here, right, you cannot use the height function because things near rate, there are two possible choices that the points close to it, you'd look at that height of this height. So for this, you'd rather solve in terms of y. That's what I was trying to tell you. Right here. You know, you want to write this as z equals f x, y. And here it is. You want to solve any of that you want to do. Why is fxy? You understand that? So it's not always going to be always in terms of height. It'll be one variable in terms of the other two. So everywhere you can make a surface patch. If you understand how to do things on surface patches, then you can understand how to work on any boundary. Alright? So this is a very special kind of parametrization. This was a very special. You keep two variables fixed and solve the third in terms of the other tool. So when you do spherical coordinates for a sphere does not this kind of parametrization. For a sphere you would do rather, for a sphere, you would, near this point, you would solve z is square root one minus x squared. Okay? If you were trying to do this as a patch, if you're trying to do a surface patch here, then here you would use z is negative square root one minus x squared minus y squared. If you're trying to do it near this point, right? You would solve for y now. This is negative y, I guess, think yeah. Right? You wouldn't want to solve for z near here because here see, firstly, you'll get zero. And then the square root function for X, Y near that point is not going to be smooth. That is also the other issue, right? You can see it even more clearly when you're doing it for a circle, right? If you're doing it for a circle, right? So near this point. If you have a patch here, you want to do y equals square root one minus x squared. But you can see that near this function, near this point, it's a problem. You have to use y equals plus there and y equals negative there. You would be there. So on this patch, you would rather solve for x. Near this point. You would rather solve for x, but it'll be, your patch will be this. You agree on that. So they may not be one way to do it for the whole surface of the whole curve. But you break it up into patches and you can make it look like a surface patch. All right, so to integrate on a circle or to integrate on a sphere, I need to understand how to integrate on surface patches. Okay? How do you differentiate on surface patches? How do you define Sobolev spaces on surface patches? Because once you understand that, because of that breakup of u into u times chi, everything else will make sense. Okay? So we need to integrate one surface patches, differentiate on the surface patches. And they find h k S. If S is a surface patch. Because then h cube delta u will be things which are just added up with surface matches. Alright? So she's already said, but I'm going to write this down now, very explicit formula for this kind of, okay, and then we'll talk about differentiation. It's a little trickier. Okay, we'll see. So f is the height function. This hill surface patches and this is your X prime, X n. This is x n equals f x prime. And its projection is this, W. Okay? You can think of this as x, y, and z. Okay? So how do we integrate on that? So firstly, basically the idea is you convert integration on that curve thing to integration on the flat things. That's what you do with parametrization. Like when you did the sphere, right? The parametrization is with a Theta and the fee, the spherical coordinates. And you convert integration on the sphere to integration on the theta fi plane. Theta goes zero to two pi, f0 goes zero to Pi, and you've done it on the flat plane. What parametrization is doing, it's converting something which is curved to an integration on a flat thing, the range in which the parameter changes. That's the whole idea. You convert everything which is curved to flat, but there is a price to pay when you fly atmosphere, right? The surface area is deformed. So what is the price that is the correction factor? It's comes from the throne of the change of variables. So for any function, so suppose you give me a function you from S to R. How do I integrate you on the surface patch S. Okay? So this is okay, how do I integrate that? And the answer is, you make it a function of, firstly, it's on that, right? So you can just make it a function of the variable x prime, because on S the Z is always f x prime. So you can integrate it on you and you want to say, okay, I've reduced it to integration in n minus one variables. This was wrong. Correct. E.g. if you was one, you would say then the surface area of S is the same thing as the area of w. Correct? If you put u equals one, right? This is integral of one on S, which is the surface area, is equal to the integral of one over W, which is the area of Wn. That's not true. Correct? When you flatten my sphere, it doesn't preserve the area, right there is a correction factor. The correction factor is that magnitude r u cross r v business. And in this case it is actually very simple. Let me rewrite it so it takes less space. You remember this? So the correction factor is ds is square root one plus z x squared plus z squared. We call this in R3, one plus grad of Z square. So it's that, that's what's going on here. So whenever you have the height function, you know how to reduce integration on that curved surface to integration on the flat. This is the correction, right? So you know this, but that's that's, that's the correction factor. So you say you okay with this. Okay, now, why this is so it's, it's very tricky. It's not so trivial to prove this by the way. Okay, I've completely avoided the question. You integrating on the surface there is a measure of there, just like there's Lebesgue measure in RM. There's a measure on the surface. I haven't even talked about what the measure is and it's not so easy to define. And I haven't shown you why this formula is true and stuff. Because before that you have to understand what is the surface measure. Okay? And if you go just one sentence but in the nodes, but it is related to something called the House staff measure. Okay, go and look up Wikipedia. It's a complicated thing. How do you define a measure on the surface is nontrivial. Okay? To integrate on a surface, you need to have a notion of a measure, just like you have a notion of integrity. A measured Lebesgue measure in our end. So how do you define a measure on a surface? It's very tricky. I say a little bit about it in the notes you read that and from that this formula, cow, but it's nontrivial to derive this Israeli non-trivial. Correct. Okay, So how does it mean for the curve, right? So in the curve case, so you have x, y, and you have a curve, right? And this parameterization, suppose y equals f x, right? So this was a to b, this is your w, right? And then if you want it, and this is your curve gamma, you want to integrate some function u, x, y on the S. It's the same thing as integrating on w, which is eight, u x fx. And then the dx ds has to be corrected, right? It's the gradient of f. Now, of course, there is, you can have a more general change of a parametrization, right? If this could be x t comma y t, then if not, instead of one, you would have the derivative of x squared plus the derivative of y squared added and stuff. So that's a more general parametrization, which is like what she was saying. There is a general formula for that too. Okay. Which is not this. Right. If you go and look in the notes, I talk about it. What this is. I've done this only for surfaces which are parametrize with the so-called height function. What if you have a more general parametrization like spherical coordinates? What's the correction factor there? Right, in three-dimensions, It's Ru plus RV. But there is no cross-product in higher dimensions. Cross-product is a very three-dimensional thing. There are other kinds of determinants and stuff you have to use. If you read the notes in that, I talk about it. What's the correction if you have a more general parametrization? But as far as we're concerned, every surface can be parameterized like this. Every patch, right? Give me any surface, no matter how complicated. Near every point. I can solve for one of the variables in terms of all the others. Okay? Near here, maybe I don't know, I'll use height here. I will maybe use the horizontal direction. Near here. Maybe I will use the vertical direction, maybe I will use the orange. Alright? So every boundary of any region can be broken up into what I call surface patches, where you solve for one of the variables in terms of all the others. So unit surface. All right, so what is my definition of a surface now, right? So I say take any bounded open set with a smooth boundary. Okay? Now what does that mean? So my definition of a smooth boundary could be that around each point I have a surface patch. That is my definition of a smooth boundary. And so when you say RN and the surface S, as we know as simple space can be over some other variables. Yeah, yeah, so this is the definition of to take an open set and you have a boundary. The boundary can have very bizarre behavior. So I'm imposing an extra condition that at each point around, at each point on the boundary, I will be able to do some open set. So that in that open set, if you look at the part of the boundary of that open set, I can solve for one of the variables in terms of the other is given like that. That's my definition of a smooth boundary. And this is, this matches with your intuitive idea of what our surfaces. That's what I'm trying to say. Professor. Yep. Is that also the idea of like irregular surface from differential geometry? So that's it. So this is actually in geometry, they talk off manifolds. Okay? So manifolds are essentially this. So you know, a surface and environmental manifold. Surface in RAM is what's called an n minus one dimensional manifold. That around each point you can draw a neighborhood that you can solve for one variable in terms of the remaining n minus one. Okay? So let's say you are in R3. You understand this fear. That's a two-dimensional manifold. Because at, around each point, you can take a neighborhood and then two of the variables can be chosen freely and you solve for the height, or you solve for the y, or you solve for the Z. So it's really two free variables. That's why it's a two dimensional manifold. Okay? If you take the open ball, that's the three-dimensional manifold. Because you can vary x, y, z arbitrarily. All three variables arbitrary in the open ball. So that's a three-dimensional manifold. If you are on a curve in R3, right? Any curve, as long as it doesn't cross itself around each point. You can solve for two of the variables in terms of third, maybe here, you can solve for, let's say this is x, y, z. You can solve for y in terms of y and z, sorry, x and z in terms of y. You can only vary one parameter. So here you can solve for x and z in terms of y. Here you can solve for maybe x and z in terms of y here, you can see you can solve for two variables in terms of the third, curves are one-dimensional manifold is only one degree of freedom. Did I answer your question, Tyler? Yes. Okay. So you can talk in manifolds in our end. So it's a subset of R n. So that around each point, there's a better definition that I'm something exactly more or less what I'm saying. You can solve for z. If it's a k dimensional manifold, you can solve for n minus k variables in terms of the fee k variable. That's like a directional microphone. Okay, so you'd need to do it around neighborhood of each point on that set. Okay? So, so a lot of these change of variables, ideas which I've talked about, it's really, it's manifolds that, or another way of talking about it. And that's what a manifold is. In differential geometry, differential topology, that's what a lot of movements. So our boundary here, if you take any open set with this food boundary of boundary is a n minus one dimensional manifold. That's what it is. Correct. Okay, so now we know how to integrate one surface patches. How to differentiate on surface patches. So this is really hard to differentiate on a manifold. That's what I'm doing. If you want to use the language of that. Let's start with something very simple in 2D. Right? So if I look at this curve gamma, this is a surface patch because at every point I can write this is my W, That's my gamma. And I can solve this for y is fx. Okay? So suppose I have a function just defined on gamma, that's it, not defined anywhere else. Okay? So I have some function of two variables, x and y. That's it. Can I take the partial derivative of u with respect to x e.g. which means I'm asking if I'm at this point. How does u change if I move parallel to the x-axis? Don't have any hope of answering the question. Meaning and say, Oh, I'm moving this direction. Tell me at what rate you is changing. Write your partial derivative is this right here at this point, x comma y. And you want to find the partial derivative. Can you make sense of this? Is this defined even know you are only defined you on that black curve. So you cannot take partial derivatives. Neither can you take the y derivative. You cannot talk of partial derivatives there. So what would differentiation are you talking about? Right? Well, the only way you can do is you, you're allowed to move on the black curve. You're only allowed to take derivatives of you in directions which are tangential to the curve. That's the only direction you are. Have a hope of differentiating. Okay? So how do we talk about that? What do you mean by that? So let me draw this other picture. I'm good. Just going to erase it so you can't talk about partial derivatives. Okay? So, but I have a function on gamma, then I have a function on w. So I define w on, this is, let's call it. It will be the value of w, where w at this point is the value of the corresponding point on the curve gamma. You remember this is a patch. Every point in x gives me a unique point on w, gives you a unique point on gamma. Okay? So I have defined W of X is. So if you give me a function on gamma, there is a corresponding function on the flat thing. Are you with me on that? Right? I want to define a function w. Its value at this point would be the value of u at that point. Okay? And I can do this in higher dimensions too. So I get a function w defined on this part here. And it's a function of one variable, I can differentiate it with respect to x, correct? Yeah. Yeah. Right. So you give me a function, real valued function on gamma, I generate a function on. W and you're not. So you can talk of dw by dx. That makes sense. And this is the same thing as u sub x plus u sub y. F prime next. Okay? I'm going to write it in this fashion. So you can take the x derivative and then you can take you, which means if u is defined on gamma, then the only derivative of u you can compute as this. You cannot conflict. This is like the derivative in that direction which is tangential. Okay? So if you want, you know, you can say D1 you or do. That's the only direction you can differentiate. Okay? You can't inject. This is basically a tangential direction. So you're differentiating in this direction which is tangential. So as you change, as on this curve, how is u, what is the rate of change as you move along this curve? That's it. That's all you can compute. How does this carryover to higher dimensions? Let's do it on a server. Once you see it on a surface, you will see how this goes higher. Hello, this is okay. Yes. Okay. Alright. Okay. So let's say I'm, I put a surface in R3, right? Remember surfaces and R3 are surface patch, I mean, the surface patches. Okay, so this is z equals f x, y, and it is sitting above. This is my S is sitting above this W. How do I differentiate on that surface S, right? So you're given a function, you from S to R is not defined anywhere else. If you are at some point here, some point P. If I ask you, can you find the partial derivative with respect to x, you will have to move away from, you move parallel to the x-axis, but that will take you away from this off the surface. And you don't have any value for you to differentiate. Same. You cannot differentiate u with respect to y or with respect to z because you'll get away from the surface. And the function is not defined there. So you can't even take the differences and do it. So which direction is can you really differentiate? Okay, Again, if you'd give me a function on S, it automatically defines a function on w of two variables. Meaningful value at this point is the value at that point. Take the value of u at that point. That's what I'm doing. Yeah. Okay. The value of W at this point is the value of u at that point. Okay? So I've managed to find a W function on a flat region. That is the whole thing. And now this is w. I can differentiate it in any direction, x and y. Okay? So if I do delta w by delta x, use the chain rule. You get u x plus f x u z. Are you okay with that chain rule? Which means you could do delta x plus f x delta z. Of you. If you do delta w by delta y, you get your y plus f sub y. Use it. So you can compute these derivatives, partial derivatives of u and any combination of them, because you can take three wx plus five W Y. So any linear combination of those. So basically what it is is, I'm going to draw a sketch here. There are these two directions. X direction is this tangential like this, and y direction goes maybe like this. You can differentiate along as you move along the x axis, parallel to the x-axis, you get a curve on the surface. That's the direction you can integrate, differentiate or that, or any combinations of those directions, right? So if I call this d1 is delta x plus f x del, and D2 is delta y plus. One can differentiate in directions which are linear combinations of D1 and D2. Your inner surface, it has a tangent plane. You can only differentiate in directions which are tangential to the surface. You cannot differentiate in other directions. In the examples that we can. Traditional sense of differences. Correct? So here, now so instead of moving horizontally, you have to take, you have, you basically take curves. Whose tangent? So what do I mean by, right? If, let's say you are in R3, right? You're not. You have a function u of x, y, z, and u at this point P, right? I say find the derivative of u at p. If you move in that direction. Correct, It's a directional derivative. Move parallel to the x-axis, y-axis, z-axis. So basically you move h units in that direction and then divide by h. Correct? You take the difference. Okay? But the question is, do you have to move only along lines? You could actually move. I want to differentiate in this direction, right? I could actually move along a curve whose tangent is in that direction. Right? So if this curve was x t y t z t, I could just study this function. A one-variable function of one variable I differentiate, I will get the rate of change of u in that direction of that, as long as it's tangent is in the green direction. So you don't always have to move straight lines. You can take any curve, which tangent? Tangent line is that? That's what I mean by moving in that direction. You don't have to take straight line curves. We're doing that. Any curve whose tangent vector is in that direction, you can work on that. I'll be okay with that. Yeah, I think absolutely. So. We are so used to saying defining the derivative as move horizontally or move in a straight line. But that actually is the more general definition. For most situations. This subsumes that definitely, if this is true, then the straight line is a particular case. Right? All you're doing is moving along a straight line curve in that direction. So whenever we are differentiating, a function in a given direction is fairly being amphipathic, right? Right. You're saying that suppose there is a temperature distribution, but there's a space craft flying, right? And the temperature is changing from point to point in space. Spacecraft is at a particular point, say, okay, if the spacecraft moved in that direction, how rapidly will the temperature change, right? You can say, oh, let's move along the straight line measure and let go to zero. Or you can say, let's take our trajectory which is tangential. And you track how the temperature is changing on the spacecraft in this curve trajectory. At that particular instant, the rate of change of temperature would be the same as if it was moving in a straight line direction at that particular instant. With an M curve as a straight line, but still we take things is correct, right. So you're moving closer and closer? Correct. But you have yes. Basically you do this, right? You have something and you reduce it to a function of one variable. And your time t and t plus h, right? Instead of moving in three space, you do t and t plus h, right? So that's what you're doing. Okay? So this is differentiation on the surface. That is why, if you recall the divergence theorem, right? You all recall the Stokes theorem. Much harder. It's really hard. Even the statement is so hard. Okay, and why is it hard? Right? Okay, let's just, let's us, this is an interesting thing to talk about for a second. The divergence theorem, right? If you have a solid D and this is true in any dimension by the way, right? It basically says that you take is wx plus d sub y plus c sub z. Dv is something on the boundary, okay? It's eight times N1 be and to the normal components where N1, N2, N3 are the normals. Feel like I'd call it, yeah, that's the statement of the divergence theorem. It's called rarely. So it says you integrate derivatives on the solid. It's equal to the value of the functions on the boundary. It's the fundamental theorem of calculus. You integrate derivative on AB is equal to the value of F and F at the end points. It's there. And that's how you prove, you prove these theorems by using the fundamental theorem of calculus, right? You convert the a sub x two boundary, b sub y two boundary, and C sub z two boundary. Okay? Alright? And this works out fine. Because at any point in this you can differentiate in the x, y, and z directions. Now what happens when you try to do this on surfaces? You have this surface and this curve gamma, right? And you say, Oh, let me take a function and some derivative s. It'll be equal to the value of f on the boundary. That's what you want to say. You want to use integration by parts. Now can you put FX and FY, FZ? Now, because your function is only defined on S, You can only differentiate in directions tangential to S. Those are the only directions which you will be able to cancel to get f on the boundary. And that is why you have this vector field. If you take f to be a, BJ's CK, you write down what they are. You will see all the derivatives involved. I forget now what they are. They're only in directions tangential to S. Okay, let me see if I can recall this. Del cross F is okay, so let's do this. Let's say I take, just take, let me take V to be zero and c to be zero. Okay? So I'm just taking one function now. So del cross F is what IJK. So you'll get zero. J is minus a y, j. Sorry, wrong. So J gives you minus a is e j minus a y k. Again, Delacroix fat. And so basically it says this is easy. Two minus a y and three. Ds is equal to something on the value of the a on the boundary of some sort. Okay? That's what the stokes theorem says. Okay? Now let's see what is this? What derivative is that? It's N2 times delta z minus n three times Delta y applied to a. Okay? So which is minus N3 times Delta y n two times delta z applied to a. I just moved to reverse the order. Okay? So this is differentiation along. There is no x component. In the y-direction, you're going minus n three. And in the z direction you're going into, that's you're doing, you're differentiating along this direction. And this vector is perpendicular. The normal vector. You can check to take the dot product minus n 3.2 and N2, N3. So this is differentiation in a direction perpendicular, so it's along the tangential direction. That's what Stokes theorem. So you can only cancel the tangential directions. The divergence theorem is a consequence. Know? All these theorems are a consequence of the fundamental theorem of calculus. Each one says, there is a general version of this in any dimension. You have k dimensional manifold and our n derivatives in directions tangential to that object can be canceled out. To give you something on the boundary of that option. You integrate on that object. This is called general Stokes theorem. Okay? And they all say you can only cancel differentiations on, in directions which are tangential to the object you're trying to integrate tomorrow. Okay? Well, I had this, this is useful for you to learn all this, but then you can begin to see in place What is why Stokes theorem so hard is because you cannot differentiate in all directions. Some of the other you have to capture the direction in which you can integrate on the surface. Those are the only ones have been kept. And that's what this captures. Del cross F dot n. That's the left-hand side of Stokes theorem. Stokes theorem is the nominal, is the tangent vector. It says, sorry, I wrote a drunk. This gives you all the derivatives of functions f is three by three functions ABC. So it's giving you a derivatives of functions agency. And so you can take two of them to be zero. Just study one at a time. Thing. Take derivatives of functions in directions which are tangential. You can cancel them out and you Colombia left on this stuff on the boundary. All these divergence theorem. So they all say this. They're all coming from the fundamental theorem of calculus. That if you integrate a derivative on the interval, just get the value of the boundary at the end points. Okay? So, well we have only 2 min, but how do you now so you've seen how to differentiate on a surface, how to integrate on a surface patch. And you basically convert everything to the flat. Given a function on a curve thing, you can do it too flat. So how do you define Sobolev spaces on a curve thing? You convert it to a flat into the sobel level on the flat. Okay, We will never really need an explicit form of how to integrate it. Do this on a surface. You just need to know it can be done. At anytime you have to prove a theorem about boundary, you convert it to flat and do it for the half-life. That's how we do this. Alright? Well, it's a lot of hand-waving today, but hopefully conceptually you understand some of these things. Some of it is stated, I didn't talk, I haven't talked about the divergence theorem and Stokes theorem, the loads. That's something separate but other things how to integrate, not differentiate on surface patches. That's in the notes. We mentioned here to help with this remains in the linear models. It was one big question that you mentioned. That does not, That's what I think because there is no derivative there. That thing is really how do you compute integrals on the surface? Now once you know how to compute integrals on a surface, then there are some relations. Write that thing which I gave you is just about how do you define integrals on a surface? Or how do you compute integral? And then once you know how to do that, then there are some relations. Derivatives on a surface in that exponential can be canceled out to give you stuff only on the boundary. And that's the Stokes theorem. Okay. Tyler, do you have any questions? No, not right now. Okay. Which I remind. Okay. I forgot this question for I don't know if you've tried it for a, if we define a bounded linear operator that requires the trace tariff. I don't know if I've said it before or not. In the inside. You need to use the trace theorem. The sales that when you restrict it to the boundary, so you know, you go from H1 to L2 or something. And that's bounded man. Parte will require that. All right. Okay. I don't think I'll give any exams in this course. I think the weird things like grades will be whatever, just try them. Then only three of you anywhere. As long as you do the work, you'll all get it. Alright. So let's just go with that. I know you're interested in some other things right here you're trying to do well, at least you write it. You're all trying to him. I could give you exams and things, but I'm not sure what purpose it will serve. I'd rather you learned and not worry about that. Right? Yeah. But what I'm saying is in a slowly, you know, you're trying to do, you're developing more advanced skills as you're working on them. So I'm happy with that. There'll be probably about a total of eight homeworks. Please make sure you do all the problems. Alright? I think I'm going to skip all exams and things and it doesn't matter, really. Great sprite nuts. You're headed towards your PhD. I think your your GPA as a graduate student of a lot matters. Let me assure you. If you do your PhD, nobody is interested in your GPA. Alright? Okay, Tyler, professor actually do have a question related to the trace theorem. So I think in the last one, let me, let me stop, pause the, let me stop the thing and then I will. Okay.
Integration and Differentiation on Surfaces Math 836 Spring 2023
From Rakesh Rakesh April 06, 2023
16 plays
16
0 comments
0
You unliked the media.