noc21 ma33 lec01

Today, we are going to talk about notation and some definitions, which you are going to frequently use in this course. So, first of all, let me begin by assuming that omega, I will write omega like this and this is an open and connected, open and connected subset of R n. So, n will be the dimension and omega, we will always assume in this course, whenever I am saying omega, just assume that omega is open connected subset of R n. So, definitely omega can be R n, but generally speaking, it is a subset of R n and it is open and connected. We also call this set as the domain. Now, with this omega, what we can do is we can define many simple notations and the first notation, which I am going to use is, you see, we are going to use del omega. This is the first notation and the second notation. So, basically we are talking about notations here. So, del omega, del omega is denoted by the boundary of omega. So, essentially what I mean is, it is, you take the closure of omega and subtract omega from it, so minus omega. So, basically sets which are in omega bar, but not in omega, that I am calling as a boundary. So, for example, let us say that if say omega is b 0 1, b 0 1, so which is set of all those mod x less than 1, then your boundary del omega will be the set of all those mod x, which is equals to 1. So, this is the notation which we are using and we will also follow this thing. So, let us say, we will say that U, so U, whenever I am writing, this will be a function, this is defined from omega to R. So, whenever I am writing omega, I am saying it is a subset of R n, open connected subset. So, U is from omega to R, I will just write and this is given by this. So, you see U of x, x I am taking from omega. So, this actually looks like this, U x 1, x 2, x of n, yes, and this x is in omega. And we say, we say U is smooth if U is infinitely differentiable. So, whenever we say infinitely differentiable and all, we mean that you can, I mean, derive this, you can take the derivative of this thing infinitely many times. Now, coming to that, what we can also say is, we denote U x i, we write U x i to be del U del x i. So, this is the partial derivative of U in the ith direction. So, this is essentially, I mean, you guys already know this is limit h tends to 0, U of, so let us say this is at the point, I mean, some point P. So, U of P plus h E i minus U of P by h. So, if the limit exists, that is, if the limit exists. So, this is the partial derivative of U, partial derivative in the ith direction. You can also define similarly, del U del gamma. So, this is, let us say at the point P. So, this will, this is the directional derivative, we call it a directional derivative of U at the point P in omega in the direction gamma. Gamma is also a direction in R n. So, essentially this, I mean, you can define it like this, limit h tends to 0, U of x plus h gamma minus, this is at the point P, this is P minus U of P by h. Obviously, if it exists, I am assuming that. So, I mean, the state of change of U in the direction gamma. So, if gamma is E i, so if gamma is E i, then of course, del U by del gamma is del U by del x i. And we also denote, denote gradient of U to be U of x 1, U of x 2, U of x n. So, you take all the partial derivatives together in a vector like this and that will be denoted by gradient of U. With this notation, del U by del gamma at the point P can be written as gradient of U at the point P dot gamma. So, this can be achieved using the definition like this. So, I mean, you can also write del U by del gamma as this. Obviously, I mean, you need regularity on U, but I mean, we will assume that for now. So, let us look at some other things. We define U plus. U plus is defined as the maximum of U and 0. And similarly, U minus is defined as the minimum of, sorry, minus minimum of U and 0. And hence, you can write U to be U plus minus U minus. Both of these are positive functions and U can be written as a, I mean, product, sorry, sum of two positive functions. Similarly, mod U will be U plus plus U minus. You can check this. It is very easy to check. So, this is another thing. And we will also define an average. So, this is the average of U, let us say, over omega. So, how we define, we write it like this, average of U over omega. So, let us say this dy. So, this, so U of y, U of y, I am not writing U of y. So, this is, this means that this is, you write the, let us say this is measure of omega. You take the integral of U dy over omega, take this whole integral and then divide it out with the measure of omega, that will give you the average of U over omega. So, this is average of U over omega. And obviously, I mean, generally we assume that the measure of omega is positive in this case. Otherwise, I mean, this will blow up. And we also write, we say that U is a Lipschitz continuous in omega if, so basically Lipschitz continuity is not in a single point, but in a omega. So, this is a non-local concept. If there exists m positive such that mod U x minus U y, this is less than equal m times mod x minus y for all x, y in omega. And note, this m is independent of x and y. So, this depends on omega of course, but I mean, it does not, it can be less than equals to, it does not depend on x and y. So, basically you can change x and y, but this m is not going to change. So, as an example, let us take an example, look at an example. Example, you can take f of x to be x square on let us say 0, 1. Of course, you can see that f x minus f of y, this is, can be written as less than equal x minus y times x plus y, which is less than equal 2 times mod x minus y and this 2, this is your m here. So, this is Lipschitz continuous. On the other hand, you see, if f of x is x square on R, so basically what I am, what I mean is x is from R to R given by f x equals to x square. So, basically f from R to R given by f x equals to x square, then what happens is one can show, show f is not Lipschitz continuous, because in that case, you cannot get a bound on this thing, here I am bounding. So, here what we are doing is mod x plus y is less than equal mod x plus mod y and this is less than equal 2 times the maximum of let us say mod z such that z is in closed 0, 1 and this is definitely 1, so this is 2, so this becomes 2 and that is why this is 2. But in this case, what happens is if you are taking your domain to be R, I do not have a control on x and y, it can be as big as possible, there cannot be a m which dominates this whole thing for any x, y. So, now we want to talk about notations for derivative. So, we, in the last part we saw that you can define the partial derivative of u and now what happens is we define a some new concept called a multi-index notation, multi-index notation. Now, what is the multi-index notation? So, we say alpha equals to alpha 1, alpha 2, alpha n, this is a vector, a vector of this form so that alpha i is always greater than equal 0 is called a multi-index, multi-index of order, of order whatever you want to call it, let us say mod alpha, mod alpha. So, mod alpha is summation alpha i, i equals to 1 to n, this is called a multi-index. So, essentially what we are talking about is any vector alpha is given by alpha 1, alpha 2, alpha n, but the condition is all the alpha i must be non-negative and that is called a multi-index essentially, a vector with this form, of this form and the order of the multi-index, the order of the multi-index is given by mod alpha, where mod alpha is the sum of alpha i, yes, right. Now, with this notation, so the question is why do we suddenly introduce a notation like this? This is introduced because let us say I want to talk something like this. So, let us say del 7 something, u pi del x1, del x2 square, something like this, let us say 5, I want to talk about this and I mean, you can understand that the more partial derivatives have the more clumsy. So, this is a very clumsy looking notation, like looking notation and you can think of it like, let us say if I am changing 7 to 10 or 11, 12, this gets more bulkier and clumsier, we do not want this thing. So, this is not recommended, that is why we introduce something called the multi-index notation. So, let us erase this thing and we come back to the multi-index notation. As we explained, multi-index is a vector which looks like this and with this, we are going to define a new thing. So, given a multi-index and alpha, define d alpha of u x, this we define as del mod alpha of u x by x1 alpha 1, xn alpha n. So, essentially what happens is, basically you are looking at the alpha partial, mod alpha partial derivative and x1 has to be alpha 1 times and xn has to be alpha n times. So, this is all linked, alpha 1 is linked with x1, alpha 2 is linked with x2 and alpha n is linked with xn. So, basically the ordering of this vector is very important and we can also write it like this, del x1 alpha 1 del xn alpha n of u of x. So, that is your multi-index. Let us take an example and see what it means. Let us say u is from R2 to R, u is from R2 to R and alpha is 1 0. Let us see what happens to d alpha of u. So, this is, you can see this is d of 1 0 of u, this should look like del u because mod alpha is 1 plus 0 which is 1 by del c. The first 1 is for x1. So, this is x1 1 and 0 is for del x2 which is 0. So, that is not there. So, we are not going to write it, this is del u del x1. Similarly, you can take another example. Let us say the same u, u is from R2 to R, let us say R2 to R and alpha is 1. In this case, d alpha of u, it should look like del 1 plus 2 is 3, del x1 del x2 square. So, it should look like this. So, that is your multi-index notation. Now, we can continue with this thing. So, let us say k is greater than equal 0 and we can write dk of u. So, what is it? This is the set of all, this is the set of all d alpha of u x such that, obviously, there is x here, I mean, it does not matter, but you can just write it like this, mod alpha is equals to k. So, basically dk of u is the alpha. So, when the order of alpha, the multi-index is k. All those partial derivatives, we are looking at it and putting it in a set that is your dk of u x. So, for example, let us say, for example, u again, I am starting with R2 to R and dk, I am talking about, I want to write what is du of x. du of x in this case will be a set which contains u of x1, u of x2, it should contain like this. And, but generally what happens is we also write the gradient of u, gradient of u, if you remember, it is gradient of u of x, we write it like gradient of u of x, this is du of x, this is du of x and that is given by, we also write it like this, u x1, u x2. So, you see, whenever we talk about gradient, we write it like a vector, u x1 and u x2, this is a vector. And we identify this, this is an identification, we identify this with d of u, which is like a set which contains u x1 and x2, this is an identification, that is all. So, we can just think of it like this. And another example, let us say, u from R2 to R and I want to talk about the second derivative. So, d2 of u x, how should it look like? It should look like u x1 x1, u x1 x2, u x2 x1 and u x2 x2. So, that is called the Hessian of u. So, this is obviously evaluated at the point x. So, this is Hessian of u at x. And you can actually define the most important operator which we are going to study in this course, which is the Laplacian of u, we write it like this, in some books it is, so note in some books this is written like this, Laplacian is given by this. This notation is wrong notation, this is the right notation. Laplacian of u, this can be defined as the trace of d2 u. So, this is essentially, what I mean by this is x1 x1 plus x2 x2. So, that is your Laplacian of u. u from R to R if it is given, d2 u will be defined like this, this is the Hessian matrix and Laplacian of u can be defined as the trace of the Hessian matrix which is given by this. Now, with this, we are going to move on and define some other notations. So, we are going to define something called the C omega, C omega are the set of all continuous functions. So, this is u, let us say from omega to R such that u is continuous. And similarly, you can also define C omega bar is a set of all those u in C omega such that u is uniformly continuous on bounded subsets of omega. And what I mean by this is, see, first of all, let us understand what I mean by a, let us say, so let us understand this, let us say x is in on the boundary, x is on the boundary of omega and we say u is continuous at x. What does that mean? I want to know. So, what it means is, it means that given epsilon greater than 0, there exists delta positive such that u of x minus u of y, this can be less than, this can be made less than epsilon for all y in omega bar intersection B x delta. So, essentially, I mean, if you just look at the ball with centre at x and radius delta and you take the intersection with the closure of omega, for all those y, u x and u y should be very close to each other. That is what I mean by a continuity of u at the point x, where x is on the boundary. Now, when we say that u is in C omega bar, what I mean is, u is a function in C omega such that it can be continuously extended till the boundary. So, this can also be rephrased as, rephrased as u C of omega bar is the set of all functions in C omega such that u admits a continuous extension up to the boundary, till the boundary, boundary, boundary. So, essentially, note what it means is, if omega is bounded, if omega is bounded, then this is the set of all uniformly continuous functions. So, then u in C omega implies u is uniformly continuous, uniformly continuous. And similarly, you can also define C k of omega, C k of omega, C k of omega is the set of all those u from omega to R such that u is k times continuously differentiable, k times continuously differentiable. And C k of omega bar is similarly, u in C k of omega such that dk of u, u is uniformly continuous on bounded subsets of omega for mod alpha equals to, for all partial derivatives of kth order. And of course, we can also define, we can also define C infinity of omega to be the intersection of C k of omega, k equals to 1 to infinity. And C infinity of omega bar can be defined as intersection k equals to, sorry, this is k equals to 0, k equals to 0, k equals to 0 to infinity, C k of omega bar. So, essentially, if you take, of course, you can understand that C of omega contains C 1 of omega and it goes on like this, this is a nested thing, nested subsets. So, you take the intersection of that thing and that will contain all functions which are infinitely differentiable and you call all those functions as smooth functions, which is C infinity omega. And C infinity omega bar are smooth functions up to the boundary. So, these are smooth functions up to boundary. We also define this thing you guys already know, but let me again define it, L p of omega. So, this we define as u from omega to R is measurable such that integral over omega mod u to the power p, this is less than infinity. So, basically you are taking the, all the functions, p-summable functions, for p greater than equal to 1, but less than infinity. So, those are L p functions. And we also define similarly, L infinity omega. So, this is the set of all essentially bounded function. What I mean by this is, this is set of all those u from omega to R is measurable, measurable such that essential supremum of u is less than infinity. So, this is L infinity function. Now, we start with something called a Cauchy inequality, Cauchy inequality. And most of you know this inequality. So, this is a b is less than a square by 2 plus b square by 2. Now, it is very easy to see that why this is true. And this is true because we know that this is for a b in R. Why this is true? This is true since a minus b whole square is always greater than equal 0 and that will give you this particular thing. So, that is your Cauchy's inequality. This is the most basic form. And there is another form which is a Cauchy's inequality, Cauchy's inequality with epsilon, with epsilon. What that says is for epsilon greater than 0, for a b positive and epsilon greater than 0, one has a b less than equal epsilon a square plus b square by 4 epsilon. This can be proved using this. So, essentially what happens is you can use Cauchy's inequality to prove this Cauchy's inequality with epsilon. You just replace. So, a b, you see a b, you can write it like this, 2 epsilon to the power half times a and dot b by 2 epsilon to the power half. Once you do that, you can see that a b can be written like this, and then you just put a square by 2 plus b square by 2, you get this inequality. So, that is Cauchy's inequality with an epsilon. Now, let us look at another inequality. So, this inequality is called Young's inequality, Young's inequality. What is Young's inequality? It says that for p between 1 and infinity and 1 by p plus 1 by q is 1, then a b less than equal, so this is a to the power p by p plus b to the power q by q. And this is obviously, we need a and b to be positive. This is very important. For a b positive, we can write a b is less than equal a to the power p by p plus b to the power q by q. So, let us look at a quick proof of this. The proof involves the convexity of exponential x. So, we know that f x equals to exponential x is convex. So, we will use this property. So, what we do is we write, let us say a b, you can write it as exponential of log a plus log b. And that can be written as exponential of 1 by p log a to the power p plus 1 by q log b to the power q. So, this can be written as 1 by p exponential of log, so this is log, this is basic, log a to the power p plus 1 by q exponential log b to the power q. So, this is true because we are using the convexity argument. So, this is given by a to the power p by p plus b to the power q by q. And you see why we are using a b positive, because otherwise I cannot define these particular things, log of a plus log of p. And this is the place where we are using a convexity. So, how are we using it? We are using it like this, e power lambda x plus 1 minus lambda y, let us say this, so f of this is less than equal, so lambda times exponential x plus 1 minus lambda times exponential y. This is what we get. So, that is your Cauchy's inequality. And in the next part, what we are going to do is we are going to look at the Cauchy, this is the Holder's inequality, so Holder's inequality. So, what is Holder's inequality? What it says is, let us say p and q between 1 and infinity and 1 by p plus 1 by q equals to 1, then if u is in Lp omega and v is in Lq omega, then integral omega u v dx, this can be dominated by norm of u Lp omega norm of v Lq omega. Now, let us look at the proof of this thing, proof is quite simple, you can use homogeneity and we can assume, so without loss of generality, let us assume that norm of u p norm of v q is 1. So, let us just prove this thing for this uv, so for the unit uv, norm of u p equals to norm of v q equals to 1 and then we can look at the other thing. So, let us say this is, if this is the case, integral uv dx over omega, this is less than equal 1 by p integral mod u to the power p dx plus 1 by q integral mod v to the power q dx and this is 1, this is 1 and then it becomes 1 by p plus 1 by q and that will be 1 and this is norm u p norm v q. So, you see this is true, so here I am using the Young's inequality and now why can we use this thing that norm u p and norm v q is 1? So, because you see, if you take, let us say u and v are any arbitrary functions in L p and L q, you can actually assume u tilde to be u by norm u p and v tilde to be v by norm v q and then you see, then u tilde p v tilde q is 1 and for this and for this u tilde and v tilde, this is true, this inequality is true. So, when you put it there, then you will get this inequality. So, that is your Holder inequality. Another very important inequality which we are going to use is called the Minkowski inequality. This I am not proving, but this is also very simple inequality to prove. It says that for 1 less than p less than infinity and u v in L p omega, we have norm u plus v L p omega less than equal norm u L p omega plus norm v L p omega. So, this is basically the final inequality in terms of norms, you can think of it like that. Yes, this can be proved using this Minkowski inequality, but I am not going to prove it right now. So, this you can just remember. Now, what we want to do is we want to calculate. So, what I mean by this, you see, generally whenever we are talking about this PDEs, we say that let us say some PDE, I do not know, maybe let us write it like f of gradient du u x, this is some relation that is equal to 0 in let us say, we say omega, some domain, domain omega, which is let us say R n and as usual, we assume that omega is open and connected. And you see, we generally assume that omega is smooth, omega is smooth or also in some books, they say del omega, which is the boundary of omega is smooth. Now, the question is this, what is smooth? What do you mean when you say omega is smooth? So, we want to put it properly in a definition. So, we say, so obviously, it is assumed that omega, now, we will assume this omega is open and bounded, bounded set of R n. So, we are giving this definition for open and bounded subset of R n. So, what we are saying is, let us say omega is open and bounded, sorry, open and bounded subset of R n, then the boundary del omega is C k, is C infinity, let us start with C infinity. C infinity is, this is what we say smooth, smooth, C infinity, if for every point, if for every point x naught on the boundary, x naught on the boundary, there exists R greater than 0 and a C infinity map, let us say gamma from R n minus 1 to R such that upon relabeling, relabeling and reorienting, reorienting the coordinate, coordinate, if necessary, we have, so U intersection B x naught R, this part of the domain is x in B x naught R such that and x is the tuple x 1, x 2, x n. So, x n should be greater than gamma of x 1, x 2, x n minus 1. Now, what this means is, what this means is, you are basically saying that, so let us draw a small maybe diagram, let us say that your domain omega and this is R n minus 1, this is R n minus 1, but it essentially says, now what is gamma, gamma is this, this is gamma, so it is saying that the graph, I mean, if this set is C infinity, so basically the boundary of the set is C infinity, it means that the boundary is smooth, it means that you take any point on the boundary, let us say this is the point on the boundary, let me put it in a different color, so this is the point in the boundary and you have a ball, you can have a ball around that, which is centered at that point, x naught, let us say this is the point x naught in the boundary, x naught and there is a ball around that such that the part, this part of the ball, this part, so the intersection part of the ball with the domain is always above the graph of a C infinity function. So, this is always above the graph of a C infinity function, so for example, let us just assume, so for example, let us say example, you see, if you take a unit ball, so unit ball, ball in R n, definitely you can see that the unit ball in R n, I mean, you can obviously always find a function gamma such that this thing happens at every point on the unit ball, so this is a C infinity smooth domain or C infinity domain, whatever you want to call it. Of course, here I am assuming open and bounded. So, another example, let us see, so let us say a domain which looks something like this, let me draw it like this, a rectangle, so is it a C infinity domain? Of course, if you take this point and if you look at a ball around this point, of course, I mean, it is a C infinity domain, because I mean, you can reorient it and maybe look at it in this way and I mean, your domain, your axis is like this and it is always above the intersection, this part, this part will always be above the graph of a C infinity function. What is the C infinity function in this case? This one, so the line, so in this case, I mean, if you reorient this line like this here, then you can understand that the C infinity function, so if you reorient this line, the axis, if you reorient the axis, you can think of this line as like this and you have a axis like this, this axis like this, this is your R n minus 1 and the above part, this part, in this case, this part is always above the graph of a C infinity function. What is the C infinity function in this case? This line, so this is a linear map and obviously, this is a C infinity and hence, this is a smooth domain, but it is smooth almost everywhere, not in any point, because you see, if you let us take this point or this point, any corner point, if you take a corner point like this, you see the intersection, so this is the intersection part and so it will look like this, intersection part, intersection part, it will look like this, so this is the axis. Now, let us say, if you reorient it, you may think of this looking like this, so you see, the graph is not, I mean, the intersection part, this part, U intersection B, this is definitely not the graph of a C infinity function, not C infinity. So, this is, this kind of domain is not a C infinity domain, you can understand that you can have the same definition by replacing C infinity with C 1 and then you just have to replace here with C 1, this is, then we call it a C 1 domain, so basically, the boundary C 1, if there is a C 1 map, which does all of this. So, you see, if you do that, then maybe you see, this is also not a C k, a C 1 function, this is also not a C 1 function, so this is mod x kind of thing, so this is Lipschitz continuous function and we say this kind of domain, the rectangular domain is a Lipschitz domain, as per this definition, Lipschitz, you just have to replace the boundary C infinity with a Lipschitz, so the boundary is Lipschitz and Lipschitz is, for every point x naught, there exists such that there exists and a Lipschitz map in that case. So, if you just replace this definition with a Lipschitz definition, then this particular domain, this particular domain is a Lipschitz domain, but Lipschitz domain, is this clear? Because the graph is above, the intersection is above a Lipschitz graph, graph of a Lipschitz continuous function, but not a C 1 continuous function, the intersection, this intersection point, the intersection region, this is above the graph of a Lipschitz continuous function, but not a C infinity continuous function or C 1 in that case. So, this is not even a C 1 domain, but it is of course a Lipschitz domain. So, that is how we deal with domain. Now, so this is what we mean when we say it is a domain, smooth domain. Now, let us come to a very important thing which we call as the Gauss Green theorem and let me change the colour. So, Gauss Green theorem, what does that theorem say? So, first of all, omega obviously is assumed to be open and bounded, please keep that in mind, omega subset of R n is open, bounded and smooth, if U C 1 of omega bar. So, essentially if omega is bounded, C 1 of omega bar consists of all uniformly continuous function. So, basically the function is uniformly continuous, the derivative, it has a derivative omega bar, the derivative is continuous and not only continuous, the derivative is also uniformly continuous. So, basically one type continuously differentiable, not only continuously differentiable, the derivative function is also uniformly continuous. Right. So, this is the case, then this happens, very, very important, this is the one thing which you should know, I mean, you do not have to know anything else in this course, you just have to know this thing. Once you know this, this is, everything is done. So, this is true, U s i over the domain omega d s, del omega U gamma i d s, this holds for i equals to 1, 2, n. So, what does that mean? What is gamma? Gamma, where gamma is gamma 1, gamma 2, gamma n is the unit outward normal to del omega. See, what he is saying is this, if you take the derivative of U, the partial derivative of U, I mean, if you are integrating it on omega, then that will give you, so you do not have to look at the whole domain to do that, you just look at the boundary of del omega and you just compute U dot gamma i d s, that is done. I mean, once you do that, you calculate this thing and this thing gives you your, I mean, evaluate this particular thing, U x i over omega and this d s, what is s? s is the surface measure as we already did. So, maybe I can just write it, del omega normal to omega and d s is the surface measure. So, basically I am integrating it with respect to omega. So, I will give you d s. Now, so this is Green-Gauss theorem, I mean, this is essentially your Green's theorem, but in a more useful form and of course, you can just use this thing to write this, integral omega divergence of U d x equals to integral over the boundary U dot gamma d s. So, this holds for all vector field U cap in C 1 of omega R m. So, this holds and a very important property, this property is this. If you replace U with U v, let us say, so U and v both are in C 1, let us say, so the product is in C 1 of omega bar. Now, if you replace U with U v, then what happens? You have the integration bypass formula. So, that gives you U x i of v d x equals to integral over omega with a negative sign U v x i d x. How are you getting this in C? If you replace U with U v, U v of x y by Lipschitz rule, U x i of v plus v x i of u, U x i of v plus v x i of u and you have a boundary term. What is the boundary term? Plus integral over the boundary U v, is replaced by U v, U v gamma i d s, gamma i d s and this is for of course, i equals to 1 to n and for all v in C 1 of omega bar. So, basically what I am doing is, I am replacing U with U v. See, if you have two arbitrary function U and v in C 1 of omega bar, then the product of U and v is also in C 1 of omega bar, I am just replacing U with U v. So, once you replace it, it is U x i of v plus v x i of u, this is what I wrote integral over omega, that is equals to integral over the boundary U v gamma i d s, that is what I wrote and this, this is the most important formula which we are going to use in PDEs. You do not know much about PDEs, does not matter. If you want to study about PDEs or the one thing which I want you to understand in this course is only this formula, this is the most important formula, integration, this is called the integration by parts formula, integration by parts. This is your usual integration by parts in one dimension, you know and this is just the n dimensional version and you can obtain this thing using Green-Gauss theorem. With this, I think we can end this lecture.