noc21 - ma33 - lec15

Today's class, we are going to talk about some existence results. So, essentially, let us recall a little bit, quite small recall, given a problem, let us solve the homogeneous problem and then we will dive into the non-homogeneous or I mean, maybe we talk about this. So, let us say you are given this problem, Laplacian of u equals to f in omega and u restricted to the boundary is 0. This is called, if you remember, this is called the Dirichlet problem. I mean, of course, it is called a Poisson equation. See, this equation is called Poisson equation, this equation. So, let me give it like this, I mean, maybe Laplacian of u equals to f, this equation, only this equation is called the Poisson equation, Poisson equation. But I mean, if you take this equation, so, but let us say this equation, Laplacian of u equals to f in omega and you are putting this initial condition, let us say u equals to g or 0, I mean, it is not a big problem, but essentially the data is given only on u. If that sort of thing is there, this problem will be called a Dirichlet problem, sorry, it is Dirichlet, Dirichlet problem. It is named after the mathematician Dirichlet. Now, and let us say you can also have this kind of problem here and the method which I am showing, you can actually modify the method to deal with that, this sort of problem also, del u del eta. So, this is outward normal on the boundary, g on del omega. So, basically delta is the outward normal at every point on the boundary. So, del u del eta that is equal to g on the boundary. So, this is called a Dirichlet problem, if the data is given on the partial derivative and not on u, then this problem is called a Neumann problem. It is called a Neumann problem. So, today what we are going to do is basically look at the Dirichlet problem. So, let us say, let us call it a g. So, minus Laplace, let us call it a, it does not matter, I mean, you can take minus and not minus, but let us just call it a minus. See, the thing is this f is arbitrary. So, I mean, if you are not convinced, you just change f to minus f. So, Laplace and u equals to some minus f which is g. Which is h. So, it does not matter, it is just a convention which we use minus Laplacian of u equals to f. Now, the point is this, we want to solve this equation. So, you remember, given an OD, what are we trying to do? So, our quest is this, our quest is to address the well-posedness, let us say the axial 1, well-posedness of 1, well-posedness of 1. Yes, and in that respect, what did we saw? We saw that, I mean, you can use maximum principle to show that the stability is there. So, for small change in the initial data, I mean, the problem, the solution changes a little bit. So, the stability, so this is the recaller, the stability question, stability or I mean, this change under the perturbation, stability can be addressed, can be addressed, addressed. Stability can be addressed using maximal principle. But please remember this thing, maximum principle, most of the maximum principles which we did are valid only for bounded domain. I mean, some are for unbounded domains, you can prove those. But the maximum principles which we proved are only for bounded domain. So, stability can be addressed using maximum principles and there is another thing called is uniqueness, uniqueness. So, uniqueness, we have seen that you can use, uniqueness is there. So, you can use energy methods for uniqueness, energy methods and what is there, moreover, you can also use maximum principles, maximum principles. So, those things are taken care of. Now, the only thing which is remaining and the most important thing is how do I find a solution for this problem here. So, let me clarify this thing more. So, essentially, what is the question now? The question is, question is existence, existence of minus Laplacian o equals to f in omega and u restricted to the boundary is g. Now, for our case now, in this case, we actually assume that f can be, f and g are smooth. So, do not worry about what is the regularity of f and g, they are smooth. And what about omega? Omega, so where, where omega is, omega is of course, is the subset of RNER, there is nothing to say. So, omega is open, bounded, very important here. So, this, what I am going to do now, for that omega has to be open, bounded and the boundary of omega, del omega is in C1, clear? Or I mean, you can also say omega is smooth. So, basically, the domain is smooth, which means that the boundary is C1, clear? So, or I mean, we also say that the smooth domain, this is basically saying that this is a smooth domain, right? Now, what I want to do, I want to find an existence of this problem. Before I go on doing whatever I am trying to do right now, let me make some small remarks. So, please keep this thing is my remark. See, what I am going to do is I am going to construct something called a Green's function. For, for a particular domain, for a given domain, for a given domain, domain, there exists a Green's function, Green's function, which can be utilized, which can be utilized, utilized to find solution of star. Is this clear? We will show that, yeah, we will do all that, do not worry about it. So, we will write down a formula of what US will look like, right? The explicit formula of US based on f and g. So, if you are given a domain for that particular, please remember this thing, for that particular domain, you can find the Green's function. And this is a theorem, right? That you, for a nice enough domain, you can actually show that there exists a Green's function, yeah? So, that is there. I mean, we will do that. We will show that. So, but in reality, but in reality. So, basically, you may say that, yeah, then for any domain, I can just find a Green's function and write down the explicit for a formula for this thing, right? Any solution of that will look like this. You do understand what I am saying? See, Green's function, once you, for a given domain, yeah, I mean, take any domain, let us just say that the domain is a ball, yeah, with center at origin and radius 1. So, for that given domain, let us say you found out a Green's function and with the help of that, you have just found out what the solution US looks like, yeah? Okay, fine. Now, and since you know that the solution is unique, you can say that is the solution which you are looking for. But the problem is this, see, for that particular domain, yeah, for the unit ball, how do you find the Green's function? That is the question. But in reality, what happens is if omega does not have nice geometric properties, okay, geometric properties, this is very important, okay? Geometric properties. So, in some places, you will see that they start constructing Green's function out of nowhere, yeah? Most of the places where you can see that, I mean, does not have any motivation of how those Green's functions are coming, yeah? And most of the times, they are like, you know, it is like, it does not, it is not based on any logic, yeah? You just, I mean, try your hand on something and get the Green's function, yeah? And most of the times, you will not find, that is the problem. See, Green's function exists for a C1 function, that is for sure, it exists, yeah? But finding an explicit Green's function is almost always impossible, you understand? Except for a few nice domains, yeah? If omega does not have a nice geometric property, and I cannot specify what sort of geometric properties I am talking about right now, but we will have some idea about what this means, yeah? Then, then finding an explicit Green's function, finding an explicit Green's function, Green's function is almost always, always impossible, okay? Now, you may, by some hook and crook for some domains, you can get it, yeah? But generally speaking, it would not be easy, I mean, not easy, I mean, it is kind of an impossible task to do, okay? So, with that in mind, please, I mean, let us start with finding at least for some problems we can find, right? And let us do that, yeah? So, first of all, let us show that there is a Green's function, yeah, which, I mean, solves our problem, okay? Let us just call that problem star, yeah? So, this is star, right? Okay, we want to solve this problem, we want to find a u x, how does u x look like, okay? So, proof, yeah? So, basically the derivation, I am doing the derivation of Green's function, Green's function, right? Derivation of Green's function. Now, so let us say, I am starting out with a u, which is in C 2 omega bar, okay? There is nothing special about this u, I am not saying that this is Laplacian, all that, nothing, yeah? It is just an arbitrary function, yeah? It is an arbitrary function. So, you start with any arbitrary function, which looks like this, clear? Okay, once you do this, then what I am going to do is, I am going to employ a similar tactic which we did in the last class, yeah? To, you know, the convolution thing which we did last class, that sort of thing. So, basically, we fix a u, we fix the x in omega, okay? And epsilon greater than 0, such that b x epsilon is contained in omega, okay? And we apply integration by parts on, let us say, we call it omega epsilon, yeah? Which is omega minus b x epsilon, okay? Why we are doing it could be clear right now. So, basically what we are doing is this, let us say that is your omega, okay? This is your omega and that is the point x, okay? Take a small neighbourhood of x, let us say this is x, yeah? That epsilon neighbourhood, yeah? I am just throwing this part out and this is your omega epsilon, right? This is your omega epsilon, omega epsilon, yeah? So, I want to work on omega epsilon right now. What I am going to do here is this. See, I am going to use integration by parts, but for, in this domain, in this domain for which function, u of y, okay? u is a function of y, okay? See, this x is fixed, I am fixing an x in omega, yeah? And for that I am using integration by parts on this u which is a function of y right now and the fundamental solution, if you remember, phi of y minus x, okay? And now I am sure you have some idea of why we are taking this deleted neighbourhood, yeah? Because y, when y is equals to x, phi blows up, right? So, we do not want to, want that thing to happen. So, in this omega epsilon, you see y, if I am starting out with a y in omega epsilon, that can never be, I mean y can never be equals to x. So, phi is well defined here, right? And of course, u is always well defined everywhere in the domain. So, it is not a problem, yeah? So, I can use the integration by parts and once you use that, what happens? Let us say omega epsilon, u of y, Laplacian of phi of y minus x minus phi of y minus x, yeah? Laplacian of u of y dy, clear? So, this is equals to integral over the boundary of omega epsilon, yeah? u of y and del phi del gamma of x minus y, okay? And minus, maybe I do not know, maybe let me write it like this, okay? And here also, let me put it like this, minus phi of x minus y, del u del gamma of y d f y, okay? So, this is just an integration by parts, I just took two parts and put it together, okay? I mean, nothing much, right? So, once I have something like this, okay? Gamma, I did not specify, gamma is the outward normal on del omega epsilon, clear? That is your gamma, okay? Now, let us look at, see, so hence, let us call this thing as a, it is a big expression, right? Let us call this for now, let us call this as a and this expression as b, okay? So, what do you have? I have omega epsilon a dy, yeah? a of y of course, a of y dy equals to, so this is a of y and this is b of y, clear? This is a of y and this is b of y and that is equals to del omega epsilon of b of y d s of y, okay? So, now let us break this part, yeah? d epsilon y, so this is, see, what is the boundary of epsilon y? Epsilon y is this interior point, interior point, right? It contains of the interior points, so the boundary is of course the boundary of omega plus the boundary of this ball, right? So, it is basically the boundary of ball with centre at x epsilon b of y, okay? Let us call it b, b tilde because ball I am writing as b, so let us say this, call it a b tilde, b tilde, b tilde is this expression, the big one. So, b tilde of y d s of y, okay? Plus the integral over the boundary del omega. If you remember, the boundary is this boundary, the boundary of omega epsilon is this one plus this one, right? So, I am just breaking it up like that and b tilde of y d s of y, clear? Now, let us look at this expression, omega epsilon del b x epsilon phi of y minus x del u del gamma of y d s y, okay? This expression I want to calculate what happens to this. See, b tilde has two expressions, right? b tilde is this one, right? So, I am just starting out with this one, right? I am not touching this one, I am just starting out with this. So, this is phi times del u del gamma and integral over the boundary, integral over just this one. I have also this part to take care of, but let us just stick with this. For that, what do you have? See, del u del gamma, u is a C 2 omega bar function, right? u is a C 2 omega bar function. So, this is continuously C, u is a C 2 omega bar function and this ball is contained in omega. So, u is continuously differentiable. So, on the boundary of the ball, there is a maximum which u is attaining and that maximum I can just take it outside. So, basically this is less than equals to the maximum of del u del gamma on b x epsilon, sorry, del b x epsilon. Yes, that I can just push it outside and once I do that, I am left out with x epsilon phi of y minus x d s y, clear? That is what I have left out with. Now, see, phi, so essentially if I am taking y from this, yes, see, I am doing this integration, yes, such that y is on the boundary of this. Where is, where does y lie? y lies on the boundary of the ball centred at x and radius epsilon and I want to integrate phi of y minus x there. So, as you guys know, E of basically y minus x is some function R of mod y minus x, right? Because phi is a radial function, clear? So, if it is a radial function, what happens is, it is basically a constant on the ball, right? I mean, y minus x, this is constant there. So, essentially I can, you can just take, take the maximum of this phi outside, yes. So, this is, if you take that particular thing, yes, so essentially what I am trying to say is phi, if you are taking y from the boundary, y minus x is something epsilon, yes, y minus x is epsilon. So, essentially what is happening, you are basically integrating 1 by, or I mean, if you want, you can just, you see, y is never equals to x, I mean, you can also think of, I mean, see, there are many ways of doing this thing, yes, you can also think of it like this. x is not equals to y, ok, because x is in the centre and y I am taking from, on the ball, on the boundary of the ball, so x is never equals to y, there is always an epsilon distance between this, ok. So, this is always defined, well defined, yes, and it is a continuous function, so you can also take the maximum outside. So, basically maximum of this and maximum of this, let us say that is your c, let us say this is your c and I am again taking maximum of phi, let us say, yes, I can take that outside also. So, again, this is on the same thing, yes, and you have the integral of del v x epsilon d x y, yes. So, if you take the integral, that is basically the surface area, right, so that will look like epsilon power n minus 1, yes, surface area, that is epsilon power n minus 1, of course, there is some constant is there and that is getting absorbed here, yes, I am not writing all that. So, this goes to 0, yes, as epsilon goes to 0, that is quite evident, is it clear, ok. Now, what we have is this, so also, also integral over del v x epsilon, ok, u of y, the other part, this is there, the other part is u of y del del gamma of phi, right, so u of y del phi del gamma of x minus, sorry, y minus x d x of y, yes, ok. Now, what exactly is this thing? If you remember from the last class, you remember, we did this exactly this sort of calculation, yes. So, what did we get? We get something like this, yes, I am not doing this part again, S y and which goes to u of x, yes. If you remember, while proving that the convolution, convolution with phi and S that solves the Poisson equation, yes, in R n, yes, when doing the last calculation, this is the last part of that calculation, yes, if you remember, please go through that if you do not remember that, yes. So, this is also true, as I said in the last course, yes, as I said in the last course. We did the exact same calculation, I am just using that, yes, I am not doing anything special here, ok. So, once you have something like this, then therefore, therefore the formula which you have is this, you see this is u of x, if you just put it together, yes, let us just put it together, this is u of x, this is going to 0. So, essentially, you see here, this term will be there, yes, this term will be there, from here you are only getting u of x, yes, because one is this part, v tilde has two parts, one is going to, this is going to 0 and this is going to u of x, yes. So, essentially, I can write it as u of x, u of x is equals to integral over the boundary, over the boundary phi of y minus x del u by del gamma, del u by del gamma of y minus u of y del phi by del gamma of y minus x, yes, and d s, d s y minus integral over omega phi of y minus x, Laplacian of u y d y. So, this is what we are getting, if you see where is it, yes, we are getting this, why? Because what happens to this term, this particular term is 0, this particular term is 0, y is never is equals to x. So, Laplacian of phi, this, this is the fundamental solution, this is always 0. So, essentially that is gone, I am only left out with the minus integral over omega epsilon phi of y minus x Laplacian of u of y d y. So, omega epsilon as x tends to 0, it becomes omega. So, integral over omega only this particular thing, yes, so that is what I wrote, where is it, you see that is what I wrote, yes. So, of course, and, and this, so let us call that as 2. So, and 2 is valid, valid for all u in C 2 omega bar and x in omega, you can do it for any x in omega, it does not matter, because for any x in omega, you can always find a ball doing that, right, because omega is open. So, here you see what is happening, if you know, let us, let us just assume that if you know what Laplacian of u is in omega, so let us say that your f, phi is already known to you, phi you already know, because phi is a fundamental solution, right, it does not change it. So, essentially you can just calculate what is integral of omega, integral of this particular thing over omega, once you know what Laplacian of u is, and that is given to you. So, basically it is given to us that Laplacian u equals to f, and u equals to 0, u equals to g equals to boundary, right. If you look at the star, see, if you look at star, where is the Laplacian u equals to f is given to you, so just replace that there, and u restricted the boundary g, it is given to you. So, let us see, this function, this holds for any u in C2, right. So, what I am going to do is, I am just going to write down here. If I put it there, you see, this Laplacian of u, this is f, minus Laplacian u equals to f, replace it by f, this already we know, so we can calculate this. Let us look here, phi we already, let us look here, u on the boundary, we know if this is g, and del phi del gamma on the boundary we can calculate, because we know what phi is, we can just calculate what del phi del gamma is. What about this one, phi we know, but del u del gamma, this is a problem, right, del u del gamma phi, we know what u does on the boundary, we do not know what del u del gamma does on the boundary. So, we have to somehow eliminate this particular expression, and then we can, I mean, go through what we are trying to do. So, this is where we are just trying to construct a Green's function. So, to do that, what we are going to do, is we are going to introduce, introduce a corrector function, a corrector function, a corrector function. So, that is given by phi of x, which is phi x of y, phi of x, which is phi x of y, and it solves, solves the boundary value problem, it solves the boundary value problem. How does it solve? It does this, Laplacian of phi x equals to 0 in omega, phi of x is equals to phi of y minus x on the boundary. So, what we are going to do is for a fixed x. So, for, so this is for a fixed x in omega, what we are going to do is this, see, we are going to fix the x in omega, and for that we are going to introduce a corrector function, which is phi x given by phi x of y. So, for every, please understand this, what we are trying to do here, we want to eliminate this del u del gamma term here. To do that, we are fixing a x in omega. So, once we fix that x, we are introducing a new function based on that fixed x, which is phi x of y, and such that Laplacian of phi x is 0 in omega, and phi x is the fundamental solution evaluated at y minus x on the boundary. Is this clear? Now, let us apply the integration by parts again. Once we apply integration by parts, we of course get this here. Let me write it down. It is phi x of y Laplacian of u y d y equals to integral over the boundary del omega u of y d phi, del phi x by del gamma of y minus phi x of y del u del gamma of y ds y. So, basically I am just using the integration by parts to write this thing. I hope you understand where this is coming from. So, this is just whatever we did earlier, this is the same integration by part. So, this is just this formula, this formula, Laplacian of phi x is 0. So, I am not writing that part and except that everything is same. I am just writing that particular formula you see. So, of course, not for that function, I mean using, I am using that formula, but for this function phi of x and q. Once I do this thing, so let us say this over del omega, this will look like u of y, del phi x del gamma of y minus phi x at y is capital Y minus x del u by del gamma y ds y. Yes, you have this. Now, you see what is our point? Our point was to eliminate this particular term del u del gamma at the point y on the boundary and this is the term, phi y minus x del u del gamma on the boundary. So, if I add this term and this term, then they actually get cancelled out. I want to eliminate that only. So, adding this thing, so maybe, so what we are going to do is now, we want to add those two terms and to do that, I mean, we want to make it a little simple, that is all. Now, we define, so now we define, you understand what I am saying? See, if I add those two terms, this term is gone. If I add this particular term and this term, I want to eliminate this term. So, basically if I add this expression and this expression together, that term is gone and I want to do that and I want to write it in a more compact form and that is why we introduce a new function. So, we define Green's function. This is what we call the Green's function, Green's function for the domain. You see, Green's function is not for anything. It is only for the domain omega is g of x, y. We always write Green's function as g of x, y. So, this is equals to, this is by definition, the fundamental solution at y minus x. Please remember, the fundamental solution is defined in whole of R n here and that is why I can just write it like this, times p x of y. So, this is what x and y is in omega and x of course, not equals to y because otherwise this is not defined. So, this is a Green's function which we defined. Now, once you define this thing, then adding, let us call it, let us call it, so that is your 2. This is your 2 and let us call it, so adding 2 and 3, 2 and 3, we get, what do we get? We get u of x is equals to minus integral over the boundary u of y del g. I am just writing it down here. Please calculate this thing. It is not very difficult to do. I mean, this is very easy to do. Please check that part, that is all. d s y minus integral over omega g of x, y Laplacian of u of y integral. So, this is for x, the omega. That is u of x. Of course, here where del g del gamma of x, y, this is given by gradient of g at the point x, y. Now, when I am writing gradient of g, I mean it is with respect to y. So, essentially g with respect to y, that is all, times gamma y. You understand? This is g with respect to y. So, dot gamma y and of course, the gamma is the unit outward normal. I am not writing all that, gamma, unit outward normal, whatever the domain is. So, in this course, I mean, please just keep this in mind. Whenever we write this del g del gamma kind of thing, you will always understand. It is always understood that this is the unit outward normal, unless otherwise written. So, that is the thing. Now, you see once, so basically what I am trying to say is once you find this g, you can just replace this g here and you can, you know, you find what u of x is and you are done. So, let us write down the representation formula, representation formula. This is called the Green's representation formula. Representation formula. So, what it says is, if u is in C 2 omega bar, solve, solve star, solve star, then u of x is written as minus integral over omega, over the boundary g of y del g del gamma of x y d s y plus integral over omega f of y g of x y d y. I am just, I just wrote this in C, Laplacian of u is f, minus Laplacian of u is f. So, that is why this is plus, plus f times g f minus integral over the boundary u of y times this. So, this is remaining unchanged, del g del gamma. I am just writing and u on the boundary is g. So, I am just writing it like this and that gives you a representation formula for any, any x in omega. And so, essentially what we did is, let me again just recall, we wanted to find the solution of a Green's function, sorry, the solution of the Foucault equation, with the Dirichlet boundary data. And to do that, what we did is, we found out a Green's function. So, the main important thing here is to find this indicator function. What is the indicator function? Yeah, to find this, sorry, the corrector function, corrector function, once you find the corrector function like this, so the important thing is this. See, if u, if omega is C 1, this is easy to find, this is always true, it is not a problem. For any u in C 2, if omega is a smooth domain, this is always true. You do not have to worry about it. The only worry is, once you find the corrector function, you have to find a corrector function for that omega. Now, it may look like not a very difficult thing to do, but it actually is the most important thing which you can, I mean, this is the difficult part, yes, so to find the corrector function. Here, I defined it like this, but in real life situations, if you have to solve the problem explicitly, you have to find the corrector function explicitly, you understand, that is going to be difficult, yeah. And once you find the corrector function, you just define a Green's function like this, I mean, there is nothing special actually, I mean, you cannot, g, I just defined it just to make it short, that is all. So, once you define this thing, you do realize that what happens to g on the boundary, it is going to be 0, right, g on the boundary of the domain is going to be 0, yeah. And so, you just have to find a g which does this and then you just write down the representation formula in terms of this function g, ok. So, one small thing, so I will remark, remark and this I want you guys to do it yourself, I mean, it is just some calculation, but I hope you can do this, this is called a symmetry, symmetry of Green's, this is Green's function, symmetry of Green's function. So, what does it say? It says that g of x comma y, this is equals to g y comma x, yeah, this holds for all x, y in omega such that x is not equals to y, is this clear? This is called a symmetry. So, essentially it is saying that if you change x to y or y to x does not matter, it does not change anything, it remains the same. So, and you guys have to check it yourself, please check it yourself, check it yourself, yourself, ok. So, please do that and this is just a property of Green's function, ok. So, once you do that you are done. Now, in the next part what we are going to do is we are going to see how to find a Green's function for a particular domain, ok. So, with this we are going to end this lecture.