 So let's look at this equation. So in some bounded domain in Rd. So the operator L, epsilon, is given by negative divergence of x. Assuming A is elliptic and one periodic, for instance. OK, thank you. No smoothness is needed for today's lecture. So you look at this equation, you actually see two scales. One is x, another is x over epsilon. This is referred as a slow variable and a fast variable. So there's a method called a formal asymptotic expansion, two-scale expansion. That is, you try to figure out the formula by expand the solution in a formal manner here. So you write u epsilon as u0 x, x over epsilon. Assuming you have two scales here, that's the first term. And then you have epsilon u1 of x, x over epsilon for the second term. And then you can have a third term u2 x, x over epsilon. And you can continue. So this is an infinite series. We're now going to concern with whether it converges. It's a formal expansion here. So then we're going to assume that each of these functions, uj of x, let me call the second variable y. So whenever you see y, you should think of x over epsilon. Is, say, one periodic in the second variable. Because that's what you have in any equation of x over epsilon whenever the coefficient is periodic in a variable there. So here, then you put this infinite series into this equation. And then you'll equal the terms with the same power of epsilon on both sides. You actually end up with, you deduce that the first term actually does not depend on the second variable. It's independent of y. This u0 actually is the limit of u epsilon. And the second term appeared in this form. So this is kaij of y and du0, the same function here, and dxj here. And this kaij of y is one periodic. One periodic and satisfy the equation. Then kaij, the operator L, I'm going to just define. So the L here, I'm going to use the variable y, divergence of A of y, gradient in y here. The divergence also in y is equal to negative L applied to the linear function yj. Again, this is a formal calculation. Well, unfortunately, I don't have time to actually show in this. Here I'm trying to do is I'm showing where this corrector is coming from. How do you find, how do you see that the corrector should satisfy that equation here? So that's how you got this equation here. So what you have here, if you put this on one side, you end up with L of kaij plus yj is equal to 0 in the whole space. So in other words, you can think this corrector kaij is a function which is used to correct the linear function. If you have a constant coefficient, like Laplacian, the linear function is a solution. But if you have a variable coefficient operator, the linear function is not a solution. And this kaij, when you add to the linear function, it becomes a solution everywhere in the space. And so that's what we're looking for here. This kaij is also one periodic. And in order to fix the one, I also ask that the mean that each periodic cell is 0 here. So in the periodic setting, the existence of the corrector is a simple manner, but it's actually one of the most difficult part if you're going beyond periodic setting. Even to the almost periodic case, the quadratic periodic case, the existence of the corrector is one of the most difficult problems here. So here is not a problem. How do you find the corrector? Show the existence uniqueness. You simply set up a binomial form on the torus. So in the notes, I use y. So y is just, you can think this is rd, modular lattice. And this is the torus here. So the h1 torus is just h1 with periodic conditions here. And then you simply using Lax Milgram to solve a problem. So you define b, let's say. The b is already defined there, phi and psi to be the integral. This is the average actually here. You don't need that. So y, a, the gradient of phi dot, gradient of psi, and dy. And you solve a use Lax Milgram theorem to solve this equation, b, chi, j. Psi is negative b, y, j, and psi. So this is true for any psi in h1 torus there. So the ellipticity gave you a corrosive binomial form. And the right-hand side gave you a bounded linear functional on h1 on the torus. So one remark here is that if you have a matrix satisfy this condition, say, dd of, say, xi, yi, aij is 0. The index i is sound. So if then the corrector actually is 0, that is because you can see the right-hand side actually give you 0 there. So in this case, if you have a matrix, each column is diverges free. You don't have a corrector. It's dandico equals 0 there. So in a lot of cases, that's a simple case there. The other thing I want to mention here is that, so here I already say that if you use a variable y here before the scale, this becomes a solution if you add yj to the corrector as a function of y. You can rescale this to see that xj plus epsilon chi j of x over epsilon is a solution in the whole space. So if you look at this original operator L sub-epsilon, you have a linear function. You add epsilon times chi j of x over epsilon to it. It becomes a solution to the operator L epsilon. And this will be useful for us later on because you'll see that it somehow indicates as far as the regularity is concerned, the best you can do is the Lipschitz. So we're actually going to prove Lipschitz estimate in lecture 3 and 4 using two different approach. OK, so these are the definition of a corrector. So once you have a corrector, we can introduce a homogenized coefficient, or effective coefficient. In the same time, that will also give you the homogenized operator or the effective operator. So the definition is that the aij hat, which is the coefficient for L0, is given by the average of aij. First term, in the second term, you have aik, the derivative of chi j in yk. The index k is summed from 1 to d. And you can actually write this in a binomial form. So it's a b of yj plus chi j comma yi plus chi there. So one simple observation is that here the coefficient for L0 is not a simple average of the coefficient for L epsilon, but rather it's kind of a nonlinear average. Because the second term involves the corrector, it's multiplied to the coefficient here. So this is a nonlinear term here. Although the first term is just the average of aij, but you need a second term to give you aij hat here. That is the homogenized coefficient, or effective coefficient. And so the first term, which is approved in the lecture notes, is that this L0 is elliptic, actually with the same lower bound. So aij hat, chi j, this is also summed. The mu is the same constant for elliptic constant for a. And by upper bound may be different. So upper bound here you have mu1 with some constant, which only depend on mu and the dimension. And so that's the ellipticity here. So that's the first term there. OK, so this is the main theorem we're going to prove today. So this gave you the homogenization of Dirichlet problem. So you're looking at a matrix, which is elliptic and periodic. You have a bounded elliptic domain. And you solve a Dirichlet problem, a typical one you see in evidence. So right hand side is h minus 1, the du of h1 0. The bounded data is h1 1 half, taking in a sense of trace, h1 1 half trace of h1. So we know the solution exists and is unique by the last milligram. For any given epsilon, actually you don't need a periodicity condition. So with the periodicity condition, what the theorem says is that as the epsilon goes to 0, u epsilon has a limit, weakly in h1. Actually, you can see a little bit more. The flux also converge. Weekly, I'm sorry, this is not right. This would be L2, weakly in L2. Second line, the first line is weakly in h1, but this is weakly in L2. And finally, the limit function u0 is a solution to the homogenized problem, which is for L0. Same domain, same f, same right hand side, same bounded data. It's just operator change from L epsilon to L0. And L0 is the one we just introduced, the homogenized or effective operator. So we know that this A hat, the homogenized coefficient, is elliptic, so this responding value problem has a unique solution, u0 there. So that's the qualitative theory for the Dirichlet problem. The same thing can be said for the Neumann problem. That is, let's say you again assumption is the same. A is elliptic, periodic. And omega is bounded, elliptic, and it's of a Neumann problem. So here I write this in maybe a slightly different form, because the right hand side has to be in the dual of some h1. So let's just try to avoid that just using f plus the divergence of g, both capital F and capital G are L2 functions. And you write in a boundary condition in this form. This is a vibrational, co-normal derivative, new epsilon. And I did not put on the board what is the co-normal derivative we're using here. So du epsilon, d new epsilon is the normal times A, the coefficient. Actually, this is the flux. That's the definition. So the co-normal derivative depends on the operator. So this is quite different from the Dirichlet problem here. So again, u epsilon, well, here to fix the solution, let's assume that the mean of u epsilon is 0 in omega. And the second line should also be L2. So u epsilon has a limit weakening in h1, and therefore strongly in L2. And u0 is the solution to this homogenized problem here. So as I said, the co-normal derivative depends on the operator. So du 0, d new 0 is the co-normal derivative associated with this operator L0. So in other words, what you have here is that d new 0, d new 0 is m dot a hat gradient of u0 there. OK, so I see I have to hurry up. Let's see, we'll be one, two. So I'm going to write down a few theorems without a proof. You can find the proof in the lecture notes and get to the main step of the theorem of the homogenization I have on the screen here. So the first one, the first result I need is the following. So I'm going to write this in B a sequence of one periodic functions and assume that the sequence L2 norm is uniformly bounded, and the average of hL has a limit C0 here. And then I'm going to choose a sequence epsilon L goes to 0. Then the conclusion is that if you look at hL, and then you rescale as x over epsilon L, and this actually has a weak limit, which is the same constant I have there, weakly in L2 omega here only need to be bounded. So in particular, if you only have one function here, if all the functions are the same, if you have, then this guy will simply weakly converge to its average weakly. And that's the consequence of this here. That's one theorem I will need. The second theorem is commonly referred as a dv-curve lemma, which will be one of the exercises for you this afternoon, dv-curve lemma. So here I have two sequence, ul and vl, be two bounded sequences in L2 omega with values in Rd. So there are vector field on omega here. And suppose that one, they have weak limit, vl weakly converge to in L2 of omega. So this problem, I mean, this lemma is trying to solve the problem that when you multiply two weakly convergent sequences, it may not converge weakly. And here is the condition which will guarantee the product also converge weakly. So the condition is that one of them, let's say the first one, is curve equal to 0 in omega. And the second condition is divergence of vl converge to f. This has to be strong, strongly in h minus 1. One has curve 0, another has divergent converges. So under these assumptions, the product of ul and vl converge weakly. Well, I'll just say that here is u dot v phi dx. For any phi in C0 infinity of omega. So ul dot vl converge in some sense weakly when you multiply to a test function there. So these are the theorem I will need in order to prove the main theorem I have there. So let me just go back to just concentrate on the Dirichlet problem here. OK, so here is the theorem. Actually here, for the later purpose of compacted smelter, I'm going to have to consider a sequence of matrix which satisfies the same condition. I'm going to call m of mu. What is m of mu? I forgot to define earlier. It's just a class of matrices which are elliptic and periodic with elliptic constant mu there. So let also take fl in h minus 1 of omega. And suppose that you have a solution to a divergence of al x yl gradient of ul equal to fl in omega. Here you realize that not only I put yl on the denominator of the variable, but I also let the matrix A self varies in the same class. And further, let's say fl converge strongly to f in h minus 1 and ul converge weakly to u in h1. And al hat, so each al will have its own effective coefficient, I'm going to denote by al hat, has a limit. It's a constant, so whatever it's a limit, I'm going to also a constant matrix. Satisfy the same condition. Then the conclusion is that A of ul converge to A0 greater of u weakly in l2 of omega. And once you have this, because each of this divergence equal to fl and fl converge strongly in l2, and this line will actually implies that the limit is a solution to this equation in omega. And here you see that we're not going to concern with the boundary data. The boundary value does not enter into the play. And this theorem, therefore, can be used to prove both Dirichlet and Neumann problems homogenization. Yes. Yes. Sorry, as all these are limit are taken as l goes to infinity. Al hat is a constant matrix. So al is in this class. Any matrix in a class has a hat, which is a constant. We don't need to concern that, because here just looking at the matrix itself, and you see that the A hat is uniquely defined whenever you have an A. Yes, you're right. So each epsilon goes to 0. It goes a hat. But here for the purpose of compactness method, we're going to use lecture three. We will allow both to change. The matrix is changing. The epsilon is also changing. All right, so how do we prove this? So you want to prove this. So I'm going to call this star. So it suffices to show that if you have a subsequence, if you have a subsequence, I'm going to call u l prime, is a subsequence such that this guy has a limit, which I weekly say in l2 for some h, then the limit has to be a0 times the gradient of u. So in some other words, whenever you have a subsequence which converge, they have to converge the same thing. And therefore, you can deduce that the whole sequence has to converge the same thing. For otherwise, you can produce two subsequences which converge to two different limits. So that's all we need to do here. So without a lot of generally, let me just assume the whole sequence converge, actually. Gradient u l converge to say h weekly. For some h, the goal is to show that this h must be given by a0 gradient of u here. So for this, we will use the divi core lemma as well as that theorem on the weak convergence of periodic functions. So what you do is that you consider the following identity, gradient of u l dot, the gradient of, here I have x k, just a linear function. Then I have a y k. Then I have a chi k l, I have a star. I will explain what the star is. y l then times phi. That's a test function. Phi is any test function here. So you can certainly remove this matrix to here. So gradient of u l dot a star l of star just means a joint. The gradient of x k plus epsilon k chi k l star x over epsilon l phi dx. This is a simple identity. I haven't proved anything. Just move the matrix a from here to here. Becomes a joint here. So what is the x l star? This is the, my notation is not very good. So this is the corrector for a l star, adjoint matrix. Adjoint matrix satisfies the same condition. And so you can also define the correctors.