Welcome back. In this segment we are going to look at a at, at another key property of the finite element method. And this is called the best approximation property. It draws in a very important way from the main result that we had in the segment preceding this one. And that result is the consistency of the finite element method. Okay? So, let's get started here. We're going to look today, look at in this segment, the best approximation property. Okay. In order to build up to this result, let's consider the following. All right? Let us denote, as we've been doing so far. Let u h, which we will recall, belongs to s h, be the finite element solution. Okay? All right. And let w h as we've been doing also. It belongs to b h, be a weighting function Okay? And note that I'm saying it is a weighting function, and the idea is that we are going to consider any waiting function living in v h. Okay? Now I'm going to introduce a code field that we have not previously considered. Okay? I'm going to denote this as capital U, h. It too belongs to s h, okay. And however it is not the finite element solution, okay. So what does this mean? This means that when we say that U h belongs to s h. We are recalling that s h finally consists of functions, right? Since capital U h is meant to denote a general member of s h which is typically different from the finite element solution, right? What that means is that u h belongs to h one on omega. And u h satisfies the Dirichlet condition. Right? The Dirichlet boundary condition. Okay. And let us suppose that we're working with the Dirichlet problem. And you know how this would work if we were doing the pure Dirichlet problem. We would have the other boundary condition also in there, okay? So this is u h. All right, if we have this, let us note that, okay. Note the following, okay? Note that any u h, okay, can be written as the finite element solution, okay, plus w h, okay? Why is this? Note that as I made a point about, w h is any waiting function, u h is a very specific member of s h. It is in fact the finite element solution. All right. Also note that u h belongs to h one and so does w h. Okay. Therefore, their sum also belongs to h one. We haven't quite proven it, but you can probably. See why that works out. It does indeed work out rigorously. Therefore, when we add u h and w h, we recover a, we recover functions which live in h one, which re, which remember, or which we recall from just out here, is also a property of u h. Okay? So take. So adding little u h and w h maintains that property of capital U h. What about satisfaction of the Dirichlet boundary condition, which capital U h has? Well, remember that since little u h is the finite element solution, it does satisfy the Dirichlet boundary condition. What about w h? Does w h not mess up that result? It doesn't, because w h is zero on the Dirichlet boundary. Okay. Okay, so note that when we write u h in this form capital U h in this form, it does indeed satisfy all the requirements we have for this new set of functions capital U h. Okay. So let me just state that. U h equals little u h. Plus w h, right, where little u h is a finite element solution and w h is any waiting function, okay? This does indeed belong to h one, right? And I would just write and satisfies. The Dirichlet boundary condition. Okay? In what we are going to do today, the main result of this segment, we will use this result in a in an important manner. Okay, so with that in place let me state the main result of this segment. It is a, it's a theorem really. Right? And than theorem is the following. If we consider the energy norm of the error. And here I'm writing the energy norm using abstract notation. In one of the segments preceding this one, we demonstrated that with abstract notation for the bilinear form as we had defined it. We do indeed get the, the energy norm of the error, if we write it in this form. All right, so. The theorem is that the energy norm of the error is bounded from above. By the energy norm of what you may also consider the error, except it is capital U h, this new field that we've introduced, minus u. Okay this is the result that we are going to prove. Why is it of interest? Note that e, the error is the finite element solution that's little minus u, okay? So what we're saying here in this theorem statement is that the energy norm of the error, 'kay, is such that it, that, it is such that the finite element method has picked a member of S, H, okay, to be the finite element solution, which gives us the lowest energy of any possible other member of S, H, 'kay? Right, so let me state the, sort of give you a plain English statement for the relevance of this theorem, 'kay? So what we are saying here is that the finite element solution minimizes, minimizes the energy known. Okay? Of U h minus u over all members. Belonging to Sh, 'kay? The space Sh consists of a number of functions, okay? One of them is the finite element solution. That finite element solution is that member of Sh that minimizes the energy norm of the difference between that function and the exact solution, 'kay? In this sense, with the energy norm as a measure, the finite element method picks the best possible approximation among all members of Sh, okay? And this is why we call this segment the best approximation property, and in fact this theorem is called the best of the, this is, this, this statement that we have here represents the best approximation property, 'kay? And this holds for all Belonging to Sh, okay? All right, we're going to set about proving it. To prove it, let us first consider the energy norm of a slightly different function. I want to consider the energy norm of e plus wh. All right? Now we know how to expand this out, right? We get a result, which works just like the expansion of a perfect square, 'kay? It is that a, this is equal to a, e, e plus 2 times a. Well let me write it out more explicitly this, this is a subsequent step, plus a(e,wh) plus a(wh,e) plus a(wh,wh). All right, and this comes about just from the expansion. If you were to go back and look at what this abstract notation actually represents for the energy norm, that using the fact that it is bilinear in its two arguments, right, with these being the two arguments. You will get this result, right? It's just like, just like the expansion of a perfect square. And then, one recognizes that because the bilinear form also is symmetric, right, we note what we get for the result in the bilinear form is symmetric. Right, we see that these two terms, right, are equal, okay? So that lets us then write it out as energy norm of e plus the bilinear form a, let me write it as (wh,e). 'Kay, recognizing that the second and third terms in the line above are equal. Plus a of (wh,wh), all right? Okay, stare at this now and try to recall whether we can say something special about this term. Right, indeed we can. In the previous segment we demonstrated that this term is equal to 0. And this term is equal to 0 is a result that we've termed consistency of the finite element method. Right, it is the result that it, it, it, it, it, it comes from consistency of the finite element method, and consistency of the finite element method itself, you recall, is the fact that the finite dimensional weak form admits the exact solution as a solution, right, it recognize the exact, recognizes the exact solution. Okay, so what this says then, is that what we started out with at the top of the slide is a of e,e plus a of wh, wh. Now, each of the two terms on the right-hand side is a, is an energy norm, right? The first is the energy norm of the error, the second is the energy norm of the waiting function. There is a property of norms, right, a fundamental property of norm. Right, almost the definition of a norm. It is that each of these has to be greater than or equal to 0, and you can go back and check from our definition of what the energy norm is, that indeed this holds, okay? All right but if this is the case, what this implies for us is that let me now just pull down the term from the left-hand side here, a of e plus wh,e plus wh equals this. Okay, what this implies for us is that a of e,e, I'm just writing the energy norm of the error, right, this term. Okay? What it implies for us is that. Well, let me ask you. How is this term related to a of e plus w h? E plus w h? What's the relation here? Right, it is lesser than or equal to it, okay? All right, simply because the sum of two nonzero terms in the right-hand side is equal to the term in the left-hand side so we get rid of one of those nonzero terms. Well, what remains must be less than or equal to what we had in the left-hand side and then just flipped around their positions. Okay, and now we're getting somewhere. In order to proceed, let me go to the next slide. I'm going to rewrite this term differently, okay, the term on the right-hand side. Note that. A, e plus w h comma e, plus w h is equal to a. Now, the error. I'm going to rewrite the error in terms of the finite element solution and the exact solution. The way we define the error was simply the finite element solution, little u h, minus the exact solution, u. And we have w h there, the same in the second slot, u h minus u, plus w h. Okay? But now I can recombine this, right? I note that when I get u and w,h, u h and w h, this is from the result we proved at the beginning of the segment, just capital U h, right? The function that we introduced which we identified as being any arbitrary function in S h. Okay? That minus u, comma, same thing on the left-hand, on, for the second slot. All right? Now, we put this together with what we had at the bottom of the previous slide, right, which is this result, in the bottom of the slide. So, combining. With the inequality. Above, right. The inequality that we came upon just a few minutes ago, we see that that inequality which is a of e comma e is lesser than or equal to this term. 'Kay? Well, that means it's also, it's, it's lesser than or equal to U h minus u comma U h minus u. This is what we set to prove out, set out to prove, okay? We're done. What we've done here is proved the best approximation property, right. And I'll restate the significance of this result. It says that the other way to write this is instead of the, you expand out the error right, which is u h minus u. Okay, so the energy norm of the error where the error is defined as a difference between the finite element solution and the exact solution is bounded from above, right, by the energy norm of any other function living in S h, right, any other function, all right, for all U h belonging to S h. But little u h, the finite element solution also belongs to S h. So what the finite element method does is that it picks for us the solution, as the solution, as the finite element solution, that member of S h, that minimizes the energy norm. All right, of the difference between the function and, and the exact solution. So it's in this sense that we have this best approximation property. Okay, so re-emphasize that, let me just state that this is what we mean by the best approximation. Property. Okay? And just remember in all of this, that what we have is little u h, is indeed the finite element solution, all right? So, the finite element method does a very efficient job once we pick our space S h, does a very efficient job of choosing that member of S h that minimizes the norm of this of this difference. 'Kay, and says that well, that is your finite element solution. Okay. All right, good. We'll end the segment here.
