Okay, welcome back. We continue now with working with the with the finite dimensional weak form or Galerkin weak form. What we saw in the last segment is was a digression to say more about what we meant by function spaces and to understand why this might be important for us, okay? Because and, and, and, you recall that we defined function spaces as being useful or, or we introduce function basis as being useful in order to give us a sense of control over the functions themselves and over their derivatives. Right? It's when we say that we have control over the functions we are saying that the functions are bounded. When we have control over the derivatives in addition to the functions we have this notion of regularity, right? That the functions are sort of smoothed and so on. And we saw that through examples. Okay, with that as background now we will actually launch into the finite element method for this 1D linear elliptic PDE. Okay? And this would be, an we are going to work off the Galerkin, or the finite dimensional weak form. In this segment, we are finally looking at the finite element method for linear elliptic PDE's in one dimension. >> Now, we're going to work off the Galerkin weak form. Okay? Recall the Galerkin of the finite dimensional weak form. One more time unto the breach. Okay, so one more time we're going to write this out. Find u h belonging to s h which is equal to s h is equal to functions u h. Now we're seeing that they belong to H1 on (0,L). Such that they satisfy our boundary condition. Now gratifyingly, we know what H 1 means. So, we already have an idea of the extent to which we've restricted our, search to solutions for this PDE. Okay? So recall all the functions we talked about in the previous segment. Those are the sorts of functions we are now, considering as candidates, okay? Okay. Actually, for this reason, u h is often also called the space of trial functions. Okay? So, often, u h is also called a trial function. Because we say we're going to try out all these functions living in s h, okay? So maybe that's a useful thing to say here. Okay. Find u h belonging to s h, such that, for all w h belonging to v h which is also drawn from H 1 Okay? Such that, for all w h belonging to v h, the following holds. Integral zero to L, w h, x sigma h A d x equals integral zero to L, w h f a d x plus w h at L, t a. Okay? That is our finite dimensional or Galerkin weak form. Now one thing I should point out here is the following. In our finite dimensional weak form, observe that we have sigma h, all right? In that finite dimensional weak form. What we imply is that sigma h is going to be obtained as E u h, x. Okay? However, f, our forcing function f, is given to us as data. So we assume, and in fact we, we, we will in fact use the fact that, in our implementation of the finite dimensional weak form. We are not going to attempt to, to approximate the, the data, right? We are going to represent the data exactly, exactly as given to us. So f, f of x is not, finite dimensional. Okay? f of x will actually be represented as it is given. Okay? In this sense, you may think of it as it's not exact but, but, but the fact that it is data is why we are not going to represent it as a finite dimensional approximation. All right? It's given to us. Okay? All right. And, and of course in this the case, does the question arise for t, the traction? Think about it. It doesn't in 1D because in 1D the traction is applied at the point. So the question of whether we approximate it, whether we write it as finite dimensional or not doesn't really arise. We go to higher dimensions and we go to three dimensional problems we will see that, there is a question to be answered for the traction. Okay? Because the traction will then be defined on the surface and so on. But, but we get to it when, when we do. Okay? So t, let me just state, is a point value. Okay? And I'm stating all these things for us to understand, why in our finite dimensional weak form, we have finite dimensional versions of w h. We also have a finite dimensional of sigma h, right? And that's this because sigma h is going to be obtained through our constitutive relation applied to the gradient of the finite dimension of trial solution, u h, okay? And on the right-hand side, again, the waiting function is finite dimensional, not so for the forcing function which is data. And it's actually not a relevant question for the traction in this one d setting, okay? So, that's what we have, all right? So that explains why we have final dimensional forms w h,and sigma h. All right, so, we can write out this finite dimensional, or Galerkin weak form, if we say, how we obtain our finite dimensional functions, okay? So, that's really the question. How do we obtain u h and w h, right? Alternately. Alternately, how do we obtain S h, and V h? We need to define those. All right, we need to say, what kinds of restrictions we're using on our finite dimensional functions, okay? In one sense, the finite element method could be thought of, as simply defining what these finite dimensional approximations are, okay? So, here, here's how we do it, okay? The way we do it is to I'll write the statement here, and then I will start sketching things. What we do is partition 0, L into finite elements, okay, which are disjoint. Sub domains of 0, L, all right? Okay. So, here's, here's how it goes. Let me just draw our bar, one more time. And here, we have our x-axis. And that is L, that is 0. Now, for brevity, as well as to make an easy transition to multi-dimensional problems, I am going to introduce notation here for this domain 0 to L. I am going to write omega is our open interval 0, L, okay? So, that is our general domain. Okay, so, the way finite elements proceeds is to partition our domain omega into sub domains, right, which are our elements. And this partitioning is done by nodes that I have, that I'm now marking here, okay? So, we look at these nodes as being defined as x we call this x sub 1, x sub 2, and so on, x sub N, right? Sorry I shouldn't call that N, I'm going to use that for another, for something else. I'm going to call this x sub, call this actually a K, big K, because I am going to call this very last node x sub N, okay? Or I think, I'll even call it, x sub number of nodes, okay? All right, now, each of these sub domains that we've thereby, defined with these nodes. These, these are going to be our elements with the finite element method are omega 1, omega 2. In general we have omega e, right, and so on, okay? Observe that what we've done here is, what the partition that we've introduced. Is the following. We've partitioned omega into sub domains omega e, all right? And the nature of this partition is such that, the total that our entire domain omega is the union of each of these sub domains omega e, all right? Furthermore, each of these sub domains omega e is an open sub domain. Okay? It's an open interval, right? So, the total domain omega is the union e equals 1 to n sub el, which for very obvious reasons is number of elements, okay? So, the domain omega is the union of each of these sub domains, omega sub e, and e here, runs from 1 to n el. Now, because omega e is open, okay, and because we have a very convenient one dimensional setting, what you will note is that each omega e is the open interval of x sub e to x sub e plus 1, right? Clearly, because omega 1 is the Omega 1, which is this one. Is the open interval, x 1 to x 2 and so on, okay? So, now, there's one more thing we need to do here for technicality. Because each of these omega es is an open interval and because we would miss out the, intervening the inter-element nodes, if we just went with this union, we need to say also the following. We need to apply closure to this union. Okay? And say that once we close that union, we also have the closure of our open interval. Okay? So this way we are making sure that we, that we are not missing any points. This is for purely technical purposes. When it comes to computing, it makes no difference because each of those points is what we call a set of zero measure. Okay? So let me just state the, the notation here. So omega bar is the closure of omega. All right? So, it is basically saying that omega bar equals, omega union the boundary points of omega which introduce even more notation is written as that, okay, partial forming, reverse the boundary forming. These are purely technical points, but it's useful to state these now so that there's no confusion later on about what we are doing with nodes, and so on. Okay? All right. So. We have, this, partition. Let me introduce, terminology which have been using already, but let me define the terminology. So. The points x e are the nodes of the partition. Right? Omega e is an element. Okay? Obvious notation, obvious, nomenclature which I'm sure you all either knew, or figured out, or anticipated or whatever, but here it is. Okay? All right, so how do we first, use this partition, how do we begin using it? As the very first step we observe that our weak form has, is stated as an integral. Okay? And what that lets us do is to write that integral as a sum over the elements, right. Over the sub domains of the partition. Okay? So, the weak form or the, the [INAUDIBLE] weak form, what this actually holds for the, for the infinite dimension weak form, as well. All right, the weak form. We will now write as follows right? Instead of writing it as a sum over zero to L, we will first write it as a sum over omega, okay? W h comma x sigma h A d x equals integral over omega, W h f A d x plus W h at, L t A. Okay? So we've gone from writing it as an integral for over zero to L to an integral over omega. Purely notation. The next thing we can do now is to write this as an integral over each element subdomain, okay? Like that, okay? But note that we need to get to all of omega, right? So here we take a sum over e, okay, where e runs over all the elements, one, two and e l. One to number of elements. Okay? And here, too, we have sum e equals one to a number of elements. Integral over omega, omega e W h f A d x plus W h at L t A. Okay? So this is the first that our, partition of the domain into element subdomains lets us [INAUDIBLE]. What that, the next step that it also makes possible for us is to now think in terms of defining our, finite dimensional functions. But observe that having reduced those integrals to integrals over smaller subdomains, really, all we need to worry about is how to represent these finite dimensional functions over the subdomains omega e. Okay, so let me just write that here. So, represent. S h and v h over. Each omega sub e. Okay? So we can really afford now to focus on what's happening within each element within each subdomain. Right. With the idea being that the union of all the subdomains are gives us the, the, the complete, the entire domain that we're interested in. And thereby we can also define our finite dimension functions over the entire domain, okay, by going to these sub domains. Okay, let me, it's useful to stop here. When we return, we will think in terms of how we define these functions over the subdomains.
