Welcome back. We'll continue with developing our finite element formulation for our linear elliptic scalar problem in 3D. Okay, as we proceed one thing that may be of use, or one thing that may be advantages to see is how to interpret the basis functions that we are using. Since we are working in three dimensions, that actually gets a little difficult. And, and by interpret, I really mean something as simple as how do you visualize them. All right, visualizing them in 3D is a little difficult because of the proliferation of dimensions. And so what I propose to do in the first few minutes now is give you a the 2D case where it's actually easier to plot up these basis functions and get more insight into what they're doing. Okay, so let's look at that, right? So, so we'll continue with the same material, but we're just doing hm, taking an aside, so to speak, here. To hopefully enhance our insight. Okay? So, the aside here is we're going to look at Lagrange polynomials. Okay in particular we're, we're looking at the linears right? All right. In 2D All right, and so properly, rather than linear, these are really bilinears therefore, okay, because of 2D. All right. So in 2D, the situation would be the following. Instead of the use of my continuum potato represented by the foam football here, I will use a true 2D domain. Okay? So this is omega. We've we have the partition. Right? Into sub-domains, and that is a typical sub-domain. Okay, we're considering now the case where our sub-domains are polyhedral. Right, in particular, we're looking at sorry, we're not looking at polyhedral, we're looking at quadrilaterals now. Okay? Right and that, that's a typical, that's a typical sub-domain. Okay, so what we've done is we've partitioned omega into quadrilateral, lateral element sub-domains. Right and these are omega es, right? As usual e equals 1 to nel. Right, and, and, and everything else holds right, so we have indeed each of these as open. So we have omega, the closure of omega is the union over e of the omega es and the closure of that union. Okay, so we have this. So we partitioned it into quadrilateral element sub-domains. And on these quadrilateral sub-domains, we are going to consider the use of bilinear basis functions. So, consider bilinear basis functions. Okay, the picture by, the picture that we use in order to help us with the construction of these basis functions is is one that continues along this theme of using an of, of using a parent sub-domain from which every element, every actual physical element is drawn. Okay, so this is omega e. And now we, we suppose that we can construct this from a parent sub-domain which is now a square. Okay, it's a square, and it lives in a space where the, where the coordinates are labeled, z and eta. Okay. Since we are working with bilinear basis functions. Or we want to work with bilinear basis functions. And we have four nodes, of course, on the physical element. And so on the pair and sub-domain as well, we have four nodes, right? And this pair and sub-domain, you recall, is omega c, all right? Same notation. The points in the z, eta domain are picked essentially as before. Accounting however for the case that, accounting however for the fact that it is now a square. Okay, so the, this domain is again a bi-unit domain. So that says that these coordinates in terms of z and eta are minus 1, minus 1, 1 minus 1. 1, 1. And 1 minus 1. Okay. All right. Again we have we number these nodes as A equals 1, 2, 3, 4, all right. Okay, and again using notation that we've employed before. We may say that that the coordinates of node A, where we, of course, we're implying this local numbering of nodes, and therefore of degrees of freedom as well, right. Okay, so the coordinates, let's, let's also make it clear here, the coordinates of local node A. Okay? Those coordinates, in general, can be denoted as zA, eta A. Okay, and they take on the values that we've put down here. Right? Minus 1, minus 1, 1 minus 1, and so on. Okay. In this setting then here are our basis functions, right? I'll go to the next slide. So, the basis functions are the following. Now as before though we know we're going to use these basis functions in the physical domain, we take advantage of this barren sub-domain, this bi-unit domain to parametrize our basis functions in terms of z and eta. So, we have here n1, z comma eta equals one-fourth, 1 minus z, 1 minus eta. N2. One-fourth, 1 plus z, 1 minus eta. N3. One-fourth, 1 plus z, 1 plus eta. And, N4. One-fourth of 1 minus z, 1 plus eta. Okay? These functions also can be constructed as you would expect using the general formula for Lagrange basis functions. Okay? And I put down that, that general formula a little later, all right? I'll, I'll give you the general formula for 3D and then you'll, you'll, it should be clear how to simplify it to, for, to the two-dimensional case. Okay. Also you note that a hint to how that general formula is going to be constructed is to notice that here we have again this idea of tensor product basis functions. If you look at any one of these basis functions, for instance this one. All right, N1, it should be clear that it is a product of a basis function in the z direction and in the eta direction. All right? And the factor of half times half gives us one-fourth here. All right? So that, that tensor product idea continues. All right. These basis functions again we observe given our definition of the specific coordinates of the nodal points in the parent domain, right? These basis functions also do satisfy the Kronecker delta property. So we have NA, zB, eta B, equals delta AB. Right, the Kronecker delta property with which we are by now, very familiar. Also, when we sum up over all the nodes in the element. We sum up the basis functions. The result is one at every single point in the, in the element really. Okay? This has been expressed of course in terms of the coordinates in the parent sub-domain. But then, because we know that we're going to construct a mapping by mapping the geometry also we using the same basis functions. Right? Everything will work out. Okay. So, the reason to go into all of this was to be able to sketch out these basis functions and that's what I'm going to do now. Okay? So in order to look at them, I'm going to give you. I'm going to attempt to give you a perspective view here, okay? So, I'm drawing the pattern sub-domain here, but because of the, the attempt to present this as a perspective it doesn't seem quite squished. Okay. So that's our sub-domain. Those are our coordinate directions. Z and eta. Right? Our local node numbers are A equals 1, 2, 3, and 4. Okay? What I'm trying to signify here is that each of these node numbers has been written on a plane that is below the plane of the, of the actual element itself, right? Of the sub-domain. All right. That is our origin. Okay? Z, NA, eta equals 0. I'm going to draw just the sh, the, the basis function corresponding to N4. All right, so. I do this by changing colors here. And let's go with green. Okay, so, if you value at N4, what you see is that at z, N eta are equal to minus 1 and 1, respectively, right? It takes on the value of 1, right, so this value is 1. Okay, so this is N4 equals 1, okay. From there, if you stick to one of the coordinate planes, it comes straight down, right? Okay, linearly, right? So it's, it's linear along eta as we just showed. And along z, okay? Once it reaches this far edge, the function stays zero. All right. And so, as a Kronecker delta property suggests, N4 equals 0 at nodes, other than node 4. Okay? Now, at this particular point, at the origin, it takes on the value of one quarter, right. And this can be checked by just substituting z, N eta equals 0. And then 4 in fact in any one of the basis functions, right, all the basis functions value it to a quarter at the origin. So we'd get, it's just about that big, right? Okay? And what I'm going to try to do is suggest the shape that it has here so it's, it's essentially going to be a. It actually falls pretty rapidly until it gets to that point one. Sorry, one quarter. And from here, it will probably form a little less rapidly. Okay? Okay, I recognize it doesn't look quite quite smooth, but anyway that, that, that's just my poor ability to draw it. Right? The important thing is that at this point right at that point in 4, equals a quarter. Okay? So, so this is what happens with N4. The same sort of thing can be checked with any other basis function, with N1, for instance. And it will look similar, it will be, it will have this sort of tent-like structure. Right? Which obtains a value of 1 at equals 1. Right? And that node ends, and goes down bilinearly to 0 at each of the other norms, okay? So, so this is, this is an attempt to provide a little more insight into what is happening with these basis functions. This tent-like shape is a way to think about them. And as you can imagine, this is, this is more difficult to represent in 3D. So we, so we just stick with the 2D representation to provide a more visual sense of what these functions are doing. All right. I think we just end the segment here, and when we come back, we will launch into we will continue with the derivation of our finite dimensional weak form.
