Welcome back. What I'm going to do in this segment and maybe one or two more is [COUGH] almost take a step back. And outline for you the main steps that are involved. Only slightly different steps that are involved in setting up two-dimensional problems. Okay? So, we're following somewhat of a deductive approach. Where we go from the more general 3D structure of, of the problem, to a 2D structure which is, of course, is a particularization of a 3D. Okay, we're going to do this in the context of the linear, lytic problem in two dimensions now, with a scalar variable. Okay? So we will call this, of course two dimensional. Linear. Elliptic. PDEs in scale, in a scalar variable. All right. So, since we are talking of linear elliptic PDEs with a scalar variable. The physical problems that this is applied to remain, heat conduction and mass diffusion in two dimensions. That's all. Okay. There's actually very little to do. Almost nothing to do. And I'm going to do just that, all right? So, here is my version of doing nothing. So, I'm going to first, we, we start out with a strong form of the problem. And, and really, all we have to observe is that everywhere in the 3D formulation where we observe that the spatial dimensions could the spatial dimension index ran from one to three. It now runs from one to two. And we need to recall for ourselves how we would construct the basis functions from over, over two dimensional elements. All right, okay. So, I'm, for, just for completeness, I'm going to put down fairly quickly, the strong form and the weak form. Okay? And in fact because I'm going to do that fairly quickly, let me straight away say that we're given the data. Okay? We're given u g. We are given j sub n. We are given f. And we are also given the constitutive Relation [SOUND] -j i. Sorry, +j i = -Kappa i j. u, j and here, we have i, j = 1, 2. Okay? This is what we're given. So, with this background, here is the strong form. Okay, the strong form is. Find u such that j i, i =, now we'd use f in our first all sort of problem. And later on, you will recall that I said, that there was reason to replace that with minus f bar. Okay? Right? This is our PDE. All right? In omega. And, and we remember that now omega is a, sorry, it's a subset. Omega is a subset of R 2. Okay? So, the domain that we're looking at would be this. Right? With the difference that instead of drawing a three dimensional basis, or three basis vectors here, I have just two. V-1, V-2. Okay. So, this is our PDE. With. Boundary conditions. U = u g on what we understand to be the Dirichlet boundary. Okay? And -j, sorry. I'm doing this all with coordinate notation. So,- j i n i = j sub n for the influx condition on, and now, I'm in boundary. And n, now, is a unit vector in two dimensions. All right? Okay, that's it. That's our strong form of the problem and I've already stated up here that i and j run over one and two. Okay? I'll go ahead and write the weak form and I will in fact write the finite dimensional weak form, right, essentially skipping a step here. So, I will right now the finite dimensional weak form. Okay. So that is find u h. Sub, sorry, it's not sub i. Find u h belonging to s h subset s where s h consists of functions u h, belonging to h 1, over the two dimensional domain omega, and satisfying the traditionally boundary condition. Okay? So find of, of this type such that for all wh belonging to Vh subset of V, okay, where now Vh consists of all weighting functions. Also, to be picked from H1 functions, and satisfying the homogeneous Dirichlet boundary condition. Okay, find such that for all wh of this type, the following holds, right? Integral over omega wh,i, jhi. Now we are in two dimensions, right? So, what we are working with here is is an integral over an area, okay? So we get dA, right? So, wh,i ji jhi equals integral over omega wh f bar dA, okay? And what we note here is that okay, that in order to get things right we will pick up a minus sign there, okay? This is, I'm just working through all those changes I made with the signs later on in the 3D problem, okay so we have this. Plus integral over the Neumann boundary wh j sub nd. Okay, now we need to be a little careful here. In two dimensions the boundary is a curve, okay? So the boundary integral here I'm going to write as dS, all right, where for us S is now a curve. Okay, and dS is the elemental curve, okay? All right, so this is what we have. And the only thing we need to recall, if you're, is as we did on the previous slide, just remember, here, that i, remember that i equals 1,2, okay? And in this goes jhi, we also know will eventually be written as capital kappa ij sorry, minus kappa ijuh,j, okay? So j and i both run over 1 and 2. All right, and you see that now those signs will cancel out, right, plus, minus and minus will cancel out and we get back everything as positive, right, the way we write it down. All right good, so, so this is what we have and you know how everything works from here, right? We are going to use elements which are subdomains. I'm not going to write all the technical stuff about the, the union of elements you know, with closure applied to them giving us the total domain and all that, right? We don't need all of that here. Let's just look at the element sub-domains, right? So, we may work with quadrilaterals here, right, of that type, okay? And just to fix idea, let's suppose that we're working with bilinear quads. All right, so this would be omega e, right? We would have A equals 1, 2, 3, 4, which we know would be constructed from a mapping from a bi-unit domain, and we studied this process. Okay, we could have then, have it constructed from a bi-unit domain, which is now expressed in terms of coordinates xi 1, xi 2, all right? And here our nodes are we, we know exactly how, what these are, right, these are minus 1, minus 1; 1, minus 1; 1, 1; and minus 1, 1, okay? Those are our nodes in the bi-unit domain, okay? Now we also remember that this mapping gives us a way to write out the geometry, okay? So we know that the geometry in a the physical domain is a function can, or can be parameterized by our coordinates in this bi-unit domain, okay? All right, so we may have this or we may choose to be working with triangles, right? And we do that or. Triangular elements. In the case of triangular elements, I just drew the very simple so then, it's the one that we looked at in a previous segment. All right, that's omega e. Once again we have nodes at the vertices, and I'm going to label them as A equals 1, 2, 3. We know this can be constructed as a mapping from a unit domain, which we also express in terms of coordinates xi 1, xi 2, and in this setting We get that point is 0, 0; 1, 0 and 0, 1, okay? And right, we also recall that in this setting our no, our degrees of freedom here are labeled 1, 2. And three. And that reminds me that I did not label the degrees off freedom in the bi-unit domain here. So let me call this one, two, three, four. Okay, but we now that already very well. Okay. And, again, we will get from here a mapping of the geometry as well xi 1, xi 2. Okay? All right, so in this setting we will get our basis functions, right? Right. There's two ways in which we would write our basis functions, depending upon whether we were working with the bilinear quad or the linear triangle. Right? In the case of the bilinear quad right? We would get bilinear basis functions right? Bilinear polynomials. Right? These would be written as n a parametrize by c1 and c2. And we know what the formula for them is right, I'm not going to rewrite it. We've written it a couple of times at least now these are the standard bilinears. We've also looked in one of the previous segments at the Generalized Lagrange polynomial formula for this, right? So these are bilinear Lagrange polynomials, right? In this case, A runs over 1, to number of nodes in the element and their number of nodes in the element is 4. Right? Okay, or we would have linear polynomials. All right? If we were working with the triangles. Right? Linear polynomials would again we written as a n parameterized now by c1, c2, and remember the way that these things are written often Using area coordinates, we would write Z, include 8, Z3. Okay. And remember that Z3 is defined, in the case of triangles as 1 minus Z1 minus Z2. Right. And we saw this as recently in E segment. In this case again A equals from 1 to number of nodes in the element. And for the case that we're considering up here on this very slide. Number of nodes on the element is equal to 3. All right? So this is what we have and and with this we go ahead and of course construct our finite dimensional functions, right? So we will go ahead and write u h over e is now sum over the number of nodes in the element. N A, right? I'm not going to write the parameters of N A because they could be different depending upon whether we are working with bilinear or linear polynomials, all right, but we're working with bilinear quads or with triangles. So I'm just going to say it's n a times d a e. And for the waiting function, right, we have w h of e equals the same sort of expansion. Okay. All right, we just need to remember that in the two cases, the number of nodes in the element could be different. All right.