Okay. So we're well into our development of the finite dimensional weak form for our linear elliptic problems in three dimensions with scalar variables. We are in fact at the point of having already decided to work with hexahedral elements. And I've stated that these will be constructed from a, by mapping from a parent domain, okay. So let's just pick up right there. So mapping from the parent domain which I've already said we are going to continue to call omega z to omega e, okay. So, here's what we have. We have omega e, and I'm somewhat purposely going to draw this as, an irregular hexahedron. So, maybe something like that. Okay. Let's see. Okay, that's reasonably irregular as a hexahedron. Okay, noted still. Okay? But now, the point is that we say that this is obtained from a mapping from a domain in which things are indeed regular. So we have a nice, cubic domain, a nice cubic, structure in the parent domain here. Okay. So this is where we are constructing this, element omega e from, right? And this is omega sub c. Now, in this domain, we, since this is three dimensional, we have coordinates c, eta, and zeta, okay. And in terms of c, eta, zeta, again this is a bi-unit domain. Okay, so omega c is also a bi-unit domain just as, we had in one d, okay. What that means is that in terms of coordinates of, the nodal points in this domain we have for these nodal points. We have the following coordinates, okay. So this point here is minus 1, minus 1, minus 1. This point is 1, minus 1, minus 1. This point is 1, 1, minus 1. And this point here is minus 1, 1, minus 1, okay? On the top surface we have 1, minus 1, minus 1. Sorry. This is minus 1, minus 1, and 1. This point here is 1, minus 1, 1. This is 1, 1, 1. And this point here is minus 1, 1, 1. All right so, so these are the points and, you note that in general each of these points can be referred to as c A, eta A, and zeta A, right? Where A, once again, runs from 1 through 8, okay? But here it matters, that we get, that we establish a numbering, okay? So, let's what I'm going to do is just so that we can, distinguish things I'm going to use a different color here to, to, mark our, nodes, okay? So I'm going to write the local node number. There are only local node numbers, of course, in the parent domain, but we're going to call them 1, 2, 3, 4, 5, 6, 7, 8. Okay? Those are the values taken on by A in the parent domain, okay. Once we have this what we need to do next is go ahead and construct our basis functions, okay? The basis functions that we have are going to be written again in terms of c, eta, and zeta. And I'm just going to list them right now, okay. So, with things set up in this manne,r in general we have this formula N A as a function of c, eta, zeta, right, is equal to, sorry, I need more room here. Okay, N A is equal to. Something expressed as a function of c, eta, zeta. Okay? And here we're already taking this approach that, though we originally introduced our basis functions in A's as being paramaterized by, physical coordinates. We are thinking of them, as, as, as physical positions in turn being parameterized by positions of this parent domain, okay. So, here is the general formula for them. N A is 1 over 8, 'kay? 1 plus xi times xi A times 1 plus eta times eta A times 1 plus zeta times zeta A. Okay. If you look at this form, what you will notice is if you also recall the way we wrote out our Lagrange basis functions, Lagrange polynomial basis functions in 1d, right in 1d, we had only one of these contributions, 'kay? Here in 3D, we have three of them. They're just multiplied together. And this particular way of constructing our basis functions is called a tens of product approach. Okay, so the NAs are, so, the NA written as a function of xi, eta, and zeta, are what are called tensor product functions, right? The idea is that you take your basic form, established in one dimension and just multiply, extended by multiplication to the other two dimensions, all right? And the factor of one-eighth will, of course account for the fact that we have one-half to the power of 3, okay, for the three dimensions. Okay perhaps for clarity it's, it's actually useful to write out the basis functions explicitly, that's what I'll do now, 'kay? So here we go. N1 function of xi eta zeta is 1 over 8 times 1 minus xi times 1 plus eta times 1 plus zeta. N2 function of xi eta zeta is 1 over 8 times 1 plus xi times 1. Sorry. These are both minus 1 minus eta, 1 minus eta [INAUDIBLE]. Okay, so it's 1 plus xi, 1 minus eta, 1 minus zeta. N3 equals 1 over 8, 1 plus xi times 1 plus eta times 1 minus zeta. N4 equals one-eighth 1 minus xi, 1 plus eta, 1 minus zeta. N5, one-eighth, 1 minus xi, 1 minus eta, 1 plus zeta. N6 equals one-eighth, 1 plus xi, 1 minus eta, 1 plus zeta. N7 is one-eighth, 1 plus xi, 1, min, 1 plus eta, 1 plus zeta. And finally, N8 is 1 over 8, 1 minus xi, 1 plus eta, 1 plus zeta. Okay, that's the whole lot of them, all right, all written as functions of xi, eta, and zeta. Now if you look at these, you, you will immediately see two properties that follow, right? From these we directly get the chronicle delta property, okay? So NA evaluated at xi B eta B zeta B, okay, is equal to delta AB, the chronicle delta. All right, and we also have this other property which is that the sum over all shape functions, right, at a point. Right at any point, indeed, is 1, all right? Properties that were identical to what we saw in 1D, 'kay the second property allows us to represent constants. The first property is the one that gives us this interpolation property also of these Lagrange polynomial basis functions. 'Kay, so I should probably state here these are the Lagrange polynomial. Basis functions. In R3, in three dimensions, okay? All right and you, and also you, you know why they're trilinear, okay? Note the trilinearity. Right? It's each of these functions is linear in each quadrant, right? For the three coordinate directions here. Okay. So these are the functions that we use to construct our finite dimensional trial solution as well as finite dimensional waiting function. Okay? And furthermore looking ahead. We also are going to use the same basis functions to interpolate the geometry. Okay? So the actual map, okay? The map. From omega C to omega E, okay? Is obtained by really interpolating. X as the function of C and here, I will use the general notation C as a vector to represent the noted coordinates C zeta, zeta. Okay? So the map is obtained by interpreting C x is a function of C, right? Where I will see, see, I'll state it the very first time and we'll use it in the future. All right? We get this interpolation by simply observing that x be e. Parameterized by C is now simply, sorry. Xe paramaterized by C, which is any point in the physical element sub domain. Right? Is obtained as an interpolation, A equals 1 to number of nodes in the element. NA dependent, depending on the coordinates in the paring subdomain. Okay? Multiplied with the actual nodal coordinates, right? In the physical subdomain, but where I've used the local numbering of nodes. Okay? All right. So these are, we just remember that these are just nodal coordinates in physical domain. All right? And that means we're dropping off. For that element, omega e. We are talking of a typical nodal coordinate as being that one, right? We interpolate over all of those and we get any point inside here xe. Okay? That's how things work and, and, and that is constructed from a mapping, from our nice regular, at least this is intended to be regular, parent subdomain. Okay? So that is the mapping that we have underlying our geometry, as well. And you remember, the term for this kind of a formulation. This is, remember, right? It's called an isoparametric. Formulation. Okay? Isoparametric, referring to the fact that to interpolate our final dimensional waiting function as well as the geometry, we're using the same basis functions. Okay? So we have isoparametric formulation. With this, we know how to go ahead now, we will be able to compute our fields including the radians. Right? And we'll end this segment here. But when we come back, we will look at how to use the isoparametric formulation to go ahead and compute gradients and so on and also do the integration. The approach is going to exactly what we did exactly like the one we followed in one dimension, except that we, we basically need to generalize all of those ideas into three dimensions. All right. Let's stop here.