Okay, so we have everything that we now need to write our integrals in terms of coordinates in the parent subdomain. So let's just go ahead and write that. And in particular the integral that we are working with is the following, right? So, so actually let me title this. Integrals in the finite dimensional Weak form. And in particular we're working with this integral All right. And where we are is a realization that this integral can now be written as minus. And that minus comes from the constitutive relation. We've agreed to pull the sum out, and write it as a sum over A and B, and for brevity we are not writing the limits that A and B run over. Right? In the general case, they run over one to number of nodes in the element, each of them. All right. We have an integral now. Okay? And inside this integral we have terms of the form, out here we have C A B, right? We don't want to forget that. Okay. The integrand is made up of terms of the following form. N A comma i C capital I comma little i kappa i, j, N B comma capital J C Capital j comma little j. And now, previously we had d, v, being the, the elemental volume in the, physical subdomain. And we just agreed at the end of the last segment, to write it as in terms of an elemental volume in the parent sub domain. But that does involve our including the determinant of j here. And let's recall that it depends upon c. Okay? D V, c. The integral here now can properly be written as an integral over omicron c. We recall that the, this is the integral which we enclose in parenthesis. And we have here, d, d, d. Okay? Alright now, if we stare at this integrand, it looks a little, a little daunting, doubtless, but we observe that there is a The Einstein summation convention is at work. And so, really what we have left is going to be an expression which takes on values depending upon the specific element degrees of freedom, capital E and capital B. Okay, the, the capital I, capital G, little I and little G are going to be so to speak contracted out, to use the terminology of densers and vectors. Okay right. So, so when we carry out this, this integral and let's assume that we do know how to carry it out. First of all observe that this integral can, can be is, is, is easy to carry, is relatively easy to carry on because our domain omicron c is a nice regular sub domain. It's a nice regular domain, it's a bi unit domain. So let me just write that little bit here. So the integral of omega C is the one thing that I want to make a difference with the way I wrote, so, so this is now a triple integral, okay? Where, C, 1 equals minus 1 to 1, C2 equals minus 1 to 1. And C3 also equals minus 1 to 1. Okay. Everything that we have in here, I'm not going to explicitly write them as depending upon c, but we do recognize that indeed, n a comma i will depend on c, because n a is trilinear. All right, and the xis. Okay, so derivative with respect to any one of those coordinates is going to leave us with something that's bilinear in general, it'll depend on two of the xis. All right? If you look at xi i comma i, and you recall how we are going to compute it It too will depend upon C. And why's that? Right. That's because we are going to compute it or we are going to identify it as being the components of the inverse of the tangent variable. Right? And the tangent map, consists of first derivatives, right? So, this thing, this, too, will be a function of c, right? Kappa ij. Well, it depends. If we have something with, uniform conductivity or uniform diffusivity, it would be a constant, right? But that does not res-, we are not restricted to that. We could have something that's a function of, this thing could. Potentially be a function of, position. Right? Could be a function of position and because we know how to write the physical position as a function of z that too would be a function of z. All right, the kappa Okay. And then we have NB, which of course, we've all ready discussed, right? We know how to write its derivatives. It's going to depend upon c as well the derivative of c itself with respect to little x little j, right? And then again, we have the determinate here. Okay? And this again is an integral over c 1, c 2, c 3. We close our parenthesis. Right, in the places multiplying d B element e. Okay? So everything in that integrand is properly a function of the xi 1, xi 2, and xi 3. That integral can be carried out. Alright, for simple cases this integral is not that difficult to carry out, analytically, and by simple cases I mean things where situations where Kappa is indeed independent of position. Okay, however we're not going to get involved right now with evaluating it analytically, because we already have a way to evaluate this, right? Do you remember how we're going to evaluate in, in the general case? Right. Using numerical integration. Okay. So we'll come back to that later We'll come back es essentially to how to do that in three dimensions, okay. But the point I want to make now is that if one looks at the integral here, okay? That can be evaluated. It's going to give us a scalar value, right? Because all the little i's, capital Is, the little js and capital Js are going to be contracted away. It's going to give us a scalar value, however. That is going to be indexed by the, by the capital A and capital B corresponding to the local degrees of freedom, okay? So I am going to denote this K, AB, sub e, okay? The AB is obvious, why it needs to be indexed by AB. It's also obvious why it needs to be indexed by little e. Okay, because this is for a specific element, okay? And so we have it. Our integral on the left is now just minus sum over A comma B, c Ae KAB, for element e, dBe, all right? And now, we take a step that we took in the one d case as well. We are going to well actually, what are we going to do? Do you recall what we do next in, in setting up our finite element equations? We get rid of our some, our explicit sum over a and d by going to matrix vector notation. Okay, let's do that on the next slide. So what we're doing is using matrix vector notation. We're using matrix vector notation for local degrees of freedom. In order to keep track of local degrees of freedom, we're going to use matrix vector notation, right, and really for their numbering. Okay? And what this means is that for element e, if we have c 1 e, c 2 e, up to c number of nodes in the element e, all right? These are the degrees of freedom. As you recall that are used to interpolate the weighting function using our basis functions, all right? Okay, so using this, right, and of course the same thing for d's. D 1 e, d 2 e, all the way up to d number of nodes in the element e, all right? Using this, what we observe is that we essentially are able to write our our integral, right? This implies that we can write integral over omega e. As now, minus we're going to write the c vector in a, as a row vector, right? And you recall how we did this in the one d case as well. Okay? We get a big matrix here. Which consists of entries K 1 1 e, all the way up to K 1 number of nodes in the element, e, okay? And, the last entry here is k number of nodes in the element time number of nodes in the element, e, okay? That is the matrix that we get. And this is multiplied by a column vector using the notation that is sometimes employed in linear algebra. Okay. All right. And, for, even for the gravity we could write this as minus c e, right? Where ce, or ce transpose is that row vector. All right, this, Ke, right, Ke is the matrix, de, okay? And by, by reference to what was done in the one d case, we would call this the element conductivity. Or diffusivity matrix, right, depending upon what problem we were solving. Depending upon what physical problem we were solving, okay? All right, clear enough hopefully. We can now follow exactly that same approach to look at the first integral on the right-hand side of our finite-dimensional weak form. Okay, so now consider integral over omega e w h f d v. With everything we've gathered together this should be relatively quick, all right? We're going to write this as integral over omega e. W h, you recall, we will right as a sum over A N A. There are no derivatives here, right? So, we have N A, C A e. I put parentheses on this to remind us that this represents w h, multiplying f, d v, okay? We can take several steps all at once now. Let's do that. Let's, pull the summation out, c, A e. We have an integral over omega e, but right, but let's plan to write that with a change of variable as an integral over omega c. All right. Inside here we continue to have N a because it does depend on position. Multiplying f and let's recall that f could be a function of position but through our mapping we have that. Okay, and any, of course, is the function of position. Okay. We're writing everything as a function of coordinates in the parent sub domain. That leaves us with just d v, right, but we know how to handle that. We know that it's a determinant of the jovian of our mapping. D, V, C. Okay. Sorry that's a little to squeezed in there. So let me write it out here more clearly. Okay. That's what we have. Right. And, of course, this is now, again, we can take several steps at once. So we can write this now as we can abandon our explicit Writing of the, of the sum over A. Right. And we know how to do that. That base, that can be done by going to this rho vector notation. With a c degrees of freedom. The integral now explicitly is an, a triple integral. C 1 equals minus 1 to 1. C 2 equals minus 1 to 1. And z 3 equals minus 1 to 1. Right? Our integrand here consists of a column vector of the shape fun, oh sorry, the basis functions. It's all, they're often called shape functions, but a term that I've been avoiding. Okay, we have the basis functions. All right, we have this. We have f, right, which could be a function, which could depend upon c through x. We know how that dependents comes about. There's an abuse of notation in, in using the same symbol f, but I will, just dock that abuse of notation. All right. And then, we have the determinant of j, function of back position. Okay. We have this and and yes, and we observe that this integrant is essentially, integral, sorry, it's integral is over D C 1. D C 2. D C 3. All right. Okay, we can compute this integral in general. Again, we will look at how we handle these types of integrals using numerical quadrature, numerical integration. And we observe that we essentially get what I will write out as being c 1 e, c 2 e, c number of nodes in the element e. Times, the column vector which I will denote as F internal. Okay? Degree of freedom one in element e. F internal, degree of freedom 2, element e. Coming down as far as f, internal Degree of freedom, in an E, element E. Okay. And finally, we could write that as C, E, transpose F Internal for element e. Okay? All very nice and concise. All right, and the, the advantages that having done it with, with the, in excruciating detail for the, 1D case we know you don't immediately have the sense to work on it. All right, we'll stop this segment here.
