Okay. Welcome back. We'll continue with now assembling our element level matrix vector representations into our global matrix vector weak form. All right at the the end of the last segment we had obtained a representation 'kay, so for the following integral, right? You probably have this on you notes, and just taking it down. I'm just copying it down from my saved slides here. Right. We said that. We, we considered a, a specific example where we said that that local degrees of freedom one, four, five, and eight were the ones that lay on the face that coincided, the face of the element that coincided with the Neumann boundary, okay. So those, so we wrote that out in the following form. And for the resulting vector that we have here, the, the column vector. The first element was non zero, the next two would be zero, the fourth would be non zero, the fifth would also be non zero. Right, the sixth and seventh would be zero and the eighth would be non zero. Now I am going to do is so, so this is, this is an example, right? This is an example where local degrees of freedom 1, 4, 5, and 8 are the ones that belong to that set, okay. What I'm going to do is just to have common notation. Observe that this can, this then can of course always be written as minus c e transpose, right? The vec, the, the role vector of all those degrees of freedom F j e. Okay, all we need to observe is that in this column vector F G, sub e, some of the contributions are 0, right? These are the ones corresponding to the degrees of freedom that do not lie on that particular element face coinciding with the Neumann boundary. The rest of them are in general non zero, okay. So the reason I'm doing this is, because this is what we want to use, when we pull together all the other integrals that we've developed over previous segments to represent sorry all the different representations we've de, developed for the, for the other integrals in the weak form. Okay, so what you want to do is use in the following, right. We now have a way to write the left-hand side, okay, as a sum over e of terms of the form c e transpose K e d e. Okay? We derived this a few segments ago, equal to sum over e, c e transpose F internal e. And now, this new term that we've developed here, right. This is again an, a, a sum over elements. But, I'm not going to write, I'm, I'm not going to I'm not going to say that this sum for the traction term extends over all the elements, right. Instead adopting the notation that we had on the, in the previous segment to designate the degrees of freedom lying on the Neumann boundary, I'm going to say that in here belongs to a set E sub n, which I think we actually already have defined, right. The elements corresponding to elements which have some part of their surface, of their faces, lying in the, lying on the Neumann boundary. Okay, so we have e belongs to this script E n c e transpose F j e, okay. The other sums run over all the elements. E going from one to nel, e going from one to nel. All right. Okay at this point I want to point I, I want to bring to your attention the fact that we can just multiply this through by minus, right, by minus 1. Changing all the signs so that we get something that we may feel a little more comfortable about. Okay. So we basically transfer the minus sign onto the forcing function. And this brings us to a, another little detail, which it is useful to bring up at this point before we go ahead to assembly, okay? And I'm referring to the fact that the way we've set things up here that term has a negative sign, okay. So I'm going to refer to that and, and I'm going to do that, in an aside. I need to make a, an, an, I need to make an Aside here. Okay. Essentially the form of the equations that we have, okay, the form of the equations that we have. Is obtained. From the strong form, right, going all the way back to the strong form. Where the strong form had the following appearance. Right. Minus divergence to flux equals, a force in function. Okay, now this particular form had been adopted by me, keep in order to be consistent. Right, to be consistent, with the way heat conduction or diffusion equations are typically written in the field of numerical methods that treats these types of problems, okay. However, it's probably useful to also recognize that this is a steady state equation. Okay arrived at from an equation of another type, right? Now, let us suppose that whatever quantity that is undergoing transport like for instance, the temperature, right? If, if one were to include the rate of the temperature, right. Let us go away from steady state but really look at how the steady state problem arises, one would get an equation of the following form. One would have c derivative time derivative of u equals minus divergence of j minus f, okay? Where this would be if you were thinking of the temperature problem for instance, right? The heat conduction problem. This left hand side would be the rate of change of temperature. All right, and the statement here would be that rate of change of temperature was driven by this term which we all ready have previously observed to be the total influx of heat into a little elemental volume. Right. Instead of total let me call it net heat influx. Okay. Right, so that is one way in which temperature rises, simply by heat being transported into a little volume. And then this term would correspond to the local heating. Local, maybe distributed Okay? What we did was, start out with this you know, what, what we did could have been arrived at and probably is arrived at, by starting out with this time dependent form, right. So this is, time dependent form, right. And seeing that well, we have steady state, and therefore we set the time rate of temperature equal to zero. Okay? And then of course that we would have been left with the right hand side equals zero, which we would have rearranged to form presented here. Okay? All right. And, and this is the form that we worked with, okay. But you observe that when viewed in the larger setting of a time dependent or transient problem. The local distributed heating or alternately, if you're looking at a diffusion problem, the local supply of mass, okay. For the way we're writing things out is properly minus of f. Okay. Right its just a matter of the way we we took the signs when we started out with the PDE in this form, all right. So what this suggests, is that if we look back to the equation that ended the previous slide, 'kay. It makes sense for us to simply flip the sign on the forcing tone, okay. And you know, we give it a different symbol, right. And we'll, that will be properly the forcing that is obtained from considerations of how the Steady State PDE can be arrived at from the transient problem? Okay, so so what we are going to do is redefine f bar equals minus f, right, and if you go back, and redo all our derivation with just this little change, what we will see then is sum over e integral w h f d v integral over omega e of this. Right? With the minus sign, 'kay. Is of course integral sum over e, integral over omega e, w h f bar, d v. Okay, and accordingly the minus sign that showed up when I wrote out the the weak form as a sum over elements, right, at the minus sign of forcing term would be absorbed, okay. So, what I'm proposing is also to write now, sum over e, C e transpose, K e, d e equals, wha, what we have so far is this sorry, I noticed I flipped the positions of this transposing element here, okay. c e transpose, K e sorry, c e transpose. F internal e plus sum over e belonging to e n, c e transpose F j e, right. Right, this is what we had at the end of two slides ago. What we're, what I'm proposing now, is to replace this with sum over e, c e transpose, F bar internal e. Okay. Right. Where the association that we're making here is that now. That that this is essentially equal to integral over omega e w h, f bar d v, okay. All right. When we do things this way, f bar turns out we be properly the forcing function that would make sense even in the case of a time dependent problem which is something that we will go to. Okay. So for our purposes there are no big changes. It's just a matter of flipping a sign on that term. Okay, and it actually makes everything consistent now, with the time dependent case. Okay. So, if we agree with if we agree about this little change we can now go on, and essentially we will now go to our step of finite element assembly. Okay. So, the assembly of global finite element equations right, in matrix vector Okay. All right and, and, you know, very well how this precedes essentially it's a matter of looking at each of our element level contributions such as this one. Right or, or, or this one and this one. And understanding how they map into the global vectors? Okay. This is relatively straightforward to do in the case of the 1D problem. It turns out to be a little more complicated in the case of the general three dimensional problem. And what we have to grapple with here is the idea of mesh connectivity? Okay so that is, what we are going to look at now? Okay.
