Great, welcome back. We continue with our development of the finite dimensional weak form. What we have got to doing at the end of the last segment was using the bar unit to mean to not only write out our basis functions. But also to set up the, the integrals that going to the weak form. Okay, so we've written out the integrals that go into the weak form into the finite dimensional weak form for the for the general element by using this this, this approach. And that's where we going to pick up today. All right. So, so the topic of the segment that we are going to work through today is the matrix vector weak form And in order to get to it I, I, essentially I'm going to continue where we were last time, 'kay. Recall that we've written out the form with the following integrals for the, for the case of the general element, 'kay. We've written out this integral. Wh comma x, sigma h Adx for the general element, okay. And it's in your notes, it's in slides corresponding to the last segment. What I'm going to start out with doing is consider this integral for the somewhat special case of element one fueled by the fact that we have a Dirichlet boundary condition on the first node of element one. Okay, so we have this and now let's write con, consider how it works out for e equals 1, okay. Now, for element e equals 1, this is the integral over omega e. W h comma x you recall for element one is written as simply, N 2 comma c, C comma x multiplying c 2 e, where e for this particular case is 1, okay? So this is our expression of W h comma x, all right. It multiplies EA, and then we have our contribution for the gradient of the trial solution. And that takes on the form, sum over B N B comma C, C comma x, d B e, and this is our representation of u h comma x. All right. Integrated over dx, all right. And now of course we take our approach of going to the by unit domain in order to write out this integral. From that we get integral over omega C the first term is, the first factor here is N 2 comma z, we have z comma x, okay. And we recall that, that z comma x is simply 2 over h e, okay. And here we have c 2 e, EA here, and for Comma X we have N B comma C, again, we get C comma X, we use 2 over h e for that as well, we have here d B e. And now, like we did in the previous segment, we're going to rewrite dx as dx we're going to write as d C dx I'm sorry, it is dx, dc, dc. All right now, dx dc we know is he over 2 dC. All right, and as we did in the segment last time, those cancel out. All right. Thus da, a similar sort of exercise for the next integral. Integral over omega e wh f A dx, leads to after going through these sorts of steps we get integral over omega C. Our representation of Wh here is simply N2 times c2e. And remember this is all for e equals 1. Which is why we don't have the contributions from the first basis function, right? We talked about that in the last segment. This multiplying f A, and now in place of dx once again we have dx, dc, right? And we write that as dx dc, dc. And we recall that this is h e divided by 2, okay? So, so we have these two integrals also written out for the special case of the very first element. And you'll recall that what this does is to leave out the use of the first basis function. All right, these integrals here involve only N 2, okay. All right, so we have this in place and what we are going to do now is essentially look at how all these terms contribute, how, how all these simplifications of these, these expansions contribute to setting, setting together the the weak form, okay? So now in order to do that let's consider the general case. Consider for a general element omega e, okay. And let's consider this, the first of these integrals, the integral that comes from the left-hand side, the weak form. That integral is integral na, we now have it written as integral over omega C sum over A, N A comma C, and we write this as C A e, E A sum over B N B comma C D B e. And the term that survives here after cancellations of d x dc, and dc dx type of factors is this one, okay. And we have here dc, all right. Now this needs to be simplified and the way we can carry out the simplification is through several steps. The first of these steps involves us recognizing that the Cs and the ds that are used here are degrees of freedom, right? The Cs is the degree of freedom that is used to interpret the weighting function. The d B sub e is the degree of freedom that's used to interpolate the trial solution, right? So what we do is to recognize that these degrees of freedom are independent. Right? They're independent of the, of the variable of integration, right. They're independent of C, right. Because these are the actual degrees of freedom that are being interpolated by basis functions that are being expressed in terms of c. As a result, we can pull them out of the integral okay, so that allows us to write this out as follows. We can now pull the the summation symbol out. And for the sake of brevity, I'm going to write this is the sum over A and B. Okay? It we get c Ae outside the integral, and within the integral we have a the integrals over omega c. 'Kay, we have here N A comma C EA N B comma C. I'm going to pull the factor of two, which comes here. For the sake of of brevity again, I'm going to put it in here and divide by hE. Okay. So have that factor 2EA over hE sitting inside there. All of this integrated over dc, and that is our integral. Okay. And within those parenthesis, and it is being multiplied by d B e, okay. Recognize any nice, and short succinct form. All right. We have this for this for this integral and let's do the same thing for [COUGH] the, the integral that involves the forcing function. Okay. That integral is integral over omega C sum over A, N A, C A e, f A, and we get here, H e over 2 dC, all right? Again observing that this is independent of C. Therefore it can be pulled out of the equation. We get sum over A, C A e, integral over omega C, N A f A, and once again I'm going to put the H e here, in this case divided by 2 dc, all right. This is what we have. And now observe that in, in the case of the first element, all right it's going to look very much like this except that the sum over A will not be required, all right. Because A will take on only the value 2. Okay? So, let me just recall this, for e equals 1 there is no sum over A, instead, we have use we, we require to use just the index A equals 2, okay. And why is it that we need to use the index only A equals 2? It's because the first basis function is not used in element one. Recognizing the fact that we have a Dirichlet boundary condition on element one, and therefore, the the weighting contribution from the degree of freedom one is not used, okay? So instead we use A equals 2, because we know that W h at 0 has to be equal to 0, okay. Implying that W h in that element, right? For e equals 1, W h is only written as N 2 c 2 e, okay? All right. So we have this and we're now going to go ahead and construct from this a form of the weak form that we call the matrix vector form. And in order to do that, we need to look at that expression, and that one. The first step to getting to a matrix vector form, of the weak form, is to, is to attempt or, or is to aim to eliminate these sums. Okay? And the way we will eliminate those sums is by writing them out using matrix vector products.