All right. We'll continue with our establishment of the matrix vector weak form. Before we, sort of plunge onward, I should just clarify some notation. So I know I used the following symbol. I've been using this symbol I probably said what it means but let me make completely clear, this symbol is since or because. All right, I realize it may be slightly archaic notation that has fallen by the wayside, but I tend to use it still. Okay, so that's where we are. Let's continue then with the matrix vector form. What we did in the previous segment was [COUGH] write out the important contributions, so, that arise from the weak form for, for it generally, right? And we consider the two main integrals that arise in the weak form from the, the integral that remains on the left hand side. As well as the one [COUGH] that the one on the left hand side which comes from the stress. And the integral that, that's on the right hand side, that arises from the force and function. We wrote these out for general elements. What I'd like to do now is, take the step of just, sort of particularizing then for the first element, right, because of the fact that it has that tertiary boundary condition. 'Kay, so note that for e equals one. Okay? We have integral over omega e w h comma x sigma h A d x. This is simply c2 e, multiplying EA over he, right? Times let me see, how do we, how does it work out now? Oh right, correct. This times minus 1 1 d1e d2e. Okay, right then and, and you can actually see this in a very simple manner by just taking the general expression for a general element e and recognizing that c1e is not present in the expansion for element one. All right? Likewise, we get integral over omega e Wh f A dx equals, in this case it's just a scalar, right. In fact, well, actually all the integrals do turn out to be scalars when we com, when we complete the matrix vector products, okay. But in this case, we don't even need to use matrices and vectors. So the forcing term is just c2e. We have f A he divided by 2 times 1, okay? So those are the contributions from element e equals 1. All right. What we're aiming to do now is pull everything together. We are going to use the same approach as we used in going from summations over the degrees of freedom for each element integral, except that we're going to now apply that idea to the sum over elements. Okay and in doing so we are talking of the following of the following complete weak form, okay? So yet again recall the finite dimension of the weak form. It's this. Sum over e integral over omega e wh comma x sigma h A dx equals sum over e integral over omega e whfAdx plus, let's not forget this term that we have not had to worry about for quite a while. Right, the term coming from the Neumann boundary condition. So, what we spent our time in doing over the last segment was expressing these integrals, right, in a more compact form using matrices and vectors. So, what we now have is the following. We have the term above implies a sum over e, sum over all the elements. Now, for the general case, we have a form which is, which involves a matrix vector product. So actually let me do one thing. Let me first write out the contribution for element 1. Okay, the special contribution for element 1. For element 1, we have c2e, but e is 1, for element 1. Okay? E A he, but that's h1, okay? Times minus 1, 1, multiplying the vector d1 1, d2 1. Okay. And recall that we write d11 and d21. The super script there refers to the degree of freedom for that element and the subscript refers to the element number. Right? So this is the contribution from the first element to this integral. Right to this integral. Okay, for the the other elements now we have the sum e equals 2 to nel, 'Kay, of c1e c2e EA over he. Times this, very simple matrix. 1 minus 1, minus 1 1. Multiplying d1e, d2e. Okay? So those two contributions, the special contribution from element one and the contributions from that sum e going from 2 to nel, give us that integral that shows up on the left hand side, involving distress. Okay the stress and the radiant of the radiant function. Now, this is equal to all the right hand side terms. And the right hand side terms, also I can write as as follows. Right? Again, as we did for the, for the integral on the left hand side, I will first write the contribution from element one. That contribution is c2 for element 1, times fAh1 over 2, where the h1 reminds us that, that is element one. Plus, a summation E equals 2, 2 number of elements, 'kay. And that summation then is okay that summation gives us c1e c2e fAhe over 2. Okay? All right? We get this contribution. And let me see. Do we oh, sorry. We have fAhe over 2, and we have fAhe over 2 again. All right? Okay. There we go. That's complete. Now the terms we've written on the righthand side are the ones that arise from that integral. Okay, and the term that we could safely not have to worry about when we were writing these matrix vector contributions, is that one. Because that appears only at a single point. It just appears at the point on which we have a Neumann boundary condition, okay? Now, we recognize that that term alone, wh at l, has a very specific, very simple representation. Can you think of what it is? It is c2 for the very last element. E equals nel. Okay? So, the term that we add on here is c2 for the very last element times the data t, which is the rhyme and condition on the traction, times the area, okay? This is it really. This is a nice compact representation of our finite dimensional weak form. This is everything. One thing that may be useful for me to point out here is related to why we get the simple representation. Okay? It's a fairly straightforward thing. And it's something that you may all already have noticed but it's useful to point it out. Okay? So, Remark. And this remark goes all the way back to our representation of in an element as being the sum of NA, let me write NA as a function, expressed as a function parametrized by c, times dAe. And also to the corresponding representation for the waiting function. Sum over b nb c c b b. All right of course it's implied here that x is I've written this on the left hand side. X it's really x parametrized by c, alright and the same thing here. The same thing here as well. Okay. So now, observe that so, so this is of course for a general element, which is written as which can be depicted as this we would have I'll make an e. Here we would have x e you would have xe plus 1. Those would be the global node numbers, right? And we recall that this comes from a bi unit domain. Which is minus 1 to 1, and that is zero. This is omega z. Right? In the one dimensional space parameterized by z. Okay. Now. Okay. So we have this picture. And what I'm going to use it to do is to ask ourselves what ha, what form these, expansions take for and wh. If we choose to evaluate these, quantities, these fields at the nodes themselves.