Okay? So let me, let me just tape that. The, the weak form is the basis of the finite element method. And because this is sort of the first time we're writing the term, I will capitalize it. Okay? It's the basis of the Finite Element Method and of other variationally based numerical methods. The term variational may be new to you. You will, before too long know what that means. Okay? Just take it from now for now. Okay? So this is doing the basis of the finite element method. Okay? The very first thing I'm going to do is demonstrate to you, the equivalence between the strong form and the weak form. Okay? So let me state that here. The claim is that the strong form, the strong form and weak form are equivalent. Okay? What that means is that one implies the other. So the strong form implies the weak form, and importantly the weak form implies the strong form. Okay? So let me, I'll write that out, strong form implies the weak form and importantly, the weak form also implies the strong form. Okay? We're going to set out to show this and it turns out that the first step is actually quite easy to show. The first step being the strong form implying the weak form. Okay? So we'll take that approach. Okay? So, consider the strong form. All right? We know the strong forms, I'm not going to write it out in great detail. I'm only going to star write out the essential ingredients of it, right? And the most essential ingredient is the PDE itself, right? So that is d dx of sigma plus f equals 0 in 0,L with the boundary conditions, B.C. for short. Now, the boundary conditions we have are the following, U at 0 equals U not, right? Now because we've, we've written out the weak form for a single Dirichlet boundary condition, only at x equals zero, that's what we have to assume for the strong form as well, right? We need to consider that strong from, we can't have different boundary conditions in the strong form and expect it to lead to the weak form that we showed. Okay? So the boundary condition at x equals L is sigma at L equals t. Right? And we need to remind ourselves that we also have the constitutive relation, right? Sigma equals EU, x. Okay? So, this is where we start out. What we are going to demonstrate now is that this leads to the weak form. And to do that, we will proceed as follows. What we are going to do, is introduce W, just like we did back there, right? W belonging to V, right? The space V, which consists of all functions W, such that W at 0 equals 0. All right? As we discussed previously, W satisfies a homogeneous Dirichlet boundary condition, right? Any place we have a Dirichlet boundary condition, W must satisfy a homogeneous Dirichlet boundary condition. Now, what we will do is the following, we're going to treat W as a waiting function, right? In fact W does get called In the context that we are developing here, W is indeed called the waiting function. Okay? And I'm going to use it indeed like a weighting function. I'm going to multiply W in to our strong form. Okay? So what I'm doing here is multiply strong form by W, and now integrate. Right? And integrate over the interval 0 to L. Okay? So integrate 0 to L, 0 to L. The right hand side, of course, stays equal to 0. Okay? Also, because, we have this we introduce the area, right? So I'm going to sneak the area also into this, multiplication here. Okay? And then I integrated it over the domain of interest, okay? Because the right hand side of zero, of course, it stays at 0. Okay? So, we have this form and now this would work. Okay? But really we haven't done anything special. What we are going to do next is use a very common technique in calculus, in integral and differential calculus, which is called integration by parts. Okay? So I'm going to say that here, integrate, by, parts. Now, in integrate by parts, what we're going to do is act upon that derivative. All right? When we do this, we get the following form. Right? When we apply integration by parts we get minus integral 0 to L, W, x sigma Adx plus w sigma A evaluated at 0 and L plus integral 0 to L, WAfdx equals 0. Okay? And integrating by parts, if it's a step that's probably familiar to all of you, we have done that. Okay? That is the common integration by parts step. Okay? And if this is something that doesn't seem terribly familiar to you, recall that integration by parts involves two steps, right? It involves, the product rule of differentiation. Right? And what is referred to in higher dimensions as the Gals divergent theorem, but in lower dimensions, it is simply the fundamental theorem of calculus. Right? All right, it's not a difficult exercise to do, it's actually quite trivial. Okay, so, we have this form now. Let's focus attention upon that term. Okay? All right? Let me re-write this, this follows let me write it now as let me do two things. Let me move this stone to the other side of the equation. Okay, right? But then let me write the other side of the equation first. So effectively what I'm getting here is the following. I get integral 0 to L W, x sigma A dx equals integral 0 to L, WfAdx plus W sigma A at 0 and L. Okay? All right? Okay, let me go to the next slide and simplify it. For doing that, let me just go back here for a second. If you look at the very last term of the last equation on this slide, we observe that it involves that term being evaluated at two limits, 0 and l. Okay? And so that gives us integral 0 to l, W, x sigma Adx equals integral 0 to L WfAdx plus W at L sigma at L A minus W at 0 sigma at 0 A. I simply expanded out that last term. Okay? Now, if we stare at this very carefully, one should be able to say something special about this very last term here. What can we say? Think about it for a few seconds. In particular what do we know about W at 0? Remember we have a homogeneous Dirichlet boundary condition on the waiting function. Right? So that factor is equal to 0. Okay? Which gives us then our final weak form, which when we also observe that in addition, we know something about sigma at L. Okay? We choose that sigma at L is equal to t. Okay? The fact that w and zero is equal to zero comes from the homogeneous Dirichlet boundary recondition on W. Okay? D-I-R is short Dirichlet. The fact that equals t comes from the Neumann boundary condition. Right? On the strong form. Okay? So, when we do all of this we get our final weak form. Integral 0 to L, W, x sigma Adx equals integral 0 to L WfAdx plus w at L t times a. Okay? This is our weak form of the linear partial differential equation of elliptic form in one dimension. Okay? And you observed that we started with the strong form and essentially obtained this weak form. Okay? It is indeed the weak form we wrote out because W, with this waiting function that we'd introduced in the weak form, does indeed satisfy all the conditions we'd assumed of it when we specified the weak form. In particular we have this homogeneous boundary condition of W. Right? And the fact that we started from the strong form, allowed us to bring in the Neumann boundary condition on sigma. Right? Okay. We'll stop the segment here.