So, essentially, what I'm going to do in this process is the following, integral over omega wi comma j sigma ij dv equals integral over omega wi fi dV plus integral over the, the entire volume, right. Wi sigma ij nj dS, all right. Now, here comes a step that is a little different from what we've done in our previous application to a three dimensional problem, okay. What we are going to use is this special decomposition of omega into. Into subset spatial omega ui, which is tertiary boundary and, Neumann boundary, okay. This holds for i equals 1, 2, 3. Okay. So what this says is that the way we're going to apply this is by regarding this product as a vector with index i. And then observing that we essentially have a, in that integrant, a dot product of w with that vector sigma n, okay? So, the way I am going to write that is the following. And we right that as integral over omega, okay, so let me write everything here. Integral over omega on the left hand side, wi comma j sigma ij db equals integral over omega wi fi dv plus. Now, because that integrand when viewed as a final, dot product of w with the vector sigma n. Right? Because that integrand has three terms in it, right? For i equal 1, 2, 3. We're going to write each one of those contributions of the dot product, right, separately, all right? So I'm going to write that as each, so each, of these decompositions of, of the boundary, all right, are going to be used, to give us the domain of integration for each integrand. Sorry, for each contribution to this integrand, right. Okay. All right, so, so the idea is you fix i, maybe i equals one, right. So you do the integral by viewing our boundary as a union of these two subsets, boundary subsets, right, but in computing that integrand you compute only for a fixed i, okay. So you fix these guys, right, these are fixed. Okay, fix everything, right. So fix all those i's. Now that gives you, the integral with just one of those components. You need to have all the components, and you do the sum. Right? So far, we've just written that integrand differently, okay? That's all we've done. Sorry, all we've done is write that integral, differently. Okay, but now we can do something nice, right? And what is that? We invoke our boundary conditions, right? Now we invoke. Boundary conditions on wi and sigma ij nj, okay. All right. And remember the way we wrote those boundary conditions, okay? Essentially, the way we wrote those boundary conditions, say that, now that we're integrating over each, special decomposition of the boundary, what we can do is this, right. So let me actually go to the next slide. So, what we can say here is the following, right. So we have integral over Omega, wi comma j Sigma ij dv equals integral over omega wi fi dv. That term has got to, got to be the most tame term we've ever encountered, right? The, the forcing function pretty much does nothing. Okay, plus. Sum i equals 1 to number of spatial dimensions integral. Now, I'm going to break up that integral, right? Because we've written it on the previous slide as an integral over the union of tertiary and Neumann boundary subsets, right? So I'm going to write it as integral over partial of omega ui of wi sigma ij nj ds plus integral over partial of omega t bar i, wi sigma ij nj ds. And I close my parentheses here, 'kay? Actually, looking at, what I've saved here from my previous slide. I'm going to go back for just a second because I realized that I did not put my elemental area, that integral, okay, but you may have caught it. All right. Here we are. All right, now is when we invoke the boundary conditions on, on wi. And what is that boundary condition on wi over the corresponding tertiary boundary? It is that it, vanishes. Right. And what is our boundary condition on sigma ij nj over that part of the Neumann boundary? It is that sigma ij nj equals at t bar i component, okay? And so we have it. We have integral over omega, wi comma j sigma ij dv equals Integral over omega w i f i d V, plus sum i going from 1, to a number of spacial dimensions. Integral over the corresponding Neumann boundary only, wi t bar i dS, okay? This is our weak form, all right? Which we've derived from the strong form. Like I said, we can go back the other way. But, it's not something we get into here. I will make one remark though. Which is that, just as we did in our little treatment of variational methods in the context of the linear, linearized elasticity problem in one leap. All right? We demonstrated there that the weak form could be obtained through these variational arguments of extremization of a free energy functional in one day. Exactly the same things holds in three dimensions as well, okay? So, the weak form can be obtained as this so-called Euler-Lagrange conditions The Euler-Lagrange conditions of a variational principle Right? And what is the variational principle? On well it's not minimization necessarily but it's really, it's extremization, it's on extremization. Of a free energy functional in three dimensions. The problem we treated a few segments back, quite a few segments back, quite a few units back, was in 1D, but essentially all those arguments hold in three dimensions. And actually, this is a very powerful approach. While in the case of the simplif, the simplest stuff, formulation of linearized elasticity, it doesn't give us anything that we can't get through other methods. Something you may be aware of is that there are so called constraint problems, right, and a common constraint on linearized elasticity is incompressibility. So if you're dealing with a linear, with a linear elastic material that's also incompressible. It emerges that the methods we work with, the standard finite element methods we work with, and that we will be developing now, do not work well. Okay? And there are fundamental pathologies which arise, and these are very well understood. And methods to circumvent that difficulty, right, are based upon defining more generalized versions of these free-energy functionals, introducing more fields and working with them. It's the domain of what are called mixed-methods, okay, and it's a very powerful approach and which, which arises from this, from the variational basis of the method. Okay, but we won't be getting into that in this entry level series of lectures. Okay, the very last thing I want to do is just go from there to putting down the finite dimensional weak form. Okay, so what we have here is will be just derived is the infinite dimensional the weak form. [NOISE] Okay. The finite dimensional weak form. Follows. All right, and now of course you're quite expert at doing this, and you know exactly what that's, what going to happen, is going to look no different except for this pacification spaces, right? But let's just put it down so that we will be able to move on with the formulation. Okay, the finite-dimensional weak form. As before, given Ugi, t bar i, fi, the constitutive relation sigma ij equals Cij KL epsilon KL. Right? That's the form of the constitutive relation. Let me state it That's a constitutive relation and the kinematic relation epsilon KL equals the neutral stuff. Right? Given all these, find, now, now here's where things become different, right? So actually let me put it on the next slide, so that we can, give it its due attention. Find sub i, right? And as, as always that superscript h reminds us that we're talking of approximations and h is going to be related to the element size. Okay,that's all it denotes. Okay, to find of i, belonging to sh, which is a subset of s. Okay. And what do we mean by sh here? sh consists of all functions. Now, we further require as we've done before that we are interested in functions that are H1 over the domain. Okay? And, of course, they must satisfy in this class of finite element methods that we're working with throughout the series of lectures. We do require that they satisfy the Dirichlet boundary condition on the relevant subset of the boundary. Okay? So, find your weight of this type such that for all wh sub i belonging to Vh subset of V, okay? And what is Vh? Vh consists of all waiting functions, also H1. Such that, Wh of i equals zero on. The corresponding Dirichlet boundary. Okay, so find Belonging to Sh such that for Wh belonging to Vh. What all is well, the weak form that we wrote before, except its called the relevant fills, replaced with their finite dimensional versions. Whi,J, sigma hij, dV. Right? And we recall here that sigma h i j is simply computed by a C i j k l epsilon h k l. Okay? And what is epsilon h k l? Well, it just obtained from this relation. With u h. Okay? All right. This is equal to integral over omega. Sorry this is over omega too. w h i f i Dv, okay? Remember that, as before, we do not consider finite dimension inversions of, f, right? The, the f is part of the data, and we take it as given. Plus sum over the spatial dimensions. Of the integral over the Neumann Boundary, of Whi, t bar i, ds. Again, because d bar is part of the data we don't try finite dimensional representations of it, okay? This is it. Just for completeness, let me squeeze in here that epsilon hkl equals that. This parentheses closes that one. And this closes that entire node. Okay. All right. We are done with this segment. When we come back we will just work with this to develop our finite element formulation
