All right, welcome back. We'll continue and we will get down to business now and we'll do so by writing out the weak form. The weak form for linearized elasticity. All right. So I'm going to take the approach we adopted when we worked with the linear elliptic PDEs with scalar variables in 3D, okay, which is that I'm going to first put down the weak form and then I will demonstrate how we obtain it from the strong form. I won't go back the other way to just avoid, just because those arguments are identical to what we did in 1B, but they are, you know, they're more subtle and so on. Okay, but however, we know how they operate and they hold. Okay, weak form of linearized elasticity. And from here onward I'm going to use coordinate notation only. Okay, Because there are details, as you can see, which become most transparent when we use coordinate notation. Okay, so given the usual data, given u_g_sub_i, t-bar_i, f_i, the constitutive relation, sigma_i_j equals c_i_j_k_l, epsilon_k_l. And the kinematic relation, epsilon_ k_l equals 1/2 partial of u_k with respect to x_l plus partial of u_l with respect to x_k. Given all of this, find u_i, which belongs to some space_s. Okay? Which consists of all u_i such that u_i equals the Dirichlet boundary data on the corresponding Dirichlet boundary reserved for that particular component. Right? Of course, what's implied here is i equals 1,2,3. Okay? Find u_i belonging to s in this, or of this type. Such that, for all w_i belonging to v, which now consists of all functions which satisfy the homogeneous boundary condition on the corresponding Dirichlet boundary, and corresponding Dirichlet boundary, right? Here, too, we imply that i equals 1,2,3. Okay? So let me read this all over. Given the data u, t-bar, u_g, t-bar, f and the constitutive relation, the constitutive relation and the kinematic relation, epsilon equals those derivatives of u, find u belong to s such that for all w belonging to b. The following holds right. All right. Now let's write out what holds. Integral over Omega. w_i, j Sigma_ij dV equals integral over omega w_i f_i dV plus, now here comes the rub. Because of the fact that our traction boundary condition needs to be specified individually for each component, we need to straight away here have a sum i going from one to number of spatial dimensions and of course we're doing this in three dimensions, right? Of integral over the corresponding traction boundary, okay, of w_i, t bar i dS. I want to emphasize here that do we imply a sum in that integrand over i? Think about it. Look at that last term. There is in fact no sum. A sum is not implied there. The sum is explicitly being carried out here but importantly we don't do that over the integrand because, why do we not do it over the integrand? It's because that domain of integration could be different for each component i, okay, so we can't quite do it as an implied sum. What we need to do instead is compute that product for each value of i, w_i, t bar i, integrated over the corresponding Nyman boundary, right, which would be different for each component. Right, we get a scalar, right, and then we sum up those scalars, right, over the three spatial dimensions or over the spatial dimensions. Okay, so that's important to note. So let me state here no sum implied for just that term, w i T-bar i. Okay. Instead we have that explicit sum. All right. This is our weak form. What I will do now is derive it from the strong form. Okay. And I'm going to take a bit of a shortcut, I'm not going to read. Well, maybe I will restate all the data very quickly. Okay. So what I'm going to state here first is that the strong form implies the weak form and the weak form also implies the strong form. But like we did for the case of the scalar 3D elliptics linear equations we'll do only the right word implication here. Right, the right word equivalence, just as a demonstration. Right. And that is the following, right. Once again given ugi t bar i, fi, the constitutive relations, sigma ij= cijkl, Epsilon kl, the kinematic relation epsilon i, epsilon let me write this as epsilon kl equals this. Okay. What we want to do is find ui such that sigma ij,g + fi = zero in omega and boundary conditions. ui = ugi on the relevant to risky boundary and sigma ij and j equals tbar i on. Okay, so this is our strong form. All right. Yet again. All right. We start out from here and now we introduce as we did before the waiting function. Right. So what we start off by saying is now let us consider wi belonging to v, which has the usual property, right which is that v consists of all wi such that wi vanishes on that Dirichlet boundary. What we will do now and this is something you may recall from our previous repeated developments of the weak form. Actually, think about it, what we do? How do we proceed now? Right. We multiply that pde with wi and integrate. Okay, so what we do is multiply the pde by wi and integrate over our domain omega, right, which is an open domain and r, r3. All right, fine. When we do this here's what we get. Get integral over omega wi sigma ig, g dv + integral over omega wi fi dv = zero. Okay, here sums are implied over i. Okay, so I'm not going to see it which means the sum is implied. Okay. So do you recall what comes next? Right. We observed that this divergence, right, we get rid of by invoking integration by parts. And integration by parts you may recall is nothing but a combination or from the product rule of differentiation and Gauss's theorem of divergence. Right, so apply to tensors now. Okay. So what we do here is integrate by parts, right? And we integrate their parts and [inaudible]. Okay. And integrating by parts remember that this is basically product rule differentiation plus the divergence theorem. Okay. And the way we do that is the following, right. We observe first that derivate one sigma can be rewritten if we just consider it to be the following. Integral over omega, the derivative of the whole product, okay, minus an integral over Omega wi, g sigma ij dv. What I've done here is apply the first of my two techniques here, right. I've applied the product rule here. Okay. And, of course, along for the ride is wi fi dv integrated over omega, the whole thing is equal to zero. Okay. The next step we invoke the divergence theorem on the first integral. And in doing so we observed that since the i index has been contracted out here, some being implied, that term in parentheses is essentially a vector, right. A vector with free index j. And we recognize that we have properly here the divergence of that vector. That vector for our purposes is something that we may probably write as w.sigma, okay, so we just had the divergence of that vector. Right. So now we apply the divergence theorem on that... Right. And in doing so what we get is integral over the boundary, okay. Wi sigma ij mj ds minus, the other terms don't get done, don't change. Minus wi, j sigma ij dv plus integral over omega wi fi dv equals zero. Okay. We have this. And what we're going to do now is two things. I'm going to rearrange things a little and the way I'm going to rearrange this is by observing that this term has a negative sign, so I'm going to move it to the other side but I'm going to change my left and right hand sides.