Okay, we'll do it here. So, the element integral. Okay, integral over omega e. W h i, rho, second time derivative. Right, this one. I'm going to go straight away into, writing it out, with our basis functions and summing over basis functions and all that sort of thing. Okay? So, we know that this is integral over omega e. Actually let me skip some more steps. Okay, because we're now such experts at this that we don't really need to write every single thing. Okay, let's do it this way. Fine, sum over a comma b. C a e i. Okay? And you recall that c a e i is each of those is the i'th degree of freedom at the a'th node of element e. Okay? This. Integral over omega e. N a, let's have the rho there. N a n b d v. And outside of this integral. Come the degrees of freedom that, are used to, construct the representation for the trial solution, right? So that will turn out to be d b element e degree of freedom i, okay? And what is implied here is a sum over our, spatial dimensions i. Okay? Now, the form that I have in parentheses on the right hand side. Okay. The everything including the integral, is what we identified previously to be the a b component of the consistent mass matrix. Okay, same thing. The same thing happens here, except for the fact that there's one thing we need account for. Which is that there is a sum over i here, okay. So let me just mention this here. Recall sum on i, okay. And what that does is lets us write this thing out as sum over a comma b, c a e i, integral over omega e. Rho n a n b d v, but now what I'm going to do is, I wrote that d v too close, I'm going to slip in here a Kronecker delta, okay? So allow me to do this and I will show you why I'm doing this. Okay? Since I've slipped in the Kronecker delta and given it indices i j. I'm going to turn the d b e i from the previous line, this one. Into d b e. J. Okay. All right. As a result, now when we put this all together what you see is that what you observe is that though the contributions to the, from the integrand are the same as we had for the, for the linear parabolic. Problem with the scalar unknown, right. What we've done here to counter the fact that we have vector unknowns, is to include this Kronecker delta right there, okay? All right, now when we put everything together, we get the following. Okay. We're going, we're now, in what I'm about to write, I'm going to abandon the explicit, writing of the sum over, over a and b. Okay? And in fact, let me give myself more room here. Just before I go ahead and write that I know that I've forgotten the second time derivatives on the d. To account for the fact that we have a second time derivative right there. Okay? Now everything's fine. Okay. So the way I'm going to write this by dropping the explicit sum over a and b is the following. I get, c one e right? Transpose okay? C two e transpose. So on up to c number of nodes in the element, e transposed. Okay, this one for instance, c two e transpose, is simply c two e one. C two e two. C two e three. Okay? All right. All of that now multiplying some big matrix which makes up our mask matrix. Okay? And out here we get our vectors, or vector of d one e double dot d to e. Double dot all the way down to d n n e. Element e. Double dot. Okay? And like I did here for c two e let me just point out that this is the little vector d two, e one, double dot, d two, e two, double dot, d two, e three. Double dot. Okay? All right, now. Note that for each combination of nodes from the c vector and the d vector, okay? We get a little matrix, okay? And that little matrix is obtained by writing out here we have here let me show you what we get here. Okay. We have here a little matrix which is row n a. N b d v over the element and let me actually make it n one n one to show you the very first combination okay? But this matrix, that integral itself, multiplies what you may call the three by three identity matrix. Okay. That's because it's this matrix that I've just written, that little sub matrix that I've just written. Which multiplies, which, sort of intercedes in the multiplication between c one e transpose and d one e double dot. Okay, this continues until we get another such matrix. Right, each of these, each of these blocks is a little matrix, right. Each of these is an integral over omega e. Of, you know, n something, n something else, integral over d v. Okay? All right? All the way down. So we have how many such blocks? All right? We have n n e such blocks in the column wise direction and n n e such blocks in the row wise direction. Okay. All right. All right. And what's important to note is that each of these little matrices is itself an identity matrix, right? And this is what our Kronecker delta does for us. Okay? And so the general term in here would be something like integral over omega e rho n a, n b, d v one zero zero zero one zero zero zero one. All right, you'd get other such block matrices everywhere. Okay. All right. And this matrix that I've written out here is our element mass matrix. Okay, in consistent form, okay, the consistent mass matrix. All right so this is an important thing to note. This is slightly different from what we saw for the linear parabolic p d, where each degree of freedom was, was, is essentially a single scalar. Unknown. Right? And, and in that case, this, the, this identity matrix essentially collapsed to the scalar one. Okay? The fact that we are doing, dealing with vector unknowns here, just, has, has essentially expanded that scalar one into a, into an identity matrix. Right? Three by three identity, that's all. Other than that, everything is the same. Okay it's probably useful for us to ask, now what are the dimensions of m, e? Okay, what will they be? Right the dimensions of m, e are number of nodes in the element, times number of spacial dimensions, squared, right? Those are the dimensions of m e. Right. Okay. That's it. This is our element mass matrix. The only thing that's different with, with regard to what we've seen before. Assembly works just as before. Assemble over degrees of freedom, common degrees of freedom across elements. Belonging to the same global degree of freedom. Implies that those corresponding matrix contributions from every. From, from, from the neighboring elements add up. Okay? All right. So let me just state that here. Assembly proceeds as before. Over global degrees of freedom. Okay? What we are left with, finally at end of the process, right, after we, we, we say that. Well this has to, our, our weak form has to hold for all c belonging to the appropriate, Euclidean space, right. What we get at the end of it, is the following set of equations, right. We get m d double dot plus k d. Equals f. Okay? When we were not doing elastodynamics, but essentially elastostatics, this first term was missing. Okay? It's just shown up now, right. Everything else is the same. We know that boundary conditions are buried inside here, right, including time dependent Dirichlet boundary conditions. That also works just as we saw earlier, right, in the case of the time dependent parabolic problem, okay. If we have Dirichlet conditions, they are time dependent. Well we know exactly what they are at every time, at, at any time, right they will go to the right hand side. What we would have here in addition are initial conditions, right. D at zero. Equals the, the vector of, well it essentially equals the vector form by taking every Dirichlet condition. At degrees of freedom lying, corresponding Dirichlet boundary. And, putting them all in a vector d nought, okay? We did this for the parabolic problem. We need another boundary condition here, sorry, another initial condition, because we are second order in time. That's d dot at time e t equals zero, which we will denote as v nought. Okay? All right? And v nought, you remember, is simply, it's simply constructed of, u dot. Sorry, it's actually I think just the vectors, the value of the function v nought, right? At points x a, right? For every degree of freedom, a, that lies, sorry, for every single degree of freedom a. Right, for the initial condition has to be specified over the entire vector D. Okay? This is essentially it. Right, just as we specified initial conditions for our parabolic problem, we have two initial conditions now for this problem. All right? Okay, so that's really all we need to know about how the matrix, the matrix equations are obtained. The process is exactly the same. We get another mass matrix. There are some detailing mass matrix, it's at, at, for corresponding to every combination of degrees of freedom, it's slightly blown up. We have that little three by three identity matrix sitting there. And we have two initial conditions, right? And we know how to specify those initial conditions. So, so really, that's what we need to know. We'll stop this segment here. When we return, we will talk about time discretization.
