Welcome back. What we did in the last segment, what we accomplished was a global matrix vector representation of the new time dependent term that arises in the unsteady form of the linear parabolic equation in three dimensions with the scalar unknown. So let's- so to speak finish the job now and we'll do it by carrying out the last remaining step in our finite element formulation which is accounting for our Dirichlet boundary conditions. Okay, so the topic of this segment is Dirichlet boundary conditions. So, for completeness, let's just- and for connections, let's just write out our finite dimensional weak form. Just the integral form and then immediately after that our matrix vector form for it. So what we have is the following: integral over omega, w h rho, partial time derivative. That's integral over omega. W h comma i kappa i j, sorry. U h comma j dV equals integral over omega, w h F dV plus integral over the influx boundary, w h j n dS. These are the terms we have. Now, we've already written- so we've already written out the matrix vector form coming from the second term on the left hand side and the two terms on the right hand side. And we know very well that after accounting for the Dirichlet conditions on those terms, we have a form that C transpose K d equals C transpose F. And this is all we would have if we were working with the steady state problem. What we've added on to this is the understanding that this term can be written as C transpose M bar, the consistent mass matrix, d bar dot. Now, let's talk about the Dirichlet boundary conditions on this time dependent term only understanding that the Dirichlet boundary conditions have already been accounted for from the remaining terms. So what we're saying is that this form follows if the Dirichlet boundary conditions from the integrals- to be really precise about this. If this follows- if the Dirichlet boundary conditions from the integrals without time derivatives have already been accounted for in this matrix vector form. In particular, note that that is what allows us to say that the key matrix is square otherwise, the presence of Dirichlet boundary conditions would make the corresponding matrix a rectangular matrix. And that's what allows us to say that we have d here and not d bar. The d here is only the final set of unknown Dirichlet conditions. And, in fact, what I'm seeing here- what we're seeing is that F already has the Dirichlet conditions accounted for and there. Some sort of jumping ahead so that's final step. All right. So this thing already has Dirichlet boundary conditions from the non time derivative terms. The non time derivative integrals. It may seem like a piecemeal way of doing it, but I think we understand why we're doing that because we've already been over that part of a steady state problem. As in fact, you could view this statement as a steady state problem in adding on the time dependent term. How will would you do it? Well, this is how. Okay, right. That's what we were doing then already the Dirichlet boundary conditions will be accounted for in the F vector, so then we have to worry only about the Dirichlet conditions which are reflected in here. So let's do that. So let's note then that C transpose M bar d bar dot can be or is indeed written as a C vector. C1 up to sum C n n e minus n d. We have that. Multiplying our M bar matrix and here on the right, we have our big d bar vector- d bar dot vector, in fact. So here we have d bar one dot. And let's suppose that here we have a d bar, a bar dot. Maybe I should make that dot in a different color. Well, never mind. That big dot is for the time derivative. A little dot are the ellipsis. So it goes on. And let's suppose that we have here a d bar with a big dot on the b bar degree of freedom and we end with our d bar n n e big dot. Now we know how this works. What we are talking about is that those degrees of freedom have Dirichlet boundary conditions. We complete this Dirichlet boundary condition on degree of freedom a bar- Global degree of freedom. On b bar. And we know how that works out because what we're seeing is that this column that we denote as M bar a bar. That's a column and, likewise, I guess we'd get an M bar b bar column here and we know what that implies. It just says that every element and that- every component in that column is multiplied by the d a bar dot component. And likewise everything in the M bar b bar column is multiplied by the D bar b bar dot degree of freedom. But those are known. So these are known. Not only are those degrees of freedom known but since they are time dependent, if they're known at every instant in time, we can indeed compute their time derivatives as well. So those components also are known. Those time derivative components all those degrees of freedom to Dirichlet degrees of freedom are know. So, what do we do? Well, we take it- take them to the right hand side because the idea is that that time dependence of those Dirichlet boundary conditions also drives the problems that we're looking at. It's a time dependent driving of the problem with Dirichlet conditions so maybe I should just state that as a remark here. So in the form that I've written things up on the previous slide, d bar a bar and d bar dot b bar drive the problem- actually, let me qualify this further- drive the initial and boundary value problem via time-dependent Dirichlet boundary conditions. So what do we do? We know very well now. Right. So, what this lets us do is to rewrite the whole problem now as C transpose M d dot plus C transpose K d equals C transpose F minus d a bar dot M bar a bar that column minus d bar dot d bar, just a component, a time-dependent degree of freedom whose time-derivative now is continuing to drive the problem multiplying this column. Now, unfortunately, I started out by calling the F that I wrote earlier to be F instead of calling it F bar or F tilde ot something. So allow me now to simply redefine this as also F. So redefine as F. So whatever F we caught from the earlier steady state problem is sort of being updated to account for this time-dependent driving with the Dirichlet conditions. So what that lets us say then is that we finally have C transpose M d dot plus K d minus F, with our redefined F including these Dirichlet conditions. All of this equals zero for all C now belonging to R of dimension n n e minus N D. And that, of course, comes from our specification of the weak form holding for all weighting functions. All right. And when we impose this we see that the final matrix vector equations, we impose to solve this, linear parabolic problem in a scalar unknown in 3D is this. So this is our semi discrete matrix vector problem. Semi discrete because we've really truly discretized only in space. And, what we really need to do now is account for how we deal with the time-dependent here. And note that all the time-dependence has now been sort of put into this time-dependent vector of degrees of freedom that we need to solve for. So we'll end the segment here. When we come back, we will focus on the time integration.