Welcome back. What we did in the last segment, what we accomplish 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 scalar, with a 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. Okay. So, for completeness, let's just and, and for connect, and for connections let's just write out our finite dimensional weak form, just, just the integral form. And then immediately after that, our matrix vector form for it, okay? So, what we have is the following. Integral over omega w h row, partial time derivative. Plus integral over omega. Wh,i, kappa ij. j sorry. U h comma j d V equals integral over omega Wh f dv, plus integral over the, the influx boundary, wh jn ds. Okay, 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 transposed k d equals c transposed f. Okay, and this is all we would have if we were working with the steady state problem. What we've added onto this is the understanding that this term can be rewritten as c transposed M bar, the consistent mass matrix, d bar dot. All right, now, let's talk the Dirichlet boundary conditions on the Stein dependent term, only understanding that the Dirichlet boundary condition have already been accounted for on the remaining terms. Okay, so. So what we're seeing is that this form follows if the Dirichlet boundary conditions From the. From the integrals to be really precise about this, okay? For if this follows, if the Dirichlet boundary from the, integrals. Without time derivatives, Have already been accounted for. In the, in this matrix vector form. [SOUND] Okay? In particular, in particular, note that that is what allows us to say that the K matrix is square. All right, otherwise the presence of Dirichlet boundary conditions makes K a, would, would make the corresponding matrix there a rectangular matrix. And that's what allows us to say that we have d here and not d bar. Right, the d here is only the final set of unknown Dirichlet conditions, and in fact, what I am saying here, what we are saying, is that F already has the Dirichlet conditions accounted for in there. Okay? So I'm sort of jumping ahead to that final step. All right. So this thing already has. The Dirichlet boundary conditions, from, the non time derivative term, right? The non time derivative integrals. Okay, 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 the steady state problem. Yes in fact, you could view this as taking the steady state problem and adding on the time dependent term, right? How would you do it? Well, this is how. Okay, right, if that's what we were doing, then already, the Dirichlet boundary conditions would be accounted for in the f vector. Okay, so then we only have to worry about the Dirichlet conditions which are reflected in here. Okay, so let's do that. Okay, so, let's note, then, that C transpose. M bar, d bar dot, okay? Can be, or, or, or is indeed written as a C vector, right? C 1 up to some, C N N E minus N D. Right? 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. Okay? So here we have, d bar 1 dot. Right? And let's suppose that here we have a d bar, d bar dot. Okay? Maybe I should make that dot in a different color. Well, never mind. That big dot is for, for the time derivative, right? The little dots are ellipses. Okay. So, it goes on and let's suppose that we have here a d bar, with a big dot, on a, on the B bar degree of freedom. And we, and with our, d bar n n e. Big dot. Okay? Now we know how this works. What we're talking about is that those degrees of freedom, have a Dirichlet boundary conditions. Okay? And we completed the Dirichlet boundary condition on degree of freedom a bar. Global degree of freedom Right on d bar, okay. And we know how that works out, because what, what we're seeing is that is that the, that this column, okay, that I'm going, that we denote as a, M bar, A bar. Right? That's a column. Right? And likewise I guess we'd get N, M bar, B bar column here. Right? And we know how that, what that implies, it just says that every element in that col, every component in that column is multiplied by the d a bar dot com-, component. And likewise, everything in the M bar B bar column is multiplied by the d B bar, sorry, t bar b bar dot, right. Degree of freedom okay, but those are known, right. So these are known, right. Not only are those, are those degrees of freedom known, but since they are time dependent, if they're known with every, at every instant in time, we can indeed compute their time derivatives as well, okay. So those components also are normal, all right. Those time derivative components, all those degrees of freedom, the Dirichlet degrees of freedom are known. 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, okay? It's a time dependent driving of the problem with Dirichlet conditions. Okay? Right. So may, maybe I should just state that as a remark here. Right? So in the form that I've written things up in the previous slide. D bar, a bar, and d bar dot b bar drive the problem actually, let, let me qualify this further, drive the initial and boundary value problem Via time-dependent, Dirichlet boundary conditions. Okay? All right so well, 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, M bar A bar, that column. Minus, d bar dot B bar, just a component, right? A time dependent or a degree of freedom, right, whose time derivative now, is dr, continuing to drive the problem, multiplying this column. Okay? Unfortunately, I started out by calling the F to the root earlier to pF instead of calling it F bar or F tilde or something. So, allow me now to simply redefine this is also F. Okay? So, redefine as F. Okay. So whatever F we caught from the earlier steady state problem has sort of been updated to account for the time-dependent driving with Dirichlet conditions, okay. All right. So what that lets us say then, is that we finally have C transposed M d dot plus k d, minus F with our redefined F, including these Dirichlet conditions, right. All of this equals 0 for all C now belonging to R of dimension n n e minus N d. Okay. And that of course comes from our, a specification in the weak form holding for all waiting functions. All right, and then we impose this, we see that the final matrix vector equations we impose to solve this linear parabolic problem in a scalar unknown in 3 D is this. Okay. So this is our semi discrete matrix vector problem. Semi discrete because we've really truly discretized only in space. All right. And what we really need to do now is account for how we, deal with the time dependent term here. And note that the all the time dependence has now been sort of put into this time dependent vector. Of degrees of freedom, right? That we need to solve for. So, so we'll end the segment here. When we come back, we will focus on the time integration.