Welcome back. We're ready now to work on our linear parabolic problems in three dimensions in scalar variables. What we did in the previous segment was setup the strong form of the problem. We'll just write it down very quickly now and, proceed on to the weak form and other things, all right? So we write, start out here with the strong form. All right. And just remember we are still talk, we are back now to talking about scalar variables. All right. Or scalar unknown, really, let me call it that. Okay, so the setting, as before, involves our domain. We have our basis vectors, e1, e2, e3. That is our domain of interest. It is omega, a point on it is x, the position vector x. And we have the setting of our Dirichlet boundary subset and the Neumann boundary subset, okay? This is the setting, what we are seeing is now given for data, right? We are given ug, jn, f. We have our, our, our old constitutive relation for these problems in 3D as well, right? Which is that minus ji equals, sorry. Plus ji equals minus kappa ij, u comma j, right? These are all the data that we use when we did the steady state problem, right? We have in addition, another coefficient, which I'm calling just rho, okay. And we made the point last time that this would be the specific heat for, unit volume, if we were doing the heat conduction problem. If we were doing mass diffusion problems, rho would typically be one, okay. Given all these data, what we want to do is find u such that. Right? The following holds. Rho time derivative with a partial time derivative of u equals minus j i comma i plus f in omega, the domain cross the time interval of interest, okay? For boundary conditions we have the same boundary, the same sort of boundary conditions that we encountered when we did the steady state problem, right? U equals u g on the Dirichlet boundary. And, right, minus ji ni equals j sub n, the influx heat, the heat influx or the mass influx, on the Neumann boundary. Additionally, we made the point that we need initial conditions, right, or an initial condition here. We have only one initial condition, because because of what? Do you, do you recall? It's because our problem has a single derivative in time, right? It's first order in time. Okay. So we need a single initial condition. And that is specified as u. Remember u is in general a function of position and time, but now we set the time equal to 0. And we say that this is some given function, u naught, suggesting the initial value, right, of u over the domain, okay? So this is what we have. And what we are faced with in this segment is setting up the, the weak form. All right, we are going to take the approach for the weak form that we'd taken before, which is, we, we have the strong form. We multiply it by a weighting function and integrate over the domain. All right, so to get to the weak form. All right. And remember this is going to be the infinite dimensional weak form. All right. In order to get to that, we see the following, right. Consider. W belonging to V, where V consists of now all functions such that w, equals 0 on partial of omega u, right, our same old weighting function, all right? We consider this, right? And we, essentially we multiply and integrate, okay? And we will do that in the next slide, okay. So what we are doing is we multiply. So we say the following, right? We have w rho, partial of u, with respect to time, equals minus w j i comma i plus wf, right? So we multiply all of this, and we integrate over the domain, right? So we integrate this over omega. So here we pick up a dv, all right? And the same thing happens here. We pick up a integral over omega. D v, and here, too, we pick up an integral, okay? That's what we have. Right, now, we proceed just as before, which is that we integrate by parts, okay. From here we integrate by parts, and just as we did in the case of the steady state problem, we integrate by parts only to have a different way to pose that, divergence theorem, right? We want to transfer that divergence theorem into something else, want to convert it into something else. And so we say all right, we see it there, and we are going to integrate by parts. Okay, now we are knowledge experts in this, so we don't need to go through all the steps. Let's just jump directly to the final form that we get on integration by parts, okay? And that is the following. It is, that, on the left-hand side now we have integral over omega w rho time derivative of u, partial time derivative of u dv equals. Now, the way that integral works is the, the way integration by parts works here is to give us two terms. One is integral over omega w comma i, j i, dV. I'll write the second volume term, which is at this point just a bystander, and in fact, indeed is a bystander through most of our, deriv, our, our, the development of our formulation. And we have, of course, the boundary term, right? We get integral over partial omega w ji ni, right? And that's in an, an integral over the surface so we have ds, all right? So nothing happening with the time dependent term on the left-hand side or the forcing function, okay? And then of course we take the usual steps which is to observe that this term, right? Is equal to integral over partial omega u, w ji ni dS minus integral over partial omega j w ji ni dS. Right, we have these two terms and then we invoke our boundary conditions, all right, on the strong form, as well as our, our homogeneous boundary condition on the weighting function, right? Given the way we've defined the weighting function, the way we've always defined the weighting function, we know that w goes to 0 on the Dirichlet boundary. So that term drops out. And here, we know that ji ni on the Neumann boundary is minus jn. All right, so, making these substitutions we arrive at integral over omega w rho partial time derivative of u, dV equals integral over omega, w comma i, ji dV plus integral over omega w f dv plus integral over the Neumann boundary, w jn dS, all right. Now, let me do just one more thing and we'll have the final weak form. I'm going to invoke the constitutive relation here, right? And we know that ji is minus kappa i j, u comma j. So, we invoke this, and then also observing that we have a minus sign in front of it, I'm going to move it to the left-hand side, okay? Right? So, what we have finally is the following. Integral over omega, w rho, partial time derivative of u, dV plus integral over omega, w comma i, kappa ij, u comma j, dV. Equals integral over omega, w f dV plus integral over the Neumann boundary, w jn dS, okay. This is everything we have, right. What, let, let me just write out now the finite dimensional weak form and we'll be ready to go, right? The finite dimensional weak form from here is obtained by just observing that any attempts to solve the, the infinite dimensional weak form are not likely to be any more easy than the strong form. So we decide to go to an approximate representation of it, all right. And what this says, is now find u h belonging to S h, all right. Which is a subset of S, okay? And what is Sh? Sh now is a collection of all functions of the type of denoted u h which, as before, we will expect to come from h1 on omega. Right, so the spatial dependence is going to be the same. Right, even in the time-dependent problem we are assuming that the kind of approximations we are going to construct will have the same dependents that we know from, from before on the spatial variable, okay? So we have this. U h equals u g on the Dirichlet boundary, okay? So find u h given this. Now we'll, let's assume all the data, right. So we know everything about the data, all right? Find u h belonging to S h such that for all w h belonging to V h subset of V, right? Where again, V h is also drawn from the space of H1 functions. Okay. Now, for all w h belonging to V h, the above weak form should hold, except that every function that is obtained from either w or u is replaced with the finite dimensional version of it, right? So we say that integral over omega w h rho partial time derivative of u h dV. Equals integral over omega w h comma i kappa i j u h, sorry, comma j. I realize I got the sign pro, I brought in the equality too early, plus sign here. DV equals integral over omega, w h f dV, plus integral over the Neumann boundary, w h j n dS, okay? This is our finite dimensional weak form for the, for the unsteady problem, right? Now, lets just stare at this for a few minutes, right? Or maybe, maybe not that long, but for a few seconds at least. Right, so now when we go through the whole process, what we expect is that just as before, right? We are going to do everything as before, right? Let's we, we, we are going to now, how do we go to these finite dimensional weak forms? Well we are, remember that, that, that these are, at this point, at this point, this problem is still finite dimensional only in space. We've done nothing about time, right? Because time is still very much of a true derivative, right? That is, that is a time derivative. We haven't done spe, anything special about approximating it yet, okay? So, as before, we will construct our finite dimensional basis by, by a partition of the domain, right? So as usual we will say partition. Omega equals union over e of each of these omegas. So e's, right, we have all that, right? Okay. And the picture is, is also the same as before. If this is omega you know, let's suppose again since we are in 3D, let us suppose that we are using our hexahedral element subdomains and. That is one of our elements, all right. This is omega sup B. Okay, all of that is the same.