Welcome back. So, we now find ourselves ready to look at the Weak Form of the linear elliptic p d e for a scalar, for scalar unknown in three dimensions, 'kay? So, let's just state the weak form and, and, and move on with the thread because we've written the strong form, and at least a couple of times, and in different ways, so All right. Here is the weak form of the problem, 'kay? I'm going to first state it, and then, obtain it, right? So, find. U belonging to S, right? Where S is equal to space of functions, right? So, equal to some functions u, such that u at u equals u g on The Dirichlet boundary, 'kay? Find u in this space from the space given everything else, right? Obviously we're given g, u g. We're given j sub n, right? We'll be given the data f of, and the constitutive relation. Our constitutive relation now is j sub i equals minus kappa i j u sub j, okay? Given all this, right? Find u such that Such that, for all w belonging to V, where V consists of space V consists of holds functions w. Where w, again, just as we saw before for the waiting function, vanishes on the Dirichlet boundary. Okay, find u belonging to S, given the data such that, for all w belonging to V, the following condition is satisfied. Integral over omega w, i j i d V, okay? Equals Integral over omega w f d V, okay? Minus integral over omega, sorry, integral over omega sub j, w times j n d S, okay? Some things to note here, first of, this is this is our weak form, 'kay? Some things to note here are that we have here the elemental volume d V, and d S here is the elemental surface area, 'kay? That should be clear because on the golf course is a subset of three dimensional space, it's volume in 3D. And therefore, it's boundary partial omega sub j, that's subset of its boundary, is indeed a surface, right? Okay, so, that's the situation we have, right? So, [COUGH] d V is the element of volume that's, that allows us to integrate over the, over this entire volume. And d S is every little elemental surface area, okay? So, this is the weak form, what I'm going to do now is get us to it, right? I'm going to show, show you how the weak form is obtained, right? As before, we are going to take, we're going to take one of the approaches that we took before which is to start out with the strong form, which you've already put down, and get to the weak form, okay? So, here's how we do it, right? So, we start out by considering Consider the strong form. Okay? We have used this stuff, right? So, we have find u Given all the data And I'm just going to put down the constitutive relation here directly without calling it a constitutive relation Right? So, given all this stuff, find u such that, st is short form for such that. We have minus j i, i equals f in omega. Because I'm a little short of space here, but I want to have it all on this slide, I'm going to put down the boundary conditions here. Dirichlet boundary condition, and our Neumann boundary condition Okay? This is what we have. Now, the approach we are going to take is the one that we took formerly, the, the approach that we took in obtaining a weak form from the strong form for the 1 d problem, okay? All right, so, what we have, is we have the strong form. What we are going to say, is consider W belonging to V, right? Where V, of course, is the, is the has the properties that we stated before. It consists of all functions that satisfied the homogeneous Dirichlet boundary condition. Right? That's what that is. Okay? Now just as we did for the 1 d problem, what do we do here? Do you recall? That's right. We multiply the pde of the strong form by w, and we integrate by parts, right? So multiply. Pde. Right? To the strong form strong form S f being short form, for strong form of pde. Okay? Multiply the pde, the strong form of the pde by w. Right? And then, we integrate by parts. Okay, that's what we do. So, let's do it in steps, right? First let's multiply the pde, and integrate it over the volume, and then we'll invoke integration by parts, okay? So, integrating it, by multiplying it by w, and then, integrating it over the volume, we have this, integral over omega minus w j i comma i d V. Right? We took the minus divergence, from the left-hand side, and multiplied it by w. Okay. And this is equal to integral on the right-hand side, integral over omega w f d V. Okay? Okay, down here is multiply the pde by w, and integrate over the volume. Okay. We're now, going to do integration by parts, and if you recall, the application of integration of parts, and functions in three dimensions, right? Where the integrations happen in three dimensions. We of course do that to the left-hand side too, right? So, this is the one that we are going to integrate. By parts. Now, when we did this in the 1 d problem, I did mention I think, that the integration by parts is really nothing other than an application of the product rule of differentiation. Combined with the, the divergence theory. Okay. That indeed holds in, in multiple dimensions as well. And to bring uh,it's useful to bring that out, because it makes very clear what how, how the integration by parts proceeds. So, we'll do that. Okay? So, recall here that when we say integration by parts, what we are talking about doing, is the product rule And Divergence theory. Okay? Firstly, the product rule tells us the following, right? It tells us that we really need to look at w multiplied by j i comma i, as being just one term, in applying the product rule of differentiation, right? So, it tells us that, that really is integral omega. W j i, the whole thing, comma i, and we should make sure not to forget our signs, right? This plus w comma i j i. Okay? So, if you expand out the first term in the integrant, you will see that it can, that one of the terms it produces because of the product rule cancels out the second term. Right? And we're left with, what we had in the first line. Okay? So, this is equal to integral over omega w f d V, okay? But then, when we stare at our left-hand side integral, we realize that we can actually, view w j i as essentially a vector, which it is. Right? Because j is a vector. Right? Soum, this is a vector. Right? Component i. And that term, is essentially the divergence of w j. And there, we invoke the divergence theorem. Okay? All right? So, this step is really just a product rule. When we do the divergence theorem, we get the following. Right? The divergence theorem tells us that this integral over omega is really an integral over, over the boundary of omega. Okay? Minus integral over the boundary of omega of w j i n i. Right? Where n i, is of course, our unit outward normal, something that we'd observed in one of the previous segments. Right? It's this, integrated over the surface of the body, plus next term remains a volume integral. And the right-hand side stays the same. Okay? And where, how we get this is through the Divergence theorem. Okay? What we are going to do next, is take the surface integral. Sorry. Take this surface integral over to the right-hand side. When we do that we get, integral over omega w comma i j i d V equals integral over omega w f d V plus. Now, that integral over the entire surface partial omega, I'm going to break up two integrals over each of the two subsets, we've identified for the boundary, right? And those subsets, you recall, are the following. One of them is the Dirichlet boundary partial omega u, right? And so, here we have w j i n i d S plus integral over the Neumann boundary w j i n i d S. Okay. Now, let's stare at the last two boundary integrals. What can we say about the first one? What can we say about the first integral? Think about it. It's over the Dirichlet boundary. It's w j i n i d S. Right, we can say that over the Dirichlet boundary because w, we are picking to live in the space V, which satisfies homogeneous Dirichlet boundary conditions. We have that, right? So, that first term, that first integral drops out. What about the second integral? We're starting with a strong form, which includes the boundary conditions, explicitly, right? And in particular, the way we wrote out the boundary conditions we specified that the heat influx vector, which is minus, it's not the heat influx vector. The heat influx minus j i n i on the boundary, right, on the Neumann boundary. Is, heat influx, if you're talking the heat conduction problem, right, and mass influx, if you're talking mass diffusion. Anyway, j i n i, we identified as being minus j sub n, okay? So, when we put these things together, we arrive at the form that I had put down, originally, right, I think at the top of the previous slide. Integral over omega, w, i j i d V equals integral over omega w f d V, okay? Minus integral over the Neumann boundary w j sub n d S, okay? This was the, what I posed originally as the weak form, okay? And so we've got there, right? We've obtained it from the strong form, right? Now, in the case of the one d problem, we also went the other way, right? We demonstrated that the weak form, and the strong form are completely equivalent, and indeed, that holds in, in this case, too, right? It holds for actually every problem, right? The strong form and the weak form of any PDE are completely equivalent, all right? And we can adopt the same approach in proving it, in this case, as well. So, we could start out with this weak form, we would have assumed the space V. We would be given the data. We would essentially do integration by parts in reverse. And then we would go through the tricky business of invoking those variational arguments, right, on how did the, on, on, on the manner in which w can be chosen, right? Or, or, or the very fact that w has to hold for, for, for all functions living in V, allows it to state, state that it also holds for certain spes, special functions, okay? And that is what would bring us to the strong form. We're not going to go through that argument here. It holds, you can try it on your own if you like. But it, it's not crucial for what we want to do, which is, of course, to work with a weak form, okay? So, we're just going to stop here, as far as deriving the weak form is concerned. Let me also, make a remark. Let me just state here, actually, it's more than just a remark, so I am going to state it in sort of the main text, so to speak. Which is to say that the weak form. What we demonstrated is that the weak form is implied by the strong form, right? This is, this is, this is what we just demonstrated in the, in the little derivation that we did. 'Kay, one can prove also that the other direction holds. Okay, it's not difficult. You can just follow the notes there and, and follow the steps we took in the one d case. Okay, we're not going to show this, this, we're not going to show the right-hand side implication. We demonstrated the left-hand side implication. Okay, well, this is it, right? So, so this is, this is our weak form, and this is what we're going to work off to, to develop the finite element method. That will however, be best done in a different segment. Before we end this segment, there's just one remark I want to make, which is actually, applicable to the, to the strong form, okay? Just a remark I want to make here, which is that, if you look at, recall the PDE of the strong form. PDE of strong form, recall the PDE of strong, the strong form, right? Which is minus j i, i equals f, right? Or in in direct notation it is minus del dot j equals f, right, in omega. I just want to recall for us the interpretation here, now, [COUGH] you know, that if we take this body, and we look at the PDE that we had the PDE really holds point wise, right? So, it holds over every little infinitesimal volume in this over this domain, right? So, if you will go inside, cut it open, and take a little volume, that PDE the, the strong form of the PDE holds over that little infinitesimal volume, 'kay? And what I just want you to do is recall for us here the interpretation of the minus divergence, all right? In this, in this, in this setting in 3D, the minus, the, the divergence itself, vector is the, is the total net outflux over that little volume, okay? So, the negative divergence is the total, is, is the net influx into that little volume. And all this PDE is saying in strong form is that their net influx into every little infinitesimal volume is driven by what we recognize to be a source term, okay? So, let me just do that little make that little argument, too. All right, so, if we have our, this is, this is our body, omega. What we're talking of doing here is considering a little volume element. Okay? And we are, the, the net influx is. Is obtained from, you know, by considering, with, with this little pill box type argument, which is sometimes presented often in the context of classical fluid mechanics. And sometimes, maybe also in the context of heat conduction, right? So, this is the net influx, right? So, so this net influx here is denoted by minus divergence of j, okay? So, what we're seeing is the net influx, right, over a little volume equals the source term, right? Or minus del dot j equals the source term, which is f, right, at every point in omega, right? So, that little pill box there represents a point in omega, right? So, we may choose to say, that okay, that has the position vector x, right? So, the point out there has the position vector x that, that little pill box has been constructed about that point x, okay? All right. I just wanted to recall this argument of this, this interpretation because I omitted to mention it. Okay, so we're done with the segment. What we we've done very quickly actually is get to the weak form. When we return, you know how this is going to proceed. We are going to set up the finite dimensional weak form. We're going to first observe that this is the infinite dimensional weak form. We'll set up the finite dimensional weak form, and develop our finite element methods. All right, we'll stop.
