Okay, we'll continue. In the previous segment, we did two things. We introduced the weak form of the partial differential equation for linear elliptic problems in one dimension. And we also did took the important step of demonstrating that the strong form of the same equation actually leads to this weak form, okay? So we demonstrated the equivalence between the strong and weak forms in one direction by first considering the form, strong form in demonstrating that it does, indeed lead, lead to the weak form that we'd specified. We are going to complete that exercise now by starting with the weak form and demonstrating that it too does lead to the strong form. Okay, so, we do the following, right so. What we're attempting to do here is show that the weak form Implies the strong form. Okay? Well, we start with the weak form then. And again, I would not write the weak form in complete gory detail except to except for the important components, okay? So the, so the weak form is the following, right. It's find u belonging to S which satisfies our Dirichlet boundary condition. Such that. You know, given all the data, and I'm not going to specify the data again in this in, in writing it here, okay? So assume that we have all the data. We have our knowledge about the boundary conditions, and constitutive relations, okay. So find u belonging to S such that for all w belonging to V, right, all waiting functions, w belonging to the space V, which consists of functions w satisfying the homogeneous Dirichlet boundary condition. Okay? The following equation holds. Integral 0 to L, w, x sigma A dx equals integral 0 to L, w A f dx plus w at L, sigma at L A. That is our weak form. All right, so we start from here and we want to get to the strong form. The approach that we take is essentially to step backwards, relative to the approach we took in the previous segment, okay? So, that means we apply integration by parts to this term. Okay? And the reason that you may already see that that should be the approach to take is that you recall that integration by parts essentially transfers derivatives from one field to another, right? And so we want to transfer the derivative from this, from w comma x onto sigma. Okay? That's the approach that we want to take here. All right, so, let's get going. Let's do that now. When apply integration by parts to that term, recalling again that integration by parts is just a combination of the product rule of differentiation and the fundamental theorem of calculus. When one is working in 1D, we get the following. We get minus integral 0 to L, w sigma comma x, which you remember this as d sigma dx, right. A dx plus w comma X, sorry it's not w comma x here. Plus w sigma A evaluated at the limits 0 and L equals, everything on the right-hand side remains the same. Because there are no, essentially because there are no derivatives on the right-hand side that we want to transfer on to other, other functions. Okay. In the next step, we are going to split out this term. Okay. We're going to split out this term, okay, into the two limits that it implies. So we get minus integral 0 to L, w sigma comma x, A dx plus w at L, sigma at L A minus w at 0, sigma at 0, A. All right, and you observe that that is essentially what we've done here, right? We've split out w sigma A, evaluated at 0 and L into the two limits. All of this is equal to, integral 0 to L w A f dx plus w at L, sigma at L, A. Okay? And now since you are already getting to be quite expert at invoking our conditions on the problem, especially our boundary conditions, it should be pretty obvious what to do in the next step, right. Think about it for a couple of seconds if you don't already know it. Paying particular focus upon the term that, I've applied an upper brace, bracket to. Right. So, what we do here is to first observe that, that goes to zero, right? And that is so because w does indeed belong to the space V, right? And we have that homogeneous Dirichlet boundary condition upon it. So that last term disappears, okay. And what I'm going to do now, is to essentially combine what remains into integrals, right, here and there, and to terms multiplying w of L here and there, okay? When I do this, I'm giving the following, I, I arrive at the following form. When I combine the integrals, right, because those integrals are over the same domain I have the following, I have w times minus sigma, x minus f. Because what I've done here, is to bring f, which was on the right-hand side in the previous equation, right? In the last line of the previous slide, right? The term, the integral involving f, I've brought to the left-hand side, okay? So, that flips its sign. I have A d x. And I'm simply going to bring everything over to the left-hand side here, okay? So, I have w at L, let me keep A here. Multiplying sigma at L, this was all the stuff on the left-hand side, and the term that I need to bring over from the right-hand side is the minus t, okay? All of this equals 0, okay? Let me just remind ourselves of what we had here. Oh, I, I noticed that when I, when I wrote this Weak Form I actually, sort of, jumped ahead of myself by writing something that I ought to have done a little differently. The sigma at L in the Weak Form, actually appears as t, right? So, I need to correct all the sigma at Ls, and replace them with t, right? So, and, that is the Weak Form. I was jumping ahead of myself by already implying the Neumann boundary conditions, but that is not how we write the Weak Form. That is how we write the Weak Form, okay? So, now, things are, things, everything's all right. Now, when I come back here, I do indeed, have, in this very last te, term sigma minus t, sig my tail minus t, okay? And this is equal to 0. Now, let's recall the conditions under which this holds, okay? This holds For all w belonging to our space V, which consists of functions w such that, w and 0 equals 0, okay. So, the statement here, the mathematical statement is that, that equation in the first line of this slide holds for all w, long as w vanishes at x equals 0, okay? Now, in particular, I'm going to claim that if this holds, the, the condition I'm going to use is that if this holds, I can consider a particular, right? So, what this implies is that it also holds for w of x equals phi of x times minus sigma, x minus f, 'kay? Where, phi of x is greater than 0 for x belonging to the open interval 0, L, okay? And phi of x equals 0 at x equals the points 0 and L, okay? So, really what we're talking about is a function of this type. If this is 0, and that is L, right? And this is our x-axis. The function phi is one that, that looks something like this, right? It may be as smooth as this, or it may be something else, 'kay? Doesn't matter, right? It's positive, right? And it right, it's positive, and it vanishes at 0 and L, okay? What this does then, is two things, right? What this implies for us is that then, it ensures that therefore, w at L equals 0, right? It also ensures Dirichlet boundary condition that w needs to satisfy, right? The fact that w has to vanish at 0 comes from the space d, right? That's, that's the condition that we're assuming on the waiting function. In addition, we've assumed a very special one, right? It's not like we've assumed them, we're saying that because this condition has to hold from all w belonging to phi. It also holds that this particular form of w, right? This one, where phi has the form that we've assumed here, okay? Well, what is that do for us now? Because we've taken a form that makes sure that w at L goes to 0, the second term in this expression here for the Weak Form vanishes, okay? That leaves us with integral 0 to L. Now, for w, we have phi of x times minus sigma, x minus f, right? All of this is our W. But then, this multiplied again, minus sigma, x minus f d x equals 0. The other term involving the boundary terms vanished by our choice of phi, okay? So, what we get from here, is now integral 0 to L phi of x minus, well, minus sigma, x minus f. The whole square d x equals 0, and I realize I'm missing an A here. Okay? Well, but if this is the case, what do we know here? We know that phi, x is greater than, or equal to 0, everywhere, right? Or 0 to L, right? In fact, its, in fact, over the, over the domain, over the interior of the domain, right? This term is indeed equal to 0. Sorry, over the interior of the domain, this term is indeed, created in 0, okay? This being the square of an expression, is also greater than, or equal to 0, okay? So, the only way this whole expression can be equal to 0, right? As an integral over the domain, is if, minus sigma, x minus f is itself equal to 0, right? In the open integral of interest 0, L, okay? But this, of course, is what? This is nothing but our, our, our the p d e, which goes into a strong form, okay? This is simply, the p d e in In the strong form. Okay? Let's continue then. So, so, what we demonstrated that having when we, when we considered the Weak form. It, since it holds for all of w, it also holds for a certain choice of w which reveals to us that particular choice of w reveals to us that the pde of the Strong form must hold. Okay? Let's continue with that exercise now. So now, let's return to this other form that we sort of massaged our weak form into, and that was integral from 0 to L, W minus sigma comma x minus f, Adx plus w with L A sigma at L minus t equals 0. Okay, for all w belonging to v. Okay, now what we've demonstrated just above in the slide, by a particular choice of w. What we've demonstrated by choosing a particular w? Is that this term actually vanishes. The pde of our Strong form must hold. That leaves us with only this last term to worry about. Okay. Okay, and this has to hold for all w belonging to v. So, in particular. It also holds. For a w of x, which is such that w at 0 equals 0, which is required by our specification of the space v, and w at L, not equal to 0, right? It also has to hold for this particular choice of w. Well, if that is the case, the only way the condition at the top of the slide can hold is, if sigma at L minus t equals 0. Okay? But after all this is nothing but the Neumann boundary condition of the Strong form. Okay so we've done two things here, we've demonstrated by starting from the Weak form that, since the Weak form has to hold for all waiting functions belong to our space v it also must hold for certain special choices of w. And from the special choices of w, we are able to demonstrate that one, the pde must hold, right? And secondly, the Neumann boundary condition must hold. Right? The pde of the strong form and the Neumann boundary condition of the strong form. It would appear that we left out one component of the Strong form here. What about the Dirichlet boundary condition on the Strong form? Where did that go? How do we make sure that that is satisfied? Think about it for a couple seconds. My claim is that it is already satisfied. Okay, so it is indeed already satisfied. That the Dirichlet boundary condition on the Strong form is already satisfied, simply from the fact that we have our Weak form here, right? On this slide, at the top of the slide, on the second line of the slide you will observe, that when I stated the Weak form, I said that we wanted to find u, belonging to S, where the space S consists of all functions that actually satisfy the specified Dirichlet boundary condition, right? Which, for this, boundary value problem that we're considering to fix ideas, and to develop our approach. The Dirichlet boundary condition in this case is that u vanishes at x equals 0. This condition, right? It's already in our Weak form. Okay? So our Weak form already has that condition, right? Let me just state that. Weak form already has the Dirichlet boundary condition on u. Okay. And that comes about because we say that u belongs to space S which of consists of all the functions u such that u at 0 equals u not. Okay the Dirichlet boundary condition so to speak is built into our Weak form. Okay? So, we've demonstrated that the Weak form implies our pde of the Strong form, it implies the Neumann boundary condition of the Strong form and it already contains the Dirichlet boundary condition of the Strong form, right? That essentially completes it, okay. So what this implies is that, the Weak form implies the pde, Neumann boundary conditions of the Strong form, and contains. Dirichlet boundary condition. Okay. So, indeed, the Weak form is completely equivalent to the Strong form. Okay, so what we demonstrated in the segment before this is that the Strong Form implies the Weak form. And we did this by starting out with the Strong form, multiplying it by the weighting function, and integrating by parts. What we've done here is start out with the Weak form, pose certain arguments makes, make the observation that since the Weak form holds for all weighting functions w belonging to the space v, it also hol, holds for certain specific forms. Right? And for it to hold those specific forms, what we've demonstrated is that certain conditions must hold. In particular, we demonstrated that the pde must hold. Right, the Strong form of the pde must hold. Our Neumann boundary must hold and the Dirichlet boundary condition is built in. So, we have that the Weak form implies the Strong form, okay? All right, so we'll stop here for this segment.
