Okay, so we'll move on with our analysis of the, more mathematical properties of the finite elements method. What we did in the last segment was establish some facts about norms. In this segment we are going to look at two properties of the finite element method, okay? Those properties are consistency, And, The best approximation property. All right, in order to do that, there's one more thing I want to do here, okay? So let's start by recalling. Recall the finite dimensions, let's just recall the weak form, okay? All right? Recall the infinite dimensional weak form, if you like. Very quickly, let me write this out. Integral over omega, w, x and now rather than write sigma, I'm going to write E u, x. We had an A here, okay? This is equal to, we said integral over omega, w, f, A, dx + w, (L) t, A. Okay? Now, note that, This is essentially is in fact our definition of a(w, u). When we defined a(w, u) we didn't include the area in that definition, but you could very well include it in the definition, right? Right? That's a(w, u). Likewise, this is w, f. Right? The way we defined it in the last segment, in this case I defined it earlier in the last segment for general mathematical functions. Here we have an a dx but then we can just say that a is sort of a part of dx and sort of lump it into the definition. What we will do now is also define this quantity to be w,t, okay? All right? At L, okay? Note that this last quantity we've just defined, (w,t) on L is also an inner product, okay? Right, it is also linear in w and it also linear in t, okay? All right. So essentially, what we are seeing is that using this notation, we can rewrite the weak form in what is called abstract notation. Okay? So in abstract notation, we have a (w,u) = (w,f) + w,t at L. Oka,y this is just abstract rotation for our week form. Now, in exactly the same sort of manner, note that we can rewrite, we can also write our finite dimensional weak form, okay? All right? Finite dimensional weak form, then you go to the next slide. Just hope that I have enough room for everything there. Integral omega, w h, x e uh,xadx. Okay, again, everything sigma h is euh,x, right? So integral over omega whfa, dx + wh at l t A. Using our notation, using our abstract notation, note that this is A. Wh, equals wh, f + wh,t at m. Okay? Abstract notation for our finite conventional weak form, okay? So we have these two weak forms in abstract notation, okay? So let's do the following, okay? So let's look at the abstract weak form in the infinite dimensional case, okay? So consider, a (w, u) = (w,f) + (w,t) at f ,okay? So this is the abstract form of the infinite dimensional weak form, okay? Now, remember what we say here for the weighting functions. We say that this weak form holds for all w belongs to v, okay? All right? But then, also recall that, w h which is our finite dimensional reading function, belongs to vh which is a subset of v, right? Right, what this says is that when we say here that the infinite dimensional weak form must hold for all w belonging to v It also holds for finite dimensional weighting functions right? because finite dimensional weighting functions also belong to the same space, okay? So what this means is that so this relation, let me call this relation a. Okay, right? That equation of the infinite dimensional weak form. Okay, so A also holds for wh, because wh itself belongs to b. Okay, what this implies is that we are able to write a(wh, u) = (wh, f) + (wh, t) at l, okay? I'm going to call this relation B. Okay? Right then, this just comes from the fact our finite dimensional weighting function also belongs to the same space v as our infinite dimensional weighting function. All right? Okay, now let me go back and label the proper finite dimensional weak form that we work with as c. Okay, all right? And now what I'm going to say is subtract, B, From, C. Okay? We're going to go to the next slide and actually do this, right? So just stare at B and C and let's subtract them from each other. Sorry, let's subtract B from C. So here is C first. a(wh, uh). I want to leave myself enough room here, okay? = (wh, f) + (wh, t) at L. This is just equation C, which is nothing but our finite dimensional weak form. I'm going to subtract from this B, which we caught by observing that the finite dimensional weighting function also satisfies, The infinite dimensional weak form, okay? All right? So what do you get here, okay? What you see on the right-hand side is 0, okay, because these forcing terms exactly canceled out. Now, recall that a is a bilinear form, right? Which means, it is linear in each of its arguments. So it's linear in the first argument as well as the second argument, right? Likewise for this. What this lets us do is to simply write what we have on the left-hand side as a (wh, u) = 0, okay? All right, so we've got as far as here, okay? Now, look carefully at that term, u. It gives us the difference between the finite element solution and u, which is the solution to the infinite dimensional weak form. The solution to the infinite dimensional weak form would be the exact solution, okay? So u is a very special, very important measure of our numerical method, what would you call it? It is the error e, okay? It's the error in the finite element, Solution and why is this? Because, u, which is the solution to the infinite dimensional weak form, has to be the exact solution, right? The infinite dimensional weak form is equivalent to the strong form. The solution to the infinite dimensional weak form has to be the solution to the strong form, which has to be the exact solution, right? All right, u is also, The exact solution. Okay? All right, so what we've just demonstrated here is a consequence of the condition that we had on the previous slide. This condition and the condition that we've marked as B, okay? So what we've seen here is a consequence of this condition. Okay, so let me state this here. This results, From, B, okay, which is called, The consistency, Condition. Okay? All right, let's look at what we have here first and then try to understand the consistency condition later. Okay, what we have here is a statement saying that a(wh, e) = 0. Okay, now this bilinear form that we have, a, can be viewed as a mathematical generalization of the idea of taking a function, wh in this case, and projecting it along a different function, e, in this case. Okay, so effectively what this is saying is that suitably defined the projection of the error, On our, Space of functions wh, right, or space of functions wh, that projection is 0, okay? What this is saying is that the projection, Of The error on, The space, Vh is zero, right? Viewed differently, it's saying that any error that exists in the finite element solution actually lies outside the space that we are even considering for our finite dimensional solution, okay? So already to saying that in some way we are actually doing very well in having chosen a space Vh in order to get our solutions, in order to look for solutions. We're doing very well, because any error that we get is not even in that space. So within that space, we are doing as best as we can, okay? All right, so it's saying that in some sense, it is saying the error, Is, Orthogonal, To, Vh, okay? Right, one can define formally and rigorously a notion of orthogonality for functions using our bi-linear form. And what it is saying is that with that definition of orthogonality, our error really lies outside of our space Vh because it has no projection left in Vh. And that's why we're seeing this result, that this projection is effectively zero. So in a sense, it's saying that we're doing very well with having chosen the space Vh, okay, that's one interpretation. Now I said that it comes from this condition B, right. So let me rewrite the condition B, okay. The condition, B states this, all right. It says that a (wh, u) = (wh, f) + (wh, t)l, okay? This is what we have. Compare this with the equation that we actually solved to get our finite element solution. So, and that equation, you recall, is the finite dimensional Weak form. In this abstract notation, the finite dimensional weak form is this. Okay, all right, this is the equation we're actually solving to get. If you look at condition B, we can regard condition B as being obtained if we simply went to our finite dimensional weak form, and replaced with u. Okay? If we simply went to our finite dimensional weak form and replaced with u, we would get condition B, right? So what is this saying? It is saying that if somehow we were to come upon the exact solution, okay, u, that exact solution also would satisfy the equation we are trying to solve, Right? So, what this is saying is the exact solution u satisfies, The finite dimensional, Weak form. Why is this important? The equation we're actually solving in our numerical method is the finite dimensional weak form. What we just demonstrated is that if our method were such or by chance, we were to, well not by chance, if we would actually pick the exact solution, right? Maybe our space was such to include the exact solution. Then, if we would take that exact solution and plug it into our solution, into the equation we are working with, we would find satisfaction of the equation, right. So our method admits the exact solution as one of the solutions it can find, that it will accept. This is not the case with certain other numerical methods, right, most prominently the finite difference method. With the finite difference method equation, the underlying equation itself is altered so that if you take the finite difference equations and substitute the exact solution, you are not guaranteed satisfaction of the finite difference equations. So in that case you're actually working with a set of equations which does not any longer admit even the exact solution as an acceptable solution. The finite element method allows the exact solution in, okay? And this is why this condition B is referred to as the consistency condition, okay? The exact solution is consistent with the finite dimension equation we are solving. The exact solution is consistent with the finite element method, okay, so this is the consistency, Condition. Okay, I should also extend this note to saying, well, we've already said the exact solution, u satisfies the finite dimensional weak form, right, so this means two things. What it means is the finite element, Method, Can recover, The exact solution. Exactly, right? And then we also mention here, this is not the case for, Other numerical methods. Right? Of course other numerical method unless they inherit this particular property wouldn't do it, most prominently finite difference methods do not have this property, okay. So this is this is important to recognize, okay. And a great deal of the strength of the finite element method comes from this result, okay. And this result consistency is what leads to the conclusion that we observed in the previous slide. That the error in the solution that we obtained, the error in our finite element solution, is orthogonal in a suitable mathematical, in a rigorous mathematical sense. It is orthogonal to the space in which we are looking for solutions, right, in the space Vh, okay? All right, let's stop the segment for here.