Welcome back. At this point, what we've managed to do with our formulation of the finite element method for these 1d linear elliptic PDEs is actually state the problem in completeness, right? We've derived the matrix vector equations for linear and quadratic basis functions explicitly. And we've also laid out the parts to be followed when we go to higher autobasis functions. And, while, in the formulations we develop, we carried out analytic integrations of all the terms that were needed. We've also looked at how we would go about carrying out numerical right? Numerical integration. And numerical integration is what would actually be done in a code, which before too long you will also be writing. What I'd like to do now is well actually observe that we are ready to move on to other problems, right? We are now ready to move on to multidimensional problems. Before we do that however, it's important to understand a little more, just a little more about the mathematical basis of the finite element method. And there are at least two things we need to do. What we will start doing with this segment, and it'll take three or four segments, or maybe a little longer. What we will do is to understand why the finite element method works, what its special properties are. And also, get a a fairly high-level view of the way finite element conversions works. Right? And we'd see the mathematical basis just to see a few of simple proofs for why the method works. Okay? All right. So that's the plan. We are going to start by talking of norms. Okay. And that, and by this, we mean mathematical norms, right? How do we estimate the, so to, so to speak, the magnitude of certain functions. 'Kay. In a manner, that makes sense for what we have to do here. Okay. Let me give you a little background before we really plunge into the subject and let us start by considering right. Consider the the finite dimensional trial solution that we are working with, okay. So consider the finite dimensional [SOUND] trial solution. Okay? One thing I'm going to start doing from now on is perhaps stop referring to this as the finite dimensional trial solution, though it is that but actually start calling it the finite element solution. Okay. And so, this is the field that we would get by carrying out the finite element solution, and maybe going back and re-interpolating from the trial solution degrees of freedom. Okay. So let me also state that. Also called The finite element solution. Okay? That's probably how we, are we refer to it from as, as we move on. All right we consider it in the context of the problem we're solving. So we have something like that, and let's suppose we have some number of elements here. Okay, what I'm going to do is consider to show that there is some generality, complete generality with what I want to state. Let's consider that we are using here quadratic basis functions. Okay? So, that would mean what are intended to be elements here, each of them has midside norms, okay? So this now is omega 1, right? That represents some general element omega e, right. I'm going to now sketch out the quadratic basis functions. All right. So we have that. We have N2. And I'm sorry, sort of straight over a little. Let me try to do better. [SOUND] We have N2. And we have N3. All right, for this element. Okay, I'm not going, well let me label them just for this one, maybe. N1, N2, and N3. Okay. For element two, let me perhaps use a different color so we can highlight a certain point. Okay. So for element two, my shape func, my basis functions are going to be, that. Okay, can, can I do better than that. Sorry. Okay, that's N1 for element two. That is N2 for element two, and that is N3. Okay? So, again, I will say N1, N2, N3. Okay? Now, of course, we also know that associated with each of these nodal points, we get trial solution, degree of freedom values, right? So let me write those. And here, I'm going to write them using global numbering, okay? So here I have d1. I have d2 here. I've realized I switched from sub to superscript in the first one. Okay. D1, d2, I'm going to write d3 here. D4, d5, and so on, okay? Now, the point I'm trying to make here is that the way you would, we would construct our feed now, right, within each element e, would be obtained by that process. Right? DAe, and we now how to go between local and global no degree of freedom numbers. Okay, observe the nature of our basis functions here. I'm going to ask you a question. What I've sketched here in red and green are the basis functions, respectively, for elements one and two. Is the solution continuous? So, the solution, the finite element that one would generate by re-interpolating from the trial solution degrees of freedom. Would that be continuous? 'Kay. Think about it. The answer is yes, it is continuous. Okay. Mathematical notation for a continuous function is C0 on omega. Okay? When we see a function of C0 on omega we're seeing, C0 simply means the function itself is continuous, right? Over the domain omega. Okay what about its derivatives? Are the function, or its derivatives continuous over omega? Think about that. Well let me ask you in step stone comma x. It turns out that it is continuous within each element. Right? So if I look at e comma x, right? Restricted to a single element, right? Within that element, e, the derivative is also continuous. Okay? So we would say that this function is C0 on omega e. Right? So the derivative is continuous within each element. Right? What about over the entire domain? Is the derivative continuous across the domain? In general, well not in general. The, the derivative indeed is not continuous across the domain. In particular, there are points of discontinuity on the, in the derivatives, right? At element edges like there. And there, right. What we will see is that comma x is discontinuous. Okay, all right. So, so comma x is discontinuous. In all of omega. Right? So now here what I am specifying is that we are not considering restricted to a single level. Right? It is across elements. Okay? So it's discontinuous in omega, and so what we say, however, is that comma x, the, the derivative is not in C0. When we take the entire domain omega into consideration, okay? So this is an important thing to notice, right? And where does this come from? Why is it that the solution right, the final element solution as itself continues, but its derivatives when considered over the entire domain omega, are discontinues. Where does that, where does that come from? Right, it comes from the basis functions, right? So what's important to notice here is that the Lagrange polynomial basis functions have been constructed. To be only C0 on omega, right. We have basis functions, which are only continuous, and that's immediately observable if you just look at this sketch that I made of the basis functions, right. Clearly the basis function that continues, but of course if you look at the inter-element nodes, it's clear that, that the derivatives are not going to be continuous. And of course, once you have that, any function that you then construct using that basis to represent the function is also going to be only continuous but not be continuous in its derivatives. Okay? So, we say that it's only C0. We say that it is not the function itself now, the function eh, or referring to basic functions themselves are not Cn on omega for n greater than 1. Sorry, n greater than 0. Okay? All right, the basis functions themselves have been constructed to only be continuous. There you see 0, okay? Their derivatives are not continuous, okay? Right, if the derivatives were continuous, we would say the functions would be Cn, right? Where n would be the number of derivatives that were continuous. Okay? So we're saying that the basis functions are not Cn, they're only C0. Okay? All right. So let me just give you an additional statement here to give you the more general case. In general, a function is in Cn omega if its derivatives Up to order n are continuous on omega. All right, our basis functions themselves are continuous. None of their derivatives are continuous. Therefore, they're only C0. 'Kay, the zero indicates no derivative continuity. 'Kay? However, remember when we talked about spaces. We introduce the spaces L2 and H1. Okay. So however, does right, the finite element solution belong to H1 on omega? Okay. All right. Remember, so what we're asking is, what this means is, if you recall, is the following. If we integrate over omega squared plus 1 over measure of omega to the power 1 over number of spatial dimensions. Okay. Sorry, this should be 2 over number of spatial dimensions. [SOUND] Okay? Times comma x squared, dx. Right. If we do this, okay, what we are asking is, is this quantity bounded? Think about it. Is not continuous, okay? So what we will find is that this term, right? The derivative term is discontinuous. However, if we integrate it, right? Discontinuities are not unintegrable, all right? They can be integrated. Okay so when we integrate it, we will see that this quantity that we put down here on the left-hand side is indeed bounded. Okay? So the answer is yes. Okay? All right, so even these continuities themselves don't pose a problem. Okay? All right. Now, okay, so does belong in H1, okay. So I'm going to use this background to say that and, and of course you know we, of course belongs in H1, because that's how we constructed right. Remember when we defined the Sh, right. Okay. So recall, belongs to Sh, which consists of, we, we said that it consists of all functions to each belonging to H1, right? On omega. And then we put in boundary conditions and so on, right? So, so by construction, it does belong to H1 and I'm, I'm just pointing out that yes, it does indeed belong to H1. Okay, with this as background, I'm going to start defining some norms.
