Okay. So we're now shaping up to get to the main result related to finite element analysis, right? An, an analysis of the conversions of the method. To get there, to, to actually get to that result, there is one more class of results that we need. So the title of this segment is going to be Sobolev estimates. Sobolev of course, refers to the type of space that we are working with and that I'll tell you what we're trying to estimate here. In order to say what we are trying to estimate let us sort of reuse our symbol U, capital U of h that we introduced before. It continues to represent a general member of the space Sh, which means it satisfies the, the rationally boundary condition and it lives in H1. Okay? For, for our particular problem, it, it lives in H1. But in the context of Sobolev estimates, which are applicable to spaces with which, which are applicable in general to Hn spaces. Let me just redefine this for now. Okay. Let me say that Belongs to Sh, which now consists of all functions. That for our, for our purposes now, which actually hold in more general settings. Okay? Consist of Hn functions. And additionally, of course, they need to satisfy the Dirichlet boundary condition. Right? And so, if we were doing the, the Dirichlet-Neumann problem, we would have At 0 equals 0. Okay. So we're re-invoking this, this same function capital. Remember, U, capital Does not need to be the, the finite element solution. It could be, right? But it does not need to be, right? So let me state that. Does not necessarily represent little Right? Which is the finite element solution. It is any other solution in Sh, right? Where Sh now has been defined to include all functions that belong, that live in Hn. Okay? All right. I want now to consider a very special member a member of this set of functions. Okay? Consider Such that. At any nodal point let me say, this as X let me just say, x sub A. Okay? So At x sub A equals little at x sub A. Right? Okay. But if A here just denotes our global numbering of degrees of freedom, then this remember, using our finite element vector notation is just the degree of freedom, d sub A. Okay? All right. All right. So what this, what this says is that A is a global degree of freedom, right? And this, this then implies that xA is a globally numbered node and dA is a globally numbered Trial solution degree of freedom. Right. Or finite element solution degree of freedom. Okay? All of that should be clear. So what we're saying is that we want to consider a special member of this class, where Is equal to the finite element solution at the node, the degrees of freedom. Right? At the nodes, okay? But you know that if we do this, we would have capital Actually equals to little everywhere, right? Because of the fact of the finite element representation. Okay? Right? So this is not, not exactly the solution we want. Not exactly the the function we want. Consider another function, U tilde h which is such, that U tilde h. Okay? At xA, now equals the exact solution evaluated at that point. Okay? The idea is that if you had the exact solution and you could evaluated it at a nodal point, right? Take that value and let's suppose that u tilde h is such that it hits the exact solution right there on the nodes. So what we're seeing here is that u tilde h, right? Expressed now as a function of parametrize by little x, right? By position. U tilde h is what is said to be nodally exact, okay? I'm going to sketch it out for us, suppose this is our domain. And for simplicity, let's just suppose we have linear basis functions. Okay? And suppose our exact solution is the following. Let me go to a different color here. Our exact solution I'm going to plot in green, okay? Let's suppose this is our exact solution, okay? All right. This is our exact solution and what we're talking of doing here is following. We are talking of we're talking of looking at these nodal points. And identifying the value of the exact solution at the nodes. Okay? And then we're talking of so, so this, so this is u, the exact solution, okay? All right, and what we are, what we want to do now is identify the u tilde h. Okay? U tilde h, remember, is a member of S h, right, which means it inherits the representation that we get from our basis functions. If we are working with linears here, u tilde h can at, can at best be linear for each element. However ,we want it to be such that actually hits the exact solution at the nodes. Okay? For many of the other elements it's, it's not going to be very different from the exact solution because perhaps the exact solution is not too different a linear, sorry, it's not too different from a linear solution elsewhere. Okay. You see that there is a difference, right? See, this is our u tilde h. And this is what we mean by saying that u tilde h is nodally exact, right, it's exactly the nodes. It hits the exact solution in the nodes, but it is a member of S h, so over an element in this particular case it is as, at, at, it is, it is, it is linear. Okay? And let me go back to my red here, okay. So this, this is just plotting up a solution up here. All right? Okay, so this is what we mean by saying u tilde h is nodally exact. Such a function, u tilde h, is what is often called an interpolate of the exact solution. Because what we are doing is taking the exact solution at the nodes, and then using the basis functions to essentially interpolate between those exact nodal values to get u tilde h, okay? U tilde h is nodally exact and is also called the interpolate. Okay? All right. Okay, and just remember that u tilde h belongs to S h, right, and that's re, reflected in the fact that it is linear over elements. In this case we are considering linears, okay? So we're considering, in this particular example only linear basis functions. All right, okay. So this is what we have. Now, here is a result. The theory of Sobolev spaces, right, or, or mathematics of Sobolev spaces, functional analysis on Sobolev spaces, gives us the following so called interpolation Error estimate. In Sobolev spaces. Okay, the interpolation error estimate is for the following quantity. It holds for the m norm, the h m norm of the difference between this interpolate and the exact solution. Okay? It's important to note why this is called the interpolation error, right? So the term on the left is my interpolation error. Interpolation error is this, okay? U tilde h minus u. The reason it's called interpolation error should be pretty obvious, right? And what is that? Think about it. What it is saying is that, suppose you had the exact solution, right, and at the nodes you were to hit the exact solution, which is what U tilde h does, right? But then because of your choice of a finite dimensional basis over the element, you are now interpolating from that exact solution. But since you are interpolating with a finite dimensional basis, you fail to hit the exact solution over an element, right, most prominently here. Therefore, this difference that we are looking at is indeed the interpolation error, okay? All right. So the Sobolev, estimate here, and, and I'm, I'm just going to state it, we're not going to prove it, is the following. It is that this interpolation error estimate is bounded from above by c as a constant. h e is our element size. Okay, and in these error estimates the we assume, or the error estimates have been derived for the case in which we have a uniform element size, okay? So the c h e to the power of a number alpha, which I will define in a little bit, okay, c h e to the power alpha times the h r norm of u, okay? So what do we have here? U tilde h minus u is the interpolation error. It's the error incurred by the fact that we're using finite dimensional basis functions even if we had the exact solution at the nodes, okay. H e, as before, is the element size. c is a constant. r is what is called the regularity of the exact solution. Okay? So note that when we're talking about u r the r norm of u, we're seeing that yes, if we were to square integrate the exact solution and also square integrate its derivatives up to r, right, up to its rth derivative, we would get a quantity which is bounded, right, and that is the r norm. Okay? All right, so let's just recall that this quantity, the r notm of our u Is a measure of smoothness, of regularity, really. r comes from regularity. Right, which is really a measure of smoothness Of u, all right? If you have the r norm of u, it means that you can take up to r derivatives of u and still have that quantity bounded, which gives you some sense of how small the u is, okay? All right, the next thing I need to tell you is about what alpha is. Right, alpha is the exponent in to which h e is raised, right, the power of h e. Alpha satisfies it, it's an exponent, obviously. And it satisfies. The following condition. Alpha is equal to the minimum of right, k plus 1 minus m, and r minus m. Well, what is k now? k, k is the order of the polynomial order of our finite dimensional basis. k is the polynomial order of the finite element, sorry, finite dimensional basis. Okay? All right. Let me do just well, let, let, let me rewrite the result here, okay. So that is h minus u interpolation error. We are computing the m norm of it, right? And this is lesser than or equal to c h e to the power alpha. Okay, so what this is telling us is that supposing in most cases, let us suppose that we have a solution just to fix ideas. Let's suppose that our exact solution is such that we can actually take derivatives of it up to a very high order and you know, we're, we're able to take derivatives of it to a very high order, which means it's very smooth, okay? What it means is that in this definition of alpha, r is going to be very big, okay? So even though we want to take, so, so, we, we are going to take m derivatives of the solution and we are interested in knowing what the m norm of it is. If r is very big, what it says is that the order of our of our exponent here, the size of our exponent is controlled by the polynomial order of the basis functions that we have. Okay? Now why does this matter? The question is, what happens with our interpolation error as we refine the mesh? Okay, if r is big enough, then alpha essentially reduces to k plus 1 minus m, right? So if r is large, okay, which means the exact solution is very smooth. Right, we have a nice, well-behaved problem. Okay? In this setting what we will see is that alpha being the minimum of those two quantities, is indeed going to come down to k plus 1 minus m, okay? And therefore, what this thing is saying, then, is that our result is that the, the interpolation error. The m norm of the interpolation error is lesser than or equal to c h e to the power k plus 1 minus m times this r norm of u. But then if u is very smooth, we expect that the r norm is not very big, right? We won't get a very large number when we integrate the functions, to integrate the function and its r derivatives up to r, okay? But that tells us then that now as as h e tends to 0, right, if our quantity k plus 1 minus m is greater than 0, what happens with the interpolation error? What happens with the norm of the interpolation error? It vanishes, right? Okay? It vanishes at the rate k plus 1 minus m, right? So it tells us that any nor, that, that this h, sorry, this m norm that we want to compute of the interpolation error also tends to 0 at the rate k plus 1 minus m. Okay? So as we refine the mesh, as we make h e smaller, right, remember h e going to 0 means we are looking at mesh refinement. We are going to smaller and smaller elements. Okay? So what the Sobolev estimate tells us is that yeah, as you refine the mesh eh, if you were to look at this interpolation error it will vanish, provided, provided what? Provided that number is greater than 0. How can we make that number greater than 0? We can make the number greater than 0 by taking a higher polynomial order, right, or making sure that we are not taking too many derivatives in computing the norm. So we don't, if we are not looking for very high m norms, right, we're not looking for taking many derivatives when we compute the m norm, right, then this holds. Right, and this is an important thing to know. This, this property, note we should note that this property comes to us directly from the Sobolev space. Right, we're not saying anything yet about the finite element solution. This result is a property of our Sobolev space S h. Okay? How is it a property of our Sobolev space S h? Well, that's what determines the polynomial order k. Right? That is determined by, by the space we have picked for S h, right? So it says as long as you pick that to be high enough, and as long as you're not looking for too many derivatives in this interpolation error it does converge. Okay? No talk yet about the finite element solution. Okay? That will come next. And that will come in the next segment.
