So, we'll proceed. What did in the last segment was look at high, the behavior of high order modes and also introduce this notion of consistency of the finite difference approach to time discretization. I ought to mention that proof of that, those results for consistency, and order of accuracy are obtained by essentially looking at the time exact solution, carrying out Taylor series expansions, and finite Taylor series expansions, and then, and then manipulating them. It's a, it's a fairly straightforward exercise but somewhat tedious, okay? So, in this segment we are going to essentially move to the end, end game for this particular topic. We are going to look at convergence. Okay, so we are going to look at convergence. Of. Time discrete. Solution. All right. In order to talk of convergence, as you may recall from our early treatment of error analysis for the finite element method itself, we need first to identify what our error is? Okay. So we define the error, and as we've done before in the context of the finite element solution itself we will define the error as the time, as the discrete solution, right? But in this case, we look at the discrete solution at n plus 1, minus the time exact solution at n plus 1. Okay? All right. Now this is what we define as e at n plus 1. Okay? Now, e n plus 1 is of course, is just another vector in the same space. Right? Right, just as d and t are sorry, your d n plus 1 n d at tn plus 1 are vectors and this rndf space. But what that means is that because we have a basis our orthonormal basis, we can expand e as well in that basis, right? So we can essentially carry out the modal decomposition. Of e at n plus 1. Right. And that model of decomposition essentially is that e at n plus 1 equals sum over m, e m for the mode at n plus 1, right, those are our scalar modal coefficients times psi m. All right, just as for any other vector. Okay, now how do we pose the question of convergence, right? We say that something has converged, right? Or we say our solution has converged, have converged. Convergence Right, is the requirement that as limit as the limit of n plus 1, tending to infinity, right, as we take more and more steps, right. What we expect to see is that e n plus 1, dotted with M e n plus 1 equals 0. Right, that limit is 0. Okay. Can you tell me why it is okay to, or can you think, why it is okay to put M inside there? Normally one would say that well, the error has to vanish, right? But why is it okay to put M in there? It's because M is positive definite. Right? All right, so let me just state this here. Since. M is positive definite, all right. What we see is that e n plus 1 dotted with M e n plus 1, right, equals 0, if and only if e n plus 1 itself equals 0. Okay? Only if, it's only when that vector itself is equal to 0 that, that quadratic product is equal to 0, since M is positive definite. And so, and this is why it means that well, yes, if, if this particular limit is tending to 0, it has to be that e itself, e n plus 1 itself is tending to 0. Okay, so this is why it is okay to consider this, as our convergence criterion. All right. But a more importantly, we are not going to work in this form. Right, we are not going to work, with e dot with M e. We're going to work in terms of modal coefficients, because our entire analysis has been in terms of modal coefficients, right. But it's okay, because I'm going show that for this quantity tending to 0. Right. This quantity that I put a brace on, okay, that quantity tends to zero, only modal coefficients themselves go to zero. Okay. So let me, I've told you what I'm going to show you. But in order to write it out, let me say the following, but not the following. Okay, e n plus 1, dotted with M, e n plus 1 equals sum over m, e m, n plus 1 psi m, dotted with M sum over l, e l n plus 1. Psi. Right, and we're going to put parentheses here. Okay. Each of those sums is for one of those e's right, in the quadratic product. All right? Okay. Well, we know how this works now. This is equal to, let me see how I want to write it, right. It is the sum over m and l, right, of, e n, n plus one, psi m dot M psi l. All right, and for no particular reason I'm going to put parentheses around that, and here I have e l n plus 1. All right? However, we know that this is delta m l. Because of the order normality of the psis with respect to m. Right? Well, if that is the case, we know that, this product that we started out with, e, n plus one, dotted with m e, n plus one, equals sum over m, e m n plus one. E m n plus one. All right, essentially what has happened there is the chronicle delta, has been used to, to turn that l index into an m index, right? So this is basically just the Euclidean norm if, with respect to the psi basis, right, so this is sum over m, e m n plus one, the whole square. Okay, so now it's clear that if this quantity is standing to zero it means that the modal coefficients, that sum itself has to tend to zero. Right, which means the modal coefficients themselves have to go to zero. Okay? So what this implies, finally, is that, limit, n plus one, tends to infinity. E, n plus one, dotted with M e, n plus one, equals, essentially limit as n plus one tends to zero, sorry, tends to infinity. Of the sum. Okay? All right? So clearly, if this, has to be equal to zero, if this limit is equal to zero, it just means that, each, every single modal coefficient, tends to zero. Okay? All right, so limit n plus 1 tends to infinity. E n plus 1. Dotted with M e n plus 1, equals zero. If and only if, each E m n plus 1, equals zero in the limit. All right? So, it means that it's okay to just look at what happens with the modal coefficients. Okay? All right, so. Study convergence. Of modal coefficients, right, E n plus one. Right and now dropping the explicit, mention of the modal, index. Right I've just dropped n. Okay? All right, well that's what we're going to do. Okay, so, how do we set it up? We set it up by going back and writing out our time-discrete, discrete equation, our modal time-discrete equation in the form that we set up in order to pose the question of consistency. All right, so here is what I say. Consider. The time discrete equation in the following form. D n plus 1 minus A d n minus L n equals zero, okay? And consider the, the exact equation also forced into this form. We know, however, that the exact equation does not cooperate, and. Give us a zero right hand side, instead we get delta t, tau at t n. Okay, right, what I'm going to do now is subtract the second equation from the first, okay, so I'm going to change signs. Okay, and now when I add those two together, I get on the left hand side, right, d n plus one, minus d t n plus one, is essentially our e, at n plus one. Right? Okay? Right? It is the, error in the corresponding modal coefficient, right? And here I get minus A e at n. All right, the Lns cancel out, and I'm left with minus delta t tau at t n. I'm going to rewrite this as an expression that will allow me to write out a recursive formula for the error. Error at n plus 1 equals A times error at e minus delta t. Tao at tn, okay? I'm going to use recursion now, and write from here e at n equals A e at n minus 1 minus delta t tau t n minus 1, okay. What that implies is that on substituting e as written here in that expression, right? I get e at n plus 1 equals A square e at n minus 1, minus A to the power 0 delta t, tau at t n minus 1, okay. Right. Minus A to the power 1 delta t tau tn minus 1, and I just realized that here it is just tn. All right, I can take yet another step, okay. What, if I were to now rewrite an equation for e n minus 1. All right, I would get A e n minus 2, minus delta t tau at t n minus 2, okay? And making the substitution, right? You see how this is going, right? What I would get is that e, sorry. E at n plus 1 equals A cube e n minus 2, minus A to the 0 delta t tau tn, minus A to the 1 delta t tau tn minus 1, minus A squared, right. Minus A squared delta t tau at tn minus 2, okay. Now one can go on with this, and I'm sure you've all done it at one stage or the other with similar expressions. When you go all the way back to, the zeroth step, right, so the initial time, we get e at n plus 1 equals A, to the power n plus 1. Okay, right. A to the power n plus 1, e at 0, okay. Minus, sum i going from 0 to n, A to the power i, delta t, tau at tn minus i, right? You can check that this is what the recursive formula reduces to, okay? All right, let's stare at this. What is e at 0? What is the error at time t equal to 0? It's equal to 0, right? Because we've made sure that our initial condition is actually obtained from the time exact solution, right? I mean, at the initial time there is no error, right, because we fixed the initial condition. [SOUND]
