Welcome back. So at the end of the last segment, we wrote out our ODE in its discretized form. And we've also introduced the the Euler family. All right, we've met the Euler family, so to speak. All right. What you're going to do now is go ahead and apply it to our ODE. And the approach we are going to take is one of revisiting this discretized form of our ODE, okay? So go back to our time-discretized ODE. All right. We write it out as M v at n plus 1 plus K d at n plus 1 equals F at n plus one, right? Where what we can say now is that we can invoke our Euler family as applied to our definition of v, okay? So what this means is that we're saying here that d at n plus one equals d at n plus delta t v at n plus alpha, okay? Where v at n plus alpha equals v n plus 1, multiplied by alpha, plus 1 minus alpha times v at n, okay? And in writing out these two equations you can see that we have essentially applied this idea of the Euler family in there, right? Okay? All of this, of course, to complete our specification of an, of even our time-discretized form of the ODE. What we are saying is that of course with all this, we are given the initial condition, all right? And that initial condition is d 0. Okay? We know what we mean by d 0. Right, in the previous segment we explicitly said what d 0 is. All right? So now with you know, in general you, you really see this idea of time stepping. The idea is that given things at d n, right? Given quantities at t n, sorry. Given quantities d n and v n at t n this is a scheme for going from n to n plus 1. Okay. And it utilizes, it uses the, the initial condition here. All right. Now, there are two sort of canonical approaches to solving this time-discrete, discretized ODE. And we'll, we'll look at both of them in turn. The first is what is called the v-Method. 'Kay? Somewhat unimaginatively. All right. Here is the v-Method. In order to app, apply the v-Method and actually to apply the other method as well, we go back and look at this. Form of the problem and look at these two equations and do the following. Okay? We first note that. Then we can rewrite now d n plus 1 as d n plus delta t, times v n plus alpha but v n plus alpha is also written up for us here. So we'll write that as alpha v n plus 1, plus 1 minus alpha, v n. Okay, all right, so that is the first step, combining the two sort of the equations that define the Euler family. Now, in the next step what we're going to do is to rewrite this as rewrite the right-hand side as d n plus 1 minus alpha delta t. V n, plus alpha delta t v n plus one. Okay. Now what we've done here, if you observe, is rewrite d n plus 1, in terms of quantities that depend only on quantities at n. Okay? So let's denote this as d tilde at n plus 1. Okay? Now, we are getting into the realm of what are called predictor-corrector methods. 'Kay? And the idea is that if we want to calculate d n plus 1, right? We have this formula for it. We predict this, right? We say that well this is our guess for what it is, okay? And in general that will not be correct and we need to correct it with this term which is the corrector. Okay? So, written differently, d n plus 1, actually not so differently, d n plus 1 is equal to the predictor d n first 1 tilde plus the corrector alpha, delta t v n plus 1. Okay, so this is a predictor-corrector method. Okay. This v-Method that I'm talking about is one where what we do is return to our time-discretized ODE. Right, which is this one, Mvn plus 1 plus Kdn plus 1 equals F at n plus 1, right? We return to this equation, and we try to rewrite it only in terms of v, okay, and the way we can do that is to essentially use for dn plus 1 here, this predictor-corrector form. Okay, so substitute. Predictor-corrector form for dn plus 1, all right? What that does then is leave us with M v at n plus 1 plus Kdn plus 1 tilde plus alpha delta t Kv at n plus 1, okay? All of this equals F at n plus 1, all right? Now, I said that the reason we call this the v method is because we're going to try and rewrite our discretized ODE entirely in terms of v. Right, entirely in terms of vn plus 1. So that is indeed what we do, okay? So we get here this implies M plus alpha delta tK vn plus 1 equals Fn plus 1 minus Kd tilde at n plus 1, okay? We solve this equation for v, for vn plus 1 and we now have the corrector, right? Because then we can add vn plus 1 from here, up there, and we get the corrected state, all right? Right, so, so basically we solve. It's this inverse acting on Fn plus 1, minus Kd tilde n plus 1, okay? Right, now I'm going to make a couple of remarks. Okay, and those remarks are the following, right? Now, the first remark is that if, instead of M, we used the, the lumped-mass, okay, and, furthermore, we use alpha equals 0, okay, this will be forward Euler. Okay? 'Kay, have this particular combination, and what we see is that for vn plus 1, which is lumped-mass inverse Fn plus 1 minus Kd tilde n plus 1, okay? Because the lumped-mass is a diagonal matrix, its inverse is trivial, right? So as a result we don't really solve equations here, right? We don't solve, we don't have to solve with a, a, a linear system of equations by, by somehow effectively generating that inverse, okay? So this is what we call an explicit method. Okay? In contrast we have if alpha is not equal to 0, okay, all right, then we have in that case even if our mass is lumped, we still have equations to solve, okay? So this, this now an implicit method. All right? Okay, so these are two, two remarks to make and what we will do is move on and proceed directly to the other canonical solution method, which is the d-method. Okay? And the d-method once again, we could look at our predictor-corrector form. But now, we sort of turn this around to eliminate v from our discretized ODE, all right? What that implies then is we write here we get vn plus 1 equals dn plus 1 minus dn plus 1, tilde, okay, and divided by alpha delta t, okay? This of course, holds only if alpha is not equal to 0, okay? All right and in this case we get v from here and then we substitute it back into our discretized form of the ODE, okay? And. Okay? So, when we substitute in our discretized form of the ODE, M will be n plus 1, we had dn plus 1 minus dn plus 1, tilde, divided by alpha delta t, right? That is our approximation for vn plus 1. Well that's our expression for vn plus 1 using the Euler family. And here we have Kdn plus 1 equals F at n plus 1. And this again we reorganize by simply multiplying through by alpha delta t, okay? So we get M plus alpha delta tK. D at n plus 1, equals alpha delta t. F at n plus 1 minus Md tilde at n plus 1, okay? Now in this form of course you realize if we can get to the final form it doesn't matter if alpha is equal to 0. Therefore this works ev, even though I, I, I wrote that in, in, in defining vn plus 1, right? Alpha even though I wrote that alpha is set equal that, alpha cannot be equal to 0, right? We still get back our we will still able to actually solve the equation and move on, okay? So this would be the v-form and now what we get here is dn plus 1 equals M plus alpha delta tK, dn plus 1, sorry. Inverse here times alpha delta t Fn plus 1 minus Md tilde, n plus 1, okay? For this sort of method if we use a lumped-mass, right, we use a lumped-mass here. What is seen is that we require less effort to set up the right-hand side, okay? All right, if we use a lumped-mass here, there are fewer operations, not less, less that limit that fewer operations. Fewer operations to form the right-hand side. Okay? But the actual equation solving effort in the two methods, the v-method and the d-method is the same, all right? So here, here we have it our you know, this, the, these are the two methods by which we can set up the the, the solution of this problem you observe that you're, you're still solving exactly the same, sol, the same serof, the same set of equations so the methods are, are, are, are really equivalent. It's just a matter of which, which variable you attempt to eliminate first, okay? [COUGH] right, so what we can begin to do now is to proceed to the to actual solution and well, sorry, we, we've already looked at the, at the solution techniques. What we can do now is to, is to set ourselves up to carry out an analysis of these methods, okay? And that may perhaps be done best in a separate segment. So when we return, we'll take up the analysis.