All right. What we're going to do in this segment is complete our stability analysis of our Time-Discrete, Single Degree of Freedom, ODE. All right? Or a single degree of freedom equation. Okay. So let's recall where, where we got. We got as far as the amplification factor. Okay. And that amplification factor gives us dn plus 1 divided by dn equals A, this is the amplification factor. And it is 1 minus 1 minus alpha delta t lambda h divided by 1 plus alpha delta t lambda h. Okay? All right. Now what we want to consider is the fact that what we're going to use in our analysis is the fact that alpha lies in the, in the closed interval, zero to one, right? Delta t, which is our time step is greater than zero. Okay? And lambda h, we've agreed is greater than or equal to zero. Okay? All right. Now the stability condition. Or the stability criteria is that the magnitude of A is lesser than or equal to 1. Okay? All right. So this is what guarantees decaying response, right? From one time step to another. Okay? By the way, the fact that we arrive at this condition is actually the, is what we call a linear stability condition. Okay? And the linear aspect of it comes from the fact that we have indeed a linear problem here. And for linear problems, one can state the stability requirement by just saying that well, if your solution does not grow from one time step to the other, you have a, you have a you have a stable problem. Okay? When we go to non-linear problems, we can't quite use that. Because the physics of non-linear problems can actually lead to solutions growing, right? Over certain regimes, right? And that, that is the right physics for those sorts of problems. So let's get back to this. So, if this is what we have let's see how we can, we can extend this. What this implies, of course, for us directly is that minus 1 should be lesser than or equal to A should be lesser than or equal to 1. Okay? So, A has to lie in that interval. All right? Okay. So let's look at this. So what that means then is that minus 1 is lesser than or equal to 1 minus 1 minus alpha delta t, lambda h divided by 1 plus alpha delta t lambda h is lesser than or equal to 1. Okay? Now because of the conditions I put up here, right? We are guaranteed that our denominator is positive, right? So that says, if we are now what, what this lets us do is safely multiply through by that by the denominator and we get here. Minus 1 plus alpha delta t lambda h is lesser than or equal to 1 minus 1 minus alpha delta t lambda h is lesser than or equal to 1 plus alpha delta t lambda h. Okay? All right. Okay. So to move on, let's just go to the next slide and let's look at the right-hand side of that inequality. Okay? So consider in form, right? 1 minus 1 minus alpha delta t lambda h is lesser than equal to 1 plus alpha delta t lambda h. Okay? We can consider this, we see that these two cancel out. Okay? And we manipulate things and what we see here is that we get alpha plus 1 minus alpha delta t lambda h has to be greater than or equal to 0. Okay? And we do this by simply moving what remains on the left-hand side of the inequality on the first line over to the right-hand side. Okay? Now those cancel out, all right? And we are left with sorry. There's a delta t here. What we are left with here is the delta t lambda h is greater than or equal to zero. Okay? That's what the right-hand equalities says. All right. But this is always satisfied. Okay? All right? Let's look now at what happens. And, and in fact, this is always satisfied regardless of alpha. We satisfied for, for all alpha, all right? Because alpha doesn't even show up in this final form. All right. Now let's consider the left-hand side inequality. Okay? And that left-hand side inequality is that minus 1, sorry, minus 1 plus alpha delta t lambda h Is lesser than or equal to 1 minus 1 minus alpha delta t lambda h. Okay? All right. So now let's work with this a little. When we work with this a little, let's, let's, let me rewrite this by moving This term over to the right-hand side and, and, and rewrite them. So when we do that, we get 1 minus 1 minus alpha delta t lambda h plus 1 minus alpha delta t lambda h is greater than or equal to 0. Okay. That is the condition which must hold and this condition you note involves alpha, delta t and lambda h. Okay? Continuing to work with it, this implies that two plus 2 minus 1 all right. Minus two alpha, hm. What's going wrong here? Alex, you may need to just give me a pause here. I've got a two here. That over back, okay. Alex, I'm going to backup, I made a mistake here. Go back to the beginning of this last equation. Okay. I note that I need a plus sign there. Okay. Put that plus sign and go ahead and now, I get 2 plus 2 alpha, right? One alpha coming from here and another alpha coming from here. 2 alpha minus 1 times delta t lambda h is greater than or equal to 0. Okay? That is the condition we need. Let's see what else we can do with this now. I could write this out as 2 has to be greater than or equal to 1 minus 2 alpha delta t lambda h. Okay? All right. All right. Now let us look at whether this can, under what conditions this may be satisfied. Okay?. Now let's save this and move on to the next slide, okay? Let's consider the following cases, okay? So case one. All right? And for case one, let us suppose that alpha is greater than or equal to one-half. Okay? What, when, when that happens, what that says, what that means is that 1 minus 2 alpha, delta t lambda h is always what? Right? It's always lesser than or equal to zero, okay? Which implies that yes, if alpha is greater than or equal to half, 2 is always greater than or equal to 1 minus 2 alpha delta t lambda h. Okay? Right? This thing always holds, holds unconditionally. Right? And by unconditionally, what we mean is that if alpha is greater than or equal to half, our stability condition holds for all delta t, right? For all delta t greater than zero, right. Okay. So we have a condition, we have a, a situation of unconditional stability of our method. Okay? So this is unconditional stability if alpha is greater than or equal to half. Okay. So, in particular, observe that alpha equals half, which is the Crank Nicholson method or the midpoint jewel as I sometimes call it, as a lot of other people call it. Right? Or even alpha equals 1, which is backwards Euler. Right? These are both unconditionally stable. You can take a time step as big as you like and you're okay. Okay? A solution may be terrible, it may be very inaccurate, but you're still stable. The solution will not blow up, right? The amplification factor will remain, the magnitude of the amplification factor will remain bound by, by one. Okay? Case two is the interesting one. Okay? Case two is that alpha belongs to zero comma half not including the half, right? Okay. All right. Which is basically zero is lesser than or equal to alpha is less than a half, okay? This is the other case. All right? So now what happens is that we have a situation of that's okay. So, so in this case, we have a, we must still have all right. So we, we are trying to see whether 2 is greater than or equal to 1 minus 2 alpha delta t lambda h. Right. This condition needs to be satisfied for alpha, such that zero is less than or equal to alpha is less than half. Okay? All right. This is what we're trying to look at, okay? Now observe that since alpha lies in this range that I've written on the right this parenthesis is always what? Is it all, what can you say about the quantity in that grid emphasis, in those parenthesis? Right? It's always positive. Okay? So now we get a condition, where we see that on, on, as far as delta t is concerned, delta t must be lesser or equal to 2 divided by 1 minus 2 alpha. We're able to divide through by 1 minus 2 alpha, because it is indeed greater than 0. Okay, so it is a positive quantity. 1 minus 2 alpha is positive. But then we see the delta t has to be lesser than or equal to 2 over 2, 1 minus 2 alpha, times lambda h. Okay? So this is a situation of conditional stability. Okay? So the algorithm is stable, right? It can be stable. But it depends upon the size of delta t, okay, conditional stability, which means conditional upon delta T. Okay? So delta T has to be bounded by something. Delta T cannot be too big. Sort of makes sense, right? You have an algorithm that is not entirely stable, that is not always stable. In order for it to stable well, take very small delta t's, right, take very small times 6. Ok, so, an example of this is of course alpha equals 0. Right, which is forward Euler. Okay. All right, now, to end this segment let us consider what else can happen here with delta t. Remember that lambda h, because of the nature of our Eigenvalue problem, lambda H if you ask yourself, what is the order of lambda h? Okay? And where does lambda h appear? Order of lambda h depends upon the Eigenvalue problem, on the Eigenvalue problem. K psi, where psi is the mode shape corresponding to this particular mode. K psi equals lambda h, n psi. Okay? Now, what this suggests, is that lambda h is of the order of, the order of lambda h is of the sort of terms that are like M inverse K. All right? Right. So it's with the order of the norm of terms in M, inverse K. All right. Now, I want us to think about how do M and K depend on the element size? Right, because these after all are obtained by the spatial discretization. Okay? So, recall M is of the order integral over omega e, and then we have the sum over e, sorry, we have the assembly over e and all that. We have the assembly over e, integral over omega e of terms of this type, right? We have, we have shape functions, right? So this is really of this is really like M A B in a particular element. Okay, so I can actually get rid of this, right. I'm just doing this element twice, sorry, I'm changing things around a lot. Okay, element twice, this is what we have, right, integral over omega E, N, A, N, B, D V. Okay, K AB from an element is of order, integral over omega e NA,X NB,X right? These are derivatives. Right? So there are derivatives involved in KAB and yes. KAB may also have a so this is xi, xj, K, kappa here may have a kappa ij. Okay? Dv, all right, and yes the mass matrix may also have a specific heat there. Right, or row could be one if you are doing diffusion. Okay? What is, what is the order of these sorts of terms? These sorts of terms, because their shape function derivatives, are of order one over h. Okay? As a result, what we're seeing is that MAB is of order may be 1, right. But KAB is of order h to the minus 2, okay. As a result, the order of lambda h, okay, is h to the minus 2, okay? In terms of the element size. Now what does that say for delta t? Delta t has to be lesser than or equal to 2 over 1 minus 2 alpha lambda h. Well, this lambda h is like h to the minus 2. Okay? As h gets smaller, our restriction on delta t gets more and more stringent. Okay? So the remark that I will make here is that restriction On delta t, gets more stringent. As h tends to 0. Essentially as we refine our finite element mesh, our spatial discretization, we see that it has an effect even on our temporal discretization if we're working with a conditionally stable method. Right? For the time integration, okay. So, this is how the finite element discretization also makes its presence felt in the time discretization when one is working with a semi discrete method and happens upon a conditionally stable problem, conditionally stable method. All right. We'll end the segment here.
