All right, so remarks. Backward Euler All right, which is alpha = 1. Damps out, I don't like doing damps, I am going to say dissipates. Dissipates, High-order modes. Okay, and this is what we call numerical dissipation. Okay, and because as I've observed this has the same high-order behavior as the exact equation it is often preferred, okay? Numerical dissipation, right? And let me also say here, it is similar, To, Time exact equation. Right? The second note here is that there really isn't much to say about forward Euler by the way. It should come as no surprise to us that the amplification factor for forward Euler tends to minus infinity is lambda h delta t gets large. We've already seen some evidence of it. And in what way have we seen evidence of it? We've seen it in the fact that if delta t gets too large, we know that the forward Euler method is no longer stable. Right, and that is also reflected in the amplification factorgetting unbounded, in this case, minus infinity, as delta t gets too large. Right, remember lambda each delta t can get large if either lambda h or delta t get large. If you have a large time step, we know that there's a tendency for forward Euler to blow up, so to speak, right? And we see that in the amplification factor tending to minus infinity, okay? So let me state that, though, forward Euler, Forward Euler which is alpha = 0. Has unbounded, Lambda has unbounded A. All of this by the way is in the context of high order modes. So I should say up here this is all for high order modes. Okay, so everything I'm writing here applies to high-order modes, okay? All right, so this should come as no surprise to us because we know that forward Euler has this tendency to sort of lose it, okay? Remark three is that the surprising thing is that the mid point rule, Or it may be a surprise, right the mid point rule which is alpha = one half, Right, has as we've seen A tending to -1, okay? What does this mean? What this means is that dn + 1 for the high-order modes, right? If this is a high-order mode I'm writing out, right? Okay, dn+ 1 is the model coefficient of a high order mode, what we're seeing is that dn+ 1 = -dn. This leads to oscillatory behavior in the high order modes. And it's not uncommon to see solutions from the midpoint rule, which, with respect to time, All right, if you're plotting certain modes with respect to time, what we will see is that if this is t1, t2, t3 and so on. All right, what we may very well see is that you get, essentially, oscillatory behavior. Okay, something that's done is to simply form a time average, okay? Right, if you take a time average you basically damp out, well you don't truly damp out, but you eliminate the effect of these oscillations in your post process solution. Right, time average the solution. Okay, to get around this. All right, this is important to know about the behavior of these methods in the high frequency limit, right, for high order modes. Okay, so we've looked at stability we've understood the behavior of high order modes. What we are going to do next is essentially prepare ourselves for talking about conversions, all right? And the way we do that is by first looking at the notion of consistency. Okay, so we look now at the idea of consistency. All right, in order to get to consistency, let's go back to our time descritized equation, okay? And that equation, as you may recall, is the following. It is dn +1, (1 + alpha) no let me back up a little here. Let me write it in fact in the following manner. Let me write it all the way back as dn+ 1- dn over delta t + lambda h, dn + alpha equals. Now up to a few seconds ago we were looking at the homogeneous problem, all right? Which we got to by turning off the forcing. We're not bringing back the forcing okay, and I'm going to write this as fn + alpha on the right-hand side. Okay, if you wonder what Fn + alpha is, Fn + alpha, right, for this particular mode is got by looking at F, Sub n + alpha, the whole vector, right. And essentially right dotting it. With whichever mode we're looking at, okay? If we were looking at mode l we would dot it with the lth mode, okay. But since we've agreed that we're going to drop the explicit mention of the model number, right? We'll just write it as Fn + alpha here. All right, now, let's multiply through by delta t and see what we get. We get dn + 1- dn + lambda h delta t. I'm going to expand out dn + alpha as alpha dn + 1 + 1- alpha dn = delta tFn + alpha. Working with this a little more, what we get is dn + 1 (1 + apha delta t lambda h)- (1- (1- alpha) delta t lambda h) dn- delta t Fn plus alpha = 0. As a final step I'm going to divide through by 1 + alpha delta t lambda h to give me dn + 1-. Now, when I divided this quantity by this I get back my amplification factor. So I'm going to jump a step and simply write that as A dn- delta t divided by 1 + alpha delta t lambda h Fn plus alpha = 0, okay? All right, and I'm going to take the final step of calling this quantity here L at n + 1. I think I'll call it. Doesn't really matter. Let me just call it L at n. Okay, all right? So this is how I want to write out my time discrete, model equation, right? Including the forcing. Now, the notion of consistency is the following. What would happen, if you were to take the time exact model equation and plug it in here, okay? So let me now write that as d at tn + 1 Is the, Time exact mode, Corresponding, To lambda h, okay? The particular value of lambda h gives us a certain mode and that's it. Okay, now the question we ask is what would happen if we were to take our time exact model solution and plug it back in here? All right, so what we are saying is that if we do that we get d(tn+1), -A d at tn, right? This is also the time exact model value at time tn, -ln. Okay, because the last term Ln does not depend upon the model values, okay? Now, this equation is equal to 0, right? The left hand side on this equation is equal to 0. My question is if we were to not plug in our exact solution here, right, as I've done. Will what I've written here on the left-hand side be equal to 0 or not? In general, it's not equal to 0, okay? In general, however, one can write it out as delta t times some quantity tau. Okay, tau, which in general depends upon the time. Okay, right? The reason we're writing it out as delta t times tau is that because we recognize that having started from this form of the equation. We've actually multiplied the ODE through by delta t. So we expect that whatever the right hand side is, it already has a factor of delta t sitting in it, right? So we choose to write it in this fashion. Okay, now this is the general form that we would get for the exact equation, right. It turns out that we can now make the following identification, okay. What we say is that, Sorry, So let me back up to the previous slide. Let me just see one more thing here. If we can show that our exact solution satisfies an equation of this type, okay? We had what we call a consistent method. Over it's more, no let me back up a little bit. It's not quite a consistent method this is a consistency condition. Okay, this is the consistency condition where we have a consistent method, if we can show further. That tau which I've written as depending upon the actual time. If you can show that tau is lesser than or equal to some constancy times delta t, to some power k, okay? Where, k is greater than 0, okay? If we can show this then we have a consistent method. Here's why. If we can show that this holds, one can say that in the limit as delta t tends to 0, right? What we will find is that d(tn + 1)- A d(tn)- Ln = delta t c delta t to the power of k, right? This thing also tends to 0. Right, this entire thing also tends to 0, okay, right? If k is greater than 0. All right and therefore what this is suggest is that yes the equation we're working with, the finite difference equation we're working with in time is one that when we plug in the exact solution though it does not immediately satisfy the exact solution, at least it does satisfy to the limit of vanishing time steps, okay? So in the limit delta t tends to 0, the time discrete equation, Are admits, The exact solution, Right and which I'm sure you will agree is a useful property for a method, right, for a numerical method. Okay, I should mention f course, that in that result, t of tau n is lesser than or equal to c delta t to the power k. c is a constant, and k is what we call k is the order of accuracy. Okay, and it turns out that k = 2, if alpha = 1half right, the mid point rule. And it's equal to 1 otherwise, Okay. All right, so the midpoint rule as you may expect because it does apply the algorithm between n and n + 1, gives you higher accuracy than any of the other methods. Backward Euler and forward Euler are down here. And any other method also has order of accuracy one. Only the midpoint rule is second order accurate. Okay, this would be a great place to end this segment.
