All right. What we saw on the previous segment was a fairly quick statement of the stability conditions for the Newmark family, right? And we saw the explicit results for some of them. We didn't derive any of them. What I'd like to do in this segment is outline the, the sort of analysis that leads to those, to those conclusions. We won't get into a completely detailed step-by-step derivation of the results. Unlike what we did for the parabolic problem, okay? But we'll sketch out the, the approach. Okay, so the stability analysis here, as in the case of the parabolic problem. Is based upon examining a particular object. Can you recall what object we examined in the case of the parabolic problem? We examined the amplification factor. Here, too, we have an amplification factor, but it is a matrix. Okay. So, the stability analysis is based on, an eigenvalue analysis of the amplification matrix, all right? Okay. And, the way we proceed with this is to, if I remember that the amplification matrix is what we denoted as A. Okay and you also recall that the amplification matrix is what showed up in this formulation of the problem as here, y n plus 1 equals Ayn plus Ln when we consider the full n homogeneous problem, okay? So what we're talking about is analyzing this, and all our detail of the Newmark family is sitting inside there, right? The particular values of gamma and beta we've chosen, and so on, okay. Here is the condition, okay? We define what is called the spectral radius. The spectral radius of A, right? And we denote that as we've used rho quite a bit, so let me use something else here. Let me just say r, okay, spectral radius A, r sub r function, okay, okay? This is defined as the maximum over i, okay? The maximum over i of lambda i of A, where those lambdas are essentially the eigenvalues of this two by two matrix. All right, and because it's a two by two matrix, of course, i just runs over one and two, okay? That's probably not even worth using an index there. Okay, so let's say that max i equals 1,2 right? Essentially it's the maximum eigenvalue, okay? Not just the maximum eigenvalue, but actually it is the magnitude of it, okay? Where we have accounting for the fact that the matrix A may not always be symmetric, and therefore, it could have complex eigenvalues, right? Accounting for that, we write our spectral radius as being defined as the max i equals 1, 2. Now, that magnitude that I wrote up there is properly the square root of the product of lambda i and its complex conjugate, which is going to be denoted as lambda i of A bar, okay? Where that bar implies the complex conjugate. Of lambda i A, okay? That is our spectral radius. Now the condition for stability requires that. All right, it requires that r, the spectral radius, should be lesser than or equal to 1. Okay, the spectral radius is defined there, should be lesser than or equal to 1. I'm going to say a little more about this condition. We have r can be lesser than or equal to 1, if lambda 1 and lambda 2 are distinct. Okay, it turns out, that if lambda 1 and lambda 2 are distinct the condition that we get that r can be lesser than or equal to 1 involves the fact that the eigenvectors of A. Are linearly independent. Okay, on the other hand r has to be strictly less than 1, right? Not lesser than or equal to 1, it has to be strictly less than 1, if lambda 1 equals lambda 2, right? We have repeated roots, okay? And in this case it turns out that the eigenvectors. Of A, are linearly dependent. Okay? I'm going to do very quick demonstration of why this is the case, okay? So, let's look at the two cases, okay. Let's first look at what happens if they are linearly independent eigenvectors. Okay. If they are linearly independent eigenvectors, then let's look at what happens for the homogeneous problem as we go from one time step to the other, okay? Essentially, what we see is that yn plus 1 equals A yn, right? But then yn is equal to A yn minus 1, and so on, right, yn minus 1 equals, tatatata, right? Goes on. So what we are seeing here is that with every step is getting multiplied by itself, right? If you just make these substitutions in here, we see that, right? Okay, so what we are seeing is that, after a certain number of steps, we're seeing that yn plus 1 equals A to the power n plus 1 times y0. All right? So, what we need to worry about is what is happening with A. All right, as it gets multiplied by itself, what are the powers of A? If we have linearly independent eigenvectors, one can show that A, okay, can be written as some matrix P, times a two by two matrix, which is lambda 1, 0, 0, lambda 2. P inverse, okay? As a result, we get from this, we get A to the power n equals P, lambda 1 to the power n, 0, 0, lambda 2 to the power n, P inverse, okay. So now you note that even if lambda 1 were equal to 1, all right this sort of a form stays well bounded, okay? All right, okay. This sort of a form stays bounded, right? So the amplification that is applied to the initial condition to get say, the nth time step solution does not get unbounded, okay? Things are different if you do have, if we have linearly dependent eigenvectors. In this case, the best we can do is write A as, let's say, some other matrix Q, two by two matrix, lambda 1, 1, lambda 2, Q inverse. Okay? This is the case of the linearly dependent eigenvectors, okay? Now, if you go through the process now, and calculate a to the power n. >> Okay, what you get is a form where you get Q, lambda 1 to the power n. You get lambda 2 to the power n, and all is looking good except for the fact that here you get n lambda, well lambda 1 is equal to lambda 2 here, okay. So it doesn't really matter. Let me just do this. In the case of linearly dependent eigenvectors, you have lambda 1 equal to lambda 2. So, you get n times lambda 1 to the power n minus 1, Q inverse. And now, do you see a problem as n gets large? Okay, what you note is that if you look at what happens here? And you have lambda 1 equal to 1. Okay? You see that the off diagonal term becomes n, okay? Right. In that case this term becomes n. Right? And then as you go to higher, and hard, higher, as you go, as you advance in time steps you have this sort of gradual sort of tendency towards unboundedness. Okay, so what happens in this case is that the off diagonal term. Diverges, as n. Okay? All right. Further analysis of this problem is based upon essentially solving for lambdas, right? So. Solving for the lambdas, right? The equation that we need to solve for the lambdas is the following. Lambda square, remember this is just the characteristic equation that you use to solve for the eigenvalues of a matrix, all right? So we will write it as lambda squared minus 2, A1, lambda plus A2 equals 0, okay. And here A1 equals one-half trace of A, and A2 equals the determinant of A. Okay? All right? With this form, it's just a simple quadratic equation, right? So we get lambda 1, lambda 2, are equal to what is it? A1 plus or minus square root of A1 squared minus A2. I believe, let me just look at that. Yeah, okay, so these are lambda 1 and lambda 2. All right. Now, here is the, sort of stability condition again written in terms of A1 and A2. Okay, because of course stability depends upon the values of lambda, but then since we have lambda 1, lambda 2 given by these conditions for A1 and A2, we can write it out in terms of A1 and A2. Okay? The conditions that we get are the following. We get minus A2 plus 1, divided by 2 is lesser than or equal to A1 is lesser than or equal to A2 plus 1 divided by 2 if the magnitude of A2 is less than 1. Okay? All right. And. Okay. And otherwise, we get minus 1 is less than A1 is less than 1 if the magnitude of A2, sorry, it's not just the magnitude of A2. It's A2 itself. Here too, it's just A2. If A2 is equal to 1. Okay, one can plot this thing up in this result up in an A1, A2 space, and here's what we see. If this is A1, and here we're plotting up A2. Okay? Let me see. I think, I'm going to mark some critical points here. These are the points 1, 1, minus 1, 1, and down here, I have the point 0 minus 1. Okay, in this, we have between 1, 1 and 0, 1, that line, and here we have that segment, okay? The first condition, this one okay, holds everywhere except for that line. Right, because that dashed line is A2 equals 1. Okay, so this sort of inverted triangle that I've drawn is the acceptable region. Okay for stability, if A2 is less than 1. Okay, if A2 equals 1, right, then what it says, is that on that line, it's got to eliminate, it's got to be, you've got to leave out those two points. Okay? That is the region of stability that we have here. Okay. Okay, we can end this segment here.
