All right. In the previous segment what we looked at was, again, a sketch of the way we would approach the stability analysis for our modal equations for linear elastodynamics, all right. What I'd like to do here is actually take a step that we also took in the case of the parabolic problem which was to go from stability. Our understanding of stability to also very quickly cover high, high order models, okay? So, we've derived stability conditions on A. All right? Our amplification matrix. And in particular, what we said was that if A1 equals one-half of the trace of A. And A2 equals determinant of A. What we found was that the conditions are based upon lambda 1, lambda 2 equals A1. Plus or minus square root of A 1 squared minus A2. Okay? This is how we determine lambda 1 and lambda 2 from our characteristic equation for this matrix. And then we impose conditions at lambda 1 and lambda 2 can be, need to be lesser than or equal to 1. We've also understood when they have to be strictly less than 1. All right. So these are the stability requirements. Right, so we also said from here r, which is [SOUND] the maximum of the square root of lambda i and lambda i bar. Right? Where the lambda i bar refers to the complex conjugate. Right? Okay, and then we say, finally, r has to be lesser than or equal to 1, right? And we've understood when it can, when it needs to be strictly less than 1. All right, so this is how we go about our stability analysis. Now you recall that when we looked at the parabolic problem, we looked at the, the amplification factor, and also used it to tell us something more about the high order modes. Okay. And we saw that the different members in that case of the Oiler family did different things to high order modes. Okay. It emerges that in the case of this problem, also the amplification factor plays a role. And in order to damp out high order modes, okay. In order to damp our high order modes, high order modes. Are decaying right? If if our eigenvalues become purely real, right? Sorry [LAUGH] it's the other way, sorry. High-order modes are non-decaying, sorry, are non-decaying, if lambda 1 and lambda 2 are purely real. Okay? And therefore in order to damp out the higher-order modes, what we need is that the discriminant of this relation, right, should be negative. Right? So what we need is that A1 square minus A2, is less than 0 for damping or digging, of high order modes. Okay, so this is really a sort of parabolic condition and if we go back and now plot out this parabolic condition on our, A1 A2 space that we introduced at the end of the last segment, here is what we'd see. That is 1, 1. This point here is minus 1, 1. And this point here is 0,1. Okay. What we did do last time was observe that our stability condition gives us this sort of triangular region. Okay. In addition to this, it emerges that our condition for damping out of higher order modes requires that there is a parabola which takes on this shape. Okay. So we need to remain within that parabola, not outside of that parabola. So, this is the region in which we get damping of high water modes and okay according to that condition. Right, so it is in that region that we get, damping of high order modes. Okay. Let me see if I can get back to my original color here. Okay, the effect this has is the following. You know, the stability condition for the condition for, for, for stability independent of time step series, right? So, so really the unconditional stability, okay, condition is the following. It is that beta is greater than or equal to. Gamma over two. Okay? So this is unconditional. Okay? It doesn't quite make sense to say stability condition and then fit an unconditional stability condition, so let me say stability. Requirement here. Stability requirement. All right, all right, so this is the unconditional stability requirement. It turns out, however that we need, beta to be, not only greater than gamma over two. But, this requirement of damping out of higher order modes, requires that beta should be greater than or equal to gamma plus half the whole squared to damp out higher order modes. Okay. And one way in which this manifests itself is in the effect on, the spectral radius. Here's what happens, right. So the effect of, of this requirement on the spectral radius is the following. Okay, turns out that if you plot out the spectral radius, here it's r, and here we look at, delta t times, 2 pi, omega h, okay, and this is just like we plotted in our, previous, study of the parabolic problem. We plotted up the amplification factor on the hori, on the vertical axis, and on the horizontal axis we plotted up lambda t times lamb, sorry, delta t times lambda h, which was the eigenvalue of that problem, all right? In this case, you're plotting up delta t times two pi omega h, which is effectively like delta t divided by the time needed of oscillation or something, okay? So for r, if this is the value 1. [SOUND] Right? What tends to happen is that as delta t over two pi omega h increases, the entity times two pi omega h increases, if you have values of gamma or, or beta that are, you know, we satisfy the unconditional stability requirement but they have a, they're not as big enough as to satisfy the second requirement, right, of being greater than gamma plus half the whole squared. Okay, here's the sort of thing that happens with the spectral radius. And along this point is where you get, real, fully real eigenvalues. Okay? So you get real eigenvalues around here. All right, and this is for, gamma by two lesser than or equal to beta, lesser than or equal to gamma plus half of the whole squared. Okay? If on the other hand you have, beta exceeding gamma plus half the whole squared, then you tend to get, a behavior which looks like this. Okay? This is for, beta greater than gamma plus half the whole squared. Okay, and this is what leads to damping of high-order modes. Okay. So, that is what we need to know about, how high-order modes depend how, how our analysis of high-order modes follows from our stability analysis. Perhaps the last thing I want to say about this particular problem is how we go about getting to the, to a notion of conversions. Okay. And before we end, and, you recall that stability and consistency are the two requirements that lead to convergence. Okay? Now, in order to understand consistency of this problem, we go back to looking at this form of the, evolution equation for the time this could rise, modal equations, okay? A is our amplification matrix. Remember y is our two vector which has the displacement of the velocity in it. Right? And Ln is what we get from the forcing. Okay, so now we're going back to the inhomogeneous problem here. Okay, again, this is exactly the way we set things up for the parabolic problem. All right. Where Ln was the effect of our putting back the forcing. Right? We, we turned off the forcing when we studied stability. We turn it back on when we study consistency. Okay. So, this is the time-discrete problem. Okay, the time-discrete modal equation. Now, for consistency, we say that well, if instead of the time-discrete, solution, we were to look at the time exact solution, and just plug it into the equation we've written above here, right? We would get y of t n plus 1 equals A, our amplification matrix, times y and tn, right? Y of tn plus one and y of tn are the corresponding exact solutions. Okay? We get a plus Ln, right? But in general, the exact, the time exact solution does not satisfy our finite difference equation. Right? So we get, in addition, a term delta t times now a vector tau depending upon tn. Its a vector tau because of course y is a two vector, right? So tau also has to be two vector. Okay? So this is the time-exact, model equation. Okay? Now, consistency requires. That tau be written as again, a, a, a 2 vector, c. You can think of it as c being two constants, c1, c2, okay. c times delta t to the power k. 'Kay, where, or we could write this as tau1, tau2. Equals both functions of t n. All right? They are equal to c1, c2, times delta t to the power of k, where k is greater than 0, and c1, c2 are constants. Okay? Right? And, when I listed the various some of the members of this family, I said that, that the, their order of accuracy was 2. Okay? That order of accuracy is given to us by k. Right? So, when a couple of segments ago, I listed a, several members of this family and said that they all had order of accuracy 2, what it means is that in this consistency condition, the order of accuracy, k, is equal to 2 for all of them. Okay? Now, when we use consistency and stability to get a convergence, right, the so called Lax Theorem, right, which is that consistency and stability give us convergence of our time integration algorithms for this problem, right? That convergence condition is dependent upon writing out our error. E again is a 2 vector, right, because we're looking at the error in the displacement and the velocity, right, for every mode, okay? So, what we get is that e at n plus 1 equals A to the power of n plus 1, times the error at, at time t equals 0, all right. And, the error at time equals 0 is also a 2 vector because it has the error in the displacement and the error in the velocity. Okay? Plus sorry, it's minus sum i going from, from 0 to n. Right, you get i going from 0 to n, delta t, A to the power i, okay? The tau vector, which is the 2 vector I wrote just above here, at tn. Okay, this is the condition that we need to evaluate, we need to work with for convergence, and, everything works out just as we saw for the parabolic problem. Right? It's just that we have to deal with a matrix vector equation and then matrix vector inequalities here, but things essentially work out. Using what we know about stability and what we have written out for consistency, we can prove that limit of en plus 1 equals 0 as delta t tends to 0. Right, its limit as delta t tends to 0 of en plus 1, right, is equal to 0, all right? You get a 0 back to there, right? So you get the error in the displacement and the velocity tending to 0. Well, that's what we need to know as far as the analysis of this problem is concerned. And, that also concludes our rather quick study of methods for linear elastodynamics in 3D.
