Welcome back. We have just embarked upon the analysis of our time integration algorithms for parabolic equations, right, as of, of, first starter in time. What we've accomplished along that direction in the last segment is a recognition that there exists this generalized eigenvalue problem that one can identify, which provides for us a basis on which to carry out a modal analysis of the problem. Okay, so let's just recall that aspect and charge ahead. Okay so we're engaged here in the analysis of our, of the time integration algorithms. All right for linear parabolic systems. Right and you will recall that this is based upon the generalized eigenvalue problem. Right, and that generalized eigenvalue problem takes on the form K, which is our stiffness matrix. Sorry, our conductivity or diffusivity matrix. Acting on a vector, that is psi M right? K psi M equals lambda M, M which is our mass matrix psi M, okay? This is the basis for it and we went ahead with this, right? What we found and I'm not going to go through the entire analysis I'm just going to recall the critical parts of it, is that one can orthonormalize the, the eigen vectors psi with respect to M. Correct, so we found that psi L dot M, psi M equals delta LM, chronicle delta, and, because of that, this result also lets us say that psi L dot K psi M equals lambda L okay? All right actually properly equals lambda M multiplied with delta LM. Okay, that's what it actually is. Okay, so it is zero, unless M equals L. Okay, the second equation, the right hand side of the second equation. Okay, and if L equals M, it's equal to psi L, which is the same as psi M. All right, and you recall that in this lambda M is our eigenvalue, right, corresponding to the eigen vector psi M. Okay, so this is orthonormalization. Okay. All right, so with this in hand, what we observed I think at the very end of the segment is that since these vectors psi form and actually span the space, they serve as a basis. Okay. So, what we have is an expansion In the psi m basis. Right, where m equals 1 up to ndf, okay? All right, and this letter say that a vector d, right? Such as the vector of global degrees of freedom that we're now working with can be expanded as sum over M d sub m psi sub m, okay? Right, and we saw that each of the d n's, the modal degrees of the, the, the modal coefficients each, right? Each d n is given by right it's equal to psi M dotted with M d. Right? This is everything that we need to know. Right and these are the modal coefficients. All right. What we are going to do now is with a summary of our analysis, of sorry, with the summary of our modal decomposition of the problem, right. Right, so this, this, this result is the modal decomposition. The modal decomposition of d. All right, so, with this, with these results, with this summary of results, we can go ahead, all right. And, what we are going to do now is, extend this idea of the modal decomposition to, the equations we are working with. All right. So, we're going to start here with a modal decomposition. Of. The time exact O D E. Right. Right, and remember everything that we were working with homogeneous equations, okay. So which equation is that? Think about it. It's this one. M d dot plus K d equals not the forcing term on the right, or well, the forcing term is equal to 0. All right we have this with this initial condition that d at times 0 equals the d 0 vector. Okay, so now we're going to do a modal decomposition of this equation itself. The way that works is the following. We already know the modal decomposition of d. Okay. That modal decomposition I'm going to write in slightly different notation, the notations not terribly different. I just want to move the m from being a subscript of the coefficient to a superscript. Okay, so that's not a power, that's a superscript. All right, and you'll see just in a bit what I want to do with the subscript. I need them in there, okay. This times psi. Let me move this also to the after the superscript. Right, the m there is also a superscript. Now remember that this d vector of ours is time dependent. Right, that's why we're able to take a time derivative. Well, we could always take a time derivative, but that's what makes sure that the time derivative isn't doesn't vanish, right. Okay, so d is a function of time. Okay, my question is where on the right hand side does that time dependence go? Does it go into the coefficient? Right, the d n's or the sines, the eigen vectors? Right, they go the time dependence shows up in the d n's. Okay, and more than just a convenience that it has to be that way. Why is it that the sines don't change with time? It's because the sines are, are the eigenvectors of the generalized eigenvalue problem. But the matrices that define the generalized eigenvalue problem, the K and the M matrices are fixed in time, okay, because we are doing a linear problem. Okay, so let's just state that, that's useful to remember, okay. So, this what, what, what's behind this particular decomposition, right, where the time go, where the time dependence is held in the, in the coefficients, right, is the fact that the set of psi m. Right? Are fixed in time. Because, our conductivity matrix K and mass matrix M are also fixed in time. Okay, that's what allows us to do this. Okay, what you see, what that also allows us to do, what that immediately lets us see is that. D dot, right? D dot is now sum over n, d dot n, function of time, psi. All right? Some of cos running from one to number of degrees of freedom. Okay? And we are going to make these, and you are going to substitute both these decompositions into our time exact O D E, right? So what this implies then is let me just say this as substituting. Okay? It implies now that M multiplying the sum over little m d dot m, psi m plus K sum over m d m, psi m equals 0. Okay what I'm going to do next is is apply linearity, right? So what this then apply, what this implies is that because M and L are matrices and because we're dealing with linear algebra there, what we can write here is that sum over M, d dot m mass matrix acting on psi m plus sum over m, d m, conductivity matrix acting on psi m equals 0. All right. Now I am going to dot this entire equation on the left, well it doesn't matter with left or right, I'm going to dot it because it's a vector equation finally. I'm going to dot it with a specific eigenvector lets say psi L, okay, so now I'm going to take psi L, okay? Dotted with the sum d m dot, psi m. Plus psi L dotted with the sum over m, d m, K psi m equals 0. Okay? Note that something has changed between those two equations, right? Something very obvious has changed. 0 here is no longer a 0 vector, it's a scalar 0. Right? Because I'm after all taking a dot product off a vector equation here, this was a vector equation, and I've taken the dot product of it with another vector of the same dimension right, psi L. Okay, but now we know how this all works out, right? Because we know exactly what happened with psi L oh I'm sorry I am missing here, sorr,y I'm missing an m here, so let me just clean this up. I'm missing here. That matrix okay, all right. Now its all right, okay. Well let's carry out these dot products right and and we know what we get from these.