We're going to start a new unit, and a new equation now, we're going to go on to hyperbolic equations. We've looked at elliptic and parabolic equations, we now move on to hyperbolic equations. The canonical example here is elastodynamics, in whatever dimensions you wish. We're going to look at it in three dimensions, because we've left 1D far behind now. Okay, so, title of this topic of this unit is methods for hyperbolic. Linear of course. Pdes in vector unknowns. All right? And the example is linear, r i, linear elastodynamics. In 3D. Okay? So the setting is the following. I don't have my lego vectors today for my bases but I'll use my fingers okay. So that's my basis and this is the body of interest. When we considered linearized steady state elasticity. We were interested, of course, in how this ball would deform or how this continuum potato would deform. We didn't, however, consider the so-called dynamic effects, or the, or the, the effects that lead to wave propagation, all right? We couldn't, for instance, for that reason, also study the problem of this ball being actually tossed through space, right, and tumbling through space, and deforming perhaps at the same time. Right. We couldn't look at the time evolution of that problem. Right. Because we were looking at the steady state problem. So now we take away restriction, and look at the full blown elastodynamics problem. Okay. So Here we have the, the setting is essentially the same, as far as our pictures are concerned here, right. So we have our figures are concerned here, e1, e2, e3. We have a body of interest, right. This is omega. As we've done before, we have a decomposition of the domain into Dirichlet, into a Dirichlet subset for that particular component of the displacement field, right? And the corresponding Neumann subset. Okay? And this holds for i equals 1, 2, 3, right. X would be a point here, right. Which will be described by its position vector. Right, everything that we've seen from before holds, right? The decomposition of the, of the boundary into. The Dirichlet. Sorry, I got the union in the wrong position. It's the union of this Dirichlet boundary and the Neumann boundary. And, of course, those are disjoined, we know that. This is the empty set, and so on, right, we have all of this. This of course holds for I equals 1, 2, 3, right, three dimensions. Okay. I'm going to straightaway put down the strong form of the problem, right. The strong form of the problem is the following. Now, given data, ugi, t bar i. F i, right? In addition, we need some more data now. We need also other functions which I'm going to denote as u i 0, okay, and v i 0. Okay, we are going to use them for initial conditions. Alright so given all of this, and of course, the constitutive relation sigma ij equals C ij kl epsilon kl. We also know that we have the kinematics, right? Epsilon kl equals one half partial of uk with respect to xl, plus partial of ul, with respect to xk. Right? We have all of that stuff. Right, the only new things that you're seeing here are these two. Functions which I'm telling you right now we've been used in initial conditions. There, there, there is one more. We do need another coefficient, which I'm going to denote again here as rho, rho here is just the mass. Sorry, the mass density. Okay? Let me get rid of these arrows from here so that it's not confusing to think that they're pointing up from density. Those are the initial conditions. All right, we have all of this. The problem that we are trying to find is the fall, is to find u i okay, now it's a function of position and time. Okay. Right, and remember that i runs over one two three. Right, so everything that I've written on the first line, the first five functions is down here. Right. These functions are all just components of vectors. Right. Okay, so we want to find u i such that. Such that the following holds. Rho, second derivative of u i, with respect to time equals sigma ig comma j, plus fi, okay. In omega cross 0, comma T. Just as we did for the time dependent parabolic problem, right. We say that our pde must hold over the spatial domain and the time interval, 0 to T. Okay? Let me leave this here and then go on to the next slide to write out boundary conditions. Boundary conditions are no different. Right? For boundary conditions we have u i, at some position x and time t equals this given function ug for component i at positions x and t, and time t. Right. Now note that we're allowing here that Dirichlet data to vary with time. Okay this just allows us to have time dependent Dirichlet boundary conditions. Just as as we had for our time dependent heat conduction or time dependent mass diffusion. Right? We allowed the possibility that the Dirichlet conditions varied with time. Okay right at any point x belonging to a point in the Dirichlet boundary, right, for that particular displacement component. Our Neumann boundary condition or our traction boundary condition also is as before. Okay. Functional position and time for a point x belonging to the corresponding Neumann boundary. And note here that I'm continuing to use t bar for the traction function where as the t here is for time. Okay so it was, it was an anticipation of this final clash of notations that I have been using t bar for the traction. Okay, initial conditions. Our elastodynamics equation our pde for elastodynamics is a second order pde in time. And therefore, how many initial conditions do we need? Two, right? So we have u i at some position x but at time t equals 0, equals u, how do I write it, u i 0. Okay, which could be a function of position. We're allowing, we, we of course need to allow the possibility that well not just the possibility. We have to allow for initial conditions so, to be defined at every point. Right. So for every point x we have an initial condition of the displacement. Right. What it basically says what is the initial configuration of the body. Okay? So, this holds of, of for all x in omega. It is second order, so we need two initial conditions. Next initial condition is for u i dot x comma 0 equals the specified distribution of velocities. Okay, what this means is that at the initial condition we are saying that not only do we start out knowing where every point on this body is, right, that is the first of those initial conditions. What we are saying we must also know what the initial velocities are. Okay. That's the second initial condition. All right, this is it. This is our strong form. Okay. I'm going to straight away go ahead and write out the weak form. Right? The weak form. Of course, this is going to be the infinite dimensional weak form, but we know that going from there to the finite dimensional weak form is not such a big thing. The weak form. Okay? Given all the data that I have just put out there, right? I'm not going to repeat the data, okay? The weak form is find. U i belonging to S, okay. Where for our purposes here S consists of all u i such that u i equals u g i on. Okay. Find this such that for all w i, belonging to V where V consists of all weighting functions. Remember, w i is our weighting, our, our weighting functions. v belongs to w i such that w i equals 0 on. That Dirichlet boundary. Okay. Such that for all w i belonging to V, the following cond, integral condition holds. Now, integral over omega, w i rho, second derivative, second time derivative of u i. Plus integral over omega, w i comma j, sigma ij, dV, plus sorry, is equal to integral over omega w i fi dV. Plus as before, the sum over spacial dimensions 1, 2, 3 now. Integral over the corresponding Neumann boundary, w i t bar i dS. Okay, that's it. Now, if you stare hard at this weak form, you should observe that it is obtained by just adding one term to our weak form for the steady state elasticity problem. Right? And that extra term is just this one. Observe furthermore that this term requires no integration by parts. Right, it's literally obtained by looking at the left hand side of our strong form, which is right here. Okay, look at the left hand side of the equation at the bottom. Multiply that by Wi, the weighting function. Integrate of the domain, right? We know that the rest of the stuff on the right hand side is what attracts integration by parts, right? Especially the divergence of sigma, the first term on the right-hand side in the strong form. Well, I know directly how the weak form arises, nothing new here, right? Just add in that, this, extra term on the left hand side. All right, we'll end the segment here. When we return we will simply write out the weak form, sorry, the finite dimensional weak form and go directly into the finite element of matrix vector recreations. Good.