Let me pull all this together and write it in a very simplified manner. Okay, so what we have here is c transpose Kd equals c transpose F. Okay. This is close to the final form of our finite element equations in matrix-vector form. Okay? But there's one other important step we need to take here. Which is, to recognize that this vector of weighting function, degrees of freedom, actually does interpolate the weighting function. Right. So recall that in each element, we had wh of e equals sum over A, NA cAe going back just briefly momentarily actually to our local element numbering system for the c degrees of freedom. Okay. Also recall that our weak form, our statement of the weak form is the following, right, in much abbreviated, in a much abbreviated version it was defined belonging to S such that, St is short for such that. For all wh belonging to V, and I'm sorry, these are both, that's Sh and Vh, right, right? For all wh belonging to Vh the following integral holds, the following equation holds, right? So this is our weak form again, yet again. Okay. What we've done now is essentially reduced that integral weak form to the equation at the top of the slide. Okay? In matrix-vector form. All right? So, and let me, yet again, state that here. All these statements, while, while a little superfluous, are repetitive definitely, are useful to drive home points. So we, we make them repeatedly. All right. So, that is our matrix-vector weak form, and the important thing that I want to point out is that this holds for all wh belong to Vh. My question to you now is that, what is it in the matrix-vector weak form that represents wh? Remember, using basis functions, finite dimension basis functions, we've gone from this integral form at the bottom of the slide, the last equation on the slide which involved fillings. To matrix-vector form which involves just degrees of freedom for matrices for vectors, and we have entries for matrices. So in that setting, what is it that represents wh? Okay? Essentially, what represents wh is the following. This equation of ours is completely equivalent to our statement at the top of the slides. C transpose Kd equals c transpose F. When we realize that this has to hold for all c, right? Belonging to this R nel space. After all is just a vector with nel entries, okay? It is when we specify that our matrix-vector weak form must hold for all c, right? Because all c belonging to this nel degree of sorry, this nel dimensional space that we are essentially imposing the same requirement as we do here. Why is this? It is because the c of degrees of freedom which interpolate the weighting functions. If our weak form has to go for all weighting functions belonging to Vh, we've already fixed the functional form of Vh by choosing particular polynomial basis functions. All right? Or by choosing certain basis functions. That degree of arbitrariness which must still hold within the space Vh is insured if we require, that the matrix-vector weak form holds for all c belonging to this nel dimensional space. Okay, well, but if this is, if this is the case, it is clear therefore that this can only hold if Kd equals F. There's a formal way of demonstrating that, and this just involves moving everything to the left-hand side, and then considering what forms c can possibly have, all the forms that c can possibly have and discovering that this must be the case. Okay. We can make that, that rigorous argument if you need to, if we need to, but I think it's pretty clear why, why this works. Okay, so, this is the final form of our final, for, for our finite element equations, okay? FE short for finite element. Right. Remarks. Okay. The first remark I want to make is that the matrix K that you have in front of you or probably in your notes is symmetric. It's positive definite. Okay. Furthermore, it has abandoned, so-called tridiagonal structure. With [SOUND] banded, tridiagonal structure. Okay. The symmetry comes from the fact. Symmetry from the fact that the term of the weak form which gave rise to our matrix K has this sort of structure. Right? Where this is our stress. Okay. So if you look at this integral as a as an object called a functional in w and u. It is bilinear, okay? Furthermore, it remains unchanged if we just swap the places of w and u, okay? Right? So the reason this thing is symmetric is because we have this term in the weak form and we could very well interchange those two positions, right? Wh and. Okay? This is what makes our final K matrix symmetric as well. If we were working with a different set of equations which, for some reason, did not give us this form of an integral where wh and could be interchanged, and still have the same meaning, right? And, and still have the same integral, then we would not have a symmetric matrix K. Okay? Oh, of course, I should also state that now K, again for obvious reasons, is called a stiffness matrix. Okay, and as you might imagine this is related to the fact that when we look at it in, in the form Kd equals F which you have before you it is reminiscent of a of a spring. Right, K being the spring stiffness, d being the displacement of the spring and f being the force on the spring. And indeed, in the early days of finite elements when there was a lot of structural mechanics work done with finite elements, this was the, well this was the right interpretation, okay? But now if you, if you imagine that you're doing a heat conduction problem, you would probably would, you would probably, properly call it something else maybe the conductivity matrix, right. But traditionally, it's called a stiffness matrix. So let me just put that in quotes. Okay now, the other bit is the positive definiteness. The positive definiteness comes from the fact that our in this particular case, our so-called material constant, E, is greater than 0. We didn't actually state that but if E and D is greater than 0, then we have a positive definite system. All right the particular banded tridiagonal structure comes from the fact that there are two first order derivatives, and that we have a linear set of basis functions. Okay, so. The bandedness is actually is, is included when I say tridiagonal. So tridiagonal from. Single derivative on wh and. All right, in this form that, that's before us in the integral, okay there's a single derivative on each of them. And from the fact that we have a linear basis functions. Okay. The next remark I want to make, and really it's the final remark of this segment, it is the fact that our force vector in this case, I'm just looking back to my notes here to make sure that I have, oh, there we go. That I have here before me, so that I don't make any errors in putting in back up here, okay. Our force vector in this case is the following. It's fAhe over 2, times 2 being all the way down to 1, plus 0 all the way down to tA, in the last entry. Plus EA over he, multiplying our Dirichlet boundary conditions u knot, and zeros all the way down. All right, there are a few things I want to point out about this. If we look at this contribution to the force, to the force term, it comes from the distributed forcing function that was specified at every point along the bar. Right, our original function, F. Right, which we assume is a constant in order to get this final form. If F were not a constant, it would be inside this it would be within this the column vector here. Okay so, so one thing to note is that if we consider the mesh and consider a stretch of the mesh with elements of that type. Okay? What we're seeing is that the contribution on a node, which has elements on either side of it, is of the form fAhe. Okay? It's only the very last element. Right? This is the point L. Okay? It's only this very last element which has a contribution, fA. So at this very last node which has a contribution fAhe over 2. Okay? And this is really because an interior node has a contribution fAhe. It sort of has the entire force on it. Because it has elements on both sides. So both those elements contribute force to it. However, this last node has a contribution only from the element to its left. Okay? This is pretty obvious. This contribution is from the traction and it's no surprise that it appears only on the very last node, because that's where the traction is indeed applied. This contribution is also very important one. You recall that this came about from the fact that Dirichlet boundary condition is known and therefore that degree of freedom in the d vector in our trial solution vector could be moved over to the right-hand side. Dirichlet b.c for boundary condition. 'Kay? So what this is giving us is the loading, right, on the problem, obtained by specifying a Dirichlet boundary condition, okay? So this is what you may call within quotes, a Dirichlet driven load, right? You, we, we all very well know that partial differential equations can be driven by either the Dirichlet or Nuemann boundary conditions. This is how the Dirichlet boundary condition itself drives the problem. Of course, in the kinds of problems, we are considering, the sort of boundary value problem we're considering, u knot was typically zero. Okay? So that actually drops out. But in the context of elasticity, what it does for us is give us the effect of the, the, the, the, the load that gets transmitted to the structure. Because of the fact that on the left end, at the left end of the structure, we've fixed the displacement to be equal to zero. Okay? So let me say just one more thing here. This Dirichlet driven load is equal to 0 in the considered case. Okay. So there we have it. The various contributions to the, to the effective forcing on the problem, coming from the distributed forcing function from the traction, the, the Neumann boundary condition, right. Since I'm calling it traction, I also ought to say that this is the Neumann boundary condition, and finally, the effect of the Dirichlet boundary condition as well. Okay, we will stop this segment here, but essentially at this point we have completed our simplest finite element problem in 1d for linear elliptic equations. Okay. Well, I should, I should also state one more thing, right. I've just said it's formulated the problem. What is the solution? Since we have Kd equals F, the solution is d equals K inverse F. Okay, that is the formal solution. Of course, you may or may not actually invoke K depending on the size of your problem, that is an entirely separate question which we don't get into here. This is just the formulation. Once we have d, of course we can then go back and reconstruct in any element as being simply the interpolation over A using the bases functions, which we know very well, times dAe. Okay? So we've recovered the actual field from the solution. Okay, here's where we definitely do stop the segment.
