Welcome back. In the previous segment we looked at how to construct quadratic basis functions for quadratic elements, and also pushed ahead to high order basis functions, right? For high order elements. The formula for higher order basis functions is obtained from the Lagrange polynomial formula. All right and we saw all of that. Just a couple of points I want to make about the Lagrange polynomial formula, and here are those points. So first of all the Lagrange polynomial formula or the Lagrange polynomials themselves. Right, they satisfy the Kronecker delta property which our basis functions have possessed. Okay, so the general formula also satisfies this Kronecker delta property and I just want to demonstrate that. So the Lagrange polynomials satisfy. This Kronecker delta property. Okay? Let's check it. So, you recall that if we have, an element with, Nne nodes we have this formula for the eight of those, basis functions, right? NE of xi we found is, the product B going from 1 to number of nodes in the element, B not equal to A. xi minus xi B divided by a product with the same limits. But, the, the, term that we're multiplying here is xi A minus xi B. Okay, so let's check this out, right. And this holds, of course, for A equals 1 to number of nodes in the element. Right. Let's check this out. So we know that from the Kronecker delta property, if we have NA, xi A, right, just substituting xi equals xi A in the formula above, we see very clearly that NA xiA, equals pi. B equals 1 to number of nodes in the element. B not equal to A. xi A minus xi B Divided by pi, B equals 1 to number of nodes in the element. B not equal to A. Again xi A minus xi B. Right, so therefore, this is, indeed, 1. Right? Which is what one would expect from the the Kronecker delta property, right? NA of xi A should be equal to 1. All right? Let's see what happens if we take a, well if we evaluate xi, at some other value of, at some other position in the, in the element in the parent domain. Okay, so, in particular let's suppose N, we evaluate NA, not at xi A, but at sum xi C, all right? Where, of course, we are saying here that C is not equal to A. Okay? So, we have the product, B equals 1 to number of nodes. Not equal to A. Here we have xi C minus xi B. Right, remember that's a product, right, of all those sorts of terms. Divided by another product, same limits. B not equal to A here. We get xi C, minus, sorry, here we get xi A minus xi B, okay. And now you see what happens right? In the numerator the product runs over B equals 1 to Nnee but B not equal to A. However, for C not equal to A what we observe is that for one of these terms does take on the value xi C itself, right, because xi B is equal to xi C. And that's what sets this u, expression to 0. Okay? So we get here 0 when C is not equal to A. All right? So the Kronecker delta property does indeed check out. It's a simple enough exercise for you to check that the other property holds. All right? So, check that, now sum over A, NA, at any value of xi, is equal to 1, okay? Using the Lagrange polynomial formula. Okay. Check this out as an exercise. All right. So, having looked at those two properties, let's move on. But now, let's try to develop our, equations, our finite element equations, our finite element formulation, for the case of quadratic basis functions. Okay? So, what we will do now is develop the finite element formulation. With quadratic. Basis functions. All right? And we will take this up directly in the finite dimensional weak form. All right, and specifically we're going to take it up in the finite dimensional weak form where we've already Introduced the notion of the partition into elements. Okay, so we have sum e going from 1 to number of elements, integral over omega e. W h comma x sigma h A dx, and we recall that sigma h is what? Sigma h is given to us through the constitutive relation which is E times u h comma x, right? This is equal to sum e equals 1 to Nel, integral over omega e, Wh f A dx plus. Let's suppose we're dealing here, and in fact let us deal with the, Dirichlet-Neumann problem. So, we get our traction term, or our, Neumann boundary term, Wh, t bar A. Okay? And I should mention that here we have the Dirichlet-Neumann problem problem. Two questions, what other kind of problem could we consider? Correct, we could consider the Dirichlet-Dirichlet problem. Another question, why do I say that this is a Dirichlet-Neumann problem? Where does that show up? Right, it shows up in the fact that this is the Neumann boundary condition. Okay? The Dirichlet condition is, of course, already embedded in our spaces. All right. So this is the problem we are looking at, and, let us now focus on this term. Okay, this left hand side integral over the element. Right? Okay. So, let's consider integral over omega e. Wh comma x we have sigma h, but let us directly use our constitutive relation here. We have E u h comma x, and I've left a little space here, because I want to introduce the area in there. Right? dx. Now, how are we going to write this out? Okay. In order to write this out we need to express our gradients. Okay? Right. And in order to express our gradients what we will do is we will recall, as we've done before, that u h comma x. It isn't element E, because we are talking of integrating only over element, over the domain omega e. But I'm going to save ourselves the writing of that extra subscript here. Okay, so I'm just going to write it as u h comma x. This is, sum A going from 1 to number of elements, but we're working with quadratics here. Right, so we have A going from 1 to 3. NA, and you'll recall that just as we did before, we are going to calculate this gradient by taking the gradient of the basis function, okay? And a similar sort of expression for the waiting function. Okay? Where we may further ask well how do we compute these gradients? The change rule is the answer. Right? So the way we would right NA comma x is NA comma xi, which is easy to do because our basis functions are indeed parametrized by xi. All right, this times xi comma x. Okay? So, having written out or having recalled how we are going to write out our gradients, right? The only difference here between with respect to what we did for the linear problem is that we need to sum over three basis functions now because we're doing quadratics. Okay, so the way we will use this is by writing this as an integral over omega e. Now for Wh comma x I'm going to write sum A going from 1 to 3, NA comma x, but we already know NA comma x is NA comma xi. xi comma x. CAe. All of this represents Wh comma x. Multiplied by EA, multiplied again by sum over B going from 1 to 3. NA comma x, but again we are going to use our change rule. It's not A, it's B here, because we are summing over B. And B comma xi, xi comma x, dBe, okay? dx. That is how we're going to compute that element integral, right? Okay then, let's see what else we can do about this. Let's, let's start out by computing each of these quantities, 'kay? The computation of xi comma x takes a little more preparation, as we know from before. And we'll postpone that for just a few minutes. All right, so let's start out here. N1 comma xi is d d xi of N1, but N1, we remember, is one-half xi times 1 minus xi. Okay? It's that derivative, all right? Which is dd xi of one-half, xi minus xi squared, okay? Which we see is one-half times 1 minus 2xi. N2 comma xi is dd xi of 1 minus xi squared, which is d, which is, well this is easy to do, I don't need another step here. This is just minus 2xi. And finally N3 comma xi equals dd xi, of one-half xi times 1 plus xi. And when we carry this out, we get one-half 1 Plus 2xi. All right? Straightforward enough. Now, what about xi comma x? What about computing this term?