In this segment we are going to extend our development of basis functions in one dimension to higher order polynomial basis functions. Okay. So the topic of this segment is going to be. Higher order polynomial Basis functions. Having already developed the linear basis functions, we will do this in a sort of deductive manner by going first to the quadratic functions, developing them explicitly and then laying down a general formula for arbitrary order polynomial. Okay. All right. So, let me write that down. We'll start with quadratic basis functions. Okay so when one is talking about quadratic basis functions in for a one dimensional problem here is the setting. This is our domain of interest. And, this is 0 that point is L. Okay? And, we have our nodes, right? So let's suppose that these are our nodes. Okay? So that would be x1, x2, x3, so on. And now let's suppose that this is element omega e. Okay? It turns out, that just as linear 1d elements have two nodes, because we need essentially two degrees of freedom in order to interpret linears, in order to write out linear equation. And so it follows that quadratic elements in 1d have how many nodes? Three. Right? So if this is omega e, if you just follow the numbering system, what you will see is that the node numbers, the global node numbers here are x2e minus 1, x2e x2e plus 1. It continues, and, essentially ends with x twice nel for number of, elements. Plus one. Okay? That's the, the global number of the last node. All right. So this is the setting we have. We are going to take this element out here and develop basis functions for it, okay. So we have here our element sort of blown up. Well let's read, renumber the nodes here. Okay, so this is the the quadratic element in it's Physical domain. As we did for the case of the linear elements and linear basis functions, we will suppose that this physical element is created for us or is obtained through a map from a bi-unit domain which is the parent domain, right. In which, on which our coordinate is labeled c. And in terms of, of c, we have this middle node with coordinate c equals 0. We have c equals minus 1, and c equals 1. Okay? And you will notice that this is very much like our the case of linear basis function, except that we have this extra node in here, right located at c equals 0. We suppose, as we did in the case of linear elements and linear basis functions, that our physical element is obtained from map, from this domain. Okay. And essentially what we are going to do here is write out our basis functions and also you know, work out what they are, what they're going to be. Now, when, when one is working with quadratic basis functions, our trial solution over element e, is going to be written as a, as essentially it's an, an expansion over the basis functions for that element, right. In this case we have e which represents the local node number or the local degree of freedom number. Running over from one to number of nodes in the element right, and in this case, equals 3, okay. So we have NA, right. We know that it's a function of x, but really it is a function of x through c. Just as we did for the case of linear basis functions. Okay. We have this and it multiplies da sub e. Okay, da sub e being the, the actual, degree of freedom value, right, a scalar, which is being, interpolated, so to speak, by our basis function, NA. All right? Also, much as we did, in the case of, linear basis functions, we will write out wh, our, weighting function, as the same sort of representation right, a equals 1 to number of nodes in the element n a x also parametrized by c through our mapping multiplying cAe okay. And the question that we are going to address over the next few minutes is just how do we construct these Na's for the case of quadratic basis functions, all right. And here's how we do it. I'm going to write out the basis functions to begin with and then we'll talk a little more about them. We have N1 and I'm going to write it now directly, parameterized by c, though we know we get to c Through x and through the mapping and so forth. Okay. N1 is c times 1 minus c divided by 2. N2 is 1 minus c squared, N3 is c times 1 plus c divided by 2. All right? You observe, of course, that these are clearly quadratic in c. It's useful to plot them up, just as we've done for the linear basis functions. And, the way I'll do it is the following. I will suppose that this is the point, c equals minus 1. C equals 0 and roughly there should be c equals 1. All right, this is of course c. All right, if here I give you the value one. All right. The very first one of them is, the following, okay? I'm drawing N1. Right? It's quadratic, okay? That's N1. To draw N2 let me try to use a different color. Just for visibility. Okay. N2 takes on the shape, Of a quadratic function. Looking like that, okay? It's meant to be symmetric. I didn't quite get it symmetric, but it is meant to be symmetric, right? And this is N2. All right, and finally we have N3, which is the following. Okay, that's N3. All right. So those are our three basis functions in the case of quadratics. Something to note about them is the following, right? If we denote c1 equals minus 1, c2 equals 0 and c3 equals 1 to be the actual values of c at the 3 nodes in the sparing domain. What you observe is that the chronicle delta property holds, right? NA cB, Na valuated at cB equals delta AB. Okay? The same chronicle delta property that our linear basis functions possess. All right. And then there is this other property, which is that sum over A, NA at some, any, at an arbitrary value of c, the sum equals 1. Okay? And this is the property that is important in order to be able to represent constant functions, okay? All right. So we have these two properties for our basis functions, just as we saw these properties in the case of linears. Right. What we will do from here is make the observation that this sort of expression, right, the expressions that we've used here to write out the quadratics, actually, are routine from a more general expression that holds for this family of Lagrange polynomials, all right? And we'll approach it by writing out the general formula for Lagrange polynomials and recovering the quadratics as a special case. Right. Okay, so higher order basis functions. To generate higher order basis functions, we fall back upon the Lagrange polynomial formula, okay? So higher order basis functions are generated by the formula for Lagrange polynomials, Lagrange polynomials. Okay? Let's suppose that we're looking at basis functions of some order, and let's denote that order is Nne. Okay. Let's suppose that we have Nne order polynomials. And you see how this works, right? Why does it work to always say that we have Nne order polynomials? Because of course, that is the number of nodes in a single element. All right, Nne. Okay. Here's how it works. So, so what we're looking at now is a single element, okay? This is all omega e, and it has some arbitrary number of nodes, okay? Right, so the local node number in here would be, local node one. And that would be local node number Nne, okay. This, this is the sort of situation we are looking at. Okay. In this setting again, we're seeing that this element comes to us from a parent domain. Where likewise we have the same number of nodes in this parent domain as well. Okay. That's node one. That's node Nne. We will denote the coordinate in that parent domain as c, right? It is a bi-unit domain again, and that is seen by the fact that c here equals minus 1. And here we have, c equals One, right. Therefore, by unit because the total length is two. All right. In this setting, here is how we write out the polynomials. NA in terms of C is the following. It is the product B=1 to number of nodes in the element. However, B can not be equal to A. Okay, where a is what we have on the left hand side. Okay, we have here c minus cB, right? And this thing is divided by another product, the limits of which are the same. But here, the expression we have is cA. Minus cB. All right, that's it. Okay. And this holds now for A=1 up to the number of nodes in the element. All right. Let's check this out in fact let's check it out not for quadratics, but actually for an even simpler class of basic functions for the linears. Okay? So let's do that now. So. Okay. In the case of Linears, let's just recall one thing before we go onto the case of linears. If you look at that element that we had. And we look at it in the, parent domain, Omega of C. Right, we have, this is equal to, this is -1, right? C=-1, C=1. All right? What we're seeing here is that when we label the c's as ca, cb, and so on, we did that right here, I just pulled up the previous slide, you see that ca and cb and so on, that we're use in our formula at the bottom of the slide, okay? Here's what they are. This is the point C one That is the point C N and E. And inside here, we may have cA, cB, right? And those would take on the actual values of C at those points. Okay, so with this setting, all right for linear's. We have just two of them, right. Just two of those node points. So in this domain, we have c1 equals minus one, c2 equals one. Right? So let's construct our polynomials from this formula. N1 is going to be the product B equals 1 to which is 2, right? Because for linears, we know that number of nodes in the element Is 2, okay? However, b cannot be equal to a. On the left-hand side, we have a=1. Right here, right? So we're saying b goes from 1 to 2, but b cannot be equal to 1, okay? And here, we have c minus cB divided by the produce over the same limits. C1 with that which was cA equals one of the left-hand side, of course, minus cB. Let's work this out in greater detail. So the numerator is, is has a single term in the product. It is c minus cB. We look at the limits here. B goes from 1 to 2 but B cannot be equal to 1 so B has to be equal to 2. And in the denominator again, we have c1-c2. Well this is c-c2? What is c2? it's right here, correct? It is one, divided by c1-c2. C1 is minus one, c2 is one. We get back one minus c over 2. Kay? Which we recognize to be N1 for the linears. Taking the same sort of approach, what you will find is that n2, c equals the product probably don't have quite enough room here, so let me go the next slide. N2, c equals product B equals 1 to 2. Now B-Not equal to 2. Here we get c minus cB. Divided by the product over the same limits. CA, which in this case is 2, minus cB. Okay? And in those limits, we recognize the only value b is allowed to take is 1. So when we work this out, we get c Minus, C-1. C-1 however is minus one. Divided by c2 which is one minus cB which is minus one. Okay. And we get one plus c over two which we also recognize to be N2. For linears. Right? So here we see that our formula works out very well for linears, it works out for quadratics, which you can check out, and it also does work for high order polynomials. Right? This is a good point to end this segment.
