Welcome back. Today we are going to start and hopefully get through most of a new unit. And that unit is going to address basis function in an in, in an, in, in a unified manner, okay? The idea is that the most standard basis functions, which are based upon Lagrange polynomials. Are constructed in two and three dimensions. Essentially through a process that is called a tensor product, representation. Right? And this is done by first constructing basis functions in 1 D and then essentially extending them to multiple dimensions. So, we're going to take that approach, and, and, and doing so present a hopefully unified picture of basis functions that we have already been using. So, all right. So we're going to look at a I call it just that, we're going to look at a unified view Of basis functions. In 1 to 3 dimensions. All right. Okay. The the approach is to start with 1 D and and, and work up from there. I should also mention that we'll start the segment by looking at this approach for Lagrange polynomials. Okay, so we're going to look at Lagrange polynomials and let's start by recalling 1 D right. In 1 D let's look at what our linear basis functions look like. And that's our element. In 1 D, we're talking about linears, so this is what we have. Of course, we construct everything in our bi-unit domain, okay? So, in this bi-unit domain we have C. I'm going to call it C 1, just to prepare ourselves for smoothly moving on to two and three dimensions, okay? So this is C 1 equals minus 1. C 1 equals 1. OK, and of course this is 0. All right? In this setting we know very well what our linear functions are. Okay. That's meant to be linear okay? So this is N 1, and we have N 2 Okay and of course they satisfy the usual properties Kronecker delta and the fact that they evaluate to one when added up at any point C 1 in the domain, okay? These are linears and you recall how they were written? They were written as N 1. Of C 1 equals 1 minus C over 2, and N 2 C 1 equals one plus C 1 over 2. Okay and then we went on to develop quadratics. Right. We have our element. C 1 is minus 1. C 1 is 0. C 1 is 1. Okay? And in this case. Our first basis function looks like This. Right? That is N 1. N 2 looks like. That, right. That's N 2 and N 3 as you will recall is that. Okay. Allright, and the expressions for them were N 1 function of xi 1 is 1 minus xi 1 times xi 1 over 2. N 2 equals 1 minus. C 1 squared, it's just one minus, C 1 squared, and then N 3, is 1 plus C 1 times C 1, divided by two. Okay all right, so we know these ve, well enough, and, the reason I put them down here is just to get us started. You recall that when we developed the finite element formulation in one d, we also presented the general Lagrange polynomial formula. Right, for, arbitrary order, right? And that formula was the following, right? So. Right? So when we develop Lagrange polynomials of order. So we going to write this out for Lagrange polynomials of order k Right? And you will recall from our treatment of problems up to now and actually from just looking at this at the sketches of our basis functions, that the order of the polynomials bears a relation to the number of degrees of freedom on the element, right? And what is that relation? Relation that the, that the order of the polynomial is the number of degrees of freedom minus 1. So, we're going to write that out. And, we, we previously called this number of nodes in the element. I'm going to use that sort of, idea. Except, I'm going to call this number of nodes in 1 D. Okay. So it's number of nodes in 1 D minus 1, which gives us the order of the polynomial. The reason I am bringing in this reference to 1 D is as you will see, we will use the one d construction to build our constructions in two and three dimensions, okay? All right. So so the polynomials of order k are the following, right? So we know by looking that at, at, at what we've written out there that we can, we have N A, right, function of C 1. Equals the product index b, running from 1 to number of nodes in 1 D, okay? Except that B. Is not equal to A. Okay. Now the product that we need to have here is the following, C 1 plus C 1 at A, divided by Z 1 at A, minus Z 1 at B. All right, and if you test this out for any of these functions I put up there for linears of quadratics, you should see that it works. Okay? So this is the setting we have. Now, we are going to generalize this right? Essentially what we are going to see and what we have seen already is that when we go to two and three dimensions, and we did look at two dimensions briefly, we're going to see that we essentially construct those bi-linear and tri-linear or bi-quadratic and tri-quadratic functions as we will also see, by forming what we call tensor products of these functions in 1 D. Okay, also what I'm going to do is that because I'm going to now use this to construct our functions in, in higher dimensions, I am going to put a tilder on everything I have written here. Okay? This is just so that I can use N itself for the actual functions in higher dimensions. Okay? Same function. I'm just putting tildes on all of it. All right, so this is what we have. So let's go, let's move on now to tensor product functions. In let's look at 2 D. Okay. Now, in 2 D, we already did see the bi-linear shape functions. Okay, so recall the bi-linears All right, the bi-linears were the following. I'll sketch one of them all right? So because we want to make use of this perspective view. C 1, C 2, okay? And we label the nodes here. Or, degrees of freedom here as A equals 1, 2, 3, 4. And, I will draw just a, and 4. Okay? And you recall that is bi-linear, so it, it, it slopes down linearly to 0 along each coordinate direction, okay? And, it also slopes down to 0 in this dimension, all right? And this dependents actually, not quite linear but looks something like that. Okay? All right, so, so this is N 4. Okay so this shape, this basis function I am showing you is N 4. Okay. And, okay, so this is what we have. I'm going to directly just write out the, the, general Lagrange polynomial formula using the stance of product, using the denser product approach. Okay. In order to do this let's note that the kind of bi-linear shape function we are constructing here has n n e degrees of freedom. Right, we're constructing four basis functions because, because this element has four degrees of freedom. And in this case, n n e equals 4. Okay? Note of course that n n e equals 4 is a number of nodes in one dimension squared for, the linear case, right? So if you take the linear basis functions, right? And well, the, the two more degrees of freedom, the bi-linears in for, for the 2 D case involve four degrees of freedom, right? So that's n n 1 D squared. What we are going to do here is, is observe that we always have some NA, Z 1, Z 2, right, are functional Z 1 and Z 2, is going to be written as a product. Of N tilde B. C 1 times N tilde C, C 2, okay? For B comma C equals 1 up to number of nodes in 1 D. Okay? This is the general formula, alright. And A of course here, A which is on the left-hand side equals 1, 2, number of nodes in the element in 2 D. Okay. It's really as simple as that. Right, and it is, and, and this sort of formula where we're really taking those basis functions we developed on the previous slide that we labeled N tilde, right? We're taking them in each dimension, right? Along each of the dimensions the, the C 1 and C 2 dimensions and simply forming a product with it, right? This sort of a product is what we mean by the denser product formula. Okay, with this sort of view, what we observe now is that N 1, C 1, C 2 is a product of N 1 tilde. C 1 N, I'm sorry, N tilde and N It's a product of N, N 1 tilde in the C 1 direction and and, two, two three four. N 1 tilde again in the C 2 direction. All right. And that becomes clear by just looking at it. So this goes on. Well actually let me write all of them. N 2. N 2 is what? In the N 1 for, in the C 1 direction. Look C 1 dimension it is N 2 tilde C 1, N 1 tilde C 2, right? This goes on and we finally write N 4, okay? N 4, C 1. C 2 is, can you work it out? It's N 2 tilde, sorry, it's N 1 tilde again. N 1 tilde, C 1. And. 2 tilde. C2. Okay, it's as simple as that. Right? Okay and you recall we did indeed look at the bi-linear case, right? And now if, when we go on to higher dim, when we go on to basis functions of higher order, all we need to do is to go back to the polynomial sorry, to the 1 D. Construction. And, figure out what order we need to order. And it send the order in 1 D that we need in order to construct these functions in 2 D is simply the square root of the degrees of freedom that we want to have, right, in 2 D. Okay, so we construct those 1 D basis functions, and then form the sort of products with it.