All right. So, again, on this light, u h, u h over element e of x is sum A going from 1 to number of nodes in the element. Na functional position d to the a e. Now, because of the way we do this, you are observing that the degrees of freedom and the basis functions which I'm drawing are fundamentally associated with nodes. Right? Right, and then here also we know that we have a degree of freedom, right, corresponding to each node. We have d1e corresponding to that degr, to that node, and we have d2e, okay. And for that reason we have often also call Na x as being the nodal basis functions. Alright and d e as being the nodal. Degrees of freedom. All right. So we've got this far with talking about how we are going to write out this expansion of our finite de, finite-dimensional trial solution over a single element. We do the same thing also for the weighting function. Right? So, similarly. For wh function of position also restrict to element e. Okay, restricted to element e because that is where we are defining it so we do the same thing. Wh of e, x equals summation e going from one to number of nodes in the element. Which in this case, because we are looking at the simplest possible finite element formulation, that number of nodes in that element is two. We have NA function of position. Now, the degrees of freedom for the weighting function will be obviously different from the degrees of freedom for the trial solution, right, these are different functions. So, though we're using the same basis functions for the weighting function and the trial solution, we have to use different degrees of freedom, if not they would exactly the same function, right? so we have c a e, okay. We are in this case again, this is the degree of freedom, dof for short of the waiting function All right, okay, lemme see. There are many things to be said about this, but first of all, this is our most elementary sort of approach to writing out a Galerkin weak form. When we have the same sort of basis function for the weighting function, as well as for the trial solution, right, this is called a Bubnov-Galerkin method. Okay, and what this means, is that, that same basis function. For Of e and Wh of e. There are other, more advanced finite element methods that for good reason, because of certain things to do with the mathematics of the particular equations being solved, do not use the same basis functions for the trial solution and the weighting function. And they're called Petrov Galerkin and we won't get to them in this, series of lectures. Okay, so all right. Let's move on here. Now, now you will note that all I've done so far is to sketch what these functions are going to be. I, I've called them linear function and so you can image what they are. You can very well construct them in terms of position over the, in terms of the x coordinate over the element. But it it turns out that we can do better, we can do something much more systematic. Okay? And here's how we do it. We first take this approach where we say that this element of interest which has coordinates Xe and Xe plus one. We regard this as being the physical representation of the element, right? Or this is the element in the physical domain. Okay? So we call this the physical domain. In order to construct our basis functions we always think of any element of open arbitrarily element has been constructed from a mapping, right, from a different domain. And in this domain the element is going to look similar, okay, except it is that. I'm not going to use X for position along this right, where, whereas I have X for positional in the physical domain, I'm not going to use X in the, in this domain. In this domain I'm going to use z. Right, as a coordinate. And this point here, the node on the right is z equals one. The node on the left is z equals -1. And the mid point is z equals 0. It's important to note that I have not introduced three nodes. The nodes that I think of are still z equals -1 and then z equals 1, okay? I just marked the midpoint, the midpoint is not a node, okay? So these are the nodes. Right in this representation. Now this is called various things. Some people may call it the mathematical domain, I call it the bi-unit domain. Bi-unit because I'm saying the length of my element in this domain is, goes to minus one to one is two, right? So it's two times one, so it's a bi-unit to me. Now we always think of our actual physical element as being constructed from a mapping, as in mapping from our bi-unit domain. okay. Why do we do this? We do this because it becomes very convenient to define things like our basis functions and later on to carry out integrations if we have this idea of the bi-unit domain. Okay, and in fact, here is what we'll do. Our basis functions, N1 of x. We will write as N1 of x, but really x, remembering that x is a mapping from this bi-unit domain, right, x is a function of z. And likewise N2 x equals N2 x mapped from the domain, from the bi-unit domain. Okay? Now, strictly speaking, there is an, an abuse of notation going on here because I cannot have a function parametrized in terms of x and then use the same symbol for the function and parametrize it in terms of z, okay? But I will nevertheless follow that abuse of notation because otherwise our notation just gets much too unwieldy. Okay. Alright, so now that we are in this bi-unit domain, it becomes very convenient to write out our basis functions. Here they are. N1 of z equals 1 minus z divided by 2 and N2 of z equals 1 plus z divided by 2. Okay? Completely clear what's going on here. We have z equals -1, z equals 1. That is our point z equals 0. If you just evaluate 1 minus z over 2 what we, you will observe is that at z equals -1, h takes on the value 1, okay? So, this is the value 1. Okay? And it goes to 0, x equals 1. Okay? This is our function and 1, sorry. That is N1. For N2, let me try and use the ability here to go to a different color. For N2 I have that. Okay? That is N2, that is again the point one, and you observed that when, where at xz equals -1, and 2 drops to 0. Okay, so I can just those facts here, okay. What we observed is that N1 at z equals -1 equals 1. N1xz equals 1 is 0. And 2 at z equals -1 is equal to 0. And, N2 at z equals 1 is 1. Okay? Now if we write these points z equals-1 and z equals 1 as let me call this point z1, and this point is z2, okay? Just corresponding to the idea that the node on the left in this bi-unit domain is node 1 for the element. And the node on the right in the bi-unit domain, this one, is node 2, okay? What that then lets me say, is that Na okay, whichever shape function, whichever basis function you want to use. By the way, basis functions are often called shape function. Okay? I just prefer the term basis function. Okay. If you cha, if you take basis function a and you evaluate it at zb, this, is, the Kronecker delta. Okay? And you know what the Kronecker delta does, right? Kronecker delta is equal to 1 if A equals B, and it's equal to 0 if A is not equal to B. Okay, the basis functions that we have described here, have the so called Kronecker delta. Property. All right? Okay so we've got as far as describing our simplest one of, you know, indeed our simplest basis functions for this finite element formulation. The simplest of finite element formulations. We will continue with this, but this is also a good place to end this segment.