So we are going to continue with our development of the Galerkin weak form, for this one dimensional linear elliptic PDE. Let's begin by recalling where we were. So we've got as far as here; if this is our domain of interest, or this is the physical problem we're looking at, we have this bar, we have our x axis, positions L0, okay? And this is our domain omega, without those endpoints. What we did already is to partition this by the use of points that we're calling nodes. We have partitioned it into subdomains, each of which is denoted as omega_sub_omega_e, indicating an element. The nodes that we have used are labeled x1, x2 and so on; in general, this would be the point X_e and that would be the point X_e+1. The point with x coordinate x=L is node, n - number of nodes, n nodes - and that indicating number of nodes and we also observe that the number of nodes is equal to our number of elements plus one in one dimension. Okay. In one dimension also, for the very specific manner in which we are partitioning omega for this simplest of finite element examples. Simplest of finite element cases. Okay. This is as far as we got and what this allows us to do is to write the Galerkin weak form as follows. We can now write it as a sum over e going from one to number of elements of an integral over omega e of wh comma x Sigma_h e d x equals sum over e going from one to number of elements integral over omega e_w h f a dx plus wh at l, which of course is our last node times t which is the traction times a_real. So this is our Galerkin weak form. Now observe that each integral really is redefined in this expression as an integral over the relevant element subdomain. What we need to do to proceed is to talk of how we are going to write out a representation for our trial solution, right? Our trial solution, uh of x and the weighting function. Okay? This is what we need to do. Now in doing this - so observe that because of the fact that we're able to use this decomposition of our integrals over these element subdomains, we really only need to worry about how to carry out this representation over an element subdomain. Okay? So let me state that. Need to represent uh and wh over omega e, right? Where, of course, e goes from one to number of elements. Right? So we have the advantage of needing to focus only upon what often gets called a local representation, right? Local being local to the element. Okay? All right. So let's look at how we can go ahead and do this now. So let's then focus upon a single element; that is xe that is xe+1 and omega e. Okay? All right. So it's over this subdomain that we need to write out our trial solution and weighting function. The way we can do this now is to define what we will call local basis functions. We defined local basis functions on omega e. Okay? All right. And what you're going to observe is as you may expect, there is going to be a finite number of these local basis functions. Okay? So we define local basis functions on omega e and let me also state that we will define a finite number of basis functions. Because we have the advantage of working just locally over the elements of domain, we're going to focus upon this element subdomain and define a finite number of basis functions over that subdomain, but then if there is a finite number of basis functions over a single subdomain and there is a finite number of subdomains, what we're going to observe, of course, is that there is a finite number of basis functions over the domain omega. Okay? So we will have a finite number of basis functions over omega e and therefore, over omega itself. So we have a finite number of basis functions. Naturally, of course, our weak form is going to be finite dimensional. Right? So that's the connection. So when it comes down to defining a finite number of basis functions, we are assured of having a finite dimensional weak form together with a weak form. Okay? That's how we want to think about it. All right. So let's talk about what these basis functions are going to be. We are going to take, in this initial entry level, very first presentation of a finite element method - we are going to take the simplest possible basis functions. We have an element and it is by choice, by the way, that I have chosen an element to have only two nodes on it. We are going to use those two nodes to write out - to expand fields of interest over the element in terms of just two basic functions. Okay? So what are we going to do is the following; we will define two basis functions over omega e. These basis functions are going to be polynomials. And since we are in one dimension, if we have two basis functions and the basis is complete, it follows that we are talking of representing functions as what type of polynomials. Think about that. If you have a complete basis where you are going to use just two basic functions in one dimension with polynomials, what kind of polynomials are we able to represent? Let me just add on that bit about complete basis. Well, you probably got the answer. What this implies is that we are essentially going to be representing functions here as linear polynomials on Omega e. Okay? This is the simplest sort of polynomial basis function we can adopt in the most standard type of finite element method. There are finite element methods in which we can even use constant polynomials, they need a little more work than we need to do - than we should be doing at this early stage of developing the finite element method. So we go with linear polynomials. Okay? All right. Let's actually try to write these down now. So the way this works is the follows; graphically, this is what we will do. We have omega e and I'm going to write the nodal coordinates here. Our basis functions are the following; they're linears, right? That one and that one. This is n1 for obvious reasons that we're going to call n2. Okay. So n1 is a function of position, and n2 is also a function of position. Okay. With these, what we will be able to do now is to write out our finite dimensional trial solution and for the very - just for this case, I'm going to specify that this is the finite dimensional trial solution over element e - restricted to e and obviously restricted to element e because it's over element e that we've defined our basis functions. So the way we write it out in this case is sum_e going from one to number of nodes in the element N_sub_n_sub_e is for a number of - the capital N is then for number, the little n is for nodes and e is element. Okay? So we have two nodes and the elements, so Nne is equal to two. So the way we write this is then N_A - function of position times some degree of freedom that is being interpolated. And we will write that degree of freedom as da_sub_e, which in detail is simply n1, functional position times a degree of freedom - sorry, that degree of freedom should be one - degree of freedom one corresponding to element e plus basis function 2 function of position times degree of freedom 2 corresponding to element e. Okay. So in this, what we have is that na of X is the basis function. All right? And da of sub e is the degree of freedom. Also this is any - is basis function A where A equals one comma two and da_sub_e is degree of freedom - A equals one comma two on element e. Okay.