So, as, as in the 1D problem, the finite. Dimensional Weak form is the basis of our finite element formulation of the problem. Okay. Now, and, and of course, what we have to do is essentially define or construct the spaces that we need. We need to define S-H and V-H. In order to define these spaces, what we will do is construct a partition of our domain, omega. So we define S-H and V-H by partitioning. Omega into. Subdomains. Omega sub e. Where I think we were working with an omega sup e actually, so let's just stick with that. Omega sup e, okay? Where, e, once again, runs between 1 and, nel. Right, nel as before, stands for number of elements in the partition. Okay? And, furthermore, since each omega e is a subset of omega, which itself is a subset of r 3, we are indeed talking of each element sub-domain omega e as being a three dimensional sub-domain. Okay, so let me just draw the picture that we have in mind and then we'll say more things about it. Okay, so this is omega, and what we are trying to say here is that we have we're thinking of partitioning this cell sub domain. Then, sorry, there's domain omega into sub domain such as this one. And maybe another one, right next to it. All right, and so on. All right, and maybe we'd call this one omega e1 We'd have omega e2, right? And so on. Okay? So mathematically, just as we had before, we see that omega is for each omega E Really is open in, in R 3. Right, so, we consider each of these formula at least to be an open subset, and therefore our total domain omega which is the union of each omega e. e running from 1 to number of elements, right? And so, and, and you recall that as in the 1D problem to make sure that we pick up, all the interface points, right? The points of the interface between two of these elements of domains. We apply closure on the left and right, okay? And that is an, an equality. All right, and of course we also know that omega E 1, intersection, omega, E 2 is the null set. Okay, since each, since each of them is open, their intersection is is definitely a null set, right? Because we we, we don't get any boundary. So, then, but but, again, that's really, a technical thing. We do indeed, imply that they are non-overlapping elements. But, what happens on their boundary is really a technical thing. Okay, so, this is the setting. What we have to do now is, talk about how, how we construct these these sub-domains. And once we do that, we will also be on our way to defining the the underlying finite element basis. Okay, there are, of course, many ways to partition these partition the domain omega into subdomains and three dimension. What we will start off with is actually not even the simplest search partition but it is it is indeed the one that is most easy to get to given what we know about the 1D problem, okay? And what that implies is that we are going to construct a partition. Where each omega E. Is a, is, is what we call a hexahedral element, okay? So, consider hexahedral. Element sub-domains. Omega e, all right? e going from 1 to, and yeah. Okay, so, what we are talking of here is the following: We have this picture We have one of these element subdomains. Right? And I'm going to denote this one omega E. That is our domain omega. Right, so that's omega e. Right? Now in order to proceed with our final element formulation let's pull this element out of there, right? Let's examine it more closely. Okay? So that is our subdomain. Right? This is omega e. Okay, the simplest case that we are going to start out looking at is where on this hexahedral element, we are going to construct basis functions that are trilinear. Okay? So we, we will consider. Trilinear. Basis functions. Okay? Now if we are talking of cons of considering trilinear basis functions, what this implies is that. We need to do this by introducing nodes over these hexahedra elements where we have one, one node at each Vertex point of the hexahedral element and no more, right, no other nodes. So, our nodes are at the hexahedral points, sorry, at the vertex points of the hexahedron. Okay, so we have. There we go, all right? This means we have eight nodes, okay? So this is when we're going to work with trilinear basis functions. We have there, elements of the means are sometimes also called eight noded elements. Right? They're sometimes called bricks, a very colloquial way of describing their, their construction. Okay. Okay, so that's the setting. Now what we will do is as before we will denote coordinates for physical coordinates for each node okay if this so this is element omega E using the idea of local Numbering of degrees of freedom and of nodes. Let us suppose that this is node a for this element. Okay, so, we are going to denote the coordinates of that point as x sub e sub A, okay? And of course, your A equals 1 through 8, okay? All right, so when I have X, A, sub E, i'm going to see that this is local, this uses a local numbering of notes, all right? Okay, and, the numbering that's followed is. Actually we get to the numbering that's followed in just a little, let's, let's, let's just stick with this now. All right, if we have if we know that a runs over one through eight, let us then use that to write out our basis function straight away, okay? So now what we have is Sub E, right? So that's In element E, is the sum, A going from 1 to number of nodes in the element, which we know by the way, is 8, right? Because of the fact that we've chosen to construct trilinear basis functions, right? With this we have our basis functions and we're going to use our, the same notation that we had before right. These basic function are general, written as, being parameterized by physical position, right? X, by position, physical space x. And, just as we had before, we are going to multiply them By decrease of freedom N D A sub E, and here too I'm using a local numbering degrees of freedom Right, and this local numbering of degrees of freedom essentially reflects the local number of nodes. All right, we have this. And then we also have w h in element e is the sum. A going from 1 to number of nodes of the element. N a function of physical position C B E. Okay. And at this point thinks look very much like thinks before except for the fact that we have a much, greater number of nodes, but we know, how these nodes are positioned right there, they're vertices of a, of, of a hexahedron. Okay, so now if you think back to the way we proceeded in 1D, we took our physical element and regarded it as being obtained as through a mapping from a so called parent domain. Right? We're going to do exactly the same here. Okay? So we will think of omega b as being obtained. By a mapping from a parent domain. Again, we're going to denote that as omega sub c. All right, having stated that this is probably even stop this segment. Because once we get into talking about pairing domain omega c it is going just difficult to extract ourselves very quickly to end the segment, so we just pick up that definition in the following segment. Okay.