With this segment, we begin our treatment of problems in multiple dimensions. And we're going to go straight away to three dimensions here. The problem we'll start out with is, in the context of mathematical description of PDEs. Of also a linear, elliptic PDE in three dimensions, with the further specification that the unknown we're solving for is a scalar. So that's what I'm going to write out, and we'll start with it, so we're now looking at, Linear, Elliptic, PDEs, In three dimensions Okay, with scalar unknowns, or scalar unknown writers, a single unknown we're solving for, So linear, elliptic, PDEs, three dimensions, scalar unknown. The canonical physical problems that are described by this sort of PDE Include heat conduction, steady state heat conduction, And also at steady state, The mass diffusion problem. All right, okay, we'll plunge right into it. We'll start out with the strong form, so strong form of the problem we're looking at is the following. So let me do this, before I start writing out the strong form, let me just sketch out the domain that we're trying to solve things on. Just as for the 1D linear elliptic problem, we sketched out this idea for bars sort of embedded in a wall, fixed in a wall, so let me do that. Now because we're doing things in 3D, we are going to make use of vector notation, so, we have here, What I would refer to as our, Basis vectors, and these will be denoted by e1 e2, e3. Note that the notation I use for vectors is to put an under bar on them. That's just the convention I follow, when we get to needing tensors, I will do the same thing. And we will simply distinguish between vectors and tensors by context. Okay, so the domain we're interested in is some arbitrary domain. And I will draw what, in the context of continuum physics, is often referred to as the continuum potato. So we have that here, the domain of interest here is labelled omega, as before, except that omega now lives in 3D. Okay, so let me write out things here, ei where i = 1, 2, and 3, the set ei where i = 1, 2, 3, constitutes, An orthonormal, Cartesian basis. Okay, do you recall what orthonormality means? The ortho refers to perpendicular, and normal refers to unit magnitude. So what that means is that if we look at ei dotted with eg, this is equal to delta ij, where delta ij is the Kronecker delta. And you remember what the particular properties of the Kronecker delta are, where delta ij = 1 for i = j, is equal to 0 for i not equal to j. And we note that this directly covers zero through normality property. Because if i = j, it says that each of these es, e1 e2 e3, is of unit magnitude. And if i is not equal to j, it tells us that they are perpendicular to each other. So that's what is implied, and here I will make it more obvious, by doing this. Okay, so these are perpendicular to each other. Cartesian, for our purposes, simply means that they are fixed, the basis vectors do not change, they're fixed in space. Let me also just, for the purpose of making this completely obvious, state that we are doing all of this in three dimensional ambient space. Okay, so this is the setting for the problem we want to consider. I have some props to help us for the rest of these lectures. That represents our basis, you can think of this as e1, e2, e3. Each of these are of unit magnitude, and they're, of course, perpendicular to each other. The domain of interest to us is this one, this is our representation of the continuum potato. It happens to be a University of Michigan Nerf, well, not Nerf, but a University of Michigan foam football. But this is the domain over which we will be describing everything happening, this is omega, for our purposes. The other thing that we will need about omega is, we will repeatedly refer to its boundary, so omega is, as before with our domains, omega's an open set. So when I talk about omega, I don't include its boundary, we will use certain notation for the boundary of omega. So let's put down that bit of notation, the boundary is going to be denoted partial of omega. That does not imply that we're taking anything like a derivative, for our purposes, it's just notation. Okay, so that's pretty much what we need to begin with. So yeah, maybe I should just say one more thing here, omega is open in R3. And if this is the first time you're encountering it, I will also say that partial of omega is the boundary, Of omega. Okay, so this setting we have, in this setting, what we're trying to do is the following. Let's state this long form, we're interested in finding some function u. Find u given, Some other quantities, so we are going to be given f again. f, as before, is a function of position, except that now position is a position vector, it's going to be denoted as x. And at the risk of going back and forth, let me just add one more thing to that figure I had on the previous slide. I'm going to say that some point x here has a position vector, so that's the position vector of point x. In the context of our props, so we are talking of the position vector from the origin of this Cartesian basis. To some arbitrary point in this domain, in our continuum potato, in omega. So we remember that this is where the basis is, we're talking of the vector from here to there. That is what we have in mind I could use this as a prop. So if the tail of this pen, the top of this pen, is where the Cartesian basis was, and we wanted to talk about this point, we have that. Sorry, that's the position vector for x. So we need that, because we want to talk about the dependence of the forcing function. So the forcing function is a function of x, it's parametrized by x, it can be defined at any point over the domain. So we're given f(x), as before, we are given u sub g, and we are given j sub n. We're also given the constitutive relation, We're given the constitutive relation, that I will initially write using what is called initial or coordinate notation. We're given a constitutive relation ji = -Kappa ij u,j, Where, again, since this is sort of the first time I'm writing it out, I'm going to tell you that i, j = 1, 2, 3. We're given all of this, so we want to find u given all this information, such that, Such that, Again, sticking with coordinate notation, -j i,i = f in omega. And we also have boundary conditions, BBC for boundary conditions, u = u sub g on partial of omega sub u, and -j dot n = j sub n on partial of omega sub j. Well, that's our strong form, obviously, I have a lot of explanation to do here, I've introduced all kinds of terms, and I need to make them clear. Let's begin with something that we already know. Let me tell you more by what I mean by decorating the boundary with the subscripts u and g. So, we have our, Bases again, this is something that I'm going to end up drawing repeatedly over the next couple of segments. We have e1, e2, e3, we have our domain omega. And now, the boundary is partial of omega, and rather than mark it as partial of omega, I'm going to tell you what partial of omega u and partial of omega j could be. Let us look at that, actually let me make that an open interval, sorry. Let me look at that interval, so that would mean all the points that lie between those two braces. If you were to just walk around the boundary, that is what I may choose to call partial of omega u. And the complement of that set is what I'm calling partial of omega j. What this simply means is that we've taken our boundary, and partitioned it into disjoined subsets. Such that they're disjoint, like I just said, so partial of omega u intersection partial of omega j is the empty set. Phi here denotes the empty set, this symbol denotes intersection, so the intersection of these two sub sets is empty. And partial of omega is always written as partial of omega u union partial omega j. So, really, if you think about the way we've marked out these boundary subsets, I've just chosen to make partial of omega u open. But that means partial of omega j is closed, in order to make sure that we don't lose the boundary points between partial of omega u and partial omega j. And of course the union gives us the total boundary, partial of omega. Just a way of splitting up the boundary. And then what we are seeing when we go back to these boundary conditions, what we are seeing is that we have, we specified u = ug on one part of the boundary. Let me take off this brace bracket. If you are specifying u = ug on some part of the boundary, what kind of a boundary condition is that? Recall from our treatment of the problem in one dimension, that's right, that is the Dirichlet boundary condition. And the other term, the other boundary condition, is our Neumann boundary condition. Okay, so this i what we mean by the boundaries, by the boundary subsets and the boundary conditions. Let's go back now and talk about the other quantity we introduced here without much fanfare, j.
