CHAPTER 10: INTRODUCTION TO NUMERICAL METHODS: Solution of 2nd Order Linear Elliptic PEDs   

In this segment I will show you how to solve elliptic partial differential equations numerically. If you want to get more information on elliptic partial differential equations just go to the website, click on keyword, and click on elliptic partial differential equations. So let's go and review what we talked about, how we can look at second order linear partial differential equations. So the differential equation which is shown to you is a second order linear partial differential equation with two independent variables. So this particular partial differential equation has two independent variables; x and y. And it has one dependent variable; u. And we can recall the criteria for this equation to be elliptic; B^2 minus 4*A*C has to be less than 0. So if we take a typical example of a Laplacian equation here which is given by this particular second order linear partial differential equation. In this case we have A equal to 1, B is 0 because there is no term of d^2T/dxdy and C is equal to 1. That gives us B^2 - 4AC is 0 - 4*1*1 which gives us minus 4 and since this number is less than 0, we can classify this equation as elliptic. So let's go and look at a physical example of an elliptic partial differential equation. So here we have a plate and it's of length L, width W. It is maintained at four different temperatures on the four different edges of this rectangle and the equation which governs the temperature is given by this Laplacian equation here. So, the temperature of inside the plate at steady state, given as a function of x and y will be determined by solving this particular second order linear partial differential equation. And let's go and see how we can do that numerically. If you take this particular rectangle which we have right here, and what we are going to do is we are going break the x-axis length, or the length of this particular plate into m segments which makes the length of each of those segments to be L/m. We do the same thing, we break up the width of this rectangle into n equal segments because delta y is equal to W/n. So basically what we have is a grid, a rectangular grid which is of which this particular rectangle plate is made of and if we look at what happens at this particular point, (x,y), the orginal second order linear partial differential equation which we have, governs the particular temperature at that point. Now, in order to be able to solve it numerically, what we are going to do is we are going to take the second derivative of temperature with respect to x and what we are going to do is we are going to write it in the central divided difference formula for it. You can derive the central divided difference formula for the second derivative of x, of second derivative of temperature with repect to x by using Taylor Series of two variables. Similarly, you will find out the second derivative of temperature with repect to y is given by a similar central divided difference formula as shown here. And as you can see here, since we are taking only the partial derivative with respect to x, only the x value changes. Since we are taking the partial derivative with respect to y, only the y value changes. But one of the things which you have got to understand is that what are these points? What is this point, what is this point, and what is this point? So if you are finding the second derivative of temperature with respect to x you are using the value of the function of the temperature at this point, the value of the temperature at this point, and the value of the temperature at this point to be able to find out what the approximation for second derivative with respect to x is. In order to find out the approximation to the second derivative of temperature with respect to y, you are using the value of the temperature at this point, the value of the temperature at this point, and the value of the temperature at that particular point. Now, let's go and write these down, these approximations for particular nodes. So, since this is a node which is at the location (x,y) what we are going to do is, we want to name the nodes, we want to number the nodes. We want to say hey, this is node i and that's, node i in the x-axis and node j in the y-axis and the surrounding nodes which we have along the x-axis which will be (i+1,j) and on the left will be (i-1,j). The node which is above the current node is (i,j+1) and the node below is (i,j-1). So we basically are replacing the actual coordinates of x and y by node numbers, that's all we are trying to do. So, we are seeing that the second derivative of temperature with respect to x. If we want to write the second derivative of temperature with respect to x at (i,j) node, rather than saying that hey, it is at, one of the terms is T at (x + deltax,y) we will say that hey, it is the value of the temperature at i+1,j which in fact is the coordinates of this particular point here are x + deltax,y so nothing has changed. It is just that we are numbering the nodes now. So similarly, this will be the (i,j)th node, this will be the (i-1,j)th node, and so we are rewriting our formula for, in terms of the node numbers. Similarly, is happens for the other second partial derivative which is respect to y. We again converting the locations of the coordinates here into node numbers, that's what we are doing here in this formula here. So now what we have is the second derivative of temperature with respect to x at the (i,j)th node. We have the second derivative of temperature with respect to y at (i,j)th node and we are going to substitute this back into the original second order linear paritial differential equation to write down the equation just in terms of the temperatures at particular nodes. So this is our original second order linear partial differential equation. We approximated this by this at the (i,j)th node. We approximated this at the same (i,j)th node by this, so we add the two. Now, rather than having partial derivatives we are writing it in terms of the temperatures at particular points and the length and the widths of the grid wich are delta x and delta y. If, for purposes of keeping thing simple, we choose delta x equal to delta y. We can multiply both sides by (delta x) squared where we will get an equation without the use of the length and the width of the grid and we will get it as follows: So what you are seeing here is that you are, you have discretized the elliptic partial differential equation. You have written the partial differential equation at node (i,j) and you have introduced 4 new unknowns in there because now your temperature is at this node, this node, and this node, and this node included. So what we have to figure out is that how can we write more equations so that we can find out the temperatures at each and every node, which is in that, at those grid points or the nodal points, how can we find the temperature of those? But the main issue which we have solved here is to see that how we have discretized out second order linear partial differential equation and find some ways of solving them. There are three different ways by which you can solve this discretized elliptic partial differential equations. One is called the Direct method, another is called the Gauss-Seidel method, and the third one's called the Lieberman method and we will be discussing each of these methods in three seperate audio-visual lectures, so that depending on whether you want to learn all three methods or just one method you'll be able to do so. And that's the end of this segment.