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.
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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.
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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.
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