[MUSIC]
Here are the governing equations for the unsteady flow.
The equation for the stream function in log polar coordinates.
So we have something that looks like the Laplacian and Cartesian coordinates.
We have these extra exponential factors.
We have a time derivative for the scale of vorticity.
We have the Reynolds number showing up in the denominator here.
Again, something that looks like Laplacian in the vorticity equation.
And then we have this nonlinear term here.
So we want to convert these now to equations on our finite difference grid.
We're going to use second order approximations for all the derivatives.
So let's start with the stream function equation.
You should know how to right down the second order method for the Laplacian.
So that will give us because we have
a minus sign here, we have 4 psi i
comma j minus psi i plus 1 comma j,
minus psi i minus 1 comma j,
minus psi ij plus 1 minus psi ij minus 1.
That's this term.
And that's equal to the right hand side.
So this term also division by h squared.
So we multiply through by h squared.
So we have an h squared, e to the 2 psi i.
And then omega at the high j point, okay?
So that's the discretization of the stream function equation.
This will be a matrix equation, will use a direct method to solve it.
Now we want to discretize the vorticity equation.
So the partial derivative of omega with respect to time
will then become a regular derivative at the ij value of the grid.
So d omega ij dt.
And then we have this term here.
So we have the Laplacian in here which will be 1 over h squared.
So we have a 2e to the minus 2 psi
i over h squared times the Reynolds number.
And this term here then will be the same as here but with a minus sign.
So it will be omega i plus
1j plus omega i minus
1j plus omega ij pluse 1
plus omega ij minus 1.
And then we have a minus 4 omega ij, okay?
So that's this Laplacian like operator and
then we have this term here.
So each of these center difference schemes for
the derivatives will have center difference divided by 2h.
So we'll have a 2h here in a 2h here in the denominator.
So that's a 4h squared.
So then we have a plus e to the minus 2
psi i over 4h squared and all that factor
multiplies these products of derivatives.
So these are centered differences.
So the derivative of psi then will
be psi i plus 1j minus psi i minus 1j.
The derivative of omega will be omega
ij plus 1 minus omega ij minus 1.
That's the derivative with respect to theta minus than
the derivative of psi with respect to theta.
So that would be psi ij plus1
minus psi ij minus 1.
And then the derivative of omega with respect to psi.
So that will be omega i plus 1j minus omega i minus 1j.
And then we close the bracket, okay?
So that should be the correct finite difference approximations.
So let me summarize, in this video then I'm
converting our couple of PDEs to differential equations on our grid.
We have the first PDE just converts to a matrix
equation for psi given if we know omega.
And the second PDE then converts to a system of ordinary
differential equations for omega on the grid,
using the spatial finite difference approximation for the spatial derivatives.
I'm Jeff Chazanov.
Thanks for watching.
And I'll see you in the next video.