We have our governing equations in polar coordinates,
but one of the problems with polar coordinates is that we
have this Laplacian operator in our equation,
and it has this somewhat complicated form r
d dr, r d dr.
It would be much easier if this was
just a second derivative of some variable.
What we want here is for
this quantity r d dr to just be d,
d some new variable which I'll call the Greek letter Ksi.
This is what we would like.
To do that, we can change variables to make it happen.
We want to try r
being a function of Ksi where this derivative then,
d, d Ksi is equal to the derivative of r
with respect to Ksi times the
partial with respect to r from the chain rule.
Now, if we compare this expression with this expression,
we see that this derivative of r with respect to
Ksi should be equal to r. We have dr,
d, Ksi, equal to r, which is just
as simple, ordinary differential equation.
The derivative of a function is
equal to the function itself.
Thus the solution to that is
the exponential function of Ksi.
Then at the same time,
we can set some initial value.
We can set that the radius when Ksi
equal to zero can be on the surface of the cylinder.
Now in the dimensionless units,
the radius of the cylinder is one.
To be on the surface of the cylinder,
this should be equal to one.
Ksi equals zero, then will designate
the surface of the cylinder.
Now this differential equation with
this initial condition is
just the simple exponential function,
so that r of Ksi is simply e to the Ksi.
The derivative of this is e to the Ksi,
so it gives us back r and that
Psi equals zero, this is one.
It's called log-polar coordinates,
because if we take the inverse of this,
then we get Ksi equals
log r. That's the natural logarithm.
What happens? Then our Del squared
operator in this coordinate system,
we have a one over r-squared,
is then an e to the minus 2 Ksi.
Then this becomes simply d squared,
d Ksi squared plus d squared, d Theta squared.
Apart from these extra exponential factors,
at least our derivatives are very clean looking
more or less like
a Cartesian coordinate system derivative.
That will make the finite-difference approximation
much easier to do.
What are our governing equations?
The first equation will be d squared Psi,
d Ksi squared plus d squared Psi, d Theta squared.
Then taking into account the exponential function,
we have minus e to the 2 Psi times the scalar vorticity.
This becomes this stream function equation.
Then the vorticity equation will become the Omega dt,
plus, now we have an extra factor e to the minus 2 Ksi.
Then we have partial of Psi with respect to Theta,
partial Omega with respect to
Ksi minus partial of Psi with respect to Ksi,
partial Omega with respect to Theta.
That's the left-hand side.
That's equal to 2 divided by the Reynolds number.
We have also a factor of e to the minus 2 Ksi.
Then we have d squared Omega,
d Ksi squared plus d squared Omega, d Theta squared.
A simple operator that looks a lot
like the Cartesian coordinate Laplacian.
Let me summarize.
Motivated by this ugly-looking Laplacian that has an r,
d, dr, r,
d, dr in it,
we would like to change coordinates from
r to Ksi to get rid of this
and replace this by partial squared with
respect to Ksi squared coordinate system.
To do that, the relevant change of
coordinate is this log-polar coordinate,
where Ksi is equal to the log of the radial variable,
or what we actually use in the calculation is
r is equal to the exponential of the new variable Ksi.
Then the Laplacian become
simpler with this extra exponential factor.
Our equation then, the derivative,
the Laplacian term in our equation,
also becomes more like
a Cartesian coordinate system and we
just need to take into account
these extra exponential factors.
These will be our equations that we solve
for our flow around a cylinder.
I'm Jeff Chasnov.
Thanks for watching,
and I'll see you in the next video.