[MUSIC]
Now we need to consider the boundary conditions on the cylinder.
There's two physical principles that we need to use.
The first is called No penetration.
The no penetration principle means that the fluid cannot flow
into the cylinder or flow out of the cylinder.
That means that the radial velocity of the fluid has to be zero.
And if we translate that into the stream function, it means the partial
derivative of the stream function with respect to theta has to equal zero.
The second physical principle is called the no slip condition.
For no slip it means that the cylinder has some friction associated with it.
So if the fluid is touching the cylinder,
it can't be moving along the direction of the cylinder.
That translates into u in the theta direction has to be zero.
Which for the stream function it means the partial of the stream
function with respect to r has to be equal to zero.
Okay, so no penetration and no slip.
What does that mean in terms of the the stream function on our grid?
We have the stream function at the first grid point in
the chipsi direction, which is on the cylinder.
What is that value?
Well, we know that the stream function is independent of theta.
The cylinder is at a fixed radius and the boundary is prescribed by moving theta.
So the fact that psi is independent of theta means that the stream
function is a constant on the cylinder.
And we know that the stream function value at the midline is zero.
So the value of the stream function on the cylinder also has to be zero.
So psi one j, so for all angles, has to be zero on the cylinder.
The second condition that the normal derivative of psi equals zero
would mean that the normal derivative whether you use r or
the r modified coordinate chipsi doesn't make any difference.
That also has to be zero.
So the second condition is that the partial derivative
of psi instead of r, now I'll use our modified coordinate
chipsi at the one j grid point also has to be zero.
So now we have a boundary condition on psi on the cylinder and
a boundary condition on the normal derivative of psi.
What we really need is a boundary condition on psi and
a boundary condition on the vorticity on the cylinder.
So in order to get the boundary condition for the vorticity,
we need to first write down the equation for psi that relates it to the vorticity.
So that equation is that the vorticity is equal to the negative
of e to the minus two chipsi times the second derivative of
the stream function with respect to chipsi plus the second
derivative of the string function with respect to theta.
But we can simplify this on a cylinder.
This equation is valid everywhere in the domain but
on the cylinder chipsi is equal to zero.
So this exponential is one.
The stream function is independent of theta.
So this second derivative is zero.
So then this reduces to negative phi squared psi phi
chipsi squared and that's true on the cylinder.
Okay, so it relates the vorticity to the stream function.
Now the problem here is that it relates to the vorticity to the second derivative of
the stream function.
But our boundary condition is on the first derivative of the stream function.
So in order to convert this derivative, this boundary condition on
the first derivative to a boundary condition on the vorticity,
we need to make use of the Taylor series expansion.
So we need to do now two Taylor series expansions.
So let me show you how that works out.
The first Taylor series expansion will be Taylor series expanding
the stream function, one grid point away from the cylinder.
So two comma j.
With Taylor series expanding the stream function one point
away from the cylinder around its value on the cylinder.
So the constant term here will be the value on the cylinder.
The next term we're one grid point away.
So it's a grid spacing h, and then we have the derivative
of psi with respect to chipsi evaluated on the cylinder.
The next term in the Taylor series expansion is
h squared over two factorial or
two times the second derivative of psi on the cylinder.
And because we want to derive our boundary conditions that
are accurate to second order,
we need to keep going in this expansion to the third power of h.
So we have h cubed over three factorial which is six.
And then we have the third derivative of psi
with respect to chipsi cubed on the cylinder.
And the final the error term here which I'll carry, so
you see that we actually are getting to second order is big O of h, to the fourth.
So that means we're dropping terms of order h to the 4th and higher.
So that's the Taylor series expansion of psi one grid point away from the cylinder.
We're going to also need the Taylor series expansion of psi two grid points
away from the cylinder.
That will be psi sub three comma j.
So again we're expanding about this value on the cylinder.
Now we're two grid points away so h becomes 2h.
And then we have the partial of psi with respect
to chipsi on the cylinder, h becomes 2h.
So this becomes 2h squared which is 4h squared divided by two is a 2h squared.
So we have a 2h squared.
The second derivative of psi with respect to chipsi on the cylinder and
now we have 2h cubed which becomes 8h cubed.
So this term I'll write as 8h cubed over six times the cubic,
the third derivative of psi on the cylinder and
the error term here is still of big O h to the fourth.
Okay these are our Taylor series expansions.
Now we want to make use of the boundary conditions.
So the value of psi on the cylinder Is zero.
So these terms are zero.
The first derivative of the first normal derivative of psi on
the cylinder is also zero.
So this term is also zero.
Okay, and then the second derivative here,
the negative of the second derivative is omega.
Which tells us that this term here,
the second derivative then is the negative of omega.
So this term here becomes negative omega on the cylinder.
So that's one comma j.
And this term then is negative omega one comma j.
So we're making use of the value of omega on the cylinder here and
then this is the h cubed term.
Okay so let me write this so we can see how this works then.
So we have psi two comma J, zero, zero.
So this is equal to h squared over two,
negative omega one comma j plus this
term which we don't know h cubed over six.
The 3rd derivative of psi with respect
to chipsi one comma j plus the error term plus big O h to the fouth.
Okay, so this first Taylor series expansion reduces to this.
The second Taylor series expansion two grid
points away from the cylinder reduces to here.
So we have now a negative two h squared,
negative two h squared, omega one comma j.
And this h cubed term now is eight H cube
over six times the third derivative at one comma j,
plus our error which is H to the fourth.
Okay now we're in a good position.
Our good position is we want to solve for the vorticity on the cylinder but
we want to eliminate this h cubed error term to get a second order method.
So to do that we need to multiply to eliminate this term.
We multiply the first term first equation by eight and
then we subtract the second equation.
Right? So when we do that we get eight times
psi two comma j, psi of three comma j.
So we have eight divided by two is four.
So that's negative four and then minus minus two.
So negative four plus two gives us a negative two.
So that's going to be equal to negative two h squared.
Omega one comma j.
And then this cubic term in h cancels, that's the whole point.
And the remainder then Is going to be terms starting with h to the 4th.
Okay now what we do then is we solve for the vorticity.
So our final result then will be omega.
I'll write it in blue, omega one J.
So we divide through by minus two h squared.
So we can use the minus sign to change the order here.
So we have psi three comma j minus eight.
Psi two comma j divided by two h squared.
And then because we divided through by h squared, the error term here is h squared.
So that gives us our second order method.
Okay, that's the boundary condition we want on the vorticity.
So we have a zero boundary condition on the stream function and
we have a boundary condition on the vorticity that now depends on the value
of the stream function, one grid point and two grid points away from the cylinder.
I think this is the most complicated mathematics in this course and
it's also extremely crucial for the numerical method.
So it gives us a source of vorticity due to
the interaction of the flow field with the cylinder.
So let me summarize to determine the cylinder boundary condition.
We use two physical principles.
No penetration condition and the no slip condition.
That gives us a boundary condition on the stream function and
on the normal derivative of the stream function.
Unfortunately, we need a boundary condition on the vorticity.
So we work a double Taylor series expansion,
one grid point away from the cylinder two grid points away from the cylinder.
And then use the stream function equation to replace the second derivative
of the stream function with the vorticity.
And that gives us then appropriate equation for
the vorticity on the cylinder and the vorticity boundary condition.
I'm Jeff Chazanov.
Thanks for watching and I'll see you in the next video