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Meurig
Jan23-12, 09:07 AM
Hi all,

I am trying to construct a numerical solution to the following linear harmonic problem posed in a wedge of interior angle 0<\alpha<pi/2

\bigtriangledown^2\phi(r,\theta), \ r>0, \ -\alpha<\theta<0

\bigtriangledown\phi\cdot\mathbf{n}=0, r>0,\ \theta=-\alpha,

\frac{\pi}{\alpha}\eta(r)-2r\eta_{r}-\frac{1}{r}\phi_{\theta}=0, r>0, \theta=0,

(1+\frac{\pi}{\alpha})\phi - 2r\phi_{r}+(1+\sigma\tan(\alpha))\eta =0, r>0, \theta=0,

In addition I have the far field boundary conditions:

\phi=r^{\frac{\pi}{2\alpha}}\sin(\frac{\pi\theta}{ 2\alpha}) as r\rightarrow\inf

\eta=\frac{\pi}{4\alpha}r^{\frac{\pi}{2\alpha}-1} as r\rightarrow\inf.

And the solution local to the tip of the wedge given by
\phi=\frac{A\alpha\sin{\alpha}(1+\sigma\tan{\alpha })}{\pi(1+\pi/\alpha)}+rA\cos(\theta+\alpha)

\eta=-\frac{A\alpha\sin\alpha}{\pi}+\eta_1 r

where A and \eta_1 can be approximated through solving the near field boundary condition
\phi_\theta +r\tan(\theta+\alpha)\phi_r=0, r=\epsilon, -\alpha<\theta<0



So far I have attempted constructing a finite difference approximation in terms of polar coordinates, but as I iterate this scheme the error increases exponentially until phi approaches infinity.

I wonder if anyone has any ideas with regards to what I should be looking to do/what I should be weary of.

Cheers,
Meurig




*edit to correct latex

meldraft
Jan24-12, 11:00 AM
How do you calculate the error, or rather, are you sure numerical error is to blame?

Where does phi become infinite? For instance, it is supposed to be infinite where a source lies.

Meurig
Feb13-12, 09:42 AM
Thanks for the reply, I think I've managed to sort that issue.

The problem I'm having now is in the implementation of the free surface boundary condition (when \bar\theta=0).

This is a graph produced using the analytical result:
https://lh4.googleusercontent.com/YGjbspVigI58cmMU02IvXSFxVreWQ4B4jrsToC--TNzEAGf3O0_CjIu6jouvfx1xXE5U6SyLdQ4f9uqKwS55rB6Zs6 3B3W8ep9A=w1600

And using my method:
https://lh5.googleusercontent.com/pwIa6Y4c4oe1P0RrazI5pgpQm5rrEc4azNujnt7HVNh9zDtO07 8kl4yQrXfZiPOjHswI0qRyc1-XcBdrA3UVFpS2ciMFAwh6G9U=w1600

The complete write up of what I'm doing is here:
https://docs.google.com/open?id=0B-bjkxDlh3fxOWQ3NTU5NjctOWZhMi00MDI1LWJhODItOTQ2NGMy NjVlY2I2

Any help would be greatly appreciated,
Meurig

fluidistic
Feb13-12, 08:23 PM
Thanks for the reply, I think I've managed to sort that issue.

The problem I'm having now is in the implementation of the free surface boundary condition (when \bar\theta=0).

This is a graph produced using the analytical result:
https://lh4.googleusercontent.com/YGjbspVigI58cmMU02IvXSFxVreWQ4B4jrsToC--TNzEAGf3O0_CjIu6jouvfx1xXE5U6SyLdQ4f9uqKwS55rB6Zs6 3B3W8ep9A=w1600

And using my method:
https://lh5.googleusercontent.com/pwIa6Y4c4oe1P0RrazI5pgpQm5rrEc4azNujnt7HVNh9zDtO07 8kl4yQrXfZiPOjHswI0qRyc1-XcBdrA3UVFpS2ciMFAwh6G9U=w1600

The complete write up of what I'm doing is here:
https://docs.google.com/open?id=0B-bjkxDlh3fxOWQ3NTU5NjctOWZhMi00MDI1LWJhODItOTQ2NGMy NjVlY2I2

Any help would be greatly appreciated,
Meurig
Unfortunately I cannot see the graphs and I guess I'm not the only one.

Meurig
Feb14-12, 04:41 AM
Apologies, thanks for bringing that to my attention.

The numerical results:
http://i39.tinypic.com/2mljn.png

The analytical results:
http://i42.tinypic.com/29wsf7q.png

As you can see, the issue lies with the values at the free-surface.

I've never dealt with any boundary conditions involving two separate functions before (here \hat\phi and \hat\eta), if anyone could point me in the direction of some reading material on this matter it would be much appreciated.

meldraft
Feb14-12, 03:23 PM
Your two domains are different. Which of the two is supposed to be the correct geometry? Also, the domains are really off since none of the dimensions match, and neither does the number of boundary segments. Unless you get the domains to have the same contours you cannot compare your results.

I am not familiar with near or far field BCs, but what I can say at this point, is that your plots definitely defer by one boundary condition. Supposing that you are plotting equipotential lines here, in your numerical plot the top horizontal line has a Neumann boundary condition (streamlines are perpendicular to the boundary). You have probably not defined that boundary condition in you analytical solution, which is why your analytical plot looks like this.