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Numerical Solutions to Laplace's equation in a wedge

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Meurig
#1
Jan23-12, 09:07 AM
P: 6
Hi all,

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

[itex]\bigtriangledown^2\phi(r,\theta), \ r>0, \ -\alpha<\theta<0[/itex]

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

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

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

In addition I have the far field boundary conditions:

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

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

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

[itex]\eta=-\frac{A\alpha\sin\alpha}{\pi}+\eta_1 r[/itex]

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



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
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meldraft
#2
Jan24-12, 11:00 AM
P: 280
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
#3
Feb13-12, 09:42 AM
P: 6
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 [itex]\bar\theta=0[/itex]).

This is a graph produced using the analytical result:


And using my method:


The complete write up of what I'm doing is here:
https://docs.google.com/open?id=0B-b...Q2NGMyNjVlY2I2

Any help would be greatly appreciated,
Meurig

fluidistic
#4
Feb13-12, 08:23 PM
PF Gold
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P: 3,225
Numerical Solutions to Laplace's equation in a wedge

Quote Quote by Meurig View Post
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 [itex]\bar\theta=0[/itex]).

This is a graph produced using the analytical result:


And using my method:


The complete write up of what I'm doing is here:
https://docs.google.com/open?id=0B-b...Q2NGMyNjVlY2I2

Any help would be greatly appreciated,
Meurig
Unfortunately I cannot see the graphs and I guess I'm not the only one.
Meurig
#5
Feb14-12, 04:41 AM
P: 6
Apologies, thanks for bringing that to my attention.

The numerical results:


The analytical results:


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 [itex]\hat\phi[/itex] and [itex]\hat\eta[/itex]), if anyone could point me in the direction of some reading material on this matter it would be much appreciated.
meldraft
#6
Feb14-12, 03:23 PM
P: 280
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.


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