- #1
Meurig
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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
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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