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#### nonsequitter

##### New member

- Feb 26, 2012

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for the Dirichlet problem in the exterior of the disc, again assuming that the solution is bounded.

So we derived the expression for the interior part of a disc in class: $u(r,\theta)=\frac{1}{2\pi}\int_0^{2\pi}g(\phi) \frac{b^2-r^2}{b^2-2brcos(\theta-\phi)+r^2} d\phi$, where b is the radius of the disc and $u(b,\theta)=g(\theta)$. It was fairly straightforward, except that we were able to just assume the form of the solution $u(r,\theta)=\frac{1}{2}\gamma_0 +\sum_{n=1}^{\infty} (\frac{r}{b})^n[\gamma_ncos(n\theta)+\delta_nsin(n\theta)]$, where $\gamma_n=\frac{1}{\pi}\int_{0}^{2\pi}g(\phi)cos(n\phi)d\phi$ and $\delta_n=\frac{1}{\pi}\int_{0}^{2\pi}g(\phi)sin(n\phi)d\phi$. We then played around with the sums of the series of sin and cos and took the real part of that to get our answer.

Now, for the exterior of the disc, I'm just not sure how to start off with the form of the solution. It seems like the $(\frac{r}{b})^n$ should really be the only difference, but I can't justify randomly picking a form. Any suggestions?