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Let $0<a<b$ where a,b are constants.

I want to solve this bad boy:

$\nabla^2 u = 0 \enspace a<r<b, \enspace 0<\theta<\pi$

$u=g(\theta) \enspace on \enspace r=a$

$u=h(\theta) \enspace on \enspace r=b$

$u(r,0)=0 \enspace , \enspace u(r,\pi)=0$

How can I solve for $u(r, \theta) $

Ok so I got a solution of :

$u(r,\theta)=\frac{1}{2}(C_0 + D_0 \ln{r}) +\sum_{n=1}^\infty (C_nr^n+D_n r^{-n})\sin{n\theta}$

And some constants:

\begin{cases} \frac{1}{2}(C_0 + D_0 \ln{a})=\frac{2}{\pi}\int_0 ^\pi g(\theta) d\theta\\ \frac{1}{2}(C_0 + D_0 \ln{b})=\frac{2}{\pi}\int_0 ^\pi h(\theta) d\theta\\ \end{cases}

\begin{cases} C_na^n+D_n a^{-n}=\frac{2}{\pi}\int_0 ^\pi g(\theta)\sin{n\theta} d\theta\\ C_nb^n+D_n b^{-n}=\frac{2}{\pi}\int_0 ^\pi h(\theta) \sin{n\theta} d\theta\\ \end{cases}

Is this as far as I can go? Is this the final answer according to the world 'solve' ? THANKS!

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Hint: Suppose $u(r, \theta)$ can be written in the form $$u(r, \theta) = \sum_{k = 1}^{\infty} v_k(r) \sin(k \theta),$$ substitute, and separate variables. (Thanks to Willie Wong for pointing out an oversight in the original solution.)

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    $\begingroup$ Unless $g(\theta)$ is proportional to $h(\theta)$, as written the ansatz will not work. But the fundamental idea is correct: postulate instead $u(r,\theta) = \sum_{k = 1}^\infty v_k(r) \sin (k\theta)$. The harmonicity requires $v_k$ be multiples of Bessel functions. The constants of proportionality are then obtained by matching boundary conditions. $\endgroup$ – Willie Wong Mar 16 '15 at 11:56
  • $\begingroup$ @WillieWong Yes, you're quite right, and cheers for pointing this out---I'd simply forgotten this when I wrote the original solution (which I've since corrected). $\endgroup$ – Travis Willse Mar 16 '15 at 13:55
  • $\begingroup$ Hmm... still kinda stuck- how can I sub this into the pde and get something useful? $\endgroup$ – sci-guy Mar 17 '15 at 1:13

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