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Solve the following PDE: $$\phi(r,\theta) = \begin{cases} \Delta \phi=0 & \quad \text{for $a \le r\le b$ }\\[8pt] \phi=V & \quad \text{for $r=b$} \\[8pt] \phi+ C \sin(n\theta)=0 & \quad \text{for $r=a$}\end{cases}$$

I know how to solve the case when the boundary condition is rotational symmetric. However, in this case, $\phi+ C \sin(n\theta)=0$ for $r=a$. The difficulty is that it is not rotational symmetric. Then I don't know how to deal with this.

Could someone kindly help? Thanks!

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    $\begingroup$ $\phi(a, \theta + 2\pi) = \phi(a,\theta )$. $\endgroup$ – Joelafrite Apr 20 '15 at 14:05
  • $\begingroup$ As @Joelafrite pointed, your boundary conditions are rotational symmetric. $\endgroup$ – user228113 Apr 20 '15 at 14:11
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In this case, it helps to solve the problem by solving for $\tilde{\phi} = \phi - V$. Write the solution as an expansion of the eigenfunctions of the spatial Laplace operator. i.e. Once you use separation of variables, you should be able to see that $$ \tilde{\phi}(r,\theta) = A_o \ln r + B_o + \sum\limits_{n=1}^{\infty} (A_n \cos(n \theta) + B_n \sin(n \theta)) (C_n r^n + D_n r^{-n}) $$ But at $r = b, \tilde{\phi} = 0 \Longrightarrow D_n = - b^{2n} C_n$. Absorbing $C_n$ into the other constants, we have

$$ \tilde{\phi}(r,\theta) = A_o \ln r + B_o + \sum\limits_{n=1}^{\infty} (A_n \cos(n \theta) + B_n \sin(n \theta)) ( r^n - b^{2n} r^{-n}) $$ Now, using the other Boundary condition, we have $A_o = 0, B_o = 0, A_m = 0 \; \; \forall m$ and $$B_n ( a^n - b^{2n} a^{-n}) = - C \quad ; \quad B_m = 0 \; \;\forall m \neq n $$

Plugging in the obtained constants we get $$ \tilde{\phi}(r,\theta) = -C \frac{( r^n - b^{2n} r^{-n})}{( a^n - b^{2n} a^{-n})} \sin(n \theta) $$

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    $\begingroup$ Thanks! Do we need to minus $V$ at $B_n ( a^n - b^{2n} a^{-n}) = - C \quad ; \quad B_m = 0 \; \;\forall m \neq n$? Since we make $\tilde{\phi} = \phi - V$? $\endgroup$ – Sherry Apr 23 '15 at 2:36
  • $\begingroup$ You are right. So you will have $B_o = -V$ $\endgroup$ – CottonTensor Apr 24 '15 at 10:58

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