# Solution of the Laplace equation in polar coordinate.

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!

• $\phi(a, \theta + 2\pi) = \phi(a,\theta )$. – Joelafrite Apr 20 '15 at 14:05
• As @Joelafrite pointed, your boundary conditions are rotational symmetric. – user228113 Apr 20 '15 at 14:11

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 $$\tilde{\phi}(r,\theta) = -C \frac{( r^n - b^{2n} r^{-n})}{( a^n - b^{2n} a^{-n})} \sin(n \theta)$$
• 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 $\tilde{\phi} = \phi - V$? – Sherry Apr 23 '15 at 2:36
• You are right. So you will have $B_o = -V$ – CottonTensor Apr 24 '15 at 10:58