Polar laplace equation

Question

I am really struggling to solve this, the initial condition $u(a,\theta)=0$ keeps throwing me off.

Answer 1

Let $u(r,\theta)=f(r)g(\theta)$. Then the Laplace equation becomes $$ f''(r)g(\theta)+\frac1r f'(r)g(\theta)+\frac1{r^2}f(r)g''(\theta)=0$$ from which one has $$ \frac{f''(r)+\frac1r f'(r)}{f(r)}=-\frac{g''(\theta)}{g(\theta)}. $$ Let $$ \frac{f''(r)+\frac1r f'(r)}{f(r)}=K \tag{1} $$ and $$-\frac{g''(\theta)}{g(\theta)}=K. \tag{2} $$ Now solving (1) and (2) gives $$ f(r)=c_1\cosh(\sqrt Kr)+c_2 i\sinh(\sqrt{K}r), g(\theta)=d_1\cos(\sqrt K\theta)+d_2\sin(\sqrt K\theta). $$ So $$ u(r,\theta)=\bigg[c_1\cosh(\sqrt Kr)+c_2 i\sinh(\sqrt{K}r)\bigg] \bigg[d_1\cos(\sqrt K\theta)+d_2\sin(\sqrt K\theta)\bigg]. $$ Let $u(r,0)=u(r,2\pi)=0$ and then one has $$ d_1=0, K=\frac{n^2}{4},n=1,2,3\cdots. $$ You can choose $K=\frac14$ or $K=1$. Let $u(a,\theta)=0,u(b,\theta)=f(\theta)$ and I think you can determine $C_1,C_2$ which is not very hard.