So I'm trying to solve this problem where there are two concentric cylinders and the voltage potential $V$ is a function of $r$ and $\phi$ but not the height $z$
I set $V$ as
$$V=f(r)g(\phi) = fg$$
so my Laplace equation would look like
$$\nabla^2V =\frac{∂^2V}{∂r^2}+\frac{1}{r}\frac{∂V}{∂r}+\frac{1}{r^2}\frac{∂^2V}{\partial\phi^2}=0$$ $$\frac{∂^2fg}{∂r^2}+\frac{1}{r}\frac{∂fg}{∂r}+\frac{1}{r^2}\frac{∂^2fg}{\partial\phi^2}=0$$ $$g\frac{∂^2f}{∂r^2}+\frac{g}{r}\frac{∂f}{∂r}+\frac{f}{r^2}\frac{∂^2g}{\partial\phi^2}=0$$ $$\frac{1}{f}\frac{∂^2f}{∂r^2}+\frac{1}{fr}\frac{∂f}{∂r}+\frac{1}{gr^2}\frac{∂^2g}{\partial\phi^2}=0$$
Then I let $$\frac{1}{f}\frac{∂^2f}{∂r^2}+\frac{1}{fr}\frac{∂f}{∂r}=-a^2$$ $$\frac{1}{gr^2}\frac{∂^2g}{\partial\phi^2}=a^2$$
I was able to solve for $g(\phi)$ which was $$g(\phi)=A \cosh(ar\phi)+B\sinh(ar\phi)$$ ...I think, but I'm not sure how to solve for $f(r)$
Any ideas?
You should totally separe $r$ and $\phi$, getting $$ \frac{r^2 f''(r)}{f(r)}+\frac{r f'(r)}{f(r)} = -\frac{g''(\phi)}{g(\phi)} $$
For this equality to hold for all $r,\phi$, both the rhs and lhs must be constant. The continuity of $g$ calls for a periodic solution, which explains why the constant should be positive and, furthermore, that $a \in \mathbb{Z}$.
The equation $-\dfrac{g''(\phi)}{g(\phi)} = a^2$ leads to $$ g(\phi) = b_1 \cos(a \phi) + b_2 \sin(a \phi). $$
Regarding $f$, it should satisfy the equation $$ r^2 f''(r) + rf'(r) - a^2 f(r) = 0, $$
which leads to $$ f(r) = c_1 \cosh(a \log r) + c_2 \sinh(a \log r). $$