If I try for a separable solution $u(r,\theta)=R(r)\Theta(\theta)$ to the equation $\Delta u=-ku$ where $k\geq0$ I run into some problems before getting to the Bessel functions part.
After separation the equation simplifies to $\frac{r^2}{R}R''+\frac{r}{R}R'+kr^2=-\frac{1}{\Theta}\Theta''$.
Now I would set both sides to be a constant $c$. But in every solution I have seen, they have set this constant to be $n^2$ (so that $c=n^2$ in my case) because $c\geq0$. But I don't understand how we can say $c\geq0$. Why is this true?
On the two-dimensional unit disk we have the parametrisation $x=r\cos\theta$ and $y=r\sin\theta$. Assuming $u$ is separable, we have $u=R\Theta$ where $\Theta$ is periodic. Setting the RHS to a constant $c$ gives $\Theta''+c\Theta=0$ so $\Theta$ is a linear combination of $\sin(\theta\sqrt c)$ and $\cos(\theta\sqrt c)$. If $c<0$ this forces $\Theta$ to be a linear combination of $i\sinh(\theta\sqrt{-c})$ and $\cosh(\theta\sqrt{-c})$ which violates periodicity. Thus $c\ge0$, and in particular, $\sqrt c$ must be an integer.