# [SOLVED]Axisymmetric case

#### dwsmith

##### Well-known member
Laplace in spherical coordinates is
$$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial u}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2u}{ \partial \varphi^2}$$

In the axisymmetric case, $\frac{\partial}{\partial\varphi} = 0$. Does that mean I can go straight to writing the equation as
$$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial u}{\partial\theta}\right)$$

I ask because I separated variables and obtained
$$\frac{2r\varphi'+r^2\varphi''}{\varphi} = \frac{\cot\theta\psi'+\psi''}{\psi}=\lambda^2$$

My radial term works out correctly but my azimuthal doesn't.
$$\varphi(r)\sim r^{\frac{-1\pm\sqrt{1+4\lambda^2}}{2}}$$

I have
$$\cot\theta\psi'+\psi''-\lambda^2\psi=0$$
However, I have that it should be
$$\cot\theta\psi'+\psi''-\left(\lambda^2 - \frac{m^2}{\sin\theta}\right)\psi=0$$