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[SOLVED] Axisymmetric case

dwsmith

Well-known member
Feb 1, 2012
1,673
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
$$
\cot\theta\psi'+\psi''-\left(\lambda^2 - \frac{m^2}{\sin\theta}\right)\psi=0
$$