Separation of variables for partial differential equations - Shallow Wave Equation Question: I've tried countless times to prove this problem, but I keep getting hung up on the indices for derivatives. Could somebody please point me in the right direction? 
Given
Solution
 A: Just remember: in the symbolism of the present problem, 
$G_\lambda = \dfrac{dG}{d\lambda}, \; F_\sigma = \dfrac{dF}{d\sigma}, \tag 0$
and so forth; there is no semantic difference 'twixt these two notations for derivatives.
From
$4\phi_{\lambda \lambda} - \dfrac{1}{\sigma}(\sigma \phi_\sigma)_\sigma = 0, \tag 1$
we obtain
$4\phi_{\lambda \lambda} - \dfrac{1}{\sigma}(\sigma \phi_\sigma)_\sigma = 4\phi_{\lambda \lambda} - \dfrac{1}{\sigma}(\phi_\sigma + \sigma \phi_{\sigma \sigma}) =  4\phi_{\lambda \lambda} - \dfrac{1}{\sigma}\phi_\sigma - \phi_{\sigma \sigma} = 0; \tag 2$
if we set
$\phi(\lambda, \sigma) = G(\lambda)F(\sigma), \tag 3$
then
$\phi_{\lambda \lambda} = G_{\lambda \lambda}F, \tag 4$
$\phi_\sigma = GF_\sigma, \tag 5$
and
$\phi_{\sigma \sigma} =  GF_{\sigma \sigma}; \tag 6$
using (3), (4), (5) and (6), (2) yields
$0 = 4\phi_{\lambda \lambda} - \dfrac{1}{\sigma}(\sigma \phi_\sigma)_\sigma = 4\phi_{\lambda \lambda} - \dfrac{1}{\sigma}\phi_\sigma - \phi_{\sigma \sigma} =  4G_{\lambda \lambda}F - \dfrac{1}{\sigma}GF_\sigma - GF_{\sigma \sigma}; \tag 7$
we may divide through by $GF$ in any region of $\lambda \sigma$-space where it is non-zero, and find
$\dfrac{4G_{\lambda \lambda}F}{FG} - \dfrac{1}{\sigma}\dfrac{GF_\sigma}{GF} - \dfrac{GF_{\sigma \sigma}}{FG} = 0; \tag 8$
cancelling the $F$s and $G$s where possible we obtain
$\dfrac{4G_{\lambda \lambda}}{G} - \dfrac{1}{\sigma}\dfrac{F_\sigma}{F} - \dfrac{F_{\sigma \sigma}}{F} = 0, \tag 9$
which it is easily seen may be re-written as
$\dfrac{4}{G}G_{\lambda \lambda} = \dfrac{1}{F}(F_{\sigma \sigma} + \sigma F_\sigma), \tag{10}$
in which the variables are neatly separated; of course (10) may also be expressed as
$\dfrac{4}{G} \dfrac{d^2G}{d\lambda^2} = \dfrac{1}{F}(\dfrac{d^2F}{d\sigma^2} + \dfrac{1}{\sigma} \dfrac{dF}{d\sigma}).  \tag{11}$
