Use the method of separation of variables to separate the differential equation into 3 ordinary differential equations. Use the method of separation of variables to separate the differential equation
$\frac{1}{\sigma^2+\tau^2}\Big[ \frac{1}{\sigma}\frac{\partial}{\partial\sigma}(\sigma\frac{\partial}{\partial\sigma} f(\sigma,\tau,\phi)\big)+\frac{1}{\tau}\frac{\partial}{\partial\tau}(\tau\frac{\partial}{\partial\tau} f(\sigma,\tau,\phi)\big)\Big] +\frac{1}{\sigma^2 \tau^2}\frac{\partial^2}{\partial \phi^2}f(\sigma,\tau,\phi)=0$
into 3 ordinary differential equations, one a function of just σ, one a
function of just τ , and one a function of just ϕ.
The method I have been taught, you assume that $f=X(\sigma)Y(\tau)Z(\phi)$. Using this method I have separated the equation into three parts, however I cannot for the life of me seem to get all three equations to be functions of only one of the variables $\sigma,\tau,\phi$. The equations I got to are:
$(1) \frac{1}{\sigma^2+\tau^2}\Big[\frac{1}{\sigma}\frac{1}{X(\sigma)}\big(\frac{\partial X(\sigma)}{\partial \sigma}+\sigma\frac{\partial^2 X(\sigma)}{\partial \sigma^2}\big)\Big]$ 
$(2) \frac{1}{\sigma^2+\tau^2}\Big[\frac{1}{\tau}\frac{1}{Y(\tau)}\big(\frac{\partial Y(\tau)}{\partial \tau}+\tau\frac{\partial^2 Y(\tau)}{\partial \tau^2}\big)\Big]$
$(3) \frac{1}{\sigma^2\tau^2}\frac{1}{Z(\phi)}\frac{\partial^2 Z(\phi)}{\partial \phi}$
Where the sum of (1), (2) and (3) is equal to zero. I have tried multiplying them through by ${\sigma^2+\tau^2}$ and this gets (1) and (2) to be only functions of one variable, but equation (3) is still a function of all three.
I think this is just algebraic manipulation, but I cannot see how to get all three equations all functions of only one variable.
I will note I do not need to solve the equations once found, just find the equations.
Thanks for any input, it's appreciated
 A: $$\frac{1}{\sigma^2+\tau^2}\Big[ \frac{1}{\sigma}\frac{\partial}{\partial\sigma}(\sigma\frac{\partial}{\partial\sigma} f(\sigma,\tau,\phi)\big)+\frac{1}{\tau}\frac{\partial}{\partial\tau}(\tau\frac{\partial}{\partial\tau} f(\sigma,\tau,\phi)\big)\Big] +\frac{1}{\sigma^2 \tau^2}\frac{\partial^2}{\partial \phi^2}f(\sigma,\tau,\phi)=0$$
$$\frac{Z\sigma^2 \tau^2}{\sigma^2+\tau^2}\Big[ \frac{Y}{\sigma}\frac{\partial}{\partial\sigma}(\sigma\frac{\partial}{\partial\sigma} X\big)+\frac{X}{\tau}\frac{\partial}{\partial\tau}(\tau\frac{\partial}{\partial\tau} Y\big)\Big] =-YX\frac{\partial^2Z}{\partial \phi^2}$$
$$\frac{\sigma^2 \tau^2}{XY(\sigma^2+\tau^2)}\Big[ \frac{Y}{\sigma}\frac{\partial}{\partial\sigma}(\sigma\frac{\partial}{\partial\sigma} X\big)+\frac{X}{\tau}\frac{\partial}{\partial\tau}(\tau\frac{\partial}{\partial\tau} Y\big)\Big] =\underbrace{-\frac 1Z\frac{\partial^2 Z}{\partial \phi^2}}_{\text {Depends only on }\phi}$$
$$\underbrace{G(\sigma,\tau)}_{\text{ depends on } \sigma \tau}=\underbrace{-\frac 1Z\frac{\partial^2 Z}{\partial \phi^2}}_{\text {Depends only on }\phi}$$
On the left you only have X,Y and $\tau$, $\sigma$ ont the right only Z
You have first equation then do the same with the XY part
$$\frac{1}{XY}\Big[ \frac{Y}{\sigma}\frac{\partial}{\partial\sigma}(\sigma\frac{\partial}{\partial\sigma} X\big)+\frac{X}{\tau}\frac{\partial}{\partial\tau}(\tau\frac{\partial}{\partial\tau} Y\big)\Big] =K\frac{(\sigma^2+\tau^2)}{\sigma^2 \tau^2}$$
$$\Big[ \frac{1}{X\sigma}\frac{\partial}{\partial\sigma}(\sigma\frac{\partial}{\partial\sigma} X\big)+\frac{1}{Y\tau}\frac{\partial}{\partial\tau}(\tau\frac{\partial}{\partial\tau} Y\big)\Big] =K\frac{(\sigma^2+\tau^2)}{\sigma^2 \tau^2}$$
It's easy to separate
$$\Big[ \frac{1}{X\sigma}\frac{\partial}{\partial\sigma}(\sigma\frac{\partial}{\partial\sigma} X\big)+\frac{1}{Y\tau}\frac{\partial}{\partial\tau}(\tau\frac{\partial}{\partial\tau} Y\big)\Big] =\frac K{\sigma^2}+\frac K{\tau^2}$$
Put the dependance on $\tau$ on one side and the other variable on the other side...
$$\ \frac{1}{X\sigma}\frac{\partial}{\partial\sigma}(\sigma\frac{\partial}{\partial\sigma} X\big)-\frac K{\sigma^2}=-\frac{1}{Y\tau}\frac{\partial}{\partial\tau}(\tau\frac{\partial}{\partial\tau} Y\big) +\frac K{\tau^2}$$
You have now
$$h(\sigma)=w(\tau)$$
You can seperate to get the second and third equation...
$$\ \frac{1}{X\sigma}\frac{\partial}{\partial\sigma}(\sigma\frac{\partial}{\partial\sigma} X\big)-\frac K{\sigma^2}=C$$
$$-\frac{1}{Y\tau}\frac{\partial}{\partial\tau}(\tau\frac{\partial}{\partial\tau} Y\big) +\frac K{\tau^2}=C$$
$$-\frac 1Z\frac{\partial^2 Z}{\partial \phi^2}=K$$
K and C are different constants they may be equal but still are constants
$$
\boxed{
\begin{array}{lc}
\text {1)} & Z''+KZ=0 \\
\text {2)} & X''+\dfrac {X'}{\sigma}-X(\dfrac K{\sigma ^2}+C)=0 \\ 
\text {3)} & Y''+\dfrac {Y'}{\tau}-Y(\dfrac K{\tau ^2}-C)=0  
\end{array}}
$$
