How to solve: $(\frac{\cot(\theta)}{r^2}\frac{\partial}{\partial\theta})\cdot(\frac{\cot(\theta)}{r^2}\frac{\partial}{\partial\theta}).$? Can someone please help me with the below? I am not sure how to go about solving this:
$$(\frac{\cot(\theta)}{r^2}\frac{\partial}{\partial\theta})\cdot(\frac{\cot(\theta)}{r^2}\frac{\partial}{\partial\theta}).$$
I am checking my answer to creeping flow around a sphere, and am stuck on the above expansion.
The governing DE I am using is:
$\left[\frac{\partial^2}{\partial r^2} + \frac{\sin(\theta)}{r^2}\frac{\partial}{\partial\theta} \left(\frac{1}{\sin(\theta)}\frac{\partial}{\partial\theta}\right)\right]^2\psi = 0$.
Likewise, I am stuck on the below expansion. I am just not sure because if I use the operator, depending on which side the () are on will give me different solutions:
$\left(\frac{\cot(\theta)}{r^2}\frac{\partial}{\partial\theta}\right)\cdot\left(\frac{1}{r^2}\frac{\partial}{\partial\theta}\frac{\partial}{\partial\theta}\right)$$
Thank you!
 A: Factor out the $\dfrac1{r^4}$. Then
$$\cot(\theta)\frac{\partial}{\partial\theta}\big(\cot(\theta)\frac{\partial}{\partial\theta}\big) = \cot(\theta)\big({-}\csc^2(\theta)\frac{\partial}{\partial\theta}+\cot(\theta)\frac{\partial^2}{\partial\theta^2}\big).$$
A: With the help of everyone who answered, commented, and some outside help, I found what I was missing. I was doing the $[a+b]^2$ operator incorrect. I did $[a+b]*[a+b]*\psi$ , foiled the bracketed terms, then operated on the $\psi$ term. This is not what the $^2$ term does. I should have done the following:
$\left[\frac{\partial^2}{\partial r^2} + \frac{\sin(\theta)}{r^2}\frac{\partial}{\partial\theta} \left(\frac{1}{\sin(\theta)}\frac{\partial}{\partial\theta}\right)\right]^2\psi$
[expand the inside]. Lets just say the expansion is equal to $\frac{\partial}{\partial E}$ for simplicity.
Then, operate that expansion on $\psi$:
$[\frac{\partial}{\partial E}]\psi$ = $\frac{\partial\psi}{\partial E}$
Then you take $\left[\frac{\partial^2}{\partial r^2} + \frac{\sin(\theta)}{r^2}\frac{\partial}{\partial\theta} \left(\frac{1}{\sin(\theta)}\frac{\partial}{\partial\theta}\right)\right]$ and operate on $\frac{\partial\psi}{\partial E}$ after expanding the inside again.
$[\frac{\partial}{\partial E}]\frac{\partial\psi}{\partial E}$ = $\frac{\partial^2\psi}{\partial E^2}$
Thank you for all the help with the comments!
-Alex
