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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!

amWhy
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Alexander Savadelis
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2 Answers2

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Factor out the $\dfrac1{r^4}$. Then $$\cot(\theta)\frac{\partial}{\partial\theta}\big(\cot(\theta)\frac{\partial}{\partial\theta}\big) = \cot(\theta)\big({-}\csc^2(\theta)\frac{\partial}{\partial\theta}+\cot(\theta)\frac{\partial^2}{\partial\theta^2}\big).$$

Ted Shifrin
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  • So I was thinking that, but then how would you do (cot(theta) * partial)(partial partial)? The order would matter. So if it was as described above, it would be cot(theta)third partial. But if you had (partial partial)(cot partial) you would have a completely different answer due to the product rule. – Alexander Savadelis Feb 09 '22 at 20:01
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    You're asking about the term in the given PDE with $(\partial_r^2 + B)^2$? Here $B=\frac{\sin(\theta)}r (\csc(\theta)\partial_\theta)\theta$ and $$B^2 = \left(\frac{\sin(\theta)}r (\csc(\theta)\partial\theta)\theta\right)\left(\frac{\sin(\theta)}r (\csc(\theta)\partial\theta)_\theta\right).$$ Just read left to write. There will be lots of terms! – Ted Shifrin Feb 09 '22 at 21:47
  • Thank you! I realized what I was missing. I was doing the [ ]^2 operator incorrect. I didn't realize how it worked, so I was foiling the brackets [ ][ ]psi which is incorrect. I should have done [ ]psi, then [ ] of that new quantity. Thank you to everyone who helped =] – Alexander Savadelis Feb 11 '22 at 14:20
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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

Alexander Savadelis
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