0
$\begingroup$

Given a function of Cartesian coordinates $(x, y)$ in $g(x, y) = \frac{d}{dx}f(x, y)-y\frac{d}{dy}f(x, y)$, I want to transform this expression to polar coordinates $(r,\phi)$. I saw a solution using built-in chain rule here: Change variables in differential expressions and a possible solution using replacement here: Transformation of Derivatives under change of coordinates. I also came across TransformedField , which seems very succinct but it would only tranform $(x, y)$ in the expression but not the derivatives, i.e. it gives me the following:

TransformedField["Cartesian" -> "Polar", 
  D[f[x, y], x] - x*D[f[x, y], y], {x, y} -> {r, \[Phi]}
] 

output

$$f^{(1,0)}(r \cos (\phi ),r \sin (\phi ))-r \cos (\phi ) f^{(0,1)}(r \cos (\phi ),r \sin (\phi ))$$.

I wonder why this is the case? Or if there is any trick to make it work with TransformedField?


Edit 1

Kuba pointed out an existing function in "MoreCalculus," which works well for simple expression. But when I use it for a more complicated expression, I'm not sure if it outputs correct result:

Input expression with $dx = \frac{d}{dx}$ and $dy = \frac{d}{dy}$ Input

Output is very long but contains an expression with prime as shown here: $$i e \left(-\frac{1}{2} B r \sin (\phi )\right)'$$

And I'm not sure what that prime denotes or why it doesn't simplify (I already used FullSimplify). Any help would be appreciated.

$\endgroup$
  • $\begingroup$ DChange[ D[f[x, y], x] - x*D[f[x, y], y], "Cartesian" -> "Polar", {x, y}, {r, \[Phi]}, f[x, y] ] seems to do the trick. Is that the answer: change of variables in differential expressions? $\endgroup$ – Kuba Jan 11 at 23:11
  • $\begingroup$ Thanks for the link Kuba! So I applied it to my more complicated expression , and one of the outputs has a prime on it (shown in the edits). Do you know what the prime means? $\endgroup$ – Histoscienology Jan 12 at 2:53
  • $\begingroup$ Please provide a copyable code for input and output. After I created that function I used it maybe twice, and it has around four unit tests so I am not surprised fails for more complicated cases ;) $\endgroup$ – Kuba Jan 12 at 20:38

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.