I have the following relations that express a change of variables

$$\begin{cases} x=ae^\theta\cos(\varphi) \\ y=ae^\theta\sin(\varphi) \end{cases} $$

And I want to use it to transform the operator: $$L=x^2 \frac {\partial^2}{\partial x^2}-2xy \frac{\partial^2}{\partial x\partial y}+y^2 \frac{\partial^2}{\partial y^2}$$

The derivatives become rather complicated and I also have trouble defining the order of regularity of the change of variables.

I would love an explicit solution as to how $$\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$$ are expressed.


1 Answer 1


I'm using subscripts for more compact notation

$$ f_\theta = f_x x_\theta + f_y y_\theta = ae^\theta\big(\cos\phi f_x + \sin\phi f_y\big) $$

$$ f_\phi = f_x x_\phi + f_y y_\phi = ae^\theta \big(-\sin\phi f_x + \cos\phi f_y\big) $$

Cross-multiply by the trig functions to eliminate one of $f_x$ or $f_y$ $$ \cos\phi f_\theta - \sin\phi f_\phi = ae^\theta f_x $$ $$ \sin\phi f_\theta + \cos\phi f_\phi = ae^\theta f_y $$

Therefore $$ f_x = \frac{e^{-\theta}}{a}\left(\cos\phi f_\theta - \sin\phi f_\phi\right) $$

$$ f_y = \frac{e^{-\theta}}{a}\left(\sin\phi f_\theta + \cos\phi f_\phi \right) $$

For the second derivatives, just apply the same operators again $$ \begin{align} f_{xx} &= \frac{e^{-\theta}}{a}\left(\cos\phi\ \partial_\theta f_x - \sin\phi \ \partial_\phi f_x \right) \\ &= \frac{e^{-\theta}}{a}\cos\phi\left(-\frac{e^{-\theta}}{a}(\cos\phi f_\theta - \sin\phi f_\phi) + \frac{e^{-\theta}}{a}(\cos\phi f_{\theta\theta} - \sin\phi f_{\theta\phi}) \right) \\ &\quad - \frac{e^{-2\theta}}{a^2}\sin\phi \left(-\sin\phi f_\theta - \cos\phi f_\phi + \cos\phi f_{\theta\phi} - \sin\phi f_{\phi\phi} \right) \\ &= \frac{e^{-2\theta}}{a^2}\left(\cos^2\phi f_{\theta\theta} + \sin^2\phi f_{\phi\phi} - \sin2\phi f_{\theta\phi} - \cos2\phi f_\theta \right) \end{align} $$

and so on...

As for the mapping, this is called the log-polar coordinate system. Curves of constant $\theta$ are circles in the $xy$ plane, with the scale being logarithmic ($\theta = 0, \ln 2, \ln 3, \dots$ have radii $a, 2a, 3a, \dots$). The only exception is the origin, which corresponds to $\theta = -\infty$, making it a singular point.

Curves of constant $\phi$ are rays starting from (but not including) the origin, similar to regular polar coordinates.


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