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Show that under the transformation $x = \rho\cos\phi$, $y = \rho\sin\phi$, the equation $$ \frac{\partial^2u}{\partial x^2} = \frac{\partial^2u}{\partial y^2} = 0$$ becomes $$\frac{\partial^2u}{\partial \rho^2} = \frac{1}{\rho}\frac{\partial u}{\partial \rho} - \frac{1}{\rho^2}\frac{\partial^2u}{\partial \phi^2} = 0$$

$u$ is function of $x$ and $y$

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What is $\,u\,$?! – DonAntonio Nov 24 '12 at 21:03
I don't see $v$ anywhere... – TonyK Nov 24 '12 at 21:07
up vote 2 down vote accepted

$\newcommand{\del}[2]{\frac{\partial #1}{\partial #2}}$ Assuming $u$ is a smooth (or at the very least continuously second differentiable) function, you can use the chain rule. You know that $$\del{u}{x}=\del{\rho}{x}\del{u}{\rho} + \del{\phi}{x}\del{u}{\phi}$$ and similarily for $y$. You can iterate the relation to derive higher order derivatives and combine your results to get your identity.

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