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 $y$