I'm currently working on how to derive the Laplacian of the Steady-state Heat Equation from the polar coordinates. In other words, I tried to show that: $\frac{\partial^{2}{u}}{{\partial r^{2}}} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^{2}} = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}$ where $x = r\cos\theta, y = r\sin\theta$
Progress so far: I used the following identities: $\frac{\partial^{2}{u}}{\partial r^{2}} = \frac{\partial^2 u}{\partial x^2}\cos^{2}{\theta} + \frac{\partial^2 u}{\partial y^2}\sin^{2}{\theta} + \frac{\partial u}{\partial x}\frac{\cos\theta}{r} +\frac{\partial u}{\partial y}\frac{\sin\theta}{r} + \frac{\partial^{2} u}{\partial x\partial y}sin{2\theta}$
$\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\cos\theta + \frac{\partial u}{\partial y}\sin\theta$
$\frac{\partial^2 u}{\partial \theta^{2}} = r^2\sin^{2}{\theta} \frac{\partial^{2} u}{\partial x^2} - r^2\sin{2\theta}\frac{\partial^2 u}{\partial x\partial y} - r\sin{\theta}\frac{\partial u}{\partial y} - r\cos{\theta}\frac{\partial u}{\partial x} + r^2\cos^{2}{\theta}\frac{\partial^2 u}{\partial y^2}$
Plugging these identities to the LHS and grouping/cancelling common terms, I end up with $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{1}{r}\frac{\partial u}{\partial r}$. In order for the equation above holds, it seems to me that I need to show $\frac{\partial u}{\partial r} = 0$. But I don't see how this is the case. I checked my calculations for the three identities above twice, and couldn't find any error, so this last step must hold! Can anyone help me resolve this last step?