How to show this property of the equation of Laplace? If $u=u(x,y)$ and $x=r\cos \theta$, $y=r \sin \theta$, show that equation of Laplace $\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=0$ turns in:
$$\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+ \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}=0$$
 A: From the link given by Emilio Novati in the comments, most of the maths is straightforward.
This bit seems difficult however, but once you can see it, the rest of the paper follows easily.
$$\dfrac{\partial^2u}{\partial r^2}= \cos\theta\dfrac{\partial}{\partial r}\dfrac{\partial u}{\partial x}+\sin\theta\dfrac{\partial}{\partial r}\dfrac{\partial u}{\partial y}$$
$$=\cos\theta\left(\dfrac{\partial}{\partial x}\dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial}{\partial y}\dfrac{\partial u}{\partial x}\dfrac{\partial y}{\partial r}\right)+
\sin\theta\left(\dfrac{\partial}{\partial y}\dfrac{\partial u}{\partial y}\dfrac{\partial x}{\partial r}+\dfrac{\partial}{\partial y}\dfrac{\partial u}{\partial y}\dfrac{\partial y}{\partial r}\right)$$
$$=\cos^2\theta\dfrac{\partial^2u}{\partial x^2}+2\cos\theta\sin\theta\dfrac{\partial^2 u}{\partial x\partial y}+\sin^2\theta\dfrac{\partial^2 u}{\partial y^2}$$
The first line differentiates $\dfrac{\partial u}{\partial r}$ through by $\dfrac{\partial}{\partial r}$. As we are dealing with partial derivatives, on the right hand side, this can be passed through the function, and only applied to components that are dependent on $r$, namely $\dfrac{\partial u}{\partial x}$ and $\dfrac{\partial u}{\partial y}$, with $\cos\theta$ and $\sin\theta$ not being affected.
The second line reapplies the chain rule to $\dfrac{\partial}{\partial r}$, and we get:
$$\dfrac{\partial}{\partial r}=\dfrac{\partial}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial}{\partial y}\dfrac{\partial y}{\partial r}$$
In the third line, we reuse $\dfrac{\partial x}{\partial r}=\cos\theta$ and $\dfrac{\partial y}{\partial r}=\sin\theta$, and also that $\dfrac{\partial}{\partial x}\dfrac{\partial u}{\partial x}=\dfrac{\partial^2u}{\partial x^2}$.
Similar techniques are used throughout the paper referenced to get the result.
