How to find out the following function f which is a function of r? If $\nabla^2 f(r) = 0$, show that $f(r) = a\log r + b$, where $a,b$ are constants and $r ^2 = x^2+y^2$.
I know $\nabla$ for cartesian coordinates. But i am not getting idea how to solve above?
 A: The Laplacian $\Delta f  = \nabla \cdot \nabla f$ in polar coordinates is,
\begin{align}
\Delta f &= \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial f}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2} \\
&= \frac{\partial^2 f}{\partial r^2} + \frac{1}{r} \frac{\partial f}{\partial r} + \frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2}.
\end{align}
which is equal to 
$$
\Delta f = f'' + \frac{1}{r} f' =0
$$
because $f = f(r)$. Denote $f'=g$, 
$$
g' + g/r =0  \implies g  =f'= \frac{C}{r} \implies f(r) = a \ln r + b
$$
$\textbf{Note :}$
You can get that form of laplacian by apply the chain rule
$$
\frac{\partial }{\partial \tilde{x}^i} = \frac{\partial x^j}{\partial \tilde{x}^i} \frac{\partial }{\partial {x}^j}
$$
to the expression of $\Delta f$ in Cartesian coordinate
$$
\Delta f = \frac{\partial^2f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}
$$
In your case $(x,y)$ and $(r,\theta)$. So
$$
\frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial }{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial }{\partial \theta}, \quad \frac{\partial}{\partial y} = \frac{\partial r}{\partial y} \frac{\partial }{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial }{\partial \theta}
$$
