How to find out the following function f which is a function of r?

Question

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?

Answer 1

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 $$ \frac{\partial}{\partial x} = \frac{\partial r}{\partial x} \frac{\partial }{\partial r} + \frac{\partial \theta}{\partial x} \frac{\partial }{\partial \theta}, \quad \frac{\partial}{\partial y} = \frac{\partial r}{\partial y} \frac{\partial }{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial }{\partial \theta} $$