Please see the attached image, containing my question and my solution.
Question: Show [using the formulas for gradient and divergence in terms of polar coordinates] that the Lapalacian of a scalar function $f(r,\theta)$ is given by $$ \nabla \cdot \nabla f = \partial_r^2 f + \frac 1r \partial_r f + \frac 1{r^2} \partial_{\theta}^2 f $$
Attached image: http://imgur.com/a/XUZ1M
I gather that the RHS is obtained by substituting the 1st result into the 2nd which then gives the answer after doing the two differentiations.
I have ajusted the notation and I have solved via a longer way, it is only that I have not looked at this area in a long time, any advice on where to start - as in like what does the initial substitution look like.
Thanks
-nomad609
Note- this question is much different to any other previous threads, I have uploaded the information to elucidate this.
Calculate the gradient first
\begin{equation} F = \nabla f = \hat{r}\frac{\partial f}{\partial r} + \hat{\theta}\frac{1}{r}\frac{\partial f}{\partial \theta} = \hat{r}F_r + \hat{\theta} F_\theta, \end{equation}
with $F_r = \partial f/\partial r$ and $F_\theta = r^{-1}\partial f/\partial \theta$. Now calculate the divergence of the field $F$
\begin{equation} \nabla \cdot \nabla f = \nabla \cdot F = \frac{1}{r}\frac{\partial }{\partial r}(r Fr) + \frac{1}{r}\frac{\partial}{\partial \theta} F_\theta = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial f}{\partial r} \right) + \frac{1}{r}\frac{\partial}{\partial \theta}\left(\frac{1}{r}\frac{\partial f}{\partial \theta} \right) \end{equation}
which can be rearranged as
\begin{equation} \nabla \cdot\nabla f = \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{equation}