Answer
We use Eq. (12) to prove that in polar coordinates:
$\Delta f = {f_{rr}} + \frac{1}{{{r^2}}}{f_{\theta \theta }} + \frac{1}{r}{f_r}$
Work Step by Step
We have the relations between rectangular coordinates and polar coordinates:
$x = r\cos \theta $, ${\ \ \ }$ $y = r\sin \theta $
In this exercise, $f$ is a function of $x$ and $y$, where $x = g\left( {r,\theta } \right)$, $y = h\left( {r,\theta } \right)$. Thus,
$f = f\left( {g\left( {r,\theta } \right),h\left( {r,\theta } \right)} \right)$
The partial derivatives of $f$ with respect to $r$ is
$\frac{{\partial f}}{{\partial r}} = {f_r} = \frac{{\partial f}}{{\partial g}}\frac{{\partial g}}{{\partial r}} + \frac{{\partial f}}{{\partial h}}\frac{{\partial h}}{{\partial r}}$
Since $x = g\left( {r,\theta } \right)$ and $y = h\left( {r,\theta } \right)$, so
${f_r} = {f_x}\cos \theta + {f_y}\sin \theta $
The partial derivatives of $x$ and $y$ are
$\frac{{\partial x}}{{\partial r}} = \cos \theta $, ${\ \ \ }$ $\frac{{\partial y}}{{\partial r}} = \sin \theta $
$\frac{{{\partial ^2}x}}{{\partial {r^2}}} = 0$, ${\ \ \ }$ $\frac{{{\partial ^2}y}}{{\partial {r^2}}} = 0$
$\frac{{\partial x}}{{\partial \theta }} = - r\sin \theta $, ${\ \ \ }$ $\frac{{\partial y}}{{\partial \theta }} = r\cos \theta $
$\frac{{{\partial ^2}x}}{{\partial {\theta ^2}}} = - r\cos \theta $, ${\ \ \ }$ $\frac{{{\partial ^2}y}}{{\partial {\theta ^2}}} = - r\sin \theta $
Using Eq. (12) we obtain for $r$
${f_{rr}} = {f_{xx}}{\left( {\frac{{\partial x}}{{\partial r}}} \right)^2} + 2{f_{xy}}\left( {\frac{{\partial x}}{{\partial r}}} \right)\left( {\frac{{\partial y}}{{\partial r}}} \right) + {f_{yy}}{\left( {\frac{{\partial y}}{{\partial r}}} \right)^2} + {f_x}\frac{{{\partial ^2}x}}{{\partial {r^2}}} + {f_y}\frac{{{\partial ^2}y}}{{\partial {r^2}}}$
(1) ${\ \ \ }$ ${f_{rr}} = {f_{xx}}{\cos ^2}\theta + 2{f_{xy}}\cos \theta \sin \theta + {f_{yy}}{\sin ^2}\theta $
Similarly, using Eq. (12) we obtain for $\theta$
${f_{\theta \theta }} = {f_{xx}}{\left( {\frac{{\partial x}}{{\partial \theta }}} \right)^2} + 2{f_{xy}}\left( {\frac{{\partial x}}{{\partial \theta }}} \right)\left( {\frac{{\partial y}}{{\partial \theta }}} \right) + {f_{yy}}{\left( {\frac{{\partial y}}{{\partial \theta }}} \right)^2} + {f_x}\frac{{{\partial ^2}x}}{{\partial {\theta ^2}}} + {f_y}\frac{{{\partial ^2}y}}{{\partial {\theta ^2}}}$
${f_{\theta \theta }} = {f_{xx}}{r^2}{\sin ^2}\theta - 2{r^2}{f_{xy}}\sin \theta \cos \theta $
${\ \ \ \ \ \ \ \ }$ $ + {f_{yy}}{r^2}{\cos ^2}\theta - {f_x}r\cos \theta - {f_y}r\sin \theta $
(2) ${\ \ \ }$ $\frac{1}{{{r^2}}}{f_{\theta \theta }} = {f_{xx}}{\sin ^2}\theta - 2{f_{xy}}\sin \theta \cos \theta $
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$ $ + {f_{yy}}{\cos ^2}\theta - \frac{{{f_x}}}{r}\cos \theta - \frac{{{f_y}}}{r}\sin \theta $
Adding equation (1) and equation (2) gives
${f_{rr}} + \frac{1}{{{r^2}}}{f_{\theta \theta }} = {f_{xx}} + {f_{yy}} - \frac{1}{r}\left( {{f_x}\cos \theta + {f_y}\sin \theta } \right)$
But ${f_r} = {f_x}\cos \theta + {f_y}\sin \theta $. So,
${f_{rr}} + \frac{1}{{{r^2}}}{f_{\theta \theta }} = {f_{xx}} + {f_{yy}} - \frac{1}{r}{f_r}$
Since $\Delta f = {f_{xx}} + {f_{yy}}$, hence we obtain Eq. (13):
$\Delta f = {f_{rr}} + \frac{1}{{{r^2}}}{f_{\theta \theta }} + \frac{1}{r}{f_r}$
