Answer
Using both rectangular and polar expressions for $\Delta f$, we verify that $f\left( {x,y} \right) = x$ and $f\left( {x,y} \right) = y$ are harmonic.
Work Step by Step
1. We have $f\left( {x,y} \right) = x$
We evaluate the Laplace operator using
a. rectangular expression
The partial derivatives are
${f_x} = 1$, ${\ \ \ }$ ${f_y} = 0$
${f_{xx}} = 0$, ${\ \ \ }$ ${f_{yy}} = 0$
Therefore, $\Delta f = {f_{xx}} + {f_{yy}} = 0$.
Since $\Delta f = {f_{xx}} + {f_{yy}} = 0$, $f$ is harmonic.
b. polar expression
In polar coordinates, we have $f\left( {r,\theta } \right) = r\cos \theta $
The partial derivatives are
${f_r} = \cos \theta $, ${\ \ \ }$ ${f_\theta } = - r\sin \theta $
${f_{rr}} = 0$, ${\ \ \ }$ ${f_{\theta \theta }} = - r\cos \theta $
Using Eq. (13) we get
$\Delta f = {f_{rr}} + \frac{1}{{{r^2}}}{f_{\theta \theta }} + \frac{1}{r}{f_r}$
$\Delta f = - \frac{1}{r}\cos \theta + \frac{1}{r}\cos \theta = 0$
Since $\Delta f = 0$, $f$ is harmonic.
2. We have $f\left( {x,y} \right) = y$
We evaluate the Laplace operator using
a. rectangular expression
The partial derivatives are
${f_x} = 0$, ${\ \ \ }$ ${f_y} = 1$
${f_{xx}} = 0$, ${\ \ \ }$ ${f_{yy}} = 0$
Therefore, $\Delta f = {f_{xx}} + {f_{yy}} = 0$.
Since $\Delta f = 0$, $f$ is harmonic.
b. polar expression
In polar coordinates, we have $f\left( {r,\theta } \right) = r\sin \theta $
The partial derivatives are
${f_r} = \sin \theta $, ${\ \ \ }$ ${f_\theta } = r\cos \theta $
${f_{rr}} = 0$, ${\ \ \ }$ ${f_{\theta \theta }} = - r\sin \theta $
Using Eq. (13) we get
$\Delta f = {f_{rr}} + \frac{1}{{{r^2}}}{f_{\theta \theta }} + \frac{1}{r}{f_r}$
$\Delta f = - \frac{1}{r}\sin \theta + \frac{1}{r}\sin \theta = 0$
Since $\Delta f = 0$, $f$ is harmonic.
