Answer
Using Clairaut's Theorem we show that if $u\left( {x,y} \right)$ is harmonic, then the partial derivatives $\frac{{\partial u}}{{\partial x}}$ and $\frac{{\partial u}}{{\partial y}}$ are harmonic.
Work Step by Step
If $u\left( {x,y} \right)$ is harmonic, then by definition
(1) ${\ \ \ \ }$ $\Delta u = {u_{xx}} + {u_{yy}} = 0$
1. Take the derivative of equation (1) with respect to $x$ gives
${u_{xxx}} + {u_{yyx}} = 0$
By Clairaut's Theorem, we can re-arrange the orders and write
${u_{xxx}} + {u_{xyy}} = 0$
${\left( {{u_x}} \right)_{xx}} + {\left( {{u_x}} \right)_{yy}} = 0$
Hence, by definition ${u_x} = \frac{{\partial u}}{{\partial x}}$ is harmonic.
2. Take the derivative of equation (1) with respect to $y$ gives
${u_{xxy}} + {u_{yyy}} = 0$
By Clairaut's Theorem, we can re-arrange the orders and write
${u_{yxx}} + {u_{yyy}} = 0$
${\left( {{u_y}} \right)_{xx}} + {\left( {{u_y}} \right)_{yy}} = 0$
Hence, by definition ${u_y} = \frac{{\partial u}}{{\partial y}}$ is harmonic.
Thus, if $u\left( {x,y} \right)$ is harmonic, then the partial derivatives $\frac{{\partial u}}{{\partial x}}$ and $\frac{{\partial u}}{{\partial y}}$ are harmonic.
