Answer
The harmonic polynomials of degree 3:
$u\left( {x,y} \right) = a{x^3} + b{x^2}y - 3ax{y^2} - \frac{1}{3}b{y^3}$
Work Step by Step
We have $u\left( {x,y} \right) = a{x^3} + b{x^2}y + cx{y^2} + d{y^3}$.
1. Take the derivatives with respect to $x$:
${u_x} = 3a{x^2} + 2bxy + c{y^2}$
${u_{xx}} = 6ax + 2by$
2. Take the derivatives with respect to $y$:
${u_y} = b{x^2} + 2cxy + 3d{y^2}$
${u_{yy}} = 2cx + 6dy$
The function $u\left( {x,y} \right)$ is harmonic if it satisfies the Laplace equation $\Delta u = 0$:
$\Delta u = {u_{xx}} + {u_{yy}} = 0$
So,
$6ax + 2by + 2cx + 6dy = 0$
$x\left( {6a + 2c} \right) + y\left( {2b + 6d} \right) = 0$
Since $x$ and $y$ are independent, the solution is
$6a + 2c = 0$ ${\ \ \ }$ and ${\ \ \ }$ $2b + 6d = 0$
Hence, $u\left( {x,y} \right)$ is a harmonic polynomials if $c = - 3a$ and $d = - \frac{1}{3}b$.
So, we can write
$u\left( {x,y} \right) = a{x^3} + b{x^2}y + cx{y^2} + d{y^3}$
$u\left( {x,y} \right) = a{x^3} + b{x^2}y - 3ax{y^2} - \frac{1}{3}b{y^3}$
