I'm trying to find the most general harmonic polynomial of form $ax^3+bx^2y+cxy^2+dy^3$. I write this polynomial as $u(x,y)$.
I calculate
$$
\frac{\partial^2 u}{\partial x^2}=6ax+2by,\qquad \frac{\partial^2 u}{\partial y^2}=2cx+6dy
$$
and conclude $3a+c=0=b+3d$. Does this just mean the most general harmonic polynomial has form $ax^3-3dx^2y-3axy^2+dy^3$ with the above condition on the coefficients? "Most general" is what my book states, and I'm not quite sure what it means.
Also, I want to find the conjugate harmonic function, say $v$. I set $\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}$ and $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$ and find
$$
v=dx^3+3ax^2y+bxy^2-ay^3+K
$$
for some constant $K$. By integrating $\frac{\partial v}{\partial x}$, finding $v$ as some polynomial in $x$ and $y$ plus some function in $y$, and then differentiating with respect to $y$ to determine that function in $y$ up to a constant. Is this the right approach?
Finally, the question asks for the corresponding analytic function. Is that just $u(x,y)+iv(x,y)$? Or something else? Thanks for taking the time to read this.
Best Answer
An alternative approach is to 'guess' the form of the final analytic function. We want $f(z)=u(x,y)+i v(x,y)$ where $z=x+i y$ and $u(x,y)$ is a homogeneous cubic polynomial. Since $u(x,y)$ is cubic, we infer that $f(z)$ should also be some cubic polynomial; since it is homogeneous, we further conclude that $f(z)=\alpha z^3$ for some $\alpha$. (If it contained $z^2$, for instance, $u(x,y)$ would have terms of degree 2 and wouldn't be homogeneous.) Writing $\alpha=a+i d$ then yields
$$f(z)=(a+i d)(x+i y)^3=(ax^3-3dx^2y-3axy^2+dy^3)+i(dx^3+3ax^2y-3dxy^2-ay^3)$$ in agreement with $u(x,y)$, $v(x,y)$ as identified in the OP.