What holomorphic functions $f = u + iv$ satisfy $u(x, y) = x^3 + xg(y)$, where $g$ is a twice continuously differentiable function?

Question

I'm trying to determine the analytic functions $f = u + iv$ for which $u(x, y) = x^3 + xg(y)$, where $g$ is a twice continuously differentiable function.

What I've tried is to figure out what v must be in terms of $x$ and $y$, and then express the resulting $f$ in terms of $z$.

First I differentiated $u$ with respect to $x$ and then integrated the result with respect to $y$. Next I applied Cauchy-Riemann to get the differential equation $6xy + h'(x) = - x g'(y)$, where $h$ is an unknown real valued function of one real variable.

This differential equation seemed too complicated to handle, and as a desperate measure to come forward I just deleted h' to get the simpler differential equation $6xy = - x g' (y)$, which I could simplify further to $6y = - g'(y)$.

The obvious solution to this differential equation is $c - 3y^2$, which means that $u = x^3 +cx - 3xy^2$.

By a technique similar to what I had used earlier, I found that the harmonic conjugate $v = 3(x^2)y + cy - y^3$, and it was easy to see that $u + iv = cz + z^3$, where $z = x + iy$.

However according to a source, the general solution is $z^3 + iaz^2 + bz + ic$, where $a, b, c ∈ \mathbb R$. So apparently I've found only some of the holomorphic functions which satisfy the requirement and not all of them. What went wrong?

Answer 1

We know that if $f = u +iv$ is holomorphic, then $u$ and $v$ are harmonic, i.e. $u_{xx} + u_{yy} = 0$ and $v_{xx} + v_{yy} = 0$. This follows from Cauchy-Riemann equations.

In particular, you can calculate $u_{xx}=6x$ and $u_{yy} = xg''(y)$, so $u_{xx}+u_{yy} = 0$ implies that $g''(y) = -6$, which gives you $g(y) = -3y^2+ay+b$.

Answer 2

Cauchy-Riemann equation:

$$\partial_x u=\partial_y v~~~~(1),~~~~~~~~~~~\partial_y u=-\partial_x v~~~~(2)$$

From (1), we get

$$\partial_y v=3x^2+g(y)\Longrightarrow v=3x^2y+\int^y g(t)dt+f(x)$$

plug into (2)

$$xg'(y)=-6xy-f'(x)\Longrightarrow \boxed{g'(y)+6y=-\frac{f'(x)}{x}}$$

The LHS is a function of $y$ and RHS is a function of $x$, this implies

$$g'(y)+6y=c_1=-\frac{f'(x)}{x}$$

hence

$$g(y)=-3y^2+c_1y+c_2$$

Answer 3

It is natural that you got some of them,not all of them as you chose a particular $f$ not the general case. Actually the equation:

$$ 6xy + f'(x)= -xg(y) $$

Is not as tricky as it looks, notice that setting for example $x=1$ you get that:

$$ 6y + f'(1) = -g'(x) \implies g(x) = -3y^2 + c_{1}y+c_{2} $$

And you should be able to finish up from here.