Existence of conjugate for $u \in C^2(\mathbb{R}^2)$ harmonic in the plane

Question

I'm looking for a proof of the following statement

Let $u \in C^2(\mathbb{R}^2)$ be a harmonic function in $\mathbb{R}^2$. Then, $u$ has an armonic conjugate.

So far, what I've noted is that this is equivalent to saying that there exists $v \in C^2(\mathbb{R}^2)$ such that $f \equiv u + iv$ is holomorphic, i.e. (by Cauchy-Riemann) that $Dv$ = $Du \cdot \begin{bmatrix}0&1\\-1&0\end{bmatrix}$. The first thing that came to my mind was $v \equiv u(\mu(x,y))$ with $\mu(x,y) = (y,-x)$, but this is not quite right, since we'd have

$Dv(x,y) = Du(y,-x)\begin{bmatrix}0&1\\-1&0\end{bmatrix}$

instead of

$Dv(x,y) = Du(x,y)\begin{bmatrix}0&1\\-1&0\end{bmatrix}$

Thoughts?

Answer 1

Consider the function $F=(g,h)^t$, where $g=-u_y$ and $h=u_x$. Since $u$ is harmonic, we have

$g_y=-u_{yy}=u_{xx}=h_x$. Hence the vecor field $F$ is conservative.

Since $ \mathbb R^2$ is simply connected, there exists a $C^1$ skalar field $v$ such that

$F= grad v=(v_x,v_y)^t$.

Therefore $u$ and $v$ satisfiy the Cauchy-Riemann differential equations.