Local existence of harmonic conjugates

Remarks

Proof idea: write $z = x + iy$ and set $g = u_x - i u_y$. The equation $u_{xx} + u_{yy} = 0$ and the symmetry $u_{xy} = u_{yx}$ are exactly the Cauchy–Riemann equations for $g$, so $g$ is holomorphic on $U$. A holomorphic function on a disc has a primitive $F$, and $F' = g$ gives $\nabla(\operatorname{Re} F) = \nabla u$. So $\operatorname{Re} F - u$ is a real constant $c$ on $D$, and $f = F - c$ works. The same proof works on every simply connected domain in place of the disc. The result is local: on $U = \C \setminus \set{0}$ the function $u(z) = \log\abs{z}$ is harmonic. On each disc that avoids $0$ it has the conjugate $v = \arg z$ for a continuous branch of the argument. It has no harmonic conjugate on all of $U$, because no continuous argument exists on a circle around $0$.

Going deeper

For a student

The theorem lets one transfer results from complex analysis to planar harmonic functions. For example, the mean value property and the maximum principle for $u$ follow from those for a local holomorphic $f$ with $\operatorname{Re} f = u$. Globally, a harmonic conjugate exists exactly when the period $\oint (-u_y \, dx + u_x \, dy)$ vanishes on each closed curve in $U$. These periods are the obstruction on multiply connected domains, as the example $\log\abs{z}$ shows.