# Harmonic functions the squares of which are harmonic as well

I have been asked to determine all such complex-valued harmonic functions whose squares are also harmonic. Clearly the set of all such functions contains as a subset the set of analytic functions (which are themselves harmonic and their squares being analytic, are again harmonic), but is that all? Or are there non-analytic harmonic functions that have harmonic squares? I need some direction to proceed with.

• $\overline {z}$ is harmonic and so is its square. Commented Jul 9, 2020 at 10:05

If $$u$$ is harmonic then $$u^{2}$$ is harmonic iff $$u_x^{2}+u_y^{2}=0$$. You can see this by computing the Laplacian. But I don't think there is a simple description of all solutions of this equation.

• Of course $u_x=\pm iu_y$ along with continuity gives $u_x \equiv iu_y$ or $u_x \equiv -iu_y$ in any open region where the partial derivatives are no zero, but zeros create a problem. Commented Jul 9, 2020 at 10:11
• Can it be deduced from your answer that a square of harmonic function is harmonic only if f is constant?
– user775699
Commented Oct 1, 2020 at 10:02
• @User Yes, that is true. Commented Oct 1, 2020 at 10:03
• sorry that was trivial, got it.
– user775699
Commented Oct 1, 2020 at 10:12

$$\partial_z u^2=2u\partial_z u$$ so $$0=\partial_{\bar z} (\partial_z u^2)=2(\partial_{\bar z}u)(\partial_z u)+2u\partial_{\bar z} (\partial_z u)=2(\partial_{\bar z}u)(\partial_z u)$$

Hence $$(\partial_{\bar z}u)(\partial_z u)=0$$ for all $$z\in \mathbb C$$

But now if $$A$$ is the set where $$\partial_zu=0$$, $$A$$ is obviously closed, so assuming $$u$$ is not analytic and $$A \ne \mathbb C$$ it follows that $$\partial_{\bar z}u=0$$ contains a non-empty open set $$B$$ and from there it obviously follows $$\partial_{\bar z}u=0$$ for all $$z \in \mathbb C$$, so $$u$$ is conjugate analytic.

(edit - per comments to explicit the above - we note that if $$u$$ harmonic, $$\partial_z u$$ is always analytic and $$\partial_{\bar z}u$$ is always conjugate analytic, so in this case, we have a conjugate-analytic function $$g=\partial_{\bar z}u$$ zero on an open non-empty set, it then follows by the identity principle for conjugate-analytic functions that $$g=0$$ on the connected component containing $$B$$; the identity principle follows from the analytic case since $$g$$ is conjugate analytic iff $$\bar g$$ is analytic and similarly $$g=0$$ iff $$\bar g=0$$))

Hence $$u$$ must be either analytic or conjugate analytic (note that the proof applies to any domain, while in general, $$u$$ is analytic or conjugate analytic on any component of the open set where it is defined, if that is not connected)

• "$\partial_{\overline{z}} u=0$ contains a non-empty open set $B$ and from there it obviously follows $\partial_{\overline{z}} u=0$ for all $z\in\mathbb{C}$."- I don't see this. Commented Jul 9, 2020 at 14:51
• note that if $u$ harmonic, $g=\partial_{\overline{z}} u$ is conjugate- analytic (that is true always since the Laplacian is $\partial_{z}\partial_{\overline{z}}$), while in this case $g=0$ on a non-empty open set, hence by the identity principle for conjugate-analytic functions, $g$ is then zero on the connected component containing $B$ (if you are comfortable with analytic functions only, note that $\bar g$ is analytic when $g$ is conjugate-analytic and $g=0$ iff $\bar g=0$) Commented Jul 9, 2020 at 15:40
• ah identity theorem! Now I get you! Thanks! Commented Jul 9, 2020 at 16:25
• happy to be of help! Commented Jul 9, 2020 at 16:26