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.
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$\begingroup$ $\overline {z}$ is harmonic and so is its square. $\endgroup$– Kavi Rama MurthyCommented Jul 9, 2020 at 10:05
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2 Answers
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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.
answered Jul 9, 2020 at 10:08
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$\begingroup$ 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. $\endgroup$ Commented Jul 9, 2020 at 10:11
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$\begingroup$ Can it be deduced from your answer that a square of harmonic function is harmonic only if f is constant? $\endgroup$– user775699Commented Oct 1, 2020 at 10:02
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$\begingroup$ @User Yes, that is true. $\endgroup$ Commented Oct 1, 2020 at 10:03
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$\begingroup$ sorry that was trivial, got it. $\endgroup$– user775699Commented Oct 1, 2020 at 10:12
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$\partial_z u^2=2u\partial_z u$ 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)
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$\begingroup$ "$\partial_{\overline{z}} u=0$ 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. $\endgroup$ Commented Jul 9, 2020 at 14:51
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$\begingroup$ 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 $\bar g=0$) $\endgroup$– ConradCommented Jul 9, 2020 at 15:40
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$\begingroup$ ah identity theorem! Now I get you! Thanks! $\endgroup$ Commented Jul 9, 2020 at 16:25
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