Showing that $u$ is constant if $u$ and $u^2$ are harmonic

Question

Let $G$ be a domain.

Let $u$ be harmonic in $G$. Show that if $u^2$ is harmonic in $G$ then $u$ is constant.

I know since $u^2$ is harmonic, that $\Delta u^2=0$, but I don't know how to advance from here.

Answer 1

We have $(u^{2})_x=2uu_x$ and $(u^{2})_{xx}=2uu_{xx}+2(u_x)^{2}$. Similarly compute $(u^{2})_{yy}$ and add. using the fact that $(u^{2})_{xx}+(u^{2})_{yy}=0$ and $u_{xx}+u_{yy}=0$ you will get $u_x^{2}+u_y^{2}=0$. Hence $u_x=u_y=0$ so $u$ is a constant.

Answer 2

Let $c:=u(0)$ and $v:=u-c$. Since $u$ and $u^2$ are harmonic, we see that $v^2$ is harmonic.

We have that $v^2 \ge 0$ on $G$ and $v^2(0)=0$. By the $\max / \min$ - principle, $v^2$ is constant $=0$, hence $v$ is constant and therefore $u$ is constant.

Answer 3

Here's a slightly more abstract proof that works in $n$ dimensions, and it's co-ordinate free!

A well-known identity from vector calculus affirms that, for vector fields $X$ and scalar functions $f$,

$\nabla \cdot (fX) = \langle X, \nabla f \rangle + f \nabla \cdot X; \tag 1$

taking

$f = u, \; X = \nabla u \tag 2$

in (1) yields

$\nabla \cdot (u \nabla u) = \langle \nabla u, \nabla u \rangle + u\nabla \cdot \nabla u = \vert \nabla u \vert^2 + u \nabla^2 u; \tag 3$

furthermore, we have

$\nabla u^2 = 2u \nabla u, \tag 4$

which when combined with (3) yields

$\nabla^2 u^2 = \nabla \cdot \nabla u^2 = \nabla \cdot (2u \nabla u) = 2\nabla \cdot (u \nabla u) = 2 \vert \nabla u \vert^2 + 2u \nabla^2 u; \tag 5$

now given that

$\nabla^2 u = \nabla^2 u^2 = 0, \tag 6$

we see via (5) that

$\vert \nabla u \vert ^2 = 0, \tag 7$

whence

$\nabla u = 0, \tag 8$

which of course implies $u$ is constant on $G$.

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