If $u$ is harmonic then $u^2$ is harmonic? If $u$ is harmonic then $u^2$ is harmonic?
My attempt:
As $u$ is harmonic then $u_{xx}+u_{yy}=0$ Moreover, $(u^2)_{xx}=2u_x.u_x+u_{xx}.2u$
and $(u^2)_{yy}=2u_y.u_y+u_{yy}.2u$
Then
$$(u^2)_{xx}+(u^2)_{yy}=2(u_x)^2+u_{xx}.2u+2(u_y)^2+2u.u_{yy}=2((u_x)^2+(u_y)^2)\not=0$$
This implies $u^2$ is not harmonic. Is correct this?
 A: Our OP Bvss12's argument is essentially correct, save fore one small detail:  the condition 
$u_x^2 + u_y^2 \ne 0 \tag 1$
is satisfied if and only if $u$ is non-constant; otherwise, if $u$ is constant, then
$u_x = u_y = 0, \tag 2$
and (1) cannot bind. The point is that for constant  $u$, $u^2$ is constant as well, and constant functions are harmonic; but for a non-constant harmonic function $u$, $u^2$ is not harmonic by virtue of (1).
This may in fact be seen in a co-ordinate free manner by means of the identity
$\nabla \cdot (u\nabla u) = \nabla u \cdot \nabla u + u \nabla^2 u;  \tag 3$
using this, we have
$\nabla^2 u^2 = \nabla \cdot \nabla u^2 = \nabla \cdot (2u\nabla u) = 2 \nabla \cdot (u \nabla u) = 2 \nabla u \cdot \nabla u + 2 u \nabla^2 u; \tag 4$
with $u$ harmonic,
$\nabla^2 u = 0, \tag 5$
(4) yields
$\nabla^2 u^2 = 2 \nabla u \cdot \nabla u \ge 0,  \tag 6$
with equality holding if and only if
$\nabla u = 0, \tag 7$
the condition that $u$ be constant.
In closing it should be observed that the conclusion that $u^2$ harmonic implies $u$ constant requires the underlying assumption that the domain $D$ of $u$ is a connected set; otherwise the most we can achieve is that $u$ must be constant on the connected components of $D$, since the constancy of $u$ is derived from path integrals of the form
$u(Q) - u(P) = \displaystyle \int_{\gamma(t)} \nabla u(\gamma(s)) \cdot \dot \gamma(s) \; ds \tag 8$
where $\gamma(t)$ is a differentiable path joining $P$ and $Q$; this integral vanishes when 
$\nabla(\gamma(t)) = 0; \tag 9$
but since $\gamma(t)$ must lie in a single component of $D$, we can only infer that $u$ is constant on such a set; $u$ may take different values on different components of $D$.
Finally, we observe that this argument extends into a any number of dimensions; there is nothing intrinsically two dimensional about what has been herein presented.
A: You are correct. Your calculation shows that $\Delta u^2=2u\Delta u+2|\nabla u|^2$ for any function $u$ whatsoever. So $u$ and $u^2$ both being harmonic implies that $2|\nabla u|^2=0$, i.e. $u$ is constant.
