$u$ and $u^2$ are harmonic. 
Let $D$ be the unit disk centered at $0$ in the complex plane,
and let $u$ be a real harmonic function on D.
Find all $u$ such that $u(0)=0$ and $u^2$ is also harmonic on $D$.

 A: If $\nabla^2 u(x,y)=0$, then 
$$\begin{align}
\nabla^2 u^2(x,y)&=2u(x,y)\nabla^2u(x,y)+2\left|\nabla u(x,y)\right|^2\\\\
&=2\left|\nabla u(x,y)\right|^2
\end{align}$$
If $\nabla^2 u^2(x,y)=0$, then $\left|\nabla u(x,y)\right|^2=0$ and therefore we find that $u(x,y)$ is a constant.  
And if $u(0,0)=0$ additionally, then $u(x,y)\equiv 0$.
A: Two other ways to the same end:


*

*The average value of a harmonic function over a ball is equal to its value at the center of the ball. Let $B(r)$ denote the ball of radius $r\in(0,1)$ and center $0$. Using the averaging property for $u^2$ and for $u$, along with Jensen's inequality, we have
$$
\eqalign{
0&=[u(0)]^2=(\pi r^2)^{-1}\int_{B(r)}[u(x)]^2\,dx\cr
&\ge \left[(\pi r^2)^{-1}\int_{B(r)}u(x)\,dx\right]^2\cr
&=0^2=0.\cr
}
$$
It follows that $\int_{B(r)} [u(x)]^2\,dx=0$ for each $r$, and so $u$ must vanish throughout $D$. [As a bonus, any strictly convex function $\varphi$ could be substituted for the square function, with the  conclusion that if $u$ and $\varphi(u)$ are both harmonic then $u$ is constant.]

*(Minimum principle) The non-negative harmonic function $u^2$ attains it minimum value $0$ at $x=0\in D$. Consequently, $[u(x)]^2=0$ for all $x\in D$, so $u(x)=0$ for all $x\in D$. [Bonus: Suppose $\psi:\Bbb R\to[0,\infty)$ vanishes only at $0$. Let $u$ be a harmonic function in $D$ such that $u(0)=0$ and $\psi(u)$ is also harmonic. Then $u(x)=0$ for all $x\in D$.]
A: Real-valued harmonic functions satisfy the minimum principle. I.e., if $v$ is real valued and harmonic in $D,$ and $v$ attains a minimum value in $D,$ then $v$ is constant. Applying this to $u^2,$ where $u$ satisfies the given condition in this problem, gives $u\equiv 0.$
