Let $B \subset \mathbb R^2$ be the unit ball centered at origin and set $B^{\pm} := B \cap \{\pm y>0\}$ and $B^0 := B \cap \{y=0\}$. Let $u = u(x,y)$ be a continuous function on $B^+ \cup B^0$ and harmonic in $B^+$.
The Schwarz reflection principle for harmonic functions asserts that if $u \equiv 0$ on $B^0$, then $u$ can be extended to a harmonic function on $B^-$ by the "odd extension", namely, $$v(x,y) := \begin{cases} u(x,y), & y \geq 0 \\ -u(x,-y), & y < 0 \end{cases}$$ is harmonic on the whole $B$.
On the other hand, can we extension $u$ to whole $B$ by "even extension", namely, $$w(x,y) := \begin{cases} u(x,y), & y \geq 0 \\ u(x,-y), & y<0 \end{cases}.$$
It is clear that if such extension is valid, we need $\frac{\partial u}{\partial y}(x,0) = 0$ since $w$ is twice differentiable.
My question: Can we extend $u$ on $B^+ \cup B^0$ harmonically on $B^-$ by even extension provided $\frac{\partial u}{\partial y}(x,0) = 0$ for all $-1 <x< 1$?
The “even extension” $w$ is harmonic in $B$ only if $u$ is identically zero.
A harmonic extension of $u$ from $B^+$ to $B$ is unique (because of the Identity principle for harmonic functions).
So if both $v$ and $w$ are harmonic extensions of $u$ then necessarily $v=w$, and that is only possible if $u$ is identically zero.