Let $u \colon \{\mbox{Im } z>0\}\subset\mathbb{C}\to \mathbb{R}$ be a positive harmonic function in the upper half plane, i.e $$ \Delta u=0,\,\, \mbox{for}\,\mbox{ Im } z>0. $$ Consider now the following mixed Dirichlet-Neumann boundary condition on the boundary $\partial\{\mbox{Im } z>0\}=\{\mbox{Im }z=0\}$ $$ u=0,\,\mbox{for }\{(x,0): -1<x<1\}\\ \frac{\partial u}{\partial y}=0,\,\mbox{for }\{(x,0): x<-1 \vee x>1\} $$ where obviously $z=x+iy$. For example $u(z)=\mbox{Re}(\sqrt{(z-1)(z+1)})$ is a solution of the previous problem, but it's unique upo to a constant? The question is how can i prove that this problem has an unique solution?
It's possible to apply a reflection to the other half plane and then apply some Liouville type result since the function turn to be bounded from below, or a maximum principle?
Every non-negative harmonic function in the upper half plane can be represented as $$ h(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{yd\rho(t)}{(t-x)^{2}+y^{2}}+Ay $$ where $A$, $D$ are positive constants and $\rho(t)$ is a non-decreasing function on $\mathbb{R}$ for which $$ \int_{-\infty}^{\infty}\frac{d\rho(t)}{1+t^{2}} < \infty. $$ The representation is unique if you add a normalization requirement to $\rho$ such that $\rho(t+0)=\rho(t)$. You can determine $\rho$ from limits of the function $h$ near the boundary. If $a$ and $b$ are points of continuity of $\rho$, then $$ \rho(b)-\rho(a) = \lim_{y\downarrow 0}\int_{a}^{b}h(x,y)dx $$ Jumps of $\rho$ are determined by $$ \frac{1}{\pi}\{\rho(t+0)-\rho(t-0)\} = \lim_{y\downarrow 0}yh(x,y) $$ And $\rho$ is differentiable at some $t$ iff the following exists $$ \rho'(t) = \lim_{y\downarrow 0}h(t,y) $$ This representation theorem gives you a unique characterization of a positive harmonic function on the upper half plane.
For uniqueness, you can also apply a version of Liouville's theorem if the function is continuous to the boundary. The details can be found in Axler's book on Harmonic Function Theory, which is a good reference to have.