I'm looking to prove the following:
Let $\Omega$ be some simply connected open domain in $\mathbb C$ with three distinct points $a,b,c \in \partial \Omega$.
If $\Omega$ has a smooth boundary, then there exists a harmonic function $h$ on $\Omega$ which is the unique solution to the following:
$\begin{cases} h = 1 & \text{at the point $a$} \\ h = 0 & \text{on the arc bc} \\ \frac{\partial h}{\partial (\nu \tau)} = 0 & \text{on the arc ab} \\ \frac{\partial h}{\partial (-\nu \tau^2)} = 0 & \text{on the arc ac} \end{cases}$
Where $\nu$ is the counter-clockwise pointing unit tangent to $\partial\Omega$, and $\tau = e^{2i\pi /3}$.
This claim is made in a paper of Stanislav Smirnov, but I have very limited background in complex analysis and PDEs in general, so I have no idea where to even start on proving that this problem has a unique solution. Any help much appreciated.
Edit: I've now been trying to prove it on the equilateral triangle and applying the Riemann mapping theorem. Specifically taking the three points to be the vertices of the triangle, and attempting to prove that if $f$ and $g$ satisfy the boundary conditions, then $f-g \equiv 0$.
I can get the uniqueness using PDE techniques. I'm going to call the tangent $T$ instead, since $\nu$ is hardwired as the normal vector in my brain. To avoid confusion I'll use $n = e^{-i \pi/2} T$ to denote the outwards unit normal vector to $\partial \Omega$.
Suppose we have two real solutions $u,v$ satisfying the conditions, and let $w = u - v$ so that we want to prove $w = 0.$ Then $w$ is a harmonic function satisfying the corresponding homogeneous boundary conditions: $$\begin{cases} \Delta w=0 & \text{on }\Omega \\ w = 0 & \text{on }\{a \} \cup bc \\ \partial w /\partial (\tau T) = 0 & \text{on }ab \\ \partial w /\partial (-\tau T^2) = 0 & \text{on }ac \end{cases}\tag{1}$$
Note that $\int_\Omega w \Delta w = 0$; so using the product rule $$\mathrm{div}(w \nabla w) = |\nabla w|^2 + w \Delta w$$ and the divergence theorem we can integrate by parts to get $$-\int_\Omega |\nabla w|^2 + \int_{\partial \Omega} w \frac{\partial w}{\partial n} = 0.$$ Breaking up the boundary integral into three pieces, the portion from $bc$ vanishes immediately; so we are left with $$\int_\Omega |\nabla w|^2 = \int_a^b w \frac{\partial w}{\partial n}+ \int_c^aw \frac{\partial w}{\partial n},$$
where by $\int_a^b$ I of course mean the line integral along the boundary curve joining $a$ to $b$. In order to extract something from the boundary condition, note that $n$ can be written as a linear combination $\alpha T + \beta V$ where $V$ is the vector field appearing in the boundary condition; i.e. $V = \tau T$ on $ab$ and $V = -\tau^2 T$ on $ca$. Note that the coefficients $\alpha,\beta$ are piecewise constant on the arcs $ab,ca$. Since $T$ is the velocity vector of the unit-speed parametrization of the boundary, we can thus write $$\int_a^b w \frac{\partial w}{\partial n} = \alpha_{ab} \int_a^b w \frac{dw}{ds} ds + \beta_{ab} \int_a^b w \frac{\partial w}{\partial V}$$ and similar for $ca.$ The second integral vanishes thanks to the boundary condition $\partial w / \partial V = 0$, while the first can be written using the fundamental theorem of calculus as $$\frac {\alpha_{ab}} 2\int_a^b \frac d{ds}(w^2) ds = \frac {\alpha_{ab}} 2 (w(b)^2 - w(a)^2),$$ which vanishes because $w(b)=w(a)=0$ from the boundary conditions. The exact same argument works for the other arc, yielding $\int_\Omega |\nabla w|^2 = 0;$ so $w = u-v$ is constant, and thus $u \equiv v$ since they have the same value at $a$.
Now, assuming instead you have a complex solution $h= f + ig$, note that the imaginary component $g$ satisfies the homogeneous system $(1)$, and thus by the argument above we know $g = 0$. Thus all solutions are real, so we have unconditional uniqueness.