I have a question on a harmonic function and the boundary behavior.


Let $\mathbb{U} \subset \mathbb{C}$ be a unit disk. We denote by $\overline{\mathbb{U}}$ the closure of $\mathbb{U}$ in $\mathbb{C}$.


We have a reflected Brownian motion $X=(\{X\_t\}\_{t \ge0}, \{P\_x\}\_{x \in \overline{U}})$ on $\overline{\mathbb{U}}$. Let $B:=B(a,r)$ be an open disk centered at $a \in \mathbb{C}$ with radius $r>0$ and $C$ a closed disk such that $C \subset \mathbb{U} \cap B$. 


$u(x):=P\_{x}(\sigma\_{C}>\tau\_{B})$ is a harmonic function with respect to $X$ on $(\overline{\mathbb{U}}\cap B) \setminus C$, which satisfies $\lim\_{x \to \partial C}u(x)=0$. Here, we define 


\begin{align\*}
\sigma\_C=\inf\{t>0 \mid X\_t \in C\},\\
\tau\_B=\inf\{t>0 \mid X\_t \notin B\}.
\end{align\*}


In other words, $u$ is a positive harmonic function on $ (\mathbb{U} \cap B) \setminus C$ with the Neumann boundary condition on $\partial \mathbb{U} \cap B$ and the Dirichlet boundary condition on $\partial C$.


**Question**


How $u(x)$ behave as $x \to \partial C$?


I am intersted in the rate of convergence of $\lim\_{x \to \partial C}u(x)=0$.


Can we construct a nice positive harmonic function on $\mathbb{U} \setminus C$ with the Neumann boundary condition on $\partial \mathbb{U}$ and the Dirichlet boundary condition on $\partial C$?? If we know the behavior near $\partial C$, we should be able to obtain the behavior of $u$ near $\partial C$ by the boundary Harnack inequality.



Harmonic functions with boundary condition
Roughly: $u(x) \approx \operatorname{dist}(x,C)$ (in the sense that the ratio is bounded), except near the corners, where $u(x) \approx |x - x\_0|^{(2\alpha / \pi) - 1} \operatorname{dist}(x,C)$, where $x\_0$ is a corner point and $\alpha$ is the interior angle at $x\_0$.


The easiest way to see this is to map your domain conformally into a square $[0,1] \times [0,1]$, so that the image $v$ of $u$ is harmonic, with Neumann boundary condition along vertical sides, Dirichlet condition $v = 0$ along the bottom side, and Dirichlet condition $v = 1$ along the top side. Then $v(x) = \operatorname{Im} x$, and everything boils down to the properties of conformal maps.


