# Harmonic functions with boundary condition

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.

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.
• Thank you for your kind reply. However, our domain is $(U \cap B) \setminus C$, right? This is not a Jordan domain. How do you map our domain to the square? Nov 11, 2019 at 19:46
• Sorry. I forgot to write the definition of $\tau_B$ and $\sigma_C$. Nov 11, 2019 at 19:55
• Ah, I thought $C$ intersects $U$ as well. Then simply $u(x) \approx \operatorname{dist}(x, C)$ by the boundary Harnack inequality, right? Nov 11, 2019 at 20:02
• Yes (assuming that $K = C$ in your comment). BHI is a local result: in your case $u$ is a positive harmonic function in a $C^{1,1}$ domain $D = B(x_0, r_2) \setminus B(x_0, r_1)$ (where $C = \overline{B}(x_0, r_1)$ and $r_2 > r_1$ is small enough), which goes to zero continuously on $\partial C = \partial B(x_0, r_1)$. Then $u$ is necessarily comparable with the distance to $\partial C$. Nov 11, 2019 at 20:32
• If we only assume that $u$ bounded by $1$ in $D$, then, by comparison principle, $u(x) \le \log(|x - x_0| / r_1) / \log(r_2 / r_1)$, and this cannot be improved. For the function $u$ in the original question: things get complicated if $\partial C$ gets too close to either $\partial B$ (then $A$ in the upper bound explodes just as above) or $\partial U$ (then a similar $A$ in the lower bound goes to zero super-fast). Nov 11, 2019 at 21:23