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 $\partial C$.


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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – sharpe
    Nov 11, 2019 at 19:46
  • $\begingroup$ Sorry. I forgot to write the definition of $\tau_B$ and $\sigma_C$. $\endgroup$
    – sharpe
    Nov 11, 2019 at 19:55
  • $\begingroup$ Ah, I thought $C$ intersects $U$ as well. Then simply $u(x) \approx \operatorname{dist}(x, C)$ by the boundary Harnack inequality, right? $\endgroup$ Nov 11, 2019 at 20:02
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    $\begingroup$ 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 $\partial C$. $\endgroup$ Nov 11, 2019 at 20:32
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    $\begingroup$ 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). $\endgroup$ Nov 11, 2019 at 21:23

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