\begin{question}

How can I prove that harmonic functions have the mean-value property using Brownian motion $B_{t} ?$
I know that I need to use the fact that $B_{t \wedge \tau}$ is a martingale where $\tau$ is a stopping time which denotes the first time the Brownian motion leaves a ball with radius $r$ around $B_{0}$

\end{question}

Fix a ball $B(x, r)$ and consider $\tau=\inf t>0,\left|B_{t}-x\right|=r$.
Now compute by Itô’s formula
$$f\left(B_{\tau}\right)=f\left(B_{0}\right)+\int_{0}^{\tau} f^{\prime}\left(B_{s}\right) d B_{s}+\frac{1}{2} \int_{0}^{\tau} \Delta f\left(B_{s}\right) d s$$
Take expectations:
$$\mathbb{E}\left[f\left(B_{\tau}\right)\right]=f\left(B_{0}\right)+\mathbb{E}\left[\int_{0}^{\tau} f^{\prime}\left(B_{s}\right) d B_{s}\right]+\frac{1}{2} \int_{0}^{\tau} \Delta f\left(B_{s}\right) d s$$
Since $\mathbb{E}\left[\int_{0}^{\tau} f^{\prime}\left(B_{s}\right) d B_{s}\right]=0$ and $\Delta f \equiv 0$ we obtain that
$$\mathbb{E}\left[f\left(B_{\tau}\right)\right]=f\left(B_{0}\right)$$
Now it suffices to take a Brownian motion starting at $x$ and to note by rotation invariance that
$$\mathbb{E}\left[f\left(B_{\tau}\right)\right]=\frac{1}{|\partial B(x, r)|} \int_{\partial B(x, r)} f(x) d S$$
and therefore you obtain the mean property
$$f(x)=\mathbb{E}\left[f\left(B_{\tau}\right)\right]=\frac{1}{|\partial B(x, r)|} \int_{\partial B(x, r)} f(x) d S$$

Remark 1. It is important to note that $\tau$ is a finite almost surely stopping time and that $\left|B_{\tau}-x\right|=r$

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 $X=\left(\left{X_{t}\right}_{t \geq 0},\left{P_{x}\right}_{x \in \bar{U}}\right)$ 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 $C \subset \mathbb{U} \cap B$.
$u(x):=P_{x}\left(\sigma_{C}>\tau_{B}\right)$ is a harmonic function with respect to $X$ on $(\overline{\mathbb{U}} \cap B) \backslash C$, which satisfies $\lim {x \rightarrow \partial C} u(x)=0$. Here, we define \begin{aligned} \sigma{C} &=\inf \left{t>0 \mid X_{t} \in C\right} \ \tau_{B} &=\inf \left{t>0 \mid X_{t} \notin B\right} \end{aligned}
In other words, $u$ is a positive harmonic function on $(\mathbb{U} \cap B) \backslash C$ 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 $x \rightarrow \partial C ?$
I am intersted in the rate of convergence of $\lim _{x \rightarrow \partial C} u(x)=0$.
Can we construct a nice positive harmonic function on $\mathbb{U} \backslash C$ 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 $u(x) \approx\left|x-x_{0}\right|^{(2 \alpha / \pi)-1} \operatorname{dist}(x, C)$, 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.

BS equation代写