I have a question on a relation between conformal mappings and diffusion processes with boundary condition.
Let $D_1$ be a smooth simply connected domain of $\mathbb{R}^2 \cong \mathbb{C}$. This may be unbounded.
We can define the normally reflecting Brownian motion $X$ on $\bar{D_1}$. We can also describe the Skorohod equation. The generator is the Laplacian $\Delta$ on $D_1$ with Neumann boundary condition.
Let $D_2 \subset \mathbb{R}^2$ be an another smooth simply connected domain. Let $\Psi:D_1 \to D_2$ be a conformal mapping. We also assume that $\Psi$ is extended to a homeomorphism from $\bar{D_1} \to \bar{D_2}$ and $\Psi(\partial D_1)=\partial D_2$ (I do not know if this assumption is necessary. Please tell me if it is unnecessary.).
By using this map, we can make the change of variable: $ D_1 \ni (\rho,z) \mapsto (r,w) \in D_2$, where $r=\text{Re}\Psi(\rho,z)$ and $z=\text{Im} \Psi(\rho,z)$.
In $(r,w)$-coordinate, $\Delta$ does not take the form $\frac{\partial^2}{\partial r^2}+\frac{\partial^2}{\partial w^2}$. In $(r,w)$-coordinate, $\Delta$ becomes an another diffusion operator. We denote by $\mathcal{L}$ the operator.
My question is as follows:
- Does the diffusion process $\Psi(X)$ correspond to the operator $\mathcal{L}$ with Neumann boundary condition on $\partial D_2$?
You will think that this is trivial. But I do not know how to justify these results.
Should I prove that the diffusion processe determined by $\mathcal{L}$ (with Neumann boundary condition) and the diffusion process $\Psi(X)$ coincide?
ADD:
$X$ satisfies the following SDE: $X_t=x+B_t+\int_{0}^{t}\nu(X_s)\,dL_s$. Here, $\{L_t\}$ denotes the boundary local time of $X$ and $\nu$ the normal inward unit vector on $\partial D_1$. I guess that $\Psi(X)$ is a time-changed reflecting Brownian motion on $D_2$.
Let $u=\text{Re}(\Psi)$, $v=\text{Im}(\Psi)$. Note that $u$ and $v$ are harmonic function on $D_1$. Then, \begin{align*} u(X_t)&=u(x)+\int_{0}^{t}\nabla u (X_s)\,dX_s +(1/2)\int_{0}^{t}\Delta u(X_s)\,ds \\ &=u(x)+\int_{0}^{t}\nabla u(X_s)\,dB_s+\int_{0}^{t}(\nabla u,\nu)(X_s)\,dL_s. \end{align*} Similarly, it holds that \begin{align*} v(X_t) &=v(x)+\int_{0}^{t}\nabla v(X_s)\,dB_s+\int_{0}^{t}(\nabla v,\nu)(X_s)\,dL_s. \end{align*} By the CR-equation, it is easy to see that \begin{align*} &<\int_{0}^{\cdot}\nabla u(X_s)\,dB_s>_t=<\int_{0}^{\cdot}\nabla v(X_s)\,dB_s>_t=\int_{0}^{t}|\Psi'(B_s)|^2\,ds \\ &<\int_{0}^{\cdot}\nabla u(X_s)\,dB_s,\int_{0}^{\cdot}\nabla v(X_s)\,dB_s>_t=0. \end{align*} Therefore, the martingale part of $\Psi(X)$ is a time-changed (two-dimensional) Brownian motion.
What the next term will be?: \begin{equation*} (\int_{0}^{t}(\nabla u,\nu)(X_s)\,dL_s, \int_{0}^{t}(\nabla u,\nu)(X_s)\,dL_s). \end{equation*}
ADD2:
I found a related result. In Chapter 5 of this book CF, the authors identify the Dirichlet form of $\Psi(X)$.
