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Let $D_1$ and $D_2$ be two bounded simply connected Jordan domains in $\mathbb{R}^2$. By Carathéodory's Theorem there exists a homeomorphism $f:\bar{D}_1 \to \bar{D}_2$ such that the restriction $f:D_1 \to D_2$ is conformal. If $E\subset \partial D_1$ is Borel measurable then the harmonic extension of the indicator function $\mathbf{1}_E$ from $\partial D_1$ to $\bar{D}_1$ is called the harmonic measure of $E$ and is denoted by $\omega(x,E;D_1)$ for $x\in \bar{D}_1$. It is well known that $f$ preserves harmonic measure in the sense that $\omega(x,E;D_1)=\omega(f(x),f(E);f(D_1))=\omega(f(x),f(E);D_2)$. Since $\omega(\cdot,E;D_1)$ is really just the solution to the BVP

    $\Delta u=0$ in $D_1$
    $u=\mathbf{1}_E$ on $\partial D_1$

we can generalize harmonic measure to other elliptic operators $\mathcal{L}$ besides the Laplacian. This generalization is called $\textit{elliptic measure}$ or $\mathcal{L}\textit{-harmonic measure}$, see Diffusions and Elliptic Operators by Bass. My question is whether given $D_1,D_2,$ and $\mathcal{L}$ as above does there exist an injective "generalized conformal map" from $D_1$ onto $D_2$ that preserves elliptic measure and obeys a first order Cauchy-Riemann type system? I'm interested in finding out whether such maps exist and what their Cauchy-Riemann systems are for constant coefficient operators of the form $\mathcal{L}u=\Delta u+\alpha\cdot \nabla u$. I believe that quasiconformal maps give an affirmative answer for a certain class of elliptic operators but I'm not sure these include what I'm looking for.

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    $\begingroup$ One way to think about defining a conformal map from a Jordan domain $D$ to the unit disk $\mathbb{D}$ is to let $u$ be a suitable multiple of the Green's function in $D$ (for Brownian motion or Laplacian depending on your perspective), let $v=v(z_0)+\int_{gamma} \frac{d}{dn} u(z)|dz|$ where $\gamma$ is a curve from $z_0$ to $z$. Then $f=e^{-(u+iv)}$ is a conformal map from $D$ to $\mathbb{D}$. For other diffusions, you still have a Green's function, but I don't know how you'd define $v$ (or if this is in any way the right way to approach things). $\endgroup$
    – ShawnD
    Mar 30, 2012 at 17:35
  • $\begingroup$ Is it obvious that conformal maps won't preserve the generalized harmonic measure! $\endgroup$
    – Brian Rushton
    Dec 7, 2012 at 2:21
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    $\begingroup$ @BrianRushton This becomes obvious with the probabilistic interpretation of elliptic measure as the exit distribution of the corresponding elliptic diffusion. Think of Brownian motion with constant drift on the unit disc. A rotation of the disc is conformal but will change the probability that the Brownian motion exits in a particular sector of the boundary. $\endgroup$
    – HMPanzo
    Feb 20, 2016 at 20:53
  • $\begingroup$ Maybe this paper is of some interest ? $\endgroup$
    – user111
    Sep 12, 2017 at 14:06

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