The continuous dependence of the Green's function on a domain Let $\Omega\in\mathbb{R}^2$ be a smooth bounded domain and $G(x,y)$ be the Green's function of $-\Delta$ in $\Omega$ with zero Dirichlet condition. Clearly $G(x,y)=-\frac{1}{2\pi}\ln|x-y|-h(x,y)$, where $h$ is a smooth function. Let $(L(x))_{2\times 2}\in C^\infty$ be a matrix-valued function.
For any $x\in X$, $L(x)$ is positive definite and there exists c>0 such that
$$c|\zeta|^2\leq \zeta L(x)\zeta, \ \ \forall \zeta\in\mathbb{R}^2, x\in X.$$
What I want to know is, whether $G(x,y)$ is continuously (smoothly) dependent of $\Omega$. In other words, let $G_{z}(x,y)$ be the Green's function of $-\Delta$ in $L(z)(\Omega):=\{L(z)x\mid x\in\Omega\}$. Is $G_z(x,y)$ continuous (smooth) about $z$?
 A: On the question of continuity, much more is true: the Green's function is continuous under convergence of domains in the sense of Carathéodory, which is the weakest reasonable topology of planar domains - weaker than e.g. Hausdorff convergence of the boundaries. See e.g., Pommerenke, Univalent functions, for the definitions.
Let me sketch a proof. Let $\Omega_n$ be a sequence of planar domains that converge in the Caratheodory sense to $\Omega$, say, simply connected or finitely connected, and $x\in\Omega$. It suffices to show that the harmonic functions $h_n(x,\cdot)$ converge to $h$ uniformly on compact subsets of $\Omega$. But they are uniformly bounded (as $\partial\Omega_n$ are eventually uniformly bounded away from $x$), thus Harnack's estimates allow one to conclude that they are equicontinuous, and by Arzela-Ascoli one can exract a convergent subsequence. Using e.g. Beurling estimate, you can show that any subsequential limit must solve the Dirichlet problem with boundary conditions $\frac{1}{2\pi}\log|x-y|$
On the question of smoothness, see Hadamard's variational formula for calculating the variational derivative of the Green's function under $C^1$ deformations of $C^1$ domains. I don't know about higher order smoothness.
