Let $\gamma:S^{1} \times [0,1] \to \mathbb{C}$ be in $C^1$ and suppose that $\gamma(.,t)$ is a Jordan curve for all $t\in [0,1]$. Call the domain bounded by $\gamma(.,t)$ as $\Omega(t)$. Assume that for all $t\in [0,1]$, $0 \in \Omega(t)$ and that there exists a $\delta>0$ such that $d(0,\partial \Omega(t) )\geq \delta$.
For each fixed $t\in [0,1]$ consider the unique Riemann map $\Phi(.,t):\mathbb{D} \to \Omega(t)$ such that $\Phi (0,t) =0 $ and $\Phi_{z}(0,t) >0$. As $\partial \Omega(t)$ is a Jordan curve we know that $\Phi(,.t)$ extends continuously to $\mathbb{D}$ by Caratheodary's theorem.
Question: Is $\Phi(z,t)$ differentiable in $t$? And if that is the case, is it true that for each fixed $t$, $\Phi_{t}(z,t)$ extends continuously to $\mathbb{D}$?
I know that $\Phi(z,t)$ is continuous in $t$ by a theorem of Rado (Proof given in the book by Pommerenke: Boundary behaviour of Conformal maps).
It seems that there is an unpublished paper (Lavrentiev curves and conformal mapping) by Coifman and Meyer on this problem where they show that the Riemann map depends analytically on the curve. However first of all I cannot find the paper anywhere and even with it I don't know how it solves the above question as the analytic dependence is with respect to the BMO norm associated with the chord arc curve (The problem is that all constants are killed by the BMO norm which corresponds to a rotation. So a lot of domains correspond to the same BMO function and hence you lose information. Also all possible Riemann maps from $\Omega$ to the upper half plane get assigned the same BMO function )
I would be very grateful if someone can answer this or give me a reference for it.
If you replace "$C^1$" by "analytic" and "$[0,1]$" by "the unit disk", this paper: http://arxiv.org/pdf/1302.2704.pdf gives several necessary and sufficient conditions.
I don't know of any results for $C^1$ regularity.