The differentiability of the distance function on asymptotically flat manifolds Let $M = \mathbb{R}^3 \setminus \overline{B_1}$ where $\overline{B_1}$ is the closed unit ball.
Let $g$ be an asymptotically flat metric of the form $g_{ij} = \delta_{ij}+h_{ij}$ in standard coordinates such that
$$ \sup_{M} ( |x|^2 |h| + |x|^3 |\partial h| + |x|^4 |\partial^2 h| ) < \epsilon$$
where $\epsilon$ is a small positive number and $|x| = \sqrt{x_1^2+ x_2^2+ x_3^2}$ and $(x_1,x_2,x_3)$ is the standard coordinates on $M$.
Then is it true that the distance function $r(\cdot) := {\rm dist}(\cdot, \partial M)$ from the boundary is differentiable on $M\setminus \partial M$ and hence defines a global foliation of spheres $S_{r_0} := \{ r=r_0 \}$?
It is clear that there exists a neighborhood $U$ of $\partial M$ such that $r$ is differentiable on $U \setminus \partial M$ since $\partial M$ is compact.
Does it suffice to show that if $r$ is differentiable on $B_R = \{x \in M| 0<r(x) < R \}$ for some $R>0$, then the mean curvature $H_r$ of the spheres $S_r$ for $r<R$ is bounded from below? I think that the mean curvature must go to $-\infty$ on $S_R$ if $r$ fails to be differentiable on $S_R$, but I am not sure if that's correct. If so, then we can study the ODE obeyed by the mean curvature $H_r$ and the traceless part of the second fundamental form $\hat A$. If the Ricci curvature falls very quickly, I think it can be shown that $H_r = \frac{1}{r} + O(\frac{1}{r^2})$.
Of course, if $g$ is the euclidean metric, then $r$ is differentiable on $M \setminus \partial M$.
Any help or references is appreciated.
 A: Here I can give a sufficient condition.
The differentiability of the distance function from $\partial M$ is basically asking whether the normal exponential map from $\partial M$ is a diffeo. (Here, the normal exponential map is $\exp: N(\partial M) \to M$ where $N(\partial M)$ is the (one-sided) normal bundle of the boundary. If you start with $p\in \partial M$ and $v$ an outward pointing normal vector to $\partial M$ and $p$, $\exp(p,v)$ is the point in $M$ obtained from evolving for time 1 along the geodesic starting at $p$ with velocity $v$.)
For the distance function to fail to be differentiable, your $\exp(p,v)$ must be singular (in the sense of having a critical point). To analyze this, you can look at Jacobi fields of the corresponding normal geodesics.
The Jacobi equation for a Jacobi field $J$ along a geodesic $\gamma$ is
$$ \ddot{J} + R(J,\dot{\gamma}, \dot{\gamma}) = 0 $$
What you know is that for the background (Euclidean) metric, the solutions orthogonal to $\dot{\gamma}$ grows linearly in $t$.
So a sufficient condition for your perturbed metric $g = \delta + h$ to exhibit similar behavior (that the normal exponential map gives a diffeomorphism) is that (i) $g$ is uniformly bounded above and below by $\delta$ (ii) the first derivative $\partial h$ is bounded by $\epsilon |x|^{-1-\gamma}$ for some $\epsilon \ll 1$ and $\gamma > 0$ (iii) and the second derivative $\partial^2 h$ to be bounded by $\epsilon |x|^{-2-\gamma}$.
