Is dynamics in GR unique?

Question

Quote:

"... a map $h$ of an open set $\Theta$ of a Bnach space $B_1$ into a Banach space $B_2$ is Lipschitz in $\Theta$ if there exists $k\in \mathbb{R}$ such that $||h(a)-h(b)||\leq k||a-b||$ for $(a,b)\in\Theta\times \Theta$".

Quote:

"0.33 The Cauchy Theorem. The differential equation $$(*)~~~~~~~~~~~~ x'=f(t,x),x(t_0)=x_0$$ has a unique continuous solution $x(t)$ if $f(t,x)$ is Lipschit in $x$ on $I\times \Omega$."

Quote:

"0.34 Remark. If $f(t,x)$ is not Lipschitz, we cannot say anything in general. But if Banach space is $\mathbb{R}^n$, then there is a solution of (*), but it may not be unique."

This made me thought about GR, with black hole, the curvature went to infinity, thus the function could not be Lipschitz. Does this imply that the dynamics around black hole was not unique?

Answer 1

This is why cosmic censorship is considered to be so important -- you are saved from this conclusion if all of the infinite curvature points are hidden behind horizons, and therefore, the exterior of the black holes can still be globally hyperbolic.

Answer 2

The question and the example are not directly related. The equations in GR are PDE and it is not just a simple fixed point argument, where you use the Lipschitz property. Also the function $f(t,x)$ in your example is part of the equation, not related to a particular solution nor initial data. So you cannot expect that some solutions of GR will not be unique because something is not Lipschitz. Even in the ODE case of your example, with a Lipschitz function, where you have existence and uniqueness of solutions, you can have some solutions that are not global i.e. they tend to infinity in finite time.

The situation in GR is that there are theorems that guarantee that there is a unique solution locally. There is also a theorem that says that there is a unique maximal solution, but only in a certain sense. It is maximal only within a class of space-times and may be extendable in a non-unique way. Here the strong cosmic censorship says that shouldn't be the case (at least not generically).