Smoothness and curvature of geodesics in a length space

Question

Let $X$ be a nice compact subset of $R^d$. Given a function $p: X \to \mathbb{R}^+$, define the length of a path $\gamma \subset X$ as $\ell(\gamma) = \int_\gamma p \, ds$, and the distance between two points $x,y \in X$ as the length of the shortest path connecting them. I'd like to prove that if $p$ is "nice" (say, continuously differentiable), then the shortest paths are smooth and have a bounded curvature. Is there an elementary way of proving this, without diving too deep into Riemannian geometry? I'd in particular like to understand what are the minimal assumptions on $p$ I need.

Answer 1

You can use variational methods. If you consider a parametrization $\alpha:[0,c] \rightarrow X$ of $\gamma$ the integral becomes $l(\gamma)=l(\alpha)=\int_0^c p(\alpha(t)) |\dot\alpha|dt$. Consider a variation $d\alpha\colon [0,c] \rightarrow \mathbb{R}^d$ with $d\alpha(0)=d\alpha(c)=0$ and $\langle d\alpha(t),\dot\alpha(t)\rangle=0$ for all $t\in [0,c]$. We get $$ l(\alpha+\epsilon d\alpha)\\ =l(\alpha)+\epsilon\int_0^c \left(\langle \nabla p(\alpha),d\alpha\rangle |\dot\alpha|+p(\alpha)\left\langle \frac{\dot\alpha}{|\dot\alpha|},\dot {d\alpha}\right\rangle\right)+O(\epsilon^2)\\ =l(\alpha)+\epsilon\int_0^c \left\langle \nabla p(\alpha)|\dot\alpha|-p(\alpha)\left(\frac{\ddot\alpha }{|\dot\alpha|}-\frac{\langle \ddot\alpha,\dot\alpha\rangle\dot\alpha}{|\dot\alpha|^3} \right) ,d\alpha\right\rangle +O(\epsilon^2). $$ As this holds for all curves $d\alpha$ we must have if $\alpha$ is a minimizer that $$ \frac{\ddot\alpha }{|\dot\alpha|^2}-\frac{\langle \ddot\alpha,\dot\alpha\rangle\dot\alpha}{|\dot\alpha|^4}=\frac{\nabla p}{p}+\lambda \dot \alpha $$ If you force $\alpha$ to be of unit arc length ($|\dot\alpha|=1$) we have $\langle \dot\alpha,\ddot\alpha\rangle=0$ and hence$$ \ddot \alpha=\frac{\nabla p}{p}+\lambda \dot\alpha\\ \Rightarrow |\ddot\alpha|^2+\lambda^2=\left|\frac{\nabla p}{p}\right|^2 \\ \Rightarrow |\ddot\alpha|\leq \left|\frac{\nabla p}{p}\right|. $$

Answer 2

Idea for an elementary proof. I'm only presenting the idea for an elementary proof because I don't have the maths to make it formal.

The idea comes from two foundations:

1. $p(x)$ represents the density of the material at point $x$. The conjecture is that if changes in density are 'nice' throughout $X$ then paths are smooth and of bounded curvature.

2. Changes in density can be treated as a changes in height if one chooses the right perspective. Imagine two sheets of paper that are treated as follows:

   • Fold the first sheet of paper, unfold it, lay it totally flat on the table, and then cover the right half in glue.

   • Fold the second sheet of paper just as the first one was folded, partially unfold it, then put on the table so that the left half lays flat on the table but the right half climbs into the air.

   Now imagine an ant crawling along the sheets of paper from left to right. If the gradient of the right-half of the second sheet of paper is chosen just right, then from a viewpoint directly above the table, the time taken by the ant to cross the sheets will be exactly the same: the impedance caused by the glue will have been replaced by the gradient from gravity.

So in the same way, I think that density function $p$ on $X \subset R^d$ can be made equivalent to constructing a surface $X' \subset R^{d'}$ where $d' \geq d$. That is, there exists a function $f : R^d \rightarrow R^{d'}$ so that for any path $\gamma$ in $X$, if $\gamma' = f \circ \gamma$ then the length of $\gamma$ in $X$ (modulated by material density $p$) is equal to the length of $\gamma'$ in $R^{d'}$ (measured across the surface $X').

Hence the conjecture is about the minimal requirements on $f$ for paths across the surface of $X'$ to be smooth. If $X$ is 'nice' then I suspect that $f$ only needs to be smooth.