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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.

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    $\begingroup$ Your formula for $\ell(\gamma)$ is meaningless. You should read any introductory book on differential geometry to find how length of a path is defined. Voting to close. $\endgroup$ – Misha May 25 '13 at 17:02
  • $\begingroup$ I don't see why it is meaningless: it is a standard conformal transformation of the standard Euclidean distance. The "introductory text books" you mention always start off with smoothness assumptions - but I am interested in whether smooothness properties and curvature bounds for the shortest paths follow from the length structure directly. $\endgroup$ – ulelux May 26 '13 at 9:35
  • $\begingroup$ In view of your comment, I think, your dx really means ds. If this is the case then you should edit your formula accordingly. $\endgroup$ – Misha May 26 '13 at 14:18
  • $\begingroup$ Now I see you point. Yes, this is a typo, I meant ds. Sorry. Here is an example: $X = [0,1]^2$, $p: X \to R$ piecewise constant, $p(x,y) = 1/2$ if $x < 1/2$ and $p(x) = 2$ if $x \geq 2$. Choose x=(0,0) and y=(1,1). It is easy to see that the shortest path between x and y (in the sense I defined it) is not smooth, it has a bend when it crosses x=1/2. So this is an example that if p is not smooth (not even continous), then the shortest path is not smooth. I suspect that if p is smooth, then so should be the shortest path, and there should be an elementary proof for this ... $\endgroup$ – ulelux May 27 '13 at 19:27
  • $\begingroup$ The question that was posed has an intuitive interpretation, "Suppose that $p(x)$ represents the density of the material at any point $x$. If changes in density are 'nice' throughout $X$, is it true that shortest paths are smooth and of bounded curvature?" The intuition is "yes" if $p(x)$ is smooth. $\endgroup$ – Patrick Hew May 30 '18 at 4:10
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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$$ \ddot \alpha=\frac{\nabla p}{p}+\lambda \dot\alpha\\ \Rightarrow |\ddot\alpha|^2+\lambda^2=\left|\frac{\nabla p}{p}\right|^2 \\ \Rightarrow |\ddot\alpha|\leq \left|\frac{\nabla p}{p}\right|. $$

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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.

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