Suppose $\mathcal M\subseteq\mathbb R^n$ is a submanifold of $\mathbb R^n$. Take $d(\cdot,\cdot):\mathcal M\times\mathcal M\to\mathbb R_+$ to be the pairwise geodesic distance function, so $d(x,y)$ is the length of the shortest path from $x\in\mathcal M$ to $y\in\mathcal M$.

Holding one coordinate fixed and varying the other, wherever $d(\cdot,\cdot)$ is differentiable it satisfies the eikonal equation $\|\nabla f(x)\|_2=1$. In particular, we can write two conditions satisfied by $d(\cdot,\cdot)$ almost everywhere: $$\|\nabla_x d(x,y)\|_2=1\qquad\textrm{and}\qquad\|\nabla_y d(x,y)\|_2=1.$$

In a sense, these two conditions are redundant. If $d(\cdot,\cdot)$ satisfies the first condition, it "looks" like a distance function from $y$ to all the other points $x$ and the other condition follows by symmetry of $d$.

Is there a single, more symmetric PDE satisfied by $d(\cdot,\cdot)$ as a function on the product manifold $\mathcal M\times\mathcal M$?

That is, if the eikonal equation is the PDE behind the single-source-all-destinations geodesic distance problem, is there a different canonical PDE that governs the pairwise geodesic distance problem? I'm hoping to identify a condition that doesn't require enforcing the eikonal condition in the $x$ and $y$ coordinates individually.


1 Answer 1


[Cross post from Twitter:]

This is not rigorous, but it may help.

Consider the Euclidean case, where $d(x,y) = |x-y|$. Then

$$\nabla d = \frac{(x-y, y-x)}{|x-y|},$$

where $\nabla$ denotes the gradient on $\mathbb{R}^{2n}$. Hence,

$$|\nabla d|^2 = \frac{|x-y|^2 + |y-x|^2}{|x-y|^2} = 2.$$

So, a necessary condition is

$$ |\nabla d|^2 = 2, $$

subject to $d(x,x) = 0$.

In the Riemannian case, we likewise have

$$\nabla d = (\nabla_x d, \nabla_y d),$$

where $\nabla_x$ and $\nabla_y$ denote the gradient with respect to the first and second arguments, respectively. Hence,

$$|\nabla d|^2 =|\nabla_x d|^2+|\nabla_y d|^2 = 1+1 = 2.$$

So then, up to a factor 2, the distance function must satisfy the usual eikonal equation on the product space, subject to Dirichlet boundary conditions

$$d(x,x) = 0.$$

Since this equation has a unique (viscosity) solution, it should characterize the distance function.

  • 1
    $\begingroup$ Awesome! Exactly what I was looking for. You rock, as always! $\endgroup$ Apr 9, 2020 at 18:05

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