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Is there a relation between a solution of Eikonal equation and curvature?

Imagine someone hands you a local solution $\phi$ of Eikonal equation, $\left\| \nabla \phi \right\|_g = 1$, on Riemannian manifold $(M,g)$. Can you say me something about the curvature?

A more precise formulation:

Let $(M,g)$ be a two dimensional Riemannian manifold. Let $p\in M$ be a point and $v\in T_pM$ a direction at $p$. Now assume that we have a neighborhood $U$ of the point $p$, a geodesic $\gamma\subset U$ going through the point $p$ in the direction $v$ and a smooth solution $\phi$ to Eikonal equation, $\left\| \nabla \phi \right\|_g = 1$, on $U$ satisfying $\phi(x) = 0$ for $x \in \gamma$. Can we say something about the Riemann curvature tensor?


Few observations:

Denote $u = \nabla \phi$. Then we know that $u \cdot v = 0$. Also by taking derivative of Eikonal equation we get $$ \nabla^2 \phi \cdot \nabla \phi = 0. $$ The second derivative of $\phi$ has the following general form $$\nabla^2 \phi = a u\otimes u + b v \otimes v + \frac{c}{2}(u\otimes v + v \otimes u)$$ where $a,b,c$ are some scalars. Combining $\nabla^2 \phi \cdot \nabla \phi = 0$ and $u \cdot v = 0$ we deduce that $a=c=0$.

What can be say about $b$? Isn't it by any chance the Gaussian curvature?

Flat case: When the manifold is just $\mathbb{R}^2$ then the geodesics are just straight lines and the solution $\phi$ is just a affine function which has zero second derivatives.


Additional question: Can you generalize the question for manifolds with arbitrary dimension? The problem is I do not know how to set up the boundary condition of the Eikonal equation.

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If $c$ is a geodesic of finite length, define $$X:=\{ \exp_{c(t)}\ \epsilon x(t)||t|\leq \epsilon_2\}$$ where $x$ is a unit parallel vector field along $c$ and $g(c'(t),x(t))=0$. Then define $$ f(p)={\rm dist}\ (X,p)-\epsilon $$

Then $f|c=0$ and $|\nabla f|=1$ And since geodesic $c$ in smooth manifold is smooth, so $X,\ f$ are smooth. Hence we do not know curvature.

(reference : https://mathoverflow.net/questions/283467/tubular-neighborhoods-of-embedded-manifolds )

[Add] high dimension : If $U_\epsilon (c)$ is $\epsilon$-tubular neighborhood, then define smooth curve $\alpha$ in the boundary s.t. $\exp_{c(t)}\ \epsilon v(t)=\alpha(t)$ where $|v(t)|=\epsilon$.

So define $f(p)={\rm dist}\ (p,\alpha)-\epsilon$ so that it is a solution.

[Add2 - dimension 2 ] Consider a variation $F(t,s)=\exp_{c(t)}\ s \nabla f(c(t))$ Since $F_s(t,0)$ is a parallel along unit speed geodesic $c$, then \begin{align*} R(c',\nabla f,\nabla f,c')&= (\nabla_t\nabla_s F_s-\nabla_s\nabla_t F_s,c') \\&= -(\nabla_s\nabla_t F_s,c') =-(\nabla_t F_s,F_t)_s+(\nabla_t F_s,\nabla_s F_t) \\&=-(\nabla_t F_s,F_t)_s\end{align*}

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  • $\begingroup$ I see. You are right. $\endgroup$ – HK Lee Jun 15 '18 at 14:48
  • $\begingroup$ @Rahul Dont you get $f(x,y)=|y-\epsilon | -\epsilon$? And you do not have problem with differentiability on $x$ axis. But I still do not understand the conclusion "Hence we do not know curvature." $\endgroup$ – tom Jun 15 '18 at 19:26
  • $\begingroup$ Also you can obtain the solution in much more straight forward way. $f(p) = dist(c,p)$ on side of the geodesic and $f(p) = - dist(c,p)$ on the other side. Note that for this it is crucial that we are in 2d and we can talk about a side from a curve. $\endgroup$ – tom Jun 15 '18 at 19:30
  • $\begingroup$ I see that your example is straight forward. And when we construct $f$, the curvature is not related. $\endgroup$ – HK Lee Jun 15 '18 at 23:50
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    $\begingroup$ Ohh sorry, my question is not too clear in that regard. I'm suspecting that there might be a relation between second derivatives of $f$ and curvature. So is there one? $\endgroup$ – tom Jun 18 '18 at 23:16

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