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.


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*}

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.