2
$\begingroup$

Let $(M,g)$ be an n dimensional smooth Riemannian manifold, possibly noncompact. Suppose there exists a harmonic function $f$ defined on $M$, then we know that $f$ is smooth. Let $S$ be the set of critical points of $f$, i.e. $$ S=\{x\in M: |\nabla f (x)|=0\}. $$ For $x\in S$, we know that the gradient flow starting from $x$ remains unchanged.

For a point $x$ such that $|\nabla f(x)|\neq 0$, is it possible that the gradient flow reaches a critical point? If it's no, for a function $f$ defined on $M$ satisfying $\Delta f=c>0$, $c$ is a constant, do we have the same property?

Improve this question
$\endgroup$
2
  • $\begingroup$ To avoid trivialities, I would definitely assume $M$ is noncompact: harmonic functions on compact manifolds are constant. If $M$ is noncompact, how about a saddle-shaped harmonic function on a disk, e.g., integrate the Poisson kernel against something like $\theta\mapsto\sin(\theta)\cos(\theta)$, and then trace the gradient flow from the boundary to the critical point? $\endgroup$
    – Neal
    Commented Sep 21, 2016 at 13:52
  • 2
    $\begingroup$ It always takes an infinite time to reach a singularity of a smooth vector field by following its flow. It may be possible to reach the singularity in finite time if the vector field admits a pole. $\endgroup$
    – Loïc Teyssier
    Commented Sep 21, 2016 at 14:17

1 Answer 1

Reset to default
6
$\begingroup$

Take $f(x,y)=x^2-y^2$ on the plane. Starting from $(0,y_0)$ leads to $(0,0)$ (in infinite time). More generally, critical points of harmonic functions on Riemannian surfaces have a stable manifold from where trajectories of the gradient flow end at the critical point.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .