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Consider a continuous 3D vector field $\vec{V}:\mathbb{R}^n\to\mathbb{R}^n$ except that $\vec{V}$ may have isolated singularities, and an open region $\Omega\subset\mathbb{R}^n$. Suppose $\vec{V}(\vec{r})$ is nonsingular for $\vec{r}\in\partial\Omega$, and let $\vec{n}(\vec{r})$ be the outward-pointing normal at a boundary point $\vec{r}\in\partial\Omega$. If the normal component of the vector field points inward at every point on the boundary,

$$\vec{V}(\vec{r})\cdot \vec{n}(\vec{r}) < 0\ \forall\ \vec{r}\in\partial\Omega.$$

Does this imply that $\vec{V}(\vec{r})$ is either zero or singular somewhere in $\Omega$? How would it be shown? I imagine this must be a reasonably well known result but I wasn't sure how to look for it.

I did find this question, which states for $\vec{V}$ defined on the unit $n$-ball $B^n$, if $\vec{V}(\vec{r})\neq 0\ \forall\ \vec{r}\in B^n$, there must be a point $\vec{r}\in\partial B^n$ at which $\vec{V}(\vec{r})$ points outward and one at which it points inward. Logically, then, if those points don't exist on $\partial B^n$ (such as if $\vec{V}$ points inward everywhere), then there must be some point in $B^n$ at which $\vec{V}(\vec{r}) = 0$, or $\vec{V}$ has a singularity in $B^n$. Intuitively I would think this should continue to hold for any region topologically equivalent to a ball, since a homotopy between $B^n$ and $\Omega$ wouldn't change whether a given zero or singularity of $V$ is inside or outside the region or on the boundary. (Is this right so far?) But what about spaces with holes? Does anything change about the argument when $\Omega$ is not homotopic to a ball?

For the curious: I was inspired to think about this by considering whether any continuous mass distribution necessarily has a point at which the gravitational forces from all different parts of the distribution cancel out. That question is easily answered in the affirmative, because the gravitational field is a gradient of a scalar function which must have some local minimum, but then I started wondering if the condition that the vector field is a gradient was really necessary.

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+1 for the physics inspiration/application – Sammy Black May 27 '13 at 2:09
I suspect that the Poincaré-Hopf index theorem is relevant. – Sammy Black May 27 '13 at 2:15
up vote 2 down vote accepted

Your instincts are right about a region equivalent to a ball. But the topology of the region definitely matters. Imagine a closed annulus (say $\{x\in \mathbb R^2: 1\le |x|\le 2\}$), with the flow spiralling in from the outside and out from the inside and a limit cycle at $|x|=3/2 cycling around. In general, see Milnor's Topology from the Differentiable Viewpoint or for the Poincaré-Hopf Theorem, even on manifolds with boundary. They typically specify that the vector field be normal and outward-pointing on the boundary (so you can glue two copies of the manifold with boundary, together with the vector field, smoothly along the boundary). Inward-pointing is just as good, and your case can be smoothly deformed to that.

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Thanks, I'll have to do some reading about this to understand it. – David Z May 29 '13 at 5:04

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