# Vanishing of a Vector Field

Suppose I have some smooth vector field $F$ defined on $\mathbb{R}^n$. Further, suppose that there is some radius $R>0$ such that for all $||x||=R$, $$\langle x , F(x) \rangle < 0$$

Ideally, I want to claim that $F$ vanishes somewhere inside the ball $||x||<R$. The motivation behind this question is to understand the dynamics of the flow under $F$, which never leaves the ball.

1. Is this true in general?
2. If not, Is this true if I assume that $\nabla \cdot F < 0$ everywhere?
3. If not, Is this true if I assume that $F(x) = -\nabla\psi(x)+Jx$ where $\psi$ is a convex function and $J$ is a skew-symmetric matrix?

I believe that the answer to 2 is affirmative, but have failed to prove the claim.

Suppose $F$ does not vanish inside $B(0,R)$ define $H:B(0,R)\rightarrow B(0,R)$ by $H(x)=R{{F(x)}\over {\|F(x)\|}}$, the Brouwer fixed point theorem implies that there exists $x$ such that $H(x)=x, x\in S^n(R)=\{x:\|x\|=R\}$ since the image of $H\subset S^n(R)$, this implies that $F(x)={{\|F(x)\|}\over R}x$ we deduce that $\langle x,F(x)\rangle=\langle x,{{\|F(x)\|\over R}}x\rangle>0$. Contradiction.
• There's a minor computation error - $H(x)=x$ should imply $RF(x)=||F(x)||x$. The proof after that basically goes word by word. Aug 25 '16 at 18:55