Fixed point involved ODE Vector field $v: R^n \to R^n$ is smooth, and $x\cdot v(x)\geq 0$ when $|x|=1$. Then consider the ODE:
$$\dot{x}(t)=-v(x(t)) \ \ t\geq 0 $$
$$x(0)=y$$
For $t>0$ fixed, the map $y\mapsto x(t,y)$ is continuous, can we claim this map has a fixed point? Any suggestions are welcome!

I can tell you the solution when $|y|\leq 1$. I think general case can be done by renormalisation. I have to find it out.
since $\frac{d}{dt}u=-f(u)$. Hence $u\frac{d}{dt}u=-uf(u)$. when $|u|=1$,
$u\frac{d}{dt}u=-uf(u) \leq 0 \implies \frac{d}{dt}|u|^2 \leq 0$.
So, if we start from $|u(0)|=|y|\leq 1$. The solution $u(t,y)$ cannot escape from the unit ball in $\mathbb R^n$.(Moment it reaches boundary, its norm begins to decrease!)
As the solution depends continuously on initial data, the map $y\rightarrow u(t,y)$ is a continuous function from $B(0,1)$ to itself. 
Use Brower Fixed point theorem to conclude that the map has a fixed point.
I shall try to solve for general case, but it looks more assumption on f is needed.
