In my work in algebraic topology I need to build a special homotopy and I came up with a construction based on some ordinary differential equation in which I am not an expert. I miss some argument to prove the continuity of the flow.

In details, $V$ is a lipschitzian vector field defined on the closed unit n-ball $B\subset \mathbb{R}^n$ and which vanishes at the origin and on the boundary of the ball but nowhere else. This induces a continuous flow $$\Phi:[0,+\infty)\times B\rightarrow B,(t,x)\mapsto \Phi_t(x) $$ characterized by $d\Phi_t(x)/dt=V(\Phi_t(x))$. I need to extend it continuously at time $t=\infty$ and I wonder whether the following criterion is correct. Assume that there exists $C>0$ such for any $x\in B$ we have $$(V(x)\cdot x)\geq C\|V(x)\|\,\|x\|$$ where on the left hand side I mean by $\cdot$ the inner product of $\mathbb{R}^n$. Is it true that then $\Phi$ would admit a continuous extension on $$ [0,+\infty]\times(B\setminus\{0\})\quad ? $$ I am not an expert in differential equations but my intuition is that the above condition requires the vector field to belong to some cone oriented in the radial direction with a uniform angle at the summit. This should imply that all trajectories, except the constant trajectory at the origin, converge to the boundary. Moreover the trajectories should belong inside a uniform "curved" cone and when we look at a point close enoughto the boundary this will ensure continuity of the asymptotic extension.

I will be happy to see a proof or have a reference if this result is classic. A counter-example is good too but will make me less happy ;-)