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 ;-)



A criterion on a vector field for its flow to extend continuously at $t=\infty$
Fix any $x\in B\setminus\{0\}$. 
Let $y:=y\_t:=\Phi\_t(x)$, $r:=r\_t:=\|y\_t\|$, $\dot{y}:=d\Phi\_t(x)/dt=V(y)$ (the velocity), $v:=\|\dot{y}\|=\|V(y)\|$ (the speed), $c:=C>0$. Then for all $t>0$ such that $0<r\_t<1$ we have 
\begin{equation\*}
 \dot r=\frac{d\|y\|}{dt}=\frac{y\cdot\dot y}{\|y\|}=\frac{y\cdot V(y)}{\|y\|}\le\|V(y)\|=v; \tag{1} 
\end{equation\*}
on the other hand, your condition $(V(x)\cdot x)\ge C\|V(x)\|\,\|x\|$ implies that 
\begin{equation\*}
 \dot r=\frac{y\cdot V(y)}{\|y\|}\ge c\|V(y)\|=cv>0, \tag{2} 
\end{equation\*}
so that $r\_t$ is increasing in $t\in[0,T)$, where 
\begin{equation\*}
 T:=\inf\{t>0\colon r\_t=1\};
\end{equation\*}
recall that $\inf\emptyset$ is defined as $\infty$. 
Thus, the limit $h:=r\_{T-}$ exists. 


Let us show that $h=1$. Indeed, suppose the contrary. Then $T=\infty$ and there is some real $t\_0>0$ such that 
\begin{equation\*}
0<h/2\le r\_t\le h<1 \text{ for all real }t\ge t\_0. \tag{3}
\end{equation\*}
Since $V$ is continuous and nonzero on 
the closed "annulus" $A:=\{z\in\mathbb R^n\colon h/2\le\|z\|\le h\}$, 
it follows that $\|V(z)\|\ge u$ for some real $u>0$ and all $z\in A$. So, for all real $t\ge t\_0$ we have $\|V(y\_t)\|\ge u$ and hence, 
by (2), 
$\dot r\ge cu>0$. This implies that for some real $t\ge t\_0$ we will have $r\_t=1$, which contradicts (3). 
Thus, 
\begin{equation\*}
 r\_{T-}=1. \tag{4}
\end{equation\*}


It also follows from (2) that for all $s$ and $t$ such that $0<s<t<T$
\begin{equation}
 r\_t-r\_s=\int\_s^t \dot r(\tau)\,d\tau\ge c\int\_s^t v(\tau)\,d\tau
 =c\int\_s^t \|\dot y(\tau)\|\,d\tau
 \ge c\|y\_t-y\_s\|. 
\end{equation}
In view of (4), we conclude that, by the Cauchy convergence criterion, the limit $y\_{T-}$ exists and is on the boundary of $B$, as desired. 




---


So far, in addition to the condition $(V(x)\cdot x)\ge C\|V(x)\|\,\|x\|$, we have only used the condition that $V$ is nonzero and continuous away from the origin and the boundary of $B$. If we also use the Lipschitz condition, we can say a bit more: that then, in fact, $T=\infty$. 
Indeed, for some real Lipschitz constant $K>0$, all $t\in[0,T)$, $y=y\_t$, and $y\_\*:=y/\|y\|$, by (1), 
$$\frac{dr}{dt}=\dot r\le \|V(y)\|\le\|V(y\_\*)\|+\|V(y)-V(y\_\*)\|\le 0+K\|y-y\_\*\|=K(1-r),
$$ 
whence $-\frac{d}{dt}\,\ln(1-r)\le K$. So, in view of (4), 
$$\infty=\lim\_{t\uparrow T}(\ln(1-r\_0)-\ln(1-r\_t))\le KT, 
$$
which does imply that $T=\infty$. 




---


If the Lipschitz condition fails to hold near the boundary of $B$, then of course $T$ may be finite. E.g., if $d=1$ and $V(y)=y\sqrt{1-|y|}$, then with $y\_0=x\in(-1,1)\setminus\{0\}$ we have
$$y\_t=\Big[1-\tanh ^2\left(\tfrac{t}{2}-\tanh ^{-1}\sqrt{1-|x|}\right)\Big]\,\text{sign} \,x
$$ 
and $y\_T=\text{sign}\,x$ for $T=2\tanh ^{-1}\sqrt{1-|x|}<\infty$.


