# A criterion on a vector field for its flow to extend continuously at $t=\infty$

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 ;-)

• Presumably you want the derivative to be $V(\Phi_t(x))$, not $V(x)$? If so, then it seems to me that you yourself proved what you need in the penultimate paragraph:) (just a little comment: you could add that when we are at least $\epsilon$ away from the boundary you move to it with the speed at least $\delta>0$ so all nonzero trajectories actually tends to the boundary) Aug 24, 2019 at 11:39

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 $$0 we have $$\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 $$\begin{equation*} 0 Since $$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 $$0 $$\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 $$T=2\tanh ^{-1}\sqrt{1-|x|}<\infty$$.

• The solution is now fully formalized, streamlined, and simplified. Aug 26, 2019 at 3:38
• thank you for the answer ! Actually I also need the continuity of the extension at $T=\infty$ to be continuous w.r.t to $x$. But I see that this follows from the inequality in the displaymath after (4), which implies that $$\|y_\infty-y_t\|\leq \frac{1-r_t}{C},$$ and from the continuity for $T<\infty$ which follows from the lipschitz hypothesis on V. So this is exactly what I needed and makes me happy :-) Aug 26, 2019 at 6:21
• @PascalLambrechts : Since you mentioned a homotopy in your post, I did suspect that actually you also needed the continuity of the extension in the initial position $x$. But, as you note, that also quickly follows from the answer. Anyhow, I am glad the answer made you happy. :-) Aug 26, 2019 at 13:17