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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 ;-)

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  • $\begingroup$ 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) $\endgroup$ – Aleksei Kulikov Aug 24 at 11:39
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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 $T=2\tanh ^{-1}\sqrt{1-|x|}<\infty$.

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  • $\begingroup$ The solution is now fully formalized, streamlined, and simplified. $\endgroup$ – Iosif Pinelis Aug 26 at 3:38
  • $\begingroup$ 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 :-) $\endgroup$ – Pascal Lambrechts Aug 26 at 6:21
  • $\begingroup$ @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. :-) $\endgroup$ – Iosif Pinelis Aug 26 at 13:17

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