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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 case $T$ 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 $T=2\tanh ^{-1}\sqrt{1-|x|}<\infty$.

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 $T=2\tanh ^{-1}\sqrt{1-|x|}<\infty$.

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*}

Let us now show that the Lipschitz condition on the vector field $V$ implies that, 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 $T=\infty$.

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 case $T$ 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 $T=2\tanh ^{-1}\sqrt{1-|x|}<\infty$.

Let $y:=y_t:=\Phi_t(x)$, $x\in B\setminus\{0\}$, $\dot{y}:=d\Phi_t(x)/dt=V(y)$, $c:=C>0$. Then your condition $(V(x)\cdot x)\ge C\|V(x)\|\,\|x\|$ means that the radial speed $v_{rad}:=v_{rad}(t):=\dot{y}\cdot y/\|y\|$ is no less than $c$ times the full speed $v:=\|\dot{y}\|$. In particular, it follows that $v_{rad}>0$ while $r:=r_t:=\|y_t\|<1$, so that $r_t$ is increasing.

Suppose that $h:=r_{\infty-}<1$. Then for some real $t_0>0$ and all real $t\ge t_0$ we have $0<h/2\le r_t\le h<1$. Since $V$ is continuous and nonzero on the closed annulus between the spheres of radii $h/2$ and $h$ centered at the origin, it follows that $\|\dot{y_t}\|=\|V(y_t)\|\ge u$ and hence $v_{rad}(t)\ge ca>0$ for some real $u>0$ and all real $t\ge t_0$. This implies that for large enough $t$ we will have $r_t>1$, which contradicts the assumption $r_{\infty-}<1$.

  So, either (i) $r_T=1$ for some real $T>0$ (which should be fine for us) or (ii) $r_{\infty-}=1$ but $r_t<1$ for all real $t\ge0$.

It remains to consider the case (ii). If $r_s=1-\delta$ for some real $s>0$ and $\delta\in(0,1)$, then the condition $v_{rad}\ge cv$ implies that, for all real $t\ge s$, the angle between the vectors $y_t$ and $y_s$ will be no greater than $\frac\delta{c(1-\delta)}$. Thus, the limit $y_{\infty-}$ will exist and be on the boundary of $B$.

Added: A detailed and improved (?) version of the latter paragraph (without using angles) is as follows: For real $t>0$ and $\Delta t\downarrow0$, we have $\Delta y_t:=y_{t+\Delta t}-y_t=V(y_t)\Delta t+o(\Delta t)$ and hence $\|\Delta y_t\|=\|V(y_t)\|\Delta t+o(\Delta t)\le\frac1c\,v_{rad}(t)\Delta t+o(\Delta t)$. Also, $\Delta r_t:=r_{t+\Delta t}-r_t=v_{rad}(t)\Delta t+o(\Delta t)$, so that $\|\Delta y_t\|\le\frac1c\,\Delta r_t+o(\Delta t)$. Integrating/telescoping this and using the triangle inequality for the norm $\|\cdot\|$, we see that for all $s$ and $t$ such that $0<s<t<\infty$ we have $\|y_t-y_s\|\le\frac1c\,(r_t-r_s)$. Since $r_{\infty-}=1$, we conclude that, by the Cauchy convergence criterion, the limit $y_{\infty-}$ exists and is on the boundary of $B$.

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 case (ii) must take place. (The condition that $V(0)=0$ is nowhere needed.) Indeed, assuming the contrary, we have case (i), by what has been proved. That is, $r_T=1$ for some real $T>0$. On the other hand, for some real Lipschitz constant $K>0$, all $t\in[0,T]$, $y=y_t$, and $y_*:=y/\|y\|$, $$\frac{dr}{dt}=v_{rad}\le v=\|V(y)\|\le\|V(y_*)\|+\|V(y)-V(y_*)\|\le 0+K\|y-y_*\|=K(1-r), $$ whence $-\frac{d}{dt}\,\ln(1-r)\le K$, so that $$\infty=\lim_{t\uparrow T}\ln\frac{1-r_0}{1-r_t}\le KT<\infty, $$ which is the sought contradiction.

If the Lipschitz condition fails to hold near the boundary of $B$, then of course case (i) may take place. 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 $T=2\tanh ^{-1}\sqrt{1-|x|}<\infty$.

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*}

Let us now show that the Lipschitz condition on the vector field $V$ implies that, 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 $T=\infty$.