If $(T_t)_{t\ge0}$ is a flow on a submanifold $\Omega$ with boundary with velocity $v$ satisfying $\langle v,\nu_{∂Ω}\rangle=0$, then $T_t(∂Ω)⊆∂Ω$

Question

Let

• $\tau>0$;
• $d\in\mathbb N$;
• $v:[0,\tau]\times\mathbb R^d\to\mathbb R^d$ be Lipschitz continuous in the second argument uniformly with respect to the first with $v(\;\cdot\;,x)\in C^0([0,\tau],\mathbb R^d)$;
• $X^x$ denote the unique element of $C^0([0,\tau],\mathbb R^d)$ with $$X^x(t)=x+\int_0^tv(s,X^x(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\tau]\tag1$$ for $x\in\mathbb R^d$ and $$T_t(x):=X^x(t)\;\;\;\text{for }x\in\mathbb R^d$$ for $t\in[0,\tau]$.

Let $\Omega$ be a $d$-dimensional embedded $C^1$-submanifold with boundary and $\nu_{\partial\Omega}$ denote the unique outer unit normal field on $\partial\Omega$. How can we show that, if $$\langle\left.v\right|_{[0,\:\tau]\times\partial\Omega},\nu_{\partial\Omega}\rangle=0\tag2,$$ then $$T_t(x)\in\partial\Omega\tag3$$ for all $x\in\partial\Omega$?

If necessary, assume that $v$ does not depend on the first argument.

I don't know if it'S relevant, but $\partial\Omega$ is a $(d-1)$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ (without boundary) and hence, for all $x\in\partial\Omega$, $$\mathbb R^d=T_x\:\partial\Omega\oplus N_x\:\partial\Omega\tag4$$ and $$N_x\:\partial\Omega=\mathbb R\nu_{\partial\Omega}(x)\tag5.$$ Moreover, it's trivially to see that $$v(0,x)\in T_x\:\partial\Omega\;\;\;\text{for all }x\in\partial\Omega\tag6.$$

I have no idea how we can show the claim, but I was able to prove the following, possibly related, results:

Proposition 1: If $x\in\mathbb R^d$ with $$v(t,x)=0\;\;\;\text{for all }t\in[0,\tau]\tag7,$$ then $$T_t(x)=x\;\;\;\text{for all }t\in[0,\tau]\tag8.$$

Proposition 2: If $B\subseteq\mathbb R^d$ with $\left.v\right|_{[0,\:\tau]\times B}=0$, then $$\left.T_t\right|_B=\operatorname{id}_B\;\;\;\text{for all }t\in[0,\tau]\tag9.$$

Proposition 3: If $B\subseteq\mathbb R^d$ is open or closed and $\left.v\right|_{[0,\:\tau]\times\partial B}=0$, then $$T_t(B)=B\;\;\;\text{for all }t\in[0,\tau]\tag{10}.$$

EDIT:

$T_t$ is a $C^1$-diffeomorphism on $\mathbb R^d$ and hence $$\partial T_t(\Omega)=T_t(\partial\Omega)\tag{11}$$ and $$\nu_{\partial T_t(\Omega)}(T_t(x))=\frac{\left({\rm D}T_t(x)^{-1}\right)^\ast\nu_{\partial\Omega}(x)}{\left\|\left({\rm D}T_t(x)^{-1}\right)^\ast\nu_{\partial\Omega}(x)\right\|}\;\;\;\text{for all }x\in\partial\Omega\tag{12}$$ for all $t\in[0,\tau]$. Does this help?

EDIT 2:

$\Omega$ being a $d$-dimensional embedded $C^1$-submanifold with boundary means that each point of $\Omega$ is locally $C^1$-diffeomorphic to $\mathbb H^d:=\mathbb R^{d-1}\times[0,\infty)$ which in turn means that for each $p\in\Omega$, there is an open neighborhood $U$ (open in $\Omega$) of $x$ and a $C^1$-diffeomorphism from $U$ onto an open subset of $\mathbb H^d$. A $C^1$-diffeomorphism between non-open subsets is a homeomorphism such that the function itself and its inverse are locally extendable to $C^1$-differentiable functions.

This particularly implies that $\partial\Omega$ is a $(d-1)$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$.

Answer 1

Let $x\in \partial \Omega$, we want to show that $T_t(x) \in \partial \Omega$.

Since $\partial\Omega$ is a $C^1$ embedded submanifold in $\mathbb R^n$, there is a $C^1$-diffeomorphism $\varphi : U \to \varphi(U) \subset \mathbb R^n$, where $x\in U$, so that \begin{align} \tag{i}\varphi(x) &= (0,0,\cdots, 0),\\ \varphi (U\cap \partial \Omega) &= \varphi (U) \cap \{ y\in \mathbb R^n: y_n = 0\}. \end{align}

Let $w(t, y)$ be the time dependent vector field on $\varphi (U)$ given by $$\tag{ii} w(t, y) = (\phi_*)_q v(t, q),\ \ \text{ where } \varphi (q) = y.$$

Note that since $\varphi$ is $C^1$ and $v$ is continuous, $w$ is a time dependent $C^0$-vector field on $\varphi (U)$.

Since $\partial \Omega\cap U$ is sent to the plane $\{ y_n = 0\}$, we have $$\tag{iii} y = (y_1, y_2, \cdots, y_{n-1}, 0)$$ whenever $\varphi (p) = y$ for some $p\in \partial \Omega \cap U$. Since $v$ is tangential to $\partial \Omega$ by (2), we have $$\tag{iv} w(t, y_1, y_2, \cdots, y_{n-1}, 0)_n = 0,$$ where $$w(t, y) = (w(t, y)_1, \cdots w(t, y)_n).$$ (To see this, for each point $y = (y_1, \cdots, y_{n-1}, 0)\in \varphi (U)$ and vector $V = (V_1, \cdots, V_{n-1}, 0)$, consider the curve $e(t) = y+tV$. We assume that $|t|$ is small so that $e(t) \in \varphi (U)$. By definition, $e(t) \in \varphi (U) \cap \{ y_n = 0\}$. By (i), $\varphi^{-1} \circ e$ is a curve on $\partial \Omega$. In particular, $$ \varphi^{-1}_* v = \frac{d}{dt} \bigg|_{t=0} \varphi^{-1} (e(t)) \in T_{\varphi^{-1}(y)} \partial \Omega.$$ Since $V$ is arbitrary, $$ (\varphi^{-1}_y)_* \{ V \in \mathbb R^n : V_n = 0\}\subset T_{\varphi^{-1}(y)}(\partial \Omega).$$ Since $\varphi^{-1}$ is a $C^1$-diffeomorphism, $\varphi^{-1}_*$ is injective. Since $\{V \in \mathbb R^n : V_n = 0\}$ is $n-1$-dimensional, we conclude that $$ (\varphi^{-1}_y)_* \{ V \in \mathbb R^n : V_n = 0\}= T_{\varphi^{-1}(y)}(\partial \Omega), $$ or $$\{ V \in \mathbb R^n : V_n = 0\}= \varphi_* T_{x}(\partial \Omega).$$ Now by (2) we have $v(z)\in T_z (\partial \Omega)$ for all $z \in \partial \Omega$. In particular, $$ w = \varphi_* v(t, \cdot) \in \{V : V_n=0\},$$ which is (iv)).

Now consider the following system of first order differential equations \begin{align} y_1' &= w_1(t, y_1(t), y_2(t), \cdots, y_{n-1}(t), 0)_1,\\ \vdots \ \ &= \qquad \qquad\vdots \\ y_{n-1}' &= w_1(t, y_1(t), y_2(t), \cdots, y_{n-1}(t), 0)_{n-1}. \end{align} with the initial conditions $$ y_1(0) = 0, \cdots, y_{n-1}(0) = 0.$$ Since $w$ is continuous, the existence theorem of ODE (see section I.1 in Hale's ODEs, which is attributed to Peano) imples that the IVP solvable for some $t\in (-\epsilon, \epsilon)$. Let $$\tilde y(t) = (\tilde y_1(t), \tilde y_2(t), \cdots, \tilde y_{n-1}(t))$$ be a solution. Let $$\tag{v}\gamma(t) = (\tilde y(t), 0).$$ Then $\gamma $ is a curve in $\varphi (U)$ and \begin{align} \gamma(0) &=(\tilde y_1(0), \tilde y_2(0), \cdots, \tilde y_{n-1}(0), 0) \\ &= (0,0,\cdots, 0, 0) = \varphi(x). \end{align} Also, \begin{align} \gamma'(t) &= (\tilde y'_1(t), \tilde y'_2(t), \cdots, \tilde y'_{n-1}(t), 0)\\ &= \big( w(t, \tilde y(t), 0)_1 , w(t,\tilde y(t),0)_2, \cdots, w(t, \tilde y(t), 0)_{n-1}, 0\big) \\ &=\big( w(t, \gamma(t))_1 , w(t,\gamma(t) )_2, \cdots, w(t, \gamma(t))_{n-1}, w(t, \gamma(t))_n\big), \end{align} where we used the definition of $\gamma(t)$ (v) and (iv). That is, $\gamma(t)$ solves the IVP \begin{align} \gamma' &= w(t, \gamma), \\ \gamma(0) &= \varphi (x). \end{align} Let $$\tag{vi} \eta(t) = \varphi^{-1} \circ \gamma(t).$$ Then $ \eta (0) = \varphi^{-1} \varphi (x) = x$ and \begin{align} \eta'(t) &= (\varphi^{-1}_*)_{\gamma(t)} \gamma'(t) \\ &= (\varphi^{-1}_*)_{\varphi (\eta(t))} w(t, \gamma(t)) \\ &= (\varphi^{-1}_*)_{\varphi (\eta(t))} (\varphi_*)_{\eta(t)} v(t, \eta(t)) \\ &= v(t, \eta(t)) \end{align} by (vi) and (ii). Thus $\eta(t)$ is a solution to the IVP \begin{align} \eta' &= v(t, \eta), \\ \eta(0) &= x. \end{align} Since $v$ satisfies the assumption for the uniqueness theorem, we find that $\eta(t)$ is the solution: Thus $\eta(t) = T_t(x)$. Since $\gamma (t)_n = 0$ by (v), we have $\eta (t) \in \partial \Omega$ by (vi). Thus $$ T_t(x) \in \partial \Omega.$$ Since $x\in \partial \Omega$ is arbitrary, we conclude $$ T_t(\partial \Omega) \subset \partial \Omega.$$

Edit: Up to now we have just proved that for all $x\in \partial \Omega$, there is $\epsilon>0$ depending on $x, \partial \Omega$ so that $T_t(x) \in \partial \Omega$ for all $|t| <\epsilon$. This is not exactly the same as $$ T_t(\partial \Omega) \subset \partial \Omega, \ \ \text{ for all } t \in [0, \tau]. $$

To show this we merely need to use the closedness of $\partial \Omega$ as a subset in $\mathbb R^n$: indeed, for each $x$, let

$$ \chi_x = \sup A, $$ where $$A = \{ \nu>0: T_t(x) \in \partial \Omega \ \ \text{ for all } |t|<\nu \}.$$ Note that the supremum exists since above set is nonempty. Now we show that $\chi_x = \tau$. If not, then $\chi_x<\tau$. First of all, the Lipschitz continuity of $v$ implies that $T_t$ is defined for all $t\in [0,\tau]$. By definition of $\chi_x$, $$ T_t(x) \in \partial \Omega, \ \ \ \text{ for all }t<\chi_x.$$ Since $\partial \Omega$ is closed, we conclude that $ T_{\chi_x}(x)\in \partial \Omega$. Then the IVP $$\begin{cases} x'= v(t, x), \\ x(\chi_x) = T_{\chi_x}(x) \end{cases}$$ is (uniquely) solvable for some $t \in (\chi_x - \delta, \chi_x + \delta)$. In particular, By the same argument we did, we have $$T_t(x) \in \partial \Omega$$ whenever $t < \chi_x + \delta'$ for some $\delta' < \delta$. But this contradicts to the definition of $\chi_x$. Thus we must have $\chi_x = \tau$ for all $x$.

Answer 2

Arctic Char's answer is perfectly fine and helped me a lot to understand the main idea. However, at least for my own reference, I post an additional answer which uses definitions of the involved concepts which I'm more familiar with.

I will slightly deviate from the notation used in the question. First of all, let $k\in\{1,\ldots,d\}$ and $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary.

We show first that, for every chart $(\Omega,\phi)$ of $M$ and $x\in\partial M$, there is a $\varepsilon>0$ with $$T_t(x)\in\partial\Omega\;\;\;\text{for all }t\in[0,\varepsilon)\tag{13}.$$ In order to do so, let $\pi$ denote the canonical projection of $\mathbb R^k$ onto $\mathbb R^{k-1}$ with $\pi(\partial\mathbb H^k)=\mathbb R^{k-1}$ and $$\tilde\phi:=\pi\circ\left.\phi\right|_{\partial\Omega}.$$ Note that $$\tilde U:=\tilde\phi(\partial\Omega)$$ is $\mathbb R^{k-1}$-open and let $$h(t,\tilde u):=T_{\tilde\phi^{-1}(\tilde u)}(\tilde\phi)v\left(t,\tilde\phi^{-1}(\tilde u)\right)\;\;\;\text{for }(t,u)\in[0,\tau]\times\tilde U.$$ Remember that $$T_{\tilde\phi^{-1}(\tilde u)}(\tilde\phi)^{-1}={\rm D}\tilde\phi^{-1}(\tilde u)={\rm D}\phi^{-1}(\iota(\tilde u))\iota=T_{\tilde\phi^{-1}(\tilde u)}(\phi)^{-1}\circ\iota\tag{14}$$ for all $\tilde u\in\tilde U$, where $\iota$ denotes the canonical embedding of $\mathbb R^{k-1}$ into $\mathbb R^k$ with $\iota\mathbb R^{k-1}=\partial\mathbb H^k$. Now let $\tilde u:=\tilde\phi(x)$. By the Peano existence theorem, there is a $\varepsilon>0,Y\in C^1([0,\varepsilon),\tilde U)$ with $$Y(t)=\tilde u+\int_0^th(s,Y(s))\:{\rm d}s\;\;\;\text{for all }t\in[0,\varepsilon)\tag{15}.$$ Let $$Z:=\tilde\phi^{-1}\circ Y.$$ By $(14)$, $$T_{Z(t)}(\tilde\phi)^{-1}={\rm D}\tilde\phi^{-1}(Y(t))\;\;\;\text{for all }t\in[0,\varepsilon)\tag{16}$$ and hence $$Z'(t)={\rm D}\tilde\phi^{-1}(Y(t))Y'(t)=\underbrace{T_{Z(t)}(\tilde\phi)^{-1}T_{Z(t)}(\tilde\phi)}_{=\:\operatorname{id}_{T_x\:\partial\Omega}}v(t,Z(t))=v(t,Z(t))\tag{17}$$ for all $t\in[0,\varepsilon)$ by the chain rule. Moreover, by definition, $Z(0)=x$ and hence, by uniqueness, $$\left.X^x\right|_{[0,\:\varepsilon)}=Z\tag{18}.$$ Clearly, $$Z([0,\varepsilon))\subseteq\tilde\phi^{-1}(\tilde U)=\partial\Omega.\tag{19}.$$

From this result, we have obtained that $$\forall x\in\partial M:\exists\varepsilon>0:\forall t\in[0,\varepsilon):T_t^{(v)}(x)\in\partial M.\tag{20}.$$ And if $\partial M$ is $\mathbb R^d$-closed$^1$, we can show (as Arctic Char did) that $$T_t(\partial M)\subseteq\partial M\;\;\;\text{for all }t\in[0,\tau].\tag{21}$$

However, how do we conclude that $(21)$ is actually an equality?

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$^1$ This is, for example, the case when $M$ is a $d$-dimensional properly embedded $C^1$-submanifold of $\mathbb R^d$ with boundary, since then $M$ is $\mathbb R^d$-closed and $\partial M$ coincides with the topological boundary of $M$.