The strategy is to work with the integral equation for the displacement $z(t) := \Phi(t,x) - \Phi(t,y)$ rather than differentiating $|z(t)|$ directly. The upper bound follows from the triangle inequality and the [Gronwall Inequality](/theorems/872). The lower bound is obtained by applying the upper bound to the reverse-time flow. The proof proceeds as follows: Step 1 derives the integral equation and the Lipschitz estimate for the velocity difference; Steps 2 and 3 establish the upper and lower bounds respectively; Step 4 extracts injectivity, the Lipschitz estimate for the inverse, and the bi-Lipschitz conclusion.
**Step 1: Integral equation and Lipschitz estimate.** Fix $x, y \in M$ and define $z: [0,T] \to \mathbb{R}^n$ by $z(t) := \Phi(t,x) - \Phi(t,y)$. Since the flow map satisfies $\Phi(t,x) = x + \int_0^t u(s, \Phi(s,x))\,d\mathcal{L}^1(s)$, we have
\begin{align*}
z(t) &= (x - y) + \int_0^t \bigl[u(s, \Phi(s,x)) - u(s, \Phi(s,y))\bigr]\,d\mathcal{L}^1(s).
\end{align*}
Write $\beta(s) := \|\nabla u(s)\|_{L^\infty}$ for brevity. Since $u \in C([0,T]; W^{1,\infty}(M; \mathbb{R}^n))$, the [mean value inequality](/theorems/328) gives
\begin{align*}
\bigl|u(s, \Phi(s,x)) - u(s, \Phi(s,y))\bigr| &\le \beta(s)\,|z(s)|
\end{align*}
for all $s \in [0,T]$.
**Step 2: Upper bound.**
[claim:Upper Bound]
$|z(t)| \le |x - y|\,\exp\!\bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\bigr)$ for all $t \in [0,T]$.
[/claim]
[proof]
Taking norms of the integral equation from Step 1 and applying the triangle inequality:
\begin{align*}
|z(t)| &\le |x - y| + \int_0^t \bigl|u(s, \Phi(s,x)) - u(s, \Phi(s,y))\bigr|\,d\mathcal{L}^1(s) \le |x - y| + \int_0^t \beta(s)\,|z(s)|\,d\mathcal{L}^1(s).
\end{align*}
The [function](/page/Function) $t \mapsto |z(t)|$ is [continuous](/page/Continuity) (since $\Phi$ is continuous in $t$ by the flow equation), and $\beta$ is [integrable](/page/Integral). The [Gronwall Inequality](/theorems/872) with $\phi(t) = |z(t)|$ and $\alpha = |x - y|$ gives
\begin{align*}
|z(t)| &\le |x - y|\,\exp\!\Bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr).
\end{align*}
[/proof]
**Step 3: Lower bound.**
[claim:Lower Bound]
$|z(t)| \ge |x - y|\,\exp\!\bigl(-\int_0^t \beta(s)\,d\mathcal{L}^1(s)\bigr)$ for all $t \in [0,T]$.
[/claim]
[proof]
Fix $t_0 \in [0,T]$. Define the reverse-time velocity field $\tilde{u}: [0,t_0] \times M \to \mathbb{R}^n$ by $\tilde{u}(s,w) := -u(t_0 - s, w)$, and let $\Psi: [0,t_0] \times M \to M$ be its flow map:
\begin{align*}
\frac{d}{ds}\Psi(s,w) &= \tilde{u}(s, \Psi(s,w)) = -u(t_0 - s, \Psi(s,w)), \quad \Psi(0,w) = w.
\end{align*}
We claim that $\Psi(s, \Phi(t_0, x)) = \Phi(t_0 - s, x)$ for all $s \in [0, t_0]$. To see this, define $\eta: [0,t_0] \to M$ by $\eta(s) := \Phi(t_0 - s, x)$. Then $\eta(0) = \Phi(t_0, x)$. Differentiating with respect to $s$ using the chain rule and the flow equation $\partial_t \Phi(t,x) = u(t, \Phi(t,x))$:
\begin{align*}
\frac{d\eta}{ds}(s) &= -\frac{\partial \Phi}{\partial t}(t_0 - s, x) = -u(t_0 - s, \Phi(t_0 - s, x)) = -u(t_0 - s, \eta(s)) = \tilde{u}(s, \eta(s)).
\end{align*}
Hence $\eta$ solves the ODE $\frac{d}{ds}\eta(s) = \tilde{u}(s, \eta(s))$ with initial condition $\eta(0) = \Phi(t_0, x)$. But $s \mapsto \Psi(s, \Phi(t_0, x))$ solves the same ODE with the same initial condition. By the uniqueness part of the [Picard Lindelof](/theorems/69) theorem, $\Psi(s, \Phi(t_0, x)) = \eta(s) = \Phi(t_0 - s, x)$ for all $s \in [0, t_0]$; in particular, $\Psi(t_0, \Phi(t_0, x)) = \Phi(0, x) = x$. Since $\|\nabla \tilde{u}(s)\|_{L^\infty} = \|\nabla u(t_0 - s)\|_{L^\infty}$, the Upper Bound applied to $\Psi$ with starting points $w_1 := \Phi(t_0, x)$ and $w_2 := \Phi(t_0, y)$ gives
\begin{align*}
|x - y| &= |\Psi(t_0, w_1) - \Psi(t_0, w_2)| \le |w_1 - w_2|\,\exp\!\Bigl(\int_0^{t_0} \|\nabla u(t_0 - s)\|_{L^\infty}\,d\mathcal{L}^1(s)\Bigr).
\end{align*}
The substitution $r = t_0 - s$ gives $\int_0^{t_0} \|\nabla u(t_0 - s)\|_{L^\infty}\,d\mathcal{L}^1(s) = \int_0^{t_0} \|\nabla u(r)\|_{L^\infty}\,d\mathcal{L}^1(r)$. Hence
\begin{align*}
|x - y| &\le |\Phi(t_0,x) - \Phi(t_0,y)|\,\exp\!\Bigl(\int_0^{t_0} \beta(s)\,d\mathcal{L}^1(s)\Bigr),
\end{align*}
and dividing both sides by the exponential (which is strictly positive) yields
\begin{align*}
|\Phi(t_0,x) - \Phi(t_0,y)| &\ge |x - y|\,\exp\!\Bigl(-\int_0^{t_0} \beta(s)\,d\mathcal{L}^1(s)\Bigr).
\end{align*}
Since $t_0 \in [0,T]$ was arbitrary, the lower bound holds for all $t \in [0,T]$.
[/proof]
**Step 4: Injectivity and bi-Lipschitz regularity.** Combining the Upper Bound and Lower Bound gives the two-sided estimate in the theorem statement. Injectivity follows from the Lower Bound: if $\Phi(t,x) = \Phi(t,y)$ then $|z(t)| = 0$, so $|x - y| \le 0$, hence $x = y$. Since $\Phi(t,\cdot)$ is injective, the inverse $\Phi(t,\cdot)^{-1}$ exists on the image of $\Phi(t,\cdot)$. Setting $X_1 := \Phi(t,x)$ and $X_2 := \Phi(t,y)$ and rearranging the Lower Bound as $|x - y| \le |z(t)|\,\exp\bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\bigr)$ gives the Lipschitz estimate for the inverse:
\begin{align*}
|\Phi(t,\cdot)^{-1}(X_1) - \Phi(t,\cdot)^{-1}(X_2)| &\le |X_1 - X_2|\,\exp\!\Bigl(\int_0^t \beta(s)\,d\mathcal{L}^1(s)\Bigr).
\end{align*}
Together with the Upper Bound, this establishes that $\Phi(t,\cdot)$ is bi-Lipschitz for all $t \in [0,T]$.
