> [!theorem] > > Let $0 \in J \subset \mathbb R$ be an open interval, $E$ be a [[Banach Space|Banach space]], $U \subset E$ be [[Open Set|open]], $f: J \times U \to E$ be a Lipschitz [[Time-Dependent Vector Field|time-dependent vector field]], and $x_0 \in U$. > > Let $r \in (0, 1)$ such that $\overline{B(x_0, 3r)} \subset U$, $L = \max(1, \text{Lip}(f))$, and $\eps < \frac{a}{L\norm{f}_u}$, then > 1. For each $x \in \overline{B(x_0, r)}$, there exists a unique [[Flow|local flow]] $\alpha: (-\eps, \eps) \times B(x, r) \to U$. If $f \in C^p$, then so is each integral curve $\alpha_x$. > 2. The mapping $\overline{B(x_0, r)} \to C((-\eps, \eps), U)$ with $y \mapsto \alpha_y$ is Lipschitz. > 3. For every pair of [[Integral Curve|integral curve]] $\alpha_1: J_1 \to U$ and $\alpha_2: J_2 \to U$ with $\alpha_1(0) = \alpha_2(0) = x_0 \in U$, $\alpha_1|_{J_1} = \alpha_2|_{J_2}$. > > *Proof*. $(1)$: Let $x \in \overline{B(x_0, r)}$, > $ > C_{x} = \bracsn{\alpha \in C((-\eps, \eps), \overline{B(x_0, 2r)}): \alpha(0) = x_0} > $ > then $C_x$ is a [[Complete Metric Space|complete metric space]] under the uniform norm. Let > $ > S_x: C_x \to C_x \quad (S\alpha)(t) = x + \int_0^tf(t, \alpha(t))dt > $ > then > $ > \begin{align*} > \abs{(S_x\alpha)(t) - x} &\le \abs{\int_0^tf(t, \alpha(t))dt} \le \int_0^t\abs{f(t, \alpha(t))}dt \\ > &\le \abs{t}\norm{f}_u \le \eps\norm{f}_u \le r > \end{align*} > $ > so $S\alpha \in S$ for all $\alpha \in C_x$. On the other hand, for any $\alpha, \beta \in C_x$, > $ > \begin{align*} > \norm{S_x\alpha - S_x\beta}_u &\le \int_0^t\abs{f(t, \alpha(t)) - f(t, \beta(t))}dt \\ > &\le \abs{t} \cdot L \cdot \norm{\alpha - \beta}_u \le \eps L \norm{\alpha - \beta}_u > \end{align*} > $ > Since $\eps L < 1$, by the [[Contractor|contraction mapping theorem]], it admits a fixed point $\alpha_x$, which is an integral curve for $f$ at $1$. > > $(2)$: Let $x, y \in \overline{B(x_0, r)}$, then > $ > \begin{align*} > \norm{\alpha_x - \alpha_y}_u &\le \sum_{k \in \nat}\normn{S^k_x\alpha_y - S_x^{k-1}\alpha_y}_u \\ > &\le \norm{S_x\alpha_y - \alpha_y}\sum_{k \in \nat}(\eps L)^k = \frac{\norm{S_x\alpha_y - \alpha_y}}{1 - \eps L} > \end{align*} > $ > where > $ > \norm{S_x\alpha_y - \alpha_y} = \norm{S_x\alpha_y - S_y\alpha_y} \le \norm{x - y} > $ > Hence $y \mapsto \alpha_y$ is Lipschitz with constant $1/(1 - \eps L)$. > > $(3)$: Let $J'$ be the set of all points in $J_1 \cap J_2$ on which $\alpha_1$ and $\alpha_2$ agree. By continuity of $\alpha_1$ and $\alpha_2$, $J'$ is relatively closed. By the local uniqueness part of $(1)$, $J'$ is open. Since $J_1 \cap J_2$ is connected, $J' = J_1 \cap J_2$.