Let $E \subset \mathbb{R}^n$ be open, let $I \subset \mathbb{R}$ be an interval, and let $f: I \times E \to \mathbb{R}^n$ be continuous. Assume that $f$ is locally Lipschitz in the state variable, locally uniformly in time, meaning that for every $(s,a) \in I \times E$ there exist a relatively open neighbourhood $T \subset I$ of $s$, an open neighbourhood $B \subset E$ of $a$, and a constant $L>0$ such that $|f(t,p)-f(t,q)| \le L|p-q|$ for all $t \in T$ and all $p,q \in B$. Let $(t_0,x_0) \in I \times E$. A relatively open interval in $I$ means a subset $J \subset I$ that is an interval and is open in the [subspace topology](/page/Subspace%20Topology) inherited from $\mathbb{R}$. A solution on a relatively open interval $J \subset I$ containing $t_0$ means a map $x: J \to E$ such that $x(t_0)=x_0$, $x$ is continuous, the relative derivative $\dot{x}(t)=\lim_{h\to 0,\ t+h\in J}(x(t+h)-x(t))/h$ exists at every $t\in J$ for which this limit is defined in the relative domain sense, and $\dot{x}(t)=f(t,x(t))$ at every such $t$. Then there exists a unique maximal solution $x: J \to E$ on a relatively open interval $J \subset I$ containing $t_0$. More precisely, $x$ is a solution on $J$, and if $y: K \to E$ is any solution of the same [initial value problem](/page/Initial%20Value%20Problem) on a relatively open interval $K \subset I$ containing $t_0$, then $K \subset J$ and $y(t)=x(t)$ for every $t \in K$.