Uniqueness of solutions for Lipschitz vector fields

Statement

Let $D$ be an open subset of $\mathbb{R} \times \mathbb{R}^n$ and let $f : D \to \mathbb{R}^n$ be continuous. Assume that $f$ is locally Lipschitz in $x$. That is, each point of $D$ has a neighbourhood $V \subseteq D$ and a constant $K \ge 0$ such that $\|f(t, x) - f(t, y)\| \le K \|x - y\|$ whenever $(t, x), (t, y) \in V$. Let $I$ be an interval, and let $\gamma_1, \gamma_2 : I \to \mathbb{R}^n$ be two solutions of $x' = f(t, x)$ whose graphs lie in $D$. If $\gamma_1(t_0) = \gamma_2(t_0)$ for one $t_0 \in I$, then $\gamma_1(t) = \gamma_2(t)$ for all $t \in I$.

Remarks

So an initial value problem for such an $f$ has at most one solution on each interval that contains $t_0$. Proof idea: the function $u(t) = \|\gamma_1(t) - \gamma_2(t)\|$ satisfies $u(t) \le K \left| \int_{t_0}^{t} u(s)\, ds \right|$ near $t_0$. Then Grönwall's inequality gives $u = 0$. A connectedness argument extends this to all of $I$. A $C^1$ function $f$ is locally Lipschitz in $x$. Without the Lipschitz condition, uniqueness can fail. See the non-uniqueness example.

Explanations

L1 human

Two solutions of a Lipschitz equation that meet at one time are the same solution. So the graphs of different solutions never cross or touch.