Proof of uniqueness of solutions for Lipschitz vector fields

Proof

We first prove agreement near a point of agreement, and then spread it along $I$ with a supremum argument.

1. Lemma. Let $c < e$ and let $h : [c, e] \to \R^n$ be continuous. Then $\norm{\int_c^e h} \le \int_c^e \norm{h}$.
2. Proof: let $v = \int_c^e h$, and assume $v \ne 0$. By linearity of the integral, $\norm{v}^2 = \int_c^e v \cdot h$. By Cauchy–Schwarz and monotonicity, this is at most $\norm{v} \int_c^e \norm{h}$.
3. Let $w = \gamma_1 - \gamma_2$ and $u = \norm{w}$, a continuous function on $I$. We may assume $I$ has more than one point.
4. Local claim: if $t_1 \in I$ and $w(t_1) = 0$, there is $\delta > 0$ with $w = 0$ on $I \cap [t_1 - \delta, t_1 + \delta]$.
5. Proof of the claim. Let $p = (t_1, \gamma_1(t_1)) = (t_1, \gamma_2(t_1)) \in D$. Take a neighbourhood $V \subseteq D$ of $p$ and $K \ge 0$ as in the Lipschitz hypothesis. $V$ contains the open ball of some radius $\rho > 0$ around $p$.
6. The curves $\gamma_1, \gamma_2$ are continuous. So there is $\delta > 0$ with $K \delta < 1$ such that $(t, \gamma_k(t))$ lies within $\rho$ of $p$ for $k = 1, 2$ and $t \in I_\delta = I \cap [t_1 - \delta, t_1 + \delta]$.
7. Shrinking $\delta$, we may assume that $I_\delta$ is a closed bounded interval. Indeed, if $t_1 + \delta' \in I$ with $\delta' > 0$, then $[t_1, t_1 + \delta'] \subseteq I$, and similarly on the left.
8. So $(t, \gamma_k(t)) \in V$ for $t \in I_\delta$. As solutions, the $\gamma_k$ have derivatives $f(t, \gamma_k(t))$, which are continuous in $t$.
9. By the fundamental theorem of calculus, componentwise, $w(t) = \int_{t_1}^{t} \left(f(s, \gamma_1(s)) - f(s, \gamma_2(s))\right) ds$ for $t \in I_\delta$, since $w(t_1) = 0$.
10. The integrand has norm at most $K u(s)$, by the choice of $V$. Let $M$ be the maximum of $u$ on $I_\delta$, which exists by the extreme value theorem.
11. By the lemma, $u(t) \le K \abs{t - t_1} M \le K \delta M$ for $t \in I_\delta$. So $M \le K \delta M$, that is, $(1 - K \delta) M \le 0$.
12. Since $K \delta < 1$ and $M \ge 0$, we get $M = 0$. This proves the local claim.
13. Right of $t_0$. Suppose $t_2 \in I$, $t_2 > t_0$ and $w(t_2) \ne 0$. Let $A$ be the set of $s \in [t_0, t_2]$ with $w = 0$ on $[t_0, s]$. It contains $t_0$ and is bounded.
14. By the completeness axiom, $s^* = \sup A$ exists, and $t_0 \le s^* \le t_2$. Each $s \in [t_0, s^*)$ lies below some element of $A$, so $w(s) = 0$.
15. By continuity of $w$, also $w(s^*) = 0$. So $s^* \in A$, and $s^* < t_2$ since $w(t_2) \ne 0$.
16. By the local claim at $s^*$, $w = 0$ on $I \cap [s^*, s^* + \delta]$. This interval contains $[s^*, \min(s^* + \delta, t_2)]$, since $t_2 \in I$.
17. So $\min(s^* + \delta, t_2) \in A$, and it is larger than $s^*$. This contradicts $s^* = \sup A$. Hence $w = 0$ on $I \cap [t_0, \infty)$.
18. Left of $t_0$. The same argument, with the infimum of the set of $s \in [t_2, t_0]$ such that $w = 0$ on $[s, t_0]$, gives $w = 0$ on $I \cap (-\infty, t_0]$.
19. So $\gamma_1(t) = \gamma_2(t)$ for all $t \in I$.