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  "body": "\n## Proof\n\nWe first prove agreement near a point of agreement, and then spread it along $I$ with a supremum argument.\n\n1. Lemma. Let $c \u003c e$ and let $h : [c, e] \\to \\R^n$ be continuous. Then $\\norm{\\int_c^e h} \\le \\int_c^e \\norm{h}$.\n2. Proof: let $v = \\int_c^e h$, and assume $v \\ne 0$. By [[linearity of the integral|080RWVVSB6]], $\\norm{v}^2 = \\int_c^e v \\cdot h$. By [[Cauchy–Schwarz|0RFWW56Y22]] and [[monotonicity|0MBWQKM8QR]], this is at most $\\norm{v} \\int_c^e \\norm{h}$.\n3. Let $w = \\gamma_1 - \\gamma_2$ and $u = \\norm{w}$, a continuous function on $I$. We may assume $I$ has more than one point.\n4. Local claim: if $t_1 \\in I$ and $w(t_1) = 0$, there is $\\delta \u003e 0$ with $w = 0$ on $I \\cap [t_1 - \\delta, t_1 + \\delta]$.\n5. 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|055VVSN00V]] hypothesis. $V$ contains the open ball of some radius $\\rho \u003e 0$ around $p$.\n6. The curves $\\gamma_1, \\gamma_2$ are continuous. So there is $\\delta \u003e 0$ with $K \\delta \u003c 1$ such that $(t, \\gamma_k(t))$ lies within $\\rho$ of $p$ for $k = 1, 2$ and $t \\in I_\\delta = I \\cap [t_1 - \\delta, t_1 + \\delta]$.\n7. Shrinking $\\delta$, we may assume that $I_\\delta$ is a closed bounded interval. Indeed, if $t_1 + \\delta' \\in I$ with $\\delta' \u003e 0$, then $[t_1, t_1 + \\delta'] \\subseteq I$, and similarly on the left.\n8. So $(t, \\gamma_k(t)) \\in V$ for $t \\in I_\\delta$. As [[solutions|04SW7WECS8]], the $\\gamma_k$ have derivatives $f(t, \\gamma_k(t))$, which are [[continuous|0NRQPQX3VA]] in $t$.\n9. By the [[fundamental theorem of calculus|0NHGY0N0JB]], 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 $w(t_1) = 0$.\n10. 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|0WS1A8E2P9]].\n11. 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, $(1 - K \\delta) M \\le 0$.\n12. Since $K \\delta \u003c 1$ and $M \\ge 0$, we get $M = 0$. This proves the local claim.\n13. Right of $t_0$. Suppose $t_2 \\in I$, $t_2 \u003e t_0$ 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.\n14. By the [[completeness axiom|0AXWET80YW]], $s^* = \\sup A$ exists, and $t_0 \\le s^* \\le t_2$. Each $s \\in [t_0, s^*)$ lies below some element of $A$, so $w(s) = 0$.\n15. By continuity of $w$, also $w(s^*) = 0$. So $s^* \\in A$, and $s^* \u003c t_2$ since $w(t_2) \\ne 0$.\n16. By the local claim at $s^*$, $w = 0$ on $I \\cap [s^*, s^* + \\delta]$. This interval contains $[s^*, \\min(s^* + \\delta, t_2)]$, since $t_2 \\in I$.\n17. So $\\min(s^* + \\delta, t_2) \\in A$, and it is larger than $s^*$. This contradicts $s^* = \\sup A$. Hence $w = 0$ on $I \\cap [t_0, \\infty)$.\n18. 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 $I \\cap (-\\infty, t_0]$.\n19. So $\\gamma_1(t) = \\gamma_2(t)$ for all $t \\in I$.\n\n## Idea\n\nNear a point where the two solutions meet, their distance $u$ satisfies $u \\le K \\delta \\max u$. On a short interval with $K \\delta \u003c 1$, this forces $u = 0$. The set where the solutions agree is therefore open in $I$, and it is closed by continuity. A supremum argument spreads the agreement to all of $I$.\n",
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