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  "body": "\n## Proof\n\nWe compare $\\|\\gamma_1 - \\gamma_2\\|$ with a slightly larger Grönwall bound and look at the first time where the comparison could fail. Here $\\|\\cdot\\|$ is the Euclidean norm on $\\mathbb{R}^n$.\n\n1. Let $\\varepsilon = \\varepsilon_1 + \\varepsilon_2$, let $e(t) = \\gamma_1(t) - \\gamma_2(t)$ and let $\\varphi(t) = \\|e(t)\\|$ for $t \\in [a, b]$. Then $\\varphi$ is continuous on $[a, b]$.\n2. For $t \\in (a, b)$, write $e'(t)$ as $\\left(\\gamma_1'(t) - f(t, \\gamma_1(t))\\right) + \\left(f(t, \\gamma_1(t)) - f(t, \\gamma_2(t))\\right) + \\left(f(t, \\gamma_2(t)) - \\gamma_2'(t)\\right)$.\n3. By the [[triangle inequality|0FJZMBX0ZZ]], the hypotheses on $\\gamma_i$ and the [[Lipschitz condition|055VVSN00V]], $\\|e'(t)\\| \\le \\varepsilon_1 + K \\varphi(t) + \\varepsilon_2 = K \\varphi(t) + \\varepsilon$.\n4. For real $\\delta', \\varepsilon'$, let $G = G_{\\delta', K, \\varepsilon'}$ be the [[Grönwall bound|0FRX8JD7QH]]. Then $G(0) = \\delta'$ and $G'(x) = K G(x) + \\varepsilon'$ for all $x$.\n5. Indeed, if $K \\ne 0$, the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential|0EX9GFJ0B3]] give $G'(x) = K \\delta' e^{Kx} + \\varepsilon' e^{Kx}$, and this equals $K G(x) + \\varepsilon'$. If $K = 0$, then $G'(x) = \\varepsilon' = K G(x) + \\varepsilon'$.\n6. Fix $\\eta \u003e 0$ and let $H = G_{\\delta + \\eta, K, \\varepsilon + \\eta}$. By step 4, $H(0) = \\delta + \\eta$ and $H'(x) = K H(x) + \\varepsilon + \\eta$.\n7. We claim that $\\varphi(t) \u003c H(t - a)$ for each $t \\in [a, b)$. Suppose not. Then the set $Z = \\{ t \\in [a, b) : \\varphi(t) \\ge H(t - a) \\}$ is non-empty. Let $t_1 = \\inf Z$.\n8. We have $\\varphi(a) \\le \\delta \u003c H(0)$. By continuity of $\\varphi$ and $H$, there is $\\rho \u003e 0$ with $\\varphi(t) \u003c H(t - a)$ for $t \\in [a, a + \\rho)$. So $t_1 \\ge a + \\rho \u003e a$.\n9. By the definition of $t_1$, $\\varphi(t) \u003c H(t - a)$ for each $t \\in [a, t_1)$. Letting $t \\to t_1$ from the left gives $\\varphi(t_1) \\le H(t_1 - a)$, since [[limits preserve non-strict inequalities|0N32V0E3XJ]].\n10. There are points $z_k \\in Z$ with $z_k \\ge t_1$ and $z_k \\to t_1$. By continuity, $\\varphi(t_1) \\ge H(t_1 - a)$. So $\\varphi(t_1) = H(t_1 - a)$. Also $t_1 \\le z_1 \u003c b$, so $t_1 \\in (a, b)$.\n11. Let $s \\in (a, t_1)$. By steps 9 and 10, $H(t_1 - a) - H(s - a) \u003c \\varphi(t_1) - \\varphi(s)$. By the triangle inequality, $\\varphi(t_1) - \\varphi(s) \\le \\|e(t_1) - e(s)\\|$.\n12. Dividing by $t_1 - s \u003e 0$ gives $\\dfrac{H(t_1 - a) - H(s - a)}{t_1 - s} \\le \\left\\| \\dfrac{e(t_1) - e(s)}{t_1 - s} \\right\\|$.\n13. Let $s \\to t_1$ from the left. The left side tends to $H'(t_1 - a)$. The quotient inside the norm tends to $e'(t_1)$, so by the reverse triangle inequality its norm tends to $\\|e'(t_1)\\|$.\n14. So $H'(t_1 - a) \\le \\|e'(t_1)\\|$. By steps 3 and 10, $\\|e'(t_1)\\| \\le K H(t_1 - a) + \\varepsilon$.\n15. By step 6, $H'(t_1 - a) = K H(t_1 - a) + \\varepsilon + \\eta$. With step 14 this gives $\\eta \\le 0$, a contradiction. This proves the claim of step 7.\n16. So $\\varphi(t) \u003c G_{\\delta + \\eta, K, \\varepsilon + \\eta}(t - a)$ for each $t \\in [a, b)$ and each $\\eta \u003e 0$.\n17. For fixed $x$, the value $G_{\\delta', K, \\varepsilon'}(x)$ is an affine function of $(\\delta', \\varepsilon')$, so it is continuous in them. Letting $\\eta \\to 0$ in step 16 gives $\\varphi(t) \\le G_{\\delta, K, \\varepsilon}(t - a)$ for $t \\in [a, b)$.\n18. Both $\\varphi$ and $x \\mapsto G_{\\delta, K, \\varepsilon}(x)$ are continuous. Letting $t \\to b$ gives the same bound at $t = b$.\n19. Hence $\\|\\gamma_1(t) - \\gamma_2(t)\\| \\le G_{\\delta, K, \\varepsilon_1 + \\varepsilon_2}(t - a)$ for each $t \\in [a, b]$.\n\n## Idea\n\nThe Grönwall bound $G$ solves $G' = K G + \\varepsilon$, while the distance $\\varphi$ grows at most at the rate $K \\varphi + \\varepsilon$. A bound that starts a little higher and grows a little faster can never be reached. At a first touching time, the distance would have to grow faster than the bound. Shrinking the margin to zero gives the estimate.\n",
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