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  "body": "\n## Statement\n\nLet $D$ be an open subset of $\\mathbb{R} \\times \\mathbb{R}^n$ and let $f : D \\to \\mathbb{R}^n$ be continuous.\nAssume 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 $(t, x), (t, y) \\in V$.\nLet $I$ be an interval, and let $\\gamma_1, \\gamma_2 : I \\to \\mathbb{R}^n$ be two [[solutions|04SW7WECS8]] of $x' = f(t, x)$ whose graphs lie in $D$.\nIf $\\gamma_1(t_0) = \\gamma_2(t_0)$ for one $t_0 \\in I$, then $\\gamma_1(t) = \\gamma_2(t)$ for all $t \\in I$.\n\n## Remarks\n\nSo an [[initial value problem|075VBY4K4X]] for such an $f$ has at most one solution on each interval that contains $t_0$.\nProof 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|07EJHQYR9A]] gives $u = 0$. A connectedness argument extends this to all of $I$.\nA $C^1$ function $f$ is locally Lipschitz in $x$.\nWithout the Lipschitz condition, uniqueness can fail. See the [[non-uniqueness example|00K7ASZ1PR]].\n\n## Explanations\n\n### L1 human\n\nTwo solutions of a Lipschitz equation that meet at one time are the same solution.\nSo the graphs of different solutions never cross or touch.\n",
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