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  "body": "\n## Counterexample\n\nFor a real number $x$, let $x^{1/3}$ be its real cube root and let $x^{2/3} = \\left(x^{1/3}\\right)^2$.\nConsider the [[initial value problem|075VBY4K4X]] $x'(t) = 3\\, x(t)^{2/3}$, $x(0) = 0$, on $\\mathbb{R}$.\nThe constant function $x(t) = 0$ is a solution.\nThe function $x(t) = t^3$ is also a solution, because its derivative is $3t^2 = 3 \\left(t^3\\right)^{2/3}$.\nMore generally, for each $c \\ge 0$, the function equal to $0$ for $t \u003c c$ and to $(t - c)^3$ for $t \\ge c$ is a solution.\nSo the problem has infinitely many solutions on every interval that contains $0$ in its interior.\n\n## Remarks\n\nThe right side $f(x) = 3 x^{2/3}$ is continuous, so the [[Peano existence theorem|098YP831QY]] applies.\nThe function $f$ is not Lipschitz on any neighbourhood of $0$: the quotient $f(x)/x = 3 x^{-1/3}$ is unbounded as $x \\to 0$. So the [[Picard–Lindelöf theorem|0P6KRD5B7A]] does not apply.\nFor an initial value $x(0) = x_0 \\neq 0$, the field is $C^1$ near $x_0$, and the solution is unique near $t = 0$.\n\n## Explanations\n\n### L1 human\n\nA solution can rest at $0$ for any length of time and then leave.\nThe field is so steep near $0$ that the equation cannot tell when the departure must happen.\nThis shows that continuity alone does not give uniqueness.\n",
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  "name": "Non-uniqueness for a non-Lipschitz equation",
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