Non-uniqueness for a non-Lipschitz equation

Counterexample

For a real number $x$, let $x^{1/3}$ be its real cube root and let $x^{2/3} = \left(x^{1/3}\right)^2$. Consider the initial value problem $x'(t) = 3\, x(t)^{2/3}$, $x(0) = 0$, on $\mathbb{R}$. The constant function $x(t) = 0$ is a solution. The function $x(t) = t^3$ is also a solution, because its derivative is $3t^2 = 3 \left(t^3\right)^{2/3}$. More generally, for each $c \ge 0$, the function equal to $0$ for $t < c$ and to $(t - c)^3$ for $t \ge c$ is a solution. So the problem has infinitely many solutions on every interval that contains $0$ in its interior.

Remarks

The right side $f(x) = 3 x^{2/3}$ is continuous, so the Peano existence theorem applies. The 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 does not apply. For an initial value $x(0) = x_0 \neq 0$, the field is $C^1$ near $x_0$, and the solution is unique near $t = 0$.

Explanations

L1 human

A solution can rest at $0$ for any length of time and then leave. The field is so steep near $0$ that the equation cannot tell when the departure must happen. This shows that continuity alone does not give uniqueness.