= Solution
For each departure time $c\in[2,h]$, define
$$
y_c(x)=\begin{cases}
(1-x/2)^2-1,&0\leq x\leq2,\\
-1,&2\leq x\leq c,\\
-1-(x-c)^2/4,&c\leq x\leq h.
\end{cases}
$$
Each branch satisfies the <ordinary differential equation>, and both joins have matching value $-1$ and matching derivative zero. Thus $y_c$ is continuously differentiable and satisfies the same <initial value problem>. If $2\leq c_1<c_2\leq h$, the solutions differ on $(c_1,c_2)$, proving \b[infinitely many solutions on every $[0,h]$ with $h>2$].
The uniqueness hypotheses fail when the solution reaches $-1$. Indeed
$$
\frac{|f(-1+\varepsilon)-f(-1)|}{\varepsilon}=\frac1{\sqrt\varepsilon}\longrightarrow\infty.
$$
No finite <Lipschitz constant> works on a rectangle containing that level in its interior. The local theorem still controls the earlier branch near zero; it cannot be extended through a point where its hypotheses fail. These are <waiting-time continuations at a non-Lipschitz equilibrium>.
\Image[/past-exam-of-the-differential-equations-1-course-of-the-university-of-oxford/2017-nonunique-continuations.png]
{title=The same initial solution admits different departure times from a non-Lipschitz equilibrium}
{description=Each curve arrives at $y=-1$ at time two, may remain there, and then decreases again. All displayed joins are continuously differentiable.}
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