$\newcommand{\R}{\mathbb{R}}$ Consider the ODE $$\begin{cases} y'(t) = f(t, y(t)) \quad \\ y(t_0) = y_0 \end{cases}$$ with $f: I \times U \to \R^n$, $t_0 \in I \subset \mathbb{R}$ an open interval and $y_0 \in U \subset \R^n$ an open domain.
By the local version of the Picard-Lindelöf (also called Cauchy-Lipschitz) Theorem, if $f$ is continuous and locally Lipschitz continuous in the second variable, a unique solution exists on a maximal interval of existence $t_0 \in I_{\max} \subset I$.
In general, the Lipschitz continuity is roughly also necessary for uniqueness (albeit slightly more moderate conditions like the Osgood Criterion exist), as illustrated by the the example $$\begin{cases} y'(t) = |y(t)|^{\alpha} \quad \\ y(0) = 0 \end{cases}$$ for $0 < \alpha < 1$. This ODE has infinitely many solutions of the form $$y_C(t)=\begin{cases} 0,&t<C, \quad \\ (1- \alpha)^{\frac{1}{1-\alpha}} (t-C)^{\frac{1}{1-\alpha}}, &t\geq C. \end{cases}$$ for $C \in [0, \infty]$ arbitrary.
Question. Why does the lacking Lipschitz continuity / Osgood condition allow for this example? What is the visual or intuitive reason that an $0<\alpha < 1$-Hölder continuous $f$ (maybe in general) allows in for multiple solutions as opposed to a Lipschitz one?
Two further remarks:
- In the above link it is emphasized that for this example the Osgood condition is violated.
- (Lastly, there is a "dual" result: For $\alpha > 1$, clearly local existence and uniqueness holds, but for $y(0) \neq 0$ the solution blows up in finite time by self-reinforcing growth due to the $\alpha > 1$ exponent.)