Finding conditions when an initial value problem has unique solution

Question

For each $\alpha>0$ find all pairs $(x_0,y_0)\in\mathbb R^2$ such that the following IVP has a unique solution in a neighbourhood of $(x_0,y_0)$ $$y'=y^\alpha,\ y(x_0)=y_0$$ I know about the Picard-Lindelof theorem which asserts uniqueness but it is only a sufficient condition. The question ask to find all such pairs.

Answer 1

The function $f(y):=y^\alpha$ is not locally Lipschitz in $y=0$ for $\alpha \in (0,1)$. If we choose $\alpha$ in this interval and $y_0\neq 0$ then we can apply the Picard-Lindelöf theorem. The same theorem applies when $\alpha \geq 1$ since $y^\alpha$ is locally Lipschitz for all $\alpha \geq 1$ in all $y\in \mathbb{R}$.

When we lose the Lipschitz continuity we can construct $2$ solutions to prove non-uniqueness. For example if $\alpha=1/2$ and $y(1)=0$ we have the solutions $y\equiv 0$ and $y=\frac{(x-1)^2}{4}$. We can easily generalize this example to all $\alpha \in (0,1)$ and $y(x_0)=0$.