ODE - Initial Value Problem

If $\alpha < 1$, then there are nonunique solutions. Consider $y(t) = 0$ and $y(t) = ((1-\alpha)t)^{1/(1-\alpha)}$, both of which are solutions to the IVP if $\alpha < 1$ (However if $\alpha \geq 1$ then the exponent is negative and coefficient $(1-\alpha)$ becomes negative, which causes the function to cease to be a solution).

Now, if $\alpha \geq 1$, then the function $f(y) = |y|^{\alpha}$ is continuously differentiable away from $0$ and the derivative near $0$ is bounded (following from $\alpha \geq 1$). Thus $f$ is locally Lipschitz, so by Picard-Lindelof, there exists a unique solution to the IVP on some interval $[0, \epsilon]$, which one can check is $y(t) = 0$. One can extend this solution repeatedly, using Picard-Lindelof, since the solution never blows up.

We elaborate on this fact. Let $\mathcal{M}$ be the set of real numbers $M$ such that there exists a unique solution to the IVP on $[0, M]$. Let $M \in \mathcal{M}$. Since uniqueness was assumed and $y(t) = 0$ is a solution on $[0, M]$, the unique solution must be $y(t) = 0$. Then one may extend this solution to a unique solution on $[0, M + \epsilon]$ by Picard-Lindelof, by considering the IVP with initial value $y(M) = 0$. Thus $M+\epsilon \in \mathcal{M}$. Furthermore, given an increasing sequence $M_1, M_2, \ldots$ of values in $\mathcal{M}$ with limit $M$, one has that $M$ in $\mathcal{M}$, since the solutions on $[0, M_i]$ can be combined to define a solution on $[0,M]$ by setting $y(M) = 0$. This solution must be unique, since any subinterval $[0, M']$ is contained in some $[0, M_i]$, where the solution is uniquely defined. Thus by real induction $\mathcal{M} = [0, \infty)$.

See http://alpha.math.uga.edu/~pete/instructors_guide_2017.pdf for an explanation of real induction.