ODE  Initial Value Problem
Let $\alpha \in \mathbb{R}, \alpha > 0$. Prove that the IVP
$y'=y^{\alpha}, \quad y(0)=0$
has a uniqe solution on $[0, \infty)$ if and only if $\alpha \geq 1$.
Answer
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 PicardLindelof, 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 PicardLindelof, 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 PicardLindelof, 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.

In the case $\alpha \geq 1$: Doesn't PicardLindelof require open intervals which $[0, \infty)$ is not? And great thanks for the link by the way :)

And can we ensure that y is nonnegative? Because if it isn't then f is not necessarily differentiable at 0 (alpha=1), or am I missing something?

You can, for example, use PicardLindelof on a closed rectangle $D = [0, \infty) \times [1,1] $ and consider $f : D \to R$ as $f(t,y) = y^\alpha$. Then $f$ is continuous in $t$ and Lipschitz continuous in $y$. The initial value problem uses the point $(0, 0) \in D$ and gives a solution $y(t)$ on $[0, \epsilon]$. The theorem requires only a closed rectangle D. This is then repeated, using the knowledge that the unique solution is actually just $0$.

The answer has been corrected to account for the nondifferentiability at $y = 0$. The point still stands that $f$ is locally Lipschitz, which can be verified by demonstrating that $f$ is continuously differentiable away from $0$ and that the derivative does not blow up near $0$. Then one can use the mean value theorem to verify Lipschitz continuity.

Thank you for your patience :)
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