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 :)
 answered
 1536 views
 $4.92
Related Questions
 Differentai equations, question 2.
 Differential equations, question 4
 Laplace transforms and initial value problems.
 Solve the twoway wave equation in terms of $u_0$
 Solve the twoway wave equation
 Solve the integral equation, (integrodifferential)
 Linearization of nonlinear differential equations near an equilibrium position
 Ordinary differential equation questions
Your deadline is too short for the level of your question. Consider extending the deadline if possible.