Solutions for the differential equation $y'=|y|^\alpha$ with $y(0)=0$ I'm studying for a final, for my first ODEs class, and I've stumbled upon this problem:
Given $\alpha > 0$ and the following ODE with initial values
$$\begin{cases}y'(t) = |y(t)|^{\alpha} \\ y(0) = 0\end{cases}$$
Determine for which $\alpha$ can we be sure that $y \equiv 0$ is the only solution to the equation, and for all the other possible values of $\alpha$, show an alternative solution. 
I could not make any progress yet, and have tried for a considerable amount of time. If I'm not mistaken, we want to determine for which $\alpha$ is $|y|^{\alpha}$ Lipschitz; because in that case, Picard's theorem guarantees that $y \equiv 0$ is the only solution. I'd really appreciate your help.   
 A: Note that $y'(t)\ge 0$ 
so $y$ is an increasing function and since $y(0)=0$ we know that $y(t)$ is the same sign as $t$.
So we can find explicit solutions because on the two intervals $]-\infty,0]$ and $[0,+\infty[$ the function $y$ is of constant sign and we can get rid of the absolute value in the equation.
Also we will suppose $y$ does not annulate elsewhere than in zero, which is equivalent to say that $y$ is strictly increasing, so we can integrate $y'/y^\alpha$. 
Unfortunately $|y|^\alpha$ is Lipschitz only for $\alpha=1$ (for other values we have either a problem in $0$ either in $\infty$), so Cauchy-Lipschitz theorem can be applied only in this case.
About regularity, since we suppose $y$ differentiable then it is continuous, and then from equation $y'$ become continuous, so if a solution exists it is at least $C^1$.

For $t\ge 0,\ y(t)\ge 0\ :\quad  y'=y^\alpha$


*

*If $\alpha=1$ then $y=Ae^t,\ $ In zero we get $A=0$ so $y=0$.

*If $\alpha>1$ let's call $\beta=(\alpha-1)>0$ then $\displaystyle{\frac1{y^\beta}=-\beta(A+t)}$ and there is no solution because of divergence in zero.

*If $\alpha<1$ let's call $\beta=(1-\alpha)\in\;]0,1]$ then $\displaystyle{y^\beta=\beta(A+t)}$. In zero we get $A=0$ so $y(t)=(\beta t)^{\frac1\beta}$



For $t\le 0,\ y(t)\le 0\ :\quad  (-y')=-(-y)^\alpha$


*

*If $\alpha=1$ then $-y=Ae^{-t},\ $ In zero we get $A=0$ so $y=0$.

*If $\alpha>1$ let's call $\beta=(\alpha-1)>0$ then $\displaystyle{\frac1{(-y)^\beta}=-\beta(A-t)}$ and there is no solution because of divergence in zero.

*If $\alpha<1$ let's call $\beta=(1-\alpha)\in\;]0,1]$ then $\displaystyle{(-y)^\beta=\beta(A-t)}$. In zero we get $A=0$ so $-y(t)=(-\beta t)^{\frac1\beta}$



conclusion :


*

*For $\alpha=1$, the only solution is $y=0$ (unicity by Cauchy-Lipchitz).

*For $\alpha>1$, $y=0$ is solution and there are no solutions strictly increasing.

*For $0<\beta=(1-\alpha)<1$, there are two solutions $y=0\ \ $ and  $\ \ y(t)=\operatorname{sgn(t)}\times(\beta|t|)^{\frac1\beta}$ 


Since unicity is not asserted except in the first case, can we find piecewise $C^1$ solutions ? 
The case $\alpha>1$ stays problematic, because $y(t)\to 0$ means $1/y^\beta\to\infty$ so it cannot be glued anywhere to $y=0$, the only other solution we know of that works in $t=0$.
For $\alpha<1$ we can glue $y=0$ on $[-\infty,t_0]$ to $y=(\beta(t-t_0))^{\frac1\beta}$ on $[t_0,+\infty]$ and this connection is $C^1$ in general since $1/\beta>1$ but it can be made $C^\infty$ for $\beta=\frac 1n$ for $n\in\mathbb N$. For other values, there is a step, where the $n^{th}-$derivative of $y$ has a vertical tangent (because $\frac 1\beta-n<0$).
So for $\alpha<1$ we can find an infinity of maximal solutions and they are even $C^\infty$ for $\alpha=\frac{n-1}{n}$ with $n\in\mathbb N$.
There are other gluing of solutions that are working, I won't explicit them all, but you can try to find them :p 
A last word to conclude, maybe there are also other hidden solutions, but since we do not have a theorem of unicity or tools to calculate them, we cannot do much more.
