Deciding whether solution to ODE is unique I have an ODE:
$$y' = y^{1/4}$$ for $y \ge 0$.
and $y' = 0$ for $y < 0$.
The initial condition is $y(a) = b$.
I'm trying to use Picard's theorem to decide whether there is a unique solution in the cases:


*

*$(a,b) = (0,0)$

*$(a,b) = (0,1)$

*$(a,b) = (1,0)$.


For the first case I can see that the function is not Lipschitz around the origin and hence we are not guaranteed a unique solution. In fact we can see that $y = (\frac{4}{3}x)^{3/4}$ is a solution, as is $y=0$.
For the second case, the ODE satisfies all the conditions of Picard's theorem (provided we choose a suitable rectangle that stays away from the origin) and hence there is a unique solution.
Now I am struggling with the third case, to me it seems like it should be exactly the same as the 1st case, however I'm not sure what happens when we alter the initial value of $x$. 
Thanks for any help

For all $C\in ]-\infty, -1]$, consider the function $\varphi_C\colon \mathbb R\to \mathbb R$ given by $$\varphi_C(x)=\begin{cases} \left(\dfrac 3 4\right)^{4/3}(x+C)^{4/3}, &\text{ if }x\ge -C\\ 0, &\text{ if }x\leq -C\end{cases}$$
These are solutions to the IVP $y'=y^{1/4}, y(1)=0$.
First check that $\varphi _C$ is differentiable (at the eventually problematic point $-C$): $$\lim \limits_{x\to -C^+}\left[\dfrac{\varphi_C(x)-\varphi_C\left(-C\right)}{x+C}\right]=\lim \limits_{x\to -C^+}\left[\left(\dfrac 3 4\right)^{4/3}\dfrac{(x+C)^{4/3}}{x+C}\right]=0.$$
Since $C\leq -1$, the initial condition is verified.
Finally check that this function is a solution to the differential equation:
$$\forall x\in \mathbb R\left[\varphi_C'(x)=\left(\dfrac 3 4\right)^{4/3}\dfrac 43(x+C)^{1/3}=\left(\dfrac 3 4\right)^{1/3}(x+C)^{1/3}=(\varphi_C(x))^{1/4}\right].$$
In fact with the abuse $C=-\infty$, these are all solutions.
Alternatively and in view of Hans Lundmark's comment take any two different solutions to the first IVP and use this answer.
