Find $a$ such that the existence and uniqueness theorem applies to the ivp I want to find all the real constants $a$ such that the existence and uniqueness theorem applies to the ivp $x'=\frac{3}{2}|x|^\frac{1}{3}, x(0)=a$.
My approach is finding for which constants $a$ the function $f(x)=\frac{3}{2}|x|^\frac{1}{3}$ is lipschitzian at $x=a$. Is this a correct approach? If yes, could you help me showing which $a$ $f(x)$ is atleast locally lipschitzian?
edit:
$f(x)$ is continuously differentiable wrt $x$ for all $x\neq0$ and there exists a ngbh at those points where it's derivative is bounded, implying $f(x)$ being lipschitzian at all points $x \neq 0$. Am i correct?
 A: All what you say is correct. For all $x \neq 0$, $f$ is continuously differentiable in a neighborhood of $x$, hence locally Lipschitz continuous in such a neighborhood of $x$. The hypothesis of Picard–Lindelöf theorem are fulfilled and you can state the existence and uniqueness of the solution to the IVP:
$$x'=\frac{3}{2}|x|^\frac{1}{3}, x(0)=a.$$
It remains to study the case $a=0$. As Picard–Lindelöf theorem only involves a material implication (and not an equivalence), you CANNOT state that (at least) a solution doesn't exist or is not unique directly.
However,$f$ is continuous at $x=0$ and you can invoke Peano existence theorem to state that a solution exists. But uniqueness is not guaranteed.
A trivial solution is the always vanishing map. If you try to solve directly the ODE by separating the variables, you'll find an infinity of other solutions. Namely
$$x(t) = \begin{cases}
0 & \text{for } t \le c\\
(t-c)^{3/2} & \text{for } t \ge c
\end{cases}$$
For all $c >0$ are also solutions. I let you discover what happens using $c < 0$.
A: Your approach is correct. It generates the answer $a>0$. 
If $a>0$, then the unique solution of the ivp is
$$x(t)=(a^{2/3}+t)^{3/2}.$$
Note that $x(t)>0$ holds for all $t$ such that the function $f$ is Lipschitz (in fact differentiable) at $x(t)$ for all $t$.
If $a=0$, then there are at least two solutions, namely $x_1(t)=0$ for all $t$ and $x_2(t)=t^{3/2}$ for all $t$.
Finally, if $a<0$, then there exist also at least two solutions, namely
$$x_1(t)=\left\{\begin{array}{cl}-(|a|^{2/3}-t)^{3/2}&\mbox{ if }t\leq |a|^{2/3}\\
0&\mbox{ if }t>|a|^{2/3},\end{array}\right.$$
and 
$$x_2(t)=\left\{\begin{array}{cl}-(|a|^{2/3}-t)^{3/2}&\mbox{ if }t\leq |a|^{2/3}\\
(t-|a|^{2/3})^{3/2}&\mbox{ if }t>|a|^{2/3},\end{array}\right.$$
