The differential equation $\dot{x} = \sqrt{|x|}$ Consider 
\begin{equation} \label{hh}
\dot{x} = \sqrt{|x|} \tag{*}
\end{equation}
a) Show that 
$$ x =  \left\{ \begin{array}{cc} 
t^2/4 & t \geq 0 \\
0 & t <0
\end{array} \right.$$
is a solution of \eqref{hh}. 
b) Are initial value solutions of \eqref{hh} unique?
c) Is it true that any solution of \eqref{hh} is monotonically increasing, decreasing, or constant?

It is part c) that I find difficult. Here are my suggestions for a) and b).
a) By inserting the given solution in \eqref{hh} it is concluded that the given x is a solution.
b) Considering the IVP with $x(0) = 0$ it is seen that besides the solution given in the exercise we also have the solution $x_2(t) = 0$, $\forall t \in \mathbb{R}$. Thus it is not true that the solutions in general are unique. This is due to the fact that $\sqrt{|x|}$ is not continous differentiable.
c) I know that the statement is true if the differential equation is continous differentiable, but here it is not. I therefore assume that the answer is no (although I'm not sure at all). Can someone help me to actually show it?
 A: You know that if a differentiable (not necessary continuously differentiable) function $f$ is such that $\dot{f}\geq 0$, then $f$ is increasing (simply write $\frac{f(y)-f(x)}{y-x}=\dot{f}(c)\geq 0$ for some $c$ in $(x,y)$). Here this result applies since $\dot{f}=\sqrt{|f|}\geq 0$.
A: By the definition of the square root, $\dot x\ge0$ and the solution is either increasing or constant.
The constant solution $x=0$ is possible, with the initial condition $x(0)=0$.
Other solutions must be increasing as of $x>0$.

For $x<0$, $\dfrac{\dot x}{\sqrt{-x}}=1$ and $-2\sqrt {-x}=t+c$, which is possible for $t+c\le0$. For $x>0$, $\dfrac{\dot x}{\sqrt{x}}=1$ and $2\sqrt {x}=t+c'$, which is possible for $t+c'\ge0$.
Hence for $x$ to be a function defined over the whole of $\mathbb R$, one must have $c=c'$*.
There are four possible modes:
$$\begin{align}x(t)&=0,\\x(t)&=-\frac{(t+c)^2}4\text{ then }0,\\x(t)&=0\text{ then }\frac{(t+c)^2}4,\\
x(t)&=-\frac{(t+c)^2}4\text{ then }\frac{(t+c)^2}4,\end{align}$$ for some $c$.

*Update: my statement is wrong, see MartinR's comment.
