When is the solution to $x' = \sqrt{|x|} +x^2$ unambiguous and global? I have a short question connected with the differential equation
$$
x'=\sqrt{|x|}+x^2,
$$
for $t>0$, $x(0)=x_0$, $x_0 \in \mathbb{R}$. Namely -- when does this equation have unambiguous solution and when does it have solution which is global in time (for all $t>0$)?
I managed to obtain the equation and it is (in implicit form)
$$
\dfrac{1}{\sqrt{3}}\arctan\left(\dfrac{\sqrt{|x|}-\frac12}{\sqrt{\frac34}}\right) +\dfrac13\ln\left(|x|-\sqrt{|x|}+1\right)-\dfrac13\ln \left| 1+|x|\sqrt{|x|} \right|=t+c,
$$
where $c$ is a constant. I do not know how to answer given question with this expression.
 A: Note that the derivative of $\sqrt{|x|} + x^2$ is not bounded at $x=0$, the uniqueness theorem is not applicable. In particular, the formula you obtained (using separation of variables) will misses a lot of other solutions.
The problem is at $x=0$. Since $x'$ is always non-negative, if $x(0) = x_0>0$, then $x$ is strictly increasing and $x(t) >0$. In particular,
$$x' = \sqrt{|x|} + x^2 > c_0 + x^2, $$
where $c_0 = \sqrt{x_0}$. Dividing by $\sqrt{|x_0|} + x^2$ and integrate from $0$ to $t$ gives
$$\frac{1}{\sqrt{c_0}} \arctan \left( \frac{x}{\sqrt{c_0}}\right) \ge t + \frac{1}{\sqrt{c_0}} \arctan \left( \sqrt{c_0}\right)
$$
which gives
$$x \ge \sqrt{c_0} \tan \left(\sqrt{c_0} t + \arctan \left( \sqrt{c_0}\right)\right)$$
and since $\tan$ blows up in finite time, the same holds for $x$ and thus $x$ cannot be defined for all $t>0$.
When $x_0 <0$, then use
$$x' = \sqrt{|x|} +x^2 > \sqrt{|x|} = \sqrt{-x}$$
whenever $x<0$. Integrating again gives
$$-2\sqrt{-x(t)} +2\sqrt{-x_0} > t\Rightarrow x(t) > -\frac{1}{4} (t-2\sqrt{-x_0})^2. $$
That is, $x(t_1) = 0$ for some $t_1 < 2\sqrt{-x_0}$. Once the solution reaches $x=0$, one can easily construct two solutions: for example
$$ x_1(t) = \begin{cases} x(t) & \text{ if } t \le t_1, \\0 & \text{ if } t > t_1\end{cases}$$
and
$$ x_2(t) = \begin{cases} x(t) & \text{ if } t \le t_1,\\ -x(2t_1-t) & \text{ if } t > t_1\end{cases}$$
The case $x_0 = 0$ can be done similarly as in the case for $x_0 <0$: you can find two solutions.
Thus the condition is that, depending $x_0$, the equation either has non-unique solution, or don't have global solution.
