# Prove that maximal solution of a Cauchy problem has domain $(-\infty,b)$

Given this Cauchy problem $$(P): \begin{cases} y'=f(y,t)=y|y|-t^2 \\ y(0)=0 \end{cases}$$ I have to prove that , called $$\overline y$$ the maximal solution of $$(P)$$, $$dom(\overline y)=(-\infty,b)$$ for some $$b>0$$. Moreover i need to demonstrate that $$\lim_{t\to-\infty}=+\infty \text{ and }\lim_{t\to b^{-}}=-\infty$$ I was able to demonstrate that the solution exist and is unique in a neighborhood of $$t=0$$ because $$\displaystyle \frac{\partial f}{\partial y}=2|y|$$ is continuous. I have found that $$\overline y$$ must be decreasing in all its domain (because $$y'\geq0 \iff y\geq|t|$$) but still I can't proceed to demonstrate none of the requests. I believe that I should use some sort of comparison but the only thing I came up with was this disequation: $$y'=y|y|-t^2\leq-y^2 \quad \text{for } \ t>0$$ and I know that a Cauchy problem with $$x'=-x^2$$ blow up in finite time but still to apply the comparison theorem I know I need to have $$x(0)\geq y(0)=0$$. The problem is that if $$x(0) \geq 0$$ the problem blow-up to $$+\infty$$ and this give me nothing about $$\overline y$$. I hope I have explained my problem well, thanks in advance for the help.

You know that $$y(t)<0$$ for $$t>0$$. Then you can use for some small time $$y'\le-t^2\implies y(t)\le-\frac{t^3}3$$ so that for instance $$y(1)\le-\frac13$$ which you can then use with your original inequality or strengthen it to $$y'\le -y^2-1 ~~\text{ for }~~ t\ge 1$$ giving a shifted tangent function as upper bound.
• Why I need to use it for some small time? Doesn't it hold for every t>0? And also how can I prove for example that the solution is defined for any $t<0$? Thank you for your help – edo1998 Jan 20 '19 at 20:01
• It does not matter where you make the cut, you get always an upper bound that proves the solution to have a pole. If the cut is made closer to zero, the bound on $b$ will be somewhat stricter. If you want really tight bounds, treat the equation as a Riccati equation with the substitution $y=\pm\frac{u'}{u}$ which then has solutions in Bessel functions or their easily calculated power series. See Riccati D.E., vertical asymptotes (and linked topics) for a similar question. – Lutz Lehmann Jan 20 '19 at 20:05