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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.

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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.

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  • $\begingroup$ 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 $\endgroup$ – edo1998 Jan 20 '19 at 20:01
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    $\begingroup$ 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. $\endgroup$ – Lutz Lehmann Jan 20 '19 at 20:05
  • $\begingroup$ Yes, I understand. I will check the Riccati equations because I don't know them $\endgroup$ – edo1998 Jan 20 '19 at 20:07

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