1
$\begingroup$

I want to show that solution of this Cauchy problem

\begin{cases} u'(t)=u(t)^2 + t \\ u(0)=0 \end{cases}

is defined for $t \in [0,\alpha]$, with $\alpha <3 $

I tried to integrate, but it's not possible because of the $t$. I can see that $u'>0$ for all $t \geq 0$ , and that's all. I don't know how to argue honestly

$\endgroup$
2
  • $\begingroup$ These equations are known as Riccatti equation $\endgroup$ – Surb Aug 23 '20 at 14:47
  • $\begingroup$ I was trying to use the substitution $u^2 = w$, as I didn't know these equations @Surb $\endgroup$ – andereBen Aug 23 '20 at 14:50
3
$\begingroup$

Note that $u(t) \geq \int_0^t sds = t^2/2$ so that $u(\sqrt{2})\ge 1$.

Since $u'(t)/u(t)^2 \ge 1$, we can integrate both sides from $\sqrt{2}$ to $t$ and we see that $-\frac{1}{u(t)}+\frac1{u(\sqrt{2})}\ge t-\sqrt{2}, $ whenever $t>\sqrt{2}$.

Thus $\frac{1}{u(t)} \le \sqrt{2}-t+\frac1{u(\sqrt{2})}$. We know $u(\sqrt{2})\geq 1$, so that $\frac{1}{u(\sqrt{2})}\leq 1$.

Thus, if $t \ge \sqrt{2}$ then $u(t)>\frac{1}{\sqrt{2}+1-t}$. Therefore the blow-up time is at most $\sqrt{2}+1<3$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks @shalop, exactly what I was looking for. So, when I am looking for the blow up time, it's convenient to try to bound from below the solution and see whether or not something goes wrong. $\endgroup$ – andereBen Aug 23 '20 at 16:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.