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Hi everyone: I need a check on the following exercise

\begin{cases} u'(t)=\sin(t)^2 - u(t)^2 \\ u(0)=-1\end{cases}

Show that the maximal solution $u:[\alpha,\beta] \rightarrow \mathbb{R}$ is such that $\beta < \infty$


First of all, the problem is well posed: the r.h.s. of the ODE is smooth and therefore we can infer there exists a unique local solution. Then, I note immediately that $$\sin(t)^2-u^2(t) > - u^2(t)$$ for all the $t$'s for which the solution is defined

Hence $$u'>-u^2$$ and integrating from $0$ to $t$ I obtain $$\frac{1}{u(t)} + 1 < t$$ and then $$u(t) > \frac{1}{t-1}$$ again for all $t$'s such that the solution is defined. But the r.h.s explodes as $t \rightarrow 1^{-}$ and therefore we can say that the $$\beta<1$$ is necessary for the solution norm not to explode


This seems to be confirmed also numerically. Is everything okay?

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    $\begingroup$ No, it doesn't work. As $t \to 1-$, $1/(t-1) \to -\infty$. So all you have shown is that $\lim_{t\to 1-} u(t) > -\infty$. $\endgroup$ – Stephen Montgomery-Smith Aug 24 '20 at 23:28
  • $\begingroup$ @StephenMontgomery-Smith But if I take the modulus, I have shown that the norm explodes, right? $\endgroup$ – andereBen Aug 24 '20 at 23:29
  • $\begingroup$ No. $u(t)$ will be negative. You cannot say $x<y$ implies $|x| < |y|$ unless you know $x \ge 0$. $\endgroup$ – Stephen Montgomery-Smith Aug 24 '20 at 23:30
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    $\begingroup$ Find an $\epsilon$ such that $v(\epsilon) > 1$. Then use $v' > v^2 - 1$. $\endgroup$ – Stephen Montgomery-Smith Aug 24 '20 at 23:56
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    $\begingroup$ I think you need to step away from stack exchange, and think long and hard about this prolem. $\endgroup$ – Stephen Montgomery-Smith Aug 25 '20 at 0:20
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At $t=0$ the right side is negative. It is easy to observe, perhaps harder to formulate, that this implies that at no point the right side can become positive, $u$ is continuously falling and thus always smaller than $-1$. Then for $t\in[0,1]$ $$ u'(t)=\sin^2t-u(t)^2\le t^2-1\implies u(t)\le -1-t+\frac{t^3}3=v(t). $$ so the solution moves decidedly away from the level $-1$ downward and $$ u(1)\le-\frac53. $$

Now use $\sin^2t\le 1\le -u(t)$, $u(t)^{-1}+1\ge0$ to derive a simple upper bound for the solution for $t\ge 1$, $$ u'\le -u-u^2\implies (u^{-1}+1)'=-u^{-2}u'\ge u^{-1}+1 \\~\\ \implies u(t)^{-1}+1\ge (u(h)^{-1}+1)e^{t-h} $$ So in the numerically simple case $h=1$ one gets for $t\ge1$ $$ u(t)^{-1}+1\ge\frac25e^{t-1} \iff u(t)\le-\frac{5}{5-2e^{t-1}} $$ This upper bound diverges to $-\infty$ at $t=1+\ln\frac52$, thus forces the solution to diverge in this direction at some time before that.

This is a rather large bound, as the numerical solution diverges shortly after $t=1$, closely following the lower bound $-\frac1{1-t}$ discussed in the question.

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