# Show that solution of a Cauchy problem is not globally defined

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?

• No, it doesn't work. As $t \to 1-$, $1/(t-1) \to -\infty$. So all you have shown is that $\lim_{t\to 1-} u(t) > -\infty$. – Stephen Montgomery-Smith Aug 24 '20 at 23:28
• @StephenMontgomery-Smith But if I take the modulus, I have shown that the norm explodes, right? – andereBen Aug 24 '20 at 23:29
• No. $u(t)$ will be negative. You cannot say $x<y$ implies $|x| < |y|$ unless you know $x \ge 0$. – Stephen Montgomery-Smith Aug 24 '20 at 23:30
• Find an $\epsilon$ such that $v(\epsilon) > 1$. Then use $v' > v^2 - 1$. – Stephen Montgomery-Smith Aug 24 '20 at 23:56
• I think you need to step away from stack exchange, and think long and hard about this prolem. – Stephen Montgomery-Smith Aug 25 '20 at 0:20

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.