I found the following exercise on a test:

Show that the maximal solutions of $x' = t^2 + x^2$ are defined on a bounded interval.

Using the preceding part:

Let $f_1, f_2$ smooth functions over $\mathbb{R}^2$ to $\mathbb{R}$ such that $f_1(t,x) < f_2 (t,x) \ \forall (t,x) \in \mathbb{R}^2$. Show that if $\varphi_i$ is solution to $x' = f_i(t,x)$ and there exist $t_0$ sucht that $\varphi_1(t_0) = \varphi_2(t_0)$ then $\varphi_1(t) \leq \varphi_2(t)$ if $t \geq t_0$ and $\varphi_1(t) \geq \varphi_2(t)$ if $t \leq t_0$.

I had no troubles with the second statement, but with the first I take $f_1(t,x) = x^2$ and $f_2(t,x) = t^2 + x^2$ and then $f_1(t,x) < f_2 (t,x)$ (it's not exactly the same that above, but works). Now, the maximal solutions to $x' = x^2$ are $$\varphi(t)=\left\{\begin{array}{lr} \frac{1}{\frac{1}{x_0} + t_0 - t} & \mbox{if } x_0 \neq 0\\ 0 & \mbox{if } x_0 = 0\\ \end{array}\right.$$ where $(t_0,x_0)$ is the initial condition and so if you take $x_0 > 0$ then the maximal solution $\varphi_0$ of $x' = t^2 + x^2$ by $(t_0,x_0)$ holds $\varphi_0 \geq \varphi$, which is defined up to $\frac{1}{x_0} + t_0 > t_0$ and this show that $\varphi_0$ is also defined up to $\frac{1}{x_0} + t_0$.

But with the other side I couldn't give a bound for the domain of $\varphi$ of the opposite side, similarly if $x_0 < 0$ and if $x_0 = 0$ I have no idea. Can somebody give an approach with this problem?

  • $\begingroup$ This does not answer your question (hence a comment), but there was a recent discussion of this ODE. $\endgroup$ – user53153 Jan 23 '13 at 22:53

If I understand correctly, you want to prove that every solution blows up in finite time, going both forward and backward in time.

Forward in time: since $x'\ge t^2$, we have $x(t)-x(t_0)\ge \frac13(t^3-t_0^3)$. In particular, the solution cannot be bounded above. Once the solution reaches a positive value, say $x(t_1)=1$, we use the comparison $x'\ge x^2$ to obtain $x(t)\ge (t_1+1-t)^{-1}$, which means $x$ blow up no later than $t_1+1$.

Backward in time: let $y(t)=-x(-t)$. Since $y'=x'(-t)=x(-t)^2+(-t)^2=y^2+t^2$, the solution $y$ blows up going forward in time. In terms of $x$, this means blow-up going backward.

  • $\begingroup$ I thought in you did for forward in time then I asked it, but what you did for backward in time it's incredible. Thank you very much. $\endgroup$ – Diego Silvera Jan 24 '13 at 0:43

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