# Second Order non-autonomous ODE - prove the solution is unbounded on [0,$\infty$)

I'm working on the problem below and have a known error

I've basically "proven" that as $t \to \infty$ the solution decouples to negative infinity. However, we have as assumption that $\phi (t) \to 0$ as $t \to \infty$

Can someone find where I went wrong?

• Can you add which textbook you use at this (if any...)? – Chip Mar 2 '16 at 2:15
• It's a problem the professor gave us. It isn't from a book as far as I know. – Phillip Hamilton Mar 2 '16 at 2:30
• I have a suggestion for another proof (unfortunately not about your proof, since I am not familiar to the Gronwall's inequality), but you have to work out the details: with $y=\exp{P}$ you get a Ricatti differential equation, for which you already know a solution. So the Ricatti then can be reduced to a first order ODE in the unknown solution (the one you want to study the properties). For the latter, you can see here the details: math.stackexchange.com/questions/462075/…. Maybe this is useful to you... – Chip Mar 2 '16 at 2:41

It is correct that $$\int_{s=0}^t a(s)e^{\int_0^t a(u)\,du}\,ds\,\le\,\int_{s=0}^t Me^{\int_0^t M\,du}\,ds,$$ but you multiply the integral with $-\varphi_0'$ which might be negative. In this case, your inequality is false.
• The integral form of Gronwall's inequality does not make assumptions as to the sign of $\alpha (t)$ en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality – Phillip Hamilton Mar 2 '16 at 3:26
• So what? You have $\int_{s=0}^t a(s)e^{\int_0^t a(u)\,du}\,ds\le\int_{s=0}^t |a(s)|e^{\int_0^t a(u)\,du}\,ds\le\int_{s=0}^t |a(s)|e^{\int_0^t |a(u)|\,du}\,ds\le\int_{s=0}^t Me^{\int_0^t M\,du}\,ds$. As I wrote, the problem is the $-\varphi_0'$ in front of the integral. Note that if $0 < s < t$ then $-5s > -5t$. – Friedrich Philipp Mar 2 '16 at 4:09