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

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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?

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  • $\begingroup$ Can you add which textbook you use at this (if any...)? $\endgroup$ – Chip Mar 2 '16 at 2:15
  • $\begingroup$ It's a problem the professor gave us. It isn't from a book as far as I know. $\endgroup$ – Phillip Hamilton Mar 2 '16 at 2:30
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    $\begingroup$ 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... $\endgroup$ – Chip Mar 2 '16 at 2:41
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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.

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  • $\begingroup$ 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 $\endgroup$ – Phillip Hamilton Mar 2 '16 at 3:26
  • $\begingroup$ 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 $-5s > -5t$. $\endgroup$ – Friedrich Philipp Mar 2 '16 at 4:09

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