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Problem: Let $y \in C^1([0,+\infty), \mathbb{R})$ and $\alpha >0$. Prove that $$ \lim_{t \to + \infty}(y'(t)+\alpha y(t))=0 \implies \lim_{t \to +\infty}y(t)=0$$

I will show two attempts I have made, both of them didn't work out for me

1st Attempt: Let $\alpha > 0$ such that $$\lim_{t \to + \infty} (y'(t) + \alpha y(t))=0 \\ \implies \forall \epsilon >0, \exists S \in \mathbb{R}: \forall t \in [0, \infty) \text{ with } t\geq S \implies y'(t)+\alpha y(t) \leq \epsilon $$

But I can easily work with $y'(t)+ \alpha y(t) \leq \epsilon$ by multiplying it with $e^{\alpha t}$ one easily finds that $$ \left(y(t)e^{\alpha t}\right)' \leq \epsilon e^{\alpha t} $$ Integration of both sides and using the fact that integration preserves the inequality I obtain that $$y(t) \leq \frac{\epsilon}{\alpha}+ C \exp(-\alpha t), \ \text{ for } C \in \mathbb{R} \\ \implies |y(t)| \leq \frac{\epsilon}{\alpha}+|C| \exp(-\alpha t) $$ Which looks promising at first, because as $t \to + \infty$ the most right term vanishes, however the fraction $\epsilon / \alpha$ doesn't provide any useful information, because possibly $\alpha, \epsilon$ are arbitrarily, so I am not sure if it is a rigorous statement to just say "$\epsilon$ can always be smaller than $\alpha$"

2nd Attempt: I was hoping that with the help of Gronwall's inequality I could get rid of the missing rigor of the above attempt

Gronwall Inequality: Assume that $y'(t) \leq f(t) + g(t)y(t)$ then we have $$ y(t) \leq y(a) \exp \left( \int_a^t g(s)ds\right) + \int_a^t f(s) \exp \left( \int_s^t g(r)dr \right) ds $$

So I did use again the only thing that I know which is that $y'(t) + \alpha y(t) \leq \epsilon $ and to apply Gronwalls Lemma I did set $f(t)= \epsilon$ and $g(t)=- \alpha$ for all $t \in [0, \infty)$

Doing the necessary integration for Gronwalls Inequality I obtain that $$ y(t) \leq y(a) \exp(-\alpha (t-a)) + \frac{\epsilon}{\alpha} \left(1-\exp(-\alpha(t-a) \right)$$ Which somehow just shows that my 1st attempt was equally good/bad as my 2nd attempt.

Any hints? Corrections?

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  • $\begingroup$ $α$ and $y$ are given constants of the problem. Thus there is no problem varying $ϵ$. -- Note that you also need to prove the lower bound. -- And your first calculation follows one of the proof strategies for Gronwalls lemma, thus it is no coincidence that the results match. $\endgroup$ – Lutz Lehmann Oct 6 '15 at 12:08
  • $\begingroup$ @LutzL I thought that my Calculations in the 1st Attempt give both because $$ |y(t)| \leq \frac{\epsilon}{\alpha}+ |C| \exp (-\alpha t) $$ from which I would have concluded, that for varying small epsilon the limit goes to zero (by definition) $\endgroup$ – Spaced Oct 6 '15 at 12:12
  • $\begingroup$ Maybe for clarification I should add that I read $$\frac{\epsilon}{\alpha}= \epsilon \frac{1}{\alpha} \approx 0 * \infty = ?$$ which might cause all the confusion $\endgroup$ – Spaced Oct 6 '15 at 12:24
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    $\begingroup$ No, you can not use this as a direct argument. However, you can exchange $y$ for $-y$ without changing the calculation, which gives the lower bound. -- And no, since $α$ is a fixed number, $1/α$ is finite. $\endgroup$ – Lutz Lehmann Oct 6 '15 at 12:24
  • $\begingroup$ @LutzL thanks for explaining that to me $\endgroup$ – Spaced Oct 6 '15 at 12:42
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let $f=y'+\alpha y$ hence $y$ is a solution of the ODE $y'+\alpha y=f$. By resulving this equation we abtain $y(t)=\lambda e^{-\alpha t}+[\int_0^tf(s)e^{\alpha s}ds]e^{-\alpha t}$.

The first terme is Ok. for the second fixe $\epsilon>0$ and $A\geq 0$ s.t $|f(s)|\leq \epsilon /2$ for all $s\geq A$. And "cut your integral " $$\int_0^t=\int_0^A+\int_A^t$$

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    $\begingroup$ This is very elegant, thanks a lot for sharing that solution with me. $\endgroup$ – Spaced Oct 6 '15 at 12:36

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