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?

  • $\begingroup$ $α$ 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
  • 1
    $\begingroup$ 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

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 " $$\int_0^t=\int_0^A+\int_A^t$$

| cite | improve this answer | |
  • 1
    $\begingroup$ This is very elegant, thanks a lot for sharing that solution with me. $\endgroup$ – Spaced Oct 6 '15 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.