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Suppose I have a harmonic oscillator of the following form:

$\ddot x(t)=-F(t)\dot x(t)-x(t), F(t)>0$ for all $t$.

From the physical perspective, the term proportional to $\dot x(t)$ represents a friction term. Hence, if $F(t)>0$ for all $t$, I would expect that $\lim_{t\to\infty} x(t)=0$. But I am not sure how to proof that, and perhaps it is not even true, and my thinking is too simplistic.

If someone had some thoughts, I would appreciate! : )

EDIT: I can probably also assume $F(t)$ to be $C^\infty$.

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    $\begingroup$ I suppose that $F$ is time depending and since this come from physics i suspect that $F$ is $C^{\infty}$ $\endgroup$
    – DINEDINE
    Commented Apr 2, 2019 at 22:05
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    $\begingroup$ It is time dependent, yes. And it most certainly $C^\infty$. I don't know too much concrete things about $F$ though. But if I could show something with the assumptions $F(t)>0$ and $C^\infty$ that would be already a good start, if not sufficient for my purposes. $\endgroup$
    – Britzel
    Commented Apr 2, 2019 at 22:11
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    $\begingroup$ That's interesting! Could you perhaps elaborate on why this is? It is not obvious to me. $\endgroup$
    – Britzel
    Commented Apr 3, 2019 at 21:38
  • $\begingroup$ @Chip This sounds like something I should try! (Would also be a chance for me to warm up some complex analysis ; ).) One question to the condition that $F(t)$ should be finite at infinity: Here you mean $t\to\infty$ (not $u\to\infty$), right? $\endgroup$
    – Britzel
    Commented Apr 4, 2019 at 10:08
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    $\begingroup$ Your problem is thoroughly analyzed in Cabot $\it{et \; al.}$, TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 361, Number 11, November 2009, Pages 5983–6017. I added some details from the paper in a comment to the answer below. $\endgroup$
    – Chip
    Commented Apr 9, 2019 at 3:20

1 Answer 1

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I will show that as long as $F(t) \geq 0$ and $\int_0^\infty F(t){\rm d}t < \infty$ then we won't have $\lim_{x\to \infty}x(t) = 0$ unless $x\equiv 0$.

Consider the energy of the oscillator, $E(t) = x^2(t) + x'^2(t)$. Using the ODE this is found to satisfy
$$E'(t) = -2F(t)x'^2(t) \geq -2F(t) E(t)$$ Since the energy is decreasing and bounded below by $0$ it will have a finite limit $E_\infty$ as $t\to\infty$. Thus the system will always settle into a circle $x^2 + x'^2 = E_\infty$ in the $(x,x')$ phase-space asymptotically and $x\to 0$ is the case only if $E_\infty = 0$. Integrating the inequality above we obtain $$E_\infty \geq E(0) e^{-2\int_0^\infty F(t){\rm d}t}$$ which is strictly positive as long as the integral in the exponential is finite and $x\not\equiv 0$. I also suspect the converse, if this integral is infinite then the limit will be zero, could be true.

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  • $\begingroup$ For keeping consistency with physical definition, I point out that the total 'energy of the oscillator' (ie, the sum of the kinetic and potential energies) reads: $E(t) = x'^2(t)/2 + x(t)^2/2$. $\endgroup$
    – Chip
    Commented Apr 8, 2019 at 5:22
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    $\begingroup$ I did some numerical simulations and they seem to confirm that if $\int_0^\infty F(t) dt$ is finite, in the $(x(t), v=x'(t))$ plane the motion approaches a limit circle that does not include $(0,0)$. $\endgroup$
    – Chip
    Commented Apr 8, 2019 at 9:43
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    $\begingroup$ Cabot et al. TRANS. AM. MATH. SOC. vol 361, 11, Nov 2009, pag. 5983–6017 gives lower and upper bounds for the energy and a detailed analysis. Interestingly, eg, for $F(t)=c/(t+1)$ one has $k/t^c \le x(t)^2+x'(t)^2 \le K/t^c$, with $k,K$ strictly positive, finite. In general, is intriguing that there is a lower ($k$ constant) / upper ($K$ constant) bound (with $F(t)$ under mild conditions given in Prop. 3.4 in paper) of the form $k e^{-\int_0^t F(s) ds}$ / $K e^{-\int_0^t F(s) ds}$, respectively. ($F, \dot F$ must go to $0$ and $\ddot F + \dot F F $ must have constant sign at $\infty$.) $\endgroup$
    – Chip
    Commented Apr 9, 2019 at 3:12
  • $\begingroup$ (continue from above): (Example 3.2 in the paper) for $F(t)=1/(1+t)^\alpha$ and $0 \lt \alpha \lt 1$, the upper and lower bounds are proportional to $e^{-t^{1-\alpha}/(1-\alpha)}$ $\endgroup$
    – Chip
    Commented Apr 9, 2019 at 3:17

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