# Harmonic oscillator with time dependent friction term

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 $$C^\infty$$.

• I suppose that $F$ is time depending and since this come from physics i suspect that $F$ is $C^{\infty}$ Commented Apr 2, 2019 at 22:05
• 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. Commented Apr 2, 2019 at 22:11
• That's interesting! Could you perhaps elaborate on why this is? It is not obvious to me. Commented Apr 3, 2019 at 21:38
• @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? Commented Apr 4, 2019 at 10:08
• 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.
– Chip
Commented Apr 9, 2019 at 3:20

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.

• 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: $E(t) = x'^2(t)/2 + x(t)^2/2$.
– Chip
Commented Apr 8, 2019 at 5:22
• 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 $(0,0)$.
– Chip
Commented Apr 8, 2019 at 9:43
• 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 $\infty$.)
– Chip
Commented Apr 9, 2019 at 3:12
• (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 $e^{-t^{1-\alpha}/(1-\alpha)}$
– Chip
Commented Apr 9, 2019 at 3:17