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 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.