I was investigating motorcycle suspensions and the velocity-squared damping formula
$$ m\ddot{x} = -kx - b\dot{x}^2 $$
which I read should more closely approximate the damping scenario in a simple spring coil-over damping shock. However, unlike the simpler linear damping formula $$ m\ddot{x} = -kx - b\dot{x} $$ which has closed-form solutions, the velocity-squared damping formula appears to only have those for specific cases: Harmonic oscillator with squared damping term.
Without finding a closed-form, I'm curious as to whether even the simpler (spring-less) but more general version of the damping equation $$m\ddot{x} = -bv^p$$
would even be guaranteed to have an asymptote -- i.e., have $\lim\limits_{t\to\infty} x\left(t\right)$ always converge to some value for any $p > 1$.
It definitely converges for $p = 1$, but in the case where $p>1$ the magnitude of the damping force $\left|bv^p\right| < \left|bv\right|$ when v < 1. So, with less damping, can $\lim\limits_{t\to\infty} x\left(t\right)$ still converge or will the object be crossing an infinite distance as $t\to\infty$ ?
$v = \dot{x}$ satisfies the first-order equation $\dot{v} = - b v^p$, which is separable and (with initial condition $v(0)=0$) has solution
$$ v(t) = (b(p-1)t + c)^{-1/(p-1)} $$
where $c = v_0^{1-p}$. Note that if $v_0 > 0$ and $p > 1$, this is defined for all $t > 0$, and $v(t) \sim k t^{-1/(p-1)}$ as $t \to \infty$ where $k$ is a positive constant. We want to know if
$$\lim_{t \to \infty} x(t) - x(0) = \int_0^\infty v(t)\; dt $$ converges. This is true if and only if $-1/(p-1) < -1$, which is equivalent to $1 < p < 2$.