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 $t\to\infty$ ?

  • $\begingroup$ It is separable but it is a second-order ODE with no analytic solution like in math.stackexchange.com/questions/60702/… . However, I'm not looking for an analytic solution, just would like to see if an asymptote is always guaranteed under my initial conditions. Also, it looks like my question received close votes for lack of context/background details, so I will add those in. $\endgroup$
    – ManRow
    Jul 19, 2020 at 20:51
  • 1
    $\begingroup$ Wolfram gives an analytical solution. $\endgroup$ Jul 19, 2020 at 20:57
  • $\begingroup$ @ManRow: You're new here so you likely do not know that you should add followup information in the question itself, not buried in comments. $\endgroup$ Jul 19, 2020 at 21:54
  • $\begingroup$ @PeterForeman Ahh, I see now. I mistakenly assumed that since math.stackexchange.com/questions/1649788/… had no closed form (except for a few specific cases), then also my simpler equation (with the squared-damping term) was unsolvable as well. This was an incorrect assumption, as you pointed out it actually does have a pretty simple closed form too -- and, as Robert Israel's answer points out, converges too for the specific case of 1 < p < 2. $\endgroup$
    – ManRow
    Jul 19, 2020 at 22:01
  • $\begingroup$ Regarding the first equation, since the system $(\dot x, \dot v) = (v, -k x/m - b v^2/m)$ is symmetric wrt the $x$-axis and the origin is a center of the linearized system, the origin is also a center of the nonlinear system. Therefore $x$ is periodic for sufficiently small $x_0$ and $v_0$. $\endgroup$
    – Maxim
    Jul 26, 2020 at 19:00

1 Answer 1


$v = \dot{x}$ 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 $1 < p < 2$.

  • $\begingroup$ Interesting! I think it should be $1 \le p < 2$, since it converges for the standard $p=1$ case as well. $\endgroup$
    – ManRow
    Jul 19, 2020 at 22:31
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    $\begingroup$ $p=1$ is a special case, where my formula for $v(t)$ doesn't work. But the OP specifically requested $p > 1$. $\endgroup$ Jul 20, 2020 at 3:15

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