It is well known that for a harmonic oscillator with linear damping,
$$\ddot x+c\dot x+x=0$$
with positive $c$, the amplitude of the oscillations decays exponentially when $c<2$. If it is higher than $2$, the system fails to oscillate at all and is said to be overdamped.
Suppose the damping is nonlinear instead, following a power law
$$\ddot x+c\lvert \dot x\rvert^{p-1}\dot x+x=0.$$
For example, $p=1$ recovers linear damping, while $p=2$ gives quadratic damping which can model aerodynamic drag. I assume that in general a closed-form solution is not possible due to the presence of the absolute value signs. What can be said about the asymptotic behaviour of the system?
Edit: While @doraemonpaul's comment and @mjqxxx's answer are very nice, I am more interested in stronger results than merely the existence or absence of overdamping. For comparison, consider a first-order nonlinear decay equation,
$$\dot x+\lvert x\rvert^{p-1}x=0.$$
The solution to this has the form $x = \pm(p-1)(t-t_0)^{1/(1-p)}$ with certain conditions on $t_0$. When $p<1$, the solution drops to zero in finite time; when $p>1$, it decays roughly as $t^{-1/(p-1)}$ which is much slower than exponential. What are the corresponding characterizations of how the amplitude of the nonlinearly damped harmonic oscillator behaves? What is the exponent of the decay when $p > 1$? Can the system come to rest in finite time if $p < 1$?
Best Answer
So, following the usual treatment of the harmonic oscillator, note that the equations of motion for $x$ and $v$ are $$ \begin{eqnarray} \dot{x} &=& +v \\ \dot{v} &=& -x - c|v|^{p}v, \end{eqnarray} $$ where the damping term is clearly small for small $v$ when $p>1$. In the two-dimensional phase space, the equations of motion become particularly simple in radial coordinates: letting $x=r\cos\theta$ and $v=r\sin\theta$, we have $$ \begin{eqnarray} \dot{r}\cos\theta - r\dot\theta\sin\theta &=& +r\sin\theta \\ \dot{r}\sin\theta + r\dot\theta\cos\theta &=& -r\cos\theta - cr^p\lvert\sin\theta\rvert^{p-1}\sin\theta, \end{eqnarray} $$ or, solving the system, $$ \begin{eqnarray} \dot{r} &=& -c r^{p}\lvert\sin\theta\rvert^{p-1}\sin^{2}\theta\\ \dot{\theta} &=&-1-cr^{p-1}\lvert\sin\theta\rvert^{p-1}\sin\theta\cos\theta. \end{eqnarray} $$ For $p>1$, the motion is always underdamped. Since $r$ is always decreasing, there comes a time where $r$ is small enough that we can guarantee that $|\dot{r}|<ar$ and $\dot{\theta}<-1+\varepsilon$ for any positive $a$ and $\varepsilon$ and for all later times, and hence $\cos\theta$ (and $x$) will change signs arbitrarily many times before $r$ reaches zero.