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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 $p < 1$?

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Doesn't the end behavior depend on the signs of the damping factor and k? You can have A LOT of cases depending on the initial conditions as well. Also, I think it would be better to write $c = |x'|^{p-1}$ which shows that the damping factor is varying. But that's jsut my take. –  jip Jul 7 '12 at 4:01
    
@jak: Fair enough. I removed the irrelevant and distracting coefficients. I didn't replace $c$ with $c\lvert\dot x\rvert^{p-1}$ because I prefer $c$ to represent a constant, but I brought the two factors closer together. –  Rahul Jul 7 '12 at 4:12
    
@jak: How many cases is "A LOT"? The phase space is just two-dimensional after all, and zero is the only attractor. I'd be surprised if there turned out to be a lot of different kinds of behaviour. But if there are, I want to know about them. –  Rahul Jul 7 '12 at 4:28
    
This question is really interesting. For example the simple pendulum motion with quadratic damping is always under-damped. See hk.knowledge.yahoo.com/question/question?qid=7010122600228 for detail explanation. –  doraemonpaul Jul 7 '12 at 4:33
    
@doraemonpaul: Thanks, I'll have to read that carefully. Does the solution there allow $x$, or rather $\theta$, to be obtained as a function of $t$? I'm interested in how $\theta$ decays as $t \to \infty$. –  Rahul Jul 7 '12 at 4:55

1 Answer 1

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.

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Thanks for this nice analysis. I'm afraid I am hoping for some stronger results, though; please see my edited question. –  Rahul Jul 7 '12 at 6:58
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@RahulNarain: Okay. Well, the $p>1$ case is still fairly simple: the adiabatic approximation applies as $r\rightarrow 0$, since $r$ is changing much more slowly than $\theta$, so you can approximate $\dot{r}$ by its average over a period of $\theta$. Then $\dot{r} \sim -c'r^{p}$ (where $c'=\langle \lvert\sin \theta\rvert^{p+1}\rangle c<c$), which gives the asymptotic behavior $r(t) \sim C t^{-1/(p-1)}$. Not sure about the $p<1$ case, though. –  mjqxxxx Jul 7 '12 at 22:28

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