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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.2012-07-07
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    @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.2012-07-07
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    @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.2012-07-07
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    This question is really interesting. For example the simple pendulum motion with quadratic damping is always under-damped. See http://hk.knowledge.yahoo.com/question/question?qid=7010122600228 for detail explanation.2012-07-07
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    @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 $t \to \infty$.2012-07-07

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