How does a harmonic oscillator with nonlinear damping behave? x¨+cx˙+x=0 with positive c, the amplitude of the oscillations decays exponentially when c&lt;2. If it is higher than 2, the system fails to oscillate at all and is said to be overdamped.

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How does a harmonic oscillator with nonlinear damping behave?  x¨+cx˙+x=0  with positive c, the amplitude of the oscillations decays exponentially when c&amp;lt;2. If it is higher than 2, the system fails to oscillate at all and is said to be overdamped.

nghodlokl · Accepted Answer

So, following the usual treatment of the harmonic oscillator, note that the equations of motion for x and v are  x˙=+v  v˙=−x−c|v|pv,  where the damping term is clearly small for small v when p&amp;gt;1. In the two-dimensional phase space, the equations of motion become particularly simple in radial coordinates: letting x=rcos⁡θ and v=rsin⁡θ, we have  r˙cos⁡θ−rθ˙sin⁡θ=+rsin⁡θ  r˙sin⁡θ+rθcos⁡θ=−rcos⁡θ−crp|sin⁡θ|p−1sin⁡θ,  or, solving the system,  r˙=−crp|sin⁡θ|p−1sin2θ  θ˙=−1−crp−1|sin⁡θ|p−1sin⁡θcos⁡θ.  For p&amp;gt;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 |r˙|&amp;lt;ar and θ˙&amp;lt;−1+ξ for any positive a and ξ and for all later times, and hence cos⁡θ (and x) will change signs arbitrarily many times before r reaches zero.

How does a harmonic oscillator with nonlinear damping behave? \ddot{x}+c \dot{x}+x=0 with

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