Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 79 / 4 / a

Use the method of multiple scales, $T=\epsilon t$, and write $x_0=A(T)\cos \psi$, $\psi =t+\theta (T)$. At first order,Removing the fundamental secular terms gives the averaged amplitude for position-dependent damping:where the brackets mean a full-period average. The initial conditions set $A(0)=1$ and $\theta (0)=0$ at leading order. For quadratic damping, $⟨{\sin }^2\psi ⟩=1/2$ and $⟨{\cos }^2\psi {\sin }^2\psi ⟩=1/8$, givingThus the Van der Pol amplitude evolution yieldsThe leading amplitude approaches the stable value two. The small initial-velocity discrepancy introduced by the slow amplitude derivative is removed by the first-order correction and does not affect this leading approximation.

For $f(x)=\sin x$, the period average $⟨\sin (A\cos \psi ){\sin }^2\psi ⟩$ vanishes: a half-period shift reverses the first factor and leaves the second unchanged. This is the zero first-order amplitude drift for odd damping. The averaged amplitude for position-dependent damping therefore has $A_T=0$, while its first-order phase equation again has ${\theta }_T=0$. With the initial conditions,The first-order forcing has no fundamental resonant component, so it produces bounded corrections rather than a secular amplitude change on this time scale. This statement concerns the leading approximation on $t=O({\epsilon }^{-1})$.

For $f(x)=|x|$ and positive amplitude $A$, the averaged amplitude for position-dependent damping givesThe average is $(4/(2\pi )){\int }_0^{\pi /2}\cos \psi {\sin }^2\psi d\psi =2/(3\pi )$. Thus the absolute-value damping amplitude law isand ${\theta }_T=0$ as in the other cases. Consequently,The dissipation makes the leading envelope decrease algebraically in slow time. The absolute value is Lipschitz continuous, so the first-order averaging argument remains valid despite its corner at zero.