= Solution
Use the <method of multiple scales>, $T=\varepsilon t$, and write $x_0=A(T)\cos\psi$, $\psi=t+\theta(T)$. At first order,
$$
(\partial_t^2+1)x_1=2A_T\sin\psi+2A\theta_T\cos\psi+A f(A\cos\psi)\sin\psi.
$$
Removing the fundamental <secular terms> gives the <averaged amplitude for position-dependent damping>:
$$
A_T=-A\langle f(A\cos\psi)\sin^2\psi\rangle,\qquad \theta_T=0,
$$
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, $\langle\sin^2\psi\rangle=1/2$ and $\langle\cos^2\psi\sin^2\psi\rangle=1/8$, giving
$$
A_T=\frac A2-\frac{A^3}{8},\qquad A(T)=\frac2{\sqrt{1+3e^{-T}}}.
$$
Thus the <Van der Pol amplitude evolution> yields
$$
\boxed{x(t;\varepsilon)=\frac{2\cos t}{\sqrt{1+3e^{-\varepsilon t}}}+O(\varepsilon),\qquad 0\leq t\leq C/\varepsilon.}
$$
The 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.
Back to article page
