Lindstedt-Poincare method to find a valid first approximation I want to use Lindstedt-Poincare method or multiple scale method to find a uniformly valid first approximation to the equation
$\frac{d^2u}{dt^2}-\epsilon (1-u^2)\frac{du}{dt}+u=0$, and show that for a large class of initial conditions, the solution approaches a limit cycle as $t\rightarrow \infty$.
I would appreciate any assistance.
 A: We want to solve the given ODE (Van der Pol Oscillator)
$$ \frac{\mathrm{d}^2 u}{\mathrm{d}t^2} - \epsilon (1-u^2)\frac{\mathrm{d}u}{\mathrm{d}t}+u=0 $$
using a multiple scales expansion with the slow time scale $\tau=\epsilon t$. Then $u=u_o(t,\tau)+\epsilon u_1(t,\tau)+o(\epsilon)$, $\frac{\mathrm{d}}{\mathrm{d}t} = \partial_t + \epsilon \partial_\tau$. At leading order,
$$\partial_t^2u_o+u_o=0$$
and hence $u_o=A(\tau)e^{it}+c.c.$ . At order $\epsilon$, 
$$ \partial_t^2 u_1+u_1 = -2i A_{\tau} e^{it}+iAe^{it}-iA^3e^{3it}-iA^2A^*e^{it} +c.c.$$
Now, since the homogeneous solutions have frequency $1$, the forcing terms with that frequency will cause resonance i.e. secular growth (prop. to $t$). Because we want to construct a solution that is uniformly valid until $t=O(\epsilon^{-1})$, we set the coefficient of $e^{it}$ on the RHS to zero to avoid that secular growth, i.e.
$$ 2A_{\tau} + (|A|^2-1)A =0$$
This is a complex ODE whose real and imaginary part read
$$2A^r_{\tau} +(|A|^2-1)A^r=0, \, 2A^i_{\tau} = (|A|^2-1)A^i$$
Multiplying by $A^r$ and $A^i$ respectively and adding gives: 
$\partial_{\tau} |A|^2 +(|A|^2-1)|A|^2 =0$ which is solved by $|A(\tau)|^2=\frac{1}{1+Be^{-\tau}}$, with $B$ a real constant. Hence $A$ satisfies
$$A_{\tau}-\frac{1}{2}\frac{1}{1+Be^{-\tau}}A=0 $$
Use the integrating factor $\sqrt{e^{\tau} +B}e^{-\tau/2}$, s.t. $\sqrt{e^{\tau} +B}e^{-\tau/2}A = C \Rightarrow A = \frac{Ce^{\tau/2}}{\sqrt{e^\tau+B}}$. By the condition that $  |A|^2 = \frac{Be^{-\tau}}{1+Be^{-\tau}} = \frac{B}{e^{\tau}+B} \Leftrightarrow  |C|^2=B$, i.e. $A=\frac{\sqrt{B}e^{\tau/2+i\phi}}{\sqrt{e^\tau+B}}$, $\phi=\arg{C}$.
$$u_o(t)= \sqrt{\frac{B}{1+Be^{-\tau}}} \cos(t+\phi) $$
This solution is uniformly valid up to $t=Ord(\epsilon^{-1})$. Now, as $t\to \infty$, then $u_o(t)\sim \sqrt{B} \cos(t+\phi)$ (where $\sim$ means asymptotic to). This is a limit cycle of radius $\sqrt{B}$ and frequency $1$.
