I'm working through some notes regarding analysis of some weakly perturbed oscillators. Among these is Duffing's oscillator which is governed by $$ \frac{d^2y}{dt^2}+y+\epsilon y^3=0, \; \; 0<\epsilon << 1 $$
Now to leading we have
\begin{align*}
\frac{d^2y_0}{dt^2}+y_0=0
\end{align*}
with general solution
$$ y_0 = a\cos t + b \sin t$$
which can be written w.l.o.g as $ y_0 = c \cos(t+\phi), \; \; c>0$.
So far so good. But now to understand the nature of the perturbation the notes make use of the Poincare-Lindstedt method, which relies on the observation that the period of the perturbed oscillator is $$2\pi(1+o(1))$$
as $\epsilon \rightarrow 0$. Here is my first question: how exactly was this observation made?
Next we define $Y(\tau; \epsilon) = y(t; \epsilon)$ where $\tau = t/\chi(\epsilon)$ and $\chi(\epsilon)$ is such that $Y$ is $2\pi$ periodic in $\tau$. Then, expanding $Y$ and $\chi$ in powers of $\epsilon$ we have
$$Y(\tau;\epsilon) \sim Y_0(\tau;\epsilon)+\epsilon Y_1(\tau;\epsilon)+…$$
and
$$\chi(\epsilon) \sim 1+\epsilon\chi_1 + \epsilon^2\chi_2+…$$
Here the $Y_i$ are $2\pi$ periodic in $\tau$. To leading order the solution has the same form, $Y_0 = a\cos(\tau + \phi)$. Going up to first order in $\epsilon$, we have
\begin{align*}
\frac{d^2Y_1}{d\tau^2} + Y_1 &= -2\chi_1Y_0-Y_0^3 \\
&= -2\chi_1a\cos(\tau+\phi) -\frac{a^3}{4}\left \{3\cos(\tau+\phi)+\cos\left(3(\tau+\phi)\right)\right\},
\end{align*}
where we have substituted for $Y_0$ and used a Fourier expansion. Then here is where I really have trouble. At this point then notes say that in order for $Y_1$ to be $2\pi$ perioduc in $\tau$ we must eliminate $\cos(\tau+\phi)$ terms, which gives us an algebraic expression for $\chi_1$. But why must we eliminate such terms in order for $Y_1$ to be $2\pi$ periodic? I don't understand this.
Best Answer
1.) The equation remains conservative for $ϵ\ne 0$ so that all solutions always lie on level surfaces of $$ E=\frac12y'^2+\frac12y^2+\fracϵ4y^4. $$ All solutions that do not lie on the level of singular points can not help but be periodic. For $ϵ>0$ there are no singular points except the origin. Any solution for $ϵ\ll 1$ will be close to a circle and at time $2\pi$ will be close to the initial point.
2.) The frequency-one-terms on the right side are in resonance to the homogeneous solution of the left side, and thus will give solution terms that are increasing in amplitude. This gives obviously only a valid approximation for a very local solution. With the perturbation approach one has introduced flexibility to shift these terms of the equation into the base frequency, or said otherwise, to compensate them with changes in the frequency. Thus by selecting $χ_0=1$ and $$ 2χ_1a+\frac{3a^3}4=0 $$ there is no longer resonance in this level of the perturbation equations, the resulting periodic solution is correct to a higher order in $ϵ$.
Essentially, if you solve the equation using a simple perturbation approach $y=y_0+ϵy_1+...$, you get the first perturbation equation as $$ y_1''+y_1=-y_0^3=-c^3\cos^3(t+ϕ)=-\frac{c^3}{4}[\cos(3t+3ϕ)+3\cos(t+ϕ)] $$ with particular solution of the form $$ y_1=A\cos(3t+3ϕ)+Bt\sin(t+ϕ) \\ \implies y_1''+y_1=-8A\cos(3t+3ϕ)+2B\cos(t+ϕ) $$ so that the coefficients are $A=\frac{c^3}{32}$ and $B=-\frac{3c^3}{8}$. In the expression for $y$ the combination of terms $$ c\cos(t+ϕ)-ϵ\frac{3c^3}{8}t\sin(t+ϕ) $$ can be interpreted as linear Taylor polynomial for $$ c\cos\left(\left(1+ϵ\frac{3c^2}{8}\right)t+ϕ\right) $$ These are equivalent approximations for $t\approx 0$, but the second one gives a result that is also globally valid for a periodic solution.