Why is $\sin^3(\tau + \phi)$ not a Secular term in the context of the van der Pol oscillator

Question

On page $223$ of Strogatz is the example of Two-Timing applied to the van der Pol oscillator $$x'' + x + \epsilon (x^2 – 1)x' = 0$$ Introducing the perturbation $x = x_0(\tau, T) + \epsilon x_1(\tau, T) + \cdots$ and time scales $\tau = t$ and $T = \epsilon t$ then we may, after substitution and appropriate grouping, obtain the $O(1)$ and $O(\epsilon)$ equations.
$$\begin{align*}
O(1): \partial_{\tau \tau}x_0 + x_0 &= 0\\
O(\epsilon): \partial_{\tau \tau}x_1 + 2\partial_{T \tau}x_0 + 2\partial_{\tau}x_0+x_1 & = 0\end{align*}$$

The solution to the $O(1)$ equation is $x_0 = r(T)\cos(\tau + \phi(T))$. Substituting appropriately and computing the derivatives they obtain equation $(39)$ (which has a tiny formatting typo I think: should be $\sin^3$ not $\sin3(\tau +\phi)$ ):

Why is $\sin^3(\tau + \phi)$ not considered a secular term? I know a secular term to be a term which grows without bound. I thought that since $\sin(\cdot)$ and $\cos(\cdot)$ were oscilitory then they were considered secular.

So why are the first two terms considered secular and the third is not?

Answer 1

I don't believe it is a typo. The reason it is not secular is because it does not resonate with the frequency of the adjoint solution.

Formally, you have a linear operator $L$ applied to the solution $x_1$ with a RHS we can call $r_1$. So $L x_1 = r_1$ for this to have a solution, recall the condition from linear algebra that says if $Ax = b$ has a solution then $A^T y = 0$ and $y^T r = 0$. This is the solvability condition.

In the context here $L = \partial_{tt} + I$, is a self-adjoint operator and the solution $y = \sin, \cos$. The solvability condition is $\langle \sin t, \sin n t \rangle$ which is 0 unless $n = 1$.

EDIT: The solution of $Lu = 0$ are sin and cos. Secular terms cause blow-up and in this context occurs because the frequencies are the same. If you are unsure. Try solving $u'' + u = \sin 3t$ vs. $u'' + u = \sin t$ and see what happens as $t \to \infty$ for both.