How to identify secular terms in multiscale expansion?

Question

How can one identify secular terms while doing multiscale expansion?

For e.g. in an initial value problem, can any term where t appears can be counted as secular term?

What about; $t e^{-t}$, is it a secular term?

Regards,

Nitin

Answer 1

Secular terms appear in the solution to a perturbation problem, and indicate that the scaling has broken down.

A perturbation expansion looks like $$ x(t)=x_0+\epsilon x_1(t)+\epsilon^2x_2(t)+\ldots,$$ where each term (i.e. each $\epsilon^ix_i$) is asymptotically smaller than the term before it. So $\epsilon x_1\ll x_0$ as $\epsilon\rightarrow0$, and $\epsilon^2x_2\ll\epsilon x_1$.

It is this ordering that allows you to separate your original equation into a series of equations that give approximate solutions.

In some cases though, your assumption that each term is asymptotically smaller than the one before it can break down. If you find a function $x_1(t)$ that grows in time, then eventually it will become larger than the $x_0$ term.

A classic example is the weakly nonlinear oscillator. Consider the IVP $$\ddot x+2\epsilon\dot x+x=0,\quad x(0)=0,\quad\dot x(0)=1.$$ Doing the normal perturbation expansion you find that $x_0=\sin(t)$, and $x_1=-t\sin(t)$. So your two-term solution is $x=\sin(t)-\epsilon t\sin(t)$. But when $t\sim1/\epsilon$, the solution looks like $x=0$ because your first and second order terms have balanced, and $x_0\sim\epsilon x_1$! So your original assumption implicit in solving for $x_0$ first, then $x_1$ and so on is not satisfied for large times.

To resolve this problem, you need to use the method of multiple scales.