 Solution

Write $A=a{e}^{i\varphi }$. The modulation equation gives $a_T=0$ and ${\varphi }_T=-5a^2/3$. Matching the leading initial conditions gives $a=1$, $\varphi (0)=0$. ThereforeThe leading amplitude remains two and has no slow growth or decay. The frequency changes from one to $1-5{ϵ}^2/3+O({ϵ}^3)$; expanding this shifted cosine recovers the second-order secular term from part (i).

There are bounded first-order waveform changes as well. If they are retained while matching $u(0)=2$, one may write $2\cos \tau +ϵ[2-(4/3)\cos \tau -(2/3)\cos 2\tau ]+\cdots$, where $\tau =(1-5{ϵ}^2/3)t$. Thus the first harmonic's coefficient is $2-4ϵ/3$, with a mean shift and a second harmonic. These bounded shape corrections are distinct from the constant leading amplitude in the zeroth-order multiple-scales solution.