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Interpret the displayed two-harmonic form as a leading term in a weakly nonlinear expansion. Write $\xi =(1+ϵ\kappa )x-t$ and ${\Psi }_0=\alpha \cos \xi +\beta \cos 2\xi$. The linear operator multiplies $\cos m\xi$ byFor $m=1,2$, this is $4ϵ\kappa +O({ϵ}^2)$ and $64ϵ\kappa +O({ϵ}^2)$, respectively. Thus detuning of a wave resonance enters at the same order as the quadratic forcing. The cosine addition formula givesProjection onto the two resonant harmonics by harmonic balance requiresThe nonzero branches are thereforeThe two signs are related by a half-period translation of $\xi$; the trivial branch $\alpha =\beta =0$ also exists.

The nonresonant first correction supplies the generated mean, third harmonic and fourth harmonic. Their linear multipliers at $ϵ=0$ are $4,40,180$, giving the nonresonant partFurther resonant amplitude corrections are determined at higher order. Consequently the nonzero answer describes an asymptotic periodic travelling wave with additional harmonics. Literally retaining only the two displayed harmonics cannot be an exact nonzero solution: their square has a positive constant term, while the linear operator applied to the two cosines has no constant term. The distinction is essential to interpreting this perturbative ansatz.