Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 336 / 3

The linear dispersion relation for a mode $e^{i(kx-\omega t)}$ is $5{\omega }^2=k^4+4$. The proposed modes share phase velocity $c$, so $5c^2{k}_j^2=k_j^4+4$. A quadratic wave interaction generates the second harmonic and the difference harmonic. With only two positive wavenumbers, phase matching for a quadratic wave interaction requires $k_2=2k_1$. Equating their phase velocities givesand thereforeThis is a two-to-one resonance of dispersive waves. Its amplitude changes accumulate on $t=O({ϵ}^{-1})$. Choose the slow time $T=ϵt/5$; this harmless constant rescaling makes the later amplitude formulas simple. With $\xi =x-t$, the leading real field is $A_0{e}^{i\xi }+B_0{e}^{2i\xi }+{\overline{A}}_0{e}^{-i\xi }+{\overline{B}}_0{e}^{-2i\xi }$.

For the fundamental, the order-$ϵ$ cross derivative in $5{\psi }_{tt}$ is $-2i(A_0{)}_T{e}^{i\xi }$, while the fundamental coefficient in ${\psi }_0{\psi }_{0x}$ is $i{\overline{A}}_0{B}_0$. For the second harmonic the corresponding terms are $-4i(B_0{)}_T{e}^{2i\xi }$ and $i{A}_0^2$. The solvability condition in the method of multiple scales removes the resonant forcing and givesThe nonresonant third and fourth harmonics enter the correction. They do not change these leading amplitude equations.

Write $A_0=R{e}^{i\theta }$, $B_0=P{e}^{i\varphi }$ and $\delta =\varphi -2\theta$. Taking real and imaginary parts gives the explosive two-to-one amplitude system in polar form:The polar phases are used where their corresponding amplitudes are nonzero. These equations implyFor the second identity, differentiate using ${\delta }_T=(R^2/(4P)+P)\sin \delta$: the terms proportional to $\sin \delta \cos \delta$ cancel. Thereforeare first integrals, and in particular

For constant phases with $\varphi =2\theta +\pi$, these real equations reduce to $R_T=RP/2$ and $P_T=R^2/4$. The initial data give $R^2-2P^2=8$, so the Riccati equation for $P$ is $2P_T=P^2+4$. Integrating and using $P(0)=2$ givesThe tangent and secant function have a pole at $T=\pi /4$, corresponding to $t=5\pi /(4ϵ)$. Both modes grow through their locked resonant interaction. This is formal finite-time blowup of the reduced amplitude equations; the weakly nonlinear expansion loses validity as the amplitudes become large, so it cannot establish a singularity of the full partial differential equation.

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.