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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.

Fordirect differentiation givesandSubstitution into the Schrodinger equation and equality of the coefficients of $x^2,x,1$ give the necessary and sufficient Riccati equation system

Let $\tau =T-t$ and setThenFor $B(z)=a-bz$ and $C(z)=c$, matching constant, linear, and quadratic coefficients givesandThe terminal condition becomes $P(0)=Q(0)=R(0)=0$. The first equation is a Riccati equation; once it is solved, the second is linear in $Q$, followed by direct integration for $P$.

The Riccati equation $b_s+b^2=-4$ with $b(0)=0$ has solutionSince ${\lambda }_s/\lambda =-b$, the initial condition $\lambda (0)=1$ gives $\lambda (s)=\sec (2s)$. ConsequentlyFinally ${\gamma }_s=-\omega$. In physical time the solution is therefore

Put $B=\xi \tan (\varphi +\alpha t)$. Matching the constant and ${\tan }^2$ coefficients in the Riccati equation requiresOne convenient choice is thereforeso

Put $v=e^S$ and $p=S^{\prime }$. The Liouville-Green exponential ansatz turns the normal-form equation into the Riccati equationWriteMatching the constant, $x^{-1}$, and $x^{-2}$ coefficients gives

For the recessive branch,soSince $y=e^{x/2}{x}^{-1/2}v$, this givesThis agrees with the first two terms of part (a),Continuing the Riccati recursion reproduces the factorial asymptotic series, so $U$ is proportional to this recessive branch.

For the dominant branch,and in factsolves the phase equation exactly. It follows thatIndeed, $e^x$ is an exact second solution of the original differential equation.

At a regular point of a second-order linear ordinary differential equation, a nontrivial solution cannot have both $x$ and $x^{\prime }$ zero, by initial-value uniqueness. Its zeros are therefore simple. Under the linearization of a Riccati equation, such a zero $t_{\ast }$ produces $y=x^{\prime }/(cx)\sim 1/[c(t-t_{\ast })]$. Zeros separate the intervals on which the Riccati equation solution is finite.