Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 65 / 3

Take the leading real field to be $A{e}^{i(k_{c}x+{\omega }_{0}t)}+B{e}^{i(-k_{c}x+{\omega }_{0}t)}+\text{complex conjugate}$. A spatial translation gives $(A,B)\mapsto (e^{i{k}_c\delta }A,e^{-i{k}_c\delta }B)$; a time translation gives $(A,B)\mapsto (e^{i\psi }A,e^{i\psi }B)$; spatial reflection exchanges $A$ and $B$.

For a monomial $A^j{\overline{A}}^k{B}^l{\overline{B}}^m$ in the first amplitude equation, covariance requires $j-k+l-m=1$ and $j-k-l+m=1$. Hence $j-k=1$, $l-m=0$. At linear order only $A$ is allowed, no quadratic monomial satisfies these conditions, and the cubic possibilities are $A^2\overline{A}$ and $AB\overline{B}$. Reflection fixes the coefficients in the second equation. Removing the small common linear frequency correction by a rotating frame makes $\mu$ real. This derives the two cubic amplitude equations, rather than merely postulating them.

Let $r=|A{|}^2$, $s=|B{|}^2$. Taking real parts givesA travelling wave has $(r,s)=(\mu /{\beta }_r,0)$, or its reflection, whenever $\mu /{\beta }_r>0$. Its phase evolves at $-{\beta }_i\mu /{\beta }_r$, shifting the physical oscillation frequency. The nonzero-amplitude radial eigenvalue is $-2\mu$, while the absent mode has real growth rate $\mu (1-{\gamma }_r/{\beta }_r)$. Therefore, on the supercritical side,The phase direction is neutral because of the symmetry, so stability means orbital stability. A branch with $\mu <0$ is radially unstable, whatever the transverse coefficient.

A standing wave has $r=s=h=\mu /({\beta }_r+{\gamma }_r)>0$. Its two phases evolve together at $-({\beta }_i+{\gamma }_i)h$; the relative phase fixes the spatial position of the standing pattern. In the two intensity directions the eigenvalues are $-2\mu$ and $-2h({\beta }_r-{\gamma }_r)$. ThusAgain the symmetry phases are neutral and the conclusion is orbital stability. Equality ${\beta }_r={\gamma }_r$ leaves a continuum of mixed intensities at cubic order; equality ${\beta }_r+{\gamma }_r=0$ removes cubic standing-wave saturation. These degeneracies require higher-order terms and cannot be classified by the cubic truncation. The zero state is linearly stable for $\mu <0$ and unstable for $\mu >0$.

Forcing at $2{\omega }_0$ preserves translation by one forcing period, which acts on the critical amplitudes as $(A,B)\mapsto (-A,-B)$. It destroys the continuous common phase symmetry. Under the remaining spatial translation, $\overline{B}$ transforms as $e^{i{k}_c\delta }\overline{B}$, just like $A$, and $\overline{A}$ transforms just like $B$. Both change sign under the remaining temporal sign reversal. They are therefore allowed linear terms in the respective amplitude equations. Spatial reflection requires their coefficients to coincide.

The resonant physical-frequency calculation gives the same result: multiplying the forcing factor $e^{2i{\omega }_{0}t}$ by the conjugate left-going mode $\overline{B}{e}^{i(k_{c}x-{\omega }_{0}t)}$ produces $\overline{B}{e}^{i(k_{c}x+{\omega }_{0}t)}$, resonant with the $A$ mode. The other term follows by reflection. Thus the leading parametrically forced counterpropagating waves obeywith the reflected equation for $B$. The forcing phase can be absorbed into a common amplitude phase so that $\nu$ is real; its magnitude measures the resonant forcing strength. Higher-order forcing corrections are beyond this leading weak-forcing truncation.

Write $\psi =\theta +\phi$, $U={\beta }_r+{\gamma }_r$ and $V={\beta }_i+{\gamma }_i$. The polar amplitude equations arewith their reflected counterparts. On the equal-amplitude standing wave branch $R=S>0$, equilibrium point requiresSquaring and adding eliminates the phase:For $U^2+V^2>0$ the candidate squared amplitudes areFor each positive root, set $\sin \psi =-Vz/\nu$ and $\cos \psi =(Uz-\mu )/\nu$. The quadratic identity ensures these define a phase. Conversely these phase equations imply the quadratic, so a nonzero phase-locked branch exists exactly when the discriminant is nonnegative and at least one root is positive.

Since $|\cos \chi |=|V|/\sqrt{U^2+V^2}$, the discriminant condition is the printed $|\mu \cos \chi /\nu |\le 1$ for $\nu \ne 0$. It is sufficient under the usual above-onset supercritical assumptions $\mu >0,U>0$, because then the plus root is positive. Without these assumptions the condition is not sufficient: $\mu =-1,U=1,V=0,\nu =1/2$ satisfies it but gives $z=-1\pm 1/2<0$. This is the positivity condition for a phase-locked standing wave, not a removable algebraic detail. The zero solution always exists and is distinct from the nonzero branch. For $\nu =0$, the undivided equations must be used; a nonzero equilibrium point in this fixed rotating frame requires $V=0$ and $z=\mu /U>0$. If $U=V=0$, nonzero equilibrium points require ${\mu }^2={\nu }^2$ and a compatible phase, with no cubic amplitude selection.

Strong enough forcing can overcome the nonlinear frequency shift, lock the sum phase to the external signal and produce a standing wave at half the forcing frequency. The difference phase remains free through spatial translation. The amplitude inequality is an existence condition, not a stability theorem: attraction requires the linearized forced amplitude equations, and nonlinear coefficients have not been restricted enough here to promise stability of every root.