Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 85 / 4

At $r=s=0$, the linear growth rate of a Fourier mode $e^{ikx}$ is $-(1-k^2{)}^2$. Its critical kernel of a linear map has $k=\pm 1$. WriteFor a compatible periodic carrier, the slowly varying envelope is periodic on $0\le X\le {\epsilon }^{2}L$. Small carrier detuning can alternatively be included in the envelope wave phase.

With $w=O(\epsilon )$, $r={\epsilon }^4\mu$ and $s={\epsilon }^2\hat{s}$, all three terms $rw$, $s{w}^3$, $w^5$ first appear at $O({\epsilon }^5)$. Balancing the time derivative thus requires $T={\epsilon }^{4}t$. Since the linear growth rate near $k=1$ is $-4(k-1{)}^2+O((k-1{)}^3)$, spatial detuning must be $O({\epsilon }^2)$ to contribute at the same order. ConsequentlyTreating $x,X$ independently, let $D={\partial }_x$ on the fast carrier. Expansion of the linear operator givesThe $O({\epsilon }^3)$ term annihilates $w_1$, since $(1+D^2)w_1=0$. At orders two through four there is no nonresonant forcing, and kernel of a linear map corrections can be absorbed into the definition of $A$, allowing $w_2=w_3=w_4=0$. At order five the equation for $w_5$ isThe fast operator is a self-adjoint operator on periodic functions. Its Fredholm alternative requires the right-hand side to have zero inner product with each kernel of a linear map mode $e^{\pm ix}$, equivalently to have zero resonant Fourier coefficients. The coefficient of $e^{ix}$ in $w_1^3$ is $3A^2\overline{A}$, and in $w_1^5$ it is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)A^3{\overline{A}}^2=10A|A{|}^4$. Therefore the solvability condition isThe conjugate condition supplies the same real equation for the negative mode. The remaining noncritical harmonics determine $w_5$ without obstruction. This is the cubic-quintic Swift–Hohenberg amplitude reduction; it retains quintic saturation because the small positive cubic coefficient makes the onset mildly subcritical.

Assume $\hat{s}>0$, as required for the subcritical case. The zero state exists for every $\mu$. For a nonzero uniform real state, put $B=A_0^2>0$. Its equilibrium point condition is $\mu +3\hat{s}B-10B^2=0$, givingBoth are positive for $-9{\hat{s}}^2/40<\mu <0$. At $\mu =0$, the lower branch reaches zero and the upper has $B=3\hat{s}/10$. For $\mu >0$ only the upper branch is positive. Each positive $B$ represents real states $A_0=\pm \sqrt{B}$, as well as their constant complex wave phase rotations.

The two nonzero branches merge atThe radial linearization is ${\lambda }_a=\mu +9\hat{s}B-50B^2=2B(3\hat{s}-20B)$. It is negative on the upper branch and positive on the lower. The zero state is stable for $\mu <0$ and unstable for $\mu >0$. Thus a subcritical pitchfork bifurcation branch leaves zero towards negative $\mu$, and its fold connects to the stable finite-amplitude branch. Constant wave phase shifts are neutral for nonzero complex states; the stated attraction is in amplitude, with spatial wave phase perturbations damped except for that uniform symmetry mode.

The linearized complex equation couples a perturbation and its conjugate, so a single complex Fourier mode by itself is not a closed perturbation ansatz. Equivalently use real amplitude and wave phase perturbations, with Fourier modes paired to make each real field. Writing $A=A_0+\alpha +i\beta$ and $B=A_0^2$ givesA real amplitude mode of slow wavenumber $\ell >0$ therefore becomes neutral whenFor a nonzero uniform state the equilibrium point condition eliminates $\mu$, leaving amplitude growth $6\hat{s}B-40B^2-4{\ell }^2$. The wave phase mode growth is simply $-4{\ell }^2$, so it cannot give a nonzero-wavenumber bifurcation from these undetuned states.

The concave quadratic has maximumThus a nonzero uniform state needs $\ell \le 3\hat{s}/(4\sqrt{10})$ to have a modulation bifurcation. The slow domain length is ${\epsilon }^{2}L$, so ${\ell }_n=2\pi n/({\epsilon }^{2}L)$. If its smallest positive wavenumber already exceeds this bound, none can become neutral. Since $s={\epsilon }^2\hat{s}$, the conclusion isThis is the modulation cutoff of a cubic-quintic uniform pattern. If the bound is reversed, candidate squared amplitudes are $B=(3\hat{s}\pm \sqrt{9{\hat{s}}^2-160{\ell }^2})/40$. They lie on the lower branch; the upper branch is already damped in amplitude and cannot undergo this stationary modulation bifurcation.

The word “nonzero” is necessary in the final claim. At $A_0=0$, the equilibrium point exists independently of $\mu$, and the mode growth is $\mu -4{\ell }^2$. It becomes neutral at $\mu =4{\ell }_n^2$ for every finite domain. For example, take $\hat{s}=1$, ${\epsilon }^{2}L=20$: then $L=20/{\epsilon }^2<8\pi \sqrt{10}/(3s)$, but the zero state has a first-sideband threshold at $\mu ={\pi }^2/25$. Therefore the printed assertion, if read to include the trivial state, is false. The derived domain-size bound is the correct statement for nonzero pattern branches.