Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 65 / 1 / a

Substitution into the Newell–Whitehead–Segel equation gives $0=(\mu -q^2-R_0^2)R_0$. Thus a nonzero roll has $R_0^2=\mu -q^2>0$; at equality the branch meets the zero solution. The two signs of $R_0$ differ by a constant phase modulation.

The sign of the transverse zigzag instability must be calculated from the differential operator. Put $D={\partial }_X+i{\partial }_Y^2/2$ and $A=R_0{e}^{iqX+i\varphi (Y)}$. The chain rule givesConsequently the change in the printed functional density isFor a Lyapunov functional one integrates this density over the transverse coordinate too, or uses a transverse spatial average, with periodic or decaying boundary conditions. A fast-$x$ average alone leaves a density depending on $Y$ and is not itself a global monotonicity statement. Indeed $D^{†}=-D$ under these boundary conditions, and the functional derivative of the resulting spatial integral is $\delta V/\delta \overline{A}=\mu A+D^{2}A-|A{|}^{2}A=A_T$. Hence $dV/dT=2⟨|A_T{|}^2⟩\ge 0$: stable rolls locally maximize this functional.

With the plus sign actually printed in $D$, $q<0$ makes both quadratic contributions to $\Delta V$ negative. These rolls are stable to long-wave transverse phase modulation. For $q>0$ a sufficiently long-wave displacement increases $V$ and is unstable. At $q=0$ the second-derivative stiffness vanishes, but the fourth-derivative and nonlinear tilt terms still decrease $V$. Linearization makes the conclusion explicit:Thus for $q>0$ the unstable band is $0<k^2<4q$, with maximum growth at $k^2=2q$; in a finite periodic cell the band must contain an allowed Fourier mode. Replacing $D$ by ${\partial }_X-i{\partial }_Y^2/2$ reverses the sign of the detuning associated with the zigzag instability. It would therefore be incorrect to import that opposite sign convention into the printed equation. Transverse stability does not exclude the distinct longitudinal Eckhaus instability.