= Solution
Substitution into the <Newell–Whitehead–Segel equation> gives $0=(\mu-q^2-R_0^2)R_0$. Thus a nonzero roll has \b[$R_0^2=\mu-q^2>0$]; 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 $A=R_0e^{iqX+i\phi(Y)}$. The <chain rule> gives
$$
DA=\left[i\left(q-\frac12\phi_Y^2\right)-\frac12\phi_{YY}\right]A.
$$
Consequently the change in the printed functional density is
$$
\Delta V=R_0^2\left\langle q\phi_Y^2-\frac14\phi_Y^4-\frac14\phi_{YY}^2\right\rangle.
$$
For 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 $D^\dagger=-D$ under these boundary conditions, and the functional derivative of the resulting spatial integral is $\delta V/\delta\overline A=\mu A+D^2A-|A|^2A=A_T$. Hence $dV/dT=2\langle|A_T|^2\rangle\ge0$: stable rolls locally maximize this functional.
With the \b[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:
$$
\phi_T=-q\phi_{YY}-\frac14\phi_{YYYY},\qquad
\boxed{\sigma(k)=qk^2-\frac14k^4.}
$$
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>.
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