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Use the translation-covariant phase expansion with $\xi =x+\varphi (Y,T)$, $Y=\epsilon y$ and $T={\epsilon }^{4}t$. Write all fast profiles at $\xi$, including the corrections. LetThe Swift–Hohenberg equation has steady equation $r{w}_0-B^2{w}_0-w_0^3=0$, and differentiation gives $L{w}_0^{\prime }=0$. On a fast periodic cell $L$ is a self-adjoint operator. We assume the translation mode is its only zero mode; otherwise one must impose the corresponding additional Fredholm solvability conditions and retain additional amplitudes.

Since ${\nabla }^2={\partial }_{\xi }^2+{\epsilon }^2{\mathcal{D}}_Y^2$, its squared operator is $B^2+2{\epsilon }^{2}B{\mathcal{D}}_Y^2+{\epsilon }^4{\mathcal{D}}_Y^4$. The time derivative first contributes at order ${\epsilon }^4$. At order ${\epsilon }^2$,The Fredholm solvability condition is orthogonality of this forcing to $w_0^{\prime }$. Periodic integration by parts givesThus the condition is $2J{\varphi }_{YY}=0$. General slowly curved rolls require $J=0$ at leading order. If $J=O({\epsilon }^2)$, this order-${\epsilon }^2$ residual first enters the order-${\epsilon }^4$ equation, so the leading order-${\epsilon }^2$ problem is solvable. In particular,is precisely the near-threshold condition. Defining $N=⟨(w_0^{\prime }{)}^2⟩>0$ and $J=-{\epsilon }^2\lambda N/2$ keeps the residual as the phase-diffusion term at the next order.

There is an important chain rule qualification in this part. The printed value of $\gamma$ follows from the expansion. Its printed value of $\alpha$ follows only if transverse differentiation of the translated correction profiles is omitted. For a consistent translation-covariant phase expansion, that omission is generally nonzero.

At the leading transverse threshold letwhere the last conditions fix the phase gauge. Put $p={\varphi }_Y$, $r_1={\varphi }_{YY}$, $s_1={\varphi }_{YYY}$ and $t_1={\varphi }_{YYYY}$, and use $w_2=C_1(\xi )r_1+C_2(\xi )p^2$. At order ${\epsilon }^4$ the Fredholm solvability condition givesThe first two contributions follow directly by differentiating:Choose the reflection-symmetric roll with $w_0$ even. Then $C_1$ is odd and $C_2$ even. Projection onto the odd translation mode eliminates $p{s}_1$, $r_1^2$ and $p^4$, leaving exactly the symmetry-allowed phase modulation equation. Its coefficients areandIf one freezes $C_j$ at the unshifted coordinate when taking transverse derivatives, ${\partial }_Y^2{w}_2=C_1{t}_1+2C_2(r_1^2+p{s}_1)$, and its projection contributes only the $\gamma$ term. The other two projected terms then give the printed expression ${\alpha }_{\mathrm{print}}=6⟨w_0^{\prime }(w_0^{\prime \prime \prime }+w_0{C}_1{C}_2)⟩/N$. But freezing these profiles is inconsistent with the translated $w_0$ in $L$ for a general finite phase displacement. The printed expression needs the extra assumption $⟨w_0^{\prime }B(C_1^{\prime \prime }+5C_2^{\prime })⟩=0$, which is not an identity.

A small-amplitude calculation demonstrates the difference. For a critical roll of leading amplitude $\eta$ and fast wavenumber $\kappa$, the steady equation and $J=0$ giveThe third harmonic coefficient in $w_0$ follows by dividing its cubic forcing $-{\eta }^3\cos (3\kappa \xi )/4$ by the fast eigenvalue $-64$. Substituting in $L{C}_j$ gives the displayed correction profiles; the near-zero first-harmonic eigenvalue in the even problem must be retained to obtain the first harmonic of $C_2$. Periodic orthogonality then givesThus the supplied nonlinear coefficient is not generally an exact consequence of the finite-amplitude translated-roll expansion. The corrected coefficient above, the printed calculation under its additional freezing assumption, and the genuine near-threshold phase modulation equation are all distinguished explicitly.