 Solution

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.