= Solution
Use the <translation-covariant phase expansion> with $\xi=x+\phi(Y,T)$, $Y=\varepsilon y$ and $T=\varepsilon^4t$. Write all fast profiles at $\xi$, including the corrections. Let
$$
B=1+\partial_\xi^2,\quad
\mathcal D_Y=\partial_Y+\phi_Y\partial_\xi,\quad
L=r-B^2-3w_0^2.
$$
The <Swift–Hohenberg equation> has steady equation $rw_0-B^2w_0-w_0^3=0$, and differentiation gives $Lw_0'=0$. 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 $\nabla^2=\partial_\xi^2+\varepsilon^2\mathcal D_Y^2$, its squared operator is $B^2+2\varepsilon^2B\mathcal D_Y^2+\varepsilon^4\mathcal D_Y^4$. The time derivative first contributes at order $\varepsilon^4$. At order $\varepsilon^2$,
$$
Lw_2=2B\mathcal D_Y^2w_0
=2\left[(w_0''''+w_0'')\phi_Y^2+(w_0'''+w_0')\phi_{YY}\right].
$$
The <Fredholm solvability condition> is orthogonality of this forcing to $w_0'$. Periodic <integration by parts> gives
$$
\langle w_0'(w_0''''+w_0'')\rangle=0,\qquad
J:=\langle w_0'(w_0'''+w_0')\rangle
=\langle (w_0')^2-(w_0'')^2\rangle.
$$
Thus the condition is $2J\phi_{YY}=0$. General slowly curved rolls require $J=0$ at leading order. If $J=O(\varepsilon^2)$, this order-$\varepsilon^2$ residual first enters the order-$\varepsilon^4$ equation, so the leading order-$\varepsilon^2$ problem is solvable. In particular,
$$
\boxed{\langle(w_0')^2+w_0'w_0'''\rangle=O(\varepsilon^2)}
$$
is precisely the near-threshold condition. Defining $N=\langle(w_0')^2\rangle>0$ and $J=-\varepsilon^2\lambda N/2$ keeps the residual as the phase-diffusion term at the next order.
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