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At $r=s=0$, the linear growth rate of a Fourier mode $e^{ikx}$ is $-(1-k^2{)}^2$. Its critical kernel of a linear map has $k=\pm 1$. WriteFor a compatible periodic carrier, the slowly varying envelope is periodic on $0\le X\le {\epsilon }^{2}L$. Small carrier detuning can alternatively be included in the envelope wave phase.

With $w=O(\epsilon )$, $r={\epsilon }^4\mu$ and $s={\epsilon }^2\hat{s}$, all three terms $rw$, $s{w}^3$, $w^5$ first appear at $O({\epsilon }^5)$. Balancing the time derivative thus requires $T={\epsilon }^{4}t$. Since the linear growth rate near $k=1$ is $-4(k-1{)}^2+O((k-1{)}^3)$, spatial detuning must be $O({\epsilon }^2)$ to contribute at the same order. ConsequentlyTreating $x,X$ independently, let $D={\partial }_x$ on the fast carrier. Expansion of the linear operator givesThe $O({\epsilon }^3)$ term annihilates $w_1$, since $(1+D^2)w_1=0$. At orders two through four there is no nonresonant forcing, and kernel of a linear map corrections can be absorbed into the definition of $A$, allowing $w_2=w_3=w_4=0$. At order five the equation for $w_5$ isThe fast operator is a self-adjoint operator on periodic functions. Its Fredholm alternative requires the right-hand side to have zero inner product with each kernel of a linear map mode $e^{\pm ix}$, equivalently to have zero resonant Fourier coefficients. The coefficient of $e^{ix}$ in $w_1^3$ is $3A^2\overline{A}$, and in $w_1^5$ it is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)A^3{\overline{A}}^2=10A|A{|}^4$. Therefore the solvability condition isThe conjugate condition supplies the same real equation for the negative mode. The remaining noncritical harmonics determine $w_5$ without obstruction. This is the cubic-quintic Swift–Hohenberg amplitude reduction; it retains quintic saturation because the small positive cubic coefficient makes the onset mildly subcritical.