= Solution
Fix $\mu=r-1-q$ and $0<q<\zeta/\sigma$. The noncritical <eigenvalues> at $\mu=0$ are $-\zeta$ and $\sigma q-\zeta<0$. Include the constant parameter as an additional centre variable. <Symmetry> permits expansions
$$
b=\frac a\zeta+b_\mu\mu a+b_3a^3+\cdots,\qquad c=c_2a^2+\cdots,
$$
where the omitted terms in the reduced equation have order $\mu^2a,\mu a^3,a^5$. At $\mu=0$, $a'$ begins at cubic order, so $c'=c_a a'$ begins at fourth order. The $c$ equation therefore gives $c_2=3/\zeta^2$.
The invariance equation for $b$ must retain $b_a a'=a'/\zeta+\cdots$. From the first equation,
$$
a'=\sigma(1-\zeta q b_\mu)\mu a-\sigma\zeta q b_3a^3+\cdots.
$$
Equating coefficients in $b_a a'=a-\zeta b-ac$ gives
$$
\frac{\sigma}{\zeta}(1-\zeta q b_\mu)=-\zeta b_\mu,\qquad
-\sigma q b_3=-\zeta b_3-\frac3{\zeta^2}.
$$
Hence
$$
b_\mu=-\frac{\sigma}{\zeta(\zeta-\sigma q)},\qquad
b_3=-\frac3{\zeta^2(\zeta-\sigma q)},
$$
and the required reduced equation is
$$
\boxed{a'=C_1\mu a+C_2a^3+\cdots,\qquad
C_1=\frac{\sigma\zeta}{\zeta-\sigma q},\quad
C_2=\frac{3\sigma q}{\zeta(\zeta-\sigma q)}.}
$$
Both coefficients are positive. Thus the <bifurcation> is a subcritical pitchfork: the no-convection state is stable for $\mu<0$ and unstable for $\mu>0$. The small nonzero branches lie on the stable-origin side,
$$
a_\pm^2=-\frac{C_1\mu}{C_2}=-\frac{\zeta^2\mu}{3q},
$$
and are unstable, since the <derivative> of the reduced <vector field> there is $-2C_1\mu>0$. Their two transverse directions remain stable, making them <saddle equilibrium> <equilibria> in the three-dimensional model. Setting $b'=c'=0$ without the invariance correction would incorrectly omit the common factor $(1-\sigma q/\zeta)^{-1}$. This is the <cubic centre-manifold reduction of porous magnetoconvection>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-56-amplitude-diagrams.png]
{title=Subcritical cubic onset and an example of bounded asymmetric quintic saturation}
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