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On the nonzero stationary branch of five-mode magnetoconvection, put $s=a^2$ and $m=(4-\varpi ){\zeta }^2/\varpi >0$. The temperature equations first give $b=a/(1+s)$ and $c=s/(1+s)$. The magnetic equations then giveDivide the steady first equation by $\sigma a$ and substitute these expressions. The exact branch isExpanding at $s=0$ givesNear $\varpi =2$ the required amplitude restriction $a^2\ll {\zeta }^2$ ensures $s\ll m$. For $\varpi \le 2$, $r_2>0$, so the paired equilibrium points emerge in the forward direction $r>1+q$. For fixed $q>0$ and small $\zeta$, the pitchfork reversal near geometric ratio two in magnetoconvection occurs atThis follows by expanding the exact zero-slope equation $q\varpi (2-\varpi )+(4-\varpi ){\zeta }^2(1+q)=0$. Above this value the branch initially runs backwards into $r<1+q$; at $q=0$ there is no reversal. This is a degenerate pitchfork bifurcation at $r_2=0$, not a conclusion that every emerging state is stable. If $B_s>0$, the center-direction cubic term has the opposite sign to $r_2$: the forward branch is locally attracting and the backward branch repelling. If $B_s<0$, an unstable transverse mode remains.

For the nearby geometry, the next coefficient is $r_4=q[(2\varpi -5)/m^2+(2-\varpi )/m]$, which is negative near the reversal. Thus a small positive $r_2$ produces a nearby maximum of $r$ at $s\simeq -r_2/(2r_4)>0$; this fold approaches the pitchfork as $r_2\downarrow 0$. Once $r_2<0$ the small-amplitude branch is backward. The exact formula eventually grows like $r\sim s$, so a backward branch must turn at a finite amplitude, potentially outside the small-amplitude expansion.

The truncated fields must be interpreted as a Galerkin method approximation: project each equation onto its retained spatial modes and discard orthogonal higher harmonics. They are not an exact pointwise solution of the full partial differential equations, because the nonlinear terms generate additional harmonics. Write $X=kx$, $Z=\pi z$, and abbreviate the printed normalizations byThen $\psi =Sa\sin X\sin Z$, $\theta =Tb\cos X\sin Z-c\sin 2Z/\pi$, and $\chi =Dd\sin X\cos Z+Ee\sin 2X$. All five amplitudes a,b,c,d,e depend on $\tau =pt$. Use the spatial Jacobian determinant $J(F,G)=F_x{G}_z-F_z{G}_x$ and the inner product over the stated rectangle. Orthogonality of the trigonometric modes makes projection equivalent to retaining their coefficients. In particular, ${\nabla }^2\psi =-p\psi$, so $J(\psi ,{\nabla }^2\psi )=0$.

For the temperature equation, direct products giveThe other temperature contribution is $-2kSac\cos X\sin Z\cos 2Z$. Since $\sin Z\cos 2Z=(\sin 3Z-\sin Z)/2$, its retained component is $kSac\cos X\sin Z$. Thus the projected temperature equation isHere the dot denotes $d/d\tau$. Since $kS/(pT)=1$ and $\phi =4{\pi }^2/p$, these equations give $\dot{b}=-b+a(1-c)$ and $\dot{c}=\phi (-c+ab)$.

Similarly, the two magnetic-advection products arewhile $J(\psi ,Ee\sin 2X)=-2\pi Sae\sin X\cos Z\cos 2X$ has retained component $\pi Sae\sin X\cos Z$. The projected induction equation is thereforeUsing $\pi S/(pD)=1$ and $4k^2/p=4-\phi$ gives $\dot{d}=-\zeta d+a(1-e)$ and $\dot{e}=-(4-\phi )\zeta e+\phi ad$.

It remains to project the Lorentz force in the vorticity equation. The linear term is $J(x,{\nabla }^2\chi )=p\pi Dd\sin X\sin Z$. The nonlinear cross term isThe Jacobian on the right is $2\pi kDEde\sin X\sin Z\cos 2X$. Its retained component is $-\pi kDEde\sin X\sin Z$, so the total retained magnetic term isMeanwhile the projected acceleration, viscous term and temperature forcing have coefficients $-p^{2}S\dot{a}$, $p^{2}Sa$, and $-RkTb$. Divide the projected vorticity equation by $-p^{2}S$ to obtain $\dot{a}=\sigma [-a+rb+\zeta qd\{(\phi -3)e-1\}]$, using $r=k^{2}R/p^3$ and $q={\pi }^{2}Q/p^2$. This derives the complete five-mode vertical-field magnetoconvection system:The Rayleigh number, Chandrasekhar number, Prandtl number and magnetic-to-thermal diffusivity ratio enter through r,q,$\sigma$ and $\zeta$ respectively.

Linearization at the zero-amplitude state leaves c and e with negative rates $-\phi$ and $-(4-\phi )\zeta$, while a,b,d have coefficient matrixA zero eigenvector satisfies $b=a$, $d=a/\zeta$ and $[-1+r-q]a=0$. Thus the stationary threshold is $\boxed{r_c=1+q}$. The reversal symmetry $(a,b,c,d,e)\mapsto (-a,-b,c,-d,e)$ makes the nontrivial steady states occur as a pair.

For these nonzero steady states put $A=a^2$. The temperature equations give $c=ab$ and $b=a(1-c)$, so $b=a/(1+A)$ and $c=A/(1+A)$. The induction equations give $\zeta d=a(1-e)$ and $(4-\phi )\zeta e=\phi ad$. With $\mu =(4-\phi ){\zeta }^2/\phi$, their solution isHere $k>0$ gives $0<\phi <4$, and positive diffusivity gives $\mu >0$. Substituting these amplitudes into the stationary a equation and dividing by a yields the stationary branch of five-mode magnetoconvection:At $a=0$ this meets $r=1+q$. Expansion in $A=a^2$ givesThe denominator of its initial slope is positive, and its numerator after multiplication by $(4-\phi ){\zeta }^2$ isTherefore $\mathcal{N}>0$ gives a supercritical pitchfork bifurcation direction, and $\mathcal{N}<0$ gives a subcritical pitchfork bifurcation direction. If $\mathcal{N}=0$, the leading slope vanishes and higher terms must be retained.

This classification describes which side of the stationary threshold contains the steady branch. It does not alone establish dynamical stability or rule out an earlier oscillatory instability. For example, at $r_c$ the characteristic polynomial factors asThe zero eigenvalue is simple unless the final coefficient vanishes; the other two eigenvalues need not both be stable for every positive parameter choice. The stationary branch formula and its direction criterion remain the requested algebraic conclusions.

Put $m=(4-\phi ){\zeta }^2/\phi$. The derivative of the stationary branch of five-mode magnetoconvection at $a^2=0$ is $r_2=1+q+q(2-\phi )/m$. For fixed positive $q$ and small $\zeta$, it changes from a forward to a backward pitchfork bifurcation direction at the displayed value. Expanding the exact numerator $q\phi (2-\phi )+(4-\phi ){\zeta }^2(1+q)$ gives this result. The expansion in amplitude requires $a^2\ll m$; branch direction alone does not determine stability against all other modes.