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At $(x,y)=(\mu ,0)$ near $\mu =1$, use $X=x-\mu$, $Y=y$, $\nu =\mu -1$. The stable direction is the $X$ axis and the centre direction is the $Y$ axis. The extended centre manifold has $X=Y^2+O(\nu Y^2,Y^4)$ and reduced flow $\dot{Y}=-\nu Y-2Y^3+\cdots$, a supercritical pitchfork bifurcation in parameter $-\nu$. At $(0,1)$ near $\mu =-1$, use $X=x$, $Y=y-1$, $\nu =\mu +1$. The centre line is $Y=-X/2$, the stable line is $X=0$, and the reduced flow is $\dot{X}=\nu X-2X^2+\cdots$, a transcritical bifurcation. The branch $x=(\mu +1)/2$, $y=\sqrt{(1-\mu )/2}$ joins the two bifurcations and is stable for $-1<\mu <1$.

A simple zero eigenvalue occurs on $c=d$, except at $(1,1)$. The odd symmetry of the vector field gives a pitchfork bifurcation. Near this curve the critical eigenvalue isFor $0<d<1$ the other two eigenvalues are stable: this is a supercritical pitchfork bifurcation as $c$ decreases through $d$. For $d>1$ it is a subcritical pitchfork bifurcation in the center direction, with an already unstable transverse direction.

For a Hopf bifurcation, factor the characteristic polynomial as $(m+\tau )(m^2+{\omega }^2)$. Comparing coefficients givesThus the trivial branch has the local bifurcation curvesThe second curve is a Hopf bifurcation with a stable third eigenvalue. The Routh-Hurwitz stability criterion gives the stable trivial region $0<d<1$, $d<c<(d-2{)}^2$. Its Hopf bifurcation is subcritical. To check the sign, normalize the critical right eigenvector by $q_x=1$ and putThe second component of the normalized left projection is $G$. There is no quadratic term at the origin, and the cubic derivative in the second component is $C_2(q,q,\bar{q})=2$. The cubic Hopf coefficient is therefore $l_1=\mathrm{Re}G/\omega =4/[\omega ((4-d{)}^2+4{\omega }^2)]>0$. A small repelling periodic orbit lies on the stable-origin side. The diagram also includes the nonzero-branch information derived next.

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.

For a flow, a hyperbolic fixed point has no eigenvalue of its linearization with zero real part; a nonhyperbolic fixed point has at least one such eigenvalue. At $(\mu ,0)$ the eigenvalues are $-\mu ,1-\mu$, while at $(0,\pm 1)$ they are $\mu +1,-2$. Thus nonhyperbolicity occurs at the two stated parameters. The following centre manifold reductions show the associated actual bifurcations.

Near $(1,0)$ put $X=x-\mu$, $Y=y$, $\nu =\mu -1$. The transformed equations areAt $\nu =0$, the non-extended stable subspace is the $X$ axis and the centre subspace is the $Y$ axis. In the extended centre manifold, write $X=h(Y,\nu )$, with $h(0,\nu )=0$. The symmetry $Y\mapsto -Y$ allows an even graph. At quadratic order its invariance equation has left side of degree at least three and right side $-h+Y^2$, so $h=Y^2+O(\nu Y^2,Y^4)$. Substitution givesThis is a supercritical pitchfork bifurcation in parameter $-\nu =1-\mu$. The central branch is stable for $\mu >1$ and unstable for $\mu <1$ near one; the two nonzero branches for $\mu <1$ are stable. At the bifurcation itself the cubic term attracts on the centre manifold.

Near $(0,1)$ put $X=x$, $Y=y-1$, $\nu =\mu +1$. ThenAt $\nu =0$, the stable subspace is $X=0$ and the centre line is $Y=-X/2$. Thus the leading extended centre graph is $Y=-X/2+O(X^2,\nu X)$, and the reduced flow isThis is a transcritical bifurcation: $X=0$ is stable for $\nu <0$, while the branch $X=\nu /2+\cdots$ is stable for $\nu >0$; the opposite branches are unstable. At the critical point the quadratic term is only one-sided attracting.

The exact additional branch in $y>0$ isIt meets $(\mu ,0)$ at $\mu =1$ and $(0,1)$ at $\mu =-1$, displaying the connecting pitchfork and transcritical branches in a quadratic-product flow. Its Jacobian has determinant $1-{\mu }^2$ and trace $(\mu -3)/2$. Hence it is stable for $-1<\mu <1$ and a saddle for $\mu <-1$. The negative-$y$ companion is needed for the pitchfork sketch near $(1,0)$; no separate analysis of $(0,-1)$ is required.

Now $G(x)=x[{\mu }^2-(x^2-a{)}^2]$. The zero point exists for all parameters and has eigenvalue $G^{\prime }(0)={\mu }^2-a^2$, so it is stable for $|\mu |<|a|$ and unstable for $|\mu |>|a|$. At every nonzero fixed point,Both outer branches are stable and both inner branches unstable.

For $a>0$, the nonzero locations are as in part (i). At $(\mu ,x)=(\pm a,0)$ the local equation is $\dot{x}=({\mu }^2-a^2)x+2a{x}^3+O(x^5)$: these are subcritical pitchfork bifurcations, with the two unstable inner branches present on the side where zero is stable. At the bifurcation itself $\dot{x}\sim 2a{x}^3$, so zero is unstable. At $(0,\pm \sqrt{a})$ the inner and outer branches cross and exchange stability in transcritical bifurcations. The positive crossing is semistable, attracting from the right; the negative crossing is semistable, attracting from the left, because the additional factor $x$ reverses the local flow there.

For $a<0$, the outer stable pair exists only for $|\mu |>|a|$. The points $(\pm |a|,0)$ are supercritical pitchfork bifurcations, since the cubic coefficient $2a$ is negative. Zero loses stability as $|\mu |$ increases across $|a|$, and a stable symmetric pair appears. At the threshold zero remains nonlinearly stable, with $\dot{x}\sim 2a{x}^3$. There are no further bifurcations. The second row of the preceding figure gives the diagrams. The plots use $a=\pm 1$; the general locations are obtained from the formulas above.

For positive diffusivity ratio $\zeta$ and $0<\phi <4$, define $\mu =(4-\phi ){\zeta }^2/\phi$. The nonzero steady branch has $b=a/(1+a^2)$, $c=a^2/(1+a^2)$, $d=\mu a/[\zeta (\mu +a^2)]$, and $e=a^2/(\mu +a^2)$. Substitution yields the displayed branch through $r=1+q$. Its initial slope in $a^2$ has the sign of $\phi q(2-\phi )+(4-\phi ){\zeta }^2(1+q)$. Positive slope is a supercritical pitchfork bifurcation direction, negative slope a subcritical pitchfork bifurcation direction. This geometric criterion alone does not assert stability against every perturbation.