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The conduction linearization has two negative eigenvalues and a cubic factor $s^3+A{s}^2+Bs+C$, where $A=\sigma +1+\zeta$, $B=\sigma (1-r)+\sigma \zeta (1+q)+\zeta$, $C=\sigma \zeta (1+q-r)$. A double zero requires $B=C=0$, giving the displayed Bogdanov–Takens bifurcation location for $0<\zeta <1$. The remaining cubic eigenvalue is $-A$. The imaginary-pair condition $AB=C$ yields $r_H=(\sigma +\zeta )(1+\zeta )/\sigma +\zeta (\sigma +\zeta )q/(\sigma +1)$ and ${\omega }^2=\sigma \zeta (1-\zeta )q/(\sigma +1)-{\zeta }^2$. Thus the oscillatory onset exists precisely for $q>q_{\ast }$ when $0<\zeta <1$.

For a reflection-symmetric vector field $\dot{u}=v+f_{30}{u}^3+f_{21}{u}^{2}v+\cdots$, $\dot{v}=g_{30}{u}^3+g_{21}{u}^{2}v+\cdots$, introduce $x=u$, $y=\dot{u}$. Differentiating gives $\dot{x}=y$ and a cubic acceleration with $x^{2}y$ coefficient $b=g_{21}+3f_{30}$. The contribution $3f_{30}$ comes from differentiating the first equation's cubic term. Further cubic near-identity transformations can remove nonresonant terms but cannot turn $b=0$ into a nonzero coefficient by a nonsingular scaling. This is a useful check before assigning a generic reflection-symmetric Bogdanov–Takens bifurcation to a calculated Taylor expansion.

The linearization of this reduction of the FitzHugh-Nagumo model at its trivial equilibrium point isA double zero eigenvalue requires both the constant and linear coefficients to vanish. Hence the codimension-two point isThere the polynomial is $m^2(m+3/2)$. The kernel of a linear map of $L$ is one-dimensional, so its zero eigenvalue has a nontrivial size-two block in Jordan normal form. This is a reflection-symmetric double-zero point of Bogdanov–Takens bifurcation type, with one stable direction; the nonlinear coefficients are computed below rather than assumed.

At conduction, $c$ and $e$ have stable eigenvalues $-\varpi$ and $-(4-\varpi )\zeta$. The remaining linearization on $(a,b,d)$ isIts characteristic polynomial is $m^3+A{m}^2+Bm+C$, whereConsequently the steady threshold is $\boxed{r_s=1+q}$. The symmetry $(a,b,d)\mapsto (-a,-b,-d)$, with $(c,e)$ unchanged, gives a generic pitchfork bifurcation, whose forward or backward direction is determined below. At threshold the other two critical-block eigenvalues satisfy $m^2+Am+B_s=0$, where $B_s=\zeta (\sigma +1)-\sigma (1-\zeta )q$. They are stable if $B_s>0$; if $B_s<0$ there is already one unstable transverse direction. Degenerate zero cubic slope requires higher-order analysis and is not a generic pitchfork.

At a Hopf bifurcation, $AB=C$, $B={\omega }^2>0$. Solving the coefficient equation givesFor physical $q\ge 0$ the oscillatory threshold therefore exists exactly when $0<\zeta <1$ and $q>q_{\ast }$. The conduction double-zero criterion for five-mode magnetoconvection givesAt this Bogdanov–Takens bifurcation, the polynomial is $m^2(m+A)$ and the zero eigenspace is one-dimensional. For $q<q_{\ast }$ conduction first loses stability at the steady threshold; for $q>q_{\ast }$ it first loses stability at $r_H<r_s$. For $\zeta \ge 1$ no positive-$q$ double-zero or oscillatory threshold exists. These conclusions also follow from the Routh-Hurwitz stability criterion, $B,C>0$ and $AB>C$. The Hopf bifurcation is transversal since ${\partial }_r(AB-C)=-\sigma (\sigma +1)\ne 0$; its nonlinear criticality is not required by this linear onset calculation.

The equilibrium points satisfy $y=0$, $x^2=\lambda$. For $\lambda <0$ there are none. For $\lambda >0$ set $a=\sqrt{\lambda }$. The Jacobian matrix isThe positive equilibrium point $(a,0)$ is always a saddle equilibrium. At $(-a,0)$ the determinant is $2a$, so it is stable for $\mu <a$ and unstable for $\mu >a$. The discriminant $(\mu -a{)}^2-8a$ distinguishes a node from a focus; its zero does not destroy hyperbolicity and is not another local bifurcation curve.

The local bifurcation curves areAt $(0,0)$ the linearization has a double zero eigenvalue and is nilpotent, giving a Bogdanov–Takens bifurcation. For $\mu \ne 0$ on $\lambda =0$, solving the local center manifold gives $\dot{x}=(\lambda -x^2)/\mu +\cdots$, displaying the saddle-node bifurcation. For $\mu <0$ the nonsaddle branch born at positive $\lambda$ is attracting; for $\mu >0$ it is repelling. On the Hopf curve the angular frequency is $\sqrt{2a}$. The permitted subcritical Hopf bifurcation creates a repelling periodic orbit on the stable-focus side $\mu <a$.

A local analysis alone does not say how far that periodic orbit extends. The global calculation below supplies the additional division of the stable-focus region. The complete nearby phase portraits and the leading global curve are shown together here. The $lambda<0$ flow has no closed periodic orbit, since a closed orbit would have index one and hence enclose an equilibrium point.

Let $q=r_{\Omega }^2$ and $h=\varpi$. The $c,e$ eigenvalues at the conductive equilibrium point are $-h$ and $-(4-h)\sigma$, both negative since $0<h<4$ and $\sigma >0$. The coupled $(a,b,d)$ Jacobian matrix has characteristic polynomialwhereA simple zero eigenvalue occurs on $\boxed{r_P=1+q}$. The reflection $(a,b,d)\mapsto (-a,-b,-d)$, with $c,e$ fixed, makes its generic steady-state bifurcation a pitchfork bifurcation.

For a nonzero imaginary pair, substitute $s=i\omega$ into $P$: the real and imaginary parts require $A_3=A_1{A}_2$ and ${\omega }^2=A_2>0$. HenceWith $0<\sigma <1$, this Hopf bifurcation curve is present only for $q>q_{TB}=(1+\sigma )/(1-\sigma )$. At its intersection with the pitchfork bifurcation curve,and $P(s)=s^2(s+1+2\sigma )$. The zero eigenvalue has geometric multiplicity one: its eigenvector has $b=a$, $d=a/\sigma$. Thus this is a reflection-symmetric Bogdanov–Takens bifurcation, rather than two independent zero modes.

The Routh-Hurwitz stability criterion gives the stable trivial state below $r_P$ for $q<q_{TB}$ and below $r_H$ for $q>q_{TB}$. The rest of the line $r_P$, beyond $q_{TB}$, is a stationary bifurcation of an already unstable trivial state. Likewise the extension of $r_H$ below $q_{TB}$ is only an algebraic line, since its putative frequency is not real. The middle panel of the bifurcation diagram in part 1(b) distinguishes the actual curves from this extension.

A codimension-three degeneracy is possible when the double-zero eigenvalue condition and the vanishing cubic restoring coefficient of the pitchfork bifurcation coincide. Equating $q_D$ and $q_{TB}$ givesEquivalently $L=\sqrt{2{\sigma }^2/(1+\sigma )}$. For every $0<\sigma <1$, this is a physically admissible aspect ratio and $0<h<4$. The other two parameters are $q=(1+\sigma )/(1-\sigma )$ and $r=2/(1-\sigma )$.

At this point the fifth-order restoring term is nonzero: the steady relation is $r-r_{TB}=k{a}^4+O(a^6)$ with $k>0$. It is therefore a degenerate reflection-symmetric Bogdanov–Takens bifurcation with three independent conditions: zero determinant, zero frequency, and zero cubic restoring coefficient. Varying the aspect ratio supplies the third parameter, in addition to $r,q$. At a fixed generic aspect ratio it is not a codimension-two event available by varying only $r,q$.

It is not a triple-zero eigenvalue: the three-mode characteristic polynomial still has the noncritical factor $s+1+2\sigma$, while the $c,e$ eigenvalues remain strictly negative. The extra degeneracy is nonlinear. A fifth-order, two-dimensional normal form is needed for its complete unfolding; the one-dimensional quintic pitchfork bifurcation used away from $q_{TB}$ cannot be substituted at the double-zero point.

For $\lambda >0$ put $s=\sqrt{\lambda }$. The equilibrium points are $(s,0)$ and $(-s,0)$, withThe positive branch is always a saddle. The negative branch is a sink for $\mu <s$ and a source for $\mu >s$; it is a node or focus according as $(\mu -s{)}^2-8s$ is positive or negative. At $\lambda =0$ the two branches meet in a saddle-node bifurcation when $\mu \ne 0$; at $(\lambda ,\mu )=(0,0)$ the double zero eigenvalue with a nontrivial Jordan block gives a Bogdanov–Takens bifurcation. There are no real equilibrium points for $\lambda <0$.

The Hopf bifurcation curve is $\mu =s>0$, equivalently $\lambda ={\mu }^2$ on its positive-$\mu$ branch. Its frequency is $\omega =\sqrt{2s}$. To determine the direction, shift $u=x+s$, $v=y$ at this threshold; thenSet $X=u$, $Y=-v/\omega$ to obtain $\dot{X}=-\omega Y$, $\dot{Y}=\omega X-X^2/\omega +XY$. A quadratic near-identity transformation, or harmonic averaging with its quadratic correction retained, gives the radial cubic coefficientwhere $g=-X^2/\omega +XY$. The radial equation is $\dot{R}=(\mu -s)R/2+{\ell }_H{R}^3+\cdots$. Hence the Hopf bifurcation is subcritical, with an unstable small periodic orbit on the stable-focus side $\mu <s$.

The saddle's outgoing separatrix can return to its incoming separatrix, creating a homoclinic orbit. This global bifurcation cannot be detected from the equilibrium point's Jacobian matrix alone. The weakly Hamiltonian scaling below locates its curve and shows how the unstable Hopf cycle grows into the saddle loop. The final diagram includes the local saddle-node and Hopf curves and this global curve; it describes the neighborhood of the double-zero point.

At $(\sigma ,\mu )=(1,0)$ the origin's matrix is $\left(\begin{array}{cc}0& 0\\ 2& 0\end{array}\right)$: it has a double zero eigenvalue and one eigenvector. Trace and determinant vary independently with $\mu$ and $\sigma$ there, so two parameters are needed. The symmetry of an odd function removes quadratic terms, giving a reflection-symmetric cubic double-zero normal form. The quadratic nonlinearity required for a generic Bogdanov–Takens bifurcation vanishes.

Use a prime for differentiation with respect to the rescaled time. Direct substitution, with all rescaled variables written without tildes, givesAt $\epsilon =0$, the equations are $u^{\prime }=-sv+v^3/2$, $v^{\prime }=2u$. Fordifferentiation gives $H^{\prime }=2u(-sv+v^3/2)+(sv-v^3/2)2u=0$. This is the Hamiltonian blow-up of a reflection-symmetric double-zero point.

For $s>0$, the equilibrium points are a centre equilibrium at $(0,0)$ and saddles at $(0,\pm \sqrt{2s})$. The centre equilibrium has eigenvalues $\pm i\sqrt{2s}$; at either saddle equilibrium they are $\pm 2\sqrt{s}$. Levels $0<H<s^2/2$ contain closed inner ovals around the centre equilibrium. The saddle equilibrium level hasThese two branches form a heteroclinic cycle connecting the saddles. The branch with $u>0$ runs from the lower saddle equilibrium to the upper one because $v^{\prime }=2u>0$; the other runs back. Outside the potential well, trajectories are open. The right panel of the preceding figure uses the coordinates $(v,u)$ to display the well and these directed connections.

Put $y=\dot{x}$. The planar autonomous differential equation isIts equilibria have $y=0$, $x^2=p$. There are none for $p<0$, one at $p=0$, and two $x_{\pm }=\pm \sqrt{p}$ for $p>0$. At an equilibrium $x_{\ast }$ the stability matrix isWriting $s=\sqrt{p}>0$, $x_{+}=s$ is always a saddle equilibrium. At $x_{-}=-s$, the determinant is $2s>0$ and the trace is $s-q$: it is attracting for $q>s$ and repelling for $q<s$.

At $p=0$ and $q\ne 0$, the eigenvalues are $0,-q$; the nonzero quadratic term and parameter unfolding give a saddle-node bifurcation. At $p>0$, the trace crosses zero at $q=s$, with eigenvalues $\pm i\sqrt{2s}$, giving a Hopf bifurcation. It is nondegenerate: around $x=-s$ the equation is $\ddot{\xi }+(q-s)\dot{\xi }+2s\xi -{\xi }^2+\xi \dot{\xi }=0$. At $q=s$, a small oscillatory expansion of leading amplitude $r$ has $⟨\xi {\dot{\xi }}^2⟩=r^4/8+O(r^5)$, so the first nonlinear damping is positive and nonzero. For example, its second-order oscillatory expansion is $\xi =r\cos (\Omega t)+r^2/(2{\Omega }^2)-r^2\cos (2\Omega t)/(6{\Omega }^2)-r^2\sin (2\Omega t)/(6\Omega )$ with ${\Omega }^2=2s$. Thus the small attracting limit cycle is on the $q<s$ side. At $(p,q)=(0,0)$ the matrix has a double zero eigenvalue and a nontrivial Jordan block, the Bogdanov–Takens bifurcation organizing the two curves. The local bifurcation curves areA change between node and focus away from these curves is not itself a local bifurcation.

For the requested global exclusion, suppose $x(t)$ were periodic with period $T$. Integrate the original equation over a period. The integrals of $\ddot{x}$, $q\dot{x}$, and $x\dot{x}=\frac{1}{2}(d/dt)x^2$ vanish. ThereforeThe left side is nonnegative and the right side is negative when $p<0$. This contradiction proves there are no periodic orbits for $p<0$, independently of $q$.