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A node equilibrium with a repeated nonzero real eigenvalue. If the planar Jacobian matrix has a one-dimensional eigenspace, its linearized solutions contain a term $t{e}^{\lambda t}$ and the node is also called improper. A negative repeated eigenvalue still gives asymptotic stability; a repeated value does not by itself indicate a bifurcation.

The FitzHugh-Nagumo model represents an excitable cell by a fast activation variable and a slower recovery variable. A common nondimensional form is $\dot{V}=V-V^3/3-W+I$, $\dot{W}=\epsilon (V+a-bW)$. Its cubic activation response and recovery feedback can produce an excitable resting state or sustained oscillations through bifurcations. Particular reductions and travelling-wave equations can have additional state variables; their signs and parameter conventions must be stated separately.

A parameter specifies a member of a family of mathematical models and is held fixed while a particular solution evolves. Varying it can change equilibrium existence, stability or long-term dynamics, producing a bifurcation. A model parameter differs from its evolving state variables and from an initial condition, even though all may be numerical inputs to a computation.

Set $z=y-\beta$. Then the autonomous planar system becomeswhich has no $\beta$ dependence. ThereforeMore precisely, replacing $\beta$ by $\beta +\delta$ sends every trajectory $(x(t),y(t))$ to $(x(t),y(t)+\delta )$, with the same time parametrization. The equilibrium points, nullclines, periodic orbits and separatrices all undergo this same translation. Their types and orientations are independent of $\beta$, so varying $\beta$ produces no bifurcation.

For $r>0$, the origin always exists. Its derivative $r-1$ makes it attracting for $0<r<1$ and repelling for $r>1$. Nonzero equilibria solveThey occur in symmetric pairs. On the first positive lobe $0<x<\pi$, the sinc function decreases from one to zero, giving one positive and one negative nonzero equilibrium for $0<r<1$. Both are repelling, since $f^{\prime }=\sin x/x-\cos x>0$ there. Near $r=1$,The unstable branches meet the stable origin in a subcritical pitchfork bifurcation. At $r=1$ the origin is still nonlinearly repelling because the leading term is $+x^3/6$. For $1<r\le 2$ it is the only equilibrium.

For each higher positive-sine lobe $2n\pi <x<(2n+1)\pi$, $n\ge 1$, the sinc function has a single positive maximum at $a_n$, whereIf $0<r<r_n$, that lobe has two roots: the smaller positive root is attracting and the larger is repelling, because $f^{\prime }=-xd(\sin x/x)/dx$. Their negative counterparts have the same stability. At $r=r_n$ the pair on each side merges in a saddle-node bifurcation; the double equilibria are semistable. For $r>r_n$ that lobe contributes none. The first such threshold is about $0.128375$.

For $n\gg 1$, put $b_n=(2n+1/2)\pi$; expanding near the sine maximum gives $a_n=b_n-b_n^{-1}+O(b_n^{-3})$ and $r_n\simeq 1/b_n$. Away from bifurcation values, the exact counting rule for $0<r<1$ and its small-$r$ estimate areAt a saddle-node threshold, count each merged positive or negative equilibrium once rather than twice. Infinitely many thresholds accumulate at $r=0$, consistent with the infinitely many equilibria in solution.

Eliminate $\dot{v}$ from the first equation before linearization:All partial derivatives below are evaluated at $(\mu ,g(\mu ))$. The Jacobian matrix isIts trace and determinant areBy linear stability of a planar equilibrium, a positive trace makes both real eigenvalues positive or gives a complex pair with positive real part. The equilibrium is consequently unstable precisely whenA negative trace gives asymptotic stability; equality leaves a purely imaginary pair and requires a separate nonlinear bifurcation calculation. In particular, the unique positive equilibrium is never a saddle equilibrium.

A steady transverse bifurcation requires a zero eigenvalue, hence $\det M=0$. This givesThe inequality is required for a positive amplitude. Since $d_R>0$, the numerator cannot vanish. The zero eigenvalue is simple provided the other eigenvalue, the trace $-2(d_R+{\beta }_R{r}_S)$, is nonzero. Equivalently,With $d,\beta$ fixed, ${\partial }_r\det M=2Q\ne 0$, so the simple zero eigenvalue crosses transversely. If this is to be the first instability of the stable pure branch, its other eigenvalue must still be negative. Comparing with $r_H=-d_R/{\beta }_R$ gives the stronger conditionFor $Q_{\ast }<Q<0$, a simple steady threshold still exists, but it lies after the Hopf threshold and the pure branch is already unstable there. At $Q=Q_{\ast }$ both transverse eigenvalues vanish, so the event is a double-zero degeneracy rather than an ordinary simple steady bifurcation.

The reduced system has the reflection symmetry $C\mapsto -C$. At a generic simple steady threshold, the new mixed states therefore occur in a reflection-related pair, with small nonzero $B$ and a corresponding correction to the $A$ magnitude and common frequency. In the original variables they are phase-locked periodic solutions involving both spatial modes: $A=a{e}^{i\tilde{\Omega }t}$, $B=b{e}^{i\tilde{\Omega }t}$, with constant nonzero complex $a,b$. The odd spatial component breaks the reflection symmetry of the pure even mode; the two branches are exchanged by $b\mapsto -b$. Determining whether the branches are supercritical or subcritical, and their stability, requires the reduced nonlinear coefficient and is not determined by this linear threshold alone.

The equilibria have $y=0$ and $x(\lambda +x^2)=0$. ThusAt $O$, the linearization has trace $\kappa$ and determinant $\lambda$. Hence $O$ is a saddle equilibrium when $\lambda <0$, a sink equilibrium when $\lambda >0,\kappa <0$, and a source equilibrium when $\lambda >0,\kappa >0$. The sink equilibrium/source equilibrium may be a node equilibrium or a focus equilibrium; the node-focus discriminant is not a bifurcation curve. At $P_{\pm }$, the trace and determinant are $\kappa +\lambda$ and $-2\lambda$. These points are sink equilibria for $\kappa <-\lambda$ and source equilibria for $\kappa >-\lambda$.

The local bifurcation curves are thereforeFor $\kappa \ne 0$ on the pitchfork curve, a leading centre-manifold equation is $\dot{x}=(\lambda x+x^3)/\kappa +\cdots$. For $\kappa <0$, with unfolding parameter $-\lambda$, the pitchfork is supercritical and creates two sink equilibria from a sink equilibrium that becomes a saddle equilibrium. For $\kappa >0$ it is subcritical in the centre direction; the nonzero branches are unstable, and the transverse eigenvalue is also positive. The origin of parameter space is a reflection-symmetric double-zero degeneration.

To determine Hopf criticality at the origin, use the energy $E=y^2/2+\lambda x^2/2+x^4/4$, whose derivative is $(\kappa -x^2)y^2$. For a small nearly harmonic oscillation $x=R\cos (\sqrt{\lambda }t)$, averaging gives $\dot{R}=\kappa R/2-R^3/8+\cdots$. Thus this Hopf bifurcation is supercritical and creates an attracting periodic orbit for $\kappa >0$. By the supplied opposite-criticality fact, the simultaneous Hopf bifurcations at the two nonzero equilibria are subcritical. A direct calculation confirms this: writing $x=\sqrt{-\lambda }+X$, $Y=-y/\sqrt{-2\lambda }$ puts the cubic radial coefficient at $1/4>0$, including the contribution from the quadratic terms.

The figure includes these local curves and the global curves derived below. All lines emerging from the double-zero point are local asymptotic sketches.

A returning orbit must have zero net change of the energy. To first order, integrate the perturbation along either unperturbed homoclinic loop:For the positive loop, $v_h=\pm u_h\sqrt{a_0^2-u_h^2/2}$; the outward and inward halves contribute equally. ThereforeandThe integrals follow by substituting $w=a_0^2-u^2/2$; for the second one use $u^2=2(a_0^2-w)$. HenceThe equality in rescaled parameters is the leading Melnikov balance, not an assertion that the entire finite-parameter bifurcation curve is exactly straight. Reflection symmetry makes both saddle equilibrium loops occur together. The global curve lies below the nonzero-equilibrium Hopf line $\kappa =-\lambda$, as shown in the parameter sketch.

The change of time is $d\tau /dt=u$, which is positive in the half-plane under consideration. Dividing the two approximate equations by $u$ givesSet $w=v-\mu$. The inhomogeneous system becomes $(u,w{)}^{\prime }=(u-w,\lambda u)$, with eigenvalues $(1\pm \sqrt{1-4\lambda })/2$. For $\lambda >1/4$, write $\omega =\sqrt{4\lambda -1}/2$. Integrating with the prescribed initial data givesandThese expressions also verify both initial values and their initial derivatives. For fixed $\mu >0$ and small $h,v_0$ on the outgoing branch, the sine coefficient in $u$ is positive and dominates the initial cosine term. The first return to small $u$ is near the first positive zero, $\omega \tau \simeq \pi$. HenceThe factor is a half-turn of the growing spiral, not its full rotation period. After this excursion the approximate trajectory approaches the small-$u$, large-$v$ region. In the full system the term $-\lambda v$, omitted during the large excursion, allows the motion to descend along that region and return.

As $\lambda \downarrow 1/4$, the rescaled excursion time diverges and the factor $e^{\tau /2}$ produces arbitrarily large excursions. At the critical value the matrix is a nontrivial Jordan block, and the limiting formula isFor the small outgoing initial data this is positive and unbounded: the unstable manifold no longer turns back. Below $1/4$ the two eigenvalues are real and positive, so there is again no oscillatory return. This quadratic escape desingularization explains loss of the periodic orbit at infinity, together with the unstable-manifold escape, as $\lambda$ decreases through $1/4$. It is not a finite homoclinic bifurcation to the origin. The divergent $\tau$ travel time must not be identified with the physical period without integrating $dt=d\tau /u$; the time change becomes singular near $u=0$.

A global bifurcation joins two mechanisms: a long passage near a hyperbolic equilibrium point and a finite excursion returning the unstable manifold to its neighborhood. Eigenvalue ratios control the local passage, while the geometry of the global return decides reinjection, folding and trapping. In the real symmetric case there are two returning branches and a power-law cusp. In the saddle-focus case rotation adds infinitely many oscillations to that cusp. The following maps display both mechanisms explicitly.

For the Lorenz configuration, enter on $z=h$ at $(x,y,h)$ and leave on $x=sh$, where $s=\mathrm{sgn}(x)$. Solving the linear flow gives a flight timePut $\gamma =-\hat{\lambda }/{\lambda }_{+}>\delta =-{\lambda }_{-}/{\lambda }_{+}$. The outgoing coordinates areThe smooth global map from each outgoing section to the incoming section has, to leading order,up to the choice of which unstable branch carries the sign of $\nu$. Reflection symmetry makes the offsets opposite and the leading coefficients equal. Since the stable transverse coordinate is multiplied by the higher power $|x{|}^{\gamma }$, it contributes only subleading terms to the leading scalar map. Absorbing powers of $h$ into a positive constant gives the Lorenz power return mapThe line $x=0$ lies on the stable manifold and has no finite return; the discontinuity is therefore a genuine missing return, not an arbitrary value to assign at zero.

If $\delta >1$, the derivative $A\delta |x{|}^{\delta -1}$ tends to zero. For $\mu <0$, there are two attracting fixed points, $x=\pm s$ with $s=-\mu +A{s}^{\delta }\simeq -\mu$. They describe two symmetry-related one-lobe attracting flow periodic orbits. For $\mu >0$, the orbit alternates between $s$ and $-s$, where $s=\mu -A{s}^{\delta }\simeq \mu$; its multiplier is $(A\delta s^{\delta -1}{)}^2<1$. This represents one attracting flow periodic orbit traversing both lobes. At $\mu =0$ the two homoclinic loops meet: this is the contracting gluing scenario, with diverging flow period at the connection.

If $\delta <1$, the local derivative is unbounded, and the small periodic orbits are repelling in the expanding return direction. For $\mu >0$, the small positive fixed point has $s\simeq (\mu /A{)}^{1/\delta }$, with a multiplier tending to infinity; there is a symmetry-related negative fixed point as well. More complex motion follows from the two branches, not merely from a large derivative at one fixed point. Choose $I=[-c\mu ,c\mu ]$ with $0<c<1$. The inverse branches areFor sufficiently small positive $\mu$, both lie inside $I$, their images are disjoint, and their derivatives are $O({\mu }^{1/\delta -1})\to 0$. Nested inverse images for every infinite sign sequence therefore define a Cantor invariant set. Iterating the forward map shifts the sequence. This constructs periodic itineraries of every symbolic period and nonperiodic itineraries; transverse contraction in the full return turns the expanding two-strip mechanism into a horseshoe.

This is the local chaotic Lorenz mechanism. A robust attracting Lorenz set additionally needs a global trapping region and the appropriate return properties. The scalar asymptote and $\delta <1$ alone do not prove that every nearby orbit is attracted to one chaotic set; periodic windows, escaping orbits and other global returns must be distinguished. Symmetry is crucial to obtaining the two homoclinic branches together in a single parameter variation.

At the no-convection equilibrium $(a,b,c)=(0,0,0)$, one eigenvalue is $-\zeta$ and the other two come fromTheir trace and determinant areA zero eigenvalue occurs at $r=1+q$. If $q<\zeta /\sigma$, the other two eigenvalues are negative, and the stationary bifurcation is the pitchfork analyzed in part (c). For $q>\zeta /\sigma$ the transverse eigenvalue is positive instead.

An imaginary pair occurs at $r=1+\zeta /\sigma$, provided $q>\zeta /\sigma$, becauseThe trace crosses zero with derivative $\sigma >0$. The cubic coupling is nondegenerate here: at criticality, putting $K=\sigma \zeta q$ reduces the oscillatory variables toFor $a=R\cos ({\omega }_H\tau )$, the forced second harmonic of $c$ has sine coefficient $9R^2\zeta {\omega }_H/[2K({\zeta }^2+4{\omega }_H^2)]$. Averaging the derivative of $(a^{\prime 2}+{\omega }_H^2{a}^2)/2$ consequently gives $-9\zeta {\omega }_H^2{R}^4/[8({\zeta }^2+4{\omega }_H^2)]<0$. Thus the Hopf bifurcation is nondegenerate and supercritical.

The two thresholds meet atHere the planar block is nonzero but has trace and determinant zero, while the third eigenvalue remains negative. It is a reflection-symmetric double-zero, codimension-two point; the ordinary simple-eigenvalue reduction of part (c) is not valid there.

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 expansionswhere the omitted terms in the reduced equation have order ${\mu }^{2}a,\mu a^3,a^5$. At $\mu =0$, $a^{\prime }$ begins at cubic order, so $c^{\prime }=c_a{a}^{\prime }$ begins at fourth order. The $c$ equation therefore gives $c_2=3/{\zeta }^2$.

The invariance equation for $b$ must retain $b_a{a}^{\prime }=a^{\prime }/\zeta +\cdots$. From the first equation,Equating coefficients in $b_a{a}^{\prime }=a-\zeta b-ac$ givesHenceand the required reduced equation isBoth 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,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 $b^{\prime }=c^{\prime }=0$ 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.

Put $a=a_1>0$, so $a_2=4a/5$ and $a_3=8a/5$. The second equilibrium equation factors as $B({\lambda }_2+a_{3}A)=0$. The resulting equilibria areThe mixed pair exists strictly when ${\lambda }_2({\lambda }_1-5{\lambda }_2/8)>0$; equality describes its merger with an axial equilibrium. At $O$ the eigenvalues of the Jacobian matrix are $({\lambda }_1,{\lambda }_2)$, and at $P$ they are $(-{\lambda }_1,{\lambda }_2-8{\lambda }_1/5)$. At a nonzero mixed equilibrium,Thus both mixed equilibria are saddle equilibria. The axial equilibria are nodes when their two real eigenvalues have the same sign and saddles when the signs differ. In particular,There are exactly three local bifurcation lines away from their common codimension-two intersection. On ${\lambda }_1=0$, $O$ and $P$ meet in a transcritical bifurcation, visible in $\dot{A}={\lambda }_{1}A+a{A}^2$ on the invariant axis $B=0$. On ${\lambda }_2=0$, the mixed pair meets $O$ in a pitchfork bifurcation. Eliminating $A$ near $O$, for fixed nonzero ${\lambda }_1$, gives $\dot{B}={\lambda }_{2}B-(a_2{a}_3/{\lambda }_1)B^3+\cdots$: this pitchfork is supercritical for ${\lambda }_1>0$ and subcritical for ${\lambda }_1<0$. On ${\lambda }_2=8{\lambda }_1/5$, the mixed pair meets $P$. Writing $A=-{\lambda }_1/a+u$ and $\eta ={\lambda }_2-8{\lambda }_1/5$, elimination gives $\dot{B}=\eta B+(a_2{a}_3/{\lambda }_1)B^3+\cdots$. This pitchfork is subcritical for ${\lambda }_1>0$ and supercritical for ${\lambda }_1<0$. A supercritical pitchfork here need not produce a stable mixed state: its other eigenvalue can already be positive. The negative mixed Jacobian determinant settles the full stability question. There is no Hopf bifurcation in this quadratic truncation.

Along ${\lambda }_2={\lambda }_1-\Delta$, the axial curves are $A=0$ and $A=-{\lambda }_1/a$. The mixed pair has the same plotted amplitude for its two signs of $B$:Consequently the mixed curves exist outside the interval whose endpoints are $\Delta$ and $-5\Delta /3$. The stable portions of the axial curves areFor $\Delta >0$, stability passes directly from $O$ to $P$ at the transcritical bifurcation ${\lambda }_1=0$; the two pitchfork bifurcations lie at $-5\Delta /3$ and $\Delta$. For $\Delta <0$, $O$ loses stability at ${\lambda }_1=\Delta$ and $P$ becomes stable only at ${\lambda }_1=-5\Delta /3$. Between these points none of the equilibria is stable. This local statement does not prescribe a bounded attractor of the quadratic truncation. All mixed portions are unstable.

The linearization at the zero solution is the self-adjoint differential operator $L=\mu -({\partial }_x^2+K{)}^2$, with domain satisfying both endpoint conditions. The Dirichlet boundary condition for $-{\partial }_x^2$ on $(-1,1)$ gives the orthogonal eigenfunctions $\cos ((n+\frac{1}{2})\pi x)$ and $\sin ((n+1)\pi x)$. Their second derivatives are multiples of themselves, so the additional condition ${\psi }_{xx}(\pm 1)=0$ is automatic. Completeness of the Dirichlet eigenfunctions also shows that these exhaust the spectrum of the polynomial operator $L$. Set $q_j=(j+1)\pi /2$; each eigenfunction has growth rate ${\sigma }_j=\mu -(K-q_j^2{)}^2$. Hence the stationary bifurcation thresholds areEquality of the first two thresholds requires $K$ to be the midpoint of ${\pi }^2/4$ and ${\pi }^2$, since these numbers are distinct. ThereforeAt this point the even and odd lowest eigenfunctions become neutral simultaneously.

Use local coordinates $q=\theta -\pi /2$, $p=\varphi$ at $P_1$. The Jacobian matrix, its trace and its determinant areConsequently $P_1$ is attracting when $T<0,D>0$, repelling when $T>0,D>0$, and a saddle equilibrium when $D<0$. A transition between a node and a focus at $T^2=4D$ changes the linear geometry but is not a bifurcation of the local topological phase portrait.

The stationary bifurcation curve is $\lambda +\kappa =2$. To identify it, write $C=\cos 2\varphi$, $S=\sin 2\varphi$. At a non-equatorial equilibrium point, eliminating $S$ givesNear $(3,-1)$, if $d=2-\lambda -\kappa >0$, these yield $p=\pm \sqrt{d}/2+O(d^{3/2})$ and $q=p+o(\sqrt{d})$. The two branches are related by the local reflection $(q,p)\mapsto (-q,-p)$. Their determinants are $-4d+o(d)$, so they are saddle equilibria. This is a pitchfork bifurcation; on the portion with $T<0$ the saddle branches lie on the side where $P_1$ is stable, giving a subcritical pitchfork bifurcation.

A Hopf bifurcation requires $T=0,D>0$, henceThe oscillation frequency is $\omega =\sqrt{-2(1+\kappa )}$. At the intersection $\boxed{(\lambda ,\kappa )=(3,-1)}$, $J\ne 0$ but $J^2=0$: there is a double zero eigenvalue with one eigenvector. The map from $(\lambda ,\kappa )$ to $(T,D)$ has nonzero Jacobian determinant, so two independent parameter conditions produce this codimension-two bifurcation. Reflection removes quadratic terms; the appropriate initial comparison is with a reflection-symmetric cubic double-zero normal form, subject to the cubic degeneracy checked below.

For completeness, the printed vector field has a subcritical, rather than supercritical, Hopf bifurcation. In the mechanical coordinates $x=p$, $y=\dot{p}$, at $\lambda =3$ its cubic acceleration has $x^{2}y$ coefficient $b=6(\kappa -3)(\kappa +1)/(\kappa -1{)}^2$ and $y^3$ coefficient $d_3=-2/(\kappa -1{)}^2$. For a small harmonic oscillation, the averaged cubic change of $H=(y^2+{\omega }^2{x}^2)/2$ is ${\omega }^2{A}^4(b+3d_3{\omega }^2)/8$, whereThus a small unstable periodic orbit lies on the attracting side of the Hopf bifurcation. This calculation matters when comparing the original equations with the later assumed normal form.

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.

At a periodic steady state, two spatial integrations of $0=(B+\delta A^2{)}_{XX}$ make $B+\delta A^2$ constant. The zero-mean constraint fixes that constant to $\delta ⟨A^2⟩$. Define $\beta =1-\delta >0$ and $\nu =\mu -\delta ⟨A^2⟩$; the stationary amplitude equation isFor a broad localized plateau, the two transition layers occupy $O(\ell )$ length while the plateau and exterior occupy $2hL$ and $2(1-h)L$. Provided both lengths are large compared with the front width $\ell$, spatial averaging gives $⟨A^2⟩=h{R}^2+O(R^2\ell /L)$.

Multiplying the stationary equation by $A_X$ gives the first integralIn the well-separated-front approximation, $A\to 0$ with $A_X\to 0$ outside, so $E=0$. On the plateau $A\to R\ne 0$, giving both $\nu +\alpha R-\beta R^2=0$ and $W(R)=0$. Substitution of the first into the second gives $R^3(\beta R/4-\alpha /6)=0$, henceThe equal-potential condition is therefore $2{\alpha }^2=-9\nu \beta$. It is a front balance, not an integration of the stationary equation without its $A_X$ multiplier.

These plateau values also produce actual isolated fronts. At this balance $W(A)=-\beta A^2(A-R{)}^2/4$, so for $\alpha >0$ the increasing front satisfies $A_X=\sqrt{\beta /2}A(R-A)$. It isFor $\alpha <0$ the sign-reversed construction gives the negative plateau. Two oppositely oriented well-separated fronts give the leading periodic mesa solution approximation; their weak interaction and tails supply finite-$L$ corrections.

Finally $\nu =\mu -\delta h{R}^2$ gives the coexistence fraction of a conserved-field amplitude mesa:The leading existence range is the interval obtained from $0<h<1$:This assumes $\alpha \ne 0$ and $L\gg \ell$, with $hL,(1-h)L\gg \ell$. At the two displayed ends a transition pair can no longer be treated as well separated, so they are asymptotic bounds rather than exact finite-period bifurcation values. For $\alpha =0$ this particular nonzero plateau balance collapses to $R=0$ and does not yield the mesa solution family.

Put $p=\dot{u}$ and extend the system by $\dot{\lambda }=\dot{\kappa }=0$. At the zero equilibrium with zero parameters, the $(u,p)$ block is the nilpotent matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, while the $v,w$ eigenvalues are $-1,-\tau$. Since $\tau >0$, these are stable. The extended centre manifold is a graph $v=h(u,p,\lambda ,\kappa )$, $w=k(u,p,\lambda ,\kappa )$ tangent to the four-dimensional centre space containing the two parameter directions. Its invariance equations areFor the quadratic jets in the ordinary total degree, the coefficient multiplying $h_p,k_p$ has no linear part. Hence $p{\partial }_u{h}_2=-h_2+u^2$ and $p{\partial }_u{k}_2=-\tau k_2+u^2/\tau$. With $h_2=a{u}^2+bup+c{p}^2$, comparison gives $a=1$, $b=-2$, $c=2$. The second graph is obtained in the same way. Terms involving parameters at quadratic order have zero forcing and vanish. Thus the quadratic centre manifold for a pair of linear filters isSubstitution into the centre equations gives the cubic termsThe last displayed cubic term is absent from the proposed second-order approximation, so its omission needs the weighted scaling rather than an ordinary-degree truncation.

Under the hinted scaling $u=\epsilon \tilde{u}$, $\kappa =\epsilon \tilde{\kappa }$, $\lambda ={\epsilon }^2\tilde{\lambda }$, $d/dt=\epsilon d/d\tilde{t}$, one has $p=O({\epsilon }^2)$ and $\dot{p}=O({\epsilon }^3)$. The invariance equations at weighted orders two and three give $v=u^2-2up+O({\epsilon }^4)$ and $w=u^2/{\tau }^2-2up/{\tau }^3+O({\epsilon }^4)$. In particular no additional parameter-dependent weighted-order-three graph terms appear. Therefore the $u{\dot{u}}^2$ term and the omitted manifold corrections affect the reduced acceleration only at order ${\epsilon }^5$, whereas $u^3$ has order ${\epsilon }^3$ and $u^2\dot{u}$ order ${\epsilon }^4$. After division by ${\epsilon }^3$, the slow equation isThis proves why the stated approximation retains the leading nonlinear restoring force and its first nonlinear damping correction. The centre manifold theorem removes the exponentially decaying transverse directions, while the retained generic coefficients determine the nearby reduced bifurcations. It is a local asymptotic description, not an exact global reduction. If either relevant coefficient vanishes, higher-order terms may be needed to resolve that degeneracy; the two extreme $s$ cases in part (b), with $\tau =1$, avoid it.

Set $A=s-1\ne 0$ and rescale $q=\sqrt{|A|}u$, $p=\dot{q}$. The reflection-symmetric cubic double-zero normal form becomesIt is invariant under $(q,p)\mapsto (-q,-p)$. Thus the large-$s$ case has $\sigma =1$ and the small-$s$ case $\sigma =-1$. At the origin the Jacobian matrix has determinant $\lambda$ and trace $\kappa$: it is a saddle for $\lambda <0$, stable for $\lambda >0,\kappa <0$, and unstable for $\lambda >0,\kappa >0$. In the latter two cases the discriminant ${\kappa }^2-4\lambda$ distinguishes nodes from foci; its zero is not an equilibrium-creation bifurcation.

The nonzero equilibria satisfy $p=0$, $q^2=\lambda /\sigma$. At either of them the determinant is $-2\lambda$ and the trace is $\kappa -2\lambda$. Therefore for $s\gg 1$, a pair of saddles exists when $\lambda >0$, while for $s\ll 1$, a pair of attracting equilibria exists when $\lambda <0,\kappa <2\lambda$, and a repelling pair when $\lambda <0,\kappa >2\lambda$. The repelling or attracting points may be nodes or foci. Their origin collision occurs on the pitchfork bifurcation line $\lambda =0$, away from $\kappa =0$. For fixed $\kappa \ne 0$, the scalar centre equation there beginsFor $\kappa <0$, the large-$s$ case is a subcritical pitchfork and the small-$s$ case a supercritical pitchfork. For $\kappa >0$, the scalar criticality reverses, but there is already an unstable transverse eigenvalue, so the full-plane branches have the stability just listed.

At the origin, $\kappa =0$, $\lambda >0$ is a Hopf bifurcation with frequency $\sqrt{\lambda }$. Averaging the leading weakly nonlinear oscillator gives the amplitude equationConsequentlyThese are respectively a supercritical Hopf bifurcation and a subcritical Hopf bifurcation, with $\lambda >0$ fixed and the cycle sufficiently small. The portraits change from a stable focus to an unstable focus surrounded by a stable cycle in the first case; in the second, a stable focus is surrounded by an unstable cycle before the focus becomes unstable and that cycle disappears.