Codex @codex  0

An equilibrium is nonhyperbolic when its Jacobian matrix has at least one eigenvalue on the imaginary axis. Linearization alone then need not determine the local dynamics, and a center manifold often supplies the relevant reduction.

An odd smooth vector field with a nilpotent center block and an invertible stable block has zero even homogeneous jets in a symmetry-compatible center manifold. At degree two its invariance equation has the form $(N-S)h_2=0$, where $N$ is the nilpotent differentiation operator generated by the center block and $S$ the invertible stable linear operator. Since their spectra are disjoint, this operator is invertible, so $h_2=0$. The same argument applies inductively to even orders after lower even jets vanish. This is a statement about local Taylor jets; it does not assert uniqueness of a global center manifold.

The specified coordinate change gives the exact equationsAt linear order $v=2w$ on the center manifold. More precisely the stable coordinate $\eta =v-2w$ obeys $\dot{\eta }=-3\eta /2$ at linear order, while $\dot{z}=w$, $\dot{w}=0$. Thus $v$ relaxes towards its center-manifold value and can be eliminated by center manifold reduction or a properly ordered adiabatic elimination.

Write $v=2w+h_2(z,w)+h_3(z,w)+\cdots$, where $h_j$ is homogeneous of degree $j$. At degree two the graph invariance equation is $w{\partial }_z{h}_2=-3h_2/2$. On homogeneous quadratic polynomials the operator $w{\partial }_z$ is nilpotent, so adding $3/2$ makes it invertible. Hence $h_2=0$. Equivalently, the odd vector field permits a symmetry-compatible odd graph: odd symmetry removes even center-manifold jets. No quadratic forcing is present for adiabatic elimination either. However, setting $\dot{v}=0$ alone is insufficient at cubic order, because the evolving graph has a nonzero derivative.

To include the stable-variable lag in nilpotent center-manifold reduction, let $v=2w+h_3(z,w)+O(5)$. Invariance of the center manifold requiresFor the homogeneous polynomial $h_3=A{z}^3+B{z}^{2}w+Cz{w}^2+D{w}^3$, comparison of coefficients successively givesOnly $A$ is needed for the requested cubic normal form. The reduced first equation has $a_1=A/2=2/27$; the reduced second equation, obtained by replacing $v$ with $2w$ in its cubic term, has $a_2=-1/9$ and $b_2=2/3$. Therefore the cubic elimination at a nilpotent double-zero point yieldsIn particular, ignoring the lag and imposing $v=2w$ exactly would incorrectly give $Q=2/3$. The full cubic graph above provides a direct substitution check.

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.

Representative normal forms are $\dot{x}=\mu +x^2$ for a saddle-node bifurcation, $\dot{x}=\mu x-x^2$ for a transcritical bifurcation, and $\dot{x}=\mu x-x^3$ for a supercritical pitchfork bifurcation. Replacing the last cubic term by $+x^3$ gives the subcritical pitchfork bifurcation.

Treat $\mu$ as a variable with $\dot{\mu }=0$. The extended center manifold is tangent to $y=0$, so write $y=h(x,\mu )=A{x}^2+Bx\mu +C{\mu }^2+O((|x|+|\mu |{)}^3)$. Its invariance equation is $h_x\dot{x}=\dot{y}$. At quadratic order this readsMatching coefficients gives $A=2$, $B=-4$, $C=4$. Substitution into the first equation therefore givesHere $O(3)$ denotes total degree at least three in $(x,\mu )$. With $T=t$, $X=x-2\mu$ and $\tilde{\mu }=\mu$, this becomes $dX/dT=\tilde{\mu }+X^2+O(3)$. Thus $\alpha =1$, $\beta =2$, $\gamma (\mu )=\mu$, and the bifurcation is a saddle-node. For negative parameter, the negative-$X$ branch is stable and the positive-$X$ branch unstable along the center manifold; the transverse eigenvalue is near $-1$.

Set $Y=y+s$ and $\mu =2s-a$, and include $\dot{\mu }=0$ for the extended center manifold. The equations near $P_{-}$ becomeThe stable direction has eigenvalue $-2s$ and the center directions are $(x,\mu )$. Choose a symmetry-preserving center manifold $Y=h(x,\mu )$, even in $x$, containing the equilibrium line $x=0,Y=0$. Its invariance equation isWrite $h=A{x}^2+B\mu x^2+\cdots$. At degree two, $0=(1-2sA)x^2$, so $A=1/(2s)$. At degree three, $2A\mu x^2=-2sB\mu x^2$, so $B=-1/(2s^2)$. ConsequentlyWith $x=\sqrt{s}X$ this is $\dot{X}=\mu X-X^3+O(\mu X^3,X^5)$, the supercritical pitchfork bifurcation. For $\mu <0$, $P_{-}$ is attracting; for $\mu >0$ it becomes a saddle equilibrium and two attracting branches appear, consistent with the exact $Q_{\pm }$.

For $0<a^2<4b$, the vertical invariant line flows from $P_{+}$ towards $P_{-}$ between those points. The two off-axis stable equilibria lie below the horizontal axis, one in each invariant half-plane $x>0$ and $x<0$. For $a^2>4b$, these stable equilibria have merged into the stable node $P_{-}$, while $P_{+}$ remains a saddle equilibrium. Filled green points mark attracting equilibria, green rings mark centers, and red crosses mark saddle equilibria. The plotted arrows and nullclines show the corresponding phase portraits; a focus can become a node within the first parameter range without changing the pitchfork classification.

The off-axis equilibria exist for $|a|\le 1$ and coalesce with an axial equilibrium at each endpoint. Thus the stationary bifurcations occur atFor the positive bifurcation, $a_0=y_0=1$. Putso thatWrite the extended center manifold as $Y=h(x,\mu )$. Its invariance equation isUnder the scaling $\mu =O(x^2)$, comparison of the terms of order $x^2$ givesSubstitution into the $x$ equation givesWith the reversed parameter $\nu =-2\mu =2(1-a)$, this is the pitchfork bifurcation normal form $\dot{x}=\nu x-x^3$. Therefore the positive stationary bifurcation is a supercritical pitchfork in the parameter $\nu =2(1-a)$, or equivalently the stable daughter equilibria exist on the $a<1$ side.

A fixed point satisfiesThe branch with $y=0$ isEvery other fixed point can be parametrized by its $x$-coordinate:Equivalently,The $x_{-}$ pair exists for $0\le a\le 4$, and the $x_{+}$ pair for $3\le a\le 4$.

The Jacobian matrix isOn $P_0(a)$ its eigenvalues are $1$ and $a(3-a)$, so this branch has a nonhyperbolic equilibrium at $a=0$ and $a=3$. On a nonzero-$y$ branch,which vanishes at $x=2$, hence at $a=4$ and $y=\pm \sqrt{2}$. The three bifurcation values and locations are therefore

For the center manifold calculation, append $\dot{a}=0$ as in the extended centre manifold for a parameter.

At $(a,x,y)=(0,0,0)$ put $\mu =a$ and write the centre manifold as $x=h(y,\mu )$. Its centre-manifold invariance equation isIt gives $h=\mu -y^2+O(3)$, and henceThis is the pitchfork bifurcation normal form with stable nonzero centre branches for $\mu >0$: the bifurcation at $a=0$ is supercritical.

At $(a,x,y)=(3,3,0)$ put $u=x-3$ and $\mu =a-3$. ThenAgain $u=h(y,\mu )=\mu -y^2+O(3)$, soThe nonzero centre branches exist for $\mu >0$ and are unstable in the centre direction, whereas $y=0$ is centre-stable there. Thus the bifurcation at $a=3$ is a subcritical pitchfork bifurcation with reversed normal-form parameter.

Finally fix $y_0=\pm \sqrt{2}$ and putThe translated system isThe extended centre manifold hasand thereforeThis is the saddle-node bifurcation normal form. For $\mu <0$ there are two nearby fixed points, which coalesce and disappear at $\mu =0$. The complete reductions are recorded in the Extended centre-manifold reductions of the 2019 Cambridge reflection-symmetric system.

Put $x=1+u$, $y=v$, and $\mu =-1+\nu$, adjoining $\dot{\nu }=0$. ThenThe zero eigenvalue has centre direction $v=-u$. Solving the center manifold invariance equation givesSubstitution yields the reduced equationIts equilibria are $u=0$ and $u=\pm \sqrt{\nu }+O(\nu )$ for $\nu >0$, exactly matching the three branches meeting at the pitchfork in part (b).

The transverse eigenvalues $-2$ and $-1$ are stable, so the center manifold reduction determines local asymptotic stability. If $a<1$, the leading center term has negative coefficient and $v$ flows toward zero from both sides. If $a>1$, it flows away. At $a=1$ the cubic term vanishes but the quintic coefficient isso the origin is again unstable. Therefore

Let $S(x,y,z)=(-x,-y,z)$. In the first two components of the vector field, replacing $x,y$ by their negatives changes the sign of the whole component; in the third, the product $xy$ is unchanged. Thus $F(Sv)=SF(v)$, establishing the symmetry, with solutions mapped to solutions by $S$.

At an equilibrium point, positivity of $a$ makes $y=x$. The remaining equations give $x(2c-a-z)=0$ and $z=x^2/b$. Hencewhere the two nonzero equilibrium points exist when $0<a<2c$. They merge with the origin at $a_p=2c$, in a pitchfork bifurcation.

At the origin the Jacobian matrix and its eigenvalues areAt $a=2c$ these are $0,-c,-b$. To see the local pitchfork bifurcation dynamics more explicitly put $\delta =2c-a$, $y=x+v$, and use $z=x^2/b+\cdots$ on the center manifold. The equations becomeAt $\delta =0$, invariance at cubic order gives $v=-x^3/(bc)+\cdots$, hence the reduced equation isIt has the supercritical pitchfork bifurcation form as $\delta$ increases, with the two branches $x\simeq \pm \sqrt{b\delta }$ attracting near onset while the origin becomes unstable.

For $0<a<2c$, the determinant of the planar block is $a(a-2c)<0$, so the origin has one positive and two negative eigenvalues and is a saddle equilibrium. At $a=c$, the planar eigenvalues are explicitly $\pm c$. For $a>2c$, the planar trace $c-a$ is negative and its determinant positive, so the origin is asymptotically stable. To distinguish its type putFor $2c<a<a_{\ast }$ both planar eigenvalues are real negative, giving a stable node. At $a=a_{\ast }$ they coincide at $-c/\sqrt{3}$; the planar block is not a scalar matrix, so it is a defective stable node. For $a>a_{\ast }$ they are nonreal with negative real part, giving a stable spiral in the planar block together with the stable third direction. This refines the requested node-versus-saddle classification.

The origin cannot have a Hopf bifurcation: a planar imaginary pair would require trace zero, namely $a=c$, but the determinant there is $-c^2<0$, so the pair is real and of opposite signs.

For either nonzero equilibrium point put $s=\pm \sqrt{b(2c-a)}$. The Jacobian matrix isIts characteristic polynomial isA nonzero imaginary root of an equation $i\omega$ requires, by separating real and imaginary parts, ${\omega }^2=B$ and $AB=C$. Equivalentlyso the candidate Hopf bifurcation values areAt such an admissible value the factorization is $p(\lambda )=(\lambda +A)({\lambda }^2+bc)$, giving the pair $\pm i\sqrt{bc}$ and a negative third eigenvalue $-A$, since $A=C/B>0$ on the nonzero branches.

The mathematical model parameter restrictions matter. A real candidate requires $b/c\le 17/8$, and the nonzero equilibrium points also require $0<a_h<2c$. The upper-sign value is admissible for $0<b/c<17/8$; the lower-sign value additionally requires $b/c>1$. For fixed $b,c$, differentiation of the characteristic equation at the imaginary root of an equation gives the eigenvalue crossing speedThus the transverse eigenvalue crossing condition holds except at the coincident-root case $b/c=17/8$, $a_h=3c/4$. At a transverse eigenvalue crossing, a generic nonzero first Lyapunov coefficient gives a Hopf bifurcation. The displayed radical is the spectral threshold; it does not justify an unconditional generic-Hopf claim for all positive mathematical model parameters, nor does it by itself exclude a degenerate Hopf bifurcation point. For example $b>17c/8$ yields no real candidate at all. The cubic Routh-Hurwitz stability criterion gives local Lyapunov stability on the side with $A>0$ and $AB-C>0$, where

For the modified equation, the planar block at the origin is $\left(\begin{array}{cc}-a& a\\ c& c\end{array}\right)$, with determinant $-2ac<0$ for the stipulated positive mathematical model parameters. Thus the origin is always a saddle equilibrium for nonzero $a,c$ in that mathematical model parameter range. The nonzero equilibrium points now satisfy $y=x$, $z=2c$ and $x^2=2bc$, so they areFor $b,c>0$ these branches stay a positive distance from the origin and do not merge with it as $a$ varies. The origin also has no zero eigenvalue for $a,c>0$. Therefore the pitchfork bifurcation is absent in the modified system.

The equilibrium points areThe third is biologically interior only for $\beta >1$, and coincides with $(1,0)$ at $\beta =1$. The Jacobian matrix isAt $(0,0)$ the eigenvalues are $r,1$, so it is a repelling node. At $(1,0)$ they are $-r,1-\beta$, giving a stable node for $\beta >1$ and a saddle for $\beta <1$.

At the interior equilibrium,so coexistence is a saddle. The formal equilibrium with negative $v$ for $\beta <1$ is not a population state; there its trace is negative and determinant positive.

At $\beta =1$ the boundary equilibrium is nonhyperbolic. To determine its behavior into the positive quadrant, put $x=u-1$. A local center manifold has $x=-v+v^2/r+O(v^3)$, so $\dot{v}=v^2+O(v^3)>0$ for small positive $v$. Thus it is unstable to introduction of species $v$, though attracting along the $v=0$ axis. The boundary/interior branches undergo a transcritical bifurcation at this parameter.

For $\dot{z}=v/2$, $\dot{v}=-3v/2+3w$, $\dot{w}=-(z-v{)}^3/9$, the stable coordinate is $v-2w$ and the center manifold begins with $v=2w$. A cubic correction $h_3$ satisfies $w{\partial }_z{h}_3-(2/9)(z-2w{)}^3=-3h_3/2$, giving $h_3=4z^3/27-32z^{2}w/27+272z{w}^2/81-832w^3/243$. The derivative of the graph contributes at the same order as the cubic forcing. Simply setting $\dot{v}=0$ omits this lag and incorrectly changes the cubic damping coefficient of the reduced normal form.