Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 51 / 1

Write $h=(h_1,h_2)$ for the homogeneous cubic correction and $A(x,y{)}^T=(y,0{)}^T$ for the nilpotent linear part. The chain rule gives the transformed cubic vector field as $f_3+Ah-DhA(x,y{)}^T$. Thus the cubic coefficients in the first equation arewhile those in the second equation areConsequently one explicit near-identity transformation isThe transformed equations have no cubic correction in $\dot{x}$, and only $x^3,x^{2}y$ survive in $\dot{y}$. The cubic coefficients areThis is the cubic elimination at a nilpotent double-zero point; the extra $3a_1$ is the nilpotent cubic damping invariant contribution from the first equation. Higher-order terms generated by the substitution are outside the cubic truncation.

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.

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 nonzero equilibrium points are paired by reflection:At either equilibrium point, the derivative of the second equation with respect to $x$ is $-1+x_{\ast }^2=2-3c/d$. The characteristic polynomial becomesIts constant term is nonzero away from the pitchfork bifurcation, so there is no further steady local bifurcation. At a Hopf bifurcation the condition $A_1{A}_2=A_3$ givesIndeed $\tau =2-d/2$ and $\tau {\omega }^2=d-c>0$ require $d<4$; within this range ${\omega }^2>0$ is equivalent to $d<1$. The apparent singularity at $d=12$ does not hide another solution of the coefficient equation. The Routh-Hurwitz stability criterion shows that both equilibrium points are stable for $0<d<1$, $c_H^{\ast }(d)<c<d$, and unstable for $c<c_H^{\ast }(d)$. For $d\ge 1$, $3c/d-d-2<1-d\le 0$, so no nonzero equilibrium point is stable.

Their Hopf bifurcations are supercritical. Here is a coefficient check that includes the effect of the quadratic term around $x_{\ast }$. Let $B_2(u,v)=2x_{\ast }{u}_x{v}_x$, $C_2(u,v,w)=2u_x{v}_x{w}_x$ and normalize $q_x=1$ as before. The inverse-linearization term has $(L^{-1}B(q,\bar{q}){)}_x=3/x_{\ast }$. AlsoThe resonant cubic Hopf coefficient isIn this simplification $x_{\ast }^2/{\omega }^2=3\tau /d$. Thus stable small periodic orbits appear around each nonzero equilibrium point on the $c<c_H^{\ast }(d)$ side. This classifies the local branches; the sketch does not infer uncomputed distant global bifurcations.

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.