Line OO — five Abelian zeros of a cubic perturbation

Status: five zeros of $I(h)$, and $H(3)\ge 5$ as a dent of 13, dropped. Kept: the degree-3 radial slice on the circle Hamiltonian, and the named family on the cubic Hamiltonian of a quadratic field. Not a dent of Li–Liu–Yang 13. Planar $H(n)$ did not move.

Imagined: an explicit cubic perturbation of a quadratic Hamiltonian has five simple zeros of $I(h)$, attaining the published local max (Han–Yang–Yu).

Two readings, both drop.

If “quadratic Hamiltonian” means $H$ of degree 2, the period annulus is circles and a degree-3 1-form has $Z(2,3)=\lfloor(3-1)/2\rfloor=1$. Five is impossible: $\tilde I(h)=c\,\mu\,h\,p(2h)$ with $\deg p\le 1$ has at most one positive zero.

If it means a quadratic field that is Hamiltonian ($H$ cubic), then a cubic perturbation is degree 3 and five zeros of $I(h)$ would be $H(3)\ge 5$, not 14. Still not a dent of Li–Liu–Yang 13. Scholarpedia records Han–Yang–Yu 2009 as a local statement: a center of a Hamiltonian quadratic system generates at most 5 cycles under cubic perturbations. That 5 is not 13, and the named closed forms below do not produce 5 zeros.

Drop immediately. No numeric hunt for 5 zeros. A closed form that does not produce 5 is a drop, not a search target.

Replay:

problems/hilbert16-limit-cycles/compute/q3/oo-five-zeros/run.sh

Python expands the identities sparsely; Rust expands them again with a BTreeMap and evaluates the residuals on an integer box. The two dumps are diffed. Exit 0. Certs: certs/core.json, certs/identities.json.

Opened this session: Scholarpedia, Han–Li–Li 2010. Weak Hilbert $Z(2,n)=\lfloor(n-1)/2\rfloor$ because $M(h)$ is a polynomial; that is the $n=3$ pointer kept below. The same article says it is proved in Han–Yang–Yu 2009 that a center of a Hamiltonian quadratic system generates at most 5 limit cycles under cubic perturbations (their [46], Hopf bifurcations for near-Hamiltonian systems, IJBC 19). Forum restatements are leads. Li–Liu–Yang 13 is still the cubic record (full text still paywalled; see RESEARCH.md).

Dropped — five zeros, and H(3) at least 5 as a dent of 13

A dent of the published cubic record would be a verified family with more than 13 isolated cubic cycles. Five simple zeros of $I(h)$ are not that. On the circle reading they contradict $Z(2,3)=1$. On the cubic-$H$ reading they would be $H(3)\ge 5$, which does not beat 13. Han–Yang–Yu’s 5 is a local cyclicity bound at one Hamiltonian quadratic center, via the expansion of $M$ at that center, not a construction of 14 and not a beat of Li–Liu–Yang. The imagined five-zero integral is not written. A numeric search of an elliptic integral is not a lower bound.

Kept — circles, degree 3, one positive zero

Take the quadratic Hamiltonian

$H=\frac{x^2+y^2}{2}.$

Its Hamiltonian field is $\dot x=y$, $\dot y=-x$. Formal differentiation gives $d(x^2+y^2)/dt\equiv 0$, so every regular level $H=h>0$ is a circle $x^2+y^2=2h$. This is the $n=3$ slice of the radial family already replayed on line JJ, not a second copy of that folder.

Perturb by $P=0$ and

$Q=\mu\, y\bigl(\alpha-(x^2+y^2)\bigr).$

Then $\deg Q=3$. On each oval, $x^2+y^2=2h$ is constant, so

$I(h)=\mu\,(\alpha-2h)\oint_{H=h} y\,dx.$

The reduced integral

$\tilde I(h)=h\,(\alpha-2h)$

is a polynomial in $Z[h,\alpha]$ with the same zeros on $h>0$ as $I(h)$. The factor $h$ is the area scaling; it vanishes only at the equilibrium $h=0$. The positive zeros are exactly the positive zeros of $\alpha-2h$. For $\alpha>0$ there is one, at $h=\alpha/2$. For $\alpha=1$,

$\tilde I(h)=h(1-2h),$

one positive zero at $h=1/2$. Matches $Z(2,3)=1$. The oval reduction $(\alpha-x^2-y^2)-(\alpha-2h)=-(x^2+y^2-2h)$ is the zero remainder.

Here $p(s)=\alpha-s$ has degree 1, so $\tilde I(h)=h\,p(2h)$ has at most one positive zero. Five positive zeros would need $\deg p\ge 5$ and $\deg Q\ge 11$. That is a degree-11 1-form, the $n=11$ attaining example of $Z(2,11)=5$, not a beat of $Z(2,3)$.

Kept — cubic Hamiltonian, quadratic field, named family

Now read “quadratic Hamiltonian” as a quadratic field that is Hamiltonian. Take

$H=\frac{y^2}{2}+\frac{x^3}{3}-\frac{x^2}{2}.$

The Hamiltonian field is $\dot x=y$, $\dot y=x-x^2$, degree 2. The certificate uses the cleared polynomial $H_6=6H=3y^2+2x^3-3x^2$, so every identity stays in $Z[x,y]$. Formal differentiation gives $dH_6/dt\equiv 0$.

Equilibria of $(y,\,x-x^2)$: $y=0$ and $x(1-x)=0$, hence $(0,0)$ and $(1,0)$. The Jacobian is

$\begin{pmatrix}0&1\\1-2x&0\end{pmatrix},$

trace 0 everywhere, determinant $2x-1$. At $(0,0)$ the determinant is $-1$ (saddle). At $(1,0)$ it is $1$ (linear center, frequency $1$). Energy: $H(0,0)=0$ and $H(1,0)=-1/6$. The potential identity

$2x^3-3x^2+1=(x-1)^2(2x+1)$

puts the well bottom at $H=-1/6$ and a second point $(-1/2,0)$ of the same energy on the unbounded left component, not a nest oval. The homoclinic through the saddle is the level $H=0$, with right tip $V(3/2)=0$. Nest ovals live in $H\in(-1/6,0)$.

Named perturbation, degree at most 3 (in fact linear):

$P=0,\qquad Q=\mu y.$

Along the perturbation, $dH/dt=\mu y^2$ and $dH_6/dt=6\mu y^2$ (block family). One-signed for $\mu\neq 0$: every oval is displaced in the same direction. The first-order integral is

$I(h)=\mu\oint_{H=h} y\,dx=\pm\mu\,\mathrm{Area}(\{H\le h\}\cap\mathrm{nest}).$

Area vanishes only at the well bottom (the oval shrinks to $(1,0)$). On a regular nest oval the area is positive: the rectangle $[3/4,5/4]\times[-1/4,1/4]$ sits inside $\{H_6\le-19/32\}\subset$ nest, because the far corner is $-19/32<0$, so that one oval already has area at least $1/4$. Scaling / monotonicity: the nest sublevels are nested, so Area is strictly increasing from $0$ at $h=-1/6$ to the homoclinic area at $h=0$. Thus $I$ has no zero on the open period interval. After the shift $\hat h=h+1/6\ge 0$, that is no positive zero, or exactly the well-bottom zero $\hat h=0$. First-order cyclicity of this family is at most 1, not 5. Regular zeros exhibited on the nest: 0. The named field after perturbation is still degree 2 ($\dot y=x-x^2+\mu y$), so it is not an $H(3)$ construction in any case.

What this is not

Not a dent of $Z(2,3)$. Not a dent of $H(3)$ or $H(n)$. Not five zeros of an Abelian integral. Not a beat of Han–Yang–Yu’s local 5, and not a replay of their expansion of $M$ at a general quadratic Hamiltonian center. Not a numeric search. The reusable lemmas are the $n=3$ radial identity $\tilde I(h)=h(\alpha-2h)$ and the one-signed $dH/dt=\mu y^2$ on the cubic Hamiltonian nest. A stranger can run run.sh and read one circle zero and zero regular nest zeros off the dump.