Line JJ — weak Hilbert number on a quadratic period annulus

Status: beat-the-formula dent dropped. Kept as a replay of the published $Z(2,n)$. Planar $H(n)$ did not move. This is the cyclicity of one Hamiltonian period annulus, not a dent of $H(2)$.

Imagined: an explicit Abelian integral with $\lfloor(n-1)/2\rfloor+1$ real zeros on the period interval, beating Scholarpedia / Han–Li–Li 2010.

Dropped: that formula is a theorem, not a conjecture. $M(h)$ is a polynomial, so the count cannot be beaten.

Replay:

problems/hilbert16-limit-cycles/compute/q3/jj-weak-hilbert/run.sh

Python expands the Hamiltonian identity, the oval reduction, and the reduced integral $\tilde I(h)=h\,p(2h)$ as sparse monomials. Rust expands the same rings with a BTreeMap and evaluates the residuals on an integer box. The two dumps are diffed. Exit 0. Cert: certs/family.json.

Opened this session: Scholarpedia, Han–Li–Li 2010. Weak Hilbert $Z(m,n)$ is the maximum number of isolated zeros of the first Melnikov function

$M(h)=\oint_{L_h}P\,dy-Q\,dx$

over polynomial Hamiltonians of degree $m$ and polynomial perturbations of degree $n$. The article states that $Z(2,n)=\lfloor(n-1)/2\rfloor$ because $M(h)$ is a polynomial. That is the record replayed here. Forum restatements are leads.

Dropped — one more zero than the formula

A dent would be a verified family with $\lfloor(n-1)/2\rfloor+1$ isolated zeros of $M(h)$ on the period interval. For a quadratic Hamiltonian the first-order integral is a polynomial (times a density that does not vanish for $h>0$). A polynomial of that degree cannot have more real zeros than the published count. The imagined extra zero is not a search target; it contradicts the theorem. The formula was not beaten.

Kept — $H=(x^2+y^2)/2$ and a radial 1-form

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 $x^2+y^2=2h$. That is the period annulus. The flow is clockwise: $x=\sqrt{2h}\,\sin t$, $y=\sqrt{2h}\,\cos t$.

Perturb by $P=0$ and

$Q=\mu\, y\, p(x^2+y^2),$

where $p$ is a univariate polynomial of degree $k$ and $\mu\neq 0$. Then $\deg Q=2k+1$. The first-order integral (either orientation) factors on each oval, because $x^2+y^2=2h$ is constant:

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

Along the Hamiltonian orientation, $\oint y\,dx=2\pi h$. The code never computes $\pi$. The reduced integral

$\tilde I(h)=h\, p(2h)$

is a polynomial in $Z[h]$ with the same zeros on $h>0$ as $I(h)$. The factor $h$ is the area scaling; it vanishes only at the equilibrium $h=0$, which is not in the period interval. The positive zeros are exactly the positive zeros of $p(2h)$.

If $p(s)=\prod_{j=1}^{k}(s-\alpha_j)$ with distinct $\alpha_j>0$, then $\tilde I$ has exactly $k$ positive zeros. The degree budget is $n=2k+1$, so

$k=\frac{n-1}{2}=\Bigl\lfloor\frac{n-1}{2}\Bigr\rfloor$

when $n$ is odd. The family attains the published $Z(2,n)$. It does not beat it.

For even $n=2k$ the same radial form cannot use a degree-$k$ factor $p$ (that would make $\deg Q=2k+1=n+1$). Taking $p$ of degree $k-1$ gives $\deg Q=2k-1\le n$ and $k-1$ positive zeros, which is again $\lfloor(n-1)/2\rfloor$. Equivalently: for every $n\ge 1$ the radial 1-form of degree $2\lfloor(n-1)/2\rfloor+1\le n$ attains exactly $Z(2,n)$ zeros and uses at most degree $n$.

A mixed term $Q=\mu\,xy\,q(x^2+y^2)$ integrates to zero on every circle (the integrand is odd in the angle). It does not add zeros and does not beat the odd-degree radial count.

Degree 3

$Z(2,3)=1$. Take $p(s)=1-s$, so $Q=\mu y\bigl(1-(x^2+y^2)\bigr)$ has degree 3. Then

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

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

Degree 5

$Z(2,5)=2$. Take $p(s)=(s-1)(s-4)=s^2-5s+4$, so $Q=\mu y(x^2+y^2-1)(x^2+y^2-4)$ has degree 5. Then

$\tilde I(h)=h(2h-1)(2h-4)=4h^3-10h^2+4h.$

Two positive zeros, at $h=1/2$ and $h=2$. Matches $Z(2,5)$. The oval reduction is $p(x^2+y^2)-p(2h)=(x^2+y^2-2h)(x^2+y^2+2h-5)$.

An extra positive zero would need $\deg\tilde I\ge 4$, hence $\deg p\ge 3$ and $\deg Q\ge 7$. That is a degree-7 1-form, which is the $n=7$ attaining example, not a beat of $Z(2,5)$.

Family upper bound, same family

On this annulus, $\tilde I(h)=h\,p(2h)$ has at most $\deg p$ zeros in $h>0$. For $Q=\mu y\,p(x^2+y^2)$ of degree at most $n$ one has $\deg p\le\lfloor(n-1)/2\rfloor$. Upper and lower match the formula on this family. That is a replay of the small-$n$ case, not an independent proof that every degree-$n$ 1-form (non-radial included) has at most that many zeros. Scholarpedia already records the general fact because $M(h)$ is a polynomial.

Enumeration $n=1,\ldots,10$ (both languages, same dump): $Z(2,n)=\lfloor(n-1)/2\rfloor$ equals the radial zero count, and the radial degree $2Z(2,n)+1$ never exceeds $n$.

Kept — this annulus is not $H(2)$

$Z(2,n)$ bounds the number of zeros of the first Melnikov function on a period annulus of a quadratic Hamiltonian. It is not a bound on the planar Hilbert number $H(2)$. The published $H(2)\ge 4$ uses a order-3 weak focus plus a surrounding cycle (Shi; Chen–Wang), or graphics, not this Hamiltonian annulus. For quadratic perturbations one has $Z(2,2)=0$: the first-order integral of this family has no positive zero. That is consistent with four quadratic cycles needing a non-Hamiltonian mechanism.

What this is not

Not a dent of $Z(2,n)$. Not a dent of $H(2)$ or $H(n)$. Not a field with $\lfloor(n-1)/2\rfloor+1$ Abelian zeros. Not a bound on cyclicity of an arbitrary quadratic period annulus beyond the published theorem. The reusable lemma is the radial attaining family: a stranger can run run.sh and read $n=3$ and $n=5$ off the dump.