Line FF — two-well cubic Hamiltonian, 14 Abelian zeros

Status: imagined 14-zero certificate dropped. Fork kept. Not a dent of H(3).

Imagined certificate. The cubic Hamiltonian field of

$H=\frac{y^2}{2}+\frac{(x^2-1)^2}{4}$

perturbed by an explicit cubic, has 14 isolated zeros of the first Abelian integral I(h) across the two small wells and the large annulus, hence H(3) ≥ 14 without Li–Liu–Yang.

Drop immediately. No 14-zero integral is written. The first Melnikov function of the named cubic perturbation is an elliptic integral on each nest; one exact sample is nonzero and the 3-jet Lyapunov quantity at each well is a nonzero multiple of μ. That proves I is not identically zero. It does not produce 14 zeros, and it does not beat Li–Liu–Yang H(3) ≥ 13.

Fork kept. The unperturbed field, the energy identities, the formula for I(h), the exact value of I on each lobe of the figure-eight, and the exact first Lyapunov quantity at each well after translation to the q1 focal form. Symbolic only; no ODE integration. Fractions as strings. Not a bound on H(n).

Replay:

problems/hilbert16-limit-cycles/compute/q3/ff-two-well/run.sh

Python expands the identities sparsely; Rust expands them again and evaluates the cleared dH/dt residuals on the integer box from −3 to 3. The two dumps are diffed. Exit 0. Certs: certs/core.json, certs/identities.json.

Dropped — fourteen zeros of I(h), hence H(3) at least 14

The fiction needs an explicit cubic perturbation whose Abelian integral has 14 isolated zeros on the three period annuli (two nests and the outer well). Picard–Fuchs dimension for this Hamiltonian is small; typical first-order counts are a handful, not 14. Harnack still caps one invariant cubic at two ovals. No closed form of I(h) with 14 roots is produced, and a numeric search of an elliptic integral is not a lower bound.

The named perturbation below is degree 3, so it is the right degree. The integral is written. The 14 roots are not.

Kept — unperturbed two-well field

The field is

$\dot x=y,\qquad \dot y=x-x^3.$

It is Hamiltonian for

$H=\frac{y^2}{2}+\frac{x^4}{4}-\frac{x^2}{2},$

which is the user’s double well shifted by a constant:

$\frac{y^2}{2}+\frac{(x^2-1)^2}{4}=H+\frac{1}{4}.$

The certificate uses the cleared polynomial H4 = 4H = 2y² + x⁴ − 2x², so every identity stays in Z[x, y]. Formal differentiation gives dH4/dt ≡ 0 (block unperturbed). Divergence vanishes identically.

Equilibria of (y, x − x³): y = 0 and x(1 − x²) = 0, hence (0, 0) and (±1, 0). The Jacobian is

$\begin{pmatrix}0&1\\1-3x^2&0\end{pmatrix},$

trace 0 everywhere, determinant 3x² − 1. At (0, 0) the determinant is −1 (saddle). At (±1, 0) it is 2 (linear centers, frequency √2). Energy: H(0, 0) = 0 and H(±1, 0) = −1/4. The figure-eight through the saddle is the level H = 0. Each nest oval lives in H ∈ (−1/4, 0).

Kept — named cubic perturbation and I(h)

Perturb by van der Pol damping on the same wells,

$\dot x=y,\qquad \dot y=x-x^3+\mu(1-x^2)y.$

Still degree 3. Along the perturbation,

$\frac{dH}{dt}=\mu\, y^2(1-x^2)$

(block perturbed; equivalently dH4/dt = 4μ y² (1 − x²)). Energy increases where |x| < 1 and decreases where |x| > 1. The first Melnikov function on a level H = h is

$I(h)=\oint_{H=h}\mu(1-x^2)\,y\,dx,$

or, on a nest oval with y = ±√(2h − x⁴/2 + x²),

$I(h)=2\mu\int_{x_{\min}}^{x_{\max}}(1-x^2)\sqrt{2h-\tfrac{1}{2}x^4+x^2}\,dx.$

No elementary antiderivative in the open wells is claimed. Two exact samples are.

At the bottom of each well the oval shrinks to a point and I(−1/4) = 0. On the figure-eight H = 0, each lobe is elementary. The right lobe has x ∈ [0, √2] and

$\sqrt{x^2-\tfrac{1}{2}x^4}=|x|\sqrt{1-\tfrac{1}{2}x^2}.$

The substitutions u = x² then w = 1 − u/2 then t = √w reduce the integral to

$\int_0^1(4t^4-2t^2)\,dt=\Bigl[\tfrac{4}{5}t^5-\tfrac{2}{3}t^3\Bigr]_0^1=\tfrac{4}{5}-\tfrac{2}{3}=\tfrac{2}{15}.$

Clearing denominators, 15 times the antiderivative is 12t⁵ − 10t³, and its derivative minus 15(4t⁴ − 2t²) is the zero polynomial (block figure_eight). The prefactors assemble to

$I_{\mathrm{right}}(0)=\frac{4\mu}{15}.$

The integrand depends only on x², so the left lobe equals the right lobe. For μ ≠ 0 this value is nonzero, so I is not the zero function on either nest (the figure-eight is the boundary of each nest). Counted isolated zeros of I on regular ovals: none exhibited. The figure-eight itself does not persist at first order.

Kept — L1 at both wells

Translate each unperturbed center to the origin and put the 3-jet in the q1 focal form