Line HH — first Melnikov on the quasi-homogeneous center
Status: 14 zeros of I(h), and the H(3) beat, dropped. Scaling and first-order cyclicity at most 1 for a named cubic family kept. Not a dent of H(3).
Imagined: a cubic (or degree-4) perturbation of the quasi-homogeneous center ẋ = 2y, ẏ = −x³ has 14 isolated zeros of the first Melnikov function I(h) on the period annulus H = x⁴ + 4y², hence H(3) ≥ 14.
Drop immediately. The unperturbed degree is already 3. A cubic perturbation stays in degree 3, so 14 zeros would be a dent of Li–Liu–Yang’s H(3) ≥ 13. Scaling on this annulus forbids it. For any cubic perturbation the first Melnikov function is of the form
which has at most two positive zeros. The named one-parameter family below has c₂ = 0 and at most one. Gavrilov–He–Xiao, arXiv:2606.22137v1, already bound the zeros of every Melnikov order for one-parameter polynomial perturbations of a quasi-homogeneous center; their §5 application is this same annulus (H = x⁴/4 + y²/2, a constant multiple of ours after rescaling y) and concludes cyclicity one for a linear unfolding. This line does not beat that paper, and it does not beat H(3) ≥ 13.
Replay:
problems/hilbert16-limit-cycles/compute/q3/hh-qh-melnikov/run.sh
Python expands the identities sparsely; Rust expands them
again and evaluates the residuals on an integer box. The two
dumps are diffed. Exit 0. Cert: certs/identities.json.
Opened this session: the abs and HTML of Gavrilov–He–Xiao, arXiv:2606.22137v1.
Dropped — 14 zeros, and H(3) at least 14
The fiction needs fourteen isolated zeros of I(h), or some other construction of fourteen isolated cubic cycles. First order on this single annulus cannot supply them. A quadratic in √h has at most two positive roots. Higher-order Melnikov functions are a different problem; Gavrilov–He–Xiao already treat that unfolding and do not produce 14. Li–Liu–Yang’s 13 is still the published cubic record (full text still paywalled; see RESEARCH.md). This line does not invent a 14-cycle cubic.
Kept — unperturbed field, degenerate center, period annulus
The field ẋ = 2y, ẏ = −x³ is Hamiltonian for
(equal to 4 times x⁴/4 + y²). Formal differentiation gives
dH/dt ≡ 0 (certificate block unperturbed). The only
equilibrium is the origin: 2y = 0 and −x³ = 0 force (x, y) =
(0, 0). The identity Q₀² + (P₀/2)² = x⁶ + y² is a sum of
even powers and vanishes only there.
The linearization at the origin is ẋ = 2y, ẏ = 0. The Jacobian is [[0, 2], [0, 0]]: trace 0, determinant 0, characteristic polynomial t². This is not a linear center. It is a nilpotent / degenerate quasi-homogeneous center. The Hessian of H at the origin is [[0, 0], [0, 8]], also degenerate.
Regular levels H = h > 0 are compact ovals. H vanishes only at the origin, H → +∞ at infinity, and ∇H = (4x³, 8y) vanishes only at the origin, so those ovals are smooth. Each is a periodic orbit. A continuum; the unperturbed field has zero isolated cycles. This is the unperturbed half already kept on line F, recorded again because the Melnikov count sits on it.
The same field is quasi-homogeneous of weighted degree 2 for weights (1, 2):
Q_0(\lambda x,\lambda^2 y)=\lambda^3 Q_0,\qquad H(\lambda x,\lambda^2 y)=\lambda^4 H.$$ Each level H = c maps to the distinct level H = λ⁴ c (block `weight`). ## Kept — first Melnikov formula Perturb by a polynomial field of degree at most 3: $$\dot x=2y+\varepsilon P,\qquad \dot y=-x^3+\varepsilon Q.$$ The first Melnikov function used here is $$I(h)=\oint_{H=h}Q\,dx-P\,dy.$$ That is the standard line integral of the perturbation 1-form ω = Q dx − P dy, matching Gavrilov–He–Xiao’s ωᵢ. Green’s theorem, counterclockwise, rewrites it as a divergence integral on the disk H ≤ h: $$I(h)=-\iint_{H\le h}(P_x+Q_y)\,dx\,dy.$$ The energy increment along the unperturbed orbit is a constant multiple of the same integral. Along the unperturbed field, dt = dx / (2y) and $$\nabla H\cdot(P,Q)=4x^3 P+8y Q.$$ When P = 0 this collapses by the polynomial identity (8y Q)/(2y) = 4Q (cleared: 8y Q − 8y Q ≡ 0), so $$\oint\nabla H\cdot(P,Q)\,dt=4\oint Q\,dx=4I(h).$$ Zeros of I and of the first-order energy increment coincide. The factor 4 is recorded; it is not a new parameter. ## Kept — Q = μ y produces no positive zero Take P = 0, Q = μ y. Then I(h) = μ ∮ y dx. Green gives ∮ y dx = −A(h), where A(h) is the area of {H ≤ h}. The change of variables x = h^{1/4} s, y = (√h / 2) t has Jacobian (1/2) h^{3/4} and sends the disk to s⁴ + t² ≤ 1. That model region has area $$4\int_0^1\sqrt{1-s^4}\,ds=B\bigl(\tfrac14,\tfrac32\bigr) =\frac{\Gamma(1/4)\Gamma(3/2)}{\Gamma(7/4)},$$ so $$A(h)=\tfrac12 h^{3/4}\,B\bigl(\tfrac14,\tfrac32\bigr).$$ (The factor is 1/2, not 2: Jacobian (1/2) h^{3/4} times Beta, not four times Beta.) Hence $$I(h)=-\mu A(h)=-\frac{\mu}{2}B\bigl(\tfrac14,\tfrac32\bigr)\,h^{3/4}.$$ This is monotone in h and vanishes only at h = 0. There is no positive zero. First-order Melnikov predicts no limit cycle from this perturbation: dH/dt = 8μ y² is one-signed for μ ≠ 0, so every oval is displaced in the same direction. The verifier does not call Gamma or Beta. It certifies the scaling A(λ⁴) = λ³ A(1) from the Jacobian identity, and that A(1) ≠ 0 by a rectangle sitting inside s⁴ + 4u² ≤ 1 (corner 5/16 < 1). A cousin with no y-factor, Q = μ(a + b x² + c y) and P = 0, is no better. The 1-forms dx and x² dx are exact, so ∮ dx = ∮ x² dx = 0 and I(h) = μ c ∮ y dx, again either identically zero or never zero. ## Kept — named family, scaling law, cyclicity at most 1 Named family, still degree 3: $$P=0,\qquad Q=\mu(\alpha-x^2)y.$$ Then div(P, Q) = μ(α − x²) and $$I(h)=\mu\bigl(\alpha J_0(h)-J_2(h)\bigr) =-\mu\bigl(\alpha M_0(h)-M_2(h)\bigr),$$ where Jₖ(h) = ∮ xᵏ y dx and Mₖ(h) = ∬_{H≤h} xᵏ dx dy, so Jₖ = −Mₖ. The extra term in dH/dt is the polynomial 8μ(α − x²) y² (block `family`). Scaling, with x = λ s, y = λ² u and h = λ⁴: the chart identity is H(λs, λ²u) = λ⁴(s⁴ + 4u²), and the Jacobian determinant is λ³ (block `scale`). The integrand xᵏ times that Jacobian is λ^{k+3} sᵏ. Therefore $$M_k(\lambda^4)=\lambda^{k+3}M_k(1).$$ In particular M₀(h) scales as h^{3/4} and M₂(h) as h^{5/4}, so the ratio is a constant times a square root: $$\frac{J_2(h)}{J_0(h)}=\frac{M_2(h)}{M_0(h)}=C\sqrt{h},\qquad C=\frac{M_2(1)}{M_0(1)}=\frac{B(3/4,3/2)}{B(1/4,3/2)}.$$ The constant C is positive: the rectangle [1/2, 3/4] × [0, 1/8] lies in s⁴ + 4u² ≤ 1 because the far corner is 97/256 < 1, and s² ≥ 1/4 there, so M₂(1) ≠ 0; the area rectangle [0, 1/2] × [0, 1/4] has far corner 5/16 < 1, so M₀(1) ≠ 0. The verifier checks those two rational comparisons and the monotonicity derivatives 4s³ and 2u. It does not evaluate C. Thus I(h) = 0 if and only if α = C √h. For fixed α that is at most one h > 0 (none if α ≤ 0). The first-order Melnikov function of this family is not identically zero unless μ = 0, because J₂/J₀ is not constant. First-order cyclicity of the period annulus, for this named family, is at most 1. That is a certified upper bound for the family, not a bound on H(3). The same scaling, applied to a general cubic (P, Q), keeps only three even moments of the quadratic divergence: ∬ 1, ∬ x², and ∬ y² (the region is symmetric in x and in y; odd monomials integrate to zero). Those scale as λ³, λ⁵, and λ⁷, so $$I(h)=h^{3/4}\bigl(c_0+c_1\sqrt{h}+c_2 h\bigr)$$ and a general cubic has at most two positive zeros of I at first order. The named family is the slice c₂ = 0. ## What this is not Not a bound on H(3). Not fourteen zeros of I(h). Not a beat of Gavrilov–He–Xiao’s application (they get cyclicity one for a linear unfolding of this annulus; we get at most one at first order for one cubic family). Not a closed-form evaluation of the Beta values in the verifier — those sit in this note as a closed form, and the replay is the scaling exponents and the two rectangle inclusions. A stranger can run `run.sh` and read the exponents 3, 5, 2 and the cyclicity 1 off the dump.