files / problems / hilbert16-limit-cycles / compute / q3 / kk-plus-one / LINE.md

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Line KK — Gasull–Santana +1 on the radial cubic

Status: two-cycle dent dropped. Kept: the translated cubic, the explicit degree-4 product, and the exact miss of the line past the translated circle. Not a dent of H(4).

Imagined certificate. Apply the Gasull–Santana construction H(n+1) ≥ H(n)+1 to the radial cubic of q1 line D (ρ² = 1/4). The result is an explicit degree-4 field with two hyperbolic isolated periodic orbits: the translated circle, plus a Hopf cycle from the new monodromic point. A constructive +1, and in particular H(4) ≥ 2.

Drop immediately. The second cycle is a Hopf they do not write term-by-term. This line does not invent a numeric (ε, δ) perturbation and call it two cycles. One explicit hyperbolic circle in degree 4, or even two, does not beat Prohens–Torregrosa H(4) ≥ 28.

Fork kept. The finite algebra of their Theorem 1, specialised to this cubic: translate the regular point (2, 0) to the origin, multiply by the line 4x − 15y, prove that line misses the translated circle, and read the Jacobian identities off the product. The n+2 field (x²+y²)X on the untranslated cubic is degree 5 and has the same unique circle; it is not a +1. Symbolic only. Fractions as strings. No ODE integration. Not a bound on H(n).

Replay:

problems/hilbert16-limit-cycles/compute/q3/kk-plus-one/run.sh

Python expands the translated field and the degree-4 product over Q. Rust expands the content-cleared integer field with a BTreeMap and evaluates the residuals on the integer box from −3 to 3. The two dumps are diffed. Exit 0. Certs: certs/core.json, certs/identities.json.

Opened this session: arXiv:2407.13465v2 and its HTML (Theorem 1 and its proof; Lemma 1; the n+2 remark in §4). Same two theorems as PAMS 153 (2025) 669–677. The introduction still records H(4) ≥ 28.

Dropped — two hyperbolic cycles, hence H(4) at least 2

The fiction needs an explicit degree-4 field with two hyperbolic isolated periodic orbits, offered as a constructive +1 and as a dent of H(4) ≥ 28.

Theorem 1 of Gasull–Santana is an existence argument, not a term-by-term field. After the product they unfold

S_{\delta}=((a+\delta)x+by)Q,$$ choose small `ε, δ` so the origin is an isolated monodromic singularity with `L₁ ≠ 0`, and perturb once more to birth a Hopf cycle. None of those perturbations is written here. A numeric integrator is not a certificate. Two cycles would not beat 28 in any case. The product field below has one proved periodic orbit: the translated circle. The origin of that product is a non-isolated equilibrium on a line of singularities, not a focus. ## Kept — translate (2, 0), then multiply The radial cubic of q1 line D, with `ρ² = 1/4`, is $$\dot x=y-x\bigl(x^2+y^2-\tfrac14\bigr),\qquad \dot y=-x-y\bigl(x^2+y^2-\tfrac14\bigr).$$ The unique periodic orbit is the circle of radius `1/2`. Polar form `ṙ = r(ρ² − r²)`, `θ̇ = −1`. The origin is an equilibrium, so the Gasull–Santana multipliers `a = −Q(0, 0)`, `b = P(0, 0)` both vanish. Multiplying the untranslated field by a line through the origin is not their construction: every such line meets the circle, because the distance from `(0, 0)` to the line is `0 < 1/2`. Take the regular point `p = (2, 0)`, outside any closed ball that contains the circle in its interior. Then $$P(2,0)=-2\bigl(4-\tfrac14\bigr)=-\tfrac{15}{2},\qquad Q(2,0)=-2.$$ Translate `p` to the origin: `P_t(x, y) = P(x+2, y)` and `Q_t(x, y) = Q(x+2, y)`. Expanding over `Q` with `ρ² = 1/4` gives $$\begin{aligned} P_t&=-x^3-xy^2-6x^2-2y^2+y-\tfrac{47}{4}x-\tfrac{15}{2},\\ Q_t&=-x^2y-y^3-4xy-x-\tfrac{15}{4}y-2. \end{aligned}$$ In particular `P_t(0, 0) = −15/2` and `Q_t(0, 0) = −2`. The Gasull–Santana line is `ax + by` with $$a=-Q_t(0,0)=2,\qquad b=P_t(0,0)=-\tfrac{15}{2},$$ so `2x − (15/2)y`. Clear content by doubling: `L = 4x − 15y`. The degree-4 field is $$R=L\,P_t,\qquad S=L\,Q_t.$$ (Equivalently work with `(2R, 2S)`, or with the integer field `(L · 4P_t, L · 4Q_t)`.) Thirteen plus ten monomials; leading term of `R` is `−4x⁴`. The whole line `L = 0` is a line of singularities. The origin is a non-isolated equilibrium. ## Kept — the line misses the translated circle The translated circle is `(x+2)² + y² = 1/4`: centre `(-2, 0)`, radius `1/2`. Distance from the centre to `4x − 15y = 0` is $$\frac{|4(-2)|}{\sqrt{16+225}}=\frac{8}{\sqrt{241}}.$$ The line misses the circle if and only if that distance exceeds the radius: $$\frac{8}{\sqrt{241}}>\frac12\iff 16>\sqrt{241}\iff 256>241 \iff 15>0.$$ Cleared comparison, integers only: `4 · 8² − 241 = 15 > 0`. The same miss as a quadratic: substituting `y = 4x/15` into the circle and clearing produces $$964x^2+3600x+3375=0,$$ discriminant `3600² − 4 · 964 · 3375 = −54000 < 0`. No real intersection. On the circle, `L ≠ 0`. Sample values: `L(-2, 0) = −8` at the centre, `L(-3/2, 0) = −6` at the rightmost point. The circle is connected and does not meet the line, so `L` has constant sign on it (negative). ## Kept — the translated circle remains a periodic orbit The original polar identity, after translation, is $$(x+2)P_t+yQ_t=\bigl((x+2)^2+y^2\bigr) \bigl(\tfrac14-(x+2)^2-y^2\bigr).$$ The right-hand side vanishes on the translated circle, so that circle is invariant for `(P_t, Q_t)`. Off the origin of the original field the angular identity is `θ̇ = −1 ≠ 0`, so the orbit is periodic. Because `L ≠ 0` on the circle, `(R, S)` is a nowhere-zero time rescaling of `(P_t, Q_t)` along that orbit. The translated circle is therefore a periodic orbit of the degree-4 field. Time is reversed (`L < 0`), so the attracting circle of the cubic becomes repelling; it stays hyperbolic. One cycle. Not two. ## Kept — Jacobian identities at the origin Write `X = (R, S) = (L P_t, L Q_t)`. At the origin `L = 0`, so $$DX(0,0)=\begin{pmatrix}L_x P_t&L_y P_t\\ L_x Q_t&L_y Q_t\end{pmatrix} =\begin{pmatrix}-30&225/2\\ -8&30\end{pmatrix}.$$ Determinant `0`, trace `0`. Their display, for the uncleared line `ax + by`, is $$DX(0,0)=\begin{pmatrix}ab&b^2\\ -a^2&-ab\end{pmatrix} =\begin{pmatrix}-15&225/4\\ -4&15\end{pmatrix},$$ again determinant `0` and trace `0`. The doubled line multiplies that matrix by `2`; the two vanishing identities are unchanged. ## Kept — n+2 on the untranslated cubic Section 4 of the paper notes that `H(n+2) ≥ H(n)+1` is easier: `Y = (x² + y²) X` is degree `n+2` and agrees with `X` off the origin. On the untranslated radial cubic this is degree 5. The polar identities become $$r\,\dot r=r^4(\rho^2-r^2),\qquad r^2\dot\theta=-r^4,$$ so `ṙ = r³(ρ² − r²)` and `θ̇ = −r²`. The only positive root of `ṙ = 0` is still `r = ρ`. Same unique circle. The origin is an extra degenerate equilibrium. No new cycle is written. This is not a `+1` in degree 4, and it is not a dent of `H(5) ≥ 37`. ## What this is not Not a bound on `H(n)`. Not a dent. Not two hyperbolic cycles. Not an explicit Hopf perturbation. Not `H(4) ≥ 2` as a published improvement. The reusable lemma is the explicit product and the miss: a stranger can run `run.sh` and read `4x − 15y` and `4 · 8² − 241 = 15` off the dump.