Line LL — cubic with an invariant line and three cycles
Status: imagined three isolated cycles dropped. Fork kept. Not a
dent of H(3).
Imagined certificate. An explicit cubic with a straight-line solution has three isolated periodic orbits.
Drop immediately. No three-cycle cubic is written. Ye / Cherkas uniqueness (at most one cycle) is a theorem for quadratics with an invariant line, not for cubics. Cubics with a line remain open. A linear Dulac factor along the line, on the named family below, does not produce three cycles: it produces zero.
Fork kept. One named cubic with a genuine invariant line (not a
line of equilibria) and a certified isolated-cycle count of 0
for that family, by Bendixson–Dulac. Symbolic only. Fractions as
strings. Not a bound on H(n).
Replay:
problems/hilbert16-limit-cycles/compute/q3/ll-invariant-line/run.sh
Python expands the identities sparsely; Rust expands them again
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:
Scholarpedia, Han–Li–Li 2010.
Lead, not a citation. The article records: Chen–Ye (class I)
and Cherkas–Zhilevich / Rychkov (class III with a = 0) give
that a quadratic with a straight-line solution has at most one
limit cycle. That is the quadratic context below. It is not
re-proved here, and it does not apply to degree 3.
Dropped — three isolated periodic orbits
The fiction needs an explicit cubic field, a genuine invariant line, and three isolated cycles. Ye / Cherkas do not forbid that: their uniqueness theorem is for degree 2. Nothing in the fetched record writes three cubic cycles on a line as a replayable field, and this line does not invent one. A numeric search, or a Poincaré–Bendixson picture without a certificate, is not a lower bound.
The named field below is degree 3 and has an invariant line. The count it certifies is 0, not 3.
Context — quadratics, at most one (Ye / Cherkas)
A quadratic planar field with an invariant straight line has at
most one limit cycle. Scholarpedia records this as the union of
Ye class I (Chen–Ye) and the Cherkas–Zhilevich / Rychkov
treatment of class III with a vanishing a. Coppel’s survey
and later expositions quote the same uniqueness. This line does
not re-prove that theorem and does not replay Cherkas’s Abel
reduction. The object below is cubic.
Kept — named cubic, invariant line y = 0
Named field, integer coefficients, degree 3:
The same orbits as the μ-family member μ = 1/16,
after a constant time rescaling (block named).
The polynomial y is a first integral of the line: the Lie
derivative is Q = 16xy = (16x) y, so the cofactor is 16x
(block line). Hence y = 0 is invariant. It is not a line of
equilibria. Restricting to y = 0 gives
The factor 16 + x² is at least 16 (block axis; as a
polynomial, 16 + x² − 16 − x² ≡ 0 and the constant 16 is
nonzero). So P(x, 0) vanishes only at the origin. Sample:
P(1, 0) = 17.
Equilibria of the named field: Q = 0 forces x = 0 or y = 0.
On x = 0 one has P = 16y, so y = 0. On y = 0 one has
P = x(16 + x²), so x = 0. The only equilibrium is the origin
(block equilibria). The Jacobian there is
trace 16, determinant 0: one eigendirection along the invariant line, as it must be. A focus cannot sit on an invariant line.
The one-parameter family, still degree 3 for μ ≠ 0,
has the same invariant line y = 0 with cofactor x, and
P(x, 0) = x(1 + μ x²). For μ ≥ 0 the axis factor 1 + μ x² is
at least 1, so again the axis is not a line of equilibria
(block family).
A degeneration that was discarded: ẋ = y,
ẏ = −y(x + μ(x² − 1)y) has Q(x, 0) ≡ 0, but also P(x, 0) ≡ 0,
so the whole x-axis is equilibria (block degeneration). That
is not a genuine invariant line in the sense of this line.
Kept — Dulac, certified count 0
Ordinary divergence of the named field is
The discriminant of that quadratic is 16² − 4 · 3 · 16 = 64
(block bendixson). It changes sign: div(0, 0) = 16 and
div(−3, 0) = −5. Bendixson’s criterion on the plane is
inconclusive. The invariant line is used.
Take the Dulac function B = 1/y, which is C¹ and nonzero on each open half-plane. The weighted divergence is
=\frac{16+3x^2}{y}.$$ The polynomial identity behind the cancellation is $$y\cdot\mathrm{div}\,X-Q=y(16+3x^2)$$ (block `dulac`; the xy terms cancel). The numerator 16 + 3x² is at least 16 and is not the zero polynomial. In {y > 0} one has div(BX) > 0; in {y < 0} one has div(BX) < 0. Each open half-plane is simply connected. Bendixson–Dulac: neither half-plane contains a closed orbit. A closed orbit of a C¹ field cannot cross an invariant line. The restricted flow on y = 0 is one-dimensional and, as above, has no periodic orbit. Therefore the named field has exactly 0 isolated periodic orbits. The same Dulac works for the whole family μ ≥ 0. There $$\mathrm{div}\,X=1+x+3\mu x^2,\qquad y\cdot\mathrm{div}\,X-Q=y(1+3\mu x^2),$$ and 1 + 3μ x² ≥ 1 (block `family`). Certified isolated-cycle count for the family: 0. When μ = 0 the field is quadratic and Ye already gives at most one; the same Dulac gives zero for this particular quadratic. That is not a proof of Ye. When 0 < μ < 1/12 the unweighted discriminant 1 − 12μ is positive, so ordinary Bendixson fails and the factor 1/y is essential. The named cubic μ = 1/16 lies in that range. The count 0 is an exact upper bound for this family, not for every cubic with a line, and not for H(3). A different cubic with a line might have cycles; this line does not produce one. ## What this is not Not a bound on H(n). Not a dent. Not three cubic cycles. Not a re-proof of Ye / Cherkas. Not a line of equilibria. The reusable lemma is the named field and the Dulac identity: a stranger can run `run.sh` and read the cycle count 0, the line y = 0, and the cancelled numerator 16 + 3x² off the JSON.