Line D — restricted exact upper bounds
Status: kept. H(n) did not move. This is not a dent of H(3).
The primary lemma is an exact uniqueness statement for one cubic family. Line C may use the same field as a seed and only needs existence of a periodic orbit. This line owns uniqueness.
Replay:
problems/hilbert16-limit-cycles/compute/q1/d-restricted-upper/run.sh
Python expands the identities sparsely; Rust expands them again
and also evaluates the polar differences on the integer box
from -3 to 3 in each variable. A degree-at-most-4 polynomial in
three variables that vanishes on that box is zero. The two dumps
are diffed. Exit 0.
Kept — radial cubic, exactly one cycle
For every real rho with , the cubic field
\dot y = -x - y(x^2+y^2-\rho^2)$$ has exactly one periodic orbit: the circle of radius $\rho$. The same identities hold in the polynomial ring Z[x, y, rho] with no restriction on rho. For every $\rho > 0$ the unique periodic orbit is still that circle. The interval (0, 1) is the sub-family named in the imagined end-state (a unit-disk seed). If rho is zero the circle collapses to the origin and there is no periodic orbit. Polar form, as polynomial identities (certificate `certs/identities.json`, block `polar`): $$xP+yQ = (x^2+y^2)(\rho^2-x^2-y^2),\qquad xQ-yP = -(x^2+y^2).$$ These are $r\,\dot r = r^2(\rho^2-r^2)$ and $r^2\dot\theta = -r^2$, i.e. $$\dot r = r(\rho^2-r^2),\qquad \dot\theta = -1.$$ Consequences, none of which needs a computer: - The only equilibrium is the origin. Off the origin the angular identity gives $xQ-yP = -r^2 \ne 0$, so (P, Q) cannot vanish. At the origin both components are zero. - The only positive root of $\dot r = 0$ is $r = \rho$. As polynomials, $r(\rho^2-r^2) = r(\rho-r)(\rho+r)$ (block `radial_speed`). - On the circle $r = \rho$ one has $\dot r = 0$ and $\dot\theta = -1 \ne 0$, so the orbit is periodic of period $2\pi$. - On $(0,\rho)$ one has $\dot r > 0$; on $(\rho,\infty)$ one has $\dot r < 0$. The circle is invariant, so no orbit crosses it. A periodic orbit cannot live in either open annulus, because r is strictly monotone there. Hence the circle is the only periodic orbit. The circle is a hyperbolic limit cycle (attracting from both sides). That is more than uniqueness; uniqueness is the claim. This is an upper bound of 1 for one documented family, not an upper bound for all cubics. It does not touch the published record H(3) at least 13. ## Kept — quadratic Hamiltonian fields, zero limit cycles Let H be a real polynomial of degree at most 3, and set $$\dot x = \partial H/\partial y,\qquad \dot y = -\partial H/\partial x.$$ The field has degree at most 2. Formal differentiation in the ring of polynomials in x, y and the cubic coefficients of H gives $$\frac{dH}{dt} = \frac{\partial H}{\partial x}\frac{\partial H}{\partial y} + \frac{\partial H}{\partial y}\Bigl(-\frac{\partial H}{\partial x}\Bigr) = 0$$ identically (certificate block `hamiltonian`, `dHdt` empty). Every orbit lies in a level of H. A periodic orbit cannot contain an equilibrium, so the gradient of H has no zero on it: the field is exactly the Hamiltonian rotation of that gradient. A regular compact connected component of a level of H in the plane is a simple closed curve, and nearby regular levels supply a continuum of nearby periodic orbits. The orbit is therefore not isolated, hence not a limit cycle. The family has periodic orbits (any nondegenerate quadratic center), but no limit cycles. Zero is sharp as an upper bound on the number of limit cycles. ## Kept — Lotka–Volterra in the open first quadrant The quadratic Kolmogorov field $$\dot x = x(a+bx+cy),\qquad \dot y = y(d+ex+fy)$$ has no limit cycle in the open first quadrant. The proof is a Dulac identity plus a parallel reduction, not a citation. Write $B = x^{\alpha-1} y^{\beta-1}$. The normalised divergence $$\frac{\mathrm{div}(BX)}{B} = (\alpha-1)\frac{P}{x}+\frac{\partial P}{\partial x} + (\beta-1)\frac{Q}{y}+\frac{\partial Q}{\partial y}$$ is the polynomial (block `lotka_volterra`) $$(\alpha a+\beta d) + \bigl((\alpha+1)b+\beta e\bigr)x + \bigl(\alpha c+(\beta+1)f\bigr)y.$$ When $\Delta = bf-ce \ne 0$, Cramer's numerators $\alpha\Delta = f(e-b)$ and $\beta\Delta = b(c-f)$ kill the coefficients of x and y (block `cramer`, both products are the zero polynomial). The remaining constant is $K = \alpha a+\beta d$. If K is nonzero, Bendixson–Dulac says there is no closed orbit in the first quadrant. If K is zero, BX is divergence-free, hence locally Hamiltonian, and periodic orbits lie in levels of a first integral, so they are not isolated. When Delta is zero the linear parts are parallel. If (b, c) is zero then $\dot x = ax$: either x is exponential (not periodic) or a is zero and motion is vertical, hence one-dimensional. The case (e, f) zero is symmetric. If (e, f) equals lambda times (b, c), the identity (block `parallel`) $$\lambda(a+bx+cy)-(d+ex+fy)=\lambda a-d$$ says that lambda log x minus log y has constant derivative lambda a minus d. If that constant is nonzero the function is strictly monotone and there is no closed orbit. If it is zero, $y/x^{\lambda}$ is a first integral and motion on each power curve is one-dimensional. In every case there is no isolated periodic orbit in the open first quadrant. A conservative Volterra center has a continuum of cycles; those are not limit cycles. ## Dropped / forked Nothing dropped. No fork. Lean was not added: the identities are short polynomial cancellations, and a mathlib `ring` lemma would repeat the Python/Rust check after a long lake fetch. ## What this is not Not a bound on H(3). Not a bound on H(2). Not a new published record. The uniqueness lemma is reusable: a stranger can replay `run.sh` and read the two polar identities off the JSON.