the one control that matters
The paper prints 0 < h < 1/100 and never says where the bound comes
from. It comes from two places: below zero the axisymmetric Liouville theorems exclude the flow,
at 1/6 the dissipation stops being integrable. Every verdict along the foot is a rational
inequality in h, decided here in integers.
what the pulses are for
The core alone does not solve Navier–Stokes. §5 leaves a
residual equal to −div(annular stress) plus a flat remainder; that stress is what the WAVES must
produce. It is producible only if it lies in a cone (4.22)/(4.23), and then Proposition 7.5 turns it
into two positive squared amplitudes whose square roots are the real wave amplitudes. Turn the
stress out of the cone and one of them goes negative: no real pair supplies it, and nothing cancels
the residual.
vs = 2 + 2/m²—
cone: |s| <—
waves reach—
The wedge is turned by its slope m rather than by the paper's vs, because
m = √(2/(vs − 2)) is what the drawing shows and vs = 2 + 2/m² is then an exact rational — so both
crossings are equalities, not limits. ts rotates the wedge (the frame change of (4.23) has
determinant 1 + ts², so it cannot fold it). The outer, faint wedge is what two real amplitudes
reach; the gap to the hatched one is the margin η
c = 1/10 of Proposition 7.5's
proof, which absorbs the errors of (7.28).
Decided in integers. The stress path itself is
drawn: a bump vanishing at both annular edges with a direction that turns across them.
The audit is next door.
the dichotomy
For this scaling family there is NO h at which an axisymmetric forced blowup from rest can work. At h = 0 the swirl is bounded and the flow is exactly type I,
which the axisymmetric Liouville theorems exclude. Above zero it is type II and escapes them —
and the swirl diverges, which the maximum principle forbids for a flow driven from rest by a
bounded force. The construction survives only by leaving the axisymmetric class, and the paper
never says so.
every scale, exactly
Rational exponents of τ in h, evaluated at the τ on screen. Decided with
BigInt, so h = 0 and h = 1/6 are equalities rather than limits.
computed here
H(Z), the one closed form in the construction: the exterior solves the
radial heat equation exactly. Residual of (A.37) at Z = 1: —.
The shear feeds the oscillation, then shortens its wavelength until
viscosity wins. Peak amplification —.
what backs each mark
Decided. every row is a rational inequality in h with integer numerators; the tab decides it with BigInt, and the boundary cases (h = 0, h = 1/6) are decided as equalities rather than approached
Decided. the cone test (4.22)/(4.23) and the two squared amplitudes of Proposition 7.5 are exact rational arithmetic in the tab — the wedge is parametrised by its slope m so that vs = 2 + 2/m² stays rational, and both cone crossings are equalities rather than limits. The stress path they are applied to is drawn. Every formula is re-derived
from the paper by instruments/navierstokes/probes/stress_cone.py, which carries six red
controls — among them that a stress just outside the cone forces a negative squared amplitude, and
that at angular mode zero the averages of Proposition 7.5 Step 1 both fail.
Computed. H(Z) = Γ(1+h)⁻¹∫₀^∞ e^{−v} v^h (1+Zv)^{−h} dv by Simpson in the tab; the residual of its differential equation (A.37) is printed beside it, and the battery checks the same integral to 25 digits
Drawn. the exponents and incompressibility are the paper’s and are exact; the radial profile shapes are chosen to be legible and claim nothing. No fluid is integrated anywhere on this page
The construction is OpenAI's,
Finite time blowup for Navier–Stokes,
166 pages, sha256 0e779481c4da…. The audit next door built its
2,486-module Lean proof on one laptop in 140 minutes and had
Comparator and a second, independently written kernel accept both theorems.
What is certified is next door.