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Eq. (49) solves Eq. (44): F˙k+Fk2+k2=0\dot F_k + F_k^2 + k^2 = 0F˙k​+Fk2​+k2=0

Proved
AKR2008.hybrid_modeF_solves_eq44

by Lucas · Sep 27, 2026 · Mathlib 0df444a (Lean v4.33.1)

mathematical-physicsquantum-gravity

Throughout, kernels of Sec. IV A are written in the real orthonormal eigenbasis f(k)f^{(k)}f(k) of −∂x2-\partial_x^2−∂x2​ (eigenvalue k2k^2k2): a kernel Axy=∑kAkfx(k)fy(k)A_{xy}=\sum_k A_k f^{(k)}_x f^{(k)}_yAxy​=∑k​Ak​fx(k)​fy(k)​ is represented by its mode coefficient AkA_kAk​, so ∫dx AyxBxz↦AkBk\int dx\,A_{yx}B_{xz}\mapsto A_kB_k∫dxAyx​Bxz​↦Ak​Bk​, ∂z2δ(y−z)↦−k2\partial_z^2\delta(y-z)\mapsto -k^2∂z2​δ(y−z)↦−k2 and ∫dx Axx↦∑kAk\int dx\,A_{xx}\mapsto\sum_k A_k∫dxAxx​↦∑k​Ak​.

For every real kkk and every time ttt with cos⁡(kt)≠0\cos(kt)\ne0cos(kt)=0, the mode coefficient Fk(t)=−ktan⁡(kt)F_k(t)=-k\tan(kt)Fk​(t)=−ktan(kt) of Eq. (49) satisfies the mode form of Eq. (44),

F˙k(t)+Fk(t)2+k2=0.\dot F_k(t) + F_k(t)^2 + k^2 = 0 .F˙k​(t)+Fk​(t)2+k2=0.

Eq. (44), F˙yz+∫dx FyxFxz−∂z2δ(y−z)=0\dot F_{yz}+\int dx\,F_{yx}F_{xz}-\partial_z^2\delta(y-z)=0F˙yz​+∫dxFyx​Fxz​−∂z2​δ(y−z)=0, is the Riccati equation for the kernel of the classical Hamilton–Jacobi functional Sc[ϕ,t]=12∬ϕyFyz(t)ϕzS^c[\phi,t]=\tfrac12\iint\phi_yF_{yz}(t)\phi_zSc[ϕ,t]=21​∬ϕy​Fyz​(t)ϕz​; it is independent of ggg and therefore also holds in the interacting solution of Sec. IV B.

Formalization Note The time derivative is taken with deriv; the hypothesis cos⁡(kt)≠0\cos(kt)\ne0cos(kt)=0 excludes the poles of tan⁡\tantan.

Preamble
import Mathlib
import Definitions.Def_AKR2008_HybridDefs
Formal statement
namespace AKR2008

theorem hybrid_modeF_solves_eq44 (k t : ℝ) (hcos : Real.cos (k * t) ≠ 0) :
    deriv (fun s => hybridModeF k s) t + hybridModeF k t ^ 2 + k ^ 2 = 0 := by sorry

end AKR2008
Source
M. Albers, C. Kiefer, M. Reginatto, Measurement analysis and quantum gravity, Phys. Rev. D 78, 064051 (2008), https://doi.org/10.1103/PhysRevD.78.064051, p. 064051-10, Sec. IV A, Eqs. (44) and (49)
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What the Lean code literally says, in plain math · Aristotle (Harmonic)

Non-blind read-back — not independent testimony. This read-back was written by the same agent that drafted the Lean statement, not by a blind auditor with a fresh context. It is not independent evidence of faithfulness: compare the Lean code against the source yourself.

For all real numbers kkk and ttt such that cos⁡(kt)≠0\cos(kt)\neq 0cos(kt)=0:

dds[−ktan⁡(ks)]s=t+(−ktan⁡(kt))2+k2=0.\frac{d}{ds}\Bigl[-k\tan(ks)\Bigr]_{s=t} + \bigl(-k\tan(kt)\bigr)^2 + k^2 = 0 .dsd​[−ktan(ks)]s=t​+(−ktan(kt))2+k2=0.

No other hypotheses; k=0k=0k=0 is allowed (then cos⁡(0)=1≠0\cos(0)=1\ne0cos(0)=1=0 and every term is 000).

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