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Eq. (51) solves Eq. (48): −12K˙k−KkFk=0-\tfrac12\dot K_k - K_kF_k = 0−21​K˙k​−Kk​Fk​=0

Proved
AKR2008.hybrid_modeK_solves_eq48

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 all real τk,k\tau_k, kτk​,k and every time ttt with cos⁡(kt)≠0\cos(kt)\ne0cos(kt)=0, the coefficients Kk(t)=τk/cos⁡2(kt)K_k(t)=\tau_k/\cos^2(kt)Kk​(t)=τk​/cos2(kt) (Eq. (51), τk\tau_kτk​ an arbitrary constant) and Fk(t)=−ktan⁡(kt)F_k(t)=-k\tan(kt)Fk​(t)=−ktan(kt) (Eq. (49)) satisfy the mode form of Eq. (48),

−12 K˙k(t)−Kk(t)Fk(t)=0.-\frac12\,\dot K_k(t) - K_k(t)F_k(t) = 0 .−21​K˙k​(t)−Kk​(t)Fk​(t)=0.

Eq. (48) governs the width kernel Kab(t)K_{ab}(t)Kab​(t) of the classical Gaussian ensemble PcP^cPc.

Preamble
import Mathlib
import Definitions.Def_AKR2008_HybridDefs
Formal statement
namespace AKR2008

theorem hybrid_modeK_solves_eq48 (τk k t : ℝ) (hcos : Real.cos (k * t) ≠ 0) :
    -(1 / 2) * deriv (fun s => hybridModeK τk k s) t - hybridModeK τk k t * hybridModeF k t = 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. (48), (49), (51)
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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 τk,k,t\tau_k,k,tτk​,k,t with cos⁡(kt)≠0\cos(kt)\ne0cos(kt)=0:

−12 dds[τkcos⁡2(ks)]s=t−τkcos⁡2(kt)⋅(−ktan⁡(kt))=0.-\frac12\,\frac{d}{ds}\Bigl[\frac{\tau_k}{\cos^2(ks)}\Bigr]_{s=t} - \frac{\tau_k}{\cos^2(kt)}\cdot\bigl(-k\tan(kt)\bigr) = 0 .−21​dsd​[cos2(ks)τk​​]s=t​−cos2(kt)τk​​⋅(−ktan(kt))=0.

No other hypotheses.

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