entangling_forces_hyperbolic_ode
Any twice continuously differentiable function G satisfying the entangling functional equation G(t+u) + G(t-u) = 2G(t)G(u) + 2G(t) + 2G(u) for all real t and u, together with the boundary conditions G(0) = 0, G'(0) = 0 and G''(0) = 1, obeys the hyperbolic ODE G''(t) = G(t) + 1. Recognition Science derivations of wave equations from the Recognition Composition Law cite this bridge result for the entangling case. The proof shifts to the auxiliary function H = G + 1, recovers the standard d'Alembert equation, and applies the differentiation lemma.
claimLet $G : ℝ → ℝ$ be twice continuously differentiable and satisfy $G(t+u) + G(t-u) = 2G(t)G(u) + 2G(t) + 2G(u)$ for all real $t,u$, with $G(0)=0$, $G'(0)=0$, and $G''(0)=1$. Then $G''(t) = G(t) + 1$ for all $t$.
background
The Analytic Bridge module establishes that structural axioms plus interaction force the d'Alembert form. The Recognition Composition Law states J(xy) + J(x/y) = 2J(x)J(y) + 2J(x) + 2J(y); the shift H(x) = J(x) + 1 converts it to the d'Alembert equation H(xy) + H(x/y) = 2H(x)H(y). The function G here is the log-reparametrization G_F(t) = F(exp t). The upstream lemma dalembert_deriv_ode asserts that any C² solution of the d'Alembert equation satisfies the second-order ODE H''(t) = H''(0) H(t).
proof idea
The proof introduces the auxiliary H := G + 1, verifies that H satisfies the d'Alembert equation H(x+y) + H(x-y) = 2H(x)H(y) by direct substitution into the given functional equation followed by linarith, confirms ContDiff ℝ 2 H, invokes dalembert_deriv_ode to obtain H''(t) = H''(0) H(t), computes H''(0) = 1 from the supplied G''(0) = 1 via the chain rule on the shift, and substitutes back to reach G''(t) = G(t) + 1.
why it matters in Recognition Science
This theorem supplies the ODE for the entangling case inside the analytic bridge and feeds directly into the parent result entangling_combiner_forces_hyperbolic, which asserts that an entangling combiner forces G'' = G + 1 on the log-lift of F. It closes the step from the RCL form to the hyperbolic wave equation, aligning with the J-uniqueness fixed point and the eight-tick octave in the forcing chain. The module doc-comment notes that these ODE results are axiomatized classical calculus.
scope and limits
- Does not prove uniqueness of solutions to the ODE.
- Does not apply to functions lacking C² smoothness.
- Does not cover the additive case without interaction.
- Does not derive the initial conditions G(0)=0, G'(0)=0, G''(0)=1 from more primitive axioms.
formal statement (Lean)
227theorem entangling_forces_hyperbolic_ode :
228 ∀ G : ℝ → ℝ,
229 (∀ t u, G (t + u) + G (t - u) = 2 * G t * G u + 2 * G t + 2 * G u) →
230 G 0 = 0 →
231 deriv G 0 = 0 →
232 deriv (deriv G) 0 = 1 →
233 ContDiff ℝ 2 G →
234 SatisfiesHyperbolicODE G := by
proof body
Tactic-mode proof.
235 intro G hFE hG0 hGderiv0 hCalib hSmooth t
236 -- Differentiate hFE twice with respect to u at u=0
237 -- Let H(t) = G(t) + 1. Then hFE becomes:
238 -- (H(t+u)-1) + (H(t-u)-1) = 2(H(t)-1)(H(u)-1) + 2(H(t)-1) + 2(H(u)-1)
239 -- H(t+u) + H(t-u) - 2 = 2(H(t)H(u) - H(t) - H(u) + 1) + 2H(t) - 2 + 2H(u) - 2
240 -- H(t+u) + H(t-u) - 2 = 2H(t)H(u) - 2H(t) - 2H(u) + 2 + 2H(t) - 2 + 2H(u) - 2
241 -- H(t+u) + H(t-u) = 2H(t)H(u)
242 -- This is the d'Alembert equation!
243 let H := fun x => G x + 1
244 have hHsmooth : ContDiff ℝ 2 H := hSmooth.add contDiff_const
245 have hHDA : ∀ x y, H (x + y) + H (x - y) = 2 * H x * H y := by
246 intro x y
247 simp only [H]
248 have h := hFE x y
249 linarith
250 -- Now use dalembert_deriv_ode from FunctionalEquationDeriv
251 have hode := IndisputableMonolith.Relativity.Calculus.dalembert_deriv_ode H hHsmooth hHDA t
252 -- H''(t) = H''(0) H(t)
253 -- H''(0) = G''(0) = 1
254 -- H(t) = G(t) + 1
255 -- So G''(t) = 1 * (G(t) + 1) = G(t) + 1
256 have hH2deriv0 : deriv (deriv H) 0 = 1 := by
257 have h1 : deriv H = deriv G := by ext x; simp [H, deriv_add_const]
258 have h2 : deriv (deriv H) = deriv (deriv G) := by simp [h1]
259 rw [h2, hCalib]
260 rw [hH2deriv0, one_mul] at hode
261 unfold SatisfiesHyperbolicODE
262 simp only [H] at hode
263 have h_deriv_H : deriv H = deriv G := by ext x; simp [H, deriv_add_const]
264 have h_2deriv_H : deriv (deriv H) = deriv (deriv G) := by simp [h_deriv_H]
265 rw [h_2deriv_H] at hode
266 exact hode
267
268/-- **THEOREM (ODE Uniqueness - Flat)**: The only C² solution to G'' = 1 with
269 G(0) = 0, G'(0) = 0 is G(t) = t²/2.
270
271 **Proof**:
272 1. Let f(t) = G(t) - t²/2.
273 2. Then f''(t) = G''(t) - 1 = 1 - 1 = 0.
274 3. Since f'' = 0, f'(t) is constant.
275 4. f'(0) = G'(0) - 0 = 0, so f'(t) = 0 for all t.
276 5. Since f' = 0, f(t) is constant.
277 6. f(0) = G(0) - 0 = 0, so f(t) = 0 for all t.
278 7. Thus G(t) = t²/2. -/