Route-D v14: the two-scale profile equation has a first integral

Phase-2 P2, Route-D leg 14. Figure: writeup/figures/fig32_route_d_v14_first_integral.png. Data: writeup/data/p2_route_d_v14_first_integral.json (rebuild the figure with writeup/4_p2_lottery/p2_route_d_v14_evidence.py; regenerate the data with experiments/p2_route_d_v14_first_integral.py, ~6 min, deterministic). Code: solver/first_integral.py + test_first_integral.py (11/11; suite 21 files green).

Rigor level: 1 (a novel numerical map) plus one exact algebraic identity. Nothing here is interval-enclosed, nothing is a certificate, and nothing is a Clay result. See the honesty section at the end.

0. The one-line summary

Thirteen legs treated

R = Ω H(Ω) − c Ω_X − a U Ω_X = 0 ,      U(X) = ∫₀^X H(Ω) dX'      (P)

as a nonlinear integro-differential equation on the whole line, to be discretized and inverted. It is not one. Writing E = c + a U for the effective transport coefficient v12 introduced, E_X = a H(Ω) exactly, so wherever Ω ≠ 0

R = 0  ⟺  Ω E_X / a = E Ω_X  ⟺  (log|Ω|)_X = (1/a)(log E)_X

and therefore |Ω| = C E^{1/a}. The amplitude gauge Ω(0) = −1 fixes C = c^{−1/a} (at X = 0, U = 0 and E = c), giving

Ω(X) = − ( E(X) / c )^{1/a} , E = c + a U , U_X = H(Ω). (FI)

That is an exact first integral of (P). Three legs' worth of structure falls out of it in one line each, and, the point of the leg, it makes the kill switch the continuation prompt has been asking for cheap enough to just run.

The kill switch passes. Posed on [0, X_c] via (FI), the approximate inverse that v12 measured diverging as J^+2.86 is flat: K^−0.0009 … K^+0.0011 over K = 48…192, at every a tested, and unchanged under the decay grading the whole-line measurement used.

1. Why it is a first integral, and what gates it

Nothing above assumes anything about Ω beyond Ω ≠ 0 and E ≠ 0 on the interval. Three independent gates:

(a) The a → 0 limit IS the exact anchor. (1 + aU/c)^{1/a} → exp(U/c), so (FI) degenerates to Ω = −exp(U/c). On the anchor U = −½ log(1+X²), c = ½, hence exp(U/c) = 1/(1+X²) and Ω = −1/(1+X²): exactly the known solution the whole project is anchored on. Measured max error over X ∈ [0,200]: 1.11e−16. The finite-a form approaches it at the predicted O(a) (fitted a^1.011).

(b) On a discretization that knows nothing about it. |Ω|/E^{1/a} should be constant. On the whole-line θ-collocation profiles of solver/collocation_newton the relative spread over the core is

The defect is an order of magnitude below the profile's own residual at every J and falls faster (×251 vs ×126 at a=0.2; ×20 vs ×9 at a=0.3). That is the signature of an exact identity evaluated on an approximate object: what is being measured is the profile, not the identity. This is the test-suite gate, deliberately written as a rate rather than a threshold, because a threshold would only have measured the profile.

(c) The reconstruction solves the original equation. Take the reduced solution, rebuild Ω, and evaluate R itself with an independent quadrature at points that are not collocation nodes: max|R| = 6.8e−7 → 4.3e−8 → 2.7e−9 → 1.9e−10 as the U quadrature refines, fitted n^−1.98, converging at the trapezoid's own second order, i.e. to zero.

2. What (FI) says without any computation

The profile ends, and it is forced rather than discovered. E is decreasing wherever H(Ω) < 0, so it reaches zero at a finite X_c; beyond it E < 0 and E^{1/a} is not real, so Ω ≡ 0. With Ω ≡ 0 outside, H(Ω) out there is still negative (the mass inside is negative), so E stays negative and the situation is self-consistent. v12 found the support edge by noticing a sign change in a measured quantity; here it is a consequence.

Caveat, stated rather than hidden: the step needs H(Ω) < 0 for X > 0, which holds on every solution found here and is what the solve returns, but proving it a priori for all admissible Ω is a separate lemma this leg did not attempt.

The zero has order 1/a, with the amplitude explicit. E vanishes linearly at X_c (E_X(X_c) = a H(Ω)(X_c) ≠ 0), so Ω ~ −A (X_c − X)^{1/a} with A = (2 s(1)/X_c)^{1/a} computable from the solution: v12 got the exponent from a leading balance and called the amplitude free; (FI) gives both. Measured edge exponents against 1/a:

(relative error ≤ 1.4e−5; v13's independent whole-line instrument got 0.02–0.6%.)

The profile is only finitely smooth. Ω ∈ C^{1/a} at the edge and no better. It is a C¹ (hence classical) solution exactly while a < 1; at a = 1, De Gregorio, the edge is a corner, and for a > 1 it is a cusp with Ω_X → ∞ there (the equation still holds in the limit, since E Ω_X ~ s^{1/a} → 0, but the solution is classical only away from ±X_c). Flagged as an observation, not a claim: whether the loss of C¹ exactly at De Gregorio is meaningful needs the literature check, not another leg.

3. The reduced system, and why it converges where the direct build did not

Scale by the radius, v = X/X_c, e(v) = E(X_c v)/c. The finite Hilbert transform is scale invariant, so X_c drops out of H and the whole problem is scalar:

c e'(v) + a X_c · Hpv` e^{1/a} ` = 0 ,   e(0) = 1 ,   e(1) = 0 ,
Hpvw = (1/π) p.v. ∫_{−1}^{1} w(|u|)/(v−u) du .                     (RS)

c enters only as a scale (dilation sends X_c → μX_c, c → μc and leaves e alone), so the module fixes c = 1 and every radius reported is the invariant X_c/c.

Two structural facts are why this works, and both are worth carrying forward:

1. e(1) = 0 goes in the ansatz, not in an extra row. e = (1−v²) s(v) with s an even Chebyshev series. The support edge is exact by construction, and the order-1/a zero of Ω = −e^{1/a} is an output, not something a grid has to resolve and not something put in by hand. (The WIP direct build solver/finite_support.py put (1−v²)^{1/a} in the ansatz by hand, which is the same thing only if you already know 1/a.)
2. The edge row is non-degenerate. The raw residual R is identically zero at X_c (every term carries Ω or Ω_X, both of which vanish) so a collocation row there carries no information, and a direct build must append an ad-hoc free-boundary condition. (RS) has c e'(1) = −a X_c Hpvw with both sides nonzero: the free boundary is priced by the equation itself.

solver/finite_support.py is the direct build. It never converged (residual 1.4 after 59 iterations at a=0.3). (RS) converges from a cold start (s ≡ 1, X_c⁰ = 10 regardless of a) in 5–10 Newton iterations to residual ~1e−14, at every a from 0.2 to 1.2. That difference is the whole content of items 1–2.

Numerics: the p.v. is taken by one subtraction against the exact p.v. ∫_{−1}^{1} du/(v−u) = log((1+v)/(1−v)), and the resulting integrand (analytic in (−1,1) with an algebraic branch point of order 1/a at each end) is integrated by composite Gauss–Legendre on geometrically graded panels. Gated against the exact airfoil family (1/π) p.v. ∫ √(1−u²) U_{n−1}(u)/(v−u) du = T_n(v): 5.3e−14 at the default rule. That gate is deliberately harsher than anything the module meets (√ is endpoint order ½; the profiles are 1/a ≥ 2, where the same rule is at machine precision by 8 grading levels).

Cross-build check. X_c/c against the whole-line collocation profile's own zero crossing of E: 3.9e−6 / 4.3e−6 / 7.3e−5 at a = 0.3 / 0.4 / 0.5. v12's two independent discretizations agreed on this invariant to 0.06–0.11%; these agree to four to five significant figures. And in its own K the reduced radius is converged to 3.7e−13 over K = 64…192.

4. THE KILL SWITCH

The question the continuation prompt posed: v13 attributed the whole-line divergence to a mode growing like (log(X/X_c))^{1/a} outside X_c, against a domain space that is a decay class, a codimension-1 range obstruction that no refinement and no bordering with a symmetry touches. The repair it named was to take the far field out of the domain. (RS) does exactly that: its unknowns are (s, X_c) and its perturbations live on [0, X_c] only, so the growing mode has nowhere to live.

‖A‖ = ‖M⁻¹‖, sup-to-sup, with the domain pushed through to function values on a fixed fine grid (a coefficient vector is not a function space) and the X_c slot measured relative to X_c:

The control, measured with the same code and the same decay grading that produced v12's number (solver/turning_point.graded_norm_by_radius, α=1.4):

which reproduces v12 (J^−0.003 at the anchor, J^+2.86 at a=0.2) closely enough to confirm it is the same object.

Two honesty items that the leg would be worth less without.

The a=0.2 full-ladder slope is K^+0.0308, not −0.0009. Both are reported. The difference is entirely the K=16 point (6.27), which is under-resolved: at a=0.2 the support radius is X_c/c = 34.3 while the core half-width stays O(1), so the reduced variable v carries a boundary layer of width ~1/X_c and needs K ≳ X_c to see it. From K=48 the row is flat to four digits. The two-scale structure of the problem shows up in the reduced formulation as a resolution requirement, and it is why small a costs more modes.

The flatness is not an artifact of choosing an unweighted norm. Repeating the ladder with the decay grading w_dom = (1+X²)^{α/2}, w_cod = (1+X²)^{(α+1)/2} at the operating α = 1.4 gives K^−0.0030 / +0.0008 / +0.0011 / +0.0013, the same verdict. It has to: on a compact interval every such weight is bounded above and below, so it can change the value but not the rate. On the whole line they are not equivalent, which is exactly why v12's ladder had to be graded. Saying this out loud matters because the unweighted whole-line norm grows like J^+0.99 even at the anchor, purely because the grid's outer radius ~4J/π grows with J, so an unweighted whole-line ladder would have "diverged" for a reason that has nothing to do with the operator.

5. Two things that changed elsewhere

The radius law, with the constants measured (recommended brick (3), delivered). E(X_c) = 0 with the far-field form U ≈ (m/π) log X + U₀, m = ∫Ω, gives

log(X_c/c) = −π( c/a + U₀ ) / m .

v12 quoted X_c ~ e^{c/a}, which is this law with the anchor's m = −π. The profiles' own m is not −π (it runs −4.55 → −2.43 over a = 0.2 … 1.0) so the coefficient of 1/a is −πc/m ≈ 0.69 at a=0.2, not 1. Measuring m and U₀ outside the support (where Ω ≡ 0 and there is no principal value) and substituting:

Agreement improves monotonically as a → 0, which is the right behaviour and attributes the residual error: the law uses a far-field expansion evaluated at X_c, and X_c grows as a falls, so the expansion gets better exactly where it is being used.

v11's fourth confirmation of a* does not survive (the other three do). v11 read a grid-refinement spread on the whole-line Newton solve (3.7e−3 at a=0.8, 1.3e−2 at a=1.0 vs 3e−5 at a=0.5) as "the solutions are continuum objects only up to a ~ 0.5", and the continuation prompt banked it as a fourth independent confirmation of the survival boundary. On (RS) the same object is grid converged in K to

i.e. eight to twelve significant figures, with Newton converging from cold start at every one. The whole-line spread was the global basis failing to represent a compactly supported profile whose edge regularity is C^{1/a} and therefore gets worse as a grows: the same instrument-artifact family as v12's Gibbs ringing (banked lesson 31).

What that retires is v11's argument, not a*. The other three confirmations (GA-, genome- and basis-convergence in §9-cont2) are about the two-scale GA problem (whether the two-scale structure of the a = 0 anchor survives continuation into a > 0, which is the domain all of this is measured on) which is a different question, and this leg says nothing about them. (It is also a different question from the published two-scale scenario: arXiv:2603.25104 scopes that to a ≤ 0.) So a* is confirmed three times, not four, and separately: the compactly supported traveling wave itself exists, as a grid-converged continuum object, well past it.

6. Gate-check against the Clay chain (answer it, don't re-paste it)

(a) Which link does this move? L1, and for the first time in three legs it moves it forward rather than reclassifying it. v12/v13 established that L1 as posed was mis-specified; v14 supplies the repair they named and shows it works: the approximate inverse exists in the continuum limit at the profile the certificate is actually about. L2, L3, L4 unmoved. Nothing here is a step whose success resolves Clay.

(b) Is another L1 leg still the best use of the next chunk? For exactly one more, yes, and the reason is new. Until now the honest answer was "the L1 machinery is priced at the wrong point"; that objection is now discharged, and the remaining work is smaller than what it replaces, because eleven legs of far-field apparatus (v3's resonance, v6's tail bound, v7–v9's estimates, v10's bracket) are simply not needed on a compact interval. The next brick is Y₀, Z₀, Z₁, Z₂ in the reduced space, i.e. brick (4) of the "what would actually be worthwhile" list, finish one certificate end-to-end, which the list ranks above another estimate leg and which is now cheap. If that does not close in float with margin, the stopping rule applies and the alternative lanes go primary.

(c) Is there a cheaper experiment that kills the whole route? The one this leg ran was it. The next equivalent is: assemble the four radii-polynomial constants in the reduced space at one a and see whether the budget closes in float. That is a day, not a leg, and it is decisive in the same way.

7. What is NOT claimed

• Not a certificate. Y₀, Z₀, Z₁, Z₂ have not been computed in the reduced space. ‖A‖ converging says the approximate inverse exists in the limit; it says nothing about whether the radii polynomial closes.
• 4.52 is not "better than 47". The v7–v9 upper bounds are for a different operator in a different space (whole line, decay-graded Hölder). The comparable quantity between the two is the slope in the discretization parameter, not the value. Quoting the value as an improvement would be the same error v10 named (lesson 28) in a new costume.
• ‖A‖ here is a measured induced norm, not an upper bound in the v6/v7 sense: it is computed exactly for the discrete matrix at each K, which is a statement about the discretization, not a continuum theorem. It is on the honest side of lesson 15 only in that it is not a family-restricted maximum: it is the exact induced norm of the object it names.
• The reduced ↔ original equivalence is for even, negative, unimodal profiles. The monotonicity that forces E to decrease is observed on every solution, not proved.
• Plain float64. Nothing is interval-enclosed. solver/interval.py still has never been pointed at any of this, correctly, because nothing has closed in float.
• ~~Novelty is unchecked. A first integral of a scalar traveling-wave equation is exactly the sort of thing that is folklore to people who work on gCLM/De Gregorio. The literature search is item (5) on the standing list and it blocks any novelty claim here as much as anywhere else.~~ [CORRECTED BY v15, 2026-08-01 (see TECHNICAL_P2_LITERATURE_SCOPE.md. The hedge above was right, and it should now be stronger than a hedge: presume the first integral is KNOWN. The search turned up arXiv:2603.25104 (Huang–Tong–Wang, March 2026), whose abstract states that the two-scale blowup's inner traveling wave) this object, has its existence established rigorously via a fixed-point method; and arXiv:2305.05895 (same group, 2023), proving existence of gCLM self-similar profiles "either smooth on the whole real line or compactly supported and smooth in the interior of their closed supports", via the fixed point of an a-dependent nonlinear map with Ω(x) = −x f(x). The natural way to build such a fixed point for Ω H(Ω) = E Ω_X is precisely this reduction. Caveat on the caveat: no paper was read (the container's network policy blocks arXiv and every publisher domain) so this is a strong lead, not a verified fact. None of the measurements in this document change. What does change is §6(b): finishing the certificate is a capability demonstration and a reproduction, not the lottery ticket.] What is not in doubt is that this project has spent thirteen legs discretizing an equation that has a closed-form reduction, which is a lesson regardless of who knew it first.

Honest ceiling. Route-D v3–v14 remain validated tooling, a no-go theorem, partial bounds at the wrong point, a diagnosis, and now a repair that passes its own kill switch. They do not climb the rigor ladder. Even the eventual success this scouts is a computer-assisted toy-model certification (Chen–Hou / Gómez-Serrano genre), not a Clay solve. 1D gCLM is a toy model of the boundary behaviour of Hou–Luo / 3D-axisymmetric Euler. Overall Clay odds ~0.05%, unchanged by this leg.