← The blow-up search

Route-T v1: bordering restores a bounded tail, and it works where the failure curve was worst

Nothing here resolves the Clay problem. This is one long-shot programme's working record, published at the confidence its own gates recorded. What this is →

Phase 2 / P2, leg 52. Code: solver/spectral_certificate.py (+5 functions), test_spectral_certificate.py, experiments/p2_route_t_v1_border.py → writeup/data/p2_route_t_v1_border.json → fig47. Working notes: PHASE2_P2_NOTES.md §41. Plan stage T (plan_of_record.py). 154 s, 6/6 pre-committed clauses.

Leg 51 measured a window that was empty by 0.606 in exponent units and answered L1's gate NO. One border row and one border column close it: the bordered tail inverse saturates at s = 0 (9.44) and s = 0.3 (11.37), both inside the band where the target profile has finite norm. The border a proof can write down achieves the optimum to three digits. It is still not a certificate, for a reason pre-committed before any number existed. No chain link moved; Clay unchanged at ~0.05%.


0. What was open

Leg 51 rebuilt the certificate in Route-E's compactified basis and got three of four terms: the operator gap vanished, Y₀ was exactly zero in rational arithmetic, Z₁ and Z₂ were rigorous and finite. The fourth term, the neglected-mode tail, had no bound in any weight class tried, because the unbounded part of an inviscid self-similar linearisation is the dilation shift, not a multiplier: the tail operator's diagonal is exactly zero.

But leg 51 also localised the failure further than "the space is wrong". The operator is not injective, and its kernel is one explicit mode: h_m ~ m^{−2.007}, the |X|^{−1} far field. A non-invertibility caused by a finite-dimensional kernel is the textbook case for bordering, which is the same move that made Route-PORT's finite block work (three gauge freedoms → three border rows). This leg measures that repair before anything is built on it.

1. T-0: the novelty pass, first, and it narrowed the claim

Five queries, three ledger entries, committed in the driver so the search is auditable rather than remembered (leg 42's failure mode was an unrecorded search). Verdict PROCEED_NARROW.

The one that matters: Breden–Desvillettes–Lessard, arXiv:1503.06315, "Rigorous numerics for nonlinear operators with tridiagonal dominant linear part." They state leg 51's problem in nearly leg 51's words (the derivative "does not have an asymptotically diagonal dominant structure", so the approximate inverse is not straightforward) and supply a construction for A in that setting.

So the general observation is not new, and leg 51's methodological claim drops to a re-derivation of something the field knows. What their paper does not obviously cover: our tail is tridiagonal and not dominant, diagonal exactly zero, and is Fredholm with a kernel. Whether their construction reaches that case was not resolved in this pass (the PDF did not extract), and it is recorded as an open question rather than as a gap.

Leg 51's prediction about the shape of the field survives the pass, the certified self-similar blow-ups using ℓ¹-Fourier machinery are dissipative (Dahne–Figueras, CGL), while the certified inviscid ones (Chen–Hou, arXiv:2210.07191 / 2305.05660) used weighted energy estimates over 145 pages instead. That is a consistency check, not evidence.

Flagged: the query naming arXiv:2604.01868 (Chen–Huang–Li, the April 2026 source of this project's target) did not resurface it, only 2021–2023 Hou–Luo work returned. A statement about the search index, not about the paper; the repo's seventh-pass record stands and a later leg should re-check.

2. T-5, which side of s = 1 the obstruction is on, stated before the ladders

In w_k = (1 + k)^s, measured on modes 65 … 3136:

kernel      h_m ~ m^(-2.0024)    ->  IN the space         iff  s < 1
cokernel    u_m ~ m^(+1.0012)    ->  functional BOUNDED   iff  s >= 1

Bordering supplies a missing range direction and annihilates a kernel. It cannot repair a cokernel functional that is not in the dual. Therefore bordering must help below s = 1 and must not help at or above it: written down before the ladders ran, and it is the prediction the rest of the leg tests.

And it re-reads leg 51's U-curve. That curve's minimum was at s = 1.00, which leg 51 recorded as the least-bad class. It is in fact the exact exponent at which the operator is marginally both failure modes at once: the kernel leaving the space precisely as the cokernel functional enters the dual. The most promising-looking point on the curve was the one point the repair could not reach.

3. T-1: the ladders

K = 64 retained modes; tail on 65 … M for M ∈ {320, 576, 1088, 2112, 3136}; weighted ℓ¹ operator norm of the inverse of

B  =  [[ T , u ],
       [ vᵀ, 0 ]]

v pinning the kernel (one extra equation), u supplying the missing range direction (one extra unknown), both normalised so the number is not an artifact of their scale.

class admissible? unbordered bordered (analytic) shape
flat s = 0 yes 4.06 → 48.76 (M^{+1.085}) 7.46 → 9.44 (M^{+0.100}) SATURATES
algebraic s = 0.3 yes 3.03 → 20.67 (M^{+0.837}) 8.09 → 11.37 (M^{+0.147}) SATURATES
algebraic s = 0.394 boundary 2.77 → 16.04 (M^{+0.765}) 8.30 → 12.16 (M^{+0.165}) SATURATES
algebraic s = 1.0 no 1.64 → 3.94 (M^{+0.379}) 9.89 → 20.51 (M^{+0.319}) still growing
algebraic s = 1.5 no 2.41 → 11.54 (M^{+0.681}) 11.44 → 31.81 (M^{+0.448}) still growing

Admissible = the target Ω ~ |X|^{−α}, α = 0.394, has finite norm in the class, i.e. s < α. s = 0.394 is the excluded boundary, reported because it saturates anyway.

The ordering inverts. Unbordered, s = 1 was the best class and s = 0 the worst. Bordered, s = 0 and s = 0.3 are bounded and s = 1 is not. Exactly T-5's prediction, and the opposite of what optimising leg 51's curve would have suggested.

Reported as a shape, not an endpoint (lesson 72). Flat increments 0.864 → 0.616 → 0.369 → 0.133; s = 0.3, 1.212 → 1.019 → 0.738 → 0.314: falling, with the fitted exponent down an order of magnitude (1.085 → 0.100). At s = 1 the increments rise across the ladder (2.44 → 2.92 → 3.25), which is what a slow divergence looks like and what a slow saturation does not.

3.1 The window is no longer empty: this is the result

leg 51:  object needs s < 0.394   |   operator wants s ~ 1   |   gap 0.606, curve never zero
leg 52:  object needs s < 0.394   |   operator BOUNDED at s = 0 and s = 0.3

The two sides overlap. Leg 51's NO was a statement about the standard construction, not about the object, which is the branch L1's own gate named in advance and did not get to take.

3.2 The kernel is one-dimensional, measured

σ_min falls 2.71e−01 → 8.54e−03 (flat class) while σ_2 stays bounded away, 2.092 → 1.652. One singular value goes to zero and the next does not, independently confirming the analytic claim that the block's first row, which involves the mode K lying outside it, kills one of the two parity chains.

4. T-2: the border a certificate can actually write down

The SVD pair is the most favourable one-dimensional bordering that exists; the analytic far-field mode and its adjoint are what a proof would use. Ratio at the top rung:

flat 1.000    s=0.3 1.007    s=0.394 1.012    |    s=1 1.130    s=1.5 1.383

In the admissible classes the explicit mode achieves the optimum to three digits. That is the clause that makes the repair usable rather than an SVD artifact, a certificate cannot border with a singular vector it computed numerically.

5. T-4, the alignment is the physics, and it fails with the repair

|cos| between the optimal border direction and the analytic far-field mode, top rung:

class flat s = 0.3 0.394 s = 1 s = 1.5
alignment 1.00000 0.99996 0.99991 0.99300 0.90209

At s = 1.5 it is also flat in M (0.90378 → 0.90209), i.e. not converging to the far field at all. Alignment and usefulness degrade together: the story predicts that and a coincidence would not. The repair is therefore not "border by something": it is add the far-field amplitude as an unknown.

6. T-3: the negative controls

border flat class, M = 320 → 3136 saturates?
analytic 7.46 → 9.44 yes
SVD (optimal) 7.46 → 9.44 yes
second singular pair 13.04 → 48.76 no
random pair 1584, 324, 379, 537, 668 no

The second-pair control lands on 48.76: exactly the unbordered value at the top rung. Bordering in the wrong direction is asymptotically worth nothing at all. The random control's first rung is an outlier of the draw; from the second onward it rises monotonically, ending 70× the analytic border. Neither saturates in any of the five classes, which is what makes "bordering fixes it" a measurement.

7. The ceiling: pre-committed as clause T-6, before any number existed

A bounded bordered tail is not a certificate. What is measured here is the weighted ℓ¹ operator norm of the inverse of the tail block plus one border row and one border column, as the number of retained modes grows.

In a certificate that border is a new unknown, the far-field amplitude, and it needs

  • its own column in the finite block,
  • its own contribution to Y₀,
  • and a matching condition between the spectral tail and the asymptotic expansion.

None of that is written here.

And the object is still the a = 0 CLM linearisation: one mode, analytic, in every class considered. Leg 51 was careful that a wall there bounds the difficulty for HL_S2_nonsymmetric from below rather than above. The same asymmetry applies to a success: this bounds it from below too. Nothing is claimed about HL_S2_nonsymmetric.

The constant is also not small: 9.44 and 11.37, against a Z₂ of 79.5 from leg 51. Whether the assembled polynomial closes with those numbers is a different question from whether the term is finite, and only the second was asked here.

8. What moved and what did not

Moved: the ℓ¹-Fourier route to a certified L1 was closed by leg 51 and is open again, with a named repair, a mechanism, a measured constant, and a prediction that was made before the measurement and held.

Did not move: L1 is still uncertified, the target object has still never been touched by this machinery, L2 and L3 are still Chen–Hou's, L4 is still out of reach by Wall 2. No link of the L1→L4 chain moved. Clay odds unchanged at ~0.05%.

9. Reproduce

.venv/bin/python -u experiments/p2_route_t_v1_border.py   # 154 s
.venv/bin/python writeup/4_p2_lottery/p2_route_t_v1_evidence.py   # fig47
.venv/bin/python test_spectral_certificate.py

Pre-committed clauses: 6/6, T0 novelty pass first and it can only narrow, T1 the gate on admissible classes, T2 analytic against optimal, T3 two negative controls diverge, T4 alignment reported at every rung, T5 the Fredholm side stated before the ladders, T6 the ceiling written before the numbers.