The wall I removed was not the only wall

Route-D v16 of a Navier–Stokes blow-up search. A negative result with a named repair, which is the useful kind. Not a certificate, not rigorous, not a Clay result, and after the last leg, not novel either.

Two legs ago I found that the equation I'd been discretizing for a year integrates once in closed form, reformulated the problem on the interval where the profile actually lives, and watched the obstruction that had stalled the previous two legs disappear. One leg ago I looked up the literature and discovered that most of this is known, that the object's existence appears to have been proved by other people in March, and that computer-assisted certification of these toy models is routine work for the groups in the area.

So this leg is the certificate, demoted to what it always should have been: a capability check. Assemble the four constants a Newton–Kantorovich argument needs, in plain floating point, and find out whether the pipeline closes before anything gets hardened into interval arithmetic. Sixteen legs have produced constants and never once a closed budget.

It doesn't close. And the reason is worth the leg.

First, the good number

The quantity a certificate needs from the candidate solution is its defect: how badly it fails to solve the equation. Not at the grid points, where a Newton solve drives it to machine zero by construction, but as a function, between them.

For five legs this project carried a defect floor of about 1e−2, from a genetic algorithm. Then a Newton solve took the nodal residual to machine zero, which sounded like a twelve-order improvement and wasn't, because the function-space defect stayed far larger. This is the first time the number that actually matters has been at machine level. The bottom row is in the table as the control: it's flat by construction, and the gap between the rows is what tells you the top row is measuring the function rather than the grid.

Then the wall

The certificate also needs a bound on the quadratic part: how much the linearization changes across the ball you're searching in. And here the whole thing falls over, twice.

The Hilbert transform is still unbounded. The velocity in this model is given by a Hilbert transform of the vorticity, and the Hilbert transform is famously unbounded on the sup norm: it turns a bounded function with a jump into one with a logarithm. Four legs ago I found this on the infinite line and spent legs 5 through 9 building a Hölder-norm apparatus to deal with it. Then leg 14 removed the far field from the problem entirely, and I think I quietly assumed that had dealt with everything.

It hadn't. Unboundedness of the Hilbert transform on sup is a local fact (it's about a jump, not about infinity) and putting the problem on a bounded interval does nothing to it. What the far-field removal killed was the decay half of the requirement. The smoothness half was never about the far field at all.

You can't discover this by sampling. Random perturbations of increasing complexity will happily report that everything is bounded, because the bad direction is a cusp in the unit ball you never stumble onto. You have to build the adversary, here, band-limited approximations of a step:

Growing linearly in the logarithm of the mode count, at 0.499 per e-fold. That's the logarithm, and it means the constant I need is infinite.

And here's the part I nearly got wrong. The obvious quick probe (take a single high-frequency mode and see what the transform does to it) also showed a divergence: 1.0, 2.9, 6.4 as the modes went up. Two independent-looking signs of the same conclusion. Very satisfying.

It was the quadrature. Refine the integration rule fourfold and the single-mode row collapses to 0.999 at every mode count, flat, while the adversary row doesn't move at all: 2.991, 3.038, 3.040, 3.041. One of those two numbers was about the operator and one was about my integrator, and they pointed the same way. If I'd only run the quick probe I'd have reached the right conclusion for a completely wrong reason, and that's the kind of thing that stays wrong for several legs.

The second wall, which is stranger

The nonlinearity in the reduced problem is e ↦ e^{1/a}, and its second derivative goes like e^{1/a − 2}. The function e vanishes linearly at the edge of the support. So the second derivative is bounded exactly when 1/a ≥ 2, that is, when

a ≤ 1/2.

Measured, by tightening how close to the support edge you look, over eight decades:

Flat to every digit up to one half, then divergent. It's a step function, and the step is at an exact rational number rather than a fitted one. It's also exactly where the profile stops being twice differentiable.

Now: this project has an independently measured "survival boundary" for the same family at a* ≈ 0.5–0.55, found three separate ways, none of which has anything to do with this calculation. (That boundary was measured at positive a, and it is the boundary of the positive-a continuation of the exact a = 0 wave, not of the published two-scale scenario, which lives at a ≤ 0.) So there's a coincidence sitting here, and I want to be careful with it in both directions.

I'm not going to claim it explains anything. Leg 14 solves the profile cleanly and grid-converged all the way to a = 1.2; the traveling wave exists well past one half. What fails at one half is my norm, not the equation. And it's fixable: put a weight on the perturbations that vanishes at the edge and the constant comes back, at the price of only allowing perturbations that vanish there too.

I'm also not going to leave it out. A previous leg found a similar-looking arithmetic coincidence at the same boundary, wrote it down, ran a control, and the control killed it. That was the right sequence. Here I haven't run a control, so it stays written down as an observation, which is how the next person gets to disprove it.

What I'm not reporting

The fourth constant (the one that handles everything outside the finite-dimensional approximation) I didn't compute at all. That's the entire content of a real computer-assisted proof, and it isn't a chunk of work, it's the work. The code returns it as None rather than zero, and deliberately refuses to assemble a budget at all. A budget built from a ledger with a hole in it, or with an infinity in it, is exactly the failure this project has spent several legs learning to avoid: five legs once reported lower bounds into a framework that needed upper bounds, and nobody was hiding it; it just never got carried to the conclusion.

Where this leaves things

The repair is named and measured. Re-run the same adversary against a Hölder norm and the divergence stops at about γ = 0.35, and a lower exponent, 0.15, still creeps, which matters, because it shows the threshold is real rather than an artefact of dividing by any seminorm at all.

The pleasing part: leg 5 found the same threshold, 0.35, on the infinite line. That's a real independent check rather than a restatement, because leg 5's norm also carried a decay grading and this one carries none. So the threshold belongs to the smoothness half, which is what you'd want to be true, and hadn't been separated before.

So the machinery from legs 5 through 9 isn't wasted. Its bounded-interval version is the next brick, and it's a smaller job than the original: no decay grading, no resonance, no matching radius, no tail bound. Roughly legs 7 through 9 with the expensive half deleted.

Whether I should do it is a different question, and after last leg the answer is probably not yet. The literature check said this whole lane is occupied territory, and the genuinely uncrowded thing on my list (searching for a discretely self-similar blow-up, the class that an old theorem of Nečas, Růžička and Šverák leaves open for Navier–Stokes) is still waiting. The certificate would be a capability. It would be a good one. It wouldn't be the point.

Figure and data: fig33_route_d_v16_rehearsal.png, writeup/data/p2_route_d_v16_rehearsal.json. Code: solver/reduced_certificate.py, test_reduced_certificate.py (16/16). Everything rebuilds from committed data.

One small correction to leg 14, while I'm here: I said the reduced solve converges from a cold start at every parameter value. At a = 0.7 it misses the basin and needs continuation from a neighbouring value. Everything else in 0.2 to 1.2 does converge cold.