The profile ends

Route-D v12 of a Navier–Stokes blow-up search. Level-1 tooling plus a structural finding. Not a certificate, not rigorous, not a Clay result.

The previous leg drove the candidate profile's residual down twelve orders of magnitude and handed over a single instruction: you measured that in the wrong coordinates. All the machinery that decides whether the argument closes (the operator norms, the constants, the budget) is written in a different, compactified discretization. Carry the profile across and measure the defect there.

I carried it across. The measurement took about an hour. Then it took the rest of the leg to understand what it was saying, and the answer was not about the defect.

The thing about a Newton solve

A Newton solve drives the residual to zero at the grid points. The certificate does not care about grid points. It cares about the residual of the underlying function, everywhere, and those are different, for a reason that is structural rather than sloppy. The nonlinear term is a product of two polynomials of degree J, so it has degree 2J, and collocation pins down only J conditions on it. The rest is invisible to the solve by construction. That leftover is exactly what the argument's first constant is made of.

So: rows the solver enforces, 1e-13 at every parameter value. The one row the gauge condition displaces, at the same moment, in the same solution: 1e-13 at a = 0, and 9e-3 at a = 0.5. Ten orders apart. Zero at the nodes is not zero as a function, and here it is not even close.

Fine: that is a known kind of problem, and it usually gets better as you refine the grid. It did not. At one parameter value the defect sat flat across a sixteen-fold refinement. And in every single case, the worst point was the outermost point of the domain.

That last detail is the tell.

The velocity has a logarithm in it

Here is the equation's far field, in one line. The profile is transported at a speed which is not the constant c in the equation, but

E(X) = c + a·U(X)

where U is the velocity the profile induces on itself. And U is an integral of a Hilbert transform, so it carries a logarithm: for large X, U ≈ (∫Ω/π)·log X, and the integral of the profile is negative.

So E decreases without bound. At some finite radius it hits zero.

At the anchor point of the whole programme, the exactly-solvable case a = 0, this cannot happen, because E ≡ c is constant. That case has a clean power-law tail, X⁻², and eleven legs of analysis were built on it: a graded space that measures how fast things decay, a resonance at exponent 2, tail bounds, far-field solution operators. All correct, all about that tail.

Turn a on and the tail is gone. Past the critical radius the transport reverses, and the profile does not decay out there, it ends. Approaching the radius from inside, the balance forces

Ω ~ (X_c − X)^{1/a}

an algebraic zero whose order is one over the advection parameter, with no fitted constant anywhere in the derivation. Beyond it, zero solves the equation exactly.

Measured, in two completely independent discretizations, at five parameter values: the critical radius agrees to 0.1% or better (once the grid resolves it at all), and the zero's order tracks 1/a to 7–9%, the residual gap being what a leading-order fit over a finite window should show.

The a = 0 anchor everything was built on is the degenerate case where that radius sits at infinity.

Why the defect measurement was misbehaving

Because a function that is identically zero outside a finite radius, represented in a global spectral basis, leaves ringing where it should be flat, and the norm the certificate uses weights the far field by a large positive power of X, which multiplies precisely that ringing. The measurement was not failing. It was reporting, accurately, that the object and the space are mismatched.

The number that looked like good news

At one parameter value the defect does converge, and at the finest grid it lands 7.7× under the budget: the first time in this project that side of the inequality has come in under target at nonzero advection.

It is not good news, and the reason is worth stating plainly, because it is the kind of error that is easy to publish by accident. Every constant in that budget was computed by linearizing at the a = 0 anchor. The certificate linearizes at the profile it certifies. So I measured the operator norm at the actual profiles:

That is not the matrix going bad, the plain condition number barely moves across that row. And one grid size cannot distinguish "large operator" from "bad grid", so I refined:

At the anchor it is flat to three decimals. At the real profile it grows like a power of the grid size. The approximate inverse that the entire argument is built around does not exist in the limit.

There is a mechanism. (Corrected in the next leg: I first wrote that linearizing about a profile with a zero of order p produces a mode blowing up like (X_c − X)^{−p}. It doesn't: that mode vanishes at the critical radius; I dropped a sign. The real obstruction is a mode that grows like (log X)^{1/a} out past the critical radius, against a space that requires decay. See BLOG_P2_ROUTED_V13.md.) The numbers above are real and the a = 0.1 value is small because that grid cannot see the problem at all.

Correct the budget for the real operator norm and the near-miss becomes a miss by three orders of magnitude. Both sides of the inequality move the wrong way, by the same mechanism.

The boundary, and a coincidence that did not survive

This project has confirmed four separate times that the two-scale structure inherited from the exact a = 0 wave stops existing somewhere around a ≈ 0.5 (that is, as a is turned up from zero, on positive a throughout) and has never had a mechanism for it. Two were available here. The critical radius shrinks steadily as a grows (at a = 0.5 it is down to about three times the profile's own width) which is a geometric story: eventually there is no room for two scales. And the zero's order is 1/a, which passes through the integer 2 at exactly a = 1/2, where the relevant transform grows a logarithm; that would be a sharp arithmetic story landing exactly on the observed boundary.

The arithmetic story predicts trouble at a = 1/3 too, where the order is 3. There is none: that point sits smoothly between its neighbours in every measurement. So the coincidence is a coincidence, and what is left is the geometric trend: smooth, with nothing special happening at 0.5.

Which is annoying, and also exactly what the earlier work said: that crossing is soft. The leg explains where the outer scale comes from and why it collapses. It does not predict the boundary.

Where this leaves the programme

Honestly? It re-specifies the target, and I would rather find that by measurement than after an attempt to make anything rigorous.

The upside is that the repair is cheaper than what it replaces. If the profile is exactly zero beyond a finite radius, a certificate can work on a finite interval, with the radius itself as an unknown, and the far field, which has absorbed eleven legs of tail bounds, resonances and graded norms, disappears, because there is nothing out there. (The reason I gave here for expecting the boundary singularity to be harmless was wrong; the next leg found there is no interior singularity to begin with, and a better reason for the same repair.)

That is untested. It is the next leg, and it comes with its own kill switch: do the free-boundary version at one parameter value and watch the operator norm as the grid refines. If the singular direction is not absorbed, the framing needs replacing rather than repairing.

Everything here is plain float64. Nothing is interval-enclosed and nothing is rigorous; this is scouting for a computer-assisted certification on a one-dimensional toy model, which is not the Clay problem and would not be mistaken for it. Code, data and the figure are in the repository; the figure rebuilds from committed data without re-running anything.