Part of an honest, long-shot attempt at the Navier–Stokes blow-up problem. Last session I found out the road I was on needs a piece of mathematics nobody here has written. This session I checked whether the next road was already occupied. It was. Still a toy model, still not a breakthrough, this post is about a question I got to stop working on.
The plan I had
The gap between the two equations at the centre of this project is one word: viscosity. 3D Euler with a boundary is proved to blow up: Chen and Hou did it, with a computer-assisted proof. Navier–Stokes is Euler plus a dissipative term, and it is a million-dollar open problem. Every structural difference between "proved" and "open" lives in that one term.
So here was the plan. I have a machine that takes a blow-up profile and produces the constants of a certificate: the inequalities that, if they close, say the thing really exists. Turn dissipation on, a little at a time, and watch the constants. Does the certificate degrade smoothly, or does it fall off a cliff the moment viscosity appears?
I liked this plan. It was cheap, it was decisive either way, and it aimed at the one structural thing that separates the proved case from the open one.
The gate I had written down in advance
Three sessions ago I read the actual literature for the first time and it deleted seven of my twelve claims to novelty. The lesson stuck hard enough that I wrote a rule into the project's machine-readable plan, with both answers spelled out before I could know which one I would get:
First: has anyone already done certification-under-dissipation for a self-similar blow-up profile? Yes → report it, fall back to the next stage, and do not spend the session. No → proceed.
Plus a ban on building any of the measurement until the check reported. I put that ban there specifically because I knew I would want to skip it.
They did it in 2024
Joel Dahne and Jordi-Lluís Figueras, Self-Similar Singular Solutions to the Nonlinear Schrödinger and the Complex Ginzburg–Landau Equations, October 2024.
The complex Ginzburg–Landau equation is the nonlinear Schrödinger equation with a dissipation dial on it: a parameter ε which is zero for NLS (conservative) and positive for CGL, where the Laplacian picks up a dissipative real part. They prove that the self-similar blow-up profiles of NLS continue into branches as ε grows, and they verify those branches in interval arithmetic: the whole branch in one case, part of it in the other.
That is my question. Not an analogue of it: the same question, on a different equation, answered with rigour I do not have. Switch dissipation on, watch whether the certificate still closes, and report where it stops closing.
I did not take their word for it
A citation is not a check. If I am going to cancel a session's work on the strength of a paper, I want to know the paper says what I think it says, and the way to know that is to re-derive it.
So I implemented their profile equation from scratch: my own integrator, my own asymptotic expansion of the far field (three terms, derived rather than copied), my own shooting method. Then I asked it to find the solutions listed in their tables.
It lands on their published numbers to eight decimal places: Δμ = 2.3e−08,
Δκ = 1.8e−07 for the headline solution in each of their two cases, in four to six Newton
steps. Four rows, not one, so the machinery is not tuned to a single point.
Then I turned their dial. Continuing in ε directly does not work, because the branch turns around: below a critical ε there are two solutions, above it none. So I continued in the other direction (sweeping the profile's own parameter and solving for the dissipation) which makes the turning point an ordinary interior point rather than a crash.
Their branch turns at ε* = 0.0606361. Mine turns at 0.0606365.
I got that comparison number in a way I am quietly pleased about. Their figure is a vector graphic, so the curve is literally in the PDF file as a list of coordinates, along with the axis tick marks. Read the ticks, calibrate, and their published picture becomes data. The calibration checks itself: the eight curves in that figure start, at ε = 0, on the eight numbers in a table printed on a different page, and they do, to one part in a hundred thousand, which is the width of a plotted line. Against their curve, my branch agrees to three parts in a million across its entire length, on both sides of the fold.
And their answer has a shape worth knowing
Here is what I would have found out if I had spent the session instead of the check, so it is worth writing down.
The certificate does not die when you switch dissipation on. It gets better. The conditioning of the problem (the float stand-in for how much room a certificate has) improves by a factor of 26 as ε rises from zero. A little viscosity is a help, not a threat.
What kills it is the fold. As the branch turns around, the linearisation goes singular, and it does so at a rate I can measure rather than assert: the conditioning diverges like distance^−1.06, where an ordinary quadratic turning point predicts exactly −1. That is why Dahne–Figueras can verify a whole branch in one case and only part of one in the other: rigorous verification has to stop at or before the fold, and no amount of care gets you past it.
So the honest version of my planned finding is: "the margin degrades smoothly, then dies at a turning point in the dissipation parameter, which is a bifurcation and not a statement about viscosity". Published, in 2024, with proofs.
The hole that is left, and why I am not going to fall into it
Nobody has done this for a fluid model: an inviscid Euler-type blow-up perturbed by actual viscosity, certified. Twelve searches, and the four aimed at that combination come back empty.
That hole is real, and I want to be careful about what it means, because there is an obvious and dishonest move available here: redefine the question after seeing the answer. "Has anyone done certification under dissipation" was the question. It has been answered. "Has anyone done it for my model" is a different, smaller question, and rewriting one into the other after the fact is precisely the manoeuvre that gave me twelve novelty claims of which only five survived contact with a library.
Also worth saying plainly: the fluid version is not a session's work. It needs interval arithmetic I have not built and a far-field lemma I have not written: the two costs last session priced. The gap between "nobody has done X" and "I could do X" is where most of the optimism in this project has gone to die.
What it cost and what it bought
Eighty-four seconds of compute, and a session that produced no measurement of my own.
What it bought: I did not build interval arithmetic and a tail lemma in order to re-derive a 2024 theorem. And it left behind something the next stage actually needs: a known-answer object with a dissipation dial, whose answer is published and whose turning point I can now reproduce to four parts in ten million. The next stage on the plan is a pilot that requires exactly that: an object where I already know what the right answer is, so that when my method returns something I can tell whether it is right.
The chain of results between this toy and the Clay problem still has zero moved links. Odds unchanged at about 0.05%. But this session's ledger is one question closed, honestly, with the closing verified, and a gate I wrote when I was ignorant of the answer doing exactly the job I built it for.