similarity coordinates · scaling laws · the axis profile · rescaling ν
core · pulse · stress · exterior · bands · solver · explorer
Every length, speed, energy and Reynolds number of the core is a power of the time remaining τ = 1 − t, and every exponent is built from one number h. The paper takes 0 < h < 1/100 (Theorem 3.1). Move h to see which laws care and which do not — and where the whole thing would break.
The change of variables the whole construction lives in (3.2): τ = q(1 − η²), z = qDη, X = r²/2q. At fixed time, q depends on height — it grows like |z|1/D away from the middle plane — so the coordinate grid in the physical r–z plane is a family of curves. Click anywhere to read (q, X, η) at that point; scrub τ to watch the grid concentrate.
The meridional flow is not a cartoon. Given the axial profile U(η), incompressibility and regularity at the axis determine the radial flux r·ur = V₀ by the identity (4.7). The paper's reference axis data is U* = 4η + j₀ with a small offset 0 < j₀ ≤ 0.05 (proof of Proposition 4.10) — the "slight upward bias" of §2.1. These are the instantaneous streamlines of that field at fixed τ, in similarity units.
For U = 4η + j₀ the radial average AX(U) equals U itself, so V₀ = (X/L)·(2Aη(4η + j₀) − 4(1 − η²)): inflow (V₀ < 0) in a band around the middle plane, outflow beyond |η| ≈ 0.7. The axial velocity vanishes on η = −j₀/4, below the middle plane.
The whole construction is done at ν = 1. One rescaling gives every viscosity: uν(x,t) = √ν·u(x/√ν, t), pν = ν·p(x/√ν, t), fν = √ν·f(x/√ν, t) — equation (10.22). Time is untouched, so the singular time stays t = 1. The periodic case (Corollary 10.6) then shrinks everything by a factor λ into the unit cube and copies it to every lattice point.
§2.1 · §3.1 · §4.1 · §10.4–10.5 of the paper
Take cylindrical coordinates (r, θ, z) about the vertical axis. The leading flow is axisymmetric — its components depend on r, z, t but not θ. In the middle of the core, fluid spirals inward toward the axis and leaves upward and downward on either side of a dividing layer close to z = 0; farther above and below there is radial outflow. Pressure falls toward the axis and supplies the centripetal force. Angular momentum carried inward spins the core up: a torque-free parcel conserves r·uθ, so moving inward raises uθ. Viscosity carries angular momentum back out; the growth of the characteristic speed is the balance of the two.
For τ > 0 the concentration scale q = q(z, τ) is the unique root of q − z²q2h = τ (Lemma 4.1: the left side is increasing in q, with derivative L = 1 − 2hη² ≥ 1 − 2h). It satisfies q ≍ τ + |z|1/D. Letting q ↓ 0 at bounded X approaches the singular point; the parameter endpoints η = ±1 with q > 0 describe t = 1 away from it. The leading field is
with E = √(2X)·F for a smooth positive F — the √X factor is the swirl vanishing linearly at the axis — and incompressibility fixing V₀ from U by (4.7). The leading radial momentum balance is the centrifugal one, ∂XΠ = E²/2X, normalised so Π → 0 at radial infinity (4.25).
On the core Cτ = {0 ≤ X ≤ Xc, |η| ≤ ηc} one has q ≍ τ, so
Two things are worth staring at. The core becomes an ever more slender column — radius shrinking faster than height — and its kinetic energy goes to zero while its speed goes to infinity. The total dissipation up to the singular time is ∫₀ τ−½−3hdτ, finite exactly when h < 1/6. The swirl Reynolds number diverges — fluid makes more and more turns per radial diffusion time — but the radial one stays bounded, so viscosity keeps competing with the inflow. Axial diffusion is weaker than radial by the factor ℓr²/ℓz² ≍ τ2h, and that ratio is the expansion parameter for the background corrections of §5 (see bands).
The lower bound on the axial speed comes from the nonzero axis datum: the paper prescribes U*(η) = 4η + j₀ at the axis with 0 < j₀ ≤ 0.05, and an azimuthal reference profile E₀ = P*(1 + η²)−1(X/XR)1/10 farther out, where P* is large (Lemma 4.8, Proposition 4.10). The asymmetry is not decoration: with exact reflection symmetry uz would vanish on z = 0 and there would be no radial shear of the axial velocity to amplify pulses near the middle plane, where the rotational mechanism is weakest (§2.1).
If (u, p, f) solves the equation at viscosity one, then uν(x,t) = √ν·u(x/√ν, t), pν = νp(x/√ν, t), fν = √ν·f(x/√ν, t) solves it at viscosity ν with the same singular time; the support becomes √ν·K and energy and dissipation scale by ν5/2 (10.22–10.23). A bounded-energy competitor at viscosity ν would rescale back to one at viscosity one, which is already excluded. For the periodic statement, choose λ > 1 with λ−1Kν inside the open cube, put t₀ = 1 − λ−2, and set ũ(x,t) = λu(λx, λ²(t − t₀)), p̃ = λ²p, f̃ = λ³f; every term of the equation picks up the same factor λ³, so ν is unchanged, and the original time one is reached at t = 1 again. Sum the integer translates: they never overlap, so the nonlinear term is exactly preserved and the pressure is periodic as the erratum to the problem statement requires (Corollary 10.6).