Corrections and completion

Cancel the leading stress, recompute everything, cancel again. Then cut off, extend the force, and compare.

After the pulses, the residual still contains the errors of the linearization, the interactions between harmonics, the terms produced by curls and cutoffs, and defects in five radial integrals. The construction removes these in a cycle whose only virtue is that it can be repeated. Each repetition buys a fixed improvement in the rate at which the residual vanishes.

The cycle

At stage $j$ the fields are $(u^{[j]},p^{[j]})$. Add a divergence-free increment $\delta u_j$ and pressure $\delta p_j$. The residual changes by the same exact identity as on the residual page, now with the current field as background:

$$\mathcal{R}(u^{[j+1]},p^{[j+1]}) = \mathcal{R}(u^{[j]},p^{[j]}) + \mathcal{L}_{u^{[j]}}(\delta u_j,\delta p_j) + \nabla\cdot(\delta u_j\otimes\delta u_j).$$

Cancelling a chosen source therefore creates a linear remainder and a quadratic self-interaction. Four operations, in a fixed order, handle the four kinds of term.

Waves. Solve the sourced amplitude equation for each nonzero angular harmonic with zero initial data. The principal operator cancels the prescribed source; the linear leftovers and the new pulses' interactions go into the next residual.
Stress increments. Add signed amplitude increments to the two leading families. Their cross covariance with the leading pulses gives a prescribed correction to the averaged stress, of either sign, because the cone condition holds with margin.
Mean flow. Correct the angularly averaged residual with zero auxiliary mean by inverting the fast auxiliary-time derivative. The axial part is realized through a vector potential, which supplies its own radial velocity.
Moments. Solve five radial moment equations. Two keep the zero angular-momentum and axial-flux integrals; three cancel the linear defects in the radial integrals of pressure and tangential momentum.

Pressure is reconstructed from the radial equation after each operation and its contribution to the axial equation enters the next residual. All products use the updated fields. The background, the leading amplitudes and the inverse operators stay fixed throughout, so the induction has a fixed vocabulary.

The exponent that improves

The residual at stage $j$ is bounded, for Cartesian derivatives of order $m$, by $q^{h\sigma_j - K_m}$ up to logarithms, with $K_m$ independent of $j$. One cycle gives

$$\sigma_0 = \tfrac15,\qquad \sigma_{j+1} = \sigma_j + \tfrac1{10},\qquad \sigma_j = \tfrac15 + \tfrac{j}{10}\longrightarrow\infty .$$

Each cycle is worth a tenth of a power of $q^{h}$. Since $h<1/100$, each cycle is worth less than a thousandth of a power of $q$. The number of cycles is unbounded, so this is enough, but it is why the construction cannot afford to be careless with any term of any order: a single dropped term of relative size $q^{0.001}$ would undo a cycle.

Summation with shrinking cutoffs

Infinitely many corrections must be added without spoiling incompressibility or the leading growth. The paper multiplies each correction's vector potential and pressure by a cutoff $\chi(a_j q)$ that equals one near $q=0$ and whose support shrinks with the stage, then takes curls. The sum is locally finite for $q>0$ and defines smooth fields for $t<1$. Comparing the residual of the sum with that of any finite stage shows that the summed residual vanishes to every order: for every derivative and every $N$,

$$\big|\partial_x^\alpha\partial_t^b\,\mathcal{R}(u,p)\big| \le C_{\alpha,b,N,X_1}\,q^{N}\qquad (0\le X\le X_1,\ q\downarrow 0).$$

This is the flatness estimate, Theorem 3.1(iii). Outside a fixed radius the residual is identically zero because the exterior is an exact heat flow. Between, the estimate is uniform on compact sets. In English: near the singular point the leftover force and all its derivatives are smaller than any power of the time remaining.

Localization

The local fields are multiplied by a smooth spatial cutoff, axisymmetric and compactly supported, and a temporal cutoff that vanishes on an initial interval. Both equal one near the singular point $(0,1)$. The cutoffs are applied to the vector potential before the curl, so the localized velocity is exactly divergence-free, has zero initial datum, and has support in a fixed compact set $K$. Differentiating the cutoffs produces extra residual terms, but they live where the cutoffs vary, away from the singular point, where every derivative of the fields has a one-sided limit at $t=1$.

Extending the force through time one

For $t<1$ set $f = \mathcal{R}(u,p)$. Flatness at the origin and the endpoint bounds away from it give every derivative of $f$ a uniform limit as $t\uparrow1$. A force on $\mathbb{R}^3\times(0,\infty)$ is then built that attains those limits from $t>1$, with support in $K\times[0,2]$. This is an application of Borel's lemma in the time variable. The result is $f\in C_c^\infty$ with $\mathcal{R}(u,p)=f$ on $[0,1)$.

Energy and the comparison

The energy identity with a bounded force gives, for $F(t)=\int_0^t\|f(s)\|_2\,ds$,

$$\|u(t)\|_2^2 + 2\int_0^t\|\nabla u(s)\|_2^2\,ds \le F(t)^2\qquad(0\le t<1),$$

so the kinetic energy is bounded and the total dissipation before time one is finite. The theorem then needs one more lemma. If some smooth solution $v$ with the same force and zero datum has bounded energy on $[0,T]$ with $T<1$, then $v=u$ there. The proof is a difference estimate on expanding balls; the pressure needs separate control because no growth at infinity has been assumed for it, and the paper recovers it through Riesz transforms. With uniqueness on every $[0,T]$, a global smooth bounded-energy solution would agree with $u$ on $[0,1)$ and inherit its growth along the path $x_\tau = (\sqrt{2X_{\rm in}\tau},0,0)$, contradicting its boundedness on a compact neighbourhood of $(0,1)$. That is Theorem 1.1.

The periodic corollary rescales the compactly supported fields into the interior of a unit cube and sums their integer translates. The supports are disjoint with a gap, so even the nonlinear term is preserved by the sum. The force is smooth on the torus with compact time support, and the same difference estimate, now without pressure or transport terms, gives uniqueness before time one.

What was hard

Nothing on this page is conceptually difficult and all of it is difficult. The 166 pages are mostly the proof that every term created by an operation is either cancelled by a later one or carried into the residual with a bound that the next cycle can absorb. The paper's own summary of what it took is Figure 6: compute the full remaining residual, correct waves and mean flow and pressure and moments, sum with shrinking cutoffs, localize and extend. The authors of the Lean formalization would say the same thing in a different language.

Next: what is settled.