The ledger of the OpenAI manuscript, version 1.1, October 1, 2026

Section 5: Correcting the base flow to every order (pp. 45 to 62)

M5.1: Expansion in the anisotropy parameter q^{2h}

Node ns-m5-1-expansion-in-the-anisotropy-parameter-q-2h, kind move, pp. 45-62.

• Statement: The formal ansatz (5.1). For n ≥ 0 set λn = 2nh, with order zero equal to the leading profile of Theorem 4.6, (E0, U0, Π0) = (E, U, Π). The coefficient fields are: - uθ,n = q^{-A+λn} En = r q^{-A-1/2+λn} ϕn/C, where En = √(2X) ϕn/C; - uz,n = q^{-A+λn} Un; - r ur,n = q^{λn} Vn; - pn = q^{-2A+λn} Πn. The spacing 2h comes from 1 - 2D = 2A - 1 = 2h (A = 1/2 + h, D = 1/2 - h). The series is formal: "convergence of the unmodified infinite series is not asserted" (p45).
• Obligation: The leading field satisfies (4.7) and (4.13). So its tangential residual is a stress divergence only after axial viscosity is removed (definition (4.12)), and its radial equation balances only pressure against centrifugal force. The omitted terms are smaller than the leading balance only by q^{2h}. Axial viscosity, for example, has size q^{-A-1+2h} = q^{-3/2+h}, which is still unbounded. It is also present near the axis and in the core, where T0 = 0 and no wave will act. These terms must be pushed to arbitrarily high order in q before a smooth force is possible.
• Mechanism: At fixed similarity coordinates, every tangential term of the leading field scales like q^{-A-1}: - the time derivative; - radial transport; - axial transport, since uz∂z costs q^{-A-D} and A + D = 1; - radial viscosity, since two radial derivatives cost q^{-1} because X = r^2/(2q). Axial viscosity costs q^{-2D} = q^{-1} q^{2h}. In r times the radial equation, pressure and centrifugal terms carry q^{-2A}, while the remaining radial terms start at q^{-1} = q^{-2A} q^{2h}. So both defects are relative corrections by the single factor q^{2h} = (ℓr/ℓz)^2; at viscosity one, ℓr = q^{1/2} and ℓz = q^D. Expanding in that factor, with similarity profiles at each order, sends each defect to the next order. Axial viscosity of order n-1 enters the tangential equations at order n. The non-centrifugal radial terms of order n-1 enter the pressure equation at order n. The azimuthal unknown is carried as ϕn, which is smooth at X = 0, so uθ,n vanishes linearly on the axis.
• Antecedent: None cited. Internal antecedents: the leading ansatz (4.3); the chain rule of Lemma 4.1, (4.2); and the remark after (4.7) that the omitted radial terms and axial viscosity carry an extra factor q^{2h}.
• Cost: A new index n and gauge λn = 2nh. Only asymptotic (not convergent) expansions can be claimed. Everything rests on the fixed h > 0 of Theorem 4.6.
• Backward question: The leading profile balances only the q^{-A-1} tangential terms and the q^{-2A} radial balance. Are all the terms it drops small in one and the same parameter, so that a single expansion can remove them?
• Checkable: Exponent fit. 1. Take h = 0.005 and the test profiles E = √(2X) e^{-X}(1+η^2)^{-1} and U = η e^{-X}, with V0 from (4.7). 2. Fix (X, η) = (1, 0.3) and take q = 1e-2 down to 1e-8. 3. At each q, lay a finite-difference stencil in physical (r, z, t). At every stencil point, compute q(z, τ) by root finding (scipy brentq) on q - z^2 q^{2h} = τ. 4. Evaluate each residual term separately in mpmath: ∂t uθ, ur(∂r + 1/r)uθ, uz∂z uθ, the radial vector Laplacian, ∂zz uθ, r∂r p, uθ^2, and r(∂t + u·∇)ur. 5. Fit log-log slopes in q. Expected slopes: -A-1 (tangential terms), -A-1+2h (axial viscosity), -2A (pressure and centrifugal), -2A+2h (other radial terms times r).
• Depends on: M4.3 (absorbs the terms deferred after (4.7), the non-cyclostrophic radial terms and axial viscosity, each smaller by exactly q^{2h}.); M4.8 (order zero (E0, U0, Π0) is the leading profile of Theorem 4.6.); M4.2 (each order copies the axis-regular ansatz (4.3): swirl q^{-A+λn}√(2X)ϕ_n/C, axial q^{-A+λn}U_n, radial flux q^{λn}V_n.); M4.1 (the spacing λ_n = 2nh comes from the exponents of (4.1), 1 - 2D = 2A - 1 = 2h, through Lemma 4.1's power counting.)
• Refs: p45 (section plan, formal expansion); p46 ((5.1), and the spacing argument citing (4.1), (4.2), (4.3), (4.7)).
• Verification: statement digest-only; hypotheses not-checked; computation checked False

M5.2: The order-n coefficient system (5.2) to (5.6), linear at positive order

Node ns-m5-2-the-order-n-coefficient-system-5-2-to-5-6-linear-at, kind move, pp. 45-62.

• Statement: Section 5, (5.2) to (5.6) (p. 46-47): substituting (5.1) and collecting q^{2nh} gives, with T_{a,n} = T_{a+λn}, Z_{a,n} = Z_{a+λn}, Z^{[2]}_{a,n} = Z_{a+λn-D}Z_{a+λn} from (4.2), b = -A - 1/2, c = -A: incompressibility (5.2), ∂_X Vn = -Z_{-A,n}Un; the swirl equation (5.3) for ϕn; the axial equation (5.4) for Un, with pressure term Z_{-2A,n}Πn; the radial balance (5.5), Πn' = C^{-2}Σ_{i+j=n}ϕiϕj - Ω_{n-1}/(2X); and the radial source Ωk (5.6). At positive order the system is linear in (ϕn, Un, Πn) with lower-order sources, and Ωk/X is regular at the axis since Vj = X vj.
• Obligation: Each order becomes solvable by linear methods, with sources built from completed lower orders. It also guarantees that the next pressure source Ωn/X is smooth at the axis.
• Mechanism: Substitute (5.1) and differentiate physically with Lemma 4.1. The second axial derivative uses Z_{a+λn-D}Z_{a+λn}, because the first axial derivative lowers the power by D. Remove the common tangential power q^{-A-1} (or q^{-2A} in r times the radial equation) and collect q^{2nh}. In the quadratic transport sums, an order-n unknown appears only in the splits (0, n) and (n, 0), multiplied by the fixed leading profile. So the order-n problem is linear. The cylindrical frame terms become: - ur(∂r + r^{-1})uθ gives Vi(ϕj' + ϕj/X); - the angular vector Laplacian gives 2(X∂XXϕn + 2∂Xϕn); - the axial scalar Laplacian gives 2(X∂XXUn + ∂XUn); - the radial operator ∂rr + r^{-1}∂r - r^{-2} applied to V/r gives 2X∂XXV/(qr). The T and Z operators contain ∂η, and Vn contains d∂ηAX(Un). So each order is a linear system of second order in X and first order in η, not an ODE.
• Antecedent: None cited. Internal: Lemma 4.1, (4.7), and Proposition 4.2 for the order-zero identities.
• Cost: The current unknowns appear under η-derivatives in three places: transport by Hc = Dη + dU0, the term ∂ηAX(Un), and d∂ηΠn in the axial pressure gradient. This forces analytic control in η (Lemma 5.1). The pressure must be solved together with ϕn and Un, because Z_{-2A,n}Πn enters (5.4).
• Backward question: If I substitute a series in q^{2h} into the residual: - Is the order-n problem linear in the order-n unknowns? - Where exactly does each omitted term land? - Does the source passed to the next order stay regular at the axis?
• Checkable: Symbolic coefficient extraction in sympy. 1. Write q = q(z, t) implicitly and use the derivative table of Lemma 4.1: q_t = -1/L, η_t = Dη/(qL), X_t = X/(qL), q_z = 2ηq^{1-D}/L, η_z = d/(q^D L), X_z = -2ηX/(q^D L). 2. Substitute the truncated sum of (5.1) for n ≤ 2 into the cylindrical axisymmetric Navier-Stokes residual, treating ε = q^{2h} as a formal symbol. 3. Extract the ε^1 and ε^2 coefficients after removing q^{-A-1} (tangential) or q^{-2A} (r times radial). 4. Compare term by term with (5.3) to (5.6). 5. Numerically, confirm that Ωk/X stays bounded as X → 0 for test Vj = X vj.
• Depends on: M5.1 (collects the q^{2nh} coefficients after substituting the ansatz (5.1) into the axisymmetric residual.); M4.1 (derivatives use the operators T_b, Z_b of (4.2), shifted by λ_n, with Z_{a+λn-D}Z_{a+λn} for axial viscosity.); M4.3 ((5.2) and (5.5) extend (4.7): incompressibility through the radial average A_X, and the pressure balance with source Ω_{n-1}.); M4.4 (order zero is governed by Proposition 4.2, whose residual identity and radial operators the positive-order rows (5.3), (5.4) reuse.)
• Refs: p46 ((5.2) to (5.6)); p47 (linearity, frame terms, regularity of Ωk/X).
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses not-checked; computation checked False

M5.3: Lemma 5.1, part 1: a local first-order system with a regular singular point at the axis

Node ns-m5-3-lemma-5-1-part-1-a-local-first-order-system-with-a, kind move, pp. 45-62.

• Statement: Lemma 5.1 says: - There is an interval 0 ≤ ξ ≤ a, with ξ = √X, reaching from the axis into the inner collar of Theorem 4.6(i), where the cone inequalities hold with a uniform margin. The endpoint a does not depend on n. - On it, each positive order has a unique solution of (5.2) to (5.6) with ϕn(0, η) = Un(0, η) = Πn(0, η) = 0. - The profiles are smooth in X and holomorphic in η near [-1, 1]. That neighborhood may depend on n and on the radial derivative order; a does not.
• Obligation: Each order must be solved on one interval that reaches strictly past Xa, for every n. The inner solution is retained on [0, X-] with X- > Xa. That is what keeps every higher-order stress supported from X- onward, strictly inside the annulus where ζ is bounded below. If the interval shrank with n, high orders would leave stress at X ≤ Xa, inside the core, where T must vanish and no wave is placed.
• Mechanism: Two features block a standard radial solve: the nonlocal average AX(Un) inherited from incompressibility, and singular coefficients at X = 0. - The auxiliary unknown Kn turns the average into the local relation ∂ξKn + 2Kn/ξ = -∂ξUn. So "the system has no unevaluated radial integral of the current unknowns." - The variable ξ = √X is proportional to r. In it, the viscous operators become radial Laplacians in dimension four (for ϕ, since uθ = rϕ/C) and dimension two (for U), and all apparent divisions by X become regular. The system has a regular singular point at ξ = 0 with exponents c = (0, 0, 2, 0, 3, 1). For regular solutions, the components with c_i > 0 vanish at the axis automatically. The components with c_i = 0 (ϕn, Un, Πn) have free axis data, and these are set to zero. The paper's stated reason: zero axis values "leave the leading traces of ϕ, U, and Π unchanged by every positive-order coefficient"; these are the axis data of Proposition B.2. All lower-order coefficients are bounded and holomorphic on Sρ = {η ∈ C : dist(η, [-1, 1]) < ρ}, a neighborhood chosen free of zeros of L. This includes their cutoffs and moment corrections, which are supported to the right of the inner rectangle.
• Antecedent: None cited. Internal: the analytic axis solution of Proposition B.2 and the common analytic region [0, Xan] of Theorem 4.6(i).
• Cost: It needs a common complex neighborhood on which all lower-order data are holomorphic and L ≠ 0. The fixed interval must lie inside the analytic region and beyond the collar that stays uncut (later X- < Xkeep < Xcut < a^2).
• Backward question: The order-n equations contain a radial average of the unknown and are singular at the axis. Can they be written as a local first-order system with a regular singular point, so that the regular solution with prescribed axis values comes from an explicit integral operator started at the axis?
• Checkable: 1. With X = ξ^2, verify (symbolically or by finite differences) that 2(X∂XX + 2∂X)f = (f_ξξ + 3f_ξ/ξ)/2 and 2(X∂XX + ∂X)f = (f_ξξ + f_ξ/ξ)/2 for a test f. 2. For U(X) = e^{-X} cos X, compute K = AX(U) - U by quadrature (scipy quad). 3. Check ∂ξK + 2K/ξ = -∂ξU at sample ξ, including small ξ, where both sides stay bounded.
• Depends on: M5.2 (recasts the order-n system (5.2) to (5.6) as the first-order system (5.7) with a regular singular point at ξ = 0.); M4.8 (the fixed interval reaches into the inner collar of Theorem 4.6(i), where all lower-order data are holomorphic in η.); MB.10 (the order-zero coefficients of (5.7) are analytic in η on the quarantined rectangle of Corollary B.6, which no later edit touches.)
• Refs: p47 (Lemma 5.1 statement, Step 1, Sρ, (5.7)); p48 (Kn relation, radial operators, pressure row).
• Verification: statement digest-only; hypotheses not-checked; computation checked False

M5.4: Lemma 5.1, part 2: nilpotent η-derivatives and a Picard series with half the derivative count

Node ns-m5-4-lemma-5-1-part-2-nilpotent-derivatives-and-a-picard, kind move, pp. 45-62.

• Statement: Lemma 5.1 (p. 48): with Hc = Dη + dU0, the η-derivative matrix A1 of (5.7) maps the first four coordinates into the last two and annihilates the last two, so A1(ξ)D0A1(s) = A1(ξ)D0∂ηA1(s) = 0 for diagonal D0. With (Gg)_i(ξ) = ∫_0^ξ (s/ξ)^{c_i}g_i(s)ds and K = G(A0 + A1∂η), k-fold compositions carry at most pk = ⌈k/2⌉ η-derivatives, so ||K^kGfn||_{Sρ'} ≤ Cn^{k+1}a^{k+1}/(k+1)! (max{1, pk/Δ})^{pk}, Δ = ρ - ρ' (5.8). Hence the Picard series Wn = Σ_k K^kGfn converges on Sρ' for every finite a, regardless of Cn, and is unique.
• Obligation: The core claim of Lemma 5.1: "the radial interval is independent of the coefficient norms at later orders." The constants of later orders are not controlled uniformly in n (see (5.17)). Any radius that depended on Cn would therefore shrink to zero along the induction.
• Mechanism: Each Picard step integrates once in ξ and may differentiate once in η. A standard Cauchy-Kovalevskaya estimate would then give a radius of order 1/Cn. Here the η-derivatives enter only the two second-order rows (for ∂ξϕn and ∂ξUn), and they act only on the undifferentiated unknowns (ϕn, Un, Kn, Πn). They come from three sources: - transport by Hc; - the ∂ηAX(Un) = ∂η(Un + Kn) part of Vn, multiplying (ϕ0 + X∂Xϕ0) or X∂XU0; - d∂ηΠn from the axial pressure gradient. G has η-independent diagonal kernels, which preserve the two coordinate blocks. So two derivative factors cannot be adjacent; an A0 factor must separate them. That caps k steps at ⌈k/2⌉ derivatives. The two sides of the estimate balance as follows. Cauchy estimates on nested strips, with the radius loss Δ split among pk derivatives, cost (pk/Δ)^{pk} ≈ (k/(2Δ))^{k/2}. The ordered integration simplex gives a^{k+1}/(k+1)!. So the kth root of the kth term decays like Cn·a·e·(2Δ)^{-1/2}·k^{-1/2}. In effect the system is second order in ξ and first order in η. The Cauchy problem from the axis therefore has solutions entire in the radial variable for η-analytic data. (Reading this as a sideways-heat structure is my gloss, not the manuscript's.)
• Antecedent: None cited; the manuscript's words are "Picard series" and "Cauchy radius loss". The argument is of Cauchy-Kovalevskaya type with nested analytic norms (Ovsyannikov style); that attribution is mine.
• Cost: The holomorphy neighborhood shrinks (ρ' < ρ) at each order and each radial derivative, so there is no analyticity uniform in n. The constants Cn stay uncontrolled in n. This is why (5.1) is not summed directly and Lemma 5.4 is needed.
• Backward question: Each order loses an η-derivative per radial integration. How many η-derivatives can actually pile up in k Picard steps? Is it few enough to solve every order on one fixed interval reaching past Xa, even though the coefficient norms grow without bound in n?
• Checkable: I ran checks (a) to (c) while digesting; all behave as stated. - (a) Linear algebra: fill the six listed entries of A1 and of ∂ηA1 with random numbers. Confirm A1 D0 A1 = 0 and A1 D0 ∂ηA1 = 0 for random diagonal D0. Also confirm A1 A0 A1 ≠ 0 for a random full A0, so derivative factors really must be separated by an A0. - (b) Evaluate the logarithm of the right side of (5.8) with lgamma, for Cn = 1e6, a = 10, Δ = 0.05 and k = 10 to 1e6. Confirm that k^{1/2} times the kth root tends to Cn·a·e·(2Δ)^{-1/2}, about 8.6e7. Equivalently, the kth root tends to zero like k^{-1/2}. - (c) Toy model with the same structure: ∂ξξ w = M ∂η w, with w(0, η) = 1/(3 - η) and ∂ξw(0, η) = 0. Its Picard series is Σ_k M^k ξ^{2k} k!/((2k)! (3 - η)^{k+1}). With mpmath, confirm the partial sums stabilize at ξ = 5 for M = 1e3 (and for larger M, with more terms). So there is no finite radius in ξ.
• Depends on: M5.3 (estimates the Picard series of the system (5.7), with the Volterra operator G built from its exponents c_i.); M5.2 (A1 collects the η-derivatives of (5.3) to (5.6): transport by H_c, ∂_η A_X(U_n) inside V_n, and d∂_ηΠ_n from the axial pressure gradient.)
• Refs: p48 (A1 entries, G, K, pk, (5.8), Picard series, uniqueness).
• Verification: statement completeness audit 2026-10-01: FRAGMENT; statement replaced from the digest; hypotheses not-checked; computation checked False

M5.5: Lemma 5.1, part 3: smoothness in X at the axis via Volterra identities and parity

Node ns-m5-5-lemma-5-1-part-3-smoothness-in-x-at-the-axis-via-volterra, kind move, pp. 45-62.

• Statement: Step 3 of Lemma 5.1: - (Gg)_i = ξ ∫_0^1 t^{c_i} g_i(tξ) dt; - ∂ξ(Gg)_i = g_i(ξ) - c_i ∫_0^1 t^{c_i} g_i(tξ) dt, which is continuous at ξ = 0. Repeated use gives every radial derivative. The integral equation extends across ξ = 0, and the equations preserve even parity of the first four coordinates and odd parity of the last two. Uniqueness forces these parities. Taylor's formula for an even smooth function of ξ then gives smoothness in X = ξ^2, to every finite order.
• Obligation: Cartesian smoothness of each coefficient field across the axis for t < 1 (Definition 3.2, via the Cartesian forms (4.4) to (4.5)). It also gives regularity of the next source Ωn/X.
• Mechanism: G never differentiates a singular expression. Its derivative identity writes ∂ξ(Gg) through g itself and a smooth average, so regularity in ξ bootstraps (with a smaller η-neighborhood when needed). But smoothness in ξ is not enough. ξ is proportional to r, and smoothness in r does not give smoothness across the axis; what is needed is smoothness in X ∝ r^2. The extended equations are invariant under ξ → -ξ with the stated parities. Uniqueness in the holomorphic class forces the solution to inherit them. An even smooth function of ξ is a smooth function of ξ^2 on the half-interval.
• Antecedent: None cited. (The even-function fact is classical and often attributed to Whitney; that attribution is mine.)
• Cost: Nothing new. It relies on the uniqueness class of Lemma 5.1 and may shrink the η-neighborhood per derivative.
• Backward question: The solution is built as a function of ξ = √X, that is, of r. How do I know it is smooth as a function of X ∝ r^2, which is what smoothness across the axis requires?
• Checkable: Take test g(ξ) (even or odd polynomials times e^{-ξ^2}) and c ∈ {1, 2, 3}. 1. Compute (Gg)(ξ) = ∫_0^ξ (s/ξ)^c g(s) ds with scipy quad and compare with ξ ∫_0^1 t^c g(tξ) dt. 2. Compare a finite-difference derivative with g(ξ) - c ∫_0^1 t^c g(tξ) dt. 3. Confirm that G maps even g to odd output (the factor ξ), matching the parity split of (5.7).
• Depends on: M5.4 (uniqueness of the Picard solution forces the even and odd parities that the extended equations preserve.); M5.3 (differentiates the Volterra operator G of the regular singular system (5.7) in ξ = √X at ξ = 0.); M4.2 (its target, smoothness in X rather than ξ ∝ r, is the axis-regularity criterion (4.4) for a smooth Cartesian field.)
• Refs: p48 to p49 (Step 3 of the proof of Lemma 5.1).
• Verification: statement digest-only; hypotheses not-checked; computation checked False

M5.6: Order-n stress primitives and the five total moments

Node ns-m5-6-order-n-stress-primitives-and-the-five-total-moments, kind move, pp. 45-62.

• Statement: Order-n stress and moments (p. 49-50): rθ,n = (R/C)[right minus left side of (5.3)], rz,n = [right minus left side of (5.4)], R = √(2X), vanish where the inner equations hold; Rj,n = q^{-A-1+λn}rj,n. The stress (5.9), physically q^{-A-1/2+λn}Tn, is Tn,θ = -R^{-2}∫_0^R ϱ²rθ,n dϱ, Tn,z = -R^{-1}∫_0^R ϱrz,n dϱ. The five moments (5.10)-(5.11), functions of η: mn,1 = ∫RUn dR, mn,2 = ∫R²En dR, mn,3 = ∫∂RΠn dR, mn,4 = ∫R²ΣUiEj dR, mn,5 = ∫(RΣUiUj - R²∂RΠn/2)dR; and Pn = (ϕn, Un, Vn/X, Πn, Fn/X), Fn = X AX(Un).
• Obligation: The order-n residual must be written as minus the cylindrical divergence of a stress supported in the annulus, the only place waves can supply it. The forward primitive from the axis vanishes wherever the inner equations hold. But beyond the source support it equals R^{-2} (or R^{-1}) times the total moment. A bare radial cutoff also leaves an exterior pressure constant and an exterior streamfunction.
• Mechanism: R^2 and R are the integrating factors of ∂R + 2/R (flux of angular momentum) and ∂R + 1/R (flux of axial momentum). So (5.9) is the primitive regular at the axis, and it vanishes past the source support exactly when ∫R^2 rθ,n dR = ∫R rz,n dR = 0. The five moments are the order-n analogs of the cumulative integrals (4.15): - mn,1 is the analog of M (radially integrated axial momentum, which is also the streamfunction at infinity); - mn,2 of I (angular momentum); - mn,3 of Cp (pressure increment from the axis to infinity); - mn,4 of J (axial transport of angular momentum); - mn,5 of S (axial momentum flux including pressure). M5.9 proves that these five conditions force both total residual integrals to vanish.
• Antecedent: None cited. Internal: - Proposition 4.2, the leading stress by radial integration; - (4.15) and Lemma 4.4; - the remark after (4.19) that, without total moment identities, a zero exterior residual could leave stresses proportional to r^{-2} and r^{-1} (Lemma A.8).
• Cost: Five scalar constraints at every order, each a function of η.
• Backward question: Integrating the order-n residual from the axis gives a stress that vanishes near the axis. Which finitely many global integrals of the order-n profile must vanish so that this stress also vanishes past the correction support, and so that pressure and streamfunction leave no exterior constants?
• Checkable: 1. Take a compactly supported test pair (rθ, rz) on 1 ≤ R ≤ 2 and compute T by (5.9) with scipy quad. 2. Verify (∂R + 2/R)Tθ = -rθ and (∂R + 1/R)Tz = -rz by finite differences. 3. Verify that for R > 2, Tθ = -R^{-2} ∫ ϱ^2 rθ dϱ and Tz = -R^{-1} ∫ ϱ rz dϱ. 4. Subtract a bump multiple from the test residual to zero each moment, and confirm the tails then vanish.
• Depends on: M4.4 (the primitives (5.9) repeat Proposition 4.2's radial integration from the axis, with integrating factors R² and R.); M5.2 (the residual coefficients r_θ,n and r_z,n are the right minus left sides of (5.3) and (5.4).); M4.5 (defines the five total moments m_n on the pattern of the cumulative integrals M, I, Cp, J, S of (4.15), taken at infinity.)
• Refs: p49 (rθ,n, rz,n, (5.9), (5.10), (5.11), Fn); p50 (mn, Pn).
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses not-checked; computation checked False

M5.7: Radial extension by a cutoff plus five reserved-patch bumps, and the affine block moment solve

Node ns-m5-7-radial-extension-by-a-cutoff-plus-five-reserved-patch, kind move, pp. 45-62.

• Statement: Lemma 5.2, Steps 1-2 (p. 50-52): fix X- < Xkeep < Xcut < a² < inf Ipos and an η-independent cutoff κ (1 on [0, Xkeep], 0 for X ≥ Xcut). Set Un = κUn^in + Σ_{j≤2}αn,j(η)b^U_j(R), En = κEn^in + Σ_{j≤3}βn,j(η)b^E_j(R) (5.14), with fixed unit-mass bumps in Jpos, and rebuild Fn, Vn, Πn by (5.15). On Ipos (U0 = 0, E0 = e∗fR^{-1-2λ}) the moments are affine in (αn, βn): mn = 0 iff B_Uαn = -d_{U,n}, B_Eβn = -d_{E,n}, with B_U, B_E moment matrices of distinct powers, invertible by Lemma A.1; so (5.16) αn = -B_U^{-1}d_{U,n}, βn = -B_E^{-1}d_{E,n}.
• Obligation: The inner solution lives only on [0, a^2]. The order-n profiles must: - be defined for all X ≥ 0, with (5.2) and (5.5) holding globally. This means exact incompressibility and exact radial balance, since the waves supply only the rθ and rz stresses and nothing in the radial equation; - equal the inner solution on [0, X-]; - keep the η-analyticity on [0, a^2] that the next order needs; - zero the five moments.
• Mechanism: κ is η-independent and the bumps vanish on [0, a^2], so the reconstructed profiles there are the analytic inner ones. On [0, X-], forward integration from zero axis data reproduces the inner solution exactly. The bumps sit on the reserved patch, where the leading profile has no axial velocity and a pure-power swirl. This decouples the moments: - U-bumps enter only mn,1 (weight R) and mn,4 (through R^2E0Un, weight R^{1-2λ}); - E-bumps enter only mn,2 (weight R^2), mn,3 (through ∂RΠn = 2E0En/R + ..., weight R^{-2-2λ}), and mn,5. - For mn,5, the pressure equation rewrites the moment as ∫(ΣUiUj - (1/2)ΣEiEj + (1/2)Ω_{n-1}) dX, so the E-bumps enter with weight R^{-2λ}. After dividing by e∗f(η), the matrices are η-independent moment matrices of distinct powers against ordered disjoint bumps. They are invertible by Lemma A.1: a nonzero combination of m distinct powers has at most m - 1 positive zeros, so the determinant integrand has one sign. Since i + j = n never pairs two order-n factors, the system is affine. It is solved for discrepancies of any size, with no smallness needed. This differs from the quadratic moment systems of Section 4, which needed Lemma A.2.
• Antecedent: No external citation. Internal: - Lemma A.1 (restated as Lemma 4.7), which the manuscript proves by Rolle's theorem and multilinearity of the determinant; - the reserved patch of Theorem 4.6(vi) and (4.30).
• Cost: - It uses the reserved interval Ipos at every order. - The coefficients αn, βn are not small and grow with n. - The bumps and the cutoff transition add order-n stress inside the annulus. - Distinct exponents require λ > 0. The inverse bound noted after Lemma A.1 degrades like O(λ^{-1}) as exponents merge, which is harmless for fixed λ.
• Backward question: Where can I add finitely many adjustable profiles so that the five moment conditions become a linear system invertible uniformly in η? It must not disturb the inner solution, its analytic region, the exterior heat flow, or the patch reserved for mean corrections.
• Checkable: 1. Choose λ, say 0.02; the paper fixes λ but does not give its value. 2. Place five standard mollifier bumps on disjoint ordered subintervals of an R-interval. 3. Compute B_U (2 by 2) and B_E (3 by 3) by quadrature and confirm the determinants are nonzero. 4. Track condition numbers as λ decreases toward 0; expect B_U to grow like λ^{-1}. 5. With test lower-order profiles, compute the five moments by quadrature for random (α, β). 6. Confirm the map is affine with Jacobian rows (B_U)_{1·}, e∗f (B_U)_{2·}, (B_E)_{1·}, 2e∗f (B_E)_{2·}, -e∗f (B_E)_{3·}, and that solving (5.16) zeros all five moments.
• Depends on: M5.6 (the five bump coefficients are fixed by zeroing the total moments m_n of (5.10), (5.11).); MA.1 (B_U and B_E are distinct-power moment matrices against ordered disjoint bumps, invertible by Lemma A.1, giving (5.16).); M5.3 (it cuts off the inner solution of Lemma 5.1, kept exactly on [0, X_keep] with its analytic region [0, a²] untouched.); M4.8 (the bumps sit on the reserved patch I_pos of Theorem 4.6(vi), where U0 = 0 and E0 is a pure power, which decouples the moments.)
• Refs: p50 (Lemma 5.2 statement, Step 1, X- < Xkeep < Xcut < a^2 < inf Ipos, κ, bumps); p51 ((5.14), (5.15), Step 2, E0 on Ipos, the mn,5 rewrite, ∂RΠn, B_U, B_E); p52 (d_{U,n}, d_{E,n}, Lemma A.1, (5.16)).
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses not-checked; computation checked False

M5.8: Support closure past X+ and exact vanishing on the mean patch

Node ns-m5-8-support-closure-past-x-and-exact-vanishing-on-the-mean, kind move, pp. 45-62.

• Statement: Lemma 5.2, Step 3 (p. 52): with X+ beyond the leading axial-velocity perturbation (hence beyond Xv), beyond every positive-order patch and before the terminal collar, the zero moments give Fn = Vn = 0 and Πn = 0 for X ≥ X+, since every Ωk term has a V factor and each ϕiϕj (i + j = n ≥ 1) a positive-order factor. Hence (5.12) supp_X Pn ⊂ [0, X+] with mn = 0; Fn/X and Vn/X are smooth at the axis; (5.17) max_{k+ℓ≤m} sup |∂X^k∂η^ℓPn| ≤ C_{n,m}; and (5.18) En = Un = Fn = Vn = 0 on the mean patch Imean for every n ≥ 1.
• Obligation: Positive orders must not touch the exterior heat flow (4.29), whose residual is identically zero; Theorem 3.1(iii) and the localization need this. They must also leave the mean patch Imean exactly at the leading power law, with zero axial velocity, for the corrections of Section 8.
• Mechanism: A radially integrated field vanishes past a support radius exactly when its total integral vanishes. The streamfunction vanishes by mn,1. The radial velocity then vanishes by (5.2). The pressure vanishes by mn,3, once its radial derivative vanishes. That radial derivative contains Ω_{n-1}, which involves the leading radial velocity V0, nonzero until Xv. The manuscript: "placing X+ beyond the leading axial-velocity perturbation is essential for the pressure support: Ω0 can remain nonzero beyond the positive-order patches." Imean lies to the right of both the inner cutoff and Ipos (sup Ipos < inf Imean). There, En = Un = 0 by construction. Fn = 0 because the axial flux has already been zeroed once Ipos is passed, and then Vn = 0 by (5.2).
• Antecedent: None cited. Internal: Theorem 4.6(v) (U = V0 = 0 on [Xv, ∞)) and Theorem 4.6(vi) (Ipos and Imean lie in (Xa, Xv), with sup Ipos < inf Imean).
• Cost: - Positive-order corrections occupy [X-, X+], a subinterval of the wave annulus. - The pressure corrections Πn need not vanish on Imean; only the velocity components are claimed to. - The constants C_{n,m} are uncontrolled in n.
• Backward question: Once the moments are zeroed, do all radially integrated fields vanish beyond a single n-independent radius, and what constrains where that radius may sit?
• Checkable: A toy order-one computation. 1. Prescribe smooth test lower-order data with U0 = V0 = 0 beyond a radius Xv. 2. Build (5.14) with coefficients from (5.16) and reconstruct (5.15) by cumulative quadrature. 3. Confirm that Fn, Vn, Πn vanish on [X+, 2X+] to quadrature tolerance. 4. With α = β = 0 instead, confirm they tend to nonzero constants: m^0_{n,1} for Fn and m^0_{n,3} for Πn.
• Depends on: M5.7 (the zeroed moments m_n,1 and m_n,3 from the bump solve make the streamfunction and pressure vanish past the patches.); M4.8 (X+ lies beyond X_v, where Theorem 4.6(v) gives U = V0 = 0, and I_mean lies right of I_pos by Theorem 4.6(vi).); M5.2 (V_n follows from A_X(U_n) by (5.2), and Π_n' from (5.5), whose Ω_{n-1} terms all carry a V factor.)
• Refs: p52 (Step 3, (5.17), (5.18), the remark on Ω0).
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses not-checked; computation checked False

M5.9: Conservative forms make the total tangential residual integrals vanish

Node ns-m5-9-conservative-forms-make-the-total-tangential-residual, kind move, pp. 45-62.

• Statement: Lemma 5.2, Step 4 (p. 53-54): for smooth axisymmetric divergence-free fields, r²Rθ and rRz are conservative (sums of ∂t, ∂r, ∂z, ∂zz of fluxes). Integrating in r, with the moment scalings (5.20) (e.g. ∫r uz,n dr = q^{1-A+λn}∫RUn dR), ∫RΠn dR = -(1/2)∫R²∂RΠn dR, and at n = 1 the identity (5.19) ∫_0^∞ r²(uθ,0 - P)dr = 0, the five vanishing moments give (5.21): ∫r²Rθ,n dr = ∫rRz,n dr = 0. Hence ∫R²rθ,n dR = ∫Rrz,n dR = 0, and (5.9) equals the backward form Tn,θ = R^{-2}∫_R^∞ ϱ²rθ,n dϱ, Tn,z = R^{-1}∫_R^∞ ϱrz,n dϱ.
• Obligation: Proves that zeroing the five moments makes the order-n stress vanish past the correction support, so the forward and backward primitives coincide.
• Mechanism: Integrating the conservative forms in r removes all radial flux terms: axis parity handles r = 0, and compact positive-order support handles infinity. What remains is ∂t and ∂z of radial integrals: - angular momentum (mn,2); - axial transport of angular momentum (mn,4); - axial flux (mn,1); - axial momentum flux including pressure (mn,5, after integrating the pressure by parts); - the order n-1 integrals hit by axial viscosity, which vanish by the previous order's moments, or at order zero by (4.28). The product power q^{...+λn} does not depend on the split i + j = n. So each integral vanishes identically in (z, t) before the physical derivatives are taken. At n = 1 the leading angular momentum integral diverges, since uθ,0 decays like r^{-1-2h}. The z-independent pure power P, which ∂zz does not see, is therefore subtracted. Differentiation under the integral is justified because beyond the terminal collar uθ,0 = K(r, t) is z-independent.
• Antecedent: None cited. Internal: (4.28), (A.45), and Lemma A.8 (the same argument at order zero).
• Cost: Order one depends on the renormalized angular moment identity of the leading construction. The resulting representation is for a signed stress; there is no positivity.
• Backward question: Which terms of the tangential residual survive integration against r^2 dr and r dr? Are five moments exactly enough to kill them all, including the pressure contribution and the axial viscosity inherited from the previous order?
• Checkable: 1. Symbolic (sympy): take ur = -∂zS/r and uz = ∂rS/r for a generic S(r, z, t), with generic uθ and p. Expand r^2 times the azimuthal residual and r times the axial residual, and confirm they equal the displayed conservative forms. 2. Quadrature: for compactly supported test En(R) and Πn(R), confirm ∫r^2uθ,n dr / (q^{3/2-A+λn}∫R^2En dR) = 1 with r = √q R. 3. Confirm ∫RΠn dR = -(1/2)∫R^2∂RΠn dR.
• Depends on: M5.8 (uses m_n = 0 and the compact support of positive orders in [0, X+] to drop the flux terms at infinity.); M5.6 (the conservative forms reduce the weighted residual integrals to derivatives of the moments (5.10), (5.11), giving the backward form of (5.9).); MA.12 (repeats Lemma A.8's order-zero argument; at n = 1 it subtracts the pure-power exterior P of (A.45) so the angular moment converges.); M4.8 (the n = 1 axial-viscosity term needs the order-zero identities (4.28) and the z-independent heat exterior (4.29).)
• Refs: p53 (conservative forms, moment scalings, (5.19), (5.20), (5.21)); p54 (removal of powers, backward representation).
• Verification: statement completeness audit 2026-10-01: TRUNCATED; statement replaced from the digest; hypotheses not-checked; computation checked False

M5.10: Order-one stress at the outer edge is flat at the rate of ζ

Node ns-m5-10-order-one-stress-at-the-outer-edge-is-flat-at-the-rate-of, kind move, pp. 45-62.

• Statement: Lemma 5.2, Step 5, and (5.13) (p. 50, 54): for n ≥ 2, supp Tn ⊂ [X-, X+] and every derivative of Tn/ζ is bounded. For n = 1 the only exterior source is -∂zz uθ,0 on the terminal collar, where uθ,0 = K fo with 1 - fo = e^{-4/δb²} times a smooth factor, δb = log(Xb/X); so supp T1 ⊂ [X-, Xb], and Lemma A.9 gives (5.22): |T1| ≤ Ce^{-4/δb²}δb^{-3}, |∂^IT1| ≤ C_Ie^{-4/δb²}δb^{-N_I}. With ζ = exp(-ca/ya² - 4/yb²) this yields (5.13): |∂^ITn| ≤ C_{n,I}ζδ^{-N_{n,I}} on Xa < X < Xb for every n ≥ 1.
• Obligation: Later sections build wave amplitudes on the weight ζ; amplitudes carry √ζ in the class W_α. They must also absorb the higher-order stress by signed corrections. So that stress must vanish at the annulus edges as fast as ζ, up to finite inverse powers of δ. Without Step 5, T1 could reach the outer edge Xb at a rate not dominated by ζ.
• Mechanism: At order one, inside the terminal collar, the leading swirl is the heat field times a terminal multiplier fo(log X). Since X = s/q depends on z, axial viscosity produces a source built from ∂y fo and ∂yy fo. These are flat at Xb at the rate e^{-4/δb^2}, the same exponent as the outer factor of ζ and as the outer rate (4.32) of T0. The backward representation of M5.9 integrates this source inward from the edge. Lemma A.9 (substitution u = δ/(1 + δ^2v)^{1/2}) gives ∫_0^δ e^{-c/u^2} u^{-j} b(u, η) du = (1/2) e^{-c/δ^2} δ^{3-j} B(δ, η), with B smooth and B(0, η) = b(0, η)/c. In words: integrating a flat factor from the edge gains δ^3. Derivatives cost only finite inverse powers of δb. Near the inner edge, T1 vanishes identically on [0, X-].
• Antecedent: None cited. Internal: Lemma A.9, the terminal multiplier (A.12), the heat profile (4.29), and the edge factorization (4.32).
• Cost: Unlike Tn for n ≥ 2, T1 reaches the outer edge Xb, with a finite inverse-power loss δ^{-N}. Later stages must accept stress weighted by ζ δ^{-N}.
• Backward question: At order one the leading exterior is not compactly supported and depends on z inside the terminal collar. Does axial viscosity acting on it leave stress at the outer edge, and does that stress vanish as fast as the weight ζ that controls the wave amplitudes?
• Checkable: 1. mpmath at 40 digits: for c = 4, j ∈ {3, 6} and b ≡ 1, compute I(δ) = ∫_0^δ e^{-4/u^2} u^{-j} du after the substitution w = u^{-2}, which removes the endpoint peak. Use δ = 0.3, 0.1, 0.05, 0.02. 2. Compare I(δ)/((1/2) e^{-4/δ^2} δ^{3-j}) with b(0)/c = 0.25; for j = 3 the ratio is exactly 1/c. I ran this while digesting: j = 3 gives 0.25, and j = 6 gives 0.2585, 0.2509, 0.2502, 0.25004. 3. Symbolically, verify the ∂zz(K fo) formula from the chain rule of Lemma 4.1 (X_z = -2ηX/(q^D L), Z_b of (4.2)), with K independent of z.
• Depends on: MA.13 (Lemma A.9 with c = 4, j = 3, 6 factors e^{-4/δ_b²} out of the backward integral of the order-one source, gaining δ_b³ in (5.22).); M5.9 (integrates the order-one residual inward from the edge with the backward stress form that the zeroed moments make valid.); M4.15 (the target weight ζ = exp(-c_a/y_a² - 4/y_b²) is defined in Step 4 of the proof of Theorem 4.6.); MA.14 (on the terminal collar u_θ,0 = K f_o with f_o' flat at the rate e^{-4/δ²}δ^{-3}, the same exponent as the outer factor of ζ.)
• Refs: p54 (Step 5, az, bz, bounds on fo, Lemma A.9, (5.22)); p50 ((5.13)); p44 (definition of ζ in the proof of Theorem 4.6).
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses not-checked; computation checked False

M5.11: Finite truncations: the residual gain grows with N, the derivative loss does not

Node ns-m5-11-finite-truncations-the-residual-gain-grows-with-n-the, kind move, pp. 45-62.

• Statement: Proposition 5.3. Define: - U^[N] = Σ_{n=0}^N (un, pn, q^{-A-1/2+λn} Tn), where the profiles include their cutoffs and moment corrections; - (5.24): Fslow(u, p, T) = R(u, p) + (∂r + 2/r)Tθ eθ + (∂r + 1/r)Tz ez; - (5.23): |V|_m = max_{|α|+b≤m} |∂x^α ∂t^b V|. Then each u^[N]_slow is divergence-free, and (5.25) holds: |Fslow(U^[N])|_m ≤ C_{N,m} q^{2h(N+1)-Km} on 0 ≤ X ≤ Xmax, -1 ≤ η ≤ 1, 0 < q ≤ 1, with Km independent of N.
• Obligation: Supplies the finite-residual hypothesis (5.33) of Lemma 5.4, with ρN = 2h(N+1) and no flat remainder. The gain must grow with N while the loss from physical differentiation stays fixed.
• Mechanism: At every retained order: - (5.2) gives exact incompressibility; - (5.5) holds globally, so the radial residual is canceled; - differentiating the primitives gives (∂R + 2/R)Tn,θ = -rθ,n and (∂R + 1/R)Tn,z = -rz,n, so the tangential residual is exactly minus the stress divergence; - order zero holds by Proposition 4.2. After truncation, every uncancelled product or shifted viscous term carries relative power at least q^{2h(N+1)}. For N = 1, for example, products of two order-one coefficients and axial viscosity on order one first enter at order two. In the similarity chart, each transverse derivative costs q^{-1/2}, each axial derivative q^{-D}, and each time derivative q^{-1}, whatever the base power b. The powers of n produced by differentiating q^{λn} enter only the constants, through (1 + |b|)^m. On the stress support X ≥ Xa > 0, r^{-1} = q^{-1/2}(2X)^{-1/2}; near the axis the smooth Cartesian representatives are used.
• Antecedent: None cited. Internal: Lemma 4.1 and Proposition 4.2.
• Cost: Bounds hold only on fixed compact profile ranges and for q ≤ 1. C_{N,m} depends on N and on Xmax. The losses Km must be carried into the summation.
• Backward question: If I stop at order N, does the residual gain grow with N while each physical derivative costs a power of q that does not depend on N?
• Checkable: A slope test of (5.26). 1. Take g(x⊥, η) = e^{-|x⊥|^2}(1 + η^2)^{-1}, b = -A + 2nh for several n, and h = 0.005. 2. Compute mixed derivatives of q^b g in Cartesian (x1, x2, z, t) by high-order finite differences in mpmath. Obtain q(z, t) by root finding on q - z^2q^{2h} = 1 - t. 3. Work at points of fixed (X, η) = (1, 0.3), for q from 1e-2 to 1e-8. 4. Fit the log-log slope and compare with b - a/2 - Dk - l. The full (5.25) needs the actual profiles. Appendices A to C construct them analytically, not in closed form.
• Depends on: M5.7 (the extended profiles satisfy (5.2) and (5.5) globally, so each truncation is divergence-free and cancels the radial residual.); M5.6 (differentiating the primitives (5.9) makes each order's tangential residual exactly minus the stress divergence.); M4.1 (the loss K_m comes from (5.26): each physical derivative costs a fixed power of q via Lemma 4.1's chain rule, at every order.); M4.4 (order zero enters through Proposition 4.2, whose tangential residual is exactly minus the divergence of the leading stress.)
• Refs: p54 (plan of subsection 5.3); p55 ((5.23), (5.24), Proposition 5.3, (5.25), proof Step 1, (5.26)); p56 (chain rule, r^{-1} factor).
• Verification: statement digest-only; hypotheses not-checked; computation checked False

M5.12: Stokes streamfunction potentials, so that cutoffs in q keep div u = 0

Node ns-m5-12-stokes-streamfunction-potentials-so-that-cutoffs-in-q, kind move, pp. 45-62.

• Statement: (5.27): - Fn = X AX(Un), Sn = q^{1-A+λn} Fn, and An = (Sn/r) eθ; - uz,n = ∂s Sn and r ur,n = -∂z Sn (s = r^2/2), so curl An = ur,n er + uz,n ez. In Cartesian form, An = (Sn/r^2)(-x2, x1, 0) = (1/2) q^{-A+λn} (Fn/X)(-x2, x1, 0). This is smooth across the axis for q > 0, because Fn/X is smooth at X = 0. The summation tuple is An = An, Bn = uθ,n eθ, pn = q^{-2A+λn} Πn, Tn = q^{-A-1/2+λn} Tn, with decay orders gn = 2nh and differential polynomial Fslow.
• Obligation: The summation cutoffs χ(cn q) depend on (z, t) through q. Multiplying the velocity coefficients by them would break div u = 0. Cutting the potential before taking the curl keeps exact incompressibility. The swirl Bn stays divergence-free under multiplication by any function of (z, t), because it is independent of θ.
• Mechanism: Axisymmetric divergence-free (ur, uz) fields are curls of azimuthal potentials (S/r)eθ, with S the Stokes streamfunction. Given uz,n = ∂s Sn, the relation -∂zSn = q^{λn}Vn is exactly (5.2), using A + D = 1 and the operator Z of (4.2). Smoothness at the axis follows because S/r^2 is a smooth function of (r^2, z, t). The first moment mn,1 = 0 makes Fn, and hence Sn, compactly supported in X. The curl of a cut potential includes the term from differentiating the cutoff, so the result stays divergence-free.
• Antecedent: None cited; the Stokes streamfunction is named without citation.
• Cost: It needs Fn/X smooth at the axis and Fn compactly supported, both from Lemma 5.2. Cutting the potential produces an extra radial velocity term, displayed as (5.45).
• Backward question: How can each order be multiplied by a cutoff depending on q = q(z, t) without destroying exact incompressibility or smoothness at the axis?
• Checkable: 1. Take Un = (1 - X) e^{-X} g(η). Then ∫_0^∞ Un dX = 0 and Fn = X e^{-X} g(η). 2. Obtain q(z, t) by root finding. 3. By finite differences in z at fixed (r, t), check that -∂zSn = q^{λn}Vn with Vn from (5.2), and that ∂sSn = q^{-A+λn}Un. 4. Check that the Cartesian divergence of curl(χ(cq)An) vanishes to finite-difference accuracy.
• Depends on: M5.2 (u_z,n = ∂_s S_n and r u_r,n = -∂_z S_n reproduce the order-n incompressibility relation (5.2), using A + D = 1.); M5.8 (F_n/X is smooth at the axis and F_n vanishes past X+ because m_n,1 = 0, so each potential is smooth and compactly supported.); M5.11 (the summation tuple carries the differential polynomial F_slow of (5.24) and the stresses of the truncations.)
• Refs: p56 (plan of subsection 5.4, potential formulas, (5.27), Cartesian form, summation tuple).
• Verification: statement digest-only; hypotheses not-checked; computation checked False

M5.13: Lemma 5.4, an abstract Borel-type summation with shrinking cutoffs

Node ns-m5-13-lemma-5-4-an-abstract-borel-type-summation-with-shrinking, kind move, pp. 57-59.

• Statement: Lemma 5.4 (Borel-type summation, pp. 57-59): let 0 < q < q0 ≤ 1, |∂z^a∂t^b q| ≤ Cq^{1-aD-b}, 0 < D < 1 (5.28); a base U0, div u0 = 0, |U0|_m ≤ C_m q^{-Km}Λlog^{Pm}, Λlog = 1 + |log q|; increments Zj = (Aj, Bj eθ, pj, Tj) with smooth zero extensions, |Zj|_m ≤ C_{j,m}q^{gj-ℓm}Λlog^{P_{j,m}}, gj ↑ ∞ (5.29)-(5.30); and a differential polynomial F (5.31)-(5.32) whose partial-sum residuals are O(q^{ρJ-K^F_m}) plus flat, ρJ → ∞ (5.33). Then there are scales aj with U = U0 + Σ_j χ(ajq)Zj (curl after cutoff) smooth, divergence-free, close to U[J] (5.35), and F(U) = O(q^N) for all N (5.36).
• Obligation: The constants of (5.17) may grow too fast for (5.1) to converge. The lemma produces genuine smooth, exactly divergence-free fields with the same asymptotics and a flat residual, in a form that the wave and mean correction sequence can reuse (Proposition 9.9).
• Mechanism: Three steps. 1. Scale invariance. On the support of χ^{(k)}(aq), 1/2 ≤ aq ≤ 1. So each factor a from the chain rule pairs with a factor q from (5.28), giving |∂z^{a'}∂t^{b'}χ(aq)| ≤ C q^{-a'D-b'} uniformly in a. Since q depends only on (z, t), curl(χ(aq)Aj) = χ(aq) curl Aj + aχ'(aq)∇q × Aj is controlled by the potential bounds, with a loss ℓ'm depending only on m and ℓ_{m+1}. 2. Diagonal choice (5.37). At step j, pick aj so large that Ĉ_{j,m} Λlog^{P̂_{j,m}} q^{gj/2} ≤ 2^{-j} on q ≤ aj^{-1}, for all m ≤ j. Half of the decay exponent pays for arbitrarily large constants. The tail beyond J is then at most 2^{-J} q^{g_{J+1}/2 - ℓ'm}. On {q ≥ δ}, only terms with aj ≤ δ^{-1} survive, which gives local finiteness. 3. Flatness. Fix (m, N), then a large J, and compare F(U) with F(U[J]) on q < 1/(2aJ), where the first J cutoffs equal one. - (5.38) bounds U[J] by q^{-Km}, with Km independent of J. - (5.39) bounds the difference by C_m q^{-Hm} |e|_{m+s}(1 + |U[J]|_{m+s} + |e|_{m+s})^{d-1}, with e = U - U[J] and Hm independent of J. - Choose J with g_{J+1}/2 - ℓ'_{m+s} ≥ N + Hm + (d - 1)(K_{m+s} + 1) and ρJ - K^F_m ≥ N + 1. Both pieces are then O(q^N).
• Antecedent: None cited. The diagonal cutoff construction is the one in the classical proof of Borel's lemma (a smooth function with a prescribed Taylor series); that attribution is mine.
• Cost: - The scales aj are not explicit; they are chosen after all constants. - The summed field equals the partial sums only on the shrinking neighborhoods q < 1/(2aJ). - The tail estimates use only half the decay exponent. - Every summand needs common supports, smooth Cartesian representatives, and smooth zero extensions, plus a common domain for all partial sums (later supplied by Lemma 9.7).
• Backward question: Given a formal expansion whose coefficient bounds may grow arbitrarily fast with the order, how do I build an actual smooth, exactly divergence-free field with the same asymptotics and a residual flat at the singular point? And how do I state it abstractly enough to reuse for the later correction cycle?
• Checkable: Mostly a pure estimate. Two ingredients are computable: - (a) For a standard smooth step χ built from e^{-1/x}, confirm numerically that sup_{q>0} |(q∂q)^j χ(aq)| is the same for a = 1, 10, 1e4. It equals sup_σ |(σ∂σ)^j χ(σ)|. - (b) A toy summation: take gj = 2jh with h = 0.005 and constants Ĉ_{j,m} = ((j + m)!)^2. Choose aj from (5.37) by root finding, then verify the tail bound (5.35) on a grid in q.
• Depends on: M5.12 (its increments are potentials A_j cut before the curl plus θ-independent swirl B_j e_θ, the device of (5.27) that keeps div u = 0.); M4.1 (the cutoffs χ(a_j q) use the concentration scale q of (4.1), whose derivative bounds (5.28) make them scale invariant.)
• Refs: p56 (summation plan, Λlog); p57 ((5.28) to (5.33), Lemma 5.4 statement, (5.34)); p58 ((5.35), (5.36), Steps 1 and 2, (5.37)); p59 (Step 3, (5.38), (5.39), choice of J, initial block, residual comparison formula, (5.40), recovery of the expansion).
• Verification: statement completeness audit 2026-10-01: INCOMPLETE; statement replaced from the digest; hypotheses astra-spot-check-2026-10-01; computation checked False; Astra spot-check: statement-rewritten

M5.14: Proposition 5.5, the realized base field and what it guarantees

Node ns-m5-14-proposition-5-5-the-realized-base-field-and-what-it, kind move, pp. 45-62.

• Statement: Proposition 5.5 (p. 60-62): there are smooth axisymmetric (uB, pB) for q > 0 and stresses Tphys supported in Xa ≤ X ≤ Xb with: div uB = 0; R(uB, pB) = -(∂r + 2/r)Tphys,θeθ - (∂r + 1/r)Tphys,zez + EB, EB = O(q^M) for all M (5.41); on [Xlo, Xhi], q^Auθ,B - E0, q^Auz,B - U0 = O(q^{2h}) and q^Aur,B = O(q^h) (5.42); |D^I(q^{A+1/2}Tphys - T0)| ≤ C_Iq^{2h}ζδ^{-N_I} on Xa < X < Xb (5.43); continuous extension to η = ±1; uB - u(0), pB - p(0) vanish for X ≥ X+, preserving (4.29); on Imean, uz,B = 0, q^Auθ,B = E0 (5.44).
• Obligation: Delivers the background every later section uses: - exact incompressibility; - residual equal to an annular stress divergence plus a flat error; - the untouched heat exterior; - the untouched mean patch; - normalized closeness to the leading field, used for the chart base velocity in Section 7 and for the growth asymptotic (3.6) at a fixed Xin ∈ (0, Xa); - weighted closeness of the stress to T0, used to choose wave amplitudes (Propositions 7.5 and 7.6); - smooth limits at t = 1 away from the origin.
• Mechanism: Apply Lemma 5.4 to the tuples of M5.12 on each compact profile range, keeping order zero uncut. By (5.26) and D < 1/2, the loss is ℓm = 2A + m, and (5.25) is (5.33) with ρN = 2h(N+1) and no remainder. Cutoffs are scale invariant in the normalized variables: (q∂q)^j χ(cn q) = (σ∂σ)^j χ(σ) at σ = cn q, bounded uniformly in n. Cutting the potential gives (5.45): r u^cut_{r,n} = q^{2nh}[χ(cn q)Vn - (2η/L)(cn q)χ'(cn q)Fn] and u^cut_{z,n} = q^{-A+2nh}χ(cn q)Un. This field is divergence-free identically, and (5.18) kills both bracket terms on Imean, which proves (5.44). Extra diagonal requirements of type (5.40) give (5.46): |D^I[χ(cn q) q^{2nh} an]| ≤ 2^{-n} q^{nh} for n ≥ max{2, |I|}. Here an is one of En, Un, Πn, Vn/√(2X), Fn/√(2X) on the enlarged rectangle, or a component of Tn/ζ for n ≥ 2. Summing gives (5.42); in the normalization q^A, the positive-order radial correction is even O(q^{3h}). Together with (5.22) for n = 1, summing also gives (5.43). The asymptotic coefficients are unchanged. Split at J = max{2N + 2, m + 1}. Keep the finitely many terms N < n < J with their original powers, which are at least q^{2h(N+1)}. Bound the tail by Σ_{n≥J} 2^{-n}q^{nh} ≤ 2^{1-J}q^{Jh} ≤ 2^{1-J}q^{2h(N+1)}. On compact sets with q bounded below, only finitely many terms survive, so their analytic η-neighborhoods intersect. Since L > 0, the one-sided extensions at η = ±1 transfer to physical variables; η = ±1 is the time t = 1 away from the origin (Section 3.1).
• Antecedent: None cited. Internal: Lemma 5.4, Proposition 5.3, Lemma 5.2, Theorem 4.6.
• Cost: - The higher-order stress T̂ - T0 is signed and bounded only by q^{2h}ζδ^{-N}. The δ^{-N} loss near the annulus edges rules out a uniform relative bound against |T0| ≥ cζ. So Section 7 keeps the fixed positive representation of T0 and treats the rest by signed corrections. - The diagonal bound gives q^{nh} rather than q^{2nh}; splitting at 2N + 2 compensates. - All estimates are uniform only on compact profile rectangles and for q > 0.
• Backward question: Does summation with shrinking cutoffs preserve exactly what later sections need, without changing any asymptotic coefficient? That means the heat exterior, the untouched mean patch, closeness to the leading field and to the stress with the ζ weight, and smooth limits at t = 1 away from the origin.
• Checkable: - (a) Symbolic: derive (5.45) as -∂z[χ(cn q)Sn], using ∂zq = 2ηq^{1-D}/L (Lemma 4.1), Sn = q^{1-A+2nh}Fn and A + D = 1. Confirm the divergence of the cut field vanishes. - (b) Numeric: for h = 0.005, N ≤ 5, m ≤ 5 and a grid of q in (0, 1], confirm Σ_{n≥J} 2^{-n}q^{nh} ≤ 2^{1-J}q^{2h(N+1)} with J = max{2N + 2, m + 1}. - (c) The quantitative bounds (5.42) and (5.43) require the constructed profiles.
• Depends on: M5.13 (Lemma 5.4 applied to the order-n tuples gives smooth, exactly divergence-free fields with a flat residual E_B and unchanged asymptotics.); M5.11 (Proposition 5.3's truncation bound (5.25) supplies the hypothesis (5.33) with ρ_N = 2h(N + 1) and an N-independent loss.); M5.10 (the ζδ^{-N} bounds (5.13), (5.22) on T_n give the weighted closeness (5.43) of the summed stress to T0.); M5.8 (positive orders vanish past X+ and, by (5.18), on I_mean, so the heat exterior and the mean patch (5.44) are untouched.)
• Refs: p60 (subsection 5.5, Proposition 5.5, (5.41) to (5.44), Step 1, ℓm = 2A + m, cutoff scale invariance); p61 (ρN, Step 2, (5.45), Step 3, (5.46), O(q^{3h}), the Step 4 split); p62 (tail bound, endpoint extension).
• Verification: statement completeness audit 2026-10-01: TRUNCATED; statement replaced from the digest; hypotheses not-checked; computation checked False