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

Appendix A: Matching radial moments and constructing the heat exterior (pp. 126 to 144)

MA.1: Distinct power weights give independent moment changes (Lemma A.1)

Node ns-ma-1-distinct-power-weights-give-independent-moment-changes, kind move, pp. 126-127.

• Statement: For distinct real α_1, ..., α_m and nonnegative nonzero smooth bumps β_j supported in the interiors of compact intervals I_1 < ... < I_m in (0, ∞), the matrix B_ij = ∫_0^∞ x^{α_i} β_j(x) dx is invertible; the same holds for exponential weights in a logarithmic coordinate, and the inverse and each fixed parameter derivative stay bounded over a smooth compact family that keeps the separations.
• Obligation: Every exact moment restoration in the paper needs the linear map from bump coefficients to moment changes to be invertible with a controlled inverse: Theorem 4.6 Step 3 (4.42), Proposition A.7, Proposition B.8, Proposition C.2, and the order-n solve (5.16) in Lemma 5.2. Without it a local edit could leave a moment discrepancy that propagates to the exterior, and Lemma 4.4(i) could not be invoked.
• Mechanism: Multilinearity in the columns gives det B = ∫_{I_1 × ... × I_m} det[x_j^{α_i}] Π_j β_j(x_j) dx. The generalized Vandermonde determinant cannot vanish on 0 < x_1 < ... < x_m: otherwise some nonzero combination of the m distinct powers would have m distinct positive zeros, whereas dividing by the lowest power and applying Rolle's theorem inductively shows it has at most m - 1. The ordered cone is connected, so the integrand has one sign and is nonzero on a set of positive measure. The substitution x = e^y gives the logarithmic version; compactness plus differentiation of B^{-1}B = Id gives the uniform bounds. The explicit 2 × 2 determinant (A.1) is proportional to the exponent gap, which quantifies the loss.
• Antecedent: None cited. The proof names Rolle's theorem (zero counting for sums of distinct powers) and multilinearity of the determinant.
• Cost: Bumps must have ordered disjoint supports and the exponents within each block must be distinct. A gap of size λ costs λ^{-1} in the inverse, so λ must be fixed before any later large parameter (the radial frequency N of Proposition C.2), or the loss must be absorbed by an exponentially small discrepancy, as in (A.15).
• Backward question: If I add finitely many fixed bumps to a profile, when do the resulting changes in finitely many power-weighted integrals span every prescribed discrepancy, and how fast does this fail as two weights coalesce?
• Checkable: Compute B for random distinct exponents and rescaled σ' bumps on ordered intervals and check that det B is nonzero with the sign of det[x_j^{α_i}] on the ordered cone; verify (A.1) symbolically; tabulate ‖B^{-1}‖ against λ for the weights (1, x^{-λ}) and (e^{(1/2-λ)y}, e^{(1/2-2λ)y}) to see the λ^{-1} growth.
• Refs: pp. 126 to 127, Lemma A.1, (A.1); restated p. 34 (Lemma 4.7).
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MA.2: Quadratic moment equations solved by contraction with C^k control (Lemma A.2)

Node ns-ma-2-quadratic-moment-equations-solved-by-contraction-with-c-k, kind move, pp. 127-128.

• Statement: Lemma A.2 (pp. 127 to 128): suppose coefficients c ∈ R^m change the required moments by exactly F_η(c) = B(η)c + Q_η(c, c) (A.2) on K = [−1, 1], with B smooth and invertible and Q smooth bilinear; let β_0 = sup_K ‖B^{−1}‖, κ_0 = sup_K ‖Q‖, d_0 = ‖d‖_{C^0(K)}. If 8β_0²κ_0 d_0 ≤ 1, then F_η(c) = d(η) has a unique solution with ‖c‖_{C^0} ≤ 2β_0 d_0, smooth in η and the limit of c ↦ B^{−1}(d − Q(c, c)) from zero; if Q = 0 no smallness is needed. If also 8β_k²κ_k‖d‖_{C^k} ≤ 1 (β_k, κ_k the C^k analogues) for a preselected k, then ‖c‖_{C^k} ≤ 2β_k‖d‖_{C^k} (A.3).
• Obligation: The moments J, S, C_p are quadratic in (U, E), so an invertible linearization alone does not restore them exactly. The lemma gives exact solutions whose finitely many prescribed η-derivatives are small, and the scaling bound turns this into small field derivatives on the patch, which is what keeps the strict cone inequalities (which involve finitely many field derivatives) intact at every correction.
• Mechanism: On the C^0 ball of radius r = 2 β_0 d_0, the map sends c to a vector of norm at most r/2 + β_0 κ_0 r^2 ≤ r and has Lipschitz constant at most 2 β_0 κ_0 r ≤ 1/2, so it is a contraction. The same argument in the Banach algebra C^k(K) yields (A.3), and because the iterates are identical, the two limits coincide. The same smallness makes B + D_c Q invertible, so the pointwise implicit function theorem gives smoothness in η, including one-sided derivatives at η = ±1; all other fixed derivatives are finite by implicit differentiation, with no simultaneous smallness over all orders. An extra compact parameter (the pulse amplitude) is carried the same way.
• Antecedent: None cited. The proof names the contraction argument in C^0 and in the Banach algebra C^k and the pointwise implicit function theorem.
• Cost: The smallness conditions 8 β_0^2 κ_0 d_0 ≤ 1 and 8 β_k^2 κ_k ‖d‖_{C^k} ≤ 1, imposed only for finitely many k; each discrepancy must therefore be made small first (by a large radius, a large frequency, or exponential decay).
• Backward question: The moment functionals are quadratic in the profile; can the finite system be solved exactly, smoothly in η, with bounds on a prescribed finite number of η-derivatives, without demanding smallness at every derivative order at once?
• Checkable: None: pure estimate. Its concrete instances are the finite solves listed under MA.3, MA.5, MA.6 and MA.11, which are checkable there.
• Depends on: M4.5 (the equations it solves are exact changes of the cumulative integrals (4.15), in which J, S, Cp are quadratic in (U, E).); MA.1 (its invertible linear part B, with bounded η-derivatives of B^{-1}, is the power-weight moment matrix of Lemma A.1.)
• Refs: pp. 127 to 128, Lemma A.2, (A.2), (A.3), scaling bound p. 128; restated pp. 34 to 35 (Lemma 4.7).
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MA.3: The five-moment Jacobian on a power-law patch (Corollary A.3)

Node ns-ma-3-the-five-moment-jacobian-on-a-power-law-patch-corollary-a, kind move, pp. 128.

• Statement: Corollary A.3 (p. 128): for the five cumulative integrals (M, I, J, S, C_p) of (4.15), under X = ρx the normalizing factors are (ρ, ρ^{3/2}, ρ^{3/2}, ρ, 1), with H = ρ^{1/2}√(2x) E (A.4). On a correction interval with U_0 = u_c(η) and E_0 = e∗ f∗(η) x^α (e∗ > 0, f∗ > 0 smooth), two additive U bumps and three additive E bumps give an invertible Jacobian of the normalized five-moment map provided α ∉ {−1/2, 1/2, 3/2}, and the moment changes have the exact quadratic form (A.2); at U_0 = 0 the three E bumps prescribe changes in (I, S, C_p) while preserving M and J.
• Obligation: Lemma 4.4(i) preserves the exterior pressure, radial velocity, Q_s, N_s, p_s and T_0 only when all five integrals agree at the joining radius. The corollary says exactly which patches let five local bumps reset all five integrals, so that each join (Proposition B.8 at α = 1/10; Theorem 4.6 Step 3, Proposition A.7 and Proposition C.2 at α = -1/2 - λ) is solvable.
• Mechanism: With u_c frozen at its base value, the row operations J → J - u_c I and S → S - 2 u_c M make the linearization block diagonal: the U block (rows M and J - u_c I, the latter with differential ∫ √(2x) E_0 δU dx) has weights 1, x^{α+1/2}, and the E block (rows I, S - 2 u_c M with differential -∫ E_0 δE dx, and C_p) has weights x^{1/2}, x^α, x^{α-1}. Distinct exponents within each block is exactly α ∉ {-1/2, 1/2, 3/2}, so Lemma A.1 applies; the original functionals have degree at most two in (U, E), so the remainder is exactly quadratic and Lemma A.2 applies (the row operations do not assume the corrected U stays constant). The two instances used: α = 1/10 (axis moment correction) with blocks (0, 3/5) and (1/2, 1/10, -9/10); α = -1/2 - λ (intermediate patches) with blocks (0, -λ) and (1/2, -1/2 - λ, -3/2 - λ), where the first block has inverse bound λ^{-1}.
• Antecedent: None cited beyond Lemma A.1.
• Cost: Correction patches must carry an exact power law with U_0 constant in X, which is why Section A.2 reserves untouched patches. On the intermediate patches the λ^{-1} inverse forces λ to be fixed before the radial frequency N.
• Backward question: On which explicit profile patches can two axial and three azimuthal bumps reset all five cumulative integrals independently, and which exponent coincidences must be avoided?
• Checkable: Assemble the 5 × 5 Jacobian of (M, I, J, S, C_p) at U_0 = u_c, E_0 = e∗ f∗ x^α with fixed bumps; check invertibility for α outside {-1/2, 1/2, 3/2} and singularity at those three values (two rows then carry the same weight); confirm the block exponents for α = 1/10 and α = -1/2 - λ and the λ^{-1} growth of the U-block inverse as λ → 0.
• Depends on: M4.5 (the map it linearizes is the five cumulative integrals (4.15), rescaled under X = ρx by the factors (A.4).); MA.1 (after row operations each block has distinct power weights against ordered disjoint bumps, so Lemma A.1 inverts it when α ∉ {-1/2, 1/2, 3/2}.); MA.2 (the full moment change is exactly quadratic, the form (A.2) that Lemma A.2 solves by contraction.)
• Refs: p. 128, Corollary A.3, (A.4); used p. 42 (4.42), pp. 139 to 140, pp. 156 to 157, p. 162.
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MA.4: The staged outer reference profile with four reserved patches (Section A.2, Proposition A.4)

Node ns-ma-4-the-staged-outer-reference-profile-with-four-reserved, kind move, pp. 129-131.

• Statement: Proposition A.4 (pp. 129 to 131): with σ of (A.5) and finite choices in the order (A.6) (M_d, T_d = e^{M_d} + 10, P∗ > e^{T_d}, 0 < λ ≪ 1, 0 < h ≪ min{λ, e^{−T_d}}, then X_R), the staged profile (reference (A.7), axial reduction, l = −λ over T_w (A.9), pulse, η-flattening, exterior transition) has U, E smooth for X > 0, E > 0, (A.8); U = M = J = 0 after the pulse, E = c∞X^{−A} after the terminal interval; the four patches keep U = 0, E ∝ f X^{−1/2−λ}; for x_− ∈ (0, 1], X_R large, the strict relaxed cone holds from x_−, the admissible cone from the first l < 0, through tail coordinate 1/2.
• Obligation: Lemma 4.8 needs one outer family that (a) fixes the axis pressure datum (MA.8), (b) ends in an η-independent power law that can be heat-replaced (MA.10, MA.11), (c) satisfies the total identities that remove stress tails (MA.12), (d) keeps the cone (MA.9), and (e) leaves untouched power-law patches for Proposition C.2 and Theorem 4.6 Step 3 (I_1), Proposition A.7 (I_2), Lemma 5.2 (I_pos = I_3) and Lemma 8.7 (I_mean = I_4). Theorem 4.6(vi) is the statement that I_3, I_4 survive with the form (4.30).
• Mechanism: Prescribing the log slope l instead of E makes every stage an explicit exponential in y, so the moment integrals and the linear equations for Q_s, N_s become explicit. The reference stage (l = 3/5, a = 4/5) is where the axis profile will later be attached. The axial reduction first flattens H (a potential vortex, a = 2) and then removes the axial velocity so slowly that |k'| ≤ e_d/(1 + y) with e_d = 4‖σ'‖_∞/M_d, keeping the axial shear b_s small. The intermediate stage l = -λ gives a = 2 + 2λ > 2 (admissible) and exact power-law patches. The pulse is the one adjustable source of positive S-moment. The interpolation (A.10) replaces the η shape f by its value 1/2 at η = ±1 without letting E increase, so the tail is η-independent, as a z-independent heat exterior requires. The steep stage (l = -1) cuts the amplitude by h^6 and resets Q_s, and the terminal factor f_o is what later produces a positive angular stress at the outer edge. Since the schedule depends only on x = X/X_R, normalized fields are X_R-independent and X_R remains a free final scale.
• Antecedent: None cited. The step (A.5) is the standard e^{-1/y^2} gluing function, used without citation.
• Cost: The ordering (A.6); auxiliary constants T_f (large), c_o (small), bump width .3; λ small enough that the four patches fit in (0, T_w); a very long logarithmic profile (roughly e^{M_d} + 13/λ + 90 log(1/λ) + O(log(1/h)) units) whose amplitude is exponentially small in 1/λ after the pulse. The inner reference branch (A.7) is not regular at the axis and must be replaced (Appendix B).
• Backward question: Can I write down, explicitly and stage by stage in log X, an outer profile that starts at the reference inner power law, ends at an η-independent power law c∞ X^{-A}, has enough independent scalar knobs to meet every total moment identity, stays inside the stress cone, and still leaves room for four later corrections?
• Checkable: Integrate (log E)' = l(y) - 1/2 along the schedule for sample parameters (work in log variables to avoid underflow) and compute the five cumulative integrals by quadrature in y. Check that k = 0 exactly on the last eleven units of [0, T_d] (since σ(log(1 + y)/M_d) = 1 iff y ≥ e^{M_d} - 1 = T_d - 11); that the patches lie in (0, T_w) iff T_w > 25; that E = c_patch f X^{-1/2-λ}, U = 0 on each patch. Digest-pass computation: ‖σ'‖_∞ = σ'(1/2) = 8, so e_d = 32/M_d, and T_f ≥ 10 (log 2) ‖σ'‖_∞ ≈ 55.5 suffices for -λ - .1 ≤ l ≤ -λ in (A.10) (there l = -λ + ϑ_f'(y)(log 2 - J_0) with ϑ_f' ≤ 0 and 0 ≤ log 2 - J_0 ≤ log 2).
• Depends on: M4.5 (its tuning targets the total cumulative integrals: U = M = J = 0 after the pulse and the identities (A.8) for S and the renormalized I.); M4.7 (it asserts the strict relaxed cone from x_- and the admissible cone from the intermediate power law onward, in Lemma 4.5's terms.); M4.4 (each stage prescribes l = D_X log H of (4.8), which fixes the shear a = 2 - 2l and makes Q_s, N_s of (4.9) explicit.)
• Refs: pp. 129 to 131, (A.5) to (A.13), Proposition A.4; used by Lemma 4.8, pp. 35 to 36.
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MA.5: Closing M = J = 0 after the axial pulse (Section A.3, first part)

Node ns-ma-5-closing-m-j-0-after-the-axial-pulse-section-a-3-first-part, kind move, pp. 130-131.

• Statement: Section A.3, closing M = J = 0 (pp. 130 to 131): on the pulse 0 ≤ y ≤ 13/λ, with l = −λ and U = E R_b, R_b = Amp(η) R_0(λy) + c_1β_1(y) + c_2β_2(y), R_0 the ramp φ_b[1 − σ(ξ_b − 10)], ξ_b = λy, and β_i identical width-.3 bumps centered at 13/λ − 3 and 13/λ − 1: given the pre-pulse bounds (A.14), e_b ≤ C_pre λ^{30} and ‖A_X(U)/E‖_{C^1_η} ≤ C_pre λ^{29}, there are end coefficients c_i, affine in Amp with ‖c_i‖_{C^1_η} ≤ C_pre e^{−c/λ}(1 + |Amp|), that give M = J = 0 after the pulse (A.15); every fixed y-derivative obeys the same bound.
• Obligation: M(∞) = J(∞) = 0 are two of the four identities (4.28). U = M = 0 beyond X_v gives V_0 = 0 by (4.7), so the exterior is purely azimuthal; M(∞) = 0 also makes the Stokes streamfunction vanish in the exterior (A = 0 for X ≥ X_ext in Theorem 3.1(iii), via (10.1)); J(∞) = 0 removes the angular-momentum transport integral in Lemma A.8.
• Mechanism: The pre-pulse debts are tiny in natural units: M is frozen after the axial reduction while XE grows like e^{(1/2-λ)y} and XHE like e^{(1/2-2λ)y} over T_w = 60 log(1/λ), and E itself drops by e^{-(1/2+λ)T_w} ≤ λ^{30}. The main pulse is cut off at ξ_b = 11 (y = 11/λ), at least 2/λ - 3 before the first bump center, so, normalized at that center, its contributions carry factors like e^{-s_i(2/λ - 3)} with s_1 = 1/2 - λ, s_2 = 1/2 - 2λ, both above .4. With E fixed, M and J are linear in U, so the correction is a linear 2 × 2 solve with weights e^{s_1 y}, e^{s_2 y}; its inverse is O(λ^{-1}) by (A.1), which the factor e^{-c/λ} absorbs.
• Antecedent: None cited.
• Cost: The gap of order 2/λ between the pulse cutoff and the end bumps; the end coefficients depend affinely on the still unknown Amp, which is fixed in MA.7.
• Backward question: After an axial pulse, how do I cancel the axial mass flux M and the axial transport of angular momentum J exactly when the only available weights are nearly degenerate (exponents differing by λ)?
• Checkable: For sample λ, form the 2 × 2 matrix of the weights e^{(1/2-λ)y}, e^{(1/2-2λ)y} against the two width-.3 bumps, check ‖B^{-1}‖ ≈ C λ^{-1}, compute the normalized contributions of the main pulse to M and J and confirm their e^{-c/λ} decay, and check e^{-(1/2+λ)·60 log(1/λ)} = λ^{30 + 60λ} ≤ λ^{30}.
• Depends on: MA.4 (works on the axial-pulse stage of the Section A.2 schedule, after the intermediate power law of length T_w = 60 log(1/λ).); MA.1 (the two end bumps solve a 2×2 system with weights e^{(1/2-λ)y}, e^{(1/2-2λ)y}, whose inverse is O(λ^{-1}) by (A.1).); M4.5 (the targets are the cumulative integrals M = ∫U and J = ∫UH of (4.15), linear in U once E is fixed.)
• Refs: pp. 130 to 131, (A.14), (A.15).
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MA.6: The renormalized angular-moment identity through the exact Q_s equation (Section A.3, second part)

Node ns-ma-6-the-renormalized-angular-moment-identity-through-the, kind move, pp. 130-132.

• Statement: Section A.3, angular identity (pp. 130 to 132): after the pulse U = 0; the interpolation (A.10) makes E η-independent, and two relative E bumps on the following 30 log(1/λ) hold impose (A.11), I = XH/(1 − λ) with ∫(E² − E²_unedited) dy = 0, coefficients O_{C^1_η}(C_pre λ^{28}). At the exterior transition Q_s = (λ − h)/(1 − λ) by (4.16); the l = −h hold runs until Q_s reaches Q_p of (A.13), and the terminal solution (A.16) vanishes from y = 3 on, so I = XH/(1 − h) beyond the tail, i.e. ∫_0^∞ (H − H_pow) dX = 0. The steep interval gives E ratio h^6 (A.17), and (A.18) bounds the release.
• Obligation: The renormalized angular identity in (A.8) and (4.28). Without it the integrated angular residual need not vanish and T_θ keeps an r^{-2} tail in the exterior, so Lemma A.8 fails. The same computation makes the reference inviscid angular stress vanish beyond the tail, and the h^6 reduction is what later makes w and T_z/T_θ small near the outer edge.
• Mechanism: The ratio r_I = I/(XH) obeys r_I' + (1 + l) r_I = 1, so on the η-independent hold it relaxes to 1/(1 - λ) at rate 1 - λ, leaving an O(λ^{28}) discrepancy after 30 log(1/λ) units; two relative bumps (I row slope 1 - λ, linearized pressure row slope -1 - 2λ, bounded inverse by (A.1)) cancel it exactly by Lemma A.2 while keeping ∫E^2 dy, hence Π_0, unchanged. Because E is now η-independent, Q_s follows an explicit scalar ODE: during l = -1 it grows linearly (Q_s' = 1 - h) to size about log(1/h), and during l = -h it decays at rate 1 - h. Since Q_p ≍ ρ_o ≍ h is smaller, the hold has a positive finite length, used as a shooting parameter so that Q_s hits Q_p exactly; the terminal source -f_o'/f_o ≤ 0 then drives Q_s to zero at y = 3. Beyond the tail H = H_pow ∝ X^{-h}, integrable at 0 since h < 1, so ∫_0^X H_pow = X H_pow/(1 - h) and I = XH/(1 - h) is the same as ∫_0^∞ (H - H_pow) dX = 0. This part uses E only and is independent of Amp.
• Antecedent: None cited (integrating-factor solution of a linear first-order ODE).
• Cost: A steep interval of length 4 log(1/h) and an h-dependent hold length; the constant c_o must make 0 ≤ f_o'/f_o < h/4.
• Backward question: How can the renormalized total angular momentum be made to vanish exactly and uniformly in η, when the moment is coupled to Q_s through an ODE in log X and the exterior must be η-independent?
• Checkable: Integrate Q' + (1 + l)Q = -l - h along the exterior schedule from Q = (λ - h)/(1 - λ), find the hold length at which Q = Q_p, verify Q_s(3) = 0 from (A.16), and confirm I(X) = XH/(1 - h) beyond the tail by direct quadrature of I. Check (A.17) as e^{2(-1)·4 log(1/h)} = h^8 and e^{(-3/2)·4 log(1/h)} = h^6. Digest-pass computation: with c_o = 0.05625 (so max f_o'/f_o = 0.225 h < h/4), Q_p/ρ_o = 7.415 for h = 10^{-3} and 7.430 for h = 10^{-6}, confirming Q_p ≍ ρ_o ≍ h.
• Depends on: MA.4 (tunes the interpolation (A.10), the hold with two bumps (A.11), and the l = -h hold length in the Section A.2 schedule.); M4.5 (with U = M = J = 0 and E η-independent, (4.16) gives Q_s = -1 + (1 - h)I/(XH), so Q_s = 0 beyond the tail means I = XH/(1 - h).); MA.5 (starts from U = M = J = 0 after the pulse, which turns Q_s into the solution of a scalar ODE in log X.); MA.2 (the two relative E bumps meet (A.11) exactly while keeping ∫E²dy, hence Π0, by Lemma A.2's contraction.)
• Refs: p. 130, (A.10) to (A.13); pp. 131 to 132, (A.16) to (A.18).
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MA.7: S(∞) = 0 by a scalar root in the pulse amplitude (Section A.3, third part)

Node ns-ma-7-s-0-by-a-scalar-root-in-the-pulse-amplitude-section-a-3, kind move, pp. 132-133.

• Statement: With M, J, I fixed independently of Amp, (A.19) holds: λ S(∞)/(X_p e_b^2 f^2) = Amp^2 K_b - (1 - e^{-26})/4 + E(Amp, η), K_b = ∫_0^{13} e^{-2ξ_b} R_0(ξ_b)^2 dξ_b, with ‖E‖_{C^1} ≤ C_pre λ(1 + log(1/λ)) on [.9, 1.2] × [-1, 1]. Since .20 < K_b ≤ 1/4, the principal expression is below -.047 at .9, above .038 at 1.2, and has amplitude derivative at least .36, so for small λ there is a unique smooth root with |∂_η Amp| ≤ C_pre λ(1 + log(1/λ)) (A.20). Both equalities of (A.8) then hold.
• Obligation: S(∞) = 0 in (A.8) and (4.28). Without it the axial momentum flux including pressure does not integrate to zero and T_z keeps an r^{-1} tail in the exterior (Lemma A.8 fails). The small η-derivative (A.20) is reused in the cone check on the pulse, where it makes the η m_η term negligible.
• Mechanism: On the pulse U^2 - E^2/2 = E^2(R_b^2 - 1/2) and XE^2 = X_p e_b^2 f^2 e^{-2λy}, so in ξ_b = λy the pulse contributes (X_p e_b^2 f^2/λ)[Amp^2 K_b - (1 - e^{-26})/4] up to exponentially small end-bump terms. Everything else (the pre-pulse debt (A.14), the interpolation and hold stages, and the release integral (A.18), all either carrying the factor e^{-26} or lacking the 1/λ length) is of relative size λ(1 + log(1/λ)). So at leading order the pulse's axial term must balance its own azimuthal term, a monotone scalar equation in Amp. The bounds on K_b follow from R_0 ≤ ξ_b (giving ∫_0^∞ ξ^2 e^{-2ξ} dξ = 1/4) and R_0 ≥ ξ_b - .02 on [.02, 9].
• Antecedent: None cited (a monotone scalar root with implicit differentiation).
• Cost: λ small enough that E(Amp, η) stays within the bracket margins; the pulse makes U/E as large as about 10·Amp, which the cone check (MA.9) must tolerate.
• Backward question: After M, J and I are fixed, which single scalar knob controls the last total moment S(∞), and is its equation monotone enough to solve uniquely with small η-derivatives?
• Checkable: Evaluate K_b by quadrature, using φ_b(ξ) = ξ - .01 for ξ ≥ .02 (because ∫_0^1 σ = 1/2). Digest-pass computation: K_b = 0.245050; principal expression -0.0515 at .9 and 0.1029 at 1.2; root Amp_0 = ((1 - e^{-26})/(4 K_b))^{1/2} = 1.01005; worst-case bracket values .81/4 - 1/4 = -.0475, 1.44(.20) - .25 = .038, slope 2(.9)(.20) = .36, matching the paper.
• Depends on: MA.5 (Amp scales the axial pulse R_b of Section A.3, whose end bumps already give M = J = 0 affinely in Amp.); MA.6 (the angular identity uses E only, so I is fixed independently of Amp and S(∞) is the one remaining moment.); M4.5 (S = ∫(U² - E²/2) of (4.15) is the moment set to zero; on the pulse U² - E²/2 = E²(R_b² - 1/2).)
• Refs: pp. 132 to 133, (A.19), (A.20).
• Verification: statement digest-only; hypotheses not-checked; computation checked False

MA.8: The analytic axis pressure datum fixed from the outer schedule (Lemma A.5)

Node ns-ma-8-the-analytic-axis-pressure-datum-fixed-from-the-outer, kind move, pp. 133-134.

• Statement: Let E_{id,sched} be the inner reference (A.7) followed by the Section A.2 outer profile with the pressure-preserving angular bumps omitted. Then Π_0(η) = -(1/2) ∫_{-∞}^{∞} E_{id,sched}(y, η)^2 dy (A.21) is independent of X_R, analytic on a complex neighborhood of [-1, 1], even, has Π_0' with the sign of η, and satisfies Π_0 ≤ -(5/2) P∗^2 f(η)^2 (A.22); for the complete reference profile, Π(X, η) = -(1/2) ∫_{log(X/X_R)}^∞ E(v, η)^2 dv (A.23).
• Obligation: The regular axis construction (Proposition B.2) needs its pressure value at X = 0 before the axis profile exists, analytic near [-1, 1] and with the sign and size properties used on p. 144 to make Z∗(η_0) ≥ c j_0 P∗^2 > 0 at the zero η_0 of H∗. The final profile must also satisfy the normalization (4.25). The lemma breaks this circularity: the datum depends only on the outer schedule, and later edits are required to preserve it.
• Mechanism: Since Π = Π_0 + C_p with C_p = (1/2)∫E^2 dy, pressure vanishing at infinity forces Π_0 = -C_p(∞). Without the angular bumps every piece of E has the form c(y) f(η)^{ϑ(y)}, 0 ≤ ϑ ≤ 1, with c and all transition lengths independent of η (ϑ = 1 through the pulse, decreasing to 0 in the interpolation, 0 beyond). On a simply connected complex neighborhood avoiding the poles and zeros of f, f^ϑ = exp(ϑ log f) is holomorphic uniformly in ϑ, and the integral converges uniformly (bounded by C e^{y/5} as y → -∞ and C e^{-(1+2h)y} as y → +∞), so Π_0 is analytic. Evenness is inherited from f. From ∂_η f^{2ϑ} = -2ϑ J_0' f^{2ϑ} with J_0' = 2η/(1 + η^2), every contribution to Π_0' has the sign of η, strictly so from the reference part, which contributes exactly -(5/2) P∗^2 f^2. No X_R enters the log-coordinate schedule, and the angular bumps change ∫E^2 dy by zero, which gives (A.23). More generally, an edit with zero total ∫(E_new^2 - E_old^2)/(2X) dX leaves the forward pressure unchanged before and after its support (not inside it), which is why pressure-neutral edits may be omitted in the backward integral (A.23).
• Antecedent: None cited (uniform integration of holomorphic functions).
• Cost: A standing constraint on every later edit of E: its total pressure increment must be zero or be restored by a five-moment match. Non-analytic edits (the heat factor) are allowed only under that constraint.
• Backward question: The axis problem needs its pressure value at X = 0 before the inner profile exists; can that datum be fixed from the outer profile alone, analytic in η, and kept invariant under every later edit?
• Checkable: Compute Π_0(η) by quadrature of the schedule; check evenness, the sign of Π_0', the bound (A.22), and that the reference piece equals -(1/2) P∗^2 f^2 ∫_{-∞}^0 e^{y/5} dy = -(5/2) P∗^2 f^2 (consistent with C_{p,0} = (5/2) P∗^2 f^2 x^{1/5} in (4.34)); evaluate at complex η near [-1, 1] to confirm analyticity.
• Depends on: MA.4 (the datum integrates E² along the Section A.2 schedule, whose pieces c(y)f(η)^{ϑ(y)} have η-independent transition lengths.); M4.5 (Π = Π(0, η) + Cp with Cp = ∫E²/(2x), so pressure vanishing at infinity forces Π0 = -Cp(∞).); MA.6 (the angular bumps of (A.11) leave ∫E²dy unchanged, so omitting them gives the same datum and (A.23) for the full profile.)
• Refs: pp. 133 to 134, Lemma A.5, (A.21) to (A.23); used p. 35 ((4.31)), p. 37, p. 144, p. 150.
• Verification: statement digest-only; hypotheses not-checked; computation checked False

MA.9: Stage-by-stage cone verification and the large-X_R scaling (Section A.5, completing Proposition A.4)

Node ns-ma-9-stage-by-stage-cone-verification-and-the-large-x-r, kind move, pp. 134-137.

• Statement: Section A.5, completing Proposition A.4 (pp. 134 to 137): with w = N_s/(E Q_s) where Q_s > 0, the sufficient test (A.24) of Lemma 4.5, a − b_s w > 0 and 2b_s w + b_s²/a + (a − 2)w² < 2, holds with uniform margins at each stage, while p_{s,1} = XQ_s/L grows with X_R. Reference (A.25): Q_s ≥ c > 0, b_s = 0, a = 4/5; axial reduction (A.26): |b_s w| ≤ C e_d, a = 2; intermediate (A.27): √λ|w| = o(1), a = 2 + 2λ; pulse (A.28) to (A.30): b_s w < .74, the quadratic < 1.68, v_s ≥ a > 2; then (A.31): |w| ≪ 1, b_s = 0, 2 < a ≤ 4. Where a ≤ 2 only the relaxed cone results.
• Obligation: The cone assertions of Proposition A.4, which become Lemma 4.8(iii) (relaxed on [e^{-5} X_R, e^{1/2} X_tail], admissible on [X_good, e^{1/2} X_tail]); Proposition 7.5 needs the admissible cone to realize the stress with positive squared wave amplitudes.
• Mechanism: Under (A.4) the shears a, b_s (logarithmic derivatives), the ratio w and Q_s are X_R-invariant, while p_{s,1} = X Q_s/L grows linearly in X_R; Lemma 4.5 then reduces the admissible cone at large p_{s,1} to (A.24) on a compact set of (a, b_s, w). Each stage is handled by solving the linear equations (4.9) for Q_s, N_s explicitly or by (4.16): on the axial reduction the slow decay of k makes b_s w small (M_d large) and P∗ > e^{T_d} makes b_s^2 small; on the intermediate interval a = 2 + 2λ > 2 and (a - 2) w^2 = 2λ w^2 = o(1); on the pulse m = A_X(U)/E solves m' + βm = R_b with β = 1/2 - λ, and a two-term expansion of the exponential convolution (A.28) together with (A.29), (A.30) gives w ≈ 2R_b - C_d ∂_{ξ_b} R_b, so, with R_b ≥ 0 and ∂_{ξ_b} R_b ≤ 1.2, both cone quantities are bounded by explicit quadratics in R = R_b; after the pulse E is exponentially small in 1/λ, and after the steep interval it is further cut by h^6, while Q_s ≥ cλ or ch, so w is negligible. Where a ≤ 2 (the reference and the first transition), only the relaxed condition results.
• Antecedent: None cited (variation of constants; Taylor's formula for the convolution remainder).
• Cost: The orderings M_d large (e_d small), P∗ > e^{T_d}, h ≪ e^{-T_d} (used in (A.26)), h/λ → 0 (used for C_d ≤ 2 + o(1)), and one final increase of X_R. The admissible condition is not obtained where a ≤ 2; that gap is left to Appendix C.
• Backward question: Along each stage of the outer profile, does the integrated inviscid stress p_s stay inside the admissible cone around the shear direction once the radial scale X_R is large, and which stages can only give the relaxed cone?
• Checkable: Verify (A.25) symbolically from (4.9) with U = 4η, E ∝ f x^{1/10} (so l = 3/5, W = -(4L - 1), Q_s = S_q/(1 + l)); check C_d → 2 at c_η = 0 as λ, h/λ → 0 and that C_d decreases in c_η. Digest-pass computation: sup_{R≥0}(-2R^2 + 2.41R) = 0.72601 < .74, sup_{R≥0}(-3.5R^2 + 4.82R) = 1.65946 < 1.68, and max ∂_ξ R_0 = 1.0, so ∂_{ξ_b} R_b ≤ 1.2 for Amp ≤ 1.2. A full check integrates (4.9) along the numerically built schedule and evaluates the gap map Ψ of (4.35) with p_{s,1} scaled by X_R.
• Depends on: M4.7 (applies Lemma 4.5's sufficient test (A.24) on compact (a, b_s, w) ranges, then makes p_s,1 = XQ_s/L large through X_R.); MA.4 (checks each stage of the schedule: reference, axial reduction, intermediate power law, pulse, interpolation, exterior transition.); M4.4 (solves (4.9) for Q_s, N_s stage by stage and uses the shear a = 2 - 2l, b_s = 2D_XU/E of (4.11).); MA.7 (on the pulse, the small η-derivative (A.20) of the amplitude makes the ηm_η term negligible in N_s/E.)
• Refs: pp. 134 to 137, (A.24) to (A.31); Proposition A.4 pp. 130 to 131; Lemma 4.5 pp. 31 to 32.
• Verification: statement completeness audit 2026-10-01: TRUNCATED; statement replaced from the digest; hypotheses not-checked; computation checked False

MA.10: The self-similar swirl heat exterior (Lemma A.6)

Node ns-ma-10-the-self-similar-swirl-heat-exterior-lemma-a-6, kind move, pp. 137-138.

• Statement: With a_K = 1 + h, the heat factor H(Z) = Γ(a_K)^{-1} ∫_0^∞ e^{-v} v^{a_K - 1} (1 + Zv)^{-h} dv, Z ≥ 0 (A.32), has H(0) = 1, is positive and smooth up to Z = 0 with every fixed derivative bounded on bounded intervals, and K(r, t) = c∞ s^{-A} H(2τ/s), s = r^2/2 (A.33), is independent of z, satisfies ∂_t K = (∂_rr + r^{-1}∂_r - r^{-2})K, and has K_r < 0; its profile E_pow(X) H(2d/X) is smooth with all η-derivatives continuous up to η = ±1.
• Obligation: The exterior must have identically zero Navier-Stokes residual (so that T_0 = 0 there and no force is needed outside the annulus) and smooth limits of every derivative as t ↑ 1 at each fixed r > 0: Theorem 3.1(iii), formula (3.5), proved on p. 116 with (A.34), and Lemma 10.2, (10.7) and (10.8). The pure power law c∞ s^{-A} is not a steady swirl solution and leaves a viscous residual outside the annulus.
• Mechanism: Inserting the self-similar ansatz with Z = 2τ/s into the swirl heat equation gives ∂_t K = -2c∞ s^{-A-1} H' and (∂_rr + r^{-1}∂_r - r^{-2})K = 2c∞ s^{-A-1}[Z^2 H'' + (2A + 1) Z H' + (A^2 - 1/4) H], which reduces to (A.37) because 2A + 1 = 2a_K and A^2 - 1/4 = a_K(a_K - 1). The Laplace-type integral solves (A.37): integrating the total derivative h ∂_v[e^{-v} v^{a_K} (1 + Zv)^{-a_K}] produces the equation with vanishing boundary terms. H(0) = 1 means the flow arrives exactly at the prescribed power law at t = 1. Differentiation under the integral, with majorant e^{-v} v^{h+m} (since (1 + Zv)^{-h-m} ≤ 1), bounds all derivatives uniformly on Z ≥ 0, which is the source of the smooth one-sided limits at τ = 0. Writing -Z H'/H as an average of hZv/(1 + Zv) against a positive density gives the monotonicity. In profile variables Z = 2(1 - η^2)/X, so η-derivatives produce only polynomials in d' = -2η, d'' = -2 times H^{(k)}(2d/X)(2/X)^k, with no division by d. By (A.35) the Taylor coefficients (h)_m (1 + h)_m/m! grow factorially, so the Taylor series at Z = 0 diverges; the paper uses only finite Taylor formulas ("no convergent Taylor series is needed"), treats the heat factor as merely smooth in η, and keeps it out of the analytic axis datum (MA.8, MA.11).
• Antecedent: None cited. The proof names the gamma integral, dominated differentiation and the finite Taylor formula.
• Cost: h > 0 ties the exterior decay X^{-A}, A = 1/2 + h, to the heat solution; the exterior profile depends on η through d = 1 - η^2 and is smooth but not analytic at η = ±1; the replacement disturbs three moments that must be restored (MA.11).
• Backward question: Is there an exact, z-independent swirl heat solution that equals the power law c∞ s^{-A} at t = 1, is smooth in time up to t = 1 at every fixed r > 0, decreases in r, and matches the profile power law to O(X^{-1}) at large X?
• Checkable: Evaluate H by quadrature and check (A.35), (A.36), (A.37), 0 ≤ -ZH'/H < h, and the swirl heat equation for K by finite differences. Digest-pass computation (mpmath, h = 0.005, 0.01, 0.2): (A.35) matches for m ≤ 3; the (A.37) residual is below 10^{-31} at Z = 10^{-3}, 0.1, 1, 10; -ZH'/H < h at all those points; (|H - 1| + |ZH'|)/(hZ) lies between 1.28 and 2.36 at Z = .01, .1, .5; the swirl heat residual of K is below 10^{-38} relative at four (r, t) points, with r K_r/(2K) in (-A, -1/2). An independent evaluation route, not stated in the manuscript and confirmed here to 12 digits: H(Z) = Z^{-1-h} U(1 + h, 2, 1/Z), with U the Tricomi confluent hypergeometric function.
• Depends on: MA.4 (replaces the schedule's η-independent exterior power law E_pow = c∞X^{-A} by a heat solution that reaches the same power at t = 1.); M4.1 (in similarity variables Z = 2τ/s = 2d/X, so the physical swirl K is the profile E_pow(X)H(2d/X), smooth up to η = ±1.)
• Refs: pp. 137 to 138, (A.32) to (A.38); used p. 33 (4.29), p. 35 (Lemma 4.8(iii)), p. 116 ((3.5)), p. 119 ((10.7), (10.8)).
• Verification: statement astra-spot-check-2026-10-01; hypotheses astra-spot-check-2026-10-01; computation checked False; Astra spot-check: correct

MA.11: Heat replacement and three-moment compensation on the second patch (Proposition A.7)

Node ns-ma-11-heat-replacement-and-three-moment-compensation-on-the, kind move, pp. 137.

• Statement: With X_K = e^{.2} X_tail and a smooth step 0 ≤ χ_K(y) ≤ 1 equal to 0 for y ≤ .2 and 1 for y ≥ .5, the replacement E_cl ↦ E_cl[1 + χ_K(y)(H(2d/X) - 1)] (A.39) can be compensated by three additive E bumps on the second reserved patch; the complete edit preserves M, J, S(∞), C_p(∞) and ∫_0^∞ (H - H_pow) dX pointwise in η, and, for large X_R, preserves E > 0 and the strict admissible cone from the patch through y = .5; the axis datum Π_0 and the profile before the patch are unchanged.
• Obligation: Lemma 4.8(iii) (the heat exterior) together with parts (i) and (ii) (the unchanged datum and exact moments). Without compensation the heat factor would shift C_p(∞) (hence Π_0, destroying the analytic datum of Appendix B and the normalization (4.25)), S(∞) and the angular moment (reintroducing stress tails).
• Mechanism: By (A.34) to (A.36), |(X∂_X)^j ∂_η^m (E - E_cl)| ≤ C e_K X_K^{-1} x^{-A-1} for x = X/X_K ≥ 1, bounded even at d = 0. Both edited regions have U = 0, so M and J are untouched; the other three discrepancies obey (A.40), (A.41), (A.42), the last using ∫_1^∞ x^{-A-1/2} dx = 1/h, finite because h is fixed before X_R. Normalized by (A.43) (e∗^2, X∗ e∗^2, X∗^{3/2} e∗) each is O_m(X_K^{-1}). On the patch E = e∗ f x∗^{-1/2-λ}, so the pressure, S and angular rows have weights f x∗^{-3/2-λ}, -f x∗^{-1/2-λ}, √2 x∗^{1/2}, with distinct exponents; Lemma A.1 gives an inverse uniform in η and Lemma A.2 gives ‖∂_η^m c‖_∞ ≤ C_m X_K^{-1} (the first two rows carry quadratic remainders, the last is linear). The normalized formulas (4.16) contain no X_R, so Q_s, N_s change by O(X_K^{-1}) at the cost of one η-derivative, and the cone margins of Proposition A.4 survive for large X_R. Zero net pressure change means forward integration from the unchanged Π_0 still gives pressure vanishing at infinity.
• Antecedent: None cited.
• Cost: A further largeness requirement on X_R. The compensation coefficients depend on η through the heat factor, which is only smooth in η; this is harmless because analyticity is required only on the axis rectangle [0, X_an] and Π_0 is unchanged.
• Backward question: If I swap the power-law tail for the exact heat flow, which cumulative moments change, by how much in natural units, and can I restore them upstream without touching the axis pressure datum or the cone?
• Checkable: Compute the three discrepancies of (A.39) by quadrature for sample (h, X_K), divide by (A.43), solve the 3 × 3 system with weights f x∗^{-3/2-λ}, -f x∗^{-1/2-λ}, √2 x∗^{1/2} plus the quadratic terms, and confirm that c = O(X_K^{-1}) and that the totals C_p(∞), S(∞), ∫(H - H_pow) dX agree with the reference profile afterward.
• Depends on: MA.10 (the heat-factor bounds (A.34) to (A.36) make the replacement's changes to Cp(∞), S(∞), and the angular moment O(X_K^{-1}).); MA.3 (three E bumps on a power-law patch with U = 0 reset (I, S, Cp) while preserving M and J, by Corollary A.3.); MA.4 (the compensation sits on the schedule's second reserved patch; the replacement acts on the tail beyond X_K = e^{.2}X_tail.); MA.9 (the stage cone margins of Proposition A.4 absorb the O(X_K^{-1}) changes in Q_s, N_s for large X_R.)
• Refs: p. 137, pp. 139 to 140, (A.39) to (A.43).
• Verification: statement digest-only; hypotheses not-checked; computation checked False

MA.12: Zero total residual integrals and the backward stress formula (Lemma A.8)

Node ns-ma-12-zero-total-residual-integrals-and-the-backward-stress, kind move, pp. 140-141.

• Statement: Suppose the leading axisymmetric field is smooth at the axis, has U = 0 and E = E_pow f_o[1 + χ_K(H(2d/X) - 1)] for y ≥ 0, and satisfies M(∞) = J(∞) = S(∞) = 0, ∫_0^∞ (H - H_pow) dX = 0 and Π(X) = -(1/2) ∫_X^∞ E(x)^2/x dx. Then ∫_0^∞ r^2 R^{(0)}_θ dr = ∫_0^∞ r R^{(0)}_z dr = 0, the stress is given by the backward integrals T_θ(r) = r^{-2} ∫_r^∞ r'^2 R^{(0)}_θ(r') dr', T_z(r) = r^{-1} ∫_r^∞ r' R^{(0)}_z(r') dr' (A.46), and it vanishes for X ≥ X_b. Restated as Lemma 4.9.
• Obligation: Theorem 4.6(ii): T_0 = 0 for X ≥ X_b. The stress is defined by integration from the axis (Proposition 4.2), so a zero exterior residual alone would still leave stresses proportional to r^{-2} (angular) and r^{-1} (axial) whenever the total weighted residual integrals are nonzero (p. 30).
• Mechanism: Integrate the conservative forms of the leading tangential residuals (those displayed in Lemma 5.2, Step 4, without the axial viscosity terms) over 0 < r < ∞. By p. 30, M, J, S are the radially integrated axial momentum, the axial transport of angular momentum, and the axial momentum flux including pressure. The time and axial derivatives fall on q^{3/2-A} ∫(H - H_pow) dX (A.45), q^{3/2-2A} J(∞), q^{1-A} M(∞) and q^{1-2A} S(∞), the last via ∫_0^∞ r p dr = -(1/2) ∫_0^∞ r' u_θ^2 dr' (Fubini), and all vanish by hypothesis. The radial boundary terms vanish by axis regularity and the exterior decay (A.44), K - K_pow = O_h(τ r^{-3-2h}), ∂_t K = O_h(r^{-3-2h}), r^2 K_r - rK = O_h(r^{-2h}): the viscous flux [r^2 ∂_r u_θ - r u_θ] tends to zero at infinity and the subtracted angular moment converges (majorant r^{-1-2h}) precisely because h > 0; the axial pressure moment converges since p = O(r^{-2-4h}). So forward and backward primitives coincide, and beyond X_b the field (0, K, 0) with centrifugal pressure has zero residual by Lemma A.6, giving T = 0 there. Once the regular axis and the moments are supplied, the stress depends only on the exterior profile.
• Antecedent: None cited (integration of conservation forms; Fubini).
• Cost: It is conditional on a regular axis and exact moment identities, which the outer profile alone cannot supply because the reference branch (A.7) is not regular; they come from Proposition B.2 and Corollary B.10 (with Proposition B.8). It also needs h > 0.
• Backward question: Once the exterior residual is zero, why should a stress defined by integrating from the axis vanish there, and which conserved totals must vanish to exclude r^{-2} and r^{-1} tails?
• Checkable: For a sample profile satisfying the hypotheses, compute R^{(0)}_θ and R^{(0)}_z (on the terminal collar they are (A.52) and p_z of (A.53)) and check numerically that the forward and backward formulas in (A.46) agree; check the scalings (A.44) from (A.36) and the tail integral ∫_{R∗}^∞ r^{-1-2h} dr = R∗^{-2h}/(2h).
• Depends on: M4.4 (the stress is Proposition 4.2's forward primitive from the axis; the lemma shows it equals the backward primitive (A.46).); M4.5 (the derivatives of the weighted residual integrals fall on M, J, S and the renormalized I, which vanish by hypothesis.); MA.11 (its hypothesis profile E = E_pow f_o[1 + χ_K(H(2d/X) - 1)] for y ≥ 0 is the heat-replaced terminal profile of Proposition A.7.); MA.10 (beyond X_b the field (0, K, 0) of Lemma A.6 has zero residual, and its heat-factor bounds give the decay (A.44) at infinity.)
• Refs: pp. 140 to 141, Lemma A.8, (A.44) to (A.46); conservative forms p. 53; restated p. 36 (Lemma 4.9).
• Verification: statement digest-only; hypotheses not-checked; computation checked False

MA.13: Factoring a flat edge weight out of a backward integral (Lemma A.9)

Node ns-ma-13-factoring-a-flat-edge-weight-out-of-a-backward-integral, kind move, pp. 141-142.

• Statement: For c > 0, a fixed integer j ≥ 0 and smooth b(u, η) on [0, δ_0] × K, ∫_0^δ e^{-c/u^2} u^{-j} b(u, η) du = (1/2) e^{-c/δ^2} δ^{3-j} B(δ, η) (A.47), with B smooth on [0, δ_0] × K, B(0, η) = b(0, η)/c, and bounded mixed derivatives of every fixed order; the same holds with any finite set of smooth compact parameters.
• Obligation: The stress vanishes to infinite order at both edges, so its direction and the weighted bounds (4.27) require factoring out the flat weight with a smooth nonvanishing remainder. Used at the outer edge (Proposition A.10, c = 4, j = 3 and j = 0), at the inner collar (p. 152, c = t_1^2, j = 0) and for the order-one outer term (5.22) in Lemma 5.2 (c = 4, j = 3, 6).
• Mechanism: The substitution u = δ/(1 + δ^2 v)^{1/2} maps (0, δ] onto v ∈ [0, ∞), turns c/u^2 into c/δ^2 + cv, and has Jacobian -(1/2) δ^3 (1 + δ^2 v)^{-3/2}, so B(δ, η) = ∫_0^∞ e^{-cv} (1 + δ^2 v)^{(j-3)/2} b(δ/(1 + δ^2 v)^{1/2}, η) dv, a Laplace-type integral whose derivatives are dominated by e^{-cv} times polynomials in v. Integrating a flat factor from the edge thus returns the same flat factor with three more powers of δ.
• Antecedent: None cited (change of variables and dominated differentiation).
• Cost: None beyond smoothness of b.
• Backward question: The stress and its sources vanish to infinite order at the outer edge; how can the relative rates of the two stress components, and hence the limiting direction, still be computed?
• Checkable: Digest-pass computation: (A.47) with b(u) = 1 + 3u + sin^2 u, c = 4, j ∈ {0, 3, 6}, δ ∈ {.05, .2} agrees to 50 digits when the left side is integrated in the variable w = c/u^2 - c/δ^2 (plain quadrature in u loses accuracy for δ ≤ .2 because the integrand is concentrated at u = δ), and B(δ) → b(0)/c as δ → 0.
• Refs: pp. 141 to 142, Lemma A.9, (A.47); used p. 54 ((5.22)), p. 143, p. 152.
• Verification: statement digest-only; hypotheses not-checked; computation checked False

MA.14: Outer-edge stress factorization and limiting direction (Proposition A.10)

Node ns-ma-14-outer-edge-stress-factorization-and-limiting-direction, kind move, pp. 142-144.

• Statement: Proposition A.10 (pp. 142 to 144): under the hypotheses of Lemma A.8, the terminal profile satisfies the admissible cone on .5 ≤ y < 3; its stress direction extends smoothly to the outer endpoint and 2 − (a − 2)(T_z/T_θ)² has a uniform positive lower bound; for δ = 3 − y small, T_{0,θ} = e^{−4/δ²} δ^{−3} b_θ(δ, η), b_θ(0, η) > 0 (A.48), T_{0,z} = e^{−4/δ²} δ³ b_z(δ, η) (A.49), T_{0,z}/T_{0,θ} = δ^6 b_z/b_θ → 0 (A.50), and |∂^I T_0| ≤ C_I e^{−4/δ²} δ^{−N_I}, |T_0| ≥ c e^{−4/δ²} δ^{−3} (A.51), constants independent of q.
• Obligation: Theorem 4.6(iii) at X = X_b (n(X_b, η) = (1, 0), b_s(X_b, η) = 0, strict margin κ in (4.26)) and the outer half of Theorem 4.6(iv) (the factor e^{-4/y_b^2} in ζ, finalized as (C.18), and the bounds (4.27)). The wave construction needs a smooth stress direction strictly inside the cone up to the edge where the stress itself tends to zero (p. 32).
• Mechanism: On y ≥ 1/2 one has u_θ = K f with f = f_o(y) and u_r = u_z = 0. Since K solves the swirl heat equation, the residual (A.52) R^{(0)}_θ = K f'/(qL) - K(f_rr + r^{-1} f_r) - 2 K_r f_r and the axial pressure gradient (A.53) come only from derivatives of f, whose argument depends on (z, t) through q (y_t = (qL)^{-1}, y_z = -2η/(q^D L)). Integrating the viscous terms by parts in (A.46) gives (A.54), a sum of three nonnegative terms because K > 0, K_r < 0, f' ≥ 0; the last yields T_θ ≥ c r K ρ_o ψ_o(y)/(qL) > 0. The axial stress comes from the pressure, quadratic in the small amplitude, so |T_z/T_θ| ≤ C q^{1-D} K = C E_pow H(2d/X) (A.55), small by the h^6 reduction; with b_s = 0 and 2 + h < a ≤ 2 + 2h (A.56), the cone reduces to (a - 2)(T_z/T_θ)^2 < 2 with fixed slack. At the endpoint ψ_o = e^{-4/δ^2} g(δ), g(δ) = 1/(e^{-(1-δ/2)^{-2}} + e^{-4/δ^2}), g(0) = e, so f' = ρ_o e^{-4/δ^2} δ^{-3}[8g(δ) + δ^3 g'(δ)]. The boundary term of (A.54), rescaled as q^{A+1/2} K f_r = 2 E_pow(X) H(2d/X) f'/√(2X), has exactly the rate e^{-4/δ^2} δ^{-3} with b_θ(0, η) = 16 ρ_o E_pow(X_b) H(2d/X_b) g(0)/√(2X_b) > 0; the two integral terms gain δ^3 by Lemma A.9 (c = 4, j = 3). For the axial component, p_z is itself an integral of K^2 f f', which the same identity turns into e^{-4/δ^2} times a smooth coefficient, and the backward integral for T_z (Lemma A.9 with j = 0) then gives e^{-4/δ^2} δ^3; the powers of q cancel as q^{A+1/2} q^{1/2-D-2A} = q^{1-D-A} = 1. The limiting direction (1, 0) is the cone axis when b_s = 0, where the normalized quadratic expression equals 2.
• Antecedent: None cited.
• Cost: Requires c_o small (f_o'/f_o < h/4), X_R large (-ZH'/H < h/4 on the collar), h small, and a positive minimum of H on [0, 2/X_b]. It fixes the exponent 4 in the outer factor of the weight ζ = exp(-t_1^2/y_a^2 - 4/y_b^2) of (C.18).
• Backward question: At the outer edge, where the stress vanishes to infinite order, what is its limiting direction, and can the terminal amplitude factor be designed so that this direction sits strictly inside the admissible cone with a positive angular component?
• Checkable: Digest-pass computation: ψ_o(3 - δ) = e^{-4/δ^2} g(δ) and -∂_y ψ_o = e^{-4/δ^2} δ^{-3}[8g + δ^3 g'] agree to 15 digits at δ = .3, .6, and g(δ) → e. A fuller check evaluates T_θ from (A.54) and T_z from (A.53) and (A.46) for sample (h, c_o, X_R, q) and fits the rates δ^{-3} and δ^3 after dividing by e^{-4/δ^2}, comparing the leading coefficient with 16 ρ_o E_pow(X_b) H(2d/X_b) e/√(2X_b) and checking that the ratio is independent of q.
• Depends on: MA.12 (evaluates on the terminal collar the backward stress integrals (A.46) that Lemma A.8 makes valid.); MA.13 (Lemma A.9 with c = 4 and j = 3, 0 factors e^{-4/δ²} out of the integral terms, giving the rates δ^{-3} and δ³.); MA.10 (K solves the swirl heat equation, so the residual comes only from derivatives of f_o; K > 0 and K_r < 0 make the angular stress positive.); MA.4 (the terminal factor f_o = 1 - ρ_o ψ_o of (A.12), with 0 ≤ f_o'/f_o < h/4, fixes the flat rate e^{-4/δ²} at the endpoint.)
• Refs: pp. 142 to 144, Proposition A.10, (A.48) to (A.56); restated p. 36 ((4.32)); used p. 44, pp. 163 to 164.
• Verification: statement completeness audit 2026-10-01: TRUNCATED; statement replaced from the digest; hypotheses not-checked; computation checked False