Other material · A dividing-plane barrier in the OpenAI forced Navier-Stokes blow-up construction
The ledger as data, version 1.0, September 30, 2026
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{
"lane": "navier-stokes-blowup",
"generated_utc": "2026-09-30T21:01:23Z",
"registration": "docs/DESIGN-BLOWUP-PULLBACK.md",
"sources": {
"manuscript": {
"title": "OpenAI 2026, Finite Time Blowup for Navier-Stokes",
"sha256": "0e779481c4da40bd28d1e642e1d8ca57447d129610df28dfa5a11e9af8ae228f",
"pages": 166
},
"euler": {
"title": "OpenAI 2026, Finite Time Blowup for the Euler Equation",
"sha256": "a0c234518e6c489e16996805023eb2e75c00b7c03455f7a3a5be2c124954bfdd",
"pages": 57
}
},
"entries": [
{
"id": "T",
"kind": "endpoint",
"name": "ns-forced-blowup-theorem-1-1",
"title": "Theorem 1.1: forced finite-time blow-up from rest at every viscosity",
"section": "1",
"pages": "1",
"refs": [
"Theorem 1.1",
"(1.1)",
"Corollary 10.6"
],
"statement": "Theorem 1.1 (OpenAI 2026): for every nu > 0 there exist a force f in C_c^infinity(R^3 x (0, infinity)), a compact set K, and smooth u, p on R^3 x [0, 1) solving Navier-Stokes (1.1) from rest, supported in K for every t < 1, with sup_t ||u(t)||_{L^2} finite and limsup_{t -> 1} ||u(t)||_{L^infinity} = infinity. Hence no global smooth bounded-energy solution for that force and datum: alternative (C); (D) on the torus by compact support (Cor. 10.6).",
"obligation": "The endpoint. It rests on the localization and comparison steps of section 10 and, through them, on the whole construction.",
"backward_question": "What is the weakest thing one must build to refute global regularity under a smooth force: a flow whose residual is smooth through the singular time while its velocity is not.",
"mechanism": "",
"antecedent": "Fefferman's problem statement [13]",
"cost": "",
"checkable": "none: the theorem statement",
"depends_on": [
"M10.2",
"M10.5",
"M10.6",
"M10.9",
"M10.12",
"M10.10",
"M10.11"
],
"constrains": [],
"reasons": {
"M10.2": "Supplies the smooth, exactly divergence-free u, p from rest with fixed compact support K, agreeing near (0, 1) with the local blowup field.",
"M10.5": "Supplies the force f ∈ C_c^∞(R³ × (0, ∞)) that extends the residual through t = 1.",
"M10.6": "Gives sup_t ‖u(t)‖_{L²} < ∞ from the equation and the smooth force.",
"M10.9": "The comparison Lemma 10.5 excludes a global smooth bounded-energy solution with the same force and datum.",
"M10.12": "Corollary 10.6 gives alternative (D) on the torus from the compact support.",
"M10.10": "the growth path: the localized velocity keeps the asymptotic (3.6) along a path where the cutoffs equal one, which is the limsup ||u||_infinity = infinity claim",
"M10.11": "the viscosity rescaling u_nu(x,t) = sqrt(nu) u(x/sqrt(nu), t) transfers the nu = 1 construction to every nu > 0 with the same singular time"
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, p. 1",
"description": "Theorem 1.1 (OpenAI 2026): for every nu > 0 there exist a force f in C_c^infinity(R^3 x (0, infinity)), a compact set K, and smooth u, p on R^3 x [0, 1) solving Navier-Stokes (1.1) from rest, supported in K for every t < 1, with sup_t ||u(t)||_{L^2} finite and limsup_{t -> 1} ||u(t)||_{L^infinity} = infinity. Hence no global smooth bounded-energy solution for that force and datum: alternative (C); (D) on the torus by compact support (Cor. 10.6). OBLIGATION: The endpoint. It rests on the localization and comparison steps of section 10 and, through them, on the whole construction. ANTECEDENT: Fefferman's problem statement [13] REFS: p. 1, Theorem 1.1; Corollary 10.6."
},
{
"id": "M4.1",
"kind": "move",
"name": "ns-m4-1-similarity-coordinates-and-the-derivative-operators",
"title": "Similarity coordinates and the derivative operators",
"section": "4",
"pages": "24-25",
"refs": [
"p. 24 to 25",
"(4.1), (4.2), Lemma 4.1",
"(3.2) on p. 7."
],
"statement": "With τ = 1 - t, the concentration scale q(z, t) > 0 is defined implicitly by τ = q(1 - η²), z = q^D η, equivalently q - z²q^{2h} = τ (4.1). For τ > 0 there is a unique root q > |z|^{1/D}, because ∂_q(q - z²q^{2h}) = 1 - 2hz²q^{2h-1} = L ≥ 1 - 2h > 0; hence |η| < 1, and η = ±1 are one-sided limits describing t = 1 away from the singular point.",
"description": "With τ = 1 - t, the concentration scale q(z, t) > 0 is defined implicitly by τ = q(1 - η²), z = q^D η, equivalently q - z²q^{2h} = τ (4.1). For τ > 0 there is a unique root q > |z|^{1/D}, because ∂_q(q - z²q^{2h}) = 1 - 2hz²q^{2h-1} = L ≥ 1 - 2h > 0; hence |η| < 1, and η = ±1 are one-sided limits describing t = 1 away from the singular point. OBLIGATION: Converts every time and axial derivative of a field of the form q^b × profile(X, η) into a profile operator times an explicit power of q. This makes the self-similar ansatz exact rather than asymptotic, supplies the power counting that separates leading terms from q^{2h}-smaller ones, and resolves the anisotropic scales ℓ_r ~ q^{1/2}, ℓ_z ~ q^D with one scale that stays regular at t = 1, z ≠ 0. MECHANISM: Implicit differentiation of the defining relation gives 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); the η_z entry uses L - 2Dη² = d, which is h + D = 1/2. The chain rule gives (4.2). ANTECEDENT: None cited (the coordinates are introduced in (3.2), Section 3.1). REFS: p. 24 to 25; (4.1), (4.2), Lemma 4.1; (3.2) on p. 7.",
"obligation": "Converts every time and axial derivative of a field of the form q^b × profile(X, η) into a profile operator times an explicit power of q. This makes the self-similar ansatz exact rather than asymptotic, supplies the power counting that separates leading terms from q^{2h}-smaller ones, and resolves the anisotropic scales ℓ_r ~ q^{1/2}, ℓ_z ~ q^D with one scale that stays regular at t = 1, z ≠ 0.",
"backward_question": "With radial scale τ^{1/2} and axial scale τ^{1/2-h}, what single scale function makes q^b × profile(r/√q, z/q^D) closed under ∂_t and ∂_z, and turns z ≠ 0 at t = 1 into a regular boundary rather than a singularity?",
"mechanism": "Implicit differentiation of the defining relation gives 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); the η_z entry uses L - 2Dη² = d, which is h + D = 1/2. The chain rule gives (4.2). Because q depends only on (z, t), radial derivatives at fixed (z, t) are (r/q)∂_X, so all physical derivatives reduce to X and η derivatives times powers of q. The exponent 1 - 2D = 2h in the defining relation is what makes the axial compression rate q^D compatible with a single time scale q.",
"antecedent": "None cited (the coordinates are introduced in (3.2), Section 3.1).",
"cost": "q is only implicit (q ≍ τ + |z|^{1/D}, Section 3.1). Every time or axial derivative carries L^{-1} and needs L > 0, hence h < 1/2 for these identities (Theorem 4.6 later takes h < 1/100). Profiles must be smooth up to η = ±1 with one-sided η-derivatives, since those endpoints are physical points at t = 1 (this appears as a hypothesis in Lemma 4.4).",
"checkable": "Exact identity. Pick h in (0, 1/2), a real b, a test f(X, η), and a point (q0, η0, X0); set r0 = √(2q0X0), z0 = η0 q0^D, t0 = 1 - q0(1 - η0²); differentiate q^b f implicitly from G(q, z, t) = q - z²q^{2h} - (1 - t) = 0 and compare with q^{b-1}T_b f and q^{b-D}Z_b f. Run for this digest in sympy at 4 random points with h up to 0.45 and b in [-3, 3]: difference exactly 0.",
"depends_on": [],
"constrains": [],
"reasons": {},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 24-25"
},
{
"id": "M4.2",
"kind": "move",
"name": "ns-m4-2-axis-regular-ansatz-for-the-leading-field",
"title": "Axis-regular ansatz for the leading field",
"section": "4",
"pages": "25",
"refs": [
"p. 25",
"(4.3), (4.4), (4.5)",
"Definition 3.2 (p. 18)."
],
"statement": "u_θ^(0) = q^{-A}E, u_z^(0) = q^{-A}U, r u_r^(0) = V0, p^(0) = q^{-2A}Π, E = C^{-1}√(2X)ϕ (4.3), with a fixed normalization C > 1 chosen in the proof of Theorem 4.6. Regularity at the axis (Definition 3.2) follows from (4.4): E = √(2X)F, V0 = Xv0, with F = ϕ/C, U, v0, Π in C^∞([0, Xc] × [-1, 1]). In Cartesian form (4.5): u1 = (v0/(2q))x1 - q^{-A-1/2}Fx2, u2 = (v0/(2q))x2 + q^{-A-1/2}Fx1, u3 = q^{-A}U, p = q^{-2A}Π.",
"description": "u_θ^(0) = q^{-A}E, u_z^(0) = q^{-A}U, r u_r^(0) = V0, p^(0) = q^{-2A}Π, E = C^{-1}√(2X)ϕ (4.3), with a fixed normalization C > 1 chosen in the proof of Theorem 4.6. Regularity at the axis (Definition 3.2) follows from (4.4): E = √(2X)F, V0 = Xv0, with F = ϕ/C, U, v0, Π in C^∞([0, Xc] × [-1, 1]). In Cartesian form (4.5): u1 = (v0/(2q))x1 - q^{-A-1/2}Fx2, u2 = (v0/(2q))x2 + q^{-A-1/2}Fx1, u3 = q^{-A}U, p = q^{-2A}Π. OBLIGATION: The velocity must be smooth in Cartesian coordinates for every t < 1, so that the residual (the eventual force) is smooth away from the singular time. The swirl must vanish linearly on the axis and the radial flux quadratically. MECHANISM: X = (x1² + x2²)/(2q) is itself a smooth Cartesian function, so any smooth function of (X, η) is smooth in Cartesian coordinates. The only nonsmooth factor is r, absorbed by r e_θ = (-x2, x1, 0) and r e_r = (x1, x2, 0): writing E = √(2X)F makes u_θ e_θ = q^{-A-1/2}F(-x2, x1, 0), and V0 = Xv0 makes u_r e_r = (v0/(2q))(x1, x2, 0). So the right condition is smoothness of F, U, v0, Π in X (not in r). ANTECEDENT: None cited. REFS: p. 25; (4.3), (4.4), (4.5); Definition 3.2 (p. 18).",
"obligation": "The velocity must be smooth in Cartesian coordinates for every t < 1, so that the residual (the eventual force) is smooth away from the singular time. The swirl must vanish linearly on the axis and the radial flux quadratically.",
"backward_question": "Which profile quantities must be smooth in X, rather than in r, for the Cartesian field to be smooth at the axis, given that X = r²/(2q) is already smooth in x1, x2?",
"mechanism": "X = (x1² + x2²)/(2q) is itself a smooth Cartesian function, so any smooth function of (X, η) is smooth in Cartesian coordinates. The only nonsmooth factor is r, absorbed by r e_θ = (-x2, x1, 0) and r e_r = (x1, x2, 0): writing E = √(2X)F makes u_θ e_θ = q^{-A-1/2}F(-x2, x1, 0), and V0 = Xv0 makes u_r e_r = (v0/(2q))(x1, x2, 0). So the right condition is smoothness of F, U, v0, Π in X (not in r).",
"antecedent": "None cited.",
"cost": "Positivity ϕ > 0 (so E > 0 for X > 0), used later to divide by F, E, H. Axisymmetry of the leading field. C becomes a free large parameter (C ≥ C0 in Proposition 4.10).",
"checkable": "Algebraic. Evaluate (4.5) against the cylindrical-to-Cartesian conversion of (4.3) at random points; for test F, U, v0, finite-difference the Cartesian components across x1 = x2 = 0 and confirm smoothness. (The divergence check is under M4.3.)",
"depends_on": [
"M4.1"
],
"constrains": [],
"reasons": {
"M4.1": "writes the leading fields as powers of the concentration scale q times profiles of the similarity variables X = r²/(2q), η of (4.1)."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 25"
},
{
"id": "M4.3",
"kind": "move",
"name": "ns-m4-3-incompressibility-and-cyclostrophic-balance-fix-v0-and",
"title": "Incompressibility and cyclostrophic balance fix V0 and Π; the q^{2h} hierarchy",
"section": "4",
"pages": "25-26",
"refs": [
"p. 25 to 26",
"(4.6), (4.7) and the paragraph after it",
"used in (5.1) to (5.6) on p. 46."
],
"statement": "With the radial average A_X(f)(X, η) = X^{-1}∫_0^X f(x, η)dx, A_X(f)(0, η) = f(0, η) (4.6): V0 = (X/L)(2ηU - 2DηA_X(U) - d∂_η A_X(U)) and Π_X = E²/(2X) (4.7). The remaining terms of the radial momentum equation, and axial viscosity relative to radial viscosity in the tangential equations, carry an additional factor at least q^{2h} at fixed profile coordinates; they are retained in the full residual and treated in Section 5, where they generate the powers q^{2nh} of (5.1).",
"description": "With the radial average A_X(f)(X, η) = X^{-1}∫_0^X f(x, η)dx, A_X(f)(0, η) = f(0, η) (4.6): V0 = (X/L)(2ηU - 2DηA_X(U) - d∂_η A_X(U)) and Π_X = E²/(2X) (4.7). The remaining terms of the radial momentum equation, and axial viscosity relative to radial viscosity in the tangential equations, carry an additional factor at least q^{2h} at fixed profile coordinates; they are retained in the full residual and treated in Section 5, where they generate the powers q^{2nh} of (5.1). OBLIGATION: Exact incompressibility of the leading field (a hard constraint throughout the paper) and removal of the leading radial momentum residual, so that only the two tangential residuals remain to be written as a stress divergence. It determines V0 and Π from the free choices E, U plus the single function Π(0, η). MECHANISM: With s = r²/2, incompressibility reads ∂_s V0 + ∂_z(q^{-A}U) = 0. Since ∂_s = q^{-1}∂_X at fixed (z, t) and A + D = 1, Lemma 4.1 gives (V0)_X = L^{-1}(2AηU - dU_η + 2ηXU_X), and integration. ANTECEDENT: None cited (classical cyclostrophic balance; the scale comparison is Sections 2.1 and 3.1). REFS: p. 25 to 26; (4.6), (4.7) and the paragraph after it; used in (5.1) to (5.6) on p. 46.",
"obligation": "Exact incompressibility of the leading field (a hard constraint throughout the paper) and removal of the leading radial momentum residual, so that only the two tangential residuals remain to be written as a stress divergence. It determines V0 and Π from the free choices E, U plus the single function Π(0, η).",
"backward_question": "Under ℓ_r ~ q^{1/2}, ℓ_z ~ q^D and velocities ~ q^{-A}, which terms of the momentum equation are leading, and what small parameter orders everything else?",
"mechanism": "With s = r²/2, incompressibility reads ∂_s V0 + ∂_z(q^{-A}U) = 0. Since ∂_s = q^{-1}∂_X at fixed (z, t) and A + D = 1, Lemma 4.1 gives (V0)_X = L^{-1}(2AηU - dU_η + 2ηXU_X), and integration from V0(0, η) = 0 gives (4.7). In the radial equation, ∂_r p^(0) and (u_θ^(0))²/r both carry q^{-2A-1/2}, with coefficients √(2X)Π_X and E²/√(2X), so the balance is Π_X = E²/(2X). The other radial terms (time derivative, transport, viscosity of u_r) scale like q^{-3/2}, a relative q^{2h}; axial versus radial diffusion is ℓ_r²/ℓ_z² = q^{2h}.",
"antecedent": "None cited (classical cyclostrophic balance; the scale comparison is Sections 2.1 and 3.1).",
"cost": "The axis pressure Π(0, η) is undetermined; it is fixed globally by normalizing Π to vanish at radial infinity ((4.25), through the exterior datum Π0 of (4.31)). The deferred q^{2h} terms create an infinite hierarchy of axisymmetric corrections (Section 5). The text states that physical derivative bounds are not proved here; these are comparisons at fixed profile coordinates.",
"checkable": "(a) With a test U, compute V0 by (4.7) and check the physical divergence (1/r)∂_r(ru_r) + ∂_z u_z. Run for this digest: exactly 0 at 4 random points. (b) Scaling. Run for this digest at fixed (X, η) = (0.7, 0.3), h = 1/10, q = 1e-2 to 1e-8: ∂_r p - u_θ²/r = 0 exactly; (axial viscosity)/(radial viscosity) of u_θ has log-slope exactly 0.2 = 2h; (remaining radial terms)/(u_θ²/r) has local slopes 0.34, 0.30, 0.25, approaching 2h from above (a q^{2h} term plus a q^{4h} term from ∂_z²u_r), consistent with \"at least q^{2h}\".",
"depends_on": [
"M4.2",
"M4.1"
],
"constrains": [],
"reasons": {
"M4.2": "imposes incompressibility and the radial momentum balance on the ansatz fields (4.3), with E = √(2X)F and r u_r = V0.",
"M4.1": "uses Lemma 4.1 to compute ∂_z(q^{-A}U) for incompressibility and to compare term sizes, finding the extra factor q^{2h}."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 25-26"
},
{
"id": "M4.4",
"kind": "move",
"name": "ns-m4-4-radial-integration-of-the-tangential-residual-into-a",
"title": "Radial integration of the tangential residual into a stress (Proposition 4.2)",
"section": "4",
"pages": "26-27",
"refs": [
"p. 26 to 27",
"(4.8) to (4.14), Proposition 4.2. The extraction of (4.13) flattens the fraction",
"the form above was confirmed by rederivation."
],
"statement": "Set H = √(2X)E (so r u_θ = q^{-h}H), F = E/√(2X), l = D_X log H, W = 1 - 2DηA_X(U) - d∂_η A_X(U), H_c = Dη + dU (4.8). Let Q_s, N_s solve D_X Q_s + (1 + l)Q_s = S_q and D_X N_s + N_s = S_n (4.9), with S_q = -Wl - h(1 - 2ηU) - H_c(log E)_η and S_n = -W D_X U - A(1 - 2ηU)U - H_c U_η - dΠ_η + 4AηΠ + 2ηD_X Π, taking the integration constants C_Q = C_N = 0 in (4.10). Define T0 = F(p_s - s), p_s = (XQ_s/L, XN_s/(LE)), s = (a, -b_s), a = 1 - 2D_X log E = 2 - 2l, b_s = 2D_X U/E (4.11).",
"description": "Set H = √(2X)E (so r u_θ = q^{-h}H), F = E/√(2X), l = D_X log H, W = 1 - 2DηA_X(U) - d∂_η A_X(U), H_c = Dη + dU (4.8). Let Q_s, N_s solve D_X Q_s + (1 + l)Q_s = S_q and D_X N_s + N_s = S_n (4.9), with S_q = -Wl - h(1 - 2ηU) - H_c(log E)_η and S_n = -W D_X U - A(1 - 2ηU)U - H_c U_η - dΠ_η + 4AηΠ + 2ηD_X Π, taking the integration constants C_Q = C_N = 0 in (4.10). Define T0 = F(p_s - s), p_s = (XQ_s/L, XN_s/(LE)), s = (a, -b_s), a = 1 - 2D_X log E = 2 - 2l, b_s = 2D_X U/E (4.11). OBLIGATION: Puts the leading residual in the only form the pulses can cancel: the divergence of a stress whose rθ and rz entries are the wave covariances ⟨w_r w_θ⟩, ⟨w_r w_z⟩ (Section 3.3, Proposition 7.5). It also gives a pointwise test for \"no stress\", the zero-stress equations (4.13), which the core near the axis must solve. MECHANISM: The inviscid angular equation is cleanest in angular momentum: (4.14) states (∂_t + u_r∂_r + u_z∂_z)(q^{-h}H) = -(q^{-h-1}/L)HS_q; the axial material derivative plus pressure gradient is -q^{-A-1}S_n/L. ANTECEDENT: None cited. REFS: p. 26 to 27; (4.8) to (4.14), Proposition 4.2. The extraction of (4.13) flattens the fraction; the form above was confirmed by rederivation.",
"obligation": "Puts the leading residual in the only form the pulses can cancel: the divergence of a stress whose rθ and rz entries are the wave covariances ⟨w_r w_θ⟩, ⟨w_r w_z⟩ (Section 3.3, Proposition 7.5). It also gives a pointwise test for \"no stress\", the zero-stress equations (4.13), which the core near the axis must solve.",
"backward_question": "Can the leading tangential residual of an arbitrary axisymmetric profile be written as a radial stress divergence, what fixes the integration constants, and why must that stress vanish near the axis (the waves act only on an annulus)?",
"mechanism": "The inviscid angular equation is cleanest in angular momentum: (4.14) states (∂_t + u_r∂_r + u_z∂_z)(q^{-h}H) = -(q^{-h-1}/L)HS_q; the axial material derivative plus pressure gradient is -q^{-A-1}S_n/L. Multiplying by the integrating factors r² and r and integrating from the axis gives FXQ_s/L and FXN_s/(LE); the ODEs (4.9) are exactly these radial integrations written with D_X (integrating factors XH and X). Radial viscosity is already a divergence: ∂_r u_θ - u_θ/r = -q^{-A-1/2}Fa and ∂_r u_z = q^{-A-1/2}Fb_s, which is the -Fs term. Regularity forces C_Q = C_N = 0, since H ≈ 2Xϕ/C near the axis turns a constant into an X^{-2} (Q_s) or X^{-1} (N_s) singularity; then Q_s(0, η) = S_q(0, η)/2 and N_s(0, η) = S_n(0, η). Substituting p_s = s (that is XQ_s/L = a, N_s/L = -2U_X) into (4.9) gives (4.13), whose viscous parts are the 4D radial Laplacian for ϕ and the 2D radial Laplacian for U in the variable X.",
"antecedent": "None cited.",
"cost": "T0 at radius X depends on E, U at all smaller radii (nonlocal), which forces the five cumulative integrals of M4.5 and complicates joining. Requires ϕ > 0. Near the axis the profile must solve (4.13), a nonlinear system whose sources contain η-derivatives of the unknowns (solved in an analytic class in Appendix B).",
"checkable": "Run for this digest in sympy (h = 7/100, C = 2, explicit non-stress-free ϕ > 0, U, Π_ax with closed-form integrals): built u, p from (4.3) and (4.7), computed R_θ^(0), R_z^(0) from (4.12) by implicit differentiation in (r, z, t), and compared with -(∂_r + 2/r)T_θ, -(∂_r + 1/r)T_z from (4.11) and (4.16): agreement to 1e-135 at 4 random points. Also verified symbolically: the shear normalization (∂_r u_θ - u_θ/r, ∂_r u_z) = -q^{-A-1/2}Fs, the reduction of p_s = s to (4.13), and, by exact series, Q_s(0) = S_q(0)/2 and N_s(0) = S_n(0).",
"depends_on": [
"M4.3",
"M4.2",
"M4.1"
],
"constrains": [],
"reasons": {
"M4.3": "uses V0 from (4.7), through the factor W, and the balance Π_X = E²/(2X), which leave only the two tangential residuals to integrate.",
"M4.2": "integrates the tangential residuals of the leading field (4.3), with swirl E = √(2X)F and angular momentum H = √(2X)E.",
"M4.1": "expresses time and axial derivatives of the q^b-scaled fields through the profile operators of Lemma 4.1 (4.2)."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 26-27"
},
{
"id": "M4.5",
"kind": "move",
"name": "ns-m4-5-the-five-cumulative-radial-integrals-lemma-4-3",
"title": "The five cumulative radial integrals (Lemma 4.3)",
"section": "4",
"pages": "28",
"refs": [
"p. 28",
"(4.15), (4.16), Lemma 4.3",
"moment interpretation on p. 30."
],
"statement": "m = (M, I, J, S, Cp) with M = ∫_0^X U, I = ∫_0^X H, J = ∫_0^X UH, S = ∫_0^X (U² - E²/2), Cp = ∫_0^X E²/(2x), and Π = Π(0, η) + Cp (4.15). Lemma 4.3 (4.16): Q_s = -W + [(1 - h)I - DηI_η - dJ_η + 2(h - D)ηJ]/(XH) and N_s = -WU + [D(M - ηM_η) + 4hηS - dS_η]/X + 4AηΠ - dΠ_η, where XW = X - 2DηM - dM_η.",
"description": "m = (M, I, J, S, Cp) with M = ∫_0^X U, I = ∫_0^X H, J = ∫_0^X UH, S = ∫_0^X (U² - E²/2), Cp = ∫_0^X E²/(2x), and Π = Π(0, η) + Cp (4.15). Lemma 4.3 (4.16): Q_s = -W + [(1 - h)I - DηI_η - dJ_η + 2(h - D)ηJ]/(XH) and N_s = -WU + [D(M - ηM_η) + 4hηS - dS_η]/X + 4AηΠ - dΠ_η, where XW = X - 2DηM - dM_η. OBLIGATION: Identifies exactly which accumulated information from smaller radii the stress depends on, so profiles on different radial intervals can be joined by matching five scalar functions of η, and so that changes in p_s are controlled by changes in profile values and integrals, with no radial derivatives. MECHANISM: Integrate the sources of (4.9) from 0 to X. The transport term -W D_X H integrates by parts using ∂_X(XW) = 1 - 2DηU - dU_η; the angular terms -dU_ηH - dUH_η combine to -d(UH)_η, and -H_c(log E)_η H = -H_c H_η; for the pressure, ∫_0^X Π = XΠ - ∫_0^X E²/2 from Π_X = E²/(2X), and the U² and E² pieces combine into S with coefficient 4h = 2(A - D). Division by XH and X gives (4.16). Physically (p. ANTECEDENT: None cited. REFS: p. 28; (4.15), (4.16), Lemma 4.3; moment interpretation on p. 30.",
"obligation": "Identifies exactly which accumulated information from smaller radii the stress depends on, so profiles on different radial intervals can be joined by matching five scalar functions of η, and so that changes in p_s are controlled by changes in profile values and integrals, with no radial derivatives.",
"backward_question": "The stress at X depends on the whole profile inside X; what is the minimal finite set of accumulated quantities that two profiles must share at a joining radius so that pressure, radial velocity, and stress agree beyond it?",
"mechanism": "Integrate the sources of (4.9) from 0 to X. The transport term -W D_X H integrates by parts using ∂_X(XW) = 1 - 2DηU - dU_η; the angular terms -dU_ηH - dUH_η combine to -d(UH)_η, and -H_c(log E)_η H = -H_c H_η; for the pressure, ∫_0^X Π = XΠ - ∫_0^X E²/2 from Π_X = E²/(2X), and the U² and E² pieces combine into S with coefficient 4h = 2(A - D). Division by XH and X gives (4.16). Physically (p. 30), M, J, S measure integrated axial momentum, axial transport of angular momentum, and axial momentum flux including pressure; I measures angular momentum.",
"antecedent": "None cited.",
"cost": "Every join must match five functions of η together with their η-derivatives, which costs finite-dimensional moment solves (Lemma 4.7) and dedicated bump intervals.",
"checkable": "Quadrature. For test profiles, compare (4.16) with (4.10) computed by quadrature of HS_q and S_n. Run for this digest: agreement to about 1e-15 at three (X, η) points (double-precision limited). Caution: (4.16) suffers X^{-2} cancellation near X = 0 in floating point; use series there. Independent hand rederivation of both lines of (4.16) also matched.",
"depends_on": [
"M4.4",
"M4.3"
],
"constrains": [],
"reasons": {
"M4.4": "integrates the sources S_q, S_n of (4.9) from the axis with C_Q = C_N = 0 to write Q_s, N_s through cumulative integrals.",
"M4.3": "uses Π_X = E²/(2X) to write Π = Π(0, η) + Cp, and A_X(U) = M/X inside W from (4.7)."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 28"
},
{
"id": "M4.6",
"kind": "move",
"name": "ns-m4-6-joining-lemma-exact-propagation-and-a-derivative-free",
"title": "Joining lemma: exact propagation and a derivative-free estimate (Lemma 4.4)",
"section": "4",
"pages": "28-30",
"refs": [
"p. 28 to 30",
"Lemma 4.4, (4.17), (4.18), (4.19)."
],
"statement": "Fix h in (0, 1/2) and two pairs (U_i, E_i), E_i > 0, sharing one axis pressure datum Π_ax(η), with Π_i = Π_ax + Cp,i, V0,i by (4.7), Q_s,i, N_s,i by (4.16), and p_s,i, a_i, b_s,i, T0,i by (4.11). (i) If (U1, E1) = (U2, E2) for X ≥ X_h and Δm(X_h, η) = 0 for all η, then m1 = m2 and Π, V0, Q_s, N_s, p_s, a, b_s, T0 all agree for X ≥ X_h.",
"description": "Fix h in (0, 1/2) and two pairs (U_i, E_i), E_i > 0, sharing one axis pressure datum Π_ax(η), with Π_i = Π_ax + Cp,i, V0,i by (4.7), Q_s,i, N_s,i by (4.16), and p_s,i, a_i, b_s,i, T0,i by (4.11). (i) If (U1, E1) = (U2, E2) for X ≥ X_h and Δm(X_h, η) = 0 for all η, then m1 = m2 and Π, V0, Q_s, N_s, p_s, a, b_s, T0 all agree for X ≥ X_h. OBLIGATION: (i) Allows gluing an inner (axis) profile to a prescribed outer profile without disturbing the outer pressure, radial velocity, stress, or heat exterior. (ii) Is the license for Step 2 of the proof: radial derivatives (hence the shear a, b_s) can change by O(1) while p_s changes only by the size of the profile and moment changes. MECHANISM: (i) Where the integrands of (4.18) agree, Δm is constant in X, hence Δm ≡ 0 beyond X_h as functions of η (so all η-derivatives vanish); the common Π_ax gives ΔΠ = ΔCp = 0, and (4.7), (4.16) contain only values, integrals, and η-derivatives. (ii) Write ΔH = √(2X)ΔE, ΔW = -(2DηΔM + dΔM_η)/X, ΔΠ = ΔCp, and use Δ(1/H) = -ΔH/(H1H2), Δ(1/E) = -ΔE/(E1E2); ANTECEDENT: None cited. REFS: p. 28 to 30; Lemma 4.4, (4.17), (4.18), (4.19).",
"obligation": "(i) Allows gluing an inner (axis) profile to a prescribed outer profile without disturbing the outer pressure, radial velocity, stress, or heat exterior. (ii) Is the license for Step 2 of the proof: radial derivatives (hence the shear a, b_s) can change by O(1) while p_s changes only by the size of the profile and moment changes.",
"backward_question": "Can the radial derivatives of a profile be changed by O(1) without changing the integrated inviscid stress p_s by more than a small amount, and can two profiles be glued exactly without the outer stress noticing?",
"mechanism": "(i) Where the integrands of (4.18) agree, Δm is constant in X, hence Δm ≡ 0 beyond X_h as functions of η (so all η-derivatives vanish); the common Π_ax gives ΔΠ = ΔCp = 0, and (4.7), (4.16) contain only values, integrals, and η-derivatives. (ii) Write ΔH = √(2X)ΔE, ΔW = -(2DηΔM + dΔM_η)/X, ΔΠ = ΔCp, and use Δ(1/H) = -ΔH/(H1H2), Δ(1/E) = -ΔE/(E1E2); on R the denominators are bounded below (X ≥ X0, L ≥ 1 - 2h, H_i ≥ √(2X0)e_min). After k η-derivatives every term contains a difference of U, E, or a component of m of order at most k + 1.",
"antecedent": "None cited.",
"cost": "One axis pressure function must be fixed before any join (so the exterior is built first). One η-derivative is lost (k + 1 to k). The full stress T0 = F(p_s - s) still depends on the shear, which (ii) does not control.",
"checkable": "(i) Take two pairs equal on X ≥ X_h, adjust the interior of one by bumps so that Δm(X_h) = 0 (a 5 × 5 solve as in (4.42)), and compare Q_s, N_s, p_s, T0 beyond X_h. (ii) Run for this digest (h = 0.005, η = 0.4, η-independent test profiles, zero-mean loop amplitudes of order one on [2, 4], evaluated at X = 3.37): with E_N = E0 exp(A(X, N log X)/N), U_N = U0 + B(X, N log X)/N, max|Δm| = 4.1e-3, 1.5e-3, 2.2e-4, 8.0e-5, 1.9e-5 for N = 10, 20, 40, 80, 160 (about N^{-2}, since the oscillation has zero mean), N·|Δp_s| stayed between 0.13 and 0.22 (Δp_s = O(N^{-1}), driven by pointwise value changes), and |a_N - a0| stayed between 0.4 and 1.0.",
"depends_on": [
"M4.5",
"M4.4",
"M4.3"
],
"constrains": [],
"reasons": {
"M4.5": "uses (4.16): Q_s, N_s depend on the profile only through values, η-derivatives, and the five cumulative integrals m.",
"M4.4": "carries the agreement of Q_s, N_s to p_s, a, b_s and T0 = F(p_s - s) through (4.11).",
"M4.3": "V0 from (4.7) and Π = Π_ax + Cp involve only values, integrals, and the shared axis datum, so they agree as well."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 28-30"
},
{
"id": "M4.7",
"kind": "move",
"name": "ns-m4-7-the-admissible-stress-cone-in-profile-variables-lemma-4-5",
"title": "The admissible stress cone in profile variables (Lemma 4.5)",
"section": "4",
"pages": "30-32",
"refs": [
"p. 30 to 32",
"(4.20) to (4.23), Lemma 4.5",
"forward use in (7.1) on p. 74."
],
"statement": "Where a > 0: t_s = -b_s/a, v_s = a(1 + t_s²), P_c = p_s,1 + t_s p_s,2, J_c = p_s,2 - t_s p_s,1 (4.20). Relaxed cone: P_c > 2 and v_s < U(P_c, J_c) := P_c + J_c²/4 - |J_c|√((P_c - 2)/2 + J_c²/16) (4.21). Admissible cone: the relaxed inequalities plus v_s > 2. Lemma 4.5: if v_s > 2, admissible is equivalent to P_c > v_s and (v_s - 2)J_c² < 2(P_c - v_s)² (4.22).",
"description": "Where a > 0: t_s = -b_s/a, v_s = a(1 + t_s²), P_c = p_s,1 + t_s p_s,2, J_c = p_s,2 - t_s p_s,1 (4.20). Relaxed cone: P_c > 2 and v_s < U(P_c, J_c) := P_c + J_c²/4 - |J_c|√((P_c - 2)/2 + J_c²/16) (4.21). Admissible cone: the relaxed inequalities plus v_s > 2. Lemma 4.5: if v_s > 2, admissible is equivalent to P_c > v_s and (v_s - 2)J_c² < 2(P_c - v_s)² (4.22). OBLIGATION: The stress must lie in the positive span of the covariance directions of two viscous wave families with real amplitudes (Proposition 7.5); this is that requirement in profile variables. The extra inequality v_s > 2 \"is required by the viscous waves\". Homogeneity lets the condition be imposed on the unit direction n = T0/|T0| where T0 → 0 at the annulus edges. MECHANISM: The polynomial 2(P_c - v)² - (v - 2)J_c² has roots v± = P_c + J_c²/4 ± |J_c|√((P_c - 2)/2 + J_c²/16), and for P_c > 2, 2 < v- ≤ P_c; ANTECEDENT: None cited in Section 4. The introduction lists the centrifugal-instability criteria of Leibovich and Stewartson [15] and of Billant and Gallaire [2, 3] as precedents for the wave dynamics this condition serves. REFS: p. 30 to 32; (4.20) to (4.23), Lemma 4.5; forward use in (7.1) on p. 74.",
"obligation": "The stress must lie in the positive span of the covariance directions of two viscous wave families with real amplitudes (Proposition 7.5); this is that requirement in profile variables. The extra inequality v_s > 2 \"is required by the viscous waves\". Homogeneity lets the condition be imposed on the unit direction n = T0/|T0| where T0 → 0 at the annulus edges.",
"backward_question": "Which stress vectors can two real wave families with positive squared amplitudes produce against the local shear, and how can this be stated so that it survives where the stress tends to zero at the annulus edges?",
"mechanism": "The polynomial 2(P_c - v)² - (v - 2)J_c² has roots v± = P_c + J_c²/4 ± |J_c|√((P_c - 2)/2 + J_c²/16), and for P_c > 2, 2 < v- ≤ P_c; so on 2 < v < P_c positivity holds exactly when v < v- = U(P_c, J_c), which is (4.22). Geometry (derived here): (1, t_s) and (-t_s, 1) have equal length, so for v_s > 2 the admissible set is the open cone about the shear direction s = a(1, t_s) with half-angle arctan√(2/(v_s - 2)), widening to a half-plane as v_s → 2+. The first inequality T0·s > 0 is positive shear production: the physical shear vector is -q^{-A-1/2}Fs and the production is -T·(shear), matching the energy-transfer identity of Section 3.3 (derived here). Section 7 (p. 74) shows v_s > 2 is positivity of the reference growth rate λ0² = 2aF0²(1 - 2/v_s), with c0² = (v_s - 2)/2. For the sufficient test, with c = 1 - b_s w/a and j = w + b_s/a one has P_c = p_s,1c, J_c = p_s,1j, and (v_s - 2)j² - 2c² = (1 + b_s²/a²)((a - 2)w² + 2b_s w + b_s²/a - 2); for large p_s,1 the fixed v_s becomes negligible, and P_K = max{(B_K + 2)/c_K, 8B_K/γ_K} works uniformly on K.",
"antecedent": "None cited in Section 4. The introduction lists the centrifugal-instability criteria of Leibovich and Stewartson [15] and of Billant and Gallaire [2, 3] as precedents for the wave dynamics this condition serves.",
"cost": "Two strict inequalities (a > 0, v_s > 2) on the whole closed annulus with a uniform margin. The join achieves only the relaxed cone on an intermediate interval, which creates the repair problem of M4.13 to M4.15.",
"checkable": "Run for this digest: random sampling of (a > 0, b_s, p_s); (4.21) plus v_s > 2 versus (4.22) gave 0 mismatches in 372,246 samples with v_s > 2 (57,267 admissible); 2 < v- ≤ P_c had 0 violations in 200,000 samples with P_c > 2; the polynomial identity in the proof and both lines of (4.23) verified symbolically.",
"depends_on": [
"M4.4",
"L.11"
],
"constrains": [],
"reasons": {
"M4.4": "builds t_s, v_s, P_c, J_c from the shear s = (a, -b_s) and the integrated vector p_s of (4.11), and writes the cone for T0 = F(p_s - s).",
"L.11": "The cone's extra inequality v_s > 2, 'required by the viscous waves', is what Duraiswami (W.8) identifies with Rayleigh's centrifugal criterion including axial shear."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 30-32"
},
{
"id": "M4.8",
"kind": "move",
"name": "ns-m4-8-the-leading-profile-theorem-theorem-4-6-the-contract",
"title": "The leading-profile theorem (Theorem 4.6): the contract handed onward",
"section": "4",
"pages": "32-34",
"refs": [
"p. 32 to 34",
"(4.24) to (4.30), Theorem 4.6."
],
"statement": "There exist fixed h in (0, 1/100), λ > 0, C > 1, 0 < X_a < X_b, and profiles E, U, Π on X ≥ 0, η in [-1, 1], such that: (i) F = ϕ/C, U, Π, V0/X are smooth on [0, R] × [-1, 1] for every finite R, ϕ > 0, and on [0, X_an] × [-1, 1] (some X_an in (X_a, X_b)) they are analytic in η on one common complex neighborhood of [-1, 1], with the inner collar [X_a, X_an] carrying the margin of (iii); Π = -∫_X^∞ E²/(2x)dx (4.25). (ii) Π_X = E²/(2X), with Π_X = F² at X = 0, and Proposition 4.2 hold exactly;",
"description": "There exist fixed h in (0, 1/100), λ > 0, C > 1, 0 < X_a < X_b, and profiles E, U, Π on X ≥ 0, η in [-1, 1], such that: (i) F = ϕ/C, U, Π, V0/X are smooth on [0, R] × [-1, 1] for every finite R, ϕ > 0, and on [0, X_an] × [-1, 1] (some X_an in (X_a, X_b)) they are analytic in η on one common complex neighborhood of [-1, 1], with the inner collar [X_a, X_an] carrying the margin of (iii); Π = -∫_X^∞ E²/(2x)dx (4.25). (ii) Π_X = E²/(2X), with Π_X = F² at X = 0, and Proposition 4.2 hold exactly; OBLIGATION: Collects everything the rest of the proof uses from the leading flow: exact incompressibility and cyclostrophic balance; a stress supported in a fixed annulus in X (the shrinking physical annulus √(2qX_a) < r < √(2qX_b)), strictly inside the wave cone with a uniform margin including the edges; flat edge behavior quantified by ζ; a residual-free heat exterior with smooth limits at every fixed r > 0 as t → 1; vanishing total moments (no stress tails); axis analyticity; and reserved power-law patches for later corrections. ANTECEDENT: None cited; the ingredients are proved in Appendices A to C. REFS: p. 32 to 34; (4.24) to (4.30), Theorem 4.6.",
"obligation": "Collects everything the rest of the proof uses from the leading flow: exact incompressibility and cyclostrophic balance; a stress supported in a fixed annulus in X (the shrinking physical annulus √(2qX_a) < r < √(2qX_b)), strictly inside the wave cone with a uniform margin including the edges; flat edge behavior quantified by ζ; a residual-free heat exterior with smooth limits at every fixed r > 0 as t → 1; vanishing total moments (no stress tails); axis analyticity; and reserved power-law patches for later corrections. Positivity of E for X > 0 is also what gives the growth u_θ = E(X*, 0)τ^{-A} on the circle η = 0 (p. 8).",
"backward_question": "What exactly must the leading profile deliver so that an axisymmetric expansion (Section 5), waves (Section 7), mean corrections (Section 8), and localization (Section 10) can all proceed with q-independent constants?",
"mechanism": "Assembled in Section 4.6 from Lemma 4.8 (exterior), Lemma 4.9 (outer edge), Proposition 4.10 (axis and inner edge), Lemma 4.11 (shear loop), Lemma 4.7 (moment solve) and Lemma 4.4 (joining); see M4.14 and M4.15.",
"antecedent": "None cited; the ingredients are proved in Appendices A to C.",
"cost": "Fixes h < 1/100 and λ, C, X_a, X_b, κ, ζ permanently. Later sections must keep the exterior (4.29) and the identities (4.28) intact, work inside the annulus with the weights ζ, δ (derivative bounds lose powers of δ), and place their radial corrections only in I_pos and I_mean. Flatness of T0 at the edges means wave amplitudes (weight √ζ in the class W_α) must vanish to infinite order there.",
"checkable": "The exterior part. Run for this digest with mpmath: K(r, τ) = c∞(r²/2)^{-A}H(4τ/r²) satisfies -∂_τK = K_rr + K_r/r - K/r² (relative residual at most 1.6e-13 at 4 points each for h = 0.005, 0.009, 0.3), K_r < 0 at all sampled points, H(0) = 1. Closed form (derived here): H(Z) = Z^{-1-h}U(1 + h, 2, 1/Z), U the Tricomi confluent hypergeometric function; agreement with the integral to 8e-26 via mpmath.hyperu. The identity q^{-A}c∞X^{-A}H(2d/X) = c∞s^{-A}H(2τ/s) follows from X = s/q, d = τ/q. The remaining assertions need the appendix profiles.",
"depends_on": [
"M4.15",
"M4.14",
"M4.7",
"M4.4",
"L.11"
],
"constrains": [],
"reasons": {
"M4.15": "Steps 3 and 4 restore the moments exactly and verify (i) to (vi), including the margin κ of (4.26) and the weight ζ of (4.27).",
"M4.14": "Steps 1 and 2 join the axis connection to the prepared exterior and make the shear admissible on the repair interval.",
"M4.7": "part (iii) is stated in the cone coordinates t_s, v_s, P_c, J_c and the admissible inequalities of Lemma 4.5.",
"M4.4": "part (ii) asserts Proposition 4.2 exactly and locates the support of the stress T0 = F(p_s - s) of (4.11).",
"L.11": "Theorem 4.6(iii) keeps v_s > 2 on the closed annulus, which Duraiswami reads as a centrifugally unstable annulus in the sense of these criteria."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 32-34"
},
{
"id": "M4.9",
"kind": "move",
"name": "ns-m4-9-moment-matrices-and-the-quadratic-moment-solve-lemma-4-7",
"title": "Moment matrices and the quadratic moment solve (Lemma 4.7)",
"section": "4",
"pages": "34-35",
"refs": [
"p. 34 to 35",
"Lemma 4.7 (Lemmas A.1, A.2)."
],
"statement": "For distinct real α1, ..., αm and nonnegative nonzero smooth β_j with ordered disjoint compact supports in (0, ∞), B_ij = ∫_0^∞ x^{α_i}β_j(x)dx is invertible, and its inverse and η-derivatives are bounded on smooth compact families preserving the hypotheses. For B(η) smooth and invertible, Q_η smooth bilinear, d in C^∞([-1, 1];",
"description": "For distinct real α1, ..., αm and nonnegative nonzero smooth β_j with ordered disjoint compact supports in (0, ∞), B_ij = ∫_0^∞ x^{α_i}β_j(x)dx is invertible, and its inverse and η-derivatives are bounded on smooth compact families preserving the hypotheses. For B(η) smooth and invertible, Q_η smooth bilinear, d in C^∞([-1, 1]; OBLIGATION: Every radial modification (heat compensation, axis continuation, loop realization) disturbs the five cumulative integrals. This lemma restores them exactly with bumps supported where the profile is a pure power law, so the prescribed exterior survives by Lemma 4.4(i). MECHANISM: By multilinearity det B = ∫det[x_j^{α_i}]∏_jβ_j(x_j)dx1...dxm; on the support x1 < ... < xm, and the integrand determinant has constant nonzero sign because a nonzero combination of m distinct powers has at most m - 1 positive zeros (induction and Rolle's theorem). ANTECEDENT: Rolle's theorem (the generalized Descartes rule of signs for sums of powers) and a contraction argument; proved as Lemmas A.1 and A.2. REFS: p. 34 to 35; Lemma 4.7 (Lemmas A.1, A.2).",
"obligation": "Every radial modification (heat compensation, axis continuation, loop realization) disturbs the five cumulative integrals. This lemma restores them exactly with bumps supported where the profile is a pure power law, so the prescribed exterior survives by Lemma 4.4(i).",
"backward_question": "If k bump functions are added to repair k moment integrals with power weights x^{α_i}, is the linear system always invertible, and does the quadratic part of the moment map spoil solvability?",
"mechanism": "By multilinearity det B = ∫det[x_j^{α_i}]∏_jβ_j(x_j)dx1...dxm; on the support x1 < ... < xm, and the integrand determinant has constant nonzero sign because a nonzero combination of m distinct powers has at most m - 1 positive zeros (induction and Rolle's theorem). The quadratic system is solved by the contraction c ↦ B^{-1}(d - Q(c, c)) on the ball of radius r_j = 2ν_j||d||_{C^j}, where 2ν_jq_jr_j ≤ 1/2; the pointwise implicit function theorem gives smoothness in η.",
"antecedent": "Rolle's theorem (the generalized Descartes rule of signs for sums of powers) and a contraction argument; proved as Lemmas A.1 and A.2.",
"cost": "Exponents must be distinct, which in Step 3 needs λ > 0: the B_U weights X^0 and X^{-λ} coalesce as λ → 0, so ||B_U^{-1}|| grows like 1/λ (derived here). The defect must be small relative to the inverse norms, which is why N is chosen after all bump shapes and radii.",
"checkable": "Run for this digest: det[x_j^{α_i}] kept one sign over 20,000 ordered random samples for α = (0, -0.3), (0.5, -0.8, -1.8), (2, -0.5, 0.7, 1.3). With the (4.42) weights, λ = 0.3, K = 1, and smooth bumps in [10, 20]: det B_U ≈ -0.030 (condition number ≈ 42), det B_E ≈ 1.8e-5 (condition number ≈ 2.3e4); det B_U/λ = -0.070, -0.143, -0.154 for λ = 0.3, 0.03, 0.003. A further check would iterate the contraction under 8ν²q||d|| ≤ 1 and confirm ||c|| ≤ 2ν||d||.",
"depends_on": [
"MA.1",
"MA.2"
],
"constrains": [],
"reasons": {
"MA.1": "its first half is Lemma A.1: distinct-power moment matrices against ordered disjoint bumps are invertible with bounded inverse.",
"MA.2": "its second half is Lemma A.2: Bc + Q_η(c, c) = d is solved by contraction, with C^k bounds on the coefficients."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 34-35"
},
{
"id": "M4.10",
"kind": "move",
"name": "ns-m4-10-the-prepared-exterior-with-an-exact-heat-tail-lemma-4-8",
"title": "The prepared exterior with an exact heat tail (Lemma 4.8)",
"section": "4",
"pages": "35-36",
"refs": [
"p. 35 to 36",
"Lemma 4.8, (4.31), (4.29)",
"Proposition A.4, Lemmas A.5, A.6, Proposition A.7."
],
"statement": "Parameters are chosen in the order M_d, P*, λ, h, with T_d = e^{M_d} + 10, P* > e^{T_d}, 0 < h < min{1/100, λ, e^{-T_d}}, and a radius R* > 0. For every X_R ≥ R* there are smooth (U_o, E_o) on (0, ∞) × [-1, 1], E_o > 0, with x = X/X_R, f(η) = (1 + η²)^{-1}: (i) U_o = 4η, E_o = P*f(η)x^{1/10} for 0 < x ≤ 1, and the pressure datum Π0(η) = -∫_0^∞ E_o²/(2X)dX (4.31) is independent of X_R, analytic on one complex neighborhood of [-1, 1], even, with Π0 ≤ -(5/2)P*²f² and ηΠ0'(η) > 0 for η ≠ 0.",
"description": "Parameters are chosen in the order M_d, P*, λ, h, with T_d = e^{M_d} + 10, P* > e^{T_d}, 0 < h < min{1/100, λ, e^{-T_d}}, and a radius R* > 0. For every X_R ≥ R* there are smooth (U_o, E_o) on (0, ∞) × [-1, 1], E_o > 0, with x = X/X_R, f(η) = (1 + η²)^{-1}: (i) U_o = 4η, E_o = P*f(η)x^{1/10} for 0 < x ≤ 1, and the pressure datum Π0(η) = -∫_0^∞ E_o²/(2X)dX (4.31) is independent of X_R, analytic on one complex neighborhood of [-1, 1], even, with Π0 ≤ -(5/2)P*²f² and ηΠ0'(η) > 0 for η ≠ 0. OBLIGATION: Supplies the exterior half of the profile: zero residual beyond X_b with a smooth limit at every fixed r > 0 as t → 1 (needed for the spatial cutoff in Section 10); the power law c∞X^{-A} that makes the exterior compatible with the self-similar scaling; the axis pressure datum Π0, fixed before the axis problem as Lemma 4.4 requires; the vanishing total moments (4.28); and spare power-law intervals for later repairs. ANTECEDENT: The classical radial heat equation for swirl; constructed in Proposition A.4, Lemma A.5, Lemma A.6, Proposition A.7. REFS: p. 35 to 36; Lemma 4.8, (4.31), (4.29); Proposition A.4, Lemmas A.5, A.6, Proposition A.7.",
"obligation": "Supplies the exterior half of the profile: zero residual beyond X_b with a smooth limit at every fixed r > 0 as t → 1 (needed for the spatial cutoff in Section 10); the power law c∞X^{-A} that makes the exterior compatible with the self-similar scaling; the axis pressure datum Π0, fixed before the axis problem as Lemma 4.4 requires; the vanishing total moments (4.28); and spare power-law intervals for later repairs.",
"backward_question": "What exterior flow has zero residual and a smooth limit at every fixed r > 0 as t → 1, yet carries the power-law tail that joins a self-similar core, and how is the axis pressure fixed before the core is solved?",
"mechanism": "The heat profile solves the radial swirl heat equation exactly and tends to c∞X^{-A} as X → ∞ at fixed d (H(0) = 1), so it is a forward heat solution that reaches the power law r^{-1-2h} at t = 1. Replacing a pure power-law tail by it changes the pressure integral and the S and renormalized I moments; the compensation on I2 (Lemma A.6, Proposition A.7) restores them. The inner branch alone contributes exactly -(5/2)P*²f² to Π0, since ∫_0^{X_R}P*²f²x^{1/5}/(2X)dX = (5/2)P*²f², and the rest adds more negative terms. The sign pattern Π0 < 0, ηΠ0' > 0 makes the leading axial pressure force on the axis point toward z = 0 (both terms of Z_{-2A}Π0 ∝ dΠ0' - 4AηΠ0 carry the sign of η; derived here), matching Section 2.1.",
"antecedent": "The classical radial heat equation for swirl; constructed in Proposition A.4, Lemma A.5, Lemma A.6, Proposition A.7.",
"cost": "A nested hierarchy with h exponentially small in e^{M_d} (h < e^{-T_d}). The cone minima may depend on X_R. Admissibility may fail on [e^{-5}X_R, X_good), where only the relaxed cone holds, and must be repaired. I2 is consumed.",
"checkable": "Run for this digest: the heat equation, K_r < 0 and H(0) = 1 (see M4.8), and the inner-branch pressure contribution -(5/2)P*²f² by symbolic integration. The intervals, moments, and cone regions need the Appendix A profile.",
"depends_on": [
"MA.4",
"MA.8",
"MA.11",
"MA.10"
],
"constrains": [],
"reasons": {
"MA.4": "the family (U_o, E_o) is the staged outer reference profile of Proposition A.4, with its four reserved patches and radii X_good, X_v, X_tail.",
"MA.8": "the axis datum Π0 of (4.31), independent of X_R, analytic, even, with Π0 ≤ -(5/2)P*²f² and ηΠ0' > 0, is Lemma A.5.",
"MA.11": "the heat replacement on the tail, compensated on the second patch, keeps Π0, the moment identities, and the cone (Proposition A.7).",
"MA.10": "the exterior E_o = c∞X^{-A}H(2d/X), whose physical swirl K solves the radial swirl heat equation with K_r < 0, is Lemma A.6."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 35-36"
},
{
"id": "M4.11",
"kind": "move",
"name": "ns-m4-11-outer-edge-stress-integrated-from-infinity-lemma-4-9",
"title": "Outer edge: stress integrated from infinity (Lemma 4.9)",
"section": "4",
"pages": "30",
"refs": [
"p. 30 (moments at infinity) and p. 36",
"Lemma 4.9, (4.32)",
"Lemma A.8, Proposition A.10."
],
"statement": "The Lemma 4.8 family can be chosen so that, for every X_R ≥ R*, any leading field that is regular at the axis, agrees with the family for X ≥ X_tail, satisfies the four moment identities (4.28), and has the canonical pressure (4.25) obeys T_θ(r) = r^{-2}∫_r^∞ r'²R_θ^(0)(r')dr' and T_z(r) = r^{-1}∫_r^∞ r'R_z^(0)(r')dr', so T0 = 0 for X ≥ X_b. On e^{1/2}X_tail ≤ X < X_b the admissible cone holds with b_s = 0, 2 + h < a ≤ 2 + 2h, T_θ > 0, 2 - (a - 2)(T_z/T_θ)² ≥ κ_o > 0;",
"description": "The Lemma 4.8 family can be chosen so that, for every X_R ≥ R*, any leading field that is regular at the axis, agrees with the family for X ≥ X_tail, satisfies the four moment identities (4.28), and has the canonical pressure (4.25) obeys T_θ(r) = r^{-2}∫_r^∞ r'²R_θ^(0)(r')dr' and T_z(r) = r^{-1}∫_r^∞ r'R_z^(0)(r')dr', so T0 = 0 for X ≥ X_b. On e^{1/2}X_tail ≤ X < X_b the admissible cone holds with b_s = 0, 2 + h < a ≤ 2 + 2h, T_θ > 0, 2 - (a - 2)(T_z/T_θ)² ≥ κ_o > 0; OBLIGATION: The stress is defined by integrating from the axis, so a zero exterior residual alone does not make the exterior stress vanish: without the moment identities, stresses proportional to r^{-2} (angular) and r^{-1} (axial) would survive (p. 30). This lemma removes those tails and fixes the outer-edge flatness and the edge direction n(X_b) = (1, 0). MECHANISM: The four total identities make the two weighted integrals ∫_0^∞ r²R_θ^(0)dr and ∫_0^∞ rR_z^(0)dr vanish (p. 30), so the forward primitive from the axis equals the backward primitive from infinity, which vanishes. ANTECEDENT: Lemma A.8 and Proposition A.10. REFS: p. 30 (moments at infinity) and p. 36; Lemma 4.9, (4.32); Lemma A.8, Proposition A.10.",
"obligation": "The stress is defined by integrating from the axis, so a zero exterior residual alone does not make the exterior stress vanish: without the moment identities, stresses proportional to r^{-2} (angular) and r^{-1} (axial) would survive (p. 30). This lemma removes those tails and fixes the outer-edge flatness and the edge direction n(X_b) = (1, 0).",
"backward_question": "The residual vanishes in the exterior, but the stress is an integral from the axis: what total integrals must vanish so that no r^{-2} and r^{-1} stress tails survive, and what is the stress direction at the outer edge?",
"mechanism": "The four total identities make the two weighted integrals ∫_0^∞ r²R_θ^(0)dr and ∫_0^∞ rR_z^(0)dr vanish (p. 30), so the forward primitive from the axis equals the backward primitive from infinity, which vanishes wherever the residual does (the exterior is purely azimuthal and independent of z, so its full residual and ∂_z²u_θ both vanish). In the terminal collar U = 0, so b_s = 0, t_s = 0, v_s = a, and the cone reduces to T_θ > 0, (a - 2)T_z² < 2T_θ². For the pure power law a = 1 + 2A = 2 + 2h; the heat factor lowers it: a = 2 + 2h + 2ZH'(Z)/H(Z) with Z = 2d/X (derived here). T0,z/T0,θ ~ y_b^6 → 0 gives the direction (1, 0).",
"antecedent": "Lemma A.8 and Proposition A.10.",
"cost": "Conditional: applies only after a regular inner extension with the same moments exists (supplied by M4.12 and the join in M4.14). Constants may depend on X_R. The flatness rate e^{-4/y_b²} is fixed here and enters ζ.",
"checkable": "Consistency in the heat region X ≥ X_b. Run for this digest: a(Z) = 2 + 2h + 2ZH'(Z)/H(Z) satisfies 2 + h < a ≤ 2 + 2h exactly when Z < 1.632 (h = 0.005), Z < 1.626 (h = 0.009), Z < 1.301 (h = 0.3), and Z = 2d/X ≤ 2/X_b is tiny because X_b > X_R = 110(CP*)^{10}. The tail mechanism is elementary: for a compactly supported residual, -r^{-2}∫_0^r s²R_θ ds equals -r^{-2}∫_0^∞ s²R_θ ds beyond the support. The collar factorization (4.32) needs the Appendix A terminal profile.",
"depends_on": [
"MA.12",
"MA.14",
"M4.10"
],
"constrains": [],
"reasons": {
"MA.12": "Lemma A.8: the moment identities and canonical pressure make the forward stress equal the backward integral, so T0 = 0 for X ≥ X_b.",
"MA.14": "the terminal-collar cone with b_s = 0, 2 + h < a ≤ 2 + 2h, the margin κ_o, and the flat factorization (4.32) are Proposition A.10.",
"M4.10": "applies to leading fields that agree with the Lemma 4.8 family for X ≥ X_tail."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 30"
},
{
"id": "M4.12",
"kind": "move",
"name": "ns-m4-12-axis-construction-and-moment-matching-proposition-4-10",
"title": "Axis construction and moment matching (Proposition 4.10)",
"section": "4",
"pages": "36-38",
"refs": [
"p. 36 to 38",
"Proposition 4.10, (4.33), (4.34)",
"(B.34), (B.35), (B.39)."
],
"statement": "Given the Lemma 4.8 data and finitely many prescribed η-derivative orders, there are axis parameters j0, δ*, σ*, Λ, a logarithmic transition length T_sh, and a threshold C0 such that every C ≥ C0 admits, with X_a = 4/Λ, X_R = 110(CP*)^{10}, x = X/X_R, X_h = X_Re^{-5}, f = (1 + η²)^{-1}, and after C the activation width t1, shear factor κ0 and remaining widths, profiles E = √(2X)ϕ/C, U, Π = Π0 + Cp on [0, X_h] × [-1, 1] with: (i) ϕ > 0 and ϕ, U, Π, V0/X smooth including X = 0, analytic in η on.",
"description": "Given the Lemma 4.8 data and finitely many prescribed η-derivative orders, there are axis parameters j0, δ*, σ*, Λ, a logarithmic transition length T_sh, and a threshold C0 such that every C ≥ C0 admits, with X_a = 4/Λ, X_R = 110(CP*)^{10}, x = X/X_R, X_h = X_Re^{-5}, f = (1 + η²)^{-1}, and after C the activation width t1, shear factor κ0 and remaining widths, profiles E = √(2X)ϕ/C, U, Π = Π0 + Cp on [0, X_h] × [-1, 1] with: (i) ϕ > 0 and ϕ, U, Π, V0/X smooth including X = 0, analytic in η on. OBLIGATION: Provides the core: a smooth, analytic, stress-free flow near the axis with the prescribed axis pressure Π0, which leaves the stress-free region with the stress entering the admissible cone (flatly), and which is continued outward to coincide with the outer reference profile with all five cumulative integrals matched, so Lemma 4.4(i) can glue it to the prepared exterior. MECHANISM: Axis axial data U* = 4η + j0 with a small upward offset 0 < j0 ≤ .05 (the asymmetric profile of Section 2.1). ANTECEDENT: Propositions B.2, B.3, B.5, Corollary B.6, Lemma B.7, Proposition B.8, Corollary B.10. REFS: p. 36 to 38; Proposition 4.10, (4.33), (4.34); (B.34), (B.35), (B.39).",
"obligation": "Provides the core: a smooth, analytic, stress-free flow near the axis with the prescribed axis pressure Π0, which leaves the stress-free region with the stress entering the admissible cone (flatly), and which is continued outward to coincide with the outer reference profile with all five cumulative integrals matched, so Lemma 4.4(i) can glue it to the prepared exterior.",
"backward_question": "Can an axis-regular, stress-free core with a prescribed axis pressure exit into the admissible cone and then be continued to match all five integrals of a prescribed outer reference? And why must the axial profile be asymmetric (j0 > 0), with its dividing layer off z = 0?",
"mechanism": "Axis axial data U* = 4η + j0 with a small upward offset 0 < j0 ≤ .05 (the asymmetric profile of Section 2.1). H* = Dη + dU* is the axial-transport coefficient H_c on the axis; it has a unique zero η0, which lies in (-j0/4, 0) (derived here), and Z* = -A(1 - 2ηU*)U* - H*U*' - dΠ0' + 4AηΠ0 (which is S_n(0, η) on the axis data, derived here) is positive there because both pressure terms are positive at η0 < 0 (Π0 < 0, ηΠ0' > 0) and dominate for large P*. The axis swirl ϕ(0, η) = exp(-Λ∫_0^η LH*/(H*² + σ*²)dw) makes -H*∂_η log ϕ(0, η) = ΛLχ, χ = H*²/(H*² + σ*²) > .99 where |Z*| ≤ δ*; then (4.13) at X = 0, -4Lϕ_X/ϕ = S_q(0, η), gives ϕ_X/ϕ = -Λχ/4 + O(1), and the axial equation gives U_X(0, η) = -Z*/(2L) (both derived here), consistent with the paper's statement that Λ sets the radial scale Y = ΛX. Propositions B.2 and B.3 give the stress-free analytic solution through Y = 4.1 and the exit inequality p_s,1 + p_s,2²/p_s,1 > 2 + c_ex at Y = 4, which is v_s > 2 + c_ex since p_s = s there. Proposition B.5 and Corollary B.6 activate the stress with width t1 and reduce the reference shear by κ0. The continuation ends at X_i = 110 with a(X_i) = 4/5, D_X U(X_i) = 0; U = G_i is held while E is matched to the reference ((B.34)); restoration of U to 4η on e^{-8} < x < e^{-7}; five moment corrections on e^{-6} < x < e^{-5} (Proposition B.8). A tolerance ε_m fixed before C keeps Q_s ≥ Q_min/2, E ≥ e*/2, .7 ≤ a ≤ .9, |N_s/(EQ_s)| ≤ w*, |b_s| ≤ .1/(1 + w*), so v_s ≤ .9 + .01/.7 < 1 and P_c = X_RxG/L > 2 with G = Q_s - b_sN_s/(aE) ≥ (6/7)Q_s: relaxed holds, admissible fails in this region. The choice X_R = 110(CP*)^{10} makes the reference E0 at X = 110 equal f/C, the same 1/C scaling as the inner E = √(2X)ϕ/C (derived here), which is why T_sh is fixed before C and x_sep = e^{T_sh}/(CP*)^{10} → 0.",
"antecedent": "Propositions B.2, B.3, B.5, Corollary B.6, Lemma B.7, Proposition B.8, Corollary B.10.",
"cost": "Many ordered parameters (j0, δ*, σ*, Λ, T_sh, C0, then t1, κ0, ω_fin). X_a = 4/Λ ties the inner annulus edge to Λ. The inner flatness rate e^{-t1²/y_a²} enters ζ (c_a = t1²). In the matching region v_s < 1, so the stretch between the first collar and X_good satisfies only the relaxed cone and must be repaired (M4.13 to M4.15).",
"checkable": "Run for this digest: (4.34) by symbolic integration of the reference pair, all five exact. Arithmetic: .9 + .01/.7 < 1 and G ≥ (6/7)Q_s ≥ 3Q_min/7 are pure inequalities. Positivity of Z* at η0 can be evaluated with any even analytic Π0 satisfying Π0 ≤ -(5/2)P*²f² and ηΠ0' > 0. The axis solve of (4.13) couples X and η derivatives (Cauchy-Kovalevskaya type); a numerical check would Taylor-expand in X with η-dependent coefficients started from ϕ(0, η) and U*, and compare the first X-slopes with -S_q(0)ϕ/(4L) and -Z*/(2L).",
"depends_on": [
"MB.14",
"MB.5",
"MB.8",
"MB.13"
],
"constrains": [],
"reasons": {
"MB.14": "Corollary B.10 is the connection it states, and Remark B.9's order j0, δ*, σ*, Λ, T_sh, C, then t1, κ0 fixes its parameter sequence.",
"MB.5": "part (i), the analytic stress-free profile solving (4.13) with T0 = 0 for X ≤ X_a = 4/Λ, is Proposition B.2.",
"MB.8": "part (ii)'s flat factorization (4.33) and admissibility on the first collar come from the shear-reduction activation of Proposition B.5.",
"MB.13": "part (iii), fields and all five integrals equal to the reference pair's values (4.34) near X_h, is Proposition B.8."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 36-38"
},
{
"id": "M4.13",
"kind": "move",
"name": "ns-m4-13-the-shear-loop-with-prescribed-mean-lemma-4-11",
"title": "The shear loop with prescribed mean (Lemma 4.11)",
"section": "4",
"pages": "38-39",
"refs": [
"p. 38 to 39",
"(4.35), (4.36), (4.37), Lemma 4.11."
],
"statement": "Let I = [X-, X+] ⋐ (0, ∞) and a, b_s, p_s smooth on I × [-1, 1] with a > 0, P_c > 2, v_s < U(P_c, J_c) throughout, and v_s > 2 near both radial endpoints. There is a smooth period-one family (a_L, -b_L)(X, η, φ), φ in R/Z, with ∫_0^1 (a_L, -b_L)dφ = (a, -b_s), equal to (a, -b_s) near both endpoints, such that every loop shear with p_s held fixed satisfies the admissible cone with margin: Ψ_i(a_L, b_L, p_s) ≥ µ_L > 0 for i = 1, ..., 4 (4.36), where Ψ(a, b, p) = (a, v - 2, c - v, 2(c - v)² - (v.",
"description": "Let I = [X-, X+] ⋐ (0, ∞) and a, b_s, p_s smooth on I × [-1, 1] with a > 0, P_c > 2, v_s < U(P_c, J_c) throughout, and v_s > 2 near both radial endpoints. There is a smooth period-one family (a_L, -b_L)(X, η, φ), φ in R/Z, with ∫_0^1 (a_L, -b_L)dφ = (a, -b_s), equal to (a, -b_s) near both endpoints, such that every loop shear with p_s held fixed satisfies the admissible cone with margin: Ψ_i(a_L, b_L, p_s) ≥ µ_L > 0 for i = 1, ..., 4 (4.36), where Ψ(a, b, p) = (a, v - 2, c - v, 2(c - v)² - (v. OBLIGATION: Removes the gap left by the join, where only the relaxed cone holds (v_s may be below 2; in the matching region of M4.12 it is below 1), without changing the integrated vector p_s, which depends only on integrals. MECHANISM: Lemma C.1 supplies a 2π-periodic ratio t_ℓ(X, η, θ′) and ρ_ℓ ≥ 0 with ⟨t_ℓ⟩ = t_s, ⟨(t_ℓ - t_s)²⟩ = ρ_ℓ/a, v_ℓ = v_s + ρ_ℓ > 2, and for every θ′, c_ℓ. ANTECEDENT: Lemma C.1 (the \"variance construction\"), with (C.8) to (C.10). No classical result is cited here; the introduction cites Daneri and Székelyhidi [10] for oscillations realizing a prescribed stress, which concerns the waves rather than this loop. REFS: p. 38 to 39; (4.35), (4.36), (4.37), Lemma 4.11.",
"obligation": "Removes the gap left by the join, where only the relaxed cone holds (v_s may be below 2; in the matching region of M4.12 it is below 1), without changing the integrated vector p_s, which depends only on integrals.",
"backward_question": "If only the relaxed cone holds on an interval, can the shear be replaced by a family whose every member satisfies the admissible cone, while keeping the same average so that the profile itself barely moves?",
"mechanism": "Lemma C.1 supplies a 2π-periodic ratio t_ℓ(X, η, θ′) and ρ_ℓ ≥ 0 with ⟨t_ℓ⟩ = t_s, ⟨(t_ℓ - t_s)²⟩ = ρ_ℓ/a, v_ℓ = v_s + ρ_ℓ > 2, and for every θ′, c_ℓ = p_s,1 + t_ℓp_s,2 > 2 and v_ℓ < U(c_ℓ, j_ℓ). Reparametrize by dφ/dθ′ = a(1 + t_ℓ²)/(2πv_ℓ) and set (a_L, -b_L) = v_ℓ(1, t_ℓ)/(1 + t_ℓ²) (4.37); direct integration gives total length 1 and mean (a, at_s) = (a, -b_s), and Lemma 4.5 gives positivity of Ψ. Geometry (derived here): v = (a² + b²)/a is convex on a > 0, and the loop runs on the circle a² + b² = v_ℓa, every point of which has v = v_ℓ > 2; the weight dφ places the barycenter at the original shear, which lies inside that circle (v_s < v_ℓ). The nonconvex constraint v > 2, the exterior of the disk (a - 1)² + b² < 1, is thus met at every loop point while failing for the average.",
"antecedent": "Lemma C.1 (the \"variance construction\"), with (C.8) to (C.10). No classical result is cited here; the introduction cites Daneri and Székelyhidi [10] for oscillations realizing a prescribed stress, which concerns the waves rather than this loop.",
"cost": "An auxiliary loop parameter and a margin µ_L that may depend on all input data. The loop is only a family of shears, not yet a profile; it is realized at a finite frequency N in M4.14.",
"checkable": "Run for this digest: a = 0.8, b_s = 0.05 (v_s = 0.803), ρ = 2.5 - v_s, t_ℓ(θ′) = t_s + √(2ρ/a)cos θ′; quadrature gives period 1 and mean (0.8, -0.05) to working precision, with v = 2.5 on the whole loop. For a given p_s, evaluating the four components of Ψ along the loop checks (4.36).",
"depends_on": [
"MC.4",
"MC.3",
"M4.7"
],
"constrains": [],
"reasons": {
"MC.4": "the period-one loop (4.37) with exact mean (a, -b_s), a uniform margin, and constancy near the endpoints is Lemma C.1's lift φ.",
"MC.3": "each member's level v = v_s + ρ, with 2 < v < U(P_c(t), J_c(t)), is set by (C.8) to (C.10).",
"M4.7": "its hypothesis is the relaxed cone (4.21), and Ψ_i > 0 in (4.36) is the admissible cone rewritten by Lemma 4.5."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 38-39"
},
{
"id": "M4.14",
"kind": "move",
"name": "ns-m4-14-proof-steps-1-and-2-join-then-realize-the-loop-at",
"title": "Proof Steps 1 and 2: join, then realize the loop at frequency N in log X",
"section": "4",
"pages": "39-42",
"refs": [
"p. 39 to 42",
"Steps 1 and 2 of the proof of Theorem 4.6",
"(4.38) to (4.41)."
],
"statement": "Step 1. Parameters satisfy 0 < C^{-1} ≪ T_sh^{-1} ≪ Λ^{-1} ≪ σ* ≪ δ* ≪ j0 ≪ ε_m ≪ h ≪ λ ≪ P*^{-1} ≪ M_d^{-1} ≪ 1, then 0 < ω_fin ≪ t1 ≪ κ0 ≪ C^{-1}, then N^{-1} ≪ ω_fin (choices right to left, Definition 3.3), with C ≥ max{C0, (R*/110)^{1/10}/P*} so X_R = 110(CP*)^{10} ≥ R*. Define (U0, E0) as the axis connection for X ≤ X_h and the prepared outer pair for X ≥ X_h; both equal (4η, P*(1 + η²)^{-1}(X/X_R)^{1/10}) near X_h.",
"description": "Step 1. Parameters satisfy 0 < C^{-1} ≪ T_sh^{-1} ≪ Λ^{-1} ≪ σ* ≪ δ* ≪ j0 ≪ ε_m ≪ h ≪ λ ≪ P*^{-1} ≪ M_d^{-1} ≪ 1, then 0 < ω_fin ≪ t1 ≪ κ0 ≪ C^{-1}, then N^{-1} ≪ ω_fin (choices right to left, Definition 3.3), with C ≥ max{C0, (R*/110)^{1/10}/P*} so X_R = 110(CP*)^{10} ≥ R*. Define (U0, E0) as the axis connection for X ≤ X_h and the prepared outer pair for X ≥ X_h; both equal (4η, P*(1 + η²)^{-1}(X/X_R)^{1/10}) near X_h. OBLIGATION: Turns the loop of shears into actual smooth profiles that satisfy the admissible cone pointwise on the whole repair interval, while keeping the profiles, the integrals, and p_s within O(1/N) of the joined pair. MECHANISM: A two-scale construction. Since D_X(N log X) = N, the φ-derivative of each primitive enters D_X log E_N and D_X U_N at order one, so the shear follows the loop pointwise (a_N ≈ a_L(X, η, N log X)), while profile values move only by A/N, B/N. ANTECEDENT: Lemma 4.4(ii), which the text says applies to the radial modulation of Proposition C.2 (the appendix version of this step, also described in Section 3.2, item 4). REFS: p. 39 to 42; Steps 1 and 2 of the proof of Theorem 4.6; (4.38) to (4.41).",
"obligation": "Turns the loop of shears into actual smooth profiles that satisfy the admissible cone pointwise on the whole repair interval, while keeping the profiles, the integrals, and p_s within O(1/N) of the joined pair.",
"backward_question": "How can a family of shears, which are radial derivatives, be realized by actual profiles when the integrated stress p_s depends only on profile values and integrals?",
"mechanism": "A two-scale construction. Since D_X(N log X) = N, the φ-derivative of each primitive enters D_X log E_N and D_X U_N at order one, so the shear follows the loop pointwise (a_N ≈ a_L(X, η, N log X)), while profile values move only by A/N, B/N. The zero mean of A and B is what keeps the profiles close (otherwise log E would drift by O(1) across I). p_s depends only on values, integrals, and η-derivatives (Lemma 4.4(ii)), and the phase N log X has no η-dependence, so p_s moves by O(1/N). The loop's strict margin µ_L absorbs these errors through a Lipschitz bound C_Ψ for Ψ on a fixed compact neighborhood. The resulting stress T0 = F(p_s - s) oscillates in log X at frequency N but lies in the cone at every point.",
"antecedent": "Lemma 4.4(ii), which the text says applies to the radial modulation of Proposition C.2 (the appendix version of this step, also described in Section 3.2, item 4).",
"cost": "A large fixed integer N, chosen last. The shear and stress have radial derivatives of size N^k (fixed constants, but large). The O(1/N) moment defect must be repaired exactly (M4.15). The margin halves.",
"checkable": "Symbolic: the a_N formula follows from (4.38) and ∂_φA = -(a_L - a0)/2 (verified for this digest in sympy); b_N the same way. Numeric: the modulation experiment under M4.6 (Δm about N^{-2}, better than the O(N^{-1}) used; Δp_s about N^{-1}; Δa of order one).",
"depends_on": [
"M4.13",
"M4.6",
"M4.12",
"M4.10"
],
"constrains": [],
"reasons": {
"M4.13": "Step 2 realizes the admissible shear loop of Lemma 4.11 through zero-mean primitives evaluated at phase N log X (4.38).",
"M4.6": "Lemma 4.4(i) carries the outer coefficients to the joined pair; Lemma 4.4(ii) bounds Δp_s by value and moment changes, giving (4.40).",
"M4.12": "the inner half of the joined pair is the axis connection of Proposition 4.10, equal to the reference pair with matched moments near X_h.",
"M4.10": "the outer half is the prepared exterior pair of Lemma 4.8, whose admissibility beyond X_good and patch I1 bound the repair interval."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 39-42"
},
{
"id": "M4.15",
"kind": "move",
"name": "ns-m4-15-proof-steps-3-and-4-exact-moment-repair-then-the-edge",
"title": "Proof Steps 3 and 4: exact moment repair, then the edge direction, margin κ, and weight ζ",
"section": "4",
"pages": "42-45",
"refs": [
"p. 42 to 45",
"Steps 3 and 4 of the proof of Theorem 4.6",
"(4.42), (4.43)."
],
"statement": "Step 3. On (Y0, Y1) ⊂ I1 the profile is the patch U = 0, E = K(η)X^{-1/2-λ}, K = c_patch(1 + η²)^{-1}. Add u_c = Σ_{i=1,2}α_i(η)β_i(X) and e_c = Σ_{j=1,2,3}ξ_j(η)γ_j(X) with ordered disjoint bumps in (Y0, Y1); the exact moment changes at Y1, ordered (M, J, I, S, Cp), are B(η)c + Q_η(c, c) = d_N(η) = -Δm_N(Y1, η) (4.42), with block-diagonal linear part B_U (weights X^0, X^{-λ}) acting on (α1, α2) and B_E (weights X^{1/2}, X^{-1/2-λ}, X^{-3/2-λ}) acting on (ξ1, ξ2, ξ3), invertible by Lemma 4.7.",
"description": "Step 3. On (Y0, Y1) ⊂ I1 the profile is the patch U = 0, E = K(η)X^{-1/2-λ}, K = c_patch(1 + η²)^{-1}. Add u_c = Σ_{i=1,2}α_i(η)β_i(X) and e_c = Σ_{j=1,2,3}ξ_j(η)γ_j(X) with ordered disjoint bumps in (Y0, Y1); the exact moment changes at Y1, ordered (M, J, I, S, Cp), are B(η)c + Q_η(c, c) = d_N(η) = -Δm_N(Y1, η) (4.42), with block-diagonal linear part B_U (weights X^0, X^{-λ}) acting on (α1, α2) and B_E (weights X^{1/2}, X^{-1/2-λ}, X^{-3/2-λ}) acting on (ξ1, ξ2, ξ3), invertible by Lemma 4.7. OBLIGATION: Restores exactly the five integrals disturbed by the modulation, so the outer stress, pressure, radial velocity, heat exterior, and moment identities are untouched; keeps the cone on the repair interval; and converts the edge factorizations (4.33), (4.32) into the uniform statements (4.26), (4.27) that later sections use as black boxes. MECHANISM: The repair bumps sit where U = 0 and E is a pure power law, so the linear moment map has power weights with distinct exponents; ANTECEDENT: Lemma 4.7; Appendix C's Proposition C.3 states the same conclusions for the profiles of Proposition C.2. REFS: p. 42 to 45; Steps 3 and 4 of the proof of Theorem 4.6; (4.42), (4.43).",
"obligation": "Restores exactly the five integrals disturbed by the modulation, so the outer stress, pressure, radial velocity, heat exterior, and moment identities are untouched; keeps the cone on the repair interval; and converts the edge factorizations (4.33), (4.32) into the uniform statements (4.26), (4.27) that later sections use as black boxes.",
"backward_question": "After the modulation leaves O(1/N) moment defects, can the five integrals be restored exactly without leaving the cone, and does the stress direction extend to the edges, where T0 vanishes, with a uniform margin?",
"mechanism": "The repair bumps sit where U = 0 and E is a pure power law, so the linear moment map has power weights with distinct exponents; the J, S, Cp rows pick up quadratic terms only (√(2X)u_ce_c, u_c² - e_c²/2, e_c²/(2X)). The defect is O(1/N), so the contraction of Lemma 4.7 applies once N is large, and the repair changes a, b_s, p_s by O(1/N), within the margin µ_R. At the edges, the factorizations isolate smooth nonvanishing direction vectors (B_a and (b_θ, y_b^6b_z)), so n extends continuously; inside, T0·(1, t_s) = F(P_c - v_s) > 0 shows T0 ≠ 0. ζ reproduces exactly the two flatness rates e^{-t1²/y_a²} and e^{-4/y_b²}, so |T0|/ζ is bounded above and below near each edge, and each derivative of an exponential factor costs only finitely many inverse powers of y_a or y_b.",
"antecedent": "Lemma 4.7; Appendix C's Proposition C.3 states the same conclusions for the profiles of Proposition C.2.",
"cost": "Consumes I1 (I2 was consumed by the heat compensation), leaving only I3, I4 for Sections 5 and 8. The margin shrinks to µ_R/4 on the repair interval. The constants C_α, m_α in (4.27) depend on derivative order. ζ and δ become the radial-edge weights of the coefficient classes ((6.23)).",
"checkable": "Solve (4.42) numerically: E0 = KX^{-1/2-λ}, U0 = 0 on (Y0, Y1), fixed bumps β_i, γ_j, and a defect vector of size 1/N; iterate c ↦ B^{-1}(d - Q(c, c)) and confirm the five integrals at Y1 match exactly with ||c|| ≤ 2ν||d||. For ζ: confirm numerically that all X-derivatives of exp(-c_a/y_a² - 4/y_b²) tend to 0 at X_a and X_b. The linear blocks were checked under M4.9.",
"depends_on": [
"M4.14",
"M4.9",
"M4.12",
"M4.11"
],
"constrains": [],
"reasons": {
"M4.14": "Step 3 repairs the O(1/N) moment defect d_N = -Δm_N(Y1) left by the modulated pair of Step 2, within its cone margin.",
"M4.9": "the repair system (4.42) on the power-law patch is solved with Lemma 4.7's invertible blocks and quadratic contraction.",
"M4.12": "the inner-edge factorization (4.33) gives n(X_a) parallel to (a, -b_s) and the factor e^{-t1²/y_a²} of ζ.",
"M4.11": "the outer-edge factorization (4.32) gives n(X_b) = (1, 0) and the factor e^{-4/y_b²} of ζ."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 42-45"
},
{
"id": "M5.1",
"kind": "move",
"name": "ns-m5-1-expansion-in-the-anisotropy-parameter-q-2h",
"title": "Expansion in the anisotropy parameter q^{2h}",
"section": "5",
"pages": "45-62",
"refs": [
"p45 (section plan, formal expansion)",
"p46 ((5.1), and the spacing argument citing (4.1), (4.2), (4.3), (4.7))."
],
"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).",
"description": "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}. 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}. REFS: p45 (section plan, formal expansion); p46 ((5.1), and the spacing argument citing (4.1), (4.2), (4.3), (4.7)).",
"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.",
"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?",
"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.",
"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",
"M4.8",
"M4.2",
"M4.1"
],
"constrains": [],
"reasons": {
"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."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 45-62"
},
{
"id": "M5.2",
"kind": "move",
"name": "ns-m5-2-the-order-n-coefficient-system-5-2-to-5-6-linear-at",
"title": "The order-n coefficient system (5.2) to (5.6), linear at positive order",
"section": "5",
"pages": "45-62",
"refs": [
"p46 ((5.2) to (5.6))",
"p47 (linearity, frame terms, regularity of Ωk/X)."
],
"statement": "Notation: - The operators of (4.2) are T_b f = L^{-1}(-b f + Dη f_η + X f_X) and Z_b f = L^{-1}(2bη f + d f_η - 2ηX f_X). - Set T_{a,n} = T_{a+λn}, Z_{a,n} = Z_{a+λn}, Z^{[2]}_{a,n} = Z_{a+λn-D} Z_{a+λn}, b = -A - 1/2, c = -A. The equations: - (5.2), incompressibility with the axis-regular constant: ∂X Vn = -Z_{-A,n} Un, and Vn/X = [2ηUn - 2η(D+λn) AX(Un) - d ∂η AX(Un)]/L. - (5.3): 2(Xϕn'' + 2ϕn') = T_{b,n}ϕn + Σ_{i+j=n}{Vi(ϕj' + ϕj/X) + Ui Z_{b,j}ϕj} - Z^{[2]}_{b,n-1}ϕ_{n-1}.",
"description": "Notation: - The operators of (4.2) are T_b f = L^{-1}(-b f + Dη f_η + X f_X) and Z_b f = L^{-1}(2bη f + d f_η - 2ηX f_X). - Set T_{a,n} = T_{a+λn}, Z_{a,n} = Z_{a+λn}, Z^{[2]}_{a,n} = Z_{a+λn-D} Z_{a+λn}, b = -A - 1/2, c = -A. The equations: - (5.2), incompressibility with the axis-regular constant: ∂X Vn = -Z_{-A,n} Un, and Vn/X = [2ηUn - 2η(D+λn) AX(Un) - d ∂η AX(Un)]/L. - (5.3): 2(Xϕn'' + 2ϕn') = T_{b,n}ϕn + Σ_{i+j=n}{Vi(ϕj' + ϕj/X) + Ui Z_{b,j}ϕj} - Z^{[2]}_{b,n-1}ϕ_{n-1}. 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}. ANTECEDENT: None cited. Internal: Lemma 4.1, (4.7), and Proposition 4.2 for the order-zero identities. REFS: p46 ((5.2) to (5.6)); p47 (linearity, frame terms, regularity of Ωk/X).",
"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.",
"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?",
"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).",
"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",
"M4.1",
"M4.3",
"M4.4"
],
"constrains": [],
"reasons": {
"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."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 45-62"
},
{
"id": "M5.3",
"kind": "move",
"name": "ns-m5-3-lemma-5-1-part-1-a-local-first-order-system-with-a",
"title": "Lemma 5.1, part 1: a local first-order system with a regular singular point at the axis",
"section": "5",
"pages": "45-62",
"refs": [
"p47 (Lemma 5.1 statement, Step 1, Sρ, (5.7))",
"p48 (Kn relation, radial operators, pressure row)."
],
"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.",
"description": "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. ANTECEDENT: None cited. Internal: the analytic axis solution of Proposition B.2 and the common analytic region [0, Xan] of Theorem 4.6(i). REFS: p47 (Lemma 5.1 statement, Step 1, Sρ, (5.7)); p48 (Kn relation, radial operators, pressure row).",
"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.",
"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?",
"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).",
"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",
"M4.8",
"MB.10"
],
"constrains": [],
"reasons": {
"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."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 45-62"
},
{
"id": "M5.4",
"kind": "move",
"name": "ns-m5-4-lemma-5-1-part-2-nilpotent-derivatives-and-a-picard",
"title": "Lemma 5.1, part 2: nilpotent η-derivatives and a Picard series with half the derivative count",
"section": "5",
"pages": "45-62",
"refs": [
"p48 (A1 entries, G, K, pk, (5.8), Picard series, uniqueness)."
],
"statement": "With Hc = Dη + dU0, the only possibly nonzero entries of A1 in (5.7) are: - (A1)51 = 2Hc/L; - (A1)52 = (A1)53 = -2d(ϕ0 + X∂Xϕ0)/L; - (A1)62 = 2(Hc - dX∂XU0)/L; - (A1)63 = -2dX∂XU0/L; - (A1)64 = 2d/L. So A1 maps the first four coordinates into the last two and annihilates the last two. For every intervening diagonal kernel D0, A1(ξ)D0A1(s) = 0 and A1(ξ)D0∂ηA1(s) = 0. Define (Gg)_i(ξ) = ∫_0^ξ (s/ξ)^{c_i} g_i(s) ds and K = G(A0 + A1∂η).",
"description": "With Hc = Dη + dU0, the only possibly nonzero entries of A1 in (5.7) are: - (A1)51 = 2Hc/L; - (A1)52 = (A1)53 = -2d(ϕ0 + X∂Xϕ0)/L; - (A1)62 = 2(Hc - dX∂XU0)/L; - (A1)63 = -2dX∂XU0/L; - (A1)64 = 2d/L. So A1 maps the first four coordinates into the last two and annihilates the last two. For every intervening diagonal kernel D0, A1(ξ)D0A1(s) = 0 and A1(ξ)D0∂ηA1(s) = 0. Define (Gg)_i(ξ) = ∫_0^ξ (s/ξ)^{c_i} g_i(s) ds and K = G(A0 + A1∂η). 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. 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. REFS: p48 (A1 entries, G, K, pk, (5.8), Picard series, uniqueness).",
"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.",
"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?",
"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.",
"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",
"M5.2"
],
"constrains": [],
"reasons": {
"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."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 45-62"
},
{
"id": "M5.5",
"kind": "move",
"name": "ns-m5-5-lemma-5-1-part-3-smoothness-in-x-at-the-axis-via-volterra",
"title": "Lemma 5.1, part 3: smoothness in X at the axis via Volterra identities and parity",
"section": "5",
"pages": "45-62",
"refs": [
"p48 to p49 (Step 3 of the proof of Lemma 5.1)."
],
"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.",
"description": "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; ANTECEDENT: None cited. (The even-function fact is classical and often attributed to Whitney; that attribution is mine.) REFS: p48 to p49 (Step 3 of the proof of Lemma 5.1).",
"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.",
"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?",
"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.",
"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",
"M5.3",
"M4.2"
],
"constrains": [],
"reasons": {
"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."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 45-62"
},
{
"id": "M5.6",
"kind": "move",
"name": "ns-m5-6-order-n-stress-primitives-and-the-five-total-moments",
"title": "Order-n stress primitives and the five total moments",
"section": "5",
"pages": "45-62",
"refs": [
"p49 (rθ,n, rz,n, (5.9), (5.10), (5.11), Fn)",
"p50 (mn, Pn)."
],
"statement": "The order-n residual coefficients are: - rθ,n = (R/C)[right side of (5.3) minus its left side]; - rz,n = [right side of (5.4) minus its left side]; - here R = √(2X). Both vanish where the inner equations hold. The physical residuals are Rj,n = q^{-A-1+λn} rj,n. The stress is (5.9), with physical stress q^{-A-1/2+λn} Tn: - Tn,θ(R) = -R^{-2} ∫_0^R ϱ^2 rθ,n dϱ; - Tn,z(R) = -R^{-1} ∫_0^R ϱ rz,n dϱ. The five total moments are (5.10) to (5.11).",
"description": "The order-n residual coefficients are: - rθ,n = (R/C)[right side of (5.3) minus its left side]; - rz,n = [right side of (5.4) minus its left side]; - here R = √(2X). Both vanish where the inner equations hold. The physical residuals are Rj,n = q^{-A-1+λn} rj,n. The stress is (5.9), with physical stress q^{-A-1/2+λn} Tn: - Tn,θ(R) = -R^{-2} ∫_0^R ϱ^2 rθ,n dϱ; - Tn,z(R) = -R^{-1} ∫_0^R ϱ rz,n dϱ. The five total moments are (5.10) to (5.11). 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. 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). REFS: p49 (rθ,n, rz,n, (5.9), (5.10), (5.11), Fn); p50 (mn, Pn).",
"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.",
"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?",
"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.",
"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",
"M5.2",
"M4.5"
],
"constrains": [],
"reasons": {
"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."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 45-62"
},
{
"id": "M5.7",
"kind": "move",
"name": "ns-m5-7-radial-extension-by-a-cutoff-plus-five-reserved-patch",
"title": "Radial extension by a cutoff plus five reserved-patch bumps, and the affine block moment solve",
"section": "5",
"pages": "45-62",
"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))."
],
"statement": "Lemma 5.2 (radii Xa < X- < X+ < Xb, independent of n), Steps 1 and 2. - Fix X- < Xkeep < Xcut < a^2 < inf Ipos, and an η-independent cutoff κ with κ = 1 on [0, Xkeep] and κ = 0 for X ≥ Xcut. - Set ϕ̃n = κϕn^in and Ũn = κUn^in. - (5.14): Un = Ũn + Σ_{j=1}^{2} αn,j(η) b^U_j(R) and En = Ẽn + Σ_{j=1}^{3} βn,j(η) b^E_j(R), with ϕn = (C/R)En. The five bumps are nonnegative with unit mass, have disjoint ordered supports in Jpos (Ipos in the R coordinate), and are fixed across orders.",
"description": "Lemma 5.2 (radii Xa < X- < X+ < Xb, independent of n), Steps 1 and 2. - Fix X- < Xkeep < Xcut < a^2 < inf Ipos, and an η-independent cutoff κ with κ = 1 on [0, Xkeep] and κ = 0 for X ≥ Xcut. - Set ϕ̃n = κϕn^in and Ũn = κUn^in. - (5.14): Un = Ũn + Σ_{j=1}^{2} αn,j(η) b^U_j(R) and En = Ẽn + Σ_{j=1}^{3} βn,j(η) b^E_j(R), with ϕn = (C/R)En. The five bumps are nonnegative with unit mass, have disjoint ordered supports in Jpos (Ipos in the R coordinate), and are fixed across orders. 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. 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). 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)).",
"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.",
"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.",
"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.",
"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",
"MA.1",
"M5.3",
"M4.8"
],
"constrains": [],
"reasons": {
"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."
},
"statement_leaks_reason": false,
"statement_leaks_answer": false,
"verified": true,
"source": "OpenAI 2026, Finite Time Blowup for Navier-Stokes, pp. 45-62"
},