The forced blowup theorem and its geometry

NS-BLOWUP N1. Independent exposition by GPT-6 Astra (OpenAI). Authored lesson text is CC0.

A fluid velocity can become arbitrarily large while its total kinetic energy stays bounded. The size of the region carrying that velocity matters. The construction studied here concentrates its fastest flow near one point, controls its full momentum residual, and turns that residual into a force smooth through the singular time.

This lesson states the exact theorem and proves the coordinate, volume, energy, stress and scaling calculations that give its mathematical picture. The full profile and correction constructions have complete proofs in the linked audit. Subsequent lessons develop those proofs in teaching order. The existing summation lesson proves the infinite-sum step in full.

Prerequisites are multivariable differentiation, integration by parts, the implicit function theorem, and the definitions of curl, divergence and the Lebesgue norms. We give the specific coordinate and integral arguments used here. The source is OpenAI’s Finite Time Blowup for Navier–Stokes, Theorem 1.1 and Sections 2–3.

N1.1. What the theorem says

Write \(x=(x_1,x_2,z)\in\mathbb R^3\). For a vector field \(u=(u_1,u_2,u_3)\), the \(i\)-th component of \((u\cdot\nabla)u\) is \(\sum_{j=1}^3u_j\partial_{x_j}u_i\); here \(x_3=z\). The Laplacian acts on each component. Fix a viscosity \(\nu>0\), and retain the full residual

A force \(f\) makes this an exact forced solution when \(\mathscr R_\nu(u,p)=f\) and \(\nabla\cdot u=0\).

The whole-space theorem. For every prescribed \(\nu>0\), there exist

and smooth velocity and pressure \(u_\nu,p_\nu\) on \(\mathbb R^3\times[0,1)\) satisfying

There is no smooth solution on all of \(\mathbb R^3\times[0,\infty)\) with the same force and initial velocity whose kinetic energy is uniformly bounded in time.

Here \(C_c^\infty\) requires one compact support in the open space–time domain and derivatives of every order. In particular the force vanishes on an initial time interval, vanishes after some finite time, and is smooth at \(t=1\). A function merely smooth at every \(t<1\) would not meet this requirement. Smoothness of \(u_\nu,p_\nu\) on \([0,1)\) means smoothness up to the initial time and on every shorter closed time interval. It does not assert finite limits of their derivatives at the singular point.

Kinetic energy is

The factor \(1/2\) is retained throughout. The theorem’s force is part of the construction; the theorem does not assert the same behavior for every force.

The periodic statement. On the particular torus \(\mathbb T^3=\mathbb R^3/\mathbb Z^3\), for every \(\nu>0\) there is a smooth force compactly supported in time and a solution \((U,P_{\rm per})\) with zero initial velocity, smooth for \(0\le t<1\), such that

Velocity and periodic pressure have support in one fixed compact subset of the interior of \((-1/2,1/2)^3\) at all times before one. No global smooth periodic solution has the same datum and force. This is original Corollary 10.6; the exact map producing it appears in N1.7.

The complete supplying proofs for these statements are the local construction, I0–I7 and compact force and final argument, F0–F8, together with their included profile and operator proofs. N1.8 identifies every part of that chain. The theorem’s statement here is a reference to that completed mathematical argument, not a new assumption used to supply its missing steps.

N1.2. The original coordinates, with their exact inverse

The profile construction first works at viscosity one; N1.7 proves the full map to each \(\nu>0\). Keep its fixed parameters

The three-dimensional cylindrical coordinates and unit vectors are

The unit vectors at \(r=0\) are not used to define the Cartesian field. Its smooth extension there will be proved below.

Retain the source’s variables \(q,X,\eta\):

Eliminating \(\eta\) gives

For \(z\ne0\), the left side is zero at \(q=|z|^{1/D}\), tends to infinity as \(q\to\infty\), and has derivative

The inequality follows because \(2h-1=-2D\), so \(z^2q^{2h-1}\le1\) on that interval. Any solution of N1.9 with \(\tau>0\) lies strictly above its left endpoint. The intermediate value theorem and strict monotonicity give exactly one positive solution. For \(z=0\), it is \(q=\tau\). The derivative at every solution is nonzero, proving smoothness by the implicit function theorem.

There are useful explicit comparisons. Put \(b=|z|^{1/D}\) and \(a=2^{1/(2D)}\). Then

For the upper bound set \(Q=a(\tau+b)\). Since \((b/Q)^{2D}\le a^{-2D}=1/2\),

Here \(a\ge2\), since \(D<1/2\). Monotonicity from N1.10 gives \(q\le Q\). The lower bounds follow directly from N1.9 and its positive left side. When \(z=0\), the bound involving \(b\) is interpreted with \(b=0\).

Let \(d=1-\eta^2\) and \(L=1-2h\eta^2\). At fixed physical coordinates, differentiation gives

For example \(q_t=-1/L\) follows by differentiating N1.9 with \(\partial_t\tau=-1\). From \(\eta=zq^{-D}\), \(\eta_z=q^{-D}(1-2D\eta^2/L)=q^{-D}d/L\), because \(L-2D\eta^2=1-\eta^2\). The remaining formulas follow by the same product rule. Every denominator stays positive, since \(L\ge1-2h\).

For a profile \(G(X,\eta)\) and any real \(b_0\), these formulas imply the complete first derivatives

These operators retain the derivative of the radial profile coordinate \(X\); treating \(X\) as fixed during physical differentiation would change the equation.

At \(\tau=0\) with \(z\ne0\), N1.9 gives

The implicit derivative remains positive. Thus the endpoints \(\eta=\pm1\) with positive \(q\) describe terminal points away from the origin. Approaching \(q=0\) on a bounded \(X\) range instead forces \(r\to0\), \(z\to0\), and \(t\to1\).

N1.3. The leading velocity and its smooth axis

Use the actual leading profiles constructed in the audit. They have the form

Their complete existence proof is G/O/H/J/Q/V, ending with V7. In particular \(E/\sqrt{2X}\), \(U\), \(V_0/X\), and \(\Pi\) extend smoothly to \(X=0\), and \(E(X,\eta)>0\) for \(X>0\). These are properties of the supplied profiles; they are not properties of arbitrary choices of four functions.

Here is the exact relation that fixes the radial component. Set

Then \(F_U=X\mathcal A_XU\), including at \(X=0\). The fixed-interval integral proves smoothness of \(\mathcal A_XU\) and each derivative. Since \(A+D=1\), N1.13 gives

Indeed the cylindrical radial contribution is \(r^{-1}\partial_r(ru_r^{(0)})=q^{-1}(V_0)_X\), and the angular contribution vanishes because the leading components are independent of \(\theta\).

Integrating the condition that N1.17 vanish, with \(V_0(0,\eta)=0\), gives

To check the coefficient of \(F_U\), use \(\int_0^XyU_y\,dy=XU-F_U\) and \(A-1=-D\). Differentiating N1.18 reproduces N1.17 exactly. The formula also proves that \(V_0/X\) is smooth at the axis.

Define \(F_E=E/\sqrt{2X}\) and \(v_0=V_0/X\), with their smooth axis values. The actual Cartesian components are

At fixed \(t<1\), \(q\ge1-t>0\), and \(X=(x_1^2+x_2^2)/(2q)\) is smooth. Thus N1.19 is smooth across \(r=0\), with zero angular and radial components on the axis. The cylindrical factor \(1/r\) in N1.15 has not produced a Cartesian singularity.

The original pressure is fixed by its radial integral:

The supplied exterior has \(E=O(X^{-A})\), so the integrand is \(O(X^{-2A-1})\) at infinity and integrable, since \(A>0\). At zero it is the smooth function \(F_E^2\). Consequently

This is the centrifugal radial balance. It cancels exactly those two terms. The time derivative, radial transport, and radial and axial viscous terms of the full residual still have to be included; the higher-order construction supplies them.

N1.4. An exact shrinking core and its energy

Let \(X_a>0\) be the supplied inner-profile radius. The local construction I6 fixes \(X_{\rm in}=X_a/2\). Choose fixed \(X_{\rm in}<X_c<X_a\) and \(0<\eta_c<1\) with \([0,X_c]\times[-\eta_c,\eta_c]\) inside that inner region. This choice also places the later growth point inside the core. At time \(t=1-\tau\), define the actual spatial region

The variables \(X,\eta\) retain N1.8; this region is not a cylinder with arbitrarily assigned height. At fixed \(\tau\),

The derivative of \(z\) with respect to \(\eta\) is

Thus each allowed \(z\) corresponds to exactly one \(\eta\). The maximal radius and the positive axial extent are exactly

Their ratio is

Both lengths go to zero, and the radial extent goes to zero faster.

The drawing shows the ordering and formulas of N1.22–N1.31. It is a schematic, with no numerical velocity samples or substituted profile constants. The editable SVG is its reproducible source.

The full volume element follows from \(dx=r\,dr\,d\theta\,dz\). At fixed \(\eta\), \(r\,dr=q\,dX\). The determinant in \((X,\eta)\mapsto(r,z)\) has no second term because \(z_X=0\). Combining N1.23–N1.24 gives

In particular

The integral is a finite strictly positive constant: its integrand is continuous and positive on a fixed compact interval. No part of that constant has been discarded. The volume power is \(1+D=3/2-h\).

Consider the kinetic energy of the leading field on this specific region. Its angular and axial contributions are exactly

Its radial contribution is

For N1.30, use \((u_r^{(0)})^2=V_0^2/(2qX)\); the energy factor \(1/2\), angular integral \(2\pi\), and denominator \(2X\) give the coefficient \(\pi/2\). The integrand at \(X=0\) extends continuously as \(Xv_0^2\). Both integrals are finite. The first is strictly positive because \(E>0\) on \(X>0\).

The two time powers are

Thus this core’s leading kinetic energy tends to zero, even though its leading angular velocity grows. This exact calculation concerns \(u^{(0)}\) on \(\mathcal C_\tau\). The full corrected velocity also contains oscillations, higher coefficients and localization terms. N1.6 proves its uniform global energy bound from the exact equation, without assuming that the leading energy calculation controls every correction.

The region \(\mathcal C_\tau\) is defined by fixed profile coordinates. It need not be a material region transported by \(u\). For a smooth incompressible velocity, a material flow map \(Y(a,t)\) satisfies \(\partial_tY=u(Y,t)\). Differentiating in \(a\) gives \(\partial_tD_aY=(D_xu)(Y,t)D_aY\); hence the determinant obeys \(\partial_t\det D_aY=(\nabla\cdot u)(Y,t)\det D_aY=0\). With initial determinant one, it remains one. This volume preservation for material sets is consistent with N1.28: the sets \(\mathcal C_\tau\) are selected by N1.22, not by a fixed collection of moving fluid particles.

N1.5. Why oscillations can carry the missing stress

At fixed \(z,t\), write the two tangential residuals as \(R_\theta(r)\) and \(R_z(r)\). The original radial stress equations are

Multiplication by \(r^2\) or \(r\), respectively, and integration from the axis give the axis-regular choices

Indeed differentiating \(r^2T_{r\theta}\) and \(rT_{rz}\) recovers N1.32, including its minus signs. The homogeneous integration constants would give \(r^{-2}\) and \(r^{-1}\) terms and are excluded by axis regularity.

Suppose a residual is zero for \(r\le r_a\) and \(r\ge r_b\), with \(0<r_a<r_b\). The corresponding stress is zero for \(r\le r_a\). For \(r\ge r_b\), it is zero if and only if the appropriate full moment vanishes:

Necessity and sufficiency both follow from the constant numerators of N1.33 outside the support. These exact moments explain why radial matching is part of the construction.

An oscillation can have zero mean velocity and nonzero mean momentum flux. To see every coefficient, let \(a,b\in\mathbb R^3\), let \(\varphi\) range over a circle, and define

Direct integration gives

The mixed integrals vanish and each square integrates to \(\pi\). Thus \(\langle w_rw_\theta\rangle=(a_ra_\theta+b_rb_\theta)/2\) and \(\langle w_rw_z\rangle=(a_ra_z+b_rb_z)/2\). Changing \(w\) to \(-w\) leaves both products unchanged.

The actual construction supplies two flux directions \(v_1,v_2\in\mathbb R^2\) and positive coefficients \(c_1,c_2\) with

At any point where \(\det(v_1,v_2)\ne0\), the coefficients have the exact formulas

Multiplying these expressions out in coordinates proves N1.37; uniqueness follows from the nonzero determinant. Positivity is the additional restriction defining the interior of the cone spanned by these ordered vectors. The profile construction proves that its actual stress satisfies this restriction, and the covariance proof, C1–C8 constructs the amplitudes, including their edge behavior.

N1.36 is an averaging identity, not by itself a solution of the momentum equation. Actual waves have moving phases, varying envelopes, viscosity, pressure and compact cutoffs. Their exact divergence-free velocities come from curls, and their derivatives contain additional terms. The pulse proof, P0–P8 and the four-operation cycle, K0–K8 retain those terms. The later mean corrections restore the exact compatibility conditions. The single actual sum then has a residual whose every fixed derivative decays to arbitrary order near the singular point.

N1.6. Smooth forcing, energy, and the surviving growth path

For any smooth divergence-free \(v\), assigning \(f=\mathscr R_\nu(v,p)\) makes the equation true at times where \(v,p\) are smooth. That identity alone gives no smooth extension of \(f\) through a singular time. The construction proves that extension for the full residual, after all corrections.

Localization also has to preserve incompressibility. If \(v=\operatorname{curl}\mathcal A\) and \(c\) is a smooth cutoff, then

The first equality is the curl product rule. The second follows by cancellation of mixed partial derivatives in divergence of a curl. The term \(\nabla c\times\mathcal A\) is part of the localized velocity and its residual. If one instead used \(cv\), its divergence would be \(\nabla c\cdot v\), which generally does not vanish. The actual local field has the potentials and smooth extensions needed for N1.39; F0–F3 proves every cutoff term, every terminal force derivative and the compact extension after time one.

Now use that actual compactly supported solution, for the original prescribed viscosity \(\nu\). At every \(T<1\), integration of the exact equation against \(u_\nu\) is legitimate on its fixed compact support. The time term is the derivative of N1.4. Since \(\nabla\cdot u_\nu=0\),

Both integrals vanish. Integrating the Laplacian by parts gives

Define \(F_\nu(t)=\int_0^t\|f_\nu(s)\|_2\,ds\). It is finite up to time one because the force is smooth with fixed compact support. For \(\delta>0\), divide N1.40 by \((\|u_\nu(t)\|_2^2+\delta^2)^{1/2}\), discard only the nonnegative dissipation in this inequality, and apply Cauchy–Schwarz. This yields

Using \(u_\nu(0)=0\), integrate and let \(\delta\downarrow0\) to obtain \(\|u_\nu(t)\|_2\le F_\nu(t)\). Return to the full identity N1.40 and retain dissipation:

Monotone convergence of the nonnegative dissipation integrals gives \(\int_0^1\|\nabla u_\nu(s)\|_2^2ds\le F_\nu(1)^2/(2\nu)\). This is a global bound for the full corrected field.

The unbounded velocity is witnessed at a particular Cartesian point. At viscosity one, the actual construction supplies \(X_{\rm in}>0\), \(e_0=E(X_{\rm in},0)>0\), and one finite constant \(C\) such that

The unchanged inner profile and the full summation remainder give this estimate in I6 and F6. All annular oscillations vanish on that path, and both localization cutoffs are one there for sufficiently small \(\tau\). At \(x_\tau\), \(\theta=0\) and \(e_\theta=(0,1,0)\), so the displayed component is \(u_2\). Choose \(\tau\) small enough that \(C\tau^{2h}\le e_0/2\). Then

Continuity implies that the essential supremum is at least this pointwise value: every smaller value is exceeded in a neighborhood of the point. Since N1.43 holds at every sufficiently small positive \(\tau\), the norm has full limit \(+\infty\), which implies the source’s stated limsup conclusion.

To exclude a different global smooth solution, one also needs a comparison theorem. For any \(T<1\), the constructed compact field is in \(L^\infty([0,T];L^2\cap L^6)\) and has \(\nabla u\in L^1([0,T];L^\infty)\). A smooth competitor with bounded kinetic energy supplies the other \(L^\infty_tL^2_x\) input. The complete whole-space comparison proof, equations (1)–(64) proves equality on \([0,T]\), without assuming decay of the competitor’s pressure. It retains the expanding spatial cutoff flux and proves its Sobolev and pressure estimates. Thus a global smooth competitor would equal \(u\) on every such interval. It would be bounded on a compact neighborhood of \((0,1)\), contradicting N1.43 along \(x_\tau\to0\). N1.7 transfers this conclusion to every viscosity.

N1.7. Every viscosity and the original periodic domain

Let \((u,p,f)\) be the constructed viscosity-one fields. For any \(\nu>0\), use the exact map

Writing \(y=x/\sqrt\nu\), the four momentum terms become

They therefore satisfy \(\mathscr R_\nu(u_\nu,p_\nu)=f_\nu\). Velocity and spatial derivative factors cancel in divergence. Time and the zero initial datum are unchanged. A compact spatial set \(\mathcal K\) maps to \(\sqrt\nu\,\mathcal K\). Each force derivative retains

For each fixed positive \(\nu\) these are finite constant multiples of the original derivatives, so smooth compact support is preserved.

The volume change is \(dx=\nu^{3/2}dy\). Hence

The full energy inequality N1.41 has the same factor \(\nu^{5/2}\) on both sides. At the point \(\sqrt\nu\,x_\tau\), the lower bound in N1.43 is multiplied by \(\sqrt\nu\). If a global competitor at viscosity \(\nu\) existed, the inverse map

would produce a viscosity-one competitor with energy \(\nu^{-5/2}\|v_\nu(t)\|_2^2\), contradicting N1.6.

For the periodic statement, fix this viscosity and choose \(\lambda>1\) so large that the common support of velocity, pressure and force, after division by \(\lambda\), lies strictly inside \(Q_0=(-1/2,1/2)^3\). Such a choice exists by compactness. Put \(t_0=1-\lambda^{-2}\), and define

The first two are used for \(t_0\le t<1\); the force continues beyond one. Extend all three by zero for \(t<t_0\). The original construction vanishes on an initial interval, so the extensions are smooth there. Every momentum term has factor \(\lambda^3\); divergence has factor \(\lambda^2\). The viscosity remains \(\nu\), and original time one is reached precisely at \(t=1\).

Set

The translates have disjoint closed supports with a positive gap. The sums are locally finite, all derivatives can be taken term by term, and all nonlinear products between distinct translates vanish. Thus

This proves the exact periodic equation, including a periodic pressure. Compact time support of the force survives N1.49. The velocity and pressure are supported in the same fixed compact subset of \(Q_0\) before time one.

At

only the zeroth translate contributes, and

The periodic comparison needs no spatial cutoff. If \(V\) is a smooth periodic competitor, put \(w=V-U\) on a shorter interval. Subtract the equations in the form

Integration against \(w\), periodic integration by parts, and divergence zero give

Multiply by \(\exp(-2\int_0^t\|\nabla U(s)\|_\infty ds)\). Its product with \(\|w(t)\|_2^2\) is nonincreasing, nonnegative, and initially zero, so \(w=0\). A global smooth competitor would therefore contradict N1.52. This proves the periodic implication from the supplied compact whole-space construction.

N1.8. How the construction supplies the theorem

The mechanism is a succession of exact constructions, not an inference from a picture or a sample of a flow. The table identifies the complete proofs available for each part; N2–N7 develop them in the corresponding lesson sequence.

The same pressure and velocity pass through these maps. In particular, the mean correction uses the reserved interval supplied by the actual profiles, the pulse inverse uses their actual shear margins, and the final localization uses the actual potential sum. The complete proof connections, Sections 2–3, check the order of choices and these receivers.

The human smooth-forcing programme provides essential context for the question. Alpöge, Buckmaster and Coiculescu’s IPM paper, arXiv:2609.16470v1, explicitly develops the Córdoba–Martínez-Zoroa approach and focuses on smoothness jointly in space and time. Its introduction and overview describe retained nonlinear residuals and oscillatory corrections. The source’s IPM theorem is a separate theorem about a transported density; it is not an input to the Navier–Stokes proof in the table. The Courant account of Alpöge and Buckmaster’s Euler and Boussinesq work identifies the companion smooth-forcing programme. Their complete mathematical treatments belong to the corresponding foundation lessons. Here the proved Navier–Stokes calculation N1.1, N1.32–N1.41 makes the relevant issue explicit: an exact residual must meet the required force class through the terminal time.

The workbench contribution used in this course is the independent derivation and comparison of these maps, including the full pressure comparison and the retained terms in the summation. Original author formulas remain identifiable. Transcription corrections restore what OpenAI wrote; editorial mathematical arguments stay in separate exposition.

N1.9. Five worked exercises

Exercise 1. Find the core’s exact horizontal section. Fix \(z\) and \(\tau>0\). Determine when this height meets \(\mathcal C_\tau\), and give the disk radius there.

Solution. The function \(\eta\mapsto\tau^D\eta(1-\eta^2)^{-D}\) is strictly increasing by N1.24. It maps \([-\eta_c,\eta_c]\) onto \([-z_{\max},z_{\max}]\). Thus the height meets the core exactly when \(|z|\le z_{\max}\). Solve that strictly monotone scalar equation for its unique \(\eta\), set \(q=\tau/(1-\eta^2)\), and use N1.22:

This is a disk, with the full factor \(2X_c\) and the same \(\eta\) determined by the height. At \(z=0\) it has radius \(\sqrt{2X_c\tau}\); at either endpoint height it has radius \(r_{\max}\). The boundary need not have the shape of a cylinder.

Exercise 2. Compare the two exact core-energy contributions. Denote the constant integrals, including their coefficients, in N1.29 and N1.30 by \(C_{\rm tan}\) and \(C_r\). Prove a formula for their ratio and show when it is zero.

Solution. The earlier proof gives \(0<C_{\rm tan}<\infty\) and \(0\le C_r<\infty\). Therefore

It tends to zero because \(h>0\). Since every factor in the radial integral is positive except \(V_0^2/X\), its value is zero precisely when \(V_0=0\) on the whole core profile rectangle. For necessity, a nonzero value at any \(X>0\) would, by continuity, give a neighborhood with positive integral. Axis continuity then includes \(X=0\). For sufficiency the integrand is identically zero. This statement concerns the leading field and does not omit corrections from the full-field energy estimate.

Exercise 3. Show why zero-mean waves can carry either sign of a cross flux. Construct two vector waves with zero mean, the same mean kinetic energy, and opposite radial–angular flux.

Solution. In the ordered components \((r,\theta,z)\), take

Both means vanish. Their mean squared lengths are \(2\langle\cos^2\varphi\rangle=1\), and their mean kinetic energies are \(1/2\). Their radial–angular fluxes are respectively \(1/2\) and \(-1/2\); both radial–axial fluxes are zero. Negating all components of either wave preserves its flux, while changing just the angular sign reverses that cross flux. These are exact circle identities. Constructing waves with the required equation, support and pressure remains the pulse construction cited in N1.5.

Exercise 4. Prove that the theorem’s force cannot be zero. Use the exact energy identity and the initial datum.

Solution. If \(f_\nu=0\), N1.41 has right side zero. Each term on its left is nonnegative, so \(\|u_\nu(t)\|_2=0\) at every \(t<1\). Smoothness makes the field identically zero: any nonzero point would have a neighborhood with positive square integral. This contradicts N1.43 and its viscosity scaling. The same argument applies on the torus, using periodic integration by parts. Thus the force in this construction is nonzero, even though it extends smoothly through the singular time.

Exercise 5. Retain every constant in periodic energy and growth. Compute the periodic \(L^2\) energy from N1.49–N1.50, and express the lower bound N1.52 in terms of \(1-\widetilde t_\tau\).

Solution. At any \(t_0\le t<1\), disjoint translates and a fundamental cube give

The kinetic energy is half this expression. The coefficient is \(\lambda^2\lambda^{-3}\), not one. Also \(\tau=\lambda^2(1-\widetilde t_\tau)\), so

The growth point is \(\widetilde x_\tau=(\sqrt{2\nu X_{\rm in}(1-\widetilde t_\tau)},0,0)\); this follows by substituting the same \(\tau\) into N1.42 and retaining both spatial scale factors. The factor \(\lambda^{-2h}\) is positive and fixed, so the lower bound diverges while the energy stays bounded.

Sources and reading scope

OpenAI’s original 166-page paper supplies Theorem 1.1, the original geometry in Sections 2–3, and Corollary 10.6. The original pages used for this lesson are 1–4, 6–10 and 125, with the full local and final constructions supplied by the pinned audit links above. N1.27–N1.30 give the full coordinate Jacobian and energy integrals behind the source’s core-size calculation. They are editorial derivations in the original variables.

For the human smooth-forcing programme, see Alpöge–Buckmaster–Coiculescu, arXiv:2609.16470v1, introduction and construction overview. This citation supplies historical and mathematical context; the present lesson concerns the forced Navier–Stokes theorem. Bibliographical details are in SOURCES.json.

The local-field construction, compact force extension and energy and comparison argument complete the proof of the theorem stated here.

From velocity profiles to stress and five exact moments

NS-BLOWUP N2.1. A leading velocity field is useful only if its full residual has the form needed by the later corrections. This lesson derives that form and proves how to alter the profiles without disturbing their exterior.

We answer three concrete questions:

• How does the velocity profile determine the radial stress, with pressure and both viscous contributions retained?
• Which five moments must be restored after changing a profile, and how can their nonlinear correction equations actually be solved?
• Which stress directions can be supplied by positive wave covariances?

The opening lesson supplies the exact target theorem and its geometry. We recall the needed coordinates here, and give the complete calculations for the stress and finite correction mechanisms. These are tools for the actual outer, axis, joining and shear constructions; the complete leading-profile existence proof is supplied by the linked construction papers and is developed further in N2. This lesson does not replace those constructions with an assumption.

The original formulas come from OpenAI's Finite Time Blowup for Navier–Stokes, Sections 4.1–4.3, Lemma 4.7 and Appendix A.1. All arguments here are editorial exposition, separate from the diplomatic transcription. Original mathematical irregularities, if found, belong in separate editorial discussion.

P0. Coordinates, regularity and incompressibility

The starting data are velocity profiles \(E,U\) and one prescribed pressure value \(\Pi_{\rm ax}(\eta)\). They determine the radial velocity and pressure by exact integration. The angular profile contains a square root at the axis, so we work with its actual smooth coefficient \(F=\phi/\mathsf C\) when testing Cartesian regularity. The factor \(\mathsf C\) remains fixed and visible.

Keep the original parameters and variables

For each \(\tau>0,z\in\mathbb R\), the equation \(q-z^2q^{2h}=\tau\) has exactly one solution \(q>|z|^{1/D}\). Indeed its left side vanishes at the lower endpoint, tends to infinity, and has derivative \(L\ge1-2h>0\) on that half-line. When \(z=0\), the endpoint is zero and the same argument applies. Thus \(|\eta|<1\). The implicit function theorem gives smooth \(q,\eta\) at every \(t<1\). Differentiation gives

The identity \(L-2D\eta^2=d\) proves the middle axial formula. For \(\mathcal D_X=X\partial_X\), any real \(b\), and a smooth profile \(f\), the chain rule therefore gives

All endpoint statements in \(\eta\) mean continuous one-sided derivatives.

Retain the fixed amplitude factor \(\mathsf C>1\) and set

The axis data are the smooth functions \(F=\phi/\mathsf C,U,v_0,\Pi\), with \(\phi>0\), on each finite closed \((X,\eta)\) rectangle. This does not require \(E\) to be smooth as a function of \(X\) at zero. In Cartesian coordinates the exact field is

These expressions prove smoothness across the axis for \(t<1\).

The radial average is \(\mathcal A_Xf=\int_0^1f(aX,\eta)\,da\). Differentiation under this fixed finite integral proves its smooth extension with value \(f(0,\eta)\). Since \(A+D=1\), incompressibility is exactly

Integrate from zero, use \(V_0(0,\eta)=0\) and \(\int_0^XxU_X\,dx=XU-\int_0^XU\,dx\). This gives the original formula, including its sign,

The second identity equates the coefficients \(\sqrt{2X}\Pi_X\) and \(E^2/\sqrt{2X}\) of \(\partial_rp^{(0)}\) and \((u_\theta^{(0)})^2/r\), both with factor \(q^{-2A-1/2}\). It cancels precisely those radial terms. The remaining radial terms are retained in P1.6 below.

P1. Exact stress, including both kinds of viscosity

The momentum residual is \(\mathcal R_\nu(u,p)=\partial_tu+(u\cdot\nabla)u-\nu\Delta u+\nabla p\). Its cylindrical components contain curvature terms. Angular momentum \(ru_\theta\) absorbs the angular transport curvature, while the radial viscous operators still retain their \(1/r\) terms. We first calculate the source's \(\nu=1\) field and then prove the full map for every original \(\nu>0\).

Define the original quantities

Using P0.3 and \(V_0L=X(2\eta U+W-1)\), direct collection of the three derivatives proves the complete material-derivative formula

Here is the collection of terms in detail. At fixed \(z,t\), \(\partial_rX=r/q\), so

Also \(A+D=1\), and hence

The time derivative contributes \(q^{b-1}(-bf+D\eta f_\eta+\mathcal D_Xf)/L\). Thus the three coefficients are

This proves P1.2 term by term, including the radial transport.

For \(b=-h,f=H\), this is \(-q^{-h-1}HS_q/L\). For \(b=-A,f=U\), adding \(\partial_zp^{(0)}=q^{-A-1}Z_{-2A}\Pi\) gives \(-q^{-A-1}S_n/L\). The equality of pressure powers uses \(2A+D=A+1\).

The two radial equations have the unique axis-regular solutions

Indeed the integrating factors are \(XH\) and \(X\). Their homogeneous constants would give \(X^{-2}\) and \(X^{-1}\) singularities, since \(H=2X\phi/\mathsf C\) and \(\phi(0,\eta)>0\). The displayed integrals are smooth and have values \(S_q(0,\eta)/2\) and \(S_n(0,\eta)\) at zero.

For \(r>0\), write \(D_\theta=\partial_r+2/r\), \(D_z=\partial_r+1/r\). The inviscid angular residual is \(-q^{-A-1}ES_q/L\), because angular momentum must first be divided by \(r\). Its negative radial primitive is

Here \(\rho\,d\rho=q\,dx\), \(\rho=\sqrt{2qx}\), and \(\int_0^X\sqrt{2x}ES_q\,dx=XHQ_s\); thus all powers and factors of two are retained. The axial calculation similarly gives \(r^{-1}\int_0^r\rho q^{-A-1}S_n/L\,d\rho=q^{-A-1/2}XN_s/(L\sqrt{2X})\).

Put

The radial viscous operators are exactly \(D_\theta(\partial_ru_\theta-u_\theta/r)\) and \(D_z\partial_ru_z\). Subtracting them from the inviscid residual proves the last line and the stress sign. In particular, \(p_s\) is the integrated inviscid contribution, not the pressure. Axial viscosity has not vanished: its exact profile is \(\partial_z^2(q^{-A}f)=q^{-A-2D}Z_{-A-D}Z_{-A}f\), for \(f=E,U\).

For completeness set \(B_r=V_0/\sqrt{2X}\), so \(u_r^{(0)}=q^{-1/2}B_r\). After the centrifugal-pressure cancellation, the full radial residual is

This follows from P1.2 and \((\partial_{rr}+r^{-1}\partial_r-r^{-2})(q^{-1/2}B_r)=q^{-3/2}(2X(B_r)_{XX}+2(B_r)_X-B_r/(2X))\). It retains every radial term. Its first power is larger than the cancelled leading power \(q^{-3/2-2h}\) by \(q^{2h}\); its axial-viscosity power differs by \(q^{4h}\). In P1.5 the axial viscosity differs from the leading tangential power \(q^{-A-1}\) by \(q^{2h}\). These are exact fixed-profile comparisons, not physical derivative estimates.

The stress vanishes exactly when \(Q_s=-2LF_X/F\) and \(N_s=-2LU_X\). Substitution in P1.3 gives

Conversely these equations make the displayed expressions for \(Q_s,N_s\) solve their radial equations and be smooth at the axis. Uniqueness in P1.3 then proves zero stress. Thus the equivalence uses the actual axis constants.

For any original viscosity \(\nu>0\), retain \(y=x_\nu/\sqrt\nu\), \(u_\nu(x_\nu,t)=\sqrt\nu u(y,t)\), \(p_\nu=\nu p(y,t)\), and \(T_\nu=\nu T(y,t)\). Direct differentiation gives \(\mathcal R_\nu(u_\nu,p_\nu)=\sqrt\nu\mathcal R_1(u,p)(y,t)\), \(D_{j,\nu}T_{j,\nu}=\sqrt\nu D_jT_j\), and \(\nu\partial_{z_\nu}^2u_{j,\nu}=\sqrt\nu\partial_z^2u_j\). P1.5 therefore retains exactly the same physical stress identity at every \(\nu\). The original coordinates are \(X_\nu=r_\nu^2/(2\nu q_\nu)\), \(\eta_\nu=z_\nu/(\sqrt\nu q_\nu^D)\), with \(q_\nu-z_\nu^2q_\nu^{2h}/\nu=\tau\). No viscosity factor is discarded.

P2. All five cumulative moments and the pressure datum

Changing a profile inside an annulus can change the stress beyond that annulus: the formulas in P1 integrate from the axis. We now identify exactly which accumulated data determine the exterior. The five entries below have different weights, and the pressure has an additional axis datum. Their distinct roles will matter when we restore them.

Use the original moments and prescribed axis pressure:

The factor \(1/2\) in \(S\) and the weight \(1/(2x)\) in \(C_p\) remain distinct. Since \(XW=X-2D\eta M-dM_\eta\), \(\partial_X(XW)=1-2D\eta U-dU_\eta\). Integration by parts in \(-WXH_X\) yields

The lower boundary vanishes because \(XWH=O(X^2)\). For the axial terms other than pressure, the identical integration by parts gives

Indeed the coefficient of \(M\) is \(1-A=D\), that of \(\int U^2\) is \(2(A-D)\eta=4h\eta\), and the two copies of \(-d\int UU_\eta\) combine to the derivative shown. No axial boundary value has been omitted: \(XWU=O(X)\) at zero.

Writing \(K_E=\int_0^XE^2/2\,dx\), integration of \(x\Pi_x=E^2/2\) gives \(\int_0^X\Pi\,dx=X\Pi-K_E\). Hence the entire pressure contribution is

The axis pressure is included in \(X\Pi\). Combining P2.2–P2.4 and dividing by the exact integrating factors proves

These are original (4.16), derived without discarding pressure or an integration constant.

P3. Exact exterior preservation and stability without radial derivative loss

There are two useful conclusions. Exact agreement of five moments preserves the exterior fields. Small differences of moments control the integrated inviscid vector without requiring small radial derivatives. The second fact allows a later construction to change the shear substantially while controlling the remaining part of the stress. Both conclusions are proved below with their domains.

For two profiles write \(\Delta f=f_2-f_1\) and \(\mathbf m_i=(M_i,I_i,J_i,S_i,C_{p,i})\). Their exact moment differences are

where every integral has measure \(dx\) and parameter \(\eta\) fixed. If the profiles agree for \(X\ge X_h>0\), all these integrands vanish there. Thus \(\Delta\mathbf m(X,\eta)=\Delta\mathbf m(X_h,\eta)\). When this latter function is zero for every \(\eta\), all its parameter derivatives vanish too. With the same \(\Pi_{\rm ax}\), P0.7 and P2.5 give equality of \(\Pi,V_0,Q_s,N_s\) on the entire exterior. Equality of the profiles on that interval also gives equality of their radial derivatives, including their common one-sided value at \(X_h\). P1.5 then proves equality of \(p_s,a,b_s,\mathcal T_0\) there. This proves the full preservation statement of Lemma 4.4(i).

The pressure datum is essential in this calculation. If instead its difference is \(g(\eta)\) while the five moments and exterior profiles agree, then

These exact terms identify what preservation requires; there is no inference that the source neglects them.

On \(\mathcal R=[X_0,X_1]\times[-1,1]\), with \(0<X_0<X_1\), set \(\|f\|_j=\max_{0\le r\le j}\sup_{\mathcal R}|\partial_\eta^rf|\), using the componentwise maximum for tuples. Assume the source bounds \(E_i\ge e_{\min}>0\) and finite common \(k+1\) norms. The exact difference expressions are

For clarity, let \(A_I=(1-h)I-D\eta I_\eta-dJ_\eta+2(h-D)\eta J\) and \(A_M=D(M-\eta M_\eta)+4h\eta S-dS_\eta\), retaining every summand. Then

This exposes the one differentiated moment in each term. To bound the other factors explicitly, use \(\partial_\eta^r(fg)=\sum_{a=0}^r\binom ra f^{(a)}g^{(r-a)}\), giving \(\|fg\|_r\le2^r\|f\|_r\|g\|_r\). If \(|f|\ge c>0\), its reciprocal derivatives are recursively

Start with \(\|f^{-1}\|_0\le c^{-1}\); this recursion gives a finite polynomial bound in \(c^{-1}\) and \(\|f\|_r\) for every \(r\). Apply it to \(E_i,H_i,L\), whose lower bounds are \(e_{\min},\sqrt{2X_0}e_{\min},1-2h\). Product differentiation of P3.4 therefore proves, with exactly the source dependence,

Here \(C_k\) depends only on \(k,h,X_0,X_1,e_{\min}\), the common input bounds, and \(\|\Pi_{\rm ax}\|_{k+1}\). No radial derivative of a difference occurs. This estimate does not control the changes of \(a,b_s\) without the corresponding radial derivatives; P1.5 retains that dependence in the full stress.

For the original joining radius \(\rho>0\), the exact comparison, not a replacement of the original objects, is

Changing the integration variable in each of P2.1 proves the five factors. Substitution in P2.5 and P1.5 cancels the same factors, so P3.6 holds in a fixed \(x\) rectangle with a constant independent of \(\rho\) whenever the listed hatted bounds are uniform. The shears obey \(a(\rho x)=\widehat a(x)\), \(b_s(\rho x)=\widehat b_s(x)\), whereas the full stress obeys

Both different powers are necessary and are retained.

P4. Invertibility and the approaching-exponent bound

To change several moments independently, place smooth corrections on several disjoint radial intervals. The resulting matrix consists of weighted bump integrals. Its invertibility follows from the zero count of a sum of distinct powers. The proof also shows why a bound can deteriorate when two exponents approach each other.

Let distinct real \(\alpha_1,\ldots,\alpha_m\) and nonnegative nonzero smooth \(\beta_j\) have supports in ordered disjoint compact subintervals \(I_1<\cdots<I_m\) of \((0,\infty)\). Define the actual matrix

A nonzero combination of \(n\) distinct powers has at most \(n-1\) distinct positive zeros. Here is the induction: divide by its least power, which does not change positive zeros. If \(n\) zeros exist, Rolle's theorem gives \(n-1\) zeros of the derivative. Unless that derivative is zero identically, it is a combination of at most \(n-1\) distinct powers and so has at most \(n-2\) zeros by induction. If it is zero identically, the divided function is a nonzero constant and has no zeros. The induction starts with one nonzero monomial. Consequently the evaluation determinant \(\det[x_j^{\alpha_i}]\) is nonzero when \(0<x_1<\cdots<x_m\). It has a constant sign on this connected domain. Expanding the determinant by permutations and applying finite Fubini gives

There is no cancellation: the product is positive on a product of open sets of positive measure. This proves the matrix assertion of Lemmas 4.7 and A.1.

For a smooth compact family preserving these hypotheses, the determinant has a positive absolute minimum. The cofactor formula bounds \(B^{-1}\); all fixed parameter derivatives follow recursively from

The assertion requires smooth dependence of the integrals, as in the source's smooth compact family. It supplies no uniform inverse at colliding exponents. For logarithmic coordinates the exact substitution is \(\int e^{\alpha_i y}\gamma_j(y)\,dy=\int x^{\alpha_i}\gamma_j(\log x)x^{-1}\,dx\). The Jacobian is part of the bump and does not change the ordering or sign.

The loss needed later is explicit. For fixed translated logarithmic bumps \(\beta(y-c_1),\beta(y-c_2)\), let \(\Delta=c_2-c_1>0\) and \(b(s)=\int e^{st}\beta(t)\,dt>0\). The original matrix and its determinant are

The matrix displayed in original (A.1) is obtained by multiplying row \(i\) by \(e^{-s_ic_1}\); recovering the original inverse multiplies by this same diagonal factor on the right. For slopes in a fixed compact interval, those factors and the positive \(b(s_i)\) have two-sided bounds. The mean value theorem gives \(|e^{s_2\Delta}-e^{s_1\Delta}|=\Delta e^{\xi\Delta}|s_2-s_1|\). Cofactors are bounded, hence the original inverse is \(O(|s_2-s_1|^{-1})\).

For weights \(1,x^{-\lambda}\) and fixed ordered bumps \(\beta_1,\beta_2\), set \(m_j=\int\beta_j>0\) and \(\mu_j=m_j^{-1}\int\log x\,\beta_j(x)\,dx\). Separation gives \(\mu_1<\mu_2\). Taylor's formula on their compact positive supports, with its bounded second derivative \((\log x)^2x^{-\lambda}\), gives

The remainder constant is finite on a fixed bounded \(\lambda\) interval. For sufficiently small positive \(\lambda\) it is at most half the magnitude of the displayed linear term, proving the required inverse bound \(O(\lambda^{-1})\). This is an analytic bound, with no numerical matrix sampling.

P5. Exact five-moment correction and the nonlinear solve

The matrix from P4 is the derivative of the moment map. The actual map also contains quadratic terms, which must be solved rather than dropped. We retain the complete original increments, give an invertible comparison map on their five rows, and solve the resulting finite equation. The parameter \(\rho\) remains in every original moment and in the support derivative estimates.

On a correction patch in \(X=\rho x\), keep the source base \(U_0=u_c(\eta)\), \(E_0=e_*f_*(\eta)x^\alpha\), where \(e_*>0\), \(f_*>0\). Use two fixed ordered axial bumps and three fixed ordered angular bumps, with additive perturbations \(u=\sum_{i=1}^2c_i\beta_i\), \(e=\sum_{j=1}^3c_{j+2}\gamma_j\). Each block has ordered disjoint supports inside the patch. The original five moment increments, integrated over the entire patch, are exactly

In particular \(ue,u^2,e^2\) remain in the actual moment map. Introduce the explicitly invertible row map \(\mathscr L_\eta\) by

Every factor is retained. Its inverse sends \(d=(d_1,\ldots,d_5)\) to

Thus solving the transformed equation is proved equivalent to restoring the original moments; it does not replace them. The linear map is block diagonal with the actual entries

The exact remainder in the same row order is

Polarization supplies a smooth bilinear map with this diagonal. P4 proves invertibility precisely when \(\alpha\notin\{-1/2,1/2,3/2\}\): the two axial exponents and the three angular exponents are then distinct. At an excluded value the indicated block actually has equal rows, so this particular additive Jacobian is singular. No conclusion about other constructions is inferred. If \(u_c=0\) and \(u=0\), then P5.1 gives exactly \(\delta M=\delta J=0\), while the three angular coefficients control \((I,S,C_p)\).

Here is the full quantitative solve of Lemmas 4.7 and A.2. In \(C^j([-1,1];\mathbb R^m)\), use the source maximum norm, let \(\nu_j\) be the multiplication norm of \(B^{-1}\), and let \(q_j\) be the bilinear norm of \(\mathscr Q\), including all product-rule constants. For a prescribed smooth discrepancy \(d\), set \(r_j=2\nu_j\|d\|_{C^j}\). The iteration

maps the closed ball of radius \(r_j\) into itself and has Lipschitz constant at most \(1/2\) whenever \(8\nu_j^2q_j\|d\|_{C^j}\le1\). In fact

The case \(r_j=0\) gives the exact zero solution. Completeness and the geometric bound on successive differences give convergence and uniqueness in the specified ball. Applying this for \(j=0\) and one preselected finite \(j=k\) gives the same limit: the iterates are literally identical and convergence in \(C^k\) implies convergence in \(C^0\). Therefore \(\|c\|_{C^k}\le2\nu_k\|d\|_{C^k}\).

Smoothness requires no simultaneous smallness of every derivative norm. At the obtained solution the exact differential is \(D_cF=B+\mathscr Q(c,\cdot)+\mathscr Q(\cdot,c)\). Its product with \(B^{-1}\) differs from the identity by an operator of norm at most \(2\nu_0q_0r_0\le1/2\); the Neumann series converges, so it is invertible. The pointwise implicit function theorem gives a smooth branch, equal to the continuous fixed-point branch by local uniqueness. One-sided endpoint derivatives follow either by smooth local extension of the finite-dimensional data or from successive differentiation and continuity up to the endpoints. The first derivative is exactly

For order \(r\), differentiate \(Bc+\mathscr Q(c,c)=d\) using the finite binomial and multinomial product rules. The terms with \(c^{(r)}\) are exactly \(D_cF\,c^{(r)}\). Every other term uses derivatives of \(c\) of order at most \(r-1\) and derivatives of the given smooth data through order \(r\). The inverse bound and induction give finite continuous derivatives of every order on the compact parameter interval. This proves the claimed smoothness without a hidden all-orders smallness assumption. When \(\mathscr Q=0\), the exact solution is \(B^{-1}d\) with no smallness requirement.

The actual support-scale derivative bound is also retained. For a fixed family \(\delta V=v_*\sum c_j(\eta)\beta_j(X/\rho)\),

For fixed-width logarithmic bumps, \(X\partial_X=\partial_{\log X}\) gives the analogous formula with the logarithmic bump derivatives. Each later preservation of a strict inequality must compare this bound with its actual positive margin. This general solve does not certify any unexamined application margin.

In the source's two applications the exponents are explicit. At \(\alpha=1/10\), the blocks have powers \((0,3/5)\) and \((1/2,1/10,-9/10)\). At \(\alpha=-1/2-\lambda\), they have powers \((0,-\lambda)\) and \((1/2,-1/2-\lambda,-3/2-\lambda)\). P4.5 gives the possible \(\lambda^{-1}\) loss in the first block; the second stays separated near zero. The source fixes positive \(\lambda\) before selecting the later radial frequency. The actual comparisons are derived in O, H and Q for the outer and joining constructions, and in V4 for the final shear correction, with positive lambda fixed before the radial frequency.

P6. The exact stress-cone test

The waves used later must have real amplitudes; their squared amplitudes are positive. This restricts the allowable stress direction. Here we prove the exact algebraic test in the original shear and stress coordinates and its connection to the covariance construction. The positive profile \(F\) and the lengths of the coordinate vectors are retained.

For \(a>0\), set \(t_s=-b_s/a\), \(v_s=a(1+t_s^2)\), \(P_c=p_{s,1}+t_sp_{s,2}\), \(J_c=p_{s,2}-t_sp_{s,1}\). These are the original coordinates, with no unit-vector factors suppressed. Define

For \(P>2\), the roots in \(v\) of \(2(P-v)^2-(v-2)J^2\) are \(v_\pm=P+J^2/4\pm|J|\sqrt{(P-2)/2+J^2/16}\). To check their location, put \(x=P-2>0\). The inequality \(\sqrt{x/2+J^2/16}\ge|J|/4\) gives \(v_-\le P\); and \((x+J^2/4)^2-J^2(x/2+J^2/16)=x^2>0\) gives \(v_->2\). The quadratic is positive below the smaller root and above the larger root. Since \(v_+\ge P\), on \(2<v<P\) positivity is exactly \(v<v_-\). This also covers \(J=0\), where both roots equal \(P\). Therefore, for \(v_s>2\), the source's admissible condition is exactly

Its map to the physical stress is exact:

Thus, where \(F>0\), P6.2 is precisely the positive stress-cone inequality used in the covariance derivation C1–C9. This proves the coordinate connection to that receiving argument, without asserting that the complete profile meets the inequality.

Finally retain the source's compact-set sufficient condition. For a nonempty compact \(K\subset\{(a,b_s,w):a>0\}\), suppose its listed pointwise quantities satisfy \(a-b_sw>0\) and \(2b_sw+b_s^2/a+(a-2)w^2<2\). Set \(c=1-b_sw/a\), \(j=w+b_s/a\), \(v=a+b_s^2/a\). Expansion gives the exact identity

Hence choose \(c_K,\gamma_K>0\), \(B_K\ge1\) with \(c\ge c_K\), \(G:=2c^2-(v-2)j^2\ge\gamma_K\), \(v\le B_K\) and \(cv\le B_K\) on \(K\). Let

For \(p_{s,1}=p\ge P_K\), \(p_{s,2}=wp\), one has \(P_c=pc\ge B_K+2>\max(2,v)\), \(J_c=pj\), and

For \(v>2\) the equivalence already proved gives admissibility. For \(v\le2\), the inequality \(\mathcal U(P_c,J_c)>2\) proves the relaxed condition directly. This completes Lemma 4.5, including its original threshold.

The figure uses the exact section \(a=3,b_s=0\) of P6.2. It is a diagram of the proved inequalities, not a numerical sample of the constructed velocity. The second panel records the five-moment correction map; its full bump integrals and support factors are given in P5.1–P5.9.

P7. Five worked exercises

Exercise 1. Identify the exact exterior stress tails. Fix the axial variable and time. Let smooth angular and axial residuals \(R_\theta(r),R_z(r)\) vanish outside \(0<r<R_*\), and suppose they vanish near the axis. Solve

with regular axis behavior. Determine exactly when the stresses vanish for \(r\ge R_*\).

Solution. Multiplication by \(r^2\) and \(r\), respectively, gives

Regularity at zero makes both integration constants zero. Therefore

They vanish near zero because the residuals do. On the exterior the integrals are the constants

Thus the exterior stresses are exactly \(-M_\theta^Rr^{-2}\) and \(-M_z^Rr^{-1}\). They both vanish there if and only if both weighted moments vanish. A nonnegative, nonzero smooth bump supported in \((R_*/3,2R_*/3)\), used as either residual, has a strictly positive corresponding moment and a nonzero tail. This proves why vanishing exterior residual alone is insufficient. The full construction arranges its total moments to eliminate these actual constants.

Exercise 2. Can a different smooth axis pressure preserve the same exterior stress? Suppose the two exterior profiles and all five moments agree as in P3, but the two axis pressure data differ by \(g\in C^\infty([-1,1])\). If their full stresses agree on a nonempty radial exterior interval for every \(\eta\), prove that \(g=0\).

Solution. P3.2, with \(X>0\) and \(L>0\), gives

Direct product differentiation, without division by \(g\), gives

Thus \(g(\eta)=c(1-\eta^2)^{-2A}\) on the connected open interval. Since \(A>0\) and \(g\) is bounded up to \(\eta=1\), the constant \(c\) is zero. Conversely \(g=0\) preserves the stress by P3.

The endpoint hypothesis matters. On the open interval, a nonzero \(c\) would give the physical pressure difference

which is spatially constant for each \(t<1\) and therefore changes no pressure gradient. Its profile fails the stipulated bounded endpoint condition. This identifies the exact pressure-gauge freedom and why it is absent from the smooth closed profile class used here.

Exercise 3. Compute the first radial coefficients at a zero-stress axis. Write

with \(\phi_0>0\) and \(\Pi(0,\eta)=\Pi_{\rm ax}(\eta)\). Derive the coefficients forced by P1.7. All remainders here are the ordinary local Taylor remainders of smooth functions on a fixed compact parameter interval.

Solution. At the axis,

Both \(XU_X\) and \(X\Pi_X\) vanish there. Hence

The factors \(4L\) and \(2L\) arise from \(X\phi_{XX}+2\phi_X\) and \(XU_{XX}+U_X\), respectively, at zero. The pressure coefficient is also fixed:

These are necessary first-coefficient identities and follow from an actual zero-stress solution. Constructing the convergent axis solution requires the full analytic argument in J0–J6; these two coefficients alone do not establish that existence.

Exercise 4. Retain the support scale in an actual moment correction. For the five coefficients in P5, suppose \(\|c\|_{C^k}\le2\nu_k\|d\|_{C^k}\). Give explicit bounds for the original increments \(u(X/\rho,\eta)\) and \(e(X/\rho,\eta)\), and determine the original change in \(J\) from transformed data \(d_2,d_3\).

Solution. The coefficients may depend on \(\eta\), but \(\rho>0\) is fixed. Repeated differentiation gives

The correction coefficients are the actual nonlinear solution, so its moment change is the target, including every quadratic contribution. Applying the inverse row map P5.3 yields

The second term cannot be discarded when \(u_c\ne0\). A finite positive margin involving \(r\) radial and \(k\) parameter derivatives is preserved by choosing the discrepancy small compared with the right sides of P7.4 and the actual margin. There is no bound uniform in \(\rho\downarrow0\) for fixed nonzero coefficients and \(r>0\); the displayed \(\rho^{-r}\) remains.

Exercise 5. Invert the stress-cone coordinates exactly. For \(a>0,F>0\), recover \(p_s\) and \(\mathcal T_0\) from \(P_c,J_c\). Then describe the section \(a=3,b_s=0\) and its boundary.

Solution. The matrix taking \(p_s\) to \((P_c,J_c)\) is

Its exact inverse gives

Since \(a=v_s/(1+t_s^2)\) and \(-b_s=at_s\), the stress is

For the stated section \(t_s=0,v_s=3,P_c=p_{s,1},J_c=p_{s,2}\). The admissible set is exactly

Its two boundary rays have \(p_{s,2}=\pm\sqrt2(p_{s,1}-3)\) for \(p_{s,1}\ge3\), with vertex \((3,0)\). The stress components there are \(F(p_{s,1}-3),Fp_{s,2}\). The boundary and vertex are excluded by the strict inequalities. Every compact subset of the open cone has positive minimum values of its two continuous defining gaps. This is the exact margin used when choosing sufficiently small later corrections.

Sources and the next construction

The original OpenAI paper, pages 24–32, 34–35 and 126–128, supplies the source formulas and arguments: the cone comparison on page 32, Lemma 4.7 and the moment identities in Appendix A. The source identity and hash are in SOURCES.json.

The complete editorial calculations are G0–G6 of the profile and moment audit. This lesson retains those proofs and develops their transport, pressure-gauge, support and cone consequences. It makes no novelty claim.

The actual next construction begins with O0–O8, which fixes the outer profile and pressure datum, and H0–H7, which inserts the exact heat exterior and restores the moments. J0–J6, Q0–Q9 and V0–V7 then provide the axis solution, join and actual admissible shear. These proofs supply the existence inputs for the full theorem; N2 continues by developing them as lessons with all their choices and estimates.

Constructing the outer profile and its heat exterior

The five moments from N2.1 give us a way to change a velocity profile while controlling its stress. We now use that method to construct a profile. We choose every radial interval, solve the moment equations, determine the pressure needed at the axis, and prove the strict stress inequalities. We then replace the distant power law by a heat flow and prove that the stress goes smoothly to zero at the outer edge.

This lesson supplies the outer part of the leading-profile construction: a complete reference family and its heat modification. Its temporary inner reference is explicitly given and has integrable boundary traces, but is not a smooth Cartesian axis field. The analytic axis, its exact join, and the shear realization have separate complete proofs linked at the end. They are the subsequent teaching components of N2. The final assembly appears in the energy and comparison lesson.

Read N1 for the original coordinates and N2.1 P0–P6 for the full stress, moments, bump-matrix inverse, nonlinear solve and cone criterion. Those proofs are included in this cumulative reader. Here \(q-z^2q^{2h}=1-t\), \(X=r^2/(2q)\), \(\eta=z/q^D\), \(A=1/2+h\), \(D=1/2-h\). We keep every original physical factor. Scalar changes of variable in integrals are always accompanied by their Jacobians and the factors needed to recover the original quantity.

The source is OpenAI, Finite Time Blowup for Navier–Stokes, original paper, pages 129–144, Proposition A.4 through Proposition A.10, equations (A.5)–(A.56). The source version and mathematical locators are in SOURCES.json. The exposition below is a separate derivation and teaching text; it does not rewrite the diplomatic transcription.

Figure. The schedule is a diagram of ordered intervals, not a plot of a sampled velocity or a scale drawing. The four reserved patches lie in the long interval before the pulse. The heat change begins at \(X_K=e^{1/5}X_{\rm tail}\); its three changed moments are restored in the second reserved patch. At \(X_b=e^3X_{\rm tail}\), both stress components vanish, while their quotient has the exact sixth-order factor proved in H6. All interval lengths and full formulas appear below.

How the construction fits together

The construction has three jobs. First, the inner reference and the gradual axial transition produce a positive angular stress. The long interval then makes the accumulated axial moments small relative to the local moment scales. A pulse and two tiny end corrections set both axial moments and the quadratic stress moment to zero. These are exact equations, not numerical approximations.

Second, after the pulse, the angular profile is made independent of \(\eta\). Two more bumps restore its angular integral without changing the total pressure increment. A steep interval reduces its amplitude enough to offset the small parameter \(h\) in the later estimates. A final holding time is chosen by an explicit formula; it makes the subtracted angular moment exactly zero.

Third, the heat integral supplies an actual exterior solution. Replacing the power law changes three moments. The three-bump nonlinear system restores them, preserves the pressure datum and keeps the stress cone strict. Backward integration then proves compact support of the stress and its direction at the outer edge. The sections follow these dependencies, so each use of a small or large parameter comes after the estimates that justify choosing it.

O0. The complete reference schedule and the order of choices

Retain \(A=1/2+h,D=1/2-h,d=1-\eta^2,L=1-2h\eta^2\), \(f=(1+\eta^2)^{-1}\), \(J_0=\log(1+\eta^2)\), \(H=\sqrt{2X}E\), \(l=X\partial_X\log H\). All profiles have \(-1\le\eta\le1\). For each interval the coordinate \(y\) starts at its left endpoint and satisfies \(dX=X\,dy\). Hence prescribing l means solving \((\log E)'=l-1/2\), with the preceding endpoint value of E. Every prime on a radial profile below is a y derivative unless otherwise indicated.

Use the source's smooth step

It is increasing and smooth, flat at its endpoints. Indeed the logarithmic derivative of the ratio of the second exponential to the first is \(-2(1-y)^{-3}-2y^{-3}<0\); endpoint smoothness follows from \(y^{-n}e^{-1/y^2}\to0\) for every n. Positive correcting bumps are fixed translates or rescalings of \(\sigma'\), with the supports stated below. Their fixed derivative norms are finite.

The parameter order is

We always take \(h<1/100\). Fixed shape constants, including the interpolation length \(T_f\) and terminal constant \(c_o\), are chosen independently of the later small parameters. Below, \(C_{\rm pre}\) denotes finite constants depending only on those shapes, \(M_d,T_d,P_*\) and a specified finite derivative order, uniformly for sufficiently small \(\lambda\) and then sufficiently small h. No such constant depends on \(X_R\) or the amplitude factor \(\mathsf C\). Constants denoted C before the choice of \(M_d\) depend only on the fixed shapes. Estimates below justify this order rather than assuming its compatibility.

The entire schedule is as follows.

1. For \(X=X_Re^y\), \(y\le0\), set \(U=4\eta\), \(E=P_*fe^{y/10}\), so \(l=3/5\). This is the temporary inner reference, not a regular physical axis.
2. For one unit of logarithmic length keep \(U=4\eta\), and set \(l=(3/5)(1-\sigma(y))\).
3. On \(0\le y\le T_d\), set \(l=0\), \(E=P_1fe^{-y/2}\), and

Here \(e^{-1/2}P_*\le P_1\le e^{1/10}P_*\). The function k is zero on the last eleven units, since it is zero for \(y\ge e^{M_d}-1=T_d-11\). 4. With \(U=0\), move l from zero to \(-\lambda\) by \(l=-\lambda\sigma(y)\) on a unit interval. Keep \(l=-\lambda\) for \(T_w=60\log(1/\lambda)\). Reserve, in this latter local coordinate, the four intervals

For sufficiently small \(\lambda\), all are nonempty, ordered and inside this interval. None of the reference-profile corrections below touches them. On them \(U=0\), \(E\) is the same positive multiple of \(fX^{-1/2-\lambda}\). 5. Let \(X_p\) be the end radius and \(E(X_p,\eta)=e_bf\). For \(0\le y\le13/\lambda\), retain \(l=-\lambda\), so \(E=e_bfe^{-(1/2+\lambda)y}\), and put

Extend \(\varphi_b\) by zero on negative arguments. The \(\beta_j\) are identical width-.3 bumps centered at \(13/\lambda-3\) and \(13/\lambda-1\). Their coefficients restore M and J; Amp restores S. These choices are made in O1–O3. 6. After the pulse put \(U=0\) permanently. Writing \(E(X_{\rm end},\eta)=e_{\rm end}f\), prescribe

Choose the fixed \(T_f\ge10\|\sigma'\|_\infty\log2\), so \(-\lambda-.1\le l\le-\lambda\). This also satisfies the source's weaker lower bound \(-\lambda-1\). Continue the resulting \(\eta\)-independent exponential for \(30\log(1/\lambda)\). Two relative E bumps of width .3, centered three and one units before that interval ends, impose

These bumps have zero endpoint values and leave all later E values unchanged. 7. At radius \(X_{\rm rel}\), with common value \(e_{\rm rel}\), move l from \(-\lambda\) to \(-1\) on a unit interval, hold \(-1\) for \(4\log(1/h)\), and move from \(-1\) to \(-h\) on a unit interval. Every transition uses \(\sigma\). Then hold \(-h\) for the finite length selected in O2. 8. On the final interval \(0\le y\le3\), use

Choose \(c_o>0\) so small that \(0\le f_o'/f_o\le h/4\). For example \(c_o\le[16(1+\|\psi_o'\|_\infty)]^{-1}\) suffices for \(h<1/100\). Extend \(f_o\) constantly on the left and by one on the right. Thereafter \(E=E_{\rm pow}=c_\infty X^{-A}\), with \(c_\infty>0\) independent of \(\eta\).

Smooth flat endpoint matching of the step functions and bumps makes this schedule smooth for \(X>0\); solving the logarithmic equation gives E positive before its small relative corrections. Positivity after those corrections is proved below. All stage lengths are finite after the parameter choices. Their endpoint radii divided by \(X_R\) are fixed.

O1. The actual axial correction and the initial moment bounds

The first correction acts on U, with E already prescribed on the pulse. Its two moment rows are linear, so we can solve them exactly before choosing the pulse amplitude. The nearby exponents cost an inverse factor \(1/\lambda\); we retain that loss and show why the long separation of supports pays for it.

Use the five original integrals \((M,I,J,S,C_p)\) from P2.1. Before the long intermediate interval all dimensionless profiles and their fixed \(\eta\) derivatives have finite bounds depending only on the preceding choices. Radial integration over this finite logarithmic interval, and over the inner reference where the powers are integrable, gives finite bounds on \(M/X_R\), \(J/X_R^{3/2}\), \(S/X_R\), and the corresponding parameter derivatives. On the intermediate interval \(U=0\), so M and J remain constant. The exact multiplicative changes in its denominators are

The entering unit transition has uniformly bounded factors. At \(T_w=60\log(1/\lambda)\), take \(\lambda<1/120\). Then \(e^{-(1/2-2\lambda)T_w}=\lambda^{30-120\lambda}\le\lambda^{29}\); the analogous M bound is no larger. The factor \(e^{2\lambda T_w}=e^{120\lambda\log(1/\lambda)}\) stays bounded as \(\lambda\downarrow0\). Integrating \(dS/dy=-XE^2/2\) over this interval and retaining the incoming S yields

Every fixed k is allowed here. The factors f and its reciprocal have bounded derivatives on [-1,1]. The preceding estimates follow from exact integrals of the stated powers; they introduce no dependence on h through a divergent exterior integral.

On the pulse, the two original axial moment rows for a relative bump \(c\beta\) are exactly

This includes the Jacobian \(dX=X_pe^y\,dy\). Let \(y_1=13/\lambda-3\), \(y_2=y_1+2\). After dividing each row by its displayed physical factor and then by \(e^{s_iy_1}\), its two entries are \(b(s_i),e^{2s_i}b(s_i)\), with \(b(s)=\int e^{st}\beta(t)\,dt>0\). The determinant is \(b(s_1)b(s_2)(e^{2s_2}-e^{2s_1})\). P4.4 proves an inverse bound \(C/\lambda\), retaining the exact row factors to recover the original moments.

The main pulse ends by \(11/\lambda\), and \(R_0\le11\) on its support. For small \(\lambda\), \(s_i\ge2/5\). Its first-center-weighted moment has magnitude at most

The incoming moments in O1.2 gain the factor \(e^{-s_iy_1}\). Multiplication by the inverse loss \(1/\lambda\) is absorbed by a smaller positive exponential rate: for every \(a>0,n\), \(\lambda^{-n}e^{-a/\lambda}\le C_{a,n}e^{-a/(2\lambda)}\), since \(v^ne^{-av/2}\) is bounded for \(v\ge1\). Therefore the exact linear solve gives affine functions of Amp with

The same statement holds for fixed radial bump derivatives; a \(\zeta_b\) derivative adds a power of \(\lambda^{-1}\), which remains exponentially small. No quadratic term was dropped: M and J are exactly linear in U for the fixed E used here.

O2. Angular moment restoration and the complete terminal schedule

The next correction acts on E after U has become zero. One equation is linear in E; the pressure equation is quadratic. We solve both, then choose the terminal holding time from the angular equation itself. This is what will make a backward stress integral possible later.

The identity \(I'=XH\) gives, for \(r_I=I/(XH)\),

Before interpolation, \(1+l\) has a fixed positive lower bound. The inner reference has \(r_I=5/8\); variation of constants bounds it and every fixed parameter derivative through the subsequent intervals. During interpolation, our choice in O0.6 keeps \(1+l\ge1-\lambda-.1>1/2\). Thus at its end \(r_I\) and its fixed parameter derivatives are bounded by \(C_{{\rm pre},k}\). On the following constant-slope interval the exact discrepancy is

At the first angular bump center \(30\log(1/\lambda)-3\), its \(C^k_\eta\) norm is at most \(C_{{\rm pre},k}\lambda^{28}\) for sufficiently small \(\lambda\).

Write the relative change as \(E\mapsto E(1+v)\), \(v=d_1\beta_1+d_2\beta_2\). The two exact rows are

Divide by \(XH\) and \(E^2\) at the first center, respectively. Their linear weights have exponents \(1-\lambda\) and \(-1-2\lambda\), separated by \(2+\lambda\). P4.4 gives a uniformly bounded inverse. The quadratic pressure term in O2.3 has uniformly bounded bilinear norms on these two fixed supports. The desired first-row discrepancy is O2.2 transported to the endpoint; equivalently, undoing that transport gives an \(O_{C^k}(C_{{\rm pre},k}\lambda^{28})\) discrepancy at the first center. The pressure discrepancy is zero. The actual finite solve P5.6–P5.8 therefore applies for sufficiently small \(\lambda\), giving

Only a specified finite k must have a small bound; smoothness of all orders follows from the same fixed branch. Taking k at least one gives \(|v|+|v'|\le C_{\rm pre}\lambda^{28}\). Thus \(1+v>0\) and the slope change \(v'/(1+v)\) is smaller than \(\lambda/2\). On the bumps, \(l\le-\lambda/2\). All earlier and later E values are unchanged. These choices involve E only and are independent of Amp.

At the exterior-transition start, \(U=M=J=0\), E and I are independent of \(\eta\), and P2.5 gives exactly

The first transition keeps a positive lower bound \(c\lambda\) when \(h\ll\lambda\). On the hold with \(l=-1\), \(Q_s'=1-h\); its length \(4\log(1/h)\) gives a value comparable to \(\log(1/h)\). The next unit transition changes that value by bounded positive factors and bounded terms, since its coefficients are bounded and its source is nonnegative. Call its final value \(Q_{\rm in}\).

For the terminal profile O0.8 define the source's exact value

On this interval \(1+l\) is bounded above and below by positive constants. Also \(\int_0^3f_o'/f_o=-\log(1-c_oh)\), between \(c_oh\) and \(2c_oh\) for the stated small h. Thus \(c h\le Q_p\le C h\), with fixed positive constants. Consequently \(Q_{\rm in}>Q_p>0\) for sufficiently small h. The additional hold at \(l=-h\) has the explicitly positive finite length

It decreases Q exactly to \(Q_p\). On the terminal interval, variation of constants with final value zero gives

Indeed differentiation gives \(Q_s'+(1+l)Q_s=-f_o'/f_o=-l-h\), and its initial value is O2.6. It is positive for every \(y<3\), vanishes at 3, and is zero thereafter. Its lower bound on \(0\le y\le1/2\) is \(ch\), because the integral over the fixed portion where \(f_o'>0\) remains present.

Beyond the terminal interval, \(H=H_{\rm pow}=\sqrt2c_\infty X^{-h}\) and \(I=XH/(1-h)\). But \(\int_0^XH_{\rm pow}\,dx=XH_{\rm pow}/(1-h)\), since \(h<1\). Therefore

This integral converges at zero and has identically zero integrand sufficiently far out. All exterior lengths and \(c_\infty\) are independent of Amp and \(\eta\).

O3. Closing S with the actual amplitude

We still have one scalar equation: the total stress moment S must vanish. The pulse amplitude is its remaining free variable. The following analytic bracket proves that an actual smooth amplitude exists, with the parameter derivatives required by the cone estimate.

The integral after the pulse must be bounded uniformly as h becomes small. On the steep hold \(l=-1\), the exact changes are

The first follows from \((\log XE^2)'=2l=-2\); the second from \((\log E)'=-3/2\), each over \(4\log(1/h)\). The first unit transition and steep hold contribute at most \(CX_{\rm rel}e_{\rm rel}^2\) to \(\int XE^2dy\). After the steep hold, \(l\le-3h/4\) at every point. Thus the rest is bounded by \(CX_{\rm rel}e_{\rm rel}^2h^8\int_0^\infty e^{-3hy/2}dy\le CX_{\rm rel}e_{\rm rel}^2h^7\). In particular the complete exterior release, regardless of its hold length, satisfies

Interpolation and the subsequent uniform interval have length \(T_f+30\log(1/\lambda)\), nonincreasing \(XE^2\) before the small bumps, and bounded relative bump factors. Their contribution, together with O3.2, is at most \(C(1+\log(1/\lambda))X_{\rm end}e_{\rm end}^2\). Fixed parameter derivatives obey the same estimate: f, its powers with exponents in [0,1], and the small angular coefficients have bounded derivatives; the release itself is independent of \(\eta\).

On the pulse the exact identity is \(XE^2=X_pe_b^2f^2e^{-2\lambda y}\). The pulse contribution to S, including its two end bumps, is therefore

To justify every part of the error bound, the incoming S gives \(O_{C^k_\eta}(C_{{\rm pre},k}\lambda(1+\log(1/\lambda)))\) by O1.2. The post-pulse E integral has the same bound by O3.2 and \(X_{\rm end}e_{\rm end}^2=X_pe_b^2e^{-26}\), with \(f^{-2}\) retained in the displayed denominator. The extra U terms are exactly \(2\operatorname{Amp}R_0(c_1\beta_1+c_2\beta_2)+(c_1\beta_1+c_2\beta_2)^2\). In fact the first term is zero because the end bumps lie after the main pulse; even its absolute bound from O1.5 is exponentially small. On the bounded amplitude bracket \([.9,1.2]\), the second term and all its fixed parameter derivatives are exponentially small too. The original coefficients are affine in Amp; differentiating them with respect to Amp uses their equally small linear coefficients. Thus

All constants are uniform for the subsequently small h. This supplies the amplitude derivative estimate as well as the parameter estimate needed to infer a unique root.

Here are exact bounds on \(K_b\). Since \(0\le R_0\le\zeta\), \(K_b\le1/4\). On \(1/50\le\zeta\le9\), \(R_0\ge\zeta-1/50\). Direct integration gives

For the strict last inequality, \(e^{-1/25}\ge24/25\) gives a first term at least \(6/25\), and \(e^{18}>18^4/24=4374\) gives a second term smaller than \(1/25\). These are exact analytic inequalities, not quadrature.

For \(F_0(a)=a^2K_b-(1-e^{-26})/4\), it follows that

The first follows from \(-19/400+e^{-26}/4<-47/1000\), using \(e^{26}>500\). The second is \((36/25)(1/5)-1/4=19/500\), and the derivative is larger than \(2(9/10)(1/5)=9/25\). Choose \(\lambda\) so that O3.4 is smaller than \(.01\), for example. Then the complete expression is negative and positive at the respective endpoints and has derivative greater than \(.35\). The intermediate value theorem and this derivative bound give exactly one root Amp in the bracket at each \(\eta\). The implicit function theorem gives a smooth root, including endpoint derivatives, with

Implicit differentiation gives finite higher derivatives, and any finite set of small derivative bounds can be secured before choosing the parameters. Consequently the actual constructed profile has \(S(\infty)=0\), as well as O2.9 and \(M=J=0\) after the pulse. The integrals converge because U is compactly supported and \(E^2\) has the exterior power \(X^{-1-2h}\).

O4. The analytic pressure supplied to the inner construction

The inner analytic construction needs its pressure before that construction starts. We therefore derive the datum from the complete outer schedule. The important identity is zero total pressure change: it lets smooth moment corrections preserve an analytic datum exactly.

Before the two pressure-preserving angular bumps, the complete schedule has the form \(E_{\rm id,sched}(y,\eta)=c(y)f(\eta)^{\vartheta(y)}\), with \(0\le\vartheta\le1\). Here use the global coordinate \(y=\log(X/X_R)\). The coefficient c and all interval lengths are independent of \(\eta\); O2 shows that the later lengths are independent of Amp as well. In the interpolation, c includes \(2^{-(1-\vartheta)}\), exactly as in O0.6. The exponent is one through the pulse, decreases to zero during interpolation, and is zero afterwards.

Choose a bounded simply connected complex neighborhood of [-1,1] whose closure avoids \(\pm i\). The function \(f(\eta)=(1+\eta^2)^{-1}\) is holomorphic and nonvanishing there, so choose an analytic logarithm. The functions \(\exp(2\vartheta\log f)\) are uniformly bounded on the closure for \(0\le\vartheta\le1\). At the two infinite ends c squared is a fixed multiple of \(e^{y/5}\) and \(e^{-(1+2h)y}\); the intervening interval is finite. Dominated integration and Morera's theorem therefore prove holomorphy on one common neighborhood of

For completeness, Morera applies because the uniform integrable majorant permits interchanging each triangular contour integral with the y integral, and each interior contour integral is zero. Continuity follows from the same majorant.

The function is even, and for real \(\eta\)

The integral in the derivative is strictly positive, since the inner reference has \(\vartheta=1,c>0\). This proves the required strict sign of \(\Pi_0'\) for \(\eta\ne0\). Its inner contribution is exactly \(-\frac12P_*^2f^2\int_{-\infty}^0e^{y/5}dy=-(5/2)P_*^2f^2\); all others are nonpositive. Independence of \(X_R\) follows from \(dX/X=dy\). Restoring the angular bumps leaves the total pressure integral exactly unchanged by O2.4. Thus for the actual reference profile

Later smooth edits need not be analytic to preserve this already fixed analytic datum: exact zero total pressure increment is the sufficient identity, proved by splitting the last integral. This proves Lemma A.5 for the actual constructed reference family.

O5. The cone through the reference and axial transitions

Having solved the moments is not enough. The stress must lie in the required cone at every radius, including inside each correction. We first prove bounds independent of the final radius. Only after those bounds hold do we enlarge \(X_R\).

Keep the source test, with \(w=N_s/(EQ_s)\),

P6 proves how these inequalities imply the relaxed cone once \(p_{s,1}=XQ_s/L\) is large enough, and the admissible cone when also \(v_s=a+b_s^2/a>2\). We first prove the two strict inequalities and \(Q_s>0\), before choosing \(X_R\).

On the reference inner interval \(\mathcal A_XU=4\eta\), \(W=1-4L\), \(l=3/5\), \((\log E)_\eta=-J_0'\). The regular radial primitive for its constant angular source gives exactly

For \(h<1/100\), \(L\ge.98\), and \(\eta J_0'=2\eta^2/(1+\eta^2)\ge0\), the numerator is at least \((3/5)(4\cdot.98-1)-.01>8/5\). Hence \(Q_s>1\). Here \(a=4/5\), \(b_s=0\), so O5.1 holds for every finite w. On any fixed \(x=X/X_R\ge x_*>0\), E is positive and all quantities are bounded.

During the first unit transition U and W retain these values, \(0\le l\le3/5\), and the angular source \(-Wl-h(1-8\eta^2)+(D+4d)\eta J_0'\) is at least \(-h\). Variation of constants in \(Q_s'+(1+l)Q_s=S_q\) gives at its end \(Q_s\ge e^{-8/5}-h>1/10\). One may use \(e^2<9\) to verify the last inequality. The same formula gives a positive bound throughout this unit. Since \(b_s=0\) and \(4/5\le a\le2\), O5.1 continues to hold.

On the axial transition, \(l=0\) and the original source is

The inequality uses \(k\ge0\) and \(\eta J_0'\ge\eta^2\). Take \(h\le e^{-T_d}/20\). The negative h term is then at most \(e^{-y}/20\). Splitting at \(y=\log2\), using \(D\ge.49\), proves for example

For \(y\le\log2\), \(\eta^2+e^{-y}\le3e^{-y}\), while the retained exponential term is \(e^{-y}/20\). For \(y\ge\log2\), the two positive terms dominate \(e^{-y}/20+.245\eta^2\). Thus the constant here is independent of \(M_d\).

We need a uniform axial-source bound to justify choosing \(M_d\) first. Write \(\mathcal A_XU=\bar k\eta\); then \(0\le k,\bar k\le4\) and \(W=1-L\bar k\) is bounded by fixed constants. Choose \(M_d\) large enough that \(e_d\le1\). The geometric portion \(-Wk'\eta-A(1-2k\eta^2)k\eta-(D+dk)\eta k\) is bounded by \(C|\eta|\), with C independent of \(M_d,T_d,P_*\). For the pressure portion use O4.3. Every unedited later E is bounded by \(E(y)e^{-(v-y)/2}\) for \(v\ge y\), and its logarithmic parameter derivative has magnitude at most \(|J_0'|\). The full future angular correction may be omitted in this backward pressure integral because its total pressure increment is zero. Thus

The last term is exactly \(\eta E^2\). At the start of the first unit transition the radial formula \(N_s=X^{-1}\int_0^XS_n\,dx\) gives \(|N_s|\le C|\eta|(1+P_*^2)\): its geometric terms are bounded odd functions of \(\eta\), and its pressure terms are bounded by \(CP_*^2|\eta|\). Integrability of the inner powers and the fixed first transition preserve this bound to the axial start. The pressure bound is independent of the later transition length, by the exponential estimate just used. Solving \(N_s'+N_s=S_n\) on the axial interval, with \(E^2=P_1^2f^2e^{-y}\), gives

We used \(f\ge1/2\), \(P_1\asymp P_*\) with fixed comparison constants, and \(P_*>e^{T_d}\). This proves the stated C can be chosen independently of \(M_d\).

Since \(b_s=2k'\eta/E\), O5.4–O5.6 yield

Here \(a=2\). Choose \(M_d\) first so \(Ce_d<1/8\), and \(P_*\) later sufficiently large that the second bound is less than \(1/4\). Then \(a-b_sw>15/8\) and \(2b_sw+b_s^2/a<3/8<2\). This proves both strict cone tests throughout the axial interval without a circular parameter choice.

O6. The intermediate interval and the pulse cone

The long waiting interval and the pulse have different roles. Waiting gives decay of the old axial stress. On the pulse, both shear components vary, so we retain the next term in the convolution formula and prove the quantitative cone margins.

The equation \(\bar k'=k-\bar k\) and the last eleven units with k zero give \(\bar k(T_d)\le4e^{-11}\), so \(L\bar k<1/2\). On the next unit transition U is zero and \(W=1-L\bar k\ge1/2\). Set

On that transition \(S_q=(-l)W-h+c_\eta\). The positive incoming value in O5.4 and variation of constants retain a positive lower bound depending only on the preceding choices, uniformly for small \(\lambda\) and h. On this finite interval w is bounded by \(C_{\rm pre}\); hence \(\sqrt\lambda|w|\) is small after choosing \(\lambda\). With \(b_s=0\) and \(2\le a\le2+2\lambda\), O5.1 follows. The admissible cone applies wherever \(l<0\), and the relaxed cone at its initial endpoint.

On the constant-slope interval let \(\alpha=1-\lambda\) and \(\bar k=\bar k_0e^{-y}\). The angular equation is exactly

The incoming value has a lower bound \(c_{\rm pre}(1+\eta^2)\). Since \(L\bar k_0<1/2\) and \(h<\lambda/4\), the right side is at least \(\lambda/4+c_\eta\). Its positive convolution proves the first of the following bounds; direct integration proves the second:

For the last line, U zero removes every geometric axial-source term. O5.5 bounds the pressure source by \(C|\eta|E^2\). Its convolution with \(e^{-y}\), and \(E^2=E(0)^2e^{-(1+2\lambda)y}\), gives \(C_{\rm pre}|\eta|(1+y)e^{-y}\) for N. Divide by the exact E. One parameter derivative gives the same bound with the \(|\eta|\) factor removed. Since \(|\eta|/(\eta^2+e^{-\alpha y})\le\frac12e^{\alpha y/2}\),

The last limit is uniform on that interval: its right side is at most \(C_{\rm pre}\sqrt\lambda(1+60\log(1/\lambda))e^{30\lambda\log(1/\lambda)}\), which tends to zero. Thus \(a=2+2\lambda,b_s=0\) satisfy both cone tests there.

For the pulse retain \(m=\mathcal A_XU/E\), \(\beta=1/2-\lambda\), and the actual R in O0.5, including both end bumps. Differentiating the radial average gives the exact equation

Extend R by zero to negative y. The main pulse is flat at its start; the later bumps have exponentially small fixed \(\zeta_b\) derivatives. Its convolution solution and Taylor's formula give

The remainder is bounded by \(\frac12\lambda^2\|\partial_{\zeta_b}^2R_b\|_\infty\int_0^\infty e^{-\beta v}v^2dv\). The same argument works after one \(\eta\) derivative. The initial m and its derivative are \(O(C_{\rm pre}\lambda^{29})\) by O1.2. The fixed point amplitude estimate O3.7 and O1.5 give \(\|m_\eta\|_\infty\le C_{\rm pre}\lambda(1+\log(1/\lambda))\).

On the pulse, \(W=1-2D\eta Em-d(Em)_\eta\). Since \((\log E)_\eta=-J_0'\), the angular source is \(\lambda-h+c_\eta+O(C_{\rm pre}E)\). The initial angular discrepancy is bounded by O6.3 at \(T_w\); \(E\le C_{\rm pre}\lambda^{30}\) throughout the pulse. Convolution with \(e^{-(1-\lambda)y}\) therefore proves

The weakened power 28 retains the source's estimate. To bound all terms of N, integrate the original S: \(dS/dy=XE^2(R_b^2-1/2)\). The ratio \(XE^2/(X_pe_b^2f^2)\) is between \(e^{-26}\) and one on the pulse. Its length is \(13/\lambda\), R and its first parameter derivative are bounded, and O1.2 controls the incoming moment. Thus

The canonical pressure and its parameter derivative are \(O(E^2)\) by O4.3, omitting the complete later pressure-preserving correction. Substitute \(M/X=Em\), \(U=ER_b\) into the complete formula P2.5. In particular \((M-\eta M_\eta)/(XE)=m-\eta m_\eta+\eta J_0'm\). The S and pressure terms are \(O(C_{\rm pre}E/\lambda)\), and \((1-W)R_b=O(C_{\rm pre}E)\). This proves the full source formula

All pressure and moment contributions have thus been bounded rather than omitted.

Divide O6.9 by O6.7. The factor \(|\eta|/(\lambda+c_\eta)\le C/\sqrt\lambda\) and the m derivative bound make its parameter term \(o(1)\). The Taylor remainder divided by Q is \(O(\lambda)\); the incoming exponential term and angular error give smaller powers. Since \(D+c_\eta-\beta=\lambda-h+c_\eta\), the coefficient of R is \((1-\lambda)/\beta=2+O(\lambda)\). Therefore, uniformly on the whole pulse,

The quotient \((D+c)/(\lambda-h+c)\) decreases for \(c\ge0\), because its derivative has numerator \(\lambda-h-D=\lambda-1/2<0\). Its maximum occurs at zero; taking \(\lambda\downarrow0\) and then \(h/\lambda\downarrow0\) gives the stated upper limit. This is compatible with O0.2.

The exact other shear is

Apart from exponentially small end bumps, R is nonnegative and \(\partial_{\zeta_b}R_b\le1.2\): the derivative of \(\varphi_b\) is at most one, and differentiating its terminal cutoff gives a nonpositive term. Both R and every fixed shape derivative have finite uniform bounds. Choose \(\lambda,h\) small enough that \(1.2C_d\le2.41\). The exact quadratic maxima then give

The rational gaps to the stated upper bounds are \(1119/80000\) and \(719/35000\). Choose the finitely many uniform o(1) terms smaller than half these gaps. This is possible by the explicit preceding bounds, involving \(\lambda\), \(\sqrt\lambda\log(1/\lambda)\), exponentially small terms and \(h/\lambda\). Negative shape derivatives only decrease the cross terms, so no lower bound on that derivative is needed. Consequently \(a-b_sw>1\), the second expression is less than 2, and \(v_s\ge a>2\). This proves the admissible cone test throughout the actual pulse, including its correcting bumps.

O7. Interpolation, the steep hold and the terminal half-interval

After the pulse, the vanishing total moments turn the remaining stress into tail integrals. The steep hold is essential: its \(h^6\) amplitude reduction compensates for the angular-stress lower bound of order h near the terminal interval.

After the pulse \(U=M=J=0\) exactly, hence \(W=1\). Before the angular bumps the interpolation source is \(-l-h+\vartheta_fc_\eta\), at least \(\lambda-h\). Its integrating coefficient \(1+l\) has fixed bounds, and the pulse endpoint Q is bounded below by \(c\lambda\). Thus \(Q_s\ge c\lambda\) through interpolation and the uniform interval. On the two bumps the perturbations of l and \((\log E)_\eta\) are \(O(C_{\rm pre}\lambda^{28})\), by O2.4, and their total logarithmic length is fixed. Variation of constants changes Q by that same order. It therefore remains at least \(c\lambda\) there, with a smaller fixed c.

The already established \(S(\infty)=0\), together with U zero, gives the exact backward identity

Before the exterior release, \(l\le-\lambda/2\), including the angular bumps, so the remaining pre-release integral is bounded by \(CXE^2/\lambda\). The release contribution is bounded by O3.2 and is no larger than another \(CXE^2\), since \(XE^2\) decreases before release up to bounded bump factors. The same proof after a parameter derivative uses the bounded logarithmic parameter derivatives and small angular-coefficient derivatives. Also O4.3 gives \(|\Pi|+|\Pi_\eta|\le CE^2\). Thus

The last statement uses \(E\le C_{\rm pre}\lambda^{30}e^{-13/(2\lambda)-13}\) at pulse end, with a nonincreasing unedited profile and small relative bumps thereafter. Here \(b_s=0\), and \(2<a\le2+2\lambda+.2+O(C_{\rm pre}\lambda^{28})\). The two cone margins are positive; \(v_s=a>2\).

From the release start onward E is independent of \(\eta\), as are its remaining moment integrals. Equations O4.3, O7.1 and P2.5 therefore give the exact formula

Every subsequent slope satisfies \(l\le-3h/4\). Hence

Thus \(|N_s|\le CE^2\) with a constant independent of h. Before and during the steep hold Q is at least \(c\lambda\); during that hold it increases by O2.5. Hence \(|w|\le Ce_{\rm rel}/\lambda=o(1)\) there. After the steep hold, its next transition retains a large positive Q, the \(-h\) hold decreases it only to \(Q_p\asymp h\), and O2.8 gives \(Q_s\ge ch\) on the first half-unit of the terminal interval. The exact \(h^6\) amplitude drop in O3.1 yields

on that latter range. This remains small even if h is much smaller than \(e_{\rm rel}\); no reversed parameter order is used. Throughout the release range in question \(b_s=0\), \(2<a\le4\), so O5.1 holds and \(v_s=a>2\). The endpoint collar after terminal coordinate \(1/2\) is intentionally not included in this conclusion: it is the source's separate heat and endpoint-stress construction.

O8. One finite parameter choice and the resulting reference profile

We now collect the inequalities in their logical order. This step proves that all choices can be satisfied by one finite profile; it does not assume that independently stated smallness conditions are automatically compatible.

Choose the fixed shapes, \(T_f\), and \(c_o\) as in O0. Select \(M_d\) to meet O5.7, define \(T_d\), then choose \(P_*\) as there. All constants denoted \(C_{\rm pre}\) are now fixed. Select \(\lambda>0\) sufficiently small to meet the finite list of moment-solve conditions O2.4, the root bracket O3.4–O3.7, the bounds O6.4, and the parts of O6.12 and O7.2 depending on \(\lambda\). All their controlling quantities tend to zero. Finally select h sufficiently small relative to this \(\lambda\) and \(e^{-T_d}\) to meet O5.4, O6.2, O6.10–O6.12 and the positive holding-length condition O2.7. Include also the finite outer heat-stress requirement of H5.5, with its constant uniform for small h and the already fixed amplitude \(e_{\rm rel}\). The h-uniform bounds in O3.2 and O7.4 ensure that making h smaller does not undo any preceding smallness condition. This proves existence of a single finite parameter choice for the entire schedule.

For any fixed \(x_*\in(0,1]\), the remaining radial range from \(X=x_*X_R\) through terminal coordinate \(1/2\) is a compact interval in \(x=X/X_R\), after these choices. The proof gives positive Q and strict O5.1 throughout it. All a, \(b_s\), w are continuous and bounded there. Their image is therefore a compact subset of the strict set in P6.4–P6.5, with positive minimum gaps. The exact comparison P3.7 proves that these quantities are independent of \(X_R\), while

Choose \(X_R\) large enough that the left side exceeds the finite threshold \(P_K\) of P6.5 everywhere. The relaxed cone follows on the entire specified compact interval. The admissible cone follows wherever \(v_s>2\), in particular throughout the intermediate power-law interval, pulse, interpolation and exterior release, and at every point with \(l<0\) in the transition into the intermediate interval. This is precisely the range in Proposition A.4. All stage boundaries divided by \(X_R\) are independent of \(X_R\) and \(\eta\).

The reference profile has positive E, smooth U and E for \(X>0\), \(M=J=U=0\) after the pulse, \(S(\infty)=0\), \(\int_0^\infty(H-H_{\rm pow})\,dX=0\), and the four unchanged reserved patches. Beyond the terminal interval it equals the stated \(\eta\)-independent power law. Its canonical pressure has the analyticity, sign, size and scale-independence properties of Lemma A.5. These conclusions establish those two source components. The heat replacement and endpoint stress are constructed next. The regular axis and its exact join remain separate components of the leading-profile theorem.

K0. The original heat integral solves the exterior equation

We now construct the exact exterior heat field from its original integral. The power law alone has a viscous residual. This calculation supplies the replacement, including the pressure and every viscosity factor. Original (A.32) defines

Here \(\Gamma(1+h)=\int_0^\infty e^{-v}v^h\,dv\); neither its argument nor the exponent of \(1+Zv\) is altered. For the full rising factorial \((b)_m=b(b+1)\cdots(b+m-1)\), with \((b)_0=1\), differentiating under this integral yields exactly

The dominating function is the displayed integrand with its last factor replaced by one; it is integrable at zero since \(h+m>-1\) and at infinity by exponential decay. Gamma integration by parts gives \(\Gamma(1+h+m)/\Gamma(1+h)=(1+h)_m\). Dominated convergence proves smooth right derivatives at zero of all orders, where equality is \(\mathcal H^{(m)}(0)=(-1)^m(h)_m(1+h)_m\). No convergent Taylor series is asserted.

Multiplying the derivative of \(e^{-v}v^{1+h}(1+Zv)^{-1-h}\) by \(h\), and expressing the result with the common factor \(e^{-v}v^h(1+Zv)^{-h-2}\), gives the polynomial \(h(1+h-v-Zv^2)\). Substituting (K0.1)–(K0.2) into the following left side gives that same integrand. Both endpoint terms vanish, so

This verifies the integral identity in original (A.37) with its boundary terms.

Retain \(s=r^2/2\) and the original positive coefficient \(c_\infty\). The physical exterior field is

All constants in this comparison remain explicit. Direct differentiation with \(Z=2\tau/s\) gives

Since \(2A+1=2a_K\) and \(A^2-1/4=a_K(a_K-1)\), (K0.3) makes these two quantities equal. Thus the exact swirl heat equation, including \(-r^{-2}K\), holds. Equation (K0.2) also gives the original uniform bound

It applies on the full nonnegative half-line.

The exterior pressure is exactly

The integral converges because \(0<\mathcal H\le1\) and its integrand is bounded by \((2^Ac_\infty)^2\rho^{-3-4h}\). Its radial derivative is \(+K(r,\tau)^2/r\), cancelling the centrifugal component \(-K^2/r\) of \((Ke_\theta\cdot\nabla)(Ke_\theta)\). The other pressure components vanish. Together with (K0.5), this proves the full momentum residual is identically zero in the exterior.

For completeness, every derivative of the pressure is controlled without discarding its endpoint. Put \(\omega=\tau/r^2\), retaining the radial exponent \(1+2h\). Differentiation gives the exact finite formula

where \(c_{0,b,0}=1\), coefficients outside \(0\le k\le a\) are zero, and \(c_{a+1,b,k}=(-1-2h-a-2b-2k)c_{a,b,k}-2c_{a,b,k-1}\). In the exterior \(X\ge X_{\rm ext}\), \(\omega=(1-\eta^2)/(2X)\le1/(2X_{\rm ext})\). Equations (K0.6)–(K0.8) bound every term. Time differentiation of (K0.7) retains the full sum \(\sum_{i=0}^b\binom bi(\partial_\tau^iK)(\partial_\tau^{b-i}K)/\rho\), bounded by \(C_b\rho^{-3-4h-2b}\); it is integrable. Each subsequent radial derivative differentiates its lower-endpoint value. This proves all fixed exterior derivative bounds and right limits for \(r\) bounded away from zero, including the pressure.

For the original viscosity, the exact exterior formulas are

The exact derivative factors are \(\partial_{r_\nu}^a\partial_\tau^bK_\nu=\nu^{(1-a)/2}(\partial_r^a\partial_\tau^bK)(r_\nu/\sqrt\nu,\tau)\) and \(\partial_{r_\nu}^a\partial_\tau^bp_{{\rm ext},\nu}=\nu^{1-a/2}(\partial_r^a\partial_\tau^bp_{\rm ext})(r_\nu/\sqrt\nu,\tau)\), and the heat equation is \(-\partial_\tau K_\nu=\nu(\partial_{r_\nu}^2+r_\nu^{-1}\partial_{r_\nu}-r_\nu^{-2})K_\nu\). The same exact centrifugal cancellation gives \(\mathscr R_\nu=0\).

H0. The original heat factor and its exact comparison with the power

Retain \(0<h<1/100\), \(A=1/2+h\), \(D=1/2-h\), \(d=1-\eta^2\), \(L=1-2h\eta^2\), and all fixed reference-profile choices. The heat factor is

The complete derivative and heat-equation proof is in K0 above. In particular

Here every factor in the rising factorial is retained and \((h)_0=(1+h)_0=1\). The same integrable majorant works at zero, so all right derivatives extend continuously there. The mean value theorem and the bound for \(m=2\) give

on a fixed bounded nonnegative interval, uniformly for \(0<h\le h_0<1/2\). For each fixed positive m, the derivative bound in H0.2 is at most \(C_mh\). This uses a finite Taylor formula, not a convergent series.

Let \(E_{\rm pow}(X)=c_\infty X^{-A}\), \(H_{\rm pow}=\sqrt{2X}E_{\rm pow}\), \(s=r^2/2\), \(\tau=1-t\). The physical heat swirl is exactly

It is independent of z. To check the sign used below, \(-Z\mathcal H'/\mathcal H\) is the average of \(hZv/(1+Zv)\) against the positive probability measure obtained by dividing the integrand of H0.1 by its integral. Consequently

In particular K is positive and strictly decreasing in r. Composition with \(Z=2(1-\eta^2)/X\) uses only \(Z\ge0\). Every parameter derivative is a finite chain-rule sum involving powers of \(2/X\), \(-2\eta\), \(-2\), and derivatives in H0.2. No division by d occurs, including at \(\eta=\pm1\).

H1. The actual heat change and its three full moment changes

The heat change is small at large X, but its integral effects have different sizes. We compute all three full discrepancies, including both quadratic terms and the angular row’s \(1/h\) factor, before solving for compensating bumps.

Use the source terminal coordinate \(y=\log(X/X_{\rm tail})\), \(X_b=e^3X_{\rm tail}\), and

The actual terminal replacement is

where \(E_{\rm cl}\) is the full reference profile, including its moment corrections. The factor in braces is a convex combination of 1 and the strictly positive heat factor; hence \(E_H>0\) everywhere. On this tail \(E_{\rm cl}=E_{\rm pow}f_o(y)\), and U is zero.

Set \(x=X/X_K\), and let \(e_K=E_{\rm cl}(X_K)\). All logarithmic derivatives of \(f_o\) and \(\chi_K\) are bounded after the preceding choices. H0.2–H0.3 and the chain rule therefore give, for every fixed \(j,m\),

For \(m=0\), the factor \(X^{-1}\) follows from \(\mathcal H-1\); every differentiated heat term also contains at least one factor \(X^{-1}\). Derivatives of the switch are supported in a fixed bounded x interval and obey the same bound. Thus H1.3 holds uniformly at both parameter endpoints.

Keep the original pressure moment \(C_p=\int E^2/(2X)\,dX\), stress moment \(S=\int(U^2-E^2/2)\,dX\), and subtracted angular moment \(\mathcal I=\int(H-H_{\rm pow})\,dX\). The exact discrepancies caused by H1.2 alone are

The subtracted power is unchanged. All quadratic terms and the factor \(1/2\) are present. Since \(\mathcal H\le1\), \(\Delta E\le0\) and \(0<E_H\le E_{\rm cl}\). Accordingly the three displayed discrepancies have signs nonpositive, nonnegative and nonpositive, respectively. The functions M and J are unchanged pointwise in X, because both their affected integrands contain U, which is zero on the edited tail.

For every fixed m, differentiating H1.4 under its absolutely convergent integrals is justified by H1.3 and its product-rule bounds. Keeping the radial factors explicitly gives

The \(1/h\) in the last row remains explicit. It is finite because h was fixed before the large-radius choice; no uniform limit as h tends to zero is used here.

H2. Three additive bumps, their exact nonlinear system and preservation

Use the second reserved patch so the heat correction has its own unchanged reference interval. The exact map below sends three bump coefficients to the three original moment changes. The nonlinear solve is the one already proved in N2.1, with its hypotheses checked here.

Use only the second reserved interval \(\mathcal I_2\). At its start \(X_*\), set \(x_*=X/X_*\). There the unmodified field is \(U=0\), \(E_0=e_*f(\eta)x_*^{-1/2-\lambda}\), \(f=(1+\eta^2)^{-1}\ge1/2\). Choose three fixed nonnegative smooth bumps \(\beta_i\) with ordered disjoint compact supports inside this patch and put

The tail and this patch are disjoint, so their moment changes add exactly. Divide the three original patch rows by \(e_*^2,X_*e_*^2,X_*^{3/2}e_*\), respectively. The resulting complete map is

Every sign, Jacobian and row factor is fixed by H1.4 applied to the patch. Recovering the original rows means multiplying by precisely those three factors; none is discarded. The linear weights are \(fx_*^{-3/2-\lambda},-fx_*^{-1/2-\lambda},\sqrt2x_*^{1/2}\). Their three exponents are distinct. P4 proves invertibility of the power-moment matrix. The row factors f, -f and \(\sqrt2\) give a uniformly invertible B on [-1,1], with bounded derivatives of each fixed order. The first two quadratic integrals remain in \(\mathscr Q\); the last row is exactly linear.

The desired vector is minus the three full discrepancies H1.4 divided by the corresponding original row factors. The ratios \(X_*/X_K\) and \(e_*/e_K\) are positive fixed numbers after the reference schedule is chosen. Thus H1.5 proves

for each fixed m. In the angular row the constant includes the retained \(1/h\) and the fixed factor \((X_K/X_*)^{3/2}(e_K/e_*)\). The pressure and S rows retain their respective fixed amplitude ratios as well.

Apply the actual iteration P5.6 with this d, this B and both quadratic rows in H2.2. Since B and \(\mathscr Q\) are independent of \(X_R\), their \(C^0\) operator norms are fixed. For sufficiently large \(X_K\), the contraction condition holds and gives the unique small solution \(c=O(X_K^{-1})\). Its differential \(B+\mathscr Q(c,\cdot)+\mathscr Q(\cdot,c)\) has a uniformly bounded inverse, by the same Neumann-series estimate. Differentiate H2.2 m times: the terms containing \(c^{(m)}\) are exactly this differential times \(c^{(m)}\). All remaining terms contain either \(d_{\rm mom}^{(m)}=O(X_K^{-1})\), one lower derivative of c times a fixed derivative of B, or a product of two lower c derivatives and fixed derivatives of \(\mathscr Q\). Induction therefore proves

A single sufficiently large radius for the \(C^0\) solve suffices for these all-fixed-order estimates, with order-dependent constants. Only the finite small bounds needed below require prescribed thresholds. No smallness of infinitely many derivative norms is assumed.

The corrected E is positive on the patch: its reference lower bound on this fixed compact \(x_*\) interval is \(e_*c_*>0\), while \(|\delta E|\le e_*C_h/X_K\). Choose the radius so the latter is smaller than \(e_*c_*/2\). It is positive elsewhere by H1.2. The solve proves exactly

In particular the actual profile still has \(M(\infty)=J(\infty)=S(\infty)=\mathcal I=0\). The pressure is integrated forward from the unchanged analytic datum \(\Pi_0\). Its total increment is unchanged, so it still vanishes at infinity. Before the compensation patch all profiles and the forward pressure are unchanged. Because U is unchanged everywhere, P0.7 also proves that \(V_0\) is unchanged everywhere.

There is an additional exact endpoint fact. At \(\eta=\pm1\), d is zero and \(\mathcal H(0)=1\), so all three tail discrepancies vanish. The zero coefficient vector solves H2.2 there, and uniqueness in the same small ball gives \(c(\pm1)=0\). Thus the entire modified E agrees with the reference E at both parameter endpoints. Moreover \(c=(1-\eta^2)\widetilde c\) with \(\widetilde c\) smooth on the closed interval: near an endpoint, the fundamental theorem of calculus writes c as its distance from that endpoint times the integral of its first derivative; the other linear factor is nonzero. The tail change has the same factor directly from

This endpoint equality does not set parameter derivatives of the changes to zero; those derivatives enter the stress and are controlled by H2.4.

H3. Preservation of the actual stress cone before the terminal collar

The three total moments now match. Inside the region where the field changed, their partial integrals need not match. We bound those partial changes and both shear variables to prove the actual cone survives everywhere in that region.

The compact interval from \(\mathcal I_2\) to terminal coordinate \(y=.5\) is fixed in \(x=X/X_R\). On it the reference E, Q and \(v_s-2\), and both sufficient cone gaps in O5.1, have positive minima. Every fixed radial and parameter derivative of the heat change and the compensating bumps is \(O_h(X_K^{-1})\) in these x coordinates. This follows from H1.3 on the tail, H2.1–H2.4 on the compact patch, and the fixed ratios of radii. In particular the needed radial shear derivatives are included.

For the cumulative moments retain exactly

Partial integrals of the absolute moment changes satisfy the same \(O_h(X_K^{-1})\) bounds after these divisions: on the patch use H2.2 with absolute values and the fixed supports, and on the tail use the majorants H1.5. Thus every required parameter derivative of \(\Delta\widehat{\mathbf m}\) is \(O_h(X_K^{-1})\). P3.4–P3.6, with the unchanged axis pressure, gives \(\Delta Q_s,\Delta N_s=O_h(X_K^{-1})\), using one additional parameter derivative. Directly differentiating \(a=1-2X\partial_X\log E\) and \(b_s=2XU_X/E\), with the positive reference E bound, gives the same bound for their differences. Division by the positive reference \(EQ_s\) gives it for \(w=N_s/(EQ_s)\) too.

These estimates preserve the positive lower bounds and strict O5.1 gaps for sufficiently large \(X_R\). The image of the perturbed \((a,b_s,w)\) lies in one fixed compact subset of those strict inequalities. P6.5 supplies a common finite threshold \(P_K\), while \(p_{s,1}=X_RxQ_s/L\) grows at least linearly with \(X_R\). Choose the radius also to exceed that threshold. Then the actual stress is admissible throughout the stated interval. This argument retains the full scaling: \(\widehat p_s=X_R^{-1}p_s\) is close, whereas an absolute small bound on \(\Delta p_s\) is not asserted. This completes the construction and cone assertion of Proposition A.7.

H4. Backward stress from the constructed profile and the boundary traces

The sign of the backward integral comes from a zero total weighted residual. We prove that identity, with every axis and infinity boundary term. The temporary axis has weaker powers than the eventual regular field, so its traces are checked directly rather than assigned regularity it does not have.

The profile now constructed has a temporary inner reference rather than a regular axis. We can nevertheless compute its backward stress directly; this avoids assuming the still-unconstructed inner field. For fixed \((z,t)\) with \(t<1\), its exact inner powers are

They follow from \(U=4\eta\), \(E=P_*f(X/X_R)^{1/10}\), P0.7 and the pressure integral. The coefficients are smooth on compact sets with \(t<1\). In particular all time and axial derivatives preserve these radial powers. They are not a claim of Cartesian smoothness at the axis.

For the physical leading tangential residuals, retaining radial viscosity and omitting axial viscosity exactly as in original (4.12), incompressibility gives the conservative identities

For example, the first transport divergence has an extra coefficient \((2ru_r+r^2\partial_ru_r+r^2\partial_zu_z)u_\theta=ru_ru_\theta\), exactly the cylindrical angular transport term. No angular term has been lost.

At zero, H4.1 gives

All four boundary traces vanish. The leading angular viscous term is a constant times \(r^{-9/5}\), whose weighted integral against \(r^2dr\) converges at zero. The other terms of H4.2 are also integrable there. Thus the forthcoming integrations are valid for the actual temporary reference. For a smooth axis field, the standard axis powers \(u_\theta,u_r=O(r)\), \(u_z=O(1)\), \(\partial_ru_z=O(r)\) give the same vanishing traces; the argument also verifies the receiving identity in Lemma A.8 once the inner field is supplied.

In the exterior \(u_\theta=K\), \(u_r=u_z=0\). Write \(K_{\rm pow}=c_\infty s^{-A}\), which is independent of t and z. H0.2–H0.3 give, locally uniformly in nonnegative bounded \(\tau\),

For \(R_*\ge1\), both the subtracted angular moment and its time derivative have the common weighted majorant

The \(r^2\) weight multiplies both terms, as in the original page. Near zero the subtracted power has weighted size \(r^{1-2h}\), integrable since \(h<1\), and its time derivative is zero. These bounds justify differentiation of the convergent subtracted moment and its change of variable:

The second integral is supported in a bounded radial region; its z derivative is justified by H4.1 at zero and local boundedness elsewhere. Exterior radial transport vanishes because \(U=M=0\) gives \(V_0=0\) exactly. H4.3–H4.6 in the first conservative identity prove \(\int_0^\infty r^2\mathscr R_\theta^{(0)}dr=0\).

For the axial identity, \(\int_0^\infty ru_zdr=q^{1-A}M(\infty)=0\). The exact pressure calculation, justified by absolute convergence or Tonelli applied to its negative, is

At infinity \(p=O(r^{-2-4h})\) and \(r^2p\to0\); near zero H4.1 suffices. The axial flux derivative is justified locally in z by the same bounds, and its exterior pressure is z independent. Radial transport and viscosity traces vanish at both ends. Thus \(\int_0^\infty r\mathscr R_z^{(0)}dr=0\).

The original forward primitives can therefore be written with the positive backward sign:

The forward primitives here equal P1.5: their lower boundary constants vanish by H4.3, and their radial integrating factors are the same \(r^2,r\). Both residuals vanish beyond \(X_b\), by the exact swirl heat equation and its z-independent canonical pressure. Hence \(\mathcal T_0=0\) for \(X\ge X_b\).

This calculation proves the outer stress for the actual reference, with its weaker axis powers. It also gives the exact transfer to the future regular inner construction: P3.1 shows that matching the five cumulative moments and common axis pressure at a radius where the profiles rejoin makes all outer \(Q_s,N_s,p_s,a,b_s,\mathcal T_0\) identical. The actual regular matching is constructed in Q0–Q9; this reference calculation uses its stated temporary axis powers.

H5. Positive angular stress and the strict cone on the terminal interval

The terminal stress is positive in its angular component. Its axial component is smaller by a controlled factor. The exact physical formulas prove both assertions and produce a strict cone margin before the stress vanishes.

On \(1/2\le y<3\), the switch is identically one. Write \(f=f_o(y)\), \(f'=\partial_y f\). Then \(u_\theta=Kf\), \(u_r=u_z=0\), and K is independent of z. The original coordinate derivatives are

Apply the product rule to the original radial swirl operator, and use its exact heat equation H0.4. Differentiating the canonical pressure at fixed r gives

For the second identity start with \(p=-\int_r^\infty K(r')^2f(y(r'))^2dr'/r'\). Its z derivative is \(4\eta(q^DL)^{-1}\int K^2ff'\,dr'/r'\), and \(dr'/r'=dv/2\) gives exactly H5.2. The integrand vanishes beyond the terminal endpoint. This also proves the differentiation directly on its bounded support.

Integrating the viscous terms of H5.2 by parts in H4.8 yields

Indeed the integral of \(-r'^2Kf_{r'r'}\) has lower boundary \(+r^2Kf_r\); its remaining coefficient is \(2r'K+r'^2K_{r'}\). Combining it with \(-r'Kf_{r'}-2r'^2K_{r'}f_{r'}\) leaves \((r'K-r'^2K_{r'})f_{r'}\). The upper boundary is zero because f is constant there. Thus the boundary sign and all three terms in H5.3 are fixed.

Each term is nonnegative by H0.5 and \(f'\ge0\). The remaining y interval has length at most \(5/2\); hence \(1\le r'/r\le e^{5/4}\). H0.5 bounds the logarithmic derivative of K between \(-1-2h\) and \(-1\), so \(K(r')/K(r)\) is bounded above and below by positive constants independent of small h and large \(X_R\) on this interval. Since \(\int_y^3f'(v)dv=\rho_o\psi_o(y)\), the last term of H5.3, with \(dr'=r'dv/2\), gives

Use \(|\eta|\le1\), \(L\ge1-2h\), \(0<f\le1\), and monotonic decrease of K and \(\psi_o\) in H5.2 and then H4.8. The finite length and radius ratio just bounded give

The last equality retains \(1-D=A\). The constants can be chosen uniformly for \(0<h\le1/100\) and sufficiently large radius. By the exact amplitude reduction O3.1 and the bounded factor \(f_o^{-1}\), the last expression is at most \(C e_{\rm rel}h^6\). Here \(e_{\rm rel}\) was fixed before h. Include the finite requirement \(Ce_{\rm rel}h^6\le1\) among the small-h choices of O8, before choosing \(X_R\). This is compatible with the earlier choices: it is one further upper bound on h, not a lower bound or a change of h after the profile is chosen. We henceforth use that common choice.

The original shear variables now obey

The reference step has \(f'/f<h/4\). H0.3 and positivity of \(\mathcal H\) give \(-Z\mathcal H'/\mathcal H<h/4\) after increasing \(X_R\). Thus \(2+h<a\le2+2h\). In P6, \(t_s=0\), \(v_s=a>2\), \(P_c-v_s=\mathcal T_{0,\theta}/F>0\), and \(J_c=\mathcal T_{0,z}/F\). Its exact cone criterion becomes

This proves admissibility at every point of the terminal interval before its outer endpoint, with fixed directional slack. It uses the actual stress calculated in H4, not an independent replacement for it.

H6. Full flat factors and the smooth limiting direction

A nonzero direction can have a smooth limit even when the vector itself becomes zero. To prove it here, we factor both components from their integral formulas. The two different powers of the distance to the edge, not a numerical plot, determine the limiting direction.

First prove the integral identity used at the edge. For \(c>0\), a fixed integer \(j\ge0\), and b smooth on \([0,\delta_0]\) times a compact parameter interval,

where the substitution \(u=\delta(1+\delta^2v)^{-1/2}\) gives exactly

The Jacobian is \(-\tfrac12\delta^3(1+\delta^2v)^{-3/2}\), and \(c/u^2=c/\delta^2+cv\). These give both the factor \(1/2\) and every power in H6.1. Each fixed mixed derivative of the last integrand is bounded by \(e^{-cv}\) times a polynomial in v, uniformly on the closed parameter rectangle: the denominator \(1+\delta^2v\) is at least one, its numerator factors are polynomially bounded in v, and all derivatives of b on its compact domain are bounded. The same holds for derivatives of its composed argument. Dominated differentiation proves smoothness through \(\delta=0\) and all fixed derivative bounds; dominated convergence gives the displayed endpoint value. The proof also permits any finite collection of extra compact smooth parameters.

Put \(\delta=3-y=\log(X_b/X)\). The specific step in O0 has, for sufficiently small nonnegative \(\delta\),

Extend \(e^{-4/\delta^2}\) by zero at zero; its derivatives of every order are a polynomial in \(1/\delta\) times the same exponential and tend to zero. The denominator defining g has a positive value at zero, so g is smooth on a closed small collar. With \(f=1-\rho_o\psi_o\) and \(\partial_y=-\partial_\delta\),

To retain every scale and integration weight, set \(R(v)=\sqrt{2X_{\rm tail}e^v}\) and \(\mathcal E(v,\eta)=E_{\rm pow}(X_{\rm tail}e^v)\mathcal H(2d/(X_{\rm tail}e^v))\). Thus \(r'=\sqrt qR(v)\) and \(K(r',t)=q^{-A}\mathcal E(v,\eta)\). Multiplying H5.3 and H4.8 by the original \(q^{A+1/2}\) gives the exact profile formulas

For the second angular term use \(K_{r'}=q^{-A-1/2}\partial_{R}\mathcal E\), \(R\partial_R\mathcal E=2\partial_v\mathcal E\), and \(f_{r'}dr'=f'dv\). For the last angular term, \(r'^2Kdr'=q^{3/2-A}R(v)^3\mathcal E\,dv/2\), while its denominator is \(qLr^2=q^2LR(y)^2\). Both have physical factor \(q^{-A-1/2}\). For the axial term, \(r'dr'=qR(v)^2dv/2\) cancels the 2 in H5.2, leaving physical factor \(q^{1/2-D-2A}\). The additional profile factor cancels it exactly because \(q^{A+1/2}q^{1/2-D-2A}=q^{1-D-A}=1\). Hence H6.5 retains the original physical factors and contains no hidden dependence on q.

All remaining coefficients of these integrals are smooth on the closed \((\delta,\eta)\) collar: \(X=X_be^{-\delta}\) is positive there, \(L\ge1-2h>0\), and the heat derivatives are smooth for the full closed interval \(-1\le\eta\le1\). The first angular term, using H6.4, is already \(e^{-4/\delta^2}\delta^{-3}\) times a smooth coefficient. Apply H6.1 with \(j=3,c=4\) to each remaining angular integral. Each becomes \(e^{-4/\delta^2}\) times a smooth function, and hence contributes \(\delta^3\) times a smooth function to that coefficient. This proves

Positivity is uniform in \(\eta\), since the heat factor has a positive minimum on \([0,2/X_b]\). Continuity and compactness give a common closed collar where \(b_\theta\) has a positive lower bound.

The inner axial integral in H6.5 has one factor f'; H6.1 with \(j=3,c=4\) writes it as \(e^{-4/u^2}\) times a smooth coefficient. The outer integral then uses H6.1 with \(j=0,c=4\). Consequently

Both coefficients are smooth on the closed collar. One can retain also the exact first axial coefficient. Set \(\mathcal E_b=E_{\rm pow}(X_b)\mathcal H(2d/X_b)\), \(R_b=\sqrt{2X_b}\). The inner integral coefficient at zero is \(\rho_og(0)\mathcal E_b^2\): its input coefficient is \(8\rho_og(0)\mathcal E_b^2\), and H6.1–H6.2 divide it by \(2c=8\). The outer integration divides by 8 once more. Thus

This is an explicit consequence of the same source integrals, not an assumption about their leading term. H6.5 also shows that the axial stress has the sign of \(\eta\) on the open collar and is zero at \(\eta=0\).

Write \(\mathcal R=\delta^6b_z/b_\theta\). Since \(\mathcal T_{0,\theta}>0\) before the edge, the unit direction is exactly

The expression extends smoothly to the closed collar. It is a direction extension; the stress itself is zero at the endpoint. H5.7 and continuity show that \(2-(a-2)\mathcal R^2\) remains uniformly positive, with endpoint value 2.

Finally, differentiating H6.6–H6.7 any fixed number of times in X and \(\eta\) introduces only finitely many inverse powers of \(\delta\). Indeed \(\delta_X=-1/X\) is smooth on the collar and the coefficient derivatives are bounded there. Therefore for every fixed mixed derivative I there are finite profile-dependent constants with

The upper bounds tend to zero to every order, proving that the stress extends smoothly by zero to \(X\ge X_b\), as H4 already requires. Neither constant depends on physical q. These calculations complete the outer-edge assertions of Proposition A.10 and the flat integral lemma A.9.

H7. The original viscosity and physical stress

For any original viscosity \(\nu>0\), use the exact spatial map, with t unchanged:

The full coordinates, with the original physical r and z now denoted \(r_\nu,z_\nu\), are

Each physical leading residual scales by \(\sqrt\nu\): time, transport, pressure gradient and \(\nu\) times the radial viscous operator all do so. The stress divergence scales by the same factor, since one radial derivative of \(\nu T(x_\nu/\sqrt\nu,t)\) contributes \(1/\sqrt\nu\). Equivalently, substitution \(r'_\nu=\sqrt\nu r'\) into the two backward integrals gives \(T_{\nu,\theta}=\nu T_\theta\) and \(T_{\nu,z}=\nu T_z\), with their respective weights \(r'^2\) and \(r'\) retained. The stress ratio and its direction therefore agree exactly with H6.7–H6.9 at the mapped point. No viscosity factor is removed from the physical fields or their bounds.

X0. Five worked exercises

These exercises make explicit the finite-derivative argument, the parameter order, the endpoint correction, the boundary traces and the edge coefficient. Each uses the constructed fields above.

Exercise 1. What exactly does Taylor's formula give for the heat factor?

For \(n\ge0\), derive a remainder valid for every \(Z\ge0\), retaining its full coefficient. Show that differentiability at zero does not justify using an infinite power series here.

Solution. K0.2 gives derivatives on the closed nonnegative half-line. Repeated use of the fundamental theorem of calculus on \([0,Z]\) therefore gives

The integral formula K0.2 shows that \(R_n\) has sign \((-1)^{n+1}\) for \(Z>0\). Thus finite even and odd truncations also give one-sided bounds with the displayed error. Every step used finitely many integrations of continuous derivatives.

For \(h>0\), let \(a_m=(h)_m(1+h)_m/m!\). All these coefficients are positive, and

The terms of the formal Taylor series therefore fail even to tend to zero at every nonzero argument. Its radius of convergence is zero. The heat factor is smooth from the right at zero and has no analytic extension through zero. This conclusion concerns this heat factor; the axis datum in O4 is a different integral and has its own proved holomorphic extension. Neither result replaces the other.

Exercise 2. Why must h be chosen before the large radius?

Calculate the angular discrepancy bound in its original units and after the correction-row division. Then show explicitly how the steep interval prevents a divergent exterior stress-moment estimate as h becomes small.

Solution. With \(A=1/2+h\), the tail integral is

Write \(r_*=X_*/X_K>0\) and \(e_*/e_K=\gamma_*>0\). These are fixed once the reference schedule is fixed. Dividing the last row of H1.5 by the original angular row factor gives

For each fixed h this tends to zero as \(X_R\), and hence \(X_K\), grows. The argument does not assert uniform smallness when h tends to zero with \(X_K\) fixed. It also retains both fixed ratios rather than treating the two patch amplitudes or radii as equal.

On the steep interval \(l=-1\), its length is \(4\log(1/h)\). Because \(d\log(XE^2)/dy=2l\) and \(d\log E/dy=l-1/2\), integration gives exactly the factors \(h^8\) and \(h^6\). After that interval \(l\le-3h/4\), so the remaining original stress-moment integral is bounded by

The earlier transition and steep interval have bound \(C X_{\rm rel}e_{\rm rel}^2\), as in O3.2. The complete bound is consequently uniform for small h. At the terminal half-interval, division of \(E\le C e_{\rm rel}h^6\) by \(Q_s\ge ch\) leaves \(C'e_{\rm rel}h^5\). Both estimates justify decreasing h after \(\lambda\), and neither requires increasing h to offset a later choice.

Exercise 3. What is preserved at the two parameter endpoints?

Prove that every heat-correction coefficient is divisible by \(1-\eta^2\) as a smooth function on the closed interval. Determine the endpoint values of the quotient and explain why the stress still needs derivative estimates.

Solution. At \(\eta=\pm1\), the heat factor is one. All three tail moment discrepancies are zero. The exact nonlinear map H2.2 has zero as a solution, and the contraction branch is unique in its fixed small ball. Hence every coefficient \(c_i\) is zero at both endpoints.

Near \(\eta=1\), the fundamental theorem of calculus gives

Dividing by \((1-\eta)(1+\eta)\) proves a smooth quotient there. The same argument at \(-1\), and ordinary division on the open interval, yields a single smooth function \(\widetilde c_i\) with

The expressions agree on the overlapping open intervals because they all equal the same quotient. The tail itself has the factor in H2.6. Its first parameter derivative can be nonzero at an endpoint: the derivative of \(1-\eta^2\) is \(-2\eta\). Thus equality of endpoint values alone gives no equality of the stress, whose moment formulas contain parameter derivatives. The complete estimates H2.4 and H3 are needed.

Exercise 4. Why may the temporary axis be used in the backward formula?

Check the four boundary terms and weighted integrability directly from H4.1. Decide whether those facts make this axis a smooth Cartesian velocity field.

Solution. Put \(\alpha=1/5\) in \(u_\theta=C_\theta r^\alpha\). Direct differentiation gives

Multiplication by the angular weight \(r^2\) leaves \(r^{1/5}\), integrable at zero. The angular transport boundary is \(C_rC_\theta r^{16/5}\). The viscous boundary is \(r^2\partial_ru_\theta-ru_\theta=-\frac45C_\theta r^{6/5}\). Both vanish. The axial boundaries are \(ru_ru_z=C_rC_zr^2\) and \(r\partial_ru_z=0\), and vanish as well. Time and axial derivatives of the smooth coefficients preserve the powers. For the subtracted exterior power, \(r^2K_{\rm pow}=2^Ac_\infty r^{1-2h}\) is integrable at zero because \(h<1\). At infinity, H4.5 supplies the integrable majorant and its time derivative. These are precisely the facts needed to integrate both conservative residual equations and obtain the positive backward sign.

They do not establish a regular axis. Along the positive Cartesian x-axis, the azimuthal velocity contributes \(u_y=C_\theta x^{1/5}\). Since \(C_\theta>0\), its derivative is \(C_\theta x^{-4/5}/5\), which is unbounded as \(x\downarrow0\). A smooth Cartesian field cannot have this restriction. The subsequent analytic inner construction replaces this temporary field and matches all five moments and the common pressure datum; P3 then preserves the outer stress exactly.

Exercise 5. Recover the exact leading direction and its viscosity dependence

Derive the first coefficient of the stress ratio at the edge directly from the double integral H6.5. Keep the original \(X_b\), L and heat amplitude, and transfer the result to viscosity \(\nu>0\).

Solution. Let \(\mathcal E_b=E_{\rm pow}(X_b)\mathcal H(2d/X_b)\), \(R_b=\sqrt{2X_b}\), and \(g(0)=e\). The first angular boundary term has coefficient

The inner axial integral has input coefficient \(8\rho_og(0)\mathcal E_b^2\). Applying H6.1 with \(c=4,j=3\) divides that coefficient by \(2c=8\) and removes its inverse cubic power. The outer integral has weight \(R_b^2\); applying H6.1 with \(c=4,j=0\) divides by 8 again and adds the factor \(\delta^3\). Multiplication by \(\eta/(LR_b)\) yields

The limit is uniform in \(\eta\), because the smooth coefficient quotient in H6.7 is continuous on a compact collar with a positive denominator. For \(\eta=0\), the axial integral is identically zero. For either nonzero sign of \(\eta\), its sign is that sign, while \(T_\theta>0\). Thus the unit direction tends smoothly to \((1,0)\) from the corresponding side.

For the original viscosity, set \(\delta_\nu=\log(X_b/X_\nu)\), with \(X_\nu,\eta_\nu\) as in H7.2. Both stresses are multiplied by the same positive factor \(\nu q_\nu^{-A-1/2}\). Their ratio and direction are therefore the same profile functions evaluated at \((X_\nu,\eta_\nu)\). In particular the coefficient in X0.7 is evaluated at \(\eta_\nu\), retaining \(X_b\), \(L(\eta_\nu)\), and \(\mathcal H(2(1-\eta_\nu^2)/X_b)\). There is no extra multiplicative viscosity factor in the ratio. The physical stresses themselves retain the full displayed factor.

What this construction supplies to the leading profile

We have constructed the full reference schedule and solved its actual moment equations. Its pressure datum is holomorphic on a common neighborhood of the closed parameter interval. The heat replacement and three-bump correction preserve that datum and every required total moment. The complete leading residual has zero weighted totals, its outer stress has compact support, and the original strict cone and smooth endpoint direction hold on the stated outer range. The stress estimates retain all physical and viscosity factors.

The remaining leading-profile components have complete separate proofs: the analytic axis J0–J7, the continuation and exact joining Q0–Q9, and the admissible shear and full leading profile V0–V7. The positive-order base Z0–Z8 receives that actual leading profile. The corresponding lessons are analytic axis, joining, admissible shear and base field.

Authored lesson text and the accompanying diagram are CC0. Mathematical source: OpenAI, Finite Time Blowup for Navier–Stokes, Appendix A and the exact coordinate and stress formulas cited above. Exposition: GPT-6 Astra (OpenAI). The original source retains its attribution and rights.

Constructing the analytic axis profile

NS-BLOWUP N2.3. The outer construction fixes an analytic pressure value at the axis. We now use that actual datum to solve the nonlinear inner equations. The solution is analytic in its scalar profile variables, gives a smooth Cartesian velocity field, has exactly zero leading tangential stress, and satisfies the strict endpoint inequality needed for its continuation.

The difficulty is concrete. The equations contain parameter derivatives as well as radial derivatives, and the radial inverse degenerates at the axis. A derivative in the parameter cannot simply be treated as a bounded operation in the coefficient norm. We prove the full mixed estimate after radial integration, construct the inverse of the leading angular operator, and then prove convergence of the actual nonlinear iteration.

The complete outer-profile lesson, O0–H7, supplies the prescribed pressure and exterior. The stress and moment lesson, P0–P6, supplies the original-coordinate equations, physical residual and exact comparison maps. Both are included in this reader. Here all original factors, including \(\Lambda\), \(\mathsf C\), the weights \(20^{-\alpha}\rho^{-\beta}\), viscosity and the pressure datum, remain explicit.

The source is OpenAI, Finite Time Blowup for Navier–Stokes, original paper, pages 26–27 and 144–150: equations (4.7)–(4.13), (B.1)–(B.21), Lemma B.1 and Propositions B.2–B.3. The source version and mathematical locators are recorded in SOURCES.json. This is a separate derivation and teaching text, not a diplomatic transcription.

Figure. The domain \(R/20+r/\rho<1\) is the exact convergence domain used for the joint coefficient bound. Its coordinates are ratios of the original radii; neither original radius is removed from the proof. The two endpoint alternatives use the original \(\chi\), \(Z_*\), \(p_1\) and \(p_2\). The last panel records the actual positive axial residual of the leading field, with its full viscosity factor, proved in Exercise 5. It is not a residual of the final corrected field.

Why the proof has this order

First choose the axis data using the pressure already supplied by the outer field. Their two complementary parameter regions will later give the endpoint inequality. Then construct a coefficient space that represents actual analytic functions and prove its multiplication and integrated derivative estimates. These estimates control every term of the original nonlinear equations.

The leading angular operator need not be small. Its repeated inverse integrations increase the radial degree, producing factorial denominators. That supplies a bounded inverse without an unsupported smallness assumption. Only the remaining nonlinear map carries \(1/\Lambda\). We choose \(\Lambda\) large enough to make that map contractive, and then choose \(\mathsf C\) large enough to control the original amplitude on a complex neighborhood. The resulting bounds hold for every subsequent admissible amplitude.

Finally, positivity permits division by the actual angular profile. We recover the smooth Cartesian field, prove the source lower bound, and check both endpoint regions. Zero leading stress retains the axial-viscosity and radial parts of the full residual; Exercise 5 computes one such term exactly.

J0. Actual axis data and their parameter domains

The pressure datum is already fixed. We choose the remaining axis data so that either the angular contribution or the axial contribution can give a strict endpoint margin. The location where \(H_*\) vanishes is central to that division of the parameter interval.

Keep the original fixed choices \(0<h<1/100\), \(A=1/2+h\), \(D=1/2-h\), \(d=1-\eta^2\), \(L=1-2h\eta^2\). The analytic pressure \(\Pi_0\) constructed in O4 is even, obeys \(\Pi_0\le-(5/2)P_*^2(1+\eta^2)^{-2}\), and has \(\eta\Pi_{0\eta}>0\) for nonzero real \(\eta\in[-1,1]\). Choose \(0<j_0\le1/20\) and set exactly

On \((-1,1)\), \(H_*/d=D\eta/d+4\eta+j_0\) has derivative \(D(1+\eta^2)/d^2+4>0\) and ranges from minus infinity to plus infinity. Its unique zero \(\eta_0\) lies in \((-1,0)\). Writing \(x=-\eta_0\), its equation is

For the bounds on x, the equation first gives \(x<j_0/4\le1/80\); its parenthesis is then less than 5 and greater than 4. At \(\eta_0\), \(H_*=0\), \(-d\Pi_{0\eta}\ge0\), and \(U_*=Dx/d\). The first term of \(Z_*\) has absolute value at most \(j_0\): use \(x<1/80\), \(D\le1/2\), \(A\le.51\), and the displayed formula for \(U_*\). The last term satisfies

The original choice \(P_*>e^{T_d}>10\) therefore gives \(Z_*(\eta_0)>\tfrac12j_0P_*^2>0\). This proves the required positivity from the actual datum.

Choose a neighborhood of \(\eta_0\) on which \(Z_*\) has a positive lower bound. A positive \(\delta_*\) smaller than that bound makes the compact set \(\{|Z_*|\le\delta_*\}\) avoid that neighborhood. On this set \(|H_*|\) has a positive minimum. Choose \(\sigma_*>0\) small enough that

If the set is empty any sufficiently small positive \(\sigma_*\) works. The strict inequalities persist on a slightly larger real compact interval \(I\supset[-1,1]\), chosen so \(H_*\) has no additional zero there. Choose a bounded simply connected complex neighborhood \(\Omega\) of I, symmetric about the real axis, with closure in a domain where \(\Pi_0\) is holomorphic and L and \(H_*^2+\sigma_*^2\) do not vanish. Compactness supplies such a neighborhood.

For \(\Lambda\ge1\) retain the original functions and amplitude order:

The holomorphic primitive exists on \(\Omega\); it is real on I, so \(\phi_*>0\) there. The amplitude \(\mathsf C\) will be chosen after \(\Lambda\). None of these choices alters the outer pressure datum.

J1. The original coefficient space and exact radial maps

The norm controls every coefficient and every parameter derivative at once. We must prove both completeness and convergence to actual functions. The variable \(Y=\Lambda X\) records a change of radial coordinate; the original \(X\) and its derivative factors are recovered explicitly.

Use the scalar radial variable \(Y=\Lambda X\). For a small fixed \(\rho>0\), retain the weights and norm

The space \(\mathcal B_\rho\) consists of smooth coefficient sequences with finite norm. It is complete. Indeed a norm-Cauchy sequence converges uniformly at each coefficient and each derivative order; the fundamental theorem of calculus identifies the successive derivative limits. Passing to the limit in every weighted inequality gives a limit in the space and norm convergence.

This coefficient space represents actual analytic functions, not just formal series. At a real center \(\eta_0\), divide the derivative estimate by \(\beta!\) and sum the joint Taylor series. For \(|Y|\le R<20\) and \(|\eta-\eta_0|\le r\) with \(R/20+r/\rho<1\), its absolute sum is bounded by

The identity follows by expanding \(\sum_{n\ge0}(R/20+r/\rho)^n\). Taylor's remainder estimate proves equality with the original smooth coefficients on the real interval. Local extensions agree on overlaps and supply a common holomorphic neighborhood, including the endpoints of I. The denominator factors in J1.1 have been retained in the norm; discarding them only in this upper bound does not change the represented function.

Write \(\mathcal I F=\int_0^YF(Y',\eta)dY'\), \(\mathcal A F=Y^{-1}\mathcal I F\), and \(\mathcal D_X=Y\partial_Y=X\partial_X\). The identity \(\mathcal A_X(F)=\int_0^1F(\Lambda aX,\eta)da=\mathcal A(F)(Y,\eta)\) proves the exact radial-average comparison. In the next radial inverse, \(\nu\in\{1,2\}\) is an integration index; the physical viscosity is restored separately in J7. For \(\nu=1,2\), the unique smooth solution of \(YG_{YY}+\nu G_Y=F\), \(G(0)=0\), is \(G=\mathcal J_\nu F\), with

One can integrate \((Y^\nu G_Y)'=Y^{\nu-1}F\); smoothness removes the singular homogeneous derivative and the zero value removes the constant. Expanding that integral gives every divisor in J1.3.

J2. Multiplication and the mixed derivative estimate

The squared denominators in the coefficient weights make the two convolutions summable. The binomial factor connects radial degree to parameter derivatives. We calculate the weight ratios before using any nonlinear estimate, including the case where a parameter derivative and a radial derivative act on different factors.

Keep the exact finite constant \(C_{\rm sq}=8\sum_{i=1}^\infty i^{-2}\). Splitting a sum at \(N/2\) gives

On the first half the ratio \((N+1)^2/(N-i+1)^2\) is at most 4; reverse the indices on the other half. The product Leibniz factor \(\binom\beta k\) cancels the two derivative factorials against \(\beta!\). The remaining binomial ratio is at most one because

The left side counts the subsets with exactly k elements in the first of two disjoint blocks of sizes \(i+k\) and \(\alpha-i+\beta-k\). Summing the resulting weight ratio in the radial and derivative indices and applying J2.1 proves

The exact ratios for increasing a radial degree are

Together with J1.3, these bound multiplication by Y, \(\mathcal I\) and \(\mathcal J_\nu\). The average divides the radial degree \(\alpha\) coefficient by \(\alpha+1\), so its norm is at most one. The second ratio, with the integration divisor \(i+1\), proves \(\|\partial_\eta\mathcal I F\|_\rho\le(80/\rho)\|F\|_\rho\).

For the differentiated product inside \(\mathcal J_\nu\), let the output radial degree be \(N=\alpha+1\), and let the differentiated F have degree i and G degree \(\alpha-i\). Allocate the added radial degree to F, writing \(I_1=i+1\), \(J_1=\alpha-i\). J2.4, the full Leibniz rule, and J2.2 bound its \(\beta\)-derivative divided by \(\mathfrak a_{N\beta}\) by

Here the numerator \(J_1\) is the logarithmic radial derivative and \(I_1\) is the parameter-derivative weight from J2.4. Since \(I_1+J_1=N\) and \(\nu\ge1\), the first fraction is at most one. Extending the first sum to all splittings of N and applying J2.1 twice proves

With an average on F its divisor \(i+1\) cancels \(I_1\). Omitting the radial derivative replaces \(J_1\) by 1, again leaving a fraction at most one. Without the parameter derivative use the first ratio of J2.4 and the radial coefficient at most \(\alpha\), divided by \((\alpha+1)(\alpha+\nu)\); J2.1 again applies. Omitting both derivatives follows directly from J2.3 and boundedness of \(\mathcal J_\nu\). Additional undifferentiated factors leave their radial indices unaltered. The same allocation of one degree to the parameter-differentiated factor and the same inverse divisor bound apply, while each additional convolution contributes \(C_{\rm sq}^2\). These arguments include zero indices and every \(\beta\). They prove the full Lemma B.1 needed below.

J3. Original nonlinear equations and every remainder

We substitute into the actual equations and display every remainder. The pressure is recovered by its original integral, so it contributes a known nonlinear function of the two unknown profiles. Complex control of the amplitude is what keeps all parameter derivatives bounded as the large parameters vary.

The original angular and axial equations for zero leading stress, and radial pressure balance, are

Their complete source formulas are P1.1. Make the exact substitutions, retaining the amplitude and radial factors,

The intended axis values are \(\Phi(0,\eta)=1\), \(\mathfrak u(0,\eta)=0\). In particular \(\Pi_X=\partial_Y\mathfrak p=g^2\Phi^2=\phi^2/\mathsf C^2\). Define

Direct substitution into J3.1 gives

with all terms

The terms contain the products \(A\eta\), not a derivative \(A_\eta\); the original page 147 has those products. To check the angular leading sign, \(H_*\zeta_*/L=-\chi\), while \(H_c\Lambda\zeta_*=-\Lambda L\chi+d\mathfrak u\zeta_*\). For the axial equation, expand its negative source before dividing by L:

Its constant part is \(-Z_*\). Expanding the first term about \(U_*\) gives exactly \(\Lambda^{-1}A(1-4\eta U_*)\mathfrak u-2A\eta\Lambda^{-2}\mathfrak u^2\); expanding \(H_cU_\eta\) gives the two linear terms and the mixed quadratic term in J3.5. Pressure contributes the last three displayed terms after multiplication by \(\Lambda\). This verifies the full substitution, rather than replacing the original equations by a different model.

Choose \(\rho\) strictly below a common complex-neighborhood distance from I to the boundary of \(\Omega\). For each \(\Lambda\), impose

The supremum is finite by the choice of closure. Then \(|g|\le1\) on \(\Omega\). Cauchy's estimate gives \(|\partial_\eta^\beta g|\le\beta!r_0^{-\beta}\) on a smaller neighborhood for some fixed \(r_0>\rho\). At radial degree zero its norm ratio is bounded by \((\beta+1)^2(\rho/r_0)^\beta\), uniformly in \(\beta,\Lambda,\mathsf C\). All fixed coefficients in J3.5 have finite norms by the same estimate. This complex bound is essential: real bounds on \(\phi_*\) alone would not control its large-parameter derivatives.

The maps \((\Phi,\mathfrak u)\mapsto\mathcal J_\nu R_i\) are consequently bounded and locally Lipschitz on any fixed ball in \(\mathcal B_\rho^2\), uniformly for \(\Lambda\ge1\) and amplitudes satisfying J3.7. Here is the full derivative check. In \(W\mathcal D_X\Phi\) and \(W\mathcal D_X\mathfrak u\), the only double differentiated products are \((\partial_\eta\mathcal A\mathfrak u)\mathcal D_X\Phi\) and \((\partial_\eta\mathcal A\mathfrak u)\mathcal D_X\mathfrak u\); J2.6 handles them. The factors B without a radial derivative use its omitted-derivative case. The terms \(H_c\Phi_\eta\), \(H_*\mathfrak u_\eta\) and \(\mathfrak u\mathfrak u_\eta\) use the same case, with extra undifferentiated factors. Finally \(\partial_\eta\mathfrak p=\partial_\eta\mathcal I(g^2\Phi^2)\) is bounded by J2.4, and \(\mathcal D_X\mathfrak p=Yg^2\Phi^2\) is bounded by multiplication and the Y operator. All other factors use J2.3. For Lipschitz bounds, subtract each monomial as a sum with one differing factor; the same estimates apply to each term. No unbounded parameter derivative is treated as a bounded operator by itself.

J4. Actual inverse, contraction and analytic solution

The operator inverse is built first, using its full sequence of powers. We then iterate one explicit nonlinear map in the complete coefficient space. Its fixed point is an actual analytic solution, and positivity is proved before dividing by the angular profile.

Let \(M_\chi\) bound multiplication by \(\chi\) in \(\mathcal B_\rho\), and let \(T=\mathcal J_2\chi/2\), with multiplication first. If the input has no radial coefficient below degree b, J2.4 gives

Multiplication by \(\chi(\eta)\) preserves the minimum degree and \(\mathcal J_2\) increases it by one, proving the second inequality by iteration. It includes all parameter derivatives through the norm. The series \(\sum_{k\ge0}(-T)^k\) converges in operator norm by its factorial bound. Multiplication of its finite partial sums by \(1+T\) leaves a remainder of norm tending to zero, on either side. Thus it is a two-sided inverse without any assumption that \(\chi\) or \(\|T\|\) is small.

The exact unperturbed solution is

Indeed the series converges everywhere and its coefficients solve \(2(zf_0''+2f_0')=-f_0\), \(f_0(0)=1\); J1.3 then gives the first identity. The axial identity follows since \(\mathcal J_1\) applied to a constant equals that constant times Y.

Use the actual map

On the closed ball of radius 1 about the center, in the maximum of the two coefficient norms, J3 supplies finite uniform bounds M and K for the numerator map and its Lipschitz constant, including \((1+T)^{-1}\). Choose \(\Lambda\ge\max(1,M,K)\), increasing it if needed. The map stays within radius \(1/2\) of the center and has Lipschitz constant at most \(1/2\). Iteration from the center is Cauchy, its limit lies in the ball by completeness, and continuity proves that it is a fixed point. The same bound proves uniqueness in the ball. It has distance at most \(M/(2\Lambda)\) from the center. All output corrections vanish at Y zero, so the intended axis values hold exactly. Choose \(\mathsf C_0(\Lambda)\) after \(\Lambda\) to include J3.7. The estimates hold for every subsequent \(\mathsf C\ge\mathsf C_0(\Lambda)\).

Choose any fixed \(4.1<R<20\) in J1.2 and then a smaller common complex neighborhood. Cauchy estimates for the actual difference from the center give, for every fixed r and s,

The constants and neighborhood are independent of sufficiently large \(\Lambda\) and of subsequent \(\mathsf C\). The original derivatives satisfy \(\partial_X^r=\Lambda^r\partial_Y^r\), so their bound is \(K_{r,s}\Lambda^{r-1}\), with the original scale retained. Reality follows either from the real iteration or from conjugation and uniqueness.

For positivity on the actual real rectangle, put \(t=z/2\) with \(0\le z\le41/10\). The alternating series after its cubic term has decreasing magnitudes and positive first remaining term. Thus

The cubic derivative is \(-[(t-4)^2+8]/48<0\), so its minimum for \(0\le t\le41/20\) is the displayed exact endpoint value. For the upper bound group the alternating series after its constant term into negative-positive pairs; magnitudes decrease from the linear term onward because their successive ratio is \(t/[(\alpha+1)(\alpha+2)]\le(41/20)/6<1\) for \(\alpha\ge1\). Since \(0\le\chi\le1\) on I, J4.4 and J4.5 give \(\Phi\ge c_0>0\) on the full real rectangle after increasing \(\Lambda\). Consequently \(\phi=\phi_*\Phi>0\), and division by \(\phi\) in the original J3.1 is valid. This proves the nonlinear equations themselves, not just their multiplied version.

The analytic scalar profiles give a smooth Cartesian axis field by P0.4–P0.7. Explicitly retain \(F=\phi/\mathsf C\), \(E=\sqrt{2X}F\), and the actual radial average in \(V_0=Xv_0\). Then

Here X is \((x_1^2+x_2^2)/(2q)\) and q is positive and smooth for \(t<1\). Every scalar coefficient is smooth in X, so these products extend smoothly across the axis. P1.3 and P1.7 identify their zero leading stress, including the zero axis integration constants. This completes Proposition B.2 on its stated interval.

J5. Positive angular source and the actual stress coefficients

The construction must control the source even where \(\chi=0\). We use a bound involving \(\sqrt\chi\) and Young’s inequality without dividing by \(\chi\). The original radial primitive then transfers that source bound to the actual stress coefficients.

The source inequalities in this section and J6 use the original real parameter interval \([-1,1]\); the larger I remains available for analytic neighborhoods and derivative bounds.

The identity

and the positivity of \(f_0\) show that \(\mathcal D_X\log f_0(Y\chi)=O(\chi)\) and \(H_*\partial_\eta\log f_0(Y\chi)=O(\chi)\). J4.4, with the positive lower bounds, therefore gives

The original logarithmic derivative contains \(\Lambda\zeta_*\), so

Also \(H_c-H_*=O(\Lambda^{-1})\), and the unweighted derivative \(\Phi_\eta/\Phi\) is uniformly bounded. The complete source \(S_q=-W(1+\mathcal D_X\log\Phi)-h(1-2\eta U)-H_c(\log\phi)_\eta\), J5.2–J5.3 and \(W-W_*=O(\Lambda^{-1})\) imply

Here \(|U|\le4.1\) follows from J4.4 for large \(\Lambda\), and hence \(|h(1-2\eta U)|\le10h\). First increase \(\Lambda\) so \(K\chi\le.025\Lambda L\chi\); then Young's inequality gives \(K_{\sigma_*}\sqrt\chi\le.025\Lambda L\chi+K'_{\sigma_*}/\Lambda\). Since \(-W_*>2.8\), \(10h\le.1\), increase \(\Lambda\) once more so the remaining constant error is at most .2. This proves the original full bound

No division by \(\chi\) has occurred, so the proof includes its zero at \(\eta_0\).

The actual axis-regular integral P1.3 reads

Its smooth extension is obtained by setting \(X'=aX\). A uniform lower bound \(c_0\) and upper bound \(C_\Phi\) for \(\Phi\), and J5.5, give \(Q_s\ge(5/4)c_0/C_\Phi\). Retain the original coefficients

The equalities with a and \(-2U_X\) follow from the actual zero-stress solution and its axis constants, as proved in P1.7. Thus, for \(0<X\le4.1/\Lambda\),

The upper bound follows from J4.4 and the positive \(\Phi\) lower bound. Each fixed parameter derivative of \(\Phi,\log\Phi,\mathfrak u,p_1,n_s\) is bounded uniformly in subsequent \(\mathsf C\). For \(\log\Phi\), the uniform derivative bound and positivity give a smaller common complex neighborhood without zeros, on which its logarithm is chosen from the positive real values. The explicit formulas and Cauchy estimates then give all these bounds. Original radial derivatives retain their powers of \(\Lambda\).

J6. The strict endpoint inequality, in both parameter regions

The endpoint proof covers every parameter value. On \(\chi>.99\), the angular ratio gives the margin. On \(\chi\le.99\), the chosen axis data force \(Z_*\) away from zero and the original amplitude makes the axial ratio large. We prove the two bounds with compatible parameter choices.

At \(X_0=4/\Lambda\), \(Y=4\). First suppose \(\chi>.99\). Put \(z=4\chi\), \(t=z/2\), so \(1.98<t\le2\). The exact series is \(f_0(z)+zf_0'(z)=\sum_{\alpha\ge0}(-t)^\alpha/(\alpha!)^2\). The alternating remainder after its fourth term is negative, with decreasing magnitudes, hence

The polynomial derivative is \(-1+t/2-t^2/12+t^3/144\). On \([1.98,2]\) its first pair is nonpositive and the other terms equal \(-t^2(1/12-t/144)<0\). Thus its maximum is at \(t=99/50\), where its exact value is \(-75535511/400000000<-.188\). Using \(0<f_0\le1\),

The actual ratio in J5.7 differs by \(O(\Lambda^{-1})\), uniformly on this region. Choose the error below .06. Then \(p_1>2.3\) at \(X_0\).

On the complementary region \(\chi\le.99\), the actual choice J0.4 gives \(|Z_*|>\delta_*\). Since \(L\le1\), J5.7 gives \(|n_s|\ge\delta_*/2\) after increasing \(\Lambda\). At the same original radius,

This retains \(E=\sqrt{2X}\phi_*\Phi/\mathsf C\) and every factor of two. For the now-fixed \(\Lambda\), let \(M_\Lambda=(\max_I\phi_*)C_\Phi\), a finite bound independent of subsequent \(\mathsf C\). Enlarge its threshold to include \(\mathsf C_0(\Lambda)\ge\sqrt{5\Lambda C_1}\,M_\Lambda/\delta_*\). J5.8 and J6.3 then give

There is no circular amplitude bound: \(C_\Phi,C_1\) came from estimates uniform for every amplitude already meeting J3.7. Taking the maximum of that threshold and the displayed one preserves all prior bounds. On the first parameter region \(p_1>2.3\); on the second \(p_2^2/p_1>2.3\), with \(p_1>0\). Both prove, for the actual analytic inner profile,

Together with J5, this proves Proposition B.3, including its parameter order and all fixed parameter-derivative assertions.

J7. What the actual axis solution supplies

The constructed profile has the original prescribed pressure at the axis, positive \(\phi\), all analytic scalar profiles, a smooth Cartesian field, exactly zero leading stress, positive \(S_q\), and the strict endpoint inequality. The maps J3.2 and J4.6 retain the original X and physical fields; \(\partial_X=\Lambda\partial_Y\) retains their derivative scales. For original viscosity \(\nu>0\), use the exact map in P1 and H7.1–H7.2: \(u_\nu=\sqrt\nu u(x_\nu/\sqrt\nu,t)\), \(p_\nu=\nu p(x_\nu/\sqrt\nu,t)\), \(T_\nu=\nu T(x_\nu/\sqrt\nu,t)=0\), with \(X_\nu=r_\nu^2/(2\nu q_\nu)\). The full momentum residual retains the axial-viscosity and radial remainder terms in P1.5–P1.6; zero leading stress does not set those terms to zero.

J8. Five worked exercises

Exercise 1. Recover the actual radial inverse kernels

Derive integral formulas for both operators \(\mathcal J_1,\mathcal J_2\), including their axis values and original radial scale.

Solution. For \(YG_{YY}+\nu G_Y=F\), multiply by \(Y^{\nu-1}\). The left side is \((Y^\nu G_Y)'\). Smoothness at zero sets the lower integration constant to zero, and \(G(0)=0\) sets the remaining constant. Hence

For \(Y>0\), the inner integral is \(O(s^\nu)\), so the integrand is bounded at zero. Set \(s=Yb\), \(t=Ya\) and change the order on \(0\le a\le b\le1\). This gives

The weights are integrable and nonnegative. Differentiation under these fixed integrals proves smoothness and \((\mathcal J_\nu F)(0)=0\), \((\mathcal J_\nu F)'(0)=F(0)/\nu\). Integrating \(a^\alpha\) against the two weights gives exactly \(1/(\alpha+1)^2\) and \(1/[(\alpha+1)(\alpha+2)]\), recovering J1.3 without discarding either divisor.

For the original \(X\)-equation with right side \(F(\Lambda X,\eta)\), put

Then \(X\partial_X^2G_X+\nu\partial_XG_X=F(\Lambda X,\eta)\), and \(G_X(0,\eta)=0\). The derivative is \(\partial_XG_X=(\mathcal J_\nu F)_Y\), while the second derivative is \(\Lambda(\mathcal J_\nu F)_{YY}\). These identities prove the complete radial scaling of the inverse.

Exercise 2. Give explicit derivative bounds on a common complex domain

Starting from the coefficient-norm bound in J4, give a concrete family of bounds for all fixed derivatives of the actual difference from the comparison pair. Then retain the original radial factors in the axial profile U.

Solution. Write the two-component difference as \(\Delta=(\Phi-\Phi^0,\mathfrak u-\mathfrak u^0)\). J4.3 gives \(\|\Delta\|_\rho\le M/(2\Lambda)\), using the maximum norm on the pair. J1.2 implies that \(|\Delta|\le4\|\Delta\|_\rho\) on every joint disk with \(|Y|\le10\) and \(|\eta-\eta_c|\le\rho/4\), where \(\eta_c\in I\). The original radius condition is \(10/20+(\rho/4)/\rho=3/4<1\).

For every real \(Y_c\in[0,4.1]\), the disk \(|Y-Y_c|\le5\) lies in \(|Y|<10\). Cauchy’s formula on the two circles therefore yields

The same domains and constants work for the whole real rectangle and every subsequent \(\mathsf C\) admitted by J3.7. This proves a concrete version of the fixed-order estimates; it does not require a single bound independent of r and s.

The original axial comparison is \(U_* -XZ_*/(2L)\), because \(\Lambda^{-1}\mathfrak u^0=-XZ_*/(2L)\). Thus the actual original-coordinate estimates are

Every bound applies on \(0\le X\le4.1/\Lambda\). The extra inverse power in the second line comes from the actual prefactor in U; it is not a change of the original profile.

Exercise 3. Compute the exact first axis derivatives

Evaluate the angular source, \(\phi_X\), \(U_X\), \(\Pi_X\) and the first stress-coefficient limits at \(X=0\). Determine which expressions depend on \(\Lambda\) and \(\mathsf C\).

Solution. At the axis, \(U=U_*\), \(\phi=\phi_*\), \(l=1\), \(W=W_*\), \(H_c=H_*\), and \((\log\phi)_\eta=\Lambda\zeta_*\). Write

This is exactly \(S_q(0,\eta)\), since \(-H_*\Lambda\zeta_*=\Lambda L\chi\). Evaluation of J3.1 at \(X=0\), with no limiting approximation, gives

For the axial identity, \(S_n(0,\eta)=Z_*\): the terms containing \(\mathcal D_XU\) or \(\mathcal D_X\Pi\) vanish, and the remaining source is precisely J0.1. Thus \(\mathfrak u_Y(0,\eta)=-Z_*/(2L)\) exactly, strengthening the approximation at this one radius.

J5.6 gives \(Q_s(0,\eta)=\mathcal S_*/2\), while \(n_s(0,\eta)=Z_*/L\). Consequently

The second equality retains \(E=\sqrt{2X}\phi/\mathsf C\). The data \(U_*,Z_*,L,\chi\) precede \(\Lambda\). The derivative \(U_X(0,\eta)\) is therefore independent of both subsequent large parameters. In contrast \(\mathcal S_*\) and \(\phi_*\) retain their \(\Lambda\)-dependence, and \(\Pi_X\) and the second coefficient limit retain their original amplitude factors.

Exercise 4. Quantify the actual inverse and fixed-point iteration

Give a tail bound for the inverse series even when \(\|T\|\) exceeds one. Also bound the error of the actual nonlinear iterates starting from the comparison pair.

Solution. Set \(a=40M_\chi\) and let \(S_N=\sum_{k=0}^N(-T)^k\). From J4.1,

If \(a/[(N+2)(N+3)]\le1/2\), each ratio of successive terms in this tail is at most \(1/2\). Hence

For every fixed finite \(a\), such an N exists. This proves the inverse approximation directly from the degree gain, without an assumption \(\|T\|<1\).

Let \(\mathcal F\) denote the full nonlinear map J4.3, \(z_0=(\Phi^0,\mathfrak u^0)\), and \(z_{n+1}=\mathcal F(z_n)\). Its invariant ball has Lipschitz constant at most \(1/2\), and \(\|z_1-z_0\|\le M/(2\Lambda)\). Induction gives \(\|z_{n+1}-z_n\|\le2^{-n}M/(2\Lambda)\). Summing all remaining increments proves

The limit z is the actual fixed point because the map is continuous on the complete ball. Both estimates hold in the full coefficient norm, including all weighted parameter derivatives. They are analytic bounds on a construction, not numerical evidence for existence.

Exercise 5. Does zero leading stress solve the full momentum equation?

Compute the axial component of the full residual at the spatial origin for the actual leading inner field. Retain the original viscosity \(\nu>0\) and identify its exact order relative to the leading equations.

Solution. First use viscosity one. On the axis \(r=0\), the actual axial velocity is

At \(z=0\), implicit differentiation or P0.2 gives

The last equality also follows from the fact that q is even in z and \(\eta\) is odd. The source data \(4\eta+j_0\) have zero second derivative in \(\eta\). Applying the product and chain rules, every term but the one containing \(q_{zz}j_0\) vanishes. Thus

The exact physical residual identity P1.5 is \(\mathcal R_1(u,p)\cdot e_z=-(\partial_r+1/r)T_z-\partial_z^2u_z\). The constructed stress is identically zero on this inner region. All fields are smooth in Cartesian coordinates, so the identity on \(r>0\) extends to the axis. Therefore

The viscosity map is \(u_\nu(x_\nu,t)=\sqrt\nu\,u(x_\nu/\sqrt\nu,t)\), \(p_\nu=\nu p(x_\nu/\sqrt\nu,t)\). Each term in the full momentum residual then has the factor \(\sqrt\nu\). At the spatial origin the exact original-viscosity identity is

Here \(2D=1-2h\). The final equality proves the precise \(q^{2h}\) relative factor at this point; the absolute residual is still nonzero. This is the leading inner field. The subsequent base coefficients and corrections address its full residual before the theorem’s final smooth forcing is obtained.

The continuation receives an actual solution

The inner profile now exists with the prescribed outer pressure datum. The complete proof gives its analytic domain, Cartesian regularity, positive angular source, zero leading stress and strict endpoint inequality, with all parameter choices in their stated order. The exercises compute exact axis derivatives, quantify the analytic estimates and identify the remaining physical residual.

The next teaching component is the continuation and exact joining Q0–Q9, including stress activation, the analytic collar, the amplitude-independent transition and all five restored moments. The shear realization V0–V7 then completes the leading profile, and the base construction Z0–Z8 treats the higher-order residual. Their lesson treatments are joining, shear and the actual base field.

The final theorem lesson, EC0–EC5, uses these constructions to prove Theorem 1.1 and Corollary 10.6.

Authored lesson text and diagram are CC0. Mathematical source: OpenAI, Finite Time Blowup for Navier–Stokes, Section 4 and Appendix B. Exposition: GPT-6 Astra (OpenAI). Original sources retain their attribution and rights.

Joining the analytic axis to the outer field

NS-BLOWUP N2.4. The inner and outer solutions now exist separately. This lesson constructs their actual smooth connection. It activates the stress at the inner edge, preserves its required direction on a first collar, continues through a region of small shear, and restores all five original radial moments before entering the outer field.

The analytic-axis lesson, J0–J8, supplies the actual inner solution with its prescribed pressure and strict endpoint inequality. The outer-profile lesson, O0–H7, supplies the complete outer field and heat exterior. The stress and moment lesson, P0–P7, proves the exact physical operators, exterior-preservation map, bump-matrix inverse and nonlinear correction estimates. All three lessons are included in this cumulative reader.

The matching problem concerns the original fields and their entire radial history. Agreeing on velocity values at one radius is insufficient: the pressure and integrated stress also depend on the five cumulative moments. We compute their exact differences, retain the original scale factors, and solve the five nonlinear equations.

The source is OpenAI, Finite Time Blowup for Navier–Stokes, original paper, pages 150–157, equations (B.22)–(B.40), Lemma B.4 through Corollary B.10. This lesson is an independent derivation and teaching text; it does not replace or amend the diplomatic transcription. Source identity and mathematical locators are recorded in SOURCES.json.

Figure. The radial positions show order rather than proportional distance. Every displayed radius, factor and endpoint is the one used below. The two bump families have disjoint supports; Exercise 4 proves the resulting exact separation of the axial equations. The lower panel gives the actual inner pressure difference from Exercise 2. The stress is zero to infinite order at its inner edge while having a nonzero limiting direction.

Why the transition length comes first

The proof has three linked tasks. First, a reference continuation gives uniform parameter bounds for an entire family of narrow transitions. These bounds let us reduce the shear without introducing an inverse power of the small shear factor into the cone estimate. The stress starts with the exact flat factor proved in Q3.

Second, the endpoint bound for \(\log(\mathsf C E)\) fixes a finite transition length before the final amplitude \(\mathsf C\). Increasing that amplitude then moves the matching scale \(X_R\) outward while keeping the earlier transition endpoint fixed. Q7 proves exactly how the five original moments transform; it does not assume their unscaled discrepancies all tend to zero.

Third, the fixed outer data determine the correction tolerance before \(X_R\) is selected. Five fixed bumps solve the actual moment equations inside that tolerance. The common pressure datum and exact moment identities then preserve the entire outer field, including its heat exterior.

Q0. Objects, norms and the reference continuation

The reference field is constructed by integrating prescribed slopes. Its cutoff is radial only, so parameter differentiation does not create inverse cutoff widths. We keep the natural axis solution and its pressure integral identifiable throughout.

Retain \(A=1/2+h\), \(D=1/2-h\), \(d=1-\eta^2\), \(L=1-2h\eta^2\), \(0<h<1/100\), and the parameter interval \(-1\le\eta\le1\). Thus \(1-2h\le L\le1\). The prescribed datum is the actual analytic \(\Pi_0\) of O4. Write the J solution as

Here \(U_*=4\eta+j_0\), \(H_*=D\eta+dU_*\), and \(W_*,Z_*,\chi\) are exactly J0.1–J0.4. The original unaccented \(\zeta_*\) and accented \(\widetilde\zeta_0\) follow (B.3). For each fixed integer \(k\), \(\|\cdot\|_k\) is the maximum of parameter derivatives through order \(k\), uniformly on the indicated radial interval. A bound uniform in \(\mathsf C\) is asserted only for \(\mathsf C\ge\mathsf C_0(\Lambda)\). Constants allowed to depend on the already fixed \(\Lambda\) will be indicated by \(K_k(\Lambda)\); they do not depend on subsequent cutoff widths.

Use exactly

The superscript distinguishes this local radius 100, called \(X_b\) in Appendix B, from the exterior stress endpoint of Appendix A. No radius has been changed. Increase \(\Lambda\) if necessary so \(X_0<100\). Fix a positive \(\bar t\) satisfying Q0.2. With the original smooth step

choose \(0<t_1\le\bar t\). Let \((U_{\rm r},\phi_{\rm r})\) equal the analytic solution for \(y\le t_1\); on \(t_1<y<2t_1\) define

Integrate from \(y=t_1\), then keep both functions constant for \(y\ge2t_1\). Positivity of \(\phi_{\rm r}\) follows by exponentiation. Flatness of the step at its endpoints makes the fields smooth at both joins. Set \(E_{\rm r}=\sqrt{2X}\phi_{\rm r}/\mathsf C\). Define \(\Pi_{\rm r},Q_{s,\rm r},N_{s,\rm r}\) by the original forward pressure and stress integrals P1.3 and P2.1, with the same \(\Pi_0\).

The J estimates give, for every fixed \(k\),

For the first two bounds integrate Q0.4 over length at most \(\bar t\): the integrands and all parameter derivatives are bounded respectively by \(K_k/\Lambda\) and \(K_k\), since \(4\le\Lambda X\le4.1\) there. Before that interval J4.4 applies, and after it the fields are constant. The radial average is exactly \(\mathcal A_Xf=\int_0^1f(aX,\eta)\,da\), a contraction in every such norm, so the first bound holds also for \(\mathcal A_XU_{\rm r}-U_*\). Finally,

The factor \(\sqrt{2X}\) in E is essential to this identity. Product differentiation and Q0.5 bound this integral through every fixed parameter order, including at zero. Exponentiation and reciprocal differentiation show that \(\mathsf C E_{\rm r}\) and \((\mathsf C E_{\rm r})^{-1}\) have bounded \(k\)-norms on \([X_0,X_i]\), with positive lower bounds there, uniformly in \(\mathsf C,t_1\). All these statements use parameter derivatives; high radial cutoff derivatives need not be bounded as \(t_1\) decreases.

Q1. The reference source and strict shear comparison

The reference continuation must retain a strict margin over the threshold two. The angular estimate covers one parameter region; the nonvanishing axial source covers the other. Both comparisons apply to the whole family before a final transition width is selected.

The weighted gradient bounds used in J5 remain valid through the cutoff. To check this directly, put \(q(Y,\eta)=\log\Phi(Y,\eta)\). Differentiating the center \(\log f_0(Y\chi)\) and using \(|H_*\chi'|\le2\|H_*'\|_\infty\chi\), together with J4.4 and positivity, gives

All denominators are bounded below by J4.5. In Q0.4 the cutoff has no parameter dependence. Integrating its parameter derivative, and using the last term of Q1.1 on the interval of length at most \(\bar t\), proves

The unweighted derivative of \(\log\phi_{\rm r}-\log\phi_*\) is also bounded by Q0.5. Since \(H_{c,\rm r}-H_*=d(U_{\rm r}-U_*)\), Q0.5 and \(|\zeta_*|\le L\sigma_*^{-1}\sqrt\chi\) give

Furthermore \(W_{\rm r}-W_*=O(\Lambda^{-1})\), by the actual radial-average formula, and \(|U_{\rm r}|\le4.1\) for large \(\Lambda\). In the exact source

Q1.2 therefore bounds the change caused by \(l_{\rm r}-1\) by \(K\chi+K/\Lambda\). Hence

For the last inequality choose \(\Lambda\) so \(K\chi\le.03\Lambda L\chi\), and use \(K_{\sigma_*}\sqrt\chi\le.03\Lambda L\chi+ K_{\sigma_*}^2/(.12\Lambda(1-2h))\). Then enlarge \(\Lambda\) so the remaining \(1/\Lambda\) terms are smaller than .3. The original bounds \(-W_*>2.8\) and \(10h\le.1\) supply the constant 2.4, with strict spare room by making the last error smaller. No division by \(\chi\) occurs. This estimate is uniform over a fixed small reference-width family.

On the natural interval, J5.7–J5.8 give \(\mathcal D_X\log\phi_{\rm nat}=-p_1/2<0\). Multiplication by the nonnegative cutoff and then setting the slope to zero proves \(l_{\rm r}\le1\).

For the axial source, retain every term of P1.1. Q0.5, \(\mathcal D_XU_{\rm r}=O(\Lambda^{-1})\), and Q0.6 yield

The two pressure derivatives are included: \(\Pi_{\rm r}-\Pi_0\) is Q0.6 and \(\mathcal D_X\Pi_{\rm r}=E_{\rm r}^2/2\). Choose \(\Lambda\), then \(\mathsf C\), so the right side for \(k=0\), after division by L, is below \(\delta_*/4\). On \(\chi\le.99\), \(|Z_*|>\delta_*\) by J0.4. The equation and its exact solution are

The analytic initial value differs from \(Z_*/L\) by \(O(\Lambda^{-1})\). Both terms in this convex average have the sign of \(Z_*\) and absolute value at least \(\delta_*/2\), after the same choice. Therefore \(|n_{s,\rm r}|\ge\delta_*/2\) on that parameter region throughout the continuation.

The positive stress integral gives \(p_{1,\rm r}>0\), and its equation is

Using \(l_{\rm r}\le1\), Q1.4 and \(X=X_0e^y\) in the integrating-factor inequality gives both exact lower comparisons

The coefficient 1.88 is \(.94\Lambda X_0/2\); the coefficient 1.2 uses \(L\le1\) and \(2.4/2\). On \(\chi>.99\), J6 gives \(p_{1,\rm r}(0)>2.3\). Replacing it by 2.3 in the first comparison gives \(1.88\chi e^y+(2.3-1.88\chi)e^{-y}\), whose derivative is positive for \(y\ge0\), because \(1.88\chi>2.3/2\). Thus \(p_{1,\rm r}>2.3\) there. The second comparison is greater than 3 for \(100\le X\le110\), since \(X_0\le4\) and \(1.2(100-16/100)>3\).

The reference pressure and field bounds, or their exact moment formulas, also give a finite upper bound \(M_0(\Lambda)\) for \(p_{1,\rm r}\), independently of subsequent \(\mathsf C,t_1\). Let \(E_{\max}(\Lambda)\) bound \(\mathsf C E_{\rm r}\). On \(\chi\le.99\),

Increase \(\mathsf C\) so this is at least \(5/2\). On the other region \(p_{1,\rm r}>2.3\). Consequently

Each fixed parameter derivative of \(p_{1,\rm r},n_{s,\rm r}\) is uniformly bounded: the moment formulas P2.5 use Q0.5–Q0.6, one additional parameter derivative, positive denominators and integrals on a fixed interval. For \(Q_{s,\rm r}\), its common amplitude factor cancels exactly between numerator and denominator. For \(N_{s,\rm r}\), Q1.5–Q1.6 apply. The lower comparison in Q1.8 is uniform and positive. Once a finite \(\mathsf C\) is fixed these bounds therefore give a finite \(V_{\max}\) bounding \(v_{\rm r}\) for every sufficiently small \(t_1\), before any one width is selected. This proves the full reference statement of Lemma B.4.

Q2. Activation and the stress cone without an inverse cutoff loss

The same positive factor multiplies both shear components. Cancel it in their ratio before estimating. This exact cancellation is why making the shear small does not make the cone errors large.

Fix a sufficiently large finite \(\mathsf C\) as above. All constants in this section may depend on it and on \(\Lambda\), but are uniform as \(t_1\) and \(0<\kappa_0<1/2\) decrease. On \(0<y<t_1\), prescribe

Integrate from the actual axis values at \(y=0\); leave the analytic solution unchanged for \(X\le X_0\). Define \(E=\sqrt{2X}\phi/\mathsf C\), and compute pressure and stresses by forward integration with the same \(\Pi_0\).

On this entire first interval the reference is the natural solution. Its zero leading stress gives the exact differences

The step is increasing, so \(\int_0^ye_a(v)\,dv\le ye_a(y)\). Parameter differentiation commutes with these integrals and never differentiates the cutoff. Thus the field differences and \(\log\phi-\log\phi_{\rm r}\) are \(O_k(ye_a)\). Exponentiating gives \(E_{\rm r}/E=1+O_k(ye_a)\). On this compact annulus the moment integrands are smooth functions of the fields; their differences vanish before \(X_0\). Their integrals obey the same bound, also with one additional parameter derivative. P3.3–P3.6 consequently prove

These are estimates on values and parameter derivatives, not bounds on radial cutoff derivatives.

Since \(p_{2,\rm r}=Xn_{s,\rm r}/E_{\rm r}\), the actual shear is exactly \(\mathfrak s=\kappa(p_{1,\rm r},p_{2,\rm r}E_{\rm r}/E)\). Divide its two components before estimating: because \(\kappa>0\),

The bounded reciprocal of \(p_{1,\rm r}\) follows from Q1.8. Every error constant is independent of \(\kappa_0^{-1}\): the common \(\kappa\) cancels exactly in \(t_s\), while \(0<\kappa\le1\) multiplies the remaining error in \(v_s\).

Choose one constant \(K\) bounding the four errors, and use \(v_{\rm r}>2+.2\), \(v_{\rm r}\le V_{\max}\). Taking \(t_1\) small enough yields positive constants \(c_1,c_2\), independent of its final smaller value, with

For example require the error in the first difference to be at most \(e_a\), the error in \(P_c\) to be at most .1, and the final displayed upper bound to be less than 2. These are finitely many width bounds on already fixed constants. P6.1–P6.2 prove the admissible cone where \(v_s>2\); where \(v_s\le2\), the fact that \(\mathcal U(P_c,J_c)>2\) proves the strict relaxed condition. Continuity from \(v_s(0,\eta)=v_{\rm r}(0,\eta)>2+.2\), uniformly on the compact parameter interval, supplies a positive first collar with \(v_s>2+.1\).

Q3. Flat stress factor and limiting direction at the inner edge

A stress that vanishes at an endpoint can still have a definite smooth direction there. We prove this by dividing by its actual flat scalar factor, computing the resulting smooth vector, and proving that vector is nonzero.

For \(y\ge0\) near zero write, exactly,

The exponential extends smoothly by zero at zero, so \(g_a\) is smooth and positive on a closed small collar. For a nonnegative integer \(m\) and a smooth \(b\), the substitution in H6.1–H6.2, now with \(c=t_1^2\), gives the full formula

Every fixed mixed derivative of the integral is dominated by \(e^{-t_1^2w}\) times a polynomial in \(w\); the chosen \(t_1>0\) is fixed here. Hence the right side is \(y^{m+3}\) times a smooth function through zero. This retains the integration factor \(1/2\), rather than only recording a vanishing order.

Apply \(m=0\) to Q2.2. Field and logarithmic differences belong to \(e_ay^3C^\infty\). Products and smooth compositions preserve this class: write, for instance, \(\exp(e_ay^3b)-1=e_ay^3b\int_0^1\exp(se_ay^3b)\,ds\). Reciprocals are handled by \((f+g)^{-1}-f^{-1}=-g/[f(f+g)]\), where the fields have nonzero denominators. Each moment and pressure difference is one further forward integral of such a difference, with smooth radial weights \(dX=X_0e^y\,dy\) and \(1/X\) on this collar. Applying Q3.2 with \(m=3\) gives \(e_ay^6C^\infty\).

Substituting these exact classes in P3.4 gives \(p_s-p_{s,\rm r}\in e_ay^3C^\infty\). The shear formula in Q2.4 then proves

Here \(F=\phi/\mathsf C\) is positive, \(p_{1,\rm r}>0\), and no division by a vanishing field was used. At the endpoint \(t_s=p_{2,\rm r}/p_{1,\rm r}\), so exactly

It follows that \(2[B_0\cdot(1,t_s)]^2-(v_s-2)[B_0\cdot(-t_s,1)]^2\) has a positive minimum at the endpoint. Smoothness and compactness preserve that minimum, and \(|B_0|>0\), on a smaller collar. Thus \(B_0/|B_0|\) is the smooth limiting unit stress direction.

Differentiating Q3.1 any fixed number of times introduces only finitely many inverse powers of \(y\). Since \(X_0>0\), \(\partial_X=X^{-1}\partial_y\) has smooth bounded coefficients here. For every fixed radial/parameter multi-index \(I\),

The upper bounds tend to zero to every order. The stress therefore extends smoothly by zero to \(X\le X_0\), while its direction extends by Q3.4. This proves the full inner endpoint assertions (B.30)–(B.31).

Q4. Continuation with small shear to the radius 110

The explicit integrated prescriptions continue the field. The small parameters are chosen in order from bounds already available for the reference family. A first-crossing argument then preserves the angular margin on the final constant-slope interval.

After \(y=t_1\), keep Q2.1 with \(\kappa=\kappa_0\) up to \(X=100\). These are explicit integrals of the already constructed reference coefficients, so they define smooth \(U,\log\phi\) and positive E. On the reference cutoff, the reference logarithmic slopes are bounded independently of its width; their integrals over length \(t_1\) are \(O_k(t_1)\). After that cutoff both reference fields are constant. Integrating the actual slopes over the fixed remaining interval gives \(O_k(\kappa_0)\). Together with Q2.2 this proves

For the last term use the exact five moment differences and P3.4 with one extra parameter derivative. Positivity and a lower bound for E follow first from the bound on the logarithm. No radial derivative of a narrow cutoff is used in this comparison.

The same common-\(\kappa\) cancellation as in Q2.4 gives

Choose \(\kappa_0\) so \(\kappa_0V_{\max}<1/4\), \(\kappa_0\sup p_{1,\rm r}<.8\), and all the errors contributed by \(\kappa_0\) in Q4.2 are below \(\min(.05,1/8)\). Then choose \(t_1>0\) so its errors satisfy the same bounds as well as Q2.5. This gives \(P_c>2+.1\) and \(v_s<1\) after the first activation interval. Since \(\mathcal U(P_c,J_c)>2\), the strict relaxed condition holds irrespective of the size of \(J_c\). The entire reference family, rather than a previously selected member, supplied the constants, so there is no circular dependence on the choice of \(t_1\).

At \(X=100\) multiply the axial prescription by a smooth \(\beta\) decreasing from 1 to 0 on a short logarithmic interval; keep \(a=\kappa_0p_{1,\rm r}\) there. The reference coefficients have been defined up to 110. For final transition lengths with sum \(\omega_{\rm fin}\), integration gives the same comparisons as Q4.1, with an additional \(K_k\omega_{\rm fin}\). On the axial cutoff,

Since \(p_{1,\rm r}>3\), \(0\le\beta\le1\), and \(\kappa_0V_{\max}<1/4\), decreasing the chosen parameters makes \(P_c>2\) and \(v_s<1\) on this cutoff. At its end \(\mathcal D_XU=0\). On the next short interval interpolate \(a\) from \(\kappa_0p_{1,\rm r}\) to .8 by \(a=(1-\theta)\kappa_0p_{1,\rm r}+.8\theta\), where \(\theta\) is a flat-ended smooth step. Keep \(\mathcal D_XU=0\). Then \(0<a\le.8\), \(b_s=0\), \(v_s=a<1\). P3.4 and the short-interval comparison with the value at 100 give \(P_c=p_1>2\) throughout. Both transitions fit strictly before 110 by requiring \(\omega_{\rm fin}<\log(110/100)\).

After the second transition keep \(a=.8\), \(l=.6\), and \(U\) constant in X. The parameter derivative of \(\log\phi\) is then constant in X. At its start it equals

To prove the stated independence from \(\mathsf C\), differentiate the integrated slopes in Q2.1 and the two final cutoffs. The only varying parameter coefficients are derivatives of \(p_{1,\rm r}\), whose bounds in Q1 are uniform in \(\mathsf C\). The constant .8 has zero parameter derivative. For U the corresponding coefficients are \(n_{s,\rm r}\), also uniformly bounded. Thus \(U-U_*=O_k(\Lambda^{-1})+O_k(\Lambda)(t_1+\kappa_0+\omega_{\rm fin})\) throughout the construction, including its radial average.

As in Q1.3, after increasing \(\Lambda\) as an axis parameter and then decreasing the final widths, the gradient term is bounded below by

The fixed-\(\Lambda\) coefficient of the last error includes multiplication by \(\Lambda\zeta_*\); it is finite and is made small by the later choices. One can simultaneously ensure \(-W\ge2.75\) and \(|U|\le4.1\). The original source then satisfies

Starting from \(p_1>2\), a first downward crossing of 2 on this interval would have \(\partial_y p_1=XS_q/L-.6p_1>X-1.2>0\). At a first crossing from above its left derivative must be nonpositive, a contradiction. Consequently \(p_1>2\) through \(X_i=110\). This completes the actual continuation of Proposition B.5, including positivity, the inner admissible collar, the later strict relaxed cone and the specified final slopes.

Q5. One analytic collar and amplitude-independent endpoint bounds

The inner parameter-analytic rectangle is protected by the direction of the radial integrals. Separately, bounds for the endpoint values use integrated cutoff values, so the transition length can be selected before the amplitude.

Choose \(0<t_c<t_1\) inside the admissible collar of Q2–Q3, with \(4e^{t_c}<4.1\). The J solution has a joint complex neighborhood of its compact radial/parameter rectangle; shrink the parameter neighborhood once so that \(\Phi\) has no zeros and L has no zeros. In the first collar the reference is the natural solution. Its \(p_{1,\rm r},n_{s,\rm r}\) and all their radial derivatives are holomorphic in that same smaller parameter neighborhood, using either J5.7 or the original forward integrals. Q2.1 then integrates holomorphic coefficients multiplied by radial-only cutoffs. Its exponential for \(\phi\) is nonzero. Differentiating these formulas radially changes cutoff derivatives but neither introduces a new parameter singularity nor changes the parameter domain.

The exact expressions

are therefore smooth in \((X,\eta)\) through \(X=0\) and holomorphic in \(\eta\) on a common neighborhood for \(0\le X\le X_0e^{t_c}\). Each fixed radial derivative has finite bounds on a smaller common neighborhood. Subsequent modifications will lie strictly to its right. Since all the integrals in Q5.1 are forward integrals, later modifications leave this rectangle unchanged. The size of \(t_c\) and the radial derivative bounds need not be uniform as the final amplitude grows or the transition widths decrease. This proves precisely Corollary B.6.

For the endpoint quantities

let \(M_k,N_k\) be the uniform bounds for \(p_{1,\rm r},n_{s,\rm r}\) from Q1. Before the final interpolation, \(a=\kappa p_{1,\rm r}\); during it a is a convex combination of \(\kappa_0p_{1,\rm r}\) and .8; afterwards it is .8. Thus \(\|a\|_k\le M_k+.8\). Integrating \(\partial_y\log\phi=-a/2\) gives the explicit finite bound

The supremum is finite by J4.4 and positivity. It depends on \(\Lambda\) and the prior axis data, but not on subsequent \(\mathsf C\) or smaller widths.

For U, the first activation contributes at most \(X_0e^{\bar t}N_kt_1/2\) to its endpoint \(k\)-norm. On the remainder, including its final cutoff, the integrand has magnitude at most \(\kappa_0X|n_{s,\rm r}|/2\); using \(dX=X\,dy\) bounds its integral by \(\kappa_0X_iN_k/2\). It is zero after the axial cutoff. Therefore

The natural constant is independent of large \(\Lambda,\mathsf C\) by J4.4, after the axis shape data have been fixed. The joining constant is independent of subsequent \(\mathsf C\) and widths. No inverse cutoff width occurs because only integrated cutoff values, not radial derivatives, were used. This proves Lemma B.7 with its full stated uniformity.

Q6. Transition length chosen before the amplitude

The interpolation changes the logarithm of the angular field. Its two endpoint parameter gradients have a common lower bound, and its radial slope remains in a fixed interval. The original source equation therefore keeps the required sign throughout the transition.

Keep \(f(\eta)=(1+\eta^2)^{-1}\). Choose a positive finite \(T_{\rm sh}\) satisfying

The right side is already fixed before the final amplitude. For \(y_i=\log(X/X_i)\ge0\), prescribe

At \(X_i\) this matches the existing constant slopes to every radial order; the step is flat at zero. E is positive by construction. Its exact slope is

Q5.3 supplies the stated interval without any growth of \(T_{\rm sh}\) with \(\mathsf C\).

The angular source remains positive on this potentially long interval. Q4.4 and J5.2 imply

where \(o(1)\) is bounded by a fixed-\(\Lambda\) constant times \(t_1+\kappa_0+\omega_{\rm fin}\). The full radial history of U stays close to \(U_*\), so this also bounds \(W-W_*\) through its actual radial average. The other endpoint gradient has the exact expression

Indeed \(\eta H_*=(D+4d)\eta^2+dj_0\eta \ge D\eta^2-j_0|\eta|\ge-j_0^2/(4D)\), and \(2|\eta|d/(1+\eta^2)\le1\). This verifies the bound at both parameter endpoints as well as near the zero of \(H_*\).

Young's inequality as in Q1 makes the right side of Q6.4 at least \(-.1\) for large \(\Lambda\) and the later small widths. Q6.5 is at least \(-.1\) for \(j_0\le.05\), large \(\Lambda\) and the same sufficiently small widths. Since the cutoff in Q6.2 is independent of \(\eta\), its parameter gradient is the convex combination of these two gradients. Also \(-W\ge2.75\) and \(|U|\le4.1\) persist. Hence the exact source satisfies

At a putative first downward crossing \(p_1=2\), the original equation gives \(\partial_{y_i}p_1=XS_q/L-lp_1>X-1.3>0\), impossible for \(X\ge110\). Thus \(p_1>2\), \(v_s=a\le.9<2\), and \(P_c=p_1>2\): the strict relaxed cone persists.

At \(X_{\rm sep}=X_ie^{T_{\rm sh}}\) the field becomes exactly

The equality retains the amplitude and every power. The step is flat at its upper endpoint, so the match is smooth. Before axial restoration the same source bound and crossing argument continue to apply to this constant-slope interval. The endpoint \(X_{\rm sep}\) is fixed before \(\mathsf C\), whereas \(X_R\) grows with it; this is the scale separation required for the moment estimates below.

Q7. All five moment errors with their original scale factors

The original moments have different powers of the radial scale. We display the invertible comparison map and carry each factor through the stress formulas. The short inner segment then has explicit small contributions in the units needed for the correction solve.

Introduce \(x=X/X_R\) only with the exact comparison

The other hatted moments are the integrals in \(dv\) of \(U,\widehat H,U\widehat H,U^2-E^2/2\), evaluated at \(X_Rv\). Changing variables in the original moments proves every factor in Q7.1. Substitution in P2.5 gives the exact original functions at \(X_Rx\):

These formulas remove \(X_R\) from the coefficient map, but not from the physical moments or from \(p_s=(X_Rx/L)(Q_s,N_s/E)\). Those factors remain in use.

The short inner portion ends at

Q0, Q5 and Q6 bound U and \(\mathsf C E\) on \(0\le X\le X_{\rm sep}\) in each fixed parameter norm, independently of later large \(\mathsf C\) and the allowed smaller widths. Near zero \(\mathsf C E=\sqrt{2X}\phi_{\rm nat}\). Therefore at \(x=x_{\rm sep}\),

For the first bounds use the bounded U and \(E=O(\mathsf C^{-1})\) in the exact integrands. For I and J integrate the retained factor \(\sqrt{2x}\) against this E bound. For pressure, change back to X: on \([0,X_0]\) its integrand is \(\mathsf C^{-2}\phi_{\rm nat}^2\); on \([X_0,X_{\rm sep}]\) it is bounded by \(K_k(\Lambda)\mathsf C^{-2}/X\), whose integral is finite because \(X_0>0\) and \(X_{\rm sep}\) is fixed. Product differentiation supplies the same estimates through every chosen finite order. No singular \(1/x\) majorant was integrated to zero.

The ideal inner part of the actual outer family has \(U_0=4\eta\), \(E_0=P_*fx^{1/10}\). Its exact moments are

In particular, at the endpoint, \(\widehat C_{p,0}(x_{\rm sep})=(5/2)f^2e^{T_{\rm sh}/5}\mathsf C^{-2}\). Every term of Q7.5, with every fixed parameter derivative, tends to zero there as \(\mathsf C\) grows. Q7.4–Q7.5 consequently make the actual differences of all five hatted moments as small as any preassigned positive tolerance, uniformly over the later width choices. Through Q7.1 this is a proved statement about the original moments, not their replacement by different objects.

After \(x_{\rm sep}\), E is exactly \(E_0\). Put \(g(\eta)=G_i-4\eta\). Choose \(\mathsf C\) so \(x_{\rm sep}<e^{-8}\), and on \(-8<\log x<-7\) restore the axial field by

Take \(\rho=1\) before that interval and \(\rho=0\) after it. The error integrands after \(x_{\rm sep}\), before any angular bumps, are exactly

Their weights \(1,x^{3/5}\) are integrable at zero. Integrating to \(e^{-5}\), and differentiating in \(\eta\), bounds the five moment discrepancies by

provided \(\|g\|_{k+1}\le1\); the extra derivative is retained for Q7.2. Here \(K_k^{\rm out}\) depends only on the fixed outer data and the fixed restoration cutoff. The error \(o_{\mathsf C}(1)\) is the sum of the explicit bounds in Q7.4–Q7.5, with fixed axis data. On \([e^{-8},e^{-5}]\), the same bound controls the fields, their logarithmic radial derivatives and the input moment histories. This follows by differentiating the fixed \(\rho\) in Q7.6; its derivative norms do not involve the small activation widths.

Q8. A fixed cone tolerance and the exact five-bump solution

The correction is a finite nonlinear construction on a fixed radial patch. Its tolerance is selected using the outer field. The cone calculation retains the factor \(X_R\), while the exact row map proves that the solved equations cancel the original moments.

Let \(\mathcal R=[e^{-8},e^{-5}]\times[-1,1]\). The ideal functions \(E_0>0\) and \(Q_{s,0}>1\) have positive minima there by O5.2; \(N_{s,0}\) is bounded by Q7.2 with Q7.5 and the fixed \(\Pi_0\). Define the actual finite constants

The reference shear is \(a=.8,b_s=0\). Apply the explicit difference and reciprocal formulas P3.3–P3.5 to Q7.2 on the fixed rectangle. They give finite Lipschitz constants for \(Q_s,N_s\) in a prescribed \(C_\eta^k\) norm, using one additional parameter derivative of the field and moment inputs. The same product and reciprocal formulas applied to \(a=1-2xE_x/E\), \(b_s=2xU_x/E\) control the shears from the fields and their first logarithmic radial derivatives. A sufficiently small positive input tolerance, depending only on these fixed outer data, therefore ensures

The last bound retains the original decimal .1 in (B.36), page 156; this factor gives the quantitative cone margin in the next calculation.

The original cone coordinates then obey

Thus \(X_R>14e^8/(3Q_{\min})\) makes \(P_c>2\) throughout \(\mathcal R\), since \(L\le1\). This proves the strict relaxed cone without making the tolerance shrink with \(X_R\).

For an actual moment correction, set \(a_0=e^{-6}\), \(b_0=e^{-5}\), \(c_j=a_0+j(b_0-a_0)/6\) for \(j=1,\ldots,5\), and \(\omega=(b_0-a_0)/20\). Use the explicit fixed positive bumps

Their supports are ordered, disjoint and strictly inside \((e^{-6},e^{-5})\). Put \(u=c_1^{\,\rm amp}(\eta)\beta_1+c_2^{\,\rm amp}(\eta)\beta_2\) and \(e=\sum_{j=3}^5c_j^{\,\rm amp}(\eta)\beta_j\). Here the superscript distinguishes the five coefficient functions from the fixed centers. The unperturbed fields on this patch already equal \(U_0=4\eta,E_0=P_*fx^{1/10}\). Replace them by \(U_0+u,E_0+e\).

The increments of the original five moments, over this whole patch, are exactly

Let \(\Delta\mathbf m\) denote the original moment discrepancy accumulated before these bumps. Apply the exact invertible row map

Its inverse is P5.3 with \(\rho=X_R,u_c=4\eta,e_*=P_*,f_*=f\). Thus solving for target \(d_{\rm tar}=-\mathscr L_\eta(\Delta\mathbf m)\) is exactly equivalent to cancelling all original moments.

The transformed linear coefficient matrix has the two actual blocks

P4 proves invertibility for these ordered nonnegative bumps and distinct powers. The factors \(P_*f\) in Q8.6 are positive and their parameter derivatives and inverses are bounded by the outer data. The exact nonlinear part, with no dropped term, is

For \(m=k+1\), let \(\nu_m\) be the operator norm of \(B^{-1}\) and \(q_m\) the bilinear norm of this quadratic map in \(C_\eta^m\), including the finite product-rule factors. Both are finite and independent of \(X_R,\mathsf C\). For

the estimates P5.6–P5.7 show that the iteration stays in the radius \(2\nu_m\|d_{\rm tar}\|_m\) ball and contracts by at most \(1/2\). It therefore converges to an actual solution of all five equations, with that norm bound. P5.8 supplies an invertible derivative in c and proves that this solution is smooth in \(\eta\), with finite derivatives of every fixed higher order. Only the selected finite norm must be small.

The same coefficient bound controls u, e, their first logarithmic radial derivatives and their partial moment integrals throughout the patch: each is a fixed finite sum or integral of the fixed bumps, with finite coefficient norms. Let \(K_{\rm patch}\) be the maximum of these finitely many operator bounds and their product-rule constants on the chosen small coefficient ball. Shrink the fixed discrepancy tolerance so that \(2K_{\rm patch}\nu_m\|d_{\rm tar}\|_m\) is below half the input tolerance of Q8.2, and so that Q8.9 holds. Require the preexisting error in Q7.8 to be below the other half. Both requirements involve only fixed outer-data constants. They guarantee Q8.2 throughout the restoration and every partial moment correction, not merely at its final endpoint.

At the right endpoint the five moment differences vanish identically as functions of \(\eta\). The bumps have compact interior supports, so the fields already agree with the ideal fields on a neighborhood of that endpoint. Continue with the actual outer family of O and H. The same \(\Pi_0\), the vanishing five moments, and P3.1–P3.4 prove exact equality thereafter of

with their actual outer-profile values, including all their derivatives. In particular the heat exterior, exact zero outer stress, smooth terminal direction and all total-moment conditions survive. This completes Proposition B.8 by an actual nonlinear moment solve.

Q9. Compatible choices and the precise completed component

The earlier estimates now give one compatible sequence of finite choices. Smallness is required in finitely many parameter norms needed by the construction. Smoothness supplies every other fixed derivative after those choices have been made.

Fix in advance the finite parameter derivative orders needed for the inequalities and finite moment solve, including the additional derivative in Q7.2. Keep the outer shape choices \(M_d,T_d,P_*,\lambda,h\) of O and H. Their constructed family has the same analytic \(\Pi_0\) for every sufficiently large \(X_R\); the heat compensation has zero pressure increment and preserves that datum.

First fix the restoration and bump functions Q7.6 and Q8.4. The compact ideal rectangle in Q8.1 then fixes the positive cone and moment-solve tolerances independently of \(X_R\). Choose \(j_0>0\), at most .05, so the contribution of the constant \(U_*-4\eta=j_0\) to Q7.8 is smaller than a chosen fraction of that tolerance. Then choose \(\delta_*,\sigma_*\) as in J0, and \(\Lambda\) large enough for J, Q1, the gradient absorptions, and \(C_k^{\rm nat}/\Lambda\) in Q5.4 to satisfy their finite inequalities. These constants are determined after \(j_0\), so small \(\sigma_*\) causes no circular choice.

The uniform reference bounds and Q5.3 now fix all required \(\mathcal B_k\), hence a positive \(T_{\rm sh}\) and \(X_{\rm sep}\), before the amplitude. Choose a finite \(\mathsf C\) large enough for the analytic construction, Q1.5 and Q1.9, and for all the following:

Here \(X_R^{\rm outer}\) is the finite threshold already obtained for the actual O–H family. The quantities required to decay in Q9.1 are explicit positive powers of \(\mathsf C^{-1}\), with constants fixed before \(\mathsf C\), so such a finite choice exists. The initial-segment bounds hold uniformly over the allowed subsequent width family by Q5.3–Q5.4; selecting smaller widths does not undo Q9.1.

With this fixed \(\mathsf C,\Lambda\), choose \(\kappa_0\), then \(t_1\), then the finite final transition widths as in Q2 and Q4. Make their additional contribution in Q5.4, Q4.4 and Q7.8 as small as the remaining fractions of the fixed tolerances require. Positivity and the reference comparisons hold for the entire smaller-width family, so decreasing a width preserves the earlier estimates. The strict inequalities before the matching patch use finite constants that may depend on \(\mathsf C,\Lambda\); their widths are chosen after those constants. The matching tolerance uses only the fixed outer data and therefore does not conflict with this order. Finally solve Q8.9. Higher derivatives exist by smoothness and the finite-dimensional inverse argument, without infinitely many simultaneous smallness requirements.

This realizes the original order (B.40):

The radial frequency for the later shear modification has not yet been chosen.

The constructed profile is smooth through the Cartesian axis by J4 and unchanged on its reserved analytic rectangle. E is positive for \(X>0\). The leading stress is zero through \(X_0\), has the exact inner flat factor and smooth nonzero limiting direction of Q3, and satisfies the admissible cone on a positive inner collar. It satisfies the strict relaxed cone thereafter through the matching point \(\log(X/X_R)=-5\). All five original radial moments, the pressure and all exterior fields match exactly there, so the already established outer-profile and heat-exterior properties remain true. The stress is nonzero between its two endpoints: throughout a strict relaxed region \(v_s<\mathcal U(P_c,J_c)\le P_c\), so its parallel component is \(F(P_c-v_s)>0\); the inner and outer collars have their proved positive direction bounds. These statements establish Lemma B.4, Proposition B.5, Corollary B.6, Lemma B.7, Proposition B.8 and Corollary B.10, with the parameter ordering of Remark B.9.

Q10. Five worked exercises

Exercise 1. Compute the coefficient after division by the flat factor

Give the exact endpoint value of the smooth coefficient in Q3.2. Apply it to both field differences in Q2.2, retaining the transition width.

Solution. Put \(c=t_1^2>0\). For every fixed nonnegative integer \(m\) and smooth \(b\), Q3.2 writes

For each fixed mixed derivative, the integrand is bounded by \(e^{-cw}\) times a polynomial in \(w\), on a sufficiently small closed collar. This is integrable because the chosen \(c\) is positive. Differentiation under the integral proves a smooth extension to zero. Since \(g_a(0)>0\) and \(\int_0^\infty e^{-cw}\,dw=1/c\), its endpoint value is exactly

The quotient notation at zero denotes this proved smooth extension.

Write \(p_e=p_{1,\rm r}(X_0,\eta)\) and \(n_e=n_{s,\rm r}(X_0,\eta)\). In the first activation interval the reference is the natural axis solution. Applying Q10.2 with \(m=0\) to the two exact integrals Q2.2 gives

Both coefficients are smooth through zero, including all fixed mixed derivatives. The original \(X_0\) comes from \(X=X_0e^y\) in the axial integrand. The angular coefficient is strictly positive because \(p_e>0\). These formulas retain the width dependence rather than claiming uniform radial bounds as \(t_1\) shrinks.

Exercise 2. Find the leading differences of all five original moments

Compare the activated field with the reference on its first collar. Compute the exact coefficients of the differences of \(M,I,J,S,C_p\) after division by \(e_a y^6\).

Solution. All moment differences vanish before \(X_0\). At the endpoint set

Each quantity is the actual endpoint value, not its leading approximation at the axis. For fixed \(X\), \(E/E_{\rm r}=H/H_{\rm r}=\exp(\log\phi-\log\phi_{\rm r})\). Q10.3 and the exact integral formula for the exponential give

Products of two field differences have zero coefficient at this order. Thus the endpoint coefficients of the original moment integrands are

The first four moments use \(dX=X_0e^y\,dy\), so their coefficients acquire another factor \(X_0/(2c)\) under the integration rule Q10.2 with \(m=3\). For the fifth moment,

Here \(dX/X=dy\); there is no additional \(X_0\) factor. The coefficient of its integrand divided by \(e_a y^3\) is \(E_e^2p_e/(4c)\). We therefore obtain the full vector

The displayed row and column notation specifies the same five ordered components. Every quotient has a smooth extension: its integrand lies in \(e_a y^3C^\infty\), and Q10.1 proves the next integration step in all fixed derivatives. The common datum \(\Pi_0\) gives \(\Delta\Pi=\Delta C_p\). In particular,

This positive difference is local to the first activation collar. The later five-bump correction cancels the complete accumulated pressure moment at the matching point; no intermediate equality of pressure was assumed.

Exercise 3. Restore the original scale in the moment-error estimates

Determine the amplitude powers in Q7.4–Q7.5 and in the corresponding original moment discrepancies. Show how a finite amplitude achieves the fixed tolerance.

Solution. With all preceding shape data fixed, put

These are the exact constants in

Let \(\Delta\widehat{\mathbf m}\) compare the actual and ideal moments at this endpoint. Q7.4 gives orders \(\mathsf C^{-10}\) for \(\widehat M,\widehat S\), \(\mathsf C^{-16}\) for \(\widehat I,\widehat J\), and \(\mathsf C^{-2}\) for \(\widehat C_p\). For the ideal moments Q7.5 gives the same orders, with the angular term of \(\widehat S_0\) carrying the still smaller factor \(\mathsf C^{-12}\). In particular its pressure term is exactly

Triangle inequalities and the finite product-rule sums prove, in every selected parameter norm, finite bounds

The constants retain the fixed \(d_{\rm sep},P_*,f,X_i\), the original inner-profile bounds and all their required derivatives. They are independent of subsequent amplitude and smaller transition widths. For a desired total error \(\varepsilon>0\), the finite choice

makes the sum of the five displayed norms at most \(\varepsilon\). Take the maximum with the other finite thresholds in Q9.1.

Restoring the original quantities by Q7.1 gives

The argument therefore does not claim that the unscaled \(M,S\) discrepancies tend to zero. The exact invertible row map Q8.6 uses their retained \(X_R\) factors and converts the proved small quantities into the actual five-equation target. Its inverse then cancels the original discrepancies exactly. The remaining axial restoration error is controlled separately by \(G_i-4\eta\), with the order of choices in Q9.

Exercise 4. Use the disjoint supports in the exact nonlinear correction

Show that the two axial bump coefficients can be solved exactly before the three angular coefficients. Retain the full nonlinear terms and prove an actual reduced solve.

Solution. Use the fixed bumps Q8.4 and write their coefficient vectors as

All five supports are disjoint. Consequently \(ue=0\) pointwise, and there are no cross terms in either square. Set

These constants are finite and positive, since the fixed supports lie inside \((e^{-6},e^{-5})\). The full quadratic remainder Q8.8 becomes exactly

For the original transformed target \(d=(d_1,\ldots,d_5)=-\mathscr L_\eta(\Delta\mathbf m)\), the first two equations therefore give

exactly. Define

The remaining three equations are precisely \(B_E\mathbf b+q_E(\mathbf b,\mathbf b)=\mathbf r\). No original moment or quadratic term has been discarded.

For a selected \(C_\eta^m\) norm, let \(\nu_U,\nu_E\) be the norms of the two actual inverse matrices, let \(q_E^*\) bound the displayed bilinear map, and let \(\gamma_m\) bound \(\mathbf a\mapsto\sum_iA_i a_i^2/(P_*f)\) by \(\gamma_m\|\mathbf a\|_m^2\). The product rule and the fixed positive \(P_*f\) make these constants finite and independent of \(X_R,\mathsf C\). Fix the target tolerance \(\varepsilon\) small enough that

Q7–Q9 achieve this fixed tolerance, together with the earlier cone tolerance, by their already proved finite choices. Then \(\|\mathbf a\|_m\le\nu_U\varepsilon\) and \(\|\mathbf r\|_m\le2\varepsilon\). On the radius \(2\nu_E\|\mathbf r\|_m\) ball, the iteration

stays in that ball and has Lipschitz constant at most \(4\nu_E^2q_E^*\|\mathbf r\|_m\le1/2\). Indeed its quadratic contribution to the radius is at most half the linear radius, by the same inequality. The complete-space contraction argument supplies the actual angular coefficients. They obey \(\|\mathbf b\|_m\le4\nu_E\varepsilon\); decreasing the fixed tolerance also meets the partial-patch bounds used in Q8.

The derivative in \(\mathbf b\) is \(B_E+2q_E(\mathbf b,\cdot)\), whose inverse follows from a norm-convergent geometric series after multiplication by \(B_E^{-1}\). Smooth dependence on \(\eta\) follows by differentiating these three finite equations and using that inverse at each fixed order. Substitution in Q8.5–Q8.6 proves exact cancellation of all five original moments. The axial coefficients are independent of the angular targets \(d_3,d_4,d_5\); their exact dependence is Q10.16. This is a property of the specified disjoint supports, not an approximation that neglects a mixed term.

Exercise 5. Determine which analyticity survives stress activation

Prove that the angular profile has the same full radial Taylor data as the natural solution at \(X_0\), yet is not real analytic in \(X\) there. Reconcile this with the common parameter-analytic collar and its exact physical coordinates.

Solution. On \(X\le X_0\), the actual and natural profiles agree. On the first activation interval, Q10.3 gives

Since \(e_a\) vanishes to infinite order and \(X=X_0e^y\) is an analytic invertible coordinate near \(y=0\), the difference and every fixed mixed derivative vanish at \(X_0\). Exponentiation proves the same equality of all derivatives for \(\phi\) and \(E\).

For every fixed real \(\eta\), Q2.2 also gives

Every integrand is positive there. The natural profile is analytic in \(X\) on a neighborhood of \(X_0\), supplied by J. If the activated \(\phi\) were analytic there as well, the equality of all Taylor coefficients would force equality on a neighborhood, contradicting Q10.21. Thus \(\phi\), and also \(E=\sqrt{2X}\phi/\mathsf C\), are smooth but not real analytic in \(X\) at \(X_0\). This statement concerns the constructed leading profile.

For parameter analyticity, Q2.1 integrates holomorphic parameter coefficients against radial-only cutoffs. Exponentiation, the nonvanishing denominators and the forward integrals Q5.1 preserve one common parameter neighborhood. Every fixed radial derivative remains holomorphic there: differentiating the cutoff changes a finite radial coefficient without changing that parameter domain. These are the exact maps that produce the regularity stated in Corollary B.6.

For the original viscosity \(\nu>0\), the physical comparison is

It follows from \(x_\nu=\sqrt\nu\,x\), with the original time and parameter data retained. For fixed \(z_\nu,t<1\), the activation edge and protected collar have the exact radii

The radial coordinate map is analytically invertible near the positive edge radius, so the failure of radial analyticity persists under this map. The collar remains smooth and parameter analytic as proved. Every subsequent modification is supported strictly outside it. The forward integrals for pressure and radial velocity use only smaller radii, so those later modifications leave all collar fields and derivatives exactly unchanged. The final matching equations separately preserve the exterior by Q8.10.

What the connected field supplies next

The actual connection now preserves the analytic inner rectangle, activates a smooth nonzero stress with its exact limiting direction, continues in the strict relaxed cone, and joins the entire outer field with exactly the same pressure and all five original moments. The finite choices and all original radial factors are part of the proof.

The next construction is the admissible-shear realization V0–V7. It preserves the protected collars and exact moments while enforcing the stronger shear condition throughout the intervening annulus. The admissible-shear lesson, V0–V8, gives the construction and its exercises. The base construction Z0–Z8 then treats the higher-order physical residual.

The final theorem lesson, EC0–EC5, uses these constructions to prove Theorem 1.1 and Corollary 10.6.

Authored lesson text and diagram are CC0. Mathematical source: OpenAI, Finite Time Blowup for Navier–Stokes, Appendix B. Exposition: GPT-6 Astra (OpenAI). Original sources retain their attribution and rights.

Realizing the admissible shear and completing the leading profile

NS-BLOWUP N2.5. The joined field now has the required axis, pressure and exterior. A part of its connecting interval still has shear below the threshold required by the later viscous waves. We construct one smooth field that satisfies the stronger cone everywhere, then prove all six conclusions of the leading-profile theorem.

The construction changes radial derivatives by an amount of order one while changing field values by an amount of order \(N^{-1}\). Here \(N\) is one sufficiently large, finite integer. A periodic family of admissible shears has exactly the original mean. Its periodic primitives produce the actual field. Five subsequent corrections restore the original cumulative moments exactly, so the axis and exterior retain their entire previous constructions.

The stress and moment lesson, P0–P7, supplies the full operators and nonlinear moment inverse. The outer lesson, O0–H7, supplies the four reserved patches, heat field and outer stress factor. The axis lesson, J0–J8, and joining lesson, Q0–Q10, supply the regular axis, exact connection and inner stress factor. All four are included in the cumulative reader.

The source is OpenAI, Finite Time Blowup for Navier–Stokes, Theorem 4.6, pages 32–34, the cone formulas on pages 31 and 38, and Appendix C, pages 157–165, (C.1)–(C.19). This is independent teaching and editorial mathematics; the diplomatic transcription continues to reproduce what OpenAI wrote. SOURCES.json records the source identity and mathematical locators.

Figure. The upper panels show the exact averaging map and actual field formulas. The radial diagram records order, with both endpoint collars protected. The lower panel displays the leading endpoint pressure-moment change proved in Exercise 2 and cancelled by the exact correction in Exercise 3. It shows no sampled fluid flow.

Why a small field change can supply the required shear

The shear uses a radial derivative; the five moments use field values. Evaluating a periodic primitive at \(N\log X\) and dividing its amplitude by \(N\) makes the field difference small while leaving the derivative of its fast phase at order one. The two shear components must be integrated with their exact different factors. In particular, changing the angular field exponentially produces an exponential denominator in the axial shear.

The periodic family must have the exact prescribed mean and depend smoothly on both profile coordinates, even where the axial stress coefficient vanishes. V1–V2 prove those properties. V3 constructs the finite-frequency field and controls every quantity needed by the cone. V4 cancels the five nonlinear moment differences. V5–V7 then carry the two endpoint factorizations through all six conclusions of Theorem 4.6.

The exercises give exact formulas at zero axial stress, sharper moment and correction estimates, explicit endpoint derivative powers and the original physical phase at every viscosity. These are proved consequences of the same construction. They do not allege an error in the source's sufficient bounds.

V0. The actual input and the interval of modification

Keep the actual joined input fixed. The radial interval below contains every place that still needs the stronger cone, and the following moment patch is already reserved by the outer construction.

Fix the entire profile constructed in Q, including its finite parameters, common axis pressure and exact exterior matching. Keep

Its inner stress endpoint is \(X_a=X_0=4/\Lambda\); its outer stress endpoint \(X_b\) is the endpoint of the actual H exterior. These are not the local continuation radius 100. The original functions are

All five cumulative integrals \(\mathbf m=(M,I,J,S,C_p)\), their integration constants and \(\Pi=\Pi_0+C_p\) remain those of P2. The input has \(a>0\), the strict relaxed cone on \((X_a,X_b)\), an admissible inner collar and the admissible outer portion beginning in the intermediate power-law interval. These properties were derived in Q9, O8 and H5, rather than assumed for a nonexistent profile.

Choose \(X_{\rm an}\) strictly inside the analytic admissible inner collar of Q5. Then choose \(X_-\) beyond \(X_{\rm an}\), still inside that collar, and \(X_+\) inside the admissible intermediate power-law interval, before the first reserved patch. This is possible by the positive lengths established in Q5 and O0.4. Set \(\mathcal I=[X_-,X_+]\). The compact interval contains every point where the input need not be admissible; the input is admissible near both endpoints. Retain shorter endpoint collars outside its support. The first reserved patch will restore moments; the third and fourth remain available for later positive-order and mean corrections. The second was already used for the heat compensation, which will be preserved exactly.

For any \(a>0,b\in\mathbb R,p\in\mathbb R^2\), keep the original gap map

It is smooth on its stated domain. P6 proves that positivity of all four components is exactly the admissible cone. With \(t_s=-b_s/a\), its c and j are \(P_c(t_s)\) and \(J_c(t_s)\), where

V1. A smooth variance family, including zero axial stress

Vary the ratio of the two shear components while keeping its average fixed. The exponential distribution makes the stress projection remain above two. Its variance can reach every finite positive target, including when the axial stress coefficient is zero.

On the compact rectangle \(\mathcal I\times[-1,1]\), choose \(0<d_0<\tfrac12\min(P_c(t_s)-2)\). This minimum is positive by the already constructed strict relaxed cone. With the separate auxiliary angle \(\theta'\), define

This auxiliary average is not the physical angular or wave-torus average. Differentiation under the integral on compact z intervals proves smoothness, and \(M_e>0\). The change \(\theta'\mapsto\theta'+\pi\) proves evenness. The numerator in the quotient vanishes at p=0. Writing it as \(p\int_0^1\partial_pG(up)du\) proves a joint smooth extension in \((\theta',\mu,p)\), with value \(t=t_s+d_0\mu\sin\theta'\) at zero. Direct integration gives

and the exact variance

The integral of the squared smoothly extended quotient is also a direct proof of smoothness at p=0.

For \(g=\log M_e\), differentiating the quotient for \(g'\) gives exactly the variance of \(\sin\theta'\) under density \(e^{z\sin\theta'}/(2\pi M_e(z))\). It is strictly positive: the density is positive everywhere and \(\sin\theta'\) is not constant on any full-measure set. Thus \(g''>0\). For p nonzero and \(\mu>0\),

For negative p both factors in brackets and outside are negative; for positive p both are positive. This proves strict increase of V for \(\mu>0\); the p=0 formula does so there as well.

The unbounded range can be proved with explicit analytic estimates. For \(|u|\le\pi\), \(1-u^2/2\le\cos u\le1-2u^2/\pi^2\). The lower bound follows by integrating \(1-\cos u\le u^2/2\); the upper follows from \(\sin(|u|/2)\ge |u|/\pi\), the chord inequality for the concave sine on \([0,\pi/2]\), in \(1-\cos u=2\sin^2(u/2)\). Centering the angle at its maximum and integrating over \(|u|\le z^{-1/2}\) for the lower bound, and over the whole real Gaussian for the upper, yields for \(z\ge1\)

Evenness supplies the same bounds for negative z with \(|z|\). Consequently the ratio in V1.3 is bounded below by a positive constant times \(\sqrt{|\mu p|}\) when \(|\mu p|\ge1\). For each fixed nonzero p this tends to infinity as \(\mu\to\infty\); for p=0 use V1.3. Given any finite target, continuity gives for each p an open neighborhood and one finite \(\mu_p\) exceeding the target there. A finite subcover of the compact p range and monotonicity give one finite \(\mu_{\max}\) such that

No uniform large-z estimate near p=0 was assumed in this compactness step.

The original upper cone bound is

P6 proves \(2<\mathcal U(P,J)\le P\) for P>2, including J=0. Continuity suffices here; no derivative of \(|J|\) is taken. The entire compact family in V1.1 with \(0\le\mu\le\mu_{\max}\) has P>2 by V1.2. Choose \(0<\delta_L<1\) below the minimum of \(\mathcal U-2\) on that family and below the positive minimum of \(v_s-2\) on fixed smaller radial boundary neighborhoods. The order is \(d_0,\mu_{\max},\delta_L\).

V2. The variance choice and the exact prescribed mean

Choose the variance only where the original shear is too small. Then change the speed with which the loop is traversed. The exact Jacobian gives both required component averages.

Using the smooth step of O0.1, take \(\zeta_L(w)=1-\sigma((w-2-\delta_L/8)/(\delta_L/8))\). It lies in [0,1], is one below \(2+\delta_L/8\) and zero above \(2+\delta_L/4\). Set exactly

Define \(\rho=0\) where the cutoff vanishes. On its possible nonzero support, \(v_*-v_s\ge\delta_L/4\); hence \(\sqrt\rho=\zeta_L(v_s)\sqrt{v_*-v_s}\), extended by zero, is smooth. Since \(v_s=a+b_s^2/a>0\) and \(v_*<5/2\), one has \(0\le\rho<3\). There is a unique \(0\le\mu<\mu_{\max}\) solving

Existence follows from V1.6 and the intermediate value theorem; strict monotonicity proves uniqueness.

Here is the smooth dependence at zero, which the ordinary derivative of V alone would not justify. Taylor expansion under the defining integral gives \(M_e(z)=1+z^2/4+O(z^4)\). Thus

where \(F_e\) is smooth and even near zero. This is an exact factorization, obtained by the twice-integrated Taylor remainder; the displayed value follows from the second derivative. In a fixed neighborhood it is positive. Therefore the signed square root \(d_0\mu\sqrt{F_e(\mu p)}\) is smooth, odd in \(\mu\), and has derivative \(d_0/\sqrt2\) at zero. Its implicit equation with smooth right side \(\sqrt\rho/\sqrt a\) gives a unique smooth \(\mu\) through zero. Away from zero use the strictly positive derivative V1.4. Local solutions agree by uniqueness, including at \(\eta=\pm1\), by smooth local extension or one-sided implicit differentiation. No cusp in \(\sqrt\rho\) is introduced.

If \(v_s\le2+\delta_L/8\), then v=v_*>2. In the transition zone, \(2<v_s\le v\le v_*\). Off the cutoff, v=v_s>2 and \(\mu=0\). Thus where the loop changes, \(2<v\le2+\delta_L/2<\mathcal U(P_c(t),J_c(t))\). Where it does not change, the original relaxed inequality and v_s>2 give the same conclusion. Every vector \(v(1,t)/(1+t^2)\) is therefore admissible.

To obtain the original average, define the lift

V1.2–V1.3 and V2.2 give \(a\langle1+t^2\rangle=a(1+t_s^2)+aV=v_s+\rho=v\). Consequently the lift increases by exactly one as \(\theta'\) increases by \(2\pi\). Its derivative is bounded below by a positive constant on the compact family, so it has a smooth parameter-dependent inverse. For completeness, the derivative of the inverse is \((\partial_{\theta'}\varphi)^{-1}\), while any parameter derivative is \(-\varphi_{\rm parameter}/\varphi_{\theta'}\); iterating these identities proves all mixed derivative bounds. The exact two averages are

They follow by substituting the full Jacobian in V2.4. On the radial boundary neighborhoods \(\zeta_L=0\), so \(\mu=0,t=t_s,v=v_s\), the lift is \(\theta'/(2\pi)\), and the loop equals the original shear. Positivity of all four components of \(\Psi(a_L,b_L,p_s)\) and compactness yield a common minimum \(\kappa_L>0\). The fixed neighborhoods yield \(\delta_\partial>0\). This establishes every assertion of Lemma C.1, with the exact period-one mean and boundary equalities.

V3. The actual radial modulation and its moment estimates

Integrate the two mean-zero shear differences in the periodic variable, then insert their primitives into the original fields. Differentiate the actual formulas before estimating; this retains the factors needed by the cone.

For any smooth period-one zero-mean f, the unique zero-mean primitive is

Its periodicity follows from \(\int_0^1f=0\), its derivative is f, and its mean is zero by construction. A constant is the only difference between primitives, proving uniqueness. Every fixed parameter derivative commutes with these bounded integrals. Define

V2.5 proves compatibility. Both primitives vanish identically on the radial boundary neighborhoods, so extension by zero is smooth. For an integer N≥1 prescribe the actual fields

Positivity of E_N and exact agreement outside \(\mathcal I\) follow immediately. Let \(\mathcal D_X=X\partial_X\) hold the independent \(\varphi\) fixed. Differentiation after substitution is \(\mathcal D_X+N\partial_\varphi\). Thus the original shears, with every amplitude factor, are exactly

In the second formula the numerator is \(2\mathcal D_XU+2\partial_\varphi\mathcal B+2\mathcal D_X\mathcal B/N=Eb_L+2\mathcal D_X\mathcal B/N\), and the denominator is \(Ee^{\mathcal A/N}\). This proves the exponential factor as well as both signs.

All coefficients and derivatives are bounded on the fixed compact radial range from \(X_-\) through the first reserved patch, with E and X bounded below. The phase is independent of \(\eta\). The identity \(e^z-1=z\int_0^1e^{sz}ds\), followed by parameter differentiation, proves for each fixed m

The norms include a radial supremum, and loop values are evaluated at the actual phase. Constants depend on the fixed loop and profile, never on N. Each radial differentiation of the composed primitives contributes at most one N, while the original perturbation has factor N^{-1}. Repeated product and chain differentiation therefore gives

Terms differentiating the exponential k times carry N^{-k} and at most N^r, with k≥1; hence they obey the same bound for N≥1. This is not smallness of radial derivatives.

Compute the actual moments from E_N,U_N, keeping the same \(\Pi_0\). For any field increments e,u, their integrand changes are exactly

in the original order (M,I,J,S,C_p). All changes vanish before \(X_-\), so their denominators are bounded away from zero wherever integrated. V3.5 and product differentiation give \(O_m(N^{-1})\) bounds for all five partial integrals, uniformly through the correction interval. The same holds for \(\Pi_N-\Pi=C_{p,N}-C_p\). P2.5 and P3.4 then give

For the last estimate use the first bound through order m+1, positive E_N, and the full multiplier X/L. The multiplier is fixed, although possibly large. None of these formulas differentiates the oscillating fields radially.

There is a compact neighborhood of the loop values with a positive lower bound for a; the smooth gap map has a finite Lipschitz constant there. V3.4–V3.8 therefore preserve every gap by at least \(\kappa_L/2\) on \(\mathcal I\) for sufficiently large N. Between \(X_+\) and the reserved patch, fields and shears equal the input, while p_s differs by \(O(N^{-1})\). The input gap has a positive minimum on this compact interval, so the same choice, enlarged if necessary, preserves it. All frequency lower bounds are finite and are chosen after the entire input profile.

V4. Restoring the five original moments on the first reserved patch

The moment mismatch reaches a patch on which the fields still have their exact original power law. The full nonlinear five-moment map is invertible there, and its small solution preserves the cone throughout the patch.

Let the first patch be \((R_-,R_+)\), with its actual radii from O0.4. There the unchanged input is

Fix five centers \(r_j=R_-+j(R_+-R_-)/6\) and width \(\omega=(R_+-R_-)/20\). Let \(\beta_j(X)=\exp[-1/(1-((X-r_j)/\omega)^2)]\) on \(|X-r_j|<\omega\), and zero otherwise. They are smooth, positive on their interiors, and have ordered disjoint compact supports inside the original patch. Put \(u=c_1\beta_1+c_2\beta_2\), \(e=\sum_{j=3}^5c_j\beta_j\), and add them to the fields after modulation.

At the patch entrance the fields themselves are exactly V4.1; the accumulated moment discrepancy is \(O_m(N^{-1})\). The correction map, in row order (M,J,I,S,C_p), has the full increments

Every integral here is in dX over the patch. At zero coefficients its matrix B is block diagonal. Its U rows have weights \(1,\sqrt2KX^{-\lambda}\); its E rows have weights \(\sqrt2X^{1/2},-KX^{-1/2-\lambda},KX^{-3/2-\lambda}\). Removing these nonzero row factors for the purpose of the determinant, while retaining them in B and its inverse, gives the exact power lists

The fixed positive \(\lambda\) makes both lists distinct. P4's determinant-integral proof applies to the specified bumps, so B is invertible on the closed parameter interval. Its inverse and every fixed parameter derivative are bounded. Its constants may depend on \(\lambda\), K and the original patch radii; no uniformity as \(\lambda\downarrow0\) is asserted or needed.

The exact quadratic remainder is

Take target d to be minus the preexisting moment discrepancy in that same row order. With \(\nu_m\) the multiplication norm of B^{-1} and q_m the quadratic bilinear norm, including product-rule constants, the iteration

is a contraction by P5.6–P5.8, provided also the same condition in C^0. Choose N so those finitely many inequalities hold. It yields smooth coefficients with the selected finite norm \(O_m(N^{-1})\), solving the original five equations exactly. All further fixed parameter derivatives exist by the invertible finite-dimensional derivative. They are also \(O_m(N^{-1})\) as N grows: differentiation of the equation gives B times the highest derivative plus terms containing c or a lower derivative of c, while every derivative of d is \(O_m(N^{-1})\) by V3.8. The inverse derivative is uniformly bounded for the small C^0 branch; induction proves the assertion. This requires no simultaneous smallness threshold for every m.

The fixed bumps bound the corrections and their first radial derivatives by a finite multiple of the selected coefficient norm. Partial moment increments obey the same bound, using V4.2 with the upper integration endpoint inside the patch. P3 therefore retains the \(O(N^{-1})\) p_s bound throughout that patch. Comparing these three finite bounds with the positive input E and the four positive input gap minima on its closed rectangle gives a further finite lower bound for N that preserves positivity and the cone everywhere, including within each bump. This comparison uses actual partial integrals, not only their endpoint values.

Let \(X_{\rm rep}=R_+\). The bumps vanish on neighborhoods of both boundaries. At its right endpoint,

The forward integrals propagate equality of all five moments, and the common \(\Pi_0\) gives equality of the complete pressure. P3 proves equality of \(V_0,Q_s,N_s,p_s\) and every parameter derivative there; field equality gives all radial derivatives and shears. This preserves the previously performed terminal compensation, heat exterior and outer endpoint. At the inner endpoint both modifications are supported later, and the same forward integration preserves the analytic rectangle and inner collar exactly. The two unused reserved patches are also unchanged. Fix one integer N satisfying the finite requirements before any physical q or later iteration stage is chosen. This proves Proposition C.2.

V5. Nonzero stress and the direction on the closed annulus

The stress vanishes at both radial edges. Divide by each proved scalar flat factor to extend its direction smoothly, then use the interior cone to obtain a uniform margin on the closed annulus.

Use E,U,Π now for the corrected fields. The leading stress vanishes on \([0,X_a]\) by J and Q3, and on \([X_b,\infty)\) by H4. These intervals are exactly preserved by V4.6 and the unchanged inner construction. In the open annulus admissibility gives \(P_c>v_s\), so \(\mathcal T_{0,\theta}+t_s\mathcal T_{0,z}=F(P_c-v_s)>0\). Thus the stress cannot vanish anywhere there.

Define \(n=\mathcal T_0/|\mathcal T_0|\) in the interior. At the inner edge Q3 supplies the exact factor

Hence n extends smoothly there as \(B_0/|B_0|\). At \(X_a\), zero leading stress gives \(p_s=(a,-b_s)=a(1,t_s)\). The limiting direction is a positive multiple of this vector. Consequently \(n_z-t_sn_\theta=0\) and \(n_\theta+t_sn_z=\sqrt{1+t_s^2}>0\) there.

At the outer edge the exact H6 factors give

and therefore

This is smooth up to the closed edge, including \(\eta=\pm1\), and b_s=0 there because U=0 in the whole outer collar. F remains positive at both edges. At the inner edge a>0 and \(v_s-2>.2\) by J6; at the outer edge H5 gives \(a=v_s\ge2+h\). Interior admissibility and compactness now give positive lower bounds for F,a,v_s−2 on the full closed annulus.

In the interior, division by the actual nonzero stress gives

The quantity \(r_n=2-(v_s-2)(n_z-t_sn_\theta)^2/u_n^2\) is positive by the cone test. Both u_n and r_n extend continuously to the edges by V5.1–V5.3, with positive u_n and r_n=2 there. Let their minima on the closed rectangle be u_>0 and r_>0. Choose

Then everywhere on that rectangle

This proves the precise directional margin of Theorem 4.6(iii), including both specified endpoint directions. A uniform raw stress-gap minimum at the zero-stress endpoints was not assumed.

V6. The common flat weight and all fixed derivatives

The common weight contains both endpoint exponentials. Near either edge the other factor is bounded above and below; on the remaining interior interval the fixed smooth field has ordinary compact bounds.

Use exactly the original weight and endpoint distances

extended by zero outside the annulus. The two positive logarithmic distances have sum \(\log(X_b/X_a)>0\). Near either endpoint the other distance is bounded below; its exponential is a smooth positive factor bounded above and below. Derivatives of the vanishing factor are its exponential times finite sums of inverse powers of the vanishing distance, because \(y_{a,X}=1/X\) and \(y_{b,X}=-1/X\) have bounded derivatives on the fixed annulus. The exponential dominates each inverse power as its distance tends to zero. This proves smoothness and flatness of ζ at both edges.

V5.1 with Q3 gives a positive constant times \(e^{-t_1^2/y_a^2}\) as a stress lower bound near the inner edge and the same exponential times finite inverse powers as each derivative upper bound. V5.2 and H6 give \(ce^{-4/y_b^2}y_b^{-3}\) as the outer lower bound and its exponential times finite inverse powers for every derivative. On a collar with y_b≤1 the lower polynomial factor is at least one. Dividing each estimate by the bounded positive factor from the opposite endpoint proves the required comparison with ζ. On the remaining closed interior interval both ζ and |T_0| have positive minima; every fixed derivative is bounded because the frequency N has already been fixed. Taking the minimum of the three positive lower constants and, for each derivative, the maximum of the three upper constants yields

The lower c is independent of α. Each m_α is a finite nonnegative integer; enlarge it if necessary to cover both endpoint powers. These estimates also prove smooth extension of the stress by zero. All constants depend on fixed profile choices and derivative orders, and none on physical q or later bands and correction stages.

V7. Every conclusion of the leading-profile theorem

The exact moment identities preserve pressure and the heat exterior. Together with the protected axis, stress direction and flat weight, they give every assertion of the leading-profile theorem.

The common analytic axis rectangle and smooth Cartesian axis follow from J4 and Q5; V3–V4 leave them unchanged. On every finite closed radial interval, the constructed fields and all fixed mixed derivatives are finite: the only rapid radial frequency is the one fixed finite N, the later corrections have fixed smooth compact supports, and the heat factor is smooth for its full nonnegative argument range by K0/H0. No analyticity of that heat factor at zero is assumed. Positivity of E on X>0 implies positivity of \(\phi=\mathsf C E/\sqrt{2X}\); at zero it follows from J. The quantities F,U,V_0/X and Π extend smoothly through the axis by Q5.1.

The original pressure identity is preserved exactly. The input pressure was canonical, and both the joining and the shear correction restored C_p with the same axis datum. Therefore

These integrals converge: near zero E=\(\sqrt{2X}F\) with smooth F, and at infinity the unchanged heat exterior decays like \(X^{-A}\), with A>0 and bounded heat factor. Differentiating gives \(\Pi_X=E^2/(2X)=F^2\), including its smooth limit at zero. P1's exact tangential residual identities apply to these actual fields; the physical leading stress is \(q^{-A-1/2}\mathcal T_0\). The full momentum residual still includes the axial-viscosity and radial remainder terms of P1.5–P1.6: completing the leading profile does not solve the entire equation. On \(0<X\le X_a\) the original equations (4.13) hold because that analytic solution is unchanged. This proves parts (i)–(ii); V5 proves (iii) and V6 proves (iv).

For part (v), all original convergent total moments remain exactly

They were established in O/H, transferred through the exact Q join, and preserved by V4.6. In particular the convergence of the subtracted angular moment and its exact value are retained, not replaced by an un-subtracted divergent integral. The original heat exterior is

K0/H0 supply its positive smooth heat solution, including Z=0. In original physical coordinates s=r²/2, its swirl is exactly \(c_\infty s^{-A}\mathcal H(2\tau/s)\); the argument equals \(4\tau/r^2\). For the earlier radius \(X_v=X_{\rm end}\) at the end of the pulse, the input has U=M=0 by O1 and O0.6. The correction patch lies before that radius and restores M exactly, so these equalities hold for all X≥X_v. P0's radial-average formula gives V_0=0 there, proving the whole outer-collar assertion.

For part (vi), retain exactly the third and fourth intervals of O0.4, in their original radii, as \(\mathcal I_{\rm pos},\mathcal I_{\rm mean}\). They are ordered, disjoint, lie in \((X_a,X_v)\), and were untouched by the Q join, the heat compensation on the second patch, and the V4 correction on the first. They therefore retain

The positive c_patch is independent of X and η, as required. These are the actual intervals receiving the later corrections, not newly chosen substitutes.

Finally, for each original viscosity \(\nu>0\), the exact H7 map retains \(u_\nu(x_\nu,t)=\sqrt\nu u(x_\nu/\sqrt\nu,t)\), \(p_\nu=\nu p\) and \(T_\nu=\nu T\) at that same mapped point. Its coordinates satisfy

All original viscosity factors remain in the physical equations; the spatial derivative contributes \(\nu^{-1/2}\), so divergence of the physical stress has the same \(\sqrt\nu\) factor as the other momentum terms. The positive scalar stress factor preserves its direction. This is the proved comparison with the source's displayed coefficient-one construction, not a change of the target viscosity.

V8. Five worked exercises

Exercise 1. Write the complete loop map at zero axial stress

At a fixed profile point with \(p_{s,2}=0\), compute the variance parameter, the loop and its lifted circle coordinate. Check its period, inverse and both component means without dividing by \(p_{s,2}\).

Solution. Retain \(a>0,t_s,v_s=a(1+t_s^2)\), the actual cutoff value \(\rho\), and \(v=v_s+\rho\) from V2. The removable value in V1 gives

The square root is smooth as a function of the profile coordinates by the specific squared cutoff in V2.1. Thus these formulas include \(\rho=0\). Integrating \(1+t^2\) exactly gives

The coefficient of \(\theta'\) is exactly \(1/(2\pi)\), since \(a(1+t_s^2+w^2/2)=v_s+\rho=v\). In particular, \(\varphi(0)=0\) and \(\varphi(\theta'+2\pi)=\varphi(\theta')+1\). Differentiating retains the positive Jacobian

On the fixed compact family both derivatives have finite positive bounds. The lift is strictly increasing and has the stated degree one, so it is a bijection of the circles. The local inverse derivative in V8.3 and its parameter derivatives give its smooth inverse globally by uniqueness.

With \(\theta'=\theta'(\varphi)\), the loop is exactly

Multiplying by the full Jacobian gives the first component average \(a\), and the second \(a t_s=-b_s\), since the integral of \(\sin\theta'\) is zero. If \(\rho=0\), the lift is \(\theta'/(2\pi)\) and the loop is the original shear. All factors \(a,v,d_0\) remain identified. The parameter \(d_0\) cancels from the zero-axial-stress loop itself, while it still determines \(\mu\) and the common compact family used in V1.

Exercise 2. Compute the leading changes of all five endpoint moments

Improve the moment estimates using the zero means of \(\mathcal A,\mathcal B\). Give the actual leading coefficient at \(X_+\) in the original order \((M,I,J,S,C_p)\), including every radial weight. State what can and cannot be improved in the integrated stress estimate.

Solution. All profiles and the loop are now fixed; only the positive integer \(N\) varies. Constants below may depend on these fixed objects and on the specified derivative orders. For a smooth period-one function \(f(X,\eta,\varphi)\) with zero periodic mean and compact radial support inside \((X_-,X_+)\), write \(f_N(X,\eta)=f(X,\eta,N\log X)\). Let \(\mathcal P\) be V3.1, and let \(\mathcal D_X=X\partial_X\) hold the independent phase fixed. Direct differentiation gives the exact formula

The lower boundary term is zero because \(f\), and therefore its periodic primitive, vanishes on a radial neighborhood of \(X_-\). The formula is also valid at either endpoint by continuity. It proves a uniform \(O_m(N^{-1})\) bound for partial integrals in every fixed \(C^m_\eta\) norm.

At \(x=X_+\) the upper boundary term vanishes too. The operator \(\mathcal K=(1+\mathcal D_X)\mathcal P\) preserves zero periodic mean and compact radial support: differentiation and the bounded phase integrals commute. Repeated use of V8.5 therefore gives, for every integer \(k\ge1\),

Every integrand on the right is a fixed smooth function on a compact set, so its integral and each of its finitely many parameter derivatives are bounded. This proves the estimate directly, with no cancellation assumption at a sampled phase.

Put \(A_N=\mathcal A(X,\eta,N\log X)\), \(B_N=\mathcal B(X,\eta,N\log X)\). The original increments are \(u=B_N/N\) and \(e=E(e^{A_N/N}-1)\). Taylor's integral identity

holds for every real \(z\). Boundedness of \(\mathcal A,\mathcal B\) and their fixed parameter derivatives makes the remainders below \(O_m(N^{-3})\), uniformly in \(X\). Substituting into all five original moment integrands yields

Here \(H=\sqrt{2X}E\), and every coefficient is integrated with the original \(dX\). For the last two rows one can check the factors directly from

Primes in this display mean derivatives in \(X\). Both first-order rows in V8.8 have zero periodic mean, as do all the other first-order rows. Formula V8.5 bounds their partial integrals by \(O_m(N^{-1})\), and the extra prefactor gives \(O_m(N^{-2})\). The quadratic terms and their partial integrals are already \(O_m(N^{-2})\). Thus the actual stronger partial-moment and pressure estimates are

At the upper endpoint use V8.6 with \(k=2\) on the first-order coefficients. On each second-order coefficient subtract its periodic mean; V8.6 with \(k=1\) bounds the remaining integral. Consequently

where the entire original coefficient vector is

The original \(M\) discrepancy is even smaller: it is \(N^{-1}\int B_N\,dX\), so V8.6 bounds it by every prescribed inverse power of \(N\). This statement concerns its endpoint value; the partial integral can have a nonzero \(N^{-2}\) boundary term in V8.5.

For each fixed \(\eta\), \(D_{C_p}\ge0\), with equality exactly when \(\mathcal A(X,\eta,\varphi)=0\) for every \(X,\varphi\). Indeed \(E^2/X>0\); a continuous square with zero weighted integral must vanish everywhere. If \(\rho>0\) at some radial point for that \(\eta\), the variance gives a nonconstant continuous \(t\) there. The function \(v/(1+t^2)\) cannot then be constant, since a nonconstant continuous real function cannot have constant square. Thus \(a_L-a\) is not identically zero in phase, \(\mathcal A\) is not identically zero, and by continuity \(D_{C_p}>0\). For that fixed parameter the uncorrected endpoint pressure difference is positive for all sufficiently large \(N\), by V8.11. V4 cancels it exactly afterward.

The bound for \(p_{s,N}-p_s\) on the modulation interval remains \(O_m(N^{-1})\): its original formulas also contain the field values \(E_N,U_N\), whose changes have that order. Between \(X_+\) and the correction patch those fields agree exactly with the input, so the improved moment estimate through order \(m+1\) gives \(O_m(N^{-2})\) for \(p_{s,N}-p_s\) there. No small radial-derivative bound has been asserted.

Exercise 3. Propagate the sharper moment estimate through the exact correction

Find the leading coefficient of the actual five bump amplitudes and prove their stronger size estimate. Preserve the row order of the correction map and every quadratic term.

Solution. Let \(\mathscr R\) denote the exact permutation

No moment changes between the end of modulation and the first patch, so V8.11 makes the actual correction target

The fixed original matrix \(B\) in V4.2–V4.3 includes its factors \(\sqrt2,K(\eta)\) and both patch radii. The actual equation is

Choose \(N\) above the finite smallness thresholds for \(C^0\) and the required \(C^m\) norm in V4.5. The same contraction then gives \(\|c_N\|_{C^m_\eta}\le2\nu_m\|d_N\|_{C^m_\eta}=O_m(N^{-2})\). For any further fixed derivative order, differentiating V8.15 gives the invertible linear operator \(B+2\mathscr Q(c_N,\cdot)\) on its highest derivative. Its inverse stays bounded on the small \(C^0\) branch; all lower derivative terms contain a derivative of \(c_N\) or of \(d_N\). Induction gives \(O_m(N^{-2})\) for every fixed \(m\), with constants and sufficiently large \(N\) allowed to depend on \(m\).

Substitute the established bound into the exact quadratic term: \(\mathscr Q(c_N,c_N)=O_{C^m_\eta}(N^{-4})\). Multiplication by the original \(B^{-1}\) in V8.15 gives

All five bump coefficients occur here in their original units. The distinct powers in V4.3 and their actual factors remain in the matrix; no bound uniform as \(\lambda\) tends to zero is used.

For the specified disjoint supports, define

All these integrals are finite and positive. With \(c_N=(a_1,a_2,b_1,b_2,b_3)^T\), the retained quadratic vector is exactly

The second component vanishes because \(ue=0\) pointwise; the square terms have the displayed signs from V4.2. This is the exact nonlinear map used in V8.15, not a linearized replacement.

Since the bumps are fixed, the field correction and every fixed radial and parameter derivative are \(O_m(N^{-2})\), with the derivative order included in the constant. Its partial moment increments have the same order by V4.2. Thus the cone estimates on the correction patch can use \(N^{-2}\) in place of the sufficient \(N^{-1}\) bound. The modulation interval still requires the \(N^{-1}\) field and shear comparisons. After the patch all five moments, the pressure and the entire exterior agree exactly, for the fixed selected \(N\), as proved in V4.6.

Exercise 4. Give explicit endpoint powers and a smooth stress amplitude

In V6.2, give a single explicit exponent for a fixed number of radial derivatives. Then prove that the scalar amplitude \(|\mathcal T_0|^{1/2}\), extended by zero, is smooth at both edges.

Solution. Put \(r=\alpha_1,m=\alpha_2\). On the inner collar the stress is \(e^{-t_1^2/y_a^2}g_aB_0\), with \(g_aB_0\) smooth and nonzero at the edge. On the outer collar it is \(e^{-4/y_b^2}y_b^{-3}(b_\theta,y_b^6b_z)\), with the vector in parentheses smooth and nonzero at the edge. Shrink the fixed collars so these nonvanishing properties hold throughout them.

Each \(y\)-derivative of \(e^{-c/y^2}\) contributes a polynomial in \(y^{-1}\) whose greatest power increases by at most three: the exponential derivative is \(2cy^{-3}e^{-c/y^2}\), while differentiating a previous inverse power increases its degree by one. Induction therefore bounds its \(j\)-th derivative by \(C_j e^{-c/y^2}y^{-3j}\) for \(0<y\le1\). The derivatives of \(y_a,y_b\) with respect to \(X\) are bounded on the fixed positive annulus. The full product and chain rules, retaining the outer factor \(y_b^{-3}\), give

Parameter derivatives fall only on smooth coefficient functions; the endpoint distances and their exponentials are independent of \(\eta\). The opposite endpoint exponential is bounded above and below on each chosen collar. On the remaining compact interval the same estimate follows by increasing \(C_{r,m}\), since \(\zeta,\delta>0\). Thus \(m_\alpha=3\alpha_1+3\) is an explicit admissible integer, independent of \(\alpha_2\).

For completeness the original weight has the exact logarithmic derivative

The second sign comes from \(y_{b,X}=-1/X\). This identity also directly verifies the power three at each differentiated endpoint.

Define \(\gamma=|\mathcal T_0|^{1/2}\) inside the annulus. Its exact collar formulas are

Every coefficient after the displayed scalar powers is smooth and positive, with bounded fixed mixed derivatives on its collar. The fractional factor \(y_b^{-3/2}\) is kept explicitly. Its \(j\)-th derivative is a constant times \(y_b^{-3/2-j}\); the exponential supplies decay exceeding every such power. The product rule gives

The lower bound is the square root of V6.2. For the upper bound the outer exponent \(3r+3/2\) is bounded by the displayed integer \(3r+2\) when \(\delta\le1\); the inner exponent is \(3r\). The interior is compact and \(\gamma>0\) there. Every derivative tends to zero at either edge, so extension by zero is smooth with all radial and mixed derivatives zero. This gives a precise smooth scalar magnitude for later amplitude constructions; it does not supply their separate wave equations or covariance identities.

Exercise 5. Recover the exact radial phase and supports at viscosity \(\nu\)

Express the actual fast phase in the original physical variables. Compute its radial, axial and time derivatives, and give its exact radial period ratio. Locate the modulation and moment correction supports.

Solution. Keep \(\nu>0\), the original physical coordinates \((r_\nu,z_\nu,t)\) with \(t<1\), and V7.5:

The coefficient of the \(q_\nu\) derivative in the implicit equation is \(L_\nu\), because \(2D=1-2h\). It is positive on the parameter domain. Differentiating the original equation at fixed other physical coordinates gives

The actual phase is \(\Phi_\nu=N\log X_\nu\), considered modulo one when evaluating the period-one primitives. Its three derivatives are therefore

In the axial identity, the exact comparison \(z_\nu q_\nu^{2h-1}/\nu=\eta_\nu/(\sqrt\nu q_\nu^D)\) uses \(D+2h-1=-D\), retaining the viscosity factor. At fixed \(z_\nu,t\), one complete increase of the phase by one changes \(X_\nu\) by the ratio \(e^{1/N}\) and \(r_\nu\) by the ratio \(e^{1/(2N)}\). Starting at radius \(r_\nu\), the exact radial distance for one phase period is

This is the period of the fast phase; the full field also has slowly varying coefficients, so it need not itself repeat after that radial shift.

The primitives vanish outside the original interval \((X_-,X_+)\) and on neighborhoods of both boundaries. Their physical support is accordingly contained in

The five moment bumps are supported in the distinct later interval

These formulas retain the fixed original endpoints, every \(\nu,q_\nu\) factor and the protected collars. The selected finite \(N\) is independent of physical \(q_\nu\), viscosity transport and later iteration stages. No infinite-frequency limiting field is used.

What the completed leading profile supplies

The five parts of N2 now construct the actual leading profile from its original moment operators through its axis, outer field, exact connection and admissible shear. V7 verifies every conclusion of Theorem 4.6, with the full source parameters and both preserved endpoint collars. The actual perturbation has one fixed finite radial frequency.

The field still has the axial-viscosity and radial residual terms retained in P1 and J8. The axis-coefficient lesson, global extension and base-field construction treat these terms. The following wave, mean-correction, iteration and terminal-force lessons complete the construction. The summation lesson proves the common summation mechanism.

Authored exposition and figure source: CC0.

Correcting the coefficients on one fixed axis interval

NS-BLOWUP N3.1. The completed leading profile leaves axial-viscosity and radial momentum terms in the full residual. This lesson derives every coefficient equation that treats those terms, constructs the actual first correction, and proves the linear solution step for each later order on the same radial interval.

The leading-profile lessons, ending in V7, supply the actual input, including one common parameter-analytic axis rectangle. The stress and moment lesson, P0–P7, gives the exact physical operators. The axis lesson, J8, computes a nonzero axial residual of the leading field. Exercise 5 below proves exactly how the first correction cancels that residual at the axis, without changing its growing axial velocity trace.

At each positive order the current unknowns satisfy a linear system with six components. One coefficient differentiates in the parameter. That derivative is unbounded on a fixed holomorphic supremum norm, so an ordinary bounded-operator argument would miss the difficulty. We prove a special matrix identity: two adjacent parameter-derivative factors vanish, even when a derivative strikes a variable coefficient. The number of surviving parameter derivatives is therefore at most half the number of integrations. The factorial from the ordered radial integrals then proves convergence for the entire fixed radial interval.

The original source is OpenAI, Finite Time Blowup for Navier–Stokes, pages 45–49, (5.1)–(5.8) and Lemma 5.1, together with the physical coordinate formulas on pages 24–26. SOURCES.json gives the bibliography and mathematical locators. This is separate teaching and editorial mathematics; the diplomatic transcription retains OpenAI's text.

Figure. The two coordinate spaces retain the original six unknowns. Their derivative map gives the exact vanishing composition used in D3. The lower panels show the convergence bound and the physical residual cancellation in Exercise 5. Radial positions indicate order, not scale; no numerical solution is plotted.

From the original residual to an actual correction

First retain the physical powers of \(q\), the pressure and both viscosity contributions. The spacing \(2h\) arises from their exact relative powers. Radial incompressibility determines the radial velocity coefficient, including its axis integration constant.

Next write the original equations as a six-component system, listing every nonzero matrix entry. Its diagonal radial singularity has an exact integral inverse. The sparse derivative structure and a shared parameter-domain loss give a convergent series and uniqueness in the stated holomorphic class.

Finally recover smoothness in \(X\), the original radial average, pressure and physical fields. The first correction uses only the already constructed leading profile. Later orders use the same solution operator after the preceding coefficients have been globally extended and their moments corrected; that global step is the next lesson. This local construction alone does not yet supply the full base field.

D0. Original physical coefficients and differential operators

The subscript now records expansion order. Keep the full leading pressure distinct from its axis trace, and include the change of physical power after each axial derivative.

Keep \(A=1/2+h,D=1/2-h,d=1-\eta^2,L=1-2h\eta^2\), with the already fixed \(0<h<1/100\). Use \(z=q^D\eta\), \(1-t=q(1-\eta^2)\), \(X=r^2/(2q)\), and the original amplitude \(\mathsf C\). For \(\lambda_n=2nh\), the physical coefficients are exactly

Here \(\Pi_0(X,\eta)\) is the full leading pressure; its axis trace is the pressure datum previously denoted \(\Pi_0(\eta)\). The subscript in this section is expansion order. No pressure constant is discarded by this notation.

For a real exponent \(\beta\), retain the original operators

so \(\partial_t(q^\beta f)=q^{\beta-1}T_\beta f\) and \(\partial_z(q^\beta f)=q^{\beta-D}Z_\beta f\). P0.2–P0.3 prove these by differentiation of the actual coordinate map. Write

The composition order in the last expression retains the change in physical power after the first derivative. Negative-index fields are zero.

Incompressibility at order n gives \(q^{\lambda_n-1}[\partial_XV_n+Z_{-A,n}U_n]=0\), because A+D=1. The regular-axis condition fixes V_n(0)=0. Integrating the displayed equation, with \(\int_0^XxU_{n,X}dx=XU_n-\int_0^XU_n dx\), proves

The integral formula retains its smooth value at X=0 and fixes the integration constant uniquely.

D1. Every term of the coefficient equations

Collect terms only after computing the full cylindrical momentum equation. The angular connection term, radial pressure equation and two different viscous order shifts all remain in the equations below.

Set \(b=-A-1/2\), \(c=-A\). For angular transport, the cylindrical connection term is part of \(u_r(\partial_r+r^{-1})u_\theta\). Substitution of D0.1 yields exactly

Indeed \((\partial_r+r^{-1})(r\phi_j)=2\phi_j+2X\phi_{j,X}\), and \(r^2=2qX\) gives the stated factor. The angular axial transport is \((r/\mathsf C)q^{b-1+\lambda_i+\lambda_j}U_iZ_{b,j}\phi_j\), because \(-A-D=-1\). The vector radial Laplacian satisfies

This follows from \(X_r=r/q\) by two ordinary derivatives, including the two cylindrical terms. Two axial derivatives introduce \(q^{-2D}=q^{-1}q^{2h}\), so they move order n−1 to order n. Equating the complete order-n angular residual to zero gives

Primes here and in D1.4–D1.6 mean \(\partial_X\) at fixed η.

For the axial component the scalar radial Laplacian is \(2q^{-1}(XU''+U')\). Radial transport contributes \(V_iU_j'\), and axial transport contributes \(U_iZ_{c,j}U_j\), each with factor \(q^{c-1+\lambda_i+\lambda_j}\). The pressure derivative has the same power because \(-2A-D=c-1\). Thus its full equation is

For the radial equation multiply the physical momentum equation by r first. The pressure and centrifugal terms at order n are \(q^{-2A+\lambda_n}[2X\Pi_n'-2X\mathsf C^{-2}\sum_{i+j=n}\phi_i\phi_j]\). The time and transport terms have power \(q^{-1+\lambda_k}\), with exact coefficients \(T_{0,k}V_k\), \(V_i(V_j'-V_j/(2X))\) and \(U_iZ_{0,j}V_j\). The vector radial Laplacian of V(X)/r is \(2XV''/(qr)\), so after multiplying by r its contribution is \(-2XV_k''\). The axial-viscosity term at that power is \(-Z_{0,k-1}^{[2]}V_{k-1}\). Since \(-1+\lambda_k=-2A+\lambda_{k+1}\), the result is exactly

where

These computations retain all physical powers and all radial and axial viscous terms in original (5.2)–(5.6).

The quotient in D1.5 is regular, with an explicit full formula. Write V_j=Xv_j using D0.4. Direct substitution in D0.2 gives \(T_\beta(Xf)=XT_{\beta-1}f\), \(Z_\beta(Xf)=XZ_{\beta-1}f\). Consequently

Every coefficient is smooth at the axis and holomorphic on a suitable parameter neighborhood whenever the preceding profiles and their needed derivatives are. The factor 1/2 in the quadratic radial term and both radial derivatives remain visible.

D2. The full six-component system

The current order is linear because every middle-index product uses previously constructed coefficients. A radial-average variable makes the system local in the radial coordinate; its defining relation will be recovered exactly.

Fix \(a>0\) with \(X_a<a^2<X_{\rm an}\), inside the common analytic rectangle proved in V7. The radius a is fixed independently of the correction order. In this paragraph use the concise symbols \(\phi=\phi_0,U=U_0,V=V_0\) for the actual leading functions and put

The W here equals the original transport coefficient in P1.1 by D0.4. The superscript on \(\pi_n^{\rm exp}\) distinguishes a scalar exponent from the pressure profile.

At positive order all middle-index terms are already of lower order. Write them, without omission, as

For n=1 these sums are empty. The factors V_i=Xv_i cancel the apparent 1/X in R_φ,n. D1.7 handles the remaining quotient. These are actual finite source expressions, not new free data.

Let \(\xi=\sqrt X\), \(K_n=\mathcal A_XU_n-U_n\), and

The use of ξ retains X=ξ² and all resulting derivative factors. D0.4 is equivalently

The second-order operators are exactly \(2(X\partial_{XX}+2\partial_X)=\tfrac12(\partial_{\xi\xi}+3\xi^{-1}\partial_\xi)\) and \(2(X\partial_{XX}+\partial_X)=\tfrac12(\partial_{\xi\xi}+\xi^{-1}\partial_\xi)\). Also \(\partial_\xi K_n+2K_n/\xi=-\partial_\xi U_n\). Thus the exact system is

Here are every nonzero entry of the matrices; all unlisted entries vanish:

The parameter-derivative matrix is

The complete source vector is

To verify these rows, the linear angular equation before the ξ conversion is

Insert D2.4 and multiply by two; \(X\phi_n'=\xi\partial_\xi\phi_n/2\) gives (A_0)55. The axial row has the same derivative coefficients with U_n, together with \(XU_{0,X}(V_n/X)+U_nZ_cU_0+Z_{-2A,n}\Pi_n+R_{U,n}\). Use the already derived pressure row \(\Pi_n'=2\phi_0\phi_n/\mathsf C^2+P_n\) inside \(Z_{-2A,n}\Pi_n=(2\pi_n^{\rm exp}\eta\Pi_n+d\Pi_{n,\eta}-2\eta X\Pi_n')/L\). This supplies (A_0)61, (A_0)64, (A_1)64 and the last term in D2.8. The first four equations then follow from the definitions and the pressure row. This proves the full matrix system, including the terms not printed explicitly in the source.

D3. Exact integral inverse and the sparse derivative structure

Invert the diagonal singular part with its actual zero-axis conditions. Then follow the two coordinate spaces through every factor to prove the vanishing composition before estimating derivatives.

For \(c_i=(C_d)_{ii}\), define

For positive ξ, differentiating the first expression gives \((Gg)_i'+c_i(Gg)_i/\xi=g_i\). The second expression extends continuously by zero at ξ=0. Conversely multiplication of a regular solution of that scalar equation by ξ^{c_i}, integration from zero, and its zero axis value recover D3.1. For c_i>0 a nonzero homogeneous integration constant is singular; for c_i=0 it violates the prescribed zero value. Thus G is the exact inverse with these boundary conditions.

Let \(E_4=\operatorname{span}(e_1,e_2,e_3,e_4)\), \(E_2=\operatorname{span}(e_5,e_6)\). D2.7 proves that A_1 maps E_4 into E_2 and annihilates E_2. Every η derivative of A_1 has the identical support pattern. All G kernels are diagonal and independent of η, so they preserve both spaces. For any such diagonal kernel D_0 and any radii ξ,s,

In particular, composing two adjacent factors of the type \(A_1\partial_\eta\) vanishes, including the term where the left derivative strikes the right coefficient. These are exact matrix identities on the original six coordinates.

Put \(\mathcal K=G(A_0+A_1\partial_\eta)\). In the expansion of \(\mathcal K^kG\), there are at most 2^k choices of the two operator types. A nonzero choice cannot contain adjacent parameter-derivative types by D3.2. Hence it has at most

parameter derivatives. This count concerns differential-operator compositions, so it already includes the product-rule terms on variable coefficients. No coefficient is treated as constant in η.

D4. Convergence on the whole fixed radial interval

Allocate the available parameter-domain loss only to the derivatives that survive. Ordered radial integration supplies a full factorial. This proves convergence without a smallness requirement on the chosen radial interval.

For the current order choose a bounded complex neighborhood \(S_\rho=\{\eta:\operatorname{dist}(\eta,[-1,1])<\rho\}\) with no zero of L, on which all current coefficients and sources are bounded and holomorphic, uniformly for \(0\le\xi\le a\). Its radius may depend on n. Such a neighborhood exists for n=1 by V7 and for the induction receiver from the finitely many lower profiles and derivatives that enter D2.2. Take a smaller radius than those available for those finitely many data, so the suprema are finite on its closure. No bound uniform in n is assumed.

For \(0<\rho'<\rho\), set \(\Delta=\rho-\rho'\). Cauchy's integral formula on a disk of radius ε inside the larger domain gives \(\|\partial_\eta f\|_{S_{r-\varepsilon}}\le\varepsilon^{-1}\|f\|_{S_r}\). In a word with p derivatives allocate loss Δ/p to each derivative; coefficient multiplications need no radius loss. For p=0 the derivative factor is one. This bounds the full operator word by a coefficient factor times \((p/\Delta)^p\), without expanding all product-rule terms separately.

Each diagonal integration kernel has norm at most one since c_i≥0 and the nested radii are ordered. Its k+1 integrations range over a simplex of volume \(\xi^{k+1}/(k+1)!\). Choose C_n≥1 bounding the matrices, source and the factor from at most 2^k words. D3.3 then gives the complete estimate

The convention at p_k=0 is one. The simplex and radius-loss estimates can be applied simultaneously: before taking suprema, each integrand has exactly the same ordered radial variables and parameter-domain losses, and all coefficient bounds are uniform in those radial variables.

The root of the right side tends to zero. For example \((k+1)!\ge((k+1)/e)^{k+1}\), from summing \(\log j\ge\int_{j-1}^j\log x\,dx\) for j≥2, while p_k≤(k+1)/2. The kth root is bounded by a fixed multiple of \((k+1)^{-1/2}\), with fixed a,C_n,Δ. Thus the actual series

converges uniformly on the whole fixed interval and every smaller parameter neighborhood. Uniform convergence preserves holomorphy in η. On a further smaller domain Cauchy's bound also gives convergence of the parameter derivative series, so bounded radial integration passes through the series. Consequently \(\mathbf W_n=G(A_0\mathbf W_n+A_1\partial_\eta\mathbf W_n+f_n)\), and D3.1 proves D2.5. No infinite-frequency or numerical approximation is used.

For uniqueness take two bounded holomorphic solutions on a common parameter neighborhood; their difference H satisfies H=\(\mathcal K^kH\) for every k. The same argument, with k integrations, bounds it by

The right side tends to zero. A common larger domain with finite suprema is obtained by first shrinking the intersection of the two solutions' domains. Hence H=0. This is exactly the stated holomorphic uniqueness class, not a uniqueness assertion for arbitrary smooth parameter dependence.

The chosen a did not enter any smallness condition. Increasing C_n or decreasing the parameter neighborhood still gives convergence by D4.1. Therefore the radial interval stays fixed at every application of this solution operator.

D5. Smoothness, parity and recovery of the original unknowns

Recover radial smoothness through a nonsingular integral identity. Reflection gives the exact parity needed to return from the square-root radial coordinate to the original \(X\).

For continuous g the inverse formula has the nonsingular derivative

It follows either by differentiating the first integral away from zero and passing to zero in the second expression, or by its scalar equation and continuous quotient. Thus continuous g gives one continuous radial derivative. Apply this to \(g=A_0\mathbf W_n+A_1\partial_\eta\mathbf W_n+f_n\); the parameter derivative is continuous on a smaller neighborhood by Cauchy's formula. Once \(\mathbf W_n\) is C^r in ξ, the same formula and product differentiation give C^{r+1} on a smaller parameter domain.

Here is the common-domain conclusion required by the next coefficient construction. At this fixed order n, take the parameter neighborhood \(S_\rho\) on which the input coefficients, source and all their radial derivatives are holomorphic. For n=1 it is supplied by the common analytic rectangle of V7; for the induction it is the intersection of the finitely many neighborhoods of the completed lower coefficients. The source formulas D2.2 use only finitely many derivatives of those fields. On any smaller closed neighborhood, each fixed radial derivative of these data is bounded uniformly on the compact radial interval. The bounds may depend on its derivative order.

Choose once \(0<\rho_2<\rho_1<\rho\). For any prescribed integer r, perform the r applications of D5.1 on the nested neighborhoods with radii \(\rho_1-j(\rho_1-\rho_2)/(r+1)\), \(j=0,\ldots,r\). Each application uses a positive radius difference, so Cauchy's formula controls the one additional parameter derivative. Radial differentiation of the integral in D5.1 introduces the factor \(t^k\) for a k-th radial derivative of its integrand; this is bounded by one on \(0\le t\le1\), so the same differentiated integral formula remains valid at the axis. Induction therefore supplies all radial derivatives through r, continuous and holomorphic on a neighborhood strictly larger than \(S_{\rho_2}\). Every occurrence is a derivative of the one D4.2 solution, by uniqueness, not a new solution chosen with r. Since r was arbitrary, all radial derivatives are holomorphic on this same fixed \(S_{\rho_2}\); their bounds can grow with r. Further parameter derivatives follow from Cauchy's formula on the positive remaining margin. This proves the common parameter neighborhood for all mixed derivatives at this fixed n. Its radius may decrease with n, while the radial interval \([0,a]\) stays unchanged.

For the parity argument extend the original coefficients to negative ξ by X=ξ². Set \(S=\operatorname{diag}(1,1,1,1,-1,-1)\). Inspection of every entry in D2.6–D2.8 gives

For example (A_0)41 is odd, (A_0)15 is even, and A_1 has only last-two-row, first-four-column entries, all even. The lower source profiles are functions of X; f_n's fourth component is odd and its last two are even. The extended G formula in D3.1 and the same absolute estimates construct a unique smooth solution on [-a,a]. Reflection \(\widetilde{\mathbf W}_n(\xi)=S\mathbf W_n(-\xi)\) obeys the same integral equation and axis data by D5.2, so uniqueness gives equality. The first four components are therefore even and the last two odd.

An even smooth function of ξ is smooth as a function of X=ξ² on the half interval. Here is the boundary check with its derivative content. For each integer M, Taylor's formula at zero has only even coefficients through degree 2M, with remainder \(\xi^{2M+2}R_M(\xi)\) whose needed derivatives are bounded. Applying \(\partial_X=(2\xi)^{-1}\partial_\xi\) k times to this remainder gives a finite sum bounded by a constant times \(\xi^{2M+2-2k}\) for M≥k. The polynomial terms give the required limits of each derivative. Taking larger M for each k proves the smooth extension; parameter derivatives and their holomorphy obey the same argument on the smaller neighborhoods already supplied above.

The first two rows recover the derivative coordinates. The third row also recovers the original radial average exactly, rather than a replacement variable. Indeed

where \(\partial_sU_n(s^2,\eta)\) denotes the derivative of the composed function. Changing variables x=s² gives \(K_n=\mathcal A_XU_n-U_n\). Formula D2.4 then reconstructs V_n and D0.4, with V_n(0)=0. The fourth row reconstructs D1.5 with \(\Pi_n(0)=0\); the last two rows, by D2.9 and the pressure substitution, reconstruct D1.3–D1.4. Thus the integral solution solves all original equations, with \(\phi_n(0)=U_n(0)=\Pi_n(0)=0\). Conversely any solution in the source's regular holomorphic class gives this six-component solution and is unique by D4.3.

D6. The actual first correction and the induction receiver

At the first order every source is already an explicit function of the leading profile. The solution series therefore constructs an actual correction. Its complete physical factors are retained when transporting to viscosity \(\nu\).

At n=1 all inputs are already supplied by the verified leading profile. Specifically,

with the full \(\Omega_0/X\) given by D1.7 using only v_0,U_0 and their derivatives; its negative-index term is zero. Put these actual functions, \(\ell=2h\), and the actual leading coefficients into D2.6–D2.8. The convergent series D4.2 is therefore an explicit exact construction of \((\phi_1,U_1,V_1,\Pi_1)\) on \([0,a^2]\), not an existence statement depending on an unspecified source. Its zero axis values preserve all leading traces. Its original physical fields are D0.1 with \(\lambda_1=2h\), and the preceding identities cancel every order-one residual on this interval.

For higher orders D2.2 displays precisely the finite receiving map from previously completed coefficients. D1.7 proves that its axis source is regular; D4 proves that its size never forces the radial interval to shrink; D5 proves the regularity needed to form the next source. This establishes the full linear solution step of original Lemma 5.1. The actual global extension in Y0–Y7 completes original Lemma 5.2: it extends each current solution, imposes the five actual residual moments and finishes that order before using it at the next. Together these constructions supply the entire global coefficient sequence; its summation into the actual base field is derived in Z0–Z8.

For the original viscosity \(\nu>0\), retain the exact map proved in P1 and H7, with \(r_\nu=\sqrt\nu r,z_\nu=\sqrt\nu z\) and time unchanged. The physical coefficient fields become

Each full physical momentum term has the same additional factor \(\sqrt\nu\): one spatial derivative contributes \(\nu^{-1/2}\), the pressure has factor ν, and the two derivatives in νΔ cancel their ν^{-1} with the original viscosity. Thus the coefficient cancellations hold at the unchanged target viscosity with every field factor retained.

D7. Five worked exercises

Exercise 1. Prove the exact composition rules for the physical operators

Explain why the two axial derivatives in D0.3 have different subscripts. Prove the corresponding mixed time–axial identity and the rules involving the original radial variable \(X\).

Solution. Let \(f(X,\eta)\) be smooth on the profile rectangle, and let \(\beta\) be a real physical exponent. D0.2 gives

Applying the same identities in the opposite order gives \(\partial_z\partial_t(q^\beta f)=q^{\beta-1-D}Z_{\beta-1}T_\beta f\). The original smooth coordinate map has \(q>0\), \(L>0\), so physical mixed derivatives commute and their common scalar factor is nonzero. Each \((X,\eta)\) with \(X\ge0,|\eta|<1\) is reached by taking any positive \(q\) and setting \(r=\sqrt{2qX},z=q^D\eta,t=1-q(1-\eta^2)\). Thus

This proves an identity on the original profile domain, not merely on one coordinate trajectory. Continuity gives the one-sided identities at \(\eta=\pm1\) and \(X=0\).

Direct substitution into the full formulas D0.2 gives

For instance, \(X\partial_X(Xf)=Xf+X^2f_X\) changes the time coefficient from \(-\beta\) to \(1-\beta\), while \(-2\eta X\partial_X(Xf)\) changes the axial coefficient from \(2\beta\eta\) to \(2(\beta-1)\eta\). Differentiating the original operators gives the last two identities by the same product rule; \(L\) has no \(X\)-dependence.

The order spacing is also exact:

The first identity accounts for axial viscosity relative to radial viscosity. The second accounts for the remaining radial momentum terms relative to centrifugal pressure. Neither permits replacing \(Z_{\beta-D}Z_\beta\) by \(Z_\beta^2\).

Exercise 2. Give a finite remainder bound for the convergent integral series

Turn D4.1 into an explicit tail estimate without shrinking \(a\). Retain both parity classes of the summation index.

Solution. Let \(z=C_na>0\), \(\Delta=\rho-\rho'>0\), and retain the actual majorants

For \(m\ge\max\{1,\lceil\Delta\rceil\}\), put \(B_m=(m/\Delta)^m\). Then

The last bound follows by integrating \(1/(1+s)\le1\) on \([0,1/m]\), giving \(m\log(1+1/m)\le1\). Hence the two exact adjacent-parity comparisons are

Choose any integer

Both ratios in D7.7 are then at most \(1/2\) for every \(m\ge M\). Summing each parity class as a geometric majorant proves the actual remainder bound

All constants in this formula are those from the already established coefficient and source bounds. Increasing \(M\) makes the right side tend to zero by D7.7. The radius \(a\), parameter margin \(\Delta\), source and physical coordinates remain unchanged. This is an analytic error estimate for the actual infinite integral series.

Exercise 3. Compute the first radial coefficients at the axis

Find the exact axis slopes of \(\phi_1,U_1,\Pi_1\), retaining the full original pressure datum. Determine the axis slopes of \(\phi_n,U_n\) at every later positive order.

Solution. For an axis trace \(f(\eta)\), define the restrictions of the original operators by

Here primes mean parameter derivatives. At \(X=0\), the term \(Xf_X\) in D0.2 vanishes for each smooth profile. This remains true after either operator is applied, so taking the axis trace of their composition gives precisely the composition in D7.10.

At each positive order all three prescribed axis traces \(\phi_n,U_n,\Pi_n\) vanish identically in \(\eta\), as do their parameter derivatives. D0.4 then gives \(V_n/X=0\) at the axis. Evaluating D1.3–D1.5 there gives

In the angular current-order terms, \(V_0(\phi_n'+\phi_n/X)\) tends to zero because \(V_0=O(X)\) and \(\phi_n=O(X)\); the term \(V_n\phi_0/X\) tends to zero because \(V_n/X=0\) at the axis. All other current-order terms vanish from their zero traces or explicit \(X\) factors. This verifies the limits rather than cancelling a singular expression formally.

For \(n=1\) use D6.1 and the actual traces \(\phi_*,U_*=4\eta+j_0\):

In particular the complete \(\Lambda\)-dependent function \(\phi_*\) is retained in the first expression.

To compute the pressure slope, retain \(g(\eta)=U_{0,X}(0,\eta)=-Z_*/(2L)\) from J8.6. The full source \(Z_*\), including the actual axis pressure \(\Pi_0(0,\eta)\), is J0.1. Expanding the exact radial-average formula D0.4 gives

Indeed \(\mathcal A_XU_0=U_*+gX/2+O(X^2)\), and \(1-D=A\). D1.7 at \(k=0\) now gives the entire finite value

The terms with an explicit \(X\) and the negative-index axial-viscosity term vanish at this endpoint. The remaining radial derivative \(-4w_*\), the quadratic factor \(1/2\) and the original pressure inside \(g\) have not been discarded.

For \(n\ge2\), all middle-index terms in \(R_{\phi,n}\) and \(R_{U,n}\) vanish at the axis: every positive-order field has zero trace and \(V_i/X\) has zero trace. The final axial-viscosity terms have the zero trace of the \((n-1)\)-st field and all its parameter derivatives. Therefore

This conclusion concerns these two slopes; it does not assert that every pressure slope or higher radial derivative vanishes.

Finally, writing \(\alpha(\eta)=U_{1,X}(0,\eta)\), the exact reconstruction gives

Every fixed parameter derivative has the same remainder order on a smaller common domain, by D5 and Taylor's integral remainder. These formulas describe the actual first correction, not a separately chosen axis polynomial.

Exercise 4. Prove a quantitative source-to-solution map

For the actual fixed matrices \(A_0,A_1\), prove linearity and a finite bound for the solution map on two parameter neighborhoods. State the norms, domains and all radius dependence.

Solution. Use the maximum norm on the original six coordinates and its induced matrix norm. Let \(\mathcal H_\rho\) be the normed space of continuous functions on \([0,a]\times S_\rho\), holomorphic in \(\eta\), with finite supremum norm \(\|\cdot\|_\rho\). We use only bounded holomorphic data; taking uniform limits makes this a Banach space. The actual matrices are bounded on this domain after the prior shrinking in D4. Put

The same root estimate as D4, or the two parity estimates in Exercise 2, proves that this series is finite for every finite \(z\ge0\) and every \(\Delta>0\).

The word proof in D3–D4 applies to any \(f\in\mathcal H_\rho\). Separate the source norm from the coefficient norms. There are at most \(2^k\) words, each coefficient multiplication costs at most \(\mathsf M\), and the full simplex has volume \(a^{k+1}/(k+1)!\). Thus

Consequently the actual integral solution operator is the bounded linear map

Linearity follows term by term and passes through uniform convergence. On an intermediate parameter neighborhood the series converges with its first parameter derivative on a smaller neighborhood. D3.1 then proves the integral equation, the differential equation and the zero-axis value. The uniqueness argument D4.3 applies to this same bounded holomorphic class.

For two source vectors in the stated space, subtracting the convergent series gives the exact difference map

The actual smooth source \(f_n\) belongs to the domain; D5 gives the extra smoothness and parity needed to recover its original fields. An arbitrary source in \(\mathcal H_\rho\) need not have that parity, so the map just proved is a six-component solution map, with original-field recovery asserted for the already verified source \(f_n\). The factor \(a\), coefficient norm and full parameter margin remain explicit. No source smallness or radial shrinking is required.

Exercise 5. Show exactly how the first correction removes the axial residual at the axis

Compute \(U_{1,X}(0,0)\) for the actual leading trace, then prove cancellation of the entire axial momentum residual at every point of the physical axis. Retain the original viscosity.

Solution. The actual leading axial trace is \(U_*=4\eta+j_0\), and \(c=-A\). Its first restricted axial derivative is

At zero the next operator \(\mathscr Z_{c-D}\) acts by parameter differentiation. The derivative of the denominator is zero there, so

This value follows from the complete coefficient equations and is independent of the later moment extensions, whose supports lie away from this axis interval.

Let \((u_\nu^{(0)},p_\nu^{(0)})\) be the actual leading physical field and \((u_{1,\nu},p_{1,\nu})\) the actual first coefficient from D6.2. Use the original physical \(q_\nu,\eta_\nu\) there. The leading tangential equations cancel time, convection, axial pressure and radial viscosity on the axis interval. Its remaining axial residual at the axis is therefore exactly

The factor comes from \(\nu\partial_{z_\nu z_\nu} [\sqrt\nu\,q_\nu^cU_*]\): each spatial derivative contributes \(\nu^{-1/2}\), leaving exactly \(\sqrt\nu\). This is the same full axial-viscosity term retained in the original residual.

The first correction has zero axial velocity and pressure traces for every \(\eta\), and \(V_1=O(X^2)\), \(\phi_1=O(X)\). In Cartesian coordinates its horizontal velocity is \(O(r_\nu^3)\) and its axial velocity is \(O(r_\nu^2)\) at fixed \(z_\nu,t\). Thus its time derivatives, axial derivatives and axial pressure derivative vanish on the axis. All extra axial convection terms vanish there too: both fields have zero horizontal velocity on the axis, and the correction's axial trace and axial derivative are zero.

Only its radial viscous term contributes to the axial residual at the axis. Since \(X=r_\nu^2/(2\nu q_\nu)\), direct radial differentiation gives

The displayed axis value is the smooth Cartesian limit. Now \(\lambda_1-1=2h-1=-2D\), and D7.12 gives twice the slope as the negative restricted second axial derivative. Combining D7.23–D7.24 in the momentum residual therefore yields the exact identity

This is cancellation for the actual first correction throughout the axis, not only cancellation of a formal coefficient at the origin.

At \(z_\nu=0\), put \(\tau=1-t\). The unchanged leading residual from J8 is \(2Aj_0\sqrt\nu\,\tau^{-A-2D}\). The first correction's original near-axis expansion is

At each fixed \(t<1,\nu>0\), the radial Laplacian of its displayed quadratic term, multiplied by \(\nu\), is exactly \(2Aj_0\sqrt\nu\,\tau^{-A-2D}\). Its negative sign in the momentum residual cancels the positive leading residual.

The velocity trace itself stays

Thus cancellation of the residual does not remove the prescribed growth of the axial trace. D7.15 also shows that later constructed positive orders have zero first radial slope and cannot change this exact axial cancellation. Residuals away from the axis, global moment conditions and the summation into a base field still require their subsequent constructions.

What this fixed-interval solution supplies

D0–D6 prove the complete coefficient equations, all matrix entries, the convergent solution operator, uniqueness, common parameter regularity, parity and recovery of the original fields. They construct the actual first positive-order correction and the full local solution step of Lemma 5.1 on the chosen interval. D7 supplies explicit remainder and stability bounds, exact axis coefficients and the physical residual cancellation.

The next lesson extends each coefficient globally and restores its five original residual moments before it is used at the next order. Those complete derivations are already supplied in the separate extension audit, followed by the actual base-field realization. The global extension lesson proves the extension and moment correction; the base-field lesson constructs their actual sum.

Authored exposition and figure source: CC0.

Extending every coefficient and cancelling its five moments

NS-BLOWUP N3.2. A local coefficient cannot simply be cut off and used at the next order. Radial integration can leave a velocity tail, a pressure constant and two stress tails. This lesson constructs the global coefficients, cancels all five original moments exactly, proves their support and retains the common parameter regularity needed for the next coefficient.

The fixed-interval lesson, D0–D7, supplies the actual inner solution. The completed leading profile, V0–V8, supplies two protected patches, the exact heat exterior and the original flat stress weight. The outer-profile lesson, O0–H7, supplies the smooth cutoff, heat integrals and convergent subtracted moment used below.

The current-order correction is an exact finite linear solve, with no small-discrepancy condition. We prove the two matrix inverses for actual smooth bumps, then prove the radial boundary terms and all integration constants vanish. The exceptional first-order outer stress receives its own endpoint calculation. The final induction constructs every positive-order coefficient; summing them into a base field is the following lesson.

The original source is OpenAI, Finite Time Blowup for Navier–Stokes, pages 49–54, (5.9)–(5.22) and the full proof of Lemma 5.2. SOURCES.json records the source identity and mathematical locators. This is separate teaching and editorial mathematics; the diplomatic transcription retains OpenAI's text.

Figure. The diagram retains the original moments, patch order and stress supports. The lower formula is the exact first-order endpoint coefficient proved in Exercise 4. Positions show radial order, not scale; no numerical field is sampled.

Y0. Fixed geometry, data and original moments

Choose the radial geometry once. The five moments have different jobs: they remove radial integration constants and the total tangential residuals. Keep their original weights and pressure derivative until compact pressure has been proved.

Keep all notation and physical fields of D0–D1: \(A=1/2+h,D=1/2-h,d=1-\eta^2,L=1-2h\eta^2\), \(\lambda_n=2nh\), \(R=\sqrt{2X}\), \(r=\sqrt qR\), \(b=-A-1/2,c=-A\), and the original amplitude \(\mathsf C\). Thus \(E_n=R\phi_n/\mathsf C\). The actual leading profile has stress endpoints \(X_a,X_b\), analytic interval \([0,X_{\rm an}]\), and ordered unused patches \(\mathcal I_{\rm pos},\mathcal I_{\rm mean}\) from V7.4. Its leading \(U_0\text{ and }V_0\) vanish after \(X_v<X_b\). Choose, once for all orders,

The first four radii are inside the positive-length analytic inner collar retained in Q5 and V7; D2 permits this same a. The outer radius can be chosen before the terminal outer collar, because the original schedule places \(X_v\) and both patches strictly before it. The subscripts here label the positive-order construction, not the distinct shear-modification interval in V0. No support radius depends on n.

Let σ be the exact smooth step O0.1 and set

This gives the required values one and zero with smooth flat transitions. At order n the five original moments are

Each sum includes both endpoint terms and all middle terms. The pressure derivative, rather than an a priori compact pressure, appears in the last moment. This distinction is retained until compactness has actually been proved.

Y1. Cutoff, five explicit bumps and reconstruction

Cutting off the two freely chosen profiles creates a smooth extension, but it does not yet close the reconstructed radial velocity and pressure. Five disjoint bumps supply the missing degrees of freedom.

Start at n=1 with the actual solution D6.1; for subsequent n use D2.2 with the preceding completed global coefficients. Write its inner solution as \(\phi_n^{\rm in},U_n^{\rm in},V_n^{\rm in},\Pi_n^{\rm in}\). The initial extension is

extended by zero beyond a². Since κ is already zero on a neighborhood of a², all radial derivatives match there.

Choose a closed interval \([R_p,R_q]\) inside \(\{R:R^2/2\in\mathcal I_{\rm pos}\}\). Set \(r_j=R_p+j(R_q-R_p)/6\), \(\omega=(R_q-R_p)/20\), and define

The spacing is \((R_q-R_p)/6\) and the sum of two widths is \((R_q-R_p)/10\), so the five compact supports are disjoint and ordered. Each is smooth and has exactly integral one; the defining integrals \(I_j\) are retained. For coefficients \(\alpha_{n,j}(\eta)\), \(\beta_{n,j}(\eta)\), put

The last quotient is smooth at X=0: every added bump vanishes near that point and the remaining quotient equals \(\widetilde\phi_n\).

Define the other fields by the exact forward integrals

\(F_n\) is the indexed streamfunction coefficient, not the leading angular ratio denoted F in V. The full expression for Ω is D1.6–D1.7; in particular \(\Omega_{n-1}\) contains only already constructed lower orders. D1.7 proves regularity of \(\Omega_{n-1}/x\) at zero. Thus these equations impose original (5.2) and (5.5) globally, with exactly the source's zero axis values.

For \(X\le X_{\rm keep}\) the original inner functions and Y1.3 coincide. Their forward equations and axis values coincide as well, so Y1.4 gives equality of all four inner profiles there. On all of \([0,a^2]\), the cutoff is independent of η and the five bumps vanish. Each fixed radial derivative therefore retains the holomorphic parameter neighborhood obtained in D5; products, finite integrals and the lower source retain it after a finite intersection of neighborhoods. Global correction coefficients need only be smooth in η because they do not enter any of these inner forward integrals.

Y2. The complete affine moment map

At one positive order, the current unknown appears only once in each product. This makes the five equations affine with all earlier orders held fixed. The original patch separates them into two constant invertible matrices.

On the unchanged positive-order patch, V7.4 gives the exact expression

The factor \(2^{1/2+\lambda}\) follows from \(X=R^2/2\) in V7.4; λ here is the fixed profile slope, distinct from \(\lambda_n=2nh\). All lower-order terms remain fixed when solving the current five equations.

The pressure equation in Y1.4 gives, with no dropped middle terms,

Using \(dX=R\,dR\) in the fifth moment gives exactly

For a current perturbation \((\delta U_n,\delta E_n)\) on the patch, the five changes are consequently

For the fourth row \(U_0\delta E_n=0\); for the fifth \(2U_0\delta U_n=0\), whereas the two E endpoints cancel the factor 1/2. No product of two current coefficients occurs: that product has order 2n rather than n. \(\Omega_{n-1}\) is unchanged. This proves affine dependence, for arbitrary finite sizes of the current discrepancies.

Keep the exact constant matrices

Let \(m_{n,j}^0\) denote Y0.3 after setting the five bump coefficients to zero and reconstructing Y1.4. The solution is the explicit pair

These divisions are only the displayed representation of the full rows Y2.4; multiplying back by \(e_*f\), \(2e_*f\) and \(-e_*f\) recovers exactly every original moment, with its sign.

Here is a direct invertibility proof. A nonzero sum of k distinct real powers of R has at most k−1 distinct positive zeros. Substitute \(R=e^t\), divide by the exponential of the least exponent, and differentiate. The derivative has at most k−1 nonzero exponential terms; induction and Rolle's theorem give the assertion, starting with one term. A zero determinant of \((R_j^{s_i})\) at k ordered positive points would contradict that assertion. Hence this determinant never vanishes on the connected region \(R_1<\cdots<R_k\), and its sign is constant there. Multilinearity and Fubini give

The ordered supports make the integrand have one fixed nonzero sign wherever the bump product is positive. The powers in both lists Y2.5 are distinct because λ>0. Thus both actual matrices are invertible, independently of n and η; their numerical values need not be estimated by sampling. The full moment-map inverse also contains the retained row factors. Since f≥1/2 and \(e_*>0\) is fixed, it and each fixed parameter derivative are bounded on [-1,1]. No bound uniform as λ or \(e_*\) tends to zero is claimed.

The discrepancies are smooth: the current \(U_n,E_n\) have compact support, all prior positive-order factors have the common support proved below, and \(\Omega_0\) has compact support because \(U_0,V_0\) do. The moments contain no current \(V_n\) or its derivatives; \(\Omega_{n-1}\) uses only completed lower orders. Thus each integral defining a discrepancy has a fixed bounded integration region with smooth integrand, regular at zero by D1.7. Differentiation under these integrals is legitimate for every fixed derivative. Equation Y2.6 then gives smooth coefficients and proves

Y3. Exact exterior closure and the reserved mean patch

Use the solved moments to prove compact support. The pressure constant disappears only after the third moment is imposed; its support can extend farther than the velocity correction patch.

For \(X\ge X_+\), the current \(U_n\) and \(E_n\) vanish by construction and

D0.4, with \(\mathcal A_XU_n=F_n/X\), gives \(V_n=0\) there, including all parameter derivatives. The leading \(U_0,V_0\) vanish there by \(X_+>X_v\). Each term of \(\Omega_{n-1}\) contains a V field or its derivative; from D1.6 these are \(T V\), \(V_i(V_j'-V_j/(2X))\), \(U_i ZV_j\), \(-2XV''\) and the lower axial viscous term. All therefore vanish in this exterior. Each product \(\phi_i\phi_j\) of positive total order has at least one positive-order factor, also zero. Equation Y1.4 now yields \(\Pi_{n,X}=0\) there. Finally

Both constants of radial integration have been removed by their exact moments, rather than by imposing an additional exterior boundary datum.

At zero,

These formulas are smooth through the axis. The resulting original tuple

has finite constants by smoothness and compactness; they may grow with n and m. The construction does not assert convergence of the unrestricted coefficient series.

The current \(E_n,U_n\) already vanish after the positive-order bump patch. Their total axial moment has vanished too, so Y3.1 and Y3.3 show that \(F_n,V_n\) vanish after that patch. In particular

This does not set \(\Pi_n\) to zero on that earlier patch: \(\Omega_0\) can extend to \(X_v\). Choosing \(X_+\) beyond \(X_v\) was necessary for the pressure-support proof. All original patch and support distinctions are retained.

Y4. Residuals, radial boundary terms and total integrals

Convert the full cylindrical equations to conservative form, keeping every radial boundary term. The five moments then cancel the integrated residuals. At first order the angular viscosity uses a convergent subtracted leading moment.

The actual order-n tangential residuals, retaining every source term, are

Primes mean \(\partial_X\) at fixed η. These are D1.3–D1.4 with the full residual signs; their physical factors are \(q^{-A-1+\lambda_n}\). They vanish on \([0,X_-]\) by the inner equations and Y1.4. The stress is defined exactly as in original (5.9):

Differentiation gives \((\partial_R+2/R)\mathcal T_{n,\theta}=-\mathfrak r_{\theta,n}\) and \((\partial_R+1/R)\mathcal T_{n,z}=-\mathfrak r_{z,n}\). The physical stress factor is \(q^{-A-1/2+\lambda_n}\), so the radial derivative contributes exactly \(q^{-1/2}\), yielding the required physical residual factor. These higher stresses are signed; no positivity is imposed on them.

To prove their total source integrals vanish, use the exact conservative equations for a smooth axisymmetric divergence-free field:

Expanding the transport derivatives and inserting \(\partial_ru_r+u_r/r+\partial_zu_z=0\) recovers the original angular connection term \(u_ru_\theta/r\) and the axial transport. Expanding the two radial viscous derivatives gives exactly \(r^2(u_{\theta,rr}+u_{\theta,r}/r-u_\theta/r^2)\) and \(r(u_{z,rr}+u_{z,r}/r)\). Thus no cylindrical term is omitted.

At the axis each \(u_{\theta,n}\) is r times a smooth function of r², \(u_{z,n}\) is smooth in r², and \(u_{r,n}\) is r times such a function, by D5 and Y3.3. Hence the radial angular transport boundary is \(O(r^4)\), the axial one \(O(r^2)\), and \(r^2\partial_ru_{\theta,n}-ru_{\theta,n}=O(r^4)\), \(r\partial_ru_{z,n}=O(r^2)\). All vanish at zero. At infinity the positive-order fields and their radial derivatives are compactly supported; the order-n transport products each contain such a factor. All radial boundary terms therefore vanish there as well.

At fixed physical z,t, the substitution \(r=\sqrt q R\) gives the exact undifferentiated moments

For the last equality \(\Pi_n\) is now known to be compact and bounded at zero. Integration by parts gives \(\int R\Pi_n dR=[R^2\Pi_n/2]_0^\infty-\frac12\int R^2\partial_R\Pi_n dR\), with both boundary values zero. This use of compact pressure follows Y3, rather than preceding its proof. All product powers depend only on i+j=n. Positive-order integrals and all their fixed physical derivatives have locally bounded compact support for t<1, which justifies their differentiation.

For axial viscosity at n≥2, the analogous order-(n−1) angular and axial moments vanish by Y2.8. At n=1 the leading axial moment vanishes by V7 and P2. The leading angular moment requires its actual subtracted form, proved in H4.6 and preserved by the exact joining and shear moments:

Near zero the subtracted power has integrable weight \(r^{1-2h}\); the actual field is regular. At infinity H0.2–H0.4 give \(K-P=O(r^{-3-2h})\) for fixed t<1, so its weighted difference is integrable because h>0. On any compact neighborhood of physical (z,t) with t<1, the exterior starts before one finite radius and \(u_{\theta,0}=K(r,t)\) is z-independent beyond it. The same is true of P. Thus the first and second z derivatives of the difference have compact exterior support and are locally bounded at the axis; differentiation under Y4.5 is justified. It proves the vanishing order-one axial-viscosity angular moment without differentiating a divergent un-subtracted integral.

Integrating Y4.3 at order n, the first two moments in Y4.4 cancel the angular time and axial transport terms, and the last two cancel the axial time and flux terms. The preceding paragraph cancels both lower axial-viscosity moments. Removing the nonzero physical factors gives

Explicitly the physical weighted factors are \(q^{1/2-A+\lambda_n}\) and \(q^{-A+\lambda_n}\), respectively. These follow from the residual factor and \(r^2dr=q^{3/2}R^2dR\) or \(r\,dr=qR\,dR\). Time differentiation loses one power of q, axial differentiation loses D, and the order-(n−1) second axial derivative uses \(\lambda_n-\lambda_{n-1}=2h=1-2D\); each term in the respective integrated equation has exactly that same power.

Y5. Full stress support and the exceptional first-order term

The backward stress integral identifies the exact support. All later orders stop before the outer collar. The first order still sees the axial derivative of the leading heat multiplier, so its full flat factor must be calculated.

Equation Y4.6 converts the exact forward integrals into backward ones:

For n≥2, every same-order exterior residual term has a positive-order factor, and the axial-viscosity term acts on order n−1≥1. All are zero beyond \(X_+\) by Y3. On the inner interval the residuals vanish and the forward integrals are zero. Consequently

At n=1 the only possible exterior term is the angular \(-\partial_{zz}u_{\theta,0}\); the leading axial field is already zero there. For \(X\ge X_b\), \(u_{\theta,0}=K(r,t)\) is independent of z. Thus

Retain the full original terminal multiplier. In \(y=\log(X/X_{\rm tail})\), with \(X_b=e^3X_{\rm tail}\) and \(\delta_b=3-y\), it is

This is O0 and H6.3, including its denominator and amplitude. On a sufficiently small closed terminal collar, g is positive and smooth, with every fixed derivative bounded. The actual physical field there is

The heat factor and all of its needed derivatives are the convergent integrals H0.1–H0.2, including the value at d=0. The factor 2 in its argument is retained.

At fixed physical r,t the coordinate identities P0 give

K has no z dependence. Differentiating \(q^{-D}a_z\) gives \(q^{-2D}b_z\), which proves the last formula and retains its exponent derivative. Both \(a_z\) and \(b_z\) and their fixed derivatives are bounded because \(L\ge1-2h>0\). For \(\delta=\delta_b\) and g as in Y5.4 the exact multiplier derivatives are

Hence the first is bounded by \(C e^{-4/\delta^2}\delta^{-3}\) and the second by \(C e^{-4/\delta^2}\delta^{-6}\) on a short collar. After multiplying the physical source by \(q^{A+1-\lambda_1}\), the power cancels exactly: \(\lambda_1=2h\) and \(1-\lambda_1-2D=0\). Thus its coefficient is the bounded smooth heat factor from Y5.5 times the two full expressions in Y5.6–Y5.7; no q-dependent constant remains.

For completeness, the radial integration gains exactly the three powers needed at this edge. For any smooth bounded \(B(s,\eta)\) on a small collar and fixed j≥0, the substitution \(s=\delta(1+\delta^2v)^{-1/2}\) gives

Its Jacobian is \(-\delta^3(1+\delta^2v)^{-3/2}/2\). Each fixed derivative of the final integrand is bounded by \(e^{-4v}\) times a polynomial in v, uniformly on the compact parameter rectangle. Dominated differentiation proves that the last integral is smooth through δ=0, with every fixed derivative finite. This also proves its endpoint value \(B(0,\eta)/4\).

In the backward angular integral Y5.1, \(R=\sqrt{2X_b}e^{-\delta/2}\) and \(dR=-R\,d\delta/2\); all these weights, including the exterior \(R^{-2}\), are smooth and bounded on the fixed collar. Applying Y5.8 to the finite terms of Y5.7, with j≤6, yields the actual smooth representation and bounds

Here \(B_1\) is smooth; no nonzero endpoint value or sign for it is required. Terms with smaller j acquire the nonnegative powers \(\delta^{6-j}\) in \(B_1\). Each radial derivative of the exponential or the factor \(\delta^{-3}\) introduces only finitely many additional inverse powers; \(\partial_X\delta=-1/X\) and its derivatives are bounded on this collar. Thus every derivative tends to zero at \(X_b\) and the stress extends smoothly by zero there.

Y6. Original flat weight, all fixed derivatives and viscosity

Combine the actual support and endpoint calculation with the same weight used by the leading profile. Restore the original viscosity in both the physical stress and the residual.

Retain V6's actual common weight and the source distance:

For n≥2, Y5.2 puts every stress and derivative in a fixed compact subinterval of \((X_a,X_b)\), where ζ has a positive minimum. Smoothness then gives Y6.2 below with \(N_{n,k,\ell}=0\). For n=1, the stress is zero on \([0,X_-]\), while near \(X_b\) the inner factor \(\exp(-t_1^2/\log(X/X_a)^2)\) and its reciprocal are smooth and bounded. Equation Y5.9 supplies Y6.2 on that outer collar. On the remaining compact middle interval, smoothness and the positive minimum of ζ supply it as well. Therefore

This proves original (5.13) for every fixed derivative; all constants may depend on n,k,ℓ and the already fixed profile parameters. No simultaneous derivative smallness or uniform bound over all orders has been used.

For the original viscosity ν>0 retain D6.2 and the stress map

The last expression is evaluated at \(X=r_\nu^2/(2\nu q),\eta=z_\nu/(\sqrt\nu q^D)\), in the original cylindrical components. Spatial differentiation contributes \(\nu^{-1/2}\); hence divergence of this stress and every term of the momentum residual have the same factor \(\sqrt\nu\). Incompressibility is preserved by the same exact derivative map. The five profile moments and all fixed profile bounds above are unchanged statements in their original variables, with these physical factors explicitly restored. This does not change the target viscosity.

Y7. Actual induction and its precise endpoint

Start with the actual first coefficient, complete its global extension, and only then form the source for the next coefficient. The construction uses the same radii, cutoff and bumps at every order.

At n=1, D6 supplies the actual inner solution using only the already constructed leading profile. Equations Y1.1–Y1.4 define its tentative global extension; Y2.6 supplies all five actual smooth correction functions without a small-discrepancy requirement. Y3–Y6 prove its complete support, moments, retained analytic interval and stress bounds, including its exceptional outer heat term.

Suppose the construction has completed the finite orders 1 through n−1. Their full sources in D2.2 are then defined and smooth; D1.7 proves axis regularity, and Y1 preserves their fixed-derivative holomorphic parameter domains on \([0,a^2]\). The D3–D5 series therefore constructs the actual next inner coefficient on that same interval. Apply the explicit Y1 and Y2 formulas to it. Their lower sources are the completed fields, and all integrals are finite by the support argument in Y2–Y3. The constant matrices are the same as before. Y3–Y6 then prove exactly the properties needed at order n. This induction constructs a single sequence at every positive integer order, with one set of radii, one cutoff and one set of bumps.

Every finite order now satisfies original (5.2) and (5.5) globally, the full equations (5.3)–(5.4) on the common inner interval, all five moment identities, the tuple support (5.12), the stress support and derivative estimate (5.13), the profile bounds (5.17), and the reserved-mean-patch equalities (5.18). The coefficient values are given by the convergent integral solution followed by finite explicit integrals and the two fixed matrix inverses. No coefficient or missing source has been assumed into existence.

Y8. Five worked exercises

Exercise 1. Prove determinant signs and explicit lower bounds for the actual bump matrices

Use the five smooth bumps in Y1.2, with their original integrals \(I_j\), to give positive lower bounds for \(|\det B_U|\) and \(\det B_E\). Retain every power of the patch endpoints and the profile slope.

Solution. Put \(H=R_q-R_p>0\) and \(g=H/15\). Consecutive support centers are separated by \(H/6\); the sum of their two widths is \(H/10\). Therefore any points in consecutive bump supports differ by at least \(g\). Every point lies between \(R_p\) and \(R_q\). Each actual density is \(b_j=w_j/I_j\) and has integral one.

For the two axial columns at \(r_1<r_2\), retain the exact determinant

The integrand is positive. Since \(\lambda>0\), its value is at least \(R_q^{-2\lambda-1}\), while \(r_1r_2\ge R_p^2\). Integrate the determinant against the two actual bump densities as in Y2.7. Their product integral is one. This proves

For the three angular columns use points \(r_1<r_2<r_3\) in their ordered supports, and put \(x_j=r_j^2\) only for this determinant calculation. Factoring the indicated powers from each column, without discarding them, gives

The cyclic row permutation has positive sign. Subtracting the first column from the other two proves the divided-difference formula displayed here. Its last factor is at least half the minimum of \(F''\) on the interval: if \(c=\min F''\), then \(F-cx^2/2\) is convex, its second divided difference is nonnegative by increasing secant slopes, and that of \(cx^2/2\) is \(c/2\). In this case

The two consecutive differences in \(x\) are at least \(2R_pg\), and the remaining difference is at least \(4R_pg\). The column factor is at least \(R_q^{-6-6\lambda}\). Integrating Y8.3 against all three actual densities therefore proves

These bounds prove invertibility for the finite-width smooth bumps themselves. They use neither point-mass replacement nor numerical determinant sampling. The factor \(\lambda\) in Y8.2 is retained; no bound uniform as \(\lambda\) tends to zero is asserted.

Exercise 2. Bound the full moment correction in every fixed parameter norm

Prove an explicit \(C^m([-1,1])\) bound for the five actual correction functions. Retain the row factors \(e_*f\), \(2e_*f\), and \(-e_*f\).

Solution. For a vector function \(v\), use \(\|v\|_{C^m}=\max_{0\le j\le m,\,|\eta|\le1}|v^{(j)}(\eta)|_\infty\), and use the induced maximum matrix norm. Since the original \(f=(1+\eta^2)^{-1}\), Y2.6 is exactly

The exact product rule, with terms of negative derivative order omitted, is

Thus \(P_m=m^2+m+2\) bounds the multiplier in every derivative through \(m\). Keeping the three rows visible gives

These are finite constants for the fixed original patch. The inverse norms can also be bounded explicitly by the adjugate formula, Y8.2 and Y8.5, since every entry of Y2.5 is bounded by the supremum of its retained power on \([R_p,R_q]\).

For example, with the exact bump derivatives

all of which are finite because their supports stay away from the axis, the actual added profiles satisfy

For \(\phi_n\), replace the angular bump in \(M_{E,k}\) by its full factor \(\mathsf C b_j^E(\sqrt{2X})/\sqrt{2X}\). Its support still stays away from zero. For two discrepancy vectors with the lower profiles fixed, subtract Y8.6: the same bounds hold for their differences. This proves a bounded linear correction map on each stated norm, with the actual row factors and supports. The discrepancies can be arbitrarily large and finite; no smallness is needed at this coefficient order.

Exercise 3. Identify the exterior defects that the five moments remove

Before imposing the moments, calculate the exterior radial velocity and pressure. After the first and third moments are zero, express both total tangential residuals in terms of the remaining moments, and recover their physical stress tails.

Solution. Keep the lower global coefficients fixed, and take a tentative smooth current \(U_n,E_n\) supported to the left of \(X_+\), with \(V_n,\Pi_n\) defined by Y1.4. The lower fields already have the support proved in the induction. The moments \(m_{n,j}\) are those of this tentative extension.

For \(X\ge X_+\), the forward formulas give exactly

Indeed \(U_n=0\), \(\mathcal A_XU_n=m_{n,1}/X\), and D0.4 gives the middle formula. The pressure derivative vanishes there because all its products have a positive-order angular factor and \(\Omega_{n-1}\) contains only the completed lower radial fields. Its axis value is zero, so its exterior constant is the third moment. Consequently the full tuple \(\mathbf P_n\) has zero exterior precisely when \(m_{n,1}=m_{n,3}=0\). Necessity uses \(F_n/X\) and \(\Pi_n\), not a claim that the equation \(V_n=0\) alone forces \(m_{n,1}=0\).

Now impose those two moments, so all current velocity and pressure fields are compactly supported. Define the original operators on functions of \(\eta\) alone,

Write \(I_{\theta,n}=\int_0^\infty R^2\mathfrak r_{\theta,n}\,dR\) and \(I_{z,n}=\int_0^\infty R\mathfrak r_{z,n}\,dR\). Apply the exact conservative equations Y4.3 and the undifferentiated moments Y4.4. The lower axial-viscosity moments are zero, including the subtracted first-order angular moment Y4.5. Hence

The angular common power is \(q^{1/2-A+\lambda_n}\); the axial common power is \(q^{-A+\lambda_n}\). In the flux terms these follow from \(A+D=1\). The second axial derivative from order \(n-1\) uses \(2h=1-2D\), exactly as in Y4.6.

Let \(X_e=X_+\) for \(n\ge2\), and \(X_e=X_b\) for \(n=1\). Beyond that radius the residual sources vanish, while their forward stress integrals need not. Their full tails are

This follows directly from Y4.2, including its negative sign. In the original viscosity variables \(R=r_\nu/\sqrt{\nu q}\), Y6.3 gives

The first and third moments remove the radial integration defects; the other three make Y8.13 zero and hence remove both stress tails. This proves the exact map from the five original moments to the exterior defects. It does not claim that five zero moments are necessary for the two differential combinations in Y8.13 alone to vanish.

Exercise 4. Compute the actual first-order outer stress coefficient

Find the endpoint value of \(B_1\) in Y5.9, retaining the full heat factor. At \(\eta=0\), determine the sharper leading term that survives when this endpoint value vanishes.

Solution. Set

These are exactly the heat factors in Y5.5, with every constant and argument retained. Put \(\delta=\log(X_b/X)\). The coefficient of the physical first-order residual on the terminal collar is

The physical power cancels because \(1-\lambda_1-2D=0\). Substituting the exact multiplier derivatives Y5.7 gives the full expression

Here \(a_z,b_z\) retain their full functions of \(\eta\) from Y5.6. The backward integral is exactly

Apply Y5.8 term by term. The coefficient of \(s^{-6}\) in the integrand, including its factor \(1/2\), is \(32\rho_o R(s)^3\mathcal E(s,\eta)a_z^2g(s)\). Y5.8 divides its endpoint value by \(8\); the outside \(R(\delta)^{-2}\) is retained. Every other term acquires a strictly positive power of \(\delta\) after extracting \(e^{-4/\delta^2}\delta^{-3}\). Thus the actual smooth coefficient obeys

It is positive for every fixed \(\eta\ne0\), including the one-sided parameter endpoints, and zero at \(\eta=0\). No factor from the heat profile or radial measure has been absorbed.

At \(\eta=0\), the exact values are \(a_z=0\), \(b_z=-2\). Y8.18 becomes \(2\rho_o\mathcal E(\delta,0)e^{-4/\delta^2}(8\delta^{-3}g+g')\). The same integral identity, now with \(j=3\) and \(j=0\), proves

The \(j=0\) term contributes three additional powers and has zero endpoint coefficient. This explains the exact improvement at the parameter midpoint without discarding the first-order outer term.

Exercise 5. Prove sharp derivative powers and the exact comparison with the leading stress

Give explicit powers in Y6.2 for \(n=1\), and compare the first correction with the original leading angular stress at the outer edge. Prove the resulting endpoint statement for every actual finite coefficient stress sum.

Solution. The representation in Y5.9 is an actual smooth factorization: \(\mathcal T_{1,\theta}=e^{-4/\delta^2}\delta^{-3}B_1(\delta,\eta)\). All fixed derivatives of \(B_1\) are bounded on a common closed collar. Since \(\partial_X\delta=-1/X\),

Each radial derivative falling on the exponential therefore costs at most three inverse powers; derivatives of a remaining power cost one, and all derivatives of \(X^{-1}\) and the smooth factors are bounded. Parameter derivatives act only on the smooth coefficient. Induction and the product rule prove

Near the outer edge, the inner exponential factor in \(\zeta\) and its reciprocal are bounded. On the remaining compact middle interval use smoothness and a positive minimum of \(\zeta\); near the inner edge the stress is zero. This proves the whole-annulus estimate with the original weight.

For fixed \(\eta\ne0\), the unique most singular derivative term is obtained by putting every radial derivative on the exponential. Every other product-rule term has a smaller inverse power. Hence

Thus the radial exponent \(3k+3\) cannot be reduced uniformly over parameters containing such a point. At \(\eta=0\), Y8.21 instead gives

This midpoint refinement concerns radial derivatives at that fixed parameter; it does not assert the same power for arbitrary parameter derivatives there.

The complete leading stress, unchanged by the joining and shear moments, satisfies H6.6–H6.8:

Both coefficients are smooth on a closed collar and \(b_\theta\) has a positive lower bound. Therefore the exact quotient of the original stresses extends smoothly, with

These are exact comparisons; Y8.20 and Y8.26 retain the full factors before taking their quotient. In particular \(\mathcal T_{1,\theta}/\zeta\) is unbounded like \(\delta^{-3}\) at every fixed nonzero \(\eta\), even though its ratio to the leading angular stress has a finite limit. The different weights must not be substituted for each other.

For any fixed \(q>0\) and integer \(N\ge1\), use the actual finite stress

On \(X>X_+\), every \(n\ge2\) term is identically zero and \(\mathcal T_{1,z}=0\). Thus throughout the outer collar

For fixed \(q\), compactness gives a common smaller collar where the angular coefficient is positive. The exact direction is

It extends smoothly to the endpoint. The original physical stress is \(\nu q^{-A-1/2}\widehat T^{[N]}\), evaluated at the original viscosity coordinates of Y6.3. Its positive scalar factor preserves this direction and retains every physical scale. These finite sums use the already constructed coefficients. Their convergence without cutoffs is not asserted; the subsequent base construction supplies the actual infinite sum.

What the global coefficient construction supplies

Y0–Y7 construct the entire actual coefficient sequence, with one fixed radial geometry, the full original equations on the protected axis interval, all five exact moments, compact coefficient support and the two required stress supports. The first-order outer term is retained with the same flat weight as the leading profile. Y8 proves quantitative matrix bounds, the exact exterior-defect map, sharp first-order edge powers and its smooth comparison with the leading stress.

The next lesson constructs the actual base field from this sequence, using cutoffs of the full vector potentials and both stress components. Its complete proof is supplied in the separate base-field audit, Z0–Z8. The actual base-field lesson, Z0–Z9, gives that construction and its consequences.

Authored lesson text and figure source: CC0.

Constructing the actual base field

NS-BLOWUP N3.3. The preceding lessons construct every coefficient. This lesson turns that sequence into an actual smooth, divergence-free velocity and its pressure and stress, while keeping the entire coefficient expansion. The remaining augmented momentum residual decays faster than every power of the original scale, with every fixed physical derivative.

The local coefficient lesson, D0–D7, and the global extension lesson, Y0–Y8, supply the actual coefficients. The leading-profile lesson, V0–V8, supplies their leading field and common stress weight. The general summation lesson gives the differential-polynomial argument; here every input is constructed for the actual base, and the full residual is written out.

The source is OpenAI, Finite Time Blowup for Navier–Stokes, pages 55–62, (5.23)–(5.46), Propositions 5.3 and 5.5 and Lemma 5.4. SOURCES.json gives the bibliographical details. This is independent teaching and editorial mathematics; the diplomatic transcription keeps OpenAI's text.

Figure. Here \(\mathcal U_B=(u_B,p_B,T_{\rm phys})\). The displayed operators and factors are the ones proved below. The final panel records the exact axis residual of the actual base, including its first cutoff. This is an operator diagram, with no numerical flow sample.

Z0. Exact Cartesian fields and the derivative loss

Begin with the original physical components. Their exact derivative powers determine how long a finite sum must be before it can control a prescribed residual derivative.

Keep the original \(q,A,D,h,\eta,X,\lambda_n=2nh\), \(R=\sqrt{2X}\) and \(\mathsf C\) from D. Define only the explicit coordinate map \(\xi=(x_1,x_2)/\sqrt q\); then \(X=|\xi|^2/2\) and every original profile below is evaluated at that same X and η. The original velocity coefficients, with every radial and angular factor, are

The factor 1/2 in the radial row follows from \(V_n/R^2=V_n/(2X)\). Y3.3 proves the smooth Cartesian representatives at the axis. On every compact profile range, all fixed mixed derivatives of these representatives are bounded by the already constructed coefficient bounds; their constants can depend on n.

On a function of \((q,\xi,\eta)\), the exact physical differentiation operators are

They follow by differentiating \(q(1-\eta^2)=1-t\) and \(q^D\eta=z\), then \(\xi=x_\perp/\sqrt q\). In particular the q and η derivatives are exactly those in P0, and the ξ terms follow from the derivative of \(q^{-1/2}\).

For a profile g(ξ,η), each time application of Z0.2 subtracts one from the q exponent, each axial application subtracts D, and each transverse application subtracts 1/2. Differentiating any coefficient involving L,η,ξ leaves bounded smooth coefficients on a compact profile range, since \(L\ge1-2h>0\). After m derivatives there are finitely many terms; each exponent differentiation contributes a factor β−c with |c|≤m. Induction on m therefore proves

Every norm here is on the specified compact profile range. Constants depend on that range and h, but the exponent loss does not depend on β or the coefficient order. Powers of \(\lambda_n\) remain in the finite coefficient constants. No coordinate, factor or field is replaced by its largest-order estimate.

Z1. All terms in the finite augmented residual

A finite sum is an actual smooth field. List its residual directly, including products beyond the truncation and both shifts of radial viscosity; then bound the resulting finite expression.

Use the original finite tuple

Here \(\mathscr R=\partial_tu-\Delta u+(u\cdot\nabla)u+\nabla p\) is the source's coefficient-one operator; the exact map back to the original ν is D6.2/Y6.3. All finite velocities are divergence-free by the independently proved order-by-order identity D0.4. Y4.1–Y4.2 cancel the two complete tangential residual coefficients at every retained positive order; the leading ones cancel by P1 and V7. The radial pressure relation D1.5 cancels the retained radial coefficients as well.

The unretained terms can be listed without an appeal to a formal infinite series. Products of two retained fields have \(i,j\le N\); tangential products have total order i+j, and their omitted terms have \(i+j>N\). Tangential axial viscosity acting on a retained order j has order j+1, so only its unretained last index \(j=N\) remains. Time, transverse viscosity and pressure differentiation otherwise retain their index in the tangential equations. In particular the exact tangential components of the remaining augmented residual are

Primes mean \(\partial_X\). All operators and profiles are the full ones in D0–D1, and \(\lambda_{i+j}=\lambda_i+\lambda_j\); the index does not denote an unconstructed product profile.

For the radial equation multiplied by r, retain the two distinct physical factors from D1.5–D1.6:

The second factor has radial coefficient order k+1 because \(-1+\lambda_k=-2A+\lambda_{k+1}\). Thus radial time and transverse viscosity at retained index j enter order j+1; radial transport products at i+j=k enter order k+1. The radial axial-viscosity term acting on j enters order j+2, because \(-2D+\lambda_j=-2A+\lambda_{j+2}\). This last shift retains the full axial viscosity of \(V_N\): it is not part of the tangential j+1 rule. The pressure/centrifugal difference has order n=i+j in its products. Every coefficient of order at most N cancels by the equations already proved, including all middle terms. All remaining orders are at least N+1 and there are only finitely many of them; radial orders can reach max(2N+1,N+2).

More explicitly the exact remaining radial component, multiplied by r, is

For N=0 the last sum has j=0; for N≥1 it has precisely \(j=N-1,N\). This includes the entire final axial-viscosity contribution. The stress pair has only tangential divergence, so it adds no radial term to this identity.

After dividing the radial equation by r, the common radial power is \(q^{-2A-1/2+\lambda_n}\); the tangential power is \(q^{-A-1+\lambda_n}\). Division by R in the radial coefficient is regular: \(\Omega_k=X\) times the smooth function D1.7, while the pressure/centrifugal bracket has factor X. Multiplication by \(e_r\) then gives \(R e_r=(\xi_1,\xi_2,0)\), with its retained numerical factors. The angular bracket in Y4.1 has its explicit R factor and hence a smooth Cartesian angular vector. The only cylindrical factors in the stress divergence occur on \([X_a,X_b]\), away from X=0, where \(R^{-1}\) and every fixed derivative are bounded. Consequently all remaining coefficient functions are smooth Cartesian functions with finite bounds on each fixed compact profile range.

Since 0<q≤1 and 2A+1/2>A+1 for h>0, the exact powers above and Z0.3 give the concrete estimate

Here \(|\cdot|_m\) is the original maximum of Cartesian space-time derivatives through order m. This estimate retains every term of Z1.2–Z1.4. Its exponent loss is independent of N, and its constant may depend on N,m and the compact profile range. It proves the finite-residual assertion of original (5.25), with an explicit allowable loss. There is no additional finite-stage remainder in this application.

Z2. The actual Stokes potential and its full cutoff curl

A scalar cutoff of a divergence-free velocity usually has nonzero divergence. Cutting its vector potential supplies the precise radial term that restores the equation.

For positive order use the exact \(F_n\) from Y1.4 and \(s=r^2/2=qX\). The physical Stokes function and vector potential are

These equalities retain the factor 1/2 and both coordinate representations. The actual potential is smooth at the Cartesian axis because \(F_n\)/X is smooth. Direct curl in cylindrical coordinates gives

The last identity follows from the full expression D0.4 and \(F_n=X\mathcal A_XU_n\); it includes the derivative of \(q^{1-A+\lambda_n}\). Thus the potential is proved to give the original radial and axial fields.

Let χ be smooth, equal to one on [0,1/2] and zero on [1,∞), and let \(c_n>0\). Cutting the potential before taking curl gives the exact fields

Indeed q is independent of r, while \(q_z=2\eta q^{1-D}/L\); multiplying \(c_n\chi' q_z\) by \(q^{1-A+\lambda_n}F_n\) produces \(q^{\lambda_n}(2\eta/L)(c_nq)\chi'F_n\) because A+D=1. The radial sign is negative from \(-\partial_z S_n\). Curl has zero divergence by commutation of Cartesian mixed derivatives, and the θ-independent angular field has zero divergence. This proves incompressibility through every cutoff transition, not merely where χ is constant.

The five coefficient entries \((\mathcal A_n,u_{\theta,n}e_\theta,p_n,\mathcal T_{n,\theta}^{\rm phys},\mathcal T_{n,z}^{\rm phys})\) have respectively the exact powers \(q^{-h+\lambda_n},q^{-A+\lambda_n},q^{-2A+\lambda_n},q^{-A-1/2+\lambda_n},q^{-A-1/2+\lambda_n}\) in their smooth Cartesian or annular profile representations. Since D<1/2, Z0.3 bounds every derivative through m by \(C_{n,m}q^{\lambda_n-(2A+m)}\). Thus the source's admissible common loss \(\ell_m=2A+m\) follows from the actual coefficient fields. The stress entries, pressure and potential are all included; no bound uniform in n is asserted.

Z3. One actual cutoff sequence for every required derivative

Choose one cutoff sequence from finite norms of already constructed coefficients. Each successive choice meets a finite list of derivative requirements, and their union covers every fixed derivative.

Choose the exact fixed step \(\chi(s)=1-\sigma(2s-1)\), with σ from O0.1. It equals one for s≤1/2 and zero for s≥1, is smooth and flat at the transition endpoints, and has values between zero and one. For a positive-order z,t derivative of χ(cq), repeated chain rule gives a finite sum of terms

Z0.2, applied repeatedly to q, gives \(|\partial_z^{a_j}\partial_t^{b_j}q|\le Cq^{1-Da_j-b_j}\). On the support of the derivative, 1/2≤cq≤1, so the factors \(c^kq^k\) are bounded independently of c. Consequently

The undifferentiated cutoff is bounded by one. These bounds hold on the whole domain because the positive derivatives vanish outside the transition.

For the full cut tuple let

All potential, direct-field, pressure and stress terms remain. A derivative through order m of its curl uses at most m+1 derivatives in total. A term with k derivatives on χ costs at most \(q^{-k}\), and the remaining at most m+1−k derivatives on the potential cost \(q^{\lambda_n-2A-(m+1-k)}\) by Z2. The full product rule therefore gives

with constants independent of c. The direct field, pressure and both stress entries obey the same bound, using fewer derivatives. All positive-order profiles have a common compact radial support, so these constants can be taken on that entire support, including its smooth zero extension. In particular the first-order stress is included by Y5.9, whose exponential beats every inverse power at its endpoint.

The additional estimates use exactly the source operators \(\mathscr D=(q\partial_q,\partial_X,\partial_\eta)\). For a multi-index I, let \(\mathscr D^I\) denote their indicated composition. They commute on functions of the independent (q,X,η). The identity

gives a uniform bound for every fixed j; the same holds with \(\chi(s)\) replaced by \(s\chi'(s)\). Each application to \(q^{2nh}\) contributes precisely 2nh. All \(\partial_X\) and \(\partial_\eta\) derivatives of the actual profiles on a fixed compact annulus are finite. Thus, for every fixed annulus, the quantities

have all \(\mathscr D^I\) bounds \(C_{n,I}q^{2nh}\) independent of c. In the last row Y5.2 places the stress in the interior \([X_-,X_+]\), so its quotient by ζ is smooth with bounded derivatives on the entire active annulus, extended by zero near both edges. This quotient estimate is not applied to n=1.

Here is one explicit sequence meeting every requirement. Set \(X_*=\max\{1,X_b\}\) and \(\mathcal K_k=[X_a/(k+1),(k+1)X_*]\times[-1,1]\), k≥1. These exhaust the positive radial coordinate and contain the active annulus. For each n, take \(M_n\ge1\) to be the maximum of the finitely many constants in Z3.4 for 0≤m≤n and in Z3.6 for |I|≤n and 1≤k≤n. Include the full bracket, not just its first term, and the entire stress quotient when n≥2. Each constant is defined by the finite product-rule bounds just proved for the already constructed profiles and the fixed χ; no unknown future coefficient enters. Define \(c_0=1\) and recursively

All \(M_n\) are finite and h>0 was fixed in the actual leading construction. Hence this gives one finite positive \(c_n\) at every order and \(c_n\ge2c_{n-1}\). For \(0<q\le1/c_n\) it implies \(M_nq^{nh}\le2^{-n}\). On larger q the n-th summand and all its derivatives vanish. Consequently

where \(B_n^{\rm sim}\) means each of the complete expressions in Z3.6 on \(\mathcal K_k\), including the full cutoff bracket and both stress quotients. The q derivative factors and the coefficients 2η/L are included in \(M_n\). This is an exact diagonal construction from finite norms; it requires no numerical estimate of the enormous possible cutoff sizes and no hypothesis of uniform coefficient growth.

Z4. Actual fields, exact divergence and their finite comparison

The chosen sequence now defines actual fields. Local finiteness proves smoothness away from the singular scale; the quantitative tail estimate controls the approach to that scale.

Define the fields by the actual locally finite series

On any compact set with \(q\ge q_*>0\), \(c_nq\ge1\) for all sufficiently large n. Thus only finitely many terms and their derivatives occur locally. Every term is smooth in Cartesian coordinates at the axis by Z0–Z2, or supported away from it for the stress. Z2.3 proves exact incompressibility, including every cutoff term, and local finiteness preserves it in the sum. All added terms extend smoothly by zero for \(q\ge1/c_1\) because χ is flat at its outer endpoint.

Every positive-order velocity and pressure term is zero when \(X\ge X_+\): its potential contains \(F_n\), its cutoff curl contains only \(F_n\) and \(V_n\), and all of these, as well as \(U_n,E_n\),\(\Pi_n\), vanish there by Y3. The stress has support \([X_a,X_b]\) because its leading and higher components do, and multiplication by χ(\(c_n\)q) does not alter the radial profile support. No conclusion that the first-order stress vanishes at \(X_+\) is used.

Fix a derivative order m. For \(J\ge\max\{1,m\}\) and \(0<q<(2c_J)^{-1}\), all first J cutoffs equal one on a neighborhood, including all their vanishing derivatives. The original tuple \(\mathcal U^{[J]}\), with its uncut potentials differentiated, is therefore the first J part of Z4.1. By Z3.8,

The last inequality uses q≤1 and the exact sum \(\sum_{n>J}2^{-n}=2^{-J}\). The common derivative loss is independent of J.

Z5. The complete augmented residual is flat

The nonlinear residual cannot be interchanged with an uncontrolled infinite series. Compare the actual field with one fixed finite sum and estimate every term of their exact difference.

For one fixed finite state set \(v=u^{[J]}\), \(p_*=p^{[J]}\), \(T_*=T^{[J]}\), and \(e=u_B-v\), \(\pi=p_B-p_*\), \(\tau=T_{\rm phys}-T_*\). Subtracting the full residual gives the exact identity

The last line contains both original tangential stress divergences. It is supported in the fixed active annulus. Write \(\partial_r=e_r\cdot\nabla_x\) there. A Cartesian derivative of order j of \(e_r,e_\theta\) is bounded by \(C_jr^{-j}\), and of 1/r by \(C_jr^{-1-j}\); their t,z derivatives vanish. Since \(r=\sqrt q R\) and \(\sqrt{2X_a}\le R\le\sqrt{2X_b}\), the product rule yields

Terms with fewer frame derivatives obey the same bound because 0<q≤1. All coefficients and supports are the original ones; no 1/r estimate is used at the axis.

For the three nonlinear terms in Z5.1, every differentiated component is a finite sum \(\partial^I(v_k\partial_ke_i)=\sum_{J'\le I}\binom I{J'}(\partial^{J'}v_k)(\partial^{I-J'}\partial_ke_i)\), and the other two products have their respective displayed factors. This proves the full estimate

In particular the quadratic new-new term and the pressure term have not been discarded.

On a fixed compact profile range, Z0.1–Z0.3 and the actual leading profile give, for every finite J,

because all coefficient increments have factor \(q^{2nh}\)≤1 and the pressure is the most singular of the undifferentiated entries. The constants may grow with J; \(K_r\) does not. The leading stress has its already proved smooth zero extension, and its support is away from the axis.

Given physical derivative order m and desired decay order M≥0, put r=m+2 and choose one finite integer J≥max{1,r} such that

Both inequalities hold for a sufficiently large finite J because h>0 and every displayed loss is independent of J. For example the ceiling of the maximum of r, \((M+\max\{K_r,H_m\}+\ell'_r)/h-1\), and \((M+2A+1/2+m)/(2h)-1\), enlarged to at least one, suffices. Keep this J fixed and restrict \(q<(2c_J)^{-1}\). Equation Z4.2 bounds e,π,τ through order r by \(2^{-J}q^\alpha\); Z5.4 bounds v by \(C_{J,r}q^{-K_r}\). The five terms in Z5.3 have powers at least α, α, \(\alpha-K_r\), 2α and \(\alpha-H_m\), all at least M by Z5.5. The finite residual Z1.5 also has power at least M. Therefore the actual field satisfies

on the chosen compact profile range as q↓0, for every m and M. The constants and neighborhood may depend on that range and on m,M. The order of choices is (m,M), then one finite J, then its q neighborhood. This uses no infinite sum of residual remainders. It proves the actual application of Lemma 5.4 to the base, including its stress variable, rather than importing unverified finite-state data.

Z6. Every original velocity and weighted stress comparison

Keep the original factors multiplying each velocity, pressure and stress component. The first stress coefficient reaches the outer endpoint and must retain its full weight.

Fix any \(0<X_{\rm lo}<X_a<X_b<X_{\rm hi}<\infty\). Choose k so its closed profile rectangle lies in \(\mathcal K_k\). For a fixed |I|=m split the positive series into a finite initial block and a tail starting at \(n_0\ge\max\{2,k,m\}\). The finite terms in Z3.6 and their \(\mathscr D^I\) derivatives are \(O(q^{2h})\), since each has its original \(q^{2nh}\) factor and the fixed cutoff derivatives are bounded. The tail is at most

Use this for each complete expression in Z3.6 and then use the exact fields Z2.3. It proves all of the following, with their original factors:

For the fourth row, the exact identity is \(q^Au_{r,B}=q^h[V_0/\sqrt{2X}+O(q^{2h})]\); the leading term is retained. The positive-order radial contribution with that same \(q^A\) factor is \(O(q^{3h})\). All derivatives of the extra cutoff term in Z2.3 are included in the third and fourth rows. The pressure statement follows from the same included coefficient row; \(\Pi_0\) here is the full leading pressure, as in D0.

For the original stress set \(\widehat T=q^{A+1/2}T_{\rm phys}\). The exact difference is

For n≥2 the quotients and all derivatives are bounded on the full active annulus by Y6, and are included in Z3.6–Z3.8. Their series and every \(\mathscr D^I\) derivative are \(O(q^{2h})\) by Z6.1 and a finite initial block. For each fixed j, the full exponential ζ of Y6.1 satisfies \(|\partial_X^j\zeta|\le C_j\zeta\delta^{-M_j}\): differentiate both exponential factors, using the bounded derivatives of \(\log(X/X_a)\), \(\log(X_b/X)\) on the fixed annulus, and retain the finite resulting inverse powers. The product rule therefore bounds the second term in Z6.3 by \(O_I(q^{2h}\zeta\delta^{-N_I})\). The first term obeys the same bound using the actual first-order stress estimate Y5.9–Y6.2 and the fixed cutoff derivatives. Taking the maximum over the finitely many product-rule exponents gives

uniformly for \(X_a<X<X_b\) and −1≤η≤1. This treats the whole first-order outer stress and both stress entries. It does not assume positivity of the higher stresses or replace them by their leading value.

Z6a. Exact outer stress coefficient of the actual base

On the terminal collar \(X>X_+\), Y5.2–Y5.3 give \(\mathcal T_n=0\) for \(n\ge2\) and \(\mathcal T_{1,z}=0\). Thus the actual sum Z6.3 is exactly

These are the full original stress components; no limiting series is interchanged here because all but one correction vanish identically on this collar. H6.6 supplies the smooth positive leading coefficient \(b_\theta\), and Y8.16–Y8.27 proves

In particular the ratio \(B_1/b_\theta\) extends smoothly through the outer endpoint, despite the unbounded first-order quotient \(\mathcal T_{1,\theta}/\zeta\) at fixed nonzero \(\eta\).

The actual cutoff satisfies \(0\le\chi\le1\). At the endpoint the angular coefficient divided by \(b_\theta\) is

For every fixed bounded interval of \(q\ge0\), continuity and compactness supply a common smaller terminal collar on which the full angular coefficient stays positive. The cutoff is zero for large \(c_1q\), so the same conclusion holds on any intended preterminal range. The exact direction is

It is smooth in \((\delta_b,\eta)\) through the endpoint for each fixed \(q>0\), with every fixed \(q\)-derivative on compact positive \(q\) intervals. The exact radial derivative bounds Y8.22–Y8.25 allow the exponent \(3k+3\) for the first-order outer stress after \(k\) radial derivatives, with arbitrary fixed parameter derivatives. This supplies an explicit exponent for that term of Z6.4; other product-rule terms retain their stated finite exponents. The actual physical stress is \(\nu q^{-A-1/2}\widehat T\) at the unchanged viscosity coordinates, so the positive scalar factor preserves the direction and all original physical factors. This proves a sharper endpoint comparison for the same base field, without altering the construction or claiming an original source error.

Z7. Formal expansion, exterior, mean patch and endpoint derivatives

The same fields recover every original coefficient order, preserve the exact heat exterior and mean-patch velocity, and have the required endpoint derivatives.

Fix a formal order N and a physical derivative order m. Take \(J\ge\max\{2N+2,m+1\}\). For sufficiently small q, the finitely many cutoffs with n≤J equal one and all their derivatives vanish. The intermediate block N<n<J then consists exactly of original coefficient terms and has powers at least 2h(N+1), with the fixed derivative losses already proved in Z0.3. The remaining tail satisfies

The constants of the finite intermediate block may depend on N,m. This proves every fixed formal order of the original tuple, including physical derivatives with an index-independent exponent loss. For the similarity derivatives the same argument applies to Z3.6, with J also exceeding the chosen annulus index k; its tail has no physical loss. The cutoff curl vanishes in the finite initial block and remains fully present in the tail bounds. Thus the asymptotic coefficients are the original \(\mathcal U^{[N]}\), with their Stokes potentials differentiated before summation as specified in the source.

One can also retain the exact component power in this comparison. For angular or axial velocity, radial velocity, pressure, and stress respectively, let \(\beta=-A,-1/2,-2A,-A-1/2\), as in Z0.1 and Z1.1. A prescribed derivative with a transverse, k axial and ℓ time derivatives has exact loss \(w=a/2+Dk+\ell\). Enlarge the finite split index J, if necessary, until \(hJ-\ell'_m\ge\beta+2h(N+1)-w\). The intermediate terms obey Z0.3 with this original β, and the tail obeys this strengthened choice of J in Z7.1. Therefore the component remainder after the original N-th truncation is \(O(q^{\beta+2h(N+1)-w})\) for that derivative. All five types of field have been retained; a larger finite J proves their finite list of requirements simultaneously. This proves the coefficient expansion with the actual physical powers as well as the common-loss estimate.

By Y3.5, \(E_n,U_n\),\(F_n,V_n\) vanish on the untouched mean patch for every positive n. Both terms in the radial bracket Z2.3 vanish there as well. Hence the exact base velocity on that patch is

There is no pressure-equality assertion on this earlier patch. In the exterior \(X\ge X_+\), Y3 gives all positive-order fields and potentials, including pressure, identically zero. Thus \(u_B=u^{(0)}\) and \(p_B=p^{(0)}\) there. For \(X\ge X_b\) this is precisely the actual heat swirl and its pressure from H0/H4 and V7, including \(U_0=V_0=0\), not just a field with the same asymptotic power.

For endpoint regularity fix a compact set with \(q\ge q_*>0\). Local finiteness leaves only finitely many positive-order terms. Each has every fixed profile derivative continuous up to η=±1 by D/Y; the leading term has that same one-sided smoothness by G/O/H/J/Q/V, including the heat factor at d=0. The cutoff and its derivatives are smooth functions of q. Formula Z0.2 transfers these limits to every fixed physical space-time derivative because \(L\ge1-2h>0\) and q is bounded below. The same applies to the pressure and stress, and then to \(\mathcal E_B\), which is their finite differential expression Z5.6. This proves the claimed continuous derivative extensions. It requires no claim of two-sided analytic extension of the terminal heat factor at d=0.

Z8. Original viscosity and the receiving construction

Restore the originally specified viscosity in the potentials, streamfunctions, velocities, pressure, stresses and residual. Their derivative factors must agree in the physical equation.

For the originally specified ν>0 use the exact map

The potential and streamfunction factors follow from \(r_\nu=\sqrt\nu r\) and \(\operatorname{curl}_\nu=\nu^{-1/2}\operatorname{curl}\); they give exactly the mapped velocity including every cutoff curl. Each momentum term has the factor \(\sqrt\nu\): one spatial derivative costs \(\nu^{-1/2}\), pressure and stress have factor ν, and \(\nu\Delta_\nu u_\nu\) has \(\nu\cdot\nu^{-1}\cdot\sqrt\nu\). Thus

Here the tangential divergence is exactly the two operators \((\partial_{r_\nu}+2/r_\nu)T_{\nu,\theta}e_\theta+(\partial_{r_\nu}+1/r_\nu)T_{\nu,z}e_z\). All residual derivative bounds gain only the fixed factors \(\nu^{(1-|\alpha|)/2}\) for |α| spatial derivatives; time is unchanged. They are therefore flat to every order in the same original \(q_\nu\). Stress support, endpoint extensions and exterior and mean-patch equalities pass through this exact coordinate map. No different viscosity or spatial scale has replaced the original problem.

Z9. Five worked exercises

Exercise 1. Find the exact divergence defect and the unique regular radial repair

Compare cutting the physical positive-order velocity directly with taking the curl of its cut potential. Prove the correction is uniquely determined among regular radial fields, and retain the original viscosity.

Solution. Write \(s=c_nq\), only as the argument of the already fixed cutoff. Since \(\operatorname{div}u_n=0\), \(q\) is independent of \(r\), and \(q_z=2\eta q^{1-D}/L\),

Here \(A+D=1\) gives the final exponent. The angular component contributes no divergence. The extra radial component in Z2.3 is

At fixed \(z,t\), every factor outside \(F_n/r\) is independent of \(r\), while \(\partial_rF_n=(r/q)U_n\). Hence

This proves the cancellation with its exact sign. Moreover \(F_n/X\) is smooth and \(X=r^2/(2q)\); thus \(w_{r,n}e_r\) is smooth in the Cartesian variables. In fact the positive-order zero axis trace gives \(F_n=O(X^2)\), so this correction is \(O(r^3)\) at fixed \(z,t\).

Any other radial solution of the same divergence equation differs by \(a(z,t)/r\). Smoothness at the axis forces \(a=0\). Equivalently, integrating \(\partial_r(rw)=-r\,\operatorname{div}[\chi u_n]\) from zero gives precisely Z9.2, because \(F_n(0)=0\). This proves uniqueness in the specified class. The common compact support and the zero mean-patch value follow from those of \(F_n\).

For the original viscosity, \(r_\nu=\sqrt\nu r\), \(w_{r,n,\nu}=\sqrt\nu w_{r,n}\), and

In the divergence, the spatial factor \(\nu^{-1/2}\) cancels the velocity factor \(\sqrt\nu\); Z9.1 and Z9.3 retain exactly their coefficient-one values at \((X_\nu,\eta_\nu,q_\nu)\). In the curl, the potential has factor \(\nu\), so its mapped curl has factor \(\sqrt\nu\), as Z8.1 requires.

Exercise 2. Compute the entire residual of the first finite sum

Set \(N=1\) in the actual tuple Z1.1. List every remaining radial and tangential term, with its physical power. In particular identify both surviving radial axial-viscosity terms.

Solution. The tangential product condition \(i+j>1\), with \(0\le i,j\le1\), leaves only \((1,1)\). The last retained axial-viscosity term has index one. Consequently

All functions retain their actual \(X,\eta\) dependence. The operators have the original shifts: \(Z_{\beta,1}^{[2]}=Z_{\beta+2h-D}Z_{\beta+2h}\).

For radial transport the condition is \(i+j+1>1\). It leaves \((0,1),(1,0),(1,1)\). The radial axial-viscosity condition \(j+2>1\) leaves both \(j=0\) and \(j=1\). Group only terms with identical original powers:

The full radial expression is

Indeed \(-1+2h=-2A+4h=-2D\), whereas \(-1+4h=-2A+6h=-2D+2h\). These identities place the two radial axial derivatives in orders two and three respectively. They also place the radial time and transverse-viscosity terms in order two. Every coefficient and both middle transport products remain in Z9.6.

Each \(G_j/X\) is smooth at \(X=0\): substitute \(V_i=Xv_i\) into every term and use the operator identities \(T_\beta(Xf)=XT_{\beta-1}f\), \(Z_\beta(Xf)=XZ_{\beta-1}f\). Thus division by \(r=\sqrt qR\) gives the smooth Cartesian radial vector with its exact factor \(X/R=R/2\). On a fixed compact profile range, the derivative estimate Z0.3 gives

This is a bound for the displayed finite residual. It does not discard the order-three term or equate a finite truncation with the actual infinite base. For viscosity \(\nu\), the entire vector in Z9.5–Z9.7 is multiplied by \(\sqrt\nu\) and evaluated at \(x=x_\nu/\sqrt\nu\); an additional spatial derivative contributes \(\nu^{-1/2}\).

Exercise 3. Compare two admissible cutoff sequences exactly

Suppose \(c_n\) and \(\widetilde c_n\) are two sequences constructed by Z3 from the same actual coefficients, each satisfying all displayed estimates. Determine their exact field difference and prove that it is flat in \(q\), including a weighted stress statement on the full active annulus.

Solution. Put \(\Delta\chi_n(q)=\chi(c_nq)-\chi(\widetilde c_nq)\). The exact difference is

Both sums are locally finite. Their difference is divergence-free by the full curl identity, and its velocity and pressure vanish outside \(X_+\). Its velocity vanishes on the unchanged mean patch. The stress difference has support in \([X_a,X_b]\).

Fix a derivative order \(m\) and decay order \(M\). Choose \(J\ge\max\{1,m\}\) with \(h(J+1)-\ell'_m\ge M\). For

both sets of first \(J\) cutoffs are identically one. Apply Z4.2 to each base relative to the same finite tuple. The exact triangle inequality gives

The constants and the neighborhood are permitted to depend on \(m,M\). This proves equality of every formal coefficient with a flat difference, rather than equality of the actual fields.

There is a stronger stress comparison after retaining its original multiplier. Define \(\Delta\widehat T=q^{A+1/2}\Delta T_{\rm phys}\). For \(q\) as in Z9.10 with \(J\ge\max\{2,|I|\}\), the first-order term vanishes exactly. Every remaining stress coefficient is supported in the fixed interior \([X_-,X_+]\); its quotient by \(\zeta\) is smooth. The two bounds Z3.8 therefore give

Choose \(J\) also so that \(h(J+1)\ge M\). This is flat uniformly through the whole active annulus and every fixed similarity derivative, with no inverse distance factor. The quotient extends by zero near both endpoints because the remaining supports lie in that fixed interior. The exact original physical relation is \(\Delta T_{{\rm phys},\nu}=\nu q_\nu^{-A-1/2}\Delta\widehat T\); no stress multiplier has been removed from the physical conclusion.

Exercise 4. Compute the exact axis residual of the actual base

Track the first cutoff through the axis cancellation in D7. Prove an exact identity for \(\mathcal E_{B,\nu}\) on the axis and retain the growing axial velocity there.

Solution. Use the actual traces and slopes already derived from the full coefficient equations:

These are D7.10–D7.15; the global moment corrections preserve the protected axis interval. At any fixed \(q_\nu>0\) only finitely many base terms occur. The axial components in Z2.3 have the factor \(\chi(c_nq_\nu)\), which is independent of \(r_\nu\). Their zero axis traces remain zero under every time and axial derivative. Their pressure derivatives in \(z_\nu\) also have zero trace.

Every positive-order horizontal velocity, including the additional radial cutoff term, is \(O(r_\nu^3)\), and the positive-order axial velocity is \(O(r_\nu^2)\). Hence all extra axial convection terms vanish at the axis. Only the transverse Laplacian of the first positive-order axial component contributes there. Its exact value is

The higher orders have zero first radial slope by Z9.13, so they contribute zero even in their cutoff transitions. The leading residual from D7.23 is \(-\sqrt\nu q_\nu^{-A-2D}\mathscr Z_{c-D}\mathscr Z_cU_*\). Since \(2h-1=-2D\), inserting Z9.13 into the negative viscous term gives

The stress and its derivatives vanish on a neighborhood of the axis because \(X_a>0\), so the augmented residual there is the ordinary momentum residual. Its horizontal components are zero: a smooth axisymmetric vector at the axis is fixed by every horizontal rotation and hence is parallel to \(e_z\). This proves the full vector identity.

In particular the residual is identically zero on the axis whenever \(0<q_\nu\le(2c_1)^{-1}\), although no open radial neighborhood with identically zero residual is claimed. At \(z_\nu=0\), write \(\tau=1-t=q_\nu\); D7.22 gives

The velocity trace is exactly

Thus exact axis cancellation preserves the prescribed growth. It supplies an identity for the constructed base, not a replacement for the later annular correction and localization arguments.

Exercise 5. Recover the full first correction and a uniform weighted remainder

Refine Z6.2–Z6.4 by retaining the first actual coefficient. State the estimates for every fixed similarity derivative on the original enlarged annulus, and treat both endpoint weights.

Solution. Keep \(X_{\rm lo},X_{\rm hi}\) and \(\mathscr D\) from Z6. For \(q<(2c_1)^{-1}\), the first cutoff is one and its derivatives vanish. Fix \(I\) and choose a tail index \(J\ge\max\{4,|I|,k\}\), where \(\mathcal K_k\) contains the chosen annulus. Every term \(2\le n<J\) has its full factor \(q^{2nh}\), hence is \(O_I(q^{4h})\); the finitely many cutoff derivatives are bounded by Z3.5. The remaining tail is bounded by \(2^{1-J}q^{Jh}\), which is also \(O(q^{4h})\). Applying this argument to every complete entry of Z3.6 yields

The entire radial cutoff bracket remains in the tail estimate; it is not set to zero there. The common \(q^A\) factor in the radial component gives the more precise expansion

Each derivative of the extra factor \(q^h\) contributes a fixed power of \(h\), so the product rule preserves this stated exponent.

For stress, subtract the full first coefficient before dividing by the original weight:

All terms on the right have support in \([X_-,X_+]\). The same finite-block and tail argument, now using the exact quotient estimates in Z3.6, proves

This quotient extends smoothly by zero near both endpoints. The exceptional first-order stress has been retained explicitly; its quotient by \(\zeta\) is not bounded at a fixed nonzero outer parameter, as Y8 proves. Differentiating Z9.20 after multiplying back by \(\zeta\) also gives the physical weighted estimate \(C_Iq^{4h}\zeta\delta^{-N_I}\) for the displayed stress remainder, with the finite product-rule exponents of the original weight.

To restore viscosity in Z9.18–Z9.21, evaluate at \((q_\nu,X_\nu,\eta_\nu)\), use \(u_{B,\nu}=\sqrt\nu u_B\), \(p_{B,\nu}=\nu p_B\), and \(T_{{\rm phys},\nu}=\nu q_\nu^{-A-1/2}\widehat T\). For example the angular comparison is for \(q_\nu^Au_{\theta,B,\nu}/\sqrt\nu\), and the pressure comparison for \(q_\nu^{2A}p_{B,\nu}/\nu\). These are exact comparisons to the original physical fields, retaining their specified viscosity.

What the actual base supplies

Z0–Z9 complete the N3 construction: the actual coefficients and their global moments now give a smooth divergence-free base, the full residual bound, every coefficient order, the required weighted stresses and endpoint derivatives, and the exact exterior and mean-patch fields. All conclusions of original Proposition 5.5 have their proofs above. The residual is flat in the shrinking scale; it is not asserted to vanish away from the axis.

The next construction supplies the oscillatory fields whose stresses cancel the retained tangential stress. It uses the auxiliary torus, separated supports, evaluated derivative operators and the local background shear. Those steps remain separate lessons.

Authored lesson text and figure source: CC0.

Separating the oscillatory supports

The actual base field has a prescribed annular stress. The next construction supplies that stress through oscillating velocities. If two unrelated velocities overlap, their product adds a term that the construction did not prescribe. We first arrange the supports so that every such product is exactly zero, even after differentiation. At the same time we must allow fast temporal variation without an equally large radial derivative.

This lesson proves the complete geometric and coefficient framework of OpenAI, Finite Time Blowup for Navier–Stokes, section 6, original pages 62–72, equations (6.1)–(6.32), Lemmas 6.1–6.3 and Proposition 6.6. The original paper supplies the objects and constants. The proofs below retain their full physical coordinate maps. The pulse equation and its particular phase are the next construction. Here the phase hypotheses are exactly the definitions of section 6, whose algebra applies to the subsequent constructed phase.

The diagram records exact maps and support implications proved in OS1–OS5. The rectangle centers are constructed in OS3 and Exercise 3. In its final panel \(u_\nu=\sum_\gamma u_{\nu,\gamma}\), as defined in Exercise 5. No flow samples are used. Editable diagram source.

OS0. Keep the physical field and every chart factor

Retain

We work where \(q>0\), initially \(1-t\ge0\), and in the fixed active shell \(0<X_a\le X\le X_b\). The base field supplies a smooth extension near \(1-t=0\). For a fixed integer band \(\ell\), set

These coordinates retain the original physical field. The chart radius is \(R=\sqrt{q/Q}\sqrt{2X}\), whereas the earlier profile radius was \(\sqrt{2X}\). On \(1/2\le q/Q\le2\),

Indeed \(|\eta|\le1\), and substitution of OS0.1 gives each bound. A sufficiently small fixed enlargement remains away from \(R=0\), with bounded \(Z,T\).

The positive viscosity is retained through the exact map

Thus \(q=q(z_\nu/\sqrt\nu,t)\), \(X=r_\nu^2/(2\nu q)\). For physical velocity, pressure, stress and augmented residual define their chart representatives by

The full receiving differential equation is derived in OS6; none of these factors is discarded.

OS1. The auxiliary variable and its evaluated derivatives

The extended field has an independent variable \(Y\in\mathbb T^2=\mathbb R^2/\mathbb Z^2\). Set

Direct multiplication gives \(J_gv_r=\Lambda_gv_r\), \(J_gv_t=T_gv_t\), \(v_r\cdot v_t=0\), and \(|v_r|^2=|v_t|^2=1+b_g^2\). Both eigenvalues exceed one and \(\det J_g=14\). Also \(\Lambda_g>2,T_g<6\) imply \(\rho_g>\log2/\log6>1/3\), so \(d_r>2(1/3-10^{-7})>0\).

Evaluate the extended field at

This is smooth on the active shell. No assertion of smoothness of \(r^{d_r}\) at the axis is needed: every auxiliary-dependent correction is supported away from the axis. Define

The chain rule proves that \(\mathfrak t f\) and \(\mathfrak r f\), evaluated at OS1.2, equal the physical derivatives of the evaluated \(f\). The derivatives \(\partial_t,\partial_r\) in OS1.3 hold \(Y\) fixed. Auxiliary periodicity does not impose periodicity on the physical coordinates.

For sufficiently large \(\ell\), choose the nonnegative integer

This is an integer covering, not a globally invertible torus coordinate change. On a band pullback, \(N_{\rm abs}=T_g^iN_i\) and \(\mathcal L_{\rm abs}=\Lambda_g^i\mathcal L_i\), where \(N_i=v_t\cdot\partial_{Y_i}\), \(\mathcal L_i=v_r\cdot\partial_{Y_i}\). Consequently

The time sign follows from \(T=(1-t)/Q\); the axial factor is \(Q^{1/2-D}=Q^h\). The floor in OS1.4 gives the two-sided bounds

For the second, raise the first covering-index bound to \(\rho_g\) and use \(d_r/2-(1+h)\rho_g=-h\kappa_s\). These are exact inequalities, including the floor losses.

The operators \(D_r,D_z,\partial_\theta,\mathfrak t_*\) commute pairwise on component functions: the only variable auxiliary coefficient depends on \(R\), and the other three operators annihilate it. However \(D_r(R^{-1})=-R^{-2}\), which must be retained when differentiating a cylindrical vector.

Both auxiliary directions have a quantitative inverse on zero-Haar-mean smooth functions. For \(n=(n_1,n_2)\in\mathbb Z^2\setminus\{0\}\), the products

are nonzero integers. If either vanished, irrationality of \(\sqrt2\) would force both coordinates to vanish. Cauchy–Schwarz bounds each conjugate by \(C_g|n|\), where \(C_g=\sqrt{1+(1+\sqrt2)^2}\). Hence

Exercise 1 constructs the inverse with this full constant and proves its derivative estimate.

OS2. Slow partitions, labels and pulse coordinates

Here is an explicit construction of the partitions used in the paper. Choose a nonnegative smooth function \(\beta\), supported in \([-1,1]\), positive on \((-1,1)\), with all derivatives zero at the endpoints; for example \(\beta(s)=e^{-1/(1-s^2)}\) on that interval and zero outside. Define

The denominator is smooth, periodic and strictly positive: some integer lies within distance \(1/2\) of every \(s\). Only three indices can meet a point in the closed supports. Every fixed derivative is bounded.

Put \(\chi_\ell(q)=b_\ell(-\log_2q)\). Its support has \(q/Q\in[1/2,2]\). In each band use the product of three such partitions at mesh \(S_*^{-3}\) in \(R,Z,T\), denoted \(\chi_{\ell,a}\). Its closed box \(B_{\ell,a}\) extends at most one mesh length from the grid center in each coordinate. These choices give

Take \(\ell_0\) sufficiently large and \(q_{\rm big}\le2^{-\ell_0}\). A band below \(\ell_0\) cannot have a nonzero cutoff at \(0<q<q_{\rm big}\). Let \(\mathcal A_\ell\) be the portion \(q<q_{\rm big}\), \(q/Q\in[1/2,2]\), \(1-t\ge0\), \(X_a\le X\le X_b\), expressed in the chart. Select

The representatives and all discrete choices stay fixed under differentiation. Write \(C_\ell(r,z,t)=(R,Z,T)\) and

The signs duplicate the cutoff: including both signs in the last sum would give two. Supports include their closure in the coordinate domain and the smooth extension near \(1-t=0\).

Every fixed derivative of these cutoffs in chart coordinates costs only a power of \(S_*\). To check the \(q\)-factor explicitly, set \(s=q/Q\). The implicit equation becomes \(s-Z^2s^{2h}=T\), whose derivative in \(s\) is \(1-2hZ^2s^{2h-1}=1-2h\eta^2\ge1-2h>0\). On the band all its derivatives and implicit derivatives are bounded. A small extension preserves a positive lower bound. Differentiating the grid factors costs at most \(S_*^{3|I|}\); the band cutoff has fixed bounds.

Assign to each label a center \(c_\gamma\in\mathbb T^2\) and a common radius \(r_0>0\), constructed in OS3. Set

with both sets taken modulo \(\mathbb Z^2\). Their absolute preimages are \(\mathcal R_\gamma^{\rm abs}=(J_g^{i(\ell)})^{-1}(\mathcal R_\gamma)\), and similarly for the enlargement. The superscript \({}^{-1}\) here denotes set preimage.

On a local lift choose \(k_{\rm copy}\in\mathbb Z^2\) and write

The dual linear functionals are \(\lambda_r(w)=v_r\cdot w/(1+b_g^2)\) and \(\lambda_t(w)=v_t\cdot w/(1+b_g^2)\). Thus \(\mathcal L_i\xi_g=N_i\eta_g=1\) and \(\mathcal L_i\eta_g=N_i\xi_g=0\). In particular

Choose nonzero \(0\le\chi_g\le1\) smoothly supported in \((-r_0,r_0)\), and a rescaled smooth time cutoff with

Such cutoffs follow from the same bump construction. Their temporal support stays strictly inside \(0<v<L_s\). They will be applied inside the velocity potentials after solving the pulse equation.

OS3. One separation construction for all labels

Enlarge slow boxes to a fixed half-width \(L\ge1\) in mesh units. Join two distinct labels when their enlarged physical boxes meet and \(|\ell-\ell'|\le4\); join the opposite signs at one box too. Meeting original slow supports implies \(|\ell-\ell'|\le2\), since both \(q/Q\) and \(q/Q'\) lie in \([1/2,2]\).

The physical mesh lengths are

For \(\ell,\ell'\ge\ell_0\), \(|\ell-\ell'|\le4\), ratios in either direction are at most

A meeting box center in each coordinate lies in an interval of length \(2L(d_{\ell,j}+d_{\ell',j})\), which contains at most \(N_{\rm mesh}\) centers at mesh \(d_{\ell',j}\). Nine possible bands and two signs give graph degree at most \(18N_{\rm mesh}^3-1\). Enumerate the countable labels and greedily assign one of \(K=18N_{\rm mesh}^3\) colors unused by already colored neighbors. A later endpoint of every edge therefore receives a different color.

The mean value theorem for \(\log\ell\) and the floor inequality give

where \(\Delta_{\max}\) is a fixed integer ceiling. Choose a center \(c_\nu\) for each color satisfying all ordered constraints

For distinct colors the excluded equality is closed with empty interior, since one center can be varied independently. For equal colors and \(\Delta>0\), \(J_g^\Delta-I\) is nonsingular: its eigenvalues are \(\Lambda_g^\Delta-1,T_g^\Delta-1>0\). Its torus kernel is finite, because on a bounded fundamental square its integer image has finitely many possibilities, each with at most one preimage. A finite union of these closed sets has empty interior: remove them successively from a nonempty open subset. The remaining open dense set contains a rational tuple. Exercise 3 provides an explicit finite selection rule.

Let \(s_0>0\) be the smallest distance of the finitely many forbidden differences to \(\mathbb Z^2\), and put \(C_e=2\sqrt{2(1+b_g^2)}\). Choose

An enlarged displacement \(e=\xi v_r+\eta_gv_t\), \(|\xi|,|\eta_g|\le2r_0\), has \(|e|\le C_er_0\). Its diameter is less than one, proving injectivity modulo the integer lattice. If absolute enlarged preimages for joined labels at levels \(i\) and \(i+\Delta\) met, their centers would satisfy

The right side has norm at most \(C_er_0(T_g^\Delta+1)<s_0/2\), since \(J_g\) is symmetric positive definite with norm \(T_g\). This contradicts the definition of \(s_0\). At \(\Delta=0\), joined labels have different colors, so that constraint was included. We have proved

Thus products of distinct-label fields supported in these slow and auxiliary sets vanish exactly. A derivative of a smooth field has support in the original closed support: on its open complement the field is identically zero. The same product identity holds for every pair of derivatives, including frame derivatives, and after physical evaluation OS1.2.

OS4. A shared torus, its sheets and its operations

Fix \((z,t)\) with \(q>0\). Choose a small neighborhood \(U\) whose band cutoffs have indices differing by at most four. This is possible because \(\log_2 q\) is continuous and the band support interval has length two in that variable. Let

Then \(0\le\Delta_\ell\le\Delta_{\max}\), and

The last map is onto: an invertible real matrix \(J_g^{i_0}\) has a real preimage of every representative of a torus point. Therefore separation of absolute preimages implies separation on this shared torus.

Writing \(\Delta=\Delta_\ell\), its sheets are

Here the real matrix inverse is used on individual lifts. Two parameter points with the same torus image have identical band coordinates by rectangle injectivity; the remaining equality identifies exactly the same coset of \(k\). Thus these sheets are disjoint. Their number is

For completeness, the lattice-index formula follows by tiling \(\mathbb R^2\) with translates of the parallelogram \(J_g^\Delta[0,1)^2\). A fundamental domain of this sublattice is also tiled, up to measure-zero boundaries, by one unit-square fundamental domain for each integer-lattice coset. Its area is both the number of cosets and \(|\det J_g^\Delta|\); finiteness follows because a bounded fundamental parallelogram contains a representative of each coset.

For a smooth band function \(f\), Fourier expansion and invertibility of \((J_g^\Delta)^T\) give

Indeed a frequency \(n\) becomes \((J_g^\Delta)^Tn\), which is zero exactly for \(n=0\). Smooth Fourier series converge absolutely, so termwise integration is justified. The same proof applies to the absolute covering. Each sheet has area \((4r_0^2(1+b_g^2))/14^\Delta\); summing its \(14^\Delta\) copies also proves preservation of the rectangle's total Haar area.

On overlaps the representatives must satisfy

in the same fixed physical band chart. Both sides equal the original extended field. At support closures the equality follows by continuity. Throughout the following operator identities \(Q\) stays fixed, even when a common-torus index changes:

The corresponding identity for \(cN\) uses \(T_g^\Delta\). Thus these operators give exactly the same evaluated derivative; no inverse torus covering is invoked.

A smooth absolute function descends to the torus of level \(i_0\) precisely when it is invariant under every deck translation \(Y\mapsto Y+a\), \(J_g^{i_0}a\in\mathbb Z^2\). Equivalently, its Fourier frequencies lie in \((J_g^{i_0})^T\mathbb Z^2\). To verify the converse direction of this frequency criterion, invariance of a character of frequency \(n\) under \(a=J_g^{-i_0}k\) for all \(k\in\mathbb Z^2\) says \((J_g^{-i_0})^Tn\in\mathbb Z^2\); this is precisely the stated lattice.

Torus translation multiplies each Fourier coefficient by a phase. Haar averaging retains the zero frequency. A directional inverse on the zero-mean subspace multiplies each nonzero frequency by a scalar. Each operation therefore preserves descent. Radial integration at fixed \((z,t)\) leaves \(q(z,t)\), every band cutoff and its covering index fixed, and preserves the same deck invariance under the integral. It can meet more grid boxes but introduces no new dyadic band.

At a point at most three band indices, three grid centers in each coordinate and two signs occur, giving at most \(3\cdot3^3\cdot2=162\) closed slow supports. On a fixed compact chart box the count is at most \(C_BS_*^9\); on a bounded radial interval at fixed \(Z,T\) it is at most \(C_IS_*^3\). These follow by counting grid points in intervals of mesh \(S_*^{-3}\); fixed enlargements change only the constants. Common-torus lifts add at most \(14^{\Delta_{\max}}\). These explicit counts prove the polynomial label losses required for radial integration.

On an individual lifted rectangle define \(\eta_g(H)=\lambda_t(J_g^\Delta H-c_\gamma-k_{\rm copy})\). The path reaching pulse coordinate \(w\in[0,L_s]\) is

Its slow and transverse coordinates are fixed, and its pulse coordinate is \(w\), by applying \(J_g^\Delta\) and then \(\lambda_r,\lambda_t\). A general source on the shared torus is evaluated at this \(H_w\) on this sheet; one does not average other sheets. Since \(\lambda_tJ_g^\Delta=T_g^\Delta\lambda_t\),

and every derivative of order at least two of these affine maps is zero. The first is bounded independently of the band; the second is bounded by a constant times \(S_*\), using OS1.6. Exercise 4 proves the exact path composition and descent.

OS5. The coefficient classes and their complete algebra

Keep the full radial weight

Extend \(\zeta\) by zero outside the active interval. Every fixed inverse power of the distance to either edge, multiplied by \(\zeta^b\) for \(b>0\), tends to zero there: after setting \(y=1/s\), the required limit is \(y^Me^{-aby^2}=0\). Differentiation produces only more finite inverse powers. Thus the same assertion holds for every fixed derivative. The quantity \(\delta\) is an estimate weight; no differentiability of its minimum is assumed. Both weights are constant along OS4.8 because \(R,Z,T\) remain fixed.

Fix a finite construction stage \(j\). A coefficient derivative is

It holds the band, label, representative and harmonic integer fixed. A mean coefficient \(f\in\mathsf M_\alpha\) is independent of \(\theta\), has the descent compatibility OS4.6, extends smoothly by zero outside the active radial shell, and satisfies for every \(I\)

Constants are uniform in the band, label, sheet and point; they may depend on fixed profile data, \(j,I\). Such a coefficient may depend on \(H\). It has neither a rectangle restriction nor a temporal envelope requirement.

A moment coefficient \(F(Z,T)\in\mathsf S_\alpha\) satisfies, for each two-variable multi-index,

For a component with \(f_* = Q^{a_f}f_{\rm phys}\), the complete radial moment map is

This follows from \(r=\sqrt Q R\), including both \(r^e\) and \(dr\). It holds before averaging too. For arbitrary viscosity and a quantity \(f_{\nu}=\nu^{b_f}f_{\rm phys}\), its full version is

For each label prescribe a phase, envelope and finite set of nonzero harmonics as in the original definition:

The actual phase and envelope are constructed with the pulse in section 7. For the algebra here, \(\varphi_\gamma\) is smooth on the enlarged lifted rectangles and the amplitude is independent of \(\theta\). The finite harmonic set is independent of the band.

Put \(x_*=(R,Z,T)\), \(K_\gamma^*=C_\ell(K_\gamma)\). The permitted domains and supports are

The enlarged open domain \(\mathcal E_\gamma^+\) replaces the closed rectangle by its open enlargement. A smooth labelled amplitude \(a_{\gamma,m}\in\mathsf W_\alpha\) on that domain has mutually compatible lift formulas and OS4.6, is independent of \(\theta\), and its support on \(\mathcal E_\gamma\) lies in \(\Omega_\gamma\). It extends smoothly by zero across slow and transverse support boundaries and both shell edges. For all fixed \(I\), on \(0\le v\le L_s\),

This is an estimate for every differentiated amplitude. We do not differentiate an inequality for \(P_v\). Before cutoff, continuation beyond \(v=0,L_s\) has no temporal envelope or zero-extension requirement. An actual cut coefficient has support in \(\Omega_\gamma^{\rm cut}\) and smooth zero extension in \(v\). No divisibility by any chosen cutoff is assumed.

For a finite or locally finite sum of real waves, collect equal label-harmonic pairs first:

Membership requires uniform OS5.9 for these collected coefficients. Mere local finiteness is not a substitute for those uniform estimates.

Leibniz' formula proves

For equal-label amplitudes the product lies in \(\mathsf W_{\alpha+\beta}\) at harmonic \(m+m'\ne0\), and in \(\mathsf M_{\alpha+\beta}\) when \(m+m'=0\), after temporal cutoff. Before cutoff the zero harmonic satisfies the mean estimates on that rectangle only. In detail, powers of \(\varepsilon,S_*,\delta^{-1}\) add in each finite Leibniz term, and

Product support is contained in the intersection. Smooth zero extensions follow from the stated support conditions and flat weight at the shell. To check an edge derivative directly, its interior limit is zero; its difference quotient across the edge is zero because the preceding derivative vanishes faster than the edge distance. Induction proves all orders. For a zero harmonic from actual cut coefficients, temporal zero extension supplies a global mean coefficient; without that extension one has only the local assertion.

For distinct labels, OS3.7 proves zero for every pair of differentiated actual fields. For one label, integer periodicity proves

This average integrates cylindrical component columns over \(0\le\theta\le2\pi\). The integral of \(e^{in\theta}\) is zero for every nonzero integer \(n\). It is a different operation from Haar averaging in \(Y\), which removes auxiliary dependence. In particular an angular mean \(f\) can have a nonzero auxiliary zero-mean part \(f^\circ=f-\langle f\rangle_Y\).

For \(\mathsf C=\mathsf M\) or \(\mathsf W\), coefficient derivatives preserve the exponent, and the evaluated operators act by

Indeed all derivatives of \(R^{d_r-1}\) are bounded on OS0.3; OS1.6 supplies the one factor \(\varepsilon^{-\kappa_s}\). Differentiating that coefficient any fixed number of times introduces no further such factor. The slow derivative term embeds in the weaker radial class because \(0<\varepsilon\le1\). The factors \(c_{i_0}\) and matrices in OS4.7 have bounded polynomial losses only. The compatibility identity OS4.6 is preserved because these are the same absolute operators at fixed \(Q\). Finite sums preserve every class. Finite products add harmonic indices, and differentiation preserves each index; at each fixed stage the resulting harmonic set is still finite.

These are amplitude maps. A full wave has the extra exact phase factors

No pulse bound has been assumed to remove these terms.

OS6. Cylindrical frames and the physical receiving equation

Use the orthonormal physical frame \((e_r,e_\theta,e_z)\). Its derivatives are \(\partial_\theta e_r=e_\theta\), \(\partial_\theta e_\theta=-e_r\), \(\partial_\theta e_z=0\). Therefore the full component transport is

The last column is obtained by differentiating the frame. It is retained for both mean and oscillatory coefficients.

Define \(\nabla_*p=(D_rp,R^{-1}\partial_\theta p,D_zp)\) and \(\Delta_s=D_r^2+R^{-1}D_r+R^{-2}\partial_\theta^2+D_z^2\). Differentiating the frame twice gives

For example \(\partial_\theta^2(u_re_r+u_\theta e_\theta)\) has radial coefficient \(\partial_\theta^2u_r-u_r-2\partial_\theta u_\theta\) and angular coefficient \(\partial_\theta^2u_\theta-u_\theta+2\partial_\theta u_r\), proving the displayed terms.

For a symmetric stress matrix \(T\) in this frame, differentiation of both frame factors gives

One can verify this directly from \(\operatorname{div}T=\sum_j(\nabla_{e_j}T)e_j\): only \(\nabla_{e_\theta}e_r=R^{-1}e_\theta\) and \(\nabla_{e_\theta}e_\theta=-R^{-1}e_r\) are nonzero frame derivatives. The derivative of the contracted frame contributes \(R^{-1}(T_{rr},T_{\theta r},T_{zr})\); the derivative of the first tensor factor contributes \(R^{-1}(-T_{\theta\theta},T_{r\theta},0)\), yielding OS6.3.

Substitute OS0.5 in \(E_\nu=\partial_tu_\nu+(u_\nu\cdot\nabla)u_\nu+\nabla p_\nu-\nu\Delta u_\nu-\operatorname{div}\mathcal T_\nu\). Since physical spatial differentiation is \((\nu Q)^{-1/2}\) times its evaluated chart counterpart, every factor gives

The time factor is \(Q^{2A+1/2-A}=Q^{1+h}\); the viscous factor is \(Q^{2A+1/2-A-1}=Q^h=\varepsilon\). The factors \(\sqrt\nu,\nu\) in OS0.5 cancel exactly as displayed, retaining the original equation for every \(\nu>0\).

The exact curl and divergence are

Inserting OS6.5 gives the six terms

The first term uses \(D_r(R^{-1})=-R^{-2}\), which cancels the additional \(R^{-1}\) in the divergence. The remaining cancellations use the commutators proved in OS1.

For a \(\theta\)-independent potential coefficient \(C_m\), retain \(n=(D_r\Phi,p/R,D_z\Phi)\). Its actual amplitude is

All cutoff derivatives inside \(C_m\) remain. In particular \(D_r\psi(v)=D_z\psi(v)=0\) by OS2.7, whereas spatial derivatives of the slow and transverse cutoffs generally remain. If the physical vector potential is \(A_\nu=\nu Q^{1/2-A}C\), then \(\nabla_{\rm phys}\times A_\nu=\sqrt\nu Q^{-A}\operatorname{curl}_*C\). This is the exact potential map corresponding to OS0.5.

OS7. Five solved exercises

Exercise 1. Invert the original torus directions

Construct the inverse of \(d\cdot\partial_H\), for \(d=v_r\) or \(v_t\), on zero-Haar-mean smooth functions. Keep the constant and prove its behavior under the covering maps.

Solution. Write \(f(H)=\sum_{n\ne0}\widehat f(n)e^{2\pi i n\cdot H}\) and set

OS1.8 bounds the multiplier by \(C_g|n|/(2\pi)\). Smooth Fourier decay then gives convergence with every derivative. With \(\|f\|_{H^s}^2=\sum_n(1+|n|^2)^s|\widehat f(n)|^2\), termwise comparison proves

The mean is zero, differentiation returns \(f\), and uniqueness follows because no nonzero frequency lies in the directional kernel. A direct \(C^k\) estimate also follows: integrate by parts \(k+4\) times in a coordinate with \(|n_j|\ge|n|/\sqrt2\); the differentiated inverse series is bounded by a constant times \(\|f\|_{C^{k+4}}\sum_{n\ne0}|n|^{-3}\), and the sum converges in two dimensions.

Let \(d=v_r\) with eigenvalue \(\lambda=\Lambda_g\), or \(d=v_t\) with \(\lambda=T_g\). A pullback frequency is \((J_g^\Delta)^Tn\), and \(d\cdot(J_g^\Delta)^Tn=\lambda^\Delta d\cdot n\). Hence

This proves the precise inverse comparison and preserves the pullback frequency lattice.

Exercise 2. Bound radial sums without losing their moments

For a fixed band suppose actual mean coefficients \(f_\gamma\) satisfy OS5.3 uniformly and are supported in the selected slow boxes. Bound a finite or locally finite sum integrated along a bounded radial interval with weight \(R^e\), retaining the number of sheets and the physical measure.

Solution. Let the interval have length \(L_R\), and let the enlarged slow-box half-width be \(L\). There are at most \(\lceil L_RS_*^3+2L\rceil+1\) radial centers, and at most \(2\lceil L\rceil+1\) centers in each of \(Z,T\). Including signs, a valid bound is

On a common torus multiply this by at most \(14^{\Delta_{\max}}\) when counting sheets separately. Haar integration already accounts for their smaller measures; counting every sheet by the whole-torus bound is a valid upper bound. At fixed derivative order, the source estimate gives a finite exponent \(d_I\), and

The interval stays away from zero and has bounded length; the edge weight bounds the entire integrand uniformly. Its differentiated version follows from the same weight estimate, with an increased finite exponent. Integrate over a fixed enclosing interval, using the smooth zero extension, so no moving support endpoint term is lost. Differentiation under the integral is justified by these uniform bounds. Therefore the Haar-averaged moment of the sum lies in \(\mathsf S_\alpha\) with at most the added factor \(14^{\Delta_{\max}}C_IS_*^3\). A finite set of overlapping bands changes only this constant and the polynomial degree. The exact physical moment is OS5.6, including \(Q^{a_f-(e+1)/2}\nu^{-b_f-(e+1)/2}\); the estimate does not remove that map.

Exercise 3. Choose all rectangle centers by a finite rational rule

Give a finite construction satisfying every constraint OS3.4 and an explicit admissible radius.

Solution. Put \(D_0=\Delta_{\max}\), \(K=18N_{\rm mesh}^3\), and form the positive integer

Each determinant is positive and nonzero by the eigenvalue calculation. Choose an integer \(N\ge\max\{7,2K(D_0+1)\}\), and choose a prime divisor \(p\) of \(BN!+1\). Such a divisor exists because this integer exceeds one. It satisfies \(p>N\) and \(p\nmid B\): a prime at most \(N\), or dividing \(B\), cannot divide a number congruent to one modulo that prime. Thus \(J_g\) and every \(J_g^\Delta-I\), \(1\le\Delta\le D_0\), are invertible over \(\mathbb F_p\).

Choose \(a_\nu\in\mathbb F_p^2\) successively. Exclude zero and, for every previously chosen \(a_\mu\), all \(J_g^\Delta a_\mu\) and \(J_g^{-\Delta}a_\mu\), \(0\le\Delta\le D_0\). At step \(\nu\) at most \(1+2(D_0+1)(\nu-1)\) points are excluded, which is less than \(p^2\). Choose the first remaining pair in lexicographic order. The exclusions impose both ordered constraints for distinct colors. For one color, \((J_g^\Delta-I)a_\nu\ne0\) because the matrix is invertible and \(a_\nu\ne0\). Therefore \(c_\nu=a_\nu/p\pmod{\mathbb Z^2}\) satisfies OS3.4.

Every forbidden difference has coordinates in \(p^{-1}\mathbb Z\) and is nonzero modulo \(\mathbb Z^2\); its distance to that lattice is at least \(1/p\). Thus the explicit choice

satisfies OS3.5 strictly. This constructs the required centers and radius by finite integer arithmetic. It establishes no need to compute a large numerical table and makes no novelty claim.

Exercise 4. Follow the same pulse sheet exactly

Prove the composition rule for OS4.8, identify its derivative, and show why descent does not require averaging sheets.

Solution. Put \(P=I-v_t\lambda_t\). Since \(\lambda_t(v_t)=1\),

Thus \(P\) is the orthogonal projection onto the original radial eigendirection, because the eigendirections are orthogonal. Write \(A_w(H)=H_w\) on one lifted rectangle. Applying \(\lambda_tJ_g^\Delta\) to OS4.8 proves \(v(A_w(H))=w\); applying \(\lambda_rJ_g^\Delta\) proves that the transverse coordinate is unchanged. Substituting in the same formula gives, for \(s,w\in[0,L_s]\),

These are identities on the chosen sheet, with its unchanged copy index. Different preimages of a band point may carry different values of a common-torus source, so the source must remain \(f(R,Z,T,A_w(H))\).

In the absolute variable the corresponding path is

Multiplying by \(J_g^{i_0}\) gives OS4.8. A deck translation preserving \(H=J_g^{i_0}Y\) translates this entire path by that deck vector and reindexes its lifted rectangle; it leaves the common source value unchanged. Consequently a uniquely specified zero-initial-data linear solution on these paths descends to the common torus: the translated solution satisfies the identical equation and initial data, and uniqueness identifies it. This is a descent statement; the quantitative pulse inverse is still to be constructed. No average over the \(14^\Delta\) sheets was taken.

Exercise 5. Compute the exact angular covariance and its physical factor

Let \(a_\gamma\) be an actual cut coefficient column and \(w_\gamma=a_\gamma e^{ik_\gamma\Phi_\gamma}+\overline{a_\gamma}e^{-ik_\gamma\Phi_\gamma}\). Compute the angular covariance of their real sum and distinguish it from auxiliary averaging.

Solution. The column tensor product is bilinear. Expanding the two factors and using OS5.13 yields

The terms at harmonics \(2,-2\) have zero angular mean. Distinct-label cross-products vanish pointwise by OS3.7, before averaging. The sum is locally finite, so

All products and averages are of cylindrical component tensors; frame derivatives in later divergence remain those of OS6.3.

For physical velocities \(u_{\nu,\gamma}=\sqrt\nu\,Q_\gamma^{-A}w_\gamma\), retain each band's factor:

Each term is evaluated using its original band chart and the same physical point. One cannot put a single band factor in front without applying the exact chart comparisons.

For \(n\in\mathbb Z^2\setminus\{0\}\), \(f(Y)=\cos(2\pi n\cdot Y)\) has angular mean \(f\) and Haar mean zero. After physical evaluation it becomes \(\cos(2\pi n\cdot(v_r r^{d_r}+v_tt))\), which is not identically zero. This proves by an explicit function why angular mean, auxiliary mean and physical evaluation must retain their separate exact maps. The covariance to be prescribed is the one in OS7.13; its actual pulse coefficients are constructed next.

What this construction supplies

We have constructed the slow partitions, all labels, one uniformly separating family of auxiliary rectangles and its pulse coordinates. We proved the common-torus maps, sheet counts, Haar identities, path derivatives, radial integration counts, complete coefficient algebra and exact physical differential operators. The next lesson constructs the particular phase, solves the pulse equation and proves the bounds needed for its amplitudes and curl potentials. This lesson does not claim that arbitrary coefficients already realize the target stress.

The exposition is based on OpenAI, section 6, and the preceding constructions cited above.

Constructing the pulses and their actual velocities

The base field supplies the shear, and the support construction supplies the slow boxes, auxiliary rectangles and physical differential operators. We now construct the oscillation inside each rectangle. Its phase and amplitude solve exact equations. Taking a curl then produces an exactly divergence-free velocity. We keep its pressure, curl correction and cutoff residual when checking the Navier–Stokes equation.

This lesson gives the local pulse construction in OpenAI, Finite Time Blowup for Navier–Stokes, Lemma 7.1, Proposition 7.2, Corollary 7.3, Lemmas 7.4 and 7.7, and Proposition 9.1. The original paper, pages 73–81, 85–87 and 101–102, supplies the original objects and formulas. Exercise 3 derives the covariance column estimate on pages 82–83. The subsequent positive stress representation and its signed inverse are separate constructions.

The diagram records the equations proved in PU1–PU8. The remainder in its last panel is retained in the actual equation. Editable diagram source.

PU0. The pulse inputs come from the constructed base

Keep \(0<h<1/100\), \(A=1/2+h\), \(D=1/2-h\), \(Q=2^{-\ell}\), \(\varepsilon=Q^h\), and \(S_*=\ell^2\). The original coordinates satisfy

The positive solution of \(q-z^2q^{2h}=1-t\) is unique for \(t<1\): the derivative of \(q^{1-2h}-(1-t)q^{-2h}\) is positive, and its endpoint limits are minus and plus infinity. At \(z=0\) the solution is \(q=1-t\). On the one-sided boundary \(t=1,z\ne0\), it is \(q=|z|^{1/D}\). Hence \(|\eta|\le1\) and \(L\ge1-2h>0\). These are the actual domains of the base field.

Put \(\xi=q/Q\in[1/2,2]\). Then

Indeed implicit differentiation gives \(\xi_Z=2\eta\xi^{1-D}/L\) and \(\xi_T=1/L\); differentiating \(X,\eta\) proves the three operators. Each fixed derivative of their coefficients is bounded on the fixed enlarged annulus, since \(\xi,L,X\) remain bounded away from zero. Repeated product rules therefore carry every fixed similarity derivative estimate of the constructed base to these coefficient variables.

Let \(R_p=\sqrt{2X}\) denote the profile radius. The exact base construction and its estimates Z4–Z6 give

These estimates hold after every fixed slow derivative. In particular \(b=O(\varepsilon)\), and the tangential differences from the leading fields are \(O(\varepsilon^2)\), because \(q^{2h}=\varepsilon^2\xi^{2h}\). The radial field here includes the derivative of every base cutoff in Z4. We have not substituted the formal profile series for the actual base.

Write

The shear construction, V5–V7, proves \(F_{\rm prof}>0\), \(a>0\), \(v_s>2\), including strict endpoint margins on the closed active annulus. Thus there are constants \(F_-,a_-,d_v>0\) and finite upper bounds \(F_+,a_+\) with \(F_-\le F^{(0)}\le F_+\), \(a_-\le a\le a_+\), and \(v_s\ge2+d_v\). The same margins hold on a fixed sufficiently small radial enlargement.

Direct differentiation using PU0.2 gives the full leading shear

Since \(a^2+b_s^2=av_s\), substitution gives

For example \(2F^{(0)}N_\theta=-2F^{(0)}a/\sqrt{av_s}\), while \(2F^{(0)}N_\theta+|g^{(0)}|=F^{(0)}a(v_s-2)/\sqrt{av_s}\); their product proves the displayed equality. In particular the positive growth rate and all frame denominators below have actual lower bounds.

The parameter \(u_*\) also comes from the constructed stress. Set \(t_s=-b_s/a\), and let \(n^T\) be the smooth unit stress direction supplied by V5, including its values at the shell edges. For the fixed \(0<\kappa_c<2\) there,

Put \(c_0=\lambda_0/(2F^{(0)}N_\theta)<0\), so \(c_0^2=(v_s-2)/2\). For \(T_0^*=\xi^{-A-1/2}\mathcal T_0\), the coordinate formulas

and PU0.7 give, in the shell interior,

At the edges use the ratio of the unit direction components. Its denominator stays away from zero by PU0.7, so the ratio extends smoothly there. If \(m_c\) is its maximum on the closed parameter rectangle, choose

This is one finite positive choice. Representative points, band labels, supports and this parameter are fixed before taking coefficient derivatives. The construction below selects one lower band threshold using finitely many value inequalities. Higher derivative orders change constants and polynomial degrees; they do not require a new domain.

PU1. Original input data and a fixed band threshold

Keep the original \(0<h<1/100\), \(A=1/2+h\), \(D=1/2-h\), \(Q=2^{-\ell}\), \(\varepsilon=Q^h\), \(S_*=\ell^2\), \(R=r/\sqrt Q\), \(Z=z/Q^D\), \(T=(1-t)/Q\), \(X=r^2/(2q)\), and the original auxiliary map \(Y=v_rr^{d_r}+v_tt\). The matrices, vectors, \(d_r\), \(\kappa_s=10^{-5}\), covering indices, chain-rule operators and weighted classes are defined with full factors in the wave-interaction audit, W1–W3. In particular,

The pulse coordinates on each lifted rectangle satisfy

Band and label choices, including the representative, are fixed before any coefficient derivative.

The source uses the smooth slow base column \((b,V,G)\), independent of the auxiliary variable, with \(b=O(\varepsilon)\) in every fixed slow derivative. The tangential base differs from its fixed order-zero profiles by \(O(\varepsilon^2)\), with all fixed slow derivatives. The boxes and their fixed enlargements have diameter at most \(C S_*^{-3}\). At the fixed representative put

Here subscript zero means the order-zero quantity at the representative; \(\lambda_0>0\). These are precisely the base properties supplied to Section 7 by Theorem 4.6 and Proposition 5.5. Their actual global construction is supplied by V and Z, and the exact base-to-chart derivative map and positive shear bounds are proved in I1–I2. The calculation below proves the pulse operation with those same input fields.

Write \(0<\lambda_-\le\lambda_0\le\lambda_+\) and \(|g_0|\le g_+\) for the fixed uniform input bounds. They already imply the relevant nonzero shear and frame factors. Indeed, with \(a=2F_0N_\theta\), the relation \(-a(a+|g_0|)>0\) gives \(-|g_0|<a<0\). Moreover

The inequality \(|g_0|\ge2\lambda_-\) follows from \(x(g-x)\le g^2/4\), whose difference from \(g^2/4\) is \((x-g/2)^2\). Thus the denominators used below have stated lower bounds, rather than implicit nonvanishing assumptions.

Keep \(u_*>0\) chosen by the original strict stress-cone condition, and retain

The fixed tie rule is not differentiated. For \(\varepsilon\le1\),

For rounding, any real number has a nonzero integer within distance one: ordinary nearest-integer rounding has error at most one half except when it selects zero, in which case one of \(1,-1\) is within distance one. This proves the displayed error without using the incorrect one-half bound near zero.

Put \(H_\Phi=pF+p_zG\). The entire phase gradient is

This follows by differentiating (PU1.5) with (PU1.1) and \(D_rv=D_zv=0\). Before rounding, at the representative,

since \(K\cdot g_0=0\). Uniform first slow derivative bounds, the box diameter, the \(O(\varepsilon^2)\) base comparison, and rounding therefore give

Consequently \(n_r=B_ss(v)-v\mathcal E_R\). The exact tangential wave-number formula gives

here \(R_0/R-1=O(S_*^{-3})\), using the positive lower bound on \(R\). The remaining axial term in (PU1.7) is bounded by \(C\varepsilon S_*\). Since \(n'=(-(H_\Phi)_R,0,-\varepsilon(H_\Phi)_Z)\), these estimates prove

once \(S_*^2(\varepsilon+\varepsilon^2+k^{-1})\le1\). This one condition holds for all sufficiently large \(\ell\): \(S_*^2=\ell^4\), \(k^{-1}\le2^{-h\ell/2}\), and every fixed power of \(\ell\) times a decreasing exponential tends to zero, as follows by taking its logarithm \(4\log\ell-(h\log2)\ell/2\).

Increase this lower band threshold finitely so that the error in (PU1.9) is at most \(B_-/2\). Then

uniformly. All fixed mixed derivatives in the slow, transverse and pulse coordinates have polynomial bounds in \(S_*\), directly from (PU1.5)–(PU1.7) and the fixed-derivative bounds of the base. Discrete data are fixed; higher derivatives do not differentiate rounding or the representative choice. Product and quotient rules, together with (PU1.10), give polynomial bounds for every smooth rational expression in these coefficients whose denominators have the displayed lower bounds. This requires no new smallness condition for a higher derivative order.

The same denominator argument applies on the enlarged lifted rectangle: its pulse coordinate ranges over \(-L_s/2<v<3L_s/2\), so \(|v|\le3L_s/2\) and \(|s(v)|\le2u_*\). The error estimates above change only their fixed constants. Include this finite enlargement in the choice of the lower band threshold. Then \(|n_{\rm tan}|\ge B_-/2\) there as well, while the base is evaluated on its prescribed smooth enlarged slow box. This supplies the smooth coefficients needed for local continuation past the pulse endpoints. The envelope and propagator bounds below are asserted on \(0\le v\le L_s\).

The full transport defect is also explicit:

Every fixed coefficient derivative of the last line has a factor \(\varepsilon\) and a polynomial in \(S_*\), because every such derivative of \(b\) is \(O(\varepsilon)\). Finally \(kp\ne0\) is an integer, so all \(e^{ikm\Phi}\), \(m\ne0\), have nonzero integer angular frequency.

Include \(2r_0S_*\ge25\) for the concentration calculation in Exercise 3. Choose a lower band index \(\ell_*\) satisfying these finitely many value conditions and set \(q_*\le\min(q_{\rm big},2^{-\ell_*-1})\). For \(q<q_*\) and \(q/Q\in[1/2,2]\), \(Q\le2q<2^{-\ell_*}\), so all relevant bands satisfy the chosen threshold. The choice has no dependence on the harmonic bound, derivative order or correction stage.

PU2. Exact moving-plane equation, frame and pressure

Retain the original matrix and projection:

Dot products here are bilinear extensions of the real Euclidean dot product; \(n\) and all these coefficients are real. In particular \(n^T\mathcal A=-(n')^T\) and \(n\cdot\Pi_nf=0\).

The three-dimensional initial-value equation

preserves the constraint exactly:

Multiplication by \(\exp(m^2\int_0^v d)\) proves that the scalar in (PU2.3) is zero. Rearranging (PU2.2) gives

Therefore the original pressure, including its complex sign, is

and multiplication by \(ikmn\) cancels the preceding normal vector because \(i^2=-1\). Thus

This includes the normal component of the source. Conversely any solution of (PU2.5) with the constraint has (PU2.4), by taking its dot product with \(n\), and then satisfies (PU2.2). Thus uniqueness for (PU2.2) proves uniqueness for the constrained amplitude and its pressure.

Define the actual frame without replacing the phase normal:

The columns of \(U\) lie in \(n^\perp\), are independent, and \((e_r^T;N_a^T)U=I_2\). Since \(c\ne0\) by (PU1.4), \(B^\ell B=I_2\), and \(BB^\ell t=t\) for \(t\in n^\perp\). We only assert this last identity on that plane: \(BB^\ell\) need not be the orthogonal projection on all of \(\mathbb C^3\). Equations (PU1.9)–(PU1.10) and (PU1.4) bound \(B,B^\ell\) uniformly and bound every fixed derivative polynomially.

At the reference normal \(n_0=B_s(s,K)\), let \(e=e_r-sK\) and write \(t=xe+yN\). The frozen \(\mathcal K_0\) uses \(F_0\) and \(g_0=|g_0|N\). Direct multiplication gives

where each pair lists its radial and tangential parts. Consequently

and \(N\cdot(-\mathcal K_0N)=0\). The normal projection contributes zero to these dot products. Since \(|e|^2=1+s^2\), \(|N|^2=1\), and \(e\cdot N=0\), the exact two-dimensional reference matrix is

Its eigenvectors are \((1,c)^T,(1,-c)^T\). Indeed, writing \(a=2F_0N_\theta\), the first component of \(L_0(1,c)^T\) is \(ac/(1+s^2)=\lambda_0/\sqrt{1+s^2}\), and its second is \(-(a+|g_0|)=\lambda_0^2/a=\lambda c\). Thus

For the actual coefficients, the algebraic projection \(-\mathcal K+nn^T\mathcal K/|n|^2\) differs from its frozen counterpart by \(O(S_*^{-1})\), by (PU1.8)–(PU1.10), the base comparison and the box diameter. The actual \(K_a,N_a,s_a\) differ from \(K,N,s\) by that order. The moving-normal term \(-n(n')^T/|n|^2\) is also \(O(S_*^{-1})\). Finally \(K_a',N_a',s_a'=O(S_*^{-1})\) by differentiating their exact quotients and using (PU1.9), while \(s'=\sigma u_*/L_s=O(S_*^{-1})\). Hence \(B'=O(S_*^{-1})\). Substituting all these differences, using the uniform matrix bounds, proves

All fixed derivatives of \(E\) are polynomially bounded; they are not asserted to be \(O(S_*^{-1})\). Similarly, with the original reference damping,

For the last inequality use \(\big||n|^2-|n_0|^2\big|\le|n-n_0|(|n|+|n_0|)\), (PU1.6), and the uniform bounds already proved.

Because \(n^T(\mathcal AB-B')=-(n')^TB-n^TB'=0\), this matrix maps into the moving plane. Consequently the exact correspondence \(t_m=Bz_m\) turns (PU2.2) into

The verified identities for \(BB^\ell\) prove both directions of this correspondence; no normal component has been removed without recovering its pressure.

PU3. Exact envelope and forward propagator

Use the original envelope

Write \(y=|s(v)|=u_*/2+u_*v/L_s\), retaining the original \(v\) and \(L_s\). Direct integration gives the exact identity

To verify it, differentiate \(\operatorname{arsinh}y\) to obtain \((1+y^2)^{-1/2}\), differentiate \(y+y^3/3\) to obtain \(1+y^2\), and use \(y'=u_*/L_s\). Both sides vanish at \(v=L_s/2\).

Let

Then \(a(L_s/2)=0\) and exactly

Since \(u_*/2\le y\le3u_*/2\), define the fixed positive constants

Each positive factor in (PU3.3) is bounded by its respective endpoint bounds, proving \(-c_+/L_s\le a'(v)\le-c_-/L_s\). Taylor's integral formula about the midpoint, applied to \(\log P\) whose value and first derivative vanish there, now gives on both sides

For the left side of the midpoint, substitute \(v=L_s/2-x\) in Taylor's integral formula; the same nonnegative triangular kernel integrates \((\log P)''\). Thus no sign is changed by integrating backward. These are analytic bounds, not numerical enclosures.

For completeness, the fundamental matrix of a continuous matrix coefficient \(A_m\) on a fixed finite interval is obtained by the series

If \(\|A_m\|\le K\) there, the \(n\)-th term is bounded by \(K^n(v-w)^n/n!\). Uniform absolute convergence, and the same estimate for the differentiated integral series, prove \(\partial_vV_m=A_mV_m\), \(V_m(w,w)=I_2\). A difference of two solutions with zero initial value satisfies \(u(v)=\int_w^v A_m(a)u(a)\,da\); iterating this identity and letting \(n\to\infty\) with the same factorial bound proves uniqueness. Thus this construction supplies the finite-interval existence and uniqueness used in PU2, rather than assuming a solution with the desired bound.

For a complex solution of \(z'=A_1z\), using the Hermitian norm only in this norm estimate,

Multiply by the exponential of minus twice the integral of the scalar coefficient to obtain a nonincreasing quantity. Taking square roots and the supremum over initial vectors proves

The scalar damping commutes with every matrix, so direct differentiation and uniqueness in (PU3.6) prove the exact identity

Since \(d>0\) and \(m\in\mathbb Z\setminus\{0\}\), its scalar factor is at most one. Therefore (PU3.8) holds for every nonzero integer \(m\), with the same constant and the same domain. This is why higher harmonics do not require another smallness threshold.

PU4. Source inverse, derivatives and the exact common-torus path

Let \(f_m\) have exactly the support, smooth-extension and fixed-derivative bounds of original Proposition 7.2: for each fixed multi-index \(I\) in \((R,Z,T,H)\),

with fixed closed slow support inside the label support, compact transverse support inside \(\operatorname{supp}\chi_g\), and zero extension at the shell edges. The full weights are

The source may depend on the common torus without descending to the finer band torus; its values on different preimages are not averaged or identified. It need not be transverse to \(n\). No divisibility by a cutoff is assumed.

The actual solution is

with pressure (PU2.4). Inserting (PU3.8) and the source bound cancels \(P(w)\) exactly. The path fixes \((R,Z,T)\), hence \(\zeta,\delta\), and its length is at most \(2r_0T_gS_*\). This gives

enlarging the constant or polynomial degree for the bounded frame factors if needed.

First differentiate in slow variables and the initial transverse coordinate, holding \(v\) fixed. The labels and \(L_s\) are constants in these derivatives. They commute with \(\partial_v\). The differentiated equation is exactly

The coefficient derivatives \(\partial^JA_m\) are bounded by \(C_{J,M}S_*^{c_J}\) for a fixed finite harmonic set \(0<|m|\le M\); the factor \(m^2\) is retained in the constant. At derivative order \(I\), Leibniz' formula bounds \(\partial^Ig_m\) by a polynomial in \(S_*\) times the same \(\varepsilon^\alpha\sqrt\zeta\delta^{-a}P\) weight. Inductively every other source in (PU4.5) has this weight too. Applying (PU3.8) again increases its polynomial degree by one from the interval length. More explicitly, if the source term for \(\partial^Ig_m\) has degree \(b_I^g\) and the previously proved solution degrees are \(b'_{I-J}\), choose

These choices prove the induction for every fixed multi-index. No additional power of \(\varepsilon\) is lost. Pulse derivatives follow by repeatedly differentiating \(z'_m=A_mz_m+g_m\); each derivative is a finite sum of products already bounded, so it has the same envelope with enlarged polynomial degree. Applying product rules to \(t_m=Bz_m\) proves its corresponding estimates. The pressure formula (PU2.4), (PU1.10), and \(|km|^{-1}\le\varepsilon^{1/2}\) give the full bounds

All source and solution derivatives have their own estimates (PU4.1); this argument does not infer derivative bounds by differentiating a pointwise inequality for the envelope.

Here is the precise map on the common torus. For \(H=J_g^{i_0}Y\), \(\Delta=i-i_0\in\{0,\ldots,\Delta_{\max}\}\), define

On a chosen lift, \(\eta_g(H)=\lambda_t(J_g^\Delta H-c_\gamma-k_{\rm copy})\), and the source in (PU4.3) is evaluated at

Because \(\lambda_tJ_g^\Delta=T_g^\Delta\lambda_t\),

The derivative of the first path map is bounded independently of the band; the derivative of \(v\) costs only a factor \(C S_*\). Differentiating a coefficient expressed in \((\xi_g,v)\) thus introduces fixed powers of these constant linear maps. Conversely the initial transverse and pulse directions in \(H\) are \(J_g^{-\Delta}v_r\) and \(T_g^{-\Delta}c_iv_t\) on this local lift; these are legitimate linear maps in the covering chart, not an asserted global inverse to a torus covering. They have bounded coefficients. These identities prove (PU4.7) in exactly the original \((R,Z,T,H)\) coefficient variables.

The path fixes the slow and transverse point. If that point is outside the prescribed support, its source is zero for the entire interval, and (PU4.3) makes the solution zero there. Its derivatives extend by zero across those boundaries by (PU4.5), or directly by solving the equation for the given smooth source extension. At shell edges, \(\sqrt\zeta\delta^{-a}\) tends to zero with every fixed derivative; the estimates in PU4 and the zero-extension argument of W3 therefore prove smooth extension for the solution and pressure. There is no claimed temporal zero extension before the pulse cutoff. Smooth coefficients on the enlarged rectangle provide a local extension of the initial-value solution past each endpoint; the bounds above concern the closed interval itself.

Deck translations preserving \(H\) translate (PU4.8) and reindex its lift by the same integer vector. The source values agree under those translations, and uniqueness proves the same for the solution. Distinct preimages of a finer band-torus point remain distinct points of \(H\), with their actual source data. On overlapping common-torus neighborhoods, pull everything to the absolute torus. The path is \(Y_w=Y+T_g^{-i}c_i(w-v(Y))v_t\) in both descriptions. Compatible source data and uniqueness then prove compatibility of the output. On compact sets with \(q>0\), uniform one-sided slow derivatives at \(\tau=0\) pass through the same differentiated equations and their integrals, proving the stated one-sided derivative property. The number of derivatives and the finite harmonic set change constants and degrees only, not the previously chosen \(q_*\).

PU5. The homogeneous pulse and its momentum flux direction

The primary stress construction needs a nonzero homogeneous solution, not the zero-data sourced solution. Use the original initial data in the growing coordinate:

All coefficients and data are real. On any interval with \(z_+>0\), put \(r=z_-/z_+\). Subtracting the two quotient terms from (PU2.11) gives exactly

Let \(c_\lambda=\lambda_-/\sqrt{1+9u_*^2/4}>0\), and retain \(|E_{ab}|\le C_E/S_*\). Choose \(K_b=2C_E/c_\lambda+1\), and include \(S_*\ge4K_b\) among the finite value conditions defining the common threshold in PU1. At either boundary \(r=\pm K_b/S_*\),

A differentiable function starting inside this interval cannot first cross its upper boundary with strictly negative derivative, or its lower boundary with strictly positive derivative. Hence \(|r|\le K_b/S_*\le1/4\) while the ratio exists. Also

so, with \(C_h=C_d+5C_E/4\),

For any fixed band, the positive continuous \(P\) has positive minimum on \([0,L_s]\). The lower bound in (PU5.4) therefore prevents \(z_+\) from reaching zero. The ratio and its bound extend to the entire interval.

In the actual \(U\) coordinates, write \(t^h=x(e_r-s_aK_a)+yN_a\). The full matrix \(M\) gives

The last expression is bounded by \((8/3)\sup|c_0\sqrt{1+s^2}|K_b/S_*\). This proves the full source direction estimate (7.21), including its sign and radial lower bound.

The value estimate also follows from \(V_1(v,0)(P(0),0)^T\), using (PU3.8). Initial data \(P(0)\) are fixed with the label, so their slow and transverse derivatives vanish. Repeating (PU4.5) for this homogeneous solution proves \(|\partial^I t^h|\le C_IS_*^{b_I}P(v)\) at every fixed derivative order; pulse derivatives again follow from the equation. Its pressure is (PU2.4) with \(f_m=0,m=1\), and has the factor \(\varepsilon^{1/2}\). These are estimates for the pulse before multiplying by the radial, transverse and slow amplitude factors used in stress realization. No shell weight or compact shell support is asserted for this unweighted homogeneous pulse itself.

Its energy identity retains the original shear and viscosity contributions. Because \(n\cdot t^h=0\), the normal term in \(\mathcal A t^h\) has zero dot product with \(t^h\). For any real \(t=(t_r,t_\theta,t_z)\), direct multiplication gives

Thus the exact homogeneous identity is

The stress map to be checked next must match both components of this radial momentum flux. Establishing (PU5.6) alone does not construct that stress map.

PU6. Actual compact velocity, pressure, and every endpoint term

Let \(0\le\psi(v)\le1\) be the original smooth pulse cutoff, equal to one on \(|v-L_s/2|\le L_s/5\), supported in \(|v-L_s/2|<L_s/3\), rescaled from a fixed function. For the sourced solution set

The exact actual velocity amplitude, rather than just its transverse part, is

Here are all component terms of the curl and its divergence. For any smooth vector column \(C\),

The original operators satisfy \([D_r,D_z]=[D_r,\partial_\theta]=[D_z,\partial_\theta]=0\), \(D_rR=1\), and \(D_zR=\partial_\theta R=0\): these follow directly from PU1.1, since its auxiliary vector field is constant and its coefficient depends only on \(R\). Therefore

For its first term use \((D_r+R^{-1})(R^{-1}f)=R^{-1}D_rf\); the three pairs then cancel. In the curl of the harmonic potential, differentiating the exponential gives

The remaining terms are exactly PU6.2. If \(a_m=\widehat t_m+\widehat r_m\), the divergence identity gives the full longitudinal relation

The class conclusion follows from one \(D_r\) loss and \(1/|km|\le\varepsilon^{1/2}\), after PU6.3 below supplies \(a_m\in\mathsf W_\alpha\). Each fixed derivative follows by the same identity and Leibniz' rule. The chain rule for the physical evaluation map preserves this exact zero divergence. Now (PU1.10) supplies the uniform denominator bound needed for this particular potential. The factor \((km)^{-1}\) increases the exponent by one half; one \(D_r\) can cost at most \(\kappa_s\). Hence, with all fixed coefficient derivatives and the original supports,

These class bounds follow by Leibniz' rule with (PU4.7); derivatives of the quotient \(n/|n|^2\) are polynomially bounded by PU1. All spatial derivatives of the temporal cutoff actually vanish here: \(D_r\psi=D_z\psi=\partial_\theta\psi=0\). Thus \(\widehat r_m=\psi r_m\) for the uncut remainder defined by (PU6.2) with \(C_m\). Time derivatives of this product still contain \(\psi' r_m\); none are deleted. Both the potential and pressure vanish on open neighborhoods of the pulse endpoints, which supplies their smooth zero extensions there. The other support boundaries were already checked in PU4.

The complete principal cutoff identity is

It follows by multiplying (PU2.5) by \(\psi\), retaining the derivative \((\psi t_m)'\), and adding \(f_m\). Where either right-hand term is nonzero, \(|v-L_s/2|\ge L_s/5\), so (PU3.5) and (PU1.2) give

Every fixed derivative of these terms has this same exponential smallness, times a permitted polynomial in \(S_*\) and shell factors, because the solution and source have their differentiated envelope bounds. The cutoff derivatives themselves obey \(|\psi^{(j)}|\le C_jL_s^{-j}\), and conversion to \(H\) multiplies only by powers of \(S_*\).

For the global assertion in original Proposition 9.1, the incoming source harmonic has its stated smooth zero extension across the entire fixed rectangle, including the temporal endpoints. This additional source property makes (PU6.4) a global residual identity after the supported fields are extended by zero. Proposition 7.2 by itself did not impose that property on an arbitrary uncut source, and the present calculation does not add it silently. The actual source containment is now proved in S1–S5, and the four updates preserving it are checked in K0–K8.

Real fields use conjugate harmonic pairs. Since \(A_m\) depends on \(m^2\) and is real, uniqueness gives \(t_{-m}=\overline{t_m}\) for \(f_{-m}=\overline{f_m}\). The factors \(i/(km)\) in the pressure and potential give \(\pi_{-m}=\overline{\pi_m}\), \(C_{-m}=\overline{C_m}\). The real cosine velocity \(t\cos(k\Phi)\) therefore has amplitudes \(t/2\) at both \(m=1,-1\), with their actual conjugate pressures and potentials. No factor two is suppressed in assembling the real velocity.

PU7. Physical factors and infinite-order cutoff decay

At viscosity one the physical potential, velocity, pressure and residual corresponding to chart fields have the factors

respectively. Their physical derivatives use the original evaluation map and chain-rule operators (PU1.1). At any fixed derivative order and finite stage, each conversion contributes only a fixed power of \(Q^{-1}\) and a polynomial in \(S_*\): the radial factor is \(Q^{-1/2}\), the axial factor \(Q^{-D}\), the slow time factor \(Q^{-1}\), and the auxiliary factors come from \(T_g^i\) and \(\Lambda_g^i\), bounded by fixed powers of \(Q^{-1}\) using their explicit covering-index formula. Derivatives of the oscillatory exponential retain their finite factors \(km\) and the displayed phase derivatives; \(k\le2Q^{-h/2}\), \(\varepsilon^{-1}=Q^{-h}\). Derivatives of the cylindrical frame introduce only finite powers of \(r^{-1}\), with \(r=\sqrt Q R\) and \(R\) bounded away from zero. All remaining fixed coefficient derivatives have already been bounded.

The edge factors \(\sqrt\zeta\delta^{-a}\) are bounded uniformly on the shell, since they vanish at both edges and are continuous on its interior. Taking any fixed physical space-time derivative of a cutoff term from (PU6.4) thus gives a bound

for fixed finite \(M,B\), allowing them to include the fixed factor \(\varepsilon^\alpha\). Since \(q\ge Q/2\), for every fixed \(N\ge0\),

The limit follows because the negative quadratic term dominates the linear and logarithmic terms; for large \(\ell\) their sum is at most \(-c\ell^2/2\). The finitely many remaining bands enlarge the bound's constant. This proves the all-order \(O(q^N)\) assertion for each fixed physical derivative, uniformly over the bands and labels. Fixed-stage label sums add at most a polynomial mesh factor, absorbed into \(S_*^B\). These residuals remain actual additive terms; flatness does not mean they are zero.

For the original arbitrary \(\nu>0\), keep the exact map of the moment audit M2.4–M2.7:

Thus the four factors in (PU7.1) become, in the same order,

The original coordinates are \(R=r_\nu/\sqrt{\nu Q}\), \(Z=z_\nu/(\sqrt\nu Q^D)\), \(T=(1-t)/Q\), with \(q_\nu-(z_\nu^2/\nu)q_\nu^{2h}=1-t\), \(X=r_\nu^2/(2\nu q_\nu)\), and \(Y=v_r(r_\nu/\sqrt\nu)^{d_r}+v_tt\). A fixed number of spatial derivatives adds a fixed power of \(\nu^{-1/2}\), absorbed only into the constant for that fixed original viscosity. In particular \(\nu\Delta_xu_\nu=\sqrt\nu\Delta_yu_1\); the factor \(\varepsilon k^2|n|^2\) in the pulse damping is unchanged by this exact map. All factors are retained, and the same all-order cutoff conclusion follows for each \(\nu>0\).

PU8. Full linearized residual after the curl

The principal amplitude-pressure operator is

On each lifted rectangle \(c_{i_0}N_{i_0}=\partial_v\) with the slow and transverse coordinates held fixed, including for sources that only descend to the common torus. The full linearization of the Navier–Stokes residual about \((b,V,G)\), on the harmonic \(ae^{ikm\Phi}\) with pressure \(\pi e^{ikm\Phi}\), has coefficient \(\mathscr L_m(a,\pi)+\mathscr L_m^{\rm rem}(a,\pi)\), where

Here is a component derivation. The two transports use the exact cylindrical formula W1.6. Their frame terms sum to \((-2Va_\theta/R,Va_r/R+ba_\theta/R,0)\). The radial derivatives of the tangential base add \((0,\partial_RV\,a_r,\partial_RG\,a_r)\); because \(\partial_RV=F+RF_R\), their sum gives the full \(\mathcal K a\) and the remaining \((b/R)a_\theta e_\theta\). The other base derivatives are exactly \(a_r(\partial_Rb)e_r+a_zD_z(b,V,G)\). Phase transport gives \(ikmE_{ik}a\) after the cancellation in (PU1.11). The pressure gradient is \(ikmn\pi+(D_r\pi,0,D_z\pi)\).

For the scalar Laplacian applied componentwise, the product rule gives its amplitude part \((D_r^2+R^{-1}D_r+D_z^2)a\), phase-quadratic part \(-k^2m^2|n|^2a\), mixed part \(2ikm(n_rD_r+n_zD_z)a\), and phase divergence \(ikm(D_rn_r+n_r/R+D_zn_z)a\). The angular derivative of \(n_\theta=p/R\) is zero. The cylindrical vector Laplacian adds \(-R^{-2}(a_r,a_\theta,0)\) and \(2R^{-2}\partial_\theta(-a_\theta,a_r,0)e^{ikm\Phi}\), which gives the last line in (PU8.2). Multiplication by the residual sign \(-\varepsilon\) places the phase-quadratic part in (PU8.1) with its positive sign. This accounts for every term in original (9.2).

For \(a\in\mathsf W_\alpha\), \(\pi\in\mathsf W_{\alpha+1/2}\), the gains over \(\alpha\), with all coefficient derivatives, are as follows:

The rules behind the table are the exact class maps of W3: one explicit \(D_r\) costs at most \(\kappa_s\), \(D_z\) adds one, \(k\) costs one half, and the displayed \(b\) or its fixed slow derivatives add one. The multiplier \(n\) has polynomial coefficient bounds, as do its derivatives before the possible one \(D_r\) loss; (PU1.11) retains the full \(\varepsilon\) in the transport defect. Every remaining factor \(R^{-1}\) is bounded on the original annulus, with bounded derivatives. Thus Leibniz' formula proves the table for every fixed coefficient derivative, retaining the shell weight and envelope from the amplitude or pressure factor.

The actual amplitude in (PU6.2) is \(a=\widehat t_m+\widehat r_m\). The exact residual after adding the prescribed source is

No curl term is absent here. The first line is the actual additive flat source from PU6–PU7. In the second line, \(c_{i_0}N_{i_0}\) preserves the coefficient exponent, \(\mathcal K\) has bounded coefficients, and \(\varepsilon k^2m^2|n|^2\) has polynomial coefficient bounds for each fixed harmonic set. Hence \(\mathscr L_m(\widehat r_m,0)\in\mathsf W_{\alpha+1/2-\kappa_s}\). The table and (PU6.3) give the same bound for the last term, because \(1-2\kappa_s\ge1/2-\kappa_s\). Consequently

The inclusion uses \(0<\varepsilon\le1\). This proves the source Proposition 9.1, and the full term table yields the stated local strengthening by \(2\kappa_s\). No improvement of the complete correction cycle is inferred; other steps still determine its gain. Supports and torus compatibility follow from the exact products and derivatives in (PU8.3), with smooth temporal extension supplied by the cutoff. The flat source is retained separately and is not fed back as a source requiring a new pulse solve. For a homogeneous supported amplitude in the stated wave class, the identical calculation applies with \(f_m=0\), leaving its actual cutoff term \(\psi't_m\).

PU9. Five worked exercises

Exercise 1. The exact frame projection and a normal source

Determine \(BB^\ell\) on all of \(\mathbb C^3\), compare it with the orthogonal projection \(\Pi_n\), and solve the zero-data inverse for a source \(f_m=\beta n\).

Solution. Put \(n_t=|n_{\rm tan}|>0\). The orthonormal real basis \(e_r,K_a,N_a\) gives \(n=n_r e_r+n_tK_a\) and \(s_a=n_r/n_t\). For \(w=w_re_r+w_KK_a+w_NN_a\), PU2.6 yields

The first map has image \(n^\perp\) and kernel \(\operatorname{span}K_a\). Its square is itself, since \(n\cdot BB^\ell w=0\). Its Euclidean operator norm is exactly \(\sqrt{1+s_a^2}\): the squared output norm is \((1+s_a^2)|w_r|^2+|w_N|^2\), bounded by \((1+s_a^2)|w|^2\), with equality at \(w=e_r\). Thus it is uniformly bounded by PU1, although it is generally oblique. When \(s_a=0\), it agrees with \(\Pi_n\).

For \(f_m=\beta n\), the projected source is zero. Uniqueness with zero initial data gives \(t_m=0\), while PU2.4 gives

This holds for any smooth coefficient \(\beta\) for which the source has the stated supports and weights. A normal source is cancelled by pressure; it has not been discarded in the projection.

Exercise 2. A quantitative gain for higher harmonics

Use the exact scalar factor in PU3.9 to bound the zero-data inverse more sharply for \(|m|\ge2\). Keep the pulse envelope and every damping factor.

Solution. By PU1.6 and PU1.10,

For \(a_m=(m^2-1)d_{\min}>0\), PU3.8–PU3.9 give

If \(G_\infty=\sup_{[0,L_s]}|g_m(w)|/P(w)\), direct integration yields

Integrating PU9.3 over \(v\) and using the nonnegative Fubini identity gives the additional bound

The frame map \(t_m=Bz_m\) multiplies these bounds by \(\sup\|B\|\). For the source weight in PU4.1, the slow factors \(\varepsilon^\alpha\sqrt\zeta\,\delta^{-a}\) are constant along the path and remain in \(G_\infty\).

The factor \((m^2-1)^{-1}\) is a value estimate for \(|m|\ge2\). It does not assert uniform bounds for every harmonic derivative: differentiating \(A_m\) retains its \(m^2\) factors, as PU4.5 shows. The fundamental harmonics \(m=\pm1\) retain the interval-length bound in PU4.4. Every estimate uses the same previously selected domain.

Exercise 3. Exact concentration bounds and the covariance columns

Derive the size and direction of the covariance of a homogeneous pulse, with both the angular factor and the auxiliary measure retained.

Solution. Write \(c_x=\tfrac34e^{-C_{h,0}}\), \(C_x=\tfrac54e^{C_{h,0}}\). Then PU5.5 gives \(c_xP\le x\le C_xP\). For one sign set

For \(L_s\ge25\), the interval \(|v-L_s/2|\le\sqrt{L_s}\) lies inside \(\psi=1\). The lower Gaussian bound in PU3.5 is at least \(e^{-c_+/2}\) there. The upper bound can be integrated over the whole real line. Consequently

For the first upper integral substitute \(w=(v-L_s/2)/\sqrt{L_s}\). The identity \(\int_{\mathbb R}e^{-c_-w^2}dw=\sqrt{\pi/c_-}\) follows by squaring the integral and using polar coordinates: \(2\pi\int_0^\infty r e^{-c_-r^2}dr=\pi/c_-\). For the second integral, the same substitution gives \(\int_{\mathbb R}|w|e^{-c_-w^2}dw=1/c_-\). The integrands are nonnegative, so these changes and products are justified by monotone integration. In particular

The finite threshold \(L_s\ge25\) is included in the initial band choice because \(L_s\ge2r_0S_*\).

The source field used for its covariance column, before the slow cutoff and scalar stress amplitude, is

Here \(\xi_g\) is the transverse coordinate of the support rectangle, distinct from \(q/Q\). The transverse cutoff has compact support, and \(t_\sigma^h\) is independent of \(\xi_g\). Define

Because \(kp\) is a nonzero integer, \((2\pi)^{-1}\int_0^{2\pi}\cos^2(kp\theta+\varphi)d\theta=1/2\). The original rectangle map has measure \(|\det(v_r,v_t)|\,d\xi_g\,d\eta_g\), with \(d\eta_g=c_i\,dv\). The support and covering proof OS2–OS4 permits computing the Haar integral on this rectangle. Thus

The integral of \(\chi_g^2\) is positive, since the prescribed transverse cutoff is not identically zero. Retain \(c_i=2r_0/L_s\). With \(D_\chi=|\det(v_r,v_t)|\int\chi_g^2\,d\xi_g\), PU9.6 gives the explicit size bounds

All constants here are the original fixed rectangle and pulse constants.

By the actual frame in PU2 and the quotient bound PU5.5,

To obtain the last bound, expand the differences of the three products, use PU1.9–PU1.10 for \(s_a-s,K_a-K,N_a-N\), and PU5.5 for \(y/x-c_0\sqrt{1+s^2}\). Every other factor is uniformly bounded. Since

the probability weight \(\psi^2x^2/J_0\) in PU9.8 gives

The square-root inequality follows from the derivative \(s/\sqrt{1+s^2}\), whose absolute value is at most one. This proves the source's covariance column estimate with its full constants and measure factors. It does not yet invert the two-column matrix or construct the stress amplitudes.

Exercise 4. The real cosine, its potential and its pressure

For a real source coefficient \(f\) and its real transverse solution \(t\) at \(m=1\), write the actual real velocity and pressure explicitly. Keep the cosine factor \(1/2\) in the harmonic coefficients.

Solution. Define the real vector and scalar

The coefficients at \(m=\pm1\) are \(t/2\), the potential coefficients are \(\pm iJ/(2k)\), and the pressure coefficients are \(\pm iH/(2k)\). After the pulse cutoff their sums are

Here \(D_r\psi=D_z\psi=\partial_\theta\psi=0\). Differentiating the sine gives \(-\psi(n\times J)\cos(k\Phi)=\psi t\cos(k\Phi)\), by \(n\cdot t=0\). Differentiating \(J\) gives the second term and all its cylindrical connection terms. The divergence of this whole velocity is exactly zero by PU6; the first cosine term alone need not be divergence-free.

At the original viscosity \(\nu>0\), evaluate these fields at

The physical potential, velocity and pressure are, respectively,

The physical curl contributes the factor \(1/\sqrt{\nu Q}\), so the first expression has exactly the curl asserted in the second. Reality has not removed the imaginary factor from each complex harmonic; it has combined the conjugate harmonics into a real sine.

Exercise 5. The exact energy supplied to the transverse pulse

Derive the integrated energy balance for \(t^h\) and for its cutoff \(\widehat t=\psi t^h\). Determine what becomes of the endpoint term.

Solution. The original shear \(g=(RF_R,G_R)\) is fixed along the pulse at fixed slow coordinates, while \(d(v)=\varepsilon k^2|n(v)|^2\) retains its full \(v\) dependence. Integration of PU5.6 gives

For the cutoff, differentiate its squared norm before integrating:

Both endpoint values of \(\widehat t\) vanish, and hence

If \(|t^h|\le C_tP\), \(0\le\psi\le1\), and \(|\psi'|\le C_\psi/L_s\), then the last term has the explicit bound

Indeed \(\psi'\) is supported where \(|v-L_s/2|\ge L_s/5\); there PU3.5 gives \(P^2\le e^{-c_-L_s/25}\), and the interval length cancels \(L_s^{-1}\).

These identities concern the transverse pulse and its stated cutoff. The actual velocity also contains the sine term in PU9.12; its equation includes the full curl residual PU8.3. Neither that term nor the cutoff term is set to zero by the energy calculation. PU7 proves the all-order physical decay of the cutoff residual while retaining it in the equation.

Sources and proof scope

OpenAI, Finite Time Blowup for Navier–Stokes, original pages 73–87, equations (7.1)–(7.40), and pages 101–102, Proposition 9.1 and equations (9.1)–(9.2). The source PDF has SHA-256 0e779481c4da40bd28d1e642e1d8ca57447d129610df28dfa5a11e9af8ae228f.

PU0 supplies the actual base inputs. PU1–PU5 prove the phase, moving-plane inverse and homogeneous pulse. PU6–PU8 give the actual compact velocity, pressure and complete linear residual. PU9 supplies five exact worked calculations, including the covariance column bounds needed for the next construction. The complete positive stress representation, signed inverse and nonlinear corrections are supplied in the separate stress covariance audit, C1–C9, and wave interaction audit, W4–W7.

Authored exposition: GPT-6 Astra (OpenAI). The authored lesson and diagram source are CC0; cited original works retain their attribution and rights.

Realizing the stress and controlling nonlinear interactions

The pulse lesson constructed actual velocities, their pressures and their complete residuals. We now choose the two pulse amplitudes to produce the prescribed two-component stress. The signed map used later in the iteration inverts the derivative of this covariance. Its finite update also has a quadratic term, which we keep throughout.

The support lesson, OS0–OS7, supplies the original slow partition, disjoint closed auxiliary supports, all covering maps and weighted coefficient classes. A wave coefficient in \(\mathsf W_\alpha\) has, at every fixed derivative order, a bound of the form \(\varepsilon^\alpha S_*^{b_I}\sqrt{\zeta}\delta^{-a_I}P(v)\), with its original support, envelope and finite nonzero harmonic set. A mean in \(\mathsf M_\mu\) has the corresponding bound \(\varepsilon^\mu S_*^{b_I}\zeta\delta^{-a_I}\), may depend on the auxiliary torus and is independent of the cylindrical angle. The actual definitions and zero extensions are those proved in OS5. Products, full coefficient derivatives and chart changes use those proofs. In particular, \(D_r\) costs at most \(\kappa_s=10^{-5}\), \(D_z\) gains one power of \(\varepsilon\), and \(k\) costs one half power. No support or weight is removed here.

The leading stress is the actual one constructed in V0–V8; the background is Z0–Z9. This lesson supplies the positive amplitude inverse, its fixed signed derivative inverse, smooth extension at both radial boundaries, full curled covariance and nonlinear transport. It also derives positive energy production using the actual shear, retaining its difference from the frozen leading shear and the viscous damping.

The diagram records exact maps proved below. It contains no sampled flow data. Editable diagram source.

Original source: OpenAI, Finite Time Blowup for Navier–Stokes, sections 7.3–7.4 and Lemma 9.2, pages 81–88 and 102–103. The source remains identifiable in the separate transcription. The actual-shear calculation is also developed in the workbench pulse derivation, identified in the accompanying source record.

The citations W2–W3 below refer to the full support and coefficient proof also supplied as OS0–OS7 in the preceding support lesson. Covariance uses \(\mathcal B_{\rm cov}\); differential transport uses \(\mathcal T_{\rm adv}\). Their definitions are displayed separately.

NI1. Original coordinates, operators and angular averaging

Keep the original parameters and coordinates

The original variables satisfy \(z=q^D\eta\), \(1-t=q(1-\eta^2)\), \(X=r^2/(2q)\). On a band, \(q/Q\in[1/2,2]\); the active radial shell has \(X_a\le X\le X_b\) with \(X_a>0\). Thus \(R^2=2(q/Q)X\), and

The bounds may be enlarged by fixed amounts inside the coordinate domain. No term involving \(R^{-1}\) or \(R^{-2}\) is suppressed.

Retain the actual auxiliary torus and its covering maps:

Only large bands with \(i\ge0\) are used. On a band representative set

The full physical chain rule is

These operators have not replaced the physical ones: \(D_r=\sqrt Q\,\mathfrak r\), \(D_z=\sqrt Q\partial_z\), and \(\mathfrak t_*=Q^{1+h}(\partial_t+N_{\rm abs})\) before evaluation, where \(\mathfrak r=\partial_r+d_rr^{d_r-1}\mathcal L_{\rm abs}\). In particular

They follow directly because the only nonconstant coefficient of the auxiliary part of \(D_r\) depends on \(R\) alone. Constant-coefficient auxiliary derivatives commute.

All vector components below are in the original orthonormal cylindrical frame \((e_r,e_\theta,e_z)\). Its derivatives are \(\partial_\theta e_r=e_\theta\), \(\partial_\theta e_\theta=-e_r\), \(\partial_\theta e_z=0\). For component columns \(u,v\), the exact transport is

The last column comes from differentiating the frame and is part of transport. The dot products of complex harmonic coefficients used below are bilinear, not Hermitian.

Here \(\langle\cdot\rangle_\theta\) averages the cylindrical component columns over \(0\le\theta\le2\pi\), exactly as in the source mean equations. It is not the operation of averaging the Cartesian components while keeping a fixed Cartesian frame. It also differs from auxiliary Haar averaging. For a fixed label with

and a \(\theta\)-independent column \(a\), direct integration gives

Indeed the integral of \(e^{in\theta}\) is \((e^{2\pi in}-1)/(in)=0\) for a nonzero integer \(n\). The frame terms in (NI1.6) have already been expressed in component columns, so this calculation applies to them too.

SC1. Exact average and the two columns

The tangential component order is \((\theta,z)\). For real cylindrical-component fields on the extended domain, let

where Haar measure has mass one. These are a quadratic map and its symmetric bilinear cross map; \(D\mathcal C(w)[v]=\mathcal B_{\rm cov}(w,v)\). The symbol \(\mathcal B_{\rm cov}\) here denotes the covariance map (original (7.23)), not the differential transport map in NI1. The averages act on cylindrical components, with the basis vectors held in that component convention.

A common torus maps onto each band torus by the integer matrix \(J^d\), where

For \(d\ge0\), partition a fundamental domain of the covering torus into the \(14^d\) inverse-image sheets of a fundamental domain of the band torus, ignoring boundaries of measure zero. Change of variables on each sheet contributes \(14^{-d}\). Their sum proves \(\int f(J^dY)\,d\mu(Y)=\int f(Y_i)\,d\mu(Y_i)\). Thus no covering degree multiplies (SC1.1). Smooth periodic integrands allow differentiation under this integral. Common-cover compatibility follows by applying this equality twice.

Fix a slow box \(\beta=(\ell,a)\). Its two labels are \((\beta,\sigma)\), \(\sigma=\pm1\), with the same slow cutoff \(\eta_\beta\). Put

With \(T_g=4+\sqrt2\), the original band choice gives

The positive constant \(r_0\) and the transverse cutoffs are fixed before bands are chosen. The unweighted, uncurled local basis waves are exactly

Here \(k=\lceil\varepsilon^{-1/2}\rceil\), \(kp\in\mathbb Z\setminus\{0\}\), and the pulse cutoff satisfies \(0\le\psi\le1\), \(\psi=1\) for \(|v-L_s/2|\le L_s/5\), with support in \(|v-L_s/2|<L_s/3\). Neither \(\eta_\beta\) nor a scalar stress amplitude is in \(b_\sigma\). Their auxiliary supports are disjoint, including the closed support neighborhoods used for derivatives, as constructed in W2.

For every fixed slow and auxiliary point, write \(k\Phi_\sigma=kp\theta+\omega\). Then

This equality remains exact after every slow derivative; differentiating its right side creates no frequency factor. On the band rectangle the area element is \(\det(v_r,v_t)d\xi_gd\eta_g=(1+b_g^2)c_i\,d\xi_gdv\). The homogeneous amplitude is independent of \(\xi_g\); write \(x_\sigma=(t^h_\sigma)_r\). Hence

The subscript \(g\) retains the cutoff belonging to the sign's own rectangle. Its squared integral is positive and bounded above and below uniformly, since these are fixed nonzero cutoffs from finitely many shapes. The determinant, \(c_i\), the cosine factor \(1/2\), and both stress components have all been retained. For real coefficients independent of both averaging variables, disjointness gives

SC2. Covariance size and direction from the exact pulse

PU3.5 and PU5.5 establish fixed constants \(c_-,c_+,c_x,C_x>0\), independent of the band, with

These bounds use the original \(B_s^2=\lambda_0/[\varepsilon k^2(1+u_*^2)^{3/2}]\). Set \(I_{0,\sigma}=\int\psi^2x_\sigma^2\,dv\) and \(I_{1,\sigma}=\int |v-L_s/2|\psi^2x_\sigma^2/L_s\,dv\), each on \([0,L_s]\). Choose the one band threshold large enough that \(L_s\ge25\). On \(|v-L_s/2|\le\sqrt{L_s}\), the cutoff equals one. Integrating the lower bound there and enlarging the upper integrals to the real line proves

For the last integral, \(2L_s^{-1}\int_0^\infty s e^{-c_-s^2/L_s}\,ds=1/c_-\). The Gaussian integral in the preceding bound follows by squaring the positive integral and using polar coordinates: \(\int_{\mathbb R^2}e^{-c_-(x^2+y^2)}dxdy=2\pi\int_0^\infty r e^{-c_-r^2}dr=\pi/c_-\). Thus these are analytic inequalities, not numerical enclosures. Equations (SC1.2), (SC1.5) and (SC2.2) yield

For clarity, \(h_-^*,h_+^*\) are fixed bounds applying to both signs, not the two different numbers \(h_-\) and \(h_+\).

In the frozen orthonormal tangential frame of PU1,

PU1 and PU5 prove, uniformly throughout the enlarged slow box,

This uses the actual formula \((t^h)_{\rm tan}=-s_axK_a+yN_a\), including its moving-frame error. The derivative of \(s\mapsto-sK+c_0\sqrt{1+s^2}N\) has norm at most \(1+|c_0|\). Since \(|s_\sigma(v)-\sigma u_*|=u_*|v-L_s/2|/L_s\), averaging (SC2.4) with the positive probability density \(\psi^2x_\sigma^2/I_{0,\sigma}\) gives the exact representation

The error is explicitly the average of \(e^{\rm loc}_\sigma\) plus the difference between the displayed reference direction and its midpoint value. Thus it includes both the moving-frame and finite-width contributions. PU1 supplies fixed \(0<A_-\le A_c\le A_+\). All fixed derivatives of \(\mathsf H\) have polynomial bounds in \(S_*\): apply Leibniz to the exact integral (SC1.5), use the PU5 bounds for each differentiated pulse, and integrate on an interval of length \(L_s\le2r_0T_gS_*\). The limits, cutoffs and \(c_i\) are fixed under slow differentiation. No derivative of a pointwise error estimate is being assumed.

SC3. The original cone margin and an explicit inverse estimate

Original (4.20) defines \(t_s=-b_s/a\) and \(v_s=a(1+t_s^2)\). Original (7.1) uses \(g=F(-a,b_s)\), with \(F,a,v_s-2\) bounded below positively. Before freezing the point,

Indeed \(2FN_\theta=-2F/\sqrt{1+t_s^2}\), \(|g|=Fa\sqrt{1+t_s^2}\), so \(\lambda_0^2=-2FN_\theta(2FN_\theta+|g|)=2aF^2(1-2/v_s)\); dividing by \((2FN_\theta)^2\) proves the last identity. Here is the full profile-to-chart comparison behind this calculation. Original (4.3)–(4.4) give \(u_\theta^{(0)}=q^{-A}E\), \(u_z^{(0)}=q^{-A}U\), \(E=\sqrt{2X}F_{\rm prof}\), \(F_{\rm prof}=\phi/\mathsf C\). With \(\xi=q/Q\), their chart values are \(V^{(0)}=\xi^{-A}E\), \(G^{(0)}=\xi^{-A}U\), and

These identities use \(X=R^2/(2\xi)\), with \(\xi,\eta\) fixed under \(\partial_R\), and the full original (4.11) definitions \(a=1-2X\partial_X\log E\), \(b_s=2XU_X/E\). Thus \(F\), \(g\), \(\lambda_0\) and the target stress each retain their displayed positive factor \(\xi^{-A-1/2}\); that factor cancels only in the explicitly computed frame and in the ratio defining \(c_0\). No amplitude normalization \(\mathsf C\) or physical stress factor has been discarded.

Write \(\kappa_c\in(0,2)\) for the constant called \(\kappa\) in original (4.26), to distinguish it from the derivative-loss constant \(\kappa_s=10^{-5}\). For the original unit stress direction \(n=\mathcal T_0/|\mathcal T_0|\), that equation is

For \(T=\mathcal T_{0,*}=(Q/q)^{A+1/2}\mathcal T_0\), (SC3.1) gives

At either edge use the smooth unit direction in (SC3.2), not the undefined quotient of two zero stress components. Its denominator stays uniformly away from zero, since \(t_s\) is bounded on the closed annulus. This proves continuity of the directional quotient there.

Let \(m_c\) be its supremum on the closed annulus. The original choice of the single constant \(u_*>0\) satisfies \(r_u=u_*/\sqrt{1+u_*^2}>m_c\). Such a choice exists because \(m_c<1\) and \(r_u\to1\) as \(u_*\to\infty\). All frozen frames and coefficients range over a compact set with the positive denominators in PU1. For each actual point of the closed shell, hold its unit stress direction fixed while comparing the frame at that point to the frame at the representative; only the frame and its growth coefficient are frozen. The frame is defined on the fixed enlarged annulus even when the representative lies outside the shell. The smooth profile-to-chart map and frame have bounded first derivatives; the enlarged slow box has diameter at most \(CS_*^{-3}\). The mean-value formula along a segment in its fixed enlarged neighborhood therefore bounds the change of the numerator and denominator of this quotient by \(C'S_*^{-3}\). For a positive denominator bounded below by \(d_0\), the quotient difference is at most \(2C'S_*^{-3}(d_0^{-1}+M_0d_0^{-2})\) once its change is at most \(d_0/2\), with \(M_0\) a fixed numerator bound. Choose the band threshold so this difference is at most \((r_u-m_c)/2\). The frozen frame then satisfies

The argument covers the shell edges using their unit directions. It is a value estimate, not smallness of arbitrarily high derivatives. One finite band threshold suffices for this margin and for the pulse estimates.

Put \(F_{NK}=[N\mid K]\); this orthogonal matrix has determinant \(+1\) in the original \((\theta,z)\) orientation. Set

Then the actual covariance matrix, including its error, factors exactly as

In the induced infinity norm, \(\|C_0^{-1}\|_\infty\le M_c=\tfrac12(A_-^{-1}+u_*^{-1})\) and \(\|E_c\|_\infty\le2C_eS_*^{-1/2}\). Enlarge the single threshold so

The geometric series \(\sum_{j\ge0}(-C_0^{-1}E_c)^j\) converges absolutely in this matrix norm and its product with \(I+C_0^{-1}E_c\) is \(I\), by cancellation of partial sums and a remainder tending to zero. This proves invertibility and the inverse norm bound \((1-\vartheta)^{-1}\).

For the target set \(s_T=-T_N/A_c>0\), \(t_T=T_K/u_*\). The reference vector is

The actual vector \(z=D_h\mathsf H^{-1}T\) equals \((I+C_0^{-1}E_c)^{-1}z^0\), whence

Also \(|T|^2=A_c^2s_T^2+u_*^2t_T^2\), so

Divide (SC3.7) by (SC2.3). This proves the positive, two-sided amplitude bounds, with every component retained:

For completeness, every complex eigenvalue of \(C_0^{-1}E_c\) has modulus at most \(\vartheta\): apply the eigenvector equation to a component of maximum modulus. Thus \(|\det(I+C_0^{-1}E_c)|\ge(1-\vartheta)^2\). Along the path multiplying this error matrix by \(t\in[0,1]\), this real determinant cannot vanish and starts at one. It is therefore positive. Consequently

This retains the orientation of both original covariance columns.

SC4. Every derivative and both shell boundaries

The exact chart variables are

Here \(R=r/\sqrt Q\), \(Z=z/Q^D\), \(T_{\rm time}=(1-t)/Q\). On the original chart, \(\xi\) is bounded above and below and \(|\eta|\le1\). The derivative in \(\xi\) of its defining equation is \(1-2h\eta^2\ge1-2h>0\). Repeated implicit differentiation, dividing only by this denominator, bounds every fixed derivative of \(\xi,X,\eta\) and \(\xi^{-A-1/2}\). Apply this chain rule to original (4.27), visually checked on page 33:

It gives \(|T|\ge c'\zeta\) and \(|\partial^IT|\le C_I\zeta\delta^{-a_I}\). These are the precise leading-profile assertions consumed here. V5–V7 construct them, and I1–I2 proves the complete receiving maps with the actual base and its fixed margins.

For every nonzero multi-index \(I\), differentiating \(\mathsf H\mathsf H^{-1}=I_2\) gives the recursion

Induction using SC2 and (SC3.9) bounds each derivative by a finite polynomial in \(S_*\). Multiplying by the full derivatives of \(T\) proves

Set \(a_\sigma=\sqrt{y_\sigma}\) in the open shell. For \(|I|>0\), repeated chain and product rules give a finite sum

Every term has weight \(\zeta^{1/2-n}\zeta^n=\sqrt\zeta\); all powers of \(S_*\) and \(\delta^{-1}\) remain in finite order-dependent bounds. For \(I=0\), take the square root of the upper bound in (SC4.2). Therefore

The fixed source weight is smooth and flat at each radial edge. Taylor's formula with integral remainder, using its vanishing jets, gives \(0\le\zeta\le C_Nd^N\) near either edge for every integer \(N\), where \(d\) is the distance to that edge. The logarithmic distances in \(\delta\) are comparable to \(d\) there, since \(0<X_a<X_b<\infty\). Given any fixed power in (SC4.4), choose \(N\) more than twice that power. Then \(\sqrt\zeta\delta^{-a'_I}\to0\). Extend every derivative by zero. Difference quotients in a smooth boundary coordinate, followed by induction on derivative order, show that these extended derivatives are derivatives of the zero-extended function. Thus \(a_\sigma\) extends smoothly by zero; no value of \(1/a_\sigma\) at an edge is being assigned. This argument uses the original flat weight and does not require replacing it by a special exponential formula.

Define

The coefficients of \(e^{\pm ik\Phi_\sigma}\) are each \(\sqrt\varepsilon a_\sigma\chi_g\psi t^h_\sigma/2\). PU5 bounds each fixed derivative of \(t^h_\sigma\) by a polynomial in \(S_*\) times the original envelope \(P(v)\). Thus the coefficients obey \(\varepsilon^{1/2}S_*^{b_I}\sqrt\zeta\delta^{-a_I}P(v)\) bounds. With the slow cutoff, \(\eta_\beta W_0\in\mathsf W_{1/2}\), including all support and common-torus requirements. Without that cutoff the same local coefficient bounds hold, and (SC1.6) gives

SC5. Signed increments and the exact retained quadratic term

Keep the positive coefficients \(a_\sigma\) just constructed fixed at all later stages. Let \(\Sigma(R,Z,T_{\rm time})\) be the real two-component stress supplied to this operator, independent of both \(\theta\) and the auxiliary variables, with the original \(\mathsf M_\alpha\) estimates

Auxiliary independence is a required input property of this particular map; membership in \(\mathsf M_\alpha\) alone does not assert it. Define in the open shell

The inverse bounds give \(|\partial^I(d_\Sigma)_\sigma|\le C_I\varepsilon^{\alpha-1}S_*^{b'_I}\zeta\delta^{-a'_I}\). Derivatives of \(y_\sigma^{-1/2}\) are finite sums of \(C y_\sigma^{-k-1/2}\prod_{j=1}^k\partial^{I_j}y_\sigma\), including \(k=0\) at order zero. Their weights are \(\zeta^{-k-1/2}\zeta^k=\zeta^{-1/2}\). The extra \(\zeta\) from \(d_\Sigma\) is therefore sufficient for every term in the product rule, giving

SC4's boundary argument proves smooth zero extension. The pulse bounds then prove \(\eta_\beta\mathcal L_\beta\Sigma\in\mathsf W_{\alpha-1/2}\). Linearity in \(\Sigma\) follows because \(\mathsf H\), \(a_\sigma\), and \(b_\sigma\) are fixed, also at the boundary after extension. No sign or relative-size condition on \(\Sigma\) has been introduced.

The coefficients can be moved outside both averages. Cross-label products are zero. Equations (SC1.5) and (SC5.1) now give

Here \(\odot\) denotes componentwise multiplication. The cosine factor \(1/2\) is already in each column of \(\mathsf H\). The cross term has the separate factor two because differentiating \(a_\sigma^2\) gives \(2a_\sigma\delta a_\sigma\). In particular the full change is

Thus the map in (SC5.1) is a right inverse of the derivative of the covariance map. It does not drop the last quadratic term. Its angularly averaged products have the weight \(\varepsilon^{2\alpha-1}\zeta\delta^{-a}P^2\), with \(P^2\le1\), and finite polynomial factors. Averaging over the mass-one auxiliary torus preserves this bound. Therefore \(\mathcal C(\eta_\beta\mathcal L_\beta\Sigma)\in\mathsf M_{2\alpha-1}\). These statements hold for every real \(\alpha\); later improvement of an iterative error additionally depends on its actual exponent.

SC6. Squared partition, chart transitions, and physical assembly

At each fixed \(q>0\), the slow partition has bounded local overlap and \(\sum_\beta\eta_\beta^2=1\), counting each box once, not once for each sign. Define exactly as in original (7.35), with \(Q_\beta=2^{-\ell_\beta}\),

Here \(\sigma(r,z,t)\) is the physical stress input, independent of the averaging variables; it is distinct from the sign label. A common torus can be chosen for all summands active at that point. Their cross-label products vanish on it by W2, even where their slow supports overlap. Then

The scale identity in the first line is \(Q^{-2A}Q^h(Q/q)^{A+1/2}=Q^{-A+h+1/2}q^{-A-1/2}=q^{-A-1/2}\), since \(A=1/2+h\). There is no missing count of two from the sign labels.

In a receiving chart of scale \(Q\), the representatives are

An overlapping band has \(|\ell_\beta-\ell|\le4\). The ratios of its scales \(Q_\beta/Q\), \(\varepsilon_\beta/\varepsilon\) and \(S_{*,\beta}/S_*\) are bounded above and below by constants independent of the band. The exact coordinate transition is

and the common-to-band torus maps are the fixed powers of \(J\) with bounded exponent difference from W2. Each fixed derivative of a coefficient after these linear maps has the same class estimate with a changed finite constant and polynomial power. The profile \(X\), its edge weight and the label's own pulse envelope are retained. Collect coefficients having the same label and harmonic before testing membership. Bounded local overlap then proves

The equalities are compatible on chart overlaps because they arise from the same physical fields in (SC6.1). Local finiteness for \(q>0\) suffices here. No extension at the singular terminal point is asserted by this assembly.

SC7. Full curls and the stress of the actual velocity

For each harmonic \(t_me^{ikm\Phi}\) of the assembled primary or signed field, \(m\ne0\), let \(n=\nabla_*\Phi\). Its full coefficient is transverse: \(n\cdot t_m=0\). PU1 proves a uniform positive lower bound for \(|n|\). The exact potential and remainder from PU6 are

The curl of \(C_me^{ikm\Phi}\) is \((t_m+r_m)e^{ikm\Phi}\), since \(ikm n\times C_m=t_m\). Its divergence is exactly zero, including the cylindrical connection terms, by NI4. For a signed amplitude in \(\mathsf W_{\alpha-1/2}\), PU6 gives \(C_m\in\mathsf W_\alpha\) and \(r_m\in\mathsf W_{\alpha-\kappa_s}\). It preserves all closed support restrictions. The \(m=+1,-1\) coefficients are conjugates, including their potentials: the sign of \(i/m\) changes on conjugation. The assembled curl field is therefore real.

In particular the initial actual wave produced by the primary amplitudes is

Its covariance in any chart is the exact expression

where the two additional terms are in \(\mathsf M_{3/2-\kappa_s}\) and \(\mathsf M_{2-2\kappa_s}\), respectively. The target in this receiving chart is the physical one from (SC6.2), not a target frozen at a different box. Since \(\kappa_s<1/2\), their sum is in \(\mathsf M_{3/2-\kappa_s}\). This supplies the primary stress wave and its actual curl defect to the iteration, using the complete actual base in Z0–Z9.

At a later stage write the actual current wave as \(U=W_0^{\rm as}+E\), with its original recorded classes \(U\in\mathsf W_{1/2}\), \(E\in\mathsf W_\rho\). For a signed increment let \(V=\mathcal L^{\rm as}\Sigma+R\), where \(R\in\mathsf W_{\alpha-\kappa_s}\) is its full curl remainder. Direct expansion, with no discarded cross term, gives

All of \(\mathcal C(\mathcal L^{\rm as}\Sigma)\), \(\mathcal B_{\rm cov}(\mathcal L^{\rm as}\Sigma,R)\) and \(\mathcal C(R)\) remain inside \(\mathcal C(V)\). The product estimates give respectively

For the last bound use \(\alpha-\kappa_s>\alpha-1/2\), so \(V\in\mathsf W_{\alpha-1/2}\). To justify these product statements at every derivative order, angular averaging of a same-label pair retains only harmonics with \(m+m'=0\), because \(kp\ne0\); their exponential phases cancel exactly. Cross-label products are identically zero. Each derivative of the remaining coefficient products has weight \(\varepsilon^{\mu+\mu'}S_*^b\zeta\delta^{-a}P^2\), and \(P^2\le1\). Integrating these finite sums over a mass-one common torus proves the displayed mean estimates with all derivatives. This verifies original (7.41)–(7.42) for the actual curl velocities.

SC8. Radial divergence after both averages

There is a small stronger estimate at this exact receiver. The three expressions in (SC7.5) have already been averaged over both the angle and the auxiliary torus. For any same-label zero-harmonic product coefficient \(f\), on a fixed common torus with index \(i_0\), holding the receiving scale \(Q\) fixed,

The second equality is integration by parts on the periodic torus; boundary faces cancel. Consequently \(D_r\langle f\rangle_H=\partial_R\langle f\rangle_H=\langle D_rf\rangle_H\). Differentiating the surviving coefficient products by ordinary \(\partial_R\) uses their coefficient bounds and introduces no \(\varepsilon^{-\kappa_s}\) factor. The derivative of the cancelled exponential is zero; it must not be estimated separately before cancellation.

For a two-component auxiliary-averaged radial stress \(S=(S_\theta,S_z)\), define its cylindrical tangential divergence

The shell has \(R\) bounded above and bounded away from zero. The zero-order factors and all their derivatives are therefore bounded. Equations (SC8.1)–(SC8.2) prove that the divergences of the three terms in (SC7.5) retain precisely the exponents displayed there. Original Corollary 7.8 allows a further loss of at most \(\kappa_s\); it is unnecessary after this full auxiliary average. The stronger claim does not apply to an angular mean still depending on the auxiliary variable, for which the fast derivative remains. It also does not remove the \(\kappa_s\) already paid when taking a curl.

SC9. All original viscosity factors

For every fixed original viscosity \(\nu>0\), let \(y=x/\sqrt\nu\), with time unchanged. The exact change of fields from M2 is

The last formula gives \(\operatorname{curl}_x\mathcal A_\nu=\sqrt\nu\operatorname{curl}_y\mathcal A_1\), because each spatial derivative contributes \(\nu^{-1/2}\). It applies to every potential in (SC7.1), including all cutoffs and remainders.

The angles and auxiliary torus measures in (SC1.1) do not change under this dilation. There is no physical-space volume integral in that definition. Throughout this note, a covariance of a field in physical units is still the average on the extended domain before evaluating the auxiliary variable at the physical phase map. No equality with the unaveraged pointwise velocity product is inferred. Thus for any two fields \(w,v\),

For a physical input \(\sigma_\nu(x,t)\), set \(\sigma_1(y,t)=\nu^{-1}\sigma_\nu(\sqrt\nu y,t)\), and define

Apply (SC6.2) and (SC9.2) to get the exact identities

Here all original coordinates are

In a chart the stress input is \(\Sigma=\nu^{-1}Q^{2A}\sigma_\nu\); physical velocity, potential and covariance carry factors \(\sqrt\nu Q^{-A}\), \(\nu Q^{1/2-A}\) and \(\nu Q^{-2A}\), respectively. Applying a physical radial divergence multiplies the last factor by \(\nu^{-1/2}Q^{-1/2}\), giving \(\sqrt\nu Q^{-2A-1/2}\), with both cylindrical terms in (SC8.2) preserved. These are exactly the physical residual factors in M2 and PU7. Thus (SC7.3)–(SC8.2) transfer to every \(\nu>0\) without dropping a viscosity factor or an auxiliary average.

SC10. Production with the actual shear and all damping

Here the flux is denoted \(T\) within this section; the time coordinate remains \(T_{\rm time}\). The constants \(c_-,c_+\) in this section bound \(|c_0|\), independently of the Gaussian constants used in SC2. The symbol \(p\) in the slow-box comparison is the slow point, not the angular phase coefficient.

For the real homogeneous amplitude \(t\), \(n_\Phi\cdot t=0\) makes the normal component of \(\mathcal A_\Phi t\) perpendicular to \(t\). Direct multiplication, keeping both cylindrical \(2F\) terms, gives

Taking \(t\cdot t'\) therefore proves the exact identity

For the flux \(T=t_r(t_\theta,t_z)=T_NN+T_KK\), the exact production is

Here \(g_0\) is the leading shear frozen at the representative; it is not identified with the actual shear even there. To keep both sources of its difference, let \(V^{(0)},G^{(0)}\) denote the exact leading chart fields, put \(\delta V=V-V^{(0)}\), \(\delta G=G-G^{(0)}\), and write

The leading and actual fields here are the constructed backgrounds. Their proved chart estimates give \(|\partial^I\delta V|\le C_{V,I}\varepsilon^2\) and \(|\partial^I\delta G|\le C_{G,I}\varepsilon^2\). Let \(R_->0\) be the minimum radius on the compact enlarged chart, \(L_g\) bound \(D_pg^{(0)}\), and \(|p-p_0|\le C_{\rm box}S_*^{-3}\), with the connecting segment in that chart. Set

The product rule and segment integration give the actual bound

At \(p_0\) only the segment term is zero. The Borel term remains. For a vector in the prescribed compact stress cone, retain its margin \(-T_N\ge c_T|T|\), \(c_T>0\), and \(|g_0|\ge g_->0\). Then Cauchy--Schwarz proves

One sufficient additional integer threshold is

Each error is then at most \(g_-c_T/4\), so the lower bound is \((g_-c_T/2)|T|\), including \(T=0\).

For the actual homogeneous pulse, positivity follows directly from its proved polarization. This requires no assertion that its instantaneous flux lies in the prescribed target cone. Retain

and choose the constants in the already proved estimates as

The quantities \(g_-,g_+,c_-,c_+\) denote positive uniform bounds \(g_-\le|g_0|\le g_+\), \(c_-\le|c_0|\le c_+\), with \(c_0<0\). Choose \(S_*\ge2E_n/B_-\). Then \(|n_{\Phi,\tan}|\ge B_-/2\). For nonzero vectors \(a,b\), addition and subtraction of \(a/|b|\), followed by \(\big||a|-|b|\big|\le|a-b|\), gives \(|a/|a|-b/|b||\le2|a-b|/|b|\). Therefore

The last estimate follows by writing \(n_{\Phi,r}-s|n_{\Phi,\tan}| =(n_{\Phi,r}-B_ss)+s(B_s-|n_{\Phi,\tan}|)\) and dividing by the actual tangential norm. Define

The exact flux and its exact remainder are

Expand the first difference as \((s_a-s)K_a+s(K_a-K)\) and the second as \((y/x-\gamma_0)N_a+\gamma_0(N_a-N)\). The preceding bounds give \(|e_T|\le C_T/S_*\), with all four terms retained. Hence

The absolute value of the last two terms is at most

Set \(W=s_{\max}+\Gamma_{\max}+C_T>0\). In addition to all previous phase and invariant-band thresholds, it suffices to choose

For \(\ell\ge1\), the last factor in \(\mathcal E_\ell\) is at most \(W\). The three error contributions are then at most \(g_-c_-/4,g_-c_-/8,g_-c_-/8\), respectively. Since \(-|g_0|\gamma_0\ge g_-c_-\), this proves

All constants concern the same selected \(h\), fixed profiles and original compact label family. The displayed thresholds are finite and are added to the finite value thresholds in PU1–PU5; no uniformity as \(h\downarrow0\) is asserted. The exact actual-frame norm is

The damping \(\varepsilon k^2|n_\Phi|^2|t|^2\) remains in the energy equation, so positive production is not a claim of positive total energy derivative. Covariance mass uses the proved positive \(x\) and the original positive integral of \(\psi^2x^2\); its argument does not omit this damping. The carrier physical wavelength is \(\sqrt Q/k\asymp Q^{1/2+h/2}\), while the leading physical wave amplitude has power \(Q^{-A}\sqrt\varepsilon=Q^{-1/2-h/2}\), up to polynomial factors in \(S_*\). Their product has power \(Q^0\); this agrees with the retained leading damping \(\varepsilon k^2\asymp1\).

NI4. The exact curl identity used by the nonlinear terms

For a smooth potential column \(C\), the full source curl is

The divergence is \((D_r+R^{-1})u_r+R^{-1}\partial_\theta u_\theta+D_zu_z\). Inserting (NI4.1) yields all six terms

For the first term we used \((D_r+R^{-1})(R^{-1}f)=R^{-1}D_rf\). The three pairs cancel by (NI1.5), including the \(R^{-1}\) contributions.

For a \(\theta\)-independent potential amplitude \(C_m\), expand the actual full curl as

All potential derivatives, including derivatives of its slow and transverse cutoffs, remain in this expression. A cutoff in \(v\) also remains inside the potential. In the original coordinates \(D_rv=D_zv=0\), so its spatial derivative terms vanish for that exact reason, not because the cutoff is omitted.

Where the source writes \(C_m=i(n\times t_m)/(km|n|^2)\), \(n\cdot t_m=0\), the first term in (NI4.3) equals

This is a pointwise identity for the original potential wherever it is defined. The uniform denominator estimate and the potential class bounds for the original phase are now derived in the pulse audit, PU1 and PU6, retaining the source base inputs. The nonlinear cancellation in this section uses (NI4.2), and therefore holds for every smooth potential coefficient, independently of that particular formula or of any pulse solution.

Expansion of the zero divergence in (NI4.2) now proves, for the full actual amplitude,

If \(a_m\in\mathsf W_\alpha\), (W3.3) gives \(\mathfrak d a_m\in\mathsf W_{\alpha-\kappa_s}\). Since \(|m|\ge1\) and \(k^{-1}\le\varepsilon^{1/2}\), division in the exact identity proves

This remains true for all fixed coefficient derivatives because \(k,m\) are constant within the label and the bound for \(\mathfrak d a_m\) holds at those orders. Using only the transverse leading amplitude while omitting the remainder in (NI4.3) would not justify (NI4.5) for the actual velocity.

Conversely, any exactly divergence-free harmonic \(a_me^{ikm\Phi}\) with a \(\theta\)-independent amplitude satisfies (NI4.5) simply by expanding its divergence. Thus the nonlinear estimates below need exact divergence, not a particular choice of vector potential. The actual source waves satisfy the identity through the full curl construction (NI4.2).

NI5. Complete mean–wave interaction and its output support

Write the actual mean increment as \(d=(d^{r},d^{\theta},d^{z})\), avoiding a replacement by its tangential leading part. It is independent of \(\theta\); it may depend on the full auxiliary torus. For \(w=ae^{ikm\Phi}\), (NI1.6) gives exactly

To obtain it, the derivative on the wave exponential contributes \(ikm n\) in all three spatial directions. The mean amplitude has no angular component derivative, while each transport contributes its own frame term. Their sum is the last line; in particular its radial coefficient is \(-2\), not \(-1\). No divergence identity is needed for (NI5.1).

The source mean increments have \(d^{\theta},d^{z}\in\mathsf M_\mu\) and \(d^{r}\in\mathsf M_{\mu+1}\). For \(a\in\mathsf W_\alpha\), each complete group in (NI5.1) has the following lower exponent, with all powers of \(S_*\), \(\delta^{-1}\), weights and fixed derivatives handled by W3:

Since \(\kappa_s=10^{-5}<1/2\) and \(\varepsilon\le1\), every group belongs to \(\mathsf W_{\alpha+\mu-1/2}\). Addition proves the first part of original Lemma 9.2, without discarding any cylindrical term.

Multiplication by a mean, regardless of the extent of its auxiliary support, cannot enlarge the wave support. The derivative of a smoothly extended wave still has its original closed support. The term differentiating the mean contains the undifferentiated wave. Every term in (NI5.1) therefore keeps the prescribed slow support, transverse support, enlarged rectangle, shell and actual temporal cutoff support. Its bounds retain the factor \(P_v\). There is no division by \(\chi_g\), \(\eta_\gamma\) or \(\psi\). The smooth mean multiplied into the smoothly extended wave has the same zero extensions. The result retains precisely the original label and nonzero integer \(m\), so (NI1.7) gives zero angular mean.

For a finite or locally finite wave sum, first collect equal \((\gamma,m)\) pairs as in Definition 6.5. Formula (NI5.1) is linear in each collected wave coefficient. Hence its bound uses the already uniform bound for that coefficient, without a factor counting its original representations. Fixed-stage finite harmonic sets remain unchanged under this operation. Local finiteness in \(q>0\) permits the equality to be summed locally. These observations prove the asserted map on the actual sums, not just on one exponential.

There is a limited strengthening of the source statement. The same output exponent holds when the radial mean belongs only to \(\mathsf M_\mu\). In the four groups involving \(d^{r}\) or its derivatives, lower the listed exponent by one. Their new lower exponents are respectively \(\alpha+\mu-1/2\), \(\alpha+\mu-\kappa_s\), \(\alpha+\mu-\kappa_s\) or \(\alpha+\mu+1\), and \(\alpha+\mu\); all are at least \(\alpha+\mu-1/2\). All other groups and support arguments are unchanged. Thus the extra radial smallness is unnecessary for this particular interaction estimate. It remains available and important in other mean equations; this calculation changes no hypothesis elsewhere in the iteration and claims no novelty.

NI6. Same-label wave interactions, with the phase contraction evaluated exactly

Take two actual harmonics of one label, \(w=ae^{ikm\Phi}\), \(w'=be^{ikm'\Phi}\), where \(m,m'\ne0\), and use their full exactly divergence-free amplitudes. Before invoking divergence, exact transport is

By (NI4.5), its complete coefficient is instead the equal expression

The factor \(m'/m\), the sign and the \(a_r/R\) term inside \(\mathfrak d a\) are retained. It is an identity, not a projection or an approximate cancellation. The coefficient of the symmetrized expression is

If the input exponents are \(\alpha,\alpha'\), every radial derivative term in (NI6.2) has exponent at least \(\alpha+\alpha'-\kappa_s\), every axial derivative term at least \(\alpha+\alpha'+1\), and every undifferentiated connection at least \(\alpha+\alpha'\). The same bounds apply to \(\mathfrak d a\) expanded as \(D_ra_r+a_r/R+D_za_z\). The finite nonzero harmonic sets bound \(|m'/m|\) by \(\max_{n\in\mathscr H_j'}|n|\), and the reverse ratio similarly. These constants may depend on the fixed stage, as the class definitions allow. This proves the exponent \(\alpha+\alpha'-\kappa_s\), with no half-power loss.

If \(m+m'\ne0\), the output has that nonzero integer harmonic and belongs to \(\mathsf W_{\alpha+\alpha'-\kappa_s}\) by W3. If \(m+m'=0\), the exponential is exactly one and the product weight is \(\zeta P_v^2\le\zeta\). The actual temporal cutoffs give a smooth field on the full auxiliary torus, so this output belongs to \(\mathsf M_{\alpha+\alpha'-\kappa_s}\), not merely to a class on an unextended rectangle. Its auxiliary Haar average may be nonzero and is retained by the mean equations. Before cutoff the corresponding assertion is only local on the labelled rectangle.

The support of every product lies in the intersection of the two input supports. Its extension and all fixed derivative bounds follow as in NI5. Distinct-label products and all their derivative terms are exactly zero by W2. Thus there are no additional cross-label frequencies, even if angular frequencies of different labels happen to coincide. Within a label the new finite harmonic set is contained in \((\mathscr H_j+\mathscr H_j')\setminus\{0\}\), with zero outputs assigned to the mean. Collect the finitely many pairs giving the same output integer; their number is at most \(|\mathscr H_j||\mathscr H_j'|\), independent of the band. This proves the map for the actual locally finite sums.

In particular the unsymmetrized transport (NI6.2) has the same bound. For a residual update \(w\mapsto w+\delta w\) the exact quadratic change is

The final term has coefficient one. Treating it as the symmetrized expression with identical arguments would introduce an erroneous factor two. Equations (NI6.2) and (NI6.3) bound the three terms in (NI6.4) with their actual multiplicities.

NI7. Receiver of the two mean updates and return to physical viscosity

Let the current full field be \(U=(b+\beta,V+v,G+\gamma)+w\). The earlier moment audit, M4–M8, and temporal audit, T3–T4, supply the actual mean increments

in Steps 3 and 4 of Proposition 9.6. In the temporal update, \(\delta\gamma=\gamma_d-\mathsf A_1\gamma_d\) is the full compactly realized axial field. The term \(-\mathsf A_1\gamma_d\), although flat after the established scaling, remains in (NI7.1) and in every occurrence of \(d^{z}\) in (NI5.1). The compact radial field \(\delta\beta=-\varepsilon\partial_Z\Psi\) also remains. The moment update has its complete supported potential and exact auxiliary-independent increments. No cutoff remainder is removed in applying this note.

For the full residual \(\mathscr R(U,P)=\mathfrak t_*U+\mathcal T_{\rm adv}(U,U)+\nabla_*P-\varepsilon\Delta_*U\), with the full cylindrical vector Laplacian, a mean update \((d,\delta P)\) has exact change

Every term except \(\mathcal T_{\rm adv}(w,d)+\mathcal T_{\rm adv}(d,w)\) is angularly constant in cylindrical components: the full vector Laplacian maps a mean column to

and the other linear terms and mean products also have no angular coefficient dependence. Thus, denoting removal of the cylindrical angular mean by \(\mathsf P_{\ne0}\),

This is the exact map between the already checked mean update and its nonzero-harmonic receiver. For the source iteration's incoming wave class \(\mathsf W_{1/2}\), NI5 gives an added nonzero source in \(\mathsf W_H\), with all the supports and envelope bounds required for the next pulse source. It has zero angular mean, so it adds no omitted term to the previously computed mean defects.

The physical conversion retains every scale. At viscosity one, \(u_*=Q^Au\) and \(\mathscr R_*=Q^{2A+1/2}\mathscr R\); hence the physical mixed transport is \(Q^{-2A-1/2}\) times the right side of (NI7.3), evaluated at the original auxiliary map. For any original \(\nu>0\), use the exact map already proved in M2.4–M2.7:

Each spatial derivative contributes \(\nu^{-1/2}\); consequently

while \(\nu\Delta_xu_\nu=\sqrt\nu\Delta_yu_1\). The physical receiver of (NI5.1) is therefore \(\sqrt\nu Q^{-2A-1/2}\mathcal Q_m(d_*,a_*)e^{ikm\Phi}\), with

The angle and time remain unchanged. A chart potential has physical factor \(\nu Q^{1/2-A}\), since its curl contributes \((\nu Q)^{-1/2}\); a chart pressure has physical factor \(\nu Q^{-2A}\). Thus exact divergence, support separation and all the displayed nonlinear identities survive this map with their full factors. This does not remove viscosity from the original theorem.

SC11. Five worked exercises

Exercise 1. An exact finite amplitude change

For the local waves in SC4–SC5, determine exactly which signed changes admit positive real squared amplitudes. Compare the finite map with its derivative and second derivative.

Solution. Keep the slow point and both averaging operations fixed. Put \(d=\mathsf H^{-1}(\Sigma/\varepsilon)\), and let

This is the intersection of two open half-lines or whole lines; it contains zero. For \(s\in I\), disjointness and the exact covariance integral give

Conversely, any field with positive real amplitudes in this fixed two-pulse basis has squared amplitudes \(\mathsf H^{-1}(T+s\Sigma/\varepsilon)=a^2+sd\), because \(\mathsf H\) is invertible. Thus \(I\) is the exact domain for positive amplitudes in this basis. Allowing a zero amplitude changes the strict inequalities to non-strict inequalities; smooth dependence at such a zero is not asserted.

At zero, direct differentiation gives

Differentiating SC11.2 twice, or substituting these expressions in the covariance integral, proves

Indeed the coefficients inside \(\varepsilon\mathsf H\) are respectively \(-d_\sigma^2/(2a_\sigma^2)\) and \(d_\sigma^2/(2a_\sigma^2)\). For a finite \(s\), rationalizing the square root gives the exact increment

On any parameter range with \(a_\sigma^2+s d_\sigma\ge c a_\sigma^2\), \(c>0\), and bounded \(s\), every differentiated numerator has the original \(\zeta\) weight and every inverse square-root derivative has weight \(\zeta^{-1/2}\) times finite powers of \(\delta^{-1}\) and \(S_*\). The chain rule in SC4 therefore proves the same smooth zero extension for this finite increment. This is an identity for the leading pulse field. Taking its actual curl adds the terms in SC7; those terms are not cancelled by SC11.2.

Exercise 2. What the signed map does to an auxiliary-dependent input

Show exactly why auxiliary independence is required in SC5. Determine the map obtained by applying the same amplitude formula to an auxiliary-dependent stress.

Solution. At a fixed slow point in the open shell define the matrix-valued function

Suppose the same formula is used with \(\delta a_\sigma(Y)=(\mathsf H^{-1}\Sigma(Y))_\sigma/(2\varepsilon a_\sigma)\). Coefficients can leave the angular average, but they cannot leave the auxiliary integral. Direct substitution gives exactly

Regarding the result as a constant function of \(Y\), \(\mathsf P\) is a projection onto the two-dimensional space of constant vectors:

The first equality follows from \(\int\mathsf K=\mathsf H\), the second applies the first to \(\mathsf P\Sigma\), and the displayed kernel is the defining equality.

This projection is not the ordinary Haar average. The two pulse supports are nonempty disjoint closed rectangles. Their union cannot be the connected torus: a union of two disjoint nonempty closed sets covering the torus would disconnect it. Its complement is a nonempty open set. Choose a nonnegative nonzero smooth bump \(\beta\) with support there and a nonzero constant vector \(c\). For \(\Sigma(Y)=\beta(Y)c\), \(\mathsf K\Sigma=0\) pointwise, so \(\mathsf P\Sigma=0\), while \(\int\Sigma\,d\mu=c\int\beta\,d\mu\ne0\). A supported slow scalar factor with the original edge weight can be included in both expressions. The actual iteration uses the auxiliary-independent stress supplied by its averaged equations; SC5 inverts precisely that domain.

Exercise 3. All terms in a finite curled covariance update

For \(U=W_0^{\rm as}+E\), put \(L=\mathcal L^{\rm as}\Sigma\) and \(V=L+R\), as in SC7. Expand the covariance of \(U+sV\) and retain the exponent of every term.

Solution. Bilinearity and symmetry give the exact polynomial

The five remainders, in their displayed order, belong to

To verify them, multiply the exponents of \(E\in\mathsf W_\rho\), \(U\in\mathsf W_{1/2}\), \(L\in\mathsf W_{\alpha-1/2}\), and \(R\in\mathsf W_{\alpha-\kappa_s}\). The two square-root edge weights multiply to \(\zeta\); the envelopes multiply to \(P^2\le1\). Opposite harmonics cancel their phases before any coefficient derivative is taken, and distinct-label products vanish. Leibniz' rule and the mass-one auxiliary integral give every derivative estimate. On a bounded interval of \(s\), its powers change only constants.

After both averages, SC8 proves that \(\mathfrak D_r\) retains each of these exponents. In physical units the covariance factor is \(\nu Q^{-2A}\), and \(\partial_{r_\nu}=(\nu Q)^{-1/2}D_r\) on the evaluated chart; the two connection coefficients scale by the same factor. Thus the physical tangential stress divergence has factor \(\sqrt\nu Q^{-2A-1/2}\), with all five remainders retained. No complete-cycle gain is inferred from this single update.

Exercise 4. The conservative mean of a real harmonic

Let \(w=ae^{ikm\Phi}+\overline a e^{-ikm\Phi}\) be an actual real divergence-free harmonic pair. Compute its full angularly averaged nonlinear transport.

Solution. Define the symmetric real matrix

The nonzero double harmonics integrate to zero because \(kp m\) is a nonzero integer. For each component \(j\), the product rule gives

The last term is zero. Add the cylindrical frame column \(R^{-1}(-w_\theta^2,w_\theta w_r,0)\), then take the angular average. Periodicity removes the angular derivative and the slow operators commute with that average. The result, component by component, is

These are the original conservative mean terms. They include every part of \(a=t+r\), hence every cutoff and curl remainder. An auxiliary average may now be taken separately; it has not been included in \(W\).

Exercise 5. The exact transport defect without incompressibility

For two same-label harmonics, determine which divergence identities are needed for the unsymmetrized estimate and retain the error when a divergence is nonzero.

Solution. For the advecting coefficient \(a\), define its actual divergence coefficient by

Multiplying this identity by \(m'/m\) gives \(ikm'n\cdot a=(m'/m)(\omega_a-\mathfrak d a)\). Substitute it into the complete transport formula NI6.1:

This identity is valid for arbitrary smooth coefficients with the stated angle independence. If \(a\in\mathsf W_\alpha\), \(b\in\mathsf W_{\alpha'}\) and \(\omega_a\in\mathsf W_\gamma\), the exact output has exponent at least

All radial derivative terms use the first exponent; the extra divergence product uses the second. The finite harmonic ratios, supports and zero-harmonic mean receiver are exactly those in NI6. If the advecting wave is divergence free, \(\omega_a=0\), the target coefficient \(b\) need not be divergence free for this unsymmetrized estimate. For the symmetrized estimate, retain also \((m/m')\omega_ba\), or impose both zero divergences.

For a transverse leading coefficient \(t\), \(n\cdot t=0\) gives \(\omega_t=\mathfrak d t\). Substitution in SC11.15 cancels the two ratio terms exactly. Its remaining transport terms obey the same class estimate, since \(\mathfrak d t\in\mathsf W_{\alpha-\kappa_s}\). This gives the exact relationship between transversality and the nonlinear bound. The constructed Navier–Stokes velocity still requires its full curl to have zero divergence; no leading field is substituted for it.

The five exercises are analytic identities and estimates for the specified objects. They introduce no numerical validation, new original-source defect or claim of novelty. The next lesson constructs the mean corrections that receive these stresses and conservative mean terms.

Radial primitives and the complete mean equations

The mean correction needs a compact pressure and an actual divergence-free velocity. A radial integral alone is insufficient: its source depends on a torus coordinate that changes along a physical radius, and its full radial integral can leave a nonzero tail. This lesson constructs the original radial operator, retains its tail exactly, proves all its derivative estimates, and derives the three scalar defects that the later mean operations must remove.

The complete component covers original section 8.1–8.4, through Corollary 8.5: the conservative equations, weighted primitives, pressure, axial potential, integrated identities and compact tangential stress. RM4 retains the exact map back to the original viscosity. The auxiliary-time inverse and the five-profile moment solve are subsequent components. The preceding pulse lesson, stress lesson and actual base construction supply their complete input fields.

This lesson gives separate mathematical exposition of the radial operators and mean equations. The original source passages and the preceding constructions are cited below.

The diagram displays the maps proved in RP1–RP8 and RM1–RM4. The editable diagram source uses the exact formulas, without numerical flow samples.

RP1. The physical map and every scale factor

Keep the original constants and coordinates

For a dyadic band and its auxiliary covering, the source uses

Only sufficiently large \(\ell\), for which \(i\geq0\), are used. The physical evaluation is

Direct multiplication gives \(J_gv_r=\Lambda_gv_r\) and \(J_gv_t=T_gv_t\). Before evaluating a coefficient at (RP1.3), its physical derivatives are consequently

Both derivatives on the right hold the independent \(Y\) fixed. Physical axial differentiation is \(\partial_z\), since \(Y\) does not depend on \(z\).

Fix a common covering index \(i_0\) on a neighborhood and write \(y=J_g^{i_0}Y\). All discrete indices and \(Q\) remain fixed during differentiation there. Put