Interior vorticity bounds on an actual annulus

We now turn the annular energy and local velocity estimates into pointwise bounds for the original velocity, vorticity and their first derivatives. Six explicit heat estimates give the vorticity maximum. A spatial continuity estimate and the complete local Newton formula then control the velocity gradient. Finally we differentiate the localized heat equation.

The construction retains a positive physical time margin. We also prove how to choose an actual annulus on which all four maxima are as small as prescribed, for unforced flow or an annulus beyond the support of an actual compact force. The full global force costs remain in that construction.

Read Selecting an annulus and controlling local velocity first. Its complete initial-data selection, heat propagation and energy estimates are the inputs here. The human comparison is Terence Tao, Quantitative bounds for critically bounded solutions to the Navier–Stokes equations, version 2, the end of the annuli-of-regularity proof. We supply the full receiving argument with original viscosity and force, including the local Newton boundary fields and point mass. The general large critical-velocity endpoint and its backpropagation argument remain unfinished.

1. The original fields, domains and physical margins

Keep the smooth original solution on \(\mathbb R^3\):

Use the actual earlier heat flow \(v(t)=H_{\nu(t-t_b)}*u(t_b)\), where \(H_s(x)=(4\pi s)^{-3/2}e^{-|x|^2/(4s)}\), and retain \(w=u-v\), \(\ell=\nabla\times v\), \(z=\nabla\times w\), and \(\omega=\nabla\times u=z+\ell\). The preceding lesson specifies the original smooth class, force decomposition and all global constants. Its actual selected interval is \(J=[t_1,t_2]\); write \(T=t_2-t_1>0\).

In particular its equations (6.1)–(6.12) give the fixed domain and full velocity bound

Here \(P_0,Q_0\) are precisely the complete cutoff quantities of that lesson, including the original velocity terms. The other actual inputs are

All derivatives use the Euclidean norm of the full ordered tensor. The constants \(h,A,B,r_-,r_+,E_*,W_0\) remain those of the selected original annulus. No equation has been rescaled.

Choose the physical ball radius \(\rho_b=h/32\), spatial step \(\epsilon=\rho_b/32=h/1024\), and time step \(\tau=T/16\). For every center \(x_0\) in the nonempty shell \(\mathcal K=\{b_-+\rho_b\leq|x_0|\leq b_+-\rho_b\}\), set

The closed ball of radius \(\rho_b\) about \(x_0\) is in \(D_{\rm out}\), by the triangle inequality. Also \(R_{10}=3\rho_b/8>0\). Estimates uniform in this center will therefore give estimates on the whole shell \(\mathcal K\). This use of fixed-radius balls avoids introducing the volume of the potentially very large selected annulus.

Let \(j\) be the exact smooth transition in equation (6.2) of the preceding lesson, with \(j=0\) on \((-\infty,0]\), \(j=1\) on \([1,\infty)\), and \(0\leq j\leq1\). Put \(b_1=\|j'\|_\infty\), \(b_2=\|j''\|_\infty\). Define, for \(1\leq k\leq9\),

The spatial cutoff is supported in the closure of \(B_{k-1}\), equals one on \(B_k\), and has its derivative band at distance at least \(\epsilon\) from \(B_k\). It is constant near the center. On its derivative band the radius is greater than \(R_{10}\), so the complete radial gradient, Laplacian and Hessian bounds are

For the Hessian, its three eigenvalues are the radial second derivative and twice the radial first derivative divided by the radius. Its Euclidean norm is at most the sum of the absolute second derivative and \(\sqrt2\) times that latter quantity; this is bounded by \(L_2\). Thus the curvature contribution has been retained. Moreover \(\chi_k(s_{k-1})=0\) and \(\Psi_k=1\) on \([s_k,t_2]\times B_k\).

Figure 1. Exact spatial radii and time margins for the displayed example \(\rho_b=1\), \(t_1=2\), \(t_2=4\). The first panel shows the physical radial sections of the nested balls. The second shows the six proved spatial exponent gains and their actual time cutoffs. The last time margin used for the derivative bound is also shown. These are cutoff geometry and exponent values; they are not sampled fluid trajectories. Equations (1.4)–(1.6) and (2.1) specify the objects. Reproducible figure source.

The initial local vorticity estimate is

Indeed the first term bounds \(z\), and \(|\ell|\leq\sqrt2|\nabla v|\) gives the second using the actual ball volume.

For a smooth original force retain its actual local datum

It is finite when \(f\) is smooth on this closed compact region, and is exactly zero for the unforced source equation. The earlier \(L^1H^1+L^2L^2\) force bounds alone do not bound (1.8). We will keep their exact heat-potential map visible, and Exercise 4 tests this distinction with complete original forced solutions.

2. Six explicit heat estimates

Set \(1/p_k=1/2-k/12\) for \(0\leq k\leq6\). Thus

Every adjacent pair loses exactly \(1/12\) in reciprocal spatial exponent. Young's convolution formula uses \(r=12/11\), since \(1+1/p_k=1/r+1/p_{k-1}\). Define the full Gaussian kernel constants

Scaling the original Gaussian gives \(\|H_{\nu a}\|_{12/11}=K_0(\nu a)^{-1/8}\) and \(\sum_j\|\partial_jH_{\nu a}\|_{12/11} =K_1(\nu a)^{-5/8}\). For a vector field \(G\), \(\nabla\times(H*G)=\sum_je_j\times(\partial_jH*G)\). Since \(|e_j\times v|\leq|v|\), these same constants bound the full vector curl. Every component is retained.

The precise time integrals needed below are

They are all finite at zero. The first two are the \(L^{4/3}\) norms of the respective kernels, for pairing with the actual \(L^4\)-time velocity.

2.1. One full step, including every cutoff and force term

Suppose the preceding step has already proved \(\sup_{[s_{k-1},t_2]}\|\omega(t)\|_{L^{p_{k-1}}(B_{k-1})} \leq X_{k-1}\). For \(k=1\), (1.7) proves exactly this statement. Set \(\Omega_k=\Psi_k\omega\). The complete localized equation from the preceding lesson is

Its initial value at \(s_{k-1}\) is zero because the actual time cutoff vanishes there. Duhamel therefore contains four non-force contributions: the curl flux, the cutoff divergence, the scalar cutoff source, and the nonlinear boundary cross product. Their bounds in \(L^\infty([s_{k-1},t_2];L^{p_k}(\mathbb R^3))\) are, in that order,

For example the first follows by Young's spatial inequality with (2.2), followed by Hölder in time with the first norm in (2.3). The second uses the factor \(2\nu\), the bound \(|\partial_j\Psi_k|\leq L_1\), and the third integral of (2.3). The third uses \(|\partial_t\Psi_k+\nu\Delta\Psi_k|\leq L_t+\nu L_2\) and the fourth integral. The last pairs the \(L^4\)-time velocity with the second norm of (2.3). Every spatial product is supported in \(B_{k-1}\), and every time is at least \(s_{k-1}\). Thus only the already proved previous norm occurs on the right side.

The two force terms have the exact combined potential

The equality is the product rule and spatial integration by parts, including the full boundary term. It holds distributionally for the original rough force classes. For the actual smooth force in (1.8), positivity and mass one make the heat semigroup an \(L^{p_k}\) contraction. The input is bounded by \(F_{\rm curl}\) on a ball of volume at most \(v_3\rho_b^3\). Hence its norm is at most \(T(v_3\rho_b^3)^{1/p_k}F_{\rm curl}\), including \(p_k=\infty\).

2.2. Completing all six steps

Define the full coefficient

For \(1\leq k\leq6\), put

All quantities are finite in the smooth-force class. Since \(\Psi_k=1\) on \([s_k,t_2]\times B_k\), the proved one-step estimate supplies the induction. It gives in particular

The last \(L^{12}\)-to-\(L^\infty\) step uses exactly the same integrable exponent \(5/8\). Thus there is no missing endpoint convolution or unspecified repetition. When \(f=0\), the full force contribution is zero and \(X_k=C^kX_0\).

For a rough force, retain instead the actual norm \(F_k=\|\mathcal H_{\Psi_k,f}\|_{L^\infty_tL^{p_k}_x}\) on \([s_{k-1},t_2]\times\mathbb R^3\). The same proof gives \(X_k=CX_{k-1}+F_k\), interpreted in the extended nonnegative reals. It does not assert that the earlier coarse force norms make all these quantities finite. The subsequent finite estimates use the unforced or the stated smooth-force input for which finiteness was proved.

3. Spatial continuity of the vorticity

Fix \(\alpha=1/4\). For an integrable differentiable scalar kernel \(K\), integration along a line and the triangle inequality give

The second inequality follows by considering which of the two positive quantities is smaller; it also holds if one is zero. The spatial Hölder seminorm \([g]_{C^\alpha}=\sup_{x\ne y}|g(x)-g(y)|/|x-y|^\alpha\) therefore obeys convolution bounds with the complete Gaussian constants

Their original heat-time factors are \((\nu a)^{-\alpha/2}\) and \((\nu a)^{-1/2-\alpha/2}\). For \(\alpha=1/4\) these are again the powers \(1/8\) and \(5/8\).

Apply the full equation (2.4) with \(k=7\), starting at \(s_6\). Its vorticity input is now the proved maximum \(X_6\); its velocity input remains the actual time norm \(\mathcal V\). Using (3.1) in each of the four non-force terms gives exactly (2.7) with \(K_0,K_1\) replaced by \(J_0,J_1\). Denote that complete expression by \(C_H\):

The force potential (2.6) has seminorm at most \((8/7)J_0\nu^{-1/8}T^{7/8}F_{\rm curl}\), using its bounded curl input and the last integral of (2.3). Thus the globally defined compact field \(\Omega_7=\Psi_7\omega\) satisfies

The maximum bound uses the actual support in \(B_6\) and \(|\Psi_7|\leq1\). It is not inferred from a seminorm. For \(t\geq s_7\), \(\Omega_7=\omega\) on \(B_7\), giving the stated spatial continuity of the original vorticity there.

4. The complete local Newton map

4.1. Recovering the velocity with its boundary fields

Use the original nonlinear velocity \(w\), whose global \(L^2\) norm is at most \(W_0\). Put \(X=\psi_8w\) and \(N(x)=1/(4\pi|x|)\). The exact compact-field identity is

Indeed \(-\Delta N=\delta_0\) and \(\nabla\times\nabla\times X-\nabla\operatorname{div}X =-\Delta X\) give \(X=\nabla\times N*(\nabla\times X) -\nabla N*(\operatorname{div}X)\). Fourier transformation verifies it off the origin. The compact \(L^2\) field has no additional constant or point-supported Fourier part. Substitute the full product curl and divergence, \(\nabla\times X=\psi_8z+\nabla\psi_8\times w\) and \(\operatorname{div}X=\nabla\psi_8\cdot w\), to obtain (4.1). This is a map for the actual cutoff field; a local curl has not been given a false global inverse.

For \(t\geq s_7\), set

The first bounds \(|z|\) on \(B_7\), and the second bounds \(\|\psi_8z(t)\|_2\) uniformly in every center. If \(x\in B_8\) and \(|y|<\epsilon\), the segment from \(x\) to \(x-y\) is in \(B_7\). The heat derivative bound there gives \(|\ell(x-y)-\ell(x)|\leq\sqrt2V_2^*|y|\). Together with (3.4) this proves \(|z(x-y)-z(x)|\leq Z_\alpha|y|^\alpha\) for these exact local increments.

The main kernel in (4.1) has magnitude \(|\nabla N(y)|=1/(4\pi|y|^2)\). Its integral over \(|y|\leq2\rho_b\) is \(2\rho_b\), which gives the bound \(2\rho_bZ_\infty\) for the main field on \(B_8\). A stronger estimate retains \(Z_2\). Split the same complete convolution at any radius \(a>0\). Its near part is at most \(aZ_\infty\), and Cauchy–Schwarz bounds the far part by \(Z_2/(2\sqrt{\pi a})\), since

If both norms are positive, choose \(a=(Z_2/(4\sqrt\pi Z_\infty))^{2/3}\). The sum is \(3(4\sqrt\pi)^{-2/3}Z_\infty^{1/3}Z_2^{2/3}\). If either norm vanishes, the actual main field is zero. The split is valid for every \(a>0\), so no restriction relative to the support radius is required.

Both boundary fields in (4.1) have their source separated from \(B_8\) by at least \(\epsilon\). Each is at most \(L_1W_0/(2\sqrt{\pi\epsilon})\), by (4.3). Keeping both and both main-field bounds gives

on \([s_7,t_2]\times B_8\).

4.2. The gradient and the Newton point mass

The full distributional Hessian is

Away from zero this is direct differentiation. To obtain the point mass, integrate by parts outside a ball of radius \(a\) against a smooth test function. The leading inner boundary term tends to its value at zero times \(-(4\pi)^{-1}\int_{\mathbb S^2}\vartheta_i\vartheta_j\,dS =-\delta_{ij}/3\). The remaining singular integral converges in principal value: its angular average is zero because that same spherical integral equals \(4\pi\delta_{ij}/3\). Taylor's first-order bound makes the subtraction of the test value integrable near zero. This proves (4.5), including its sign and point mass.

The ordinary Hessian kernel has eigenvalues proportional to \(2,-1,-1\), and its full tensor magnitude is \(\sqrt6/(4\pi|y|^3)\). For the derivative of the main curl term in (4.1), split its principal value at \(\epsilon\). On the inner ball, \(\psi_8=1\) at both evaluation points and the zero angular average permits the increment in (4.2). Its bound is \(\sqrt6 Z_\alpha\epsilon^\alpha/\alpha\). On \(\epsilon\leq|y|\leq2\rho_b\), the bound is \(\sqrt6Z_\infty\log(2\rho_b/\epsilon)\). Beyond \(2\rho_b\) the actual source vanishes for \(x\in B_8\). The point mass in (4.5) gives a cross-product matrix whose Euclidean norm is at most \(\sqrt2Z_\infty/3\).

The differentiated boundary fields have no point source at \(x\), because their supports are separated. The full far Hessian norm is

Thus their sum is at most \(\sqrt{2/\pi}L_1W_0\epsilon^{-3/2}\). For the curl term, each row of the Hessian acts by a cross product, bounded by its row norm times the vector norm; summing the row squares introduces no further factor. Consequently

Adding the actual heat field proves

The complete local map retains the original nonlinear velocity at the boundary, the Newton point mass and the original heat field.

5. The first vorticity derivative

Use (2.4) with \(k=9\), starting at \(s_8\). The compact field \(V=\psi_9u\) is defined on the full space and satisfies \(\|V\|_\infty\leq U_{\rm loc}\) and \(\|\nabla V\|_\infty\leq L_1U_{\rm loc}+G_u\). Its smooth zero extension has that same global derivative bound. Integrating along line segments, as in (3.1), gives

For times at least \(s_8\), \(\Omega_7=\omega\) on the support of \(\psi_9\). Hence \(\psi_9(u\times\omega)=V\times\Omega_7\) globally. Both terms of the product difference must be included. They give

For the second line, the compact field \(\partial_j\psi_9\) has maximum at most \(L_1\) and derivative maximum at most \(L_2\); multiply its full difference bound by that of \(\Omega_7\). The time factors are at most one.

Every second Gaussian derivative has integral zero. Thus its convolution with a bounded Hölder field \(g\) is exactly its integral against \(g(x-y)-g(x)\). Define the finite complete moments

Their original heat-time factors are \((\nu a)^{-1+\alpha/2}\) and \((\nu a)^{-1/2}\). Differentiate the curl-flux and cutoff-divergence potentials in (2.4), use (5.2) and their cancellation, and integrate \(\int_0^Ta^{-1+\alpha/2}\,da=2T^{\alpha/2}/\alpha\). Their sum is bounded by \((2/\alpha)J_{2,\alpha}\nu^{-1+\alpha/2}T^{\alpha/2} (A_{\rm nl}+2\nu A_{\rm cut})\).

The remaining non-force source has maximum at most \((L_t+\nu L_2+L_1U_{\rm loc})X_6\). One Gaussian gradient integrates to \(2J_\nabla\nu^{-1/2}\sqrt T\). The full force potential (2.6) obeys the same gradient bound with input \(F_{\rm curl}\). We obtain

The differentiated integrals are justified first with a positive upper time gap, then by the displayed integrable moments and dominated convergence. On this final region the cutoff is identically one, so its derivative does not alter \(\nabla\omega\). The initial term at \(s_8\) was exactly zero.

All constants are uniform in the center \(x_0\). Taking the value at each such center gives the full original-annulus bounds

No hidden higher-derivative norm of \(u\) occurs on the right side. All full force and boundary terms have explicit receiving bounds.

6. Choosing an annulus with arbitrarily small maxima

Boundedness alone is insufficient when a later weighted estimate requires small coefficients. We now choose them from the actual construction. Fix the original solution, \(\nu\), interval, selected time \(t_1\), and global input bounds before varying \(h\). Fix \(h_0,a_0>0\) and take \(h\geq h_0\). In the preceding lesson's selection formulas choose the smaller heat target

The symbol \(\alpha_h\) is the numerical heat target; the Hölder exponent stays exactly \(\alpha=1/4\). Use (6.1) everywhere that target enters the earlier formulas, including the initial tolerance and Gaussian distance. Every energy inequality remains valid: decreasing the target decreases its three heat-budget terms, and the shell-count and radius proofs allow the smaller tolerances and larger radius. In particular \(V_j^*\leq a_0h_0^2h^{-2}\) for \(j=0,1,2\).

This section applies to \(f=0\), and to an actual smooth force supported in \(\{|x|\leq R_f\}\) throughout \(J\). In the latter case add \(\Lambda^{m-2}(R_f+h_0)\) to the maximum defining the original \(R_{\min}\). Then \(A\geq R_f+h_0\), so the whole receiving annulus has actual \(\nabla\times f=0\). Thus \(F_{\rm curl}=0\) here. All earlier global contributions of the force to \(W_0,M_\delta\), total speed and shell selection stay unchanged. The original PDE still contains its force.

6.1. Full constants fixed before the height is chosen

Keep the preceding lesson's constants \(\lambda_5,\lambda_8,\lambda_9,C_2,L=t_2-t_b\). Define

To verify these constants, the earlier energy coefficient satisfies \(\mathcal A_0(h)\leq\overline A\), and its budget \(e_0(h)\leq E_*/8\). Hence its actual local energies obey \(Z_0^2\leq E_c/h\) and \(Z_1^2\leq D_c/h\). The earlier annular cutoff length is \(d=A/8\geq h/8\), and its lower radius is at least \(2A\). Its gradient and Laplacian bounds are consequently at most \(g_1/h\) and \(g_2/h^2\).

Insert these full bounds in the earlier formulas for \(P_0^2,P_1^2,Q_0^2\), giving

For example the gradient-cutoff term of \(Q_0^2\) is at most \(12g_1^2TP_{1c}h^{-3}\), bounded by \(12g_1^2TP_{1c}h_0^{-2}h^{-1}\); its zeroth-order term is treated with \(h^{-4}\leq h_0^{-3}h^{-1}\). Every contribution appears in (6.2) before this comparison. Equations (1.2) and (1.7) now give \(\mathcal V\leq\overline Vh^{-1/2}\) and \(X_0\leq\overline X_0h^{-1/2}\).

The physical ball cutoffs (1.5) have exactly

Define \(\overline C\) by the complete expression (2.7) with \(\mathcal V,L_1,L_2\) replaced by \(\overline Vh_0^{-1/2},k_1/h_0,k_2/h_0^2\). Define \(\overline C_H\) by the same substitutions in (3.3). These are explicit finite expressions retaining every displayed summand, kernel constant and viscosity power. Put

Because every coefficient is nonnegative, (2.9) and (3.4) give \(X_6\leq\overline Xh^{-1/2}\) and \(H_7\leq\overline Hh^{-1/2}\). Also \(Z_\infty\leq\overline Zh^{-1/2}\) and \(Z_2=\sqrt{E_c}h^{-1/2}\).

6.2. The actual velocity and derivative powers

The retained \(L^2\) term of (4.4) gives

The first term is the complete optimized Newton estimate. The second retains its boundary term of order \(h^{-3/2}\), and the third retains the original heat field of order \(h^{-2}\). Their stated bounds use \(h\geq h_0\). Using only the other main-field bound \(2\rho_bZ_\infty\) would lose this decay.

In (4.7) the logarithm is exactly \(\log64\). With \(\alpha=1/4\), let

Substituting \(Z_\alpha=H_7+\sqrt2V_2^*\epsilon^{1-\alpha}\) in (4.7)–(4.8) proves \(G_u\leq\overline Gh^{-1/4}\). The five displayed terms receive, respectively, the vorticity Hölder increment, the original heat derivative increment, the Newton far integral and point mass, both boundary fields, and the added heat gradient. Before comparison their height powers are \(h^{-1/4}\), \(h^{-1}\), \(h^{-1/2}\), \(h^{-5/2}\), and \(h^{-2}\). Thus (6.7) gives every required \(h_0\) factor explicitly.

For the full products in (5.2), define

Then \(A_{\rm nl}\leq B_{\rm nl}h^{-15/16}\), \(A_{\rm cut}\leq B_{\rm cut}h^{-3/2}\), and \((L_t+\nu L_2+L_1U_{\rm loc})X_6 \leq B_{\rm rem}h^{-1/2}\). For example the mixed nonlinear term has exponents \(3/8+1/16+1/2=15/16\): these come from its velocity, gradient and vorticity factors. The faster-decaying partners are retained with their \(h_0\) factors in (6.8). Consequently define

All these constants are fixed before \(h\) is selected. For a prescribed positive tolerance \(\varepsilon_*\), take the actual finite height

Run the original annulus selection with this \(h\), the smaller target (6.1), and the additional force-support entry when needed. It produces a finite \(R\) with the explicit upper radius of the preceding lesson. The estimates just proved give

This is actual smallness derived from the original solution, with the full receiving geometry and time margin. Different tolerances for the four quantities can be used in the four respective entries of (6.10). Nothing in this construction changes the original viscosity, coordinates or force.

7. Source comparison and five solved exercises

The comparison passage is Tao, arXiv:1908.04958v2, original author article.tex lines 859–892. It lists local gradient-energy estimates and successive regularity gains in its unforced, unit-viscosity setting. Here every gain, cutoff, Newton boundary field, point mass and time margin is proved with the original parameters. The additional force receiver is identified explicitly. The source's displayed closed initial-time bounds are not imported as estimates from an initial spatial norm.

The original argument and the preceding lessons guide this construction; no novelty is asserted. These finite annular estimates still need their complete receiving maps into the weighted heat estimates and frequency backpropagation argument. The general endpoint theorem is not proved by local annular regularity alone.

Exercise 1: verify the exponent gains and their time condition

Check all six spatial steps in (2.1), including the last one. More generally, determine an explicit finite number of equal steps from \(L^2\) to \(L^\infty\) when the local velocity is in \(L^q_tL^\infty_x\) with \(q>2\), and prove the time integrability of each differentiated heat convolution.

Solution. For our sequence, \(1/p_{k-1}-1/p_k=1/12\) for every \(1\leq k\leq6\). Thus the kernel space is \(L^{12/11}\), and its derivative exponent is \(1/2+(3/2)(1/12)=5/8\). Hölder pairs the velocity in time with exponent \(q'=4/3\); the resulting power is \(5/6<1\). This includes \(p_5=12,p_6=\infty\) without change.

For general \(q>2\), choose any integer \(n>3q/(2(q-2))\), set \(1/p_k=1/2-k/(2n)\), and let \(r^{-1}=1-1/(2n)\). Each derivative kernel has exponent

It has time norm

Young's exact spatial relation and Hölder therefore give each step as in Section 2, with its actual Gaussian \(L^r\) derivative constant, viscosity factor \(\nu^{-a}\), and the full norm (7.2). The zeroth derivative kernel has the smaller exponent \(3/(4n)\); it and the ordinary time integrals for cutoff terms are also finite. This proves the stated finite exponent mechanism. The other hypotheses, initial local vorticity norm and actual force potentials remain required exactly as in the full argument; they have not been manufactured by changing \(q\).

Exercise 2: why the boundary fields cannot be omitted

Construct a smooth compact divergence-free field that is a nonzero constant on a ball. Apply (4.1) with its cutoff supported within that ball. Identify which exact terms recover the velocity even though the local curl is zero.

Solution. Fix a nonzero vector \(a\), a radius \(R>0\), and a smooth radial cutoff \(\eta\) that is one on \(B(0,2R)\) and zero outside \(B(0,3R)\). It is smooth at the origin because it is constant there. Define the compact vector potential \(A(x)=\tfrac12\eta(x)(a\times x)\) and

The component product rule gives the equality, since \(\nabla\times(a\times x)=2a\). As a curl, \(w\) has zero divergence. It is identically \(a\) on \(B(0,2R)\), hence \(z=\nabla\times w=0\) there.

Choose a smooth cutoff \(\psi\) supported in \(B(0,R)\), equal to one on \(B(0,R/2)\). The full identity (4.1), applied to this actual compact field, has zero main source \(\psi z\). For \(x\in B(0,R/2)\) it gives exactly

The Fourier/Poisson proof of (4.1) establishes this identity including both boundary fields. Omitting them would assert \(a=0\), contrary to its choice. The example demonstrates the actual map from boundary data to the locally curl-free velocity; it does not assert that such a field has zero global curl.

Exercise 3: independent tolerances and the full Newton split

Prove the optimized main-field constant in (4.4), including zero inputs. Then give the exact height choice when the four desired tolerances are \(\varepsilon_u,\varepsilon_{\nabla u}, \varepsilon_\omega,\varepsilon_{\nabla\omega}>0\).

Solution. The complete split at \(a>0\) bounds the main field by \(aZ_\infty+Z_2/(2\sqrt{\pi a})\). Its derivative with respect to \(a\) is \(Z_\infty-Z_2/(4\sqrt\pi a^{3/2})\). For positive norms it vanishes at the radius stated after (4.3); the second term there is twice the first. Their sum is \(3(4\sqrt\pi)^{-2/3}Z_\infty^{1/3}Z_2^{2/3}\). If either bounding norm is zero, the actual main source is zero almost everywhere and the field is zero; the same bound holds.

All constants in Section 6 were fixed before selecting \(h\). Thus choose

Each of the four proved height bounds is then at most its own tolerance. Repeat the actual shell selection with that height and target (6.1). The entire construction still has a finite explicit radius. Independent physical tolerances change neither the original objects nor their equations.

Exercise 4: the original force classes do not supply a curl maximum

Construct smooth compactly forced solutions from rest whose velocities have bounded \(L^\infty_tL^3_x\) norm and whose \(L^2_tL^2_x\) forces tend to zero, but whose vorticity derivatives are unbounded. Compute the actual force and its curl at a chosen time. Explain precisely what this tests.

Solution. Choose a nonzero smooth compact divergence-free vector field \(b\). For example take \(b=\nabla\times(\phi e_3)\), where \(\phi(x)=\beta(x_1)\beta(x_2)\beta(x_3)\), and \(\beta(r)=e^{-1/(1-r^2)}\) for \(|r|<1\), zero otherwise. This gives a nonzero smooth compact field. Its curl is nonzero: otherwise the full Fourier div-curl identity would give \(\nabla b=0\), forcing the compact field to vanish. Similarly \(\nabla\nabla\times b\) is nonzero, since a compact constant curl would be zero.

Let \(t_c\) lie strictly inside a fixed time interval \(I\), and set \(\eta(s)=s e^{-1/(1-s^2)}\) for \(|s|<1\), zero otherwise. It is smooth, \(\eta(0)=0\), and \(\eta'(0)=e^{-1}\). For sufficiently large positive integers \(N\), define in the original coordinates

Direct differentiation verifies (1.1) with this complete force and the original \(\nu\). The velocity is divergence free. Its time support is strictly inside \(I\), so it starts from rest and returns to rest; every field is smooth and compact in space and time. Use \(f_a=0\), \(f_b=f_N\).

The original substitutions in the integrals give exactly \(\|u_N\|_{L^\infty_tL^3_x} =N^{-1}\|\eta\|_\infty\|b\|_3\). For the force, retain all three terms and use the triangle inequality:

Indeed the space-time norm contributes \(N^{-5/2}\), while the three differential terms carry \(N^2,N^2,N\). Thus the displayed force norm tends to zero. But \(\|\nabla\omega_N\|_\infty =N^2\|\eta\|_\infty\|\nabla\nabla\times b\|_\infty\) diverges. At \(t=t_c\), both terms carrying \(\eta\) vanish exactly, so

This tests the proposed uniform bound of a pointwise derivative or force curl by the stated coarse norms; that bound is false. It does not give a finite-time singularity or a counterexample for one fixed force. The actual curl datum in (1.8) records the growing contribution exactly, and an annulus beyond these compact force supports remains covered by Section 6.

Exercise 5: every physical scaling factor in the receiver

For \(\lambda>0\), compare the original solution with \(u^\lambda(t,x)=\lambda u(\lambda^2t,\lambda x)\) on the correspondingly mapped interval. Derive the original force, vorticity and derivative factors. Check the scaling of the full recurrence (2.8), the Newton receiver and the final derivative bound while keeping \(\nu\).

Solution. Set \(p^\lambda(t,x)=\lambda^2p(\lambda^2t,\lambda x)\) and \(f^\lambda(t,x)=\lambda^3f(\lambda^2t,\lambda x)\). Every term of (1.1) has factor \(\lambda^3\), so the viscosity remains exactly \(\nu\). Direct derivatives give

The actual heat flow and its remainder have the same velocity factor. All spatial radii and lengths become their original values divided by \(\lambda\), and all time margins become their original values divided by \(\lambda^2\). Thus

The norm identities follow by substitution in the original integrals. For \(W_0\), both its nonlinear heat term and its full force integral have the stated factor. Every summand of (2.7) has factor \(\lambda^{1/4}\). Since \(3/p_{k-1}-3/p_k=1/4\), its product with the preceding norm has exactly the new factor. The complete force term in (2.8) has factor \(\lambda^{-2}\lambda^{-3/p_k}\lambda^4 =\lambda^{2-3/p_k}\) as well.

The Hölder input has factor \(\lambda^{2+\alpha}\), while \(Z_2\) has factor \(\lambda^{1/2}\). Both main terms of (4.4) therefore have factor \(\lambda\); so does its boundary term. Each term of (4.7) has factor \(\lambda^2\), including the point mass. In (5.2), \(A_{\rm nl}\) has factor \(\lambda^{3+\alpha}\), and \(A_{\rm cut}\) the same factor. The first coefficient of (5.4) has factor \(\lambda^{-\alpha}\). The remaining bracket has factor \(\lambda^4\) and its time coefficient \(\lambda^{-1}\). Thus every term of \(G_\omega\) has exactly factor \(\lambda^3\), agreeing with (7.9). This verifies the full map while retaining all force, boundary, time and viscosity factors. The original calculation remains the one used in the lesson; this is its proved comparison.

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