Vorticity, strain and continuation

Vorticity measures local rotation. Strain measures stretching and contraction. Both come from the same velocity gradient, but they retain different parts of that matrix. This lesson constructs their exact connecting maps, proves the enstrophy identity, and uses it to decide when the strong solution can continue. The force and the original positive viscosity remain in every estimate.

The middle strain eigenvalue enters through a spatial integral identity. That identity does not prescribe the vorticity direction at each point. A compact forced flow at the end of the lesson shows the difference explicitly.

1. Original equation and solution class

Let \(\Omega=\mathbb R^3\) or let

All integrals use the original Lebesgue measure, including the volume \(V\) on \(Q\). We use the solution constructed in Strong solutions and continuation:

The maximal solution interval is \([0,T_*)\), with \(T_*\leq T_f\). On each compact subinterval it satisfies

Here and below a time norm is over the interval under discussion. A finite endpoint \(T_*<T_f\) requires an unbounded \(H^1\) norm, by the complete construction and continuation proof in that chapter. In particular the force is prescribed beyond any endpoint to which our continuation assertion applies.

Write the original gradient, symmetric part and skew part as

Matrix norms are Frobenius norms, and matrix pairings sum all entries. Thus \(\operatorname{tr}S=0\), and direct component calculation gives

For example \(W_{12}=-\omega_3/2\), \(W_{23}=-\omega_1/2\), and \(W_{31}=-\omega_2/2\); the other entries follow by skew symmetry. The vector identity \(\omega\times(\omega\times z) =\omega(\omega\cdot z)-|\omega|^2z\) proves the square formula.

We will use the already proved Sobolev constants

The inequality holds for scalar, vector and matrix fields with these same constants. The periodic lower-order term is retained; the whole-space gradient inequality is the stronger inequality proved in the preceding chapters, applied to the same original field.

2. Exact reconstruction and the strain constraint

Our whole-space Fourier transform has kernel \(e^{-2\pi i x\cdot\xi}\). On \(Q\), put \(\kappa_k=(k_1/L_1,k_2/L_2,k_3/L_3)\), and use coefficients \(u_k=V^{-1}\int_Q u(x)e^{-2\pi i\kappa_k\cdot x}\,dx\). For every nonzero frequency, solenoidality and the curl formula give

The second identity follows by taking \(\xi\times\) of the first and using \(\xi\cdot\widehat u=0\). The identical formulas hold with \(\xi,\widehat u,\widehat\omega\) replaced by \(\kappa_k,u_k,\omega_k\) for \(k\ne0\).

The derivative coefficients at zero vanish. On the periodic domain the missing velocity coefficient is exactly

This follows by integrating (1.2), since the transport, pressure and Laplacian terms have zero integral. On \(\mathbb R^3\), an additional distribution supported at frequency zero would be a polynomial velocity; no nonzero such polynomial belongs to the specified \(L^2\) class.

Parseval and \(|\xi\times\widehat u|^2=|\xi|^2|\widehat u|^2\) therefore give the exact identities

To check the factor for \(S\), its Fourier matrix is \(\pi i(\widehat u\otimes\xi+\xi\otimes\widehat u)\). The two summands have zero Frobenius cross term and equal squared norm \(|\xi|^2|\widehat u|^2\). Multiplication by \(2\pi|\xi|\) proves the derivative version. On \(Q\) all these integrals become \(V\sum_k\); that factor occurs on both sides and is not changed.

There is also an exact reconstruction from a candidate strain matrix. Let \(T\) be a symmetric matrix-valued \(L^2\) field. For \(\xi\ne0\) define

The matrix \(\widehat{\Pi T}\) is symmetric and has trace zero because \(\xi\cdot v=0\). It satisfies \((\widehat{\Pi T})\xi=v\), so \(\Pi^2=\Pi\). For any \(w\perp\xi\), its error is orthogonal to \(\xi\otimes w+w\otimes\xi\): indeed the Frobenius pairing is

Consequently \(\Pi\) is the orthogonal projection onto the strain range, and

The gradient convention in the middle line is (1.4).

On \(\mathbb R^3\), (2.4) defines an actual tempered distribution: near zero its size is bounded by \(|\widehat T|/(\pi r)\), and \(r^{-1}\) is locally square integrable in three dimensions. At high frequency its pairing with a Schwartz function is integrable by Cauchy–Schwarz. Cutting off to compact frequency annuli gives a Cauchy sequence in the gradient norm and hence, by (1.6), in \(L^6\). Its \(L^6\) limit is the same distribution. Thus the target of \(T\mapsto u_T\) is the actual \(L^6\) realization of \(\dot H^1_\sigma\), with norm squared \(2\|\Pi T\|_2^2\). It belongs to \(H^1_\sigma\) precisely when

This condition retains the original velocity's low frequencies.

On \(Q\), use the same formulas at \(\kappa_k\ne0\), put \((\Pi T)_0=0\), and choose \((u_T)_0=0\) for this reconstruction. The positive gap \(\min_{k\ne0}|\kappa_k|>0\) and (2.6) imply \(u_T\in H^1_\sigma(Q)\). Adding the actual mean (2.2) recovers the original velocity. The constant strain error is retained in \(\|T-\Pi T\|_2^2\), including its factor \(V\).

For completeness, the strain range has an exact differential description:

Here \((\operatorname{div}T)_i=\sum_j\partial_jT_{ij}\). At nonzero frequency the second equation is \(r^2\widehat T=\xi\otimes(\widehat T\xi) +(\widehat T\xi)\otimes\xi\). Taking its trace gives \(\xi\cdot\widehat T\xi=0\), and then the equation is exactly \(\widehat T=\widehat{\Pi T}\). Conversely (2.4) has trace zero and satisfies that equation. On \(\mathbb R^3\) this proves the equivalence in \(L^2\); on \(Q\) one must also require \(T_0=0\). Exercise 3 identifies the missing constant matrices if that last condition is omitted. Equations (2.4)–(2.8) give the reconstruction, its error, and all its kernels and ranges.

3. Full strain equation and pressure pairing

Differentiate the original momentum equation using (1.4). The product rule gives \(\partial_j(u_k\partial_k u_i) =u_k\partial_kG_{ij}+G_{ik}G_{kj}\). Thus

The symmetric part of \(G^2\) is \(S^2+W^2\), since \(SW+WS\) is skew symmetric. Substituting (1.5) proves

Taking the trace recovers the original pressure equation:

Taking curl in (1.2), or taking the skew part of (3.1) component by component, gives

These identities hold distributionally at the solution regularity (1.3).

The pressure is the potential already reconstructed in the previous lessons. In particular it has a nonlinear part \(p_N\) and a force part \(p_F\), with

On \(Q\) the nonzero modes use \(\kappa_k\), the pressure mean is chosen zero, and \(P_0=I_3\); thus no mean force is assigned to pressure. On \(\mathbb R^3\) the distributional low-frequency potential is the one in Pressure and the divergence-free projection.

At an \(H^2\) time, \(u\otimes u\in H^2\). To verify this claim, use \(u\in L^\infty\), \(\nabla u\in L^4\), and \(\nabla^2u\in L^2\): every second product derivative is a sum of \(u_i\partial_{jk}u_l\), \(\partial_j u_i\partial_k u_l\), and the corresponding terms with the factors reversed. These belong to \(L^2\); the undifferentiated and first derivatives do also. The previous chapters prove the required embeddings. The multiplier in (3.5) is bounded, so \(p_N\in H^2\).

For \(f_b\in L^2\), however, (3.5) gives only \(\nabla p_{F,b}\in L^2\) and \(\operatorname{Hess}p_{F,b}\in H^{-1}\). That is enough. Since \(\operatorname{div}S=\Delta u/2\), the exact dual pairing is

The first equality vanishes by the orthogonality of the gradient and solenoidal Fourier ranges, not by assuming an extra derivative of the force. For \(f_a\in H^1\) the same statements have ordinary \(L^2\) matrix pairings. The nonlinear pressure has zero pairing with \(S\) by the same integration by parts and orthogonality. All time pairings are legitimate: \(f_a\) is integrable in \(H^1\), \(S\) is continuous in \(L^2\), and \(f_b\) and \(S\) have the dual square-integrable classes just used.

There are two specific source corrections. In Evan Miller's middle-eigenvalue paper, version 4, the forced strain display immediately before Definition ExtForce (author TeX lines 557–558) prints a plus sign before \(|\omega|^2I_3/4\). Its earlier equation NavierStrain and its mild strain formula have the sign consistent with (3.2). The incorrect display differs from (3.2) by \(|\omega|^2I_3/2\), whose trace is \(3|\omega|^2/2\). Exercise 5 gives an actual compact forced solution on which this difference is nonzero.

In the proof Strn, author TeX lines 692–696 claim that the pressure Hessian is \(L^2\) under the allowed \(L^2\) force assumption. Exercise 4 gives a source-class solution for which it is not. Formula (3.6) supplies the needed orthogonality in the correct dual spaces and preserves the enstrophy argument. Miller writes the gradient as \(G^{\mathsf T}\): his symmetric part is exactly \(S\), and his antisymmetric part is \(-W\). Transposition preserves the trace, Frobenius norm and symmetric square contribution, so this convention difference does not change either correction.

4. Enstrophy and the cubic integral identity

Pair (3.4) with \(\omega\). The transport term vanishes by \(\operatorname{div}u=0\). On the whole space a cutoff at radius \(R\) bounds its boundary error by \(C R^{-1}\|u\|_\infty\|\omega\|_2^2\), which tends to zero. Curl is self-adjoint under the integral pairing, and \(\nabla\times\omega=-\Delta u\). Thus the full identity is

There is no differentiation of \(f_b\) in its final pairing.

Here is the time justification for the actual solution (1.3). Subtract the proved \(L^2\) energy identity from the full \(H^1\) energy identity in the strong-solution chapter. Their difference is the gradient energy identity, with the same two force summands. The spatial nonlinear identity leading to (4.1) follows by solenoidal \(H^2\) approximation. At almost every time \(u\) is bounded and \(\omega\in H^1\cap L^4\). On a compact time interval, the transport pairing is integrable from \(u\in L^2L^\infty\), \(\nabla\omega\in L^2L^2\), and \(\omega\in L^\infty L^2\). The stretching term is integrable since

The last time factor is integrable by Hölder on a finite interval. Both force pairings in (4.1) are integrable in their stated classes. Consequently \(Y\) is absolutely continuous on every compact solution interval, and (4.1) integrates between every pair of its endpoints.

The cubic identity requires the spatial strain constraint; it is not an identity for three arbitrary matrix fields. For a smooth solenoidal field write repeated indices as sums from 1 to 3. Successive integrations by parts give

The first integration has no term from \(\partial_j\partial_k u_j\); the second has no term from \(\partial_i u_i\). In the penultimate step, interchanging the dummy indices \(j,k\) shows that the two product-rule terms have the same integral. The last integral is zero by solenoidality. Whole-space cutoff errors are bounded by a constant times \(R^{-1}\|u\|_\infty\|G\|_2^2\). The remaining terms are bounded by \(\|u\|_\infty\|G\|_2\|\nabla G\|_2\). Approximation in \(H^2\) gives convergence of \(G\) in \(L^3\) and proves (4.3) for every field needed in (4.1).

Expansion of the trace, using symmetry and skew symmetry, gives

Terms with an odd single skew factor have zero trace; \(\operatorname{tr}W^3=0\). The \(-|\omega|^2I_3/4\) contribution in (1.5) has zero pairing with \(S\) because \(\operatorname{tr}S=0\).

Let \(\lambda_1\leq\lambda_2\leq\lambda_3\) be the eigenvalues of \(S\), counted with multiplicity. Their sum is zero, and expanding \((\lambda_1+\lambda_2+\lambda_3)^3\), or substituting \(\lambda_3=-\lambda_1-\lambda_2\), proves \(\operatorname{tr}(S^3)=3\det S\). Thus

and (4.1) is equivalently

Both representations retain every force contribution and the original dissipation. In particular, a growing enstrophy in a forced flow need not come from a positive stretching integral.

5. Three continuation criteria

The trace constraint improves an elementary matrix estimate. If \(S\) is symmetric with trace zero, then

Indeed, for each eigenvalue \(\lambda_i\), the other two sum to \(-\lambda_i\), so the sum of their squares is at least \(\lambda_i^2/2\). Hence \(|S|^2\geq3\lambda_i^2/2\). The matrix with eigenvalues \((2c,-c,-c)\) attains this algebraic constant. This is a sharp matrix inequality; no sharpness claim for a whole-flow continuation threshold follows from that observation.

From (2.3), (5.1), and Cauchy–Schwarz,

This retains a factor lost by estimating the operator norm with the full matrix norm without using its zero trace.

For a second criterion, define the actual directional stretching by

Here \(r_+=\max(r,0)\). Since \(I_\omega=\int\alpha|\omega|^2\), we have \(I_\omega\leq hY\). The Rayleigh quotient also gives \(\alpha\leq\lambda_3\) where \(\omega\ne0\); hence \(h\leq\|\lambda_3{}_+\|_\infty\).

For the third criterion use (4.5). If \(\lambda_2\geq0\), then

Indeed \(-2\lambda_1\lambda_3=|S|^2-2\lambda_2^2\), by the zero trace. If \(\lambda_2<0\), the two lowest eigenvalues are negative and the highest is positive, so \(-\det S\leq0\). Consequently the full remainder has the useful bound

All terms before the last upper bound are kept in (4.6) and (5.5). The upper bound permits the continuation estimate; it does not assert that the negative cubic term vanishes.

Put

Let \(k(t)\) be any one of \(\|\omega\|_\infty/\sqrt3\), \(h(t)\), or \(\|\lambda_2{}_+\|_\infty\). The proved upper bound is \(I_\omega\leq kY\). The two exact scalar inequalities

then give

If \(H(t)=\int_0^tK(s)\,ds\), multiplication by \(e^{-H}\) and integration prove

The second line follows from the same integrated inequality by retaining the nonnegative terminal term before discarding it.

If \(T_*<T_f\) and any one of

is finite, (5.9) bounds \(Y\) up to that endpoint. The original \(L^2\) energy estimate also stays finite for the force in (1.2). By (2.3) this bounds the full \(H^1\) norm and contradicts its continuation alternative. Thus each integral in (5.10) must be infinite at such a finite maximal endpoint. One may start these estimates at any earlier time of the strong solution instead of zero.

These are viscous theorems for (1.2). The proof uses \(\nu>0\) in (5.7) and the viscous \(H^1\) continuation theorem. Setting \(\nu=0\) does not prove the Euler theorem of Beale, Kato and Majda. That theorem requires a separate high-regularity continuation argument. No pointwise bound of the strain by the pointwise vorticity has been assumed here: the exact connections used are (2.3), (4.5) and the proved integral bounds.

6. Mixed norms and finite sums

We can weaken the uniform spatial bound by allowing a time-space norm. There are two versions. In the vorticity version decompose the actual vector field as \(\omega=\sum_{j=1}^m z_j\), and set \(d=2/\sqrt3\). In the middle-eigenvalue version decompose the actual scalar function as \(\lambda_2{}_+=\sum_{j=1}^m z_j\), and set \(d=2\). Summands need not be positive, divergence-free, or solutions of any equation; this is a decomposition used only inside the exact stretching integral.

For each summand choose

For a finite \(q=q_j\), let

Hölder interpolation between \(L^2\) and \(L^6\) gives \(\|F\|_{r_q}\leq\|F\|_2^{1-\theta}\|F\|_6^\theta\): raise to the \(r_q\) power and apply Hölder with reciprocal exponents \((1-\theta)r_q/2\) and \(\theta r_q/6\), whose sum is one. Using (1.6) and (2.3) gives, with \(E=D+\delta_\Omega Y\),

For the vector version, substitute one copy of \(\omega=\sum_jz_j\) in \(I_\omega\), apply (5.1), and then Hölder with exponents \(q,r_q,r_q\). For the scalar version use \(I_\omega\leq2\int\lambda_2{}_+|S|^2\) and the same exponents. In both cases twice the stretching contribution of a finite summand is bounded by

The \(q_j=\infty\) summand instead contributes at most \(d k_jY\).

All absorption constants can be kept explicit. For \(0<\theta<1\) and \(\varepsilon>0\), maximization of \(c x^\theta-\varepsilon x\) over \(x\geq0\) gives

For \(c,Y>0\) the maximum occurs at \(x=(\theta c/\varepsilon)^{1/(1-\theta)}\) after putting \(x=D/Y\). The cases \(cY=0\) follow directly or by continuity. Also \((D+\delta_\Omega Y)^\theta\leq D^\theta+\delta_\Omega Y^\theta\), by concavity and \(\delta_\Omega\in\{0,1\}\).

Let \(N\) be the number of finite \(q_j\), and choose \(\varepsilon=\nu/(2\max(N,1))\). The sum of the absorbed terms in (6.5) is at most \(\nu D/2\). This time use \(2b\sqrt D\leq(\nu/2)D+2b^2/\nu\) for the force. Equation (4.1) then proves

If \(N=0\), the unused dissipation is nonnegative and the same inequality follows by discarding it. For finite \(T_*\), the assumed \(L^{p_j}\) membership also implies \(L^1\) membership by Hölder; the \(\infty\)-space summands have \(p_j=1\). Thus \(K,R\) are integrable. Formula (5.9), with these \(K,R\), proves continuation. This proves the finite-sum criterion with every constant, original domain, viscosity and force contribution retained.

At the spatial endpoint there is a proved smallness variant. Let the actual observable, either \(\omega\) or \(\lambda_2{}_+\), be \(z+g\), with

Now (6.3) uses \(r_q=6\), \(\theta=1\). Twice the first stretching contribution is at most \(d C_\Omega^2\|z\|_{3/2}(D+\delta_\Omega Y)\). Absorb at most \(\nu D/2\), retain its lower-order contribution, and use the force estimate from (6.6). This gives (6.6) with

Again (5.9) proves continuation. In particular one may choose an integrable threshold \(H(t)\geq0\), put \(z=\omega\,1_{|\omega|>H(t)}\), \(g=\omega\,1_{|\omega|\leq H(t)}\), or use the corresponding decomposition of \(\lambda_2{}_+\). The small-tail hypothesis is exactly (6.7), while \(\|g(t)\|_\infty\leq H(t)\). No assertion about an arbitrary large \(L^\infty L^{3/2}\) observable is being substituted for this proved endpoint statement.

7. Five exercises with complete solutions

Exercise 1: the critical exponents of the actual observables

For \(R>0\), send \((x,t)\) to \((x/R,t/R^2)\), and define

Find the exact maps on all quantities used in the continuation estimates.

Solution. The time derivative, convection, pressure gradient and viscous Laplacian each acquire \(R^{-3}\), so (1.2) holds with the same \(\nu\), on the mapped interval and domain. On the box the new lengths are \(RL_j\), its volume is \(R^3V\), and its velocity mean is \(R^{-1}u_0(t/R^2)\). The strain, vorticity, eigenvalues and directional stretching acquire \(R^{-2}\):

The eigenvalue ordering is preserved because \(R^{-2}>0\); the zero case in (5.3) maps to itself. Thus each observable has mixed norm

The usual zero reciprocal is used for an infinite exponent. This proves the exponent relation in (6.1) directly from the original fields. Furthermore

The time derivative of \(Y_R\) has factor \(R^{-3}\). The two force pairings have that factor too: curl of \(f_{a,R}\) has \(R^{-4}\), vorticity has \(R^{-2}\), and spatial measure has \(R^3\); \(f_{b,R}\) and \(-\Delta u_R\) each have \(R^{-3}\). Thus every term of (4.1) transforms identically. The original mean and lower-order norm terms remain as specified, not as derivatives of a rescaled mean-zero replacement.

Exercise 2: an exact forced helical wave

On the original box take \(n\geq1\), \(k=2\pi n/L_3\), a real smooth function \(A(t)\), and

Find the pressure, force, strain eigenvalues and complete enstrophy balance.

Solution. The field has zero mean and zero divergence. Its third component is zero and its first two components depend only on \(x_3\), so \((u\cdot\nabla)u=0\). Direct differentiation gives \(\nabla\times u=ku\) and \(\Delta u=-k^2u\). Hence the exact solution has

Its symmetric gradient is

The vector \(e\) is in its zero eigenspace, and the remaining two eigenvalues are \(\pm|kA|/2\). Thus \(\lambda_2=0\), \(S\omega=0\), \(\alpha=0\), and \(\det S=0\). The actual norms and force work are

Substitution gives \(Y'/2+\nu D=Vk^2AA'+\nu Vk^4A^2\), exactly (4.1) with zero stretching. Likewise the kinetic energy balance is \(VAA'+\nu Vk^2A^2=(f,u)\). The amplitude can grow because the prescribed force supplies work. The wave is not evidence that force work disappears from the strain criterion.

Exercise 3: the periodic strain defect at zero frequency

Suppose a symmetric \(L^2(Q)\) tensor satisfies (2.8). Describe its relation to the strain of a periodic velocity without assuming its mean vanishes.

Solution. Every nonzero coefficient lies in the range of (2.4), by the proof of (2.8). Its constant coefficient \(T_0\) is symmetric and trace-free but otherwise arbitrary. Formula (2.4), with zero velocity mean, constructs \(u_T\in H^1_\sigma(Q)\) with

An integral of a periodic derivative is zero, so \(T_0\ne0\) cannot be the strain of any periodic velocity. The map sending \(T\) to \(T_0\) is onto the five-dimensional space of real symmetric trace-free constant matrices: each such constant tensor satisfies (2.8). Its kernel is precisely the periodic strain range. The inclusion of constant tensors splits this map, and (7.9) proves the orthogonal direct sum. The velocity kernel is separately the three-dimensional space of constant vectors; adding (2.2) supplies the actual original-flow mean. These two constant spaces have different maps and dimensions, all explicitly retained.

Exercise 4: a force with a non-square-integrable pressure Hessian

On \(\mathbb R^3\), define a real scalar distribution \(\phi\) by

Take a nonzero real \(\gamma\in C_c^\infty(0,T)\). Verify that \(u=0\), \(p=\gamma(t)\phi\), \(f=\gamma(t)\nabla\phi\) is an actual solution in the source force class, and compute the relevant pressure regularity.

Solution. The transform in (7.10) is real and even, so its inverse distribution is real. Radial integration and the original Fourier factors give

The last norm is the Frobenius norm, using \(\sum_{ij}\xi_i^2\xi_j^2=|\xi|^4\). With the full \(H^{-1}\) weight its squared norm is instead

The force belongs to \(L^2_tL^2_x\). Since \(\nabla p=f\), the original momentum equation holds with zero velocity for every \(\nu>0\). The pressure formula (3.3) is exactly \(-\Delta p=-\operatorname{div}f\); there is no low-frequency ambiguity because (7.10) vanishes near zero. The mild velocity equation also holds: its integrand \(-\nabla p+f\) is identically zero. At every time when \(\gamma\ne0\), the pressure Hessian fails to be \(L^2\). In particular setting \(\nu=1\) gives a solution in Definition ExtForce of the cited source. The correct \(H^{-1},H^1\) pairing (3.6) remains valid. In the full strain equation the pressure Hessian and symmetric force gradient cancel as distributions.

Exercise 5: zero total stretching without middle-eigenvector alignment

Construct a compact smooth forced flow for which both integrals in (4.5) vanish, but vorticity is aligned with the largest strain eigenvalue on one open ball and the smallest on another.

Solution. Keep arbitrary constants \(a>0\), \(\ell>0\), and \(\nu>0\). Put

Define an exact smooth cutoff by

The denominator is positive for every real \(s\). All derivatives of \(b\) vanish at zero from either side: for \(s>0\), each derivative is a polynomial in \(s^{-1}\) times \(e^{-1/s}\), which tends to zero. Hence \(\chi\) is smooth, is one for \(s\leq1/4\), and is zero for \(s\geq1\). Since \(\operatorname{tr}M=0\), the vector identity for curl of a cross product gives \(\nabla\times(y\times My)=-3My\). Thus \(v=My\) on \(|y|<\ell/2\), and \(v\) is supported in \(|y|\leq\ell\).

Here are the full transition formulas, to keep the annular force visible. Let \(r^2=|y|^2\), \(s=r^2/\ell^2\), \(q=y\cdot My\), and \(B=r^2My-qy\). Primes below are derivatives of \(\chi\) at \(s\). The product and vector triple-product rules give

Consequently every gradient column and the complete Laplacian are

To verify the last coefficients, note that \(\Delta B=10My-4S_My\) and \(y\cdot\nabla B=3B\); apply the Laplacian product rule to both terms of (7.15). These formulas include the entire transition region.

Now put \(y_+=x-x_+\), \(y_-=-x-x_+\), and

This is the curl of \(A(y_+)-A(y_-)\); the minus sign compensates for the derivative of \(y_-\). It is therefore divergence-free. Its two support balls, centered at \(x_+\) and \(-x_+\), are disjoint. It is even in \(x\), so \(G,S,\omega\) are odd. In particular \(\det S\) and \(\omega\cdot S\omega\) are odd integrable functions and each has integral zero.

On the ball of radius \(\ell/2\) about \(x_+\), the exact data are

On the corresponding ball about \(-x_+\), they are

The middle eigenspace is \(\mathbb Re_2\) in both balls. Vorticity is along \(e_3\), the largest eigenspace in (7.18) and the smallest in (7.19). The disagreement occurs on open sets of positive volume.

To make this an actual stationary solution of the original viscous equation, use the full compact smooth force

The two supports are disjoint, so there are no omitted cross-products. The chain rule gives \(G_u=G_v(y_+)-G_v(y_-)\) and \(\Delta u=\Delta v(y_+)+\Delta v(y_-)\). Equations (7.15)–(7.16) then verify exactly that \(f=(u\cdot\nabla)u-\nu\Delta u\), including the annuli. This force is in both classes in (1.2) on every finite time interval. Integration gives \((f,u)=\nu\|\nabla u\|_2^2\) and \((f,-\Delta u)=\nu D\): the transport contribution to the latter is \(-I_\omega=0\), by (4.1) or its proved spatial identity. Thus both full energy balances agree with stationarity.

The figure takes \(a=\ell=1\). The left panel is the exact \(x_2=0\) section of the two support balls and their affine inner balls; the other panels show the exact spectra in (7.18)–(7.19), with each eigenvalue labelled by its original coordinate direction. It depicts neither an approximate solution nor a numerical proof of cancellation: the cancellation is the parity calculation above. The original figure source is included in the course download.

This example refutes a pointwise alignment inference from zero total stretching. It does not refute an independently measured statistical tendency. The alignment discussion following VortStretch in Miller's version 4 is therefore read with that precise limitation. The source integral identity itself is retained and proved in (4.3)–(4.5). For the other source correction in Section 3, set \(\nu=1\) in this same solution: the incorrect forced strain display has residual \(|\omega|^2I_3/2=a^2I_3/2\) on both inner balls. This verifies the sign correction on an actual smooth compact source-class solution.

8. Sources and the next questions

Evan Miller's middle-eigenvalue paper, arXiv:1710.05569v4 provides the strain constraint, isometries, enstrophy identity and middle-eigenvalue criterion compared here. The exact source locators are StrainSpace, TensorIsometry, Strn, DET2, EigenBound and RegCrit. Sections 2–6 give independent proofs on the original whole space and arbitrary rectangular periodic boxes, with the full force of (1.2). The source-specific corrections and examples are identified in Sections 3 and 7 rather than silently substituted into the author's text.

Miller's Navier–Stokes regularity criteria in sum spaces, arXiv:2007.02023v1, particularly StrainSumSpace, EigenEndpoint and VortSumSpace, is the primary-source comparison for Section 6. Its introduction credits Jiří Neustupa and Patrick Penel with the earlier middle-eigenvalue criterion. The two-summand and small-tail arguments there motivate the finite-sum calculation here. Our vector estimate also retains the trace-free matrix gain (5.1); the source's Cauchy–Schwarz step in VortSumSpace retains only the strain–vorticity isometry. We make no novelty claim for these elementary consequences.

The historical Euler reference is J. T. Beale, T. Kato and A. Majda, “Remarks on the breakdown of smooth solutions for the 3-D Euler equations,” Communications in Mathematical Physics 94 (1984), 61–66. That original paper has not been used as a read proof here. The cited Miller source and Terence Tao's localization paper, arXiv:1108.1165v4 supply the historical routing references. The Euler theorem, the general large velocity \(L^\infty_tL^3_x\) endpoint, and the later forced-flow constructions require their own complete arguments. The present chapter proves exactly the viscous criteria and source corrections stated above.