Intermittent fields and exact periodic product estimates

The waves of lesson 21 have an exact average tensor, but their viscous cost grows with frequency. This chapter constructs the source's intermittent amplitudes: large values on small regions with an exactly fixed second norm. It proves their original-period covering map, all derivative bounds and the product estimates needed when their coefficients vary in space and time.

The original pressure and projection conventions remain those of lesson 2. The labels IB, LD and DC identify the three connected parts of the proof. References to RS and BG mean the complete maps and rational frames of lesson 21. Every original factor in the kernel and physical frequency is retained, including the period, integer covering degree and temporal parameter.

The human source is Tristan Buckmaster and Vlad Vicol, Nonuniqueness of weak solutions to the Navier–Stokes equation, arXiv:1709.10033v4. The original author source was read at lines 474–672 and 1533–1592 for these kernels and product estimates, and at 776–836 and 1341–1387 for the temporal correction and its receiving pair. The exposition and proofs here are independently written.

The source's product lemma omits control of the zeroth-order amplitude. A constant amplitude gives a counterexample to that printed statement. DC13–DC17 retain the missing term; DC18–DC22 prove a stronger estimate using only three derivative orders in three dimensions, for all exponents from one to infinity. This is a local source correction and proved strengthening; it does not assert failure or completion of the source's whole iteration. No novelty or independent review is claimed.

The left panel is the exact one-dimensional section \(D_r(y,0,0)\) of the three-dimensional source kernel, not its spatial average. The middle panel plots the proved upper bound IB13 with the actual source period \(L=2\pi\); it does not plot sampled field norms. The last panel displays the pair-invariant signed vector in IB24. Proof: IB1–IB6, IB13 and IB24. Reproducible figure source.

IB1. The complete original finite sums

Let \(r\geq1\) be an integer and keep \(M=2r+1\). The one-dimensional and three-dimensional source kernels are

For \(y\notin2\pi\mathbb Z\), multiplying the finite geometric sum by \(1-e^{iy}\) and using \(1-e^{iz}=-2ie^{iz/2}\sin(z/2)\) gives

The second formula is the actual finite sum at each removable point. Opposite indices prove reality. On \([-\pi,\pi]\), the triangle inequality and the concavity bound \(\sin(|y|/2)\geq |y|/\pi\) give

For the lower bound, every real summand \(\cos(jy)\) is at least \(\cos(1/2)\), because \(|jy|\leq r/M<1/2\), and the imaginary summands cancel. The upper value \(M\) is attained at zero.

For \(1<p<\infty\), split the upper integral at \(\pi/M\) and the lower integral at \(M^{-1}\). Direct integration gives

At \(p=1\), the same split instead gives

Orthogonality of the original exponentials gives the exact, unweighted integral \(\int|d_r|^2=2\pi M\). Thus Fubini with the full prefactor in IB1 proves

These are the original source kernels and their actual measures.

IB2. The integer map on the original torus

Keep the original cube \(\mathbb T_L^3\), \(L>0\), a constructed BG family, its common integer \(N_\Lambda\), a positive integer \(\lambda\) divisible by \(N_\Lambda\), and

These requirements and the later smallness condition have actual simultaneous choices. With the same \(r,N_\Lambda,c_\Lambda\), take

Then \(m=1\), \(\lambda\) is the required positive multiple, \(\sigma r\leq c_\Lambda/(10N_\Lambda)\), and \(\lambda<\mu<\lambda^2\), since \(\lambda>1\). Thus the source's later temporal interval is available. The finite formulas below in fact hold for every \(\mu>0\). Choose one representative from each opposite pair and call that set \(\Lambda^+\); let \(\Lambda^-=-\Lambda^+\). For a representative \(\xi\), keep its original polarization \(A_\xi\) and \(C_\xi=\xi\times A_\xi\). Define

The rows of \(O_\xi\) are an oriented orthonormal basis because \(\xi\times A_\xi=C_\xi\). Therefore \(\det O_\xi=1\), \(\det \mathsf M_\xi=(mN_\Lambda)^3\ne0\). Each matrix entry is integer by BG16 and BG18. The actual angular map is

Its integer matrix makes it well-defined. It is onto: given any real representative of a target angle, apply the actual inverse matrix \(\mathsf M_\xi^{-1}\) and multiply by \(L/(2\pi)\). The different preimages are exactly indexed by \(\mathbb Z^3/\mathsf M_\xi\mathbb Z^3\). This lattice quotient has \(|\det \mathsf M_\xi|\) elements: unit lattice cells, one for each coset, form a fundamental region for \(\mathsf M_\xi\mathbb Z^3\), while its parallelepiped has volume \(|\det \mathsf M_\xi|\); both regions tile by the same lattice, hence have equal volume. In particular, the map has multiplicity \((mN_\Lambda)^3\), and its local Jacobian is \(\beta^3\). It is generally not a one-to-one rotation of the torus.

The exact integral relation for every continuous periodic \(H\) is

Here is a direct proof that also avoids assuming a rational rotation separately descends to the torus. For \(H(y)=e^{ik\cdot y}\), the left side is zero unless \(\mathsf M_\xi^Tk=0\), because it is an original integer Fourier mode in \(x\). The matrix is invertible, so only \(k=0\) survives. Both sides agree on every trigonometric polynomial. Such polynomials approximate each continuous periodic \(H\) uniformly: convolve with the product of the three finite kernels

Each is nonnegative, has integral \(2\pi\), and is a finite trigonometric polynomial. At distance at least \(\delta>0\) from \(2\pi\mathbb Z\), the geometric sum gives \(F_q(z)\leq[(q+1)\sin^2(\delta/2)]^{-1}\). The mass outside a three-dimensional \(\delta\) box therefore tends to zero, by the union bound and the integrals of the other two factors. Inside that box uniform continuity bounds the convolution error by the modulus of continuity of \(H\). This proves uniform convergence and then IB10 by integration.

The original kernel has spatial period \(L/m\) in each coordinate: that translation changes its arguments by \(2\pi N_\Lambda O_\xi e_j\), an integer multiple of \(2\pi\). Minimality of this period is not asserted. At \(L=2\pi\), the source period is \(2\pi/(\lambda\sigma)\); the source's displayed quotient by \(2\pi\lambda\sigma\mathbb Z\) reverses that reciprocal. The equality of averages in IB10 comes from this integer covering, with its full multiplicity and Jacobian. The individual rational rotation and dilation need not separately preserve the same torus.

IB3. Original-volume norm bounds and exact derivative integrals

Apply IB10 to \(H=|D_r|^p\), then use IB4–IB6. For \(1<p<\infty\),

The original-volume endpoint statements are

Surjectivity of IB9 proves the exact maximum. These assertions hold at every actual time; the translating angle does not change the average. For fixed \(L\), the ratio of the last upper bound to the exact \(L^2\) norm tends to zero as \(r\) tends to infinity. For example, \((1+\log M)^3/M^{3/2}\to0\) by three applications of the elementary derivative rule to \((1+\log x)^3/x^{3/2}\).

The complete expansion is

The original lattice vectors are distinct because the frame is a basis. Parseval therefore retains every summand separately. For nonnegative integers \(a,b\), define \(\nabla^a\) to contain all ordered spatial partial derivatives of length \(a\), with their Euclidean square sum. Direct differentiation gives

For \(a=0\) or \(b=0\), the corresponding zeroth power in this finite sum is one, including at the zero index. The finite identity \(\sum_{j=-r}^r j^2=r(r+1)(2r+1)/3\) follows by summing \((j+1)^3-j^3=3j^2+3j+1\) from \(j=0\) to \(r\). Thus

The \(A_\xi\) and \(C_\xi\) directional derivatives have the same squared integral as the \(\xi\) derivative.

IB4. The actual carriers and both signed directions

For \(\zeta\in\Lambda^+\cup\Lambda^-\), retain

The complete product is the source intermittent field; its prefactor is the one in IB1. It is not assumed divergence-free. Since \(|B_\zeta|=1\), every norm of \(\mathbb W_\zeta\) equals the corresponding norm of \(\eta_\zeta\). For \(\zeta=\varepsilon\xi\), \(\xi\in\Lambda^+\), its full spatial frequencies are

If the original parameter inequality \(\sigma r\leq c_\Lambda/(10N_\Lambda)\) holds, put \(\delta_r=\sqrt3\,\sigma N_\Lambda r\). Then every frequency in IB18 has length between \(\varkappa(1-\delta_r)\) and \(\varkappa(1+\delta_r)\), with \(\delta_r\leq\sqrt3c_\Lambda/10<1/2\). For two nonopposite directions their product has carriers at \(\varkappa(\zeta+\zeta')\) and perturbation of length at most \(2\delta_r\varkappa\), so its full support lies in

This proves, in particular, the source intervals \([\varkappa/2,2\varkappa]\) and \([c_\Lambda\varkappa,4\varkappa]\), retaining the tighter actual margins. The argument uses only finite Minkowski sums, so no projection cutoff convention is silently assumed.

Differentiating IB8, and using its definition at the negative direction, gives the full signed identity

The product rule and the original complex polarization give

As \(\eta_\zeta\) is real, the squared first magnitude is one half the sum of the squared derivatives in the two transverse directions. The squared second magnitude is the full gradient square minus that half sum. Applying IB16 yields the exact defects

For every \(a,b\geq0\), the corresponding full derivative identity is

It follows directly from the distinct original frequencies IB18. No original \(N_\Lambda\), volume or physical frequency is omitted.

IB5. The exact opposite pair and its sign

Opposite coefficients still give a real full field because \(\eta_{-\zeta}=\eta_\zeta\) and \(B_{-\zeta}=\overline{B_\zeta}\). BG23 proves the pointwise tensor and its divergence:

Indeed the constant tensor on the first line gives \(\nabla\eta_\zeta^2-\zeta(\zeta\cdot\nabla)\eta_\zeta^2\); IB20 then gives the second line. Replacing \(\zeta\) by \(-\zeta\) leaves both sides unchanged, since \(\varepsilon_{-\zeta}(-\zeta)=\varepsilon_\zeta\zeta\). The source equation eq:useful:2 has the second line without \(\varepsilon_\zeta\). It is correct for the chosen positive representatives, and the formula for the whole opposite-closed family is IB24.

This is not a vacuous sign difference. In the finite expansion of \(\eta_\xi^2\), the frequency with indices \((2r,2r,2r)\) has exactly one contributing ordered pair of index triples. Its coefficient is \(M^{-3}e^{i2r\beta\mu t}\). Its time derivative is nonzero because \(r,\beta,\mu>0\). Thus omitting the sign fails at the negative direction.

The complete average follows from IB13:

Grouping the two orientations and using the complete BG15 identity proves the final equality. Here \(R\) is any actual matrix in the proved positive domain, including the enlarged trace slice of lesson 21 Exercise 2. The nonopposite products have no zero frequency by IB19, so the full constant-coefficient sum has this same average tensor.

LD1. A finite kernel that reproduces the original Fourier modes

For a positive integer \(m\), keep the full finite kernel

Expansion of the square proves the coefficient at \(j\): there are exactly \(m-|j|\) ordered index pairs with that difference. Integration leaves its constant coefficient one. The geometric-sum identity gives

No singularity is introduced at a removable point.

For the actual positive integer \(r\), put

For \(|j|<r\), the complete coefficient is \(2(1-|j|/(2r))-(1-|j|/r)=1\). At \(|j|=r\) the second term is zero and the first is one. Outside \(2r\), both are zero. Hence every original polynomial \(f(y)=\sum_{|j|\leq r}a_je^{ijy}\) satisfies

This follows by multiplying the finite Fourier coefficients. It is an exact map on these original modes, with the full convolution factor retained.

LD2. Every derivative of the kernel, with explicit constants

On \(0<|y|\leq\pi\), set \(h(y)=\sin(y/2)^{-2}\). Define the finite constants

Every one is a specified positive real number depending only on the nonnegative integer order. We claim

For order zero this is the original sine lower bound of IB3. To prove the higher orders, set \(g(y)=\sin^2(y/2)=(1-\cos y)/2\), so \(gh=1\). The full derivatives satisfy

Differentiate \(gh=1\) exactly \(a\) times, solve for the term with \(h^{(a)}\), and apply induction:

This proves LD6 for every order without dropping any derivative from the reciprocal.

The numerator \(n_m(y)=\sin^2(my/2)\) has \(|n_m^{(\ell)}(y)|\leq m^\ell\) for all \(\ell\geq0\): at order zero it is at most one; at positive order its exact trigonometric derivative is at most \(m^\ell/2\). Using the larger displayed bound for all orders, Leibniz in LD2 gives, whenever \(|y|\geq m^{-1}\),

On the complementary interval the full Fourier formula LD1 has nonnegative coefficients with sum \(m\). Therefore

For \(a=0\), the zero-frequency zeroth power is one. Since \(m^{-1}\leq\pi\), direct integration of both regions yields

The last line retains both kernels and their coefficients: it is \(2\|T_{2r}^{(a)}\|_1+\|T_r^{(a)}\|_1\). These constants are finite for every order and independent of the original frequency parameter.

LD3. All \(L^p\) derivatives of the original three-dimensional kernel

For every \(1\leq p\leq\infty\), convolution with an integrable kernel \(K\) on the original \(2\pi\)-periodic circle obeys

If \(\|K\|_1=0\) this is immediate. Otherwise, for finite \(p\geq1\), the triangle inequality followed by Jensen for the probability measure \(|K|/\|K\|_1\) gives the pointwise \(p\)-th power bound by \(\|K\|_1^{p-1}\int|K(z)||f(y-z)|^p\,dz\). Integrating and translating the full periodic integral proves LD12. At \(p=\infty\), use the pointwise supremum in that same integral. The argument also covers complex-valued \(f\).

Define

LD4 and LD11–LD12 prove for every original polynomial of degree at most \(r\),

For \(a=0\) this is the exact identity. For positive order, the factor \(1/(2\pi)\) in LD4 gives exactly LD13.

Retain the full \(D_r=M^{-3/2}d_r(y_1)d_r(y_2)d_r(y_3)\) of IB1. For a multi-index \(\alpha=(\alpha_1,\alpha_2,\alpha_3)\), Fubini and the independent variable derivatives give

At finite \(p\), take the product of the three integrals before the \(p\)-th root. At \(p=\infty\), take the product of their suprema. The original prefactor is on both sides throughout.

For nonnegative integers \(N,K\), put

The ordered derivative tensor of IB15 contains \(N!/(\alpha_1!\alpha_2!\alpha_3!)\) copies of \(\partial^{\alpha+(K,0,0)}D_r\). Its pointwise Euclidean norm is bounded by the sum of the norms of those multi-index blocks. Applying LD15 and the triangle inequality in \(L^p\), including its endpoints, gives

Thus all tensor multiplicities are accounted for explicitly.

LD4. The actual original fields and their complete Leibniz terms

Use the actual \(L,\varkappa,\beta,\mu\) and original covering of IB7–IB10. Each spatial derivative of \(\eta_\zeta\) is \(\beta\) times the corresponding row transformation of a derivative in its original kernel arguments. The complete ordered derivative tensor transforms by the \(N\)-fold tensor product of the orthogonal frame matrix. That tensor product is an isometry: expanding its Euclidean square norm contracts each pair of matrix factors to an identity. Each time derivative contributes \(\beta\mu\,\partial_{y_1}\) for the common representative. Applying the full covering integral therefore gives the exact norm map

At \(p=\infty\), the volume factor is one and surjectivity proves equality of the suprema.

For concise display of the full proved original bounds define the explicit scalar

This is exactly the earlier IB12–IB13 upper bound, with all volume factors and finite-\(r\) contributions displayed. Equations LD17–LD18 now prove

For the complete field \(\mathbb W_\zeta=\eta_\zeta B_\zeta e^{i\varkappa\zeta\cdot x}\), every original ordered derivative has the exact expansion

The empty derivative product is the identity; the empty scalar product is one. No cross term has been omitted. For fixed \(|E|=a\), its full derivative array is a permutation of the tensor product of \(a\) amplitude derivatives with \(N-a\) copies of \(i\varkappa\zeta\) and the fixed vector \(B_\zeta\). The Euclidean norm of a tensor product is the product of the norms; both \(|\zeta|\) and \(|B_\zeta|\) are one. There are exactly \(\binom Na\) subsets of that size. The triangle inequality with LD20 proves

The second line substitutes the actual equality \(\beta=\varkappa\sigma N_\Lambda\); the first line retains every contributing derivative order. Under the proved original support constraint, its final finite sum is at most

This recovers the source derivative powers, including the full \(N_\Lambda\), physical period and temporal factors, with an explicit constant independent of a later iteration. The \(p=1\) case has the full logarithmic factor in LD19. For \(p=2\), the sharper exact sums IB15 and IB23 remain available; this upper bound does not replace those identities.

DC1. Original cubes and the exact periodic mass

Keep dimension \(n\geq1\), original cube period \(L>0\), a positive integer \(m\), and the actual length and diameter

Their interiors are disjoint and they cover the original cube, apart from their measure-zero faces. Let \(g\in L^p(\mathbb T_L^n)\) have period \(h\) in every coordinate. Translation by \(h\ell\) gives for finite \(p\)

At \(p=\infty\), every cell has the same essential supremum. These assertions also hold for a vector-valued \(g\), with its Euclidean magnitude.

For any smooth finite-dimensional tensor field \(H\) on one closed cell, integration of the line segment from \(y\) to \(x\) gives the exact formula

The segment remains in the closed cell. The Hilbert–Schmidt norm bounds the contraction by \(|DH|\,|x-y|\), and \(|x-y|\leq d\). Hence

For a smooth scalar or vector amplitude \(f\), let \(\nabla^j f\) be the full ordered derivative tensor, with its Euclidean norm. Applying DC4 successively to those actual tensors gives, for every positive integer \(M\),

This follows by finite substitution, retaining each term; it does not require a convergent infinite expansion. The full zeroth-order average remains present.

DC2. A direct \(L^p\) product map

For finite \(1\leq p<\infty\), multiply DC5 by \(|g|\), integrate its \(p\)-th power over the cells, and use DC2. The triangle inequality for the resulting finite \(\ell^p\) norm gives

The second inequality is Jensen on each cell: the sum of the averaged tensor \(p\)-th powers is at most \(h^{-n}\|\nabla^jf\|_p^p\). The exact product \(m^{-n/p}h^{-n/p}=L^{-n/p}\) retains the original volume. For vector \(f\) and vector \(g\), interpret \(fg\) as their tensor product, whose Euclidean magnitude is \(|f||g|\); the same proof then applies componentwise through that norm.

At \(p=\infty\), DC6 also holds, with every volume exponent zero: its \(j=0\) term is already the exact product supremum bound, and all remaining terms are nonnegative. No squaring of the amplitude is needed at \(p=2\), or at any other exponent.

DC3. An explicit original-period bound for the remainder

For a smooth one-dimensional periodic tensor function \(H\), the fundamental theorem of calculus, averaged over \(y\), gives at every \(x\in[0,L]\)

To verify the kernel, write \(H(x)-L^{-1}\int H(y)\,dy=L^{-1}\int\int_y^xH'(s)\,ds\,dy\). For \(s<x\), the allowed \(y\) occupy length \(s\); for \(s>x\), the integral orientation is negative and the allowed \(y\) occupy length \(L-s\). This proves precisely the displayed \(b_x\), including its sign.

Apply DC7 in each of the \(n\) original coordinates. The operators in different coordinates commute by Fubini. Expanding all \(2^n\) choices gives the full representation

An empty product is one. There are no omitted boundary terms: DC7 is an identity on the whole closed interval and was applied before integration. The factor \(L^{-1}\) is used exactly once for each coordinate in which the derivative choice was not taken.

The triangle inequality, \(|b_x|\leq1\), and Hölder on the original volume \(L^n\) now give, for all \(1\leq p\leq\infty\),

For \(p=\infty\), the integral is bounded by volume times the supremum; for \(p=1\), no volume factor comes from Hölder. This proves the needed embedding directly, with its full period dependence and zero-order contribution.

For \(H=\nabla^Mf\), every component of \(\partial^\epsilon\nabla^Mf\) occurs in the full ordered tensor \(\nabla^{M+|\epsilon|}f\). Fixing the first \(|\epsilon|\) derivative indices gives an injective inclusion of arrays; hence its Euclidean norm is bounded by that of the full higher tensor. Grouping the \(\binom ns\) subsets with size \(s\) proves

DC4. The complete finite bound and the source comparison

Substitute DC10 into the unchanged original DC6:

This holds for each smooth periodic amplitude \(f\), every actual small-period \(g\), all positive integers \(M,m\), and every \(1\leq p\leq\infty\). In particular, it is already a complete product estimate without an assumed frequency hierarchy.

If the actual finite amplitude costs satisfy \(\|\nabla^jf\|_p\leq C_f\Lambda^j\) for \(0\leq j\leq M+n\), put \(q=d\Lambda\). The full right side is then bounded by

The polynomial in the second term is exactly the binomial sum in DC11. For \(0\leq q<1\), the finite geometric sum is \((1-q^M)/(1-q)\); its infinite upper bound is \(1/(1-q)\). Both keep the actual remainder.

If only the source's positive-order costs are available, retain the missing zeroth-order norm explicitly instead:

The sum is empty when \(M=1\). This proves the exact repair with the source's existing derivative information rather than assuming away the missing amplitude.

For the source choices \(n=3\), \(L=2\pi\), \(\Lambda=\lambda\geq1\), \(m=\kappa\),

the actual positive-order contribution obeys

Thus DC13 gives the complete source-domain bound

If the missing zeroth-order estimate is also given, replace \(\|f\|_p\) by \(C_f\), obtaining the source type of conclusion with explicit constant \((2\pi)^{-3/p}[3/2+(1+2\pi)^3]\). Only derivatives through order \(M+3\) were used, one fewer than the source's stated \(M+4\), and the estimate holds for the entire interval of \(p\), including infinity.

The omitted amplitude cannot be inferred from the original positive-order assumptions. For example, take \(\lambda=1\), \(M=1\), and any integer \(\kappa\geq6\pi\sqrt3\); then DC14 holds. Let \(C_f=1\), \(f(x)=A\) with arbitrary \(A>0\), and \(g=1\). Every positive derivative of \(f\) is zero, satisfying all stated bounds, while

has no bound by a universal multiple of \(C_f\|g\|_p\) as \(A\) increases. The complete estimate DC16 keeps \(\|f\|_p=A(2\pi)^{3/p}\) and remains valid. This local statement and its repair do not determine the validity of an iteration that also controls its actual amplitudes; those receiving costs must be checked separately.

DC5. A stronger cellwise bound with only mixed first derivatives

The same exact averaging map gives a stronger finite estimate. Keep each original cell, with lower corner \(a=h\ell\). On its \(i\)-th interval replace the kernel of DC7 by

The proof of DC7 uses only the fundamental theorem of calculus, so it applies on this interval without a periodicity assumption on the restricted amplitude. Applying it in every coordinate and expanding all choices gives, for \(x\) in the closed cell,

The empty product is one and all boundary contributions are already in the signed kernels. Hölder on the actual cell yields

For finite \(p\), multiply this bound by the common cell mass of \(|g|^p\), sum, and take the \(p\)-th root. Minkowski on the finite cell index set, followed by disjointness, proves

The volume factor is again exactly \(m^{-n/p}h^{-n/p}=L^{-n/p}\). The final inequality includes each mixed derivative in the full ordered tensor of its order. For \(p=\infty\), the zeroth term alone gives the required product bound, so both inequalities remain true. The same tensor product interpretation as DC6 covers vector amplitudes and factors.

Thus no derivative order beyond \(n\) is needed. No parameter hierarchy is required for DC21 itself. In the source's three dimensions, positive derivative bounds through order three give

The additional source condition involving \(M\) is unnecessary for this estimate. DC11–DC16 remain proved, with all of their terms retained; DC21 supplies a shorter-derivative receiver. The counterexample DC17 still requires the zeroth-order term.

The left panel displays the actual functions \(f(x)=2+0.4\sin x\) and \(g(x)=1+\cos(4x)\) on \([0,2\pi]\), with cell length \(h=\pi/2\). Every cell has the same integral of \(g\); the horizontal segments are the exact cell averages of \(f\). These are example amplitudes, not fluid velocities. The right panel displays the signed kernel DC7 at \(x=L/3\), including its downward jump of one. DC18–DC21 apply the same identity on each original small cell. The plots illustrate the proved maps; sampled curves do not establish the estimates. Reproducible figure source. Human comparison: Buckmaster–Vicol, arXiv:1709.10033v4, Lemma lem:Lp:independence and Appendix app:Lp:product.

Five solved exercises

Exercise 1: compute the exact fourth norm and the squared factor

Compute the fourth norm of the original field, retaining the finite parameter \(M=2r+1\) and original volume \(L^3\).

Solution. Every coefficient of \(d_r^2\) counts ordered pairs with a fixed sum. Thus

The sum formula \(\sum_{j=1}^{k}j^2=k(k+1)(2k+1)/6\) follows by induction: subtract its value at \(k-1\) and obtain \(k^2\), with the value zero at \(k=0\). Parseval gives the middle equality, with the full original circle measure. Taking the three independent integrals and the fourth power of \(M^{-3/2}\), then applying the exact covering IB10, gives

Consequently the actual squared factor \(\theta=\eta_\xi^2\) has

The last identity expands the complete square and uses \(\int\theta=L^3\). This is the exact cost received by the temporal correction in Exercise 5.

Exercise 2: retain every fourth-order spatial and time contribution

Compute the norms of the full ordered second spatial derivative, the second time derivative, and the mixed space–time derivative.

Solution. Define the exact one-index moments

For the fourth power, the polynomial \(P(k)=k(k+1)(2k+1)(3k^2+3k-1)/30\) has \(P(0)=0\) and \(P(k)-P(k-1)=k^4\), as expansion of the five factors verifies. Hence the positive-index sum is \(P(r)\); symmetry and division by \(M\) give EX4. The same argument with the square-sum polynomial gives \(s_2\).

In the exact finite sum IB15 expand every factor:

The three indices range independently, so

The factors six and two retain all mixed-index contributions. For example, the last sum is the average of \((j^2+k^2+\ell^2)j^2\), with one fourth-power term and two independent square products. Substitution of EX4 gives every coefficient as an explicit polynomial in the original \(r\); neither time frequency nor volume is removed.

Exercise 3: find an exact energy map and its boundary resonance

Show that an amplitude whose Fourier frequencies have magnitude strictly below \(\beta/2\) has an exact product-energy identity. Test the boundary with an actual periodic amplitude.

Solution. The complete square of the original finite kernel has the expansion

It follows by multiplying the three exact one-dimensional sums of EX1. Its constant coefficient is one. Orthogonality of the frame proves that every nonzero frequency has magnitude at least \(\beta\), with equality on the six coordinate indices.

Let \(a\) be any real trigonometric polynomial on the original torus whose physical frequencies \(k\in(2\pi/L)\mathbb Z^3\) satisfy \(|k|<\beta/2\). Every frequency of \(a^2\) is a sum of two such frequencies and has magnitude strictly below \(\beta\). In the integral of \(a^2\theta\), a nonzero frequency from EX6 therefore cannot cancel any frequency of \(a^2\). Integration leaves exactly

This proves an equality, stronger than the product bound for this precisely specified class of original amplitudes.

To test its boundary, choose the integer \(m=\lambda\sigma\) even. Since \(N_\Lambda\xi\in\mathbb Z^3\), \(a(x)=\cos((\beta/2)\xi\cdot x)\) is genuinely periodic on \(\mathbb T_L^3\). Such parameters satisfy IB7 by taking \(\lambda\) a sufficiently large multiple of \(N_\Lambda\) and \(\sigma=m/\lambda\), so the example is within the original support restrictions. The coefficients at the two indices \((\pm1,0,0)\) in EX6 are \((M-1)M^{-1}e^{\pm i\beta\mu t}\). Since \(a^2=(1+\cos(\beta\xi\cdot x))/2\), exact integration gives

The equality EX7 fails at this boundary, for example at \(t=0\). Thus the strict radius \(\beta/2\) cannot be enlarged uniformly over the permitted periods and parameters.

Exercise 4: keep different original periods and cell lengths

Extend the stronger cell estimate to an original rectangular torus without replacing its periods by a common length.

Solution. Keep \(L_i>0\), positive integers \(m_i\), \(h_i=L_i/m_i\), volume \(V=\prod_i L_i\), and cell volume \(v=\prod_i h_i\). Suppose \(g\) has period \(h_i\) in coordinate \(i\). Every one of the \(\prod_i m_i\) cells has the same \(|g|^p\) integral, equal to \((v/V)\|g\|_p^p\). In coordinate \(i\), DC18 now has its own denominator \(h_i\). Expanding all coordinate choices and applying Hölder on the original cell gives

Minkowski on the cell index set and the full mass factor \((v/V)^{1/p}\) yield

This includes \(p=1\); at \(p=\infty\) the zeroth term already proves the product estimate. Every original period and mixed derivative is present, with no replacement by a largest or smallest length. The proof also applies to vector fields with the tensor product convention of DC6.

Exercise 5: construct the temporal correction and its whole pair residual

For one positive representative \(\xi\), take a smooth real amplitude \(a(x,t)\), put \(A=a^2\) and \(\theta=\eta_\xi^2\). Construct its temporal correction, pressure and exact residual, including all amplitude derivatives and constant modes.

Solution. Let \(P\) be the original Leray projection from lesson 2, and let \(P_0v=v-\langle v\rangle\). On the original torus these commute. The inverse Laplacian has its zero mode set to zero. Define

The exact Fourier multipliers of the projections prove \(\operatorname{div}z=0\) and \(\langle z\rangle=0\), and commute with time differentiation of these smooth fields. By the positive-representative transport IB20, \(\xi\cdot\nabla\theta=\mu^{-1}\partial_t\theta\). Expanding the divergence and time derivative gives

Here the exact identity \(I-PP_0=\nabla\Delta^{-1}\operatorname{div}+\langle\cdot\rangle\) accounts for both the pressure part and the constant mode. There is no assumed vanishing mean of a product. Instead, periodic integration by parts proves

Set the actual pressure and symmetric trace-free stress to

The complete inverse divergence in lesson 21 has \(\operatorname{div}\mathcal Rv=P_0v\). Equations EX12–EX14 therefore establish the exact pair equation

In particular, if \(A\) is constant, \(b=0\) and this entire pair residual vanishes. The sign of \(z\) in EX11 is positive. This agrees with Buckmaster–Vicol's temporal corrector e:temporal_corrector, original source lines 805–809, whose sum is explicitly over positive representatives.

The stronger product estimate gives concrete receiving bounds. Write \(h=L/(\lambda\sigma)\) and \(E_M=((2M^2+1)/(3M))^{3/2}\), keeping its whole expression. Orthogonality makes \(PP_0\) an \(L^2\) contraction; the sharp original-period inverse bound RS13 is \(C_L=\sqrt{2(1+(L/(2\pi))^2)}\) from \(L^2\) to \(H^1\). Using DC21 and EX3 gives

The factors \(L^{-3/2}\) in DC21 and \(L^{3/2}\) in EX3 cancel exactly, while the physical period remains in \(h\) and \(C_L\). These are bounds for the explicit fields EX11 and EX14, with their actual amplitude costs.

Equation EX15 is a complete map for the paired quadratic tensor. To turn it into a fluid velocity correction, the principal field, its incompressibility correction, the original viscosity term, transport by the old velocity and all quadratic cross terms must also enter the stress equation. Those terms are the next calculation; EX15 alone makes no claim of an infinite construction or a new fluid solution.

What the next correction must contain

The complete finite kernels, exact fourth norms and temporal pair map are now available. Exercise 5 inserts the stronger product estimate into its actual velocity and stress bounds, including the original mean and pressure. The finite results are proved for the periods, indices and fields stated here.

The next lesson must add the full incompressibility correction, transport by the old velocity, the original positive viscosity, all cross terms, and the true varying-amplitude oscillation stress. The remaining frequency estimates must apply to those actual terms before any infinite iteration is concluded. The later Albritton–Brué–Colombo, Alpöge–Buckmaster, OpenAI and workbench constructions keep their own equations and precise hypotheses.