The bounded critical-velocity endpoint

A bounded \(L^3\) velocity norm prevents a finite singular endpoint for the unforced Navier–Stokes equation on three-dimensional space, in the energy classes defined in the earlier lessons. This chapter proves that statement from the frequency estimate, then transfers it from smooth solutions to the actual strong and energy weak solutions.

The original viscosity stays positive and explicit. We keep the full heat field and every cross term in the nonlinear vorticity identity. The smoothing argument uses the exact time norm of lesson 5, and the final identification uses the full energy comparison of lesson 6. The frequency and mass providers are lesson 12, lesson 15, and lesson 19. No additional regularity is assumed in the weak-solution conclusion.

The human source is Terence Tao, Quantitative bounds for critically bounded solutions to the Navier–Stokes equations, arXiv:1908.04958v2, original author article.tex, Applications, lines 1422–1490. Its displayed enstrophy identity lists five terms, although its following definitions and estimates include a sixth. Section 2 derives the full identity and keeps that term visibly. This is independent course exposition and author self-check; no novelty or independent review is claimed.

1. The full identity in the original equation

Keep a smooth unforced solution of the original equation

The original velocity and its derivatives have bounded \(L^2\) norms on each compact time interval under consideration, and \(\sup_t\|u(t)\|_3\leq U\). For an actual earlier time \(t_b\), let

The Gaussian is exactly \(H_s(x)=(4\pi s)^{-3/2}e^{-|x|^2/(4s)}\). Since \(\ell_t=\nu\Delta\ell\), taking the original curl gives

Every sign follows from \(\omega_t-\nu\Delta\omega=-(u\cdot\nabla)\omega+ (\omega\cdot\nabla)u\); pressure has zero curl because it is the original scalar pressure. No replacement pressure is introduced.

2. All six contributions and the global curl map

Multiplying 1.3 by \(z\) and integrating proves

where

The label \(Y_1\) in the source is its unit-viscosity dissipation; here it is retained as \(\nu D\). To justify the sole transport cancellation, use compact spatial cutoffs \(\chi_R\) equal to one on the radius-\(R\) ball, with \(|\nabla\chi_R|\leq C/R\). Divergence freedom gives \(-\int\chi_R z\cdot(u\cdot\nabla)z= (1/2)\int(u\cdot\nabla\chi_R)|z|^2\). Its absolute value is at most \(C\|u\|_\infty\|z\|_2^2/(2R)\), which tends to zero. The diffusion term converges to \(-\nu D\) by its \(H^1\) integrability and the same cutoff limit. All remaining products are integrable in the stated smooth class and by the bounds below. This proves 2.1 with the missing \(Y_6\) present.

The global Fourier identity for the solenoidal \(w\) is

Indeed the original Fourier multiplier is \(2\pi i\xi\):

The last summand is zero by the actual divergence condition. Parseval gives 2.3. This identity is used on the whole space, not on an unproved restricted curl inverse.

3. Bounds for all four heat cross terms

Write \(s=t-t_b>0\). The original Gaussian bounds from lesson 13 (1.5) retain

All tensors carry their full Euclidean norms. The curl inequalities give \(\|\ell\|_p\leq\sqrt2\|\nabla v\|_p\) and \(\|\nabla\ell\|_p\leq\sqrt2\|\nabla^2v\|_p\). Hölder, 2.3 and 3.1 therefore prove separately

For \(Y_6\), the two original heat powers are \(-1/2\) and \(-3/4\), totaling \(-5/4\). For \(Y_2\), the second derivative in \(L^6\) has that same power. Their two constants remain distinct.

For every \(A\geq0,E\geq0,s>0\), the square \((s^{-1/2}\sqrt E-s^{1/2}A)^2\geq0\) proves \(2A\sqrt E\leq s^{-1}E+sA^2\). Use \(A=(C_{2,6}+C_{1,3}C_{1,6})U^2(\nu s)^{-5/4}\) in the sum of the first two estimates. The complete differential consequence is

Both source heat terms, both mixed-gradient terms, the original viscosity and the actual positive time gap are present. 3.3 is a proved estimate; its nonlinear term \(Y_3\) has not been assumed small or omitted.

4. Original projections and the high-frequency bound

First let \(u,p\) be a smooth unforced solution on \([a,a+T]\times\mathbb R^3\), \(T>0\), of

Use the smooth energy class of lesson 19, with bounded spatial \(L^2\) derivative norms on this closed interval. Their sizes will not enter the estimates. If \(U=0\), the velocity is zero by continuity and all assertions below follow directly. Hence assume \(U>0\). Keep the actual heat field and nonlinear remainder \(v(t)=H_{\nu(t-a)}*u(a)\), \(w=u-v\), \(\ell=\nabla\times v\), \(z=\nabla\times w\), and \(E=\|z\|_2^2/2\), \(D=\|\nabla z\|_2^2\).

The original radial functions are

The denominator is positive. Its two summands are oppositely monotone where both are nonzero, so \(\chi\) is nonincreasing. Consequently \(0\leq p\leq1\). It vanishes for \(|\xi|\leq1/4\) and \(|\xi|\geq1\), including the two boundary spheres, and \(\widetilde p=1\) on its support. For any fixed original frequency \(N_{\rm ref}>0\), the grid \(\mathcal D=N_{\rm ref}2^{\mathbb Z}\) therefore has \(\sum_{N\in\mathcal D}p(\xi/N)=1\) for every \(\xi\ne0\), by the complete telescoping identity. We do not set \(N_{\rm ref}\) or any physical frequency to one.

Define the finite positive constants

Smooth compact Fourier support makes these kernels Schwartz: repeated integration by parts gives arbitrarily high polynomial decay. Nonconstancy of the companion makes \(\kappa_\partial>0\). This \(\kappa_p\) is distinct from the quantity \(c_p=\|p\|_2\) in lesson 12.

Retain the exact \(b_0>0\) of lesson 15 (2.3), and choose

In \(\mathcal Q\), all symbols are the complete lesson 19 (7.1) constants constructed at this chosen \(b\). In particular the \(K\) there is the longer-time receiver constant. Here \(K_{\rm heat}\) is exactly the full kernel constant of lesson 15 (1.4), and \(c_{\rm heat}=\pi^2/8\); those are distinct constants. No later frequency or time occurs in any of these choices.

For every \(t\in[a+T/2,a+T]\), the lesson 19 (7.1) result applies on the original interval \([a,t]\). Its length is at least \(T/2\). Thus \(N\geq N_c\) cannot have a point with \(|P_Nu(t,x)|\geq bN\): such an event would imply \((t-a)N^2<\mathcal Q\), contradicting 4.4. The original heat-kernel estimate also gives

If \(K_{\rm heat}U/b\leq1\), the second estimate follows without a positive logarithm. Otherwise 4.4 supplies exactly the needed exponential factor. The low-frequency bound follows from the full convolution kernel \(N^3\mathcal F^{-1}p(Nx)\), whose \(L^{3/2}\) norm is \(\kappa_p N\), and \(\|w\|_3\leq\|u\|_3+\|v\|_3\leq2U\). This explicitly includes the heat part of the high-frequency nonlinear field, which cannot be removed from \(P_Nw\).

5. The full trilinear term and its frequency count

At any of those times set \(z_N=P_Nz\), \(W_N=\nabla P_Nw\), and \(a_N=\|z_N\|_2\). The global curl identity gives

Indeed \(P_N\widetilde P_N=P_N\). Each original derivative therefore has kernel \(N^4\mathcal F^{-1}(2\pi i\xi_j\widetilde p)(Nx)\), of \(L^1\) norm equal to \(N\) times the stated one. For the gradient tensor its Euclidean norm is bounded by the sum of the three derivative-vector norms. For the curl, \(\nabla\times f=\sum_j e_j\times\partial_jf\), and each cross product has operator norm one. This proves both bounds without losing a tensor component.

The actual term is

Here is the exact support test. Put \(H=\max(N_1,N_2,N_3)\), let \(M\) be the middle frequency, and \(L\) the least. Where the three Fourier factors can be nonzero, their vectors have sum zero and the highest vector has magnitude strictly greater than \(H/4\), while the other two have magnitudes strictly less than \(M\) and \(L\leq M\). Hence \(H/4<2M\), or \(H<8M\). On the original dyadic grid this leaves only

In particular the putative ratio \(H/M=8\) does not contribute. One may prove the vanishing integral first for compact Fourier approximants and then by \(L^2\) convergence and the bounded third factor; the supports never meet zero in the convolution when \(H\geq8M\). This also handles boundary spheres.

The pointwise contraction in 5.2 is bounded by the product of the two vorticity-vector norms and the full gradient-tensor norm. For each of at most six orderings, place the two larger-frequency factors in \(L^2\), and the smaller in \(L^\infty\). Whichever original role occupies each position, 5.1 gives the same bounds. Ties may be counted more than once, which only increases this nonnegative upper bound. From 4.5–5.1 and the complete geometric sums,

For every positive upper endpoint \(R\), the sum of \(L^2\) over grid frequencies at most \(R\) is at most \(4R^2/3\): take the largest such grid frequency and sum \(\sum_{k\geq0}4^{-k}=4/3\). This proof does not require \(N_c\) to belong to the grid.

Cauchy–Schwarz and the bijection \(H\mapsto2^{-j}H\) give

The second line applies Cauchy to \((Ha_H)((2^{-j}H)a_{2^{-j}H})\), retaining the extra factor \(2^{-j}\). Furthermore Parseval and \(0\leq p\leq1\) give

The first two inequalities use \(\sum p^2\leq\sum p=1\) and the original support relation \(N\leq4|\xi|\) wherever \(p(\xi/N)\ne0\). Thus the factor \(4/\pi^2\) includes the full gradient multiplier \(2\pi i\xi\). 4.4 proves the final absorption.

To justify the infinite expansion, truncate the same dyadic decomposition in all three original fields. The telescoping multipliers converge in each of their finite Sobolev norms. In the stated smooth class this gives \(L^2\) convergence of both differentiated fields and \(L^\infty\) convergence of the factor needed in the integral; Fourier Cauchy–Schwarz proves the latter from \(H^2\). The trilinear integrals therefore converge to \(Y_3\). The sum of their absolute values is finite by 5.4–5.6: \(E,D<\infty\), and both discrete Cauchy estimates are valid for finite sets before their monotone limit. This proves 5.2 and 5.6 with all low and high frequencies present.

6. Original enstrophy and gradient bounds

Insert 5.6 into the complete 3.3 inequality. Define

Since \(s=t-a\geq T/2\), the original equation gives on \(I=[a+T/2,a+T]\)

Both terms containing the source \(Y_6\) remain in \(B_c\). The heat-gradient contributions and the Young-square coefficient remain in \(A_c\).

For the actual earlier interval take \(\delta=T/4\). The complete nonlinear energy estimate of lesson 12 gives

The initial nonlinear energy term is bounded by \(W_0^2\) before evaluation; both original square-root endpoints are shown. There is no force contribution in the original unforced equation. The final integral uses the global identity \(\|z\|_2^2=\|\nabla w\|_2^2\).

For each \(t\in[a+3T/4,a+T]\), the actual preceding interval \([t-T/4,t]\) lies in \(I\). Its integral bound gives a time \(r\) in that interval with \(E(r)\leq2M_c/\sqrt T\). Continuity justifies the pointwise choice. Integrating 6.2 with its exact integrating factor gives

The last estimate integrates 6.2, uses the already proved initial and maximum energies on that quarter interval, and keeps the nonnegative final energy before bounding it below by zero. In particular every constant here is independent of the auxiliary higher norms.

For the whole original velocity retain the original \(L^2\) energy as well. Put \(J_\nabla=\|\nabla H_1\|_1\). The exact unforced energy identity and the Gaussian convolution give, on that last quarter,

This adds back the full heat field, rather than identifying the nonlinear gradient with the full velocity gradient. The square retains its mixed term. The original \(H^1\) norm is \(\|u\|_2^2+\|\nabla u\|_2^2\).

7. Positive-time smoothness of the H1 solution

To apply the preceding calculation to the solution constructed in lesson 5, we supply its actual positive-time smoothing. For an integer \(m\geq1\) define

The ordered derivative tensors give the second equality. For \(m=1,2\) these are precisely the original norms of lesson 5; no lower-order contribution is removed. On an interval \(I\) give \(X_m\) the actual norm \(\max\{\sup_I\|u(t)\|_{H^m}, (\int_I\|u(t)\|_{H^{m+1}}^2\,dt)^{1/2}\}\). This is the exact time norm of lesson 5 at \(m=1\). The viscosity remains in the original equation and estimates. Let \(S_6=4/\sqrt3\), so the proved whole-space Sobolev inequality is \(\|f\|_6\leq S_6\|\nabla f\|_2\).

For an integer \(s\geq2\) put

The integral is finite by polar coordinates at zero and infinity. The \(s=2\) value follows from the full evaluation integral in lesson 13 (2.4). Fourier Cauchy–Schwarz gives \(\|\widehat f\|_1\leq\kappa_s\|f\|_{H^s}\) and hence \(\|f\|_\infty\leq\kappa_s\|f\|_{H^s}\). The weight inequality \(\langle\xi\rangle^s\leq 2^{s-1}(\langle\eta\rangle^s+\langle\xi-\eta\rangle^s)\), where \(\langle\xi\rangle=(1+4\pi^2|\xi|^2)^{1/2}\), and Young's convolution inequality prove

The weight inequality follows from the Euclidean triangle inequality followed by the convex power inequality. Apply it to the full Fourier product convolution, estimate each of its two terms by \(L^2*L^1\), and use 7.2. Approximation proves the formula for all indicated Sobolev inputs.

At the lower product order needed for \(m=2\), the exact product gradient and the original embeddings give

The first inequality uses the triangle inequality in the combined value-and-gradient \(L^2\) norm. It also holds for scalar-vector products with their Euclidean norms.

Thus for \(m\geq2\)

Here \(C_m\) is a product constant local to this section; it is not the stretching constant in the earlier annulus. The factor three comes from summing the three original ordered velocity components. For \(m\geq3\), the sharper first factor \(\|a\|_{H^{m-1}}\) also follows from 7.3. Here is the complete map to the earlier linear estimate. The Fourier multiplier \(T_m=(1-\Delta)^{(m-1)/2}\) maps \(C H^m\cap L^2H^{m+1}\) isometrically onto \(C H^1\cap L^2H^2\), with the actual time norms above. Its inverse has multiplier \((1+4\pi^2|\xi|^2)^{-(m-1)/2}\). Both commute with the original heat semigroup and Leray projection, and \(T_m\) maps the force \(L^2H^{m-1}\) isometrically to \(L^2L^2\). Applying lesson 5 to the transformed linear equation and pulling it back therefore proves

Every original viscosity factor is retained in this bound. Its nonlinear map consequently has norm at most \(L_{\nu,\tau}\nu^{-1/2}C_m\sqrt\tau\) times the two \(X_m\) norms. This tends to zero with \(\tau\). The complete successive-approximation argument of lesson 5 therefore constructs an \(X_m\) solution from each \(H^m\) datum. Its uniqueness is the already proved \(H^1\) uniqueness. The same restart argument yields a blowup alternative in \(H^m\).

Now take an existing unforced \(X_1\) solution on a compact interval strictly inside its maximal interval. Almost every positive time has \(H^2\) data. Start the \(X_2\) construction at any such time. It agrees with the \(X_1\) solution on the overlap. Twice differentiating the original equation and pairing with its complete ordered second derivative gives

The differentiated nonlinearity has the three remaining products \((\partial_a u_j)\partial_j\partial_bu_i\), \((\partial_b u_j)\partial_j\partial_au_i\), and \((\partial_a\partial_bu_j)\partial_ju_i\); the original \(u\cdot\nabla\partial_a\partial_bu_i\) cancels by divergence freedom. Tensor Cauchy–Schwarz bounds each remaining contraction by \(|\nabla u|\,|\nabla^2u|^2\), giving exactly the three. The pressure pairing vanishes by the differentiated divergence condition. The original force is zero. Sobolev and Young give the displayed coefficients. Fourier truncation and the same integrable product bounds justify the energy inequality for the \(X_2\) solution. More explicitly, 7.5 gives \(u_t\in L^2H^1\) on a bounded \(X_2\) interval, since \(\Delta u\in L^2H^1\) and the projected nonlinearity belongs to \(L^2H^1\). Thus \(\nabla^2u\in L^2H^1\) has time derivative in \(L^2H^{-1}\). Apply the time-and-Fourier smoothing argument of the linear energy identity, pair these two spaces, and pass to the limit. The cancellation and three products above are legitimate in that pairing: their absolute integral is bounded by the right side of 7.6, integrable on every \(X_2\) interval. The coefficient in the last line is integrable on the entire compact interval, because the existing \(X_1\) solution has bounded \(H^1\) norm and square-integrable \(H^2\) norm. Gronwall therefore prevents an \(H^2\) endpoint there and gives the stated \(L^2H^3\) bound as well.

For \(m\geq3\), suppose the \(X_{m-1}\) conclusion is known. Choose an actual later time with \(H^m\) datum. The \(X_m\) solution again agrees with \(u\). Pairing the equation with its full Bessel weight gives

Indeed the force-free nonlinear pairing is at most \(\|(u\cdot\nabla)u\|_{H^{m-1}}\|u\|_{H^{m+1}}\). Use the sharper 7.5 bound and Young. The exact identity \(\|u\|_{H^{m+1}}^2=\|u\|_{H^m}^2+\|\nabla u\|_{H^m}^2\) explains the retained \(+\nu\) term. Gronwall on every compact interval gives persistence and the next integrated derivative. Induction proves the result for all finite \(m\). The starting times may be chosen in successive subintervals below any prescribed positive time; their lengths can be a geometric sequence with finite sum. Thus each compact interval separated from the original initial time has bounded norms of every spatial order. The original projected equation and its pressure Fourier formula then give successive time derivatives; Sobolev embedding at arbitrarily high orders makes them continuous. This proves actual positive-time smoothness of the original \(X_1\) solution, not an additional regularity assumption in the application below.

8. Continuation of the original H1 solution

Let the unforced original whole-space \(H^1\) solution have a finite maximal time \(T_*\), and suppose \(\sup_{0\leq t<T_*}\|u(t)\|_3\leq U<\infty\). For \(U=0\) it is the zero solution. Otherwise fix an actual \(a\in(0,T_*)\), and set \(L_*=T_*-a>0\). For every \(t_n\uparrow T_*\) with \(t_n-a\geq L_*/2\), positive-time smoothing proves all the smooth-class hypotheses on \([a,t_n]\). Use 6.5 with the original duration \(T_n=t_n-a\). All constants \(E_c\) were constructed from the same fixed upper bound \(U\), original \(\nu\) and kernels, so they are independent of \(n\). Thus

The complete \(H^1\) restart theorem of lesson 5 now constructs solutions for one common positive time length from every \(u(t_n)\). Its construction uses the original \(\nu\) and zero force; its contraction constant tends to zero with that length and its linear trajectory is bounded by \(L_{\nu,1}R_*\). For large \(n\), the restarted interval crosses \(T_*\). Uniqueness glues it to the original solution, contradicting maximality. We have proved

The proof also applies to a bounded \(L^3\) norm on a final subinterval, since \(a\) can be chosen inside that subinterval. It does not assert that the norm tends to infinity at every time. No periodic or forced endpoint theorem is inferred by removing their additional terms.

9. The actual energy weak solution

Let \(u\) be an unforced whole-space energy weak solution on \([0,T]\), with the exact class and all-ending-time energy inequality of lesson 6, and assume

For any \(\varepsilon>0\), choose \(a\in(0,\varepsilon)\) in the common full-measure set of permissible energy starts and times with \(u(a)\in H^1_\sigma\). Such times exist because the energy class has \(u\in L^2(0,T;H^1)\). Start the original \(H^1\) solution \(v\) at this actual datum. The complete weak–strong comparison in lesson 6 gives \(u=v\) at every time of their common interval. The essential \(L^3\) upper bound of \(u\) transfers to \(v\) at almost every time. Its \(H^1\) continuity, interpolation and the original whole-space Sobolev embedding give continuity in \(L^3\), so the same upper bound holds at every time of that interval. 8.2 prevents its maximal endpoint at or before \(T\). The strong solution therefore continues through \(T\), and the comparison remains valid up to that original endpoint. We obtain

For \(H^1\) initial data the conclusion includes the initial strong interval and continuation through \(T\). The pressure is recovered by the same original Leray/Fourier map; on overlaps its gradient agrees with that of the weak equation. No new choice of force, viscosity or initial trace is used. These statements establish the bounded-\(L^3\) endpoint in the unforced whole-space energy class.

Figure: Sections 5–8. The vanishing support endpoints leave three possible ratios between the two largest frequencies. The lower panel shows the actual nested physical time intervals used in the enstrophy bound and restart. The original equation is unchanged. Reproducible figure source.

10. A smaller frequency-event constant

Keep exactly the amplitude \(b\) chosen in 4.4. Evaluate the complete annulus and the stronger covering coefficient at that amplitude, before making any radius choice. Set

All constants are the original ones. In particular \(0<d<A_0<B_0\) and \(m-gs\geq m/3>0\). Lesson 19 (8.18)–(8.19) proves that \(\Phi\) has an attained minimum on this interval. Let \(s_*\) be any minimizer and define

Here is the strict comparison, including its geometric reason. At \(s=s_0=r_c^*\), the original optimized mass is \(L_{s_0}=L_c^*\). The old shell endpoints, before multiplication by \(2\sqrt K\), were \(A_0-d\) and \(B_0+\Lambda^{-2}\overline R_c/4\). The actual bump endpoints are \(A_0-s_0\) and \(B_0+s_0\). The inner one is at least the old one. The outer one is strictly less than the old one, because \(s_0\leq d\) and

Both \(\mathsf B\) and \(\epsilon_a\) are positive and \(\Lambda>1\). Thus \(q_{s_0}<q_{\rm shell}\), and \(\Phi(s_*)\leq\Phi(s_0)<\log(q_{\rm shell})/L_c^*\). Since \(U>0\), exponentiation proves the strict comparison in 10.2. The original disjoint-shell proof applies to each radius and gives its first inequality, including empty and singleton scale intervals. No uniqueness of \(s_*\) is needed, since the minimum value is unique.

11. The complete heat threshold and enstrophy estimate

Retain the original frequency kernels and heat field of 4.1–4.5. Write the full heat threshold separately:

The inequality is strict exactly when \(H_0<2\mathcal Q\). Indeed, if \(H_0\geq2\mathcal Q\), both maxima equal \(H_0\). Otherwise both \(2\mathcal Q_{\rm opt}\) and \(H_0\) are strictly less than \(2\mathcal Q\), which is the old maximum. This checks the actual heat contribution instead of assuming that an improved frequency-event constant must improve every later coefficient.

For \(t-a\geq T/2\), the new event bound excludes \(|P_Nu(t,x)|\geq bN\) at \(N\geq N_{c,{\rm opt}}\). The unchanged full exponential heat bound excludes a heat contribution larger than \(bN\) at the same frequencies. Consequently the complete trilinear calculation 5.1–5.6 gives

No coefficient of the source sixth term has been removed. The original nonlinear energy coefficient is still \(M_c=(16\kappa_0^2+9C_{0,6}^2)U^4\nu^{-5/2}>0\). The actual preceding-quarter selection and full integrating factor therefore give

12. Exact monotonicity and the original velocity

For every \(A>0\),

The identity follows by integrating the exponential; differentiating on this finite interval is justified by its continuous derivative. Thus \(E_{\rm opt}\leq E_c\), strictly precisely when \(\mathcal N_{\rm opt}^2<\mathcal N^2\). The dissipation coefficient is also strictly increasing in \(A\): its derivative is \((2/\nu)[(1+A/4)E'(A)+E(A)/4]>0\). It follows that \(D_{\rm opt}\leq D_c\), with the same exact strictness condition.

Add back the complete original heat field and \(L^2\) energy:

Here \(J_\nabla=\|\nabla H_1\|_1\) and \(L_*=T_*-a\) are exactly 6.5 and 8.1. Both mixed squares are retained. The endpoint restart uses \(R_{*,{\rm opt}}\) in the same proved local existence theorem. Every smoothing and weak-solution comparison remains the original 7.1–9.2 argument. The smaller shell therefore gives a strictly smaller event constant in every nonzero case, and a strictly smaller enstrophy, dissipation and velocity bound exactly when the old threshold was not already set by the original heat field.

The argument retains the precise unforced whole-space hypotheses. It extends no forced or periodic theorem and makes no novelty claim. Human comparison remains Terence Tao, arXiv:1908.04958v2, original author article.tex 1422–1490; the improved radius and envelope arguments are explicitly the course consequences in lesson 19, Exercises 4–5.

Figure: 10.1–12.2. The upper panel displays both original envelopes with their common physical length factor shown. Spacing is schematic; the two inner endpoints may coincide. The lower panel gives both exact heat-threshold cases. The original velocity, equation and time coordinates remain unchanged. Reproducible figure source.

13. Five exercises with complete solutions

Exercise 1: optimize the heat cross-term weight

In Section 3, replace the Young weight \(1/s\) by \(\lambda/s\), where \(\lambda>0\). Keep the original amplitude, improved frequency cutoff and all heat constants. Prove that the last-quarter energy coefficient has a unique best value of \(\lambda\).

Solution. The complete nonnegative square is

Use the same actual \(A=(C_{2,6}+C_{1,3}C_{1,6})U^2(\nu s)^{-5/4}\). Both original heat terms and both mixed-gradient terms remain. On the later half interval put

These \(A_0,B_0\) name the coefficients in this exercise; they are distinct from the radial endpoints used in Sections 10–12. The earlier integrated energy remains \(M_c\sqrt T/2\). Thus on the final quarter the complete coefficient is

The first term tends to infinity as \(\lambda\to\infty\). As \(\lambda\downarrow0\), the integral is at least \(1/4\), so the second term tends to infinity. Continuity gives an attained minimum on a compact subinterval of \((0,\infty)\).

To prove uniqueness, differentiate every retained term:

The integrands and their derivatives are continuous on each compact positive \(\lambda\) interval, which justifies these differentiations. Strict convexity makes the minimizer unique. It is the unique zero of the complete first line of (13.4). Taking \(\lambda=1\) recovers every coefficient in Section 11. The optimized \(F\) is therefore no larger than its proved energy coefficient there. No heat summand is discarded in choosing the weight.

Exercise 2: optimize the actual preceding time window

Suppose \(A,B,M_c>0\) and the actual estimates on \([a+T/2,a+T]\) are

For a fixed \(\rho\in(1/2,1)\), find the best uniform bound on \([a+\rho T,a+T]\) obtainable by choosing a preceding window of length \(\sigma T\).

Solution. Every \(0<\sigma\leq\rho-1/2\) keeps the actual interval \([t-\sigma T,t]\) in the known domain. Its integral supplies an actual time \(r\) there with \(E(r)\leq M_c/(2\sigma\sqrt T)\). The original integrating factor gives

The quadratic in parentheses is strictly increasing for \(\sigma>0\), starts negative and tends to infinity. Its unique positive root, retaining all coefficients, is

Both displayed forms follow by multiplication by the full conjugate; no coefficient is absorbed into a changed parameter. The derivative changes from negative to positive, proving the exact optimum on the permitted original time windows.

Integration of (13.5), retaining its final nonnegative energy until the last inequality, gives

The original heat field is bounded at the actual time gap \(\rho T\). The full velocity consequently satisfies

The original choice \(\rho=3/4,\sigma=1/4\) recovers Section 6. Its coefficient improves strictly when the free root is below \(1/4\); otherwise that original window is already optimal within this family.

Exercise 3: give a common restart length explicitly

Let the original unforced whole-space \(H^1\) data have norm at most \(R>0\), with viscosity \(\nu>0\). Give a positive local existence time common to all such data, and use it in the actual endpoint contradiction.

Solution. Use exactly the time norm and linear constant of lesson 5. Put

The original free heat trajectory has norm at most \(d_R=L_{\nu,1}R\) on this interval, since \(L_{\nu,\tau_R}\leq L_{\nu,1}\). The exact ordered bilinear estimate of lesson 5 gives

The fixed-point map on the actual ball of radius \(2d_R\) therefore maps that ball into itself and has contraction constant at most \(1/2\). Its convergent iteration, original equation recovery and uniqueness are the complete construction of lesson 5. There is no change of viscosity, data or force. If \(R=0\), the data are zero and the unforced zero solution exists for every positive time.

Apply (13.10) to \(R=R_{*,{\rm opt}}\) from Section 12. Every selected endpoint datum \(u(t_n)\) has at least this same existence time. For \(t_n>T_*-\tau_R\), its solution crosses \(T_*\); uniqueness on the actual overlap gives the contradiction. This supplies a numerical formula for the previously constructed common length, with all its original parameter dependence.

Exercise 4: use both actual projection bands

The bound in Section 5 used the support inequality \(N\leq4|\xi|\) separately at every frequency. Retain the actual smooth projection and compute its two possibly nonzero weights at each nonzero Fourier point. Prove a stronger dissipation estimate without changing the amplitude or any preceding event constant.

Solution. Let \(r=|\xi|>0\) and choose the unique grid frequency \(N\) with \(2r\leq N<4r\). Only \(N\) and \(N/2\) can have nonzero projection weight: a nonzero weight requires \(r<N'<4r\), and the boundary weights vanish. Put

Here \(\tau\) is a Fourier-radius variable in this exercise, distinct from the local existence time in Exercise 3. Since \(\phi(\xi/N)=1\) and \(\phi(4\xi/N)=0\), the exact two weights are

This includes the grid-boundary case \(\tau=1\), where \(h=0\). The cutoff is the original one of Section 4; in particular it is nonincreasing and \(\chi(5/8)=1/2\), since the two exponential arguments \(1-5/8\) and \(5/8-1/4\) both equal \(3/8\).

For \(1/4<\tau\leq5/8\), we have \(1/2\leq h\leq1\). The convex quadratic \(4(1-h)^2+h^2\) is at most the larger endpoint value \(5/4\). Its quotient by \(\tau\) is therefore at most \(5\). For \(5/8\leq\tau\leq1\), we have \(0\leq h\leq1/2\); the same endpoint comparison bounds the numerator by \(4\) and its quotient by \(32/5\). Hence, retaining the full Fourier gradient multiplier,

We assert this proved bound, not sharpness of its constant. The complete trilinear sum of Section 5 now gives

The second line uses the unchanged original \(b\leq\pi^2\nu/(224\kappa_\partial)\). Thus every event constant, heat cutoff and energy coefficient in Sections 10–12 remains valid. The complete integrated dissipation coefficient becomes

All original projection weights are retained throughout; their exact pairing improves the estimate.

Figure: Exercise 4, equations (13.12)–(13.16). The first panel keeps the exact original smooth cutoff; the second retains both weights in their complete squared-frequency sum. The sampled curve illustrates the formula. The two-piece analytical estimate proves its bound, including every Fourier gradient factor. Reproducible figure source.

Exercise 5: combine the improved weights and time windows

Use the optimized spatial radius of Section 10, the Young weight of Exercise 1, the actual window of Exercise 2, and the full two-band dissipation of Exercise 4. For fixed \(\rho\in(1/2,1)\), prove that the joint energy coefficient has a unique attained minimum. Give its receiving dissipation and original velocity bounds.

Solution. Keep the complete \(A_0,B_0,M_c\) of (13.2) and the original nonlinear energy estimate. For \(\lambda>0\) and \(0<\sigma\leq\sigma_{\max}:=\rho-1/2\), define the full expression

The preceding proofs use the same original fields, so they give \(E(t)\leq\mathcal F(\lambda,\sigma)/\sqrt T\) on the entire interval \([a+\rho T,a+T]\).

Here is a complete attainment argument. Set \(C_*=\mathcal F(1,\sigma_{\max})>0\). On the nonempty sublevel set \(\mathcal F\leq C_*\), the first positive summand gives \(\sigma\geq\sigma_0:=M_c/(2C_*)>0\). The second summand is at least \(B_0\sigma/\lambda\), so \(\lambda\geq B_0\sigma_0/C_*>0\). The first summand is also at least \([M_c/(2\sigma_{\max})]e^{2\lambda\sigma_0}\), whence

The logarithm is positive because the sublevel contains the point \((1,\sigma_{\max})\). These are closed finite bounds separated from both zero endpoints. Continuity makes the sublevel compact. The minimum is therefore attained at some \((\lambda_*,\sigma_*)\), with value \(F_*=\mathcal F(\lambda_*,\sigma_*)>0\).

For each fixed \(\lambda\), (13.7) gives the unique best \(\sigma\), with \(A=A_\lambda,B=B_\lambda\). For fixed \(\sigma\), the second \(\lambda\)-derivative is

Thus its best \(\lambda\) is the unique root

In fact the joint minimizer is unique. Here is the complete comparison that proves this stronger conclusion. The map

is a bijection with inverse \((\log\lambda,\log\sigma)\). It changes only the coordinates of the optimization parameters; the original equation, fields, time windows and full objective (13.17) remain fixed. In its original heat integral put \(r=e^yq\), retaining \(dr=e^y\,dq\) and both endpoints. The exact pullback is

Every original coefficient and Jacobian appears in this identity. For a direction \(v=(\alpha,\beta)\), differentiation gives

Since \(A_0>0\), the first line is positive for every nonzero \(v\): vanishing would force first \(\beta=0\) and then \(\alpha=0\). The second line is nonnegative for every \(q\in[0,1]\). The full Hessian is consequently

On every compact parameter set the integrand and its first two derivatives are continuous and bounded on the entire integration interval, which justifies the differentiation. The domain in (13.20a) is convex. Restricting \(G\) to the segment between any two distinct points gives a function with positive second derivative. Integrating that derivative proves strict convexity, so two different minimum points cannot exist. The already proved attainment and the bijection \(\Psi\) therefore give exactly one original pair \((\lambda_*,\sigma_*)\).

The two one-variable conditions are now also sufficient. Equation (13.20) gives \(\partial_xG=\lambda\partial_\lambda \mathcal F=0\). At the best window in (13.7), \(\partial_yG=\sigma\partial_\sigma\mathcal F\) is zero in the interior and is nonpositive at the upper boundary. Thus its inner product with every feasible displacement is nonnegative. Integrating the nonnegative second derivative along that displacement proves the convex supporting inequality and hence global minimality. Strict convexity makes this the unique pair. This proves the characterization without replacing joint minimization by two unproved separate choices.

Put \(A_*=A_{\lambda_*}\), \(B_*=B_{\lambda_*}\). With the complete remaining dissipation \(4\nu/5\), integration on the actual final interval gives

Both original heat and energy contributions remain in the square. At \(\rho=3/4\), the admissible pair \((1,1/4)\) recovers the energy coefficient of Section 11, so \(F_*\leq E_{\rm opt}\). The displayed dissipation estimate is the bound at this energy-minimizing pair; no separate minimum of that coefficient is assumed.

Finally the endpoint sequence of Section 8 has \(T_n\geq L_*/2\), so the actual common bound is

Use this exact number in Exercise 3 and in the unchanged weak–strong comparison. This propagates all the proved improvements through the endpoint conclusion while preserving the original equation, solution, pressure and viscosity.

14. Scope and source comparison

The complete endpoint theorem applies to the unforced whole-space energy classes stated in Sections 8–9. The original source sixth term remains visible, and the higher-order smoothing uses the actual earlier time norm with its complete isometry and inverse. The two exact frequency counts, smaller envelopes and optimized weights are proved additions to the receiving calculation.

A finite maximal strong endpoint therefore requires an unbounded critical velocity supremum on every final interval. The theorem here does not claim a pointwise limit of that norm along every approaching time. The periodic endpoint and the later forced constructions retain their own domains, forces and required arguments. The remaining series develops those assigned constructions and compares the human, OpenAI and workbench proofs in their actual scope.