Original clocks, weighted transport and exact corrections

Chen and Hou's stability construction changes both time and spatial scale, controls outgoing transport with weights, and corrects an approximate evolution so that its residual vanishes to the required order at the origin. This lesson proves those maps with every original factor, coupling, endpoint and correction term retained.

Part 1 calculates the physical clock and its exact remaining-time integral. Part 2 constructs a complete annular damping weight and keeps its cost in the nonlocal operator. Part 3 gives the full scalar and three-component analytic corrections, their time tails, and the return from independent components to a temperature field. Part 4 solves the original frozen transport model, including every mixed derivative and both homogeneous and inhomogeneous Sobolev norms.

Five solved exercises calculate a clock with a logarithmic correction, an explicit annular damping exponent, a nonzero analytic correction tail, every derivative norm of a quadratic Gaussian, and all three forced correction coefficients with their actual gradient defect.

Each part identifies the original author-source version and precise passages. Corrections to the explanatory note and frozen model remain visible. These comparisons do not establish a failure of the main blowup theorem or certify its full computational stability argument.

Part 1. Exact physical time and every original field scale

This calculation receives the original dynamic scales in Jiajie Chen and Thomas Y. Hou, arXiv:2210.07191v4, original TeX lines710–735 and815–850. The stated profile and rate errors are at lines799–809. The source explicitly calls the later rate formulas informal and does not assert convergence to an exact profile. Its displayed decimal scaling parameters at1495 are rounded, as stated at1478–1482; none is used here as an exact enclosure.

The arguments below prove the clock and field maps on a fully specified class of rate functions. They do not supply the source's still-unread nonlinear stability estimates or validated computational constants. The original half-plane, time origin and initial scale are retained.

QC1. The actual clock and its complete inverse

Write the original spatial point as \(\boldsymbol x=(x,y)\) in \(\mathbb H=\mathbb R\times[0,\infty)\), and keep \(\tau\) for rescaled time and \(t\) for physical time. Fix the source constants \(\bar c_l>0\), \(\bar c_\omega<0\), the actual initial vorticity scale \(A=C_\omega(0)>0\), and \(\delta\) with \(0\leq\delta<-\bar c_\omega\). Define both positive rate endpoints without replacing the original rate:

Take any continuous original rate \(c_\omega\) satisfying QC1. The full source scales and time are

Here \(C_l(0)=1\) is the original source value. The initial factor \(A\), and hence \(A^2\), are precisely the source's extension at825. Integrating QC1 over \([0,\tau]\), exponentiating and retaining \(A\) gives

The two exponential integrals are finite because both endpoints are strictly positive. The time derivative is exactly \(t'(\tau)=C_\omega(\tau)>0\). Thus \(t\) is a strictly increasing \(C^1\) bijection from \([0,\infty)\) onto \([0,T)\): continuity gives every intermediate value, strict increase gives uniqueness, and the defining improper integral gives the limit \(T\). Its inverse \(\tau(t)\) is \(C^1\) at every \(t<T\), with derivative \(1/C_\omega(\tau(t))\). In particular the complete space-time map and its inverse are

The determinant follows from the block triangular derivative: its spatial block is \(C_l I_2\), its lower-left block is zero and its last diagonal entry is \(C_\omega\). The moving spatial column \(C_l'(\tau)\boldsymbol y\) is retained and does not alter that determinant. Positive dilation preserves both the half-plane and its boundary. No point at \(t=T\) is in the inverse map's domain.

QC2. The exact remaining-time factor

For every \(\tau\geq0\), define

The integrand lies between \(e^{-a_+u}\) and \(e^{-a_-u}\). Changing variables \(s=\tau+u\) in the tail of the original time integral, and using \(C_\omega(\tau+u)=C_\omega(\tau) \exp(\int_\tau^{\tau+u}c_\omega)\), proves

The last identity follows by differentiating the first identity, whose two factors are \(C^1\); it requires no differentiation under an improper integral. Every constant in QC6 is attained by the continuous constant endpoint rates \(c_\omega=-a_+\) and \(c_\omega=-a_-\), respectively. The original scale therefore obeys the sharp two-sided remaining-time estimate

The upper and lower sides retain different constants. Replacing both by \(-\bar c_\omega\) is not an exact identity for the rate class QC1.

QC3. Spatial scales and the entire accumulated rate error

Put \(\Delta_t=T-t(\tau)>0\). Integrating the exponential envelopes from \(\tau\) to infinity yields

Since \(C_l=e^{-\bar c_l\tau}\), all quantities are positive and raising each side to the indicated positive power gives

and hence

The cap at one uses \(\tau\geq0\). The lower base in QC9 is at most one because \(\Delta_t\leq T\leq A/a_-\); thus these endpoint bounds are consistent at \(\tau=0\) as well.

There is also a full identity retaining the deviation of the original rate rather than only its bound. Set

The logarithm of QC2 is \(\log(C_\omega/A)=\bar c_\omega\tau+B(\tau)\). Multiplying this equality by \(\gamma\), and retaining the entire \(B\) term, gives

For \(\delta=0\), the only rate is \(c_\omega=\bar c_\omega\). Then \(B=0\), \(q=1/(-\bar c_\omega)\), and every displayed clock formula specializes to its exact constant-rate value.

For \(\delta>0\), the two permitted constant rates \(c_\omega=\bar c_\omega+\delta=-a_-\) and \(c_\omega=\bar c_\omega-\delta=-a_+\) give, respectively,

Since \(a_-<-\bar c_\omega<a_+\), dividing the first expression by \(\Delta_t^\gamma\) tends to zero as \(\Delta_t\downarrow0\), and dividing the second by it tends to infinity. This proves exactly why a bounded rate error alone does not give a uniform multiplicative comparison with the central power. These examples are clocks in QC1, not asserted PDE solutions or counterexamples to the source theorem.

Even the factor \(q\) need not converge in this rate class. For the specified rate \(c_\omega(\tau)=\bar c_\omega+\delta\sin\tau\),

The integrand makes \(q\) continuous and \(2\pi\)-periodic by dominated convergence, with dominating function \(e^{-a_-u}\). If \(q\) were constant, QC6 would force \(c_\omega=-1/q\) to be constant, contrary to \(\delta>0\). A nonconstant continuous periodic function cannot have a limit at infinity: two different values repeat along two sequences of times tending to infinity. This proves the asserted nonconvergence directly.

QC4. Every field, derivative and measure factor

Keep the original source fields \(\widetilde\omega,\widetilde\theta,\widetilde{\boldsymbol u}\) in the rescaled half-plane. The identity \(c_\theta=\bar c_l+2c_\omega\), together with the actual initial factors in QC2, gives

At \(\boldsymbol y=C_l(\tau)^{-1}\boldsymbol x\), the full inverse source transformation is therefore

For any original multi-index \(\alpha=(\alpha_1,\alpha_2)\) of order \(k=\alpha_1+\alpha_2\), spatial differentiation at fixed time multiplies each field by \(C_l^{-k}\). Thus the exact derivative formulas retain factors \(C_\omega^{-1}C_l^{-k}\), \(C_\omega^{-2}C_l^{1-k}\) and \(C_\omega^{-1}C_l^{1-k}\), in the order of QC16. In particular the original temperature gradient has the full factor \(C_\omega^{-2}\); its spatial scale cancels only after both original factors are displayed.

For \(1\leq p<\infty\), the substitution \(\boldsymbol x=C_l\boldsymbol y\) has Jacobian \(C_l^2\). Taking the \(p\)-th root of the resulting integral proves

All time-dependent factors on the right are evaluated at \(\tau\). The identities also hold with infinite values when the fields are not integrable. For \(p=\infty\), the bijective spatial dilation preserves the essential supremum, so the three factors are respectively \(C_\omega^{-1}C_l^{-k}\), \(C_\omega^{-2}C_l^{1-k}\), and \(C_\omega^{-1}C_l^{1-k}\). No finite-energy conclusion is inferred from bounded vorticity alone.

QC5. Exact receiving estimate for the source's stated error sizes

Let \(M=\|\bar\omega\|_{L^\infty(\mathbb H)}<\infty\) and suppose a rescaled field has the actual error \(\|\widetilde\omega(\tau)-\bar\omega\|_\infty\leq\epsilon_\omega\) at every time. The triangle inequality and its reverse give \((M-\epsilon_\omega)_+\leq\|\widetilde\omega(\tau)\|_\infty \leq M+\epsilon_\omega\), where \((r)_+=\max\{r,0\}\). Multiplying QC16 by \(\Delta_t\), using QC6 and retaining both endpoints, gives

This is a direct algebraic receiving estimate. When \(M>\epsilon_\omega\), its left coefficient is positive and it gives the exact lower rate for the supplied field. Without that strict inequality it does not claim a positive lower bound.

At time zero \(\widetilde\omega(0)=A\omega_{\rm phy}(0)\). Writing \(M_0=\|\widetilde\omega(0)\|_\infty\), any nonzero initial field satisfies \(A=M_0/\|\omega_{\rm phy}(0)\|_\infty\). Substitution into QC3 proves the more precise initial-data form

There is no division by a zero initial norm: QC19 is stated only for a nonzero initial field, while QC3 holds in every case in QC1–QC2.

The source's stated \(E_*=5\cdot10^{-6}\) gives exactly

Thus QC18–QC19 supply the exact comparison receiving those stated errors, with the original \(\bar c_\omega\) still present. Checking the actual profile enclosure and the nonlinear estimates that produce these errors remains part of the continuing source audit. The proof above has not replaced either calculation by the rounded value at1495.

For completeness, for each component \(j=1,2\) put \(M_{\theta,j}=\|\partial_j\bar\theta\|_\infty\) and use an actual bound \(\|\partial_j\widetilde\theta- \partial_j\bar\theta\|_\infty\leq\epsilon_{\theta,j}\). The gradient factor in QC16 and the square of QC6 prove

The source asserts the same \(1/1000\) error for each of its two gradient components. QC21 keeps them componentwise and does not silently replace two component bounds by a Euclidean vector bound.

The figure uses the exact specified clock \(A=5\), \(\bar c_\omega=-2\), \(\bar c_l=3\) and \(c_\omega=-2+\tfrac14\sin\tau\), with \(a_-=7/4\), \(a_+=9/4\). The integral curve is sampled numerically; the two proved bounds are QC6 and QC9. These are clock examples, not the source's numerical profile. The complete reproducible figure source and numerical samples retain the original factors. Human source and exact proof locators are given above; no interval certificate is claimed for plotted samples.

The next calculation is the complete original weighted transport system and the analytic correction of its near-origin jets. QC1–QC21 give exact clock consequences, not the missing stability proof.

Part 2. The full original weighted transport system and an explicit annular weight

Human source: Jiajie Chen and Thomas Y. Hou, arXiv:2210.07191v4, original TeX1559–1632 and1690–1799. This calculation retains the complete three-component equation; it does not delete the nonlocal terms, profile residuals or nonlinear terms. It constructs the bounded annular factor described at1744–1758 with explicit constants. The full nonlinear and computational stability estimates remain to be read and checked.

WT1. All components of the original equation

Use the source's perturbation notation, in which \(\omega,\theta\) are perturbations of \(\bar\omega,\bar\theta\), and \(c_\omega=u_x(0)\) is the perturbation of the rate. Retain the source's fixed derivative anchors, so the perturbation of \(c_l\) is zero and that of \(c_\theta\) is \(2u_x(0)\). On the original half-plane let

The stream function and boundary conditions are those of the source; the velocity here is its exact Biot–Savart velocity, not the separately represented numerical stream function. The full equation is

with every matrix entry retained:

The nonlocal part is the following full vector, not a numerical bound:

The full nonlinear and original residual vectors are

To verify WT2, insert the full barred-plus-perturbation fields into the source equations. The vorticity equation has transport \(-b\cdot\nabla\omega\), the local terms \(\bar c_\omega\omega+\eta\), the first entry of WT4, the first entry of WT5 and the barred residual. These exhaust its products. For the temperature equation differentiate once in \(x\): the derivative of transport contributes \(-\bar c_l\eta-\bar u_x\eta-\bar v_x\xi\), and \(\bar c_\theta\eta-\bar c_l\eta=2\bar c_\omega\eta\). Differentiating its perturbation transport and rate products gives exactly the second entries of WT4–WT5. Differentiate instead in \(y\): the corresponding local entries are \(-\bar u_y\eta+(\bar c_\theta-\bar c_l-\bar v_y)\xi\). The exact divergence identity \(\bar v_y=-\bar u_x\) gives the third row of WT3; all other differentiated products are the third entries of WT4–WT5. Thus no term from the source's two spatial directions is discarded.

WT2. The complete componentwise weight map

On any region excluding the origin let \(\varphi_i>0\) be \(C^1\) time-independent weights, and put \(D_\varphi=\operatorname{diag}(\varphi_1,\varphi_2,\varphi_3)\), \(Z=D_\varphi W\). Multiplication is invertible there. The product rule proves the exact conjugated equation

In particular the three off-diagonal coefficients are exactly \(\varphi_1/\varphi_2\), \(-\bar v_x\varphi_2/\varphi_3\) and \(-\bar u_y\varphi_3/\varphi_2\). They cannot be ignored when choosing different weights for the components.

WT3. A constructed weight with the source's exact plateaus

Retain \(0<R_1<R_2\) and work in the original positive quadrant \(x,y\geq0\), away from its origin. Let

and, for \(r=|\boldsymbol x|\), set

The oriented integral in WT8 is zero for \(r\leq a\), since its integrand vanishes there. It equals \(Q\) for \(r\geq d\). The positive integrand on \((a,d)\) gives \(0<Q<d-a\), and consequently

All derivatives of the exponential in WT7 tend to zero at either endpoint: each derivative is that exponential times a finite rational function of \(r-a\) and \(d-r\); an exponential of a negative reciprocal dominates every such power. Thus \(h\), its integral and \(\varphi_0\) are smooth. The last function is constant in a full ball about the origin, so the radial definition is smooth there too. For \(R_1\leq r\leq R_2\) and \(x_i>0\), WT9 has \(\partial_i\varphi_0<0\). At the corresponding coordinate axis that derivative is zero, which is the exact boundary behavior.

Retain actual positive outgoing constants \(c_1,c_2\) in \(b_1(\boldsymbol x)\geq c_1x\) and \(b_2(\boldsymbol x)\geq c_2y\), and set \(c_*=\min(c_1,c_2)>0\). The constants are not replaced by the source's rounded illustrative decimals. Then

This retains the two individual coordinate contributions before taking their lower bound. On the closed annulus \(R_1\leq r\leq R_2\), the concave quadratic \((r-a)(d-r)\) attains its minimum at an endpoint. The second endpoint value minus the first is \((R_2-R_1)(R_2-R_1/2)>0\). Therefore

The complete integral \(Q\) remains in the stronger coefficient; the following explicit lower bound uses only \(Q<d-a\). The inequality is valid at the axes as well, because WT10 uses the radial dot product rather than strict negativity of both partials.

WT4. Strong local damping without deleting the remaining equation

For fixed \(\alpha_i<0\) and real \(\beta_i\), retain the original two-term radial factors

Differentiation gives \(d_i=d_{s,i}+m d_0\). The common bounded factor cancels in every same-point ratio \(\varphi_i/\varphi_j=\psi_i/\psi_j\), so the off-diagonal matrix does not acquire an unrecorded dependence on \(m\). The original weighted spaces are unchanged as sets, with the explicit norm comparison

This follows directly by multiplying \(2^{-m}\leq\varphi_0^m\leq1\). In particular the factor is identically one near the origin. It does not repair a missing vanishing jet required by a singular \(\psi_i\).

On the closed quadrant annulus define the exact three row quantities

When the barred velocity is \(C^1\) on this annulus, all functions are continuous and the denominators are positive. Compactness makes \(K\) finite. For any prescribed local damping \(\kappa>0\), the fully specified choice

makes every diagonal-plus-absolute-off-diagonal row sum of the exact matrix in WT6 at most \(-\kappa\) throughout the annulus. Indeed, that row sum is \(\rho_i+m d_0\leq K-mC_*\); WT15 gives \(-\kappa\) when \(K+\kappa\geq0\), and when it is negative, \(m=0\) and \(K<-\kappa\). This proves the full local matrix statement with unequal base weights and without discarding the \(\eta\)-to-\(\omega\) or either cross-gradient coupling.

Here is its precise receiving estimate. Let \(X(t)\) be an actual characteristic, \(X'=b(X)\), which stays in the annulus on \([t_0,t_1]\). For a differentiable solution of the complete WT2, put \(z(t)=Z(X(t),t)\). Its equation is exactly WT6 evaluated on this path. The upper right derivative of \(n(t)=\max_i|z_i(t)|\) satisfies

To prove this finite-dimensional statement, at a nonzero maximizing component multiply its differential equation by its sign. The diagonal term is its diagonal coefficient times \(n\), and each off-diagonal term is at most its absolute coefficient times \(n\). At a maximizing zero all components vanish, so the derivative of its absolute value is bounded by the forcing norm. For a maximum of finitely many differentiable absolute values, its upper right derivative is at most the maximum of these derivatives among the maximizing components: any nonmaximizing component has a strict value gap and cannot become a maximizer over sufficiently short times. WT15 now gives WT16. Multiplication by \(e^{\kappa t}\) and integration yields

Equivalently, the integrating-factor comparison follows by integrating the almost-everywhere differential inequality for the locally Lipschitz function \(n\). Every original forcing term from WT4–WT5 remains inside this estimate. The interval restriction matters: WT17 does not assert that a characteristic remains in the annulus forever.

The same-point cancellation in WT12 does not extend to nonlocal kernels. For an actual integral term \((KW)_i(\boldsymbol x)=\sum_j\int K_{ij}(\boldsymbol x,\boldsymbol y) W_j(\boldsymbol y)\,d\boldsymbol y\), whenever that integral exists, the exact conjugated kernel is

Both extremal ratios occur between the two exact plateaus of WT9. The kernel identity follows by substituting \(W=D_\varphi^{-1}Z\) before integration. For a principal value, the identical substitution holds at each original cutoff before taking its limit; no existence of that limit is inferred from WT18. This is the exact additional weight factor that the remaining nonlocal estimate must handle.

The specified illustration uses \(R_1=1\), \(R_2=3\), the outgoing field \(b=(x/2,3y)\), and the original ray \(x=y\geq0\). On that ray \((b\cdot\boldsymbol x)/r=(7/4)r\), so the plotted damping is exactly WT10 with that substitution. The shaded region is the actual annulus, and the dashed lower bound is WT11 with the full \(Q=\int_{1/2}^{6}h(s)\,ds\). Numerical quadrature samples the explicit functions; the inequalities are proved above, not inferred from the plot. The reproducible source and samples are retained. This field is a specified illustration of the outgoing class, not the source approximate Euler/Boussinesq profile.

WT1–WT18 prove the complete local transport map and a constructive annular damping mechanism. The next source work is to prove the required weighted nonlocal bounds and the analytic corrections of near-origin jets; stronger local damping alone has not supplied them.

Part 3. Exact local corrections, the complete residual and its original receiving map

Human sources: Jiajie Chen and Thomas Y. Hou, the analytic correction note, arXiv:2607.15256v1, original TeX1128–1525 and1545–1921; Part I, arXiv:2210.07191v4, original TeX2230–2340 and15795–15860; and Part II, arXiv:2305.05660v3, original TeX927–985,1035–1090,1106–1360,6163–6181,6314–6317 and6474–6498. These exact versions matter: the note's displayed error estimate omits a time-derivative term which Part I v4 explicitly retains. Part II retains it through the correction ODE before estimating the complete residual. The calculations below prove that comparison; they do not certify all the remaining stability and numerical estimates.

JC1. The original quadratic defect and its exact projection

Work on \(\mathbb H=\mathbb R\times[0,\infty)\) with original coordinates \(\boldsymbol x=(x,y)\), \(r=\sqrt{x^2+y^2}\). Derivatives at the boundary origin are derivatives of functions smooth up to that boundary. Retain the source operator

The functional is used on its actual domain of convergence. In the source model it is bounded on the receiving space. The local coefficients obey

The last expansion follows from the source's oddness in \(x\), its quadratic vanishing, and Taylor's formula. Keep all coefficients and assume the differentiability needed in each following formula; \(C^{3,1}\) represented spatial functions suffice for the residual estimates, as the note itself observes at1518–1523.

Let \(\mathcal Y\) consist of \(C^{2,1}\) functions whose derivatives of order less than two vanish at zero and whose only possibly nonzero second derivative there is the mixed derivative. Set

Every \(f\in\mathcal Y\) has \(f=Jf\,xy+O(r^3)\), and every \(f\in\mathcal X\) is \(O(r^3)\). Choose an actual compactly supported \(\chi\in\mathcal Y\) with \(J\chi=1\). Then

These identities follow by applying \(J\chi=1\), not by discarding a small coefficient. The exact maps \(f\mapsto(\Pi f,Jf)\) and \((h,z)\mapsto h+z\chi\) are inverse linear maps between \(\mathcal Y\) and \(\mathcal X\oplus\mathbb R\). Thus the quadratic obstruction is the actual one-dimensional quotient \(\mathcal Y/\mathcal X\), with its supplied section and projection.

For \(f\in\mathcal Y\) of sufficient regularity, Leibniz's rule gives

For example, \(\partial_{xy}(\overline V_1 f_x)(0)\) has the four terms \(\overline V_{1,xy}f_x\), \(\overline V_{1,x}f_{xy}\), \(\overline V_{1,y}f_{xx}\), and \(\overline V_1f_{xxy}\). Only the second survives, giving \(c_1Jf\). The four terms from \(\overline V_2f_y\) similarly give \(c_2Jf\). The four terms from \(\bar c f\) give \(\bar c(0)Jf\), and the functional is constant in space, so the final term gives \(\alpha P(f)\). The same product rules show that \(\mathcal L f\) has zero value, zero gradient and zero pure second derivatives at the origin. These claims use the diagonal first derivative of \(\overline V\) in JC2.

The full operator \(\mathcal L_0=\mathcal L-\alpha\chi P\) consequently satisfies \(J\mathcal L_0f=\lambda Jf\). For an adequately regular solution of \(\partial_tW_1=\mathcal L_0W_1-R\) with all derivatives through order two of \(R\) zero at the origin, the lower derivatives evolve by a homogeneous finite linear system; zero initial derivatives remain zero by uniqueness for that system. After those derivatives vanish, the mixed derivative obeys \((JW_1)'=\lambda JW_1\). Thus initial cubic vanishing is preserved. This is an exact statement about the regular evolution; existence and global stability of the source PDE are not inferred from this jet identity.

JC2. A constructed correction with its full stream function

Take an even \(C^\infty\) compactly supported cutoff \(\kappa\), equal to one in a neighborhood of zero. Write \(K(x,y)=\kappa(x)\kappa(y)\) and define

The complete expression, including every transition term, is

Near zero \(\chi=xy\) exactly. It is odd in \(x\), compactly supported and smooth up to the original boundary. Its original velocity, with \(\nabla^\perp=(-\partial_y,\partial_x)\), is

This identification with the original Dirichlet inverse can be proved directly. For two points \(\boldsymbol x,\boldsymbol z\) in the open half-plane, its Green function is \((2\pi)^{-1}\log(|\boldsymbol x-\boldsymbol z^*|/ |\boldsymbol x-\boldsymbol z|)\), where \(\boldsymbol z^*=(z_1,-z_2)\). The identity \(-\Delta_{\boldsymbol z}[-(2\pi)^{-1}\log|\boldsymbol x-\boldsymbol z|] =\delta_{\boldsymbol x}\) follows from its radial derivative \(-1/(2\pi\rho)\) on a circle of radius \(\rho\) about \(\boldsymbol x\). Its outward flux is \(-1\), so the negative Laplacian has mass one. On the inner boundary of the punctured half-plane the outward normal points toward the center; that reversal gives the corresponding positive term in Green's identity. The reflected term is harmonic inside the half-plane and cancels the boundary value. Integrate Green's identity against the compactly supported \(\Phi\): the outer boundary contributes zero, and on \(y=0\) both \(\Phi\) and the Green function vanish. Removing a small circle and taking its radius to zero gives \((-\Delta)^{-1}(-\Delta\Phi)=\Phi\). This proves JC8 for the original inverse, with its boundary condition fixed.

In particular, for the source functional \(P(f)=u_x(f)(0)=-\partial_{xy}(-\Delta)^{-1}f(0)\), JC8 gives

The functional has an explicit bounded domain adequate for the example below. Let \(f\) be a smooth function supported in the closed half-ball of radius \(R\), whose extension to the whole plane is odd in each coordinate and smooth, and set \(\|f\|_{2,R}=\sup_{0<r\leq R}|f|/r^2\). Then

To justify the derivative, extend \(f\) oddly across the boundary. The whole-plane fundamental solution is \(N(\boldsymbol z)=-(2\pi)^{-1}\log|\boldsymbol z|\). In \(\partial_{xy}(N*f)(0)=\int N\partial_{xy}f\), integrate by parts twice outside a circle of radius \(\rho\). The omitted interior integral is \(O(\rho^2|\log\rho|)\). The two boundary terms are bounded by constants times \(\rho^2|\log\rho|\) and \(\rho^2\), since \(f=O(r^2)\), \(\nabla f=O(r)\), \(N=O(|\log\rho|)\) and \(\nabla N=O(\rho^{-1})\). All tend to zero. The resulting kernel is \(\partial_{xy}N=z_1z_2/(\pi|\boldsymbol z|^4)\), which is absolutely integrable against \(f\) at zero. Odd reflection gives the Dirichlet potential on the half-plane; oddness in both coordinates makes the whole-plane integral four times the quadrant integral. The original velocity adds its minus sign. Finally, polar coordinates give \(\int_0^{\pi/2}\cos\vartheta\sin\vartheta\,d\vartheta=1/2\) and \(\int_0^Rr\,dr=R^2/2\), proving the bound in JC9a. Thus the model example uses the actual bounded Biot–Savart functional, including its sign and domain, rather than an unspecified scalar map.

The correction note's footnote at1413 instead prints \(-\Delta(+xy^3K/6)\), which equals \(-xy\) near zero and has mixed derivative \(-1\). It cannot satisfy the subsequent required mixed derivative \(+1\). The sign in JC6 is the one explicitly present in Part I v4 at2260 and Part II v3 at1150. The note's later elliptic decomposition at1885–1891 also uses this correct sign. The note at1921 also prints the opposite sign in its formula for \(u_x(\varepsilon)(0)\). The original convention \(u=-\phi_y\) gives the minus sign in JC9a; Part I v4 at9069 and Part II v3 at6173–6174 retain it. This local sign discrepancy does not change the zero value for the particular correction JC6.

JC3. Both analytic corrections and the full evolution residual

Let \(g^{(0)}(t,x,y)\) be the actual globally represented function, odd in \(x\), with sufficient spatial regularity and locally Lipschitz, piecewise smooth time dependence. Let \(\chi_1=x+O(r^3)\) be a fixed smooth compactly supported odd function. The first correction is

Oddness eliminates the constant and pure \(y\) derivatives and the pure second \(x\) derivative at zero. JC10 eliminates the remaining linear derivative, so \(g^{(1)}(t)\in\mathcal Y\). Every derivative here is the exact derivative of the supplied basis representation. For example, if \(g^{(0)}=\sum_{ij}a_{ij}(t)B_i(x)C_j(y)\), then \(\partial_xg^{(0)}(t,0)=\sum_{ij}a_{ij}(t)B_i'(0)C_j(0)\). No second rounding of that expression is part of JC10.

Keep the original target \(\chi_\dagger\in\mathcal Y\) with \(J\chi_\dagger=1\), and put

The lower jets of \(E_0,E\) vanish by JC5 and the preceding product rules. For the actual \(\chi\) in JC6, define

The integral is an exact ordinary integral of the represented defect; bounded measurable \(e\) gives an absolutely continuous \(a\), with the differential identity almost everywhere. For piecewise smooth inputs it holds classically away from their time breakpoints. Initial matching gives \(J(g^{(2)}(0)-\chi_\dagger)=e_0+a(0)=0\), and JC5, JC9 and JC12 give \(JE_2=e+a'-\lambda a=0\). All lower jets remain zero, hence

These are three different statements. The note's prose at1483 says \(g^{(2)}\) itself vanishes cubically. At initial time its mixed derivative is one, so that prose is false. The correct conditions are the note's own display at1299–1300, reproduced exactly in JC13.

The derivative term \(a'\chi\) in JC12 must be retained. The note omits it from the estimate at1635–1638 and1664–1666. It also omits \(e\) in the first equality at1454 and restricts the second condition at1440 to \(s=0\). The full ODE is the identity at every time in JC12. Part I v4 explicitly contains the full conditions at2287–2306 and the \(a'\) terms at15819 and15849.

There is an exact alternative that keeps the cancellation before estimating. With \(\Pi\) from JC4,

Both parentheses separately belong to \(\mathcal X\), by JC5. This identity incorporates \(a'\) through its full ODE; it does not drop it. It is the scalar version of Part II's residual decomposition at1288–1343.

The exact cancellation itself does not require \(P(\chi)=0\). For any supplied \(\chi\in\mathcal Y\) with \(J\chi=1\), retain instead \(\Lambda=J\mathcal L\chi=\lambda+\alpha P(\chi)\) throughout the ODE, integral and JC14. The same algebra proves all three conclusions in JC13. The source's construction has the useful additional property \(P(\chi)=0\), so its correction rate remains the original \(\lambda\); without it the actual extra rate term must be estimated and may affect temporal decay.

JC4. Complete weighted estimates and the infinite time tail

For a scalar function with zero derivatives through order two at zero, Taylor's integral formula on the original half-ball gives

The third differential contains all four original terms \(x^3f_{xxx}+3x^2yf_{xxy}+3xy^2f_{xyy}+y^3f_{yyy}\). The integral of \((1-s)^2/2\) is \(1/6\). For \(C^{2,1}\) functions use the second-order integral remainder and the Lipschitz constant of the Hessian; it gives the same coefficient. The constant \(1/6\) is attained by \(f=x^3/6\) on the positive \(x\)-axis when the norm of its third differential is one.

Writing \(M_3(f;R)=\sup_{r\leq R}\|D^3f\|_{\rm op}\), JC12–JC15 give both complete bounds

The first keeps the derivative explicitly; the second retains its entire quadratic correction inside \(\Pi E\). Away from zero the pointwise bound is

with the same three terms.

For an actual negative \(\lambda=-\gamma<0\), a defect supported in \([0,T_0]\), and \(|e(t)|\leq\varepsilon\), the full original integral in JC12 gives

Each formula follows by integrating the positive exponential kernel on \([0,\min(t,T_0)]\). The correction is not truncated at \(T_0\). No numerical value such as the source's claimed rate below \(-5\) is assumed here in place of its still-unchecked parameter enclosure.

JC5. A counterexample to the omitted-term bound

The missing term is not harmless in the displayed estimate as a general statement. The following example keeps all the source model structure. Let \(w(s)=e^{-1/s}\) for \(s>0\) and zero otherwise, and \(\vartheta(s)=w(s)/(w(s)+w(1-s))\). The denominator is strictly positive; \(\vartheta\) is smooth, zero for \(s\leq0\) and one for \(s\geq1\). Choose

Thus \(\kappa=1\) on \(|s|\leq1/8\), zero on \(|s|\geq1/4\), and the JC6 correction has support in \(r\leq\sqrt2/4<1/2\). Also \(H=xy\) on \(r\leq1\), with compact support in \(r\leq2\). Use the actual Biot–Savart functional from JC9 and set

All original model hypotheses hold, with \(\alpha=1\) and \(P(\chi)=0\). The functions \(\chi,\mathcal L\chi,H\) are smooth, odd in both coordinates and supported in \(r\leq2\), so JC9a supplies \(|P(f)|\leq4\|f\|_{2,2}/\pi\) on their receiving space. For an arbitrary \(\varepsilon>0\), let \(\rho(t)=\vartheta(2-t)\) on \(t\geq0\) and supply the finite global representation

It is smooth, compactly supported in time and space, odd in \(x\) and quadratic at the origin. It can itself be the raw representation \(g^{(0)}\), since its first derivative at the origin is zero. Since \(\rho=1\) on \([0,1]\), direct differentiation at zero gives

On \(r\leq1\), every third derivative of \(E(0)=\varepsilon xy\) is zero. Both terms retained in the note's right side at1637–1638 therefore vanish: its other term has factor \(|a(0)|=0\). But at the actual point \((x,y)=(3/(4\sqrt2),3/(4\sqrt2))\), the radius is \(3/4\), \(\chi=0\) and \(H=xy=9/32\). Consequently

No implicit finite constant can bound this positive value by that zero right side. The corrected JC16 retains \(|a'|M_3(\chi;1)\), or equivalently \(M_3(\Pi E;1)\), and has no such contradiction. This is a counterexample to the note's omitted-term estimate, not to the source's main theorem. Part I v4 already includes the missing term.

The figure fixes the actual \(\varepsilon=1/1000\) and follows the original ray \(x=y=r/\sqrt2\) at \(t=0\). The shaded interval has exact cancellation because both cutoff factors are one. Every transition term in JC7 contributes to the plotted correction. At \(r=3/4\), JC22 gives the exact value \(1/1500\), while the incomplete bound is zero. The curves are numerical samples of these specified functions; the strict witness is proved in JC18–JC22. The figure source and sample values are retained. This is an illustration of the model counterexample, not the source's computed approximate profile.

JC6. The full Duhamel receiving map

Let \(W_1\) be a regular function for which the following integrals and applications of the original operator exist, and retain \(\alpha=J\overline F\). Define the actual corrected response

For the supplied finite basis representations, differentiation and integration are justified term by term on compact time intervals. The same calculation holds for locally absolutely continuous functions with integrable bounds on the displayed terms. Differentiating the upper endpoint and retaining the whole residual JC12 gives

Every spatial jet through degree two of \(\mathcal R\) is zero by JC13, whenever differentiation under the integral is licensed by the same finite representation or uniform local derivative bounds. If \(W_1\) satisfies the exact modified equation \(\partial_t W_1=(\mathcal L-\alpha\chi_\dagger P)W_1- \mathcal R(W_1)\), addition to JC24 proves \(\partial_t(W_1+\widehat W_2)=\mathcal L(W_1+\widehat W_2)\) with its original initial data. This proves the explicit map back to the original equation, including the initial mismatch and time tail. It does not claim that the weighted norm of \(\widehat W_2\) is cubic; only the main component and its residual receive that norm.

There is also an exact norm estimate. If \(|P(f)|\leq C_P\|f\|_X\) on the stated receiving space, and both corrected residual terms are in \(X\), the triangle inequality gives

No small constant is asserted without estimating the actual norms. The complete JC16–JC17 quantities are the inputs for that estimate.

JC7. Exact relation to the source's three-component correction

Part II retains the actual three correction components \(B\boldsymbol a=(a_1\chi_{12},a_2\chi_{22},a_3\chi_{32})\), where \(\Phi_*=-xy^3K_*/6\), \(\chi_{12}=-\Delta\Phi_*\), \(\chi_{22}=xyK_*\), \(\chi_{32}=x^2K_*/2\). Here the actual source cutoff is \(K_*(x,y)=\kappa_*(x)\kappa_*(y)\), not the particular cutoff chosen for JC18. Retain its full definition on the original nonnegative coordinate and its even symmetry extension:

These are the source's parameters \(a=1/3\), \(b=3/2\). The exponential cutoff is flat at both endpoints, by the same exponential-versus-power argument used for JC18. The full rational factor remains: \(\kappa_*(s)=1-81s^4+O(s^8)\) up to a flat term, so \(\kappa_*-1\) has zero derivatives through degree three. Consequently \(\Phi_*=-xy^3/6+O(r^8)\), \(\chi_{12}=xy+O(r^6)\), \(\chi_{22}=xy+O(r^6)\) and \(\chi_{32}=x^2/2+O(r^6)\), with these orders justified by their smooth Taylor expansions. JC7–JC8 apply with every \(\kappa\) replaced by \(\kappa_*\), preserving all transition terms. The exact velocity is \(\nabla^\perp\Phi_*\), and \(P(\chi_{12})=0\) still holds. Define \(J\boldsymbol f=(\partial_{xy}f_1(0), \partial_{xy}f_2(0),\partial_{xx}f_3(0))^T\). Then \(JB=I_3\) exactly, including the factor \(1/2\) in the third correction. These components belong to the source's enlarged linear system at951: \(\eta,\xi\) are independent there. Their relation to an actual temperature gradient is proved below.

For the complete linear part of WT1–WT5, its action on these actual corrections has jet matrix

Indeed, the velocity induced by \(\chi_{12}\) is cubic by JC8–JC9, so its product with \(\nabla\bar\omega\) is cubic and its derivative products with \(\nabla\bar\theta=O(r)\) are cubic. Its rate functional is zero. Near zero the barred velocity is \((\bar u_x(0)x,-\bar u_x(0)y)+O(r^2)\); this follows from the source symmetry and no-flow boundary, with divergence zero. Acting on \(xy\), the two linear transport coefficients add to \(2\bar c_l\). The original \(+\eta\) term gives the entry one in the first row. In the second row the local growth remains \(2\bar c_\omega-\bar u_x(0)\). In the third row, acting on \(x^2/2\) gives twice \(\bar c_l+\bar u_x(0)\), while the local growth is \(2\bar c_\omega+\bar u_x(0)\). Their difference is the third diagonal in JC26. Finally \(\bar v_x(0)=\bar u_y(0)=0\), so neither cross-gradient coupling contributes a quadratic jet. All these couplings remain in the full functions \(\mathcal L B\).

For the actual first-corrected approximation \(\widehat W^{(1)}\), let \(\boldsymbol E=(\partial_t-\mathcal L)\widehat W^{(1)}\), \(\boldsymbol F=J\boldsymbol E\), and set \(\boldsymbol a'=\Gamma\boldsymbol a-\boldsymbol F\), with initial value chosen to match the three target jets. Direct expansion gives

This is exactly the algebra performed in Part II at1284–1303, before its split into the numerical velocity, velocity error and profile-velocity error. In particular it accounts for \(\boldsymbol a'\); it does not use the note's incomplete bound. The full original functions in each parenthesis are still needed for the quantitative estimates. Their jets alone are not those estimates.

Keeping the source relation \(\bar c_\omega=\bar c_l/2+\bar u_x(0)\), put \(r_1=-2\bar c_l+\bar c_\omega\) and \(r_2=-2\bar c_l+2\bar c_\omega-\bar u_x(0)\). Then \(r_2-r_1=\bar c_l/2\), with all original rates retained, and

Differentiate the integral entry to obtain \(h'=r_1h+e^{r_2t}\), \(h(0)=0\). This verifies the full matrix equation and identity initial value, proving JC28. The forced solution is exactly \(\boldsymbol a(t)=e^{\Gamma t}\boldsymbol a(0) -\int_0^t e^{\Gamma(t-s)}\boldsymbol F(s)\,ds\), so the first component retains both \(F_1\) and the transfer from \(F_2\). For \(\bar c_l>0\) and \(\bar c_\omega<0\), the original formulas give \(r_1<0\) and \(r_2=-3\bar c_l/2+\bar c_\omega<0\). This proves decay of this finite correction matrix in that class. It neither estimates the full infinite-dimensional operator nor supplies the source's numerical claim that its particular rates are below \(-5\).

JC8. The exact return from independent components to a temperature

Retain the original \(b=\bar c_l(x,y)+\bar{\boldsymbol u}\), \(\bar c_\theta=\bar c_l+2\bar c_\omega\), \(c_\omega=u_x(0)\) and \(h=-\boldsymbol u\cdot\nabla\bar\theta+2c_\omega\bar\theta\). The last two linear equations in Part II935–944, including all nonlocal terms, are precisely

For example, \(h_x=-\boldsymbol u_x\cdot\nabla\bar\theta -\boldsymbol u\cdot\nabla\bar\theta_x+2c_\omega\bar\theta_x\); the \(y\) derivative retains the corresponding three terms. Thus the displayed compact form preserves the original full functions.

Define the actual incompatibility \(D=\partial_x\xi-\partial_y\eta\). Differentiate the second equation in \(x\) and subtract the \(y\) derivative of the first. The transport differentiation contributes

Differentiating the matrix products contributes

The forcing contributes \(h_{xy}-h_{xy}=0\). Hence

The characteristic formula holds on each existing characteristic interval beginning at zero; it follows by a scalar integrating factor. Here \(\operatorname{div}b=2\bar c_l\) uses the exact divergence of the original barred velocity. Thus the independent-component evolution preserves the class of compatible gradients. No choice of a correction coefficient is presumed to do so by itself: for the source's exact correction,

The local leading term is \((a_3-a_2)x\); the two distinct cutoff derivatives also remain. Even \(a_2=a_3\) does not cancel them throughout the support. The exact Duhamel sum, with compatible initial data, obeys the full linear equations and therefore JC30. This proves its return to the gradient class without asserting compatibility of every auxiliary component.

For completeness construct the original temperature, rather than only showing zero incompatibility. On the half-plane, for \(D=0\) put

The oriented first integral includes negative \(x\). Direct differentiation gives \(\theta_y=\xi\) and \(\theta_x=\eta(t,x,0)+\int_0^y\xi_x(t,x,s)ds=\eta\). Applying both derivatives to \(Q=\theta_t+b\cdot\nabla\theta-\bar c_\theta\theta-h\) and using JC29 gives \(Q_x=Q_y=0\). The half-plane is connected, so \(Q\) is constant in space. The original symmetries give \(b(0,0)=0\); at the origin the ODE for \(q\) gives \(Q=0\). This proves the full original linearized temperature equation with its specified initial constant and boundary origin. Two reconstructed temperatures with the same gradient differ by a spatial constant, whose evolution is exactly \(q'=\bar c_\theta q\); fixing \(q(0)\) therefore gives uniqueness. This is the actual inverse map from the compatible enlarged system to temperature, including its constant.

The nonlinear terms in WT5 have the same exact consequence. Replace \(b\) by \(b+\boldsymbol u\), the scalar growth by \(\bar c_l+2\bar c_\omega+2c_\omega(t)\), and \(h\) by \(h+\bar F_\theta\). Their divergence is still \(2\bar c_l\), and the same expansion gives

The integrating factor is the exponential of the full time integral of this coefficient. This propagates compatibility through the complete perturbation equation without discarding its profile residual.

The local omissions in the correction note have exact repairs already visible in the current main papers. The remaining work is to check their actual represented profiles, full nonlocal estimates and numerical residual bounds. No main-theorem failure follows from the local omissions established here.

Part 4. Exact derivatives, singular weights and both Sobolev norms in the original transport model

Human sources: Jiajie Chen and Thomas Y. Hou, Part I, arXiv:2210.07191v4, original TeX1851–1938, and the analytic correction note, arXiv:2607.15256v1, original TeX315–358. Part I1935 and the note354 say that decay of the pure \(x\) derivative in the displayed model needs order at least fourteen. The exact calculation below gives order eight for strict \(L^2\) decay in Part I's specified model, with order seven neutral. It also proves the full mixed-derivative and coupled equations, and the exact distinction between a homogeneous derivative norm and the inhomogeneous Sobolev norm. These are statements about the original displayed constant-coefficient model; no nonlocal term of the full Euler/Boussinesq problem is thereby controlled.

LT1. Complete original solutions and their supports

Retain the original quadrant \(Q=(0,\infty)^2\), coordinates \((x,y)\), time \(t\geq0\), and coefficients \(a_1,a_2>0\) and real \(a_3\). The source's model is

The triple on the right is exactly the model printed at1855; it is not an exact enclosure for the full source profile. We retain general coefficients while deriving the maps and then substitute that triple. Take \(\omega_0,\eta_0\in C_c^\infty(Q)\). Extending oddly in \(x\) gives the source symmetry without imposing an artificial boundary value at \(y=0\) on the full PDE. This initial class already contains nonzero examples sufficient to test every asserted necessity below. The same explicit formulas apply to other initial functions whenever the indicated norms and derivatives exist.

Let

Both maps preserve the quadrant, and each is the inverse of the other at fixed time. Their Jacobians are respectively \(e^{(a_1+a_2)t}\) and \(e^{-(a_1+a_2)t}\). Along the first map the second equation is \(\dot\eta=a_3\eta\); the first then has the integrating factor \(e^t\). Therefore the unique smooth solution with these initial values is

Uniqueness follows along every characteristic from the two scalar ODEs, whose initial footpoints cover the quadrant. Direct differentiation also verifies the full two equations, including the \(+\eta\) coupling. The support of each transported initial field is precisely the image of its support under \(X(t;\cdot)\). The vorticity support is contained in the union of its two such images; cancellation can make that inclusion strict. At every finite time the supports remain compact and avoid both axes. No boundary integral or far-field contribution is lost in subsequent integrations by parts.

LT2. Every mixed derivative and original measure factor

For integers \(i,j\geq0\), differentiation at fixed time gives

No mixed derivative is omitted. In the differentiated equations the transport contributes exactly \(-ia_1-ja_2\) to each scalar growth coefficient; its constant diagonal gradient generates no coupling between distinct derivative indices. Changing variables using the actual Jacobian in LT2 proves, for \(1\leq p<\infty\),

For \(p=\infty\) the same formulas have measure contribution zero: composition with the bijection preserves the supremum. For any nonzero derivative the first equality proves strict decay exactly when its exponent is negative, neutral norm when it is zero, and strict growth when it is positive. A zero derivative stays identically zero and is not a witness for any strict threshold.

Specialize to the original triple in LT1 and \(p=2\). The exact formulas are

In particular \(k=7\) is neutral and every \(k\geq8\) decays. For a concrete nonzero function in the stated original class let \(w(s)=e^{-1/s}\) for \(s>0\) and zero otherwise, and set \(b(s)=w(s-1)w(2-s)\), \(\eta_0(x,y)=b(x)b(y)\). This is smooth and supported in \([1,2]^2\). Its \(k\)-th pure \(x\) derivative is nonzero: otherwise \(b\) would be a polynomial of degree at most \(k-1\) on the connected positive half-line, and its vanishing on \((2,\infty)\) would force it to be zero everywhere, a contradiction. For \(k=0\) nonvanishing is immediate. Thus LT6 gives actual examples at every threshold, including strict decay with \(k=8<14\).

LT3. The entire homogeneous derivative norm and the coupled system

Specify the order-\(k\) homogeneous derivative seminorm by

This formula fixes every term and coefficient in the convention used here. On the compactly supported smooth class it is a norm: if all order-\(k\) derivatives vanish, its pure \(x\) derivative vanishes, and the polynomial argument just given on each horizontal line forces \(f=0\). For \(k=0\) it is the ordinary \(L^2\) norm. For \(a_2\geq a_1>0\), LT5 with \(p=2\) gives the full identity and bound

All mixed derivatives have been retained in the sum. If \(a_2>a_1\) and \(\eta_0\ne0\), division by the largest exponential and the finite sum give

The positivity follows on some horizontal slice from the same compact-support polynomial argument, then by continuity on a neighborhood of that slice. Thus the exponent in LT8 cannot be improved uniformly in time by introducing a fixed finite prefactor. For the source model it is \((7-k)/2\), and all homogeneous norms of order \(k\geq8\) decay. The mixed derivatives do not create an additional difficulty in this frozen model.

For the coupled vorticity put \(d_k=(a_1+a_2)/2-ka_1\). Apply LT4 termwise in the finite direct sum of \(L^2\) spaces. Every dilation factor is at most \(e^{d_kt}\); the triangle inequality in that direct sum gives

Here \(H(t)\geq0\) for \(t\geq0\), directly from its integral, including the resonant value in LT2. For the exact triple in LT1, \(d_k=3-k/2\) and \(H(t)=\tfrac23(e^{t/2}-e^{-t})\). Consequently

For \(k\geq8\), both right sides tend exponentially to zero, with the exact transfer coefficient \(2/3\) retained. In particular, their sum is at most \(e^{(7-k)t/2}(|\omega_0|_{\dot H^k} +\tfrac53|\eta_0|_{\dot H^k})\). This proves decay for the full two-component homogeneous norm without replacing the vorticity equation by the temperature-derivative equation.

LT4. Why this is not decay in the full unweighted Sobolev norm

Define the inhomogeneous norm by retaining all its original orders:

For the source model the exact formula is

Indeed, the zero-order term has exponent seven and every other term has a strictly smaller exponent. There are only finitely many terms, so passage to the displayed limit is immediate term by term. For every nonzero compactly supported initial \(\eta_0\), this norm grows at the exact leading rate \(e^{7t/2}\) for every finite \(k\). Increasing the derivative order cannot remove the zero-order term in an inhomogeneous norm. For the coupled system the sum of the two inhomogeneous norms is at least this growing temperature-derivative norm, so it cannot decay either.

The source's surrounding caution about establishing a full stability estimate remains relevant. The corrected statement is precise: the model's top homogeneous derivatives decay starting at order eight; its full unweighted inhomogeneous Sobolev norm does not decay for nonzero \(\eta_0\). Neither statement estimates the variable coefficients or nonlocal operators in the full source PDE.

LT5. Every energy-weight factor and the exact derivative correspondence

Retain arbitrary real \(\alpha,\beta\) and the original weight \(\phi(x,y)=x^{-\alpha}y^{-\beta}\) on \(Q\). For a compactly supported initial field as above every weighted integral is finite. Changing variables in LT3 proves the full energy identity

The factors arise separately from the square of the field, the Jacobian, and the two powers of the weight. Integrating by parts gives the same result: the \(x\)-transport contributes \(a_1(1-\alpha)\int|\eta|^2\phi\) to the energy derivative, the \(y\)-transport contributes \(a_2(1-\beta)\int|\eta|^2\phi\), and growth contributes \(2a_3\int|\eta|^2\phi\). Compact supports make the boundary terms zero. For broader classes those boundary limits must actually vanish before this integration-by-parts proof is used.

For the exact model, the exponent in LT14 is \(7-\alpha/2-11\beta/2\). With \(\beta=0\), the weighted energy is nonincreasing for \(\alpha\geq14\) and strictly decays for \(\alpha>14\), with neutrality at fourteen. Thus the printed number fourteen does arise for this energy-weight exponent. It is not the corresponding derivative order. The exact comparison is as follows.

For \(i,j\geq0\) set \(T_{ij}f=x^iy^j\partial_x^i\partial_y^jf\). The Euler transport operator \(L_b=a_1x\partial_x+a_2y\partial_y\) satisfies

For the first identity, \([x\partial_x,\partial_x]=-\partial_x\) and induction gives the factor \(-i\); the \(y\) factor is proved in the same way, and cross-coordinate derivatives commute. Differentiating the multiplying monomial contributes \((ia_1+ja_2)x^iy^j\), canceling exactly that commutator. The norm identity is pointwise multiplication on \(Q\), followed by integration; no equivalence constant or omitted factor is involved. Hence \(T_{ij}\eta\) obeys the original scalar transport equation with growth \(a_3\), and LT14 with \(\alpha=2i,\beta=2j\) is exactly LT5 squared. This explains the factor of two while proving the actual intertwining map.

The image of this map can also be specified exactly. In one dimension let \(n\geq1\) and \(F\in C_c^\infty(0,\infty)\). Define

Differentiating under the integral \(n\) times, with the lower endpoint contribution retained on the last step, gives \(\partial_x^n I_nF=F\). Above the support the integral is zero. Below its support expand the entire polynomial \((s-x)^{n-1}=\sum_{m=0}^{n-1}\binom{n-1}{m}s^m(-x)^{n-1-m}\). Thus \(I_nF\) vanishes below the support exactly when \(\int_0^\infty s^mF(s)ds=0\) for every \(0\leq m<n\). These conditions are also necessary for \(F=\partial_x^nf\) with compactly supported \(f\), by integration by parts \(n\) times. The difference \(I_n\partial_x^nf-f\) has zero \(n\)-th derivative and vanishes above both supports; it is a polynomial that vanishes on an interval, hence zero. This proves both inverse identities on the specified image.

Apply this argument separately in \(x\) and \(y\). The map \(T_{ij}:C_c^\infty(Q)\to C_c^\infty(Q)\) is a bijection onto the subspace of functions \(G\) for which \(F(x,y)=x^{-i}y^{-j}G(x,y)\) obeys

at every value of the other coordinate. A family of conditions is empty when its index is zero. The inverse is \(I_i^xI_j^yF\), where \(I_0\) denotes the identity. If \(i,j\geq1\), its full formula is

Fubini is valid on the finite supports. The moment conditions make the result vanish below each support as well as above it, so it lies in \(C_c^\infty(Q)\). The two derivatives recover \(F\), and multiplication recovers \(G\). Applying the one-dimensional inverse identities to an original \(f\) recovers that same \(f\), proving injectivity as well as surjectivity onto the stated subspace. This supplies the precise objects and inverse map behind LT15.

For a general field nonzero at \(y=0\), a factor \(y^{-\beta}\) in its squared energy does require \(\beta<1\) for local integrability near any boundary point where its value stays away from zero. Indeed, continuity supplies a positive lower bound on a small rectangle, and \(\int_0^d y^{-\beta}dy\) diverges when \(\beta\geq1\). In LT15 the weighted field is \(T_{ij}f\), which contains the actual factor \(y^j\); its square cancels \(y^{-2j}\) exactly. The boundary statement for the original undifferentiated field therefore does not forbid the derivative norm. This uses the proved map, not an identification of two different fields.

LT6. Original radial weights and exact anisotropic dilation

For completeness retain the actual original radial weight \(r^{-\gamma}\), \(r=\sqrt{x^2+y^2}\), with \(\gamma\geq0\). At \((x,y)=X(t;s,r_0)\), LT3 gives

The symbol \(r_0\) is the second original coordinate of the initial point here, not its radius. Both anisotropic contributions remain in the denominator. If \(a_2\geq a_1\), it follows that

For the original triple, strict scalar damping holds for \(\gamma>1\); retaining only quadratic vanishing permits \(1<\gamma\leq2\) near the origin, exactly as stated in the source. The full angular coefficient is \(a_3-\gamma(a_1x^2+a_2y^2)/(x^2+y^2)\), obtained by differentiating the original radial weight, with no averaged replacement of the two coordinates.

The first bound in LT20 has the exact operator norm on the compact smooth class used here. To prove the lower bound, fix \(s=1\) and let \(r_0>0\) decrease to zero. The multiplier in LT19 tends to \(e^{(a_3-a_1\gamma)t}\). Choose a smooth nonnegative bump \(\psi\) supported in a ball in \(Q\), equal to one at that point, and bounded by one; then \(\eta_0(s,r_0)=(s^2+r_0^2)^{\gamma/2}\psi(s,r_0)\) has weighted norm one and realizes the exact multiplier at the selected point. Its support avoids zero, so every real \(\gamma\) in the stated range yields a smooth field. Odd extension in \(x\) preserves the source symmetry. Taking the limit proves the lower bound, while LT20 proves the upper bound. No value on the excluded axis is needed to attain the supremum as a limit.

These computations complete the original frozen model, its full derivative system, all measure factors, the singular-weight comparison and the exact norm thresholds. The next source calculation must retain the variable-coefficient and nonlocal operators. Their commutators and quantitative bounds are not supplied by the constant-coefficient model.

Five solved exercises

These exercises use the complete proofs QC1–QC21, WT1–WT18, JC1–JC32 and LT1–LT20. Their human source comparisons are Jiajie Chen and Thomas Y. Hou, Part I v4, Part II v3, and the analytic correction note v1. The numerical constants below specify actual mathematical examples; they do not replace the source's approximate-profile coefficients by rounded values or certify its full nonlinear theorem.

Exercise 1. An exact clock with a logarithmic spatial correction

Use the original clock definitions with \(A=3\), \(\bar c_l=5\), \(\bar c_\omega=-2\) and \(c_\omega(\tau)=-2-(1+\tau)^{-1}\). Calculate the entire physical clock and both field scales, give an exact remaining-time formula, and determine the leading length scale as physical time approaches its endpoint. Preserve the original initial factors.

Solution. The rate is continuous and lies in \([-3,-2]\), so it also lies in the QC1 interval with \(\delta=1<2\). Its full accumulated deviation is \(B(\tau)=-\log(1+\tau)\). Direct integration gives

Define the actual exponential integral on its positive real domain by \(E_1(z)=\int_z^\infty e^{-v}v^{-1}\,dv\), \(z>0\). Its integral converges because its lower endpoint is positive and the tail has exponential decay. Set \(v=2(1+s)\) in the original time integral and retain \(ds/(1+s)=dv/v\). Then

Differentiation of the defining integral gives \(E_1'(z)=-e^{-z}/z\); hence \(t'=C_\omega>0\), verifying the original clock directly. For \(u\geq0\) the exact ratio \(C_\omega(\tau+u)/C_\omega(\tau)\) gives

For the lower bound use \(\log(1+u/(1+\tau))\leq u/(1+\tau)\), proved by integrating \(1/(1+s)\leq1\) from zero to the nonnegative argument. The upper bound uses the ratio at most one. The limit also follows from these two bounds; no asymptotic formula for the special function is required.

The original length scale therefore has the exact identity

Indeed, \(\Delta=3e^{-2\tau}q(\tau)/(1+\tau)\); solving for \(e^{-2\tau}\) and raising to \(5/2\) proves EX4. For large \(\tau\), \(\Delta<1\) and \(\log(1/\Delta)=2\tau+\log(1+\tau)-\log(3q(\tau))\). The final logarithm is bounded by EX3, and \(\log(1+\tau)/\tau\to0\), for instance by comparison of its derivative with \(\tau^{-1/2}\) beyond one and integration. Thus \(\log(1/\Delta)/\tau\to2\), and EX4 gives

Every constant comes from the original \(A=3\) and the exact limit of \(q\). Finally, the physical field multipliers from QC16 are \(e^{2\tau}(1+\tau)/3\) for vorticity, \(e^{-\tau}(1+\tau)^2/9\) for temperature, and \(e^{-3\tau}(1+\tau)/3\) for velocity, all at the original point \(e^{5\tau}\boldsymbol x\). This is a fully specified clock example; its rate alone is not an Euler/Boussinesq solution.

Exercise 2. A complete annular weight for the original local system

Use the exact two-component model LT1 with \(b=(x/2,11y/2)\). Construct an annular weight for \(1\leq r\leq3\) with inner plateau one and outer plateau one-half, and combine it with \(\psi=r^{-2}+r\). Give an exact exponent ensuring a row damping bound of one for both equations. Calculate the time interval during which the characteristic starting at \((1/\sqrt2,1/\sqrt2)\) receives that bound.

Solution. Keep the full WT7 bump

The WT7–WT11 proof gives both exact plateaus and, on the annulus, \(d_0=(b\cdot\nabla\varphi_0)/\varphi_0\leq-C_*\), where

Here \(c_*=1/2\), \(R_1=1\) and \((R_1/2)(2R_2-R_1)=5/2\); none of these factors is omitted. The original radial logarithmic derivative for \(\psi\) is

On \(1\leq r\leq3\), the first factor increases from \(-1/2\) to \(25/28\), since its derivative is \(9r^2/(1+r^3)^2>0\). The second factor lies in \([1/2,11/2]\). Where the first factor is negative, their product is negative. Where it is nonnegative, it is at most \((25/28)(11/2)=275/56\), attained at \((0,3)\) in the closed quadrant and approached from the open quadrant. Thus its exact supremum is \(275/56\).

The complete local matrix for \((\omega,\eta)\) is \(M=\left(\begin{smallmatrix}-1&1\\0&1/2\end{smallmatrix}\right)\). With the common weight \(\varphi=\psi\varphi_0^m\), the two row quantities before the added bounded factor are \(d_s\) and \(1/2+d_s\). Their joint maximum is \(303/56\). Therefore the exact choice

makes both row sums at most \(-1\) throughout the annulus. Along any characteristic that stays there, WT16–WT17 consequently give

There is no forcing in this specified original local model. For the full perturbation equation every forcing term in WT17 remains; EX10 does not discard it on that different equation.

The stated characteristic is \(X(t)=(e^{t/2}/\sqrt2,e^{11t/2}/\sqrt2)\), so its radius squared is \((e^t+e^{11t})/2\). This is strictly increasing from one to infinity. Its unique exit time is

Existence and uniqueness follow because the polynomial is continuous, strictly increasing on \([1,\infty)\), negative at one and tends to infinity. The time bounds follow from \(e^t\leq|X(t)|^2\leq e^{11t}\). Thus EX10 applies on the entire interval \([0,T_{\rm exit}]\), and does not assert this annular estimate after the trajectory leaves. The norm comparison still has its complete cost \(2^{-m}\|\psi W\|_\infty \leq\|\varphi W\|_\infty\leq\|\psi W\|_\infty\); the exponent in EX9 is not hidden in a constant.

Exercise 3. The exact nonlocal coefficient and the entire correction tail

For the actual scalar counterexample JC18–JC22, calculate \(P(H)\), \(P(\mathcal L\chi)\), the full jet defect \(e(t)\), and the correction coefficient \(a(t)\), including its tail after the represented approximation vanishes. Use \(\varepsilon=1/1000\) to show that the tail is nonzero.

Solution. Retain precisely the JC18 cutoff \(\vartheta\), \(H(x,y)=xy\vartheta((4-x^2-y^2)/3)\), the JC6 correction \(\chi=-\Delta(-xy^3\kappa(x)\kappa(y)/6)\), \(\rho(t)=\vartheta(2-t)\) and \(\mathcal L=-x\partial_x-y\partial_y-1+\chi P\). The actual quadrant kernel is the one proved in JC9a, so polar coordinates give

Indeed, \(\cos^2\theta\sin^2\theta=(1-\cos4\theta)/8\), so the angular integral is \(\pi/16\). For the radial integral put \(z=(4-r^2)/3\), retaining \(r\,dr=-3\,dz/2\). The result is \((3/2)\int_0^{4/3}\vartheta(z)dz\). The exact defining formula gives \(\vartheta(z)+\vartheta(1-z)=1\), so its integral over \([0,1]\) is \(1/2\); it equals one on \([1,4/3]\). The radial integral is therefore \((3/2)(1/2+1/3)=5/4\).

Write the full scalar kernel as \(p(x,y)=-(4/\pi)xy/(x^2+y^2)^2\). It is homogeneous of degree \(-2\), hence \(\operatorname{div}(p(x,y)(x,y))=2p+(x,y)\cdot\nabla p=0\) away from zero. Integrating \(p(x,y)(x,y)\cdot\nabla\chi\) over a punctured quadrant gives only boundary terms. The axes have zero normal component of \((x,y)\), the outer boundary lies beyond the support, and the inner-circle contribution is \(O(r^2)\) because \(\chi=O(r^2)\), \(p=O(r^{-2})\), and \((x,y)\cdot n\,ds=O(r^2)\). Thus \(P((x,y)\cdot\nabla\chi)=0\). Since \(P(\chi)=0\),

Here \(Jf=f_{xy}(0)\), and the middle equality uses the actual jet formula JC5. Put \(q_0(t)=1+(-3+\varepsilon)t\). Then \(Jg^{(1)}=\rho q_0\). Applying \(J\mathcal Lf=-3Jf+P(f)\), with the full nonlocal term, gives

The coefficient \(53/16\) is the sum of \(3\) and \(5/16\); the latter comes from the original nonlocal functional and cannot be removed. The initial mismatch is zero, so \(a(0)=0\). For \(z(t)=Jg^{(2)}(t)=\rho q_0+a\), the exact correction ODE therefore becomes \(z'=-3z-(5\varepsilon/16)\rho t\), \(z(0)=1\). Integration yields the entire coefficient

For \(0\leq t\leq1\), \(\rho=1\), and the exact integral is \(e^{3t}(t/3-1/9)+1/9\). Thus

For \(t\geq2\), the original approximation is zero, but its exact correction is

For \(\varepsilon=1/1000\), using \(0\leq\rho\leq1\), \(\rho=1\) on \([0,1]\), and the strict transition, gives

The last bound uses \(e<3\). One elementary proof is

because \(n!\geq2^{n-1}\) and the inequality is strict for \(n\geq3\). Thus the tail is rigorously nonzero. This makes the time-support distinction concrete: the supplied approximation vanishes after time two; the analytic correction continues with the full positive coefficient in EX17.

Exercise 4. Every derivative norm of an explicit quadratic Gaussian

Solve the original frozen system with \(\eta_0(x,y)=xy e^{-x^2-y^2}\) and \(\omega_0=0\) on the positive quadrant. Compute every mixed \(L^2\) derivative norm, the entire order-eight homogeneous and inhomogeneous norms, and the coupled vorticity norms. Compare their exact large-time behavior.

Solution. The initial function is smooth, odd in each coordinate, quadratically vanishing at zero and rapidly decreasing. All integrations below converge. LT3 gives the actual fields

The factor \(e^{-6t}\) from the two initial coordinates is retained before multiplication by \(e^{t/2}\) or by \(\tfrac23(e^{t/2}-e^{-t})\). Direct substitution into LT1 checks both equations and the original initial data.

Define the Hermite polynomials by \(H_n(x)=(-1)^ne^{x^2}(d/dx)^ne^{-x^2}\). Their generating function is \(e^{2xs-s^2}=\sum_{n\geq0}H_n(x)s^n/n!\), obtained by Taylor expansion of \(e^{-(x-s)^2}\). Completing the square gives

For bounded \(s,z\), derivatives of the integrand are polynomials times a Gaussian with a fixed integrable majorant, so its coefficients may be extracted under the integral. The coefficient of \(s^nz^n\) on the right comes only from degree \(2n\): it is \((2n)!/[2^n(n!)^3]\) times \(\sqrt{\pi/2}\). Multiplication by \((n!)^2\) therefore proves

The square is even, so the positive-half-line integral is exactly half of this expression. Since \(xe^{-x^2}=-\tfrac12(d/dx)e^{-x^2}\),

The coefficient \(\pi/128\) retains the squared \(1/4\), both half-line factors, and both Gaussian integrals. The time factor is the complete LT6 measure-and-derivative factor. Hence the two requested full norms are exactly

These finite sums retain every mixed derivative, including each lower order in the second norm. In particular,

All other terms have strictly smaller exponents, proving both limits directly. EX19 multiplies every spatial derivative of \(\eta\) by the same nonnegative scalar \(\tfrac23(1-e^{-3t/2})\), so the vorticity's two squared norms equal the corresponding expressions in EX23 times \(\tfrac49(1-e^{-3t/2})^2\). Its two limits are therefore the limits in EX24 multiplied by \(4/9\).

The first two panels use the exact LT6 exponents. The last panel evaluates both complete finite sums EX23, divided by their respective initial norms, and the exact zero-order norm ratio. A temporarily decreasing inhomogeneous norm still has the proved eventual growth EX24. Every coefficient is specified; the plots are numerical evaluations of proved formulas. The reproducible source and samples are retained.

Exercise 5. All correction coefficients and the actual gradient defect

Use the full source correction matrix JC26 with \(\bar c_l=3\), \(\bar c_\omega=-1\), \(\bar u_x(0)=-5/2\), initial coefficient vector \((1,-1,2)^T\), and forcing \(\boldsymbol F(t)=(1,2,3)^Te^{-4t}\). These specified rates satisfy the original relation \(\bar c_\omega=\bar c_l/2+\bar u_x(0)\). Solve the entire forced correction system. Determine whether equal second and third coefficients at a time make the correction a temperature gradient, using the actual source cutoff \(K_*\).

Solution. The original diagonal rates are \(r_1=-7\) and \(r_2=-11/2\). Thus

The last two integrating factors give their complete formulas. Their substitution into the first equation, including the forcing through \(a_2\), yields

For example, the first convolution with \(a_2\) contributes

and the direct forcing contributes \(-(1/3)(e^{-4t}-e^{-7t})\). Adding the original \(e^{-7t}\) initial term gives exactly EX26. Differentiation verifies all three equations, and substitution of zero gives all three original initial values.

The local leading incompatibility is \((a_3-a_2)x=(\tfrac{11}{3}e^{-11t/2} -\tfrac23e^{-4t})x\). It vanishes at the unique nonnegative time

Uniqueness follows from strict decrease of \(e^{-3t/2}\) from one to zero. However, the actual complete defect is JC31, with both cutoff derivatives. At the original point \(x=y=s=1/6\), using \(a=a_2(t_*)=a_3(t_*)\), it is

To prove strict nonvanishing from the full source formula, retain \(\kappa_*(s)=(1-\chi_e(2s/3))/(1+81s^4)\). At \(s=1/6\), the argument of \(\chi_e\) is \(1/9\), strictly between its two endpoints. On that interval \(\chi_e=(1+e^p)^{-1}\) with \(p'=-(z^{-2}+(z-1)^{-2})<0\), so \(\chi_e'=-e^pp'/(1+e^p)^2>0\). The numerator of \(\kappa_*\) is positive with a negative derivative; the positive denominator has a positive derivative for \(s>0\). The quotient therefore has \(\kappa_*(s)>0\) and \(\kappa_*'(s)<0\). Along with \(a\ne0\), this proves EX28.

Thus agreement of the leading quadratic coefficients is insufficient for the auxiliary correction to be a gradient on its full support. The exact return map is JC29–JC32: the complete reconstructed solution of the original linear system preserves zero compatibility defect from its initial data, and its temperature is the stated path integral with the original time-dependent constant. That proved map explains how the source can use these auxiliary components without identifying each one with a separate physical temperature.

Continue with the profile and stability arguments

The exact clocks, weighted transport, analytic corrections and frozen model now have complete receiving formulas. The remaining work is to follow the source's quantitative nonlocal estimates, actual represented profiles, supplementary calculations and validated computational errors. Those arguments supply information that these local comparisons alone cannot provide. The series then continues through the remaining human constructions, Alpöge–Buckmaster, OpenAI and retained workbench results.