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The Kármán-Howarth equation evolves two-point correlations in homogeneous isotropic turbulence. In trace form it is ${\partial }_{t}C=r^{-2}{\partial }_r[r^{-1}{\partial }_r(r^4{u}^{3}K)]+2\nu r^{-2}{\partial }_r(r^{2}C\prime )$. Integrating with radial volume weight reduces conservation of the Saffman integral to vanishing nonlinear and viscous boundary fluxes.

For unforced homogeneous turbulence with isotropic turbulence statistics, put $F=u^{2}f=u^2-S_2/2$ and $S_3=6u^{3}K$. The longitudinal Kármán-Howarth equation is $F_t=r^{-4}{\partial }_r[r^4(S_3/6-\nu S_2^{\prime })]$. Since ${\partial }_t{u}^2=-2ϵ/3$, regularity at zero givesNeglecting the last term in a locally equilibrated small-scale range gives the displayed equation. The Kolmogorov four-fifths law follows where the viscous term is negligible as well.

The signed third-order longitudinal structure function has this leading inertial range form under homogeneity, isotropy and the usual local-equilibrium scale separation. Its coefficient follows from the Kármán-Howarth equation and the mean kinetic energy budget, without assuming the Kolmogorov two-thirds law. Its negative sign encodes a forward energy cascade; it does not establish every higher-order similarity exponent.

Put $W=⟨{\omega }^2⟩$, $C=⟨|\nabla \times \mathbf{\omega }{|}^2⟩$ and $M=⟨G^3⟩$, where $G={\partial }_x{u}_x$. The small-$r$ Taylor series giveIn the Kármán-Howarth equation, the coefficient of $r^2$ on the left is $-\dot{W}/30$. On the right, the triple-correlation term contributes $7M/6$ and the viscous term contributes $\nu C/15$. HenceComparing with the mean enstrophy balance identifies the mean vortex stretching:Also $S_2=2(u^2-F)=W{r}^2/15+\cdots$, whereas differentiability gives $S_2=⟨G^2⟩r^2+\cdots$. Thus $⟨G^2⟩=W/15$. Homogeneity makes $⟨G⟩=0$, so the longitudinal velocity-gradient skewness is $S_0=M/(W/15{)}^{3/2}$. Substitution givesPositive mean vortex stretching therefore requires negative skewness. A centered Gaussian distribution has zero third moment, so the derivative statistics cannot be Gaussian. This conclusion uses the physically positive mean stretching, not homogeneity alone.

The probability density function has mean zero but a longer or stronger negative tail, representing intermittent compression. The illustration uses a standardized Gaussian mixture distribution solely to sketch this sign of skewness; it is not turbulence simulation data or a fitted universal distribution.

Let $u^2$ denote one-component velocity variance, $F=u^{2}f$ the longitudinal velocity correlation, and $S_p=⟨(\Delta v{)}^p⟩$ the signed longitudinal structure functions. Homogeneity gives $F=u^2-S_2/2$, while the third-order convention here is $u^{3}K=S_3/6$. Thus the Kármán-Howarth equation becomesThe mean kinetic energy per unit mass is $3u^2/2$, so its decay gives ${\partial }_t{u}^2=-2ϵ/3$. In the universal equilibrium range, the small-scale variation ${\partial }_t{S}_2$ is negligible at leading order, hence ${\partial }_{t}F\simeq -2ϵ/3$. Integrating from zero and using regularity givesThe Kolmogorov equation for structure functions is thereforeFor finite local unsteadiness the right side has the additional term $-3r^{-4}{\int }_0^r{s}^4{\partial }_t{S}_2(s,t)ds$; dropping it is the local-equilibrium approximation used here.

In the inertial range the viscous term is also negligible, leaving the Kolmogorov four-fifths law, $S_3(r)=-4ϵr/5$. Its coefficient and sign follow from the exact energy balance and Kármán-Howarth equation, rather than from dimensional similarity. It supplies a firm third-order constraint on Kolmogorov 1941 theory and measures the forward energy cascade, while leaving the second- and higher-order exponents open to internal intermittency corrections.

Put $F(r)=r^4{u}^{3}K(r)$ and retain the scalar trace correlation $C(r)$. For radial functions the Laplacian is ${\nabla }^{2}C=r^{-2}(r^2{C}^{\prime }{)}^{\prime }$. Multiplying the trace Kármán-Howarth equation by the radial volume element and integrating givesRegularity at the origin removes the lower endpoints. The stated triple-correlation tail gives $F=u^3(a+b/r+\cdots )$ and hence $F^{\prime }/r\to 0$. Rapid enough decay of the trace correlation removes the viscous endpoint. ThereforeFor the Loitsyansky integral, multiplying by $-4\pi r^4$ gives the nonlinear contributionThe given tail removes $r{F}^{\prime }$ but permits a nonzero limit of $F$. The viscous contribution isHere the stronger weighted endpoints vanish under the assumed sufficiently rapid correlation decay. Combining the terms yieldsThe conserved Saffman integral measures the density of large-volume momentum fluctuations established in part (i). Internal interactions redistribute momentum without changing that leading extensive variance when forcing and far-field momentum fluxes are absent. This is a statistical expression of linear momentum conservation, not the assertion that kinetic energy is conserved during viscous decay.

Multiply the Kármán-Howarth equation for $C(r,t)$ by $4\pi r^2$ and integrate in $r$. Integration by parts gives the exact boundary-flux expressionAt the origin smooth isotropic correlations give $K=O(r)$ or smaller and $C^{\prime }=O(r)$, so both boundary terms vanish. At infinity the usual regular power-law decay $K=O(r^{-4})$, with the corresponding derivative decay, makes the first term tend to zero. A sufficiently smooth integrable trace correlation makes $r^2{C}^{\prime }\to 0$ as well. Under these physical far-field conditions,The exact required conditions are the limits of these two boundary fluxes. A bare big-$O$ condition on $K$ without control of its derivatives is not by itself a rigorous sufficient condition: the smooth tail $K(r)=r^{-4}\sin (r^2)$ is $O(r^{-4})$, but $r^{-1}(r^{4}K{)}^{\prime }=2\cos (r^2)$ has no zero limit. The intended power-law assumption excludes such increasingly rapid far-field oscillations; this qualification is needed in a literal proof.

Physically, $L$ is the large-volume momentum variance divided by volume. Internal advective and pressure forces exchange momentum among eddies but do not provide a net impulse to the whole unforced field. For a large spherical region, the variance balance is driven by its boundary momentum flux. The stated correlation decay makes that contribution negligible per unit volume as the region grows. Thus the conservation law preserves momentum-fluctuation density, not kinetic energy. kinematic viscosity can remove kinetic energy while leaving this integral invariant.