= Solution
Let $u^2$ denote one-component <velocity> <variance>, $F=u^2f$ the <longitudinal velocity correlation>, and $S_p=\langle(\Delta v)^p\rangle$ the signed <longitudinal structure functions>. Homogeneity gives $F=u^2-S_2/2$, while the third-order convention here is $u^3K=S_3/6$. Thus the <Kármán-Howarth equation> becomes
$$
\partial_tF=\frac1{r^4}\partial_r\left[r^4\left(\frac{S_3}{6}-\nu S_2'\right)\right].
$$
The mean <kinetic energy> per unit mass is $3u^2/2$, so its decay gives $\partial_tu^2=-2\epsilon/3$. In the <universal equilibrium range>, the small-scale variation $\partial_tS_2$ is negligible at leading order, hence $\partial_tF\simeq-2\epsilon/3$. Integrating from zero and using regularity gives
$$
r^4\left(\frac{S_3}{6}-\nu S_2'\right)
=-\frac{2\epsilon}{3}\frac{r^5}{5}.
$$
The <Kolmogorov equation for structure functions> is therefore
$$
\boxed{S_3(r)-6\nu S_2'(r)=-\frac45\epsilon r}.
$$
For finite local unsteadiness the right side has the additional term $-3r^{-4}\int_0^r s^4\partial_tS_2(s,t)\,ds$; 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>, \b[$S_3(r)=-4\epsilon r/5$]. 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.
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