 Solution

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.