Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 73 / 3 / v

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.