Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 87 / 1

Let $C(r)=⟨\mathbf{u}(\mathbf{x})\cdot \mathbf{u}(\mathbf{x}+\mathbf{r})⟩$, with zero mean velocity, and normalize the turbulent energy spectrum by ${\int }_0^{\infty }E(k)dk=⟨|\mathbf{u}{|}^2⟩/2$. Use the Fourier transform pairIsotropic turbulence makes the transform radial, while the energy normalization gives $E(k)=2\pi k^2\hat{C}(k)$. ThusAssume, for example, $\int (1+r^4)|C(r)|d^{3}r<\infty$, sufficient to control the remainder. Expanding the radial kernel gives $\sin (kr)/(kr)=1-k^2{r}^2/6+O(k^4{r}^4)$. With the Saffman integral $L=\int C(r)d^{3}r$ and Loitsyansky integral $I=-\int r^{2}C(r)d^{3}r$, this provesThe original PDF defines the two distinct integrals; the TeX aid accidentally repeats the definition of $I$ in place of $L$.

For a growing volume $V$, let ${\mathbf{P}}_V={\int }_V\mathbf{u}dV$ be its momentum per unit density. Statistical homogeneity givesFor volumes whose boundary-to-volume ratio tends to zero and integrable $C$, divide by $|V|$ and use dominated convergence:If the volume contains many weakly dependent eddies with uncompensated random impulses, the central limit theorem suggests root-mean-square total momentum proportional to $|V{|}^{1/2}$; its variance is extensive and $L$ is nonzero. This is a physical expectation, not a theorem for all turbulent fields: cancellation between cells can eliminate the extensive term.

In particular $L=0$ for an ensemble built from locally momentum-compensated eddies with zero net impulse, or when the complete incompressible velocity correlation tensor decays rapidly enough to have no longitudinal $r^{-3}$ tail. Merely imposing zero total momentum in one finite periodic box is insufficient to prove $L=0$ in an infinite-volume limit: the finite-box constraint fixes one global mode. Rapid decay of the trace $C$ alone is also insufficient; the individual tensor components can retain algebraic tails whose trace cancels.

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.

Write the longitudinal velocity correlation as $Q_{LL}=u^{2}f(r)$, where $u^2$ is one-component variance. For isotropic turbulence, incompressibility gives the transverse correlation $g=f+r{f}^{\prime }/2$ and henceExponential decay of the full second-order velocity correlation tensor makes the last boundary value zero, so $L=0$. For invariance of $I$ we additionally need $r^{4}K\to 0$, as part (ii) shows. This follows if the premise of rapid decay of all two-point correlations includes the two-position triple correlation and persists during evolution; exponential second-order decay at one initial time alone does not establish it. Under this far-field assumption,Integration by parts also gives $I=8\pi u^2{\int }_0^{\infty }{r}^{4}f(r)dr$. Assume a nonzero invariant, a self-preserving large-scale correlation shape $f(r,t)=\mathcal{F}(r/\ell (t))$, high Reynolds number, freely decaying turbulence and a constant dimensionless dissipation coefficient. Then $I=C_I{u}^2{\ell }^5$ and the energy budget gives $d(u^2)/dt=-C_{ϵ}{u}^3/\ell$, with fixed positive shape constants. Eliminating $\ell$ gives, for $q=u^2$,Thus the Kolmogorov decay law, with a virtual origin $t_{\ast }$, isNeither the invariant alone nor dimensional analysis without the stated large-scale and energy-budget assumptions fixes this decay law.

Now let $\mathbf{J}={\int }_V\mathbf{x}\times \mathbf{u}dV$. Integrate the supplied identity over both position variables. For a genuinely confined incompressible flow with $\mathbf{u}\cdot \mathbf{n}=0$ on the boundary, each divergence produces a zero surface term, because its flux contains the normal velocity at the corresponding point. Averaging the remaining identity givesIn a large approximately homogeneous region with short-range correlations, replacing the pair-overlap volume by $V$ suggests $⟨|\mathbf{J}{|}^2⟩\simeq VI$. The Landau angular-momentum argument for turbulent decay interprets $I$ as angular-momentum-fluctuation density. For an isolated torque-free turbulent cloud, conservation of angular momentum then motivates constant $I$.

The weaknesses concern precisely the passage to an infinite homogeneous field. A fixed large subvolume is not an isolated cloud: it exchanges momentum and angular momentum by advection and fluid pressure forces. Impermeable walls eliminate the stated velocity-flux terms but need not eliminate viscous or fluid pressure torques. Moreover nonlocal fluid pressure can generate long-range correlations, invalidating the assumed decay, and boundary terms with large lever arms require control before taking a volume limit. Thus the exact integrated identity is sound under its boundary conditions, but the inferred conservation law requires additional hypotheses.

The concern is that the rapid-decay assumptions need not be preserved by the dynamics. In a constant-density incompressible flow, the divergence of the Navier-Stokes equation gives the pressure Poisson equation for incompressible flow, with $\pi =p/\rho$:For a localized eddy, inversion with the free-space Green function and two integrations by parts yieldAt large separation the differentiated kernel has leading form $(3n_i{n}_j-{\delta }_{ij})/r^3$. Thus an anisotropic stress moment of an individual eddy produces a quadrupolar fluid pressure of order $r^{-3}$, even if the velocity source is spatially localized. Correlating that fluid pressure with local $u_x^2$ allows $⟨u_x^2{p}^{\prime }⟩=O(r^{-3})$. Isotropy may cancel a mean multipole but does not automatically cancel such a mixed fourth-order statistic.

The triple-correlation evolution contains a pressure-gradient term of the form $-⟨u_x^2{\partial }_{x^{\prime }}{\pi }^{\prime }⟩$, so these long-range pressure correlations in turbulence can generate a longitudinal triple-correlation tail $K\sim a{r}^{-4}$. With $L=0$, part (ii) then gives $\dot{I}=8\pi u^{3}a$, which need not vanish. This is the substance of the objection to deducing exact conservation from initially short velocity correlations. Batchelor and Proudman's original long-range calculation concerned anisotropic turbulence; extending it to isotropic decay requires the appropriate higher-order correlations and allows cancellations. A permitted tail is not a proof that its coefficient stays appreciable in fully developed turbulence.

This possibility appealed to workers constructing heuristic turbulence closures. A closure that generated a changing low-wavenumber $k^4$ coefficient and a decay exponent different from $10/7$ could then be viewed as representing a pressure-mediated effect, rather than as automatically violating a universal invariant. The closure's far-field coefficient still needs physical validation; closure-generated nonconservation does not itself prove the true dynamics nonconserving.

For evidence available before this exam, the 2006 simulations of Ishida, Davidson and Kaneda found that $I$ approached an approximately constant value after a transient when the box was much larger than the integral scale of turbulence and the Reynolds number was large. The energy decay was correspondingly close to $u^2\sim t^{-10/7}$. The observed conclusion was approximate late-time conservation in that regime, indicating weak net long-range triple-correlation transfer, rather than exact conservation for every initial condition, box size or Reynolds number.