Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 46 / 1 / i / Solution

Let ${\sigma }_v^2=⟨v(0{)}^2⟩$, assumed finite. The displacement is $\Delta X(t)={\int }_0^{t}v(s)ds$. By stationarity, its two-time covariance is $⟨v(s)v(r)⟩={\sigma }_v^2\rho (s-r)$, with $\rho$ even for a real scalar process. Integrating over the two triangles of the square gives the Taylor turbulent dispersion identityA sufficient condition for ordinary long-time variance growth is ${\int }_0^{\infty }|\rho (s)|ds<\infty$ with ${\int }_0^{\infty }\rho (s)ds>0$. The dominated convergence theorem then givesMore generally the necessary asymptotic condition for a finite positive variance-based diffusion coefficient is convergence of ${\sigma }_v^2{\int }_0^t(1-s/t)\rho (s)ds$ to that coefficient. Absolute integrability is sufficient, not necessary. A zero integral does not give nondegenerate ordinary diffusion. Moreover a second-moment calculation alone does not establish a normal distribution or an entire diffusive scaling limit; those require additional probabilistic assumptions.

For the specified positive autocorrelation, direct integration gives, when $\alpha \ne 1,2$,Thus the power-law velocity-correlation dispersion has the following regimes. If $\alpha >1$, the Lagrangian integral time is $1/(\alpha -1)$ and ordinary variance growth holds, with $D={\sigma }_v^2/(\alpha -1)$. At $\alpha =2$ the exact expression is $2{\sigma }_v^2[t-\log (1+t)]$. If $\alpha =1$,For $0<\alpha <1$,The last two regimes spread faster than ordinary diffusion and have no finite long-time diffusion coefficient. Their slowly decaying velocity memory is responsible. At short times all these correlations instead give the ballistic law $⟨\Delta X^2⟩\sim {\sigma }_v^2{t}^2$.