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

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 $⟨\Delta X^2⟩\sim {\sigma }_v^2{t}^2$.

Take $\kappa >0$ and initially $\omega \ne 0$. Write $m_j(y,t)={\int }_{\mathbb{R}}{x}^j\chi (x,y,t)dx$, and set $k=\pi /L$, $a=\kappa k^2$. The advection-diffusion equation is ${\chi }_t+u{\chi }_x=\kappa ({\chi }_{xx}+{\chi }_{yy})$. Integration by parts in the unbounded coordinate gives the longitudinal moment hierarchy for shear transportAll three moments obey Neumann boundary conditions at the walls. The initial values are $m_0=L$, $m_1=0$, $m_2=L^3/12$. These manipulations require finite moments and vanishing boundary terms at $x=\pm \infty$; diffusion of the compact initial distribution supplies the required decay for positive time.

The zeroth moment remains exactly $L$. Since $\sin (ky)$ is a Neumann eigenfunction, put $m_1=Lg(t)\sin (ky)$. Then $g^{\prime }+ag=U\cos (\omega t)$ and $g(0)=0$, givingFor the second moment, $2u{m}_1=LUg\cos (\omega t)[1-\cos (2ky)]$. Thus $m_2=A(t)+B(t)\cos (2ky)$, withThe equation for $B$ has positive damping and bounded forcing, so $B$ remains bounded. In the equation for $A$, the decaying term in $g$ contributes only a bounded accumulated transient. Over a complete nonzero-frequency cycle, $⟨{\cos }^2(\omega t){⟩}_t=1/2$ and $⟨\sin (\omega t)\cos (\omega t){⟩}_t=0$. ConsequentlyThis proves oscillating shear dispersion in an insulating channel, including the local-in-$y$ leading term, not just a cross-channel average. The instantaneous derivative retains bounded periodic oscillations; its cycle-averaged slope is $2K{m}_0$. Thus $2Kt{m}_0$ is the leading second moment, rather than a literal rate proportional to $t$.

The asymptotic statement holds for fixed positive diffusivity and fixed nonzero frequency, after times long compared with $a^{-1}$ and the oscillation period. If $|\omega |L^2/\kappa \ll 1$, transverse diffusion adjusts rapidly to the changing shear flow, andThis is the cycle average of the instantaneous steady-shear enhancement $U^2{\cos }^2(\omega t)/(2a)$. At exactly $\omega =0$, there is no cycle average: $g\to U/a$ and $K=\kappa +U^2/(2a)$. Hence the zero-frequency and long-time limits do not commute. In the intermediate range $a^{-1}\ll t\ll |\omega {|}^{-1}$, the shear is approximately steady at its initial amplitude and that latter enhancement is appropriate.