Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 334 / 2

For a unidirectional, $z$-independent velocityincompressibility holds automatically. The axial Stokes flow equation isNo external pressure gradient is imposed, so $p_z=0$ andThus the axial velocity is a harmonic function in the disk.

The boundary data are $U\mathrm{sgn}(\sin \theta )$, whose Fourier series isRegular harmonic modes in a disk are $(r/R{)}^n\sin (n\theta )$ and $(r/R{)}^n\cos (n\theta )$. Matching the odd boundary data givesEvery term is regular at $r=0$, and the series approaches the prescribed values at every boundary point away from the two jump discontinuities.

For $|z|<1$,Set $z=(r/R)e^{i\theta }$ and take the imaginary part. Since $r<R$, the real part of the relevant denominator is positive, and the result isAs $r\to R^{-}$ this tends to $U$ for $0<\theta <\pi$ and to $-U$ for $-\pi <\theta <0$.

The transverse and longitudinal Péclet numbers areThe Taylor dispersion regime requires transverse diffusion to act within a longitudinal advection time,while longitudinal molecular diffusion is slow on that advection time,Together,

The advection-diffusion equation isWritewhere the bar is the cross-gap average. In the long, late-time Taylor regime, $c^{\prime }$ adjusts rapidly across the gap while $\bar{c}$ varies slowly along the cell. The leading fluctuation balance isIts scaling isThis final inequality is precisely the transverse-equilibration condition from part i.

Put $c^{\prime }=\chi (y){\bar{c}}_x$. The balance and reflecting boundary conditions giveIntegration yieldsAveraging the full transport equation givesNowbecause $\overline{y^2}=h^2/3$ and $\overline{y^4}=h^4/5$. Thus the flow-induced Taylor dispersion coefficient and total effective diffusivity are