Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 81 / 4 / b / Solution

First distinguish concentration conventions. If $C$ is cell number density, the excess mixture density is $v\Delta \rho C$. If $\phi =vC$ is cell volume fraction, it is $\Delta \rho \phi$. The printed definition of $R$ is dimensionless for number density, whereas the prose calls ${\hat{n}}_0$ a volume fraction. Use $C_0$ for mean number density and ${\phi }_0=v{C}_0$ for mean volume fraction. Then the consistent Rayleigh number isThe dimensionless concentration $n=C/C_0=\phi /{\phi }_0$ is the same in both conventions. The extra factor $v$ must be removed if the source symbol is literally a volume fraction.

Remove the hydrostatic basic pressure and use a dilute Boussinesq approximation. With ${\hat{C}}_e=\hat{C}-C_0\bar{n}$ and perturbation fluid velocity $\hat{\mathbf{u}}$, the linear incompressible Newtonian fluid equations areThe cell conservation equation linearizes toAt this order the vertical orientation perturbation is zero, because $|\mathbf{p}|=1$. The instantaneous-orientation approximation from the introductory calculation givesIncompressibility implies ${\hat{\nabla }}_{\perp }\cdot (\hat{\mathbf{\omega }}\times {\mathbf{e}}_3{)}_{\perp }=-\hat{\Delta }\hat{w}$. This identity fixes the sign of the gyrotactic term.

Apply the length, viscous-time, velocity, concentration and pressure scales in the paper. Writing $n_e=n-\bar{n}$, the dimensionless equations arewhere $S=\nu /D$ and $G=V_{s}B/(2H)$. The factor $S$ appears in the cell equation because its diffusive time differs from the viscous time used for $t$.

For a normal mode $[n_e,w]=[N,W]e^{\sigma t+i\kappa x}$, set $\mathcal{L}=d^2/d{z}^2-{\kappa }^2$. Taking the divergence of the momentum equation gives $\mathcal{L}{P}_e=-R{N}^{\prime }$. Applying $\mathcal{L}$ to its vertical component eliminates pressure and yieldsThe concentration equation becomesThese reproduce the requested modal equations, with the physically consistent sign of the orientation balance and concentration convention made explicit.

At both plates, the no-slip boundary condition gives $w=0$ and horizontal perturbation velocity zero. For a nonzero horizontal wavenumber, incompressibility then gives $W^{\prime }=0$. The normal cell flux is $\bar{n}w+hN-N^{\prime }$; the orientation correction has no vertical component at this order. Impermeability therefore givesIf orientation lag is retained rather than the given instantaneous balance, ${\mathbf{p}}_{\perp }$ acquires the relaxation response $(1+B{\partial }_{\hat{t}}{)}^{-1}$ and the modal gyrotactic term changes accordingly. The displayed equations are the quasistatic-orientation model requested here.