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

Write ${\ell }_g$ for the dimensional offset of the centre of mass, reserving $h$ below for the dimensionless swimming parameter. The weight $-mg\mathbf{k}$ acts at $-{\ell }_g\mathbf{p}$, so its torque about the geometric centre is $mg{\ell }_g\mathbf{p}\times \mathbf{k}$. Buoyancy acts through the geometric centre and has no such torque. For a spherical cell, the ambient angular velocity is $\mathbf{\omega }/2$ and the rotational resistance is ${\zeta }_r=\rho \nu {\alpha }_{\perp }v$. Neglecting angular inertia, torque balance givesSince $\dot{\mathbf{p}}={\mathbf{\Omega }}_{\mathrm{cell}}\times \mathbf{p}$, the bottom-heavy spherical-cell orientation dynamics isThus a steady orientation satisfiesThe printed algebraic equation has the opposite sign on its right side, if vorticity means the usual $\nabla \times \mathbf{u}$. Its later concentration equation is consistent with the corrected sign. In particular, linearizing the corrected equation about $\mathbf{p}=\mathbf{k}$ gives ${\mathbf{p}}_{\perp }=(B/2)(\mathbf{\omega }\times \mathbf{k}{)}_{\perp }$, whereas the printed sign would reverse that response. Merely changing the definition of vorticity would also require changing the ensuing fluid and orientation calculations.

The gyrotactic reorientation time $B$ measures gravitational restoration against rotational viscous resistance. In still fluid a tilt angle obeys $\dot{\theta }=-\sin \theta /B$, soFor small tilt it is precisely the exponential relaxation time. Applying the algebraic orientation balance to a time-dependent disturbance additionally assumes that this relaxation is fast enough to neglect orientation lag. Keeping the full dynamic orientation equation would introduce an additional relaxation factor into the perturbation problem.

With zero fluid velocity and all mean orientations vertical, the steady cell conservation equation isThe plates are impermeable to cells, so the constant vertical flux is zero. Hence $D{\hat{n}}_{\hat{z}}=V_s\hat{n}$, giving $\hat{n}=C\exp (V_s\hat{z}/D)$. Determine $C$ by the prescribed depth average:With $z=\hat{z}/H$ and $h=V_{s}H/D$, the result isThe same normalized profile describes cell number density or cell volume fraction, because cell volume is constant. Its weak-stratification expansion iswhich is depth-normalized at each displayed order and will be used in part (c). Upward swimming makes concentration larger near the upper plate, creating the density stratification responsible for bioconvection when cells are heavier than the fluid.

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.

Set $\sigma =0$ under the assumed stationary-onset hypothesis, and let $\kappa =h{\kappa }^{\prime }$ with fixed nonzero ${\kappa }^{\prime }$. Substitute $W=h{W}_0+h^2{W}_1+\cdots$, $N=N_0+h{N}_1+h^2{N}_2+\cdots$, and $R_c=h^{-1}(R_0+h{R}_1+\cdots )$ into the modal equations and plate boundary conditions.

At leading order the concentration equation is $N_0^{\prime \prime }=0$, with $N_0^{\prime }=0$ at both plates. Normalize its nonzero constant to $N_0=1$. The first nonzero momentum equation isIntegrating four times and imposing all four clamped boundary conditions givesAt the next concentration order, $\bar{n}=1+O(h)$ and ${\bar{n}}^{\prime }=h+O(h^2)$, so the vertical advection of the basic concentration does not yet contribute. ThusIntegration gives $N_1=-G{W}_0+z+C_1$. The constant $C_1$ is an amplitude-normalization freedom; set it to zero, obtainingThere is no need to require zero depth average of this concentration eigenfunction: for a nonzero horizontal wavenumber the horizontal average is already zero.

Use $\bar{n}=1+h(z+1/2)+O(h^2)$ and ${\bar{n}}^{\prime }=h+O(h^2)$. At order $h^2$, the concentration equation isThe corresponding cell-flux boundary conditions are $N_2^{\prime }=N_1$ at both plates. Integrate the equation from $-1$ to zero. Its left derivative term satisfiesThe $N_1^{\prime }$ terms therefore cancel. The no-slip conditions imply $\int W_1^{\prime \prime }dz=0$. Also integration by parts givesConsequently the solvability condition is simplysince ${\int }_{-1}^0{z}^2(z+1{)}^{2}dz=1/30$. For every fixed nonzero ${\kappa }^{\prime }$, the long-wave bioconvection threshold in weak stratification is thereforeThe cancellation shows why $G$ does not enter this leading coefficient; neither does $S$ at stationary onset. In dimensional terms, the leading critical mean number density and volume fraction areThese assume heavier-than-fluid cells, $\Delta \rho >0$. Exactly ${\kappa }^{\prime }=0$ makes the integrated equation identically zero and corresponds to a spatially uniform change of conserved cell number, not a convective mode; the statement independent of ${\kappa }^{\prime }$ is interpreted for nonzero long waves or their limit. This is the leading threshold within the specified long-wave, real-eigenvalue assumptions. It does not by itself rule out a different finite-wavelength or oscillatory instability outside those assumptions.