Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 81 / 4

Let $f$ be probability density per unit solid angle, with ${\int }_{S^2}fd\Omega =1$ and isotropic initial condition $f(\mathbf{p},0)=1/(4\pi )$. In still fluid, bottom-heavy spherical-cell orientation dynamics gives $\dot{\mathbf{p}}=B^{-1}[\mathbf{k}-(\mathbf{k}\cdot \mathbf{p})\mathbf{p}]$. Adding isotropic rotational diffusion gives the gyrotactic orientation Fokker-Planck equationThe derivatives are tangential to the unit sphere. Rotational symmetry about $\mathbf{k}$ is preserved by both the equation and the initial condition, so $f$ depends only on $\theta ,t$. Since $\dot{\theta }=-\sin \theta /B$, the polar probability flux is $j_{\theta }=-\sin \theta f/B-D_r{f}_{\theta }$. Expanding ${\partial }_{t}f+{\sin }^{-1}\theta {\partial }_{\theta }(\sin \theta j_{\theta })=0$ yieldsRegular solutions have no probability flux through the poles. The mean swimming velocity is ${\mathbf{V}}_c=V_s⟨\mathbf{p}⟩$, with only a vertical component by symmetry.

For $D_r>0$, the steady polar flux satisfies ${\partial }_{\theta }(\sin \theta j_{\theta })=0$. Regularity at both poles makes its integration constant zero. ThusNormalize with $x=\cos \theta$:The zero-flow steady gyrotactic orientation distribution is thereforeDifferentiating the logarithm of its normalization integral gives the Langevin function meanAs $\Lambda \to 0$, $f_0\to 1/(4\pi )$ and the mean tends to zero; as $\Lambda \to \infty$, the distribution concentrates at the upward pole and the mean tends to $V_s\mathbf{k}$. At zero diffusivity the pole-concentrated steady state is a probability measure, not a regular density of this formula.

With $x=\cos \theta$ and $D_r=0$, the Fokker-Planck equation becomes the probability-transport equationIts characteristic obeys $\dot{x}=(1-x^2)/B$. Writing $s=t/B$ and $q=\tanh s$ givesConservation of orientation probability pushes the initially uniform density forward by this Jacobian. ThusSince $q=(1-\alpha )/(1+\alpha )$ with $\alpha =e^{-2t/B}$, the deterministic alignment of an initially isotropic orientation distribution isAt $t=0$, $\alpha =1$ and $f=1/(4\pi )$; the characteristic Jacobian preserves $2\pi {\int }_{-1}^{1}fdx=1$ at every finite time. Its reciprocal is quadratic in $x$, as in the suggested substitution.

To compute the mean, integrate explicitly:ConsequentlyThe apparent singularity at $t=0$ is removable: $⟨x⟩=2t/(3B)+O((t/B{)}^3)$, so the initial mean is zero. At late times,Except for the exactly downward initial orientation, which has zero probability in the initial smooth distribution, every characteristic aligns upward. The infinite-time orientation distribution is a point mass at the upward pole.

From part (i), the steady mean has $\coth \Lambda -1/\Lambda =1-1/\Lambda +O(e^{-2\Lambda })$. Hence $B{D}_r={\Lambda }^{-1}\to 0$ gives ${\mathbf{V}}_c\to V_s\mathbf{k}$. Part (ii) gives the same mean as $t\to \infty$. ThusBoth limiting distributions concentrate weakly at the upward pole. A common limiting mean does not identify their finite-time distributions or the order in which the limits are taken.

Time scales must be compared at a specified accuracy. The exact deterministic solution approaches the limiting mean with deficit $(4t/B-2)e^{-2t/B}$: its alignment scale is $B$, with exponential late scale $B/2$. For weak but nonzero diffusion, linearize about the upward direction using the two transverse tilt components $\mathbf{q}$. They obey the Ornstein-Uhlenbeck processThe stationary angular variance is $2B{D}_r$, and nearby axisymmetric variance perturbations decay on $B/2$ to leading weak-noise order. This is weak-noise relaxation of gyrotactic alignment: the aligning torque, not the pure rotational-diffusion time $D_r^{-1}$, controls the weak-noise alignment rate.

Nevertheless the full relaxation times are not identical. Finite diffusion leaves mean deficit approximately $B{D}_r$, whereas deterministic alignment has no such floor. Starting from isotropy, reducing the transient deficit to that tiny floor requires $(4t/B-2)e^{-2t/B}\sim B{D}_r$, orwith smaller logarithmic corrections. The two limits have the same mean and the same order-$B$ weak-noise alignment mechanism, but not identical transients or accuracy-dependent settling times. No finite time achieves the exact deterministic point-mass limit; positive diffusion never gives that exact point mass at all.