Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 81 / 3

Fix the sense of rotation by taking the fluid velocity to be $\mathbf{u}=\Omega (-x_3,0,x_1)$, with $\Omega >0$. Its vorticity is $-2\Omega {\mathbf{e}}_2$. In the convention where $B$ is the gyrotactic reorientation time appearing with aligning rate $1/B$, the bottom-heavy spherical-cell orientation dynamics isFor the stable orientation in the transverse plane, write $\mathbf{p}=(\sin \vartheta ,0,\cos \vartheta )$. The angle equation is $\dot{\vartheta }=-\sin \vartheta /B-\Omega$. If $B\Omega <1$, its stable fixed point satisfiesThe other planar fixed point has negative $p_3$ and is unstable. The positive-$p_3$ orientation also damps a small out-of-plane component because ${\dot{p}}_2=-p_3{p}_2/B$.

Once the orientation has relaxed to ${\mathbf{p}}_{\ast }$, the transverse trajectory obeysDefine the shifted axis byThen $(x_1-x_{1A},x_3-x_{3A})$ rotates rigidly with angular speed $\Omega$, so its squared length is constant. These are the gyrotactic circular paths in a rotating cylinder, andHere $\alpha$ is the acute angle between the plane through the two parallel axes and the horizontal plane; the directed displacement $OA$ in the chosen cross-section points to the lower-left. Reversing the sense of rotation reverses the horizontal displacement. Arbitrary initial orientations have an orientation transient before the exact circular trajectories; neglecting randomness alone does not remove that transient. The bulk result also assumes that the trajectory does not intersect the cylinder wall.

Use the rotation convention of part (a), so the orientation drift is $B^{-1}({\mathbf{e}}_3-p_3\mathbf{p})-\Omega {\mathbf{e}}_2\times \mathbf{p}$. The steady gyrotactic orientation Fokker-Planck equation on the unit sphere isLet $\eta =(B\Omega {)}^{-1}$, $\lambda =(B{D}_R{)}^{-1}$, and introduce the rotation generator $\mathcal{A}=({\mathbf{e}}_2\times \mathbf{p})\cdot {\nabla }_p$. Dividing by $\Omega$ givesThe leading isotropic density is $f_0=1/(4\pi )$. At first order, writing $f=f_0+\eta f_1+\cdots$, rotational diffusion makes no contribution because ${\Delta }_p{f}_0=0$. The tangential drift is the spherical gradient of $p_3$, whose divergence is ${\Delta }_p{p}_3=-2p_3$. ThereforeIn the polar convention of the paper, $p_1=\sin \theta \cos \varphi$, and $\mathcal{A}{p}_1=p_3$. Thus, in the specified first-harmonic form,The correction integrates to zero, preserving normalization. The first correction has no dependence on $\lambda$ because the isotropic leading density has zero spherical Laplacian; $\lambda$ enters the next-order equations.

Spherical symmetry gives $\int p_i{p}_{j}d\Omega =4\pi {\delta }_{ij}/3$. Consequently the rapid-rotation mean gyrotactic orientation isIt is this small horizontal swimming bias, rather than a mean vertical orientation at leading order, that enters the concentration equation of part (c).

Let $C=N_{0}n$ be the cell number density. The steady cell conservation equation with isotropic translational diffusion isUsing the rapid-rotation mean gyrotactic orientation, put $a_s=2V_s/(3B\Omega )$, so the swimming drift is $-a_s{\mathbf{e}}_1$. In the transverse plane, $x_1=Rr\cos {\theta }^{\prime }$ and $x_3=Rr\sin {\theta }^{\prime }$. Thus $\mathbf{u}\cdot \nabla =\Omega {\partial }_{{\theta }^{\prime }}$ andMultiplying the conservation equation by $-R^2/D$ giveswith ${\beta }^2=\Omega R^2/D$ and $\epsilon =a_{s}R/D=2V_{s}R/(3B\Omega D)$. The base fluid velocity is tangent to the wall. Zero normal cell flux there is $-a_{s}n\cos {\theta }^{\prime }-D{n}_r/R=0$, orNormalization requires the cross-sectional mean of $n$ to be one.

For $\epsilon \ll 1$, set $n=1+\epsilon \mathrm{Re}[a(r)e^{i{\theta }^{\prime }}]+O({\epsilon }^2)$. The drift term acting on the leading constant vanishes, and the first angular harmonic satisfiesWriting $q=e^{i\pi /4}\beta$ turns this into the Modified Bessel differential equation of order one. Its regular solution iswhere $I_1$ is the Modified Bessel function of the first kind. The singular second solution is excluded at the axis, and the displayed coefficient enforces the wall boundary condition. The angular harmonic has zero area average, so the normalization is satisfied at this order.

When $\beta \gg 1$, the gyrotactic concentration layer at a rotating-cylinder wall has stretched coordinate $s=\beta (1-r)$. The leading radial equation is $a_{ss}-ia=0$. The solution decaying into the interior is proportional to $e^{-e^{i\pi /4}s}$; the wall condition fixes its coefficient to $-1/q$. HenceThis also follows from the large-argument asymptotic expansion of $I_1$, which gives $a\sim -r^{-1/2}{e}^{-q(1-r)}/q$ near the wall; $r^{-1/2}=1+O({\beta }^{-1})$ within the layer. Its thickness is $R/\beta =\sqrt{D/\Omega }$. The quoted exponential is a wall-layer approximation, not an exact solution all the way to $r=0$; the Modified Bessel function of the first kind formula supplies the regular continuation through the axis.