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For a bottom-heavy spherical swimmer in axial flow $u_z=W_0(-1+r^2+w)$, the bottom-heavy spherical-cell orientation dynamics with aligning rate $1/B$ gives $\sin \theta =-\lambda (2r+w^{\prime })$, $\lambda =B{W}_0/(2R)$. On the upward-stable branch $\cos \theta >0$, the cells tilt inward in a downward Poiseuille flow. Zero radial cell flux with scalar diffusivity gives $n^{\prime }=\chi n\sin \theta$, $\chi =V_{c}R/D$. For $w=0$, integration and the average normalization $2{\int }_0^{1}nrdr=1$ give the displayed concentration. A steady deterministic orientation requires $|\lambda (2r+w^{\prime })|<1$; beyond this shear threshold the stationary orientation closure fails.

Let $\hat{u}(\hat{r})=-W_0(1-{\hat{r}}^2/R^2)+\hat{w}(\hat{r})$ and $\mathbf{p}=\sin \theta {\mathbf{e}}_r+\cos \theta \mathbf{k}$, with $\cos \theta >0$ for the upward-stable orientation. Cell conservation in a swimming suspension gives radial fluxRegularity at the axis and impermeability at the wall imply ${\hat{j}}_r=0$, hence $D{\hat{n}}^{\prime }=V_C\hat{n}\sin \theta$. Axial advection has no divergence in this fully developed, axially uniform state.

The fluid is incompressible. Use dynamic stresslet strength $S_d$ and dynamic viscosity ${\mu }_d$ first, so the swimming contribution to its stress is $S_d\hat{n}(\mathbf{pp}-I/3)$. The isotropic part can be absorbed into the effective pressure. With the ordinary fluid hydrostatic pressure removed, the axial momentum equation isThe final term is the downward excess weight of the cells. Sedimentation is neglected in their swimming flux, not in this buoyancy force. Fluid convective inertia vanishes for the specified fully developed velocity even before a low-Reynolds-number approximation is invoked.

There is a dimensional convention to make explicit. The printed definition of $\gamma$ contains the extra fluid density $\rho$. All the printed coefficients are consistent if this question uses density-normalized pressure, kinematic viscosity and stresslet strength:Then ${\hat{p}}^{\prime }=4\mu W_0/R^2$ generates the stated background Poiseuille flow. If $\mu$ and $S$ instead mean dynamic viscosity and a physical force-dipole strength, the correct buoyancy parameter is $g\Delta \rho \nu N_0{R}^2/({\mu }_d{W}_0)$, without a further density denominator. The following derivation retains the consistent density-normalized interpretation of the printed parameters.

The bottom-heavy spherical-cell orientation dynamics isIts scalar tilt equation is $\dot{\theta }=-\sin \theta /B-{\hat{u}}^{\prime }/2$. Steady gravitational-viscous torque balance therefore gives $\sin \theta =-B{\hat{u}}^{\prime }/2$. This equilibrium is stable on the $\cos \theta >0$ branch and exists only when $|B{\hat{u}}^{\prime }/2|<1$.

Put $r=\hat{r}/R$, $w=\hat{w}/W_0$, $n=\hat{n}/N_0$. Subtract the background pressure-driven balance and divide by $\mu W_0/R^2$. The governing system iswhereThe last parameter contains $B$, as in the PDF, not the diffusivity substituted by the converted TeX. Boundary and normalization conditions areThe wall cell-flux condition is already incorporated in $n^{\prime }=\chi n\sin \theta$; there is no prescribed wall concentration or additional fixed-volume-flux condition. The radial divergence of the stresslet stress can be balanced by a radial pressure depending only on $r$. It neither produces radial flow in this ansatz nor changes the imposed axial pressure gradient, so it is eliminated from the axial problem, not assumed to vanish.

For $\gamma \ll 1$, $\sigma =\gamma {\sigma }_1$, let $w=\gamma w_1+O({\gamma }^2)$, $n=n_0+\gamma n_1+O({\gamma }^2)$. The leading tilt hasThe supplied $B{W}_0/R<1$ ensures this leading orientation remains on the stable branch. Put $a=\lambda \chi$. Integrating the concentration equation and imposing the average concentration gives the gyrotactic focusing in a downward pipe flow profileAt $a=0$ the continuous limit is $n_0=1$.

At first order the axial equation isIntegrate from the regular axis, where both boundary terms vanish:Since $\sin 2{\theta }_0=-4\lambda r\sqrt{1-4{\lambda }^2{r}^2}$, the stresslet correction to gyrotactic pipe flow iswith its regular limiting value at $r=0$, andThe orientation and concentration corrections, if needed to complete the expansion, follow without another differential boundary problem:These enforce $2{\int }_0^1{n}_{1}rdr=0$.

For $\lambda >0$, write the two positive integrand coefficients as $F(s)=A(1-e^{-a{s}^2})/(2as)$ and $H(s)=4\lambda s{n}_0(s)\sqrt{1-4{\lambda }^2{s}^2}$. ThenThus $w_1<0$ throughout the interior for ${\sigma }_1\ge 0$, and at any given radius a positive correction is possible only ifA simple sufficient condition for negativity is ${\sigma }_1\ge -1/(8\lambda )$: indeedWhen $\lambda =0$, the stresslet contribution vanishes and the buoyancy correction is negative regardless of ${\sigma }_1$.

Physically, downward shear tilts upward swimmers toward the axis, concentrating excess weight there and enhancing the downward flow. Positive $S$ in the specified $+Sn\mathbf{pp}$ stress convention reinforces this effect; sufficiently negative $S$ opposes it. A positive $w$ means reduced downward velocity, not necessarily reversal of the total pipe flow. The deterministic orientation closure fails at excessive shear, and scalar diffusivity neglects directional and shear-dependent dispersion.

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.

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.

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.

Close to the upward pole, the two Cartesian tilt components of bottom-heavy spherical-cell orientation dynamics obey the displayed Ornstein-Uhlenbeck process. Thus $d⟨|\mathbf{q}{|}^2⟩/dt=-2⟨|\mathbf{q}{|}^2⟩/B+4D_r$. The stationary spread is $2B{D}_r$, and late axisymmetric variance perturbations relax on $B/2$ in this weak-noise approximation. The zero-noise mean from deterministic alignment of an initially isotropic orientation distribution has the same exponential scale but a different transient prefactor. Reaching a tiny diffusion-supported mean-deficit floor $B{D}_r$ from an initially isotropic population takes order $B\log [(B{D}_r{)}^{-1}]$, with logarithmic corrections. Equality of limiting mean directions does not make the full transients identical, and the alignment time is not automatically the pure rotational-diffusion time $D_r^{-1}$.