Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 81 (The Fluid Dynamics of Swimming Organisms)

Take $x$ tailward, so forward swimming has laboratory velocity $-U\mathbf{i}$. For an inextensible travelling waveform, material motion in the swimmer frame is $V\mathbf{i}-c\mathbf{t}$: a material point passes through the wave along its tangent while the wave pattern translates at $V$. Thus the relative velocity of otherwise stationary fluid isPut $q=V-U$ and choose $\mathbf{t}=(X_s,Y_s)$, $\mathbf{n}=(-Y_s,X_s)$, with $X_s^2+Y_s^2=1$. ThenOn the forward-swimming branch $0<U<V<c$ and $X_s>0$, the tangential slip is positive. A law for reversed tangential slip would also need its sign, not just its absolute magnitude.

The quadratic normal law represents inertial cross-flow drag: dynamic pressure scales as fluid density times $w_n^2$, while $|w_n|w_n$ chooses the force direction. Its coefficient contains the cross-sectional size and an empirical drag coefficient. The $|w_t{|}^{3/2}$ exponent is characteristic of an attached laminar boundary layer: its thickness scales as $\delta \sim (\nu s/|w_t|{)}^{1/2}$ and its shear as $\mu |w_t|/\delta$. However, this gives local shear proportional to $|w_t{|}^{3/2}{s}^{-1/2}$, not $s^{1/2}$. The positive power of $s$ is genuinely printed in this paper and repeated in its target formulas. Below it is treated as the prescribed phenomenological weight, rather than misidentified as standard local laminar skin-friction scaling along a slender swimmer.

The normal axial force on an element isThe prescribed tangential force gives $d{F}_{T,x}=K_T(c-q{X}_s{)}^{3/2}{X}_s{s}^{1/2}ds$. Zero mean axial force therefore gives the nonlinear resistive-force balance for a travelling filamentThe absolute values are present in the original PDF. In a periodically beating, constant-mean-speed model, phase-average the integrals where necessary; the leading approximation below is phase independent. Local force laws neglect nonlocal hydrodynamic interactions and assume nearly steady local response.

For the sinusoidal displacement, let $\epsilon =\beta k\ll 1$. Then $Y_s=\epsilon \cos ks$ and $X_s=(1-Y_s^2{)}^{1/2}=1+O({\epsilon }^2)$. Over one wavelength,Replace $X_s$ by one in the leading tangential term and balance the forces. After multiplication by $3/2$,This replacement also requires $c{\epsilon }^2\ll c-V+U$ for a uniformly small relative error inside the fractional power. If the tangential slip is itself $O(c{\epsilon }^2)$, its variation along the body must be retained. Periodicity gives $V/c=⟨X_s⟩=1-{\epsilon }^2/4+O({\epsilon }^4)$, so small geometric slope alone does not guarantee that additional slip condition.

For Lighthill elongated-body theory, put $h(x,t)\simeq \beta \sin [k(x-Vt)]$ and $w_{\perp }=h_t+U{h}_x$. The transverse fluid kinetic energy per unit length is $M_A{w}_{\perp }^2/2$. In the usual tail-flux approximation, with negligible leading-end contribution, the lateral work flux is $M_{A}U⟨h_t{w}_{\perp }⟩$, whereas kinetic energy carried into the wake is $M_{A}U⟨w_{\perp }^2⟩/2$. Their difference is the useful propulsive power $TU$. ConsequentlyBalancing this against the drag of the printed tangential model, $D=(2/3)K_T{L}^{3/2}(c-V+U{)}^{3/2}$, givesFor a genuine local laminar skin-friction estimate, write $dD=C_T{u}_t^{3/2}{s}^{-1/2}ds$, where $C_T$ has the appropriate different dimensions. Its integrated drag is $2C_T{u}_t^{3/2}\sqrt{L}$, and the corresponding elongated-body balance is instead $M_A{\beta }^2{k}^2(V^2-U^2)/8=C_T\sqrt{L}(c-V+U{)}^{3/2}$. The printed $L^{3/2}$ expression is therefore a result of its stated force model, not of that standard boundary-layer law.

The two thrust models describe different mechanisms: cross-flow resistive-force theory uses dissipative local drag, while elongated-body theory uses reactive acceleration of added mass. Both suppress details of the wake, body-body interactions, finite slenderness, skin-friction unsteadiness, separation and possible turbulent transition. The tail-only reactive result also requires a negligible leading-end flux, usually supplied by a nose taper or small leading-end motion; a blunt constant-cross-section body waving equally at both ends cannot automatically omit its leading-end contribution. Speed fluctuations and recoil may matter, and a planar constant-amplitude waveform is an idealization of a real sea snake.

Choose the helix handedness so its unit tangent is $\mathbf{t}=\cos \alpha \mathbf{i}+\sin \alpha {\mathbf{e}}_{\varphi }$. Let $q={\Omega }_0-\Omega$; the body rotates at $+\Omega \mathbf{i}$, while the flagellum rotates at $-q\mathbf{i}$ in the laboratory, since its relative motor rotation is $-{\Omega }_0\mathbf{i}$. Both translate at $-U\mathbf{i}$. Thus the relative fluid velocity on a flagellar element is $\mathbf{w}=U\mathbf{i}+bq{\mathbf{e}}_{\varphi }$.

Low-Reynolds-number resistive-force theory gives force per unit length on the flagellumIts axial and azimuthal components areDefine the integrated axial resistance matrix of a slender helix coefficientsThe sphere has translational resistance $A_b=6\pi \mu a$ by the Stokes drag law and rotational resistance $H_b=8\pi \mu a^3$. Its hydrodynamic force is $+A_{b}U\mathbf{i}$ and its torque $-H_b\Omega \mathbf{i}$. Integration of $f_x$ and $b{f}_{\varphi }$ gives the zero total force and torque conditionsDividing the first by $L$ and expanding the coefficients gives exactly the two requested balances. The internal motor torque cancels in the total torque balance; it is not an external applied torque.

Eliminate $U$ and use $q+\Omega ={\Omega }_0$. The effective flagellar rotational resistance is $H_e=H-C^2/(A+A_b)$, soThe determinant of helical resistance in resistive-force theory is $AH-C^2=b^2{L}^2{K}_N{K}_T>0$, hence the denominator is positive. Opposite handedness changes the sign of the translation-rotation coupling and reverses the swimming direction.

For $K_T=K$, $K_N=2K$, $\alpha =\pi /4$ and $a=b$, the coefficients are $A=3KL/2$, $C=KaL/2$, $H=3K{a}^{2}L/2$. ThereforeEquivalently, with $h=KL/(\pi \mu a)$, $U=a{\Omega }_{0}4h/(2h^2+21h+48)$. These are positive speed magnitudes for the selected handedness; the translational velocity is $-U\mathbf{i}$.

This helical microswimmer with a spherical head approximation neglects head-tail hydrodynamic interactions as permitted, treats the flagellum as slender and rigid, and averages away transverse forces associated with finite helical ends. The flow is quasistatic Stokes flow, so neither body inertia nor fluid inertia enters the balance.

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.

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.