Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 81 (The Fluid Dynamics of Swimming Organisms)

The reaction sign and the meaning of recoil translation matter here. In Lighthill elongated-body theory, put $\mathcal{D}={\partial }_t+U{\partial }_x$. The transverse momentum imparted to the fluid per unit length is $m\mathcal{D}h$, so the lateral force on the fish is $-\mathcal{D}(m\mathcal{D}h)$. Thus body mass and positive added mass enter effective inertia with the same sign. The periodic trailing-edge thrust, to quadratic order in slope, isThe reaction sign and this small-amplitude thrust formula can also be checked in the primary derivation of the elongated-body force, equations (2.6), (3.2) and the discussion after (3.5). For the long-body limit below, the recoiling leading edge contributes only at relative order ${\alpha }^{-2}$.

The requested recoil and thrust formulas cannot all hold for the literal fish described. In particular, the printed thrust would remain $-m{U}^2/4$ when $h_1=0$, although a straight, unperturbed fish has $h_t=h_x=0$ and zero reactive thrust. Its negative term is missing an amplitude-squared slope factor. The stated effective translational coefficient $M^{\prime }-1$ also has the wrong sign for positive added mass, and the literal centre of mass translation cannot be equated to the translation of an arbitrarily chosen undeformed body reference line. The PDF defines $I^{\prime }=I/[m(\alpha L{)}^3]$; the converted TeX incorrectly uses the square.

Here are the force balance and angular momentum calculations for a consistent interpretation of the stated uniform fish. They give a corrected recoil solution, as well as a direct test of the printed equations. Defineand let $R$ be actual centre of mass displacement. The prescribed deformation has mean $a(t)=h_{1}f(t)/[2(\alpha +1)]$. Its mass-weighted mean must be removed in the recoil correction in elongated-body theory:Indeed $\int hdx=\ell R$, since $\int \xi dx=0$ and $\int h_{0}dx=\ell a$. Introduce the two geometric integralsFor uniform mass per unit length, $I={\mu }_{b}J$ at leading small-amplitude order. Retaining the symbol $I$ makes the rotational momentum balance transparent.

The needed integrals of transverse acceleration and convective derivatives areThe second spatial derivative includes the hinge's slope jump, as a distributional derivative; discarding it would incorrectly remove the $U^2$ terms. Newton's lateral equation is $M\ddot{R}=-m\int {\mathcal{D}}^{2}hdx$. The angular momentum of the deforming fish is $I\dot{\theta }+{\mu }_{b}C\dot{f}$, so its time derivative equals $-m\int \xi {\mathcal{D}}^{2}hdx$. Consequently the consistent uniform-added-mass recoil equations areThese follow directly from the local Lighthill elongated-body theory model with the physical reaction sign. An alternative reference-line translation replaces $R$ by $R-a$, and changes the shape forcing, but cannot change $M+m\ell$ to a difference of two positive inertias. Leading-edge details of a uniform-depth idealization can require an end model; they do not alter this reaction-sign test or the zero-stroke counterexample.

For $f=\cos \omega t$, write $R=\mathrm{Re}(r{e}^{i\omega t})$ and $\theta =\mathrm{Re}(\tau e^{i\omega t})$, choosing the periodic response and omitting free rigid-motion constants. The two balances give the explicit recoil amplitudesAt the tail the displacement and slope amplitudes are $H_L=h_1+r+\ell \tau /2-h_1/[2(\alpha +1)]$ and $S_L=h_1/L+\tau$. Thus a corrected, directly calculable mean thrust iswithin the trailing-edge approximation, with any retained leading-edge contribution treated consistently. This expression has the correct dimensions and vanishes quadratically with stroke amplitude.

A particularly simple independent check is the limit $U=0$. For the uniform fish, $I={\mu }_{b}J$, the balances give $\ddot{R}=0$ and $\theta =-Cf/J$ for the periodic response. The tail displacement amplitude is then exactlyHence, to the requested relative order,The leading-edge contribution is only $O({\alpha }^{-2})$ here. This recoil reduction contradicts the printed positive correction, for example for $M^{\prime }=2$, $I^{\prime }=1/6$. It also shows explicitly why treating the prescribed shape's moving mass as stationary would give a different model.

For completeness, the positive correction quoted in the question has a simple conditional origin if its two displayed recoil equations are accepted as formal equations rather than derived physical balances. Put $J^{\prime }=I^{\prime }+1/6$ and keep $U/(\omega L)$ fixed as $\alpha \to \infty$, away from $M^{\prime }=1$. Adding those equations and then using the first gives their leading periodic responseUnder the additional reference-line interpretation $h=h_0+R+\xi \theta$, the in-phase tail amplitude is $h_1[1+1/(2\alpha (M^{\prime }-1))+1/(8\alpha J^{\prime })]$. Squaring it in the trailing-edge thrust gives the printed frequency correction. Even this formal route, however, gives the amplitude-consistent resultnot a term $-m{U}^2/4$ independent of $h_1$. The quadrature part of $R$ first changes the squared displacement at relative order ${\alpha }^{-2}$, while recoil changes the tail slope at that order as well. This conditional calculation explains the intended algebra without presenting the flawed starting equations as a physical theorem.

Reactive thrust is positive when the tail's lateral-motion term exceeds its streamwise slope penalty; for an unrecoiling hinged tail this requires $\omega L>U$. A real steady swimming speed additionally balances the thrust against resistive drag force, absent from this inertial idealization. Recoil can alter either term and is not universally thrust-enhancing. A pole at $M^{\prime }=1$ is not a physical negative-added-mass resonance: the genuine combined inertias remain positive.

Write $\omega =kc$, $q=k(1+i)$, and $E=e^{qL}$. The local resistive-force theory force per unit length on the filament iswhere $\mathbf{t}=(X_s,Y_s)$ is the unit tangent and $\mathbf{v}=(X_t-U,Y_t-V)$ the material velocity relative to the fluid. Inextensibility gives $X_s=(1-Y_s^2{)}^{1/2}=1-Y_s^2/2+\cdots$. With $X(0,t)=0$, $X_t=-{\int }_0^s{Y}_s{Y}_{st}d{s}^{\prime }+\cdots$ is second order and has zero time average for a periodic prescribed stroke. The actual small-slope condition is $k|H|e^{kL}\ll 1$ for fixed $kL$; $kH\ll 1$ alone is not uniform when $kL$ grows.

At first order, the transverse force is $f_y=-\mu K_N(Y_t-V)$. Adding the spherical-head drag force $\mu K_{N}L\delta V$ and imposing zero total force givesEquivalently $V(t)=\mathrm{Re}[-i\omega H(E-1)e^{-i\omega t}/(qL(1+\delta ))]$. This form is useful for the subsequent time averages.

At second order, the axial resistive-force theory force isAdding the head drag force gives the head-drag correction to planar flagellar propulsion:since $\overline{X_t}=0$. Directly, $Y_s=kH{e}^{ks}(\cos \vartheta -\sin \vartheta )$ and $Y_t=kcH{e}^{ks}\sin \vartheta$, with $\vartheta =k(s-ct)$. HenceAlso ${\int }_0^L{Y}_{s}ds=Y(L,t)-Y(0,t)$, whose complex amplitude is $H(E-1)$. Using $\overline{\mathrm{Re}(a{e}^{-i\omega t})\mathrm{Re}(b{e}^{-i\omega t})}=\mathrm{Re}(a\overline{b})/2$, together with $\mathrm{Re}(-i\omega /q)=-c/2$, givesAs $|E-1{|}^2=e^{2kL}-2e^{kL}\cos kL+1$, the requested mean speed follows:Anisotropic resistance is essential: $\gamma =1$ removes propulsion at this order. The mean motion is opposite to the direction of wave propagation when the bracket is positive, since the head's axial velocity is $-U$.

For the basal bending moment of a planar flagellum, take positive moment to be the positive out-of-plane torque exerted on the distal filament at its base. The leading hydrodynamic torque on that filament is $-\mu K_N{\int }_0^{L}s(Y_t-V)ds$, so balance requiresAn explicit unsimplified time-dependent answer isThe first term follows from ${\int }_0^{L}s{e}^{qs}ds=[E(qL-1)+1]/q^2$. This is the total internal bending moment required by the hydrodynamic load. It equals the contractile moment when passive elasticity is neglected. If a specified passive moment $M_{\mathrm{pass}}$ also acts, the required active moment is $M_0-M_{\mathrm{pass}}$; no elastic modulus is supplied here.

The prescribed base coordinate needs a geometrical qualification. Literally, a coordinate measured from the attachment point must have $Y(0,t)=0$, whereas the printed waveform has $Y(0,t)=H\cos \omega t$. The preceding calculation reproduces the printed speed by treating that waveform as the prescribed coordinate, including its moving transverse offset. If the intention is instead a fixed attachment at the origin, replace it by $\tilde{Y}=H[e^{ks}\cos k(s-ct)-\cos \omega t]$. ThenThese follow by replacing $Y_t$ with $Y_t-Y_t(0,t)$ in the same force balance. The basal moment likewise uses ${\tilde{Y}}_t$ and $\tilde{V}$. Thus the printed kinematics and its claim that the origin is the actual attachment cannot both be imposed without this correction.

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.

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.