Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 81 / 2 / Solution

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.