Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 77 / 2

In resistive-force theory, the flagellum is slender and moves at a small Reynolds number, so local viscous resistance is linear in its velocity and fluid and body inertia are neglected. Normal and tangential resistance coefficients are treated as positive constants, usually ${\kappa }_N>{\kappa }_T$. The approximation neglects distant hydrodynamic interactions along the filament, head-tail interactions, nearby boundaries and self-contact. The centreline is an inextensible filament with prescribed shape. Its local resistance tensor isSince the filament velocity relative to distant fluid is $-\mathbf{w}$, the fluid force on a filament element is $K\mathbf{w}ds$. The equal and opposite force on the fluid has total axial component $T$. Projecting onto $\mathbf{i}$ givesThis establishes the thrust expression with its force-on-fluid convention.

For a uniform travelling wave, write the moving centreline as $\mathbf{r}(s,t)={\mathbf{r}}_{\mathrm{wave}}(s-ct)+(V-U)t\mathbf{i}$, where $s$ is material arc length and $|{\mathbf{r}}_{\mathrm{wave}}^{\prime }|=1$. Differentiating at fixed material label gives ${\mathbf{r}}_t=(V-U)\mathbf{i}-c\mathbf{t}$. HenceLet $q=\mathbf{t}\cdot \mathbf{i}=X_s$, $\alpha =⟨q⟩$ and $\beta =⟨q^2⟩$, where these brackets are contour-length averages. The axial wavelength divided by contour wavelength is $\alpha =V/c$. For a filament comprising whole wave periods, or after the usual period averaging, $c{\int }_0^{L}qds=VL$. This averaging is needed to avoid end-phase corrections for an arbitrary fraction of a wavelength. Substitution of $\mathbf{w}$ into the thrust expression now givesFor $\gamma ={\kappa }_T/{\kappa }_N$ and $r=1-\gamma$, this is $T/({\kappa }_{N}L)=Vr(1-\beta )-U(1-r\beta )$. The swimming body is force-free, so the reaction thrust on the head-tail assembly balances its head drag. Thus $T=\delta {\kappa }_{N}LU$, and linear resistive-force propulsion of a travelling filament givesWe henceforth assume $0<\gamma <1$, $\delta \ge 0$ and $0<\beta <1$. Isotropic resistance, $\gamma =1$, gives no propulsion in this model.

The flagellar contribution to viscous dissipation is the positive quadratic form $P_f=\int \mathbf{w}\cdot K\mathbf{w}ds$. Expanding it, and using $c⟨q⟩=V$, givesThere is also head dissipation $\delta {\kappa }_{N}L{U}^2$. Equivalently, the work of the flagellar deformation relative to the translating head is $P_f+UT$; the force balance makes $UT$ precisely the head dissipation. Consequently, with $\alpha =V/c$,This derives the rate of working, rather than merely inferring it from the thrust.

The Cauchy-Schwarz inequality gives ${\alpha }^2=⟨q{⟩}^2\le ⟨q^2⟩=\beta$. Equality requires $q$ to be constant almost everywhere. For a planar periodic filament this is an ideal sawtooth with alternating transverse tangent and fixed axial tangent, or a limit with negligibly rounded corners. It is not an identity for a general sinusoidal wave, and the ideal calculation omits the elastic cost of sharp corners.

For the stipulated equality ${\alpha }^2=\beta$, put $d=1+\delta$ and eliminate $V$ at fixed $U$ using $V/U=(d-r\beta )/[r(1-\beta )]$. Direct simplification yieldsThe derivative isWriting $A=r({\delta }^2-\gamma )$ and $B=\gamma d^2$, the numerator is $A{\beta }^2+2B\beta -B$. It is negative at zero and equals $(\delta +\gamma {)}^2>0$ at one. Its derivative is positive throughout $(0,1)$: this is immediate if $A\ge 0$; if $A<0$, its minimum derivative is $2(A+B)=2(\delta +\gamma {)}^2>0$. Since $F$ diverges at both endpoints, the unique stationary point is the global fixed-speed energy optimum for a planar flagellum:This form includes the case $A=0$, when ${\beta }_{\ast }=1/2$, without a removable division by zero. With no head drag it reduces to ${\beta }_{\ast }=1/(1+\sqrt{\gamma })$.

The printed final optimality formula does not follow from the preceding thrust and power equations. For a concrete check, $\gamma =1/2$, $\delta =0$ gives $F=(2-\beta )/[\beta (1-\beta )]$. The derived minimum is at $\beta =2-\sqrt{2}$. The printed formula instead gives ${\beta }^2=\sqrt{6}-2$, hence $\beta \approx 0.67044$, where $F^{\prime }\ne 0$ and the expenditure is greater. Thus the thrust and working formulas can both be established as above, but the last printed conclusion must be replaced by the derived optimum.