Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 355 (Biological Physics and Fluid Dynamics)

In the Monge representation, $ds=dx+O(h_x^2)$ and the signed curvature is $\kappa =h_{xx}+O(h_x^2{h}_{xx})$. To quadratic order,Two integrations by parts giveThus a filament with free ends has zero bending moment and shear force,and the local Stokesian dynamics of an elastic filament isWith $X=x/L$, $H=h/L$, and $\tau =At/(\zeta L^4)$, the dimensionless problem isat $X=0,1$.

In the dark, energy minimization gives $h_{xx}=-H_0$. Translation and rotation are zero-energy freedoms; one representative is $h_{-}(x)=-H_0{x}^2/2$. After illumination, choose the equivalent light-adapted equilibriumThen $u=h-h_{+}$ has homogeneous free-end conditions $u_{xx}=u_{xxx}=0$, and its initial valueis orthogonal to the two rigid zero modes $1$ and $x$.

Let $q_n>0$ solve the free--free biharmonic eigenvalue equationand chooseThese are orthogonal eigenfunctions of the biharmonic operator with ${\varphi }_n^{\prime \prime }={\varphi }_n^{\prime \prime \prime }=0$ at both ends. The shape iswhereThe omitted zero modes would only translate or rotate the whole filament.

The transverse force density is $f=-A{h}_{xxxx}=\zeta h_t$. Its resultant isby the free-end shear conditions. To linear order, its torque about $x=0$ isbecause both end curvatures equal the same imposed ${\kappa }_0$. Thus every decaying mode is force-free and torque-free, consistently leaving the translational and rotational zero modes unchanged.

In resistive-force theory, the force per unit length is local and linear in the filament velocity relative to the fluid:The distinct parallel and perpendicular drag coefficients encode the anisotropic resistance of a slender filament. Neglecting nonlocal hydrodynamic interactions makes this a leading logarithmic approximation to slender-body theory.

Choose the handedness for which the helix tangent is $\mathbf{t}=\cos \theta {\mathbf{e}}_z+\sin \theta {\mathbf{e}}_{\varphi }$. Its local rigid velocity is $\mathbf{u}=\tilde{U}{\mathbf{e}}_z+R\tilde{\Omega }{\mathbf{e}}_{\varphi }$, so $\mathbf{u}\cdot \mathbf{t}=\tilde{U}\cos \theta +R\tilde{\Omega }\sin \theta$. Integrating the axial force and torque from the drag law gives the symmetric hydrodynamic resistance matrix in the question, withReversing helical handedness reverses the sign of $B$ but leaves $A$ and $D$ unchanged.

Force and torque freedom of the cell--helix pair giveWritingsolution of this linear system yieldsThe body counter-rotates relative to the motor, and the swimming direction reverses with the helix handedness through $B$.

The helix exerts on the fluid the torqueOverall torque balance also gives $N=-D_0\Omega$. HenceFor the usual choice $\omega >0$, this is positive when the resistance matrix is positive definite.

Substitution into $ϵ=A_0{U}^2/(N\omega )$ givesWhen $B^2\ll AD$ and $D\ll D_0$,After scaling body size so that $A_0\mapsto \lambda A_0$, differentiation with respect to $\lambda A_0$ shows that the optimum impedance match is $\lambda A_0=A$. Therefore

For a nearly planar membrane, the quadratic Helfrich energy of a Fourier mode isThe equipartition theorem and $A-A_{\mathrm{proj}}=\int |\nabla h{|}^2{d}^{2}x/2$ giveUsing the stated molecular and system-size cutoffs, $q_{\max }=\pi /a$ and $q_{\min }=\pi /\sqrt{A}$, gives

Subtracting the two fluctuation areas givesWhen both tensions are much smaller than ${\pi }^2{k}_c/a^2$, the molecular-scale factors cancel to leading order, leavingIncreasing tension suppresses thermal membrane undulations, releasing their hidden excess area into the tether.

Set ${\sigma }_0=0$. Equating the cylindrical tether area $2\pi rL$ to $A\Delta \alpha$ givesorBecause a cylinder has $2H=1/r$, the energy becomesThe two stationarity equations implyEliminating $r$ gives the implicit entropic membrane-tether force--extension relationAs tether length grows, fluctuation area is depleted, tension rises, the tether narrows, and the required force increases.

For a membrane connected to a reservoir at fixed tension, its energy per length is $\pi k_c/r+2\pi \sigma r-F$. Minimization instead givesThe reservoir supplies area without changing $\sigma$, so the force is independent of extension; the finite vesicle has an entropic, strain-stiffening response.