Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 355 (Biological Physics and Fluid Dynamics)

The higher-order Euler-Lagrange equation for a functional depending on $h_x$ and $h_{xx}$ givesWith the membrane elastic lengththe general free-membrane profile isTwo integrations by parts, followed by use of the field equation, turn the energy of any free interval $[a,b]$ into the boundary term

Take the undeformed membrane to have $h,h_x\to 0$ at infinity. In the small-contact approximation, the cylindrical profile isContinuity of height and slope at the right contact point and exponential decay then givewith its reflected copy on the left. The two free tails carryWithin the contact, $h_{xx}=-1/R$. Since ${\delta }_o\ll \xi$, the tension contribution there is smaller than the bending contribution by $O({\delta }_o^2/{\xi }^2)$, soWriting the dimensionless adhesion strength as $U=R^2\mathcal{U}/k_c$, the total energy isIts stationary point isIt is an admissible bound state only for $U>1/2$; otherwise the constrained minimum is the unbound state ${\delta }_o=0$. Substitution gives

Reflection symmetry makes the two outer free tails identical. Their combined energy isBetween the cylinders the free profile is even. Up to an irrelevant additive height it has the formSlope matching at $x=L-{\delta }_i$ givesDirect integration, or the boundary expression from part (a), yieldsAt the retained order ${\delta }_i/\xi \ll 1$, replace the argument by $L/\xi$. The four contact halves contribute bending plus adhesion energyThusIndependent minimization givesand henceAt infinite separation the energy is twice the one-cylinder minimum. The membrane-mediated interaction potential is thereforeIt is positive and decreases monotonically to zero, so the two cylinders repel. The physical cause is the overlap of their exponentially relaxing membrane deformations.

For a cylinder of radius $r$ and length $L$, the mean curvature is $H=1/(2r)$ and the area is $2\pi rL$. Its Helfrich membrane energy per unit length is thereforeThe first term favors a narrow tube and the second penalizes its curvature. Setting the derivative with respect to $r$ to zero givesThe positive second derivative confirms a minimum.

In the high-tension, narrow-gap regime, the almost spherical membrane has area $4\pi (R+\delta {)}^2$. Its spherical bending energy $8\pi k_c$ is independent of radius, while Helfrich repulsion over area $4\pi R^2$ costsUp to terms independent of $\delta$, a suitable free energy is consequentlyIts stationary point satisfiesUsing $\delta /R\ll 1$ gives the fluctuation-supported membrane--particle gapThe inverse-square entropic repulsion prevents contact, while membrane tension limits the area gained by opening the gap.

Let $z$ measure distance across the narrow gap and let $\theta$ be polar angle about the tube axis. In lubrication theory, radial velocity and radial pressure variation are negligible. Axisymmetric incompressible flow is obtained frombecause ${\partial }_{\theta }(u_{\theta }\sin \theta )=0$. The tangential Stokes equation then separates:and hence, after choosing an irrelevant pressure constant,

In the sphere frame, $u_z=-u_{\theta }\sin \theta =-\varphi$. The no-slip boundary condition gives $\varphi (0)=0$ on the sphere and $\varphi (\delta )=U$ on the membrane translating backward relative to it. The sphere-frame volume flux inherited from the narrow remote tube is $-\pi r_0^{2}U$, soUnder the asymptotic condition $r_0^2\ll R\delta$, the right-hand side is negligible at leading order. Solving the quadratic profile subject to the two wall values and zero leading-order integral givesChanging the chosen positive tube direction reverses both signs but leaves the drag magnitude unchanged.

The pressure scale is $p\sim \mu UR/{\delta }^2$, whereas the viscous shear scale is $\tau \sim \mu U/\delta$. After multiplication by comparable areas, pressure drag exceeds shear drag by $R/\delta \gg 1$, an instance of lubrication pressure dominates shear stress. Put $c=\cos \Delta \theta$. The axial pressure force isAs $\Delta \theta \to 0$, the bracket tends to $-2$, and thereforeThe resulting confined-sphere drag coefficient isThus it exceeds the free Stokes drag law coefficient by $2R^2/{\delta }^2$. The Stokes–Einstein relation then gives

Let the unit normal $\mathbf{n}$ point from the fluid toward the free surface, let $\mathbf{t}$ lie along the surface, and write the swimmer direction asReflecting the swimmer position across the plane and reflecting its orientation toconstructs the free-surface image of a force dipole. At the surface the two dipoles have equal tangential velocity and opposite normal velocity. Their sum therefore satisfies the no-penetration boundary condition; tangential velocity is even across the plane and normal velocity is odd, so the tangential traction vanishes and the stress-free boundary condition is also satisfied.

The vector from the image to the swimmer is $\mathbf{r}=-2h\mathbf{n}$, for which $({\mathbf{p}}^{\ast }\cdot \mathbf{r}{)}^2=4h^2{\sin }^2\theta$. Evaluating the image force-dipole flow at the swimmer givesThere is no tangential image velocity at the swimmer in this point-dipole approximation. With the convention that $\mathcal{P}>0$ is an extensile pusher microswimmer, a nearly parallel pusher is attracted toward the surface, whereas a nearly parallel contractile puller microswimmer is repelled. For $|\sin \theta |>1/\sqrt{3}$ the normal drift reverses because the image samples the axial rather than equatorial part of the dipolar flow.

For a fieldonly the gradient of the scalar prefactor contributes to its vorticity. At the swimmer,The prescribed rotation law has ${\mathbf{\Omega }}_f={\mathbf{\omega }}^{\ast }/2$. Since $\mathbf{b}\times \mathbf{p}=-\partial \mathbf{p}/\partial \theta$, comparison with $\dot{\mathbf{p}}=(\partial \mathbf{p}/\partial \theta )\dot{\theta }$ givesA pusher has $\mathcal{P}>0$ and rotates toward the stable parallel orientation $\theta =0$; there its image flow attracts it to the free surface. A puller has $\mathcal{P}<0$ and rotates toward a normal orientation $|\theta |=\pi /2$. Depending on which way its polar swimming direction points, it then swims away from the surface or meets it head-on; the dipolar image drift at the exactly normal orientation is toward the surface for this puller sign convention.