Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 355 / 1

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.