Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 355 / 1

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.