In the <Monge representation>, $ds=dx+O(h_x^2)$ and the signed curvature is $\kappa=h_{xx}+O(h_x^2h_{xx})$. To quadratic order,
$$
E[h]=\frac A2\int_0^L(h_{xx}-\kappa_0)^2dx.
$$
Two integrations by parts give
$$
\delta E=A\left[(h_{xx}-\kappa_0)\delta h_x-h_{xxx}\delta h\right]_0^L
+A\int_0^Lh_{xxxx}\delta h\,dx.
$$
Thus a filament with free ends has zero bending moment and shear force,
$$
\boxed{h_{xx}(0,t)=h_{xx}(L,t)=\kappa_0,
\qquad h_{xxx}(0,t)=h_{xxx}(L,t)=0,}
$$
and the local <Stokesian dynamics of an elastic filament> is
$$
\boxed{\zeta h_t=-Ah_{xxxx}.}
$$
With $X=x/L$, $H=h/L$, and $\tau=At/(\zeta L^4)$, the dimensionless problem is
$$
H_\tau=-H_{XXXX},
\qquad H_{XX}=\kappa_0L,
\qquad H_{XXX}=0
$$
at $X=0,1$.

Solution

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

In local Stokesian filament dynamics, viscous drag balances the variational elastic force without inertia. A small-slope filament of bending modulus $A$ and transverse drag $\zeta$ obeys $\zeta h_t=-Ah_{xxxx}$.


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 Incoming links: Stokesian dynamics of an elastic filament