 Solution

Balancing the viscous and bending terms gives $\zeta \omega h\sim Ah/{\ell }^4$, so the elastohydrodynamic penetration length isInserting $h=h_0\mathrm{Re}[e^{i\omega t}F(x/\ell )]$ into the linear equation yields $F^{\prime \prime \prime \prime }=-iF$, with $F(0)=1$, $F^{\prime \prime }(0)=0$ and decay at infinity. Write $C=\cos (\pi /8)$, $S=\sin (\pi /8)$. Of the four roots of ${\lambda }^4=-i$, precisely two have negative real part:The two decaying solutions are $e^{{\lambda }_L\eta }$ and $e^{{\lambda }_R\eta }$, and the general decaying response is their linear combination. Both coefficients are nonzero: if one vanished, $F^{\prime \prime }(0)=0$ would force the other to vanish, contradicting $F(0)=1$.

Their real parts in the time-dependent response are proportional toConstant phases move respectively with velocities $-\omega \ell /S$ and $\omega \ell /C$, so these are damped travelling waves moving left and right. Since $S<C$, the right-going wave has the longer attenuation length and dominates at large $x$. Thus the less strongly damped, outward-moving wave dominates the far-field response. This is the oscillatory bending of a moment-free semi-infinite filament; the two boundary conditions determine its amplitudes without needing them explicitly here.