= Solution
Balancing the viscous and bending terms gives $\zeta\omega h\sim Ah/\ell^4$, so the <elastohydrodynamic penetration length> is
$$
\boxed{\ell=(A/(\zeta\omega))^{1/4}.}
$$
Inserting $h=h_0\operatorname{Re}[e^{i\omega t}F(x/\ell)]$ into the linear equation yields $F''''=-iF$, with $F(0)=1$, $F''(0)=0$ 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:
$$
\lambda_L=-C+iS,\qquad\lambda_R=-S-iC.
$$
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, $F''(0)=0$ would force the other to vanish, contradicting $F(0)=1$.
Their real parts in the time-dependent response are proportional to
$$
e^{-C\eta}\cos(\omega t+S\eta+\phi_L),\qquad
e^{-S\eta}\cos(\omega t-C\eta+\phi_R).
$$
Constant 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 \b[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.
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