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Set $F=e^{k\eta }$. The characteristic polynomial is $k^4=-i$, whose roots have arguments $-\pi /8$, $3\pi /8$, $7\pi /8$, and $11\pi /8$. Only the last two have negative real part and decay at infinity. Write $c=\cos (\pi /8)$ and $s=\sin (\pi /8)$; the admissible roots areThus $F=a_{\ell }{e}^{k_{\ell }\eta }+a_r{e}^{k_r\eta }$. Since $k_r=i{k}_{\ell }$, the boundary equations are $a_{\ell }+a_r=1$ and $k_{\ell }^2(a_{\ell }-a_r)=0$. ThereforeTaking real parts gives the oscillatory bending of a moment-free semi-infinite filamentThe first phase is constant along paths with $dx/dt=-\omega \ell /s$, so that contribution travels left. The second has $dx/dt=\omega \ell /c$ and travels right. Both are forced, attenuating patterns in an overdamped medium; they are not undamped inertial waves. Their amplitudes are equal at the driven end, but $c\simeq 0.924>s\simeq 0.383$ gives decay lengths $\ell /c\simeq 1.08\ell$ and $\ell /s\simeq 2.61\ell$. Hence the right-travelling contribution dominates away from the end, because it decays more slowly; its amplitude relative to the left-travelling one grows as $e^{(c-s)x/\ell }$.

The envelopes in the sketch show this amplitude comparison independently of the instantaneous phase. At particular points a cosine can vanish or the two terms can interfere, so dominance refers to envelopes and the far-field response, not to a strict pointwise inequality at every time.

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.

For the long filament, take the time-periodic solution of the dimensionless bending equation, after transients have decayed. Write $y=\mathrm{Re}[Y(x)e^{it}]$. Then $Y^{(4)}=-iY$, so the spatial exponents satisfy ${\lambda }^4=-i$. Define $C=\cos (\pi /8)$ and $S=\sin (\pi /8)$. The two roots with negative real part are ${\lambda }_1=-C+iS$ and ${\lambda }_2=-S-iC$; the other roots violate decay at infinity.

Hence $Y=A{e}^{{\lambda }_{1}x}+B{e}^{{\lambda }_{2}x}$. The driven displacement gives $A+B=y_0$, while the zero bending moment gives ${\lambda }_1^{2}A+{\lambda }_2^{2}B=0$. Since ${\lambda }_2^2=-{\lambda }_1^2$, $A=B=y_0/2$. The oscillatory bending of a moment-free semi-infinite filament isHolding each wave phase constant gives its dimensionless phase velocity:The first travels toward the actuator and attenuates over length $1/C$; the second travels away and attenuates over the longer length $1/S$. These are spatially damped phase patterns in an overdamped bending equation, not two undamped inertial beam waves. Their combination satisfies both displacement and zero-moment conditions at the driven end.