 Solution

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.