= Solution
For one transverse displacement $\psi$ the density is
$$
\mathcal L=\frac\rho2\psi_t^2-\frac F2\psi_x^2-\frac B2\psi_{xx}^2.
$$
Its variation is $\delta S=\int dt\,dx[\rho\psi_t\delta\psi_t-F\psi_x\delta\psi_x-B\psi_{xx}\delta\psi_{xx}]$. Integrating once in time and twice in space gives the bulk coefficient
$$
-\rho\psi_{tt}+F\psi_{xx}-B\psi_{xxxx}=0.
$$
The spatial boundary contribution is $[(-F\psi_x+B\psi_{xxx})\delta\psi-B\psi_{xx}\delta\psi_x]$ integrated over time. Thus
$$
\boxed{\rho\psi_{tt}=F\psi_{xx}-B\psi_{xxxx}.}
$$
Insert $\psi\propto e^{i(kx-\omega t)}$ to obtain
$$
\boxed{\omega^2=\frac{Fk^2+Bk^4}{\rho}=\frac{k^2}{\rho}(F+Bk^2).}
$$
For compression $F<0$, sufficiently long waves have $\omega^2<0$ and grow exponentially. Positive bending rigidity stabilizes short wavelengths; without it the compressed small-slope model would have no short-wavelength regularization.
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