= Solution
Linearize about the resting state $\eta=0$, $\phi=0$. Introduce a dimensionless small parameter $\epsilon$ and write
$$
\eta=\epsilon\eta_1+O(\epsilon^2),\qquad
\phi=\epsilon\phi_1+O(\epsilon^2),
$$
holding the depth, <bending stiffness> and variation scales fixed. Thus both boundary displacement and induced <velocity> are small. For any first-order derivative $D\phi$, a <Taylor expansion> at the moving surface gives
$$
(D\phi)(x,\eta,t)
=\epsilon(D\phi_1)(x,0,t)+O(\epsilon^2).
$$
The displacement correction to the leading field is $\epsilon^2\eta_1\partial_z(D\phi_1)(x,0,t)$, so evaluating it on $z=0$ changes it only at second order. Also $\phi_x\eta_x=O(\epsilon^2)$ and $|\nabla\phi|^2=O(\epsilon^2)$. The first-order <kinematic boundary condition> and dynamic boundary condition are therefore
$$
(\eta_1)_t=(\phi_1)_z(x,0,t),\qquad
(\phi_1)_t(x,0,t)+g\eta_1+\frac B\rho(\eta_1)_{xxxx}=0.
$$
Dropping the subscript, the <linearized boundary condition> pair is
$$
\boxed{\eta_t=\phi_z,\qquad
\phi_t+g\eta+\frac B\rho\eta_{xxxx}=0\quad\text{at }z=0}.
$$
The fixed-bed condition remains $\phi_z(x,-h,t)=0$. The source's shorthand $|\eta|\ll1$ requires a length scale: for the subsequent <plane wave>, small surface slope $k|\alpha|\ll1$ and displacement small compared with depth are sufficient conditions. Small height alone, with arbitrarily rapid spatial variation or a finite background disturbance, does not justify discarding the products.
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