= Solution
The <long-wave approximation> makes $u$ independent of $z$ at leading order. <Mass conservation> and symmetry about $z=0$ then give
$$
u_x+w_z=0,
\qquad
\boxed{w=-zu_x}.
$$
For an incompressible Newtonian fluid,
$$
\sigma_{xx}=-p+2\mu u_x,
\qquad
\sigma_{zz}=-p+2\mu w_z=-p-2\mu u_x.
$$
The leading normal-stress balance on either surface is
$$
\boxed{\sigma_{zz}=-p_a+\gamma h_{xx}}.
$$
Since $\sigma_{xx}-\sigma_{zz}=4\mu u_x$, it follows that
$$
\boxed{\sigma_{xx}=-p_a+\gamma h_{xx}+4\mu u_x}.
$$
For a slice of length $\delta x$, the axial forces are the integrated normal stresses $2h\sigma_{xx}$ on its vertical ends, ambient pressure on the varying end height, and the horizontal components of <surface tension> on its two sloping faces. Expanding their difference to first order in $\delta x$ cancels the uniform $p_a$ terms and the lower-order capillary terms, leaving
$$
\boxed{
\frac\partial{\partial x}
\left(4\mu h\frac{\partial u}{\partial x}\right)
+\gamma h\frac{\partial^3h}{\partial x^3}=0}.
$$
The kinematic condition on $z=h(x,t)$ is $h_t+uh_x=w$. Substituting $w=-hu_x$ gives the second required relation,
$$
\boxed{h_t+(uh)_x=0}.
$$
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