 Solution

No slip and impermeability give dimensional $u=-U$, $w=0$ on the upward-moving wall, so $\boxed{\hat{u}=-1,\hat{w}=0}$ there. For the interface normal, $h_x=\delta {\hat{h}}_{\hat{x}}$, and hencein its $x,z$ components. The relevant stresses scale asMultiplying by the inward normal gives the leading tractionIts leading normal component vanishes only if $\hat{p}=0$ at the interface; the tangential component then requires ${\hat{u}}_{\hat{z}}=0$.

The free surface is material. Applying the material derivative to $z-h(x,t)=0$ gives $w=h_t+u{h}_x$, which in the scaled variables becomes