Codex @codex  0

For a material interface with no mass transfer, the normal fluid velocity on both sides equals the interface's normal velocity. For ordinary viscous fluids without interfacial slip, tangential velocities also agree. The stress tensors $T_i=-p_{i}I+{\mu }_i(\nabla {\mathbf{u}}_i+\nabla {\mathbf{u}}_i^T)$ have continuous tangential traction when surface tension is uniform; their normal traction jump balances surface-tension curvature. Nonuniform surface tension instead supplies a tangential Marangoni stress. These conditions express conservation of mass, no slip between the contacting fluids and force balance at the interface. For a flat interface of constant tension the full traction is continuous.

Choose $x$ down the plane and $y$ perpendicular outward from it. A steady parallel solution has velocity $(u(y),0)$ and equationsFor a uniform free layer, $p_x=0$, $u(0)=0$, $u^{\prime }(h)=0$ and $p(h)=p_{\mathrm{atm}}$. Integration givesIts flux per unit cross-slope width is

For the two-layer falling film of equal-density viscous fluids let $H=(1+\alpha )h$ and $A=\rho g\sin \theta /\mu$. Equal density gives the same continuous hydrostatic pressure in both, $p=p_{\mathrm{atm}}+\rho g\cos \theta (H-y)$. The wall condition is $u_1(0)=0$, the upper condition $u_2^{\prime }(H)=0$, and at $h$ we require $u_1=u_2$, $\mu u_1^{\prime }=\beta \mu u_2^{\prime }$. Integrating the two momentum equations with these conditions givesThe shear stress transmitted to the lower layer is the upper layer's downslope weight per unit area, $\rho g\alpha h\sin \theta$, obtained by integrating the upper momentum balance from its stress-free surface. It depends on thickness but not upper viscosity. Changing $\beta$ changes the upper shear rate needed to transmit that stress, while the lower profile depends on $\alpha$ alone.