= Solution
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 $T_i=-p_iI+\mu_i(\nabla\mathbf u_i+\nabla\mathbf u_i^T)$ 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 equations
$$
0=-p_x+\mu u''+\rho g\sin\theta,\qquad0=-p_y-\rho g\cos\theta.
$$
For a uniform free layer, $p_x=0$, $u(0)=0$, $u'(h)=0$ and $p(h)=p_{\rm atm}$. <Integration> gives
$$
\boxed{p=p_{\rm atm}+\rho g\cos\theta(h-y),\quad u=\frac{\rho g\sin\theta}{\mu}\left(hy-\frac{y^2}{2}\right).}
$$
Its flux per unit cross-slope width is
$$
\boxed{\int_0^h u(y)dy=\frac{\rho g h^3\sin\theta}{3\mu}.}
$$
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, $p=p_{\rm atm}+\rho g\cos\theta(H-y)$. The wall condition is $u_1(0)=0$, the upper condition $u_2'(H)=0$, and at $h$ we require $u_1=u_2$, $\mu u_1'=\beta\mu u_2'$. Integrating the two <momentum> equations with these conditions gives
$$
\boxed{u_1(y)=A\left(Hy-\frac{y^2}{2}\right),\quad0\leq y\leq h,}
$$
$$
\boxed{u_2(h+z)=Ah^2\left(\alpha+\frac12\right)+\frac A\beta\left(\alpha h z-\frac{z^2}{2}\right),\quad0\leq z\leq\alpha h.}
$$
The <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.
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