A layer of incompressible fluid of density $\rho$ and viscosity $\mu$ flows steadily down a plane inclined at an angle $\theta$ to the horizontal. The layer is of uniform thickness $h$ measured perpendicular to the plane and the viscosity of the overlying air can be neglected. Using coordinates $x$ parallel to the plane (in steepest downwards direction) and $y$ normal to the plane, write down the equations of motion and the boundary conditions on the plane and on the free top surface. Determine the pressure and velocity fields and show that the volume flux down the plane is

$$\frac{\rho g h^{3} \sin \theta}{3 \mu}$$

Consider now the case where a second layer of fluid, of uniform thickness $\alpha h$, viscosity $\beta \mu$ and density $\rho$, flows steadily on top of the first layer. Explain why one of the appropriate boundary conditions between the two fluids is

$$\mu \frac{\partial}{\partial y} u\left(h_{-}\right)=\beta \mu \frac{\partial}{\partial y} u\left(h_{+}\right),$$

where $u$ is the component of velocity in the $x$ direction and $h_{-}$and $h_{+}$refer to just below and just above the boundary respectively. Determine the velocity field in each layer.