Viscous fluid with dynamic viscosity μ flows with velocity (ux,uy,uz)≡(uH,uz) (in cartesian coordinates x,y,z) in a shallow container with a free surface at z=0. The base of the container is rigid, and is at z=−h(x,y). A horizontal stress S(x,y) is applied at the free surface. Gravity may be neglected.
Using lubrication theory (conditions for the validity of which should be clearly stated), show that the horizontal volume flux q(x,y)≡∫−h0uHdz satisfies the equations
∇⋅q=0,μq=−31h3∇p+21h2S
where p(x,y) is the pressure. Find also an expression for the surface velocity u0(x,y)≡ uH(x,y,0) in terms of S,q and h.
Now suppose that the container is cylindrical with boundary at x2+y2=a2, where a≫h, and that the surface stress is uniform and in the x-direction, so S=(S0,0) with S0 constant. It can be assumed that the correct boundary condition to apply at x2+y2=a2 is q⋅n=0, where n is the unit normal.
Write q=∇ψ(x,y)×z^, and show that ψ satisfies the equation
∇⋅(h31∇ψ)=−2μh2S0∂y∂h
Deduce that if h=h0 (constant) then q=0. Find u0 in this case.
Now suppose that h=h0(1+ϵy/a), where ϵ≪1. Verify that to leading order in ϵ,ψ=ϵC(x2+y2−a2) for some constant C to be determined. Hence determine u0 up to and including terms of order ϵ.
[Hint: ∇×(A×z^)=z^⋅∇A−z^∇⋅A for any vector field A.]