Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 39A (Fluid Dynamics II)

For an incompressible Newtonian fluid with negligible inertia and no body force, the Stokes equations areThey are linear because any linear combination of solutions corresponds to the same linear combination of boundary data and applied forces. They are reversible because replacing all imposed velocities and forces by their negatives sendsFluid particles therefore retrace their paths under reversal. This is kinematic reversibility of Stokes flow.

The geometry and boundary velocities are axisymmetric and linear in the axial pseudovector $\mathbf{\Omega }$. The pressure is a true scalar. No nonzero parity-even scalar can be formed linearly from $\mathbf{\Omega }$ and the radial vector $\mathbf{x}$: $\mathbf{\Omega }\cdot \mathbf{x}$ is a pseudoscalar. Hence the flow-induced pressure is spatially constant.

Equivalently, taking the divergence of the Stokes momentum equation gives the Harmonic pressure and vorticity in Stokes flow resultand symmetry plus the homogeneous boundary data excludes every nonconstant harmonic pressure mode. Since pressure is defined up to an additive constant, we choose

Both candidate fields are divergence-free. For solid-body rotation,each component is linear in $\mathbf{x}$, so ${\nabla }^2\mathbf{u}=0$. Also,is harmonic for $r>0$, because ${\nabla }^2(1/r)=0$ there and constant-coefficient derivatives commute with the Laplacian. Thus both solve the Stokes equations with $p=0$.

The rotationally symmetric solution is consequentlyThe no-slip boundary condition givesThereforeandThis is the Rotational Stokes flow between concentric spheres.

WriteThe velocity gradient of the solid-rotation part is antisymmetric, so it drops out of the Newtonian fluid stress tensor. Sinceand $p=0$, the stress isEquivalently, with $\mathbf{n}=\mathbf{x}/r$,

On $r=c$, the traction $\mathbf{\sigma }\mathbf{n}$ exerted by the outer fluid on the inner fluid isThe inner fluid exerts the opposite traction on the outer fluid. Its couple is thereforeUsingand the supplied spherical integral,we obtainThus the radius cancels:This is the Torque in rotational Stokes flow between concentric spheres.

When $a\ll b$,the rotational drag torque for a sphere in an effectively unbounded fluid. If $h=b-a\ll a$, thensothe inverse-gap growth expected from a thin Couette flow.