Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 39D (Fluid Dynamics)

At zero Reynolds number, an incompressible Newtonian fluid obeys the Stokes equationsTaking the divergence gives ${\nabla }^{2}p=0$. Taking the curl and writing $\mathbf{\omega }=\nabla \times \mathbf{u}$ givesThus both pressure and vorticity are harmonic in the fluid.

In the laboratory frame, the boundary conditions areand, on $r=a$,The first is impermeability of the solid surface and the second is viscous no slip. Together they say $\mathbf{u}=\mathbf{U}$ on the sphere.

For the stated solution, putThen $u_j=A{U}_j+Bs{x}_j$, andwhere

Taking the curl of $A\mathbf{U}$ gives $\nabla A\times \mathbf{U}$, whileTherefore

For $r>0$,andThe pressure and vorticity can be writtenConstant-coefficient derivatives commute with the Laplacian, so both fields are harmonic away from the sphere.

Taking the trace of the velocity gradient givesDirect substitution yieldssoThis verifies incompressibility directly.

For a Newtonian fluid,On $r=a$, with $\mathbf{n}=\mathbf{x}/a$,The surface velocity gradient is thereforeContracting the stress with $n_j$, the normal part of the viscous stress cancels the pressure contribution, leaving the uniform traction

Integrating the traction over the sphere givesHence the force exerted by the fluid on the sphere isThis is Stokes' drag law; the sign is opposite to the sphere's velocity.