Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 37B (Fluid Dynamics II)

The steady Stokes equations are $-\nabla p+\mu {\nabla }^2\mathbf{u}=0$, $\nabla \cdot \mathbf{u}=0$. Taking the curl gives ${\nabla }^2\mathbf{\omega }=0$. For an axisymmetric azimuthal component $\omega {\mathbf{e}}_{\varphi }$, the vector Laplacian's azimuthal component is $({\nabla }^2-1/(R^2{\sin }^2\theta ))\omega$. Expanding the derivatives showsWith $\omega =-D^2\Psi /(R\sin \theta )$, the vorticity equation consequently gives $D^4\Psi =0$. This derives the streamfunction equation from momentum rather than treating the scalar vorticity component as an ordinary scalar harmonic function.

For $\Psi =f(R){\sin }^2\theta$, the angular operator gives $D^2\Psi =(f^{\prime \prime }-2f/R^2){\sin }^2\theta$. If $f=AR+B/R$, then $D^2\Psi =-2A{\sin }^2\theta /R$, which is annihilated by $D^2$, verifying the differential equation. The supplied curl formula givesBoth decay at infinity. At $R=a$, no slip requires $u_R=U\cos \theta$, $u_{\theta }=-U\sin \theta$, so $2(A/a+B/a^3)=U$ and $A/a-B/a^3=U$. ThereforeThis is the decaying translating-sphere solution in the laboratory frame; in the body frame one would additionally include the uniform flow at infinity.

Substitution givesRotational symmetry about the translation axis forbids a transverse force. Linearity of the Stokes equations and boundary conditions makes the force linear in $U$; dissipation makes the force on the translating sphere oppose its motion. Thus the drag is axial and proportional to $U$, without computing the stress integral. Dimensional analysis further gives scale $\mu aU$; the exact single-sphere coefficient is $6\pi$.

For side-by-side spheres, reflection in the vertical mid-plane exchanges the spheres while preserving their common velocity, so their axial drags are equal. For vertically aligned spheres, reflection in the horizontal mid-plane exchanges them and reverses velocity; applying linearity to reverse all velocities back restores the original problem and again gives equal axial drag. These are symmetries of the exact two-sphere Stokes problem.

At a distant side-by-side centre, the leading flow produced by one sphere is $3aU/(4b)$ in the direction of translation. On its axis it is $3aU/(2b)$. This ambient entrainment lowers the relative velocity of the other sphere. Using Stokes drag law with that relative velocity gives the mutual drag reduction of two distant translating spheres:The feedback of the corrected force into the other sphere's flow enters at the next order. If the coefficient $6\pi$ is needed directly from the preceding solution, its pressure anomaly is $p=3\mu Ua\cos \theta /(2R^2)$. At the surface the traction components are $t_R=-3\mu U\cos \theta /(2a)$ and $t_{\theta }=3\mu U\sin \theta /(2a)$; their axial projection is the constant $-3\mu U/(2a)$. Integrating over $4\pi a^2$ gives the single-sphere force $-6\pi \mu aU{\mathbf{e}}_z$, as used above.