Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 86 / 3

The leading disturbance in sphere Stokes flow has size $u\sim Ua/r$. Consequently viscous diffusion has size $\nu u/r^2$, background advection has size $Uu/r$, and their ratio is $Ur/\nu$. Even if the sphere Reynolds number $Ua/\nu$ is small, this ratio becomes order one at $r\sim \nu /U$. Thus the Stokes approximation is nonuniform at distances of order $\nu /U$. At that outer distance the disturbance itself is small relative to $U$, so self-convection is smaller than background advection and the Oseen approximation is appropriate:Write $\mathbf{u}=\nabla \varphi +\nu \nabla \chi -\mathbf{U}\chi$. It is sufficient outside the origin to impose $\Delta \varphi =0$ and $(\nu \Delta -\mathbf{U}\cdot \nabla )\chi =0$: these give zero divergence, and substitution into momentum gives $p=-\rho \mathbf{U}\cdot \nabla \varphi$. With $\chi =h{e}^{\mathbf{U}\cdot \mathbf{x}/(2\nu )}$, differentiation yieldsA decaying radial choice is $h=C{e}^{-Ur/(2\nu )}/r$. ThereforeThe exponent is nonpositive, so this scalar decays at infinity, including as $1/r$ directly downstream. In the overlap $a\ll r\ll \nu /U$,Choosing $\varphi =-\nu C/r$ cancels the first term's gradient in $\nabla \varphi +\nu \nabla \chi$. The remaining leading velocity isMatching fixes $C=3a/2$. Thus the potential-source and wake decomposition of sphere Oseen flow isIts pressure is $p=-3\mu a(\mathbf{U}\cdot \mathbf{x})/(2r^3)$, the sign appropriate to a fixed sphere in an incident flow. The velocity decays in every direction and matches the requested leading inner disturbance; the sphere's potential-dipole term enters a higher matching order.

The scalar potential is a point source: $\nabla \varphi =(3a\nu /2)\mathbf{x}/r^3$, so integration over a centred sphere givesTo see the fluid wake, take $z$ downstream along $\mathbf{U}$ and transverse radius $R_{\perp }$. For $z\gg \nu /U$, $r-z\simeq R_{\perp }^2/(2z)$, givingThe fluid wake is concentrated within $R_{\perp }=O(\sqrt{\nu z/U})$. Its integrated mass deficit is $\rho U\int \chi dA\simeq 6\pi \mu a$, balancing the outward potential-source mass flux. The corresponding momentum deficit iswhich equals the leading sphere drag. The Gaussian integral used here is ${\int }_0^{\infty }{e}^{-U{R}_{\perp }^2/(4\nu z)}2\pi R_{\perp }d{R}_{\perp }=4\pi \nu z/U$.

The PDF labels $6\pi \mu aU$ as the mass-source strength, but this has dimensions of force. The actual mass-source strength is $6\pi \mu a$; $6\pi \mu aU$ is the momentum deficit or drag. The decomposition above supplies both quantities and identifies the printed normalization error.