Past exam of the flows, fluctuations and complexity course of the University of Oxford / 2016 / 1 / d

On the surface, $u_r=0$ and $u_{\theta }=(3v/2)\sin \theta$. The only surface stagnation points are thereforeThere are no other stagnation points in the exterior: away from the axis the polar component is nonzero, and on the axis the radial component vanishes only at $r=R$.

Near the front point put $(X,Y,Z)=(x,y,z-R)$. Expanding the potential from solution givesso the linear velocity field is $(3vX/(2R),3vY/(2R),-3vZ/R)$. This agrees with solution for $W=3v/(4R)$.

At the rear point, using $Z=z+R$ instead, the quadratic term has the opposite sign: $W=-3v/(4R)$. Incoming axial flow and tangential outflow at the front become tangential inflow and axial outflow at the back. The local streamline geometry is unchanged while its direction reverses.