Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 37E (Fluid Dynamics II)

Incompressibility gives $r^{-1}(ru{)}_r+r^{-1}{v}_{\varphi }=0$, permitting a stream function with $u={\psi }_{\varphi }/r$, $v=-{\psi }_r$. Its axial vorticity is $-\Delta \psi$. Taking the curl of the Stokes flow momentum equation eliminates pressure and gives $\Delta$ of that vorticity zero, hence ${\Delta }^2\psi =0$.

At an impermeable constant-angle plane, $v=0$ means $\psi$ is constant along that boundary. Fix that constant to zero. A rigid boundary also requires ${\psi }_{\varphi }=0$. The tangential stress is $\mu (-{\psi }_{rr}+{\psi }_r/r+{\psi }_{\varphi \varphi }/r^2)$; along an impermeable boundary its first two terms vanish, so a stress-free boundary requires ${\psi }_{\varphi \varphi }=0$.

Write $\psi =(S/\mu )r^{2}F(\varphi )$. The biharmonic equation becomes $F^{\prime \prime \prime \prime }+4F^{\prime \prime }=0$. The boundary conditions are $F(-\alpha )=F^{\prime }(-\alpha )=F(0)=0$ and $F^{\prime \prime }(0)=1$. ConsequentlyThe requested $f$ is $F/\mu$. Evaluating $u(r,0)=(Sr/\mu )F^{\prime }(0)$ givesIts small-angle ratio is $\alpha /4+O({\alpha }^3)$, so $U$ and $S$ have the same sign for sufficiently small positive angle. At the first nonzero critical root, $2{\alpha }_c\simeq 4.493409$, the denominator vanishes while the numerator does not. The pure $r^2$ similarity solution is then resonant and cannot satisfy the forcing; logarithmic radial terms or additional global boundary data are needed. Just beyond this angle, and up to $\alpha <\pi$, the denominator is negative and numerator positive, so the formal surface flow reverses relative to $S$. This is not a contradiction of viscous dissipation: the wedge is radially unbounded and its radial-end energy flux has not been prescribed. A physical finite-domain flow needs those outer conditions, and the Stokes approximation also fails sufficiently far out when the velocity grows with $r$.