Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 38B / b

Conservation of volume flux through a circular arc of radius $r$ givesso the outer flow is the sink flow in a sectorThis is a potential flow and already satisfies the no-penetration boundary condition on the radial walls. It has nonzero tangential speed $Q/r$ there, however, and therefore violates the no-slip boundary condition. Thin vorticity-bearing boundary layers are needed to reduce that speed to zero at each wall.

Along $y=0$, put $x=r$ and $U(x)=-Q/x$. If $\delta (x)$ is the layer thickness, thenThe boundary-layer thickness from an advection-diffusion balance therefore givesand henceup to a constant factor absorbed into the similarity coordinate.

LetUsing the Cartesian streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$ givesAt fixed $y$,The terms proportional to $\eta f^{\prime }{f}^{\prime \prime }$ cancel, leavingSince $U=-Q/x$, $U{U}^{\prime }=-Q^2/x^3$, while $\nu u_{yy}=Q^2{f}^{\prime \prime \prime }/x^3$. The boundary-layer equation therefore becomesChoosing the wall value of the streamfunction as zero, imposing no slip, and matching to $U=-Q/x$ give

SetWriting $C=\cosh z$, direct differentiation givesOn the other hand,Thus $F^{\prime \prime }=1-F^2$, which is precisely $f^{\prime \prime \prime }=1-(f^{\prime }{)}^2$. The far-field condition is automatic because $F\to -1$. No slip requiressoFor $c=+\mathrm{arcosh}5$, $F$ decreases directly from $0$ to $-1$, matching flow everywhere toward the sink. For the negative root, $F$ initially becomes positive and creates a reverse-flow region next to the wall. The physically likely choice is therefore