Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 39C (Fluid Dynamics II)

The Reynolds number isWhen it is large, viscosity is negligible in most of the flow, but the inviscid outer solution cannot generally satisfy the no-slip boundary condition at a rigid wall. A thin boundary layer resolves this mismatch: tangential gradients remain of outer scale $\mathcal{L}$, normal gradients have the much shorter scale $\delta$, and the small viscous coefficient is offset by $u_{yy}=O(\mathcal{U}/{\delta }^2)$. Matching a viscous inner solution to an inviscid outer solution retains the leading viscous effects without solving the full equations uniformly everywhere.

The wall speed setsUsing $x$ as the local streamwise length giveswhich is large when $x\gg \sqrt{\nu /\alpha }$. The boundary-layer scalingthen givesso the stretching-sheet layer has constant thickness to leading order.

With zero imposed pressure gradient, the two-dimensional boundary-layer equations areIntroduce the stream function byThe boundary conditions areThe additive constant in $\psi$ may be chosen so that $\psi (x,0)=0$.

PutThenSubstitution in the boundary-layer momentum equation, using $\nu /{\delta }^2=\alpha$, givesThe wall and far-field conditions becomeThese are the similarity equations for the boundary layer over a linearly stretching sheet.

Set $f=A+B{e}^{-C\eta }$ with $C>0$. The differential equation reduces toso $A=C$. The two wall conditions giveHence $C^2=1$, and positivity of $C$ yieldsor

As $\eta \to \infty$,Thus the tangential velocity matches the quiescent outer fluid, but the normal velocity tends toThe stretching sheet entrains fluid toward itself at a constant leading-order speed. Consequently a globally stationary exterior flow is impossible: the outer flow must provide this normal influx and complete the mass balance away from the local boundary-layer approximation.