Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 331 / 2 / b

Normalize $U$ and $c$ by $U_{\max }$ and write $h={\zeta }_L\in (0,1)$, $\alpha =kL>0$. In each open region $U^{\prime \prime }=0$, so the Rayleigh equation for inviscid shear flow reduces to ${\hat{w}}^{\prime \prime }-{\alpha }^2\hat{w}=0$ for the nonreal modes of interest. Incorporating the wall conditions from the start, the bounded piecewise-linear shear layer hasThese are four independent regional amplitudes before matching. The same expressions extend to noncritical neutral modes and their limits.

Impermeability gives $\hat{w}(\pm 1)=0$. At $\zeta =\pm h$, the base velocity is continuous but its derivative jumps. The normal velocity and pressure must remain continuous. From incompressibility and the streamwise linearized momentum equation,Therefore, with derivatives now taken in $\zeta$, the vorticity-jump matching for an inviscid shear flow isAt the upper jump $[U^{\prime }]=-1/h$ and at the lower jump $[U^{\prime }]=1/h$, where brackets mean the limit from larger $\zeta$ minus that from smaller $\zeta$. Equivalently $(U-c)[{\hat{w}}^{\prime }]=[U^{\prime }]\hat{w}$ away from $U=c$. Setting $[{\hat{w}}^{\prime }]=0$ would discard the vorticity jumps and give the wrong eigenproblem.

Let $q=\alpha h$, $d=\alpha (1-h)$, and use the four regional amplitudes above. Continuity of $\hat{w}$ at the two jumps givesPressure continuity, using vorticity-jump matching for an inviscid shear flow, gives the other two equations:These are the required homogeneous four equations. As a useful reduction, put $X=\tanh q$, $Y=\tanh d$, $H=q(1+XY)$ and $J=q(X+Y)$. Eliminating $A,D$ leavesThe vanishing determinant gives $c^2=(H-Y)(J-XY)/(HJ)$, which expands to the printed dispersion relation.