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

First suppose $c_i\ne 0$ and put $F=\hat{w}/(U-c)$. The wall condition gives $F(\pm L)=0$. Substitution into the Rayleigh equation for inviscid shear flow givesMultiply by $\overline{F}$ and use integration by parts. The resulting weighted identity for Howard's semicircle theorem isIts imaginary part gives $\int UQ=c_{r}I$, and its real part then gives $\int U^{2}Q=(c_r^2+c_i^2)I$. Set $a=U_{\min }$ and $b=U_{\max }$. Since $(U-a)(b-U)\ge 0$,Completing the square proves Howard's semicircle theorem:In particular every growing normal mode lies in the upper semicircle for $k>0$. If $c$ is real and outside $[a,b]$, $F$ is again regular, but the same identity has a strictly positive integrand unless the mode vanishes. Such a neutral mode is impossible. Consequently neutral regular modes also satisfy the bound; no division by $c_i$ is needed for this last conclusion. The nonreal proof avoids critical-level singularities altogether.

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.

For fixed $h\in (0,1)$, $X=\alpha h+O({\alpha }^3)$ and $Y=\alpha (1-h)+O({\alpha }^3)$. The factored dispersion relation therefore giveswhere the latter follows from $X,Y\to 1$ and $c^2\sim (1-1/(2\alpha h){)}^2$. A real negative $c^2$ yields a growing branch $c=i\sqrt{-c^2}$ for positive $k$. Thus $h<1/2$ gives instability at sufficiently small nonzero wavenumber.

For the converse, monotonicity toward the limiting value one would suffice, as permitted in the question. In fact a direct sign argument avoids this extra assumption. With $q=\alpha h>0$, $X=\tanh q<q$, so $J-XY=qX+(q-X)Y>0$. If $h\ge 1/2$, then $q\ge d=\alpha (1-h)$ and $Y=\tanh d<d\le q$. Hence $H-Y=q(1+XY)-Y>0$, so $c^2>0$ for every $\alpha >0$. We obtain the exact instability threshold for a bounded piecewise-linear shear layer:At ${\zeta }_L=1/2$ the long-wave limiting value is zero, but every finite positive wavenumber has $c^2>0$.