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Let $\zeta =z/L\in [-1,1]$, $h=L_s/L\in (0,1)$, and normalize velocity by its maximum magnitude. The profile is $U=\zeta /h$ for $|\zeta |<h$ and $U=\mathrm{sgn}\zeta$ outside. Each region has exponential vertical-velocity modes; the two walls and vorticity-jump matching for an inviscid shear flow determine their coupling. With $\alpha =kL$, $q=\alpha h$, $X=\tanh q$, $Y=\tanh [\alpha (1-h)]$, $H=q(1+XY)$ and $J=q(X+Y)$, elimination gives $c^2=(H-Y)(J-XY)/(HJ)$, with $c$ normalized by the velocity scale.

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.

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.