= Solution
Impermeability gives $\widehat w(\pm1)=0$. 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,
$$
\widehat p=\frac{\rho}{ik}\bigl[(U-c)\widehat w_z-U_z\widehat w\bigr].
$$
Therefore, with derivatives now taken in $\zeta$, the <vorticity-jump matching for an inviscid shear flow> is
$$
\boxed{[\widehat w]=0,\qquad[(U-c)\widehat w'-U'\widehat w]=0\quad\text{at }\zeta=\pm h.}
$$
At the upper jump $[U']=-1/h$ and at the lower jump $[U']=1/h$, where brackets mean the limit from larger $\zeta$ minus that from smaller $\zeta$. Equivalently $(U-c)[\widehat w']=[U']\widehat w$ away from $U=c$. Setting $[\widehat w']=0$ would discard the vorticity jumps and give the wrong eigenproblem.
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