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For a lower uniformly stratified half-space and an upper well-mixed half-space, the vertical velocity amplitude is continuous and its derivative jumps by $[W^{\prime }]=-k^2\Delta bW/{\omega }^2$. A wave incident upwards with vertical phase factor $e^{-imz}$ therefore has upper evanescent amplitude $T/W_i=2im/[k(1-S)+im]$, where $m=k\sqrt{N^2/{\omega }^2-1}$ and $S=\Delta bk/{\omega }^2$. Its magnitude peaks at $S=1$, with amplitude twice the incident amplitude. Lossless reflection carries all incident energy back down; the evanescent upper field carries no mean outgoing vertical energy flux.

In the lubrication regime the thin region has longitudinal scale $\ell \sim \sqrt{ab}$, since the height increase $x^2/(2a)$ becomes comparable with $b$ there. The characteristic shear is $\mu V/b$, so force per axial length scales as $\mu V\ell /b\sim \mu V\sqrt{a/b}$. Negligible inertia, no slip, no cylinder rotation, and full wetting are understood.

Use a frame translating with the cylinder. The wall moves with velocity $-V$, the cylinder is stationary, and the gap is $h(x)=b+x^2/(2a)$. Lubrication momentum gives $\mu u_{yy}=p_x$, $p_y=0$, soContinuity makes $q$ constant. Equal ambient pressures at both infinities give $\int p_{x}dx=0$, hence $q=-(V/2)(\int h^{-2}dx)/(\int h^{-3}dx)=-2Vb/3$. ThereforeThe vertical velocity, with the wall condition $w(0)=0$, follows explicitly from incompressibility:It gives $w(h)=0$, consistently with a horizontally translating cylinder in the lab frame.

The wall shear is $\mu u_y(0)=4\mu V(1/h-b/h^2)$. With $x=\sqrt{2ab}s$, its integral isForce balance equates this wall force to the required cylinder force at leading order; the pressure and viscous stresses on remote boundaries make no leading contribution. This is the cylinder translating parallel to a wall in a thin gap result. The asymptotic sign denotes the approximation implicit in the parabolic gap, rather than an exact finite-gap cylinder formula.

Let $\hat{\eta }$ be the displacement of a material interface and use $[f{]}_{-}^{+}=f_{+}-f_{-}$. On each side, the kinematic boundary condition isThere is one displaced material interface, so its displacement is continuous. Thus the first jump condition isIn particular, vertical velocity need not be continuous when the background velocity jumps.

With no surface tension, pressure continuity applies on the displaced interface, rather than at its undisplaced height. Linearizing gives $[\hat{p}+\hat{\eta }{\overline{p}}^{\prime }{]}_{-}^{+}=0$. Since ${\overline{p}}^{\prime }=-g\overline{\rho }$, this becomes $[\hat{p}-g\overline{\rho }\hat{\eta }{]}_{-}^{+}=0$. The horizontal momentum calculation in part (a) givesSubstitute this and $\hat{\eta }=\hat{w}/[ik(\overline{U}-c)]$ to obtain the second jump condition:These jump conditions for stratified inviscid shear flow use one-sided values of ${\overline{U}}^{\prime }$ and apply to jumps of mass density, vorticity, or velocity. If $\overline{U}$ is continuous, the first condition reduces to continuity of $\hat{w}$; a jump of vorticity generally still prevents continuity of ${\hat{w}}^{\prime }$. This derivation avoids multiplying singular distributional derivatives of a discontinuous velocity in the Taylor–Goldstein equation.

The floor contributes gravitational potential $mgz=m{\omega }^2(x_1^2+y_1^2)/2$. Neglecting vertical velocity as specified, for the nonrotating cup orientation $\varphi =-\Omega$ the Lagrangian isFor general $\alpha =\Omega +\varphi$, replace the velocity components by ${\dot{x}}_1-\alpha y_1+R\Omega \sin \varphi t$ and ${\dot{y}}_1+\alpha x_1+R\Omega \cos \varphi t$. The potential remains the same because the paraboloid is axisymmetric.