Put $Z=\epsilon\zeta$ and seek the slowly attenuating <wave envelope> $A(Z)$ in
$$
\psi=A(Z)e^{i(k\xi-\omega t)}.
$$
At order $\epsilon^0$, the equation gives the <internal gravity wave> <dispersion relation>
$$
\omega^2=N^2\sin^2\theta.
$$
At order $\epsilon^1$, retaining one derivative of the slowly varying amplitude gives
$$
2i\omega k^4A+2iN^2k\sin\theta\cos\theta\,A_Z=0.
$$
Using $\omega=N\sin\theta$ therefore yields
$$
A_Z=-\frac{k^3}{N\cos\theta}A.
$$
Hence the leading <viscous attenuation of an internal-wave beam> is
$$
\boxed{\widetilde\psi(\zeta)=A_0
\exp\left(-\frac{\epsilon k^3}{N\cos\theta}\zeta\right)
=A_0e^{-k\zeta/\operatorname{Re}},}
\qquad
\boxed{\operatorname{Re}=\frac{N\cos\theta}{\epsilon k^2}.}
$$
The wave <energy density>, being quadratic in the amplitude, decays twice as rapidly in the exponent.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 345 / 1 / c / Solution

The incident <internal-wave ray> has equation $z=-rx$. Intersecting it with the bottom $z=-H_0+sx$ gives
$$
x_1=\frac{H_0}{r+s}.
$$
Its distance to the slope is consequently
$$
\boxed{\ell=\frac{x_1}{\cos\theta}
=\frac{H_0}{\sin\theta+s\cos\theta}.}
$$
The subcritically reflected ray rises through the same vertical distance at angle $\theta$, so its next free-surface reflection is at
$$
\boxed{x_2=2x_1=\frac{2H_0}{r+s}.}
$$

On the first leg, <viscous attenuation of an internal-wave beam> multiplies the <energy density> by $e^{-2k\ell/\operatorname{Re}}$. The bottom reflection multiplies it by $\gamma^2$ and changes the <wavenumber> to $\gamma k$. Since the corresponding <Reynolds number> is $\operatorname{Re}/\gamma^2$, attenuation on the second leg contributes $e^{-2\gamma^3k\ell/\operatorname{Re}}$. Ignoring boundary-layer enhancement as requested,
$$
\boxed{\overline E_2=\gamma^2\overline E_0
\exp\left[-\frac{2k\ell}{\operatorname{Re}}(1+\gamma^3)\right].}
$$

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 345 / 1 / e / Solution

When <kinematic viscosity> and <mass diffusivity> are equal to $\epsilon$, a monochromatic internal-wave beam with <wavenumber> $k$ and ray angle $\theta$ has leading stream-function amplitude
$$
A(\zeta)=A(0)\exp\left(-\frac{\epsilon k^3\zeta}{N\cos\theta}\right)
=A(0)e^{-k\zeta/\operatorname{Re}},
\qquad
\operatorname{Re}=\frac{N\cos\theta}{\epsilon k^2}.
$$
The cubic dependence on $k$ makes slope-focused, short internal waves dissipate especially rapidly.

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