= Solution
Use the attenuation coefficient read from the PDF,
$$
\lambda=\frac{\nu k^3}{2N\sin\theta},\qquad A(\xi)=U_0e^{-\lambda\xi},\qquad U=A(\xi)\cos\omega t,
$$
where here $k$ is the beam wave-number magnitude and the angle convention has $\sin\theta>0$ for the chosen forward attenuation coordinate. The omitted denominator in the TeX is essential both physically and dimensionally. Work in the prescribed along-beam kinematic model, with small parcel excursion compared with $\lambda^{-1}$.
The first-order along-beam <fluid displacement>, initially zero, is $\delta\xi=A\sin(\omega t)/\omega$. Therefore the leading displacement correction to the along-beam velocity is
$$
\boxed{u_s(\xi,t)=\delta\xi\,\partial_\xi U
=-\frac{\lambda U_0^2}{2\omega}e^{-2\lambda\xi}\sin(2\omega t)
=-\frac{\nu k^3U_0^2}{4N\omega\sin\theta}e^{-\nu k^3\xi/(N\sin\theta)}\sin(2\omega t).}
$$
This <oscillatory drift of an attenuated internal-wave beam> is oscillatory, so
$$
\boxed{\int_0^{2\pi/\omega}u_s\,dt=0.}
$$
Indeed the scalar parcel equation $\dot\Xi=U_0e^{-\lambda\Xi}\cos\omega t$ can be integrated exactly:
$$
e^{\lambda\Xi(t)}=e^{\lambda\Xi(0)}+\frac{\lambda U_0}{\omega}\sin\omega t.
$$
For excursions small enough that the right side stays positive, the parcel returns to its initial along-beam coordinate after each period. This demonstrates the zero net drift in this supplied model. It is not a claim that all components of a viscous beam, or a separately generated Eulerian mean flow, vanish; the question specifies the along-beam component only.
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