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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-345/1/c/solution