Define the <wave-action density>
$$
W=\frac{\partial\overline L^{(0)}}{\partial\omega}
=\frac12A^2\omega.
$$
The local dispersion relation implies
$$
c_g=\frac{\partial\omega}{\partial k}
=\frac{c^2k}{\omega},
$$
which is the <group velocity>. Moreover,
$$
\frac{\partial\overline L^{(0)}}{\partial k}
=-\frac12A^2c^2k=-c_gW.
$$
Substitution into the phase-variation equation gives the <wave-action conservation law>
$$
\boxed{
\partial_TW+\partial_X(c_gW)=0.}
$$
On the two nondispersive branches $\omega=\pm ck$, the group velocities are $c_g=\pm c$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 336 / 1 / b / iii / Solution

{title2=$W$}

For an averaged Lagrangian, the wave-action density is $W=\partial\overline L/\partial\omega$, up to the sign convention used for the fast phase.


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