Let
$$
I=\int\widehat w^2\,dz.
$$
Because the <Rayleigh quotient> is stationary with respect to first-order changes of its eigenfunction, only the explicit dependence of $l^2$ on $c=\omega/k$ contributes when the quotient is differentiated. Thus
$$
2k\,dk
=\frac1I\int
\frac{\partial l^2}{\partial c}
\widehat w^2\,dz\,dc,
$$
where
$$
\frac{\partial l^2}{\partial c}
=\frac{2N^2}{(\overline u-c)^3}
-\frac{\overline u_{zz}}{(\overline u-c)^2}.
$$
It follows that
$$
\frac{dc}{dk}
=\frac{2kI}
{\displaystyle\int
(\partial l^2/\partial c)\widehat w^2\,dz}.
$$
Since $\omega=kc$, the horizontal <group velocity> is
$$
\boxed{
\frac{\partial\omega}{\partial k}
=c+
\frac{2k^2I}
{\displaystyle\int
\left[
\frac{2N^2}{(\overline u-c)^3}
-\frac{\overline u_{zz}}{(\overline u-c)^2}
\right]\widehat w^2\,dz}}.
$$

For a stationary ridge wave, $c=0$. Assume $\overline u>0$, so downstream is the positive $x$ direction. The stated inequality is
$$
l^2> -\frac{N^2}{\overline u^2}
\quad\Longleftrightarrow\quad
\frac{2N^2}{\overline u^3}
-\frac{\overline u_{zz}}{\overline u^2}>0.
$$
The denominator in the group-velocity formula is then positive, and
$$
\boxed{\partial\omega/\partial k>0}.
$$
The <atmospheric internal gravity waves> generated by the ridge consequently carry their wave packet downstream.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 333 / 3 / f / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-333/3/f/solution