Put $l=\pi/L$. For the specified <quasi-geostrophic streamfunction>, the complex amplitudes of the disturbance fields are
$$
\widehat u=-l\widehat\psi\cos(ly),
\qquad
\widehat v=ik\widehat\psi\sin(ly).
$$
The product of these two amplitudes is purely imaginary after one is conjugated, so its zonal mean vanishes:
$$
\overline{u'v'}
=\frac12\operatorname{Re}(\widehat u\widehat v^*)=0.
$$
Consequently
$$
\boxed{\overline F^{(y)}=0}.
$$

<Geostrophic balance> and <hydrostatic pressure> give
$$
\widehat\rho
=-\frac{\rho_0f_0}{g}
\widehat\psi_z\sin(ly).
$$
Therefore
$$
\overline{\rho'v'}
=-\frac{\rho_0f_0k}{2g}
\operatorname{Im}
(\widehat\psi_z\widehat\psi^*)
\sin^2(ly).
$$
Since $d\rho_s/dz=-\rho_0N^2/g$, the vertical <Eliassen–Palm flux> is
$$
\boxed{
\overline F^{(z)}
=\frac{f_0^2}{N^2}
\frac k2
\operatorname{Im}
(\widehat\psi_z\widehat\psi^*)
\sin^2\frac{\pi y}{L}}.
$$
Thus it has the stated form
$$
\boxed{
\overline F^{(z)}=F_0\Theta(z)
\sin^2\frac{\pi y}{L}},
\qquad
F_0=\frac{f_0^2}{N^2},
$$
with
$$
\boxed{
\Theta(z)=\frac k2
\operatorname{Im}
(\widehat\psi_z\widehat\psi^*)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 333 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-333/4/c/solution