Dot the momentum equation with $\mathbf u$. The <Coriolis acceleration> does no work, while the buoyancy-variable equation gives
$$
\frac D{Dt}\frac12N^2\theta^2=N^2\theta u_z,
$$
which cancels the buoyancy work $-N^2\theta u_z$. Thus
$$
\frac{DE}{Dt}=-\frac1{\rho_0}\mathbf u\cdot\nabla P
+3\Omega^2xu_x.
$$
Using incompressibility,
$$
\mathbf u\cdot\nabla P=\nabla\cdot(P\mathbf u),
\qquad
3\Omega^2xu_x=\nabla\cdot\left(\frac32\Omega^2x^2\mathbf u\right).
$$
Therefore
$$
\boxed{\partial_tE+\nabla\cdot\mathbf F=0,
\qquad
\mathbf F=\mathbf u\left(E+\frac P{\rho_0}-\frac32\Omega^2x^2\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 321 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-321/3/b/solution