Apply $\widehat{\mathbf z}\mathbin\cdot\nabla\times$ to the linear momentum equation. The pressure and buoyancy terms vanish, while incompressibility gives $\widehat{\mathbf z}\cdot\nabla\times(\widehat{\mathbf z}\times\mathbf u')=-D W$. Hence
$$
\boxed{(\partial_t-\sigma\nabla^2)\omega-\lambda DW=0}.
$$
Applying $\widehat{\mathbf z}\cdot\nabla\times\nabla\times$ and using the identity supplied in the question gives
$$
\boxed{
(\partial_t-\sigma\nabla^2)\nabla^2W
+\lambda D\omega
=\sigma Ra\,\nabla_H^2\theta'}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 331 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-331/1/c/solution