Differentiate horizontal momentum with respect to $z$ and vertical momentum with respect to $x$. When the two equations are subtracted, the terms proportional to $\overline u_z(u'_x+w'_z)$ vanish by <incompressible flow>. The pressure derivatives also cancel, leaving
$$
\boxed{
D_t(u'_z-w'_x)+\overline u_{zz}w'+\sigma'_x=0}.
$$

Differentiate this equation with respect to $x$. Since continuity gives
$$
(u'_z-w'_x)_x
=-(w'_{xx}+w'_{zz}),
$$
one obtains
$$
-D_t\nabla^2w'
+\overline u_{zz}w'_x+\sigma'_{xx}=0.
$$
Apply $D_t$ and use the buoyancy equation $D_t\sigma'=-N^2w'$. Multiplication by $-1$ then gives
$$
\boxed{
D_t^2(w'_{xx}+w'_{zz})
-\overline u_{zz}D_tw'_x
+N^2w'_{xx}=0}.
$$

Solution

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

 Solution

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