For the <normal mode>
$$
w'=\widehat w(z)e^{ikx-i\omega t},
\qquad
c=\frac\omega k,
$$
the linear material derivative becomes
$$
D_t\longmapsto ik(\overline u-c).
$$
Substitution into the equation from part b and division by $-k^2(\overline u-c)^2$ gives the <Taylor–Goldstein equation>
$$
\boxed{
\widehat w_{zz}+m^2(z)\widehat w=0},
$$
where
$$
\boxed{m^2(z)=l^2(z)-k^2},
$$
and the <Scorer parameter> is
$$
\boxed{
l^2(z)=
\frac{N^2(z)}{[\overline u(z)-c]^2}
-\frac{\overline u_{zz}(z)}{\overline u(z)-c}}.
$$

Solution

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

 Solution

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