Elimination of $\widehat u$ and $\widehat\phi$ gives
$$
\widehat v_{yy}-\frac{\beta^2B}{c^2A}y^2\widehat v
+\frac{i\beta k}{A}\widehat v=0.
$$
Choose $\sigma^2=(\beta/c)(B/A)^{1/2}$ with $\operatorname{Re}\sigma^2>0$ and set $Y=\sigma y$. The Hermite eigenvalue condition gives, for $n=1,2,\ldots$,
$$
\boxed{\widehat v=V_n(\sigma y)=H_n(\sigma y)e^{-\sigma^2y^2/2},}
$$
$$
\boxed{(\gamma-i\omega)(\alpha-i\omega)+\frac{c^2k^2}{(2n+1)^2}=0.}
$$
Squaring the eigenvalue condition introduces an extraneous branch, so one must also impose $ikc/\sqrt{AB}=2n+1$ and $\operatorname{Re}\sigma^2>0$. For real $k$, trapped propagating roots exist when
$$
\boxed{|k|>\frac{(2n+1)|\alpha-\gamma|}{2c},}
$$
and the accepted root propagates westward, $k\operatorname{Re}\omega<0$. The other root has an outward-growing meridional structure.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 333 / 4 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-333/4/ii/solution