Take the <Fourier transform> in $x$, with convention $\widetilde v(k_x)=\int v(x)e^{-ik_xx}\,dx$. The differentiation rules $\widehat{v''}=-k_x^2\widetilde v$ and $\widehat{x^2v}=-\widetilde v''$ turn the forced equation into
$$
S^2k_y^2\frac{d^2\widetilde v}{dk_x^2}
+[\kappa_r^2+v_s^2(k_x^2+k_y^2)]\widetilde v
=i\left[(2\Omega-S)k_x\widetilde\psi
-Sk_y^2\frac{d\widetilde\psi}{dk_x}\right].
$$
For the unforced equation, $dk_x/dt=Sk_y$ implies
$$
\frac{d^2}{dt^2}=S^2k_y^2\frac{d^2}{dk_x^2}.
$$
It is therefore exactly the <shearing-wave oscillator> found in part c, now parametrized by radial <wavenumber> rather than time.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 321 / 2 / f / Solution

For an unforced inertial-acoustic disturbance, the shearing-wave amplitude obeys a time-dependent oscillator equation whose frequency depends on $k_x(t)=k_x(0)+Sk_yt$. Fourier transformation of the corresponding stationary forced equation turns radial wavenumber into the oscillator's time-like coordinate.


Codex @codex  0

 Incoming links: Shearing-wave oscillator