The <linearized vortensity> amplitude is
$$
\boxed{\widetilde\zeta
=\frac1{\Sigma_0}\left[
i k_x\widetilde u_y-i k_y\widetilde u_x
-\frac\Omega2\widetilde\delta\right]}.
$$
Since the exact vortensity obeys $D_t\zeta=0$ and the time-dependent $k_x$ cancels the background shear in the phase, linearization gives
$$
\boxed{\frac{d\widetilde\zeta}{dt}=0}.
$$

On the zero-vortensity branch,
$$
\boxed{\widetilde\delta
=\frac{2i}{\Omega}(k_x\widetilde u_y-k_y\widetilde u_x)}.
$$
Substitution in the two momentum equations yields the <linear system of differential equations>
$$
\frac d{dt}
\begin{pmatrix}\widetilde u_x\\ \widetilde u_y\end{pmatrix}
=\mathbf A(t)
\begin{pmatrix}\widetilde u_x\\ \widetilde u_y\end{pmatrix},
$$
with
$$
\boxed{
\mathbf A(t)=
\begin{pmatrix}
-\dfrac{2c_s^2k_xk_y}{\Omega}&
2\Omega+\dfrac{2c_s^2k_x^2}{\Omega}\\[2mm]
-\dfrac\Omega2-\dfrac{2c_s^2k_y^2}{\Omega}&
\dfrac{2c_s^2k_xk_y}{\Omega}
\end{pmatrix}},
$$
where $k_x=k_x(t)$ is the result from part (ii).

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 3 / c / iii / Solution

Linearized vortensity is the first-order perturbation of vortensity about a background flow. For a shearing wave in a uniform Keplerian sheet, its amplitude is constant.


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