= Solution
Apply the <shearing-wave ansatz> to each disturbance. Acting on its phase gives
$$
D_0e^{i[k_x(t)x+k_y(t)y]}=i[(\dot k_x-Sk_y)x+\dot k_y y]e^{i\mathbf k\cdot\mathbf x}.
$$
Thus a spatially uniform amplitude system exists when
$$
\boxed{\dot{\mathbf k}=Sk_y\mathbf e_x,\qquad k_y=\text{constant},\qquad k_x(t)=k_x(0)+Sk_yt.}
$$
For a nonzero horizontal <wavenumber> $k=|\mathbf k|$, the <razor-thin disk Poisson kernel> follows from $(\partial_z^2-k^2)\widetilde\Phi_d'=4\pi G\widetilde\Sigma'\delta(z)$. The potential decays above and below the sheet as $\widetilde\Phi_{d,m}'e^{-k|z|}$. Integrating across the midplane gives the jump $-2k\widetilde\Phi_{d,m}'=4\pi G\widetilde\Sigma'$, hence
$$
\widetilde\Phi_{d,m}'=-\frac{2\pi G}{k}\widetilde\Sigma'.
$$
Substitute this potential and the phase evolution into the linearized equations. The <barotropic shearing-wave gravitational amplitude system> is
$$
\boxed{\begin{aligned}\dot{\widetilde\Sigma}'&=-i\Sigma_0(k_x\widetilde v_x+k_y\widetilde v_y),\\\dot{\widetilde v}_x-2\Omega\widetilde v_y&=-ik_x\left(c_s^2-\frac{2\pi G\Sigma_0}{k}\right)\frac{\widetilde\Sigma'}{\Sigma_0},\\\dot{\widetilde v}_y+(2\Omega-S)\widetilde v_x&=-ik_y\left(c_s^2-\frac{2\pi G\Sigma_0}{k}\right)\frac{\widetilde\Sigma'}{\Sigma_0}.\end{aligned}}
$$
The self-gravity sign is attractive and opposes the <pressure> response. The uniform $k=0$ <mass density> shift requires its own gravitational reference and is not obtained by dividing by $k$.
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