= Solution
Initially every particle has $A=B=0$, so its <Jacobi energy in a shearing sheet> is $-3\Omega_0^2x_0^2/8$. During an <inelastic collision>, positions are fixed at the instant of impact, while <momentum conservation> preserves the sum of the tangential velocities. Hence the sum of $p_y=\dot y+2\Omega_0x$, and therefore the sum of the <epicyclic guiding center> positions $x_0=2p_y/\Omega_0$, is unchanged. Between collisions, these quantities are individually conserved.
The collision dissipates kinetic energy without changing the instantaneous tidal potential. Consequently the total <Jacobi energy in a shearing sheet> decreases. At any later time, writing $x_{0j}$ for the particles' current <epicyclic guiding center> positions gives
$$
E=E_{\rm osc}-\frac38\Omega_0^2\sum_jx_{0j}^2,\qquad
E_{\rm osc}=\frac12\Omega_0^2\sum_j(|A_j|^2+|B_j|^2)\geq0.
$$
If $D\geq0$ is the accumulated energy dissipated in the <inelastic collisions>, comparison with the initial circular orbits gives
$$
\boxed{\sum_jx_{0j}^2
=\sum_jx_{0j,\rm initial}^2+\frac{8}{3\Omega_0^2}(D+E_{\rm osc})}.
$$
Thus the mean guiding-center position stays fixed, while its variance grows. The initial ensemble has no preference for positive or negative $x$, and the local equations and collision law preserve the symmetry $(x,y,z)\mapsto(-x,-y,z)$. The spreading is therefore symmetric in the ensemble average: <angular momentum transport> moves some particles inward and others outward. A particular finite random realization need not be exactly symmetric. This is the microscopic energy argument for <dissipative spreading of a planetary ring>.
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