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For either scalar equation, the complementary solution isso initial velocity transients decay when $t\gg (ϵ\Omega {)}^{-1}$. In the exact corrected forced dust epicycle with aerodynamic drag, $v_{x,p}^{\prime }=u\cos \theta$ is a particular solution, and the original radial equation then gives $v_{y,p}^{\prime }=u\sin \theta /2$. ConsequentlyThis is resonant dust entrainment by an epicyclic gas wave: in the adopted long-wavelength gas approximation, the forcing is a free epicycle of the dust too, so matching the gas makes the drag zero. The convergence holds for every $ϵ>0$ in this linear model; at $ϵ=0$ transients never decay, so the zero-drag and late-time limits do not commute.

If instead the printed scalar equation is treated literally for finite $ϵ$, its particular solution isIt has amplitude $|u|/\sqrt{1+{ϵ}^2/4}$ and phase offset $\arctan (ϵ/2)$, not exact phase/amplitude agreement. Agreement in that truncated equation means only leading order as $ϵ\to 0$. This distinguishes the intended approximation from a false exact reading of the printed assertion.

There is a further order-of-limits condition: a real finite-wavelength gas wave has $\omega -\Omega \ne 0$. Leading resonant entrainment over the damping time requires $|\omega -\Omega |\ll ϵ\Omega$, or $(k{c}_s/\Omega {)}^2\ll ϵ$. Otherwise solve the dust transfer equations at the actual $\omega$; detuning can reduce the dust response even though $|k|c_s/\Omega \ll 1$.

The pressureless dust fluid in a shearing sheet has no pressure restoration and is driven by drag force toward the prescribed gas velocity; gas backreaction is omitted in the given model. After linearizing about the common background shear, axisymmetry givesApply $D_{ϵ}={\partial }_t+ϵ\Omega$ to the first equation and use the second to eliminate $v_y^{\prime }$. The exact linear scalar equation isFor the supplied leading-order epicycle, set $\theta =kx-\Omega t$. Since ${\partial }_t{u}_x^{\prime }=\Omega u\sin \theta$ and $u_y^{\prime }=u\sin \theta /2$, this becomesThe final cosine term is missing from the printed equation. Dropping it is justified at leading order in $0<ϵ\ll 1$, recovering the intended forcing, but the printed equation retains an order-${ϵ}^2$ frequency term on the left and is not an exact derivation at that order. A direct countercheck is ${\mathbf{v}}^{\prime }={\mathbf{u}}^{\prime }$: the drag vanishes and the pair solves both dust equations exactly, while $v_x^{\prime }=u\cos \theta$ leaves residual ${ϵ}^2{\Omega }^{2}u\cos \theta$ in the printed scalar equation. This is the forced dust epicycle with aerodynamic drag; its damping time is $(ϵ\Omega {)}^{-1}$.

If the prescribed gas velocity is itself a free epicyclic motion solution, subtract its two momentum equations from the forced dust epicycle with aerodynamic drag equations. The velocity difference obeys a damped epicycle and decays as $e^{-ϵ\Omega t}$. Therefore dust approaches the gas velocity with identical phase and amplitude for every positive drag rate. At zero drag the difference does not decay, so the late-time limit is nonuniform as drag vanishes. A finite-wavelength inertial-acoustic wave is detuned from the free epicycle; matching its phase and amplitude in the weak-drag limit additionally requires $|\omega -\Omega |\ll ϵ\Omega$. For a gas density wave of frequency $\omega$, the exact radial transfer amplitude isobtained by solving the two dust equations and using $U_y=-i\Omega U_x/(2\omega )$. At $\omega =\Omega$ it is exactly one.