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Let ${\Sigma }_{d0}$ be the uniform dust background and put ${\delta }_d={\Sigma }_d^{\prime }/{\Sigma }_{d0}$. The linearized continuity equation is ${\partial }_t{\delta }_d=-{\partial }_x{v}_x^{\prime }$ because background shear has no divergence or axisymmetric advection. Using the late entrained velocity $v_x^{\prime }=u\cos \theta$ givesThus the coherent travelling wave has maximum fractional excess $|ku|/\Omega$, and its crest density is ${\Sigma }_{d0}[1+|ku|/\Omega ]$ when the stationary part is absent. This is the density contrast of an entrained dust wave, and requires $|ku|/\Omega \ll 1$ in the present linear approximation. Its fractional oscillating density amplitude is the same as the gas's long-wavelength amplitude, so the coherent wave produces no additional dust-to-gas concentration ratio at this order.

Velocity damping alone does not erase the time-independent $C(x)$; initial density and the history of the forcing are needed to fix it. The printed data therefore determine the travelling-wave contrast, rather than a unique total late-time concentration for arbitrary initial histories. As an explicit example, switch on the prescribed gas epicycle at $t=0$ with initially uniform dust and ${\mathbf{v}}^{\prime }=0$. The corrected velocity pair gives $v_x^{\prime }=u(1-e^{-ϵ\Omega t})\cos \theta$. Integrating continuity then leavesat late times. It is smaller by order $ϵ$ in the stipulated weak-coupling regime, and confirms the stated leading wave-crest estimate while showing why arbitrary stationary density structure is not removed by the velocity transient.