= Solution
Write $\Delta\mathbf u=(U,W,0)$. The background shear gives
$$
\mathbf u_0\mathbin\cdot\nabla\mathbf u_0
=-\frac32\Omega_KU\mathbf e_y.
$$
The constant radial and azimuthal components of the dust equation are therefore
$$
2\Omega_KW-\frac U\tau=0,
\qquad
\frac12\Omega_KU+\frac{W-V}{\tau}=0.
$$
With the <Stokes number> $\mathrm{St}=\tau\Omega_K$, their solution is
$$
\boxed{\Delta\mathbf u
=\frac{2\mathrm{St}\,V}{1+\mathrm{St}^2}\mathbf e_x
+\frac{V}{1+\mathrm{St}^2}\mathbf e_y}.
$$
The gas is slower than a Keplerian particle, so the particle feels an aerodynamic headwind. <Drag force> removes its <angular momentum>, and it drifts radially inward because $U<0$. Its radial speed is
$$
|U|=2|V|\frac{\mathrm{St}}{1+\mathrm{St}^2}.
$$
Differentiation with respect to $\mathrm{St}>0$ shows that the unique maximum occurs at
$$
\boxed{\tau\Omega_K=1},
$$
where $|U|=|V|$.
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