= Solution
Define the Eulerian-mean streamfunction by
$$
(\overline v_a,\overline w_a)
=(\overline\chi_{a,z},-\overline\chi_{a,y}).
$$
With
$$
A=\frac{\overline{\rho'v'}}{d\rho_s/dz}
=\frac{\overline F^{(z)}}{f_0},
$$
the transformation in part a gives
$$
\overline v_a^*
=(\overline\chi_a-A)_z,
\qquad
\overline w_a^*
=-(\overline\chi_a-A)_y.
$$
Hence, up to an irrelevant additive constant,
$$
\boxed{
\overline\chi_a
=\overline\chi_a^*
+\frac{\overline F^{(z)}}{f_0}}.
$$
For the step-profile flux,
$$
\boxed{
\overline\chi_a
=\overline\chi_a^*
+\frac{F_0}{f_0}\Theta(z)
\sin^2\frac{\pi y}{L}}.
$$
The residual streamfunction $\overline\chi_a^*$ vanishes on the rigid boundaries, decays away from the absorption level, and has opposite-signed values immediately below and above $z=H$. Its vertical derivative gives a zonal acceleration $\overline u_t$ concentrated around $H$ and largest near the channel center. Its meridional derivative gives a dipolar density tendency: $\overline\rho_t$ changes sign across the channel center and reverses vertical structure across the absorption level.
The eddy term in $\overline\chi_a$ has a compensating downward jump at $H$, so the Eulerian-mean streamfunction is continuous even though $\overline\chi_a^*$ jumps in the idealized step limit. Below the critical layer, the Eulerian view contains a broad circulation associated with the eddy density flux; the transformed view subtracts that reversible eddy-induced motion and isolates the residual circulation forced where the waves dissipate.
In the Eulerian density budget, vertical advection by $\overline w_a$ and the divergence of $\overline{\rho'v'}$ can be individually large and largely cancel. The <transformed Eulerian mean> combines them into advection by $\overline w_a^*$, making the irreversible mean response to <wave-activity deposition> much clearer.
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