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Use the meridional overturning streamfunction conventionDifferentiating geostrophic and hydrostatic balance gives the thermal wind relation $f_0{\overline{u}}_z=(g/{\rho }_0){\overline{\rho }}_y$. Differentiate it in time and use the mean momentum and density equations. The resulting diagnostic elliptic boundary value problem isFor the transformed circulation it becomes the Eliassen equation for residual circulation, with signs appropriate to the convention above:At the rigid meridional walls, impermeability requires each streamfunction to be constant along each wall. Set those constants to zero, choosing the solution with no imposed net vertical throughflow. Since the prescribed density flux vanishes at both walls, the same conditions apply to ${\mathcal{X}}^{\ast }$.

There is a genuine data limitation: prescribing $F$ alone does not prescribe $R$. The term $-f_0{R}_{yz}$ remains in both circulation equations. To give the usual explicit step-flux circulation, assume vanishing meridional momentum-flux divergence, $R_y=0$, with no independently imposed circulation. The formulas below are conditional on that assumption, not a unique consequence of the density flux alone.

Let $l=\pi /L$, $\kappa =N_{0}l/|f_0|$ and $A_0=-1/S>0$, using the flux unit in which its lower plateau is $-1$. The bounded solution has $\mathcal{X}=\sin (ly)\chi (z)$. Its vertical equation isAt vertical infinity require bounded Eulerian flow and vanishing residual flow. Requiring the Eulerian vertical velocity to vanish also at lower infinity would be inconsistent with the continuing lower-region flux divergence. At $z=0$, continuity of $\chi$ and ${\chi }^{\prime }$ follows because the forcing has a step but no delta function. The step-flux residual circulation in a stratified channel is thereforeHere ${\mathcal{X}}^{\ast }=\sin (ly){\chi }^{\ast }$, because $F/S=A_0\sin (ly)$ below zero and vanishes above it. The residual streamfunction decays at both infinities, whereas the Eulerian streamfunction tends to $-A_0\sin (ly)$ below and to zero above.

For $z\ne 0$, the Eulerian velocity isThe flow descends near the $y=0$ wall and rises near $y=L$, with a negative meridional return flow concentrated near $z=0$. Below the dissipation level its vertical flow balances the continuing divergence of eddy density transport. There is no independently added throughflow, but opposing vertical transports at lower infinity are required by this idealized forcing.

Away from the discontinuity the residual meridional velocity equals the Eulerian value, whileThus the residual mean circulation forms two oppositely rotating cells, confined within vertical distance $O(1/\kappa )$ of the termination of the wave flux. Its streamfunction jumps by $-A_0\sin (ly)$ at zero. Distributionally this supplies the horizontal return transportThe Dirac delta sheet is a consequence of the discontinuous imposed flux, not an extra boundary condition. Smoothing the flux termination replaces it by a thin finite return current. Streamlines must not be drawn as continuously crossing the jump without that transport.

The physical Eulerian mean flow is continuous. In the transformed momentum equation, the sheet contribution to $f_0{v}^{\ast }$ cancels the sheet in the flux divergence, leaving the smooth mean accelerationThus termination of the vertical Eliassen–Palm flux produces a finite column-integrated wave drag, redistributed over depth by the residual mean circulation.

For completeness, the general bounded response can be calculated without setting $R_y$ to zero. Expand $R_{yz}$ in wall-compatible sine modes, with coefficients $B_n(z)$, $l_n=n\pi /L$ and ${\kappa }_n=N_0{l}_n/|f_0|$. Add to either displayed streamfunction the same correctionfor decaying momentum-flux forcing. The Green function sign follows from $({\partial }_z^2-{\kappa }_n^2)e^{-{\kappa }_n|z-z^{\prime }|}=-2{\kappa }_n\delta (z-z^{\prime })$.

An explicit counterexample to uniqueness is the additional momentum flux $R=ϵ[1-\cos (2ly)]e^{-a|z|}$, with $a>0$. It vanishes at both walls and leaves the prescribed density flux unchanged. For $a\ne 2\kappa$ it addsto both circulations. This nonzero bounded, wall-impermeable solution verifies that the original mean equations need an additional momentum-flux specification to select a unique sketch.