= Solution
Steady axisymmetric <mass conservation> is $\nabla\cdot(\rho\mathbf u_p)=0$, permitting the <axisymmetric hydrodynamic mass flux function> representation
$$
u_R=-\frac{\psi_z}{\rho R},\qquad u_z=\frac{\psi_R}{\rho R},\qquad
\mathbf u_p\cdot\nabla\psi=0.
$$
Thus stream surfaces are regular level surfaces of $\psi$. Adiabaticity gives $\mathbf u\cdot\nabla s=0$, and axisymmetry removes the azimuthal derivative, so the <specific entropy> is constant along their meridional streamlines. The azimuthal <Euler equations for an inviscid fluid> has no <pressure> or gravitational torque:
$$
\mathbf u_p\cdot\nabla u_\phi+u_Ru_\phi/R=0
\quad\Longrightarrow\quad
\mathbf u_p\cdot\nabla(Ru_\phi)=0.
$$
The conserved <specific angular momentum> is $\ell=Ru_\phi$. Taking the scalar product of the steady <Crocco form of the unsteady Euler equation> with $\mathbf u$ also gives
$$
\mathbf u\cdot\nabla\varepsilon=T\mathbf u\cdot\nabla s=0,\qquad
\varepsilon=u^2/2+\Phi+w.
$$
Consequently, on each connected regular family of stream surfaces,
$$
\boxed{s=s(\psi),\qquad \ell=\ell(\psi),\qquad
\varepsilon=\varepsilon(\psi).}
$$
These functions express entropy, angular momentum and the <Bernoulli function> carried by a <stellar wind>. Stagnation regions and disconnected branches with identical flux labels require their own matching information; the argument establishes the functions locally where the poloidal flow is nonzero.
To derive the <hydrodynamic transfield equation for an axisymmetric wind>, decompose the <vorticity>. The poloidal part comes from the toroidal <velocity>:
$$
\boldsymbol\omega_p=\nabla(Ru_\phi)\times\nabla\phi
=\ell'(\psi)\nabla\psi\times\nabla\phi=\rho\ell'\mathbf u_p.
$$
The azimuthal part comes from the poloidal <velocity>:
$$
\begin{aligned}
\omega_\phi&=\partial_z u_R-\partial_Ru_z\\
&=-\partial_z\left(\frac{\psi_z}{\rho R}\right)
-\partial_R\left(\frac{\psi_R}{\rho R}\right)
=-R\nabla\cdot\left(\frac{\nabla\psi}{\rho R^2}\right).
\end{aligned}
$$
The final divergence is the full axisymmetric cylindrical divergence, including its radial $1/R$ factor.
Since $\mathbf u_p=\nabla\psi\times\mathbf e_\phi/(\rho R)$, the meridional part of $\boldsymbol\omega\times\mathbf u$ is
$$
\boldsymbol\omega_p\times u_\phi\mathbf e_\phi
+\omega_\phi\mathbf e_\phi\times\mathbf u_p
=\left(-\frac{\ell\ell'}{R^2}+\frac{\omega_\phi}{\rho R}\right)\nabla\psi.
$$
The remaining steady <Euler equations for an inviscid fluid> is therefore
$$
\left(-\frac{\ell\ell'}{R^2}+\frac{\omega_\phi}{\rho R}\right)\nabla\psi
=(Ts'-\varepsilon')\nabla\psi.
$$
Canceling $\nabla\psi$ on a regular stream surface and inserting $\omega_\phi$ gives
$$
\boxed{\frac1\rho\nabla\cdot\left(\frac{\nabla\psi}{\rho R^2}\right)
=\frac{d\varepsilon}{d\psi}-T\frac{ds}{d\psi}
-\frac{\ell}{R^2}\frac{d\ell}{d\psi}.}
$$
The sign of the angular-momentum term follows from $\mathbf u_p\times\mathbf e_\phi=-\nabla\psi/(\rho R)$. To determine the wind, this transverse force equation is supplemented by its <Bernoulli function> relation
$$
\varepsilon(\psi)=\frac{|\nabla\psi|^2}{2\rho^2R^2}
+\frac{\ell(\psi)^2}{2R^2}+\Phi+w(\rho,s(\psi)),
$$
and the <equation of state>, with the <gravitational potential> supplied externally or by the <Poisson equation>.
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