= Solution
In the narrow polar <flux tube>, the <magnetic field lines> are almost radial and the transverse width is much smaller than the radial scale. The <Lorentz force density> has no component along the <magnetic field>, so the longitudinal <magnetically channelled accretion> is hydrodynamic. Let $v(r)>0$ denote inward speed. <Mass conservation>, the <Euler equations for an inviscid fluid>, and the <polytropic equation of state> give
$$
\rho vA=\dot M_{\rm tube},\qquad
v\frac{dv}{dr}=-\frac1\rho\frac{dp}{dr}-\frac{GM_\star}{r^2},\qquad
p=K\rho^\gamma.
$$
Using the <adiabatic sound speed> $c^2=\gamma K\rho^{\gamma-1}$ and $A\propto r^3$,
$$
\frac{\rho'}\rho=-\frac{v'}v-\frac3r,
\qquad
\boxed{\left(v-\frac{c^2}{v}\right)v'=\frac{3c^2}{r}-\frac{GM_\star}{r^2}.}
$$
The <Bernoulli equation> is
$$
\boxed{\frac{v^2}{2}+\frac{c^2}{\gamma-1}-\frac{GM_\star}{r}=\frac{c_0^2}{\gamma-1},}
$$
where the right-hand side comes from matching to a nearly stationary reservoir with negligible <gravitational potential>. The fixed <magnetic field> provides transverse confinement; neglecting magnetic forces in this one-dimensional equation concerns their longitudinal component.
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