= Solution
Write $H=z^+$ and $B_+=B_x^+$, and assume $B_z\ne0$ and $q>0$. In a <steady state>, the <horizontally invariant magnetized shearing-sheet equations> give
$$
v_x'=0,\qquad
B_y'=\frac{\mu_0\rho}{B_z}(2-q)\Omega v_x.
$$
Since $B_y$ vanishes at both boundaries, $v_x=0$ for $q\ne2$, and then $B_y=0$ throughout. The other two equations are
$$
v_y=-\frac{B_z}{2\mu_0\rho\Omega}B_x',\qquad
v_y'=\frac{q\Omega}{B_z}B_x.
$$
Eliminating $v_y$ gives the <harmonic oscillator equation>
$$
B_x''+K^2B_x=0,\qquad
K^2=\frac{2q\mu_0\rho\Omega^2}{B_z^2}.
$$
The midplane-symmetric <magnetic bending in an incompressible disk> has odd $B_x$ and even $v_y$. Applying $B_x(\pm H)=\pm B_+$ gives
$$
\boxed{B_x=B_+\frac{\sin(Kz)}{\sin(KH)},\qquad B_y=0,\qquad v_x=0},
$$
$$
\boxed{v_y=-\frac{B_zKB_+}{2\mu_0\rho\Omega\sin(KH)}\cos(Kz)}.
$$
For nonzero imposed inclination $B_+$, this equilibrium exists only when $\sin(KH)\ne0$. More generally, $B_x=A\sin(Kz)+D\cos(Kz)$; the <boundary conditions> require $A\sin(KH)=B_+$ and $D\cos(KH)=0$. Thus the displayed solution is unique away from these resonances. At $\cos(KH)=0$, an additional even homogeneous solution is possible unless midplane symmetry is imposed. At $q=2$, an arbitrary constant $v_x$ is also allowed because its coefficient in the azimuthal equation vanishes; choosing $v_x=0$ gives the same symmetric equilibrium.
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