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For nonzero imposed surface inclination, magnetic bending in an incompressible disk is nonmonotonic when $KH>\pi /2$: the first maximum of $|B_x|$ occurs at $z=\pi /(2K)<H$, followed by a decrease. This is exactly the finite-thickness magnetorotational instability criterion for the same slab. The zero-inclination equilibrium can still be unstable, so nonmonotonic bending is a property of the forced profile, rather than a condition on every possible unstable equilibrium.

Write $H=z^{+}$ and $B_{+}=B_x^{+}$, and assume $B_z\ne 0$ and $q>0$. In a steady state, the horizontally invariant magnetized shearing-sheet equations giveSince $B_y$ vanishes at both boundaries, $v_x=0$ for $q\ne 2$, and then $B_y=0$ throughout. The other two equations areEliminating $v_y$ gives the harmonic oscillator equationThe midplane-symmetric magnetic bending in an incompressible disk has odd $B_x$ and even $v_y$. Applying $B_x(\pm H)=\pm B_{+}$ givesFor nonzero imposed inclination $B_{+}$, this equilibrium exists only when $\sin (KH)\ne 0$. 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.