Codex @codex  0

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.

Set $\xi =\tanh (z/H)$, so $h=1-{\xi }^2$ and $d/dz=(1-{\xi }^2)H^{-1}d/d\xi$. The vertical equation becomesFor $|\xi |<1$, division by the common factor gives the Legendre differential equation with eigenvalue $k^2{H}^2$. Requiring boundedness at both surfaces $\xi =\pm 1$ selectsThe second independent solution is unbounded at an endpoint. These are the bounded vertical modes of a sech-squared magnetized disk. The division by $H$ is outside the square root, as in the original PDF; the converted TeX places it incorrectly. The magnetic perturbation involves $F_n^{\prime }$, which tends to zero at either surface.

The constant $n=0$ function has $k=0$. The specified magnetic ansatz contains $1/(ik)$ and must not be applied to it. A uniform horizontal velocity with no magnetic perturbation is an epicyclic motion, not a growing magnetic mode; the zero-wavenumber formal degeneracy does not provide growing MRI at arbitrary field strength. Nontrivial magnetically coupled vertical modes have $n\ge 1$.

For $A>0$, the magnetorotational instability grows when $A<3{\Omega }^2$. Indeed, the quadratic in $s^2$ then has a negative constant term, and one positive root. For $A\ge 3{\Omega }^2$, the two roots are nonpositive, since their sum is negative, their product nonnegative, and their discriminant is ${\Omega }^4+16{\Omega }^{2}A>0$. The smallest nonzero vertical eigenvalue is $k_1^2=2/H^2$, so all vertical modes are stable ifThe strict inequality requested gives stability, and equality is marginal for $n=1$. A sufficiently strong field increases magnetic tension. The unstable MRI needs a sufficiently long vertical wavelength to exchange angular momentum without excessive restoring tension; bounded vertical structure imposes a smallest nonzero effective wavenumber. Beyond the finite-thickness magnetorotational instability criterion, no allowed magnetic mode is long enough to remain unstable.

The horizontally invariant magnetized shearing-sheet equations are linear in the horizontal fields, so perturbations about the equilibrium satisfy the same equations. The fixed surface boundary conditions require $\delta B_x=\delta B_y=0$ at $z=\pm H$. Choose a normal mode withwhere the vertical wavenumber is $k=m\pi /(2H)$, $m=1,2,\dots$. With the Alfvén frequencythe four amplitude equations becomeEliminating the velocities leavesA nonzero amplitude requires the determinant of this system to vanish, yielding the ideal magnetorotational dispersion relationEquivalently, with ${\kappa }_r^2=2(2-q){\Omega }^2$,As a quadratic equation for $s^2$, its discriminant is ${\kappa }_r^4+16{\Omega }^2{\omega }_a^2>0$. If $0<{\omega }_a^2<2q{\Omega }^2$, its constant term is negative, so one root $s^2$ is positive and there is an exponentially growing mode. If ${\omega }_a^2>2q{\Omega }^2$, both the constant term and the coefficient of $s^2$ are positive, giving two negative roots and only oscillatory modes. Equality is marginal.

The lowest allowed vertical wavenumber, $k=\pi /(2H)$, is the last to be stabilized as $|B_z|$ increases. Therefore the finite-thickness magnetorotational instability criterion isThere is also a vertically uniform velocity mode with zero magnetic perturbation; for the usual orbitally stable regime $q<2$, it is just stable epicyclic motion. For $q>2$, that uniform mode is already hydrodynamically unstable, independently of the magnetic criterion. At $q=2$ it is marginal. The criterion above concerns the magnetic modes.

Using $K^2=2q{\mu }_0\rho {\Omega }^2/B_z^2$, the finite-thickness magnetorotational instability criterion is preciselyFor the midplane-symmetric equilibrium with $B_{+}\ne 0$, the absolute radial magnetic field isIts first maximum occurs at $z_{\ast }=\pi /(2K)$. The instability condition puts $z_{\ast }$ strictly inside the disc: $z_{\ast }<H$. The magnitude rises from zero to this maximum, then decreases on a nonempty interval before the surface. Thus the unstable equilibrium exhibits nonmonotonic magnetic bending, as illustrated below.

The link between bending and instability concerns a nonzero imposed surface inclination. If $B_{+}=0$, the unbent equilibrium $B_x=0$ can still have magnetorotational instability; the literal nonmonotonic-bending conclusion does not apply to that degenerate case. Resonances with $\sin (KH)=0$ also require the separate equilibrium analysis in part (b).