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In the steady state of the horizontally invariant magnetized shearing-sheet equations, let $B_x(\pm H)=\pm B_{+}$ and $B_y(\pm H)=0$. For $q>0$ and $B_z\ne 0$, the symmetric equilibrium hasThis follows by combining azimuthal induction with radial magnetic tension balance to obtain $B_x^{\prime \prime }+K^2{B}_x=0$. Nonzero $B_{+}$ requires $\sin (KH)\ne 0$; even homogeneous additions at $\cos (KH)=0$ are excluded by imposing midplane symmetry.

In the rotating shearing sheet, write the background orbital shear as ${\mathbf{U}}_0=-q\Omega x{\mathbf{e}}_y$. The rotating momentum equation contains Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{u}$ and the shearing-sheet tidal potential; the background shear balances the radial tidal acceleration.

SetGauss's law for magnetism gives ${\partial }_z{B}_z=0$, and the vertical ideal magnetohydrodynamic induction equation gives ${\partial }_t{B}_z=0$. Horizontal invariance and $v_z=0$ remove the nonlinear horizontal advection. The remaining background-shear term is $(\mathbf{v}\cdot \nabla ){\mathbf{U}}_0=-q\Omega v_x{\mathbf{e}}_y$. The horizontal magnetic tension is $B_z{\partial }_z{\mathbf{B}}_h/{\mu }_0$, while horizontal pressure gradients vanish. Subtracting background balance yieldsFor ideal magnetohydrodynamics, the ideal magnetohydrodynamic induction equation is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}$. Its horizontal components giveThese horizontally invariant magnetized shearing-sheet equations are exact within the stated local, incompressible ansatz, even for finite horizontal amplitudes. The vertical equation determines the pressure needed to balance vertical gravity and magnetic pressure; it does not add another horizontal evolution equation.

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.

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.