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The additional conditions are $\boxed{\nabla \cdot \mathbf{u}=0,\nabla \cdot \mathbf{B}=0}$: incompressible flow and the solenoidal magnetic field constraint. The scalar $\Psi$ combines pressure per unit density with magnetic pressure per unit density and any conservative gravitational and centrifugal potentials. In particular, the Lorentz force identity separates magnetic tension from $-\nabla (B^2/2{\mu }_0\rho )$. Adding this latter term to the rotating-flow pressure potential gives the pressure appearing in the stated equation; the Coriolis force remains explicit.

Work in magnetic velocity units $\mathbf{B}/\sqrt{{\mu }_0\rho }$, and write $\mathbf{U}=-2Ax{\mathbf{e}}_y$ for the background shearing sheet flow. Its advective acceleration is zero, and $2\Omega {\mathbf{e}}_z\times \mathbf{U}=4\Omega Ax{\mathbf{e}}_x=-\nabla (-2\Omega A{x}^2)$, so the stated quadratic background pressure potential gives an exact balance. A spatially uniform mean magnetic field must evolve asThis is stretching of radial field into azimuthal field by differential rotation.

For the shearing wave phase $\Theta =\mathbf{k}(t)\cdot \mathbf{x}$, advection by the background givesIt vanishes identically ifThe changing wavevector describes the shear of initially planar wavefronts. The mean-field stretching and wavevector evolution cancel in the Alfvén frequency:Hence $\boxed{{\omega }_A=\mathbf{k}\cdot \bar{\mathbf{b}}\text{is constant}}$.

These solutions are exact finite-amplitude disturbances, not merely a linearization. To see this, let ${\mathbf{u}}^{\prime }=\mathrm{Re}(\mathbf{v}{e}^{i\Theta })$ and ${\mathbf{B}}^{\prime }=\mathrm{Re}(\mathbf{b}{e}^{i\Theta })$ in magnetic velocity units. The divergence constraints give $\mathbf{k}\cdot \mathbf{v}=\mathbf{k}\cdot \mathbf{b}=0$. Thus both real disturbances are pointwise perpendicular to k, whereas all their spatial derivatives are along k. Every quadratic term ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }$, ${\mathbf{B}}^{\prime }\cdot \nabla {\mathbf{B}}^{\prime }$, ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{B}}^{\prime }$ and ${\mathbf{B}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }$ vanishes, including terms involving conjugate amplitudes. This extends the exact incompressible shearing wave to an exact magnetic shearing wave. The gas pressure can compensate quadratic magnetic-pressure terms even though the combined $\Psi$ contains only its stated background and wave harmonic.

The remaining wave terms are therefore preciselyThe shear terms arise from perturbations acting on $\nabla \mathbf{U}$, the magnetic terms from the uniform field acting on the wave, and the last terms from diffusion. Pressure enforces the moving-wavevector velocity constraint: differentiating it givesUsing this pressure leaves $\mathbf{k}\cdot \mathbf{v}=0$ invariant. Differentiating $\mathbf{k}\cdot \mathbf{b}$ in the induction equation similarly leaves it zero. Thus the evolution is consistent with both constraints for initially transverse amplitudes.

Now take positive $\Omega$, so the stipulated condition implies $0<A<2\Omega$, and take magnetic Prandtl number $\nu /\eta =1$, with $\nu =\eta >0$. For a vertical wavevector $\mathbf{k}=k{\mathbf{e}}_z$, use the equal-diffusivity magnetorotational channel formThe two horizontal momentum equations, after adding them, give $\dot{V}=(A-\nu k^2)V$. Either component then gives $i{\omega }_{A}B=(A-2\Omega )V$. The first induction equation gives $({\partial }_t+\nu k^2)B=i{\omega }_{A}V$. These equations are simultaneously solved byThe second induction component gives the same relation, since $B_y=-B_x$. Exponential growth occurs when $\boxed{\nu k^2<A}$. The quarter-cycle phase between magnetic and velocity perturbations is required by the factor i in the Fourier equations.

For Keplerian rotation, expanding the local shear gives $2A=-(r{\Omega }^{\prime })=3\Omega /2$, hence $A=3\Omega /4$. Therefore ${\omega }_A^2=15{\Omega }^2/16$. With $v_{Az}={\bar{B}}_z/\sqrt{{\mu }_0\rho }$ and ${\omega }_A=k{v}_{Az}$, the growth condition for this prescribed polarization becomesA weaker field would require a shorter wavelength to retain this mode's selected Alfvén frequency, making diffusive damping too strong. The required wavelength must also fit any actual vertical domain; no vertical boundary conditions are specified here.

The field bound is for the requested equal-component mode, rather than a universal necessary condition for all magnetorotational instability. This distinction matters if all vertical wavelengths are admitted. Indeed, let $s=\gamma +\nu k^2$ for a general horizontal polarization. Eliminating its two magnetic amplitudes gives $(s+{\omega }_A^2/s)v_x-2\Omega v_y=0$ and $(s+{\omega }_A^2/s)v_y+[2(\Omega -A)-2A{\omega }_A^2/s^2]v_x=0$. Their determinant givesFor a Keplerian disk, ${\kappa }^2={\Omega }^2$. As a concrete counterexample to a universal reading of the bound, in consistent dimensionless units take $\Omega =1$, $A=3/4$, $\nu =\eta =1$, $v_{Az}=0.1$, and $k=0.01$. The positive root gives $s\simeq 0.00173205$ and $\gamma =s-0.0001>0$, even though $v_{Az}^2=0.01<1.25$. This mode has a different polarization and a much longer wavelength. Real finite-height discs restrict such wavelengths; the PDF supplies no such restriction.

Finally, the horizontally averaged radial flux of azimuthal momentum is the sum of Reynolds stress and magnetic tension stress:Multiplication by the local radius gives an outward angular-momentum flux. Equal radial and azimuthal velocities give the largest positive product at fixed horizontal kinetic energy, while opposite magnetic components give the largest positive $-B_x{B}_y$ at fixed horizontal magnetic energy. Both follow from $2|ab|\le |a{|}^2+|b{|}^2$. This stress-maximizing magnetorotational channel polarization therefore makes both contributions reinforce outward angular momentum transport. It extracts shear energy at rate $-T_{xy}{\partial }_x{U}_y=2A{T}_{xy}>0$, explaining why these correlations are especially effective in driving accretion.