= Solution
Choose a reference <circular orbit> at radius $r_0$, rotate the frame with $\boldsymbol\Omega=\Omega(r_0)\mathbf e_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\phi-\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading <differential rotation>. The background <velocity> relative to the rotating frame is
$$
\mathbf u_0=r_0\Omega'(r_0)x\mathbf e_y=-2Ax\mathbf e_y,\qquad A=-\frac12r_0\Omega'(r_0).
$$
Thus $A=3\Omega/4$ for a <Keplerian disk>. The <shearing sheet> retains <Coriolis acceleration> and the linear tidal acceleration. The radial <shearing-sheet tidal potential> is $-2\Omega Ax^2$, whose force $4\Omega Ax\mathbf e_x$ balances the background <Coriolis acceleration>. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background <pressure> $p_0(z)$ and disappears when that equilibrium is subtracted.
Write the total <velocity> as $\mathbf u=\mathbf u_0+\mathbf v$. In the rotating-frame <magnetohydrodynamic momentum equation>, background advection is $(\mathbf u_0\cdot\nabla)\mathbf v=-2Ax\partial_y\mathbf v$ and advection of the background by the perturbation is $(\mathbf v\cdot\nabla)\mathbf u_0=-2Av_x\mathbf e_y$. The uniform-viscosity <diffusion> of the linear background shear is zero. For a solenoidal <magnetic field> the <Lorentz force> per unit <mass> is
$$
\frac{(\nabla\times\mathbf B)\times\mathbf B}{\mu_0\rho}=\frac{\mathbf B\cdot\nabla\mathbf B}{\mu_0\rho}-\nabla\frac{B^2}{2\mu_0\rho}.
$$
At constant <density>, absorb the magnetic-pressure gradient into $\psi=(p-p_0)/\rho+B^2/(2\mu_0\rho)$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf v\cdot\nabla\mathbf v$, <Coriolis acceleration> $2\boldsymbol\Omega\times\mathbf v$, <magnetic tension> and viscous <diffusion>.
For constant finite <electrical conductivity> $\sigma_e$, put $\eta=(\mu_0\sigma_e)^{-1}$. The <resistive induction equation> with uniform <magnetic diffusivity> is $\partial_t\mathbf B+\mathbf u\cdot\nabla\mathbf B=\mathbf B\cdot\nabla\mathbf u+\eta\nabla^2\mathbf B$. Substituting the background shear supplies the term $-2AB_x\mathbf e_y$ and the same background-advection operator as in the momentum equation. The two further conditions are
$$
\boxed{\nabla\cdot\mathbf v=0,\qquad\nabla\cdot\mathbf B=0.}
$$
They express <incompressibility> and the absence of <magnetic monopoles>, respectively.
Now take real <plane waves> with common phase $\vartheta=\mathbf k(t)\cdot\mathbf x$ and real <wavevector>. The solenoidal conditions become
$$
\mathbf k\cdot\widetilde{\mathbf v}=0,\qquad \mathbf k\cdot\widetilde{\mathbf B}=0.
$$
Every field depends on position only through $\vartheta$. Hence for either transverse <vector> field $\mathbf P$ and either wave field $\mathbf Q$, $(\mathbf P\cdot\nabla)\mathbf Q=(\mathbf P\cdot\mathbf k)\partial_\vartheta\mathbf Q=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate <amplitudes>. These waves are therefore exact solutions at finite <amplitude>, not merely a <linearization>.
Acting with $D_0=\partial_t-2Ax\partial_y$ on the phase yields
$$
D_0\vartheta=(\dot k_x-2Ak_y)x+\dot k_y y+\dot k_z z.
$$
Canceling its position-dependent part requires
$$
\boxed{\dot k_x=2Ak_y,\qquad\dot k_y=\dot k_z=0,\qquad k_x(t)=k_{x0}+2Ak_yt.}
$$
The remaining <amplitude> equations are
$$
\boxed{\dot{\widetilde{\mathbf v}}-2A\widetilde v_x\mathbf e_y+2\boldsymbol\Omega\times\widetilde{\mathbf v}=-i\mathbf k\widetilde\psi-\nu k^2\widetilde{\mathbf v},\qquad\dot{\widetilde{\mathbf B}}=-2A\widetilde B_x\mathbf e_y-\eta k^2\widetilde{\mathbf B}.}
$$
Differentiating the <velocity> constraint determines the <pressure> <amplitude>:
$$
i k^2\widetilde\psi=4Ak_y\widetilde v_x-2\mathbf k\cdot(\boldsymbol\Omega\times\widetilde{\mathbf v}).
$$
The <resistive induction equation> preserves its constraint because $\dot{\mathbf k}\cdot\widetilde{\mathbf B}=2Ak_y\widetilde B_x$ cancels the corresponding shear term in $\mathbf k\cdot\dot{\widetilde{\mathbf B}}$. The specified <magnetic field> has no spatially uniform part. Thus there is no background <magnetic field> producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified <pressure>; the physical <pressure> can include a constant and a second spatial harmonic from $-B^2/(2\mu_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact <zero-mean magnetic shearing wave>.
For decay, take the real part of the <scalar product> of the velocity-amplitude equation with its <complex conjugate>. <Pressure> does no work because $\mathbf k\cdot\widetilde{\mathbf v}=0$, and the real part of Coriolis work is zero. Therefore
$$
\frac{d}{dt}|\widetilde{\mathbf v}|^2=4A\operatorname{Re}(\widetilde v_x\widetilde v_y^*)-2\nu k^2|\widetilde{\mathbf v}|^2\le(2|A|-2\nu k^2)|\widetilde{\mathbf v}|^2.
$$
The <resistive induction equation> similarly gives
$$
\frac{d}{dt}|\widetilde{\mathbf B}|^2=-4A\operatorname{Re}(\widetilde B_x\widetilde B_y^*)-2\eta k^2|\widetilde{\mathbf B}|^2\le(2|A|-2\eta k^2)|\widetilde{\mathbf B}|^2,
$$
using $2|u_xu_y|\le|\mathbf u|^2$. Integrating these differential inequalities, define
$$
J(t)=\int_0^t k(s)^2\,ds=(k_{x0}^2+k_y^2+k_z^2)t+2Ak_{x0}k_yt^2+\frac43A^2k_y^2t^3.
$$
Then
$$
|\widetilde{\mathbf v}(t)|^2\le|\widetilde{\mathbf v}(0)|^2e^{2|A|t-2\nu J(t)},\qquad |\widetilde{\mathbf B}(t)|^2\le|\widetilde{\mathbf B}(0)|^2e^{2|A|t-2\eta J(t)}.
$$
A non-axisymmetric wave has $k_y\ne0$. For nonzero differential shear and positive $\nu,\eta$, the negative cubic terms dominate the linear upper bound on shear work. Consequently
$$
\boxed{\widetilde{\mathbf v}(t)\longrightarrow0,\qquad\widetilde{\mathbf B}(t)\longrightarrow0\quad(t\to\infty).}
$$
This <cubic-exponent decay of a nonaxisymmetric shearing wave> allows transient <energy> growth before the winding to short radial wavelengths makes <diffusion> dominant. The magnetic <amplitudes> can also be written explicitly as
$$
\widetilde{\mathbf B}(t)=e^{-\eta J(t)}\big(\widetilde B_x(0),\ \widetilde B_y(0)-2At\widetilde B_x(0),\ \widetilde B_z(0)\big),
$$
which confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical <velocity> <amplitude> can remain constant, and an ideal vertical magnetic <amplitude> can remain constant when $\eta=0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.
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