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Write $\mathbf{u}={\partial }_t\mathbf{\xi }$. In the uniform equilibrium, the linearized ideal magnetohydrodynamic equations reduce toIntegrating from the undisplaced reference state, and using the uniform background magnetic field, givesThese are the perturbations induced by the fluid displacement; independent time-independent changes to the reference state are excluded. The linear Lorentz force density is $(\nabla \times \delta \mathbf{B})\times \mathbf{B}/{\mu }_0$. Its magnetic pressure and magnetic tension parts giveFor a Fourier mode, put $s=\mathbf{k}\cdot \mathbf{\xi }$, $q=\mathbf{B}\cdot \mathbf{\xi }$ and $K=\mathbf{k}\cdot \mathbf{B}$. Then $\delta p=-i\gamma ps$ and $\delta \mathbf{B}=i(K\mathbf{\xi }-\mathbf{B}s)$. Substitution givesIf $s=q=0$, the fluid displacement is perpendicular to both the wave vector and the magnetic field. Only magnetic tension restores it, andThis is the Alfvén wave. The Alfvén velocity is the field-directed propagation vector: $\omega =\pm \mathbf{k}\cdot {\mathbf{v}}_a$. The signed phase velocity normal to the wavefront is $\pm v_a\cos \theta \hat{\mathbf{k}}$; the group velocity is $\pm {\mathbf{v}}_a$. The distinction matters for oblique propagation. For generic directions the Alfvén wave has one transverse polarization, with $\delta \rho =\delta p=0$.

To obtain the other magnetohydrodynamic waves, take the scalar products with $\mathbf{k}$ and $\mathbf{B}$:The determinant condition for a nonzero pair $(s,q)$ isThus the fast magnetosonic wave and slow magnetosonic wave haveTheir fluid displacements lie in the plane of $\mathbf{k}$ and $\mathbf{B}$ and are generally compressive. In the fast magnetosonic wave, gas and magnetic pressure provide the stronger restoring combination; in the slow magnetosonic wave, their perturbations oppose one another. The phase speeds depend on direction. They satisfy $v_{\mathrm{sl}}\le \min (c_s,v_a)$ and $v_{\mathrm{f}}\ge \max (c_s,v_a)$. For parallel propagation the two speeds are $c_s$ and $v_a$, with a degeneracy between a transverse branch and the Alfvén wave. For perpendicular propagation $v_{\mathrm{f}}^2=c_s^2+v_a^2$ and $v_{\mathrm{sl}}=0$; the latter is a nonpropagating limiting disturbance. These special directions require interpreting the polarizations by continuity rather than assuming three distinct nonzero frequencies.

For $\bar{\nu }=C{r}^a{\Sigma }^b$, the response is $q=b+1$. A nonaccreting background has ${\bar{\nu }}_0{\Sigma }_0\propto r^{-1/2}$, while ${\Sigma }_0\propto r^{-p}$, soConsequently ${\bar{\nu }}_0\propto r^{p-1/2}$. On this steady background, introduce the square-root-radius diffusion transformSince $q$ is constant and ${\partial }_r=(2x{)}^{-1}{\partial }_x$, the linear equation becomesThus $A(x)\propto x^{2p-3}$, and the required choice isFor a Fourier mode $g\propto e^{\lambda t+i{k}_{x}x}$, $\lambda =-A{k}_x^2$. Positive kinematic viscosity makes $A$ have the sign of $q$, so $q<0$ gives growing modes whose rate increases with $k_x^2$. The formal equation is a backward heat equation and predicts arbitrarily rapid small-scale amplification. Physically the thin disc diffusion closure applies only to wavelengths sufficiently larger than the thickness and stress-relaxation scales; this formal limit identifies the need for a cutoff, rather than a finite fastest wavelength absent from the model. The boundary case $q=0$ has vanishing linear transport response.

The zero Dirichlet boundary conditions permit an exact finite-interval calculation. Let $T=\mathrm{tridiag}(1,-2,1)$ on the $M$ interior grid points. The Dirichlet discrete Laplacian has an orthogonal eigenvector basisConsequently the update matrix $G=(I-rT{)}^{-1}$ has modal amplification factorsFor every physical $r\ge 0$, all denominators are at least one. Orthogonal diagonalization therefore provesuniformly in $M$, $r$ and $n$. The same bound controls initial perturbations; a forcing increment is propagated by contraction, so successive increments accumulate at most by their sum. The highest-order consistent method is unconditionally stable for all $r\ge 0$. This proof incorporates the boundary conditions, whereas a periodic Fourier mode calculation alone would not do so.

For completeness, if “range” is interpreted algebraically to include negative $r$ on a fixed grid, the exact power-stable range isIndeed, for $r<0$ every denominator is less than one, so $|g_j|\le 1$ requires $1+4r{s}_j\le -1$ for every $j$. The strongest restriction comes from $s_1$. Equality gives a simple amplification factor $-1$ and is allowed because the update is orthogonally diagonalizable. Other negative values either amplify a mode or make the update singular. This extra branch describes backward time stepping on a fixed spatial grid; it is not a positive-time diffusion discretization, and no fixed negative $r$ remains in that branch as $M\to \infty$.