Eigenmode equations using $(P_1, \xi _{\psi }, \xi _s, \nabla \cdot \xi )$ as variables

Using Eqs. (121) and (122) to eliminate $Q_{\psi}$, $Q_s$, Eqs. (124) and (125) are written, respectively, as

$$\nabla \Psi \cdot \nabla P_1 = \omega^2 \rho_0 \xi_{\psi} +{\textmu}_0^{- 1} \vert \nabla \Psi \... ... \frac{\mathbf{B}_0 \cdot \nabla \xi_{\psi}}{\vert \nabla \Psi \vert^2} \right)$$

$$+ ({\textmu}_0^{- 1} \vert \nabla \Psi \vert^2 S - B_0^2 \sigma) \f... ...rt \nabla \Psi \vert^2}{B^2_0} (\mathbf{B}_0 \cdot \nabla \xi_s - S \xi_{\psi})$$

$$+ 2{\textmu}_0^{- 1} \kappa_{\psi} Q_b .$$ (127)

and

$$- \omega^2 \rho_0 \vert \nabla \Psi \vert^2 \xi_s = - (\mathbf{B}_0 \times \nabla \Psi) \cdot \nabla P_1 + B_0^2 \sig... ...a \Psi \vert^2}{B^2_0} (\mathbf{B}_0 \cdot \nabla \xi_s - S \xi_{\psi}) \right]$$

$$+ 2{\textmu}_0^{- 1} B_0^2 \kappa_s Q_b$$ (128)

Following Ref. [3], we will express the final eigenmodes equations in terms of the following four variables: $P_1$, $\xi_{\psi}$, $\xi_s$, and $\nabla \cdot \ensuremath{\boldsymbol{\xi}}$. In order to achieve this, we need to eliminate unwanted variables. Since we will use $P_1 \equiv p_1 +\mathbf{B}_0 \cdot \mathbf{B}_1 /{\textmu}_0$ instead of $p_1$ as one of the variables. we write the equation (120) for the perturbed pressure in terms of $P_1$ variable, which gives

$$P_1 -{\textmu}_0^{- 1} Q_b + p_0' \xi_{\psi} = - \gamma p_0 \nabla \cdot \ensuremath{\boldsymbol{\xi}}.$$ (129)

Using (129) to eliminate $Q_b$ in Eq. (128), we obtain

$$- \frac{\omega^2 \rho_0 \vert \nabla \Psi \vert^2}{B_0^2} \xi_s = - \left( \frac{\mathbf{B}_0 \times \nabla \Psi}{B_0^2} \right) \c... ...a \Psi \vert^2}{B^2_0} (\mathbf{B}_0 \cdot \nabla \xi_s - S \xi_{\psi}) \right]$$

$$+ 2 \kappa_s \gamma p_0 \nabla \cdot \ensuremath{\boldsymbol{\xi}}+ 2 \kappa_s P_1 + 2 \kappa_s p_0' \xi_{\psi}$$ (130)

Using pressure equation (120) to eliminate $p_1$ in Eq. (126), we obtain

$$\omega^2 \rho_0 \xi_b = - \gamma p_0 \mathbf{B}_0 \cdot \nabla (\nabla \cdot \ensuremath{\boldsymbol{\xi}}),$$ (131)

which can be used in Eq. (123) to eliminate $\xi_b$, yielding

$$Q_b = \xi_{\psi} \left( - \frac{2 B^2_0}{\vert \nabla \Psi \vert^... ...}_0 \cdot \nabla (\nabla \cdot \ensuremath{\boldsymbol{\xi}})}{B^2_0} \right) .$$ (132)

Using Eq. (132) to eliminate $Q_b$ in Eq. (129), we obtain

$$P_1 +{\textmu}_0^{- 1} \left[ \frac{2 B^2_0}{\vert \nabla \Psi \v... ...^2_0} \right) \right] = - \gamma p_0 \nabla \cdot \ensuremath{\boldsymbol{\xi}}$$ (133)

Equations (130) and (133), which involves only surface derivative operators, can be put in the following matrix form:

$$\left(\begin{array}{cc} E_{11} & E_{12}\\ E_{21} & E_{22} \end{ar... ...} \end{array}\right) \left(\begin{array}{c} P_1\\ \xi_{\psi} \end{array}\right)$$ (134)

with the matrix elements given by

$$E_{11} = - \frac{\omega^2 \rho_0 \vert \nabla \Psi \vert^2}{B_0^2... ...left( \frac{\vert \nabla \Psi \vert^2}{B^2_0} \mathbf{B}_0 \cdot \nabla \right)$$ (135)

$$E_{12} = - 2 \kappa_s \gamma p_0$$ (136)

$$E_{21} = 2{\textmu}_0^{- 1} \kappa_s$$ (137)

$$E_{22} = \frac{{\textmu}_0^{- 1} B_0^2 + \gamma p_0}{B_0^2} +{\te... ...mathbf{B}_0 \cdot \nabla \left( \frac{\mathbf{B}_0 \cdot \nabla}{B^2_0} \right)$$ (138)

$$F_{11} = 2 \kappa_s - \left( \frac{\mathbf{B}_0 \times \nabla \Psi}{B_0^2} \right) \cdot \nabla$$ (139)

$$F_{12} = \sigma \mathbf{B}_0 \cdot \nabla -{\textmu}_0^{- 1} \mat... ...c{\vert \nabla \Psi \vert^2}{B^2_0} S \right) + 2 \kappa_s \frac{d p_0}{d \Psi}$$ (140)

$$F_{21} = - \frac{1}{B_0^2}$$ (141)

$$F_{22} = -{\textmu}_0^{- 1} \frac{2}{\vert \nabla \Psi \vert^2} \kappa_{\psi} .$$ (142)

Equations (134)-(142) agree with equations (25), (28), and (29) of Ref [3].

Using Eq. (129) to eliminate $Q_b$ in Eq. (127), we obtain

$$\nabla \Psi \cdot \nabla P_1 = \omega^2 \rho_0 \xi_{\psi} +{\textmu}_0^{- 1} \vert \nabla \Psi \... ... \frac{\mathbf{B}_0 \cdot \nabla \xi_{\psi}}{\vert \nabla \Psi \vert^2} \right)$$

$$+ ({\textmu}_0^{- 1} \vert \nabla \Psi \vert^2 S - B_0^2 \sigma) \f... ...rt \nabla \Psi \vert^2}{B^2_0} (\mathbf{B}_0 \cdot \nabla \xi_s - S \xi_{\psi})$$

$$+ 2 \kappa_{\psi} [P_1 + \gamma p_0 \nabla \cdot \ensuremath{\boldsymbol{\xi}}+ \xi_{\psi} p_0']$$ (143)

The equation for the divergence of $\ensuremath{\boldsymbol{\xi}}$ [Eq. (95)] is written

$$\nabla \Psi \cdot \nabla \xi_{\psi} = \vert \nabla \Psi \vert^2 \... ...}_0 \cdot \nabla (\nabla \cdot \ensuremath{\boldsymbol{\xi}})}{B^2_0} \right) .$$ (144)

Equations (143) and (144) can be put in the following matrix form

$$\nabla \Psi \cdot \nabla \left(\begin{array}{c} P_1\\ \xi_{\psi} ... ...rray}{c} \xi_s\\ \nabla \cdot \ensuremath{\boldsymbol{\xi}} \end{array}\right),$$ (145)

with the matrix elements given by

$$C_{11} = 2 \kappa_{\psi}$$ (146)

$$C_{12} = \omega^2 \rho_0 +{\textmu}_0^{- 1} \vert \nabla \Psi \ve... ...frac{\vert \nabla \Psi \vert^2}{B^2_0} S + 2 \kappa_{\psi} \frac{d p_0}{d \Psi}$$ (147)

$$D_{11} = ({\textmu}_0^{- 1} \vert \nabla \Psi \vert^2 S - B_0^2 \sigma) \frac{\vert \nabla \Psi \vert^2}{B^2_0} \mathbf{B}_0 \cdot \nabla$$ (148)

$$D_{12} = 2 \gamma p_0 \kappa_{\psi}$$ (149)

$$C_{21} = 0$$ (150)

$$C_{22} = - \vert \nabla \Psi \vert^2 \nabla \cdot \left( \frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \right)$$ (151)

$$D_{21} = - \vert \nabla \Psi \vert^2 \frac{(\mathbf{B}_0 \times \nabla \Psi)}{B^2_0} \cdot \nabla + 2 \vert \nabla \Psi \vert^2 \kappa_s$$ (152)

$$D_{22} = \vert \nabla \Psi \vert^2 + \frac{\gamma p_0 \vert \nabl... ...thbf{B}_0 \cdot \nabla \left( \frac{\mathbf{B}_0 \cdot \nabla}{B^2_0} \right) .$$ (153)

Equations (146)-(153) agree with Equations (26) and (27) in Ref.[3]. (It took me several months to manage to put the equations in the matrix form given here.)