Discrete form of elements of matrix $C$, $D$, $E$, and $F$

The elements of matrix $C$, $D$, $E$, and $F$ are two dimensional differential operators about $(\theta, \zeta)$, which can be called surface differential operators. As discussed above, we use Fourier expansion to treat the differential with respect to $\theta$ and $\zeta$. In this method, we need to take inner product between different Fourier harmonics. Noting this, we recognize that it is useful to define the following inner product operator:

$$\langle W \rangle_{n' n m' m} \equiv \frac{1}{(2 \pi)^2} \int_0^{... ... [i (n' \zeta - m' \theta)] W \exp [i (m \theta - n \zeta)] d \theta d \zeta,$$

where $W$ is a surface differential operator of the following form

$$W = a (\psi, \theta) \frac{\partial}{\partial \theta} + b (\psi, \theta) \frac{\partial}{\partial \zeta} .$$ (175)

Because both of the coefficients in expression (175) are independent of $\zeta$, it is ready to see that, for $n' \neq n$, $\langle W \rangle_{n' n m' m} = 0$. This indicates that $G_{n m}$ with different $n$ are decoupled with each other.

For notation ease, $\langle W \rangle_{n' n m' m}$ is denoted by $\langle W \rangle_{m' m}$ when $n' = n$, i.e.,

$$\langle W \rangle_{m' m} = \frac{1}{2 \pi} \int_0^{2 \pi} \exp [i (n \zeta - m' \theta)] W \exp [i (m \theta - n \zeta)] d \theta,$$

[For the special case that $W$ is an algebra operator $W = W (\psi, \theta)$, equation (5) can be reduced to

$$\langle W \rangle_{m' m} = \frac{1}{2 \pi} \int_0^{2 \pi} e^{i (m - m') \theta} W d \theta,$$ (176)

where $W (\psi, \theta)$ is independent of $\zeta$ because, as we will see below, $W$ is determined by equilibrium quantities (for example, $W (\psi, \theta) = \frac{\mathcal{J}^{- 2}}{B_0^2}$), which is axisymetrical. Expression (176) is a Fourier integration over the interval $[0, 2 \pi]$, which can be efficiently calculated by using the FFT algorithm (details are given in Chapter 13.9 of Ref. [5]).] After using the Fourier harmonics expansion and taking the inner product, every element of the matrices $C, D, E$, and $F$ becomes a $L \times L$ matrix, where $L$ is the total number of the Fourier harmonics included in the expansion. Taking the matrix $\overline{E}$ as an example, it is discretized as

$$\left(\begin{array}{cc} \overline{E}_{11} & \overline{E}_{12}\\ \... ...erline{E}_{22}^{(L 1)} & \ldots & \overline{E}_{22}^{(L L)} \end{array}\right),$$ (177)

where $\overline{E}_{11}^{(m' m)} \equiv \langle \overline{E}_{11} \rangle_{m' m}$, $\overline{E}_{12}^{(m' m)} \equiv \langle \overline{E}_{12} \rangle_{m' m}$, $\overline{E}_{21}^{(m' m)} \equiv \langle \overline{E}_{21} \rangle_{m' m}$, $\overline{E}_{22}^{(m' m)} \equiv \langle \overline{E}_{22} \rangle_{m' m}$. Next, let us derive the expressions of $\overline{E}_{11}^{(m' m)}$, $\overline{E}_{12}^{(m' m)}$, $\overline{E}_{21}^{(m' m)}$, $\overline{E}_{22}^{(m' m)}$. The goal of he derivation is to perform the surface differential operators so that all the inner products take the form of the Fourier integration given by Eq. (176). For the convenience of reference, the expression of matrix $\overline{E}$ is repeated here:

$$\overline{E} = \left(\begin{array}{cc} - \frac{2 \overline{\omeg... ...a \left( \frac{\mathbf{B}_0 \cdot \nabla}{B^2_0} \right) \end{array}\right) .$$

Then $\overline{E}_{11}^{(m' m)}$ is written as

$$\overline{E}_{11}^{m' m} \equiv \langle \overline{E}_{11} \rangle_{m' m}$$

$$= - \frac{2 \overline{\omega}^2 \overline{\rho}_0}{R_a^2} \left\lan... ...bla \Psi \vert^2}{B^2_0} \mathbf{B}_0 \cdot \nabla \right) \right\rangle_{m' m}$$ (178)

Making use of Eq. (173), equation (178) is written as

$$\overline{E}_{11}^{m' m} = - \frac{2 \overline{\omega}^2 \overlin... ...rt \nabla \Psi \vert^2 \mathcal{J}^{- 2}}{B_0^2} \right\rangle_{m' m} \right] .$$ (179)

Note that all the operators within the inner operator $\langle \ldots \rangle_{m' m}$ of the above equation are algebra operators. Therefore the calculation of the inner product $\langle \ldots \rangle_{m' m}$ reduces to the calculation of the Fourier integration (176), which can be efficiently calculated by using the FFT algorithm (it is thus implemented in GTAW). Similarly, the discrete form of the other matrix elements are written respectively as:

$$\overline{E}_{12}^{m' m} \equiv \langle \overline{E}_{12} \rangle_{m' m}$$

$$= - 2 \gamma \overline{p}_0 \langle \kappa_s \rangle_{m' m} .$$ (180)

$$\overline{E}_{21}^{m' m} \equiv \langle \overline{E}_{21} \rangle_{m' m}$$

$$= \frac{4}{B_a^2} \langle \kappa_s \rangle_{m' m}$$ (181)

$$\overline{E}_{22}^{m' m} = \langle \overline{E}_{22} \rangle_{m' m}$$

$$= \frac{2}{B_a^2} \langle 1 \rangle_{m' m} + \gamma \overline{p}_0 ... ...bla \left( \frac{\mathbf{B}_0 \cdot \nabla}{B^2_0} \right) \right\rangle_{m' m}$$

$$= \frac{2}{B_a^2} \langle 1 \rangle_{m' m} + \gamma \overline{p}_0 ... ... q)^2 \left\langle \frac{\mathcal{J}^{- 2}}{B_0^2} \right\rangle_{m' m} \right]$$ (182)

Next, consider the discrete form of the normalized $F$ matrix, which is given by

$$\left(\begin{array}{cc} F_{11} & \overline{F}_{12}\\ F_{21} & \ov... ...ac{4}{B_a^2} \frac{\kappa_{\psi}}{\vert \nabla \Psi \vert^2} \end{array}\right)$$ (183)

Using Eqs. (139) and (159), we obtain

$$F_{11}^{m' m} = \langle F_{11} \rangle_{m' m}$$

$$= 2 \langle \kappa_s \rangle_{m' m} - \left\langle \left( \frac{\mathbf{B}_0 \times \nabla \Psi}{B_0^2} \right) \cdot \nabla \right\rangle_{m' m}$$

$$= 2 \langle \kappa_s \rangle_{m' m} - i (m - n q) \Psi' g \left\lan... ...c{\mathcal{J}^{- 1}}{B_0^2} \right\rangle_{m' m} + i n \langle 1 \rangle_{m' m}$$ (184)

$$\overline{F}_{12}^{m' m} = \langle \overline{F}_{12} \rangle_{m' m}$$

$$= \left\langle \frac{2}{B_a^2} {\textmu}_0 \sigma \mathbf{B}_0 \cdo... ...^2_0} \right) + 2 \kappa_s \frac{d \overline{p}_0}{d \Psi} \right\rangle_{m' m}$$

$$= \frac{2}{B_a^2} \langle {\textmu}_0 \sigma \mathbf{B}_0 \cdot \na... ...ngle_{m' m} + 2 \frac{d \overline{p}_0}{d \Psi} \langle \kappa_s \rangle_{m' m}$$

$$= - \Psi' \frac{2}{B_a^2} i (m - n q) \langle {\textmu}_0 \sigma \m... ... m} \right] + 2 \frac{d \overline{p}_0}{d \Psi} \langle \kappa_s \rangle_{m' m}$$ (185)

$$\langle F_{21} \rangle_{m' m} = - \left\langle \frac{1}{B_0^2} \right\rangle_{m' m}$$ (186)

$$\langle \overline{F}_{22} \rangle_{m' m} = - \frac{4}{B_a^2} \left\langle \frac{\kappa_{\psi}}{\vert \nabla \Psi \vert^2} \right\rangle_{m' m},$$ (187)

Next, we derive the discrete form of matrix $C$ and $D$. Before doing this, we examine matrix equation (161), which can be written as

$$\left(\begin{array}{cc} \nabla \Psi \cdot \nabla & 0\\ 0 & \nabla... ...ray}{c} \xi_s\\ \nabla \cdot \ensuremath{\boldsymbol{\xi}} \end{array}\right) .$$ (188)

Using the expression of the operator $\nabla \Psi \cdot \nabla$, i.e.,

$$\nabla \Psi \cdot \nabla = \Psi' \vert \nabla \psi \vert^2 \frac{... ...heta} + \Psi' (\nabla \zeta \cdot \nabla \psi) \frac{\partial}{\partial \zeta},$$ (189)

equation (188) is written as

$$\left(\begin{array}{cc} \Psi' \vert \nabla \psi \vert^2 & 0\\ 0 &... ...array}{c} \xi_s\\ \nabla \cdot \ensuremath{\boldsymbol{\xi}} \end{array}\right)$$ (190)

Define the first matrix on the r.h.s of the above equation as $\mathcal{H}$, then $\mathcal{H}_{12} =\mathcal{H}_{21} = 0$, and $\mathcal{H}_{11}$ and $\mathcal{H}_{22}$ are given by

$$\mathcal{H}_{11} =\mathcal{H}_{22} = - \Psi' \nabla \theta \cdot ... ...eta} - \Psi' (\nabla \zeta \cdot \nabla \psi) \frac{\partial}{\partial \zeta} .$$ (191)

Then

$$\langle \mathcal{H}_{11} \rangle_{m' m} = \langle \mathcal{H}_{22... ...rangle_{m' m} + i n \Psi' \langle \nabla \zeta \cdot \nabla \psi \rangle_{m' m}$$ (192)

$$\langle C_{11} \rangle_{m' m} = 2 \langle \kappa_{\psi} \rangle_{m' m}$$ (193)

$$\langle \overline{C}_{12} \rangle = \frac{2}{R_a^2} \overline{\omega}^2 \overline{\rho}_0 \langle 1 \... ...{m' m} + 2 \langle \kappa_{\psi} \rangle_{m' m} \frac{d \overline{p}_0}{d \Psi}$$

$$= \frac{2}{R_a^2} \overline{\omega}^2 \overline{\rho}_0 \langle 1 \... ...{m' m} + 2 \langle \kappa_{\psi} \rangle_{m' m} \frac{d \overline{p}_0}{d \Psi}$$

$$\langle C_{22} \rangle_{m' m} = - \left\langle \vert \nabla \Psi ... ...eft( \frac{\nabla \Psi}{\vert \nabla \Psi \vert^2} \right) \right\rangle_{m' m}$$ (194)

The formula for calculating the right-hand side of Eq. (194) is given in Sec. 8.6.

$$\langle \overline{D}_{11} \rangle_{m' m} = - i (m - n q) \frac{2}{B_a^2} \Psi' \left\langle (\vert \nabla \P... ... \frac{\mathcal{J}^{- 1} \vert \nabla \Psi \vert^2}{B^2_0} \right\rangle_{m' m}$$ (195)

$$\langle \overline{D}_{12} \rangle_{m' m} = 2 \gamma \overline{p}_0 \langle \kappa_{\psi} \rangle_{m' m} .$$ (196)

$$\langle D_{21} \rangle_{m' m} = - i (m - n q) \Psi' g \left\langle \frac{\vert \nabla \Psi \vert^... ...^2 \rangle_{m' m} + 2 \langle \vert \nabla \Psi \vert^2 \kappa_s \rangle_{m' m}$$ (197)

$$\langle D_{22} \rangle_{m' m} = \langle \vert \nabla \Psi \vert^2 \rangle_{m' m} + \frac{\gamma \... ...bla \left( \frac{\mathbf{B}_0 \cdot \nabla}{B^2_0} \right) \right\rangle_{m' m}$$

$$= \langle \vert \nabla \Psi \vert^2 \rangle_{m' m} + (\Psi')^2 \fra... ...vert \nabla \Psi \vert^2 \mathcal{J}^{- 2}}{B_0^2} \right\rangle_{m' m} \right]$$ (198)