Grad-Shafranov equation with prescribed safety factor profile (to be finished)

In the GS equation, $g (\Psi)$ is one of the two free functions which can be prescribed by users. In some cases, we want to specify the safety factor profile $q (\Psi)$, instead of $g (\Psi)$, in solving the GS equation. Next, we derive the form of the GS equation that contains $q (\Psi)$, instead of $g (\Psi)$, as a free function. The safety factor defined in Eq. (182) can be written

$$q (\psi) = - \frac{1}{2 \pi} \frac{g}{\Psi'} \int_0^{2 \pi} \frac{\mathcal{J}}{R^2} d \theta$$ (469)

$$= - \frac{1}{2 \pi} \frac{g}{\Psi'} \frac{\int_0^{2 \pi} R^{- 2} \m... ... \theta}{\int_0^{2 \pi} \mathcal{J}d \theta} \int_0^{2 \pi} \mathcal{J}d \theta$$

$$= - \frac{1}{2 \pi} \frac{g}{\Psi'} \langle R^{- 2} \rangle \int_0^{2 \pi} \mathcal{J}d \theta$$ (470)

$$= - \ensuremath{\operatorname{sgn}} (\mathcal{J}) \frac{V'}{(2 \pi)^2} \frac{\langle R^{- 2} \rangle}{\Psi'} g.$$ (471)

Equation (471) gives the relation between the safety factor $q$ and the toroidal field function $g$. This relation can be used in the GS equation to eliminate $g$ in favor of $q$, which gives

$$g = - \ensuremath{\operatorname{sgn}} (\mathcal{J}) \frac{(2 \pi)^2}{V' \langle R^{- 2} \rangle} \Psi' q.$$ (472)

$$\Rightarrow \frac{d g}{d \Psi} g = \frac{(2 \pi)^2}{V' \langle R^... ...e} q \left[ \frac{(2 \pi)^2}{V' \langle R^{- 2} \rangle} q \Psi' \right]_{\psi}$$ (473)

Multiplying Eq. (468) by $R^{- 2}$ gives

$$\frac{1}{\mathcal{J}} \left[ \left( \Psi' \frac{\mathcal{J}}{R^2}... ...{\theta} \right] + \mu_0 \frac{d p}{d \Psi} + R^{- 2} \frac{d g}{d \Psi} g = 0.$$ (474)

Surface-averaging the above equation, we obtain

$$\frac{2 \pi}{V'} \int d \theta \left[ \left( \Psi' \frac{\mathcal... ... + \mu_0 \frac{d p}{d \Psi} + \langle R^{- 2} \rangle \frac{d g}{d \Psi} g = 0,$$ (475)

$$\Rightarrow \frac{2 \pi}{V'} \int d \theta \left( \Psi' \frac{\ma... ... + \mu_0 \frac{d p}{d \Psi} + \langle R^{- 2} \rangle \frac{d g}{d \Psi} g = 0,$$ (476)

$$\Rightarrow \frac{2 \pi}{V'} \left[ \Psi' \int d \theta \left( \m... ... + \mu_0 \frac{d p}{d \Psi} + \langle R^{- 2} \rangle \frac{d g}{d \Psi} g = 0,$$ (477)

$$\Rightarrow \frac{1}{V'} \left[ V' \Psi' \left\langle \frac{\vert... ... + \mu_0 \frac{d p}{d \Psi} + \langle R^{- 2} \rangle \frac{d g}{d \Psi} g = 0.$$ (478)

$$\Rightarrow \left[ V' \Psi' \left\langle \frac{\vert \nabla \psi ... ..._0 V' \frac{d p}{d \Psi} + \langle R^{- 2} \rangle V' \frac{d g}{d \Psi} g = 0,$$ (479)

Substitute Eq. (473) into the above equation to eliminate $g d g / d \Psi$, we obtain

$$\Rightarrow \left[ V' \Psi' \left\langle \frac{\vert \nabla \psi ... ...(2 \pi)^4 \left[ \frac{q \Psi'}{V' \langle R^{- 2} \rangle} \right]_{\psi} = 0,$$ (480)

Eq. (480) agrees with Eq. (5.55) in Ref. [9].

$$\Rightarrow V' \left\langle \frac{\vert \nabla \psi \vert^2}{R^2}... ... (2 \pi)^4 \left[ \frac{q}{V' \langle R^{- 2} \rangle} \right]_{\psi} \Psi' = 0$$ (481)

$$\Rightarrow \Psi'' = - \frac{1}{V' D} \left\{ \left[ V' \left\lan... ...le R^{- 2} \rangle} \right]_{\psi} \Psi' + \mu_0 V' \frac{d p}{d \Psi} \right\}$$ (482)

where

$$D = \left\langle \frac{\vert \nabla \psi \vert^2}{R^2} \right\rangle + \frac{(2 \pi)^4}{\langle R^{- 2} \rangle} \left( \frac{q}{V'} \right)^2$$ (483)

The GS equation is

$$\triangle^{\star} \Psi = - \mu_0 R^2 \frac{d p}{d \Psi} - g \frac{d g}{d \Psi}$$ (484)

$$\Rightarrow \triangle^{\star} \Psi = - \mu_0 R^2 \frac{d p}{d \Ps... ...{- 2} \rangle} \left[ \frac{q \Psi'}{V' \langle R^{- 2} \rangle} \right]_{\psi}$$ (485)

$$\Rightarrow \triangle^{\star} \Psi = - \mu_0 R^2 \frac{d p}{d \Ps... ...si} \Psi' + \left[ \frac{q}{V' \langle R^{- 2} \rangle} \right] \Psi'' \right\}$$ (486)

Using Eq. (482) to eliminate $\Psi''$ in the above equation, the coefficients before ( $- \mu_0 d p / d \Psi$) is written as

$$B = R^2 - \frac{q (2 \pi)^4}{V' \langle R^{- 2} \rangle} \left\{ \left[ \frac{q}{V' \langle R^{- 2} \rangle} \right] \frac{1}{V' D} V' \right\}$$

$$= \frac{1}{D} \left\{ R^2 D - \left( \frac{q}{V'} \right)^2 \frac{(2 \pi)^4}{\langle R^{- 2} \rangle^2} \right\} .$$ (487)

Substituting the expression of $D$ into the above equation, we obtain

$$B = \frac{1}{D} \left\{ R^2 \left\langle \frac{\vert \nabla \psi \ver... ...ft( \frac{q}{V'} \right)^2 \frac{(2 \pi)^4}{\langle R^{- 2} \rangle^2} \right\}$$

$$= \frac{1}{D} \left\{ R^2 \left\langle \frac{\vert \nabla \psi \ver... ...{- 2} \rangle} \left( R^2 - \frac{1}{\langle R^{- 2} \rangle} \right) \right\},$$ (488)

which is equal to the expression (5.58) in Ref. [9]. The coefficient before $\Psi'$ is written as

$$A = - \frac{q (2 \pi)^4}{V' \langle R^{- 2} \rangle} \left\{ \lef... ...4 \left[ \frac{q}{V' \langle R^{- 2} \rangle} \right]_{\psi} \right] \right\}$$

Define

$$\alpha = \left\langle \frac{\vert \nabla \psi \vert^2}{R^2} \right\rangle,$$

$$\beta = \frac{q}{V' \langle R^{- 2} \rangle} .$$

$$A = - (2 \pi)^4 \beta \left\{ \beta_{\psi} - \beta \frac{1}{V' D} [(V' \alpha)_{\psi} + q (2 \pi)^4 \beta_{\psi}] \right\}$$

$$\Longrightarrow \frac{D}{- (2 \pi)^4 \beta} A = D \beta_{\psi} - \beta \frac{1}{V'} [(V' \alpha)_{\psi} + q (2 \pi)^4 \beta_{\psi}]$$ (489)

$$\Longrightarrow \frac{D}{- \beta (2 \pi)^4} A = D \beta_{\psi} - \beta \frac{1}{V'} [V'' \alpha + V' \alpha_{\psi} + q (2 \pi)^4 \beta_{\psi}]$$ (490)

Using

$$D = \alpha + (2 \pi)^4 \langle R^{- 2} \rangle \beta^2$$ (491)

Eq. () is written as

$$\left. \Longrightarrow \frac{D}{- \beta (2 \pi)^4} A = \alpha \be... ...c{1}{V'} V' \alpha_{\psi} - \beta \frac{1}{V'} q (2 \pi)^4 \beta_{\psi} \right]$$ (492)

$$\left. \Longrightarrow \frac{D}{- \beta (2 \pi)^4} A = \alpha \be... ...pha - \beta \alpha_{\psi} - \beta \frac{1}{V'} q (2 \pi)^4 \beta_{\psi} \right]$$ (493)

$$\frac{A}{- \beta (2 \pi)^4} D = \alpha \beta_{\psi} + (2 \pi)^4 \... ... - \beta \alpha_{\psi} - \beta^2 \langle R^{- 2} \rangle (2 \pi)^4 \beta_{\psi}$$ (494)

$$\Longrightarrow \frac{A}{- \beta (2 \pi)^4} D = \alpha \beta_{\psi} - \beta \frac{1}{V'} V'' \alpha - \beta \alpha_{\psi}$$ (495)

$$\Longrightarrow \frac{A}{- \beta (2 \pi)^4} D = - \beta^2 \left( \frac{\alpha}{\beta} \right)_{\psi} - \beta \frac{1}{V'} V'' \alpha$$ (496)

$$\Longrightarrow \frac{A}{- \beta (2 \pi)^4} D = - \beta^2 \left( ... ...c{\alpha}{\beta} \right)_{\psi} - \beta^2 \frac{1}{V'} V'' \frac{\alpha}{\beta}$$ (497)

$$\Longrightarrow \frac{A}{- \beta (2 \pi)^4} D = - \beta^2 \left[ ... ...c{\alpha}{\beta} \right)_{\psi} + \frac{1}{V'} V'' \frac{\alpha}{\beta} \right]$$ (498)

$$\Longrightarrow \frac{A}{- \beta (2 \pi)^4} D = - \beta^2 \frac{1... ...left( \frac{\alpha}{\beta} \right)_{\psi} V' + V'' \frac{\alpha}{\beta} \right]$$ (499)

$$\Longrightarrow \frac{A}{- \beta (2 \pi)^4} D = - \beta^2 \frac{1}{V'} \left( \frac{\alpha}{\beta} V' \right)_{\psi}$$ (500)

$$\Longrightarrow A = \frac{(2 \pi)^4}{D} \left( \frac{q}{V' \langl... ...^2} \right\rangle \frac{\langle R^{- 2} \rangle}{q} V^{\prime 2} \right]_{\psi}$$ (501)

But the expression of $A$ is slightly different from that given in Ref. [9] [Eq. (5.57)]. Using the above coefficients, the GS equation with the $q$-profile held fixed is written as

$$\left( \Delta^{\star} - A \frac{\partial}{\partial \psi} \right) \Psi = - B \mu_0 \frac{d p}{d \Psi} .$$ (502)