Solov'ev Equilibrium

Axisymmetric Solov'ev equilibra in (R/R₀,Z/R₀,ϕ) coordinates. Based on Cerfon & Freidberg, Physics of Plasmas 17, 032502, 2010, and Freidberg, Ideal Magnetohydrodynamics, 2014.

The covariant components of the vector potential are given by

\[A (x, y, \phi) = \left( \frac{B_0 R_0}{2} \, \frac{y}{x} , \, - \frac{B_0 R_0}{2} \, \ln x , \, \psi(x,y) \right)^T ,\]

with $x = R/R_0$ and $y = Z/R_0$. The normalised poloidal flux $\psi$ is given by

\[\psi (x,y) = \psi_0 + \sum \limits_{i=1}^{7} c_i \psi_i (x,y) ,\]

with

\[\begin{aligned} \psi_{0} &= \frac{x^4}{8} + \alpha \left( \frac{1}{2} x^2 \, \ln x - \frac{x^4}{8} \right) , \\ \psi_{1} &= 1 , \\ \psi_{2} &= x^2 , \\ \psi_{3} &= y^2 - x^2 \, \ln x , \\ \psi_{4} &= x^4 - 4 x^2 y^2 , \\ \psi_{5} &= 2 y^4 9 y^2 x^2 + 3 x^4 \, \ln x - 12 x^2 y^2 \, \ln x , \\ \psi_{6} &= x^6 - 12 x^4 y^2 + 8 x^2 y^4 , \\ \psi_{7} &= 8 y^6 - 140 y^4 x^2 + 75 y^2 x^4 - 15 x^6 \, \ln x + 180 x^4 y^2 \, \ln x - 120 x^2 y^4 \, \ln x . \end{aligned}\]

This formula describes exact solutions of the Grad-Shafranov equation with up-down symmetry. The constants $c_i$ are determined from boundary constraints on $\psi$, that are derived from the following analytic model for a smooth, elongated "D" shaped cross section:

\[\begin{aligned} x &= 1 + \epsilon \, \cos (\tau + \delta_0 \, \sin \tau) , \\ y &= \epsilon \kappa \, \sin (\tau) , \end{aligned}\]

where $0 \leq \tau < 2 \pi$, $\epsilon = a / R_0$ is the inverse aspect ratio, $\kappa$ the elongation, and $\sin \delta_0 = \delta$ is the triangularity.

Defining three test points, namely

• the high point $(1 - \delta \epsilon, \kappa \epsilon)$,
• the inner equatorial point $(1 - \epsilon, 0)$,
• and the outer equatorial point $(1 + \epsilon, 0)$,

the following geometric constraints can be posed on the solution:

\[\begin{aligned} \psi (1 + \epsilon, 0) &= 0 , \\ \psi (1 - \epsilon, 0) &= 0 , \\ \psi (1 - \delta \epsilon, \kappa \epsilon) &= 0 , \\ \psi_{x} (1 - \delta \epsilon, \kappa \epsilon) &= 0 , \\ \psi_{yy} (1 + \epsilon, 0) &= - N_1 \psi_{x} (1 + \epsilon, 0) , \\ \psi_{yy} (1 - \epsilon, 0) &= - N_2 \psi_{x} (1 - \epsilon, 0) , \\ \psi_{xx} (1 - \delta \epsilon, \kappa \epsilon) &= - N_3 \psi_y (1 - \delta \epsilon, \kappa \epsilon) . \end{aligned}\]

The first three equations define the three test points, the fourth equations enforces the high point to be a maximum, and the last three equations define the curvature at the test points.

The coefficients $N_j$ can be found from the analytic model cross section as

\[\begin{aligned} N_1 &= \left[ \frac{d^2 x}{dy^2} \right]_{\tau = 0} = - \frac{(1 + \delta_0)^2}{\epsilon \kappa^2} , \\ N_2 &= \left[ \frac{d^2 x}{dy^2} \right]_{\tau = \pi} = \hphantom{-} \frac{(1 - \delta_0)^2}{\epsilon \kappa^2} , \\ N_3 &= \left[ \frac{d^2 x}{dy^2} \right]_{\tau = \pi/2} = - \frac{\kappa}{\epsilon \, \cos^2 \delta_0} . \end{aligned}\]

For a given value of the constant $a$ above conditions reduce to a set of seven linear inhomogeneous algebraic equations for the unknown $c_i$, which can easily be solved.

Parameters:

• R₀: position of magnetic axis
• B₀: B-field at magnetic axis
• ϵ: inverse aspect ratio
• κ: elongation
• δ: triangularity
• α: free constant, determined to match a given beta value

Axisymmetric Solov'ev equilibra with X-point in (R/R₀,Z/R₀,phi) coordinates. Based on Cerfon & Freidberg, Physics of Plasmas 17, 032502, 2010, and Freidberg, Ideal Magnetohydrodynamics, 2014.

The covariant components of the vector potential are given by

\[A (x, y, \phi) = \left( \frac{B_0 R_0}{2} \, \frac{y}{x} , \, - \frac{B_0 R_0}{2} \, \ln x , \, \psi(x,y) \right)^T ,\]

with $x = R/R_0$ and $y = Z/R_0$. The normalised poloidal flux $\psi$ is given by

\[\psi (x,y) = \psi_0 + \sum \limits_{i=1}^{12} c_i \psi_i (x,y) ,\]

with

\[\begin{aligned} \psi_{0} &= \frac{x^4}{8} + \alpha \left( \frac{1}{2} x^2 \, \ln x - \frac{x^4}{8} \right) , \\ \psi_{1} &= 1 , \\ \psi_{2} &= x^2 , \\ \psi_{3} &= y^2 - x^2 \, \ln x , \\ \psi_{4} &= x^4 - 4 x^2 y^2 , \\ \psi_{5} &= 2 y^4 9 y^2 x^2 + 3 x^4 \, \ln x - 12 x^2 y^2 \, \ln x , \\ \psi_{6} &= x^6 - 12 x^4 y^2 + 8 x^2 y^4 , \\ \psi_{7} &= 8 y^6 - 140 y^4 x^2 + 75 y^2 x^4 - 15 x^6 \, \ln x + 180 x^4 y^2 \, \ln x - 120 x^2 y^4 \, \ln x , \\ \psi_{8} &= y , \\ \psi_{9} &= y x^2 , \\ \psi_{10} &= y^3 - 3 y x^2 \, \ln x , \\ \psi_{11} &= 3 y x^4 - 4 y^3 x^2 , \\ \psi_{12} &= 8 y^5 - 45 y x^4 - 80 y^3 x^2 \, \ln x + 60 y x^4 \, \ln x . \end{aligned}\]

This formula describes exact solutions of the Grad-Shafranov equation with up-down asymmetry. The constants $c_i$ are determined from boundary constraints on $\psi$, that are derived from the following analytic model for a smooth, elongated "D" shaped cross section:

\[\begin{aligned} x &= 1 + \epsilon \, \cos (\tau + \arcsin \delta \, \sin \tau) , \\ y &= \epsilon \kappa \, \sin (\tau) , \end{aligned}\]

where $0 \leq \tau < 2 \pi$, $\epsilon = a / R_0$ is the inverse aspect ratio, $\kappa$ the elongation, and $\sin \delta_0 = \delta$ is the triangularity.

Defining four test points, namely

• the high point $(1 - \delta \epsilon, \kappa \epsilon)$,
• the inner equatorial point $(1 - \epsilon, 0)$,
• and the outer equatorial point $(1 + \epsilon, 0)$,
• the position of the X-point $(x_{\mathrm{sep}}, y_{\mathrm{sep}})$,

the following geometric constraints can be posed on the solution:

\[\begin{aligned} \psi (1 + \epsilon, 0) &= 0 , \\ \psi (1 - \epsilon, 0) &= 0 , \\ \psi (1 - \delta \epsilon, \kappa \epsilon) &= 0 , \\ \psi (x_{\mathrm{sep}}, y_{\mathrm{sep}}) &= 0 , \\ \psi_{y} (1 + \epsilon, 0) &= 0 , \\ \psi_{y} (1 - \epsilon, 0) &= 0 , \\ \psi_{x} (1 - \delta \epsilon, \kappa \epsilon) &= 0 , \\ \psi_{x} (x_{\mathrm{sep}}, y_{\mathrm{sep}}) &= 0 , \\ \psi_{y} (x_{\mathrm{sep}}, y_{\mathrm{sep}}) &= 0 , \\ \psi_{yy} (1 + \epsilon, 0) &= - N_1 \psi_{x} (1 + \epsilon, 0) , \\ \psi_{yy} (1 - \epsilon, 0) &= - N_2 \psi_{x} (1 - \epsilon, 0) , \\ \psi_{xx} (1 - \delta \epsilon, \kappa \epsilon) &= - N_3 \psi_y (1 - \delta \epsilon, \kappa \epsilon) . \end{aligned}\]

The first four equations define the four test points, the fifth and sixth equations define the up-down symmetry, the seventh equations enforces the high point to be a maximum, the eighth and ninth eqaution set the $x$- and $y$-components of the magnetic field at the X-point to zero, and the last three equations define the curvature at the first three test points.

The coefficients $N_j$ can be found from the analytic model cross section as

\[\begin{aligned} N_1 &= \left[ \frac{d^2 x}{dy^2} \right]_{\tau = 0} = - \frac{(1 + \delta_0)^2}{\epsilon \kappa^2} , \\ N_2 &= \left[ \frac{d^2 x}{dy^2} \right]_{\tau = \pi} = \hphantom{-} \frac{(1 - \delta_0)^2}{\epsilon \kappa^2} , \\ N_3 &= \left[ \frac{d^2 x}{dy^2} \right]_{\tau = \pi/2} = - \frac{\kappa}{\epsilon \, \cos^2 \delta_0} . \end{aligned}\]

For a given value of the constant $a$ above conditions reduce to a set of seven linear inhomogeneous algebraic equations for the unknown $c_i$, which can easily be solved.

Parameters:

• R₀: position of magnetic axis
• B₀: B-field at magnetic axis
• ϵ: inverse aspect ratio
• κ: elongation
• δ: triangularity
• α: free constant, determined to match a given beta value
• xsep: x position of the X point
• ysep: y position of the X point