For a <barotropic fluid>, $\nabla p=\rho\nabla h$. Magnetostatic force balance and $L\alpha=F(\beta)$ give
$$
\rho\nabla(\Phi+h)=\mathbf F_m
=\frac{L\beta-FF'}{\mu_0r^2}\nabla\beta.
$$
The left side is a gradient multiplied by $\rho$, so taking the curl shows that $(L\beta-FF')/(r^2\rho)$ is constant on each $\beta$ surface. Absorbing the fixed factor $\mu_0$ into an arbitrary function $G$ gives the <Axisymmetric magnetostatic Grad-Shafranov system>
$$
\boxed{L\beta=F(\beta)\frac{dF}{d\beta}+r^2\rho G(\beta)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 314 / 3 / d / Solution

{c}

For an axisymmetric vector potential $\mathbf A=\nabla\alpha\times\nabla\phi+\beta\nabla\phi$, define
$$
L=-r\partial_r(r^{-1}\partial_r)-\partial_z^2.
$$
Magnetostatic azimuthal force balance gives $L\alpha=F(\beta)$. In a barotropic equilibrium the remaining balance integrates to
$$
L\beta=F(\beta)F'(\beta)+r^2\rho G(\beta)
$$
for flux functions $F$ and $G$.


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