Spherical magnetostatic pressure balance (${\Delta }_{\ast }\chi +\psi {\psi }^{\prime }=-{\mu }_0{r}^2{\sin }^2\theta p^{\prime }$)

For an axisymmetric magnetic field $\mathbf{B}=\psi {\mathbf{e}}_{\varphi }/(r\sin \theta )+\nabla \times [\chi {\mathbf{e}}_{\varphi }/(r\sin \theta )]$, the poloidal magnetic flux function gives $B_r={\chi }_{\theta }/(r^2\sin \theta )$ and $B_{\theta }=-{\chi }_r/(r\sin \theta )$. Force balance implies locally $p=p(\chi )$ and $\psi =\psi (\chi )$ on regular connected flux surfaces. Direct calculation gives $(\nabla \times \mathbf{B})\times \mathbf{B}=-({\Delta }_{\ast }\chi +\psi {\psi }^{\prime })\nabla \chi /(r^2{\sin }^2\theta )$, with ${\Delta }_{\ast }\chi ={\chi }_{rr}+\sin \theta {\partial }_{\theta }({\chi }_{\theta }/\sin \theta )/r^2$. Equating this to ${\mu }_0\nabla p$ proves the displayed magnetostatic equilibrium equation, including its sign and dimensions.

Set $\psi =\alpha \chi$, $p=p_0+P\chi$ and $\chi =F(r){\sin }^2\theta$. A regular family is $F=A[\cos (\alpha r)-\sin (\alpha r)/(\alpha r)]+c{r}^2$, where $P=-{\alpha }^{2}c/{\mu }_0$. A zero magnetic field at radius a requires $F(a)=F^{\prime }(a)=0$. For nonzero A this yields the displayed condition with $x=\alpha a$ and $c=-A[\cos x-\sin x/x]/a^2$. The boundary pressure is $p_0$, independent of angle. The trigonometric pair is tied by regularity and the radial differential equation; its coefficients are not independent.