15.4 Pressure and toroidal field function profile

The function p(Ψ) and g(Ψ) in the GS equation are free functions which must be specified by users before solving the GS equation. Next, we discuss one way to specify the free functions. Following Ref. [16],  we take P(Ψ) and g(Ψ) to be of the forms

P(Ψ) = P0 − (P0 − Pb)ˆp(Ψ),
(415)

1       1         --
2g2(Ψ) = 2 g20(1 − γ ˆg(Ψ )),
(416)

with ˆp and ĝ chosen to be of polynomial form:

ˆp(Ψ) = Ψ-α,
(417)

  --  --β
ˆg(Ψ) = Ψ ,
(418)

where

--  Ψ − Ψa
Ψ ≡ Ψ-−-Ψ--,
     b    a
(419)

with Ψb the value of Ψ on the boundary, Ψa the value of Ψ on the magnetic axis, α, β, γ, P0, Pb, and g0 are free parameters. Using the profiles of P and g given by Eqs. (415) and (416), we obtain

dP-(Ψ-)= − 1-(P − P )αΨα−1
  dΨ      Δ   0   b
(420)

where Δ = Ψb Ψa, and

                --
g dg-= − 1g20γ 1-βΨβ−1.
  dΨ     2   Δ
(421)

Then the term on the r.h.s (nonlinear source term) of the GS equation is written

      dP    dg        [ 1          -α−1]   1   1  -β−1
− μ0R2--- − --g = μ0R2  --(P0 − Pb)α Ψ    +  -g20γ--βΨ   .
       dΨ   dΨ          Δ                  2   Δ
(422)

The value of parameters P0, Pb, and g0 in Eqs. (415) and (416), and the value of α and β in Eqs. (417) and (418) can be chosen arbitrarily. The parameter γ is used to set the value of the total toroidal current. The toroidal current density is given by Eq. (548), i.e.,

Jϕ = RdP- + 1-dg-g-,
      dΨ    μ0dΨ R
(423)

which can be integrated over the poloidal cross section within the boundary magnetic surface to give the total toroidal current,

    ∫
Iϕ =   JϕdS
    ∫  (  dP   1  dgg )
  =     R ---+ -------  dRdZ
    ∫  [ (dΨ  )μ0dΨ R                        ]
  =     R  −-1  (P0 − Pb)ˆp′(Ψ) −-1--1g2γ 1-ˆg′(Ψ) dRdZ          (424)
            Δ                  μ0R 2 0 Δ
Using
        1
dRdZ = --𝒥 dψd𝜃,
       R
(425)

Eq. (424) is written as

    ∫ [(   1)           --    1  1    1   --]
Iϕ =      −--  (P0 − Pb)ˆp′(Ψ)− ---2- g02γ --ˆg′(Ψ) 𝒥 dψd𝜃,
          Δ                  μ0R 2    Δ
(426)

from which we solve for γ, giving

                    ∫ ′ --
γ = −-ΔIϕ1− (P0∫ [−1Pb)-[pˆ](Ψ)]𝒥-dψd𝜃.
        2μ0g20   R2ˆg′(Ψ) 𝒥dψd 𝜃
(427)

If the total toroidal current Iϕ is given, Eq. (427) can be used to determine the value of γ.