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A.15 Large aspect ratio expansion

Consider the case that the boundary flux surface is circular with radius r = a and the center of the cirle at (R = R0,Z = 0). Consider the case 𝜀 = r∕R0 → 0. Expanding Ψ in the small parameter 𝜀,

Ψ = Ψ0 + Ψ1
(591)

where Ψ0 ∼ O(𝜀0), Ψ1 ∼ O(𝜀1). Substituting Eq. (591) into Eq. (590), we obtain

1 r∂ ∂r-r∂Ψ0 -∂r-+1 r∂ ∂r-r∂Ψ1 -∂r-+ 1 r2∂2Ψ0 -∂𝜃2+ 1 r2∂2Ψ1 ∂𝜃2-− 1 R0 +-rcos𝜃(∂ Ψ0 ∂ Ψ01 ) -∂r-cos𝜃− -∂𝜃-r sin𝜃− 1 R0-+-rcos𝜃( ∂Ψ1 ∂Ψ1 1 ) -∂r-cos𝜃− -∂𝜃-r sin𝜃 = −μ0(R0+r cos𝜃)2P′(Ψ 0+Ψ1)−g′(Ψ0+Ψ1)g(Ψ0+Ψ1)

Multiplying the above equation by R02, we obtain

( ) ( ) 21 ∂--∂Ψ0- 2 1∂--∂Ψ1- 21-∂2Ψ0 2-1∂2Ψ1 ----R20---- ∂Ψ0- ∂Ψ0-1 ----R20---- ∂Ψ1- ∂Ψ1-1 2 2 ′ 2′ R0r ∂rr ∂r +R 0 r∂rr ∂r +R 0r2∂ 𝜃2 +R 0r2 ∂𝜃2 −R0 + rcos𝜃 ∂r cos𝜃− ∂𝜃 r sin𝜃 − R0 + rcos𝜃 ∂r cos𝜃 − ∂𝜃 r sin 𝜃 = − μ0R 0(R0+r cos𝜃) P (Ψ0+ Ψ1)−
(592)

Further assume the following orderings (why?)

∂-Ψ −1 R0 ∂r ∼ O(𝜀 )Ψ,
(593)

and

∂Ψ ∂𝜃-∼ O (𝜀0)Ψ.
(594)

Using these orderings, the order of the terms in Eq. (592) can be estimated as

21 ∂ ∂Ψ0 Ψ0 −2 R0r∂r-r∂r--∼ 𝜀2-∼ O(𝜀 )
(595)

R201-∂-r∂Ψ1-∼ Ψ1-∼ O(𝜀−1) r∂r ∂r 𝜀2
(596)

1 ∂2Ψ0 1 R20r2-∂𝜃2 ∼ 𝜀2Ψ0 ∼ O(𝜀−2)
(597)

1 ∂2Ψ1 Ψ1 R20r2-∂𝜃2 ∼ 𝜀2-∼ O(𝜀−1)
(598)

----R0---- ≈ 1− -r-cos𝜃 R0 + rcos𝜃 R0
(599)

∂ Ψ0 Ψ0 R0-∂r-cos𝜃 ∼ 𝜀-∼ O (𝜀−1)
(600)

∂Ψ01 Ψ0 −1 R0 ∂𝜃-r sin 𝜃 ∼-𝜀-∼ O (𝜀 )
(601)

R0 ∂Ψ1-cos 𝜃 ∼ Ψ1-= O (𝜀0) ∂r 𝜀
(602)

∂ Ψ11 Ψ1 R0-∂𝜃-r sin𝜃 ∼ 𝜀-= O(𝜀0)
(603)

The leading order (𝜀−2 order) balance is given by the following equation:

2 R201-∂-r∂Ψ0+ R20 1-∂-Ψ0 = − μ0R40P′(Ψ0) − R20g′(Ψ0)g(Ψ0), r∂r ∂r r2 ∂𝜃2
(604)

It is reasonable to assume that Ψ0 is independent of 𝜃 since Ψ0 corresponds to the limit a∕R → 0. (The limit a∕R → 0 can have two cases, one is r → 0, another is R →∞. In the former case, Ψ must be independent of 𝜃 since Ψ should be single-valued. The latter case corresponds to a cylinder, for which it is reasonable (really?) to assume that Ψ0 is independent of 𝜃.) Then Eq. (604) is written

1∂-r∂Ψ0-= − μ0R2P ′(Ψ0)− g′(Ψ0)g(Ψ0). r∂r ∂r 0
(605)

(My remarks: The leading order equation (605) does not corresponds strictly to a cylinder equilibrium because the magnetic field B = ∇Ψ0 ×∇ϕ + g∇ϕ depends on 𝜃.) The next order (𝜀−1 order) equation is

R021 r-∂- ∂rr∂-Ψ1 ∂r+R0212 r 2 ∂-Ψ12 ∂ 𝜃−R0∂Ψ0- ∂rcos𝜃 = −μ0R022R 0r cos𝜃P′(Ψ0)−μ0R04P′′(Ψ 0)Ψ1−R02[g′(Ψ 0)g(Ψ0)]′Ψ1

2 R21 ∂-r∂Ψ1+R2 1-∂-Ψ1+{μ0R4 P′′(Ψ0)+R2[g′(Ψ0)g(Ψ0)]′}Ψ1 = − μ0R22R0rcos𝜃P ′(Ψ0)+R0 ∂Ψ0cos𝜃 0r ∂r ∂r 0 r2 ∂𝜃2 0 0 0 ∂r
(606)

2 R201 ∂-r∂Ψ1+R20 12-∂-Ψ21+ d-{μ0R40P ′(Ψ0)+R20g′(Ψ0)g(Ψ0)}-drΨ1 = − μ0R202R0rcos𝜃P′(Ψ0)+R0 ∂Ψ0cos𝜃 r ∂r ∂r r ∂𝜃 dr dΨ0 ∂r
(607)

It is obvious that the simple poloidal dependence of cos𝜃 will satisfy the above equation. Therefore, we consider Ψ1 of the form

dΨ0(r) Ψ1 = Δ(r) dr cos𝜃,
(608)

where Δ(r) is a new function to be determined. Substitute this into the Eq. (), we obtain an equation for Δ(r),

( ) 21 d- dΨ0dΔ- d2Ψ0 2 1- dΨ0(r) d- 4 ′ 2′ -dr- dΨ0(r) 2 ′ dΨ0- R0r dr rdr dr + rΔ dr2 − R0 r2 Δ dr + dr{μ0R0P (Ψ0)+R 0g (Ψ0)g(Ψ0)}dΨ0Δ dr = − μ0R02R0rP (Ψ0)+R0 dr
(609)

21 d ( dΨ0dΔ ) 21 d ( d2Ψ0 ) 21 dΨ0(r) d 4 ′ 2 ′ 2 ′ dΨ0 R0r dr rdr--dr +R 0r dr rΔ-dr2 − R 0r2Δ--dr--+ dr{μ0R 0P (Ψ0)+R 0g(Ψ0)g(Ψ0)}Δ = − μ0R02R0rP (Ψ0)+R0 dr--
(610)

1 d ( dΨ dΔ ) 1 d ( d2Ψ ) d2Ψ dΔ 1 dΨ (r) d dΨ R20- -- r--0--- + ΔR20--- r---02 +R20--20---− R20-2Δ --0---+--{μ0R40P′(Ψ0)+R20g′(Ψ0)g(Ψ0)}Δ = − μ0R202R0rP ′(Ψ0)+R0--0- r dr dr dr rdr dr dr dr r dr dr dr
(611)

( ) [ ( 2 ) ] 2 R201 d- rdΨ0dΔ- + ΔR20 1 d- rd-Ψ0 − 1-dΨ0(r) +R20d-Ψ0dΔ-+ d-{μ0R40P′(Ψ0)+R20g′(Ψ0)g(Ψ0)}Δ = − μ0R202R0rP′(Ψ0)+R0 dΨ0 r dr dr dr r dr dr2 r2 dr dr2 dr dr dr
(612)

Using the identity

d- dr[ ( )] 1 d- r dΨ0- r dr dr = 1 r-d dr( 2 ) rd-Ψ02 dr−-12 rdΨ0(r)- dr,

equation () is written as

( ) [ ( ) ] 2 R201 d- rdΨ0dΔ- + ΔR20 d 1-d rdΨ0- +R20d-Ψ20dΔ-+ d-{μ0R40P ′(Ψ0)+R20g′(Ψ0)g(Ψ0)}Δ = − μ0R202R0rP′(Ψ0)+R0 dΨ0 r dr dr dr dr rdr dr dr dr dr dr
(613)

Using the leading order equation (), we know that the second and fourth term on the l.h.s of the above equation cancel each other, giving

( ) 21-d dΨ0-dΔ- 2d2Ψ0dΔ- 2 ′ dΨ0- R 0rdr r dr dr + R 0dr2 dr = − μ0R02R0rP (Ψ0)+ R0 dr
(614)

1 d ( dΨ dΔ ) d2Ψ dΔ 1 dΨ ⇒ --- r--0---- + --20---= − μ02R0rP′(Ψ0)+ -----0- rdr dr dr dr dr R0 dr
(615)

Using the identity

[ ] 1 dr d ( dΨ0)2 dΔ 1d ( dΨ0dΔ ) d2Ψ0 dΔ rdΨ-dr r dr-- dr- = rdr r-dr-dr + -dr2 dr , 0
(616)

equation (615) is written

[ ( )2 ] 1-dr-d- r dΨ0- dΔ- = − μ02R0rP ′(Ψ0)+ -1-dΨ0, rdΨ0 dr dr dr R0 dr
(617)

[ ] 1 d ( dΨ0)2 dΔ dP 1 ( dΨ0)2 ⇒ rdr r dr-- dr- = − μ02R0rdr-+ R0- dr-- ,
(618)

Using

1 dΨ B𝜃0 = ----0- R0 dr
(619)

equation (618) is written

1 d [ dΔ ] 1 dP B2 - -- rB2𝜃0--- = − μ02--r---+ --𝜃0. r dr dr R0 dr R0
(620)

d [ dΔ] r ( dP (Ψ ) ) ⇒ -- rB2𝜃0--- = --- − 2μ0r----0- + B2𝜃0 , dr dr R0 dr
(621)

which agrees with equation (3.6.7) in Wessson’s book[27].

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