Zeroth-Order Solution

To zeroth order in $\epsilon$, Equation (11.72) yields

$$\partial_X^{\,2}{\mit\Psi}_0 = 1.$$ (11.100)

Solving this equation subject to the boundary condition (11.83), we obtain

$${\mit\Psi}_0 = {\mit\Omega}(X,\zeta)\equiv \frac{X^2}{2} + \cos\zeta.$$ (11.101)

Thus, we again conclude that, to lowest order in our expansion, the magnetic flux-surfaces in the island region have the constant-$\psi$ structure pictured in Figure 5.7. The island O-points correspond to ${\mit\Omega}=-1$ and $\zeta=(2k-1)\,\pi$ (where $k$ is an integer), the X-points correspond to ${\mit\Omega}=+1$ and $\zeta = 2k\,\pi$, and the magnetic separatrix corresponds to ${\mit\Omega}=+1$.

To zeroth order in $\epsilon$, Equation (11.71) yields

$$\{{\cal N}_0, {\mit\Omega}\}= 0,$$ (11.102)

where use has been made of Equation (11.101). Given that ${\cal N}$ is an odd function of $X$, it follows that

$${\cal N}_0(X,\zeta,T) = \varsigma\, {\cal N}_{(0)}({\mit\Omega},T).$$ (11.103)

By symmetry, ${\cal N}_{0}=0$ inside the magnetic separatrix of the island chain. Let

$$L({\mit\Omega},T) = \frac{\partial{\cal N}_{(0)}}{\partial{\mit\Omega}}.$$ (11.104)

Note that $L({\mit\Omega}\leq 1,T)=0$. Equations (11.84) and (11.101) imply that

$$L({\mit\Omega}\rightarrow\infty,T) = \frac{1}{\sqrt{2\,{\mit\Omega}}}.$$ (11.105)

To zeroth order in $\epsilon$, Equation (11.68) yields

$$\{{\mit\Phi}_0,{\mit\Omega}\}=0,$$ (11.106)

where use has been made of Equations (11.101) and (11.103). Given that ${\mit\Phi}$ is an odd function of $X$, it follows that

$${\mit\Phi}_0(X,\zeta,T) = \varsigma\, {\mit\Phi}_{(0)}({\mit\Omega},T).$$ (11.107)

By symmetry, ${\mit\Phi}_{0}=0$ inside the magnetic separatrix of the island chain. Let

$$M({\mit\Omega},T) = \frac{\partial{\mit\Phi}_{(0)}}{\partial{\mit\Omega}}.$$ (11.108)

Note that $M({\mit\Omega}\leq 1,T)=0$. Equations (11.85) and (11.101) imply that

$$M({\mit\Omega}\rightarrow\infty,T) = \frac{v(T)}{\sqrt{2\,{\mit\Omega}}}+ v'(T).$$ (11.109)

To zeroth order in $\epsilon$, Equation (11.69) yields

$$\{{\cal V}_0,{\mit\Omega}\} = 0,$$ (11.110)

where use has been made of Equations (11.101), (11.103), and (11.107). Given that ${\cal V}$ is an even function of $X$, we can write

$${\cal V}_0(X,\zeta,T) = {\cal V}_{(0)}({\mit\Omega},T).$$ (11.111)

Finally, to zeroth order in $\epsilon$, Equation (11.70) gives

$$\{{\cal J}_0,{\mit\Omega}\} =-\zeta_g\,\{L\,\vert X\vert,{\mit\Omega}\}+\frac{1}{2}\left\{\pa... ...\Omega}\!\left[M\left(M+\frac{L}{1+\tau}\right)\right]X^2, {\mit\Omega}\right\}$$

$$\phantom{=} -\frac{\epsilon_\theta}{\epsilon_q}\,X\,\partial_{\mit\Omega}\!\left(\xi^{-1}\,{\cal V}_{(0)}-\vert X\vert\,F\right),$$ (11.112)

where

$$F({\mit\Omega}, T) = M({\mit\Omega}, T) + \frac{1}{1+\tau}\left(1-\alpha_\theta\,\frac{\eta_i}{1+\eta_i}\right)L({\mit\Omega},T),$$ (11.113)

and use has been made of Equations (11.101), (11.103), (11.104), (11.107), and (11.108). Moreover, $\partial_{\mit\Omega}\equiv \partial/\partial{\mit\Omega}$. Note that $F({\mit\Omega}\leq 1,T)=0$.

Let us write

$${\cal J}_0({\mit\Omega},\zeta,T) = {\cal J}_0^{(c)}({\mit\Omega},\zeta,T) + {\cal J}_0^{(s)}({\mit\Omega},\zeta,T) ,$$ (11.114)

where ${\cal J}_0^{(c)}$ has the symmetry of $\cos\zeta$, whereas ${\cal J}_0^{(s)}$ has the symmetry of $\sin\zeta$. It follows from Equations (11.112) and (11.114) that

$$\{{\cal J}_0^{(c)},{\mit\Omega}\} =-\zeta_g\,\{L\,\vert X\vert,{\... ...Omega}\!\left[M\left(M+\frac{L}{1+\tau}\right)\right]X^2, {\mit\Omega}\right\},$$ (11.115)

which implies that

$${\cal J}_0^{(c)}({\mit\Omega},\zeta,T)= \overline{\cal J}_0({\mit... ...ial_{\mit\Omega}\!\left[M\left(M+\frac{L}{1+\tau}\right)\right]\widetilde{X^2},$$ (11.116)

where the $\widetilde{\phantom x}$ operator is defined in Section 8.6, and $\overline{{\cal J}}_0({\mit\Omega},T)$ is an undetermined flux-surface function.

Equations (11.112) and (11.114) also yield

$$\{{\cal J}^{(s)}_0,{\mit\Omega}\} =-\frac{\epsilon_\theta}{\epsil... ...\partial_{\mit\Omega}\!\left(\xi^{-1}\,{\cal V}_{(0)} - \vert X\vert\,F\right).$$ (11.117)

The flux-surface average (see Section 8.6) of the previous equation gives

$$\langle X\,\partial_{\mit\Omega}(\xi^{-1}\,{\cal V}_{(0)} - \vert X\vert\,F)\rangle =0.$$ (11.118)

Inside the separatrix of the magnetic island chain, recalling that $F({\mit\Omega}\leq 1,T)=0$, the previous equation reduces to

$$\langle X\rangle\, \partial_{\mit\Omega} {\cal V}_{(0)} = 0,$$ (11.119)

whereas outside the separatrix, it gives

$$\partial_{\mit\Omega}\!\left(\xi^{-1}\,\langle\vert X\vert\rangle \,{\cal V}_{(0)} - \langle X^{\,2}\rangle\, F\right)= 0,$$ (11.120)

where use has been made of the easily proved identity $\langle X\,\partial_{\mit\Omega} G\rangle \equiv \partial_{\mit\Omega}\langle X\,G\rangle$. Thus, we conclude that

$${\cal V}_{(0)} ({\mit\Omega}, T)= c_0(T)$$ (11.121)

inside the separatrix, and

$${\cal V}_{(0)}({\mit\Omega}, T) = \xi\,\langle X^{\,2}\rangle \,F({\mit\Omega},T) + c_1(T)$$ (11.122)

outside the separatrix. Here, use has been made of $\langle\vert X\vert\rangle =1$ outside the separatrix. However, the boundary condition (11.86), combined with Equations (11.101), (11.105), (11.109), and (11.113), implies that $c_1(T)= 0$. Finally, Equations (11.117), (11.121), and (11.122) yield

$$\{{\cal J}^{(s)},{\mit\Omega}\} =- \frac{\epsilon_\theta}{\epsilo... ...l_{\mit\Omega}\!\left[\left(\langle X^{\,2}\rangle-\vert X\vert\right)F\right].$$ (11.123)