Solution in Presence of Resistive Wall

Suppose, now, that the wall at $r=r_w$ possesses non-zero electrical resistivity, but is surrounded by a perfectly conducting wall located at radius $r_c>r_w$. The most general solution to the cylindrical tearing mode equation, (3.60), in the outer region can be written

$$\delta\psi(r,t) = {\mit\Psi}_s(t)\,\hat{\psi}_s(r) + {\mit\Psi}_w(t)\,\hat{\psi}_w(r),$$ (3.76)

where the real function $\hat{\psi}_s(r)$ is specified in Section 3.8, and the real function $\hat{\psi}_w(r)$ is a solution of

$$\frac{d^2\hat{\psi}_w}{dr^2} + \frac{1}{r}\,\frac{d\hat{\psi}_w}{dr}-\frac{m^2}{r^2}\,\hat{\psi}_w - \frac{J_z'\,\hat{\psi}_w}{r\,(1/q-n/m)}= 0$$ (3.77)

that satisfies

$$\hat{\psi}_w(r\leq r_s) =0,$$ (3.78)

$$\hat{\psi}_w(r_w) =1,$$ (3.79)

$$\hat{\psi}_w(r\geq r_c) = 0.$$ (3.80)

Again, Equation (3.80) ensures that the perturbed magnetic field associated with the tearing mode cannot penetrate the perfectly conducting wall. It is easily seen that

$$\hat{\psi}_w(r_w< r < r_c) = \frac{(r/r_c)^{-m} - (r/r_c)^m}{(r_w/r_c)^{-m} - (r_w/r_c)^m}.$$ (3.81)

In general, $\delta\psi$ is continuous across the resistive wall (in accordance with Maxwell's equations), whereas $\partial\delta\psi/\partial r$ is discontinuous. The discontinuity in $\partial\delta\psi/\partial r$ is caused by a helical current sheet induced in the wall. The complex quantity

$${\mit\Psi}_w(t)=\delta\psi(r_w,t),$$ (3.82)

determines the amplitude and phase of the perturbed magnetic flux that penetrates the resistive wall. The complex quantity

$${\mit\Delta\Psi}_w(t) = \left[r\,\frac{\partial\delta\psi}{\partial r}\right]_{r_{w-}}^{r_{w+}}$$ (3.83)

parameterizes the amplitude and phase of the helical current sheet flowing in the wall. Simultaneously matching the outer solution (3.76) across the rational surface and the resistive wall yields [4]

$${\mit\Delta\Psi}_s = E_{ss}\,{\mit\Psi}_s + E_{sw}\,{\mit\Psi}_w,$$ (3.84)

$${\mit\Delta\Psi}_w = E_{ws}\,{\mit\Psi}_s+ E_{ww}\,{\mit\Psi}_w.$$ (3.85)

Here,

$$E_{ww} = \left[r\,\frac{d\hat{\psi}_w}{dr}\right]_{r_{w-}}^{r_{w+}},$$ (3.86)

$$E_{sw} =\left[r\,\frac{{d\hat\psi}_w}{dr}\right]_{r=r_{s+}},$$ (3.87)

$$E_{ws} =-\left[r\,\frac{{d\hat\psi}_s}{dr}\right]_{r=r_{w-}}$$ (3.88)

are real quantities determined by the solutions of Equations (3.64) and (3.77) in the outer region.

Equations (3.64) and (3.77) can be combined to give

$$\frac{d}{dr}\!\left(\hat{\psi}_s\,r\,\frac{d\hat{\psi}_w}{dr} - \hat{\psi}_w\,r\,\frac{d\hat{\psi}_s}{dr}\right) = 0.$$ (3.89)

If we integrate the previous equation from $r=r_{s+}$ to $r=r_{w-}$, making use of Equations (3.66), (3.67), (3.78), (3.79), (3.87), and (3.88), then we obtain [4]

$$E_{sw} = E_{ws}.$$ (3.90)