Solution in Presence of Resistive Wall
Suppose, now, that the wall at
possesses non-zero electrical resistivity, but is surrounded by a perfectly conducting wall
located at radius
.
The most general solution to the cylindrical tearing mode equation, (3.60), in the outer region can be written
![$\displaystyle \delta\psi(r,t) = {\mit\Psi}_s(t)\,\hat{\psi}_s(r) + {\mit\Psi}_w(t)\,\hat{\psi}_w(r),$](img1558.png) |
(3.76) |
where the real function
is specified in Section 3.8, and the real function
is a solution of
![$\displaystyle \frac{d^2\hat{\psi}_w}{dr^2} + \frac{1}{r}\,\frac{d\hat{\psi}_w}{dr}-\frac{m^2}{r^2}\,\hat{\psi}_w - \frac{J_z'\,\hat{\psi}_w}{r\,(1/q-n/m)}= 0$](img1560.png) |
(3.77) |
that satisfies
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
![$\displaystyle \hat{\psi}_w(r_w< r < r_c) = \frac{(r/r_c)^{-m} - (r/r_c)^m}{(r_w/r_c)^{-m} - (r_w/r_c)^m}.$](img1564.png) |
(3.81) |
In general,
is continuous across the resistive wall (in accordance with Maxwell's equations), whereas
is discontinuous. The discontinuity in
is caused by a helical current sheet induced in the wall. The complex quantity
![$\displaystyle {\mit\Psi}_w(t)=\delta\psi(r_w,t),$](img1566.png) |
(3.82) |
determines the amplitude and
phase of the perturbed magnetic flux that penetrates the resistive wall. The complex quantity
![$\displaystyle {\mit\Delta\Psi}_w(t) = \left[r\,\frac{\partial\delta\psi}{\partial r}\right]_{r_{w-}}^{r_{w+}}$](img1567.png) |
(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]
Here,
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
![$\displaystyle \frac{d}{dr}\!\left(\hat{\psi}_s\,r\,\frac{d\hat{\psi}_w}{dr} - \hat{\psi}_w\,r\,\frac{d\hat{\psi}_s}{dr}\right) = 0.$](img1578.png) |
(3.89) |
If we integrate the previous equation from
to
, making use of Equations (3.66), (3.67), (3.78), (3.79), (3.87),
and (3.88), then we obtain [4]
![$\displaystyle E_{sw} = E_{ws}.$](img1581.png) |
(3.90) |