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Appendix A: Derivation of kink mode equation
We give a short derivation of the basic equations, starting from Eqs. (3)–(5). The solutions are obtained in the three regions, inside the flux tube, outside the flux tube and in the transition layer.
A.1. Internal solution
Using the thin tube (or long wavelength) limit, the internal solutions can be
expressed as 
(A.1)Hence, at
r = R − l/2 = R(1 − ϵ/2),
we have 
(A.2)
A.2. External solution
Again using the thin tube limit, we have
(A.3)Hence, at
r = R + l/2 = R(1 + ϵ/2),
we have 
(A.4)Since both
ξr and P are
continuous across the transition layer as ϵ → 0, we can state
ξe = η + δξr, Pe = Pi + δP,
where both δξr and
δP tend to zero as ϵ → 0. Using (A.4), we have, correct to
O(ϵ2), 
(A.5)Rearranging
Eq. (A.5), the final equation for
the propagating kink mode is 
(A.6)Note that the
right hand side of (A.6) is of
O(ϵ). It is the leading order expressions for
δξr and
δP that we now need to calculate and this is done from the
transition layer solutions.
A.3. Transition layer solution
Integrating Eq. (3) across the thin
transition layer, we have 
(A.7)Integrating
(4) we have
(A.8)Remembering that
η is independent of r,
is l times
the average value of the operator ℒ and, for the linear density profile, the average
is ℒk, thus, 
(A.9)since
ℒkη = O(ϵ). Hence, for
the linear density profile 
(A.10)
We substitute Eq. (A.7) into
Eq. (A.6) and, using both Eq. (A.10),
ℒi + ℒe = 2ℒk and that again
ℒkη = O(ϵ), this
results in the propagating kink mode equation 
(A.11)
A.4. Integration of ξϕ across the transition layer
In this section we evaluate
(A.12)Using the solution
for ξϕ given by (26), the integral across the transition
layer is made up of four terms. These are evaluated in turn.
A.4.1. Term 1
Now the integral of the first term on the RHS of Eq. (26), due to the radial profile of the driving boundary condition
of ξϕ, is
For
large kz, this is proportional to
(kz)-1. The influence of the choice of boundary
condition does becomes less important after several wavelengths.
For small values of kz, we can expand the result in a series to
show that the first term is 
For small
κ, Term 1 can be expressed as
where we have
defined 
(A.13)
A.4.2. Term 2
The second term on the RHS integrates to give 
where
Ci(x) and Si(x) are the Cosine integral
and Sine integral respectively, defined by
where γ = 0.57721... is Euler’s constant and
Again this term is
proportional to (kz)-1 for large kz.
For small values of kz, it is easier to start from the integral
expression. Hence, the first two terms in the Taylor series are
For
small values of κ, term 2 can be shown to reduce to
where
Z = κkz/2, as above.
A.4.3. Term 3
The third term is 
The
expansion of the coefficient of 
for small kz gives to leading order
The expansion for
small κ gives, where
Z = κkz/2,
A.4.4. Term 4
Consider the final term, 
The
expansion for small κ gives 
A.5. Final expression
We can now bring together the expressions for all four terms to rewrite the kink mode
equation, (A.11), as
Expressing
η as 
, our
final equation is 
(A.14)where
,
,
ℒ1 = d2/dz2 − 2ikd/dz
and
ℒe = −k2κ + ℒk.
In Eq. (A.14), the operator,
ℒ1, acting on the final terms in the curly brackets on the right hand
side, results in terms that are small for κ ≪ 1. In fact, the terms
remain small even for κ ≤ 1/2. Hence, we will
neglect them and the comparison with the numerical results confirms this is a valid
assumption (see Sect. 5).
Equation (A.14) is an inhomogeneous,
integro-differential equation for 
,
the slowly varying amplitude function that describes the damping of the kink mode.
© ESO, 2013