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 Issue A&A Volume 551, March 2013 A39 14 The Sun https://doi.org/10.1051/0004-6361/201220617 18 February 2013

## Online material

### 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.