Issue 
A&A
Volume 551, March 2013



Article Number  A39  
Number of page(s)  14  
Section  The Sun  
DOI  https://doi.org/10.1051/00046361/201220617  
Published online  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 ξ_{r} and P are continuous across the transition layer as ϵ → 0, we can state ξ_{e} = η + δξ_{r}, P_{e} = P_{i} + δP, where both δξ_{r} and δP tend to zero as ϵ → 0. Using (A.4), we have, correct to O(ϵ^{2}), (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 δξ_{r} 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 ℒ_{k}, thus, (A.9)since ℒ_{k}η = 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), ℒ_{i} + ℒ_{e} = 2ℒ_{k} and that again ℒ_{k}η = O(ϵ), this results in the propagating kink mode equation (A.11)
A.4. Integration of ξ_{ϕ} across the transition layer
In this section we evaluate (A.12)Using the solution for ξ_{ϕ} 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 ξ_{ϕ}, 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 , , ℒ_{1} = d^{2}/dz^{2} − 2ikd/dz and ℒ_{e} = −k^{2}κ + ℒ_{k}. In Eq. (A.14), the operator, ℒ_{1}, 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, integrodifferential equation for , the slowly varying amplitude function that describes the damping of the kink mode.
© ESO, 2013
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