m. s. ruderman, m. goossens and j. andries- nonlinear propagating kink waves in thin magnetic tubes

8/3/2019 M. S. Ruderman, M. Goossens and J. Andries- Nonlinear propagating kink waves in thin magnetic tubes

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Nonlinear propagating kink waves in thin magnetic tubes

M. S. Ruderman,1,a M. Goossens,2 and J. Andries2,31Department of Applied Mathematics, University of Sheffield, Hicks Building,

Hounsfield Road, Sheffield S3 7RH, United Kingdom2Centrum voor Plasma Astrofysica, Katholieke Universiteit Leuven, Leuven B-3001, Belgium

3Centre for Stellar and Planetary Astrophysics, Monash University, Victoria 3800, Australia

Received 5 January 2010; accepted 21 June 2010; published online 18 August 2010

The propagation of nonlinear nonaxisymmetric waves along a magnetic tube in an incompressibleplasma embedded in a magnetic-free plasma is studied. The plasma and magnetic parameters in the

tube core as well as plasma parameters in the external plasma are constant. Between the tube core

and the magnetic-free plasma there is a thin annulus where the Alfvn speed monotonically

decreases to zero. In this annulus there is a cylindrical surface where the phase speed of the global

wave matches the local Alfvn speed. In the vicinity of this surface there is an efficient conversion

of the global wave energy in the energy of local Alfvn waves. This results in the resonant

absorption of the global wave and, as a consequence, in the global wave damping. The wave

amplitude is assumed to be small and used as a small parameter in the singular perturbation method

that is used to derive the nonlinear governing equation for nonaxisymmetric waves. This equation

accounts both for nonlinearity and wave damping due to resonant absorption. A particular class of

solutions of this equation in the form of helical waves is studied numerically. The main result

obtained in this study is that nonlinearity accelerates the wave damping. It also distorts the shape of

the tube boundary due to nonlinear generation of fluting modes. 2010 American Institute of

Physics. doi:10.1063/1.3464464

I. INTRODUCTION

This work was originally motivated by the recent sug-

gestion by Ruytova et al.1

and Ruytova and Hogenaar2

that

the moving magnetic features around sunspots can be de-

scribed by solitonlike and shocklike kink perturbations of

magnetic flux tubes in the penumbral regions. These pertur-

bations are excited by the negative energy wave NEW in-

stability. The model itself is very interesting and deservesattention of both solar and plasma physicists. However, there

is one very serious problem related to this model. To describe

the excitation and subsequent nonlinear evolution of kink

perturbations, the Kortewegde Vries KdV equation withadditional terms describing the wave damping due to dissi-

pation and wave excitation due to NEW instability is used

see Eq. 7 in Ruytova et al.1 and Eq. 1 in Ruytova andHogenaar

2. It is claimed that the KdV equation for nonlinearkink waves was derived by Ruytova and Sakai.

3First, we

should note that Ruytova and Sakai3

derived not the KdV but

the modified Kortewegde Vries mKdV equation for thekink waves. And second, there is an error in their derivation.

A simple qualitative analysis shows that the nonlinear

evolution of kink waves in a magnetic tube cannot be de-

scribed by a one-dimensional equation. In addition to the

kink mode, there is an infinite set of fluting modes, all of

them propagating with the same phase speed Ck in the thin-

tube approximation. This implies that there is a very strong

nonlinear resonant interaction between the kink and fluting

modes that provides a quadratic nonlinearity. It is also clear

that the kink and all fluting modes have to be described by

one equation; they cannot be separated. Since the kink and

each fluting mode have different dependence on in cylin-

drical coordinates r, , and z, with the z-axis coinciding with

the tube axis, the equation that describes their simultaneous

nonlinear evolution has to be a two-dimensional equation

with and z being independent variables.

In their derivation of the mKdV equation for the kink

mode, Ruytova and Sakai3

imposed the boundary conditions

at the unperturbed boundary see their Eqs. 6 and 7,while in the nonlinear theory, the boundary perturbation has

to be taken into account. The neglect of the boundary pertur-

bation is equivalent to the assumption that the tube cross-

section remains circular during the tube motion, which elimi-

nates all fluting modes. It is the boundary perturbation that

provides the interaction between the kink and fluting modes.

Thus, Ruytova and Sakai3

eliminated the main nonlinear

effect and then took into account much weaker nonlinear

interaction between the kink mode and the longitudinal

motion.

In this paper, we aim to derive a nonlinear equation de-

scribing the simultaneous evolution of the kink and flutingmodes in a thin magnetic tube. We use the approximation of

an incompressible plasma, which is, although reasonable, not

ideal for the applications to the solar photosphere. However,

we are encouraged by the fact that in the linear approxima-

tion, the kink and fluting modes are not affected by com-

pressibility at all. We hope that this property is approxi-

mately valid even in the nonlinear regime, at least when we

consider perturbations with moderate amplitudes. We assume

that the tube is inhomogeneous with the inhomogeneity con-

fined to a thin transitional layer near the tube boundary. TheaElectronic mail: [email protected].

PHYSICS OF PLASMAS 17, 082108 2010

1070-664X/2010/178 /082108/23/$30.00 2010 American Institute of Physic17, 082108-1

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presence of inhomogeneity results in the resonant absorption

of the wave energy. The decrement due to resonant absorp-

tion is proportional to l /R, where R is the tube radius and l is

the thickness of the inhomogeneous layer. The wave disper-

sion is proportional to R /L2, where L is the characteristicwavelength. This means that under the assumption that

R /L l /R, wave damping due to resonant absorption

strongly dominates wave dispersion. This observation en-

courages us to neglect the wave dispersion in our analysis.Although, as we have already stated, this work was

originally motivated by publications by Ryutova with

co-authors,13

it has much wider applications. It is well

known that the solar atmosphere is strongly magnetically

structured. The most common elements of the magnetic

structuring are magnetic flux tubes. These tubes are formed

by the magnetic flux concentrations in the solar photosphere.

In the corona, they are formed by thin regions of enhanced

plasma density elongated in the magnetic field direction

which are called coronal loops. Waves and oscillations are

routinely observed in the magnetic structures. We have al-

ready mentioned moving magnetic features in vicinities ofsunspots. Another example is standing kink oscillations that

have been observed for more than decade see, e.g., the re-view by Aschwanden

4. Recently, Farahani et al.5 suggestedinterpreting transverse waves observed in solar coronal hot

jets, discovered by Hinode, in terms of magnetohydrody-

namic kink waves.

While the analysis of this paper is definitely relevant for

the interpretation of moving magnetic features in vicinities

of sunspots and transverse waves in solar coronal hot jets, it

cannot be directly applied to kink oscillations of coronal

loops. The reason is that we consider propagating waves,

while the transverse coronal loop oscillations are standing

waves. Since very often the observed kink oscillations ofcoronal loops have sufficiently large amplitudes, nonlinearity

can strongly affect these oscillations. However, at present,

only the linear theory of standing kink oscillations of coronal

loops has been developed see, e.g., the review by Rudermanand Erdly

6. Hence, the development of nonlinear theory ofcoronal loop kink oscillations is very much on the agenda. It

is well known that the nonlinear theory of standing waves is

more complicated than that of propagating waves. The de-

velopment of nonlinear theory of propagating waves is an

absolutely essential step in the development of nonlinear

theory of standing waves.

The application of theory of nonlinear kink waves propa-gating along magnetic flux tube is not restricted to the solar

physics. Another important application is the interpretation

of propagating disturbances in astrophysical jets.

The paper is organized as follows. In the next section we

introduce the magnetohydrodynamic MHD equations inLagrangian variables. In Sec. III, we formulate the problem

describing the equilibrium, write down the MHD equations

in cylindrical coordinates, and state main assumptions. In

Sec. IV, we combine the reductive perturbation method and

the method of matched asymptotic expansions to derive the

model equation describing the propagation of nonaxisym-

metric waves along a thin magnetic tube. In Sec. V, we

present the results of the numerical solution of the model

equation. Section VI contains the summary of the results

obtained in this paper and our conclusions.

II. MHD EQUATIONS IN LAGRANGIAN VARIABLES

In what follows, we study the wave propagation in a

magnetic tube with a thin transitional layer. The amplitude of

the tube boundary displacement is of the order of the thick-ness of this layer. This fact causes serious complications be-

cause, when solving the nonlinear MHD equations in the

transitional layer, we have to solve them in a region with

undetermined boundaries. A similar problem was encoun-

tered by Ruderman and Goossens7

when they studied the

nonlinear surface wave propagation on a magnetic interface

with a thin transitional layer. To solve this problem, they

considered the magnetic flux function as an independent

variable instead of the Cartesian coordinate in the direction

perpendicular to the unperturbed interface. The values of the

flux function at the boundaries of the transitional region re-

main the same at all moments of time. As a result, in the new

variables, the transitional layer has well-defined boundaries.The method used by Ruderman and Goossens

7was

based on the fact that the problem that they studied was

two-dimensional, so that it was possible to express the mag-

netic field in terms of the flux function. Studying propagation

of nonlinear kink waves in a magnetic tube results in an

essentially three-dimensional problem, so that the magnetic

field cannot be expressed in terms of the flux function. Now

the best way to overcome difficulties related to the moving

boundaries of the transitional layer is to use the Lagrangian

description. In Lagrangian variables, the equations determin-

ing the transitional layer boundaries are independent of time,

so that the transition layer has well-defined boundaries.

Since the Lagrangian MHD equations are not as well

known as their Eulerian counterparts, we will give a brief

derivation of these equations in this section. When doing so

we will first use Cartesian coordinates. We will write the

equations in the invariant form containing only vector quan-

tities and the operator . After that they can be used in any

curvilinear coordinates.

Let us introduce the Lagrangian variables a

= a1 , a2 , a3, the position vector x= x1 ,x2 ,x3, and the dis-placement vector u=xa=u1 , u2 , u3. Here x=xt,a andx0 ,a =a, so that u0 ,a =0. We assume that the plasma isincompressible, so that the plasma density is independent

of time =a

. The only dissipative process that we takeinto account is viscosity. Hence, the magnetic field is de-scribed by the ideal induction equation.

We start our derivation from recalling that in the

Lagrangian description, the ideal induction equation can be

integrated thus giving the expression for the magnetic induc-

tion, B, in term of the plasma displacement see, e.g.,Moffatt

8. For an incompressible plasma, this expressionreads

B = B0 + b = B0jx

aj= B0 + B0 u, 1

where B0 =B0 ,a and =/a1 ,/a2 ,/a3.

082108-2 Ruderman, Goossens, and Andries Phys. Plasmas 17, 082108 2010



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The momentum equation in Lagrangian variables has the

form e.g., Temam and Miranville9

2ui

t2=

ji

aj, 2

where ji is the nominal or first PiolaKirchhoff stress ten-

sor. In an incompressible medium, this tensor is related to the

Cauchy stress tensor Tji

used in Eulerian description by

ji = Tkiaj

xk, 3

where we have used the Einstein summation rule with re-

spect to repeating indices. Note that the PiolaKirchhoff

stress tensor introduced in Ref. 9 is equal to T, where the

superscript T indicates a transposed tensor and, from now on,

we denote tensors by bold capital letters with the hat.

The Cauchy stress tensor is the sum of three tensors, the

pressure tensor, the Maxwell electromagnetic tensor, and the

viscosity tensor,

Tki = pki +1

01

2B2ki + BkBi + Wki , 4

where p is the plasma pressure, ki is the Kronecker delta-

symbol, 0 is the magnetic permeability of free space, is

the dynamic viscosity, and

Wki =vk

xi+

vi

xk, 5

vi being the components of the velocity v related to the dis-

placement by v=u /t. The components of the viscous force

are given by

Wki

xk=

2vk

xi xk+

xk vi

xk =

xk vi

xk ,

where we have taken into account that v=0 in an incom-

pressible medium. This result implies that we can substitute

Wki =vi

xk=

2ui

tal

al

xk6

for Wki in Eq. 4.Let us introduce the equilibrium total pressure plasma

plus magnetic P0 and the total pressure perturbation P,

P0 = p0 +B0

2

20, P = p + 1

0B0 b +

120

b2, 7

where p0 is the equilibrium plasma pressure and p =pp0.

Then we can write the expression for the Cauchy tensor in

the invariant form as

T = T0 PI+

1

0B0b + bB0 + bb + W , 8

where the components of the modified viscosity tensor W

are given by Eq. 6 and the unperturbed Cauchy tensor T0 isgiven by

T0 = P0I+

1

0B0B0. 9

In the equilibrium u= 0, = T0 so that T0 satisfies the equi-librium condition

T0 = 0. 10

I is the unit tensor with the components ji and bb is the

dyadic product of two vectors.

In an incompressible medium, the determinant of matrix

with the components xi /aj is equal to unity. If we identify

a second order tensor with the matrix of its components, then

we can write this condition as

detI+ u = 1. 11

For what follows, it is convenient to introduce tensor A

with the components Aij =ai /xj. This tensor satisfies the

equation

I+ uT A = I. 12

Now Eq. 3 can be rewritten as

= A T 13

and tensor W is given by

W = AT u

t. 14

Using Eq. 1 yields

ai

xjBj = B0k

xj

ak

ai

xj= B0kki = B0i . 15

With the aid of this result and Eq. 1, we obtain

A B0B0 + B0b + bB0 + bb

= A BB = B0B0 + B0 u. 16

With the aid of Jacobis formula for the derivatives of the

determinant of a nonsingular matrix M depending on a pa-

rameter ,

ddetM

d= detM trdM

dM 1 ,

where tr denotes the trace of a matrix, we can prove the

identity

ajaj

xi = 0 ,

valid for an incompressible medium. Using this identity and

Eqs. 7, 8, 10, 13, 14, and 16, we rewrite Eq. 2 inthe invariant form

082108-3 Nonlinear propagating kink waves in thin magnetic tubes Phys. Plasmas 17, 082108 2010



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2u

t2= P0 A

T P0 + P +1

0 B0B0 u

+ AT AT ut

. 17Equations 11 and 17 constitute the closed system of equa-tions describing the plasma motion in Lagrangian variables.

Although it was derived in Cartesian coordinates, it can be

used in any curvilinear coordinates because it is written in

the invariant form.

III. PROBLEM FORMULATION

In what follows we consider propagation of nonlinear

kink waves in a magnetic tube. In the equilibrium state, there

is a magnetic tube of radius R surrounded by an annulus of

thickness l. The annulus is assumed to be thin, lR. We

introduce cylindrical coordinates r, , and z so that a1 = r,

a2 =, and a3 =z. The unit vectors of this coordinate system

in the r, , and z-directions are er, e, and ez, respectively.The z-axis coincides with the tube axis. The equilibrium

magnetic field is everywhere parallel to the tube axis

B0 =B0ez. The density and the equilibrium magnetic field are

constant inside the tube =i =const; B0 =Bi =const for

rR. Outside the annulus, the density is also constant and

the magnetic field is zero =e =const; B0 =0 for rR + l. In

the annulus, monotonically varies from i to e and B0 is

monotonically decreasing from Bi to zero. For this equilib-

rium, the equilibrium condition 10 takes the form

P0 = p0 +1

20B0

2 = const. 18

An important quantity in our analysis is the Alfvn speed

VA =B001/2. It is constant inside the tube VA = VAi

=Bi0i1/2 for rR. We assume that VA monotonically

decreases from VAi to zero in the annulus.

Let us now write down the governing equations in cy-

lindrical coordinates. To do this, we need formulae for dif-

ferent expressions containing the operator written in cylin-

drical coordinates. They can be found elsewhere e.g.,Goedbloed and Poedts

10. The only problem is to find theexpression for u. Usually, only the expression for w u is

given. However, it is easy to obtain the expression for u

using the expression for w u. For this, it is enough to take

wequal to the first, second, and third unit vector of thecurvilinear coordinate system. In this way, we obtain that the

matrix of tensor u is given in cylindrical coordinates by

u = ur

r

u

r

uz

r

1

r

ur

u

r

1

r

u

+

ur

r

1

r

uz

ur

z

u

z

uz

z

. 19Using this result, we immediately write down the incom-

pressibility Eq. 11 in the explicit form as

1 +

ur

r

u

r

uz

r

1

r

ur

u

r1 +

1

r

u

+

ur

r

1

r

uz

ur

z

u

z1 +

uz

z

= 1. 20The best way to evaluate the third term on the right-hand

side of Eq. 17 is to calculate it in Cartesian coordinates. Asa result, we obtain

1

0 B0B0 u = VA

2 2u

z2. 21

Using this result, we write the components of Eq. 17 incylindrical coordinates as

2ur

t2= VA

2 2ur

z2 A11

P

r

A21

r

P

A31

P

z+ fr, 22

2u

t2= V

A

2 2u

z2 A

12

P

r

A22

r

P

A

32

P

z+ f

,

23

2uz

t2= VA

2 2uz

z2 A13

P

r

A23

r

P

A33

P

z+ fz . 24

Here A11 , . . . ,A33 are the components of tensor A, given by

Eqs. A1A9 in Appendix A, and

f= AT AT ut

. 25Equations 20 and 2224 will be used in the next section

to derive the nonlinear governing equation for kink waves.

IV. DERIVATION OF NONLINEAR GOVERNINGEQUATION

In this section, we derive the nonlinear governing equa-

tion for kink waves in the magnetic tube. To do this we use

the reductive perturbation method.1113

We assume that the

wave is launched by a driver at z =0 and then propagates in

the positive z-direction. We also assume that the wave am-

plitude is small, u /R =O, 1 for rR and rR + lwe will see in what follows that u can be of the order of Rin the inhomogeneous annulus. In that case, the nonlinear

effects become pronounced at a nonlinearity distance of theorder of 1L, where L is the characteristic wavelength. It

follows from the linear theory of resonant absorption that in

the initial value problem, the ratio of the wave period to the

damping time is of the order of l /R e.g., Hollweg andYang,

14Goossens et al.,

15and Ruderman and Roberts

16.Translated to the boundary problem studied in this paper, this

result implies that the ratio of the damping distance to the

wavelength is of the order of R / 1. Our aim is to derive an

equation that describes the competition between the damping

and nonlinearity, so that we consider the situation where the

nonlinear and damping distances are of the same order. In

accordance with this, we take l /R =. We also restrict our




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rr

3rurm1

r2 m2

2urm

1

2= 0. 38

In what follows, we assume that the driving at z =0 starts at

some moment of time, and before that, the plasma was at

rest. This means that all perturbations decay as . The

solution to Eq. 38 satisfying this condition and regular atr=0 is

urm1 = Um,r

m1 , 39

where Um, is an arbitrary function. Now we easily findthe expressions for the other variables

um1 = iUmr

m1 sgn m , 40

Pm1 = i

C2 VAi2

mC2

2Um

2rm. 41

In the second order approximation, we obtain from Eqs.

2830, with the aid of Eq. 32, the following equations:

rur2

r +

u2

= rur

1

r 2

+

u1

r ur

1

u1

, 42

i1 VAi2

C22ur2

2+

P2

r

= 2iVAi

2

C

2ur

1

+

ur1

r

P1

r+

1

r

u1

r

P1

, 43

i1 VAi2

C22u2

2+

1

r

P2

= 2iVAi

2

C

2u

1

+1

rur

1

u1

P1

r

1

r

ur1

r

P1

. 44

Let us recall the formula for the coefficients of the expansion

of the product of two functions in the Fourier series

fgm = k=

fkfmk. 45

Using this formula and Eqs. 3941, we rewrite Eqs.4244 in the transformed form as

rurm2

r

+ imum2 =

k=

r2km3mk

k 1m k 1UkUmk, 46

i1 VAi2

C22urm2

2+

Pm2

r

= 2iVAi

2

C

2Um

rm1

iC2 VAi

2

C2

k=

r2km3mkk 1Uk

2Umk

2, 47

i1 VAi2

C22um2

2+

im

rPm

2

= 2iVAi

2

C

2Um

rm1 sgn m

iiC2 VAi

2

C2

k=

r2km3mkk sgn kUk

2Umk

2, 48

where

mk = 1 sgnkm k . 49

When deriving Eqs. 4648, we have used the fact thatmk0 only when km k0 and, when the latter inequal-ity is satisfied, k + m k = 2k m. Now we eliminate allvariables in favor of u

rm

2in this system of equations. It is

shown in Appendix B that, as a result, we arrive at

rr

2rurm2

r m2

urm2

= 2 k=

r2km3mk k 1m k 1

mUmkUk

sgn k+ m k 1

UkUmk .

50

The general solution to this equation is the sum of a particu-

lar solution and the complimentary function regular at r= 0,

so that we obtain

urm2

=1

2

k=

r2km3mk

mUmkUk

sgn k+ k 1

UkUmk

+Um

rm1 , 51

where Um, is an arbitrary function. When deriving thisequation, we have used the identity

2k m 22 m2 = 4k 1m k 1 52

valid for km k0. It is shown in Appendix B that the

expression for Pm2 can be written in the form

Pm2 =

2iVAi2

mC

2Um

rm

iC2 VAi

2

2mC2

k=

r2km2mkk sgn k 22

UkUmk

+ mUk

2Umk

2 i C2 VAi

2

mC2

2Um

2rm. 53

We do not give the expression for um

2because it is not used

in what follows.




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B. Solution in external region

The external region is defined by the inequality

rR + l. We split this region in two parts: the inner external

region and the outer external region. To do this, we introduce

a small parameter 1 satisfying the condition R /L11.

The inner external region is determined by the inequality

R + lr11R and the outer external region by the in-

equality r11R. Since both regions contain points with

r11R, they overlap.

In the outer external region, r can be of the order of or

even larger than L, so that, in this region, the characteristic

scale in the radial direction is L, i.e., it is the same as the

characteristic scale in the z-direction. This implies that we

have to introduce the stretching variable in the radial direc-

tion similar to z1. This variable is r1 =r. We consider per-

turbations decaying when r with the characteristic scale

of decay near the tube equal to R. This means that the wave

amplitude is small in the outer external region and the wave

motion can be described by the linearized system of equa-

tions. We obtain this system of equations linearizing Eqs.

20 and 2224, neglecting the dissipative terms andtransforming the obtained equations to the stretching vari-ables t1, r1, and z1. As a result, we arrive at

1

r1

r1ur

r1+

1

r1

u

+

uz

z1= 0 , 54

e

2ur

t12

= P

r1, 55

e

2u

t12 =

1

r1

P

, 56

e

2uz

t12

= P

z1. 57

Recall that B0 =0 in the external region. Expanding all vari-

ables in the Fourier series with respect to and eliminating

urm, uzm, and Pm, we obtain the equation for um

1

r12

r1r1

r1um

r1

m2

r12

um +

2um

z12

= 0. 58

The Fourier coefficients of other quantities are given in terms

of um

by

urm = ir1

m

um

r1, uzm =

ir1

m

um

z1,

59

Pm =ir1e

m

2um

t12 .

We assume that all functions are given at z1 = 0. In particular,

um = um0 r1 , t1 at z1 =0. In what follows, we assume that

um0 r1 , t1 vanishes sufficiently fast as r1. Let us intro-

duce um = um um0 ecz1, where c is an arbitrary positive

constant. Substituting this expression in Eq. 58 yields

1

r12

r1r1

r1um

r1

m2

r12 um +

2um

z12

= gr1,t1ecz1, 60

where

gr1,t1 = 1

r12

r1r1

r1um0

r1+

m2

r12 um

0 c2um0 . 61

Since um

=0 at z1

=0, we can apply the sinus Fourier trans-

form to this function. We also apply this transform to ecz1,

so that

umr1,z1,t1 = 0

umr1,,t1sinz1d,

62

ecz1 =2

0

sinz1

c2 + 2d.

Note that the second equation in Eq. 62 is only valid forz10. Substituting Eq. 62 in Eq. 60, we obtain

1

r1

r1r1

r

1um

r1

m2

r1um 2r1u

m =

2r1

gr

1,t

1c2 + 2 .

63

The homogeneous counterpart of this equation is the modi-

fied Bessel equation for function r1um, so that the compli-

mentary function is a linear combination of the modified

Bessel functions of the first kind Imr1 and second kindKmr1, divided by r1. Then, using the standard method ofvariation of arbitrary constants, it is straightforward to find

the solution to Eq. 63 vanishing as r1. It is given by

um =2

r1c2

+ 2

r1

r2gr,t1ImrKmr1

Imr1Kmrdr + Cm,t1Kmr1 , 64

where Cm, t1 is an arbitrary function. In what follows, weassume that mCm, t1 is bounded as 0 in order tohave the last term in Eq. 64 regular at =0. SubstitutingEq. 61 in Eq. 64, after simple algebra, we rewrite Eq. 64as

um = 2um

0 r1,t1

c2 + 2

2

r1

r1

r2um0 r,t1

ImrKmr1 Imr1Kmrdr

+ Cm,t1Kmr1 . 65

Now we make the inverse Fourier transform and use the

second equation in Eq. 62 to obtain

um = um + um0

ecz1

=0

Cm,t1Kmr1sinz1d

2

r1

0

sinz1dr1

r2um0 r,t1

ImrKmr1 Imr1Kmrdr. 66




8/23

The solution in the outer external region is only needed to

obtain the boundary conditions for the solution in the inner

external region. To do this, we use the method of matched

asymptotic expansions e.g., Bender and Orszag19. In accor-dance with this method, we have to obtain the inner expan-

sion, which is the solution in the inner external region, and

the outer expansion, which is the solution in the outer exter-

nal region. To match the inner and outer expansions, we need

to obtain the so-called inner expansion of the outer expan-sion and the outer expansion of the inner expansion. To ob-

tain the first of the two expansions, we substitute r1 =r in

the outer expansion determined by Eqs. 66 and 59 andexpand it in power series of . Using the asymptotic

formula20

Kmx 2m1m 1!xm

valid for x+0, we obtain

urm urmrm1, um umr

m1 ,

67

uzm uzmrm, Pm Pmr

m,

where urm, um, uzm, and Pm are functions oft1 and z1. We do

not give the expressions for them because they are not used

in what follows. The outer expansion of the inner expansion

is obtained by substituting r=1r1 in the inner expansion

and expanding it in power series of . The matching condi-

tion is that the outer expansion of the inner expansion coin-

cides with the inner expansion of the outer expansion when

r1 =r is substituted in the former or, equivalently, r=1r1

is substituted in the latter. It is obvious from Eq. 67 thatthe necessary condition for satisfying the matching condition

is that the solution in the inner external region vanishes

as r. Hence, in the rest of this subsection, we are looking

for the solution in the inner external region vanishing atinfinity.

The analysis in the rest of this subsection parallels to the

analysis in Sec. IV A. Once again we are looking for the

solution to Eqs. 20, 22, and 23 that can be reduced toEqs. 2830, and then the solution is looked for in theform of expansions given by Eq. 31. In the first order ap-proximation, we obtain Eqs. 3234 and solve them usingthe expansions in the Fourier series with respect to . Once

again we eliminate all variables in favor of urm

1to obtain Eq.

38. However, now we look not for the solution to this equa-tion regular at r=0 but for the solution vanishing as r.

This solution is given by

urm1 = Wm,r

m1 , 68

where Wm, is an arbitrary function. The expressions forthe other variables are given by

um1 = iWmr

m1 sgn m , 69

Pm1 =

e

m

2Wm

2rm. 70

In the second order approximation, we once again obtain

Eqs. 4244. Using Eqs. 45 and 6870, we transformthese equations to the form very similar to Eqs. 4648.

Solving these transformed equations is almost exact repeti-

tion of solving Eqs. 4648, so that we omit all the detailsand present only the final results. The radial displacement

and the total pressure perturbation in the second order ap-

proximation are given in the inner external region by

urm2

=

1

2

k=

r2km3mk

mWmkWk

sgn k+ k + 1

WkWmk

+Wm

rm1 , 71

Pm2 =

e

2m

k=

r2km2mk

mWk2Wmk2

+ k + 1

2

2WkWmksgn k

+ e

m2W

m

2rm, 72

where Wm, is an arbitrary function.

C. Solution in the annulus

In this subsection, we are looking for the solution in the

annulus determined by RrR + l. The Alfvn resonant sur-

face r= rA, with rA determined by the equation VArA = C,plays an important role in our analysis. In a thin dissipative

layer embracing this surface, the resonant interaction be-

tween the global wave motion and the local Alfvn oscilla-

tions of the individual magnetic field lines occurs. This in-teraction causes the oscillation amplitude in the resonant

layer to grow at the expense of the global wave motion. As a

result, the global wave motion damps.

The interaction between the global wave motion and the

local Alfvn oscillations near the Alfvn resonant surface

starts immediately after the propagating wave is launched at

z =0 by the driver. After the initial phase of growth, the am-

plitude of the wave motion in the resonant layer reaches its

limiting value and does not grow any longer. However, in

general, the ratio of the oscillation amplitude in the resonant

layer and the amplitude of the global wave may continue to

increase due to the damping of the global wave. Another

important process that occurs in the resonant layer is phasemixing. This process occurs because, due to the radial depen-

dence of the Alfvn velocity, the oscillation frequencies of

neighboring magnetic field lines are slightly different. As a

result, the phase difference between oscillations of two

neighboring magnetic field lines linearly grows with time,

which causes the linear growth of gradients in the radial

direction. Eventually, this process is stopped by dissipation

e.g., viscosity or resistivity and after that, the physical en-ergy damping starts.

In linear MHD, the motion in the resonant layer was

studied for the initial value problem.2124

However, the re-

sults of this study are easily translated to the boundary prob-




9/23

lem considered in this paper if we swap the time and

z-coordinate. To discuss these results, it is convenient to in-

troduce the viscous Reynolds number Re= iRC/. In accor-

dance with the analysis of motion in the resonant layer, there

are two different regimes of this motion. The first regime

occurs when Re1/3l /R2 =O2. In this case, the thick-ness of the dissipative layer is of the order of Re 1/3lRL1/3,which is, in turn, of the order of Re1/3R l2 /R =Ol.25,26

This regime can be called quasistationary because the wavemotion in the dissipative layer is almost exactly the same as

that in stationary resonant layers that exist in the driven

problem. When Re1/3 =O, the thickness of the dissipativelayer becomes of the order of l. In that case, we cannot use

the term resonant absorption anymore; rather, we have nor-

mal viscous wave damping. So the wave damping due to

resonant absorption occurs only when Re1/3.

The second regime is nonstationary.22,23

It occurs when

Re1/3l /R2 =O2. In this case, the thickness of the dis-sipative layer is of the order of l2 /R =Ol. The wave mo-tion in the dissipative layer is qualitatively different from

that in the driven problem. An important result is that the

thickness of the dissipative layer is independent of theReynolds number. Another extremely important result is that

the damping decrement due to resonant absorption is inde-

pendent of the Reynolds number. It is the same in both

regimes.

To make our analysis valid for both regimes, we take

Re1/3 =2 /, where . Then 1 corresponds to the sta-

tionary and 1 to the nonstationary regime. The dissipa-

tive layer is determined by the inequality r rAl /,where =min1 ,, and the part of the annulus outside thedissipative layer is determined by r rAl /. Since themotion in the dissipative layer and outside the dissipative

layer are qualitatively different, we obtain the solutions in

these two regions separately and then match them in the

overlap region determined by r rAl /.

1. Solution outside dissipative layer

In the annulus outside the dissipative layer, the charac-

teristic scale in the radial direction is l. Since l /R =O, it isconvenient to introduce a new stretching variable in the an-

nulus =1rR. Since uz and P have to be continuous atthe annulus boundaries, the estimates 27 remain valid inthe annulus. Viscosity is only important in the dissipative

layer, while it can be neglected outside the dissipative layer.Then, using the estimates 27, we rewrite Eqs. 20, 22,and 23 in the new variables as

R + ur

+

u

+ ur +

ur

u

+ ur

u

ur

u =O3, 73

31 VA2C2

2ur2

+P

1

ur

P

1

R

u

P

= O5 , 74

21 VA2C2

2u2

+ 23VA

2

C

2u

+

1

R +

P

1

R + ur

uP

ur

P

= O5. 75

Once again we are looking for the solution to Eqs. 7375in the form of expansions given by Eq. 31. In the first order

approximation, we obtain from Eqs. 73 and 74ur

1

= 0,

P1

= 0 , 76

so that ur

1and P1 are independent of . Using this result,

we obtain from Eq. 75 in the first order approximation

1 VA2C2

2u12

+1

R

P1

= 0. 77

In the second order approximation, we obtain from Eqs. 73and 74

Rur2

= ur

1 u1

+ u

1

ur

1

u

1 , 78

P2

=

1

R

u1

P1

1 VA2

C22ur1

2. 79

It follows from Eq. 77 that u

1has a singularity at

r= rA. It becomes of the order of 1 when r rA, thus

violating the assumption that u=O. This observationgives an additional ground to our statement that the motion

in the dissipative layer has to be considered separately from

the motion outside the dissipative layer.

2. Solution in dissipative layer

Since the thickness of the dissipative layer is of the order

ofl /=O2R /, we introduce the new stretching variable=2r rA in the dissipative layer. If =O1, then thethickness of the dissipative layer is of the order ofl. Then it

follows from Eq. 77 that u

1/R =O1 at the boundary of

the dissipative layer, i.e., u, is of the order of unity in this

layer. Although the motion in the dissipative layer is mainly

in the azimuthal direction, so that a displaced plasma particle

remains at almost the same distance from the tube axis, ur is

of the order of unity as soon as u

is of the order of unity.

This is clearly seen from Fig. 1. If the initial radial coordi-

nate of the particle is r, then its radial coordinate after the

perturbation is ur+ r2 + u

21/2. Since the variation of theradial position of any particle is of the order of , ur+ r

2

+ u2 = r2 +OR. In addition, r= rA1 +O

2 for any pointin the dissipative layer. This analysis inspires us to introduce

the new variables and h, such that

ur = rA + hcos r, u= rA + hsin . 80

The characteristic scale of variation of function VA2r is l.

Since the thickness of the dissipative layer is much smaller

than l, we use in this layer the approximate expression




10/23

C2 VA2r = 2/, =

dVA2

d=

dVA2

dr. 81

For

, we have the estimate

=OC2 /R.Let us now obtain an approximate expression for the

viscous force f. When doing so, we keep only the largest

terms. First of all we note that the first estimate in Eq. 27,uz /R =O

2, remains valid in the dissipative layer. Usingthese estimates and Eqs. A1A9, we obtain

AT = 2er1 + 1r

u

+

ur

r

1

r

u

e

r ur

u

ur

. 82

When deriving this expression, we neglect all terms of the

order or smaller than 2. Let us introduce =+ andthen use =and as the independent variables instead of

and . We also introduce two unit vectors see Fig. 1,

er = er cos + e sin , e = er sin + e cos .

83

Then we rewrite Eq. 82 in a very compact form,

2/AT = er

. 84

Using Eqs. 80 and 83, and once again retaining only thelargest terms, we obtain

u

t=

ur

ter +

u

te= rA

te . 85

The derivatives of the unit vectors are given by10

er

= e,

e

= er,

with all other derivatives being zero. With the aid of these

formulae, it is straightforward to show that

er

=

e

= 0. 86

Then

2/AT u

t= rA

2

tere . 87

Since Re=36, it is convenient to introduce the scaled dy-

namic viscosity =36=iRC. Then, using Eqs. 84,86, and 87, we eventually arrive at

f= 323rAe

2

. 88With the aid of Eqs. A1A9 and the first estimate in

Eq. 27, we write the mass conservation Eq. 20 in the newvariables as

ur

r+ u

+ ur u

ur

u +

2

u

+ ur

= O3/. 89

Substituting Eq. 80 in Eq. 89 and using the new variables and , we reduce Eq. 89 to a very simple form

rA + h

h

= /rA +O

2/. 90

Once again using the estimate uz /R =O2 and taking

into account Eq. 81, we reduce Eqs. 22 and 23 to

5 C2

2ur

2+ 2

2ur

2fr +

rr+ u

+ ur

+O3 P

1

ru

+ O3 P

= O6/,

91

5 C2

2u

2+ 2

2u

2f

r ur

u

+ O3 P

+1

rur

+ 2 + O3 P

= O6/.

92

Now we substitute Eqs. 80 and 88 in Eqs. 91 and 92and use the variables and as the independent variables

instead of and . As a result, we obtain two equations for

and P. Multiplying the first equation by cos , the second

equation by sin , and adding the results, we obtain

r

e

e

u

u

u

r

r

e

e

r

FIG. 1. The sketch of the particle displacement. It is clearly seen that al-

though the variation of the radial distance of a particle is small, both ur and

u are large.




11/23

P

= O5/. 93

In a similar way, multiplying the first equation by sin , the

second equation by cos , and adding the results, we arrive at

3rA2

C2

+

+ 2

+

1 +

+ P

323rA2

2

= O4/. 94

When deriving Eq. 93, we have used the estimate=O that follows from Eqs. 77 and 80. To derive Eq.94 we have used Eqs. 90 and 93. Equations 90, 93,and 94 constitute a closed system of equations for variablesh, , and P.

When studying propagation of surface waves on a thin

transitional layer in the linear approximation Mok and

Einaudy21 made an assumption that 1. Later,Ruderman and Goossens

7made the same assumption when

they extended the analysis by Mok and Einaudy to include

the effect of nonlinearity. Ruderman et al.22

relaxed the as-

sumption made by Mok and Einaudy21

and developed the

linear theory of surface waves on a thin transitional layer

valid for 1, i.e., for arbitrary . However, an attempt

to extend their analysis to the nonlinear case results in a very

complicated nonlinear Eq. 94. It seems improbable that itcan be solved analytically. To make analytical progress, we

introduce the same restriction as in Mok and Einaudy21

and

Ruderman and Goossens7

and take 1. Since =O,we can now neglect all nonlinear terms in Eq. 94 in thelowest order approximation with respect to . In particular,

= 1 +

1.

Then, noting that now = and introducing P=3P, we

obtain from Eq. 94 in the lowest order approximation withrespect to and ,

A

C2

2

2

3

2=

rA2

P

, 95

where A =rA. When deriving this equation, we have

noted that the ratio of the second term in the first squarebrackets in Eq. 94 to the first one is of the order of , sothat this second term can be neglected, and we have also

used the fact that =. In accordance with Eq. 93, we can

consider P in Eq. 95 as in dependent of.To solve Eq. 95, we introduce the Fourier transform

with respect to ,

f=

feid, f =1

2

feid.

Usually, this transform is used only for functions that decay

at infinity. However, it also can be applied to periodic func-

tions if we allow the Fourier transform to be a generalized

function. In that case, the Fourier transform of a periodic

function f has the form

f= 2n=

fn+ n0 ,

where fn are the coefficients in the expansion of f in the

Fourier series, 2/0 is the period, and is the Diracdelta-function.

Applying the Fourier transform to Eq. 95, we obtain

A2 iC

2

2

2=C2

rA2

P

. 96

This equation was extensively studied in the linear theory of

stationary dissipative layers that are present in the driven

problem. In particular, Goossens et al.26

showed that its so-

lution vanishing as can be written as

= iC2

rA2A

2

P

F

,

97

where the F-function and the parameter are given by

= C2

1/3

, Fx = 0

expisx sgn s3/3ds .

98

Let us expand P in the power series of given by Eq.

31. Then, in accordance with Eq. 93, we obtain

P = 3P1,, + 4P2,, + . . . . 99

Using the expansion h = h1

+ /h2

+... and Eqs. 80 and90, we arrive at

ur = rAcos 1 + h1,,cos + 2

h2,,, / + . . . . 100

With the aid of the formula

=

1 = 1

1 , 101

we obtain from Eq. 90

h2

=

1

1

. 102Now it follows from Eqs. 97 and 102 that

h2 /= C2

rA2A

2

2P

2G , 103

where the G-function is given by26

Gx = 0

es

3/3

sexpisx sgn 1ds 104

and we have neglected an arbitrary constant as unimportant.




12/23

3. Calculations of jumps of total pressureand displacement across the annulus

The final goal of Sec. IV C is to calculate the jumps of urand P across the annulus determined by

ur = urR + l urR, P = PR + l PR. 105

These jumps will be then used to connect the solution in the

internal and external regions. To calculate these jumps, weneed to match in the overlap regions the solution in the dis-

sipative layer and in the two regions outside the dissipative

layer. The overlap regions are defined by the inequality

1 1, or, which is the same, by / R1.The inner expansions of P and ur are given by Eqs. 99 and100. To obtain the outer expansion of the inner expansion,

we substitute =1with =1rA R in the inner ex-

pansion and then re-expand it with respect to at fixed .

Using the integration by parts and Eq. 99, we easily obtainfrom Eq. 97

= 1 + C2

rA2A

2

P

1

+ . . . . 106

An important result that follows from this equation is that

=O. Then it follows from the relation /=// and Eq. 101 that we can substituteP

1/ for P

1/ in Eq. 106. Let us introduce function

P1 defined by

2P1

2= P1. 107

Since we assume that perturbations of all quantities vanish as

, function P1 is defined uniquely. Now it follows

from Eq. 106 that

1 = C2

rA2A

P1

. 108

Using Eqs. 101, 106, and 108 we rewrite Eq. 99 as

P = 4C2

rA2A

P1

P1

+ 3P1,, + P2

,, + . . . , 109

where the dots indicate terms with the positive powers of.

This is the outer expansion of the inner expansion for P.

To obtain the outer expansion of the inner expansion for

ur, we use the asymptotic formula26

Gx = lnx 1

3E+ ln 3 +

i

2sgnx + . . . , 110

where E 0.577 is Eulers constant. Using this formula andEqs. 99, 103, 106, and 108, we obtain from Eq. 100that the outer expansion of the inner expansion of ur is given

by

ur =2C2

rA2A

P1

C2

2rAA2

P1

+

1

h1

+

2P1

2ln + E+ ln 3

3 + 2

2 + h1 + . . . ,

111

where now, h1 is a function of , , and , the dots once

again stand for terms with positive powers of, and function

=,, is determined by its Fourier transform

= P1ln1 + i

2sgn . 112

Now we proceed to calculating the inner expansion of

the outer expansion. We start from transforming Eqs. 78and 79. Using Eqs. 76 and 77, we rewrite Eqs. 78 and79 as

ur2

=

C2

R2C2 VA2

P1

ur

1

+C2

2RC2 VA2

P1

ur

1

R+

C2

R2

2P1

2

d

C2 VA2

+ ur2,, ,

113

P2 = C2

R2C2 VA2

P1

P1

2ur

1

2

1 VA2C2

d + P2,,, 114where = 0, + =

1l, ur

2,, and P

2,, are arbi-trary functions, and the and + subscripts indicate the

regions at the left and the right of the dissipative layer. Now

we substitute =0 +/with 0 =1rA R in the solution

outside the dissipative layer and then expand the obtained

expressions with respect to , keeping constant. After that,

to compare with the outer expansion of the inner expansion

given by Eqs. 109 and 111, we substitute back=/

0 = /. As a result, we obtain the inner expansion

of the outer expansion written in terms of

ur = 2

C2

rA2A

2

P

1

2ln

/

P

1

C22rAA

2

P1

+

1

ur1

+ ur11 0

R

+ 2ur2 +

2C2

R2

2P1

2

0 1C2 VA

2

1

A 0

d + . . . ,115




13/23

P = 4C2

rA2A

P1

P1

+ 3P

1 + P2 + . . . ,

116

where =0, the dots indicate terms with positive pow-

ers of , and now we systematically use the and + sub-

scripts to distinguish quantities in the inner expansion of theouter expansion from those in the outer expansion of the

inner expansion. When deriving these equations, we have

taken into account that R = rA +O. Comparing Eqs. 109and 116 yields

P1 = P1, P

2 = P2 . 117

Comparing Eqs. 111 and 115 results in

ur1 = h1 , 118

ur2 =

C2

rA2A

13

2P1

2E+ ln3

3 +

2

2+ h1

0

R

C2

R2

2P1

2

0 1C2 VA2 1

A 0

d. 119Now it follows from Eqs. 76, 117, and 118 that

ur1

= 0, P1

= 0 , 120

so that there are no jumps of the radial displacement and the

total pressure across the annulus in the first order approxi-

mation. Using Eqs. 115 and 117, we obtain

P2 = 2

eR2

P1

P1

R

lC2

2ur1

2

R

R+l

C2 VA2dr. 121

When deriving this equation we used the approximate rela-tion C Ck. Finally, with the aid of Eqs. 114, 118, and119, we arrive at

ur2 =

2

eR2

P1

ur1

+

C2

lR

2P1

2P

R

R+l dr

C2 VA2

ur1 +

C2

R2A

2

2R + l R , 122

where P indicates the principal Cauchy part of the integral.

Since sgn =sgn, it follows from Eq. 112 that

R + l R = iP1 sgnsgn. 123

Let us introduce the Hilbert transform with respect to ,2729

Lf =1

P

fd

. 124

Then, using the relation between the Fourier and Hilbert

transforms

Lf= i sgnf, 125

we eventually rewrite Eq. 122 as

ur2 =

2

eR2

P1

ur1

C2

R2AL 2P1

2 ur1

+C2

lR

2P1

2P

R

R+l dr

C2 VA2

. 126

When deriving this equation, we once again used the ap-

proximate relation C Ck. Equations 120, 121, and 126will be used in Sec. IV D to connect the solutions in the

internal rR and external rR + l regions.

D. Matching solutions in the internal and externalregions

In this subsection, we use the expressions for the jumps

of the radial displacement and the perturbation of the total

pressure across the annulus to connect the solutions in the

internal and external regions. For this we calculate the jumpsof ur and P using these solutions and then compare the re-

sults with the jumps given by Eqs. 120, 121, and 126.In the first order approximation, we use Eqs. 39, 41,

68, and 70 to obtain

urm1 = WmR + l

m1 UmRm1 , 127

Pm1 =

e

m

2Wm

2R + lm + i

C2 VAi2

mC2

2Um

2Rm.

128

Since, in accordance with Eq. 120, the left-hand sides ofEqs. 127 and 128 are equal to zero, we obtain from theseequations

Wm = R2m1 + l/Rm+1Um, 129

C2 =iVAi

2

i + e1 + l/R Ck

21 el/Ri + e

= Ck2 + O.130

Using these results and Eqs. 51, 53, 71, and 72, weobtain in the second order approximation




14/23

urm

2 = k=

R2km3sgn k sgnm k

mUmkUk

+ k

UkUmk

+Wm

R + lm1

Um

Rm1 , 131

Pm2 =

2Ci + e

m

2Um

Rm

e

m

k=

R2km21 sgnkm k

mUk2Umk2

+ k

2

2UkUmk

+e

m

2Wm

2R + lm + i

C2 VAi2

mC2

2Um

2Rm.

132When deriving these equations, we have used the approxi-

mation R1 + l /RR.To compare Eqs. 131 and 132 with Eqs. 121 and

126, we need to express in terms ofU the functions ur

1and

P1 calculated in the annulus that are present on the right-

hand sides of Eqs. 121 and 126. Since both these func-tions are independent of r in the annulus, they are equal to

their values at r=R. Hence, the coefficients of the expansion

in the Fourier series of ur

1are given by Eq. 39 and it

follows from Eq. 41 that

Pm1 = i

C2 VAi2

mC2RmUm. 133

Then the comparison of Eqs. 131 and 132 with Eqs. 121and 126 results in

Wm

R + lm

Um

Rm1 + l/R = R

m

+ k=

msgn k sgnm k mkk

+ ksgn k+ sgnm k

kmk

eRC2m

l

m

P

R

R+l dr

C2 VA2

+eC

2m

ALm

, 134

2Wm

2R + lm

2Um

2Rm1 + l/R

=2RCi + e

e

2m

+

k=

m1 + sgnkm kk2mk2

+ k sgn m1 sgnkm k

2

2kmk

mR

elC2

2m

2

R

R+l

C2 VA2 dr, 135

where m =Rm1Um. It follows from Eq. 39 that ,, =ur

1R ,,,, i.e., gives the radial displacement of the tubesurface divided by in the first order approximation. When deriving Eq. 135, we have used Eq. 130.

Differentiating Eq. 134 with respect to and subtracting the result from Eq. 135, we arrive at

2RCi + e

e

2m

+

k=

sgn k+ sgnm k mk2mk2

sgn k k

2

2kmk sgn k sgnm k

m

mkk

k 22

kmksgn m + R2m2

eC

2m

AL 2m

2 + mR

l

2m

2

PR

R+l eC2C2 VA

2C2 VA

2

eC2 dr. 136




15/23

It is shown in Appendix C how to simplify the sum in this

equation. Using Eq. C12, we rewrite Eq. 136 as

C

2m

+

q

2

2m

2+

qm

R

k=

sgn m sgn k

mkk

k

mk

sgn k

+qm

l

2m

2 L 2m

2 = 0 , 137

where

q =e

i + e, =

eC2

2A, =

=

dVA2

dr,

138

=1

2P

R

R+l eC2C2 VA

2C2 VA

2

eC2 dr.

Let us now introduce the Hilbert transform with respect to

,2729

H =1

2

0

2

cot

2d. 139

Then, using the expression for the Fourier coefficients of the

Hilbert-transformed function27

Hm = im sgn m , 140

and returning to the original variables, we transform Eq.

137 to recall that = l /R

C 2

tz

+ 1 + lq2R2

t2

+ qR

3

t2HL

=q

R

tH

t H

t

tH

t .

141

When deriving this equation, we have made the substitution

=1 and then dropped the tilde. As a result, the coeffi-

cient at the first term of the expansion for ur given by Eq.

31 is canceled out and is equal to the radial displacementat the tube boundary in the first order approximation.

Linearizing Eq. 141 and taking =expim+ k,where is real, we obtain the dispersion relation

k= 1/C0 m+ im , 142

where C0 = Ck1 lq /R +Ol2/R2. We see that the wave

propagates from the driver at z =0 in the positive z-direction

without dispersion with the phase speed close to the kink

speed Ck note that =Ol /R. There is the small correctionto this phase speed proportional to l /R, the correction de-

pending on m. There is also spatial damping described bythe imaginary part of k. Since =Ol /R, the characteristicdamping length is equal to the wavelength times R / l. It is

also inversely proportional to m.

The temporal damping of nonaxisymmetric magnetic

tube oscillations was studied by Goossens at al.15

In particu-

lar, they gave the expression for the decrement in the thin-

tube approximation. If we substitute the zero external mag-

netic field in their formula, then we obtain for the temporal

decrement the expression

=2e

2Ck

3m

ARi + eL,

where L is the wavelength. The wave amplitude decreases by

e times after the time 1. During this time, the wave runs

the distance Ck1, so that its amplitude decreases by e times

at the distance Ck1. This means that the spatial decrement

is equal to / Ck. Taking into account that = 2Ck/L, it isstraightforward to verify that / Ck=m1 +Ol /R.Hence, our results agree very well with the results obtained

by Goossens et al.

15

Ruderman30

derived an equation describing propagation

of nonaxisymmetric surface waves in a magnetic tube with a

sharp boundary see his Eq. 2.29. He took weak dispersionrelated to the finiteness of the tube radius into account. It is

interesting to compare his equation with Eq. 141. For thiswe first neglect dispersive terms in Rudermans Eq. 2.29and then take l0 in Eq. 141, so that ==0. Then weinterchange t and z in Eq. 141. For this we transform Eq.141 to the new variables, z= ct and t=z / c. After that, thetwo equations, one obtained from Rudermans Eq. 2.29 andthe other obtained from Eq. 141, exactly coincide.

Another interesting comparison can be made if we sim-

plify Eq. 141 by considering its particular solutions in theform of helical waves. These solutions depend on z and

=t, where is a constant. After that, Eq. 141 re-duces to an equation that has only two independent variables,

z and . Interestingly, by a simple rescaling of variables, this

equation can be reduced to the equation describing the

propagation of long nonlinear waves on finite-thickness mag-

netic interface7

or to the equation describing the propagation

of long nonlinear waves on a sharp magnetic interface in a

plasma with strongly anisotropic viscosity.31

We are now in a position to discuss qualitatively the

nonlinearity effect on the propagation of kink waves along

the tube. If a kink wave with the frequency 0 is launched at

z =0, then, during its propagation in the positive z-direction,

the quadratic nonlinearity present in Eq. 141 will generatehigher harmonics with respect to and , which are the

fluting waves m1 with frequencies multiple to 0.Since, in accordance with Eq. 142, the damping rate isproportional both to m and , this will result in the accel-eration of the wave damping. This process will be studied

numerically in Sec. V.

It is shown in Appendix D that the wave energy flux

integrated over the time is equal to CR2i +eE, where E isgiven by




16/23

E=

dt0

2Ht

0

tdd. 143

It is also shown in Appendix D that E0, it satisfies the

equation

dE

dz=

dt0

2

2

t2L

td, 144

and the right-hand side of this equation is negative. We see

that E is a conserved quantity when there is no wave damp-

ing due to resonant absorption = 0. We also can see thatthe wave energy flux decreases due to resonant absorption as

was expected.

V. NUMERICAL RESULTS

In this section, we present the results of numerical study

of the nonlinearity effect on the wave damping. To simplify

the analysis, we assume that C2 VA2 is a symmetric func-

tion, which means that rA =R + l /2 and C2 VA

2 is an oddfunction of the variable r rA. Then = 0.

Equation 141 is a complicated integrodifferential equa-tion. Its numerical solution is a difficult problem. It would be

simplified very much if we consider solutions periodic with

respect to time because in that case it could be reduced to an

infinite system of ordinary differential equations. This sys-

tem could be then truncated and easily solved numerically

using, for example, the RungeKutta method.

Unfortunately, this approach does not work. The reason

is the following. In Sec. IV, we assumed that the tube starts

to be driven at a finite moment of time, and before that the

plasma was at rest. This, in particular, implies that =0 for

t t0

, where t0

is a constant. Obviously, no periodic function

of time satisfies this condition.

The condition ur

1=0 as was used in Sec. IV to

eliminate two arbitrary functions of r and that appear after

integrating Eq. 38 twice with respect to , and to obtain thesolution to this equation given by Eq. 39. We can get thesame result if, instead of the condition u

r

1=0 as , we

impose the condition that ur

1is a periodic function ofwith

the zero mean value over the period. However, this condition

is incompatible with Eq. 141. Really, the assumption thatu

r

1is a periodic function of implies that is a periodic

function of t. But then it follows that the mean values of all

terms but the last one in Eq. 141 are zeros, while the mean

value of the last term is nonzero.However, there is one exception. This is the case of he-

lical perturbations. Let us look for solutions to Eq. 141 thatdepend not on t and separately but on their linear combi-

nation ht. Obviously, is a periodic function of ht

with the period 2. In addition, we assume that the mean

value of over the period with respect to ht is zero.

The wave is driven at z =0. The driver is determined by

the boundary condition

= A sin . 145

Since we assume that the perturbation amplitude is of the

order of, it follows that A /R =O.

In order to solve Eq. 141 numerically, we introducenew dimensionless variables

= ht+zh

C1 + lq

2R, Z= qhz

RC, =

A. 146

Transforming Eq. 141 to these variables, integrating it oncewith respect to , and using the condition that the mean

value of over the period is zero, we reduce it to

Z=

2

2

N

H

H

H

, 147

where the nonlinearity parameter is given by N=A /. When

deriving Eq. 147, we have used the fact that now

H = L

=1

2

0

2

cot

2d, 148

and the identity Eq. D7. By a simple rescaling of variables,this equation can be reduced to the equations describing the

nonlinear wave propagation on magnetic interfaces.7,31

It is

straightforward to see that the condition that is a periodic

function of with the zero mean value is compatible with

Eq. 147. Really, using the identity Eq. D14, we immedi-ately obtain that the mean value of the right-hand side of Eq.

147 is zero. This implies that if the mean value of is zeroat Z=0, then it is zero at any Z0.

To solve Eq. 147, we expand in the Fourier serieswith respect to . The formulae

27

Hsin m = sgn m cos m,

149Hcos m = sgn m sin m,

imply that the Hilbert transform of an even function is an

odd function and vice verse. Then it follows that, if a solu-

tion of Eq. 147 is an odd function at Z=0, it remains oddfor any Z0. Hence, we can write the Fourier series for

as

= m=1

mZsin m. 150

The system of equations for the Fourier coefficients was de-

rived in Ref. 7. In our notation it reads

dm

dZ= Nm

n=1

nnm+n 1

2n=1

m1

nm nnmn m2m . 151

Equation 151 is an infinite system of ordinary differentialequations. It follows from Eq. 145 that the initial condi-tions at Z=0 are given by

1 = 1, m = 0 for m 1. 152

For the numerical solution of system Eq. 151 with theinitial conditions Eq. 152, we truncated expansion Eq.




17/23

150 and put m =0 for mM. In this way, we reduced Eq.151 to the system of M ordinary differential equationswhich can be easily solved numerically.

Let us transform the expression Eq. 143 for the quan-tity E proportional to the energy flux integrated over the

time. Since now is a periodic function oft, the limits of the

integral with respect to t has to be equal to 0 and 2/h.

Then, using Eqs. 149 and 150, we obtain

E= 22A2hE, E= m=1

mm2 . 153

Equation 144 reduces to

dE

dZ=

m=1

m3m2 . 154

This equation was used to verify the accuracy of solution of

Eq. 151. For this we calculated Efirst using the solution ofEq. 151 and then by the direct integration of Eq. 154, andcompared the results.

Figure 2 shows the dependence of E on the distance Z

for different values of the nonlinearity parameter N. When

N=0, i.e., in the linear approximation, E decays exponen-

tially, so that ln E is a linear function of Z. This result is in acomplete agreement with the linear theory of wave damping

due to resonant absorption. It is clearly seen in Fig. 2 that

increasing N results in faster energy dissipation. This is an

intuitively expected result. Nonlinearity transfers energy

from the fundamental harmonic to higher harmonics, which

damp faster. The larger the nonlinearity parameter N, the

more efficient the nonlinear energy transfer.

Here we have to make one comment. When deriving the

nonlinear governing equation, we have used the long-

wavelength approximation, which implies that we only con-

sidered perturbations with the characteristic spatial scale

much larger than the tube radius. The nonlinear generation of

higher harmonics causes the wave steepening and the de-

crease of the characteristic spatial scale. The larger the N, the

smaller spatial scale is formed by the nonlinear steepening.

This means that we can consider only moderate values of N,

while the theory breaks down for very large values of N. This

situation is typical for all nonlinear long-wavelength theo-

ries. For example, the shallow water theory describing

propagation of long waves on water surface breaks down

when the nonlinearity parameter, this time characterizing the

relative importance of nonlinearity and dispersion, becomes

very large.

In accordance with Eq. 153, the total dimensionless

energy flux E is equal to the sum of energies of separateharmonics, the energy of the mth harmonic being equal to

mm2 . Figures 35 show the dependence of the harmonic

energy on Z for N=5, 10, and 50, respectively. We see that

the amplitudes of overtones are much smaller than the am-

plitude of the fundamental harmonic even for the large value

of the nonlinearity parameter.

0.001

0.01

0.1

1

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

E

Z

FIG. 2. Color online The dependence of the total dimensionless energy Eon the dimensionless distance Z for different values of the nonlinearity

parameter N. The full, dashed, long dashed, and dotted curves correspond to

values N of 0, 5, 10, and 50, respectively.

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5

E

Z

FIG. 3. Color online The dependence of the total dimensionless energy Efull curve and the energies of the first four harmonics dashed, shortdashed, dotted, and dashed dotted on the dimensionless distance Z for

N=5 .

0

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5

E

Z

FIG. 4. Color online The same as Fig. 3 but for N=10.




18/23

The internal radius of the unperturbed tube is equal to R.

Let us take a point M on the unperturbed boundary corre-sponding to the azimuthal angle . Its position vector is Rer.

In the first order approximation with respect to , the dis-

placement of a point at the tube boundary in the radial direc-

tion is A, while, in accordance with Eqs. 39, 40, and140, the displacement in the azimuthal direction is AH.Hence the displacement vector of point M is

Aer + AHe,

so that the new position vector of point M is

Rer + Aer + AHe.

Then the Cartesian coordinates of M are

x = R cos + A cos AHsin ,

155y = R sin + A sin + AHcos .

These equations define the cross-section of the perturbed

boundary of the tube by the plane Z=const as a parametric

curve with being the parameter. Using Eqs. 149 and150, we rewrite Eq. 155 as

x = R cos + A m=1

mZsinm ,

156

y = R sin + A m=1

mZcosm .

Since depends on t the shape of the cross-section of per-

turbed boundary of the tube by the plane, Z=const varies

with time. However, it is possible to get read off the time

dependence. Introduce new coordinates

x = x cos 0 + y sin 0,

157y = x sin 0 + y cos 0 ,

where

0

0.2

0.4

0.6

0.8

1

0 0.05 0.1 0.15 0.2 0.25

E

Z

FIG. 5. Color online The same as Fig. 3 but for N=50.

-2

1.5

-1

0.5

0

0.5

1

1.5

2

-2 - 1.5 - 1 -0.5 0 0.5 1 1.5 2

Z= 0.0000

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-2 - 1.5 - 1 -0.5 0 0.5 1 1.5 2

Z= 0.0400

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-2 - 1.5 - 1 -0.5 0 0.5 1 1.5 2

Z= 0.0800

-2

1.5

-1

0.5

0

0.5

1

1.5

2

-2 - 1.5 - 1 -0.5 0 0.5 1 1.5 2

Z= 0.1600

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-2 - 1.5 - 1 -0.5 0 0.5 1 1.5 2

Z= 0.2400

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

-2 - 1.5 - 1 -0.5 0 0.5 1 1.5 2

Z= 0.3200

FIG. 6. Color online Deformation of the shape of the tube boundary.Parameters are N=10 and A /R =0.5. The other parameters are adjusted such

that the helical shape spirals once around the axis in a distance Z=0.25. The

different frames correspond to different positions Z enhanced online.

URL: http://dx.doi.org/10.1063/1.3464464.1


http://dx.doi.org/10.1063/1.3464464.1http://dx.doi.org/10.1063/1.3464464.1http://dx.doi.org/10.1063/1.3464464.1http://dx.doi.org/10.1063/1.3464464.1


19/23

0 = htzh

C1 + lq

2R .

At fixed Z, the new coordinate system rotates with respect to

the old one with the angular velocity h. Now Eq. 156 aretransformed to

x = R cos + Am=1

m+1Zsinm,

158

y = R sin + A m=0

m+1Zcosm.

These equations define the cross-section of the perturbed

boundary of the tube by the plane Z=const as a parametric

curve with being the parameter. Now the shape of this

curve is independent of time. In Fig. 6, the cross-section of

the perturbed tube boundary by the plane Z=Z0 is shown for

N=10, A /R =0.5, and six different values ofZ0. Note that the

governing equation Eq. 141 was derived under assumptionthat the perturbation amplitude is small. In particular, this

implies that A /R1. Otherwise the parameter A /R can be

chosen arbitrarily because the solution of the dimensionless

Eq. 147 depends only on N, while it is independent ofA /R.When producing Fig. 6, we chose an artificially large value

of A /R to make the boundary deformation visible more

clearly.

VI. SUMMARY AND CONCLUSIONS

In this paper, we have studied the propagation of non-

axisymmetric waves on a magnetic tube in an incompressible

plasma. The plasma outside the tube is magnetic-free andhomogeneous. The tube consists of a core with homogeneous

magnetic field and density surrounded by an annulus where

the magnetic field magnitude decreases to zero, and the

plasma density varies from its value in the core to its value

outside the tube. It is assumed that the Alfvn speed is a

monotonically decreasing function in the annulus.

The wave amplitude is considered to be small and used

as a small parameter in the singular perturbation method.

This method is used to derive the nonlinear equation describ-

ing propagation of long waves excited at one end of the tube

by an external driver along the tube see Eq. 141. Thephase speed of the waves is equal to the kink speed. Inside

the annulus, there is a cylindrical surface where the Alfvnspeed matches the kink speed. As a result, there is a strong

conversion of the global wave mode in the local Alfvn os-

cillations that causes resonant absorption of energy of the

global wave.

A particular class of solutions of the nonlinear governing

equation in the form of helical waves is studied numerically.

Nonlinearity converts the energy of long waves in the energy

of shorter waves that are subject to stronger resonant absorp-

tion. As a result, nonlinearity accelerates the wave damping

due to resonant absorption. Nonlinearity also causes the dis-

tortion of the tube boundary related to the generation of flut-

ing modes by the kink mode.

ACKNOWLEDGMENTS

This work was carried out when M.S. Ruderman was a

guest in the Katholieke Universiteit Leuven. He acknowl-

edges the financial support from the Onderzoekfonds K.U.

Leuven during his visit and he is grateful to the Centre

Plasma Astrophysica and M. Goossens for the warm hospi-

tality. J. Andries acknowledges the support by the Interna-

tional Outgoing Marie Curie Fellowship within the SeventhEuropean Community Framework Programme. J. Andries is

Postdoctoral Fellow of the National Fund for Scientific Re-

search FWO-Vlaanderen, Flanders, Belgium.

APPENDIX A: CALCULATION OF COMPONENTS

OF TENSOR A

In this appendix, we calculate the components of tensor

A. It follows from Eq. 12 that the matrix of coefficients ofthis tensor is inverse to the matrix of coefficients of tensor

I+uT. In accordance with Eqs. 18 and 19, the ele-

ments of matrix of coefficients of tensor I

+u coincide withthe elements of the determinant on the right-hand side of Eq.

19. Now, taking into account that the determinant of the

matrix of coefficients of tensor I+uT is equal to unity, it

is straightforward to calculate the components of tensor A,

A11 = 1 +1

r

u

+

uz

z+

ur

r+

1

r

u

uz

z

u

z

uz

+ ur

uz

z , A1

A12 = 1r

ur

u+

ur

uz

z

ur

z

uz

u

uz

z , A2

A13 = ur

z

1

r ur

z

u

u

z

ur

+ ur

ur

z+ u

u

z ,A3

A21 = u

r

u

r

uz

z+

u

z

uz

r, A4

A22 = 1 +u

rr +

uz

z +u

rr

uz

z u

rz

uz

r , A5

A23 = u

z

ur

r

u

z+

ur

z

u

r, A6

A31 = uz

r

1

r uz

r

u

uz

u

r+ ur

uz

r , A7

A32 = 1

r uz

+

ur

r

uz

ur

uz

r+ u

uz

r , A8




20/23

A33 = 1 +ur

r+

1

r

u

+

ur

r+

1

r

urr

u

ur

u

r+ ur

ur

r+ u

u

r . A9

APPENDIX B: DERIVATION OF EXPRESSIONS

FOR RADIAL DISPLACEMENT AND PRESSUREIN SECOND ORDER APPROXIMATION

In this appendix, we derive Eqs. 50 and 53. Eliminat-ing u

m

2from Eqs. 46 and 48, we obtain

Pm2 =

ri

m21 VAi2

C23rurm2

r2

2iVAi2

mC

2Um

rm

+iC

2 VAi2

C2m2

k=

r2km2mkk 1

m k 1

2

2

UkUmk

mUk

2Umk

2sgn k . B1

Substituting Eq. B1 in Eq. 47 yields

rr

3rurm2

r2 m2

urm2

2

= k=

r2km3mkk 1

2k m 2k m 1

2

2UkUmk

m2k m 2sgn k mUk

2Umk

2 . B2

Using the identity

2k m 2sgn k m = 2m k 1sgn k B3

valid for km k0 we rewrite this equation as

rr

3rurm2

r2 m2

urm2

2

= k=

r2km3mkk 1 k m 1

2k m 2 22

UkUmk

2mUk

2Umk

2sgn k . B4

Making the substitution k= m kand then dropping the tilde,

we obtain

k=

r2kmmkk 1k m 1Uk

2Umk

2sgn k

= k=

r2kmmk

k 1k m 1Umk

2Uk

2sgn k. B5

When deriving this equation, we have taken into account that

sgn k=sgnm k when mk0. It follows from Eq. B5that

2 k=

r2kmmkk 1k m 1Uk

2Umk

2sgn k

= k=

r2kmmkk 1k m 1

UkUmk

Umk

Uk

sgn k. B6

Substituting Eq. B6 in Eq. B4 and using the identity

UkUmk

Umk

Uk

=

2

2UkUmk 2Umk

Uk

B7

and Eq. B3, we eventually arrive at Eq. 50.Let us now proceed to the derivation of Eq. 53. Sub-

stituting Eq. 51 in Eq. B1, we obtain

Pm2 =

2iVAi2

mC

2Um

rm i

C2 VAi2

mC2

2Um

2rm

iC2

VAi2

2C2m k=

r2km2mk sgn k

k 1 22

UkUmk + 2k 1Uk

2Umk

2

+ 2k m 2

UmkUk

. B8

When deriving this equation, we have used the identity

2m k 2k m = m sgn k, B9

valid when km k0. Making the substitution k= m k,

dropping the tilde, and recalling that sgnm k =sgn kwhen mk0, we obtain

k=

r2kmmk sgn k2k m 2

UmkUk

= k=

r2kmmk sgn k2k m 2

Umk2Uk

2+

Uk

Umk

. B10

Using the same method, we show that




21/23

k=

r2kmmk sgn k2k m 2Uk

Umk

= 0. B11

Substituting Eqs. B10 and B11 in Eq. B8 and using Eq.B9 with m k substituted for k, we eventually arrive at Eq.53.

APPENDIX C: EVALUATION OF THE SUMIN EQUATION 135

In this appendix, we show how to simplify the sum in

Eq. 136. Let us consider

S1 = k=

ksgn k+ sgnm kkmk. C1

The substitution m kk transforms this expression to

S1 = k=

m ksgn k+ sgnm kkmk. C2

Adding Eqs. C1 and C2 yields

S1 =m

2

k=

sgn k+ sgnm kkmk. C3

Next we consider

S2 = k=

ksgn k sgnm kkmk. C4

The substitution m kk transforms it to

S2 = k=

m ksgn k sgnm kkmk. C5

Adding Eqs. C1 and C2 and using the identity

k m k = m sgn k, C6

valid for km k0 we obtain

S2 =m

2

k=

1 sgnkm kkmk. C7

Using Eqs. C3 and C7 and the symmetry with respect tothe substitution m kk, we rewrite the sum in Eq. 136 as

S = m

k=

1 + sgnkm kk

2mk

2sgn m

sgn kmk2k

2 k

2mk

2

1

22 sgn k 1 sgnkm ksgn m

2

2kmk . C8

Once again, using the symmetry with respect to the substitu-

tion m kk, it i

m. s. ruderman, m. goossens and j. andries- nonlinear propagating kink waves in thin magnetic tubes

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