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8/3/2019 M. S. Ruderman, M. Goossens and J. Andries- Nonlinear propagating kink waves in thin magnetic tubes
1/23
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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http://dx.doi.org/10.1063/1.3464464http://dx.doi.org/10.1063/1.3464464http://dx.doi.org/10.1063/1.3464464http://dx.doi.org/10.1063/1.34644648/3/2019 M. S. Ruderman, M. Goossens and J. Andries- Nonlinear propagating kink waves in thin magnetic tubes
2/23
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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3/23
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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8/3/2019 M. S. Ruderman, M. Goossens and J. Andries- Nonlinear propagating kink waves in thin magnetic tubes
4/23
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
082108-4 Ruderman, Goossens, and Andries Phys. Plasmas 17, 082108 2010
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8/3/2019 M. S. Ruderman, M. Goossens and J. Andries- Nonlinear propagating kink waves in thin magnetic tubes
6/23
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.
082108-6 Ruderman, Goossens, and Andries Phys. Plasmas 17, 082108 2010
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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
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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-
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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
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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.
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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.
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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
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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
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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
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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
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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.
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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.
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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
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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
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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
082108-20 Ruderman, Goossens, and Andries Phys. Plasmas 17, 082108 2010
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8/3/2019 M. S. Ruderman, M. Goossens and J. Andries- Nonlinear propagating kink waves in thin magnetic tubes
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