Nonlinear Waves
Roger Grimshaw, Gennady El, Karima Khusnutdinova
Lecture 10
Sine-Gordon Equation
February 24, 2010
Lecture 10: Sine-Gordon equation
by Karima Khusnutdinova
Introduction
The sine-Gordon (SG) equation
φtt − φxx = − sin φ
(1)
has many applications:
I
it appeared in 19th century in the differential geometry of surfaces
with constant negative Gaussian curvature (Bour 1862, Bianchi,
Enneper)
I
as a model of dislocations in a crystal (Frenkel and Kontorova, 1939)
I
as a model for elementary particles (Perring and Skyrme, 1962)
I
as a model of transverse electromagnetic waves on a
superconducting ‘strip-line’ transmission system (long Josephson
junctions, Scott and Johnson, 1969)
I
etc.
The equation is integrable by the Inverse Scattering Transform (IST)
(Takhtajan and Faddeev, AKNS 1974).
10.1 Sine-Gordon equation: Bäcklund Transformations
and integrability
The SG equation (1) can be written in the characteristic form
φξη = sin φ,
where
ξ=
x −t
x +t
, η=
.
2
2
Then, it appears as a compatibility condition of the following (AKNS)
linear problem:
ψ1 ξ = −iζψ1 −
ψ2 ξ = iζψ2 +
φξ
ψ2 ,
2
φξ
ψ1 ;
2
and
i
[cos φ ψ1 + sin φ ψ2 ],
4ζ
i
=
[sin φ ψ1 − cos φ ψ2 ];
4ζ
(2)
ψ1 η =
ψ2 η
providing the basis for the IST.
(3)
10.1 Sine-Gordon equation: Bäcklund Transformations
and integrability
Bäcklund transformations (BT) for the SG equation were devised in
1880s in differential geometry and are attributed to Bianchi and
Bäcklund. They have the form:
1
φ − φ̃
(φ + φ̃)ξ = α sin
,
2
2
1
φ + φ̃
1
(φ − φ̃)η = sin
,
2
α
2
(4)
where both φ and φ̃ are solutions of the SG equation (1), and can be
viewed as a transformation of the SG equation into itself. BT allow one
to construct hierarchies of solutions, starting from some simple known
solutions. Are there any other Klein-Gordon equations φξη = F (φ),
admitting Bäcklund transformation of the form:
φ − φ̃
1
(φ + φ̃)ξ = f (
),
2
2
1
φ + φ̃
(φ − φ̃)η = g (
)?
2
2
(5)
(6)
10.1 Sine-Gordon equation: Bäcklund Transformations
and integrability
Differentiating (5) with respect to η and (6) with respect to ξ, we obtain
φ − φ̃
φ + φ̃
1
(φ + φ̃)ξη = f 0 (
)g (
),
2
2
2
φ + φ̃
φ − φ̃
1
(φ − φ̃)ξη = g 0 (
)f (
),
2
2
2
(7)
(8)
which implies
φ − φ̃
φ + φ̃
φ + φ̃
φ − φ̃
)g (
) + g 0(
)f (
), (9)
2
2
2
2
φ − φ̃
φ + φ̃
φ + φ̃
φ − φ̃
)g (
) − g 0(
)f (
). (10)
= F (φ̃) = f 0 (
2
2
2
2
φξη = F (φ) = f 0 (
φ̃ξη
Introducing functions v = (φ + φ̃)/2 and w = (φ − φ̃)/2, we rewrite the
previous equations as
F (v + w ) = g (v )f 0 (w ) + g 0 (v )f (w ),
0
0
F (v − w ) = g (v )f (w ) − g (v )f (w ).
(11)
(12)
10.1 Sine-Gordon equation: Bäcklund Transformations
and integrability
Differentiating either of these equations with respect to v and then w ,
and either adding or subtracting the results, we obtain
f 00 (w )
g 00 (v )
=
= λ = const.
g (v )
f (w )
Thus,
g 00 = λg ,
f 00 = λf ,
and λ can be assumed to be −1, 0, 1. For λ = −1 we find
g (v ) = β sin v ,
f (w ) = α sin w .
(13)
Substituting (13) into (11) and (12), we obtain F (φ) = sin φ for
β = α−1 , giving us the BT for the sine-Gordon equation. The choice of
λ = 1 yields the sinh-Gordon equation, λ = 0 - the linear Klein-Gordon
equation. Thus, the SG equation is a very special equation (existence of
the BT for it is related to the integrability of this equation by the IST).
10.1 Sine-Gordon equation: Bäcklund Transformations
and integrability
Derivation of BTs from AKNS problem?
(Following H-H Chen, Phys. Rev. Lett. 33 (1974) 925-928)
I
ψ1
Having AKNS problem for ψ1 and ψ2 , introduce G = ψ
and derive
2
equations for Gξ and Gη (turn out to be Riccati equations).
I
Eliminating φ (solution of the SG equation) from these equations,
one can obtain an equation for G only.
I
This equation turns out to be invariant under the transformation
(G , ζ) → (−G , −ζ), indicating that there is another solution φ̃ of
the SG equation, related to (−G , −ζ). This gives a second pair of
equations for Gξ and Gη , containing φ̃.
I
BTs (relating φ and φ̃) can be derived from the two pairs of
equations for Gξ and Gη .
10.2 Symmetries of the SG equation
In the following we use the following continuous and discrete symmetries,
admitted by the SG equation (1):
t → t + t0 ,
x → x,
φ→φ
(shift in t),
t → t,
x → x + x0 ,
φ→φ
(shift in x),
t → t,
x → x,
φ → φ + 2πn,
n is any integer,
(discrete shifts in φ),
t → −t,
t → t,
t → t,
x → x,
φ → φ (reflection in t),
x → −x, φ → φ (reflection in x),
x → x, φ → −φ (reflection in φ),
t − cx
x − ct
t→√
, x→√
,
1 − c2
1 − c2
(Lorenz transformation).
φ → φ,
For example, if φ is a solution, then −φ is a solution as well, etc.
10.3 Bäcklund Transformations: Kink and Antikink
Given a solution φ0 of the SG equation, BT (4) allows one to find a
2-parameter family of solutions. Consider the trivial solution φ = 0.
Substituting it into (4), we obtain
φ̃ξ = −2α sin
φ̃
,
2
φ̃η = −
2
φ̃
sin .
α
2
Integrating,
Z
φ̃
2αξ = −
d φ̃
φ̃
2
sin
2
η=−
α
Z
φ̃
d φ̃
sin
φ̃
2
= −2 ln (tan
φ̃
) + p(η),
4
= −2 ln (tan
φ̃
) + q(ξ).
4
Therefore, tan φ̃4 = exp (aξ + 1a η + δ), where a = −α, δ = const.
Thus, the trivial solution φ = 0 has generated a new solution:
1
φ1 = φ̃ = 4 tan−1 exp(aξ + η + δ).
a
(14)
10.3 Bäcklund Transformations: Kink and Antikink
Reverting to t and x:
φ1 = φ̃ = 4 tan
where v =
1−a2
1+a2 ,
−1
x − vt
exp σ √
+δ ,
1 − v2
σ = ±1 (we can assume δ = 0, up to shifts in x, t).
Σ= 1 HkinkL, v = 0.5, t = 0
Σ= -1 HantikinkL, v = 0.5, t = 0
6
6
5
5
4
4
j1 3
j1 3
2
2
1
1
0
-10
-5
0
x
5
10
0
-10
-5
0
x
5
10
Figure: Kink and antikink for v = 0.5 at t = 0.
Remarks. Note that derivatives of these solutions are hump-shaped, like
a solitary wave.
Also note that 2πn, where n is an integer can be added to any solution of
the SG equation due to invariance of the equation with respect to this
transformation.
10.4 Addition theorem
Now we can again use the BT (4), so that φ1 generates a new solution
φ2 . To do this, put φ = φ1 into the BT (4), and integrate to find a new
solution. This new solution can then also be used to find another
solution, etc. However, the direct integration is not easy, and the actual
way to use the BTs for finding new solutions is based on the fact that
BTs commute, which is shown on the diagram below.
Figure: http://homepages.tversu.ru/∼s000154/collision/main.html
Remark. To be precise, BTs can be made commutative by adjusting the
integration constants (e.g., McLaughlin and Scott 1973, Seeger et al.
1953).
10.4 Addition theorem
Let us take a sequence of pairs (φ0 , φ11 ), (φ11 , φ2 ), (φ0 , φ21 ), (φ21 , φ2 ).
Then, we have the following BTs (follow the diagram):
1
(φ0 + φ11 )ξ
2
1
(φ0 − φ11 )η
2
1
(φ0 + φ21 )ξ
2
1
(φ0 − φ21 )η
2
φ0 − φ11
,
2
1
φ0 + φ11
=
sin
;
α1
2
φ11 − φ2
1
(φ11 + φ2 )ξ = α2 sin
,
2
2
1
1
φ11 + φ2
(φ11 − φ2 )η =
sin
;
2
α2
2
φ0 − φ21
= α2 sin
,
2
1
φ0 + φ21
=
sin
;
α2
2
1
φ21 − φ2
(φ21 + φ2 )ξ = α1 sin
,
2
2
1
1
φ21 + φ2
(φ21 − φ2 )η =
sin
.
2
α1
2
= α1 sin
(15)
(16)
(17)
(18)
(19)
(20)
(21)
(22)
10.4 Addition theorem
Taking (15) - (17) - (19) + (21) and using the trigonometric formula
α−β
sin α + sin β = 2 sin α+β
2 cos 2 , one can obtain
α1 sin
φ0 − φ11 + φ21 − φ2
φ0 + φ11 − φ21 − φ2
= α2 sin
,
4
4
which yields, by using sin(α ± β) = sin α cos β ± sin β cos α with
α = φ0 − φ2 and β = φ21 − φ11 ,
tan
φ0 − φ2
α2 + α1
φ21 − φ11
=
.
tan
4
α2 − α1
4
(23)
The same formula follows from the second group of equations, (16),
(18), (20) and (22).
The relation (23) is called the addition theorem for the Bäcklund
transformations. This relation allows one to obtain new solution φ2
algebraically, from the known solutions φ0 , φ11 , φ21 , without any
integration. Repeatedly applying this theorem, one can obtain
successively new solutions by means of purely algebraic manipulations.
10.5 Two-wave interactions (kink-kink, kink-antikink, etc.)
Let φ0 = 0. Then φ11 and φ21 are (solutions are parametrized by
ai = −αi ):
2
ai + 1
1 − ai2
−1
φi1 = 4 tan exp θi (+2πni ), θi =
(x −
t) + δi
2ai
1 + ai2
Using the addition theorem (23) and the formula
tan α−tan β
tan(α − β) = 1+tan
α tan β , we find the two-soliton solution:
φ2 = 4 tan
−1
a1 + a2 e θ2 − e θ1
·
.
a1 − a2 1 + e θ1 +θ2
Choosing different values of parameters a1 and a2 , one can obtain
solutions describing various interaction scenarios (kink-kink,
kink-antikink, etc.)
(24)
10.5 Two-wave interactions (kink-kink, kink-antikink, etc.)
Example 1 (worked example in the lecture). The symmetric kink-kink
solution (head-on collision, same velocities) is obtained when
a2 = −1/a1 , a1 > 0 and δ1 = δ2 = 0. It can be written as
"
#
x
v sinh √1−v
2
1 − a12
−1
.
(25)
φ2 = 4 tan
, where v =
vt
cosh √1−v 2
1 + a12
Considering the asymptotics of (24) as t → −∞ (and then x → −∞
and x → +∞) and as t → +∞√(and then x → −∞ and x → +∞), one
can find the phase shift to be 2 v 2 − 1 log v (the only “evidence” of
interaction).
5
10
j2 0
5
-5
-10
0x
-5
t
-5
0
5
10
-10
Figure: Head-on collision of two kinks for v = 0.6.
10.5 Two-wave interactions (kink-kink, kink-antikink, etc.)
Example 2. The symmetric kink-antikink solution (head-on collision,
same velocities) is obtained when a2 = 1/a1 , a1 < 0 and δ1 = δ2 = 0. It
can be written as
#
"
vt
sinh √1−v
2
1 − a12
−1
.
(26)
,
where
v
=
φ2 = 4 tan
x
v cosh √1−v
1 + a12
2
10
5
t 0
-5
-10
5
j2 0
-5
10
5
0
x
-5
-10
Figure: Head-on collision of kink and antikink for v = 0.6.
10.6 Breather
The symmetric kink-antikink solution (26) takes an interesting form if
the velocity parameter v is allowed to be imaginary. Setting
iω
v=√
, ω < 1,
1 − ω2
one obtains the breather
"√
#
2
sin
ωt
1
−
ω
√
·
φ2 = 4 tan−1
,
(27)
ω
cosh 1 − ω 2 x
which is a localized, but oscillating in time solution.
4
2
j2 0
-2
-4
10
20
10
0 t
5
0
x
-10
-5
-10
-20
Figure: A stationary breather for ω = 0.5.
10.6 Breather
The solution (27) describes a stationary breather. Using the Lorentz
symmetry of the SG equation:
x − ct
x→√
,
1 − c2
t − cx
t→√
,
1 − c2
one can obtain a moving breather:

√
√
2
sin[ ω(t−cx)
]
2
1
−
ω
1−c
.
√
·
φ2 = 4 tan−1 
2 (x−ct)
ω
√
cosh [ 1−ω
]
1−c 2
Here, an envelope velocity c is equal to the reciprocal of its carrier
velocty c −1 .
10.7 Soliton “ladder”
The procedure, described above, can be continued, resulting in the
so-called soliton “ladder”, shown below:
Figure: http://homepages.tversu.ru/∼s000154/collision/main.html
10.8 Soliton interactions
I
Interactions of any SG solitons are accompanied by phase shifts only
(elastic interactions).
I
Interactions have a two-particle nature: when several solitons collide,
a shift of any soliton involved into the interaction is equal to the
sum of the shifts caused by its independent interactions with other
solitons.
10.8 Soliton interactions
Visit: http://homepages.tversu.ru/∼s000154/collision/main.html
to see some movies, showing kink-kink, kink-antikink, kink-breather, etc.
interactions.
Lecture 10 References
From the list of main references:
1. Ablowitz, M.J. & Segur, H. 1981 Solitons and the Inverse Scattering
Transform, SIAM.
2. Dodd, R.K., Eilbeck, J.C., Gibbon, J.D., & Morris, H.C. 1982 Solitons
and Nonlinear Wave Equations, Academic Press Inc.
3. Drazin, P.G. and Johnson, R.S. 1989 Solitons: an Introduction,
Cambridge University Press, London.
4. Scott, A. 1999 Nonlinear Science: Emergence and Dynamics of
Coherent Structures, Oxford University Press Inc., New York.
Additional references:
1. Lamb, G.M. 1980 Elements of Soliton Theory, Wiley, New York.
2. Chen, H-H, General Derivation of Bäcklund Transformations from
Inverse Scattering Problems, Phys. Rev. Lett. 33 (1974) 925-928.
3. http://homepages.tversu.ru/∼s000154/collision/main.html