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4

The sine-Gordon equationon time scales

Jan L. Cieslinski∗, Tomasz Nikiciuk†

Uniwersytet w Bia lymstoku, Wydzia l Fizyki,

ul. Lipowa 41, 15-424 Bia lystok, Poland

Kamil Waskiewicz‡,

Instytut Geofizyki PAN,

ul. Ks. Janusza 64, 01-452 Warszawa, Poland

Abstract

We formulate and discuss integrable analogue of the sine-Gordonequation on arbitrary time scales. This unification contains the sine-Gordon equation, discrete sine-Gordon equation and the Hirota equa-tion (doubly discrete sine-Gordon equation) as special cases. Wepresent the Lax pair, check compatibility conditions and constructthe Darboux-Backlund transformation. Finally, we obtain a solitonsolution on arbitrary time scale. The solution is expressed by the socalled Cayley exponential function.

Keywords: integrable systems, solitons, time scales, discretizationPACS numbers: 02.30.Ik, 05.45.YvMSC 2010: 34N05, 35Q51, 39A12

∗e-mail: janek@ alpha.uwb.edu.pl†e-mail: niki@ alpha.uwb.edu.pl‡e-mail: kamilwas@igf.edu.pl

1

1 Introduction

The sine-Gordon equation φ,xy = sinφ is one of the classical soliton equa-tions with numerous applications in many fields [2, 14, 23], from differentialgeometry [4, 7, 25, 28] to applied physics, including relativistic field theory[29], Josephson junctions [30, 33], propagation of deformations along DNAdouble helix [36], dispalcements in crystals [16], domain walls in ferroelectricand ferromagnetic materials [22], mechanical transmission lines [30, 32], andmany others. In spite of the long history the sine-Gordon equation and itsnumerous extensions and generalizations still attracts attention of researchers[13, 15, 24, 31, 35]. In this paper we formulate and study an integrable ex-tension of the sine-Gordon equation on arbitrary time scales, which includes,as special cases, the discrete case [27] and doubly discrete case (the Hirotaequation) [20].

A time scale T is just any (non-empty) closed subset of R [18], includingspecial cases like T = R, T = Z, T = hZ and q-calculus. Time scales (ormeasure chains) were introduced in order to unify continuous and discretecalculus [19]. We recall several notions which will be used throughout thepaper (here we assume t ∈ T, note that in the rest of the paper we have twotime scales with elements denoted respectively by x, y). The forward jumpoperator is defined as

tσ := inf{s ∈ T : s > t}. (1.1)

we denote also

φσ(t) := φ(tσ). (1.2)

In the case T = hZ the forward jump is a shift. Graininess is defned by

µ(t) := tσ − t . (1.3)

In the case T = hZ we have µ = h (time step). In the case T = R, we have,obviously, µ = 0. The delta derivative is defined by

f∆(t) := lims→t

f(tσ)− f(s)

tσ − s. (1.4)

Note that

f + µf∆ = fσ. (1.5)

2

Delta exponential function ea(t), see [6, 18], satisfies initial value problem

e∆a = aea, ea(0) = 1, (1.6)

where, in general, a = a(t). We also have:

eσa = (1 + aµ)ea (1.7)

In some applications (e.g., in trigonometry) another definition of the ex-ponential function is much more convenient [11, 12]. This is the Cayleyexponential function which satisfies

E∆a =

1

2a(Ea + Eσ

a ), Ea(0) = 1, Eσa =

1 + aµ

2

1− aµ

2

Ea . (1.8)

In the continuous case (T = R) exponential functions become identical andfor a = const we have: ea(t) = Ea(t) = eat.

Dynamic equations on time scales are counterparts of differential equa-tions. They unify continuous and discrete dynamical systems [1, 5, 21]. Somesoliton equations on time scales were already studied within the Hamiltonianframework [3, 17, 34]. This paper is a continuation of [8], where the Darboux-Backlund transformation for a class of 2×2 linear problems was constructed.We will construct this transformation for the integrable analogue of the sine-Gordon equation on time scales and compute a soliton solution expressed bythe Cayley exponential function.

2 Lax pair

A standard Lax pair for the sine-Gordon equation φ,xy = sinφ:

Ψ,x =

(

iζ −12φ,x

12φ,x −iζ

)

≡ iζσ3 −1

2iφ,x σ2 ,

Ψ,y =1

4iζ

(

cosφ sinφsinφ − cos φ

)

≡1

4iζ(cosφσ3 + sinφσ1) ,

(2.1)

where σ1, σ2, σ3 are Pauli matrices. We postulate the following ime scaleanalogue for this Lax pair:

Ψ∆x = UΨ , U = iζσ3 −sin µxφ

∆x

2

µx

iσ2 −1− cos µxφ

∆x

2

µx

Ψ∆y = VΨ , V =1

4iζ

(

cosφσy + φ

2σ3 + sin

φσy + φ

2σ1

)

(2.2)

3

We make important assumption throughout this paper:

µx = µx(x) , µy = µy(y) , (2.3)

i.e., we consider time scales T = T1 ×T2 with graininess µ = (µx(x), µy(y)).If µ → (0, 0), then the Lax pair (2.2) becomes (2.1). It is convenient todenote (compare [4]):

δ :=1

2φ∆x , α :=

1

2(φ+ φσy) . (2.4)

Then, the Lax pair (2.2) takes the form

Ψ∆x =

(

−ζe3 +sinµxδ

µx

e2 −1− cosµxδ

µx

)

Ψ ,

Ψ∆y =1

4ζ(e3 cosα + e1 sinα)Ψ,

(2.5)

where e1 = −iσ1, e2 = −iσ2, e3 = −iσ3. We have

e1e2 = −e2e1 = e3, e2e3 = −e3e2 = e1,

e3e1 = −e1e3 = e2, e21 = e22 = e23 = −1.(2.6)

Lemma 2.1

α∆x = δσy + δ , µx(δ + δσy) = ασx − α. (2.7)

Proof: Using (2.4) we get

α∆x =1

2

(

φ∆x + φσ2∆x)

= δ + δσy . (2.8)

The second equality follows by (1.5). ✷

Proposition 2.2 Compatibility conditions, U∆y −V ∆x +UσyV −V σxU = 0,for (2.2) are given by

4

µxµy

sinµxµy φ

∆x∆y

4= sin

φσxσy + φσx + φσy + φ

4, (2.9)

where for µx = 0 or µy = 0 the left-hand side becomes φ∆x∆y (as limitsµx → 0 or µy → 0 suggest).

4

Proof: Compatibility conditions yield a system of four scalar equations given bythe decomposition in the basis 1, e1, e2, e3. Coefficients by e1, e3 yield

(sinα)∆x = sinµxδσy cosα−(1−cos µxδ

σy ) sinα+cosασx sinµxδ+(1−cos µxδ) sinασx

µx,

(cosα)∆x = − sinµxδσy sinα−(1−cos µxδ

σy ) cosα−sinασx sinµxδ+(1−cos µxδ) cosασx

µx.

(2.10)

If µx 6= 0, then we use (1.5) to transform these equations into

sinασx cosµxδ − cosασx sinµxδ = sinα cosµxδσy + cosα sinµxδ

σy ,

cosασx cosµxδ + sinασx sinµxδ = cosα cosµxδσy − sinα sinµxδ

σy .(2.11)

Using well known trigonometric identities, we get

sin(ασx − µxδ) = sin(µxδσy + α),

cos(ασx − µxδ) = cos(µxδσy + α),

(2.12)

which is identically satisfied by (1.5) and Lemma 2.1. If µx = 0, then (2.10) reduceto

α∆x cosα = (δσy + δ) cosα, α∆x sinα = (δσy + δ) sinα, (2.13)

which is satisfied by Lemma 2.1, as well.Dealing with coefficients by 1 and e2 we will consider three cases separately.

In the first case we assume µx 6= 0, µy 6= 0. Then coefficients by 1 and e2 yield

cosµxδσy = cosµxδ + 1

4µxµy(cosασx − cosα),

sinµxδσy = sinµxδ + 1

4µxµy(sinασx + sinα),(2.14)

Hence, using trigonometric identities, we get

sin µx(δσy+δ)2 sin µx(δσy−δ)

2 = 14µxµy sin ασx−α

2 sin ασx+α2 ,

cos µx(δσy+δ)2 sin µx(δσy−δ)

2 = 14µxµy cos ασx−α

2 sin ασx+α2 .

(2.15)

By virtue od Lemma 2.1 these equations reduce to the following single equation

sinµx(δσy − δ)

2=

1

4µxµy sin

ασx + α

2. (2.16)

Using (2.4) and taking into account that δσy = δ + µyδ∆y we get (2.9).

5

In the second case we assume µx 6= 0 and µy = 0 (hence, in particular, δσy = δ

and α = φ). Then coefficients by 1 and e2 yield

−δ∆y sinµxδ = 14(cosασx − cosα),

δ∆y cosµxδ = 14(sinασx + sinα), .

(2.17)

In this case Lemma 2.1 implies µxδ = 12 (ασx − α). Therefore

sin ασx−α2

(

δ∆y − 12 sin ασx+α

2

)

= 0,

cos ασx−α2

(

δ∆y − 12 sin ασx+α

2

)

= 0,(2.18)

hence

δ∆y =1

2sin

ασx + α

2, (2.19)

and, substituting δ = 12α

∆x , α = φ, we obtain

α∆x∆y = sinασx + α

2−→ φ∆x∆y = sin

φσx + φ

2, (2.20)

which can be considered as a special case of (2.9).In the third case we assume µx = 0. Now the Lax pair essentially simplifies

Ψ∆x = (iζσ3 − iσ2δ)Ψ , Ψ∆y =1

4iζ(σ3 cosα + σ1 sinα) Ψ (2.21)

Compatibility conditions yield three equations (the term proportional to the unitmatix vanishes):

δ∆y = 12 sinα ,

(−α∆x + δσy + δ) cosα = 0 ,

(α∆x − δσy − δ) sinα = 0 ,

(2.22)

The last two equations are identically satisfied by virtue of Lemma 2.1. Then,using (2.4), we obtain

φ∆x∆y = sinφσy + φ

2, (2.23)

which can be considered as a special case of (2.9). ✷

6

Corollary 2.3 The equation (2.9) will be called the sine-Gordon equationon time scales. For µx 6= 0, µy 6= 0 it coincides with the Hirota equation(doubly discrete sine-Gordon equation) [20, 26] and for µy = 0 it coincideswith the discrete sine-Gordon equation [27].

The equation (2.9) can be rewritten as

φ∆1∆2 = F

(

1

4µxµy sin〈〈φ〉〉

)

sin〈〈φ〉〉, (2.24)

where

〈〈φ〉〉 =1

4(φσxσy + φσx + φσy + φ) , (2.25)

and F (x) := x−1 arcsin x, F (0) = 1.

3 Darboux-Backlund transformation

The Darboux-Backlund transformation on time scales was formulated andderived in [8]. This is a gauge-like transformation preserving the structure ofthe Lax pair, see [9]. Let Ψ = DΨ (D is called “Darboux matrix”). Thenmatrices U, V of the Lax pair (2.2) transform as follows

U = D∆1D−1 +Dσ1UD−1 ,

V = D∆2D−1 +Dσ2V D−1 .(3.1)

In this paper we confine ourselves to the binary Darboux matrix which adds(as a kind of nonlinear superposition) a soliton onto a given background.Fortunatelly, the Darboux matrix has exactly the same form in all cases(continuous, discrete and time scales) [8]:

D =λ− κ1A

λ− iκ1

. (3.2)

where A = i(I − 2P ) and P is a Hermitean projector (P 2 = P , P † = P )defined by

kerP = Ψ(iκ1)~c1 , ImP = Ψ(−iκ1)~c⊥1 , (3.3)

7

where ~c1 is a constant vector. Moreover, the Lax pair (2.2) satisfies:

U(−ζ) = e−12 U(ζ)e2, V (−ζ) = e−1

2 V (ζ)e2, (3.4)

which implies the following constraint on the projector P [8]:

e2P = (I − P )e2. (3.5)

Hence [8]:

P =1

2(I + σ1 sin θ + σ3 cos θ) =

1

2

(

1 + cos θ sin θsin θ 1− cos θ

)

. (3.6)

Taking into account well known trigonometric identities, we can represent Pas

P =

(

cos θ2

sin θ2

)

(

cos θ2, sin θ

2

)

. (3.7)

Likewise, we have

I − P =1

2

(

1− cos θ − sin θ− sin θ 1 + cos θ

)

, (3.8)

and

I − P =

(

− sin θ2

cos θ2

)

(

− sin θ2, cos θ

2

)

. (3.9)

Therefore,

imP = f

(

cos θ2

sin θ2

)

, kerP = g

(

− sin θ2

cos θ2

)

, (3.10)

where f, g are arbitrary functions. The function θ, parameterizing the pro-jector, can be uniquely derived from (3.3), provided that Ψ(ζ) is known.Alternatively, we can solve dynamic equations for θ.

8

Proposition 3.1 Dynamic equations for θ parameterizing the projector P

read

1µx

sin(

µxδ −θσx−θ

2

)

= κ1 sinθσx+θ

2, (for µx 6= 0),

1µy

sin θσy−θ2

= 14κ1

sin(

α− θ+θσy

2

)

(for µy 6= 0),

δ − 12θ,x = κ1 sin

θσx+θ2

, (for µx = 0),

θ,y =1

2κ1sin(α− θ), (for µy = 0).

(3.11)

Proof: We confine ourselves to the case µx 6= 0, µy 6= 0. Other cases can beconsidered in exactly the same way (all intermediate steps can be obtain as aformal limit). From (3.3) it follows that

(kerP )∆x = U(iκ1)kerP, (kerP )∆y = V (iκ1)imP. (3.12)

Analogical equations for imP can be omitted because in this case imP is uniquelydetemined by kerP . Indeed, imP = (kerP )⊥ and imP = e2kerP . Note that U(iκ1)and V (iκ1) are real matrices, provided that all coefficients and fields are real:

U(iκ1) = κ1

(

−1 00 1

)

+ sinµxδµx

(

0 −11 0

)

− 1−cosµxδµx

(

1 00 1

)

,

V (iκ1) = − 14κ1

(

cosα sinα

sinα − cosα

)

.

(3.13)

Substituting (3.10) into (3.12) and dividing by, respectively, f and g, we obtain:

(sin θ2 )∆x + f∆x

fsin θσx

2 = −κ1 sin θ2 + sinµxδ

µxcos θ

2 −1−cosµxδ

µxsin θ

2 ,

(cos θ2 )∆x + f∆x

fcos θσx

2 = κ1 cos θ2 −

sinµxδµx

sin θ2 − 1−cosµxδ

µxcos θ

2 ,

(sin θ2 )∆y + g∆y

gsin θσy

2 = − 14κ1

(

cosα cos θ2 − sinα cos θ

2

)

,

(cos θ2 )∆y + g∆y

gcos θσy

2 = 14κ1

(

sinα sin θ2 + cosα cos θ

2

)

,

(3.14)

Eliminating f and g from (3.14) we get two equations

1µx

sin θσx−θ2 = −κ1 sin θσx+θ

2 + sinµxδµx

cos θσx−θ2 + 1−cosµxδ

µxsin θσx−θ

2 ,

1µy

sin θ−θσy

2 = 14κ1

sin(

θ+θσy

2 − α)

and, after elementary calculation, we get (3.11). ✷

9

Proposition 3.2 ([8]) If D is given by (3.2), (3.3) and the Lax pair is ofthe form

U = u0 + ζu1 , V = v0 +1

ζv1 , (3.15)

where u0, u1, v0, v1 are linear combinations of I, e1, e2, e3 with real coefficientsand (3.4) is satisfied, then the Darboux-Backlund transformation is given by

u1 = u1 , u0 = u0 + κ1

(

u1A− Aσxu1

)

,

v0 = v0 , v1 = −Aσyv1A−1 ,

(3.16)

where A = i(I − 2P ) = e3 cos θ + e1 sin θ.

Proposition 3.2 can be applied directly to the Lax pair (2.2). Two equa-tions are trivially satisfied (u1 = u1 = −e3 and v0 = v0 = 0). The equationfor u0 splits into two equations:

sin µxδ

µx

=sin µxδ

µx

− κ1(sin θσx + sin θ),

1− cosµxδ

µx

=1− cosµxδ

µx

+ κ1(cos θσx − cos θ),

(3.17)

for µx 6= 0. In the case µx = 0 we have, instead,

δ = δ − 2κ1 sin θ ,

12δ2 = 1

2δ2 − κ1θ,x sin θ.

(3.18)

The last equation of (3.16) will be simplified using an elegant approach ofClifford numbers instead of cumbersome straightforward calculations.

Lemma 3.3 For any scalar function β = β(x, y) we have

cos β + e1e3 sin β = exp(βe1e3),

e3 cos β + e1 sin β = e3(cos β + e1e3 sin β) = e3 exp(βe1e3),

e3 exp(βe1e3) = exp(−βe1e3)e3

(3.19)

10

Proof: We use (2.6) and observe that (e1e3)2 = −1. The first equation can be

proved using power series, like the famous formula cos β + i sin β = exp(iβ). ✷

Taking into account properties listed in Lemma 3.3 we can easily trans-form the last equation of (3.16) into

e3 exp(αe1e3) = −e33 exp(e1e3θσx) exp(−e1e3α) exp(e1e3θ). (3.20)

Because e33 = −e3 and exponents of all exponential functions commute, wehave

e3 exp(αe1e3) = exp e1e3(θσx + θ − α), (3.21)

hence we finally get: α = −α + θσx + θ.

Proposition 3.4 The Darboux-Backlund transformation (3.16) for the Laxpair (2.2) is equivalent to the system:

α = −α + θσy + θ,

δ + δ = θ∆x ,(3.22)

which can be reduced to one equation for φ:

φ = 2θ − φ , (3.23)

where θ is computed either from (3.3) or from Proposition 3.1.

Proof: The first equation of (3.22) was already obtained from (3.21). Consideringthe second equation, we first assume µx 6= 0. The system (3.17) can be rewrittenas

1µx

sin µx(δ−δ)2 cos µx(δ+δ)

2 = −κ1 sin θσx+θ2 cos θσx−θ

2 ,

1µx

sin µx(δ−δ)2 sin µx(δ+δ)

2 = −κ1 sin θσx+θ2 sin θσx−θ

2 ,

(3.24)

Dividing these equations side by side, we get

1µx

tanµx(δ+δ)

2 = 1µx

tan θσx−θ2 , (3.25)

which implies the first equation of (3.22). The remaining equation,

1µx

sin µx(δ−δ)2 = −κ1 sin θσx+θ

2 , (3.26)

11

after substituting δ = θ∆x − δ, reduces to the first equation of (3.11). The caseµx = 0 is simpler. Substituting δ − δ from the first equation of (3.18) into thesecond equation of (3.18) we get the second equation of (3.22).

Finally, substituting (2.4) into (3.22), we get(

φ + φ− 2θ)∆x

= 0 ,

(

φ + φ− 2θ)σy

= −(φ + φ− 2θ) .

(3.27)

Hence, φ + φ − 2θ = χ, where χ is a function of y such that χσy = −χ. Without

loss of generality we can take χ = 0 because transformation φ → φ + χ does not

change the Lax pair (2.2). ✷

4 Solitons on an arbitrary time scale

Pure soliton solutions are obtained, as usual, by applying the Darboux-Backlund transformation to the simplest (“vacuum”) background φ = 0.In this case the Lax pair reduces to

Ψ∆x = iζσ3Ψ , Ψ∆y =1

4iζσ3Ψ, (4.1)

which can be easily solved (assuming initial condition Ψ(0, 0) = I):

Ψ =

eiζ(x)e 1

4iζ(y) 0

0 e−iζ(x)e− 1

4iζ(y)

, (4.2)

where e denotes the delta exponential function (1.6). From the previoussection we know that φ = 2θ − φ. Therefore one-soliton solution is given by

φsol = 2θ, (4.3)

where θ is associated with the simplified linear system (4.2).

Proposition 4.1 One-soliton solution to the sine-Gordon equation on anarbitrary time scale (equation (2.9)) is given by

φsol = 4 arctan(

c E−2κ1(x)E− 1

2κ1

(y))

, (4.4)

where E denotes the Cayley-exponential function (1.8) and c, κ1 are real pa-rameters.

12

Proof: In the case α = δ = 0 equations (3.11) read

1µx

sin θσx−θ2 = −κ1 sin θσx+θ

2 , (for µx 6= 0)

1µy

sin θσy−θ2 = − 1

4κ1sin θσy+θ

2 , (for µy 6= 0)

θ,x = −2κ1 sin θ, (for µx = 0),

θ,y = − 12κ1

sin θ (for µy = 0).

(4.5)

Equations (4.5) can be easily transformed into:

tan θσx

2 = 1−µxκ1

1+µxκ1tan θ

2 , (for µx 6= 0)

tan θσy

2 =1−

µy

4κ1

1+µy

4κ1

tan θ2 , (for µy 6= 0)

(

tan θ2

)

,x = −2κ1 tan θ2 , (for µx = 0),

(

tan θ2

)

,y = − 12κ1

tan θ2 , (for µy = 0)

(4.6)

which yields (4.4) by virtue of (1.8) (the continuous case agrees with this solution,

as well). Another method to get (4.4), even more straightforward, consists in using

(3.3) and (3.10). ✷

In the continuous and discrete case the formula (4.4) reduces to knownexpressions:

φsol−cont = 4 arctan exp(−2κ1x− 12κ1

y + c0) ,

φsol−discr = 4 arctan

(

c(

1−κ1hx

1+κ1hx

)xhx

(

1−hy

4κ1

1+hy

4κ1

)y

hy

)

,(4.7)

where hx = µx and hy = µy. The discrete soliton has, in fact, a similar shapeas the smooth soliton. Indeed,

φsol−discr = 4 arctan exp(−2κ1x−1

2κ1y + c0), (4.8)

where c = exp c0 and

κ1 =1

2hx

ln

(

1 + κ1hx

1− κ1hx

)

, κ−11 =

1

2hy

ln

(

1 + hy

4κ1

1− hy

4κ1

)

. (4.9)

13

Therefore, on two most popular time scales (T = R, T = hZ) solitons aresolitary waves. In the discrete case velocity and shape depend on hx and hy.However, in general, one-soliton solution (4.4) is not a solitary wave.

5 Summary

We formulated and studied an integrable analogue of the sine-Gordon equa-tion on an arbitrary time scale. In particular, we provided parallel compu-tations for the discrete and continuous case. The Darboux-Backlund trans-formation has the same form on any time scale. Soliton solutions of thesine-Gordon equation on time scales are expressed by the Cayely exponentialfunction. Therefore, like in the continuous case, soliton solutions of the non-linear equation are expressed in terms of exponential functions which wereintroduced as solutions to linear equations.

Acknowledgement. The first author (J.L.C.) is partly supported by theNational Science Centre (NCN) grant no. 2011/01/B/ST1/05137.

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