Darboux Transformation and Exact Solutions of the Continuous Heisenberg Spin Chain Equation

Darboux Transformation and Exact Solutions of the Continuous Heisenberg Spin Chain Equation

Ai-Hua Chen^aand Fan-Fan Wang^b

aCollege of Science, University of Shanghai for Science and Technology, Shanghai, 200093, People’s Republic of China

bDepartment of Mathematics, East China University of Science and Technology, Shanghai, 200237, People’s Republic of China

Reprint requests to F.-F. W.; E-mail:ffwang@ecust.edu.cn Z. Naturforsch.69a,9 – 16 (2014) / DOI: 10.5560/ZNA.2013-0067

Received March 11, 2013 / revised September 10, 2013 / published online October 30, 2013 In this paper, we give theN-fold Darboux transformation (DT) for the continuous Heisenberg spin chain which describes the motion of the isotropic ferromagnets in the complex case. By using this DT, we getN-soliton solutions and a new exact solution of the spin chain from a trivial seed solution and a plane wave seed solution, respectively.

Key words:Darboux Transformation; Heisenberg Spin Chain; Breather Solution; Rational Solution.

PACS numbers:02.30.Ik; 02.30.Jr

1. Introduction

In the 1970s, the continuous Heisenberg spin chain which describes the motion of the magnetization vec- tor of the isotropic ferromagnets has attracted much attention [1–4]. In [3], the explicit single-soliton so- lutions in the isotropic case were presented. In [4], the inverse scattering method was applied to the continu- ous Heisenberg spin chain and its Lax representation was obtained. In [5], the N-soliton solution was ob- tained by applying the Wadati gauge transformation to the inverse scattering problem of a nonlinear evolu- tion equation. In [6], theN-soltion solution was given explicitly according to the bilinear equations obtained by Hirota. Since the 1970s, the continuous Heisenberg spin chain also received much attention. In [7], the higher-order Heisenberg spin chain equations were de- duced by Chen and Li and they proved that these equa- tions are equivalent to the evolution equations of the Ablowitz–Kaup–Newell–Segur (AKNS) By using the nonlinearization method, Qiao [8] presented a finite- dimensional integrable system and the involutive so- lutions of the higher-order Heisenberg spin chain. In [9], the relation among the different systems associated with the Heisenberg magnetic equation was studied by the reduction procedure.

The Darboux transformation (DT) as a useful method to get explicit solutions of nonlinear partial differential equations has been utilized to obtain so- lutions of the continuous Heisenberg spin chain. In [10,11], for the spectral problem related to the con- tinuous Heisenberg spin chain, explicit one-fold Dar- boux transformations were constructed to obtain its soliton solutions. In [12], Cie´sli´nski and Czarnecka constructed the Darboux–Bäcklund transformation for the two-dimensional Heisenberg chain. In [13], Saleem and Hassan constructed the Darboux transformation for the generalized Heisenberg magnet model and ob- tained multi-soliton solutions in terms of quasidetermi- nants.

In [11], the soliton solutions were obtained for the continuous Heisenberg spin chain equation by the one- fold DT, and the iterative relationship between theNth solution and the (N+1)st solution can be derived by applying the one-fold DTNtimes. However, the rela- tionships between the new solutions and the seed solu- tions are not given. In this paper, according to the de- terminant representation of DT [14–17], we directly construct the N-fold Darboux transformation for the continuous Heisenberg spin chain by using a similar method as in [18]. By using of theN-fold DT, we can obtain the relationships between the new solutions and

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the seed solutions without tedious and complicated it- erations.

In this paper, we study the continuous Heisenberg spin chain introduced in [4], where Takhtajan consid- ered an infinite linear chain with spin-densitys_j(x,t) for j=1,2,3 and length of the spins equal to 1, i.e.

s²₁(x,t) +s²₂(x,t) +s²₃(x,t) =1. The motion equation of the continuous Heisenberg spin chain is

S_t= 1

2i(SS_xx−S_xxS), (1)

whereS=s₁σ₁+s₂σ₂+s₃σ₃, andσ_j(j=1,2,3)are the Pauli matrices

σ1=

0 1 1 0

, σ2=

0 −i

i 0

,

σ₃=

1 0 0 −1

.

(2)

If we letw=s3andu=s1−is₂, then from (1), we get u_t=i(uw_xx−wu_xx), w_t= i

2(u^∗u_xx−uu^∗_xx), (3) where w is a real-valued function, u is a complex- valued function, andu^∗denotes the complex conjugate ofu. According to the length constraint condition, we have

|u|²+w²=1. (4)

In Section2, according to the spectral problems in [7–11], we give the Lax pair of (3) and (4). Then we construct the N-fold Darboux transformation for the Heisenberg spin chain (1). In Section3, using different seed solutions, we get exact solutions of (1). From the trivial seed solution, we get breather solutions for the first two components and soliton solution for the third component of the spin chain. These solutions are sim- ilar with those in [10,11]. From the plane wave seed solution, we get a new exact solution of the spin chain.

In Section4, we make our conclusion.

2. Darboux Transformation

The spectral problem of the system (3) and (4) (cf.

[7–11]) is given by φx=Uφ=iλ

w u u^∗ −w

φ with |u|²+w²=1,

(5)

and the corresponding auxiliary spectral problem is given by

φt=Vφ (6)

= 2iwλ²+¹₂(uu^∗_x−u^∗ux)λ 2iuλ²+ (wux−uwx)λ 2iu^∗λ²+ (u^∗w_x−wu^∗_x)λ −2iwλ²−¹₂(uu^∗_x−u^∗u_x)λ

! φ,

whereφ= (φ₁,φ2)^T.λis a constant spectral parameter, uandware functions ofxandt.

We want to find a transformation ¯φ=Tφ, such that the Lax pair

φx=Uφ, φt=Vφ (7) is transformed into

φ¯x=U¯φ¯, φ¯t=V¯φ¯, (8) where ¯U and ¯V have the same forms asU andV re- spectively, which are obtained by replacingu,u^∗,w, u_x,u^∗_x,w_x with ¯u, ¯u^∗, ¯w, ¯u_x, ¯u^∗_x, ¯w_x, respectively, inU andV. Then the matrixT satisfies

T_x+TU=U T¯ , T_t+TV=V T¯ . (9) According to the zero curvature equation, if(u,w)sat- isfies (3),(u,¯ w)¯ also satisfies (3). The transformation φ¯=Tφ is a Darboux transformation for the Lax pair (5) and (6).

In the following, we first construct the Darboux transformation for the Lax pair (5) and (6). Then we give the relationship between(u,w)and(u,¯w).¯

Firstly, the Lax pair (5) and (6) satisfies the loop group conditions

U(λ^∗) =σ2U^∗(λ)σ₂, V(λ^∗) =σ2V^∗(λ)σ₂, (10) whereσ2 is the Pauli matrix defined in (2). Equation (10) implies thatT satisfiesT(λ^∗) =σ₂T^∗(λ)σ₂(con- straints imposed by reduction groups, see [19–21]).

Then from (9), we have to demandT(0) =const. in order to preserve the conditionsU(0) =V(0) =0 and U(0) =¯ V¯(0) =0. For simplicity, we supposeT(0) =I.

So we let

T =









 1+

N

∑

k=1

A_kλ^k

N

∑

k=1

B_kλ^k

−

N

∑

k=1

B^∗_kλ^k 1+

N

∑

k=1

A^∗_kλ^k











(11)

whereA_kandB_k(1≤k≤N)are functions ofxandt. Let Φ^(j) = (ϕ₁(λ_j),ϕ2(λ_j))^T and Ψ^(j) = (ψ₁(λ_j),ψ₂(λ_j))^T be two basic solutions of (5) and (6) withλ =λj (j=1,2, . . . ,N). ThenΦ∗^(j) = (−ϕ₂^∗(λ_j),ϕ₁^∗(λ_j))^T and Ψ∗^j = (−ψ₂^∗(λ_j),ψ₁^∗(λ_j))^T be two basic solutions of (5) and (6) with λ =λ^∗_j (j=1,2, . . . ,N). A_k and B_k (1≤k≤N) satisfy the following two algebraic systems:

N

∑

k=1

(A_k+η_jB_k)λ_j^k=−1, (12)

N k=1

∑

(−B_k+η^∗_jA_k)λ^∗_j^k=−η^∗_j, (13)

where

ηj=ϕ2(λ_j)−r_jψ2(λ_j)

ϕ₁(λ_j)−r_jψ₁(λ_j), 1≤ j≤N. (14) Note that complex λ_j and real r_j should be suitably chosen such that the determinants of the coefficient matrices of systems (12) and (13) are nonzero and thus A_kandB_k(1≤k≤N)can be uniquely determined.

From the form ofT in (11), we find that detT(λ) = (|A_N|²+|B_N|²)

N

∏

(λ−λj)(λ−λ_j^∗), (15) which means that λ_j andλ^∗_j are roots of detT(λ) = 0. At the same time, from T(0) = I, we know that detT(0) =1= (|A_N|²+B²_N)|λ₁|²|λ₂|²· · · |λ_N|², i.e.

|A_N|²+|B_N|²= 1

|λ₁|²|λ₂|²· · · |λ_N|². (16) In the following theorem, we give the relationship between(u,w)and(u,¯ w).¯

Theorem 1. If(u,w)is a solution of (3) and (4),(u,¯ w)¯ with

¯

u=uA²_N−u^∗B²_N−2wA_NB_N

|A_N|²+|B_N|² ,u[N], w¯=w(|A_N|²− |B_N|²) +u^∗A^∗_NB_N+uA_NB^∗_N

|A_N|²+|B_N|² ,w[N]

(17)

is a new solution of (3) and (4), where A_N and B_N are determined by (12) and (13).

Proof. From (17), it can be verified by a straightfor- ward calculation that|u|¯²+w¯²=1. In order to prove that(u,¯ w)¯ is a new solution of (3), we need to verify the validity of (9).

Step 1: We verify the validity of the first part of (9), i.e.T_x+TU=U T¯ .

LetT⁻¹=adj(T)/detT and (T_x+TU)·adj(T) =

f₁₁(λ) f₁₂(λ) f₂₁(λ) f₂₂(λ)

,F(λ). (18) A direct calculation shows that λ⁻¹f₁₁(λ), λ⁻¹f12(λ),λ⁻¹f21(λ), andλ⁻¹f22(λ)are all 2Nth- order polynomials inλ, andF(λ^∗) =σ2F^∗(λ)σ₂. By using (5), (12), (13), and (14), we get

η_jx=iu^∗λ_j−2iwλ_jη_j−iuλ_jη²_j, 1≤j≤N. (19) We can verify thatλjandλ^∗_j (1≤j≤N)are roots of

f_kl(λ) =0 fork,l=1,2. Therefore, we have

(T_x+TU)·adj(T) =detT·P(λ) (20) withP(λ^∗) =σ2P^∗(λ)σ₂, i.e.

P(λ) = p⁽¹⁾₁₁λ p⁽¹⁾₁₂λ

−p⁽¹⁾₁₂^∗λ p⁽¹⁾₁₁

∗

λ

!

, (21)

wherep⁽¹⁾₁₁ andp⁽¹⁾₁₂ are independent ofλ. We can fur- ther rewrite (20) as

T_x+TU=P(λ)T. (22)

Comparing the coefficients ofλ^N+1in (22), we find iwA_N+iu^∗B_N=p⁽¹⁾₁₁A_N−p⁽¹⁾₁₂B^∗_N,

iuA_N−iwB_N=p⁽¹⁾₁₁B_N+p⁽¹⁾₁₂A^∗_N.

(23) From (17) and (23), we have

p⁽¹⁾₁₁ =i ¯w, p⁽¹⁾₁₂ =i ¯u. (24) Then we obtainT_x+TU=U T¯ .

Step 2: As in Step 1, we verify the validity of the second part of (9), i.e.T_t+TV=V T¯ similarly.

LetT⁻¹=adj(T)/detT and (T_t+TV)·adj(T) =

g11(λ) g12(λ) g21(λ) g22(λ)

,G(λ). (25) A direct calculation shows that λ⁻¹g₁₁(λ), λ⁻¹g12(λ), λ⁻¹g21(λ), and λ⁻¹g22(λ) are all (2N + 1)st-order polynomials in λ and G(λ^∗) = σ₂G^∗(λ)σ₂. By using (6), (12), (13), and (14), we result in

η_jt=2iu^∗λ_j²+ (u^∗w_x−wu^∗_x)λ_j

−

4iwλ²_j + (uu^∗_x−u^∗u_x) η_j

−

2iuλ²_j+ (wu_x−uw_x)λ_j

η²_j, 1≤j≤N. (26)

We can verify thatλ_jandλ_j^∗(1≤j≤N)are roots of g_kl(λ) =0 fork,l=1,2. Therefore, we have

(T_t+TV)·adj(T) =detT·Q(λ) (27) withQ(λ^∗) =σ₂Q^∗(λ)σ₂, i.e.

Q(λ) = (28)

q⁽²⁾₁₁λ²+q⁽¹⁾₁₁λ q⁽²⁾₁₂λ²+q⁽¹⁾₁₂λ

−q⁽²⁾₁₂^∗λ²−q⁽¹⁾₁₂^∗λ q⁽²⁾₁₁^∗λ²+q⁽¹⁾₁₁^∗λ

! ,

whereq^(k)₁₁ andq^(k)₁₂ (k=1,2)are independent ofλ. We can further rewrite (27) as

T_t+TV=Q(λ)T. (29)

Comparing the coefficients ofλ^N+2in (29), we get 2iwA_N+2iu^∗B_N =q⁽²⁾₁₁A_N−q⁽²⁾₁₂B^∗_N, 2iuA_N−2iwB_N=q⁽²⁾₁₁B_N+q⁽²⁾₁₂A^∗_N.

(30) From (17) and (30), we obtain

q⁽²⁾₁₁ =2i ¯w, q⁽²⁾₁₂ =2i ¯u. (31) If we further compare the coefficients ofλ^N+1of (29), by using (31), we have

1

2(uu^∗_x−u_xu^∗)A_N+2iwAN−1+ (u^∗wx−wu^∗_x)B_N (32) +2iu^∗BN−1=q⁽¹⁾₁₁A_N+2i ¯wAN−1−q⁽¹⁾₁₂B^∗_N−2i ¯uB^∗_N−1,

−1

2(uu^∗_x−u_xu^∗)B_N−2iwBN−1+ (wu_x−w_xu)A_N +2iuAN−1=q⁽¹⁾₁₁B_N+2i ¯wBN−1+q⁽¹⁾₁₂A^∗_N+2i ¯uA^∗_N−1.

According to step 1, by comparing the coefficients of λ^NinT_x+TU=U T¯ , we find

ANx+iwAN−1+iu^∗BN−1=i ¯wAN−1−i ¯uB^∗_N−1, BNx+iuA_N−1−iwB_N−1=i ¯wBN−1+i ¯uA^∗_N−1. (33)

By using (32), (33), and the expressions of ¯u and

¯

w, through tedious calculation (which can also be done easily by using the Mathematica software), we achieve

q⁽¹⁾₁₁ =1

2(u¯u¯^∗_x−u¯^∗u¯_x), q⁽¹⁾₁₂ =1

2(w¯u¯_x−u¯w¯_x). (34) Then we have T_t+TV =V T¯ . The proof is com- plete.

From Theorem1and the transformationw=s3,u= s₁+is₂, we find that

s¯1= (u¯+u¯^∗)/2, s¯2=i(u¯−u¯^∗)/2, s¯3=w¯ (35) is a new real-valued solution of the Heisenberg equa- tion (1).

In the following section, we will obtain exact so- lutions of (3) and (4) by the use of Theorem1. Then through (35), we will obtain real-valued exact solutions of the Heisenberg equation (1).

3. Exact Solutions

In this section, we will take a trivial solution and a plane wave solution as the seed solution, respectively, to obtain exact solutions of the Heisenberg equation (1).

3.1. Trivial Seed Solution

In this subsection, by using the N-fold Darboux transformation and taking a trivial solution as seed so- lution, we give the breather and soliton solutions of (1).

It is easy to see that(u,w) = (0,1)is a trivial solution of (3) and (4), and we take it as the seed solution. From (16) and (17), the solution of (3) and (4) is given by

u[N] =−2|λ₁|²|λ₂|²· · · |λ_N|²A_NB_N,

w[N] =1−2|λ₁|²|λ₂|²· · · |λ_N|²|B_N|². (36)

From (35), we can obtain a series of solutions of the Heisenberg equation (1) which are given by

s₁[N] = u[N] +u[N]^∗

2 =−|λ₁|²|λ₂|²· · · |λ_N|²

·(A_NB_N+A^∗_NB^∗_N), s₂[N] = i(u[N]−u[N]^∗)

2 =i|λ₁|²|λ₂|²· · · |λ_N|²

·(A^∗_NB^∗_N−A_NB_N),

s3[N] =w[N] =1−2|λ₁|²|λ₂|²· · · |λ_N|²|B_N|². (37) We choose two basic solutions of (5) and (6) as

ϕ(λ_j) = (e^{i(λjx+2λ2jt)},0)^T, ψ(λ_j) = (0,e^{−i(λjx+2λ2jt)})^T.

(38) For simplicity we letr_j=−1(j=1,2, . . . ,N)in (14).

Then we have

ηj=e^{−2i(λjx+2λ2jt)}, j=1,2, . . . ,N. (39) From the linear system (12), (13) and the expression of (39), we find that ifλ_j(j=1,2, . . . ,N)are real, the obtained solutions are trivial. To get non-trivial solu- tions, we chooseλj(j=1,2, . . . ,N)as complex.

WhenN=1, takingλ1=τ1+iτ2, whereτ1andτ2

are real, the solution of (3) and (4) is u[1] =−e^{2i(τ1x+2(τ12−τ22)t)}

τ₁²+τ₂²

·h

2τ₂²tanh(2τ₂(x+4τ₁t))−2iτ₁τ₂ i

·sech(2τ₂(x+4τ₁t)), w[1] =1− 2τ₂²

τ₁²+τ₂²

sech²(2τ₂(x+4τ₁t)).

(40)

-2 -1

0 1

2 x

-0.4 -0.2

0 0.2

0.4

t -11-20

2

s1 1 -2

-1 0 x 1

(a)

-3 -2

-1 0 x

0 0.2

0.4 0.6

0.81 t -11-20

2

s2 1 -3

-2 x -1

(b)

-0.5 0 x 0.5

-0.2 -0.1

0 0.1

0.2

t -11-20

s3 1 -0.5

0 x 0.5

(c)

Fig. 1 (colour online). Plots of solution (41) of Heisenberg equation (1) whenτ1=1 andτ2=2.

Then a solution of the Heisenberg equation (1) is s1[1] = −1

τ₁²+τ₂² h

2τ₂²tanh(2τ₂(x+4τ₁t))

·cos(2τ₁x+4(τ₁²−τ₂²)t) +2τ₁τ2sin(2τ₁x+4(τ₁²−τ₂²)t)i

·sech(2τ₂(x+4τ1t)), s₂[1] = 1

τ₁²+τ₂² h

2τ₂²tanh(2τ₂(x+4τ₁t)) (41)

·sin(2τ₁x+4(τ₁²−τ₂²)t)

−2τ₁τ2cos(2τ₁x+4(τ₁²−τ₂²)t)i

·sech(2τ₂(x+4τ₁t)), s₃[1] =1− 2τ₂²

τ₁²+τ₂²

sech²(2τ₂(x+4τ₁t)).

For given values of τ1=1 and τ2=2, we give the plots of s₁[1], s₂[1], and s₃[1] in Figure 1. We find that s₃[1] is a bell-shaped single-soliton solu- tion, and s₁[1] and s₂[1] are single-soliton solutions with exchange interactions, i.e. they are breather so- lutions.

WhenN=2, for simplicity, taking λ1=τ1+iτ2, λ2=−τ₁+iτ₂, whereτ1andτ2are real, the solution of (3) and (4) is

u[2] =−4τ₁τ2

θ₁² h

θ2+2iτ₁τ2(e^2α1−e^2α2)i

(θ₃+iθ₄),

w[2] =1−8τ₁²τ₂² θ₁²

·

(e^α1+e^α1+2α2)²+ (e^α2+e^2α1+α2)²−2θ₅ (42)

-2 0

x 2 -0.5

0 0.5 t -2

0 2 s1 2

-2 0 x 2

(a)

-2 0

x 2 -0.5

0 0.5 t -2

0 2 s2 2

-2 0 x 2

(b)

-1 0

x 1 -0.2

0 0.2 t -11-20

2 s3 2

-1 0 x 1

(c)

Fig. 2 (colour online). Plots of solution (43) of Heisenberg equation (1) whenτ₁=1 andτ₂=2.

and the solution of (1) is s₁[2] =4τ₁τ2

θ₁²

−θ2θ3+2τ₁τ2(e^2α1−e^2α2)θ₄ , s₂[2] =4τ₁τ2

θ₁²

θ2θ4+2τ₁τ2(e^2α1−e^2α2)θ₃ ,

s3[2] =1−8τ₁²τ₂² θ₁²

(e^α1+e^α1+2α2)² + (e^α2+e^2α1+α2)²−2θ₅

,

(43) where

θ1= (e^2α1+1)(e^2α2+1)τ₁²−(2 e^α1+α2cos(β₁−β2)

−e^2α1−e^2α2)τ₂²,

θ2= (e^2α1+1)(e^2α2+1)τ₁²+ (2 e^α1+α2cos(β₁−β2)

−e^2α1−e^2α2)τ₂²,

θ₃= (e^α2+e^2α1+α2)sinβ₂−(e^α1+e^α1+2α2)sinβ₁, θ4= (e^α2+e^2α1+α2)cosβ₂−(e^α1+e^α1+2α2)cosβ1, θ₅= (e^α1+e^α1+2α2)(e^α2+e^2α1+α2)cos(β₁−β2) with

α1=2τ2(x+4τ1t), β1=−2(τ₁x+2 τ₁²−τ₂²)t , α₂=2τ₂(x−4τ₁t), β₂=2(τ₁x−2(τ₁²−τ₂²)t).

For given values ofτ1=1 andτ2=2, we give the plots of (43) in Figure2. We find thats1[2]ands2[2]

are two head-on breather solutions, ands₃[2]is a bell- shaped two-soliton solution.

3.2. Plane Wave Seed Solution

In this subsection, by the modified Darboux trans- formation method [22–24], we can get rational solu- tions of (3) and (4) by the one-fold Darboux transfor- mation. Then, new exact solutions of (1) can be ob- tained.

In order to get rational solutions, we take a plane wave solution as the seed solution. It is easy to see that(u,w) = (be^i(2x+4at),a)satisfies (3), wherea and b(b6=0)are real. To satisfy (4), we havea²+b²=1.

Taking(u,w)as the seed solution, we suppose that the basic solutions of (5) and (6) have the following forms:

φ1(λ),φ1(x,t)

= (a₁x+b1t+c1xt+d₁)e^i(x+2at), φ2(λ),φ2(x,t)

= (a₂x+b₂t+c₂xt+d₂)e^−i(x+2at),

(44)

wherea_i,b_i,c_i, andd_i(i=1,2)are complex, andλ= a+ib. Substituting (44) into the Lax pair (5) and (6) and comparing all the coefficients, we have

a1=ia₂,b1= (−2b+4ia)a₂, b2= (4a+2ib)a₂,

c₁=c₂=0,d₂=a−ib

b a₂−id₁.

(45)

In order to preserve the positivity of|φ₁|²+|φ₂|²and get some solutions with simple forms, we choosea₂= i,d₁=a−ib

2b . Then, φ1(x,t) =h

−x−(4a+2ib)t+a−ib 2b

i

e^i(x+2at), φ2(x,t) =h

ix−(2b−4ia)t+b+ia 2b

i

e^−i(x+2at). (46)

For simplicity, we letr1=0 in (14), and we get η1(λ),η1(x,t) =φ2(x,t)

φ₁(x,t)

=

ix−(2b−4ia)t+b+ia 2b

−x−(4a+2ib)t+a−ib 2b

e^−2i(x+2at). (47)

-2 -1

0 1

2

x -5

-2.5 0 52.5

t -1-2210 s1 1

-2 -1

0 1

2

x -1-210

(a)

-2 0

2 x -5

-2.5 0 52.5

t -1-2210 s2 1

-2 0

2 x -1-210

(b)

-1 0

1 x -1

-0.5 0 0.5 1

t -1

0 1 s3 1

-1 0

1 x -1

0 1

(c)

Fig. 3 (colour online). Plots of solution (51) of Heisenberg equation (1).

Then from the linear systems (12) and (13), we ob- tain

A₁=−a−ib+ (a+ib)|η₁|²

1+|η1|² , B₁= 2ibη₁^∗

1+|η1|². (48) From (16) and (17), the solution of (3) and (4) is given by

u[1] =uA²₁−u^∗B²₁−2wA₁B₁,

w[1] =w(|A₁|²− |B₁|²) +u^∗A^∗₁B1+uA1B^∗₁. (49)

Now from (35), we obtain the solution of the Heisen- berg equation (1) which is given by

s₁[1] =1 2 h

u(A²₁−B^∗₁²) +u^∗(A^∗₁²−B²₁)

−2w(A₁B₁+A^∗₁B^∗₁)i , s₂[1] =i

2 h

u(A²₁+B^∗₁²)−u^∗(A^∗₁²+B²₁)

−2w(A₁B₁−A^∗₁B^∗₁)i ,

s₃[1] =w(|A₁|²− |B₁|²) +u^∗A^∗₁B₁+uA₁B^∗₁.

(50)

If we further leta=0, b=1, the solution of the Heisenberg equation (1) is

s₁[1] =− h

(4x²+16t²+1)²−32x²i

cos 2x−8x(4x²+16t²−1)sin 2x (4x²+16t²+1)² , s₂[1] =

8x(4x²+16t²−1)cos 2x+h

(4x²+16t²+1)²−32x² i

sin 2x (4x²+16t²+1)² , s₃[1] =− 64xt

(4x²+16t²+1)².

(51)

The plots of (51) are given in Figure3. Note that s₃[1]is a rational solution. To the best of our knowledge, (51) is a new solution of the Heisenberg equation (1).

Since the seed solution is related toλ, we cannot ap- ply theN-fold DT whenN=2,3, . . .. Although we can iterate the one-fold DT to get new solutions, it would be much more complicated.

4. Conclusions

In this paper, we construct the N-fold DT of the Heisenberg equation (1). By the use of the obtained DT, from the trivial seed solution (u,w) = (0,1), we get breather solutions for the first two components and

soliton solution for the third component of the spin chain (s₁,s₂,s₃). From the plane wave seed solution (u,w) = (be^i(2x+4at),a)witha²+b²=1, we get a new exact solution of the spin chain(s₁,s₂,s₃). Particularly, the third component of the spin chain is a rational so- lution.

Acknowledgements

We are most grateful to the anonymous referee for pointing out much relevant literature and crucial help in improving the original manuscript. The work de- scribed in this paper was supported by the Fundamen- tal Research Funds for the Central Universities.

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