Darboux transformation and soliton solutions of the semi-discrete massive Thirring model
aa r X i v : . [ n li n . S I] J u l DARBOUX TRANSFORMATION AND SOLITON SOLUTIONSOF THE SEMI-DISCRETE MASSIVE THIRRING MODEL
TAO XU AND DMITRY E. PELINOVSKYA
BSTRACT . A one-fold Darboux transformation between solutions of the semi-discrete massive Thirring modelis derived using the Lax pair and dressing methods. This transformation is used to find the exact expressions forsoliton solutions on zero and nonzero backgrounds. It is shown that the discrete solitons have the same propertiesas solitons of the continuous massive Thirring model.
1. I
NTRODUCTION
The massive Thirring model (MTM) in laboratory coordinates is an example of the nonlinear Dirac equationarising in two-dimensional quantum field theory [23], optical Bragg gratings [7], and diatomic chains withperiodic couplings [1]. This model received much of attention because of its integrability [17] which was usedto study the inverse scattering [13–16, 21, 27, 28], soliton solutions [2–4, 20], spectral and orbital stability ofsolitons [6, 12, 22], and construction of rogue waves [8].Several integrable semi-discretizations of the MTM in characteristic coordinates were proposed in the litera-ture [18, 19, 24–26] by discretizing one of the two characteristic coordinates. These semi-discretizations are notrelevant for the time-evolution problem related to the MTM in laboratory coordinates. It was only recently [11]when the integrable semi-discretization of the MTM in laboratory coordinates was derived. The correspondingsemi-discrete MTM is written as the following system of three coupled equations:(1) dU n dt + Q n +1 + Q n + 2i h ( R n +1 − R n ) + U n ( ¯ R n + ¯ R n +1 ) − U n ( | Q n +1 | + | Q n | + | R n +1 | + | R n | ) − i h U n ( ¯ Q n +1 − ¯ Q n ) = 0 , − h ( Q n +1 − Q n ) + 2 U n − | U n | ( Q n +1 + Q n ) = 0 ,R n +1 + R n − U n + i h | U n | ( R n +1 − R n ) = 0 , where h is the lattice spacing of the spatial discretization and n is the discrete lattice variable. ¯ R n and ¯ Q n denote the complex conjugate of R n and Q n respectively. Only the first equation of the system (1) representsthe time evolution problem, whereas the other two equations represent the constraints which define componentsof { R n } n ∈ Z and { Q n } n ∈ Z in terms of { U n } n ∈ Z instantaneously in time t .In the continuum limit h → , the slowly varying solutions to the system (1) can be represented by U n ( t ) = U ( x = hn, t ) , R n ( t ) = R ( x = hn, t ) , Q n ( t ) = Q ( x = nh, t ) , where the continuous variables satisfy the following three equations:(2) ∂U∂t + i ∂R∂x + Q + U ¯ R − U ( | Q | + | R | ) = 0 , − i ∂Q∂x + U − | U | Q = 0 ,R − U = 0 . he system (2) in variables U ( x, t ) = u ( x, t − x ) and Q ( x, t ) = v ( x, t − x ) yields the continuous MTM systemin the form:(3) i (cid:18) ∂u∂t + ∂u∂x (cid:19) + v = | v | u, i (cid:18) ∂v∂t − ∂v∂x (cid:19) + u = | u | v. It is shown in [11] that the semi-discrete MTM system (1) is the compatibility condition(4) ddt N n ( λ ) = P n +1 ( λ ) N n ( λ ) − N n ( λ ) P n ( λ ) , of the following Lax pair of two linear equations: Φ n +1 ( λ ) = N n ( λ )Φ n ( λ ) , N n ( λ ) = λ + hλ (cid:18) i2 h | U n | − i2 h | U n | (cid:19) U n − i2 h | U n | U n − i2 h | U n | hλ − λ (cid:18) i2 h | U n | − i2 h | U n | (cid:19) , (5a) ddt Φ n ( λ ) = P n ( λ )Φ n ( λ ) , P n ( λ ) = i2 (cid:18) λ − | R n | λR n − Q n λ − λ ¯ R n − ¯ Q n λ − | Q n | − λ − (cid:19) , (5b)where Φ n ( λ ) ∈ C is defined for n ∈ Z and λ is a spectral parameter.Because the passage from the discrete system (1) to the continuum limit (3) involves the change of the coor-dinates U ( x, t ) = u ( x, t − x ) and Q ( x, t ) = v ( x, t − x ) , the initial-value problem for the semi-discrete MTMsystem (1) does not represent the initial-value problem for the continuous MTM system (3) in time variable t . Inaddition, numerical explorations of the semi-discrete system (1) are challenging because the last two constraintsin the system (1) may lead to appearance of bounded but non-decaying sequences { R n } n ∈ Z and { Q n } n ∈ Z inresponse to the bounded and decaying sequence { U n } n ∈ Z . On the other hand, since the semi-discrete MTMsystem (1) has the Lax pair of linear equations (5), it is integrable by the inverse scattering transform methodwhich implies existence of infinitely many conserved quantities, exact solutions, transformations between dif-ferent solutions, and reductions to other integrable equations [10]. These properties of integrable systems werenot explored for the semi-discrete MTM system (1) in the previous work [11].The purpose of this work is to derive the one-fold Darboux transformation between solutions of the semi-discrete MTM system (1). We employ the Darboux transformation in order to generate one-soliton and two-soliton solutions on zero background in the exact analytical form. By looking at the continuum limit h → , weshow that the discrete solitons share many properties with their continuous counterparts. We also construct one-soliton solutions on a nonzero constant background. Further properties of the model, e.g. conserved quantitiesand solvability of the initial-value problem, are left for further studies.The following theorem represents the main result of this work. Theorem 1.
Let Φ n ( λ ) = ( f n , g n ) T be a nonzero solution of the Lax pair (5) with λ = λ and ( U n , R n , Q n ) be a solution of the semi-discrete MTM system (1). Another solution of the semi-discrete MTM system (1) isgiven by U [1] n = − λ | f n | + λ | g n | ) U n − h | λ | ( λ | f n | + ¯ λ | g n | ) U n + 2 i( λ − ¯ λ ) f n ¯ g n λ | f n | + ¯ λ | g n | ) − h | λ | (¯ λ | f n | + λ | g n | ) + h ( λ − ¯ λ ) ¯ f n g n U n , (6a) R [1] n = − (cid:0) ¯ λ | f n | + λ | g n | (cid:1) R n + (cid:0) λ − ¯ λ (cid:1) f n ¯ g n λ | f n | + ¯ λ | g n | , (6b) Q [1] n = − | λ | (cid:0) λ | f n | + ¯ λ | g n | (cid:1) Q n + (cid:0) λ − ¯ λ (cid:1) f n ¯ g n | λ | (¯ λ | f n | + λ | g n | ) . (6c) heorem 1 is proven in Section 2 using the Lax pair (5) and the dressing methods. One-soliton and two-soliton solutions on zero background are obtained in Section 3. One-soliton solutions on a nonzero constantbackground are constructed in Section 4. Both zero and nonzero constant backgrounds are modulationallystable in the evolution of the semi-discrete MTM system (1). A summary and further directions are discussedin Section 5. 2. P ROOF OF THE ONE - FOLD D ARBOUX TRANSFORMATION
The one-fold Darboux transformation takes an abstract form (see, e.g., [9]): Φ [1] ( λ ) = T ( λ )Φ( λ ) , (7)where T ( λ ) is the Darboux matrix, Φ( λ ) is a solution to the system (5), whereas Φ [1] ( λ ) is a solution of thetransformed system Φ [1] n +1 ( λ ) = N [1] n ( λ )Φ [1] n ( λ ) , ddt Φ [1] n ( λ ) = P [1] n ( λ )Φ [1] n ( λ ) , (8)with N [1] n ( λ ) and P [1] n ( λ ) having the same form as N n ( λ ) and P n ( λ ) except that the potentials (cid:0) U n , Q n , R n (cid:1) are replaced by (cid:0) U [1] n , Q [1] n , R [1] n (cid:1) . By substituting (7) into the linear equations (8) and using the linear equations(5), we obtain the following system of equations for the Darboux matrix T ( λ ) : T n +1 ( λ ) N n ( λ ) = N [1] n ( λ ) T n ( λ ) , (9a) ddt T n ( λ ) + T n ( λ ) P n ( λ ) = P [1] n ( λ ) T n ( λ ) . (9b)Since N n ( λ ) and P n ( λ ) in (5) contain both the positive and negative powers of λ , we take the one-fold Dar-boux matrix T ( λ ) in the following form (used in [29] in the context of the semi-discrete nonlocal nonlinearSchr¨odinger equation):(10) T n ( λ ; t ) = P l = − a l,n ( t ) λ l P l = − b l,n ( t ) λ l P l = − c l,n ( t ) λ l P l = − d l,n ( t ) λ l , where the coefficients are to be determined. Before further work, we shall simplify the Darboux matrix in (10)by using some constraints following from the system (9). Expanding Eq. (9b) in powers of λ and equating thecoefficients of λ and λ − to , we verify that b ,n = c ,n = b − ,n = c − ,n = 0 . (11)Collecting coefficients of other powers of λ yields the following system of equations: λ : a ,n R n − b ,n − d ,n R [1] n = 0 , (12a) λ : a ,n ¯ R [1] n − c ,n − d ,n ¯ R n = 0 , (12b) λ − : a − ,n Q n + b ,n − d − ,n Q [1] n = 0 , (12c) λ − : a − ,n ¯ Q [1] n + c ,n − d − ,n ¯ Q n = 0 , (12d) λ : a ,n R n − d ,n R [1] n = 0 , (12e) λ : − a ,n ¯ R [1] n + d ,n ¯ R n = 0 , (12f) λ : (cid:0) | R [1] n | − | R n | (cid:1) a ,n + ¯ R n b ,n − R [1] n c ,n − da ,n dt = 0 , (12g) : R n c ,n − ¯ R [1] n b ,n + (cid:0) | Q n | − | Q [1] n | (cid:1) d ,n − dd ,n dt = 0 , (12h) λ − : a ,n Q n − d ,n Q [1] n = 0 , (12i) λ − : a ,n ¯ Q [1] n − d ,n ¯ Q n = 0 , (12j) λ − : a − ,n (cid:0) | R [1] n | − | R n | ) − b ,n ¯ Q n + c ,n Q [1] n − da − ,n dt = 0 , (12k) λ − : d − ,n (cid:0) | Q n | − | Q [1] n | (cid:1) − c ,n Q n + b ,n ¯ Q [1] n − dd − ,n dt = 0 , (12l) λ : a ,n (cid:0) | R [1] n | − | R n | (cid:1) − da ,n dt = 0 , (12m) λ : d ,n (cid:0) | Q n | − | Q [1] n | (cid:1) − dd ,n dt = 0 , (12n) λ : b ,n (cid:0) | Q n | + | R [1] n | (cid:1) − a ,n Q n + a − ,n R n + d ,n Q [1] n − d − ,n R [1] n − db ,n dt = 0 , (12o) λ : c ,n (cid:0) | Q [1] n | + | R n | (cid:1) − a ,n ¯ Q [1] n + a − ,n ¯ R [1] n + d ,n ¯ Q n − d − ,n ¯ R n + 2i dc ,n dt = 0 . (12p)It follows from Eqs. (12e), (12f), (12i) and (12j) that if ( | Q [1] | , | R [1] | ) = ( | Q | , | R | ) , then a ,n = d ,n = 0 , afterwhich Eqs. (12m) and (12n) are identically satisfied. Solving Eqs. (12a), (12b), (12c), and (12d) yields b ,n = a ,n R n − d ,n R [1] n = d − ,n Q [1] n − a − ,n Q n , (13a) c ,n = a ,n ¯ R [1] n − d ,n ¯ R n = d − ,n ¯ Q n − a − ,n ¯ Q [1] n . (13b)Plugging (13) into Eqs. (12g) and (12l) gives da ,n dt = dd − ,n dt = 0 . (14)All constraints of the system (12) are satisfied except for Eqs. (12h), (12k), (12o), and (12p). It is howeverdifficult to compute relations between the new and old potentials from these four equations. Therefore, we willobtain the relations between ( R n , Q n ) and ( R [1] n , Q [1] n ) by using dressing methods from Appendix A in [5].Expanding Eq. (9a) in powers of λ and equating the coefficients of λ and λ − to , we verify that a ,n +1 = a ,n , d − ,n +1 = d − ,n . (15)Combining Eqs. (14) and (15), we conclude that a ,n ( t ) and d − ,n ( t ) are constants both in t and n . For nor-malization purposes, we set a ,n ( t ) = 1 and d − ,n ( t ) = | λ | . We also re-enumerate the remaining coefficientsas follows: a − ,n ( t ) = a n ( t ) | λ | , b ,n ( t ) = b n ( t ) , c ,n ( t ) = c n ( t ) , and d ,n ( t ) = d n ( t ) . The Darboux matrix T [1] n given previously by (10) is now rewritten in the simplified form: T n ( λ ) = λ + a n | λ | λ b n c n d n λ + | λ | λ ! . (16)In order to determine a n ( t ) , b n ( t ) , c n ( t ) , and d n ( t ) , we use the symmetry properties of the Lax pair (5). Thisallows us to find simultaneously both the coefficients of T ( λ ) and the transformations between the potentials (cid:0) U, Q, R (cid:1) and (cid:0) U [1] , Q [1] , R [1] (cid:1) . Lemma 2.
Let Φ( λ ) = (cid:0) f, g ) T be a nonzero solution of the Lax pair (5) at λ = λ . Then, (17) [Φ(¯ λ )] n = Ω n (cid:18) − ¯ g n ¯ f n (cid:19) , [Φ( − λ )] n = ( − n (cid:18) − f n g n (cid:19) , [Φ( − ¯ λ )] n = ( − n Ω n (cid:18) ¯ g n ¯ f n (cid:19) , re solutions of the Lax pair (5) at λ = ¯ λ , λ = − λ , and λ = − ¯ λ respectively, where Ω n ( t ) satisfies: Ω n +1 = − S n Ω n , S n := 1 + i2 h | U n | − i2 h | U n | , (18a) d Ω n dt = M n Ω n , M n := i2 (cid:0) ¯ λ − ¯ λ − + | Q n | − | R n | (cid:1) . (18b) Proof.
It follows from (5a) that components of Φ( λ ) satisfy the system of difference equations: f n +1 = (cid:16) λ + hλ S n (cid:17) f n + U n − i2 h | U n | g n ,g n +1 = U n − i2 h | U n | f n + (cid:16) hλ − λ S n (cid:17) g n , (19)whereas components of Φ(¯ λ ) satisfy the system of difference equations: Ω n +1 ¯ g n +1 = (cid:16) ¯ λ + h ¯ λ S n (cid:17) Ω n ¯ g n − U n − i2 h | U n | Ω n ¯ f n , Ω n +1 ¯ f n +1 = − U n − i2 h | U n | Ω n ¯ g n + (cid:16) h ¯ λ − ¯ λ S n (cid:17) Ω n ¯ f n . (20)Dividing (20) by Ω n +1 and taking the complex conjugation yields (19) if and only if Ω satisfies the differenceequation (18a). Similarly, it follows from (5b) that components of Φ( λ ) satisfy the time evolution equations: (cid:26) df n dt = i2 (cid:2) ( λ − | R n | ) f n + ( λ R n − λ − Q n ) g n (cid:3) , dg n dt = i2 (cid:2) ( λ ¯ R n − λ − ¯ Q n ) f n + ( − λ − + | Q n | ) g n (cid:3) , (21)whereas components of Φ(¯ λ ) satisfy the time evolution equations: ( d Ω n dt ¯ g n + Ω n d ¯ g n dt = i2 (cid:2) (¯ λ − | R n | )Ω n ¯ g n − (¯ λ R n − ¯ λ − Q n )Ω n ¯ f n (cid:3) , d Ω n dt ¯ f n + Ω n d ¯ f n dt = i2 (cid:2) − (¯ λ ¯ R n − ¯ λ − ¯ Q n )Ω n ¯ g n + ( − ¯ λ − + | Q n | )Ω n ¯ f n (cid:3) , (22)Taking the complex conjugation of (22) yields (21) if and only if Ω satisfies the time evolution equation (18b).The other two solutions in (17) are obtained by the symmetry of the system (5) with respect to the reflection λ → − λ . Lemma 3.
Let Φ( λ ) = ( f, g ) T be in the kernel of the Darboux matrix T ( λ ) and Φ(¯ λ ) = Ω (cid:0) − ¯ g, ¯ f (cid:1) T be inthe kernel of T (¯ λ ) . Then, the coefficients of T ( λ ) in (16) are given by (23) a n = − ¯∆ n ∆ n , b n = − (cid:0) λ − ¯ λ (cid:1) f n ¯ g n ∆ n , c n = (cid:0) λ − ¯ λ (cid:1) ¯ f n g n ∆ n , d n = − ¯∆ n ∆ n , where ∆ n := ¯ λ | f n | + λ | g n | . Furthermore, Φ( − λ ) and Φ( − ¯ λ ) in (17) are in the kernel of T ( − λ ) and T ( − ¯ λ ) respectively.Proof. We rewrite the linear equations for T ( λ )Φ( λ ) = 0 and T (¯ λ )Φ(¯ λ ) = 0 in the following explicitform:(24) ( λ + a n ¯ λ ) f n + b n g n = 0 ,c n f n + ( d n λ + ¯ λ ) g n = 0 , − (¯ λ + a n λ )¯ g n + b n ¯ f n = 0 , − c n ¯ g n + ( d n ¯ λ + λ ) ¯ f n = 0 , where the scalar factor Ω has been canceled out. Solving the linear system (24) with Cramer’s rule yields (23).Then, it follows from (16) and (23) that T n ( λ ) can be written in the form: T n ( λ ) = ( λ − λ )( λ − ¯ λ )2 λ ∆ n ˆ T n ( λ ) , (25) here ˆ T n ( λ ) = 1 λ − λ (cid:18) ¯ g n ¯ f n (cid:19) (cid:0) g n − f n (cid:1) + 1 λ + λ (cid:18) − ¯ g n ¯ f n (cid:19) (cid:0) g n f n (cid:1) + 1 λ − ¯ λ (cid:18) f n − g n (cid:19) (cid:0) ¯ f n ¯ g n (cid:1) + 1 λ + ¯ λ (cid:18) f n g n (cid:19) (cid:0) − ¯ f n ¯ g n (cid:1) . It follows from (25) that T n ( λ ) (cid:18) f n g n (cid:19) = (cid:18) (cid:19) , T n ( − λ ) (cid:18) − f n g n (cid:19) = (cid:18) (cid:19) ,T n (¯ λ ) (cid:18) − ¯ g n ¯ f n (cid:19) = (cid:18) (cid:19) , T n ( − ¯ λ ) (cid:18) ¯ g n ¯ f n (cid:19) = (cid:18) (cid:19) , hence T ( ± λ )Φ( ± λ ) = 0 and T ( ± ¯ λ )Φ( ± ¯ λ ) = 0 . Lemma 4.
Let the Darboux matrix T ( λ ) be in the form (16) with the coefficients given by Eqs. (23). Then, thedeterminant of T ( λ ) is given by det T n ( λ ) = − ( λ − λ )( λ − ¯ λ ) λ ¯∆ n ∆ n . (26) Proof.
Expanding det T n ( λ ) given by (16) yields det T n ( λ ) = d n λ + a n d n | λ | − b n c n + | λ | + a n | λ | λ − . (27)Since ± λ and ± ¯ λ are the roots of det T ( λ ) , we obtain (26). Alternatively, substituting (23) into (27) yields(26).For λ = ± λ and λ = ± ¯ λ , we define(28) ad T n ( λ ) = det T n ( λ )[ T n ( λ )] − = d n λ + | λ | λ − b n − c n λ + a n | λ | λ ! . and obtain ad T n ( λ ) from (16) and (23) in the form: ad T n ( λ ) = ( λ − λ )( λ − ¯ λ )2 λ ∆ n ad ˆ T n ( λ ) , (29)where ad ˆ T n ( λ ) = 1 λ − λ (cid:18) f n g n (cid:19) (cid:0) − ¯ f n ¯ g n (cid:1) + 1 λ + λ (cid:18) f n − g n (cid:19) (cid:0) ¯ f n ¯ g n (cid:1) + 1 λ − ¯ λ (cid:18) ¯ g n − ¯ f n (cid:19) (cid:0) − g n − f n (cid:1) + 1 λ + ¯ λ (cid:18) ¯ g n ¯ f n (cid:19) (cid:0) g n − f n (cid:1) . New potentials N [1] n ( λ ) and P [1] n ( λ ) are derived from Eqs. (9) by using the Darboux matrix T ( λ ) . Assuming λ = ± λ and λ = ± ¯ λ , we obtain from (9) and (29) that N [1] n ( λ ) = 1det T n ( λ ) T n +1 ( λ ) N n ( λ )ad T n ( λ )= − λ n T n +1 ( λ ) N n ( λ )ad ˆ T n ( λ ) (30)and P [1] n ( λ ) = 1det T n ( λ ) (cid:20) ddt T n ( λ ) + T n ( λ ) P n ( λ ) (cid:21) ad T n ( λ ) − λ n (cid:20) ddt T n ( λ ) + T n ( λ ) P n ( λ ) (cid:21) ad ˆ T n ( λ ) , (31)where the expressions (26) and (29) have been used.First, we compute the products in the right-hand side of Eq. (30). By Lemma 2 and direct computations, weobtain N n ( λ ) (cid:18) f n g n (cid:19) = (cid:18) f n +1 g n +1 (cid:19) + ( λ − λ ) (cid:18) − hλλ S n − hλλ − S n (cid:19) (cid:18) f n g n (cid:19) , (32a) N n ( λ ) (cid:18) f n − g n (cid:19) = (cid:18) − f n +1 g n +1 (cid:19) + ( λ + λ ) (cid:18) hλλ S n hλλ − S n (cid:19) (cid:18) f n − g n (cid:19) , (32b) N n ( λ ) (cid:18) ¯ g n − ¯ f n (cid:19) = − S n (cid:18) ¯ g n +1 − ¯ f n +1 (cid:19) + ( λ − ¯ λ ) − hλ ¯ λ S n − hλ ¯ λ − S n ! (cid:18) ¯ g n − ¯ f n (cid:19) , (32c) N n ( λ ) (cid:18) ¯ g n ¯ f n (cid:19) = S n (cid:18) ¯ g n +1 ¯ f n +1 (cid:19) + ( λ + ¯ λ ) hλ ¯ λ S n hλ ¯ λ − S n ! (cid:18) ¯ g n ¯ f n (cid:19) , (32d)where S n is defined in Eq. (18a). By using this table, we compute the first product in (30): N n ( λ )ad ˆ T n ( λ ) = 1 λ − λ (cid:18) f n +1 g n +1 (cid:19) (cid:0) − ¯ f n ¯ g n (cid:1) + 1 λ + λ (cid:18) − f n +1 g n +1 (cid:19) (cid:0) ¯ f n ¯ g n (cid:1) + 1 λ − ¯ λ S n (cid:18) ¯ g n +1 − ¯ f n +1 (cid:19) (cid:0) g n f n (cid:1) + 1 λ + ¯ λ S n (cid:18) ¯ g n +1 ¯ f n +1 (cid:19) (cid:0) g n − f n (cid:1) + 4i hλ | λ | (cid:18) S n ∆ n − ¯∆ n (cid:19) . By Lemma 3 and direct computations, we obtain T n ( λ ) (cid:18) f n g n (cid:19) = ( λ − λ ) − a n ¯ λ λ d n − ¯ λ λ ! (cid:18) f n g n (cid:19) , (33a) T n ( λ ) (cid:18) f n − g n (cid:19) = ( λ + λ ) a n ¯ λ λ d n + ¯ λ λ ! (cid:18) f n − g n (cid:19) , (33b) T n ( λ ) (cid:18) ¯ g n − ¯ f n (cid:19) = ( λ − ¯ λ ) (cid:18) − a n λ λ d n − λ λ (cid:19) (cid:18) ¯ g n − ¯ f n (cid:19) , (33c) T n ( λ ) (cid:18) ¯ g n ¯ f n (cid:19) = ( λ + ¯ λ ) (cid:18) a n λ λ d n + λ λ (cid:19) (cid:18) ¯ g n ¯ f n (cid:19) . (33d)By using this table, we compute the second product in (30): T n +1 ( λ ) N n ( λ )ad ˆ T n ( λ ) = 2 (cid:18) − (cid:0) f n +1 ¯ f n − S n ¯ g n +1 g n (cid:1) − a n +1 λ (cid:0) ¯ λ f n +1 ¯ g n + S n λ ¯ g n +1 f n (cid:1) λ (cid:0) ¯ λ g n +1 ¯ f n + S n λ ¯ f n +1 g n (cid:1) d n +1 (cid:0) g n +1 ¯ g n − S n ¯ f n +1 f n (cid:1) (cid:19) + 4i hλ | λ | λ + a n +1 | λ | λ b n +1 c n +1 d n +1 λ + | λ | λ ! (cid:18) S n ∆ n − ¯∆ n (cid:19) . Substituting this expression into (30), we finally obtain N [1] n ( λ ) = (cid:18) δ λ + hλ δ δ δ hλ − δ λ (cid:19) , (34) here δ = ¯ f n f n +1 − S n g n ¯ g n +1 ¯∆ n − h S n ∆ n | λ | ¯∆ n ,δ = − a n +1 S n ∆ n ¯∆ n ,δ = a n +1 ¯ λ f n +1 ¯ g n + S n λ ¯ g n +1 f n ¯∆ n + 2 i b n +1 h | λ | ,δ = − ¯ λ g n +1 ¯ f n + S n λ ¯ f n +1 g n ¯∆ n − c n +1 S n ∆ n h | λ | ¯∆ n ,δ = − d n +1 h | λ | + d n +1 g n +1 ¯ g n − S n ¯ f n +1 f n ¯∆ n . It follows from substitution of (19) and (20) for f n +1 , g n +1 , ¯ f n +1 and ¯ g n +1 that ¯ f n f n +1 − S n g n ¯ g n +1 = ¯∆ n + 2 i S n ∆ n h | λ | and g n +1 ¯ g n − S n ¯ f n +1 f n = − S n ∆ n + 2 i ¯∆ n h | λ | . As a result, we verify that δ = 1 and δ = δ . We represent N [1] n ( λ ) in (34) in the same form as N n ( λ ) in (5a),therefore, we write δ = 1 + i2 hW n − i2 hW n , δ = 2 Y n − i2 hW n , δ = 2 Z n − i2 hW n (35)for some Y n , Z n , and W n . Using Eqs. (23) for a n +1 , b n +1 , and c n +1 and solving Eq. (35) for W n , Y n , and Z n yield W n = 2 i( ¯∆ n ∆ n +1 − S n ¯∆ n +1 ∆ n ) h ( ¯∆ n ∆ n +1 + S n ¯∆ n +1 ∆ n ) , (36a) Y n = − h | λ | ¯∆ n +1 (cid:0) λ S n f n ¯ g n +1 + ¯ λ f n +1 ¯ g n (cid:1) + 2 i( λ − ¯ λ ) ¯∆ n f n +1 ¯ g n +1 h | λ | ( ¯∆ n ∆ n +1 + S n ¯∆ n +1 ∆ n ) , (36b) Z n = − h | λ | ∆ n +1 (cid:0) λ S n ¯ f n +1 g n + ¯ λ ¯ f n g n +1 (cid:1) + 2 i( λ − ¯ λ ) S n ∆ n ¯ f n +1 g n +1 h | λ | ( ¯∆ n ∆ n +1 + S n ¯∆ n +1 ∆ n ) . (36c)Substituting Eqs. (19) and (20) into Eqs. (36b)–(36c) simplifies Y n and Z n to the form: Y n = h | λ | ¯∆ n U n − (cid:0) λ − ¯ λ (cid:1) f n ¯ g n − n U n h (cid:0) λ − ¯ λ (cid:1) ¯ f n g n U n − h | λ | ∆ n + 2i ¯∆ n , (37a) Z n = h | λ | ∆ n ¯ U n − (cid:0) λ − ¯ λ (cid:1) g n ¯ f n + 2i ¯∆ n ¯ U n h (cid:0) ¯ λ − λ (cid:1) f n ¯ g n ¯ U n − h | λ | ¯∆ n − n . (37b)It follows from Eqs. (37) that Y n = ¯ Z n . We have checked with the aid of Wolfram’s MATHEMATICA fromEq. (36a) that W n = Y n Z n is satisfied. As a result, we conclude that N [1] n ( λ ) in (34) is the same as that of N n ( λ ) in (5a) with the correspondence: U [1] n = Y n , U [1] n = Z n = ¯ Y n , and | U [1] n | = W n = | Y n | . Thus,Eq. (6a) follows from the transformation formula (37a). ext, we prove Eq. (31) and derive the transformations for R n and Q n in Eqs. (6b) and (6c). Again, usingLemma 2 and direct computations, we obtain P n ( λ ) (cid:18) f n g n (cid:19) = (cid:18) f n,t g n,t (cid:19) + ( λ − λ ) H ( λ ) (cid:18) f n g n (cid:19) , (38a) P n ( λ ) (cid:18) f n − g n (cid:19) = (cid:18) f n,t − g n,t (cid:19) + ( λ + λ ) H ( λ ) (cid:18) f n − g n (cid:19) , (38b) P n ( λ ) (cid:18) ¯ g n − ¯ f n (cid:19) = (cid:18) ¯ g n,t − ¯ f n,t (cid:19) + M n (cid:18) ¯ g n − ¯ f n (cid:19) + ( λ − ¯ λ ) H ( λ ) (cid:18) ¯ g n − ¯ f n (cid:19) , (38c) P n ( λ ) (cid:18) ¯ g n ¯ f n (cid:19) = (cid:18) ¯ g n,t ¯ f n,t (cid:19) + M n (cid:18) ¯ g n ¯ f n (cid:19) + ( λ + ¯ λ ) H ( λ ) (cid:18) ¯ g n ¯ f n (cid:19) , (38d)where M n is defined in Eq. (18b) and matrices H , , , ( λ ) are given by H ( λ ) = i2 λ + λ R n + λλ Q n ¯ R n + λλ ¯ Q n λ + λ λ λ ! ,H ( λ ) = i2 λ − λ R n − λλ Q n ¯ R n − λλ ¯ Q n λ − λ λ λ ! ,H ( λ ) = i2 λ + ¯ λ R n + λ ¯ λ Q n ¯ R n + λ ¯ λ ¯ Q n λ +¯ λ λ ¯ λ ! ,H ( λ ) = i2 λ − ¯ λ R n − λ ¯ λ Q n ¯ R n − λ ¯ λ ¯ Q n λ − ¯ λ λ ¯ λ ! . Based on the results in Eq. (38), the product in the right-hand side of Eq. (31) can be obtained as (cid:20) ddt T n ( λ ) + T n ( λ ) P n ( λ ) (cid:21) ad ˆ T n ( λ )= 1 λ − λ (cid:20) T n ( λ ) (cid:18) f n g n (cid:19)(cid:21) t (cid:0) − ¯ f n ¯ g n (cid:1) + 1 λ + λ (cid:20) T n ( λ ) (cid:18) f n − g n (cid:19)(cid:21) t (cid:0) ¯ f n ¯ g n (cid:1) + 1 λ − ¯ λ (cid:20) T n ( λ ) (cid:18) ¯ g n − ¯ f n (cid:19)(cid:21) t (cid:0) − g n − f n (cid:1) + 1 λ + ¯ λ (cid:20) T n ( λ ) (cid:18) ¯ g n ¯ f n (cid:19)(cid:21) t (cid:0) g n − f n (cid:1) + T n ( λ ) H ( λ ) (cid:18) f n g n (cid:19) (cid:0) − ¯ f n ¯ g n (cid:1) + T n ( λ ) H ( λ ) (cid:18) f n − g n (cid:19) (cid:0) ¯ f n ¯ g n (cid:1) + T n ( λ ) H ( λ ) (cid:18) ¯ g n − ¯ f n (cid:19) (cid:0) − g n − f n (cid:1) + T n ( λ ) H ( λ ) (cid:18) ¯ g n ¯ f n (cid:19) (cid:0) g n − f n (cid:1) + M n (cid:20) λ − ¯ λ T n ( λ ) (cid:18) ¯ g n − ¯ f n (cid:19) (cid:0) − g n − f n (cid:1) + 1 λ + ¯ λ T n ( λ ) (cid:18) ¯ g n ¯ f n (cid:19) (cid:0) g n − f n (cid:1)(cid:21) . Expanding the above equation and substituting it into (31) gives P [1] n ( λ ) = 1¯∆ n (cid:18) − ¯ λ ¯ f n ( a n f n ) t − λ g n ( a n ¯ g n ) t λ ( f n ¯ g n,t − f n,t ¯ g n ) λ d n (cid:0) ¯ f n g n,t − ¯ f n,t g n (cid:1) λ f n ¯ f n,t + ¯ λ ¯ g n g n,t (cid:19) i2 λ + | λ | a n + b n | λ | (cid:16) ∆ n ¯∆ n ¯ Q n − λ − ¯ λ | λ | ¯∆ n ¯ f n g n (cid:17) − (cid:16) a n λ + λ | λ | (cid:17) Q n − b n λ | λ | λc n + (cid:16) λ + λd n | λ | (cid:17) (cid:16) ∆ n ¯∆ n ¯ Q n − λ − ¯ λ | λ | ¯∆ n g n ¯ f n (cid:17) − λ − ( d n + c n Q n ) | λ | + M n λ ∆ n | g n | λ ¯∆ n f n ¯ g nλ ∆ n ¯ f n g n λ ¯∆ n | f n | ! , (39)where we have used Eq. (23) in obtaining the last term. Thus, P [1] n can be formally written in the form P [1] n ( λ ) = i2 (cid:18) λ − π π π λ − π λ − π λ − π λ − π π − λ − (cid:19) , (40)Comparing Eqs. (39) and (40) and using Eqs. (23) together with (21), we can express π i ’s ( ≤ i ≤ ) as π = − ∆ n R n + (cid:0) λ − ¯ λ (cid:1) f n ¯ g n ¯∆ n , (41a) π = − | λ | ¯∆ n Q n + (cid:0) λ − ¯ λ (cid:1) f n ¯ g n | λ | ∆ n , (41b) π = − ¯∆ n ¯ R n + (cid:0) ¯ λ − λ (cid:1) ¯ f n g n ∆ n , (41c) π = − | λ | ∆ n ¯ Q n + (cid:0) ¯ λ − λ (cid:1) ¯ f n g n | λ | ¯∆ n , (41d)where Wolfram’s MATHEMATICA has been used for simplification. It is obvious from (41) that ¯ π = π and ¯ π = π . As a result, we conclude that P [1] n ( λ ) in (39) is the same as that of P n ( λ ) in (5b) with thecorrespondence: R [1] n = π and Q [1] n = π . Thus, Eqs. (6b)–(6c) follow from the transformation formulas(41a)–(41b). Theorem 1 is proven with the algorithmic computations.3. S OLITON SOLUTIONS ON ZERO BACKGROUND
Here we use the one-fold Darboux transformation of Theorem 1 and construct soliton solutions on zerobackground. Hence we take zero potentials ( U, R, Q ) = (0 , , in the transformation formula (6) and obtain U [1] n = − λ − ¯ λ ) f n ¯ g n λ | f n | + ¯ λ | g n | ) − h | λ | (¯ λ | f n | + λ | g n | ) , (42a) R [1] n = − ( λ − ¯ λ ) f n ¯ g n λ | f n | + ¯ λ | g n | , (42b) Q [1] n = − ( λ − ¯ λ ) f n ¯ g n | λ | (¯ λ | f n | + λ | g n | ) , (42c)where Φ n ( λ ) = ( f n , g n ) T is a nonzero solution of the Lax pair (5) with λ = λ at the zero background. First,we prove that the zero background is linearly stable in the semi-discrete MTM system (1). Next, we constructJost solutions of the Lax pair (5) at the zero background. At last, we obtain and study the exact expressions forone-soliton and two-soliton solutions. .1. Stability of zero background.
Linearization of the semi-discrete MTM system (1) at the zero backgroundis written as the linear system(43) du n dt + q n +1 + q n + 2i h ( r n +1 − r n ) = 0 ,q n +1 − q n + i hu n = 0 ,r n +1 + r n − u n = 0 . Thanks to the linear superposition principle, we use the discrete Fourier transform on the lattice,(44) u n = 12 π Z π − π ˆ u ( θ ) e i nθ dθ, n ∈ Z , invert the second and third equations of the differential-difference system (43), and obtain the following differ-ential equation with parameter θ ∈ ( − π, π ) \{ } :(45) h d ˆ udt = (cid:18) h e i θ + 1 e i θ − − e i θ − e i θ + 1 (cid:19) ˆ u. Separating variables in ˆ u = ˆ u ( θ ) e − i tω ( θ ) yields the dispersion relation for the Fourier mode ˆ u ( θ ) :(46) ω ( θ ) = 1 h sin θ (cid:20)(cid:18) h (cid:19) + (cid:18) h − (cid:19) cos θ (cid:21) , θ ∈ ( − π, π ) \{ } . Since ω ( θ ) ∈ R for every θ ∈ ( − π, π ) \{ } , the zero background is linearly stable. Note however that | ω ( θ ) | →∞ as θ → and θ → ± π . Divergences of the dispersion relation in (46) as θ → and θ → ± π are related toinversion of the second and third difference equations in the linear system (43).3.2. Solutions of the Lax pair (5) at zero background.
Lax pair (5) at the zero background is decoupled intotwo systems which admit the following two linearly independent solutions:(47) [Φ + ( λ )] n ( t ) = α (cid:18) (cid:19) µ n + e i λ t , [Φ − ( λ )] n ( t ) = β (cid:18) (cid:19) µ n − e − i2 λ t , where α, β ∈ C \{ } are parameters and µ ± ( λ ) := 2 i hλ ± λ. We say that Φ( λ ) is the Jost function if λ ∈ C yields either | µ + ( λ ) | = 1 or | µ − ( λ ) | = 1 , in which caseone of the two fundamental solutions in (47) is bounded in the limit | n | → ∞ . Constraints | µ ± ( λ ) | = 1 for λ = | λ | e i θ/ in the polar form are equivalent to the following equation:(48) | λ | ± h sin( θ ) + 4 h | λ | = 1 . Roots of Eq. (48) in the complex plane for λ ∈ C are shown on Fig. 1 for h < (left) and h > (right). Forevery λ on each curve of the Lax spectrum, there exists one Jost function in (47) which remains bounded in thelimit | n | → ∞ . On the other hand, thanks to the time dependence in (47), Jost functions remain bounded alsoin the limit | t | → ∞ if and only if λ ∈ R . No such Jost functions exist for h < as is seen from the left panelof Fig. 1. In other words, all Jost functions diverge exponentially either as t → −∞ or as t → + ∞ if h < . e( λ ) I m ( λ ) −2 −1 0 1 2−2−1012 Re( λ ) I m ( λ ) −2 −1 0 1 2−2−1012 F IGURE
1. Solutions to the transcendental equation (48) in the complex plane for h = 2 (left)and h = 6 (right). Each curve encloses a point λ where either µ + ( λ ) = 0 or µ − ( λ ) = 0 .3.3. One-soliton solutions.
Fix λ ∈ C such that µ ± ( λ ) = 0 and λ / ∈ R . Taking a general solution for Φ( λ ) = ( f, g ) T , we write f and g in the form:(49) f n ( t ) = α e ξ ,n ( t ) , g n ( t ) = β e η ,n ( t ) , where(50) ξ ,n ( t ) = n log (cid:16) λ + 2i hλ (cid:17) + i2 λ t, η ,n ( t ) = n log (cid:16) − λ + 2i hλ (cid:17) − i t λ , and α , β ∈ C \{ } are parameters. Without loss of generality, we set λ = δ e i θ / with some δ > and θ ∈ (0 , π ) . Substituting Eq. (49) into Eqs. (42) yields the exact one-soliton solution in the form: U [1] n = − δ α ¯ β sin θ e i θ / | β | (2 + i hδ e i θ ) e η ,n − ξ ,n + | α | (2 e i θ + i hδ ) e − ¯ η ,n +¯ ξ ,n , (51a) R [1] n = − δ α ¯ β sin θ e i θ / | β | e η ,n − ξ ,n + | α | e − ¯ η ,n +¯ ξ ,n +i θ , (51b) Q [1] n = − α ¯ β sin θ e i θ / δ ( | β | e η ,n − ξ ,n +i θ + | α | e − ¯ η ,n +¯ ξ ,n ) , (51c)where ξ ,n ( t ) − η ,n ( t ) = n log (cid:16) − i hδ e i θ hδ e i θ (cid:17) + i2 (cid:0) δ + 1 δ (cid:1) cos θ t − (cid:0) δ − δ (cid:1) sin θ t. Fig. 2(a)–2(c) presents the one-soliton solutions (51) for α = 1 , β = 1 + i , λ = 2 e π i , and h = 1 .Let us check that the discrete solitons (51) recover solitons of the continuous MTM system (2). In orderto simplify the computations, we set δ = 1 , which corresponds to the case of stationary solitons [6, 22]. Bydefining x n = hn , n ∈ Z and taking the limit h → , we obtain for δ = 1 : U [1] n → U ( x, t ) = − α ¯ β sin θ e i cos θ ( t − x ) | α | e sin θ x +i θ / + | β | e − sin θ x − i θ / , (52a) R [1] n → R ( x, t ) = − α ¯ β sin θ e i cos θ ( t − x ) | α | e sin θ x +i θ / + | β | e − sin θ x − i θ / , (52b) Q [1] n → Q ( x, t ) = − α ¯ β sin θ e i cos θ ( t − x ) | α | e sin θ x − i θ / + | β | e − sin θ x +i θ / , (52c) a) (b) (c) F IGURE
2. An example of the one-soliton solutions (51). The following components areshown: | U n | (left), | R n | (middle), and | Q n | (right).which agree with the MTM solitons in the continuous system (2). Parameters α , β ∈ C \{ } determinetranslations in space and rotation in time, whereas θ ∈ (0 , π ) determines the frequency ω := cos θ ∈ ( − , of the continuous MTM solitons. In the limit ω → ( θ → ), the MTM soliton (52) degenerates to the zerosolution, whereas in the limit ω → − ( θ → π ) and α = β , it becomes the algebraic solitons:(53) U ( x, t ) → U a ( x, t ) = − e − i( t − x ) x − i / . Discrete solitons (51) enjoy the same properties as the continuous solitons. In particular, let us recover thediscrete algebraic soliton for the case α = β and δ = 1 in the limit θ → π . By setting θ = π − ǫ andexpanding to the first order in ǫ , we obtain from (51a) U [1] n = 4( ǫ + O ( ǫ )) e − i t (2 − i h − ǫh + O ( ǫ )) (cid:16) − i h +i ǫ (2+i h ) / O ( ǫ )2+i h +i ǫ (2 − i h ) / O ( ǫ ) (cid:17) n − (2 − i h − ǫ + O ( ǫ )) (cid:16) − i h − i ǫ (2+i h ) / O ( ǫ )2+i h − i ǫ (2 − i h ) / O ( ǫ ) (cid:17) n . This expression yields in the limit ǫ → the discrete algebraic soliton(54) U [1] n → [ U a ] n = − e − i t nh (2 − i h )4+ h −
2i + h (cid:18) h − i h (cid:19) n . If x n = hn , n ∈ Z , the discrete algebraic soliton (54) reduces in the limit h → to the continuous algebraicsoliton (53). Similarly, one can prove that the discrete soliton (51) degenerates to the zero solution in the limit θ → .3.4. Two-soliton solutions.
In order to construct the two-soliton solutions, one needs to use the one-foldDarboux transformation (6) twice. Fix λ , λ ∈ C \{ } such that µ ± ( λ , ) = 0 , λ , / ∈ R , λ = ± λ , and λ = ± ¯ λ . A general solution of the Lax pair (5) with λ = λ and λ = λ at zero background is written in theform(55) [Φ( λ )] n ( t ) = (cid:18) α e ξ ,n ( t ) β e η ,n ( t ) (cid:19) , [Φ( λ )] n ( t ) = (cid:18) α e ξ ,n ( t ) β e η ,n ( t ) (cid:19) , where ξ j,n and η j,n with j = 1 , are given by (50) for λ , , and α , , β , ∈ C \{ } are parameters.By using the one-fold Darboux transformation (6) with zero potentials, λ = λ , and Φ = Φ( λ ) , weobtain the one-soliton solutions ( U [1] n , R [1] n , Q [1] n ) in the form (51). The transformed eigenfunction Φ [1] ( λ ) = T [1] ( λ )Φ( λ ) satisfies the Lax pair (5) with the potentials ( U [1] n , R [1] n , Q [1] n ) and λ = λ . By using the one-foldDarboux transformation (6) with ( U n , R n , Q n ) replaced by ( U [1] n , R [1] n , Q [1] n ) , λ replaced by λ , and Φ( λ ) eplaced by Φ [1] ( λ ) , we obtain the two-soliton solutions ( U [2] n , R [2] n , Q [2] n ) in the explicit form (which is notgiven here).Fig. 3(a)–3(c) shows the two-soliton solutions for α = 1 , β = 1 + i , α = 1 , β = 1 , λ = √ e i π/ , λ = √ e i arctan 22 , and h = 1 . The two-soliton solutions feature elastic collisions of two individual solitons withpreservation of their shapes. Such collisions are very common in integrable equations including the continuousMTM system (3). (a) (b) (c) F IGURE
3. An example of the two-soliton solutions.4. S
OLITON SOLUTIONS ON NONZERO CONSTANT BACKGROUND
Here we use the one-fold Darboux transformation of Theorem 1 and construct soliton solutions on nonzeroconstant background ( U, R, Q ) = ( ρ, ρ, ρ − ) , where ρ > is a real parameter. Similarly to Section 3, we provethat the nonzero constant background is linearly stable in the semi-discrete MTM system (1) for every ρ > ,construct Jost solutions of the Lax pair (5) at nonzero constant background, and then finally obtain the exactexpressions for one-soliton solutions.4.1. Stability of nonzero constant background.
Linearization of the semi-discrete MTM system (1) at thenonzero constant background ( U, R, Q ) = ( ρ, ρ, ρ − ) with ρ > yields the linear system of equations:(56) du n dt + 2 (cid:18) ρ − ρ (cid:19) u n + (cid:18) hρ (cid:19) (cid:18) h r n +1 − ¯ q n +1 (cid:19) − (cid:18) − i hρ (cid:19) (cid:18) h r n + ¯ q n (cid:19) = 0 , (cid:18) hρ (cid:19) ¯ q n +1 − (cid:18) − i hρ (cid:19) ¯ q n + i hu n = 0 , (cid:18) hρ (cid:19) r n +1 + (cid:18) − i hρ (cid:19) r n − u n = 0 . By using the discrete Fourier transform on the lattice (44), we close the linear system (56) at the followingdifferential equation with parameter θ ∈ ( − π, π ) :(57) i h d ˆ udt + h (cid:18) ρ − ρ (cid:19) ˆ u + h θ − hρ sin θ sin θ + hρ cos θ − sin θ + hρ cos θ cos θ − hρ sin θ ! ˆ u = 0 . The dispersion relation following from linear equation (57) is purely real, which implies that the nonzeroconstant background is linearly stable for every ρ > . Note that the linear equation (57) does not reduce toequation (45) in the limit ρ → because the nonzero constant background ( U, R, Q ) = ( ρ, ρ, ρ − ) is singularin this limit, hence the variable q in the linearized system (43) is replaced by ¯ q in the system (56). ote that ( u, v ) = ( ρ, ρ − ) is also the nonzero constant solution of the continuous MTM system (3). How-ever, computations similar to those in (56) and (57) show that the nonzero constant background for any ρ > is modulationally unstable. This is different from the conclusion on the nonzero constant background in thesemi-discrete MTM system (1).4.2. Solutions of the Lax pair (5) at nonzero constant background.
Solving Lax pair (5) with the potentials ( U, R, Q ) = ( ρ, ρ, ρ − ) , we have two linearly independent solutions:(58) [Φ + ( λ )] n ( t ) = α (cid:18) ρ − λ (cid:19) µ n + e i2 (cid:16) ρ − ρ (cid:17) t , [Φ − ( λ )] n ( t ) = β (cid:18) λρ (cid:19) µ n − e i2 (cid:16) λ − λ (cid:17) t , where α, β ∈ C \{ } are parameters and µ + ( λ ) := (cid:18) hλ − λ (cid:19) i hρ − i hρ , µ − ( λ ) := 2 i hλ + λ. Similarly to the case of zero potentials, we say that Φ( λ ) is a Jost function if λ ∈ C yields either | µ + ( λ ) | = 1 or | µ − ( λ ) | = 1 . Interestingly, the constraints | µ ± ( λ ) | = 1 with λ = | λ | e i θ/ yield the same equation (48).Hence, any point on each curve of the Lax spectrum shown on Fig. 1 gives one Jost function in (58) whichremains bounded in the limit | n | → ∞ . The function of Φ + ( λ ) is always bounded in the limit | t | → ∞ since ρ > . On the other hand, Φ − ( λ ) is bounded as | t | → ∞ if and only if λ ∈ R , and no such Jost functionsexist for Φ − ( λ ) if h < .4.3. One-breather solutions.
Fix λ ∈ C such that µ ± ( λ ) = 0 and λ / ∈ R . Let Φ( λ ) = ( f, g ) T be thegeneral solution of Lax pair (5) with ( U, R, Q ) = ( ρ, ρ, ρ − ) and λ = λ . We write f and g in the form(59) f ,n = α ρ e µ ,n ( t ) + β λ e ν ,n ( t ) , g ,n = − α λ e µ ,n ( t ) + β ρ e ν ,n ( t ) , with µ ,n ( t ) = n log "(cid:18) hλ − λ (cid:19) i hρ − i hρ + i2 (cid:18) ρ − ρ (cid:19) t,ν ,n ( t ) = n log (cid:18) λ + 2i hλ (cid:19) + i2 (cid:18) λ − λ (cid:19) t, where α , β ∈ C \{ } are parameters. Substituting Eq. (59) into Eqs. (6), we obtain the one-breather solutionsat nonzero constant background as follows: U [1] n = − | α | ρ ¯ λ h λ ¯ χ e Θ ,n + | β | ρλ h ¯ λ χ e − Θ ,n + ¯ α β | λ | h ρ ( λ − ¯ λ ) e − i Ξ ,n | α | λ h λ ¯ χ e Θ ,n + | β | ¯ λ h ¯ λ χ e − Θ ,n − ¯ α β ρh ρ ( λ − ¯ λ ) e − i Ξ ,n , (60a) R [1] n = − | α | ρ ¯ λ ¯ χ e Θ ,n + | β | ρλ χ e − Θ ,n − ¯ α β | λ | ( λ − ¯ λ ) e − i Ξ ,n | α | λ ¯ χ e Θ ,n + | β | ¯ λ χ e − Θ ,n + ¯ α β ρ ( λ − ¯ λ ) e − i Ξ ,n , (60b) Q [1] n = − | α | ¯ λ χ e Θ ,n + | β | λ ¯ χ e − Θ ,n + α ¯ β ρ ( λ − ¯ λ ) e i Ξ ,n ρ | λ | (cid:2) | α | ¯ λ χ e Θ ,n + | β | λ ¯ χ e − Θ ,n − α ¯ β ρ ( λ − ¯ λ ) e i Ξ ,n (cid:3) , (60c)where Θ ,n = Re( µ ,n − ν ,n ) , Ξ ,n = Im( µ ,n − ν ,n ) ,χ = ρ + λ , ¯ χ = ρ + ¯ λ ,h λ = −
2i + hλ , h ¯ λ = −
2i + h ¯ λ , h ρ = 2i + hρ . ue to the presence of the oscillatory terms e i Ξ ,n and e − i Ξ ,n , solutions (60), in general, exhibit the localizedbreathers which oscillate periodically both in n and t . Fig. 4(a)–4(c) illustrates the one-breather solutions (60)at the constant background for α = 1 , β = 1 + i , ρ = 1 , λ = 2 e i π/ , and h = 3 / . (a) (b) (c) F IGURE
4. An example of the one-breather solutions (60) at the nonzero background.No periodic oscillations occur in the one-breather solutions (60) if and only if Ξ ,n = 0 . In this case,solutions (60) describe one-solitons illustrated on Fig. 5(a)–5(c) for α = 1 , β = 1 + i , ρ = / √ , λ = q e π i , and h = √ . (a) (b) (c) F IGURE
5. An example of the one-breather solutions (60) without periodic oscillations.We show that the one-breather solutions (60) feature no periodic oscillations if the modulus and argument of λ are given by | λ | = 1 ρ r h , arg( λ ) = 12 arccos (cid:18) h − ρ − h ρ (cid:19) (61)in the two regions described byeither h > ρ , ρ < , or < h < ρ , ρ > . (62)Note that the two regions intersect at ρ = 1 , h = 2 , for which | λ | = 1 whereas arg( λ ) is not determined. Infact, we show that arg( λ ) ∈ ( π , π ) . The existence region for non-oscillating one-soliton solutions (60) on the ( h, ρ ) plane is displayed in Fig. 6. h ρ F IGURE
6. Region on the ( h, ρ ) plane given by (62).In order to verify (61), we note that the condition Ξ ,n = 0 is equivalent to the system of two equations ( ρ − ρ − ¯ λ − λ + λ + λ = 0 , ρ h | λ | − h | λ | ρ + h (cid:16) ¯ λ λ + λ ¯ λ (cid:17) (cid:16) − h ρ (cid:17) = 0 , (63)subject to the constraint(64) (cid:18) h | λ | − | λ | (cid:19) (cid:18) − h ρ (cid:19) − ρ (cid:18) ¯ λ λ + λ ¯ λ (cid:19) > . By using the polar form λ = δ e i θ / with δ > and θ ∈ (0 , π ) , we rewrite the constraints (63)–(64) in theform: ( ρ − ρ + (cid:16) δ − δ (cid:17) cos θ = 0 ,δ h ρ + δ (cid:0) h ρ − (cid:1) cos θ − ρ = 0 , (65)subject to the constraint(66) (cid:0) δ h − (cid:1) (cid:0) h ρ − (cid:1) δ h − ρ cos θ > . Let us first assume that δ = 1 , in which case the first equation in (65) gives a unique solution for θ :(67) cos θ = ρ − ρ − δ − − δ . Substituting (67) into the second equation in (65) yields the following equation δ h ρ − δ ( h ρ + 4) + 4 ρ = 0 with two roots δ = ρ and δ h ρ = 4 . Since δ = ρ implies cos θ = − in (67), which is not admissible,we only have one positive root for δ given by(68) δ = √ ρ √ h , which implies(69) cos θ = 2 h − ρ − h ρ hanks to (67). Solutions (68) and (69) are equivalent to (61). The constraint (66) with the solutions (68)–(69)is rewritten in the form (1 − ρ )( h ρ + 4) hρ ( h ρ − > , from which the two regions in (62) follow. In the exceptional case, δ = 1 , we have from the first equation in(65) that ρ = 1 whereas cos θ is not determined. Then, the second equation in (65) implies that h = 2 since cos θ = − is not admissible. The constraint (66) yields cos θ < so that θ ∈ (cid:0) π , π (cid:1) .5. C ONCLUSION
We have derived the one-fold Darboux transformation between solutions of the semi-discrete MTM systemusing the Lax pair and the dressing methods. When one solution of the semi-discrete MTM system is either zeroor nonzero constant, the one-fold Darboux transformation generates one-soliton solution on the zero or nonzeroconstant background respectively. When the one-fold Darboux transformation is used recursively, it also allowsus to construct two-soliton solutions and generally multi-soliton solutions. We have showed that properties ofthe discrete solitons in the semi-discrete MTM system are very similar to properties of the continuous MTMsolitons.Among further problems related to the semi-discrete MTM system, we mention construction of conservedquantities which may clarify orbital stability of the discrete MTM solitons, similar to the work [22]. Anotherdirection is to develop the inverse scattering transform for solutions of the Cauchy problem associated withthe semi-discrete MTM system, similar to the work [21]. Since numerical simulations of the semi-discreteMTM system (1) present serious challenges, it may be interesting to look for another version of the integrablesemi-discretization of the continuous MTM system (3).
Acknowledgement.
The authors thank Leeor Greenblat for collaboration on numerical exploration of thesemi-discrete MTM system during an undergraduate research project. The work of TX was partially supportedby the National Natural Science Foundation of China (No. 11705284) and the program of China Scholar-ship Council (No. 201806445009). TX also appreciates the hospitality of the Department of Mathematics &Statistics at McMaster University during his visit in 2019. The work of DEP is supported by the State taskprogram in the sphere of scientific activity of Ministry of Education and Science of the Russian Federation(Task No. 5.5176.2017/8.9) and from the grant of President of Russian Federation for the leading scientificschools (NSH-2685.2018.5). R
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EPARTMENT OF M ATHEMATICS AND S TATISTICS , M C M ASTER U NIVERSITY , H
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