Nonlinearization of the Lax pairs for discrete Ablowitz–Ladik hierarchy

J. Math. Anal. Appl. 327 (2007) 829–853 www.elsevier.com/locate/jmaa

Nonlinearization of the Lax pairs for discrete Ablowitz–Ladik hierarchy Xianguo Geng a,∗ , H.H. Dai b a Department of Mathematics, Zhengzhou University, Zhengzhou, Henan 450052, People’s Republic of China b Department of Mathematics, City University of Hong Kong, 83 Tat Chee Avenue,

Kowloon, Hong Kong, People’s Republic of China Received 23 December 2005 Available online 30 May 2006 Submitted by Steven G. Krantz

Abstract The discrete Ablowitz–Ladik hierarchy with four potentials and the Hamiltonian structures are derived. Under a constraint between the potentials and eigenfunctions, the nonlinearization of the Lax pairs associated with the discrete Ablowitz–Ladik hierarchy leads to a new symplectic map and a class of finitedimensional Hamiltonian systems. The generating function of the integrals of motion is presented, by which the symplectic map and these finite-dimensional Hamiltonian systems are further proved to be completely integrable in the Liouville sense. Each member in the discrete Ablowitz–Ladik hierarchy is decomposed into a Hamiltonian system of ordinary differential equations plus the discrete flow generated by the symplectic map. © 2006 Elsevier Inc. All rights reserved. Keywords: Discrete Ablowitz–Ladik hierarchy; Nonlinearization of the Lax pairs

1. Introduction During the recent decades, the study of soliton equations is one of the most prominent events in the field of nonlinear sciences. Fairly satisfied understanding has been got for nonlinear integrable models. Quite a few methods of solution of such soliton equations are well established, such as the inverse scattering transformation, the bilinear transformation methods of Hirota, the * Corresponding author.

E-mail address: [email protected] (X. Geng). 0022-247X/$ – see front matter © 2006 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2006.04.033

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dressing method, the Bäcklund and the Darboux transformation, the algebraic curve method, and so on [1–5]. A systemic approach, so-called nonlinearization of eigenvalue problems or Lax pairs, was developed in Refs. [6–8]. This approach offers an effective way to generate new finite-dimensional integrable Hamiltonian systems from soliton equations. It is always interesting to decompose a nonlinear partial differential equation into a pair of systems of ordinary differential equations both from theoretical point and practical point of view. The nonlinearization approach can decompose soliton equations into temporal and spatial finite-dimensional Hamiltonian systems, which also makes it very natural to compute solutions of soliton equations numerically, for instance, by the symplectic method. Some physically important solutions of soliton equations, including soliton solutions, periodic solutions and quasi-periodic solutions, may be engendered through solving the compatible system of ordinary differential equations [9–12]. Therefore it may give out a direct way to observe evolution behavior of nonlinear phenomena. Subsequently the nonlinearization approach has been applied to discrete soliton equations [13–18], for example, Toda lattice, Kac–van Moerbeke lattice, and others [9–22]. The main aim of the present paper is that the nonlinearization approach is developed and applied to the discrete Ablowitz–Ladik hierarchy with four potentials. We first derive the discrete Ablowitz–Ladik hierarchy with four potentials associated with a spectral problem and their Hamiltonian structures. Then we propose a constraint between the potentials and eigenfunctions. The nonlinearization of the Lax pairs for the discrete Ablowitz–Ladik hierarchy leads to a new integrable symplectic map and a class of finite-dimensional integrable Hamiltonian systems. As an application, solutions of the discrete Ablowitz–Ladik hierarchy are decomposed into solving compatible Hamiltonian systems of ordinary differential equations plus the discrete flow generated by the symplectic map. The outline of the paper is as follows. In Section 2, with the aid of the decomposition of zero-curvature equation, we shall construct the discrete Ablowitz–Ladik hierarchy with four potentials. In Section 3, we derive the Lenard gradients and establish Hamiltonian structures of the discrete Ablowitz–Ladik hierarchy. In Section 4, we introduce the Bargmann constraint between the potentials and eigenfunctions, from which a new symplectic map and a class of finitedimensional Hamiltonian systems are obtained. The generating function approach is used to calculate the involutivity of integrals, by which the symplectic map and these finite-dimensional Hamiltonian systems are further proved to be completely integrable in the Liouville sense. In Section 5, the discrete Ablowitz–Ladik hierarchy is decomposed into the integrable symplectic map and the compatible Hamiltonian system of ordinary differential equations. 2. The hierarchy of differential-difference equations In this section, we shall derive the discrete Ablowitz–Ladik hierarchy associated with a discrete spectral problem with four potentials [23]   λ + Rn Sn Qn + λ−1 Sn 1 , (2.1) χ(n + 1) = Un χ(n), Un = γn πn Rn + λTn λ−1 + Qn Tn where Qn , Rn , Sn and Tn are four potentials, λ is a constant spectral parameter,    T γn = 1 − Qn Rn , πn = 1 − Sn Tn , χ(n) = χ1 (n), χ2 (n) . A number of research on the discrete Ablowitz–Ladik hierarchy has been conducted. For example, its inverse scattering transform, soliton solutions, the Hamiltonian structures, Darboux

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transformation, and others have been discussed in Refs. [23–27]. In what follows, we first give solutions of the stationary discrete zero-curvature equation. Then we construct the discrete Ablowitz–Ladik hierarchy. Proposition 1. Let two matrices Vn and Vˆn satisfy Vˆn Un(1) − Un(1) Vn = 0, where

(z2

= λ)

Un(1) =

1 γn



z zRn

Vn+1 Un(2) − Un(2) Vˆn = 0,

z−1 Qn z−1

 Un(2) =

,

1 πn



(2.2) z−1 Sn z−1

z zTn

 .

(2.3)

Then Vn satisfies the discrete stationary zero-curvature equation Vn+1 Un − Un Vn = 0,

(2.4)

and det Vn and det Vˆn are constants independent of n. (2)

(1)

Proof. Noticing Un = Un Un , we have     Vn+1 Un − Un Vn = Vn+1 Un(2) − Un(2) Vˆn Un(1) + Un(2) Vˆn Un(1) − Un(1) Vn = 0. By (2.2), we arrive at det Vn+1 = det Vˆn and det Vˆn = det Vn , which imply that det Vn and det Vˆn are constants independent of n. The proof is completed. 2 Assume that solutions Vn and Vˆn of (2.2) take form     An Bn an bn ˆ , Vn = . Vn = Cn −An cn −an

(2.5)

Then (2.2) can be written as an − An + Rn bn − λ−1 Qn Cn = 0, bn − λBn + Qn (an + An ) = 0, cn − λ−1 Cn − Rn (an + An ) = 0, An − an + Qn cn − λRn Bn = 0, An+1 − an + Tn Bn+1 − λ

−1

(2.6)

Sn cn = 0,

Bn+1 − λbn + Sn (an + An+1 ) = 0, Cn+1 − λ−1 cn − Tn (an + An+1 ) = 0, an − An+1 − Tn bn λ + Sn Cn+1 = 0.

(2.7)

Let An =

∞ 

An λ−j , (j )

Bn =

j =0

an =

∞  j =0

∞ 

Bn λ−j , (j )

Cn =

j =0

an λ−j , (j )

bn =

∞  j =0

∞ 

Cn λ−j , (j )

j =0

bn λ−j , (j )

cn =

∞  j =0

cn λ−j (j )

(2.8)

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with the condition A2n + Bn Cn = α02 ,

an2 + bn cn = α02 ,

(2.9)

where α0 is a constant. Substituting (2.8) into (2.6), (2.7) and (2.9) and comparing the coefficients of the same power for λ, we obtain A(0) n = α0 ,

Bn(0) = 0,

an(0) = α0 ,

bn(0) = 0,

Cn(0) = 2α0 Tn−1 , cn(0) = 2α0 Rn ,

(2.10)

(j ) (j ) (j ) (j −1) an − An + Rn bn − Qn Cn = 0,  (j ) (j +1) (j ) (j )  bn − Bn + Qn an + An = 0,  (j ) (j ) (j −1) (j )  c n − Cn − Rn an + An = 0, (j ) (j ) (j ) (j +1) An − an + Qn cn − Rn Bn = 0, (j ) (j ) (j ) (j −1) An+1 − an + Tn Bn+1 − Sn cn = 0,

(2.11)

 (j ) (j )  + Sn an + An+1 = 0,  (j ) (j ) (j −1) (j )  − Tn an + An+1 = 0, Cn+1 − cn (j +1)

(j )

Bn+1 − bn (j )

(j +1)

(j )

an − An+1 − Tn bn j 

(j −k)

A(k) n An

+

k=0 j 

j 

(j )

+ Sn Cn+1 = 0, (j −k)

Bn(k) Cn

(2.12)

= 0,

k=0 (j −k)

an(k) an

+

k=0

j 

(j −k)

bn(k) cn

= 0 (j  1).

(2.13)

k=0

(j )

(j )

(j )

(j )

(j )

(j )

Then An , Bn , Cn , an , bn , cn are uniquely determined by the recursion relations (2.10)– (2.13). It is easy to see that A(1) n = −2α0 Qn Tn−1 ,

Bn(1) = 2α0 Qn ,

Cn(1) = 2α0 Rn−1 − 2α0 Tn−1 (Qn Tn−1 + Rn−1 Sn−1 ), A(2) n = −2α0 (Qn Rn−1 + Sn Tn−1 ) + 2α0 Qn Tn−1 (Rn Sn + Qn Tn−1 + Rn−1 Sn−1 ), Bn(2) = 2α0 Sn − 2α0 Qn (Qn Tn−1 + Rn Sn ), an(1) = −2α0 Rn Sn ,

bn(1) = 2α0 Sn ,

cn(1) = 2α0 Tn−1 − 2α0 Rn (Qn Tn−1 + Rn Sn ), an(2) = −2α0 (Qn+1 Rn + Sn Tn−1 ) + 2α0 Rn Sn (Qn+1 Tn + Qn Tn−1 + Rn Sn ), bn(2) = 2α0 Qn+1 − 2α0 Sn (Qn+1 Tn + Rn Sn ).

(2.14)

On the other hand, we consider the following assumption: An =

∞ 

(j ) Aˆ n λj ,

Bn =

j =0

an =

∞  j =0

∞ 

(j ) Bˆ n λj ,

Cn =

j =0 (j )

aˆ n λj ,

bn =

∞  j =0

(j ) bˆn λj ,

∞ 

(j ) Cˆ n λj ,

j =0

cn =

∞  j =0

(j )

cˆn λj

(2.15)

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833

with the condition A2n + Bn Cn = β02 ,

an2 + bn cn = β02 ,

(2.16)

where β0 is a constant. Substituting (2.15) into (2.6), (2.7) and (2.16), we obtain Bˆ n(0) = −2β0 Sn−1 , bˆn(0) = −2β0 Qn ,

Aˆ (0) n = β0 , aˆ n(0) = β0 ,

Cˆ n(0) = 0, cˆn(0) = 0,

(2.17)

(j ) (j ) (j ) (j +1) aˆ n − Aˆ n + Rn bˆn − Qn Cˆ n = 0,   (j ) (j −1) (j ) (j ) + Qn aˆ n + Aˆ n = 0, bˆn − Bˆ n

 (j ) (j ) (j +1) (j )  cˆn − Cˆ n − Rn aˆ n + Aˆ n = 0,

(j ) (j ) (j ) (j −1) Aˆ n − aˆ n + Qn cˆn − Rn Bˆ n = 0, (j ) (j ) (j ) (j +1) ˆ ˆ = 0, An+1 − aˆ n + Tn Bn+1 − Sn cˆn  (j )  (j ) (j −1) (j ) + Sn aˆ n + Aˆ n+1 = 0, Bˆ n+1 − bˆn  (j ) (j ) (j +1) (j )  − Tn aˆ n + Aˆ n+1 = 0, Cˆ n+1 − cˆn

(2.18)

(j ) (j ) (j −1) (j ) + Sn Cˆ n+1 = 0, aˆ n − Aˆ n+1 − Tn bˆn j 

ˆ (j −k) + Aˆ (k) n An

k=0 j 

j 

(2.19)

(j −k) = 0, Bˆ n(k) Cˆ n

k=0 (j −k)

aˆ n(k) aˆ n

+

k=0

j 

(j −k) = 0 (j  1). bˆn(k) cˆn

(2.20)

k=0

Then Aˆ n , Bˆ n , Cˆ n , aˆ n , bˆn , cˆn are uniquely determined by the recursion relations (2.17)– (2.20), and (j )

(j )

(j )

(j )

(j )

(j )

Bˆ n(1) = −2β0 Qn−1 + 2β0 Sn−1 (Rn Sn−1 + Qn−1 Tn−1 ), Aˆ (1) n = −2β0 Rn Sn−1 , Cˆ n(1) = −2β0 Rn , Cˆ n(2) = −2β0 Tn + 2β0 Rn (Rn Sn−1 + Qn Tn ), Aˆ (2) n = −2β0 (Qn−1 Rn + Sn−1 Tn ) + 2β0 Rn Sn−1 (Qn Tn + Rn Sn−1 + Qn−1 Tn−1 ), (1) aˆ n = −2β0 Qn Tn , bˆn(1) = −2β0 Sn−1 + 2β0 Qn (Rn Sn−1 + Qn Tn ), cˆn(1) = −2β0 Tn ,

cˆn(2) = −2β0 Rn+1 + 2β0 Tn (Rn+1 Sn + Qn Tn ),

aˆ n(2) = −2β0 (Qn Rn+1 + Sn−1 Tn ) + 2β0 Qn Tn (Rn+1 Sn + Qn Tn + Rn Sn−1 ). (m)

Proposition 2. Let two matrices Vn

and Vˆn

(1) Untm = Vˆn(m) Un(1) − Un(1) Vn(m) , (m)

which imply that Vn (m)

(2)

be solutions of the equations

(2) (m) Untm = Vn+1 Un(2) − Un(2) Vˆn(m) ,

(2.22)

satisfies the discrete zero-curvature equation

Untm = Vn+1 Un − Un Vn(m) . (1)

(m)

(2.21)

Here Un and Un are defined by (2.3).

(2.23)

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Proof. By using (2.2), a direct calculation shows that   (2) (m) (m) Untm − Vn+1 Un + Un Vn(m) = Untm − Vn+1 Un(2) + Un(2) Vˆn(m) Un(1)  (1)  + Un(2) Untm − Vˆn(m) Un(1) + Un(1) Vn(m) = 0. The proof is completed.

2

and Vˆn of (2.22) take form   ˜ (m)  (m) B˜ n(m) An a˜ n (m) ˆ Vn(m) = , V = (m) (m) (m) n C˜ n −A˜ n cn (m)

(m)

Let Vn

 b˜n(m) (m)

−a˜ n

,

(2.24)

with m 

A˜ (m) n =

(j )

An λm−j +

j =0

B˜ n(m) =

j =0

m 

(j )

Bn λm−j +

j =0

C˜ n(m) =

m−1 

b˜n(m) =

m 

(j )

Cn λm−j +

m 

m 

j =0

j =0

m 

(j )

m−1 

j =0

(j ) Bˆ n λj −m ,

(j ) Cˆ n λj −m ,

(2.25)

j =0

an λm−j +

bn λm−j +

j =0

c˜n(m) =

m−1 

(j )

m−1 

 1 (j ) ˆ (m) , Aˆ n λj −m − A(m) n + An 2

j =0

j =0

a˜ n(m) =

m 

aˆ n λj −m − (j )

 1  (m) an + aˆ n(m) , 2

(j ) bˆn λj −m ,

j =0 (j )

cn λm−j +

m 

cˆn λj −m . (j )

Then (2.22) can be written as   −1 ˜ (m) ˜ (m) γn γn−1 t = a˜ n(m) − A˜ (m) n + Rn bn − λ Qn Cn , m  −1    Qntm + Qn γn γn t = b˜n(m) − λB˜ n(m) + Qn a˜ n(m) + A˜ (m) , n m  −1    Rntm + Rn γn γn t = c˜n(m) − λ−1 C˜ n(m) − Rn a˜ n(m) + A˜ (m) , n m  −1  γn γn t = A˜ (m) ˜ n(m) + Qn c˜n(m) − λRn B˜ n(m) , n −a m  −1  (m) (m) πn πn t = A˜ n+1 − a˜ n(m) + Tn B˜ n+1 − λ−1 Sn c˜n(m) , m     (m) Sntm + Sn πn πn−1 t = B˜ n+1 , − λb˜n(m) + Sn a˜ n(m) + A˜ (m) n+1 m  −1   (m) (m)  Tntm + Tn πn πn t = C˜ n+1 − λ−1 c˜n(m) − Tn a˜ n(m) + A˜ n+1 , m   (m) (m) πn πn−1 t = a˜ n(m) − A˜ n+1 − λTn b˜n(m) + Sn C˜ n+1 , m

that is

(2.26)

j =0

(2.27)

(2.28)

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835

   1 ˆ (m) + Rn bn(m) − Qn Cn(m−1) , γn γn−1 t = an(m) + aˆ n(m) − A(m) n − An m 2  −1   1  ˆ (m) , Qntm + Qn γn γn t = bn(m) − Bˆ n(m−1) + Qn an(m) + aˆ n(m) + A(m) n + An m 2  −1    1 ˆ (m) , Rntm + Rn γn γn t = cˆn(m) − Cn(m−1) − Rn an(m) + aˆ n(m) + A(m) n + An m 2    1 (m) ˆ (m) γn γn−1 t = A(m) ˆ n(m) + Qn cˆn(m) − Rn Bˆ n(m−1) , (2.29) n + An − an − a m 2    1  (m) (m) (m) πn πn−1 t = An+1 + Aˆ n+1 − an(m) − aˆ n(m) + Tn Bn+1 − Sn cn(m−1) , m 2   1  (m) (m) (m)  Sntm + Sn πn πn−1 t = Bn+1 − bˆn(m−1) + Sn an(m) + aˆ n(m) + An+1 + Aˆ n+1 , m 2  −1   1 (m) (m) (m)  (m−1) Tntm + Tn πn πn t = Cˆ n+1 − cn − Tn an(m) + aˆ n(m) + An+1 + Aˆ n+1 , m 2  −1   1  (m) (m) (m) (m) πn πn t = an + aˆ n(m) − An+1 − Aˆ n+1 − Tn bˆn(m−1) + Sn Cˆ n+1 . (2.30) m 2 A direct calculation shows that (2.29) and (2.30) are equivalent to the following differentialdifference equations:   (m−1) ˆ (m) , Qntm = γn2 bn(m) − Bˆ n(m−1) + Qn A(m) n + An + Qn Cn  (m)  2 (m) (m−1) (m) (m−1) Rntm = γn cˆn − Cn , − Rn An + Aˆ n − Rn Bˆ n   (m) Sntm = πn2 Bn+1 − bˆn(m−1) + Sn an(m) + aˆ n(m) + Sn cn(m−1) ,   (m) Tntm = πn2 Cˆ n+1 − cn(m−1) − Tn an(m) + aˆ n(m) − Tn bˆn(m−1) , m  0. (2.31) The first three systems in (2.31) read Qnt0 = (α0 + β0 )Qn , Snt0 = (α0 + β0 )Sn ,

Rnt0 = −(α0 + β0 )Rn , Tnt0 = −(α0 + β0 )Tn ,

(2.32)

Qnt1 = 2(1 − Qn Rn )(α0 Sn + β0 Sn−1 ), Rnt1 = −2(1 − Qn Rn )(β0 Tn + α0 Tn−1 ), Snt1 = 2(1 − Sn Tn )(α0 Qn+1 + β0 Qn ), Tnt1 = −2(1 − Sn Tn )(β0 Rn+1 + α0 Rn ),  Qnt2 = 2α0 (1 − Qn Rn ) Qn+1 − Sn (Qn+1 Tn + Rn Sn + Qn Tn−1 )  + 2β0 (1 − Qn Rn ) Qn−1 − Sn−1 (Qn Tn + Rn Sn−1 + Qn−1 Tn−1 ) ,  Rnt2 = 2α0 (1 − Qn Rn ) −Rn−1 + Tn−1 (Rn Sn + Qn Tn−1 + Rn−1 Sn−1 )  + 2β0 (1 − Qn Rn ) −Rn+1 + Tn (Rn+1 Sn + Qn Tn + Rn Sn−1 ) ,  Snt2 = 2α0 (1 − Sn Tn ) Sn+1 − Qn+1 (Qn+1 Tn + Rn+1 Sn+1 + Rn Sn )  + 2β0 (1 − Sn Tn ) Sn−1 − Qn (Rn+1 Sn + Rn Sn−1 + Qn Tn ) ,  Tnt2 = 2α0 (1 − Sn Tn ) −Tn−1 + Rn (Qn+1 Tn + Qn Tn−1 + Rn Sn )  + 2β0 (1 − Sn Tn ) −Tn+1 + Rn+1 (Qn+1 Tn+1 + Rn+1 Sn + Qn Tn ) .

(2.33)

(2.34)

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In particular, letting Rn = ∓Qn = un , Sn = ∓Tn = vn , and α0 = −β0 = 12 , (2.33) is reduced to the self-dual network [23–28]     unt1 = 1 ± u2n (±vn−1 ∓ vn ), vnt1 = 1 ± vn2 (±un ∓ un+1 ). (2.35) Similarly, we let Rn = ∓Q∗n , Sn = ∓Tn∗ , with α0 = β0 = − 12 i, (2.33) is reduced to a discretized second order in time nonlinear Schrödinger equation [23]       ∗ ∗ , Snt1 = i 1 ± Sn Sn∗ ±Rn+1 (2.36) ± Rn∗ . Rnt1 = i 1 ± Rn Rn∗ ∓Sn∗ ∓ Sn−1 Alternatively, if Qn = yn , Rn = 0, Sn = 1 − xn , Tn = 1, with α0 = −β0 = 12 , we have from (2.33) that xnt1 = xn (yn − yn+1 ),

ynt1 = xn−1 − xn ,

(2.37)

which is the Toda lattice equation. 3. The Lenard gradients and the Hamiltonian structures In this section, we shall derive the Lenard gradients and the Hamiltonian structures of differential-difference equations (2.31). To this end, we introduce a transformation λ−1 Cn + Rn An = γn2 gn(1) ,

λBn − Qn An = γn2 gn(2) ,

λ−1 cn + Tn an = πn2 gn(3) ,

λbn − Sn an = πn2 gn(4) .

(3.1)

Then (2.6) and (2.7) can be written as     Rn Sn an + πn2 gn(4) = λ γn2 Qn gn(1) + γn2 An − an ,   Sn an + πn2 gn(4) = λ γn2 gn(2) − Qn an ,   Rn an + γn2 gn(1) = λ πn2 gn(3) − Tn an ,   an − γn2 An + γn2 Rn gn(2) = λQn πn2 gn(3) − Tn an ,    (2)  2 Tn Qn+1 An+1 + γn+1 gn+1 = λ πn2 an + πn2 Sn gn(3) − An+1 ,   (2) 2 Qn+1 An+1 + γn+1 gn+1 = λ πn2 gn(4) − Sn An+1 ,   2 (1) gn+1 − Rn+1 An+1 , Tn An+1 + πn2 gn(3) = λ γn+1  (1)  2 πn2 an − An+1 − πn2 Tn gn(4) = λSn Rn+1 An+1 − γn+1 gn+1 .

(3.2)

(3.3)

Proposition 3. Under the transformation (3.1), the matrices Vn and Vˆn determined by (2.5) satisfy (2.2) if and if only the function gλ (n) satisfies the discrete Lenard equation Kn gλ (n) = λJn gλ (n)

(3.4)

(1) (2) (3) (4) where gλ (n) = (gn , gn , gn , gn , an , An )T and two matrix ⎞ ⎛ 0 0 0 πn2 Sn 0 ⎜ γ2 0 0 0 Rn 0 ⎟ ⎟ ⎜ n ⎟ ⎜ ⎜ 0 Eγn2 0 0 0 EQn ⎟ ⎟, Kn = ⎜ ⎜ 0 0 πn2 0 0 Tn E ⎟ ⎟ ⎜

⎜ ⎝ Qn 0

−Rn

0

0

−1

0

Sn

−Tn

1

⎟ 1 ⎠

−E

operators

X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

⎛

0

γn2

⎜ 0 0 ⎜ ⎜ ⎜ 0 0 Jn = ⎜ ⎜ Eγ 2 0 ⎜ n ⎜ ⎝ 0 0 0 0 Here the shift operator E

0

0

−Qn

837

⎞

0

πn2 0 −Tn 0 ⎟ ⎟ ⎟ 2 0 −Sn E ⎟ 0 πn ⎟. 0 0 0 −ERn ⎟ ⎟ ⎟ 0 0 0 0 ⎠ 0 0 0 0 is defined as Ef (n) = f (n + 1).

Proof. From (3.2) and (3.3), we have Qn gn(1) − Rn gn(2) − an + An = 0, Sn gn(3) − Tn gn(4) + an − An+1 = 0,

(3.5)

which, together with the second and third expressions of (3.2) and (3.3), imply (3.4). Contrarily, a direct calculation shows that (3.2) and (3.3) hold in view of (3.4). The proof is completed. 2 Substituting (2.8) and (2.15) into (3.1), respectively, we obtain two solutions of (3.4): g˜ λ (n) =

∞ 

g˜ n λ−j , (j )

j =0

gˆ λ (n) =

∞ 

(3.6)

j =0

with the Lenard gradient sequence (j ) ⎞ ⎛ −2 (j −1) γn (Cn + Rn An ) ⎜ γ −2 (B (j +1) − Q A(j ) ) ⎟ ⎜ n n n ⎟ n ⎜ ⎟ ⎜ π −2 (c(j −1) + T a (j ) ) ⎟ n n n ⎜ n ⎟ (j ) g˜ n = ⎜ ⎟, ⎜ π −2 (bn(j +1) − Sn an(j ) ) ⎟ ⎜ n ⎟ ⎜ ⎟ (j ) ⎝ ⎠ an (j )

⎛

(j )

gˆ n λj

An −2 ⎞

Rn γ n ⎜ Q γ −2 ⎟ ⎜ n n ⎟ ⎜ ⎟ ⎜ Tn πn−2 ⎟ (0) ⎜ ⎟, g˜ n = α0 ⎜ −2 ⎟ ⎜ Sn πn ⎟ ⎜ ⎟ ⎝ 1 ⎠ 1 It is easy to calculate that ⎛ ⎞ Tn−1 ⎜ ⎟ Sn ⎜ ⎟ ⎜ ⎟ ⎜ Rn ⎟ ⎟, g˜ n(1) = 2α0 ⎜ ⎜ Q ⎟ n+1 ⎟ ⎜ ⎜ ⎟ ⎝ −Rn Sn ⎠ −Sn Tn−1

⎛

(j +1) (j ) ⎞ γn−2 (Cˆ n + Rn Aˆ n ) ⎜ −2 ˆ (j −1) (j ) ⎟ ⎜ γn (Bn − Qn Aˆ n ) ⎟ ⎜ ⎟ ⎜ −2 (j +1) (j ) ⎟ + Tn aˆ n ) ⎟ ⎜ πn (cˆn (j ) gˆ n = ⎜ ⎟, ⎜ π −2 (bˆ (j −1) − S aˆ (j ) ) ⎟ ⎜ n ⎟ n n n ⎜ ⎟ (j ) ⎝ ⎠ aˆ n (j ) ˆ An

⎛

⎞ Rn γn−2 ⎜ Q γ −2 ⎟ ⎜ n n ⎟ ⎜ ⎟ ⎜ Tn πn−2 ⎟ (0) ⎜ ⎟. gˆ n = −β0 ⎜ −2 ⎟ ⎜ Sn πn ⎟

j  1,

(3.7)

(3.8)

⎜ ⎟ ⎝ −1 ⎠ −1 ⎛

⎜ ⎜ ⎜ ⎜ gˆ n(1) = −2β0 ⎜ ⎜ ⎜ ⎜ ⎝

Tn Sn−1 Rn+1 Qn Qn Tn Rn Sn−1

⎞ ⎟ ⎟ ⎟ ⎟ ⎟. ⎟ ⎟ ⎟ ⎠

(3.9)

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Utilizing the transformation (3.1), (2.5) can be put in the equivalent forms   (2) λ−1 (γn2 gn + Qn An ) An   , Vn gλ (n) = (1) −An λ(γn2 gn − Rn An )   (4) λ−1 (πn2 gn + Sn an ) an   ˆ Vn gλ (n) = . (3) −an λ(πn2 gn − Tn an )

(3.10) (3.11)

Noting (2.9), (2.16) and (3.6), we arrive at the following fact. Proposition 4.     det Vn g˜ λ (n) = det Vˆn g˜ λ (n) = −α02 ,     det Vn gˆ λ (n) = det Vˆn gˆ λ (n) = −β02 .

(3.12)

For the sake of convenience, we introduce an inverse operator by ⎞ ⎛ 0 γn2 0 0 −Qn 0 ⎜ 0 0 πn2 0 −Tn 0 ⎟ ⎟ ⎜ ⎟ ⎜ 2 ⎟ ⎜ 0 0 0 π 0 −S E n n ⎟. ⎜ ¯ Jn = ⎜ 2 0 0 0 0 −ERn ⎟ ⎟ ⎜ Eγn ⎟ ⎜ ⎝ Qn −Rn 0 0 −1 1 ⎠ 0 0 Sn −Tn 1 −E With the help of (3.4) and (3.6), we arrive at (j )

(j +1)

Kn g˜ n = Jn g˜ n (j +1) Kn gˆ n

Jn g˜ n(0) = 0,

(3.13)

Kn gˆ n(0) = 0,

(3.14)

,

J¯n g˜ n(0) = 0,

(3.15)

(j ) = J¯n gˆ n ,

Kn gˆ n(0) = 0.

(3.16)

,

(j ) = Jn gˆ n ,

which are equivalent to (j +1)

Kn g˜ n = J¯n g˜ n (j )

(j +1) Kn gˆ n

Utilizing (2.11) and (2.12), we have 2 (m−1) γn2 bn(m) = Bn(m+1) − 2Qn A(m) , n − Qn Cn 2 ˆ (m−1) , γn2 cˆn(m) = Cˆ n(m+1) + 2Rn Aˆ (m) n − Rn Bn (m)

πn2 Bn+1 = bn(m+1) − 2Sn an(m) − Sn2 cn(m−1) , (m) πn2 Cˆ n+1 = cˆn(m+1) + 2Tn aˆ n(m) − Tn2 bˆn(m−1) .

Resorting to (3.7), (3.15) and (3.16), Eqs. (2.31) can be written as   (Qntm , Rntm , Sntm , Tntm )T = Xm (n) = Jn L g˜ n(m) − gˆ n(m) , m  0,

(3.17)

(3.18)

where L is the projective map f = (f (1) , f (2) , f (3) , f (4) , f (5) , f (6) )T → (f (1) , f (2) , f (3) , f (4) )T , and Jn is a skew-symmetric operator

X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

⎛

0

γn2

0

0

0 0 0

0 0 −πn2

⎜ −γ 2 ⎜ Jn = ⎜ n ⎝ 0

0

839

⎞

0 ⎟ ⎟ ⎟. 2 πn ⎠ 0

To establish Hamiltonian structures of the discrete Ablowitz–Ladik hierarchy (2.31), we need the following quantities, which are easy to calculate (Vn = Vn Un−1 )  −1  (λ + Qn Tn )An − (λTn + Rn )Bn (λ + Rn Sn )Bn − (Qn + λ−1 Sn )An 1 , Vn = γn πn (λTn + Rn )An + (λ−1 + Qn Tn )Cn −(Qn + λ−1 Sn )Cn − (λ + Rn Sn )An    1  ∂Un = 2 2λ−1 An − Rn Bn + λ−2 Qn Cn , tr Vn (3.19) ∂λ γn      1  1 ∂Un ∂Un = 2 λ−1 Cn + Rn An , = 2 (λBn − Qn An ), tr Vn tr Vn (3.20) ∂Qn ∂Rn γn γn   1  ∂Un = 2 2 2λ−1 Rn + Tn (1 + Qn Rn ) An tr Vn ∂Sn γn πn    − Rn (Rn + λTn )Bn + λ−1 λ−1 + Qn Tn Cn ,   1   ∂Un = 2 2 − 2λQn + Sn (1 + Qn Rn ) An tr Vn ∂Tn γn πn    − Qn Qn + λ−1 Sn Cn + λ(λ + Rn Sn )Bn . (3.21) From (2.6), we have an γn2 = (1 + Qn Rn )An − λRn Bn + λ−1 Qn Cn , cn γn2 = λ−1 Cn − λRn2 Bn + 2Rn An , bn γn2 = λBn − λ−1 Q2n Cn − 2Qn An ,

  2λ−1 An − Rn Bn + λ−2 Qn Cn = λ−1 2γn2 An − 2Qn Rn an − Rn bn + Qn cn , which, together with (3.21), (3.19), implies      1  1 ∂Un ∂Un = 2 λ−1 cn + Tn an , = 2 (λbn − Sn an ), tr Vn tr Vn ∂Sn ∂Tn πn πn    λ−1  ∂Un = 2 2γn2 An − 2Qn Rn an − Rn bn + Qn cn . tr Vn ∂λ γn

(3.22)

(3.23)

Using (3.1), (2.8), (2.15), (3.6) and the trace identity [29], we arrive at (δ/δQn , δ/δRn , δ/δSn , δ/δTn )T H˜ j = Lg˜ n , (j )

(j ) (δ/δQn , δ/δRn , δ/δSn , δ/δTn )T Hˆ j = Lgˆ n ,

j  1,

(3.24)

where 1  (j ) (j ) (j ) (j )  2Qn Rn an + Rn bn − Qn cn − 2γn2 An , H˜ j = 2 j γn 1  2 ˆ (j ) (j ) (j ) (j )  2γn An − 2Qn Rn aˆ n − Rn bˆn + Qn cˆn . Hˆ j = 2 j γn

(3.25)

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X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

Therefore, we deduce that the desired Hamiltonian form of the Ablowitz–Ladik hierarchy (3.18) (Qntm , Rntm , Sntm , Tntm )T = Jn (δ/δQn , δ/δRn , δ/δSn , δ/δTn )T Hm ,

m  1,

(3.26)

with Hj = H˜ j − Hˆ j . Specially, (2.33) can be written as the Hamiltonian form (Qnt1 , Rnt1 , Snt1 , Tnt1 )T = Jn (δ/δQn , δ/δRn , δ/δSn , δ/δTn )T H1

(3.27)

with the Hamiltonian H1 = 2α0 (Qn Tn−1 + Rn Sn ) + 2β0 (Qn Tn + Rn Sn−1 ). 4. A new integrable symplectic map For simplicity, in what follows we write f (n) = f, f (n + k) = E k f , and so on. Assume that λj and χ = (pj , qj )T (1  j  N ) are N distinct eigenparameters and the associated eigenfunctions for the spectral problem (2.1). Then N copies of the discrete spectral problem (2.1) can be written as πEpj = zj pˆ j + zj−1 S qˆj ,

πEqj = zj T pˆ j + zj−1 qˆj ,

γ pˆ j = zj pj + zj−1 Qqj ,

γ qˆj = zj Rpj + zj−1 qj ,

1  j  N.

It is easy to calculate that ⎛ ⎞ ⎛ −γ −2 (λ−1 q 2 + Rpj qj ) ⎞ j j δλj /δQ ⎟ 2 −2 ⎜ δλ /δR ⎟ ⎜ γ (λj pj + Qpj qj ) ⎟ ⎜ j ⎟ ⎜ ⎟ ⎜ ⎟=⎜ ⎟ −1 ⎝ δλj /δS ⎠ ⎜ ⎝ −π −2 (λj qˆj2 + T pˆ j qˆj ) ⎠ δλj /δT π −2 (λ pˆ 2 + S pˆ qˆ ) j j

(4.1)

(4.2)

j j

up to a constant factor, which is extended to ⎛ ⎞ 2 −γ −2 (λ−1 j qj + Rpj qj ) ⎜ −2 ⎟ ⎜ γ (λj pj2 + Qpj qj ) ⎟ ⎜ ⎟ ⎜ 2 + T pˆ qˆ ) ⎟ ⎜ −π −2 (λ−1 ⎟ q ˆ j j j j ∇λj = ⎜ ⎟. ⎜ ⎟ 2 −2 ⎜ π (λj pˆ j + S pˆ j qˆj ) ⎟ ⎜ ⎟ ⎝ ⎠ −pˆ j qˆj −pj qj

(4.3)

Noticing the identities Q

δλj δλj −R + pˆ j qˆj − pj qj = 0, δQ δR

S

δλj δλj −T − pˆ j qˆj + Epj qj = 0, δS δT

(4.4)

it is not difficult to verify that K∇λj = λj J¯∇λj .

(4.5)

Consider the Bargmann constraint [6] N  (0)   (0) L g˜ + gˆ = L∇λj , j =1

(4.6)

X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

841

which is equivalent to Q = α1 Λp, p, ˆ p, ˆ S = αˆ 1 Λp,

  R = β1 Λ−1 q, q ,   T = βˆ1 Λ−1 q, ˆ qˆ

(4.7)

with 1 , α0 − β0 − p, q 1 , αˆ 1 = α0 − β0 − p, ˆ q ˆ α1 =

1 , β0 − α0 − p, q 1 βˆ1 = , β0 − α0 − p, ˆ q ˆ

β1 =

(4.8) (4.9)

where ·,· stands for the canonical inner product in RN , Λ = diag(λ1 , . . . , λN ), p = (p1 , . . . , pN )T , q = (q1 , . . . , qN )T , pˆ = (pˆ 1 , . . . , pˆ N )T , qˆ = (qˆ1 , . . . , qˆN )T . Substituting (4.7) into (4.1) yields Ep =  pˆ = 

Z pˆ + αˆ 1 Λp, ˆ pZ ˆ −1 qˆ −1 q, 1 − αˆ 1 βˆ1 Λp, ˆ pΛ ˆ ˆ q ˆ

Zp + α1 Λp, pZ −1 q 1 − α1 β1 Λp, pΛ−1 q, q

,

,

Eq =  qˆ = 

βˆ1 Λ−1 q, ˆ qZ ˆ pˆ + Z −1 qˆ −1 q, 1 − αˆ 1 βˆ1 Λp, ˆ pΛ ˆ ˆ q ˆ

,

β1 Λ−1 q, qZp + Z −1 q

(4.10)

1 − α1 β1 Λp, pΛ−1 q, q

with Z = diag(z1 , . . . , zN ). Equation (4.10) is equivalently written as follows         p pˆ pˆ p E =S , =S , q qˆ qˆ q where

    ξ ξ S: → = η η

1



1 − α˜ 1 β˜1 Λξ, ξ Λ−1 η, η

Zξ + α˜ 1 Λξ, ξ Z −1 η β˜1 Λ−1 η, ηZξ + Z −1 η

(4.11)  (4.12)

with α˜ 1 =

1 , α0 − β0 − ξ, η

β˜1 =

1 . β0 − α0 − ξ, η

Proposition 5. S determined by (4.12) is a symplectic map in (R2N , dp ∧ dq). Proof. Through tedious calculations, we obtain that dξ ∧ dη = dξ ∧ dη.

2

Proposition 6. Let p and q be solutions of (4.10). Then p, q and p, ˆ q ˆ are conserved integrals of the n-flow, and p, q = p, ˆ q. ˆ This implies that αˆ 1 = α1 , βˆ1 = β1 . Proof. Using (4.10) and (4.7), we arrive at     γ 2 p, ˆ q ˆ = α1 + β1 + α1 β1 p, q Λp, p Λ−1 q, q + p, q   = α1−1 + β1−1 + p, q QR + p, q = γ 2 p, q, which gives rise to p, q = p, ˆ q, ˆ αˆ 1 = α1 , βˆ1 = β1 . The latter two expressions of (4.7) can be written as

842

X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

S=

α1 (Λ2 p, p + 2α1 Λp, pΛp, q + α12 Λp, p2 q, q) , 1 − α1 β1 Λp, pΛ−1 q, q

T=

β1 (Λ−2 q, q + 2β1 Λ−1 q, qΛ−1 p, q + β12 Λ−1 q, q2 p, p) . 1 − α1 β1 Λp, pΛ−1 q, q

(4.13)

Therefore, we have γ 2 π 2 (E − 1)p, q    = T Λ2 p, p + 2α1 Λp, pΛp, q + α12 Λp, p2 q, q       2  + S Λ−2 q, q + 2β1 Λ−1 q, q Λ−1 p, q + β12 Λ−1 q, q p, p   + R(1 + ST )Λp, p + Q(1 + ST ) Λ−1 q, q + 2(QR + ST )p, q     = γ 2 ST α −1 + β −1 + QR(1 + ST ) α −1 + β −1 + 2(QR + ST )p, q = 0. The proof is finished.

2

The first two expressions of (4.7) and Eq. (4.13) can be denoted as ⎛ ⎞ Q ⎜R⎟ ⎝ ⎠ = fS (p, q). S T In order to prove the integrability of the symplectic map S, we introduce two functions ⎛ ⎞ [−2α1 Pλ (q, q) + R(1 − 2α1 Pλ (Λp, q))]/(2α1 γ 2 ) ⎜ [2α P (Λ2 p, p) + Q(1 + 2α P (Λp, q))]/(2α γ 2 ) ⎟ 1 λ 1 λ 1 ⎜ ⎟ ⎜ 2) ⎟ ⎜ [−2αˆ 1 Pλ (q, ⎟ ˆ q) ˆ + T (1 − 2 α ˆ P (Λ p, ˆ q))]/(2 ˆ α ˆ π 1 λ 1 ⎟, Gλ = ⎜ ⎜ [2αˆ P (Λ2 p, 2 ˆ p) ˆ + S(1 + 2αˆ 1 Pλ (Λp, ˆ q))]/(2 ˆ αˆ 1 π ) ⎟ 1 λ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ [1 − 2αˆ 1 Pλ (Λp, ˆ q)]/(2 ˆ αˆ 1 ) ⎛

(4.14)

(4.15)

(4.16)

[1 − 2α1 Pλ (Λp, q)]/(2α1 )

⎞ [−2β1 Pλ (Λ−2 q, q) − R(1 + 2β1 Pλ (Λ−1 p, q))]/(2β1 γ 2 ) ⎜ [2β P (p, p) − Q(1 − 2β P (Λ−1 p, q))]/(2β γ 2 ) ⎟ 1 λ 1 λ 1 ⎜ ⎟ ⎜ ⎟ ˆ1 Pλ (Λ−1 p, ˆ1 π 2 ) ⎟ ⎜ [−2βˆ1 Pλ (Λ−2 q, ˆ q) ˆ − T (1 + 2 β ˆ q))]/(2 ˆ β ⎜ ⎟, ˆ Gλ = ⎜ ˆ p) ˆ − S(1 − 2βˆ1 Pλ (Λ−1 p, ˆ q))]/(2 ˆ βˆ1 π 2 ) ⎟ ⎜ [2βˆ1 Pλ (p, ⎟ ⎜ ⎟ ⎝ ⎠ [1 − 2βˆ1 Pλ (Λ−1 p, ˆ q)]/(2 ˆ βˆ1 )

(4.17)

[1 − 2β1 Pλ (Λ−1 p, q)]/(2β1 ) where bilinear functions Pλ (ς, η) and Pλ (ς, η) on RN and their partial-fraction expansions and Laurent expansions are defined as N ∞       ςj ηj Pλ (ς, η) = (λ − Λ)−1 ς, η = = λ−m−1 Λm ς, η , λ − λj j =1

−1   Pλ (ς, η) = λ−1 − Λ−1 ς, η =

m=0

N  j =1

ςj ηj λ−1

− λ−1 j

=

∞  m=0

  λm+1 Λ−m ς, η .

(4.18)

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843

The first expansion and second one in (4.18) are convergent absolutely outside |λ| > max{|λ1 |, . . . , |λN |} and inside |λ| < min{|λ1 |, . . . , |λN |}, respectively. Now we introduce two matrices Vλ and Vλ by   1 Pλ (Λp, p) 2α1 − Pλ (Λp, q) Vλ = , (4.19) −λPλ (q, q) − 2α1 1 + Pλ (Λp, q)  1  −1 λ−1 Pλ (p, p) 2β1 − Pλ (Λ p, q) Vλ = . (4.20) −Pλ (Λ−1 q, q) − 2β1 1 + Pλ (Λ−1 p, q) Proposition 7. The matrices Vλ and Vλ given by (4.19) and (4.20) satisfy Eq. (2.2), respectively. Proof. With the aid of (4.1) and the identities Ep, p + β1−1 S = 0,

Eq, q + α1−1 T = 0,

q, ˆ q ˆ + α1−1 R = 0, p, ˆ p ˆ + β1−1 Q = 0,  l  l−1    l−1  Pλ Λ ς, η = λPλ Λ ς, η − Λ ς, η ,       Pλ Λl ς, η = λ−1 Pλ Λl+1 ς, η − Λl+1 ς, η ,

(4.21) (4.22)

a direct calculation shows that the desired result holds. The proof is completed.

2

Therefore, the generating functions Fλ = det Vλ and Fλ = det Vλ are independent constants of n-flow. This means that Fλ and Fλ are invariant under the action of the symplectic map S. Then we have    Pλ (Λp, p) Pλ (p, q)    + (α0 − β0 )Pλ (Λp, q) − 1 Fλ = λ  Pλ (Λp, q) Pλ (q, q)  4α12 =

∞ 

λ−m−1 Fm ,

m=−1

  Pλ (p, p) Fλ = λ−1  Pλ (p, q) =

∞ 

(4.23)    Pλ (Λ−1 p, q)  1 + (β0 − α0 )Pλ Λ−1 p, q − 2  −1 Pλ (Λ q, q) 4β1

λm+1 Fm ,

(4.24)

m=−1

where 2 1 α0 − β0 − p, q , 4  m    Λj +1 p, p Λm−j p, q   m+1    Fm = p, p ,  Λj +1 p, q Λm−j q, q  + (α0 − β0 ) Λ F−1 = −

m  0,

(4.25)

j =0

2 1 β0 − α0 − p, q , 4  m    Λ−j p, p Λ−m+j −1 p, q   −m−1    Fm = p, p ,  Λ−j p, q Λ−m+j −1 q, q  + (β0 − α0 ) Λ j =0 F−1 = −

m  0.

(4.26)

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In order to prove the involutivity of {Fm } and {Fm }, we introduce the generating function method, which is convenient in a series of later calculations. Denote the variables of Fλ -flow by τλ and Fλ -flow by τ˜λ , respectively. Let us consider Fλ and Fλ as two Hamiltonians in the symplectic space (R2N , dp ∧ dq). A direct calculation gives the canonical equations of the Fλ flow and Fλ -flow:       pk −∂Fλ /∂qk pk d = I ∇k F λ = = W (λ, λk ) , (4.27) dτλ qk ∂Fλ /∂pk qk and d d τ˜λ



pk qk



 = I ∇k F λ =

−∂Fλ /∂qk ∂Fλ /∂pk



 = W(λ, λk )

pk qk

 (4.28)

,

where 1 W (λ, μ) = λ−μ



(λ + μ)[Pλ (Λp, q) −

1 2α1 ]

−(λ + μ)[Pλ (Λp, q) −

2λμPλ (q, q)

1 W(λ, μ) = −1 λ − μ−1  −1 (λ + μ−1 )[Pλ (Λ−1 p, q) − × 2λ−1 Pλ (Λ−1 q, q)

1 2β1 ]



−2λPλ (Λp, p) 1 2α1 ]

,

(4.29)



−2λ−1 μ−1 Pλ (p, p) −(λ−1 + μ−1 )[Pλ (Λ−1 p, q) −

1 2β1 ]

.

(4.30) The Poisson bracket of two smooth functions h1 and h2 in the symplectic space (R2N , dp ∧ dq) is defined as      N   ∂h1 ∂h2 ∂h1 ∂h2 ∂h1 ∂h2 ∂h1 ∂h2 = , − , . − {h1 , h2 } = ∂qj ∂pj ∂pj ∂qj ∂q ∂p ∂p ∂q j =1

Theorem 8. The matrices Vμ and Vμ satisfy the Lax equations alone the τλ -flow and τ˜λ -flow:  d Vμ = W (λ, μ), Vμ , dτλ  d Vμ = W(λ, μ), Vμ , d τ˜λ

(4.31) (4.32)

which imply {Fμ , Fλ } = 0,

{Fμ , Fλ } = 0,

∀λ, μ ∈ C,

(4.33)

{Fj , Fk } = 0,

{Fj , Fk } = 0,

∀j, k  −1.

(4.34)

Proof. Using (4.27) and (4.28), we have d p, q = 0, dτλ

d p, q = 0. d τ˜λ

Notice the identities (4.22) and

(4.35)

X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

845

   1   l Pλ Λ ς, η − Pμ Λl ς, η , μ−λ  −1      l     1 −1 −1 −1 −1 −1 l μ −Λ λ −Λ Pλ Λ ς, η − Pμ Λl ς, η , Λ ς, η = −1 −1 μ −λ (4.36)



 (μ − Λ)−1 (λ − Λ)−1 Λl ς, η =

a direct calculation shows that (4.31) and (4.32) hold. As a consequence of the Lax equations (4.31) and (4.32) we obtain 0=

dFμ d det Vμ = = {Fμ , Fλ }, dτλ dτλ

0=

dFμ d det Vμ = = {Fμ , Fλ }. d τ˜λ d τ˜λ

Substituting the expansions (4.23) and (4.24) into (4.33) gives (4.34) by comparing the same power of λ, μ. 2 From (4.23), it is easy to see that N   11 2 12 21 − Vλ Vλ = Fλ = − Vλ

Γj r(λ) 1 w(λ) =− 2 , − 2 =− 2 λ − λj 4α1 4α1 u(λ) 4α1 u2 (λ) j =1

Vλ12 = Pλ (Λp, p) = where



Γj = λj

n(λ) , u(λ)

(4.37) (4.38)

 N    λk (pj qk − pk qj )2 2 α0 − β0 − p, q pj qj + q, qpj + , λj − λk

(4.39)

k=1 k =j

u(λ) =

N 

(λ − λj ),

w(λ) =

j =1

r(λ) = u(λ)w(λ) =

N 

(λ − λj +N ),

j =1 2N 

(λ − λj ),

n(λ) =

j =1

N −1 

(λ − μj ).

(4.40)

j =1

Here the roots {μj } are defined as elliptic variables. Proposition 9. The elliptic coordinates satisfy the evolution equation along the τλ flow: dμj α1 λn(λ) , =− u(λ)(λ − μj )n (μj ) 2 r(μj ) dτλ 

1  j  N − 1.

(4.41)

Moreover, N −1  j =1

−1−l α1 μN dμj λN −l j  , =− u(λ) 2 r(μj ) dτλ

1  l  N − 1.

Proof. Substituting λ = μj in (4.37), we have  r(μj ) 11 . Vμj = 2α1 u(μj )

(4.42)

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In the second component of (4.31), d 12 2  2λVμ11 Vλ12 − (λ + μ)Vλ11 Vμ12 , Vμ = dτλ λ−μ let μ = μj . After some calculation, we obtain (4.41). By means of the interpolation formula for polynomials, it is easy to see that (4.42) holds. 2 For fixed λ0 , we introduce the quasi-Abel–Jacobi coordinates: ρl =

μ N −1  j  j =1 λ

α1 μN −1−l dμ , √ 2 r(μ)

1  l  N − 1.

(4.43)

0

Then we have dρl λN −l . =− dτλ u(λ)

(4.44)

It is not difficult to verify that the coefficient of the expansion ∞

 1 = ωk λ−k (1 − λ1 λ−1 ) · · · (1 − λN λ−1 )

(4.45)

k=0

can be determined recursively by ω0 = 1,

ω1 = s1 ,

   1 sk + ωk = si ωj , k

(4.46)

i+j =k i,j 1

where sk = λk1 + · · · + λkN . Denote the variable of Fk -flow by τk (k  −1). Noticing the definition of the Poisson bracket and comparing the coefficients of λ−k−1 in the expansion of (4.44) give dρl = {ρl , Fk } = −ωk+1−l , dτk

1  l  N − 1,

with supplementary definition ω−j = 0 (j = 1, 2, . . .). Thus we have ⎛ 1 ω1 ω2 ⎜ 1 ω1 ⎜   ⎜ dρ dρ dρ dρ .. ⎜ = −⎜ = 0, , ,..., . ⎜ dτ−1 dτ0 dτ1 dτN −2 ⎜ ⎝

(4.47)

··· ··· .. . .. .

⎞ ωN −2 ωN −3 ⎟ ⎟ .. ⎟ ⎟ . ⎟, ⎟ ⎟ ω1 ⎠ 1

(4.48)

where ρ = (ρ1 , . . . , ρN −1 )T . Proposition 10. The conserved integrals {F−1 , F0 , . . . , FN −2 } and {F−1 , F0 , . . . , FN −2 } are functionally independent, respectively. Proof. We only prove the linear independence of the differentials dF−1 , dF0 , . . . , dFN −2 . need −2 Suppose N k=−1 κk dFk = 0. We have

X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853 N −2 

0=

κj ω2 (IdFj , Idρl ) =

j =−1

N −2 

κj {ρl , Fj } =

j =−1

N −2 

κj

j =0

dρl , dτj

847

l = 1, . . . , N − 1.

Thus κ0 = · · · = κN −2 = 0 because of (4.48). Thus κ−1 dF−1 = 0, which implies that κ−1 = 0 since dF−1 =

N  1 α0 − β0 − p, q (qj dpj + pj dqj ) = 0 2 j =1

as long as α0 − β0 − p, q = 0. Similarly, we can prove the linear independence of the differentials dF−1 , dF0 , . . . , dFN −2 . 2 The invariance of Fλ and Fλ under the symplectic map S implies the invariance of Fm −2 N −2 and Fm . The integrals {Fk }N k=−1 or {Fk }k=−1 are in involution in pair and functionally independent. Therefore, we have the following assertion [30,31]. Theorem 11. The symplectic map S defined by (4.12) and the finite-dimensional Hamiltonian systems     −∂Fm /∂p d p , 0  m  N − 2, (4.49) = dτm q ∂Fm /∂q and d d τˆm

    −∂Fm /∂p p , = q ∂Fm /∂q

0  m  N − 2,

(4.50)

are completely integrable in the Liouville sense. 5. Decomposition of the discrete Ablowitz–Ladik hierarchy In this section, we shall give a decomposition of the discrete Ablowitz–Ladik hierarchy. With the aid of Proposition 6 and (4.6), we have N 

∇λj = g˜ (0) + gˆ (0) + ε0 σ0 ,

(5.1)

j =1

where ε0 = −α0 − β0 − p, q, σ0 = (0, 0, 0, 0, 1, 1)T . It is easy to see that     ker J¯ = 0 g˜ (0) |α0 =1 : ∀0 , ker K = ˆ0 gˆ (0) |β0 =1 : ∀ˆ0 , 1 (1) 1 (1) g˜ + ker J¯, K −1 J¯σ0 = gˆ + ker K. J¯−1 Kσ0 = 2α0 2β0 Acting with J¯−1 K and K −1 J¯ upon (5.1) k times, respectively, we obtain   N  ε0 g˜ (k) + 1 g˜ (k−1) + · · · + k−1 g˜ (1) + k g˜ (0) , λkj ∇λj = 1 + 2α0 j =1

N  j =1

λ−k j ∇λj

 ε0 gˆ (k) + ˆ1 gˆ (k−1) + · · · + ˆk−1 g˜ (1) + ˆk g˜ (0) , = 1+ 2β0

k  1,

(5.2)



k  1,

(5.3)

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in view of (3.15), (3.16) and (4.5), where j and ˆj are constants. Using (4.16), (4.3) and noting Proposition 6, we have   N ∞ N    λj ∇λj 1 ε0 (0) (0) −k g˜ + Gλ = g˜ + = 1+ λ λkj ∇λj 2α0 α1 λ − λj 2α0 

j =1



= 1+ 

ε0 g˜ (0) + 2α0

ε0 = 1+ + 2α0

∞ 

∞ 

λ−k

k=1

k λ

−k

k=1





∞ 

k=1

j =1

 ε0 1+ g˜ (k) + 1 g˜ (k−1) + · · · + k−1 g˜ (1) + k g˜ (0) 2α0

λ−k g˜ (k) = λ g˜ λ

(5.4)

k=0

with ∞

λ = 1 +

 ε0 + k λ−k . 2α0

(5.5)

k=1

In a similar way, we obtain from (4.17), (4.3) and (5.3) that ˆ λ = ˆλ gˆ λ , G

∞

 ε0 ˆλ = 1 + + ˆk λk . 2β0

(5.6)

k=1

According to Proposition 4 and noticing (5.5), (5.6), we arrive at Fλ = −α02 2λ ,

Fλ = −β02 ˆλ2 .

(5.7)

Let us introduce two sets of integrals {Hk }, {Hk } by 1 H0 = − α0 p, q, 2

Hk = α02 k ,

k  1,

(5.8)

1 H0 = − β0 p, q, Hk = β02 ˆk , 2 which are put in the equivalent forms

k  1,

(5.9)

and

2 Hλ , α0 2 2β0 ˆλ = β0 − α0 + Hλ β0 2α0 λ = α0 − β0 +

(5.10) (5.11)

with the aid of the generating functions Hλ =

∞  k=0

Hk λ−k ,

Hλ =

∞ 

(5.12)

k=0

Using (5.7), (5.10) and (5.11), we have 2  1 2 α0 − β0 + Hλ , −Fλ = 4 α0 which imply

Hk λk .

2  1 2 β0 − α0 + Hλ , −Fλ = 4 β0

(5.13)

X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

1 H0 = − α0 p, q, 2

849

H1 = −α0 α1 F0 ,

Hk+1 = −α0 α1 Fk − α1 α0−1

k 

Hj Hk+1−j ,

k  1,

(5.14)

k  1,

(5.15)

j =1

and 1 H0 = − β0 p, q, 2

H1 = −β0 β1 F0 ,

Hk+1 = −β0 β1 Fk − β1 β0−1

k 

Hj Hk+1−j ,

j =1

in view of (4.23) and (4.24). The involutivity of {Hk } and {Hk } is respectively based on the equalities α2 {Fμ , Fλ } = 0, {Hμ , Hλ } =  0 4 Fλ Fμ β2 {Fμ , Fλ } = 0. {Hμ , Hλ } =  0 4 Fλ Fμ

(5.16)

Denote the variables of the Hλ -flow and Hλ -flow by t˜λ and tˆλ , respectively. By the Leibniz rule of the Poisson bracket we get {h, Fλ } = −2λ {h, Hλ },

{h, Fλ } = −2ˆλ {h, Hλ }

(5.17)

for any smooth function h. Therefore, we have d 1 d =− , 2λ dτλ d t˜λ

d 1 d =− . d tˆλ 2˜λ d τ˜λ

(5.18)

From (4.27) and (4.28), we arrive at       Pλ (Λ2 p, p) + Q( 2α1 1 + Pλ (Λp, q)) Q α1 Λp, pτλ  d = −2 =2 , (5.19) dτλ R β1 Λ−1 q, qτλ  Pλ (q, q) − R( 2α1 1 − Pλ (Λp, q))       Pλ (p, p) − Q( 2β1 1 − Pλ (Λ−1 p, q)) Q α1 Λp, pτ˜λ  d =2 =2 . d τ˜λ R β1 Λ−1 q, qτ˜λ  Pλ (Λ−2 q, q) + R( 2β1 1 + Pλ (Λ−1 p, q)) (5.20) Resorting to the second expression of (4.10) and (4.27), we have γ q = Z qˆ − RZ p, ˆ γp = Z −1 pˆ − QZ −1 q, ˆ 11 −1 ˆ − 2λVˆλ12 (λ − Λ)−1 Λ(qˆ − R p), ˆ γ Zpτλ = Vˆλ (λ − Λ) (λ + Λ)(pˆ − Qq)

(5.21)

γ Zqτλ = 2λVˆλ21 (λ − Λ)−1 Λ(pˆ − Qq) ˆ − Vˆλ11 (λ − Λ)−1 (λ + Λ)(qˆ − R p), ˆ

(5.22)

where 1 , Vˆλ11 = Pλ (Λp, q) − 2α1 Then we obtain by (4.10) that

Vˆλ12 = Pλ (Λp, p),

Vˆλ21 = Pλ (q, q).

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 1 1 Zpτλ + Qτλ Z −1 q + QZ −1 qτλ + (QR)τλ pˆ , pˆ τλ = γ 2γ  1 1 Rτλ Zp + RZpτλ + Z −1 qτλ + (QR)τλ qˆ . qˆτλ = γ 2γ

(5.23)

Using (5.19) and (5.21)–(5.23), a direct calculation gives that  γ2 Sτλ = γ 2 Λp, ˆ pˆ τλ  = 2 (1 + QR)Vˆλ11 + QVˆλ21 + λR Vˆλ12 Pλ (Λ2 p, ˆ p) ˆ 2αˆ 1  − 2λ 2QVˆλ11 + λVˆλ12 + Q2 Vˆλ21 Pλ (Λp, ˆ q) ˆ  1 1 ˆ p ˆ + (1 + QR)Vˆλ11 + 2QVˆλ21 + QRτλ − RQτλ Λp, 2 2  + 2QVˆλ11 + 2λVˆλ12 + Qτλ Λp, ˆ q, ˆ (5.24)    γ2 Tτλ = γ 2 Λ−1 q, ˆ qˆτλ = −2 (1 + QR)Vˆλ11 + QVˆλ21 + λR Vˆλ12 Pλ (q, ˆ q) ˆ ˆ 2β1  ˆ q) ˆ + 2 2R Vˆλ11 + λR 2 Vˆλ12 + Vˆλ21 Pλ (p,    1 1 − (1 + QR)Vˆλ11 + 2QVˆλ21 + QRτλ − RQτλ Λ−1 q, ˆ qˆ 2 2  −1   11 21 ˆ qˆ . (5.25) + 2R Vˆλ + 2Vˆλ + Rτλ Λ p, Noticing (4.22) and (5.19), it is easy to calculate that  1 11 21 12 2 ˆ ˆ ˆ , (1 + QR)Vλ + QVλ + λR Vλ = γ Pλ (Λp, ˆ q) ˆ − 2α1 2QVˆλ11 + λVˆλ12 + Q2 Vˆλ21 = γ 2 Pλ (Λp, ˆ p), ˆ  1 1 1 11 21 2 ˆ ˆ (1 + QR)Vλ + 2QVλ + QRτλ − RQτλ = γ Pλ (Λp, , ˆ q) ˆ − 2 2 2α1 2QVˆλ11 + 2λVˆλ12 + Qτλ = 0, 2R Vˆλ11 + Vˆλ21 + λR 2 Vˆλ12 = λγ 2 Pλ (q, ˆ q), ˆ 11 21 ˆ ˆ 2R Vλ + 2Vλ + Rτλ = 0. Substituting (5.26) into (5.24) and (5.25) yields     Pλ (Λ2 p, ˆ p) ˆ + S( 2α1ˆ + Pλ (Λp, ˆ q)) ˆ S d 1 = −2 , dτλ T ˆ q) ˆ − T ( 2α1ˆ − Pλ (Λp, ˆ q)) ˆ Pλ (q,

(5.26)

(5.27)

1

where α1 = αˆ 1 and β1 = βˆ1 are used. Similarly, we obtain  −1  −1  γ Zpτ˜λ = Vˆ λ11 λ−1 − Λ−1 λ + Λ−1 (pˆ − Qq) ˆ   −1 ˆ 12 −1 −1 −1 − 2λ Vλ λ − Λ (qˆ − R p), ˆ   −1 ˆ γ Zqτ˜λ = 2λ−1 Vˆ λ21 λ−1 − Λ−1 Λ−1 (pˆ − Qq)  −1 −1  −1  11 −1 −1 − Vˆ λ λ − Λ λ + Λ (qˆ − R p) ˆ with

(5.28)

X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

    1 Vˆ λ11 = Pλ Λ−1 p, q − , Vˆ λ12 = Pλ (p, p), Vˆ λ21 = Pλ Λ−1 q, q , 2β1  γ2 Sτ˜λ = 2 (1 + QR)Vˆ λ11 + R Vˆ λ12 + λ−1 QVˆ λ21 Pλ (p, ˆ p) ˆ 2αˆ 1  − 2 2QVˆ λ11 + Vˆ λ12 + λ−1 Q2 Vˆ λ21 Pλ (p, ˆ q) ˆ  1 1 ˆ p ˆ + (1 + QR)Vˆ λ11 + 2R Vˆλ12 + QRτ˜λ − RQτ˜λ Λp, 2 2  + Qτ˜ − 2QVˆ λ11 − 2Vˆ λ12 Λp, ˆ q, ˆ λ

γ2 2βˆ1

(5.29)

   Tτ˜λ = −2 (1 + QR)Vˆ λ11 + R Vˆ λ12 + λ−1 QVˆ λ21 Pλ Λ−2 q, ˆ qˆ    ˆ qˆ + 2 2R Vˆ λ11 + R 2 Vˆλ12 + λ−1 Vˆ λ21 Pλ Λ−2 p,    1 1 − (1 + QR)Vˆ λ11 + 2R Vˆ λ12 + QRτ˜λ − RQτ˜λ Λ−1 q, ˆ qˆ 2 2  −1   11 2 ˆ 12 ˆ ˆ qˆ , + 2R Vλ + 2R Vλ + Rτ˜ Λ p, λ

and

851

   1 , ˆ qˆ − (1 + QR)Vˆ λ11 + R Vˆ λ12 + λ−1 QVˆ λ21 = γ 2 Pλ Λ−1 p, 2β1 2QVˆ λ11 + Vˆ λ12 + λ−1 Q2 Vˆ λ21 = λ−1 γ 2 Pλ (p, ˆ p), ˆ    1 1 1 11 12 2 ˆ ˆ , (1 + QR)Vλ + 2R Vλ + QRτ˜λ − RQτ˜λ = γ Pλ Λ−1 p, ˆ qˆ − 2 2 2β1 Qτ˜λ − 2QVˆ λ11 − 2Vˆ λ12 = 0,   2R Vˆ λ11 + R 2 Vˆλ12 + λ−1 Vˆ λ21 = γ 2 Pλ Λ−1 q, ˆ qˆ ,   2R Vˆ λ11 + 2R 2 Vˆ λ12 + Rτ˜λ = 2γ 2 P Λ−1 q, ˆ qˆ .

Therefore, we arrive at     Pλ (p, ˆ p) ˆ − S( 1ˆ − Pλ (Λ−1 p, ˆ q)) ˆ S d 2β1 =2 . d τ˜λ T Pλ (Λ−2 q, ˆ q) ˆ + T ( 1 + Pλ (Λ−1 p, ˆ q)) ˆ

(5.30)

(5.31)

(5.32)

2βˆ1

From (5.19), (5.20), (5.27) and (5.32), we have ⎛ ⎞ ⎛ ⎞ Q Q d ⎜R⎟ d ⎜R⎟ ˆ λ. ⎝ ⎠ = −2J LGλ , ⎝ ⎠ = 2J LG dτλ S d τ˜λ S T T Substituting (5.4) and (5.6) into (5.33) yields ⎛ ⎞ Q ∞  d ⎜R⎟ = J L g ˜ = J L g˜ (j ) λ−j , ⎝ ⎠ λ d t˜λ S j =0 T

(5.33)

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X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

⎛

⎞ Q ∞  d ⎜R⎟ = −J L g ˆ = −J L gˆ (j ) λj (5.34) ⎝ ⎠ λ d tˆλ S j =0 T in view of (5.18) and (3.6). Assume that t˜j and tˆj are the variables of Hj -flow and Hj -flow, respectively. Therefore we get from (5.34) that ⎛ ⎞ ⎛ ⎞ Q Q d ⎜R⎟ d ⎜R⎟ (5.35) ⎝ ⎠ = J Lg˜ j , ⎝ ⎠ = −J Lgˆ j , j  0. d t˜j S d tˆj S T T Theorem 12. Let (p(n, tm ), q(n, tm ))T be a compatible solution of the discrete S 2 -flow and H¯ m -flow     p −∂ H¯ m /∂p d , m  0, = dtm q ∂ H¯ m /∂q where H¯ m = Hm + Hm , and tm stands for the variable of H¯ m -flow. Then ⎛ ⎞ Q   ⎜R⎟ ⎝ ⎠ = fS p(n, tm ), q(n, tm ) S T determined by (4.15) solves the mth discrete Ablowitz–Ladik equation (3.18) ⎛ ⎞ Q ∂ ⎜R⎟ ⎝ ⎠ = Xm , m  0. ∂tm S T

(4.10)

(5.36)

(5.37)

(5.38)

According to Theorem 12, let (p 0 (tm ), q 0 (tm ))T be the solution of the initial-value problem ⎧     p −∂ H¯ m /∂p d ⎪ ⎪ , m  0, = ⎪ ⎨ dtm q ∂ H¯ /∂q    0 m (5.39) p p ⎪ ⎪ ⎪ . = ⎩ q tm =0 q0 Then the algorithm

⎞ Qn (tm )  0    ⎟ p (tm ) S 2n p(n, tm ) fS ⎜ ⎜ Rn (tm ) ⎟ −→ −→ ⎟ ⎜ ⎝ Sn (tm ) ⎠ q 0 (tm ) q(n, tm ) Tn (tm ) yields a solution to the mth discrete Ablowitz–Ladik equation (3.18). ⎛

Acknowledgments The work described in this paper was supported by a grant from City University of Hong Kong (Project No. 7001645). One of the authors (X.G.G.) acknowledges the support by National Natural Science Foundation of China (Project 10471132) and the Special Funds for Chinese Major State Basic Research Project “Nonlinear Science.”

X. Geng, H.H. Dai / J. Math. Anal. Appl. 327 (2007) 829–853

853

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