- Email: [email protected]
Notes on “Soliton solutions by Darboux transformation and some reductions for a new Hamiltonian lattice hierarchy” [Phys. Scr. 82 (2010) 015008]
Results in Physics 6 (2016) 982–984
Contents lists available at ScienceDirect
Results in Physics journal homepage: www.journals.elsevier.com/results-in-physics
Microarticle
Notes on ‘‘Soliton solutions by Darboux transformation and some reductions for a new Hamiltonian lattice hierarchy” [Phys. Scr. 82 (2010) 015008] Xi-Xiang Xu College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao 266590, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 24 September 2016 Received in revised form 8 November 2016 Accepted 11 November 2016 Available online 16 November 2016
We demonstrate that the Darboux transformation in the paper ‘‘Soliton solutions by Darboux transformation and some reductions for a new Hamiltonian lattice hierarchy” [Phys. Scr. 82 (2010) 015008] is incorrect, and establish a correct Darboux transformation. Ó 2016 The Author. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
Keywords: Integrable lattice equation Lax pair Gauge transformation Darboux transformation
It is well-known that the Darboux transformation (DT) provide us a purely algebraic, powerful tool to generate solutions for the soliton equations. It is not only applicable to continuous integrable systems [1,2], but also can be applied to discrete integrable systems [3–5]. In [6] the authors introduce a discrete matrix spectral problem
Ewn ¼ U n ðun ; kÞwn ;
U n ðun ; kÞ ¼
0
rn
sn
k þ pkn
! ð1Þ
ð1Þ U n;t ¼ ðEV ð1Þ n ÞU n U n V n
derive an integrable lattice equation:
8 > < r nt ¼ r n ðpn r n1 sn þ r n snþ1 Þ; snt ¼ sn ðpn þ rn1 sn r n sn1 Þ; > : pnt ¼ pn ðr n1 sn r n snþ1 Þ:
Eqs. (1) and (2) form a Lax pair of Eq. (4). In Ref. [6], for two different values k1 ; k2 , we have two solution T
and the auxiliary spectral problems associated with the spectral problem (1):
wn;t ¼ V nð1Þ ðun ; kÞwn ;
V ð1Þ n ðun ; kÞ ¼
12 k2 þ r n1 sn
rn1 k 1 2 k 2
sn k
rn1 sn
! ; ð2Þ
in
Eqs.
(1)
and
(2),
r n ; sn
and
pn
are
potentials,
r n ¼ rðn; tÞ; sn ¼ sðn; tÞ; pn ¼ pðn; tÞ; un ¼ ðrn ; sn ; pn ÞT . Let f ¼ f ðnÞ be a lattice function. The shift operator E, the inverse of E are defined by
ðEf ÞðnÞ ¼ f ðn þ 1Þ;
ðE1 f ÞðnÞ ¼ f ðn 1Þ:
ð3Þ
The compatibility condition between Eqs. (1) and (2), i.e., the discrete zero curvature equation
ð4Þ
T
wn ¼ ðw1n ; w2n Þ ; /n ¼ ð/1n ; /2n Þ of Eqs. (1) and (2). Let us denote
gi ðnÞ ¼
/2n ðki Þ di w2n ðki Þ /1n ðki Þ di w1n ðki Þ
;
i ¼ 1; 2;
ð5Þ
where di ; i ¼ 1; 2, are properly selected. In Ref. [6] authors introduce a gauge transformation of the spectral problems (1) and (2),
!
~ n ¼ T n un ; u
un ¼
u1n ; Tn ¼ u2n
k þ t 11 ðnÞ
t 12
ðnÞt 21 ðnÞ
k þ t22kðnÞ
where 2 g1 ðnÞ t11 ðnÞ ¼ k1 gg2 ðnÞk ; ðnÞg ðnÞ 1
t21 ðnÞ ¼
2
ðg1 ðnÞg2 ðnÞÞðk22 k21 Þ ; k1 g2 ðnÞk2 g1 ðnÞ
E-mail address: [email protected] http://dx.doi.org/10.1016/j.rinp.2016.11.020 2211-3797/Ó 2016 The Author. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
t 12 ðnÞ ¼ g
k2 k1
1 ðnÞg2 ðnÞ
t22 ðnÞ ¼
;
k1 k2 ðk1 g1 ðnÞk2 g2 ðnÞÞ : k1 g2 ðnÞk2 g1 ðnÞ
!
:
983
X.-X. Xu / Results in Physics 6 (2016) 982–984
In what follows, it is easy to verify that T n , which is presented as the gauge transformation matrix in Ref. [6], is not suitable. Owing to the following expression is presented
ðk k1 Þðk k2 Þðg1 ðnÞðk1 þ kÞ g2 ðk2 þ kÞÞ Det½T n ¼ : kðg1 ðnÞ g2 ðnÞÞ
with
nn ¼
n12
ð0Þ
ð1Þ
;
ð0Þ
n22 k þ n22 =k
n21 ð0Þ
!
ð0Þ
0
ð0Þ
ðiÞ
where n12 ; n21 ; n22 ; i ¼ 0; 1 are all independent of k. Thus we have
Pnþ1 U n ¼ nn Pn :
This implies
Det½T n – ðk k1 Þðk k2 Þ:
ð11Þ
and the Eqs. (4.14) and (4.17) in the Ref. [6]
By comparing the coefficients of ki ; i ¼; 1; 0; 1 , in both sides of Eq. (11), we obtain that
T nþ1 U n T n ¼ Det½TnP n ;
n12 ¼ r n þ Bnþ1 ¼ ~r n ;
ð0Þ
ðT n;t þ T n V ð1Þ n Þ ¼ ðDet½T n ÞQ n
are not true. Hence the proofs in Ref. [6] are incorrect. In what follows, we will briefly give right proof process.It is the key problem to select the suitable gauge transformation matrix [3–5]. Let us consider gauge transformation matrix
Pn ¼
k 2 þ An
Bn k
Cn k
k2 þ Dn
!
:
An ¼ Cn ¼
g1 ðnÞg2 ðnÞðk21 k22 Þ ; k1 g2 ðnÞk2 g1 ðnÞ
ð0Þ
~n : n22 ¼ pn þ rn C nþ1 sn Bn Bn C n þ Dnþ1 Dn ¼ p e n . The proof According to Eqs. (8) and (11), we obtain that nn ¼ U is completed. h
V ð1Þ n in the Eq. (2) under the transformation (9), i.e.
Bn ¼
e ð1Þ ¼ V n
k22 k21 ; k1 g1 ðnÞk2 g2 ðnÞ
1 ðnÞk2 g2 ðnÞÞ Dn ¼ k1 kk21ðkg1 gðnÞk : 2 g ðnÞ 2
1
Det½Pn ¼ ðk2 k21 Þðk2 k22 Þ:
ð6Þ
We introduce the gauge transformation
~ n ¼ Pn w ; w n
12 k2 þ ~r n1~sn
ð7Þ
~sn k
1 2 k 2
!
:
~r n1~sn
gi;t ðnÞ ¼ sn ki þ ðk2 2rn1 sn Þgi ðnÞ rn1 ki ðgi ðnÞÞ2 ; i ¼ 1; 2: Let us denote ðPn;t þ Pn V ð1Þ n ÞPn ¼
g 11 ðk; nÞ g 12 ðk; nÞ : g 21 ðk; nÞ g 22 ðk; nÞ
We get by direct calculation that g 11 ðk; nÞ and g 22 ðk; nÞ are 6thorder polynomial in k; g 12 ðk; nÞ and g 21 ðk; nÞ are 5th-order polynomial in k, and
which transform two eigenvalue problems (1) and (2) into
e ð1Þ w ~n ~ n;t ¼ V w n
g 11 ðki ; nÞ;
with
e n ¼ Pnþ1 U n P1 ; V e ð1Þ ¼ ðPn;t þ Pn V ð1Þ ÞP1 : U n n n n
~r n1 k
Proof. According to the Eqs. (2) and (5), we know
Through a expatiatory direct computation or by a computer algebra system, for example, Mathematica etc. We can obtain that
e nw ~ n; ~n ¼ U Ew
ð1Þ
n22 ¼ 1;
e ð1Þ Proposition 2. The matrix V n defined by (8) has the same form as
where k1 k2 ðk1 g2 ðnÞk2 g1 ðnÞÞ ; k1 g1 ðnÞk2 g2 ðnÞ
ð0Þ
n21 ¼ sn C n ¼ ~sn ;
ð8Þ
g 12 ðki ; nÞ;
g 21 ðki ; nÞ;
g 22 ðki ; nÞ;
i ¼ 1; 2:
are all zero.Then we have ðPn;t þ P:V ð1Þ n ÞPn ¼ Det½Pn gn
e n defined by (8) has the same form as U n Proposition 1. The matrix U in the Eq. (1), in which the old potentials rn ; sn ; pn are mapped into ~n on the basis of new potentials ~r n ; ~sn ; p
8 > < ~r n ¼ r n þ Bnþ1 ; ~sn ¼ sn C n ; > :~ pn ¼ pn þ r n C nþ1 sn Bn Bn C n þ Dnþ1 Dn :
ð9Þ
Proof. Let us set Pn is adjoint matrix of Pn , from Eqs. (1), (2) and (5), we have
gi ½n þ 1 ¼
sn þ pkni gi ½n r n gi ½n
;
i ¼ 1; 2;
ð10Þ
and we obtain that
Pnþ1 U n Pn ¼
f 11 ðk; nÞ f 12 ðk; nÞ : f 21 ðk; nÞ f 22 ðk; nÞ
We find that f 11 ðk; nÞ is a 3th-order polynomial in k; f 12 ðk; nÞ and f 21 ðk; nÞ, are all 4th-order polynomials in k; kf 22 ðk; nÞ is 6th-order polynomial in k, and through a tedious but direct computation or by computer algebra system, we get that f 11 ðk; nÞ ¼ 0; k1 and k2 are all roots of f 12 ðk; nÞ; f 21 ðk; nÞ and kf 22 ðk; nÞ. Thus, we have n
Pnþ1 U n P ¼ Det½Pn nn
with
vn ¼
ð0Þ 2 gð2Þ gð1Þ 11 k þ g11 12 k ð2Þ 2 ð0Þ gð1Þ k g k þ g22 21 22 ð1Þ
ð1Þ
ðiÞ
!
ðiÞ
where g12 ; g21 ; g11 ; g22 ; i; j ¼ 1; 2, are all independent of k. Thus we obtain
Pn;t þ P:V ð1Þ n ¼ gn Pn :
ð12Þ j
By comparing the coefficients of k ; j ¼ 0; 1; 2, in both sides of Eq. (12), we have
1 2
ð2Þ ð0Þ ð0Þ gð2Þ ; g11 ¼ g22 ¼ ðr n þ Bnþ1 Þðsn þ C n Þ 11 ¼ g22 ¼
¼ ~r n~sn1 ;
ð0Þ ð0Þ v12 ¼ ^r n ; v21 ¼ ^sn1 :
b ð1Þ On the basis of the Eqs. (8) and (12), we get that gn ¼ V n . The proof is completed. h The transformations (6) and (9):
~ n ; ~r n ; ~sn Þ ðwn ; r n ; sn Þ > ðw constitute a Darboux transformation of Eq. (4). Based on the Propositions 1 and 2, we can obtain the following theorem.
984
X.-X. Xu / Results in Physics 6 (2016) 982–984
Theorem 1. Every solution ðr n ; sn ÞT of Eq. (4) is mapped into a new T solution ð~r n ; ~sn Þ of Eq. (4) under the transformation (9). Acknowledgement This work is supported by the Nature Science Foundation of Shandong Province of China (Grant No. ZR2014AM001).
References [1] Ma WX. Lett. Math. Phys. 1997;39:33–49. [2] VB Matveev, MA Salle. Darboux Transformations and Solitons. Verlag, Berlin: Springer; 1991. [3] Wu YT, Geng XG. J. Phys. A: Math.Gen. 1998;31. L677-84. [4] Xu XX, Yang HX, Sun YP. Mod. Phys. Lett. B. 2006;20:641–8. [5] Xu XX. Appl. Math. Comput. 2015;251:275–83. [6] Tian SF, Zhang HQ. Phys. Scr. 2010;82:015008.
Contents lists available at ScienceDirect
Results in Physics journal homepage: www.journals.elsevier.com/results-in-physics
Microarticle
Notes on ‘‘Soliton solutions by Darboux transformation and some reductions for a new Hamiltonian lattice hierarchy” [Phys. Scr. 82 (2010) 015008] Xi-Xiang Xu College of Mathematics and Systems Science, Shandong University of Science and Technology, Qingdao 266590, China
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 24 September 2016 Received in revised form 8 November 2016 Accepted 11 November 2016 Available online 16 November 2016
We demonstrate that the Darboux transformation in the paper ‘‘Soliton solutions by Darboux transformation and some reductions for a new Hamiltonian lattice hierarchy” [Phys. Scr. 82 (2010) 015008] is incorrect, and establish a correct Darboux transformation. Ó 2016 The Author. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
Keywords: Integrable lattice equation Lax pair Gauge transformation Darboux transformation
It is well-known that the Darboux transformation (DT) provide us a purely algebraic, powerful tool to generate solutions for the soliton equations. It is not only applicable to continuous integrable systems [1,2], but also can be applied to discrete integrable systems [3–5]. In [6] the authors introduce a discrete matrix spectral problem
Ewn ¼ U n ðun ; kÞwn ;
U n ðun ; kÞ ¼
0
rn
sn
k þ pkn
! ð1Þ
ð1Þ U n;t ¼ ðEV ð1Þ n ÞU n U n V n
derive an integrable lattice equation:
8 > < r nt ¼ r n ðpn r n1 sn þ r n snþ1 Þ; snt ¼ sn ðpn þ rn1 sn r n sn1 Þ; > : pnt ¼ pn ðr n1 sn r n snþ1 Þ:
Eqs. (1) and (2) form a Lax pair of Eq. (4). In Ref. [6], for two different values k1 ; k2 , we have two solution T
and the auxiliary spectral problems associated with the spectral problem (1):
wn;t ¼ V nð1Þ ðun ; kÞwn ;
V ð1Þ n ðun ; kÞ ¼
12 k2 þ r n1 sn
rn1 k 1 2 k 2
sn k
rn1 sn
! ; ð2Þ
in
Eqs.
(1)
and
(2),
r n ; sn
and
pn
are
potentials,
r n ¼ rðn; tÞ; sn ¼ sðn; tÞ; pn ¼ pðn; tÞ; un ¼ ðrn ; sn ; pn ÞT . Let f ¼ f ðnÞ be a lattice function. The shift operator E, the inverse of E are defined by
ðEf ÞðnÞ ¼ f ðn þ 1Þ;
ðE1 f ÞðnÞ ¼ f ðn 1Þ:
ð3Þ
The compatibility condition between Eqs. (1) and (2), i.e., the discrete zero curvature equation
ð4Þ
T
wn ¼ ðw1n ; w2n Þ ; /n ¼ ð/1n ; /2n Þ of Eqs. (1) and (2). Let us denote
gi ðnÞ ¼
/2n ðki Þ di w2n ðki Þ /1n ðki Þ di w1n ðki Þ
;
i ¼ 1; 2;
ð5Þ
where di ; i ¼ 1; 2, are properly selected. In Ref. [6] authors introduce a gauge transformation of the spectral problems (1) and (2),
!
~ n ¼ T n un ; u
un ¼
u1n ; Tn ¼ u2n
k þ t 11 ðnÞ
t 12
ðnÞt 21 ðnÞ
k þ t22kðnÞ
where 2 g1 ðnÞ t11 ðnÞ ¼ k1 gg2 ðnÞk ; ðnÞg ðnÞ 1
t21 ðnÞ ¼
2
ðg1 ðnÞg2 ðnÞÞðk22 k21 Þ ; k1 g2 ðnÞk2 g1 ðnÞ
E-mail address: [email protected] http://dx.doi.org/10.1016/j.rinp.2016.11.020 2211-3797/Ó 2016 The Author. Published by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
t 12 ðnÞ ¼ g
k2 k1
1 ðnÞg2 ðnÞ
t22 ðnÞ ¼
;
k1 k2 ðk1 g1 ðnÞk2 g2 ðnÞÞ : k1 g2 ðnÞk2 g1 ðnÞ
!
:
983
X.-X. Xu / Results in Physics 6 (2016) 982–984
In what follows, it is easy to verify that T n , which is presented as the gauge transformation matrix in Ref. [6], is not suitable. Owing to the following expression is presented
ðk k1 Þðk k2 Þðg1 ðnÞðk1 þ kÞ g2 ðk2 þ kÞÞ Det½T n ¼ : kðg1 ðnÞ g2 ðnÞÞ
with
nn ¼
n12
ð0Þ
ð1Þ
;
ð0Þ
n22 k þ n22 =k
n21 ð0Þ
!
ð0Þ
0
ð0Þ
ðiÞ
where n12 ; n21 ; n22 ; i ¼ 0; 1 are all independent of k. Thus we have
Pnþ1 U n ¼ nn Pn :
This implies
Det½T n – ðk k1 Þðk k2 Þ:
ð11Þ
and the Eqs. (4.14) and (4.17) in the Ref. [6]
By comparing the coefficients of ki ; i ¼; 1; 0; 1 , in both sides of Eq. (11), we obtain that
T nþ1 U n T n ¼ Det½TnP n ;
n12 ¼ r n þ Bnþ1 ¼ ~r n ;
ð0Þ
ðT n;t þ T n V ð1Þ n Þ ¼ ðDet½T n ÞQ n
are not true. Hence the proofs in Ref. [6] are incorrect. In what follows, we will briefly give right proof process.It is the key problem to select the suitable gauge transformation matrix [3–5]. Let us consider gauge transformation matrix
Pn ¼
k 2 þ An
Bn k
Cn k
k2 þ Dn
!
:
An ¼ Cn ¼
g1 ðnÞg2 ðnÞðk21 k22 Þ ; k1 g2 ðnÞk2 g1 ðnÞ
ð0Þ
~n : n22 ¼ pn þ rn C nþ1 sn Bn Bn C n þ Dnþ1 Dn ¼ p e n . The proof According to Eqs. (8) and (11), we obtain that nn ¼ U is completed. h
V ð1Þ n in the Eq. (2) under the transformation (9), i.e.
Bn ¼
e ð1Þ ¼ V n
k22 k21 ; k1 g1 ðnÞk2 g2 ðnÞ
1 ðnÞk2 g2 ðnÞÞ Dn ¼ k1 kk21ðkg1 gðnÞk : 2 g ðnÞ 2
1
Det½Pn ¼ ðk2 k21 Þðk2 k22 Þ:
ð6Þ
We introduce the gauge transformation
~ n ¼ Pn w ; w n
12 k2 þ ~r n1~sn
ð7Þ
~sn k
1 2 k 2
!
:
~r n1~sn
gi;t ðnÞ ¼ sn ki þ ðk2 2rn1 sn Þgi ðnÞ rn1 ki ðgi ðnÞÞ2 ; i ¼ 1; 2: Let us denote ðPn;t þ Pn V ð1Þ n ÞPn ¼
g 11 ðk; nÞ g 12 ðk; nÞ : g 21 ðk; nÞ g 22 ðk; nÞ
We get by direct calculation that g 11 ðk; nÞ and g 22 ðk; nÞ are 6thorder polynomial in k; g 12 ðk; nÞ and g 21 ðk; nÞ are 5th-order polynomial in k, and
which transform two eigenvalue problems (1) and (2) into
e ð1Þ w ~n ~ n;t ¼ V w n
g 11 ðki ; nÞ;
with
e n ¼ Pnþ1 U n P1 ; V e ð1Þ ¼ ðPn;t þ Pn V ð1Þ ÞP1 : U n n n n
~r n1 k
Proof. According to the Eqs. (2) and (5), we know
Through a expatiatory direct computation or by a computer algebra system, for example, Mathematica etc. We can obtain that
e nw ~ n; ~n ¼ U Ew
ð1Þ
n22 ¼ 1;
e ð1Þ Proposition 2. The matrix V n defined by (8) has the same form as
where k1 k2 ðk1 g2 ðnÞk2 g1 ðnÞÞ ; k1 g1 ðnÞk2 g2 ðnÞ
ð0Þ
n21 ¼ sn C n ¼ ~sn ;
ð8Þ
g 12 ðki ; nÞ;
g 21 ðki ; nÞ;
g 22 ðki ; nÞ;
i ¼ 1; 2:
are all zero.Then we have ðPn;t þ P:V ð1Þ n ÞPn ¼ Det½Pn gn
e n defined by (8) has the same form as U n Proposition 1. The matrix U in the Eq. (1), in which the old potentials rn ; sn ; pn are mapped into ~n on the basis of new potentials ~r n ; ~sn ; p
8 > < ~r n ¼ r n þ Bnþ1 ; ~sn ¼ sn C n ; > :~ pn ¼ pn þ r n C nþ1 sn Bn Bn C n þ Dnþ1 Dn :
ð9Þ
Proof. Let us set Pn is adjoint matrix of Pn , from Eqs. (1), (2) and (5), we have
gi ½n þ 1 ¼
sn þ pkni gi ½n r n gi ½n
;
i ¼ 1; 2;
ð10Þ
and we obtain that
Pnþ1 U n Pn ¼
f 11 ðk; nÞ f 12 ðk; nÞ : f 21 ðk; nÞ f 22 ðk; nÞ
We find that f 11 ðk; nÞ is a 3th-order polynomial in k; f 12 ðk; nÞ and f 21 ðk; nÞ, are all 4th-order polynomials in k; kf 22 ðk; nÞ is 6th-order polynomial in k, and through a tedious but direct computation or by computer algebra system, we get that f 11 ðk; nÞ ¼ 0; k1 and k2 are all roots of f 12 ðk; nÞ; f 21 ðk; nÞ and kf 22 ðk; nÞ. Thus, we have n
Pnþ1 U n P ¼ Det½Pn nn
with
vn ¼
ð0Þ 2 gð2Þ gð1Þ 11 k þ g11 12 k ð2Þ 2 ð0Þ gð1Þ k g k þ g22 21 22 ð1Þ
ð1Þ
ðiÞ
!
ðiÞ
where g12 ; g21 ; g11 ; g22 ; i; j ¼ 1; 2, are all independent of k. Thus we obtain
Pn;t þ P:V ð1Þ n ¼ gn Pn :
ð12Þ j
By comparing the coefficients of k ; j ¼ 0; 1; 2, in both sides of Eq. (12), we have
1 2
ð2Þ ð0Þ ð0Þ gð2Þ ; g11 ¼ g22 ¼ ðr n þ Bnþ1 Þðsn þ C n Þ 11 ¼ g22 ¼
¼ ~r n~sn1 ;
ð0Þ ð0Þ v12 ¼ ^r n ; v21 ¼ ^sn1 :
b ð1Þ On the basis of the Eqs. (8) and (12), we get that gn ¼ V n . The proof is completed. h The transformations (6) and (9):
~ n ; ~r n ; ~sn Þ ðwn ; r n ; sn Þ > ðw constitute a Darboux transformation of Eq. (4). Based on the Propositions 1 and 2, we can obtain the following theorem.
984
X.-X. Xu / Results in Physics 6 (2016) 982–984
Theorem 1. Every solution ðr n ; sn ÞT of Eq. (4) is mapped into a new T solution ð~r n ; ~sn Þ of Eq. (4) under the transformation (9). Acknowledgement This work is supported by the Nature Science Foundation of Shandong Province of China (Grant No. ZR2014AM001).
References [1] Ma WX. Lett. Math. Phys. 1997;39:33–49. [2] VB Matveev, MA Salle. Darboux Transformations and Solitons. Verlag, Berlin: Springer; 1991. [3] Wu YT, Geng XG. J. Phys. A: Math.Gen. 1998;31. L677-84. [4] Xu XX, Yang HX, Sun YP. Mod. Phys. Lett. B. 2006;20:641–8. [5] Xu XX. Appl. Math. Comput. 2015;251:275–83. [6] Tian SF, Zhang HQ. Phys. Scr. 2010;82:015008.