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Hamiltonian forms of the two new integrable systems and two kinds of Darboux transformations
Applied Mathematics and Computation 244 (2014) 261–273
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Hamiltonian forms of the two new integrable systems and two kinds of Darboux transformations Baiying He, Liangyun Chen ⇑ School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China
a r t i c l e
i n f o
a b s t r a c t g a new Lax Based on the Lie algebra glð2Þ, taking a kind of corresponding loop algebra glð2Þ, integrable hierarchy can be obtained. Then, by means of the quadratic-form identity, the corresponding bi-Hamiltonian structure was worked out. Expanding Lie algebra glð2Þ, and making use of the new zero curvature equation Zhang (2008) [9], we obtain an integrable hierarchy and its Hamiltonian structure. At last, two kinds of Darboux transformations of the equation are generated. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Matrix Lie algebra Quadratic-form identity Bi-Hamiltonian structure Darboux transformation
1. Introduction By using some matrix Lie algebras, Guizhang [1] once introduced a powerful tool for generating integrable Hamiltonian systems [3–6], which was called Tu scheme. By employing the corresponding loop algebras, some interesting soliton hierarchies of evolution equations were worked out. However, we have found that some solitary hierarchy can be obtained by a kind of vector loop algebra [8]. P P ~ Let G be a s-dimensional Lie algebra with basis e1 ; e2 ; . . . ; es . Take a ¼ sk¼1 ak ek ; b ¼ sk¼1 bk ek 2 G. The loop algebra G generated by G has the basis ek ðmÞ ¼ ek km ; 1 6 k 6 s; m 2 Z, the commuting operations read ½ek ðmÞ; ej ðnÞ ¼ ½ek ; ej kmþn . ~ is given by The column vector form of G
~ ¼ a ¼ ða1 ; . . . ; as Þ; G
ak ¼
X ak;m km ;
½a; b ¼ c ¼ ðc1 ; . . . ; cs ÞT :
m
~ is as follows The linear isospectral problem established by G
w@ ¼ ½U; w; wt ¼ ½V; w;
kt ¼ 0;
ð1Þ
P where @ ¼ nk¼1 ak @x@ , ak are arbitrary constants, w@ denotes the derivative sum of w with aspect to xk ; k ¼ 1; 2; . . . ; n. k The compatibility condition of (1) is the zero curvature equation
U t V @ þ ½U; V ¼ 0; its stationary zero curvature equation reads
V @ ¼ ½U; V: ⇑ Corresponding author. E-mail address: [email protected] (L. Chen). http://dx.doi.org/10.1016/j.amc.2014.07.006 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
ð2Þ
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
~ u ¼ ðu1 ; . . . ; up ÞT , assume rankU 0 ¼ rankðui U i Þ ¼ a; 1 6 i 6 p, then U is called Take U ¼ Uðk; uÞ ¼ U 0 þ Rpi¼1 ui U i ; U i 2 G; the same-rank, denote by
@ ¼ a; rankðUÞ ¼ rankð@Þ ¼ rank @xk
1 6 k 6 n:
ð3Þ
Let two same-rank solutions V 1 and V 2 satisfy the relation V 1 ¼ cV 2 ; c ¼ constant, then we see that T ~ the constant matrix Theorem 1. Let (3) hold. Two same-rank solutions of (2) possess V 1 ¼ cV 2 . Set ½a; b ¼ aT RðbÞ; a; b 2 G, F ¼ ðfij Þss meets T
F ¼ FT ;
RðbÞF ¼ ðRðbÞFÞ :
ð4Þ
Define a quadratic functional as follows
fa; bg ¼ aT Fb;
~ 8a; b 2 G;
then the following identity holds
d @ @U ; kc V; fV; U k g ¼ kc dui @k @ui
i ¼ 1; . . . ; p;
ð5Þ
where c is a constant to be determined, V is a same-rank solution of 2. 5 is called the quadratic-form identity. Zhang and Fan [2] proposed the Lie algebra glð2Þ ¼ spanfe1 ; e2 ; e3 g, where
e1 ¼
1
0
0 1
;
e2 ¼
0
0
1 0
;
e3 ¼
1 2 0
1
;
M1 ¼
1 0 1 1
;
along with the commutative relation
½e1 ; e2 ¼ 2e2 ; ½e1 ; e3 ¼ 2e2 þ 2e3 ; ½e2 ; e3 ¼ 2e1 ; ½ei ; ej ¼ ei M 1 ej ej M 1 ei ; ei ; ej 2 glð2Þ:
g a new Lax In this paper, we consider the known Lie algebra glð2Þ, and take a kind of corresponding loop algebra glð2Þ, integrable hierarchy can be obtained. Then, by means of the quadratic-form identity, the corresponding bi-Hamiltonian structure was worked out. In Section 3, we want to expand the Lie algebra glð2Þ into the Lie algebra G1 , and take a kind of loop algebra G~1 . Next, we start from an isospectral problem to obtain the integrable hierarchy (26), by making use of the zero curvature equation [9]. In addition, by using the quadratic-form identity, we can obtain a Hamiltonian structure. Study of the algebraic properties of the equations is an important aspect in soliton theory. Some ways for generating Darboux transformations of nonlinear soliton equations by starting from isospectral problems [10]. In Section 4, when n ¼ 2; b ¼ 2, the integrable hierarchy (26) can be reduced to a new Eq. (31). In [11] the Darboux transformations for a Lax pair integrable systems are investigated in detail. Based on this, we obtain two kinds of Darboux transformations of the new Eq. (31). 2. A new integrable hierarchy and its bi-Hamiltonian structure g Based on the known Lie algebra glð2Þ, we construct the following loop algebra glð2Þ:
8 ek ði; nÞ ¼ ek k2nþi ; k—spectral parameter; > > > > > ½e1 ði; mÞ; e2 ðj; nÞ ¼ 2e2 ðdij ; m þ n þ qij Þ; > > > > > > ½e1 ði; mÞ; e3 ðj; nÞ ¼ 2e2 ðdij ; m þ n þ qij Þ þ 2e3 ðdij ; m þ n þ qij Þ; > > > > > ½e < 2 ði; mÞ; e3 ðj; nÞ ¼ 2e1 ðdij ; m þ n þ qij Þ; i þ j; i þ j < 2; > > d ¼ i;j > > 0; i þ j ¼ 2; > > > > > 0; i þ j < 2; > > > q ¼ > > i;j 1; i þ j ¼ 2; > > : degðek ði; nÞÞ ¼ 2n þ i; k ¼ 1; 2; 3; i; j 2 f0; 1g: Consider an isospectral problem
ux ¼ U u; ut ¼ V u; kt ¼ 0;
where
U ¼ e1 ð1; 0Þ þ u1 e1 ð1; 1Þ þ u2 e2 ð0; 0Þ þ u3 e2 ð1; 1Þ þ u4 e3 ð0; 0Þ þ u5 e3 ð1; 1Þ
ð6Þ
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
and
V¼
X a0m e1 ð0; mÞ þ b0m e2 ð0; mÞ þ c0m e3 ð0; mÞ þa1m e1 ð1; mÞ þ b1m e2 ð1; mÞ þ c1m e3 ð1; mÞ
mP0
:
ð7Þ
The stationary zero curvature representation
V x ¼ ½U; V; gives
8 a0mx ¼ 2u2 c0m þ 2u3 c1m 2u4 b0m 2u5 b1m ; > > > > > a1mþ1x ¼ 2u2 c1mþ1 þ 2u3 c0m 2u4 b1mþ1 2u5 b0m ; > > > < b0mx ¼ 2b1mþ1 þ 2c1mþ1 2u1 b1m 2u1 c1m þ 2u2 a0m þ 2u3 a1m þ 2u4 a0m þ 2u5 a1m ; > b1mþ1x ¼ 2b0mþ1 þ 2c0mþ1 2u1 b0m 2u1 c0m þ 2u2 a1mþ1 þ 2u3 a0m þ 2u4 a1mþ1 þ 2u5 a0m ; > > > > > c > 0mx ¼ 2c1mþ1 þ 2u1 c1m 2u4 a0m 2u5 a1m ; > : c1mþ1x ¼ 2c0mþ1 þ 2u1 c0m 2u4 a1mþ1 2u5 a0m :
ð8Þ
Take the initial data as
a00 ¼ a;
b00 ¼ c00 ¼ a10 ¼ b10 ¼ c10 ¼ 0:
ð9Þ
From the recursion relation (8), we have
b11 ¼ au2 ;
a11 ¼ 0;
c11 ¼ au4 ;
a
b01 ¼ ðu2x þ u4x Þ au3 2
ð10Þ
and
c01 ¼
a 2
a
a01 ¼ au2 u4 þ u24 : 2
u4x au5 ;
ð11Þ
Denoting
8 ðnÞ n > < V þ ¼ Rm¼0 ða0m e1 ð0; n mÞ þ b0m e2 ð0; n mÞ þ c0m e3 ð0; n mÞ þa1m e1 ð1; n mÞ þ b1m e2 ð1; n mÞ þ c1m e3 ð1; n mÞÞ; > : ðnÞ V þ þ V ðnÞ ¼ k2n V: The stationary zero curvature representation V x ¼ ½U; V can be expressed as ðnÞ ðnÞ ðnÞ V ðnÞ þx þ ½U; V þ ¼ V x ½U; V :
A direct calculation shows ðnÞ V ðnÞ þx þ ½U; V þ ¼ ða1nþ1x 2u2 c 1nþ1 þ 2u4 b1nþ1 Þe1 ð1; 1Þ þ ð2b1nþ1 2c 1nþ1 Þe2 ð0; 0Þ þ ðb1nþ1x 2b0nþ1 2c0nþ1
2u2 a1nþ1 2u4 a1nþ1 Þe2 ð1; 1Þ þ ð2c1nþ1 Þe3 ð0; 0Þ þ ðc1nþ1x þ 2c0nþ1 þ 2u4 a1nþ1 Þe3 ð1; 1Þ: Taking V
Ut
ðnÞ
¼
V ðnÞ x
V ðnÞ þ ,
the zero curvature equation
þ ½U; V ðnÞ ¼ 0;
leads to the following Lax integrable hierarchy 1 0 1 0 u1 @ a1nþ1x þ 2u2 c1nþ1 2u4 b1nþ1 C B B C B 0 2b1nþ1 þ 2c1nþ1 C B B u2 C B C B B C B C B B C B ut ¼ B u3 C ¼ B b1nþ1x þ 2b0nþ1 þ 2c0nþ1 þ 2u2 a1nþ1 þ 2u4 a1nþ1 C ¼ B 2u2 þ 2u4 C B B C B C B B C B 0 2c1nþ1 A @ @ u4 A @ 0
u5
c1nþ1x 2c0nþ1 2u4 a1nþ1
t
2u4
1 0 1 0 u1 0 2u3 c0n 2u5 b0n C B B C B 0 2b1nþ1 þ 2c1nþ1 C B B u2 C B C B B C B C B B C B ¼ J1 P 1nþ1 ¼ B u3 C ¼ B 2u3 a0n þ 2u5 a0n 2u1 b0n 2u1 c0n C ¼ B 2u3 þ 2u5 C B B C B C B B C B 0 2c1nþ1 A @ @ u4 A @ 0
u5
2u5 a0n þ 2u1 c0n
t
2u5
0 2u2 2u4 0
0
0
@
0
2
2
@
0 2u3 2u5 0
0
0
0
2
0
0
2u1
1 a1nþ1 CB C 0 2 CB c0nþ1 C CB C CB C 2 @ CB c1nþ1 C CB C CB C 0 0 A@ b0nþ1 c0nþ1 A 0 2u4
0
10
0
b1nþ1 c1nþ1
1 a0n CB C 2 0 CB c1nþ1 C CB C CB C 0 2u1 CB c0n C ¼ J 2 Q 1nþ1 ; CB C CB C 0 0 A@ b1nþ1 c1nþ1 A 0 2u5
0
0
10
b0n c0n ð12Þ
where J 1 and J 2 satisfy with
J i
¼ J i ; i ¼ 1; 2. So J 1 and J 2 are the Hamiltonian operators.
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
From (8), we can obtain a recurrence operator
0
0 B B l21 B L1 ¼ B B 0 B @ 0
l12 l22
l51
l13 l23
l14 l24
1
0
0
0
0
1
l52
l53
l54
1 l15 C l25 C C 0 C C; C 0 A l55
where
l12 ¼ 2@ 1 u2 2@ 1 u4 ; l15 l23
l13 ¼ 2@ 1 u3 2@ 1 u5 ;
l14 ¼ 2@ 1 u4 ; @ ¼ 2@ 1 u5 ; l21 ¼ u5 ; l22 ¼ 2u4 ð@ 1 u2 þ @ 1 u4 Þ þ ; 2 ¼ 2u4 ð@ 1 u3 þ @ 1 u5 Þ u1 ; l24 ¼ 2u4 @ 1 u4 ; l25 ¼ 2u4 @ 1 u5 ; l52 ¼ 2ðu2 þ u4 Þð@ 1 u3 þ @ 1 u5 Þ; l53 ¼ u1 þ 2ðu2 þ u4 Þ@ 1 u5 ; @ ¼ 2ðu2 þ u4 Þ@ 1 u4 þ ; l55 ¼ u1 þ 2ðu2 þ u4 Þ@ 1 u5 : 2
l51 ¼ u3 þ u5 ; l54
After calculations, we get P 1nþ1 ¼ L1 Q 1nþ1 . When u1 ¼ u3 ¼ u5 ¼ 0, the hierarchy (12) can be reduce to
ut ¼
u2
u4
¼
2b1nþ1 þ 2c1nþ1
2c1nþ1
t
¼
0
2
2 0
c1nþ1 b1nþ1 c1nþ1
:
ð13Þ
From (8), we have
b1nþ1 þ c1nþ1
c1nþ1
¼ L2
b0n þ c0n c0n
;
ð14Þ
where
L2 ¼
A
B
C
D
and
A¼
@ þ 2ðu2 þ u4 Þ@ 1 u4 ; 2
C ¼ 2u4 @ 1 u4 ;
B¼
@ þ 2ðu2 þ u4 Þ@ 1 ðu2 þ u4 Þ; 2
@ D ¼ 2u4 @ 1 ðu2 þ u4 Þ: 2
From (6) and (7), it is easy to see that
u1 u3 u5 T U ¼ k þ ; u2 þ ; u4 þ ; k k k
T
V ¼ ða0 þ ka1 ; b0 þ kb1 ; c0 þ kc1 Þ ;
P P P where ai ¼ mP0 aði; mÞk2m ; bi ¼ mP0 bði; mÞk2m ; ci ¼ mP0 cði; mÞk2m ; i ¼ 0; 1. Now we set a ¼ a1 e1 þ a2 e2 þ a3 e3 ; b ¼ b1 e1 þ b2 e2 þ b3 e3 ; a; b 2 glð2Þ. To construct Hamiltonian structures of the obtained integrable hierarchy, we need to transform the Lie algebra glð2Þ into a vector form through the mapping
f : glð2Þ ! R3 ;
a ! ða1 ; a2 ; a3 ÞT :
The mapping f induces a Lie algebraic structure on R3 , isomorphic to the matrix Lie algebra glð2Þ. The corresponding commutator ½:; : on R3 is given by T
½a; b ¼ aT R1 ðbÞ;
a ¼ ða1 ; a2 ; a3 ÞT ;
T
b ¼ ðb1 ; b2 ; b3 Þ 2 R3 :
So, we have T
½a; b ¼ ð2a2 b3 2a3 b2 ; 2a1 b2 2a1 b3 þ 2a2 b1 þ 2a3 b1 ; 2a1 b3 2a3 b1 Þ ¼ aT R1 ðbÞ; where
0
0
B R1 ðbÞ ¼ @ 2b3 2b2
2b2 2b3 2b1 2b1
2b3
1
C 0 A: 2b1
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
The matrix equation (4) gives a system of linear equations on the elements of F 1 . Solving the resulting system, we obtain
0
1
B F1 ¼ @ 0
0 0
0
1
C 1 A:
0 1 1 After calculations we have
fV; U u1 g ¼ a1 þ
a0 ; k
fV; U u2 g ¼ c0 þ kc1 ;
fV; U u5 g ¼ b1 c1
b0 þ c 0 ; k
fV; U u3 g ¼ c1 þ
c0 ; k
fV; U u4 g ¼ b0 c0 kb1 kc1 ;
u1 u3 u5 fV; U k g ¼ 1 2 ða0 þ ka1 Þ þ 2 ðc0 þ kc1 Þ þ 2 ðb0 þ c0 þ kb1 þ kc1 Þ: k k k
Inserting the above results into the quadratic-form identity (5), we get
0 d du
ða0 þ ka1 Þ uk21 ða0 þ ka1 Þ þ uk23 ðc0 þ kc1 Þ
!
þ uk25 ðb0 þ c0 þ kðb1 þ c1 ÞÞ
c
¼k
a1 þ ak0
1
C B c0 þ kc1 C B C @ cB c0 C: B c1 þ k kB C @k B C @ b0 c0 kðb1 þ c1 Þ A
ð15Þ
0 b1 c1 b0 þc k
By comparing the coefficients of k2n3 and k2n2 in both sides of (15), we see that
d ða1nþ2 u1 a1nþ1 þ u3 c1nþ1 þ u5 ðb1nþ1 þ c1nþ1 ÞÞ ¼ ð2n 2 þ cÞP1nþ1 du
ð16Þ
d ða0nþ1 u1 a0n þ u3 c0n þ u5 ðb0n þ c0n ÞÞ ¼ ð2n 1 þ cÞQ 1nþ1 : du
ð17Þ
and
From (8)–(11), one infers c ¼ 0. Thus, the Eq. (16) and (17) give
a1nþ2 u1 a1nþ1 þ u3 c1nþ1 þ u5 ðb1nþ1 þ c1nþ1 Þ ; 2n þ 2
P1nþ1 ¼
dH1n ; du
H1n ¼
Q 1nþ1 ¼
dH2n ; du
H2n ¼
a0nþ1 u1 a0n þ u3 c0n þ u5 ðb0n þ c0n Þ : 2n þ 1
The integrable hierarchy (12) can be written as
ut ¼ J 1
dH1n dH2n ¼ J2 : du du
Therefore, the hierarchy (18) has bi-Hamiltonian structure. It is easy to verify that J 1 L1 ¼ the integrable in the Liouville sense.
ð18Þ L1 J 1
¼ J 2 . So, the system (18) is
3. The generalized zero curvature equation and the application of the Lie algebra G In terms of the bilinear operator m n n 0 0 0 Dm t Dx f g ¼ ð@ t @ t Þ ð@ x @ x0 Þ f ðt; xÞgðt ; x Þjt 0 ¼t;x0 ¼x ;
where Dt , Dx are differential operators, f ðt; xÞ, gðt; xÞ are differentiable functions on variations t; x. We consider the isospectral problems in [9] as follows:
ðDx þ Dt Þu 1 ¼ U u; ðDx Dt Þu 1 ¼ V u:
ð19Þ
From (19) and the properties of the bilinear operator, it follows that
ð@ x þ @ t Þu 1 ¼ U u; ð@ x @ t Þu 1 ¼ V u:
Eq. (20) is equivalent to
ux þ ut ¼ U u; ux ut ¼ V u:
ð20Þ
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
Based on their compatibility conditions utx ¼ uxt , we have the zero curvature equation
U x U t ðV x þ V t Þ þ ½V; U ¼ 0:
ð21Þ
By using of the zero curvature Eq. (21), we can obtain some new integrable hierarchies. Consider the matrix Lie algebra G1 , which is expanded by glð2Þ
G1 ¼ spanfg 1 ; g 2 ; g 3 ; g 4 ; g 5 ; g 6 g; where
0
1
0
0
0
1
0
B 0 1 0 0 C B C g1 ¼ B C; @0 0 1 0 A 0 0 0 1 1 0 0 1 0 B 0 0 0 1 C C B g4 ¼ B C; @0 0 0 0 A 0
0 0 0
0 0
0 0
1
0
B1 0 0 0C B C g2 ¼ B C; @0 0 0 0A 0 0 1 0 1 0 0 0 0 B0 0 1 0C C B g5 ¼ B C; @0 0 0 0A 0
0
0 0
1 2 0
B0 B g3 ¼ B @0 0
1 0 0
0
0 C C C; 1 2 A 0 1 0
1 0 0 1 2 B0 0 0 1 C C B g6 ¼ B C; @0 0 0 0 A 0
0 0
0 0 0
1
1 1 0 0 0 B1 1 0 0C C B M2 ¼ B C; @0 0 1 0A
0
0
0
0 1 1
along with the following commutative relations
8 ½g 1 ; g 2 ¼ 2g 2 ; ½g 1 ; g 3 ¼ 2g 2 þ 2g 3 ; ½g 1 ; g 4 ¼ 0; ½g 1 ; g 5 ¼ 2g 5 ; > > > > > > < ½g 1 ; g 6 ¼ 2g 5 þ 2g 6 ; ½g 2 ; g 3 ¼ 2g 1 ; ½g 2 ; g 4 ¼ 2g 5 ; ½g 2 ; g 5 ¼ 0; ½g 2 ; g 6 ¼ 2g 4 ; ½g 3 ; g 4 ¼ 2g 5 2g 6 ; ½g 3 ; g 5 ¼ 2g 4 ; ½g 3 ; g 6 ¼ 0; > > > > ½g 4 ; g 5 ¼ ½g 4 ; g 6 ¼ ½g 5 ; g 6 ¼ 0; > > : ½g i ; g j ¼ g i M 2 g j g j M 2 g i ; g i ; g j 2 G1 : We can construct the following loop algebra G~1 ¼ spanfg 1 ðnÞ; g 2 ðnÞ; g 3 ðnÞ; g 4 ðnÞ; g 5 ðnÞ; g 6 ðnÞg, where g i ðnÞ ¼ g i kn ; i ¼ 1; 2; 3; 4; 5; 6, along with ½g i ðmÞ; g j ðnÞ ¼ ½g i ; g j kmþn ; i; j ¼ 1; 2; 3; 4; 5; 6. Consider the following isospectral problem
ðDx þ Dt Þu 1 ¼ U u; U ¼ g 1 ð1Þ þ u1 g 2 ð0Þ þ u2 g 3 ð0Þ þ u3 g 5 ð0Þ þ u4 g 6 ð0Þ:
ð22Þ
Set
V¼
X
ðam g 1 ðmÞ þ bm g 2 ðmÞ þ cm g 3 ðmÞ þ dm g 4 ðmÞ þ em g 5 ðmÞ þ fm g 6 ðmÞÞ:
mP0
The stationary zero curvature equation
V x þ V t ¼ ½V; U;
ð23Þ
gives rise to
8 amz ¼ 2u2 bm 2u1 cm ; > > > > > bmz ¼ 2u1 am 2u2 am þ 2bmþ1 þ 2cmþ1 ; > > > > > > < cmz ¼ 2u2 am 2cmþ1 ; dmz ¼ 2u4 bm 2u3 cm þ 2u2 em 2u1 fm ; > > > > emz ¼ 2u3 am 2u4 am 2u1 dm 2u2 dm þ 2emþ1 þ 2f mþ1 ; > > > > > fmz ¼ 2u4 am þ 2u2 dm 2f mþ1 ; > > :@ @ ¼ @x þ @t@ : @z
ð24Þ
Take the initial data as
a0 ¼ b;
b0 ¼ c0 ¼ d0 ¼ e0 ¼ f0 ¼ 0:
ð25Þ
Then the recursion relation (24) uniquely determines the sequence of sets of ai ; bi ; ci ; di ; ei ; f i , i P 1. The first few sets are listed as follows:
a1 ¼ 0; b1 ¼ bu1 ; b a2 ¼ bu1 u2 þ u22 ; 2
c1 ¼ bu2 ; b2 ¼
d1 ¼ 0;
b ðu1z þ u2z Þ; 2
e1 ¼ bu3 ;
f 1 ¼ bu4 :
b c2 ¼ u2z ; 2
B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
d2 ¼ b u1 u4 þ u2 u3 þ u2 u4 ;
b ðu3z þ u4z Þ; 2
e2 ¼
267
b f 2 ¼ u4z : 2
Denoting n V ðnÞ þ ¼ Rm¼0 ðam g 1 ðn mÞ þ bm g 2 ðn mÞ þ c m g 3 ðn mÞ þ dm g 4 ðn mÞ þ em g 5 ðn mÞ þ fm g 6 ðn mÞÞ:
A direct calculation reads ðnÞ ðnÞ ðV ðnÞ þx þ V þt Þ þ ½V þ ; U ¼ ð2bmþ1 2c mþ1 Þg 2 ð0Þ þ ð2c mþ1 Þg 3 ð0Þ þ ð2emþ1 2f mþ1 Þg 5 ð0Þ þ ð2f mþ1 Þg 6 ð0Þ:
Taking V ðnÞ ¼ V ðnÞ þ , the zero curvature equation ðnÞ ðnÞ U x U t ðV ðnÞ ; U ¼ 0; x þ V t Þ þ ½V
leads to the following Lax integrable hierarchy
1 u1 Bu C B 2C B C @ u3 A
1 1 0 2bnþ1 þ 2cnþ1 cnþ1 fnþ1 C C B B 2cnþ1 C B B bnþ1 cnþ1 enþ1 fnþ1 C ¼B C ¼ J3 B C ¼ J 3 P 2nþ1 ; A @ 2enþ1 þ 2f nþ1 A @ cnþ1
0
u4
0
bnþ1 cnþ1
2f nþ1
xt
ð26Þ
where
0
0
0
2
1
2 0
0 2
C C C: A
2
0
0
B0 0 B J3 ¼ B @ 0 2 2
0
From (24), we are easy to obtain the recurrence relation
0
0
l11
B l0 B P2nþ1 ¼ L3 P2n ¼ B 21 @ 0 0
l12
0
l13
0
l22 0
0
l23 0 l33
0
l43
0
l14
1
0
0 l24 C C 0 CP 2n l34 A
0
l44
0
and
@ 0 0 l11 ¼ l33 ¼ 2u2 @ 1 u1 2u2 @ 1 u2 ; 2 0
0
0
0
l21 ¼ l43 ¼ 2u1 @ 1 u1 2u2 @ 1 u1 l22 ¼ l44 ¼
0
0
l12 ¼ l34 ¼ 2u2 @ 1 u2 ;
@ 2u1 @ 1 u2 2u2 @ 1 u2 ; 2
@ þ 2u1 @ 1 u2 þ 2u2 @ 1 u2 ; 2
0
l14 ¼ 2u4 @ 1 u2 þ 2u2 @ 1 u4 ;
0
l13 ¼ 2u4 @ 1 u1 2u2 @ 1 u3 2u4 @ 1 u2 2u2 @ 1 u4 ; 0
l23 ¼ 2u3 @ 1 u1 2u1 @ 1 u1 2u1 @ 1 u3 2u2 @ 1 u3 2u3 @ 1 u2 2u4 @ 1 u2 2u1 @ 1 u4 2u2 @ 1 u4 ; 0
l24 ¼ 2u3 @ 1 u2 þ 2u4 @ 1 u2 þ 2u1 @ 1 u4 þ 2u2 @ 1 u4 : Therefore, the system (26) can be written as
1 u1 Bu C B 2C ¼B C @ u3 A 0
uxt
u4
1 ðu2 þ u4 Þb B ðu þ u þ u þ u Þb C 1 2 3 4 C B ¼ J3 Ln3 B C: A @ u2 b 0
ðu1 þ u2 Þb
xt
When u3 ¼ u4 ¼ 0, then hierarchy (26) can reduce to
uxt ¼
u1 u2
¼ xt
2bnþ1 þ 2cnþ1 2cnþ1
¼
0
2
2 0
cnþ1 bnþ1 cnþ1
:
ð27Þ
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
From (24), we have
bnþ1 þ cnþ1 cnþ1
@ 2
¼
þ 2ðu1 þ u2 Þ@ 1 u2
@ 2
þ 2ðu1 þ u2 Þ@ 1 ðu1 þ u2 Þ
2u2 @ 1 u2
@ 2
!
þ 2u2 @ 1 ðu1 þ u2 Þ
bn þ c n : cn
ð28Þ
This form is similar to the AKNS hierarchy, but not the AKNS hierarchy. Based on the above calculations, it is not difficult to show the following theorem: Theorem 2. Suppose that xðkÞ is the polynomial of degree n, which is independent of x. If k is not rely on t, that is kt ¼ 0, and the matrix V meets the following boundary condition
0 2cnþ1 B @V B 2bnþ1 þ 2cnþ1 ¼B @z ðu1 ;u2 ;u3 ;u4 Þ¼ð0;0;0;0Þ @ 0
2f nþ1
2cnþ1 0
2enþ1 þ 2f nþ1 2cnþ1
2f nþ1 C C C: 4cnþ1 A
0
2bnþ1 þ 2cnþ1
2cnþ1
0
4f nþ1
1
4cnþ1
Then from (21), (22), (23) and (25), we can only determine the matrix V, and the isospectral hierarchy
1 u1 Bu C B 2C B C @ u3 A
1 ðu2 þ u4 Þb B ðu þ u þ u þ u Þb C 1 2 3 4 C B ¼ xðL3 ÞB C; A @ u2 b
0
u4
0
ðu1 þ u2 Þb
xt
where L3 is a recurrence operator. Next, we will establish the Hamiltonian structure of the integrable system by using the quadratic-form identity. Set
a ¼ a1 g 1 þ a2 g 2 þ a3 g 3 þ a4 g 4 þ a5 g 5 þ a6 g 6 ;
b ¼ b1 g 1 þ b2 g 2 þ b3 g 3 þ b4 g 4 þ b5 g 5 þ b6 g 6 ;
a; b 2 G1
To construct Hamiltonian structures of the obtained integrable hierarchy, we need to transform the Lie algebra G1 into a rector form through the mapping
a ! ða1 ; a2 ; a3 ; a4 ; a5 ; a6 ÞT :
f2 : G1 ! R6 ;
The mapping f2 induces a Lie algebraic structure on R6 , isomorphic to the matrix Lie algebra G1 . The corresponding commutator [.,.] on R6 is given by T
½a; b ¼ aT R2 ðbÞ;
a ¼ ða1 ; . . . ; a6 ÞT ;
T
b ¼ ðb1 ; . . . ; b6 Þ 2 R6
and T
½a; b ¼ð2a2 b3 2a3 b2 ;
2a1 b2 2a1 b3 þ 2a2 b1 þ 2a3 b1 ;
2a1 b3 2a3 b1 ;
2a1 b5 2a1 b6 þ 2a2 b4 þ 2a3 b4 2a4 b2 2a4 b3 þ 2a5 b1 þ 2a6 b1 ;
2a2 b6 2a3 b5 þ 2a5 b2 2a6 b2 ; T
2a1 b6 2a3 b4 þ 2a4 b3 2a6 b1 Þ ¼ aT R2 ðbÞ;
where
0
0
2b2 2b3
2b3
0
2b5 2b6
2b1
0
2b6
2b4
2b1
2b1
2b5
2b4
0
0
0
2b2 2b3
0 0
0 0
2b3 2b2
2b1 2b1
B 2b 3 B B B 2b2 R2 ðbÞ ¼ B B 0 B B @ 0 0
2b6
1
C C C 2b4 C C: 2b3 C C C 0 A 0
2b1
From [7], we know that R2 ðbÞ is required to satisfy with
F 2 ¼ F T2 ;
T
R2 ðbÞF 2 ¼ F 2 ðR2 ðbÞÞ :
This matrix equation gives a system of linear equations on the elements of F 2 . Solving the resulting system, we obtain
0
1 0 0 B 0 0 1 B B B 0 1 1 F2 ¼ B B1 0 0 B B @ 0 0 1 0 1 1
1
0
0
1
1 C C C 0 1 1 C C: 0 0 0 C C C 0 0 0 A 0 0 0 0
0
In order to the Hamiltonian structure of the Lax integrable system, we define a linear functional fa; bg as follows: fa; bg ¼ aT F 2 b.
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
By a direct calculation, we have
@U V; ¼ c f ; @r
@U V; ¼ b c e f ; @u1
V;
@U @u3
¼ c;
@U V; ¼ b c; @u4
V;
@U @k
¼ a þ d:
Inserting the above result into the quadratic-form identity (5), we get
0
c f B b c e f d @ B ða þ dÞ ¼ kc kc B du @k @ c
1 C C C: A
ð29Þ
b c By comparing the coefficients of kn2 in (29), we see that
0
cn fn
1
B b c e f C d n n nC B n ðanþ2 þ dnþ2 Þ ¼ ðn 1 þ cÞB C ¼ ðn 1 þ cÞP2n : @ A du cn bn cn It is easy to see that c ¼ 0, thus we have
P2nþ1 ¼
dH3nþ1 ; du
H3nþ1 ¼
anþ2 þ dnþ2 : nþ1
The integrable hierarchy (26) can be written as
uxt ¼ J 3
dH3nþ1 : du
ð30Þ
Which is the Hamiltonian structure of the hierarchy (26). After tedious computations, we have J 3 L3 ¼ L3 J 3 . So, the hierarchy (30) is integrable in the Liouville sense. 4. Two kinds of Darboux transformations of the Eq. (31) Let b ¼ 2; n ¼ 2, the hierarchy (26) reduces to a new equation as follows
8 u1;xt > > > > < u2;xt > > u3;xt > > : u4;xt
¼ 2b ðu1zz þ u2zz Þ þ 2bu21 u2 þ 3bu1 u22 þ bu32 ; ¼ 2b u2zz 2bu1 u22 bu32 ; ¼ 2b ðu3zz þ u4zz Þ þ 4bu1 u2 u3 þ 3bu3 u22 þ 6bu1 u2 u4 þ 3bu22 u4 þ 2bu21 u4 ;
ð31Þ
¼ 2b u4zz 4bu1 u2 u4 3bu22 u4 2bu22 u3 ;
whose Lax pair matrices present that
ux þ ut ¼ U u; ux ut ¼ V u;
where
0 B B U¼B @
1
k þ u2
2u2
u4
2u4
u1
k þ u2
u3
u4
0
0
k þ u2
2u2
0
0
u1
k þ u2
C C C A
and
0
V 11 BV B 21 V ¼B @ 0 0
1
V 12
V 13
V 14
V 22
V 23
0
V 11
V 24 C C C; V 12 A
0
V 21
V 22
V 11 ¼ 2k2 þ 2u1 u2 þ u22 þ 2ku2 u2z ; V 22 ¼ 2k2 2u1 u2 u22 þ 2ku2 u2z ;
ð32Þ
V 12 ¼ 4ku2 þ 2u2z ;
V 21 ¼ 2ku1 þ u1z þ u2z ;
V 23 ¼ 2ku3 þ u3z þ u4z ;
V 13 ¼ 2ðu1 u4 þ u2 u3 þ u2 u4 Þ þ 2ku4 u4z ;
V 14 ¼ 4ku4 þ 2u4z ;
V 24 ¼ 2ðu1 u4 þ u2 u3 þ u2 u4 Þ þ 2ku4 u4z :
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
In this section, we will investigate the Dardoux transformations of the Eq. (31). Consider a Darboux transformation
/0 ¼ T/ and require /0 and / satisfying the spectral problem
/x ¼ U/;
ð33Þ
/0x ¼ U 0 /0
and the spectral problem (22), where U 0 has the same form as U expect replacing u1 ; u2 ; u3 ; u4 by u01 ; u02 ; u03 ; u04 . It is easy to see that T meets
T x þ TU ¼ U 0 T:
ð34Þ
In what follows, we discuss two kinds of forms of matrix T. Assume that
0 T ¼ kT 1 þ T 0 ;
1
0
a1
b1
e1
f1
Bc B 1 T1 ¼ B @0
d1
s1
0
a1
t1 C C C; b1 A
0
0
c1
d1
a
b
e
1
f
Bc d s t C B C T0 ¼ B C; @0 0 a bA 0 0
c
d
where a1 ; b1 ; c1 ; d1 ; e1 ; f 1 ; s1 ; t1 and a; b; c; d; e; f ; s; t are functions in x and t. From the spectral problem (22), we have
0
k þ u2
2u2
u4
2u4
1
0
0
1
u1 0
k þ u2 0
u3 k þ u2
u4 2u2
C B 0 1 0 C B C ¼ kB A @0 0 1
0 0
0
0
u1
k þ u2
0
k þ u02
2u02
u04
2u04
u01
k þ u02
u03
0
0
k þ u02
0
0
u01
B B U¼B @
0
1
0
0
0
u2
1
2u2
u4
2u4
C Bu C B 1 CþB A @0
u2 0
u3 u2
u4 C C C ¼ kE0 þ U 0 2u2 A
0
0
u1
0 1
u2
and
0 B B U0 ¼ B @
1
u04 C C 0 C ¼ kE0 þ U 0 : 2u02 A k þ u02
Based on the Eq. (34), we obtain 0
a1x b1x Bc d B 1x 1x T x ¼ kB @ 0 0 0 0 0
a1 B 2 B c1 k B @0 0
e1x s1x a1x c1x b1 d1 0 0
e1 s1 a1 c1
1 0 f1x ax bx ex B t 1x C C B cx dx sx CþB b1x A @ 0 0 ax d1x 0 0 cx 1 0 f1 a 1 b1 C Bc d t1 C B 1 1 CE0 kB @0 0 b1 A d1 0 0
1 0 fx a 1 b1 C Bc d tx C B 1 1 2 C ¼ k E0 B @0 0 bx A dx 0 0 e1 s1 a1 c1
e1 s1 a1 c1
1 0 1 0 f1 a b e f a1 C B C B t1 C Bc d s t C 0 B c1 C þ kE0 B C þ kU B @0 0 a bA @0 b1 A d1 0 0 0 c d
1 0 1 0 f1 a b e f a C B C Bc t1 C Bc d s t C B CU kB CE B @0 0 a bA 0 @0 b1 A d1 0 0 c d 0
b d 0 0
e s a c
b1 d1 0 0
e1 s1 a1 c1
1 0 1 f1 a b e f C B C t1 C 0B c d s t C CþU B C @0 0 a bA b1 A d1 0 0 c d
1 f tC C CU: bA d ð35Þ
j
Comparing the coefficients of k ðj ¼ 0; 1; 2Þ in Eq. (35). As for the case of j ¼ 2, we get that
a1 a1 ¼ 0;
b1 þ b1 ¼ 0;
e1 e1 ¼ 0;
f 1 þ f1 ¼ 0;
c1 c1 ¼ 0; s1 s1 ¼ 0;
d1 þ d1 ¼ 0; t 1 þ t 1 ¼ 0:
So we can obtain that
b1 ¼ 0;
c1 ¼ 0;
f 1 ¼ 0;
s1 ¼ 0:
ð36Þ
The case of j ¼ 1 leads to the following equations
8 a1x ¼ a þ u02 a1 2u02 c1 u2 a1 u1 b1 a; > > > > > b1x ¼ b þ u02 b1 2u02 d1 þ 2u2 a1 u2 b1 þ b; > > > > > c1x ¼ c þ u01 a1 þ u02 c1 u2 c1 u1 d1 c; > > > < d1x ¼ d þ u01 b1 þ u02 d1 þ 2u2 c1 u2 d1 þ d; > e1x ¼ e þ u02 e1 2u02 s1 þ u04 a1 2u04 c1 u4 a1 u3 b1 u2 e1 u1 f1 e; > > > > 0 0 0 0 > > f1x ¼ f þ u2 f1 2u2 t1 þ u4 b1 2u4 d1 þ 2u4 a1 u4 b1 þ 2u2 e1 u2 f1 þ f ; > > > 0 0 0 0 > s ¼ s þ u e þ u s þ u a þ u > 1x 1 1 2 1 3 1 4 c 1 u4 c 1 u3 d1 u2 s1 u1 t 1 s; > : 0 0 0 0 t1x ¼ t þ u1 f1 þ u2 t 1 þ u3 b1 þ u4 d1 þ 2u4 c1 u4 d1 þ 2u2 s1 u2 t 1 þ t
ð37Þ
B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
271
And the case of j ¼ 0 leads to the following equations
8 ax ¼ u02 a 2u02 c u2 a u1 b; > > > > > bx ¼ u02 b 2u02 d þ 2u2 a u2 b; > > > > > cx ¼ u01 a þ u02 c u2 c u1 d; > > > < d ¼ u0 b þ u0 d þ 2u c u d; x 2 2 1 2 > ex ¼ u02 e 2u02 s þ u04 a 2u04 c u4 a u3 b u2 e u1 f ; > > > > > fx ¼ u02 f 2u02 t þ u04 b 2u04 d þ 2u4 a u4 b þ 2u2 e u2 f ; > > > > > > sx ¼ u01 e þ u02 s þ u03 a þ u04 c u4 c u3 d u2 s u1 t; > : t x ¼ u01 f þ u02 t þ u03 b þ u04 d þ 2u4 c u4 d þ 2u2 s u2 t: Substituting (36) into (37) gives
(
a1 ¼ d1 ; f ¼
ðu04
e1 ¼ t1 ;
u4 Þa1 þ
c ¼ 12 ðu01 u1 Þa1 ;
b ¼ ðu02 u1 Þa1 ;
ðu02
u2 Þe1 ;
s¼
1 ½ðu03 2
ð38Þ
u3 Þa1 þ ðu01 u1 Þe1 :
4.1. The first Darboux transformation Set a1 ¼ d1 ¼ 0; e1 ¼ t1 ¼ 1. From (38), it is easy to see that b ¼ 0; c ¼ 0; f ¼ u02 u2 ; s ¼ 12 ðu01 u1 Þ. The matrix T can be written to the following form
0
0 0 1 0
1
0
a
0
B0 0 0 1C B0 d C B B T ¼ kB CþB @0 0 0 0A @0 0 0 0 0
0
0 0
u02 u2
e 1 ðu01 2
u1 Þ
t
a
0
0
d
1
C C C: A
0
Since / and / of (33) are 4 4 matrices, we need that det/0 ¼ 0. There exists a constant k ¼ k1 and a solution / ¼ ð/1 ; /2 ; /3 ; /4 ÞT of the spectral problem (22) which satisfy that
8 a/3 ¼ 0; > > > > < d/4 ¼ 0;
) a ¼ 0; ) d ¼ 0;
k1 /3 þ a/1 þ e/3 þ ðu02 u2 Þ/4 ¼ 0; ) e ¼ k1 ðu02 u2 Þ //43 ; > > > > : k / þ d/ þ 1 ðu0 u Þ/ þ t/ ¼ 0; ) t ¼ k 1 ðu0 u Þ /3 : 1 4 1 1 1 / 2 3 4 1 1 2 2 4 We are easy to get the Darboux transformation T:
1 00 0 0 1 0 B0 0 0 1C B 0 C B B T ¼ k1 B CþB @0 0 0 0A B @0 0 0 0 0 0 0
0 k1 ðu02 u2 Þ //43 0 0 0
1 ðu01 2
u1 Þ 0 0
u02 u2
1
C k1 12 ðu01 u1 Þ //34 C C: C A 0 0
4.2. The second Darboux transformation Set a1 ¼ d1 ¼ 1; e1 ¼ t 1 ¼ 0. We have the following form of the Darboux transformation T:
T ¼ kI S;
ð39Þ
where I is a 4 4 unit matrix. From the Eq. (34), we have
Sx þ ðkI SÞU ¼ U 0 ðkI SÞ;
) Sx þ ðkI SÞðkE0 þ U 0 Þ ¼ ðkE0 þ U 00 ÞðkI SÞ:
ð40Þ
j
Comparing the coefficients of k ðj ¼ 0; 1; 2Þ in Eq. (40), we get
8 2 > < k : E0 ¼ E0 ; k1 : U 0 SE0 ¼ U 00 E0 S; ) U 00 ¼ U 0 SE0 þ E0 S; > : 0 k : Sx ¼ SU 0 U 00 S: ) Sx ¼ U 00 S SU 0 :
ð41Þ
From (41), we have
Sx ¼ ðU 0 SE0 þ E0 SÞS SU 0 ¼ U 0 S SE0 S þ E0 S2 SU 0 :
ð42Þ
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
Now, we need to find the matrix S, which can be expressed by the eigenfunction of (33). Assume that H is the solution of (33), that is
Hx ¼ UH ¼ E0 H ^ þU 0 H;
ð43Þ
where ^ ¼ diagðk1 ; k1 ; k1 ; k1 Þ. Set
S ¼ H ^ H1 :
ð44Þ
From (43), we have
Sx ¼ Hx ^ H1 H ^ H1 Hx H1 ¼ ðE0 H ^ þU 0 HÞ ^ H1 SðE0 H ^ þU 0 HÞH1 ¼ E0 S2 þ U 0 S SE0 S SU 0 :
ð45Þ
It is easy to see that (45) is equivalent to (42). Next, we shall prove again that the transformations (39) map equation
/t ¼ V/; into equation
/0t ¼ V 0 /0 ; where V is presented in (32), and V 0 has the same form as V except replacing u1 ; u2 ; u3 ; u4 ; u1z ; u2z , u3z ; u4z by u01 ; u02 ; u03 ; u04 ; u01z ; u02z ; u03z ; u04z . To find a new solution of the soliton equations, we should prove
T t þ TV ¼ V 0 T:
ð46Þ
Proof. From (32), we get
V ¼ 2k2 E0 þ 2kU 0 þ V 0 ; where
0
v 11 v 12 v 13 Bv B 21 v 22 v 23 V0 ¼ B @ 0 0 v 11 0 0 v 21
v 14 1 v 24 C C C; v 12 A v 22
v 11 ¼ 2u1 u2 þ u22 u2z ; v 12 ¼ 2u2z ; v 13 ¼ 2ðu1 u4 þ u2 u3 þ u2 u4 Þ u4z ; v 21 ¼ u1z þ u2z ; v 14 ¼ 2u4z ; v 22 ¼ 2u1 u2 u22 u2z ; v 23 ¼ u3z þ u4z ; v 24 ¼ 2ðu1 u4 þ u2 u3 þ u2 u4 Þ u4z : After the Darboux transformation /0 ¼ T/, we have
/0t ¼ V 0 /0 ¼ ð2k2 E0 þ 2kU 00 þ V 00 Þ/0 ; Assume H satisfies that
Hx ¼ UH; , we are easy to see that Ht ¼ VH;
Ht ¼ VH ¼ 2E0 H^2 þ 2U 0 H ^ þV 0 H: From (46), we get
St þ ðkI SÞð2k2 E0 þ 2kU 0 þ V 0 Þ ¼ ð2k2 E0 þ 2kU 00 þ V 00 ÞðkI SÞ:
ð47Þ
j
Comparing the coefficients of k ðj ¼ 0; 1; 2; 3Þ in Eq. (47), we get
8 3 > k > > > < k2 > k1 > > > : 0 k
: 2E0 ¼ 2E0 ; : 2U 0 2SE0 ¼ 2U 00 2E0 S; : V 0 2SU 0 ¼ V 00 2U 00 S; : St SV 0 ¼ V 00 S:
) U 00 ¼ U 0 SE0 þ E0 S; ) V 00 ¼ V 0 2SU 0 þ 2U 00 S;
) St ¼ V 00 S SV 0 :
From (44) and (48), we have
St ¼ Ht ^ H1 H ^ H1 Ht H1 ¼ ð2E0 H^2 þ 2U 0 H ^ þV 0 HÞ ^ H1 Sð2E0 H^2 þ 2U 0 H ^ þV 0 HÞH1 ¼ 2E0 S3 þ 2U 0 S2 þ V 0 S 2SE0 S2 2SU 0 S SV 0 And
St ¼ V 00 S SV 0 ¼ ðV 0 2SU 0 þ 2U 00 SÞS SV 0 ¼ V 0 S 2SU 0 S þ 2ðU 0 SE0 þ E0 SÞS2 SV 0 ¼ V 0 S 2SU 0 S þ 2U 0 S2 2SE0 S2 þ 2E0 S3 SV 0 :
ð48Þ
B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
273
5. Conclusion Based on a pair of the isospectral problem
ux ¼ U u ut ¼ V u , in terms of the compatibility conditions: uxt ¼ utx , we are usually
easy to obtain the zero curvature equation U t V x þ ½U; V ¼ 0. Then making use of the corresponding stationary zero curvature equation and the commutative operation, the Lax integrable hierarchy can be obtained. In addition, taking advantage of the trace identity or the quadratic-form identity, the Hamiltonian structure can be worked out. At last, we get two kinds of Darboux transformations of Eq. (31). In this article, according to the properties of the bilinear operator and the zero curvature Eq. (21), we consider the isospectral problem (22), integrable hierarchy (26) can be obtained. Ultimately, by making use of the quadratic-form identity, we obtain the Hamiltonian structure of the hierarchy (26). From Sections 2 and 3 of this paper, we find that Eq. (13) and (14) are equivalent to Eq. (27) and (28). In other words, the hierarchy (12) and hierarchy (26) can reduce to the same hierarchy. In Section 4, we let b ¼ 2; n ¼ 2, the integrable hierarchy (26) can be reduced to Eq. (31). In terms of the investigation of Darboux transformations [11], we get two kinds of Darboux transformations of Eq. (31). Acknowledgments The authors would like to thank the referee for valuable comments and suggestions on this article. This work was supported by the National Natural Science Foundation of China (No. 11171055), the Natural Science Foundation of Jilin province (Nos. 201115006 and 20130101068JC), Scientific Research Foundation for Returned Scholars Ministry of Education of China. References [1] Tu. Guizhang, The trace identity, a powerful tool for constructing the Hamiltonian structure of integrable systems, J. Math. Phys. 30 (2) (1989) 330– 338. [2] Honwah Tam, Yufeng Zhang, Spectral radius analysis of matrices and the associated with integrable systems, Int. J. Modern Phys. B 23 (24) (2009) 4855–4879. [3] Zhang Yufeng, Liu Jing, Induced Lie algebras of a six-dimensional matrix Lie algebra, Commun. Theor. Phys. (Beijing, China) 50 (2008) 289–294. [4] Yufeng Zhang, Qingyou Yan, Integrable couplings of the hierarchies of evolution equations, Chaos Solitons Fractals 16 (2003) 263–269. [5] Wen-Xiu Ma, Multi-component bi-Hamiltonian Dirac integrable equations, Chaos Solitons Fractals 39 (2009) 282–287. [6] Wen-Xiu Ma, A multi-component Lax integrable hierarchy with Hamiltonian structure, Pac. J. Appl. Math. 1 (2) (2008) 191–197. [7] Fu-Kui Guo, Huan-He Dong, A new loop algebra and corresponding computing formula of constant in quadratic-form identity, Commun. Theor. Phys. (Beijing) 47 (6) (2007) 981–986. [8] Fukui Guo, Yufeng Zhang, The integrable coupling of the AKNS hierarchy and its Hamiltonian structure, Chaos Solitons Fractals 32 (2007) 1898–1902. [9] Yufeng Zhang, A complex higher-dimensional Lie algebra with real and imaginary structure constants as well as its decomposition, Commun. Theor. Phys. (Beijing) 50 (5) (2008) 1021–1026. [10] Yufeng Zhang, Zhong Han, Hon-Wah Tam, An integrable hierarchy and Darboux transformations, bilinear Backlund transformations of a reduced equation, Appl. Math. Comput. 219 (2013) 5837–5848. [11] Li Yishen, Solitons and Integrable Systems, Shanghai Science and Technology Education Publishing House, 1999.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Hamiltonian forms of the two new integrable systems and two kinds of Darboux transformations Baiying He, Liangyun Chen ⇑ School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China
a r t i c l e
i n f o
a b s t r a c t g a new Lax Based on the Lie algebra glð2Þ, taking a kind of corresponding loop algebra glð2Þ, integrable hierarchy can be obtained. Then, by means of the quadratic-form identity, the corresponding bi-Hamiltonian structure was worked out. Expanding Lie algebra glð2Þ, and making use of the new zero curvature equation Zhang (2008) [9], we obtain an integrable hierarchy and its Hamiltonian structure. At last, two kinds of Darboux transformations of the equation are generated. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Matrix Lie algebra Quadratic-form identity Bi-Hamiltonian structure Darboux transformation
1. Introduction By using some matrix Lie algebras, Guizhang [1] once introduced a powerful tool for generating integrable Hamiltonian systems [3–6], which was called Tu scheme. By employing the corresponding loop algebras, some interesting soliton hierarchies of evolution equations were worked out. However, we have found that some solitary hierarchy can be obtained by a kind of vector loop algebra [8]. P P ~ Let G be a s-dimensional Lie algebra with basis e1 ; e2 ; . . . ; es . Take a ¼ sk¼1 ak ek ; b ¼ sk¼1 bk ek 2 G. The loop algebra G generated by G has the basis ek ðmÞ ¼ ek km ; 1 6 k 6 s; m 2 Z, the commuting operations read ½ek ðmÞ; ej ðnÞ ¼ ½ek ; ej kmþn . ~ is given by The column vector form of G
~ ¼ a ¼ ða1 ; . . . ; as Þ; G
ak ¼
X ak;m km ;
½a; b ¼ c ¼ ðc1 ; . . . ; cs ÞT :
m
~ is as follows The linear isospectral problem established by G
w@ ¼ ½U; w; wt ¼ ½V; w;
kt ¼ 0;
ð1Þ
P where @ ¼ nk¼1 ak @x@ , ak are arbitrary constants, w@ denotes the derivative sum of w with aspect to xk ; k ¼ 1; 2; . . . ; n. k The compatibility condition of (1) is the zero curvature equation
U t V @ þ ½U; V ¼ 0; its stationary zero curvature equation reads
V @ ¼ ½U; V: ⇑ Corresponding author. E-mail address: [email protected] (L. Chen). http://dx.doi.org/10.1016/j.amc.2014.07.006 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
ð2Þ
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
~ u ¼ ðu1 ; . . . ; up ÞT , assume rankU 0 ¼ rankðui U i Þ ¼ a; 1 6 i 6 p, then U is called Take U ¼ Uðk; uÞ ¼ U 0 þ Rpi¼1 ui U i ; U i 2 G; the same-rank, denote by
@ ¼ a; rankðUÞ ¼ rankð@Þ ¼ rank @xk
1 6 k 6 n:
ð3Þ
Let two same-rank solutions V 1 and V 2 satisfy the relation V 1 ¼ cV 2 ; c ¼ constant, then we see that T ~ the constant matrix Theorem 1. Let (3) hold. Two same-rank solutions of (2) possess V 1 ¼ cV 2 . Set ½a; b ¼ aT RðbÞ; a; b 2 G, F ¼ ðfij Þss meets T
F ¼ FT ;
RðbÞF ¼ ðRðbÞFÞ :
ð4Þ
Define a quadratic functional as follows
fa; bg ¼ aT Fb;
~ 8a; b 2 G;
then the following identity holds
d @ @U ; kc V; fV; U k g ¼ kc dui @k @ui
i ¼ 1; . . . ; p;
ð5Þ
where c is a constant to be determined, V is a same-rank solution of 2. 5 is called the quadratic-form identity. Zhang and Fan [2] proposed the Lie algebra glð2Þ ¼ spanfe1 ; e2 ; e3 g, where
e1 ¼
1
0
0 1
;
e2 ¼
0
0
1 0
;
e3 ¼
1 2 0
1
;
M1 ¼
1 0 1 1
;
along with the commutative relation
½e1 ; e2 ¼ 2e2 ; ½e1 ; e3 ¼ 2e2 þ 2e3 ; ½e2 ; e3 ¼ 2e1 ; ½ei ; ej ¼ ei M 1 ej ej M 1 ei ; ei ; ej 2 glð2Þ:
g a new Lax In this paper, we consider the known Lie algebra glð2Þ, and take a kind of corresponding loop algebra glð2Þ, integrable hierarchy can be obtained. Then, by means of the quadratic-form identity, the corresponding bi-Hamiltonian structure was worked out. In Section 3, we want to expand the Lie algebra glð2Þ into the Lie algebra G1 , and take a kind of loop algebra G~1 . Next, we start from an isospectral problem to obtain the integrable hierarchy (26), by making use of the zero curvature equation [9]. In addition, by using the quadratic-form identity, we can obtain a Hamiltonian structure. Study of the algebraic properties of the equations is an important aspect in soliton theory. Some ways for generating Darboux transformations of nonlinear soliton equations by starting from isospectral problems [10]. In Section 4, when n ¼ 2; b ¼ 2, the integrable hierarchy (26) can be reduced to a new Eq. (31). In [11] the Darboux transformations for a Lax pair integrable systems are investigated in detail. Based on this, we obtain two kinds of Darboux transformations of the new Eq. (31). 2. A new integrable hierarchy and its bi-Hamiltonian structure g Based on the known Lie algebra glð2Þ, we construct the following loop algebra glð2Þ:
8 ek ði; nÞ ¼ ek k2nþi ; k—spectral parameter; > > > > > ½e1 ði; mÞ; e2 ðj; nÞ ¼ 2e2 ðdij ; m þ n þ qij Þ; > > > > > > ½e1 ði; mÞ; e3 ðj; nÞ ¼ 2e2 ðdij ; m þ n þ qij Þ þ 2e3 ðdij ; m þ n þ qij Þ; > > > > > ½e < 2 ði; mÞ; e3 ðj; nÞ ¼ 2e1 ðdij ; m þ n þ qij Þ; i þ j; i þ j < 2; > > d ¼ i;j > > 0; i þ j ¼ 2; > > > > > 0; i þ j < 2; > > > q ¼ > > i;j 1; i þ j ¼ 2; > > : degðek ði; nÞÞ ¼ 2n þ i; k ¼ 1; 2; 3; i; j 2 f0; 1g: Consider an isospectral problem
ux ¼ U u; ut ¼ V u; kt ¼ 0;
where
U ¼ e1 ð1; 0Þ þ u1 e1 ð1; 1Þ þ u2 e2 ð0; 0Þ þ u3 e2 ð1; 1Þ þ u4 e3 ð0; 0Þ þ u5 e3 ð1; 1Þ
ð6Þ
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
and
V¼
X a0m e1 ð0; mÞ þ b0m e2 ð0; mÞ þ c0m e3 ð0; mÞ þa1m e1 ð1; mÞ þ b1m e2 ð1; mÞ þ c1m e3 ð1; mÞ
mP0
:
ð7Þ
The stationary zero curvature representation
V x ¼ ½U; V; gives
8 a0mx ¼ 2u2 c0m þ 2u3 c1m 2u4 b0m 2u5 b1m ; > > > > > a1mþ1x ¼ 2u2 c1mþ1 þ 2u3 c0m 2u4 b1mþ1 2u5 b0m ; > > > < b0mx ¼ 2b1mþ1 þ 2c1mþ1 2u1 b1m 2u1 c1m þ 2u2 a0m þ 2u3 a1m þ 2u4 a0m þ 2u5 a1m ; > b1mþ1x ¼ 2b0mþ1 þ 2c0mþ1 2u1 b0m 2u1 c0m þ 2u2 a1mþ1 þ 2u3 a0m þ 2u4 a1mþ1 þ 2u5 a0m ; > > > > > c > 0mx ¼ 2c1mþ1 þ 2u1 c1m 2u4 a0m 2u5 a1m ; > : c1mþ1x ¼ 2c0mþ1 þ 2u1 c0m 2u4 a1mþ1 2u5 a0m :
ð8Þ
Take the initial data as
a00 ¼ a;
b00 ¼ c00 ¼ a10 ¼ b10 ¼ c10 ¼ 0:
ð9Þ
From the recursion relation (8), we have
b11 ¼ au2 ;
a11 ¼ 0;
c11 ¼ au4 ;
a
b01 ¼ ðu2x þ u4x Þ au3 2
ð10Þ
and
c01 ¼
a 2
a
a01 ¼ au2 u4 þ u24 : 2
u4x au5 ;
ð11Þ
Denoting
8 ðnÞ n > < V þ ¼ Rm¼0 ða0m e1 ð0; n mÞ þ b0m e2 ð0; n mÞ þ c0m e3 ð0; n mÞ þa1m e1 ð1; n mÞ þ b1m e2 ð1; n mÞ þ c1m e3 ð1; n mÞÞ; > : ðnÞ V þ þ V ðnÞ ¼ k2n V: The stationary zero curvature representation V x ¼ ½U; V can be expressed as ðnÞ ðnÞ ðnÞ V ðnÞ þx þ ½U; V þ ¼ V x ½U; V :
A direct calculation shows ðnÞ V ðnÞ þx þ ½U; V þ ¼ ða1nþ1x 2u2 c 1nþ1 þ 2u4 b1nþ1 Þe1 ð1; 1Þ þ ð2b1nþ1 2c 1nþ1 Þe2 ð0; 0Þ þ ðb1nþ1x 2b0nþ1 2c0nþ1
2u2 a1nþ1 2u4 a1nþ1 Þe2 ð1; 1Þ þ ð2c1nþ1 Þe3 ð0; 0Þ þ ðc1nþ1x þ 2c0nþ1 þ 2u4 a1nþ1 Þe3 ð1; 1Þ: Taking V
Ut
ðnÞ
¼
V ðnÞ x
V ðnÞ þ ,
the zero curvature equation
þ ½U; V ðnÞ ¼ 0;
leads to the following Lax integrable hierarchy 1 0 1 0 u1 @ a1nþ1x þ 2u2 c1nþ1 2u4 b1nþ1 C B B C B 0 2b1nþ1 þ 2c1nþ1 C B B u2 C B C B B C B C B B C B ut ¼ B u3 C ¼ B b1nþ1x þ 2b0nþ1 þ 2c0nþ1 þ 2u2 a1nþ1 þ 2u4 a1nþ1 C ¼ B 2u2 þ 2u4 C B B C B C B B C B 0 2c1nþ1 A @ @ u4 A @ 0
u5
c1nþ1x 2c0nþ1 2u4 a1nþ1
t
2u4
1 0 1 0 u1 0 2u3 c0n 2u5 b0n C B B C B 0 2b1nþ1 þ 2c1nþ1 C B B u2 C B C B B C B C B B C B ¼ J1 P 1nþ1 ¼ B u3 C ¼ B 2u3 a0n þ 2u5 a0n 2u1 b0n 2u1 c0n C ¼ B 2u3 þ 2u5 C B B C B C B B C B 0 2c1nþ1 A @ @ u4 A @ 0
u5
2u5 a0n þ 2u1 c0n
t
2u5
0 2u2 2u4 0
0
0
@
0
2
2
@
0 2u3 2u5 0
0
0
0
2
0
0
2u1
1 a1nþ1 CB C 0 2 CB c0nþ1 C CB C CB C 2 @ CB c1nþ1 C CB C CB C 0 0 A@ b0nþ1 c0nþ1 A 0 2u4
0
10
0
b1nþ1 c1nþ1
1 a0n CB C 2 0 CB c1nþ1 C CB C CB C 0 2u1 CB c0n C ¼ J 2 Q 1nþ1 ; CB C CB C 0 0 A@ b1nþ1 c1nþ1 A 0 2u5
0
0
10
b0n c0n ð12Þ
where J 1 and J 2 satisfy with
J i
¼ J i ; i ¼ 1; 2. So J 1 and J 2 are the Hamiltonian operators.
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
From (8), we can obtain a recurrence operator
0
0 B B l21 B L1 ¼ B B 0 B @ 0
l12 l22
l51
l13 l23
l14 l24
1
0
0
0
0
1
l52
l53
l54
1 l15 C l25 C C 0 C C; C 0 A l55
where
l12 ¼ 2@ 1 u2 2@ 1 u4 ; l15 l23
l13 ¼ 2@ 1 u3 2@ 1 u5 ;
l14 ¼ 2@ 1 u4 ; @ ¼ 2@ 1 u5 ; l21 ¼ u5 ; l22 ¼ 2u4 ð@ 1 u2 þ @ 1 u4 Þ þ ; 2 ¼ 2u4 ð@ 1 u3 þ @ 1 u5 Þ u1 ; l24 ¼ 2u4 @ 1 u4 ; l25 ¼ 2u4 @ 1 u5 ; l52 ¼ 2ðu2 þ u4 Þð@ 1 u3 þ @ 1 u5 Þ; l53 ¼ u1 þ 2ðu2 þ u4 Þ@ 1 u5 ; @ ¼ 2ðu2 þ u4 Þ@ 1 u4 þ ; l55 ¼ u1 þ 2ðu2 þ u4 Þ@ 1 u5 : 2
l51 ¼ u3 þ u5 ; l54
After calculations, we get P 1nþ1 ¼ L1 Q 1nþ1 . When u1 ¼ u3 ¼ u5 ¼ 0, the hierarchy (12) can be reduce to
ut ¼
u2
u4
¼
2b1nþ1 þ 2c1nþ1
2c1nþ1
t
¼
0
2
2 0
c1nþ1 b1nþ1 c1nþ1
:
ð13Þ
From (8), we have
b1nþ1 þ c1nþ1
c1nþ1
¼ L2
b0n þ c0n c0n
;
ð14Þ
where
L2 ¼
A
B
C
D
and
A¼
@ þ 2ðu2 þ u4 Þ@ 1 u4 ; 2
C ¼ 2u4 @ 1 u4 ;
B¼
@ þ 2ðu2 þ u4 Þ@ 1 ðu2 þ u4 Þ; 2
@ D ¼ 2u4 @ 1 ðu2 þ u4 Þ: 2
From (6) and (7), it is easy to see that
u1 u3 u5 T U ¼ k þ ; u2 þ ; u4 þ ; k k k
T
V ¼ ða0 þ ka1 ; b0 þ kb1 ; c0 þ kc1 Þ ;
P P P where ai ¼ mP0 aði; mÞk2m ; bi ¼ mP0 bði; mÞk2m ; ci ¼ mP0 cði; mÞk2m ; i ¼ 0; 1. Now we set a ¼ a1 e1 þ a2 e2 þ a3 e3 ; b ¼ b1 e1 þ b2 e2 þ b3 e3 ; a; b 2 glð2Þ. To construct Hamiltonian structures of the obtained integrable hierarchy, we need to transform the Lie algebra glð2Þ into a vector form through the mapping
f : glð2Þ ! R3 ;
a ! ða1 ; a2 ; a3 ÞT :
The mapping f induces a Lie algebraic structure on R3 , isomorphic to the matrix Lie algebra glð2Þ. The corresponding commutator ½:; : on R3 is given by T
½a; b ¼ aT R1 ðbÞ;
a ¼ ða1 ; a2 ; a3 ÞT ;
T
b ¼ ðb1 ; b2 ; b3 Þ 2 R3 :
So, we have T
½a; b ¼ ð2a2 b3 2a3 b2 ; 2a1 b2 2a1 b3 þ 2a2 b1 þ 2a3 b1 ; 2a1 b3 2a3 b1 Þ ¼ aT R1 ðbÞ; where
0
0
B R1 ðbÞ ¼ @ 2b3 2b2
2b2 2b3 2b1 2b1
2b3
1
C 0 A: 2b1
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
The matrix equation (4) gives a system of linear equations on the elements of F 1 . Solving the resulting system, we obtain
0
1
B F1 ¼ @ 0
0 0
0
1
C 1 A:
0 1 1 After calculations we have
fV; U u1 g ¼ a1 þ
a0 ; k
fV; U u2 g ¼ c0 þ kc1 ;
fV; U u5 g ¼ b1 c1
b0 þ c 0 ; k
fV; U u3 g ¼ c1 þ
c0 ; k
fV; U u4 g ¼ b0 c0 kb1 kc1 ;
u1 u3 u5 fV; U k g ¼ 1 2 ða0 þ ka1 Þ þ 2 ðc0 þ kc1 Þ þ 2 ðb0 þ c0 þ kb1 þ kc1 Þ: k k k
Inserting the above results into the quadratic-form identity (5), we get
0 d du
ða0 þ ka1 Þ uk21 ða0 þ ka1 Þ þ uk23 ðc0 þ kc1 Þ
!
þ uk25 ðb0 þ c0 þ kðb1 þ c1 ÞÞ
c
¼k
a1 þ ak0
1
C B c0 þ kc1 C B C @ cB c0 C: B c1 þ k kB C @k B C @ b0 c0 kðb1 þ c1 Þ A
ð15Þ
0 b1 c1 b0 þc k
By comparing the coefficients of k2n3 and k2n2 in both sides of (15), we see that
d ða1nþ2 u1 a1nþ1 þ u3 c1nþ1 þ u5 ðb1nþ1 þ c1nþ1 ÞÞ ¼ ð2n 2 þ cÞP1nþ1 du
ð16Þ
d ða0nþ1 u1 a0n þ u3 c0n þ u5 ðb0n þ c0n ÞÞ ¼ ð2n 1 þ cÞQ 1nþ1 : du
ð17Þ
and
From (8)–(11), one infers c ¼ 0. Thus, the Eq. (16) and (17) give
a1nþ2 u1 a1nþ1 þ u3 c1nþ1 þ u5 ðb1nþ1 þ c1nþ1 Þ ; 2n þ 2
P1nþ1 ¼
dH1n ; du
H1n ¼
Q 1nþ1 ¼
dH2n ; du
H2n ¼
a0nþ1 u1 a0n þ u3 c0n þ u5 ðb0n þ c0n Þ : 2n þ 1
The integrable hierarchy (12) can be written as
ut ¼ J 1
dH1n dH2n ¼ J2 : du du
Therefore, the hierarchy (18) has bi-Hamiltonian structure. It is easy to verify that J 1 L1 ¼ the integrable in the Liouville sense.
ð18Þ L1 J 1
¼ J 2 . So, the system (18) is
3. The generalized zero curvature equation and the application of the Lie algebra G In terms of the bilinear operator m n n 0 0 0 Dm t Dx f g ¼ ð@ t @ t Þ ð@ x @ x0 Þ f ðt; xÞgðt ; x Þjt 0 ¼t;x0 ¼x ;
where Dt , Dx are differential operators, f ðt; xÞ, gðt; xÞ are differentiable functions on variations t; x. We consider the isospectral problems in [9] as follows:
ðDx þ Dt Þu 1 ¼ U u; ðDx Dt Þu 1 ¼ V u:
ð19Þ
From (19) and the properties of the bilinear operator, it follows that
ð@ x þ @ t Þu 1 ¼ U u; ð@ x @ t Þu 1 ¼ V u:
Eq. (20) is equivalent to
ux þ ut ¼ U u; ux ut ¼ V u:
ð20Þ
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
Based on their compatibility conditions utx ¼ uxt , we have the zero curvature equation
U x U t ðV x þ V t Þ þ ½V; U ¼ 0:
ð21Þ
By using of the zero curvature Eq. (21), we can obtain some new integrable hierarchies. Consider the matrix Lie algebra G1 , which is expanded by glð2Þ
G1 ¼ spanfg 1 ; g 2 ; g 3 ; g 4 ; g 5 ; g 6 g; where
0
1
0
0
0
1
0
B 0 1 0 0 C B C g1 ¼ B C; @0 0 1 0 A 0 0 0 1 1 0 0 1 0 B 0 0 0 1 C C B g4 ¼ B C; @0 0 0 0 A 0
0 0 0
0 0
0 0
1
0
B1 0 0 0C B C g2 ¼ B C; @0 0 0 0A 0 0 1 0 1 0 0 0 0 B0 0 1 0C C B g5 ¼ B C; @0 0 0 0A 0
0
0 0
1 2 0
B0 B g3 ¼ B @0 0
1 0 0
0
0 C C C; 1 2 A 0 1 0
1 0 0 1 2 B0 0 0 1 C C B g6 ¼ B C; @0 0 0 0 A 0
0 0
0 0 0
1
1 1 0 0 0 B1 1 0 0C C B M2 ¼ B C; @0 0 1 0A
0
0
0
0 1 1
along with the following commutative relations
8 ½g 1 ; g 2 ¼ 2g 2 ; ½g 1 ; g 3 ¼ 2g 2 þ 2g 3 ; ½g 1 ; g 4 ¼ 0; ½g 1 ; g 5 ¼ 2g 5 ; > > > > > > < ½g 1 ; g 6 ¼ 2g 5 þ 2g 6 ; ½g 2 ; g 3 ¼ 2g 1 ; ½g 2 ; g 4 ¼ 2g 5 ; ½g 2 ; g 5 ¼ 0; ½g 2 ; g 6 ¼ 2g 4 ; ½g 3 ; g 4 ¼ 2g 5 2g 6 ; ½g 3 ; g 5 ¼ 2g 4 ; ½g 3 ; g 6 ¼ 0; > > > > ½g 4 ; g 5 ¼ ½g 4 ; g 6 ¼ ½g 5 ; g 6 ¼ 0; > > : ½g i ; g j ¼ g i M 2 g j g j M 2 g i ; g i ; g j 2 G1 : We can construct the following loop algebra G~1 ¼ spanfg 1 ðnÞ; g 2 ðnÞ; g 3 ðnÞ; g 4 ðnÞ; g 5 ðnÞ; g 6 ðnÞg, where g i ðnÞ ¼ g i kn ; i ¼ 1; 2; 3; 4; 5; 6, along with ½g i ðmÞ; g j ðnÞ ¼ ½g i ; g j kmþn ; i; j ¼ 1; 2; 3; 4; 5; 6. Consider the following isospectral problem
ðDx þ Dt Þu 1 ¼ U u; U ¼ g 1 ð1Þ þ u1 g 2 ð0Þ þ u2 g 3 ð0Þ þ u3 g 5 ð0Þ þ u4 g 6 ð0Þ:
ð22Þ
Set
V¼
X
ðam g 1 ðmÞ þ bm g 2 ðmÞ þ cm g 3 ðmÞ þ dm g 4 ðmÞ þ em g 5 ðmÞ þ fm g 6 ðmÞÞ:
mP0
The stationary zero curvature equation
V x þ V t ¼ ½V; U;
ð23Þ
gives rise to
8 amz ¼ 2u2 bm 2u1 cm ; > > > > > bmz ¼ 2u1 am 2u2 am þ 2bmþ1 þ 2cmþ1 ; > > > > > > < cmz ¼ 2u2 am 2cmþ1 ; dmz ¼ 2u4 bm 2u3 cm þ 2u2 em 2u1 fm ; > > > > emz ¼ 2u3 am 2u4 am 2u1 dm 2u2 dm þ 2emþ1 þ 2f mþ1 ; > > > > > fmz ¼ 2u4 am þ 2u2 dm 2f mþ1 ; > > :@ @ ¼ @x þ @t@ : @z
ð24Þ
Take the initial data as
a0 ¼ b;
b0 ¼ c0 ¼ d0 ¼ e0 ¼ f0 ¼ 0:
ð25Þ
Then the recursion relation (24) uniquely determines the sequence of sets of ai ; bi ; ci ; di ; ei ; f i , i P 1. The first few sets are listed as follows:
a1 ¼ 0; b1 ¼ bu1 ; b a2 ¼ bu1 u2 þ u22 ; 2
c1 ¼ bu2 ; b2 ¼
d1 ¼ 0;
b ðu1z þ u2z Þ; 2
e1 ¼ bu3 ;
f 1 ¼ bu4 :
b c2 ¼ u2z ; 2
B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
d2 ¼ b u1 u4 þ u2 u3 þ u2 u4 ;
b ðu3z þ u4z Þ; 2
e2 ¼
267
b f 2 ¼ u4z : 2
Denoting n V ðnÞ þ ¼ Rm¼0 ðam g 1 ðn mÞ þ bm g 2 ðn mÞ þ c m g 3 ðn mÞ þ dm g 4 ðn mÞ þ em g 5 ðn mÞ þ fm g 6 ðn mÞÞ:
A direct calculation reads ðnÞ ðnÞ ðV ðnÞ þx þ V þt Þ þ ½V þ ; U ¼ ð2bmþ1 2c mþ1 Þg 2 ð0Þ þ ð2c mþ1 Þg 3 ð0Þ þ ð2emþ1 2f mþ1 Þg 5 ð0Þ þ ð2f mþ1 Þg 6 ð0Þ:
Taking V ðnÞ ¼ V ðnÞ þ , the zero curvature equation ðnÞ ðnÞ U x U t ðV ðnÞ ; U ¼ 0; x þ V t Þ þ ½V
leads to the following Lax integrable hierarchy
1 u1 Bu C B 2C B C @ u3 A
1 1 0 2bnþ1 þ 2cnþ1 cnþ1 fnþ1 C C B B 2cnþ1 C B B bnþ1 cnþ1 enþ1 fnþ1 C ¼B C ¼ J3 B C ¼ J 3 P 2nþ1 ; A @ 2enþ1 þ 2f nþ1 A @ cnþ1
0
u4
0
bnþ1 cnþ1
2f nþ1
xt
ð26Þ
where
0
0
0
2
1
2 0
0 2
C C C: A
2
0
0
B0 0 B J3 ¼ B @ 0 2 2
0
From (24), we are easy to obtain the recurrence relation
0
0
l11
B l0 B P2nþ1 ¼ L3 P2n ¼ B 21 @ 0 0
l12
0
l13
0
l22 0
0
l23 0 l33
0
l43
0
l14
1
0
0 l24 C C 0 CP 2n l34 A
0
l44
0
and
@ 0 0 l11 ¼ l33 ¼ 2u2 @ 1 u1 2u2 @ 1 u2 ; 2 0
0
0
0
l21 ¼ l43 ¼ 2u1 @ 1 u1 2u2 @ 1 u1 l22 ¼ l44 ¼
0
0
l12 ¼ l34 ¼ 2u2 @ 1 u2 ;
@ 2u1 @ 1 u2 2u2 @ 1 u2 ; 2
@ þ 2u1 @ 1 u2 þ 2u2 @ 1 u2 ; 2
0
l14 ¼ 2u4 @ 1 u2 þ 2u2 @ 1 u4 ;
0
l13 ¼ 2u4 @ 1 u1 2u2 @ 1 u3 2u4 @ 1 u2 2u2 @ 1 u4 ; 0
l23 ¼ 2u3 @ 1 u1 2u1 @ 1 u1 2u1 @ 1 u3 2u2 @ 1 u3 2u3 @ 1 u2 2u4 @ 1 u2 2u1 @ 1 u4 2u2 @ 1 u4 ; 0
l24 ¼ 2u3 @ 1 u2 þ 2u4 @ 1 u2 þ 2u1 @ 1 u4 þ 2u2 @ 1 u4 : Therefore, the system (26) can be written as
1 u1 Bu C B 2C ¼B C @ u3 A 0
uxt
u4
1 ðu2 þ u4 Þb B ðu þ u þ u þ u Þb C 1 2 3 4 C B ¼ J3 Ln3 B C: A @ u2 b 0
ðu1 þ u2 Þb
xt
When u3 ¼ u4 ¼ 0, then hierarchy (26) can reduce to
uxt ¼
u1 u2
¼ xt
2bnþ1 þ 2cnþ1 2cnþ1
¼
0
2
2 0
cnþ1 bnþ1 cnþ1
:
ð27Þ
268
B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
From (24), we have
bnþ1 þ cnþ1 cnþ1
@ 2
¼
þ 2ðu1 þ u2 Þ@ 1 u2
@ 2
þ 2ðu1 þ u2 Þ@ 1 ðu1 þ u2 Þ
2u2 @ 1 u2
@ 2
!
þ 2u2 @ 1 ðu1 þ u2 Þ
bn þ c n : cn
ð28Þ
This form is similar to the AKNS hierarchy, but not the AKNS hierarchy. Based on the above calculations, it is not difficult to show the following theorem: Theorem 2. Suppose that xðkÞ is the polynomial of degree n, which is independent of x. If k is not rely on t, that is kt ¼ 0, and the matrix V meets the following boundary condition
0 2cnþ1 B @V B 2bnþ1 þ 2cnþ1 ¼B @z ðu1 ;u2 ;u3 ;u4 Þ¼ð0;0;0;0Þ @ 0
2f nþ1
2cnþ1 0
2enþ1 þ 2f nþ1 2cnþ1
2f nþ1 C C C: 4cnþ1 A
0
2bnþ1 þ 2cnþ1
2cnþ1
0
4f nþ1
1
4cnþ1
Then from (21), (22), (23) and (25), we can only determine the matrix V, and the isospectral hierarchy
1 u1 Bu C B 2C B C @ u3 A
1 ðu2 þ u4 Þb B ðu þ u þ u þ u Þb C 1 2 3 4 C B ¼ xðL3 ÞB C; A @ u2 b
0
u4
0
ðu1 þ u2 Þb
xt
where L3 is a recurrence operator. Next, we will establish the Hamiltonian structure of the integrable system by using the quadratic-form identity. Set
a ¼ a1 g 1 þ a2 g 2 þ a3 g 3 þ a4 g 4 þ a5 g 5 þ a6 g 6 ;
b ¼ b1 g 1 þ b2 g 2 þ b3 g 3 þ b4 g 4 þ b5 g 5 þ b6 g 6 ;
a; b 2 G1
To construct Hamiltonian structures of the obtained integrable hierarchy, we need to transform the Lie algebra G1 into a rector form through the mapping
a ! ða1 ; a2 ; a3 ; a4 ; a5 ; a6 ÞT :
f2 : G1 ! R6 ;
The mapping f2 induces a Lie algebraic structure on R6 , isomorphic to the matrix Lie algebra G1 . The corresponding commutator [.,.] on R6 is given by T
½a; b ¼ aT R2 ðbÞ;
a ¼ ða1 ; . . . ; a6 ÞT ;
T
b ¼ ðb1 ; . . . ; b6 Þ 2 R6
and T
½a; b ¼ð2a2 b3 2a3 b2 ;
2a1 b2 2a1 b3 þ 2a2 b1 þ 2a3 b1 ;
2a1 b3 2a3 b1 ;
2a1 b5 2a1 b6 þ 2a2 b4 þ 2a3 b4 2a4 b2 2a4 b3 þ 2a5 b1 þ 2a6 b1 ;
2a2 b6 2a3 b5 þ 2a5 b2 2a6 b2 ; T
2a1 b6 2a3 b4 þ 2a4 b3 2a6 b1 Þ ¼ aT R2 ðbÞ;
where
0
0
2b2 2b3
2b3
0
2b5 2b6
2b1
0
2b6
2b4
2b1
2b1
2b5
2b4
0
0
0
2b2 2b3
0 0
0 0
2b3 2b2
2b1 2b1
B 2b 3 B B B 2b2 R2 ðbÞ ¼ B B 0 B B @ 0 0
2b6
1
C C C 2b4 C C: 2b3 C C C 0 A 0
2b1
From [7], we know that R2 ðbÞ is required to satisfy with
F 2 ¼ F T2 ;
T
R2 ðbÞF 2 ¼ F 2 ðR2 ðbÞÞ :
This matrix equation gives a system of linear equations on the elements of F 2 . Solving the resulting system, we obtain
0
1 0 0 B 0 0 1 B B B 0 1 1 F2 ¼ B B1 0 0 B B @ 0 0 1 0 1 1
1
0
0
1
1 C C C 0 1 1 C C: 0 0 0 C C C 0 0 0 A 0 0 0 0
0
In order to the Hamiltonian structure of the Lax integrable system, we define a linear functional fa; bg as follows: fa; bg ¼ aT F 2 b.
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By a direct calculation, we have
@U V; ¼ c f ; @r
@U V; ¼ b c e f ; @u1
V;
@U @u3
¼ c;
@U V; ¼ b c; @u4
V;
@U @k
¼ a þ d:
Inserting the above result into the quadratic-form identity (5), we get
0
c f B b c e f d @ B ða þ dÞ ¼ kc kc B du @k @ c
1 C C C: A
ð29Þ
b c By comparing the coefficients of kn2 in (29), we see that
0
cn fn
1
B b c e f C d n n nC B n ðanþ2 þ dnþ2 Þ ¼ ðn 1 þ cÞB C ¼ ðn 1 þ cÞP2n : @ A du cn bn cn It is easy to see that c ¼ 0, thus we have
P2nþ1 ¼
dH3nþ1 ; du
H3nþ1 ¼
anþ2 þ dnþ2 : nþ1
The integrable hierarchy (26) can be written as
uxt ¼ J 3
dH3nþ1 : du
ð30Þ
Which is the Hamiltonian structure of the hierarchy (26). After tedious computations, we have J 3 L3 ¼ L3 J 3 . So, the hierarchy (30) is integrable in the Liouville sense. 4. Two kinds of Darboux transformations of the Eq. (31) Let b ¼ 2; n ¼ 2, the hierarchy (26) reduces to a new equation as follows
8 u1;xt > > > > < u2;xt > > u3;xt > > : u4;xt
¼ 2b ðu1zz þ u2zz Þ þ 2bu21 u2 þ 3bu1 u22 þ bu32 ; ¼ 2b u2zz 2bu1 u22 bu32 ; ¼ 2b ðu3zz þ u4zz Þ þ 4bu1 u2 u3 þ 3bu3 u22 þ 6bu1 u2 u4 þ 3bu22 u4 þ 2bu21 u4 ;
ð31Þ
¼ 2b u4zz 4bu1 u2 u4 3bu22 u4 2bu22 u3 ;
whose Lax pair matrices present that
ux þ ut ¼ U u; ux ut ¼ V u;
where
0 B B U¼B @
1
k þ u2
2u2
u4
2u4
u1
k þ u2
u3
u4
0
0
k þ u2
2u2
0
0
u1
k þ u2
C C C A
and
0
V 11 BV B 21 V ¼B @ 0 0
1
V 12
V 13
V 14
V 22
V 23
0
V 11
V 24 C C C; V 12 A
0
V 21
V 22
V 11 ¼ 2k2 þ 2u1 u2 þ u22 þ 2ku2 u2z ; V 22 ¼ 2k2 2u1 u2 u22 þ 2ku2 u2z ;
ð32Þ
V 12 ¼ 4ku2 þ 2u2z ;
V 21 ¼ 2ku1 þ u1z þ u2z ;
V 23 ¼ 2ku3 þ u3z þ u4z ;
V 13 ¼ 2ðu1 u4 þ u2 u3 þ u2 u4 Þ þ 2ku4 u4z ;
V 14 ¼ 4ku4 þ 2u4z ;
V 24 ¼ 2ðu1 u4 þ u2 u3 þ u2 u4 Þ þ 2ku4 u4z :
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
In this section, we will investigate the Dardoux transformations of the Eq. (31). Consider a Darboux transformation
/0 ¼ T/ and require /0 and / satisfying the spectral problem
/x ¼ U/;
ð33Þ
/0x ¼ U 0 /0
and the spectral problem (22), where U 0 has the same form as U expect replacing u1 ; u2 ; u3 ; u4 by u01 ; u02 ; u03 ; u04 . It is easy to see that T meets
T x þ TU ¼ U 0 T:
ð34Þ
In what follows, we discuss two kinds of forms of matrix T. Assume that
0 T ¼ kT 1 þ T 0 ;
1
0
a1
b1
e1
f1
Bc B 1 T1 ¼ B @0
d1
s1
0
a1
t1 C C C; b1 A
0
0
c1
d1
a
b
e
1
f
Bc d s t C B C T0 ¼ B C; @0 0 a bA 0 0
c
d
where a1 ; b1 ; c1 ; d1 ; e1 ; f 1 ; s1 ; t1 and a; b; c; d; e; f ; s; t are functions in x and t. From the spectral problem (22), we have
0
k þ u2
2u2
u4
2u4
1
0
0
1
u1 0
k þ u2 0
u3 k þ u2
u4 2u2
C B 0 1 0 C B C ¼ kB A @0 0 1
0 0
0
0
u1
k þ u2
0
k þ u02
2u02
u04
2u04
u01
k þ u02
u03
0
0
k þ u02
0
0
u01
B B U¼B @
0
1
0
0
0
u2
1
2u2
u4
2u4
C Bu C B 1 CþB A @0
u2 0
u3 u2
u4 C C C ¼ kE0 þ U 0 2u2 A
0
0
u1
0 1
u2
and
0 B B U0 ¼ B @
1
u04 C C 0 C ¼ kE0 þ U 0 : 2u02 A k þ u02
Based on the Eq. (34), we obtain 0
a1x b1x Bc d B 1x 1x T x ¼ kB @ 0 0 0 0 0
a1 B 2 B c1 k B @0 0
e1x s1x a1x c1x b1 d1 0 0
e1 s1 a1 c1
1 0 f1x ax bx ex B t 1x C C B cx dx sx CþB b1x A @ 0 0 ax d1x 0 0 cx 1 0 f1 a 1 b1 C Bc d t1 C B 1 1 CE0 kB @0 0 b1 A d1 0 0
1 0 fx a 1 b1 C Bc d tx C B 1 1 2 C ¼ k E0 B @0 0 bx A dx 0 0 e1 s1 a1 c1
e1 s1 a1 c1
1 0 1 0 f1 a b e f a1 C B C B t1 C Bc d s t C 0 B c1 C þ kE0 B C þ kU B @0 0 a bA @0 b1 A d1 0 0 0 c d
1 0 1 0 f1 a b e f a C B C Bc t1 C Bc d s t C B CU kB CE B @0 0 a bA 0 @0 b1 A d1 0 0 c d 0
b d 0 0
e s a c
b1 d1 0 0
e1 s1 a1 c1
1 0 1 f1 a b e f C B C t1 C 0B c d s t C CþU B C @0 0 a bA b1 A d1 0 0 c d
1 f tC C CU: bA d ð35Þ
j
Comparing the coefficients of k ðj ¼ 0; 1; 2Þ in Eq. (35). As for the case of j ¼ 2, we get that
a1 a1 ¼ 0;
b1 þ b1 ¼ 0;
e1 e1 ¼ 0;
f 1 þ f1 ¼ 0;
c1 c1 ¼ 0; s1 s1 ¼ 0;
d1 þ d1 ¼ 0; t 1 þ t 1 ¼ 0:
So we can obtain that
b1 ¼ 0;
c1 ¼ 0;
f 1 ¼ 0;
s1 ¼ 0:
ð36Þ
The case of j ¼ 1 leads to the following equations
8 a1x ¼ a þ u02 a1 2u02 c1 u2 a1 u1 b1 a; > > > > > b1x ¼ b þ u02 b1 2u02 d1 þ 2u2 a1 u2 b1 þ b; > > > > > c1x ¼ c þ u01 a1 þ u02 c1 u2 c1 u1 d1 c; > > > < d1x ¼ d þ u01 b1 þ u02 d1 þ 2u2 c1 u2 d1 þ d; > e1x ¼ e þ u02 e1 2u02 s1 þ u04 a1 2u04 c1 u4 a1 u3 b1 u2 e1 u1 f1 e; > > > > 0 0 0 0 > > f1x ¼ f þ u2 f1 2u2 t1 þ u4 b1 2u4 d1 þ 2u4 a1 u4 b1 þ 2u2 e1 u2 f1 þ f ; > > > 0 0 0 0 > s ¼ s þ u e þ u s þ u a þ u > 1x 1 1 2 1 3 1 4 c 1 u4 c 1 u3 d1 u2 s1 u1 t 1 s; > : 0 0 0 0 t1x ¼ t þ u1 f1 þ u2 t 1 þ u3 b1 þ u4 d1 þ 2u4 c1 u4 d1 þ 2u2 s1 u2 t 1 þ t
ð37Þ
B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
271
And the case of j ¼ 0 leads to the following equations
8 ax ¼ u02 a 2u02 c u2 a u1 b; > > > > > bx ¼ u02 b 2u02 d þ 2u2 a u2 b; > > > > > cx ¼ u01 a þ u02 c u2 c u1 d; > > > < d ¼ u0 b þ u0 d þ 2u c u d; x 2 2 1 2 > ex ¼ u02 e 2u02 s þ u04 a 2u04 c u4 a u3 b u2 e u1 f ; > > > > > fx ¼ u02 f 2u02 t þ u04 b 2u04 d þ 2u4 a u4 b þ 2u2 e u2 f ; > > > > > > sx ¼ u01 e þ u02 s þ u03 a þ u04 c u4 c u3 d u2 s u1 t; > : t x ¼ u01 f þ u02 t þ u03 b þ u04 d þ 2u4 c u4 d þ 2u2 s u2 t: Substituting (36) into (37) gives
(
a1 ¼ d1 ; f ¼
ðu04
e1 ¼ t1 ;
u4 Þa1 þ
c ¼ 12 ðu01 u1 Þa1 ;
b ¼ ðu02 u1 Þa1 ;
ðu02
u2 Þe1 ;
s¼
1 ½ðu03 2
ð38Þ
u3 Þa1 þ ðu01 u1 Þe1 :
4.1. The first Darboux transformation Set a1 ¼ d1 ¼ 0; e1 ¼ t1 ¼ 1. From (38), it is easy to see that b ¼ 0; c ¼ 0; f ¼ u02 u2 ; s ¼ 12 ðu01 u1 Þ. The matrix T can be written to the following form
0
0 0 1 0
1
0
a
0
B0 0 0 1C B0 d C B B T ¼ kB CþB @0 0 0 0A @0 0 0 0 0
0
0 0
u02 u2
e 1 ðu01 2
u1 Þ
t
a
0
0
d
1
C C C: A
0
Since / and / of (33) are 4 4 matrices, we need that det/0 ¼ 0. There exists a constant k ¼ k1 and a solution / ¼ ð/1 ; /2 ; /3 ; /4 ÞT of the spectral problem (22) which satisfy that
8 a/3 ¼ 0; > > > > < d/4 ¼ 0;
) a ¼ 0; ) d ¼ 0;
k1 /3 þ a/1 þ e/3 þ ðu02 u2 Þ/4 ¼ 0; ) e ¼ k1 ðu02 u2 Þ //43 ; > > > > : k / þ d/ þ 1 ðu0 u Þ/ þ t/ ¼ 0; ) t ¼ k 1 ðu0 u Þ /3 : 1 4 1 1 1 / 2 3 4 1 1 2 2 4 We are easy to get the Darboux transformation T:
1 00 0 0 1 0 B0 0 0 1C B 0 C B B T ¼ k1 B CþB @0 0 0 0A B @0 0 0 0 0 0 0
0 k1 ðu02 u2 Þ //43 0 0 0
1 ðu01 2
u1 Þ 0 0
u02 u2
1
C k1 12 ðu01 u1 Þ //34 C C: C A 0 0
4.2. The second Darboux transformation Set a1 ¼ d1 ¼ 1; e1 ¼ t 1 ¼ 0. We have the following form of the Darboux transformation T:
T ¼ kI S;
ð39Þ
where I is a 4 4 unit matrix. From the Eq. (34), we have
Sx þ ðkI SÞU ¼ U 0 ðkI SÞ;
) Sx þ ðkI SÞðkE0 þ U 0 Þ ¼ ðkE0 þ U 00 ÞðkI SÞ:
ð40Þ
j
Comparing the coefficients of k ðj ¼ 0; 1; 2Þ in Eq. (40), we get
8 2 > < k : E0 ¼ E0 ; k1 : U 0 SE0 ¼ U 00 E0 S; ) U 00 ¼ U 0 SE0 þ E0 S; > : 0 k : Sx ¼ SU 0 U 00 S: ) Sx ¼ U 00 S SU 0 :
ð41Þ
From (41), we have
Sx ¼ ðU 0 SE0 þ E0 SÞS SU 0 ¼ U 0 S SE0 S þ E0 S2 SU 0 :
ð42Þ
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B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
Now, we need to find the matrix S, which can be expressed by the eigenfunction of (33). Assume that H is the solution of (33), that is
Hx ¼ UH ¼ E0 H ^ þU 0 H;
ð43Þ
where ^ ¼ diagðk1 ; k1 ; k1 ; k1 Þ. Set
S ¼ H ^ H1 :
ð44Þ
From (43), we have
Sx ¼ Hx ^ H1 H ^ H1 Hx H1 ¼ ðE0 H ^ þU 0 HÞ ^ H1 SðE0 H ^ þU 0 HÞH1 ¼ E0 S2 þ U 0 S SE0 S SU 0 :
ð45Þ
It is easy to see that (45) is equivalent to (42). Next, we shall prove again that the transformations (39) map equation
/t ¼ V/; into equation
/0t ¼ V 0 /0 ; where V is presented in (32), and V 0 has the same form as V except replacing u1 ; u2 ; u3 ; u4 ; u1z ; u2z , u3z ; u4z by u01 ; u02 ; u03 ; u04 ; u01z ; u02z ; u03z ; u04z . To find a new solution of the soliton equations, we should prove
T t þ TV ¼ V 0 T:
ð46Þ
Proof. From (32), we get
V ¼ 2k2 E0 þ 2kU 0 þ V 0 ; where
0
v 11 v 12 v 13 Bv B 21 v 22 v 23 V0 ¼ B @ 0 0 v 11 0 0 v 21
v 14 1 v 24 C C C; v 12 A v 22
v 11 ¼ 2u1 u2 þ u22 u2z ; v 12 ¼ 2u2z ; v 13 ¼ 2ðu1 u4 þ u2 u3 þ u2 u4 Þ u4z ; v 21 ¼ u1z þ u2z ; v 14 ¼ 2u4z ; v 22 ¼ 2u1 u2 u22 u2z ; v 23 ¼ u3z þ u4z ; v 24 ¼ 2ðu1 u4 þ u2 u3 þ u2 u4 Þ u4z : After the Darboux transformation /0 ¼ T/, we have
/0t ¼ V 0 /0 ¼ ð2k2 E0 þ 2kU 00 þ V 00 Þ/0 ; Assume H satisfies that
Hx ¼ UH; , we are easy to see that Ht ¼ VH;
Ht ¼ VH ¼ 2E0 H^2 þ 2U 0 H ^ þV 0 H: From (46), we get
St þ ðkI SÞð2k2 E0 þ 2kU 0 þ V 0 Þ ¼ ð2k2 E0 þ 2kU 00 þ V 00 ÞðkI SÞ:
ð47Þ
j
Comparing the coefficients of k ðj ¼ 0; 1; 2; 3Þ in Eq. (47), we get
8 3 > k > > > < k2 > k1 > > > : 0 k
: 2E0 ¼ 2E0 ; : 2U 0 2SE0 ¼ 2U 00 2E0 S; : V 0 2SU 0 ¼ V 00 2U 00 S; : St SV 0 ¼ V 00 S:
) U 00 ¼ U 0 SE0 þ E0 S; ) V 00 ¼ V 0 2SU 0 þ 2U 00 S;
) St ¼ V 00 S SV 0 :
From (44) and (48), we have
St ¼ Ht ^ H1 H ^ H1 Ht H1 ¼ ð2E0 H^2 þ 2U 0 H ^ þV 0 HÞ ^ H1 Sð2E0 H^2 þ 2U 0 H ^ þV 0 HÞH1 ¼ 2E0 S3 þ 2U 0 S2 þ V 0 S 2SE0 S2 2SU 0 S SV 0 And
St ¼ V 00 S SV 0 ¼ ðV 0 2SU 0 þ 2U 00 SÞS SV 0 ¼ V 0 S 2SU 0 S þ 2ðU 0 SE0 þ E0 SÞS2 SV 0 ¼ V 0 S 2SU 0 S þ 2U 0 S2 2SE0 S2 þ 2E0 S3 SV 0 :
ð48Þ
B. He, L. Chen / Applied Mathematics and Computation 244 (2014) 261–273
273
5. Conclusion Based on a pair of the isospectral problem
ux ¼ U u ut ¼ V u , in terms of the compatibility conditions: uxt ¼ utx , we are usually
easy to obtain the zero curvature equation U t V x þ ½U; V ¼ 0. Then making use of the corresponding stationary zero curvature equation and the commutative operation, the Lax integrable hierarchy can be obtained. In addition, taking advantage of the trace identity or the quadratic-form identity, the Hamiltonian structure can be worked out. At last, we get two kinds of Darboux transformations of Eq. (31). In this article, according to the properties of the bilinear operator and the zero curvature Eq. (21), we consider the isospectral problem (22), integrable hierarchy (26) can be obtained. Ultimately, by making use of the quadratic-form identity, we obtain the Hamiltonian structure of the hierarchy (26). From Sections 2 and 3 of this paper, we find that Eq. (13) and (14) are equivalent to Eq. (27) and (28). In other words, the hierarchy (12) and hierarchy (26) can reduce to the same hierarchy. In Section 4, we let b ¼ 2; n ¼ 2, the integrable hierarchy (26) can be reduced to Eq. (31). In terms of the investigation of Darboux transformations [11], we get two kinds of Darboux transformations of Eq. (31). Acknowledgments The authors would like to thank the referee for valuable comments and suggestions on this article. This work was supported by the National Natural Science Foundation of China (No. 11171055), the Natural Science Foundation of Jilin province (Nos. 201115006 and 20130101068JC), Scientific Research Foundation for Returned Scholars Ministry of Education of China. References [1] Tu. Guizhang, The trace identity, a powerful tool for constructing the Hamiltonian structure of integrable systems, J. Math. Phys. 30 (2) (1989) 330– 338. [2] Honwah Tam, Yufeng Zhang, Spectral radius analysis of matrices and the associated with integrable systems, Int. J. Modern Phys. B 23 (24) (2009) 4855–4879. [3] Zhang Yufeng, Liu Jing, Induced Lie algebras of a six-dimensional matrix Lie algebra, Commun. Theor. Phys. (Beijing, China) 50 (2008) 289–294. [4] Yufeng Zhang, Qingyou Yan, Integrable couplings of the hierarchies of evolution equations, Chaos Solitons Fractals 16 (2003) 263–269. [5] Wen-Xiu Ma, Multi-component bi-Hamiltonian Dirac integrable equations, Chaos Solitons Fractals 39 (2009) 282–287. [6] Wen-Xiu Ma, A multi-component Lax integrable hierarchy with Hamiltonian structure, Pac. J. Appl. Math. 1 (2) (2008) 191–197. [7] Fu-Kui Guo, Huan-He Dong, A new loop algebra and corresponding computing formula of constant in quadratic-form identity, Commun. Theor. Phys. (Beijing) 47 (6) (2007) 981–986. [8] Fukui Guo, Yufeng Zhang, The integrable coupling of the AKNS hierarchy and its Hamiltonian structure, Chaos Solitons Fractals 32 (2007) 1898–1902. [9] Yufeng Zhang, A complex higher-dimensional Lie algebra with real and imaginary structure constants as well as its decomposition, Commun. Theor. Phys. (Beijing) 50 (5) (2008) 1021–1026. [10] Yufeng Zhang, Zhong Han, Hon-Wah Tam, An integrable hierarchy and Darboux transformations, bilinear Backlund transformations of a reduced equation, Appl. Math. Comput. 219 (2013) 5837–5848. [11] Li Yishen, Solitons and Integrable Systems, Shanghai Science and Technology Education Publishing House, 1999.