Using Vector-Product Loop Algebra to Generate Integrable Systems

Abstract

A new three-dimensional Lie algebra and its loop algebra are proposed by us, whose commutator is a vector product. Based on this, a positive flow and a negative flow are obtained by introducing a new kind of spectral problem expressed by the vector product, which reduces to a generalized KdV equation, a generalized Schrödinger equation, a sine-Gordon equation, and a sinh-Gordon equation. Next, the well-known Tu scheme is generalized for generating isospectral integrable hierarchies and non-isospectral integrable hierarchies. It is important that we make use of the variational method to create a new vector-product trace identity for which the Hamiltonian structure of the isospectral integrable hierarchy presented in the paper is worded out. Finally, we further enlarge the three-dimensional loop algebra into a six-dimensional loop algebra so that a new isospectral integrable hierarchy which is a type of extended integrable model is produced whose bi-Hamiltonian structure is also derived from the vector-product trace identity. This new approach presented in the paper possesses extensive applications in the aspect of generating integrable hierarchies of evolution equations.

1. Introduction

Searching for new integrable systems has been an important aspect in the integrable system field, and zero-curvature equations play a prominent role in constructing integrable equations (see e.g., [1,2,3,4]). The Hamiltonian structure can be furnished for the resulting integrable equations by applying the trace identity [1] and the quadratic-type identity [5,6]. Tu Guizhang [1] applied the finite-dimensional matrix Lie algebras to introduce isospectral problems for some integrable hierarchies of equations and the resultant Hamiltonian structure generated was called the Tu scheme by Ma Wen-Xiu [5]. Later, Zhang et al. [7] extended the Tu scheme to the case of generated non-isospectral integrable hierarchies. By using the Tu scheme and its applications for isospectral and nonisospectral integrable systems, many integrable equations and their some properties (such as Hamiltonian structures, Backlund transformations, and so on) are followed (see, for example, [8,9,10,11,12]). In the paper, we want to apply a Lie algebra which consists of vectors to introduce isospectral and non-isospectral Lax pairs, whose compatibility conditions give rise to isospectral and non-isospectral zero-curvature equations for which two integrable hierarchies are obtained. In order to search the Hamiltonian structure of the isospectral integrable hierarchy by using the variational method, a new vector-product trace identity is generated. In the following, we recall some basic notations and computing formulae based on any standard university courses in linear algebra or any analytic geometry of space. It is noteworthy that the vectors in the following Definitions 1 and 2 and Lemmas 1 and 2 as well as Theorem 1 are all three-dimensional, which belong to the space $R^3$.

Definition 1.

The inner product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ in the space $R^3$ denoted by $\overrightarrow{a}\ast \overrightarrow{b}$ is defined as

$$\overrightarrow{a}\ast \overrightarrow{b}=|\overrightarrow{a}\left|\right|\overrightarrow{b}|cos〈(\overrightarrow{a},\overrightarrow{b})〉,$$

where $〈(\overrightarrow{a},\overrightarrow{b})〉$ stands for the angle of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, and $0\le 〈(\overrightarrow{a},\overrightarrow{b})〉\le \pi$.

For arbitrary vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$, the inner product has the following properties

(1) 

$\overrightarrow{a}\ast \overrightarrow{b}=\overrightarrow{b}\ast \overrightarrow{a},$

(2) 

$(\lambda \overrightarrow{a})\ast \overrightarrow{b}=\lambda \ast (\overrightarrow{a}\ast \overrightarrow{b}),$

(3) 

$(\overrightarrow{a}+\overrightarrow{b})\ast \overrightarrow{c}=\overrightarrow{a}\ast \overrightarrow{c}+\overrightarrow{b}\ast \overrightarrow{c}.$

Definition 2.

The outer product (again called vector product) of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ in the space $R^3$ is noted by $\overrightarrow{a}\times \overrightarrow{b}$ whose length is given by

$$|\overrightarrow{a}\times \overrightarrow{b}|=|\overrightarrow{a}\left|\right|\overrightarrow{b}|sin〈(\overrightarrow{a},\overrightarrow{b})〉,$$

whose direction is required to be vertical to the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, further satisfies the right-handed coordinate system. For arbitrary vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$, the vector product possesses the following properties

(1) 

$\overrightarrow{a}\times \overrightarrow{b}=-(\overrightarrow{b}\times \overrightarrow{a}),$

(2) 

$(\lambda \overrightarrow{a})\times \overrightarrow{b}=\lambda (\overrightarrow{a}\times \overrightarrow{b}),$

(3) 

$\overrightarrow{a}\times (\overrightarrow{b}+\overrightarrow{c})=\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c},(\overrightarrow{a}+\overrightarrow{b})\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{c}.$

Assume ${\overrightarrow{e}}_1,{\overrightarrow{e}}_2$ and ${\overrightarrow{e}}_3$ are unit coordinate vectors in the right-handed vectical coordinate system $\left\{0;{\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3\right\}$, then for arbitrary vectors $\overrightarrow{a}=(a_1,a_2,a_3),\overrightarrow{b}=(b_1,b_2,b_3)\in \left\{0;{\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3\right\},$ where $a_i,b_i$ are coordinates of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, respectively, the inner product reads that

$$\overrightarrow{a}\ast \overrightarrow{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3.$$

(1)

The vector product of the vectors $\overrightarrow{a},\overrightarrow{b}$ is given by

$$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}{\overrightarrow{e}}_1& {\overrightarrow{e}}_2& {\overrightarrow{e}}_3\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|.$$

(2)

Lemma 1.

For arbitrary three vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},$ the following identity holds

$$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}\ast \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\ast \overrightarrow{b})\overrightarrow{c}.$$

(3)

Proof. 

It is enough to prove (3) under the right-handed vertical coordinate system $\left\{0;{\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3\right\}$. Assume $\overrightarrow{a}=(a_1,a_2,a_3),\overrightarrow{b}=(b_1,b_2,b_3),\overrightarrow{c}=(c_1,c_2,c_3)$, then

$$\begin{array}{cc}\hfill & \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(a_1{\overrightarrow{e}}_1+a_2{\overrightarrow{e}}_2+a_3{\overrightarrow{e}}_3)\times \left[(b_1{\overrightarrow{e}}_1+b_2{\overrightarrow{e}}_2+b_3{\overrightarrow{e}}_3)\times (c_1{\overrightarrow{e}}_1+c_2{\overrightarrow{e}}_2+c_3{\overrightarrow{e}}_3)\right]\hfill \\ \hfill =& (a_1{\overrightarrow{e}}_1+c_2{\overrightarrow{e}}_2+c_3{\overrightarrow{e}}_3)\times \left[(b_2{c}_3-b_3{\overrightarrow{e}}_2){\overrightarrow{e}}_1+(b_3{c}_1-b_1{c}_3){\overrightarrow{e}}_2+(b_1{c}_2-b_2{c}_1){\overrightarrow{e}}_3\right]\hfill \\ \hfill =& \left[a_1(b_3{c}_1-b_1{c}_3)-a_2(b_2{c}_3-b_3{c}_2)\right]{\overrightarrow{e}}_1\times {\overrightarrow{e}}_2+\left[a_2(b_1{c}_2-b_2{c}_1)-a_3(b_3{c}_1-b_1{c}_3)\right]{\overrightarrow{e}}_2\times {\overrightarrow{e}}_3\hfill \\ & +\left[a_3(b_2{c}_3-b_3{c}_2)-a_1(b_1{c}_2-b_2{c}_1)\right]{\overrightarrow{e}}_3\times {\overrightarrow{e}}_1\hfill \\ \hfill =& (a_1{b}_3{c}_1-a_1{b}_1{c}_3-a_2{b}_2{c}_3+a_2{b}_3{c}_2){\overrightarrow{e}}_1\times {\overrightarrow{e}}_2+(a_2{b}_1{c}_2-a_2{b}_2{c}_1-a_3{b}_3{c}_1+a_3{b}_1{c}_3){\overrightarrow{e}}_2\times {\overrightarrow{e}}_3\hfill \\ & +(a_3{b}_2{c}_3-a_3{b}_3{c}_2-a_1{b}_1{c}_2+a_1{b}_2{c}_1){\overrightarrow{e}}_3\times {\overrightarrow{e}}_1\hfill \\ \hfill =& \left[b_1(a_2{c}_2+a_3{c}_3)-c_1(a_2{b}_2+a_3{b}_3)\right]{\overrightarrow{e}}_1+\left[b_2(a_3{c}_3+a_1{c}_1)-c_2(a_3{b}_3+a_1{b}_1)\right]{\overrightarrow{e}}_2\hfill \\ \hfill & +\left[b_3(a_1{c}_1+a_2{c}_2)-c_3(a_1{b}_1+a_2{b}_2)\right]{\overrightarrow{e}}_3.\hfill \end{array}$$

It is easy to compute that

$$\begin{array}{cc}\hfill \overrightarrow{a}\ast \overrightarrow{c}=& a_1{c}_1+a_2{c}_2+a_3{c}_3,\hfill \\ \hfill \overrightarrow{a}\ast \overrightarrow{b}=& a_1{b}_1+a_2{b}_2+a_3{b}_3,\hfill \end{array}$$

$$\begin{array}{cc}\hfill (\overrightarrow{a}\ast \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\ast \overrightarrow{b})\overrightarrow{c}=& (a_1{c}_1+a_2{c}_2+a_3{c}_3)(b_1{\overrightarrow{e}}_1+b_2{\overrightarrow{e}}_2+b_3{\overrightarrow{e}}_3)-(a_1{b}_1+a_2{b}_2+a_3{b}_3)(c_1{\overrightarrow{e}}_1\hfill \\ & +c_2{\overrightarrow{e}}_2+c_3{\overrightarrow{e}}_3)\hfill \\ \hfill =& [b_1(a_1{c}_1+a_2{c}_2+a_3{c}_3)-c_1(a_1{b}_1+a_2{b}_2+a_3{b}_3){\overrightarrow{e}}_1]+[b_2(a_1{c}_1+a_2{c}_2\hfill \\ & +a_3{c}_3)-c_2(a_1{b}_1+a_2{b}_2+a_3{c}_3)]{\overrightarrow{e}}_2+[b_3(a_1{c}_1+a_2{c}_2+a_3{c}_3)-c_3(a_1{b}_1\hfill \\ & +a_2{b}_2+a_3{b}_3)]{\overrightarrow{e}}_3.\hfill \end{array}$$

Hence, formula (3) is proved. □

Lemma 2.

For arbitrary vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$, the Jacobi identity holds

$$\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})+\overrightarrow{b}\times (\overrightarrow{c}\times \overrightarrow{a})+\overrightarrow{c}\times (\overrightarrow{a}\times \overrightarrow{b})=\overrightarrow{0}.$$

(4)

Proof. 

By applying Lemma 1, we find that

$$\begin{array}{cc}\hfill & \overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(\overrightarrow{a}\ast \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{a}\ast \overrightarrow{b})\overrightarrow{c},\hfill \\ \hfill & \overrightarrow{b}\times (\overrightarrow{c}\times \overrightarrow{a})=(\overrightarrow{b}\ast \overrightarrow{a})\overrightarrow{c}-(\overrightarrow{b}\ast \overrightarrow{c})\overrightarrow{a},\hfill \\ \hfill & \overrightarrow{c}\times (\overrightarrow{a}\times \overrightarrow{b})=(\overrightarrow{c}\ast \overrightarrow{b})\overrightarrow{a}-(\overrightarrow{c}\ast \overrightarrow{a})\overrightarrow{b}.\hfill \end{array}$$

In terms of the properties of inner product and outer product, the Jacobi identity is completed to prove.

Denote a vector space by $\mathcal{S}$, that is

$$\mathcal{S}=span\left\{{\overrightarrow{e}}_1,{\overrightarrow{e}}_2,{\overrightarrow{e}}_3\right\},$$

then an arbitrary row vector space $\mathcal{W}$ with three dimensions is isomorphic to the space $\mathcal{S}$. Therefore, for arbitrary vectors $\overrightarrow{a},\overrightarrow{b}\in \mathcal{W}$, we define a commutation operation by vector product

$${[\overrightarrow{a},\overrightarrow{b}]}_{\mathcal{V}}=\overrightarrow{a}\times \overrightarrow{b}.$$

(5)

It can be verified that $\mathcal{W}$ is a Lie algebra. That is, □

Theorem 1.

For three dimensional row-vector space $\mathcal{W}$, if $\overrightarrow{a},\overrightarrow{b}\in \mathcal{W}$ satisfy (5), then $\mathcal{W}$ is a Lie algebra.

Proof. 

According to the properties of vector product, we have

$${[\overrightarrow{a},\overrightarrow{b}]}_{\mathcal{V}}=-{[\overrightarrow{b},\overrightarrow{a}]}_{\mathcal{V}},$$

$${[\alpha \overrightarrow{a}+\beta \overrightarrow{b},\overrightarrow{c}]}_{\mathcal{V}}=\alpha {[\overrightarrow{a},\overrightarrow{c}]}_{\mathcal{V}}+\beta {[\overrightarrow{b},\overrightarrow{c}]}_{\mathcal{V}}.$$

Again by using Lemma 2, the Jacobi identity holds. Hence, $\mathcal{W}$ is a Lie algebra with the commutator (5). □

Remark 1.

We are all familiar with following two types of 2-order matrix Lie algebras:

$$g_1=span\{\overline{e_1},\overline{e_2},\overline{e_3}\},$$

where

$$\overline{e_1}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),\overline{e_2}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),\overline{e_3}=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),$$

along with the following commutators:

$$[\overline{e_1},\overline{e_2}]=2\overline{e_2},[\overline{e_1},\overline{e_3}]=-2\overline{e_3},[\overline{e_2},\overline{e_3}]=\overline{e_1};$$

$$g_2=span\{\widetilde{e_1},\widetilde{e_2},\widetilde{e_3}\},$$

where

$$\widetilde{e_1}=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),\widetilde{e_2}=\frac{1}{2}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),\widetilde{e_3}=\frac{1}{2}\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),$$

equiped wtih the commutators:

$$[\widetilde{e_1},\widetilde{e_2}]=\widetilde{e_3},[\widetilde{e_1},\widetilde{e_3}]=\widetilde{e_1},[\widetilde{e_2},\widetilde{e_3}]=-\widetilde{e_1}.$$

By using the above Lie algebras, one has obtained many important integrable equations by the use of the Tu scheme(see, examples [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17]). It is easy to find that the Lie algebra $\mathcal{S}$ is different from the Lie algebras $g_1$ and $g_2$ on real field R, that is, the commutative relations in $\mathcal{S}$ are different from those of the Lie algebras $g_1$ and $g_2$, which indicates that applying the Lie algebra $\mathcal{S}$ could produce new integrable equations. The following examples presented in the next section will verify this fact. Only on the complex filed $iR$, the Lie algebra $\mathcal{S}$ is the same with the Pauli matrix Lie algebra

$${\sigma }_1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\sigma }_2=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),{\sigma }_3=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),$$

with commutative relations

$[{\sigma }_1,{\sigma }_2]=2i{\sigma }_3,[{\sigma }_2,{\sigma }_3]=2i{\sigma }_1,[{\sigma }_3,{\sigma }_1]=2i{\sigma }_2.$

The corresponding loop algebra of Lie algebra $\mathcal{S}$ denoted by

$\tilde{\mathcal{S}}=span\{\overrightarrow{e_1}(n),\overrightarrow{e_2}(n),\overrightarrow{e_3}(n)\},$ where $\overrightarrow{e_i}(n)=\overrightarrow{e_i}{\lambda }^n,n\in N;i=1,2,3.$

Theorem 2.

For arbitrary vectors $\overrightarrow{U},\overrightarrow{V},\overrightarrow{\varphi }\in \mathcal{W}$, suppose that

$${\overrightarrow{\varphi }}_x=\overrightarrow{U}\times \overrightarrow{\varphi },{\overrightarrow{\varphi }}_t=\overrightarrow{V}\times \overrightarrow{\varphi },$$

(6)

then the compatibility condition of (6) leads to a new form of zero-curvature equation

$${\overrightarrow{U}}_t-{\overrightarrow{V}}_x+\overrightarrow{U}\times \overrightarrow{V}=\overrightarrow{0}.$$

(7)

Proof. 

A direct calculation reads by using Lemma 1 that

$$\begin{array}{cc}\hfill & {\overrightarrow{\varphi }}_{xt}={\overrightarrow{U}}_t\times \overrightarrow{\varphi }+\overrightarrow{U}\times {\overrightarrow{\varphi }}_t={\overrightarrow{U}}_t\times \overrightarrow{\varphi }+\overrightarrow{U}\times (\overrightarrow{V}\times \overrightarrow{\varphi })={\overrightarrow{U}}_t\times \overrightarrow{\varphi }+(\overrightarrow{U}\ast \overrightarrow{\varphi })\overrightarrow{V}-(\overrightarrow{U}\ast \overrightarrow{V})\overrightarrow{\varphi },\hfill \\ \hfill & {\overrightarrow{\varphi }}_{tx}={\overrightarrow{V}}_x\times \overrightarrow{\varphi }+\overrightarrow{V}\times {\overrightarrow{\varphi }}_x={\overrightarrow{V}}_x\times \overrightarrow{\varphi }+\overrightarrow{V}\times (\overrightarrow{U}\times \overrightarrow{\varphi })={\overrightarrow{V}}_x\times \overrightarrow{\varphi }+(\overrightarrow{V}\ast \overrightarrow{\varphi })\overrightarrow{U}-(\overrightarrow{V}\ast \overrightarrow{U})\overrightarrow{\varphi },\hfill \\ \hfill & {\overrightarrow{\varphi }}_{xt}={\overrightarrow{\varphi }}_{tx}\Rightarrow {\overrightarrow{U}}_t\times \overrightarrow{\varphi }-{\overrightarrow{V}}_x\times \overrightarrow{\varphi }-(\overrightarrow{U}\ast \overrightarrow{V}-\overrightarrow{V}\ast \overrightarrow{U})\overrightarrow{\varphi }+(\overrightarrow{U}\times \overrightarrow{V})\times \overrightarrow{\varphi }=\overrightarrow{0},\hfill \end{array}$$

that is, since $\overrightarrow{\varphi }$ is arbitrary, hence

$${\overrightarrow{U}}_t-{\overrightarrow{V}}_x+\overrightarrow{U}\times \overrightarrow{V}=\overrightarrow{0}.$$

We call Equation (7) a vector-product zero-curvature equation. □

2. A Positive Flow and a Negative Flow

In order to display the action of (7) different from that of the standard zero-curvature equation with Lie bracket

$$U_t-V_x+[U,V]=0,$$

(8)

where $[U,V]=UV-VU$, U and V are square matrices, we first recall the AKNS equations.

In Ref. [18], let $U=\left(\begin{array}{cc}-i\lambda & q\\ r& i\lambda \end{array}\right),V=\left(\begin{array}{cc}A& B\\ C& -A\end{array}\right),$ then the zero-curvature Equation (8) gives rise to

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{A}_x=qC-rB,\hfill \\ q_t=B_x+2i\lambda B+2qA,\hfill \\ r_t=C_x-2i\lambda C-2rA.\hfill \end{array}\right.\end{array}\end{array}$$

(9)

Case 1: Assume

$$A={\displaystyle \sum _{j=0}^3}{a}_j{\lambda }^j,B={\displaystyle \sum _{j=0}^3}{b}_j{\lambda }^j,C={\displaystyle \sum _{j=0}^3}{c}_j{\lambda }^j.$$

(10)

Substituting (10) into (9) and taking $a_3=a_3^0$(const), $a_2=a_2^0$(const) yields the following nonlinear integrable equations (positive flow):

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_t=-\frac{1}{4}i{a}_3^0(q_{xxx}-6qr{q}_x)-\frac{1}{2}{a}_2^0(q_{xx}-2q^{2}r)+i{a}_1^0{q}_x+2a_0^{0}q,\hfill \\ r_t=-\frac{1}{4}i{a}_3^0(r_{xxx}-6qr{r}_x)+\frac{1}{2}{a}_2^0(r_{xx}-2q{r}^2)+i{a}_1^0{r}_x-2a_0^{0}r,\hfill \end{array}\right.\end{array}\end{array}$$

(11)

where $a_1^0$ is an integral constant of $a_1$, and $a_0^0$ is an integral constant of $a_0$.

Let $a_0^0=a_1^0=a_3^0=0,a_2^0=-2i,r=\mp q^{\ast }$, (11) reduces to the nonlinear Schrödinger equation

$$i{q}_t+q_{xx}\pm 2q^2{q}^{\ast }=0.$$

(12)

Taking $a_0^0=a_1^0=a_2^0=0,a_3^0=-4i,r=1,$ (11) reduces to the standard KdV equation

$$q_t+6q^2{q}_x+q_{xxx}=0.$$

(13)

Case 2: Assume

$$A=\frac{a}{\lambda },B=\frac{b}{\lambda },C=\frac{c}{\lambda },$$

(14)

and insert (14) into (9), one obtains (negative flow)

$$a_x=\frac{i}{2}{(qr)}_t,q_{xt}=-4iaq,r_{xt}=-4iar.$$

(15)

Let $a=\frac{i}{4}cosu,b=c=\frac{i}{4}sinu,q=-r=-\frac{u_x}{2},$ (15) becomes

$$u_{xt}=sinu,$$

(16)

which is called a sine-Gordon equation.

Set $a=\frac{i}{4}coshu,-b=c=\frac{i}{4}sinhu,q=r=\frac{u_x}{2}$, then (15) reduces to

$$u_{xt}=sinhu,$$

(17)

which is called a sinh-Gordon equation.

In what follows, we want to make use of new zero-curvature Equation (7) going on the generation of positive flow and negative flow by following the above AKNS spectral problem. Assume $\overrightarrow{U}=(i\lambda ,q,r),\overrightarrow{V}=(A,B,C),$ and substitute them into (7), we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_t-B_x+rA-i\lambda C=0,\hfill \\ r_t-C_x+i\lambda B-qA=0,\hfill \\ -A_x+qC-rB=0.\hfill \end{array}\right.\end{array}\end{array}$$

(18)

It is easy to find that (18) is different from (9).

Case 1’: Taking $A,B,C$ as the form of (10), then (18) reduces to

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_t-b_{0x}+r{a}_0=0,\hfill \\ r_t-c_{0x}-q{a}_0=0,\hfill \end{array}\right.\end{array}\end{array}$$

(19)

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{a}_{jx}=q{c}_j-r{b}_j,j=0,1,2,\hfill \\ -c_{jx}+i{b}_j-q{a}_j=0,j=1,2,3,\hfill \\ -b_{jx}+r{a}_j-i{c}_j=0,j=1,2,3.\hfill \\ a_{3x}=0.\hfill \end{array}\right.\end{array}\end{array}$$

(20)

Set $a_3=a_3^0$(const),$a_2=a_2^0$(const), then from (20) we obtain

$$\begin{array}{cc}\hfill & b_2=-i{a}_3^{0}q,c_2=-i{a}_3^{0}r,\hfill \\ \hfill & b_1=-a_3^0{r}_x-i{a}_2^{0}q,c_1=a_3^0{q}_x-i{a}_2^{0}r,\hfill \\ \hfill & b_1=-a_3^0{r}_x-i{a}_2^{0}q,c_1=a_3^0{q}_x-i{a}_2^{0}r,\hfill \\ \hfill & a_1=\frac{1}{2}{a}_3^0{q}^2+\frac{1}{2}{a}_3^0{r}^2+a_1^0,\hfill \\ \hfill & b_0=-i{a}_3^0{q}_{xx}-a_2^0{r}_x-\frac{i}{2}{a}_3^0{q}^3-\frac{i}{2}{a}_3^{0}q{r}^2-i{a}_1^{0}q,\hfill \\ \hfill & c_0=-i{a}_3^0{r}_{xx}+a_2^0{q}_x-\frac{i}{2}{a}_3^{0}r{q}^2-\frac{i}{2}{a}_3^0{r}^3-i{a}_1^{0}r,\hfill \\ \hfill & a_0=i{a}_3^0(r{q}_x-r_{x}q)+\frac{1}{2}{a}_2^0{q}^2+\frac{1}{2}{a}_2^0{r}^2+a_0^0.\hfill \end{array}$$

Thus, (19) becomes the following positive order flow:

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_t=-i{a}_3^0{q}_{xxx}-a_2^0{r}_{xx}-\frac{3i}{2}{a}_3^0{q}^2{q}_x-\frac{3i}{2}{a}_3^0{r}^2{q}_x-\frac{1}{2}{a}_2^{0}r{q}^2-\frac{1}{2}{a}_2^0{r}^3-a_0^{0}r-i{a}_1^0{q}_x,\hfill \\ r_t=-i{a}_3^0{r}_{xxx}+a_2^0{q}_{xx}-\frac{3i}{2}{a}_3^0{r}_x{q}^2-\frac{3i}{2}{a}_3^0{r}^{2}r+\frac{1}{2}{a}_2^0{q}^3+\frac{1}{2}{a}_2^{0}q{r}^2+a_0^{0}q-i{a}_1^0{r}_x.\hfill \end{array}\right.\end{array}\end{array}$$

(21)

Obviously, (21) differs from (11). Set $a_2^0=0,a_0^0=0,a_3^0=i,a_2^0=-2,a_1^0=i,r=1,$ (21) reduces to a generalized KdV equation

$$q_t=q_{xxx}+\frac{3}{2}{q}^2{q}_x+\frac{5}{2}{q}_x+q^2,$$

which is different from the mKdV equation presented by Li [18] which was obtained by using the standard zero-curvature equation with the Lie bracket:

$$q_t+6q^2{q}_x+q_{xxx}=0.$$

Taking $a_3^0=0,r=ϵq^{\ast }$, then (21) becomes

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_t=\frac{1}{{ϵ}^{\ast }}{a}_2^{0^{\ast }}{q}_{xx}^{\ast }+\frac{1}{2{ϵ}^{\ast }}{a}_2^{0^{\ast }}{q}^{\ast 3}+\frac{1}{2}{a}_2^{0^{\ast }}{ϵ}^{\ast }{q}^{\ast }{q}^2+\frac{1}{{ϵ}^{\ast }}{a}_0^{0^{\ast }}{q}^{\ast }+i{a}_1^{0^{\ast }}{q}_x,\hfill \\ q_t=-ϵa_2^0{q}_{xx}^{\ast }-\frac{1}{2}ϵa_2^0{q}^{\ast }{q}^2-\frac{1}{2}{a}_2^0{ϵ}^3{q}^{\ast 3}-a_0^0ϵq^{\ast }-i{a}_1^0{q}_x.\hfill \end{array}\right.\end{array}\end{array}$$

(22)

In order to be compatible with (22), we let

$$\frac{1}{{ϵ}^{\ast }}=-ϵ,\frac{1}{{ϵ}^{\ast }}=-{ϵ}^3,{ϵ}^{\ast }=-ϵ,a_1^{0^{\ast }}=-a_1^0,$$

Hence, taking $ϵ=i,a_1^0=i,$ we obtain a nonlinear integrable equation

$$q_t=-i{a}_2^0{q}_{xx}^{\ast }+\frac{1}{2}i{a}_2^0{q}^{\ast }{q}^2+\frac{1}{2}{a}_2^{0}i{q}^{\ast 3}-i{a}_0^0{q}^{\ast }+q_x,$$

(23)

which is different from the standard NLS Equation (12). Therefore, we find that the new vector product equation can really generate new integrable equations.

Case 2’: Taking $A,B,C$ as the form (14), then (18) reduces to

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_t-\frac{b_x-ra}{\lambda }-ic=0,\hfill \\ r_t-\frac{c_x+qa}{\lambda }+ib=0,\hfill \\ a_x=qc-rb.\hfill \end{array}\right.\end{array}\end{array}$$

Let $b_x=ra,c_x=-qa,$ then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \left\{\begin{array}{c}{q}_t=ic,\hfill \\ r_t=-ib.\hfill \end{array}\right.\hfill \end{array}\end{array}$$

$$\begin{array}{cc}\hfill & a_x=\frac{1}{2i}{(q^2+r^2)}_t.\hfill \end{array}$$

(24)

Hence,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_{xt}=i{c}_x=-iqa,\hfill \\ r_{xt}=-ira,\hfill \end{array}\right.\end{array}\end{array}$$

(25)

Set $q=r$, (25) becomes

$$q_{xt}=-iqa,$$

(26)

(24) reduces to

$$a_x=-2iq{q}_t.$$

(27)

If $a=\frac{1}{2i}cosu,q=-\frac{1}{2i}{u}_x,$ then (26) or (27) turns to

$$u_{xt}=sinu,$$

which is just right (16). If $a=\frac{1}{2i}coshu,q=\frac{i}{2}{u}_x,$ then (27) gives rise to

$$u_{xt}=sinhu,$$

which is Equation (17). Hence, some integrable equations obtained by the usual zero-curvature Equation (8) can be generated by our new vector product Equation (7).

If $q^2+r^2=v_x\ge 0,a=2i{v}_t,$ then we can set $q=\sqrt{v_x}sin\theta ,r=\sqrt{v_x}cos\theta$, $\theta$ is independent of $x,t$. Thus, (26) can lead to a new integrable equation as follows:

$$v_x^2{v}_t=\frac{1}{2}{v}_{xxt}{v}_x-\frac{1}{4}{v}_{xx}{v}_{xt}.$$

If $s=\sqrt{v_x},$ then the second equation in (25) leads to the following rational integrable equation

$${(\frac{s_{xt}}{4s})}_x=s{s}_t.$$

(28)

From the above discussions, we find that the vector-product Equation (7) is able to produce new integrable equations.

3. Isospectral and Non-Isospectral Integrable Hierachies

3.1. Isospectral Integrable Hierarchy

In the section, we want to extend the Tu scheme to the case where we make use of the vector product Equation (7) replacing the zero-curvature Equation (8).

First of all, solving the stationary equation

$${\overrightarrow{V}}_x=\overrightarrow{U}\times \overrightarrow{V},$$

(29)

where $\overrightarrow{U}=(i\lambda ,q,r),\overrightarrow{V}=(A,B,C)={\displaystyle \sum _{j\ge 0}}(a_j{\overrightarrow{e}}_1(-j)+b_j{\overrightarrow{e}}_2(-j)+c_j{\overrightarrow{e}}_3(-j),{\overrightarrow{e}}_k(-j)={\overrightarrow{e}}_k{\lambda }^{-j},k=1,2,3,$ yields

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{A}_x=qC-rB,\hfill \\ B_x=rA-i\lambda C,\hfill \\ C_x=i\lambda B-qA,\hfill \end{array}\right.\end{array}\end{array}$$

which can be written as

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{a}_{jx}=q{c}_j-r{b}_j,\hfill \\ c_{j+1}=\frac{1}{i}(r{a}_j-b_{jx}),\hfill \\ b_{j+1}=\frac{1}{i}(c_{jx}+q{a}_j).\hfill \end{array}\right.\end{array}\end{array}$$

(30)

Taking $a_0={\alpha }_0(t),b_0=c_0=0,$ then from (30) we obtain

$$\begin{array}{cc}\hfill a_1=& {\alpha }_1(t),b_1=\frac{1}{i}{\alpha }_{0}q,c_1=\frac{1}{i}{\alpha }_{0}r,\hfill \\ \hfill a_2=& \frac{{\alpha }_0}{2}(q^2+r^2)+{\alpha }_2(t),b_2=\frac{1}{i}{\alpha }_{1}q-{\alpha }_0{r}_x,\hfill \\ \hfill c_2=& \frac{1}{i}{\alpha }_{1}r+{\alpha }_0{q}_x,\hfill \\ \hfill a_3=& \frac{{\alpha }_1}{2}(q^2+r^2)+\frac{{\alpha }_0}{i}(q{r}_x-r{q}_x)+{\alpha }_3(t),\hfill \\ \hfill b_3=& -{\alpha }_1{r}_x+\frac{{\alpha }_0}{i}{q}_{xx}+\frac{{\alpha }_0}{2i}q(q^2+r^2)+\frac{{\alpha }_2}{i}q,\hfill \\ \hfill c_3=& \frac{{\alpha }_0}{2i}r(q^2+r^2)+\frac{{\alpha }_2}{i}r+{\alpha }_1{q}_x+\frac{{\alpha }_0}{i}{r}_{xx},\hfill \\ \hfill b_4=& -\frac{{\alpha }_0}{2}{(r{q}^2+r^3)}_x-{\alpha }_2{r}_x+\frac{{\alpha }_1}{i}{q}_{xx}-{\alpha }_0{r}_{xxx}+\frac{{\alpha }_1}{2i}q(q^2+r^2)-{\alpha }_{0}q(q{r}_x-r{q}_x)+\frac{{\alpha }_3}{i}q,\hfill \\ \hfill c_4=& \frac{{\alpha }_1}{2i}r(q^2+r^2)-{\alpha }_{0}r(q{r}_x-r{q}_x)+\frac{{\alpha }_3}{i}r+\frac{{\alpha }_1}{i}{r}_{xx}+{\alpha }_0{q}_{xxx}+\frac{{\alpha }_0}{2}{(q^3+q{r}^2)}_x+{\alpha }_2{q}_x,\hfill \\ \hfill & \cdots \hfill \end{array}$$

Next, taking

$${\overrightarrow{V}}_{+}^{(n)}={\displaystyle \sum _{j=0}^n}(a_j{\overrightarrow{e}}_1(n-j)+b_j{\overrightarrow{e}}_2(n-j)+c_j{\overrightarrow{e}}_3(n-j))={\lambda }^n\overrightarrow{V}-{\overrightarrow{V}}_{-}^{(n)},$$

then (29) can be decomposed into the following form

$$-{({\overrightarrow{V}}_{+}^{(n)})}_x+\overrightarrow{U}\times {\overrightarrow{V}}_{+}^{(n)}={({\overrightarrow{V}}_{-}^{(n)})}_x-\overrightarrow{U}\times {\overrightarrow{V}}_{-}^{(n)}.$$

(31)

It is easy to see that the degree of the left-hand side is more than 0, while that of the right-hand side is less than 0. Therefore, the degree of (31) is taken to be zero. Hence, we have

$$-{({\overrightarrow{V}}_{+}^{(n)})}_x+\overrightarrow{U}\times {\overrightarrow{V}}_{+}^{(n)}=i{c}_{n+1}{\overrightarrow{e}}_2(0)-i{b}_{n+1}{\overrightarrow{e}}_3(0).$$

Let ${\overrightarrow{V}}^{(n)}={\overrightarrow{V}}_{+}^{(n)},$ we obtain the following theorem:

Theorem 3.

Assume that $\overrightarrow{U}=(i\lambda ,q,r),{\overrightarrow{V}}^{(n)}={\displaystyle \sum _{j\ge 0}^n}(a_j{\overrightarrow{e}}_1(n-j)+b_j{\overrightarrow{e}}_2(n-j)+c_j{\overrightarrow{e}}_3(n-j)),$ then the improved form of vector product Equation (7)

$${\overrightarrow{U}}_t-{\overrightarrow{V}}_x^{(n)}+\overrightarrow{U}\times {\overrightarrow{V}}^{(n)}=\overrightarrow{0},$$

(32)

admits an isospectral integrable hierarchy

$${\left(\begin{array}{c}q\\ r\end{array}\right)}_{t_n}=i\left(\begin{array}{c}-c_{n+1}\\ b_{n+1}\end{array}\right)=iJ\left(\begin{array}{c}{b}_{n+1}\\ c_{n+1}\end{array}\right),$$

(33)

where $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ is a Hamiltonian operator.

In terms of (30) we obtain

$$\left(\begin{array}{c}{b}_{n+1}\\ c_{n+1}\end{array}\right)=\frac{1}{i}L\left(\begin{array}{c}{b}_n\\ c_n\end{array}\right)+\frac{{\alpha }_n}{i}\left(\begin{array}{c}q\\ r\end{array}\right).$$

Hence, (33) can be written as

$$\begin{array}{cc}\hfill {\left(\begin{array}{c}q\\ r\end{array}\right)}_{t_n}=& J\left[L\left(\begin{array}{c}{b}_n\\ c_n\end{array}\right)+{\alpha }_n\left(\begin{array}{c}q\\ r\end{array}\right)\right]=\cdots =\frac{1}{i^{n-1}}J{L}^n\left(\begin{array}{c}{b}_1\\ c_1\end{array}\right)+\frac{1}{i^{n-2}}J{L}^{n-1}\left(\begin{array}{c}{b}_2\\ c_2\end{array}\right)+\cdots \hfill \\ & +\frac{1}{i^2}{\alpha }_{n-2}J{L}^2\left(\begin{array}{c}q\\ r\end{array}\right)+\frac{{\alpha }_{n-1}}{i}JL\left(\begin{array}{c}q\\ r\end{array}\right)+{\alpha }_n\left(\begin{array}{c}q\\ r\end{array}\right),\hfill \end{array}$$

where $L=\left(\begin{array}{cc}-q{\partial }^{-1}r& \partial +q{\partial }^{-1}q\\ -\partial -r{\partial }^{-1}r& r{\partial }^{-1}q\end{array}\right),$ and ${\alpha }_k={\alpha }_k(t)(k=0,1,2,\cdots )$ are integrable constants. Some reductions of (33) are presented as follows.

Taking $n=2$, one infers that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_{t_2}=-\frac{{\alpha }_0}{2}r(q^2+r^2)-{\alpha }_{2}r-i{\alpha }_1{q}_x-{\alpha }_0{r}_{xx},\hfill \\ r_{t_2}=-i{\alpha }_1{r}_x+{\alpha }_0{q}_{xx}+\frac{{\alpha }_0}{2}q(q^2+r^2)+{\alpha }_{2}q,\hfill \end{array}\right.\end{array}\end{array}$$

which is a general equation system of (23).

Taking $n=3$, we have

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_{t_3}=-\frac{{\alpha }_1}{2}r(q^2+r^2)+i{\alpha }_{0}r(q{r}_x-r{q}_x)-{\alpha }_{3}r-{\alpha }_1{r}_{xx}-i{\alpha }_0{q}_{xxx}-\frac{i{\alpha }_0}{2}{(q^3+q{r}^2)}_x-i{\alpha }_2{q}_x,\hfill \\ r_{t_3}=-\frac{i{\alpha }_0}{2}{(r{q}^2+r^3)}_x-i{\alpha }_2{r}_x+{\alpha }_1{q}_{xx}-i{\alpha }_0{r}_{xxx}+\frac{{\alpha }_1}{2}q(q^2+r^2)-i{\alpha }_{0}q(q{r}_x-r{q}_x)+{\alpha }_{3}q,\hfill \end{array}\right.\end{array}\end{array}$$

which is a general equation system of the generalized KdV equation.

3.2. Non-Isospectral Integrable Hierarchy

Ma [19] and Qiao [20] once generated some non-isospectral integrable hierarchies of evolution equations when the spectral parameter $\lambda$ has the evolution ${\lambda }_t=\frac{\partial \lambda }{\partial t}=\alpha (x,t){\lambda }^n,n=1,2,\cdots .$ Li [18] considered the spectral evolution form ${\lambda }_t={\displaystyle \sum _{j=0}^m}{k}_j(t){\lambda }^{m-j}.$ Obviously, the works finished by Ma and Qiao are the special case of Li. Zhang, et al. [7] applied the Tu scheme for the generation of non-isospectral integrable hierarchies under the framework of matrix zero-curvature Equation (8). In this section, we would like to extend the Tu scheme to the case where the non-isospectral integrable hierarchies are produced by the vector-product Equation (7) as follows:

$$\frac{\partial \overrightarrow{U}}{\partial u}{u}_t+\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_t-{\overrightarrow{V}}_x+\overrightarrow{U}\times \overrightarrow{V}=\overrightarrow{0}.$$

(34)

In fact, set ${\overrightarrow{\varphi }}_x=\overrightarrow{U}\times \overrightarrow{\varphi },{\overrightarrow{\varphi }}_t=\overrightarrow{V}\times \overrightarrow{\varphi },\overrightarrow{U}=\overrightarrow{U}(u,\lambda ),$ then

$$\begin{array}{cc}\hfill & {\overrightarrow{\varphi }}_{xt}=(\frac{\partial \overrightarrow{U}}{\partial u}{u}_t+\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_t)\times \overrightarrow{\varphi }+\overrightarrow{U}\times (\overrightarrow{V}\times \overrightarrow{\varphi }),\hfill \\ \hfill & {\overrightarrow{\varphi }}_{tx}={\overrightarrow{V}}_x\times \overrightarrow{\varphi }+\overrightarrow{V}\times {\overrightarrow{\varphi }}_x={\overrightarrow{V}}_x\times \overrightarrow{\varphi }+\overrightarrow{V}\times (\overrightarrow{U}\times \overrightarrow{\varphi }),\hfill \\ \hfill & {\overrightarrow{\varphi }}_{xt}={\overrightarrow{\varphi }}_{tx}\Rightarrow \hfill \\ \hfill & (\frac{\partial \overrightarrow{U}}{\partial u}{u}_t+\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_t)\times \overrightarrow{\varphi }+(\overrightarrow{U}\ast \overrightarrow{\varphi })\overrightarrow{V}-(\overrightarrow{U}\ast \overrightarrow{V})\overrightarrow{\varphi }={\overrightarrow{V}}_x\times \overrightarrow{\varphi }+(\overrightarrow{V}\ast \overrightarrow{\varphi })\overrightarrow{U}-(\overrightarrow{V}\ast \overrightarrow{U})\overrightarrow{\varphi },\hfill \\ \hfill & (\frac{\partial \overrightarrow{U}}{\partial u}{u}_t+\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_t)\times \overrightarrow{\varphi }-{\overrightarrow{V}}_x\times \overrightarrow{\varphi }+(\overrightarrow{V}\ast \overrightarrow{U}-\overrightarrow{U}\ast \overrightarrow{V})\overrightarrow{\varphi }+(\overrightarrow{U}\ast \overrightarrow{\varphi })\overrightarrow{V}-(\overrightarrow{V}\ast \overrightarrow{\varphi })\overrightarrow{U}=\overrightarrow{0}.\hfill \end{array}$$

It is easy to find that

$$(\overrightarrow{U}\ast \overrightarrow{\varphi })\overrightarrow{V}-(\overrightarrow{V}\ast \overrightarrow{\varphi })\overrightarrow{U}=-\overrightarrow{\varphi }\times (\overrightarrow{U}\times \overrightarrow{V}),$$

that is,

$$(\overrightarrow{U}\times \overrightarrow{V})\times \overrightarrow{\varphi }=(\overrightarrow{U}\ast \overrightarrow{\varphi })\overrightarrow{V}-(\overrightarrow{V}\ast \overrightarrow{\varphi })\overrightarrow{U}.$$

Hence, we have

$$(\frac{\partial \overrightarrow{U}}{\partial u}{u}_t+\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_t)\times \overrightarrow{\varphi }-{\overrightarrow{V}}_x\times \overrightarrow{\varphi }+(\overrightarrow{U}\times \overrightarrow{V})\times \overrightarrow{\varphi }=\overrightarrow{0}.$$

Since $\overrightarrow{\varphi }$ is arbitrary, (34) holds.

Theorem 4.

Assume $\overrightarrow{U}=(i\lambda ,q,r),\overrightarrow{V}={\displaystyle \sum _{j\ge 0}}({\overline{a}}_j{e}_1(-j)+{\overline{b}}_j{e}_2(-j)+{\overline{c}}_j{e}_3(-j)),{\lambda }_t={\displaystyle \sum _{j\ge 0}}{k}_j(t){\lambda }^{-j},$ then one infers that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{\overline{a}}_{jx}=i{k}_j(t)+q{\overline{c}}_j-r{\overline{b}}_j,\hfill \\ {\overline{c}}_{j+1}=\frac{1}{i}(r{\overline{a}}_j-{\overline{b}}_{jx}),\hfill \\ {\overline{b}}_{j+1}=\frac{1}{i}({\overline{c}}_{jx}+q{\overline{a}}_j).\hfill \end{array}\right.\end{array}\end{array}$$

(35)

Proof. 

Since $\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_t=i{\displaystyle \sum _{j\ge 0}}{k}_j(t){\overrightarrow{e}}_1(-j),$ the stationary vector-product equation

$${\overrightarrow{V}}_x=\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_t+\overrightarrow{U}\times \overrightarrow{V},$$

(36)

directly leads to the set of Equations (35).

From (35), one infers that

$${\overline{a}}_{j+1}=-\frac{1}{i}{\int }^x(q{\overline{b}}_{jx}+r{\overline{c}}_{jx})d_x+i{k}_{j+1}(t)x.$$

Taking ${\overline{b}}_0={\overline{c}}_0=0,{\overline{a}}_0=i{k}_0(t)x,$ then from (35) one obtains that

$$\begin{array}{cc}\hfill & {\overline{b}}_1=k_{0}xq,{\overline{c}}_1=k_{0}xr,{\overline{a}}_1=i{k}_{1}x,\hfill \\ \hfill & {\overline{c}}_2=k_{1}xr+i{k}_0{(xq)}_x,\hfill \\ \hfill & {\overline{b}}_2=-i{k}_0{(xr)}_x+k_{1}xq,\hfill \\ \hfill & {\overline{a}}_2=-\frac{1}{2i}{k}_0{\partial }_x^{-1}(q^2+r^2)-\frac{1}{2i}{k}_{0}x(q^2+r^2)+i{k}_{2}x,\hfill \\ \hfill & {\overline{b}}_3=k_0{(xq)}_{xx}-i{k}_1{(xr)}_x+\frac{1}{2}{k}_{0}q{\partial }^{-1}(q^2+r^2)+\frac{1}{2}{k}_{0}xq(q^2+r^2)+k_{2}xq,\hfill \\ \hfill & {\overline{c}}_3=\frac{1}{2}{k}_{0}r{\partial }_x^{-1}(q^2+r^2)+\frac{1}{2}{k}_{0}xr(q^2+r^2)+k_{2}xr+k_0(xr)xx+i{k}_1{(xq)}_x,\hfill \\ \hfill & \cdots ,\hfill \end{array}$$

(37)

where $k_i=k_i(t),i=0,1,2,\cdots .$

Denoting ${\overrightarrow{V}}_{+}^{(m)}={\displaystyle \sum _{j=0}^n}({\overline{a}}_j{\overrightarrow{e}}_1(m-j)+{\overline{b}}_j{\overrightarrow{e}}_2(m-j)+{\overline{c}}_j{\overrightarrow{e}}_3(m-j))={\lambda }^m\overrightarrow{V}-{\overrightarrow{V}}_{-}^{(m)},{\lambda }_{+t}^{(m)}={\displaystyle \sum _{j=0}^m}{k}_j(t){\lambda }^{m-j},$ then (36) can be decomposed into the following

$$-{({\overrightarrow{V}}_{+}^{(m)})}_x+\overrightarrow{U}\times {\overrightarrow{V}}_{+}^{(m)}+\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_{+t}^{(m)}={({\overrightarrow{V}}_{-}^{(m)})}_x-\overrightarrow{U}\times {\overrightarrow{V}}_{-}^{(m)}-\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_{-t}^{(m)}.$$

Hence,

$$-{({\overrightarrow{V}}_{+}^{(m)})}_x+\overrightarrow{U}\times {\overrightarrow{V}}_{+}^{(m)}+\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_{+t}^{(m)}=i{c}_{m+1}{\overrightarrow{e}}_2(0)-i{b}_{m+1}{\overrightarrow{e}}_3(0).$$

Thus, we obtain the following consequence: □

Theorem 5.

Assume $\overrightarrow{U}=(i\lambda ,q,r),\overrightarrow{V}=(A,B,C),{\lambda }_{+t}^{(m)}={\displaystyle \sum _{j=0}^m}{k}_j(t){\overrightarrow{e}}_1(m-j),$ then the improved vector product equation

$${\overrightarrow{U}}_t-{({\overrightarrow{V}}_{+}^{(m)})}_x+\overrightarrow{U}\times {\overrightarrow{V}}_{+}^{(m)}+\frac{\partial \overrightarrow{U}}{\partial \lambda }{\lambda }_{+t}^{(m)}=\overrightarrow{0},$$

admits a non-isospectral integrable hierarchy

$${\left(\begin{array}{c}q\\ r\end{array}\right)}_{t_m}=iJ\left(\begin{array}{c}{\overline{b}}_{m+1}\\ {\overline{c}}_{m+1}\end{array}\right).$$

From (35), we find that

$$\left(\begin{array}{c}{\overline{b}}_{m+1}\\ {\overline{c}}_{m+1}\end{array}\right)=\frac{1}{i}L\left(\begin{array}{c}{\overline{b}}_m\\ {\overline{c}}_m\end{array}\right)+k_m(t)\left(\begin{array}{c}xq\\ xr\end{array}\right),$$

where L is the same as that in Theorem 3.

When $m=1$, (37) reduces to

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_{t_1}=-i{\overline{c}}_2=k_0{(xq)}_x-i{k}_{1}xr,\hfill \\ r_{t_1}=i{\overline{b}}_2=k_0{(xr)}_x+i{k}_{1}xq,\hfill \end{array}\right.\end{array}\end{array}$$

when $m=2$, (37) gives that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \left\{\begin{array}{c}{q}_{t_2}=-i{\overline{c}}_3=-\frac{i}{2}{k}_{0}r{\partial }_x^{-1}(q^2+r^2)-\frac{i}{2}{k}_{0}xr(q^2+r^2)-i{k}_{2}xr-i{k}_0{(xr)}_{xx}+k_1{(xq)}_x,\hfill \\ r_{t_2}=i{\overline{b}}_3=i{k}_0{(xq)}_{xx}+k_1{(xr)}_x+\frac{i}{2}{k}_{0}q{\partial }_x^{-1}(q^2+r^2)+\frac{i}{2}{k}_{0}xq(q^2+r^2)+i{k}_{2}xq,\hfill \end{array}\right.\end{array}\end{array}$$

(38)

which is a nonlocal integrable system. Taking $r=q^{\ast }$, then (38) reduces to a new nonlocal Schrödinger-type equation

$$q_t=-i{k}_0{(x{q}^{\ast })}_{xx}+k_1{(xq)}_x-\frac{i}{2}{k}_0{q}^{\ast }{\partial }_x^{-1}(q^{\ast 2}+q^2)-\frac{i}{2}{k}_{0}x{q}^{\ast }(q^{\ast 2}+q^2)-i{k}_{2}x{q}^{\ast }.$$

(39)

4. A Vector-Product Trace Identity

The trace identity [1] proposed by Tu Guizhang is a powerful tool for generating a Hamiltonian structure of integrable hierarchies produced by the zero-curvature Equation (8). A generalized identity called the quadratic-form identity was presented by Guo and Zhang [5]. In this section, we follow the idea for deriving the quadratic-form identity to deduce a new vector-product identity for generating a Hamiltonian structure of the integrable hierarchies obtained by the vector-product Equation (7).

Assume $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\in \mathcal{W},$ a linear functional ${\left\{\overrightarrow{a},\overrightarrow{b}\right\}}_{\mathcal{V}}=:\overrightarrow{a}F{\overrightarrow{b}}^T,$ where F is a square matrix with constant entries. Two requirements are imposed on the functional ${\left\{\overrightarrow{a},\overrightarrow{b}\right\}}_{\mathcal{V}}$ which are listed as follows

$$\begin{array}{cc}\hfill & {\left\{\overrightarrow{a},\overrightarrow{b}\right\}}_{\mathcal{V}}={\left\{\overrightarrow{b},\overrightarrow{a}\right\}}_{\mathcal{V}},\hfill \end{array}$$

(40)

$$\begin{array}{cc}& {\left\{\overrightarrow{a},{[\overrightarrow{b},\overrightarrow{c}]}_{\mathcal{V}}\right\}}_{\mathcal{V}}={\left\{{[\overrightarrow{a},\overrightarrow{b}]}_{\mathcal{V}},\overrightarrow{c}\right\}}_{\mathcal{V}}.\hfill \end{array}$$

(41)

Equation (40) requires that F is symmetry, i.e., $F^T=F$. Since the vector product of $\overrightarrow{a}$ and $\overrightarrow{b}$ can be written as

$$\begin{array}{cc}\hfill {[\overrightarrow{a},\overrightarrow{b}]}_{\mathcal{V}}& =\overrightarrow{a}\times \overrightarrow{b}=(a_2{b}_3-a_3{b}_2,a_3{b}_1-b_3{a}_1,a_1{b}_2-b_1{a}_2)\hfill \\ & =(a_1,a_2,a_3)\left(\begin{array}{ccc}0& -b_3& b_2\\ b_3& 0& b_1\\ -b_2& b_1& 0\end{array}\right)=:\overrightarrow{a}R(b),\hfill \end{array}$$

(42)

where $\overrightarrow{a}=(a_1,a_2,a_3),\overrightarrow{b}=(b_1,b_2,b_3).$ Therefore, (41) requires that

$$R(b)F=-{(R(b)F)}^T$$

(43)

A linear functional is introduced in the following

$$Q={\left\{\overrightarrow{V},{\overrightarrow{U}}_{\lambda }\right\}}_{\mathcal{V}}+{\left\{\overrightarrow{\Lambda },{\overrightarrow{V}}_x-{[\overrightarrow{U},\overrightarrow{V}]}_{\mathcal{V}}\right\}}_{\mathcal{V}},$$

(44)

where ${\overrightarrow{U}}_{\lambda }$ denotes the partial derivative with respect to spectral parameter $\lambda$, and $\overrightarrow{V},\overrightarrow{\Lambda }\in \mathcal{W}$ are two vectors to be determined. We again present the variational derivative

$$\frac{\delta }{\delta u}={\displaystyle \sum _{k=0}^{\infty }}{(-D)}^k\frac{\partial }{\partial (D^{k}u)},D=\frac{d}{dx},$$

and apply it to (44), two variational equations are obtained

$$\begin{array}{cc}\hfill & \frac{\delta }{\delta \overrightarrow{V}}Q={\overrightarrow{U}}_{\lambda }-{\overrightarrow{\Lambda }}_x+{[\overrightarrow{U},\overrightarrow{\Lambda }]}_{\mathcal{V}},\hfill \\ \hfill & \frac{\delta }{\delta \overrightarrow{\Lambda }}Q={\overrightarrow{V}}_x-{[\overrightarrow{U},\overrightarrow{V}]}_{\mathcal{V}}.\hfill \end{array}$$

(45)

Given the following constrained conditions

$$\frac{\delta }{\delta \overrightarrow{V}}Q=0\Rightarrow {\overrightarrow{U}}_{\lambda }-{\overrightarrow{\Lambda }}_x+{[\overrightarrow{U},\overrightarrow{\Lambda }]}_{\mathcal{V}}=\overrightarrow{0}.$$

(46)

$$\frac{\delta }{\delta \overrightarrow{\Lambda }}Q=0\Rightarrow {\overrightarrow{V}}_x-{[\overrightarrow{U},\overrightarrow{V}]}_{\mathcal{V}}=\overrightarrow{0}.$$

(47)

By using (47) and (44), we have

$$\frac{\delta }{\delta u}\left\{\overrightarrow{V},{\overrightarrow{U}}_{\lambda }\right\}=\frac{\delta Q}{\delta u}={\left\{\overrightarrow{V},\frac{\partial {\overrightarrow{U}}_{\lambda }}{\partial u}\right\}}_{\mathcal{V}}+{\left\{{[\overrightarrow{\Lambda },\overrightarrow{V}]}_{\mathcal{V}},\frac{\partial \overrightarrow{U}}{\partial u}\right\}}_{\mathcal{V}}.$$

(48)

By using (46), (47) and the Jacobi identity, one infers that

$${[\overrightarrow{\Lambda },\overrightarrow{V}]}_{\mathcal{V}}={[{\overrightarrow{U}}_{\lambda }+{[\overrightarrow{U},\overrightarrow{\Lambda }]}_{\mathcal{V}}]}_{\mathcal{V}}+{[\overrightarrow{\Lambda },{[\overrightarrow{U},\overrightarrow{V}]}_{\mathcal{V}}]}_{\mathcal{V}}={[{\overrightarrow{U}}_{\lambda },\overrightarrow{V}]}_{\lambda }+{[\overrightarrow{U},{[\overrightarrow{\Lambda },\overrightarrow{V}]}_{\mathcal{V}}]}_{\mathcal{V}}.$$

(49)

Again from (47), we have

$${\overrightarrow{V}}_{\lambda ,x}={\overrightarrow{V}}_{x,\lambda }={[{\overrightarrow{U}}_{\lambda },\overrightarrow{V}]}_{\mathcal{V}}+{[\overrightarrow{U},{\overrightarrow{V}}_{\lambda }]}_{\mathcal{V}}.$$

Hence, ${[\overrightarrow{\Lambda },\overrightarrow{V}]}_{\mathcal{V}}-{\overrightarrow{V}}_{\lambda }=:\overrightarrow{P}$ is a solution to the equation

$${\overrightarrow{V}}_x={[\overrightarrow{U},\overrightarrow{V}]}_{\mathcal{V}}.$$

(50)

As we all know, if ${\overrightarrow{V}}_1,{\overrightarrow{V}}_2$ are solutions of (50), then there exists a constant $\gamma$ so that

$${\overrightarrow{V}}_1=\gamma {\overrightarrow{V}}_2.$$

(51)

Therefore, by using (51) and $rank(\overrightarrow{P})=rank({\overrightarrow{V}}_{\lambda })=rank(\frac{1}{\lambda }\overrightarrow{V}),$ we have

$$\overrightarrow{P}={[\overrightarrow{\Lambda },\overrightarrow{V}]}_{\mathcal{V}}-{\overrightarrow{V}}_{\lambda }=\frac{\gamma }{\lambda }\overrightarrow{V}.$$

(52)

Since $\frac{1}{\lambda }\overrightarrow{V}$ is also a solution of (50), we find that (48) can be written as follows:

$$\frac{\delta }{\delta u}{\left\{\overrightarrow{V},{\overrightarrow{U}}_{\lambda }\right\}}_{\mathcal{V}}=\frac{\partial }{\partial \lambda }{\left\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial u}\right\}}_{\mathcal{V}}+({\lambda }^{-\gamma }\frac{\partial }{\partial \lambda }{\lambda }^{\gamma }){\left\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial u}\right\}}_{\mathcal{V}}={\lambda }^{-\gamma }\frac{\partial }{\partial \lambda }{\lambda }^{\gamma }{\left\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial u}\right\}}_{\mathcal{V}}.$$

(53)

Summarizing the above discussion, we obtain the following consequence.

Theorem 6

(a vector-product trace identity). Let $\mathcal{W}$ be a Lie algebra with row vectors, $\overrightarrow{U}=\overrightarrow{U}(u,\lambda )\in \mathcal{W}$ be homogeneous in rank and ${\left\{·,·\right\}}_{\mathcal{V}}$ denote a non-degenerate symmetric form invariant under the Lie algebra $\mathcal{W}$. Then we have the following vector-product trace identity:

$$\frac{\delta }{\delta u}{\left\{\overrightarrow{V},{\overrightarrow{U}}_{\lambda }\right\}}_{\mathcal{V}}={\lambda }^{-\gamma }\frac{\partial }{\partial \lambda }{\lambda }^{\gamma }{\left\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial u}\right\}}_{\mathcal{V}}.$$

(54)

In what follows, we apply (54) to deduce the Hamiltonian structure of the integrable hierarchy (33). In terms of (43), a direct calculation gives

$$F=\left(\begin{array}{ccc}\rho & 0& 0\\ 0& \rho & 0\\ 0& 0& \rho \end{array}\right),$$

where ρ is a constant. Noting $\overrightarrow{V}=(a,b,c),{\overrightarrow{U}}_{\lambda }=(i,0,0),$ then we have

$${\left\{\overrightarrow{V},{\overrightarrow{U}}_{\lambda }\right\}}_{\mathcal{V}}=\overrightarrow{V}F{\overrightarrow{U}}_{\lambda }^T=i\rho a,$$

$${\left\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial q}\right\}}_{\mathcal{V}}=\rho b,{\left\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial r}\right\}}_{\mathcal{V}}=\rho c.$$

Substituting the above computations into (54) yields

$$\frac{\delta }{\delta u}(i\rho a)={\lambda }^{-\gamma }\frac{\partial }{\partial \lambda }{\lambda }^{\gamma }\left(\begin{array}{c}\rho b\\ \rho c\end{array}\right),$$

where $a={\displaystyle \sum _{j\ge 0}}{a}_j{\lambda }^{-j},b={\displaystyle \sum _{j\ge 0}}{b}_j{\lambda }^{-j},c={\displaystyle \sum _{j\ge 0}}{c}_j{\lambda }^{-j}.$ Comparing the coefficients of ${\lambda }^{-n-1},$ we obtain

$$\frac{\delta }{\delta u}\left(\begin{array}{c}i{a}_{n+1}\\ -n+\gamma \end{array}\right)=\left(\begin{array}{c}{b}_n\\ c_n\end{array}\right).$$

(55)

Inserting some initial values $b_1,c_1,a_2$ into (55), we find $\gamma =0.$ Therefore, one obtains

$$\left(\begin{array}{c}{b}_n\\ c_n\end{array}\right)=\frac{\delta H_n}{\delta u},H_n=-\frac{a_{n+1}}{n}.$$

Hence, the Hamiltonian form (33) is given by

$${\left(\begin{array}{c}q\\ r\end{array}\right)}_{t_n}=iJL\left(\begin{array}{c}{b}_n\\ c_n\end{array}\right)=iJL\frac{\delta H_n}{\delta u}.$$

(56)

5. An Enlarged Loop Algebra of the Loop Algebra $\tilde{\mathcal{S}}$ and Its Applications

In the section, we shall enlarge the loop algebra $\tilde{\mathcal{S}}$ by enlarging the degrees of spectral parameter $\lambda$ of elements in $\tilde{\mathcal{S}}$, that is, an enlarged loop algebra is given by

$G=span\{\overrightarrow{e_1}(0,n),\overrightarrow{e_1}(1,n),\overrightarrow{e_2}(0,n),\overrightarrow{e_2}(1,n),\overrightarrow{e_3}(0,n),\overrightarrow{e_3}(1,n)\}$,

where

$\overrightarrow{e_1}(0,n)=\overrightarrow{e_1}{\lambda }^{2n},\overrightarrow{e_1}(1,n)=\overrightarrow{e_1}{\lambda }^{2n+1},\overrightarrow{e_2}(0,n)=\overrightarrow{e_2}{\lambda }^{2n},$

$\overrightarrow{e_2}(1,n)=\overrightarrow{e_2}{\lambda }^{2n+1},\overrightarrow{e_3}(0,n)=\overrightarrow{e_3}{\lambda }^{2n},\overrightarrow{e_3}(1,n)=\overrightarrow{e_3}{\lambda }^{2n+1}.$

The degrees of every element in G are defined by

$\mathrm{deg}(\overrightarrow{e_1}(0,n))=\mathrm{deg}(\overrightarrow{e_2}(0,n))=\mathrm{deg}(\overrightarrow{e_3}(0,n))=2n,$

$\mathrm{deg}(\overrightarrow{e_1}(1,n))=\mathrm{deg}(\overrightarrow{e_2}(1,n))=\mathrm{deg}(\overrightarrow{e_3}(1,n))=2n+1.$

The commutative relations of the loop algebra G are as follows:

$[\overrightarrow{e_1}(0,m),\overrightarrow{e_2}(0,n)]=\overrightarrow{e_3}(0,m+n),[\overrightarrow{e_1}(0,m),\overrightarrow{e_2}(1,n)]=\overrightarrow{e_3}(1,m+n),$

$[\overrightarrow{e_1}(0,m),\overrightarrow{e_3}(0,n)]=-\overrightarrow{e_2}(0,m+n),[\overrightarrow{e_1}(0,m),\overrightarrow{e_3}(1,n)]=-\overrightarrow{e_3}(1,m+n),$

$[\overrightarrow{e_1}(1,m),\overrightarrow{e_2}(0,n)]=\overrightarrow{e_3}(1,m+n),[\overrightarrow{e_1}(1,m),\overrightarrow{e_2}(1,n)]=\overrightarrow{e_3}(0,m+n+1),$

$[\overrightarrow{e_1}(1,m),\overrightarrow{e_3}(0,n)]=-\overrightarrow{e_2}(1,m+n),[\overrightarrow{e_1}(1,m),\overrightarrow{e_3}(1,n)]=-\overrightarrow{e_2}(0,m+n+1),$

$[\overrightarrow{e_2}(0,m),\overrightarrow{e_3}(0,n)]=\overrightarrow{e_1}(0,m+n),[\overrightarrow{e_2}(0,m),\overrightarrow{e_3}(1,n)]=\overrightarrow{e_1}(1,m+n),$

$[\overrightarrow{e_2}(1,m),\overrightarrow{e_3}(0,n)]=\overrightarrow{e_1}(1,m+n),[\overrightarrow{e_2}(1,m),\overrightarrow{e_3}(1,n)]=\overrightarrow{e_1}(0,m+n+1).$

Now, we make use of the loop algebra G to introduce the following isospectral problem:

$${\overrightarrow{\psi }}_x=\overrightarrow{U}\times \overrightarrow{\psi },{\overrightarrow{\psi }}_t=\overrightarrow{V}\times \overrightarrow{\psi },{\lambda }_t=0,\overrightarrow{\psi }={({\psi }_1,{\psi }_2)}^T,$$

(57)

where

$\overrightarrow{U}=\overrightarrow{e_1}(1,0)+u_1\overrightarrow{e_2}(0,0)+u_2\overrightarrow{e_3}(0,0)+u_3\overrightarrow{e_2}(1,-1)+u_4\overrightarrow{e_3(}1,-1)+u_5\overrightarrow{e_1}(1,-1),$

$\overrightarrow{V}={\displaystyle \sum _{m\ge 0}}(a(0,m)\overrightarrow{e_1}(0,-m)+a(1,m)\overrightarrow{e_1}(1,-m)+b(0,m)\overrightarrow{e_2}(0,-m)+b(1,m)\overrightarrow{e_2}(1,-m)$

$+c(0,m)\overrightarrow{e_3}(0,-m)+c(1,m)\overrightarrow{e_3}(1,-m).$

The compatibility condition of (57) is of the form (7) whose stationary zero-curvature equation

$${\overrightarrow{V}}_x=\overrightarrow{U}\times \overrightarrow{V}$$

(58)

leads to the following set of partial differential equations

$$\begin{array}{cc}\hfill & a_x(0,m)=u_{1}c(0,m)-u_{2}b(0,m)+u_{3}c(1,m)-u_{4}b(1,m),\hfill \\ \hfill & a_x(1,m+1)=u_{1}c(1,m+1)-u_{2}b(1,m+1)+u_{3}c(0,m)-u_{4}b(0,m),\hfill \\ \hfill & b_x(0,m)=-c(1,m+1)+u_{2}a(0,m)+u_{4}a(1,m)-u_{5}c(1,m),\hfill \\ \hfill & b_x(1,m+1)=-c(0,m+1)+u_{2}a(1,m+1)+u_{4}a(0,m)-u_{5}c(0,m),\hfill \\ \hfill & c_x(0,m)=b(1,m+1)-u_{1}a(0,m)-u_{3}a(1,m)+u_{5}b(1,m),\hfill \\ \hfill & c_x(1,m+1)=b(0,m+1)-u_{1}a(1,m+1)-u_{3}a(0,m)+u_{5}b(0,m).\hfill \end{array}$$

(59)

Denoting

${\overrightarrow{V}}_{+}^{(n)}={({\lambda }^n\overrightarrow{V})}_{+}={\displaystyle \sum _{m=0}^n}(a(0,m)\overrightarrow{e_1}(0,-m)+a(1,m)\overrightarrow{e_1}(1,-m)+(b(0,m)\overrightarrow{e_2}(0,-m)+b(1,m)\overrightarrow{e_2}(1,-m)+c(0,m)\overrightarrow{e_3}(0,-m)+c(1,m)\overrightarrow{e_3}(1,-m)){\lambda }^n,$

${\overrightarrow{V}}_{-}^{(n)}={\lambda }^n\overrightarrow{V}-{\overrightarrow{V}}_{+}^{(n)},$

then Equation (58) can be decomposed into the following form:

$$-{({\overrightarrow{V}}_{+}^{(n)})}_x+\overrightarrow{U}\times {\overrightarrow{V}}_{+}^{(n)}={({\overrightarrow{V}}_{-}^{(n)})}_x-\overrightarrow{U}\times {\overrightarrow{V}}_{-}^{(n)},$$

(60)

which is far different from decomposed equation presented by the stationary equation of (8). The degrees of the left-hand side of (60) are more than $-1$, while those of the right-hand side are less than 0. Therefore, the degrees of both sides are taken as $-1,0$. Thus, we obtain that

$$\begin{gathered} -{({\overrightarrow{V}}_{+}^{(n)})}_x+\overrightarrow{U}\times {\overrightarrow{V}}_{+}^{(n)}=c(1,n+1)\overrightarrow{e_2}(0,0)-b(1,n+1)\overrightarrow{e_3}(0,0)+[b_x(1,n+1)+c(0,n+1)-u_{2}a(1,n+1)]\overrightarrow{e_2}(1,-1)+ \\ [-b(0,n+1)+u_{1}a(1,n+1)+c_x(1,n+1)]\overrightarrow{e_3}(1,-1)+[a_x(1,n+1)-u_{1}c(1,n+1)+u_{2}b(1,n+1)]\overrightarrow{e_1}(1,-1). \end{gathered}$$

Let ${\overrightarrow{V}}^{(n)}={\overrightarrow{V}}_{+}^{(n)},$ then the vector-product zero-curvature equation

$${\overrightarrow{U}}_t-{\overrightarrow{V}}_x^{(n)}+\overrightarrow{U}\times {\overrightarrow{V}}^{(n)}=\overrightarrow{0}$$

gives rise to the following isospectral integrable hierarchy of equations

$$\begin{array}{ccc}& & u_t={\left(\begin{array}{c}{u}_1\\ u_2\\ u_3\\ u_4\end{array}\right)}_t=\left(\begin{array}{c}-c(1,n+1)\\ b(1,n+1)\\ -b_x(1,n+1)-c(0,n+1)+u_{2}a(1,n+1)\\ b(0,n+1)-u_{1}a(1,n+1)-c_x(1,n+1)\\ -a_x(1,n+1)+u_{1}c(1,n+1)-u_{2}b(1,n+1)\end{array}\right)\hfill \\ & & =\left(\begin{array}{ccccc}0& 0& 0& -1& 0\\ 0& 0& -1& 0& 0\\ 0& 1& -{\partial }_x& 0& u_2\\ 1& 0& 0& -{\partial }_x& -u_1\\ 0& 0& -u_2& u_1& -{\partial }_x\end{array}\right)\left(\begin{array}{c}b(0,n+1)\\ c(0,n+1)\\ b(1,n+1)\\ c(1,n+1)\\ a(1,n+1)\end{array}\right)=:J_1\left(\begin{array}{c}b(0,n+1)\\ c(0,n+1)\\ b(1,n+1)\\ c(1,n+1)\\ a(1,n+1)\end{array}\right)\hfill \\ & & =\left(\begin{array}{c}-c(1,n+1)\\ b(1,n+1)\\ -u_{4}a(0,n)+u_{5}c(0,n)\\ u_{3}a(0,n)-u_{5}b(0,n)\\ -u_{3}c(0,n)+u_{4}b(0,n)\end{array}\right)=\left(\begin{array}{ccccc}0& -1& 0& 0& 0\\ 1& 0& 0& 0& 0\\ 0& 0& 0& u_5& -u_4\\ 0& 0& -u_5& 0& u_3\\ 0& 0& u_4& -u_3& 0\end{array}\right)\left(\begin{array}{c}b(1,n+1)\\ c(1,n+1)\\ b(0,n)\\ c(0,n)\\ a(0,n)\end{array}\right)\hfill \\ & & =:J_2\left(\begin{array}{c}b(1,n+1)\\ c(1,n+1)\\ b(0,n)\\ c(0,n)\\ a(0,n)\end{array}\right).\hfill \end{array}$$

(61)

In what follows, we investigate the Hamiltonian structure by using the vector-product trace identity (54). Noting that

$\overrightarrow{V}=(a(0)+\lambda a(1),b(0)+\lambda b(1),c(0)+\lambda c(1)),\overrightarrow{U}=(\lambda +\frac{u_5}{\lambda },u_1+\frac{u_3}{\lambda },u_2+\frac{u_4}{\lambda }),$

it is easy to compute that

${\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial u_1}\}}_v=b(0)+\lambda b(1),{\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial u_2}\}}_v=c(0)+\lambda c(1),{\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial u_3}\}}_v=b(1)+\frac{b(0)}{\lambda },$

${\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial u_4}\}}_v=c(1)+\frac{c(0)}{\lambda },{\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial u_5}\}}_v=a(1)+\frac{a(0)}{\lambda },$

${\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial \lambda }\}}_v=(1-\frac{u_5}{{\lambda }^2})(a(0)+\lambda a(1))-\frac{u_3}{{\lambda }^2}(b(0)+\lambda b(1))-\frac{u_4}{{\lambda }^2}(c(0)+\lambda c(1)),$

where $a(0)={\displaystyle \sum _{m\ge 0}}a(0,m){\lambda }^{-2m},a(1)={\displaystyle \sum _{m\ge 0}}a(1,m){\lambda }^{-2m},\dots$

Substituting the above consequences into the formula (54) reads that

$$\frac{\delta }{\delta u}{\{\overrightarrow{V},\frac{\partial \overrightarrow{U}}{\partial \lambda }\}}_v={\lambda }^{-\gamma }\frac{\partial }{\partial \lambda }{\lambda }^{\gamma }\left(\begin{array}{c}b(0)+\lambda b(1)\\ c(0)+\lambda c(1)\\ b(1)+\frac{b(0)}{\lambda }\\ c(1)+\frac{c(0)}{\lambda }\\ a(1)+\frac{a(0)}{\lambda }\end{array}\right).$$

(62)

Comparing the coefficients of ${\lambda }^{-2n-3}$ in (62) yields that

$$\frac{\delta }{\delta u}[a(1,n+2)-u_{5}a(1,n+1)-u_{3}b(1,n+1)-u_{4}c(1,n+1)]=(-2n-2+\gamma )\left(\begin{array}{c}b(0,n+1)\\ c(0,n+1)\\ b(1,n+1)\\ c(1,n+1)\\ a(1,n+1)\end{array}\right).$$

Comparing the coefficients of ${\lambda }^{-2n-2}$ gives that

$$\frac{\delta }{\delta u}[a(0,n+1)-u_{5}a(0,n)-u_{3}b(0,n)-u_{4}c(0,n)]=(-2n-1+\gamma )\left(\begin{array}{c}b(1,n+1)\\ c(1,n+1)\\ b(0,n)\\ c(0,n)\\ a(0,n)\end{array}\right).$$

Taking $n=0,$ we find that $\gamma =0.$ Therefore, we have that

$$\left(\begin{array}{c}b(0,n+1)\\ c(0,n+1)\\ b(1,n+1)\\ c(1,n+1)\\ a(1,n+1)\end{array}\right)=\frac{\delta H(1,2n+2)}{\delta u},$$

where

$H(1,2n+2)=\frac{1}{2n+2}[-a(1,n+2)+u_{5}a(1,n+1)+u_{3}b(1,n+1)+u_{4}c(1,n+1)].$

$$\left(\begin{array}{c}b(1,n+1)\\ c(1,n+1)\\ b(0,n)\\ c(0,n)\\ a(0,n)\end{array}\right)=\frac{\delta H(2,2n+1)}{\delta u},$$

where

$H(2,2n+1)=\frac{1}{2n+1}[-a(0,n+1)+u_{5}a(0,n)+u_{3}b(0,n)+u_{4}c(0,n)].$

Thus, we produce the Hamiltonian structure of the isospectral integrable hierarchy (61) as follows

$$u_t=J_1\frac{\delta H(1,2n+2)}{\delta u}=J_2\frac{\delta H(2,2n+1)}{\delta u},n\ge 0.$$

(63)

From (59), we obtain the recursion operator:

$$L=\left(\begin{array}{ccccc}-u_1{\partial }_x^{-1}{u}_2& {\partial }_x+u_1{\partial }_x^{-1}{u}_1& -u_1{\partial }_x^{-1}{u}_4-u_5& u_1{\partial }_x^{-1}{u}_3& u_3\\ -{\partial }_x-u_2{\partial }_x^{-1}{u}_2& u_2{\partial }_x^{-1}{u}_1& -u_2{\partial }_x^{-1}{u}_4& u_2{\partial }_x^{-1}{u}_3-u_5& u_4\\ 1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ -{\partial }_x^{-1}{u}_2& {\partial }_x^{-1}{u}_1& -{\partial }_x^{-1}{u}_4& {\partial }_x^{-1}{u}_3& 0\end{array}\right),$$

which satisfies that

$$\begin{array}{cc}\hfill & \left(\begin{array}{c}b(0,n+1)\\ c(0,n+1)\\ b(1,n+1)\\ c(1,n+1)\\ a(1,n+1)\end{array}\right)=L\left(\begin{array}{c}b(1,n+1)\\ c(1,n+1)\\ b(0,n)\\ c(0,n)\\ a(0,n)\end{array}\right),\left(\begin{array}{c}b(1,n+1)\\ c(1,n+1)\\ b(0,n)\\ c(0,n)\\ a(0,n)\end{array}\right)=L\left(\begin{array}{c}b(0,n)\\ c(0,n)\\ b(1,n+1)\\ c(1,n)\\ a(1,n)\end{array}\right),\hfill \\ \hfill & J_{1}L=L^{\ast }{J}_1=J_2.\hfill \end{array}$$

(64)

As long as we can verify that $J_1$ and $J_2$ are a Hamiltonian pair, we conclude that (63) is a bi-Hamiltonian structure. In fact, for arbitrary constants $c_1$ and $c_2$, it is enough that we only verify $J=:c_1{J}_1+c_2{J}_2$ is a Hamiltonian operator identically. After complicated calculation, we find that J is really a Hamiltonian operator; here we omit the computation.

Remark 2.

When $u_3=u_4=u_5=0,$ the hierarchy (59) reduces to the integrable hierarchy (33) except for the difference complex number i. Therefore, (61) is a type of extended integrable model of (33), that is, (33) is a special case.

6. Conclusions

In the paper, we adopted the vector-product Lie algebra to show a method for generating integrable hierarchies. In order to illustrate the efficient roles of the method, we take the AKNS spectral problem(ZS spectral problem) as an example to really generate new integrable equations compared with the old matrix method by using the zero-curvature Equation (8). In addition, we derived a new computing formula for Hamilton structure by using the variational approach. It may be interesting work to investigate symmetries and their Lie algebras of the equations obtained in the paper, which will be further discussed by following the ideas in [21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45] in the forthcoming days.