# Conservation law and Lie symmetry analysis of the (1+1) dimensional
dispersive long-wave equation

Long Ju1 Faiza Afzal1 and Yufeng Zhang1,2,
1School of Mathematics,China University of Mining and Technology, Xuzhou,
221116, China;
2Jiangsu Center for Applied Mathematics (CUMT), Xuzhou, Jiangsu 221116, China.
Corresponding author<EMAIL_ADDRESS>
###### Abstract

In this paper, we mainly study the integrability of (1+1)-dimensional
dispersive long-wave equation. Firstly, the Lie symmetry analysis of the
equation is carried out in the first part. And the optimal system of the
equation is obtained according to the symmetry, and the invariant solution and
the reduced form of the target equation are solved according to the results.
Secondly, we use different methods to solve the conservation law of the target
equation. To begin with, we give the adjoint determination equation and
adjoint symmetry of the (1+1)-dimensional dispersive long-wave equation, and
use the adjoint symmetry as the equation multiplier to find several
conservation laws. Then we get a Lie bracket by using the relationship between
the symmetry of the equation and the adjoint symmetry. Next its strict self-
adjoint property is verified, and its conservation laws are solved by
Ibragimov’s method. Finally, the conservation laws of the target equation are
solved by Noether’s theorem. Thirdly we calculate some exact solutions of the
target equation by three different methods. In the end of the paper, the
Hamiltonian structure of the target equation, the generalized pre-symplectic
that maps symmetries into adjoint-symmetries and some of its soliton solutions
are calculated. In conclusion, we use the direct construction of conservation
law method, Ibragimov’s method and so on to solve some new conservation laws
of (1+1)-dimensional dispersive long-wave equation, use the relationship
between symmetry and adjoint symmetry to construct the corresponding Lie
brackets, and obtain some linear soliton solutions according to the
conservation law of the equation.
Keywords: conservation law; extended Kudryashov method; Hamiltonian structure;
Ibragimov method; Noether’s theorem; (1+1)-dimensional dispersive long-wave
equation
PACS numbers: 05.45.Yv, 02.30.Jr, 02.30.Ik

##  1 Introduction

Lie symmetry has been applied in most fields of mathematics and has played a
crucial role, such as differential equations [1, 2], Lie algebra [3, 4] and
classical mechanics [5]. Meanwhile, the Lie symmetry method is also considered
to be one of the most effective methods to derive the explicit solutions of
nonlinear partial differential equations. At the same time, on the basis of
symmetry, the integrability of differential equations can be considered. From
an algebraic viewpoint, the infinitesimal symmetries of a PDE are the
solutions of the linearisation (Frechet derivative) equation holding on the
space of solutions to the partial differential equation. Solutions of the
adjoint linearisation equation, holding on the space of solutions to the
partial differential equation, are called adjoint-symmetries and provide a
direct link to conservation laws. In particular, adjoint-symmetries that
satisfy a certain variational condition represent multipliers which yield
conservation laws.

In the past study of differential equations, the conservation law is
undoubtedly a very important part, especially in terms of integrability and
linearization [6, 7, 8]. How to solve the conservation law of the equation has
become a first problem to be faced. Conservation laws are indispensable in the
study of mathematical physics equations. At present, there are many methods to
construct the conservation laws of mathematical physics equations, such as
Noether theorem, known conservation laws of symmetric action, the variational
derivative method, the recursive formula method and so on. In this study, we
choose three methods to solve the conservation laws of partial differential
equations, which are direct construction of conservation laws [9, 10, 11, 12],
adjoint equation method [13, 14, 15, 16, 17] and using Noether’s theorem to
solve conservation laws.

This article manly deals with the (1+1) dimensional dispersive long-wave
equation [19] of the form

$\displaystyle u_{t}+uu_{x}+v_{x}=0,$ (1) $\displaystyle
v_{t}+u_{x}v+uv_{x}+\frac{1}{3}u_{xxx}=0,$

where $v$ is the elevation of water and $u$ is the surface velocity of water
along $x$-direction. System (1) can be traced back to the works of Broer [20],
Kaup [21], Martinez [22] and Kupershmidt [23]. It is very useful for coastal
and civil engineers to comprehend the solutions of the nonlinear water wave
model in harbor and coastal designs. In this paper, we mainly use three ways
to solve the conservation law of equation (1) and analyze its Lie symmetry. At
the same time, we use the relationship between symmetry and adjoint symmetry
[24, 25, 26, 27, 28] to obtain a new bracket of the target equation, and solve
its Hamiltonian structure and linear soliton solution. Finally, we calculate
some exact solutions [29, 30, 31, 32, 33] of the target equation by the
extended Kudryashov method and seek a few solutions in terms of hyperbolic
tangent functions and $\exp(\xi)$.

##  2 Lie symmetries analysis

###  2.1 Lie point symmetries

In this section, Lie symmetry analysis on the (1+1) dimension dispersive long-
wave equation was performed alongside considering one Lie group of
transformations parameter

$\displaystyle t\rightarrow t+\epsilon\xi^{1}(x,t,u,v),x\rightarrow
x+\epsilon\xi^{2}(x,t,u,v),$ (2) $\displaystyle u\rightarrow
u+\epsilon\xi^{3}(x,t,u,v),v\rightarrow v+\epsilon\eta^{1}(x,t,u,v),$

with a small parameter $\epsilon\leq 1$. And the corresponding generator of
the Lie algebra of the form

$X=\xi^{1}(t,x,u,v)\frac{\partial}{\partial
t}+\xi^{2}(t,x,u,v)\frac{\partial}{\partial
x}+\eta^{1}(t,x,u,v)\frac{\partial}{\partial
u}+\eta^{2}(t,x,u,v)\frac{\partial}{\partial v}.$ (3)

Thus the third prolongation $pr^{(3)}X$ is

$pr^{(3)}X=X+\eta^{1}_{t}\frac{\partial}{\partial
u_{t}}+\eta^{1}_{x}\frac{\partial}{\partial
u_{x}}+\eta^{2}_{t}\frac{\partial}{\partial
v_{t}}+\eta^{2}_{x}\frac{\partial}{\partial
v_{x}}+\eta^{1}_{xx}\frac{\partial}{\partial
u_{xx}}+\eta^{1}_{xy}\frac{\partial}{\partial u_{xy}}+\cdots.$ (4)

where
$\eta^{\alpha}_{i}=D_{i}\eta^{\alpha}-(D_{i}\xi^{j})u_{j},i=1,2,\alpha=1,2.$
$\eta^{\alpha}_{i_{i}i_{2}\cdots
i_{k}}=D_{i_{k}}\eta^{\alpha}_{i_{1}i_{2}\cdots
i_{k-1}}-(D_{i_{k}}\xi^{j})u_{i_{1}i_{2}\cdots i_{k-1}j},i_{l}=1,2,\alpha=1,2$
for $l=1,2,\dots k$ with $k=2,3,\dots.$ and the operators $D_{x}$ and $D_{t}$
are the total derivatives with respect to $x$ and $t$. The determining
equations of Eq. (1) arises from the following invariance condition

$pr^{(3)}X(G)|_{G=0}=0.$ (5)

Then we obtain the overdetermined system of the partial differential equations

$\displaystyle\xi^{1}_{tt}=0,\xi^{2}_{tt}=0,\xi^{1}_{u}=0,\xi^{1}_{v}=0,\xi^{1}_{x}=0,\xi^{2}_{u}=0,$
(6)
$\displaystyle\xi^{2}_{v}=0,\xi^{2}_{x}=\frac{\xi^{1}_{t}}{2},\eta^{1}=\frac{u\xi^{1}_{t}}{2}+\xi^{2}_{t},\eta^{2}=v\xi^{1}_{t}.$

Solving the system, one can get

$\displaystyle\xi^{1}=C_{1}t+C_{2},\xi^{2}=\frac{C_{1}}{2}x+C_{3}t+C_{4},\eta^{1}=\frac{C_{1}u}{2}+C_{3},\eta^{2}=C_{1}v,$
(7)

where $C_{i},i=1,\cdots,4.$ are arbitrary constants. Hence the infinitesimal
symmetries of Eq. (1) form the four dimensional Lie algebra $L_{6}$ spanned by
the following independent operators:

$\displaystyle X_{1}=\frac{\partial}{\partial
t},X_{2}=\frac{\partial}{\partial x},X_{3}=t\frac{\partial}{\partial
x}+\frac{\partial}{\partial u},$ (8) $\displaystyle
X_{4}=\frac{x}{2}\frac{\partial}{\partial x}+t\frac{\partial}{\partial
t}-\frac{u}{2}\frac{\partial}{\partial u}-v\frac{\partial}{\partial v}.$

By solving the following system of ordinary differential equations with
initial condition

$\displaystyle\frac{dt^{*}}{d\epsilon}=\xi^{1}(t^{*},x^{*},u^{*},v^{*})|_{\epsilon=0}=t,\frac{dx^{*}}{d\epsilon}=\xi^{2}(t^{*},x^{*},u^{*},v^{*})|_{\epsilon=0}=x,$
(9)
$\displaystyle\frac{du^{*}}{d\epsilon}=\eta^{1}(t^{*},x^{*},u^{*},v^{*})|_{\epsilon=0}=u,\frac{dv^{*}}{d\epsilon}=\eta^{2}(t^{*},x^{*},u^{*},v^{*})|_{\epsilon=0}=v.$

We can obtain the group transformation which is generated by the infinitesimal
generator $X_{i}(i=1,\dots,5)$, respectively

$\displaystyle G_{1}:(t,x,u,v)\rightarrow(t+\epsilon,x,u,v),$ (10)
$\displaystyle G_{2}:(t,x,u,v)\rightarrow(t,x+\epsilon,u,v),$ $\displaystyle
G_{3}:(t,x,u,v)\rightarrow(t,x+t\epsilon,u+\epsilon,v),$ $\displaystyle
G_{4}:(t,x,u,v)\rightarrow(te^{\epsilon},xe^{\frac{\epsilon}{2}},u=ue^{-\frac{\epsilon}{2}},v=ve^{-\epsilon}).$

Theorem 1. If $(u(x,t),v(x,t))$ is a solution of Heisenberg equation, so are
the functions

$\displaystyle G_{1}(\epsilon)\cdot
u(x,t)=u(x,t-\epsilon),G_{1}(\epsilon)\cdot v(x,t)=v(x,t-\epsilon),$ (11)
$\displaystyle G_{2}(\epsilon)\cdot
u(x,t)=u(x-\epsilon,t),G_{1}(\epsilon)\cdot v(x,t)=v(x-\epsilon,t),$
$\displaystyle G_{3}(\epsilon)\cdot
u(x,t)=u(x-t\epsilon,t)-\epsilon,G_{3}(\epsilon)\cdot
v(x,t)=v(x-t\epsilon,t),$ $\displaystyle G_{4}(\epsilon)\cdot
u(x,t)=u(xe^{-\frac{\epsilon}{2}},te^{-\epsilon})e^{\frac{\epsilon}{2}},G_{4}(\epsilon)\cdot
v(x,t)=v(xe^{-\frac{\epsilon}{2}},te^{-\epsilon})e^{\epsilon}.$

###  2.2 Optimal system of one-dimensional subalgebras

In 2010, Ibragimov et al. presented a concise method to get the optimal
system, which only relies on the commutator table. Then Abdulwahhab
constructed an optimal system of the 2-dimensional Burgers equations. In this
part, we try to find the optimal system of (1+1)-dimensional dispersive long-
wave equation. First, we construct a Lie bracket $L_{5}$. Their algebra has
the non-zero commutators

$[X_{1},X_{3}]=X_{2},[X_{1},X_{4}]=X_{1},[X_{2},X_{4}]=\frac{1}{2}X_{2},[X_{3},X_{4}]=-\frac{1}{2}X_{3}.$
(12)

We take any operator $X=l^{1}X_{1}+l^{2}X_{2}+l^{3}X_{3}+l^{4}X_{4}.$ To find
the linear transformations of the vector $l=(l^{1},l^{2},l^{3},l^{4})$, we can
use the following generators

$E_{i}=c_{ij}^{k}l^{j}\frac{\partial}{\partial l^{k}},i=1,2,3,4,$ (13)

where $c_{ij}^{k}$ is defined by the formula $[X_{i}.X_{j}]=c_{ij}^{k}X^{k}.$
So after calculation, $E_{1},E_{2},E_{3},E_{4}$ has the following forms:

$\displaystyle E_{1}=l^{3}\frac{\partial}{\partial
l_{2}}+l^{4}\frac{\partial}{\partial
l^{1}},E_{2}=\frac{1}{2}l^{4}\frac{\partial}{\partial l^{2}},$ (14)
$\displaystyle E_{3}=-l^{1}\frac{\partial}{\partial
l^{2}},E_{4}=-l^{1}\frac{\partial}{\partial
l^{1}}-\frac{1}{2}l^{2}\frac{\partial}{\partial
l^{2}}-\frac{1}{2}l^{3}\frac{\partial}{\partial l^{3}}.$

Let us find the transformations provided by the generators (14). For the
generator $E_{1},E_{2},E_{3},E_{4}$, the Lie equations with the parameter
$a_{1},a_{2},a_{3},a_{4}$ are written

$\displaystyle\frac{d\widetilde{l}^{1}}{a_{1}}=\widetilde{l}^{4},\frac{d\widetilde{l}^{2}}{a_{1}}=\widetilde{l}^{3},\frac{d\widetilde{l}^{3}}{a_{1}}=0,\frac{d\widetilde{l}^{4}}{a_{1}}=0,$
(15)
$\displaystyle\frac{d\widetilde{l}^{1}}{a_{2}}=0,\frac{d\widetilde{l}^{2}}{a_{2}}=\frac{1}{2}\widetilde{l}^{4},\frac{d\widetilde{l}^{3}}{a_{2}}=0,\frac{d\widetilde{l}^{4}}{a_{2}}=0,$
$\displaystyle\frac{d\widetilde{l}^{1}}{a_{3}}=0,\frac{d\widetilde{l}^{2}}{a_{3}}=-\widetilde{l}^{1},\frac{d\widetilde{l}^{3}}{a_{3}}=-\frac{1}{2}\widetilde{l}^{4},\frac{d\widetilde{l}^{4}}{a_{3}}=0,$
$\displaystyle\frac{d\widetilde{l}^{1}}{a_{4}}=-\widetilde{l}^{1},\frac{d\widetilde{l}^{2}}{a_{4}}=-\frac{1}{2}\widetilde{l}^{2},\frac{d\widetilde{l}^{3}}{a_{4}}=-\frac{1}{2}\widetilde{l}^{3},\frac{d\widetilde{l}^{4}}{a_{4}}=0.$

The solutions of these equations give the transformations

$\displaystyle
T_{1}:\widetilde{l}^{1}=l_{1}+a_{1}l^{4},\widetilde{l}^{2}=l^{2}+a_{1}l^{3},\widetilde{l}^{3}=l^{3},\widetilde{l}^{4}=l^{4},$
(16) $\displaystyle
T_{2}:\widetilde{l}^{1}=l^{1},\widetilde{l}^{2}=l^{2},\widetilde{l}^{3}=l^{3}+\frac{1}{2}a_{2}l^{4},\widetilde{l}^{4}=l^{4},$
$\displaystyle
T_{3}:\widetilde{l}^{1}=l^{1},\widetilde{l}^{2}=l^{2}-a_{3}l^{1},\widetilde{l}^{3}=l^{3}-\frac{1}{2}a_{3}l^{4},\widetilde{l}^{4}=l^{4},$
$\displaystyle
T_{4}:\widetilde{l}^{1}=e^{-a_{4}}l^{1},\widetilde{l}^{2}=e^{-\frac{1}{2}a_{4}}l^{2},\widetilde{l}^{3}=e^{-\frac{1}{2}a_{4}}l^{3},\widetilde{l}^{4}=l^{4}.$

The construction of the optimal system requires a simplification of the vector
$l=(l^{1},l^{2},l^{3},l^{4})$ by use of $T_{1},T_{2},T_{3},T_{4}$. And the
construction will be divided into two cases.
Case 1. $l^{1}\neq 0.$ By taking $a_{3}=\frac{l^{2}}{l^{1}}$ in $T_{3}$, we
can make $\widetilde{l}^{2}=0$. The vector is reduced to the form
$l=(l^{1},0,l^{3},l^{4})$.
1.1 $l^{4}\neq 0$ By taking $a_{2}=-2\frac{l^{3}}{l^{4}}$ in $T_{2}$, we can
make $\widetilde{l}^{3}=0$. The vector is reduced to the form
$l=(l^{1},0,0,l^{4})$.
Then by taking $a_{1}=-\frac{l^{1}}{l^{4}}$.The vector is reduced to the form
$l=(0,0,0,l^{4})$.
we obtain the following representatives

$X_{4}.$

1.2 $l^{4}\neq 0$ The vector is reduced to the form $l=(l^{1},0,l^{3},0)$.
Taking all the possible combinations, we obtain the following representatives

$X_{1},X_{1}\pm X_{3}.$

Case 2. $l^{1}=0.$ The vector is reduced to the form
$l=(0,l^{2},l^{3},l^{4})$.
2.1 $l^{3}\neq 0$. By taking $a_{1}=-\frac{l^{2}}{l^{3}}$ in $T_{1}$, we can
make $\widetilde{l}^{2}=0$. The vector is reduced to the form
$l=(0,0,l^{3},l^{4})$.
Then if we set $l^{4}=0$, we obtain the following representatives

$X_{3}.$

If $l^{4}\neq 0$, by taking $a_{3}=\frac{2l^{3}}{l^{4}}$ in $T_{3}$. The
vector is reduced to the form $l=(0,0,0,l^{4})$. we obtain the following
representatives

$X_{4}.$

2.2 $l^{3}=0$ The vector is reduced to the form $l=(0,l^{2},0,l^{4})$. Taking
all the possible combinations, we obtain the following representatives

$X_{2},X_{4},X_{2}\pm X_{3}.$

Theorem 2. The following operators provide an optimal system of one-
dimensional subalgebras of the Lie algebra spanned by Eq. (1):

$X_{1},X_{2},X_{3},X_{4},X_{1}\pm X_{3},X_{2}\pm X_{4}$

.

###  2.3 Reductions and the invariant solutions

After we calculate the optimal system of subalgebras, we can investigate the
symmetry reductions of Eq. (1) by integrating the characteristic equations.
Next, we analyze the reduced order and invariant solution of the Eq. (1).

####  2.3.1 Solutions through $X_{1}+X_{3}$

For the generator

$X_{1}+X_{3}=\frac{\partial}{\partial t}+t\frac{\partial}{\partial
x}+\frac{\partial}{\partial u},$

the characteristic equation is written as

$\frac{dt}{1}=\frac{dx}{t}=\frac{du}{1},$ (17)

We can get a similar variables according to the characteristic equation:

$Q=\frac{t^{2}}{2}-x,u=t+f(Q),v=g(Q).$

After bringing them into Eq.(1), we can obtain the reduced form of it:

$\displaystyle 1+f^{{}^{\prime}}t+(t+f)(-f^{{}^{\prime}})-g^{{}^{\prime}}=0,$
(18) $\displaystyle
tg^{{}^{\prime}}-f^{{}^{\prime}}g+(t+f)(-g^{{}^{\prime}})-\frac{f^{{}^{\prime\prime\prime}}}{3}=0.$

By solving it, We can get the form of the solution of the Eq. (1)

$f=c_{1},g=Q+c_{2}=\frac{t^{2}}{2}-x+c_{2}.$

So we obtain the invaint solution of Eq. (1)

$u=t+c_{1},v-\frac{t^{2}}{2}-x+c_{2},$ (19)

where $c_{1}$ and $c_{2}$ are arbitrary constants.

####  2.3.2 Solutions through $X_{2}+X_{4}$

For the generator

$X_{2}+X_{4}=t\frac{\partial}{\partial
t}+(\frac{x}{2}+1)\frac{\partial}{\partial
x}-\frac{u}{2}\frac{\partial}{\partial u}-v\frac{\partial}{\partial v},$

the characteristic equation is written as

$\frac{dx}{1+\frac{x}{2}}=\frac{dt}{t}=\frac{du}{-\frac{u}{2}}=\frac{dv}{-v}.$
(20)

By calculation, we can get the expressions of similar variables and $u,v$:

$R=\frac{x+2}{\sqrt{t}},u=\frac{f(R)}{\sqrt{t}},v=\frac{g(R)}{t}.$

Then we can get the reduced form of Eq. (1)

$\displaystyle-R\frac{f^{{}^{\prime}}}{2}-\frac{f}{2}+ff^{{}^{\prime}}+g^{{}^{\prime}}=0,$
(21) $\displaystyle-
g-R\frac{g^{{}^{\prime}}}{2}+gf^{{}^{\prime}}+fg^{{}^{\prime}}+\frac{f^{{}^{\prime\prime\prime}}}{3}=0.$

By solving the system (21), we can obtain the solution of it

$f=R,g=c_{1}.$

So we get the invarint solution of Eq. (1)

$u=\frac{1}{\sqrt{t}}\cdot\frac{x+2}{\sqrt{t}}=\frac{x+2}{t},v=\frac{c_{1}}{t},$
(22)

where $c_{1}$ is an arbitrary constant.

##  3 Adjoint symmetries analysis

###  3.1 Constructing conservation laws by adjoint symmetries

In this section, we try to use the adjoint determination equation of the
equation to solve its adjoint symmetry, and use the adjoint symmetry to solve
the conservation law of the target equation. First we consider a system of
partial differential equations with $N$ independent variables
$\bm{u}=(u^{1},\dots,u^{N})$ and $n+1$ independent variables $(t,\bm{x})$, the
forms are as follows:

$G^{i}=\frac{\partial u^{j}}{\partial
t}+g^{j}(t,\bm{x},\bm{u},\partial_{\bm{x}}\bm{u},\dots,\partial^{m}_{\bm{x}}\bm{u})=0,j=1,\dots,N,$
(23)

with $\bm{x}$ derivatives of $\bm{u}$ up to some order $m$. And we can use
$\partial_{\bm{x}}\bm{u},\partial_{\bm{x}}^{2}\bm{u}$, etc. to represent all
the derivatives of $u^{j}$ with respect to ${x^{i}}$. We denote partial
derivatives $\frac{\partial}{\partial t}$ and $\frac{\partial}{\partial
x^{i}}$ with $\bm{x}$ derivatives of $\bm{u}$ up to some order $m$. We denote
partial derivatives $\frac{\partial}{\partial t}$ and
$\frac{\partial}{\partial x^{i}}$ by subscripts $t$ and $i$ respectively. And
likewise, $D_{t}$ and $D_{i}$ represent the total derivatives with respect to
$x^{i}$ and $t$.
Definition 4 Multipliers for PDE system (23) are a set of expressions

${\lambda_{1}(t,\bm{x},\bm{u},\partial_{\bm{x}}\bm{u},\dots,\partial^{q}_{\bm{x}}\bm{u}),\cdots,\lambda_{N}(t,\bm{x},\bm{u},\partial_{\bm{x}}\bm{u},\dots,\partial^{q}_{\bm{x}}\bm{u})}$

satisfying

$(u^{j}_{t}+g^{j})\lambda_{j}=D_{t}\phi^{t}+D_{i}\phi^{i},$ (24)

for some expression
$\phi^{t}(t,\bm{x},\bm{u},\partial_{\bm{x}}\bm{u},\dots,\partial^{k}_{\bm{x}}\bm{u})$
and
$\phi^{i}(t,\bm{x},\bm{u},\partial_{\bm{x}}\bm{u},\dots,\partial^{k}_{\bm{x}}\bm{u})$
for all functions $\bm{u}(t,\bm{x}).$
Theorem 3 For the Cauchy-Kovalevskaya PDE system (23), every nontrivial
conservation law in normal form
$D_{t}\phi^{t}(\bm{x},t,\partial_{\bm{x}}\bm{u},\cdots)+D_{i}\phi^{i}(\bm{x},t,\partial_{\bm{x}}\bm{u},\cdots)=0$
is uniquely characterized by a set of multipliers $\Lambda_{\sigma}$ with no
dependence on ut and differential consequences, satisfying the relations
$D_{t}\phi^{t}+D_{i}(\phi^{i}-\Gamma^{i})=(u_{t}^{\sigma}+g^{\sigma})\Lambda_{\sigma}$
and $\Lambda_{\sigma}=\widehat{E}_{u}^{\sigma}(\phi^{t}),\sigma=1,\cdots,N.$
holding for all functions $\bm{u}(t,\bm{x})$, where $\Gamma^{i}$ is given by
an expression proportional to $u_{t}^{\sigma}+g^{\sigma}$ (and differential
consequences), $\widehat{E}_{u}^{\sigma}=\partial_{u^{\sigma}}-D_{i}\partial
u^{\sigma}_{i}+D_{i}D_{j}\partial_{u_{ij}^{\sigma}}+\cdots.$

The standard determining condition for multiplier expressions
$\Lambda_{\sigma}(t,\bm{x},\bm{u},\partial_{\bm{x}}\bm{u},\cdots)$ arises from
the definition (24) by the wellknown result that divergence expressions are
characterized by annihilation under the full Euler operator

$E_{u^{\sigma}}=\partial_{u^{\sigma}}-D_{i}\partial_{u_{i}^{\sigma}}-D_{t}\partial_{u_{t}^{\sigma}}+D_{i}D_{j}\partial_{u_{ij}^{\sigma}}+D_{t}D_{j}\partial_{u_{ij}^{\sigma}}+\cdots.$

This yields

$E_{u^{\sigma}}(u_{t}^{\rho}\Lambda_{\rho}+g^{\rho}\Lambda_{\rho})=0.$ (25)

To carry out the splitting of $D_{t}\Lambda_{\sigma}$, we ues the identity

$D_{t}=\partial_{t}+(u_{t}^{\rho}+g^{\rho})\partial_{u^{\rho}}+(u_{ti}^{\rho}+D_{i}g^{\rho})\partial_{u^{\rho}_{i}}+\cdots.$

Consequently, the non-leading terms in Eq. (25) are as follows

$\displaystyle-\frac{\partial\Lambda_{\sigma}}{\partial
t}+(\frac{\partial\Lambda_{\sigma}}{\partial
u^{\rho}}g^{\rho}+\frac{\partial\Lambda_{\sigma}}{\partial
u^{\rho}_{i}}D_{i}g^{\rho}+\cdots+\frac{\partial\Lambda_{\sigma}}{\partial
u^{\rho}_{i_{1}\cdots i_{p}}}D_{i}\cdots D_{i_{p}}g^{\rho})$ (26)
$\displaystyle+\frac{g^{\rho}}{\partial
u^{\sigma}}\Lambda_{\rho}-D_{i}(\frac{g^{\rho}}{\partial
u^{\sigma}_{i}}\Lambda_{\rho})+\cdots+(-1)^{m}D_{i_{1}}\cdots
D_{i_{m}}(\frac{g^{\rho}}{\partial u^{\sigma}_{i_{1}\cdots
i_{m},}}\Lambda_{\rho})=0,\sigma=1,\dots,N.$

It is obviously the adjoint determining equation of Eq. (23), when
$\bm{u}(t,\bm{x})$ is the solution of Eq.(23). Therefore, we know that the
multiplier of the conservation law of the target equation is actually its
special adjoint symmetry. Next, we try to use the adjoint symmetry of Eq. (1)
to solve its conservation law. First, we can write the adjoint determining
equation of the Eq. (1)

$\displaystyle-
D_{t}Q_{1}-uD_{x}Q_{1}-vD_{x}Q_{2}-\frac{1}{3}D_{x}^{3}Q_{2}=0,$ (27)
$\displaystyle-D_{t}Q_{2}-uD_{x}Q_{2}-D_{x}Q_{1}=0,$

where $Q_{i},i=1,2.$ It has the expression
$Q_{i}(x,t,\bm{u},\partial_{x}\bm{u},\partial_{x^{2}}\bm{u})$. By expanding
the formula (27) and separating the coefficients, we can get an overdetermined
equation system about $Q_{i}$. After solving the overdetermined equation
system, we can obtain the adjoint symmetry of Eq. (1)

$\displaystyle
Q_{1}=(v_{xx}+\frac{9}{4}v^{2}+\frac{9}{4}u^{2}v+\frac{6uu_{xx}+3u_{x}^{2}}{4},u_{xx}+\frac{3u^{3}}{4}+\frac{9}{2}uv),$
(28) $\displaystyle
Q_{2}=(tv,tu-x),Q_{3}=(tv,\frac{u^{2}}{2}+v),Q_{4}=(v,u),Q_{5}=(1,0),Q_{6}=(0,1).$

By verifying the folumar (25), we can find that $Q_{i},i=1,\dots$ is the
multiplier of the equation. Therefore, we can naturally obtain the
conservation law corresponding to each adjoint symmetry.

$\displaystyle\phi_{t}^{1}=u_{x}v_{x}+vu_{xx}+uv_{xx}+\frac{3}{4}u^{3}v+\frac{9}{4}v^{2}u+\frac{3}{4}uu_{x}^{2}+\frac{3}{4}u^{2}u_{xx},$
(29)
$\displaystyle\phi_{x}^{1}=\frac{27}{8}u^{2}v^{2}+\frac{v_{x}^{2}}{2}+\frac{3}{4}v^{3}+\frac{u_{xx}^{2}}{6}-uv_{xt}+\frac{3}{4}u^{4}v+\frac{3}{8}u^{2}u_{x}^{2}+\frac{u^{3}u_{xx}}{4}+\frac{3uvu_{xx}}{2}+uu_{x}v_{x}-\frac{u_{x}^{2}v}{4}-\frac{3u^{2}u_{xt}}{4}.$

for $Q_{1}$.

$\displaystyle\phi_{t}^{2}=tuv-xv,$ (30)
$\displaystyle\phi_{x}^{2}=\frac{tv^{2}}{2}-xuv+tu^{2}v+\frac{t}{3}uu_{xx}-\frac{t}{6}u_{x}^{2}-\frac{xu_{xx}}{3}+\frac{u_{x}}{3},$

for $Q_{2}$.

$\displaystyle\phi_{t}^{3}=v^{2}u+\frac{u^{3}v}{2}+\frac{u^{2}u_{xx}}{6}+\frac{vu_{xx}}{3}+\frac{u_{x}u_{t}}{3},$
$\displaystyle\phi_{x}^{3}=\frac{v^{2}}{2}+\frac{u^{2}v}{2}-\frac{u_{x}^{2}}{6},$
(31)

for $Q_{3}$.

$\displaystyle\phi_{t}^{4}=uv,$ (32)
$\displaystyle\phi_{x}^{4}=\frac{v^{2}}{2}+vu^{2}+\frac{uu_{xx}}{3}-\frac{u_{x}^{2}}{6},$

for $Q_{4}$.

$\displaystyle\phi_{t}^{5}=u+v$
$\displaystyle\phi_{x}^{5}=\frac{u^{2}}{2}+v+uv+\frac{u_{xx}}{3},$ (33)

for $Q_{5}+Q_{6}$.

###  3.2 Symmetry actions and brackets for adjoint-symmetries

In this section, we will use the symmetry of the target equation to act on its
adjoint symmetry and obtain a Lie algebra of it. First, we introduce some
theorems.
Theorem 4: For any regular partial differential system, there are two actions
of symmetries:

$\displaystyle
Q_{A}\overset{X_{p}}{\longrightarrow}Q^{{}^{\prime}}(P)_{A}+R_{p}^{*}(Q)_{A},$
(34) $\displaystyle
Q_{A}\overset{X_{p}}{\longrightarrow}R_{p}^{*}(Q)_{A}-R_{Q}^{*}(P)_{A},$ (35)

on the linear space of adjoint-symmetries. The second symmetry action (35)
maps adjoint-symmetries into conservation law multipliers. The difference of
the first and second actions yields the linear mapping

$Q_{A}\overset{X_{p}}{\longrightarrow}Q^{{}^{\prime}}(P)_{A}+R_{Q}^{*}(P)_{A},$
(36)

where $R_{p},R_{Q}$ mean the operator from the determining equation of the
symmetry and adjoint sysmetry of the target equation
$R_{p}(G)=G^{{}^{\prime}}(p),R_{Q}(G)=G^{{}^{\prime*}}(Q)$ and $P$ and $Q$ is
symmetry and adjoint symmetry of the PDE system.

The action (36) will be trivial when the adjoint-symmetry is a conservation
law multiplier, as follows from the relation (36) which holds under certain
mild conditions on the form of the PDE system $G^{A}=0$ and the functions
$Q_{A}$.
Theorem 5: Fix an adjoint-symmetry $Q_{A}$ in AdjSymmG, and let $S_{Q}$ be the
dual linear operator $S_{Q}(P)_{A}:=S_{p}(Q)_{A}$ associated with a symmetry
action $S_{p}$ on AdjSymmG. If the kernel of $S_{Q}$ is an ideal in SymmG,
then

${}^{Q}\left[Q_{1},Q_{2}\right]_{A}:=S_{Q}([S_{Q}^{-1}Q_{1},S_{Q}^{-1}Q_{2}])_{A},$
(37)

defines a bilinear bracket on the linear space $S_{Q}(Symm_{G})\in
AdjSymm_{G}.$ This bracket can be expressed as

${}^{Q}\left[Q_{1},Q_{2}\right]_{A}=Q_{2}^{{}^{\prime}}(S_{Q}^{-1}Q_{1})-Q_{1}^{{}^{\prime}}(S_{Q}^{-1}Q_{2})-S_{Q}^{{}^{\prime}}(S_{Q}^{-1}Q_{2})(S_{Q}^{-1}Q_{1})+S_{Q}^{{}^{\prime}}(S_{Q}^{-1}Q_{1})(S_{Q}^{-1}Q_{2}),$
(38)

where $S_{Q}^{{}^{\prime}}$ denotes the Frechet derivative of $S_{Q}$. In the
previous part, we have solved a series of symmetries

$\displaystyle
P_{1}=(-u_{t},-v_{t}),P_{2}=(-u_{x},-v_{x}),P_{3}=(1-tu_{x},-tv_{x}),P_{4}=(-\frac{u}{2}-\frac{x}{2}u_{x}-tu_{t},-v-\frac{x}{2}v_{x}-tv_{t}),$
(39)

and adjoint symmetries of the target equation

$\displaystyle
Q_{1}=(v_{xx}+\frac{9}{4}v^{2}+\frac{9}{4}u^{2}v+\frac{6uu_{xx}+3u_{x}^{2}}{4},u_{xx}+\frac{3u^{3}}{4}+\frac{9}{2}uv),$
(40) $\displaystyle
Q_{2}=(tv,tu-x),Q_{3}=(tv,\frac{u^{2}}{2}+v),Q_{4}=(v,u),Q_{5}=(1,0),Q_{6}=(0,1).$

Their algebra has the non-zero commutators

$\left[P_{1},P_{3}\right]=P_{4},\left[P_{1},P_{4}\right]=P_{1},\left[P_{2},P_{4}\right]=\frac{P_{2}}{2},\left[P_{3},P_{4}\right]=-\frac{P_{3}}{2}.$
(41)

Since the above six adjoint equations are all multipliers of the target
equation, action (36) is trivial for every $Q_{i}$. And We can find that a is
an evolution equation system, so we have a conclusion
$Q^{{}^{\prime}}=-R_{p}^{*}$ and action (34) is the same as action (35). Then
we choose action (34) to analysis. We can get $R_{p_{i}}$ by caluation

$\displaystyle R_{p_{1}}=\left(\begin{array}[]{cc}-D_{t}&0\\\ 0&-D_{t}\\\
\end{array}\right),R_{p_{2}}=\left(\begin{array}[]{cc}-D_{x}&0\\\ 0&-D_{x}\\\
\end{array}\right),$ (42) $\displaystyle
R_{p_{3}}=\left(\begin{array}[]{cc}-tD_{x}&0\\\ 0&-tD_{x}\\\
\end{array}\right),R_{p_{4}}=\left(\begin{array}[]{cc}-\frac{3}{2}-\frac{x}{2}D_{x}-tD_{t}&0\\\
0&-\frac{x}{2}D_{x}-tD_{t}-2\\\ \end{array}\right).$

Table 1:  Symmetry action (34) on adjoint-symmetries | $P_{1}$ | $P_{2}$ | $P_{3}$ | $P_{4}$
---|---|---|---|---
$Q_{1}$ | 0 | 0 | $\frac{9}{2}Q_{3}$ | $-2Q_{1}$
$Q_{2}$ | $Q_{4}$ | $-Q_{6}$ | 0 | 0
$Q_{3}$ | 0 | 0 | $Q_{4}$ | $-\frac{3}{2}Q_{3}$
$Q_{4}$ | 0 | 0 | $Q_{6}$ | $-Q_{4}$
$Q_{5}$ | 0 | 0 | 0 | 0
$Q_{6}$ | 0 | 0 | 0 | 0

According to Theorem 5, we can know that the condition for the existence of
bracket is kernel of $S_{Q}$ is an ideal for an fixed adjoint symmetry.
Therefore, after verification, we get $Q_{i}$ satisfies the condition when
$i=1,3,\cdots,6$. However, the resulting bracket is trivial when $i=5,6$. For
example, when we fix $Q_{1}$, we get
$ker(S_{Q_{1}})=\left\\{0,P_{1},P_{2}\right\\}$. For any $P_{j}\in
Symm_{G},\left[P_{i},P_{j}\right]\in ker(S_{Q_{1}})$, where $P_{i}\in
ker(S_{Q_{1}})$. Therefore, we can find the following commutators after the
action of the adjoint symmetry by the commutators between the symmetries from
table 1:

$\displaystyle{}^{Q_{1}}\left[Q_{1},Q_{3}\right]=S_{Q_{1}}\left[S_{Q_{1}}^{-1}Q_{1},S_{Q_{1}}^{-1}Q_{3}\right]=S_{Q_{1}}\left[-\frac{1}{2}P_{4},\frac{2}{9}P_{3}\right]=S_{Q_{1}}(-\frac{P_{3}}{18})=-\frac{Q_{3}}{4},$
(43)
$\displaystyle{}^{Q_{3}}\left[Q_{3},Q_{4}\right]=S_{Q_{3}}\left[S_{Q_{3}}^{-1}Q_{3},S_{Q_{3}}^{-1}Q_{4}\right]=S_{Q_{3}}\left[-\frac{2}{3}P_{4},P_{3}\right]=S_{Q_{1}}(\frac{P_{3}}{3})=\frac{Q_{4}}{3},$
$\displaystyle{}^{Q_{4}}\left[Q_{4},Q_{6}\right]=S_{Q_{4}}\left[S_{Q_{4}}^{-1}Q_{4},S_{Q_{4}}^{-1}Q_{6}\right]=S_{Q_{4}}\left[-P_{4},P_{3}\right]=S_{Q_{4}}(\frac{P_{3}}{2})=\frac{Q_{6}}{2}.$

##  4 Compute conservation laws by Noether’s theorem

In this part, we try to use Noether’s theorem to solve the conservation laws
of the (1+1)-dimensional dispersive long-wave equation. First, we introduce
the operator

$E_{\gamma}=\frac{\partial}{\partial u^{\gamma}}-D_{i}\frac{\partial}{\partial
u_{i}^{\gamma}}+D_{i}D_{j}\frac{\partial}{\partial
u_{ij}^{\gamma}}+\cdots+(-1)^{k}D_{i_{1}}D_{i_{2}}\cdots
D_{i_{k}}\frac{\partial}{\partial u_{i_{1}i_{2}\cdots i_{k}}^{\gamma}}.$

Then let

$\displaystyle W^{i}[u,v]=$ $\displaystyle v^{\gamma}[\frac{\partial
L}{\partial u_{i}^{\gamma}}+\cdots+(-1)^{k-1}D_{i_{1}}D_{i_{2}}\cdots
D_{i_{k-1}}\frac{\partial L}{\partial u_{ii_{1}i_{2}\cdots
i_{k-1}}^{\gamma}}]+(D_{i_{1}}v^{\gamma})[\frac{\partial L}{\partial
u_{i_{1}i}^{\gamma}}+\cdots$ (44)
$\displaystyle+(-1)^{k-2}D_{i_{2}}D_{i_{3}}\cdots D_{i_{k-1}}\frac{\partial
L}{\partial u_{ii_{1}i_{2}\cdots
i_{k-1}}^{\gamma}}]+\cdots+(D_{i_{1}}D_{i_{2}}\cdots
D_{i_{k-1}}v^{\gamma})\frac{\partial L}{\partial u_{ii_{1}i_{2}\cdots
i_{k-1}}}.$

Definition 3 A transformation
$x^{*}=x,u^{*}=u+\epsilon\eta(x,u,u_{(1)},\dots,u_{(p)})$ is a variation
symmetry of the action intergral $J[u]$ if for any $u(x)$ there exists some
vector function

$A(x,u,u_{(1)},\dots,u_{(r)})=(A^{1},A^{2},\dots,A^{n}),$

of $x,u$ and its derivatives to some finite order $r$, such that

$U^{k}L=D_{i}A^{i},$ (45)

where $U$ represents the symmetric infinitesimal operator of the target
equation and $L$ represents Euler-Lagrangian equation of the equation, i.e.
$E_{r}L=G$.
Theorem 4 Let $U=\eta^{\gamma}\frac{\partial}{\partial u^{\gamma}}$ be the
infinitesimal generator of transformations
$x^{*}=x,u^{*}=u+\epsilon\eta(x,u,u_{(1)},\dots,u_{(p)})$ and let $U^{(k)}$ be
its $k$th extension. If $U$ is the infinitesimal generator of a variational
symmetry of an action integral so that $U^{(k)}=D_{i}A^{i}$ holds for any
$u(x)$, then the conservation law

$D_{i}(W^{i}[u,\eta]-A^{i})=0,$ (46)

holds for any solution $u(\bm{x})$ of the Euler-Lagrange equations
$E_{\gamma}(L)=0,\gamma=1,2,\dots,m$. Then for the (1+1) dimension dispersive
long-wave equation

$\displaystyle u_{t}+uu_{x}+v_{x}=0,$ (47) $\displaystyle
v_{t}+u_{x}v+v_{x}u+\frac{1}{3}u_{xxx}=0.$

It has no Lagrangian $L$. However if we make the substitution
$q=u_{x},r=v_{x}$ in (47) then the Eq. (47) becomes

$\displaystyle q_{xt}+q_{x}q_{xx}+r_{xx}=0,$ (48) $\displaystyle
r_{xt}+q_{xx}r_{x}+q_{x}r_{xx}+\frac{1}{3}q_{xxxx}=0.$

is the Euler-Lagrange equation for

$L=-q_{x}r_{t}-\frac{r_{x}^{2}}{2}-\frac{q_{x}^{2}r_{x}}{2}+\frac{1}{6}q_{xx}^{2}.$
(49)

Eq. (47) admits an infinite sequence of Lie-Backlund symmetries. The first few
infinitesimal generators are:

$\displaystyle U_{1}=u_{x}\frac{\partial}{\partial
u}+v_{x}\frac{\partial}{\partial v},U_{2}=u_{t}\frac{\partial}{\partial
u}+v_{t}\frac{\partial}{\partial v},U_{3}=(1-tu_{x})\frac{\partial}{\partial
u}-tv_{x}\partial v,$ (50) $\displaystyle
U_{4}=(\frac{u}{2}+\frac{xu_{x}}{2}+tu_{t})\frac{\partial}{\partial
u}+(v+\frac{x}{2}v_{x}+tv_{t})\frac{\partial}{\partial v}.$

Since $q=\int udx,r=\int vdx$, the corresponding infinitesimal generators for
(are obtained by integrating with respect to $x$) the coefficients of the
infinitesimal generators (50) and then replacing $u$ and $v$ by $q_{x}$ and
$r_{x}$:

$\displaystyle V_{1}=q_{x}\frac{\partial}{\partial
q}+r_{x}\frac{\partial}{\partial r},V_{2}=q_{t}\frac{\partial}{\partial
q}+r_{t}\frac{\partial}{\partial r},V_{3}=(x-tq_{x})\frac{\partial}{\partial
q}-tr_{x}\frac{\partial}{\partial r},$ (51) $\displaystyle
V_{4}=(tq_{t}+\frac{x}{2}q_{x})\frac{\partial}{\partial
q}+(\frac{r}{2}+tr_{t}+\frac{x}{2}r_{x})\frac{\partial}{\partial r}.$

Firstly, we calculate $W^{i}[u,\eta]$ by the formula

$V^{(k)}L=E_{\gamma}(L)\eta+D_{i}W^{i}.$ (52)

We use the general formula $V=\eta_{1}\frac{\partial}{\partial
q}+\eta_{2}\frac{\partial}{\partial r}$. Then the Eq. (52) becomes

$\displaystyle D_{i}W^{i}=$ $\displaystyle
U^{(k)}L-E_{\gamma}(L)\eta=-r_{t}D_{x}\eta_{1}-q_{x}D_{t}\eta_{2}-r_{x}D_{x}\eta_{2}-\frac{q_{x}^{2}}{2}D_{x}\eta_{2}-q_{x}r_{x}D_{x}\eta_{1}+\frac{1}{3}q_{xx}D_{x}^{2}\eta_{1}$
(53)
$\displaystyle-\eta_{1}(r_{xt}+q_{xx}r_{x}+q_{x}r_{xx}+\frac{1}{3}q_{xxxx})-\eta_{2}(q_{xt}+q_{x}q_{xx}+r_{xx})$
$\displaystyle=$ $\displaystyle
D_{x}(-\eta_{1}r_{t}-r_{x}\eta_{2}-\frac{q_{x}^{2}\eta_{2}}{2}-q_{x}r_{x}\eta_{1}-\frac{\eta_{1}q_{xxx}}{3}+\frac{q_{xx}}{3}D_{x}\eta_{1})+D_{t}(-q_{x}\eta_{2}).$

Therefore we can obtain
$W^{1}=-\eta_{1}r_{t}-r_{x}\eta_{2}-\frac{q_{x}^{2}\eta_{2}}{2}-q_{x}r_{x}\eta_{1}-\frac{\eta_{1}q_{xxx}}{3}+\frac{q_{xx}}{3}D_{x}\eta_{1},W^{2}=-q_{x}\eta_{2}.$Then
the conservation laws $D_{x}f^{1}+D_{y}f^{2}=0$ where
$f^{1}=W^{1}-A^{1},f^{2}=W^{2}-A^{2}$ are obtained as follows:

(i) From $V_{1}$:$V_{1}L=D_{x}L$. Hence $A^{1}=L,A^{2}=0$;

$\displaystyle
f^{1}_{1}=-\frac{r_{x}^{2}}{2}-q_{x}^{2}r_{x}-\frac{q_{x}q_{xxx}}{3}+\frac{q_{xx}^{2}}{6},$
(54) $\displaystyle f^{2}_{1}=-q_{x}r_{x}.$

We can simply verify it:

$D_{x}f^{1}_{1}+D_{t}f^{2}_{1}=-r_{x}r_{xx}-2q_{x}r_{x}q_{xx}-q_{x}^{2}r_{xx}-\frac{q_{x}q_{xxxx}}{3}-q_{xt}r_{x}-q_{x}r_{xt}=0,$

where $(q,r)$ is a solution of Eq. (48).

(ii) From $V_{2}$:$V_{2}L=D_{t}L$. Hence $A^{1}=0,A^{2}=L$;

$\displaystyle
f^{1}_{2}=-q_{t}r_{t}-r_{t}r_{x}-\frac{q_{x}^{2}r_{t}}{2}-q_{t}q_{x}r_{x}-\frac{q_{t}q_{xxx}}{3}+\frac{q_{xx}q_{xt}}{3},$
(55) $\displaystyle
f^{2}_{2}=\frac{r_{x}^{2}}{2}+\frac{q_{x}^{2}r_{x}}{2}-\frac{q_{xx}^{2}}{6}.$

We can simply verify it:

$D_{x}f^{1}_{2}+D_{t}f^{2}_{2}=-q_{xt}r_{t}-q_{t}r_{xt}-r_{t}r_{xx}-r_{t}q_{x}q_{xx}-q_{t}q_{xx}r_{x}-q_{t}q_{x}r_{xx}-\frac{q_{t}q_{xxxx}}{3}=0,$

where $(q,r)$ is a solution of Eq. (48).

(iii) From
$V_{3}$:$V_{3}L=D_{x}(tq_{x}r_{t}+\frac{t}{2}r_{x}^{2}+\frac{tq_{x}^{2}r_{x}}{2}-\frac{t}{6}q_{xx}^{2})+D_{t}(-r)$.
Hence
$A^{1}=tq_{x}r_{t}+\frac{t}{2}r_{x}^{2}+\frac{tq_{x}^{2}r_{x}}{2}-\frac{t}{6}q_{xx}^{2}$,$A^{2}=-r$;

$\displaystyle
f^{1}_{3}=-xr_{t}+\frac{tr_{x}^{2}}{2}-q_{x}r_{x}x+tq_{x}^{2}r_{x}-\frac{xq_{xxx}}{3}+\frac{tq_{x}q_{xxx}}{3}+\frac{q_{xx}}{3}-\frac{tq_{xx}^{2}}{6},$
(56) $\displaystyle f^{2}_{3}=tq_{x}r_{x}+r.$

We can simply verify it:

$\displaystyle D_{x}f^{1}_{3}+D_{t}f^{2}_{3}=$ $\displaystyle-
xr_{xt}+tq_{x}r_{xt}+tr_{x}r_{xx}+2tq_{x}r_{x}q_{xx}-q_{xx}r_{x}x-q_{x}r_{xx}x+tq_{x}^{2}r_{xx}$
$\displaystyle-\frac{x}{3}q_{xxxx}+\frac{tq_{x}q_{xxxx}}{3}+q_{xt}tr_{x}=0.$

where $(q,r)$ is a solution of Eq. (48).

(iv) From $V_{4}$: After using $V_{4}$ to act on $L$, we find that it can not
be reduced to a conserved form. So $V_{4}$ is not the variation symmetry.

##  5 Solving conservation law by Ibragimov method

In this part we try to solve the equation by Ibragimov method to obtain the
conservation law. First, we introduce a theorem:
Theorem 1. Any Lie point, Lie-Bäcklund, and nonlocal symmetry

$X=\xi^{i}(x,u,u_{(1)},\ldots)\frac{\partial}{\partial
x^{i}}+\eta^{\alpha}(x,u,u_{(1)},\ldots)\frac{\partial}{\partial u^{\alpha}}$

leads to the conservation law $D_{i}(C^{i})=0$, i.e.

$\displaystyle C^{i}=$
$\displaystyle\xi^{i}\mathscr{L}+W^{\alpha}[\frac{\partial\mathscr{L}}{\partial
u^{\alpha}_{i}}-D_{j}(\frac{\partial\mathscr{L}}{\partial
u^{\alpha}_{ij}})+D_{j}D_{j}(\frac{\partial\mathscr{L}}{\partial
u^{\alpha}_{ijk}})-\cdots]$ (57)
$\displaystyle+D_{j}(W^{\alpha})[\frac{\partial\mathscr{L}}{\partial
u^{\alpha}_{ij}}-D_{k}(\frac{\partial\mathscr{L}}{\partial
u^{\alpha}_{ijk}})+\cdots]+D_{j}D_{k}(W^{\alpha})[\frac{\partial\mathscr{L}}{\partial
u^{\alpha}_{ijk}}-\cdots]+\cdots,$

where

$\mathscr{L}=\sum\limits_{i=1}^{m}v^{i}F_{i}(x,u,u_{(1)},\ldots,u_{(s)}),W^{\alpha}=\eta^{\alpha}-\xi^{j}u_{j}^{\alpha},\alpha=1,...,m.$

Then we start with a system of $m$ differential equations with the following
forms

$F_{\alpha}(x,u,u_{(1)},...,u_{(s)})=0,\alpha=1,2,\ldots,m,$ (58)

which admits the adjoint equations

$F_{\alpha}^{*}(x,u^{1},u^{2},u^{1}_{(1)},u^{2}_{(1)},\ldots,u^{1}_{(s)},u^{2}_{(s)})=0,\alpha=1,2,\ldots,m,$
(59)

where

$\displaystyle
F_{\alpha}^{*}(x,u^{1},u^{2},u^{1}_{(1)},u^{2}_{(1)},\ldots,u^{1}_{(s)},u^{2}_{(s)})=\frac{\delta\mathcal{L}}{\delta
u^{\alpha}},$ (60) $\displaystyle
u^{1}_{(1)}={u^{\alpha}_{i}},u^{1}_{(s)}={u^{\alpha}_{i_{1}\cdots
i_{s}}},u^{2}_{(1)}={v^{\alpha}_{i}},u^{2}_{(s)}={v^{\alpha}_{i_{1}\cdots
i_{s}}},$

where $\mathcal{L}$ means the formal Lagrangian for Eq. (98), i.e.

$\mathcal{L}=\sum\limits_{i=1}^{m}v^{i}F_{i}(x,u,u_{(1)},\ldots,u_{(s)}),$
(61)

$\frac{\delta}{\delta u^{\alpha}}$ means the variational derivative with the
following forms

$\frac{\delta}{\delta u^{\alpha}}=\frac{\partial}{\partial
u^{\alpha}}+\sum\limits^{\infty}_{s=1}(-1)^{s}D_{i_{i}}\cdots
D_{i_{s}}\frac{\partial}{\partial u^{\alpha}_{i_{1}\cdots i_{s}}}.$

Definition 2. We say that the differential Eq.(58) is strictly self-adjoint if
the adjoint Eq.(59) becomes equivalent to the original Eq.(60) upon the
substitution

$v^{\alpha}=u^{\alpha},\alpha=1,2,\ldots,m.$

It means that the equation

$F_{\alpha}^{*}(x,u,v,u_{(1)},v_{(1)},\ldots,u_{(s)},v_{(s)})=F_{\alpha}(x,u,v,u_{(1)},v_{(1)},\ldots,u_{(s)},v_{(s)}).$
(62)

Next we begin to study the Eq. (1). We note
$\mathcal{L}=w^{1}(v_{t}+u_{x}v+uv_{x}+\frac{1}{3}u_{xxx})+w^{2}(u_{t}+uu_{x}+v_{x})$,
where $w^{1},w^{2}$ have the expressions $w^{1}(x,t),w^{2}(x,t)$.
Theorem 2. The (1+1)-dimensional dispersive long-wave equation

$\displaystyle v_{t}+u_{x}v+uv_{x}+\frac{1}{3}u_{xxx}=0,$ (63) $\displaystyle
u_{t}+uu_{x}+v_{x}=0.$

is strictly self-adjoint.

Proof: The formal Lagrangian of the (1+1)-dimensional dispersive long-wave
equation reads

$\mathcal{L}=w^{1}(v_{t}+u_{x}v+uv_{x}+\frac{1}{3}u_{xxx})+w^{2}(u_{t}+uu_{x}+v_{x}),$
(64)

Then we can get

$F^{*}=\left(\begin{array}[]{c}\frac{\delta\mathcal{L}}{\delta u}\\\
\frac{\delta\mathcal{L}}{\delta v}\\\
\end{array}\right)=\left(\begin{array}[]{c}-vD_{x}w^{1}-\frac{1}{3}D_{x}^{3}w^{1}-D_{t}w^{2}-uD_{x}w^{2}\\\
-vD_{x}w^{1}-uD_{x}w^{1}-D_{x}w^{2}\\\ \end{array}\right),$ (65)

If we make $w^{1}=u,w^{2}=v$,

$\begin{cases}\frac{\delta\mathcal{L}}{\delta
u}=-v_{t}-vu_{x}-uv_{x}-\frac{1}{3}u_{xxx}\\\ \frac{\delta\mathcal{L}}{\delta
v}=-u_{t}-uu_{x}-v_{x}\end{cases}=-F.$ (66)

Therefore, according to the Definition 2, the Eq. (1) is nonlinear self-
adjoint.
Next, we use Theorem 1 and the symmetry of the Eq. (1) to solve the
corresponding conservation law. It has symmetries:

$\displaystyle X_{1}=\frac{\partial}{\partial
t},X_{2}=\frac{\partial}{\partial x},X_{3}=t\frac{\partial}{\partial
x}+\frac{\partial}{\partial u},$ (67) $\displaystyle
X_{4}=\frac{x}{2}\frac{\partial}{\partial x}+t\frac{\partial}{\partial
t}-\frac{u}{2}\frac{\partial}{\partial u}-v\frac{\partial}{\partial v}.$

According to the formula (57), we have

$W^{1}_{1}=-u_{x},W^{1}_{2}=-v_{x},$

for $X_{1}$. Then we obtain

$\displaystyle C_{x}^{1}=$
$\displaystyle\xi^{x}\mathcal{L}+W^{1}_{1}[\frac{\partial\mathcal{L}}{\partial
u_{x}}+D_{x}^{2}(\frac{\partial\mathcal{L}}{\partial
u_{xxx}})]+W^{1}_{2}\frac{\partial\mathcal{L}}{\partial
v_{x}}+D_{x}W^{1}_{1}(-D_{x})(\frac{\partial\mathcal{L}}{\partial
u_{xxx}})+D_{x}^{2}W^{1}_{1}\frac{\partial\mathcal{L}}{\partial u_{xxx}}$ (68)
$\displaystyle=$ $\displaystyle
w^{1}v_{t}+w^{2}u_{t}-\frac{u_{x}w^{1}_{xx}}{3}+\frac{w^{1}_{x}u_{xx}}{3},$
$\displaystyle C_{t}^{1}=$
$\displaystyle\xi^{t}L+W^{1}\frac{\partial\mathcal{L}}{\partial
u_{t}}+W^{2}\frac{\partial\mathcal{L}}{\partial v_{t}}$ $\displaystyle=$
$\displaystyle-u_{x}w^{2}-v_{x}w^{1}.$

Then we have $W^{2}_{1}=-u_{t},W^{2}_{2}=-v_{t}$ for $X_{2}$, and the
corresponding conservation law is

$\displaystyle
C_{x}^{2}=-u_{t}vw^{1}-uu_{t}w^{2}-\frac{u_{t}w^{1}_{xx}}{3}-uv_{t}w^{1}-v_{t}w^{2}+\frac{u_{xt}w^{1}_{x}}{3}-\frac{u_{xxt}w^{1}}{3},$
(69) $\displaystyle
C_{t}^{2}=w^{1}u_{x}v+w^{1}v_{x}u+\frac{w^{1}u_{xxx}}{3}+uw^{2}u_{x}+w^{2}v_{x}.$

For $X_{3}$, we get $W^{3}_{1}=1-tu_{x},W^{3}_{2}=-tv_{x}$ and the
conservation law is

$\displaystyle
C_{x}^{3}=tw^{1}v_{t}+tw^{2}u_{t}+vw^{1}+uw^{2}+\frac{w^{1}_{xx}}{3}-\frac{tu_{x}w^{1}_{xx}}{3}+\frac{tu_{xx}w^{1}_{x}}{3},$
(70) $\displaystyle C_{t}^{3}=w^{2}-w^{2}tu_{x}-tv_{x}w^{1}.$

For $X_{4}$, we have
$W^{4}_{1}=-\frac{u}{2}-\frac{xu_{x}}{2}-tu_{t},W^{4}_{2}=-v-\frac{xv_{x}}{2}-tv_{t}$.
Then the corresponding conservation law is

$\displaystyle C_{x}^{4}=$ $\displaystyle-
tvu_{t}w^{1}-tu_{t}uw^{2}-tv_{t}uw^{1}-tv_{t}w^{2}+\frac{w^{1}_{x}xu_{xx}}{6}+\frac{w^{1}_{x}tu_{xt}}{3}-\frac{w^{1}tu_{xxt}}{3}+\frac{xw^{1}v_{t}}{2}+\frac{xw^{2}u_{t}}{2}-\frac{3uvw^{1}}{2}$
(71)
$\displaystyle-\frac{xu_{x}w^{1}_{xx}}{6}-\frac{tu_{t}w^{1}_{xx}}{3}-\frac{w^{2}u^{2}}{2}-\frac{uw^{1}_{xx}}{6}-vw^{2}+\frac{w^{1}_{x}u_{x}}{3}-\frac{w^{1}u_{xx}}{2},$
$\displaystyle C_{t}^{4}=$ $\displaystyle
tw^{1}u_{x}v+tw^{1}v_{x}u+\frac{tw^{1}}{3}u_{xxx}+tw^{2}uu_{x}+tw^{2}v_{x}-\frac{uw^{2}}{2}-\frac{w^{2}xu_{x}}{2}-vw^{1}-\frac{w^{1}xv_{x}}{2}.$

##  6 Hamiltonian Structure and Line Soliton Solution

For the (1+1) dimensional dispersive long-wave equation equation

$\displaystyle u_{t}+uu_{x}+v_{x}=0,$ (72) $\displaystyle
v_{t}+u_{x}v+uv_{x}+\frac{1}{3}u_{xxx}=0.$

We notice that it has a Hamiltonian formulation
$u_{t}=-\mathcal{D}(\frac{\delta H}{\delta u})$

$\left(\begin{array}[]{c}u\\\ v\\\
\end{array}\right)_{t}=-{\mathcal{D}}\left(\begin{array}[]{c}\frac{\delta
H}{\delta u}\\\ \frac{\delta H}{\delta v}\\\
\end{array}\right)=-\mathcal{D}\left(\begin{array}[]{c}uv+\frac{u_{xx}}{3}\\\
v+\frac{u^{2}}{2}\\\ \end{array}\right),$ (73)

where $H=\int\frac{v^{2}}{2}+\frac{u^{2}v}{2}-\frac{u_{x}^{2}}{6}dx$ is the
Hamiltonian functional, and $\mathcal{D}=\left(\begin{array}[]{cc}0&D_{x}\\\
D_{x}&0\end{array}\right)$ is a Hamiltonian operator. Then since $\mathcal{D}$
is a Hamiltonian operator, it can map adjoint-symmetries into symmetries, so
$\mathcal{J}=\left(\begin{array}[]{cc}0&D_{x}^{-1}\\\
D_{x}^{-1}&0\end{array}\right)$ can map symmetries into adjoint-symmetries.
And we can use the above symmetry to get the adjoint symmetry of some
objective equations. Applying this latter operator to the scaling symmetry, we
obtain the adjoint-symmetry:

$\displaystyle Q_{1}=\mathcal{J}\left(\begin{array}[]{c}-u_{t}\\\
-v_{t}\end{array}\right)=\left(\begin{array}[]{c}-r_{t}\\\
-q_{t}\end{array}\right),Q_{2}=\mathcal{J}\left(\begin{array}[]{c}-u_{x}\\\
-v_{x}\end{array}\right)=\left(\begin{array}[]{c}v\\\ u\end{array}\right),$
(74) $\displaystyle Q_{3}=\mathcal{J}\left(\begin{array}[]{c}1-tu_{x}\\\
-tv_{x}\end{array}\right)=\left(\begin{array}[]{c}-tv\\\
x-tu\end{array}\right),Q_{4}=\mathcal{J}\left(\begin{array}[]{c}-\frac{u}{2}-\frac{xu_{x}}{2}-tu_{t}\\\
-v-\frac{xv_{x}}{2}-tv_{t}\end{array}\right)=\left(\begin{array}[]{c}-r-tr_{t}-\frac{xv}{2}+r\\\
-\frac{q}{2}-\frac{xu}{2}+q-tq_{t}\end{array}\right),$

where $u=q_{x},v=r_{x}$.
Next, we try to solve the soliton solution by using the conservation law
constructed by the adjoint symmetry. A line soliton is a solitary traveling
wave $u=U(x-\mu t),v=V(x-\mu t)$ in one dimension where the parameter $\mu$
means the speed of the wave. Then we study the conservation laws of the ZK
equation $\phi^{t},\ \phi^{x}$ which doesn’t contain the variables $t,x.$ Then
the conservation law is obtained by reduction

$D_{t}|_{u=U(\xi)}=-\mu\frac{d}{d\xi},D_{x}|_{u=U(\xi)}=\frac{d}{d\xi},\xi=x-\mu
t,$ (75)

yielding

$\frac{d}{d\xi}(\phi^{x}-\mu\phi^{t})=0.$ (76)

So $(\phi^{x}-\mu\phi^{t})=C.$ Then we begin with the equation (1). Using the
transformation $u(x,t)=U(\xi),v(x,t)=V(\xi),$ we can obtain the nonlinear
ordinary differential equation:

$\displaystyle-\mu U^{{}^{\prime}}+UU^{{}^{\prime}}+V^{{}^{\prime}}=0,$ (77)
$\displaystyle-\mu
V^{{}^{\prime}}+UV^{{}^{\prime}}+VU^{{}^{\prime}}+\frac{U^{{}^{\prime\prime\prime}}}{3}=0.$

for $U(\xi),V(\xi)$. Conservation laws (29), (31), (32) and (33) do not
contain the variables $t,x$. When the first integral formula
$(\phi^{x}-\mu\phi^{t})=C$ is applied to these three conservation laws, we
obtain

$\displaystyle C_{1}=$
$\displaystyle\frac{27U^{2}V^{2}}{8}+\frac{(V^{{}^{\prime}})^{2}}{2}+\frac{3V^{3}}{4}+\frac{(U^{{}^{\prime\prime}})^{2}}{6}+\mu
UV^{{}^{\prime\prime}}+\mu
VU^{{}^{\prime\prime}}+\frac{3U^{4}V}{4}+\frac{3U^{2}(U^{{}^{\prime}})^{2}}{8}+\frac{U^{3}U^{{}^{\prime\prime}}}{4}$
$\displaystyle+\frac{3UVU^{{}^{\prime\prime}}}{2}+UU^{{}^{\prime}}V^{{}^{\prime}}-\frac{V(U^{{}^{\prime}})^{2})}{4}+\frac{3\mu
U^{2}U^{{}^{\prime\prime}}}{4}-\mu(U^{{}^{\prime}}V^{{}^{\prime}}+VU^{{}^{\prime\prime}}+V^{{}^{\prime\prime}}U+\frac{3U^{3}V}{4}+\frac{9V^{2}U}{4}$
$\displaystyle+\frac{3U(U^{{}^{\prime}})^{2}}{4}+\frac{3U^{2}U^{{}^{\prime\prime}}}{4})$
(78) $\displaystyle=$
$\displaystyle\frac{3U^{4}V}{4}-\frac{3U^{3}V\mu}{4}+\frac{U^{3}U^{{}^{\prime\prime}}}{4}+\frac{27U^{2}V^{2}}{8}+\frac{3U^{2}(U^{{}^{\prime}})^{2}}{8}-\frac{9UV^{2}\mu}{4}-\frac{3U(U^{{}^{\prime}})^{2}\mu)}{4}+\frac{3UVU^{{}^{\prime\prime}}}{2}$
$\displaystyle+UU^{{}^{\prime}}V^{{}^{\prime}}+\frac{3V^{3}}{4}-\frac{V(U^{{}^{\prime}})^{2}}{4}-U^{{}^{\prime}}V^{{}^{\prime}}\mu+\frac{(U^{{}^{\prime\prime}})^{2}}{6}+\frac{(V^{{}^{\prime}})^{2}}{2}=0,$
$\displaystyle C_{2}=$
$\displaystyle\frac{U^{3}V}{2}-\frac{U^{2}V\mu}{2}+\frac{U^{2}U^{{}^{\prime\prime}}}{6}+V^{2}U-\frac{V^{2}\mu}{2}-\frac{\mu(U^{{}^{\prime}})^{2}}{6}+\frac{VU^{{}^{\prime\prime}}}{3}=0,$
(79) $\displaystyle C_{3}=$
$\displaystyle\frac{V^{2}}{2}+U^{2}V+\frac{UU^{{}^{\prime\prime}}}{3}-\frac{(U^{{}^{\prime}})^{2}}{6}-\mu
UV=0,$ (80) $\displaystyle C_{4}=$ $\displaystyle
UV+\frac{U^{{}^{\prime\prime}}}{3}+\frac{U^{2}}{2}+V-\mu U-\mu V=0.$ (81)

We impose the asymptotic conditions
$U,U^{{}^{\prime}},U^{{}^{\prime\prime}},U^{{}^{\prime\prime\prime}}\rightarrow
0$ as $|\xi|\rightarrow\infty.$ Then we combine the formulas (78), (79), (80)
and (81), then we can calculate its general line soliton solutions:

$\displaystyle U_{1}=$
$\displaystyle\frac{2e^{C_{1}\mu\sqrt{3}}\mu}{e^{\mu\xi\sqrt{3}}(-1+\frac{e^{C_{1}\mu\sqrt{3}}}{e^{\mu\xi\sqrt{3}}})},V_{1}=-\frac{2e^{\mu\xi\sqrt{3}}\mu^{2}e^{C_{1}\mu\sqrt{3}}}{(e^{C_{1}\mu\sqrt{3}}-e^{\mu\xi\sqrt{3}})^{2}},$
(82) $\displaystyle U_{2}=$
$\displaystyle\frac{2e^{\xi\mu\sqrt{3}}\mu}{e^{C_{1}\mu\sqrt{3}}(-1+\frac{e^{\xi\mu\sqrt{3}}}{e^{\mu
C_{1}\sqrt{3}}})},V_{2}=-\frac{2e^{\mu\xi\sqrt{3}}\mu^{2}e^{C_{1}\mu\sqrt{3}}}{(e^{C_{1}\mu\sqrt{3}}-e^{\mu\xi\sqrt{3}})^{2}}.$
(83)

where $C_{1}$ is an arbitrary constant.

Figure 1: 3D plots and density plots of the $u(x,y,t)$ given by Eq. (138) for
parameters $a_{1}=1,t=1$ and $a_{1}=-1,t=1$.

Figure 2: 3D plots and density plots of the $u(x,y,t)$ given by Eq. (138) for
parameters $a_{1}=1,t=1$ and $a_{1}=-1,t=1$.

##  7 Exact solutions of (1+1) dimensional dispersive long-wave equation

###  7.1 The improved extended Kudryashov method

In this section we will try to use the extended Kudrtashov method to find some
general solutions of the Eq. (1). To obtain the exact solutions of (1+1)
dimensional dispersive long-wave equation, we set

$\displaystyle u(x,t)=$ $\displaystyle
A_{0}+{\sum\limits_{K=1}\limits^{M}}{\sum\limits_{i+j=K}}A_{ij}{\phi}^{i}(\xi){\psi}^{j}(\eta)+{\sum\limits_{K=1}\limits^{M}}B_{K}{\psi}^{-K}(\eta)+{\sum\limits_{j=1}\limits^{M}}C_{j}{\psi}^{-j}(\eta),$
(84) $\displaystyle v(x,t)=$ $\displaystyle
a_{0}+{\sum\limits_{K=1}\limits^{N}}{\sum\limits_{i+j=K}}a_{ij}{\phi}^{i}(\xi){\psi}^{j}(\eta)+{\sum\limits_{K=1}\limits^{N}}b_{K}{\psi}^{-K}(\eta)+{\sum\limits_{j=1}\limits^{N}}c_{j}{\psi}^{-j}(\eta).$

Firstly, we need to determine the values of $M,N$ by homogeneous balance
method. We can obtain $M=1,N=2$ by caluation. Therefore, we set the following
general solution

$\displaystyle
u(x,t)=A_{0}+A_{10}{\phi}(\xi)+A_{01}{\psi}(\eta)+\frac{B_{1}}{{\psi}(\eta)}+\frac{C_{1}}{\phi(\xi)},$
(85) $\displaystyle
v(x,t)=a_{0}+a_{10}{\phi}(\xi)+a_{01}{\psi}(\eta)+a_{11}+\frac{b_{1}}{{\psi}(\eta)}+\frac{c_{1}}{\phi(\xi)},$

where
$\xi=\kappa_{1}x+\omega_{1}t+r_{1}y,\eta=\kappa_{2}x+\omega_{2}t+r_{2}y$, and
$\kappa_{1}$,$\kappa_{2}$,$\omega_{1},$$\omega_{2}$,$r_{1}$,$r_{2}$,$A_{0}$$,A_{10}$$,A_{01}$$,B_{1}$,$C_{1}$
are underdetermined constants and the functions ${\phi(\xi)}$ and
${\psi(\eta)}$ satisfy the Bernoulli and the Riccati equations. Substitute
(85) into (1), then we will obtain a system of algebraic equations by solving
these algebraic equations by Maple. We will get the following traveling wave
solution of Eq. (1), then we can get the exact solution expression of $u$ and
$v$ by solving these equations together. Some of the results are shown as
follows:

$\displaystyle u_{1}=$ $\displaystyle
A_{0}+C_{1}\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}},$ (86) $\displaystyle
v_{1}=$ $\displaystyle
a_{0}-\frac{C_{1}(A_{0}\kappa_{1}+\omega_{1})}{\kappa_{1}}\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}}+c_{2}(\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}})^{2},$

where $u=\kappa_{1}x+\omega_{1}t,v=\kappa_{2}x+\omega_{2}t$.

$\displaystyle u_{2}=$ $\displaystyle A_{0},$ (87) $\displaystyle v_{2}=$
$\displaystyle
a_{0}+b_{1}(\frac{R_{2}c_{1}\kappa_{1}}{b_{1}\kappa_{2}}\eta+\eta_{0})+c_{1}\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}}+c_{2}(\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}})^{2},$

where $u=\kappa_{1}x+\omega_{1}t,v=\kappa_{2}x+\omega_{2}t$.

$\displaystyle u_{3}=$ $\displaystyle
A_{0}+C_{1}\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}},$ (88) $\displaystyle
V_{3}=$ $\displaystyle
a_{0}+b_{1}(\frac{2S_{2}}{\frac{C_{1}R_{2}\kappa_{1}}{\omega_{2}}-\sqrt{\mu}\tanh(\frac{\sqrt{\mu}}{2}\eta+\eta_{0})})-\frac{C_{1}(A_{0}\kappa_{1}+\omega_{1})}{\kappa_{1}}\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}}+c_{2}(\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}})^{2},$

where $u=\kappa_{1}x+\omega_{1}t,v=\kappa_{2}x+\omega_{2}t$.

$\displaystyle u_{4}=$ $\displaystyle
A_{0}+C_{1}\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}},$ (89) $\displaystyle
v_{4}=$ $\displaystyle
a_{0}+b_{1}(\frac{R_{2}C_{1}\omega_{1}}{b_{1}\kappa_{2}}\eta+\eta_{0})+c_{1}\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}}+c_{2}(\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}})^{2},$

where $u=\kappa_{1}x+\omega_{1}t,v=\kappa_{2}x+\omega_{2}t$.

$\displaystyle u_{5}=$ $\displaystyle A_{0},$ (90) $\displaystyle v_{5}=$
$\displaystyle
a_{0}-a_{01}\frac{1}{\frac{R_{2}c_{1}\kappa_{1}}{a_{01}\kappa_{2}}\eta+\eta_{0}}+c_{1}\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}}+c_{2}(\frac{R_{2}+R_{1}e^{R_{1}\xi+\xi_{0}}}{R_{1}})^{2}.$

where $u=\kappa_{1}x+\omega_{1}t,v=\kappa_{2}x+\omega_{2}t$.

###  7.2 Hyperbolic tangent method

In this section we shall seek a solution to Eq. (1) in terms of hyperbolic
tangent functions of the following form

$\displaystyle u(\xi)=$ $\displaystyle a_{0}+a_{1}\tanh\xi,$ (91)
$\displaystyle v(\xi)=$ $\displaystyle
b_{0}+b_{1}\tanh\xi+b_{2}(\tanh\xi)^{2}$

where $\xi=x-\mu t$ and $a_{0},a_{1},b_{0},b_{1},b_{2},\mu$ are constants to
be determined. Bring (91) into Eq. (1) and set the coefficients of the same
power of $\tanh\xi$ equal to zero the following algebraic equations are
obtained

$\displaystyle-\frac{64}{3}a_{1}-8\mu
b_{1}+8a_{1}b_{0}+8b_{1}a_{0}-24a_{1}b_{2}=0,$ (92) $\displaystyle-4\mu
b_{1}-8\mu
b_{2}+4a_{1}b_{0}+8a_{1}b_{1}+12a_{1}b_{2}+4a_{0}b_{1}+8b_{2}a_{0}+\frac{16}{3}a_{1}=0,$
$\displaystyle-4\mu b_{1}+8\mu
b_{2}+4a_{1}b_{0}-8a_{1}b_{1}+12a_{1}b_{2}+4a_{0}b_{1}-8b_{2}a_{0}+\frac{16}{3}a_{1}=0,$
$\displaystyle 4a_{0}a_{1}-4\mu a_{1}+4b_{1}+4a_{1}^{2}+8b_{2}+4a_{1}^{2}=0,$
$\displaystyle 4a_{0}a_{1}-4a_{1}^{2}+4b_{1}-4\mu a_{1}-8b_{2}=0.$

By solving the above system, we get exact solutions of Eq. (1)

$\displaystyle u_{1}=\mu-\frac{2\sqrt{3}\tanh(\mu t-x)}{3},$ (93)
$\displaystyle v_{1}=\frac{2}{3}-\frac{2(\tanh(\mu t-x))^{2}}{3}.$

Figure 3: 3D plots and density plots of the $u(x,y,t)$ given by Eq. (138) for
parameters $a_{1}=1,t=1$ and $a_{1}=-1,t=1$.

Figure 4: 3D plots and density plots of the $u(x,y,t)$ given by Eq. (138) for
parameters $a_{1}=1,t=1$ and $a_{1}=-1,t=1$.

###  7.3 Rational functions in $\exp(\xi)$

In this section we shall seek a rational function type of solution to Eq. (1),
in terms of $\exp(\xi)$ and we set the following form:

$\displaystyle u(\xi)=$ $\displaystyle a_{0}+\frac{a_{1}}{1+e^{\xi}},$ (94)
$\displaystyle v(\xi)=$ $\displaystyle
b_{0}+\frac{b_{1}}{1+e^{\xi}}+b_{2}(\frac{1}{1+e^{\xi}})^{2},$

where $\xi=x-\mu t$ and $a_{0},a_{1},b_{0},b_{1},b_{2},\omega$ are constants
to be determined. Bring (94) into Eq. (1) and set the coefficients of the same
power of $e^{\xi}$ equal to zero the following algebraic equations are
obtained

$\displaystyle-a_{1}a_{0}-a_{1}^{2}+\mu a_{1}-b_{1}-2b_{2}=0,$ (95)
$\displaystyle-a_{1}a_{0}+\mu a_{1}-b_{1}=0,$ $\displaystyle\mu b_{1}+2\mu
b_{2}-a_{1}b_{0}-2a_{1}b_{1}-3a_{1}b_{2}-b_{1}a_{0}-2b_{2}a_{0}-\frac{1}{3}a_{1}=0,$
$\displaystyle 2\mu b_{1}+2\mu
b_{2}-2a_{1}b_{0}-2a_{1}b_{1}-2b_{1}a_{0}-2b_{2}a_{0}+\frac{4}{3}a_{1}=0,$
$\displaystyle\mu b_{1}-a_{1}b_{0}-b_{1}a_{0}-\frac{1}{3}a_{1}=0.$

Then the solution may be given as follows

$\displaystyle
u_{1}=a_{0}+\frac{2\sqrt{3}}{3(1+e^{x-(a_{0}+\frac{\sqrt{3}}{3})t})},$ (96)
$\displaystyle
v_{1}=\frac{2}{3(1+e^{x-(a_{0}+\frac{\sqrt{3}}{3})t})}-\frac{2}{3(1+e^{x-(a_{0}+\frac{\sqrt{3}}{3})t})^{2}}.$

##  8 Conclusion

In this paper, we first analyze the Lie symmetry of the (1+1) dimensional
dispersive long-wave equation, and obtain some Lie point symmetries, invariant
solutions and a reduced equation. At the same time, three different methods
are used to solve the conservation laws of the ZK equation. The first is to
use the direct construction method to solve the multiplier of the target
equation. The adjoint symmetry of the equation is used to substitute the
symmetry in a partial differential equation that lacks variational symmetry.
At this time, the adjoint symmetry satisfies the linear adjoint symmetry
determining equations. The symmetry invariant condition is replaced by the
adjoint symmetry invariant condition, and an adjoint symmetry formula is
presented. The second method is the adjoint equation method. The adjoint
equation method proposed by Ibragimov can use the symmetry of the equation to
calculate the conservation law through the explicit formula. It is convenient
to calculate and does not require complex analysis. The third method is to use
Noether’s theorem. The conservation law is solved by the variational symmetry
of the target equation.

In fact, the three methods used in this paper to calculate the conservation
law of equations have different advantages. The adjoint equation method
proposed by Ibragimov and Noether’s theorem can use the symmetry of the
equation to calculate the conservation law through the explicit formula. The
former is convenient to calculate and does not require complex analysis. It
has a wide range of applications, but the results are directly affected by the
symmetry of the equation. And the limitation of using the Noether’s theorem to
solve conservation laws is that sometimes the Euler equation of the target
equation cannot be obviously obtained. The advantage of constructing
conservation laws directly is that it is not necessary to use the variational
symmetry of the equation. For a partial differential equation without
variational symmetry, the adjoint symmetry of the equation is used to replace
the symmetry. At this time, the adjoint symmetry satisfies the linear adjoint
symmetry determining equations. The symmetry invariant condition is replaced
by the adjoint symmetry invariant condition, and a formula using adjoint
symmetry is given. However, this method is computationally complex and does
not apply to any type of equation or equation system. The three methods can be
naturally applied to higher dimensional differential equations and
differential equation systems.

At the same time, we use the symmetry and adjoint symmetry of the obtained
target equation to calculate a Lie bracket on the set of adjoint-symmetries
given by the range of a symmetry action. Then we use the improved Extended
Kudryashov method to solve some exact solutions of (1+1) dimensional
dispersive long-wave equation and seek a few solutions in terms of hyperbolic
tangent functions and $\exp(\xi)$. In the last part, we analyze the
Hamiltonian structure of the target equation, calculate its generalized pre-
symplectic operator and solve the linear soliton solution of the target
equation.

## Acknowledgments

This work is supported by the National Natural Science Foundation of China
(Grant No. 12371256 & No. 11971475).

## Conflict of interest

The authors declare that they have no known competing financial interests.

## Data availability

No data was used for the research described in the article.

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