# Lie Symmetry Analysis and Some New Exact Solutions to the KP-BBM Equation

Arindam<EMAIL_ADDRESS>Sarit
<EMAIL_ADDRESS>

Dept. of Maths.
National Institute of Technology Durgapur, India

###### Abstract

This paper is aimed to study the KP-BBM equation, which was proposed by Abdul
Majid Wazwaz [Wazwaz. A.M.: Applied Mathematics and Computation, 169 (2005),
700–712.]. To check its integrability Painlevé test has been performed. Lie
Symmetry analysis has been done and point symmetry generators are obtained.
The invariants of the Lie algebra are found and the one dimensional optimal
system for subalgebras of the obtained Lie algebra is constructed by using the
Hu-Li-Chen algorithm. Three similarity reductions and corresponding exact
solutions are derived. Also Homogeneous balance method, $Tanh$ method are used
to find exact solutions. Solitary wave like solutions are obtained and plotted
for some suitable values of the parameters involved. The effects of the
nonlinear coefficient and dispersion coefficient on the obtained solitary
waves are discussed.

Keywords: KP-BBM equation, Painlevé test, Lie Symmetry, Exact solutions,
Solitary wave solution.
MSC 2020: 35B06, 35C08.

##  1 Introduction:

Nonlinear differential equations (NDE) have become one of the most essential
tools to study many problems of different branches of physics and mathematics
[1, 2]. It is used to formulate different models of astrophysics, cosmology,
hydrodynamics, plasma physics, nonlinear optics, mathematical biology [2, 3].
Integrability provides significant information for physical phenomena
described by NDEs [4]. A robust and profound tool for checking the
integrability of a differential equation is the Painlevé test [4, 5]. Though
it was introduced by Painlevé and his followers on ordinary differential
equations, the concept was extended in the context of PDE by Weiss, Tabor and
Carnevale [6]. The importance of integrability, Painlevé property in the study
of differential equations are well discussed in [7]. On the other hand exact
solutions of NDEs provides a lot of quantitative information to these
problems. In last few decades mathematicians proposed numerous techniques in
this direction. Among them homogeneous balance method [8, 9], Lie symmetry
method [10, 11], $Tanh$ method [12], sub ODE method [13], Hirota’s bilinear
method [14], First integral method [15] are widely used for finding exact
solutions. Homogeneous balance (HB) method is introduced by Mingling Wang
[16]. Wang et al. obtained exact solutions of some well known nonlinear
partial differential equations by using HB method in [8]. An extended version
of this method is reported by Maitra et al. while obtaining a system of
solutions of the SIDV equation [9]. The $Tanh$ method, first introduced by
Malfliet and Hereman [12], is widely applied to find $Sech^{2}$ solutions of
NDEs. The First integral method, another useful method, first introduced by
Zhaosheng Feng [17] based on theory of commutative algebra. A vivid and
rigorous description with application of this method is given in [15] by Ghosh
et al.

Lie group of symmetry method, developed by Sophus Lie, is very useful and one
of the classic techniques of solving differential equations [10].
Corresponding to each subgroup of the Lie group, the differential equation can
be reduced to a differential equation with less independent variables which
possesses group invariant solutions. As almost always there exists infinite
number of such subgroups [18], it is not possible to list all such group
invariant solutions. For this reason the necessity of finding group invariant
solutions that are inequivalent to each other came into account which leads to
the concept of optimal system of subalgebras [18]. In application, to each
member of the optimal system group invariant solutions are found. By using the
invariants of the Lie algebra, Hu-Li-Chen proposed an effective algorithm for
finding optimal system of subalgebras [19]. Beside finding exact solutions,
conservation laws and many other important properties of differential
equations can be constructed by using the knowledge of symmetry [11].

In this work we are going to study the KP-BBM (Kadomtsev-Petviashvili-
Benjamin-Bona-Mahony) equation, in the context of Lie symmetry analysis, exact
solutions, Painlevé test:

$\displaystyle(u_{t}+u_{x}+a(u^{2})_{x}+bu_{xxt})_{x}+ku_{yy}=0$ (1)
$\displaystyle\Rightarrow$ $\displaystyle
u_{xt}+u_{xx}+2a(u_{x})^{2}+2auu_{xx}+bu_{xxxt}+ku_{yy}=0,$ (2)

which was introduced by Abdul Majid Wazwaz [20], where $a,b$ and $k$ are the
coefficients of nonlinear, dispersion and dissipation terms respectively.
Wazwaz derived this equation from the BBM (Benjamin-Bona-Mahony) equation
[20], which is used as a model for propagation of long waves. He found
compactons, solitons and periodic solutions. Zhou et al. found some exact
travelling wave solutions by generalized sub ODE method in [13]. Using Hirota
bilinear method Manafian et al. have obtained periodic wave solutions [21].

The contents of this article is arranged as follows: in Section 2 and 3
Painlevé test and Lie symmetry analysis have been performed on the KP-BBM
equation respectively. Exact solutions are obtained in Section 4 by
homogeneous balance method, in Section 5 by $Tanh$ method. Ultimately the
outcomes of this work is discussed in Section 7.

##  2 Painlevé Test:

A partial differential equation is said to have Painlevé property if its
solution is single valued about its movable singularity manifold [6]. So, if
$\phi(x,y,t)=0$ is the singularity manifold of (1), ($\phi(x,y,t)$ is analytic
about the singularity manifold) and $u=u(x,y,t)$ is its solution then we
assume that

$\displaystyle u=u(x,y,t)=\phi^{\alpha}\Sigma_{j=0}^{\infty}u_{j}\phi^{j},$
(3)

where $u_{j}=u_{j}(x,y,t)$ and $\phi=\phi(x,y,t)$ are analytic near the
singularity manifold $M=\left\\{(x,y,t):\phi(x,y,t)=0\right\\}.$

By leading order analysis we get $\alpha=-2$. Therefore,

$\displaystyle u=\Sigma_{j=0}^{\infty}u_{j}\phi^{j-2}.$ (4)

Now for the recursion relation we collect the coefficients of $\phi^{j-6}$
from (2) after substituting (4). For $j=0$ we find that

$\displaystyle u_{0}=-\frac{6b}{a}\phi_{x}\phi_{t}$ (5)

For $j\neq 0$ we obtain the recursion relation as

$\displaystyle(j+1)(j-4)(j-5)(j-6)bu_{j}\phi_{x}^{3}\phi_{t}=-[u_{j-4,xt}+(j-5)u_{j-3,x}\phi_{t}+(j-5)u_{j-3,t}\phi_{x}$
$\displaystyle+(j-4)(j-5)u_{j-2}\phi_{x}\phi_{t}+(j-5)u_{j-3}\phi_{xt}+u_{j-4,xx}+2(j-5)u_{j-3,x}\phi_{x}+(j-4)(j-5)u_{j-2}\phi_{x}^{2}$
$\displaystyle+(j-5)u_{j-3}\phi_{xx}+b\\{u_{j-4,xxxt}+(j-5)u_{j-3,xxx}\phi_{t}+3(j-5)u_{j-3,xxt}\phi_{x}+3(j-4)(j-5)u_{j-2,xx}\phi_{x}\phi_{t}$
$\displaystyle+3(j-5)u_{j-3,xx}\phi_{xt}+3(j-4)(j-5)u_{j-2,xt}\phi_{x}^{2}+3(j-3)(j-4)(j-5)u_{j-1,x}\phi_{x}^{2}\phi_{t}$
$\displaystyle+6(j-4)(j-5)u_{j-2,x}\phi_{x}\phi_{xt}+2(j-5)u_{j-3,xt}\phi_{xx}+3(j-4)(j-5)u_{j-2,x}\phi_{xx}\phi_{t}+3(j-5)u_{j-3,x}\phi_{xxt}$
$\displaystyle+(j-3)(j-4)(j-5)u_{j-1,t}\phi_{x}^{3}+3(j-3)(j-4)(j-5)u_{j-1}\phi_{x}^{2}\phi_{xt}+3(j-4)(j-5)u_{j-2,t}\phi_{x}\phi_{xx}$
$\displaystyle+3(j-3)(j-4)(j-5)u_{j-1}\phi_{x}\phi_{xx}\phi_{t}+3(j-4)(j-5)u_{j-2}\phi_{xt}\phi_{xx}+3(j-4)(j-5)u_{j-2}\phi_{x}\phi_{xxt}$
$\displaystyle+(j-5)u_{j-3,xt}\phi_{xx}+(j-5)u_{j-3,t}\phi_{xxx}+(j-4)(j-5)u_{j-2}\phi_{xxx}\phi_{t}+(j-5)u_{j-3}\phi_{xxxt}\\}+k\\{u_{j-4,yy}$
$\displaystyle+2(j-5)u_{j-3,y}\phi_{y}+(j-4)(j-5)u_{j-2}\phi_{y}^{2}+(j-5)u_{j-3}\phi_{yy}\\}$
$\displaystyle+2a\Sigma_{r=1}^{j-1}\\{u_{j-r,x}u_{r-2,x}+(j-r-2)u_{j-r}u_{r-1,x}\phi_{x}+(r-3)u_{j-r,x}u_{r-1}\phi_{x}+(j-r-2)(r-2)u_{j-r}u_{r}\phi_{x}^{2}$
$\displaystyle+u_{j-r}u_{r-2,xx}+2(r-3)u_{j-r}u_{r-1,x}\phi_{x}+(r-2)(r-3)u_{j-r}u_{r}\phi_{x}^{2}+(r-3)u_{j-r}u_{r-1}\phi_{xx}\\}$
$\displaystyle+2a\\{u_{0,x}u_{j-2,x}-2u_{0}u_{j-1,x}\phi_{x}+(j-3)u_{0,x}u_{j-1}\phi_{x}+u_{0}u_{j-2,xx}+2(j-3)u_{0}u_{j-1,x}\phi_{x}$
$\displaystyle+(j-3)u_{0}u_{j-1}\phi_{xx}\\}]$ (6)

From the recursion relation (6) we find the resonances as $j=-1,4,5,6.$ the
resonance at $j=-1$ indicates the arbitrariness of the singularity manifold.
Resonances at $j=4,5,6$ indicates that $u_{4},u_{5},u_{6}$ must be arbitrary
and the right hand side of (6) must have to vanish identically at $j=4,5,6.$
But by using the software Mathematica we found non zero right hand side for
$j=4,5,6$. i.e the compatibility conditions for Painlevé test for these
resonances are not satisfied. Thus the KP-BBM equation fails to pass the
Painlevé test and hence this equation does not possess the Painlevé property.

##  3 Lie Point Symmetry Analysis:

Let us define the following point transformations:
$(x,y,t,u)\rightarrow(\hat{x},\hat{y},\hat{t},\hat{u})$ where

$\displaystyle\hat{x}=\hat{x}(x,y,t,u(x,y,t);\epsilon),\hat{y}=\hat{y}(x,y,t,u(x,y,t);\epsilon),\hat{t}=\hat{t}(x,y,t,u(x,y,t);\epsilon),$
$\displaystyle\hat{u}=\hat{u}(x,y,t,u(x,y,t);\epsilon).$ (7)

Tangent vectors $(\xi,\gamma,\tau,\eta)$ at
$(\hat{x},\hat{y},\hat{t},\hat{u})$ is defined by

$\displaystyle\frac{d\hat{x}}{d\epsilon}=\xi(\hat{x},\hat{y},\hat{t},\hat{u}),\frac{d\hat{y}}{d\epsilon}=\gamma(\hat{x},\hat{y},\hat{t},\hat{u}),\frac{d\hat{t}}{d\epsilon}=\tau(\hat{x},\hat{y},\hat{t},\hat{u}),\frac{d\hat{u}}{d\epsilon}=\eta(\hat{x},\hat{y},\hat{t},\hat{u})$
(8)

satisfying the conditions

$\displaystyle(\hat{x},\hat{y},\hat{t},\hat{u})|_{\epsilon=0}=(x,y,t,u)$ (9)

Therefore, the Taylor’s series of the Lie group action are

$\displaystyle\hat{x}=x+\epsilon\xi(x,y,t,u)+o(\epsilon^{2})$
$\displaystyle\hat{y}=y+\epsilon\gamma(x,y,t,u)+o(\epsilon^{2})$
$\displaystyle\hat{t}=t+\epsilon\tau(x,y,t,u)+o(\epsilon^{2})$
$\displaystyle\hat{u}=u+\epsilon\eta(x,y,t,u)+o(\epsilon^{2})$ (10)

The differential coefficients become:

$\displaystyle\hat{u}_{\hat{x}}=u_{x}+\epsilon\eta^{x}(x,y,t,u,u_{t},u_{y},u_{x})+o(\epsilon^{2})$
$\displaystyle\hat{u}_{\hat{x}\hat{x}}=u_{xx}+\epsilon\eta^{xx}(x,y,t,u,u_{t},u_{y},u_{x},...)+o(\epsilon^{2})$
$\displaystyle\hat{u}_{\hat{y}}=u_{y}+\epsilon\eta^{y}(x,y,t,u,u_{t},u_{y},u_{x})+o(\epsilon^{2})$
$\displaystyle\hat{u}_{\hat{y}\hat{y}}=u_{yy}+\epsilon\eta^{yy}(x,y,t,u,u_{t},u_{y},u_{x},...)+o(\epsilon^{2})$
$\displaystyle\hat{u}_{\hat{x}\hat{t}}=u_{xt}+\epsilon\eta^{xt}(x,y,t,u,u_{t},u_{y},u_{x},...)+o(\epsilon^{2})$
$\displaystyle\hat{u}_{\hat{x}\hat{x}\hat{x}\hat{t}}=u_{xxxt}+\epsilon\eta^{xxxt}(x,y,t,u,u_{t},u_{y},u_{x},...)+o(\epsilon^{2})$
(11)

where

$\displaystyle\eta^{x}=D_{x}\eta-u_{x}D_{x}\xi-u_{y}D_{x}\gamma-
u_{t}D_{x}\tau$ $\displaystyle\eta^{xx}=D_{x}\eta^{x}-u_{xx}D_{x}\xi-
u_{xy}D_{x}\gamma-u_{xt}D_{x}\tau$ $\displaystyle\eta^{y}=D_{y}\eta-
u_{x}D_{y}\xi-u_{y}D_{y}\gamma-u_{t}D_{y}\tau$
$\displaystyle\eta^{yy}=D_{y}\eta^{y}-u_{xy}D_{y}\xi-u_{yy}D_{y}\gamma-
u_{yt}D_{y}\tau$ $\displaystyle\eta^{xt}=D_{t}\eta^{x}-u_{xx}D_{t}\xi-
u_{xy}D_{t}\gamma-u_{xt}D_{t}\tau$
$\displaystyle\eta^{xxxt}=D_{t}\eta^{xxx}-u_{xxxx}D_{t}\xi-
u_{xxxy}D_{t}\gamma-u_{xxxt}D_{t}\tau$ (12)

and
$D_{x}=\frac{\partial}{\partial x}+u_{x}\frac{\partial}{\partial
u}+u_{xx}\frac{\partial}{\partial u_{x}}+u_{xy}\frac{\partial}{\partial
u_{y}}+u_{xt}\frac{\partial}{\partial u_{t}}+u_{xxt}\frac{\partial}{\partial
u_{xt}}+...$ ,
$D_{y}=\frac{\partial}{\partial y}+u_{y}\frac{\partial}{\partial
u}+u_{xy}\frac{\partial}{\partial u_{x}}+u_{yy}\frac{\partial}{\partial
u_{y}}+u_{yt}\frac{\partial}{\partial u_{t}}+u_{xyt}\frac{\partial}{\partial
u_{xt}}+...$ ,
$D_{t}=\frac{\partial}{\partial t}+u_{t}\frac{\partial}{\partial
u}+u_{xt}\frac{\partial}{\partial u_{x}}+u_{ty}\frac{\partial}{\partial
u_{y}}+u_{tt}\frac{\partial}{\partial u_{t}}+u_{xtt}\frac{\partial}{\partial
u_{xt}}+...$ .
The transformations (7) will be a symmetry transformation for (1) if it
satisfies the following condition

$\displaystyle\hat{u}_{\hat{x}\hat{t}}+\hat{u}_{\hat{x}\hat{x}}+2a(\hat{u}_{\hat{x}})^{2}+2a\hat{u}\hat{u}_{\hat{x}\hat{x}}+b\hat{u}_{\hat{x}\hat{x}\hat{x}\hat{t}}+k\hat{u}_{\hat{y}\hat{y}}=0$
$\displaystyle\Rightarrow$ $\displaystyle
u_{xt}+\epsilon\eta^{xt}+u_{xx}+\epsilon\eta^{xx}+2a(u_{x}+\epsilon\eta^{x})^{2}+2a(u+\epsilon\eta)(u_{xx}+\epsilon\eta^{xx})+b(u_{xxxt}+\epsilon\eta^{xxxt})$
(13) $\displaystyle+k(u_{yy}+\epsilon\eta^{yy})=0$

Collecting the coefficients of $\epsilon$ from (13) we get

$\displaystyle\eta^{xt}+\eta^{xx}+4au_{x}\eta^{x}+2au_{xx}\eta+2au\eta^{xx}+b\eta^{xxxt}+k\eta^{yy}=0,$
(14)

which gives the symmetry condition (linearized) for (1).

We calculate $\eta^{x},\eta^{xt},\eta^{xx}...$ etc from (12) and use them in
(14). The functions $\xi,\gamma,\tau,\eta$ are free from the terms
$u_{x},u_{y},u_{t},u_{xx},u_{xy},......$ . So by equating the coefficients of
the various derivative terms of $u$ to 0, a set of equations is obtained, from
which we derive the following

$\displaystyle\xi=c_{1}x+c_{2},\gamma=c_{1}y+c_{4},\tau=-2c_{1}t+c_{3},\eta=c_{1}u+\frac{c_{1}}{2a};$
(15)

$c_{1},\hskip 2.84526ptc_{2},\hskip 2.84526ptc_{3},\hskip 2.84526ptc_{4}$ are
any constants.

The infinitesimal symmetry generator for (1) is

$\displaystyle\Gamma$ $\displaystyle=\xi\frac{\partial}{\partial
x}+\gamma\frac{\partial}{\partial y}+\tau\frac{\partial}{\partial
t}+\eta\frac{\partial}{\partial u}$ (16)
$\displaystyle=c_{1}\left(x\frac{\partial}{\partial
x}+y\frac{\partial}{\partial y}-2t\frac{\partial}{\partial
t}+(u+\frac{1}{2a})\frac{\partial}{\partial
u}\right)+c_{2}\frac{\partial}{\partial x}+c_{3}\frac{\partial}{\partial
t}+c_{4}\frac{\partial}{\partial y}$

Thus a Lie algebra $V$ is obtained which is generated by

$\displaystyle\Gamma_{1}=x\frac{\partial}{\partial
x}+y\frac{\partial}{\partial y}-2t\frac{\partial}{\partial
t}+\left(u+\frac{1}{2a}\right)\frac{\partial}{\partial
u},\Gamma_{2}=\frac{\partial}{\partial x},\Gamma_{3}=\frac{\partial}{\partial
t},\Gamma_{4}=\frac{\partial}{\partial y}.$ (17)

The commutator table is given below:

Table 1: Commutator table $[\Gamma_{i},\Gamma_{j}]$ | $\Gamma_{1}$ | $\Gamma_{2}$ | $\Gamma_{3}$ | $\Gamma_{4}$
---|---|---|---|---
$\Gamma_{1}$ | 0 | -$\Gamma_{2}$ | 2$\Gamma_{3}$ | $-\Gamma_{4}$
$\Gamma_{2}$ | $\Gamma_{2}$ | 0 | 0 | 0
$\Gamma_{3}$ | -2$\Gamma_{3}$ | 0 | 0 | 0
$\Gamma_{4}$ | $\Gamma_{4}$ | 0 | 0 | 0

The first derived [22] Lie algebra $V^{1}$ is defined by
$V^{1}=[V,V]=\\{\Gamma_{2},\Gamma_{3},\Gamma_{4}\\}$. Similarly the second
derived Lie algebra is given by $V^{2}=[V^{1},V^{1}]=0,$ which shows that the
Lie algebra $V$ is solvable and hence the related differential equation can be
solved. [A Lie algebra $V$ is solvable iff $V^{n+1}=[V^{n},V^{n}]=0$ for some
natural $n$. ]

###  3.1 Construction of Adjoint matrix:

To compute the adjoint relation we consider the Lie series [19]

$\displaystyle
Ad_{exp(\epsilon\Gamma_{i})}\Gamma_{j}=\Gamma_{j}-\epsilon[\Gamma_{i},\Gamma_{j}]+\frac{\epsilon^{2}}{2!}[\Gamma_{i},[\Gamma_{i},\Gamma_{j}]]-\frac{\epsilon^{3}}{3!}[\Gamma_{i},[\Gamma_{i},[\Gamma_{i},\Gamma_{j}]]]+....$
(18)

The adjoint table is given below

Table 2: Adjoint table $Ad$ | $\Gamma_{1}$ | $\Gamma_{2}$ | $\Gamma_{3}$ | $\Gamma_{4}$
---|---|---|---|---
$\Gamma_{1}$ | $\Gamma_{1}$ | $e^{\epsilon}\Gamma_{2}$ | $e^{-2\epsilon}\Gamma_{3}$ | $e^{\epsilon}\Gamma_{4}$
$\Gamma_{2}$ | $(\Gamma_{1}-\epsilon\Gamma_{2})$ | $\Gamma_{2}$ | $\Gamma_{3}$ | $\Gamma_{4}$
$\Gamma_{3}$ | $(\Gamma_{1}+2\epsilon\Gamma_{3})$ | $\Gamma_{2}$ | $\Gamma_{3}$ | $\Gamma_{4}$
$\Gamma_{4}$ | $(\Gamma_{1}-\epsilon\Gamma_{4})$ | $\Gamma_{2}$ | $\Gamma_{3}$ | $\Gamma_{4}$

The adjoint action of $\Gamma_{1}$ on any vector $\Gamma\in V$ is given by

$\displaystyle
Ad_{exp(\epsilon\Gamma_{1})}\Gamma=d_{1}\Gamma_{1}+d_{2}\Gamma_{2}+d_{3}\Gamma_{3}+d_{4}\Gamma_{4},$
(19)

which can be expressed in the matrix form as

$\displaystyle
Ad_{exp(\epsilon\Gamma_{1})}\Gamma=[d_{1},d_{2},d_{3},d_{4}]A_{1}[\Gamma_{1},\Gamma_{2},\Gamma_{3},\Gamma_{4}].$

where

$\displaystyle A_{1}=\begin{bmatrix}1&0&0&0\\\ 0&e^{\epsilon_{1}}&0&0\\\
0&0&e^{-2\epsilon_{1}}&0\\\ 0&0&0&e^{\epsilon_{1}}\end{bmatrix}$

Similarly we obtain

$\displaystyle A_{2}=\begin{bmatrix}1&-\epsilon_{2}&0&0\\\ 0&1&0&0\\\
0&0&1&0\\\ 0&0&0&1\\\
\end{bmatrix},A_{3}=\begin{bmatrix}1&0&2\epsilon_{3}&0\\\ 0&1&0&0\\\
0&0&1&0\\\ 0&0&0&1\\\
\end{bmatrix},A_{4}=\begin{bmatrix}1&0&0&-\epsilon_{4}\\\ 0&1&0&0\\\
0&0&1&0\\\ 0&0&0&1\\\ \end{bmatrix}$

the global adjoint matrix $A$ is given by

$\displaystyle A=\begin{bmatrix}1&-\epsilon_{2}&2\epsilon_{3}&-\epsilon_{4}\\\
0&e^{\epsilon_{1}}&0&0\\\ 0&0&e^{-2\epsilon_{1}}&0\\\
0&0&0&e^{\epsilon_{1}}\end{bmatrix}.$ (20)

The general adjoint transformation equation is given by

$\displaystyle(\tilde{a}_{1},\tilde{a}_{2},\tilde{a}_{3},\tilde{a}_{4})=(a_{1},a_{2},a_{3},a_{4})A,$
(21)

by which the vector
$a_{1}\Gamma_{1}+a_{2}\Gamma_{2}+a_{3}\Gamma_{3}+a_{4}\Gamma_{4}$ is
transformed into the vector
$\tilde{a}_{1}\Gamma_{1}+\tilde{a}_{2}\Gamma_{2}+\tilde{a}_{3}\Gamma_{3}+\tilde{a}_{4}\Gamma_{4}$
by adjoint action.

###  3.2 Calculation of the Invariants of the Lie Algebra

In this section we are going to find the invariants of the obtained Lie
algebra by a method proposed by Hu et. al. in [19]. Let $U,W$ be two elements
of the Lie algebra spanned by ${\Gamma_{1},\Gamma_{2},\Gamma_{3},\Gamma_{4}}$.
Then $U,W$ can be expressed as $U=\Sigma_{i=1}^{4}a_{i}\Gamma_{i}$ and
$W=\Sigma_{j=1}^{4}b_{j}\Gamma_{j}$.

Now

$\displaystyle Ad_{exp(\epsilon W)}U$
$\displaystyle=U-\epsilon[W,U]+\frac{\epsilon^{2}}{2!}[W,[W,U]]-....$
$\displaystyle=U-\epsilon[W,U]+o(\epsilon^{2})$
$\displaystyle=(a_{1}\Gamma_{1}+a_{2}\Gamma_{2}+a_{3}\Gamma_{3}+a_{4}\Gamma_{4})$
$\displaystyle\hskip
28.45274pt-\epsilon[(b_{1}\Gamma_{1}+b_{2}\Gamma_{2}+b_{3}\Gamma_{3}+b_{4}\Gamma_{4}),(a_{1}\Gamma_{1}+a_{2}\Gamma_{2}+a_{3}\Gamma_{3}+a_{4}\Gamma_{4})]+o(\epsilon^{2})$
$\displaystyle=(a_{1}\Gamma_{1}+a_{2}\Gamma_{2}+a_{3}\Gamma_{3}+a_{4}\Gamma_{4})-\epsilon((\theta_{1}\Gamma_{1}+\theta_{2}\Gamma_{2}+\theta_{3}\Gamma_{3}+\theta_{4}\Gamma_{4}))+o(\epsilon^{2}),$

where $\theta_{i}(a_{1},...,a_{4},b_{1},...,b_{4})$ can easily be obtained
from the commutator table. Thus the action of $Ad_{exp(\epsilon W)}$ on $U$
can be written as

$(a_{1},a_{2},a_{3},a_{4})\rightarrow(a_{1}-\epsilon\theta_{1},a_{2}-\epsilon\theta_{2},a_{3}-\epsilon\theta_{3},a_{4}-\epsilon\theta_{4})+o(\epsilon^{2})$.
Let $\phi$ be an invariant of the Lie algebra. Then

$\displaystyle\phi(a_{1},a_{2},a_{3},a_{4})=\phi(a_{1}-\epsilon\theta_{1}+o(\epsilon^{2}),a_{2}-\epsilon\theta_{2}+o(\epsilon^{2}),a_{3}-\epsilon\theta_{3}+o(\epsilon^{2}),a_{4}-\epsilon\theta_{4}+o(\epsilon^{2}))$
(23)

By applying Taylor’s theorem, the right hand side of (23) becomes

$\displaystyle\phi(a_{1}-\epsilon\theta_{1}+o(\epsilon^{2}),a_{2}-\epsilon\theta_{2}+o(\epsilon^{2}),a_{3}-\epsilon\theta_{3}+o(\epsilon^{2}),a_{4}-\epsilon\theta_{4}+o(\epsilon^{2}))=\phi(a_{1},a_{2},a_{3},a_{4})$
$\displaystyle\hskip
142.26378pt-\epsilon\left(\theta_{1}\frac{\partial\phi}{\partial
a_{1}}+\theta_{2}\frac{\partial\phi}{\partial
a_{2}}+\theta_{3}\frac{\partial\phi}{\partial
a_{3}}+\theta_{4}\frac{\partial\phi}{\partial a_{4}}\right)+o(\epsilon^{2}).$
(24)

Since $\phi$ is an invariant we must have

$\displaystyle\theta_{1}\frac{\partial\phi}{\partial
a_{1}}+\theta_{2}\frac{\partial\phi}{\partial
a_{2}}+\theta_{3}\frac{\partial\phi}{\partial
a_{3}}+\theta_{4}\frac{\partial\phi}{\partial a_{4}}=0.$ (25)

We calculate $\theta_{i}$ from the commutator table as follows

$\theta_{1}=0,\hskip 5.69054pt\theta_{2}=-b_{1}a_{2}+b_{2}a_{1},\hskip
5.69054pt\theta_{3}=2b_{1}a_{3}-2b_{3}a_{1},\hskip
5.69054pt\theta_{4}=-b_{1}a_{4}+b_{4}a_{1}$.

Then (25) becomes

$\displaystyle(b_{2}a_{1}-b_{1}a_{2})\frac{\partial\phi}{\partial
a_{2}}+2(b_{1}a_{3}-b_{3}a_{1})\frac{\partial\phi}{\partial
a_{3}}+(b_{4}a_{1}-b_{1}a_{4})\frac{\partial\phi}{\partial a_{4}}=0$ (26)

Equating the coefficients of each $b_{i}$ to 0, we get the following system of
partial differential equations

$\displaystyle a_{1}\frac{\partial\phi}{\partial a_{4}}=0,$ $\displaystyle
a_{1}\frac{\partial\phi}{\partial a_{3}}=0,$ $\displaystyle
a_{1}\frac{\partial\phi}{\partial a_{2}}=0,$ $\displaystyle
a_{2}\frac{\partial\phi}{\partial a_{2}}-2a_{3}\frac{\partial\phi}{\partial
a_{3}}+a_{4}\frac{\partial\phi}{\partial a_{4}}=0,$ (27)

Solving the above system we get $\phi(a_{1},a_{2},a_{3},a_{4})=F(a_{1})$, an
arbitrary function of $a_{1}$.

###  3.3 One Dimensional Optimal System of Lie Sub-algebras

In this section, we are looking for a one dimensional optimal system of
subalgebras of the Lie algebra $V$ by using Hu-Li-Chen algorithm [19]. A list
of one dimensional subalgebras is said to form an one dimensional optimal
system if each subalgebra of the Lie algebra is equivalent to a definite
member of this list [23].

According to the Hu-Li-Chen algorithm at first the basic invariant(s) is (are)
to be scaled to ’-1,1,0’ or ’1,0’ respectively as they are even or odd. Then
for each case one representative element
$\tilde{v}=\tilde{a}_{1}\Gamma_{1}+\tilde{a}_{2}\Gamma_{2}+\tilde{a}_{3}\Gamma_{3}+\tilde{a}_{4}\Gamma_{4}$
has to be selected for which the adjoint transformation equation (21) can be
solved for $\epsilon_{1},\epsilon_{2},\epsilon_{3},\epsilon_{4}$. At the time
of scaling it must be kept in mind that once an invariant is scaled to a
nonzero integer then the other invariants (if there be any) can not be
adjusted.

By using (20) in (21), the adjoint transformation equation can be explicitly
written as

$\displaystyle(\tilde{a}_{1},\tilde{a}_{2},\tilde{a}_{3},\tilde{a}_{4})=(a_{1},-\epsilon_{2}a_{1}+e^{\epsilon_{1}}a_{2},2\epsilon_{3}a_{1}+e^{-2\epsilon_{1}}a_{3},-\epsilon_{4}a_{1}+e^{\epsilon_{1}}a_{4}).$
(28)

The invariant of this Lie algebra is found to be
$\phi(a_{1},a_{2},a_{3},a_{4})=F(a_{1})$, where $F$ is an arbitrary function
of $a_{1}$. $a_{1}$ is a basic invariant whose degree is one, an odd integer.
So, by Hu-Li-Chen algorithm we have to consider two cases: $a_{1}=1$ and
$a_{1}=0$.
Case 1: $a_{1}=1$
Let us choose the representative element as $\tilde{v}=\Gamma_{1}$ i.e.
$\tilde{a}_{2}=\tilde{a}_{3}=\tilde{a}_{4}=0$ and $\tilde{a}_{1}=1$.
Under such choice the adjoint transformation equation (28) can be solved for
$\epsilon_{i}$ $(i=1,2,3,4)$ as $\epsilon_{1}=0$, $\epsilon_{2}=a_{2}$,
$\epsilon_{3}=-\frac{1}{2}a_{3}$, $\epsilon_{4}=a_{4}$ which indicates that
the selected representative element is correct. Therefore, all the elements of
the form $\Gamma_{1}+a_{2}\Gamma_{2}+a_{3}\Gamma_{3}+a_{4}\Gamma_{4}$ are
equivalent to $\Gamma_{1}$.
Case 2: $a_{1}=0$
For $a_{1}=0$, the system of equations (27) reduces to

$a_{2}\frac{\partial\phi}{\partial a_{2}}-2a_{3}\frac{\partial\phi}{\partial
a_{3}}+a_{4}\frac{\partial\phi}{\partial a_{4}}=0$

solving which we get

$\phi(a_{1},a_{2},a_{3},a_{4})=F(a_{2}^{2}a_{3},a_{4}^{2}a_{3})$

where $F$ is an arbitrary function of two variables. Thus we have two basic
invariants $\Delta_{1}=a_{2}^{2}a_{3}$ and $\Delta_{2}=a_{4}^{2}a_{3}$ both of
which are odd degree polynomials. According to Hu-Li-Chen algorithm we have
three sub-cases

$\\{\Delta_{1}=1,\Delta_{2}=c\\},\hskip
8.53581pt\\{\Delta_{1}=0,\Delta_{2}=1\\},\hskip
8.53581pt\\{\Delta_{1}=0,\Delta_{2}=0\\}$

Subcase 2.1: $\\{\Delta_{1}=a_{2}^{2}a_{3}=1,\Delta_{2}=a_{4}^{2}a_{3}=c\\}$

Obviously $a_{2}\neq 0$, $a_{3}>0$. The following subcases under this subcase
2.1 are to be considered:

Subsubcase 2.1.1: $a_{2}>0$, $a_{3}>0$. Then from (28) we have

$\tilde{a}_{1}=0,\hskip 5.69054pte^{\epsilon_{1}}a_{2}=\tilde{a}_{2},\hskip
5.69054pte^{-2\epsilon_{1}}a_{3}=\tilde{a}_{3},\hskip
5.69054pte^{\epsilon_{1}}a_{4}=\tilde{a}_{4}$

which give

$\epsilon_{1}=\log\frac{\tilde{a}_{2}}{a_{2}}=-\frac{1}{2}\log\frac{\tilde{a}_{3}}{a_{3}}=\log\frac{\tilde{a}_{4}}{a_{4}},\hskip
8.53581pt\tilde{a}_{2}>0,\tilde{a}_{3}>0$

which implies

$\tilde{a}_{2}^{2}\tilde{a}_{3}=a_{2}^{2}a_{3}=1,\hskip
8.53581pt\tilde{a}_{4}^{2}\tilde{a}_{3}=a_{4}^{2}a_{3}=c$

Considering all of the above restrictions on
$\tilde{a}_{1},\tilde{a}_{2},\tilde{a}_{3},\tilde{a}_{4}$ we can take

$\tilde{a}_{1}=0,\tilde{a}_{2}=1,\tilde{a}_{3}=1,\tilde{a}_{4}=\sqrt{c}$

Therefore, the corresponding representative element is

$\Gamma_{2}+\Gamma_{3}+\sqrt{c}\Gamma_{4}$

.

Subsubcase 2.1.2: $a_{2}<0$, $a_{3}>0$. In this case the representative
element can be chosen as

$-\Gamma_{2}+\Gamma_{3}+\sqrt{c}\Gamma_{4}$

Subcase 2.2: $\\{\Delta_{1}=a_{2}^{2}a_{3}=0,\Delta_{2}=a_{4}^{2}a_{3}=1\\}$

Clearly here $a_{2}=0,\hskip 2.84526pta_{3}>0$ and either $a_{4}>0$ or
$a_{4}<0$. Thus we have to consider two subcases under this subcase 2.2:

Subsubcase 2.2.1: $a_{2}=0$, $a_{3}>0$, $a_{4}>0$. Proceeding in similar
fashion we found the representative $\Gamma_{3}+\Gamma_{4}$.

Subsubcase 2.2.2: $a_{2}=0$, $a_{3}>0$, $a_{4}<0$. The corresponding
representative is found to be $\Gamma_{3}-\Gamma_{4}$.

Subcase 2.3: $\\{\Delta_{1}=a_{2}^{2}a_{3}=0,\Delta_{2}=a_{4}^{2}a_{3}=0\\}$
which can be divided into following subcases:

Subsubcase 2.3.1: $a_{2}=0$, $a_{3}\neq 0$, $a_{4}=0$. There are two
possibilities $a_{3}>0$ and $a_{3}<0$, correspondingly representative elements
are found to be $\Gamma_{3}$ and $-\Gamma_{3}$ which are equivalent to each
other. So in this Subsubcase 2.3.1 we choose $\Gamma_{3}$ as representative
member in the optimal list.

Subsubcase 2.3.2: $a_{3}=0$. All the possibilities and the corresponding
representatives are listed below:
$(i)$ $a_{1}=0$, $a_{2}>0$, $a_{3}=0$, $a_{4}=0$ $\Rightarrow$ representative
$\Gamma_{2}$
$(ii)$ $a_{1}=0$, $a_{2}<0$, $a_{3}=0$, $a_{4}=0$ $\Rightarrow$ representative
$-\Gamma_{2}$
$(iii)$ $a_{1}=0$, $a_{2}=0$, $a_{3}=0$, $a_{4}>0$ $\Rightarrow$
representative $\Gamma_{4}$
$(iv)$ $a_{1}=0$, $a_{2}=0$, $a_{3}=0$, $a_{4}<0$ $\Rightarrow$ representative
$-\Gamma_{4}$
$(v)$ $a_{1}=0$, $a_{2}>0$, $a_{3}=0$, $a_{4}>0$ $\Rightarrow$ representative
$\Gamma_{2}+\Gamma_{4}$
$(vi)$ $a_{1}=0$, $a_{2}>0$, $a_{3}=0$, $a_{4}<0$ $\Rightarrow$ representative
$\Gamma_{2}-\Gamma_{4}$
$(vii)$ $a_{1}=0$, $a_{2}<0$, $a_{3}=0$, $a_{4}>0$ $\Rightarrow$
representative $-\Gamma_{2}+\Gamma_{4}$
$(viii)$ $a_{1}=0$, $a_{2}<0$, $a_{3}=0$, $a_{4}<0$ $\Rightarrow$
representative $\Gamma_{2}-\Gamma_{4}$

Thus in this case we get four distinct inequivalent representative vectors as

$\Gamma_{2},\hskip 5.69054pt\Gamma_{4},\hskip
5.69054pt\Gamma_{2}+\Gamma_{4},\hskip 5.69054pt\Gamma_{2}-\Gamma_{4}$

.

Therefore, by listing all the above selected representatives we have the
following optimal system of sub-algebras:

$\\{\Gamma_{1}\\},\hskip 5.69054pt\\{\Gamma_{2}\\},\hskip
5.69054pt\\{\Gamma_{3}\\},\hskip 5.69054pt\\{\Gamma_{4}\\},\hskip
5.69054pt\\{\Gamma_{2}+\Gamma_{4}\\},\hskip
5.69054pt\\{\Gamma_{2}-\Gamma_{4}\\},\hskip
5.69054pt\\{\Gamma_{3}+\Gamma_{4}\\},\hskip
5.69054pt\\{\Gamma_{3}-\Gamma_{4}\\},$ $\hskip
5.69054pt\\{\Gamma_{2}+\Gamma_{3}+\sqrt{c}\Gamma_{4}\\},\hskip
5.69054pt\\{-\Gamma_{2}+\Gamma_{3}+\sqrt{c}\Gamma_{4}\\}$

###  3.4 Similarity Reductions and Exact solutions

The invariant surface condition for (1) is given by

$\displaystyle\eta-\xi\frac{\partial u}{\partial x}-\gamma\frac{\partial
u}{\partial y}-\tau\frac{\partial u}{\partial t}=0,$ (29)

which is a quasilinear equation in $u(x,y,t)$ and its auxiliary equations are

$\displaystyle\frac{dx}{\xi}=\frac{dy}{\gamma}=\frac{dt}{\tau}=\frac{du}{\eta}$
(30)

Let us find the similarity reduction corresponding to the subalgebra
$\\{\Gamma_{2}+\Gamma_{3}+\sqrt{c}\Gamma_{4}\\}$ with taking the constant
$c=1$. The auxiliary equation (30) becomes

$\displaystyle\frac{dx}{1}=\frac{dy}{1}=\frac{dt}{1}=\frac{du}{0}.$

We find the general solution of (29) in the following forms:

$\displaystyle u(x,y,t)=F(x-y,y-t)$ (31) $\displaystyle u(x,y,t)=G(x-y,x-t)$
(32) $\displaystyle u(x,y,t)=H(x-t,y-t)$ (33)

By putting (31) in (1) we get the following similarity reduction with the
similarity variables $\alpha=x-y,\beta=y-t$

$\displaystyle-
F_{\alpha\beta}+F_{\alpha\alpha}+a(F^{2})_{\alpha\alpha}-bF_{\alpha\alpha\alpha\beta}+k(F_{\alpha\alpha}-2F_{\alpha\beta}+F_{\beta\beta})=0,$
(34)

solving which we get the following exact solution of (1)

$\displaystyle
u(x,y,t)=\frac{3(\lambda-1)(k\lambda-k-1)}{2a}Sech^{2}\left\\{\frac{1}{2}\sqrt{\frac{(\lambda-1)(k\lambda-k-1)}{b\lambda}}(x+(\lambda-1)y-\lambda
t)\right\\},$ (35)

where $\lambda$ is an arbitrary constant.

By putting (32) in (1) we get another similarity reduction with the similarity
variables $\alpha=x-y,\beta=x-t$

$\displaystyle(1+k)G_{\alpha\alpha}+G_{\alpha\beta}+a(G^{2})_{\alpha\alpha}+2a(G^{2})_{\alpha\beta}+a(G^{2})_{\beta\beta}-bG_{\alpha\alpha\alpha\beta}-3bG_{\alpha\alpha\beta\beta}-3bG_{\alpha\beta\beta\beta}-bG_{\beta\beta\beta\beta}=0,$
(36)

solving which we get the following exact solution of (1)

$\displaystyle
u(x,y,t)=-\frac{3(2+k+\lambda-\lambda^{2})}{2a(1+\lambda)^{2}}Sech^{2}\left\\{\frac{1}{2}\sqrt{\frac{2+k+\lambda-\lambda^{2}}{b\lambda(1+\lambda)^{3}}}((1+\lambda)x-y-\lambda
t)\right\\},$ (37)

where $\lambda(\neq-1)$ is any constant other than -1.

Now by putting (33) in (1) we get one another similarity reduction with the
similarity variables $\alpha=x-t,\beta=y-t$

$\displaystyle
kH_{\beta\beta}-H_{\alpha\beta}+a(H^{2})_{\alpha\alpha}-bH_{\alpha\alpha\alpha\alpha}-bH_{\alpha\alpha\alpha\beta}=0,$
(38)

solving which we get another exact solution of (1) as

$\displaystyle
u(x,y,t)=\frac{3(k\lambda^{2}-\lambda)}{2a}Sech^{2}\left\\{\frac{1}{2}\sqrt{\frac{k\lambda^{2}-\lambda}{b(\lambda+1)}}(x+\lambda
y-(\lambda+1)t)\right\\},$ (39)

where $\lambda(\neq-1)$ is an any constant other than -1.

Amplitude of each solitary wave (35), (37), (39) decreases with respect to the
nonlinear coefficient $a$ and the widths increase with respect to the
dispersion coefficient $b$. $a$, $b$ don’t have any effect on the velocity of
the above solitary waves.

Figure 1: Effect of the dispersion coefficient $b$ on the width and velocity
of the solitary wave (39), taking $k=1,\lambda=2,a=1,y=0$.

The similarity reductions and corresponding exact solutions for each member of
the optimal system can be obtained similarly.

##  4 Solution by Homogeneous Balance Method:

Here we are going to solve (1) by considering

$\displaystyle u(x,y,t)=\frac{\partial^{p}f(\phi(x,y,t))}{\partial
x^{p}}+u_{1}(x,y,t),$ (40)

in which $\phi,u_{1}$, $f$ are functions of $x,y,t$ to be found. We choose
$\phi$ in such a way that its $x-$derivatives and $t-$derivatives are same.
i.e. $\phi_{x}=\phi_{t},\phi_{xt}=\phi_{xx},\phi_{xxxt}=\phi_{xxxx}$ etc.

Balancing the highest power of $\phi_{x}$ we find that $p=2$. Therefore,

$\displaystyle u(x,y,t)=\frac{\partial^{2}f(\phi(x,y,t))}{\partial
x^{2}}+u_{1}(x,y,t)=\frac{d^{2}f}{d\phi^{2}}\left(\frac{d\phi}{dx}\right)^{2}+\frac{df}{d\phi}\frac{d^{2}\phi}{dx^{2}}+u_{1},$
(41)

Using (41) in (1) we get,

$\displaystyle(2af^{\prime\prime\prime
2}+2af^{\prime\prime}f^{iv}+bf^{vi})\phi_{x}^{6}+(24af^{\prime\prime}f^{\prime\prime\prime}+2af^{\prime}f^{iv}+15bf^{v})\phi_{x}^{4}\phi_{xx}+(2f^{iv}\phi_{x}^{4}+24af^{\prime\prime
2}\phi_{x}^{2}\phi_{xx}^{2}$
$\displaystyle+4af^{\prime}f^{\prime\prime\prime}\phi_{x}^{3}\phi_{xxx}+8af^{\prime\prime
2}\phi_{x}^{3}\phi_{xxx}+12af^{\prime}f^{\prime\prime\prime}\phi_{x}^{2}\phi_{xx}^{2}+2af^{iv}\phi_{x}^{4}u_{1}+45bf^{iv}\phi_{x}^{2}\phi_{xx}^{2}+20bf^{iv}\phi_{x}^{3}\phi_{xxx}$
$\displaystyle+kf^{iv}\phi_{x}^{2}\phi_{y}^{2})+(12f^{\prime\prime\prime}\phi_{x}^{2}\phi_{xx}+20af^{\prime}f^{\prime\prime}\phi_{x}\phi_{xx}\phi_{xxx}+4af^{\prime\prime\prime}\phi_{x}^{3}u_{1x}+2af^{\prime}f^{\prime\prime}\phi_{x}^{2}\phi_{xxxx}+6af^{\prime}f^{\prime\prime}\phi_{xx}^{3}$
$\displaystyle+12af^{\prime\prime\prime}\phi_{x}^{2}\phi_{xx}u_{1}+15bf^{\prime\prime\prime}\phi_{xx}^{3}+60bf^{\prime\prime\prime}\phi_{x}\phi_{xx}\phi_{xxx}+15bf^{\prime\prime\prime}\phi_{x}^{2}\phi_{xxxx}+4kf^{\prime\prime\prime}\phi_{x}\phi_{xy}\phi_{y}+kf^{\prime\prime\prime}\phi_{x}^{2}\phi_{yy}$
$\displaystyle+kf^{\prime\prime\prime}\phi_{xx}\phi_{y}^{2})+(6f^{\prime\prime}\phi_{xx}^{2}+8f^{\prime\prime}\phi_{x}\phi_{xxx}+2af^{\prime
2}\phi_{xxx}^{2}+12af^{\prime\prime}\phi_{x}\phi_{xx}u_{1x}+2af^{\prime\prime}\phi_{x}^{2}u_{1xx}$
$\displaystyle+2af^{\prime
2}\phi_{xx}\phi_{xxxx}+6af^{\prime\prime}\phi_{xx}^{2}u_{1}+8af^{\prime\prime}\phi_{x}\phi_{xxx}u_{1}+10bf^{\prime\prime}\phi_{xxx}^{2}+15bf^{\prime\prime}\phi_{xx}\phi_{xxxx}+6bf^{\prime\prime}\phi_{x}\phi_{xxxxx}$
$\displaystyle+2kf^{\prime\prime}\phi_{xy}^{2}+2kf^{\prime\prime}\phi_{x}\phi_{xyy}+2kf^{\prime\prime}\phi_{xxy}\phi_{y}+kf^{\prime\prime}\phi_{xx}\phi_{yy})+(2f^{\prime}\phi_{xxxx}+4af^{\prime}\phi_{xxx}u_{1x}+2af^{\prime}\phi_{xx}u_{1xx}$
$\displaystyle+2af^{\prime}\phi_{xxxx}u_{1}+bf^{\prime}\phi_{xxxxxx}+kf^{\prime}\phi_{xxyy})+(u_{1xt}+u_{1xx}+2au_{1x}^{2}+2au_{1}u_{1xx}+bu_{1xxxt}+ku_{1yy})$
$\displaystyle=0,$ (42)

where
$f^{\prime}=\frac{df}{d\phi},f^{\prime\prime}=\frac{d^{2}f}{d\phi^{2}},......$
.

Equating coefficient of $\phi_{x}^{6}$ in (42) to zero we get

$\displaystyle 2af^{\prime\prime\prime 2}+2af^{\prime\prime}f^{iv}+bf^{vi}=0$
(43)

$f=ln\phi$ is a solution of (43) if $a=6b.$ So, let us take

$\displaystyle f=ln\phi$ (44) $\displaystyle a=6b$ (45)

Then $f^{\prime\prime}f^{\prime\prime\prime}=-\frac{1}{12}f^{v}$,
$f^{\prime}f^{iv}=-\frac{1}{4}f^{v}$, $f^{\prime\prime 2}=-\frac{1}{6}f^{iv}$,
$f^{\prime}f^{\prime\prime\prime}=-\frac{1}{3}f^{iv}$,
$f^{\prime}f^{\prime\prime}=-\frac{1}{2}f^{\prime\prime\prime}$, $f^{\prime
2}=-f^{\prime\prime}.$ Using the above results in (42) and then equating the
coefficients of $f^{iv},f^{\prime\prime\prime},f^{\prime\prime},f^{\prime}$
and $f$ free terms to zero we get a system of differential equations in
$\phi$, which has a solution of the form

$\displaystyle\phi(x,y,t)=1+exp(\alpha x+\beta y+\alpha t+\theta_{0}).$ (46)

[remember that $x-$derivatives and $t-$derivatives of $\phi$ are same].
Using (44), (45), (46) in (42) we obtain the following system of algebraic
equations:

$\displaystyle
2\alpha^{4}+12b\alpha^{4}u_{1}+b\alpha^{6}+k\alpha^{2}\beta^{2}=0,$ (47)
$\displaystyle
12\alpha^{4}+72b\alpha^{4}u_{1}+6b\alpha^{6}+6k\alpha^{2}\beta^{2}+24b\alpha^{3}u_{1x}=0,$
(48) $\displaystyle
14\alpha^{4}+84b\alpha^{4}u_{1}+7b\alpha^{6}+7k\alpha^{2}\beta^{2}+72b\alpha^{3}u_{1x}+12b\alpha^{2}u_{1xx}=0,$
(49) $\displaystyle
2\alpha^{4}+12b\alpha^{4}u_{1}+b\alpha^{6}+k\alpha^{2}\beta^{2}+24b\alpha^{3}u_{1x}+12b\alpha^{2}u_{1xx}=0,$
(50) $\displaystyle
u_{1xt}+u_{1xx}+2au_{1x}^{2}+2au_{1}u_{1xx}+bu_{1xxxt}+ku_{1yy}=0.$ (51)

From (47) we see that $u_{1}$ is a constant and

$\displaystyle\beta=\pm\sqrt{-\frac{2\alpha^{2}+12b\alpha^{2}u_{1}+b\alpha^{4}}{k}},$
(52)

Using (44), (46) in (41) we find that

$\displaystyle u(x,y,t)=\frac{\alpha^{2}}{2+2cosh(\alpha x+\beta y+\alpha
t+\theta_{0})}+u_{1}$ (53)

gives a solution of (1) with $a=6b$; where $\beta$ is given by (52) and
$\alpha,u_{1},\theta_{0}$ are arbitrary constants.

t=0 t=10 t=15

Figure 2: Movement of the wave (53) with
$\alpha=\gamma=k=1,b=1,u_{1}=-1,\theta_{0}$

##  5 Solution by Tanh method:

To solve equation (1) by Tanh method, we transform the partial differential
equation (1) to the following ordinary differential equation by using the
transformation $u(x,y,t)=f(z)$ where $z=x-\lambda y-\omega t$:

$\displaystyle(1-\omega+k\lambda^{2})f^{\prime\prime}+a(f^{2})^{\prime\prime}-b\omega
f^{iv}=0$ $\displaystyle\Rightarrow$
$\displaystyle(1-\omega+k\lambda^{2})f+af^{2}-b\omega f^{\prime\prime}=0.$
(54)

(integrating twice and taking the integration constant 0).
Let us impose the following boundary condition on (54)

$\displaystyle f(z)\rightarrow 0\hskip 14.22636ptand\hskip
14.22636ptf^{n}(z)\rightarrow 0\hskip 14.22636pt(n\in\mathbb{N})\hskip
14.22636ptas\hskip 14.22636ptz\rightarrow\pm\infty,$ (55)

$f^{n}(z)$ denotes the $n$-th derivative of $f$ with respect to $z$.

For finding the solution in Tanh method we suppose that

$\displaystyle u(x,y,t)=f(z)=S(Y)=\Sigma_{n=0}^{J}c_{n}Y^{n},$ (56)

where $Y=tanh\hskip 2.84526ptz=tanh\hskip 2.84526pt(x-\lambda y-\omega t)$.
Then

$\displaystyle
f^{\prime}(z)=\frac{df}{dz}=\frac{df}{dY}\frac{dY}{dz}=(1-Y^{2})\frac{dS}{dY},$
(57) $\displaystyle
f^{\prime\prime}(z)=\frac{d^{2}f}{dz^{2}}=\frac{d}{dz}\left(\frac{df}{dz}\right)=\frac{d}{dY}\left((1-Y^{2})\frac{dS}{dY}\right)\frac{dY}{dz}$
$\displaystyle\hskip
31.29802pt=-2Y(1-Y^{2})\frac{dS}{dY}+(1-Y^{2})^{2}\frac{d^{2}S}{dY^{2}}$ (58)

By using (57), (58) equation (54) becomes:

$\displaystyle(1-\omega+k\lambda^{2})S(Y)+aS^{2}(Y)+2b\omega
Y(1-Y^{2})\frac{dS}{dY}-b\omega(1-Y^{2})^{2}\frac{d^{2}S}{dY^{2}}=0$ (59)

By leading order analysis from (56) and (59) it is found that $J=2$.
The boundary conditions (55) becomes

$\displaystyle S(Y)\rightarrow 0\hskip 8.53581ptas\hskip
8.53581ptY\rightarrow\pm 1.$ (60)

Considering the case $Y\rightarrow 1$, the solution takes the following form

$\displaystyle S(Y)=d_{0}(1-Y)(1+d_{1}Y)$ (61)

We directly substitute (61) in (59) and make equal the coefficients of various
degrees of $Y$ to 0. A set of algebraic equations is thus obtained, solving
which we get

$\displaystyle d_{0}=-\frac{6b\omega}{a},\hskip 5.69054ptd_{1}=1,\hskip
5.69054pt\omega=\frac{1+k\lambda^{2}}{1+4b},\hskip
5.69054pt(b\neq-\frac{1}{4}).$ (62)

Thus from (56) and (61) we obtain a solitary wave solution of (1) with
$b\neq-\frac{1}{4}$ as

$\displaystyle
u(x,y,t)=-\frac{6b(1+k\lambda^{2})}{a(1+4b)}Sech^{2}\left(x-\lambda
y-\frac{1+k\lambda^{2}}{1+4b}t\right),$ (63)

where $\lambda$ is an arbitrary constant.

The amplitude decreases with respect to the nonlinear coefficient $a$, though
$a$ has no effect on the velocity. The amplitude increases and the velocity
decreases with respect to the dispersion coefficient $b$. That is the waves
with greater amplitude moves slower than the waves with less amplitude. Though
amplitude increases with respect to $b$, it has an upper limit
$\frac{3(1+k\lambda^{2})}{2a}$ as $b\rightarrow\infty$.

Figure 3: Effect of the dispersion coefficient $b$ on the amplitude and
velocity of the solitary wave (63), taking $k=1,\lambda=1,a=-1,y=0$.

t=0 t=10 t=15

Figure 4: Movement of the solitary wave (63) with $\lambda=b=k=1,a=-1$

##  6 Conclusion:

This research article is aimed to study the Painlevé property, Lie point
symmetry analysis and finding exact solutions of the KP-BBM equation. Painlevé
test is performed by using WTC method [6] and the equation is found to fail
the Painlevé test which indicates its nonintegrability. The Lie point symmetry
generators are obtained explicitly and the obtained Lie algebra is found to be
solvable. The general adjoint transformation matrix, invariants of the Lie
algebra are obtained and one dimensional optimal system for subalgebras is
constructed by Hu-Li-Chen algorithm. Three similarity reductions are obtained
and solved analytically. We got three solitary wave solutions from these
similarity reductions and it is seen that the amplitudes of these solitary
waves are inversely proportional with the nonlinear parameter $a$ and the
dispersion parameter $b$ increases their widths. We successfully apply HB
method and $Tanh$ method to derive exact solutions, that once again prove the
efficiency of these methods. Solitary wave like solution are obtained by the
methods of HB and $Tanh$ and are sketched with suitable parametric values.
Figure 2 and Figure 4 depict the time evolution of the solitary waves in (53)
and (63). The dispersion parameter $b$ has an important effect on the solution
(63), interestingly it is noticed that the waves with greater amplitude moves
slower than the waves with less amplitude. The obtained solutions are also
checked by using Mathematica software and to the best of our knowledge they
are not reported earlier in any other research work. The newly obtained
solutions may shed light which are governed by KP-BBM equation.

Statements and Declarations:
Acknowledgement: The authors are grateful to NIT Durgapur, India for their
research support.
Data Availability Statement: The data that supports the findings of this study
are available within the article.
Conflict of Interest: The authors declare that they have no conflict of
interest.

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