Abstract— A brief report on the different methods for finding exact traveling wave solutions of nonlinear evolution equations is presented. In a very recently work it had been shown that most of methods that exist in the literature are equivalent to the “generalized mapping method” that unifies them. This method is extended here to find a class of formal exact solutions to Burgers equation with space-dependent-coefficients.
Index Term--
Exact solutions, Extended Unified method, Burgers equation, variable coefficients
1. INTRODUCTION We consider the equation the following equation
,
1
,
0
)
,...,
,
,
,
,
,
,
(
22 2
m
x
u
x
u
t
x
u
x
u
t
u
u
t
x
f
mm
(1)
where the function
f
is a polynomial in the arguments. When the equation (1) does not depend explicitly onx
and
t
, it can be reduced to a subclass of ordinary differential equations by using the Lie groups for partial differential equations [1] or by using similarity transformations. Among these equations, the traveling wave has the form,
,
,
0
)
,...,
,
,
(
( )z
x
c
t
dz
du
u
u
u
u
u
g
m
(2)
Which results due to the translational symmetry of (1). The Painlevé analysis is used to test the integrability of partial differential equations that had been developed in [2]. Auto-Bäcklund transformation deals with the exact solutions that were obtained for integrable forms of (2) by truncating the Painleve' expansion [3-9]. Recently Auto-Bäcklund transformation that was extrapolated in [10-14] and the homogenous balance method in [15-19] assert a solution for evolution equations with variable coefficients in the form
), , ( ) ) ( ( )
,
( (0)
2 2
t x u a
x t x
u x
m m
Where
is the base function.2. EXTENDED UNIFIED METHOD
Explicit solutions of evolution equations of type (2) are, in fact, particular solutions. In this respect, these solutions are mapped
to other solutions that are given in terms of known elementary or special functions. Recently in [20] the class of these solutions was obtained by the generalized mapping method (GMM). This method generalizes the results as polynomial or rational function solutions. In the present paper, we extend this method to handle equations of type (1).
2.1- Polynomial Solutions
In this section, we search for polynomial solutions of
equations (2) in
)},
(
)
(
),
(
,
:
{
),
(
1
xk p x t
k q t S
P
P
K
R
R
S
R
C
11
0 0
)
,
(
)
,
(
)
(
),
,
(
)
,
(
)
(
k
i
s
i
i i
x k i
i t
k
b
x
t
x
t
P
c
x
t
x
t
P
.
Indeed the set
S
contains elementary or elliptic functions for some particular values ofq
,
p
,
k
andk
1. The mapping method asserts that there exists a positive integern
and a mapping
si
i i
S
S
t
x
t
x
a
v
v
R
C
M
0
}
),
,
(
)
,
(
,
{
,
)
(
:
Such that
M
(
u
)
P
n(
)
and satisfies the properties.
))
(
(
)
(
,
))
(
(
)
(
),
(
)
(
)
(
u
1u
2M
u
1M
u
2M
u
tM
u
tM
u
xM
u
xM
Thus
M
is a ring homomorphism that conserves differentiation. By the former conditions, we find that
(
)
,
(
)
(
)
)
(
( 1 1
( 1 )
x k n x t
k n
t
P
M
u
P
u
M
By using the properties of
M
and the last results and as,...)
,
,
,
(
x
t
u
u
tf
f
is a polynomial in its arguments, we find that
M
(
f
)
is a polynomial and there existss
0
s
such that
(
)
)
(
0
s
P
f
M
.It is worthy to notice that all theseExact Solutions of Burgers Equation with
Space Dependent Coefficients by the Extended
Unified Method
Nasser S. Elazab
Department of Mathematics, Faculty of Science, Cairo University, Giza- Egypt.
Nasser.Elazab@yahoo.com
,
)
(
)
(
)
(
1u
1 2u
2 1M
u
1 2M
u
2polynomials have different coefficients. More simply the mapping
M
assigns tou
andf
gives two auxiliary equations, the polynomialsP
n(
)
and(
)
0
s
P
respectively.In case of equations (1)
s
0
n
m
m
k
. The utility of theabove presentation helps us to give arguments to the statements of the conditions in lemmas 2.1 and 2.2. Also, we think that it allows for constructing more generalization and it is more appropriate when (1) is a vector equation.
We substitute for
m m
x t
x
u
u
u
u
,....,
,
,
as polynomials in
, sothat the function
f
is a polynomial in
, together with two auxiliary equations. In the applications we may write directly)
(
),
(
0
sn
f
P
P
u
. From the previous analysis we may write
ni
i i
x
t
a
u
0
)
,
(
,(3)
Where for instance, we assume that
k
1
k
, so that theauxiliary equations are
ki
i i
x k
i
i i
t
b
x
t
c
x
t
0 0
)
,
(
,
)
,
(
,(4)
Together with the compatibility equation
(5)
When substituting from (3) and (4) into (1) we find that it is transformed to ( )
(
)
0
0
f s
P
, that gives rise to
0
0 0
2 1
0
0
,
0
))
,
(
),
,
(
),
,
(
),
,
(
),
,
(
(
s
i
i r
r r
r r
i
a
x
t
b
x
t
c
x
t
a
x
t
a
x
t
h
x
t
(6)
n
r
0
0
,
1
,...,
andr
1,
r
2
0
,
1
,...,
k
By equating the coefficients of
i,i
0
,
1
,...,
s
0 to zero, we get a set of (s
0
1
) algebraic (or differential) equations,namely the principle equations, in the functions
a
i,
b
i,
c
i. On the other hand the equations that result from compatibility equation (5) count:2
k
1
,
k
2
.We mention that these later unknown functions count:
3
2
k
n
.In equation (1), if
u
ju
x andm m
x
u
are the highest nonlinear
and the highest order derivative terms respectively, then we get
the balancing conditions as
k
m
m
n
k
n
j
n
s
0
1
. Thus by solving forn
, we find that it depends onm
,
j
andk
. The last result and thenumber of compatibility equations namely
2
k
1
,
k
2
determine if the equations to be solved are over-determined or under-determined. The number of the determining equations,balances, namely
)
3
2
(
)
1
2
(
)
1
(
n
m
mk
k
n
k
is zero or greater than zero or less than zero respectively. From this last conditions we may determine a consistency condition that will be identified in the lemmas. In what follows necessary conditions for the existence of polynomial solutions will be stated.Lemma 2.1. For polynomial- Solutions of (1) (as a polynomial in
) to exist it is necessary that(i)
(
m
1
)(
k
1
)
/
j
(:
n
)
is a positive integer(ii)
m
(
k
1
)
3
m
when the equation (1) in the absence ofx
andt
passes the Painlevé test. Otherwisem
is replaced by 2.We notice that the first and the second conditions in lemma 2.1 are the balancing and the consistency conditions respectively. For details see [20]
2.2 – The rational function solutions
Here, also we search for solution of equation (1) in
C
S(
R
)
. For rational function solutions of equation (1), we consider the space of functions}
,
)
(
)
(
,
{
S
Q
Pn
v
v
r
R
andQ
r(
)
has nozeros in
K
R
.The definitions in the above and the GMM for rational function solutions assert that there exists a mapping
S
Q
P
u
M
R
C
M
r n R
R S
R
,
)
(
)
(
)
(
,
)
(
:
.
The properties of these mapping are the same properties of the mapping
M
(
u
)
in section 2.1. By bearing in mind these properties and from equations (4) and (5), we find that)
(
)
(
)
(
,
)
(
)
(
)
(
1) 1 ( , 2 2
) 1 ( , 1
mr r k n x
R r
r k n t
R
Q
P
u
M
Q
P
u
M
Thus,
i
m
x
u
M
i Ri
R
(
)
,
1
,
2
,....,
. By using the
properties of
M
R and the last results, we getM
R(
f
)
R and there exists
1
s
such that)
(
)
(
)
(
1 1
mr R s R
Q
P
f
M
. Indeeds
1 depends onm
k
r
n
,
,
,
, where in the case mentioned in the abover
m
k
m
m
n
s
1
. Simply, we write,
txxt
r i i i n i i id
a
u
0 0
. (7)So the equation (1) is transformed to
(
)
0
1R
s
P
.Equivalently, the last identity becomes
1 3 1 2 0 0,
0
))
,
(
),
,
(
),
,
(
),
,
(
(
s i i r r r ri
a
x
t
d
x
t
b
x
t
c
x
t
h
(8)
n
r
0
0
,
1
,...,
andr
1,
r
3
0
,
1
,...,
k
, alsor
2
0
,
1
,....,
r
.
In (8), by equating the coefficients of
i,i
0
,
1
,...,
s
1 to zero, we get a set of (s
1
1
) equations, that determine the functionsa
i,
b
i,
c
i andd
i. We mention that these later functions countn
2
k
r
3
. By using the same assumptions on equation (1) as in section 2.1, the balancing condition is
j
m
s
r
m
k
m
m
n
j
m
r
j
n
j
m
m
j
r
r
m
k
m
m
n
r
k
n
j
n
1
,
)
)
1
((
1
)),
1
(
(
1
1 (9)Now by solving (9) for
n
, we find that it depends onk
r
j
n
,
,
,
and, in both two cases, we get the same equation forr
n
. Hereafter, we distinguish between the two cases mentioned in (9).From the last results and whenj
m
1
, the number of the determining equations, balances the number of unknowns, is over-determined or is under-determined whenthe difference, namely
)
3
2
(
)
1
2
(
)
1
(
n
m
m
k
r
m
k
n
k
r
is zero or greater than zero or less than zero. But when1
m
j
this difference is)
3
2
(
)
1
2
(
)))
1
(
(
1
(
n
m
m
k
r
m
r
j
m
k
n
k
r
. From these last
Conditions, we may determine the consistency condition that will be identified in the following lemma.
Lemma 2.2. For solitary wave – rational solutions of equation (2) to exist it is necessary that
)
(:
/
)
1
)(
1
(
)
(
i
m
k
j
n
r
is an integer1
,2
2
)
1
(
)
2
(
1
,
3
)
1
(
)
1
(
)
(
ii
r
m
k
m
m
j
m
or
r
j
k
m
k
j
m
in the case when equation (1) passes the Painleve' test. For details see [20].
3. EXACT SOLUTIONS OF SPACE DEPENDENT BURGERS EQUATION
Here we extend the method to the variable coefficient Burgers equation
0
x
0,
,
0
)
(
)
(
f
x
u
g
x
u
u
t
u
t xx x , (10)Where
f
andg
are arbitrary functions inx
.We mention that (10) is a fundamental partial differential equations from fluid mechanics. It occurs in various areas of applied Mathematics, such as modeling of gas dynamics and traffic flow. In fact, differential equations with variable coefficients are more suitable to practical models. Many exact solutions are obtained when the coefficients are time dependent [21].
3.1 The polynomial function solutions
In what follows we shall derive a polynomial solution of equation (10) from the condition (i) in lemma 2.1, we have two cases:
n
1
whenk
2
andn
2
, whenk
3
, where in (ii), the consistency condition holds in these two cases.I. When
k
2
,
n
1
, by substituting into (3) and (4) and (10), we get six principle equations. We mention that calculations are carried out by using MATHEMATICA where standard functions in calculus and algebra were only needed. When solving the principle equations, namely those arising from (6) we get four equations, where these equations are solved explicitly to)
,
(
)
(
2
)
,
(
22
x
t
h
x
c
x
t
a
, (11))
(
)
,
(
)
(
)
(
)
,
(
)
,
(
)
,
(
)
,
(
)
,
(
0 2 1 2 22
x
r
t
x
c
x
h
x
h
t
x
c
t
x
c
t
x
c
t
x
a
t
x
b
x, (12)
)
13
(
))),
,
(
)
(
))
,
(
)
(
)(
,
(
2
)
,
(
)
(
2
))
,
(
)
(
)
,
(
)
,
(
)(
,
(
)
,
(
)
(
)
,
(
)
,
(
)(
(
))
,
(
)
(
)
,
(
)
,
(
)
(
)
,
(
)
,
(
)(
,
(
(
)
(
)
(
)
,
(
1
)
,
(
2 2 1 2 1 0 2 2 2 2 1 2 2 2 1 0 2 1t
x
c
x
h
t
x
c
x
h
t
x
c
t
x
c
x
h
t
x
c
x
h
t
x
a
t
x
h
t
x
c
t
x
c
x
r
t
x
c
t
x
c
x
h
t
x
c
x
h
t
x
h
t
x
c
x
h
t
x
c
t
x
c
t
x
a
x
r
x
h
t
x
c
t
x
b
xx x x x x x t x
where)
(
1
)
(
,
)
(
)
(
)
(
x
g
x
r
x
g
x
f
x
h
. Similar equation for)
,
(
0
x
t
b
holds but they are too lengthy to be written here. Hence all the principle equations solved. Now, we consider the compatibility equations
xt
tx which are given formally by.
0
)
,
(
)
,
(
)
,
(
)
,
(
)
,
(
)
,
(
,
0
)
,
(
)
,
(
)
,
(
)
,
(
2
)
,
(
)
,
(
2
,
0
)
,
(
)
,
(
)
,
(
)
,
(
)
,
(
)
,
(
2 2 2 1 1 2 1 1 0 2 2 0 0 0 0 1 1 0
t
x
b
t
x
c
t
x
c
t
x
b
t
x
c
t
x
b
t
x
b
t
x
c
t
x
c
t
x
b
t
x
c
t
x
b
t
x
b
t
x
c
t
x
c
t
x
b
t
x
c
t
x
b
x t x t x t (14)To simplify computations, we make the transformation
),
,
(
)
,
(
)
,
(
),
,
(
)
,
(
)
,
(
2 1 12
x
t
p
x
t
c
x
t
c
x
t
p
x
t
C
x
t
where
C
1(
x
,
t
)
is an arbitrary function. To evaluate)
,
(
0
x
t
a
we solve last equation in (12) to finda
0(
x
,
t
)
, hence)
)
(
)
(
(
)
(
)
(
)
)
(
)
(
)(
,
(
)(
(
)
,
(
10
x
r
x
h
x
h
x
r
x
r
x
h
t
x
C
x
h
t
x
a
,(15)
For convenience, we use the transformation
)
,
(
4
)
,
(
4
)
,
(
)
,
(
2
)
,
(
2
0 2
1 1
0
t
x
c
t
x
C
t
x
C
t
x
C
t
x
c
x
,(16)
As the computations are too lengthy in the general case, we consider a power law functions
h
(
x
)
h
0x
n,
r
(
x
)
r
0x
m, in the original variablem n
m
r
x
h
x
f
x
r
x
g
0 0
0
)
(
,
1
)
(
.Now, to find
C
0(
x
,
t
)
we use the following steps of computations:The second and first equation in (14) can be solved to find
)
,
(
),
,
(
00
x
t
C
x
t
C
x t respectively. CalculateC
0tx(
x
,
t
)
andbalance with
C
0xt(
x
,
t
)
, we get an equation, which can besolved to find
C
0(
x
,
t
)
, hence,
)
(
4
)
,
(
)
,
(
2 20
x
n
m
n
m
L
t
x
C
(17) Where
)
3
13
12
(
)
4
3
(
)
8
12
(
)
3
7
(
)
,
(
x
t
m
4m
3n
m
2n
n
2n
2n
n
2mn
n
n
2L
.
Balance
C
0x(
x
,
t
)
from equation (17) withC
0x from second equation in (14), we get an algebraic equation inm
,
n
. This equation is0
))
3
10
8
(
)
2
11
12
(
)
4
7
(
)(
1
(
n
m
3
m
2
n
m
n
n
2
n
n
n
2
m
, (18)
The solution of equation (18) leads to
2
5
8
4
2
,
3
4
m
m
m
2n
m
n
.In follows we fin the solution of equation (10):
Case (1) when
3
4
m
n
, by solving the first auxiliaryequation in (4) (we get
(
x
,
t
)
) and substituting into second auxiliary equation in (4) to find the arbitrary time dependentfunction of integration, we get
5
8
m
, so5 8
0 5
4
0
,
(
)
)
(
h
x
r
x
r
x
x
h
and in the original variable0 5 8
5 12
0 0
)
(
,
)
(
r
x
x
g
x
r
h
x
f
Finally we get
(
x
,
t
)
as)
,
(
)
4
)
5
2
(
25
(
10
)
,
(
)
4
)
5
2
(
25
(
5
28
)
35
6
(
25
)
,
(
2 5 7
0 5 2
0
1 5 7
0 5 2
0 5 2
0 5 2
0
t
x
c
x
x
t
h
x
r
t
x
C
x
t
h
x
x
r
x
t
h
x
r
t
x
, (19)
By substituting from (11-13) and the value of
a
0(
x
,
t
)
into (3) we get5 1 5 2
0 5 2
0
5 2
0 5
2
0 0
)
4
)
5
2
(
25
(
5
12
)
15
2
(
25
(
2
)
,
(
x
x
t
h
x
r
x
t
h
x
r
h
t
x
u
,(20)
It is worth noticing that one can verify that the solution given by (20) satisfies (10).
Case (2) when
2
5
8
4
2
m
m
m
2n
, we getinfinite number of solutions for each value of
m
.For simplicity, we take
m
7
hence2
305
5
n
in away similar to the above, we get
7 0 2
305 5
0
,
(
)
)
(
x
h
x
r
x
r
x
h
and in the original
variable
7 0 2
305 19
0
0
1
)
(
,
)
(
x
r
x
g
x
r
h
x
f
also the
auxiliary equations solve to
)
21
(
),
)
305
57
271
(
305
563
10725
(
2
1
(
)
305
23
(
2
)(
,
(
)
)
305
57
271
(
305
563
10725
(
2
1
(
)
305
11
197
(
2
0
)
305
25
[(
)
,
(
1
)
,
(
0 0 2
305
2 23
1
0 2
305
2 23
r
t
h
x
x
t
x
xC
r
t
h
x
x
t
x
Y
t
x
where
),
)
305
57
271
(
305
563
10725
(
2
1
(
)
305
23
(
2
)(
,
(
2
)
,
(
0 0 2
305 2
23
2
r
t
h
x
x
t
x
xc
t
x
Y
We get the solution of (10) as
)
,
(
)
,
(
)
,
(
t
x
M
t
x
W
t
x
u
,(22) where
))
)
305
113
1687
(
28
)
305
563
10725
((
)
305
1691
29885
(
56
(
14
)
,
(
2 305
2 23
0 0 0
x
x
r
x
t
h
h
t
x
W
))
)
305
335
9641
(
7
)
305
121
7473
((
)
305
7333
123395
(
14
(
)
,
(
2 305
2 23 2
305
0 0 2
7
x
x
x
r
t
h
x
t
x
M
Again, the solution (22) verifies the equation (10).
II. When
k
3
,
n
2
, the equations (3) and (4) becomes,
)
,
(
20
i
i i
x
t
a
u
30 3
0
)
,
(
,
)
,
(
i
i i
x i
i i
t
b
x
t
c
x
t
As the calculations above, we get two solutions for equation (10), the first solution is
0 0
2 3 0
2
1
2
)
,
(
r t h
e
x
x
h
t
x
u
,(23)
where,
2 0 2
0
0
,
(
)
1
)
(
x
r
x
g
x
r
h
x
f
.The second solution is
)
2
1
(
)
2
12
17
(
)
2
17
24
(
2
))
)
2
7
10
(
2
)
2
17
24
((
)
2
12
17
(
4
(
)
,
(
0 0
0 0 2
1 0
x
r
t
h
x
r
t
h
x
h
t
x
u
, (24)
where,
2 1 0 0
0
1
)
(
,
)
(
x
r
x
g
x
r
h
x
f
.4. CONCLUSION
In this paper, we suggested an extended unified method for finding exact solutions to evolution equations with variable coefficients. A wide class of exact solutions to Burgers equation with space-dependent-coefficients had been obtained. The method and the solutions that we obtain here are completely new.
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