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Appl. Appl. Math.
ISSN: 1932-9466
Applications and Applied
Mathematics:
An International Journal
(AAM)
Vol. 6, Issue 1 (June 2011) pp. 384 – 395
(Previously, Vol. 6, Issue 11, pp. 2125– 2136)
Exact Soliton Solutions for Second-Order Benjamin-Ono Equation
Nasir Taghizadeh, Mohammad Mirzazadeh and Foroozan Farahrooz
Department of Mathematics
Faculty of Mathematics
University of Guilan
P.O.Box 1914 Rasht, Iran
taghizadeh@guilan.ac.ir; mirzazadehs2@guilan.ac.ir; f.farahrooz@yahoo.com
Received: July 15, 2010; Accepted: February 3, 2011
Abstract
The homogeneous balance method is proposed for seeking the travelling wave solutions
of the second-order Benjamin-Ono equation. Many exact traveling wave solutions of
second-order Benjamin-Ono equation, which contain soliton like and periodic-like solutions are
successfully obtained. This method is straightforward and concise, and it may also be applied to
other nonlinear evolution equations.
Keywords: Homogeneous balance method; second-order Benjamin-Ono equation; Riccati
equation; Soliton-like solution; Periodic-like solution.
MSC 2000: 34B15; 47E05; 35G25.
1. Introduction
It is well known that the nonlinear partial differential equations (NPDEs) are widely used to
describe complex phenomena in various fields of sciences, such as physics, biology, chemistry,
etc. Exact solutions of these equations are therefore very important and significant in the
nonlinear sciences.
In recent years, Wang [(1995), (1996)] and Khalafallah (2009) presented a useful homogeneous
balance method for finding exact solutions of certain nonlinear partial differential equations. Fan
(2000) used the homogeneous balance method to search for the Backlund transformation and
similarity reductions of nonlinear partial differential equations.
The aim of this paper is to find exact soliton solutions of the second-order Benjamin-Ono
Equation, using the homogeneous balance method.
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2. Second-Order Benjamin-Ono Equation
For the second-order Benjamin-Ono equation [Hereman et al. (1986)]:
u tt   (u 2 ) xx   u xxxx  0,
(1)
where  and  are nonnegative constants,
Let us consider the traveling wave solutions
u (x , t )  u ( ),   k (x  lt )   0 ,
(2)
Where k , l and  0 are constants. Then Eq. (1) becomes
l 2u   2 (u ) 2  2uu    k 2u   0.
(3)
We now seek the solutions of Eq. (3) in the form
m
u   qi  i ,
(4)
i 0
where q i are constants to be determined later and

satisfy the following Riccati equation
   a 2  b  c
(5)
where a , b and c are constants.
Balancing the highest order derivative term with nonlinear term in Eq. (3) gives leading order
m  2. We may therefore choose
u  q 0  q1  q 2 2 ,
where q 0 , q1 and q 2 are constants to be determined and  satisfy Eq. (5).
Substituting (6) and (5) into Eq. (3), we have
(6)
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Nasir Taghizadeh et al.
l 2u   2 (u ) 2  2uu    k 2u   l 2 (q 0  q1  q 2 2 )
2 ((q 0  q1  q 2 2 )) 2  2 (q 0  q1  q 2 2 )(q 0  q1
q 2 2 )   k 2 (q 0  q1  q 2 2 )  {20 a 2q 22  120 k 2a 4q 2 } 6
{24 a 2q1q 2  24  k 2a 4q1  336  k 2a 3bq 2  36 abq 22 } 5
{6 a 2q12  60  k 2a 3bq1  (240  k 2ca 3  330  k 2b 2a 2  6l 2a 2 )q 2
12 a 2q 0q 2  42 abq1q 2  (32 ac  16b 2 )q 22 } 4  {10 abq12 
(40  k 2ca 3  2l 2a 2  50  k 2b 2a 2 )q1  4 a 2q 0q1  (36ca  18b 2 )q1q 2
 (10l 2ab  440  k 2bca 2  130  k 2b 3a )q 2  20 abq 0q 2  28bcq 22 } 3
{(8 ac  4b 2 )q12  (60  k 2bca 2  15 k 2b 3a  3l 2ba )q1  6baq 0q1
30bcq1q 2  (8l 2ac  4l 2b 2  136  k 2c 2a 2  232  k 2cb 2a  16  k 2b 4 )q 2
 (16 ac  8b 2 )q 0q 2  12ac 2q 22 } 2  {6cbq12  (16  k 2c 2a 2  2al 2c
22  ak 2b 2c   k 2b 4  l 2b 2 )q1  12c 2q1q 2  (4 ac  2b 2 )q 0q1
12bcq 0q 2  (6l 2bc  120  k 2bc 2a  30  k 2b 3c )q 2 }  2c 2q12
2bcq 0q1  (8 k 2ac 2b   k 2b 3c  l 2bc )q1  (16  k 2c 3a
2l 2c 2  14  k 2b 2c 2 )q 2  4c 2q 0q 2  0.
Setting the coefficients  i (i  0,1, 2,3, 4,5, 6) to zero yields the following set of algebraic
equations:
20 a 2q 22  120  k 2a 4q 2  0,
24 a 2q1q 2  24  k 2a 4q1  336  k 2a 3bq 2  36 abq 22  0,
6 a 2q12  60  k 2a 3bq1  (240 k 2ca 3  330 k 2b 2a 2  6l 2a 2 )q 2
12 a 2q 0q 2  42 abq1q 2  (32 ac  16b 2 )q 22  0,
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10 abq12  (40  k 2ca 3  2l 2a 2  50  k 2b 2a 2 )q1
4 a 2q 0q1  (36ca  18b 2 )q1q 2  (10l 2ab 
(7)
440  k 2bca 2  130  k 2b 3a )q 2  20 abq 0q 2
28bcq 22  0,
(8 ac  4b 2 )q12  (60  k 2bca 2  15 k 2b 3a  3l 2ba )q1
6baq 0q1  30bcq1q 2  (8l 2ac  4l 2b 2  136  k 2c 2a 2
232  k 2cb 2a  16  k 2b 4 )q 2  (16 ac  8b 2 )q 0q 2
12ac 2q 22  0,
6cbq12  (16 k 2c 2a 2  2al 2c
22 ak 2b 2c   k 2b 4  l 2b 2 )q1  12c 2q1q 2  (4 ac  2b 2 )q 0q1
12bcq 0q 2  (6l 2bc  120 k 2bc 2a  30 k 2b 3c )q 2  0,
2c 2q12  2bcq 0q1  (8 k 2ac 2b   k 2b 3c  l 2bc )q1
(16  k 2c 3a  2l 2c 2  14  k 2b 2c 2 )q 2  4c 2q 0q 2  0.
For which, with the aid of Maple, we obtain the following solution of the above set of
algebraic equations:
q0  
 k 2b 2  8 k 2ca  l 2
,
2
q1  
6  k 2ab

,
q2  
6  k 2a 2

.
(8)
For the Riccati equation (5), we can solve it by using the homogeneous balance method as
follows:
m
Case: I. Let    bi tanh i  . Balancing   with  2 leads to
i 0
  b0  b1 tanh  .
Substituting (9) into (5), we obtain the following solution of (5):
b2
1
   (b  2 tanh  ), ac   1.
4
2a
(9)
(10)
From (8), (10) and (6), we have the following traveling wave solution of second-order BenjaminOno equation (1):
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u (x , t ) 
2  k 2b 2  8 k 2ca  l 2 6k 2 

tanh 2 [k (x  lt )   0 ],
2

(11)
where
b2
ac 
 1.
4
m
Similarly, let    bi coth i  , then we obtain the following new traveling wave soliton
i 0
solutions of second-order Benjamin-Ono equation (1):
2  k 2b 2  8 k 2ca  l 2 6k 2 

coth 2 [k (x  lt )  0 ],
u (x , t ) 
2

(12)
where
ac 
b2
 1.
4
Case: II. From (Zhao, X. Q.; Tang, D. B.; 2002), when a=1, b=0, the Riccati Eq. (5) has the
following solutions:
 =- -c tanh( c  ), c  0,
1
   , c  0,

(13)
  c tan( c  ), c  0.
From (6), (8) and (13), we have the following traveling wave solutions of second-order
Benjamin-Ono equation (1) :
When c  0, we have
u (x , t )  

6 k 2a 2c

 k 2b 2  8 k 2ca  l 2 6  k 2ab c

tanh[ c (k (x  lt )   0 )]

2
tanh [ c (k (x  lt )   0 )].
2
When c  0, we have
(14)
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u (x , t )  
 k 2b 2  8 k 2ca  l 2
6  k 2ab

 (k (x  lt )  0 )
2
(15)
6 k 2a 2
.

 (k (x  lt )  0 ) 2
When c  0, we have
u (x , t )  

 k 2b 2  8 k 2ca  l 2 6 k 2ab c

tan[ c (k (x  lt )  0 )]
2

6 k 2a 2c

(16)
tan [ c (k (x  lt )  0 )].
2
Case: III. We suppose that the Riccati equation (5) has the following solutions of the form
m
  A 0   (A i f i  B i f
i 1
(17)
g ),
i 1
with
f 
1
sinh 
, g
,
cosh   r
cosh   r
which satisfy
f ( )  f ( ) g ( ), g ( )  1  g 2 ( )  rf ( ),
g 2 ( )  1  2rf ( )  (r 2  1)f 2 ( ).
Balancing   with  2 leads to
  A 0  A1f  B1 g .
(18)
Substituting (18) into (5), collecting the coefficient of the same power
f i ( ) g i ( )(i  0,1, 2; j  0,1) and setting each of the obtained coefficients to zero yield the
following set of algebra equations
aA 02  aB 12  bA 0  c  0,
2aA 0 A1  2arB 12  rB 1  bA1  0,
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aA12  a ( r 2  1)B 12  ( r 2  1) B 1  0,
2aA 0 B 1  bB 1  0,
2aA1B 1  A1  0,
which have solutions
A0  
b
,
2a
A1  
(r 2  1)
,
4a 2
B1  
1
,
2a
c
b 2 1
.
4a
(19)
From (18), (19) we obtain
sinh   (r 2  1)
1
   (b 
).
2a
cosh   r
(20)
Also from (6), (8) and (20), we obtain the new solutions of second-order Benjamin-Ono equation
(1):
u (x , t ) 
2 k 2b 2  8 k 2ca  l 2
2
(21)
3k 2  sinh(k (x  lt )  0 )  (r 2  1) 2

(
).
2
cosh(k (x  lt )  0 )  r
Case: IV. We take  in the Riccati equation (5) being of the form
  e p   (z )  p 4 ( ),
(22)
1
where
z  e p 2  p 3 ,
where p1 , p 2 and p 3 are constants to be determined.
Substituting (22) into (5) we find that when c 
 
p1e p1
p b
 1
.
p1
2a
a (e  p 3 )
If p 3  1 in (23), we have
 p12  b 2
, we have
4a
(23)
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391
 
p1
1
b
tanh( p1 )  .
2a
2
2a
(24)
If p 3  1 in (23), we have
 
p1
1
b
coth( p1 )  .
2a
2
2a
(25)
From (6), (8) and (23), we obtain the following new traveling wave solutions of second-order
Benjamin-Ono equation (1):
 k 2b 2  8 k 2ca  l 2
u (x , t )  
2
p ( k ( x  lt )  )
2
pe
p b
p e p ( k ( x  lt )  )
p b
6 k
( p 1( k ( x  lt )  )
)( p 1( k ( x  lt )  )
).

 1
 1
 e
2
2
e
 p3
 p3
1
1
0
0
1
1
0
(26)
0
When p3 = 1, we obtain the following traveling wave (soliton-like) solutions of second-order
Benjamin-Ono equation (1):
u (x , t ) 
p
2 k 2b 2  8 k 2ca  l 2 3k 2  p12

tanh 2 [ 1 (k (x  lt )  0 )].
2
2
2
(27)
When p 3  1, we have the following traveling wave (periodic-like) solutions of equal width
wave equation (1):
p
2 k 2b 2  8 k 2ca  l 2 3k 2  p12

(28)
u (x , t ) 
coth 2 [ 1 (k (x  lt )  0 )].
2
2
2
Case: V. We suppose that the Riccati equation (5) have the following solutions of the form:
m
  A 0   sinh i 1 (A i sinh w  B i cosh w ),
(29)
i 1
dw
dw
 sinh w or
 coshw . It is easy to find that m  1 by balancing   and  2 .
d
d
So we choose
where
  A 0  A1 sinh w  B 1 coshw ,
(30)
dw
dw
 sinh w , we substitute (30) and
 sinhw , into (5) and set the coefficients of
d
d
sinh i w cosh i w (i  0,1, 2; j  0,1) to zero. A set of algebraic equations is obtained as follows:
when
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aA 02  aB 12  bA 0  c  0,
2aA 0 A1  bA1  0,
aA12  aB 12  B 1  0,
2aA 0 B 1  bB 1  0,
2aA1B 1  A1  0,
for which, we have the following solutions:
A0  
b
,
2a
1
B1  ,
a
A1  0,
(31)
where
c
b2 4
,
4a
and
A0  
b
,
2a
A1  
1
,
2a
B1 
1
,
2a
(32)
where
b 2 1
c
.
4a
From
dw
 sinh w , we have
d
sinh w   csc h  , cosh w   coth  .
(33)
Also (31) − (33), give
 
where
b  2 coth 
,
2a
(34)
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c
b2 4
,
4a
and
 
b  csc h   coth 
,
2a
(35)
where
c
b 2 1
.
4a
From (6), (8), (34) and (35) we get the following traveling wave solutions of second-order
Benjamin-Ono equation (1):
u (x , t ) 
2 k 2b 2  8 k 2ca  l 2 6k 2 

coth 2 [k (x  lt )  0 ],
2

(36)
where
ac 
b2
 1.
4
and
u (x , t ) 
2 k 2b 2  8 k 2ca  l 2
2
3k 2 

(coth(k (x  lt )  0 )  csc h (k (x  lt )   0 )) 2 ,
2
(37)
where
c
b 2 1
.
4a
dw
 coshw , we obtain the following traveling wave (periodic-like)
d
solutions of second-order Benjamin-Ono equation (1):
Similarly, when
u (x , t ) 
2 k 2b 2  8 k 2ca  l 2 6k 2 

cot 2 [k (x  lt )  0 ],
2

(38)
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Nasir Taghizadeh et al.
where
ac 
b2
 1.
4
and
u (x , t ) 
2 k 2b 2  8 k 2ca  l 2
2
3k 2 

(cot(k (x  lt )  0 )  csc(k (x  lt )  0 ))2 ,
2
(39)
where
c
b 2 1
.
4a
In summary we have used the homogeneous balance method to obtain many traveling wave
solutions of second-order Benjamin-Ono equation.
We now summarize the key steps as follows:
Step 1: For a given nonlinear evolution equation
F (u ,u t ,u x ,u xt ,u tt ,...)  0,
(40)
we consider its traveling wave solutions u (x , t )  u ( ),   k ( x  lt )  0 then Eq. (40) is
reduced to an nonlinear ordinary differential equation
Q (u , u , u , u ,...)  0,
(41)
d
.
d
Step 2: For a given ansatz equation (for example, the ansatz equation is    a 2  b  c in this
paper), the form of u is decided and the homogeneous balance method is used on Eq. (41) to
find the coefficients of u .
where a prime denotes
Step 3: The homogeneous balance method is used to solve the ansatz equation.
Step 4: Finally, the traveling wave solutions of Eq. (40) are obtained by combining steps 2 and
3.
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3. Conclusion
The second-order Benjamin-Ono equation is soluble using the homogeneous balance method.
The efficiency of this method was demonstrated. New exact solution of the second-order
Benjamin-Ono equation was obtained. The solutions obtained may be significant and important
for the explanation of some practical physical problems. The method may also be applied to
other nonlinear partial differential equations.
REFERENCES
Fan, E.G. (2000). Two new applications of the homogeneous balance method. Phys. Lett. A.
Vol. 265, pp. 353-357.
Hereman, W., Banerjee, P. P., Korpel, A., Assanto, G., Van Immerzeele, A. and Meerpole, A.
(1986). Exact solitary wave solutions of nonlinear evolution and wave equations using a
direct algebraic method. J. Phys. A. Math. Gen. Vol. 19, No. 5, PP. 607-628.
Khalafallah, M. (2009). New exact traveling wave solutions of the (3+1) dimensional
Kadomtsev-Petviashvili (KP) equation. Commun. Nonlinear Sci. Numer. Simul. Vol. 14,
pp. 1169-1175.
Khalafallah, M. (2009). Exact traveling wave solutions of the Boussinesq-Burger equation.
Math. Comput. Model. Vol. 49, pp. 666-671.
Wang, M.L. (1995). Solitary wave solution for variant Boussinesq equation. Phys. Lett. A. Vol.
199, pp. 169-172.
Wang, M.L. (1996). Application of homogeneous balance method to exact solutions of nonlinear
equation in mathematical physics. Phys. Lett. A. Vol. 216, pp. 67-75.
Zhao, X.Q. and Tang, D.B. (2002). A new note on a homogeneous balance method. Phys. Lett.
A. Vol. 297, pp. 59-67.