Download Double Exp-Function Method for Multisoliton Solutions of The

Acta Mathematicae Applicatae Sinica, English Series
Vol. 32, No. 2 (2016) 461–468
DOI: 10.1007/s10255-016-0572-y
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Acta Mathemacae Applicatae Sinica,
English Series
The Editorial Office of AMAS &
Springer-Verlag Berlin Heidelberg 2016
Double Exp-Function Method for Multisoliton Solutions
of The Tzitzeica-Dodd-Bullough Equation
Alaattin ESEN, N. Murat YAGMURLU† , Orkun TASBOZAN
Department of Mathematics, Faculty of Science and Art, Inönü University, Malatya 44280, Turkey
(† E-mail: [email protected])
Abstract In this work, it is aimed to find one- and two-soliton solutions to nonlinear Tzitzeica-Dodd-Bullough
(TDB) equation. Since the double exp-function method has been widely used to solve several nonlinear evolution
equations in mathematical physics, we have also used it with the help of symbolic computation for solving the
present equation. The method seems to be easier and more accurate thanks to the recent developments in the
field of symbolic computation.
Keywords
double exp-function method; one-soliton; two-soliton; Tzitzeica-Dodd-Bullough equation; solitary
waves
2000 MR Subject Classification
1
35Q51; 74J35; 33F10
Introduction
The travelling wave solutions of nonlinear partial differential equations and their investigation
have an important place in the study of nonlinear physical phnomena. Numerical modellings
of many phenomena in scientific and engineering fields such as fluid mechanics, plasma physcis,
optical fibers, biology, solid state physics, chemical kinematics, chemical physics, and geochemistry result in nonlinear phenomena. Thus, obtaining their exact and numerical solutions have
become very important in recent years. Many authors have used many different approaches
and techniques to develop nonlinear dispersive and dissipative problems. They include the
inverse scattering transform method[2] , the Bäcklund transform[21] , Darboux transforms[6,16],
the parameter-expansion method[17,22] , the tanh method[15] , extended tanh method[4,7] , sinecosine method[23] , exp-function method[5,11] , F -expansion method[14,24] and (G′ /G)-expansion
method[13,18] .
In the present study, double exp-function method has been proposed and succesfully applied
to find the multisoliton solutions of nonlinear the TDB equation.
2
The Double-Exp Function Method
The exp-function method first introduced by Wu and He in order to solve differential equations[19] ,
and it has been systematically investigated by many authors. Then multiple exp-function was
first proposed by He[12] and used by many authors. Later, the double exp-function method and
its application have been presented by Fu and Dai[8] . The method has been applied by many
authors to solve different kinds of equations.
In this study, the general nonlinear PDE given in the following form
P (u, ut , ux , utt , utx , uxx , · · ·) = 0
Manuscript received October 19, 2011. Revised November 5, 2012.
(1)
462
A. ESEN, N.M. YAGMURLU, O. TASBOZAN
will be considered. It is assumed that the two-wave solution to Eq.(1) can be stated in terms
of fractional form as
f (x, t)
,
(2)
u=
g(x, t)
where f (x, t) and g(x, t) are ansatz functions of the two-soliton forms
f (x, t) = 1 + eξ + eη + A1 eξ+η ,
g(x, t) = 1 + eξ + eη + A2 eξ+η .
(3)
(4)
If we put Eqs.(3) and (4) into Eq.(2), then we easily obtain
u=
f (x, t)
e−(ξ+η)/2 + e(ξ−η)/2 + e(η−ξ)/2 + A1 e(ξ+η)/2
= −(ξ+η)/2
.
g(x, t)
e
+ e(ξ−η)/2 + e(η−ξ)/2 + A1 e(ξ+η)/2
This newly obtained equation is a special case of the general ansatz of the exp-function
method with N wave velocities and N frequencies, which was proposed by He[9] and then
was used for the search for the solitary solutions of nonlinear differential equations, nonlinear
differential-difference equations, and nonlinear fractional differential equations[21] . Particularly,
a two-soliton ansatz of the exp-function method can be defined as follows
u=
a−2 e−η + a−1 e−ξ + a0 + a1 eξ + a2 eη
,
b−2 e−η + b−1 e−ξ + b0 + b1 eξ + b2 eη
where ξ = c1 x + c2 t, η = c3 x + c4 t and ai , bj (−2 ≤ i, j ≤ 2) are constants and will be
determined.
3
The TBD Equation
In this section, we consider the TBD equation[22]
uxt = e−u + e−2u .
(5)
The equation plays an important role in many scientific applications such as solid state physics,
nonlinear optics, dusty plasma, plasma physics, fluid dynamics, mathematical biology, dislocations in crystals, kink dynamics, chemical kinetics and quantum field theory[3] . The equation
has been solved by many authors using different methods and techniques. Among others,
Wazwaz[20] has applied tanh method to the Dod-Bullough-Mikhailov and the Tzitzeica-DodBullough equations to derive solitons and periodic solutions for these equations and Abazari[1]
has applied (G′ /G)-expansion method for constructing more general exact solutions of the TDB
equation. Using the transformation
v(x, t) = e−u
Eq.(5) becomes
−vvxt + vx vt − v 3 − v 4 = 0.
(6)
Applying double exp-function method, we assume that
v=
a1 eξ + a2 e−ξ + a5 + a3 eη + a4 e−η
,
k1 eξ + k2 e−ξ + k5 + k3 eη + k4 e−η
(7)
where ξ = c1 x + c2 t and η = c3 x + c4 t. If we substitute (7) into (6) and set all of the coefficients
of e(iξ+jη) to zero, then we result in a system of algebraic equations. If this system is solved
with the help of any mathematical software, many abundant new solutions are obtained. After
Double Exp-Function Method for Multisoliton Solutions of The Tzitzeica-Dodd-Bullough Equation
463
some simple computations, we finally obtained one- and two-soliton solutions for 22 different
cases.
(I) One-soliton Solutions:
Case 1.
1
a3 = a4 = 0, a1 = −k1 , a2 = −k2 , a5 = −k5 , k3 = 0, c1 = c2 = 0, c3 = − ,
c4
k1 + k2 + k5
.
u1 (x, t) = − ln −
k1 + k2 + k5 + k4 e−(c3 x+c4 t)
(8)
If k1 + k2 + k5 = 1 and k4 = 1 are taken, using Eq.(8) results in the following solitary wave
solution
1
1
1
u(x, t) = − ln
tanh − (c3 x + c4 t) −
.
2
2
2
If k1 + k2 + k5 = 1 and k4 = −1 are taken, using Eq.(8) results in the following solitary
wave solution
1
1
1
u(x, t) = − ln
coth − (c3 x + c4 t) −
.
2
2
2
Case 2.
a3 = a4 = 0, a1 = −k1 , a2 = −k2 , a5 = −k5 , k4 = 0, c1 = c2 = 0, c3 = −
u2 (x, t) = − ln
1
,
c4
−(k1 + k2 + k5 )
.
k1 + k2 + k5 + k3 e(c3 x+c4 t)
(9)
If k1 + k2 + k5 = 1 and k3 = 1 are taken, using Eq.(9) results in the following solitary wave
solution
1
1
1
u(x, t) = − ln
tanh
(c3 x + c4 t) −
.
2
2
2
If k1 + k2 + k5 = 1 and k3 = −1 are taken, using Eq.(10) results in the following solitary
wave solution
1
1
1
u(x, t) = − ln
coth
(c3 x + c4 t) −
.
2
2
2
Case 3.
a2 = a4 = a5 = 0, a1 = −k1 , a3 = −k3 , k5 = 0, c1 = c3 = −
u3 (x, t) = − ln
1
, c2 = c 4 ,
4c4
−k1 e(c1 x+c2 t) − k3 e(c3 x+c4 t)
.
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + k3 e(c3 x+c4 t) + k4 e−(c3 x+c4 t)
(10)
If k1 + k3 = 1 and k2 + k4 = 1 are taken, using Eq.(10) results in the following solitary wave
solution
1
1
u(x, t) = − ln
tanh(−(c3 x + c4 t)) −
.
2
2
If k1 + k3 = 1 and k2 + k4 = −1 are taken, using Eq.(10) results in the following solitary
wave solution
1
1
u(x, t) = − ln
coth(−(c3 x + c4 t)) −
.
2
2
Case 4.
1
1
, c2 = −c4 , c3 = −
,
4c4
4c4
−k1 e(c1 x+c2 t) − k4 e−(c3 x+c4 t)
u4 (x, t) = − ln
.
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + k3 e(c3 x+c4 t) + k4 e−(c3 x+c4 t)
a2 = a3 = a5 = 0, a1 = −k1 , a4 = −k4 , k5 = 0, c1 =
(11)
464
A. ESEN, N.M. YAGMURLU, O. TASBOZAN
If k1 + k4 = 1 and k2 + k3 = 1 are taken, using Eq.(11) results in the following solitary wave
solution
1
1
tanh(c3 x + c4 t) −
.
u(x, t) = − ln
2
2
If k1 + k4 = 1 and k2 + k3 = −1 are taken, using Eq.(11) results in the following solitary
wave solution
1
1
u(x, t) = − ln
coth(c3 x + c4 t) −
.
2
2
Case 5.
c1 c4 − 1
k1 k2
, c3 = 0, c2 =
,
k4
c1
−k1 e(c1 x+c2 t) − k4 e−c4 t
u5 (x, t) = − ln
.
(12)
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + kk1 k4 2 ec4 t + k4 e−c4 t
a2 = a3 = a5 = 0, a1 = −k1 , a4 = −k4 , k5 = 0, k3 =
If k4 = 1 and k2 = 1 are taken, using Eq.(12) results in the following solitary wave solution
u(x, t) = − ln
1
1
1
tanh − (c1 x + (c2 − c4 )t) −
.
2
2
2
If k4 = 1 and k2 = −1 are taken, using Eq.(12) results in the following solitary wave solution
u(x, t) = − ln
1
1
1
coth − (c1 x + (c2 − c4 )t) −
.
2
2
2
Case 6.
1 + c1 c4
k1 k2
, c3 = 0, c2 = −
,
k4
c1
−k2 e−(c1 x+c2 t) − k4 e−c4 t
u6 (x, t) = − ln
.
(13)
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + k3 ec4 t + k4 e−c4 t
a1 = a3 = a5 = 0, a2 = −k2 , a4 = −k4 , k5 = 0, k3 =
If k4 = 1 and k1 = 1 are taken, using Eq.(13) results in the following solitary wave solution
u(x, t) = − ln
1
2
tanh
1
1
(c1 x + (c2 + c4 )t) −
.
2
2
If k4 = 1 and k1 = −1 are taken, using Eq.(13) results in the following solitary wave solution
u(x, t) = − ln
1
2
coth
1
1
(c1 x + (c2 + c4 )t) −
.
2
2
Case 7.
a2 = a3 = a5 = 0, a1 = −k1 , a4 = −k4 , k5 = 0, k3 =
u7 (x, t) = − ln
(c1 x+c2 t)
−c3 x
k1 k2
c2 c3 − 1
, c4 = 0, c1 =
,
k4
c2
−k1 e
− k4 e
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + kk1 k4 2 ec3 x + k4 e−c3
.
x
(14)
(15)
If k4 = 1 and k2 = 1 are taken, using Eq.(15) results in the following solitary wave solution
u(x, t) = − ln
1
2
tanh
1
1
((c3 − c1 )x − c2 t) −
.
2
2
Double Exp-Function Method for Multisoliton Solutions of The Tzitzeica-Dodd-Bullough Equation
465
If k4 = 1 and k2 = −1 are taken, using Eq.(15) results in the following solitary wave solution
u(x, t) = − ln
1
2
coth
1
1
((c3 − c1 )x − c2 t) −
.
2
2
Case 8.
a1 = a2 = a3 = 0,
u8 (x, t) = − ln
a4 = −k4 ,
k1 = k2 = 0,
k5 = −
k4 k3 + a25
,
a5
c3 = −
1
,
c4
a5 − k4 e−(c3 x+c4 t)
.
(c
x+c
t)
−(c
x+c
t)
k5 + k3 e 3 4 + k4 e 3 4
(16)
If a5 = −1 and k1 = 1 are taken, using Eq.(16) results in the following solitary wave solution
u(x, t) = − ln
1
2
tanh
1
1
(c3 x + c4 t) −
.
2
2
If a5 = −1 and k1 = −1 are taken, using Eq.(16) results in the following solitary wave
solution
1
1
1
coth
(c3 x + c4 t) −
.
u(x, t) = − ln
2
2
2
Case 9.
a2 = a3 = a4 = 0,
u9 (x, t) = − ln
a1 = −k1 ,
k3 = k4 = 0,
k5 = −
k1 k2 + a25
,
a5
c1 = −
1
,
c2
−k1 e(c1 x+c2 t) + a5
.
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + k5
(17)
If a5 = −1 and k2 = 1 are taken, using Eq.(17) results in the following solitary wave solution
u(x, t) = − ln
1
1
1
tanh − (c1 x + c2 t) −
.
2
2
2
If a5 = −1 and k2 = −1 are taken, using Eq.(17) results in
solution
1
1
u(x, t) = − ln
coth − (c1 x + c2 t) −
2
2
the following solitary wave
1
.
2
Case 10.
a2 (1 + c1 c2 )
a1 = a2 = a3 = a4 = 0, k3 = k4 = 0, k1 = 5 2 2
,
4c1 c2 k2
a5
u10 (x, t) = − ln
.
(c
x+c
t)
−(c
x+c
t)
k1 e 1 2 + k2 e 1 2 + k5
k5 =
a5
,
c1 c2
(II) Two-soliton Solutions
Case 11.
a1 = a3 = a4 = a5 = 0,
u11 (x, t) = − ln
a2 = −k2 ,
k1 = k4 = k5 = 0,
−k2 e−(c1 x+c2 t)
.
k2 e−(c1 x+c2 t) + k3 e(c3 x+c4 t)
c1 = −
1 + c2 c3 + c3 c4
,
c2 + c4
466
A. ESEN, N.M. YAGMURLU, O. TASBOZAN
Case 12.
a1 = a3 = a4 = a5 = 0,
u12 (x, t) = − ln
a2 = −k2 ,
k1 = k3 = k5 = 0,
c1 =
−1 + c2 c3 − c3 c4
,
c2 − c4
c1 =
−1 + c2 c3 − c3 c4
,
c2 − c4
−k2 e−(c1 x+c2 t)
.
k2 e−(c1 x+c2 t) + k4 e−(c3 x+c4 t)
Case 13.
a1 = a2 = a4 = a5 = 0,
u13 (x, t) = − ln
a3 = −k3 ,
k2 = k4 = k5 = 0,
−k3 e(c3 x+c4 t)
.
k1 e(c1 x+c2 t) + k3 e(c3 x+c4 t)
Case 14.
a2 = a3 = a4 = a5 = 0,
u14 (x, t) = − ln
a1 = −k1 ,
k2 = k3 = k5 = 0,
c1 = −
1 + c2 c3 + c3 c4
,
c2 + c4
−k1 e(c1 x+c2 t)
.
k1 e(c1 x+c2 t) + k4 e−(c3 x+c4 t)
Case 15.
a2 = a3 = a4 = a5 = 0,
u15 (x, t) = − ln
a1 = −k1 ,
k2 = k4 = k5 = 0,
c1 =
−1 + c2 c3 − c3 c4
,
c2 − c4
−k1 e(c1 x+c2 t)
.
k1 e(c1 x+c2 t) + k3 e(c3 x+c4 t)
Case 16.
k3 k4
k3 k4
1 + c1 c2 − c1 c4
, a4 = −k4 , k5 = 0, k1 =
, c3 =
,
k2
k2
c2 − c4
a1 e(c1 x+c2 t) − k4 e−(c3 x+c4 t)
u16 (x, t) = − ln
.
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + k3 e(c3 x+c4 t) + k4 e−(c3 x+c4 t)
a2 = a3 = a5 = 0, a1 = −
Case 17.
a1 = a2 = a4 = a5 = 0,
u17 (x, t) = − ln
a3 = −k3 ,
k1 = k4 = k5 = 0,
c1 = −
1 + c2 c3 + c3 c4
,
c2 + c4
−k3 e(c3 x+c4 t)
.
k2 e−(c1 x+c2 t) + k3 e(c3 x+c4 t)
Case 18.
k3 k4
1 + c1 c2 − c1 c4
k3 k4
, a3 = −k3 , k5 = 0, k2 =
, c3 =
,
k1
k1
c2 − c4
a2 e−(c1 x+c2 t) − k3 e(c3 x+c4 t)
u18 (x, t) = − ln
.
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + k3 e(c3 x+c4 t) + k4 e−(c3 x+c4 t)
a1 = a4 = a5 = 0, a2 = −
Case 19.
k3 k4
k3 k4
1 + c1 c2 + c1 c4
, a4 = −k4 , k5 = 0, k2 =
, c3 = −
,
k1
k1
c2 + c4
a2 e−(c1 x+c2 t) − k4 e−(c3 x+c4 t)
u19 (x, t) = − ln
.
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + k3 e(c3 x+c4 t) + k4 e−(c3 x+c4 t)
a1 = a3 = a5 = 0, a2 = −
467
Double Exp-Function Method for Multisoliton Solutions of The Tzitzeica-Dodd-Bullough Equation
Case 20.
a1 = a2 = a3 = a5 = 0,
u20 (x, t) = − ln
a4 = −k4 ,
k2 = k3 = k5 = 0,
c1 = −
1 + c2 c3 + c3 c4
,
c2 + c4
−k4 e−(c3 x+c4 t)
.
k1 e(c1 x+c2 t) + k4 e−(c3 x+c4 t)
Case 21.
k3 k4
k3 k4
1 + c1 c2 + c1 c4
, a3 = −k3 , k5 = 0, k1 =
, c3 = −
,
k2
k2
c2 + c4
a1 e(c1 x+c2 t) − k3 e(c3 x+c4 t)
u21 (x, t) = − ln
.
k1 e(c1 x+c2 t) + k2 e−(c1 x+c2 t) + k3 e(c3 x+c4 t) + k4 e−(c3 x+c4 t)
a2 = a4 = a5 = 0, a1 = −
Case 22.
a1 = a2 = a3 = a5 = 0,
u22 (x, t) = − ln
a4 = −k4 ,
k1 = k3 = k5 = 0,
c1 =
c2
,
c4 (c4 − c2 )
c3 = −
1
,
c4
−k4 e−(c3 x+c4 t)
.
k2 e−(c1 x+c2 t) + k4 e−(c3 x+c4 t)
As seen from the conclusions, the method seems to be easier and faster with the usage
of advanced symbolic computation systems. Our solutions are new soliton solutions of the
Tzitzeica-Dod-Bullough equation. In this study, we have obtained a wide range of solutions of
the problem, thus those solutions will be more meaningful and helpful in the examination of
physical problems related to the equation.
4
Conclusions
In this study, analytical solutions of the TDB equation has been successfully obtained and presented using symbolic computation systems. In the solution process, the double exp-function
method has been preferred due to its widespread usage in handling well-known evolution equations, discrete nonlinear equations and nonlinear equation systems. The solutions obtained in
the present study are more extensive compared with those available in the literature. Thus, we
are in the opinion that the presented results will be helpful for the applications in mathematical
physics and applied mathematics, particularly numerical simulations. In conclusion, we assert
that the exp-function method is plain and can be applied to a wider range of several nonlinear
evolution equations.
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