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IJRRAS 18 (2) ● February 2014 www.arpapress.com/Volumes/Vol18Issue2/IJRRAS_18_2_04.pdf
132
NEW EXACT SOLUTIONS OF SOME NONLINEAR PARTIAL
DIFFERENTIAL EQUATIONS VIA THE IMPROVED EXP-FUNCTION
METHOD
M. F. El-Sabbagh, R. Zait and R. M. Abdelazeem
Mathematics Department, Faculty of Science, MiniaUniversity, Egypt.
Corresponding e-mail: [email protected]
ABSTRACT In this paper, we establish new exact solutions of some nonlinear partial differential equations (PDEs) of interest
such as the Kaup–Kupershmidt, the generalized shallow water, the Boussinesq equations via the improved Exp–
function method. Also the method is used to construct periodic and solitary wave solutions for the considered
equations as well.
Keywords: Nonlinear PDEs, Exact solutions, The improved Exp–function method.
1. INTRODUCTION The nonlinear evolution equations (NLEEs) are widely used as models to describe complex physical phenomena in
various field of science, particularly in fluid mechanics, solid state physics, plasma waves and chemical physics.
Nonlinear equations covers also the following subjects: surface wave in compressible fluid, hydro magnetic waves
in cold plasma, acoustic waves in un harmonic crystal, ect. . The wide applicability of these equations is the main
reason for they have attracted so much attention from mathematicians in the last decades. The investigation of the
exact solutions of non linear partial differential equations (PDEs) plays an important role in the study of non-linear
physical phenomena. When we want to understand the physical mechanism of phenomena in nature, described by
non linear PDEs, exact solutions have to be explored. The study of nonlinear PDEs becomes one of the most
important topics in mathematical physics. Recently there are many new methods to obtain exact solutions of
nonlinear PDEs such as sine-cosine function method [1-5], tanh function method [6-8], (𝐺/
𝐺 )-expansion method
[9-13 ], extended Jacobi elliptic function method [14, 15]. He and Wu [16], proposed a straightforward and concise
method, called Exp-function method [17-24], to obtain generalized solitary wave solutions of nonlinear PDEs. Ali
[25] improved this method and obtained new exp-function solutions and periodic solutions as well.
In this paper, we use the improved Exp-function method [25, 26 ] to search for new solitary wave solutions,
compact like solutions and periodic solutions of some nonlinear PDEs, such as the Kaup–Kupershmidt equation
[27-35], the generalized shallow water equation [36, 37], and the Boussinesq equation [38-42].
2. THE IMPROVED EXP-FUNCTION METHOD:
We present the improved Exp-function method [25] in the following steps:
1- Consider the following nonlinear PDE with two independent variables 𝑥, 𝑡 and dependant variable 𝑢:
𝑁(𝑢,𝑢𝑡 ,𝑢𝑥 ,𝑢𝑥𝑥 ,𝑢𝑥𝑡 ,𝑢𝑡𝑡 ,… ) = 0 (1),
where 𝑁 is in general a polynomial function of its argument and the subscripts denote the partial derivatives.
2- We seek a traveling wave solution of Eq. (1) in the form
𝑢 𝑥, 𝑡 = 𝑢 𝜉 , 𝜉 = 𝑘𝑥 + 𝜔𝑡 2 ,
where 𝑘 and 𝜔 are constants to be determined
3- Using the transformation (2), Eq. (1) can be reduced to an ordinary differential equation (ODE):
𝐺 𝑢,𝑢′,𝑢′′,… = 0 3 ,
where 𝐺 is a polynomial of 𝑢 and its derivatives.
4- Through this method, we express the solution of the nonlinear PDE (1) in the form:
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𝑢 𝜂 = 𝐴𝑗𝑒𝑥𝑝(𝑗𝜉)𝑚
𝑗=0
𝐵𝑖𝑒𝑥𝑝(𝑖𝜉)𝑛𝑖=0
4 ,
where 𝑚 and 𝑛 are positive integers that could be freely chosen.
5- To determine 𝑢 𝑥, 𝑡 explicitly, one may apply the following producer:
(i) Substitute Eq. (4) into Eq. (3), then the lift hand side of Eq. (3) is converted into a polynomial in
𝑒𝑥𝑝(𝜉). Setting all coefficients of 𝑒𝑥𝑝(𝜉) to zero yield a system of algebraic equations for
𝐴0,𝐴1 ,𝐴2,… .𝐴𝑚 , 𝐵0 ,𝐵1 ,𝐵2 ,… .𝐵𝑛 , 𝑘 𝑎𝑛𝑑 𝜔.
(ii) Solve these algebraic equations to obtain A0, A1, A2,… , Am , 𝐵0 ,𝐵1 ,𝐵2 ,… ,𝐵𝑛 , 𝑘 𝑎𝑛𝑑 𝜔.
3. APPLICATIONS In order to illustrate the effectiveness of the above method, examples of mathematical and physical interests are
chosen as follows:
3.1The Kaup–Kupershmidt equation [27-35]
The Kaup–Kupershmidt equation is the nonlinear fifth-order partial differential equation; It is the first equation in a
hierarchy of integrable equations with Lax operator 𝜕3
𝜕𝑥3 + 2𝑢𝜕
𝜕𝑥+
𝜕𝑢
𝜕𝑥. It has properties similar (but not identical) to
those of the better known KdV hierarchy in which the Lax operator has order two.
In the present paper we introduce new exact solutions of the Kaup–Kupershmidt equation via the improved Exp-
function method as follows:
Consider the Kaup–Kupershmidt equation is given as:
𝑢𝑡 = 𝑢𝑥𝑥𝑥𝑥𝑥 − 20𝑢𝑢𝑥𝑥𝑥 − 50𝑢𝑥𝑢𝑥𝑥 + 80𝑢2𝑢𝑥 (5)
Using the transformation:
𝑢 = 𝑢 𝜂 , 𝜂 = 𝑥 − 𝑐𝑡 6 ,
where 𝑐 is a constant to be determined later. Substituting Eq. (6) into Eq. (5) we get
𝑐𝑢′ − 𝑢′′′′′ + 20𝑢𝑢′′′ + 50𝑢′𝑢′′ − 80𝑢2𝑢′ = 0 7 ,
where primes denote derivatives with respect to 𝜂. Now we study the following cases:
𝑪𝒂𝒔𝒆 𝟏: 𝒎 = 𝟐,𝒏 = 𝟐:
According to the improved Exp-function method, the travelling wave solution of Eq. (5) in this case can be written
as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜂) + 𝐴2𝑒𝑥𝑝(2𝜂)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜂) + 𝐵2𝑒𝑥𝑝(2𝜂) (8)
In case 𝐵2 ≠ 0, Eq. (8) can be simplified as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜂) + 𝐴2𝑒𝑥𝑝(2𝜂)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜂) + 𝑒𝑥𝑝(2𝜂) (9)
Substituting Eq. (9) into Eq. (7), and using the Maple, equating to zero the coefficients of all powers of
𝑒𝑥𝑝(𝜂)yields a set of algebraic equations for 𝐴0, 𝐴1, 𝐴2, 𝐵0 ,𝐵1 and c. Solving these system of algebraic equations,
with the aid of Maple, we obtain families of solutions as the following:
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𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:
𝐵0 = 1,𝐴1 = −5,𝐴0 =1
2, 𝑐 = −11,𝐴2 =
1
2,𝐵1 = 2.
Thus the following solution is
𝑢1 𝑥, 𝑡 =
12− 5𝑒𝑥𝑝(𝑥 + 11𝑡) +
12𝑒𝑥𝑝(2𝑥 + 22𝑡)
1 + 2𝑒𝑥𝑝(𝑥 + 11𝑡) + 𝑒𝑥𝑝(2𝑥 + 22𝑡) (10)
𝑭𝒊𝒈𝒖𝒓𝒆 𝟏. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟓 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏𝟎 .
𝑭𝒂𝒎𝒊𝒍𝒚 𝟐:
𝐵0 = 1,𝐴1 =−5
8,𝐴0 =
1
16, 𝑐 =
−1
16,𝐴2 =
1
16,𝐵1 = 2.
Thus the following solution is
𝑢2 𝑥, 𝑡 =
116
−58𝑒𝑥𝑝(𝑥 +
𝑡16
) +1
16𝑒𝑥𝑝(2𝑥 +
𝑡8
)
1 + 2𝑒𝑥𝑝(𝑥 +𝑡
16) + 𝑒𝑥𝑝(2𝑥 +
𝑡8
) (11)
𝑭𝒊𝒈𝒖𝒓𝒆 𝟐. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟓 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏𝟏 .
𝑪𝒂𝒔𝒆 𝟐: 𝒎 = 𝟑,𝒏 = 𝟑:
According to the improved Exp-function method, the travelling wave solution of Eq. (5) in this case can be written
as:
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𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜂) + 𝐴2𝑒𝑥𝑝(2𝜂) + 𝐴3𝑒𝑥𝑝(3𝜂)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜂) + 𝐵2𝑒𝑥𝑝(2𝜂) + 𝐵3𝑒𝑥𝑝(3𝜂) (12)
In case 𝐵3 ≠ 0, Eq. (12) can be simplified as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜂) + 𝐴2𝑒𝑥𝑝(2𝜂) + 𝐴3𝑒𝑥𝑝(3𝜂)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜂) + 𝐵2𝑒𝑥𝑝(2𝜂) + 𝑒𝑥𝑝(3𝜂) 13
Substituting Eq. (13) into Eq. (7), and using the Maple, equating to zero the coefficients of all powers of
𝑒𝑥 𝑝 𝜂 yields a set of algebraic equations for 𝐴0, 𝐴1, 𝐴2, 𝐴3, 𝐵0, 𝐵1 ,𝐵2 and c. Solving the system of algebraic
equations given above, with the aid of Maple, we obtain:
𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:
𝐴0 = 0,𝐴1 =1
8𝐵2
2 ,𝐴2 =−5
2𝐵2 ,𝐴3 =
1
2,𝐵0 = 0,𝐵1 =
1
4𝐵2
2 ,𝐵2 = 𝐵2 , 𝑐 = −11.
Thus the following solution is
𝑢5 𝑥, 𝑡 =
14𝐵2
2𝑒𝑥𝑝(𝑥 + 11𝑡) − 5𝐵2𝑒𝑥𝑝(2𝑥 + 22𝑡) + 𝑒𝑥𝑝(3𝑥 + 33𝑡)
12𝐵2
2𝑒𝑥𝑝(𝑥 + 11𝑡) + 2𝐵2𝑒𝑥𝑝(2𝑥 + 22𝑡) + 2𝑒𝑥𝑝(3𝑥 + 33𝑡) (14)
If 𝐵2 = 1, then
𝑢5 𝑥, 𝑡 =
14𝑒𝑥𝑝(𝑥 + 11𝑡) − 5𝑒𝑥𝑝(2𝑥 + 22𝑡) + 𝑒𝑥𝑝(3𝑥 + 33𝑡)
12𝑒𝑥𝑝(𝑥 + 11𝑡) + 2𝑒𝑥𝑝(2𝑥 + 22𝑡) + 2𝑒𝑥𝑝(3𝑥 + 33𝑡)
(15)
𝑭𝒊𝒈𝒖𝒓𝒆 𝟑. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟓 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏𝟓 ,𝑩𝟐 = 𝟏.
𝑭𝒂𝒎𝒊𝒍𝒚 𝟐:
𝐴0 = 0,𝐴1 =1
64𝐵2
2 ,𝐴2 =−5
16𝐵2 ,𝐴3 =
1
16,𝐵0 = 0,𝐵1 =
1
4𝐵2
2 ,𝐵2 = 𝐵2 , 𝑐 = −1
16.
Thus the following solution is
𝑢6 𝑥, 𝑡 =
14𝐵2
2𝑒𝑥𝑝(𝑥 +𝑡
16) − 5𝐵2𝑒𝑥𝑝(2𝑥 +
𝑡8
) + 𝑒𝑥𝑝(3𝑥 +3𝑡16
)
4(𝐵22𝑒𝑥𝑝(𝑥 +
𝑡16
) + 4𝐵2𝑒𝑥𝑝(2𝑥 +𝑡8
) + 4𝑒𝑥𝑝(3𝑥 +3𝑡16
)) (16)
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If 𝐵2 = 1, then
𝑢6 𝑥, 𝑡 =
14𝑒𝑥𝑝(𝑥 +
𝑡16
) − 5𝑒𝑥𝑝(2𝑥 +𝑡8
) + 𝑒𝑥𝑝(3𝑥 +3𝑡16
)
4(𝑒𝑥 𝑝 𝑥 +𝑡
16 + 4𝑒𝑥 𝑝 2𝑥 +
𝑡8 + 4𝑒𝑥 𝑝 3𝑥 +
3𝑡16
) (17)
𝑭𝒊𝒈𝒖𝒓𝒆 𝟒. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟓 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝟏𝟕 ,𝑩𝟐 = 𝟏.
3.2 The generalized shallow water equation: [36, 37]
The shallow water wave equations describe the evolution of incompressible flow, neglecting density change along
the depth. The shallow water wave equations are applicable to cases where the horizontal scale of the flow is much
bigger than the depth of fluid. The shallow water equations have been extensively used for a wide variety of coastal
phenomena, such as tide-currents, pollutant- dispersion storm-surges, tsunami-wave propagation, etc..
In the present paper we introduce new exact solutions of the generalized shallow water equation via the improved
Exp-function Method as follows:
Consider the generalized shallow water equation:
𝑢𝑥𝑥𝑥𝑡 + 𝛼𝑢𝑥𝑢𝑥𝑡 + 𝛽𝑢𝑡𝑢𝑥𝑥 − 𝑢𝑥𝑡 − 𝑢𝑥𝑥 = 0 18 ,
where 𝛼 and 𝛽 are arbitrary nonzero constants.
Using the transformation:
𝑢 = 𝑢 𝜂 , 𝜂 = 𝑥 − 𝑐𝑡 19 ,
where 𝑐 is a constant to be determined later. Substituting Eq. (19) into Eq. (18) we get
−𝑐𝑢′′′′ − 𝑐𝛼𝑢′𝑢′′ − 𝑐𝛽𝑢′𝑢′′ − 𝑐𝑢′′ − 𝑢′′ = 0 (20),
where the prime denotes the differential with respect to 𝜂.
Now we study the following cases:
𝑪𝒂𝒔𝒆 𝟏: 𝒎 = 𝟐,𝒏 = 𝟐:
According to the improved Exp-function method, the travelling wave solution of Eq. (18) in this case can be written
as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜂) + 𝐴2𝑒𝑥𝑝(2𝜂)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜂) + 𝐵2𝑒𝑥𝑝(2𝜂) (21)
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In case 𝐵2 ≠ 0, Eq.(21) can be simplified as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜂) + 𝐴2𝑒𝑥𝑝(2𝜂)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜂) + 𝑒𝑥𝑝(2𝜂) 22
Substituting Eq. (22) into Eq. (20), and using the Maple, equating to zero the coefficients of all powers of
𝑒𝑥𝑝(𝜂)yields a set of algebraic equations for 𝐴0, 𝐴1, 𝐴2,𝐵0, 𝐵1 and c. Solving the system of algebraic equation
given above, with the aid of Maple, we obtain:
𝐴0 =𝐵0(−24+𝐴2𝛽+𝐴2𝛼)
𝛼+𝛽,𝐴1 = 0,𝐴2 = 𝐴2,𝐵0 = 𝐵0 ,𝐵1 = 0, 𝑐 =
−1
3.
Thus the following solution is
𝑢1 𝑥, 𝑡 =
𝐵0(−24 + 𝐴2𝛽 + 𝐴2𝛼)𝛼 + 𝛽
+ 𝐴2𝑒𝑥𝑝(2𝑥 +2𝑡3
)
𝐵0 + 𝑒𝑥𝑝(2𝑥 +2𝑡3
) (23)
If 𝐴2 = 𝐵0 = 1, 𝛼 = 𝛽 = 2, then
𝑢1 𝑥, 𝑡 =−5 + 𝑒𝑥𝑝(2𝑥 +
2𝑡3
)
1 + 𝑒𝑥𝑝(2𝑥 +2𝑡3
) (24)
𝑭𝒊𝒈𝒖𝒓𝒆 𝟓. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟏𝟖 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟐𝟒 ,𝑨𝟐 = 𝑩𝟎 = 𝟏, 𝜶 = 𝜷 = 𝟐.
𝑪𝒂𝒔𝒆 𝟐: 𝒎 = 𝟑,𝒏 = 𝟑:
According to the improved Exp-function method, the travelling wave solution of Eq. (18) in this case can be written
as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜂) + 𝐴2𝑒𝑥𝑝(2𝜂) + 𝐴3𝑒𝑥𝑝(3𝜂)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜂) + 𝐵2𝑒𝑥𝑝(2𝜂) + 𝐵3𝑒𝑥𝑝(3𝜂) (25)
In case 𝐵3 ≠ 0, Eq.(25) can be simplified as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜂) + 𝐴2𝑒𝑥𝑝(2𝜂) + 𝐴3𝑒𝑥𝑝(3𝜂)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜂) + 𝐵2𝑒𝑥𝑝(2𝜂) + 𝑒𝑥𝑝(3𝜂) 26
Substituting Eq. (26) into Eq. (20), and using the Maple, equating to zero the coefficients of all powers of 𝑒𝑥𝑝(𝜂)
yields a set of algebraic equations for𝐴0, 𝐴1, 𝐴2, 𝐴3, 𝐵0, 𝐵1 ,𝐵2 and c. Solving the system of algebraic equation
given above, with the aid of Maple, we obtain:
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𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:
𝐴0 = 0,𝐴1 =𝐵1(−24 + 𝐴3𝛽 + 𝐴3𝛼)
𝛼 + 𝛽,𝐴2 = 0,𝐴3 = 𝐴3,𝐵0 = 0,𝐵1 = 𝐵1 ,𝐵2 = 𝐵2 , 𝑐 =
−1
3
Thus the following solution is
𝑢2 𝑥, 𝑡 =
𝐵1(−24 + 𝐴3𝛽 + 𝐴3𝛼)𝛼 + 𝛽
𝑒𝑥𝑝(𝑥 +𝑡3
) + 𝐴3𝑒𝑥𝑝(3𝑥 + 𝑡)
𝐵1𝑒𝑥𝑝(𝑥 +𝑡3
) + 𝑒𝑥𝑝(3𝑥 + 𝑡) (27)
If 𝐵1 = 𝐴3 = 2, 𝛼 = 𝛽 = 1, then
𝑢2 𝑥, 𝑡 =−20𝑒𝑥𝑝(𝑥 +
𝑡3
) + 2𝑒𝑥𝑝(3𝑥 + 𝑡)
2𝑒𝑥𝑝(𝑥 +𝑡3
) + 𝑒𝑥𝑝(3𝑥 + 𝑡) (28)
𝑭𝒊𝒈𝒖𝒓𝒆 𝟔. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟏𝟖 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟐𝟖 ,𝑨𝟑 = 𝑩𝟏 = 𝟐, 𝜶 = 𝜷 = 𝟏.
𝑭𝒂𝒎𝒊𝒍𝒚 𝟐:
𝐴0 =𝐵0(𝐴3𝛽 + 𝐴3𝛼 − 36)
𝛼 + 𝛽,𝐴1 = 0,𝐴2 = 0,𝐴3 = 𝐴3,𝐵0 = 𝐵0 ,𝐵1 = 0,𝐵2 = 0, 𝑐 =
−1
8
Thus the following solution is
𝑢3 𝑥, 𝑡 =
𝐵0(𝐴3𝛽 + 𝐴3𝛼 − 36)𝛼 + 𝛽
+ 𝐴3𝑒𝑥𝑝(3𝑥 +3𝑡8
)
𝐵0 + 𝑒𝑥𝑝(3𝑥 +3𝑡8
) (29)
If 𝐵0 = 𝐴3 = 2, 𝛼 = 𝛽 = 1, then
𝑢3 𝑥, 𝑡 =−32 + 2𝑒𝑥𝑝(3𝑥 +
3𝑡8
)
2 + 𝑒𝑥𝑝(3𝑥 +3𝑡8
) (30)
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𝑭𝒊𝒈𝒖𝒓𝒆 𝟕. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟏𝟖 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟑𝟎 ,𝑨𝟑 = 𝑩𝟎 = 𝟐,𝜶 = 𝜷 = 𝟏.
3.3 The Boussinesq equation: [38-42]
The Boussinesq-type equations, which include the lowest-order effects of nonlinearity and frequency dispersion as
additions to the simplest non-dispersive linear long wave theory, provide a sound and increasingly well-tested basis
for the simulation of wave propagation in coastal regions. The standard Boussinesq equations for variable water
depth were first derived by Peregrine (1967), who used depth-averaged velocity as a dependent variable.
In the present paper we introduce new exact solutions of the Boussinesq equation via the improved Exp-function
method as follows:
Consider the Boussinesq equation:
𝑢𝑡𝑡 − 𝑢𝑥𝑥 − 𝑢𝑥𝑥𝑥𝑥 − 6(𝑢𝑥)2 − 6𝑢𝑢𝑥𝑥 = 0 (31)
Using the transformation:
𝑢 = 𝑢 𝜂 , 𝜉 = 𝑘𝑥 + 𝜔𝑡 32 ,
where 𝑘,𝜔 are constants to be determined later. Substituting Eq. (32) into Eq. (31) we get
𝜔2𝑢′′ − 𝑘2𝑢′′ − 𝑘4𝑢′′′′ − 6𝑘2(𝑢′)2 − 6𝑘2𝑢𝑢′′ = 0 33 ,
where the prime denotes the differential with respect to 𝜉. Now we study the following cases:
𝑪𝒂𝒔𝒆 𝟏: 𝒎 = 𝟐,𝒏 = 𝟑:
According to the improved Exp-function method, the travelling wave solution of Eq. (31) in this case can be written
as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜉) + 𝐴2𝑒𝑥𝑝(2𝜉)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜉) + 𝐵2𝑒𝑥𝑝(2𝜉) + 𝐵3𝑒𝑥𝑝(3𝜉) (34)
In case 𝐵3 ≠ 0, Eq.(34) can be simplified as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥𝑝(𝜉) + 𝐴2𝑒𝑥𝑝(2𝜉)
𝐵0 + 𝐵1𝑒𝑥𝑝(𝜉) + 𝐵2𝑒𝑥𝑝(2𝜉) + 𝑒𝑥𝑝(3𝜉) 35
Substituting Eq. (35) into Eq. (33), and using the Maple, equating to zero the coefficients of all powers of
𝑒𝑥 𝑝 𝜉 yields a set of algebraic equations for 𝐴0, 𝐴1, 𝐴2, 𝐵0, 𝐵1 ,𝐵2 , 𝑘 𝑎𝑛𝑑 𝜔. Solving the system of algebraic
equation given above, with the aid of Maple, we obtain:
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𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:
𝐴0 = 0,𝐴1 = 0,𝐴2 = 𝑘2𝐵2 ,𝐵0 = 0,𝐵1 =𝐵2
2
4,𝐵2 = 𝐵2 ,𝜔 = 𝑘 1 + 𝑘2
Thus the following solution is
𝑢1 𝑥, 𝑡 =𝑘2𝐵2𝑒𝑥𝑝(2𝑘𝑥 + 2𝑘𝑡 1 + 𝑘2)
𝐵22
4𝑒𝑥𝑝(𝑘𝑥 + 𝑘𝑡 1 + 𝑘2) + 𝐵2𝑒𝑥𝑝(2𝑘𝑥 + 2𝑘𝑡 1 + 𝑘2) + 𝑒𝑥𝑝(3𝑘𝑥 + 3𝑘𝑡 1 + 𝑘2)
(36)
If 𝑘 = 1, 𝐵2 = 2, then
𝑢1 𝑥, 𝑡 =2𝑒𝑥𝑝(2𝑥 + 2𝑡 2)
𝑒𝑥𝑝(𝑥 + 𝑡 2) + 2𝑒𝑥𝑝(2𝑥 + 2𝑡 2) + 𝑒𝑥𝑝(3𝑥 + 3𝑡 2) (37)
𝑭𝒊𝒈𝒖𝒓𝒆 𝟖. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟑𝟏 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟑𝟕 ,𝑩𝟐 = 𝟐,𝒌 = 𝟏.
𝑭𝒂𝒎𝒊𝒍𝒚 𝟐:
𝐴0 = 0,𝐴1 = 0,𝐴2 = 𝑘2𝐵2 ,𝐵0 = 0,𝐵1 =𝐵2
2
4,𝐵2 = 𝐵2 ,𝜔 = −𝑘 1 + 𝑘2
Thus the following solution is
𝑢2 𝑥, 𝑡 =𝑘2𝐵2𝑒𝑥𝑝(2𝑘𝑥 − 2𝑘𝑡 1 + 𝑘2)
𝐵22
4𝑒𝑥𝑝(𝑘𝑥 − 𝑘𝑡 1 + 𝑘2) + 𝐵2𝑒𝑥𝑝(2𝑘𝑥 − 2𝑘𝑡 1 + 𝑘2) + 𝑒𝑥𝑝(3𝑘𝑥 − 3𝑘𝑡 1 + 𝑘2)
(38)
If 𝐵2 = 2, 𝑘 = 1, then
𝑢2 𝑥, 𝑡 =2𝑒𝑥 𝑝 2𝑥 − 2𝑡 2
𝑒𝑥𝑝(𝑥 − 𝑡 2) + 2𝑒𝑥𝑝(2𝑥 − 2𝑡 2) + 𝑒𝑥𝑝(3𝑥 − 3𝑡 2) (39)
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𝑭𝒊𝒈𝒖𝒓𝒆 𝟗. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟑𝟏 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟑𝟗 ,𝑩𝟐 = 𝟐,𝒌 = 𝟏.
𝑪𝒂𝒔𝒆 𝟐: 𝒎 = 𝟐,𝒏 = 𝟒:
According to the improved Exp-function method, the travelling wave solution of Eq. (31) in this case can be written
as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥 𝑝 𝜉 + 𝐴2𝑒𝑥 𝑝 2𝜉
𝐵0 + 𝐵1𝑒𝑥 𝑝 𝜉 + 𝐵2𝑒𝑥 𝑝 2𝜉 + 𝐵3𝑒𝑥 𝑝 3𝜉 + 𝐵4𝑒𝑥 𝑝 4𝜉 (40)
In case 𝐵4 ≠ 0, Eq.(40) can be simplified as:
𝑢 𝑥, 𝑡 = 𝐴0 + 𝐴1𝑒𝑥 𝑝 𝜉 + 𝐴2𝑒𝑥 𝑝 2𝜉
𝐵0 + 𝐵1𝑒𝑥 𝑝 𝜉 + 𝐵2𝑒𝑥 𝑝 2𝜉 + 𝐵3𝑒𝑥 𝑝 3𝜉 + 𝑒𝑥 𝑝 4𝜉 41
Substituting Eq. (41) into Eq. (33), and using the Maple, equating to zero the coefficients of all powers of
𝑒𝑥 𝑝 𝜉 yields a set of algebraic equations for 𝐴0, 𝐴1, 𝐴2, 𝐵0 , 𝐵1 ,𝐵2 ,𝐵3 , 𝑘 and 𝜔. Solving the system of algebraic
equation given above, with the aid of Maple, we obtain:
𝑭𝒂𝒎𝒊𝒍𝒚 𝟏:
𝐴0 = 0,𝐴1 = 0,𝐴2 = 𝐴2,𝐵0 =𝐴2
2
64𝑘4,𝐵1 = 0,𝐵2 =
𝐴2
4𝑘2,𝐵3 = 𝐵3 ,𝜔 = 𝑘 1 + 4𝑘2
Thus the following solution is
𝑢3 𝑥, 𝑡 =𝐴2𝑒𝑥𝑝(2𝑘𝑥 + 2𝑘𝑡 1 + 4𝑘2)
𝐴22
64𝑘4 +𝐴2
4𝑘2 𝑒𝑥𝑝(2𝑘𝑥 + 2𝑘𝑡 1 + 4𝑘2) + 𝑒𝑥𝑝(4𝑘𝑥 + 4𝑘𝑡 1 + 4𝑘2)
(42)
If 𝐴2 = 2, 𝑘 = 1, then
𝑢3 𝑥, 𝑡 =2𝑒𝑥𝑝(2𝑥 + 2𝑡 5)
116
+12𝑒𝑥𝑝(2𝑥 + 2𝑡 5) + 𝑒𝑥𝑝(4𝑥 + 4𝑡 5)
(43)
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𝑭𝒊𝒈𝒖𝒓𝒆 𝟏𝟎. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟑𝟏 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟒𝟑 ,𝑨𝟐 = 𝟐,𝒌 = 𝟏.
𝑭𝒂𝒎𝒊𝒍𝒚 𝟐:
𝐴0 == 0,𝐴1 = 0,𝐴2 = 𝐴2,𝐵0 =𝐴2
2
64𝑘4,𝐵1 = 0,𝐵2 =
𝐴2
4𝑘2,𝐵3 = 𝐵3 ,𝜔 = −𝑘 1 + 4𝑘2
Thus the following solution is
𝑢4 𝑥, 𝑡 =𝐴2𝑒𝑥𝑝(2𝑘𝑥 − 2𝑘𝑡 1 + 4𝑘2)
𝐴22
64𝑘4 +𝐴2
4𝑘2 𝑒𝑥𝑝(2𝑘𝑥 − 2𝑘𝑡 1 + 4𝑘2) + 𝑒𝑥𝑝(4𝑘𝑥 − 4𝑘𝑡 1 + 4𝑘2)
(44)
If 𝐴2 = 2, 𝑘 = 1, then
𝑢4 𝑥, 𝑡 =2𝑒𝑥𝑝(2𝑥 − 2𝑡 5)
116
+12𝑒𝑥𝑝(2𝑥 − 2𝑡 5) + 𝑒𝑥𝑝(4𝑥 − 4𝑡 5)
(45)
𝑭𝒊𝒈𝒖𝒓𝒆 𝟏𝟏. 𝑻𝒓𝒂𝒗𝒆𝒍𝒊𝒏𝒈 𝒘𝒂𝒗𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑬𝒒. 𝟑𝟑 𝒇𝒐𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑬𝒒. 𝟒𝟓 ,𝑨𝟐 = 𝟐,𝒌 = 𝟏.
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4. CONCULSION
In this paper, the improved Exp-function method has been successfully applied to obtain new solutions of three
nonlinear partial differential equations. Thus, the improved Exp-function method can be extended to solve the
problems of nonlinear partial differential equations which arising in the theory of solitons and other areas.
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