Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 767380, 16 pages
http://dx.doi.org/10.1155/2013/767380
Research Article
New Exact Solutions for New Model Nonlinear Partial
Differential Equation
A. Maher,1 H. M. El-Hawary,1 and M. S. Al-Amry2
1
2
Department of Mathematics, Faculty of Science, Assiut University, Assiut 71516, Egypt
Department of Mathematics, Faculty of Science and Education, Aden University, Aden, Yemen
Correspondence should be addressed to A. Maher; a maher1969@yahoo.com
Received 10 February 2013; Accepted 20 April 2013
Academic Editor: Anjan Biswas
Copyright © 2013 A. Maher et al. This is an open access article distributed under the Creative Commons Attribution License, which
permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
In this paper we propose a new form of Padé-II equation, namely, a combined Padé-II and modified Padé-II equation. The mapping
method is a promising method to solve nonlinear evaluation equations. Therefore, we apply it, to solve the combined PadéII and modified Padé-II equation. Exact travelling wave solutions are obtained and expressed in terms of hyperbolic functions,
trigonometric functions, rational functions, and elliptic functions.
1. Introduction
In recent years, directly searching for exact solutions of nonlinear partial differential equations (PDEs) has become more
and more attractive field in different branches of physics and
applied mathematics. These equations appear in condensed
matter, solid state physics, fluid mechanics, chemical kinetics,
plasma physics, nonlinear optics, propagation of fluxions in
Josephson junctions, theory of turbulence, ocean dynamics,
biophysics star formation, and many others.
In order to get exact solutions directly, many powerful
methods have been introduced such as the (𝐺󸀠 /𝐺)-expansion
method [1], inverse scattering method [2, 3], Hirota’s bilinear
method [4, 5], the tanh method [6, 7], the sine-cosine
method [8, 9], Bäcklund transformation method [10, 11], the
homogeneous balance [12, 13], Darboux transformation [14],
and the Jacobi elliptic function expansion method [15].
Recently, Peng [16] introduced a new approach, namely,
the mapping method for a reliable treatment of the nonlinear
wave equations. The useful mapping method is then widely
used by many authors [17, 18].
2. Description of the Method
Consider the general nonlinear partial differential equations
(PDEs); say, in two variables,
𝑃 (𝑢, 𝑢𝑥 , 𝑢𝑡 , 𝑢𝑥𝑥 , 𝑢𝑥𝑡 , . . .) = 0.
(1)
Let 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝜇(𝑥 − 𝑐𝑡); then (1) reduces to a
nonlinear ordinary differential equation (ODE)
𝑄 (𝑢, 𝑢󸀠 , 𝑢󸀠󸀠 , . . .) = 0.
(2)
Assume the solution of (2) takes the form
𝑚
𝑖
−𝑖
𝑢 (𝑥, 𝑡) = 𝑢 (𝜉) = 𝑎0 + ∑𝑎𝑖 (𝑓(𝜉)) + 𝑏𝑖 (𝑓(𝜉)) ,
(3)
𝑖=1
where the coefficients 𝑎𝑖 (𝑖 = 0, 1, 2, . . . , 𝑚), 𝜇, and 𝑐 are
constants to be determined, and 𝑓 = 𝑓(𝜉) satisfies a nonlinear
ordinary differential equation
𝑑𝑓 (𝜉)
1
= √ 𝑝𝑓2 (𝜉) + 𝑞𝑓4 (𝜉) + 𝑟,
𝑑𝜉
2
𝑝, 𝑞, 𝑟 ∈ 𝑅,
(4)
where the coefficients 𝑎0 , 𝑎𝑖 , 𝑏𝑖 (𝑖 = 1, 2, . . . 𝑚), 𝜇, and 𝑐 are
constants to be determined and 𝑓 = 𝑓(𝜉) satisfies (4); the
parameter 𝑚 will be found by balancing the highest-order
nonlinear terms with the highest-order partial derivative
term in the given equation. Substituting (3) into (2), using
(4) repeatedly and setting the coefficients of the each order
of 𝑓𝑖 (𝜉), 𝑓𝑖 (𝜉)√𝑝𝑓2 (𝜉) + (1/2)𝑞𝑓4 (𝜉) + 𝑟 to zero, we obtain
a set of nonlinear algebraic equations for 𝑎0 , 𝑎𝑖 , 𝑏𝑖 (𝑖 =
1, 2, . . . 𝑛), 𝜇, and 𝑐. With the aid of the computer program
Maple, we can solve the set of nonlinear algebraic equations
2
Journal of Applied Mathematics
and obtain all the constants 𝑎0 , 𝑎𝑖 , 𝑏𝑖 (𝑖 = 1, 2, . . . 𝑛), 𝜇, and 𝑐.
The ODE (4) has the following solutions:
(1) 𝑓(𝜉) = sech(𝜉), 𝑝 = 1, 𝑞 = −2, 𝑟 = 0,
3. Application
In this section, we present our proposed equation, namely, a
combined Padé-II and modified Padé-II equation, as the form
(2) 𝑓(𝜉) = tanh(𝜉), 𝑝 = −2, 𝑞 = 2, 𝑟 = 1,
𝑢𝑡 (𝑥, 𝑡) + 𝑢𝑥 (𝑥, 𝑡) + 𝑃 (𝑢) 𝑢𝑥 (𝑥, 𝑡)
(3) 𝑓(𝜉) = (1/2) tanh(2𝜉), (1/2) coth(2𝜉), 𝑝 = −8, 𝑞 =
32, 𝑟 = 1,
(4) 𝑓(𝜉) = (1/2) tan(2𝜉), −(1/2)cot(2𝜉), 𝑝 = 8, 𝑞 =
32, 𝑟 = 1,
(5) 𝑓(𝜉) = sn 𝜉, 𝑝 = −(𝑘2 + 1), 𝑞 = 2𝑘2 , 𝑟 = 1,
(6) 𝑓(𝜉) = ns 𝜉, 𝑝 = −(𝑘2 + 1), 𝑞 = 2, 𝑟 = 𝑘2 ,
(7) 𝑓(𝜉) = cd 𝜉, 𝑝 = −(𝑘2 + 1), 𝑞 = 2𝑘2 , 𝑟 = 1,
+ 𝑎𝑢𝑥𝑥𝑥 (𝑥, 𝑡) + 𝑏𝑢𝑥𝑥𝑡 (𝑥, 𝑡) = 0,
where 𝑃(𝑢) = 𝑢(𝑥, 𝑡)+𝑢2 (𝑥, 𝑡), 𝑎, and 𝑏 are real numbers [19].
Now, we apply the mapping method to solve our equation.
Consequently we get the original solutions for our new
equation, as the follows.
Substituting 𝑢(𝑥, 𝑡) = 𝑢(𝜉), 𝜉 = 𝜆(𝑥 − 𝑐𝑡) in (5) and
integrating once yield
(8) 𝑓(𝜉) = dc 𝜉, 𝑝 = −(𝑘2 + 1), 𝑞 = 2 , 𝑟 = 𝑘2 ,
(9) 𝑓(𝜉) = cn 𝜉, 𝑝 = 2𝑘2 − 1, 𝑞 = −2𝑘2 , 𝑟 = 1 − 𝑘2 ,
(1 − 𝑐) 𝑢 (𝜉) +
(10) 𝑓(𝜉) = nc 𝜉, 𝑝 = 2𝑘2 − 1, 𝑞 = 2(1 − 𝑘2 ), 𝑟 = −𝑘2 ,
2
3
(6)
󸀠󸀠
Balancing the order of the nonlinear term 𝑢3 with the highest
derivative 𝑢󸀠󸀠 gives 3𝑚 = 𝑚 + 2 that gives 𝑚 = 1. Thus, the
solution of (6) has the form
(12) 𝑓(𝜉) = nd 𝜉, 𝑝 = 2 − 𝑘2 , 𝑞 = 2(𝑘2 − 1), 𝑟 = −1,
(13) 𝑓(𝜉) = cs 𝜉, 𝑝 = 2 − 𝑘2 , 𝑞 = 2, 𝑟 = 1 − 𝑘2 ,
(14) 𝑓(𝜉) = sc 𝜉, 𝑝 = 2 − 𝑘2 , 𝑞 = 2(1 − 𝑘2 ), 𝑟 = 1,
1
(15) 𝑓(𝜉) = ds 𝜉, 𝑝 = −1 + 2𝑘2 , 𝑞 = 2, 𝑟 = −𝑘2 (1 − 𝑘2 ),
2
2
(𝑢 (𝜉))
(𝑢 (𝜉))
+
2
3
+ 𝜆 (𝑎 − 𝑏𝑐) 𝑢 (𝜉) = 0.
(11) 𝑓(𝜉) = dn 𝜉, 𝑝 = 2 − 𝑘2 , 𝑞 = −2, 𝑟 = −(1 − 𝑘2 ),
2
(5)
𝑢 (𝜉) = ∑𝑎𝑖 𝑓(𝜉)𝑖 = 𝑎0 + 𝑎1 𝑓 (𝜉) + 𝑏1 𝑓(𝜉)−1 ,
(16) 𝑓(𝜉) = sd 𝜉, 𝑝 = −1 + 2𝑘 , 𝑞 = 2𝑘 (𝑘 − 1), 𝑟 = 1,
(17) 𝑓(𝜉) = sc 𝜉 ± nc 𝜉, 𝑝 = (1 + 𝑘2 )/2, 𝑞 = (1 − 𝑘2 )/2, 𝑟 =
(1 − 𝑘2 )/4,
(18) 𝑓(𝜉) = sn 𝜉/(1 ± dn 𝜉), 𝑝 = (𝑘2 − 2)/2, 𝑞 = 𝑘2 /2, 𝑟 =
1/4,
(19) 𝑓(𝜉) = dn 𝜉/(1 ± 𝑘 sn 𝜉), 𝑝 = (𝑘2 + 1)/2, 𝑞 = (𝑘2 −
1)/2, 𝑟 = (1 − 𝑘2 )/4,
(20) 𝑓(𝜉) = 𝑘 cn 𝜉 ± dn 𝜉, 𝑝 = (𝑘2 + 1)/2, 𝑞 = −1/2, 𝑟 =
2
−(1 − 𝑘2 ) /4,
(21) 𝑓(𝜉) = cn 𝜉/(1 ± sn 𝜉), 𝑝 = (𝑘2 + 1)/2, 𝑞 = (1 −
𝑘2 )/2, 𝑟 = (1 − 𝑘2 )/4,
2
(7)
𝑖=0
2
where
𝑑𝑓 (𝜉)
1
= √ 𝑝𝑓2 (𝜉) + 𝑞𝑓4 (𝜉) + 𝑟,
𝑑𝜉
2
𝑝, 𝑞, 𝑟 ∈ 𝑅.
(8)
Substituting (7) in (6) and using (8), collecting the
coefficients of each power of 𝑓𝑖 , 0 ≤ 𝑖 ≤ 6, setting each
coefficient to zero, and solving the resulting system, we obtain
the following sets of solutions:
(1) 𝑎0 = 0, 𝑎1 = 𝑏1 = 0, 𝑐 = 𝑐, 𝜆 = 𝜆,
(2) 𝑎0 = 𝑎0 , 𝑎1 = 𝑏1 = 0, 𝑐 = 𝑐, 𝜆 = 𝜆,
(22) 𝑓(𝜉) = 𝑘 sn 𝜉 ± 𝑖 dn 𝜉, 𝑝 = (1 − 2𝑘 )/2, 𝑞 = 1/2, 𝑟 =
𝑘2 /4,
(3) 𝑎0 = −1/2, 𝑎1 = √−𝑞/4𝑝, 𝑏1 = 0, 𝑐 = 5/6, 𝜆 =
±√1/(12𝑎𝑝 − 10𝑏𝑝),
(23) 𝑓(𝜉) = 𝑘 sn 𝜉 ± 𝑖 cn 𝜉, 𝑝 = (𝑘2 − 2)/2, 𝑞 = 𝑘2 /2, 𝑟 =
𝑘2 /4,
(4) 𝑎0 = −1/2, 𝑎1 = −√−𝑞/4𝑝, 𝑏1 = 0, 𝑐 = 5/6, 𝜆 =
±√1/(12𝑎𝑝 − 10𝑏𝑝),
(24) 𝑓(𝜉) = ns 𝜉 ± ds 𝜉, 𝑝 = (𝑘2 − 2)/2, 𝑞 = 1/2, 𝑟 = 𝑘4 /4,
(25) 𝑓(𝜉) = ns 𝜉 − cs 𝜉, 𝑝 = (1 − 2𝑘2 )/2, 𝑞 = 1/2, 𝑟 = 1/4,
(5) 𝑎0 = −1/2, 𝑎1 = 0, 𝑏1 = √−𝑟/2𝑝, 𝑐 = 5/6, 𝜆 =
±√1/(12𝑎𝑝 − 10𝑏𝑝),
(26) 𝑓(𝜉) = cn 𝜉/(√1 − 𝑘2 sn 𝜉 ± dn 𝜉), 𝑝
2𝑘2 )/2, 𝑞 = 1/2, 𝑟 = 1/4,
(6) 𝑎0 = −1/2, 𝑎1 = 0, 𝑏1 = −√−𝑟/2𝑝, 𝑐 = 5/6, 𝜆 =
±√1/(12𝑎𝑝 − 10𝑏𝑝),
=
(1 −
(27) 𝑓(𝜉) = sn 𝜉/(cn 𝜉 ± dn 𝜉), 𝑝 = (1 + 𝑘2 )/2, 𝑞 =
2
(1 − 𝑘2 ) /2, 𝑟 = 1/4,
(28) 𝑓(𝜉) = cn 𝜉/(√1 − 𝑘2 ± dn 𝜉), 𝑝 = (𝑘2 − 2)/2, 𝑞 =
𝑘2 /2, 𝑟 = 1/4,
(29) 𝑓(𝜉) = −1/√𝑐/2𝜉, 𝑝 = 0, 𝑞 = 𝑐, 𝑟 = 0,
𝜉
(30) 𝑓(𝜉) = 𝑒 , 𝑝 = 1, 𝑞 = 0, 𝑟 = 0.
(7) 𝑎0 = −1/2, 𝑎1 = (1/2)√(𝑝𝑞 + 3𝑞√2𝑟𝑞)/(18𝑟𝑞 − 𝑝2 ),
𝑏1 = −(1/2)((6𝑟𝑞 + 𝑝√2𝑟𝑞)/
√(𝑝𝑞 + 3𝑞√2𝑟𝑞)/(18𝑟𝑞 − 𝑝2 )(𝑝2 − 18𝑟𝑞)),
𝑐 = 5/6, 𝜆 = ±(1/√2)
×√(6𝑝2 𝑎 − 108𝑎𝑟𝑞 − 5𝑝2 𝑏 + 90𝑟𝑏𝑞)(𝑝 + 3√2𝑟𝑞)/
(6𝑝2 𝑎 − 108𝑎𝑟𝑞 − 5𝑝2 𝑏 + 90𝑟𝑏𝑞),
Journal of Applied Mathematics
3
(8) 𝑎0 = −1/2, 𝑎1 = −(1/2)√(𝑝𝑞 + 3𝑞√2𝑟𝑞)/(18𝑟𝑞 − 𝑝2 ),
𝑏1 = (1/2)((6𝑟𝑞 + 𝑝√2𝑟𝑞)/
√(𝑝𝑞 + 3𝑞√2𝑟𝑞)/(18𝑟𝑞 − 𝑝2 )(𝑝2 − 18𝑟𝑞)),
𝑐 = 5/6, 𝜆 = ±(1/√2)
𝑢9,10 (𝑥, 𝑡)
5
1 1
1
(𝑥 − 𝑡)) ,
= − ± coth (
2 2
6
2√−6𝑎 + 5𝑏
𝑢11,12 (𝑥, 𝑡)
×√(6𝑝2 𝑎 − 108𝑎𝑟𝑞 − 5𝑝2 𝑏 + 90𝑟𝑏𝑞)(𝑝 + 3√2𝑟𝑞)/
1 1
1
5
= − ± tanh (
(𝑥 − 𝑡))
2 4
6
2√2√−6𝑎 + 5𝑏
(6𝑝2 𝑎 − 108𝑎𝑟𝑞 − 5𝑝2 𝑏 + 90𝑟𝑏𝑞),
(9) 𝑎0 = −1/2, 𝑎1 = (1/2)√(𝑝𝑞 − 3𝑞√2𝑟𝑞)/(18𝑟𝑞 − 𝑝2 ),
𝑏1 = −(1/2)((−6𝑟𝑞 + 𝑝√2𝑟𝑞)/
√(𝑝𝑞 − 3𝑞√2𝑟𝑞)/(18𝑟𝑞 − 𝑝2 ) (𝑝2 − 18𝑟𝑞)),
𝑐 = 5/6, 𝜆 = ±(1/√2)
±
𝑢13,14 (𝑥, 𝑡)
×√(−10𝑝2 𝑏+180𝑟𝑏𝑞+12𝑝2 𝑎−216𝑎𝑟𝑞)(𝑝−3√2𝑟𝑞)/
(6𝑝2 𝑎 − 108𝑎𝑟𝑞 − 5𝑝2 𝑏 + 90𝑟𝑏𝑞),
5
1
1
1
(𝑥 − 𝑡))
tan (
=− ±
√
2 2√2
6
√
2 2 −6𝑎 + 5𝑏
±
(10) 𝑎0 = −1/2, 𝑎1 = −(1/2)√(𝑝𝑞 − 3𝑞√2𝑟𝑞)/(18𝑟𝑞 − 𝑝2 ),
𝑏1 = (1/2)((−6𝑟𝑞 + 𝑝√2𝑟𝑞)/
√(𝑝𝑞 − 3𝑞√2𝑟𝑞)/(18𝑟𝑞 − 𝑝2 ) (𝑝2 − 18𝑟𝑞)),
𝑐 = 5/6, 𝜆 = ±(1/√2)
(6𝑝2 𝑎 − 108𝑎𝑟𝑞 − 5𝑝2 𝑏 + 90𝑟𝑏𝑞).
Using (7), the solution of (8) when 𝑝 = 1, 𝑞 = −2, and
𝑟 = 0, and the sets of solutions (1)–(10), we get
𝑢2 (𝑥, 𝑡) = 𝑎0 ,
∀𝑎0 ∈ 𝑅,
(9)
(10)
𝑢17,18 (𝑥, 𝑡)
5
1 1
1
(𝑥 − 𝑡)) ,
= − ± 𝑖cot (
√
2 2
6
2 6𝑎 − 5𝑏
(13)
5
1
1
(𝑥 − 𝑡)) ,
± 𝑖cot (
√
4
6
√
2 2 6𝑎 − 5𝑏
𝑢21,22 (𝑥, 𝑡)
∓
𝑢5,6 (𝑥, 𝑡)
1
5
𝑖
coth (
(𝑥 − 𝑡)) .
6
2√2
2√2√6𝑎 − 5𝑏
(11)
Using (7), the solution of (8) when 𝑝 = −2, 𝑞 = 2, and
𝑟 = 1, and the sets of solutions (3)–(10), we get for 𝑎 < (5/6)𝑏
5
1 1
1
(𝑥 − 𝑡)) ,
= − ± tanh (
2 2
6
2√−6𝑎 + 5𝑏
5
1 1
1
(𝑥 − 𝑡)) ,
= − ± 𝑖 tan (
√
2 2
6
2 6𝑎 − 5𝑏
5
1
1
𝑖
(𝑥 − 𝑡))
tanh (
=− ±
√
√
2 2 2
6
2√2 6𝑎 − 5𝑏
for 𝑎 < (5/6)𝑏
𝑢7,8 (𝑥, 𝑡)
𝑢15,16 (𝑥, 𝑡)
5
1 1
1
(𝑥 − 𝑡))
= − ± 𝑖 tan (
√
2 4
6
√
2 2 6𝑎 − 5𝑏
𝑢3,4 (𝑥, 𝑡)
5
1
1
1
(𝑥 − 𝑡)) .
sec (
=− ±
2 √2
6
√2√−6𝑎 + 5𝑏
(12)
𝑢19,20 (𝑥, 𝑡)
for 𝑎 > (5/6)𝑏
5
1
1
1
(𝑥 − 𝑡)) ,
sech (
=− ±
2 √2
6
√2√6𝑎 − 5𝑏
5
1
1
(𝑥 − 𝑡)) .
cot (
√
√
6
√
2 2
2 2 −6𝑎 + 5𝑏
For 𝑎 > (5/6)𝑏
×√(−10𝑝2 𝑏+180𝑟𝑏𝑞+12𝑝2 𝑎−216𝑎𝑟𝑞)(𝑝−3√2𝑟𝑞)/
𝑢1 (𝑥, 𝑡) = 0,
5
1
1
(𝑥 − 𝑡)) ,
coth (
√
4
6
√
2 2 −6𝑎 + 5𝑏
Using (7), the solution of (8) when 𝑝 = 8, 𝑞 = 32,
and 𝑟 = 1, and the sets of solutions (3)–(10), we get,
[𝑢7,8 (𝑥, 𝑡), 𝑢9,10 (𝑥, 𝑡), . . . , 𝑢21,22 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = −8, 𝑞 =
32, and 𝑟 = 1, and the sets of solutions (3)–(10), we get,
[𝑢7,8 (𝑥, 𝑡), 𝑢9,10 (𝑥, 𝑡), . . . , 𝑢21,22 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = −(𝑘2 + 1), 𝑞 =
2
2𝑘 , and 𝑟 = 1, and the sets of solutions (3)–(10), we get
𝑢23,24,...,30 (𝑥, 𝑡) = 𝑎0 + 𝑎1 sn 𝜉 + 𝑏1 ns 𝜉, where 𝑎0 , 𝑎1 , and 𝑏1 are
defined in the sets of solutions (3)–(10).
4
Journal of Applied Mathematics
Note that, when 𝑘 → 1, we obtain [𝑢7,8 (𝑥, 𝑡),
𝑢9,10 (𝑥, 𝑡), . . . , 𝑢21,22 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain for
𝑎 > (5/6)𝑏
𝑢31,32 (𝑥, 𝑡) = −
±
1
2
5
1
𝑖
(𝑥 − 𝑡)) ,
csch (
√2
6
√2√6𝑎 − 5𝑏
(14)
for 𝑎 < (5/6)𝑏
1
2
(15)
1
1
5
csc (
±
(𝑥 − 𝑡)) .
√2
6
√2√−6𝑎 + 5𝑏
𝑢33,34 (𝑥, 𝑡) = −
Using (7), the solution of (8) when 𝑝 = −(𝑘2 + 1), 𝑞 =
2𝑘 , and 𝑟 = 1, and the sets of solutions (3)–(10), we get
𝑢35,36,...,42 (𝑥, 𝑡) = 𝑎0 + 𝑎1 cd 𝜉 + 𝑏1 dc 𝜉, where 𝑎0 , 𝑎1 and 𝑏1 are
defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1 we obtain constant solutions,
when 𝑘 → 0 we obtain, [𝑢3,4 (𝑥, 𝑡) and 𝑢5,6 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = −(𝑘2 + 1), 𝑞 =
2, and 𝑟 = 𝑘2 , and the sets of solutions (3)–(10), we get
𝑢43,44,...,50 (𝑥, 𝑡) = 𝑎0 + 𝑎1 ns 𝜉 + 𝑏1 sn 𝜉, where 𝑎0 , 𝑎1 , and 𝑏1
are defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain, [𝑢7,8 (𝑥, 𝑡),
→
0, we obtain
𝑢9,10 (𝑥, 𝑡), . . . , 𝑢21,22 (𝑥, 𝑡)], when 𝑘
[𝑢31,32 (𝑥, 𝑡) and 𝑢33,34 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = −(𝑘2 + 1), 𝑞 =
2, and 𝑟 = 𝑘2 , and the sets of solutions (3)–(10), we get
𝑢51,52,...,57 (𝑥, 𝑡) = 𝑎0 + 𝑎1 dc 𝜉 + 𝑏1 cd 𝜉, where 𝑎0 , 𝑎1 , and 𝑏1
are defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain constant solution, and
when 𝑘 → 0, we obtain [𝑢3,4 (𝑥, 𝑡) and 𝑢5,6 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = 2𝑘2 − 1, 𝑞 =
−2𝑘2 , and 𝑟 = 1 − 𝑘2 , and the sets of solutions (3)–(10), we get
𝑢58,59,...,65 (𝑥, 𝑡) = 𝑎0 + 𝑎1 cn 𝜉 + 𝑏1 nc 𝜉, where 𝑎0 , 𝑎1 , and 𝑏1 are
defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢3,4 (𝑥, 𝑡) and
𝑢5,6 (𝑥, 𝑡)], when 𝑘 → 0, we obtain [𝑢3,4 (𝑥, 𝑡) and 𝑢5,6 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = 2𝑘2 − 1, 𝑞 =
2(1 − 𝑘2 ), and 𝑟 = −𝑘2 , and the sets of solutions (3)–(10), we
get 𝑢66,67,...,73 (𝑥, 𝑡) = 𝑎0 + 𝑎1 nc 𝜉 + 𝑏1 cn 𝜉, where 𝑎0 , 𝑎1 , and 𝑏1
are defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢3,4 (𝑥, 𝑡) and
𝑢5,6 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain [𝑢3,4 (𝑥, 𝑡) and
𝑢5,6 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = 2 − 𝑘2 , 𝑞 = −2,
and 𝑟 = −(1 − 𝑘2 ), and the sets of solutions (3)–(10), we get
𝑢74,75,...,81 (𝑥, 𝑡) = 𝑎0 + 𝑎1 dn 𝜉 + 𝑏1 nd 𝜉, where 𝑎0 , 𝑎1 , and 𝑏1 are
defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢3,4 (𝑥, 𝑡) and
𝑢5,6 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain constant solutions.
Using (7), the solution of (8) when 𝑝 = 2 − 𝑘2 , 𝑞 =
2
2(𝑘 − 1), and 𝑟 = −1, and the sets of solutions (3)–(10), we
get 𝑢82,83,...,89 (𝑥, 𝑡) = 𝑎0 + 𝑎1 nd 𝜉 + 𝑏1 dn 𝜉, where 𝑎0 , 𝑎1 , and
𝑏1 are defined in the sets of solutions (3)–(10).
2
Note that, when 𝑘 → 1, we obtain [𝑢3,4 (𝑥, 𝑡) and
𝑢5,6 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain constant solutions.
Using (7), the solution of (8) when 𝑝 = 2 − 𝑘2 , 𝑞 = 2,
and 𝑟 = 1 − 𝑘2 , and the sets of solutions (3)–(10), we get
𝑢90,91,...,97 (𝑥, 𝑡) = 𝑎0 + 𝑎1 cs 𝜉 + 𝑏1 sc 𝜉, where 𝑎0 , 𝑎1 , and 𝑏1 are
defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢31,32 (𝑥, 𝑡) and
𝑢33,34 (𝑥, 𝑡)], when 𝑘 → 0, we obtain [𝑢7,8 (𝑥, 𝑡), 𝑢9,10 (𝑥, 𝑡), . . .,
𝑢21,22 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = 2 − 𝑘2 , 𝑞 =
2(1 − 𝑘2 ), and 𝑟 = 1, and the sets of solutions (3)–(10), we
get 𝑢98,99,...,105 (𝑥, 𝑡) = 𝑎0 + 𝑎1 sc 𝜉 + 𝑏1 cs 𝜉, where 𝑎0 , 𝑎1 , and 𝑏1
are defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢31,32 (𝑥, 𝑡) and
𝑢33,34 (𝑥, 𝑡)], when 𝑘 → 0, we obtain [𝑢7,8 (𝑥, 𝑡), 𝑢9,10 (𝑥, 𝑡), . . .,
𝑢21,22 (𝑥, 𝑡)].
Using (7), the solution of (8), when 𝑝 = −1 + 2𝑘2 , 𝑞 = 2,
and 𝑟 = −𝑘2 (1 − 𝑘2 ), and the sets of solutions (3)–(10), we get
𝑢106,107,...,113 (𝑥, 𝑡) = 𝑎0 + 𝑎1 ds 𝜉 + 𝑏1 sd 𝜉, where 𝑎0 , 𝑎1 , and 𝑏1
are defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢31,32 (𝑥, 𝑡) and
𝑢33,34 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain also [𝑢31,32 (𝑥, 𝑡)
and 𝑢33,34 (𝑥, 𝑡)].
Using (7), the solution of (8), when 𝑝 = −1 + 2𝑘2 , 𝑞 =
2 2
2𝑘 (𝑘 − 1), and 𝑟 = 1, and the sets of solutions (3)–(10), we
get 𝑢114,115,...,121 (𝑥, 𝑡) = 𝑎0 + 𝑎1 sd 𝜉 + 𝑏1 ds 𝜉, where 𝑎0 , 𝑎1 , and
𝑏1 are defined in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢31,32 (𝑥, 𝑡) and
𝑢33,34 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain also [𝑢31,32 (𝑥, 𝑡) and
𝑢33,34 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = (1+𝑘2 )/2, 𝑞 = (1−
2
𝑘 )/2, and 𝑟 = (1−𝑘2 )/4, and the sets of solutions (3)–(10), we
get 𝑢122,123,...,129 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (sc 𝜉 ± nc 𝜉) + 𝑏1 (1/(sc 𝜉 ± nc 𝜉)),
where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the sets of solutions (3)–
(10).
Note that, when 𝑘 → 1, we obtain constant solutions,
and when 𝑘 → 0, we obtain [𝑢7,8 (𝑥, 𝑡), 𝑢9,10 (𝑥, 𝑡), . . .,
𝑢21,22 (𝑥, 𝑡)], and for 𝑎 > (5/6)𝑏
𝑢130,131 (𝑥, 𝑡)
1 𝑖
=− +
2 2
× (tan (
1
5
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
± sec (
𝑢132,133 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))) ,
√6𝑎 − 5𝑏
6
1 𝑖
=− −
2 2
× (tan (
1
5
(𝑥 − 𝑡))
√6a − 5𝑏
6
± sec (
5
1
(𝑥 − 𝑡))) ,
√6𝑎 − 5𝑏
6
Journal of Applied Mathematics
𝑢134,135 (𝑥, 𝑡)
=
5
𝑢144,145 (𝑥, 𝑡)
5
1
1
(𝑥 − 𝑡))
(−1 + 𝑖 × (tan (
√6𝑎 − 5𝑏
2
6
± sec (
−1
5
1
(𝑥 − 𝑡))) ) ,
√6𝑎 − 5𝑏
6
1
1
=− ±
2 2√2
× (𝑖 tanh (
𝑢136,137 (𝑥, 𝑡)
=
− sech (
5
1
1
(𝑥 − 𝑡))
(−1 − 𝑖 × (tan (
√6𝑎 − 5𝑏
2
6
± sec (
−1
5
1
(𝑥 − 𝑡))) ) ,
√6𝑎 − 5𝑏
6
𝑢138,139 (𝑥, 𝑡)
5
1
1
1
1
(𝑥 − 𝑡))
(𝑖 tanh (
=− ±
√2 √6𝑎 − 5𝑏
2 2√2
6
+ sech (
±
5
1
1
(𝑥 − 𝑡)))
√2 √6𝑎 − 5𝑏
6
5
1
1
1
(𝑥 − 𝑡))
× (𝑖 tanh (
√2 √6𝑎 − 5𝑏
6
2√2
+ sech(
−1
5
1
1
(𝑥 − 𝑡)) ,
√2 √6𝑎 − 5𝑏
6
±
+ sech (
1 𝑖
=− +
2 4
× (tan (
1
1
5
(𝑥 − 𝑡))
2 √6𝑎 − 5𝑏
6
± sec (
5
1
1
(𝑥 − 𝑡))
√2 √6𝑎 − 5𝑏
6
1
5
1
(𝑥 − 𝑡)))
√2 √6𝑎 − 5𝑏
6
5
1
1
1
(𝑥 − 𝑡))
× (𝑖 tanh (
√2 √6𝑎 − 5𝑏
6
2√2
− sech (
𝑢142,143 (𝑥, 𝑡)
+ 𝑖 × (4 (tan (
1
5
1
(𝑥 − 𝑡))
2 √6𝑎 − 5𝑏
6
−1
1
5
1
(𝑥 − 𝑡)))) ,
2 √6𝑎 − 5𝑏
6
−1
5
1
1
(𝑥 − 𝑡))) ,
√2 √6𝑎 − 5𝑏
6
1 𝑖
=− −
2 4
× (tan (
1
5
1
(𝑥 − 𝑡))
2 √6𝑎 − 5𝑏
6
± sec (
1
5
1
(𝑥 − 𝑡)))
2 √6𝑎 − 5𝑏
6
− 𝑖 × (4 (tan (
1
5
1
(𝑥 − 𝑡))
2 √6𝑎 − 5𝑏
6
∓ sec (
1
1
=− ±
2 2√2
−1
1
5
1
(𝑥 − 𝑡)))) .
2 √6𝑎 − 5𝑏
6
(16)
5
1
1
(𝑥 − 𝑡))
√2 √6𝑎 − 5𝑏
6
+ sech (
±
1
5
1
(𝑥 − 𝑡)))
2 √6𝑎 − 5𝑏
6
𝑢148,149 (𝑥, 𝑡)
− sech (
× (𝑖 tanh (
−1
5
1
1
(𝑥 − 𝑡))) ,
√2 √6𝑎 − 5𝑏
6
𝑢146,147 (𝑥, 𝑡)
∓ sec (
1
1
=− ±
2 2√2
× (𝑖 tanh (
5
1
1
(𝑥 − 𝑡)))
√2 √6𝑎 − 5𝑏
6
5
1
1
1
(𝑥 − 𝑡))
× (𝑖 tanh (
√2 √6𝑎 − 5𝑏
6
2√2
𝑢140,141 (𝑥, 𝑡)
±
5
1
1
(𝑥 − 𝑡))
√2 √6𝑎 − 5𝑏
6
1
5
1
(𝑥 − 𝑡)))
√2 √6𝑎 − 5𝑏
6
1
1
1
5
× (𝑖 tanh (
(𝑥 − 𝑡))
√2 √6𝑎 − 5𝑏
6
2√2
− sech (
−1
1
1
5
(𝑥 − 𝑡))) ,
√2 √6𝑎 − 5𝑏
6
For 𝑎 < (5/6)𝑏
𝑢150,151 (𝑥, 𝑡)
5
1 1
1
(𝑥 − 𝑡))
= − ± tanh (
√−6𝑎 + 5𝑏
2 2
6
5
1
𝑖
(𝑥 − 𝑡)) ,
± sech (
√−6𝑎 + 5𝑏
2
6
6
Journal of Applied Mathematics
𝑢152,153 (𝑥, 𝑡)
𝑢161 (𝑥, 𝑡)
5
1 1
1
(𝑥 − 𝑡))
= − ± tanh (
√−6𝑎 + 5𝑏
2 2
6
1
1
5
1
1
(tan (
(𝑥 − 𝑡))
=− +
√2 √−6𝑎 + 5𝑏
2 2√2
6
𝑖
1
5
∓ sech (
(𝑥 − 𝑡)) ,
√−6𝑎 + 5𝑏
2
6
𝑢154,155 (𝑥, 𝑡)
5
1
1
(𝑥 − 𝑡))
= (−1 + 1 × (tanh (
√−6𝑎 + 5𝑏
2
6
−1
5
1
(𝑥 − 𝑡))) ) ,
± 𝑖 sech (
√−6𝑎 + 5𝑏
6
𝑢156,157 (𝑥, 𝑡)
=
− sec (
+
+ sec (
1
1
=− −
2 2√2
−1
5
1
(𝑥 − 𝑡))) ) ,
± 𝑖 sech (
√−6𝑎 + 5𝑏
6
𝑢158,159 (𝑥, 𝑡)
5
1
1
1
1
(𝑥 − 𝑡))
(tan (
=− +
√2 √−6𝑎 + 5𝑏
2 2√2
6
± sec (
5
1
1
(𝑥 − 𝑡)))
√2 √−6𝑎 + 5𝑏
6
5
1
1
1
(𝑥 − 𝑡))
× (tan (
√2 √−6𝑎 + 5𝑏
6
2√2
± sec (
𝑢160 (𝑥, 𝑡)
+
5
1
1
(𝑥 − 𝑡)))
√2 √−6𝑎 + 5𝑏
6
5
1
1
1
(𝑥 − 𝑡))
(tan (
√2 √−6𝑎 + 5𝑏
6
2√2
± sec (
−1
5
1
1
(𝑥 − 𝑡))) ,
√2 √−6𝑎 + 5𝑏
6
𝑢164 (𝑥, 𝑡)
1
1
=− −
2 2√2
× (tan (
+ sec (
−
5
1
1
(𝑥 − 𝑡))
√2 √−6𝑎 + 5𝑏
6
5
1
1
(𝑥 − 𝑡)))
√2 √−6𝑎 + 5𝑏
6
1
1
5
1
× (tan (
(𝑥 − 𝑡))
√2 √−6𝑎 + 5𝑏
6
2√2
− sec (
−1
5
1
1
(𝑥 − 𝑡))) ,
√2 √−6𝑎 + 5𝑏
6
𝑢165 (𝑥, 𝑡)
1
5
1
(𝑥 − 𝑡))
√2 √−6𝑎 + 5𝑏
6
+ sec (
−
−1
5
1
1
(𝑥 − 𝑡))) ,
√2 √−6𝑎 + 5𝑏
6
1
1
=− +
2 2√2
× (tan (
5
1
1
(𝑥 − 𝑡))
√2 √−6𝑎 + 5𝑏
6
± sec (
5
1
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
−1
5
1
1
(𝑥 − 𝑡))) ,
√2 √−6𝑎 + 5𝑏
6
𝑢162,163 (𝑥, 𝑡)
1
(−1 − 1
2
+
1
5
1
1
(𝑥 − 𝑡))
(tan (
√2 √−6𝑎 + 5𝑏
6
2√2
× (tan (
× (tanh (
5
1
1
(𝑥 − 𝑡)))
√2 √−6𝑎 + 5𝑏
6
1
1
=− −
2 2√2
× (tan (
5
1
1
(𝑥 − 𝑡)))
√2 √−6𝑎 + 5𝑏
6
1
1
5
1
× (tan (
(𝑥 − 𝑡))
√2 √−6𝑎 + 5𝑏
6
2√2
− sec (
−1
1
5
1
(𝑥 − 𝑡))) ,
√2 √−6𝑎 + 5𝑏
6
1
1
5
(𝑥 − 𝑡))
√2 √−6𝑎 + 5𝑏
6
− sec (
−
1
5
1
(𝑥 − 𝑡)))
√2 √−6𝑎 + 5𝑏
6
5
1
1
1
(𝑥 − 𝑡))
× (tan (
√
√
√
6
2 2
2 −6𝑎 + 5𝑏
+ sec (
−1
5
1
1
(𝑥 − 𝑡))) ,
√2 √−6𝑎 + 5𝑏
6
Journal of Applied Mathematics
7
𝑢166,167 (𝑥, 𝑡)
𝑢184,185 (𝑥, 𝑡)
1 1
=− +
2 4
5
1
1
(𝑥 − 𝑡)))
(1 ± sech (
1 2
√−6𝑎 + 5𝑏
6
=− −
,
5
1
2
(𝑥 − 𝑡))
tanh (
√−6𝑎 + 5𝑏
6
× (tanh (
1
5
1
(𝑥 − 𝑡))
2 √−6𝑎 + 5𝑏
6
± 𝑖 sech (
1
1
5
(𝑥 − 𝑡)))
√
2 −6𝑎 + 5𝑏
6
+ 1 × (4 (tanh (
𝑢186,187 (𝑥, 𝑡)
1
5
1
1
(𝑥 − 𝑡)))
(tanh (
1 4
2 √−6𝑎 + 5𝑏
6
=− +
1
1
5
2
1 ± sech (
(𝑥 − 𝑡))
2 √−6𝑎 + 5𝑏
6
1
5
1
(𝑥 − 𝑡))
2 √−6𝑎 + 5𝑏
6
± 𝑖 sech (
−1
1
5
1
(𝑥 − 𝑡)))) ,
2 √−6𝑎 + 5𝑏
6
1
1
5
1
(𝑥 − 𝑡)))
(1 ± sech (
4
2 √−6𝑎 + 5𝑏
6
+
,
1
5
1
(𝑥 − 𝑡))
tanh (
2 √−6𝑎 + 5𝑏
6
𝑢168,169 (𝑥, 𝑡)
1 1
=− −
2 4
× (tanh (
± 𝑖 sech (
1
1
5
(𝑥 − 𝑡))
2 √−6𝑎 + 5𝑏
6
𝑢188,189 (𝑥, 𝑡)
1
5
1
1
(𝑥 − 𝑡)))
(tanh (
1 4
2 √−6𝑎 + 5𝑏
6
=− −
1
5
1
2
(𝑥 − 𝑡))
1 ± sech (
2 √−6𝑎 + 5𝑏
6
1
1
5
(𝑥 − 𝑡)))
2 √−6𝑎 + 5𝑏
6
− 1 × (4 (tanh (
1
5
1
(𝑥 − 𝑡))
2 √−6𝑎 + 5𝑏
6
−1
1
5
1
(𝑥 − 𝑡)))) .
± 𝑖 sech (
2 √−6𝑎 + 5𝑏
6
(17)
Using (7), the solution of (8) when 𝑝 = (𝑘2 − 2)/2, 𝑞 =
𝑘 /2, and 𝑟 = 1/4, and the sets of solutions (3)–(10), we
get 𝑢170,171,...,177 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (sn 𝜉/(1 ± dn 𝜉)) + 𝑏1 ((1 ±
dn𝜉)/sn𝜉), where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the sets of
solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain for 𝑎 < (5/6)𝑏
2
𝑢178,179 (𝑥, 𝑡)
5
1
1
(𝑥 − 𝑡))
tanh (
1
√−6𝑎 + 5𝑏
2
6
=− +
,
5
1
2
(𝑥 − 𝑡))
1 ± sech (
√−6𝑎 + 5𝑏
6
1
1
5
1
(𝑥 − 𝑡)))
(1 ± sech (
4
2 √−6𝑎 + 5𝑏
6
−
,
1
5
1
(𝑥 − 𝑡))
tanh (
2 √−6𝑎 + 5𝑏
6
𝑢190 (𝑥, 𝑡)
1
5
1
1
(𝑥 − 𝑡)))
(tanh (
1 4
2 √−6𝑎 + 5𝑏
6
=− +
1
5
1
2
(𝑥 − 𝑡))
1 − sech (
2 √−6𝑎 + 5𝑏
6
1
5
1
1
(𝑥 − 𝑡)))
(1 + sech (
4
2 √−6𝑎 + 5𝑏
6
,
−
1
5
1
(𝑥 − 𝑡))
tanh (
2 √−6𝑎 + 5𝑏
6
𝑢180,181 (𝑥, 𝑡)
𝑢191 (𝑥, 𝑡)
5
1
1
(𝑥 − 𝑡))
(tanh
1
√−6𝑎 + 5𝑏
2
6
=− −
,
1
5
2
1 ± sech (
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
1
1
1
5
(tanh (
(𝑥 − 𝑡)))
1 4
2 √−6𝑎 + 5𝑏
6
=− −
1
5
1
2
(𝑥 − 𝑡))
1 + sech (
2 √−6𝑎 + 5𝑏
6
𝑢182,183 (𝑥, 𝑡)
1
1
5
(1 ± sech (
(𝑥 − 𝑡)))
1 2
√−6𝑎 + 5𝑏
6
=− +
,
5
1
2
(𝑥 − 𝑡))
tanh (
√−6𝑎 + 5𝑏
6
1
1
5
1
(𝑥 − 𝑡)))
(1 − sech (
4
2 √−6𝑎 + 5𝑏
6
+
.
1
5
1
(𝑥 − 𝑡))
tanh (
2 √−6𝑎 + 5𝑏
6
(18)
8
Journal of Applied Mathematics
𝑢205 (𝑥, 𝑡)
For 𝑎 > (5/6)𝑏
1
5
1
1
(𝑥 − 𝑡)))
(tan (
1
4
2 √6𝑎 − 5𝑏
6
=− −𝑖
1
5
1
2
(𝑥 − 𝑡))
1 + sec (
2 √6𝑎 − 5𝑏
6
𝑢192,193 (𝑥, 𝑡)
5
1
1
(𝑥 − 𝑡))
tan (
1
√6𝑎 − 5𝑏
2
6
=− +𝑖
,
5
1
2
(𝑥 − 𝑡))
1 ± sec (
√6𝑎 − 5𝑏
6
𝑢194,195 (𝑥, 𝑡)
5
1
1
(𝑥 − 𝑡))
(tan
1
√6𝑎 − 5𝑏
2
6
=− −
,
5
1
2
(𝑥 − 𝑡))
1 ± sec (
√6𝑎 − 5𝑏
6
𝑢196,197 (𝑥, 𝑡)
5
1
1
(𝑥 − 𝑡)))
(1 ± sec (
1 2
√6𝑎 − 5𝑏
6
=− +
,
5
1
2
(𝑥 − 𝑡))
𝑖 tan (
√6𝑎 − 5𝑏
6
𝑢198,199 (𝑥, 𝑡)
5
1
1
(𝑥 − 𝑡)))
(1 ± sec (
1 2
√6𝑎 − 5𝑏
6
=− −
,
5
1
2
(𝑥 − 𝑡))
𝑖 tan (
√6𝑎 − 5𝑏
6
𝑢200,201 (𝑥, 𝑡)
1
5
1
1
(𝑥 − 𝑡)))
(tan (
1
4
2 √6𝑎 − 5𝑏
6
=− +𝑖
1
5
1
2
(𝑥 − 𝑡))
1 ± sec (
2 √6𝑎 − 5𝑏
6
1
1
5
1
(𝑥 − 𝑡)))
(1 ± sec (
4
2 √6𝑎 − 5𝑏
6
+
,
1
5
1
(𝑥 − 𝑡))
𝑖 tan (
2 √6𝑎 − 5𝑏
6
𝑢202,203 (𝑥, 𝑡)
1
5
1
1
(𝑥 − 𝑡)))
(tan (
1
4
2 √6𝑎 − 5𝑏
6
=− −𝑖
1
5
1
2
(𝑥 − 𝑡))
1 ± sec (
2 √6𝑎 − 5𝑏
6
1
1
1
5
(1 ± sec (
(𝑥 − 𝑡)))
4
2 √6𝑎 − 5𝑏
6
−
,
1
1
5
𝑖 tan (
(𝑥 − 𝑡))
2 √6𝑎 − 5𝑏
6
𝑢204 (𝑥, 𝑡)
1
5
1
1
(𝑥 − 𝑡)))
(tan (
1
4
2 √6𝑎 − 5𝑏
6
=− +𝑖
1
5
1
2
(𝑥 − 𝑡))
1 − sec (
2 √6𝑎 − 5𝑏
6
1
1
5
1
(𝑥 − 𝑡)))
(1 + sec (
4
2 √6𝑎 − 5𝑏
6
−
,
1
5
1
(𝑥 − 𝑡))
𝑖 tan (
2 √6𝑎 − 5𝑏
6
1
1
5
1
(𝑥 − 𝑡)))
(1 − sec (
4
2 √6𝑎 − 5𝑏
6
+
.
1
5
1
(𝑥 − 𝑡))
𝑖 tan (
2 √6𝑎 − 5𝑏
6
(19)
When 𝑘 → 0, we obtain [𝑢31,32 (𝑥, 𝑡) and 𝑢33,34 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = (𝑘2 + 1)/2, 𝑞 =
(𝑘 − 1)/2, and 𝑟 = (1 − 𝑘2 )/4, and the sets of solutions (3)–
(10), we get 𝑢206,207,...,213 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (dn 𝜉/(1 ± 𝑘 sn 𝜉)) +
𝑏1 ((1 ± 𝑘 sn 𝜉)/dn 𝜉), where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the
sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain constant solutions,
and when 𝑘 → 0, we obtain constant solutions.
Using (7), the solution of (8) when 𝑝 = (𝑘2 + 1)/2, 𝑞 =
2
−1/2, and 𝑟 = −(1 − 𝑘2 ) /4, and the sets of solutions (3)–
(10), we get 𝑢214,215,...,221 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (𝑘 cn 𝜉 ± dn 𝜉) +
𝑏1 /(𝑘 cn 𝜉 ± dn 𝜉), where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the sets
of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢3,4 (𝑥, 𝑡) and
𝑢5,6 (𝑥, 𝑡)] and when 𝑘 → 0, we obtain constant solution.
2
Using (7), the solution of (8) when 𝑝 = (𝑘2 + 1)/2, 𝑞 =
(1 − 𝑘2 )/2, and 𝑟 = (1 − 𝑘2 )/4, and the sets of solutions
(3)–(10), we get 𝑢222,223,...,229 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (cn𝜉/(1 ± sn𝜉)) +
𝑏1 ((1 ± sn𝜉)/cn𝜉), where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the sets
of solutions (3)–(10).
Note that, when 𝑘 → 1, we obtain constant solution, and
when 𝑘 → 0, we obtain for 𝑎 > (5/6)𝑏
𝑢230,231 (𝑥, 𝑡)
1 𝑖
=− +
2 2
5
1
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
,
5
1
(𝑥 − 𝑡))
1 ± sin (
√6𝑎 − 5𝑏
6
cos (
𝑢232,233 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
1 𝑖
√6𝑎 − 5𝑏
6
=− −
,
1
5
2 2
1 ± sin (
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
cos (
𝑢234,235 (𝑥, 𝑡)
1 𝑖
=− +
2 2
1
5
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
,
5
1
(𝑥 − 𝑡))
cos (
√6𝑎 − 5𝑏
6
1 ± sin (
Journal of Applied Mathematics
9
𝑢236,237 (𝑥, 𝑡)
For 𝑎 < (5/6)𝑏
𝑢246,247 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
1 𝑖
√6𝑎 − 5𝑏
6
=− −
,
5
1
2 2
(𝑥 − 𝑡))
cos (
√6𝑎 − 5𝑏
6
1 ± sin (
5
1
(𝑥 − 𝑡))
1 1
√−6𝑎 + 5𝑏
6
=− +
,
5
1
2 2
(𝑥 − 𝑡))
𝑖 ± sinh (
√−6𝑎 + 5𝑏
6
𝑢248,249 (𝑥, 𝑡)
cosh (
𝑢238,239 (𝑥, 𝑡)
1
5
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
,
5
1
(𝑥 − 𝑡))
𝑖 ± sinh (
√−6𝑎 + 5𝑏
6
𝑢250,251 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
6
1
1
√2√6𝑎 − 5𝑏
=− +
1
5
2 2√2
1 ± 𝑖 sinh (
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
cosh (
1
+
2√2
1 1
=− −
2 2
5
1
(𝑥 − 𝑡))
√
6
√2 6𝑎 − 5𝑏
,
5
1
(𝑥 − 𝑡))
cosh (
6
√2√6𝑎 − 5𝑏
1 ± 𝑖 sinh (
1 1
=− +
2 2
5
1
(𝑥 − 𝑡))
1 1
√−6𝑎 + 5𝑏
6
=− −
,
5
1
2 2
(𝑥 − 𝑡))
cosh (
√−6𝑎 + 5𝑏
6
𝑢254,255 (𝑥, 𝑡)
𝑖 ± sinh (
1
5
cosh (
(𝑥 − 𝑡))
6
1
1
√2√6𝑎 − 5𝑏
=− −
5
1
2 2√2
(𝑥 − 𝑡))
1 ± 𝑖 sinh (
√
6
√2 6𝑎 − 5𝑏
1
2√2
5
1
(𝑥 − 𝑡))
√
6
1
1
√2 −6𝑎 + 5𝑏
=− +
5
1
2 2√2
(𝑥 − 𝑡))
1 ± sin (
6
√2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡))
1 ± sin (
6
1
√2√−6𝑎 + 5𝑏
+
,
5
1
2√2 cos (
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
,
5
1
(𝑥 − 𝑡))
cosh (
6
√2√6𝑎 − 5𝑏
cos (
1 ± 𝑖 sinh (
𝑢242,243 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
6
1 1
√2√6𝑎 − 5𝑏
=− ±
5
1
2 4
(𝑥 − 𝑡))
𝑖 + sinh (
√
6
√2 6𝑎 − 5𝑏
cosh (
1
±
4
𝑢256,257 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
cos (
6
1
1
√2√−6𝑎 + 5𝑏
=− −
5
1
2 2√2
(𝑥 − 𝑡))
1 ± sin (
6
√2√−6𝑎 + 5𝑏
1
5
1 ± sin (
(𝑥 − 𝑡))
6
1
√2√−6𝑎 + 5𝑏
−
,
5
1
2√2 cos (
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
𝑢258,259 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
,
5
1
(𝑥 − 𝑡))
cosh (
6
√2√6𝑎 − 5𝑏
𝑖 − sinh (
𝑢244,245 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
6
1 1
√2√6𝑎 − 5𝑏
=− ±
5
1
2 4
(𝑥 − 𝑡))
𝑖 − sinh (
6
√2√6𝑎 − 5𝑏
cosh (
1
±
4
5
1
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
,
5
1
(𝑥 − 𝑡))
cosh (
√−6𝑎 + 5𝑏
6
𝑖 ± sinh (
𝑢252,253 (𝑥, 𝑡)
𝑢240,241 (𝑥, 𝑡)
−
cosh (
5
1
(𝑥 − 𝑡))
6
1 𝑖
√2√−6𝑎 + 5𝑏
=− +
5
1
2 4
(𝑥 − 𝑡))
1 ± sin (
6
√2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡))
1 ∓ sin (
6
𝑖
√2√−6𝑎 + 5𝑏
−
,
1
5
4
cos (
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
cos (
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
.
5
1
(𝑥 − 𝑡))
cosh (
6
√2√6𝑎 − 5𝑏
𝑖 + sinh (
(20)
10
Journal of Applied Mathematics
𝑢260,261 (𝑥, 𝑡)
1 𝑖
=− −
2 4
𝑖
+
4
𝑢290,291 (𝑥, 𝑡)
1
=− +1
2
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡))
1 ± sin (
6
√2√−6𝑎 + 5𝑏
cos (
× (2 (coth (
± csch (
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
.
5
1
(𝑥 − 𝑡))
cos (
6
√2√−6𝑎 + 5𝑏
1 ∓ sin (
(21)
Using (7), the solution of (8) when 𝑝 = (1 − 2𝑘2 )/2, 𝑞 =
1/2, and 𝑟 = 𝑘2 /4, and the sets of solutions (3)–(10), we get
𝑢262,263,...269 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (𝑘 sn 𝜉 ± 𝑖 dn 𝜉) + 𝑏1 (1/(𝑘 sn 𝜉 ±
𝑖 dn 𝜉)), where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the sets of solutions
(3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢130,131 (𝑥, 𝑡),
𝑢132,133 (𝑥, 𝑡), . . . , 𝑢168,169 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain
constant solutions.
Using (7), the solution of (8) when 𝑝 = (𝑘2 − 2)/2, 𝑞 =
2
𝑘 /2, and 𝑟 = 𝑘2 /4, and the sets of solutions (3)–(10), we get
𝑢270,271,...,277 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (𝑘 sn 𝜉 ± 𝑖 cn 𝜉) + 𝑏1 (1/(𝑘 sn 𝜉 ±
𝑖 cn 𝜉)), where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the sets of solutions
(3)–(10).
Note that, when 𝑘 → 1, we obtain [𝑢130,131 (𝑥, 𝑡),
𝑢132,133 (𝑥, 𝑡), . . . , 𝑢168,169 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain
constant solutions.
Using (7), the solution of (8) when 𝑝 = (𝑘2 − 2)/2, 𝑞 =
1/2, and 𝑟 = 𝑘4 /4, and the sets of solutions (3)–(10), we get
𝑢278,279,...,285 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (ns 𝜉 ± ds 𝜉) + 𝑏1 (1/(ns 𝜉 ± ds 𝜉)),
where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the sets of solutions (3)–
(10).
Note that, when 𝑘 → 1, we obtain [𝑢7,8 (𝑥, 𝑡),
𝑢9,10 (𝑥, 𝑡), . . . , 𝑢21,22 (𝑥, 𝑡)], and for 𝑎 < (5/6)𝑏
𝑢286,287 (𝑥, 𝑡)
1 1
=− +
2 2
× (coth (
𝑢292,293 (𝑥, 𝑡)
1
=− −1
2
× (2 (coth (
−1
5
1
(𝑥 − 𝑡)))) ,
√−6𝑎 + 5𝑏
6
5
1
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
± csch (
−1
5
1
(𝑥 − 𝑡)))) ,
√−6𝑎 + 5𝑏
6
𝑢294,295 (𝑥, 𝑡)
1 1
=− +
2 4
5
1
(𝑥 − 𝑡))
6
2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡)))
± csch (
6
2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡))
+ 1 × (4 (coth (
6
2√−6𝑎 + 5𝑏
× (coth (
± csch (
−1
5
1
(𝑥 − 𝑡)))) ,
6
2√−6𝑎 + 5𝑏
𝑢296,297 (𝑥, 𝑡)
1 1
=− −
2 4
5
1
(𝑥 − 𝑡))
6
2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡)))
± csch (
6
2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡))
− 1 × (4 (coth (
6
2√−6𝑎 + 5𝑏
× (coth (
± csch (
5
1
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
5
1
(𝑥 − 𝑡))) ,
± csch (
√−6𝑎 + 5𝑏
6
𝑢288,289 (𝑥, 𝑡)
1 1
=− −
2 2
× (coth (
5
1
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
5
1
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
± csch (
5
1
(𝑥 − 𝑡))) ,
√−6𝑎 + 5𝑏
6
−1
1
5
(𝑥 − 𝑡)))) ,
6
2√−6𝑎 + 5𝑏
𝑢298,299 (𝑥, 𝑡)
1 √2
=− +
2
4
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡)))
± csc (
6
√2√−6𝑎 + 5𝑏
× (cot (
+ √2 × (4 (cot (
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
± csc (
−1
5
1
(𝑥 − 𝑡)))) ,
6
√2√−6𝑎 + 5𝑏
Journal of Applied Mathematics
11
𝑢300,301 (𝑥, 𝑡)
𝑢308,309 (𝑥, 𝑡)
1 √2
=− −
2
4
1 𝑖
=− −
2 2
× (cot (
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
± csc (
5
1
(𝑥 − 𝑡)))
6
√2√−6𝑎 + 5𝑏
− √2 × (4 (cot (
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
± csc (
−1
1
5
(𝑥 − 𝑡)))) ,
6
√2√−6𝑎 + 5𝑏
× (cot (
5
1
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
± csc (
𝑢310,311 (𝑥, 𝑡)
1
=− +𝑖
2
× (2 (cot (
𝑢302,303 (𝑥, 𝑡)
1
5
× (cot (
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡)))
6
√2√−6𝑎 + 5𝑏
+ √2 × (4 (cot (
𝑢304,305 (𝑥, 𝑡)
−1
5
1
(𝑥 − 𝑡)))) ,
6
√2√−6𝑎 + 5𝑏
1 √2
=− −
2
4
± csc (
5
1
(𝑥 − 𝑡)))
6
√2√−6𝑎 + 5𝑏
− √2 × (4 (cot (
× (2 (cot (
−1
5
1
(𝑥 − 𝑡)))) .
6
√2√−6𝑎 + 5𝑏
(22)
For 𝑎 > (5/6)𝑏
1 𝑖
=− +
2 2
1 𝑖
=− +
2 4
5
1
(𝑥 − 𝑡))
√
6
2 6𝑎 − 5𝑏
± csc (
± csc (
5
1
(𝑥 − 𝑡))) ,
√6𝑎 − 5𝑏
6
5
1
(𝑥 − 𝑡)))
√
6
2 6𝑎 − 5𝑏
− 𝑖 × (4 (cot (
5
1
(𝑥 − 𝑡))
6
2√6𝑎 − 5𝑏
± csc (
−1
5
1
(𝑥 − 𝑡)))) ,
6
2√6𝑎 − 5𝑏
𝑢316,317 (𝑥, 𝑡)
1 𝑖
=− −
2 4
5
1
(𝑥 − 𝑡))
6
2√6𝑎 − 5𝑏
± csc (
1
5
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
−1
5
1
(𝑥 − 𝑡)))) ,
√6𝑎 − 5𝑏
6
𝑢314,315 (𝑥, 𝑡)
× (cot (
𝑢306,307 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
± csc (
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
∓ csc (
× (cot (
1
=− −𝑖
2
× (cot (
1
5
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
−1
5
1
(𝑥 − 𝑡)))) ,
√6𝑎 − 5𝑏
6
𝑢312,313 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
∓ csc (
× (cot (
1
5
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
± csc (
1 √2
=− +
2
4
± csc (
5
1
(𝑥 − 𝑡))) ,
√6𝑎 − 5𝑏
6
5
1
(𝑥 − 𝑡)))
√
6
2 6𝑎 − 5𝑏
+ 𝑖 × (4 (cot (
5
1
(𝑥 − 𝑡))
√
6
2 6𝑎 − 5𝑏
± csc (
−1
5
1
(𝑥 − 𝑡)))) ,
6
2√6𝑎 − 5𝑏
12
Journal of Applied Mathematics
𝑢318,319 (𝑥, 𝑡)
1 √2𝑖
=− +
2
4
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
5
1
(𝑥 − 𝑡)))
± csch (
6
√2√6𝑎 − 5𝑏
5
1
(𝑥 − 𝑡))
− √2𝑖 × (4 (coth (
6
√2√6𝑎 − 5𝑏
× (coth (
−1
5
1
(𝑥 − 𝑡)))) ,
± csch (
6
√2√6𝑎 − 5𝑏
𝑢320,321 (𝑥, 𝑡)
1 √2𝑖
=− −
2
4
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
5
1
(𝑥 − 𝑡)))
± csch (
6
√2√6𝑎 − 5𝑏
5
1
(𝑥 − 𝑡))
+ √2𝑖 × (4 (coth (
6
√2√6𝑎 − 5𝑏
−1
5
1
(𝑥 − 𝑡)))) ,
± csch (
6
√2√6𝑎 − 5𝑏
𝑢322,323 (𝑥, 𝑡)
× (coth (
1 √2𝑖
=− +
2
4
Using (7), the solution of (8) when 𝑝 = (1 − 2𝑘2 )/2, 𝑞 =
1/2, and 𝑟 = 1/4, and the sets of solutions (3)–(10), we
get 𝑢334,335,...,341 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (cn𝜉/(√1 − 𝑘2 sn 𝜉 ± dn 𝜉)) +
𝑏1 ((√1 − 𝑘2 sn 𝜉 ± dn 𝜉)/cn 𝜉), where 𝑎0 , 𝑎1 , and 𝑏1 are defined
in the sets of solutions (3)–(10).
Note that, when 𝑘 → 1 we obtain constant solutions,
when 𝑘 → 0 we obtain [𝑢230,231 (𝑥, 𝑡), 𝑢232,233 (𝑥, 𝑡), . . .,
𝑢260,261 (𝑥, 𝑡)].
Using (7), the solution of (8) when 𝑝 = (1 + 𝑘2 )/2, 𝑞 =
2
(1 − 𝑘2 ) /2, and 𝑟 = 1/4, and the sets of solutions (3)–(10), we
get 𝑢342,343,...,349 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (sn 𝜉/(cn 𝜉 ± dn 𝜉)) + 𝑏1 ((cn 𝜉 ±
dn 𝜉)/sn 𝜉), where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the sets of
solutions (3)–(10).
Note that, when 𝑘 → 1, we get [𝑢31,32 (𝑥, 𝑡) and
𝑢33,34 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain for 𝑎 > (5/6)𝑏
𝑢350,351 (𝑥, 𝑡)
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
5
1
(𝑥 − 𝑡)))
± csch (
6
√2√6𝑎 − 5𝑏
5
1
(𝑥 − 𝑡))
− √2𝑖 × (4 (coth (
6
√2√6𝑎 − 5𝑏
× (coth (
∓ csch (
−1
5
1
(𝑥 − 𝑡)))) ,
6
√2√6𝑎 − 5𝑏
𝑢324,325 (𝑥, 𝑡)
1 𝑖
=− +
2 2
5
1
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
×(
),
5
1
(𝑥 − 𝑡)) ± 1
cos (
√6𝑎 − 5𝑏
6
sin (
𝑢352,353 (𝑥, 𝑡)
1 𝑖
=− −
2 2
1 √2𝑖
=− −
2
4
1
5
× (coth (
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
± csch (
Using (7), the solution of (8) when 𝑝 = (1 − 2𝑘2 )/2, 𝑞 =
1/2, and 𝑟 = 1/4, and the sets of solutions (3)–(10), we get
𝑢326,327,...,333 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (ns 𝜉 − cs 𝜉) + 𝑏1 (1/(ns 𝜉 − cs 𝜉)),
where 𝑎0 , 𝑎1 , and 𝑏1 are defined in the sets of solutions (3)–
(10).
Note that, when 𝑘 → 1, we obtain [𝑢7,8 (𝑥, 𝑡),
𝑢9,10 (𝑥, 𝑡), . . . , 𝑢21,22 (𝑥, 𝑡)] and [𝑢286,287 (𝑥, 𝑡), 𝑢288,289 (𝑥, 𝑡), . . .,
𝑢324,325 (𝑥, 𝑡)], and when 𝑘 → 0, we obtain [𝑢7,8 (𝑥, 𝑡),
𝑢9,10 (𝑥, 𝑡), . . . , 𝑢21,22 (𝑥, 𝑡)] and [𝑢286,287 (𝑥, 𝑡), 𝑢288,289 (𝑥, 𝑡), . . .,
𝑢324,325 (𝑥, 𝑡)].
1
5
(𝑥 − 𝑡)))
6
√2√6𝑎 − 5𝑏
+ √2𝑖 × (4 (coth (
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
∓ csch (
−1
5
1
(𝑥 − 𝑡)))) .
6
√2√6𝑎 − 5𝑏
(23)
When 𝑘 → 0, we obtain [𝑢31,32 (𝑥, 𝑡) and 𝑢33,34 (𝑥, 𝑡)].
5
1
(𝑥 − 𝑡))
√6𝑎 − 5𝑏
6
×(
),
5
1
(𝑥 − 𝑡)) ± 1
cos (
√6𝑎 − 5𝑏
6
sin (
𝑢354,355 (𝑥, 𝑡)
1 𝑖
=− +
2 2
1
5
(𝑥 − 𝑡)) ± 1
√6𝑎 − 5𝑏
6
×(
),
5
1
(𝑥 − 𝑡))
sin (
√6𝑎 − 5𝑏
6
cos (
Journal of Applied Mathematics
13
𝑢356,357 (𝑥, 𝑡)
𝑢364,365 (𝑥, 𝑡)
1 𝑖
=− −
2 2
1
1
=− +
2 2√2
5
1
(𝑥 − 𝑡)) ± 1
√6𝑎 − 5𝑏
6
×(
),
5
1
(𝑥 − 𝑡))
sin (
√6𝑎 − 5𝑏
6
cos (
𝑢358,359 (𝑥, 𝑡)
1
1
=− +
2 2√2
5
1
(𝑥 − 𝑡)) ± 1
6
√2√6𝑎 − 5𝑏
×(
5
1
(𝑥 − 𝑡))
sinh (
6
√2√6𝑎 − 5𝑏
cosh (
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
+
),
5
1
(𝑥 − 𝑡)) ± 1
cosh (
6
√2√6𝑎 − 5𝑏
sinh (
𝑢360,361 (𝑥, 𝑡)
1
1
=− −
2 2√2
5
1
(𝑥 − 𝑡)) ± 1
cosh (
√
6
√2 6𝑎 − 5𝑏
×(
5
1
(𝑥 − 𝑡))
sinh (
6
√2√6𝑎 − 5𝑏
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
+
),
5
1
(𝑥 − 𝑡)) ± 1
cosh (
6
√2√6𝑎 − 5𝑏
sinh (
𝑢362,363 (𝑥, 𝑡)
1
1
=− +
2 2√2
5
1
(𝑥 − 𝑡)) + 1
6
√2√6𝑎 − 5𝑏
×(
5
1
(𝑥 − 𝑡))
sinh (
√
6
√2 6𝑎 − 5𝑏
cosh (
5
1
(𝑥 − 𝑡))
√
6
√2 6𝑎 − 5𝑏
+
),
5
1
(𝑥 − 𝑡)) − 1
cosh (
6
√2√6𝑎 − 5𝑏
sinh (
5
1
(𝑥 − 𝑡)) − 1
6
√2√6𝑎 − 5𝑏
×(
5
1
(𝑥 − 𝑡))
sinh (
√
6
√2 6𝑎 − 5𝑏
cosh (
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
+
),
5
1
(𝑥 − 𝑡)) + 1
cosh (
6
√2√6𝑎 − 5𝑏
sinh (
𝑢366,367 (𝑥, 𝑡)
1
1
=− −
2 2√2
5
1
(𝑥 − 𝑡)) + 1
6
√2√6𝑎 − 5𝑏
×(
5
1
(𝑥 − 𝑡))
sinh (
6
√2√6𝑎 − 5𝑏
cosh (
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
+
),
5
1
(𝑥 − 𝑡)) − 1
cosh (
6
√2√6𝑎 − 5𝑏
sinh (
𝑢368,369 (𝑥, 𝑡)
1
1
=− −
2 2√2
5
1
(𝑥 − 𝑡)) − 1
6
√2√6𝑎 − 5𝑏
×(
5
1
(𝑥 − 𝑡))
sinh (
6
√2√6𝑎 − 5𝑏
cosh (
5
1
(𝑥 − 𝑡))
6
√2√6𝑎 − 5𝑏
+
),
5
1
(𝑥 − 𝑡)) + 1
cosh (
6
√2√6𝑎 − 5𝑏
sinh (
𝑢370,371 (𝑥, 𝑡)
1 𝑖
=− +
2 4
5
1
(𝑥 − 𝑡)) ± 1
6
2√6𝑎 − 5𝑏
×(
5
1
(𝑥 − 𝑡))
sin (
6
2√6𝑎 − 5𝑏
cos (
5
1
(𝑥 − 𝑡))
6
2√6𝑎 − 5𝑏
−
),
5
1
(𝑥 − 𝑡)) ± 1
cos (
6
2√6𝑎 − 5𝑏
sin (
14
Journal of Applied Mathematics
𝑢372,373 (𝑥, 𝑡)
𝑢382,383 (𝑥, 𝑡)
1 𝑖
=− −
2 4
1
1
=− +
2 2√2
5
1
(𝑥 − 𝑡)) ± 1
6
√2√−6𝑎 + 5𝑏
×(
5
1
(𝑥 − 𝑡))
sin (
√
6
√2 −6𝑎 + 5𝑏
5
1
(𝑥 − 𝑡)) ± 1
√
6
2 6𝑎 − 5𝑏
×(
5
1
(𝑥 − 𝑡))
sin (
6
2√6𝑎 − 5𝑏
cos (
cos (
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
+
),
5
1
(𝑥 − 𝑡)) ± 1
cos (
6
√2√−6𝑎 + 5𝑏
1
5
(𝑥 − 𝑡))
6
2√6𝑎 − 5𝑏
+
).
5
1
(𝑥 − 𝑡)) ± 1
cos (
6
2√6𝑎 − 5𝑏
sin (
sin (
(24)
𝑢384,385 (𝑥, 𝑡)
1
1
=− −
2 2√2
For 𝑎 < (5/6)𝑏
𝑢374,375 (𝑥, 𝑡)
1 1
=− +
2 2
5
1
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
×(
),
5
1
(𝑥 − 𝑡)) ± 1
cosh (
√−6𝑎 + 5𝑏
6
sinh (
5
1
(𝑥 − 𝑡)) ± 1
6
√2√−6𝑎 + 5𝑏
×(
5
1
(𝑥 − 𝑡))
sin (
6
√2√−6𝑎 + 5𝑏
cos (
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
+
),
5
1
(𝑥 − 𝑡)) ± 1
cos (
6
√2√−6𝑎 + 5𝑏
sin (
𝑢375,377 (𝑥, 𝑡)
𝑢386,387 (𝑥, 𝑡)
1 1
=− −
2 2
1
1
=− +
2 2√2
5
1
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
×(
),
5
1
(𝑥 − 𝑡)) ± 1
cosh (
√−6𝑎 + 5𝑏
6
sinh (
𝑢378,379 (𝑥, 𝑡)
1 1
=− +
2 2
5
1
(𝑥 − 𝑡)) ± 1
√−6𝑎 + 5𝑏
6
×(
),
1
5
sinh (
(𝑥 − 𝑡))
√−6𝑎 + 5𝑏
6
cosh (
𝑢380,381 (𝑥, 𝑡)
1 1
=− −
2 2
1
5
(𝑥 − 𝑡)) ± 1
√−6𝑎 + 5𝑏
6
×(
),
5
1
(𝑥 − 𝑡))
sinh (
√−6𝑎 + 5𝑏
6
cosh (
5
1
(𝑥 − 𝑡)) + 1
6
√2√−6𝑎 + 5𝑏
×(
5
1
(𝑥 − 𝑡))
sin (
6
√2√−6𝑎 + 5𝑏
cos (
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
+
),
5
1
(𝑥 − 𝑡)) − 1
cos (
6
√2√−6𝑎 + 5𝑏
sin (
𝑢388,389 (𝑥, 𝑡)
1
1
=− +
2 2√2
5
1
(𝑥 − 𝑡)) − 1
6
√2√−6𝑎 + 5𝑏
×(
5
1
(𝑥 − 𝑡))
sin (
6
√2√−6𝑎 + 5𝑏
cos (
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
+
),
5
1
(𝑥 − 𝑡)) + 1
cos (
6
√2√−6𝑎 + 5𝑏
sin (
Journal of Applied Mathematics
15
𝑢390,391 (𝑥, 𝑡)
Using (7), the solution of (8) when 𝑝 = (𝑘2 − 2)/2, 𝑞 =
𝑘 /2, and 𝑟 = 1/4, and the sets of solutions (3)–(10), we
get 𝑢398,399,...,405 (𝑥, 𝑡) = 𝑎0 + 𝑎1 (cn 𝜉/(√1 − 𝑘2 ± dn 𝜉)) +
𝑏1 ((√1 − 𝑘2 ± dn 𝜉)/cn 𝜉), where 𝑎0 , 𝑎1 , and 𝑏1 are defined in
the sets of solutions (3)–(10).
Note that, when 𝑘 → 1, we get constant solutions, and
when 𝑘 → 0, we obtain, [𝑢3,4 (𝑥, 𝑡) and 𝑢5,6 (𝑥, 𝑡)].
2
1
1
=− −
2 2√2
5
1
(𝑥 − 𝑡)) + 1
6
√2√−6𝑎 + 5𝑏
×(
5
1
(𝑥 − 𝑡))
sin (
6
√2√−6𝑎 + 5𝑏
cos (
4. Conclusion
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
+
),
1
5
cos (
(𝑥 − 𝑡)) − 1
6
√2√−6𝑎 + 5𝑏
sin (
In this paper, the mapping method has been successfully
implemented to find new traveling wave solutions for our
new proposed equation, namely, a combined Padé-II and
modified Padé-II equation. The results show that this method
is a powerful mathematical tool for obtaining exact solutions
for our equation. It is also a promising method to solve other
nonlinear partial differential equations.
𝑢392,393 (𝑥, 𝑡)
1
1
=− −
2 2√2
References
5
1
(𝑥 − 𝑡)) − 1
6
√2√−6𝑎 + 5𝑏
×(
5
1
(𝑥 − 𝑡))
sin (
6
√2√−6𝑎 + 5𝑏
cos (
󸀠
5
1
(𝑥 − 𝑡))
6
√2√−6𝑎 + 5𝑏
+
),
5
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6
√2√−6𝑎 + 5𝑏
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𝑢394,395 (𝑥, 𝑡)
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6
2 −6𝑎 + 5𝑏
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1 1
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2 4
5
1
(𝑥 − 𝑡)) ± 1
6
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×(
5
1
(𝑥 − 𝑡))
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√
6
2 −6𝑎 + 5𝑏
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5
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6
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−
).
5
1
(𝑥 − 𝑡)) ± 1
cosh (
6
2√−6𝑎 + 5𝑏
sinh (
(25)
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