New Jacobi Elliptic Function Solutions for the Kudryashov-Sinelshchikov Equation Using Improved F-Expansion Method

Abstract

Based on the F-expansion method with a new subequation, an improved F-expansion method is introduced. As illustrative examples, some new exact solutions expressed by the Jacobi elliptic function of the Kudryashov-Sinelshchikov equation are obtained. When the modulus m of the Jacobi elliptic function is driven to the limits 1 and 0, some exact solutions expressed by hyperbolic function and trigonometric function can also be obtained. The method is straightforward and concise and is promising and powerful for other nonlinear evolution equations in mathematical physics.

1. Introduction

It has recently become more interesting to obtain exact solutions of nonlinear partial differential equations. These equations are mathematical models of complex physical phenomena that arise in engineering, applied mathematics, chemistry, biology, mechanics, physics, and so forth. Thus, the investigation of the traveling wave solutions to nonlinear evolution equations (NLEEs) plays an important role in mathematical physics. A lot of physical models have supported a wide variety of solitary wave solutions.

In 2010, Kudryashov and Sinelshchikov [1] introduced the following equation: 𝑢_𝑡+𝛾𝑢𝑢_𝑥+𝑢_𝑥𝑥𝑥−𝜀𝑢𝑢_𝑥𝑥𝑥−𝜅𝑢_𝑥𝑢_𝑥𝑥−𝜈𝑢_𝑥𝑥−𝛿𝑢𝑢_𝑥𝑥=0, (1) where𝛾, 𝜀, 𝜅, 𝜈, and𝛿are real parameters. Equation (1) describes the pressure waves in the liquid with gas bubbles taking into account the heat transfer and viscosity. When𝜀=𝜅=𝛿=0 and 𝜀=𝜅=𝜈=𝛿=0, (1) becomes the BKdV equation and the KdV equation, respectively. So, (1) can be considered as the generalization of KdV equation. Therefore, the study to (1) is more meaningful than KdV equation and BKdV equation. We call this equation the Kudryashov-Sinelshchikov equation.

Equation (1) was studied by many researchers in various methods. In the case of 𝜈=𝛿=0, it was studied by Ryabov, using a modification of the truncated expansion method [2], by Randrüüt in a more straightforward manner [3], by Li et al., using the bifurcation method of dynamical systems [4–6], by Nadjafikhah and Shirvani-Sh, using the Lie symmetry method [7], and by He, using𝐺^′/𝐺-expansion method [8]. In the case of𝜈≠0, 𝛿≠0, (1) was studied by Efimova using the modified simplest equation method [9], by Mirzazadeh and Eslami, using first integral method [10]. And they obtained some results when𝛽 took special values.

We noticed that the Jacobi elliptic function solutions of (1) are only reported in [8] with special case 𝛽=−3 and 𝛽=−4. Our aim is to find some new solutions expressed by the Jacobi elliptic function making use of improved F-expansion method.

The organization of the paper is as follows: in Section 2, a brief description of the improved F-expansion for finding traveling wave solutions of nonlinear equations is given. In Sections 3 and 4, we will study, respectively, the Kudryashov-Sinelshchikov equation with the situation 𝜈=𝛿=0 and 𝜈≠0, 𝛿≠0 by the improved F-expansion methods. Finally conclusions are given in Section 5.

2. Description of the Improved Methods

Based on F-expansion method [11–13], the main procedures of the improved F-expansion method are as follows.

Step 1. Consider a general nonlinear PDE in the form 𝐹𝑢,𝑢_𝑥,𝑢_𝑡,𝑢_𝑥𝑥,𝑢_𝑥𝑡,…=0. (2) Using 𝑢(𝑥,𝑡)=𝑈(𝜉), 𝜉=𝑥−𝑐𝑡, we can rewrite (2) as the following nonlinear ODE: 𝐹𝑈,𝑈^′,𝑈^′′,…=0, (3) where the prime denotes differentiation with respect to 𝜉.

Step 2. Suppose that the solution of ODE (3) can be written as follows: 𝑈(𝜉)=𝐴₀+^𝑛_𝑖=1𝐴_𝑖𝐹^𝑖(𝜉)+𝐵_𝑖𝐹^−𝑖(𝜉) (4) or 𝑈(𝜉)=𝐴₀+^𝑛_𝑖=1𝐴_𝑖𝐹^𝑖(𝜉)+𝐵_𝑖𝐹^𝑖−1𝐹^′(𝜉), (5) where 𝐴_𝑖,𝐵_𝑖(𝑖=1,2,…,𝑛) are constants to be determined later and 𝑛 is a positive integer that is given by the homogeneous balance principle. And 𝐹(𝜉) satisfies the following equation 𝐹^′(𝜉)²=𝑞𝐹³(𝜉)+𝑝𝐹²(𝜉)+𝑟𝐹(𝜉), (6) where 𝑟, 𝑝, and 𝑞 are constant.

Step 3. Substituting (4) or (5) along with (6) into (3) and then setting all the coefficients of 𝐹^𝑗(𝜉)𝐹^′𝑘(𝜉)(𝑗=1,2,…,𝑘=0,1) of the resulting system to zero yield a set of overdetermined nonlinear algebraic equations for 𝐴₀,𝐴_𝑖, and 𝐵_𝑖(𝑖=1,2,…𝑛).

Step 4. Assuming that the constants 𝐴₀, 𝐴_𝑖, and 𝐵_𝑖(𝑖=1,2,…𝑛) can be obtained by solving the algebraic equations in Step 3, then by substituting these constants and the solutions of (6) that can be found in Table 1 into (4), we can obtain the explicit solutions of (2) immediately.

Table 1 

Relations between values of 𝑟, 𝑝, 𝑞, and corresponding 𝐹(𝜉) in 𝐹^′2=𝑞𝐹³+𝑝𝐹²+𝑟𝐹.

3. Exact Solutions of the Kudryashov-Sinelshchikov Equation in the Case of 𝛿=𝜈=0

Using scale transformation 𝑥=𝑥^′,𝑡=𝑡^′𝑢,𝑢=^′𝜀, (7) we can write the Kudryashov-Sinelshchikov equation (1) in the form 𝑢_𝑡+𝛼𝑢𝑢_𝑥+𝑢_𝑥𝑥𝑥−𝑢𝑢_𝑥𝑥𝑥−𝛽𝑢_𝑥𝑢_𝑥𝑥−𝜈𝑢_𝑥𝑥−𝜇𝑢𝑢_𝑥𝑥=0, (8) where𝛼=𝛾/𝜀,𝛽=𝜅/𝜀, and𝜇=𝛿/𝜀 (primes are omitted). When𝛿=𝜈=0, (8) becomes 𝑢_𝑡+𝛼𝑢𝑢_𝑥+𝑢_𝑥𝑥𝑥−𝑢𝑢_𝑥𝑥𝑥−𝛽𝑢_𝑥𝑢_𝑥𝑥=0. (9)

Let 𝑢(𝑥,𝑡)=1−𝜑(𝜉),𝜉=𝑥−𝑐𝑡, (10) where 𝑐 is the wave speed. Under this transformation, (9) can be reduced to the following ordinary differential equation (ODE): 𝑐𝜑^′−𝛼1−𝜑𝜑^′−𝜑^′′′+1−𝜑𝜑^′′′−𝛽𝜑^′𝜑^′′=0. (11) Integrating (11) once with respect to𝜉and setting the constant of integration to 𝑅, we have 12𝛼𝜑²+(𝑐−𝛼)𝜑−𝜑𝜑^′′−12𝛽𝜑^′2+𝑅=0. (12)

Balancing 𝜑𝜑^′′ with (𝜑^′)² in (12) we find that 𝑛+𝑛+2=2(𝑛+1), so𝑛is an arbitrary positive integer. For simplicity, we take 𝑛=2. Suppose that (12) owns the solutions in the form 𝜑(𝜉)=𝐴₀+𝐴₁𝐹(𝜉)+𝐴₂𝐹²(𝜉)+𝐵₁𝐹(𝜉)+𝐵₂𝐹²(𝜉). (13)

Substituting (13) and (6) into (12) and then setting all the coefficients of 𝐹^𝑘(𝑘=−5,…,5) of the resulting system to zero, we can obtain the following results: 𝐴₁=𝐴₂=𝐵₂=0,𝑐=2𝐴₀𝐵₁𝑝−𝐵²₁𝑞+3𝑟𝐴₀−𝐵₁𝑝−3𝑟𝐴²₀𝐵₁,𝛼=3𝑟𝐴₀−𝐵₁𝑝𝐵₁𝐴,𝛽=−3,₂=𝐵₁=𝐵₂=0,𝑐=2𝐴₀𝐴₁𝑝−𝐴₁𝑝+3𝑞𝐴₀−𝐴²₁𝑟−3𝑞𝐴²₀𝐴₁,𝛼=3𝑞𝐴₀−𝐴₁𝑝𝐴₁𝐴,𝛽=−3,₂=0,𝐵₁=𝐴₁𝑟𝑞,𝐵₂=0,𝑐=4𝐴²₁𝑟+2𝐴₀𝐴₁𝑝−𝐴₁𝑝+3𝑞𝐴₀−3𝑞𝐴²₀𝐴₁,𝛼=3𝑞𝐴₀−𝐴₁𝑝𝐴₁𝐴,𝛽=−3,₀=𝐴₂2𝑤𝑝−𝑟3𝑞,𝐴₁=2𝑤𝐴₂,𝐵₁=𝐵₂=0,𝑐=−−18𝑞²𝑤+2𝐴₂𝑝²𝑤−6𝑤𝐴₂𝑟𝑞+6𝑝𝑞−𝐴₂𝑝𝑟,53𝑞𝛼=−2𝑝+6𝑤𝑞,𝛽=−2, (14)(15)(16)(17) where 𝑤=(𝑝±𝑝²−3𝑟𝑞)/3𝑞, 𝑞≠0. Consider the following: 𝐴₀=2𝐴₂2𝑤𝑝+5𝑟3𝑞,𝐴₁=4𝐴₂𝐵𝑤,₁=4𝐴₂𝑤𝑟𝑞,𝐵₂=𝐴₂𝑟²𝑞²,𝑐=−2−18𝑤𝑞²+2𝐴₂𝑝²𝑤+24𝐴₂𝑤𝑟𝑞+3𝑝𝑞−16𝐴₂𝑝𝑟×3𝑞⁻¹,5𝛼=−2𝑝+12𝑤𝑞,𝛽=−2, (18) where 𝑤=(𝑝±𝑝²−3𝑟𝑞)/3𝑞, 𝑞≠0. 𝐴₀=𝐵₂2𝑤₁𝑝−𝑞,𝐴3𝑟₁=𝐴₂=0,𝐵₁=2𝑤₁𝐵₂,𝑐=−−18𝑤₁𝑟²+2𝐵₂𝑝²𝑤₁−6𝑤₁𝐵₂𝑞𝑟+6𝑝𝑟−𝐵₂𝑝𝑞,3𝑟𝛼=−2𝑝+6𝑤₁5𝑟,𝛽=−2, (19) where 𝑤₁=(𝑝±𝑝²−3𝑟𝑞)/3𝑟, 𝑟≠0.

Substituting (14)–(19) into (13) with (10), we obtain, respectively, the following formal solution of (9): 𝑢(𝑥,𝑡)=1−𝐴₀−𝐵₁𝐹(𝜉), (20) where𝜉=𝑥−((2𝐴₀𝐵₁𝑝−𝐵²₁𝑞+3𝑟𝐴₀−𝐵₁𝑝−3𝑟𝐴²₀)/𝐵₁)𝑡, 𝛼=(3𝑟𝐴₀−𝐵₁𝑝)/𝐵₁, and 𝛽=−3. Consider the following: 𝑢(𝑥,𝑡)=1−𝐴₀−𝐴₁𝐹(𝜉), (21) where𝜉=𝑥−((2𝐴₀𝐴₁𝑝−𝐴₁𝑝+3𝑞𝐴₀−𝐴²₁𝑟−3𝑞𝐴²₀)/𝐴₁)𝑡, 𝛼=(3𝑞𝐴₀−𝐴₁𝑝)/𝐴₁, and 𝛽=−3. Consider the following: 𝑢(𝑥,𝑡)=1−𝐴₀−𝐴₁𝐹(𝜉)−𝐴₁𝑟(𝜉),𝑞𝐹 (22) where𝜉=𝑥−((4𝐴²₁𝑟+2𝐴₀𝐴₁𝑝−𝐴₁𝑝+3𝑞𝐴₀−3𝑞𝐴²₀)/𝐴₁)𝑡, 𝛼=(3𝑞𝐴₀−𝐴₁𝑝)/𝐴₁, and 𝛽=−3. Consider the following: 𝑢()𝐴𝑥,𝑡=1−²2𝑤𝑝−𝑟3𝑞−2𝑤𝐴₂𝐹(𝜉)−𝐴₂𝐹²(𝜉), (23) where𝜉=𝑥+((−18𝑞²𝑤+2𝐴₂𝑝²𝑤−6𝑤𝐴₂𝑟𝑞+6𝑝𝑞−𝐴₂𝑝𝑟)/3𝑞)𝑡, 𝛼=−2𝑝+6𝑤𝑞, and 𝛽=−5/2.Consider the following: 𝑢(𝑥,𝑡)=1−2𝐴₂2𝑤𝑝+5𝑟3𝑞−4𝑤𝐴₂𝐹(𝜉)−𝐴₂(𝐹(𝜉))²−4𝑤𝐴₂𝑟(𝜉)−𝐴𝑞𝐹²𝑟²𝑞²𝐹²(𝜉), (24) where𝜉=𝑥+(2(−18𝑤𝑞²+2𝐴₂𝑝²𝑤+24𝐴₂𝑤𝑟𝑞+3𝑝𝑞−16𝐴₂𝑝𝑟)/3𝑞)𝑡, 𝛼=−2𝑝+12𝑤𝑞, and 𝛽=−5/2. Consider the following: 𝑢()𝐵𝑥,𝑡=1−²𝑤₁−3𝑟2𝐵₂𝑤₁𝐹(𝜉)−𝐵₂𝐹²(𝜉), (25) where𝜉=𝑥+((−18𝑤₁𝑟²+2𝐵₂𝑝²𝑤₁−6𝑤₁𝐵₂𝑞𝑟+6𝑝𝑟−𝐵₂𝑝𝑞)/3𝑟)𝑡, 𝛼=−2𝑝+6𝑤₁𝑟, and 𝛽=−5/2.

Combining (20)–(25) with Table 1, many exact solutions of (9) can be obtained. For simplicity, we just give out case 1 of Table 1, the other cases can be discussed similarly.

When𝑟=4, 𝑝=−4(𝑚²+1), and 𝑞=4𝑚², the solution of elliptic Equation (6) is𝐹(𝜉)=sn²(𝜉,𝑚). Substituting it into (20)–(24), we can obtain the following solutions of (9).

From (20), one has 𝑢₁(𝑥,𝑡)=1−𝐴₀−𝐵₁ns²(𝜉,𝑚), (26) where𝜉=𝑥+(4(2𝐴₀𝐵₁𝑚²+2𝐴₀𝐵₁+𝐵²₁𝑚²−3𝐴₀−𝐵₁𝑚²−𝐵₁+3𝐴²₀)/𝐵₁)𝑡, 𝛼=(4(3𝐴₀+𝐵₁𝑚²+𝐵₁)/𝐵₁), and 𝛽=−3.

When𝑚→1 and sn(𝜉,𝑚)→tanh(𝜉), (26) becomes 𝑢(𝑥,𝑡)=1−𝐴₀−𝐵₁coth²(𝜉), (27) where𝜉=𝑥+(4(4𝐴₀𝐵₁−3𝐴₀−2𝐵₁+𝐵²₁+3𝐴²₀)/𝐵₁)𝑡, 𝛼=4(3𝐴₀+2𝐵₁)/𝐵₁, and 𝛽=−3.

When𝑚→0 and sn(𝜉,𝑚)→sin(𝜉), (26) becomes 𝑢(𝑥,𝑡)=1−𝐴₀−𝐵₁sec²(𝜉), (28) where𝜉=𝑥+(4(2𝐴₀𝐵₁−3𝐴₀−𝐵₁+3𝐴²₀)/𝐵₁)𝑡, 𝛼=(4(3𝐴₀+𝐵₁)/𝐵₁), and 𝛽=−3.

From (21), we have 𝑢(𝑥,𝑡)=1−𝐴₀−𝐴₁sn²(𝜉,𝑚), (29) where𝜉=𝑥+4((2𝐴₀𝐴₁𝑚²+2𝐴₀𝐴₁+3𝑚²𝐴²₀+𝐴²₁−𝐴₁𝑚²−𝐴₁−3𝑚²𝐴₀)/𝐴₁)𝑡, 𝛼=4((𝐴₁𝑚²+𝐴₁+3𝑚²𝐴₀)/𝐴₁), and 𝛽=−3.

When𝑚→1, (29) becomes a hyperbolic function solution, 𝑢(𝑥,𝑡)=1−𝐴₀−𝐵₁tanh²(𝜉), (30) where𝜉=𝑥+(4(4𝐴₀𝐵₁−3𝐴₀−2𝐵₁+𝐵²₁+3𝐴²₀)/𝐵₁)𝑡, 𝛼=(4(3𝐴₀+2𝐵₁)/𝐵₁), and 𝛽=−3.

When𝑚→0, (29) becomes a trigonometric function solution, 𝑢(𝑥,𝑡)=1−𝐴₀−𝐵₁sin²(𝜉), (31) where𝜉=𝑥+4(2𝐴₀+𝐴₁−1)𝑡, 𝛼=4, and 𝛽=−3.

From (22), we have 𝑢(𝑥,𝑡)=1−𝐴₀−𝐴₁sn²()−𝐴𝜉,𝑚¹𝑟𝑞ns²(𝜉,𝑚), (32) where𝜉=𝑥+(4(2𝐴₀𝐴₁𝑚²+2𝐴₀𝐴₁+3𝑚²𝐴²₀−4𝐴²₁−𝐴₁𝑚²−𝐴₁−3𝑚²𝐴₀)/𝐴₁)𝑡, 𝛼=(4(𝐴₁𝑚²+𝐴₁+3𝑚²𝐴₀)/𝐴₁), and 𝛽=−3.

When𝑚→1, (32) becomes a hyperbolic function solution, 𝑢(𝑥,𝑡)=1−𝐴₀−𝐴₁tanh²(𝜉)−𝐴₁𝑟𝑞coth²(𝜉), (33) where𝜉=𝑥+(4(4𝐴₀𝐴₁+3𝐴²₀−4𝐴²₁−2𝐴₁−3𝐴₀)/𝐴₁)𝑡, 𝛼=(4(2𝐴₁+3𝐴₀)/𝐴₁), and 𝛽=−3.

From (23), we have 𝑢()𝐴𝑥,𝑡=1+²2𝑤𝑚²+1+13𝑚²−2𝑤𝐴₂sn²(𝜉,𝑚)−𝐴₂sn⁴(𝜉,𝑚), (34) where𝑤=(−𝑚²−1±√𝑚⁴−𝑚²+1)/3𝑚², 𝜉=𝑥+(4(−2𝑤𝐴₂𝑚²+2𝑤𝐴₂𝑚⁴+2𝑤𝐴₂−18𝑤𝑚⁴+𝐴₂𝑚²+𝐴₂−6𝑚⁴−6𝑚²)/3𝑚²)𝑡, 𝛼=8(3𝑤𝑚²+𝑚²+1), and 𝛽=−5/2.

When𝑚→1, (34) becomes a hyperbolic function solution, 𝑢()1𝑥,𝑡=1−𝐴12²𝑤−2𝑤𝐴₂tanh²(𝜉)−𝐴₂tanh⁴(𝜉,𝑥), (35) where 𝑤=(−2±1)/3, 𝜉=8(3𝑤−(1/3)𝐴₂𝑤+2−(1/3)𝐴₁)𝑡, 𝛼=16+24𝑤, and 𝛽=−5/2.

From (24), we have 𝑢(𝑥,𝑡)=1+2𝐴₂2𝑤𝑚²+1−53𝑚²−4𝑤𝐴₂sn²(𝜉,𝑚)−𝐴₂sn⁴()−𝜉,𝑚4𝑤𝐴₂𝑚²ns²()−𝐴𝜉,𝑚²𝑚⁴ns⁴(𝜉,𝑚), (36) where 𝑤=(−𝑚²−1±√𝑚⁴+14𝑚²+1)/6𝑚², 𝜉=𝑥+(8(−18𝑤𝑚⁴+2𝑤𝐴₂𝑚⁴+28𝑤𝐴₂𝑚²+2𝑤𝐴₂−3𝑚⁴−3𝑚²+16𝐴₂𝑚²+16𝐴₂)/3𝑚²)𝑡, 𝛼=8(𝑚²+1+6𝑤𝑚²), and 𝛽=−5/2.

When𝑚→1, (36) becomes a hyperbolic function solution, 𝑢()2𝑥,𝑡=1+3𝐴₂(4𝑤−5)−4𝑤𝐴₂tanh²(𝜉)−𝐴₂tanh⁴(𝜉)−4𝑤𝐴₂coth²(𝜉)−𝐴₂coth⁴(𝜉), (37) where𝑤=(−1±2)/3,𝜉=𝑥−(48𝑤−(256/3)𝑤𝐴₂+16−(256/3)𝐴₂)𝑡,𝛼=16+48𝑤, and 𝛽=−5/2.

From (25), we have 𝑢()1𝑥,𝑡=1+3𝐵₂2𝑤𝑚²+1+𝑚²−2𝑤𝐵₂ns²(𝜉,𝑚)−𝐵₂ns⁴(𝜉,𝑚), (38) where𝑤=(−𝑚²−1±√𝑚⁴−𝑚²+1)/3,𝜉=𝑥+((8/3)𝑤𝐵₂𝑚⁴+(8/3)𝑤𝐵₂𝑚²−(8/3)𝑤𝐵₂+24𝑤+8𝑚²+8−(4/3)𝐵₂𝑚⁴−(4/3)𝐵₂𝑚²)𝑡, 𝛼=8(3𝑤+𝑚²+1), and 𝛽=−5/2.

When𝑚→1, (38) becomes a hyperbolic function solution, 𝑢()1𝑥,𝑡=1+3𝐵₂(4𝑤+1)−2𝑤𝐵₂coth²(𝜉)−𝐵₂coth⁴(𝜉), (39) where𝑤=(−2±1)/3, 𝜉=𝑥−(24𝑤−(8/3)𝑤𝐵₂+16−(8/3)𝐵₂)𝑡, 𝛼=16+24𝑤, and 𝛽=−5/2.

When𝑚→0, (38) becomes a triangle function solution, 𝑢()2𝑥,𝑡=1+3𝑤𝐵₂−2𝑤𝐵₂csc²(𝜉)−𝐵₂csc⁴(𝜉), (40) where𝑤=(−1±1)/3, 𝜉=𝑥−(24𝑤+8−(8/3)𝐵₂𝑤)𝑡, 𝛼=24𝑤+8, and 𝛽=−5/2.

We notice that when𝛽=−3, some Jacobi elliptic function solutions have been given in [8]. Part of our results may be the same with them. However, when 𝛽=−(5/2), the Jacobi elliptic function solutions of (9) have not been reported in the related literatures, so we believe that our solutions (34), (36), and (38) are new.

4. Exact Solutions of the Kudryashov-Sinelshchikov Equation in the Case of 𝛿≠0,𝜇≠0

In this case, by similar process, (1) can be changed into the following PDE: 12𝛼𝜑²+(𝑐−𝛼)𝜑−𝜑𝜑^′′−12𝛽𝜑^′2+𝜈𝜑^′+𝜇1−𝜑𝜑^′+𝑅=0, (41) where𝜑^′=𝑑𝜑/𝑑𝜉, 𝜉=𝑥−𝑐𝑡, and 𝑅is an integral constant.

Suppose that (41) owns the solutions in the form 𝜑(𝜉)=𝐴₀+𝐴₁𝐹(𝜉)+𝐴₂𝐹²(𝜉)+𝐵₁𝐹(𝜉)^′+𝐵₂𝐹(𝜉)𝐹(𝜉)^′. (42)

Substituting (42) and (6) into (41) and then setting all the coefficients of𝐹^𝑗(𝜉)𝐹^′𝑘(𝜉)(𝑗=0,1,…,6,𝑘=0,1) of the resulting system to zero, we can obtain the following results: 𝐴₀=0,𝐴₁=𝑤𝐵₁,𝐴₂=𝐵₂5=0,𝑐=−3𝑤𝐵₁4𝑟−34𝑝,𝛼=−38𝑝,𝛽=−32,𝜆=3𝐵₁1𝑟+31𝑤,𝜇=−3𝑤, (43) where𝑤=±√𝑝. Consider the following: 𝐴₀=𝐵₂𝑤2𝑝𝑤²−2𝑝²+3𝑟𝑞9𝑞²,𝐴₁=2𝐵₂𝑤𝑤−𝑝,𝐴3𝑞₂=−𝐵₂𝑤,𝐵₁𝐵=−²𝑤²−𝑝,3𝑞𝑐=−𝑤14𝑤²𝐵₂𝑝²−42𝑤²𝐵₂𝑟𝑞+63𝐵₂𝑟𝑝𝑞+108𝑤𝑞²−14𝐵₂𝑝³×45𝑞²⁻¹,𝛼=−125𝑤²,𝛽=−125,𝜆=−2𝑤²𝐵₂𝑝²−2𝐵₂𝑝³+9𝐵₂𝑟𝑝𝑞−6𝑤²𝐵₂𝑟𝑞+9𝑤𝑞²×45𝑞²⁻¹,1𝜇=5𝑤 (44) where𝑤=±⁴𝑝²−3𝑟𝑞.

Substituting (43) and (44) along with (10) into (42), we obtain, respectively, the following formal solution of (8): 𝑢(𝑥,𝑡)=1−𝑤𝐵₁𝐹(𝜉)−𝐵₁𝐹^′(𝜉), (45) where 𝑤=±√𝑝,𝜉=𝑥+((5/3)𝑤𝐵₁𝑟+(4/3)𝑝)𝑡,𝛼=−(4/3)𝑝,𝛽=−8/3,𝜆=(2/3)𝐵₁𝑟+(1/3)𝑤, and 𝜇=−(1/3)𝑤. Consider the following: 𝑢()𝐵𝑥,𝑡=1−²𝑤2𝑝𝑤²−2𝑝²+3𝑟𝑞9𝑞²−2𝐵₂𝑤𝑤²−𝑝𝐹(𝜉)3𝑞+𝑤𝐵₂𝐹²(𝜉)+𝐵₂𝑤²−𝑝𝐹^′(𝜉)3𝑞−𝐵₂𝐹(𝜉)𝐹^′(𝜉), (46) where𝑤=±⁴𝑝²−3𝑟𝑞, 𝜉=𝑥+(𝑤(14𝑤²𝐵₂𝑝²−42𝑤²𝐵₂𝑟𝑞+63𝐵₂𝑟𝑝𝑞+108𝑤𝑞²−14𝐵₂𝑝³)/45𝑞²)𝑡, 𝛼=−(12/5)𝑤², 𝛽=−12/5, 𝜆=−((2𝑤²𝐵₂𝑝²−2𝐵₂𝑝³+9𝐵₂𝑟𝑝𝑞−6𝑤²𝐵₂𝑟𝑞+9𝑤𝑞²)/45𝑞²), and 𝜇=(1/5)𝑤.

Combining (45) and (46) with Table 1, many exact solutions of (8) can be obtained. For simplicity, we just give out case 3 of Table 1, and the other cases can be discussed similarly.

When𝑟=−4(1−𝑚²), 𝑝=4(2−𝑚²), 𝑞=−4, and the solution of elliptic Equation (6) is𝐹(𝜉)=dn²(𝜉,𝑚). Substituting it into (45) and (46), we can obtain the following Jacobi elliptic function solutions of (8): 𝑢(𝑥,𝑡)=1−𝐵₁𝑤dn²(𝜉,𝑚)−2𝑚²sn(𝜉,𝑚)cn(𝜉,𝑚)dn(𝜉,𝑚), (47) where𝑤=±2√2−𝑚², 𝜉=𝑥−((20/3)𝑤𝐵₁−(20/3)𝑤𝐵₁𝑚²−(32/3)+(16/3)𝑚²)𝑡, 𝛼=(16/3)(𝑚²−2), 𝛽=−8/3,𝜈=(1/3)(−8𝐵₁+8𝐵₁𝑚²+𝑤), 𝜇=−(1/3)𝑤, and 𝑢()𝐵𝑥,𝑡=1−²𝑤2𝑝𝑤²−2𝑝²+3𝑟𝑞9𝑞²−2𝐵₂𝑤𝑤²−𝑝dn²(𝜉,𝑚)3𝑞+𝑤𝐵₂dn⁴()−𝜉,𝑚2𝑚²𝐵₂𝑤²−𝑝dn(𝜉,𝑚)cn(𝜉,𝑚)sn(𝜉,𝑚)3𝑞+2𝑚²𝐵₂dn³(𝜉,𝑚)cn(𝜉,𝑚)sn(𝜉,𝑚), (48) where𝑤=±2√⁴1−𝑚²−𝑚⁴,𝜉=𝑥+(2𝑤/45)(7𝑤²𝐵₂−7𝑤²𝐵₂𝑚²+28𝐵₂−42𝐵₂𝑚²−42𝐵₂𝑚⁴+7𝑤²𝐵₂𝑚⁴+28𝐵₂𝑚⁶+54𝑤)𝑡, 𝛼=−(12/5)𝑤²,𝛽=−(12/5), 𝜈=−(2/45)𝑤²𝐵₂+(2/45)𝑤²𝐵₂𝑚²−(2/45)𝑤²𝐵₂𝑚⁴−(8/45)𝐵₂+(4/15)𝐵₂𝑚²+(4/15)𝐵₂𝑚⁴−(8/45)𝐵₂𝑚⁶−(1/5)𝑤, and 𝜇=(1/5)𝑤.

Similarly, when 𝑚→1 we can obtain hyperbolic function solutions of (8). Here, we omit them.

We indicate that these results with 𝛽=−12/5 and 𝛽=−8/3 are new.

5. Conclusions

The Jacobi elliptic function solutions of (1) are only reported in [8] with special case of 𝛽=−3 and𝛽=−4. In the present work, we successfully obtained some new Jacobi elliptic function solutions of the Kudryashov-Sinelshchikov equation with 𝛽=−3,𝛽=−5/2,𝛽=−8/3, and 𝛽=−12/5 using the improved F-expansion method. When the modulus m of the Jacobi elliptic function is driven to the limits 1 and 0, some exact solutions expressed by hyperbolic function and trigonometric function can also be obtained. All the results we obtained have been verified. The related results are enriched.

Acknowledgments

This research is supported by the Natural Science Foundation of China (11161020), the Natural Science Foundation of Yunnan (2013FZ117), and Research Foundation of Honghe University (10XJY120).