Classification of Exact Solutions for Generalized Form of 𝐾(𝑚,𝑛) Equation

Abstract

The classification of exact solutions, including solitons and elliptic solutions, to the generalized 𝐾(𝑚,𝑛) equation by the complete discrimination system for polynomial method has been obtained. From here, we find some interesting results for nonlinear partial differential equations with generalized evolution.

1. Introduction

In science and engineering applications, it is often very difficult to obtain analytical solutions of partial differential equations. Recently, many exact solutions of partial differential equations have been examined by the use of trial equation method. Also there are a lot of important methods that have been defined such as Hirota method, tanh-coth method, sine-cosine method, the trial equation method, and the extended trial equation method [1–15] to find exact solutions to nonlinear partial differential equations. There are a lot of nonlinear evolution equations that are solved by the use of various mathematical methods. Soliton solutions, singular solitons, and other solutions have been found by using these approaches. These obtained solutions have been encountered in various areas of applied mathematics and are very important.

In Section 2, we introduce an extended trial equation method for nonlinear evolution equations with higher order nonlinearity. In Section 3, as applications, we procure some exact solutions to nonlinear partial differential equations such as the generalized form of 𝐾(𝑚,𝑛) equation [16–18]: 𝑞^𝑙_𝑡+𝑎𝑞^𝑚𝑞_𝑥+𝑏𝑞^𝑛_𝑥𝑥𝑥=0, (1) where 𝑎,𝑏∈𝑅 are constants since 𝑙,𝑚, and 𝑛∈𝑍⁺. Here, the first term is the generalized evolution term, while the second term represents the nonlinear term and the third term is the dispersion term. This equation is the generalized form of the KdV equation, where, in particular, the case 𝑙=𝑚=𝑛=1 leads to the KdV equation. The Korteweg de Vries equation is one of the most important equations in applied mathematics and physics. There have been several kinds of solutions, such as compactons, that are studied in the context of 𝐾(𝑚,𝑛) equation, for various situations. We now offer a more general trial equation method for discussion as follows.

2. The Extended Trial Equation Method

Step 1. For a given nonlinear partial differential equation 𝑃𝑢,𝑢_𝑡,𝑢_𝑥,𝑢_𝑥𝑥,…=0, (2) take the general wave transformation 𝑢𝑥₁,𝑥₂,…,𝑥_𝑁,𝑡=𝑢𝜂,𝜂=𝜆^𝑁_𝑗=1𝑥_𝑗−𝑐𝑡, (3) where 𝜆≠0 and 𝑐≠0. Substituting (3) into (2) yields a nonlinear ordinary differential equation: 𝑁𝑢,𝑢^′,𝑢^′′,…=0. (4)

Step 2. Take the finite series and trial equation as follows: 𝑢=^𝛿_𝑖=0𝜏_𝑖Γ^𝑖, (5) where Γ^′2(Γ)=Φ(Γ)=ΛΨ(Γ)=𝜉_𝜃Γ^𝜃+⋯+𝜉₁Γ+𝜉₀𝜁_𝜖Γ^𝜖+⋯+𝜁₁Γ+𝜁₀. (6) Using (5) and (6), we can write 𝑢^′2=Φ(Γ)Ψ(Γ)^𝛿_𝑖=0𝑖𝜏_𝑖Γ^𝑖−12,𝑢^′′=Φ^′(Γ)Ψ(Γ)−Φ(Γ)Ψ^′(Γ)2Ψ²(Γ)^𝛿_𝑖=0𝑖𝜏_𝑖Γ^𝑖−1+Φ(Γ)Ψ(Γ)^𝛿_𝑖=0𝑖(𝑖−1)𝜏_𝑖Γ^𝑖−2, (7) where Φ(Γ) and Ψ(Γ) are polynomials. Substituting these relations into (4) yields an equation of polynomial Ω(Γ) of Γ: Ω(Γ)=𝜚_𝑠Γ^𝑠+⋯+𝜚₁Γ+𝜚₀=0. (8) According to the balance principle, we can find a relation of 𝜃, 𝜖, and 𝛿. We can calculate some values of 𝜃, 𝜖, and 𝛿.

Step 3. Letting the coefficients of Ω(Γ) all be zero will yield an algebraic equations system: 𝜚_𝑖=0,𝑖=0,…,𝑠. (9) Solving this system, we will determine the values of 𝜉₀,…,𝜉_𝜃; 𝜁₀,…,𝜁_𝜖; and 𝜏₀,…,𝜏_𝛿.

Step 4. Reduce (6) to the elementary integral form ±𝜂−𝜂₀=𝑑ΓΛ(Γ)=Ψ(Γ)Φ(Γ)𝑑Γ. (10) Using a complete discrimination system for polynomial to classify the roots of Φ(Γ), we solve (10) and obtain the exact solutions to (4). Furthermore, we can write the exact traveling wave solutions to (2), respectively.

3. Application to the Generalized Form of 𝐾(𝑚,𝑛) Equation

In order to look for travelling wave solutions of (1), we make the transformation 𝑞(𝑥,𝑡)=𝑢(𝜂), 𝜂=𝑥−𝑐𝑡, where 𝑐 is the wave speed. Therefore it can be converted to the ODE −𝑐𝑢^𝑙𝜂^′+𝑎𝑢𝑚+1^𝑚+1𝜂^′+𝑏𝑢^𝑛𝜂^′′′=0, (11) where prime denotes the derivative with respect to 𝜂. Then, integrating this equation with respect to 𝜂 one time and setting the integration constant to zero, we obtain −𝑐𝑢^𝑙𝜂+𝑎𝑢𝑚+1^𝑚+1𝜂+𝑏𝑢^𝑛𝜂^′′=0. (12) Let 𝑙=𝑛, applying balance and using the following transformation: 𝑢=𝑣^{1/(𝑚−𝑛+1)}. (13) Equation (12) turns into the following equation: −𝑐(𝑚+1)(𝑚+1−𝑛)²𝑣²+𝑎(𝑚+1−𝑛)²𝑣³+𝑏𝑛(𝑚+1)(2𝑛−𝑚−1)𝑣^′2+𝑏𝑛(𝑚+1)(𝑚+1−𝑛)𝑣𝑣^′′=0. (14) Substituting (7) into (14) and using balance principle yield 𝜃=𝜖+𝛿+2. (15) After this solution procedure, we obtain the results as follows.

Case 1. If we take 𝜖=0, 𝛿=1, and 𝜃=3, then 𝑣^′2=𝜏₁²𝜉₃Γ³+𝜉₂Γ²+𝜉₁Γ+𝜉₀𝜁₀,𝑣^′′=𝜏₁3𝜉₃Γ²+2𝜉₂Γ+𝜉₁2𝜁₀, (16) where 𝜉₃≠0 and 𝜁₀≠0. Respectively, solving the algebraic equation system (9) yields 𝜉₀𝜉=−²¹(3+3𝑚−5𝑛)(1+𝑚+𝑛)16𝜉₂(1+𝑚−2𝑛)²,𝜉₁=𝜉₁,𝜉₂=𝜉₂,𝜉₃=−8𝜉²₂(1+𝑚−2𝑛)(1+𝑚−𝑛)𝜉₁(1+𝑚+𝑛)²,𝜏₀=𝜏₀,𝜏₁4()𝜉=−1+𝑚−2𝑛₂𝜏₀(1+𝑚+𝑛)𝜉₁,𝜁₀=−𝑏𝑛𝜉₂(1+𝑚)𝑎(1+𝑚−𝑛)𝜏₀,()𝜏𝑐=𝑎𝑛5+5𝑚−7𝑛₀().1+𝑚)(1+𝑚−𝑛)(1+𝑚+𝑛 (17) Substituting these results into (6) and (10), we have±𝜂−𝜂₀=𝐴2∫𝑑ΓΓ³−𝜉₁(1+𝑚+𝑛)²8𝜉₂()Γ1+𝑚−2𝑛)(1+𝑚−𝑛²−𝜉²₁(1+𝑚+𝑛)²8𝜉²₂()𝜉1+𝑚−2𝑛)(1+𝑚−𝑛Γ+³¹(3+3𝑚−5𝑛)(1+𝑚+𝑛)³128𝜉²₂(1+𝑚−2𝑛)³(),1+𝑚−𝑛 (18)where 𝐴=𝑏𝑛𝜉₁(1+𝑚)(1+𝑚+𝑛)²2𝑎𝜉₂𝜏₀(1+𝑚−𝑛)²().1+𝑚−2𝑛 (19) Integrating (18), we obtain the solutions to (1) as follows: ±𝜂−𝜂₀𝐴=−Γ−𝛼₁,±𝜂−𝜂₀=𝐴√𝛼₂−𝛼₁arctanΓ−𝛼₂𝛼₂−𝛼₁,𝛼₂>𝛼₁,±𝜂−𝜂₀=𝐴√𝛼₁−𝛼₂lnΓ−𝛼₂−√𝛼₁−𝛼₂Γ−𝛼₂+√𝛼₁−𝛼₂,𝛼₁>𝛼₂,±𝜂−𝜂₀𝐴=−√𝛼₁−𝛼₃𝐹𝜑,𝑙,𝛼₁>𝛼₂>𝛼₃, (20) where 𝐹𝜑,𝑙=^𝜑₀𝑑𝜓1−𝑙²sin²𝜓,𝜑=arcsinΓ−𝛼₃𝛼₂−𝛼₃,𝑙²=𝛼₂−𝛼₃𝛼₁−𝛼₃. (21) Also 𝛼₁, 𝛼₂, and 𝛼₃ are the roots of the polynomial equation Γ³+𝜉₂𝜉₃Γ²+𝜉₁𝜉₃𝜉Γ+⁰𝜉₃=0. (22) Substituting solutions (20) into (5) and (13), we have 𝑢()=𝜏𝑥,𝑡₀+𝜏₁𝛼₁+𝐴²𝜏₁𝑥−𝑎𝑛(5+5𝑚−7𝑛)𝜏₀×((1+𝑚)(1+𝑚−𝑛)(1+𝑚+𝑛))⁻¹×𝑡−𝜂₀^{2−11/(𝑚−𝑛+1)},𝑢(𝑥,𝑡)=𝜏₀+𝜏₁𝛼₁+𝜏₁𝛼₂−𝛼₁×sech²√𝛼₂−𝛼₁𝐴𝑥−𝑎𝑛(5+5𝑚−7𝑛)𝜏₀×)×(((1+𝑚)(1+𝑚−𝑛1+𝑚+𝑛))⁻¹×𝑡−𝜂₀^{1/(𝑚−𝑛+1)},𝑢()=𝜏𝑥,𝑡₀+𝜏₁𝛼₁+𝜏₁𝛼₁−𝛼₂×cosech²√𝛼₁−𝛼₂()𝜏2𝐴𝑥−𝑎𝑛5+5𝑚−7𝑛₀×)×(((1+𝑚)(1+𝑚−𝑛1+𝑚+𝑛))⁻¹×𝑡−𝜂₀^{1/(𝑚−𝑛+1)},𝑢()=𝜏𝑥,𝑡₀+𝜏₁𝛼₁+𝜏₁𝛼₂−𝛼₁×𝑠𝑛²±√𝛼₂−𝛼₁𝐴×()𝜏𝑥−𝑎𝑛5+5𝑚−7𝑛₀(1+𝑚)(1+𝑚−𝑛)(1+𝑚+𝑛)×𝑡−𝜂₀,𝛼₁−𝛼₃𝛼₁−𝛼₂^{−11/(𝑚−𝑛+1)}. (23) If we take 𝜏₀=−𝜏₁𝛼₁ and 𝜂₀=0, then solutions (23) can reduce to rational function solution 𝑢()=𝐴𝑥,𝑡𝑥−𝑐𝑡^{2/(𝑚−𝑛+1)}, (24) 1-soliton wave solution 𝑢()=𝐵𝑥,𝑡cosh^{2/(𝑚−𝑛+1)}(𝐵(,𝑥−𝑐𝑡)) (25) singular soliton solution 𝑢()=𝐶𝑥,𝑡sinh^{2/(𝑚−𝑛+1)}(𝐶(,𝑥−𝑐𝑡)) (26) and elliptic soliton solution 𝑢()=𝐵𝑥,𝑡𝑠𝑛^{2/(𝑚−𝑛+1)},𝜑,𝑙 (27) where 𝐴=𝐴√𝜏₁, 𝐵=(𝜏₁(𝛼₂−𝛼₁))^{1/(𝑚−𝑛+1)}, 𝐵=√𝛼₂−𝛼₁/𝐴, 𝐶=(𝜏₁(𝛼₁−𝛼₂))^{1/(𝑚−𝑛+1)}, 𝐶=√𝛼₁−𝛼₂/2𝐴, 𝜑=±(√𝛼₂−𝛼₁/𝐴)(𝑥−𝑐𝑡), 𝑙²=(𝛼₁−𝛼₃)/(𝛼₁−𝛼₂), and 𝑐=𝑎𝑛(5+5𝑚−7𝑛)𝜏₁𝛼₁/(1+𝑚)(1+𝑚−𝑛)(1+𝑚+𝑛). Here, 𝐵 and 𝐶 are the amplitudes of the solitons, while 𝐵 and 𝐶 are the inverse widths of the solitons and 𝑐 is the velocity. Thus, we can say that the solitons exist for 𝜏₁>0.

Remark 1. If we choose the corresponding values for some parameters, solution (25) is in full agreement with solution (21) mentioned in [17].

Case 2. If we take 𝜖=0, 𝛿=2, and 𝜃=4, then 𝑣^′2=𝜏₁+2𝜏₂Γ²𝜉₄Γ⁴+𝜉₃Γ³+𝜉₂Γ²+𝜉₁Γ+𝜉₀𝜁₀, (28) where 𝜉₄≠0 and 𝜁₀≠0. Respectively, solving the algebraic equation system (9) yields 𝜉₀=𝜉₀,𝜉₁=𝜉₁,𝜉₂=𝜉²₁3𝜉₀,𝜉₃=𝜉³₁24𝜉²₀,𝜉₄=𝜉⁴₁576𝜉³₀,𝜁₀()𝜉=−𝑏𝑛𝑚+1)(𝑚+𝑛+1³₁24𝑎(𝑚−𝑛+1)²𝜉²₀𝜏₁,𝜏₀=2𝜉₀𝜏₁𝜉₁,𝜏₁=𝜏₁,𝜏₂=𝜉₁𝜏₁12𝜉₀,𝑐=−2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁. (29) Substituting these results into (6) and (10), we get ±𝜂−𝜂₀=2𝐴₁()×Γ𝑑Γ₄+24𝜉₀𝜉₁Γ³+192𝜉²₀𝜉²₁Γ²+576𝜉³₀𝜉³₁Γ+576𝜉⁴₀𝜉⁴₁^−1/2, (30) where 𝐴₁=−6𝑏𝑛𝜉₀(1+𝑚)(1+𝑚+𝑛)/𝑎𝜉₁𝜏₁(1+𝑚−𝑛)². Integrating (30), we obtain the solutions to (1) as follows: ±𝜂−𝜂₀=−2𝐴₁Γ−𝛼₁,±𝜂−𝜂₀=4𝐴₁𝛼₁−𝛼₂Γ−𝛼₂Γ−𝛼₁,𝛼₁>𝛼₂,±𝜂−𝜂₀=2𝐴₁𝛼₁−𝛼₂lnΓ−𝛼₁Γ−𝛼₂,±𝜂−𝜂₀=4𝐴₁𝛼₁−𝛼₂𝛼₁−𝛼₃×lnΓ−𝛼₂𝛼₁−𝛼₃−Γ−𝛼₃𝛼₁−𝛼₂Γ−𝛼₂𝛼₁−𝛼₃+Γ−𝛼₃𝛼₁−𝛼₂,𝛼₁>𝛼₂>𝛼₃,±𝜂−𝜂₀=4𝐴₁𝛼₁−𝛼₄𝛼₂−𝛼₃𝐹,𝛼𝜑,𝑙₁>𝛼₂>𝛼₃>𝛼₄, (31) where 𝜑₁=arcsinΓ−𝛼₂𝛼₁−𝛼₄Γ−𝛼₁𝛼₂−𝛼₄,𝑙²₁=𝛼₁−𝛼₃𝛼₂−𝛼₄𝛼₂−𝛼₃𝛼₁−𝛼₄. (32) Also 𝛼₁, 𝛼₂, 𝛼₃, and 𝛼₄ are the roots of the polynomial equation Γ⁴+𝜉₃𝜉₄Γ³+𝜉₂𝜉₄Γ²+𝜉₁𝜉₄𝜉Γ+⁰𝜉₄=0. (33) Substituting solutions (31) into (5) and (13), we have 𝑢()=𝜏𝑥,𝑡₀+𝜏₁𝛼₁±2𝜏₁𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁𝑡−𝜂₀⁻¹+𝜏₂𝛼₁±2𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁×𝑡−𝜂₀^{−121/(𝑚−𝑛+1)},𝑢()=𝜏𝑥,𝑡₀+𝜏₁𝛼₁+16𝐴²₁𝛼₂−𝛼₁𝜏₁×16𝐴²₁−𝛼₁−𝛼₂×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁𝑡−𝜂₀²⁻¹+𝜏₂𝛼₁+16𝐴²₁𝛼₂−𝛼₁×16𝐴²₁−𝛼₁−𝛼₂×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁×𝑡−𝜂₀^{2−121/(𝑚−𝑛+1)},𝑢()=𝜏𝑥,𝑡₀+𝜏₁𝛼₂+𝛼₂−𝛼₁𝜏₁×𝛼exp¹−𝛼₂2𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁𝑡−𝜂₀−1⁻¹+𝜏₂𝛼₂+𝛼₂−𝛼₁×𝛼exp¹−𝛼₂2𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁×𝑡−𝜂₀−1^{−121/(𝑚−𝑛+1)},𝑢()=𝜏𝑥,𝑡₀+𝜏₁𝛼₁+𝛼₁−𝛼₂𝜏₁×𝛼exp¹−𝛼₂2𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁×(𝑚+1)(𝑚+𝑛+1)𝜉₁⁻¹×𝑡−𝜂₀−1⁻¹+𝜏₂𝛼₁+𝛼₁−𝛼₂×𝛼exp¹−𝛼₂2𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁×(𝑚+1)(𝑚+𝑛+1)𝜉₁⁻¹×𝑡−𝜂₀−1^{−121/(𝑚−𝑛+1)},𝑢()=𝜏𝑥,𝑡₀+𝜏₁𝛼₁−2𝛼₁−𝛼₂𝛼₁−𝛼₃𝜏₁×2𝛼₁−𝛼₂−𝛼₃+𝛼₃−𝛼₂×cosh𝛼₁−𝛼₂𝛼₁−𝛼₃2𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁×𝑡−𝜂₀⁻¹+𝜏₂𝛼₁−2𝛼₁−𝛼₂𝛼₁−𝛼₃×2𝛼₁−𝛼₂−𝛼₃+𝛼₃−𝛼₂×cosh𝛼₁−𝛼₂𝛼₁−𝛼₃2𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁×𝑡−𝜂₀^{−121/(𝑚−𝑛+1)},𝑢()=𝜏𝑥,𝑡₀+𝜏₁𝛼₂+𝛼₁−𝛼₂𝛼₄−𝛼₂𝜏₁×𝛼₁−𝛼₄×𝑠𝑛²𝛼₁−𝛼₃𝛼₂−𝛼₄4𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁×𝑡−𝜂₀,𝛼₂−𝛼₃𝛼₁−𝛼₄𝛼₁−𝛼₃𝛼₂−𝛼₄+𝛼₄−𝛼₂⁻¹+𝜏₂𝛼₂+𝛼₁−𝛼₂𝛼₄−𝛼₂𝜏₁×𝛼₁−𝛼₄×𝑠𝑛²𝛼₁−𝛼₃𝛼₂−𝛼₄4𝐴₁×𝑥+2𝑎𝑛𝜉₀𝜏₁(𝑚+1)(𝑚+𝑛+1)𝜉₁×𝑡−𝜂₀,𝛼₂−𝛼₃𝛼₁−𝛼₄𝛼₁−𝛼₃𝛼₂−𝛼₄+𝛼₄−𝛼₂^{−121/(𝑚−𝑛+1)}. (34)
For simplicity, if we take 𝜂₀=0, then we can write solutions (34) as follows: 𝑢(𝑥,𝑡)=²_𝑖=0𝜏_𝑖𝛼₁±2𝐴₁𝑥−𝑐𝑡^{𝑖1/(𝑚−𝑛+1)},𝑢()=𝑥,𝑡_2𝑖=0𝜏_𝑖𝛼₁+16𝐴²₁𝛼₁−𝛼₂×16𝐴²₁−𝛼₁−𝛼₂(𝑥−𝑐𝑡)^{2−1𝑖1/(𝑚−𝑛+1)},𝑢(𝑥,𝑡)=²_𝑖=0𝜏_𝑖𝛼₂+𝛼₂−𝛼₁exp𝐵₁(𝑥−𝑐𝑡)−1^{𝑖1/(𝑚−𝑛+1)},𝑢(𝑥,𝑡)=²_𝑖=0𝜏_𝑖𝛼₁+𝛼₁−𝛼₂exp𝐵₁(𝑥−𝑐𝑡)−1^{𝑖1/(𝑚−𝑛+1)},𝑢(𝑥,𝑡)=²_𝑖=0𝜏_𝑖𝛼₁−2𝛼₁−𝛼₂𝛼₁−𝛼₃×2𝛼₁−𝛼₂−𝛼₃+𝛼₃−𝛼₂×cosh𝐶₁(𝑥−𝑐𝑡)^{−1𝑖1/(𝑚−𝑛+1)},𝑢(𝑥,𝑡)=²_𝑖=0𝜏_𝑖𝛼₂+𝛼₁−𝛼₂𝛼₄−𝛼₂×𝛼₁−𝛼₄𝑠𝑛²𝜑,𝑙+𝛼₄−𝛼₂^{−1𝑖1/(𝑚−𝑛+1)}, (35) where 𝐵₁=(𝛼₁−𝛼₂)/2𝐴₁, 𝐶₁=(𝛼₁−𝛼₂)(𝛼₁−𝛼₃)/2𝐴₁, 𝜑₁=((𝛼₁−𝛼₃)(𝛼₂−𝛼₄)/4𝐴₁)(𝑥−𝑐𝑡), 𝑙²₁=(𝛼₂−𝛼₃)(𝛼₁−𝛼₄)/(𝛼₁−𝛼₃)(𝛼₂−𝛼₄), and 𝑐=−2𝑎𝑛𝜉₀𝜏₁/(𝑚+1)(𝑚+𝑛+1)𝜉₁. Here, 𝐴₁ is the amplitude of the soliton, while 𝑐 is the velocity and 𝐵₁ and 𝐶₁ are the inverse widths of the solitons. Thus, we can say that the solitons exist for 𝜏₁>0.

Case 3. If we take 𝜖=0, 𝛿=3, and 𝜃=5, then 𝑣^′2=𝜏₁+2𝜏₂Γ+3𝜏₃Γ²²×𝜉₅Γ⁵+𝜉₄Γ⁴+𝜉₃Γ³+𝜉₂Γ²+𝜉₁Γ+𝜉₀×𝜁₀⁻¹, (36) where 𝜉₅≠0 and 𝜁₀≠0. Respectively, solving the algebraic equation system (9) yields 𝜉₀=𝜉₅𝜏²₂−4𝜏₁𝜏₃2𝜏³₂−9𝜏₁𝜏₂𝜏₃+2𝜏²₂−3𝜏₁𝜏₃³81𝜏⁵₃,𝜉₁𝜉=−⁵4𝜏⁴₂+9𝜏₁𝜏²₂𝜏₃−108𝜏²₁𝜏²₃+4𝜏₂𝜏²₂−3𝜏₁𝜏₃³81𝜏⁴₃,𝜉₂=𝜉₅−11𝜏³₂+63𝜏₁𝜏₂𝜏₃−2𝜏²₂−3𝜏₁𝜏₃³27𝜏³₃,𝜉₃=𝜉₅𝜏²₂+7𝜏₁𝜏₃3𝜏²₃,𝜉₄=5𝜉₅𝜏₂3𝜏₃,𝜁₀()𝜉=−9𝑏𝑛𝑚+1)(𝑚+𝑛+1₅2𝑎𝜏₃(𝑚−𝑛+1)²,𝜏₀=−2𝜏³₂−9𝜏₁𝜏₂𝜏₃+2𝜏²₂−3𝜏₁𝜏₃³27𝜏²₃,𝜏₁=𝜏₁,𝜏₂=𝜏₂,𝜏₃=𝜏₃,𝜉₅=𝜉₅,𝑐=−8𝑎𝑛𝜏²₂−3𝜏₁𝜏₃³27(𝑚+1)(𝑚+𝑛+1)𝜏²₃. (37) Substituting these results into (6) and (10), we get ±𝜂−𝜂₀=3𝐴₂()×Γ𝑑Γ₅+𝜉₄𝜉₅Γ⁴+𝜉₃𝜉₅Γ³+𝜉₂𝜉₅Γ²+𝜉₁𝜉₅𝜉Γ+⁰𝜉₅^−1/2, (38) where 𝐴₂=−𝑏𝑛(1+𝑚)(1+𝑚+𝑛)/2𝑎𝜏₃(1+𝑚−𝑛)². Integrating (38), we obtain the solutions to (1) as follows: ±𝜂−𝜂₀=−2𝐴₂Γ−𝛼₁³,±𝜂−𝜂₀=3𝐴₂arctanhΓ−𝛼₂/𝛼₁−𝛼₂𝛼₁−𝛼₂^3/2−3𝐴₂Γ−𝛼₂𝛼₁−𝛼₂Γ−𝛼₁,𝛼₁>𝛼₂,±𝜂−𝜂₀=−6𝐴₂arctanΓ−𝛼₁/𝛼₁−𝛼₂𝛼₁−𝛼₂^3/2−6𝐴₂Γ−𝛼₁𝛼₁−𝛼₂,±𝜂−𝜂₀=6𝐴₂arctanhΓ−𝛼₃/𝛼₂−𝛼₃𝛼₁−𝛼₂×1√𝛼₂−𝛼₃−1√𝛼₁−𝛼₃,𝛼₁>𝛼₂>𝛼₃,±𝜂−𝜂₀=−6𝐴₂Γ−𝛼₁𝛼₁−𝛼₂𝛼₁−𝛼₃×Γ−𝛼₂Γ−𝛼₃+𝑖𝐸𝜑,𝑙−𝐹𝜑,𝑙, (39) where 𝐸𝜑,𝑙=^𝜑₀1−𝑙²sin²𝜑𝜓𝑑𝜓,₂=−arcsinΓ−𝛼₁𝛼₂−𝛼₁,𝑙²₂=𝛼₁−𝛼₂𝛼₁−𝛼₃,±𝜂−𝜂₀=−6𝑖𝐴₂√𝛼₂−𝛼₃𝛼₁−𝛼₂𝐹,𝛼𝜑,𝑙−𝜋𝜑,𝑛,𝑙₁>𝛼₂>𝛼₃>𝛼₄, (40) where 𝜑₃=−arcsin𝛼₃−𝛼₂Γ−𝛼₂,𝑙²₃=𝛼₂−𝛼₄𝛼₂−𝛼₃,𝛼𝑛=²−𝛼₁𝛼₂−𝛼₃. (41) Also 𝛼₁, 𝛼₂, 𝛼₃, 𝛼₄, and 𝛼₅ are the roots of the polynomial equation Γ⁵+𝜉₄𝜉₅Γ⁴+𝜉₃𝜉₅Γ³+𝜉₂𝜉₅Γ²+𝜉₁𝜉₅𝜉Γ+⁰𝜉₅=0. (42)

Case 4. If we take 𝜖=1, 𝛿=1, and 𝜃=4, then 𝑣^′2=𝜏²₁𝜉₄Γ⁴+𝜉₃Γ³+𝜉₂Γ²+𝜉₁Γ+𝜉₀𝜁₀+𝜁₁Γ, (43) where 𝜉₄≠0 and 𝜁₁≠0. Respectively, solving the algebraic equation system (9) yields 𝜉₀=𝜁₀𝜏²₀𝑀+2𝑎(1+𝑚−𝑛)²2𝜁₁𝜏₀+𝜁₀𝜏₁𝑏𝑛(1+𝑚)(1+𝑚+𝑛)𝜁₁𝜏²₁,𝜉₃=𝜉₃,𝜉₄()=−2𝑎1+𝑚−𝑛²𝜁₁𝜏₁(),𝜉𝑏𝑛1+𝑚)(1+𝑚+𝑛₁=𝜏₀4𝑎(1+𝑚−𝑛)²𝜁²₁𝜏²₀+2𝜁₀𝜏₁𝑀+2𝑎(1+𝑚−𝑛)²𝜁₀𝜏₁+𝜁₁𝜏₀𝑀+8𝑎(1+𝑚−𝑛)²𝜁₀𝜏₁×𝑏𝑛(1+𝑚)(1+𝑚+𝑛)𝜁₁𝜏²₁⁻¹,𝜉₂=6𝑎(1+𝑚−𝑛)²𝜁²₁𝜏²₀+2𝜁₁𝜏₀𝑀+2𝑎(1+𝑚−𝑛)²𝜁₀𝜏₁+𝜁₀𝜏₁𝑀+2𝑎(1+𝑚−𝑛)²𝜁₀𝜏₁×𝑏𝑛(1+𝑚)(1+𝑚+𝑛)𝜁₁𝜏²₁⁻¹,𝜁₀=𝜁₀,𝜁₁=𝜁₁,𝜏₀=𝜏₀,𝜏₁=𝜏₁,𝑛()𝑐=𝑀+2𝑎1+𝑚−𝑛²3𝜁₁𝜏₀+𝜁₀𝜏₁(1+𝑚)(1+𝑚+𝑛)(1+𝑚−𝑛)²𝜁₁, (44) where 𝑀=𝑏𝑛(1+𝑚)(1+𝑚+𝑛)𝜉₃. Substituting these results into (6) and (10), we get ±𝜂−𝜂₀=𝐴₃𝜁Γ+⁰𝜁₁×Γ⁴+𝜉₃𝜉₄Γ³+𝜉₂𝜉₄Γ²+𝜉₁𝜉₄𝜉Γ+⁰𝜉₄^−11/2𝑑Γ, (45) where 𝐴₃=𝑏𝑛(1+𝑚)(1+𝑚+𝑛)/2𝑎𝜏₁(1+𝑚−𝑛)². Integrating (45), we obtain the solution to (1) as follows: ±𝜂−𝜂₀=−𝐴₃𝜁₁𝜁₀+𝜁₁𝛼₁×arctanh𝜁₀+𝜁₁Γ𝜁₀+𝜁₁𝛼₁−𝐴₃Γ−𝛼₁𝜁₀+𝜁₁Γ𝜁₁,±𝜂−𝜂₀=2𝐴₃𝛼₁−𝛼₂×−𝜁₀+𝜁₁𝛼₁𝜁₁×arctanh𝜁₀+𝜁₁Γ𝜁₀+𝜁₁𝛼₁+𝜁₀+𝜁₁𝛼₂×arctanh𝜁₀+𝜁₁Γ𝜁₀+𝜁₁𝛼₂,±𝜂−𝜂₀=2𝐴₃×Γ−𝛼₁𝜁₀+𝜁₁Γ𝜁₁Γ−𝛼₂²+𝑖√𝛼₁−𝛼₂𝐸𝜑,𝑙−𝐹𝜑,𝑙, (46) where 𝜑₄=−arcsin𝜁₁𝛼₁−Γ𝜁₀+𝜁₁𝛼₁,𝑙²₄=𝜁₀+𝜁₁𝛼₁𝜁₁𝛼₁−𝛼₂,±𝜂−𝜂₀=−2𝐴₃√𝛼₂−𝛼₁𝐸𝜑,𝑙, (47) where 𝜑₅=arcsin𝛼₂−𝛼₁Γ−𝛼₁,𝑙²₅=𝜁₀+𝜁₁𝛼₁𝜁₁𝛼₁−𝛼₂,±𝜂−𝜂₀=−2𝑖𝐴₃𝛼₁−𝛼₂𝜁₁𝜁₀+𝜁₁𝛼₂×𝜁₀𝐹𝜑,𝑙−𝜋𝜑,𝑛,𝑙+𝜁₁𝛼₂𝐹𝜑,𝑙−𝛼₂𝜋𝜑,𝑛,𝑙, (48) where 𝜑₆=−arcsin𝜁₀+𝜁₁𝛼₂𝜁₁𝛼₂,𝑙−Γ₂₆=𝜁₁𝛼₂−𝛼₃𝜁₀+𝜁₁𝛼₂,𝑛₁=𝜁₁𝛼₂−𝛼₁𝜁₀+𝜁₁𝛼₂. (49)

Case 5. If we take 𝜖=1, 𝛿=2, and 𝜃=5, then (𝑣^′)²=𝜏₁+2𝜏₂Γ²𝜉₅Γ⁵+𝜉₄Γ⁴+𝜉₃Γ³+𝜉₂Γ²+𝜉₁Γ+𝜉₀𝜁₀+𝜁₁Γ, (50) where 𝜉₅≠0 and 𝜁₁≠0. Respectively, solving the algebraic equation system (9) yields 𝜉₀=𝜏²₀−2𝜉₅𝜏₁+𝜉₄𝜏₂𝜏³₂,𝜉₁=𝜏₀2𝜉₄𝜏₁𝜏₂+𝜉₅−4𝜏²₁+𝜏₀𝜏₂𝜏³₂,𝜉₂=𝜉₄𝜏₂𝜏²₁+2𝜏₀𝜏₂−2𝜉₅𝜏³₁+𝜏₀𝜏₁𝜏₂𝜏³₂,𝜉₃=−3𝜉₅𝜏²₁+2𝜏₂𝜉₅𝜏₀+𝜉₄𝜏₁𝜏²₂,𝜁₀=−2𝑏𝑛(1+𝑚)(1+𝑚+𝑛)𝜉₄𝜏₂−2𝜉₅𝜏₁𝑎(1+𝑚−𝑛)²𝜏²₂,𝜁₁=−2𝑏𝑛(1+𝑚)(1+𝑚+𝑛)𝜉₅𝑎(1+𝑚−𝑛)²𝜏²₂,𝜉₄=𝜉₄,𝜉₅=𝜉₅,𝜏₀=𝜏₀,𝜏₁=𝜏₁,𝜏₂=𝜏₂,𝜏𝑐=−𝑎𝑛²₁−4𝜏₀𝜏₂2(1+𝑚)(1+𝑚+𝑛)𝜏₂. (51) Substituting these results into (6) and (10), we get ±𝜂−𝜂₀=𝐴₄×𝜁Γ+⁰𝜁₁×Γ⁵+𝜉₄𝜉₅Γ⁴+𝜉₃𝜉₅Γ³+𝜉₂𝜉₅Γ²+𝜉₁𝜉₅𝜉Γ+⁰𝜉₅^−11/2𝑑Γ, (52) where 𝐴₄=−2𝑏𝑛(1+𝑚)(1+𝑚+𝑛)/𝑎𝜏²₂(1+𝑚−𝑛)². Integrating (52), we obtain the solution to (1) as follows: ±𝜂−𝜂₀=−2𝐴₄3𝜁₁𝜁₀+𝜁₁𝛼₁𝜁₀+𝜁₁ΓΓ−𝛼₁^3/2,±𝜂−𝜂₀=−𝐴₄𝜁₀+𝜁₁𝛼₂2𝛼₁−𝛼₂^3/2𝜁₁𝜁₀+𝜁₁𝛼₁×lnΓ−𝛼₁×𝜁₀Γ+𝛼₁−2𝛼₂+2𝜁₀+𝜁₁Γ𝜁₀+𝜁₁𝛼₁Γ−𝛼₂𝛼₁−𝛼₂+𝜁₁2Γ𝛼₁−𝛼₂Γ+𝛼₁⁻¹−𝐴₄𝛼₁−𝛼₂Γ−𝛼₁𝜁₀+𝜁₁ΓΓ−𝛼₂𝜁₁,𝛼₁>𝛼₂,±𝜂−𝜂₀=−2𝐴₄𝛼₁−𝛼₂𝜁₀+𝜁₁Γ𝜁₁Γ−𝛼₁−2𝐴₄𝛼₁−𝛼₂^3/2𝜁₀+𝜁₁𝛼₂𝜁₁×arctanΓ−𝛼₁𝜁₀+𝜁₁𝛼₂𝛼₁−𝛼₂𝜁₀+𝜁₁Γ,±𝜂−𝜂₀=−𝐴₄𝛼₁−𝛼₃𝜁₀+𝜁₁𝛼₂𝜁₁𝛼₂−𝛼₃×ln𝛼₂×𝜁−Γ₀Γ+𝛼₂−2𝛼₃+2𝜁₀+𝜁₁Γ𝜁₀+𝜁₁𝛼₂Γ−𝛼₃𝛼₂−𝛼₃+𝜁₁2Γ𝛼₂−𝛼₃Γ+𝛼₂⁻¹−𝐴₄𝛼₁−𝛼₃×𝜁₀+𝜁₁𝛼₁𝜁₁𝛼₁−𝛼₃×ln𝜁₀Γ+𝛼₁−2𝛼₃+2𝜁₀+𝜁₁Γ𝜁₀+𝜁₁𝛼₁Γ−𝛼₃𝛼₁−𝛼₃+𝜁₁2Γ𝛼₁−𝛼₃Γ+𝛼₁×Γ−𝛼₂⁻¹,𝛼₁>𝛼₂>𝛼₃,±𝜂−𝜂₀=−2𝐴₄𝛼₁−𝛼₃×𝜁₀+𝜁₁𝛼₃𝜁₁𝛼₁−𝛼₂𝐸,𝛼𝜑,𝑙₁>𝛼₂>𝛼₃, (53) where 𝜑₇=arcsinΓ−𝛼₃𝛼₂−𝛼₁Γ−𝛼₁𝛼₂−𝛼₃,𝑙²₇=𝛼₃−𝛼₂𝜁₀+𝜁₁𝛼₁𝛼₁−𝛼₂𝜁₀+𝜁₁𝛼₃,±𝜂−𝜂₀=2𝐴₄𝛼₂−𝛼₄𝛼₁−𝛼₂𝛼₃−𝛼₄𝜁₁𝛼₂−𝛼₃𝜁₀+𝜁₁𝛼₄×𝜁₀+𝜁₁Γ𝛼₃−𝛼₄𝛼₁−𝛼₂𝜋+𝜁𝜑,𝑛,𝑙₀+𝜁₁𝛼₂𝛼₄−𝛼₃𝛼₂−𝛼₄𝐹𝜑,𝑙, (54) where 𝜑₈=arcsinΓ−𝛼₃𝛼₂−𝛼₁Γ−𝛼₁𝛼₂−𝛼₃,𝑙²₈=𝛼₃−𝛼₂𝜁₀+𝜁₁𝛼₁𝛼₁−𝛼₂𝜁₀+𝜁₁𝛼₃,𝑛₂𝛼=−¹−𝛼₂𝛼₃−𝛼₄𝛼₂−𝛼₃𝛼₁−𝛼₄,𝛼₁>𝛼₂>𝛼₃>𝛼₄. (55)

4. Discussion

Thus we introduce a more general extended trial equation method for nonlinear partial differential equations as follows.

Step 1. Extended trial equation (6) can be reduced to the following more general form: 𝐴(Γ)𝑢=𝐵(Γ)=∑^𝛿_𝑖=0𝜏_𝑖Γ^𝑖∑^𝜇_𝑗=0𝜔_𝑗Γ^𝑗, (56) where Γ^′2(Γ)=Φ(Γ)=ΛΨ(Γ)=𝜉_𝜃Γ^𝜃+⋯+𝜉₁Γ+𝜉₀𝜁_𝜖Γ^𝜖+⋯+𝜁₁Γ+𝜁₀. (57) Here, 𝜏_𝑖(𝑖=0,…,𝛿), 𝜔_𝑗(𝑗=0,…,𝜇), 𝜉_𝜍(𝜍=0,…,𝜃), and 𝜁_𝜎(𝜎=0,…,𝜖) are the constants to be specified.

Step 2. Taking trial equations (56) and (57), we derive the following equations: 𝑢^′2=Φ(Γ)Ψ(Γ)𝐴^′(Γ)𝐵(Γ)−𝐴(Γ)𝐵^′(Γ)²𝐵⁴(Γ),𝑢^′′=𝐴^′(Γ)𝐵(Γ)−𝐴(Γ)𝐵^′(Γ)×Φ^′(Γ)Ψ(Γ)−Φ(Γ)Ψ^′(Γ)𝐵(Γ)−4Φ(Γ)Ψ(Γ)𝐵^′(Γ(Γ)Ψ(Γ)𝐵(Γ)𝐴)+2Φ^′′(Γ)𝐵(Γ)−𝐴(Γ)𝐵^′′(Γ)×2𝐵³(Γ)Ψ²(Γ)⁻¹ (58)(59) and other derivation terms such as 𝑢^′′′.

Step 3. Substituting 𝑢^′, 𝑢^′′, and other derivation terms into (5) yields the following equation: Ω(Γ)=𝜚_𝑠Γ^𝑠+⋯+𝜚₁Γ+𝜚₀=0. (60) According to the balance principle we can determine a relation of 𝜃, 𝜖, 𝛿, and 𝜇.

Step 4. Letting the coefficients of Ω(Γ) all be zero will yield an algebraic equations system 𝜚_𝑖=0(𝑖=0,…,𝑠). Solving this equations system, we will determine the values 𝜏₀,…𝜏_𝛿; 𝜔₀,…,𝜔_𝜇; 𝜉₀,…,𝜉_𝜃; and 𝜁₀,…,𝜁_𝜖.

Step 5. Substituting the results obtained in Step 4 into (57) and integrating (57), we can find the exact solutions of (3).

5. Conclusions and Remarks

In this study, we proposed an extended trial equation method and used it to obtain some soliton and elliptic function solutions to the generalized 𝐾(𝑚,𝑛) equation. Otherwise, we discussed a more general trial equation method. The proposed method can also be applied to other nonlinear differential equations with nonlinear evolution.