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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 478531, 16 pages
http://dx.doi.org/10.1155/2012/478531
Research Article

Classification of Exact Solutions for Some Nonlinear Partial Differential Equations with Generalized Evolution

1Department of Mathematics, Faculty of Science, Bozok University, 66100 Yozgat, Turkey
2Department of Mathematics, Faculty of Science, Ege University, 35100 Bornova-Izmir, Turkey
3Department of Mathematics, Faculty of Science, Gazi University, 06500 Teknikokullar-Ankara, Turkey

Received 13 March 2012; Accepted 17 May 2012

Academic Editor: Ravshan Ashurov

Copyright © 2012 Yusuf Pandir 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.

Abstract

We obtain the classification of exact solutions, including soliton, rational, and elliptic solutions, to the one-dimensional general improved Camassa Holm KP equation and KdV equation by the complete discrimination system for polynomial method. In discussion, we propose a more general trial equation method for nonlinear partial differential equations with generalized evolution.

1. Introduction

To construct exact solutions to nonlinear partial differential equations, some important methods have been defined such as Hirota method, tanh-coth method, the exponential function method, (𝐺/𝐺)-expansion method, the trial equation method, [115]. There are a lot of nonlinear evolution equations that are integrated using the various mathematical methods. Soliton solutions, compactons, singular solitons, and other solutions have been found by using these approaches. These types of solutions are very important and appear in various areas of applied mathematics.

In Section 2, we give a new trial equation method for nonlinear evolution equations with higher-order nonlinearity. In Section 3, as applications, we obtain some exact solutions to two nonlinear partial diffeential equations such as the one-dimensional general improved Camassa Holm KP equation [16]: 𝑢𝑡+2𝑘𝑢𝑥𝑢𝑥𝑥𝑡+𝑎𝑢𝑛(𝑢𝑛)𝑥𝑥+𝑢𝑦𝑦=0,(1.1) the dimensionless form of the generalized KdV equation [17]: 𝑢𝑡+𝑎𝑢𝑛𝑢𝑥+𝑏𝑢2𝑛𝑢𝑥+𝛿𝑢𝑥𝑥𝑥=0.(1.2) In discussion, we propose a more general trial equation method.

2. The Extended Trial Equation Method

Step 1. For a given nonlinear partial differential equation, 𝑃𝑢,𝑢𝑡,𝑢𝑥,𝑢𝑥𝑥,=0,(2.1) take the general wave transformation: 𝑢𝑥1,𝑥2,,𝑥𝑁,𝑡=𝑢(𝜂),𝜂=𝜆𝑁𝑗=1𝑥𝑗,𝑐𝑡(2.2) where 𝜆0 and 𝑐0. Substituting (2.2) into (2.1) yields a nonlinear ordinary differential equation: 𝑁𝑢,𝑢,𝑢,=0.(2.3)

Step 2. Take the finite series and trial equation as follows: 𝑢=𝛿𝑖=0𝜏𝑖Γ𝑖,(2.4) where Γ2=Λ(Γ)=Φ(Γ)=𝜉Ψ(Γ)𝜃Γ𝜃++𝜉1Γ+𝜉0𝜁𝜖Γ𝜖++𝜁1Γ+𝜁0.(2.5) Using (2.4) and (2.5), we can write 𝑢2=Φ(Γ)Ψ(Γ)𝛿𝑖=0𝑖𝜏𝑖Γ𝑖12,𝑢=Φ(Γ)Ψ(Γ)Φ(Γ)Ψ(Γ)2Ψ2(Γ)𝛿𝑖=0𝑖𝜏𝑖Γ𝑖1+Φ(Γ)Ψ(Γ)𝛿𝑖=0𝑖(𝑖1)𝜏𝑖Γ𝑖2,(2.6) where Φ(Γ) and Ψ(Γ) are polynomials. Substituting these relations into (2.3) yields an equation of polynomial Ω(Γ) of Γ: Ω(Γ)=𝜚𝑠Γ𝑠++𝜚1Γ+𝜚0=0.(2.7) According to the balance principle, we can find a relation of 𝜃, 𝜖, and 𝛿. We can compute some values of 𝜃, 𝜖, and 𝛿.

Step 3. Let the coefficients of Ω(Γ) all be zero will yield an algebraic equations system: 𝜚𝑖=0,𝑖=0,,𝑠.(2.8) Solving this system, we will determine the values of 𝜉0,,𝜉𝜃, 𝜁0,,𝜁𝜖, and 𝜏0,,𝜏𝛿.

Step 4. Reduce (2.5) to the elementary integral form: ±𝜂𝜂0=𝑑Γ=Λ(Γ)Ψ(Γ)Φ(Γ)𝑑Γ.(2.9) Using a complete discrimination system for polynomial to classify the roots of Φ(Γ), we solve (2.9) and obtain the exact solutions to (2.3). Furthermore, we can write the exact traveling wave solutions to (2.1), respectively.

3. Applications

Example 3.1 (Application to the Camassa Holm KP equation). In order to look for travelling wave solutions of (1.1), we make the transformation 𝑢(𝑥,𝑦,𝑡)=𝑢(𝜂),𝜂=𝑚(𝑥+𝑦𝑐𝑡), where 𝑚 and 𝑐 are arbitrary constants. Then, integrating this equation with respect to 𝜂 twice and setting the integration constant to zero, we obtain 𝑎(2𝑘+1𝑐)𝑢+2𝑢2𝑛+𝑚2𝑐𝑢=0.(3.1) We use the following transformation: 𝑢=𝑣1/(2𝑛1).(3.2) Equation (3.1) turns into the equation 𝑚2𝑐(2𝑛1)𝑣𝑣+2𝑚2𝑐(1𝑛)𝑣2+(2𝑘+1𝑐)(2𝑛1)2𝑣2+𝑎2(2𝑛1)2𝑣3=0.(3.3) Substituting (2.6) into (3.3) and using balance principle yield 𝜃=𝜖+𝛿+2.(3.4) After this solution procedure, we obtain the results as follows.

Case 1. If we take 𝜖=0, 𝛿=1, and 𝜃=3, then 𝑣2=𝜏12𝜉3Γ3+𝜉2Γ2+𝜉1Γ+𝜉0𝜁0,(3.5) where 𝜉30, 𝜁00. Respectively, solving the algebraic equation system (2.8) yields 𝜉1=𝑎(12𝑛)2𝜁0𝜏30+𝑚2𝜉0𝜏212+4𝑘+4𝑛+8𝑘𝑛+3𝑎𝜏0𝑚2𝜏0𝜏1𝑞+𝑎𝜏0,𝜉2=2𝑎(12𝑛)2𝜁0𝜏30+𝑚2𝜏21𝑞+3𝑎𝜏0𝑚2𝜏20𝑞+𝑎𝜏0,𝜉3=𝑎𝜏1(12𝑛)2𝜁0𝜏20+𝑚2𝜉0𝜏21𝑚2𝜏20𝑞+𝑎𝜏0,𝜉0=𝜉0,𝜁0=𝜁0,𝜏0=𝜏0,𝜏1=𝜏1,𝑐=(12𝑛)2𝜁0𝜏20𝑞+𝑎𝜏0(1+2𝑛)(12𝑛)2𝜁0𝜏20𝑚𝜉0𝜏21,(3.6) where 𝑞=(1+2𝑘)(1+2𝑛). Substituting these results into (2.5) and (2.9), we have ±𝜂𝜂0=𝑚𝐴𝑑ΓΓ3+2𝑎(12𝑛)2𝜁0𝜏30+𝑚2𝜏21𝑞+3𝑎𝜏0𝑎𝜏1(12𝑛)2𝜁0𝜏20+𝑚2𝜉0𝜏21Γ2𝑚+𝒜Γ+2𝜉0𝜏20𝑞+𝑎𝜏0𝑎𝜏1(12𝑛)2𝜁0𝜏20+𝑚2𝜉0𝜏21,(3.7)
where 𝒜 denotes by ((𝑎(12𝑛)2𝜁0𝜏40+𝑚2𝜉0𝜏0𝜏21(2+4𝑘+4𝑛+8𝑘𝑛+3𝑎𝜏0))/𝑎𝜏21((12𝑛)2𝜁0𝜏20+𝑚2𝜉0𝜏21))), and 𝐴=𝜁0𝜏20(𝑞+𝑎𝜏0)/𝑎𝜏1((12𝑛)2𝜁0𝜏20+𝑚2𝜉0𝜏21). Integrating (3.7), we obtain the solutions to the (1.1) as follows: ±𝜂𝜂0=2𝑚𝐴1Γ𝛼1,±(3.8)𝜂𝜂0=2𝑚𝐴𝛼2𝛼1arctanΓ𝛼2𝛼2𝛼1,𝛼2>𝛼1,±(3.9)𝜂𝜂0=𝑚𝐴𝛼1𝛼2|||||lnΓ𝛼2𝛼1𝛼2Γ𝛼2+𝛼1𝛼2|||||,𝛼1>𝛼2±,(3.10)𝜂𝜂0=2𝑚𝐴𝛼1𝛼3𝐹(𝜑,𝑙),𝛼1>𝛼2>𝛼3,(3.11) where 𝜁𝐴=0𝜏20𝑞+𝑎𝜏0𝑎𝜏1(12𝑛)2𝜁0𝜏20+𝑚2𝜉0𝜏21,𝐹(𝜑,𝑙)=𝜑0𝑑𝜓1𝑙2sin2𝜓,(3.12)𝜑=arcsinΓ𝛼3𝛼2𝛼3,𝑙2=𝛼2𝛼3𝛼1𝛼3.(3.13) Also 𝛼1, 𝛼2, and 𝛼3 are the roots of the polynomial equation Γ3+𝜉2𝜉3Γ2+𝜉1𝜉3𝜉Γ+0𝜉3=0.(3.14) Substituting the solutions (3.8)–(3.10) into (2.4) and (3.2), we have 𝜏𝑢(𝑥,𝑦,𝑡)=0+𝜏1𝛼1+4𝜏1𝐴𝑥+𝑦𝑡𝜂0/𝑚21/(2𝑛1),𝜏𝑢(𝑥,𝑦,𝑡)=0+𝜏1𝛼2+𝜏1𝛼1𝛼2tanh212𝛼1𝛼2𝐴𝜂𝑥+𝑦𝑡+0𝑚1/(2𝑛1),𝜏𝑢(𝑥,𝑦,𝑡)=0+𝜏1𝛼1+𝜏1𝛼1𝛼2cosech212𝛼1𝛼2𝐴(𝑥+𝑦𝑡)1/(2𝑛1),(3.15) where denote by ((12𝑛)2𝜁0𝜏20(𝑞+𝑎𝜏0)/(1+2𝑛)((12𝑛)2𝜁0𝜏20𝑚𝜉0𝜏21)).
If we take 𝜏0=𝜏1𝛼1 and 𝜂0=0, then the solutions (3.15) can reduce to rational function solution: 2𝑢(𝑥,𝑦,𝑡)=𝐴𝑥+𝑦(12𝑛)2𝛼21𝑞𝑎𝜏1𝛼1/(1+2𝑛)(12𝑛)2𝜁0𝛼21𝑚𝜉0𝑡2/(2𝑛1),(3.16) 1-soliton solution: 𝐴𝑢(𝑥,𝑦,𝑡)=1cosh2/(2𝑛1)[𝐵],(𝑥+𝑦𝑣𝑡)(3.17) and singular soliton solution: 𝐴𝑢(𝑥,𝑦,𝑡)=2sinh2/(2𝑛1)[𝐵],(𝑥+𝑦𝑣𝑡)(3.18) where 𝐴=𝐴𝜏1, 𝐴1=(𝜏1(𝛼2𝛼1))1/(2𝑛1), 𝐴2=(𝜏1(𝛼1𝛼2))1/(2𝑛1), 𝐵=(1/2)(𝛼1𝛼2)/𝐴, and 𝑣=(12𝑛)2𝜁0𝛼21(𝑞𝑎𝜏1𝛼1)/(1+2𝑛)((12𝑛)2𝜁0𝛼21𝑚𝜉0). Here, 𝐴1 and 𝐴2 are the amplitudes of the solitons, while 𝑣 is the velocity and 𝐵 is the inverse width of the solitons. Thus, we can say that the solitons exist for 𝜏1>0.

Case 2. If we take 𝜖=0, 𝛿=2 and 𝜃=4, then 𝑣2=𝜏1+2𝜏2Γ2𝜉4Γ4+𝜉3Γ3+𝜉2Γ2+𝜉1Γ+𝜉0𝜁0,(3.19) where 𝜉40, 𝜁00. Respectively, solving the algebraic equation system (2.8) yields as follows.
Subcase 2.1. It holds that 𝜉0=𝜉0,𝜉1=𝜉4𝜏314𝜏32+4𝜉0𝜏2𝜏1,𝜉2=5𝜉4𝜏214𝜏22+4𝜉0𝜏22𝜏21,𝜉3=2𝜉4𝜏1𝜏2,𝜉4=𝜉4,𝜏1=𝜏1,𝜏2=𝜏2,𝜁0𝑚=216𝑎𝜉0𝜏42+𝜉4𝜏21𝑎𝜏21+4𝜏2𝑞𝑎(12𝑛)2𝜏21𝜏22,𝜏0=𝜏214𝜏2,𝑎𝜉𝑐=1+2𝑘+4𝜏4116𝜉0𝜏424(1+2𝑛)𝜉4𝜏21𝜏2,(3.20) where 𝑞=(1+2𝑘)(1+2𝑛). Substituting these results into (2.5) and (2.9), we get ±𝜂𝜂0=𝑚16𝑎𝜉0𝜏42𝜉4𝜏41𝑎𝜏21+4𝑞𝜏2𝑎(12𝑛)2𝜉4𝜏21𝜏22×𝑑ΓΓ4+2𝜏1/𝜏2Γ3+5𝜏21/4𝜏22+4𝜉0𝜏22/𝜏21Γ2+𝜏31/4𝜏32+4𝜉0𝜏2/𝜏1𝜉Γ+0/𝜉4.(3.21) Integrating (3.21), we obtain the solutions to (1.1) as follows: ±𝜂𝜂0=𝑚𝐵Γ𝛼1,±(3.22)𝜂𝜂0=2𝑚𝐵𝛼1𝛼2Γ𝛼2Γ𝛼1,𝛼1>𝛼2,±(3.23)𝜂𝜂0=𝑚𝐵𝛼1𝛼2||||lnΓ𝛼1Γ𝛼2||||±,(3.24)𝜂𝜂0=𝑚𝐵𝛼1𝛼2𝛼1𝛼3||||||lnΓ𝛼2𝛼1𝛼3Γ𝛼3𝛼1𝛼2Γ𝛼2𝛼1𝛼3+Γ𝛼3𝛼1𝛼2||||||,𝛼1>𝛼2>𝛼3,±(3.25)𝜂𝜂0=2𝑚𝐵𝛼1𝛼3𝛼2𝛼4𝐹(𝜑,𝑙),𝛼1>𝛼2>𝛼3>𝛼4,(3.26) where 𝐵=16𝑎𝜉0𝜏42𝜉4𝜏41𝑎𝜏21+4𝑞𝜏2𝑎(12𝑛)2𝜉4𝜏21𝜏22,𝐹(𝜑,𝑙)=𝜑0𝑑𝜓1𝑙2sin2𝜓,(3.27)𝜑=arcsinΓ𝛼1𝛼2𝛼4Γ𝛼2𝛼1𝛼4,𝑙2=𝛼2𝛼3𝛼1𝛼4𝛼1𝛼3𝛼2𝛼4.(3.28) Also 𝛼1, 𝛼2, 𝛼3, and 𝛼4 are the roots of the polynomial equation: Γ4+𝜉3𝜉4Γ3+𝜉2𝜉4Γ2+𝜉1𝜉4𝜉Γ+0𝜉4=0.(3.29) Substituting the solutions (3.22)–(3.25) into (2.4) and (3.2), we have 𝜏𝑢(𝑥,𝑦,𝑡)=0+𝜏1𝛼1±𝜏1𝐵𝜂𝒞0/𝑚+𝜏2𝛼1±𝐵𝜂𝒞0/𝑚21/(2𝑛1),𝜏𝑢(𝑥,𝑦,𝑡)=0+𝜏1𝛼1+4𝐵2𝛼2𝛼1𝜏14𝐵2𝛼1𝛼2𝒞𝜂02+𝜏2𝛼1+4𝐵2𝛼2𝛼14𝐵2𝛼1𝛼2𝒞𝜂0221/(2𝑛1),𝜏𝑢(𝑥,𝑦,𝑡)=0+𝜏1𝛼2+𝛼2𝛼1𝜏1𝛼exp1𝛼2/𝐵𝒞𝜂01+𝜏2𝛼2+𝛼2𝛼1𝛼exp1𝛼2/𝐵𝒞𝜂0121/(2𝑛1),𝜏𝑢(𝑥,𝑦,𝑡)=0+𝜏1𝛼1+𝛼1𝛼2𝜏1𝛼exp1𝛼2/𝐵𝒞𝜂01+𝜏2𝛼1+𝛼1𝛼2𝛼exp1𝛼2/𝐵𝒞𝜂0121/(2𝑛1),𝜏𝑢(𝑥,𝑦,𝑡)=0+𝜏1𝛼12𝛼1𝛼2𝛼1𝛼3𝜏12𝛼1𝛼2𝛼3+𝛼3𝛼2cosh𝛼1𝛼2𝛼1𝛼3/𝐵(𝒞)+𝜏2𝛼12𝛼1𝛼2𝛼1𝛼32𝛼1𝛼2𝛼3+𝛼3𝛼2cosh𝛼1𝛼2𝛼1𝛼3/𝐵(𝒞)21/(2𝑛1),(3.30) where 𝒞 denotes by 𝑥+𝑦(1+2𝑘+(𝑎(𝜉4𝜏4116𝜉0𝜏42)/4(1+2𝑛)𝜉4𝜏21𝜏2))𝑡.
For simplicity, we can write the solutions (3.30) as follows: 𝑢(𝑥,𝑦,𝑡)=2𝑖=0𝜏𝑖𝛼1±𝐵𝜂𝒞0/𝑚𝑖1/(2𝑛1),𝑢(𝑥,𝑦,𝑡)=2𝑖=0𝜏𝑖𝛼1+4𝐵2𝛼2𝛼14𝐵2𝛼1𝛼2𝒞𝜂02𝑖1/(2𝑛1),𝑢(𝑥,𝑦,𝑡)=2𝑖=0𝜏𝑖𝛼2+𝛼2𝛼1𝛼exp1𝛼2/𝐵𝒞𝜂01𝑖1/(2𝑛1),𝑢(𝑥,𝑦,𝑡)=2𝑖=0𝜏𝑖𝛼1+𝛼1𝛼2𝛼exp1𝛼2/𝐵𝒞𝜂01𝑖1/(2𝑛1),𝑢(𝑥,𝑦,𝑡)=2𝑖=0𝜏𝑖𝛼12𝛼1𝛼2𝛼1𝛼32𝛼1𝛼2𝛼3+𝛼3𝛼2cosh𝛼1𝛼2𝛼1𝛼3/𝐵(𝒞)𝑖1/(2𝑛1),(3.31)Subcase 2.2. It holds that 𝜉0=𝜉0,𝜉1=𝜉3=0,𝜉2=2𝜉0𝜉4,𝜉4=𝜉4,𝜏1=0,𝜏2=𝜏2,𝜁04=𝑞𝑚2𝜉4+𝑎𝑚2𝜉0𝜉4𝜏2𝑎(12𝑛)2𝜏2,𝜏0=𝜉0𝜏2𝜉4𝑎,𝑐=1+2𝑘+𝜉0𝜏2(1+2𝑛)𝜉4,(3.32) where 𝑞=(1+2𝑘)(1+2𝑛). Substituting these results into (2.5) and (2.9), we get ±𝜂𝜂0=𝑚4𝑞𝜉4𝑎𝜉0𝜉4𝜏2𝑎(12𝑛)2𝜉4𝜏2𝑑ΓΓ4+2𝜉0𝜉4/𝜉4Γ2+𝜉0/𝜉4.(3.33) Integrating (3.33), we obtain the solutions to the (1.1) as follows.
If we denote 𝐹(Γ)=Γ4+2𝜉0𝜉4𝜉4Γ2+𝜉0𝜉4=𝑅2+𝑑1𝑅+𝑑0,(3.34) where Γ2=𝑅, 𝐹(Γ)=𝐺(𝑅), then we can write complete discrimination system of 𝐺(𝑅) as follows: Δ=𝑑214𝑑0.(3.35) Correspondingly, there are the following two cases to be discussed.(1) If Δ>0,then we have𝐹(Γ)=(Γ𝛼1)(Γ+𝛼1)(Γ𝛼2)(Γ+𝛼2),𝛼1𝛼2.Therefore, the solution is given by ±𝜂𝜂0=𝑚𝐶𝛼1𝐹(𝜑,𝑙),(3.36)
where 𝐶=4𝑞𝜉4𝑎𝜉0𝜉4𝜏2𝑎(12𝑛)2𝜉4𝜏2,𝐹(𝜑,𝑙)=𝜑0𝑑𝜓1𝑙2sin2𝜓,Γ(3.37)𝜑=arcsin𝛼1,𝑙2=𝛼1𝛼2,𝛼2>𝛼1.(3.38)(2) If Δ=0, then we have 𝐹(Γ)=(Γ𝛼1)2(Γ+𝛼1)2. From here, the solutions can be found as ±𝜂𝜂0=𝑚𝐶𝛼1Γarctanh𝛼1𝜏,(3.39)𝑢(𝑥,𝑦,𝑡)=0+𝜏1𝛼1±tanh𝛼1𝐶𝑎𝑥+𝑦1+2𝑘+𝜉0𝜏2(1+2𝑛)𝜉4𝜂𝑡0𝑚+𝜏2𝛼1tanh2±𝛼1𝐶𝑎𝑥+𝑦1+2𝑘+𝜉0𝜏2(1+2𝑛)𝜉4𝜂𝑡0𝑚1/(2𝑛1).(3.40)
For simplicity, we can write (3.40) as follows: =𝑢(𝑥,𝑦,𝑡)2𝑖=0𝜏𝑖𝛼1±tanh𝛼1𝐶𝑎𝑥+𝑦1+2𝑘+𝜉0𝜏2(1+2𝑛)𝜉4𝜂𝑡0𝑚𝑖1/(2𝑛1).(3.41)

Example 3.2 (Application to the generalized KdV equation). Using a complex variation 𝜂 defined as 𝜂=𝑘𝑥+𝑤𝑡, we can convert (1.2) into ordinary different equation, which reads 𝑤𝑢+𝑎𝑘𝑢𝑛𝑢+𝑏𝑘𝑢2𝑛𝑢+𝛿𝑘3𝑢=0,(3.42) where the prime denotes the derivative with respect to 𝜂. Integrating (3.42), and setting the constant of integration to be zero, we obtain 𝑤𝑢+𝑎𝑘𝑢𝑛+1𝑛+1+𝑏𝑘𝑢2𝑛+12𝑛+1+𝛿𝑘3𝑢=0.(3.43) By the using of the transformation 𝑢=𝑣1/𝑛, (3.43) reduces to 𝛿𝑘3𝑛(𝑛+1)(2𝑛+1)𝑣𝑣+𝛿𝑘31𝑛2𝑣(2𝑛+1)2+𝑏𝑘𝑛2(𝑛+1)𝑣4+𝑎𝑘𝑛2(2𝑛+1)𝑣3+𝑛2(𝑛+1)(2𝑛+1)𝑤𝑣2=0.(3.44) Substituting (2.6) into (3.44) and using balance principle yield 𝜃=𝜖+2𝛿+2.(3.45) If we take 𝜃=4, 𝜖=0, and 𝛿=1, then 𝑣2=𝜏21𝜉4Γ4+𝜉3Γ3+𝜉2Γ2+𝜉1Γ+𝜉0𝜁0,(3.46) where 𝜉40, 𝜁00. Respectively, solving the algebraic equation system (2.8) yields 𝜉0=𝒟2𝜏31,𝜉1=𝜏31,𝜉2=𝜉2,𝜉3=𝜉3,𝜉4=𝑏(2+𝑛)𝜉34𝜏12𝑎+2𝑎𝑛+2𝑏(2+𝑛)𝜏0,𝜁0𝑘=2(1+𝑛)(2+𝑛)(1+2𝑛)𝛿𝜉32𝑛2𝑎+2𝑎𝑛+2𝑏(𝑛+2)𝜏0𝜏1,𝜏0=𝜏0,𝜏1=𝜏1,𝑤=2𝑘3𝜉3𝜏0𝑎+2𝑎𝑛+𝑏(2+𝑛)𝜏0𝜉2𝜏1𝑎+2𝑎𝑛+𝑏(2+𝑛)𝜏0𝜉32+7𝑛+7𝑛2+2𝑛3.(3.47) where 𝒟 denotes by 𝜏20((𝜉3𝜏0(4𝑎(1+2𝑛)+5𝑏(2+𝑛)𝜏0)/(𝑎+2𝑎𝑛+2𝑏(2+𝑛)𝜏0))+2𝜉2𝜏1) and denote by 𝜏0((𝜉3𝜏0(3𝑎(1+2𝑛)+4𝑏(2+𝑛)𝜏0)/(𝑎+2𝑎𝑛+2𝑏(2+𝑛)𝜏0))+2𝜉2𝜏1). Substituting these results into (2.5) and (2.9), we can write ±𝜂𝜂0=𝑘𝜏1(1+𝑛)(1+2𝑛)𝛿4𝑛2𝑏×𝑑ΓΓ4+𝒩2𝑏(2+𝑛)𝜏1Γ3+𝜉2(𝒩)2𝑏(2+𝑛)𝜉2𝜏1Γ2+(𝒩)2𝑏(2+𝑛)𝜉3𝜏41Γ+𝒟(𝒩)4𝑏(2+𝑛)𝜉3𝜏41,(3.48) where 𝒩 denotes by 𝑎+2𝑎𝑛+𝑏(2+𝑛)𝜏0. Integrating (3.48), we obtain the solutions to (1.2) as follows: ±𝜂𝜂0=𝑘𝐵Γ𝛼1,±(3.49)𝜂𝜂0=2𝑘𝐵𝛼1𝛼2Γ𝛼2Γ𝛼1,𝛼2>𝛼1,±(3.50)𝜂𝜂0=𝑘𝐵𝛼1𝛼2||||lnΓ𝛼1Γ𝛼2||||±,(3.51)𝜂𝜂0=𝑘𝐵𝛼1𝛼2𝛼1𝛼3||||||lnΓ𝛼2𝛼1𝛼3Γ𝛼3𝛼1𝛼2Γ𝛼2𝛼1𝛼3+Γ𝛼3𝛼1𝛼2||||||,𝛼1>𝛼2>𝛼3,±(3.52)𝜂𝜂0=2𝑘𝐵𝛼1𝛼3𝛼2𝛼4𝐹(𝜑,𝑙),𝛼1>𝛼2>𝛼3>𝛼4,(3.53) where 1𝐵=𝜏1(1+𝑛)(1+2𝑛)𝛿4𝑛2𝑏,𝐹(𝜑,𝑙)=𝜑0𝑑𝜓1𝑙2sin2𝜓,(3.54)𝜑=arcsinΓ𝛼1𝛼2𝛼4Γ𝛼2𝛼1𝛼4,𝑙2=𝛼2𝛼3𝛼1𝛼4𝛼1𝛼3𝛼2𝛼4.(3.55) Also 𝛼1, 𝛼2, 𝛼3, and 𝛼4 are the roots of the polynomial equation: Γ4+𝜉3𝜉4Γ3+𝜉2𝜉4Γ2+𝜉1𝜉4𝜉Γ+0𝜉4=0.(3.56) Substituting the solutions (3.49)–(3.52) into (2.4) and (3.2), we find 𝜏𝑢(𝑥,𝑡)=0+𝜏1𝛼1±𝜏1𝐵𝜂0/𝑘1/𝑛,𝜏𝑢(𝑥,𝑡)=0+𝜏1𝛼1+4𝐵2𝛼2𝛼1𝜏14𝐵2𝛼1𝛼2𝜂0/𝑘21/𝑛,𝜏𝑢(𝑥,𝑡)=0+𝜏1𝛼2+𝛼2𝛼1𝜏1𝛼exp1𝛼2𝜂/𝐵0/𝑘11/𝑛,𝜏𝑢(𝑥,𝑡)=0+𝜏1𝛼1+𝛼1𝛼2𝜏1𝛼exp1𝛼2𝜂/𝐵0/𝑘11/𝑛,𝜏𝑢(𝑥,𝑡)=0+𝜏1𝛼12𝛼1𝛼2𝛼1𝛼3𝜏12𝛼1𝛼2𝛼3+𝛼3𝛼2cosh𝛼1𝛼2𝛼1𝛼3/𝐵1/𝑛,(3.57) where denotes by 𝑥+((2[3𝜉3𝜏0(𝑎+2𝑎𝑛+𝑏(2+𝑛)𝜏0)𝜉2𝜏1(𝑎+2𝑎𝑛+𝑏(2+𝑛)𝜏0)])/𝜉3(2+7𝑛+7𝑛2+2𝑛3))𝑡.
If we take 𝜏0=𝜏1𝛼1 and 𝜂0=0, then the solutions (3.57) can reduce to rational function solutions: ±𝐵𝑢(𝑥,𝑡)=𝑥+2𝜏1𝑎+2𝑎𝑛𝑏(2+𝑛)𝜏1𝛼13𝜉3𝛼1+𝜉2/𝜉32+7𝑛+7𝑛2+2𝑛3𝑡1/𝑛,4𝐵𝑢(𝑥,𝑡)=2𝛼2𝛼1𝜏14𝐵2𝛼1𝛼2𝑥+2𝜏1𝑎+2𝑎𝑛𝑏(2+𝑛)𝜏1𝛼13𝜉3𝛼1+𝜉2/𝜉32+7𝑛+7𝑛2+2𝑛3𝑡21/𝑛,(3.58) traveling wave solutions: 𝛼𝑢(𝑥,𝑡)=2𝛼1𝜏12𝛼1coth1𝛼22𝐵𝑥+2𝜏1𝑎+2𝑎𝑛𝑏(2+𝑛)𝜏1𝛼13𝜉3𝛼1+𝜉2𝜉32+7𝑛+7𝑛2+2𝑛3𝑡1/𝑛,(3.59) and soliton solution: 𝐴𝑢(𝑥,𝑡)=3𝐵𝐷+cosh1(𝑥𝑣𝑡)1/𝑛,(3.60) where 𝐵=𝐵𝜏1, 𝐴3=(2(𝛼1𝛼2)(𝛼1𝛼3)𝜏1/(𝛼3𝛼2))1/𝑛, 𝐵1=(𝛼1𝛼2)(𝛼1𝛼3)/𝐵, 𝐷=(2𝛼1𝛼2𝛼3)/(𝛼3𝛼2), and 𝑣=2𝜏1(𝑎+2𝑎𝑛𝑏(2+𝑛)𝜏1𝛼1)(3𝜉3𝛼1+𝜉2)/𝜉3(2+7𝑛+7𝑛2+2𝑛3). Here, 𝐴3 is the amplitude of the soliton, while 𝑣 is the velocity and 𝐵1 is the inverse width of the soliton. Thus, we can say that the solitons exist for 𝜏1<0.

4. Discussion

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

Step 1. The extended trial equation (2.4) can be reduced to the following more general form: 𝐴𝑢=(Γ)=𝐵(Γ)𝛿𝑖=0𝜏𝑖Γ𝑖𝜇𝑗=0𝜔𝑗Γ𝑗,(4.1) where Γ2=Λ(Γ)=Φ(Γ)=𝜉Ψ(Γ)𝜃Γ𝜃++𝜉1Γ+𝜉0𝜁𝜖Γ𝜖++𝜁1Γ+𝜁0.(4.2) Here, 𝜏𝑖(𝑖=0,,𝛿), 𝜔𝑗(𝑗=0,,𝜇), 𝜉𝜍(𝜍=0,,𝜃), and 𝜁𝜎(𝜎=0,,𝜖) are the constants to be determined.

Step 2. Taking trial equations (4.1) and (4.2), we derive the following equations: 𝑢2=Φ(Γ)𝐴Ψ(Γ)(Γ)𝐵(Γ)𝐴(Γ)𝐵(Γ)2𝐵4,𝑢(Γ)=𝐴(Γ)𝐵(Γ)𝐴(Γ)𝐵Φ(Γ)(Γ)Ψ(Γ)Φ(Γ)Ψ(Γ)𝐵(Γ)4Φ(Γ)Ψ(Γ)𝐵(Γ)2𝐵3(Γ)Ψ2+𝐴(Γ)2Φ(Γ)Ψ(Γ)𝐵(Γ)(Γ)𝐵(Γ)𝐴(Γ)𝐵(Γ)2𝐵3(Γ)Ψ2,(Γ)(4.3) and other derivation terms such as 𝑢, and so on.

Step 3. Substituting 𝑢, 𝑢 and other derivation terms into (2.3) yields the following equation: Ω(Γ)=𝜚𝑠Γ𝑠++𝜚1Γ+𝜚0=0.(4.4) 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 𝜚𝑖=0(𝑖=0,,𝑠). Solving this equations system, we will determine the values 𝜏0,𝜏𝛿; 𝜔0,,𝜔𝜇; 𝜉0,𝜉𝜃, and 𝜁0,,𝜁𝜖.

Step 5. Substituting the results obtained in Step 4 into (4.2) and integrating equation (4.2), we can find the exact solutions of (2.1).

5. Conclusions and Remarks

In this study, we proposed a new trial equation method and used it to obtain some soliton and elliptic function solutions to the Camassa Holm KP equation and the one-dimensional general improved KdV equation. Otherwise, we discussed a more general trial equation method. We think that the proposed method can also be applied to other nonlinear differential equations with nonlinear evolution.

Also, the convergence analysis of obtained elliptic solutions is given as follows: 𝐹(𝜙,𝑙)=𝜙0𝑑𝜙1𝑙2sin2𝜙,(5.1) where sin𝜙=Γ𝛼3𝛼2𝛼3,𝑙=𝛼2𝛼3𝛼1𝛼3.(5.2) Especially, 𝜙=𝜋/2, we have 𝐹𝜋2=,𝑙0𝜋/2𝑑𝜙1𝑙2sin2𝜙1=𝜙+2𝑘2𝑣2++1.3(2𝑛1)𝑘2.4(2𝑛)2𝑛𝑣2𝑛+,(5.3) where 𝑣2𝑛=𝜙0sin2𝑛𝜙𝑑𝜙. Taking the value 𝜙=𝜋/2, we have 𝑣2𝑛=((1.3(2𝑛1))/(2.4(2𝑛)))(𝜋/2). Therefore, if we take Γ(𝑡)=𝛼2 in (3.13), Γ(𝑡)=𝛼4 in (3.28) and (3.55), Γ(𝑡)=𝛼1 in (3.38), for each 𝑡, then we have 𝐹𝜋2=𝜋,𝑙2𝑛=0(2𝑛)!22𝑛(𝑛!)22𝑙2𝑛.(5.4) By the using radius of convergence of power series: 1𝑅=lim𝑛𝑎𝑛+1/𝑎𝑛,(5.5) where 𝑎𝑛=((2𝑛)!/22𝑛(𝑛!)2)2. We have the radius of convergence of power series 𝑅=1. We can say that power series converges for each 0<𝑙<1, diverges for each 𝑙>1. Consequently, the inequalities in (3.11), (3.26)–(3.53), and (3.38) are obtained by using 0<𝑙<1, respectively.

Acknowledgment

The research has been supported by the Scientific and Technological Research Council of Turkey (TUBITAK) and Yozgat University Foundation.

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