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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, (???/??)-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]: ?????+2??????-????????+??????(????)?????+??????=0,(1.1) the dimensionless form of the generalized KdV equation [17]: ????+??????????+????2??????+??????????=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, ?????,????,????,???????,=0,(2.1) take the general wave transformation: ?????1,??2,,??????,??=??(??),??=???????=1?????,-????(2.2) where ???0 and ???0. Substituting (2.2) into (2.1) yields a nonlinear ordinary differential equation: ?????,???,?????,=0.(2.3)

Step 2. Take the finite series and trial equation as follows: ??=?????=0????G??,(2.4) where ?G??2=?(G)=F(G)=???(G)??G??+?+??1G+??0????G??+?+??1G+??0.(2.5) Using (2.4) and (2.5), we can write ?????2=F(G)??(G)?????=0??????G??-1?2,????=F?(G)?(G)-F(G)??(G)2?2(?G)?????=0??????G??-1?+F(G)??(G)?????=0??(??-1)????G??-2?,(2.6) where F(G) and ?(G) are polynomials. Substituting these relations into (2.3) yields an equation of polynomial O(G) of G: O(G)=????G??+?+??1G+??0=0.(2.7) According to the balance principle, we can find a relation of ??, ??, and ??. We can compute some values of ??, ??, and ??.

Step 3. Let the coefficients of O(G) all be zero will yield an algebraic equations system: ????=0,??=0,,??.(2.8) Solving this system, we will determine the values of ??0,,????, ??0,,????, and ??0,,????.

Step 4. Reduce (2.5) to the elementary integral form: ±???-??0?=???Gv=???(G)?(G)F(G)??G.(2.9) Using a complete discrimination system for polynomial to classify the roots of F(G), 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 ??(??,??,??)=??(??),??=??(??+??-????), where ?? and ?? are arbitrary constants. Then, integrating this equation with respect to ?? twice and setting the integration constant to zero, we obtain ??(2??+1-??)??+2??2??+??2??????=0.(3.1) We use the following transformation: ??=??1/(2??-1).(3.2) Equation (3.1) turns into the equation ??2??(2??-1)??????+2??2??(1-??)???2+(2??+1-??)(2??-1)2??2+??2(2??-1)2??3=0.(3.3) Substituting (2.6) into (3.3) and using balance principle yield ??=??+??+2.(3.4) After this solution procedure, we obtain the results as follows.

Case 1. If we take ??=0, ??=1, and ??=3, then ?????2=???1?2???3G3+??2G2+??1G+??0???0,(3.5) where ??3?0, ??0?0. Respectively, solving the algebraic equation system (2.8) yields ??1=-??(1-2??)2??0??30+??2??0??21?2+4??+4??+8????+3????0???2??0??1???+????0?,??2=-2??(1-2??)2??0??30+??2??21???+3????0???2??20???+????0?,??3=????1?-(1-2??)2??0??20+??2??0??21???2??20???+????0?,??0=??0,??0=??0,??0=??0,??1=??1,??=(1-2??)2??0??20???+????0??(1+2??)(1-2??)2??0??20-????0??21?,(3.6) where ??=(1+2??)(1+2??). Substituting these results into (2.5) and (2.9), we have ±???-??0?v=???????G????G3+-2??(1-2??)2??0??30+??2??21???+3????0?????1?-(1-2??)2??0??20+??2??0??21?G2??+??G+2??0??20???+????0?????1?-(1-2??)2??0??20+??2??0??21?,(3.7)
where ?? denotes by ((-??(1-2??)2??0??40+??2??0??0??21(2+4??+4??+8????+3????0))/????21(-(1-2??)2??0??20+??2??0??21))), and ??=??0??20(??+????0)/????1(-(1-2??)2??0??20+??2??0??21). Integrating (3.7), we obtain the solutions to the (1.1) as follows: ±???-??0?v=-2????1vG-??1,±?(3.8)??-??0??=2??????2-??1?arctanG-??2??2-??1,??2>??1,±?(3.9)??-??0??=??????1-??2|||||vlnG-??2-v??1-??2vG-??2+v??1-??2|||||,??1>??2±?,(3.10)??-??0??=2??????1-??3??(??,??),??1>??2>??3,(3.11) where ????=0??20???+????0?????1?-(1-2??)2??0??20+??2??0??21??,??(??,??)=??0????v1-??2sin2??,?(3.12)??=arcsinG-??3??2-??3,??2=??2-??3??1-??3.(3.13) Also ??1, ??2, and ??3 are the roots of the polynomial equation G3+??2??3G2+??1??3??G+0??3=0.(3.14) Substituting the solutions (3.8)–(3.10) into (2.4) and (3.2), we have ?????(??,??,??)=0+??1??1+4??1?????+??-B??-??0?/??2?1/(2??-1),?????(??,??,??)=0+??1??2+??1???1-??2?tanh2?12???1-??2???????+??-B??+0?????1/(2??-1),?????(??,??,??)=0+??1??1+??1???1-??2?cosech2?12???1-??2??(??+??-B??)??1/(2??-1),(3.15) where B denote by ((1-2??)2??0??20(??+????0)/(1+2??)((1-2??)2??0??20-????0??21)).
If we take ??0=-??1??1 and ??0=0, then the solutions (3.15) can reduce to rational function solution: ?2v??(??,??,??)=??????+??-(1-2??)2??21???-????1??1??/(1+2??)(1-2??)2??0??21-????0?????2/(2??-1),(3.16) 1-soliton solution: ????(??,??,??)=1cosh2/(2??-1)[??],(??+??-????)(3.17) and singular soliton solution: ????(??,??,??)=2sinh2/(2??-1)[??],(??+??-????)(3.18) where ???=????1, ??1=(??1(??2-??1))1/(2??-1), ??2=(??1(??1-??2))1/(2??-1), v??=(1/2)(??1-??2)/??, and ??=(1-2??)2??0??21(??-????1??1)/(1+2??)((1-2??)2??0??21-????0). Here, ??1 and ??2 are the amplitudes of the solitons, while ?? is the velocity and ?? is the inverse width of the solitons. Thus, we can say that the solitons exist for ??1>0.

Case 2. If we take ??=0, ??=2 and ??=4, then ?????2=???1+2??2G?2???4G4+??3G3+??2G2+??1G+??0???0,(3.19) where ??4?0, ??0?0. Respectively, solving the algebraic equation system (2.8) yields as follows.
Subcase 2.1. It holds that ??0=??0,??1=??4??314??32+4??0??2??1,??2=5??4??214??22+4??0??22??21,??3=2??4??1??2,??4=??4,??1=??1,??2=??2,??0??=-2?-16????0??42+??4??21?????21+4??2??????(1-2??)2??21??22,??0=??214??2,???????=1+2??+4??41-16??0??42?4(1+2??)??4??21??2,(3.20) where ??=(1+2??)(1+2??). Substituting these results into (2.5) and (2.9), we get ±???-??0?????=???16????0??42-??4??41?????21+4????2????(1-2??)2??4??21??22×???G?G4+?2??1/??2?G3+??5??21/4??22?+?4??0??22/??21G??2+????31/4??32?+?4??0??2/??1?????G+0/??4?.(3.21) Integrating (3.21), we obtain the solutions to (1.1) as follows: ±???-??0?=-????G-??1,±?(3.22)??-??0?=2??????1-??2?G-??2G-??1,??1>??2,±?(3.23)??-??0?=??????1-??2||||lnG-??1G-??2||||±?,(3.24)??-??0?=????????1-??2????1-??3?||||||?ln?G-??2????1-??3?-??G-??3????1-??2???G-??2????1-??3?+??G-??3????1-??2?||||||,??1>??2>??3,±?(3.25)??-??0?=2????????1-??3????2-??4???(??,??),??1>??2>??3>??4,(3.26) where ??????=?16????0??42-??4??41?????21+4????2????(1-2??)2??4??21??22?,??(??,??)=??0????v1-??2sin2??,????(3.27)??=arcsin?G-??1????2-??4??G-??2????1-??4?,??2=???2-??3????1-??4????1-??3????2-??4?.(3.28) Also ??1, ??2, ??3, and ??4 are the roots of the polynomial equation: G4+??3??4G3+??2??4G2+??1??4??G+0??4=0.(3.29) Substituting the solutions (3.22)–(3.25) into (2.4) and (3.2), we have ????????(??,??,??)=0+??1??1±??1???????-0?/??+??2???1±???????-0??/??2????1/(2??-1),?????(??,??,??)=0+??1??1+4??2???2-??1???14??2-????1-??2????-??0??2+??2???1+4??2???2-??1?4??2-????1-??2????-??0??2?2?????1/(2??-1),?????(??,??,??)=0+??1??2+???2-??1???1??exp???1-??2?/??????-??0??-1+??2???2+???2-??1???exp???1-??2?/??????-??0???-12?????1/(2??-1),?????(??,??,??)=0+??1??1+???1-??2???1??exp???1-??2?/??????-??0??-1+??2???1+???1-??2???exp???1-??2?/??????-??0???-12?????1/(2??-1),???????????(??,??,??)=0+??1??1-2???1-??2????1-??3???12??1-??2-??3+???3-??2??cosh?????1-??2????1-??3???/??(??)+??2????????1-2???1-??2????1-??3?2??1-??2-??3+???3-??2??cosh?????1-??2????1-??3?????????/??(??)2???????1/(2??-1),(3.30) where ?? denotes by ??+??-(1+2??+(??(??4??41-16??0??42)/4(1+2??)??4??21??2))??.
For simplicity, we can write the solutions (3.30) as follows: ???(??,??,??)=2???=0???????1±???????-0??/?????1/(2??-1),???(??,??,??)=2???=0???????1+4??2???2-??1?4??2-????1-??2????-??0??2????1/(2??-1),???(??,??,??)=2???=0???????2+???2-??1???exp???1-??2?/??????-??0???-1???1/(2??-1),???(??,??,??)=2???=0???????1+???1-??2???exp???1-??2?/??????-??0???-1???1/(2??-1),????????(??,??,??)=2???=0????????????1-2???1-??2????1-??3?2??1-??2-??3+???3-??2??cosh?????1-??2????1-??3?????????/??(??)????????1/(2??-1),(3.31)Subcase 2.2. It holds that ??0=??0,??1=??3=0,??2v=2??0??4,??4=??4,??1=0,??2=??2,??04?=-????2??4+????2v??0??4??2???(1-2??)2??2,??0=v??0??2v??4??v,??=1+2??+??0??2v(1+2??)??4,(3.32) where ??=(1+2??)(1+2??). Substituting these results into (2.5) and (2.9), we get ±???-??0?????=??-4????4v-????0??4??2??(1-2??)2??4??2???G?G4+?2v??0??4/??4?G2+???0/??4?.(3.33) Integrating (3.33), we obtain the solutions to the (1.1) as follows.
If we denote ??(G)=G4+2v??0??4??4G2+??0??4=??2+??1??+??0,(3.34) where G2=??, ??(G)=??(??), then we can write complete discrimination system of ??(??) as follows: ?=??21-4??0.(3.35) Correspondingly, there are the following two cases to be discussed.(1) If ?>0,then we havev??(G)=(G-??1v)(G+??1v)(G-??2v)(G+??2),??1???2.Therefore, the solution is given by ±???-??0?v=??????1??(??,??),(3.36)
where ??????=-4????4v-????0??4??2??(1-2??)2??4??2?,??(??,??)=??0????v1-??2sin2??,?G(3.37)??=arcsinv??1?,??2=??1??2,??2>??1.(3.38)(2) If ?=0, then we have v??(G)=(G-??1)2v(G+??1)2. From here, the solutions can be found as ±???-??0?=????v??1?Garctanhv??1????,(3.39)??(??,??,??)=0+??1v??1?±vtanh??1??????v??+??-1+2??+??0??2(v1+2??)??4?????-0????+??2??1tanh2?±v??1??????v??+??-1+2??+??0??2v(1+2??)??4?????-0?????1/(2??-1).(3.40)
For simplicity, we can write (3.40) as follows: =???(??,??,??)2???=0?????v??1?±vtanh??1??????v??+??-1+2??+??0??2(v1+2??)??4?????-0????????1/(2??-1).(3.41)

Example 3.2 (Application to the generalized KdV equation). Using a complex variation ?? defined as ??=????+????, we can convert (1.2) into ordinary different equation, which reads ?????+???????????+??????2?????+????3?????=0,(3.42) where the prime denotes the derivative with respect to ??. Integrating (3.42), and setting the constant of integration to be zero, we obtain ????+????????+1??+1+??????2??+12??+1+????3????=0.(3.43) By the using of the transformation ??=??1/??, (3.43) reduces to ????3??(??+1)(2??+1)??????+????3?1-??2????(2??+1)??2+??????2(??+1)??4+??????2(2??+1)??3+??2(??+1)(2??+1)????2=0.(3.44) Substituting (2.6) into (3.44) and using balance principle yield ??=??+2??+2.(3.45) If we take ??=4, ??=0, and ??=1, then ?????2=??21???4G4+??3G3+??2G2+??1G+??0???0,(3.46) where ??4?0, ??0?0. Respectively, solving the algebraic equation system (2.8) yields ??0=??2??31,??1=F??31,??2=??2,??3=??3,??4=??(2+??)??34??12???+2????+2??(2+??)??0?,??0??=-2(1+??)(2+??)(1+2??)????32??2???+2????+2??(??+2)??0???1,??0=??0,??1=??1?,??=-2??3??3??0???+2????+??(2+??)??0?-??2??1???+2????+??(2+??)??0????3?2+7??+7??2+2??3?.(3.47) where ?? denotes by ??20(-(??3??0(4??(1+2??)+5??(2+??)??0)/(??+2????+2??(2+??)??0))+2??2??1) and F denote by ??0(-(??3??0(3??(1+2??)+4??(2+??)??0)/(??+2????+2??(2+??)??0))+2??2??1). Substituting these results into (2.5) and (2.9), we can write ±???-??0?=????1?-(1+??)(1+2??)??4??2??×???G?G4+??2??(2+??)??1G3+??2(??)2??(2+??)??2??1G2+F(??)2??(2+??)??3??41G+??(??)4??(2+??)??3??41,(3.48) where ?? denotes by ??+2????+??(2+??)??0. Integrating (3.48), we obtain the solutions to (1.2) as follows: ±???-??0?=-????G-??1,±?(3.49)??-??0?=2??????1-??2?G-??2G-??1,??2>??1,±?(3.50)??-??0?=??????1-??2||||lnG-??1G-??2||||±?,(3.51)??-??0?=????????1-??2????1-??3?||||||?ln?G-??2????1-??3?-??G-??3????1-??2???G-??2????1-??3?+??G-??3????1-??2?||||||,??1>??2>??3,±?(3.52)??-??0?=2????????1-??3????2-??4???(??,??),??1>??2>??3>??4,(3.53) where 1??=??1?-(1+??)(1+2??)??4??2???,??(??,??)=??0????v1-??2sin2??,????(3.54)??=arcsin?G-??1????2-??4??G-??2????1-??4?,??2=???2-??3????1-??4????1-??3????2-??4?.(3.55) Also ??1, ??2, ??3, and ??4 are the roots of the polynomial equation: G4+??3??4G3+??2??4G2+??1??4??G+0??4=0.(3.56) Substituting the solutions (3.49)–(3.52) into (2.4) and (3.2), we find ?????(??,??)=0+??1??1±??1?????M-0??/??1/??,?????(??,??)=0+??1??1+4??2???2-??1???14??2-????1-??2?????M-0/?????2?1/??,?????(??,??)=0+??1??2+???2-??1???1??exp???1-??2????/????M-0?/?????-11/??,?????(??,??)=0+??1??1+???1-??2???1??exp???1-??2????/????M-0?/?????-11/??,???????????(??,??)=0+??1??1-2???1-??2????1-??3???12??1-??2-??3+???3-??2??cosh?????1-??2????1-??3??M????????/??1/??,(3.57) where M denotes by ??+((-2[3??3??0(??+2????+??(2+??)??0)-??2??1(??+2????+??(2+??)??0)])/??3(2+7??+7??2+2??3))??.
If we take ??0=-??1??1 and ??0=0, then the solutions (3.57) can reduce to rational function solutions: ?±?????(??,??)=???+2??1???+2????-??(2+??)??1??1??3??3??1+??2?/??3?2+7??+7??2+2??3?????1/??,?????4?????(??,??)=2???2-??1???1?4??2-????1-??2?????+2??1???+2????-??(2+??)??1??1??3??3??1+??2?/??3?2+7??+7??2+2??3??????2??????1/??,(3.58) traveling wave solutions: ??????(??,??)=2-??1???12?????1±coth1-??2??2????+2??1???+2????-??(2+??)??1??1??3??3??1+??2???3?2+7??+7??2+2??3???????1/??,(3.59) and soliton solution: ????(??,??)=3??????+cosh1(??-????)??1/??,(3.60) where ???=????1, ??3=(2(??1-??2)(??1-??3)??1/(??3-??2))1/??, ??1=v(??1-??2)(??1-??3)/??, ??=(2??1-??2-??3)/(??3-??2), and ??=-2??1(??+2????-??(2+??)??1??1)(3??3??1+??2)/??3(2+7??+7??2+2??3). Here, ??3 is the amplitude of the soliton, while ?? is the velocity and ??1 is the inverse width of the soliton. Thus, we can say that the solitons exist for ??1<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: ????=(G)=???(G)????=0????G???????=0????G??,(4.1) where ?G??2=?(G)=F(G)=???(G)??G??+?+??1G+??0????G??+?+??1G+??0.(4.2) Here, ????(??=0,,??), ????(??=0,,??), ????(??=0,,??), and ????(??=0,,??) are the constants to be determined.

Step 2. Taking trial equations (4.1) and (4.2), we derive the following equations: ?????2=F(G)????(G)?(G)??(G)-??(G)????(G)2??4,??(G)??=????(G)??(G)-??(G)???F(G)????(G)?(G)-F(G)???(G)??(G)-4F(G)?(G)????(G)2??3(G)?2+???(G)2F(G)?(G)??(G)??(G)??(G)-??(G)?????(G)2??3(G)?2,(G)(4.3) and other derivation terms such as ?????, and so on.

Step 3. Substituting ???, ???? and other derivation terms into (2.3) yields the following equation: O(G)=????G??+?+??1G+??0=0.(4.4) According to the balance principle, we can determine a relation of ??, ??, ?? and ??.

Step 4. Letting the coefficients of O(G) all be zero will yield an algebraic equations system ????=0(??=0,,??). Solving this equations system, we will determine the values ??0,????; ??0,,????; ??0,????, and ??0,,????.

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: ???(??,??)=??0??????1-??2sin2???,(5.1) where ?sin??=G-??3??2-??3?,??=??2-??3??1-??3.(5.2) Especially, ??=??/2, we have ?????2?=?,??0??/2??????1-??2sin2???1=??+2??2??2+?+1.3?(2??-1)??2.4?(2??)2????2??+?,(5.3) where ??2??=???0sin2????????. Taking the value ??=??/2, we have ??2??=((1.3?(2??-1))/(2.4?(2??)))(??/2). Therefore, if we take G(??)=??2 in (3.13), G(??)=??4 in (3.28) and (3.55), vG(??)=??1 in (3.38), for each ??, then we have ?????2?=??,??28???=0?(2??)!22??(??!)2?2??2??.(5.4) By the using radius of convergence of power series: 1??=lim???8?????+1/?????,(5.5) where ????=((2??)!/22??(??!)2)2. We have the radius of convergence of power series ??=1. We can say that power series converges for each 0<??<1, diverges for each ??>1. Consequently, the inequalities in (3.11), (3.26)–(3.53), and (3.38) are obtained by using 0<??<1, respectively.

Acknowledgment

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

References

  1. R. Hirota, “Exact solution of the korteweg-de vries equation for multiple Collisions of solitons,” Physical Review Letters, vol. 27, no. 18, pp. 1192–1194, 1971. View at Publisher · View at Google Scholar · View at Scopus
  2. W. Malfliet and W. Hereman, “The tanh method. I. Exact solutions of nonlinear evolution and wave equations,” Physica Scripta, vol. 54, no. 6, pp. 563–568, 1996. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  3. H. X. Wu and J. H. He, “Exp-function method and its application to nonlinear equations,” Chaos, Solitons and Fractals, vol. 30, pp. 700–708, 2006. View at Google Scholar
  4. E. Misirli and Y. Gurefe, “Exp-function method for solving nonlinear evolution equations,” Mathematical & Computational Applications, vol. 16, no. 1, pp. 258–266, 2011. View at Google Scholar · View at Zentralblatt MATH
  5. Y. Gurefe and E. Misirli, “Exp-function method for solving nonlinear evolution equations with higher order nonlinearity,” Computers & Mathematics with Applications, vol. 61, no. 8, pp. 2025–2030, 2011. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  6. M. Wang, X. Li, and J. Zhang, “The (G/G)-expansion method and travelling wave solutions of nonlinear evolution equations in mathematical physics,” Physics Letters. A, vol. 372, no. 4, pp. 417–423, 2008. View at Publisher · View at Google Scholar
  7. Y. Gurefe and E. Misirli, “New variable separation solutions of two-dimensional Burgers system,” Applied Mathematics and Computation, vol. 217, no. 22, pp. 9189–9197, 2011. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  8. Y. Gurefe, A. Sonmezoglu, and E. Misirli, “Application of the trial equation method for solving some nonlinear evolution equations arising in mathematical physics,” Pramana, vol. 77, pp. 1023–1029, 2011. View at Google Scholar
  9. Y. Gurefe, A. Sonmezoglu, and E. Misirli, “Application of an irrational trial equation method to high-dimensional nonlinear evolution equations,” Journal of Advanced Mathematical Studies, vol. 5, pp. 41–47, 2012. View at Google Scholar
  10. C. S. Liu, “Trial equation method and its applications to nonlinear evolution equations,” Acta Physica Sinica, vol. 54, no. 6, pp. 2505–2509, 2005. View at Google Scholar · View at Zentralblatt MATH
  11. C. S. Liu, “Trial equation method for nonlinear evolution equations with rank inhomogeneous: mathematical discussions and applications,” Communications in Theoretical Physics, vol. 45, pp. 219–223, 2006. View at Google Scholar
  12. C. S. Liu, “A new trial equation method and its applications,” Communications in Theoretical Physics, vol. 45, pp. 395–397, 2006. View at Google Scholar
  13. C. Y. Jun, “Classification of traveling wave solutions to the Vakhnenko equations,” Computers & Mathematics with Applications, vol. 62, no. 10, pp. 3987–3996, 2011. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  14. C. Y. Jun, “Classification of traveling wave solutions to the modified form of the Degasperis-Procesi equation,” Mathematical and Computer Modelling, vol. 56, pp. 43–48, 2012. View at Google Scholar
  15. C.-S. Liu, “Applications of complete discrimination system for polynomial for classifications of traveling wave solutions to nonlinear differential equations,” Computer Physics Communications, vol. 181, no. 2, pp. 317–324, 2010. View at Publisher · View at Google Scholar · View at Zentralblatt MATH
  16. M. M. Kabir, “Analytic solutions for a nonlinear variant of the (2+1) dimensional Camassa Holm KP equation,” Australian Journal of Basic and Applied Sciences, vol. 5, no. 12, pp. 1566–1577, 2011. View at Google Scholar
  17. W. Zhang, Q. Chang, and B. Jiang, “Explicit exact solitary-wave solutions for compound KdV-type and compound KdV-Burgers-type equations with nonlinear terms of any order,” Chaos, Solitons and Fractals, vol. 13, no. 2, pp. 311–319, 2002. View at Publisher · View at Google Scholar · View at Zentralblatt MATH