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

Authors: Hasan Bulut
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The classification of exact solutions, including solitons and elliptic solutions, to the generalizedK(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.


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## Body

## 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 Section2, 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]:
(1)(ql)t+aqmqx+b(qn)xxx=0,
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(2)P(u,ut,ux,uxx,…)=0,
take the general wave transformation
(3)u(x1,x2,…,xN,t)=u(η),η=λ(∑j=1N‍xj-ct),
where λ≠0 and c≠0. Substituting (3) into (2) yields a nonlinear ordinary differential equation:
(4)N(u,u′,u′′,…)=0.Step 2.
Take the finite series and trial equation as follows:(5)u=∑i=0δτiΓi,
where
(6)(Γ′)2=Λ(Γ)=Φ(Γ)Ψ(Γ)=ξθΓθ+⋯+ξ1Γ+ξ0ζϵΓϵ+⋯+ζ1Γ+ζ0.
Using (5) and (6), we can write
(7)(u′)2=Φ(Γ)Ψ(Γ)(∑i=0δ‍iτiΓi-1)2,u′′=Φ′(Γ)Ψ(Γ)-Φ(Γ)Ψ′(Γ)2Ψ2(Γ)(∑i=0δ‍iτiΓi-1)+Φ(Γ)Ψ(Γ)(∑i=0δ‍i(i-1)τiΓi-2),
where Φ(Γ) and Ψ(Γ) are polynomials. Substituting these relations into (4) yields an equation of polynomial Ω(Γ) of Γ:
(8)Ω(Γ)=ϱsΓs+⋯+ϱ1Γ+ϱ0=0.
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 ofΩ(Γ) all be zero will yield an algebraic equations system:
(9)ϱi=0,i=0,…,s.
Solving this system, we will determine the values of ξ0,…,ξθ; ζ0,…,ζϵ; and τ0,…,τδ.Step 4.
Reduce (6) to the elementary integral form
(10)±(η-η0)=∫‍dΓΛ(Γ)=∫‍Ψ(Γ)Φ(Γ)dΓ.
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 ofK(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
(11)-c(ul(η))′+am+1(um+1(η))′+b(un(η))′′′=0,
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
(12)-cul(η)+am+1um+1(η)+b(un(η))′′=0.
Let l=n, applying balance and using the following transformation:
(13)u=v1/(m-n+1).
Equation (12) turns into the following equation:
(14)-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.
Substituting (7) into (14) and using balance principle yield
(15)θ=ϵ+δ+2.
After this solution procedure, we obtain the results as follows.Case 1.
If we takeϵ=0, δ=1, and θ=3, then
(16)(v′)2=(τ1)2(ξ3Γ3+ξ2Γ2+ξ1Γ+ξ0)ζ0,v′′=τ1(3ξ3Γ2+2ξ2Γ+ξ1)2ζ0,
where ξ3≠0 and ζ0≠0. Respectively, solving the algebraic equation system (9) yields
(17)ξ0=-ξ12(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,τ1=-4(1+m-2n)ξ2τ0(1+m+n)ξ1,ζ0=-bnξ2(1+m)a(1+m-n)τ0,c=an(5+5m-7n)τ0(1+m)(1+m-n)(1+m+n).
Substituting these results into (6) and (10), we have(18)±(η-η0)=A2∫dΓΓ3-ξ1(1+m+n)28ξ2(1+m-2n)(1+m-n)Γ2-ξ12(1+m+n)28ξ22(1+m-2n)(1+m-n)Γ+ξ13(3+3m-5n)(1+m+n)3128ξ22(1+m-2n)3(1+m-n),where
(19)A=bnξ1(1+m)(1+m+n)22aξ2τ0(1+m-n)2(1+m-2n).
Integrating (18), we obtain the solutions to (1) as follows:
(20)±(η-η0)=-AΓ-α1,±(η-η0)=Aα2-α1arctanΓ-α2α2-α1,α2>α1,±(η-η0)=Aα1-α2ln|Γ-α2-α1-α2Γ-α2+α1-α2|,α1>α2,±(η-η0)=-Aα1-α3F(φ,l),α1>α2>α3,
where
(21)F(φ,l)=∫0φ‍dψ1-l2sin2ψ,φ=arcsinΓ-α3α2-α3,l2=α2-α3α1-α3.
Also α1, α2, and α3 are the roots of the polynomial equation
(22)Γ3+ξ2ξ3Γ2+ξ1ξ3Γ+ξ0ξ3=0.
Substituting solutions (20) into (5) and (13), we have
(23)u(x,t)=[×t-η0)2)-1(A2τ1)τ0+τ1α1+(×t-η0)2)-1)A2τ1(×t-η0×t-η0)2)-1)((1+m)(1+m-n)(1+m+n))-1)2(×t-η0)2)-1)((1+m)(1+m-n)(1+m+n))-1x-(an(5+5m-7n)τ0)×((1+m)(1+m-n)(1+m+n))-1×t-η0×t-η0)2)-1)((1+m)(1+m-n)(1+m+n))-1)2)-1)]1/(m-n+1),u(x,t)=[(α2-α1A(x-an(5+5m-7n)τ0(1+m)(1+m-n)(1+m+n)t-η0,))τ0+τ1α1+τ1(α2-α1)×sech2(α2-α1A(α2-α1Ax-(an(5+5m-7n)τ0)(α2-α1A(α2-α1Allll×((1+m)(1+m-n)(α2-α1A(α2-α1Allll×(1+m+n))-1(α2-α1A(α2-α1Allll×t-η0α2-α1A))(α2-α1A]1/(m-n+1),u(x,t)=[α1-α22Aτ0+τ1α1+τ1(α1-α2)×cosech2(α1-α22A((α1-α22Ax-(an(5+5m-7n)τ0)(α1-α22Alllllllllllll×((1+m)(1+m-n)(α1-α22Alllllllllllll×(1+m+n))-1(α1-α22Alllllllllllll×t-η0(α1-α22A))]1/(m-n+1),u(x,t)=[×α1-α3α1-α2))-1τ0+τ1α1+(τ1(α2-α1))×(sn2(±α2-α1Aα1-α3α1-α2)×(x-an(5+5m-7n)τ0(1+m)(1+m-n)(1+m+n)sn2(α2-α1A×t-η0an(5+5m-7n)τ0(1+m)(1+m-n)(1+m+n)),α1-α3α1-α2)α1-α3α1-α2))-1]1/(m-n+1).
If we take τ0=-τ1α1 and η0=0, then solutions (23) can reduce to rational function solution
(24)u(x,t)=(A~x-ct)2/(m-n+1),
1-soliton wave solution
(25)u(x,t)=B~cosh2/(m-n+1)(B(x-ct)),
singular soliton solution
(26)u(x,t)=C~sinh2/(m-n+1)(C(x-ct)),
and elliptic soliton solution
(27)u(x,t)=B~sn2/(m-n+1)(φ,l),
where A~=Aτ1, 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 takeϵ=0, δ=2, and θ=4, then
(28)(v′)2=(τ1+2τ2Γ)2(ξ4Γ4+ξ3Γ3+ξ2Γ2+ξ1Γ+ξ0)ζ0,
where ξ4≠0 and ζ0≠0. Respectively, solving the algebraic equation system (9) yields
(29)ξ0=ξ0,ξ1=ξ1,ξ2=ξ123ξ0,ξ3=ξ1324ξ02,ξ4=ξ14576ξ03,ζ0=-bn(m+1)(m+n+1)ξ1324a(m-n+1)2ξ02τ1,τ0=2ξ0τ1ξ1,τ1=τ1,τ2=ξ1τ112ξ0,c=-2anξ0τ1(m+1)(m+n+1)ξ1.
Substituting these results into (6) and (10), we get
(30)±(η-η0)=2A1∫((576ξ03ξ13)Γ+(576ξ04ξ14))-1/2(dΓ)×(Γ4+(24ξ0ξ1)Γ3+(192ξ02ξ12)Γ2+(576ξ03ξ13)Γ+(576ξ04ξ14))-1/2(576ξ03ξ13)Γ+(576ξ04ξ14))-1/2),
where A1=-6bnξ0(1+m)(1+m+n)/aξ1τ1(1+m-n)2. Integrating (30), we obtain the solutions to (1) as follows:
(31)±(η-η0)=-2A1Γ-α1,±(η-η0)=4A1α1-α2Γ-α2Γ-α1,α1>α2,±(η-η0)=2A1α1-α2ln|Γ-α1Γ-α2|,±(η-η0)=4A1(α1-α2)(α1-α3)×ln|(Γ-α2)(α1-α3)-(Γ-α3)(α1-α2)(Γ-α2)(α1-α3)+(Γ-α3)(α1-α2)|,4A1(α1-α2)(α1-α3)llllllllllllllllllllllllα1>α2>α3,±(η-η0)=4A1(α1-α4)(α2-α3)F(φ,l),lllllllllllllllllllα1>α2>α3>α4,
where
(32)φ1=arcsin(Γ-α2)(α1-α4)(Γ-α1)(α2-α4),l12=(α1-α3)(α2-α4)(α2-α3)(α1-α4).
Also α1, α2, α3, and α4 are the roots of the polynomial equation
(33)Γ4+ξ3ξ4Γ3+ξ2ξ4Γ2+ξ1ξ4Γ+ξ0ξ4=0.
Substituting solutions (31) into (5) and (13), we have
(34)u(x,t)=[×t-η0)-1)22anξ0τ1(m+1)(m+n+1)ξ1×t-η02anξ0τ1(m+1)(m+n+1)ξ1)-1)2τ0+τ1α1±(2τ1A1)×(x+2anξ0τ1(m+1)(m+n+1)ξ1t-η0)-1+τ2(2anξ0τ1(m+1)(m+n+1)ξ1×t-η02anξ0τ1(m+1)(m+n+1)ξ1)-1)2α1±(2A1)×(x+2anξ0τ1(m+1)(m+n+1)ξ1×t-η02anξ0τ1(m+1)(m+n+1)ξ1)-1)22anξ0τ1(m+1)(m+n+1)ξ1]1/(m-n+1),u(x,t)=[×t-η0)]2-1)2×t-η0×t-η0)]2-1)2)]2-1)2τ0+τ1α1+(16A12(α2-α1)τ1)×(t-η0×t-η0)]2-1)2)16A12-[t-η0×t-η0)]2-1)2)(α1-α2)×(x+2anξ0τ1(m+1)(m+n+1)ξ1t-η0)]2)-1+τ2(t-η0×t-η0)]2-1)2)×t-η0×t-η0)]2-1)2)]2)-1α1+(16A12(α2-α1))×(×t-η0)]2-1)216A12-[×t-η0)]2-1)2(α1-α2)×(x+2anξ0τ1(m+1)(m+n+1)ξ1×t-η0×t-η0)]2-1)2)]2)-1)2]1/(m-n+1),u(x,t)=[t-η0)((α1-α2)(2A1))]-1))-1)2τ0+τ1α2+((α2-α1)τ1)×(exp[α1-α22A1×(x+2anξ0τ1(m+1)(m+n+1)ξ1t-η0)]-1)-1+τ2(t-η02anξ0τ1(m+1)(m+n+1)ξ1)((α1-α2)(2A1))]-1)-12anξ0τ1(m+1)(m+n+1)ξ1α2+(α2-α1)×(exp[α1-α22A1×(x+2anξ0τ1(m+1)(m+n+1)ξ1×t-η02anξ0τ1(m+1)(m+n+1)ξ1)((α1-α2)(2A1))]-1)-1)2]1/(m-n+1),u(x,t)=[((α1+((α1-α2))t-η0)]-1))-1)2t-η0)((α1-α2)(2A1))]-1))-1)2τ0+τ1α1+((α1-α2)τ1)×(t-η0)α1-α22A1]exp[α1-α22A1×(x+(2anξ0τ1)×((m+1)(m+n+1)ξ1)-1×t-η0)α1-α22A1α1-α22A1]-1)-1+τ2(t-η0)α1-α22A1]-1)-1α1+(α1-α2)×(t-η0)α1-α22A1]exp[α1-α22A1τ0+τ1α1llll×(x+(2anξ0τ1)τ0+τ1α1llll×((m+1)(m+n+1)ξ1)-1τ0+τ1α1llll×t-η0)α1-α22A1]-1)-1)2]1/(m-n+1),u(x,t)=[(α1-α2)(α1-α3)2A1τ0+τ1α1-(2(α1-α2)(α1-α3)τ1)×(t-η0)(α1-α2)(α1-α3)2A1])2α1-α2-α3+(α3-α2)×cosh[(α1-α2)(α1-α3)2A1×(x+2anξ0τ1(m+1)(m+n+1)ξ1×t-η02anξ0τ1(m+1)(m+n+1)ξ1)(α1-α2)(α1-α3)2A1])-1+τ2(t-η0(α1-α2)(α1-α3)2A1)])-1)2α1-(2(α1-α2)(α1-α3))×(t-η0(α1-α2)(α1-α3)2A1)])-1)22α1-α2-α3+(α3-α2)×cosh[(α1-α2)(α1-α3)2A1×(t-η0(α1-α2)(α1-α3)2A1)])-1)2x+2anξ0τ1(m+1)(m+n+1)ξ1×t-η0t-η0(α1-α2)(α1-α3)2A1)])-1)2)]t-η0(α1-α2)(α1-α3)2A1)])-1)22α1-α2-α3+(α3-α2))-1)2(α1-α2)(α1-α3)2A1]1/(m-n+1),u(x,t)=[(α1-α3)(α2-α4)4A1+α4-α2(α1-α3)(α2-α4)4A1)-1)2τ0+τ1α2+((α1-α2)(α4-α2)τ1)×((α1-α3)(α2-α4)4A1(α1-α4)×sn2((α1-α3)(α2-α4)4A1×(x+2anξ0τ1(m+1)(m+n+1)ξ1×t-η02anξ0τ1(m+1)(m+n+1)ξ1),(α2-α3)(α1-α4)(α1-α3)(α2-α4)(α1-α3)(α2-α4)4A1)+α4-α2(α1-α3)(α2-α4)4A1)-1+τ2((α1-α3)(α2-α4)4A1α2+((α1-α2)(α4-α2)τ1)×((α1-α3)(α2-α4)4A1(α1-α4)×sn2((α1-α3)(α2-α4)4A1×((α1-α3)(α2-α4)4A1x+2anξ0τ1(m+1)(m+n+1)ξ1×t-η0(α1-α3)(α2-α4)4A1),(α2-α3)(α1-α4)(α1-α3)(α2-α4)(α1-α3)(α2-α4)4A1)+α4-α2(α1-α3)(α2-α4)4A1)-1)2]1/(m-n+1).
For simplicity, if we takeη0=0, then we can write solutions (34) as follows:
(35)u(x,t)=[∑i=02‍τi(α1±2A1x-ct)i]1/(m-n+1),u(x,t)=[∑i=02‍τi((16A12-[(α1-α2)(x-ct)]2)-1α1+(16A12(α1-α2))×(16A12-[(α1-α2)(x-ct)]2)-1)i∑i=02‍]1/(m-n+1),u(x,t)=[∑i=02‍τi(α2+α2-α1exp[B1(x-ct)]-1)i]1/(m-n+1),u(x,t)=[∑i=02‍τi(α1+α1-α2exp[B1(x-ct)]-1)i]1/(m-n+1),u(x,t)=[∑i=02‍τi(×cosh[C1(x-ct)])-1α1-(2(α1-α2)(α1-α3))×(2α1-α2-α3+(α3-α2)×cosh[C1(x-ct)])-1)i∑i=02‍]1/(m-n+1),u(x,t)=[∑i=02‍τi(+α4-α2(α1-α4)sn2(φ,l))-1α2+((α1-α2)(α4-α2))llllll×((α1-α4)sn2(φ,l)lllllllll+α4-α2(α1-α4)sn2(φ,l))-1)i∑i=02‍]1/(m-n+1),
where B1=(α1-α2)/2A1, C1=(α1-α2)(α1-α3)/2A1, φ1=((α1-α3)(α2-α4)/4A1)(x-ct), l12=(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 takeϵ=0, δ=3, and θ=5, then
(36)(v′)2=(τ1+2τ2Γ+3τ3Γ2)2×(ξ5Γ5+ξ4Γ4+ξ3Γ3+ξ2Γ2+ξ1Γ+ξ0)×(ζ0)-1,
where ξ5≠0 and ζ0≠0. Respectively, solving the algebraic equation system (9) yields
(37)ξ0=ξ5(τ22-4τ1τ3)(2τ23-9τ1τ2τ3+2(τ22-3τ1τ3)3)81τ35,ξ1=-ξ5(4τ24+9τ1τ22τ3-108τ12τ32+4τ2(τ22-3τ1τ3)3)81τ34,ξ2=ξ5(-11τ23+63τ1τ2τ3-2(τ22-3τ1τ3)3)27τ33,ξ3=ξ5(τ22+7τ1τ3)3τ32,ξ4=5ξ5τ23τ3,ζ0=-9bn(m+1)(m+n+1)ξ52aτ3(m-n+1)2,τ0=-2τ23-9τ1τ2τ3+2(τ22-3τ1τ3)327τ32,τ1=τ1,τ2=τ2,τ3=τ3,ξ5=ξ5,c=-8an(τ22-3τ1τ3)327(m+1)(m+n+1)τ32.
Substituting these results into (6) and (10), we get
(38)±(η-η0)=3A2∫(+ξ2ξ5Γ2+ξ1ξ5Γ+ξ0ξ5)1/2(dΓ)×(Γ5+ξ4ξ5Γ4+ξ3ξ5Γ3lllll+ξ2ξ5Γ2+ξ1ξ5Γ+ξ0ξ5)-1/2),
where A2=-bn(1+m)(1+m+n)/2aτ3(1+m-n)2. Integrating (38), we obtain the solutions to (1) as follows:
(39)±(η-η0)=-2A2(Γ-α1)3,±(η-η0)=3A2arctanh[(Γ-α2)/(α1-α2)](α1-α2)3/2-3A2Γ-α2(α1-α2)(Γ-α1),α1>α2,±(η-η0)=-6A2arctan[(Γ-α1)/(α1-α2)](α1-α2)3/2-6A2Γ-α1(α1-α2),±(η-η0)=6A2arctanh[(Γ-α3)/(α2-α3)]α1-α2×(1α2-α3-1α1-α3),aaaα1>α2>α3,±(η-η0)=-6A2Γ-α1(α1-α2)(α1-α3)×[(Γ-α2)(Γ-α3)+i(E(φ,l)-F(φ,l))],
where
(40)E(φ,l)=∫0φ‍1-l2sin2ψdψ,φ2=-arcsinΓ-α1α2-α1,l22=α1-α2α1-α3,±(η-η0)=-6iA2α2-α3(α1-α2)(F(φ,l)-π(φ,n,l)),-6iA2α2-α3(α1-α2)lllllllllllllα1>α2>α3>α4,
where
(41)φ3=-arcsinα3-α2Γ-α2,l32=α2-α4α2-α3,n=α2-α1α2-α3.
Also α1, α2, α3, α4, and α5 are the roots of the polynomial equation
(42)Γ5+ξ4ξ5Γ4+ξ3ξ5Γ3+ξ2ξ5Γ2+ξ1ξ5Γ+ξ0ξ5=0.Case 4.
If we takeϵ=1, δ=1, and θ=4, then
(43)(v′)2=τ12(ξ4Γ4+ξ3Γ3+ξ2Γ2+ξ1Γ+ξ0)ζ0+ζ1Γ,
where ξ4≠0 and ζ1≠0. Respectively, solving the algebraic equation system (9) yields
(44)ξ0=ζ0τ02(M+2a(1+m-n)2(2ζ1τ0+ζ0τ1))bn(1+m)(1+m+n)ζ1τ12,ξ3=ξ3,ξ4=-2a(1+m-n)2ζ1τ1bn(1+m)(1+m+n),ξ1=(τ0(4a(1+m-n)2ζ12τ02+2ζ0τ1(M+2a(1+m-n)2ζ0τ1)+ζ1τ0(M+8a(1+m-n)2ζ0τ1)))×(bn(1+m)(1+m+n)ζ1τ12)-1,ξ2=(6a(1+m-n)2ζ12τ02+2ζ1τ0(M+2a(1+m-n)2ζ0τ1)+ζ0τ1(M+2a(1+m-n)2ζ0τ1))×(bn(1+m)(1+m+n)ζ1τ12)-1,ζ0=ζ0,ζ1=ζ1,τ0=τ0,τ1=τ1,c=n(M+2a(1+m-n)2(3ζ1τ0+ζ0τ1))(1+m)(1+m+n)(1+m-n)2ζ1,
where M=bn(1+m)(1+m+n)ξ3. Substituting these results into (6) and (10), we get
(45)±(η-η0)=A3∫((Γ+ζ0ζ1)×(Γ4+(ξ3ξ4)Γ3+(ξ2ξ4)Γ2+(ξ1ξ4)Γ+(ξ0ξ4))-1)1/2dΓ,
where A3=bn(1+m)(1+m+n)/2aτ1(1+m-n)2. Integrating (45), we obtain the solution to (1) as follows:
(46)±(η-η0)=-A3ζ1ζ0+ζ1α1×arctanh[ζ0+ζ1Γζ0+ζ1α1]-A3Γ-α1ζ0+ζ1Γζ1,±(η-η0)=2A3α1-α2×(-ζ0+ζ1α1ζ1×arctanh[ζ0+ζ1Γζ0+ζ1α1]+ζ0+ζ1α2×arctanh[ζ0+ζ1Γζ0+ζ1α2]),±(η-η0)=2A3×((Γ-α1)(ζ0+ζ1Γ)ζ1(Γ-α2)2+iα1-α2(E(φ,l)-F(φ,l))(Γ-α1)(ζ0+ζ1Γ)ζ1(Γ-α2)2),
where
(47)φ4=-arcsinζ1(α1-Γ)ζ0+ζ1α1,l42=ζ0+ζ1α1ζ1(α1-α2),±(η-η0)=-2A3α2-α1E(φ,l),
where
(48)φ5=arcsin[α2-α1Γ-α1],l52=ζ0+ζ1α1ζ1(α1-α2),±(η-η0)=-2iA3(α1-α2)ζ1(ζ0+ζ1α2)×(ζ0(F(φ,l)-π(φ,n,l))+ζ1(α2F(φ,l)-α2π(φ,n,l))),
where
(49)φ6=-arcsinζ0+ζ1α2ζ1(α2-Γ),l62=ζ1(α2-α3)ζ0+ζ1α2,n1=ζ1(α2-α1)ζ0+ζ1α2.Case 5.
If we takeϵ=1, δ=2, and θ=5, then
(50)(v′)2=(τ1+2τ2Γ)2(ξ5Γ5+ξ4Γ4+ξ3Γ3+ξ2Γ2+ξ1Γ+ξ0)ζ0+ζ1Γ,
where ξ5≠0 and ζ1≠0. Respectively, solving the algebraic equation system (9) yields
(51)ξ0=τ02(-2ξ5τ1+ξ4τ2)τ23,ξ1=τ0(2ξ4τ1τ2+ξ5(-4τ12+τ0τ2))τ23,ξ2=ξ4τ2(τ12+2τ0τ2)-2ξ5(τ13+τ0τ1τ2)τ23,ξ3=-3ξ5τ12+2τ2(ξ5τ0+ξ4τ1)τ22,ζ0=-2bn(1+m)(1+m+n)(ξ4τ2-2ξ5τ1)a(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,c=-an(τ12-4τ0τ2)2(1+m)(1+m+n)τ2.
Substituting these results into (6) and (10), we get
(52)±(η-η0)=A4×∫(+(ξ2ξ5)Γ2+(ξ1ξ5)Γ+(ξ0ξ5))-1(Γ+ζ0ζ1)×(Γ5+ξ4ξ5Γ4+ξ3ξ5Γ3+ξ2ξ5Γ2+ξ2ξ5Γ2+ξ1ξ5Γ+ξ0ξ5)-1+(ξ2ξ5)Γ2+(ξ1ξ5)Γ+(ξ0ξ5))-1)1/2dΓ,
where A4=-2bn(1+m)(1+m+n)/aτ22(1+m-n)2. Integrating (52), we obtain the solution to (1) as follows:
(53)±(η-η0)=-2A43ζ1(ζ0+ζ1α1)(ζ0+ζ1ΓΓ-α1)3/2,±(η-η0)=-A4(ζ0+ζ1α2)2(α1-α2)3/2ζ1(ζ0+ζ1α1)×ln|+ζ1(2Γα1-α2(Γ+α1)))-1(Γ-α1)×(ζ0(Γ+α1-2α2)+2(ζ0+ζ1Γ)(ζ0+ζ1α1)(Γ-α2)(α1-α2)+ζ1(2Γα1-α2(Γ+α1)))-1|-A4(α1-α2)(Γ-α1)(ζ0+ζ1Γ)(Γ-α2)ζ1,α1>α2,±(η-η0)=-2A4(α1-α2)ζ0+ζ1Γζ1(Γ-α1)-2A4(α1-α2)3/2ζ0+ζ1α2ζ1×arctan[(Γ-α1)(ζ0+ζ1α2)(α1-α2)(ζ0+ζ1Γ)],±(η-η0)=-A4α1-α3ζ0+ζ1α2ζ1(α2-α3)×ln|+ζ1((2Γα1-α3(Γ+α1))2Γα2-α3(Γ+α2)))-1(α2-Γ)×(ζ0(Γ+α2-2α3)+2(ζ0+ζ1Γ)(ζ0+ζ1α2)(Γ-α3)(α2-α3)+ζ1((2Γα1-α3(Γ+α1))2Γα2-α3(Γ+α2)))-1|-A4α1-α3×ζ0+ζ1α1ζ1(α1-α3)×ln|(Γ-α2)-1(ζ0(Γ+α1-2α3)+2(ζ0+ζ1Γ)(ζ0+ζ1α1)(Γ-α3)(α1-α3)+ζ1(2Γα1-α3(Γ+α1)))×(Γ-α2)-1|,α1>α2>α3,±(η-η0)=-2A4α1-α3×ζ0+ζ1α3ζ1(α1-α2)E(φ,l),α1>α2>α3,
where
(54)φ7=arcsin(Γ-α3)(α2-α1)(Γ-α1)(α2-α3),l72=(α3-α2)(ζ0+ζ1α1)(α1-α2)(ζ0+ζ1α3),±(η-η0)=2A4(α2-α4)(α1-α2)(α3-α4)ζ1(α2-α3)(ζ0+ζ1α4)×((ζ0+ζ1Γ)(α3-α4)α1-α2π(φ,n,l)+(ζ0+ζ1α2)(α4-α3)α2-α4F(φ,l)),
where
(55)φ8=arcsin(Γ-α3)(α2-α1)(Γ-α1)(α2-α3),l82=(α3-α2)(ζ0+ζ1α1)(α1-α2)(ζ0+ζ1α3),n2=-(α1-α2)(α3-α4)(α2-α3)(α1-α4),α1>α2>α3>α4.


## 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:
(56)u=A(Γ)B(Γ)=∑i=0δτiΓi∑j=0μωjΓj,
where
(57)(Γ′)2=Λ(Γ)=Φ(Γ)Ψ(Γ)=ξθΓθ+⋯+ξ1Γ+ξ0ζϵΓϵ+⋯+ζ1Γ+ζ0.
Here, τi(i=0,…,δ), ωj(j=0,…,μ), ξς(ς=0,…,θ), and ζσ(σ=0,…,ϵ) are the constants to be specified.Step 2.
Taking trial equations (56) and (57), we derive the following equations:
(58)(u′)2=Φ(Γ)Ψ(Γ)(A′(Γ)B(Γ)-A(Γ)B′(Γ))2B4(Γ),(59)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)}+2Φ(Γ)Ψ(Γ)B(Γ)(A′′(Γ)B(Γ)-A(Γ)B′′(Γ))×(2B3(Γ)Ψ2(Γ))-1
and other derivation terms such as u′′′.Step 3.
Substitutingu′, u′′, and other derivation terms into (5) yields the following equation:
(60)Ω(Γ)=ϱsΓs+⋯+ϱ1Γ+ϱ0=0.
According to the balance principle we can determine a relation of θ, ϵ, δ, and μ.Step 4.
Letting the coefficients ofΩ(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 Step4 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 generalizedK(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.



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*Source: 742643-2013-11-24.xml*