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

Authors: Yusuf Pandir; Yusuf Gurefe; Ugur Kadak; Emine Misirli
Journal: Abstract and Applied Analysis
 (2012)

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## 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.


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## Body

## 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, [1–15]. 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 Section2, 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]:
(1.1)(ut+2kux-uxxt+aun(un)x)x+uyy=0,
the dimensionless form of the generalized KdV equation [17]:
(1.2)ut+aunux+bu2nux+δuxxx=0.
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,(2.1)P(u,ut,ux,uxx,…)=0,
take the general wave transformation:
(2.2)u(x1,x2,…,xN,t)=u(η),η=λ(∑j=1N‍xj-ct),
where λ≠0 and c≠0. Substituting (2.2) into (2.1) yields a nonlinear ordinary differential equation:
(2.3)N(u,u′,u′′,…)=0.Step 2.
Take the finite series and trial equation as follows:(2.4)u=∑i=0δ‍τiΓi,
where
(2.5)(Γ′)2=Λ(Γ)=Φ(Γ)Ψ(Γ)=ξθΓθ+⋯+ξ1Γ+ξ0ζϵΓϵ+⋯+ζ1Γ+ζ0.
Using (2.4) and (2.5), we can write
(2.6)(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 (2.3) yields an equation of polynomial Ω(Γ) of Γ:
(2.7)Ω(Γ)=ϱsΓs+⋯+ϱ1Γ+ϱ0=0.
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 ofΩ(Γ) all be zero will yield an algebraic equations system:
(2.8)ϱi=0,i=0,…,s.
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:
(2.9)±(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 (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+y-ct), 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
(3.1)(2k+1-c)u+a2u2n+m2cu′′=0.
We use the following transformation:
(3.2)u=v1/(2n-1).
Equation (3.1) turns into the equation
(3.3)m2c(2n-1)vv′′+2m2c(1-n)v′2+(2k+1-c)(2n-1)2v2+a2(2n-1)2v3=0.
Substituting (2.6) into (3.3) and using balance principle yield
(3.4)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
(3.5)(v′)2=(τ1)2(ξ3Γ3+ξ2Γ2+ξ1Γ+ξ0)ζ0,
where ξ3≠0, ζ0≠0. Respectively, solving the algebraic equation system (2.8) yields
(3.6)ξ1=-a(1-2n)2ζ0τ03+m2ξ0τ12(2+4k+4n+8kn+3aτ0)m2τ0τ1(q+aτ0),ξ2=-2a(1-2n)2ζ0τ03+m2τ12(q+3aτ0)m2τ02(q+aτ0),ξ3=aτ1(-(1-2n)2ζ0τ02+m2ξ0τ12)m2τ02(q+aτ0),ξ0=ξ0,ζ0=ζ0,τ0=τ0,τ1=τ1,c=(1-2n)2ζ0τ02(q+aτ0)(1+2n)((1-2n)2ζ0τ02-mξ0τ12),
where q=(1+2k)(1+2n). Substituting these results into (2.5) and (2.9), we have
(3.7)±(η-η0)=mA∫‍dΓΓ3+-2a(1-2n)2ζ0τ03+m2τ12(q+3aτ0)aτ1(-(1-2n)2ζ0τ02+m2ξ0τ12)Γ2+𝒜Γ+m2ξ0τ02(q+aτ0)aτ1(-(1-2n)2ζ0τ02+m2ξ0τ12),
where𝒜 denotes by ((-a(1-2n)2ζ0τ04+m2ξ0τ0τ12(2+4k+4n+8kn+3aτ0))/aτ12(-(1-2n)2ζ0τ02+m2ξ0τ12))), and A=ζ0τ02(q+aτ0)/aτ1(-(1-2n)2ζ0τ02+m2ξ0τ12). Integrating (3.7), we obtain the solutions to the (1.1) as follows:
(3.8)±(η-η0)=-2mA1Γ-α1,(3.9)±(η-η0)=2mAα2-α1arctanΓ-α2α2-α1,α2>α1,(3.10)±(η-η0)=mAα1-α2ln|Γ-α2-α1-α2Γ-α2+α1-α2|,α1>α2,(3.11)±(η-η0)=2mAα1-α3F(φ,l),α1>α2>α3,
where
(3.12)A=ζ0τ02(q+aτ0)aτ1(-(1-2n)2ζ0τ02+m2ξ0τ12),F(φ,l)=∫0φ‍dψ1-l2sin2ψ,(3.13)φ=arcsinΓ-α3α2-α3,l2=α2-α3α1-α3.
Also α1, α2, and α3 are the roots of the polynomial equation
(3.14)Γ3+ξ2ξ3Γ2+ξ1ξ3Γ+ξ0ξ3=0.
Substituting the solutions (3.8)–(3.10) into (2.4) and (3.2), we have
(3.15)u(x,y,t)=[τ0+τ1α1+4τ1A(x+y-ℬt-η0/m)2]1/(2n-1),u(x,y,t)=[τ0+τ1α2+τ1(α1-α2)tanh2(12α1-α2A(x+y-ℬt+η0m))]1/(2n-1),u(x,y,t)=[τ0+τ1α1+τ1(α1-α2)cosech2(12α1-α2A(x+y-ℬt))]1/(2n-1),
where ℬ denote by ((1-2n)2ζ0τ02(q+aτ0)/(1+2n)((1-2n)2ζ0τ02-mξ0τ12)).
If we takeτ0=-τ1α1 and η0=0, then the solutions (3.15) can reduce to rational function solution:
(3.16)u(x,y,t)=[2A~x+y-((1-2n)2α12(q-aτ1α1)/(1+2n)((1-2n)2ζ0α12-mξ0))t]2/(2n-1),
1-soliton solution:
(3.17)u(x,y,t)=A1cosh2/(2n-1)[B(x+y-vt)],
and singular soliton solution:
(3.18)u(x,y,t)=A2sinh2/(2n-1)[B(x+y-vt)],
where A~=Aτ1, A1=(τ1(α2-α1))1/(2n-1), A2=(τ1(α1-α2))1/(2n-1), B=(1/2)(α1-α2)/A, and v=(1-2n)2ζ0α12(q-aτ1α1)/(1+2n)((1-2n)2ζ0α12-mξ0). 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 takeϵ=0, δ=2 and θ=4, then
(3.19)(v′)2=(τ1+2τ2Γ)2(ξ4Γ4+ξ3Γ3+ξ2Γ2+ξ1Γ+ξ0)ζ0,
where ξ4≠0, ζ0≠0. Respectively, solving the algebraic equation system (2.8) yields as follows.
Subcase 2.1. It holds that
(3.20)ξ0=ξ0,ξ1=ξ4τ134τ23+4ξ0τ2τ1,ξ2=5ξ4τ124τ22+4ξ0τ22τ12,ξ3=2ξ4τ1τ2,ξ4=ξ4,τ1=τ1,τ2=τ2,ζ0=-m2(-16aξ0τ24+ξ4τ12(aτ12+4τ2q))a(1-2n)2τ12τ22,τ0=τ124τ2,c=1+2k+a(ξ4τ14-16ξ0τ24)4(1+2n)ξ4τ12τ2,
where q=(1+2k)(1+2n). Substituting these results into (2.5) and (2.9), we get
(3.21)±(η-η0)=m(16aξ0τ24-ξ4τ14(aτ12+4qτ2))a(1-2n)2ξ4τ12τ22×∫‍dΓΓ4+(2τ1/τ2)Γ3+((5τ12/4τ22)+(4ξ0τ22/τ12))Γ2+((τ13/4τ23)+(4ξ0τ2/τ1))Γ+(ξ0/ξ4).
Integrating  (3.21), we obtain the solutions to (1.1) as follows:
(3.22)±(η-η0)=-mBΓ-α1,(3.23)±(η-η0)=2mBα1-α2Γ-α2Γ-α1,α1>α2,(3.24)±(η-η0)=mBα1-α2ln|Γ-α1Γ-α2|,(3.25)±(η-η0)=mB(α1-α2)(α1-α3)ln|(Γ-α2)(α1-α3)-(Γ-α3)(α1-α2)(Γ-α2)(α1-α3)+(Γ-α3)(α1-α2)|,α1>α2>α3,(3.26)±(η-η0)=2mB(α1-α3)(α2-α4)F(φ,l),α1>α2>α3>α4,
where
(3.27)B=(16aξ0τ24-ξ4τ14(aτ12+4qτ2))a(1-2n)2ξ4τ12τ22,F(φ,l)=∫0φ‍dψ1-l2sin2ψ,(3.28)φ=arcsin(Γ-α1)(α2-α4)(Γ-α2)(α1-α4),l2=(α2-α3)(α1-α4)(α1-α3)(α2-α4).
Also α1, α2, α3, and α4 are the roots of the polynomial equation:
(3.29)Γ4+ξ3ξ4Γ3+ξ2ξ4Γ2+ξ1ξ4Γ+ξ0ξ4=0.
Substituting the solutions (3.22)–(3.25) into (2.4) and (3.2), we have
(3.30)u(x,y,t)=[τ0+τ1α1±τ1B𝒞-(η0/m)+τ2(α1±B𝒞-(η0/m))2]1/(2n-1),u(x,y,t)={τ0+τ1α1+4B2(α2-α1)τ14B2-[(α1-α2)(𝒞-η0)]2+τ2(α1+4B2(α2-α1)4B2-[(α1-α2)(𝒞-η0)]2)2}1/(2n-1),u(x,y,t)={τ0+τ1α2+(α2-α1)τ1exp[((α1-α2)/B)(𝒞-η0)]-1+τ2(α2+(α2-α1)exp[((α1-α2)/B)(𝒞-η0)]-1)2}1/(2n-1),u(x,y,t)={τ0+τ1α1+(α1-α2)τ1exp[((α1-α2)/B)(𝒞-η0)]-1+τ2(α1+(α1-α2)exp[((α1-α2)/B)(𝒞-η0)]-1)2}1/(2n-1),u(x,y,t)={τ0+τ1α1-2(α1-α2)(α1-α3)τ12α1-α2-α3+(α3-α2)cosh[((α1-α2)(α1-α3)/B)(𝒞)]+τ2(α1-2(α1-α2)(α1-α3)2α1-α2-α3+(α3-α2)cosh[((α1-α2)(α1-α3)/B)(𝒞)])2}1/(2n-1),
where 𝒞 denotes by x+y-(1+2k+(a(ξ4τ14-16ξ0τ24)/4(1+2n)ξ4τ12τ2))t.
For simplicity, we can write the solutions (3.30) as follows:
(3.31)u(x,y,t)=[∑i=02‍τi(α1±B𝒞-(η0/m))i]1/(2n-1),u(x,y,t)=[∑i=02‍τi(α1+4B2(α2-α1)4B2-[(α1-α2)(𝒞-η0)]2)i]1/(2n-1),u(x,y,t)=[∑i=02‍τi(α2+(α2-α1)exp[((α1-α2)/B)(𝒞-η0)]-1)i]1/(2n-1),u(x,y,t)=[∑i=02‍τi(α1+(α1-α2)exp[((α1-α2)/B)(𝒞-η0)]-1)i]1/(2n-1),u(x,y,t)=[∑i=02‍τi(α1-2(α1-α2)(α1-α3)2α1-α2-α3+(α3-α2)cosh[((α1-α2)(α1-α3)/B)(𝒞)])i]1/(2n-1),Subcase 2.2. It holds that
(3.32)ξ0=ξ0,ξ1=ξ3=0,ξ2=2ξ0ξ4,ξ4=ξ4,τ1=0,τ2=τ2,ζ0=-4(qm2ξ4+am2ξ0ξ4τ2)a(1-2n)2τ2,τ0=ξ0τ2ξ4,c=1+2k+aξ0τ2(1+2n)ξ4,
where q=(1+2k)(1+2n). Substituting these results into (2.5) and (2.9), we get
(3.33)±(η-η0)=m-4qξ4-aξ0ξ4τ2a(1-2n)2ξ4τ2∫‍dΓΓ4+(2ξ0ξ4/ξ4)Γ2+(ξ0/ξ4).
Integrating (3.33), we obtain the solutions to the (1.1) as follows.
If we denote(3.34)F(Γ)=Γ4+2ξ0ξ4ξ4Γ2+ξ0ξ4=R2+d1R+d0,
where Γ2=R, F(Γ)=G(R), then we can write complete discrimination system of G(R) as follows:
(3.35)Δ=d12-4d0.
Correspondingly, there are the following two cases to be discussed.(1)
IfΔ>0,then we haveF(Γ)=(Γ-α1)(Γ+α1)(Γ-α2)(Γ+α2),α1≠α2.Therefore, the solution is given by
(3.36)±(η-η0)=mCα1F(φ,l),
where(3.37)C=-4qξ4-aξ0ξ4τ2a(1-2n)2ξ4τ2,F(φ,l)=∫0φ‍dψ1-l2sin2ψ,(3.38)φ=arcsin(Γα1),l2=α1α2,α2>α1.(2)
IfΔ=0, then we have F(Γ)=(Γ-α1)2(Γ+α1)2. From here, the solutions can be found as
(3.39)±(η-η0)=mCα1arctanh(Γα1),(3.40)u(x,y,t)={τ0+τ1α1tanh[±α1C(x+y-(1+2k+aξ0τ2(1+2n)ξ4)t-η0m)]+τ2α1tanh2[±α1C(x+y-(1+2k+aξ0τ2(1+2n)ξ4)t-η0m)]}1/(2n-1).
For simplicity, we can write (3.40) as follows:
(3.41)u(x,y,t)={∑i=02‍τi{α1tanh[±α1C(x+y-(1+2k+aξ0τ2(1+2n)ξ4)t-η0m)]}i}1/(2n-1).Example 3.2 (Application to the generalized KdV equation).
Using a complex variationη defined as η=kx+wt, we can convert (1.2) into ordinary different equation, which reads
(3.42)wu′+akunu′+bku2nu′+δk3u′′′=0,
where the prime denotes the derivative with respect to η. Integrating (3.42), and setting the constant of integration to be zero, we obtain
(3.43)wu+akn+1un+1+bk2n+1u2n+1+δk3u′′=0.
By the using of the transformation u=v1/n, (3.43) reduces to
(3.44)δk3n(n+1)(2n+1)vv′′+δk3(1-n2)(2n+1)(v′)2+bkn2(n+1)v4+akn2(2n+1)v3+n2(n+1)(2n+1)wv2=0.
Substituting  (2.6) into (3.44) and using balance principle yield
(3.45)θ=ϵ+2δ+2.
If we take θ=4, ϵ=0, and δ=1, then
(3.46)(v′)2=τ12(ξ4Γ4+ξ3Γ3+ξ2Γ2+ξ1Γ+ξ0)ζ0,
where ξ4≠0, ζ0≠0. Respectively, solving the algebraic equation system (2.8) yields
(3.47)ξ0=𝒟2τ13,ξ1=ℱτ13,ξ2=ξ2,ξ3=ξ3,ξ4=b(2+n)ξ34τ12(a+2an+2b(2+n)τ0),ζ0=-k2(1+n)(2+n)(1+2n)δξ32n2(a+2an+2b(n+2)τ0)τ1,τ0=τ0,τ1=τ1,w=-2k[3ξ3τ0(a+2an+b(2+n)τ0)-ξ2τ1(a+2an+b(2+n)τ0)]ξ3(2+7n+7n2+2n3).
where 𝒟 denotes by τ02(-(ξ3τ0(4a(1+2n)+5b(2+n)τ0)/(a+2an+2b(2+n)τ0))+2ξ2τ1) and ℱ 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
(3.48)±(η-η0)=kτ1-(1+n)(1+2n)δ4n2b×∫‍dΓΓ4+𝒩2b(2+n)τ1Γ3+ξ2(𝒩)2b(2+n)ξ2τ1Γ2+ℱ(𝒩)2b(2+n)ξ3τ14Γ+𝒟(𝒩)4b(2+n)ξ3τ14,
where 𝒩 denotes by a+2an+b(2+n)τ0. Integrating (3.48), we obtain the solutions to (1.2) as follows:
(3.49)±(η-η0)=-kBΓ-α1,(3.50)±(η-η0)=2kBα1-α2Γ-α2Γ-α1,α2>α1,(3.51)±(η-η0)=kBα1-α2ln|Γ-α1Γ-α2|,(3.52)±(η-η0)=kB(α1-α2)(α1-α3)ln|(Γ-α2)(α1-α3)-(Γ-α3)(α1-α2)(Γ-α2)(α1-α3)+(Γ-α3)(α1-α2)|,α1>α2>α3,(3.53)±(η-η0)=2kB(α1-α3)(α2-α4)F(φ,l),α1>α2>α3>α4,
where
(3.54)B=1τ1-(1+n)(1+2n)δ4n2b,F(φ,l)=∫0φ‍dψ1-l2sin2ψ,(3.55)φ=arcsin(Γ-α1)(α2-α4)(Γ-α2)(α1-α4),l2=(α2-α3)(α1-α4)(α1-α3)(α2-α4).
Also α1, α2, α3, and α4 are the roots of the polynomial equation:
(3.56)Γ4+ξ3ξ4Γ3+ξ2ξ4Γ2+ξ1ξ4Γ+ξ0ξ4=0.
Substituting the solutions (3.49)–(3.52) into (2.4) and (3.2), we find
(3.57)u(x,t)=[τ0+τ1α1±τ1Bℳ-(η0/k)]1/n,u(x,t)={τ0+τ1α1+4B2(α2-α1)τ14B2-[(α1-α2)(ℳ-(η0/k))]2}1/n,u(x,t)={τ0+τ1α2+(α2-α1)τ1exp[((α1-α2)/B)(ℳ-(η0/k))]-1}1/n,u(x,t)={τ0+τ1α1+(α1-α2)τ1exp[((α1-α2)/B)(ℳ-(η0/k))]-1}1/n,u(x,t)={τ0+τ1α1-2(α1-α2)(α1-α3)τ12α1-α2-α3+(α3-α2)cosh[((α1-α2)(α1-α3)/B)ℳ]}1/n,
where ℳ 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 takeτ0=-τ1α1 and η0=0, then the solutions (3.57) can reduce to rational function solutions:
(3.58)u(x,t)=[±B~x+(2τ1(a+2an-b(2+n)τ1α1)(3ξ3α1+ξ2)/ξ3(2+7n+7n2+2n3))t]1/n,u(x,t)={4B~2(α2-α1)τ1[4B2-((α1-α2)(x+(2τ1(a+2an-b(2+n)τ1α1)(3ξ3α1+ξ2)/ξ3(2+7n+7n2+2n3))t))2]}1/n,
traveling wave solutions:
(3.59)u(x,t)={(α2-α1)τ12{1∓coth[(α1-α2)2B(x+2τ1(a+2an-b(2+n)τ1α1)(3ξ3α1+ξ2)ξ3(2+7n+7n2+2n3)t)]}}1/n,
and soliton solution:
(3.60)u(x,t)=A3(D+cosh[B1(x-vt)])1/n,
where B~=Bτ1, A3=(2(α1-α2)(α1-α3)τ1/(α3-α2))1/n, B1=(α1-α2)(α1-α3)/B, D=(2α1-α2-α3)/(α3-α2), and v=-2τ1(a+2an-b(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:
(4.1)u=A(Γ)B(Γ)=∑i=0δ‍τiΓi∑j=0μ‍ωjΓj,
where
(4.2)(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 determined.Step 2.
Taking trial equations (4.1) and (4.2), we derive the following equations:
(4.3)(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)}2B3(Γ)Ψ2(Γ)+2Φ(Γ)Ψ(Γ)B(Γ)(A′′(Γ)B(Γ)-A(Γ)B′′(Γ))2B3(Γ)Ψ2(Γ),
and other derivation terms such as u′′′, and so on.Step 3.
Substitutingu′, u′′ and other derivation terms into (2.3) yields the following equation:
(4.4)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 (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:(5.1)F(ϕ,l)=∫0ϕ‍dϕ(1-l2sin2ϕ),
where
(5.2)sinϕ=Γ-α3α2-α3,l=α2-α3α1-α3.
Especially, ϕ=π/2, we have
(5.3)F(π2,l)=∫0π/2‍dϕ(1-l2sin2ϕ)=ϕ+12k2v2+⋯+1.3⋯(2n-1)2.4⋯(2n)k2nv2n+⋯,
where v2n=∫0ϕ‍sin2nϕdϕ. Taking the value ϕ=π/2, we have v2n=((1.3⋯(2n-1))/(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
(5.4)F(π2,l)=π2∑n=0∞‍((2n)!22n(n!)2)2l2n.
By the using radius of convergence of power series:
(5.5)R=1limn→∞(an+1/an),
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.



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*Source: 478531-2012-08-27.xml*