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Open Physics

formerly Central European Journal of Physics

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Volume 16, Issue 1

Issues

Volume 13 (2015)

Lie symmetry analysis and conservation laws for the time fractional simplified modified Kawahara equation

Dumitru Baleanu
  • Corresponding author
  • Cankaya University, Department of Mathematics, Öǧretmenler Cad. 1406530, Ankara, Türkiye
  • Institute of Space Sciences, Magurele, Bucharest, Romania
  • Other articles by this author:
  • De Gruyter OnlineGoogle Scholar
/ Mustafa Inc / Abdullahi Yusuf / Aliyu Isa Aliyu
Published Online: 2018-05-30 | DOI: https://doi.org/10.1515/phys-2018-0042

Abstract

In this work, Lie symmetry analysis for the time fractional simplified modified Kawahara (SMK) equation with Riemann-Liouville (RL) derivative, is analyzed. We transform the time fractional SMK equation to nonlinear ordinary differential equation (ODE) of fractional order using its Lie point symmetries with a new dependent variable. In the reduced equation, the derivative is in the Erdelyi-Kober (EK) sense. We solve the reduced fractional ODE using a power series technique. Using Ibragimov’s nonlocal conservation method to time fractional partial differential equations, we compute conservation laws (Cls) for the time fractional SMK equation. Some figures of the obtained explicit solution are presented.

Keywords: time fractional SMK; Lie symmetry; exact solutions; conservation laws

PACS: 02.20.Sv; 02.20.Hj; 02.20.Qs; 02.60.Cb

1 Introduction

Symmetry analysis has many applications in the field of science and engineering. Lie’s method is one of the global and efficient methods for investigating analytical solutions and symmetry properties of nonlinear partial differential equations (NLPDEs) [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17]. Fractional calculus has been successfully used to explain many complex nonlinear phenomena and dynamic processes in physics, engineering, electromagnetics, viscoelasticity, and electrochemistry [18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34].

Generally, physical phenomenon might depend on its current state and on its historical states, which can be modelled successfully by applying the theory of derivatives and integrals of fractional order [35, 36]. Due to this, several analytical techniques are used to derive exact, explicit, and numerical solutions of nonlinear fractional partial differential equations (FPDEs) [30, 31, 32, 33, 34]. We find very few studies of symmetry analysis for FPDEs and their group properties are not plainly understood [37, 38, 39, 40, 41].

In other words, Cls are universally known to possess an important role in the analysis of NLPDEs from a physical viewpoint [42] . If the considered system has Cls, then its integrability will be possible [43, 44]. Noether theorem supplies us with a strategic idea for constructing Cls of NLPDEs so long as the Noether symmetry associated with the Lagrangian is known for Euler-Lagrange equations [45]. Nevertheless, there are some techniques in the literature for obtaining the Cls of the NLPDEs, that do not have the Lagrangian [46, 47].

Time fractional NLPDEs come from classical NLPDEs by replacing its time derivative with a fractional derivative. In the present work, we study Lie symmetry analysis, explicit solution using the power series technique and Ibragimov’s nonlocal Cls [48] for the time fractional SMK equation given by

αutα+βu2ux+γuxxxxx=0,(1)

in Eq. (1), 0 < α ≤ 1, and β and γ are arbitrary constants, and α is the order of the fractional time derivative. If α = 1, Eq. (1) reduces to the classical SMK equation which was considered for exact travelling wave solutions and Cls in [49, 50, 51]. Moreover, one can find more details on the construction of analytical, exact, numerical solutions, and other information for classical NLPDEs, in [52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69].

2 Preliminaries

Consider the RL fractional derivative [70, 71] given by

Dαf(t)=dnfdtn,α=n,ddtnInαf(t),0n1<α<n,

where n is a natural number and Iμf(t) is the RL fractional integral of order μ given by

Iμf(t)=1Γ(μ)0t(ts)μ1f(s)ds,μ>0Iμf(t)=f(t)(2)

and Γ(z) represents Gamma function.

Consider time-fractional PDEs as below

tαu=F(t,x,u,ux,uxx,uxxx,...),(0<α<1).(3)

Given a one-parameter Lie group of infinitesimal transformations of the form

t¯=t+ϵξ2(t,x,u)+O(ϵ2),x¯=x+ϵξ1(t,x,u)+O(ϵ2),u¯=u+ϵη(t,x,u)+O(ϵ2),αu¯t¯=αutα+ϵηα0(t,x,u)+O(ϵ2),u¯x¯=ux+ϵηx(t,x,u)+O(ϵ2),2u¯x¯2=2ux2+ϵηxx(t,x,u)+O(ϵ2),3u¯x¯3=3ux3+ϵηxxx(t,x,u)+O(ϵ2),4u¯x¯4=4ux4+ϵηxxxx(t,x,u)+O(ϵ2),5u¯x¯5=5ux5+ϵηxxxxx(t,x,u)+O(ϵ2),(4)

where

ηx=Dx(η)uxDx(ξ1)utDt(ξ2),ηxx=Dx(ηx)uxtDx(ξ1)uxxDt(ξ2),ηxxx=Dx(ηxx)uxxtDx(ξ1)uxxxDt(ξ2),ηxxxx=Dx(ηxxx)uxxxtDx(ξ1)uxxxxDt(ξ2),ηxxxxx=Dx(ηxxxx)uxxxxtDx(ξ1)uxxxxxDt(ξ2),(5)

In Eq. (5), Dx is the total differential operator defined by

Dx=x+uxu+uxxux+....

The corresponding Lie algebra of symmetries consists of a set of vector fields of the form

X=ξ1x+ξ2t+ηu.(6)

The vector field Eq. (6) is a Lie point symmetry of Eq. (3) provided

Pα,5rX()|=0=0.(7)

Also, the invariance condition yields [72] gives

ξ2(t,x,u)|t=0=0,(8)

and the αth extended infinitesimal related to RL fractional time derivative with Eq. (8) is given by [54, 55].

ηα0=αηtα+(ηuαDt(ξ2))αutαuαηutα+μn=1αnDtn(ξ1)Dtαn(ux)+n=1[αnαηutααn+1Dtn+1(ξ2)]Dtαn(u),(9)

in Eq. (9),

μ=n=2m=2nk=2mr=0k1αnnmkr1k!tnαΓ(n+1α)×[u]rmtm[ukr]nm+ktnmuk.(10)

It is worth noting that, μ = 0 if the infinitesimal η is linear in u, due to the presence of ηkuk, where k ≥ 2 in Eq. (10).

Definition 2.1

The function u = Θ(x, t) is an invariant solution of Eq. (3) corresponding to the infinitesimal generator Eq. (6) provided that

  1. u = Θ(x, t) satisfies Eq. (3).

  2. u = Θ(x, t) is an invariant surface of Eq. (5), that is to say

ξ2(x,t,Θ)Θt+ξ1(x,t,Θ)Θx=η(x,t,Θ).

3 Lie symmetries and reduction for Eq. (1)

Suppose that Eq. (1) is an invariant under Eq. (5), we have that

u¯t¯α+βu¯2u¯x¯+γu¯x¯x¯x¯x¯x¯=0,(11)

so that, u = u(x, t) satisfies Eq. (1). Using Eq. (5) in Eq. (11), we get the invariant equation

ηα0+(2βuux)η+(βu2)ηx+γηxxxxx=0.(12)

Putting the values of ηα0, ηx and ηxxxxx from Eq. (5) and Eq. (9) into Eq. (12) and isolating coefficients in partial derivatives with respect to x and power of u, we get

tαηutαηuβu2ηxγηxxxxx=0,

αntn(η)αn+1Dtn+1(ξ2)=0,n=1,2,...

ξu1=ξu2=ξt1=ξx2=ηuu=0,5ξx1αξt2=0.

Solving these equations, we get:

ξ1=c1+xαc2,ξ2=5tc2,η=2αuc2,

where c1 and c2 are arbitrary constants. Thus infinitesimal symmetry group for Eq. (1) is spanned by the two vector fields

X1=x,X2=xαx+5tt2uαu.(13)

The similarity variables for the infinitesimal generator X2 can be obtained by solving the following equations

dxαx=dt5t=du2αu.

Solving the above equations, we get

z1=xtα5,z2=ut2α5.(14)

Hence, from the symmetry X2, we get the group-invariant solution

u=t2α5f(ξ),ξ=xtα5,(15)

in Eq. (15), f is an arbitrary function of ξ. Using Eq. (15), Eq. (1) is transformed to a special nonlinear ODE of fractional order.

Consider the following theorem

Theorem 3.1

The similarity transformation Eq. (15) reduces Eq. (1) to the nonlinear ODE of fractional order as below:

(P5α17α5,αf)(ξ)+βf2fξ+γfξξξξξ=0(16)

with the EK fractional differential operator [22]

(Pβξ2,αf)=Πj=0n1(ξ2+j1βddξ)(Kβξ2+α,nαf)(ξ),(17)

n={[α]+1,αN,α,αN,(18)

where

(Kβξ2,αf)(ξ)=1Γ(α)1(u1)α1u(ξ2+α)f(ξu1β)du,α>0,f(ξ),α=0,(19)

is the EK fractional integral operator [74, 75].

Proof

Let n − 1 < α < 1, n = 1, 2, 3, … Based on the RL fractional derivative in Eq. (15), we get

αutα=ntn[1Γ(nα)1t(ts)nα1sα5f(xs(α5))ds].(20)

Let v=ts,ds=tv2dv. Thus, Eq. (20) becomes

αutα=ntn[tn7α51Γ(nα)1(v1)nα1v(n+17α5)f(ξvα5)dv].(21)

Applying EK fractional integral operator Eq. (19) in Eq. (21), we get

αutα=ntn[tn7α5(K5α12α5,nαf)(ξ)].(22)

We simplify the right hand side of Eq. (22). Consider ξ = xtα5, ϕ ∈ (0, ∞), we acquire

ttϕ(ξ)=tx(α5)tα51ϕ(ξ)=α5ξξϕ(ξ).(23)

Hence,

ntn[tn7α5(K5α12α5,nαf)(ξ)]=n1tn1[t(tn7α5(K5α12α5,nαf)(ξ))]=n1tn1[tn7α51(n7α5α5ξξ)(K5α12α5,nαf)(ξ)].(24)

Repeating n − 1 times, we have

ntn[tn7α5(K5α12α5,nαf)(ξ)]=n1tn1[t(tn7α5(K5α12α5,nαf)(ξ))]=n1tn1[tn7α51(n7α5α5ξξ)×(K5α12α5,nαf)(ξ)]...=t7α5Πj=0n1[(17α5+jα5ξξ)×(K5α12α5,nαf)(ξ)].(25)

Applying EK fractional differential operator Eq. (17) in Eq. (25), we get

nutn[(tn7α5(K5α1+α5,nαf)(ξ))]=t7α5(P5α15α6,αf)(ξ)(26)

Substituting Eq. (26) into Eq. (22), we get

αutα=t7α5(P5α17α5,αf)(ξ)(27)

Thus, Eq. (1) can be reduced into a fractional order ODE

(P5α17α5,αf)(ξ)+βf2fξ+γfξξξξξ=0(28)

The proof of the theorem is completed. □

4 Conservation laws

We now construct the Cls for Eq. (1). We start with some definitions. The RL left-sided time-fractional derivative given by

oDtαu=Dtn(oInαu),(29)

where Dt is the total differential operator with respect to t, n = [α] + 1, and oInαu represents the left sided time-fractional integral of nα order given by

(oInαu)(x,t)=1Γ(nα)0tu(θ,x)(tθ)1n+αdθ.(30)

In Eq. (30), Γ(z) represents Gamma function.

A Cls for Eq. (1) is represented as

Dt(Ct)+Dx(Cx)=0,(31)

where Ct = Ct(x, t, u,…), Cx = Cx (x, t, u, …), and Eq. (31) holds for all solutions u(x, t) of the Eq. (1).

We now apply Ibragimov method [48] for constructing the Cls of Eq. (1). Lagrangian for Eq. (1) can be presented as

L=v(x,t)(αutα+βu2ux+γuxxxxx)(32)

where v(x, t) is another dependent variable. The Euler-Lagrange operator [45, 46] is

δδu=u+(Dtα)DtαuDxux+Dxx2uxxDxxx3uxxx+Dxxxx4uxxxxDxxxxx5uxxxxx,(33)

where (Dtα) is the adjoint operator of (Dtα). The adjoint equation to Eq. (1) is given by [48]

δLδu=0.(34)

Consider two independent variables x, t and one dependent variable u(x, t), we have that

X¯+Dt(ξ2)l+Dx(ξ1)l=Wδδu+DtNt+DxNx,(35)

in Eq. (35), l represent the identity operator, δδu is the Euler-Lagrangian operator, Nt and Nx represent the Noether operation, is defined by

X¯=ξ2t+ξ1x+ηu+ηα0Dtαu+ηxux+ηxxuxx+ηxxxuxxx+ηxxxxuxxxx+ηxxxxxuxxxxx,(36)

and the Lie characteristic function W is given by

W=ηξ2utξ1ux.(37)

When RL time-fractional derivative is used in Eq. (1), Nt is defined by [45, 46]

Nt=ξ2l+k=0n1(1)koDα1k(W)DtkoDtαu(1)n×J(W,DtnoDtαu).(38)

With J given by

J(f,g)=1Γ(nα)0ttTf(ξ2,x)g(μ,x)(μξ2)α+1ndμdt.(39)

For Eq. (1), the operator Nx is given by

Nx=ξ1l+W(uxDxuxx+Dx2uxxxDx3uxxxx+Dx4uxxxxx)+Dx(W)(uxxDxuxxx+Dx2uxxxxDx3uxxxxx)+Dx2(W)(uxxxDxuxxxx+Dx2uxxxxx)+Dx3(W)(uxxxxDxuxxxxx)+Dx4(Wi)uxxxxx(40)

The invariance condition for any given generator X of Eq. (1) and its solutions reads

(X¯L+Dt(ξ2)L+Dx(ξ1)L)|Eq.(1)=0,(41)

and consequently the Cls of Eq. (1) can be written as

Dt(NtL)+Dx(NxL)=0.(42)

Now, we present the Cls for Eq. (1) using the basic definitions presented above. We consider two cases corresponding to the order of α.

  • Case 1

    When α ∈ (0, 1), with the help of Eq. (38) and Eq. (39), the components of the conserved vectors are

    Cit=ξ2L+(1)0oDtα1(Wi)Dt0L(oDtαu)(1)1×J(Wi,Dt1L(oDtαu))=voDtα1(Wi)+J(Wi,vt),

    Cix=ξ1l+Wi(uxDxuxx+Dx2uxxxDx3uxxxx+Dx4uxxxxx)+Dx(Wi)(uxxDxuxxx+Dx2uxxxxDx3uxxxxx)+Dx2(Wi)(uxxxDxuxxxx+Dx2uxxxxx)+Dx3(Wi)(uxxxxDxuxxxxx)+Dx4(Wi)uxxxxx

    =Wi(βu2v+Dx4γv)+Dx(Wi)(Dx3γv)+Dx2(Wi)(Dx2γv)γvxDx3(Wi)+γvDx4(Wi),

    where i = 1, 2 and the functions Wi are given by

    W1=ux,W2=2uα5tutαxux.

  • Case 2

    When α ∈ (1, 2), with the help of Eq. (38) and Eq. (39), the components of the conserved vectors are

    Cit=ξ2L+(1)0oDtα1(Wi)Dt0L(oDtαu)(1)1J(Wi,Dt1L(oDtαu))+(1)1oDtα2(Wi)Dt1L(oDtαu)(1)1J(Wi,Dt1L(oDtαu))

    =voDtα1(Wi)+J(Wi,vt)vtoDtα2(Wi)J(Wi,vtt)

    Cix=ξ1l+Wi(uxDxuxx+Dx2uxxxDx3uxxxx+Dx4uxxxxx)+Dx(Wi)(uxxDxuxxx+Dx2uxxxxDx3uxxxxx)+Dx2(Wi)(uxxxDxuxxxx+Dx2uxxxxx)+Dx3(Wi)(uxxxxDxuxxxxx)+Dx4(Wi)uxxxxx

    =Wi(βu2v+Dx4γv)+Dx(Wi)(Dx3γv)+Dx2(Wi)(Dx2γv)γvxDx3(Wi)+γvDx4(Wi),

    where i = 1, 2 and the functions Wi are given by

    W1=ux,W2=2uα5tutαxux.

5 Explicit power series solutions

Here, we investigate the exact analytic solutions via power series method [76] and symbolic computations [77] for Eq. (28). Set

f(ξ)=n=0anξn,(43)

from Eq. (43), we can have

f=n=0nanξn1,f(5)=n=0n(n1)(n2)(n3)(n4)anξn5.(44)

Substituting Eqs. (44) into Eq. (28), we obtain

n=0Γ(212α5+nα5)Γ(27α5+nα)5anξn+βn=0n=0anξn×n=0(n+1)an+1ξn+γn=0(n+5)(n+4)(n+3)×(n+2)(n+1)an+5ξn=0.(45)

Comparing coefficients in Eq. (45) when n = 0, we obtain

a5=1120γ(Γ(212α5)Γ(27α5)a0+βa02a1),(46)

when n ≥ 1, we have

an+5=1(n+5)(n+4)(n+3)(n+2)(n+1)γ×{Γ(212α5+nα5)Γ(27α5+nα)5an+βk=0nj=0kaj(nk+1)ank+1}.

Thus, the power series solution for Eq. (28) can be represented in the form:

f(ξ)=a0+a1ξ+a2ξ2+a3ξ3+a4ξ4+a5ξ5+n=1an+5ξn+5=a0+a1ξ+a2ξ2+a3ξ3+a4ξ4+1120γ(Γ(212α5)Γ(27α5)a0+βa02a1)ξ5+n=11(n+5)(n+4)(n+3)(n+2)(n+1)γ×{Γ(212α5+nα5)Γ(27α5+nα)5an+βk=0nj=0kaj×(nk+1)ank+1}ξn+5

Consequently, we acquire the exact power series solution for Eq. (28) as

u(x,t)=a0t2α5+a1xt3α5+a2x2t4α5+a3x3tα+a4x4t6α5+1120γ(Γ(212α5)Γ(27α5)a0+βa02a1)x5t7α5+n=11(n+5)(n+4)(n+3)(n+2)(n+1)γ×{Γ(212α5+nα5)Γ(27α5+nα)5an+βk=0nj=0kaj×(nk+1)ank+1}xn+5t(n+7α)5(47)

6 Physical interpretation of the power series solution for Eqs. (47)

In order to have clear and proper understanding of the physical properties of the power series solution, the 3-D, 2-D, and contour plots for the solution Eqs. (47), are plotted in Figures 1-4 by using suitable parameter values.

3D plot of (47) a0 = a1 = a2 = 1, a3 = 0.5, a4 = 1.7, β = 2, γ = 1, α = 0.5, Γ = 0.85
Figure 1

3D plot of (47) a0 = a1 = a2 = 1, a3 = 0.5, a4 = 1.7, β = 2, γ = 1, α = 0.5, Γ = 0.85

Contour plot of (47) a0 = a1 = a2 = 1, a3 = 0.5, a4 = 1.7, β = 2, γ = 1, α = 0.5, Γ = 0.85
Figure 2

Contour plot of (47) a0 = a1 = a2 = 1, a3 = 0.5, a4 = 1.7, β = 2, γ = 1, α = 0.5, Γ = 0.85

3D plot of (47) a0 = a1 = a2 = 0.8, a3 = a4 = 1, β = 1.2, γ = 3, α = 0.9,Γ = 0.1
Figure 3

3D plot of (47) a0 = a1 = a2 = 0.8, a3 = a4 = 1, β = 1.2, γ = 3, α = 0.9,Γ = 0.1

Contour plot of (47) a0 = a1 = a2 = 0.8, a3 = a4 = 1, β = 1.2, γ = 3, α = 0.9,Γ = 0.1
Figure 4

Contour plot of (47) a0 = a1 = a2 = 0.8, a3 = a4 = 1, β = 1.2, γ = 3, α = 0.9,Γ = 0.1

7 Concluding remarks

In this research, we analyzed time fractional SMK by means of Lie symmetry analysis using the RL derivative. We reduced the governing equation to a nonlinear ODE of fractional order. The obtained fractional ODE was solved using a power series technique. Ibragimov’s nonlocal conservation theorem was applied to establish Cls for the governing equation. Some 3-D, 2-D, and contour plots were also presented.

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About the article

Received: 2017-08-26

Accepted: 2018-02-10

Published Online: 2018-05-30


Citation Information: Open Physics, Volume 16, Issue 1, Pages 302–310, ISSN (Online) 2391-5471, DOI: https://doi.org/10.1515/phys-2018-0042.

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© 2018 Dumitru Baleanu et al.. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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