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Physics & Astronomy International Journal

Review Article Volume 2 Issue 5

The unified sub-equation method and its applications to conformable space-time fractional fourth-order pochhammer-chree equation

Zayed EME, Shohib RMA

Mathematics Department, Zagazig University, Egypt

Correspondence: Elsayed ME Zayed, Department of Mathematics, Faculty of Sciences, Zagazig University, Zagazig, Egypt

Received: January 24, 2018 | Published: October 9, 2018

Citation: Zayed EME, Shohib RMA. The unified sub-equation method and its applications to conformable space-time fractional fourth-order pochhammer-chree equation. Phys Astron Int J. 2018;2(5):451-464. DOI: 10.15406/paij.2018.02.00124

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Abstrat

In this article, we apply the unified sub-equation method proposed by Lu Bin and Zhang Hong Qing to construct many new Jacobi elliptic function solutions, solitons and other solutions for the conformable space-time fractional fourth-order Pochhammer-Chree equation. This method is direct and more powerful than the projective Riccati equation method. The solitons and other solutions of this equation can be found from the Jacobi elliptic solutions when its modulasm1 or m0 respectively. Comparing our new results with the well-known results is given.

Keywords: unified sub-equation method, jacobi elliptic function solutions, dark, singular and bright solitons, periodic solutions, the conformable space-time fractional fourth-order pochhammer-chree equation

PACS

02.03.Jr; 04.20.Jb; 05.45.Yv.

Introduction

When the nonlinear partial differential equations (PDEs) are analyzed, one of the most important equation is the construction of the exact solutions of those equations. Searching for the exact solutions of those equations plays an important role in the study of nonlinear physical phenomena. Nonlinear wave phenomena appear in various scientific and engineering fields, such as fluid mechanics, plasma physics, optical fibers, biology, solid state physics, chemical kinematics, chemical physics, geochemistry, thermodynamics, soli mechanics, civil engineering, and non-Newtonian fluids to the natural sciences including population acology, inflectious disease epidemiology, natural networks and so on. Throughout the past few decades, a particular attention has been given to the problem of finding the exact solutions of these nonlinear PDEs. By virtue of these solutions, one may give better insight into the physical aspects of the nonlinear models studies. In recent years, quite a few methods for constructing explicit and solitary wave solutions of the nonlinear PDEs have been presented. A variety of powerful methods, such as the inverse scattering method,1 the Hirota method,2 the Bäcklund transform method,3,4 the Painlevé expansion method,5 the exp-function method,6,7 the sub-ODE method,8-10 the Jacobi elliptic function method,11,12 the sine-cosine function method,13,14 the (G'/G) -expansion method,15-17 the modified simple equation method,18,19 the Kudryashov method,20,21 the multiple exp–function method,22,23 the homogeneous balance method,24 the auxiliary equation method,25,26 the extended auxiliary equation method,27-29 the soliton ansatz method,30-33 the new mapping method,34,35 the first integral method,36,37 the (G'/G,1/G) -expansion method38,39 and an unified sub-equation method,40 the projective Riccati equation method41,42 and so on. Motivated by the projective Riccati equation method proposed by Conte & Musette43 and developed by Yan,44 we present a new direct algebraic method proposed by Bin L & Hong-Qing Z40 to obtain many new double periodic wave solutions of nonlinear PDEs which cannot be acquired by using the projective Riccati equation method.

The objective of this article is to apply a unified sub-equation method combined with the conformable space-time fractional derivatives40 for finding many new Jacobi elliptic function solutions, solitons and other solutions of the following nonlinear conformable space-time fractional fourth-order Pochhammer-Chree equation:

2αut2α2αt2α(2βux2β)2βx2β(α1u+β1un+1+γ1u2n+1)=0,n1, (1.1)

Where 0<α,β1 and u(x,t) is a real function, while α1 , β1 and γ1 are arbitrary constants. Equation (1.1) represents a nonlinear model of longitudinal wave propagation of elastic rods.45,46 Here the exponent n1 is the power law nonlinearity parameter. When α=β=1, Equation (1.1) has been discussed in42 using the generalized projective Riccati equation method, in47 using the extended (G'/G) -expansion method, in48 using the (G'/G) -expansion method, in49 using the tanh-coth and the sine-cosine methods, and in50 using the exp-function method.

This article is organized as follows: In Section 2, the description of conformable fractional derivative is given. In Section 3, the description of the unified sub-equation method combined with the conformable space-time fractional derivatives is obtained. In Section 4, we apply this method to the conformable space-time fractional fourth-order Pochhammer-Chree equation (1.1). In Section 5, we present the graphical representations for some solutions of Equation (1.1). In Section 6, conclusions are obtained. To the best of our knowledge. Equation (1.1) has not been previously considered in literature using the method of Section 3.

Description of the conformable fractional derivative

Khalil et al.51 introduced a novel definition of fractional derivative named the conformable fractional derivative, which can rectify the deficiencies of the other definitions.

Definition1: Suppose f:[0,)R is a function. Then, the conformable fractional derivative of f of order α is defined as

Tα(f)(t)=limτ0f(t+τt1α)f(t)τ, (2.1)

For all t>0 and α(0,1]. Several properties of the conformable fractional derivative are given below as in51−53

Thereom 1: Suppose α(0,1], and f and g are α differentiable at t>0. Then

Tα(af+bg)=aTα(f)+bTα(g),a,bR. (2.2)

Tα(tμ)=μtμα,μR. (2.3)

Tα(fg)=fTα(g)+gTα(f), (2.4)

Tα(fg)=(gTα(f)fTα(g)g2), (2.5)

Furthermore, if f is differentiable; then

Tα(f)(t)=t1αdfdt(t). (2.6)

Thereom 2: Suppose f:[0,)R is a differentiable function and also α differentiable. Let g be a function defined in the range of f and also differentiable. Then

Tα(f0g)(t)=t1αg'(t)f'(g(t)). (2.7)

Description of the unified sub-equation method combined with the conformable space-time fractional derivatives

Consider the following nonlinear PDE:

F(u,αutα,βuxβ,2αut2α,2βux2β,.....)=0,0<α,β1 (3.1)

Where F is a polynomial in u(x,t) and its partial derivatives, in which the highest order derivatives and the nonlinear terms are involved. In the following, we give the main steps of this method:

Step 1: We use the conformable space-time wave transformation:

u(x,t)=u(ξ),ξ=xββc1tαα, (3.2)

Where c1 is a constant, to reduce Equation (3.1) to the following ODE:

P(u,u',u'',.....)=0, (3.3)

Where P is a polynomial in u(ξ) and its total derivatives, such that '=ddξ .

Step 2: We assume that Equation (3.3) has the formal solution:

u(ξ)=a0+Ni=1fi1(ξ)[aif(ξ)+big(ξ)], (3.4)

Where a0 , ai , bi (i=1,.....,N) are constants to be determined later, such that aN0 or bN0, while f(ξ) and g(ξ) satisfy the auxiliary ODEs:

f'(ξ)=f(ξ)g(ξ), (3.5)

g'(ξ)=q+g2(ξ)+rf2(ξ), (3.6)

g2(ξ)=[q+r2f2(ξ)+cf2(ξ)] (3.7)

Where q , r and c are constants.

Step 3: We determine the positive integer N in (3.4) by using the homogeneous balance between the highest order derivatives and the nonlinear terms in Equation (3.3). More precisely, we define the degree of u(ξ) as D[u(ξ)]=N , which gives rise to the degree of other expressions as follows:

D[up1(ξ)(dq1u(ξ)dξq1)s1]=NP1+s1(q1+N). (3.8)

From (3.8) we can get the value of N in (3.4). In some nonlinear equations, the balance number N is not a positive integer. In this case, we make the following transformations:

(a) When N=q1p1, where q1p1 is a fraction in the lowest terms, we let

u(ξ)=[v(ξ)]q1p1, (3.9)

When N is a negative number, we let

u(ξ)=[v(ξ)]N, (3.10)

And substitute (3.9) or (3.10) into Equation (3.3) to get a new equation in terms of the function v(ξ) with a positive integer balance number.

Step 4: We substitute (3.4) along with (3.5)-(3.7) into Equation (3.3) and collect all terms of the same order of fi(ξ) gj(ξ) (i,j=0,1,.....) and set them to zero, yield a set of algebraic equations which can be solved by using the Maple or Mathematical to find a0 , ai , bi , c1 , q , r , c .

Step 5: It is well-known40 that (3.5), (3.6) have the following Jacobi elliptic function solutions:

  1. If q=(1+m2), r=2m2, c=1,  then
  2. f1(ξ)=1sn(ξ,m),g1(ξ)=cn(ξ,m)dn(ξ,m)sn(ξ,m).

  3. If q=(12m2), r=2m2, c=(m21), then
  4. f2(ξ)=1cn(ξ,m),g2(ξ)=sn(ξ,m)dn(ξ,m)cn(ξ,m).

  5. If q=(2+m2), r=2, c=(1m2), then
  6. f3(ξ)=1dn(ξ,m),g3(ξ)=m2sn(ξ,m)cn(ξ,m)dn(ξ,m).

  7. If q=(1+m2), r=2, c=m2, then
  8. f4(ξ)=sn(ξ,m),g4(ξ)=cn(ξ,m)dn(ξ,m)sn(ξ,m).

  9. If q=(12m2), r=(2+2m2),c=m2, then
  10. f5(ξ)=cn(ξ,m),g5(ξ)=sn(ξ,m)dn(ξ,m)cn(ξ,m).

  11. If q=(2+m2), r=(22m2), c=1, then
  12. f6(ξ)=dn(ξ,m),g6(ξ)=m2sn(ξ,m)cn(ξ,m)dn(ξ,m).

  13. If q=(2+m2), r=(2+2m2), c=1, then
  14. f7(ξ)=cn(ξ,m)sn(ξ,m),g7(ξ)=dn(ξ,m)sn(ξ,m)cn(ξ,m).

  15. If q=(12m2), r=(2m22m4), c=1, then
  16. f8(ξ)=dn(ξ,m)sn(ξ,m),g8(ξ)=cn(ξ,m)sn(ξ,m)dn(ξ,m).

  17. If q=(2+m2), r=2, c=(1+m2), then
  18. f9(ξ)=sn(ξ,m)cn(ξ,m),g9(ξ)=dn(ξ,m)sn(ξ,m)cn(ξ,m).

  19. If q=(1+m2), r=2m2, c=1,then
  20. f10(ξ)=dn(ξ,m)cn(ξ,m),g10(ξ)=(1m2)sn(ξ,m)cn(ξ,m)dn(ξ,m).

  21. If q=(12m2), r=2, c=(m2m4),then
  22. f11(ξ)=sn(ξ,m)dn(ξ,m),g11(ξ)=cn(ξ,m)sn(ξ,m)dn(ξ,m).

  23. If q=12(1+2m2), r=12, c=14,then
  24. f12(ξ)=cn(ξ,m)±1sn(ξ,m),g12(ξ)=ds(ξ,m).

  25. If q=12(m2+1), r=12(1m2), c=14(1m2), then
  26. f13(ξ)=dn(ξ,m)msn(ξ,m)±1,g13(ξ)=mcd(ξ,m).

  27. If q=12(1+m2) ,r=12(m21)2 , c=14, then
  28. f14(ξ)=mcn(ξ,m)+dn(ξ,m),g14(ξ)=msn(ξ,m).

  29. If q=12(1+m2), r=12(m21)2,c=14,  then
  30. f15(ξ)=cn(ξ,m)±dn(ξ,m)sn(ξ,m),g15(ξ)=ns(ξ,m).

  31. If q=(m26m+1), r=2,c=4m (m1)2,then
  32. f16(ξ)=sn(ξ,m)msn2(ξ,m)1,g16(ξ)=cs(ξ,m)dn(ξ,m)[msn2(ξ,m)+1msn2(ξ,m)1].

  33. If q=(m2+6m+1), r=2, c=4m,(m+1)2, then
  34. f17(ξ)=sn(ξ,m)msn2(ξ,m)+1,g17(ξ)=cs(ξ,m)dn(ξ,m)[msn2(ξ,m)1msn2(ξ,m)+1].

  35. If q=12(1+m2), r=12(m21),c=14(m21),then
  36. f18(ξ)=cn(ξ,m)sn(ξ,m)±1,g18(ξ)=dc(ξ,m).

  37. If q=12(2m2), r=m42,c=14,then
  38. f19(ξ)=dn(ξ,m)±1sn(ξ,m),g19(ξ)=cs(ξ,m).

  39. If q=12(2m21), r=12, c=14,then
  40. f20(ξ)=sn(ξ,m)1±cn(ξ,m),g20(ξ)=±ds(ξ,m).

  41. If q=12(1+m2), r=12, c=14(m21)2, then

f21(ξ)=sn(ξ,m)cn(ξ,m)±dn(ξ,m),g21(ξ)=±ns(ξ,m).

Where sn(ξ,m), cn(ξ,m) and dn(ξ,m) are Jacobi elliptic sine function, Jacobi elliptic cosine function, Jacobi elliptic function of the third kind respectively, and m denotes the modulus of Jacobi elliptic functions, where 0m1 . It is well-known54−56 that the Jacobi-elliptic functions satisfy the following relations:

cn2(ξ,m)=1sn2(ξ,m),dn2(ξ,m)=1m2sn2(ξ,m),

sn'(ξ,m)=cn(ξ,m)dn(ξ,m),cn'(ξ,m)=sn(ξ,m)dn(ξ,m),

dn'(ξ,m)=m2sn(ξ,m)cn(ξ,m).

ns(ξ,m)=1sn(ξ,m),nc(ξ,m)=1cn(ξ,m),nd(ξ,m)=1dn(ξ,m),

sc(ξ,m)=sn(ξ,m)cn(ξ,m),sd(ξ,m)=sn(ξ,m)dn(ξ,m),cs(ξ,m)=cn(ξ,m)sn(ξ,m),

cd(ξ,m)=cn(ξ,m)dn(ξ,m),ds(ξ,m)=dn(ξ,m)sn(ξ,m),dc(ξ,m)=dn(ξ,m)cn(ξ,m),

The Jacobi elliptic functions degenerate into hyperbolic functions when m1 as follows:

sn(ξ,1)tanh(ξ),cn(ξ,1)sech(ξ),dn(ξ,1)sech(ξ),ns(ξ,1)coth(ξ),dc(ξ,1)1,

ds(ξ,1)cosech(ξ),sc(ξ,1)sinh(ξ),sd(ξ,1)sinh(ξ),cs(ξ,1)cosech(ξ),

And into trigonometric functions when m0 as follows:

sn(ξ,0)sin(ξ),cn(ξ,0)cos(ξ),dn(ξ,0)1,ns(ξ,0)cosec(ξ),cs(ξ,0)cot(ξ),

ds(ξ,0)cosec(ξ),sc(ξ,0)tan(ξ),sd(ξ,0)sin(ξ),dc(ξ,0)sec(ξ)

Step 6: We substitute the values a0 , ai bi and the solutions (1)−(21) given in step 5 into (3.4), to get the exact solutions of Equation (3.1).

Applications

In this section, we apply the method of section 3, to solve Equation (1.1).To this aim, we first use the conformable space-time wave transformation:

u(x,t)=u(ξ),ξ=xββc1tαα, (4.1)

Where c1 is a non zero constant and 0<α,β1 , to reduce Equation (1.1) into the following nonlinear ordinary differential equation (ODE):

c21u''c21u''''(α1u+β1un+1+γ1u2n+1)''=0, (4.2)

Integrating Equation (4.2) twice with respect to ξ and vanshing the constants of integration, we get

(c21α1)uc21u''β1un+1γ1u2n+1=0, (4.3)

Balancing u'' with u2n+1 , we get N=1n . According to (3.9) we use the transformation:

u(ξ)=v1n(ξ), (4.4)

Where v(ξ) is a new function of ξ , to reduce Equation (4.3) into the new ODE:

(c21α1)n2v2c21nvv''c21(1n)v'2β1n2v3γ1n2v4=0, (4.5)

Balancing vv'' with v4 in Equation (4.5), we get N=1 . According to the form (3.4), Equation (4.5) has the formal solution:

v(ξ)=a0+a1f(ξ)+b1g(ξ), (4.6)

Where a0 , a1 , b1 are constants to be determined, such that a10 or b10 . Substituting (4.6) along with (3.5)−(3.7) into Equation (4.5) and collecting all terms of the same order of fi(ξ) , gj(ξ),(i,j=0,1,2,...) and setting them to zero, we have the following algebraic equations:

{f4(ξ):6n2γ1a21b21c+c21a21cc21b21c2+nc21a21cn2γ1a41n2γ1b41c2c21nb21c2=0,f4(ξ):c21r2b21+c21nr2b21+n2γ1r2b41=0,f3(ξ):n2β1a31+3n2β1a1b21c+2nc21a0a1c+12n2γ1a0a1b21c4n2γ1a0a31=0,f2(ξ):c21a21qn2c21b21c+6n2γ1a20b21c6n2γ1a20a21+n2α1b21c3n2β1a0a21+n2c21a21n2α1a21:2n2γ1b41qc+6n2γ1a21b21q2nc21b21qc+3n2β1a0b21c=0,f(ξ):3n2β1a1qb214n2γ1a30a1+nc21a0qa1+2n2c21a1a0+12n2γ1a0a1qb21:3n2β1a20a12n2α1a0a1=0,f1(ξ):3n2β1a1rb21+12n2γ1a0a1rb21=0,(4.7)f0(ξ):2c21rb21c2n2c21b21qnc21a21r+2n2α1b21q2n2γ1b41q26nc21rb21c+12n2γ1a20b21q:6n2β1a0b21q+6n2γ1a21b21r2n2γ1b41rc+2n2c21a202n2α1a202n2β1a302n2γ1a40+c21a21r=0,f2(ξ)g(ξ):2nc21a0b1c3n2β1a21b1+n2β1b31c+4n2γ1a0b31c12n2γ1a0a21b1=0,f2(ξ)g(ξ):2nc21a0rb1+n2β1rb31+4n2γ1a0rb31=0,f3(ξ)g(ξ):2nc21a1b1c+2c21a1b1c4n2γ1a31b1+4n2γ1a1b31c=0,g(ξ):2n2c21a0b1+n2β1b31q2n2α1a0b14n2γ1a30b1+4n2γ1a0b31q3n2β1a20b1=0,f(ξ)g(ξ):2n2α1a1b1+2n2c21a1b1+4n2γ1a1b31q6n2β1a0a1b112n2γ1a20a1b1+nc21b1a1q=0,f1(ξ)g(ξ):2nc21a1rb1+2n2γ1a1rb31c21a1rb1=0,}

According to Step5 of Section 3, we have the following results:

Result 1: If we substitute q=(1+m2),r=2m2,c=1 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,β1=0,a0=0,a1=0,b1=2α1γ1(1+2m2),c1=α1(1+2m2),α1<0,γ1<0} (4.8)

or

{m=m,n=2,α1=1564β21(m21)γ1m2,a0=3β18γ1,a1=3β18mγ1,b1=0,c1=β14m3γ1,γ1<0} (4.9)

Substituting (4.8) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=2α1γ1(1+2m2)(cn(ξ,m)dn(ξ,m)sn(ξ,m)),α1<0,γ1<0 (4.10)

Where ξ=xββ(α1(1+2m2))tαα . If m0 , then we have the periodic solution:

u(x,t)=2α1γ1[cot(xββα1tαα)], (4.11)

while if m1 , then we have the solitary wave solution:

u(x,t)=2α13γ1[coth(xββα13tαα)tanh(xββα13tαα)]. (4.12)

Substituting (4.9) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1(11msn(ξ,m))}12,γ1<0,β1>0 (4.13)

where ξ=xββ(β14m3γ1)tαα . If m1 , then we have the singular soliton solution:

u(x,t)={3β18γ1[1coth(xβββ143γ1tαα)]}12. (4.14)

Result 2: If we substitute q=(12m2), r=2m2, c=(m21) into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=2,a0=3β18γ1,a1=3β18γ1,c1=β143γ1(m21),α1=3β21(4m21)64γ1(m21),b1=0,γ1(m21)>0,} (4.15)

Substituting (4.15) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1(11cn(ξ,m))}12,γ1β1<0, (4.16)

where ξ=xββ(β143γ1(m21))tαα. If m0 , then we have the periodic solution:

u(x,t)={3β18γ1[1sec(xβββ143γ1tαα)]}12,β1γ1<0. (4.17)

Result 3: If we substitute q=(2+m2), r=2, c=(1m2) into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,β1=0,a0=0,a1=0,b1=2α1γ1(2m25),c1=α1(2m25),α1>0,γ1<0,} (4.18)

Substituting (4.18) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=m22α1γ1(2m25)(sn(ξ,m)cn(ξ,m)dn(ξ,m)),α1>0,γ1<0 (4.19)

where ξ=xββ(α1(2m25))tαα. If m1 , then we have the dark soliton solution:

u(x,t)=2α13γ1[tanh(xββα13tαα)]. (4.20)

Result 4: If we substitute q=(1+m2), r=2, c=m2 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=2,a0=3β18γ1,b1=0,a1=3β18γ1,α=1564β21(m21)γ1m2,c1=β14m3γ1,γ1<0} (4.21)

Substituting (4.21) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1(1+sn(ξ,m))}12,γ1<0,β1>0 (4.22)

where ξ=xββ(β14m3γ1)tαα. If m1 , then we have the dark soliton solution:

u(x,t)={3β18γ1[1+tanh(xβββ143γ1tαα)]}12. (4.23)

Result 5: If we substitute q=(12m2), r=(2+2m2),c=m2 into the algebraic equations (4.7) and use the Maple, we have

{n=1,a0=β13γ1,a1=0,b1=β13γ12m21,c1=β1312γ1(2m21),α1=β21(8m25)18γ1(2m21),γ1(2m21)<0.} (4.24)

Substituting (4.24) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=β13γ1[112m21(sn(ξ,m)dn(ξ,m)cn(ξ,m))] (4.25)

where ξ=xββ(β1312γ1(2m21))tαα, γ1>0 or γ1<0. If m1 , then we have the dark soliton solution:

u(x,t)=β13γ1[1tanh(xβββ1312γ1tαα)].γ1<0. (4.26)

Result 6: If we substitute q=(2+m2), r=(22m2), c=1 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=2,α1=364β21(m2+8)γ1,a0=3β18γ1,a1=3β18γ1,b1=0,c1=β143γ1,γ1>0} (4.27)

Substituting (4.27) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1(1+dn(ξ,m))}12,γ1>0,β1<0. (4.28)

where ξ=xββ(β143γ1)tαα. If m1 , then we have the bright soliton solution:

u(x,t)={3β18γ1[1+sech(xβββ143γ1tαα)]}12. (4.29)

Result 7: If we substitute q=(2+m2), r=(2+2m2), c=1 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=2,b1=3β18m2γ1(1+1m2),a0=3β18γ1,c1=β14m23γ1(m22+21m2),α1=3β2116m4γ1((m22+21m2)(m43m2+4+(m24)1m2)(1+1m2)2),a1=0,(m22+21m2)γ1>0,} (4.30)

Substituting (4.30) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1[1+(1+1m2m2)(dn(ξ,m)sn(ξ,m)cn(ξ,m))]}12,γ1β1<0. (4.31)

where ξ=xββ(β14m23γ1(m22+21m2))tαα. If m1 , then we have the same solitary wave solution (4.14).

Result 8. If we substitute q=(12m2), r=(2m22m4), c=1 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,β1=0,a0=0,b1=0,a1=α1γ1(m21),c1=α12(m21),α1(m21)<0,γ1<0} (4.32)

Substituting (4.32) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=α1γ1(m21)(dn(ξ,m)sn(ξ,m)), (4.33)

where ξ=xββ(α12(m21))tαα. If m0 , then we have the periodic solution:

u(x,t)=α1γ1[cosec(xββα12tαα)]. (4.34)

Result 9: If we substitute q=(2+m2), r=2, c=(1+m2) into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,β1=0,a0=0,a1=0,b1=2α1γ1(2m25),c1=α1(2m25),α1>0,γ1<0} (4.35)

Substituting (4.35) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=2α1γ1(2m25)(dn(ξ,m)sn(ξ,m)cn(ξ,m)),α1>0,γ1<0 (4.36)

where ξ=xββ(α1(2m25))tαα. If m1 , then we have the singular soliton solution:

u(x,t)=2α13γ1[coth(xββα13tαα)], (4.37)

while if m0 , then we have the periodic solution:

u(x,t)=2α15γ1[cosec(xββα15tαα)sec(xββα15tαα)]. (4.38)

Result 10: If we substitute q=(1+m2), r=2m2, c=1 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=2,α1=3β2164γ1(m21),a0=3β18γ1,a1=3β18γ1,b1=0,c1=β143γ1,γ1<0} (4.39)

Substituting (4.39) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1[1+dn(ξ,m)cn(ξ,m)]}12,β1>0,γ1<0 (4.40)

where ξ=xββ(β143γ1)tαα. If m0 , then we have the periodic solution:

u(x,t)={3β18γ1[1+sec(xβββ143γ1tαα)]}12. (4.41)

Result 11: If we substitute q=(12m2), r=2, c=(m2m4) into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,β1=0,a0=0,a1=0,b1=2α1γ1(4m21),c1=α1(4m21),(4m21)α1>0,γ1<0,} (4.42)

Substituting (4.42) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=2α1γ1(4m21)[cn(ξ,m)sn(ξ,m)dn(ξ,m)],ξ=xββ(α1(4m21))tαα, (4.43)

If m0 , then we have the same periodic solution (4.11), while if m1 , then we have the same singular soliton solution (4.37).

Result 12: If we substitute q=12(1+2m2),r=12,c=14 into the algebraic equations (4.7) and use the Maple, we have:

{m=m,n=1,a0=0,a1=0,β1=0,b1=α1γ1(m21),c1=α12(m21),α1(m21)<0,γ1<0} (4.44)

Substituting (4.44) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=[α1γ1(m21)ds(ξ,m)], (3.45)

where ξ=xββ(α12(m21))tαα. If m0 , then we have the same periodic solution (4.34).

Result 13: If we substitute q=12(m2+1),r=12(1m2),c=14(1m2) into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,a0=0,b1=0,β1=0,a1=α1γ1,c1=2α1(m21),α1(m21)<0,γ1α1>0} (4.46)

Substituting (4.46) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=α1γ1[dn(ξ,m)msn(ξ,m)±1], (4.47)

where ξ=xββ(2α1(m21))tαα.

Result 14: If we substitute q=12(1+m2), r=12(m21)2,c=14 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,α=β21(1+2m2)9γ1(m2+1),a0=β13γ1,a1=0,b1=β13γ12(m2+1),c1=β131γ1(m2+1),γ1<0} (4.48)

Substituting (4.48) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=β13γ1[1m2(m2+1)sn(ξ,m)], (4.49)

where ξ=xββ(β131γ1(m2+1))tαα. If m1 , then we have the same dark soliton solution (4.26).

Result 15: If we substitute q=12(1+m2),r=12(m21)2,c=14 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=2,α1=3β21(m21)64γ1,a0=3β18γ1,a1=0,b1=3β18γ1,c1=β143γ1,γ1<0} (4.50)

Substituting (4.50) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1[1ns(ξ,m)]}12,β1>0,γ1<0. (4.51)

where ξ=xββ(β143γ1)tαα. If m1, then we have the singular soliton solution:

u(x,t)={3β18γ1[1coth(xβββ143γ1tαα)]}12, (4.52)

while if m0, then we have the periodic solutions.

u(x,t)={3β18γ1[1cosec(xβββ143γ1tαα)]}12. (4.53)

Result 16: If we substitute q=(m26m+1), r=2, c=4m (m1)2 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,β1=0,a0=0,a1=0,b1=2α1γ1(2m212m+1),c1=α1(2m212m+1),α1(2m212m+1)<0,γ1<0} (4.54)

Substituting (4.54) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=2α1γ1(2m212m+1)[cs(ξ,m)dn(ξ,m)(msn2(ξ,m)+1msn2(ξ,m)1)], (4.55)

wher ξ=xββ(α1(2m212m+1))tαα. If m0 , then we have the same periodic solution (4.11), while if m1, then we have the solitary wave solutions:

u(x,t)=2α19γ1[coth(xββα13tαα)+tanh(xββα13tαα)]. (4.56)

Result 17: If we substitute q=(m2+6m+1), r=2, c=4m(m+1)2 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=2,a0=3β18γ1,a1=3β18γ1(m+1),b1=0,α1=3β21(m218m+5)256mγ1,c1=β183mγ1,γ1<0} (4.57)

Substituting (4.57) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1[1+(m+1)sn(ξ,m)msn2(ξ,m)+1]}12,β1>0,γ1<0 (4.58)

where ξ=xββ(β183mγ1)tαα. If m1 , then we have the solitary wave solution:

u(x,t)={3β18γ1[1+2tanh(xβββ183γ1tαα)1+tanh2(xβββ183γ1tαα)]}12. (4.59)

Result 18: If we substitute q=12(1+m2), r=12(m21), c=14(m21) into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=2,a0=3β18γ1,b1=3β18γ1,a1=0,α1=3β21(m21)64γ1,c1=β143γ1,γ1<0} (4.60)

Substituting (4.60) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1[1dc(ξ,m)]}12,β1>0,γ1<0 (4.61)

where ξ=xββ(β143γ1)tαα. If m0 , then we have the same periodic solutions (4.17) and (4.41) respectively.

Result 19: If we substitute q=12(2m2),r=12m4,c=14 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=2,a0=3β18γ1,a1=3β18γ1m2(1+1m2),c1=β12m23m2+61m26γ1α1=3β2116γ1m4(((10m2)1m2+m4+6m210)(m22+21m2)(1+1m2)2),b1=0,(3m2+61m26)γ1>0,} (4.62)

Substituting (4.62) into (4.4), (4.6),we have the Jacobi elliptic solutions:

u(x,t)={3β18γ1[1(1+1m2m2)(dn(ξ,m)±1sn(ξ,m))]}12,β1γ1<0 (4.63)

where ξ=xββ(β12m23m2+61m26γ1)tαα. If m1 , then we have the solitary wave solutions:

u(x,t)={3β18γ1[1+cosech(xβββ123γ1tαα)±coth(xβββ123γ1tαα)]}12. (4.64)

Result 20: If we substitute q=12(2m21), r=12, c=14 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,a0=0,b1=0,β1=0,a1=α1γ1(2m2+1),c1=2α1(2m2+1),α1>0,γ1<0} (4.65)

Substituting (4.65) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=α1γ1(2m2+1)(sn(ξ,m)1±cn(ξ,m)),α1>0,γ1<0 (4.66)

where ξ=xββ(2α1(2m2+1))tαα. If m1 , then we have the solitary wave solution:

u(x,t)=α13γ1[tanh(xββ2α13tαα)1±sech(xββ2α13tαα))], (4.67)

while if m0 , then we have the periodic solution:

u(x,t)=α1γ1[tan(xββ2α1tαα)sec(xββ2α1tαα)±1]. (4.68)

Result 21: If we substitute q=12(1+m2), r=12, c=14(m21)2 into the algebraic equations (4.7) and use the Maple, we have

{m=m,n=1,a0=β13γ1,a1=0,b1=β13γ12(1+m2),c1=β131γ1(1+m2),α1=β21(1+2m2)9γ1(1+m2),γ1<0,} (4.69)

Substituting (4.69) into (4.4), (4.6), we have the Jacobi elliptic solutions:

u(x,t)=β13γ1[1±2(1+m2)ns(ξ,m)], (4.70)

where ξ=xββ(β131γ1(1+m2))tαα. If m1 , then we have the singular soliton solution:

u(x,t)=β13γ1[1±coth(xβββ1312γ1tαα)], (4.71)

while if m0 , then we have the periodic solution:

u(x,t)=β13γ1[1±2cosec(xβββ131γ1tαα)]. (4.72)

The graphical representations of some solutions

In this section, we present some graphs of the solitons and other solutions of Eq.(1.1). Let us now examine Figures (1 - 12) as it illustrates some of our solutions obtained in this article. To this aim, we select some special values of the parameters obtained for example, in some of the solutions of (4.10), (4.12), (4.22), (4.23), (4.28), (4.37), (4.41), (4.56), (4.58), (4.64), (4.67) and (4.72) of the conformable space-time fractional fourth-order Pochhammer-Chree equation (1.1). For more convenience the graphical representations of these solutions are shown in the following figures.

From the above Figures, one can see that the obtained solutions possess the Jacobi elliptic solutions, the conformable fractional solitary wave solution, the Jacobi elliptic conformable fractional solution and the solitary wave solutions . Also, these Figures expressing the behaviour of these solutions which give some perspective readers how the behaviour solutions are produced.

Conclusion

We have derived many Jacobi elliptic function solutions, the solitary wave solutions, singular solitary wave solutions and the trigonometric function solutions of the conformable space-time fractional fourth-order Pochhammer-Chree equation (1.1) using the unified sub-equation method combined with the conformable space-time fractional derivatives described in Sec. 3. On comparing our results in this article with that obtained in42–50 using different methods, we conclude that the Jacobi elliptic solutions obtained in our article are new, while some solitary wave solutions, singular solitary wave solutions and the trigonometric function solutions obtained in our article are equivalent to that obtained in42–50. From these discussions, we conclude that the proposed method of Sec.3, is direct, concise and effective powerful mathematical tools for obtaining the exact solutions of other nonlinear evolution equations. Finally, our results in this article have been checked using the Maple by putting them back into the original equation (1.1).57

Acknowledgements

None.

Conflict of interest

The author declares that there is no conflict of interest.

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