# New Traveling Wave Solutions of the Higher Dimensional Nonlinear Partial Differential Equation by the Exp-Function Method

Authors: Hasibun Naher; Farah Aini Abdullah; M. Ali Akbar
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We construct new analytical solutions of the (3+1)-dimensional modified KdV-Zakharov-Kuznetsev equation by the Exp-function method. Plentiful exact traveling wave solutions with arbitrary parameters are effectively obtained by the method. The obtained results show that the Exp-function method is effective and straightforward mathematical tool for searching analytical solutions with arbitrary parameters of higher-dimensional nonlinear partial differential equation.


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## Body

## 1. Introduction

Nonlinear partial differential equations (NLPDEs) play a prominent role in different branches of the applied sciences. In recent time, many researchers investigated exact traveling wave solutions of NLPDEs which play a crucial role to reveal the insight of complex physical phenomena. In the past several decades, a variety of effective and powerful methods, such as variational iteration method [1–3], tanh-coth method [4], homotopy perturbation method [5–7], Fan subequation method [8], projective Riccati equation method [9], differential transform method [10], direct algebraic method [11], first integral method [12], Hirota’s bilinear method [13], modified extended direct algebraic method [14], extended tanh method [15], Backlund transformation [16], bifurcation method [17], Cole-Hopf transformation method [18], sech-tanh method [19], (G′/G)-expansion method [20–22], modified (G′/G)-expansion method [23], multiwave method [24], extended (G′/G)-expansion method [25, 26], and others [27–33] were used to seek exact traveling wave solutions of the nonlinear evolution equations (NLEEs).Recently, He and Wu [34] have presented a novel method called the Exp-function method for searching traveling wave solutions of the nonlinear evolution equations arising in mathematical physics. The Exp-function method is widely used to many kinds of NLPDEs, such as good Boussinesq equations [35], nonlinear differential equations [36], higher-order boundary value problems [37], nonlinear problems [38], Calogero-Degasperis-Fokas equation [39], nonlinear reaction-diffusion equations [40], 2D Bratu type equation [41], nonlinear lattice differential equations [42], generalized-Zakharov equations [43], (3 + 1)-dimensional Jimbo-Miwa equation [44], modified Zakharov-Kuznetsov equation [45], Brusselator reaction diffusion model [46], nonlinear heat equation [47], and the other important NLPDEs [48–51].In this article, we apply the Exp-function method [34] to obtain the analytical solutions of the nonlinear partial differential equation, namely, (3 + 1)-dimensional modified KdV-Zakharov-Kuznetsev equation.


## 2. Description of the Exp-Function Method

Consider the general nonlinear partial differential equation(2.1)P(u,ut,ux,uy,uz,utt,uxt,uxx,uxy,uyy,uyt,uzz,uzt,uzx,uzy,…)=0.
The main steps of the Exp-function method [34] are as follows.Step 1.
Consider a complex variable as(2.2)u(x,y,z,t)=u(η),η=x+y+z-Vt.
Now using (2.2), (2.1) converts to a nonlinear ordinary differential equation for u(η)(2.3)Q(u,u′,u′′,u′′′,…)=0,
where primes denote the ordinary derivative with respect to η.Step 2.
We assume that the traveling wave solution of (2.3) can be expressed in the form [34]
(2.4)u(η)=∑n=-cdanexp⁡(nη)∑m=-pqbmexp⁡(mη)=a-cexp⁡(-cη)+⋯+adexp⁡(dη)b-pexp⁡(-pη)+⋯+bqexp⁡(qη),
where c,d,p, and q are positive integers to be determined later, and an and bm are unknown constants. Equation (2.4) can be rewritten in the following equivalent form:
(2.5)u(η)=acexp⁡(cη)+⋯+a-dexp⁡(-dη)bpexp⁡(pη)+⋯+b-qexp⁡(-qη).Step 3.
In order to determine the values ofc and p, we balance the highest order linear term with the highest order nonlinear term, and, determining the values of d and q, we balance the lowest order linear term with the lowest order nonlinear term in (2.3). Thus, we obtain the values of c,d,p,  and q.Step 4.
Substituting the values ofc,d,p, and q into (2.5), and then substituted (2.5) into (2.3) and simplifying, we obtain
(2.6)∑iCiexp⁡(±iη)=0,i=0,1,2,3,….
Then each coefficient Ci=0 is to set, yields a system of algebraic equations for ac′s and bp′s.Step 5.
We assume that the unknownac′s and bp′s can be determined by solving the system of algebraic equations obtained in Step 4. Putting these values into (2.5), we obtain exact traveling wave solutions of the (2.1).


## 3. Application of the Method

In this section, we apply the method to construct the traveling wave solutions of the (3 + 1)-dimensional modified KdV-Zakharov-Kuznetsev equation. The obtained solutions will be displayed in Figures1, 2, 3, 4, 5, and 6 by using the software Maple 13.Figure 1
Periodic solution forα=-1, y=0 and z=0.Figure 2
Periodic solution forβ=2, y=0 and z=0.Figure 3
Solitons solution fory=0 and z=0.Figure 4
Solitons solution fory=0 and z=0.Figure 5
Solitons solution fory=0 and z=0.Figure 6
Solitons solution fory=0 and z=0.We consider the (3 + 1)-dimensional modified KdV-Zakharov-Kuznetsev equation(3.1)ut+αu2ux+uxxx+uxyy+uxzz=0,
where α is a nonzero constant.Zayed [52] solved (3.1) using the (G′/G)-expansion method. Later, in article [53], he solved same equation by the generalized (G′/G)-expansion method.Here, we will solve this equation by the Exp-function method.Now, we use the transformation (2.2) into (3.1), which yields(3.2)-Vu′+αu2u′+3u′′′=0,
where primes denote the derivatives with respect to η.According to Step2, the solution of (3.2) can be written in the form of (2.5). To determine the values of c and p, according to Step 3, we balance the highest order linear term of u′′′ with the highest order nonlinear term of u2u′ in (3.2), that is, u′′′  and  u2u′.  Therefore, we have the following:(3.3)u′′′=c1exp⁡[(3p+c)η]+⋯c2exp⁡[4pη]+⋯,u2u′=c3exp⁡[(p+3c)η]+⋯c4exp⁡[4pη]+⋯,
where cj are coefficients only for simplicity; from (3.3), we obtain that
(3.4)3p+c=p+3c,whichleadsp=c.
To determine the values of d and q, we balance the lowest order linear term of u′′′ with the lowest order nonlinear term of u2u′  in (3.2). We have(3.5)u′′′=⋯+d1exp⁡[-(d-q)η]⋯+d2exp⁡[-4qη],u2u′=⋯+d3exp⁡[-3(d-q)η]⋯+d4exp⁡[-4qη],
where dj are determined coefficients only for simplicity; from (3.5), we obtain
(3.6)-(d-q)=-3(d-q),whichleadsq=d.
Any real values can be considered for c and d, since they are free parameters. But the final solutions of (3.1) do not depend upon the choice of c and d.Case 1.
We setp=c=1 and q=d=1.
For this case, the trial solution (2.5) reduces to
(3.7)u(η)=a1eη+a0+a-1e-ηb1eη+b0+b-1e-η.
Since, b1≠0, (3.7) can be simplified
(3.8)u(η)=a1eη+a0+a-1e-ηeη+b0+b-1e-η.
By substituting (3.8) into (3.2) and equating the coefficients of exp(±nη),n=0,1,2,3,…, with the aid of Maple 13, we obtain a set of algebraic equations in terms of a-1,a0,a1,b-1,b0, and V(3.9)1A(C3e3η+C2e2η+C1eη+C0+C-1e-η+C-2e-2η+C-3e-3η)=0.
And, setting each coefficient of exp(±nη),n=0,1,2,3,…,  to zero, we obtain
(3.10)C3=0,C2=0,C1=0,C0=0,C-1=0,C-2=0,C-3=0.
For determining unknowns, we solve the obtained system of algebraic (3.10) with the aid of Maple 13, and we obtain four different sets of solutions.Set 1.
We obtain that(3.11)b-1=b-1,a-1=∓6b-1-2α,a0=0,a1=±6-2α,b0=0,V=-6,
where b-1 is free parameter.Set 2.
We obtain that(3.12)a0=a0,b0=b0,a-1=∓112-2α(2αa02+9b02),a1=±3-2α,b-1=118αa02+14b02,V=-32,
where a0 and b0 are free parameters.Set 3.
We obtain that(3.13)a1=a1,b0=b0,a-1=b02(2αa12+9)8αa1,a0=b0(αa12+9)αa1,b-1=b02(2αa12+9)8αa12,V=3+αa12,
where a1 and b0 are free parameters.Set 4.
We obtain that(3.14)a0=a0,a-1=0,a1=0,b-1=172αa02,b0=0,V=3,
where a0 is free parameter.
Now, substituting (3.11) into (3.8), we obtain traveling wave solution
(3.15)u(η)=±6eη∓6b-1e-η-2α(eη+b-1e-η).
Equation (3.15) can be simplified as
(3.16)u(η)=±6-2α[1-2b-1(cosh⁡η-sinh⁡η)(1+b-1)cosh⁡η+(1-b-1)sinh⁡η],
where η=x+y+z+6t.
Ifb-1=1 from (3.16), we obtain
(3.17)u(η)=±6i2αtanhη.
Substituting (3.12) into (3.8) and simplifying, we get traveling wave solution
(3.18)u(η)=±3-2α[1+12(±a0-2α+3b0)-2(2αa02+9b02)(cosh⁡η-sinh⁡η)(36+2αa02+9b02)cosh⁡η+(36-2αa02-9b02)sinh⁡η+36b0],
where η=x+y+z+(3/2)t.
Ifα is negative, that is, α=-β,  β>0,  b0=2 and a0=0, then from (3.18), we obtain
(3.19)u(η)=±32βtanhη2.
Substituting (3.13) into (3.8) and simplifying, we obtain
(3.20)u(η)=a1[1+72b0{8αa12+b02(2αa12+9)}cosh⁡η+{8αa12-b02(2αa12+9)}sinh⁡η+8αa12b0],
where η=x+y+z-(3+αa12)t.
Ifb0=1,α=6, and a1=1/2, (3.20) becomes
(3.21)u(η)=12+31+2cosh⁡η.
Substituting (3.14) into (3.8) and simplifying, we obtain
(3.22)u(η)=72a0(72+αa02)cosh⁡η+(72-αa02)sinh⁡η,
where η=x+y+z-3t.
Ifa0=3 and α=8, (3.22) becomes
(3.23)u(η)=32sechη.Case 2.
We setp=c=2 and q=d=1.
For this case, the trial solution (2.5) reduces to
(3.24)u(η)=a2e2η+a1eη+a0+a-1e-ηb2e2η+b1eη+b0+b-1e-η.
Since, there are some free parameters in (3.24), for simplicity, we may consider that b2=1 and b-1=0.  Then the solution (3.24) is simplified as
(3.25)u(η)=a2e2η+a1eη+a0+a-1e-ηe2η+b1eη+b0.
Performing the same procedure as described in Case 1, we obtain four sets of solutions.Set 1.
We obtain that(3.26)b0=b0,a-1=0,a0=∓6b0-2α,a1=0,a2=±6-2α,b1=0,V=-6,
where b0 is free parameter.Set 2.
We obtain that(3.27)a1=a1,b1=b1,a-1=0,a0=∓112-2α(2αa12+9b12),a2=±3-2α,b0=118αa12+14b12,V=-32,
where a1 and b1 are free parameters.Set 3.
We obtain that(3.28)a2=a2,b1=b1,a-1=0,a0=b12(2αa22+9)8αa2,a1=b1(αa22+9)αa2,b0=b12(2αa22+9)8αa22,V=αa22+3,
where a2 and b1 are free parameters.Set 4.
We obtain that(3.29)a1=a1,a-1=0,a0=0,a2=0,b0=αa1272,b1=0,V=3,
where a1 is a free parameter.
Using (3.26) into (3.25) and simplifying, we obtain that
(3.30)u(η)=±6-2α[1-2b0(cosh⁡η-sinh⁡η)(1+b0)cosh⁡η+(1-b0)sinh⁡η].
If b0=1, from (3.30), we obtain that
(3.31)u(η)=±6i2αtanhη,
where η=x+y+z+6t.
Substituting (3.27) into (3.25) and simplifying, we obtain that
(3.32)u(η)=±3-2α[1+12(±a1-2α+3b1)-2(2αa12+9b12)(cosh⁡η-sinh⁡η)(36+2αa12+9b12)cosh⁡η+(36-2αa12-9b12)sinh⁡η+36b1].
If α is negative, that is,  α=-β,  β>0,  b1=2 and a1=0, (3.32) can be simplified as
(3.33)u(η)=±32βtanhη2,
where η=x+y+z+(3/2)t.
Substituting (3.28) into (3.25) and simplifying, we obtain that
(3.34)u(η)=a2[1+72b1{8αa22+b12(2αa22+9)}cosh⁡η+{8αa22-b12(2αa22+9)}sinh⁡η+8αa22b1].
If b1=1,α=6, and a2=1/2, (3.34) becomes
(3.35)u(η)=12+31+2cosh⁡η,
where η=x+y+z-(3+αa22)t.
Using (3.29) into (3.25) and simplifying, we obtain that
(3.36)u(η)=72a1(72+αa12)cosh⁡η+(72-αa12)sinh⁡η.
If a1=3, and α=8, (3.36) becomes
(3.37)u(η)=32sechη,
where η=x+y+z-3t.Case 3.
We setp=c=2 and q=d=2.
For this case, the trial solution (2.5) reduces to
(3.38)u(η)=a2e2η+a1eη+a0+a-1e-η+a-2e-2ηb2e2η+b1eη+b0+b-1e-η+b-2e-2η.
Since, there are some free parameters in (3.38), we may consider b2=1,a-2=0,b-2=0, and b-1=0  so that the (3.38) reduces to the (3.25). This indicates that the Case 3 is equivalent to the Case 2. Equation (3.38) can be rewritten as
(3.39)u(η)=a2eη+a1+a0e-η+a-1e-2η+a-2e-3ηb2eη+b1+b0e-η+b-1e-2η+b-2e-3η.
If we puta-2=0,a-1=0,b2=1,b-2=0, and b-1=0  into (3.39), we obtain the solution form as (3.8). This implies that the Case 3 is equivalent to the Case 1.
Also, if we considerp=c=3 and q=d=3,  it can be shown that this Case is also equivalent to the Cases 1 and 2.
Therefore, we think that no need to find the solutions again.It is noted that the solution (3.17) and (3.31) are identical, solution (3.19) and (3.33) are identical, solution (3.21) and (3.35) are identical, and solution (3.23) and (3.37) are identical.Beyond Table1, Zayed [52] obtained another solution (3.39). But, we obtain two more new solutions (3.21) and (3.23).Table 1
Comparison between Zayed [52] solutions and our solutions.
Zayed [52] solutionsOur solutions(i) Ifλ=2,μ=0, equation (3.40) becomes u(η)=±6i2αtanhη.(i) Solution (3.17) is u(η)=±6i2αtanhη.(ii) If1+λ2=4μ and α is replaced with β, equation (3.38) becomes u(η)=±3i2βtanη2.(ii) Ifη is replaced with iη, solution (3.19) becomes u(η)=±3i2βtanη2.Graphical Representations of the Solutions
The above solutions are shown with the aid of Maple 13 in the graphs.


## 4. Conclusions

Using the Exp-function method, with the aid of symbolic computation software Maple 13, new exact traveling wave solutions of the (3 + 1)-dimensional modified KdV-Zakharov-Kuznetsev equation are constructed. It is important that some of the obtained solutions are identical to the solutions available in the literature and some are new. These solutions can be used to describe the insight of the complex physical phenomena.



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*Source: 575387-2012-01-24.xml*