Journal of Applied Mathematics

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Journal of Applied Mathematics / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 575387 | 14 pages | https://doi.org/10.1155/2012/575387

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

Hasibun Naher ,1,2 Farah Aini Abdullah,1 and M. Ali Akbar1,3

1School of Mathematical Sciences, Universiti Sains Malaysia, 11800 Penang, Malaysia

2Department of Mathematics and Natural Sciences, Brac University, 66 Mohakhali, Dhaka 1212, Bangladesh

3Department of Applied Mathematics, University of Rajshahi, Rajshahi, Bangladesh

Academic Editor: A. A. Soliman
Received12 Jul 2011
Revised04 Oct 2011
Accepted04 Oct 2011
Published24 Jan 2012

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.

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 equation𝑃𝑢,𝑢𝑡,𝑢𝑥,𝑢𝑦,𝑢𝑧,𝑢𝑡𝑡,𝑢𝑥𝑡,𝑢𝑥𝑥,𝑢𝑥𝑦,𝑢𝑦𝑦,𝑢𝑦𝑡,𝑢𝑧𝑧,𝑢𝑧𝑡,𝑢𝑧𝑥,𝑢𝑧𝑦,…=0.(2.1) The main steps of the Exp-function method [34] are as follows.

Step 1. Consider a complex variable as 𝑢(𝑥,𝑦,𝑧,𝑡)=𝑢(𝜂),𝜂=𝑥+𝑦+𝑧−𝑉𝑡.(2.2) Now using (2.2), (2.1) converts to a nonlinear ordinary differential equation for 𝑢(𝜂)𝑄𝑢,𝑢,𝑢,𝑢,…=0,(2.3) 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] ∑𝑢(𝜂)=𝑑𝑛=−𝑐𝑎𝑛exp(𝑛𝜂)∑𝑞𝑚=−𝑝𝑏𝑚=𝑎exp(𝑚𝜂)−𝑐exp(−𝑐𝜂)+⋯+𝑎𝑑exp(𝑑𝜂)𝑏−𝑝exp(−𝑝𝜂)+⋯+𝑏𝑞exp(𝑞𝜂),(2.4) 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: 𝑎𝑢(𝜂)=𝑐exp(𝑐𝜂)+⋯+𝑎−𝑑exp(−𝑑𝜂)𝑏𝑝exp(𝑝𝜂)+⋯+𝑏−𝑞exp(−𝑞𝜂).(2.5)

Step 3. In order to determine the values of 𝑐 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 of 𝑐,𝑑,𝑝, and 𝑞 into (2.5), and then substituted (2.5) into (2.3) and simplifying, we obtain 𝑖𝐶𝑖exp(±𝑖𝜂)=0,𝑖=0,1,2,3,….(2.6) 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 unknown 𝑎𝑐′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 Figures 1, 2, 3, 4, 5, and 6 by using the software Maple 13.


Figure 1 
Periodic solution for 𝛼 = − 1 , 𝑦 = 0 and 𝑧 = 0 .

Figure 2 
Periodic solution for 𝛽 = 2 , 𝑦 = 0 and 𝑧 = 0 .

Figure 3 
Solitons solution for 𝑦 = 0 and 𝑧 = 0 .

Figure 4 
Solitons solution for 𝑦 = 0 and 𝑧 = 0 .

Figure 5 
Solitons solution for 𝑦 = 0 and 𝑧 = 0 .

Figure 6 
Solitons solution for 𝑦 = 0 and 𝑧 = 0 .

We consider the (3 + 1)-dimensional modified KdV-Zakharov-Kuznetsev equation𝑢𝑡+𝛼𝑢2𝑢𝑥+𝑢𝑥𝑥𝑥+𝑢𝑥𝑦𝑦+𝑢𝑥𝑧𝑧=0,(3.1) 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−𝑉𝑢+𝛼𝑢2𝑢+3𝑢=0,(3.2) where primes denote the derivatives with respect to 𝜂.

According to Step 2, 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:𝑐𝑢′′′=1[]exp(3𝑝+𝑐)𝜂+⋯𝑐2[],𝑢exp4𝑝𝜂+⋯2𝑐𝑢′=3[(]exp𝑝+3𝑐)𝜂+⋯𝑐4[],exp4𝑝𝜂+⋯(3.3) where 𝑐𝑗 are coefficients only for simplicity; from (3.3), we obtain that 3𝑝+𝑐=𝑝+3𝑐,whichleads𝑝=𝑐.(3.4) 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 have𝑢′′′=⋯+𝑑1[]exp−(𝑑−𝑞)𝜂⋯+𝑑2[],𝑢exp−4𝑞𝜂2𝑢′=⋯+𝑑3[]exp−3(𝑑−𝑞)𝜂⋯+𝑑4[],exp−4𝑞𝜂(3.5) where 𝑑𝑗 are determined coefficients only for simplicity; from (3.5), we obtain −(𝑑−𝑞)=−3(𝑑−𝑞),whichleads𝑞=𝑑.(3.6) 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 set 𝑝=𝑐=1 and 𝑞=𝑑=1.
For this case, the trial solution (2.5) reduces to 𝑎𝑢(𝜂)=1𝑒𝜂+𝑎0+𝑎−1𝑒−𝜂𝑏1𝑒𝜂+𝑏0+𝑏−1𝑒−𝜂.(3.7) Since, 𝑏1≠0, (3.7) can be simplified 𝑎𝑢(𝜂)=1𝑒𝜂+𝑎0+𝑎−1𝑒−𝜂𝑒𝜂+𝑏0+𝑏−1𝑒−𝜂.(3.8) 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 𝑉1𝐴𝐶3𝑒3𝜂+𝐶2𝑒2𝜂+𝐶1𝑒𝜂+𝐶0+𝐶−1𝑒−𝜂+𝐶−2𝑒−2𝜂+𝐶−3𝑒−3𝜂=0.(3.9) And, setting each coefficient of exp(±𝑛𝜂),𝑛=0,1,2,3,…, to zero, we obtain 𝐶3=0,𝐶2=0,𝐶1=0,𝐶0=0,𝐶−1=0,𝐶−2=0,𝐶−3=0.(3.10) 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 𝑏−1=𝑏−1,𝑎−1=∓6𝑏−1√−2𝛼,𝑎0=0,𝑎16=±√−2𝛼,𝑏0=0,𝑉=−6,(3.11) where 𝑏−1 is free parameter.Set 2. We obtain that 𝑎0=𝑎0,𝑏0=𝑏0,𝑎−11=∓√12−2𝛼2𝛼𝑎20+9𝑏20,𝑎13=±√−2𝛼,𝑏−1=118𝛼𝑎20+14𝑏203,𝑉=−2,(3.12) where 𝑎0 and 𝑏0 are free parameters.Set 3. We obtain that 𝑎1=𝑎1,𝑏0=𝑏0,𝑎−1=𝑏202𝛼𝑎21+98𝛼𝑎1,𝑎0=𝑏0𝛼𝑎21+9𝛼𝑎1,𝑏−1=𝑏202𝛼𝑎21+98𝛼𝑎21,𝑉=3+𝛼𝑎21,(3.13) where 𝑎1 and 𝑏0 are free parameters.Set 4. We obtain that 𝑎0=𝑎0,𝑎−1=0,𝑎1=0,𝑏−1=172𝛼𝑎20,𝑏0=0,𝑉=3,(3.14) where 𝑎0 is free parameter.
Now, substituting (3.11) into (3.8), we obtain traveling wave solution 𝑢(𝜂)=±6𝑒𝜂∓6𝑏−1𝑒−𝜂√𝑒−2𝛼𝜂+𝑏−1𝑒−𝜂.(3.15) Equation (3.15) can be simplified as 𝑢(𝜂)=±6√−2𝛼1−2𝑏−1(cosh𝜂−sinh𝜂)1+𝑏−1cosh𝜂+1−𝑏−1sinh𝜂,(3.16) where 𝜂=𝑥+𝑦+𝑧+6𝑡.
If 𝑏−1=1 from (3.16), we obtain 𝑢(𝜂)=±6𝑖√2𝛼tanh𝜂.(3.17) Substituting (3.12) into (3.8) and simplifying, we get traveling wave solution 𝑢(𝜂)=±3√⎡⎢⎢⎣−2𝛼1+12±𝑎0√−2𝛼+3𝑏0−22𝛼𝑎20+9𝑏20(cosh𝜂−sinh𝜂)36+2𝛼𝑎20+9𝑏20cosh𝜂+36−2𝛼𝑎20−9𝑏20sinh𝜂+36𝑏0⎤⎥⎥⎦,(3.18) where 𝜂=𝑥+𝑦+𝑧+(3/2)𝑡.
If 𝛼 is negative, that is, 𝛼=−𝛽, 𝛽>0, 𝑏0=2 and 𝑎0=0, then from (3.18), we obtain 𝑢(𝜂)=±3√𝜂2𝛽tanh2.(3.19) Substituting (3.13) into (3.8) and simplifying, we obtain 𝑢(𝜂)=𝑎11+72𝑏08𝛼𝑎21+𝑏202𝛼𝑎21+9cosh𝜂+8𝛼𝑎21−𝑏202𝛼𝑎21+9sinh𝜂+8𝛼𝑎21𝑏0,(3.20) where 𝜂=𝑥+𝑦+𝑧−(3+𝛼𝑎21)𝑡.
If 𝑏0=1,𝛼=6, and 𝑎1=1/2, (3.20) becomes 1𝑢(𝜂)=2+31+2cosh𝜂.(3.21) Substituting (3.14) into (3.8) and simplifying, we obtain 𝑢(𝜂)=72𝑎072+𝛼𝑎20cosh𝜂+72−𝛼𝑎20sinh𝜂,(3.22) where 𝜂=𝑥+𝑦+𝑧−3𝑡.
If 𝑎0=3 and 𝛼=8, (3.22) becomes 3𝑢(𝜂)=2secℎ𝜂.(3.23)

Case 2. We set 𝑝=𝑐=2 and 𝑞=𝑑=1.
For this case, the trial solution (2.5) reduces to 𝑎𝑢(𝜂)=2𝑒2𝜂+𝑎1𝑒𝜂+𝑎0+𝑎−1𝑒−𝜂𝑏2𝑒2𝜂+𝑏1𝑒𝜂+𝑏0+𝑏−1𝑒−𝜂.(3.24) 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 𝑎𝑢(𝜂)=2𝑒2𝜂+𝑎1𝑒𝜂+𝑎0+𝑎−1𝑒−𝜂𝑒2𝜂+𝑏1𝑒𝜂+𝑏0.(3.25) Performing the same procedure as described in Case 1, we obtain four sets of solutions.

Set 1. We obtain that𝑏0=𝑏0,𝑎−1=0,𝑎0=∓6𝑏0√−2𝛼,𝑎1=0,𝑎26=±√−2𝛼,𝑏1=0,𝑉=−6,(3.26) where 𝑏0 is free parameter.

Set 2. We obtain that 𝑎1=𝑎1,𝑏1=𝑏1,𝑎−1=0,𝑎01=∓√12−2𝛼2𝛼𝑎21+9𝑏21,𝑎23=±√−2𝛼,𝑏0=118𝛼𝑎21+14𝑏21,𝑉=−32,(3.27) where 𝑎1 and 𝑏1 are free parameters.

Set 3. We obtain that 𝑎2=𝑎2,𝑏1=𝑏1,𝑎−1=0,𝑎0=𝑏212𝛼𝑎22+98𝛼𝑎2,𝑎1=𝑏1𝛼𝑎22+9𝛼𝑎2,𝑏0=𝑏212𝛼𝑎22+98𝛼𝑎22,𝑉=𝛼𝑎22+3,(3.28) where 𝑎2 and 𝑏1 are free parameters.

Set 4. We obtain that𝑎1=𝑎1,𝑎−1=0,𝑎0=0,𝑎2=0,𝑏0=𝛼𝑎2172,𝑏1=0,𝑉=3,(3.29) where 𝑎1 is a free parameter.
Using (3.26) into (3.25) and simplifying, we obtain that 𝑢(𝜂)=±6√−2𝛼1−2𝑏0(cosh𝜂−sinh𝜂)1+𝑏0cosh𝜂+1−𝑏0sinh𝜂.(3.30) If 𝑏0=1, from (3.30), we obtain that 𝑢(𝜂)=±6𝑖√2𝛼tanh𝜂,(3.31) where 𝜂=𝑥+𝑦+𝑧+6𝑡.
Substituting (3.27) into (3.25) and simplifying, we obtain that 𝑢(𝜂)=±3√⎡⎢⎢⎣−2𝛼1+12±𝑎1√−2𝛼+3𝑏1−22𝛼𝑎21+9𝑏21(cosh𝜂−sinh𝜂)36+2𝛼𝑎21+9𝑏21cosh𝜂+36−2𝛼𝑎21−9𝑏21sinh𝜂+36𝑏1⎤⎥⎥⎦.(3.32) If 𝛼 is negative, that is, 𝛼=−𝛽, 𝛽>0, 𝑏1=2 and 𝑎1=0, (3.32) can be simplified as 𝑢(𝜂)=±3√𝜂2𝛽tanh2,(3.33) where 𝜂=𝑥+𝑦+𝑧+(3/2)𝑡.
Substituting (3.28) into (3.25) and simplifying, we obtain that 𝑢(𝜂)=𝑎21+72𝑏18𝛼𝑎22+𝑏212𝛼𝑎22+9cosh𝜂+8𝛼𝑎22−𝑏212𝛼𝑎22+9sinh𝜂+8𝛼𝑎22𝑏1.(3.34) If 𝑏1=1,𝛼=6, and 𝑎2=1/2, (3.34) becomes 1𝑢(𝜂)=2+31+2cosh𝜂,(3.35) where 𝜂=𝑥+𝑦+𝑧−(3+𝛼𝑎22)𝑡.
Using (3.29) into (3.25) and simplifying, we obtain that 𝑢(𝜂)=72𝑎172+𝛼𝑎21cosh𝜂+72−𝛼𝑎21sinh𝜂.(3.36) If 𝑎1=3, and 𝛼=8, (3.36) becomes 3𝑢(𝜂)=2secℎ𝜂,(3.37) where 𝜂=𝑥+𝑦+𝑧−3𝑡.

Case 3. We set 𝑝=𝑐=2 and 𝑞=𝑑=2.
For this case, the trial solution (2.5) reduces to 𝑎𝑢(𝜂)=2𝑒2𝜂+𝑎1𝑒𝜂+𝑎0+𝑎−1𝑒−𝜂+𝑎−2𝑒−2𝜂𝑏2𝑒2𝜂+𝑏1𝑒𝜂+𝑏0+𝑏−1𝑒−𝜂+𝑏−2𝑒−2𝜂.(3.38) 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 𝑎𝑢(𝜂)=2𝑒𝜂+𝑎1+𝑎0𝑒−𝜂+𝑎−1𝑒−2𝜂+𝑎−2𝑒−3𝜂𝑏2𝑒𝜂+𝑏1+𝑏0𝑒−𝜂+𝑏−1𝑒−2𝜂+𝑏−2𝑒−3𝜂.(3.39)
If we put 𝑎−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 consider 𝑝=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 Table 1, Zayed [52] obtained another solution (3.39). But, we obtain two more new solutions (3.21) and (3.23).

Zayed [52] solutionsOur solutions
(i) If 𝜆 = 2 , 𝜇 = 0 , equation (3.40) becomes 𝑢 ( 𝜂 ) = ± 6 𝑖 √ 2 𝛼 t a n h 𝜂 .(i) Solution (3.17) is 𝑢 ( 𝜂 ) = ± 6 𝑖 √ 2 𝛼 t a n h 𝜂 .
(ii) If 1 + 𝜆 2 = 4 𝜇 and 𝛼 is replaced with 𝛽 , equation (3.38) becomes 𝑢 ( 𝜂 ) = ± 3 𝑖 √ 𝜂 2 𝛽 t a n 2 .(ii) If 𝜂 is replaced with 𝑖 𝜂 , solution (3.19) becomes 𝑢 ( 𝜂 ) = ± 3 𝑖 √ 𝜂 2 𝛽 t a n 2 .
Table 1 
Comparison between Zayed [52] solutions and our solutions.

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.

Acknowledgment

This paper is supported by USM short-term Grant no. 304/PMATHS/6310072, and the authors would like to express their thanks to the School of Mathematical Sciences, USM for providing related research facilities. The authors are also grateful to the referee(s) for their valuable comments and suggestions.

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