Journal of Applied Mathematics

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

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 equation𝑃𝑢,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)𝑄𝑢,ğ‘¢î…ž,ğ‘¢î…žî…ž,ğ‘¢î…žî…žî…žî€¸,…=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=âˆ’ğ‘ğ‘Žğ‘›exp(𝑛𝜂)âˆ‘ğ‘žğ‘š=−𝑝𝑏𝑚=ğ‘Žexp(𝑚𝜂)−𝑐exp(−𝑐𝜂)+⋯+ğ‘Žğ‘‘exp(𝑑𝜂)𝑏−𝑝exp(−𝑝𝜂)+⋯+ğ‘ğ‘žexp(ğ‘žğœ‚),(2.4) where 𝑐,𝑑,𝑝, and ğ‘ž are positive integers to be determined later, and ğ‘Žğ‘› and 𝑏𝑚 are unknown constants. Equation (2.4) can be rewritten in the following equivalent form: ğ‘Žğ‘¢(𝜂)=𝑐exp(𝑐𝜂)+⋯+ğ‘Žâˆ’ğ‘‘exp(−𝑑𝜂)𝑏𝑝exp(𝑝𝜂)+⋯+ğ‘âˆ’ğ‘žexp(âˆ’ğ‘žğœ‚).(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 ğ‘ž, 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 ğ‘ž.

Step 4. Substituting the values of 𝑐,𝑑,𝑝, and ğ‘ž 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 ğ‘Žğ‘â€²ğ‘  and 𝑏𝑝′𝑠.

Step 5. We assume that the unknown ğ‘Žğ‘â€²ğ‘  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.

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 yieldsâˆ’ğ‘‰ğ‘¢î…ž+𝛼𝑢2ğ‘¢î…ž+3ğ‘¢î…žî…žî…ž=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 ğ‘ž, we balance the lowest order linear term of ğ‘¢î…žâ€²â€² with the lowest order nonlinear term of 𝑢2𝑢′ in (3.2). We have𝑢′′′=⋯+𝑑1[]exp−(ğ‘‘âˆ’ğ‘ž)𝜂⋯+𝑑2[],𝑢exp−4ğ‘žğœ‚2𝑢′=⋯+𝑑3[]exp−3(ğ‘‘âˆ’ğ‘ž)𝜂⋯+𝑑4[],exp−4ğ‘žğœ‚(3.5) where 𝑑𝑗 are determined coefficients only for simplicity; from (3.5), we obtain −(ğ‘‘âˆ’ğ‘ž)=−3(ğ‘‘âˆ’ğ‘ž),whichleadsğ‘ž=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 ğ‘ž=𝑑=1.
For this case, the trial solution (2.5) reduces to ğ‘Žğ‘¢(𝜂)=1𝑒𝜂+ğ‘Ž0+ğ‘Žâˆ’1𝑒−𝜂𝑏1𝑒𝜂+𝑏0+𝑏−1𝑒−𝜂.(3.7) Since, 𝑏1≠0, (3.7) can be simplified ğ‘Žğ‘¢(𝜂)=1𝑒𝜂+ğ‘Ž0+ğ‘Žâˆ’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 ğ‘Žâˆ’1,ğ‘Ž0,ğ‘Ž1,𝑏−1,𝑏0, and 𝑉1𝐴𝐶3𝑒3𝜂+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,ğ‘Žâˆ’1=∓6𝑏−1√−2𝛼,ğ‘Ž0=0,ğ‘Ž16=±√−2𝛼,𝑏0=0,𝑉=−6,(3.11) where 𝑏−1 is free parameter.Set 2. We obtain that ğ‘Ž0=ğ‘Ž0,𝑏0=𝑏0,ğ‘Žâˆ’11=∓√12−2𝛼2ğ›¼ğ‘Ž20+9𝑏20,ğ‘Ž13=±√−2𝛼,𝑏−1=118ğ›¼ğ‘Ž20+14𝑏203,𝑉=−2,(3.12) where ğ‘Ž0 and 𝑏0 are free parameters.Set 3. We obtain that ğ‘Ž1=ğ‘Ž1,𝑏0=𝑏0,ğ‘Žâˆ’1=𝑏202ğ›¼ğ‘Ž21+98ğ›¼ğ‘Ž1,ğ‘Ž0=𝑏0î€·ğ›¼ğ‘Ž21+9ğ›¼ğ‘Ž1,𝑏−1=𝑏202ğ›¼ğ‘Ž21+98ğ›¼ğ‘Ž21,𝑉=3+ğ›¼ğ‘Ž21,(3.13) where ğ‘Ž1 and 𝑏0 are free parameters.Set 4. We obtain that ğ‘Ž0=ğ‘Ž0,ğ‘Žâˆ’1=0,ğ‘Ž1=0,𝑏−1=172ğ›¼ğ‘Ž20,𝑏0=0,𝑉=3,(3.14) where ğ‘Ž0 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−1cosh𝜂+1−𝑏−1sinh𝜂,(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+12Â±ğ‘Ž0√−2𝛼+3𝑏0−22ğ›¼ğ‘Ž20+9𝑏20(cosh𝜂−sinh𝜂)36+2ğ›¼ğ‘Ž20+9𝑏20cosh𝜂+36−2ğ›¼ğ‘Ž20−9𝑏20sinh𝜂+36𝑏0⎤⎥⎥⎦,(3.18) where 𝜂=𝑥+𝑦+𝑧+(3/2)𝑡.
If 𝛼 is negative, that is, 𝛼=−𝛽, 𝛽>0, 𝑏0=2 and ğ‘Ž0=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)=ğ‘Ž11+72𝑏08ğ›¼ğ‘Ž21+𝑏202ğ›¼ğ‘Ž21+9cosh𝜂+8ğ›¼ğ‘Ž21−𝑏202ğ›¼ğ‘Ž21+9sinh𝜂+8ğ›¼ğ‘Ž21𝑏0,(3.20) where 𝜂=𝑥+𝑦+𝑧−(3+ğ›¼ğ‘Ž21)𝑡.
If 𝑏0=1,𝛼=6, and ğ‘Ž1=1/2, (3.20) becomes 1𝑢(𝜂)=2+31+2cosh𝜂.(3.21) Substituting (3.14) into (3.8) and simplifying, we obtain 𝑢(𝜂)=72ğ‘Ž072+ğ›¼ğ‘Ž20cosh𝜂+72âˆ’ğ›¼ğ‘Ž20sinh𝜂,(3.22) where 𝜂=𝑥+𝑦+𝑧−3𝑡.
If ğ‘Ž0=3 and 𝛼=8, (3.22) becomes 3𝑢(𝜂)=2secâ„Žğœ‚.(3.23)

Case 2. We set 𝑝=𝑐=2 and ğ‘ž=𝑑=1.
For this case, the trial solution (2.5) reduces to ğ‘Žğ‘¢(𝜂)=2𝑒2𝜂+ğ‘Ž1𝑒𝜂+ğ‘Ž0+ğ‘Žâˆ’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 ğ‘Žğ‘¢(𝜂)=2𝑒2𝜂+ğ‘Ž1𝑒𝜂+ğ‘Ž0+ğ‘Žâˆ’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,ğ‘Žâˆ’1=0,ğ‘Ž0=∓6𝑏0√−2𝛼,ğ‘Ž1=0,ğ‘Ž26=±√−2𝛼,𝑏1=0,𝑉=−6,(3.26) where 𝑏0 is free parameter.

Set 2. We obtain that ğ‘Ž1=ğ‘Ž1,𝑏1=𝑏1,ğ‘Žâˆ’1=0,ğ‘Ž01=∓√12−2𝛼2ğ›¼ğ‘Ž21+9𝑏21,ğ‘Ž23=±√−2𝛼,𝑏0=118ğ›¼ğ‘Ž21+14𝑏21,𝑉=−32,(3.27) where ğ‘Ž1 and 𝑏1 are free parameters.

Set 3. We obtain that ğ‘Ž2=ğ‘Ž2,𝑏1=𝑏1,ğ‘Žâˆ’1=0,ğ‘Ž0=𝑏212ğ›¼ğ‘Ž22+98ğ›¼ğ‘Ž2,ğ‘Ž1=𝑏1î€·ğ›¼ğ‘Ž22+9ğ›¼ğ‘Ž2,𝑏0=𝑏212ğ›¼ğ‘Ž22+98ğ›¼ğ‘Ž22,𝑉=ğ›¼ğ‘Ž22+3,(3.28) where ğ‘Ž2 and 𝑏1 are free parameters.

Set 4. We obtain thatğ‘Ž1=ğ‘Ž1,ğ‘Žâˆ’1=0,ğ‘Ž0=0,ğ‘Ž2=0,𝑏0=ğ›¼ğ‘Ž2172,𝑏1=0,𝑉=3,(3.29) where ğ‘Ž1 is a free parameter.
Using (3.26) into (3.25) and simplifying, we obtain that 𝑢(𝜂)=±6√−2𝛼1−2𝑏0(cosh𝜂−sinh𝜂)1+𝑏0cosh𝜂+1−𝑏0sinh𝜂.(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+12Â±ğ‘Ž1√−2𝛼+3𝑏1−22ğ›¼ğ‘Ž21+9𝑏21(cosh𝜂−sinh𝜂)36+2ğ›¼ğ‘Ž21+9𝑏21cosh𝜂+36−2ğ›¼ğ‘Ž21−9𝑏21sinh𝜂+36𝑏1⎤⎥⎥⎦.(3.32) If 𝛼 is negative, that is, 𝛼=−𝛽, 𝛽>0, 𝑏1=2 and ğ‘Ž1=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)=ğ‘Ž21+72𝑏18ğ›¼ğ‘Ž22+𝑏212ğ›¼ğ‘Ž22+9cosh𝜂+8ğ›¼ğ‘Ž22−𝑏212ğ›¼ğ‘Ž22+9sinh𝜂+8ğ›¼ğ‘Ž22𝑏1.(3.34) If 𝑏1=1,𝛼=6, and ğ‘Ž2=1/2, (3.34) becomes 1𝑢(𝜂)=2+31+2cosh𝜂,(3.35) where 𝜂=𝑥+𝑦+𝑧−(3+ğ›¼ğ‘Ž22)𝑡.
Using (3.29) into (3.25) and simplifying, we obtain that 𝑢(𝜂)=72ğ‘Ž172+ğ›¼ğ‘Ž21cosh𝜂+72âˆ’ğ›¼ğ‘Ž21sinh𝜂.(3.36) If ğ‘Ž1=3, and 𝛼=8, (3.36) becomes 3𝑢(𝜂)=2secâ„Žğœ‚,(3.37) where 𝜂=𝑥+𝑦+𝑧−3𝑡.

Case 3. We set 𝑝=𝑐=2 and ğ‘ž=𝑑=2.
For this case, the trial solution (2.5) reduces to ğ‘Žğ‘¢(𝜂)=2𝑒2𝜂+ğ‘Ž1𝑒𝜂+ğ‘Ž0+ğ‘Žâˆ’1𝑒−𝜂+ğ‘Žâˆ’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,ğ‘Žâˆ’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 ğ‘Žğ‘¢(𝜂)=2𝑒𝜂+ğ‘Ž1+ğ‘Ž0𝑒−𝜂+ğ‘Žâˆ’1𝑒−2𝜂+ğ‘Žâˆ’2𝑒−3𝜂𝑏2𝑒𝜂+𝑏1+𝑏0𝑒−𝜂+𝑏−1𝑒−2𝜂+𝑏−2𝑒−3𝜂.(3.39)
If we put ğ‘Žâˆ’2=0,ğ‘Žâˆ’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 ğ‘ž=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 .

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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Copyright © 2012 Hasibun Naher et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


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