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New analytical solutions for the coupled nonlinear Maccari’s system
Alexandria Engineering Journal (2016) 55, 2839–2847
H O S T E D BY
Alexandria University
Alexandria Engineering Journal www.elsevier.com/locate/aej www.sciencedirect.com
ORIGINAL ARTICLE
New analytical solutions for the coupled nonlinear Maccari’s system Ahmad Neirameh Department of Mathematics, Faculty of Sciences, Gonbad Kavous University, Gonbad, Iran Received 10 May 2016; revised 6 July 2016; accepted 11 July 2016 Available online 30 July 2016
KEYWORDS Coupled nonlinear Maccari’s system; Analytical method; Exact solution
Abstract In this present study using the new analytical method to the coupled nonlinear Maccari’s system in following form 8
iQt þ Qxx þ RQ ¼ 0; > > > < iS þ S þ RS ¼ 0; t
xx
> iNt þ Nxx þ RN ¼ 0; > > : Rt þ Ry þ ðjQ þ S þ Nj2 Þx ¼ 0
ð1Þ
We obtain new class of exact traveling wave solutions for this system. This work shows that the new extension of the (G0 /G)-expansion method is sufficient, effective and suitable for solving other nonlinear evolution equations; it deserves further applying and studying as well. To our knowledge, the solutions obtained in this paper have not been reported in the literature so far.
Ó 2016 Faculty of Engineering, Alexandria University. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
1. Introduction It is significant to looking for the exact traveling wave solutions for nonlinear partial differential equations (PDEs) in nonlinear sciences. These equations are widely used to describe many important phenomena and dynamic processes in physics, chemistry, biology, fluid dynamics and plasma. As mathematical models of the phenomena, the investigation of exact solutions of these equations will help us to understand these phenomena better. In the recent decades, various effective approaches have been developed to construct the exact traveling wave solutions of these equations. Therefore, exact traveling wave solution methods of PDEs have become Peer review under responsibility of Faculty of Engineering, Alexandria University.
more and more important resulting in methods such as the Kudryashov Method [1–3], the homotopy perturbation method [4–10], the (G0 /G) -expansion method [11–15], the Exp-function method [16–18], the modified simple equation method [19–22], and Hirota’s bilinear transformation method [23,24]. Exact solutions allow researchers to design and run experiments, by creating appropriate natural conditions, to determine these parameters or functions. Therefore, investigating exact traveling wave solutions is becoming successively attractive in nonlinear sciences day by day. However, not all equations posed of these models are solvable. The objective of this article was to apply new extension of the (G0 /G)-expansion method [25] to construct the exact traveling wave solutions for NLEEs in mathematical physics via the coupled nonlinear Maccari’s system. We assume the solution
http://dx.doi.org/10.1016/j.aej.2016.07.007 1110-0168 Ó 2016 Faculty of Engineering, Alexandria University. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
2840
A. Neirameh
P of NLEEs is of the form uðnÞ ¼ ni¼0 ai ðm þ FðnÞÞi þ Pn i where FðnÞ ¼ G0 =G, and G ¼ GðnÞ satisfies i¼1 bi ðm þ FðnÞÞ the ordinary differential equation G00 ðnÞ þ kG0 ðnÞþ lGðnÞ ¼ 0;, where k and l are arbitrary constants. From our observation we found that if we set m ¼ 0 and leave out the portion Pn i in our solution, then our solutions coini¼1 bi ðm þ FðnÞÞ cide with the solution introduced by Wang et al. [11]. Hence we conclude that the basic (G0 /G)-expansion method established by Wang et al. [11] is the particular case of our new extension of the (G0 /G)-expansion method. The paper is organized as follows. In Section 2, the enhanced new extension of the (G0 /G)-expansion method is discussed. In Section 3, we apply this method to the coupled nonlinear Maccari’s system and some conclusions are given in Section 4.
Suppose the general nonlinear partial differential equation, Pðu; ut ; ux ; utt ; uxx ; . . .Þ ¼ 0;
ð2Þ
where u ¼ uðx; tÞ is an unknown function, P is a polynomial in uðx; tÞ and its partial derivatives in which the highest order partial derivatives and the nonlinear terms are involved. The main steps of new extension of ðG0 =GÞ-expansion method combined with the algebra expansion are as follows: Step 1: The traveling wave variable ansatz, uðx; tÞ ¼ uðnÞ;
ð3Þ
where x 2 R f0g is the speed of the traveling wave, and permits us to transform Eq. (2) into the following ODE: Qðu; u0 ; u00 ; . . .Þ ¼ 0;
n n X X ai ðm þ FðnÞÞi þ bi ðm þ FðnÞÞi i¼0
F5 ¼
B k ; A þ Bn 2
ð11Þ
Step 3: The positive integer n can be determined by considering the homogeneous balance between the highest order derivatives and the nonlinear terms appearing in Eq. (2) or Eq. (4). Moreover precisely, we define the degree of uðnÞ as DðuðnÞÞ ¼ n which gives rise to the degree of other expression as follows:
dq u ¼ n þ q; q dn
q s du D up ¼ np þ sðn þ qÞ: dnq
ð12Þ
Therefore we can find the value of n in Eq. (5), using Eq. (12). Step 4: Substituting Eq. (5) along with Eq. (6) into Eq. (4) together with the value of n obtained in step 3, we obtain polynomials in F i and F i ði ¼ 1; 2; 3; . . .Þ, and then setting each coefficient of the resulted polynomial to zero yields a system of algebraic equations for an ; bn and x. Step 5: Suppose the values of the constants an ; bn and x can be determined by solving the system of algebraic equations obtained in step 4. Since the general solutions of Eq. (6) are known, substituting an ; bn and x into Eq. (5), we obtain some exact traveling wave solutions of the nonlinear evolution Eq. (2).
ð4Þ
where the superscripts stand for the ordinary derivatives with respect to n. Step 2: Suppose the traveling wave solution of Eq. (4) can be expressed by a polynomial in F ðnÞ as follows: uðnÞ ¼
ð10Þ
When X ¼ k2 4l ¼ 0,
D
2. An analytical method
n ¼ x xt;
pffiffiffiffi pffiffiffiffi ! X X k F4 ¼ tan A n ; 2 2 2
ð5Þ
i¼1
where FðnÞ ¼ G0 =G; an and bn are not zero simultaneously. Also G ¼ GðnÞ satisfies the ordinary differential equation, G00 ðnÞ þ kG0 ðnÞ þ lGðnÞ ¼ 0;
ð6Þ
where k and l are arbitrary constants to be determined later. The solutions of Eq. (6) can be written as follows: When X ¼ k2 4l > 0, pffiffiffiffi pffiffiffiffi ! k X X coth A þ n ; F1 ¼ ð7Þ 2 2 2 pffiffiffiffi pffiffiffiffi ! k X X tanh A þ n ; F2 ¼ 2 2 2 When X ¼ k2 4l < 0, pffiffiffiffi pffiffiffiffi ! k X X cot A þ n ; F3 ¼ 2 2 2
ð8Þ
3. Application to the coupled nonlinear Maccari’s system In order to seek exact solutions of system (1), we suppose Qðx; y; tÞ ¼ uðx; y; tÞeikðkxþayþktþlÞ
Nðx; y; tÞ ¼ wðx; y; tÞeikðkxþayþktþlÞ where k; a; l and k are constants to be determined later, l is an arbitrary constant. Substituting Eqs. (12) into system (1) yields 8 iðut þ 2kux Þ þ uxx ð1 þ k2 Þu þ uR ¼ 0; > > > > > > < iðvt þ 2kvx Þ þ vxx ð1 þ k2 Þv þ vR ¼ 0; > > iðwt þ 2kwx Þ þ wxx ð1 þ k2 Þw þ wR ¼ 0; > > > > : Rt þ Ry þ ðu2 Þx ¼ 0:
ð13Þ
Using the transformation u ¼ uðnÞ; v ¼ vðnÞ; n ¼ x þ by 2kt
ð9Þ
ð12Þ
Sðx; y; tÞ ¼ vðx; y; tÞeikðkxþayþktþlÞ
w ¼ wðnÞ;
R ¼ RðnÞ
where b is a constant, system (13) becomes the following
ð14Þ
The coupled nonlinear Maccari’s system 8 > u00 ðk þ k2 Þu þ uR ¼ 0; > > > > > > < v00 ðk þ k2 Þv þ vR ¼ 0; : 2 00 > > > w ðk þ k Þw þ wR ¼ 0; > > > > 0 : ðb 2kÞR0 þ ððu þ v þ wÞ2 Þ ¼ 0
2841 Set 2: a1 ¼ 0 ð15Þ
1 ðu þ v þ wÞ2 : b 2k
ð16Þ
Substituting Eq. (16) into other segments of (15) yields 8 00 1 u ðk þ k2 Þu b2k ðu þ v þ wÞ2 u ¼ 0; > > > > < 1 ðu þ v þ wÞ2 v ¼ 0; : v00 ðk þ k2 Þv b2k ð17Þ > > > > : 00 1 ðu þ v þ wÞ2 w ¼ 0; w ðk þ k2 Þw b2k Of course, we cannot solve system (17) directly; thus, we should give a simple relation between u, v and w. Here we set v ¼ r1 u;
w ¼ r2 u
ð18Þ
where r1 and r2 are arbitrary constants. Substituting (18) into the system (17), we have u00 ðk þ k2 Þu
ð1 þ r1 þ r2 Þ2 3 u ¼ 0; b 2k
ð19Þ
Balancing the highest order derivative u and nonlinear term u3 from Eq. (19), we obtain 3n ¼ n þ 2, which gives n ¼ 1. So u ¼ a0 þ a1 ðm þ FÞ þ b1 ðm þ FÞ1
ð20Þ
Now substituting Eq. (20) along with Eq. (6) into Eq. (19), we get a polynomial in FðnÞ. Equating the coefficient of same power of F i ðnÞði ¼ 0; 1; 2; . . .Þ, we attain the system of algebraic equations, and by solving these obtained system of equations for a0 ; a1 ; b1 ; m and R, and by solving obtained system we get two sets of values in following form: Set 1: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ a1 ¼ 1 þ r1 þ r2 b1 ¼
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ a0 ¼ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
ðb2kÞðlm2 lkmþl2 Þ 3ð1þr1 þr2 Þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2
a0 ¼ ð1þr11 þr2 Þ
where prime denotes the differential with respect to n. Integrating the fourth segment of system (15) with respect to n and taking the integration constant as zero yield R¼
b1 ¼
ðb2kÞðkm k mþlkkkÞ 3
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðb2kÞðkm2 k2 mþlkkkÞ 3
Now by using these sets of solutions for a0 ; a1 ; b1 ; m; R and by using (17) along with Eqs. (6)–(10) we have following solutions for coupled nonlinear Maccari’s system: 3.1. Hyperbolic function solutions When X ¼ k2 4l > 0, we get the following solutions. Family 1: By using set 1 and Eq. (20) along with (7) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ k m u1 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi ½18kb36kk14mbþ28mkþ 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 196l98k2 168kmþ59k2 98k þ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l coth Aþ ðxþby2ktÞ mþ 2 2 2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ: 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k þ
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l coth Aþ ðxþby2ktÞ mþ 2 2 2 So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ r1 2ðb 2kÞ k m v1 ¼ 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mkþ 6ð2k bÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ 2 2 196l 98k 168km þ 59k 98k þ 1 þ r1 þ r2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi 2 2 k k 4l k 4l coth A þ ðx þ by 2ktÞ mþ 2 2 2 r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k þ
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi k k2 4l k2 4l coth A þ ðx þ by 2ktÞ mþ 2 2 2 And
2842 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ r2 2ðb2kÞ k m w1 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 42 b2k ð1þ2r 1 þ2r2 þr1 þ2r1 r2 þr1 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ 196l98k2 168kmþ59k2 98k þ 1þr1 þr! 2 ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l coth Aþ ðxþby2ktÞ mþ 2 2 2 r2 ð1þr1 þr2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 49 b2k ð1þ2r 1 þ2r2 þr1 þ2r1 r2 þr1 Þ " pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k ! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l coth Aþ ðxþby2ktÞ mþ 2 2 2 Now from Eq. (16) we have
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ ð1 þ r1 þ r2 Þ2 k m 2k b 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2k ð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mkþ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 196l 98k2 168km þ 59k2 98k þ 1 þ r1 þ r2 ! ! ffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi p k2 4l k2 4l k mþ coth A þ ðx þ by 2ktÞ 2 2 2 ð1 þ r1 þ r2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 49 h b 2kð1 þ 2r1 þ 2r2 þ r1 þ 2rp1 rffiffiffi2 þ r1 Þ 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
R1 ¼
! !1 92 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi = k k2 4l k2 4l mþ coth A þ ðx þ by 2ktÞ ; 2 2 2
Family 2: By using set 1 and Eq. (20) along with (8) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 42 b2kð1þ2r1 þ2r2 þr21 þ2r p1ffiffirffi 2 þr1 Þ ½18kb36kk14mbþ28mkþ 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 196l98k2 168kmþ59k2 98k þ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2 ð1þr1 þr2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 1 r2 þr1 Þ h49 b2kð1þ2r1 þ2r2 þr1 þ2rp ffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k ! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2
u2 ¼
A. Neirameh So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb2kÞ r1 2ðb2kÞ k m v2 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ r1 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2
þ
r1
r1 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2 And w2 ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ r2 2ðb2kÞ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ þ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2 r2
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2
The coupled nonlinear Maccari’s system Now from Eq. (16) we have 2
ð1 þ r1 þ r2 Þ 2k b ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ
R2 ¼
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi ½18kb 36kk 14mb þ 28mkþ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 1 þ r1 þ r2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi k k2 4l k2 4l tanh A þ ðx þ by 2ktÞ mþ 2 2 2
þ
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k þ
! !1 92 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi = k k2 4l k2 4l tanh A þ ðx þ by 2ktÞ mþ ; 2 2 2
3.2. Trigonometric function solutions Family 3: By using set 1 and Eq. (20) along with (9) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ k m u3 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
þ
ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k ! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
2843 So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb2kÞ r1 2ðb2kÞ k m v3 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ r1 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
þ
r1
r1 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2 And w3 ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ r2 2ðb2kÞ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
þ
r2
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
2844
A. Neirameh
Now from Eq. (16) we have 2
ð1 þ r1 þ r2 Þ 2k b ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ
R3 ¼
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mkþ 6ð2k bÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 1 þ r1 þ r2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot A þ ðx þ by 2ktÞ mþ 2 2 2 þ
ð1 þ r1 þ r2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k ! !1 92 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi = k k2 4l k2 4l mþ cot A þ ðx þ by 2ktÞ ; 2 2 2
Family 4: By using set 1 and Eq. (20) along with (10) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ u4 ¼ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ r1 2ðb 2kÞ k m v4 ¼ 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi ½18kb 36kk 14mb þ 28mkþ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ þ 1 þ r1 þ r2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi k k2 4l k2 4l tan A ðx þ by 2ktÞ mþ 2 2 2 r1
r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ
þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
mþ And
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ r2 2ðb2kÞ k m w4 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ þ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tan A ðxþby2ktÞ mþ 2 2 2
þ
ð1þr1 þr2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tan A ðxþby2ktÞ mþ 2 2 2
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi k k2 4l k2 4l tan A ðx þ by 2ktÞ 2 2 2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tan A ðxþby2ktÞ mþ 2 2 2 r2
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tan A ðxþby2ktÞ mþ 2 2 2 Now from Eq. (16) we have
The coupled nonlinear Maccari’s system ð1þr1 þr2 Þ2 2kb ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ
R4 ¼
ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k2 4l k2 4l k mþ tan A ðxþby2ktÞ 2 2 2
þ
ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k ! !1 92 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi = k2 4l k2 4l k mþ tan A ðxþby2ktÞ ; 2 2 2
3.3. Rational function solutions Family 5: By using set 1 and Eq. (20) along with (11) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ k m u5 ¼ 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ
2845 So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ r1 2ðb 2kÞ k m v5 ¼ 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ B k þ mþ A þ Bðx þ by 2ktÞ 2 1 þ r1 þ r2 r1 ð1 þ r1 þ r2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k 1 B k mþ A þ Bðx þ by 2ktÞ 2 And w5 ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb 2kÞ r2 2ðb 2kÞ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ r2 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ ½18kb 36kk 14mb þ 28mk qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi þ 6ð2k bÞ 196l 98k2 168km þ 59k2 98k þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb 2kÞ B k mþ 1 þ r1 þ r2 A þ Bðx þ by 2ktÞ 2
þ
r2 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ
½18kb 36kk 14mb þ 28mk qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi þ 6ð2k bÞ 196l 98k2 168km þ 59k2 98k mþ
1 B k A þ Bðx þ by 2ktÞ 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ B k mþ þ A þ Bðx þ by 2ktÞ 2 1 þ r1 þ r2 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ
þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
mþ
1 B k A þ Bðx þ by 2ktÞ 2
Figure 1 Soliton solution of u4 for k ¼ 1; k ¼ 0; b ¼ 1; l ¼ 0; r1 ¼ r2 ¼ 0 in region x ¼ 10 . . . 10; y ¼ 10::10.
2846
A. Neirameh
Figure 2 Soliton solution of u3 for k ¼ 1; k ¼ 0; b ¼ 1; l ¼ 0; r1 ¼ r2 ¼ 0 in region x ¼ 10 . . . 10; y ¼ 10::10. Figure 3 Soliton solution of u2 for k ¼ 1; k ¼ 0; b ¼ 1; l ¼ 0; r1 ¼ r2 ¼ 0 in region x ¼ 10 . . . 10; y ¼ 10::10.
Now from Eq. (16) we have ð1 þ r1 þ r2 Þ2 R5 ¼ 2k b
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ
½18kb 36kk 14mb þ 28mk qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi þ 6ð2k bÞ 196l 98k2 168km þ 59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ B k mþ þ A þ Bðx þ by 2ktÞ 2 1 þ r1 þ r2 ð1 þ r1 þ r2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ 18kb 36kk 14mb þ 28mk þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 6ð2k bÞ 196l 98k2 168km þ 59k2 98k
1 )2 B k mþ A þ Bðx þ by 2ktÞ 2
Figure 4 Soliton solution of u1 for k ¼ 1; k ¼ 0; b ¼ 1; l ¼ 0; r1 ¼ r2 ¼ 0 in region x ¼ 10 . . . 10; y ¼ 10::10.
Like above we can obtain the new class of exact solutions from set 2 similar set 1 (see Figs. 1–4). Acknowledgment 4. Conclusion In this paper, we successfully use the new extension of the (G0 / G)-expansion method for solving the coupled nonlinear Maccari’s system. We have successfully obtained some new exact traveling wave solutions of the coupled nonlinear Maccari’s system with parameters. We have given some figures expressing the behavior of the obtained solutions of Eq. (1) which give some perspective readers how the behavior solutions are produced.
The authors wish to thank the referees for their comments on this paper. References [1] A. Neirameh, Exact analytical solutions for 3D-Gross-Pitaevskii equation with periodic potential by using the Kudryashov method, J. Egypt. Math. Soc. (2016), http://dx.doi.org/10.1016/ j.joems.2014.11.004 (in press).
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2847 [14] H. Kim, R. Sakthivel, New exact travelling wave solutions of some nonlinear higher dimensional physical models, Z. Naturforsch. A – Phys. Sci. 65a (2010) 197–202. [15] H. Kim, R. Sakthivel, New exact traveling wave solutions of some nonlinear higher-dimensional physical models, Rep. Math. Phys. 70 (1) (2012) 39–50. [16] A. Bekir, A. Boz, Exact solutions for nonlinear evolution equations using Exp function method, Phys. Lett. A 372 (2008) 1619–1625. [17] M.A. Akbar, N.H.M. Ali, Exp-function method for Duffing equation and new solutions of (2+1) dimensional dispersive long wave equations, Prog. Appl. Math. 1 (2) (2011) 30–42. [18] H. Naher, A.F. Abdullah, M.A. Akbar, The Exp-function method for new exact solutions of the nonlinear partial differential equations, Int. J. Phys. Sci. 6 (29) (2011) 6706–6716. [19] A.J.M. Jawad, M.D. Petkovic, A. Biswas, Modified simple equation method for nonlinear evolution equations, Appl. Math. Comput. 217 (2010) 869–877. [20] E.M.E. Zayed, A note on the modified simple equation method applied to Sharma–Tasso–Olver equation, Appl. Math. Comput. 218 (2011) 3962–3964. [21] K. Khan, M.A. Akbar, Exact and solitary wave solutions for the Tzitzeica– Dodd-Bullough and the modified KdV–Zakharov– Kuznetsov equations using the modified simple equation method, Ain Shams Eng. J. (2013), http://dx.doi.org/10.1016/j. asej.2013.01.010. [22] K. Khan, M.A. Akbar, Exact solutions of the (2+1)dimensional cubic Klein– Gordon equation and the (3+1)dimensional Zakharov-Kuznetsov equation using the modified simple equation method, J. Assoc. Arab. Univ. Basic Appl. Sci. (2014), http://dx.doi.org/10.1016/j.jaubas.2013.05.001. [23] R. Hirota, Exact envelope soliton solutions of a nonlinear wave equation, J. Math. Phys. 14 (1973) 805–810. [24] R. Hirota, J. Satsuma, Soliton solutions of a coupled KDV equation, Phys. Lett. A 85 (1981) 404–408. [25] M.S. Islam et al, An analytical method for finding exact solutions of modified Korteweg–de Vries equation, Results Phys. (2015), http://dx.doi.org/10.1016/j.rinp.2015.01.007.
H O S T E D BY
Alexandria University
Alexandria Engineering Journal www.elsevier.com/locate/aej www.sciencedirect.com
ORIGINAL ARTICLE
New analytical solutions for the coupled nonlinear Maccari’s system Ahmad Neirameh Department of Mathematics, Faculty of Sciences, Gonbad Kavous University, Gonbad, Iran Received 10 May 2016; revised 6 July 2016; accepted 11 July 2016 Available online 30 July 2016
KEYWORDS Coupled nonlinear Maccari’s system; Analytical method; Exact solution
Abstract In this present study using the new analytical method to the coupled nonlinear Maccari’s system in following form 8
iQt þ Qxx þ RQ ¼ 0; > > > < iS þ S þ RS ¼ 0; t
xx
> iNt þ Nxx þ RN ¼ 0; > > : Rt þ Ry þ ðjQ þ S þ Nj2 Þx ¼ 0
ð1Þ
We obtain new class of exact traveling wave solutions for this system. This work shows that the new extension of the (G0 /G)-expansion method is sufficient, effective and suitable for solving other nonlinear evolution equations; it deserves further applying and studying as well. To our knowledge, the solutions obtained in this paper have not been reported in the literature so far.
Ó 2016 Faculty of Engineering, Alexandria University. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
1. Introduction It is significant to looking for the exact traveling wave solutions for nonlinear partial differential equations (PDEs) in nonlinear sciences. These equations are widely used to describe many important phenomena and dynamic processes in physics, chemistry, biology, fluid dynamics and plasma. As mathematical models of the phenomena, the investigation of exact solutions of these equations will help us to understand these phenomena better. In the recent decades, various effective approaches have been developed to construct the exact traveling wave solutions of these equations. Therefore, exact traveling wave solution methods of PDEs have become Peer review under responsibility of Faculty of Engineering, Alexandria University.
more and more important resulting in methods such as the Kudryashov Method [1–3], the homotopy perturbation method [4–10], the (G0 /G) -expansion method [11–15], the Exp-function method [16–18], the modified simple equation method [19–22], and Hirota’s bilinear transformation method [23,24]. Exact solutions allow researchers to design and run experiments, by creating appropriate natural conditions, to determine these parameters or functions. Therefore, investigating exact traveling wave solutions is becoming successively attractive in nonlinear sciences day by day. However, not all equations posed of these models are solvable. The objective of this article was to apply new extension of the (G0 /G)-expansion method [25] to construct the exact traveling wave solutions for NLEEs in mathematical physics via the coupled nonlinear Maccari’s system. We assume the solution
http://dx.doi.org/10.1016/j.aej.2016.07.007 1110-0168 Ó 2016 Faculty of Engineering, Alexandria University. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).
2840
A. Neirameh
P of NLEEs is of the form uðnÞ ¼ ni¼0 ai ðm þ FðnÞÞi þ Pn i where FðnÞ ¼ G0 =G, and G ¼ GðnÞ satisfies i¼1 bi ðm þ FðnÞÞ the ordinary differential equation G00 ðnÞ þ kG0 ðnÞþ lGðnÞ ¼ 0;, where k and l are arbitrary constants. From our observation we found that if we set m ¼ 0 and leave out the portion Pn i in our solution, then our solutions coini¼1 bi ðm þ FðnÞÞ cide with the solution introduced by Wang et al. [11]. Hence we conclude that the basic (G0 /G)-expansion method established by Wang et al. [11] is the particular case of our new extension of the (G0 /G)-expansion method. The paper is organized as follows. In Section 2, the enhanced new extension of the (G0 /G)-expansion method is discussed. In Section 3, we apply this method to the coupled nonlinear Maccari’s system and some conclusions are given in Section 4.
Suppose the general nonlinear partial differential equation, Pðu; ut ; ux ; utt ; uxx ; . . .Þ ¼ 0;
ð2Þ
where u ¼ uðx; tÞ is an unknown function, P is a polynomial in uðx; tÞ and its partial derivatives in which the highest order partial derivatives and the nonlinear terms are involved. The main steps of new extension of ðG0 =GÞ-expansion method combined with the algebra expansion are as follows: Step 1: The traveling wave variable ansatz, uðx; tÞ ¼ uðnÞ;
ð3Þ
where x 2 R f0g is the speed of the traveling wave, and permits us to transform Eq. (2) into the following ODE: Qðu; u0 ; u00 ; . . .Þ ¼ 0;
n n X X ai ðm þ FðnÞÞi þ bi ðm þ FðnÞÞi i¼0
F5 ¼
B k ; A þ Bn 2
ð11Þ
Step 3: The positive integer n can be determined by considering the homogeneous balance between the highest order derivatives and the nonlinear terms appearing in Eq. (2) or Eq. (4). Moreover precisely, we define the degree of uðnÞ as DðuðnÞÞ ¼ n which gives rise to the degree of other expression as follows:
dq u ¼ n þ q; q dn
q s du D up ¼ np þ sðn þ qÞ: dnq
ð12Þ
Therefore we can find the value of n in Eq. (5), using Eq. (12). Step 4: Substituting Eq. (5) along with Eq. (6) into Eq. (4) together with the value of n obtained in step 3, we obtain polynomials in F i and F i ði ¼ 1; 2; 3; . . .Þ, and then setting each coefficient of the resulted polynomial to zero yields a system of algebraic equations for an ; bn and x. Step 5: Suppose the values of the constants an ; bn and x can be determined by solving the system of algebraic equations obtained in step 4. Since the general solutions of Eq. (6) are known, substituting an ; bn and x into Eq. (5), we obtain some exact traveling wave solutions of the nonlinear evolution Eq. (2).
ð4Þ
where the superscripts stand for the ordinary derivatives with respect to n. Step 2: Suppose the traveling wave solution of Eq. (4) can be expressed by a polynomial in F ðnÞ as follows: uðnÞ ¼
ð10Þ
When X ¼ k2 4l ¼ 0,
D
2. An analytical method
n ¼ x xt;
pffiffiffiffi pffiffiffiffi ! X X k F4 ¼ tan A n ; 2 2 2
ð5Þ
i¼1
where FðnÞ ¼ G0 =G; an and bn are not zero simultaneously. Also G ¼ GðnÞ satisfies the ordinary differential equation, G00 ðnÞ þ kG0 ðnÞ þ lGðnÞ ¼ 0;
ð6Þ
where k and l are arbitrary constants to be determined later. The solutions of Eq. (6) can be written as follows: When X ¼ k2 4l > 0, pffiffiffiffi pffiffiffiffi ! k X X coth A þ n ; F1 ¼ ð7Þ 2 2 2 pffiffiffiffi pffiffiffiffi ! k X X tanh A þ n ; F2 ¼ 2 2 2 When X ¼ k2 4l < 0, pffiffiffiffi pffiffiffiffi ! k X X cot A þ n ; F3 ¼ 2 2 2
ð8Þ
3. Application to the coupled nonlinear Maccari’s system In order to seek exact solutions of system (1), we suppose Qðx; y; tÞ ¼ uðx; y; tÞeikðkxþayþktþlÞ
Nðx; y; tÞ ¼ wðx; y; tÞeikðkxþayþktþlÞ where k; a; l and k are constants to be determined later, l is an arbitrary constant. Substituting Eqs. (12) into system (1) yields 8 iðut þ 2kux Þ þ uxx ð1 þ k2 Þu þ uR ¼ 0; > > > > > > < iðvt þ 2kvx Þ þ vxx ð1 þ k2 Þv þ vR ¼ 0; > > iðwt þ 2kwx Þ þ wxx ð1 þ k2 Þw þ wR ¼ 0; > > > > : Rt þ Ry þ ðu2 Þx ¼ 0:
ð13Þ
Using the transformation u ¼ uðnÞ; v ¼ vðnÞ; n ¼ x þ by 2kt
ð9Þ
ð12Þ
Sðx; y; tÞ ¼ vðx; y; tÞeikðkxþayþktþlÞ
w ¼ wðnÞ;
R ¼ RðnÞ
where b is a constant, system (13) becomes the following
ð14Þ
The coupled nonlinear Maccari’s system 8 > u00 ðk þ k2 Þu þ uR ¼ 0; > > > > > > < v00 ðk þ k2 Þv þ vR ¼ 0; : 2 00 > > > w ðk þ k Þw þ wR ¼ 0; > > > > 0 : ðb 2kÞR0 þ ððu þ v þ wÞ2 Þ ¼ 0
2841 Set 2: a1 ¼ 0 ð15Þ
1 ðu þ v þ wÞ2 : b 2k
ð16Þ
Substituting Eq. (16) into other segments of (15) yields 8 00 1 u ðk þ k2 Þu b2k ðu þ v þ wÞ2 u ¼ 0; > > > > < 1 ðu þ v þ wÞ2 v ¼ 0; : v00 ðk þ k2 Þv b2k ð17Þ > > > > : 00 1 ðu þ v þ wÞ2 w ¼ 0; w ðk þ k2 Þw b2k Of course, we cannot solve system (17) directly; thus, we should give a simple relation between u, v and w. Here we set v ¼ r1 u;
w ¼ r2 u
ð18Þ
where r1 and r2 are arbitrary constants. Substituting (18) into the system (17), we have u00 ðk þ k2 Þu
ð1 þ r1 þ r2 Þ2 3 u ¼ 0; b 2k
ð19Þ
Balancing the highest order derivative u and nonlinear term u3 from Eq. (19), we obtain 3n ¼ n þ 2, which gives n ¼ 1. So u ¼ a0 þ a1 ðm þ FÞ þ b1 ðm þ FÞ1
ð20Þ
Now substituting Eq. (20) along with Eq. (6) into Eq. (19), we get a polynomial in FðnÞ. Equating the coefficient of same power of F i ðnÞði ¼ 0; 1; 2; . . .Þ, we attain the system of algebraic equations, and by solving these obtained system of equations for a0 ; a1 ; b1 ; m and R, and by solving obtained system we get two sets of values in following form: Set 1: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ a1 ¼ 1 þ r1 þ r2 b1 ¼
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ a0 ¼ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
ðb2kÞðlm2 lkmþl2 Þ 3ð1þr1 þr2 Þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2
a0 ¼ ð1þr11 þr2 Þ
where prime denotes the differential with respect to n. Integrating the fourth segment of system (15) with respect to n and taking the integration constant as zero yield R¼
b1 ¼
ðb2kÞðkm k mþlkkkÞ 3
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðb2kÞðkm2 k2 mþlkkkÞ 3
Now by using these sets of solutions for a0 ; a1 ; b1 ; m; R and by using (17) along with Eqs. (6)–(10) we have following solutions for coupled nonlinear Maccari’s system: 3.1. Hyperbolic function solutions When X ¼ k2 4l > 0, we get the following solutions. Family 1: By using set 1 and Eq. (20) along with (7) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ k m u1 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi ½18kb36kk14mbþ28mkþ 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 196l98k2 168kmþ59k2 98k þ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l coth Aþ ðxþby2ktÞ mþ 2 2 2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ: 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k þ
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l coth Aþ ðxþby2ktÞ mþ 2 2 2 So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ r1 2ðb 2kÞ k m v1 ¼ 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mkþ 6ð2k bÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ 2 2 196l 98k 168km þ 59k 98k þ 1 þ r1 þ r2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi 2 2 k k 4l k 4l coth A þ ðx þ by 2ktÞ mþ 2 2 2 r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k þ
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi k k2 4l k2 4l coth A þ ðx þ by 2ktÞ mþ 2 2 2 And
2842 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ r2 2ðb2kÞ k m w1 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 42 b2k ð1þ2r 1 þ2r2 þr1 þ2r1 r2 þr1 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ 196l98k2 168kmþ59k2 98k þ 1þr1 þr! 2 ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l coth Aþ ðxþby2ktÞ mþ 2 2 2 r2 ð1þr1 þr2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 49 b2k ð1þ2r 1 þ2r2 þr1 þ2r1 r2 þr1 Þ " pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k ! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l coth Aþ ðxþby2ktÞ mþ 2 2 2 Now from Eq. (16) we have
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ ð1 þ r1 þ r2 Þ2 k m 2k b 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2k ð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mkþ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 196l 98k2 168km þ 59k2 98k þ 1 þ r1 þ r2 ! ! ffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi p k2 4l k2 4l k mþ coth A þ ðx þ by 2ktÞ 2 2 2 ð1 þ r1 þ r2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 49 h b 2kð1 þ 2r1 þ 2r2 þ r1 þ 2rp1 rffiffiffi2 þ r1 Þ 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
R1 ¼
! !1 92 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi = k k2 4l k2 4l mþ coth A þ ðx þ by 2ktÞ ; 2 2 2
Family 2: By using set 1 and Eq. (20) along with (8) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 42 b2kð1þ2r1 þ2r2 þr21 þ2r p1ffiffirffi 2 þr1 Þ ½18kb36kk14mbþ28mkþ 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 196l98k2 168kmþ59k2 98k þ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2 ð1þr1 þr2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 1 r2 þr1 Þ h49 b2kð1þ2r1 þ2r2 þr1 þ2rp ffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k ! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2
u2 ¼
A. Neirameh So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb2kÞ r1 2ðb2kÞ k m v2 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ r1 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2
þ
r1
r1 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2 And w2 ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ r2 2ðb2kÞ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ þ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2 r2
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tanh Aþ ðxþby2ktÞ mþ 2 2 2
The coupled nonlinear Maccari’s system Now from Eq. (16) we have 2
ð1 þ r1 þ r2 Þ 2k b ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ
R2 ¼
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi ½18kb 36kk 14mb þ 28mkþ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 1 þ r1 þ r2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi k k2 4l k2 4l tanh A þ ðx þ by 2ktÞ mþ 2 2 2
þ
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k þ
! !1 92 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi = k k2 4l k2 4l tanh A þ ðx þ by 2ktÞ mþ ; 2 2 2
3.2. Trigonometric function solutions Family 3: By using set 1 and Eq. (20) along with (9) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ k m u3 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
þ
ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k ! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
2843 So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb2kÞ r1 2ðb2kÞ k m v3 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ r1 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
þ
r1
r1 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2 And w3 ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ r2 2ðb2kÞ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
þ
r2
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot Aþ ðxþby2ktÞ mþ 2 2 2
2844
A. Neirameh
Now from Eq. (16) we have 2
ð1 þ r1 þ r2 Þ 2k b ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ
R3 ¼
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mkþ 6ð2k bÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 1 þ r1 þ r2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l cot A þ ðx þ by 2ktÞ mþ 2 2 2 þ
ð1 þ r1 þ r2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k ! !1 92 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi = k k2 4l k2 4l mþ cot A þ ðx þ by 2ktÞ ; 2 2 2
Family 4: By using set 1 and Eq. (20) along with (10) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ u4 ¼ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ r1 2ðb 2kÞ k m v4 ¼ 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi ½18kb 36kk 14mb þ 28mkþ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ þ 1 þ r1 þ r2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi k k2 4l k2 4l tan A ðx þ by 2ktÞ mþ 2 2 2 r1
r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ h pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ
þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
mþ And
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb2kÞ r2 2ðb2kÞ k m w4 ¼ 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ þ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tan A ðxþby2ktÞ mþ 2 2 2
þ
ð1þr1 þr2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tan A ðxþby2ktÞ mþ 2 2 2
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ffi k k2 4l k2 4l tan A ðx þ by 2ktÞ 2 2 2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tan A ðxþby2ktÞ mþ 2 2 2 r2
r2 ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k
! !1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k k2 4l k2 4l tan A ðxþby2ktÞ mþ 2 2 2 Now from Eq. (16) we have
The coupled nonlinear Maccari’s system ð1þr1 þr2 Þ2 2kb ( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 2ðb2kÞ k m 2ð1þr1 þr2 Þ 3ð1þr1 þr2 Þ
R4 ¼
ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb2kÞ 1þr1 þr2 ! ! pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi k2 4l k2 4l k mþ tan A ðxþby2ktÞ 2 2 2
þ
ð1þr1 þr2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b2kð1þ2r1 þ2r2 þr21 þ2r1 r2 þr21 Þ h pffiffiffi 18kb36kk14mbþ28mkþ 6ð2kbÞ
þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l98k2 168kmþ59k2 98k ! !1 92 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi = k2 4l k2 4l k mþ tan A ðxþby2ktÞ ; 2 2 2
3.3. Rational function solutions Family 5: By using set 1 and Eq. (20) along with (11) we have solutions of Eq. (19) as follows pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ k m u5 ¼ 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ
2845 So from (18) we have pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ r1 2ðb 2kÞ k m v5 ¼ 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ r1 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r1 2ðb 2kÞ B k þ mþ A þ Bðx þ by 2ktÞ 2 1 þ r1 þ r2 r1 ð1 þ r1 þ r2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k 1 B k mþ A þ Bðx þ by 2ktÞ 2 And w5 ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb 2kÞ r2 2ðb 2kÞ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ r2 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ ½18kb 36kk 14mb þ 28mk qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi þ 6ð2k bÞ 196l 98k2 168km þ 59k2 98k þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2 2ðb 2kÞ B k mþ 1 þ r1 þ r2 A þ Bðx þ by 2ktÞ 2
þ
r2 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ
½18kb 36kk 14mb þ 28mk qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi þ 6ð2k bÞ 196l 98k2 168km þ 59k2 98k mþ
1 B k A þ Bðx þ by 2ktÞ 2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ B k mþ þ A þ Bðx þ by 2ktÞ 2 1 þ r1 þ r2 ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ pffiffiffi 18kb 36kk 14mb þ 28mk þ 6ð2k bÞ
þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 196l 98k2 168km þ 59k2 98k
mþ
1 B k A þ Bðx þ by 2ktÞ 2
Figure 1 Soliton solution of u4 for k ¼ 1; k ¼ 0; b ¼ 1; l ¼ 0; r1 ¼ r2 ¼ 0 in region x ¼ 10 . . . 10; y ¼ 10::10.
2846
A. Neirameh
Figure 2 Soliton solution of u3 for k ¼ 1; k ¼ 0; b ¼ 1; l ¼ 0; r1 ¼ r2 ¼ 0 in region x ¼ 10 . . . 10; y ¼ 10::10. Figure 3 Soliton solution of u2 for k ¼ 1; k ¼ 0; b ¼ 1; l ¼ 0; r1 ¼ r2 ¼ 0 in region x ¼ 10 . . . 10; y ¼ 10::10.
Now from Eq. (16) we have ð1 þ r1 þ r2 Þ2 R5 ¼ 2k b
( pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ 2ðb 2kÞ k m 2ð1 þ r1 þ r2 Þ 3ð1 þ r1 þ r2 Þ
ð1 þ r1 þ r2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 42 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ
½18kb 36kk 14mb þ 28mk qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi þ 6ð2k bÞ 196l 98k2 168km þ 59k2 98k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ðb 2kÞ B k mþ þ A þ Bðx þ by 2ktÞ 2 1 þ r1 þ r2 ð1 þ r1 þ r2 Þ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 49 b 2kð1 þ 2r1 þ 2r2 þ r21 þ 2r1 r2 þ r21 Þ 18kb 36kk 14mb þ 28mk þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 6ð2k bÞ 196l 98k2 168km þ 59k2 98k
1 )2 B k mþ A þ Bðx þ by 2ktÞ 2
Figure 4 Soliton solution of u1 for k ¼ 1; k ¼ 0; b ¼ 1; l ¼ 0; r1 ¼ r2 ¼ 0 in region x ¼ 10 . . . 10; y ¼ 10::10.
Like above we can obtain the new class of exact solutions from set 2 similar set 1 (see Figs. 1–4). Acknowledgment 4. Conclusion In this paper, we successfully use the new extension of the (G0 / G)-expansion method for solving the coupled nonlinear Maccari’s system. We have successfully obtained some new exact traveling wave solutions of the coupled nonlinear Maccari’s system with parameters. We have given some figures expressing the behavior of the obtained solutions of Eq. (1) which give some perspective readers how the behavior solutions are produced.
The authors wish to thank the referees for their comments on this paper. References [1] A. Neirameh, Exact analytical solutions for 3D-Gross-Pitaevskii equation with periodic potential by using the Kudryashov method, J. Egypt. Math. Soc. (2016), http://dx.doi.org/10.1016/ j.joems.2014.11.004 (in press).
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