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Exact single traveling wave solutions to generalized (2+1)-dimensional Gardner equation with variable coefficients
Journal Pre-Proof Exact single traveling wave solutions to generalized (2+1)-dimensional Gardner equation with variable coefficients Yue Kai, Bailin Zheng, Nan Yang, Wenlong Xu PII: DOI: Reference:
S2211-3797(19)31630-4 https://doi.org/10.1016/j.rinp.2019.102527 RINP 102527
To appear in:
Results in Physics
Received Date: Accepted Date:
24 May 2019 18 July 2019
Please cite this article as: Kai, Y., Zheng, B., Yang, N., Xu, W., Exact single traveling wave solutions to generalized (2+1)-dimensional Gardner equation with variable coefficients, Results in Physics (2019), doi: https://doi.org/ 10.1016/j.rinp.2019.102527
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The Authors
JOURNAL PRE-PROOF
Exact single traveling wave solutions to generalized (2+1)-dimensional Gardner equation with variable coefficients
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July 23, 2019
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Yue Kai, Bailin Zheng*, Nan Yang, Wenlong Xu School of aerospace engineering and applied mechanics Tongji University Shanghai 200092, China Email: [email protected] ∗
Abstract
Introduction
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We firstly apply the trial equation method to generalized (2+1)-dimensional Gardner equation to reduce the nonlinear partial differential equation into ordinary equation. Then by the complete discrimination system for polynomial method, the classification of the exact single traveling wave solutions is presented. This is the first time that the two methods are applied to integral-differential equation. In the classification, we can see that the equation has rational function type solutions, solitary wave solutions, triangle function type periodic solutions and Jacobian elliptic functions solutions with double periodic, which are impossible to be obtained by other methods. Moreover, to ensure the existences of these solutions, concrete examples are also constructed with specific parameters. To the best of our knowledge, the results presented in the paper are brand new and can not be found in any other papers. Keywords: Exact traveling wave solutions; complete discrimination system for polynomial; trial equation method; generalized (2+1) dimensional Gardner equation.
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Nonlinear partial differential equations are arised from many scientific fields, such as physics, biology and so on. For hundreds of years, scientists devoted themselves to formulate methods to find the solutions to the nonlinear problems[1-4]. For example, Wenxiu Ma initially applied the reduced rational function method to Kolmogorov-PetrovskiiPiskunov equation and obtained exact traveling wave solutions to the equation[5]. For the reason that the numerical solutions are not exact, the exact solutions are appeared to be even more important in some special problems such as chaos, and they could also be helpful to understand the physical phenomena and models better. So constructing the exact solutions to the nonlinear differential equations is of great significance. Recently, lump solutions have also attracted a lot attention and many important results have been achieved[6-9]. Other new and important developments for searching for analytical traveling wave solutions for partial differential equations can be seen in [10-24]. In this paper, the complete discrimination system for polynomial method[25-30] and trail ∗ This work was supported by the National Natural Science Foundation of China under Grant No. 11872280.
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equation method[31-36] are applied to find the exact traveling wave solutions to generalized (2+1)-dimensional Gardner equation with variable coefficients which is merely solvable by other methods. To our knowledge, this is the first time that the methods are applied to integral-differential equation and the results in the present paper are all new. The concrete examples under the specific parameters are also represented to ensure the existences of each solutions. The KdV equation is of great importance and has broad applications in hydrodynamics[3739]. However, in some special situations, the KdV equation is helpless when we encounter with the situations that we must consider the cubic nonlinear terms, for example, the fluid in the vicinity of the critical velocity or at the critical density, and by taking the high-order terms into consideration, the modified KdV(mKdV) equation is formulated. However,sometimes the high order nonlinear terms also need to be considered, which leads us to combine the KdV equation with the mKdV equation, and as a result, the Gardner equation ut + αuux + βuxxx + γu2 ux = 0, (1)
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is obtained, where the coefficients in the equation are all constants. Gardner equation is widely used in various area in plasma and other fields of science [40-45]. Some special kind of exact solutions such as soliton solutions to the equation could be seen in Ref. [46]. Recently, scientists found that by generalizing the constant-coefficient equation into variable coefficient[47-79], then the equation has a wider range of application, for example, some special kinds of the equations with the distance-dependent coefficients in hydrodynamics[50-55]. In this paper, we consider generalized (2+1)-dimensional Gardener equation with variable coefficients given as follows [56] Z Z ut +α(x, y, t)uux +β(x, y, t)uxxx +γ(x, y, t)u2 ux +µ(x, y, t) uyy dx+λ(x, y, t)ux uy dx = 0,
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(2) where x, y are scaled space coordinates, and t is the time coordinate, the amplitude of the wave is u(x, y, t), and the variable coefficients such as α(x, y, t) are all given functions. Due to the significance of the equation(see [56] and references therein), studying the integrability and constructing the exact solutions to this problem are significant tasks. But unfortunately, Eq. (2) is not solvable in the general case, and the integrability of a simplified case is shown in [56]. The trial equation method is initially proposed by Liu to obtain the single traveling wave solutions to nonlinear partial differential equations, and this method is not only efficient to constant-coefficients equations, but also to variable-coefficients equations[32]. According to Ref. [32], we can see that by taking the following transformation u(x, y, t) = u(ξ), ξ = k1 (t)x + k2 (t)y + ... + ω(t),
(3)
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then we can reduce the variable coefficients equation M (t, x, y, u, ux , uy , ...) = 0,
(4)
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into ordinary equation with constant coefficients N (t, x, y, k1 , k2 , ω, u, u0 , u00 , ...) = 0.
(5)
(u0 )2 = F (u),
(6)
Then by taking and substituting Eq. (6) into Eq. (5), a simplified ordinary equation can be gotten, and the trial equation F (u) could be any kind of function. Upon obtaining the trail equation, the original equation can be written into the integral form as Z du p ±(ξ − ξ0 ) = , (7) F (u) 2
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Exact solutions to generalized (2+1) dimensional Gardner equation with variable coefficients
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where ξ0 is an integral constant. In particular, if F (u) is a polynomial function, then the complete discrimination system for polynomial method can be applied to obtain the classification of the single traveling wave solutions to the original equation. Recent results about the complete discrimination system for polynomial method can be seen in [57-64]. From all above, we can summarize the four steps of applying the two methods to obtain the exact traveling wave solutions to the nonlinear equations with variable-coefficients as follows: step 1 Take the variable coefficients traveling wave transformation to reduce the original equation into ordinary equation. step 2 Choose proper trial equation to obtain the ordinary equations system. step 3 Solve the equations system, and substitute it into the trial equation. Usually, we must impose some restrictions on the equation so that the equations system is solvable. step 4 Using the complete discrimination system for polynomial method to deal with the integral form of the original equation (7), and give the classification of single traveling wave solutions to the original equation.
To get the exact solutions to Eq. (2), we first rewrite it as
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ut + α(x, y, t)uux + β(x, y, t)uxxx + γ(x, y, t)u2 ux + µ(x, y, t)wy + λ(x, y, t)ux w = 0, (8) uy = wx ,
(9)
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and then take the following transformation
u(x, y, t) = u(ξ), w(x, y, t) = w(ξ), ξ = k1 (t)x + k2 (t)y + ω(t),
(10)
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where ki (t)(i = 1, 2, 3) are real functions, k2 (t)/k1 (t) = a and a is a constant. Substituting (10) into Eq. (9) yields w0 = au0 , (11) with the prime standing for d/dξ, then we have
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w = au,
(12)
Substituting (10) and (12) into (8), we can get dk1 dk2 dw 0 x+ y+ )u + k1 α(x, y, t)uu0 + β(x, y, t)k13 u000 + k1 γu2 u0 + dt dt dt ak2 µ(x, y, t)u0 + aλk1 uu0 = 0.
(13)
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(
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Now, we use trial equation method to deal with Eq. (13). Let (u0 )2 = an un + an−1 un−1 + ... + a1 u + a0 .
(14)
Then by the balance principle, we get n = 4. And from Eq. (13), three ordinary differential equations can be obtained dk1 dk2 dw x+ y+ + ak2 µ + a2 βk13 = 0, dt dt dt
(15)
αk1 + aλk1 + 3a3 βk13 = 0,
(16)
k1 γ + 6a4 βk13 = 0.
(17)
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JOURNAL PRE-PROOF Actually, this equations system is not integratable. So we must put restrictions on it. Setting a3 = 2b1 a4 , a2 = 6b2 a4 , where bi are constants, bi 6= 0(i = 1, 2) and µ = f (t)y,
(18)
α + aλ = b1 γ,
(19)
dk1 dk2 dw x+ y+ + a2 k2 µ = b2 k1 γ, dt dt dt γ = f0 (t)x + f1 (t)y + f2 (t),
(20) (21)
Z and
g(t)e−a2
R
Z
w(t) = Ab2
where g(t) = Ab2 f1 (t)e is reduced into
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f0 (t)dt
f (t)dt
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k2 (t) =
R
f0 (t)dt
,
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k1 (t) = Ae
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where f (t) and fi (t)(i = 0, 1, 2) are the given time-dependent functions. Then Eqs. (15-17) become dk1 = b2 k1 f0 (t), (22) dt dk2 + a2 k2 f (t) = b2 k1 f1 (t), (23) dt and dw = b2 k1 f2 (t). (24) dt From the equations above yields
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f2 (t)e
dtea2
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f0 (t)dt
f (t)dt
,
dt,
(25) (26)
(27)
and A is an integral constant. Then the original equation
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(u0 )2 = a4 u4 + a3 u3 + a2 u2 + a1 u + a0 ,
(28)
where a3 = 2b1 a4 , a2 = 6b2 a4 , and a1 , a0 are arbitrary constants. Let 1
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ϕ = (a4 ) 4 (u +
1 b1 ), ξ1 = (a4 ) 4 ξ, 2
(29)
and combine (29) with Eq. (28), we have
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where
ϕ2ξ1 = H(ϕ) = ϕ4 + pϕ2 + qϕ + r,
(30)
1 a2 a3 a2 a3 p = √ , q = ( 32 − + a1 )(a4 )− 4 , a4 8a4 2a4
(31)
and r=−
3a43 a2 a23 a1 a3 + − + a0 . 3 256a4 16a24 4a4
(32)
We denote D1 = 4, D2 = −p, D3 = −2p3 + 8pr − 9q 2 , 27 D4 = −p2 q 2 + 4p4 q + 36pq 2 r − 32p2 q 2 − q 4 + 64r3 , E2 = 9p2 − 32pr. 4 According to Ref. [26], we discuss nine cases in the following.
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(33)
JOURNAL PRE-PROOF Case 1 D2 < 0, D3 = 0 and D4 = 0, H(ϕ) has a pair of double conjugate complex roots, namely H(ϕ) = (ϕ2 + s2 )2 , (34) where s > 0. By use of Eq. (30), we can get ϕ = s tan(s(ξ1 − ξ0 )).
(35)
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For example if p = 1, q = 0 and r = 1, then we have s = 1 and the solution of Eq. (28) is given by ϕ = tan(ξ1 − ξ0 ). (36)
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Case 2 When D2 = D3 = D4 = 0. H(ϕ) has a real root zero of multiplicities four, which gives H(ϕ) = ϕ4 , (37)
dϕ 1 = − −1 . ϕ2 ϕ
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Z ξ1 − ξ0 = For example, when p = q = r = 0, we have
1 . ξ1 − ξ0
(38)
(39)
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ϕ=−
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by using Eq. (30), we have
if µ > ν, we have
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Case 3 When D2 > 0, D3 = D4 = 0 and E2 > 0. H(ϕ) has two double distinct real roots such that H(ϕ) = (ϕ − µ)2 (ϕ − ν)2 , (40)
Z
±(ξ1 − ξ0 ) =
dϕ 1 ϕ−µ = ln | |. (ϕ − µ)(ϕ − ν) µ−ν ϕ−ν
(41)
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For ϕ > µ or ω < ν, by use of (41) we have ϕ=
ν−µ ν−µ (µ − ν)(ξ1 − ξ0 ) +ν = [coth − 1] + ν, 2 2 e(µ−ν)(ξ1 −ξ0 ) − 1
(42)
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when ν < ϕ < µ, we have ϕ=
ν−µ ν−µ (µ − ν)(ξ1 − ξ0 ) +ν = [tanh − 1] + ν. 2 2 −e(µ−ν)(ξ1 −ξ0 ) − 1
(43)
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We can see that Eqs. (42-43) are two solitary wave solutions. For example, when w = 1 1 9 2 , β = − 8 , α = − 2 , we have p = −2, r = 0, and q = 1, then we can get the solution of original equation as 1 ϕ−1 ±(ξ1 − ξ0 ) = ln | |. (44) 2 ϕ+1 Case 4 D2 > 0, D3 = D4 = E2 = 0. H(ϕ) = (ϕ − l)3 (ϕ + 3l), then by integrating Eq. (30), we have s Z dϕ 1 ϕ + 3l p ±(ξ1 − ξ0 ) = = . (45) l ϕ−l (ϕ − l) (ϕ − l)(ϕ + 3l) When l > 0, ϕ > l or ϕ < −3l, we have ϕ=
4l + l, l2 (ξ1 − ξ0 )2 − 1 5
(46)
JOURNAL PRE-PROOF which is a rational solution. When l < 0, ϕ < l or ϕ > −3l, the corresponding solution is the same as (46). For example, when p = −6, q = 8 and r = −3, then we have l = 1, namely r Z dϕ ϕ+3 p ±(ξ1 − ξ0 ) = = , (47) ϕ−1 (ϕ − 1) (ϕ − 1)(ϕ + 3)
µ= √
3l , + s2
4l2
and p
4l2 + s2 − √
correspondingly, we have √
(50)
√
4l2 +s2 (ξ1 −ξ0 ) −µ+ √ (e± 4l2 +s2 (ξ1 −ξ0 )
4l2 + s2 (2 − µ) . − µ)2 − 1
(51)
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ϕ=
e±
3l2 , 4l2 + s2
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δ=
(49)
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where
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and the corresponding solution is given by (46). Case 5 D2 D3 < 0, and D4 = 0. H(ϕ) = (ϕ − l)2 [(ϕ + l)2 + s2 ] with the constants l and s real. By using Eq. (30), we have p Z µϕ + δ − (ϕ + l)2 + s2 dϕ 1 p ±(ξ1 − ξ0 ) = =√ ln | |, (48) ϕ−l 4l2 + s2 (ϕ − l) (ϕ + l)2 + s2
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(51) is also a solitary wave solution. For example, when p = −1, q = −2, and r = 2, we have l = 1, s = 1, then the solution of Eq. (13) is given by √ e±3(ξ1 −ξ0 ) − 1 + 4l2 + s2 ϕ= . (52) (e±3(ξ1 −ξ0 ) − 1)2 − 1 Case 6 Di > 0(i = 2, 3, 4). H(ω) has four distinct real roots, namely
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H(ϕ) = (ϕ − α1 )(ϕ − α2 )(ϕ − α3 )(ϕ − α4 ),
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where α1 > α2 > α3 > α4 . Then we have Z dϕ p ξ1 − ξ0 = . (ϕ − α1 )(ϕ − α2 )(ϕ − α3 )(ϕ − α4 )
(54)
(55)
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When α4 > 0, if ϕ > α1 or ϕ < α4 . By using Eq. (54), we can obtain √ (α1 −α3 )(α2 −α4 ) α2 (α1 − α4 ) sn2 ( (ξ1 − ξ0 ), m) − α1 (α2 − α4 ) 2 √ ϕ= , (α −α )(α −α ) 1 3 2 4 (α1 − α4 ) sn2 ( (ξ − ξ ), m) − (α − α ) 1 0 2 4 2
(53)
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and if α3 < ϕ < α2 , then we have √ (α1 −α3 )(α2 −α4 ) α4 (α2 − α3 ) sn2 ( (ξ1 − ξ0 ), m) − α3 (α2 − α4 ) 2 √ ω= , (α1 −α3 )(α2 −α4 ) (α2 − α3 ) sn2 ( (ξ − ξ ), m) − (α − α ) 1 0 2 4 2 1 −α4 )(α2 −α3 ) where m2 = (α (α1 −α3 )(α2 −α4 ) . For a4 < 0, if α1 > ϕ > α2 , similarly we can get √ (α1 −α3 )(α2 −α4 ) α3 (α1 − α2 ) sn2 ( (ξ1 − ξ0 ), m) − α2 (α1 − α3 ) 2 √ ϕ= , (α −α )(α −α ) 1 3 2 4 (α1 − α2 ) sn2 ( (ξ − ξ ), m) − (α − α ) 1 0 1 3 2
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(56)
(57)
JOURNAL PRE-PROOF and when α4 < ϕ < α3 , we have √ 2
ϕ=
α1 (α3 − α4 ) sn (
√
(α3 − α4 ) sn2 (
(α1 −α3 )(α2 −α4 ) (ξ1 2 (α1 −α3 )(α2 −α4 ) (ξ1 2
− ξ0 ), m) − α4 (α3 − α1 )
,
(58)
− ξ0 ), m) − (α3 − α1 )
√
9 (ξ1 √2 9 2 4 sn ( 2 (ξ1
− ξ0 ), 89 ) − 6 − ξ0 ), 89 ) − 3
.
(59)
Case 7 D2 D3 ≥ 0 and D4 < 0. H(ϕ) is given by
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H(ϕ) = (ϕ − µ)(ϕ − ν)((ϕ − l)2 + s2 ),
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ϕ=
4 sn2 (
F
1 −α2 )(α3 −α4 ) where m2 = (α (α1 −α3 )(α2 −α4 ) . The expressions (55-58) are elliptic functions solutions with double periodic. For instance, when p = −5, q = 0, r = 4 and ϕ > 2, then we have α1 = 2, α2 = 1, α3 = −1, α4 = −2, and ϕ is given by
Let
1 (µ − ν)d, 2 1 (µ − ν)c, 2 s c=µ−l− , m1
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1 (µ + ν)c − 2 1 b = (µ + ν)d − 2
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where µ > ν and s > 0. Then the following formula can be gotten Z dω p ±(ξ1 − ξ0 ) = . (ϕ − µ)(ϕ − ν)((ϕ − l)2 + s2 )
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a=
d = µ − l − sm1 ,
s2 + (µ − l)(ν − l) , s(µ − ν) p m1 = E + E 2 + 1.
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E=
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Then we can get the solutions to Eq. (13) as √ ∓2sm1 (µ−ν) a cn( (ξ1 − ξ0 ), m) + b √ 2mm1 ϕ= , ∓2sm1 (µ−ν) c cn( (ξ1 − ξ0 ), m) + d 2mm1
(60)
(61)
(62) (63) (64) (65) (66) (67)
(68)
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1 where m2 = 1+m 2 , and (68) is also an elliptic function solution. For instance, when p = 3, 1 q = −4, then µ = 1, ν = −1 and l = 0, s = 2, then
a = −d = 3, b = c = 0, E =
3 1 , m1 = 2, m = , 4 5
by taking them into (68), we can obtain the solution as p ∓2sm1 (µ − ν) ϕ = − cn( (ξ1 − ξ0 ), m), 2mm1
(69)
(70)
Case 8 D2 D3 ≤ 0 and D4 > 0. H(ϕ) has two pairs of conjugate complex roots, i.e. H(ϕ) = ((ϕ − l1 )2 + s21 )((ϕ − l1 )2 + s22 ),
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(71)
JOURNAL PRE-PROOF where s1 ≥ s2 > 0. Let a = l1 c + s1 d,
(72)
b = l1 d − s1 c, s2 c = −s1 − , m1
(73)
d = l1 − l2 ,
(75)
(74)
(l1 − l2 )2 + s21 + s22 , 2s1 s2 p m1 = E + E 2 − 1.
(76)
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E=
(77)
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Then we have
a sn(η(ξ1 − ξ0 ), m) + b cn(η(ξ1 − ξ0 ), m) , (78) c sn(η(ξ1 − ξ0 ), m) + d cn(η(ξ1 − ξ0 ), m) √ s2 (c2 +d2 )(m21 c2 +d2 ) m21 −1 2 where m = m2 and η = . (78) is also an elliptic functions solution. c2 +d2 1 For example, when p = 5, q = 4, we have l1 = l2 = 0, s1 = 1, s2 = 2, then we get that
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ϕ=
a = d = 0, b = c = −2, η = 4, m = so ϕ is as follows
cn(4(ξ1 − ξ0 ), 34 ) . sn(4(ξ1 − ξ0 ), 43 )
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ϕ=
3 , 4
(79)
(80)
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Case 9 D2 , D3 > 0, and D4 = 0. H(ϕ) has two single real roots and a real roots with multiplicities two, i.e. H(ϕ) = (ϕ − α1 )(ϕ − α2 )(ϕ − α3 )2 . (81) where α1 > α2 , and αi is unequal to each other. Let α1 − α2 α1 + α2 ( − α3 ). 2 2
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c=
(82)
From Eq. (30), we have
Z
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±(ξ1 − ξ0 ) =
dϕ p . (ϕ − α3 ) ϕ − α1 )(ϕ − α2 )
(83)
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Because of the relations between α1 , α2 and α3 , c 6= ±1. If c2 − 1 > 0, by using Eq. (81), we have 1 y − c1 (ξ1 − ξ0 ) = − √ ln | |, (84) 2 y + c1 c −1
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and when c2 − 1 < 0, we can obtain
where c1 =
q
c+1 c−1
p c+1 (ξ1 − ξ0 ) = − (1 − c2 ) arctan y, (85) 1−c q α1 −α2 and y = 1 − α1 +α2 α1 −α2 . For example, when p = −1, q = 0, (ϕ−
2
)+
2
we have α1 = 1, α2 = −1, and α3 = 0, which yields ϕ=
1 . cos 2(ξ1 − ξ0 )
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(86)
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Conclusion
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In this paper, we apply two methods, namely the trial equation method and the complete discrimination system for polynomial method, to solve generalized (2+1) dimensional Gardner equation with variable coefficients. The classification of the solution is given, including solitary solutions, rational solution and elliptic double periodic functions solution, which are impossible to get by other methods. To the best of our knowledge, our results are all new and we haven’t seen any similar conclusions in the published papers. Considering the significance of the generalized (2+1) dimensional Gardner equation in mathematics and physics, it is an important job to give the exact solutions to this equation as being done in the present paper.
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S2211-3797(19)31630-4 https://doi.org/10.1016/j.rinp.2019.102527 RINP 102527
To appear in:
Results in Physics
Received Date: Accepted Date:
24 May 2019 18 July 2019
Please cite this article as: Kai, Y., Zheng, B., Yang, N., Xu, W., Exact single traveling wave solutions to generalized (2+1)-dimensional Gardner equation with variable coefficients, Results in Physics (2019), doi: https://doi.org/ 10.1016/j.rinp.2019.102527
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The Authors
JOURNAL PRE-PROOF
Exact single traveling wave solutions to generalized (2+1)-dimensional Gardner equation with variable coefficients
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July 23, 2019
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Yue Kai, Bailin Zheng*, Nan Yang, Wenlong Xu School of aerospace engineering and applied mechanics Tongji University Shanghai 200092, China Email: [email protected] ∗
Abstract
Introduction
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We firstly apply the trial equation method to generalized (2+1)-dimensional Gardner equation to reduce the nonlinear partial differential equation into ordinary equation. Then by the complete discrimination system for polynomial method, the classification of the exact single traveling wave solutions is presented. This is the first time that the two methods are applied to integral-differential equation. In the classification, we can see that the equation has rational function type solutions, solitary wave solutions, triangle function type periodic solutions and Jacobian elliptic functions solutions with double periodic, which are impossible to be obtained by other methods. Moreover, to ensure the existences of these solutions, concrete examples are also constructed with specific parameters. To the best of our knowledge, the results presented in the paper are brand new and can not be found in any other papers. Keywords: Exact traveling wave solutions; complete discrimination system for polynomial; trial equation method; generalized (2+1) dimensional Gardner equation.
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Nonlinear partial differential equations are arised from many scientific fields, such as physics, biology and so on. For hundreds of years, scientists devoted themselves to formulate methods to find the solutions to the nonlinear problems[1-4]. For example, Wenxiu Ma initially applied the reduced rational function method to Kolmogorov-PetrovskiiPiskunov equation and obtained exact traveling wave solutions to the equation[5]. For the reason that the numerical solutions are not exact, the exact solutions are appeared to be even more important in some special problems such as chaos, and they could also be helpful to understand the physical phenomena and models better. So constructing the exact solutions to the nonlinear differential equations is of great significance. Recently, lump solutions have also attracted a lot attention and many important results have been achieved[6-9]. Other new and important developments for searching for analytical traveling wave solutions for partial differential equations can be seen in [10-24]. In this paper, the complete discrimination system for polynomial method[25-30] and trail ∗ This work was supported by the National Natural Science Foundation of China under Grant No. 11872280.
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equation method[31-36] are applied to find the exact traveling wave solutions to generalized (2+1)-dimensional Gardner equation with variable coefficients which is merely solvable by other methods. To our knowledge, this is the first time that the methods are applied to integral-differential equation and the results in the present paper are all new. The concrete examples under the specific parameters are also represented to ensure the existences of each solutions. The KdV equation is of great importance and has broad applications in hydrodynamics[3739]. However, in some special situations, the KdV equation is helpless when we encounter with the situations that we must consider the cubic nonlinear terms, for example, the fluid in the vicinity of the critical velocity or at the critical density, and by taking the high-order terms into consideration, the modified KdV(mKdV) equation is formulated. However,sometimes the high order nonlinear terms also need to be considered, which leads us to combine the KdV equation with the mKdV equation, and as a result, the Gardner equation ut + αuux + βuxxx + γu2 ux = 0, (1)
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is obtained, where the coefficients in the equation are all constants. Gardner equation is widely used in various area in plasma and other fields of science [40-45]. Some special kind of exact solutions such as soliton solutions to the equation could be seen in Ref. [46]. Recently, scientists found that by generalizing the constant-coefficient equation into variable coefficient[47-79], then the equation has a wider range of application, for example, some special kinds of the equations with the distance-dependent coefficients in hydrodynamics[50-55]. In this paper, we consider generalized (2+1)-dimensional Gardener equation with variable coefficients given as follows [56] Z Z ut +α(x, y, t)uux +β(x, y, t)uxxx +γ(x, y, t)u2 ux +µ(x, y, t) uyy dx+λ(x, y, t)ux uy dx = 0,
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(2) where x, y are scaled space coordinates, and t is the time coordinate, the amplitude of the wave is u(x, y, t), and the variable coefficients such as α(x, y, t) are all given functions. Due to the significance of the equation(see [56] and references therein), studying the integrability and constructing the exact solutions to this problem are significant tasks. But unfortunately, Eq. (2) is not solvable in the general case, and the integrability of a simplified case is shown in [56]. The trial equation method is initially proposed by Liu to obtain the single traveling wave solutions to nonlinear partial differential equations, and this method is not only efficient to constant-coefficients equations, but also to variable-coefficients equations[32]. According to Ref. [32], we can see that by taking the following transformation u(x, y, t) = u(ξ), ξ = k1 (t)x + k2 (t)y + ... + ω(t),
(3)
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then we can reduce the variable coefficients equation M (t, x, y, u, ux , uy , ...) = 0,
(4)
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into ordinary equation with constant coefficients N (t, x, y, k1 , k2 , ω, u, u0 , u00 , ...) = 0.
(5)
(u0 )2 = F (u),
(6)
Then by taking and substituting Eq. (6) into Eq. (5), a simplified ordinary equation can be gotten, and the trial equation F (u) could be any kind of function. Upon obtaining the trail equation, the original equation can be written into the integral form as Z du p ±(ξ − ξ0 ) = , (7) F (u) 2
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Exact solutions to generalized (2+1) dimensional Gardner equation with variable coefficients
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2
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where ξ0 is an integral constant. In particular, if F (u) is a polynomial function, then the complete discrimination system for polynomial method can be applied to obtain the classification of the single traveling wave solutions to the original equation. Recent results about the complete discrimination system for polynomial method can be seen in [57-64]. From all above, we can summarize the four steps of applying the two methods to obtain the exact traveling wave solutions to the nonlinear equations with variable-coefficients as follows: step 1 Take the variable coefficients traveling wave transformation to reduce the original equation into ordinary equation. step 2 Choose proper trial equation to obtain the ordinary equations system. step 3 Solve the equations system, and substitute it into the trial equation. Usually, we must impose some restrictions on the equation so that the equations system is solvable. step 4 Using the complete discrimination system for polynomial method to deal with the integral form of the original equation (7), and give the classification of single traveling wave solutions to the original equation.
To get the exact solutions to Eq. (2), we first rewrite it as
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ut + α(x, y, t)uux + β(x, y, t)uxxx + γ(x, y, t)u2 ux + µ(x, y, t)wy + λ(x, y, t)ux w = 0, (8) uy = wx ,
(9)
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and then take the following transformation
u(x, y, t) = u(ξ), w(x, y, t) = w(ξ), ξ = k1 (t)x + k2 (t)y + ω(t),
(10)
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where ki (t)(i = 1, 2, 3) are real functions, k2 (t)/k1 (t) = a and a is a constant. Substituting (10) into Eq. (9) yields w0 = au0 , (11) with the prime standing for d/dξ, then we have
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w = au,
(12)
Substituting (10) and (12) into (8), we can get dk1 dk2 dw 0 x+ y+ )u + k1 α(x, y, t)uu0 + β(x, y, t)k13 u000 + k1 γu2 u0 + dt dt dt ak2 µ(x, y, t)u0 + aλk1 uu0 = 0.
(13)
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(
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Now, we use trial equation method to deal with Eq. (13). Let (u0 )2 = an un + an−1 un−1 + ... + a1 u + a0 .
(14)
Then by the balance principle, we get n = 4. And from Eq. (13), three ordinary differential equations can be obtained dk1 dk2 dw x+ y+ + ak2 µ + a2 βk13 = 0, dt dt dt
(15)
αk1 + aλk1 + 3a3 βk13 = 0,
(16)
k1 γ + 6a4 βk13 = 0.
(17)
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JOURNAL PRE-PROOF Actually, this equations system is not integratable. So we must put restrictions on it. Setting a3 = 2b1 a4 , a2 = 6b2 a4 , where bi are constants, bi 6= 0(i = 1, 2) and µ = f (t)y,
(18)
α + aλ = b1 γ,
(19)
dk1 dk2 dw x+ y+ + a2 k2 µ = b2 k1 γ, dt dt dt γ = f0 (t)x + f1 (t)y + f2 (t),
(20) (21)
Z and
g(t)e−a2
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Z
w(t) = Ab2
where g(t) = Ab2 f1 (t)e is reduced into
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f0 (t)dt
f (t)dt
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k2 (t) =
R
f0 (t)dt
,
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k1 (t) = Ae
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F
where f (t) and fi (t)(i = 0, 1, 2) are the given time-dependent functions. Then Eqs. (15-17) become dk1 = b2 k1 f0 (t), (22) dt dk2 + a2 k2 f (t) = b2 k1 f1 (t), (23) dt and dw = b2 k1 f2 (t). (24) dt From the equations above yields
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f2 (t)e
dtea2
R
f0 (t)dt
f (t)dt
,
dt,
(25) (26)
(27)
and A is an integral constant. Then the original equation
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(u0 )2 = a4 u4 + a3 u3 + a2 u2 + a1 u + a0 ,
(28)
where a3 = 2b1 a4 , a2 = 6b2 a4 , and a1 , a0 are arbitrary constants. Let 1
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ϕ = (a4 ) 4 (u +
1 b1 ), ξ1 = (a4 ) 4 ξ, 2
(29)
and combine (29) with Eq. (28), we have
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where
ϕ2ξ1 = H(ϕ) = ϕ4 + pϕ2 + qϕ + r,
(30)
1 a2 a3 a2 a3 p = √ , q = ( 32 − + a1 )(a4 )− 4 , a4 8a4 2a4
(31)
and r=−
3a43 a2 a23 a1 a3 + − + a0 . 3 256a4 16a24 4a4
(32)
We denote D1 = 4, D2 = −p, D3 = −2p3 + 8pr − 9q 2 , 27 D4 = −p2 q 2 + 4p4 q + 36pq 2 r − 32p2 q 2 − q 4 + 64r3 , E2 = 9p2 − 32pr. 4 According to Ref. [26], we discuss nine cases in the following.
4
(33)
JOURNAL PRE-PROOF Case 1 D2 < 0, D3 = 0 and D4 = 0, H(ϕ) has a pair of double conjugate complex roots, namely H(ϕ) = (ϕ2 + s2 )2 , (34) where s > 0. By use of Eq. (30), we can get ϕ = s tan(s(ξ1 − ξ0 )).
(35)
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For example if p = 1, q = 0 and r = 1, then we have s = 1 and the solution of Eq. (28) is given by ϕ = tan(ξ1 − ξ0 ). (36)
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Case 2 When D2 = D3 = D4 = 0. H(ϕ) has a real root zero of multiplicities four, which gives H(ϕ) = ϕ4 , (37)
dϕ 1 = − −1 . ϕ2 ϕ
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Z ξ1 − ξ0 = For example, when p = q = r = 0, we have
1 . ξ1 − ξ0
(38)
(39)
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ϕ=−
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by using Eq. (30), we have
if µ > ν, we have
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Case 3 When D2 > 0, D3 = D4 = 0 and E2 > 0. H(ϕ) has two double distinct real roots such that H(ϕ) = (ϕ − µ)2 (ϕ − ν)2 , (40)
Z
±(ξ1 − ξ0 ) =
dϕ 1 ϕ−µ = ln | |. (ϕ − µ)(ϕ − ν) µ−ν ϕ−ν
(41)
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For ϕ > µ or ω < ν, by use of (41) we have ϕ=
ν−µ ν−µ (µ − ν)(ξ1 − ξ0 ) +ν = [coth − 1] + ν, 2 2 e(µ−ν)(ξ1 −ξ0 ) − 1
(42)
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when ν < ϕ < µ, we have ϕ=
ν−µ ν−µ (µ − ν)(ξ1 − ξ0 ) +ν = [tanh − 1] + ν. 2 2 −e(µ−ν)(ξ1 −ξ0 ) − 1
(43)
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We can see that Eqs. (42-43) are two solitary wave solutions. For example, when w = 1 1 9 2 , β = − 8 , α = − 2 , we have p = −2, r = 0, and q = 1, then we can get the solution of original equation as 1 ϕ−1 ±(ξ1 − ξ0 ) = ln | |. (44) 2 ϕ+1 Case 4 D2 > 0, D3 = D4 = E2 = 0. H(ϕ) = (ϕ − l)3 (ϕ + 3l), then by integrating Eq. (30), we have s Z dϕ 1 ϕ + 3l p ±(ξ1 − ξ0 ) = = . (45) l ϕ−l (ϕ − l) (ϕ − l)(ϕ + 3l) When l > 0, ϕ > l or ϕ < −3l, we have ϕ=
4l + l, l2 (ξ1 − ξ0 )2 − 1 5
(46)
JOURNAL PRE-PROOF which is a rational solution. When l < 0, ϕ < l or ϕ > −3l, the corresponding solution is the same as (46). For example, when p = −6, q = 8 and r = −3, then we have l = 1, namely r Z dϕ ϕ+3 p ±(ξ1 − ξ0 ) = = , (47) ϕ−1 (ϕ − 1) (ϕ − 1)(ϕ + 3)
µ= √
3l , + s2
4l2
and p
4l2 + s2 − √
correspondingly, we have √
(50)
√
4l2 +s2 (ξ1 −ξ0 ) −µ+ √ (e± 4l2 +s2 (ξ1 −ξ0 )
4l2 + s2 (2 − µ) . − µ)2 − 1
(51)
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ϕ=
e±
3l2 , 4l2 + s2
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δ=
(49)
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where
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and the corresponding solution is given by (46). Case 5 D2 D3 < 0, and D4 = 0. H(ϕ) = (ϕ − l)2 [(ϕ + l)2 + s2 ] with the constants l and s real. By using Eq. (30), we have p Z µϕ + δ − (ϕ + l)2 + s2 dϕ 1 p ±(ξ1 − ξ0 ) = =√ ln | |, (48) ϕ−l 4l2 + s2 (ϕ − l) (ϕ + l)2 + s2
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(51) is also a solitary wave solution. For example, when p = −1, q = −2, and r = 2, we have l = 1, s = 1, then the solution of Eq. (13) is given by √ e±3(ξ1 −ξ0 ) − 1 + 4l2 + s2 ϕ= . (52) (e±3(ξ1 −ξ0 ) − 1)2 − 1 Case 6 Di > 0(i = 2, 3, 4). H(ω) has four distinct real roots, namely
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H(ϕ) = (ϕ − α1 )(ϕ − α2 )(ϕ − α3 )(ϕ − α4 ),
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where α1 > α2 > α3 > α4 . Then we have Z dϕ p ξ1 − ξ0 = . (ϕ − α1 )(ϕ − α2 )(ϕ − α3 )(ϕ − α4 )
(54)
(55)
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When α4 > 0, if ϕ > α1 or ϕ < α4 . By using Eq. (54), we can obtain √ (α1 −α3 )(α2 −α4 ) α2 (α1 − α4 ) sn2 ( (ξ1 − ξ0 ), m) − α1 (α2 − α4 ) 2 √ ϕ= , (α −α )(α −α ) 1 3 2 4 (α1 − α4 ) sn2 ( (ξ − ξ ), m) − (α − α ) 1 0 2 4 2
(53)
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and if α3 < ϕ < α2 , then we have √ (α1 −α3 )(α2 −α4 ) α4 (α2 − α3 ) sn2 ( (ξ1 − ξ0 ), m) − α3 (α2 − α4 ) 2 √ ω= , (α1 −α3 )(α2 −α4 ) (α2 − α3 ) sn2 ( (ξ − ξ ), m) − (α − α ) 1 0 2 4 2 1 −α4 )(α2 −α3 ) where m2 = (α (α1 −α3 )(α2 −α4 ) . For a4 < 0, if α1 > ϕ > α2 , similarly we can get √ (α1 −α3 )(α2 −α4 ) α3 (α1 − α2 ) sn2 ( (ξ1 − ξ0 ), m) − α2 (α1 − α3 ) 2 √ ϕ= , (α −α )(α −α ) 1 3 2 4 (α1 − α2 ) sn2 ( (ξ − ξ ), m) − (α − α ) 1 0 1 3 2
6
(56)
(57)
JOURNAL PRE-PROOF and when α4 < ϕ < α3 , we have √ 2
ϕ=
α1 (α3 − α4 ) sn (
√
(α3 − α4 ) sn2 (
(α1 −α3 )(α2 −α4 ) (ξ1 2 (α1 −α3 )(α2 −α4 ) (ξ1 2
− ξ0 ), m) − α4 (α3 − α1 )
,
(58)
− ξ0 ), m) − (α3 − α1 )
√
9 (ξ1 √2 9 2 4 sn ( 2 (ξ1
− ξ0 ), 89 ) − 6 − ξ0 ), 89 ) − 3
.
(59)
Case 7 D2 D3 ≥ 0 and D4 < 0. H(ϕ) is given by
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H(ϕ) = (ϕ − µ)(ϕ − ν)((ϕ − l)2 + s2 ),
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ϕ=
4 sn2 (
F
1 −α2 )(α3 −α4 ) where m2 = (α (α1 −α3 )(α2 −α4 ) . The expressions (55-58) are elliptic functions solutions with double periodic. For instance, when p = −5, q = 0, r = 4 and ϕ > 2, then we have α1 = 2, α2 = 1, α3 = −1, α4 = −2, and ϕ is given by
Let
1 (µ − ν)d, 2 1 (µ − ν)c, 2 s c=µ−l− , m1
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1 (µ + ν)c − 2 1 b = (µ + ν)d − 2
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where µ > ν and s > 0. Then the following formula can be gotten Z dω p ±(ξ1 − ξ0 ) = . (ϕ − µ)(ϕ − ν)((ϕ − l)2 + s2 )
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a=
d = µ − l − sm1 ,
s2 + (µ − l)(ν − l) , s(µ − ν) p m1 = E + E 2 + 1.
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E=
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Then we can get the solutions to Eq. (13) as √ ∓2sm1 (µ−ν) a cn( (ξ1 − ξ0 ), m) + b √ 2mm1 ϕ= , ∓2sm1 (µ−ν) c cn( (ξ1 − ξ0 ), m) + d 2mm1
(60)
(61)
(62) (63) (64) (65) (66) (67)
(68)
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1 where m2 = 1+m 2 , and (68) is also an elliptic function solution. For instance, when p = 3, 1 q = −4, then µ = 1, ν = −1 and l = 0, s = 2, then
a = −d = 3, b = c = 0, E =
3 1 , m1 = 2, m = , 4 5
by taking them into (68), we can obtain the solution as p ∓2sm1 (µ − ν) ϕ = − cn( (ξ1 − ξ0 ), m), 2mm1
(69)
(70)
Case 8 D2 D3 ≤ 0 and D4 > 0. H(ϕ) has two pairs of conjugate complex roots, i.e. H(ϕ) = ((ϕ − l1 )2 + s21 )((ϕ − l1 )2 + s22 ),
7
(71)
JOURNAL PRE-PROOF where s1 ≥ s2 > 0. Let a = l1 c + s1 d,
(72)
b = l1 d − s1 c, s2 c = −s1 − , m1
(73)
d = l1 − l2 ,
(75)
(74)
(l1 − l2 )2 + s21 + s22 , 2s1 s2 p m1 = E + E 2 − 1.
(76)
F
E=
(77)
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Then we have
a sn(η(ξ1 − ξ0 ), m) + b cn(η(ξ1 − ξ0 ), m) , (78) c sn(η(ξ1 − ξ0 ), m) + d cn(η(ξ1 − ξ0 ), m) √ s2 (c2 +d2 )(m21 c2 +d2 ) m21 −1 2 where m = m2 and η = . (78) is also an elliptic functions solution. c2 +d2 1 For example, when p = 5, q = 4, we have l1 = l2 = 0, s1 = 1, s2 = 2, then we get that
PR
O
ϕ=
a = d = 0, b = c = −2, η = 4, m = so ϕ is as follows
cn(4(ξ1 − ξ0 ), 34 ) . sn(4(ξ1 − ξ0 ), 43 )
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ϕ=
3 , 4
(79)
(80)
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Case 9 D2 , D3 > 0, and D4 = 0. H(ϕ) has two single real roots and a real roots with multiplicities two, i.e. H(ϕ) = (ϕ − α1 )(ϕ − α2 )(ϕ − α3 )2 . (81) where α1 > α2 , and αi is unequal to each other. Let α1 − α2 α1 + α2 ( − α3 ). 2 2
AL
c=
(82)
From Eq. (30), we have
Z
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±(ξ1 − ξ0 ) =
dϕ p . (ϕ − α3 ) ϕ − α1 )(ϕ − α2 )
(83)
U
Because of the relations between α1 , α2 and α3 , c 6= ±1. If c2 − 1 > 0, by using Eq. (81), we have 1 y − c1 (ξ1 − ξ0 ) = − √ ln | |, (84) 2 y + c1 c −1
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and when c2 − 1 < 0, we can obtain
where c1 =
q
c+1 c−1
p c+1 (ξ1 − ξ0 ) = − (1 − c2 ) arctan y, (85) 1−c q α1 −α2 and y = 1 − α1 +α2 α1 −α2 . For example, when p = −1, q = 0, (ϕ−
2
)+
2
we have α1 = 1, α2 = −1, and α3 = 0, which yields ϕ=
1 . cos 2(ξ1 − ξ0 )
8
(86)
JOURNAL PRE-PROOF
3
Conclusion
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F
In this paper, we apply two methods, namely the trial equation method and the complete discrimination system for polynomial method, to solve generalized (2+1) dimensional Gardner equation with variable coefficients. The classification of the solution is given, including solitary solutions, rational solution and elliptic double periodic functions solution, which are impossible to get by other methods. To the best of our knowledge, our results are all new and we haven’t seen any similar conclusions in the published papers. Considering the significance of the generalized (2+1) dimensional Gardner equation in mathematics and physics, it is an important job to give the exact solutions to this equation as being done in the present paper.
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