Applied Mathematics and Computation 280 (2016) 39–45
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On solutions of generalized modified Korteweg–de Vries equation of the fifth order with dissipation Nikolay A. Kudryashov∗ National Research Nuclear University MEPhI (Moscow Engineering Physics Institute), 31 Kashirskoe Shosse, Moscow 115409, Russian Federation
a r t i c l e
i n f o
a b s t r a c t
Keywords: Korteweg–de Vries equation of the fifth order Painlevé test Painlevé property Elliptic solution Exact solution
The generalized modified Korteweg–de Vries equation of the fifth order with dissipation is considered. The Painlevé test is applied for studying integrability of this equation. It is shown that the generalized modified Korteweg–de Vries equation of the fifth order does not pass the Painlevé test in the general case but has the expansion of the solution in the Laurent series. As a consequence the equation can have some exact solutions at additional conditions on the parameters of the equation. We present the effective modification of methods for finding of solitary wave and elliptic solutions of nonlinear differential equations. Solitary wave and elliptic solutions of the generalized modified Korteweg–de Vries equation of the fifth order are found by means of expansion for solution in the Laurent series. These solutions can be used for description of nonlinear waves in the medium with dissipation, dispersion. © 2016 Elsevier Inc. All rights reserved.
1. Introduction In this paper we consider the equation in the form
ut − 10 u2 uxxx − 40 u ux uxx − 10 u3x + uxxxxx + 30 u4 ux + δ uxxxx + β uxxx + α uxx = 0,
(1)
where α , β , and δ are constant parameters of equation. Eq. (1) is the generalized modified Korteweg–de Vries equation of the fifth order with processes of a dissipation, dispersion and instability. At α = 0, β = 0 and δ = 0 Eq. (1) is the famous modified fifth-order Korteweg–de Vries equation. We have obtained from Eq. (1) by multiplying equation on u, and integration with respect to x from −∞ to ∞ the following equality
1 d 2 dt
∞
−∞
u2 dx = α
∞
−∞
u2x dx − δ
∞
−∞
u2xx dx.
(2)
One can see from (2) that at α < 0 and δ > 0 we have the description of the dissipative process but in the case of α > 0 and δ < 0 we obtain the active process in Eq. (1). It is known that the modified Korteweg–de Vries equation of the fifth order is used for description of nonlinear wave processes on water [1]. We can assume that the generalized modified Korteweg–de Vries equation can also be applied for description of nonlinear waves when we need to take into account the dissipation and instability. ∗
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http://dx.doi.org/10.1016/j.amc.2016.01.032 0096-3003/© 2016 Elsevier Inc. All rights reserved.
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The aim of this paper is to investigate the analytical properties of Eq. (1). First of all we are going to study the integrability of this equation using the Painlevé test. In fact we can guess that Eq. (1) is not integrable in the general case, but the Painlevé test may allow us to obtain additional conditions for existence of some exact solutions for Eq. (1). The outline of this manuscript is the following. In Section 2 we apply the Painlevé test to analyze Eq. (1). Solitary wave solutions of this equation are found in Section 3. In Section 4 we present a method for finding doubly periodic solution of Eq. (1) and try to look for elliptic solutions of this equation. 2. Painlevé analysis of the generalized modified Korteweg–de Vries equation of the fifth order Let us study Eq. (1) using the traveling wave solution
u(x, t ) = w(z ),
z = x + C0 t.
(3)
Substituting (3) into Eq. (1) and integrating expression with respect to z, we get the nonlinear differential equation in the form
wzzzz − 10ww2z − 10w2 wzz + 6w5 + C0 w + C1 + α wz + β wzz + δ wzzz = 0,
(4)
where C1 is a constant of integration. The equation with the leading members is found from Eq. (4). It has the form
wzzzz − 10ww2z − 10w2 wzz + 6w5 = 0.
(5)
Substituting
w (z ) =
a0 . zp
(6)
into all terms of Eq. (5) we obtain that there are four branches of the expansion of the solution in the Laurent series. They are
a0(1) , p = (1, 1 ), a0(3) , p = (2, 1 ),
a0(2) , p = (−1, 1 ),
a0(4) , p = (−2, 1 ).
(7)
To find the Fuchs indices of expansions, we substitute the expression
w (z ) =
a0(1,2) + w j z j−1 . z
(8)
into Eq. (5) again. Equating expressions at w j to zero, we obtain the Fuchs indices in the form:
j1 = −1,
j2 = 2,
j3 = 3,
j4 = 6.
(9)
Using the expression
w (z ) =
a0(3,4) + w j z j−1 . z
(10)
in Eq. (5), we have the Fuchs indices in the form:
j1 = −1,
j2 = −3,
j3 = 6,
j4 = 8.
(11)
Taking these Fuchs indices into consideration, we see that Eq. (4) can pass the Painlevé test. However we have to check the third step of the Painlevé test for Eq. (4). With this aim we substitute the truncated expansion for solution of Eq. (4) in the Laurent series in the form:
w=
1 + a1 + a2 z + a3 z2 + a4 z3 + a5 z4 + a6 z5 + · · · z
(12)
One can note that there is symmetry of Eq. (4). It is determined by transformation w(z ) → −w(z ). As a result of this symmetry we can consider further only two branches of solutions with a0(1 ) and a0(3 ) . Substituting expansion (12) into Eq. (4) we have:
3 , 10
(13)
δ 5 = 0,
(14)
a1 = −
β+ 81 50
9 5
δ 3 + α + 30 δ a2 = 0,
a4 =
5 2 6 a2 − 2 5
δ a3 +
1 81 C0 + 20 100
(15)
δ 2 a2 +
243 20000
δ4,
(16)
N. A. Kudryashov / Applied Mathematics and Computation 280 (2016) 39–45
a5 = −
17 12
δ a2 2 +
1 18
β a3 +
1 36
α a2 +
5 81 a2 a3 − 3 200000
δ5 −
4 3 729 243 7 C0 δ 2 + δ C1 + β C0 + 20 δ a2 a3 + β δ4 + 50 3 10 10000 50 1107 5589 + 3 δ 3 a3 + δ 4 a2 − 41 δ 2 a2 2 + δ 6 = 0. 250 50000
1 1 9 C0 δ + C1 − 120 36 100
β δ 2 a2 +
4 3
δ α a2 −
68 15
41
δ 3 a2 +
2 5
δ 2 a3 ,
(17)
δ β a3 + 2 α a3 + 15 β a2 2 (18)
Taking the Fuchs indices into account, we hope to obtain the coefficients a2 , a3 and a6 as arbitrary constants but this is not the case. We can see that there is no integrability of Eq. (4) because we cannot take these coefficients in the form of arbitrary constants. However, we can have the partial integrability and as a result we can have some exact solutions. It is possible when
9 5
β = − δ2.
(19)
In this case the coefficient a2 can be taken as arbitrary constant. However the coefficient a3 is not arbitrary in this case. For arbitrary constant a3 we can take the coefficient a2 in the form:
a2 = −
27 500
δ2 −
1 30
α . δ
(20)
Taking the conditions (19) and (20) into account, we have the condition for the coefficient a6 in the form:
2 δ2 252 4δ C1 − C0 + 3 5 25
δ 3 a3 +
4 3
α a3 +
108 243 δ 6 − 15625 625
δ3α −
3 α2 = 0. 25
(21)
From (21) we find the value of coefficient a3 in the form:
a3 =
18750 C0 δ 2 − 62500 δ C1 + 5625 α 2 − 729 δ 6 + 8100 δ 3 α . 2500 (189 δ 3 + 25 α )
(22)
In fact we can think that a2 is an arbitrary coefficient at condition (19). The coefficient a3 is arbitrary at condition (20) and the coefficient a6 is arbitrary at condition (22). The main result of the Painlevé test is that there is the expansion for the solution of equation in the Laurent series. So, at conditions (19), (20) and (22) we obtain the expansion of the general solution of Eq. (4) in the Laurent series. The necessary condition for existence of meromorphic solution is satisfied, and we can have some exact solutions of Eq. (4). Using the same approach we can consider the second branch of the solution too. Substituting the truncated expansion in the form
w (z ) =
2 + b1 + b2 z + b3 z 2 + b4 z 3 + b5 z 4 + b6 z 5 + b7 z 6 + b8 z 7 + · · · z
(23)
into Eq. (4) we have:
b1 =
3 δ, 70
b2 = − b3 =
β 90
(24)
−
11 δ 2 , 4900
23 δ β 199 δ 3 α + + , 1029000 180 18900
b4 = −
587 δ 4 11 δ α 199 δ 2 β 23 β 2 C0 − − − − . 140 24010000 14700 1029000 113400
(25)
(26)
(27)
The coefficients b5 and b7 are also found but their form is cumbersome and we do not give these expressions. The coefficients b6 can be taken arbitrary at additional condition:
C1 =
193 δ 2 α 314 α β 398003 δ 5 57 C0 δ 1289 δ β 2 44837 δ 3 β α2 + + + + + . + 99225 5250 4725 420175000 245 15 δ 5402250
(28)
However for an arbitrary value of the coefficient b8 we can find the value of the coefficient b6 . We do not present this expression here. It is clear that there is the expansion of solution in the Laurent series for the second branch. Using the results of the Painlevé test, we can do the following conclusion. The generalized fifth-order modified Korteweg– de Vries equation can have solitary wave solutions expressed via the hyperbolic functions at condition (19) corresponding to two branches. At condition α = β = δ = 0 we obtain the well-known integrable case of the modified Korteweg–de Vries equation. The coefficients a2 , a3 , a6 for the first branch and the coefficients b6 and b8 for the second branch are arbitrary.
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3. Solitary wave solutions of Eq. (4) Let us look for the solitary wave solutions of Eq. (4). At the present time there are many approaches for finding solitary wave solutions of nonlinear differential equations. To look for the exact solutions of nonintegrable equations we can use the method of singular manifold [2,3], the tanh - method [4–6], the method of the simplest equation [7–9], the method of G /G expansion [10,11] and the method of the logistic function [12–14]. However all methods mentioned above are only suitable for finding solitary wave solutions. In this paper we use the effective modification of the method presented in [12,15]. The essence of our approach is the following. We look for solution of Eq. (4) using the logistic function in the form:
w(z ) = A + BQ (z ),
Q (z ) =
1 , 1 − e−k z+k z0
z = x + C0 t.
(29)
Solution has the first order pole at z = z0 which corresponds to the pole order of the general solution on complex domain of solution (4). Our aim is to find the coefficients A, B of solution (29), and conditions on parameters of Eq. (4). We are doing the following steps. Calculating the expansion of function (29) in the Laurent series at z = 0. We can omit z0 because Eq. (4) is autonomous. We have the expansion in the form:
w (z ) =
B Bk Bk3 3 B Bk5 5 Bk7 +A+ + z− z + z − z7 + · · · kz 2 12 720 30240 1209600
(30)
Substituting (30) into Eq. (4), we get three values of the coefficient B in the form
B(1,2) = ± k,
B(3,4) = ± 2 k
(31) (i )
These values of coefficients Bi correspond to the values of a0 at the Painlevé analysis of Eq. (4). In the case B = B(1,2 ) we have solutions
w(1,2) (z ) = ∓
3δ 1 k∓ ± k Q (z ) 2 10
(32)
These solutions are possible at the following conditions
9 δ (125 k2 + 81 δ 2 ), 5 50 3 27 2 2 243 4 729 k δ − C0 = − k4 − δ , C1 = − 8 20 1000 12500
β = − δ2, α = −
δ5 +
9 20
δ k4 .
(33)
At B = B(3,4 ) , we obtain the solitary solutions in the form
w(3,4) (z ) = ∓ k ±
3δ ± 2 k Q ( z ). 70
(34)
These solutions satisfy Eq. (4) at conditions on the parameters of the equation
99 2 23 2 81 3 k δ+ δ , α= δ , 490 7 8575 27 k2 δ 2 729 δ 5 243 δ 4 36 δ k4 C0 = −6 k4 − − , C1 = − . 245 2401000 210087500 35
β = −15 k2 −
(35)
To obtain the kinks defined by formulae (32) and (34) we have to take z0 = i π + z0 , where z0 is a new arbitrary constant. Solutions (32) and (34) are kinks at arbitrary value of the wave number. We demonstrate solutions w(1,2 ) (z ) in Fig. 1. Solutions w(3,4 ) (z ) are similar to the form of solutions w(1,2 ) (z ). 4. Elliptic solutions of Eq. (4) We can note that the necessary conditions for existing of the doubly periodic solution for Eq. (4) are satisfied because we have
a0(1) + a0(2) = 0,
a0(3) + a0(4) = 0.
(36)
Consequently we can try to look for some elliptic solutions of Eq. (4). To search for the doubly periodic solution of Eq. (4) we use the modification of the method given in works [16–18]. The essence of this modified approach is the following. It is known that the meromorphic solutions of the autonomous nonlinear ordinary differential equations are expressed via rational, trigonometric and elliptic solutions. The form of these solutions can be written taking into consideration the quantity of branches and the pole order for the general solution of nonlinear differential equation. The main theorem and all details of this choice can be found in papers [16–18]. The most simple solution for nonlinear differential equation can be found if we take into account only one branch of expansion in the Laurent series. In the case of two and more branches of solution we have more difficult structure.
N. A. Kudryashov / Applied Mathematics and Computation 280 (2016) 39–45
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Fig. 1. Solution w(1,2) (z ) (left, right) of Eq. (4) at δ = −5 and at wave numbers k = 1, k = 3, k = 5, k = 7, k = 9.
Let us search for the elliptic solution of Eq. (4). The general solution of this equation has the first pole order and there are four branches of the expansion for the solution in the Laurent series. Two branches can be taken into account for finding exact solutions. So, we can look for the elliptic solution at condition (19) of the fifth-order differential equation in the form:
w (z ) = H +
A + B℘z (z − z0 , g2 , g3 ) , C + ℘(z − z0 , g2 , g3 )
(37)
where ℘(z − z0 , g2 , g3 ) is the Weierstrass function, z0 is an arbitrary constant, which we do not use at calculations but we keep it in mind, H, A, B and C are constants depending on parameters of equation. We need to look for these constants. To find them, we use the Laurent series of expression (37). Using the MAPLE one can obtain the expansion of function (37) in the Laurent series in the form
2B + H + 2 B C z + A z2 + z 1 A g2 z6 + · · · + A C2 − 20
w (z ) = −
1 5
B g2 − 2 B C 2 z3 − A C z4 +
3
14
B g3 −
1 1 B C g2 − B C g2 + 2 B C 3 z5 10 5 (38)
Substituting the Laurent series into Eq. (4) and equating expressions at different power of z to zero, we obtain the system of algebraic equations. Solving this system we find the values for parameters of solution H, B, C, g2 , g3 and conditions on parameters of equation. Assuming α = β = δ = 0 we obtain the elliptic solutions of the equation in the form:
w (1 ) ( z ) =
℘z (z − z0 , g2 , g3 ) + 2 A 2℘(z − z0 , g2 , g3 ) +
C1 12 A
,
(39)
where the invariants g2 and g3 are determined by the formulae:
g2 = −
5 C1 2 C0 − , 2 192 A2
g3 = −4 A2 −
19 C1 3 C0 C1 . − 48 A 13824 A3
(40)
Another elliptic solution can be written at α = β = δ = 0 in the form:
w (2 ) ( z ) =
℘z (z − z0 , g2 , g3 ) , ℘(z − z0 , g2 , g3 )
(41)
where the following conditions are realized
g2 =
C0 , 28
g3 = 0,
C1 = 0.
(42)
We demonstrate the solutions w(1,2 ) (z ) in Fig. (2). The solution w(1 ) (z ) is given at C0 = 10, A = −0.1, C1 = 10. The solution w(2 ) (z ) is presented at C0 = 28 and g3 = 0.
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N. A. Kudryashov / Applied Mathematics and Computation 280 (2016) 39–45
Fig. 2. Elliptic solution w(1,2) (z ) (left, right) of Eq. (4).
Solving algebraic equations at α = 0 β = 0 δ = 0 we obtain the solution at β = − 59 δ 2 in the form:
w (3 ) ( z ) =
3δ + 10
℘z (z, g2 , g3 ) , + 15αδ + ℘(z, g2 , g3 )
27 δ 2 250
(43)
where
1701 4 27 1 α2 C0 δ + δα− − , 6250 125 60 δ 2 2 C0 α 617463 δ 6 51 α 2 39 C0 δ 2 11529δ 3 α α3 − + − , − + g3 = 3 50000 180 δ 1000 31250000 625000 27000δ 81 δ 2 α 9 α2 729 δ 5 3 − − C0 δ. C1 = − 62500 625 100 δ 10 g2 =
(44)
However solution (43) is degenerated, because g32 − 27 g23 = 0. Assuming the conditions
9 5
β = − δ2,
α=
27 25
δ3
(45)
we also have solution
w (4 ) ( z ) =
50℘z (z − z0 , g2 , g3 ) 3δ + , 10 9 δ 2 + 100℘(z − z0 , g2 , g3 )
(46)
where
g2 =
243 2500
δ 4 , g3 =
729 125000
δ 6 , C0 =
486 625
δ 4 , C1 = −
However solution (46) is degenerated too. We can also obtain the solution at β = − 95 δ 2 and α =
w (5 ) ( z ) = −
9800℘z (z − z0 , g2 , g3 ) 3δ + , 70 9800℘(z − z0 , g2 , g3 )−87 δ 2
1233 3430
1458 3125
δ5.
(47)
δ 3 in the form: (48)
where
22707 δ 4 658503 δ 6 , g3 = − , 24010000 117649000000 959607 4 2451627 C0 = − δ , C1 = δ5. 12005000 210087500 g2 =
(49)
However solution (46) is degenerated too. We obtain that all the elliptic solutions of Eq. (4) at α = 0 and δ = 0 are degenerated. The elliptic solutions are found only at α = 0 and δ = 0.
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5. Conclusion In this paper we have studied the generalized modified Korteweg–de Vries equation. We have applied the Painlevé test to this equation and have obtained that it does not pass the Painlevé test. However we have found that there are four branches of the expansion for the general solution of the equation. Taking into account the necessary condition for existence of an exact solution, we have found some exact solutions of Eq. (4) expressed via the logistic function and the Weierstrass elliptic function. We have obtained some elliptic and solitary wave solutions of the generalized Korteweg–de Vries equation but it turned out that elliptic solutions at α = 0 and γ = 0 are degenerated. We have obtained only two elliptic solutions for the fifth-order modified Korteweg–de Vries equation. Acknowledgments This research was supported by Russian Science Foundation Grant no. 14-11-00258. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]
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