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Sine–Cosine method for finding the soliton solutions of the generalized fifth-order nonlinear equation
Chaos, Solitons and Fractals 33 (2007) 1610–1617 www.elsevier.com/locate/chaos
Sine–Cosine method for finding the soliton solutions of the generalized fifth-order nonlinear equation Qasem M. Al-Mdallal *, Muhammad I. Syam Department of Mathematical Sciences, College of Science, P.O. Box 17551, UAE University, Al-Ain, United Arab Emirates Accepted 6 March 2006
Abstract In this paper, we use a modified form of the Sine–Cosine method for obtaining exact soliton solutions of the generalized fifth-order nonlinear evolution equation. Analysis for this method is presented. The present method shows that the solutions involve either sec2 or sech2 under certain conditions. General forms of those conditions are determined for the first time. Exact solutions for special cases of this problem such as the Sawada-Kotera and Lax equations are determined and found to be compared well with the previous studies. Ó 2006 Published by Elsevier Ltd.
1. Introduction The generalized fifth-order nonlinear equation of the form ut þ auuxxx þ bux uxx þ cu2 ux þ uxxxxx ¼ 0; covers many well-known equations as the real constants a, b, and c take different values. For example, Kaup–Kupershmidt (KK) equation [6,8–10] when (a, b, c) = (15, 75/2,45) and Sawada-Kotera equations (SK-I) and (SK-II) [4,5] with (a, b, c) = (15, 15,45) and (a, b, c) = (5, 5, 5), respectively. In addition to those equations we have the Lax equation [11] with (a, b, c) = (10, 20, 30). Those four equations have been widely studied over the past two decades and found to be integrable. The SK and Lax equation have N–soliton solutions and infinite set of conservation laws. The KK equation is in integrable and have bilinear representations, but the explicit form of its N-solitons solutions are not known [8]. Recently, several investigations on the solutions of the SK and KK equations are done. Hereman and Nuseir [6] constructed multi soliton solutions for the KK equation using the Hirota’ method. Ablowitz and Clrkson [1] used the inverse scattering transform method to deal with the nonlinear equations where solitons solutions were developed. Fan [7] approximated the soliton solution of the KK equation using the adomian decomposition method (ADM)
*
Corresponding author. E-mail addresses: [email protected] (Q.M. Al-Mdallal), [email protected] (M.I. Syam).
0960-0779/$ - see front matter Ó 2006 Published by Elsevier Ltd. doi:10.1016/j.chaos.2006.03.039
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[2,3,12–14]. This technique gives the solution with direct transformation. However, it needs to calculate many integrals to get good approximation for the solution. In this paper, we will use the Sine–Cosine method [15–17]. The mechanism of this method in solving a nonlinear time dependent partial differential equation with one variable x in the form Gðu; ut ; ux ; u2x ; u3x ; u4x ; . . . ; unx Þ;
ð1Þ
m
where umx denotes ddxmu and n 2 N, is described below. Applying the wave variable n = (x ct) we may transform the partial differential Eq. (4) into an ordinary differential equation given by GðU ; U n ; U 2n ; U 3n ; U 4n ; . . . ; U nn Þ;
ð2Þ
where Un = u(t, x). It is noted that Eq. (2) could be integrable with respect to n, hence integrating both sides of Eq. (2) with respect to n and using the constant of integration to be zero we obtain an easier form with lower order. The next crucial step is expressing the solution in the Cosine form U ðnÞ ¼ d þ k cosm ðlnÞ;
ð3Þ
where d, k, m and l are constants need to be determined. The direct substitution of (7) into Eq. (2) gives a trigonometric equation of cosine terms with coefficients involve the unknowns constants, d, k, m and l. It is worth mentioning that collecting all terms in the equation with same power in cos should be done carefully so we can determine the correct values for the unknown constants. In practice, the value for m should be determined first so we can reduce the number of algebraic equations related to the coefficients of cosk, where k is a positive integer belongs to the set of all powers in the trigonometric equation. Furthermore, equating the coefficients of the same power to zero produces an algebraic system of equations that can be solved by Maple or Mathematica. The paper will be organized as follows. In Section 2, analysis of the Cosine method is given while some examples are presented in Section 3.
2. Analysis of the Method Consider the generalized fifth-order nonlinear evolution equation ut þ auuxxx þ bux uxx þ cu2 ux þ uxxxxx ¼ 0;
ð4Þ
where a, b and c are real numbers. Introducing the wave variable n = x ct, where c is a constant, into the differential Eq. (4) gives the following ordinary differential equation cun þ auunnn þ bun unn þ cu2 un þ unnnnn ¼ 0; where un ¼ du and dn on u dn u ¼ oxn dnn
and
ou du ¼ c ; ot dn
n ¼ 1; 2; 3; 4.
For simplicity, the above equation can be written in the form 1 1 cun þ aðuunn Þn þ ðb aÞðu2n Þn þ cðu3 Þn þ unnnnn ¼ 0. 2 3
ð5Þ
Integrating Eq. (5) with respect to n and using the constant of integration to be zero, one obtain 1 1 cu þ auunn þ ðb aÞu2n þ cu3 þ unnnn ¼ 0. 2 3
ð6Þ
Assume that the solution of Eq. (6) is given by uðnÞ ¼ d þ k cosm ðlnÞ;
ð7Þ
where d, k, m and l are constants need to be determined. Most users of the cosine method assume that d = 0, while the present investigation shows that assuming a nonzero value for the constant d produces some other solutions for Eq. (4). Since we are interesting in a non constant solution, assume that k, l 5 0. The direct substitution of the general solution (7) into Eq. (6) gives
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d 2 1 ck3 ðcd 3cÞ þ k2 ml2 ðma þ mb 2aÞ cos2m2 ðlnÞ þ kml4 ðm 1Þðm 2Þ ðm 3Þ cosm4 ðlnÞ þ cos3m ðlnÞ 3 2 3 k2 ðm2 l2 a þ m2 l2 b 2dcÞ cos2m ðlnÞ þ ml2 kðm 1Þð2m2 l2 þ 4ml2 4l2 þ adÞ cosm2 ðlnÞ þ kðam2 l2 d 2 m4 l4 þ c cd2 Þ cosm ðlnÞ ¼ 0.
ð8Þ
The following Lemma is needed to simplify the last equation. Lemma 1. For any nonzero integer m, we have (i) (ii) (iii) (iv)
3m > max{2m 2, m 4, m 2, 2m, m} if m is a positive integer. 3m < min{2m 2, m 4, m 2, 2m, m} if m is a negative integer less than 2. 3m = m 2 62 {2m 2, m 4, 2m, m} if m = 1. 3m = 2m 2 = m 4 62 {m 2, 2m, m} if m = 2.
Now, the term cos3m(ln) in Eq. (8) leads to the conclusion that this term should be added to another terms of this equations. Consequently, from Lemma 1 m should equal to either 1 or 2. If m = 1, then Eq. (8) becomes 2 d 2 1 2 2 ck 4 4 5 4 2 ðcd 3cÞ þ k l ða þ b þ 2aÞ cos ðlnÞ þ 24kl cos ðlnÞ þ k 20l þ 2l ad cos3 ðlnÞ 3 2 3
k2 2 ðl a þ l2 b 2dcÞ cos2 ðlnÞ kðal2 d l4 þ c cd2 Þ cos1 ðlnÞ ¼ 0. 2
Therefore, the coefficient of cos5(ln) must be zero which is conflict with the assumption k, l > 0. Hence we assume that the value of m is 2. By replacing m by 2 in Eq. (8), we obtain d 2 ðcd 3cÞ kðc þ 4al2 d cd2 16l4 Þ cos2 ðlnÞ 3 þ kð2akl2 2kl2 b þ kcd þ 6al2 d 120l4 Þ cos4 ðlnÞ k 2 ðck þ 12akl2 þ 6kl2 b þ 360l4 Þ cos6 ðlnÞ ¼ 0. 3
ð9Þ
Equating the coefficients of the trigonometric functions in Eq. (9) to zero together with using the assumption k 5 0, we obtain a system of four algebraic equations with three unknowns; those equations are dðcd2 3cÞ ¼ 0;
ð10Þ
2
2
4
ðc þ 4al d cd 16l Þ ¼ 0; 2
2
ð11Þ 2
4
ð2akl 2kl b þ kcd þ 6al d 120l Þ ¼ 0; 2
2
2
4
ðck þ 12akl þ 6kl b þ 360l Þ ¼ 0.
ð12Þ ð13Þ
Eq. (10) gives three values for d which are sffiffiffiffiffi sffiffiffiffiffi 3c 3c ; and d ¼ d ¼ 0; d ¼ c c
ð14Þ
provided that c 5 0. On the other hand, adding Eq. (12) to Eq. (13) and then factorizing the resulting equation gives ðk þ 3dÞðck þ 6al2 Þ ¼ 0; 2
. In the remaining of this section, we deal with the following five cases. which implies that either k = 3d or k ¼ 6al c 2 Case One: When d = 0 and k ¼ 6al c The direct substitution of those values into Eq. (11) gives c 16l4 ¼ 0.
ð15Þ
Hence, Eq. (12) implies the following condition on a, b, and c, ða2 þ ab 10cÞ ¼ 0
ð16Þ
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or b¼
10c a2 . a
ð17Þ
On the other hand, Eq. (15) gives four possible values for l given by l¼
ffiffiffi 1p 4 c; 2
l¼
ffiffiffi 1p 4 c; 2
1 pffiffiffi l ¼ i 4 c; 2
1 pffiffiffi l¼ i4c 2
and, therefore, the corresponding values for k are given by pffiffiffi pffiffiffi pffiffiffi pffiffiffi 3a c 3a c 3a c 3a c k¼ ; k¼ ; k¼ ; k¼ . 2c 2c 2c 2c Consequently, this case shows the existence of two possible solutions given by pffiffiffi ffiffiffi 3a c 1p 4 cðx ctÞ sec2 uðt; xÞ ¼ 2 2c and pffiffiffi ffiffiffi 3a c 1p 4 sech2 cðx ctÞ . 2c 2 qffiffiffi 2 Case Two: When d ¼ 3cc and k ¼ 6al c uðt; xÞ ¼
Following exactly the same procedure presented in previous subsection, we get the following restriction on a, b, and c, b¼
10c a2 . a
The possible values for l and the corresponding values for k are given by pffiffiffipffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffipffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4 c a 3c þ 3a2 c 8c2 2 4 c a 3c þ 3a2 c 8c2 l¼ ; l¼ ; 4 4 c c pffiffiffipffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffipffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4 c a 3c 3a2 c 8c2 2 4 c a 3c 3a2 c 8c2 ; l¼ ; l¼ 4 4 c c pffiffiffi pffiffiffi 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c 8c2 ; k¼ k ¼ a 3c þ 3a a 3c þ 3a2 c 8c2 ; 2 2 4c 4c pffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3a c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c 8c2 ; k¼ k ¼ a 3c 3a a 3c 3a2 c 8c2 . 2 2 4c 4c Consequently, this case produces two possible solutions given by 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ffiffiffi p 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c þ 3a 2 c uðt; xÞ ¼ ðx ctÞA a 3c þ 3a2 c 8c2 sec2 @ c 4c2 4 c and
0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ffiffiffi p 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c 3a 2 c 2 uðt; xÞ ¼ ðx ctÞA. a 3c 3a2 c 8c2 sec @ 4c2 4 c c qffiffiffi 2 Case Three: When d ¼ 3cc and k ¼ 6al . c The restriction on a, b, and c in this case is given by b¼
10c a2 a
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which is similar to the previous two cases. The possible solutions for this case are 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffi ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c þ 3a 2 c þ ðx ctÞA uðt; xÞ ¼ a 3c 3a2 c 8c2 sec2 @ c c 4 4c2 and
0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffi ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c 3a 2 c 2 uðt; xÞ ¼ ðx ctÞA. a 3c 3a2 c 8c2 sec @ c 4 c 4c2
We can summarize the previous three cases in the following theorem. 2
Theorem 2. The generalized fifth-order nonlinear evolution Eq. (1) has six solutions when b ¼ 10ca which are a pffiffiffi ffiffi ffi p 3a c 1 4 sec2 cðx ctÞ ; u1 ðt; xÞ ¼ 2c 2 pffiffiffi ffiffiffi 3a c 1p 4 cðx ctÞ ; u2 ðt; xÞ ¼ sech2 2 2c 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi p ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c þ 3a 2 c ðx ctÞA; a 3c þ 3a2 c 8c2 sec2 @ u3 ðt; xÞ ¼ c 4c2 4 c 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffi ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c 3a 2 c u4 ðt; xÞ ¼ ðx ctÞA; a 3c 3a2 c 8c2 sec2 @ c c 4 4c2 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi p ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c þ 3a 2 c 2 u5 ðt; xÞ ¼ þ ðx ctÞA a 3c 3a2 c 8c2 sec @ c 4c2 4 c and
0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi p ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c 3a 2 c u6 ðt; xÞ ¼ ðx ctÞA. a 3c 3a2 c 8c2 sec2 @ c 4c2 4 c qffiffiffi Case Four: When d ¼ 3cc and k = 3d Substitution of those values into Eqs. (11) and (12) we obtain pffiffiffiffiffiffiffi 6 pffiffiffiffiffiffiffi c 3cc þ 6cal2 8 3ccl4 ¼ 0 c
ð18Þ
and
pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 9 12acl2 þ 6cl2 b 3c 3cc 40 3ccl4 ¼ 0. c
ð19Þ
It is noted that the form of Eq. (19) is different than that of Eq. (16) where an implicit relation between both l and b can easily captured. Therefore we use Eq. (18) to determine the value of l then to figure out the restriction on a, b, and c those values are found to be vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffipffiffiffi u pffiffiffipffiffiffi u 2 2 u u 6 4 c t 3 3a þ 9a 24c 6 4 c t 3 3a þ 9a 24c l¼ ; l¼ ; pffiffiffi pffiffiffi 12 12 c c vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffipffiffiffi u pffiffiffipffiffiffi u 2 u 3 3a þ 9a2 24c u 4 4 6 ct 6 c t 3 3a þ 9a 24c ; l¼ ; l¼ pffiffiffi pffiffiffi 12 12 c c with the corresponding restrictions on b depending on the value of l
Q.M. Al-Mdallal, M.I. Syam / Chaos, Solitons and Fractals 33 (2007) 1610–1617
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c þ 9a 3a2 8c pffiffiffi ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a þ 3a2 8c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c 9a 3a2 8c pffiffiffi b¼ ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a 3a2 8c b¼
1615
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c þ 9a 3a2 8c pffiffiffi ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a þ 3a2 8c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c 9a 3a2 8c pffiffiffi b¼ . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a 3a2 8c b¼
Therefore, this case produces the following solutions vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi sffiffiffiffiffi sffiffiffiffiffi pffiffiffipffiffiffi u u 3 3a þ 9a2 24c 4 3c 3c 2 B 6 c t C ðx ctÞA 3 sec @ uðt; xÞ ¼ pffiffiffi c c 12 c and
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi sffiffiffiffiffi sffiffiffiffiffi pffiffiffipffiffiffi u u 3 3a þ 9a2 24c 4 3c 3c 2 B 6 c t C uðt; xÞ ¼ 3 sec @ ðx ctÞA. pffiffiffi c c c 12
qffiffiffi Case Five: When d ¼ 3cc and k = 3d Following the same methodology presented in case four, we found that the possible values for l are vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffipffiffiffi u pffiffiffipffiffiffi u u 3 3a þ 9a2 24c u 3 3a þ 9a2 24c 4 4 6 ct 6 ct l¼ ; l¼ ; pffiffiffi pffiffiffi c c 12 12 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffipffiffiffi u pffiffiffipffiffiffi u u 3 3a þ 9a2 24c u 3 3a þ 9a2 24c 4 4 6 ct 6 ct ; l¼ ; l¼ pffiffiffi pffiffiffi 12 12 c c with the corresponding restrictions on b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 9 3a2 þ 8 3c þ 9a 3a2 8c 9 3a2 þ 8 3c þ 9a 3a2 8c pffiffiffi ; b ¼ ; b¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3 3a þ 3a2 8c 3 3a þ 3a2 8c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c 9a 3a2 8c 9 3a2 8 3c 9a 3a2 8c pffiffiffi pffiffiffi b¼ ; b¼ . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a 3a2 8c 3 3a 3a2 8c Consequently, the possible solutions related to this case are vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi sffiffiffiffiffi sffiffiffiffiffi pffiffiffipffiffiffi u 2 u 3c 3c 2 B 6 4 c t 3 3a þ 9a 24c C uðt; xÞ ¼ þ3 sec @ ðx ctÞA pffiffiffi c 12 c c and
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi sffiffiffiffiffi sffiffiffiffiffi pffiffiffipffiffiffi u 2 u 3c 3c 2 B 6 4 c t 3 3a þ 9a 24c C þ3 sec @ ðx ctÞA. uðt; xÞ ¼ pffiffiffi c c 12 c
3. Examples In this section, we shall apply the method developed in Section 2 to some well-known differential equations and give their exact solutions. Example 1. One form for the Sawada-Kotera equation is given by ut 15uuxxx 15ux uxx þ 45u2 ux þ uxxxxx ¼ 0.
ð20Þ
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Following the methodology presented in Section 2, we see that the possible exact solutions are given by p pffiffiffi ffiffiffi 4 c c sec2 ðx ctÞ ; u1 ðt; xÞ ¼ 2 2 p pffiffiffi ffiffiffi 4 c c u2 ðt; xÞ ¼ sech2 ðx ctÞ ; 2 2 rffiffiffiffiffi pffiffiffi p ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 c pffiffiffi pffiffiffiffiffi c pffiffiffiffiffi pffiffiffiffiffi 3c u3 ðt; xÞ ¼ 7 15 sech2 60 28ðx ctÞ ; þ 4 4 15 rffiffiffiffiffi pffiffiffi p ffi ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3c c pffiffiffi pffiffiffiffiffi c pffiffiffiffiffi pffiffiffiffiffi 7 þ 15 sech2 60 þ 28ðx ctÞ ; u4 ðt; xÞ ¼ 15 4 4 rffiffiffiffiffi pffiffiffi pffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3c c pffiffiffi pffiffiffiffiffi 2 4 c pffiffiffiffiffi pffiffiffiffiffi u5 ðt; xÞ ¼ 7 þ 15 sec 60 þ 28ðx ctÞ ; þ 15 4 4 rffiffiffiffiffi pffiffiffi p ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3c c pffiffiffiffiffi pffiffiffi c pffiffiffiffiffi pffiffiffiffiffi u6 ðt; xÞ ¼ 15 7 sec2 60 28ðx ctÞ . þ 15 4 4
ð21Þ ð22Þ ð23Þ ð24Þ ð25Þ ð26Þ
Example 2. Consider the following Sawada-Kotera equation (SK) ut þ 5uuxxx þ 5ux uxx þ 5u2 ux þ uxxxxx ¼ 0; Following the methodology presented in Section 2, we see that the possible exact solutions are given by p ffiffiffi pffiffiffi 4 c 3 c u1 ðt; xÞ ¼ ðx ctÞ ; sec2 2 2 p ffiffiffi pffiffiffi 4 c 3 c u2 ðt; xÞ ¼ ðx ctÞ ; sech2 2 2 rffiffiffiffiffi p ffi pffiffiffi pffiffiffiffiffi ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 pffiffiffi 3c 3 c c pffiffiffiffiffi pffiffiffiffiffi 2 þ 15 7 sech 60 28ðx ctÞ ; u3 ðt; xÞ ¼ 5 4 4 rffiffiffiffiffi p ffi pffiffiffi ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3c 3 c pffiffiffiffiffi pffiffiffi c pffiffiffiffiffi pffiffiffiffiffi u4 ðt; xÞ ¼ 60 þ 28ðx ctÞ ; þ 15 þ 7 sech2 5 4 4 rffiffiffiffiffi p ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 4 c pffiffiffiffiffi pffiffiffiffiffi 3c 3 c pffiffiffiffiffi pffiffiffi 2 u5 ðt; xÞ ¼ þ 60 28ðx ctÞ ; 15 7 sech 5 4 4 rffiffiffiffiffi p ffi pffiffiffi ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3c 3 c pffiffiffiffiffi pffiffiffi c pffiffiffiffiffi pffiffiffiffiffi 60 þ 28ðx ctÞ . 15 þ 7 sech2 u6 ðt; xÞ ¼ þ 4 4 5
ð27Þ
ð28Þ ð29Þ ð30Þ ð31Þ ð32Þ ð33Þ
Example 3. The Lax equation is given by ut þ 10uuxxx þ 20ux uxx þ 30u2 ux þ uxxxxx ¼ 0. Following the methodology presented in Section 2, we see that the possible exact solutions are given by pffiffiffi ffiffiffi c 1p 4 cos2 cðx ctÞ ; u1 ðt; xÞ ¼ 2 2 pffiffiffi p ffiffiffi i c 4 cos2 u2 ðt; xÞ ¼ cðx ctÞ ; 2 2 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi c 1 pffiffiffiffiffi 10 90c þ 1800c 2 ðx ctÞA; u3 ðt; xÞ ¼ 10 90c þ 1800c cos2 @ 30 4 10 120 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi c 1 pffiffiffiffiffi 10 90c 1800c 2 u4 ðt; xÞ ¼ 10 90c 1800c cos2 @ ðx ctÞA; 10 120 30 4
ð34Þ
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0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi c 1 pffiffiffiffiffi 10 90c þ 1800c 2 @ 2 þ 10 90c 1800c cos ðx ctÞA; u5 ðt; xÞ ¼ 10 120 30 4 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi c 1 pffiffiffiffiffi 10 90c 1800c 2 @ 2 u6 ðt; xÞ ¼ 10 90c 1800c cos ðx ctÞA. 10 120 30 4
References [1] Ablowitz MJ, Clarkson PA. Solitons, nonlinear evolution equations and inverse scattering. Cambridge: Cambridge University Press; 1991. [2] Adomian G. Solving frontier problems of physics: the decomposition methods. Boston (MA): Kluwer; 1994. [3] Adomian G. A review of the decomposition method in applied mathematics. J Math Anal Appl 1988;135(2):501–44. [4] Aiyer RN, Fuchssteiner B, Oevel W. Solitons and discrete eigen-functions of the recursion operator of non-linear evolution equations: I. The Caudrey–Dodd–Gibbon–Sawada-Kotera Equations. J Phys A: Math Gen 1986;19:3755–70. [5] Chen B, Xie Y. An auto-Bcklund transformation and exact solutions of stochastic Wick-type Sawada-Kotera equations. Chaos, Solitons & Fractals 2005;23(1):243–8. [6] Hereman W, Nuseir A. Symbolic methods to construct exact solution of nonlinear differential equations. Math Comput Simulat 1997;43(1):13–27. [7] Fan E. Uniformly constructing a series of explicit exact solutions to nonlinear equations in mathematical physics. Chaos, Solitons & Fractals 2003;16(5):819–39. [8] Inc M. On numerical soliton solution of the Kaup–Kupershmidt equation and convergences analysis of the decomposition method. Appl Math Comput 2006;172(1):72–85. [9] Musette M, Verhoeven C. Nonlinear superposition formula for the Kaup–Kupershmidt partial differential equation. Physica D 2000;144(1–2):211–20. [10] Parker A. On soliton solutions of the Kaup–Kupershmidt equation. II. ‘Anomalous’ N-soliton solutions. Physica D: Nonlinear Phenom 2000;137(1–2):34–48. [11] Soliman AA. A numerical simulation and explicit solutions of KdV-Burgers’ and Lax’s seventh-order KdV equations. Chaos, Solitons & Fractals 2006;29(2):294–302. [12] Syam MI. Adomian decomposition method for approximating the solution of the Korteweg–deVries equation. Appl Math Comput 2005;162(3):1465–73. [13] Wazwaz AM. A Reliable algorithm for obtaining positive solutions for nonlinear boundary value problems. Int J Comput Math Appl 2001;41:1237–44. [14] Wazwaz AM. Approximate solutions to boundary value problems of higher order by the modified decomposition method. Comput Math Appl 2000;40:679–91. [15] Wazwaz AM. Exact solutions for the generalized sine-Gordon and the generalized sinh-Gordon equations. Chaos, Solitons & Fractals 2006;28(1):127–35. [16] Wazwaz AM. The tanh method and a variable separated ODE method for solving double sine-Gordon equation. Phys Lett A 2006;350(5–6):367–70. [17] Wazwaz AM, Helal MA. Nonlinear variants of the BBM equation with compact and noncompact physical structures. Chaos, Solitons & Fractals 2005;26(3):767–76.
Sine–Cosine method for finding the soliton solutions of the generalized fifth-order nonlinear equation Qasem M. Al-Mdallal *, Muhammad I. Syam Department of Mathematical Sciences, College of Science, P.O. Box 17551, UAE University, Al-Ain, United Arab Emirates Accepted 6 March 2006
Abstract In this paper, we use a modified form of the Sine–Cosine method for obtaining exact soliton solutions of the generalized fifth-order nonlinear evolution equation. Analysis for this method is presented. The present method shows that the solutions involve either sec2 or sech2 under certain conditions. General forms of those conditions are determined for the first time. Exact solutions for special cases of this problem such as the Sawada-Kotera and Lax equations are determined and found to be compared well with the previous studies. Ó 2006 Published by Elsevier Ltd.
1. Introduction The generalized fifth-order nonlinear equation of the form ut þ auuxxx þ bux uxx þ cu2 ux þ uxxxxx ¼ 0; covers many well-known equations as the real constants a, b, and c take different values. For example, Kaup–Kupershmidt (KK) equation [6,8–10] when (a, b, c) = (15, 75/2,45) and Sawada-Kotera equations (SK-I) and (SK-II) [4,5] with (a, b, c) = (15, 15,45) and (a, b, c) = (5, 5, 5), respectively. In addition to those equations we have the Lax equation [11] with (a, b, c) = (10, 20, 30). Those four equations have been widely studied over the past two decades and found to be integrable. The SK and Lax equation have N–soliton solutions and infinite set of conservation laws. The KK equation is in integrable and have bilinear representations, but the explicit form of its N-solitons solutions are not known [8]. Recently, several investigations on the solutions of the SK and KK equations are done. Hereman and Nuseir [6] constructed multi soliton solutions for the KK equation using the Hirota’ method. Ablowitz and Clrkson [1] used the inverse scattering transform method to deal with the nonlinear equations where solitons solutions were developed. Fan [7] approximated the soliton solution of the KK equation using the adomian decomposition method (ADM)
*
Corresponding author. E-mail addresses: [email protected] (Q.M. Al-Mdallal), [email protected] (M.I. Syam).
0960-0779/$ - see front matter Ó 2006 Published by Elsevier Ltd. doi:10.1016/j.chaos.2006.03.039
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[2,3,12–14]. This technique gives the solution with direct transformation. However, it needs to calculate many integrals to get good approximation for the solution. In this paper, we will use the Sine–Cosine method [15–17]. The mechanism of this method in solving a nonlinear time dependent partial differential equation with one variable x in the form Gðu; ut ; ux ; u2x ; u3x ; u4x ; . . . ; unx Þ;
ð1Þ
m
where umx denotes ddxmu and n 2 N, is described below. Applying the wave variable n = (x ct) we may transform the partial differential Eq. (4) into an ordinary differential equation given by GðU ; U n ; U 2n ; U 3n ; U 4n ; . . . ; U nn Þ;
ð2Þ
where Un = u(t, x). It is noted that Eq. (2) could be integrable with respect to n, hence integrating both sides of Eq. (2) with respect to n and using the constant of integration to be zero we obtain an easier form with lower order. The next crucial step is expressing the solution in the Cosine form U ðnÞ ¼ d þ k cosm ðlnÞ;
ð3Þ
where d, k, m and l are constants need to be determined. The direct substitution of (7) into Eq. (2) gives a trigonometric equation of cosine terms with coefficients involve the unknowns constants, d, k, m and l. It is worth mentioning that collecting all terms in the equation with same power in cos should be done carefully so we can determine the correct values for the unknown constants. In practice, the value for m should be determined first so we can reduce the number of algebraic equations related to the coefficients of cosk, where k is a positive integer belongs to the set of all powers in the trigonometric equation. Furthermore, equating the coefficients of the same power to zero produces an algebraic system of equations that can be solved by Maple or Mathematica. The paper will be organized as follows. In Section 2, analysis of the Cosine method is given while some examples are presented in Section 3.
2. Analysis of the Method Consider the generalized fifth-order nonlinear evolution equation ut þ auuxxx þ bux uxx þ cu2 ux þ uxxxxx ¼ 0;
ð4Þ
where a, b and c are real numbers. Introducing the wave variable n = x ct, where c is a constant, into the differential Eq. (4) gives the following ordinary differential equation cun þ auunnn þ bun unn þ cu2 un þ unnnnn ¼ 0; where un ¼ du and dn on u dn u ¼ oxn dnn
and
ou du ¼ c ; ot dn
n ¼ 1; 2; 3; 4.
For simplicity, the above equation can be written in the form 1 1 cun þ aðuunn Þn þ ðb aÞðu2n Þn þ cðu3 Þn þ unnnnn ¼ 0. 2 3
ð5Þ
Integrating Eq. (5) with respect to n and using the constant of integration to be zero, one obtain 1 1 cu þ auunn þ ðb aÞu2n þ cu3 þ unnnn ¼ 0. 2 3
ð6Þ
Assume that the solution of Eq. (6) is given by uðnÞ ¼ d þ k cosm ðlnÞ;
ð7Þ
where d, k, m and l are constants need to be determined. Most users of the cosine method assume that d = 0, while the present investigation shows that assuming a nonzero value for the constant d produces some other solutions for Eq. (4). Since we are interesting in a non constant solution, assume that k, l 5 0. The direct substitution of the general solution (7) into Eq. (6) gives
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d 2 1 ck3 ðcd 3cÞ þ k2 ml2 ðma þ mb 2aÞ cos2m2 ðlnÞ þ kml4 ðm 1Þðm 2Þ ðm 3Þ cosm4 ðlnÞ þ cos3m ðlnÞ 3 2 3 k2 ðm2 l2 a þ m2 l2 b 2dcÞ cos2m ðlnÞ þ ml2 kðm 1Þð2m2 l2 þ 4ml2 4l2 þ adÞ cosm2 ðlnÞ þ kðam2 l2 d 2 m4 l4 þ c cd2 Þ cosm ðlnÞ ¼ 0.
ð8Þ
The following Lemma is needed to simplify the last equation. Lemma 1. For any nonzero integer m, we have (i) (ii) (iii) (iv)
3m > max{2m 2, m 4, m 2, 2m, m} if m is a positive integer. 3m < min{2m 2, m 4, m 2, 2m, m} if m is a negative integer less than 2. 3m = m 2 62 {2m 2, m 4, 2m, m} if m = 1. 3m = 2m 2 = m 4 62 {m 2, 2m, m} if m = 2.
Now, the term cos3m(ln) in Eq. (8) leads to the conclusion that this term should be added to another terms of this equations. Consequently, from Lemma 1 m should equal to either 1 or 2. If m = 1, then Eq. (8) becomes 2 d 2 1 2 2 ck 4 4 5 4 2 ðcd 3cÞ þ k l ða þ b þ 2aÞ cos ðlnÞ þ 24kl cos ðlnÞ þ k 20l þ 2l ad cos3 ðlnÞ 3 2 3
k2 2 ðl a þ l2 b 2dcÞ cos2 ðlnÞ kðal2 d l4 þ c cd2 Þ cos1 ðlnÞ ¼ 0. 2
Therefore, the coefficient of cos5(ln) must be zero which is conflict with the assumption k, l > 0. Hence we assume that the value of m is 2. By replacing m by 2 in Eq. (8), we obtain d 2 ðcd 3cÞ kðc þ 4al2 d cd2 16l4 Þ cos2 ðlnÞ 3 þ kð2akl2 2kl2 b þ kcd þ 6al2 d 120l4 Þ cos4 ðlnÞ k 2 ðck þ 12akl2 þ 6kl2 b þ 360l4 Þ cos6 ðlnÞ ¼ 0. 3
ð9Þ
Equating the coefficients of the trigonometric functions in Eq. (9) to zero together with using the assumption k 5 0, we obtain a system of four algebraic equations with three unknowns; those equations are dðcd2 3cÞ ¼ 0;
ð10Þ
2
2
4
ðc þ 4al d cd 16l Þ ¼ 0; 2
2
ð11Þ 2
4
ð2akl 2kl b þ kcd þ 6al d 120l Þ ¼ 0; 2
2
2
4
ðck þ 12akl þ 6kl b þ 360l Þ ¼ 0.
ð12Þ ð13Þ
Eq. (10) gives three values for d which are sffiffiffiffiffi sffiffiffiffiffi 3c 3c ; and d ¼ d ¼ 0; d ¼ c c
ð14Þ
provided that c 5 0. On the other hand, adding Eq. (12) to Eq. (13) and then factorizing the resulting equation gives ðk þ 3dÞðck þ 6al2 Þ ¼ 0; 2
. In the remaining of this section, we deal with the following five cases. which implies that either k = 3d or k ¼ 6al c 2 Case One: When d = 0 and k ¼ 6al c The direct substitution of those values into Eq. (11) gives c 16l4 ¼ 0.
ð15Þ
Hence, Eq. (12) implies the following condition on a, b, and c, ða2 þ ab 10cÞ ¼ 0
ð16Þ
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1613
or b¼
10c a2 . a
ð17Þ
On the other hand, Eq. (15) gives four possible values for l given by l¼
ffiffiffi 1p 4 c; 2
l¼
ffiffiffi 1p 4 c; 2
1 pffiffiffi l ¼ i 4 c; 2
1 pffiffiffi l¼ i4c 2
and, therefore, the corresponding values for k are given by pffiffiffi pffiffiffi pffiffiffi pffiffiffi 3a c 3a c 3a c 3a c k¼ ; k¼ ; k¼ ; k¼ . 2c 2c 2c 2c Consequently, this case shows the existence of two possible solutions given by pffiffiffi ffiffiffi 3a c 1p 4 cðx ctÞ sec2 uðt; xÞ ¼ 2 2c and pffiffiffi ffiffiffi 3a c 1p 4 sech2 cðx ctÞ . 2c 2 qffiffiffi 2 Case Two: When d ¼ 3cc and k ¼ 6al c uðt; xÞ ¼
Following exactly the same procedure presented in previous subsection, we get the following restriction on a, b, and c, b¼
10c a2 . a
The possible values for l and the corresponding values for k are given by pffiffiffipffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffipffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4 c a 3c þ 3a2 c 8c2 2 4 c a 3c þ 3a2 c 8c2 l¼ ; l¼ ; 4 4 c c pffiffiffipffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffipffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4 c a 3c 3a2 c 8c2 2 4 c a 3c 3a2 c 8c2 ; l¼ ; l¼ 4 4 c c pffiffiffi pffiffiffi 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c 8c2 ; k¼ k ¼ a 3c þ 3a a 3c þ 3a2 c 8c2 ; 2 2 4c 4c pffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3a c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c 8c2 ; k¼ k ¼ a 3c 3a a 3c 3a2 c 8c2 . 2 2 4c 4c Consequently, this case produces two possible solutions given by 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ffiffiffi p 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c þ 3a 2 c uðt; xÞ ¼ ðx ctÞA a 3c þ 3a2 c 8c2 sec2 @ c 4c2 4 c and
0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ffiffiffi p 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c 3a 2 c 2 uðt; xÞ ¼ ðx ctÞA. a 3c 3a2 c 8c2 sec @ 4c2 4 c c qffiffiffi 2 Case Three: When d ¼ 3cc and k ¼ 6al . c The restriction on a, b, and c in this case is given by b¼
10c a2 a
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which is similar to the previous two cases. The possible solutions for this case are 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffi ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c þ 3a 2 c þ ðx ctÞA uðt; xÞ ¼ a 3c 3a2 c 8c2 sec2 @ c c 4 4c2 and
0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffi ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c 3a 2 c 2 uðt; xÞ ¼ ðx ctÞA. a 3c 3a2 c 8c2 sec @ c 4 c 4c2
We can summarize the previous three cases in the following theorem. 2
Theorem 2. The generalized fifth-order nonlinear evolution Eq. (1) has six solutions when b ¼ 10ca which are a pffiffiffi ffiffi ffi p 3a c 1 4 sec2 cðx ctÞ ; u1 ðt; xÞ ¼ 2c 2 pffiffiffi ffiffiffi 3a c 1p 4 cðx ctÞ ; u2 ðt; xÞ ¼ sech2 2 2c 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi p ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c þ 3a 2 c ðx ctÞA; a 3c þ 3a2 c 8c2 sec2 @ u3 ðt; xÞ ¼ c 4c2 4 c 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p pffiffiffi ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c 3a 2 c u4 ðt; xÞ ¼ ðx ctÞA; a 3c 3a2 c 8c2 sec2 @ c c 4 4c2 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi p ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c þ 3a 2 c 2 u5 ðt; xÞ ¼ þ ðx ctÞA a 3c 3a2 c 8c2 sec @ c 4c2 4 c and
0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ffi sffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi p ffiffiffi 4 2 c 8c2 3c 3a c pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a 3c 3a 2 c u6 ðt; xÞ ¼ ðx ctÞA. a 3c 3a2 c 8c2 sec2 @ c 4c2 4 c qffiffiffi Case Four: When d ¼ 3cc and k = 3d Substitution of those values into Eqs. (11) and (12) we obtain pffiffiffiffiffiffiffi 6 pffiffiffiffiffiffiffi c 3cc þ 6cal2 8 3ccl4 ¼ 0 c
ð18Þ
and
pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 9 12acl2 þ 6cl2 b 3c 3cc 40 3ccl4 ¼ 0. c
ð19Þ
It is noted that the form of Eq. (19) is different than that of Eq. (16) where an implicit relation between both l and b can easily captured. Therefore we use Eq. (18) to determine the value of l then to figure out the restriction on a, b, and c those values are found to be vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffipffiffiffi u pffiffiffipffiffiffi u 2 2 u u 6 4 c t 3 3a þ 9a 24c 6 4 c t 3 3a þ 9a 24c l¼ ; l¼ ; pffiffiffi pffiffiffi 12 12 c c vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffipffiffiffi u pffiffiffipffiffiffi u 2 u 3 3a þ 9a2 24c u 4 4 6 ct 6 c t 3 3a þ 9a 24c ; l¼ ; l¼ pffiffiffi pffiffiffi 12 12 c c with the corresponding restrictions on b depending on the value of l
Q.M. Al-Mdallal, M.I. Syam / Chaos, Solitons and Fractals 33 (2007) 1610–1617
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c þ 9a 3a2 8c pffiffiffi ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a þ 3a2 8c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c 9a 3a2 8c pffiffiffi b¼ ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a 3a2 8c b¼
1615
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c þ 9a 3a2 8c pffiffiffi ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a þ 3a2 8c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c 9a 3a2 8c pffiffiffi b¼ . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a 3a2 8c b¼
Therefore, this case produces the following solutions vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi sffiffiffiffiffi sffiffiffiffiffi pffiffiffipffiffiffi u u 3 3a þ 9a2 24c 4 3c 3c 2 B 6 c t C ðx ctÞA 3 sec @ uðt; xÞ ¼ pffiffiffi c c 12 c and
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi sffiffiffiffiffi sffiffiffiffiffi pffiffiffipffiffiffi u u 3 3a þ 9a2 24c 4 3c 3c 2 B 6 c t C uðt; xÞ ¼ 3 sec @ ðx ctÞA. pffiffiffi c c c 12
qffiffiffi Case Five: When d ¼ 3cc and k = 3d Following the same methodology presented in case four, we found that the possible values for l are vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffipffiffiffi u pffiffiffipffiffiffi u u 3 3a þ 9a2 24c u 3 3a þ 9a2 24c 4 4 6 ct 6 ct l¼ ; l¼ ; pffiffiffi pffiffiffi c c 12 12 vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffipffiffiffi u pffiffiffipffiffiffi u u 3 3a þ 9a2 24c u 3 3a þ 9a2 24c 4 4 6 ct 6 ct ; l¼ ; l¼ pffiffiffi pffiffiffi 12 12 c c with the corresponding restrictions on b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 9 3a2 þ 8 3c þ 9a 3a2 8c 9 3a2 þ 8 3c þ 9a 3a2 8c pffiffiffi ; b ¼ ; b¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 3 3a þ 3a2 8c 3 3a þ 3a2 8c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 9 3a2 8 3c 9a 3a2 8c 9 3a2 8 3c 9a 3a2 8c pffiffiffi pffiffiffi b¼ ; b¼ . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 3a 3a2 8c 3 3a 3a2 8c Consequently, the possible solutions related to this case are vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi sffiffiffiffiffi sffiffiffiffiffi pffiffiffipffiffiffi u 2 u 3c 3c 2 B 6 4 c t 3 3a þ 9a 24c C uðt; xÞ ¼ þ3 sec @ ðx ctÞA pffiffiffi c 12 c c and
vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi sffiffiffiffiffi sffiffiffiffiffi pffiffiffipffiffiffi u 2 u 3c 3c 2 B 6 4 c t 3 3a þ 9a 24c C þ3 sec @ ðx ctÞA. uðt; xÞ ¼ pffiffiffi c c 12 c
3. Examples In this section, we shall apply the method developed in Section 2 to some well-known differential equations and give their exact solutions. Example 1. One form for the Sawada-Kotera equation is given by ut 15uuxxx 15ux uxx þ 45u2 ux þ uxxxxx ¼ 0.
ð20Þ
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Following the methodology presented in Section 2, we see that the possible exact solutions are given by p pffiffiffi ffiffiffi 4 c c sec2 ðx ctÞ ; u1 ðt; xÞ ¼ 2 2 p pffiffiffi ffiffiffi 4 c c u2 ðt; xÞ ¼ sech2 ðx ctÞ ; 2 2 rffiffiffiffiffi pffiffiffi p ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 c pffiffiffi pffiffiffiffiffi c pffiffiffiffiffi pffiffiffiffiffi 3c u3 ðt; xÞ ¼ 7 15 sech2 60 28ðx ctÞ ; þ 4 4 15 rffiffiffiffiffi pffiffiffi p ffi ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3c c pffiffiffi pffiffiffiffiffi c pffiffiffiffiffi pffiffiffiffiffi 7 þ 15 sech2 60 þ 28ðx ctÞ ; u4 ðt; xÞ ¼ 15 4 4 rffiffiffiffiffi pffiffiffi pffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3c c pffiffiffi pffiffiffiffiffi 2 4 c pffiffiffiffiffi pffiffiffiffiffi u5 ðt; xÞ ¼ 7 þ 15 sec 60 þ 28ðx ctÞ ; þ 15 4 4 rffiffiffiffiffi pffiffiffi p ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3c c pffiffiffiffiffi pffiffiffi c pffiffiffiffiffi pffiffiffiffiffi u6 ðt; xÞ ¼ 15 7 sec2 60 28ðx ctÞ . þ 15 4 4
ð21Þ ð22Þ ð23Þ ð24Þ ð25Þ ð26Þ
Example 2. Consider the following Sawada-Kotera equation (SK) ut þ 5uuxxx þ 5ux uxx þ 5u2 ux þ uxxxxx ¼ 0; Following the methodology presented in Section 2, we see that the possible exact solutions are given by p ffiffiffi pffiffiffi 4 c 3 c u1 ðt; xÞ ¼ ðx ctÞ ; sec2 2 2 p ffiffiffi pffiffiffi 4 c 3 c u2 ðt; xÞ ¼ ðx ctÞ ; sech2 2 2 rffiffiffiffiffi p ffi pffiffiffi pffiffiffiffiffi ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 pffiffiffi 3c 3 c c pffiffiffiffiffi pffiffiffiffiffi 2 þ 15 7 sech 60 28ðx ctÞ ; u3 ðt; xÞ ¼ 5 4 4 rffiffiffiffiffi p ffi pffiffiffi ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3c 3 c pffiffiffiffiffi pffiffiffi c pffiffiffiffiffi pffiffiffiffiffi u4 ðt; xÞ ¼ 60 þ 28ðx ctÞ ; þ 15 þ 7 sech2 5 4 4 rffiffiffiffiffi p ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi 4 c pffiffiffiffiffi pffiffiffiffiffi 3c 3 c pffiffiffiffiffi pffiffiffi 2 u5 ðt; xÞ ¼ þ 60 28ðx ctÞ ; 15 7 sech 5 4 4 rffiffiffiffiffi p ffi pffiffiffi ffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 3c 3 c pffiffiffiffiffi pffiffiffi c pffiffiffiffiffi pffiffiffiffiffi 60 þ 28ðx ctÞ . 15 þ 7 sech2 u6 ðt; xÞ ¼ þ 4 4 5
ð27Þ
ð28Þ ð29Þ ð30Þ ð31Þ ð32Þ ð33Þ
Example 3. The Lax equation is given by ut þ 10uuxxx þ 20ux uxx þ 30u2 ux þ uxxxxx ¼ 0. Following the methodology presented in Section 2, we see that the possible exact solutions are given by pffiffiffi ffiffiffi c 1p 4 cos2 cðx ctÞ ; u1 ðt; xÞ ¼ 2 2 pffiffiffi p ffiffiffi i c 4 cos2 u2 ðt; xÞ ¼ cðx ctÞ ; 2 2 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi c 1 pffiffiffiffiffi 10 90c þ 1800c 2 ðx ctÞA; u3 ðt; xÞ ¼ 10 90c þ 1800c cos2 @ 30 4 10 120 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi c 1 pffiffiffiffiffi 10 90c 1800c 2 u4 ðt; xÞ ¼ 10 90c 1800c cos2 @ ðx ctÞA; 10 120 30 4
ð34Þ
Q.M. Al-Mdallal, M.I. Syam / Chaos, Solitons and Fractals 33 (2007) 1610–1617
1617
0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi c 1 pffiffiffiffiffi 10 90c þ 1800c 2 @ 2 þ 10 90c 1800c cos ðx ctÞA; u5 ðt; xÞ ¼ 10 120 30 4 0pffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 pffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi c 1 pffiffiffiffiffi 10 90c 1800c 2 @ 2 u6 ðt; xÞ ¼ 10 90c 1800c cos ðx ctÞA. 10 120 30 4
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