Optik - Milivoj R. Belic

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Optik

journal homepage: www.elsevier.com/locate/ijleo

Original research article

Optical solitons and other solutions to Kudryashov’s equation withthree innovative integration norms

Elsayed M.E. Zayeda, Reham M.A. Shohiba, Anjan Biswasb,c,d,e, Mehmet Ekicif,*,Houria Trikig, Abdullah Kamis Alzahranic, Milivoj R. Belich

aMathematics Department, Faculty of Science, Zagazig University, Zagazig, EgyptbDepartment of Physics, Chemistry and Mathematics, Alabama A&M University, Normal, AL 35762-4900, USAcDepartment of Mathematics, King Abdulaziz University, Jeddah 21589, Saudi ArabiadDepartment of Mathematics and Statistics, Tshwane University of Technology, Pretoria 0008, South AfricaeDepartment of Applied Mathematics, National Research Nuclear University, 31 Kashirskoe Shosse, Moscow 115409, Russian FederationfDepartment of Mathematics, Faculty of Science and Arts, Yozgat Bozok University, 66100 Yozgat, Turkeyg Radiation Physics Laboratory, Department of Physics, Faculty of Sciences, Badji Mokhtar University, P. O. Box 12, 23000 Annaba, Algeriah Science Program, Texas A&M University at Qatar, PO Box 23874 Doha, Qatar

A R T I C L E I N F O

OCIS:060.2310060.4510060.5530190.3270.190.4370Keywords:SolitonsIntegrabilityJacobi's elliptic functions

A B S T R A C T

This paper applies three different integration algorithms, namely, unified auxiliary equationscheme, new mapping method and extended auxiliary equation criteria to Kudryashov's equationthat studies optical solitons with four nonlinear terms. These schemes reveal Jacobi's ellipticfunctions, solitons and other solutions to the model. These are classified as bright, dark, singularand combo optical solitons.

1. Introduction

One of the most innovative laws of refractive index that led to a new model, during 2018, for addressing the soliton pulsepropagation across intercontinental distances is Kudryashov's equation (KE) [1–4]. This model comes with four nonlinear forms thatare bound together. The special cases of KE is the well known and well familiar nonlinear Schrödinger's equation (NLSE) that is themost common form visible all across the literature. The two immediate special cases of KE are NLSE with power law and dual-powerlaw nonlinearity, whose further special cases are NLSE with Kerr law and parabolic law nonlinear forms. Thus KE, is the latest andmost generalized version of NLSE as it now stands. An important note in this context would be that the model is not yet experi-mentally proven. It is purely a theoretical form of refractive index as of now.

KE, thus being the latest model to address soliton dynamics through optical fibers, needs to be studied from its integrabilitystandpoint. Sure enough, it will not pass Painleve test of integrability. Thus inverse scattering transform is an epic failure based onPainleve conjecture. However, it is not the end of the road. A few integration schemes have led to the retrieval of soliton solutions tothe model. These are the method of undetermined coefficients, F -expansion scheme and the method of extended trial function [1–4].There exists a wide variety of algorithms that are applicable to nonlinear optics models. These include variational iteration scheme,mapping method, traveling waves, first integral approach and several others [5–15].

https://doi.org/10.1016/j.ijleo.2020.164431Received 6 January 2020; Accepted 16 February 2020

⁎ Corresponding author.E-mail address: [email protected] (M. Ekici).

Optik - International Journal for Light and Electron Optics 211 (2020) 164431

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It is about time to revisit the model one more time with a fresher perspective. Thus paper thus addresses the model with threeapproaches that have not been applied to KE in the past. These are unified auxiliary equation scheme, new mapping method andextended auxiliary equation criteria. These yield soliton solutions to the model along with several other solutions that emerge as abyproduct. The solutions are expressed in terms of Jacobi's elliptic function which, in the limiting case when the modulus of ellipticityapproaches zero or unity, gives soliton solutions. These are all enlisted in the rest of the paper after drawing a quick pen–picture tothese schemes.

1.1. Governing model

KE that has been proposed during 2019 reads as follows [4]:

⎜ ⎟+ + ⎛⎝

+ + + ⎞⎠

= = −bq

bq

b q b q q iiq aq| | | |

| | | | 0, 1 ,t n nn n

xx12

23 4

2

(1)

where a, b1, b2, b3 and b4 are non zero constants, while the independent variables x and t represent spatial and temporal variables,respectively. The dependent variable q x t( , ) is the complex valued describing the pulse profile. Here, a group velocity dispersion(GVD), while the coefficients b b b, ,1 2 3and b4 give self-phase modulation (SPM). The first term in Eq. (1) accounts for linear temporalevolution. When = =b b 01 2 , the model (1) transforms to the dual-power law of refractive index, while if = = =b b b 01 2 3 , it convertsto the power law. When =n 1, and = = =b b b 01 2 3 , it reduces to the well-known NLSE. When =n 2, along with =b 02 , it reduces toNLSE with anti–cubic nonlinearity.

Eq. (1) has been discussed using the traveling wave reduction, F–expansion method and undetermined coefficients where its exactsolutions have been located [1–3]. The objective of the current work is to construct the optical solitons and other solutions for Eq. (1)using three strategic integration algorithms mentioned in the abstract. To our best knowledge, Eq. (1) was not studied elsewhereemploying the unified auxiliary equation method, the new mapping method and the extended auxiliary equation method.

2. Mathematical preliminaries

In this section, we assume that Eq. (1) possesses solution in the form:

= − + + = −q x t φ ξ i ωt θ ξ x υt( , ) ( ) exp[ ( kx )], ,0 (2)

where υ, k, ω and θ0 are all non-zero constants to be determined. The parameter υ represents the soliton velocity. From the phasecomponent, k gives the frequency of the soliton, while ω is the wave number and θ0 is the phase constant. Here, φ ξ( ) is a real valuedfunction which stands for the pulse shape. Inserting (2) into (1) and distinguishing the real and imaginary parts, one has

″ − + + + + + =− − + +aφ ω φ b φ b φ b φ b φ( ak ) 0,n n n n21

1 22

13

14

1 2 (3)

and

= −υ 2ak, (4)

where (4) gives the velocity of the soliton. Setting

=φ ξ H ξ( ) ( ),n1

(5)

where H ξ( ) is a new function of ξ , we get the equation

″ + − ′ + + − + + + =a n H n b n b H n ω H n b H n b HanHH (1 ) ( ak ) 0.2 21

22

2 2 2 23

3 24

4 (6)

Balancing ″HH with H4 in (6), yields the balancenumber =N 1. The problem now is to solve Eq. (6) using the following threeintegration schemes.

3. Unified auxiliary equation method

According to this method, we assume that Eq. (6) has the formal solution:

= + +H ξ L L f ξ R g ξ( ) ( ) ( ),0 1 1 (7)

where L0, L1 and R1 are constants to be determined, such that ≠L 01 or ≠R 01 , while the functions f ξ( ) and g ξ( ) satisfy the auxiliaryODEs:

′ =f ξ f ξ g ξ( ) ( ) ( ), (8)

′ = + + −g ξ q g ξ r f ξ( ) ( ) ( ),12

12 (9)

= −⎡⎣

+ + ⎤⎦

−g ξ q r f ξ c f ξ( )2

( ) ( ) ,21

1 2 2(10)

where q1, r1and c are constants. Substituting (7) along with (8)–(10) into Eq. (6) and collecting all terms of the same order of f ξ g ξ( ) ( )

E.M.E. Zayed, et al. Optik - International Journal for Light and Electron Optics 211 (2020) 164431

2

and setting them to zero, we get the following system of algebraic equations:

+ − + − + =− − − + + =

+ + − − − −+ − + + − =− − − − + +− + =

+ + − − − + + + +

− + + + + − + =

− + =

⎡⎣

⎛⎝

− + + − ⎞⎠

+ ⎤⎦

=

+ + =

− ⎡⎣

− − + + + − − ⎤⎦

=

− + + − − + =

− ⎡⎣

− + + ⎤⎦

=

− + + − =

− ⎡⎣

+ − ⎤⎦

=

− ⎡⎣

+ + ⎤⎦

=

−

−

−

−

−

f ξ n L n b c R R Lf ξ n L R b L L R b n L L b n L bf ξ q R b R n R n R b n q L R b n L L R b

n L L b L n cωR n L L b n ωLf ξ n q L L R b L L n L L n q L R b n L L b n L L b

n ωL L n L b

f ξ n b R q R q q L b r L b q L b q ω b L b L b

n L ω n b L L L

f ξ L n r R L b b

f ξ R n b q R b L ω b L

f ξ R r n R b a n

g ξ n R b L L b L ω q b R b q R b

f ξ g ξ R n q R b ω L b L b

f ξ g ξ n L L b b

f ξ g ξ L R n R b c n n L b

f ξ g ξ L R r n R b a n

f ξ g ξ R L b b

( ): ac( 1)(cR ) ( 6cL ) 0,( ): 12 cL 2acnL 3 cn 4 0,( ): 2 cn 2acnq ack 6 cL 6 ak 3 cn

6 aq 3 0,( ): 12 anq 2ak 3 4 3

2 0,

( ): 12

[2 ( cr ) (2ak 12 6 6 2 ) 2 2 2 ]

( ak ) 12

anr ( 6cR ) 12

ar ( 2cR ) 0,

( ): 6 ( 14

) 0,

( ): nr 3 12

( ak ) 32

aq 0,

( ): 14

[ ( 1)] 0,

( ): 2 2 32

( ak 2 ) 12

( ) 0,

( ) ( ): nL [ (4 2( ak ) 12 6 ) aq ] 0,

( ) ( ): 2nR 2 (cR 3 )( 14

) acL 0,

( ) ( ): 2 [2 ac( 1) 2 ] 0,

( ) ( ): 2 ( 12

) 0,

( ) ( ): nr 2nR ( 14

) aL 0.

412

12 2

42

14

12

12

14

3 20 1 1

24 0 1

21 1

23

20 1

34

213

32 2

1 14

4 1 12 2 2

12 2

02

12

42

1 12

12

42 2

12 2

0 12

32

02

12

4 1 12 2

12 2

0 12

32

12

21 0 1 1

24 1 0 1

2 20 1

21 1 1

23

203

1 42

02

1 32

0 12

1 2

0 24 1

412

1 12 2

1 1 02

4 1 12

4 1 0 3 1 4 04

3 03

1

202 2 2

2 0 1 12

12

1 12

12

11

21 1

20 4 3

21 1

24 1 1

24 0

2 23 0 1

412

12 2

12

4

21 4 0

302

3 02

1 4 12

2 1 12

3

1 1 1 12

42

02

4 0 3 1

21 1

212

0 4 3 0

31 1

212

42

12

4

11 1 1

212

4

21 1 1

20 4 3 0 (11)

On solving the above algebraic equations (11) with the aid of Maple or Mathematica to get the following two cases:Case-1

= =+ − − +

+= − +

+= ± − + =

=− + − + − + + − +

+

=+ − + − +

+

n n ωn q n k n b n

n b nL n b

n bR a n

n bL

ba b n n q n n n b b n n b n

n n b

bn n b b n n b n

n b n

,2ab ( 2) (2 ) 3 ( 1)

2 ( 2), ( 1)

2( 2), ( 1) , 0,

16 ( 1)( 2) (2cr ) ( 1)( 1) [8aq ( 2) ( 1)]16 ( 2)

,

( 1)( 2) [4aq ( 2) ( 1)]4 ( 2)

.

42

12 2 2

32

24

2 03

41 2

41

1

242 2 4

1 12 2 2

32

1 42 2

32

4 443

23 1 4

2 232

242 3 (12)

From (2), (5), (7) and (12), then we have the solutions:

= ⎧⎨⎩

− ++

± − + ⎫⎬⎭

− + +q x t n bn b

a nn b

g ξ( , ) ( 1)2( 2)

( 1) ( ) e ,n

i ωt θ3

42

4

1

( kx )0

(13)

provided <b b 03 4 and <ab 04 .Case-2

= = −+ + + + + + +

+= − +

+

= ± + = = −− + + + + + +

+

= −+ − + + +

+

n n ωn b n b b n q k b q n

n b nL n b

n b

L nn b

R bn a b n q b b n n n n b

n n b

bn n b b n n b n

n b n

,2ak (8ak 3 ) [2ab ( 4 ) 3 ] 8ab ( 1)

2 ( 2), ( 1)

2( 2),

ac( 1) , 0,( 1)[8 cr ( 2) 4an ( 1)( 2) ( 1) ]

16 ( 2),

( 1)( 2) [2aq ( 2) ( 1)]4 ( 2)

.

2 44

3 24 3

2 24 1

232

4 12

42 0

3

4

1 24

1 1

2 21 4

2 4 21 3

24

2 4 234

4 443

23 1 4

2 232

242 3 (14)

From (2), (5), (7) and (14), then we have the solutions:

= ⎧⎨⎩

− ++

± + ⎫⎬⎭

− + +q x t n bn b

nn b

f ξ( , ) ( 1)2( 2)

ac( 1) ( ) e ,n

i ωt θ3

42

4

1

( kx )0

(15)

provided acb 04 .


3

3.1. Solutions to Eq. (1) with case-1

For case 1, Eq. (1) has the following families of Jacobi elliptic function solutions:Family-1. If = −q m(1 2 )1

2 , =r m212, = −c m( 1)2 , then,

= =f ξ ξ m g ξξ m ξ m

ξ m( ) nc( , ), ( )

sn( , ) dn( , )cn( , )

.(16)

From (13) and (16), we have

⎜ ⎟= ⎧⎨⎩

− ++

± − + ⎛⎝

+ ++

⎞⎠

⎫⎬⎭

− + +q x t n bn b

a nn b

x m x mx m

( , ) ( 1)2( 2)

( 1) sn[ 2akt, ] dn[ 2akt, ]cn[ 2akt, ]

e .n

i ωt θ3

42

4

1

( kx )0

(17)

In particular, if → −m 1 , then we have the dark soliton solutions:

= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x( , ) ( 1)2( 2)

( 1) tanh[ 2akt] e ,n

i ωt θ3

42

4

1

( kx )0

(18)

while if → +m 0 , then we have the periodic solutions:

= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x( , ) ( 1)2( 2)

( 1) tan[ 2akt] e .n

i ωt θ3

42

4

1

( kx )0

(19)

Note that, the solutions (18) and (19) are equivalent to the solutions (46) and (61) in the article [1], respectively.Family-2. If = − +q m( 2 )1

2 , =r 21 , = −c m(1 )2 , then,

= =f ξ ξ m g ξm ξ m ξ m

ξ m( ) nd( , ), ( )

sn( , ) cn( , )dn( , )

.2

(20)


= ⎧⎨⎩

− ++

± − + ⎡⎣⎢

+ ++

⎤⎦⎥

⎫⎬⎭

− + +q x t n bn b

a nn b

m x m x mx m

( , ) ( 1)2( 2)

( 1) sn[ 2akt, ]cn[ 2akt, ]dn[ 2akt, ]

e .n

i ωt θ3

42

4

21

( kx )0

(21)

In particular, if → −m 1 , then we have the same dark soliton solutions (18).Family-3. If = −q m(1 2 )1

2 , = − + =r m c m( 2 2 ),12 2, then,

= =−

f ξ ξ m g ξξ m ξ m

ξ m( ) cn( , ), ( )

sn( , ) dn( , )cn( , )

.(22)


= ⎧⎨⎩

− ++

± − + ⎡⎣⎢

+ ++

⎤⎦⎥

⎫⎬⎭

− + +q x t n bn b

a nn b

x m x mx m

( , ) ( 1)2( 2)

( 1) sn[ 2akt, ] dn[ 2akt, ]cn[ 2akt, ]

e .n

i ωt θ3

42

4

1

( kx )0

(23)

In particular, if → −m 1 , then we have the dark soliton solutions (18), while if → +m 0 , then we have the same periodic solutions (19).Family-4. If = − +q m( 2 )1

2 , = − +r m( 2 2 )12 , = −c 1, then,

= =−

f ξ ξ m g ξξ m

ξ m ξ m( ) cs( , ), ( )

dn( , )sn( , )cn( , )

.(24)


= ⎧⎨⎩

− ++

± − + ⎡⎣⎢

++ +

⎤⎦⎥

⎫⎬⎭

− + +q x t n bn b

a nn b

x mx m x m

( , ) ( 1)2( 2)

( 1) dn[ 2akt, ]sn[ 2akt, ]cn[ 2akt, ]

e .n

i ωt θ3

42

4

1

( kx )0

(25)

In particular, if → −m 1 , then we have the singular soliton solutions:

= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x( , ) ( 1)2( 2)

( 1) coth[ 2akt] e ,n

i ωt θ3

42

4

1

( kx )0

(26)


= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x( , ) ( 1)2( 2)

4 ( 1) csc[2( 2akt)] e .n

i ωt θ3

42

4

1

( kx )0

(27)

Note that, the solution (26) is equivalent to the solution (48) in the article [1].


4

Family-5. If = −q m(1 2 )12 , = −r m m(2 2 )1

2 4 , = −c 1, then,

= =−

f ξ ξ m g ξξ m

ξ m ξ m( ) ds( , ), ( )

cn( , )sn( , ) dn( , )

.(28)


= ⎧⎨⎩

− ++

± − + ⎡⎣⎢

++ +

⎤⎦⎥

⎫⎬⎭

− + +q x t n bn b

a nn b

x mx m x m

( , ) ( 1)2( 2)

( 1) cn[ 2akt, ]sn[ 2akt, ]dn[ 2akt, ]

e .n

i ωt θ3

42

4

1

( kx )0

(29)

In particular, if → −m 1 , then we have the same singular soliton solutions (26), while if → +m 0 , then we have the periodic solutions:

= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x( , ) ( 1)2( 2)

( 1) cot[ 2akt] e .n

i ωt θ3

42

4

1

( kx )0

(30)

Note that, the solution (30) is equivalent to the solution (62) in the article [1].Family-6. If = +q m(1 )1

2 , = −r m212, = −c 1, then,

= =−

f ξ ξ m g ξm ξ m

ξ m ξ m( ) dc( , ), ( )

(1 ) sn( , )cn( , ) dn( , )

.2

(31)


= ⎧⎨⎩

− ++

± − + ⎡⎣⎢

− ++ +

⎤⎦⎥

⎫⎬⎭

− + +q x t n bn b

a nn b

m x mx m x m

( , ) ( 1)2( 2)

( 1) (1 ) sn[ 2akt, ]cn[ 2akt, ]dn[ 2akt, ]

e .n

i ωt θ3

42

4

21

( kx )0

(32)

In particular, if → +m 0 , then we have the same periodic solutions (19).Family-7. If = − + = − = −q m r c( 1 2 ), ,1

12

21

12

14 , then,

=±

= ∓f ξξ m

ξ mg ξ ξ m( )

cn( , ) 1sn( , )

( ) ds( , ).(33)


= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x m( , ) ( 1)2( 2)

( 1) ds[ 2akt, ] e .n

i ωt θ3

42

4

1

( kx )0

(34)

In particular, if → −m 1 , then we have the singular soliton solutions:

= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x( , ) ( 1)2( 2)

( 1) csch[ 2akt] e ,n

i ωt θ3

42

4

1

( kx )0

(35)


= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x( , ) ( 1)2( 2)

( 1) csc[ 2akt] e .n

i ωt θ3

42

4

1

( kx )0

(36)

Note that, the solutions (35) and (36) are equivalent to the solutions (49) and (59) in the article [1], respectively.Family-8. If = − + = − = −q m r m c m( 1), (1 ), (1 )1

12

21

12

2 14

2 , then,

=±

= ∓f ξξ m

m ξ mg ξ m ξ m( )

dn( , )sn( , ) 1

, ( ) cd( , ).(37)


= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a n mn b

x m( , ) ( 1)2( 2)

( 1) cd[ 2akt, ] e .n

i ωt θ3

4

2

24

1

( kx )0

(38)

Family-9. If = − +q m m( 6 1)12 , = −r 21 , =c m4 −m( 1)2, then,

=−

= − ⎡⎣⎢

+−

⎤⎦⎥

f ξξ m

m ξ mg ξ ξ m ξ m

m ξ mm ξ m

( )sn( , )

sn ( , ) 1( ) cs( , ) dn( , )

sn ( , ) 1sn ( , ) 1

.2

2

2 (39)


⎜ ⎟= ⎧⎨⎩

− ++

± − + ⎡⎣⎢

+ + ⎛⎝

+ ++ −

⎞⎠

⎤⎦⎥

⎫⎬⎭

− + +q x t n bn b

a nn b

x m x m m x mm x m

( , ) ( 1)2( 2)

( 1) cs[ 2akt, ] dn[ 2akt, ] sn [ 2akt, ] 1sn [ 2akt, ] 1

e .n

i ωt θ3

42

4

2

2

1

( kx )0

(40)


5

In particular, if → −m 1 , then we have dark-singular combo solitons:

= ⎧⎨⎩

− ++

± − + + + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x x( , ) ( 1)2( 2)

( 1) (coth[ 2akt] tanh[ 2akt]) e ,n

i ωt θ3

42

4

1

( kx )0

(41)

while if → +m 0 , then we have the same periodic solutions (30).Family-10. If = + +q m m( 6 1)1

2 , = −r 2,1 = −c m4 +m( 1)2, then,

=+

= − ⎡⎣⎢

−+

⎤⎦⎥

f ξξ m

m ξ mg ξ ξ m ξ m

m ξ mm ξ m

( )sn( , )

sn ( , ) 1( ) cs( , ) dn( , )

sn ( , ) 1sn ( , ) 1

.2

2

2 (42)


⎜ ⎟= ⎧⎨⎩

− ++

± − + ⎡⎣⎢

+ + ⎛⎝

+ −+ +

⎞⎠

⎤⎦⎥

⎫⎬⎭

− + +q x t n bn b

a nn b

x m x m m x mm x m

( , ) ( 1)2( 2)

( 1) cs[ 2akt, ] dn[ 2akt, ] sn [ 2akt, ] 1sn [ 2akt, ] 1

e .n

i ωt θ3

42

4

2

2

1

( kx )0

(43)

In particular, if → −m 1 , then we have the solitary solutions:

⎜ ⎟= ⎧⎨⎩

− ++

± − + ⎛⎝

++ + +

⎞⎠

⎫⎬⎭

− + +q x t n bn b

a nn b

xx x

( , ) ( 1)2( 2)

( 1) sech [ 2akt]tanh[ 2akt] tanh [ 2akt]

e ,n

i ωt θ3

42

4

4

3

1

( kx )0

(44)

while if → +m 0 , then we have the same periodic solutions (30).Family-11. If = −q m(2 1)1

12

2 , = −r112 , = −c 1

4 , then,

=±

= ±f ξξ m

ξ mg ξ ξ m( )

sn( , )1 cn( , )

( ) ds( , ).(45)


= ⎧⎨⎩

− ++

± − + + ⎫⎬⎭

− + +q x t n bn b

a nn b

x m( , ) ( 1)2( 2)

( 1) ds[ 2akt, ] e .n

i ωt θ3

42

4

1

( kx )0

(46)

In particular, if → −m 1 , then we have the same singular soliton solutions (35), while if → +m 0 , then we have the same periodicsolutions (36).

Similarly, we can find many other solutions of Eq. (1) using (15) which are omitted here for simplicity.

4. New mapping method

According to the new mapping method, Eq. (6) has the following formal solution:

= + +H ξ α α F ξ α F ξ( ) ( ) ( ),0 1 22 (47)

where α α,0 1 and α2 are constants to be determined, such that ≠α 02 , while the function F ξ( ) satisfies the following first orderequation

′ = + + +F ξ r p F ξq

F ξ s F ξ( ) ( )2

( )3

( ),22

2 2 4 6(48)

where r p q s, , ,2 2 are constants, such that ≠s 0. Substituting (47) along with (48) into Eq. (6), collecting the coefficients of each power′ = … =F ξ F ξ e a b h j( ) ( ( )) , ( 0, , , , , 0, 1)e j ,and setting these coefficients to zero, we have the following algebraic equations as follows:


6

+ + =

+ + =

+ + + + + + + =

+ + + + + =

⎛⎝

− + + ⎞⎠

+ ⎡⎣

+ + ⎤⎦

+ +

+ + =

− + + + + + + + + =

⎡⎣

+ − + + ⎤⎦

+ ⎡⎣

+ − + ⎤⎦

+ −

+ + =

− ⎡⎣

⎛⎝

− − + + − ⎞⎠

− − − ⎤⎦

=

+ − + + + + − + =

F ξ α n n b α

F ξ α α n n b α α

F ξ α n α n nα α n α α b b n α α b

F ξ α α n nα α nα b n α α α b α b

F ξ n α α b ω α b n α α α b b α b n α α α

a pα q α

F ξ n α α ω α b α b n α α b b aα n pα q α α α

F ξ n α α b α b α ω b n α α b α b ω aα n α p r α

a pα r α

F ξ α n α b α b α ω b pα r α aα r

F ξ n b α b α α ω b α b α α α α

( ): 43

as ( 1) 0,

( ): 13

as (4 7 ) 4 0,

( ): aq ( 2) 13

as( (1 2 ) 8 ) (4 ) 6 0,

( ): 2aq ( 1) (as 12 ) (4 3 ) 0,

( ): 2 13

( ak ) (4 ) 13

aq ( 16

)

16

(8 ) 0,

( ): ( 2( ak ) 12 6 ) 4 ( 14

) ( ) 4ap 0,

( ): 34

12

( ak ) 14

34

2 13

( ak ) ( 12

)

14

( 4 ) 0,

( ): 2 2 32

( ak ) 12

12

an( 2 ) 2 0,

( ): [ ( ak ) ] 2anr ( 12

) ar 0.

822 2

4 24

71 2

24 1 2

3

62 2

212

0 22

23

0 4 32

12

22

4

52 1 2 0 1 2

24

21 2 1

24 2 3

4 222

02

42

0 32

12

2 0 4 3 14

4 2 0 2 12

22

2 12

3 21 2

202

4 0 32

13

0 4 3 1 2 2 0 1 2

2 22 0

34 0

23 0

22

212

02

4 0 32

2 0 2 2

12

2 22

12

03

4 02

3 02

2 0 2 2 2 2

0 24 0

43 0

302 2

2 0 1 2 0 2 12

2 12

(49)

With reference to [15], there are three types of solutions of the algebraic equations (49) as follows:

Type1. Substituting =s qp

316

22, =r p

q21627

2

2into the algebraic equations (49) and solving them by Maple or Mathematica, we have the

following result:

= =+ + + + − +

+ +− = − + + +

+ +

=− + − − + − + − + + − +

+ +

= −− + − + − + − − + + + − +

+ +

= = = −+ +

+ + +

n n ωb n α α b b n n b n

b n np n n α b b n

n n

bα α b n n α b b n n α b b n n n b n n

b n n

bα b n n α b b n n n α b b n n n b n n

b n n

α α α αn n

n n α b b n

,2 ( 2) 2 ( 2)( 1) ( 1)

( 2) ( 1)ak , [2( 2) ( 1)]

4ab ( 2) ( 1),

[5 ( 1)( 2) 6 ( 1)( 2) 3 ( 2)( 1)( 1) 4 ( 1)( 1) ]27 ( 1)( 2)

,

16 ( 2)( 2) 3 ( 1)( 2)( 2) 15 ( 2)( 2)( 1) 2 ( 2)( 1)27 ( 1)( 2)

,

, 0,aq ( 3 2)

[2( 2) ( 1)].

42 2

02

0 3 4 32 2

42

22

0 4 32

42

10 0

343 3

02

3 42 2 2

0 32

42

33 3

42 3

203

43 3

02

3 42 2

0 32

42

33 3

42 3

0 0 1 22

2

20 4 3

(50)

From (2), (5), (47), (50) and −F ξ F ξ( ) ( )1 4 of step 5 in [15], we obtain the soliton solutions:

=

⎧

⎨⎪

⎩⎪

− + + ++

⎡

⎣

⎢⎢⎢

⎛⎝

+ ⎞⎠

+ ⎛⎝

+ ⎞⎠

⎤

⎦

⎥⎥⎥

⎫

⎬⎪

⎭⎪

+ + ++ +

+ + ++ +

− + +q x t α n α b b nb n

x t

x t( , ) 4[2( 2) ( 1)]

3 ( 2)

tanh ϵ ( 2ak )

3 tanh ϵ ( 2ak )e ,

n n α b b nn n

n n α b b nn n

n

i ωt θ0

0 4 3

4

2 [2( 2) ( 1)]12ab ( 2) ( 1)

2 [2( 2) ( 1)]12ab ( 2) ( 1)

1

( kx )

2 0 4 3 2

4 2

2 0 4 3 2

4 2

0

(51)

and

=

⎧

⎨⎪

⎩⎪

− + + ++

⎡

⎣

⎢⎢⎢

⎛⎝

+ ⎞⎠

+ ⎛⎝

+ ⎞⎠

⎤

⎦

⎥⎥⎥

⎫

⎬⎪

⎭⎪

+ + ++ +

+ + ++ +


x t

x t( , ) 4[2( 2) ( 1)]

3 ( 2)

coth ϵ ( 2ak )

3 coth ϵ ( 2ak )e ,

n n α b b nn n

n n α b b nn n

n

i ωt θ0

0 4 3

4

2 [2( 2) ( 1)]12ab ( 2) ( 1)

2 [2( 2) ( 1)]12ab ( 2) ( 1)

1

( kx )

2 0 4 3 2

4 2

2 0 4 3 2

4 2

0

(52)

provided + + + ab 04 . Also, the periodic solutions:

=

⎧

⎨⎪

⎩⎪

+ + + ++

⎡

⎣

⎢⎢⎢⎢

⎛⎝

− + ⎞⎠

− ⎛⎝

− + ⎞⎠

⎤

⎦

⎥⎥⎥⎥

⎫

⎬⎪

⎭⎪

+ + ++ +

+ + ++ +


x t

x t( , ) 4[2( 2) ( 1)]

3 ( 2)

tan ϵ ( 2ak )

3 tan ϵ ( 2ak )e ,

n n α b b nn n

n n α b b nn n

n

i ωt θ0

0 4 3

4

2 [2( 2) ( 1)]12ab ( 2) ( 1)

2 [2( 2) ( 1)]12ab ( 2) ( 1)

1

( kx )

2 0 4 3 2

4 2

2 0 4 3 2

4 2

0

(53)

and


7

=

⎧

⎨⎪

⎩⎪

+ + + ++

⎡

⎣

⎢⎢⎢⎢

⎛⎝

− + ⎞⎠

− ⎛⎝

− + ⎞⎠

⎤

⎦

⎥⎥⎥⎥

⎫

⎬⎪

⎭⎪

+ + ++ +

+ + ++ +


x t

x t( , ) 4[2( 2) ( 1)]

3 ( 2)

cot ϵ ( 2ak )

3 cot ϵ ( 2ak )e ,

n n α b b nn n

n n α b b nn n

n

i ωt θ0

0 4 3

4

2 [2( 2) ( 1)]12ab ( 2) ( 1)

2 [2( 2) ( 1)]12ab ( 2) ( 1)

1

( kx )

2 0 4 3 2

4 2

2 0 4 3 2

4 2

0

(54)

provided + + + >b n α b b n[2( 2) ( 1)] 04 0 4 3 , <ab 04 and = ±ϵ 1.

Type2. Substituting =s qp

316

22, =r 02 into the algebraic equations (49) and solving them by Maple or Mathematica, we have the

following result:

= =+ + + + − +

+ +− = − + + +

+ +

= −− + + + − + − +

+ +

= − + + −+

= = = −+ +

+ + +

n n ωα b n α b b n n b n

n n bp n n α b b n

n n

bα α b n n α b b n n b n n

n n b

b α b n α b n n bb n

α α α αn n

n n α b b n

,2 ( 2) 2 ( 2)( 1) ( 1)

( 1)( 2)ak , [2( 2) ( 1)]

4ab ( 2) ( 1),

[ ( 1)( 2) 2 ( 2)( 1) ( 1)( 1) ]( 1)( 2)

,

[( 4) ( 1)( 2) ]( 2)

, , 0,aq ( 1)( 2)

[2( 2) ( 1)].

02

42 2

0 3 4 32 2

24

22

0 4 32

42

102

02

42 2

0 3 42

32 2

24

20 3

20 4 3

42 0 0 1 2

22

0 4 3 (55)

From (2), (5), (47), (55) and F ξ F ξ( ), ( )5 6 of step 5 in [15], we obtain the dark soliton solution:

⎜ ⎟=⎧⎨⎩

− + + ++

⎡

⎣⎢ ± ⎛

⎝− + + +

+ ++ ⎞

⎠

⎤

⎦⎥

⎫⎬⎭


n n α b b nn n

x t( , ) [2( 2) ( 1)]2 ( 2)

1 tanh ϵ [2( 2) ( 1)]4ab ( 2) ( 1)

( 2ak ) e ,n

i ωt θ0

0 4 3

4

20 4 3

2

42

1

( kx )0

(56)

and the singular soliton solution

⎜ ⎟=⎧⎨⎩

− + + ++

⎡

⎣⎢ ± ⎛

⎝− + + +

+ ++ ⎞

⎠

⎤

⎦⎥

⎫⎬⎭


n n α b b nn n

x t( , ) [2( 2) ( 1)]2 ( 2)

1 coth ϵ [2( 2) ( 1)]4ab ( 2) ( 1)

( 2ak ) e .n

i ωt θ0

0 4 3

4

20 4 3

2

42

1

( kx )0

(57)

provided + + + <b n α b b n[2( 2) ( 1)] 04 0 4 3 and <ab 04 .Type3. Substituting =r 02 into the algebraic equations (49) and solving them by Maple or Mathematica, we have the following

result:

= =+ + − −

+− = =

= − + =− ⎡

⎣− + + − ⎤

⎦+

= −− ⎡

⎣− + + − ⎤

⎦+

=+ ⎡

⎣− − ⎤

⎦+

+

+

+

+

n n ωn α b n sα b

n s nα α α

α nn b

bα n α b n n sα b

n s n

bα n α b n n sα b

n s n

bb n q α

n

,4aps( 1) 3nq 6

( 1)ak , , 0,

4as( 1)3

,( 1) nq 4aps( 1)

( 1),

( 2) 3 nq 8aps( 1) 4

2 ( 1),

( 2) 4ns

2ns( 1),

nb

nb

nb

nb

2 0 43as( 1) 2

02

4

22

0 0 1

2 24

1

02

0 2 43as( 1) 2

02

4

2

2

0 0 2 43as( 1) 2

02

4

2

3

4 23as( 1)

0

4

4

4

4

(58)

provided <sab 04 . From (2), (5), (47), (58) and −F ξ F ξ( ) ( )7 20 of step 5 in [15], we obtain the soliton solutions:

= ⎧⎨⎩

− − + ⎡

⎣⎢

+− + +

⎤

⎦⎥

⎫⎬⎭

− + +q x t α nn b

p x tq p x t

( , ) 6pq 4as( 1)3

sech [ ( 2ak )]3 4ps(1 ϵ tanh[ ( 2ak )])

e ,n

i ωt θ0 2 2

4

2

22 2

1

( kx )0

(59)

provided > <p q0, 02 and <s 0.

= ⎧⎨⎩

+ − + ⎡

⎣⎢

+− + +

⎤

⎦⎥

⎫⎬⎭

− + +q x t α nn b

p x tq p x t

( , ) 6pq 4as( 1)3

csch [ ( 2ak )]3 4ps(1 ϵ coth[ ( 2ak )])

e ,n

i ωt θ0 2 2

4

2

22 2

1

( kx )0

(60)

provided > >p q0, 02 and <s 0.

= ⎧⎨⎩

− − + ⎡⎣⎢

++ +

⎤⎦⎥

⎫⎬⎭

− + +q x t α p nn b

p x tq p x t

( , ) 6 4as( 1)3

sech [ ( 2ak )]3 4ϵ 3ps tanh[ ( 2ak )]

e ,n

i ωt θ0 2

4

2

2

1

( kx )0

(61)

provided > s 0.


8

= ⎧⎨⎩

+ − + ⎡⎣⎢

++ +

⎤⎦⎥

⎫⎬⎭


p x tq p x t

( , ) 6 4as( 1)3

csch [ ( 2ak )]3 4ϵ 3ps coth[ ( 2ak )]

e ,n

i ωt θ0 2

4

2

2

1

( kx )0

(62)

provided > >p q0, 02 and >s 0. We obtain the bright soliton solutions:

= ⎧⎨⎩

+ − + ⎡

⎣⎢ + − +

⎤

⎦⎥

⎫⎬⎭

− + +q x t α p nn b M p x t M q

( , ) 12 4as( 1)3

12 cosh [ϵ ( 2ak )] ( 3 )

e ,n

i ωt θ0 2

42

2

1

( kx )0

(63)

= ⎧⎨⎩

+ − + ⎡

⎣⎢ + −

⎤

⎦⎥

⎫⎬⎭

− + +q x t α p nn b M p x t q

( , ) 12 4as( 1)3

1ϵ cosh[2 ( 2ak )] 3

e ,n

i ωt θ0 2

4 2

1

( kx )0

(64)

provided > M q(9 48ps) 022 . We obtain the singular soliton solutions:

= ⎧⎨⎩

+ − + ⎡

⎣⎢ + + −

⎤

⎦⎥

⎫⎬⎭

− + +q x t α p nn b M p x t M q

( , ) 12 4as( 1)3

12 sinh [ϵ ( 2ak )] ( 3 )

e ,n

i ωt θ0 2

42

2

1

( kx )0

(65)

provided > M 0.

= ⎧⎨⎩

+ − + ⎡

⎣⎢ − + −

⎤

⎦⎥

⎫⎬⎭


( , ) 12 4as( 1)3

1ϵ sinh[2 ( 2ak )] 3

e ,n

i ωt θ0 2

4 2

1

( kx )0

(66)

provided >p 0, <q 02 and <M 0. and = ±ϵ 1. Also, the periodic solutions:

= ⎧⎨⎩

− − + ⎡⎣⎢

− ++ − − +

⎤⎦⎥

⎫⎬⎭


p x tq p x t

( , ) 6 4as( 1)3

sec [ ( 2ak )]3 4ϵ 3ps tan[ ( 2ak )]

e ,n

i ωt θ0 2

4

2

2

1

( kx )0

(67)

= ⎧⎨⎩

− − + ⎡⎣⎢

− ++ − − +

⎤⎦⎥

⎫⎬⎭


p x tq p x t

( , ) 6 4as( 1)3

csc [ ( 2ak )]3 4ϵ 3ps cot[ ( 2ak )]

e ,n

i ωt θ0 2

4

2

2

1

( kx )0

(68)

provided q 02 and >s 0.

= ⎧⎨⎩

− − + ⎡

⎣⎢

− +− − − +

⎤

⎦⎥

⎫⎬⎭


p x tM M q p x t

( , ) 12 4as( 1)3

sec [ϵ ( 2ak )]2 ( 3 )sec [ϵ ( 2ak )]

e ,n

i ωt θ0 2

4

2

22

1

( kx )0

(69)

= ⎧⎨⎩

+ − + ⎡

⎣⎢

− +− + − +

⎤

⎦⎥

⎫⎬⎭


p x tM M q p x t

( , ) 12 4as( 1)3

csc [ϵ ( 2ak )]2 ( 3 )csc [ϵ ( 2ak )]

e ,n

i ωt θ0 2

4

2

22

1

( kx )0

(70)

provided M 0.

= ⎧⎨⎩

+ − + ⎡

⎣⎢ − + −

⎤

⎦⎥

⎫⎬⎭


( , ) 12 4as( 1)3

1ϵ cos[2 ( 2ak )] 3

e ,n

i ωt θ0 2

4 2

1

( kx )0

(71)

= ⎧⎨⎩

+ − + ⎡

⎣⎢ − + −

⎤

⎦⎥

⎫⎬⎭


( , ) 12 4as( 1)3

1ϵ sin[2 ( 2ak )] 3

e ,n

i ωt θ0 2

4 2

1

( kx )0

(72)

provided M 0.

5. Extended auxiliary equation

According to this method, we assume that Eq. (6) has the same formal solution (47), while the function F ξ( ) satisfies the followingfirst order equation:

′ = + + +F ξ c c F ξ c F ξ c F ξ( ) ( ) ( ) ( ),20 2

24

46

6 (73)

where =c j d( 0, 2, , 6)j are constants to be determined. It is well known that Eq. (73) has the following solution:

= ⎡⎣⎢

− ± ⎤⎦⎥

F ξ cc

f ξ( ) 12

(1 ( )) ,4

6

12

(74)

where f ξ( ) could be expressed through the Jacobi elliptic functions ξ m ξ msn( , ), cn( , ), ξ mdn( , ) and so on. Here < <m0 1 is the


9

modulus of the Jacobi elliptic functions. Substituting (47) along with (73) into Eq. (6), collecting the coefficients of each power′ = … =F τ F τ l j( )( ( )) , ( 0, 1, 2, ,8, 0, 1)l j , and setting these coefficients to zero, we have the following set of algebraic equations:

+ + =+ + =+ + + + + + + =

+ + + + + =

⎛⎝

− + + ⎞⎠

+ ⎡⎣

+ + ⎤⎦

+ +

+ + =

− + + + + + + + + =

⎡⎣

+ − + + ⎤⎦

+ ⎡⎣

+ − + ⎤⎦

+ −

+ + =

− ⎡⎣

⎛⎝

− − + + − ⎞⎠

− − − ⎤⎦

=

+ − + + + + − + =

F ξ α n n b αF ξ α α n n b α αF ξ α n α n nα α n α α b b n α α bF ξ α α n nα α nα b n α α α b α b

F ξ n α α b ω α b n α α α b b α b n α α α

a c α c α

F ξ n α α ω α b α b n α α b b aα n c α c α α α

F ξ n α α b α b α ω b n α α b α b ω aα n α c c α

a c α c α

F ξ α n α b α b α ω b c α c α aα c

F ξ n b α b α α ω b α b α α α α

( ): 4ac ( 1) 0,( ): ac (4 7 ) 4 0,( ): 2ac ( 2) ac ( (1 2 ) 8 ) (4 ) 6 0,( ): 4ac ( 1) (3ac 12 ) (4 3 ) 0,

( ): 2 13

( ak ) (4 ) 13

2ac ( 16

)

16

(8 2 ) 0,

( ): ( 2( ak ) 12 6 ) 4 ( 14

) ( 2 ) 4ac 0,

( ): 34

12

( ak ) 14

34

2 13

( ak ) ( 12

)

14

( 4 ) 0,

( ): 2 2 32

( ak ) 12

12

an( 2 ) 2 0,

( ): [ ( ak ) ] 2anc ( 12

) ac 0.

86 2

2 24 2

4

76 1 2

24 1 2

3

64 2

26 1

20 2

223

0 4 32

12

22

45

4 1 2 0 1 6 22

42

1 2 12

4 2 3

4 222

02

42

0 32

12

2 0 4 3 14

4 4 0 2 12

2 22

4 12

3 21 2

202

4 0 32

13

0 4 3 1 2 2 4 0 2 1 2

2 22 0

34 0

23 0

22

212

02

4 0 32

2 0 2 0 2

2 12

0 22

12

03

4 02

3 02

2 2 0 0 2 2 0

0 24 0

43 0

302 2

2 0 1 0 0 2 12

0 12

(75)

By solving the system (75), the following solution set is recovered:

= =+ + + + − + + + +

+ +

=− − + + + + + +

+ +

= −− + − + + − + −

+ +

= = = − + + ++ +

= −+

= = =

n n ωn n n α b n n n n n α b n

n n n

bα n a α c α c n n n α b n n α b n

n n n

ba n n α c α c n α b n n n α b n

n n n

c c c c c n α α b n b na n n

cn α ba n

α α α α α

,4ac ( 2)( 1) 6 ( 2) ak ( 1)( 2) 6 ( 1)

( 3 2),

( 1)[4 ( )( 1)( 2) 3 ( 2) 2 ( 1)]( 3 2)

,

2 ( 4)( 1)(2 ) 3 ( 1)( 2) 4 ( 4)( 3 2)

,

, , [2 ( 2) ( 1)]2 ( 3 2)

,4 ( 1)

, , 0, .

22

02

42 2 2

0 32 2

10 0 2 2 0

203

42

02

32 2

2

20 2 2 0

202

32

03

42

2 2

0 0 2 2 42

2 0 4 32 6

222

40 0 1 2 2

(76)

From (2), (5), (47), (74) and (76), then we have the solutions:

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

f ξ( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

( ) e ,n

i ωt θ3

4

4 0 3

4

1

( kx )0

(77)

provided c 06 , then

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

+ ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nb n n

x t m( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

sn [2 ( 2) ( 1)]4am ( 1)( 2)

( 2ak ), e ,n

i ωt θ3

4

4 0 3

4

20 4 3

2

24

2

1

( kx )0

(78)

or

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

+ ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b m

n α b n b nb n n

x t m( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

ns [2 ( 2) ( 1)]4am ( 1)( 2)

( 2ak ), e ,n

i ωt θ3

4

4 0 3

4

20 4 3

2

24

2

1

( kx )0

(79)

provided <ab 04 . In particular, if → −m 1 , then we have the dark soliton solution:

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

+ ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nn n

x t( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

tanh [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ) e ,n

i ωt θ3

4

4 0 3

4

20 4 3

2

42

1

( kx )0

(80)

as well as the singular soliton solution:


10

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

+ ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nn n

x t( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

coth [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ) e .n

i ωt θ3

4

4 0 3

4

20 4 3

2

42

1

( kx )0

(81)

Note that, when = ⎡⎣

− − ⎤⎦

+ ++α a n

n bn bn b0

( 1) ( 1)2( 2)2 4

34, the solutions (80) and (81) are equivalent to the solutions (18) and (26), respectively.

Family 2. If = −c c mc0

(1 )32

43 2

62 , = −c c m

c2(5 )16

42 2

6, >c 06 , then

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

+ ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

m n b α n bn b

n α b n b nn n

x t m( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

sn [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ), e

,

ni ωt θ3

4

4 0 3

4

20 4 3

2

42

1

( kx )0

(82)

or

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

+ ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nn n

x t m( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

ns [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ), e ,n

i ωt θ3

4

4 0 3

4

20 4 3

2

42

1

( kx )0

(83)

provided <ab 04 . In particular, if → −m 1 , then we have the same dark soliton solution and singular soliton solution (80) and (81),respectively, while if → +m 0 , then we have the periodic solutions:

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

+ ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nn n

x t( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

csc [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ) e .n

i ωt θ3

4

4 0 3

4

20 4 3

2

42

1

( kx )0

(84)


− − ⎤⎦

+ ++α a n

n bn bn b0

( 1) ( 1)2( 2)2 4

34, the solution (84) is equivalent to the solution (36).

Family 3. If =c c

m c0 3243

262 , = +c c m

c m2(4 1)16

42 2

6 2 , <c 06 , then

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢

+ + ++ +

+ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nm n n

x t m( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

cn [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ), e ,n

i ωt θ3

4

4 0 3

4

20 4 3

2

42 2

1

( kx )0

(85)

or

⎜ ⎟= ⎧⎨⎩

− ++

± − ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢

+ + ++ +

+ ⎤

⎦⎥

⎫⎬⎭

− + +

q x t n bn b

m n b α n bn b

n α b n b nm n n

x t m( , ) ( 1)2( 2)

1 2( 2) ( 1)2( 2)

sd [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ),

e ,

n

i ωt θ

3

4

2 4 0 3

4

20 4 3

2

42 2

1

( kx )0 (86)

provided >ab 04 . In particular, if → −m 1 , then we have the bright soliton solution:

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢

+ + ++ +

+ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nn n

x t( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

sech [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ) e .n

i ωt θ3

4

4 0 3

4

20 4 3

2

42

1

( kx )0

(87)


− ⎤⎦

+ ++α a n m

n bn bn b0

( 1) ( 1)2( 2)

22 4

34, the solutions (85) and (87) are equivalent to the solutions (16) and (47) in the article

[1], respectively.

Family 4. If =−

c c m

c m0 32 ( 1)43 2

62 2 , = −

−c c m

c m2(5 4)

16 ( 1)42 2

6 2 , <c 06 , then

=⎧

⎨⎩

− ++

±−

⎡

⎣⎢ − + + +

− + ++ ⎤

⎦⎥

⎫

⎬⎭

+ + ++ − + +

( )q x t n b

n b mn α b n b n

m n nx t m( , ) ( 1)

2( 2) 1dn [2 ( 2) ( 1)]

4ab ( 1)( 1)( 2)( 2ak ), e ,

n b α n bn b

n

i ωt θ3

4

2( 2) ( 1)2( 2)

2

20 4 3

2

42 2

1

( kx )

4 0 34 0

(88)

or

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

− + ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nm n n

x t m( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

nd [2 ( 2) ( 1)]4ab ( 1)( 1)( 2)

( 2ak ), e ,n

i ωt θ3

4

4 0 3

4

20 4 3

2

42 2

1

( kx )0

(89)

provided − <mab ( 1) 042 .


11

Family 5. If =−

c c

c m0 32 (1 )43

62 2 , = −

−c c m

c m2(4 5)

16 ( 1)42 2

6 2 , >c 06 , then

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

− + ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nm n n

x t m( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

nc [2 ( 2) ( 1)]4ab (1 )( 1)( 2)

( 2ak ), e ,n

i ωt θ3

4

4 0 3

4

20 4 3

2

42 2

1

( kx )0

(90)

or

=⎧

⎨⎩

− ++

±−

⎡

⎣⎢ − + + +

− + ++ ⎤

⎦⎥

⎫

⎬⎭

+ + ++ − + +

( )q x t n b

n b mn α b n b n

m n nx t m( , ) ( 1)

2( 2) 1ds [2 ( 2) ( 1)]

4ab (1 )( 1)( 2)( 2ak ), e ,

n b α n bn b

n

i ωt θ3

4

2( 2) ( 1)2( 2)

2

20 4 3

2

42 2

1

( kx )

4 0 34 0

(91)

provided <ab 04 . In particular, if → +m 0 , then we have the same periodic solutions (84) and the periodic solutions:

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢ − + + +

+ ++ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nn n

x t( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

sec [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ) e .n

i ωt θ3

4

4 0 3

4

20 4 3

2

42

1

( kx )0

(92)


− − ⎤⎦

+ ++α a n

n bn bn b0

( 1) ( 1)2( 2)2 4

34, the solution (92) is equivalent to the solution (60) in the article [1].

Family 6. If =−

c c

c m0 32 (1 )43

62 2 , = −

−c c m

c m2(4 5)

16 ( 1)42 2

6 2 , >c 06 , then

⎜ ⎟= ⎧⎨⎩

− ++

± ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢

+ + ++ +

+ ⎤

⎦⎥

⎫⎬⎭

− + +q x t n bn b

n b α n bn b

n α b n b nn n

x t m( , ) ( 1)2( 2)

2( 2) ( 1)2( 2)

dn [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ), e ,n

i ωt θ3

4

4 0 3

4

20 4 3

2

42

1

( kx )0

(93)

or

⎜ ⎟= ⎧⎨⎩

− ++

± − ⎛⎝

+ + ++

⎞⎠

⎡

⎣⎢

+ + ++ +

+ ⎤

⎦⎥

⎫⎬⎭

− + +

q x t n bn b

m n b α n bn b

n α b n b nn n

x t m( , ) ( 1)2( 2)

1 2( 2) ( 1)2( 2)

nd [2 ( 2) ( 1)]4ab ( 1)( 2)

( 2ak ),

e ,

n

i ωt θ

3

4

2 4 0 3

4

20 4 3

2

42

1

( kx )0 (94)

provided >ab 04 . In particular, if → −m 1 , then we have the same bright soliton solutions (87).

6. Conclusions

This paper implements three innovative integration schemes to the KE that serves as the latest model to address soliton dynamicsin nonlinear optical fibers. While the model is at its infancy, the soliton solutions reported in this paper as well as earlier serve as astrong footing to advance further along with KE. One of the immediate project that needs to be talked about is the formulation of themodel for birefringent fibers and with DWDM topology. The model will also be applied to additional optoelectronic devices such asoptical couplers, magneto-optic waveguides, PCF, matamaterials and such. Those studies are being currently undertaken. Theirexpected results are on the horizon and are soon to be visible.

Conflict of interest

The authors also declare that there is no conflict of interest.

Acknowledgements

The research work of the seventh author (MRB) was supported by the grant NPRP 11S-1246-170033 from QNRF and he isthankful for it.

References

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Optik - Milivoj R. Belic

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