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  • Research Article𝐹-Expansion Method and New Exact Solutions ofthe Schrödinger-KdV Equation

    Ali Filiz,1 Mehmet Ekici,2 and Abdullah Sonmezoglu2

    1 Department of Mathematics, Faculty of Science and Arts, Adnan Menderes University, 09010 Aydin, Turkey2Department of Mathematics, Faculty of Science and Arts, Bozok University, 66100 Yozgat, Turkey

    Correspondence should be addressed to Ali Filiz; [email protected]

    Received 17 August 2013; Accepted 27 October 2013; Published 29 January 2014

    Academic Editors: A. K. Sharma and C. Yiu

    Copyright © 2014 Ali Filiz et al. This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

    𝐹-expansionmethod is proposed to seek exact solutions of nonlinear evolution equations.With the aid of symbolic computation, wechoose the Schrödinger-KdV equation with a source to illustrate the validity and advantages of the proposed method. A number ofJacobi-elliptic function solutions are obtained including theWeierstrass-elliptic function solutions.When themodulusm of Jacobi-elliptic function approaches to 1 and 0, soliton-like solutions and trigonometric-function solutions are also obtained, respectively.The proposed method is a straightforward, short, promising, and powerful method for the nonlinear evolution equations inmathematical physics.

    1. Introduction

    Nonlinear evolution equations are widely used to describecomplex phenomena in many scientific and engineeringfields, such as fluid dynamics, plasma physics, hydrodynam-ics, solid state physics, optical fibers, and acoustics.Therefore,finding solutions of such nonlinear evolution equations isimportant. However, determining solutions of nonlinearevolution equations is a very difficult task and only in certaincases one can obtain exact solutions. Recently,many powerfulmethods to obtain exact solutions of nonlinear evolutionequations have been proposed, such as the inverse scatteringmethod [1], the Bäcklund transformation method [2, 3], theHirota bilinear scheme [4, 5], the Painlev expansion [6],the homotopy perturbation method [7, 8], the homogenousbalance method [9], the variational method [10–12], thetanh function method [13–16], the trial function and thesine-cosine method [17], (𝐺/𝐺)-expansion method [18, 19],the trial equation method [20–28], the auxiliary equationmethod [29], the Jacobian-elliptic function method [30–33], the 𝐹-expansion method [34–38], and the Exp-functionmethod [39–42].

    In the present research, we shall apply the the 𝐹-expan-sion method to obtain 52 types of exact solution: six for

    the Weierstrass-elliptic function solutions and the rest forJacobian-elliptic function solutions of the Schrdinger-KdVequation:

    𝑖𝑢𝑡= 𝑢

    𝑥𝑥+ 𝑢V, V

    𝑡+ 6VV

    𝑥+ V

    𝑥𝑥𝑥= (|𝑢|

    2)𝑥. (1)

    Among themethodsmentioned above, the auxiliary equationmethod [29] is based on the assumption that the travellingwave solutions are in the form

    𝑢 (𝜂) =

    𝑛

    𝑖=0

    𝑎𝑖𝑧𝑖(𝜂) , 𝜂 = 𝛼 (𝑥 − 𝛽𝑡) , (2)

    where 𝑧(𝜂) satisfies the following auxiliary ordinary differen-tial equation:

    (𝑑𝑧

    𝑑𝜂)

    2

    = 𝑎𝑧2(𝜂) + 𝑏𝑧

    3(𝜂) + 𝑐𝑧

    4(𝜂) , (3)

    where 𝑎, 𝑏, and 𝑐 are real parameters. Although many exactsolutions were obtained in [29] via the auxiliary equation (3),all these solutions are expressed only in terms of hyperbolicand trigonometric functions. In this paper, we want togeneralize the work in [29]. We propose a new auxiliary

    Hindawi Publishing Corporatione Scientific World JournalVolume 2014, Article ID 534063, 14 pageshttp://dx.doi.org/10.1155/2014/534063

  • 2 The Scientific World Journal

    equation which has more general exact solutions in termsof Jacobian-elliptic and the Weierstrass-elliptic functions.Moreover, many exact solutions in terms of hyperbolic andtrigonometric functions can be also obtained when themodulus of Jacobian-elliptic functions tends to one and zero,respectively.

    The rest of the paper is arranged as follows. In Section 2,we briefly describe the auxiliary equation method (𝐹-expansion method) for nonlinear evolution equations. Byusing the method proposed in Section 2, Jacobian-ellipticand theWeierstrass-elliptic functions solutions are presentedin Sections 3 and 4, respectively. Soliton-like solutions andtrigonometric-function solutions are listed in Sections 5 and6, respectively. Some conclusions are given in Section 7. Thepaper is ended by Appendices A–D which play an importantrole in obtaining the solutions.

    2. Description of the 𝐹-Expansion Method

    Consider a nonlinear partial differential equation (PDE) withindependent variables 𝑥, 𝑡 and dependent variable 𝑢:

    𝑁(𝑢, 𝑢𝑡, 𝑢𝑥, 𝑢𝑥𝑥, . . .) = 0. (4)

    Assume that 𝑢(𝑥, 𝑡) = 𝑢(𝜉), where the wave variable 𝜉 =𝑥 + 𝑐𝑡. By this, the nonlinear PDE (4) reduces to an ordinarydifferential equation (ODE):

    𝑁(𝑢, 𝑐𝑢, 𝑢, 𝑢, . . .) = 0. (5)

    Then we seek its solutions in the form

    𝑢 (𝜉) =

    𝑚

    𝑖=0

    𝑎𝑖𝐹𝑖

    (𝜉) , (6)

    where 𝑎𝑖, 𝑖 = 0, 1, 2, . . . , 𝑚, are constants to be determined,𝑚

    is a positive integer which can be evaluated by balancing thehighest order nonlinear term(s) and the highest order partialderivative of 𝑢 in (4), and𝐹(𝜉) satisfies the following auxiliaryequation:

    𝐹

    (𝜉) = 𝜎√𝑃𝐹4(𝜉) + 𝑄𝐹

    2(𝜉) + 𝑅, (7)

    where 𝜎 = ±1 and𝑃,𝑄, and𝑅 are constants.The last equationhence holds for 𝐹(𝜉):

    𝐹= 2𝑃𝐹

    3+ 𝑄𝐹,

    𝐹= (6𝑃𝐹

    2+ 𝑄)𝐹

    ,

    𝐹(4)= 24𝑃

    2𝐹5+ 20𝑃𝑄𝐹

    3+ (12𝑃𝑅 + 𝑄

    2) 𝐹,

    𝐹(5)= (120𝑃

    2𝐹4+ 60𝑃𝑄𝐹

    2+ 12𝑃𝑅 + 𝑄

    2) 𝐹

    ...

    (8)

    In Appendices A and B, we present 52 types of exact solutionfor (7) (see [34–37, 43] for details). In fact, these exactsolutions can be used to construct more exact solutions for(1).

    3. New Exact Jacobian-Elliptic FunctionSolutions of the Schrödinger-KdV Equation

    The coupled Schrödinger-KdV equation

    𝑖𝑢𝑡− 𝑢

    𝑥𝑥− 𝑢V = 0, V

    𝑡+ 6VV

    𝑥+ V

    𝑥𝑥𝑥− (|𝑢|

    2)𝑥= 0 (9)

    is known to describe various processes in dusty plasma, suchas Langmuir, dust-acoustic wave, and electromagnetic waves[44–47]. Exact solution of (9) was studied by many authors[48–51]. Here the 𝐹-expansion method is applied to system(9) and gives some new solutions. Let

    𝑢 = 𝑒𝑖𝜃𝑈 (𝜉) , V = 𝑉 (𝜉) ,

    𝜃 = 𝛼𝑥 + 𝛽𝑡, 𝜉 = 𝑥 + 𝑐𝑡,

    (10)

    where 𝛼, 𝛽, and 𝑐 are constants.Substituting (10) into (9), we find that 𝑐 = 2𝛼 and 𝑉, 𝑈

    satisfy the following coupled nonlinear ordinary differentialsystem:

    𝑈+ (𝛽 − 𝛼

    2)𝑈 + 𝑈𝑉 = 0,

    2𝛼𝑉+ 6𝑉𝑉

    + 𝑉

    − (𝑈

    2)

    = 0.

    (11)

    Balancing the highest nonlinear terms and the highest orderderivative terms in (11), we find 𝑚 = 2 and 𝑛 = 2. Therefore,we suppose that the solution of (11) can be expressed by

    𝑈 (𝜉) = 𝑎0+ 𝑎

    1𝐹 (𝜉) + 𝑎

    2𝐹2

    (𝜉) ,

    𝑉 (𝜉) = 𝑏0+ 𝑏1𝐹 (𝜉) + 𝑏

    2𝐹2

    (𝜉) ,

    (12)

    where 𝑎0, 𝑎1, 𝑎2, 𝑏0, 𝑏1, and 𝑏

    2are constants to be determined

    later and 𝐹(𝜉) is a solution of ODE (7). Inserting (12) into(11) with the aid of (7), the left-hand side of (11) becomespolynomials in𝐹(𝜉) if canceling𝐹 and setting the coefficientsof the polynomial to zero yields a set of algebraic equations,𝑎0, 𝑎1, 𝑎2, 𝑏0, 𝑏1, and 𝑏

    2. Solving the system of algebraic

    equations with the aid of Mathematica, we obtain

    𝑎0= 0, 𝑎

    1= ±2√𝑃 (𝑄 − 𝛼 − 3𝛼2 + 3𝛽), 𝑎

    2= 0,

    𝑏0= 𝛼

    2− 𝛽 − 𝑄, 𝑏

    1= 0, 𝑏

    2= −2𝑃.

    (13)

    Substituting these results into (12), we have the followingformal solution of (11):

    𝑈 = ±2√𝑃 (𝑄 − 𝛼 − 3𝛼2 + 3𝛽)𝐹 (𝜉) ,

    𝑉 = 𝛼2− 𝛽 − 𝑄 − 2𝑃𝐹

    2

    (𝜉) , where 𝜉 = 𝑥 + 𝑐𝑡.(14)

    With the aid of Appendix A and from the formal solution of(14) along with (10), one can deduce more general combinedJacobian-elliptic function solutions of (1). Hence, the follow-ing exact solutions are obtained.

  • The Scientific World Journal 3

    Case 1. 𝑃 = 𝑚2, 𝑄 = −(1 + 𝑚2), 𝑅 = 1, 𝐹(𝜉) = 𝑠𝑛𝜉,

    𝑢1= 𝑒

    𝑖𝜃{±2𝑚√−1 − 𝑚2 − 𝛼 − 3𝛼2 + 3𝛽𝑠𝑛𝜉} ,

    V1= 𝛼

    2− 𝛽 + 1 + 𝑚

    2− 2𝑚

    2𝑠𝑛2𝜉.

    (15)

    Case 2. 𝑃 = 𝑚2, 𝑄 = −(1 + 𝑚2), 𝑅 = 1, 𝐹(𝜉) = 𝑐𝑑𝜉,

    𝑢2= 𝑒

    𝑖𝜃{±2𝑚√−1 − 𝑚2 − 𝛼 − 3𝛼2 + 3𝛽𝑐𝑑𝜉} ,

    V2= 𝛼

    2− 𝛽 + 1 + 𝑚

    2− 2𝑚

    2𝑐𝑑2𝜉.

    (16)

    Case 3. 𝑃 = −𝑚2, 𝑄 = 2𝑚2 − 1, 𝑅 = 1 − 𝑚2, 𝐹(𝜉) = 𝑐𝑛𝜉,

    𝑢3= 𝑒

    𝑖𝜃{±2𝑚√−2𝑚2 + 1 + 𝛼 + 3𝛼2 − 3𝛽𝑐𝑛𝜉} ,

    V3= 𝛼

    2− 𝛽 − 2𝑚

    2+ 1 + 2𝑚

    2𝑐𝑛2𝜉.

    (17)

    Case 4. 𝑃 = −1, 𝑄 = 2 − 𝑚2, 𝑅 = 𝑚2 − 1, 𝐹(𝜉) = 𝑑𝑛𝜉,

    𝑢4= 𝑒

    𝑖𝜃{±2√−2 + 𝑚2 + 𝛼 + 3𝛼2 − 3𝛽𝑑𝑛𝜉} ,

    V4= 𝛼

    2− 𝛽 − 2 + 𝑚

    2+ 2𝑑𝑛

    2𝜉.

    (18)

    Case 5. 𝑃 = 1, 𝑄 = −(1 + 𝑚2), 𝑅 = 𝑚2, 𝐹(𝜉) = 𝑛𝑠𝜉,

    𝑢5= 𝑒

    𝑖𝜃{±2√−1 − 𝑚2 − 𝛼 − 3𝛼2 + 3𝛽𝑛𝑠𝜉} ,

    V5= 𝛼

    2− 𝛽 + 1 + 𝑚

    2− 2𝑛𝑠

    2𝜉.

    (19)

    Case 6. 𝑃 = 1, 𝑄 = −(1 + 𝑚2), 𝑅 = 𝑚2, 𝐹(𝜉) = 𝑑𝑐𝜉,

    𝑢6= 𝑒

    𝑖𝜃{±2√−1 − 𝑚2 − 𝛼 − 3𝛼2 + 3𝛽𝑑𝑐𝜉} ,

    V6= 𝛼

    2− 𝛽 + 1 + 𝑚

    2− 2𝑑𝑐

    2𝜉.

    (20)

    Case 7. 𝑃 = 1 − 𝑚2, 𝑄 = 2𝑚2 − 1, 𝑅 = −𝑚2, 𝐹(𝜉) = 𝑛𝑐𝜉,

    𝑢7= 𝑒

    𝑖𝜃{±2√(1 − 𝑚2) (2𝑚2 − 1 − 𝛼 − 3𝛼2 + 3𝛽)𝑛𝑐𝜉} ,

    V7= 𝛼

    2− 𝛽 − 2𝑚

    2+ 1 − 2 (1 − 𝑚

    2) 𝑛𝑐

    2𝜉.

    (21)

    Case 8. 𝑃 = 𝑚2 − 1, 𝑄 = 2 − 𝑚2, 𝑅 = −1, 𝐹(𝜉) = 𝑛𝑑𝜉,

    𝑢8= 𝑒

    𝑖𝜃{±2√(𝑚2 − 1) (2 − 𝑚2 − 𝛼 − 3𝛼2 + 3𝛽)𝑛𝑑𝜉} ,

    V8= 𝛼

    2− 𝛽 − 2 + 𝑚

    2− 2 (𝑚

    2− 1) 𝑛𝑑

    2𝜉.

    (22)

    Case 9. 𝑃 = 1 − 𝑚2, 𝑄 = 2 − 𝑚2, 𝑅 = 1, 𝐹(𝜉) = 𝑠𝑐𝜉,

    𝑢9= 𝑒

    𝑖𝜃{±2√(1 − 𝑚2) (2 − 𝑚2 − 𝛼 − 3𝛼2 + 3𝛽)𝑠𝑐𝜉} ,

    V9= 𝛼

    2− 𝛽 − 2 + 𝑚

    2− 2 (1 − 𝑚

    2) 𝑠𝑐

    2𝜉.

    (23)

    Case 10. 𝑃 = −𝑚2(1 − 𝑚2), 𝑄 = 2𝑚2 − 1, 𝑅 = 1, 𝐹(𝜉) = 𝑠𝑑𝜉,

    𝑢10= 𝑒

    𝑖𝜃{±2𝑚√(𝑚2 − 1) (2𝑚2 − 1 − 𝛼 − 3𝛼2 + 3𝛽)𝑠𝑑𝜉} ,

    V10= 𝛼

    2− 𝛽 − 2𝑚

    2+ 1 + 2𝑚

    2(1 − 𝑚

    2) 𝑠𝑑

    2𝜉.

    (24)

    Case 11. 𝑃 = 1, 𝑄 = 2 − 𝑚2, 𝑅 = 1 − 𝑚2, 𝐹(𝜉) = 𝑐𝑠𝜉,

    𝑢11= 𝑒

    𝑖𝜃{±2√2 − 𝑚2 − 𝛼 − 3𝛼2 + 3𝛽𝑐𝑠𝜉} ,

    V11= 𝛼

    2− 𝛽 − 2 + 𝑚

    2− 2𝑐𝑠

    2𝜉.

    (25)

    Case 12. 𝑃 = 1, 𝑄 = 2𝑚2 − 1, 𝑅 = −𝑚2(1 − 𝑚2), 𝐹(𝜉) = 𝑑𝑠𝜉,

    𝑢12= 𝑒

    𝑖𝜃{±2√2𝑚2 − 1 − 𝛼 − 3𝛼2 + 3𝛽𝑑𝑠𝜉} ,

    V12= 𝛼

    2− 𝛽 − 2𝑚

    2+ 1 − 2𝑑𝑠

    2𝜉.

    (26)

    Case 13. 𝑃 = 1/4,𝑄 = (1−2𝑚2)/2, 𝑅 = 1/4, 𝐹(𝜉) = 𝑛𝑠𝜉±𝑐𝑠𝜉,

    𝑢13= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝑚

    2− 2𝛼 − 6𝛼

    2+ 6𝛽

    2(𝑛𝑠𝜉 ± 𝑐𝑠𝜉)

    }

    }

    }

    ,

    V13=1

    2{2𝛼

    2− 2𝛽 − 1 + 2𝑚

    2− (𝑛𝑠𝜉 ± 𝑐𝑠𝜉)

    2

    } .

    (27)

    Case 14. 𝑃 = (1 − 𝑚2)/4, 𝑄 = (1 + 𝑚2)/2, 𝑅 = (1 − 𝑚2)/4,𝐹(𝜉) = 𝑛𝑐𝜉 ± 𝑠𝑐𝜉,

    𝑢14= 𝑒

    𝑖𝜃{

    {

    {

    ±√(1 − 𝑚

    2) (1 + 𝑚

    2− 2𝛼 − 6𝛼

    2+ 6𝛽)

    2

    × (𝑛𝑐𝜉 ± 𝑠𝑐𝜉)

    }

    }

    }

    ,

    V14=1

    2{2𝛼

    2− 2𝛽 − 1 − 𝑚

    2− (1 − 𝑚

    2) (𝑛𝑐𝜉 ± 𝑠𝑐𝜉)

    2

    } .

    (28)

    Case 15. 𝑃 = 1/4,𝑄 = (𝑚2−2)/2,𝑅 = 𝑚2/4,𝐹(𝜉) = 𝑛𝑠𝜉±𝑑𝑠𝜉,

    𝑢15= 𝑒

    𝑖𝜃{

    {

    {

    ±√𝑚2− 2 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(𝑛𝑠𝜉 ± 𝑑𝑠𝜉)

    }

    }

    }

    ,

    V15=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2+ 2 − (𝑛𝑠𝜉 ± 𝑑𝑠𝜉)

    2

    } .

    (29)

    Case 16. 𝑃 = 𝑚2/4, 𝑄 = (𝑚2 − 2)/2, 𝑅 = 𝑚2/4, 𝐹(𝜉) = 𝑠𝑛𝜉 ±𝑖𝑐𝑛𝜉,

    𝑢16= 𝑒

    𝑖𝜃{

    {

    {

    ±𝑚√𝑚2− 2 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(𝑠𝑛𝜉 ± 𝑖𝑐𝑛𝜉)

    }

    }

    }

    ,

    V16=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2+ 2 − 𝑚

    2

    (𝑠𝑛𝜉 ± 𝑖𝑐𝑛𝜉)2

    } .

    (30)

  • 4 The Scientific World Journal

    Case 17. 𝑃 = 𝑚2/4, 𝑄 = (𝑚2 − 2)/2, 𝑅 = 𝑚2/4, 𝐹(𝜉) =√1 − 𝑚2𝑠𝑑𝜉 ± 𝑐𝑑𝜉,

    𝑢17= 𝑒

    𝑖𝜃{

    {

    {

    ±𝑚√𝑚2− 2 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    × (√1 − 𝑚2𝑠𝑑𝜉 ± 𝑐𝑑𝜉)

    }

    }

    }

    ,

    V17=1

    2{2𝛼

    2−2𝛽 − 𝑚

    2+2 − 𝑚

    2(√1 − 𝑚2𝑠𝑑𝜉 ± 𝑐𝑑𝜉)

    2

    } .

    (31)

    Case 18. 𝑃 = 1/4, 𝑄 = (1 − 𝑚2)/2, 𝑅 = 1/4, 𝐹(𝜉) = 𝑚𝑐𝑑𝜉 ±𝑖√1 − 𝑚2𝑛𝑑𝜉,

    𝑢18= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 𝑚

    2− 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    × (𝑚𝑐𝑑𝜉 ± 𝑖√1 − 𝑚2𝑛𝑑𝜉)

    }

    }

    }

    ,

    V18=1

    2{2𝛼

    2− 2𝛽 − 1 + 𝑚

    2

    − (𝑚𝑐𝑑𝜉 ± 𝑖√1 − 𝑚2𝑛𝑑𝜉)

    2

    } .

    (32)

    Case 19. 𝑃 = 1/4, 𝑄 = (1 − 2𝑚2)/2, 𝑅 = 1/4, 𝐹(𝜉) = 𝑚𝑠𝑛𝜉 ±𝑖𝑑𝑛𝜉,

    𝑢19= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝑚

    2− 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    × (𝑚𝑠𝑛𝜉 ± 𝑖𝑑𝑛𝜉)

    }

    }

    }

    ,

    V19=1

    2{2𝛼

    2− 2𝛽 − 1 + 2𝑚

    2− (𝑚𝑠𝑛𝜉 ± 𝑖𝑑𝑛𝜉)

    2

    } .

    (33)

    Case 20. 𝑃 = 1/4, 𝑄 = (1 − 𝑚2)/2, 𝑅 = 1/4, 𝐹(𝜉) =√1 − 𝑚2𝑠𝑐𝜉 ± 𝑖𝑑𝑐𝜉,

    𝑢20= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 𝑚

    2− 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    × (√1 − 𝑚2𝑠𝑐𝜉 ± 𝑖𝑑𝑐𝜉)

    }

    }

    }

    ,

    V20=1

    2{2𝛼

    2−2𝛽 − 1+ 𝑚

    2−(√1 − 𝑚2𝑠𝑐𝜉 ± 𝑖𝑑𝑐𝜉)

    2

    } .

    (34)

    Case 21. 𝑃 = (𝑚2 − 1)/4, 𝑄 = (𝑚2 + 1)/2, 𝑅 = (𝑚2 − 1)/4,𝐹(𝜉) = 𝑚𝑠𝑑𝜉 ± 𝑛𝑑𝜉,

    𝑢21= 𝑒

    𝑖𝜃{

    {

    {

    ±√(𝑚

    2− 1) (𝑚

    2+ 1 − 2𝛼 − 6𝛼

    2+ 6𝛽)

    2

    × (𝑚𝑠𝑑𝜉 ± 𝑛𝑑𝜉)

    }

    }

    }

    ,

    V21=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2− 1 − (𝑚

    2− 1)

    × (𝑚𝑠𝑑𝜉 ± 𝑛𝑑𝜉)2

    } .

    (35)

    Case 22. 𝑃 = 𝑚2/4,𝑄 = (𝑚2 −2)/2, 𝑅 = 1/4, 𝐹(𝜉) = 𝑠𝑛𝜉/(1±𝑑𝑛𝜉),

    𝑢22= 𝑒

    𝑖𝜃{

    {

    {

    ±𝑚√𝑚2− 2 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    𝑠𝑛𝜉

    1 ± 𝑑𝑛𝜉

    }

    }

    }

    ,

    V22=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2+ 2 − 𝑚

    2(

    𝑠𝑛𝜉

    1 ± 𝑑𝑛𝜉)

    2

    } .

    (36)

    Case 23. 𝑃 = −1/4, 𝑄 = (𝑚2 + 1)/2, 𝑅 = (1 − 𝑚2)2/4, 𝐹(𝜉) =𝑚𝑐𝑛𝜉 ± 𝑑𝑛𝜉,

    𝑢23=𝑒𝑖𝜃{

    {

    {

    ±√−𝑚

    2− 1 + 2𝛼 + 6𝛼

    2− 6𝛽

    2(𝑚𝑐𝑛𝜉 ± 𝑑𝑛𝜉)

    }

    }

    }

    ,

    V23=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2− 1 + (𝑚𝑐𝑛𝜉 ± 𝑑𝑛𝜉)

    2

    } .

    (37)

    Case 24. 𝑃 = (1 − 𝑚2)2/4, 𝑄 = (𝑚2 + 1)/2, 𝑅 = 1/4, 𝐹(𝜉) =𝑑𝑠𝜉 ± 𝑐𝑠𝜉,

    𝑢24= 𝑒

    𝑖𝜃{

    {

    {

    ±(1 − 𝑚2)√

    𝑚2+ 1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    × (𝑑𝑠𝜉 ± 𝑐𝑠𝜉)

    }

    }

    }

    ,

    V24=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2− 1 − (1 − 𝑚

    2)2

    (𝑑𝑠𝜉 ± 𝑐𝑠𝜉)2

    } .

    (38)

  • The Scientific World Journal 5

    Case 25. 𝑃 = 𝑚4(1 −𝑚2)/2(2 −𝑚2),𝑄 = 2(1 −𝑚2)/(𝑚2 − 2),𝑅 = (1 − 𝑚

    2)/2(2 − 𝑚

    2), 𝐹(𝜉) = 𝑑𝑐𝜉 ± √1 − 𝑚2𝑛𝑐𝜉,

    𝑢25

    = 𝑒𝑖𝜃{±

    √2𝑚2

    2 − 𝑚2

    × √(𝑚2 − 1) [2 (1 − 𝑚2) + (𝑚2 − 2) (−𝛼 − 3𝛼2 + 3𝛽)]

    × (𝑑𝑐𝜉 ± √1 − 𝑚2𝑛𝑐𝜉) } ,

    V25

    =1

    𝑚2 − 2{ (𝑚

    2− 2) (𝛼

    2− 𝛽) + (1 − 𝑚

    2)

    × [−2 + 𝑚4(𝑑𝑐𝜉 ± √1 − 𝑚2𝑛𝑐𝜉)

    2

    ]} .

    (39)

    Case 26. 𝑃 > 0, 𝑄 < 0, 𝑅 = 𝑚2𝑄2/(1 + 𝑚2)2𝑃, 𝐹(𝜉) =√−𝑚2𝑄/(1 + 𝑚2)𝑃𝑠𝑛(√−𝑄/(1 + 𝑚2)𝜉),

    𝑢26= 𝑒

    𝑖𝜃{

    {

    {

    ±2𝑚√𝑄(−𝑄 + 𝛼 + 3𝛼

    2− 3𝛽)

    1 + 𝑚2

    × 𝑠𝑛(√−𝑄

    1 + 𝑚2𝜉)

    }

    }

    }

    ,

    V26= 𝛼

    2− 𝛽 − 𝑄 +

    2𝑚2𝑄

    𝑚2 + 1𝑠𝑛2(√

    −𝑄

    1 + 𝑚2𝜉) .

    (40)

    Case 27. 𝑃 < 0, 𝑄 > 0, 𝑅 = (1 − 𝑚2)𝑄2/(𝑚2 − 2)2𝑃, 𝐹(𝜉) =√−𝑄/(2 − 𝑚2)𝑃𝑑𝑛(√𝑄/(2 − 𝑚2)𝜉),

    𝑢27= 𝑒

    𝑖𝜃{

    {

    {

    ±2√𝑄(−𝑄 + 𝛼 + 3𝛼

    2− 3𝛽)

    2 − 𝑚2

    × 𝑑𝑛(√𝑄

    2 − 𝑚2𝜉)

    }

    }

    }

    ,

    V27= 𝛼

    2− 𝛽 − 𝑄 +

    2𝑄

    2 − 𝑚2𝑑𝑛2(√

    𝑄

    2 − 𝑚2𝜉) .

    (41)

    Case 28. 𝑃 < 0, 𝑄 > 0, 𝑅 = 𝑚2(𝑚2 − 1)𝑄2/(2𝑚2 − 1)2𝑃,𝐹(𝜉) = √−𝑚2𝑄/(2𝑚2 − 1)𝑃𝑐𝑛(√𝑄/(2𝑚2 − 1)𝜉),

    𝑢28= 𝑒

    𝑖𝜃{

    {

    {

    ±2𝑚√𝑄(−𝑄 + 𝛼 + 3𝛼

    2− 3𝛽)

    2𝑚2 − 1𝑐𝑛

    × (√𝑄

    2𝑚2 − 1𝜉)

    }

    }

    }

    ,

    V28= 𝛼

    2− 𝛽 − 𝑄 +

    2𝑚2𝑄

    2𝑚2 − 1𝑐𝑛2(√

    𝑄

    2𝑚2 − 1𝜉) .

    (42)

    Case 29. 𝑃 = 1, 𝑄 = 2 − 4𝑚2, 𝑅 = 1, 𝐹(𝜉) = 𝑠𝑛𝜉𝑑𝑛𝜉/𝑐𝑛𝜉,

    𝑢29= 𝑒

    𝑖𝜃{±2√2 − 4𝑚2 − 𝛼 − 3𝛼2 + 3𝛽

    𝑠𝑛𝜉𝑑𝑛𝜉

    𝑐𝑛𝜉} ,

    V29= 𝛼

    2− 𝛽 − 2 + 4𝑚

    2−2𝑠𝑛

    2𝜉𝑑𝑛

    2𝜉

    𝑐𝑛2𝜉.

    (43)

    Case 30. 𝑃 = 𝑚2, 𝑄 = 2, 𝑅 = 1, 𝐹(𝜉) = 𝑠𝑛𝜉𝑐𝑛𝜉/𝑑𝑛𝜉,

    𝑢30= 𝑒

    𝑖𝜃{±2𝑚√2 − 𝛼 − 3𝛼2 + 3𝛽

    𝑠𝑛𝜉𝑐𝑛𝜉

    𝑑𝑛𝜉} ,

    V30= 𝛼

    2− 𝛽 − 2 −

    2𝑚2𝑠𝑛2𝜉𝑐𝑛

    2𝜉

    𝑑𝑛2𝜉.

    (44)

    Case 31. 𝑃 = 1, 𝑄 = 𝑚2 + 2, 𝑅 = 1 − 2𝑚2 + 𝑚4, 𝐹(𝜉) =𝑑𝑛𝜉𝑐𝑛𝜉/𝑠𝑛𝜉,

    𝑢31= 𝑒

    𝑖𝜃{±2√𝑚2 + 2 − 𝛼 − 3𝛼2 + 3𝛽

    𝑑𝑛𝜉𝑐𝑛𝜉

    𝑠𝑛𝜉} ,

    V31= 𝛼

    2− 𝛽 − 𝑚

    2− 2 −

    2𝑑𝑛2𝜉𝑐𝑛

    2𝜉

    𝑠𝑛2𝜉.

    (45)

    Case 32. 𝑃 = 𝐴2(𝑚 − 1)2/4, 𝑄 = (𝑚2 + 1)/2 + 3𝑚, 𝑅 = (𝑚 −1)2/4𝐴

    2, 𝐹(𝜉) = 𝑑𝑛𝜉𝑐𝑛𝜉/𝐴(1 + 𝑠𝑛𝜉)(1 + 𝑚𝑠𝑛𝜉),

    𝑢32= 𝑒

    𝑖𝜃{

    {

    {

    ± (𝑚 − 1)√𝑚2+ 6𝑚 + 1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    ×𝑑𝑛𝜉𝑐𝑛𝜉

    (1 + 𝑠𝑛𝜉) (1 + 𝑚𝑠𝑛𝜉)

    }

    }

    }

    ,

    V32=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2− 6𝑚 − 1

    −(𝑚 − 1)

    2𝑑𝑛2𝜉𝑐𝑛

    2𝜉

    (1 + 𝑠𝑛𝜉)2

    (1 + 𝑚𝑠𝑛𝜉)2} .

    (46)

    Case 33. 𝑃 = 𝐴2(𝑚 + 1)2/4, 𝑄 = (𝑚2 + 1)/2 − 3𝑚, 𝑅 = (𝑚 +1)2/4𝐴

    2, 𝐹(𝜉) = 𝑑𝑛𝜉𝑐𝑛𝜉/𝐴(1 + 𝑠𝑛𝜉)(1 − 𝑚𝑠𝑛𝜉),

    𝑢33= 𝑒

    𝑖𝜃{

    {

    {

    ± (𝑚 + 1)√𝑚2− 6𝑚 + 1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    ×𝑑𝑛𝜉𝑐𝑛𝜉

    (1 + 𝑠𝑛𝜉) (1 − 𝑚𝑠𝑛𝜉)

    }

    }

    }

    ,

    V33=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2+ 6𝑚 − 1

    −(𝑚 + 1)

    2𝑑𝑛2𝜉𝑐𝑛

    2𝜉

    (1 + 𝑠𝑛𝜉)2

    (1 − 𝑚𝑠𝑛𝜉)2} .

    (47)

  • 6 The Scientific World Journal

    Case 34. 𝑃 = −4/𝑚, 𝑄 = 6𝑚 −𝑚2 − 1, 𝑅 = −2𝑚3 + 𝑚4 + 𝑚2,𝐹(𝜉) = 𝑚𝑐𝑛𝜉𝑑𝑛𝜉/(𝑚𝑠𝑛

    2𝜉 + 1),

    𝑢34= 𝑒

    𝑖𝜃{±4√𝑚2 − 6𝑚 + 1 + 𝛼 + 3𝛼2 − 3𝛽

    √𝑚𝑐𝑛𝜉𝑑𝑛𝜉

    𝑚𝑠𝑛2𝜉 + 1} ,

    V34= 𝛼

    2− 𝛽 + 𝑚

    2− 6𝑚 + 1 +

    8𝑚𝑐𝑛2𝜉𝑑𝑛

    2𝜉

    (𝑚𝑠𝑛2𝜉 + 1)2.

    (48)

    Case 35. 𝑃 = 4/𝑚, 𝑄 = −6𝑚 − 𝑚2 − 1, 𝑅 = 2𝑚3 + 𝑚4 + 𝑚2,𝐹(𝜉) = 𝑚𝑐𝑛𝜉𝑑𝑛𝜉/(𝑚𝑠𝑛

    2𝜉 − 1),

    𝑢35= 𝑒

    𝑖𝜃{ ± 4√−𝑚2 − 6𝑚 − 1 − 𝛼 − 3𝛼2 + 3𝛽

    ×√𝑚𝑐𝑛𝜉𝑑𝑛𝜉

    𝑚𝑠𝑛2𝜉 − 1} ,

    V35= 𝛼

    2− 𝛽 + 𝑚

    2+ 6𝑚 + 1 −

    8𝑚𝑐𝑛2𝜉𝑑𝑛

    2𝜉

    (𝑚𝑠𝑛2𝜉 − 1)2.

    (49)

    Case 36. 𝑃 = 1/4,𝑄 = (1 − 2𝑚2)/2, 𝑅 = 1/4, 𝐹(𝜉) = 𝑠𝑛𝜉/(1 ±𝑐𝑛𝜉),

    𝑢36= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝑚

    2− 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    𝑠𝑛𝜉

    1 ± 𝑐𝑛𝜉

    }

    }

    }

    ,

    V36=1

    2{2𝛼

    2− 2𝛽 − 1 + 2𝑚

    2−

    𝑠𝑛2𝜉

    (1 ± 𝑐𝑛𝜉)2} .

    (50)

    Case 37. 𝑃 = (1 − 𝑚2)/4, 𝑄 = (1 + 𝑚2)/2, 𝑅 = (1 − 𝑚2)/4,𝐹(𝜉) = 𝑐𝑛𝜉/(1 ± 𝑠𝑛𝜉),

    𝑢37= 𝑒

    𝑖𝜃{

    {

    {

    ±√(1 − 𝑚

    2) (1 + 𝑚

    2− 2𝛼 − 6𝛼

    2+ 6𝛽)

    2

    ×𝑐𝑛𝜉

    1 ± 𝑠𝑛𝜉} ,

    V37=1

    2{2𝛼

    2− 2𝛽 − 1 − 𝑚

    2−

    (1 − 𝑚2) 𝑐𝑛

    2𝜉

    (1 ± 𝑠𝑛𝜉)2} .

    (51)

    Case 38. 𝑃 = 4𝑚1, 𝑄 = 2 + 6𝑚

    1− 𝑚

    2, 𝑅 = 2 + 2𝑚1− 𝑚

    2,𝐹(𝜉) = 𝑚

    2𝑠𝑛𝜉𝑐𝑛𝜉/(𝑚

    1− 𝑑𝑛

    2𝜉),

    𝑢38= 𝑒

    𝑖𝜃{ ± 4√𝑚

    1(2 + 6𝑚

    1− 𝑚2 − 𝛼 − 3𝛼2 + 3𝛽)

    ×𝑚2𝑠𝑛𝜉𝑐𝑛𝜉

    𝑚1− 𝑑𝑛2𝜉

    } ,

    V38= 𝛼

    2− 𝛽 − 2 − 6𝑚

    1+ 𝑚

    2−8𝑚

    1𝑚4𝑠𝑛2𝜉𝑐𝑛

    2𝜉

    (𝑚1− 𝑑𝑛2𝜉)

    2.

    (52)

    Case 39. 𝑃 = −4𝑚1, 𝑄 = 2 − 6𝑚

    1− 𝑚

    2, 𝑅 = 2 − 2𝑚1− 𝑚

    2,𝐹(𝜉) = −𝑚

    2𝑠𝑛𝜉𝑐𝑛𝜉/(𝑚

    1+ 𝑑𝑛

    2𝜉),

    𝑢39= 𝑒

    𝑖𝜃{ ± 4√𝑚

    1(−2 + 6𝑚

    1+ 𝑚2 + 𝛼 + 3𝛼2 − 3𝛽)

    ×𝑚2𝑠𝑛𝜉𝑐𝑛𝜉

    𝑚1+ 𝑑𝑛2𝜉

    } ,

    V39= 𝛼

    2− 𝛽 − 2 + 6𝑚

    1+ 𝑚

    2+8𝑚

    1𝑚4𝑠𝑛2𝜉𝑐𝑛

    2𝜉

    (𝑚1+ 𝑑𝑛2𝜉)

    2.

    (53)

    Case 40. 𝑃 = (2−𝑚2 −2𝑚1)/4,𝑄 = 𝑚2/2−1−3𝑚

    1, 𝑅 = (2−

    𝑚2−2𝑚

    1)/4, 𝐹(𝜉) = 𝑚2𝑠𝑛𝜉𝑐𝑛𝜉/(𝑠𝑛2𝜉+ (1+𝑚

    1)𝑑𝑛𝜉−1−𝑚

    1),

    𝑢40

    = 𝑒𝑖𝜃{

    {

    {

    ±√(2 − 𝑚

    2− 2𝑚

    1) (𝑚

    2− 2 − 6𝑚

    1− 2𝛼 − 6𝛼

    2+ 6𝛽)

    2

    ×𝑚2𝑠𝑛𝜉𝑐𝑛𝜉

    𝑠𝑛2𝜉 + (1 + 𝑚1) 𝑑𝑛𝜉 − 1 − 𝑚

    1

    }

    }

    }

    ,

    V40=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2+ 2 + 6𝑚

    1

    −𝑚4(2 − 𝑚

    2− 2𝑚

    1) 𝑠𝑛

    2𝜉𝑐𝑛

    2𝜉

    [𝑠𝑛2𝜉 + (1 + 𝑚1) 𝑑𝑛𝜉 − 1 − 𝑚

    1]2} .

    (54)

    Case 41. 𝑃 = (2−𝑚2 +2𝑚1)/4,𝑄 = 𝑚2/2 − 1+ 3𝑚

    1, 𝑅 = (2−

    𝑚2+2𝑚

    1)/4,𝐹(𝜉) = 𝑚2𝑠𝑛𝜉𝑐𝑛𝜉/(𝑠𝑛2𝜉+(−1+𝑚

    1)𝑑𝑛𝜉−1−𝑚

    1),

    𝑢41

    = 𝑒𝑖𝜃{

    {

    {

    ±√(2 − 𝑚

    2+ 2𝑚

    1) (𝑚

    2− 2 + 6𝑚

    1− 2𝛼 − 6𝛼

    2+ 6𝛽)

    2

    ×𝑚2𝑠𝑛𝜉𝑐𝑛𝜉

    𝑠𝑛2𝜉 + (−1 + 𝑚1) 𝑑𝑛𝜉 − 1 − 𝑚

    1

    }

    }

    }

    ,

    V41=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2+ 2 − 6𝑚

    1

    −𝑚4(2 − 𝑚

    2+ 2𝑚

    1) 𝑠𝑛

    2𝜉𝑐𝑛

    2𝜉

    [𝑠𝑛2𝜉 + (−1 + 𝑚1) 𝑑𝑛𝜉 − 1 − 𝑚

    1]2} .

    (55)

    Case 42. 𝑃 = (𝐶2𝑚4−(𝐵2+𝐶2)𝑚2+𝐵2)/4,𝑄 = (𝑚2+1)/2,𝑅 =(𝑚2− 1)/4(𝐶

    2𝑚2− 𝐵

    2), 𝐹(𝜉) = (√(𝐵2 − 𝐶2)/(𝐵2 − 𝐶2𝑚2) +

    𝑠𝑛𝜉)/(𝐵𝑐𝑛𝜉 + 𝐶𝑑𝑛𝜉),

  • The Scientific World Journal 7

    𝑢42

    = 𝑒𝑖𝜃

    {{

    {{

    {

    ±√[𝐶2𝑚4− (𝐵

    2+ 𝐶

    2)𝑚

    2+ 𝐵

    2] (𝑚

    2+ 1 − 2𝛼 − 6𝛼

    2+ 6𝛽)

    2

    ×

    √(𝐵2− 𝐶

    2) / (𝐵

    2− 𝐶

    2𝑚2) + 𝑠𝑛𝜉

    𝐵𝑐𝑛𝜉 + 𝐶𝑑𝑛𝜉

    }}

    }}

    }

    ,

    V42=1

    2

    {{

    {{

    {

    2𝛼2− 2𝛽 − 𝑚

    2− 1 − [𝐶

    2𝑚4− (𝐵

    2+ 𝐶

    2)𝑚

    2+ 𝐵

    2]

    × (

    √(𝐵2− 𝐶

    2) / (𝐵

    2− 𝐶

    2𝑚2) + 𝑠𝑛𝜉

    𝐵𝑐𝑛𝜉 + 𝐶𝑑𝑛𝜉

    )

    2

    }}

    }}

    }

    .

    (56)

    Case 43. 𝑃 = (𝐵2 + 𝐶2𝑚2)/4, 𝑄 = 1/2 − 𝑚2, 𝑅 =1/4(𝐵

    2+ 𝐶

    2𝑚2), 𝐹(𝜉) = (√(𝐶2𝑚2 + 𝐵2 − 𝐶2)/(𝐵2 + 𝐶2𝑚2) +

    𝑐𝑛𝜉)/(𝐵𝑠𝑛𝜉 + 𝐶𝑑𝑛𝜉),

    𝑢43= 𝑒

    𝑖𝜃

    {{

    {{

    {

    ±√(𝐵2+ 𝐶

    2𝑚2) (1 − 2𝑚

    2− 2𝛼 − 6𝛼

    2+ 6𝛽)

    2

    ×

    √(𝐶2𝑚2 + 𝐵2 − 𝐶2) / (𝐵2 + 𝐶2𝑚2) + 𝑐𝑛𝜉

    𝐵𝑠𝑛𝜉 + 𝐶𝑑𝑛𝜉

    }}

    }}

    }

    ,

    V43=1

    2{2𝛼

    2− 2𝛽 − 1 + 2𝑚

    2− (𝐵

    2+ 𝐶

    2𝑚2)

    ×(√(𝐶2𝑚2 + 𝐵2 − 𝐶2)/(𝐵2 + 𝐶2𝑚2) + 𝑐𝑛𝜉

    𝐵𝑠𝑛𝜉 + 𝐶𝑑𝑛𝜉)

    2

    } .

    (57)

    Case 44. 𝑃 = (𝐵2 +𝐶2)/4,𝑄 = 𝑚2/2 − 1, 𝑅 = 𝑚4/4(𝐵2 +𝐶2),𝐹(𝜉) = (√(𝐵2 + 𝐶2 − 𝐶2𝑚2)/(𝐵2 + 𝐶2)+𝑑𝑛𝜉)/(𝐵𝑠𝑛𝜉+𝐶𝑐𝑛𝜉),

    𝑢44= 𝑒

    𝑖𝜃{

    {

    {

    ±√(𝐵2+ 𝐶

    2) (𝑚

    2− 2 − 2𝛼 − 6𝛼

    2+ 6𝛽)

    2

    ×

    √(𝐵2 + 𝐶2 − 𝐶2𝑚2) / (𝐵2 + 𝐶2) + 𝑑𝑛𝜉

    𝐵𝑠𝑛𝜉 + 𝐶𝑐𝑛𝜉

    }}

    }}

    }

    ,

    V44=1

    2{2𝛼

    2− 2𝛽 − 𝑚

    2+ 2 − (𝐵

    2+ 𝐶

    2)

    × (√(𝐵2 + 𝐶2 − 𝐶2𝑚2)/(𝐵2 + 𝐶2) + 𝑑𝑛𝜉

    𝐵𝑠𝑛𝜉 + 𝐶𝑐𝑛𝜉)

    2

    } .

    (58)

    Case 45. 𝑃 = −(𝑚2 + 2𝑚 + 1)𝐵2, 𝑄 = 2𝑚2 + 2, 𝑅 = (2𝑚 −𝑚2− 1)/𝐵

    2, 𝐹(𝜉) = (𝑚𝑠𝑛2𝜉 − 1)/𝐵(𝑚𝑠𝑛2𝜉 + 1),

    𝑢45= 𝑒

    𝑖𝜃{ ± 2 (𝑚 + 1)√−2𝑚

    2 − 2 + 𝛼 + 3𝛼2 − 3𝛽

    ×𝑚𝑠𝑛

    2𝜉 − 1

    𝑚𝑠𝑛2𝜉 + 1} ,

    V45= 𝛼

    2− 𝛽 − 2𝑚

    2− 2 + 2(𝑚 + 1)

    2

    × (𝑚𝑠𝑛

    2𝜉 − 1

    𝑚𝑠𝑛2𝜉 + 1)

    2

    .

    (59)

    Case 46. 𝑃 = −(𝑚2 − 2𝑚 + 1)𝐵2, 𝑄 = 2𝑚2 + 2, 𝑅 = −(2𝑚 +𝑚2+ 1)/𝐵

    2, 𝐹(𝜉) = (𝑚𝑠𝑛2𝜉 + 1)/𝐵(𝑚𝑠𝑛2𝜉 − 1),

    𝑢46= 𝑒

    𝑖𝜃{ ± 2 (𝑚 − 1)√−2𝑚

    2 − 2 + 𝛼 + 3𝛼2 − 3𝛽

    ×𝑚𝑠𝑛

    2𝜉 + 1

    𝑚𝑠𝑛2𝜉 − 1} ,

    V46= 𝛼

    2− 𝛽 − 2𝑚

    2− 2 + 2(𝑚 − 1)

    2

    × (𝑚𝑠𝑛

    2𝜉 + 1

    𝑚𝑠𝑛2𝜉 − 1)

    2

    .

    (60)

    We note that there is much duplication in the list of 46solutions in terms of Jacobian-elliptic functions. Here aresome examples; using the well-known identities relatingJacobian-elliptic functions (see 121.00, 129.01, 129.02, and129.03 in [52], e.g.) reveals that 𝑢

    1, 𝑢3, and 𝑢

    4are identical;

    V1, V3, and V

    4are identical; 𝑢

    2, 𝑢8, and 𝑢

    10are identical; V

    2,

    V8, and V

    10are identical; 𝑢

    5, 𝑢11, and 𝑢

    12are identical; V

    5, V11,

    and V12

    are identical; 𝑢6, 𝑢7, and 𝑢

    9are identical; V

    6, V7, and

    V9are identical. Use of 162.01 in [52] reveals that 𝑢

    27and 𝑢

    28

    are equivalent and V27and V

    28are equivalent.

    4. The New Weierstrass-Elliptic FunctionSolutions of the Schrödinger-KdV Equation

    On using the solutions given in [43], mentioned inAppendix B, and from the formal solution (14) along with(10), we get then the following exact solutions.

    Case 47. 𝑔2= (4/3)(𝑄

    2− 3𝑃𝑅), 𝑔

    3= (4𝑄/27)(−2𝑄

    2+ 9𝑃𝑅),

    𝐹(𝜉) = √(1/𝑃)[℘(𝜉; 𝑔2, 𝑔3) − (1/3)𝑄],

    𝑢47= 𝑒

    𝑖𝜃{ ± 2√𝑃 (𝑄 − 𝛼 − 3𝛼2 + 3𝛽)

    ×√1

    𝑃[℘ (𝜉; 𝑔

    2, 𝑔3) −

    1

    3𝑄]} ,

    V47= 𝛼

    2− 𝛽 − 𝑄 − 2 [℘ (𝜉; 𝑔

    2, 𝑔3) −

    1

    3𝑄] .

    (61)

  • 8 The Scientific World Journal

    Case 48. 𝑔2= (4/3)(𝑄

    2− 3𝑃𝑅), 𝑔

    3= (4𝑄/27)(−2𝑄

    2+ 9𝑃𝑅),

    𝐹(𝜉) = √3𝑅/(3℘(𝜉; 𝑔2, 𝑔3) − 𝑄),

    𝑢48= 𝑒

    𝑖𝜃{ ± 2√𝑃 (𝑄 − 𝛼 − 3𝛼2 + 3𝛽)

    × √3𝑅

    3℘ (𝜉; 𝑔2, 𝑔3) − 𝑄

    } ,

    V48= 𝛼

    2− 𝛽 − 𝑄 −

    6𝑃𝑅

    3℘ (𝜉; 𝑔2, 𝑔3) − 𝑄

    .

    (62)

    Case 49. 𝑔2= −(5𝑄𝐷 + 4𝑄

    2+ 33𝑃𝑄𝑅)/12, 𝑔

    3= (21𝑄

    2

    𝐷 − 63𝑃𝑅𝐷 + 20𝑄3− 27𝑃𝑄𝑅)/216, 𝐹(𝜉) =

    √12𝑅℘(𝜉; 𝑔2, 𝑔3) + 2𝑅(2𝑄 + 𝐷)/(12℘(𝜉; 𝑔

    2, 𝑔3) + 𝐷),

    𝑢49= 𝑒

    𝑖𝜃

    {{

    {{

    {

    ± 2√𝑃 (𝑄 − 𝛼 − 3𝛼2 + 3𝛽)

    ×

    √12𝑅℘ (𝜉; 𝑔2, 𝑔3) + 2𝑅 (2𝑄 + 𝐷)

    12℘ (𝜉; 𝑔2, 𝑔3) + 𝐷

    }}

    }}

    }

    ,

    V49= 𝛼

    2− 𝛽 − 𝑄

    −4𝑃𝑅 [6℘ (𝜉; 𝑔

    2, 𝑔3) + 2𝑄 + 𝐷]

    [12℘ (𝜉; 𝑔2, 𝑔3) + 𝐷]

    2.

    (63)

    Case 50. 𝑔2= (1/12)𝑄

    2+ 𝑃𝑅, 𝑔

    3= (1/216)𝑄(36𝑃𝑅 − 𝑄

    2),

    𝐹(𝜉) = √𝑅[6℘(𝜉; 𝑔2, 𝑔3) + 𝑄]/3℘

    (𝜉; 𝑔

    2, 𝑔3),

    𝑢50= 𝑒

    𝑖𝜃{ ± 2√𝑃𝑅 (𝑄 − 𝛼 − 3𝛼2 + 3𝛽)

    ×6℘ (𝜉; 𝑔

    2, 𝑔3) + 𝑄

    3℘ (𝜉; 𝑔2, 𝑔3)} ,

    V50= 𝛼

    2− 𝛽 − 𝑄 −

    2𝑃𝑅[6℘ (𝜉; 𝑔2, 𝑔3) + 𝑄]

    2

    9[℘ (𝜉; 𝑔2, 𝑔3)]2

    .

    (64)

    Case 51. 𝑔2= (1/12)𝑄

    2+ 𝑃𝑅, 𝑔

    3= (1/216)𝑄(36𝑃𝑅 − 𝑄

    2),

    𝐹(𝜉) = 3℘(𝜉; 𝑔

    2, 𝑔3)/√𝑃[6℘(𝜉; 𝑔

    2, 𝑔3) + 𝑄],

    𝑢51= 𝑒

    𝑖𝜃{±2√𝑄 − 𝛼 − 3𝛼2 + 3𝛽

    3℘(𝜉; 𝑔

    2, 𝑔3)

    6℘ (𝜉; 𝑔2, 𝑔3) + 𝑄

    } ,

    V51= 𝛼

    2− 𝛽 − 𝑄 −

    18[℘(𝜉; 𝑔

    2, 𝑔3)]2

    [6℘ (𝜉; 𝑔2, 𝑔3) + 𝑄]

    2.

    (65)

    Case 52. 𝑅 = 5𝑄2/36𝑃, 𝑔2= 2𝑄

    2/9, 𝑔

    3= 𝑄

    3/54, 𝐹(𝜉) =

    𝑄√−15𝑄/2𝑃℘(𝜉; 𝑔2, 𝑔3)/(3℘(𝜉; 𝑔

    2, 𝑔3) + 𝑄),

    𝑢52= 𝑒

    𝑖𝜃{ ± 2√𝑃 (𝑄 − 𝛼 − 3𝛼2 + 3𝛽)

    ×𝑄√−15𝑄/2𝑃℘ (𝜉; 𝑔

    2, 𝑔3)

    3℘ (𝜉; 𝑔2, 𝑔3) + 𝑄

    } ,

    V52= 𝛼

    2− 𝛽 − 𝑄 +

    15𝑄3℘2(𝜉; 𝑔

    2, 𝑔3)

    [3℘ (𝜉; 𝑔2, 𝑔3) + 𝑄]

    2.

    (66)

    It should be noted that any solution that can be expressed interms of aWeierstrass-elliptic function can be also convertedinto a solution in terms of a Jacobian-elliptic function (formore details, see [53]). Consequently, Cases 47–52 are alreadycovered in Cases 1–46. For example, using 1031.01 in [52]reveals that, with the 𝑃, 𝑄, and 𝑅 values for Case 1, 𝑢

    1and

    𝑢48are identical and V

    1and V

    48are identical.

    5. New Soliton-Like Solutions of theSchrödinger-KdV Equation

    Some soliton-like solutions of (1) can be obtained in thelimited case when the modulus 𝑚 → 1 (see Appendix C),as follows:

    𝑢1= 𝑒

    𝑖𝜃{±2√−2 − 𝛼 − 3𝛼2 + 3𝛽tanh𝜉} ,

    V1= 𝛼

    2− 𝛽 + 2sech2𝜉,

    𝑢3= 𝑒

    𝑖𝜃{±2√−1 + 𝛼 + 3𝛼2 − 3𝛽sech𝜉} ,

    V3= 𝛼

    2− 𝛽 − 1 + 2sech2𝜉,

    𝑢5= 𝑒

    𝑖𝜃{±2√−2 − 𝛼 − 3𝛼2 + 3𝛽coth𝜉} ,

    V5= 𝛼

    2− 𝛽 − 2csch2𝜉,

    𝑢11= 𝑒

    𝑖𝜃{±2√1 − 𝛼 − 3𝛼2 + 3𝛽csch𝜉} ,

    V11= 𝛼

    2− 𝛽 − 1 − 2csch2𝜉,

    𝑢13= 𝑒

    𝑖𝜃{

    {

    {

    ±√−1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(coth 𝜉 ± csch𝜉)

    }

    }

    }

    ,

    V13=1

    2{2𝛼

    2− 2𝛽 + 1 − (coth 𝜉 ± csch𝜉)2} ,

    𝑢16= 𝑒

    𝑖𝜃{

    {

    {

    ±√−1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(tanh𝜉 ± 𝑖sech𝜉)

    }

    }

    }

    ,

    V16=1

    2{2𝛼

    2− 2𝛽 + 1 − (tanh𝜉 ± 𝑖𝑠ech𝜉)2} ,

  • The Scientific World Journal 9

    𝑢22= 𝑒

    𝑖𝜃{

    {

    {

    ±√−1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    tanh𝜉1 ± sech𝜉

    }

    }

    }

    ,

    V22=1

    2{2𝛼

    2− 2𝛽 + 1 − (

    tanh𝜉1 ± sech𝜉

    )

    2

    } ,

    𝑢23= 𝑒

    𝑖𝜃{±2√−1 + 𝛼 + 3𝛼2 − 3𝛽sech𝜉} ,

    V23= 𝛼

    2− 𝛽 − 1 + 2sech2𝜉,

    𝑢26= 𝑒

    𝑖𝜃{

    {

    {

    ±2√𝑄(−𝑄 + 𝛼 + 3𝛼

    2− 3𝛽)

    2tanh(√−𝑄

    2𝜉)

    }

    }

    }

    ,

    V26= 𝛼

    2− 𝛽 − 𝑄sech2 (√−𝑄

    2𝜉) ,

    𝑢27= 𝑒

    𝑖𝜃{±2√𝑄 (−𝑄 + 𝛼 + 3𝛼2 − 3𝛽)sech (√𝑄𝜉)} ,

    V27= 𝛼

    2− 𝛽 − 𝑄 [1 − 2sech2 (√𝑄𝜉)] ,

    𝑢30= 𝑒

    𝑖𝜃{±2√2 − 𝛼 − 3𝛼2 + 3𝛽tanh𝜉} ,

    V30= 𝛼

    2− 𝛽 − 2 (2 − sech2𝜉) ,

    𝑢31= 𝑒

    𝑖𝜃{±2√3 − 𝛼 − 3𝛼2 + 3𝛽sech𝜉csch𝜉} ,

    V31= 𝛼

    2− 𝛽 − 3 − 2(sech𝜉csch𝜉)2,

    𝑢34= 𝑒

    𝑖𝜃{±4√−4 + 𝛼 + 3𝛼2 − 3𝛽

    sech2𝜉1 + tanh2𝜉

    } ,

    V34= 𝛼

    2− 𝛽 − 4 +

    8sech4𝜉

    (1 + tanh2𝜉)2,

    𝑢40= 𝑒

    𝑖𝜃{

    {

    {

    ±√(−1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2)1 + sech𝜉1 − sech𝜉

    }

    }

    }

    ,

    V40=1

    2{2𝛼

    2− 2𝛽 + 1 −

    1 + sech𝜉1 − sech𝜉

    } ,

    𝑢41= 𝑒

    𝑖𝜃{

    {

    {

    ±√(−1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2)1 − sech𝜉1 + sech𝜉

    }

    }

    }

    ,

    V41=1

    2{2𝛼

    2− 2𝛽 + 1 −

    1 − sech𝜉1 + sech𝜉

    } ,

    𝑢43= 𝑒

    𝑖𝜃{

    {

    {

    ±√−1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    𝐵 + √𝐵2 + 𝐶2sech𝜉𝐵tanh𝜉 + 𝐶sech𝜉

    }

    }

    }

    ,

    V43=1

    2{2𝛼

    2− 2𝛽 + 1 − (

    𝐵 + √𝐵2 + 𝐶2sech𝜉𝐵tanh𝜉 + 𝐶sech𝜉

    )

    2

    } .

    (67)

    Here, it should be noted that each exact solution given in (67)can be split into two solutions if one chooses the (+ve) and(−ve) signs, respectively, but they have not been calculated.Also, all the exact solutions given by (67) can be verifiedby substitution. The main feature for some of these exactsolutions is the inclusion of the free parameters 𝑄, 𝐵, and 𝐶.

    6. New Trigonometric-Function Solutions ofthe Schrödinger-KdV Equation

    Some trigonometric-function solutions of (1) can be obtainedin the limited case when the modulus𝑚 → 0. For example,

    𝑢5= 𝑒

    𝑖𝜃{±2√−1 − 𝛼 − 3𝛼2 + 3𝛽csc𝜉} ,

    V5= 𝛼

    2− 𝛽 + 1 − 2csc2𝜉,

    𝑢6= 𝑒

    𝑖𝜃{±2√−1 − 𝛼 − 3𝛼2 + 3𝛽sec𝜉} ,

    V6= 𝛼

    2− 𝛽 + 1 − 2sec2𝜉,

    𝑢9= 𝑒

    𝑖𝜃{±2√2 − 𝛼 − 3𝛼2 + 3𝛽tan𝜉} ,

    V9= 𝛼

    2− 𝛽 − 2sec2𝜉,

    𝑢11= 𝑒

    𝑖𝜃{±2√2 − 𝛼 − 3𝛼2 + 3𝛽cot𝜉} ,

    V11= 𝛼

    2− 𝛽 − 2csc2𝜉,

    𝑢13= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(csc𝜉 ± cot𝜉)

    }

    }

    }

    ,

    V13=1

    2{2𝛼

    2− 2𝛽 − 1 − (csc𝜉 ± cot𝜉)2} ,

    𝑢14= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(sec𝜉 ± tan𝜉)

    }

    }

    }

    ,

    V14=1

    2{2𝛼

    2− 2𝛽 − 1 − (sec𝜉 ± tan 𝜉)2} ,

    𝑢24= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(csc𝜉 ± cot𝜉)

    }

    }

    }

    ,

    V24=1

    2{2𝛼

    2− 2𝛽 − 1 + (csc𝜉 ± cot𝜉)2} ,

    𝑢32= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(sec𝜉 − tan𝜉)

    }

    }

    }

    ,

    V32= 𝛼

    2− 𝛽 − sec𝜉 (sec𝜉 − tan𝜉) ,

    𝑢36= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(csc𝜉 ± cot𝜉)

    }

    }

    }

    ,

  • 10 The Scientific World Journal

    V36=1

    2{2𝛼

    2− 2𝛽 − 1 − (csc𝜉 ± cot𝜉)2} ,

    𝑢37= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2(sec𝜉 ± tan𝜉)

    }

    }

    }

    ,

    V37=1

    2{2𝛼

    2− 2𝛽 − 1 − (sec𝜉 ± tan𝜉)2} ,

    𝑢42= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    √𝐵2 − 𝐶2 + 𝐵sin𝜉𝐵cos𝜉 + 𝐶

    }

    }

    }

    ,

    V42=1

    2{2𝛼

    2− 2𝛽 − 1 − (

    √𝐵2 − 𝐶2 + 𝐵sin𝜉𝐵cos𝜉 + 𝐶

    )

    2

    } ,

    𝑢43= 𝑒

    𝑖𝜃{

    {

    {

    ±√1 − 2𝛼 − 6𝛼

    2+ 6𝛽

    2

    √𝐵2 − 𝐶2 + 𝐵cos𝜉𝐵sin𝜉 + 𝐶

    }

    }

    }

    ,

    V43=1

    2{2𝛼

    2− 2𝛽 − 1 − (

    √𝐵2 − 𝐶2 + 𝐵cos𝜉𝐵sin𝜉 + 𝐶

    )

    2

    } ,

    𝑢44= 𝑒

    𝑖𝜃{±√−1 − 𝛼 − 3𝛼2 + 3𝛽

    2√𝐵2 + 𝐶2

    𝐵sin𝜉 + 𝐶cos𝜉} ,

    V44= 𝛼

    2− 𝛽 + 1 −

    2 (𝐵2+ 𝐶

    2)

    (𝐵sin𝜉 + 𝐶cos𝜉)2.

    (68)

    Here, we note also that each trigonometric-function solutionobtained in this section can split into two solutions if wechoose the (+ve) and (−ve) signs, respectively. Besides, allthese solutions can be verified by direct substitution. Also, themain feature for some of these exact solutions is the inclusionof the free parameters 𝑄, 𝐵, and 𝐶.

    7. Conclusion

    In this paper, the 𝐹-expansion method has been applied toconstruct 52 types of exact solution of the the Schrödinger-KdV equation. The main advantage of this method overother methods is that it possesses all types of exact solu-tion, including those of Jacobian-elliptic and Weierstrass-elliptic functions. Moreover, the soliton-like solutions andtrigonometric-function solutions have been also obtained asthe modulus 𝑚 of Jacobi-elliptic function approaches to 1and 0. It can be said that the results in this paper providegood supplements to the existing literature and are useful fordescribing certain nonlinear phenomena.Thismethod can beapplied to many other nonlinear evolution equations. Finally,it is worthwhile to mention that the proposed method is alsoa straightforward, short, promising, and powerful methodfor other nonlinear evolution equations in mathematicalphysics.

    Appendices

    A. Relations between Values of (𝑃, 𝑄, 𝑅) andCorresponding 𝐹(𝜉) in (7)

    Relations between values of (𝑃,𝑄,𝑅) and corresponding𝐹(𝜉)in (7), where 𝐴, 𝐵, and 𝐶 are arbitrary constants and 𝑚

    1=

    √1 − 𝑚2. As shown in Table 1.

    B. The Weierstrass-Elliptic Function Solutionsfor (7)

    The Weierstrass-elliptic function solutions for (7), where𝐷 = (1/2)(−5𝑄 ± √9𝑄2 − 36𝑃𝑅) and ℘(𝜉; 𝑔

    2, 𝑔3) =

    𝑑℘(𝜉; 𝑔2, 𝑔3)/𝑑𝜉. As shown in Table 2.

    C. Relations between Jacobian-EllipticFunctions and Hyperbolic Functions

    The Jacobian-elliptic functions degenerate into hyperbolicfunctions when𝑚 → 1 as follows:𝑠𝑛𝜉 → tanh𝜉, 𝑐𝑛𝜉 → sech𝜉, 𝑑𝑛𝜉 → sech𝜉,

    𝑠𝑐𝜉 → sinh𝜉, 𝑠𝑑𝜉 → sinh𝜉, 𝑐𝑑𝜉 → 1,

    𝑛𝑠𝜉 → coth 𝜉, 𝑛𝑐𝜉 → cosh𝜉, 𝑛𝑑𝜉 → cosh𝜉,

    𝑐𝑠𝜉 → csch𝜉, 𝑑𝑠𝜉 → csch𝜉, 𝑑𝑐𝜉 → 1.(C.1)

    The Jacobian-elliptic functions degenerate into trigono-metric functions when𝑚 → 0 as follows:𝑠𝑛𝜉 → sin𝜉, 𝑐𝑛𝜉 → cos𝜉, 𝑑𝑛𝜉 → 1,

    𝑠𝑐𝜉 → tan𝜉, 𝑠𝑑𝜉 → sin𝜉, 𝑐𝑑𝜉 → cos 𝜉,

    𝑛𝑠𝜉 → csc𝜉, 𝑛𝑐𝜉 → sec𝜉, 𝑛𝑑𝜉 → 1,

    𝑐𝑠𝜉 → cot𝜉, 𝑑𝑠𝜉 → csc𝜉, 𝑑𝑐𝜉 → sec𝜉.

    (C.2)

    D. Some Trigonometric and HyperbolicIdentities

    Consider the following:

    coth 𝜃 − csch𝜃 = tanh 𝜃2, csc𝜃 − cot𝜃 = tan 𝜃

    2,

    coth 𝜃 + csch𝜃 = coth 𝜃2, csc𝜃 + cot𝜃 = cot𝜃

    2,

    tanh 𝜃 + 𝑖sech𝜃 = tanh [12(𝜃 +

    𝑖𝜋

    2)] ,

    sec𝜃 + tan 𝜃 = tan [12(𝜃 +

    𝜋

    2)] ,

    tanh 𝜃 − 𝑖sech𝜃 = coth [12(𝜃 +

    𝑖𝜋

    2)] ,

    sec𝜃 − tan 𝜃 = cot [12(𝜃 +

    𝜋

    2)] .

    (D.1)

  • The Scientific World Journal 11

    Table 1

    Case 𝑃 𝑄 𝑅 𝐹(𝜉)

    1 𝑚2

    −(1 + 𝑚2) 1 𝑠𝑛𝜉

    2 𝑚2

    −(1 + 𝑚2) 1 𝑐𝑑𝜉 =

    𝑐𝑛𝜉

    𝑑𝑛𝜉

    3 −𝑚2

    2𝑚2− 1 1 − 𝑚

    2𝑐𝑛𝜉

    4 −1 2 − 𝑚2

    𝑚2− 1 𝑑𝑛𝜉

    5 1 −(1 + 𝑚2) 𝑚

    2𝑛𝑠𝜉 = (𝑠𝑛𝜉)

    −1

    6 1 −(1 + 𝑚2) 𝑚

    2𝑑𝑐𝜉 =

    𝑑𝑛𝜉

    𝑐𝑛𝜉

    7 1 − 𝑚2

    2𝑚2− 1 −𝑚

    2𝑛𝑐𝜉 = (𝑐𝑛𝜉)

    −1

    8 𝑚2− 1 2 − 𝑚

    2−1 𝑛𝑑𝜉 = (𝑑𝑛𝜉)

    −1

    9 1 − 𝑚2

    2 − 𝑚2

    1 𝑠𝑐𝜉 =𝑠𝑛𝜉

    𝑐𝑛𝜉

    10 −𝑚2(1 − 𝑚

    2) 2𝑚

    2− 1 1 𝑠𝑑𝜉 =

    𝑠𝑛𝜉

    𝑑𝑛𝜉

    11 1 2 − 𝑚2

    1 − 𝑚2

    𝑐𝑠𝜉 =𝑐𝑛𝜉

    𝑠𝑛𝜉

    12 1 2𝑚2− 1 −𝑚

    2(1 − 𝑚

    2) 𝑑𝑠𝜉 =

    𝑑𝑛𝜉

    𝑠𝑛𝜉

    131

    4

    1 − 2𝑚2

    2

    1

    4𝑛𝑠𝜉 ± 𝑐𝑠𝜉

    141 − 𝑚

    2

    4

    1 + 𝑚2

    2

    1 − 𝑚2

    4𝑛𝑐𝜉 ± 𝑠𝑐𝜉

    151

    4

    𝑚2− 2

    2

    𝑚2

    4𝑛𝑠𝜉 ± 𝑑𝑠𝜉

    16𝑚2

    4

    𝑚2− 2

    2

    𝑚2

    4𝑠𝑛𝜉 ± 𝑖𝑐𝑛𝜉

    17𝑚2

    4

    𝑚2− 2

    2

    𝑚2

    4

    √1 − 𝑚2𝑠𝑑𝜉 ± 𝑐𝑑𝜉

    181

    4

    1 − 𝑚2

    2

    1

    4𝑚𝑐𝑑𝜉 ± 𝑖√1 − 𝑚2𝑛𝑑𝜉

    191

    4

    1 − 2𝑚2

    2

    1

    4𝑚𝑠𝑛𝜉 ± 𝑖𝑑𝑛𝜉

    201

    4

    1 − 𝑚2

    2

    1

    4

    √1 − 𝑚2𝑠𝑐𝜉 ± 𝑑𝑐𝜉

    21𝑚2− 1

    4

    𝑚2+ 1

    2

    𝑚2− 1

    4𝑚𝑠𝑑𝜉 ± 𝑛𝑑𝜉

    22𝑚2

    4

    𝑚2− 2

    2

    1

    4

    𝑠𝑛𝜉

    1 ± 𝑑𝑛𝜉

    23 −1

    4

    𝑚2+ 1

    2

    (1 − 𝑚2)2

    4𝑚𝑐𝑛𝜉 ± 𝑑𝑛𝜉

    24(1 − 𝑚

    2)2

    4

    𝑚2+ 1

    2

    1

    4𝑑𝑠𝜉 ± 𝑐𝑠𝜉

    25𝑚4(1 − 𝑚

    2)

    2(2 − 𝑚2)

    2(1 − 𝑚2)

    𝑚2 − 2

    1 − 𝑚2

    2(2 − 𝑚2)𝑑𝑐𝜉 ± √1 − 𝑚2𝑛𝑐𝜉

    26 𝑃 > 0 𝑄 < 0𝑚2𝑄2

    (1 + 𝑚2)2

    𝑃

    √−𝑚

    2𝑄

    (1 + 𝑚2)𝑃𝑠𝑛(√

    −𝑄

    1 + 𝑚2𝜉)

  • 12 The Scientific World Journal

    Table 1: Continued.

    Case 𝑃 𝑄 𝑅 𝐹(𝜉)

    27 𝑃 < 0 𝑄 > 0(1 − 𝑚

    2)𝑄2

    (𝑚2 − 2)2

    𝑃

    √−𝑄

    (2 − 𝑚2)𝑃𝑑𝑛(√

    𝑄

    2 − 𝑚2𝜉)

    28 𝑃 < 0 𝑄 > 0𝑚2(𝑚2− 1)𝑄

    2

    (2𝑚2 − 1)2

    𝑃

    √−𝑚2𝑄

    (2𝑚2 − 1)𝑃𝑐𝑛(√

    𝑄

    2𝑚2 − 1𝜉)

    29 1 2 − 4𝑚2

    1𝑠𝑛𝜉𝑑𝑛𝜉

    𝑐𝑛𝜉

    30 𝑚4

    2 1𝑠𝑛𝜉𝑐𝑛𝜉

    𝑑𝑛𝜉

    31 1 𝑚2+ 2 1 − 2𝑚

    2+ 𝑚

    4𝑑𝑛𝜉𝑐𝑛𝜉

    𝑠𝑛𝜉

    32𝐴2(𝑚 − 1)

    2

    4

    𝑚2+ 1

    2+ 3𝑚

    (𝑚 − 1)2

    4𝐴2

    𝑑𝑛𝜉𝑐𝑛𝜉

    𝐴(1 + 𝑠𝑛𝜉)(1 + 𝑚 𝑠𝑛𝜉)

    33𝐴2(𝑚 + 1)

    2

    4

    𝑚2+ 1

    2− 3𝑚

    (𝑚 + 1)2

    4𝐴2

    𝑑𝑛𝜉𝑐𝑛𝜉

    𝐴(1 + 𝑠𝑛𝜉)(1 − 𝑚 𝑠𝑛𝜉)

    34 −4

    𝑚6𝑚 − 𝑚

    2− 1 −2𝑚

    3+ 𝑚

    4+ 𝑚

    2𝑚𝑐𝑛𝜉𝑑𝑛𝜉

    𝑚𝑠𝑛2𝜉 + 1

    354

    𝑚−6𝑚 − 𝑚

    2− 1 2𝑚

    3+ 𝑚

    4+ 𝑚

    2𝑚𝑐𝑛𝜉𝑑𝑛𝜉

    𝑚𝑠𝑛2𝜉 − 1

    361

    4

    1 − 2𝑚2

    2

    1

    4

    𝑠𝑛𝜉

    1 ± 𝑐𝑛𝜉

    371 − 𝑚

    2

    4

    1 + 𝑚2

    2

    1 − 𝑚2

    4

    𝑐𝑛𝜉

    1 ± 𝑠𝑛𝜉

    38 4𝑚1

    2 + 6𝑚1− 𝑚

    22 + 2𝑚

    1− 𝑚

    2𝑚2𝑠𝑛𝜉𝑐𝑛𝜉

    𝑚1− 𝑑𝑛2𝜉

    39 −4𝑚1

    2 − 6𝑚1− 𝑚

    22 − 2𝑚

    1− 𝑚

    2−𝑚2𝑠𝑛𝜉𝑐𝑛𝜉

    𝑚1+ 𝑑𝑛2𝜉

    402 − 𝑚

    2− 2𝑚

    1

    4

    𝑚2

    2− 1 − 3𝑚

    1

    2 − 𝑚2− 2𝑚

    1

    4

    𝑚2𝑠𝑛𝜉𝑐𝑛𝜉

    𝑠𝑛2𝜉 + (1 + 𝑚1)𝑑𝑛𝜉 − 1 − 𝑚

    1

    412 − 𝑚

    2+ 2𝑚

    1

    4

    𝑚2

    2− 1 + 3𝑚

    1

    2 − 𝑚2+ 2𝑚

    1

    4

    𝑚2𝑠𝑛𝜉𝑐𝑛𝜉

    𝑠𝑛2𝜉 + (−1 + 𝑚1)𝑑𝑛𝜉 − 1 + 𝑚

    1

    42𝐶2𝑚4− (𝐵

    2+ 𝐶

    2)𝑚2+ 𝐵

    2

    4

    𝑚2+ 1

    2

    𝑚2− 1

    4(𝐶2𝑚2 − 𝐵2)

    √((𝐵2 − 𝐶2)/(𝐵2 − 𝐶2 𝑚2)) + 𝑠𝑛𝜉

    𝐵𝑐𝑛𝜉 + 𝐶𝑑𝑛𝜉

    43𝐵2+ 𝐶

    2𝑚2

    4

    1

    2− 𝑚

    21

    4(𝐶2𝑚2 + 𝐵2)

    √((𝐶2𝑚2 + 𝐵2 − 𝐶2)/(𝐵2 + 𝐶2𝑚2)) + 𝑐𝑛𝜉

    𝐵𝑠𝑛𝜉 + 𝐶𝑑𝑛𝜉

    44𝐵2+ 𝐶

    2

    4

    𝑚2

    2− 1

    𝑚4

    4(𝐶2 + 𝐵2)

    √((𝐵2 + 𝐶2 − 𝐶2𝑚2)/(𝐵2 + 𝐶2)) + 𝑑𝑛𝜉

    𝐵𝑠𝑛𝜉 + 𝐶𝑐𝑛𝜉

    45 −(𝑚2+ 2𝑚 + 1)𝐵

    22𝑚

    2+ 2

    2𝑚 − 𝑚2− 1

    𝐵2

    𝑚𝑠𝑛2𝜉 − 1

    𝐵(𝑚𝑠𝑛2𝜉 + 1)

    46 −(𝑚2− 2𝑚 + 1)𝐵

    22𝑚

    2+ 2 −

    2𝑚 + 𝑚2+ 1

    𝐵2

    𝑚𝑠𝑛2𝜉 + 1

    𝐵(𝑚𝑠𝑛2𝜉 − 1)

  • The Scientific World Journal 13

    Table 2

    Case 𝑔2

    𝑔3

    𝐹(𝜉)

    474

    3(𝑄2− 3𝑃𝑅)

    4𝑄

    27(−2𝑄

    2+ 9𝑃𝑅) √

    1

    𝑃(℘(𝜉; 𝑔

    2, 𝑔3) −

    1

    3𝑄)

    484

    3(𝑄2− 3𝑃𝑅)

    4𝑄

    27(−2𝑄

    2+ 9𝑃𝑅) √

    3𝑅

    3℘(𝜉; 𝑔2, 𝑔3) − 𝑄

    49 −5𝑄𝐷 + 4𝑄

    2+ 33𝑃𝑄𝑅

    12

    21𝑄2𝐷 − 63𝑃𝑅𝐷 + 20𝑄

    3− 27𝑃𝑄𝑅

    216

    √12𝑅℘(𝜉; 𝑔2, 𝑔3) + 2𝑅(2𝑄 + 𝐷)

    12℘(𝜉; 𝑔2, 𝑔3) + 𝐷

    501

    12𝑄2+ 𝑃𝑅

    1

    216𝑄(36𝑃𝑅 − 𝑄

    2)

    √𝑅[6℘(𝜉; 𝑔2, 𝑔3) + 𝑄]

    3℘(𝜉; 𝑔2, 𝑔3)

    511

    12𝑄2+ 𝑃𝑅

    1

    216𝑄(36𝑃𝑅 − 𝑄

    2)

    3℘(𝜉; 𝑔

    2, 𝑔3)

    √𝑃[6℘(𝜉; 𝑔2, 𝑔3) + 𝑄]

    522𝑄

    2

    9

    𝑄3

    54

    𝑄√−15𝑄/2𝑃℘(𝜉; 𝑔2, 𝑔3)

    3℘(𝜉; 𝑔2, 𝑔3) + 𝑄

    , 𝑅 = 5𝑄2

    36𝑃

    Conflict of Interests

    The authors declare that there is no conflict of interests re-garding the publication of this paper.

    References

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