Research article

Forming localized waves of the nonlinearity of the DNA dynamics arising in oscillator-chain of Peyrard-Bishop model

  • Received: 28 October 2019 Accepted: 24 February 2020 Published: 09 March 2020
  • MSC : 65D19, 65H10, 35A20, 35A24, 35C08, 35G50

  • In this article, the mathematical modeling of DNA vibration dynamics has been considered that describes the nonlinear interaction between adjacent displacements along with the Hydrogen bonds with utilizing five techniques, namely, the improved tan(φ/2)-expansion method (ITEM), the exp(-Ω(η))-expansion method (EEM), the improved exp(-Ω(η))-expansion method (IEEM), the generalized (G'/G)-expansion method (GGM), and the exp-function method (EFM) to get the new exact solutions. This model of the equation is analyzed using the aforementioned schemes. The different kinds of traveling wave solutions: solitary, topological, periodic and rational, are fall out as a by-product of these schemes. Finally, the existence of the solutions for the constraint conditions is also shown.

    Citation: Jalil Manafian, Onur Alp Ilhan, Sizar Abid Mohammed. Forming localized waves of the nonlinearity of the DNA dynamics arising in oscillator-chain of Peyrard-Bishop model[J]. AIMS Mathematics, 2020, 5(3): 2461-2483. doi: 10.3934/math.2020163

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  • In this article, the mathematical modeling of DNA vibration dynamics has been considered that describes the nonlinear interaction between adjacent displacements along with the Hydrogen bonds with utilizing five techniques, namely, the improved tan(φ/2)-expansion method (ITEM), the exp(-Ω(η))-expansion method (EEM), the improved exp(-Ω(η))-expansion method (IEEM), the generalized (G'/G)-expansion method (GGM), and the exp-function method (EFM) to get the new exact solutions. This model of the equation is analyzed using the aforementioned schemes. The different kinds of traveling wave solutions: solitary, topological, periodic and rational, are fall out as a by-product of these schemes. Finally, the existence of the solutions for the constraint conditions is also shown.


    Traveling wave and soliton solutions are one of the most interesting and fascinating areas of research in different fields of engineering and physical sciences. These models are basic ingredients of sciences in which play important roles in numerous areas such as biology, physics, chemistry, fluid mechanics and many engineering and science applications among others [1,2,3,4,5]. Furthermore, the approaches to solving these types of equations alongside nonlinear PDEs ranging from analytical to numerical methods are very important in many engineering and sciences applications. Some of these methods include finding the exact solutions by using the special techniques in which can be manifested to new works with vigorous references. Consequently, it is imperative to address the dynamics of these soliton pulses from a mathematical aspect. This will lead to a deeper understanding of the engineering perspective of these solutions [6,7,8,9,10,11,12,13].

    In this paper, we will study the different kinds of traveling wave solutions in mathematical model in DNA dynamics from a purely mathematical viewpoint. Therefore, the importance of this paper will be to extract the exact traveling wave solution for the nonlinear model. This model is described for first by Peyrard-Bishop, that takes into consideration the inclusion of nonlinear interaction between adjacent displacements along with the Hydrogen bonds [14]. There are several integration tools available to solve the model. Many such nonlinear equations as DNA dynamics have been examined with regards to soliton theory, where complete integrability was emphasized by various analytical techniques.

    For investigating the appearance of solitonic structures of the oscillator-chain of Peyrard-Bishop model has been analyzed by [14,15]. The balance between weak nonlinearity and dispersion in DNA dynamic model with linear dispersion and nonlinear dispersion arise in works of Dusuel et al. [16] and Alvarez et al. [17]. Treatment of mathematical and physical modeling of equations of DNA dynamics show that those can be reduced to a significant nonlinear formations. The nonlinearity of the DNA dynamic model arises in localized waves in which have a few considerable features, as example in transporting energy without dissipation. A few methods in which physical properties of DNA dynamics have been investigated by the numerous authors [18,19,20,21,22,23]. There are techniques usually used in biological systems such as the discrete derivative operator (DDO) technique applying to long-range interactions systems [24,25], the semi-discrete approximation [26,27,28], the solitary perturbation technique [29,30], the modified extended tanh function method [31,32,33].

    Author of [34] made use of the Hirota bilinear method of the bidirectional Sawada-Kotera equation to obtain new lump-type solutions and interaction phenomenon. In [35], author found the lump soliton and novel solitary wave solutions for the (3+1)-dimensional extended Jimbo-Miwa equations. Manafian and co-author found the interaction phenomenon to the (2+1)-dimensional Breaking Soliton equation [36]. Moreover, Ilhan et al. determined lump wave solutions and the interactions between lump solutions for a variable-coefficient Kadomtsev-Petviashvili equation [37]. Authors of [38] obtained the stationary solutions of various nonlinear Schrödinger equations. Younas and co-authors [38] studied the nonlinear chirp solitons for the model of Schröodinger-Hirota equation with concluding the bright, dark and singular solitons. Ali and co-workers [39] utilized the extended trial equation method and retrieved Jacobi elliptic, periodic, bright and singular solitons for paraxial nonlinear Schröodinger equation. In [40], the first and second-order rogue wave solutions were gained for the coupled Schrödinger equations. Arif et al. investigated the solitons and lump wave solutions to the graphene thermophoretic motion system [41]. Structures of this paper as follows, the nonlinear DNA dynamics model has been summarized in section 2. In sections 3–7, an overview of the integration schemes are given along with the analysis of the model including the improved tan(ϕ/2)-expansion method, exp(Ω(η))-expansion method, improved exp(Ω(η))-expansion method, generalized (G'/G)-expansion method, and exp-function method, respectively. The next section gives the discussions about the model. In the last section, the conclusions have been given.

    It is popular that DNA molecule is a double helix. This means that it consists of two complementary polymeric chains twisted around each other [42]. The B-form DNA in theWatsonCrick model is a double helix, which contains of two strands. The masses of nucleotides do not vary too much which means that one can assume a homogeneous crystal structure. The strands are coupled to each other through the hydrogen bonds, so that these bonds are weak while the harmonic longitudinal are strong the PB model neglects all the displacements beside the transversal [43]. The Hamiltonian model of Peyrard and Bishop [43], and the equations found in the literature, is modeled by the Morse potential as

    VM(unvn)=D[ea(unvn)1]2, (2.1)

    in which un and vn are the displacements of the nucleotides. Also, the Hamiltonian for the DNA chain was described by Zdravković [43]. Moreover, the improved version of the PB model, introduced by Dauxois [44]. The Hamiltonian for describing the strand aperture the hydrogen bonds can be stated as [45]

    H(u)=12mq2n+k12Δ2un+k24Δ4un+δ(e2aun1)2,     Δun=un+1un, (2.2)

    in which k1 and k2 denote the strength for the linear and nonlinear couplings respectively and qn=m˙un is the momentum for the displacement un. Searching Starting with the hamiltonian (2.2) the equation of motion in the continuum limit can be stated by the following form

    2ut2(l1+3l22ux2)22aD eau(eau1), (2.3)

    with l1=k1md2,  l2=k2md4,  D=δm,  α2a and being d the inter-site nucleotide distance in the DNA ladder ([46,47,48]). In this paper, consider the Peyrard-Bishop DNA dynamic model equation as follows

    utt(l1+3l2u2x)uxx2αΩeαu(eαu1)=0, (2.4)

    where l1,l2,α and Ω=D are constants. By make the following transformations

    u(x,t)=u(ξ),ξ=xβt, (2.5)

    then the Eq. (2.4), can be reduced to the ordinary differential equation as

    β2u(l1+3l2(u)2)u2αΩeαu(eαu1)=0. (2.6)

    By multiplying the Eq. (2.6) by u and integrating once with respect to ξ, we get

    (β2l1)2(u)234l2(u)4+Ωeαu(eαu2)+R=0. (2.7)

    By starting hypothesis is taken to be

    v(ξ)=eαu(ξ). (2.8)

    By appending (2.8) into Eq. (2.7), the nonlinear equation is achieved as follows

    (β2l1)2α2v2(v)234α4l2(v)4+Ωv5(v2)+Rv4=0. (2.9)

    In this section, the improved tan(ϕ/2)-expansion method [8,9] has been summarized to obtain the solutions of nonlinear partial differential equations (NPDEs). Hence, consider the NPDEs in the following way:

    N(u,ux,ut,uxx,utt,...)=0, (3.1)

    where N is a polynomial of u and its partial derivatives in which the relationship of higher order derivatives and nonlinear terms. To find the traveling wave solutions, we outline the following sequence of steps towards the extended tanh method:

    Step 1. Firstly, by using traveling wave transformation

    ξ=xβt, (3.2)

    where β is non-zero arbitrary constant, permits to reduce Eq. (6.1) to an ODE of u=u(ξ) in the following form

    Q(u,u,βu,u,β2u,...)=0. (3.3)

    Step 2. Assuming that the solution of Eq. (6.1) can be expressed by the following ansatz:

    u(ξ)=S(ϕ)=mk=0Ak[tan(ϕ/2)]k, (3.4)

    where Ak(0km) are the parameters to be determined and Am0 and ϕ=ϕ(ξ) satisfies in the ordinary differential equation as follows:

    ϕ(ξ)=asin(ϕ(ξ))+bcos(ϕ(ξ))+c. (3.5)

    The particular solutions of Eq. (3.5) will be read as:

    Family 1: When Δ=a2+b2c2<0 and bc0, then ϕ(ξ)=2tan1[abcΔbctan(Δ2¯ξ)].

    Family 2: When Δ=a2+b2c2>0 and bc0, then ϕ(ξ)=2tan1[abc+Δbctanh(Δ2¯ξ)].

    For see the rest seventeen families refer to Ref. [8,9]. Also, ¯ξ=ξ+C,p,Ak,Bk(k=1,2,...,m),a,b and c are constants to be determined later.

    Step 3. To determine the positive integer m, we usually balance linear terms of the highest order in the resulting equation with the highest order nonlinear terms appearing in equation (3.3).

    Step 4. We collect all the terms with the same order of tan(ϕ/2)k,(k=0,1,2,...) together. Equate each coefficient of the polynomials to zero, yields the set of algebraic equations for A0,Ak(k=1,2,...,m),a,b and c with the aid of the Maple.

    Step 5. Solving the algebraic equations in Step 4, then substituting A0,A1,...,Bm,a,b,c in (3.4).

    Consider the homogeneous balance principle between the highest order derivatives (v)4 and nonlinear terms v6, and get

    A4m+4tan4m+4(ϕ/2)(v)4=v6A6mtan6m(ϕ/2)4m+4=6mm=2.

    Therefore, the equation (3.4) takes the form

    v(ξ)=2k=0Aktank(ϕ/2). (3.6)

    Substitute equation (3.6) and its derivatives into equation (2.9). Algebraic equations set can be obtained after equating the coefficients of tanp(ϕ/2) for p=0,1,...,12, and setting equal to zero. After solving the nonlinear algebraic equations, the following values of a,b,c,β,A0,A1,A2 can be obtained:

    Set Ⅰ.

    β=2l1+23Ωl23Rl2,  a=Ξ13A2l2,   b=4108ΩA22l23α3l2,c=c,Δ=α2A2(β22l1)3l2(bc)23A2l2, (3.7)
    A0=2Ωα4A22+3l2(bc)32Ωα4A2,   A1=Ξ1(bc)32Ωα4A22,   A2=A2,
    Ξ1=3A2l2(α2β2A22α2A2l13b2A2l2+3c2A2l23b2l2+6bcl23c2l2).

    By utilizing of Family 1, the trigonometric function solution becomes

    u1(x,t)=1αln[2Ωα4A22+3l2(bc)32Ωα4A2Ξ1(bc)32Ωα4A22[abcΔbctan(Δ2¯ξ)] (3.8)
    +A2[abcΔbctan(Δ2¯ξ)]2],    ¯ξ=x2l1+23Ωl23Rl2t+C.

    The existence of the solution for the constraint condition is as A2(β22l1)3l2<(4108ΩA22l23α3l2cα)2.

    By utilizing of Family 2, the hyperbolic function solution becomes

    u1(x,t)=1αln[2Ωα4A22+3l2(bc)32Ωα4A2Ξ1(bc)32Ωα4A22[abcΔbctan(Δ2¯ξ)] (3.9)
    +A2[abcΔbctan(Δ2¯ξ)]2],    ¯ξ=x2l1+23Ωl23Rl2t+C.

    The existence of the solution for the constraint condition is as A2(β22l1)3l2>(4108ΩA22l23α3l2cα)2.

    This section elucidates a systematic explanation of the exp(Ω(η))-expansion method to obtain the solutions of nonlinear partial differential equations (NPDEs). Hence, take the NPDEs in the following way:

    N(u,ux,ut,uxx,utt,...)=0, (4.1)

    where N is a polynomial of u and its partial derivatives in which the relationship of higher order derivatives and nonlinear terms. To find the traveling wave solutions, we outline the following sequence of steps towards the extended tanh method:

    Step 1. Firstly, by utilizing the traveling wave transformation

    ξ=xβt, (4.2)

    where β is non-zero arbitrary constant, permits to reduce equation (4.1) to an ODE of u=u(ξ) in the following form

    Q(u,u,βu,u,β2u,...)=0, (4.3)

    Step 2. Assuming that the solution of equation (4.1) can be expressed by the following ansatz:

    U(ξ)=mj=0AjFj(ξ), (4.4)

    where F(η)=exp(Φ(ξ)) and Aj(0jm), are the parameters to be determined Am0, and, Φ=Φ(ξ) satisfying the ODE given below

    Φ=μF1(ξ)+F(ξ)+λ. (4.5)

    The particular solutions of equation (4.5) will be read as:

    Solution-1: When μ0 and λ24μ>0, therefore we attain Φ(η)=ln(λ24μ2μtanh(λ24μ2(ξ+E))λ2μ).

    Solution-2: When μ0 and λ24μ<0, therefore we attain Φ(η)=ln(λ2+4μ2μtan(λ2+4μ2(ξ+E))λ2μ).

    Solution-3: When μ=0, λ0, and λ24μ>0, therefore we attain Φ(η)=ln(λexp(λ(ξ+E))1).

    Solution-4: When μ0, λ0, and λ24μ=0, therefore we attain Φ(η)=ln(2λ(ξ+E)+4λ2(ξ+E)).

    Solution-5: When μ=0, λ=0, and λ24μ=0, therefore we attain Φ(η)=ln(ξ+E), where Aj(0jm), E,λ and μ are also the constants to be explored later.

    Step 3. To determine the positive integer m, we usually balance the linear terms of the highest order in the resulting equation with the highest order nonlinear terms appearing in equation (4.3).

    Step 4. We collect all the terms with the same order of F(ξ)k,(k=0,1,2,...) together. Equate each coefficient of the polynomials of F(ξ)k to zero, yields the set of algebraic equations for A0,Ak(k=1,2,...,m),λ and μ with the aid of the Maple.

    Step 5. Solving the algebraic equations in Step 4, then substituting A0,A1,...,Am,λ,μ in (4.4).

    Consider the homogeneous balance principle between the highest order derivatives (v)4 and nonlinear terms v6, we obtain 4m+4=6m, then m=2. The exact solution can be expressed in the following form

    v(ξ)=2k=0AkFk(η), (4.6)

    Substitute equation (4.6) and its derivatives into equation (2.9). The algebraic equations set can be obtained after equating the coefficients of F(ξ) for p = 0, 1, ..., 20, and setting equal to zero. After solving the nonlinear algebraic equations, the following values of λ,μ,β,A0,A1,A2 can be obtained:

    Set Ⅰ.

    Σ1=125Ω5α3de3Ξ3Ξ31(λ24μ)250Ω4Rα4de2Ξ1Ξ2Ξ31250Ω5Rα5e3Ξ3Ξ21+5000Ω6α6d(45Ω4R)Ξ31 (4.7)
    6α2l1l2Ξ429l22Ξ3Ξ32(λ24μ)+5000Ω5α6l1Ξ413Ωαl2Ξ2(50ΩRα2eΞ2Ξ3+2αd(2025Ω2+65ΩR+R2)Ξ22+
    5ΩeΞ1Ξ3(225Ω2αe+5ΩRαe+2025Ω2d+115ΩRd+R2d)(λ24μ)),   
    Σ2=25000Ω6Rα6dΞ31+1250Ω5Rα5e3Ξ3Ξ21+15Ω2αdel2Ξ1Ξ3Ξ22(λ24μ)
    18l22Ξ32(225Ω2λ2+5ΩRλ2+11250Ω2μ+670ΩRμ+6R2μ)+25Ω2α3eΞ3(5Ω3de2Ξ31(λ24μ)+6l2RΞ22)
    75Ω2α2l2Ξ2(Ωe2Ξ3Ξ21)(λ24μ)4RdΞ2+250Ω4α4Ξ1(30Ωl2Ξ31(λ2+8μ)+Rde2Ξ2Ξ3),
    Ξ1=45Ω+R,   Ξ2=2025Ω2+115ΩR+R2,   Ξ3=1575Ω2+105ΩR+R2,  β=(2500Ω5α4Ξ413l2Ξ42)Σ1α(2500Ω5α4Ξ413l2Ξ42),
    d=3Ωl2Ω,  e=412Ω3l2Ω,   A0=Σ26α2Ωd(2500Ω5α4Ξ413l2Ξ42),  A1=2λdα2,  A2=2dα2.

    By utilizing of Family 1, the hyperbolic function solution becomes

    u1(x,t)=1αln[A0+2λdα2(λ24μ2μtanh(λ24μ2(η+E))λ2μ)1 (4.8)
    +2dα2(λ24μ2μtanh(λ24μ2(η+E))λ2μ)2],   η=xβt.

    The existence of the solution for the constraint condition is as l2(α4(Ω2α2e212Rl2)eα3Ω)>0.

    By utilizing of Family 2, the trigonometric function solution becomes

    u2(x,t)=1αln[A0+2λdα2(λ2+4μ2μtan(λ2+4μ2(η+E))λ2μ)1 (4.9)
    +2dα2(λ2+4μ2μtan(λ2+4μ2(η+E))λ2μ)2],   η=xβt.

    The existence of the solution for the constraint condition is as l2(α4(Ω2α2e212Rl2)eα3Ω)<0.

    By utilizing of Family 3, the hyperbolic function solution becomes

    u3(x,t)=1αln[A0+2λdα2(λexp(λ(η+E))1)+2dα2(λexp(λ(η+E))1)2],  η=xβt. (4.10)

    The existence of the solution for the constraint condition is as l2(α4(Ω2α2e212Rl2)eα3Ω)>0, and

    λ=3l2(eα3Ω+12μl2+Ω2α6e212Rα4l2)3l2,  λ24μ=eα3Ω+α4(Ω2α2e212Rl2)3l2.

    In this section the improved exp(Ω(η))-expansion method is utilized to obtain the solutions of nonlinear partial differential equations (NPDEs). Hence, consider the NPDEs in the following way:

    N(u,ux,ut,uxx,utt,...)=0, (5.1)

    where N is a polynomial of u and its partial derivatives in which the relationship of higher order derivatives and nonlinear terms. To find the traveling wave solutions, we outline the following sequence of steps towards the extended tanh method:

    Step 1. Firstly, by using the traveling wave transformation

    ξ=xβt, (5.2)

    where β is non-zero arbitrary constant, permits to reduce equation (4.1) to an ODE of u=u(ξ) in the following form

    Q(u,u,βu,u,β2u,...)=0, (5.3)

    Step 2. Assuming that the solution of equation (5.1) can be expressed by the following ansatz:

    U(ξ)=mj=0AjFj(ξ)+mj=1BjFj(ξ), (5.4)

    where F(η)=exp(Φ(ξ)) and Aj(0jm),Bj(1jm), are the parameters to be determined Am0, and, Φ=Φ(ξ) satisfying the ODE given below

    Φ=μF1(ξ)+F(ξ)+λ. (5.5)

    The particular solutions of equation (5.5) will be read like before section.

    Step 3. To determine the positive integer m, we usually balance the linear terms of the highest order in the resulting equation with the highest order nonlinear terms appearing in equation (4.3).

    Step 4. We collect all the terms with the same order of F(ξ)k,(k=0,1,2,...) together. Equate each coefficient of the polynomials of F(ξ)k to zero, yields the set of algebraic equations for A0,Ak,Bk(k=1,2,...,m),λ and μ with the aid of the Maple.

    Step 5. Solving the algebraic equations in Step 4, then substituting A0,A1,B1,...,Am,Bm,λ,μ in (5.4).

    The exact solution will be the same as the previous section as

    v(ξ)=2k=0AkFk(η)+2k=1BkFk(η), (5.6)

    Substitute equation (5.6) and its derivatives into equation (2.9). The algebraic equations set can be obtained after equating the coefficients of F(ξ) for p=0,1,...,20, and setting equal to zero. After solving the nonlinear algebraic equations, the following values of λ,μ,β,A0,A1,B1,A2,B2 can be obtained:

    Set Ⅰ.

    Σ1=5000Ω5α6Ξ31(45Ω2d4ΩRd+45Ωl1+Rl1)1250Ω5Rα5e3Ξ3Ξ21250Ω4Rα4de2Ξ1Ξ2Ξ3 (5.7)
    25Ω2α3eΞ2(5Ω3de2Ξ31(λ24μ)+6Rl2Ξ22)3α2l2Ξ2(25Ω3e2Ξ3Ξ21(λ24μ)+
    2Ξ22(2025Ω3d+65Ω2Rd+ΩR2d+2025Ω2l1+115ΩRl1+R2l1))15Ω2αdel2Ξ1Ξ3Ξ22(λ24μ)
    9l22Ξ3Ξ32(λ24μ),   d=3Ωl2Ω,  e=412Ω3l2Ω,   β=(2500Ω5α4Ξ413l2Ξ42)Σ1α(2500Ω5α4Ξ413l2Ξ42),
    A0=Σ26α2Ωd(2500Ω5α4Ξ413l2Ξ42),  A1=2λdα2,  A2=2dα2,    B1=0,   B2=Σ2360dl2Σ3,
    Σ2=25000Ω6Rα6dΞ31+1250Ω5Rα5e3Ξ3Ξ21+15Ω2αdel2Ξ1Ξ3Ξ22(λ24μ)+25Ω2α3eΞ3(5Ω3de2Ξ31(λ24μ)+6l2RΞ22)
    18l22Ξ32(225Ω2λ2+5ΩRλ2+11250Ω2μ+670ΩRμ+6R2μ)+75Ω2α2l2Ξ2(Ωe2Ξ3Ξ21(λ24μ)4RdΞ22)
    +250Ω4α4Ξ1(30Ωl2Ξ31(λ2+8μ)+Rde2Ξ2Ξ3),
    Σ3=6250000Ω10α8Ξ813l2(Ξ42)(5000Ω5α4Ξ413l2Ξ42),
    Ξ1=45Ω+R,   Ξ2=2025Ω2+115ΩR+R2,   Ξ3=1575Ω2+105ΩR+R2.

    By utilizing of Family 1, the hyperbolic function solution becomes

    u1(x,t)=1αln[A0+2λdα2(λ24μ2μtanh(λ24μ2¯η)λ2μ)1 (5.8)
    +2dα2(λ24μ2μtanh(λ24μ2¯η)λ2μ)2+Σ2360dl2Σ3(λ24μ2μtanh(λ24μ2¯η)λ2μ)2].

    The existence of the solution for the constraint condition is as l2(α4(Ω2α2e212Rl2)eα3Ω)>0.

    By utilizing of Family 2, the trigonometric function solution becomes

    u2(x,t)=1αln[A0+2λdα2(λ2+4μ2μtan(λ2+4μ2¯η)λ2μ)1 (5.9)
    +2dα2(λ2+4μ2μtan(λ2+4μ2¯η)λ2μ)2+Σ2360dl2Σ3(λ2+4μ2μtan(λ2+4μ2¯η)λ2μ)2].

    The existence of the solution for the constraint condition is as l2(α4(Ω2α2e212Rl2)eα3Ω)<0.

    By utilizing of Family 3, the kink-soliton solution becomes

    u3(x,t)=1αln[A0+2λdα2(λexp(λ¯η)1)+2dα2(λexp(λ¯η)1)2+Σ2360dl2Σ3(λexp(λ¯η)1)2]. (5.10)

    The existence of the solution for the constraint condition is as l2(α4(Ω2α2e212Rl2)eα3Ω)>0 and

    ¯η=xβt+E,  λ=3l2(eα3Ω+12μl2+Ω2α6e212Rα4l2)3l2,  λ24μ=eα3Ω+α4(Ω2α2e212Rl2)3l2.

    Set Ⅱ.

    Σ1=5000Ω5α6Ξ31(45Ω2d4ΩRd+45Ωl1+Rl1)1250μΩ5Rα5e3Ξ3Ξ21250μ2Ω4Rα4de2Ξ1Ξ2Ξ3 (5.11)
    25Ω2α3eμΞ3(5Ω3de2Ξ31(λ24μ)+6Rl2μ2Ξ22)3α2l2μ2Ξ2(25Ω3e2Ξ3Ξ21(λ24μ)+
    2μ2Ξ22(2025Ω3d+65Ω2Rd+ΩR2d+2025Ω2l1+115ΩRl1+R2l1))15μ3Ω2αdel2Ξ1Ξ3Ξ22(λ24μ)
    9μ4l22Ξ3Ξ32(λ24μ),  d=3Ωl2Ω,  e=412Ω3l2Ω,  β=(2500Ω5α4Ξ413μ4l2Ξ42)Σ1α(2500Ω5α4Ξ413μ4l2Ξ42),
    Σ2=1250μΩ5Rα5e3Ξ3Ξ21+15μ3Ω2αdel2Ξ1Ξ3Ξ22(λ24μ)+250Ω4α4Ξ1(30Ωl2Ξ31(λ2+8μ)+Rde2μ2Ξ2Ξ3)
    18μ4l22Ξ32(225Ω2λ2+5ΩRλ2+11250Ω2μ+670ΩRμ+6R2μ)+25μΩ2α3eΞ3(5Ω3de2Ξ31(λ24μ)+6μ2l2RΞ22)+
    75μ2Ω2α2l2Ξ2(Ωe2(1575Ω2+105ΩR+R2)Ξ21(λ24μ)4μ2RdΞ2)+25000Ω6Rα6dΞ31,
    A0=Σ26α2Ωd(2500Ω5α4Ξ413μ4l2Ξ42),  A1=0,  A2=0,  B1=2dλμα2,  B2=2dμ2α2,
    Ξ1=45Ω+R,   Ξ2=2025Ω2+115ΩR+R2,   Ξ3=1575Ω2+105ΩR+R2.

    By utilizing of Family 1, the hyperbolic function solution becomes

    u1(x,t)=1αln[A0+2dλμα2(λ24μ2μtanh(λ24μ2¯η)λ2μ)+ (5.12)
    2μ2dα2(λ24μ2μtanh(λ24μ2¯η)λ2μ)2],   ¯η=xβt+E.

    The existence of the solution for the constraint condition is as l2μ(α4(Ω2α2e212μ2Rl2)eα3Ω)>0.

    By utilizing of Family 2, the trigonometric function solution becomes

    u2(x,t)=1αln[A0+2dλμα2(λ2+4μ2μtan(λ2+4μ2¯η)λ2μ)+ (5.13)
    2μ2dα2(λ2+4μ2μtan(λ2+4μ2¯η)λ2μ)2],   ¯η=xβt+E.

    The existence of the solution for the constraint condition is as l2μ(α4(Ω2α2e212μ2Rl2)eα3Ω)<0.

    By utilizing of Family 3, the exponential function solution becomes

    u3(x,t)=1αln[A0+2μλdα2(exp(λ¯η)λ)+2dμα2(exp(λ¯η)λ)2],   ¯η=xβt+E. (5.14)
    λ=3l2(eα3Ω+12μl2+Ω2α6e212Rα4l2)3l2,  λ24μ=eα3Ω+α4(Ω2α2e212Rl2)3l2.

    The existence of the solution for the constraint condition is as l2μ(α4(Ω2α2e212μRl2)eα3Ω)>0.

    As the fourth method, the generalized (G'/G)-expansion method has been summarized to obtain the solutions of NPDEs. Hence, consider the NPDEs of in the following way:

    N(u,ux,ut,uxx,utt,...)=0, (6.1)

    where N is a polynomial of u and its partial derivatives in which the relationship of higher order derivatives and nonlinear terms. To find the traveling wave solutions, we outline the following sequence of steps towards the GGM:

    Step 1. Firstly, by using traveling wave transformation

    ξ=xβt, (6.2)

    where β is non-zero arbitrary constant, permits to reduce equation (6.1) to an ODE of u=u(ξ) in the following form

    Q(u,u,βu,u,β2u,...)=0, (6.3)

    Step 2. Assuming that the solution of equation (6.1) can be expressed by the following ansatz:

    u(ξ)=S(Φ(ξ))=mk=0AkΦ(ξ)k, (6.4)

    where, Ak(0km) are constants to be determined, such that Am0, and Φ(ξ)=G(ξ)/G(ξ) satisfies the following ODE:

    k1GGk2GGk3(G)2k4G2=0. (6.5)

    The particular solutions of equation (6.5) will be read as:

    Family 1: When k20, f=k1k3 and s=k22+4k4(k1k3)>0, then Φ(ξ)=k22f+s2fC1sinh(s2k1ξ)+C2cosh(s2k1ξ)C1cosh(s2k1ξ)+C2sinh(s2k1ξ).

    Family 2: When k20, f=k1k3 and s=k22+4k4(k1k3)<0, then Φ(ξ)=k22f+s2fC1sin(s2k1ξ)+C2cos(s2k1ξ)C1cos(s2k1ξ)+C2sin(s2k1ξ).

    Family 3: When k20, f=k1k3 and s=k22+4k4(k1k3)=0, then Φ(ξ)=k22f+C2C1+C2ξ.

    Family 4: When k2=0, f=k1k3 and g=fk4>0, then Φ(ξ)=gfC1sinh(gk1ξ)+C2cosh(gk1ξ)C1cosh(gk1ξ)+C2sinh(gk1ξ).

    Family 5: When k2=0, f=k1k3 and g=fk4<0, then Φ(ξ)=gfC1sin(gk1ξ)+C2cos(gk1ξ)C1cos(gk1ξ)+C2sin(gk1ξ).

    Family 6: When k4=0 and f=k1k3, then Φ(ξ)=C1k22exp(k2k1ξ)fk1+C1k1k2exp(k2k1ξ).

    Family 7: When k20 and f=k1k3=0, then Φ(ξ)=k4k2+C1exp(k2k1ξ),

    Family 8: When k1=k3, k2=0 and f=k1k3=0, then Φ(ξ)=C1+k4k1ξ,

    Family 9: When k3=2k1, k2=0 and k4=0, then Φ(ξ)=1C1+(k3k11)ξ, where d0,dj,ej(j=1,...,m),k1,k2,k3 and k4 are constants to be determined later.

    Step 3. To determine the positive integer m, we usually balance linear terms of the highest order in the resulting equation with the highest order nonlinear terms appearing in equation (3.3).

    Step 4. We collect all the terms with the same order of Φ(ξ)k,(k=0,1,2,...) together. Equate each coefficient of the polynomials of i to zero, yields the set of algebraic equations for A0,Ak(k=1,2,...,m),k1,k2,k3, and k4 with the aid of the Maple.

    Step 5. Solving the algebraic equations in Step 4, then substituting A0,A1,...,Bm,k1,k2,k3,k4 in (6.4).

    By processing the generalized G/G-expansion method and considering the homogeneous balance principle, we get the exact solution in the following form

    u(ξ)=A0+A1Φ(ξ)+A1Φ(ξ)2. (6.6)

    Solving the nonlinear algebraic equations, we have the following sets of coefficients for the solutions of (6.6) as given below:

    Subset Ⅰ.

    β=A2(2α2A2k12l112k12l2ϵ1(ϵ12)(A01)3l2(4A0k12A2ϵ324k12))k1αA2, (6.7)
    ε1=123/44ΩA22l23α12l2,  ε2=3l2k1(4ΩA22l23123/4α+12l2)6l2,
    ε3=13A2l26A2l2((A01)(6ϵ2l2(ϵ22k1)+6k12l2)+ε4,
    ε4=36ϵ2l22(ϵ22k1)(ϵ222ϵ2k1+2k12)(A01)23α4A22k14l2(ΩA022ΩA0+R)+36k14l22(A01)2,
    A1=12ε3l2(ε11)3k1Ωα4A2,  A2=A2,  k1=k1,  k2=ε3,  k3=ε1,  k4=Ωα4A0A2k112l2(ϵ133ϵ12+3ϵ11),
    s=k22+4k4(k1k3)=ϵ32Ωα4A0A2k123(ϵ11)2l2.

    Based on the Family 1, the exact soliton solution can be written as

    u1(x,t)=1αln[A0+12ε3l2(ε11)3k1Ωα4A2{k22f+s2fC1sinh(s2k1ξ)+C2cosh(s2k1ξ)C1cosh(s2k1ξ)+C2sinh(s2k1ξ)}+ (6.8)
    A2{k22f+s2fC1sinh(s2k1ξ)+C2cosh(s2k1ξ)C1cosh(s2k1ξ)+C2sinh(s2k1ξ)}2],

    in which ξ=xA2(2α2A2k12l112k12l2ϵ1(ϵ12)(A01)3l2(4A0k12A2ϵ324k12))k1αA2t. The existence of the solution for the constraint condition is as |ε3|>α2|k1||ε11|A0A2Ω3l2.

    Based on the Family 2, the exact periodic solution can be written as

    u2(x,t)=1αln[A0+12ε3l2(ε11)3k1Ωα4A2{k22f+s2fC1sin(s2k1ξ)+C2cos(s2k1ξ)C1cos(s2k1ξ)+C2sin(s2k1ξ)}+ (6.9)
    A2{k22f+s2fC1sin(s2k1ξ)+C2cos(s2k1ξ)C1cos(s2k1ξ)+C2sin(s2k1ξ)}2],

    in which ξ=xA2(2α2A2k12l112k12l2ϵ1(ϵ12)(A01)3l2(4A0k12A2ϵ324k12))k1αA2t. The existence of the solution for the constraint condition is as |ε3|<α2|k1||ε11|A0A2Ω3l2.

    Based on the Family 3, the exact singular solution can be written as

    u3(x,t)=1αln[A0+(2Ω2α4A04k14+3ΩA02l26ΩA0l2+3Rl2)ε3(ε11)32Ω2A02k1α4(ϵ2k1)2(A01){k22f+C2C1+C2ξ}+ (6.10)
    24ΩA02l2(ϵ2k1)2(A01)4Ω2α4A04k14+3ΩA02l26ΩA0l2+3Rl2{k22f+C2C1+C2ξ}2],

    in which ξ=xA2(2α2A2k12l112k12l2ϵ1(ϵ12)(A01)3l2(4A0k12A2ϵ324k12))k1αA2t and A2=24ΩA02l2(ϵ2k1)2(A01)4Ω2α4A04k14+3ΩA02l26ΩA0l2+3Rl2.

    Based on the Family 6, the exact kink solution can be written as

    u4(x,t)=1αln[12ε3l2(ε11)3k1Ωα4A2{C1k22exp(k2k1ξ)fk1+C1k1k2exp(k2k1ξ)}+A2{C1k22exp(k2k1ξ)fk1+C1k1k2exp(k2k1ξ)}2], (6.11)

    in which ξ=xA2(2α2A2k12l1+12k12l2ϵ1(ϵ12)3l2(A2ϵ324k12))k1αA2t.

    We first consider the nonlinear equation of form

    N(u,ut,ux,uxx,utt,utx,...)=0, (7.1)

    and introduce a transformation as

    u(x,t)=u(η),ξ=xβt, (7.2)

    where β is constant to be determined later. Therefore the Eq. (7.1) is reduced to an ODE as follows

    M(u,βu,u,u,...)=0. (7.3)

    The EFM is based on the assumption that the travelling wave solutions can be expressed in the form

    u(ξ)=dn=canexp(nξ)qm=pbmexp(mξ), (7.4)

    where c, d, p and q are positive integers which could be freely chosen, an's and bm's are unknown constants to be determined.

    We apply the Exp-function method to Eq. (2.9). In order to determine values of c and p, we balance the terms (v)4 and v6 in Eq. (2.9) along with Eq. (7.4), then we get

    (v)4=c1exp(4(c+p)ξ)+...c2exp(8pξ)+...,v6=c3exp((6c+2p)ξ)+...c4exp(8pξ)+..., (7.5)

    respectively. Balancing highest order of the Exp–function in (7.5) and get 4c+4p=6c+2p, which leads to the result c=p. Similarly, to find values of d and q, for the terms (v)4 and v6 in Eq. (2.9) by simple calculation, we attain

    (v)4=d1exp(4(d+q)ξ)+...d2exp(8qξ)+...,v6=d3exp((6d+2q)ξ)+...d4exp(8qξ)+..., (7.6)

    respectively. Balancing lowest order of the Exp–function in (7.6), we achieve d=q.

    Case Ⅰ: p=c=1 and q=d=1.

    For simplicity, we set a1=0,b1=1, p=c=1 and d=q=1. Then Eq. (7.4) reduces to

    v(ξ)=a1exp(ξ)+a0exp(ξ)+b0+b1exp(ξ). (7.7)

    Substituting (7.7) into Eq. (2.9), we get an equation in the following form

    ([b1exp(ξ)+b0+exp(ξ)]4)14n=4Cnexp(nξ)=0, (7.8)

    where Cn(4n4) are polynomial expressions in terms of a1,a0,a1,b0,b1 and β. Thus, solving the resulting system Cn=0(4n4) simultaneously, we acquire the following set as

    (Ⅰ) The first set is:

    a1=0,a0=b0(4α2R+β2l1)2α2Ω,b0=b0,b1=14b20,R=3l2+2α2(l1β2)4α4,β=±1α3l2+α2(l1±23l2Ω), (7.9)
    v(ξ)=2b0(4α2R+β2l1)α2Ω(2eξ/2+b0eξ/2)2,ξ=x1α3l2+α2(l1±23l2Ω)t, (7.10)

    then the solution equation (2.4) will be as

    u1(x,t)=1αln[2b0(4α2R+β2l1)α2Ω(2e12[x1α3l2+α2(l1±23l2Ω)t]+b0e12[x1α3l2+α2(l1±23l2Ω)t])2]. (7.11)

    If we choose b0=2 and b0=2, then the solution equation (7.11), respectively, give:

    u2(x,t)=1αln[b0(4α2R+β2l1)8α2Ωsech2(x212α3l2+α2(l1±23l2Ω)t)], (7.12)
    u3(x,t)=1αln[b0(4α2R+β2l1)8α2Ωcsch2(x212α3l2+α2(l1±23l2Ω)t)], (7.13)

    (Ⅱ) The second set is:

    a1=Ω±Ω2ΩRΩ,a0=b0(a1(R2Ω)+R)Ra1Ω,b0=b0,b1=0,R=R,β=β, (7.14)
    v(ξ)=a0+a1eξb0+eξ,ξ=xβt, (7.15)

    then the solution equation (2.4) will be as

    u4(x,t)=1αln[b0(a1(R2Ω)+R)Ra1Ω+Ω±Ω2ΩRΩexβtb0+exβt]. (7.16)

    Case Ⅱ: p=c=2 and q=d=2.

    Since the values of c and d can be freely chosen, we set p=c=2 and d=q=2 and then the trial function (7.4) becomes

    u(ξ)=a2exp(2ξ)+a1exp(ξ)+a0+a1exp(ξ)+a2exp(2ξ)b2exp(2ξ)+b1exp(ξ)+b0+b1exp(ξ)+b2exp(2ξ). (7.17)

    There are some free parameters in (7.17), we set b2=1 and a1=a1=a2=b1=b1=0 for simplicity, the trial function, (7.17) is simplified as follows

    u(ξ)=a0+a2exp(2ξ)exp(2ξ)+b0+b2exp(2ξ). (7.18)

    Substituting (7.18) into Eq. (2.9), we get an equation in the following form

    ([b2exp(2ξ)+b0+exp(2ξ)]8)18n2=4Cnexp(nξ)=0, (7.19)

    where Cn(8n16) are polynomial expressions in terms of a2,a0,a1,b0,b2 and β. Thus, solving the resulting system Cn=0(8n16) simultaneously, we obtain the following set of algebraic equations

    (Ⅰ) The first set is:

    a2=0,a0=2b0(α2R+β2l1)α2Ω,b0=b0,b1=14b20,R=6l2+α2(l1β2)α4,β=±1α3ΩΩ(36l22+α4Ω(α2l16l2)), (7.20)
    v(ξ)=8b0(α2R+β2l1)α2Ω(2eξ+b0eξ)2,ξ=x1α3ΩΩ(36l22+α4Ω(α2l16l2))t, (7.21)

    then the solution equation (2.4) will be as

    u5(x,t)=1αln[8b0(α2R+β2l1)α2Ω(2e[x1α3ΩΩ(36l22+α4Ω(α2l16l2))t]+b0e[x1α3ΩΩ(36l22+α4Ω(α2l16l2))t])2]. (7.22)

    If we choose b0=2 and b0=2, then the solution equation (7.22), respectively, give:

    u6(x,t)=1αln[b0(α2R+β2l1)2α2Ωsech2(x1α3ΩΩ(36l22+α4Ω(α2l16l2))t)], (7.23)
    u7(x,t)=1αln[b0(α2R+β2l1)2α2Ωcsch2(x1α3ΩΩ(36l22+α4Ω(α2l16l2))t)], (7.24)

    (Ⅱ) The second set is:

    a2=Ω±Ω2ΩRΩ,a0=b0(a1(R2Ω)+R)Ra1Ω,b0=b0,b2=0,R=R,β=β, (7.25)
    v(ξ)=a0+a2e2ξb0+e2ξ,ξ=xβt, (7.26)

    then the solution equation (2.4) will be as

    u8(x,t)=1αln[b0(a2(R2Ω)+R)Ra2Ω+Ω±Ω2ΩRΩe2x2βtb0+e2x2βt]. (7.27)

    This paper finds many novel hyperbolic, trigonometric, kink, and kink-singular soliton solutions to governing model. With the help of some calculations, surfaces of results reported have been observed in Figures 15. These figures are depended on the family conditions which are of important physically. It has been investigated that all figures plotted have symbolized the nonlinear DNA dynamics. These mathematical properties come from trigonometric and hyperbolic function properties. In this sense, from the mathematical and physical points of views, these results take play an important role in explaining waves propagation in nonlinear dispersion. Hence, we consider surfaces plotted in this paper have proved such physical meaning of the solutions. In Figure 1 and Figure 2, we have depicted the 3D, 2D, contour, and density schematic representation of the analytical and numerical solutions at few space positions for three different waves at x=1, x=0, and x=1 by taking l1=1,l2=1,α=3,Ω=1,R=1,μ=1.5 for (5.8) and l1=1,l2=2,α=3,Ω=10,R=5,μ=1.5 for (5.9). We observe that the breathe soliton wave move in direction (x,t) and increases with move of negative (x,t) to positive (x,t). Also, the periodic wave solution for (6.9) by taking A0=1,A2=2,k1=2,l1=3,l2=2,α=2,Ω=5,C1=2,C2=3,R=5 is presented in Figure 3. Moreover, the rational kink wave solution for the DNA dynamics (6.10) by taking A0=1,A2=1,k1=2,l1=3,l2=2,α=2,Ω=5,C1=2,C2=3,R=5, is offered in Figure 4. Likewise, the DNA dynamics for (6.11) by taking A0=0,A2=1,k1=2,l1=3,l2=2,α=2,Ω=5,R=5, along with 3D plot, density plot, contour plot, and 2D plot with at spaces at x=1, x=0, and x=1 are plotted in Figure 5.

    Figure 1.  Plot of DNA dynamics (5.8) by taking l1=1,l2=1,α=3,Ω=1,R=1,μ=1.5 and (a) 3D plot, (b) density plot, (c) contour plot and (d) 2D plot with at space (a) red x=1, blue x=0, and green x=1.
    Figure 2.  Plot of DNA dynamics (5.9) by taking l1=1,l2=2,α=3,Ω=10,R=5,μ=1.5 and (a) 3D plot, (b) density plot, (c) contour plot and (d) 2D plot with at space (a) red x=1, blue x=0, and green x=1.
    Figure 3.  Plot of DNA dynamics (6.9) by taking A0=1,A2=2,k1=2,l1=3,l2=2,α=2,Ω=5,C1=2,C2=3,R=5 and (a) 3D plot, (b) density plot, (c) contour plot and (d) 2D plot with at space (a) red x=1, blue x=0, and green x=1.
    Figure 4.  Plot of DNA dynamics (6.10) by taking A0=1,A2=1,k1=2,l1=3,l2=2,α=2,Ω=5,C1=2,C2=3,R=5, and (a) 3D plot, (b) density plot, (c) contour plot and (d) 2D plot with at space (a) red x=1, blue x=0, and green x=1.
    Figure 5.  Plot of DNA dynamics (6.11) by taking A0=0,A2=1,k1=2,l1=3,l2=2,α=2,Ω=5,R=5, and (a) 3D plot, (b) density plot, (c) contour plot and (d) 2D plot with at space (a) red x=1, blue x=0, and green x=1.

    The article obtains, the traveling wave solutions of different kinds, which are solitary, topological, singular, periodic and rational solutions to the model for DNA dynamics. The integration mechanisms that are adopted, are improved tan(ϕ/2)-expansion scheme, exp(Ω(η))-expansion scheme, improved exp(Ω(η))-expansion scheme, generalized (G'/G)-expansion scheme, and exp-function scheme. It is quite visible that these integration schemes has its limitations. Thus, this paper are provides a lot of encouragement for future research in DNA dynamics. Afterwards extra solution methods will be applied to obtain lump and singular soliton solutions to the nonlinear model. In addition to, this model will be considered with other forms of nonlinear media. The constructed results may be helpful in explaining the physical meaning of the studied models and other related nonlinear phenomena models. Results are beneficial to the study of the nonlinear DNA dynamics. All calculations in this paper have been made quickly with the aid of the Maple.

    The authors would like to thank the given comments and valuable recommendations by respected Editor and the reviewers provided to improve the paper.

    The authors have declared no conflict of interest.



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