# Bifurcation Analysis and Exact Wave Solutions for the Double-Chain Model of DNA

Authors: Taher S. Hassan; A.A. Elmandouh; Adel A. Attiya; Ahmed Y. Khedr
Journal: Journal of Mathematics
 (2022)

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## Abstract

This work aims to study analytically the nonlinear model for deoxyribonucleic acid (DNA). Based on the complete discrimination and direct method, some new wave solutions are introduced. These solutions are sorted into solitary, periodic, kink (antikink), and singular solutions. Moreover, a part of them is illustrated graphically. Based on Hamilton concepts, we study the bifurcation and phase portrait for the Hamilton system corresponding to the model under consideration.


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## Body

## 1. Introduction

Deoxyribonucleic acid (DNA) molecule carries the information for living beings that is required to live and propagate themselves. The nonlinear model for deoxyribonucleic acid (DNA) is attractive for the study because its properties can be investigated precisely by experiments gathering both the physical methods and biological tools [1]. In 1953, Watson and Crick [2] initially discovered the double helix construction of DNA; in spite of that, it is not easy until now to find a specific mathematical model that involved all its characteristics. The reason is its complicated structure and the existence of several motions such as the torsional, transverse, and longitudinal motions [3]. However, the existence of several motions of the DNA on thoroughly distinct time scales becomes aidable to model a few of these that predominate in the given time scale range. The idea that was introduced by Davydov [4] in his pioneer works related to the theoretical studies of the nonlinear characteristics of DNA has been firstly utilized by Englander and coauthors in 1980 to investigate dynamics of DNA regarding the nitrogen rational motion [5]. This idea has been further developed in several subsequent works. For instance, Yomosa proposed a model of the dynamics plane of the base rotator [6], and this study was followed by Takeno and Homma who got better this model by considering the degree of freedom describing base rotations in the plane perpendicular to the helical axis around the structure of the backbone [7]. The denaturation process in which the base transverse motion along the hydrogen bond is regarded was investigated by Peyrard and Bishop [8]. Two kinds of internal motions, which have been proposed by Muto et al., contribute mainly to the denaturation process of DNA. These motions are longitudinal motions over the backbone and transverse motions over the hydrogen bond [9]. This model has been developed and improved in several works in order to study distinct motions and construct solitary wave-type solutions [10–14]. These waves acquire their significance from their ability to transmit energy without losing, i.e., the energy is conserved [8, 15], and moreover, they explicate the long-range interaction of kink solitons in the double chain [16, 17] and transcription regulation [18].Taking into account some acceptable approximations from the point of biological science, the two equations describe a DNA model with double chains consisting of elastic two long homogeneous strands. These strands characterize two polynucleotide chains of DNA molecules attached by an elastic membrane which represents the hydrogen bonds between the base pair of two chains. The dynamical nonlinear system characterizing the double-chain model of DNA takes the form [19, 20](1)Itt−e12Ixx=a1I+b1IJ+c1I3+d1IJ2,Jtt−e22Ixx=a2J+b2I2+c2I2J+d2J3+m0,where I refers to the longitudinal displacement difference between the top and bottom wires, while J indicates the transverse displacements between the upper and lower strands, and ei,ai,bi,ci,di,andm0, i = 1, 2, are constants given by(2)e1=±R1ρ,e2=±R2ρ,a1=±2μρσhh−l0,a2=±2μρσ,b1=2b2=22μl0ρσh2,c1=c2=−2μl0h3ρσ,d1=d2=4μl0h3ρσ,m0=μ2ρσh−l0,where ρ,σ,R1,andR2 refer to density of mass, area of cross section, Young’s modulus, and density of the tension of each strand, μ indicates the stiffness of the elastic membrane, h is the distance between the two strands, and l0 is the height of membrane in the equilibrium.To transform the nonlinear system (1) into a single partial differential equation, we present(3)J=αI+β,where αandβ are two arbitrary constants, and furthermore, we assume β=h/2 and R1=R2. Thus, the linear system (1) is reduced to(4)Itt−e12Ixx=f3I3+f2I2+f1I,where fi are arbitrary constants introduced for suitability, and they are given by(5)f3=ω0h34α2−2,f2=62αω0h2,f1=6ω0h−2ω0l0,ω0=l0μσρ.The nonlinear model (1) has been investigated in several works. Riccati parameterized factorization method has been applied in [20] to construct some solitary wave solutions for the DNA model (1). The Φ6− model expansion method has been utilized in [21] to construct some solutions which are assorted into solitary, kink, and singular waves. Some solitary wave solutions for the double-chain model of DNA have been introduced and discussed [19]. Some exact wave solutions of this model have been constructed by using Conte’s Painlevé truncation expansion and Pickering’s truncation expansion [22]. Some bounded wave solutions for this model such as bell-shaped solitary waves and periodic waves have been formulated based on the method of the dynamical systems [23]. The generalized exponential rational function method has been applied to introduce exact form solutions and solitonic structures for this model [24].Despite the wide variety of methods used to find wave solutions to nonlinear PDEs, the problem under study has a simple history, for instance, the bifurcation analysis [25–34], sine-Gordon expansion method [35, 36], Hirota bilinear technique [37], G′/G method [38–40], differential transform method (DTM), homotopy perturbation method (HPM) [41], Lie symmetry method [42], homotopic analysis method [43], trigonometric function series method [44], modified mapping method and extended mapping method [45], modified trigonometric function series method [46], tanh-coth expansion method and Jacobi function expansion method [47], and Jacobi elliptic function expansion method [48], and for more different techniques, see [49–53].In this work, we are interested in constructing some traveling wave solutions for the nonlinear model (1) which is equivalent to building a wave solution for the reduced equation (1). We apply the complete discriminant system in addition to determining the intervals of permitted real propagations. The significance of finding these intervals enables us to construct only real wave solutions, and furthermore, for the same constraints on the system’s parameters, there are several intervals of possible real wave propagations. Hence, the missing of such study in previous works leads to missing some wave solutions and the appearance of complex solutions. The bifurcation analysis is introduced which plays an important role in determining the types of the solutions before constructing them. We are also interested in studying the influence of the system’s parameters on the solutions.This work is organized as follows: Section2 involves the reduction of the DNA model to ordinary differential equations and using the complete discrimination method to construct some traveling wave solutions for equation (1). Section 3 contains the study of some dynamical properties of equation (1) by employing the complete discrimination and Hamiltonian concepts. Section 4 is a graphic representation of some of the obtained solutions. Furthermore, it examines the influence of the one-parameter changing on the solutions keeping the other parameters fixed. Section 4 is a collection and the summary of the obtained results.


## 2. Exact Wave Solutions

Applying the wave transformationIx,t=uζ,ζ=kx−ωt, to equation (4), we obtain(6)u′′=f3ω2−k2e12u3+f2ω2−k2e12u2+f1ω2−ka2e12u,where k is a constant that specifies the cosine of angle of propagation with ζ−axis and ω is an arbitrary constant that characterizes the speed of the wave and ω2−k2e12≠0. For simplicity, we insert(7)uζ=pζ−f23f3,into equation (6), and we get(8)p″ζ=2n4p3ζ+n2pζ+n12,where ′ indicates derivatives with respect to ζ and ni,i=2,3,4, are constants introduced for suitability and they are given by(9)n4=f32ω2−k2e12,n2=f22−3f3f13f3k2e12−ω2,n1=2f39f1f2−2f2227f32k2e12−ω2.Integrating both sides of equation (8) with respect to p, we obtain(10)p′2=n4p4+γ2p2+γ1p+γ0,where γi are arbitrary parameters. Separating the variables, we obtain the differential form(11)dζ=dpn4F4p,where(12)F4p=p4+γ2p2+γ1p+γ0,in which(13)γ2=n2n4,γ1=n1n4.To integrate both sides of equation (11), the range of the parameters is required to be determined. The cause for this is that different values of the parameters imply different solutions to the integral. Hence, the key steps are to find the range of these parameters and consequently integrate both sides of equation (11). There are many tools utilized to find these ranges of parameters. In this work, we will apply a complete discrimination system for a polynomial. This method is a natural generalization of the discrimination Δ=b2−4ac for the quadratic polynomial ax2+bx+c, but it becomes difficult to calculate it for the higher degree polynomials. This problem had been solved with aid of computer algebra programs by Yang et al. by presenting an algorithm to compute the complete discrimination system for polynomial [54]. The complete discrimination system for the quartic polynomial F4p=p4+γ2p2+γ1p+γ0 is given in [55], and it admits the form(14)D1=4,D2=−γ2,D3=−2γ23+8γ2γ0−9γ12,E2=9γ22−32γ2γ0,D4=−γ23γ12+4γ0γ24+36γ2γ12−32γ22γ02−274γ14+64γ04.We study eight cases that describe different types of the roots for polynomial (12). To avoid confounding, we collect the classification of all the different types of the roots of the polynomial F4p by utilizing the discriminant system in Table 1. Furthermore, we integrate only on certain intervals for p in which n4F4p is positive in order to get real solutions.Table 1
Types of the roots of the polynomialF4p.
No.Conditions on the discriminant systemTypes of the roots forF4p1D2=D3=D4=0All roots are equal to zero2D3=D4=E2=0Two real roots: one is triple and the other is simple3D3=D4=0,E2>0Two double roots4D2>0,D3>0,D4>0Four real roots5D4=0,D2D3<0One double root and two complex conjugate roots6D4<0,D2D3>0Two real roots and two complex conjugate roots7D2D3≤0,D4>0Two conjugate complex roots8D2>0,D3>0,D4=0Four real roots: one double and others simple


### 2.1. Case 1

Polynomial (12) has four real roots which are equal to zero if D2=D3=D4=0. Hence, it is written as F4p=p4. Assuming −∞<p<0 and pζ0=−∞ and integrating (11), we obtain(15)Ix,t=−1n4kx+ωt−ζ0−2ah2a2−1,and consequently, we have(16)Jx,t=α−1n4kx+ωt−ζ0−2ah2a2−1+β.Solutions (15) and (16) are singular solutions, and their singularity points lie on the plane kx+ct−ζ0=0.


### 2.2. Case 2

IfD3=D4=E2=0, then the quartic polynomial (12) has two real roots: one is simple and the other is triple. Hence, it can be expressed as F4p=p−p13p+3p1, where p1 is assumed to be positive, i.e., p1>0. We consider two subcases according to n4 is positive or negative:(i)
Forn4>0, the possible interval for real propagation is p∈−∞,−3p1∪p1,∞. Thus, if we choose p<−3p1, assume pζ0=−∞, and integrate both sides of equation (11), we obtain(17)n4∫ζ0ζdζ=∫−∞pdpp−p1p−p1p+3p1.It follows(18)pζ=p1−p1p1n4ζ−ζ0+1+1n4ζ−ζ0.Thus, the solution of equation (1) becomes(19)Ix,t=p1−p1p1n4ζ−ζ0+1+1n4ζ−ζ0−f23f3,Jx,t=αp1−p1p1n4ζ−ζ0+1+1n4ζ−ζ0−f23f3+β.Similarly, we can calculate the solution ifp∈p1,∞.(ii)
Ifn4<0, the possible interval for p to obtain real propagation is p∈−3p1,p1. Thus, we assume pζ0=−3p1 and integrate both sides of equation (11), and we get(20)Ix,t=p1−4p11+4n4p12ζ−ζ02−f23f3,Jx,t=αp1−4p11+4n4p12ζ−ζ02−f23f3+β.Both solutions (19) and (20) are singular solutions for equation (1).


### 2.3. Case 3

The polynomialF4p has two double real zeros, namely, ±p1, where p1>0, if D3=D4=0, E2>0,D2>0. Hence, it can be introduced as F4p=p2−p122. We consider the following two cases in which n4 is either positive or negative.(i)
Ifn4>0, the intervals for real propagation are p<−p1,−p1<p<p1, and p>p1. If we consider the case in which p<−p1 and assume pζ0=−∞, equation (11) gives(21)n4∫ζ0ζdζ=∫−∞pdpp2−p12.It follows(22)p=−p1cothp1n4ζ−ζ0.Using equations (7) and (3), we obtain a wave solution for equation (1) in the form(23)Ix,t=−p1cothp1n4ζ−ζ0−f23f3,Jx,t=α−p1cothp1n4ζ−ζ0−f23f3+β.Whenp>p1, equation (1) has the same solution shown in equation (23) if p1⟶−p1. Similarly, if we choose p∈−p1,p1 and assume pζ0=0, equation (1) has a solution in the form(24)Ix,t=−p1tanhp1n4ζ−ζ0−f23f3,Jx,t=α−p1tanhp1n4ζ−ζ0−f23f3+β.(ii)
The case in whichn4<0 is excluded since n4F4p<0 for all p∈ℝ.


### 2.4. Case 4

The polynomialF4p has four real zeros, namely, p1,p2,p3,−p1+p2+p3, where we assumed 0<p1<p2<p3 if D2>0,D3>0,D4>0. Therefore, it takes the form F4p=p−p1p−p2p−p3p+p1+p2+p3. Now, we consider the two cases n4>0 and n4>0, individually:(i)
Whenn4>0, the possible intervals of p for real propagation are p<−p1+p2+p3,p1<p<p2,andp>p3. If we choose p<−p1+p2+p3 and assume pζ0=−p1+p2+p3, equation (11) becomes(25)n4∫ζ0ζdζ=∫−p1+p2+p3pdpp−p1p−p2p−p3p+p1+p2+p3.It implies to(26)p=p1−p1−p32p1+p2+p3p1−p3+p1+p2+2p3sn2Ω1ζ−ζ0,k1,ζ0<ζ<ζ1,whereΩ1=1/2n4p3−p1p1+2p2+p3, k1=p2−p1p1+p2+2p3/p3−p1p1+2p2+p3, and ζ1=1/Ω1Kk1. Kk1 is a complete elliptic integral of the first type [56]. Using equations (7) and (3), we obtain a new traveling wave solution for equation (1) in the form(27)Ω1x,t=p1−p1−p32p1+p2+p3p1−p3+p1+p2+2p3sn2Ω1ζ−ζ0,k1−f23f3,Jx,t=αp1−p1−p32p1+p2+p3p1−p3+p1+p2+2p3sn2Ω1ζ−ζ0,k1−f23f3+β.If we selectp1<p<p2, postulate pζ0=p1, and follow the same procedures, we will obtain a new traveling wave solution for equation (1) in the form(28)Ix,t=−p1−p2−2p3+p1+2p2+p32p1+p2+2p3p1+2p2+p3+p1−p2sn2Ω1ζ−ζ0,k1−f23f3,Jx,tα−p1−p2−2p3+p1+2p2+p32p1+p2+2p3p1+2p2+p3+p1−p2sn2Ω1ζ−ζ0,k1−f23f3+β,whereζ0<ζ<ζ1. Also, if we elect p>p3 and suppose pζ0=p3, we will get a new wave solution for equation (1) in the form(29)Ix,t=p2+p2−p3p1+2p2+p3p1+p2+2p3sn2Ω1ζ−ζ0,k1−f23f3,Jx,t=αp2+p2−p3p1+2p2+p3p1+p2+2p3sn2Ω1ζ−ζ0,k1−f23f3+β,whereζ0<ζ<ζ1.(ii)
Ifn4<0, the allowed intervals of p for real propagation are −p1−p2−p3<p<p1 or p2<p<p3. Thus, if we chose p1<p<p and assume pζ0=−p1−p2−p3, equation (11) takes the form(30)−n4∫ζ0ζdζ=∫−p1−p2−p3pdp−p−p1p2−pp3−pp+p1+p2+p3.It gives(31)p=p3+p1−p3p1+p2+2p3p3−p1+2p1+p2+p3sn2Ω2ζ−ζ0,k2,ζ0<ζ<ζ2,whereΩ2=1/2−n4p3−p12p2+p1+p3, k2=p3−p22p1+p2+p3/p3−p12p2+p1+p3, and ζ2=1/Ω2Kk2. Utilizing equations (7) and (3), we obtain a new solution for equation (1):(32)Ix,t=p3+p1−p3p1+p2+2p3p3−p1+2p1+p2+p3sn2Ω2ζ−ζ0,k2−f23f3,Jx,t=αp3+p1−p3p1+p2+2p3p3−p1+2p1+p2+p3sn2Ω2ζ−ζ0,k2+β.Similarly, if we selectp2<p<p3 and assume pζ0=p2, we present a new wave solution for equation (1) in the form(33)Ix,t=p1−p1−p2p1−p3p1−p3+p3−p2sn2Ω2ζ−ζ0,k2−f23f3,Jx,t=αp1−p1−p2p1−p3p1−p3+p3−p2sn2Ω2ζ−ζ0,k2−f23f3+β.


### 2.5. Case 5

The polynomialF4p has one double real root and two conjugate complex roots if D4=0 and D2D3<0. Therefore, it takes the form F4p=p−p12p−p2p−p2∗, where ∗ refers to the complex conjugate and p1=−Rep2. We consider the case in which n4>0, and sequentially, the allowed intervals for real propagation are p<p1 or p>p1. Choosing p>p1, assuming pζ0=∞, and integrating both sides of equation (11), we obtain(34)p=p1+4p12+ρ2−2p1+ρsinhn44p12+ρ2ζ−ϵ,where ϵ=ζ0−sinh−12p1/ρ/n44p12+ρ2 is a new constant which is introduced for suitability and ρ=Imp2. Employing equations (7) and (3), we obtain a solution for equation (1) in the form(35)Ix,t=p1+4p12+ρ2−2p1+ρsinhn44p12+ρ2ζ−ϵ−f23f3,Jx,t=αp1+4p12+ρ2−2p1+ρsinhn44p12+ρ2ζ−ϵ−f23f3+β.It can be noted that the case in whichn4 is negative does not work because F4p≥0 for all p∈ℝ.


### 2.6. Case 6

The polynomialF4P has two real roots and two complex conjugate roots if D4<0 and D2D3>0. Hence, it can be written in the form F4p=p−p1p−p2p−p3p−p3∗, where p1<p2 and Rep3=1/2p1+p2. We consider the following:(i)
Ifn4>0, then the permitted intervals for real propagation are p>p2 and p<p1. Selecting p<p2 and pζ0=p2 and integrating both sides of equation (11), we obtain(36)p=p2B1−p1A1+p2B1+p1A1cnn4A1B1ζ−ζ0,k2A1+B1cnn4A1B1ζ−ζ0,k2−A1−B1,wherek2=A1+B12−p2−p12/4A1B1 and A12=B12=1/4p1−p22+Im2p3. Taking into account equations (7) and (3), we obtain a novel wave solution for equation (1) in the form(37)Ix,t=p2B1−p1A1+p2B1+p1A1cnn4A1B1kx+ωt−ζ0,k2A1+B1cnn4A1B1ζ−ζ0,k2−A1−B1−f23f3,Jx,t=αp2B1−p1A1+p2B1+p1A1cnn4A1B1kx+ωt,k2A1+B1cnn4A1B1ζ−ζ0,k2−A1−B1−f23f3+β.(ii)
Ifn4<0, then the allowed intervals of possible propagation are p1<p<p2. Assuming pζ0=p1 and integrating both sides of equation (11), we obtain(38)p=p2B2+p1A2+p1A2−p2B2cn−n4A2B2ζ−ζ0,k3B2+A2−B2−A2cn−n4A2B2ζ−ζ0,k3,wherek3=p2−p12−A2−B22/4A2B2, A22=1/4p1+3p22+Im2p3, and B22=1/43p1+p22+Im2p3. Utilizing equations (7) and (3), we construct a novel wave solution for equation (1) in the form(39)Ix,t=p2B2+p1A2+p1A2−p2B2cn−n4A2B2kx+ωt−ζ0,k3B2+A2−B2−A2cn−n4A2B2kx+ωt−ζ0,k3−f23f3,Jx,t=αp2B2+p1A2+p1A2−p2B2cn−n4A2B2kx+ωt−ζ0,k3B2+A2−B2−A2cn−n4A2B2kx+ωt−ζ0,k3−f23f3+β.


### 2.7. Case 7

The polynomialF4p has two conjugate complex roots, namely, p1,p1∗,p2,p2∗, if D2D3≤0 and D4>0. Therefore, it is expressed as F4p=p−p1p−p1∗p−p2p−p2∗, where Re.p1=−Re.p2. The permitted interval for real propagation for the case n4>0 is p∈ℝ. Thus, we follow similar steps as above and obtain a new traveling wave solution for equation (1) as(40)Ix,t=Rep1+Imp1δ+Imp1+Rep1δtnn4/2A2+B2kx+ωt−ζ0,k4δ+tnn4/2A2+B2kx+ωt−ζ0,k4−f23f3,Jx,t=αRep1+Imp1δ+Imp1+Rep1δtnn4/2A2+B2kx+ωt−ζ0,k4δ+tnn4/2A2+B2kx+ωt−ζ0,k4−f23f3+β,where A22=Imp1−Imp22,B22=Imp1−Imp22+4Re2p1, δ2=4ℜp12−A2−B22/A2+B22−4ℜp12, and k4=2A2B2/A2+B2. It can be noted that the case in which n4<0 does not work because F4p>0 for all p∈ℝ.


### 2.8. Case 8

The polynomialF4p has four real roots in which one of them is double and the others are simple if D2>0,D3>0, and D4=0. Hence, it takes the form F4p=p−p12p−p2p−p3, where p1<p2<p3 and p3=−2p1+p2. We consider the two cases in which n4 is either positive or negative:(i)
Ifn4>0, then the possible intervals for real propagation are p<p1,p>p3, and p1<p<p2. With similar computations as in previous cases, we present the solution of equation (1) directly.If we choosep>−2p1+p2 and assume pζ0=−2p1+p2, we have a new wave solution for equation (1) in the form(41)Ix,t=−2p1−p2+4p2−p1sechn4−3p1−p2kx+ωt−ζ0−f23f3,Jx,t=α−2p1−p2+4p2−p1sechn4−3p1−p2kx+ωt−ζ0−f23f3+β.In similar calculations, we can calculate the wave solution forp<p1 and p1<p<p2.(ii)
Ifn4<0, the allowed interval for real propagation is p∈p2,p3, and postulating pζ0=p2, we obtain a new wave solution for equation (1) in the form(42)Ix,t=p1+p2−p13p1+p22p1+p1+p2cosh−n43p1+p2p1−p2kx−ωt−ζ0−f23f3,Jx,t=αp1+p2−p13p1+p22p1+p1+p2cosh−n43p1+p2p1−p2kx−ωt−ζ0−f23f3+β.


## 2.1. Case 1

Polynomial (12) has four real roots which are equal to zero if D2=D3=D4=0. Hence, it is written as F4p=p4. Assuming −∞<p<0 and pζ0=−∞ and integrating (11), we obtain(15)Ix,t=−1n4kx+ωt−ζ0−2ah2a2−1,and consequently, we have(16)Jx,t=α−1n4kx+ωt−ζ0−2ah2a2−1+β.Solutions (15) and (16) are singular solutions, and their singularity points lie on the plane kx+ct−ζ0=0.


## 2.2. Case 2

IfD3=D4=E2=0, then the quartic polynomial (12) has two real roots: one is simple and the other is triple. Hence, it can be expressed as F4p=p−p13p+3p1, where p1 is assumed to be positive, i.e., p1>0. We consider two subcases according to n4 is positive or negative:(i)
Forn4>0, the possible interval for real propagation is p∈−∞,−3p1∪p1,∞. Thus, if we choose p<−3p1, assume pζ0=−∞, and integrate both sides of equation (11), we obtain(17)n4∫ζ0ζdζ=∫−∞pdpp−p1p−p1p+3p1.It follows(18)pζ=p1−p1p1n4ζ−ζ0+1+1n4ζ−ζ0.Thus, the solution of equation (1) becomes(19)Ix,t=p1−p1p1n4ζ−ζ0+1+1n4ζ−ζ0−f23f3,Jx,t=αp1−p1p1n4ζ−ζ0+1+1n4ζ−ζ0−f23f3+β.Similarly, we can calculate the solution ifp∈p1,∞.(ii)
Ifn4<0, the possible interval for p to obtain real propagation is p∈−3p1,p1. Thus, we assume pζ0=−3p1 and integrate both sides of equation (11), and we get(20)Ix,t=p1−4p11+4n4p12ζ−ζ02−f23f3,Jx,t=αp1−4p11+4n4p12ζ−ζ02−f23f3+β.Both solutions (19) and (20) are singular solutions for equation (1).


## 2.3. Case 3

The polynomialF4p has two double real zeros, namely, ±p1, where p1>0, if D3=D4=0, E2>0,D2>0. Hence, it can be introduced as F4p=p2−p122. We consider the following two cases in which n4 is either positive or negative.(i)
Ifn4>0, the intervals for real propagation are p<−p1,−p1<p<p1, and p>p1. If we consider the case in which p<−p1 and assume pζ0=−∞, equation (11) gives(21)n4∫ζ0ζdζ=∫−∞pdpp2−p12.It follows(22)p=−p1cothp1n4ζ−ζ0.Using equations (7) and (3), we obtain a wave solution for equation (1) in the form(23)Ix,t=−p1cothp1n4ζ−ζ0−f23f3,Jx,t=α−p1cothp1n4ζ−ζ0−f23f3+β.Whenp>p1, equation (1) has the same solution shown in equation (23) if p1⟶−p1. Similarly, if we choose p∈−p1,p1 and assume pζ0=0, equation (1) has a solution in the form(24)Ix,t=−p1tanhp1n4ζ−ζ0−f23f3,Jx,t=α−p1tanhp1n4ζ−ζ0−f23f3+β.(ii)
The case in whichn4<0 is excluded since n4F4p<0 for all p∈ℝ.


## 2.4. Case 4

The polynomialF4p has four real zeros, namely, p1,p2,p3,−p1+p2+p3, where we assumed 0<p1<p2<p3 if D2>0,D3>0,D4>0. Therefore, it takes the form F4p=p−p1p−p2p−p3p+p1+p2+p3. Now, we consider the two cases n4>0 and n4>0, individually:(i)
Whenn4>0, the possible intervals of p for real propagation are p<−p1+p2+p3,p1<p<p2,andp>p3. If we choose p<−p1+p2+p3 and assume pζ0=−p1+p2+p3, equation (11) becomes(25)n4∫ζ0ζdζ=∫−p1+p2+p3pdpp−p1p−p2p−p3p+p1+p2+p3.It implies to(26)p=p1−p1−p32p1+p2+p3p1−p3+p1+p2+2p3sn2Ω1ζ−ζ0,k1,ζ0<ζ<ζ1,whereΩ1=1/2n4p3−p1p1+2p2+p3, k1=p2−p1p1+p2+2p3/p3−p1p1+2p2+p3, and ζ1=1/Ω1Kk1. Kk1 is a complete elliptic integral of the first type [56]. Using equations (7) and (3), we obtain a new traveling wave solution for equation (1) in the form(27)Ω1x,t=p1−p1−p32p1+p2+p3p1−p3+p1+p2+2p3sn2Ω1ζ−ζ0,k1−f23f3,Jx,t=αp1−p1−p32p1+p2+p3p1−p3+p1+p2+2p3sn2Ω1ζ−ζ0,k1−f23f3+β.If we selectp1<p<p2, postulate pζ0=p1, and follow the same procedures, we will obtain a new traveling wave solution for equation (1) in the form(28)Ix,t=−p1−p2−2p3+p1+2p2+p32p1+p2+2p3p1+2p2+p3+p1−p2sn2Ω1ζ−ζ0,k1−f23f3,Jx,tα−p1−p2−2p3+p1+2p2+p32p1+p2+2p3p1+2p2+p3+p1−p2sn2Ω1ζ−ζ0,k1−f23f3+β,whereζ0<ζ<ζ1. Also, if we elect p>p3 and suppose pζ0=p3, we will get a new wave solution for equation (1) in the form(29)Ix,t=p2+p2−p3p1+2p2+p3p1+p2+2p3sn2Ω1ζ−ζ0,k1−f23f3,Jx,t=αp2+p2−p3p1+2p2+p3p1+p2+2p3sn2Ω1ζ−ζ0,k1−f23f3+β,whereζ0<ζ<ζ1.(ii)
Ifn4<0, the allowed intervals of p for real propagation are −p1−p2−p3<p<p1 or p2<p<p3. Thus, if we chose p1<p<p and assume pζ0=−p1−p2−p3, equation (11) takes the form(30)−n4∫ζ0ζdζ=∫−p1−p2−p3pdp−p−p1p2−pp3−pp+p1+p2+p3.It gives(31)p=p3+p1−p3p1+p2+2p3p3−p1+2p1+p2+p3sn2Ω2ζ−ζ0,k2,ζ0<ζ<ζ2,whereΩ2=1/2−n4p3−p12p2+p1+p3, k2=p3−p22p1+p2+p3/p3−p12p2+p1+p3, and ζ2=1/Ω2Kk2. Utilizing equations (7) and (3), we obtain a new solution for equation (1):(32)Ix,t=p3+p1−p3p1+p2+2p3p3−p1+2p1+p2+p3sn2Ω2ζ−ζ0,k2−f23f3,Jx,t=αp3+p1−p3p1+p2+2p3p3−p1+2p1+p2+p3sn2Ω2ζ−ζ0,k2+β.Similarly, if we selectp2<p<p3 and assume pζ0=p2, we present a new wave solution for equation (1) in the form(33)Ix,t=p1−p1−p2p1−p3p1−p3+p3−p2sn2Ω2ζ−ζ0,k2−f23f3,Jx,t=αp1−p1−p2p1−p3p1−p3+p3−p2sn2Ω2ζ−ζ0,k2−f23f3+β.


## 2.5. Case 5

The polynomialF4p has one double real root and two conjugate complex roots if D4=0 and D2D3<0. Therefore, it takes the form F4p=p−p12p−p2p−p2∗, where ∗ refers to the complex conjugate and p1=−Rep2. We consider the case in which n4>0, and sequentially, the allowed intervals for real propagation are p<p1 or p>p1. Choosing p>p1, assuming pζ0=∞, and integrating both sides of equation (11), we obtain(34)p=p1+4p12+ρ2−2p1+ρsinhn44p12+ρ2ζ−ϵ,where ϵ=ζ0−sinh−12p1/ρ/n44p12+ρ2 is a new constant which is introduced for suitability and ρ=Imp2. Employing equations (7) and (3), we obtain a solution for equation (1) in the form(35)Ix,t=p1+4p12+ρ2−2p1+ρsinhn44p12+ρ2ζ−ϵ−f23f3,Jx,t=αp1+4p12+ρ2−2p1+ρsinhn44p12+ρ2ζ−ϵ−f23f3+β.It can be noted that the case in whichn4 is negative does not work because F4p≥0 for all p∈ℝ.


## 2.6. Case 6

The polynomialF4P has two real roots and two complex conjugate roots if D4<0 and D2D3>0. Hence, it can be written in the form F4p=p−p1p−p2p−p3p−p3∗, where p1<p2 and Rep3=1/2p1+p2. We consider the following:(i)
Ifn4>0, then the permitted intervals for real propagation are p>p2 and p<p1. Selecting p<p2 and pζ0=p2 and integrating both sides of equation (11), we obtain(36)p=p2B1−p1A1+p2B1+p1A1cnn4A1B1ζ−ζ0,k2A1+B1cnn4A1B1ζ−ζ0,k2−A1−B1,wherek2=A1+B12−p2−p12/4A1B1 and A12=B12=1/4p1−p22+Im2p3. Taking into account equations (7) and (3), we obtain a novel wave solution for equation (1) in the form(37)Ix,t=p2B1−p1A1+p2B1+p1A1cnn4A1B1kx+ωt−ζ0,k2A1+B1cnn4A1B1ζ−ζ0,k2−A1−B1−f23f3,Jx,t=αp2B1−p1A1+p2B1+p1A1cnn4A1B1kx+ωt,k2A1+B1cnn4A1B1ζ−ζ0,k2−A1−B1−f23f3+β.(ii)
Ifn4<0, then the allowed intervals of possible propagation are p1<p<p2. Assuming pζ0=p1 and integrating both sides of equation (11), we obtain(38)p=p2B2+p1A2+p1A2−p2B2cn−n4A2B2ζ−ζ0,k3B2+A2−B2−A2cn−n4A2B2ζ−ζ0,k3,wherek3=p2−p12−A2−B22/4A2B2, A22=1/4p1+3p22+Im2p3, and B22=1/43p1+p22+Im2p3. Utilizing equations (7) and (3), we construct a novel wave solution for equation (1) in the form(39)Ix,t=p2B2+p1A2+p1A2−p2B2cn−n4A2B2kx+ωt−ζ0,k3B2+A2−B2−A2cn−n4A2B2kx+ωt−ζ0,k3−f23f3,Jx,t=αp2B2+p1A2+p1A2−p2B2cn−n4A2B2kx+ωt−ζ0,k3B2+A2−B2−A2cn−n4A2B2kx+ωt−ζ0,k3−f23f3+β.


## 2.7. Case 7

The polynomialF4p has two conjugate complex roots, namely, p1,p1∗,p2,p2∗, if D2D3≤0 and D4>0. Therefore, it is expressed as F4p=p−p1p−p1∗p−p2p−p2∗, where Re.p1=−Re.p2. The permitted interval for real propagation for the case n4>0 is p∈ℝ. Thus, we follow similar steps as above and obtain a new traveling wave solution for equation (1) as(40)Ix,t=Rep1+Imp1δ+Imp1+Rep1δtnn4/2A2+B2kx+ωt−ζ0,k4δ+tnn4/2A2+B2kx+ωt−ζ0,k4−f23f3,Jx,t=αRep1+Imp1δ+Imp1+Rep1δtnn4/2A2+B2kx+ωt−ζ0,k4δ+tnn4/2A2+B2kx+ωt−ζ0,k4−f23f3+β,where A22=Imp1−Imp22,B22=Imp1−Imp22+4Re2p1, δ2=4ℜp12−A2−B22/A2+B22−4ℜp12, and k4=2A2B2/A2+B2. It can be noted that the case in which n4<0 does not work because F4p>0 for all p∈ℝ.


## 2.8. Case 8

The polynomialF4p has four real roots in which one of them is double and the others are simple if D2>0,D3>0, and D4=0. Hence, it takes the form F4p=p−p12p−p2p−p3, where p1<p2<p3 and p3=−2p1+p2. We consider the two cases in which n4 is either positive or negative:(i)
Ifn4>0, then the possible intervals for real propagation are p<p1,p>p3, and p1<p<p2. With similar computations as in previous cases, we present the solution of equation (1) directly.If we choosep>−2p1+p2 and assume pζ0=−2p1+p2, we have a new wave solution for equation (1) in the form(41)Ix,t=−2p1−p2+4p2−p1sechn4−3p1−p2kx+ωt−ζ0−f23f3,Jx,t=α−2p1−p2+4p2−p1sechn4−3p1−p2kx+ωt−ζ0−f23f3+β.In similar calculations, we can calculate the wave solution forp<p1 and p1<p<p2.(ii)
Ifn4<0, the allowed interval for real propagation is p∈p2,p3, and postulating pζ0=p2, we obtain a new wave solution for equation (1) in the form(42)Ix,t=p1+p2−p13p1+p22p1+p1+p2cosh−n43p1+p2p1−p2kx−ωt−ζ0−f23f3,Jx,t=αp1+p2−p13p1+p22p1+p1+p2cosh−n43p1+p2p1−p2kx−ωt−ζ0−f23f3+β.


## 3. Dynamic Properties

The aim of this section is to investigate some dynamic properties for equation (4) by investigating the bifurcation and phase portrait for the traveling wave system corresponding to equation (8) which takes the form(43)p′=z,z′=2n4p3+γ22p+γ14.System (43) is a Hamiltonian system with one degree of freedom related to Hamilton function:(44)12z2+Vp=h,where h is an arbitrary constant and(45)Vq=−n42p4+γ2p2+γ1p,is the potential function. It is well known that the equilibrium points for the Hamilton system (43) are also critical points for the potential function (45), i.e., they are the roots of(46)dVdp=−2n4p3+γ22p+γ14.Thus, we use the discrimination of (46) to determine the number of equilibrium points. The discrimination of (46) is(47)Δ=−18γ128+γ2327.Now, let us determine the number of equilibrium points for the Hamilton system (43) and study the properties of its phase space. Thus, we need to define energy curve corresponding to(48)Ch=p,z∈ℝ2:z2=2h−Vp.It is well known that any orbit for the Hamilton system (43) is an energy curve on a certain level of the energy.Case 1.
The dynamical system (43) has a unique equilibrium point ifdV/dp=0has a unique real root. This happens in two cases which are studied individually:(i)
IfΔ=0andγ2=0, thendV/dp=0has one triple real root, i.e.,dV/dp=−2n4p3. This shows0,0is a unique equilibrium point for system(43)which is saddle ifn4>0, and it is center ifn4<0. The phase space for this case is outlined by Figures1(a)and1(b). The value of the energy at the equilibrium point0,0ish1=V0,0=0. The following proposition describes Figure 1.(ii)
IfΔ<0, thendV/dp=0has one real zero and two complex conjugate roots, i.e.,(49)dVdp=−2n4p−ap+a22+m2.It is clear that the pointa,0is a unique equilibrium point for the Hamilton system(43)and it is saddle ifn4>0and center ifn4<0. The phase space is clarified by Figures2(a) and 2(b).Figure 1
Phase portrait for the Hamilton system (43) for γ1=0andγ2=0. (a) n4=1 and (b) n4=−1.
(a)(b)Figure 2
Phase portrait for the Hamilton system (43) for Δ<0. (a) n4=1,γ1=−16,andγ2=−4. (b) n4=−1,γ1=−16,andγ2=−4.
(a)(b)Proposition 1.
The Hamiltonian system (43) has a unique equilibrium point 0,0 if γ1=γ2=0. If n4>0, it is a saddle point and all the orbits are unbounded, see Figure 1(a). If n4<0, system (43) has a family of bounded periodic orbits Ch:h>h1 about the center point 0,0 as outlined by Figure 1(b). A similar conclusion can be presented to describe Figure 2.Case 2.
IfΔ=0andγ2<0, thendV/dp=0has two real roots: one is simple and the other is double. Thus, we writedV/dp=−2n4p−a2p+2aand soa,0and−2a,0are two equilibrium points for system (43). It is clear thata,0is a cusp while−2a,0is a center ifn4<0and saddle ifn4>0. The phase portrait for this case is outlined inFigure3. The following proposition gives a short description for the phase portrait for this case.Figure 3
Phase portrait for the Hamilton system (43) for Δ=0andγ2<0. (a)n4=1,γ1=8,andγ2=−6. (b)n4=−1,γ1=8,andγ2=−6.
(a)(b)Proposition 2.
The Hamiltonian system (43) has two equilibrium points E1=a,0 and E2−2a,0. If n4>0, then E1 is cusp point while E2 is saddle, and furthermore, all the phase space orbits are unbounded as outlined in Figure 3(a). While if n4<0, E1 is a cusp and E2 is a center. The Hamilton system (43) has two bounded families of periodic orbits which are illustrated in green and blue and separated by the phase curve Ch:h=Va,0 in red.Case 3.
IfΔ>0andγ2<0, thendV/dphas three real roots, i.e., it can be written asdV/dp=−2n4p−ap−bp+a+b, where we assumedb>a>0. Consequently, the dynamical system(43)has three equilibrium pointsa,0,b,0,and−a−b,0. Ifn4>0, thena,0is center andb,0and−a−b,0are saddle points. While ifn4<0,a,0is saddle point and the other two equilibrium points are center. The phase portrait for this case is described in Figures4(a) and 4(b). We describe the phase portrait for the Hamiltonian system in the following proposition.Figure 4
Phase portrait for the Hamilton system (43) for Δ>0andγ2<0. (a)n4=1,γ1=8,andγ2=−6. (b)n4=−1,γ1=8,andγ2=−6.
(a)(b)Proposition 3.
IfΔ>0andγ2<0, then the Hamilton system (43) has three equilibrium points a,0,b,0,and−a−b,0. If n4>0, there are two families of orbits Ch:h∈Va,Vb in green in which one of them is bounded and surrounded by the homoclinic orbit Ch:h=V2a while the other family is unbounded. Moreover, all the other orbits are unbounded, see Figure 4(a) for more clarification. If n4<0, system (43) has three equilibrium points in which one is a saddle and the others are centers. It has three bounded families of periodic orbits. Two of them in blue and green are periodic orbits around the two center points 2a,0and−a−b,0, and they are separated by the homoclinic orbit in red Ch:h=Va. The third one is a family of superperiodic orbits in brown Ch:h>Va around the two centers points and lies outside the homoclinic orbit in red. For more details about superperiodic orbits, see, for example, [57].The investigation of the type of the phase space orbits is helpful in determining the types of the solutions. For instance, the existence of periodic orbits, homoclinic orbits, and heteroclinic orbits for the traveling wave system (43) indicates the existence of periodic wave solutions, solitary, and kink solutions for equation (1). Furthermore, this analysis can be employed to construct the traveling wave solution by introducing the constraints on the coefficients of function (12). Consequently, we can prove the following theorem.Theorem 1.
LetIx,t=ukx−ωtandJx,t=αIx,t+βbe a solution for the double-chain model of DNA (1), then(i)
It is a periodic solution if(a)
n4<0,h>0,Δ=0γ1=γ2=0,(b)
n4<0,h>Va,0,Δ<0,(c)
n4<0,Δ=0,γ2<0,h∈V−2a,Va∪Va,∞,(d)
n4>0,Δ>0,γ2>0,h∈Va,Vb,(e)
n4<0,Δ>0,γ2>0,h∈Va,∞,(f)
n4<0,Δ>0,γ2>0,h∈Va,∞∪Vb,Va∪Vc,Vb.(ii)
It is a solitary wave solution ifn4<0,Δ>0,γ2<0,andh=Va.(iii)
It is a kink (antikink) solution ifn4>0,Δ>0,γ2<0,andh=Vb.


## 4. Graphic Interpretations

This section aims to illustrate some of the obtained solutions graphically. Moreover, we study the influence of the physical parameters on the obtained solutions by considering two types of solutions: one of them is kink solution and the other is periodic.Figure5 and 6 illustrates the kink solution (24) for different aspects. Figures 5(a) and 5(b) clarify the 3D and contour representation for the kink solution (24) when k=1,ω=2,σ=0.001,ρ=0.1,μ=0.0001,h=0.002,R1=0.1, and l0=0.002. Now, we illustrate graphically the influence of some parameters on the kink solution while the other parameters are fixed. Figure 6(a) illustrates the effect of the change of the distance between the two strands. It is remarkable the amplitude of solution (24) is decreased when the distance between the two strands is increased. Figure 6(b) clarifies the amplitude of the solution is increased when the stiffness of the elastic membrane is increased. Figure 6(c) clarifies the amplitude of the solution is decreased when the area of the cross section of each strand is increased. Figure 6(d) outlines the amplitude of the kink solution (24) is increased when the height of the membrane in the equilibrium is increased (see Figure 7).Figure 5
Graphic representation of solution (24) for R1=0.1,h=0.002,k=1,l0=0.002,μ=0.0001,ω=2,ρ=0.1,andσ=0.001. (a) 3D graphic and (b) 2D contour plot.
(a)(b)Figure 6
Parameters affecting solution (24): (a) changes in h for R1=0.1,k=1,l0=0.001,μ=0.002,ρ=0.1,andσ=0.0001, (b) changes in μ for R1=0.1,h=0.002,k=1,l0=0.001,ω=2,ρ=0.1,andσ=0.0001, (c) changes in σ for R1=0.1,h=0.002,k=1,l0=0.001,μ=0.0001,ω=0.1,andρ=0.1, and (d) changes in l0 for R1=0.1,h=0.002,k=1,σ=0.001,ρ=0.1,μ=0.0001,andω=2.
(a)(b)(c)(d)Figure 7
Graphic representation of solution (28) for R1=0.1,h=0.002,k=1,l0=0.002,μ=0.0001,ω=0.2,ρ=0.1,andσ=0.001. (a) 3D graphic and (b) 2D contour plot.
(a)(b)Now, we are going to clarify solution (28) graphically and study the influence of parameter changes on solution (28). Solution (28) is periodic as outlined in Figure 8(a), and its contour is illustrated in Figure 8(b). If the distance between the two strands is increased and the other parameters are fixed, the amplitude of the solution is unchanged but the width of the solution is increased as outlined in Figure 8(a). The amplitude of solution (28) is not affected by the changes in the stiffness of the elastic membrane, while the width of the solution is decreased when the stiffness of the elastic membrane is increased as clarified by Figure 8(d). If the area of the cross section of each strand is increased, then the amplitude keeps fixed while the width is increased, see Figure 8(d). Figure 9(b) outlines the amplitude of the periodic solution (28) is kept unchanged while its width is decreased when the height of the membrane in the equilibrium is increased. Figure 9(a) illustrates the influence of the superperiodic wave solution (39) due the changes in the stiffness of the elastic membrane. If the stiffness of the elastic membrane increases, the amplitude is kept unchanged while the width decreases. Figure 9(b) clarifies the influence of distance between the two strands on the superwave solution (39). If distance between the two strands is increased, then the amplitude and the width of superwave solution (39) are increased.Figure 8
Parameters affecting solution (28): (a) changes in h for R1=0.1,k=1,l0=0.001,μ=0.002,ω=2,ρ=0.1,andσ=0.0001, (b) changes in μ for R1=0.1,h=0.002,k=1,l0=0.001,ω=2,ρ=0.1,andσ=0.0001, (c) changes in σ for R1=0.1,h=0.002,k=1,l0=0.001,μ=0.0001,ω=2,andρ=0.1, and (d) changes in l0 for R1=0.1,h=0.0002,k=1,μ=0.0001,ω=2,ρ=0.1,andσ=0.001.
(a)(b)(c)(d)Figure 9
Parameters affecting the superwave solution (39): (a) changes in μ for R1=0.01,α=0.8,h=0.006,k=1,l0=0.001,ρ=0.00002,andσ=0.01 and (b) changes in h for R1=0.01,α=0.8,h=0.006,k=1,l0=0.001,ρ=0.00002,andσ=0.01.
(a)(b)It is worth mentioning that we can make the same study for the remaining obtained solutions. However, we only give the 3D graphic and the 2D counter for some solutions. The 3D graphic representation and the singularity plane are outlined in Figure10(a), while Figure 10(b) illustrates the 2D contour of solution (15). Solution (37) is illustrated in Figure 11.Figure 10
Graphic representation for the singular wave solution (15). (a) 3D graphic and (b) 2D contour.
(a)(b)Figure 11
Graphic representation for the singular wave solution (37). (a) 3D graphic and (b) 2D contour.
(a)(b)


## 5. Conclusion

This work is aimed to study analytically the double-chain model for deoxyribonucleic acid (DNA). A certain wave transformation has been applied to equation (1) to transform it into an ordinary differential equation. The integration of this equation reacquired some studies on the parameters. This study has been performed by applying the complete discrimination of the polynomial F4p. Moreover, we have determined the possible interval of real propagations. Such study is more significant because the missing of such study implies to loss some solutions and also, give rise to complex solutions which are undesirable in real problems. For instance, there are several solutions corresponding to the same conditions on the discriminant system as outlined in Case 4. We have introduced new waves’ solutions for equation (1). Let us compare the results obtained in the present article with the well-known results obtained by other authors using different methods as follows: our results in a new double-chain model of DNA are new and different from those obtained in references [19–24]. We have studied the influence of some parameters such as the distance between the two strands, the stiffness of the elastic membrane, the area of the cross section of each strand, and the height of the membrane in the equilibrium. We have considered two types of solutions: one is kink (24) and the other is periodic (28). We have shown graphically the amplitude of the kink solution is decreased when the distance between the two strands or the area of the cross section of each strand is increased, while it is increased when the stiffness of the elastic membrane or the height of the membrane in the equilibrium is increased. For more clarification, see Figure 6. The amplitude of the periodic solution remains approximately unchanged when these physical parameters are changed, but the width has been affected. The width is increased due to the increase of the distance between the two strands or the area of the cross section of each strand, while it is decreased as a result of increasing the stiffness of the elastic membrane or the height of the membrane in the equilibrium. For more illustrations, see Figure 10. From another point of view, this ODE has been expressed as a one-dimensional Hamiltonian system that describes the physical motion of a particle with one degree of freedom under the action of potential function Vp given by (45). Based on the Hamiltonian concepts, we have studied some qualitative analyses such as phase portrait and bifurcation. The description of phase space has been presented through Propositions 1, 2, and 3. Moreover, these propositions contain the conditions for the existence of periodic and solitary wave solutions.



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