# The Modified Rational Jacobi Elliptic Functions Method for Nonlinear Differential Difference Equations

Authors: Khaled A. Gepreel; Taher A. Nofal; Ali A. Al-Thobaiti
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We modified the rational Jacobi elliptic functions method to construct some new exact solutions for nonlinear differential difference equations in mathematical physics via the lattice equation, the discrete nonlinear Schrodinger equation with a saturable nonlinearity, the discrete nonlinear Klein-Gordon equation, and the quintic discrete nonlinear Schrodinger equation. Some new types of the Jacobi elliptic solutions are obtained for some nonlinear differential difference equations in mathematical physics. The proposed method is more effective and powerful to obtain the exact solutions for nonlinear differential difference equations.


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## Body

## Introduction

In recent years, the study of difference equations has acquired a new significance, due in large part to their use in the formulation and analysis of discrete-time systems, the numerical integration of differential equation by finite-difference schemes and the study of deterministic chaos.Since the work of Fermi et al. in the 1960s [1], DDEs have been the focus of many nonlinear studies. On the other hand, a considerable number of well-known analytic methods are successfully extended to nonlinear DDEs by researchers [2–17]. However, no method obeys the strength and the flexibility for finding all solutions to all types of nonlinear DDEs.  Zhang et al. [18] and Aslan [19] used the (G'/G)-expansion function method to some physically important nonlinear DDEs. Qiong and Bin [12] constructed the Jacobi elliptic solutions for nonlinear DDEs.  Recently, Zhang [20] and Gepreel [21] and Gepreel and Shehata [22] have used the Jacobi elliptic function method for constructing new and more general Jacobi elliptic function solutions of the nonlinear differential difference equations. The main objective of this paper is to modify the rational Jacobi elliptic functions method to obtain the exact wave solutions for nonlinear DDEs. We use this method to calculate the exact wave solutions for some nonlinear DDEs in mathematical physics via the lattice equation, the discrete nonlinear Schrodinger equation with a saturable nonlinearity, the discrete nonlinear Klein-Gordon equation, and the quintic discrete nonlinear Schrodinger equation.


## 2. Description of the Modified Rational Jacobi Elliptic Functions Method

In this section, we would like to outline an algorithm for using the modified rational Jacobi elliptic functions method to solve nonlinear DDEs. For a given nonlinear DDEs,(2.1)Δ(un+p1(x),…,un+pk(x),un+p1′(x),…,un+pk′(x),…,un+p1(r)(x),…,un+pk(r)(x),vn+p1(x),…,vn+pk(x),vn+p1′(x),…,vn+pk′(x),…,vn+p1(r)(x),…,vn+pk(r)(x),….)=0,
where Δ=(Δ1,...,Δg),x=(x1,x2,...,xm),n=(n1,...,nQ), and g,m,Q,p1,...,pk are integers,  ui(r), vi(r) denotes the set of all rth order derivatives of ui, vi with respect to x.The main steps of the algorithm for the modified rational Jacobi elliptic functions method to nonlinear DDEs are outlined as follows.Step 1.
We seek the traveling wave solutions of the following form:(2.2)un(x)=U(ξn),vn(x)=V(ξn),…,
where
(2.3)ξn=∑i=1Qdini+∑j=1mcjxj+ξ0,
where di(i=1,…,Q), cj(j=1,…,m) and the phase ξ0 are constants to be determined later. The transformations (2.2) are reduced (2.1) to the following ordinary differential difference equations:
(2.4)Ω(U(ξn+p1),…,U(ξn+pk),U′(ξn+p1),…,U′(ξn+pk),…,U(r)(ξn+p1),…,U(r)(ξn+pk),V(ξn+p1),…,V(ξn+pk),V′(ξn+p1),…,V′(ξn+pk),…,Vn+p1(r)(ξn+p1),…,Vn+pk(r)(ξn+pk),…)=0,
where Ω=(Ω1,…,Ωg).Step 2.
We suppose the modified rational Jacobi elliptic functions solutions of (2.4) in the following  form:
(2.5)U(ξn)=∑i=0Kαi(F′(ξn)F(ξn))i+∑i=1Kβi(F(ξn)F′(ξn))i,V(ξn)=∑j=0Lγj(F′(ξn)F(ξn))j+∑j=1Lλj(F(ξn)F′(ξn))j,…,
where αi, βi(i=1,2,…,K), γj, λj(j=1,2,…,L) are constants to be determined later while F(ξn) satisfies  a discrete  Jacobi elliptic differential equation:
(2.6)F′2(ξn)=e0+e1F2(ξn)+e2F4(ξn),
and  e0, e1, e2  are arbitrary constants.Step 3.
Since the general solution of the proposed equation (2.6) is difficult to obtain and so the iteration relations corresponding to the general exact solutions we discuss the solutions  of  the  proposed discrete  Jacobi elliptic differential equation (2.6) at some special cases to e0, e1, and  e2, to cover all the Jacobi elliptic functions as follows.Type 1.
Ife0=1, e1=-(1+m2), e2=m2.   In this case, (2.6) has the solution F(ξn)=sn(ξn,m),  where sn(ξn,m) is the  Jacobi elliptic sine function, and m is the modulus.  In this case from using the properties of Jacobi elliptic functions (see [22]),  the  series expansion solutions (2.5) take the following  form:
(2.7)U(ξn)=∑i=0Kαi(cn(ξn,m)dn(ξn,m)sn(ξn,m))i+∑i=1Kβi(sn(ξn,m)cn(ξn,m)dn(ξn,m)),V(ξn)=∑i=0Lγi(cn(ξn,m)dn(ξn,m)sn(ξn,m))i+∑i=1Lλi(sn(ξn,m)cn(ξn,m)dn(ξn,m)),….
Further using the properties of Jacobi elliptic functions, the iterative relations can be written in the following form:
(2.8)U(ξn±p)=∑i=0Kαi(F′(ξn±p)F(ξn±p))i+∑i=1Kβi(F(ξn±p)F′(ξn±p))i,V(ξn±p)=∑i=0Lγi(F′(ξn±p)F(ξn±p))i+∑i=1Lλi(F(ξn±p)F′(ξn±p))i,…,
where
(2.9)F'(ξn±p)F(ξn±p)=1M1{±cn(d,m)cn(ξn,m)dn(ξn,m)dn(d,m)±m2sn(d,m)sn(ξn,m)∓2m2sn(d,m)sn3(ξn,m)∓2m2sn3(d,m)sn(ξn,m)±m2sn3(d,m)sn3(ξn,m)+sn(d,m)sn(ξn,m)±m4sn3(d,m)sn3(ξn,m)∓m2sn2(d,m)sn2(ξn,m)dn(ξn,m)dn(d,m)cn(d,m)cn(ξn,m)},M1=-cn(ϕ,m)dn(ϕ,m)sn(ξn,m)∓sn(ϕ,m)dn(ξn,m)(ϕ)cn(ξn,m)+m2sn3(ξn,m)sn2(ϕ,m)cn(ϕ,m)dn(ϕ,m)±m2sn2(ξn,m)sn3(ϕ,m)cn(ξn,m)dn(ξn,m),d=ps1d1+ps2d2+⋯+psQdQ,  psj is the jth   component  of shift vector ps.Type 2.
Ife0=1-m2, e1=2m2-1, and e2=-m2.   In this case, (2.6) has the solution F(ξn)=cn(ξn,m).  From using the properties of  Jacobi elliptic functions,  the  series expansion solutions (2.5) take the following  form:
(2.10)U(ξn)=∑i=0Kαi(-sn(ξn,m)dn(ξn,m)cn(ξn,m))i+∑i=1Kβi(-cn(ξn,m)sn(ξn,m)dn(ξn,m))i,V(ξn)=∑i=0Lγi(-sn(ξn,m)dn(ξn,m)cn(ξn,m))i+∑i=1Lλi(-cn(ξn,m)sn(ξn,m)dn(ξn,m))i,….Type 3.
Ife0=m2-1, e1=2-m2, and e2=-1.   In this case, (2.6) has the solution F(ξn)=dn(ξn,m). From using the properties of Jacobi elliptic functions,  the  series expansion solutions (2.5) take the  following  form:
(2.11)U(ξn)=∑i=0Kαi(-m2sn(ξn,m)cn(ξn,m)dn(ξn,m))i+∑i=1Kβi(-dn(ξn,m)m2sn(ξn,m)cn(ξn,m))i,V(ξn)=∑i=0Lγi(-m2sn(ξn,m)cn(ξn,m)dn(ξn,m))i+∑i=1Lλi(-dn(ξn,m)m2sn(ξn,m)cn(ξn,m))i,….
From the properties of the  Jacobi elliptic functions, we can deduce  the iterative relation to the above kind of solutions from Types 2 and 3  as we show in Type 1.Step 4.
Determining the degreeK,L,… of  (2.5)  by balancing the  nonlinear term(s) and the highest order derivatives of U(ξn),V(ξn),… in (2.4).  It should be noted that the leading terms U(ξn±p),V(ξn±p),…,p≠0 will not affect the balance because we are interested in balancing the terms of F′(ξn)/F(ξn).Step 5.
SubstitutingU(ξn),V(ξn),… in each type form (2.1)–(2.4) and the given values of K,L,…  into (2.4).  Cleaning  the denominator and collecting all terms with the same degree of sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the  left-hand side of (2.4) is converted into a  polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m). Setting each coefficient of this polynomial to zero, we derive a set of algebraic equations for αi, βi, di, γi, λi, and ci.Step 6.
Solving the over determined system of nonlinear algebraic equations by using  Maple or Mathematica. We end up with explicit expressions forαi, βi, di, γi, λi, and cj.Step 7.
Substitutingαi, βi, di, γi, λi, and ci  into U(ξn), V(ξn),… in the corresponding type from (2.1)–(2.4), we can finally obtain exact solutions for (2.1).


## 3. Applications

In this section, we use the proposed method to construct the rational Jacobi elliptic wave solutions for the nonlinear DDEs via the  lattice equation, the discrete  nonlinear Schrodinger equation with a saturable  nonlinearity, the discrete nonlinear Klein-Gordon equation, and the quintic discrete nonlinear Schrodinger  equation, which are very important in mathematical physics and have been paid attention to by many  researchers.


### 3.1. Example 1.  The  Lattice Equation

In this section, we study the  lattice equation which takes the following form [23–26]:
(3.1)dun(t)dt=(α+βun+γun2)(un-1-un+1),
where α, β, and γ are nonzero constants. This equation contains hybrid lattice equation, mKdV lattice equation, modified Volterra lattice equation, and Langmuir chain equation for some special values of α, β, γ.  According to the above steps, to seek traveling wave solutions of (3.1), we construct  the transformation
(3.2)un(t)=U(ξn),ξn=dn+c1t+ξ0,
where d, c1, and ξ0 are constants. The transformation (3.2) permits us converting equation (3.1) into the following form:
(3.3)c1U′(ξn)=[α+βU(ξn)+γU2(ξn)][U(ξn-d)-U(ξn+d)],
where ′=d/dξn. Considering the homogeneous balance between the highest order derivative and the nonlinear term in (3.3), we get  K=1.  Thus, the solution of (3.3) has the following form:
(3.4)U(ξn)=α0+α1(F′(ξn)F(ξn))+β1(F(ξn)F′(ξn)),
where α0, α1, and β1 are constants to be determined later and F(ξn) satisfies  a discrete Jacobi elliptic ordinary differential equation (2.6). When we discuss the solutions of (2.6), we get the following types.Type 1.
Ife0=1, e1=-(1+m2), and e2=m2.  In this case, the series expansion solution of (3.3) has the form:
(3.5)U(ξn)=α0+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+β1sn(ξn,m)cn(ξn,m)dn(ξn,m).
With the help of Maple, we substitute (3.5) and (2.8)  into (3.3). Cleaning  the denominator and collecting all terms with the same degree of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the  left-hand side of (3.3) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m).  Setting each coefficient of this polynomial to be zero, we derive a set of algebraic equations for α0, α1, d, β1, γ, β, α, and c1.  Solving the set of algebraic equations by using Maple or Mathematica software package, we have the following.Family 1.
(3.6)
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where   α1, d, γ, β, and m are arbitrary constants.
From  (3.5) and (3.6), the solution  of (3.3) takes the following form:
(3.7)U(ξn)=-β2γ+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+α1(m2-1)sn(ξn,m)cn(ξn,m)dn(ξn,m),
where ξn=dn+(2γα12[cn2(d,m)-sn2(d,m)dn2(d,m)]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.  Figure 1 illustrates the behavior of the exact solution (3.7).(a) represents the Jacobi elliptic solution (3.7) when β=m=0.5, γ=1, α1=1.5, d=0.3, n=2. (b) represents the Jacobi elliptic solution (3.7) when β=m=0.5, γ=1, α1=1.5, d=0.3, t=2.
(a)
(b)Family 2.
(3.8)
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where   α1, d, γ, β, and m are arbitrary constants.
From  (3.5) and (3.8), the solution  of (3.3) has the following form:
(3.9)U(ξn)=-β2γ+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)-α1(m2-1)sn(ξn,m)cn(ξn,m)dn(ξn,m),
where ξn=dn+(2γα12[dn2(d,m)-m2sn2(d,m)cn2(d,m)]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.Type 2.
Ife0=1-m2, e1=2m2-1, and e2=-m2.   In this case, the solution of (3.3) has the form:
(3.10)U(ξn)=α0-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-β1cn(ξn,m)sn(ξn,m)dn(ξn,m).
With the help of Maple, we substitute (3.10) into (3.3). Cleaning  the denominator and collecting all terms with the same degree of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the  left-hand side of (3.3) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m).   Setting each coefficient of this polynomial to zero, we derive a set of algebraic equations in α0, α1, d, β1, γ, β, α, and c1. Solving the set of algebraic equations by using Maple or Mathematica software package, we get the following.Family 1.
(3.11)
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where   α1, d, γ, β, and m are arbitrary constants.
In this case, the solution of (3.3) takes  the following form:
(3.12)U(ξn)=-β2γ-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)+α1cn(ξn,m)sn(ξn,m)dn(ξn,m),
where   ξn=dn+(2γα12[cn2(d,m)-sn2(d,m)dn2(d,m)]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.
The Jacobi elliptic functions could be generated into the hyperbolic functions whenm tends to one in the other hand, they are generated into trigonometrical functions when m tends to zero.
Whenm=0, the trigonometrical solution (3.12) takes  the following form:
(3.13)U(ξn)=-β2γ+2α1cot(2ξn),whereξn=dn+4γα12cot(2d)t+ξ0.
Also if m=1, the hyperbolic solution (3.12) takes  the following form:
(3.14)U(ξn)=-β2γ+2α1sinh(2ξn),whereξn=dn+4γα12sinh(2d)t+ξ0.Family 2.
(3.15)
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where   α1, d,γ, β, and m  are arbitrary constants.
In this case, the solution of (3.3) takes  the following form:
(3.16)U(ξn)=-β2γ-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-α1cn(ξn,m)sn(ξn,m)dn(ξn,m),
where    ξn=dn-(2γα12[m2sn4(d,m)-1]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.
Whenm=0, the trigonometrical solution (3.16) takes  the following form:
(3.17)U(ξn)=-β2γ-2α1sin(2ξn),whereξn=dn+4γα12sin(2d)t+ξ0.
When m=1, the hyperbolic solution (3.16) takes  the following form:
(3.18)U(ξn)=-β2γ-2α1coth(2ξn),whereξn=dn+4γα12coth(2d)t+ξ0.Type 3.
Ife0=m2-1, e1=2-m2, and e2=-1. In this case, the series expansion solution of (3.3) has the form:
(3.19)U(ξn)=α0-m2α1sn(ξn,m)cn(ξn,m)dn(ξn,m)-β1dn(ξn,m)m2sn(ξn,m)cn(ξn,m).
Consequently, using the Maple or Mathematica  we get the following results.Family 1.
(3.20)
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where   α1, d,γ, β, and m are arbitrary constants.
In this case, the solution of (3.3) takes the following form:
(3.21)U(ξn)=-β2γ-α1m2sn(ξn,m)cn(ξn,m)dn(ξn,m)+α1dn(ξn,m)sn(ξn,m)cn(ξn,m),
where    ξn=dn+(2γα12[dn2(d,m)-m2sn2(d,m)cn2(d,m)]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.Family 2.
(3.22)
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)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   α1, d,γ, β, and m are arbitrary constants.
In this case, the solution of (3.3) takes the following form:
(3.23)U(ξn)=-β2γ-α1m2sn(ξn,m)cn(ξn,m)dn(ξn,m)-α1dn(ξn,m)sn(ξn,m)cn(ξn,m),
where       ξn=dn-(2γα12[m2sn4(d,m)-1]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.


### 3.2. Example 2.  The Discrete Nonlinear Schrodinger Equation

The discrete nonlinear Schrodinger equation (DNSE) is one of the most fundamental nonlinear lattice models [8].  Its arise  in nonlinear optics as a model of infinite wave guide arrays [27] and has been recently implemented to describe Bose-Einstein condensates in optical lattices.  The class of DNSE model with saturable  nonlinearity is also of particular interest in their own right, due to a feature first unveiled in [28].  In this section, we study the DNSE with a saturable  nonlinearity [29, 30] form:
(3.24)i∂ψn∂t+(ψn+1+ψn-1-2ψn)+ν|ψn|21+μ|ψn|2ψn=0,
which describes optical pulse propagations in various doped fibers, ψn is a complex valued wave function at sites n while ν and μ are real parameters. We make the transformation:
(3.25)ψn=ϕ(ξn)e-i(σt+ρ),ξn=αn+β,
where σ, ρ, α, and β are arbitrary constants. The transformation (3.25) permits us converting equation (3.24) into the following nonlinear difference equation:
(3.26)(σ-2)φ(ξn)+φ(ξn+α)+φ(ξn-α)+νφ3(ξn)1+μφ2(ξn)=0.
We assume that (3.26) has  a solution of the form:
(3.27)ϕ(ξn)=α1(F′(ξn)F(ξn))+β1(F(ξn)F′(ξn))+α0,
where α0, α1, and β1 are constants to be determined later and F(ξn) satisfies  a discrete Jacobi elliptic differential equation (2.6). When we discuss the solutions of  (3.26), we have the following types.Type 1.
Ife0=1, e1=-(1+m2), and e2=m2.  In this case, the series expansion solution of (3.26) has the form:
(3.28)ϕ(ξn)=α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+β1sn(ξn,m)cn(ξn,m)dn(ξn,m)+α0.
With the help of Maple, we substitute (3.28) and (2.8) into (3.26), cleaning  the denominator and collecting all terms with the same order of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the  left-hand side of (3.26) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m). Setting each coefficient of this polynomial to zero, we derive a set of algebraic equations for α0, α1, σ, β1, ρ, α, and β. Solving the set of algebraic equations by using Maple or Mathematica software package, we obtain the following.Family 1.
(3.29)
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
s
n
2
(
α
,
m
)
+
1
]
,
β
1
=
(
m
2
-
1
)
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
s
n
2
(
α
,
m
)
+
1
]
,
μ
<
0
,
ν
=
-
2
μ
[
m
4
s
n
8
(
α
,
m
)
-
2
m
4
s
n
6
(
α
,
m
)
+
2
m
2
s
n
2
(
α
,
m
)
-
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
2
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
α
,
m
)
-
4
s
n
2
(
α
,
m
)
+
1
,
α
0
=
0
,
σ
=
4
s
n
2
(
α
,
m
)
[
m
4
s
n
6
(
α
,
m
)
-
(
m
4
+
2
m
2
)
s
n
4
(
α
,
m
)
+
(
m
2
+
2
)
s
n
2
(
α
,
m
)
+
m
2
-
2
]
m
4
s
n
8
(
α
,
m
)
-
4
m
2
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
α
,
m
)
-
4
s
n
2
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form: (3.30)ψn=(sn(α,m)cn(α,m)dn(α,m)cn(ξn,m)dn(ξn,m)-μ[m2sn4(α,m)-2sn2(α,m)+1]sn(ξn,m)+(m2-1)sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)-μ[m2sn4(α,m)-2sn2(α,m)+1]cn(ξn,m)dn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔄m4sn8(α,m)-4m2sn6(α,m)+(2m2+4)sn4(α,m)-4sn2(α,m)+1+ρ]}),where 𝔄 denotes [m4sn6(α,m)-(m4+2m2)sn4(α,m)+(m2+2)sn2(α,m)+m2-2] and ξn=αn+β. Figure 2 illustrates the behavior exact solution (3.30).(a) and (b) represent the Jacobi elliptic solution (3.30) when m=0.5, β=1.5, α=0.2, ρ=0.4, n=2. (c) and (d) represent the Jacobi elliptic solution (3.30) when m=0.5, β=1.5, α=0.2, ρ=0.4, t=1.5.
(a)
(b)
(c)
(d)Family 2.
(3.31)
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
m
2
s
n
2
(
α
,
m
)
+
1
]
,
β
1
=
-
(
m
2
-
1
)
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
m
2
s
n
2
(
α
,
m
)
+
1
]
,
μ
<
0
,
ν
=
-
2
μ
[
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
6
(
α
,
m
)
+
2
s
n
2
(
α
,
m
)
-
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
4
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
m
4
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
1
,
α
0
=
0
,
σ
=
4
s
n
2
(
α
,
m
)
[
m
4
s
n
6
(
α
,
m
)
-
(
2
m
4
+
m
2
)
s
n
4
(
α
,
m
)
+
(
2
m
4
+
m
2
)
s
n
2
(
α
,
m
)
-
2
m
2
+
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
4
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
m
4
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form: (3.32)ψn=(sn(α,m)cn(α,m)dn(α,m)cn(ξn,m)dn(ξn,m)-μ[m2sn4(α,m)-2m2sn2(α,m)+1]sn(ξn,m)-(m2-1)sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)-μ[m2sn4(α,m)-2m2sn2(α,m)+1]cn(ξn,m)dn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔚m4sn8(α,m)-4m4sn6(α,m)+(2m2+4m4)sn4(α,m)-4m2sn2(α,m)+1+ρ]}),where 𝔚 denotes [m4sn6(α,m)-(2m4+m2)sn4(α,m)+(2m4+m2)sn2(α,m)-2m2+1] and ξn=αn+β.Type 2.
Ife0=1-m2, e1=2m2-1, and e2=-m2.  In this case, the solution of (3.26) has the form:
(3.33)ϕ(ξn)=α0-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-β1cn(ξn,m)sn(ξn,m)dn(ξn,m).
Consequently, using Maple or Mathematica we get the following results.Family 1.
(3.34)
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
s
n
2
(
α
,
m
)
+
1
]
,
β
1
=
-
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
s
n
2
(
α
,
m
)
+
1
]
,
μ
<
0
,
ν
=
-
2
μ
[
m
4
s
n
8
(
α
,
m
)
-
2
m
4
s
n
6
(
α
,
m
)
+
2
m
2
s
n
2
(
α
,
m
)
-
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
2
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
α
,
m
)
-
4
s
n
2
(
α
,
m
)
+
1
,
α
0
=
0
,
σ
=
4
s
n
2
(
α
,
m
)
[
m
4
s
n
6
(
α
,
m
)
-
(
m
4
+
2
m
2
)
s
n
4
(
α
,
m
)
+
(
m
2
+
2
)
s
n
2
(
α
,
m
)
+
m
2
-
2
]
m
4
s
n
8
(
α
,
m
)
-
4
m
2
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
α
,
m
)
-
4
s
n
2
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form:
(3.35)ψn=(-sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)dn(ξn,m)-μ[m2sn4(α,m)-2sn2(α,m)+1]cn(ξn,m)+sn(α,m)cn(α,m)dn(α,m)cn(ξn,m)-μ[m2sn4(α,m)-2sn2(α,m)+1]sn(ξn,m)dn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔄m4sn8(α,m)-4m2sn6(α,m)+(2m2+4)sn4(α,m)-4sn2(α,m)+1+ρ]}),
where ξn=αn+β.
In the special case, whenm=0, the trigonometrical solution (3.35) takes  the following form:
(3.36)ψn=tan(2α)cot(2ξn)-μ(Exp{-i[-2ttan2(2α)+ρ]}),whereξn=αn+β.
Also ifm=1, the hyperbolic solution (3.35) takes  the following form:
(3.37)ψn=(sinh(2α)-μsinh(2ξn))(Exp{i[4tsinh2(α)-ρ]}),whereξn=αn+β.Family 2.
(3.38)
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
1
]
,
β
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
1
]
,
μ
<
0
,
ν
=
2
μ
[
m
4
s
n
8
(
α
,
m
)
-
2
(
m
4
+
m
2
)
s
n
6
(
α
,
m
)
+
6
m
2
s
n
4
(
α
,
m
)
-
(
2
+
2
m
2
)
s
n
2
(
α
,
m
)
+
1
]
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
4
(
α
,
m
)
+
1
,
σ
=
4
s
n
2
(
α
,
m
)
[
(
m
4
+
m
2
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
m
2
+
1
]
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
4
(
α
,
m
)
+
1
,
α
0
=
0
.
In this case, the solution of (3.24) takes the following form:
(3.39)ψn=(-sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)dn(ξn,m)-μ[m2sn4(α,m)-1]cn(ξn,m)-sn(α,m)cn(α,m)dn(α,m)cn(ξn,m)-μ[m2sn4(α,m)-1]sn(ξn,m)dn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔛m4sn8(α,m)-2m2sn4(α,m)+1+ρ]}),
where 𝔛 denotes [(m4+m2)sn4(α,m)-4m2sn2(α,m)+m2+1] and ξn=αn+β.
Ifm=0, the trigonometrical solution (3.39) takes  the following form:
(3.40)ψn=sin(2α)-μsin(2ξn)(Exp{-i[4tsin2(α)+ρ]}),whereξn=αn+β.
Also ifm=1, the hyperbolic solution (3.39) takes  the following form:
(3.41)ψn=tanh(2α)coth(2ξn)-μ(Exp{-i[2ttanh2(2α)+ρ]}),whereξn=αn+β.Type 3.
ife0=m2-1, e1=2-m2, and e2=-1. In this case, the series expansion solution of (3.26)  has the form:
(3.42)ϕ(ξn)=α0-m2α1sn(ξn,m)cn(ξn,m)dn(ξn,m)-β1dn(ξn,m)m2sn(ξn,m)cn(ξn,m).
Consequently, using Maple or Mathematica we get the following results.Family 1.
(3.43)
α
0
=
0
,
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
m
2
s
n
2
(
α
,
m
)
+
1
]
,
β
1
=
-
m
2
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
m
2
s
n
2
(
α
,
m
)
+
1
]
,
μ
<
0
,
ν
=
-
2
μ
(
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
6
(
α
,
m
)
+
2
s
n
2
(
α
,
m
)
-
1
)
m
4
s
n
8
(
α
,
m
)
-
4
m
4
s
n
6
(
α
,
m
)
+
(
4
m
4
+
2
m
2
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
1
,
σ
=
4
s
n
2
(
α
,
m
)
[
m
4
s
n
6
(
α
,
m
)
-
(
2
m
4
+
m
2
)
s
n
4
(
α
,
m
)
+
(
2
m
4
+
m
2
)
s
n
2
(
α
,
m
)
-
2
m
2
+
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
4
s
n
6
(
α
,
m
)
+
(
4
m
4
+
2
m
2
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form:
(3.44)ψn=(-m2sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)cn(ξn,m)-μ[m2sn4(α,m)-2m2sn2(α,m)+1]dn(ξn,m)+sn(α,m)cn(α,m)dn(α,m)dn(ξn,m)-μ[m2sn4(α,m)-2m2sn2(α,m)+1]sn(ξn,m)cn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔚m4sn8(α,m)-4m4sn6(α,m)+(4m4+2m2)sn4(α,m)-4m2sn2(α,m)+1+ρ]})
where ξn=αn+β.Family 2.
(3.45)
α
0
=
0
,
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
1
]
,
β
1
=
m
2
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
1
]
,
μ
<
0
,
ν
=
2
μ
[
(
m
4
s
n
8
(
α
,
m
)
-
2
(
m
4
+
m
2
)
s
n
6
(
α
,
m
)
+
6
m
2
s
n
4
(
α
,
m
)
-
(
2
+
2
m
2
)
s
n
2
(
α
,
m
)
+
1
)
]
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
4
(
α
,
m
)
+
1
,
σ
=
4
s
n
2
(
α
,
m
)
[
(
m
4
+
m
2
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
m
2
+
1
]
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
4
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form:
(3.46)ψn=(-m2sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)cn(ξn,m)-μ[m2sn4(α,m)-1]dn(ξn,m)-sn(α,m)cn(α,m)dn(α,m)dn(ξn,m)-μ[m2sn4(α,m)-1]sn(ξn,m)cn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔛m4sn8(α,m)-2m2sn4(α,m)+1+ρ]}),
where ξn=αn+β.


### 3.3. Example 3.  The Discrete Nonlinear Klein-Gordon Equation

In this section, we consider the following discrete nonlinear Klein-Gordon equation [31]:
(3.47)d2un(t)dt2=g(un)(un+1+un-1-2sun).
The nonconstant (in contrast to the standard models of harmonic coupling and linear dispersion [32]) function g(un) ensures the presence of nonlinear dispersion, which is critical for the existence of compactly supported solutions and  s  can take values in the interval [-1,1]. Kevrekidis and Konotop [31] have obtained some exact compaction solutions and claim that this DDE does not have the traveling compact solution.If we setg(un)=A+Cun2 where A and C are arbitrary constants and take the traveling transformation:
(3.48)un(t)=U(ξn),ξn=dn+c1t+ξ0,
where d, c1, and ξ0 are constants. The transformation (3.48) permits us converting equation (3.47) into the following form:
(3.49)c12U′′(ξn)=[A+CU2(ξn)][U(ξn+d)+U(ξn-d)-2sU(ξn)],
where  ′=d/dξn. Considering the homogeneous balance between the highest-order derivative and the nonlinear term in (3.49), we get K=1. Thus, the solutions of (3.49) have the following form:
(3.50)U(ξn)=α0+α1(F′(ξn)F(ξn))+β1(F(ξn)F′(ξn)),
where  α0, α1, and β1 are constants to be determined later and F(ξn) satisfies a discrete Jacobi elliptic ordinary differential equation (2.6), we have the following types.Type 1.
Ife0=1, e1=-(1+m2), and e2=m2.  In this case, the series expansion solution of (3.49) has the form:
(3.51)U(ξn)=α0+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+β1sn(ξn,m)cn(ξn,m)dn(ξn,m).
With the help of Maple, we substitute (3.51) and (2.8)  into (3.49). Cleaning  the denominator and collecting all terms with the same degree  of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the left-hand side of (3.49) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m). Setting each coefficient of this polynomial to zero, we derive a set of algebraic equations for α0, α1, d, β1, A, C, s, and c1.  Solving the set of algebraic equations by using Maple or Mathematica software package, we have the following.Family 1.
(3.52)
β
1
=
(
m
2
-
1
)
c
1
L
1
,
α
0
=
0
,
s
=
-
1
C
L
1
2
,
α
1
=
c
1
L
1
,
A
=
-
C
L
1
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
4
m
2
s
n
6
(
d
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
d
,
m
)
-
4
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants and
(3.53)L1=m4sn4(d,m)cn4(d,m)+dn4(d,m)C[m4sn8(d,m)-2m4sn6(d,m)+2m2sn2(d,m)-1].
In this case, the solution of (3.47) takes the following form:
(3.54)U(ξn)=c1L1cn(ξn,m)dn(ξn,m)sn(ξn,m)+(m2-1)c1L1sn(ξn,m)cn(ξn,m)dn(ξn,m),
where   ξn=dn+c1t+ξ0.
Figure3 illustrates the behavior of the exact solution (3.54).(a) represents the Jacobi elliptic solution (3.54) when m=0.5, c1=0.4, C=-0.6, d=0.3, n=2. (b) represents the Jacobi elliptic solution (3.54) when m=0.5, c1=0.4, C=-0.6, d=0.3, t=1.5.
(a)
(b)Family 2.
(3.55)
β
1
=
-
(
m
2
-
1
)
c
1
L
2
,
α
0
=
0
,
s
=
-
1
C
L
2
2
,
α
1
=
c
1
L
2
,
A
=
-
C
L
2
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
4
m
4
s
n
6
(
d
,
m
)
+
(
4
m
4
+
2
m
2
)
s
n
4
(
d
,
m
)
-
4
m
2
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants and
(3.56)L2=sn4(d,m)dn4(d,m)+cn4(d,m)C[m4sn8(d,m)-2m2sn6(d,m)+2sn2(d,m)-1].
In this case, the solution of (3.47) takes the following form:
(3.57)U(ξn)=c1L2cn(ξn,m)dn(ξn,m)sn(ξn,m)-(m2-1)c1L2sn(ξn,m)cn(ξn,m)dn(ξn,m),
where   ξn=dn+c1t+ξ0.Type 2.
Ife0=1-m2, e1=2m2-1, and e2=-m2.   In this case, the series expansion solution of (3.49) has the form:
(3.58)U(ξn)=α0-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-β1cn(ξn,m)sn(ξn,m)dn(ξn,m).
Consequently, using Maple or Mathematica we get the following results.Family 1.
(3.59)
β
1
=
-
c
1
L
1
,
α
0
=
0
,
s
=
-
1
C
L
1
2
,
α
1
=
c
1
L
1
,
A
=
-
C
L
1
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
4
m
2
s
n
6
(
d
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
d
,
m
)
-
4
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants.
In this case, the solution of (3.47) takes the following form:
(3.60)U(ξn)=-c1L1sn(ξn,m)dn(ξn,m)cn(ξn,m)+c1L1cn(ξn,m)sn(ξn,m)dn(ξn,m),
where   ξn=dn+c1t+ξ0.
In the special case, whenm=0, the trigonometrical solution (3.60) takes  the following form:
(3.61)U(ξn)=2c1cot(2ξn)-C,whereξn=dn+c1t+ξ0.
Also whenm=1, the hyperbolic solution (3.60) takes  the following form:
(3.62)U(ξn)=2c1coth(2ξn)[tanh4(d)+1]C[tanh4(d)-1],whereξn=dn+c1t+ξ0.Family 2.
(3.63)
β
1
=
c
1
L
3
,
α
0
=
0
,
s
=
-
1
C
L
3
2
,
α
1
=
c
1
L
3
,
A
=
-
C
L
3
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
2
s
n
4
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants, while
(3.64)L3=m4sn8(d,m)-2(m4+m2)sn6(d,m)+2(m4+m2+1)sn4(d,m)-2(m2+1)sn2(d,m)+1-C[m4sn8(d,m)-2(m4+m2)sn6(d,m)+6m2sn4(d,m)-2(m2+1)sn2(d,m)+1].
In this case, the solution of (3.47) takes the following form:
(3.65)U(ξn)=-c1L3sn(ξn,m)dn(ξn,m)cn(ξn,m)-c1L3cn(ξn,m)sn(ξn,m)dn(ξn,m),
where   ξn=dn+c1t+ξ0.
Whenm=0, the trigonometrical solution (3.65) takes  the following form:
(3.66)U(ξn)=-2c1sin(2ξn)-sin2(2d)+2-2Ccos(2d),whereξn=dn+c1t+ξ0.
Also ifm=1, the hyperbolic solution (3.65) takes  the following form:
(3.67)U(ξn)=-2c1coth(2ξn)-C,whereξn=dn+c1t+ξ0.Type 3.
Ife0=m2-1, e1=2-m2, and e2=-1. In this case, the series expansion solution of (3.49) has the form:
(3.68)U(ξn)=α0-m2α1sn(ξn,m)cn(ξn,m)dn(ξn,m)-β1dn(ξn,m)m2sn(ξn,m)cn(ξn,m).
Consequently, using Maple or Mathematica we get the following results.Family 1.
(3.69)
β
1
=
-
c
1
L
2
m
2
,
α
0
=
0
,
s
=
-
1
C
L
2
2
,
α
1
=
c
1
L
2
,
A
=
-
C
L
2
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
4
m
4
s
n
6
(
d
,
m
)
+
(
4
m
4
+
2
m
2
)
s
n
4
(
d
,
m
)
-
4
m
2
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants and
(3.70)L2=sn4(d,m)dn4(d,m)+cn4(d,m)C[m4sn8(d,m)-2m2sn6(d,m)+2sn2(d,m)-1].
In this case, the solution of (3.47) takes the following form:
(3.71)U(ξn)=-c1L3m2sn(ξn,m)cn(ξn,m)dn(ξn,m)+c1L3dn(ξn,m)sn(ξn,m)cn(ξn,m),
where   ξn=dn+c1t+ξ0.Family 2.
(3.72)
β
1
=
m
2
c
1
L
3
,
α
0
=
0
,
s
=
-
1
C
L
3
2
,
α
1
=
c
1
L
3
,
A
=
-
C
L
3
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
2
s
n
4
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants and
(3.73)L3=m4sn8(d,m)-2(m4+m2)sn6(d,m)+2(m4+m2+1)sn4(d,m)-2(m2+1)sn2(d,m)+1-C[m4sn8(d,m)-2(m4+m2)sn6(d,m)+6m2sn4(d,m)-2(m2+1)sn2(d,m)+1].
In this case, the solution of (3.47) takes the following form:
(3.74)U(ξn)=-m2c1L3sn(ξn,m)cn(ξn,m)dn(ξn,m)-c1L3dn(ξn,m)sn(ξn,m)cn(ξn,m),
where   ξn=dn+c1t+ξ0.


### 3.4. Example 4. The Quintic Discrete Nonlinear Schrodinger  Equation

We discuss the quintic discrete nonlinear Schrodinger (QDNLS) equation [33]:
(3.75)idψndt+α(ψn+1-2ψn+ψn-1)+β|ψn|2ψn+γ|ψn|2(ψn+1+ψn-1)+δ|ψn|4(ψn+1+ψn-1)=0,
which describes the propagation of discrete self-trapped beams in an array of weakly coupled nonlinear  optical waveguides. Equation (3.75) was presented  for the first time in [33], together with its localized solutions. We are looking for solutions of the form:
(3.76)ψn(t)=ϕne-iωt,
where ϕn is a real function with respect to the discrete variable n. Substitution of equation (3.76) into (3.75) yields the corresponding symmetric equation as  follows:
(3.77)ϕn+1+ϕn-1=(2α-ω)ϕn-βϕn3α+γϕn2+δϕn4.
We take the traveling wave of transformation
(3.78)ϕn=U(ξn),ξn=dn+k,
where d and k are constants. The transformation (3.78)  permits us converting equation (3.77) into the following form:
(3.79)U(ξn+d)+U(ξn-d)=(2α-ω)U(ξn)-βU3(ξn)α+γU2(ξn)+δU4(ξn).
We suppose that the solutions of (3.79) have the following form:
(3.80)U(ξn)=α0+α1(F′(ξn)F(ξn))+β1(F(ξn)F′(ξn)),
where α0, α1, and β1 are constant to be determined later and F(ξn) satisfies a discrete Jacobi elliptic ordinary differential equation (2.6), we have the following types.Type 1.
Ife0=1, e1=-(1+m2), and e2=m2.  In this case, the series expansion solution of (3.79) has the form:
(3.81)U(ξn)=α0+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+β1sn(ξn,m)cn(ξn,m)dn(ξn,m).
With the help of Maple, we substitute (3.81) and (2.8)  into (3.79). Cleaning  the denominator and collecting all terms with the same degree  of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the left-hand side of (3.79) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m). Setting each coefficient of this polynomial to be zero, we derive a set of algebraic equations for α0, α1, d, β1, α, ω, β, γ, and δ. Solving the set of algebraic equations by using Maple or Mathematica software package, we have the following.Family 1.
(3.82)
β
1
=
α
1
(
m
2
-
1
)
,
α
0
=
0
,
β
=
-
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
4
s
n
6
(
d
,
m
)
+
2
m
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
4
α
1
2
[
H
1
δ
α
1
2
+
H
2
γ
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
4
(
d
,
m
)
,
α
=
-
α
1
2
[
H
3
δ
α
1
2
+
H
4
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
where   α1, δ, and d are arbitrary constants and
(3.83)H1=m8sn12(d,m)-6m6sn10(d,m)+(14m4+m6)sn8(d,m)-(16m2+4m4)sn6(d,m)+(8-m4+8m2)sn4(d,m)+(2m2-8)sn2(d,m)-m2+2,H2=m6sn10(d,m)-(m6+3m4)sn8(d,m)+(4m2+2m4)sn6(d,m)+(m4-3m2-2)sn4(d,m)+(2-m2)sn2(d,m),H3=m8sn16(d,m)-8m6sn14(d,m)+(4m6sn12+24m4)sn12(d,m)-(32m2+24m4)sn10(d,m)+(16+48m2+6m4)sn8(d,m)-(24m2+32)sn6(d,m)+(24+4m2)sn4(d,m)-8sn2(d,m)+1,H4=m6sn14(d,m)-(m6+5m4)sn12(d,m)+(7m4+8m2)sn10(d,m)-(14m2+2m4+4)sn8(d,m)+(7m2+8)sn6(d,m)-(5+m2)sn4(d,m)+sn2(d,m).
In this case, the solution of (3.75) takes the following form:
(3.84)ψn(t)=α1[cn2(ξn,m)dn2(ξn,m)+(m2-1)sn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)×Exp{-i[4α12[H1δα12+H2γ]sn2(d,m)cn2(d,m)dn4(d,m)]t},
where   ξn=dn+k.
Figure4 illustrates the behavior of the exact solution (3.84).(a) and (b) represent the Jacobi elliptic solution (3.84) when m=0.5, α1=0.6, k=0.2, δ=0.4, γ=0.1, d=0.3, n=2. (c) and (d) represent the Jacobi elliptic solution (3.84) when m=0.5, α1=0.6, k=0.2, δ=0.4, γ=0.1, d=0.3, t=1.5.
(a)
(b)
(c)
(d)Family 2.
(3.85)
β
1
=
-
α
1
(
m
2
-
1
)
,
α
0
=
0
,
β
=
-
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
2
s
n
6
(
d
,
m
)
+
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
4
α
1
2
[
H
5
δ
α
1
2
+
H
6
γ
]
s
n
2
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
2
(
d
,
m
)
,
α
=
-
α
1
2
[
H
7
δ
α
1
2
+
H
8
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
where   α1, δ, and d are arbitrary constants and
(3.86)H5=m6sn12(d,m)-6m6sn10(d,m)+(14m6+m4)sn8(d,m)-(16m6+4m4)sn6(d,m)+(8m4-m2+8m6)sn4(d,m)+(2m2-8m4)sn2(d,m)+2m2-1,H6=m4sn10(d,m)-(m2+3m4)sn8(d,m)+(2m2+4m4)sn6(d,m)+(1-3m2-2m4)sn4(d,m)+(2m2-1)sn2(d,m),H7=m8sn16(d,m)-8m8sn14(d,m)+(24m8+4m6)sn12(d,m)-(24m6+32m8)sn10(d,m)+(6m4+48m6+16m8)sn8(d,m)-(24m4+32m6)sn6(d,m)+(24m4+4m2)sn4(d,m)-8m2sn2(d,m)+1,H8=m6sn14(d,m)-(5m6+m4)sn12(d,m)+(7m4+8m6)sn10(d,m)-(14m4+2m2+4m6)sn8(d,m)+(8m4+7m2)sn6(d,m)-(5m2+1)sn4(d,m)+sn2(d,m).
In this case, the solution of (3.75) takes the following form:
(3.87)ψn(t)=α1[cn2(ξn,m)dn2(ξn,m)-(m2-1)sn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)×Exp{-i[4α12[H5δα12+H6γ]sn2(d,m)cn4(d,m)dn2(d,m)]t},
where   ξn=dn+k.Type 2.
Ife0=1-m2, e1=2m2-1, and e2=-m2.   In this case, the solution of (3.79) has the form:
(3.88)U(ξn)=α0-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-β1cn(ξn,m)sn(ξn,m)dn(ξn,m).
Consequently, using the Maple or Mathematica we get the following results.Family 1.
(3.89)
β
1
=
-
α
1
,
α
0
=
0
,
β
=
-
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
4
s
n
6
(
d
,
m
)
+
2
m
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
4
α
1
2
[
H
1
δ
α
1
2
+
H
2
γ
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
4
(
d
,
m
)
,
α
=
-
α
1
2
[
H
3
δ
α
1
2
+
H
4
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
where   α1, δ, and d are arbitrary constants.
In this case, the solution of (3.75) takes the following form:
(3.90)ψn(t)=α1[cn2(ξn,m)-sn2(ξn,m)dn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)Exp{-i[4α12[H1δα12+H2γ]sn2(d,m)cn2(d,m)dn4(d,m)]t},
where   ξn=dn+k.
In the special case, whenm=0, the trigonometrical solution (3.90) takes  the following form:
(3.91)ψn(t)=2α1cot(2ξn)Exp{-i[32α14δ+32α14δsin2(2d)+8α12γ]t},whereξn=dn+k.
Whenm=1, the hyperbolic solution (3.90) takes  the following form:
(3.92)ψn(t)=2α1sinh(2ξn)Exp{-i[4α12[h1δα12+h2γ]tanh2(d)sech6(d)]t},
where
(3.93)h1=tanh12(d)-6tanh10(d)+15tanh8(d)-20tanh6(d)+15tanh4(d)-6tanh2(d)+1,h2=tanh10(d)-4tanh8(d)+6tanh6(d)-4tanh4(d)+tanh2(d),
and  ξn=dn+k.Family 2.
(3.94)
β
=
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
(
m
4
+
m
2
)
s
n
6
(
d
,
m
)
+
6
m
2
s
n
4
(
d
,
m
)
-
2
(
m
2
+
1
)
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
-
4
α
1
2
[
H
9
δ
α
1
2
+
H
10
γ
]
s
n
2
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
α
=
-
α
1
2
[
H
11
δ
α
1
2
+
H
12
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
β
1
=
α
1
,
α
0
=
0
,
where   α1, δ, and d are arbitrary constants and
(3.95)H9=(m8+m6)sn12(d,m)-4m6sn10(d,m)-(m6+m4)sn8(d,m)+8m4sn6(d,m)-(m4+m2)sn4(d,m)-4m2sn2(d,m)+m2+1,H10=(m6+m4)sn10(d,m)-(m6+6m4+m2)sn8(d,m)+6(m4+m2)sn6(d,m)-(m4+6m2+1)sn4(d,m)+(m2+1)sn2(d,m),H11=m8sn16(d,m)-4m6sn12(d,m)+6m4sn8(d,m)-4m2sn4(d,m)+1,H12=m6sn14(d,m)-(m6+m4)sn12(d,m)-m4sn10(d,m)+2(m4+m2)sn8(d,m)-m2sn6(d,m)-(m2+1)sn4(d,m)+sn2(d,m).
In this case, the solution of (3.75) takes the following form:
(3.96)ψn(t)=-α1[cn2(ξn,m)+sn2(ξn,m)dn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)Exp{-i[-4α12[H9δα12+H10γ]sn2(d,m)cn4(d,m)dn4(d,m)]t},
where   ξn=dn+k.
Whenm=0, the trigonometrical solution (3.96) takes  the following form:
(3.97)ψn(t)=-2α1sin(2ξn)Exp{-i[-4α12[δα12+sin2(d)cos2(d)γ]sin2(d)cos4(d)]t},
where ξn=dn+k.
Whenm=1, the hyperbolic solution (3.96) takes  the following form:
(3.98)ψn(t)=-2α1coth(2ξn)Exp{-i[-4α12[h9δα12+h10γ]tanh2(d)sech8(d)]t},
where
(3.99)h9=2tanh12(d)-4tanh10(d)-2tanh8(d)+8tanh6(d)-2tanh4(d)-4tanh2(d)+2,h10=2tanh10(d)-8tanh8(d)+12tanh6(d)-8tanh4(d)+2tanh2(d),
and  ξn=dn+k.Type 3.
Ife0=m2-1, e1=2-m2, and e2=-1. In this case, the series expansion solution of (3.79) has the form:
(3.100)U(ξn)=α0-m2α1sn(ξn,m)cn(ξn,m)dn(ξn,m)-β1dn(ξn,m)m2sn(ξn,m)cn(ξn,m).
Consequently, using the Maple or Mathematica we get the following results:Family 1.
(3.101)
β
1
=
-
α
1
m
2
,
α
0
=
0
,
β
=
-
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
2
s
n
6
(
d
,
m
)
+
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
4
α
1
2
[
H
5
δ
α
1
2
+
H
6
γ
]
s
n
2
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
2
(
d
,
m
)
,
α
=
-
α
1
2
[
H
7
δ
α
1
2
+
H
8
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
where   α1, δ, and d are arbitrary constants.
In this case, the solution of (3.75) takes the following form:
(3.102)ψn(t)=α1[dn2(ξn,m)-m2sn2(ξn,m)cn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)Exp{-i[4α12[H5δα12+H6γ]sn2(d,m)cn4(d,m)dn2(d,m)]t},
where   ξn=dn+k.Family 2.
(3.103)
β
=
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
(
m
4
+
m
2
)
s
n
6
(
d
,
m
)
+
6
m
2
s
n
4
(
d
,
m
)
-
2
(
m
2
+
1
)
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
-
4
α
1
2
[
H
9
δ
α
1
2
+
H
10
γ
]
s
n
2
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
α
=
-
α
1
2
[
H
11
δ
α
1
2
+
H
12
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
β
1
=
m
2
α
1
,
α
0
=
0
,
where   α1, δ, and d are arbitrary constants.
In this case, the solution of (3.75) takes the following form:
(3.104)ψn(t)=-α1[m2sn2(ξn,m)cn2(ξn,m)+dn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)×Exp{-i[-4α12[H9δα12+H10γ]sn2(d,m)cn4(d,m)dn4(d,m)]t},
where   ξn=dn+k.Remark 3.1.
Wang and Ma [34] constructed the Jacobi elliptic function solution to some nonlinear DDEs in terms of  Jacobi elliptic functions sn or cn or dn. In Zhang [20], an algorithm is devised to derive exact traveling wave solutions of nonlinear DDEs by means of  Jacobi elliptic functions. Gepreel and Shehata [22] put a direct method to construct the rational Jacobi elliptic solution for nonlinear DDEs. In this paper, we modified the direct method which was disscussed in [22]. When we are constructing the solutions of nonlinear DDEs we neglected the case βi=0   because if  βi=0, we get the same solution which was discussed  by [22]. The Jacobi elliptic functions could be generated into the hyperbolic functions when m tends to one in the other hand, they are generated into trigonometrical functions when m tends to zero as shown in Table 1.Table 1
The relation between the Jacobi elliptic functions and the trigonometrical functions, hyperbolic functions.
m
s
n (ξ,m)
c
n (ξ,m)
d
n (ξ,m)
n
s (ξ,m)
c
s (ξ,m)
d
s (ξ,m)
0
sin
(
ξ
)
cos (ξ)
1
csc (ξ)
cot (ξ)
csc (ξ)
1
tanh(ξ)
sech(ξ)
sech(ξ)
coth(ξ)
csch(ξ)
csch(ξ)


## 3.1. Example 1.  The  Lattice Equation

In this section, we study the  lattice equation which takes the following form [23–26]:
(3.1)dun(t)dt=(α+βun+γun2)(un-1-un+1),
where α, β, and γ are nonzero constants. This equation contains hybrid lattice equation, mKdV lattice equation, modified Volterra lattice equation, and Langmuir chain equation for some special values of α, β, γ.  According to the above steps, to seek traveling wave solutions of (3.1), we construct  the transformation
(3.2)un(t)=U(ξn),ξn=dn+c1t+ξ0,
where d, c1, and ξ0 are constants. The transformation (3.2) permits us converting equation (3.1) into the following form:
(3.3)c1U′(ξn)=[α+βU(ξn)+γU2(ξn)][U(ξn-d)-U(ξn+d)],
where ′=d/dξn. Considering the homogeneous balance between the highest order derivative and the nonlinear term in (3.3), we get  K=1.  Thus, the solution of (3.3) has the following form:
(3.4)U(ξn)=α0+α1(F′(ξn)F(ξn))+β1(F(ξn)F′(ξn)),
where α0, α1, and β1 are constants to be determined later and F(ξn) satisfies  a discrete Jacobi elliptic ordinary differential equation (2.6). When we discuss the solutions of (2.6), we get the following types.Type 1.
Ife0=1, e1=-(1+m2), and e2=m2.  In this case, the series expansion solution of (3.3) has the form:
(3.5)U(ξn)=α0+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+β1sn(ξn,m)cn(ξn,m)dn(ξn,m).
With the help of Maple, we substitute (3.5) and (2.8)  into (3.3). Cleaning  the denominator and collecting all terms with the same degree of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the  left-hand side of (3.3) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m).  Setting each coefficient of this polynomial to be zero, we derive a set of algebraic equations for α0, α1, d, β1, γ, β, α, and c1.  Solving the set of algebraic equations by using Maple or Mathematica software package, we have the following.Family 1.
(3.6)
α
0
=
-
β
2
γ
,
β
1
=
(
m
2
-
1
)
α
1
,
c
1
=
2
γ
α
1
2
[
c
n
2
(
d
,
m
)
-
s
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
]
s
n
(
d
,
m
)
c
n
(
d
,
m
)
d
n
(
d
,
m
)
,
α
=
β
2
4
γ
+
α
1
2
γ
[
-
m
4
s
n
8
(
d
,
m
)
+
4
m
2
s
n
6
(
d
,
m
)
-
(
4
+
2
m
2
)
s
n
4
(
d
,
m
)
+
4
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   α1, d, γ, β, and m are arbitrary constants.
From  (3.5) and (3.6), the solution  of (3.3) takes the following form:
(3.7)U(ξn)=-β2γ+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+α1(m2-1)sn(ξn,m)cn(ξn,m)dn(ξn,m),
where ξn=dn+(2γα12[cn2(d,m)-sn2(d,m)dn2(d,m)]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.  Figure 1 illustrates the behavior of the exact solution (3.7).(a) represents the Jacobi elliptic solution (3.7) when β=m=0.5, γ=1, α1=1.5, d=0.3, n=2. (b) represents the Jacobi elliptic solution (3.7) when β=m=0.5, γ=1, α1=1.5, d=0.3, t=2.
(a)
(b)Family 2.
(3.8)
α
0
=
-
β
2
γ
,
β
1
=
-
(
m
2
-
1
)
α
1
,
c
1
=
2
γ
α
1
2
[
d
n
2
(
d
,
m
)
-
m
2
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
]
s
n
(
d
,
m
)
c
n
(
d
,
m
)
d
n
(
d
,
m
)
,
α
=
β
2
4
γ
+
α
1
2
γ
[
-
m
4
s
n
8
(
d
,
m
)
+
4
m
4
s
n
6
(
d
,
m
)
-
(
4
m
4
+
2
m
2
)
s
n
4
(
d
,
m
)
+
4
m
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   α1, d, γ, β, and m are arbitrary constants.
From  (3.5) and (3.8), the solution  of (3.3) has the following form:
(3.9)U(ξn)=-β2γ+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)-α1(m2-1)sn(ξn,m)cn(ξn,m)dn(ξn,m),
where ξn=dn+(2γα12[dn2(d,m)-m2sn2(d,m)cn2(d,m)]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.Type 2.
Ife0=1-m2, e1=2m2-1, and e2=-m2.   In this case, the solution of (3.3) has the form:
(3.10)U(ξn)=α0-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-β1cn(ξn,m)sn(ξn,m)dn(ξn,m).
With the help of Maple, we substitute (3.10) into (3.3). Cleaning  the denominator and collecting all terms with the same degree of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the  left-hand side of (3.3) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m).   Setting each coefficient of this polynomial to zero, we derive a set of algebraic equations in α0, α1, d, β1, γ, β, α, and c1. Solving the set of algebraic equations by using Maple or Mathematica software package, we get the following.Family 1.
(3.11)
α
0
=
-
β
2
γ
,
β
1
=
-
α
1
,
c
1
=
2
γ
α
1
2
[
c
n
2
(
d
,
m
)
-
s
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
]
s
n
(
d
,
m
)
c
n
(
d
,
m
)
d
n
(
d
,
m
)
,
α
=
β
2
4
γ
+
α
1
2
γ
[
-
m
4
s
n
8
(
d
,
m
)
+
4
m
2
s
n
6
(
d
,
m
)
-
(
4
+
2
m
2
)
s
n
4
(
d
,
m
)
+
4
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   α1, d, γ, β, and m are arbitrary constants.
In this case, the solution of (3.3) takes  the following form:
(3.12)U(ξn)=-β2γ-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)+α1cn(ξn,m)sn(ξn,m)dn(ξn,m),
where   ξn=dn+(2γα12[cn2(d,m)-sn2(d,m)dn2(d,m)]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.
The Jacobi elliptic functions could be generated into the hyperbolic functions whenm tends to one in the other hand, they are generated into trigonometrical functions when m tends to zero.
Whenm=0, the trigonometrical solution (3.12) takes  the following form:
(3.13)U(ξn)=-β2γ+2α1cot(2ξn),whereξn=dn+4γα12cot(2d)t+ξ0.
Also if m=1, the hyperbolic solution (3.12) takes  the following form:
(3.14)U(ξn)=-β2γ+2α1sinh(2ξn),whereξn=dn+4γα12sinh(2d)t+ξ0.Family 2.
(3.15)
α
0
=
-
β
2
γ
,
β
1
=
α
1
,
c
1
=
-
2
γ
α
1
2
[
m
2
s
n
4
(
d
,
m
)
-
1
]
s
n
(
d
,
m
)
c
n
(
d
,
m
)
d
n
(
d
,
m
)
,
α
=
β
2
4
γ
+
α
1
2
γ
[
-
m
4
s
n
8
(
d
,
m
)
+
2
m
2
s
n
4
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   α1, d,γ, β, and m  are arbitrary constants.
In this case, the solution of (3.3) takes  the following form:
(3.16)U(ξn)=-β2γ-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-α1cn(ξn,m)sn(ξn,m)dn(ξn,m),
where    ξn=dn-(2γα12[m2sn4(d,m)-1]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.
Whenm=0, the trigonometrical solution (3.16) takes  the following form:
(3.17)U(ξn)=-β2γ-2α1sin(2ξn),whereξn=dn+4γα12sin(2d)t+ξ0.
When m=1, the hyperbolic solution (3.16) takes  the following form:
(3.18)U(ξn)=-β2γ-2α1coth(2ξn),whereξn=dn+4γα12coth(2d)t+ξ0.Type 3.
Ife0=m2-1, e1=2-m2, and e2=-1. In this case, the series expansion solution of (3.3) has the form:
(3.19)U(ξn)=α0-m2α1sn(ξn,m)cn(ξn,m)dn(ξn,m)-β1dn(ξn,m)m2sn(ξn,m)cn(ξn,m).
Consequently, using the Maple or Mathematica  we get the following results.Family 1.
(3.20)
α
0
=
-
β
2
γ
,
β
1
=
-
m
2
α
1
,
c
1
=
2
γ
α
1
2
[
d
n
2
(
d
,
m
)
-
m
2
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
]
s
n
(
d
,
m
)
c
n
(
d
,
m
)
d
n
(
d
,
m
)
,
α
=
β
2
4
γ
+
α
1
2
γ
[
-
m
4
s
n
8
(
d
,
m
)
+
4
m
4
s
n
6
(
d
,
m
)
-
(
4
m
4
+
2
m
2
)
s
n
4
(
d
,
m
)
+
4
m
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   α1, d,γ, β, and m are arbitrary constants.
In this case, the solution of (3.3) takes the following form:
(3.21)U(ξn)=-β2γ-α1m2sn(ξn,m)cn(ξn,m)dn(ξn,m)+α1dn(ξn,m)sn(ξn,m)cn(ξn,m),
where    ξn=dn+(2γα12[dn2(d,m)-m2sn2(d,m)cn2(d,m)]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.Family 2.
(3.22)
α
0
=
-
β
2
γ
,
β
1
=
α
1
m
2
,
c
1
=
-
2
γ
α
1
2
[
m
2
s
n
4
(
d
,
m
)
-
1
]
s
n
(
d
,
m
)
c
n
(
d
,
m
)
d
n
(
d
,
m
)
,
α
=
β
2
4
γ
+
α
1
2
γ
[
-
m
4
s
n
8
(
d
,
m
)
+
2
m
2
s
n
4
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   α1, d,γ, β, and m are arbitrary constants.
In this case, the solution of (3.3) takes the following form:
(3.23)U(ξn)=-β2γ-α1m2sn(ξn,m)cn(ξn,m)dn(ξn,m)-α1dn(ξn,m)sn(ξn,m)cn(ξn,m),
where       ξn=dn-(2γα12[m2sn4(d,m)-1]/sn(d,m)cn(d,m)dn(d,m))t+ξ0.


## 3.2. Example 2.  The Discrete Nonlinear Schrodinger Equation

The discrete nonlinear Schrodinger equation (DNSE) is one of the most fundamental nonlinear lattice models [8].  Its arise  in nonlinear optics as a model of infinite wave guide arrays [27] and has been recently implemented to describe Bose-Einstein condensates in optical lattices.  The class of DNSE model with saturable  nonlinearity is also of particular interest in their own right, due to a feature first unveiled in [28].  In this section, we study the DNSE with a saturable  nonlinearity [29, 30] form:
(3.24)i∂ψn∂t+(ψn+1+ψn-1-2ψn)+ν|ψn|21+μ|ψn|2ψn=0,
which describes optical pulse propagations in various doped fibers, ψn is a complex valued wave function at sites n while ν and μ are real parameters. We make the transformation:
(3.25)ψn=ϕ(ξn)e-i(σt+ρ),ξn=αn+β,
where σ, ρ, α, and β are arbitrary constants. The transformation (3.25) permits us converting equation (3.24) into the following nonlinear difference equation:
(3.26)(σ-2)φ(ξn)+φ(ξn+α)+φ(ξn-α)+νφ3(ξn)1+μφ2(ξn)=0.
We assume that (3.26) has  a solution of the form:
(3.27)ϕ(ξn)=α1(F′(ξn)F(ξn))+β1(F(ξn)F′(ξn))+α0,
where α0, α1, and β1 are constants to be determined later and F(ξn) satisfies  a discrete Jacobi elliptic differential equation (2.6). When we discuss the solutions of  (3.26), we have the following types.Type 1.
Ife0=1, e1=-(1+m2), and e2=m2.  In this case, the series expansion solution of (3.26) has the form:
(3.28)ϕ(ξn)=α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+β1sn(ξn,m)cn(ξn,m)dn(ξn,m)+α0.
With the help of Maple, we substitute (3.28) and (2.8) into (3.26), cleaning  the denominator and collecting all terms with the same order of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the  left-hand side of (3.26) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m). Setting each coefficient of this polynomial to zero, we derive a set of algebraic equations for α0, α1, σ, β1, ρ, α, and β. Solving the set of algebraic equations by using Maple or Mathematica software package, we obtain the following.Family 1.
(3.29)
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
s
n
2
(
α
,
m
)
+
1
]
,
β
1
=
(
m
2
-
1
)
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
s
n
2
(
α
,
m
)
+
1
]
,
μ
<
0
,
ν
=
-
2
μ
[
m
4
s
n
8
(
α
,
m
)
-
2
m
4
s
n
6
(
α
,
m
)
+
2
m
2
s
n
2
(
α
,
m
)
-
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
2
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
α
,
m
)
-
4
s
n
2
(
α
,
m
)
+
1
,
α
0
=
0
,
σ
=
4
s
n
2
(
α
,
m
)
[
m
4
s
n
6
(
α
,
m
)
-
(
m
4
+
2
m
2
)
s
n
4
(
α
,
m
)
+
(
m
2
+
2
)
s
n
2
(
α
,
m
)
+
m
2
-
2
]
m
4
s
n
8
(
α
,
m
)
-
4
m
2
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
α
,
m
)
-
4
s
n
2
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form: (3.30)ψn=(sn(α,m)cn(α,m)dn(α,m)cn(ξn,m)dn(ξn,m)-μ[m2sn4(α,m)-2sn2(α,m)+1]sn(ξn,m)+(m2-1)sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)-μ[m2sn4(α,m)-2sn2(α,m)+1]cn(ξn,m)dn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔄m4sn8(α,m)-4m2sn6(α,m)+(2m2+4)sn4(α,m)-4sn2(α,m)+1+ρ]}),where 𝔄 denotes [m4sn6(α,m)-(m4+2m2)sn4(α,m)+(m2+2)sn2(α,m)+m2-2] and ξn=αn+β. Figure 2 illustrates the behavior exact solution (3.30).(a) and (b) represent the Jacobi elliptic solution (3.30) when m=0.5, β=1.5, α=0.2, ρ=0.4, n=2. (c) and (d) represent the Jacobi elliptic solution (3.30) when m=0.5, β=1.5, α=0.2, ρ=0.4, t=1.5.
(a)
(b)
(c)
(d)Family 2.
(3.31)
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
m
2
s
n
2
(
α
,
m
)
+
1
]
,
β
1
=
-
(
m
2
-
1
)
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
m
2
s
n
2
(
α
,
m
)
+
1
]
,
μ
<
0
,
ν
=
-
2
μ
[
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
6
(
α
,
m
)
+
2
s
n
2
(
α
,
m
)
-
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
4
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
m
4
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
1
,
α
0
=
0
,
σ
=
4
s
n
2
(
α
,
m
)
[
m
4
s
n
6
(
α
,
m
)
-
(
2
m
4
+
m
2
)
s
n
4
(
α
,
m
)
+
(
2
m
4
+
m
2
)
s
n
2
(
α
,
m
)
-
2
m
2
+
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
4
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
m
4
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form: (3.32)ψn=(sn(α,m)cn(α,m)dn(α,m)cn(ξn,m)dn(ξn,m)-μ[m2sn4(α,m)-2m2sn2(α,m)+1]sn(ξn,m)-(m2-1)sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)-μ[m2sn4(α,m)-2m2sn2(α,m)+1]cn(ξn,m)dn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔚m4sn8(α,m)-4m4sn6(α,m)+(2m2+4m4)sn4(α,m)-4m2sn2(α,m)+1+ρ]}),where 𝔚 denotes [m4sn6(α,m)-(2m4+m2)sn4(α,m)+(2m4+m2)sn2(α,m)-2m2+1] and ξn=αn+β.Type 2.
Ife0=1-m2, e1=2m2-1, and e2=-m2.  In this case, the solution of (3.26) has the form:
(3.33)ϕ(ξn)=α0-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-β1cn(ξn,m)sn(ξn,m)dn(ξn,m).
Consequently, using Maple or Mathematica we get the following results.Family 1.
(3.34)
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
s
n
2
(
α
,
m
)
+
1
]
,
β
1
=
-
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
s
n
2
(
α
,
m
)
+
1
]
,
μ
<
0
,
ν
=
-
2
μ
[
m
4
s
n
8
(
α
,
m
)
-
2
m
4
s
n
6
(
α
,
m
)
+
2
m
2
s
n
2
(
α
,
m
)
-
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
2
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
α
,
m
)
-
4
s
n
2
(
α
,
m
)
+
1
,
α
0
=
0
,
σ
=
4
s
n
2
(
α
,
m
)
[
m
4
s
n
6
(
α
,
m
)
-
(
m
4
+
2
m
2
)
s
n
4
(
α
,
m
)
+
(
m
2
+
2
)
s
n
2
(
α
,
m
)
+
m
2
-
2
]
m
4
s
n
8
(
α
,
m
)
-
4
m
2
s
n
6
(
α
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
α
,
m
)
-
4
s
n
2
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form:
(3.35)ψn=(-sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)dn(ξn,m)-μ[m2sn4(α,m)-2sn2(α,m)+1]cn(ξn,m)+sn(α,m)cn(α,m)dn(α,m)cn(ξn,m)-μ[m2sn4(α,m)-2sn2(α,m)+1]sn(ξn,m)dn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔄m4sn8(α,m)-4m2sn6(α,m)+(2m2+4)sn4(α,m)-4sn2(α,m)+1+ρ]}),
where ξn=αn+β.
In the special case, whenm=0, the trigonometrical solution (3.35) takes  the following form:
(3.36)ψn=tan(2α)cot(2ξn)-μ(Exp{-i[-2ttan2(2α)+ρ]}),whereξn=αn+β.
Also ifm=1, the hyperbolic solution (3.35) takes  the following form:
(3.37)ψn=(sinh(2α)-μsinh(2ξn))(Exp{i[4tsinh2(α)-ρ]}),whereξn=αn+β.Family 2.
(3.38)
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
1
]
,
β
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
1
]
,
μ
<
0
,
ν
=
2
μ
[
m
4
s
n
8
(
α
,
m
)
-
2
(
m
4
+
m
2
)
s
n
6
(
α
,
m
)
+
6
m
2
s
n
4
(
α
,
m
)
-
(
2
+
2
m
2
)
s
n
2
(
α
,
m
)
+
1
]
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
4
(
α
,
m
)
+
1
,
σ
=
4
s
n
2
(
α
,
m
)
[
(
m
4
+
m
2
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
m
2
+
1
]
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
4
(
α
,
m
)
+
1
,
α
0
=
0
.
In this case, the solution of (3.24) takes the following form:
(3.39)ψn=(-sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)dn(ξn,m)-μ[m2sn4(α,m)-1]cn(ξn,m)-sn(α,m)cn(α,m)dn(α,m)cn(ξn,m)-μ[m2sn4(α,m)-1]sn(ξn,m)dn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔛m4sn8(α,m)-2m2sn4(α,m)+1+ρ]}),
where 𝔛 denotes [(m4+m2)sn4(α,m)-4m2sn2(α,m)+m2+1] and ξn=αn+β.
Ifm=0, the trigonometrical solution (3.39) takes  the following form:
(3.40)ψn=sin(2α)-μsin(2ξn)(Exp{-i[4tsin2(α)+ρ]}),whereξn=αn+β.
Also ifm=1, the hyperbolic solution (3.39) takes  the following form:
(3.41)ψn=tanh(2α)coth(2ξn)-μ(Exp{-i[2ttanh2(2α)+ρ]}),whereξn=αn+β.Type 3.
ife0=m2-1, e1=2-m2, and e2=-1. In this case, the series expansion solution of (3.26)  has the form:
(3.42)ϕ(ξn)=α0-m2α1sn(ξn,m)cn(ξn,m)dn(ξn,m)-β1dn(ξn,m)m2sn(ξn,m)cn(ξn,m).
Consequently, using Maple or Mathematica we get the following results.Family 1.
(3.43)
α
0
=
0
,
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
m
2
s
n
2
(
α
,
m
)
+
1
]
,
β
1
=
-
m
2
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
2
m
2
s
n
2
(
α
,
m
)
+
1
]
,
μ
<
0
,
ν
=
-
2
μ
(
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
6
(
α
,
m
)
+
2
s
n
2
(
α
,
m
)
-
1
)
m
4
s
n
8
(
α
,
m
)
-
4
m
4
s
n
6
(
α
,
m
)
+
(
4
m
4
+
2
m
2
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
1
,
σ
=
4
s
n
2
(
α
,
m
)
[
m
4
s
n
6
(
α
,
m
)
-
(
2
m
4
+
m
2
)
s
n
4
(
α
,
m
)
+
(
2
m
4
+
m
2
)
s
n
2
(
α
,
m
)
-
2
m
2
+
1
]
m
4
s
n
8
(
α
,
m
)
-
4
m
4
s
n
6
(
α
,
m
)
+
(
4
m
4
+
2
m
2
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form:
(3.44)ψn=(-m2sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)cn(ξn,m)-μ[m2sn4(α,m)-2m2sn2(α,m)+1]dn(ξn,m)+sn(α,m)cn(α,m)dn(α,m)dn(ξn,m)-μ[m2sn4(α,m)-2m2sn2(α,m)+1]sn(ξn,m)cn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔚m4sn8(α,m)-4m4sn6(α,m)+(4m4+2m2)sn4(α,m)-4m2sn2(α,m)+1+ρ]})
where ξn=αn+β.Family 2.
(3.45)
α
0
=
0
,
α
1
=
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
1
]
,
β
1
=
m
2
s
n
(
α
,
m
)
c
n
(
α
,
m
)
d
n
(
α
,
m
)
-
μ
[
m
2
s
n
4
(
α
,
m
)
-
1
]
,
μ
<
0
,
ν
=
2
μ
[
(
m
4
s
n
8
(
α
,
m
)
-
2
(
m
4
+
m
2
)
s
n
6
(
α
,
m
)
+
6
m
2
s
n
4
(
α
,
m
)
-
(
2
+
2
m
2
)
s
n
2
(
α
,
m
)
+
1
)
]
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
4
(
α
,
m
)
+
1
,
σ
=
4
s
n
2
(
α
,
m
)
[
(
m
4
+
m
2
)
s
n
4
(
α
,
m
)
-
4
m
2
s
n
2
(
α
,
m
)
+
m
2
+
1
]
m
4
s
n
8
(
α
,
m
)
-
2
m
2
s
n
4
(
α
,
m
)
+
1
.
In this case, the solution of (3.24) takes the following form:
(3.46)ψn=(-m2sn(α,m)cn(α,m)dn(α,m)sn(ξn,m)cn(ξn,m)-μ[m2sn4(α,m)-1]dn(ξn,m)-sn(α,m)cn(α,m)dn(α,m)dn(ξn,m)-μ[m2sn4(α,m)-1]sn(ξn,m)cn(ξn,m))×(Exp{-i[4tsn2(α,m)𝔛m4sn8(α,m)-2m2sn4(α,m)+1+ρ]}),
where ξn=αn+β.


## 3.3. Example 3.  The Discrete Nonlinear Klein-Gordon Equation

In this section, we consider the following discrete nonlinear Klein-Gordon equation [31]:
(3.47)d2un(t)dt2=g(un)(un+1+un-1-2sun).
The nonconstant (in contrast to the standard models of harmonic coupling and linear dispersion [32]) function g(un) ensures the presence of nonlinear dispersion, which is critical for the existence of compactly supported solutions and  s  can take values in the interval [-1,1]. Kevrekidis and Konotop [31] have obtained some exact compaction solutions and claim that this DDE does not have the traveling compact solution.If we setg(un)=A+Cun2 where A and C are arbitrary constants and take the traveling transformation:
(3.48)un(t)=U(ξn),ξn=dn+c1t+ξ0,
where d, c1, and ξ0 are constants. The transformation (3.48) permits us converting equation (3.47) into the following form:
(3.49)c12U′′(ξn)=[A+CU2(ξn)][U(ξn+d)+U(ξn-d)-2sU(ξn)],
where  ′=d/dξn. Considering the homogeneous balance between the highest-order derivative and the nonlinear term in (3.49), we get K=1. Thus, the solutions of (3.49) have the following form:
(3.50)U(ξn)=α0+α1(F′(ξn)F(ξn))+β1(F(ξn)F′(ξn)),
where  α0, α1, and β1 are constants to be determined later and F(ξn) satisfies a discrete Jacobi elliptic ordinary differential equation (2.6), we have the following types.Type 1.
Ife0=1, e1=-(1+m2), and e2=m2.  In this case, the series expansion solution of (3.49) has the form:
(3.51)U(ξn)=α0+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+β1sn(ξn,m)cn(ξn,m)dn(ξn,m).
With the help of Maple, we substitute (3.51) and (2.8)  into (3.49). Cleaning  the denominator and collecting all terms with the same degree  of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the left-hand side of (3.49) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m). Setting each coefficient of this polynomial to zero, we derive a set of algebraic equations for α0, α1, d, β1, A, C, s, and c1.  Solving the set of algebraic equations by using Maple or Mathematica software package, we have the following.Family 1.
(3.52)
β
1
=
(
m
2
-
1
)
c
1
L
1
,
α
0
=
0
,
s
=
-
1
C
L
1
2
,
α
1
=
c
1
L
1
,
A
=
-
C
L
1
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
4
m
2
s
n
6
(
d
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
d
,
m
)
-
4
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants and
(3.53)L1=m4sn4(d,m)cn4(d,m)+dn4(d,m)C[m4sn8(d,m)-2m4sn6(d,m)+2m2sn2(d,m)-1].
In this case, the solution of (3.47) takes the following form:
(3.54)U(ξn)=c1L1cn(ξn,m)dn(ξn,m)sn(ξn,m)+(m2-1)c1L1sn(ξn,m)cn(ξn,m)dn(ξn,m),
where   ξn=dn+c1t+ξ0.
Figure3 illustrates the behavior of the exact solution (3.54).(a) represents the Jacobi elliptic solution (3.54) when m=0.5, c1=0.4, C=-0.6, d=0.3, n=2. (b) represents the Jacobi elliptic solution (3.54) when m=0.5, c1=0.4, C=-0.6, d=0.3, t=1.5.
(a)
(b)Family 2.
(3.55)
β
1
=
-
(
m
2
-
1
)
c
1
L
2
,
α
0
=
0
,
s
=
-
1
C
L
2
2
,
α
1
=
c
1
L
2
,
A
=
-
C
L
2
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
4
m
4
s
n
6
(
d
,
m
)
+
(
4
m
4
+
2
m
2
)
s
n
4
(
d
,
m
)
-
4
m
2
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants and
(3.56)L2=sn4(d,m)dn4(d,m)+cn4(d,m)C[m4sn8(d,m)-2m2sn6(d,m)+2sn2(d,m)-1].
In this case, the solution of (3.47) takes the following form:
(3.57)U(ξn)=c1L2cn(ξn,m)dn(ξn,m)sn(ξn,m)-(m2-1)c1L2sn(ξn,m)cn(ξn,m)dn(ξn,m),
where   ξn=dn+c1t+ξ0.Type 2.
Ife0=1-m2, e1=2m2-1, and e2=-m2.   In this case, the series expansion solution of (3.49) has the form:
(3.58)U(ξn)=α0-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-β1cn(ξn,m)sn(ξn,m)dn(ξn,m).
Consequently, using Maple or Mathematica we get the following results.Family 1.
(3.59)
β
1
=
-
c
1
L
1
,
α
0
=
0
,
s
=
-
1
C
L
1
2
,
α
1
=
c
1
L
1
,
A
=
-
C
L
1
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
4
m
2
s
n
6
(
d
,
m
)
+
(
2
m
2
+
4
)
s
n
4
(
d
,
m
)
-
4
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants.
In this case, the solution of (3.47) takes the following form:
(3.60)U(ξn)=-c1L1sn(ξn,m)dn(ξn,m)cn(ξn,m)+c1L1cn(ξn,m)sn(ξn,m)dn(ξn,m),
where   ξn=dn+c1t+ξ0.
In the special case, whenm=0, the trigonometrical solution (3.60) takes  the following form:
(3.61)U(ξn)=2c1cot(2ξn)-C,whereξn=dn+c1t+ξ0.
Also whenm=1, the hyperbolic solution (3.60) takes  the following form:
(3.62)U(ξn)=2c1coth(2ξn)[tanh4(d)+1]C[tanh4(d)-1],whereξn=dn+c1t+ξ0.Family 2.
(3.63)
β
1
=
c
1
L
3
,
α
0
=
0
,
s
=
-
1
C
L
3
2
,
α
1
=
c
1
L
3
,
A
=
-
C
L
3
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
2
s
n
4
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants, while
(3.64)L3=m4sn8(d,m)-2(m4+m2)sn6(d,m)+2(m4+m2+1)sn4(d,m)-2(m2+1)sn2(d,m)+1-C[m4sn8(d,m)-2(m4+m2)sn6(d,m)+6m2sn4(d,m)-2(m2+1)sn2(d,m)+1].
In this case, the solution of (3.47) takes the following form:
(3.65)U(ξn)=-c1L3sn(ξn,m)dn(ξn,m)cn(ξn,m)-c1L3cn(ξn,m)sn(ξn,m)dn(ξn,m),
where   ξn=dn+c1t+ξ0.
Whenm=0, the trigonometrical solution (3.65) takes  the following form:
(3.66)U(ξn)=-2c1sin(2ξn)-sin2(2d)+2-2Ccos(2d),whereξn=dn+c1t+ξ0.
Also ifm=1, the hyperbolic solution (3.65) takes  the following form:
(3.67)U(ξn)=-2c1coth(2ξn)-C,whereξn=dn+c1t+ξ0.Type 3.
Ife0=m2-1, e1=2-m2, and e2=-1. In this case, the series expansion solution of (3.49) has the form:
(3.68)U(ξn)=α0-m2α1sn(ξn,m)cn(ξn,m)dn(ξn,m)-β1dn(ξn,m)m2sn(ξn,m)cn(ξn,m).
Consequently, using Maple or Mathematica we get the following results.Family 1.
(3.69)
β
1
=
-
c
1
L
2
m
2
,
α
0
=
0
,
s
=
-
1
C
L
2
2
,
α
1
=
c
1
L
2
,
A
=
-
C
L
2
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
4
m
4
s
n
6
(
d
,
m
)
+
(
4
m
4
+
2
m
2
)
s
n
4
(
d
,
m
)
-
4
m
2
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants and
(3.70)L2=sn4(d,m)dn4(d,m)+cn4(d,m)C[m4sn8(d,m)-2m2sn6(d,m)+2sn2(d,m)-1].
In this case, the solution of (3.47) takes the following form:
(3.71)U(ξn)=-c1L3m2sn(ξn,m)cn(ξn,m)dn(ξn,m)+c1L3dn(ξn,m)sn(ξn,m)cn(ξn,m),
where   ξn=dn+c1t+ξ0.Family 2.
(3.72)
β
1
=
m
2
c
1
L
3
,
α
0
=
0
,
s
=
-
1
C
L
3
2
,
α
1
=
c
1
L
3
,
A
=
-
C
L
3
2
c
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
2
s
n
4
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
where   d, c1, and C are arbitrary constants and
(3.73)L3=m4sn8(d,m)-2(m4+m2)sn6(d,m)+2(m4+m2+1)sn4(d,m)-2(m2+1)sn2(d,m)+1-C[m4sn8(d,m)-2(m4+m2)sn6(d,m)+6m2sn4(d,m)-2(m2+1)sn2(d,m)+1].
In this case, the solution of (3.47) takes the following form:
(3.74)U(ξn)=-m2c1L3sn(ξn,m)cn(ξn,m)dn(ξn,m)-c1L3dn(ξn,m)sn(ξn,m)cn(ξn,m),
where   ξn=dn+c1t+ξ0.


## 3.4. Example 4. The Quintic Discrete Nonlinear Schrodinger  Equation

We discuss the quintic discrete nonlinear Schrodinger (QDNLS) equation [33]:
(3.75)idψndt+α(ψn+1-2ψn+ψn-1)+β|ψn|2ψn+γ|ψn|2(ψn+1+ψn-1)+δ|ψn|4(ψn+1+ψn-1)=0,
which describes the propagation of discrete self-trapped beams in an array of weakly coupled nonlinear  optical waveguides. Equation (3.75) was presented  for the first time in [33], together with its localized solutions. We are looking for solutions of the form:
(3.76)ψn(t)=ϕne-iωt,
where ϕn is a real function with respect to the discrete variable n. Substitution of equation (3.76) into (3.75) yields the corresponding symmetric equation as  follows:
(3.77)ϕn+1+ϕn-1=(2α-ω)ϕn-βϕn3α+γϕn2+δϕn4.
We take the traveling wave of transformation
(3.78)ϕn=U(ξn),ξn=dn+k,
where d and k are constants. The transformation (3.78)  permits us converting equation (3.77) into the following form:
(3.79)U(ξn+d)+U(ξn-d)=(2α-ω)U(ξn)-βU3(ξn)α+γU2(ξn)+δU4(ξn).
We suppose that the solutions of (3.79) have the following form:
(3.80)U(ξn)=α0+α1(F′(ξn)F(ξn))+β1(F(ξn)F′(ξn)),
where α0, α1, and β1 are constant to be determined later and F(ξn) satisfies a discrete Jacobi elliptic ordinary differential equation (2.6), we have the following types.Type 1.
Ife0=1, e1=-(1+m2), and e2=m2.  In this case, the series expansion solution of (3.79) has the form:
(3.81)U(ξn)=α0+α1cn(ξn,m)dn(ξn,m)sn(ξn,m)+β1sn(ξn,m)cn(ξn,m)dn(ξn,m).
With the help of Maple, we substitute (3.81) and (2.8)  into (3.79). Cleaning  the denominator and collecting all terms with the same degree  of  sn(ξn,m), cn(ξn,m), dn(ξn,m) together, the left-hand side of (3.79) is converted into polynomial in sn(ξn,m), cn(ξn,m), dn(ξn,m). Setting each coefficient of this polynomial to be zero, we derive a set of algebraic equations for α0, α1, d, β1, α, ω, β, γ, and δ. Solving the set of algebraic equations by using Maple or Mathematica software package, we have the following.Family 1.
(3.82)
β
1
=
α
1
(
m
2
-
1
)
,
α
0
=
0
,
β
=
-
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
4
s
n
6
(
d
,
m
)
+
2
m
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
4
α
1
2
[
H
1
δ
α
1
2
+
H
2
γ
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
4
(
d
,
m
)
,
α
=
-
α
1
2
[
H
3
δ
α
1
2
+
H
4
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
where   α1, δ, and d are arbitrary constants and
(3.83)H1=m8sn12(d,m)-6m6sn10(d,m)+(14m4+m6)sn8(d,m)-(16m2+4m4)sn6(d,m)+(8-m4+8m2)sn4(d,m)+(2m2-8)sn2(d,m)-m2+2,H2=m6sn10(d,m)-(m6+3m4)sn8(d,m)+(4m2+2m4)sn6(d,m)+(m4-3m2-2)sn4(d,m)+(2-m2)sn2(d,m),H3=m8sn16(d,m)-8m6sn14(d,m)+(4m6sn12+24m4)sn12(d,m)-(32m2+24m4)sn10(d,m)+(16+48m2+6m4)sn8(d,m)-(24m2+32)sn6(d,m)+(24+4m2)sn4(d,m)-8sn2(d,m)+1,H4=m6sn14(d,m)-(m6+5m4)sn12(d,m)+(7m4+8m2)sn10(d,m)-(14m2+2m4+4)sn8(d,m)+(7m2+8)sn6(d,m)-(5+m2)sn4(d,m)+sn2(d,m).
In this case, the solution of (3.75) takes the following form:
(3.84)ψn(t)=α1[cn2(ξn,m)dn2(ξn,m)+(m2-1)sn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)×Exp{-i[4α12[H1δα12+H2γ]sn2(d,m)cn2(d,m)dn4(d,m)]t},
where   ξn=dn+k.
Figure4 illustrates the behavior of the exact solution (3.84).(a) and (b) represent the Jacobi elliptic solution (3.84) when m=0.5, α1=0.6, k=0.2, δ=0.4, γ=0.1, d=0.3, n=2. (c) and (d) represent the Jacobi elliptic solution (3.84) when m=0.5, α1=0.6, k=0.2, δ=0.4, γ=0.1, d=0.3, t=1.5.
(a)
(b)
(c)
(d)Family 2.
(3.85)
β
1
=
-
α
1
(
m
2
-
1
)
,
α
0
=
0
,
β
=
-
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
2
s
n
6
(
d
,
m
)
+
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
4
α
1
2
[
H
5
δ
α
1
2
+
H
6
γ
]
s
n
2
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
2
(
d
,
m
)
,
α
=
-
α
1
2
[
H
7
δ
α
1
2
+
H
8
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
where   α1, δ, and d are arbitrary constants and
(3.86)H5=m6sn12(d,m)-6m6sn10(d,m)+(14m6+m4)sn8(d,m)-(16m6+4m4)sn6(d,m)+(8m4-m2+8m6)sn4(d,m)+(2m2-8m4)sn2(d,m)+2m2-1,H6=m4sn10(d,m)-(m2+3m4)sn8(d,m)+(2m2+4m4)sn6(d,m)+(1-3m2-2m4)sn4(d,m)+(2m2-1)sn2(d,m),H7=m8sn16(d,m)-8m8sn14(d,m)+(24m8+4m6)sn12(d,m)-(24m6+32m8)sn10(d,m)+(6m4+48m6+16m8)sn8(d,m)-(24m4+32m6)sn6(d,m)+(24m4+4m2)sn4(d,m)-8m2sn2(d,m)+1,H8=m6sn14(d,m)-(5m6+m4)sn12(d,m)+(7m4+8m6)sn10(d,m)-(14m4+2m2+4m6)sn8(d,m)+(8m4+7m2)sn6(d,m)-(5m2+1)sn4(d,m)+sn2(d,m).
In this case, the solution of (3.75) takes the following form:
(3.87)ψn(t)=α1[cn2(ξn,m)dn2(ξn,m)-(m2-1)sn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)×Exp{-i[4α12[H5δα12+H6γ]sn2(d,m)cn4(d,m)dn2(d,m)]t},
where   ξn=dn+k.Type 2.
Ife0=1-m2, e1=2m2-1, and e2=-m2.   In this case, the solution of (3.79) has the form:
(3.88)U(ξn)=α0-α1sn(ξn,m)dn(ξn,m)cn(ξn,m)-β1cn(ξn,m)sn(ξn,m)dn(ξn,m).
Consequently, using the Maple or Mathematica we get the following results.Family 1.
(3.89)
β
1
=
-
α
1
,
α
0
=
0
,
β
=
-
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
4
s
n
6
(
d
,
m
)
+
2
m
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
4
α
1
2
[
H
1
δ
α
1
2
+
H
2
γ
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
4
(
d
,
m
)
,
α
=
-
α
1
2
[
H
3
δ
α
1
2
+
H
4
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
where   α1, δ, and d are arbitrary constants.
In this case, the solution of (3.75) takes the following form:
(3.90)ψn(t)=α1[cn2(ξn,m)-sn2(ξn,m)dn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)Exp{-i[4α12[H1δα12+H2γ]sn2(d,m)cn2(d,m)dn4(d,m)]t},
where   ξn=dn+k.
In the special case, whenm=0, the trigonometrical solution (3.90) takes  the following form:
(3.91)ψn(t)=2α1cot(2ξn)Exp{-i[32α14δ+32α14δsin2(2d)+8α12γ]t},whereξn=dn+k.
Whenm=1, the hyperbolic solution (3.90) takes  the following form:
(3.92)ψn(t)=2α1sinh(2ξn)Exp{-i[4α12[h1δα12+h2γ]tanh2(d)sech6(d)]t},
where
(3.93)h1=tanh12(d)-6tanh10(d)+15tanh8(d)-20tanh6(d)+15tanh4(d)-6tanh2(d)+1,h2=tanh10(d)-4tanh8(d)+6tanh6(d)-4tanh4(d)+tanh2(d),
and  ξn=dn+k.Family 2.
(3.94)
β
=
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
(
m
4
+
m
2
)
s
n
6
(
d
,
m
)
+
6
m
2
s
n
4
(
d
,
m
)
-
2
(
m
2
+
1
)
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
-
4
α
1
2
[
H
9
δ
α
1
2
+
H
10
γ
]
s
n
2
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
α
=
-
α
1
2
[
H
11
δ
α
1
2
+
H
12
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
β
1
=
α
1
,
α
0
=
0
,
where   α1, δ, and d are arbitrary constants and
(3.95)H9=(m8+m6)sn12(d,m)-4m6sn10(d,m)-(m6+m4)sn8(d,m)+8m4sn6(d,m)-(m4+m2)sn4(d,m)-4m2sn2(d,m)+m2+1,H10=(m6+m4)sn10(d,m)-(m6+6m4+m2)sn8(d,m)+6(m4+m2)sn6(d,m)-(m4+6m2+1)sn4(d,m)+(m2+1)sn2(d,m),H11=m8sn16(d,m)-4m6sn12(d,m)+6m4sn8(d,m)-4m2sn4(d,m)+1,H12=m6sn14(d,m)-(m6+m4)sn12(d,m)-m4sn10(d,m)+2(m4+m2)sn8(d,m)-m2sn6(d,m)-(m2+1)sn4(d,m)+sn2(d,m).
In this case, the solution of (3.75) takes the following form:
(3.96)ψn(t)=-α1[cn2(ξn,m)+sn2(ξn,m)dn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)Exp{-i[-4α12[H9δα12+H10γ]sn2(d,m)cn4(d,m)dn4(d,m)]t},
where   ξn=dn+k.
Whenm=0, the trigonometrical solution (3.96) takes  the following form:
(3.97)ψn(t)=-2α1sin(2ξn)Exp{-i[-4α12[δα12+sin2(d)cos2(d)γ]sin2(d)cos4(d)]t},
where ξn=dn+k.
Whenm=1, the hyperbolic solution (3.96) takes  the following form:
(3.98)ψn(t)=-2α1coth(2ξn)Exp{-i[-4α12[h9δα12+h10γ]tanh2(d)sech8(d)]t},
where
(3.99)h9=2tanh12(d)-4tanh10(d)-2tanh8(d)+8tanh6(d)-2tanh4(d)-4tanh2(d)+2,h10=2tanh10(d)-8tanh8(d)+12tanh6(d)-8tanh4(d)+2tanh2(d),
and  ξn=dn+k.Type 3.
Ife0=m2-1, e1=2-m2, and e2=-1. In this case, the series expansion solution of (3.79) has the form:
(3.100)U(ξn)=α0-m2α1sn(ξn,m)cn(ξn,m)dn(ξn,m)-β1dn(ξn,m)m2sn(ξn,m)cn(ξn,m).
Consequently, using the Maple or Mathematica we get the following results:Family 1.
(3.101)
β
1
=
-
α
1
m
2
,
α
0
=
0
,
β
=
-
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
m
2
s
n
6
(
d
,
m
)
+
2
s
n
2
(
d
,
m
)
-
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
4
α
1
2
[
H
5
δ
α
1
2
+
H
6
γ
]
s
n
2
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
2
(
d
,
m
)
,
α
=
-
α
1
2
[
H
7
δ
α
1
2
+
H
8
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
where   α1, δ, and d are arbitrary constants.
In this case, the solution of (3.75) takes the following form:
(3.102)ψn(t)=α1[dn2(ξn,m)-m2sn2(ξn,m)cn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)Exp{-i[4α12[H5δα12+H6γ]sn2(d,m)cn4(d,m)dn2(d,m)]t},
where   ξn=dn+k.Family 2.
(3.103)
β
=
2
δ
α
1
2
[
m
4
s
n
8
(
d
,
m
)
-
2
(
m
4
+
m
2
)
s
n
6
(
d
,
m
)
+
6
m
2
s
n
4
(
d
,
m
)
-
2
(
m
2
+
1
)
s
n
2
(
d
,
m
)
+
1
]
s
n
2
(
d
,
m
)
c
n
2
(
d
,
m
)
d
n
2
(
d
,
m
)
,
ω
=
-
4
α
1
2
[
H
9
δ
α
1
2
+
H
10
γ
]
s
n
2
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
α
=
-
α
1
2
[
H
11
δ
α
1
2
+
H
12
γ
]
s
n
4
(
d
,
m
)
c
n
4
(
d
,
m
)
d
n
4
(
d
,
m
)
,
β
1
=
m
2
α
1
,
α
0
=
0
,
where   α1, δ, and d are arbitrary constants.
In this case, the solution of (3.75) takes the following form:
(3.104)ψn(t)=-α1[m2sn2(ξn,m)cn2(ξn,m)+dn2(ξn,m)]sn(ξn,m)cn(ξn,m)dn(ξn,m)×Exp{-i[-4α12[H9δα12+H10γ]sn2(d,m)cn4(d,m)dn4(d,m)]t},
where   ξn=dn+k.Remark 3.1.
Wang and Ma [34] constructed the Jacobi elliptic function solution to some nonlinear DDEs in terms of  Jacobi elliptic functions sn or cn or dn. In Zhang [20], an algorithm is devised to derive exact traveling wave solutions of nonlinear DDEs by means of  Jacobi elliptic functions. Gepreel and Shehata [22] put a direct method to construct the rational Jacobi elliptic solution for nonlinear DDEs. In this paper, we modified the direct method which was disscussed in [22]. When we are constructing the solutions of nonlinear DDEs we neglected the case βi=0   because if  βi=0, we get the same solution which was discussed  by [22]. The Jacobi elliptic functions could be generated into the hyperbolic functions when m tends to one in the other hand, they are generated into trigonometrical functions when m tends to zero as shown in Table 1.Table 1
The relation between the Jacobi elliptic functions and the trigonometrical functions, hyperbolic functions.
m
s
n (ξ,m)
c
n (ξ,m)
d
n (ξ,m)
n
s (ξ,m)
c
s (ξ,m)
d
s (ξ,m)
0
sin
(
ξ
)
cos (ξ)
1
csc (ξ)
cot (ξ)
csc (ξ)
1
tanh(ξ)
sech(ξ)
sech(ξ)
coth(ξ)
csch(ξ)
csch(ξ)


## 4. Conclusion

In this paper, we modified the rational Jacobi elliptic functions method to calculate some new exact solutions  for the nonlinear difference  differential equations via the   lattice equation, the discrete  nonlinear Schrodinger equation with a saturable  nonlinearity, the discrete nonlinear Klein-Gordon equation, and the quintic discrete nonlinear Schrodinger  equation.  As a result, many new and more rational Jacobi elliptic solutions are obtained, from which hyperbolic function solutions and trigonometric function solutions are derived when the modulim→1 and m→0.



---

*Source: 427479-2012-10-17.xml*