# On a Theory for Analysing Second-Order Systems of Ordinary Discrete Equations

Authors: J. J. H. Bashingwa; A. H. Kara
Journal: Journal of Mathematics
 (2019)

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## Abstract

We present geometric based methods for solving systems of discrete or difference equations and introduce a technique for finding conservation laws for such systems.


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## Body

## 1. Introduction

Depending on a model being studied, some physical laws may be described by differential equations (DEs). Lie group theory provides us with powerful tools for obtaining analytical solutions of such equations [1]. Over the last 30 years, a considerable amount of work has been invested into applying Lie’s theory to solve and classify difference equations (ΔEs) (see [2–6] and references therein).The use of symmetry methods for ordinary difference equations(OΔEs) has been introduced by Maeda [2]. He showed that the resulting linearized symmetry condition (LSC) amounts to a set of functional equations which is hard to solve in general. Hydon introduced a technique to solve LSC and obtain symmetries in closed form by repetitive differentiations [5]. For example, a full classification of second-order OΔEs according to their point symmetries exists in the literature [6]. However, for systems of difference equations (SΔEs), most of results are based on induction methods (see [7–9] and references therein). In this paper, we present a method for solving SΔEs using their underlying symmetry.It has been proved in [5] that every second-order linear homogeneous OΔE has an eight-dimension Lie algebra isomorphic to sl(3). This is not valid for second-order system of difference equations (SΔEs). We shall prove this in Section 3.


## 2. Groundwork

Let us consider anN-th order system of rΔEs(1)xn+Ni=ωin,xn1,…,xnr,xn+11,…,xn+1r,…,xn+N-11,…,xn+N-1r,i=1,…,r.We assume that for each ωi there exists at least one xnj(i,j=1,…,r) such that ∂ωi/∂xnj≠0.Consider a point transformation(2)Γϵ:X↦X^X,ϵwhere X=(xn1,…,xnr) are continuous variables. Γ will be called one-parameter Lie group of transformations if it satisfies the following properties:(i)
Γ0 is the identity map, i.e., X^=X for ϵ=0.(ii)
ΓμΓν=Γμ+ν for every μ and ν close to 0.(iii)
Eachx^ni can be expanded as a Taylor series in a neighbourhood of ϵ=0.Therefore, we have(3)x^n+ji=xn+ji+ϵSjQin,xn1,…,xnr,xn+11,…,xn+1r,…,xn+N-11,…,xn+N-1r+Oϵ2where Qi are continuous functions which we shall refer to as characteristics, i=1,…,r,j=1,…,N, and S is the “shift” operator. It is defined as follows:(4)S:n↦n+1,Skxni=xn+kiWe define the discrete differentiation operator as follows:(5)Δ=S-Idwhere Id is the identity operator.The symmetry condition for theSΔEs (1) is(6)x^n+Ni=ωin,x^n1,…,x^nr,x^n+11,…,x^n+1r,…,x^n+N-11,…,x^n+N-1r,i=1,…,r.whenever (1) holds.Lie symmetries are obtained by linearizing the symmetry condition (6) about the identity. We have the following system of linearized symmetry condition (SLSC):(7)SNQi-Xωi=0,i=1,…,r.where the symmetry generator X is given by(8)X=∑j=0N-1∑i=1rSjQi∂∂xn+jiDefinition 1.
A functionwn is invariant function under the Lie group of transformations Γ if (9)Xwn=0,wherewn can be found by solving the characteristic equation(10)dxn1Q1=⋯=dxnrQr=dxn+11SQ1=⋯=dxn+1rSQr=⋯.=dxn+N-11SN-1Q1=⋯=dxn+N-1rSN-1Qr=wn0Theorem 2.
The discrete differential operatorΔ in (5) and the generator of symmetry X in (8) commute.Proof.
We prove the theorem forN=1 any generalisation is straightforward.(11)X,ΔFn,Xn=XΔFn,Xn-ΔXFn,xni,Xn=x1,x2,…,xr=XFn+1,Xn+1-Fn,Xn-Δ∑i=1rQi∂∂xniFn,Xn=∑i=1rSQi∂∂xn+1iFn+1,Xn+1-Qi∂∂xniFn,Xn-∑i=1rSQi∂∂xniFn,Xn-Qi∂∂xniFn,Xn=0Corollary 3.
For each invariantwn, Swn is also an invariant.Proof.
We haveX{Δ(wn)}=Δ{Xwn}=0.
Equivalently,X{(S-Id)wn}=X{S(wn)}-Xwn=0
orX(Swn)=0.We shall use this corollary for reductions in Section3.A first integral for the system (1) is a quantity ϕ(n,xn,yn,xn+1,yn+1) such that(12)Δϕn,xn,yn,xn+1,yn+1=0whenever (1) holds.In Section5, we shall use the condition (12) to develop a constructive technique for obtaining first integrals.Remark 4.
In this paper, we shall consider Lie point symmetry; i.e, the characteristics are given byQi(n,xn1,…,xnr).We refer the reader to [1] for more information on symmetry methods for differential equations.


## 3. Symmetries and Reductions

### 3.1. Finding Characteristics

Consider a second-order system of 2ΔEs(13)xn+2=ω1n,xn,yn,xn+1,yn+1,yn+2=ω2n,xn,yn,xn+1,yn+1.We assume that ∂ω1/∂xn≠0 or ∂ω1/∂yn≠0 and ∂ω2/∂xn≠0 or ∂ω2/∂yn≠0, so the system is of second order.The SLCS (7) reduces to(14)S2Q1-Q1ω1,xn-Q2ω1,yn-SQ1ω1,xn+1-SQ2ω1,yn+1=0,(15)S2Q2-Q1ω2,xn-Q2ω2,yn-SQ1ω2,xn+1-SQ2ω2,yn+1=0.where g,x=∂g/∂x,Q1=Q1(n,xn,yn), and Q2=Q2(n,xn,yn).The functional equations (14) and (15) contain functions Q1 and Q2 with different pairs of arguments making them difficult to solve. For concreteness, if, for instance, the discrete variable n stands for “state” in physics, Q1(n,xn,yn) and S(Q1)≡Q1(n+1,xn+1,yn+1) belong to two different states.To overcome this, we proceed as follows.Step 1 (elimination ofS2(Q1) and S2(Q2)).
We differentiate (total differentiation) (14) and (15) with respect to xn and yn, respectively, keeping ω1 and ω2 fixed. Here, we take xn+1 as function of xn,yn,yn+1,ω1,ω2 and yn+1 as function of xn,yn,xn+1,ω1,ω2.The total derivative operators are given by(16)ddxn=∂∂xn+∂xn+1∂xn∂∂xn+1+∂yn+1∂xn∂∂yn+1+⋯ddyn=∂∂yn+∂xn+1∂yn∂∂xn+1+∂yn+1∂yn∂∂yn+1+⋯In this case, this is simplified to(17)ddxn=∂∂xn-ω1,xnω1,xn+1+ω2,xnω2,xn+1∂∂xn+1-ω1,xnω1,yn+1+ω2,xnω2,yn+1∂∂yn+1(18)ddyn=∂∂yn-ω1,ynω1,xn+1+ω2,ynω2,xn+1∂∂xn+1-ω1,ynω1,yn+1+ω2,ynω2,yn+1∂∂yn+1So we apply the operator (17) to (14) and (18) to (15) keeping ω1 and ω2 fixed. This leads to the determining system(19)Q1ω1,xn+Q2ω1,yn,xn+SQ1ω1,xn+1,xn+SQ2ω1,yn+1,xn-ω1,xnω1,xn+1+ω2,xnω2,xn+1Q1ω1,xn+Q2ω1,yn,xn+1-ω1,xnω1,xn+1+ω2,xnω2,xn+1SQ1ω1,xn+1,xn+1+SQ2ω1,yn+1,xn+1+SQ1,xn+1ω1,xn+1+SQ2,xn+1ω1,yn+1-ω1,xnω1,yn+1+ω2,xnω2,yn+1Q1ω1,xn+Q2ω1,ynyn+1-ω1,xnω1,yn+1+ω2,xnω2,yn+1SQ1ω1,xn+1,yn+1+SQ2ω1,yn+1,yn+1+SQ1,yn+1ω1,xn+1+SQ2,yn+1ω1,yn+1=0,(20)Q1ω2,xn+Q2ω2,yn,yn+SQ1ω2,xn+1,yn+SQ2ω2,yn+1,yn-ω1,ynω1,xn+1+ω2,ynω2,xn+1Q1ω2,xn+Q2ω2,yn,xn+1-ω1,ynω1,xn+1+ω2,ynω2,xn+1SQ1ω2,xn+1,xn+1+SQ2ω2,yn+1,xn+1+SQ1,xn+1ω2,xn+1+SQ2,xn+1ω2,yn+1-ω1,ynω1,yn+1+ω2,ynω2,yn+1Q1ω2,xn+Q2ω2,ynyn+1-ω1,ynω1,yn+1+ω2,ynω2,yn+1SQ1ω2,xn+1,yn+1+SQ2ω2,yn+1,yn+1+SQ1,yn+1ω2,xn+1+SQ2,yn+1ω2,yn+1=0.Step 2 (elimination ofS(Q1) and S(Q2)).
We now differentiate (19) and (20) with respect to xn and yn, respectively, keeping xn+1 and yn+1 fixed. This means that we apply the operator ∂/∂xn on (19) and ∂/∂yn on (20). For a second-order SΔEs we need at most to differentiate four times. After separating with respect to xn+1 and yn+1 the resulting equations, we obtain a system of DEs in Q1 and Q2 which is solvable by hand or by using a computer algebra package.Step 3 (explicit form of constant of integration).
When integrating in Step2 to obtain the characteristics Q1 and Q2, we have constant of integration which appears to be functions of n. To obtain their explicit form, we need to substitute the results obtained in Step 2 in (19) and (20). If we do not succeed in obtaining all the constant of integration, we need further substitution in the SLSC (14) and (15).


### 3.2. Reductions

Consider a second-orderSΔEs(21)xn+2=ω1n,xn,yn,xn+1,yn+1,yn+2=ω2n,xn,yn,xn+1,yn+1and its symmetry generator(22)X=Q1∂∂xn+Q2∂∂yn+SQ1∂∂xn+1+SQ2∂∂yn+1The method of characteristics for partial differential equations (PDEs)(23)dxnQ1=dynQ2=dxn+1Q1=dyn+1Q2=Wn0leads to three independent constants of integration K1,K2,K3. Each invariant under X is function of those constant. Wn=f(K1,K2,K3).For second-order systems, two invariants suffice to do reduction of the systems.Let(24)un=f1n,xn,yn,xn+1,yn+1,vn=f2n,xn,yn,xn+1,yn+1be the invariants functions under X. We choose them in a way that the Jacobian is nonzero.(25)∂f1,f2∂xn+1,yn+1≠0That is, (24) can be inverted as follows:(26)xn+1=g1n,xn,yn,un,vn,yn+1=g2n,xn,yn,un,vnBy Corollary 3, Svn and Sun are also invariant functions. Therefore, the solution of (21) satisfies(27)un+1=Ω1un,vn,vn+1=Ω2un,vn(27) is a first-order SΔEs which can be solved by further reductions or by using computer algebra software (maple, Mathematica,…) for linear systems. Note that there exist some first-order systems which cannot be solved analytically.The general solution is(28)un=un,C1,C2,vn=vn,C1,C2for some constant C1,C2.So the second-order system (21) is equivalent to the first-order system obtained by substituting (28) in (26):(29)xn+1=g1n,xn,yn,un,C1,C2,vn,C1,C2,yn+1=g2n,xn,yn,un,C1,C2,vn,C1,C2(29) also admits the symmetries generated by X. The best way to integrate any first-order analytic ΔE is to use its canonical coordinates [10],(30)Tn=Tn,xn,yn,which satisfy(31)XTn=1The obvious choice of canonical coordinates is (see [10])(32)tn=∫dxnQ1n,xn,yn,g1n,xn,yn,un,C1,C2,vn,C1,C2,g2n,xn,yn,un,C1,C2,vn,C1,C2,sn=∫dynQ2n,xn,yn,g1n,xn,yn,un,C1,C2,vn,C1,C2,g2n,xn,yn,un,C1,C2,vn,C1,C2


### 3.3. Applications

#### 3.3.1. Example 1

Consider the most general homogeneous second-order linear system of difference equations(33)xn+2=a1nxn+a2nyn+a3nxn+1+a4nyn+1,yn+2=b1nxn+b2nyn+b3nxn+1+b4nyn+1where ai(n),bi(n),i=1,…,4 are arbitrary functions.One can readily verify that the determining system (19) and (20) amounts to(34)Q1,xnxn=Q2,xnxn=0,Q1,ynyn=Q2,ynyn=0Therefore(35)Q1n,xn,yn=C1xn+C2yn+F1n,Q2n,xn,yn=C3xn+C4yn+F2nwhere Ci,i=1,…,4 are constants.The characteristics in (35) must satisfy the SLSC (14) and (15). Hence, we have(36)F1n+2-a1nF1n+a2nF2n+a3nF1n+1+a4nF2n+1=0,F2n+2-b1nF1n+b2nF2n+b3nF1n+1+b4nF2n+1=0.and(37)C1=C4,C2=C3=0So, (35) is simplified to(38)Q1n,xn,yn=C1xn+F1n,Q2n,xn,yn=C1yn+F2nThe first generator of symmetry for a second-order homogeneous linear system (36) is the scaling symmetry given by(39)X=xn∂xn+yn∂ynThe system (36), which governs the remaining generators of the Lie point symmetry for the system (33), is of second order in F1 and F2. Its general solution is(40)F1n=g1n,K1,K2,K3,K4,F2n=g2n,K1,K2,K3,K4where K1,i=1,…,4 are constants.So, the most large Lie algebra of symmetry generators which can be obtained from a homogeneous second-order system of 2 difference equations has dimension five.For clarification let us considera 1 ( n ) = a 3 ( n ) = a 4 ( n ) = 0 ,   a 2 ( n ) = 1 and b2(n)=b3(n)=b4(n)=0,b1(n)=1. The system (33) becomes (41)xn+2=yn,yn+2=xnThe system which governs the remaining generators of the Lie point symmetry in this case is given by(42)F1n+2-F2n=0,F2n+2-F1n=0The general solutions for this system will be(43)F1n=C11+-1n+in+-in4+C21--1n-iin+i-in4+C31+-1n-in--in4C41--1n+iin-i-in4F2n=C31+-1n+in+-in4+C41--1n-iin+i-in4+C11+-1n-in--in4C21--1n+iin-i-in4Therefore we have 5 generators of the Lie point symmetry spanned by(44)X0=xn∂xn+yn∂ynX1=1+-1n+in+-in4∂xn+1+-1n-in--in4∂ynX2=1--1n-iin+i-in4∂xn+1--1n+iin-i-in4∂ynX3=1+-1n-in--in4∂xn+1+-1n+in+-in4∂ynX4=1--1n+iin-i-in4∂xn+1--1n-iin+i-in4∂yn


#### 3.3.2. Example 2

Consider the system(45)xn+2=xnyn+1+1xn+yn+1yn+2=ynxn+1+1yn+xn+1(45) is a special case of systems investigated in [11], where the author looked at the stability of the systems.We choose the ansatzQ1(n,xn),Q2(n,yn).The determining system (19) and (20) amounts to(46)-Q2,ynxn+12yn2+SQ1,xn+1xn+12yn2+2Q2xn+12yn-2SQ1yn2xn+1+Q2,ynxn+12+Q2,ynyn2-SQ1,xn+1xn+12-SQ1,xn+1yn2-2Q2yn+2SQ1xn+1-Q2,yn+SQ1,xn+1=0(47)-Q1,xnxn2yn+12+SQ2,yn+1xn2yn+12+2Q1yn+12xn-2SQ2xn2yn+1+Q1,xnxn2+Q1,xnyn+12-SQ2,yn+1xn2-SQ2,yn+1yn+12-2Q1xn+2SQ2yn+1-Q1,xn+SQ2,yn+1=0Differentiating twice (46) with respect to xn and twice (47) with respect to yn keeping xn+1 and yn+1 fixed we obtain, after separating with respect to xn+1 and yn+1, the following system of Des:(48)Q1′′+xnQ1′′′+2Q1xn2-2Q1′xn+Q1′′xn2-Q1′′′xn=0Q2′′+ynQ2′′′+2Q2yn2-2Q2′yn+Q2′′yn2-Q2′′′yn=0whose most general solutions are(49)Q1n,xn=F1nxn+F2nxn2-1ln⁡xn+1xn-1+F3nxn2-1Q2n,yn=F4nyn+F5nyn2-1ln⁡yn+1yn-1+F6nyn2-1To obtain the nature of functions F1,…,F6 we substitute (49) in (46) and (47). After separating with respect to xn,xn+1,yn and yn+1 we get the following SΔEs:(50)-4F2n-2F4n+1+4F5n+1-2F1n=04F2n-2F4n+1-4F5n+1+2F1n=0-4F5n-2F1n+1+4F2n+1-2F4n=04F5n-2F1n+1-4F2n+1+2F4n=0whose solutions are(51)F1n=F4n=0F2n=C1+-1nC2F5n=C1--1nC2The remaining unknown functions F3(n) and F6(n) are determined by substituting (51) and (49) into the SLSC (14) and (15). This leads to the SΔEs(52)F3n-F3n+2+F6n+1=0F6n+F3n+1-F6n+2=0The general solutions to (52) are given by(53)F3n=1+-1n5-1+5n-1+1+5n-1C32n--1+5n-1+5nC62n+1-1+-1n5-1+5n-1-1+5n-1C52n--1+5n+1+5nC42n+1F6n=1+-1n5-1+5n-1+1+5n-1C52n--1+5n-1+5nC42n+1-1+-1n5-1+5n-1-1+5n-1C32n--1+5n+1+5nC62n+1where C1,…,C6 are arbitrary constants. It follows that the characteristics are given by(54)Q1=C1+-1nC2xn2-1ln⁡xn+1xn-1+1+-1n5-1+5n-1+1+5n-1C32n--1+5n-1+5nC62n+1-1+-1n5-1+5n-1-1+5n-1C52n--1+5n+1+5nC42n+1xn2-1Q2=C1--1nC2yn2-1ln⁡yn+1yn-1+1+-1n5-1+5n-1+1+5n-1C52n--1+5n-1+5nC42n+1-1+-1n5-1+5n-1-1+5n-1C32n--1+5n+1+5nC62n+1yn2-1Therefore, we have six generators of Lie point symmetry(55)X1=xn2-1ln⁡xn+1xn-1∂xn+yn2-1ln⁡yn+1yn-1∂ynX2=-1nxn2-1ln⁡xn+1xn-1∂xn--1nyn2-1ln⁡yn+1yn-1∂ynX3=1+-1n5-1+5n-1+1+5n-12nxn2-1∂xn+-1+-1n5-1+5n-1-1+5n-12nyn2-1∂ynX4=-1+-1n5-1+5n+1+5n2n+1xn2-1∂xn+1+-1n5-1+5n-1+5n2n+1yn2-1∂ynX5=-1+-1n5-1+5n-1-1+5n-12nxn2-1∂xn+1+-1n5-1+5n-1+1+5n-12nyn2-1∂ynX6=1+-1n5-1+5n-1+5n2n+1xn2-1∂xn+-1+-1n5-1+5n+1+5n2n+1yn2-1∂ynEach generator in (55) can be used to reduce the order of (45).Let us considerX1. By the characteristic method for Partial Differential Equations, the invariants are given by following equation:(56)dxnxn2-1lnxn+1/xn-1=dynyn2-1ln⁡yn+1/yn-1=dxn+1xn+12-1lnxn+1+1/xn+1-1=dyn+1yn+12-1ln⁡yn+1+1/yn+1-1=Vn0We get(57)C1=ln⁡xn+1/xn-1ln⁡yn+1+1/yn+1-1,C2=ln⁡xn+1/xn-1ln⁡yn+1/yn-1,C3=ln⁡xn+1/xn-1ln⁡xn+1+1/xn+1-1,Vn=fC1,C2,C3where C1,C2,C3 are constants.If we choosef(C1,C2,C3)=C1, we have(58)un=ln⁡xn+1/xn-1ln⁡yn+1+1/yn+1-1and if we choose f(C1,C2,C3)=C3/C2, we have(59)vn=ln⁡yn+1/yn-1ln⁡xn+1+1/xn+1-1From (58) and (59)we deduce(60)un+1=11+vn,vn+1=11+unLet us now consider the generator X3. The resulting invariants are(61)vn=xn-1/xn+1αnyn+1-1/yn+1+1,un=yn-1/yn+1βnxn+1-1/xn+1+1where(62)αn=1+5n--1+5n21+5n-1+-1+5n-1,βn=1+5n+-1+5n21+5n-1--1+5n-1Note also the relationship between them(63)αn+1-1=1βn,βn+1-1=1αnFrom (61), we deduce the following relation:(64)vn+1=1un1/βn,un+1=1vn1/αnOne can readily check that the general solution to (64) is given by(65)un=1+-1n2u0∏k=0n-2/21/β2kα2k+1+1--1n2v0∏k=0n-1/21/α2k∏k=0n-3/21/β2k+1-1vn=1+-1n2v0∏k=0n-2/21/α2kβ2k+1+1--1n2u0∏k=0n-1/21/β2k∏k=0n-3/21/α2k+1-1where αn and βn are defined in (62).From (61) we obtain(66)vnyn+1-1yn+1+1=xn-1xn+1αn,unxn+1-1xn+1+1=yn-1yn+1βnwhich is a first-order system after substitution of un,vn by the results given in (65). Its solutions can be obtained by using the following canonical coordinates:(67)sn=ln⁡xn-1xn+1,tn=ln⁡yn-1yn+1This leads to the the following linear system with variable coefficients:(68)tn+1=αnsn+γnsn+1=βntn+φnwhere -γn=ln⁡un and -φn=ln⁡vn.The latter is a linear first-order system with variable coefficients. Its general solution is(69)s2n=∏k=0n-1α2kβ2k+1s0+∑r=0n-1γ2r∏j=r+1n-1α2j∏j=rn-1β2j+1+∑r=0n-1φ2r+1∏j=r+1n-1α2jβ2j+1t2n=∏k=0n-1β2kα2k+1t0+∑r=0n-1φ2r∏j=r+1n-1β2j∏j=rn-1α2j+1+∑r=0n-1γ2r+1∏j=r+1n-1β2jα2j+1t2n+1=∏k=0nα2k∏k=0n-1β2k+1s0+∑r=0nγ2r∏j=r+1nα2j∏j=rn-1β2j+1+∑r=0n-1φ2r+1∏j=r+1nα2j∏j=r+1n-1β2j+1s2n+1=∏k=0nβ2k∏k=0n-1α2k+1t0+∑r=0nφ2r∏j=r+1nβ2j∏j=rn-1α2j+1+∑r=0n-1γ2r+1∏j=r+1nβ2j∏j=r+1n-1α2j+1The general solution of (45) is obtained by substituting (69) into (67).


## 3.1. Finding Characteristics

Consider a second-order system of 2ΔEs(13)xn+2=ω1n,xn,yn,xn+1,yn+1,yn+2=ω2n,xn,yn,xn+1,yn+1.We assume that ∂ω1/∂xn≠0 or ∂ω1/∂yn≠0 and ∂ω2/∂xn≠0 or ∂ω2/∂yn≠0, so the system is of second order.The SLCS (7) reduces to(14)S2Q1-Q1ω1,xn-Q2ω1,yn-SQ1ω1,xn+1-SQ2ω1,yn+1=0,(15)S2Q2-Q1ω2,xn-Q2ω2,yn-SQ1ω2,xn+1-SQ2ω2,yn+1=0.where g,x=∂g/∂x,Q1=Q1(n,xn,yn), and Q2=Q2(n,xn,yn).The functional equations (14) and (15) contain functions Q1 and Q2 with different pairs of arguments making them difficult to solve. For concreteness, if, for instance, the discrete variable n stands for “state” in physics, Q1(n,xn,yn) and S(Q1)≡Q1(n+1,xn+1,yn+1) belong to two different states.To overcome this, we proceed as follows.Step 1 (elimination ofS2(Q1) and S2(Q2)).
We differentiate (total differentiation) (14) and (15) with respect to xn and yn, respectively, keeping ω1 and ω2 fixed. Here, we take xn+1 as function of xn,yn,yn+1,ω1,ω2 and yn+1 as function of xn,yn,xn+1,ω1,ω2.The total derivative operators are given by(16)ddxn=∂∂xn+∂xn+1∂xn∂∂xn+1+∂yn+1∂xn∂∂yn+1+⋯ddyn=∂∂yn+∂xn+1∂yn∂∂xn+1+∂yn+1∂yn∂∂yn+1+⋯In this case, this is simplified to(17)ddxn=∂∂xn-ω1,xnω1,xn+1+ω2,xnω2,xn+1∂∂xn+1-ω1,xnω1,yn+1+ω2,xnω2,yn+1∂∂yn+1(18)ddyn=∂∂yn-ω1,ynω1,xn+1+ω2,ynω2,xn+1∂∂xn+1-ω1,ynω1,yn+1+ω2,ynω2,yn+1∂∂yn+1So we apply the operator (17) to (14) and (18) to (15) keeping ω1 and ω2 fixed. This leads to the determining system(19)Q1ω1,xn+Q2ω1,yn,xn+SQ1ω1,xn+1,xn+SQ2ω1,yn+1,xn-ω1,xnω1,xn+1+ω2,xnω2,xn+1Q1ω1,xn+Q2ω1,yn,xn+1-ω1,xnω1,xn+1+ω2,xnω2,xn+1SQ1ω1,xn+1,xn+1+SQ2ω1,yn+1,xn+1+SQ1,xn+1ω1,xn+1+SQ2,xn+1ω1,yn+1-ω1,xnω1,yn+1+ω2,xnω2,yn+1Q1ω1,xn+Q2ω1,ynyn+1-ω1,xnω1,yn+1+ω2,xnω2,yn+1SQ1ω1,xn+1,yn+1+SQ2ω1,yn+1,yn+1+SQ1,yn+1ω1,xn+1+SQ2,yn+1ω1,yn+1=0,(20)Q1ω2,xn+Q2ω2,yn,yn+SQ1ω2,xn+1,yn+SQ2ω2,yn+1,yn-ω1,ynω1,xn+1+ω2,ynω2,xn+1Q1ω2,xn+Q2ω2,yn,xn+1-ω1,ynω1,xn+1+ω2,ynω2,xn+1SQ1ω2,xn+1,xn+1+SQ2ω2,yn+1,xn+1+SQ1,xn+1ω2,xn+1+SQ2,xn+1ω2,yn+1-ω1,ynω1,yn+1+ω2,ynω2,yn+1Q1ω2,xn+Q2ω2,ynyn+1-ω1,ynω1,yn+1+ω2,ynω2,yn+1SQ1ω2,xn+1,yn+1+SQ2ω2,yn+1,yn+1+SQ1,yn+1ω2,xn+1+SQ2,yn+1ω2,yn+1=0.Step 2 (elimination ofS(Q1) and S(Q2)).
We now differentiate (19) and (20) with respect to xn and yn, respectively, keeping xn+1 and yn+1 fixed. This means that we apply the operator ∂/∂xn on (19) and ∂/∂yn on (20). For a second-order SΔEs we need at most to differentiate four times. After separating with respect to xn+1 and yn+1 the resulting equations, we obtain a system of DEs in Q1 and Q2 which is solvable by hand or by using a computer algebra package.Step 3 (explicit form of constant of integration).
When integrating in Step2 to obtain the characteristics Q1 and Q2, we have constant of integration which appears to be functions of n. To obtain their explicit form, we need to substitute the results obtained in Step 2 in (19) and (20). If we do not succeed in obtaining all the constant of integration, we need further substitution in the SLSC (14) and (15).


## 3.2. Reductions

Consider a second-orderSΔEs(21)xn+2=ω1n,xn,yn,xn+1,yn+1,yn+2=ω2n,xn,yn,xn+1,yn+1and its symmetry generator(22)X=Q1∂∂xn+Q2∂∂yn+SQ1∂∂xn+1+SQ2∂∂yn+1The method of characteristics for partial differential equations (PDEs)(23)dxnQ1=dynQ2=dxn+1Q1=dyn+1Q2=Wn0leads to three independent constants of integration K1,K2,K3. Each invariant under X is function of those constant. Wn=f(K1,K2,K3).For second-order systems, two invariants suffice to do reduction of the systems.Let(24)un=f1n,xn,yn,xn+1,yn+1,vn=f2n,xn,yn,xn+1,yn+1be the invariants functions under X. We choose them in a way that the Jacobian is nonzero.(25)∂f1,f2∂xn+1,yn+1≠0That is, (24) can be inverted as follows:(26)xn+1=g1n,xn,yn,un,vn,yn+1=g2n,xn,yn,un,vnBy Corollary 3, Svn and Sun are also invariant functions. Therefore, the solution of (21) satisfies(27)un+1=Ω1un,vn,vn+1=Ω2un,vn(27) is a first-order SΔEs which can be solved by further reductions or by using computer algebra software (maple, Mathematica,…) for linear systems. Note that there exist some first-order systems which cannot be solved analytically.The general solution is(28)un=un,C1,C2,vn=vn,C1,C2for some constant C1,C2.So the second-order system (21) is equivalent to the first-order system obtained by substituting (28) in (26):(29)xn+1=g1n,xn,yn,un,C1,C2,vn,C1,C2,yn+1=g2n,xn,yn,un,C1,C2,vn,C1,C2(29) also admits the symmetries generated by X. The best way to integrate any first-order analytic ΔE is to use its canonical coordinates [10],(30)Tn=Tn,xn,yn,which satisfy(31)XTn=1The obvious choice of canonical coordinates is (see [10])(32)tn=∫dxnQ1n,xn,yn,g1n,xn,yn,un,C1,C2,vn,C1,C2,g2n,xn,yn,un,C1,C2,vn,C1,C2,sn=∫dynQ2n,xn,yn,g1n,xn,yn,un,C1,C2,vn,C1,C2,g2n,xn,yn,un,C1,C2,vn,C1,C2


## 3.3. Applications

### 3.3.1. Example 1

Consider the most general homogeneous second-order linear system of difference equations(33)xn+2=a1nxn+a2nyn+a3nxn+1+a4nyn+1,yn+2=b1nxn+b2nyn+b3nxn+1+b4nyn+1where ai(n),bi(n),i=1,…,4 are arbitrary functions.One can readily verify that the determining system (19) and (20) amounts to(34)Q1,xnxn=Q2,xnxn=0,Q1,ynyn=Q2,ynyn=0Therefore(35)Q1n,xn,yn=C1xn+C2yn+F1n,Q2n,xn,yn=C3xn+C4yn+F2nwhere Ci,i=1,…,4 are constants.The characteristics in (35) must satisfy the SLSC (14) and (15). Hence, we have(36)F1n+2-a1nF1n+a2nF2n+a3nF1n+1+a4nF2n+1=0,F2n+2-b1nF1n+b2nF2n+b3nF1n+1+b4nF2n+1=0.and(37)C1=C4,C2=C3=0So, (35) is simplified to(38)Q1n,xn,yn=C1xn+F1n,Q2n,xn,yn=C1yn+F2nThe first generator of symmetry for a second-order homogeneous linear system (36) is the scaling symmetry given by(39)X=xn∂xn+yn∂ynThe system (36), which governs the remaining generators of the Lie point symmetry for the system (33), is of second order in F1 and F2. Its general solution is(40)F1n=g1n,K1,K2,K3,K4,F2n=g2n,K1,K2,K3,K4where K1,i=1,…,4 are constants.So, the most large Lie algebra of symmetry generators which can be obtained from a homogeneous second-order system of 2 difference equations has dimension five.For clarification let us considera 1 ( n ) = a 3 ( n ) = a 4 ( n ) = 0 ,   a 2 ( n ) = 1 and b2(n)=b3(n)=b4(n)=0,b1(n)=1. The system (33) becomes (41)xn+2=yn,yn+2=xnThe system which governs the remaining generators of the Lie point symmetry in this case is given by(42)F1n+2-F2n=0,F2n+2-F1n=0The general solutions for this system will be(43)F1n=C11+-1n+in+-in4+C21--1n-iin+i-in4+C31+-1n-in--in4C41--1n+iin-i-in4F2n=C31+-1n+in+-in4+C41--1n-iin+i-in4+C11+-1n-in--in4C21--1n+iin-i-in4Therefore we have 5 generators of the Lie point symmetry spanned by(44)X0=xn∂xn+yn∂ynX1=1+-1n+in+-in4∂xn+1+-1n-in--in4∂ynX2=1--1n-iin+i-in4∂xn+1--1n+iin-i-in4∂ynX3=1+-1n-in--in4∂xn+1+-1n+in+-in4∂ynX4=1--1n+iin-i-in4∂xn+1--1n-iin+i-in4∂yn


### 3.3.2. Example 2

Consider the system(45)xn+2=xnyn+1+1xn+yn+1yn+2=ynxn+1+1yn+xn+1(45) is a special case of systems investigated in [11], where the author looked at the stability of the systems.We choose the ansatzQ1(n,xn),Q2(n,yn).The determining system (19) and (20) amounts to(46)-Q2,ynxn+12yn2+SQ1,xn+1xn+12yn2+2Q2xn+12yn-2SQ1yn2xn+1+Q2,ynxn+12+Q2,ynyn2-SQ1,xn+1xn+12-SQ1,xn+1yn2-2Q2yn+2SQ1xn+1-Q2,yn+SQ1,xn+1=0(47)-Q1,xnxn2yn+12+SQ2,yn+1xn2yn+12+2Q1yn+12xn-2SQ2xn2yn+1+Q1,xnxn2+Q1,xnyn+12-SQ2,yn+1xn2-SQ2,yn+1yn+12-2Q1xn+2SQ2yn+1-Q1,xn+SQ2,yn+1=0Differentiating twice (46) with respect to xn and twice (47) with respect to yn keeping xn+1 and yn+1 fixed we obtain, after separating with respect to xn+1 and yn+1, the following system of Des:(48)Q1′′+xnQ1′′′+2Q1xn2-2Q1′xn+Q1′′xn2-Q1′′′xn=0Q2′′+ynQ2′′′+2Q2yn2-2Q2′yn+Q2′′yn2-Q2′′′yn=0whose most general solutions are(49)Q1n,xn=F1nxn+F2nxn2-1ln⁡xn+1xn-1+F3nxn2-1Q2n,yn=F4nyn+F5nyn2-1ln⁡yn+1yn-1+F6nyn2-1To obtain the nature of functions F1,…,F6 we substitute (49) in (46) and (47). After separating with respect to xn,xn+1,yn and yn+1 we get the following SΔEs:(50)-4F2n-2F4n+1+4F5n+1-2F1n=04F2n-2F4n+1-4F5n+1+2F1n=0-4F5n-2F1n+1+4F2n+1-2F4n=04F5n-2F1n+1-4F2n+1+2F4n=0whose solutions are(51)F1n=F4n=0F2n=C1+-1nC2F5n=C1--1nC2The remaining unknown functions F3(n) and F6(n) are determined by substituting (51) and (49) into the SLSC (14) and (15). This leads to the SΔEs(52)F3n-F3n+2+F6n+1=0F6n+F3n+1-F6n+2=0The general solutions to (52) are given by(53)F3n=1+-1n5-1+5n-1+1+5n-1C32n--1+5n-1+5nC62n+1-1+-1n5-1+5n-1-1+5n-1C52n--1+5n+1+5nC42n+1F6n=1+-1n5-1+5n-1+1+5n-1C52n--1+5n-1+5nC42n+1-1+-1n5-1+5n-1-1+5n-1C32n--1+5n+1+5nC62n+1where C1,…,C6 are arbitrary constants. It follows that the characteristics are given by(54)Q1=C1+-1nC2xn2-1ln⁡xn+1xn-1+1+-1n5-1+5n-1+1+5n-1C32n--1+5n-1+5nC62n+1-1+-1n5-1+5n-1-1+5n-1C52n--1+5n+1+5nC42n+1xn2-1Q2=C1--1nC2yn2-1ln⁡yn+1yn-1+1+-1n5-1+5n-1+1+5n-1C52n--1+5n-1+5nC42n+1-1+-1n5-1+5n-1-1+5n-1C32n--1+5n+1+5nC62n+1yn2-1Therefore, we have six generators of Lie point symmetry(55)X1=xn2-1ln⁡xn+1xn-1∂xn+yn2-1ln⁡yn+1yn-1∂ynX2=-1nxn2-1ln⁡xn+1xn-1∂xn--1nyn2-1ln⁡yn+1yn-1∂ynX3=1+-1n5-1+5n-1+1+5n-12nxn2-1∂xn+-1+-1n5-1+5n-1-1+5n-12nyn2-1∂ynX4=-1+-1n5-1+5n+1+5n2n+1xn2-1∂xn+1+-1n5-1+5n-1+5n2n+1yn2-1∂ynX5=-1+-1n5-1+5n-1-1+5n-12nxn2-1∂xn+1+-1n5-1+5n-1+1+5n-12nyn2-1∂ynX6=1+-1n5-1+5n-1+5n2n+1xn2-1∂xn+-1+-1n5-1+5n+1+5n2n+1yn2-1∂ynEach generator in (55) can be used to reduce the order of (45).Let us considerX1. By the characteristic method for Partial Differential Equations, the invariants are given by following equation:(56)dxnxn2-1lnxn+1/xn-1=dynyn2-1ln⁡yn+1/yn-1=dxn+1xn+12-1lnxn+1+1/xn+1-1=dyn+1yn+12-1ln⁡yn+1+1/yn+1-1=Vn0We get(57)C1=ln⁡xn+1/xn-1ln⁡yn+1+1/yn+1-1,C2=ln⁡xn+1/xn-1ln⁡yn+1/yn-1,C3=ln⁡xn+1/xn-1ln⁡xn+1+1/xn+1-1,Vn=fC1,C2,C3where C1,C2,C3 are constants.If we choosef(C1,C2,C3)=C1, we have(58)un=ln⁡xn+1/xn-1ln⁡yn+1+1/yn+1-1and if we choose f(C1,C2,C3)=C3/C2, we have(59)vn=ln⁡yn+1/yn-1ln⁡xn+1+1/xn+1-1From (58) and (59)we deduce(60)un+1=11+vn,vn+1=11+unLet us now consider the generator X3. The resulting invariants are(61)vn=xn-1/xn+1αnyn+1-1/yn+1+1,un=yn-1/yn+1βnxn+1-1/xn+1+1where(62)αn=1+5n--1+5n21+5n-1+-1+5n-1,βn=1+5n+-1+5n21+5n-1--1+5n-1Note also the relationship between them(63)αn+1-1=1βn,βn+1-1=1αnFrom (61), we deduce the following relation:(64)vn+1=1un1/βn,un+1=1vn1/αnOne can readily check that the general solution to (64) is given by(65)un=1+-1n2u0∏k=0n-2/21/β2kα2k+1+1--1n2v0∏k=0n-1/21/α2k∏k=0n-3/21/β2k+1-1vn=1+-1n2v0∏k=0n-2/21/α2kβ2k+1+1--1n2u0∏k=0n-1/21/β2k∏k=0n-3/21/α2k+1-1where αn and βn are defined in (62).From (61) we obtain(66)vnyn+1-1yn+1+1=xn-1xn+1αn,unxn+1-1xn+1+1=yn-1yn+1βnwhich is a first-order system after substitution of un,vn by the results given in (65). Its solutions can be obtained by using the following canonical coordinates:(67)sn=ln⁡xn-1xn+1,tn=ln⁡yn-1yn+1This leads to the the following linear system with variable coefficients:(68)tn+1=αnsn+γnsn+1=βntn+φnwhere -γn=ln⁡un and -φn=ln⁡vn.The latter is a linear first-order system with variable coefficients. Its general solution is(69)s2n=∏k=0n-1α2kβ2k+1s0+∑r=0n-1γ2r∏j=r+1n-1α2j∏j=rn-1β2j+1+∑r=0n-1φ2r+1∏j=r+1n-1α2jβ2j+1t2n=∏k=0n-1β2kα2k+1t0+∑r=0n-1φ2r∏j=r+1n-1β2j∏j=rn-1α2j+1+∑r=0n-1γ2r+1∏j=r+1n-1β2jα2j+1t2n+1=∏k=0nα2k∏k=0n-1β2k+1s0+∑r=0nγ2r∏j=r+1nα2j∏j=rn-1β2j+1+∑r=0n-1φ2r+1∏j=r+1nα2j∏j=r+1n-1β2j+1s2n+1=∏k=0nβ2k∏k=0n-1α2k+1t0+∑r=0nφ2r∏j=r+1nβ2j∏j=rn-1α2j+1+∑r=0n-1γ2r+1∏j=r+1nβ2j∏j=r+1n-1α2j+1The general solution of (45) is obtained by substituting (69) into (67).


## 3.3.1. Example 1

Consider the most general homogeneous second-order linear system of difference equations(33)xn+2=a1nxn+a2nyn+a3nxn+1+a4nyn+1,yn+2=b1nxn+b2nyn+b3nxn+1+b4nyn+1where ai(n),bi(n),i=1,…,4 are arbitrary functions.One can readily verify that the determining system (19) and (20) amounts to(34)Q1,xnxn=Q2,xnxn=0,Q1,ynyn=Q2,ynyn=0Therefore(35)Q1n,xn,yn=C1xn+C2yn+F1n,Q2n,xn,yn=C3xn+C4yn+F2nwhere Ci,i=1,…,4 are constants.The characteristics in (35) must satisfy the SLSC (14) and (15). Hence, we have(36)F1n+2-a1nF1n+a2nF2n+a3nF1n+1+a4nF2n+1=0,F2n+2-b1nF1n+b2nF2n+b3nF1n+1+b4nF2n+1=0.and(37)C1=C4,C2=C3=0So, (35) is simplified to(38)Q1n,xn,yn=C1xn+F1n,Q2n,xn,yn=C1yn+F2nThe first generator of symmetry for a second-order homogeneous linear system (36) is the scaling symmetry given by(39)X=xn∂xn+yn∂ynThe system (36), which governs the remaining generators of the Lie point symmetry for the system (33), is of second order in F1 and F2. Its general solution is(40)F1n=g1n,K1,K2,K3,K4,F2n=g2n,K1,K2,K3,K4where K1,i=1,…,4 are constants.So, the most large Lie algebra of symmetry generators which can be obtained from a homogeneous second-order system of 2 difference equations has dimension five.For clarification let us considera 1 ( n ) = a 3 ( n ) = a 4 ( n ) = 0 ,   a 2 ( n ) = 1 and b2(n)=b3(n)=b4(n)=0,b1(n)=1. The system (33) becomes (41)xn+2=yn,yn+2=xnThe system which governs the remaining generators of the Lie point symmetry in this case is given by(42)F1n+2-F2n=0,F2n+2-F1n=0The general solutions for this system will be(43)F1n=C11+-1n+in+-in4+C21--1n-iin+i-in4+C31+-1n-in--in4C41--1n+iin-i-in4F2n=C31+-1n+in+-in4+C41--1n-iin+i-in4+C11+-1n-in--in4C21--1n+iin-i-in4Therefore we have 5 generators of the Lie point symmetry spanned by(44)X0=xn∂xn+yn∂ynX1=1+-1n+in+-in4∂xn+1+-1n-in--in4∂ynX2=1--1n-iin+i-in4∂xn+1--1n+iin-i-in4∂ynX3=1+-1n-in--in4∂xn+1+-1n+in+-in4∂ynX4=1--1n+iin-i-in4∂xn+1--1n-iin+i-in4∂yn


## 3.3.2. Example 2

Consider the system(45)xn+2=xnyn+1+1xn+yn+1yn+2=ynxn+1+1yn+xn+1(45) is a special case of systems investigated in [11], where the author looked at the stability of the systems.We choose the ansatzQ1(n,xn),Q2(n,yn).The determining system (19) and (20) amounts to(46)-Q2,ynxn+12yn2+SQ1,xn+1xn+12yn2+2Q2xn+12yn-2SQ1yn2xn+1+Q2,ynxn+12+Q2,ynyn2-SQ1,xn+1xn+12-SQ1,xn+1yn2-2Q2yn+2SQ1xn+1-Q2,yn+SQ1,xn+1=0(47)-Q1,xnxn2yn+12+SQ2,yn+1xn2yn+12+2Q1yn+12xn-2SQ2xn2yn+1+Q1,xnxn2+Q1,xnyn+12-SQ2,yn+1xn2-SQ2,yn+1yn+12-2Q1xn+2SQ2yn+1-Q1,xn+SQ2,yn+1=0Differentiating twice (46) with respect to xn and twice (47) with respect to yn keeping xn+1 and yn+1 fixed we obtain, after separating with respect to xn+1 and yn+1, the following system of Des:(48)Q1′′+xnQ1′′′+2Q1xn2-2Q1′xn+Q1′′xn2-Q1′′′xn=0Q2′′+ynQ2′′′+2Q2yn2-2Q2′yn+Q2′′yn2-Q2′′′yn=0whose most general solutions are(49)Q1n,xn=F1nxn+F2nxn2-1ln⁡xn+1xn-1+F3nxn2-1Q2n,yn=F4nyn+F5nyn2-1ln⁡yn+1yn-1+F6nyn2-1To obtain the nature of functions F1,…,F6 we substitute (49) in (46) and (47). After separating with respect to xn,xn+1,yn and yn+1 we get the following SΔEs:(50)-4F2n-2F4n+1+4F5n+1-2F1n=04F2n-2F4n+1-4F5n+1+2F1n=0-4F5n-2F1n+1+4F2n+1-2F4n=04F5n-2F1n+1-4F2n+1+2F4n=0whose solutions are(51)F1n=F4n=0F2n=C1+-1nC2F5n=C1--1nC2The remaining unknown functions F3(n) and F6(n) are determined by substituting (51) and (49) into the SLSC (14) and (15). This leads to the SΔEs(52)F3n-F3n+2+F6n+1=0F6n+F3n+1-F6n+2=0The general solutions to (52) are given by(53)F3n=1+-1n5-1+5n-1+1+5n-1C32n--1+5n-1+5nC62n+1-1+-1n5-1+5n-1-1+5n-1C52n--1+5n+1+5nC42n+1F6n=1+-1n5-1+5n-1+1+5n-1C52n--1+5n-1+5nC42n+1-1+-1n5-1+5n-1-1+5n-1C32n--1+5n+1+5nC62n+1where C1,…,C6 are arbitrary constants. It follows that the characteristics are given by(54)Q1=C1+-1nC2xn2-1ln⁡xn+1xn-1+1+-1n5-1+5n-1+1+5n-1C32n--1+5n-1+5nC62n+1-1+-1n5-1+5n-1-1+5n-1C52n--1+5n+1+5nC42n+1xn2-1Q2=C1--1nC2yn2-1ln⁡yn+1yn-1+1+-1n5-1+5n-1+1+5n-1C52n--1+5n-1+5nC42n+1-1+-1n5-1+5n-1-1+5n-1C32n--1+5n+1+5nC62n+1yn2-1Therefore, we have six generators of Lie point symmetry(55)X1=xn2-1ln⁡xn+1xn-1∂xn+yn2-1ln⁡yn+1yn-1∂ynX2=-1nxn2-1ln⁡xn+1xn-1∂xn--1nyn2-1ln⁡yn+1yn-1∂ynX3=1+-1n5-1+5n-1+1+5n-12nxn2-1∂xn+-1+-1n5-1+5n-1-1+5n-12nyn2-1∂ynX4=-1+-1n5-1+5n+1+5n2n+1xn2-1∂xn+1+-1n5-1+5n-1+5n2n+1yn2-1∂ynX5=-1+-1n5-1+5n-1-1+5n-12nxn2-1∂xn+1+-1n5-1+5n-1+1+5n-12nyn2-1∂ynX6=1+-1n5-1+5n-1+5n2n+1xn2-1∂xn+-1+-1n5-1+5n+1+5n2n+1yn2-1∂ynEach generator in (55) can be used to reduce the order of (45).Let us considerX1. By the characteristic method for Partial Differential Equations, the invariants are given by following equation:(56)dxnxn2-1lnxn+1/xn-1=dynyn2-1ln⁡yn+1/yn-1=dxn+1xn+12-1lnxn+1+1/xn+1-1=dyn+1yn+12-1ln⁡yn+1+1/yn+1-1=Vn0We get(57)C1=ln⁡xn+1/xn-1ln⁡yn+1+1/yn+1-1,C2=ln⁡xn+1/xn-1ln⁡yn+1/yn-1,C3=ln⁡xn+1/xn-1ln⁡xn+1+1/xn+1-1,Vn=fC1,C2,C3where C1,C2,C3 are constants.If we choosef(C1,C2,C3)=C1, we have(58)un=ln⁡xn+1/xn-1ln⁡yn+1+1/yn+1-1and if we choose f(C1,C2,C3)=C3/C2, we have(59)vn=ln⁡yn+1/yn-1ln⁡xn+1+1/xn+1-1From (58) and (59)we deduce(60)un+1=11+vn,vn+1=11+unLet us now consider the generator X3. The resulting invariants are(61)vn=xn-1/xn+1αnyn+1-1/yn+1+1,un=yn-1/yn+1βnxn+1-1/xn+1+1where(62)αn=1+5n--1+5n21+5n-1+-1+5n-1,βn=1+5n+-1+5n21+5n-1--1+5n-1Note also the relationship between them(63)αn+1-1=1βn,βn+1-1=1αnFrom (61), we deduce the following relation:(64)vn+1=1un1/βn,un+1=1vn1/αnOne can readily check that the general solution to (64) is given by(65)un=1+-1n2u0∏k=0n-2/21/β2kα2k+1+1--1n2v0∏k=0n-1/21/α2k∏k=0n-3/21/β2k+1-1vn=1+-1n2v0∏k=0n-2/21/α2kβ2k+1+1--1n2u0∏k=0n-1/21/β2k∏k=0n-3/21/α2k+1-1where αn and βn are defined in (62).From (61) we obtain(66)vnyn+1-1yn+1+1=xn-1xn+1αn,unxn+1-1xn+1+1=yn-1yn+1βnwhich is a first-order system after substitution of un,vn by the results given in (65). Its solutions can be obtained by using the following canonical coordinates:(67)sn=ln⁡xn-1xn+1,tn=ln⁡yn-1yn+1This leads to the the following linear system with variable coefficients:(68)tn+1=αnsn+γnsn+1=βntn+φnwhere -γn=ln⁡un and -φn=ln⁡vn.The latter is a linear first-order system with variable coefficients. Its general solution is(69)s2n=∏k=0n-1α2kβ2k+1s0+∑r=0n-1γ2r∏j=r+1n-1α2j∏j=rn-1β2j+1+∑r=0n-1φ2r+1∏j=r+1n-1α2jβ2j+1t2n=∏k=0n-1β2kα2k+1t0+∑r=0n-1φ2r∏j=r+1n-1β2j∏j=rn-1α2j+1+∑r=0n-1γ2r+1∏j=r+1n-1β2jα2j+1t2n+1=∏k=0nα2k∏k=0n-1β2k+1s0+∑r=0nγ2r∏j=r+1nα2j∏j=rn-1β2j+1+∑r=0n-1φ2r+1∏j=r+1nα2j∏j=r+1n-1β2j+1s2n+1=∏k=0nβ2k∏k=0n-1α2k+1t0+∑r=0nφ2r∏j=r+1nβ2j∏j=rn-1α2j+1+∑r=0n-1γ2r+1∏j=r+1nβ2j∏j=r+1n-1α2j+1The general solution of (45) is obtained by substituting (69) into (67).


## 4. Conservation Laws

In Section2, we have defined a first integral associated with a second-oreder SΔEs. It is given by (12)(70)ϕn,xn,yn,xn+1,yn+1=ϕn+1,xn+1,yn+1,ω1,ω2Let(71)P1=∂ϕ∂xn,P2=∂ϕ∂xn+1,Q1=∂ϕ∂yn,Q2=∂ϕ∂yn+1,By differentiating (70) with respect to xn,yn,xn+1 and yn+1 we obtain(72)P1=SP2ω1,xn+SQ2ω2,xnQ1=SP2ω1,yn+SQ2ω2,ynand(73)P2=SP1+SP2ω1,xn+1+SQ2ω2,xn+1Q2=SQ1+SP2ω1,yn+1+SQ2ω2,yn+1The substitution of (72) in (73) leads to the following second-order system of functional equations:(74)S2P2Sω1,xn+S2Q2Sω2,xn+SP2ω1,xn+1+SQ2ω2,xn+1-P2=0S2P2Sω1,yn+S2Q2Sω2,yn+SP2ω1,yn+1+SQ2ω2,yn+1-Q2=0As for SLSC, we differentiate repeatedly to obtain a system of DEs for P2 and Q2. Given the solutions P2,Q2 of (74) we easily construct P1,Q1. For consistency of our solutions, we must check the integrability conditions(75)∂P1∂xn+1=∂P2∂xnand(76)∂Q1∂yn+1=∂Q2∂ynThe first integral is then given by(77)ϕ=∫P1dxn+P2dxn+1+Q1dyn+Q2dyn+1+FnThe constant of integration F(n) which is a function depending on n is determined by substituting (77) in (70).


### 4.1. Applications

Let us consider the second-order SΔEs(78)xn+2=anyn,yn+2=bnxnBy choosing the ansatz P2(n,xn,yn) and Q2(n,xn,yn) one can readily check that the determining system (74) is simplified to(79)Q2n+2,ω1,ω2bn+1-P2n,xn,yn=0,P2n+2,ω1,ω2an+1-Q2n,xn,yn=0.where ω1 and ω2 denote the right-hand side of (78).Differentiating (79) with respect to xn and yn leads to(80)P2=xnψ1n+ynψ2n+ψ3n,Q2=xnψ4n+ynψ5n+ψ6nThus, we have from (72)(81)P1=bnxn+1ψ4n+1+yn+1ψ5n+1+ψ6n+1Q1=anxn+1ψ1n+1+yn+1ψ2n+1+ψ3n+1Substituting (80) in (79) and separating with respect to xn and yn we obtain the system(82)anbn+1ψ4n+2-ψ2n=0,bnan+1ψ2n+2-ψ4n=0,bnbn+1ψ5n+2-ψ1n=0,anan+1ψ1n+2-ψ5n=0,bn+1ψ6n+2-ψ3n=0,an+1ψ3n+2-ψ6n=0.The solutions to (82) will provide us with the explicit form of ψi,i=1,…,6The first integral is then given by(83)ϕ=∫bnxn+1ψ4n+1+yn+1ψ5n+1+ψ6n+1dxn+xnψ1n+ynψ2n+ψ3ndxn+1anxn+1ψ1n+1+yn+1ψ2n+1+ψ3n+1dyn+xnψ4n+ynψ5n+ψ6ndyn+1+Kifor some constants Ki.For clarification, let us considera(n)=b(n)=1, that is,(84)xn+2=yn,yn+2=xnThe solutions to (82) will be(85)ψ1n=C11+-1n+in+-in4+C21--1n-iin+i-in4+C31+-1n-in--in4+C41--1n+iin-i-in4ψ5n=C31+-1n+in+-in4+C41--1n-iin+i-in4+C11+-1n-in--in4+C21--1n+iin-i-in4ψ2n=C51+-1n+in+-in4+C61--1n-iin+i-in4+C71+-1n-in--in4+C81--1n+iin-i-in4ψ4n=C71+-1n+in+-in4+C81--1n-iin+i-in4+C51+-1n-in--in4+C61--1n+iin-i-in4ψ3n=C91+-1n+in+-in4+C101--1n-iin+i-in4+C111+-1n-in--in4+C121--1n+iin-i-in4ψ6n=C111+-1n+in+-in4+C121--1n-iin+i-in4+C91+-1n-in--in4+C101--1n+iin-i-in4where Ci,i=1,…,12 are constants. We have twelve solutions for P2 and Q2. That is,(1)
If(86)P2=1+-1n+in+-in4xn,Q2=1+-1n-in--in4yn,then(87)Q1=1--1n+iin-i-in4xn+1,P1=1--1n-iin+i-in4yn+1(2)
If(88)P2=1--1n-iin+i-in4xn,Q2=1--1n+iin-i-in4yn,then(89)Q1=1+-1n+in+-in4xn+1,P1=1+-1n-in--in4yn+1(3)
If(90)P2=1+-1n-in--in4)xn,Q2=1+-1n+in+-in4yn,then(91)Q1=1--1n-iin+i-in4xn+1,P1=1--1n+iin-i-in4yn+1(4)
If(92)P2=1--1n+iin-i-in4xn,Q2=1--1n-iin+i-in4yn,then(93)Q1=1+-1n-in--in4xn+1,P1=1+-1n+in+-in4yn+1(5)
If(94)P2=1+-1n+in+-in4yn,Q2=1+-1n-in--in4xn,then(95)Q1=1--1n+iin-i-in4yn+1,P1=1--1n-iin+i-in4xn+1(6)
If(96)P2=1--1n-iin+i-in4yn,Q2=1--1n+iin-i-in4xn,then(97)Q1=1+-1n+in+-in4yn+1,P1=1+-1n-in--in4xn+1(7)
If(98)P2=1+-1n-in--in4)yn,Q2=1+-1n+in+-in4xn,then(99)Q1=1--1n-iin+i-in4yn+1,P1=1--1n+iin-i-in4xn+1(8)
If(100)P2=1--1n+iin-i-in4yn,Q2=1--1n-iin+i-in4xn,then(101)Q1=1+-1n-in--in4yn+1,P1=1+-1n+in+-in4xn+1(9)
If(102)P2=1+-1n+in+-in4,Q2=1+-1n-in--in4,then(103)Q1=1--1n+iin-i-in4,P1=1--1n-iin+i-in4(10)
If(104)P2=1--1n-iin+i-in4,Q2=1--1n+iin-i-in4,then(105)Q1=1+-1n+in+-in4,P1=1+-1n-in--in4(11)
If(106)P2=1+-1n-in--in4),Q2=1+-1n+in+-in4,then(107)Q1=1--1n-iin+i-in4,P1=1--1n+iin-i-in4(12)
If(108)P2=1--1n+iin-i-in4,Q2=1--1n-iin+i-in4,then(109)Q1=1+-1n-in--in4,P1=1+-1n+in+-in4Therefore, we obtain twelve conservation laws for the system (84). They are given by(110)ϕ1=1+-1n+in+-in4xnxn+1+1+-1n-in--in4ynyn+1+1--1n+iin-i-in4xn+1yn+1--1n-iin+i-in4yn+1xn+K1ϕ2=1--1n-iin+i-in4xnxn+1+1--1n+iin-i-in4ynyn+1+1+-1n+in+-in4xn+1yn+1+-1n-in--in4yn+1xn+K2ϕ3=1+-1n-in--in4)xnxn+1+1+-1n+in+-in4ynyn+1+1--1n-iin+i-in4xn+1yn+1--1n+iin-i-in4yn+1xn+K3ϕ4=1--1n+iin-i-in4xnxn+1+1--1n-iin+i-in4ynyn+1+1+-1n-in--in4xn+1yn+1+-1n+in+-in4yn+1xn+K4ϕ5=1+-1n+in+-in4ynxn+1+1+-1n-in--in4xnyn+1+1--1n+iin-i-in4yn+1yn+1--1n-iin+i-in4xn+1xn+K5ϕ6=1--1n-iin+i-in4ynxn+1+1--1n+iin-i-in4xnyn+1+1+-1n+in+-in4yn+1yn+1+-1n-in--in4xn+1xn+K6ϕ7=1+-1n-in--in4)ynxn+1+1+-1n+in+-in4xnyn+1+1--1n-iin+i-in4yn+1yn+1--1n+iin-i-in4xn+1xn+K7ϕ8=1--1n+iin-i-in4ynxn+1+1--1n-iin+i-in4xnyn+1+1+-1n-in--in4yn+1yn+1+-1n+in+-in4xn+1xn+K8ϕ9=1+-1n+in+-in4xn+1+1+-1n-in--in4yn+1+1--1n+iin-i-in4yn+1--1n-iin+i-in4xn+K9ϕ10=1--1n-iin+i-in4xn+1+1--1n+iin-i-in4yn+1+1+-1n+in+-in4yn+1+-1n-in--in4xn+K10ϕ11=1+-1n-in--in4)xn+1+1+-1n+in+-in4yn+1+1--1n-iin+i-in4yn+1--1n+iin-i-in4xn+K11ϕ12=1--1n+iin-i-in4xn+1+1--1n-iin+i-in4yn+1+1+-1n-in--in4yn+1+-1n+in+-in4xn+K12


## 4.1. Applications

Let us consider the second-order SΔEs(78)xn+2=anyn,yn+2=bnxnBy choosing the ansatz P2(n,xn,yn) and Q2(n,xn,yn) one can readily check that the determining system (74) is simplified to(79)Q2n+2,ω1,ω2bn+1-P2n,xn,yn=0,P2n+2,ω1,ω2an+1-Q2n,xn,yn=0.where ω1 and ω2 denote the right-hand side of (78).Differentiating (79) with respect to xn and yn leads to(80)P2=xnψ1n+ynψ2n+ψ3n,Q2=xnψ4n+ynψ5n+ψ6nThus, we have from (72)(81)P1=bnxn+1ψ4n+1+yn+1ψ5n+1+ψ6n+1Q1=anxn+1ψ1n+1+yn+1ψ2n+1+ψ3n+1Substituting (80) in (79) and separating with respect to xn and yn we obtain the system(82)anbn+1ψ4n+2-ψ2n=0,bnan+1ψ2n+2-ψ4n=0,bnbn+1ψ5n+2-ψ1n=0,anan+1ψ1n+2-ψ5n=0,bn+1ψ6n+2-ψ3n=0,an+1ψ3n+2-ψ6n=0.The solutions to (82) will provide us with the explicit form of ψi,i=1,…,6The first integral is then given by(83)ϕ=∫bnxn+1ψ4n+1+yn+1ψ5n+1+ψ6n+1dxn+xnψ1n+ynψ2n+ψ3ndxn+1anxn+1ψ1n+1+yn+1ψ2n+1+ψ3n+1dyn+xnψ4n+ynψ5n+ψ6ndyn+1+Kifor some constants Ki.For clarification, let us considera(n)=b(n)=1, that is,(84)xn+2=yn,yn+2=xnThe solutions to (82) will be(85)ψ1n=C11+-1n+in+-in4+C21--1n-iin+i-in4+C31+-1n-in--in4+C41--1n+iin-i-in4ψ5n=C31+-1n+in+-in4+C41--1n-iin+i-in4+C11+-1n-in--in4+C21--1n+iin-i-in4ψ2n=C51+-1n+in+-in4+C61--1n-iin+i-in4+C71+-1n-in--in4+C81--1n+iin-i-in4ψ4n=C71+-1n+in+-in4+C81--1n-iin+i-in4+C51+-1n-in--in4+C61--1n+iin-i-in4ψ3n=C91+-1n+in+-in4+C101--1n-iin+i-in4+C111+-1n-in--in4+C121--1n+iin-i-in4ψ6n=C111+-1n+in+-in4+C121--1n-iin+i-in4+C91+-1n-in--in4+C101--1n+iin-i-in4where Ci,i=1,…,12 are constants. We have twelve solutions for P2 and Q2. That is,(1)
If(86)P2=1+-1n+in+-in4xn,Q2=1+-1n-in--in4yn,then(87)Q1=1--1n+iin-i-in4xn+1,P1=1--1n-iin+i-in4yn+1(2)
If(88)P2=1--1n-iin+i-in4xn,Q2=1--1n+iin-i-in4yn,then(89)Q1=1+-1n+in+-in4xn+1,P1=1+-1n-in--in4yn+1(3)
If(90)P2=1+-1n-in--in4)xn,Q2=1+-1n+in+-in4yn,then(91)Q1=1--1n-iin+i-in4xn+1,P1=1--1n+iin-i-in4yn+1(4)
If(92)P2=1--1n+iin-i-in4xn,Q2=1--1n-iin+i-in4yn,then(93)Q1=1+-1n-in--in4xn+1,P1=1+-1n+in+-in4yn+1(5)
If(94)P2=1+-1n+in+-in4yn,Q2=1+-1n-in--in4xn,then(95)Q1=1--1n+iin-i-in4yn+1,P1=1--1n-iin+i-in4xn+1(6)
If(96)P2=1--1n-iin+i-in4yn,Q2=1--1n+iin-i-in4xn,then(97)Q1=1+-1n+in+-in4yn+1,P1=1+-1n-in--in4xn+1(7)
If(98)P2=1+-1n-in--in4)yn,Q2=1+-1n+in+-in4xn,then(99)Q1=1--1n-iin+i-in4yn+1,P1=1--1n+iin-i-in4xn+1(8)
If(100)P2=1--1n+iin-i-in4yn,Q2=1--1n-iin+i-in4xn,then(101)Q1=1+-1n-in--in4yn+1,P1=1+-1n+in+-in4xn+1(9)
If(102)P2=1+-1n+in+-in4,Q2=1+-1n-in--in4,then(103)Q1=1--1n+iin-i-in4,P1=1--1n-iin+i-in4(10)
If(104)P2=1--1n-iin+i-in4,Q2=1--1n+iin-i-in4,then(105)Q1=1+-1n+in+-in4,P1=1+-1n-in--in4(11)
If(106)P2=1+-1n-in--in4),Q2=1+-1n+in+-in4,then(107)Q1=1--1n-iin+i-in4,P1=1--1n+iin-i-in4(12)
If(108)P2=1--1n+iin-i-in4,Q2=1--1n-iin+i-in4,then(109)Q1=1+-1n-in--in4,P1=1+-1n+in+-in4Therefore, we obtain twelve conservation laws for the system (84). They are given by(110)ϕ1=1+-1n+in+-in4xnxn+1+1+-1n-in--in4ynyn+1+1--1n+iin-i-in4xn+1yn+1--1n-iin+i-in4yn+1xn+K1ϕ2=1--1n-iin+i-in4xnxn+1+1--1n+iin-i-in4ynyn+1+1+-1n+in+-in4xn+1yn+1+-1n-in--in4yn+1xn+K2ϕ3=1+-1n-in--in4)xnxn+1+1+-1n+in+-in4ynyn+1+1--1n-iin+i-in4xn+1yn+1--1n+iin-i-in4yn+1xn+K3ϕ4=1--1n+iin-i-in4xnxn+1+1--1n-iin+i-in4ynyn+1+1+-1n-in--in4xn+1yn+1+-1n+in+-in4yn+1xn+K4ϕ5=1+-1n+in+-in4ynxn+1+1+-1n-in--in4xnyn+1+1--1n+iin-i-in4yn+1yn+1--1n-iin+i-in4xn+1xn+K5ϕ6=1--1n-iin+i-in4ynxn+1+1--1n+iin-i-in4xnyn+1+1+-1n+in+-in4yn+1yn+1+-1n-in--in4xn+1xn+K6ϕ7=1+-1n-in--in4)ynxn+1+1+-1n+in+-in4xnyn+1+1--1n-iin+i-in4yn+1yn+1--1n+iin-i-in4xn+1xn+K7ϕ8=1--1n+iin-i-in4ynxn+1+1--1n-iin+i-in4xnyn+1+1+-1n-in--in4yn+1yn+1+-1n+in+-in4xn+1xn+K8ϕ9=1+-1n+in+-in4xn+1+1+-1n-in--in4yn+1+1--1n+iin-i-in4yn+1--1n-iin+i-in4xn+K9ϕ10=1--1n-iin+i-in4xn+1+1--1n+iin-i-in4yn+1+1+-1n+in+-in4yn+1+-1n-in--in4xn+K10ϕ11=1+-1n-in--in4)xn+1+1+-1n+in+-in4yn+1+1--1n-iin+i-in4yn+1--1n+iin-i-in4xn+K11ϕ12=1--1n+iin-i-in4xn+1+1--1n-iin+i-in4yn+1+1+-1n-in--in4yn+1+-1n+in+-in4xn+K12


## 5. Conclusion and Discussions

We have presented a method for obtaining nontrivial symmetries and how to use them for solving a second-order SΔEs. Each symmetry can be used to reduce the order. However, different symmetries lead to different reductions (see (60) and (64)), but the same solution. We also proposed a technique to construct first integral associated to second-order systems of difference equations.



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*Source: 8256867-2019-05-12.xml*