Symmetry Analysis and Conservation Laws of some third-order Difference Equations

TL;DR

This paper presents a method to find Lie symmetries of third-order difference equations, enabling order reduction and solution derivation, along with a technique for obtaining their first integrals.

Contribution

It introduces a new approach for symmetry analysis and solution methods specifically for third-order difference equations.

Findings

Lie symmetries can be systematically found for third-order difference equations.

Symmetries facilitate order reduction and solution derivation.

A technique for obtaining first integrals of these equations is proposed.

Abstract

We derive a method for finding Lie Symmetries for third-order difference equations. We use these symmetries to reduce the order of the difference equations and hence obtain the solutions of some third-order difference equations. We also introduce a technique for obtaining their first integrals.

Equations

$$u_{n+N}=\omega (n,u_n,u_{n+1},...,u_{n+N-1}),$$

$${\Gamma }_{ϵ}:x\mapsto \tilde{x}(x;ϵ),$$

$$\widetilde{x_i}(x:ϵ)=x_i+ϵ{\xi }_i(x)+O({ϵ}^2),i=1,...,p.$$

$$\tilde{n}=n;{\tilde{u}}_n\simeq u_n+ϵQ(n,u_n)$$

$$X=Q(n,u_n)\partial u_n+SQ\partial u_{n+1}+S^{2}Q(n,u_n)\partial u_{n+2}+...+S^{N-1}Q(n,u_n)\partial u_{n+N-1},$$

$$S:n\mapsto n+1$$

$${\tilde{u}}_{n+N}=\omega (n,{\tilde{u}}_n,{\tilde{u}}_{n+1},...,{\tilde{u}}_{n+N-1})$$

$$S^{(N)}Q-X\omega =0,$$

$$(S-id)\varphi (n,u_n,u_{n+1},...,u_{n+N-1})=0,$$

$$u_{n+3}=\omega (n,u_n,u_{n+1},u_{n+2}).$$

$$S^{3}Q-X\omega =0,$$

$$X=Q(n,u_n)\partial u_n+SQ(n,u_n)\partial u_{n+1}+S^{2}Q(n,u_n)\partial u_{n+2},$$

$$Q(n+3,\omega )-\frac{\partial \omega }{\partial u_n}Q(n,u_n)-\frac{\partial \omega }{\partial u_{n+1}}SQ(n,u_n)-\frac{\partial \omega }{\partial u_{n+2}}{S}^{2}Q(n,u_n),$$

$$\begin{split}&\frac{\omega_{,u_{n}}}{\omega_{,u_{n+1}}}\big{[}\omega_{,u_{n}u_{n+1}}Q(n,u_{n})+\omega_{,u_{n+1}}SQ^{\prime}(n,u_{n})+\omega_{,u_{n+1}u_{n+1}}SQ(n,u_{n})\\ &+\omega_{,u_{n+1}u_{n+2}}S^{2}Q(n,u_{n})\big{]}-\omega_{,u_{n}}Q^{\prime}(n,u_{n})-\omega_{,u_{n}u_{n}}Q(n,u_{n})\\ &-\omega_{,u_{n}u_{n+1}}SQ(n,u_{n})-\omega_{,u_{n}u_{n+2}}S^{2}Q(n,u_{n})=0,\end{split}$$

$$\begin{array}{rl}& \frac{{\omega }_{,u_n}}{{\omega }_{,u_{n+2}}}\left[{\omega }_{,u_n{u}_{n+2}}Q(n,u_n)+{\omega }_{,u_{n+2}}{S}^2{Q}^{\prime }(n,u_n)+{\omega }_{,u_{n+2}{u}_{n+2}}{S}^{2}Q(n,u_n)\right]\\ & -{\omega }_{,u_n}{Q}^{\prime }(n,u_n)-{\omega }_{,u_n{u}_n}Q(n,u_n)-{\omega }_{,u_n{u}_{n+2}}{S}^{2}Q(n,u_n)=0.\end{array}$$

$$\varphi =\varphi (n,u_n,u_{n+1},u_{n+2})$$

$$S\varphi =\varphi ,$$

$$\varphi (n+1,u_{n+1},u_{n+2},\omega )=\varphi (n,u_n,u_{n+1},u_{n+2}).$$

$$P_0=\frac{\partial \varphi }{\partial u_n},P_1=\frac{\partial \varphi }{\partial u_{n+1}},P_2=\frac{\partial \varphi }{\partial u_{n+2}}.$$

$$\begin{array}{rl}{P}_0& =\left(\frac{\partial \omega }{\partial u_n}\right)\cdot S(P_2)\\ P_1& =S(P_0)+\left(\frac{\partial \omega }{\partial u_{n+1}}\right)\cdot S(P_2)\\ P_2& =S(P_1)+\left(\frac{\partial \omega }{\partial u_{n+2}}\right)\cdot S(P_2).\end{array}$$

$$S^2({\omega }_{,u_n})S^3(P_2)+S({\omega }_{,u_{n+1}})S^2(P_2)+{\omega }_{,u_{n+2}}S(P_2)-P_2=0.$$

$$u_{n+3}=\frac{a{u}_{n+1}{u}_{n+2}}{u_{n+2}+b{n}_n},$$

$$\begin{array}{rl}& \frac{-u_{n+1}}{b{u}_n+u_{n+2}}\left[-\frac{b{u}_{n+2}}{(b{u}_n+u_{n+2}{)}^2}Q+\frac{u_{n+2}}{b{u}_n+u_{n+2}}S{Q}^{\prime }+\frac{b{u}_n}{(b{u}_n+u_{n+2}{)}^2}{S}^{2}Q\right]\\ & +\frac{u_{n+1}{u}_{n+2}}{(b{u}_n+u_{n+2}{)}^2}{Q}^{\prime }-\frac{2b{u}_{n+1}{u}_{n+2}}{(u_{n+2}+b{u}_n{)}^3}Q+\frac{u_{n+2}}{(b{u}_n+u_{n+2}{)}^2}SQ\\ & +\frac{u_{n+1}(b{u}_n-u_{n+2})}{(b{u}_n+u_{n+2}{)}^3}=0.\end{array}$$

$$b{Q}^{(2)}(n,u_n)+(b{u}_n+u_{n+2})Q^{(3)}=0.$$

$$Q^{(2)}=0,$$

$$Q(n,u_n)=c_1(n)+u_n{c}_2(n).$$

$$c_1(n)=0c_2(n)=k_0+k_1{b}^{\frac{n}{2}}+(-\sqrt{b}{)}^n{k}_2,$$

$$Q(n,u_n)=u_n{k}_1+(-1{)}^n{u}_n{k}_2.$$

$$\begin{array}{rl}{X}_1& =u_n\partial u_n+u_{n+1}\partial u_{n+1}+u_{n+2}\partial u_{n+2}\\ X_2& =u_n(-1{)}^n\partial u_n+u_{n+1}(-1{)}^{n+1}\partial u_{n+1}+u_{n+2}(-1{)}^{n+2}\partial u_{n+2}\end{array}$$

$$\frac{d{u}_n}{u_n}=\frac{d{u}_{n+1}}{u_{n+1}}=\frac{d{u}_{n+2}}{u_{n+2}}=\frac{d{v}_n}{0}.$$

Full text

Key words: Difference equation; symmetry; reduction; group invariant solutions

1 Introduction

The concept of Lie Symmetry analysis applied onto difference equations ($\Delta$E’s) was studied by P. Hydon [1] and others see [2, 3, 4, 5]. In [1], the author introduced a method for obtaining symmetries and first integrals of second-order difference equations. In this paper, we extend these findings and introduce a technique for obtaining symmetries and first integrals for third-order $\Delta$E’s.

1.1 Mathematical tools to be used

The definitions and notations used in this paper follow from the ones adopted by Hybon in [1]. Assume we have an $Nth-order$ difference equation of the form

[TABLE]

for some function $\omega$. We will assume that $\frac{\partial\omega}{\partial u_{n}}\neq 0$ and that the point transformations are given by

[TABLE]

where $x=x_{i},$ $i=1,\dots,p$ are continuous variables. We then refer to $\Gamma$ as one-parameter Lie group of transformations if it satisfies the following transformations (see [6])

• •

$\Gamma_{0}$ is the identity map if $\tilde{x}=x$ when $\epsilon=0$

• •

$\Gamma_{a}\Gamma_{b}=\Gamma_{a+b}$ for every $a$ and $b$ sufficiently close to 0

• •

Each $\tilde{x_{i}}$ can be represented as a Taylor series (in a neighbourhood of $\epsilon=0$ that is determined by $x$), and therefore

[TABLE]

We shall assume that the Lie point symmetries in this article take the following form

[TABLE]

and that the analogous infinitesimal symmetry generator takes the form

[TABLE]

where $S$ is defined as the shift operator, i.e.

[TABLE]

and $Q(n,u_{n})$ is referred to as the characteristic. We shall define the symmetry condition as

[TABLE]

each time (1) is true. Substituting the Lie point symmetries (4) into the symmetry condition (7) yields linearized symmetry condition

[TABLE]

whenever (1) is true. We then define the conservation laws of (1) as follows

[TABLE]

where $id$ is the identity function each time (1) holds.

2 Finding symmetries and first integrals

2.1 Symmetries

Consider the following third-order difference equation

[TABLE]

The linearized symmetry condition then amounts to

[TABLE]

where

[TABLE]

is the analogous symmetry generator. In order to obtain the characteristic $Q(n,u_{n}),$ we apply (11) to (10) to obtain

[TABLE]

where $\omega$ represents the left hand side of (10). We then differentiate (13) with respect to $u_{n}$, keeping $\omega$ and taking $u_{n+1}$ as a function of $u_{n},u_{n+1},u_{n+2}$ and $\omega.$ We end up with the following determining equation

[TABLE]

where $g_{,x}$ denotes the derivative of $g$ with respect to $x.$ Note that (10) may sometimes be independent of $u_{n+1}$ or $u_{n+2},$ and so the above determining equation changes. For an example if (10) is independent of $u_{n+1},$ the resulting determining equation is

[TABLE]

2.2 Conservation Laws

We now introduce the general case for finding first integrals of third-order $\Delta$E’s. Assume that

[TABLE]

is the first integral of (10). According to the equation

[TABLE]

we have that

[TABLE]

We then define

[TABLE]

It then follows that

[TABLE]

We then join (20) to form a single equation involving only $P_{2},$ which will be the determining equation for computing first integrals. The equation is as follows

[TABLE]

We can then solve for $P_{2}$ in (21), hence solve (20) and integrate to obtain the first integrals.

3 Applications

We are going to use the determining equation (14) to obtain the symmetries of two third-order $\Delta$E’s.

3.1 Symmetries and exact solutions

3.1.1 Example 1

Consider the third-order difference equation

[TABLE]

where $a$ and $b$ are arbitrary constants. The determining equation (14) becomes

[TABLE]

Multiplying by $(bu_{n}+u_{n+2})^{3}$ throughout and differentiating twice with respect to $u_{n}$ we get the following differential equation

[TABLE]

Separating (24) with respect to $u_{n+2}$ we get that

[TABLE]

whose most general solution is

[TABLE]

To obtain the nature of the constants $c_{1},$ and $c_{2},$ which are functions of $n,$ we substitute (26) back into (13) and (23). We find that

[TABLE]

where $k_{1}$ and $k_{2}$ are arbitrary constants. It then follows after setting $b=1$ without loss of generosity that

[TABLE]

We therefore have that the symmetries of (22) are

[TABLE]

We note that $X_{1}$ is a symmetry for any value of $b$. From $X_{1}$ in (29), the resulting invariants are

[TABLE]

• •

Working with the first $\left(\frac{du_{n}}{u_{n}}\right)$ and second $\left(\frac{du_{n+1}}{u_{n+1}}\right)$ invariants, first $\left(\frac{du_{n}}{u_{n}}\right)$ and third $\left(\frac{du_{n+2}}{u_{n+2}}\right)$ invariants we get that

[TABLE]

where $K_{1}$ and $K_{2}$ are constants.

We choose $f(K_{1},K_{2})=K_{1}\cdot K_{2}$, therefore

[TABLE]

hence

[TABLE]

Solving (33) results in

[TABLE]

We have reduced a third-order difference equation (10) into a second-order difference equation (34a).

Case a=b: Combining (32) and (34b), we have

[TABLE]

Therefore, the solution to (22), when $a=b$, is given by

[TABLE]

where $v_{k_{1}}$ in given in (34b) and $(a)_{n}$ is the Pochhammer symbol (rising factorial.)

Case $a\neq b$: Combining (32) and (34a), we have

[TABLE]

Therefore, the solution to (22) is given by

[TABLE]

where $v_{k_{1}}$ in given in (34a).

• -

Suppose that we choose $f(K_{1},K_{2})=K_{1},$ following the same method above, we end up with $v_{n}=\frac{u_{n}}{u_{n+1}}$, i.e.,

[TABLE]

One can solve for $v_{n}$ from (39) and hence solve for $u_{n}.$

• •

Working with the first $\left(\frac{du_{n}}{u_{n}}\right)$ and third invariant $\left(\frac{du_{n+2}}{u_{n+2}}\right),$ second invariant $\left(\frac{du_{n+1}}{u_{n+1}}\right)$ and third invariant $\left(\frac{du_{n+2}}{u_{n+2}}\right),$ we get that

[TABLE]

Choosing $f(K_{1},K_{2})=K_{2}/K_{1}$ we get that

[TABLE]

and therefore

[TABLE]

The solution of ((42)) is given in [7]( put $A=a/b$ and B=1/b) by

[TABLE]

with $-1/((1/b)v_{0}v_{1})\notin\{1,1+(a/b),\dots,\sum_{i=0}^{n-2}(a/b)^{i}\}$. Combining (41) and (43) we get that

[TABLE]

where $v_{n}$ is given in (43), is also a solution to (22).

3.1.2 Example 2

We now shift our attention to focus on a second third-order difference equation

[TABLE]

where $a$ and $c$ are arbitrary functions of $n$. We note that (45) is independent of $u_{n+1},$ so we use determining equation (15). One can easily establish that the solution to the determining equation is

[TABLE]

where $k_{1}$ is a constant and $k_{2}$ is a function of $n.$ Substituting (45) into (11) we get that

[TABLE]

where we notice that $k_{2}(n)$ satisfies the original equation (45). Assuming that $U_{1},$ $U_{2}$ and $U_{3}$ are solutions of (47), the resulting symmetries are

[TABLE]

The characteristic equation, using $X_{0}$, is given by

[TABLE]

The invariant $V_{n}$ is then given by

[TABLE]

Assume $g(x,y)=x$, we have

[TABLE]

and this implies that

[TABLE]

On the other hand, by assuming that $g(x,y)=1/x$, we have

[TABLE]

and this implies that

[TABLE]

With these change of variables we have obtained two different reduced forms (in terms of order), given by ((50)) and ((51)), of (45).

3.2 Conservation Laws

3.2.1 Example 1

Consider the equation (45), assume that one required to obtain the first integrals. By letting $P_{2}=P_{2}(n,u_{n}),$ one can find that the solution to the determining equation (21) is

[TABLE]

Substituting (54), we find that $B(n)$ satisfies the equation

[TABLE]

We can then shown that

[TABLE]

where $B_{i}^{\prime}s,$ $i=1,...3$ are solutions to (55). We can therefore deduce that the first integrals of such equations (45) are

[TABLE]

for some functions $G_{i},i=1,2,3$, to be determined. By substituting (60) into (17), we were able to find that $G_{i}(n+1)=G_{i}(n)=K,$ where $K$ is a constant. For simplicity, we shall assume that $K$ is zero. It has to be noted that now that we know the symmetries and first integrals of (45), we can create new first integrals. Below are new first integrals of (45):

[TABLE]

Also, from

[TABLE]

we obtain new integrals

[TABLE]

3.2.2 Example 2

Now consider the equation (22), i.e.,

[TABLE]

Assuming one required to compute the first integrals, one would hence use (21) to come up with the equation

[TABLE]

Due to the fact that in the above equation, $P_{2}$ takes up different arguments, we first differentiate with respect to $u_{n}$ keeping $\omega=u_{n+3}$ fixed and by assuming $u_{n+2}$ is a function of $u_{n}$ and $\omega.$ We differentiate the resulting equation with respect to $u_{n}$ six times and we then split it with respect to $u_{n+1}$ and $u_{n+2}$. to get to equation

[TABLE]

After solving the differential equation (67) and substituting the result back into (66) we found that $P_{2}=0,$ hence $P_{1}=0$ and $P_{0}=0.$ This implies that the first integral of (22) is of a trivial nature and it can be written in the form

[TABLE]

where $K$ is a constant.

4 Conclusion

We have explored a way of obtaining symmetries and first integrals of third-order difference equations. We have also demonstrated one can use the symmetries to reduce the order of the difference equations. What we have to note is that when reducing the order, we used only one of the symmetries, i.e., $u_{n}$ whilst we could have also used another symmetry which could have also worked the same way but yielded a different result.