
In this paper, we obtain the form of the solutions of the following rational systems of difference equations
xn+1=yn−1znzn±xn−2,yn+1=zn−1xnxn±yn−2, zn+1=xn−1ynyn±zn−2,
with initial values are non-zero real numbers.
Citation: E. M. Elsayed, Q. Din, N. A. Bukhary. Theoretical and numerical analysis of solutions of some systems of nonlinear difference equations[J]. AIMS Mathematics, 2022, 7(8): 15532-15549. doi: 10.3934/math.2022851
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In this paper, we obtain the form of the solutions of the following rational systems of difference equations
xn+1=yn−1znzn±xn−2,yn+1=zn−1xnxn±yn−2, zn+1=xn−1ynyn±zn−2,
with initial values are non-zero real numbers.
This paper is devoted to study the expressions forms of the solutions and periodic nature of the following third-order rational systems of difference equations
xn+1=yn−1znzn±xn−2,yn+1=zn−1xnxn±yn−2, zn+1=xn−1ynyn±zn−2, |
with initial conditions are non-zero real numbers.
In the recent years, there has been great concern in studying the systems of difference equations. One of the most important reasons for this is a exigency for some mechanization which can be used in discussing equations emerge in mathematical models characterizing real life situations in economic, genetics, probability theory, psychology, population biology and so on.
Difference equations display naturally as discrete peer and as numerical solutions of differential equations having more applications in ecology, biology, physics, economy, and so forth. For all that the difference equations are quite simple in expressions, it is frequently difficult to realize completely the dynamics of their solutions see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19] and the related references therein.
There are some papers dealed with the difference equations systems, for example, The periodic nature of the solutions of the nonlinear difference equations system
An+1=1Cn,Bn+1=BnAn−1Bn−1,Cn+1=1An−1, |
has been studied by Cinar in [7].
Almatrafi [3] determined the analytical solutions of the following systems of rational recursive equations
xn+1=xn−1yn−3yn−1(±1±xn−1yn−3),yn+1=yn−1xn−3xn−1(±1±yn−1xn−3). |
In [20], Khaliq and Shoaib studied the local and global asymptotic behavior of non-negative equilibrium points of a three-dimensional system of two order rational difference equations
xn+1=xn−1ε+xn−1yn−1zn−1,yn+1=yn−1ζ+xn−1yn−1zn−1, zn+1=zn−1η+xn−1yn−1zn−1. |
In [9], Elabbasy et al. obtained the form of the solutions of some cases of the following system of difference equations
xn+1=a1+a2yna3zn+a4xn−1zn, yn+1=b1zn−1+b2znb3xnyn+b4xnyn−1,zn+1=c1zn−1+c2znc3xn−1yn−1+c4xn−1yn+c5xnyn. |
In [12], Elsayed et al. have got the solutions of the systems of rational higher order difference equations
An+1=1An−pBn−p,Bn+1=An−pBn−pAn−qBn−q, |
and
An+1=1An−pBn−pCn−p,Bn+1=An−pBn−pCn−pAn−qBn−qCn−q,Cn+1=An−qBn−qCn−qAn−rBn−rCn−r. |
Kurbanli [25,26] investigated the behavior of the solutions of the following systems
An+1=An−1An−1Bn−1,Bn+1=Bn−1Bn−1An−1, Cn+1=1CnBn,An+1=An−1An−1Bn−1,Bn+1=Bn−1Bn−1An−1, Cn+1=Cn−1Cn−1Bn−1. |
In [32], Yalçınkaya has obtained the conditions for the global asymptotically stable of the system
An+1=BnAn−1+aBn+An−1,Bn+1=AnBn−1+aAn+Bn−1. |
Zhang et al. [39] investigated the persistence, boundedness and the global asymptotically stable of the solutions of the following system
Rn=A+1Qn−p, Qn=A+Qn−1Rn−rQn−s. |
Similar to difference equations and systems were studied see [21,22,23,24,27,28,29,30,31,32,33,34,35,36,37,38].
In this section, we obtain the expressions form of the solutions of the following three dimension system of difference equations
xn+1=yn−1znzn+xn−2,yn+1=zn−1xnxn+yn−2, zn+1=xn−1ynyn+zn−2, | (1) |
where n∈N0 and the initial conditions are non-zero real numbers.
Theorem 1. We assume that {xn,yn,zn} are solutions of system (1).Then
x6n−2=ak3nn−1∏i=0(a+(6i)k)(a+(6i+2)k)(a+(6i+4)k),x6n−1=bf3nn−1∏i=0(g+(6i+1)f)(g+(6i+3)f)(g+(6i+5)f),x6n=c3n+1n−1∏i=0(d+(6i+2)c)(d+(6i+4)c)(d+(6i+6)c),x6n+1=ek3n+1(a+k)n−1∏i=0(a+(6i+3)k)(a+(6i+5)k)(a+(6i+7)k), |
x6n+2=f3n+2(g+2f)n−1∏i=0(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f),x6n+3=hc3n+2(d+c)(d+3c)n−1∏i=0(d+(6i+5)c)(d+(6i+7)c)(d+(6i+9)c), |
y6n−2=dc3nn−1∏i=0(d+(6i)c)(d+(6i+2)c)(d+(6i+4)c),y6n−1=ek3nn−1∏i=0(a+(6i+1)k)(a+(6i+3)k)(a+(6i+5)k),y6n=f3n+1n−1∏i=0(g+(6i+2)f)(g+(6i+4)f)(g+(6i+6)f),y6n+1=hc3n+1(d+c)n−1∏i=0(d+(6i+3)c)(d+(6i+5)c)(d+(6i+7)c),y6n+2=k3n+2(a+2k)n−1∏i=0(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k),y6n+3=bf3n+2(g+f)(g+3f)n−1∏i=0(g+(6i+5)f)(g+(6i+7)f)(g+(6i+9)f), |
and
z6n−2=gf3nn−1∏i=0(g+(6i)f)(g+(6i+2)f)(g+(6i+4)f),z6n−1=hc3nn−1∏i=0(d+(6i+1)c)(d+(6i+3)c)(d+(6i+5)c),z6n=k3n+1n−1∏i=0(a+(6i+2)k)(a+(6i+4)k)(a+(6i+6)k),z6n+1=bf3n+1(g+f)n−1∏i=0(g+(6i+3)f)(g+(6i+5)f)(g+(6i+7)f), |
z6n+2=c3n+2(d+2c)n−1∏i=0(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c),z6n+3=ek3n+2(a+k)(a+3k)n−1∏i=0(a+(6i+5)k)(a+(6i+7)k)(a+(6i+9)k), |
where x−2=a, x−1=b, x0=c, y−2=d, y−1=e, y0=f, z−2=g, z−1=h and z0=k.
Proof. For n=0 the result holds. Now assume that n>1 and that our assumption holds for n−1, that is,
x6n−8=ak3n−3n−2∏i=0(a+(6i)k)(a+(6i+2)k)(a+(6i+4)k),x6n−7=bf3n−3n−2∏i=0(g+(6i+1)f)(g+(6i+3)f)(g+(6i+5)f),x6n−6=c3n−2n−2∏i=0(d+(6i+2)c)(d+(6i+4)c)(d+(6i+6)c),x6n−5=ek3n−2(a+k)n−2∏i=0(a+(6i+3)k)(a+(6i+5)k)(a+(6i+7)k),x6n−4=f3n−1(g+2f)n−2∏i=0(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f),x6n−3=hc3n−1(d+c)(d+3c)n−2∏i=0(d+(6i+5)c)(d+(6i+7)c)(d+(6i+9)c), |
y6n−8=dc3n−3n−2∏i=0(d+(6i)c)(d+(6i+2)c)(d+(6i+4)c),y6n−7=ek3n−3n−2∏i=0(a+(6i+1)k)(a+(6i+3)k)(a+(6i+5)k),y6n−6=f3n−2n−2∏i=0(g+(6i+2)f)(g+(6i+4)f)(g+(6i+6)f), |
y6n−5=hc3n−2(d+c)n−2∏i=0(d+(6i+3)c)(d+(6i+5)c)(d+(6i+7)c),y6n−4=k3n−1(a+2k)n−2∏i=0(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k),y6n−3=bf3n−1(g+f)(g+3f)n−2∏i=0(g+(6i+5)f)(g+(6i+7)f)(g+(6i+9)f), |
and
z6n−8=gf3n−3n−2∏i=0(g+(6i)f)(g+(6i+2)f)(g+(6i+4)f),z6n−7=hc3n−3n−2∏i=0(d+(6i+1)c)(d+(6i+3)c)(d+(6i+5)c),z6n−6=k3n−2n−2∏i=0(a+(6i+2)k)(a+(6i+4)k)(a+(6i+6)k),z6n−5=bf3n−2(g+f)n−2∏i=0(g+(6i+3)f)(g+(6i+5)f)(g+(6i+7)f),z6n−4=c3n−1(d+2c)n−2∏i=0(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c),z6n−3=ek3n−1(a+k)(a+3k)n−2∏i=0(a+(6i+5)k)(a+(6i+7)k)(a+(6i+9)k). |
It follows from Eq (1) that
x6n−2=y6n−4z6n−3z6n−3+x6n−5=(k3n−1(a+2k)n−2∏i=0(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k) )(ek3n−1(a+k)(a+3k)n−2∏i=0(a+(6i+5)k)(a+(6i+7)k)(a+(6i+9)k) )(ek3n−1(a+k)(a+3k)n−2∏i=0(a+(6i+5)k)(a+(6i+7)k)(a+(6i+9)k) )+(ek3n−2(a+k)n−2∏i=0(a+(6i+3)k)(a+(6i+5)k)(a+(6i+7)k) )=(k3n(a+2k)n−2∏i=0(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k))(a+3k)n−2∏i=0(a+(6i+9)k)[(k(a+3k)n−2∏i=0(a+(6i+9)k))+(1n−2∏i=0(a+(6i+3)k))]=(k3n(a+2k)n−2∏i=0(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k))[k+((a+3k)n−2∏i=0(a+(6i+9)k)n−2∏i=0(a+(6i+3)k))]=(k3n(a+2k)n−2∏i=0(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k))[k+(a+(6n−3)k)]=ak3na(a+2k)(a+(6n−2)k)n−2∏i=0(a+(6i+4)k)(a+(6i+6)k)(a+(6i+8)k). |
Then we see that
x6n−2=k3nn−1∏i=0(a+(6i)k)(a+(6i+2)k)(a+(6i+4)k). |
Also, we see from Eq (1) that
y6n−2=z6n−4x6n−3x6n−3+y6n−5=(c3n−1(d+2c)n−2∏i=0(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c) )(hc3n−1(d+c)(d+3c)n−2∏i=0(d+(6i+5)c)(d+(6i+7)c)(d+(6i+9)c) )(hc3n−1(d+c)(d+3c)n−2∏i=0(d+(6i+5)c)(d+(6i+7)c)(d+(6i+9)c) )+(hc3n−2(d+c)n−2∏i=0(d+(6i+3)c)(d+(6i+5)c)(d+(6i+7)c) )=(c3n(d+2c)n−2∏i=0(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c))(d+3c)n−2∏i=0(d+(6i+9)c)[(c(d+3c)n−2∏i=0(d+(6i+9)c))+(1n−2∏i=0(d+(6i+3)c))]=(c3n(d+2c)n−2∏i=0(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c))[c+d+(6n−3)c]=c3n[d+(6n−2)c](d+2c)n−2∏i=0(d+(6i+4)c)(d+(6i+6)c)(d+(6i+8)c). |
Then
y6n−2=dc3nn−1∏i=0(d+(6i)c)(d+(6i+2)c)(d+(6i+4)c). |
Finally from Eq (1), we see that
z6n−2=x6n−4y6n−3y6n−3+z6n−5=(f3n−1(g+2f)n−2∏i=0(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f) )(bf3n−1(g+f)(g+3f)n−2∏i=0(g+(6i+5)f)(g+(6i+7)f)(g+(6i+9)f) )(bf3n−1(g+f)(g+3f)n−2∏i=0(g+(6i+5)f)(g+(6i+7)f)(g+(6i+9)f) )+(bf3n−2(g+f)n−2∏i=0(g+(6i+3)f)(g+(6i+5)f)(g+(6i+7)f) )=(f3n(g+2f)n−2∏i=0(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f))(g+3f)n−2∏i=0(g+(6i+9)f)[(f(g+3f)n−2∏i=0(g+(6i+9)f))+(1n−2∏i=0(g+(6i+3)f))]=(f3n(g+2f)n−2∏i=0(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f))[f+((g+3f)n−2∏i=0(g+(6i+9)f)n−2∏i=0(g+(6i+3)f))]=(f3n(g+2f)n−2∏i=0(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f))[f+(g+(6n−3)f)]=f3n(g+(6n−2)f)(g+2f)n−2∏i=0(g+(6i+4)f)(g+(6i+6)f)(g+(6i+8)f). |
Thus
z3n−2=gf3nn−1∏i=0(g+(6i)f)(g+(6i+2)f)(g+(6i+4)f). |
By similar way, one can show the other relations. This completes the proof.
Lemma 1. Let {xn,yn,zn} be a positive solution of system (1), then all solution of (1) is bounded and approaching to zero.
Proof. It follows from Eq (1) that
xn+1=yn−1znzn+xn−2≤yn−1, yn+1=zn−1xnxn+yn−2≤zn−1,zn+1=xn−1ynyn+zn−2≤xn−1, |
we see that
xn+4≤yn+2, yn+2≤zn, zn≤xn−2, ⇒ xn+4<xn−2,yn+4≤zn+2, zn+2≤xn, xn≤yn−2, ⇒ yn+4<yn−2,zn+4≤xn+2, xn+2≤yn, yn≤zn−2, ⇒ zn+4<zn−2, |
Then all subsequences of {xn,yn,zn} (i.e., for {xn} are {x6n−2}, {x6n−1}, {x6n}, {x6n+1}, {x6n+2}, {x6n+3} are decreasing and at that time are bounded from above by K,L and M since K=max{x−2,x−1,x0,x1,x2,x3}, L=max{y−2,y−1,y0,y1,y2,y3} and M=max{z−2,z−1,z0,z1,z2,z3}.
Example 1. We assume an interesting numerical example for the system (1) with x−2=−.22,x−1=−.4, x0=.12,y−2=−.62, y−1=4, y0=.3,z−2=.4,z−1=.53 andz0=−2. (See Figure 1).
In this section, we get the solution's form of the following system of difference equations
xn+1=yn−1znzn+xn−2,yn+1=zn−1xnxn+yn−2, zn+1=xn−1ynyn−zn−2, | (2) |
where n∈N0 and the initial values are non-zero real numbers with x−2≠±z0,≠−2z0, z−2≠y0,≠2y0,≠3y0 and y−2≠2x0,≠±x0.
Theorem 2. Assume that {xn,yn,zn} are solutions of (2). Then for n=0,1,2,...,
x6n−2=(−1)nk3na2n−1(a+2k)n, x6n−1=(−1)nbf3n(f−g)2n(3f−g)n, x6n=(−1)nc3n+1d2n(2c−d)n,x6n+1=ek3n+1(a−k)n(a+k)2n+1, x6n+2=(−1)nf3n+2gn(2f−g)2n+1, x6n+3=(−1)nhc3n+2(c−d)2n+1(c+d)n+1, |
y6n−2=(−1)nc3nd2n−1(2c−d)n, y6n−1=ek3n(a−k)n(a+k)2n, y6n=(−1)nf3n+1gn(2f−g)2n,y6n+1=(−1)nhc3n+1(c−d)2n(c+d)n+1, y6n+2=(−1)nk3n+2a2n(a+2k)n+1, y6n+3=(−1)nbf3n+2(f−g)2n+1(3f−g)n+1, |
and
z6n−2=(−1)nf3ngn−1(2f−g)2n, z6n−1=(−1)nhc3n(c−d)2n(c+d)n, z6n=(−1)nk3n+1a2n(a+2k)n,z6n+1=(−1)nbf3n+1(f−g)2n+1(3f−g)n, z6n+2=(−1)n+1c3n+2d2n+1(2c−d)n, z6n+3=−ek3n+2(a−k)n(a+k)2n+2, |
where x−2=a, x−1=b, x0=c, y−2=d, y−1=e, y0=f, z−2=g, z−1=h and z0=k.
Proof. The result is true for n=0. Now suppose that n>0 and that our claim verified for n−1. That is,
x6n−8=(−1)n−1k3n−3a2n−3(a+2k)n−1, x6n−7=(−1)n−1bf3n−3(f−g)2n−2(3f−g)n−1, x6n−6=(−1)n−1c3n−2d2n−2(2c−d)n−1,x6n−5=ek3n−2(a−k)n−1(a+k)2n−1, x6n−4=(−1)n−1f3n−1gn−1(2f−g)2n−1, x6n−3=(−1)n−1hc3n−1(c−d)2n−1(c+d)n, |
y6n−8=(−1)n−1c3n−3d2n−3(2c−d)n−1, y6n−7=ek3n−3(a−k)n−1(a+k)2n−2, y6n−6=(−1)n−1f3n−2gn−1(2f−g)2n−2,y6n−5=(−1)n−1hc3n−2(c−d)2n−2(c+d)n, y6n−4=(−1)n−1k3n−1a2n−2(a+2k)n, y6n−3=(−1)n−1bf3n−1(f−g)2n−1(3f−g)n, |
and
z6n−8=(−1)n−1f3n−3gn−2(2f−g)2n−2, z6n−7=(−1)n−1hc3n−3(c−d)2n−2(c+d)n−1, z6n−6=(−1)n−1k3n−2a2n−2(a+2k)n−1,z6n−5=(−1)n−1bf3n−2(f−g)2n−1(3f−g)n−1, z6n−4=(−1)nc3n−1d2n−1(2c−d)n−1, z6n−3=−ek3n−1(a−k)n−1(a+k)2n. |
Now from Eq (2), it follows that
x6n−2=y6n−4z6n−3z6n−3+x6n−5=((−1)n−1k3n−1a2n−2(a+2k)n)(−ek3n−1(a−k)n−1(a+k)2n)(−ek3n−1(a−k)n−1(a+k)2n)+(ek3n−2(a−k)n−1(a+k)2n−1)=((−1)nk3na2n−2(a+2k)n)(−k+a+k)=(−1)nk3na2n−1(a+2k)n,y6n−2=z6n−4x6n−3x6n−3+y6n−5=((−1)nc3n−1d2n−1(2c−d)n−1)((−1)n−1hc3n−1(c−d)2n−1(c+d)n)((−1)n−1hc3n−1(c−d)2n−1(c+d)n)+((−1)n−1hc3n−2(c−d)2n−2(c+d)n)=((−1)nc3nd2n−1(2c−d)n−1)c+c−d=(−1)nc3nd2n−1(2c−d)n,z6n−2=x6n−4y6n−3y6n−3−z6n−5=((−1)n−1f3n−1gn−1(2f−g)2n−1)((−1)n−1bf3n−1(f−g)2n−1(3f−g)n)((−1)n−1bf3n−1(f−g)2n−1(3f−g)n)−((−1)n−1bf3n−2(f−g)2n−1(3f−g)n−1)=((−1)n−1f3ngn−1(2f−g)2n−1)(f−3f+g)=(−1)nf3ngn−1(2f−g)2n. |
Also, we see from Eq (2) that
x6n−1=y6n−3z6n−2z6n−2+x6n−4=((−1)n−1bf3n−1(f−g)2n−1(3f−g)n)((−1)nf3ngn−1(2f−g)2n)((−1)nf3ngn−1(2f−g)2n)+((−1)n−1f3n−1gn−1(2f−g)2n−1)=((−1)nbf3n(f−g)2n−1(3f−g)n)(−f+2f−g)=(−1)nbf3n(f−g)2n(3f−g)n,y6n−1=z6n−3x6n−2x6n−2+y6n−4=(−ek3n−1(a−k)n−1(a+k)2n)((−1)nk3na2n−1(a+2k)n)((−1)nk3na2n−1(a+2k)n)+((−1)n−1k3n−1a2n−2(a+2k)n)=(ek3n(a−k)n−1(a+k)2n)−k+a=ek3n(a−k)n(a+k)2n,z6n−1=x6n−3y6n−2y6n−2−z6n−4=((−1)n−1hc3n−1(c−d)2n−1(c+d)n)((−1)nc3nd2n−1(2c−d)n)((−1)nc3nd2n−1(2c−d)n)−((−1)nc3n−1d2n−1(2c−d)n−1)=((−1)n−1hc3n(c−d)2n−1(c+d)n)c−(2c−d)=(−1)nhc3n(c−d)2n(c+d)n. |
Also, we can prove the other relations.
Example 2. See below Figure 2 for system (2) with the initial conditions x−2=11,x−1=5, x0=13,y−2=6, y−1=7, y0=3,z−2=14, z−1=9 andz0=2.
Here, we obtain the form of solutions of the system
xn+1=yn−1znzn+xn−2,yn+1=zn−1xnxn−yn−2, zn+1=xn−1ynyn+zn−2, | (3) |
where n∈N0 and the initial values are non-zero real numbers with x−2≠±z0,≠2z0, z−2≠±y0,≠−2y0 and y−2≠x0,≠2x0,≠3x0.
Theorem 3. If {xn,yn,zn} are solutions of system (3) where x−2=a, x−1=b, x0=c, y−2=d, y−1=e, y0=f, z−2=g, z−1=h and z0=k. Then for n=0,1,2,...,
x6n−2=k3na2n−1(a−2k)n, x6n−1=(−1)nbf3n(f−g)n(f+g)2n, x6n=(−1)nc3n+1dn(d−2c)2n,x6n+1=(−1)nek3n+1(a−k)2n(a+k)n+1, x6n+2=(−1)nf3n+2g2n(2f+g)n+1, x6n+3=(−1)nhc3n+2(c−d)2n+1(3c−d)n+1, |
y6n−2=(−1)nc3ndn−1(d−2c)2n, y6n−1=(−1)nek3n(a−k)2n(a+k)n, y6n=(−1)nf3n+1g2n(2f+g)n,y6n+1=(−1)nhc3n+1(c−d)2n+1(3c−d)n, y6n+2=−k3n+2a2n+1(a−2k)n, y6n+3=(−1)nbf3n+2(f−g)n(f+g)2n+2, |
and
z6n−2=(−1)nf3ng2n−1(2f+g)n, z6n−1=(−1)nhc3n(c−d)2n(3c−d)n, z6n=k3n+1a2n(a−2k)n,z6n+1=(−1)nbf3n+1(f−g)n(f+g)2n+1, z6n+2=(−1)nc3n+2dn(2c−d)2n+1, z6n+3=(−1)n+1ek3n+2(a−k)2n+1(a+k)n+1. |
Proof. As the proof of Theorem 2 and so will be left to the reader.
Example 3. We put the initials x−2=8,x−1=15, x0=13,y−2=6,y−1=7, y0=3,z−2=14,z−1=19 andz0=2, for the system (3), see Figure 3.
The following systems can be treated similarly.
In this section, we deal with the solutions of the following system
xn+1=yn−1znzn−xn−2,yn+1=zn−1xnxn+yn−2, zn+1=xn−1ynyn+zn−2, | (4) |
where n∈N0 and the initial values are non-zero real with x−2≠z0,≠2z0,≠3z0, z−2≠±y0,≠2y0 and y−2≠±x0,≠−2x0.
Theorem 4. The solutions of system (4) are given by
x6n−2=(−1)nk3nan−1(a−2k)2n, x6n−1=(−1)nbf3n(f−g)2n(f+g)n, x6n=(−1)nc3n+1d2n(d+2c)n,x6n+1=−ek3n+1(a−k)2n+1(a−3k)n, x6n+2=(−1)n+1f3n+2g2n+1(2f−g)n, x6n+3=(−1)n+1hc3n+2(c−d)n(c+d)2n+2, |
y6n−2=(−1)nc3nd2n−1(d+2c)n, y6n−1=ek3n(a−k)2n(a−3k)n, y6n=(−1)nf3n+1g2n(2f−g)n,y6n+1=(−1)nhc3n+1(c+d)2n+1(c−d)n, y6n+2=−k3n+2an(a−2k)2n+1, y6n+3=(−1)nbf3n+2(f−g)2n+1(f+g)n+1, |
and
z6n−2=(−1)nf3ng2n−1(2f−g)n, z6n−1=(−1)nhc3n(c+d)2n(c−d)n, z6n=(−1)nk3n+1an(a−2k)2n,z6n+1=(−1)nbf3n+1(f−g)2n(f+g)n+1, z6n+2=(−1)nc3n+2d2n(2c+d)n+1, z6n+3=ek3n+2(a−k)2n+1(a−3k)n+1, |
where x−2=a, x−1=b, x0=c, y−2=d, y−1=e, y0=f, z−2=g, z−1=h and z0=k.
Example 4. Figure 4 shows the behavior of the solution of system (4) with x−2=18,x−1=−15, x0=3,y−2=6, y−1=.7, y0=−3, z−2=4,z−1=−9 andz0=5.
In this section, we obtain the solutions of the difference system
xn+1=yn−1znzn−xn−2,yn+1=zn−1xnxn−yn−2, zn+1=xn−1ynyn−zn−2, | (5) |
where the initials are arbitrary non-zero real numbers with x−2≠z0, z−2≠y0 and y−2≠x0.
Theorem 5. If {xn,yn,zn} are solutions of system (5) where x−2=a, x−1=b, x0=c, y−2=d, y−1=e, y0=f, z−2=g, z−1=h and z0=k. Then
x6n−2=k3na3n−1, x6n−1=bf3n(f−g)3n, x6n=c3n+1d3n,x6n+1=ek3n+1(k−a)3n+1, x6n+2=f3n+2g3n+1, x6n+3=hc3n+2(c−d)3n+2, |
y6n−2=c3nd3n−1, y6n−1=ek3n(k−a)3n, y6n=f3n+1g3n,y6n+1=hc3n+1(c−d)3n+1, y6n+2=k3n+2a3n+1, y6n+3=bf3n+2(f−g)3n+2, |
and
z6n−2=f3ng3n−1, z6n−1=hc3n(c−d)3n, z6n=k3n+1a3n,z6n+1=bf3n+1(f−g)3n+1, z6n+2=c3n+2d3n+1, z6n+3=ek3n+2(k−a)3n+2. |
Example 5. Figure 5 shows the dynamics of the solution of system (5) with x−2=18,x−1=−15,x0=3,y−2=6,y−1=.7, y0=−3,z−2=4,z−1=−9 andz0=5.
This paper discussed the expression's form and boundedness of some systems of rational third order difference equations. In Section 2, we studied the qualitative behavior of system xn+1=yn−1znzn+xn−2,yn+1=zn−1xnxn+yn−2, zn+1=xn−1ynyn+zn−2, first we have got the form of the solutions of this system, studied the boundedness and gave numerical example and drew it by using Matlab. In Section 3, we have got the solution's of the system xn+1=yn−1znzn+xn−2,yn+1=zn−1xnxn+yn−2, zn+1=xn−1ynyn−zn−2, and take a numerical example. In Sections 4–6, we obtained the solution of the following systems respectively, xn+1=yn−1znzn+xn−2,yn+1=zn−1xnxn−yn−2, zn+1=xn−1ynyn+zn−2, xn+1=yn−1znzn−xn−2,yn+1=zn−1xnxn+yn−2, zn+1=xn−1ynyn+zn−2, and xn+1=yn−1znzn−xn−2,yn+1=zn−1xnxn−yn−2, zn+1=xn−1ynyn−zn−2. Also, in each case we take a numerical example to illustrates the results.
This project was funded by the Deanship of Scientific Research (DSR) at King Abdulaziz University, Jeddah, under grant no. (G: 233–130–1441). The authors, therefore, acknowledge with thanks DSR for technical and financial support.
All authors declare no conflicts of interest in this paper.
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