Raafat Abo-Zeid

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Kyungpook Mathematical Journal 2016; 56(2): 507-516

Published online June 1, 2016

Copyright © Kyungpook Mathematical Journal.

Behavior of Solutions of a Fourth Order Difference Equation

Raafat Abo-Zeid

Department of Basic Science, The Valley Higher Institute for Engineering & Technology, Cairo, Egypt

Received: November 5, 2015; Accepted: February 5, 2016

In this paper, we introduce an explicit formula for the solutions and discuss the global behavior of solutions of the difference equation xn+1=axn-3b-cxn-1xn-3,         n=0,1,

where a, b, c are positive real numbers and the initial conditions x−3, x−2, x−1, x0 are real numbers.

Keywords: difference equation, periodic solution, convergence.

Difference equations have played an important role in analysis of mathematical models of biology, physics and engineering. Recently, there has been a great interest in studying properties of nonlinear and rational difference equations. One can see [3, 7, 9, 10, 11, 12, 13, 14, 16, 17] and the references therein.

In [8], E.M. Elsayed determined the solutions to some difference equations. He obtained the solution to the difference equation

xn+1=xn-31-xn-1xn-3,         n=0,1,

where the initial conditions are arbitrary nonzero positive real numbers. But he didn’t point to any constraints on the initial conditions.

In fact, if we start with initial conditions x0 = 2, x−1 = 1, x−2 = 1, x−3 = 0.5 in equation (1.1), then undefined value for x3 will be obtained. Therefore, additional information about the initial conditions must be given for any solution of equation (1.1) to be well-defined.

In [4], M. Aloqeili discussed the stability properties and semicycle behavior of the solutions of the difference equation

xn+1=xn-1a-xnxn-1,         n=0,1,

with real initial conditions and positive real number a.

In [1], we have discussed the oscillation, boundedness and the global behavior of all admissible solutions of the difference equation

xn+1=Axn-1B-Cxnxn-2,         n=0,1,

where A, B, C are positive real numbers.

In [2], we have also discussed the oscillation, periodicity, boundedness and the global behavior of all admissible solutions of the difference equation

xn+1=Axn-2r-1B-Ci=lkxn-2i,         n=0,1,

where A, B, C are positive real numbers.

In [5], the authors investigated the asymptotic behavior of solutions of the equation

xn+1=axn-1b+cxnxn-1,         n=0,1,

with positive parameters a and c, negative parameter b and nonnegative initial conditions.

In [6], they also used the explicit formula for the solutions of the equation

xn+1=axn-1b+cxnxn-1,         n=0,1,

with positive parameters and nonnegative initial conditions in investigating their behavior.

In [15], H. Sedaghat determined the global behavior of all solutions of the rational difference equations

xn+1=axn-1xnxn-1+b,         xn+1=axnxn-1xn+bxn-2,         n=0,1,

where a, b > 0.

In this paper, we introduce an explicit formula and discuss the global behavior of solutions of the difference equation

xn+1=axn-3b-cxn-1xn-3,         n=0,1,

where a, b, c are positive real numbers and the initial conditions x−3, x−2, x−1, x0 are real numbers.

We define αi = x−2+ix−4+i, i = 1, 2.

Theorem 2.1

Let x−3, x−2, x−1and x0be real numbers such that for any i ∈ {1, 2}, αibck=0n(ab)kfor all n ∈ ℕ. If ab, then the solution {xn}n=-3 ofequation (1.2) is

xn={x-3j=0n-14(ba)2jθ1-c(ba)2j+1θ1-c,n=1,5,9,x-2j=0n-24(ba)2jθ2-c(ba)2j+1θ2-c,n=2,6,10,x-1j=0n-34(ba)2j+1θ1-c(ba)2j+2θ1-c,n=3,7,11,x0j=0n-44(ba)2j+1θ2-c(ba)2j+2θ2-c,n=4,8,12,

where θi=a-b+cαiαi, αi = x−2+ix−4+i, and i = 1, 2.

Proof

We can write the given solution as

x4m+1=x-3j=0m(ba)2jθ1-c(ba)2j+1θ1-c,         x4m+2=x-2j=0m(ba)2jθ2-c(ba)2j+1θ2-c,x4m+3=x-1j=0m(ba)2j+1θ1-c(ba)2j+2θ1-c,         x4m+4=x0j=0m(ba)2j+1θ2-c(ba)2j+2θ2-c,         m=0,1,

It is easy to check the result when m = 0. Suppose that the result is true for m > 0. Then

x4(m+1)+1=ax4m+1b-cx4m+1x4m+3=ax-3j=0m(ba)2jθ1-c(ba)2j+1θ1-cb-cx-3j=0m(ba)2jθ1-c(ba)2j+1θ1-cx-1j=0m(ba)2j+1θ1-c(ba)2j+2θ1-c=ax-3j=0m(ba)2jθ1-c(ba)2j+1θ1-cb-cx-3(j=0m(ba)2jθ1-c)x-1j=0m1(ba)2j+2θ1-c=ax-3j=0m(ba)2jθ1-c(ba)2j+1θ1-cb-cx-1x-3(θ1-c)(1(ba)2m+2θ1-c)=ax-3((ba)2m+2θ1-c)j=0m(ba)2jθ1-c(ba)2j+1θ1-cb((ba)2m+2θ1-c)-cα1(θ1-c)=ax-3((ba)2m+2θ1-c)j=0m(ba)2jθ1-c(ba)2j+1θ1-cb((ba)2m+2θ1-c)-c(a-b)=x-3((ba)2m+2θ1-c)j=0m(ba)2jθ1-c(ba)2j+1θ1-cba((ba)2m+2θ1-c)-ca(a-b)=x-3(ba)2m+2θ1-c)((ba)2m+3θ1-c)j=0m(ba)2jθ1-c(ba)2j+1θ1-c=x-3j=0m+1(ba)2jθ1-c(ba)2j+1θ1-c.

Similarly we can show that

x4(m+1)+2=x-2j=0m+1(ba)2jθ2-c(ba)2j+1θ2-c,         x4(m+1)+3=x-1j=0m+1(ba)2j+1θ1-c(ba)2j+2θ1-c

and

x4(m+1)+4=x0j=0m+1(ba)2j+1θ2-c(ba)2j+2θ2-c.

This completes the proof.

In this section, we investigate the global behavior of equation (1.2) with ab, using the explicit formula of its solution.

We can write the solution of equation (1.2) as

x4m+2t+i=x-4+2t+ij=0mζ(j,t,i),

where ζ(j,t,i)=(ba)2j+tθi-c(ba)2j+t+1θi-c, t ∈ {0, 1} and i ∈ {1, 2}.

Theorem 3.1

Let {xn}n=-3 be a solution of equation (1.2) such that for any i ∈ {1, 2}, αi-bck=0n(ab)kfor all n ∈ ℕ. If αi=b-ac for all i ∈ {1, 2}, then {xn}n=-3 is periodic with prime period 4.

Proof

Assume that αi=b-ac for all i ∈ {1, 2}. Then θi = 0 for all i ∈ {1, 2}.

Therefore,

x4m+2t+i=x-4+2t+ij=0mζ(j,t,i)=x-4+2t+i,         m=0,1,

This completes the proof.

In the following Theorem, suppose that αib-ac for all i ∈ {1, 2}.

Theorem 3.2

Let {xn}n=-3 be a solution of equation (1.2) such that for any i ∈ {1, 2}, αibck=0n(ab)kfor all n ∈ ℕ. Then the following statements are true.

  • If a < b, then {xn}n=-3 converges to 0.

  • If a > b, then {xn}n=-3 converges to a period-4 solution.

Proof
  • If a < b, then ζ(j, t, i) converges to ab<1 as j → ∞, for all t ∈ {0, 1} and i ∈ {1, 2}. So, for every pair (t, i) ∈ {0, 1} × {1, 2} we have for a given 0 < ε < 1 that, there exists j0(t, i) ∈ ℕ such that, | ζ(j, t, i) |< ε for all j ≥ j0(t, i). If we set j0 = max0≤t≤1,1≤i≤2j0(t, i), then for all t ∈ {0, 1} and i ∈ {1, 2} we get x4m+2t+i=x-4+2t+ij=0mζ(j,t,i)=x-4+2t+ij=0j0-1ζ(j,t,i)j=j0mζ(j,t,i)<x-4+2t+ij=0j0-1ζ(j,t,i)ɛm-j0+1.

    As m tends to infinity, the solution {xn}n=-3 converges to 0.

  • If a > b, then ζ(j, t, i) → 1 as j → ∞, t ∈ {0, 1} and i ∈ {1, 2}. This implies that, for every pair (t, i) ∈ {0, 1} × {1, 2}, there exists j1(t, i) ∈ ℕ such that ζ(j, t, i) > 0 for all t ∈ {0, 1} and i ∈ {1, 2}. If we set j1 = max0≤t≤1,1≤i≤2j1(t, i), then for all t ∈ {0, 1} and i ∈ {1, 2} we get x4m+2t+i=x-4+2t+ij=0mζ(j,t,i)=x-4+2t+ij=0j1-1ζ(j,t,i)exp(j=j1mln(ζ(j,t,i))).

We shall test the convergence of the series j=j1ln(ζ(j,t,i)).

Since for all t ∈ {0, 1} and i ∈ {1, 2} we have limjln(ζ(j+1,t,i)ln(ζ(j,t,i)=00, using L’Hospital’s rule we obtain

limjln ζ(j+1,t,i)ln ζ(j,t,i)=(ba)2<1.

It follows from the ratio test that the series j=j1ln ζ(j,t,i) is convergent. This ensures that there are four positive real numbers μti, t ∈ {0, 1} and i ∈ {1, 2} such that

limmx4m+2t+i=μti,         t{0,1}         and         i{1,2}

where

μti=x-4+2t+ij=0(ba)2j+tθi-c(ba)2j+t+1θi-c,         t{0,1}         and         i{1,2}.

This completes the proof.

Example (1)

Figure 1. shows that if a = 2, b = 3, c = 1 (a < b), then the solution {xn}n=-3 of equation (1.2) with initial conditions x−3 = 0.2, x−2 = 2, x−1 = −2 and x0 = 0.4 converges to 0.

Example (2)

Figure 2. shows that if a = 3, b = 1, c = 0.8 (a > b), then the solution {xn}n=-3 of equation (1.2) with initial conditions x−3 = 0.2, x−2 = 2, x−1 = −2 and x0 = 0.4 converges to a period-4 solution.

a = b

In this section, we investigate the behavior of the solution of the difference equation

xn+1=axn-3a-cxn-1xn-3,         n=0,1,

Theorem 4.1

Let x−3, x−2, x−1and x0be real numbers such that for any i ∈ {1, 2}, αiac(n+1)for all n ∈ ℕ. Then the solution{xn}n=-3ofequation (3.1) is

xn={x-3j=0n-14a-(2j)cα1a-(2j+1)cα1,n=1,5,9,x-2j=0n-24a-(2j)cα2a-(2j+1)cα2,n=2,6,10,x-1j=0n-34a-(2j+1)cα1a-(2j+2)cα1,n=3,7,11,x0j=0n-44a-(2j+1)cα2a-(2j+2)cα2,n=4,8,12,
Proof

The proof is similar to that of Theorem (2.1) and will be omitted.

We can write the solution of equation (3.1) as

x4m+2t+i=x-4+2t+ij=0mγ(j,t,i),

where γ(j,t,i)=a-(2j+t)cαia-(2j+t+1)cαi, t ∈ {0, 1} and i ∈ {1, 2}.

Theorem 4.2

Let{xn}n=-3be a nontrivial solution of equation (3.1) such that for any i ∈ {1, 2}, αiac(n+1)for all n ∈ ℕ. If αi = 0 for all i ∈ {1, 2}, then{xn}n=-3is periodic with prime period 4.

Proof

Assume that αi = 0 for all i ∈ {1, 2}. Then γ(j, t, i) = 1 for all t ∈ {0, 1} and i ∈ {1, 2}. Therefore,

x4m+2t+i=x-4+2t+ij=0mγ(j,t,i)=x-4+2t+i,         m=0,1,

This completes the proof.

In the following Theorem, suppose that αi ≠ 0 for all i ∈ {1, 2}.

Theorem 4.3

Let{xn}n=-3be a solution of equation (3.1) such that for any i ∈ {1, 2}, αiac(n+1)for all n ∈ ℕ. Then{xn}n=-3converges to 0.

Proof

It is clear that γ(j, t, i) → 1 as j → ∞, t ∈ {0, 1} and i ∈ {1, 2}. This implies that, for every pair (t, i) ∈ {0, 1}×{1, 2} there exists j2(t, i) ∈ ℕ such that, γ(j, t, i) > 0 for all t ∈ {0, 1} and i ∈ {1, 2}. If we set j2 = max0≤t≤1,1≤i≤2j2(t, i), then for all t ∈ {0, 1} and i ∈ {1, 2} we get

x4m+2t+i=x-4+2t+ij=0mγ(j,t,i)=x-4+2t+ij=0j2-1γ(j,t,i)exp(-j=j2mln1γ(j,t,i)).

We shall show that j=j2ln1γ(j,t,i)=j=j2lna-(2j+t+1)cαia-(2j+t)cαi=, by considering the series j=j2-cαia-(2j+t)cαi. As

limjln(1/γ(j,t,i))-cαi/(a-(2j+t))cαi=limjln((a-(2j+t+1)cαi)/(a-(2j+t)cαi))-cαi/(a-(2j+t)cαi)=1,

using the limit comparison test, we get j=j2ln1γ(j,t,i)=. Then

x4m+2t+i=x-4+2t+ij=0j2-1γ(j,t,i)exp(-j=j2mln1γ(j,t,i))

converges to 0 as m→∞. Therefore, {xn}n=-3 converges to 0.

a = b = c

In this section, we investigate the behavior of the solution of the difference equation

xn+1=xn-31-xn-1xn-3,         n=0,1,

Theorem 5.1

Let x−3, x−2, x−1and x0be real numbers such that for any i ∈ {1, 2}, αi1n+1for all n ∈ ℕ. Then the solution{xn}n=-3ofequation (3.3) is

xn={x-3j=0n-141-(2j)α11-(2j+1)α1,n=1,5,9,x-2j=0n-241-(2j)α21-(2j+1)α2,n=2,6,10,x-1j=0n-341-(2j+1)α11-(2j+2)α1,n=3,7,11,x0j=0n-441-(2j+1)α21-(2j+2)α2,n=4,8,12,
Proof

The proof is similar to that of Theorem (2.1) and will be omitted.

Theorem 5.2

Let{xn}n=-3be a nontrivial solution of equation (3.3) such that for any i ∈ {1, 2}, αi-bck=0n(ab)kfor all n ∈ ℕ. If αi = 0 for all i ∈ {1, 2}, then{xn}n=-3is periodic with prime period 4.

Proof

Assume that αi = 0 for all i ∈ {1, 2}. Then

x4m+2t+i=x-4+2t+i,         m=0,1,

This completes the proof.

In the following Theorem, suppose that αi ≠ 0 for all i ∈ {1, 2}.

Theorem 5.3

Let{xn}n=-3be a solution of equation (3.3) such that for any i ∈ {1, 2}, αi1n+1for all n ∈ ℕ. Then{xn}n=-3converges to 0.

Example (3)

Figure 3. shows that if a = b = 1, c = 1.5, then the solution {xn}n=-3 of equation (3.1) with initial conditionsx−3 = 5, x−2 = −1, x−1 = 1.3 and x0 = −1.1 converges to 0.

Example (4)

Figure 4. shows that if a = b = c, then the solution {xn}n=-3 of equation (3.3) with initial conditions x−3 = 5, x−2 = 1, x−1 = 1.3 and x0 = −1.1 converges to 0.

Fig. 1. xn+1=2xn-33-xn-1xn-3
Fig. 2. xn+1=3xn-31-0.8xn-1xn-3
Fig. 3. xn+1=xn-31-1.5xn-1xn-3
Fig. 4. xn+1=xn-31-xn-1xn-3
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