# On a Higher-Order Nonlinear Difference Equation

Authors: Bratislav D. Iričanin
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

This paper shows that all positive solutions of a higher-order nonlinear difference equation are bounded, extending some recent results in the
literature.


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## Body

## 1. Introduction

There is a considerable interest in studying nonlinear difference equations nowadays; see, for example, [1–40] and numerous references listed therein.The investigation of the higher-order nonlinear difference equation(1.1)xn=A+xn-mpxn-kr,n∈ℕ0,
where A,r>0 and  p≥0, and k,m∈ℕ, k≠m, was suggested by Stević at numerous talks and in papers (see, e.g., [20, 28, 30, 34–38] and the related references therein).In this paper we show that under some conditions on parametersA,r, and p all positive solutions of the difference equation  (1.2)xn=A+xn-1pxn-kr,n∈ℕ0,
where k∈ℕ∖{1}, are bounded. To do this we modify some methods and ideas from Stević's papers [30, 35–37]. Our motivation stems from these four papers.The reader can find results for some particular cases of (1.2), as well as on some closely related equations treated in, for example, [1, 2, 5–11, 18–20, 26, 30, 33–35, 38, 40].


## 2. Main Result

Here we investigate the boundedness of the positive solutions to  (1.2) for the case 0<p<(rkk/(k-1)k-1)1/k. The following result completely describes the boundedness of positive solutions to (1.2) in this case. The result is an extension of one of the main results in [35].Theorem 2.1.
Assume,p,r>0 and  k∈ℕ∖{1}. Then every positive solution of (1.2) is bounded if
(2.1)0<p<(rkk(k-1)k-1)1/k.Proof.
First note that from (1.2)  it directly follows that
(2.2)xn>A,forn∈ℕ0.
Using (1.2), it follows that
(2.3)xn=A+xn-1pxn-kr=A+(xn-1xn-kr/p)p=A+(Axn-kr/p+xn-2pxn-kr/pxn-k-1r)p=A+(Axn-kr/p+(xn-2xn-kr/p2xn-k-1r/p)p)p=A+(Axn-kr/p+(Axn-kr/p2xn-k-1r/p+xn-3pxn-kr/p2xn-k-1r/pxn-k-2r)p)p=A+(Axn-kr/p+(Axn-kr/p2xn-k-1r/p+(xn-3xn-kr/p3xn-k-1r/p2xn-k-2r/p)p)p)p.
After k steps we obtain the following formula
(2.4)xn=A+(Axn-kr/p+(Axn-kr/p2xn-k-1r/p+(Axn-kr/p3xn-k-1r/p2xn-k-2r/p+⋯+(Axn-kr/pk-1xn-k-1r/pk-2⋯xn-(2k-2)r/p+xn-kpxn-kr/pk-1xn-k-1r/pk-2⋯xn-(2k-2)r/pxn-(2k-1)r)p⋯)p=A+((Axn-kr/pk-1xn-k-1r/pk-2⋯xn-(2k-2)r/p+xn-kp-(r/pk-1)xn-k-1r/pk-2⋯xn-(2k-2)r/pxn-(2k-1)r)pAxn-kr/p+(Axn-kr/p2xn-k-1r/p+(Axn-kr/p3xn-k-1r/p2xn-k-2r/p+⋯+(Axn-kr/pk-1xn-k-1r/pk-2⋯xn-(2k-2)r/p+xn-kp-(r/pk-1)xn-k-1r/pk-2⋯xn-(2k-2)r/pxn-(2k-1)r)p⋯)p.
Two subcases can be considered now.
Case 1 (r≥pk).
Ifr≥pk, then by (2.2) equality (2.4) implies that
(2.5)xn<A+(AAr/p+(AAr/p2+r/p+(AAr/p3+r/p2+r/p+⋯+(AAr/pk-1+r/pk-2+⋯+r/p+1Ar/pk-1+r/pk-2+⋯+r/p+r-p)p⋯)p<∞,
for n≥2k-1. This means that (xn) is a bounded sequence.Case 2 (pk>r).
In this case we have(2.6)p-rpk-1>0.
From (2.4) and (1.2) we further obtain
(2.7)xn=A+(xn-kp-(r/pk-1)xn-k-1r/pk-2⋯xn-(2k-2)r/pxn-(2k-1)rAxn-kr/p+(Axn-kr/p2xn-k-1r/p+(Axn-kr/p3xn-k-1r/p2xn-k-2r/p+⋯+(Axn-kr/pk-1xn-k-1r/pk-2⋯xn-(2k-2)r/p+xn-kp-r/pk-1xn-k-1r/pk-2⋯xn-(2k-2)r/pxn-(2k-1)r)p⋯)p=A+(xn-kp-z0(0)(∏j=1k-2xn-k-jz0(j))xn-(2k-1)rAxn-kr/p+(Axn-kr/p2xn-k-1r/p+(Axn-kr/p3xn-k-1r/p2xn-k-2r/p+⋯+(A∏j=0k-2xn-k-jz0(j)+xn-kp-z0(0)(∏j=1k-2xn-k-jz0(j))xn-(2k-1)r)p⋯)p=A+((A∏j=0k-2xn-k-1-jz1(j)+xn-k-1p-z1(0)∏j=1k-2xn-k-1-jz1(j)xn-2kr)p(p-z0(0))A∏j=0k-2xn-k-jz0(j)Axn-kr/p+(Axn-kr/p2xn-k-1r/p+(Axn-kr/p3xn-k-1r/p2xn-k-2r/p+⋯+(A∏j=0k-2xn-k-jz0(j)+(xn-k(∏j=1k-2xn-k-jz0(j)/p-z0(0))xn-(2k-1)r/(p-z0(0)))p(p-z0(0)))p⋯)p=A+(Axn-kr/p+(A∏j=0k-2xn-k-1-jz1(j)+xn-k-1p-z1(0)∏j=1k-2xn-k-1-jz1(j)xn-2kr)p(p-z0(0))((A∏j=0k-2xn-k-1-jz1(j)+xn-k-1p-z1(0)∏j=1k-2xn-k-1-jz1(j)xn-2kr)p(p-z0(0))Axn-kr/p2xn-k-1r/p+(Axn-kr/p3xn-k-1r/p2xn-k-2r/p+⋯+(A∏j=0k-2xn-k-1-jz1(j)+xn-k-1p-z1(0)∏j=1k-2xn-k-1-jz1(j)xn-2kr)p(p-z0(0)))p⋯)p=⋯=A+(Axn-kr/p+(A∏j=0k-2xn-k-m-jzm(j)+xn-k-mp-zm(0)(∏j=1k-2xn-k-m-jzm(j))xn-2k+1-mr)p∏i=0m-1(p-zi(0))((A∏j=0k-2xn-k-m-jzm(j)+xn-k-mp-zm(0)(∏j=1k-2xn-k-m-jzm(j))xn-2k+1-mr)p∏i=0m-1(p-zi(0))Axn-kr/p2xn-k-1r/p+(Axn-kr/p3xn-k-1r/p2xn-k-2r/p+⋯+(A∏j=0k-2xn-k-m-jzm(j)+xn-k-mp-zm(0)(∏j=1k-2xn-k-m-jzm(j))xn-2k+1-mr)p∏i=0m-1(p-zi(0)))p⋯)p,
for each k∈ℕ∖{1} and every n≥2k+m-1, where the sequences (zm(j)),j=0,1,…,k-2, satisfy the system
(2.8)zm+1(0)=zm(1)p-zm(0),zm+1(1)=zm(2)p-zm(0),…,zm+1(k-3)=zm(k-2)p-zm(0),zm+1(k-2)=rp-zm(0),
and the initial values are given by
(2.9)z0(j)=rpj+1-k,j=0,1,…,k-2.
Note thatpk>r implies that z0(0)<p. Assume zm(0)<p for every m∈ℕ0.
By a direct calculation it follows thatz0(j)<z1(j), j=0,1,…,k-2, which, along with (2.8) implies that (zm(j)),j=0,1,…,k-2, are strictly increasing sequences.
From system (2.8), we have,
(2.10)zm+1(0)=r(p-zm(0))(p-zm-1(0))⋯(p-zm-k+2(0)),m≥k-2.
If it werezm(0)<p, m∈ℕ0, then there was
(2.11)limm→∞zm(0)=z∈(0,p].
Clearly z is a solution of the equation
(2.12)f(x)=x(p-x)k-1-r=0.
Since(2.13)f(0)=f(p)=-r,
and
(2.14)f'(x)=(p-x)k-2(p-kx),
we see that the function  f attains its maximum at the point  x=p/k.
Further, by  assumption (2.1) we get
(2.15)f(pk)=(k-1)k-1kk(pk-rkk(k-1)k-1)<0,
which along with (2.13) implies that (2.12) does not have solutions on (0,p], arriving at a contradiction.
This implies that there is a fixed indexm0∈ℕ such that
(2.16)zm0-1(0)<p,zm0(0)≥p.
From this, inequality (2.2), and identity (2.7) with m=m0, it follows that
(2.17)xn=A+((A∏j=0k-2xn-k-m-jzm(j)+xn-k-mp-zm(0)(∏j=1k-2xn-k-m-jzm(j))xn-2k+1-mr)p∏i=0m-1(p-zi(0))Axn-kr/p+(A∏j=0k-2xn-k-m-jzm(j)+xn-k-mp-zm(0)(∏j=1k-2xn-k-m-jzm(j))xn-2k+1-mr)p∏i=0m-1(p-zi(0))(+⋯+(A∏j=0k-2xn-k-m0-jzm0(j)+xn-k-m0p-zm0(0)(∏j=1k-2xn-k-m0-jzm0(j))xn-2k+1-m0r)p∏i=0m0-1(p-zi(0))Axn-kr/p2xn-k-1r/p+(Axn-kr/p3xn-k-1r/p2xn-k-2r/p+⋯+(A∏j=0k-2xn-k-m0-jzm0(j)+xn-k-m0p-zm0(0)(∏j=1k-2xn-k-m0-jzm0(j))xn-2k+1-m0r)p∏i=0m0-1(p-zi(0)))p⋯)p≤A+(+…+(AA∑j=0k-2zm0(j)+1Ar+zm0(0)-p+∑j=1k-2zm0(j))p∏i=0m0-1(p-zi(0))AAr/p++…+(AA∑j=0k-2zm0(j)+1Ar+zm0(0)-p+∑j=1k-2zm0(j))p∏i=0m0-1(p-zi(0))(+⋯+(AA∑j=0k-2zm0(j)+1Ar+zm0(0)-p+∑j=1k-2zm0(j))p∏i=0m0-1(p-zi(0))AAr/p2+r/p+(AAr/p3+r/p2+r/p+⋯+(AA∑j=0k-2zm0(j)+1Ar+zm0(0)-p+∑j=1k-2zm0(j))p∏i=0m0-1(p-zi(0)))p⋯)p<∞
for n≥2k+m0-1.
From (2.17) the boundedness of the sequence (xn) directly follows, as desired.



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*Source: 418273-2010-07-04.xml*