Selçuk J. Appl. Math. Selçuk Journal of
Vol. 6. No. 2. pp. 59-68, 2005 Applied Mathematics
Global Asymptotic Stability in a Rational Di¤erence Equation
Ibrahim Yalç¬nkaya
Department of Mathematics, Education Faculty of Selçuk University, Konya, Turkey; e-mail:iyalcinkaya@ selcuk.edu.tr
Summary. In this paper a su¢ cient condition is obtained for the global asmp-totic stability of the following rational di¤erence equation
xn+1= 1 + xa
n 1+ xnxan 2 xn+ xan 1+ xan 2
; n = 0; 1; 2; :::
where a 2 [0; 1) and the initial values x 2; x 1; x0 (0; 1):
Key words: Rational di¤erence equation; Locally asymptotic stability, Global asymptotic stability; Semicycle; Equilibrium point.
1. Introduction
In [1] Amleh, Kruse and Ladas proved that the positive equilibrium x = 1 of the di¤erence equations
xn+1= xn+ xn 1xn 2 xnxn 1+ xn 2 ; n = 0; 1; 2; ::: xn+1= xn 1+ xnxn 2 xnxn 1+ xn 2 ; n = 0; 1; 2; ::: xn+1= xn+ xn 1xn 2 xnxn 2+ xn 1 ; n = 0; 1; 2; :::
with positive initial conditions x 2; x 1; x0, is globally asymptotically stable. In [3] Papaschinopoluos and Schinas proved that the unique equilibrium x = 1 of the family of di¤erence equations
xn+1= ( k X i=o i6=j;j 1 xn i+ xn j+1xn j+ 1)= k X i=0 xn i; j = 1; 2; :::; k
with positive initial conditions x0; x 1; :::; x k (n f0; 1; :::g and k f1; 2; :::g); is globally asymptotically stable and proved the oscillatory behaviour of the positive solutions of the family of di¤erence equations.
Moreover, in [4] Xianyi and Deming proved that the unique equilibrium of the di¤erence equation
xn+1=
xnxn 1+ 1 xn+ xn 1
; n = 0; 1; 2; :::
with positive initial values x 1; x0, is globally asymptotically stable.
In [6] Kruse and Nasemann studied the global asymptotic stability of the unique positive equilibrium of a discrete dynamical system and as a special result they proved that the unique positive equilibrium x = 1 of the Putnam di¤erence equation
xn+1=
xn+ xn 1+ xn 2xn 3 xnxn 1+ xn 2+ xn 3
; n = 0; 1; 2; :::
is globally asymptotically stable, where the initial values x 3; x 2; x 1; x0 are positive real numbers.
Also, in [5] Xianyi and Deming proved that the equilibrium of the di¤erence equation xn+1= xnxbn 1+ xbn 2+ a xb n 1+ xnxbn 2+ a ; n = 0; 1; 2; :::
is globally asymptotically stable, where a; b [0; 1) and the initial values x 2; x 1; x0 (0; 1):
In this paper, we consider the following rational di¤erence equation,
(1) xn+1=
1 + xan 1+ xnxan 2 xn+ xan 1+ xan 2
where a [0; 1) and the initial values x 2; x 1; x0 (0; 1): Our main aim is to investigate the global asymptotic stability of its solutions.
Here, we review some results which will be useful in our investigation of the behaviour of solutions of Eq.(1). Let I be some interval of real numbers and let f C1[Ik+1; I]: Let x I be an equilibrium point of di¤erence equation
(2) xn+1= f (xn; :::; xn k); n = 0; 1; 2; ::: that is
x = f (x; :::; x): De…nition 1Let x be an equilibrium point of Eq.(2).
(a) The equilibrium x is called locally stable if for every > 0, there exists > 0 such that x0; :::; x k I and jx0 xj + ::: + jx k xj < , then
jxn xj < ; for all n k :
(b) The equilibrium x is called locally asymptotically stable if it is locally stable and if there exists > 0 such that if x0; :::; x k I and jx0 xj + ::: + jx k xj < , then
lim n!1xn=x
(c) The equilibrium x is called golobal attractor if for every x0; :::; x k I we have
lim n!1xn=x:
(d) The equilibrium x is called globally asymptotically stable if it is locally stable and is a global attractor.
De…nition 2Now assume k=2 in Eq.(2) that is, consider the equation xn+1= f (xn; xn 1; xn 2); n = 0 ; 1 ; 2 ; ::: Let r =@f @u(x;x;x) and s = @f @v(x;x;x); t = @f @w(x;x;x)
denote the partial derivates of f (u; v; w) evaluated at an equilibrium x: Then the equation
yn+1= pyn+ syn 1+ tyn 2 n = 0; 1; 2; :::
is called the linearized equation associated with Eq.(2) about the equilibrium point x. Its characteristic equation is
The following result is well-known. [2] Theorem 1Linearized Stability
(a) If all roots of Eq.(3) lie inside of open unit disk j j < 1; then the equilibrium x is locally asymptotically stable.
(b) If at least one of the roots of Eq.(3) has absolute value greater than one, then the equilibrium x is unstable.
(c) A necessary and su¢ cient condition for all roots of Eq.(3) to lie inside the open unit disk j j < 1; is
jr + tj < 1 s;
jr 3tj < 3 + s;
t2 s rt < 1 :
In this case, the locally asymptotically stable equilibrium x is also called a sink. De…nition 3Let x be positive equilibrium point of Eq.(2).
A positive semicycle of a solution fxng of Eq.(2) consists of a "string" of terms fxl; xl+1; :::; xmg, all greater than or equal to the equiblirium x, with l 2 and m 1 such that
either l = 2 or l > 2 and xl 1<x and
either m = 1 or m < 1 and xm+1<x
A negative semicycle of a solution fxng of Eq.(2) consists of a "string" of terms fxl; xl+1; :::; xmg, all less than x, with l 2 and m 1 such that
either l = 2 or l > 2 and xl 1 x and
either m = 1 or m < 1 and xm+1 x:
It is easy to see that Eq.(1) has a unique positive equilibrium x = 1. Eq.(1) is interesting in its own right. To the best of our knowledge, whereas, Eq.(1) has not been investigated so far. Therefore, theoretically, it is meaningful to study the qualitative properties of Eq.(1).
Theorem 2The positive equilibrium point x of Eq.(1) is globally asymptotically stable.
2. Several Lemmas
Lemma 1 A positive solution fxng1n= 2 of Eq.(1) is eventually equal to 1 if and only if
(4) (x0 1)(x 1 1)(x 2 1) = 0:
Proof. Assume that Eq.(4) holds. Then according to Eq.(1), it is easy to see that the following conclusions are true:
(a) if x0= 1, then xn= 1 for n 0; (b) if x 1= 1, then xn= 1 for n 2; (c) if x 2= 1, then xn = 1 for n 1: Conversely, assume that
(5) (x0 1)(x 1 1)(x 2 1) 6= 0:
Then we show
xn6= 1 for any n 1
For the sake of contradiction, assume that for some N 1;
(6) xN = 1 and that xn6= 1 for 2 n N 1 Clearly
1 = xN = 1 + xa
N 2+ xN 1xaN 3 xN 1+ xaN 2+ xaN 3 which implies xN 1= 1 or xN 3= 1, This contradicts Eq.(6).
Remark 1If the initial conditions do not satisfy the equality (4), then xn6= 1 for n 2 and xn6= xn 1for n 0
Lemma 2Let fxng1n= 2 be positive solution of Eq.(1) which is not eventually equal to 1. Then the following conclusions are valid:
(i) (xn+1 xn)(xn 1) < 0 for n 0;
Proof. In view of Eq.(1), we obtain xn+1 xn= (1 xn)(1 + xn+ xan 1) xn+ xan 1+ xan 2 ; n = 0; 1; 2; ::: and xn+1 1 = (xn 1)(xan 2 1) xn+ xan 1+ xan 2 ; n = 0; 1; 2; ::: From this, Inequalities (i) and (ii) follow. The proof is complated. Lemma 3For Eq.(1), the following statements are true:
(A) There exists a positive semicycle of Eq.(1) which has an in…nite number of terms and monotonically tends the positive equilibrium point x;
(B)(i) Every positive semicycle consists of one or two terms; (ii) Every negative semicycle consists of one or three terms;
(iii) Every positive semicycle of length two is followed by a negative semicycle of length three;
(iv) Every positive semicycle of length one is followed by a negative semicycle of length one;
(v) Every negative semicycle of length three is followed by a positive semicycle of length one;
(vii) Every negative semicycle of length one is followed by a positive semicycle of length two.
Proof. (A) If x 2> 1; x 1> 1 and x0 > 1; then in terms of Lemma 2 (ii), it follows that xn > 1; n 2; i.e. this positive semicycle has in…nitive number of terms. Furthermore, according to Lemma 2 (i), we know that xn is strictly decreasing for n 0: So, the limit lim
n!1xn= l exists and is …nite. Taking limit on both sides of Eq.(1), we have
l =l
a+1+ la+ 1 l + 2la and thus l = 1:
We have just seen that a positive semicycle of a solution which is not eventually greater than or equal to 1 has one or two terms and a negative semicycle has at most three terms. Clearly, one of the following cases must occur for some N 0:
(B) CASE1: xN 2> 1; xN > 1 and xN 1< 1 In view of inequalitly (ii) of Lemma 2
xN +1 > 1; xN +2< 1; xN +3 < 1; xN +4< 1; xN +5 > 1; xN +6 < 1; xN +7 > 1; xN +8> 1; xN +9< 1; xN +10< 1; xN +11< 1 and xN +12> 1:
Therefore positive semicycle of length 2 is followed by a negative semicycle of length 3 which in turn is followed by a positive semicycle with exactly one term. CASE2: xN 2< 1; xN 1< 1 and xN < 1
In view of inequalitly (ii) of Lemma 2
xN +1 > 1; xN +2< 1; xN +3 > 1; xN +4> 1; xN +5 < 1; xN +6 < 1; xN +7 < 1; xN +8 > 1; xN +9 < 1; xN +10 > 1; xN +11 > 1 ; xN +12 < 1; xN +13 < 1 and xN +14< 1:
Therefore positive semicycle of length 1 is followed by a negative semicycle of length 1 which in turn is followed by a positive semicycle with exactly two terms. 3. Main Result
Theorem 3The positive equilibrium point x of Eq.(1) is globally asymptotically stable.
Proof. We must show that the positive equilibrium point x of Eq.(1) is both locally asymptotically stable and global attractor. The linearized equation of Eq.(1) about the positive equilibrium point x is
zn+1= 0:zn+ 0:zn 1+ 0:zn 2
and so it is clear from Theoerem 1 that positive equilibrium point x of Eq.(1) is locally asymptotically stable. It remains to verify that every positive solution fxng1n= 2 of Eq.(1) converges to x as n! 1: Namely, We want to prove
(7) lim
n!1xn = x = 1
If the solution fxng1n= 2 of Eq.(1) is nonoscillatory about the positive equi-librium point x of Eq.(1), the according to Lemmas 1 and 3, respectively, we know that the solution is either eventually equal to 1 or an eventually positive one which has an in…nite number of terms and monotonically tends the positive equilibrium point x of Eq.(1) and so Eq.(7) holds. Therefore, it is su¢ cient to prove that Eq.(7) holds for strictly oscillatory solutions. Let now fxng to be strictly oscillatory about the positive equilibrium point x of Eq.(1). By virtue of Lemmas 2 (ii) and 3 one can see that every positive semicycle of this solu-tion has one or two terms and every negative semicycle except perhaps for the
…rst has exactly one or three terms. Every positive semicycle of length one is followed by a negative semicycle of length one and every positive semicycle of length two is followed by a negative semicycle of length three. For the conve-nience of statement, without loss of generality, we use the following denotation. We denote by xp the term of a positive semicycle of length one, followed by xp+1 which is the term of a negative semicycle of length one. Afterwards, there are positive semicycles: xp+2; xp+3, followed by xp+4; xp+5; xp+6 which are the terms of a negative semicycle of length three, in turn followed by the positive semicycle : xp+7 and so on.
Therefore, we have the following sequences consisting positive and negative semi-cycles:
fxp+7ng1n=0; fxp+7n+1gn=01 ; fxp+7n+2; xp+7n+3g1n=0; fxp+7n+4; xp+7n+5; xp+7n+6g1n=0; n = 0; 1; :::
We now have the following a¢ rmations:
(i) xp+7n+2> xp+7n+3; xp+7n+4< xp+7n+5< xp+7n+6;
(ii) xp+7n+1xp+7n+2< 1; xp+7n+3xp+7n+4> 1; xp+7n+6xp+7n+7< 1; (iii) xp+7n+7xp+7n+8> 1:
In fact, the inequality (i) immediately follows from Lemma 2(i) xp+7n+2= 1+xap+7n+xp+7n+1xap+7n 1 xp+7n+1+xap+7n+xap+7n 1 < 1+x a p+7n+xp+7n+1xap+7n 1 xp+7n+1+xp+7n+1xap+7n+x2p+7n+1xap+7n 1 = 1 xp+7n+1; xp+7n+4= 1+xap+7n+2+xp+7n+3xap+7n+1 xp+7n+3+xap+7n+2+xap+7n+1 > 1+x a p+7n+2+xp+7n+3xap+7n+1 xp+7n+3+xp+7n+3xap+7n+2+x2p+7n+3xap+7n+1 = 1 xp+7n+3; and xp+7n+7= 1+xa p+7n+5+xp+7n+6xap+7n+4 xp+7n+6+xap+7n+5+xap+7n+4 < 1+x a p+7n+5+xp+7n+6xap+7n+4 xp+7n+6+xp+7n+6xap+7n+5+x2p+7n+6xap+7n+2 = 1 xp+7n+6;
one can see that (ii) is valid.
As for (iii) immediately obtained from xp+7n+8= 1+xa p+7n+6+xp+7n+7xap+7n+5 xp+7n+7+xap+7n+6+xap+7n+5 > 1+x a p+7n+6+xp+7n+7xap+7n+5 xp+7n+7+xp+7n+7xap+7n+6+x2p+7n+7xap+7n+5 = 1 xp+7n+7; n = 0; 1; 2; ::::
Combining the above observations, we derive xp+7n+1 < 1 xp+7n+2 < 1 xp+7n+3 < xp+7n+4< xp+7n+5< xp+7n+6 (8) < 1 xp+7n+7 < xp+7n+8; n = 0; 1; 2; :::
From Eq.(8), one can see that fxp+7n+1g1n=0is increasing with upper bound 1. So the limit lim
n!1xp+7n+1= L exists and is …nite. Accordingly, by view of Eq. (8), we obtain lim n!1xp+7n+4= limn!1xp+7n+5= limn!1xp+7n+6= L and lim n!1xp+7n+2= limn!1xp+7n+3= limn!1xp+7n+7= 1=L: It su¢ ces to verify that L = 1: To this end, note that
xp+7n+6=
1 + xap+7n+4+ xp+7n+5xap+7n+3 xp+7n+5+ xap+7n+4+ xap+7n+3 Take the limits on both sides of the above equality and obtain
L = 1 + L
a+ L(1=L)a L + La+ (1=L)a and thus, L = 1:
We have shown that lim
n!1xp+7n+m = 1 for m f1; 2; 3; 4; 5; 6; 7g
Thus, we complete the proof of this theorem.
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