Applied Mathematics and Computation 135 (2003) 377–382 www.elsevier.com/locate/amc
On the global attractivity of systems of nonlinear difference equations H. El-Owaidy, H.Y. Mohamed
*
Mathematics Department, Faculty of Science, Al Azhar University, Nasr-City, Egypt
Abstract We study two systems of nonlinear difference equations, the first system of the form Xnþ1 ¼ AXn þ F ðXnk Þ, where A is an m m matrix and F 2 C½½0; mÞm ; ð0; mÞm , and the second Xnþ1 ¼ GðXn ; . . . ; Xnk Þ;
n ¼ 0; 1 . . . ;
where G 2 C½ð0; mÞmðkþ1Þ ; ð0; mÞm . We study the global attractivity for the two systems under some sufficient conditions. Ó 2002 Published by Elsevier Science Inc. Keywords: Global attractivity; System of nonlinear difference equations
1. Introduction Our aim in this paper is to establish that every positive solutions of the systems of equations: Xnþ1 ¼ AXn þ F ðXnk Þ; Xnþ1 ¼ GðXn ; . . . ; Xnk Þ;
n ¼ 0; 1; . . . ; n ¼ 0; 1; . . . ;
ð1:1Þ ð1:2Þ
where G 2 C½ð0; 1Þmðkþ1Þ ; ð0; 1Þ ; k is a positive integer, A is an m m matrix m m and F 2 C½½0; 1Þ ; ð0; 1Þ , are a global attractor under some sufficient
*
Corresponding author. E-mail address:
[email protected] (H.Y. Mohamed).
0096-3003/02/$ - see front matter Ó 2002 Published by Elsevier Science Inc. PII: S 0 0 9 6 - 3 0 0 3 ( 0 1 ) 0 0 3 3 8 - 1
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conditions. We extend the results in [1,2] on the scalar difference equations to the systems of equations (1.1) and (1.2). First, we need these definitions to use it for the general case. T
Definition 1.1. We say that the vector X P Y , such that X ¼ ðX 1 ; . . . ; X m Þ , T m Y ¼ ðY 1 ; . . . ; Y m Þ , X ; Y 2 ½0; 1Þ if and only if X i P Y i for all i ¼ 1; . . . ; m. m
Lemma 1.1. ð½0; 1Þ ; PÞ is partially ordered relation. Proof. The proof is clear. Definition 1.2. We say that the operator H is nondecreasing if and only if m m X P Y implies H ðxÞ P H ðY Þ, where H 2 C½½0; mÞ ; ð0; mÞ . Definition 1.3. We say that the operator H is nonincreasing if and only if m m X P Y implies H ðX Þ 6 H ðY Þ, where H 2 C½½0; mÞ ; ð0; mÞ and so we can define increasing and decreasing functions similarly. Definition 1.4. We say that lim Xn ¼ C if and only if lim Xni ¼ C i . Also we can n!1 n!1 define lim supXn and lim inf Xn similarly. n!1
n!1
Lemma 1.2. The following statements are true: (a) lim inf Xn 6 lim supXn . n!1
n!1
(b) If lim supXn ¼ lim inf Xn , then lim Xn exists and n!1
n!1
n!1
lim Xn ¼ lim sup Xn ¼ lim inf Xn :
n!1
n!1
n!1
(c) If A is an m m matrix and A P 0ðaij P 0Þ nonnegative matrix, then A is nondecreasing. 1 (d) If A is nonnegative and kAk < 1, then ðI AÞ is nondecreasing. Proof. The proof of a, b and c is clear, and we shall prove (d). We can write ðI AÞ1 as ðI AÞ
1
¼ I þ A þ A2 þ
since kAk < 1. From (c), A is nondcreasing, so if X P Y , then we have by induction n X i¼0
An X P
n X i¼0
An Y :
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379
By taking n tends to infinity we get, 1
1
ðI AÞ X P ðI AÞ Y :
Lemma 1.3. If Xnþ1 6 AXn þ C;
n ¼ 0; 1; . . . ;
m
C 2 ½0; 1Þ , C is a constant vector and kAk < 1, A P 0. Then lim sup Xn 6 ðI AÞ1 C: n!1
Proof. By induction we find Xnþ1 6 An X0 þ ðI þ A þ þ An1 ÞC: So lim supXn 6 ðI AÞ1 C, since lim An X0 ¼ 0, and this completes the n!1
n!1
proof.
2. Global attractivity of Xnþ1 ¼ AXn þ FðXnk Þ Now, we extent the results in [1] on the scalar difference equations to the system of equations (open problem in [3]) Xnþ1 ¼ AXn þ F ðXnk Þ:
ð2:1Þ m
m
Theorem 2.1. If Xnþ1 ¼ AXn þ F ðXnk Þ, where F 2 C½½0; 1Þ ; ð0; 1Þ and the following conditions are satisfied: ðH1 Þ F ðU Þ is decreasing in U. ðH2 Þ A P 0 and kAk < 1 or (the spectral radius qðAÞ < 1; qðAÞ ¼ maxfjkj: k is an eigenvalue of Ag). ðH3 Þ Assume that the system U ¼ ðI AÞ1 F ðLÞ
and
L ¼ ðI AÞ1 F ðU Þ
ð2:2Þ
has exactly one solution fL; U g and L; U > 0. ðH4 Þ The initial conditions, Xn > 0 for n ¼ k; . . . ; 0. Then Eq. (2.1) has a unique positive equilibrium X , U ¼ L ¼ X and every solution of Eq. (2.1) is attracted to X ; that is lim Xn ¼ X . n!1
Proof. (i) Existence. Eq. (2.1) has at least one positive solution because if we set 1
uðX Þ ¼ X ðI AÞ F ðX Þ: Then we have uð0Þ ¼ ðI AÞ1 F ð0Þ < 0
and
uð1Þ ¼ 1:
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Then Eq. (2.1) has at least one positive solution. (ii) Uniqueness. Clearly X is a positive equilibrium of (2.1) if and only if X is a positive solution of the equation U ¼ ðI AÞ1 F ðU Þ: Since, the system (2.2) has exactly one solution, then from this and (i), Eq. (2.1) has a unique positive equilibrium X . (iii) Global attractivity. From (2.1) and H1 we have Xnþ1 6 AXn þ F ðXnk Þ 6 AXn þ F ð0Þ and so by Lemma 1.3 1
lim sup Xn 6 ðI AÞ F ð0Þ: n!1
Now we can define, U1 ¼ ðI AÞ1 F ð0Þ
and
L1 ¼ ðI AÞ1 F ðU1 Þ
and for r ¼ 1; 2; . . . we have 1
Urþ1 ¼ ðI AÞ F ðLr Þ
and
1
Lrþ1 ¼ ðI AÞ F ðUrþ1 Þ:
Now we can see by induction that fUr g is a decreasing sequence, and fLr g is an increasing sequence and that for r ¼ 1; 2; . . .. We have Lr 6 lim inf Xn 6 lim sup Xn 6 Ur ; n!1
L ¼ lim Lr and U ¼ lim Ur : r!1
n!1
r!1
Then fU ; Lg satisfies 1
U ¼ ðI AÞ F ðLÞ;
1
L ¼ ðI AÞ F ðU Þ:
From ðH3 Þ we have L ¼ U ¼ X , then lim Xn ¼ X and the proof is comn!1 plete. Corollary 2.2. Assume that H1 , H2 and H4 are satisfied, suppose that F has a unique fixed point X > 0 and that X is a global attractor of all positive solutions of the first-order difference equations: 1
Ynþ1 ¼ ðI AÞ F ðYn Þ;
n ¼ 0; 1; . . . ;
ð2:3Þ
where Y0 > 0. Then X is a global attractor of all positive solutions of the system of Eq. (2.1).
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3. Global attractivity of Xnþ1 ¼ GðXn ; . . . ; Xnk Þ We extent the results in [2] on the scalar difference to the system of equations, Xnþ1 ¼ GðXn ; . . . ; Xnk Þ mðkþ1Þ
n ¼ 0; 1; . . . ;
ð3:1Þ
m
where G 2 C½ð0; 1Þ ; ð0; 1Þ is increasing in each of its arguments and Xk . . . X0 are positive [1–3]. Theorem 3.1. Assume that Eq. (3.1) has a unique positive equilibrium X and suppose that the function H is defined by H ðX Þ ¼ GðX ; . . . ; X Þ satisfies X > X if and only if H ðX Þ < X , and X < X if and only if H ðX Þ > X . Then X is a global attractor of all positive solutions of Eq. (3.1). Proof. Set m¼ M¼
m 1 m T 1 minfXk ; . . . ; X01 ; X g; . . . ; min Xk ; . . . ; X0m ; X ;
m 1 m T 1 maxfXk ; . . . ; X01 ; X g; . . . ; max Xk ; . . . ; X0m ; X :
Clearly, 0 < m 6 X 6 M < 1 and for n ¼ k; . . . ; 0, we have ð3:2Þ
m 6 Xn 6 M: We shall prove it for all n P k. Let (3.2) hold for all n 6 N . Then XN þ1 ¼ GðXN ; . . . XN k Þ 6 GðM . . . ; MÞ ¼ hðMÞ 6 M
and by induction, we can get Xn 6 M. Similarly, we have Xn P m for all n P k. Now, we can set k ¼ lim inf Xn ; l ¼ lim sup Xn : n!1
n!1
Then, k ¼ lim inf GðXn ; . . . :Xnk Þ P Gðk; . . . kÞ ¼ H ðkÞ: n!1
So k P X . Similarly l 6 X . Then k ¼ l ¼ X , and the proof is complete.
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