Electronic Journal of Qualitative Theory of Differential Equations 2013, No. 47, 1-18;http://www.math.u-szeged.hu/ejqtde/
ON A HIGHER-ORDER SYSTEM OF DIFFERENCE EQUATIONS
STEVO STEVI ´C∗, MOHAMMED A. ALGHAMDI, ABDULLAH ALOTAIBI, AND NASEER SHAHZAD
Abstract. Here we study the following system of difference equations xn=f−1 cnf(xn−2k)
an+bnQk
i=1g(yn−(2i−1))f(xn−2i)
! ,
yn=g−1 γng(yn−2k) αn+βnQk
i=1f(xn−(2i−1))g(yn−2i)
! ,
n∈N0,wheref andgare increasing real functions such thatf(0) =g(0) = 0, and coefficientsan, bn,cn,αn,βn,γn,n∈N0, and initial valuesx−i,y−i, i∈ {1,2, . . . ,2k} are real numbers. We show that the system is solvable in closed form, and study asymptotic behavior of its solutions.
1. Introduction
Difference equations and systems of difference equations attract lots of attention (see, e.g. [1]–[49] and references therein). Among numerous topics in this area of mathematics, studying systems of difference equations is one of some recent interest [7, 9, 11, 15, 16, 17, 18, 19, 21, 23, 35, 36, 39, 40, 41, 42, 44, 45, 46, 47, 48], while solving difference equations and applying them in other areas of sciences re- attracted some attention quite recently (see, for example, [1, 2, 6, 7, 22, 28, 29, 32, 33, 35, 36, 37, 39, 40, 42, 43, 44, 45, 46, 47, 48]). Among others, the attention was trigged off by note [28] where an equation is solved in an elegant way. Some old methods for solving difference equations can be found, e.g., in [14].
In [44], S. Stevi´c studied the following system of difference equations xn= cnxn−4
an+bnyn−1xn−2yn−3xn−4, yn= γnyn−4
αn+βnxn−1yn−2xn−3yn−4, n∈N0, (1) with real coefficientsan, bn,cn,αn,βn,γn,n∈N0, and initial valuesx−i,y−i,i∈ {1,2,3,4}, such thatcn 6= 0,γn 6= 0,n∈N0. He showed that system (1) is solvable in closed form, and described behavior of all well-defined solutions of the system for constant coefficientsan, bn,cn,αn,βnandγn. Paper [44] is a natural continuation of his previous investigations in [7, 28, 35, 36, 37, 39, 40, 43, 45, 46, 47, 48], where related difference equations and systems of difference equations were considered.
2010Mathematics Subject Classification. Primary 39A10.
Key words and phrases. Solvable system, system of difference equations, asymptotic behavior.
∗Corresponding author.
EJQTDE, 2013 No. 47, p. 1
Motivated by this line of investigations, here we study the following system of difference equations
xn =f−1 cnf(xn−2k) an+bnQk
i=1g(yn−(2i−1))f(xn−2i)
! ,
yn =g−1 γng(yn−2k) αn+βnQk
i=1f(xn−(2i−1))g(yn−2i)
!
, n∈N0,
(2)
wheref andg are increasing real functions, such that
f(0) =g(0) = 0, (3)
and coefficients an, bn, cn, αn, βn, γn, n ∈ N0, and initial values x−i, y−i, i ∈ {1,2, . . . ,2k}are real numbers.
We show that system (2) is also solvable in closed form, and study the behavior of well-defined solutions of the system when the sequencesan, bn, cn,αn, βn and γn are constant.
Recall that solution (xn, yn)n≥−2k, of system (2) is periodic with periodp, if xn+p=xn and yn+p=yn, n≥ −2k.
For some results on the periodicity or asymptotic periodicity see, e.g., [4, 5, 10, 11, 12, 13, 14, 23, 24, 26, 27, 31, 34, 38, 41, 49].
2. Solvability of system (2) in closed form
Assume thatx−i 6= 0,y−i 6= 0, i ∈ {1,2, . . . ,2k}. Then (2), the monotonicity of f and g and conditions f(0) = g(0) = 0, imply that xn 6= 0 and yn 6= 0, for every n ∈ N0. Then in this case the following change of variables along with the invertibility of functionsf andg
un= 1
Qk−1
i=0 f(xn−2i)g(yn−2i−1), vn= 1 Qk−1
i=0 g(yn−2i)f(xn−2i−1), n≥ −1, (4) transforms system (2) into the next system of linear difference equations
un =an
cn
vn−1+bn
cn
, vn= αn
γn
un−1+βn
γn
, n∈N0. (5)
From (5) we have that
un= anαn−1
cnγn−1un−2+anβn−1 cnγn−1 +bn
cn
, vn= αnan−1
γncn−1vn−2+αnbn−1
γncn−1 +βn
γn
, n∈N,
EJQTDE, 2013 No. 47, p. 2
from which we get (for details see [44]) u2n=u0
n
Y
j=1
a2jα2j−1 c2jγ2j−1 +
n
X
i=1
a2iβ2i−1 c2iγ2i−1 +b2i
c2i n
Y
s=i+1
a2sα2s−1
c2sγ2s−1, (6) u2n−1=u−1
n
Y
j=1
a2j−1α2j−2
c2j−1γ2j−2 +
n
X
i=1
a2i−1β2i−2
c2i−1γ2i−2 +b2i−1
c2i−1 n
Y
s=i+1
a2s−1α2s−2
c2s−1γ2s−2, (7) v2n=v0
n
Y
j=1
α2ja2j−1
γ2jc2j−1 +
n
X
i=1
α2ib2i−1
γ2ic2i−1 +β2i
γ2i
n Y
s=i+1
α2sa2s−1
γ2sc2s−1, (8) v2n−1=v−1
n
Y
j=1
α2j−1a2j−2 γ2j−1c2j−2
+
n
X
i=1
α2i−1b2i−2 γ2i−1c2i−2
+β2i−1 γ2i−1
n Y
s=i+1
α2s−1a2s−2 γ2s−1c2s−2
. (9) From (4) we have that
f(x2km+i) = v2km+i−1 u2km+i
f(x2k(m−1)+i), i∈ {0, . . . ,2k−1}, m∈N0,and
g(y2km+i) = u2km+i−1
v2km+i g(y2k(m−1)+i), i∈ {0, . . . ,2k−1},
for 2km+i ≥ 0, from which along with the invertibility of functions f and g it follows that for everym∈N0 and eachi∈ {0, . . . ,2k−1}
x2km+i=f−1
f(xi)
m
Y
j=1
v2kj+i−1
u2kj+i
, (10)
y2km+i=g−1
g(yi)
m
Y
j=1
u2kj+i−1 v2kj+i
. (11) Using (6)–(9) in (10) and (11) we get solutions of system (2) in closed form.
3. System (2) with constant coefficients Let
an= ˆa, bn= ˆb, cn = ˆc, αn= ˆα, βn= ˆβ and γn= ˆγ, n∈N0, then we have
xn=f−1 ˆcf(xn−2k) ˆ
a+ ˆbQk
i=1g(yn−(2i−1))f(xn−2i)
! ,
yn=g−1 γg(yˆ n−2k) ˆ
α+ ˆβQk
i=1f(xn−(2i−1))g(yn−2i)
!
, n∈N0.
(12)
If ˆc = 0, then sincef(0) = f−1(0) = 0 we have that xn = 0, n ∈ N0, so that g(yn) =αˆγˆg(yn−2k) forn∈Nand consequently
y2km−i=g−1 γˆ
ˆ α
m g(y−i)
,
EJQTDE, 2013 No. 47, p. 3
for everym∈N0andi∈ {0,1, . . . ,2k−1}.
If ˆγ = 0, then since g(0) =g−1(0) = 0 we have that yn = 0, n∈N0, implying f(xn) = ˆaˆcf(xn−2k) forn∈Nand consequently
x2km−i=f−1 ˆc
ˆ a
m f(x−i)
, for everym∈N0andi∈ {0,1, . . . ,2k−1}.
From now on we will assume that ˆc 6= 0 and ˆγ 6= 0. Note that in this case, system (12) can be written in the following form
xn=f−1 f(xn−2k) a+bQk
i=1g(yn−(2i−1))f(xn−2i)
! ,
yn=g−1 g(yn−2k)
α+βQk
i=1f(xn−(2i−1))g(yn−2i)
!
, n∈N0,
(13)
where a = ˆa/ˆc, b = ˆb/ˆc, α = ˆα/ˆγ and β = ˆβ/ˆγ. Therefore we will study system (13) instead of system (12).
Assumex−i6= 0 and y−i6= 0 for everyi∈ {1,2, . . . ,2k}. System (5) becomes un =avn−1+b, vn=αun−1+β, n∈N0, (14) which implies that
un =aαun−2+aβ+b, (15)
vn =aαvn−2+αb+β, n∈N. (16)
From (15) and (16) (or (6)–(9)) we obtain
u2n−l=u−l(aα)n+ (aβ+b)1−(aα)n 1−aα
= aβ+b+ (aα)n(u−l(1−aα)−aβ−b)
1−aα , (17)
n∈N0,l∈ {0,1}, ifaα6= 1, or
u2n−l=u−l+ (aβ+b)n, n∈N0, l∈ {0,1}, (18) ifaα= 1,and
v2n−l=v−l(aα)n+ (αb+β)1−(aα)n 1−aα
= αb+β+ (aα)n(v−l(1−aα)−αb−β)
1−aα , (19)
n∈N0,l∈ {0,1}, ifaα6= 1, or
v2n−l=v−l+ (αb+β)n, n∈N0, l∈ {0,1}, (20) ifaα= 1.
From relations (17)–(20) we easily obtain the following formulae for solutions of system (13).
EJQTDE, 2013 No. 47, p. 4
Case aα= 1. In this case we have that x2km+2s=f−1
f(x2s)
m
Y
j=1
v2kj+2s−1
u2kj+2s
=f−1
f(x2s)
m
Y
j=1
v−1+ (αb+β)(kj+s) u0+ (aβ+b)(kj+s)
, (21)
x2km+2s+1=f−1
f(x2s+1)
m
Y
j=1
v2kj+2s u2kj+2s+1
=f−1
f(x2s+1)
m
Y
j=1
v0+ (αb+β)(kj+s) u−1+ (aβ+b)(kj+s+ 1)
, (22)
y2km+2s=g−1
g(y2s)
m
Y
j=1
u2kj+2s−1
v2kj+2s
=g−1
g(y2s)
m
Y
j=1
u−1+ (aβ+b)(kj+s) v0+ (αb+β)(kj+s)
, (23)
y2km+2s+1=g−1
g(y2s+1)
m
Y
j=1
u2kj+2s v2kj+2s+1
=g−1
g(y2s+1)
m
Y
j=1
u0+ (aβ+b)(kj+s) v−1+ (αb+β)(kj+s+ 1)
, (24) for everym∈N0ands∈ {0,1, . . . , k−1}.
Case aα6= 1. We have x2km+2s=f−1
f(x2s)
m
Y
j=1
v2kj+2s−1 u2kj+2s
=f−1
f(x2s)
m
Y
j=1
(αb+β+ (aα)kj+s(v−1(1−aα)−αb−β)) (aβ+b+ (aα)kj+s(u0(1−aα)−aβ−b))
, (25) x2km+2s+1=f−1
f(x2s+1)
m
Y
j=1
v2kj+2s
u2kj+2s+1
=f−1
f(x2s+1)
m
Y
j=1
(αb+β+ (aα)kj+s(v0(1−aα)−αb−β)) (aβ+b+ (aα)kj+s+1(u−1(1−aα)−aβ−b))
, (26) EJQTDE, 2013 No. 47, p. 5
y2km+2s=g−1
g(y2s)
m
Y
j=1
u2kj+2s−1 v2kj+2s
=g−1
g(y2s)
m
Y
j=1
(aβ+b+ (aα)kj+s(u−1(1−aα)−aβ−b)) (αb+β+ (aα)kj+s(v0(1−aα)−αb−β))
, (27)
y2km+2s+1=g−1
g(y2s+1)
m
Y
j=1
u2kj+2s
v2kj+2s+1
=g−1
g(y2s+1)
m
Y
j=1
(aβ+b+ (aα)kj+s(u0(1−aα)−aβ−b)) (αb+β+ (aα)kj+s+1(v−1(1−aα)−αb−β))
, (28) for everym∈N0ands∈ {0,1, . . . , k−1}.
4. Behavior of solutions of system(13)
Prior to proving the main results on behavior of solutions of system (13) we present the following extension of Lemma 1 in [44] which guarantees the existence of 2kand 4kperiodic solutions of system (13).
Lemma 1. Assume that aα6= 1, f, g:R→R are increasing functions satisfying the conditions in (3). Then the following statements are true.
(a) Ifαb+β=aβ+b, then system (13)has2k-periodic solutions.
(b) Ifαb+β=−(aβ+b), andf andg are odd, then system (13)has4k-periodic solutions.
Proof. It is easy to see that system (14) has a unique equilibrium solution un = ¯u= aβ+b
1−aα 6= 0, vn= ¯v= αb+β
1−aα 6= 0, n≥ −1.
This along with (4) implies that
f(xn) = 1−aα
(aβ+b)g(yn−2k+1)Qk−1
j=1g(yn−2j+1)f(xn−2j)
=1−aα
aβ+bvn−1f(xn−2k) = αb+β
aβ+bf(xn−2k), n∈N0, (29) and
g(yn) = 1−aα
(αb+β)f(xn−2k+1)Qk−1
j=1f(xn−2j+1)g(yn−2j)
=1−aα
αb+βun−1g(yn−2k) = aβ+b
αb+βg(yn−2k), n∈N0. (30) (a) Since αb+β = aβ+b, from (29) and (30) we get f(xn) = f(xn−2k) and g(yn) =g(yn−2k), from which it follows thatxn =xn−2k andyn =yn−2k that is, there is a 2k-periodic solution of system (13).
(b)Sinceαb+β =−(aβ+b), from (29), (30), and sincef andgare odd functions, we get f(xn) = −f(xn−2k) = f(−xn−2k) and g(yn) = −g(yn−2k) = g(−yn−2k) EJQTDE, 2013 No. 47, p. 6
which implies that xn =−xn−2k andyn =−yn−2k, and consequently xn =xn−4k andyn =yn−4k, that is, there is a 4k-periodic solution of system (13).
Theorem 1. Assume that aα = 1, f, g :R →R are continuous, odd, increasing functions satisfying the conditions in (3), and (xn, yn)n≥−2k is a well-defined so- lution of system (13) such that x−i 6= 06=y−i, i= 1, . . . ,2k. Then the following statements are true.
(a) If|αb+β|<|aβ+b|, thenxn→0 and|yn| →g−1(+∞), asn→ ∞.
(b) If|αb+β|>|aβ+b|, thenyn→0 and|xn| →f−1(+∞), asn→ ∞.
(c) If αb+β = aβ+b 6= 0 and v−1αb+β−u0 > 0, then |x2km+2s| → f−1(+∞), s ∈ {0,1, . . . , k−1}, asm→ ∞.
(d) Ifαb+β=aβ+b6= 0andv−1αb+β−u0 <0, then|x2km+2s| →0,s∈ {0,1, . . . , k−1}, asm→ ∞.
(e) If αb+β = aβ +b 6= 0 and v−1 = u0, then the sequences x2km+2s, s ∈ {0,1, . . . , k−1}, are convergent.
(f ) If αb+β = aβ +b 6= 0 and v0αb+β−u−1 > 1, then |x2km+2s+1| → f−1(+∞), s∈ {0,1, . . . , k−1}, asm→ ∞.
(g) Ifαb+β =aβ+b6= 0and v0αb+β−u−1 <1, then|x2km+2s+1| →0,s∈ {0,1, . . . , k−
1}, asm→ ∞.
(h) Ifαb+β=aβ+b6= 0 andv0=u−1+αb+β, then the sequencesx2km+2s+1, s∈ {0,1, . . . , k−1}, are convergent.
(i) If αb+β = aβ +b 6= 0 and uαb+β−1−v0 > 0, then |y2km+2s| → g−1(+∞), s ∈ {0,1, . . . , k−1}, asm→ ∞.
(j) Ifαb+β=aβ+b6= 0and u−1αb+β−v0 <0, theny2km+2s→0,s∈ {0,1, . . . , k−1}, asm→ ∞.
(k) If αb+β = aβ +b 6= 0 and u−1 = v0, then the sequences y2km+2s, s ∈ {0,1, . . . , k−1}, are convergent.
(l) If αb+β = aβ +b 6= 0 and uαb+β0−v−1 > 1, then |y2km+2s+1| → g−1(+∞), s∈ {0,1, . . . , k−1}, asm→ ∞.
(m) Ifαb+β=aβ+b6= 0and uαb+β0−v−1 <1, theny2km+2s+1→0,s∈ {0,1, . . . , k− 1}, asm→ ∞.
(n) If αb+β=aβ+b6= 0 andu0=v−1+αb+β, then the sequencesy2km+2s+1, s∈ {0,1, . . . , k−1}, are convergent.
(o) If αb+β = −(aβ +b) 6= 0 and v−1αb+β+u0 > 0, then |x2km+2s| → f−1(+∞), s∈ {0,1, . . . , k−1}, asm→ ∞.
(p) Ifαb+β=−(aβ+b)6= 0and v−1αb+β+u0 <0, then|x2km+2s| →0,s∈ {0,1, . . . , k−
1}, asm→ ∞.
(q) If αb+β = −(aβ+b) 6= 0 andv−1 = −u0, then the sequences x4km+2s and x4km+2k+2s,s∈ {0,1, . . . , k−1}, are convergent.
(r) If αb+β = −(aβ+b) 6= 0 and v0αb+β+u−1 >1, then |x2km+2s+1| → f−1(+∞), s∈ {0,1, . . . , k−1}, asm→ ∞.
(s) If αb+β = −(aβ +b) 6= 0 and v0αb+β+u−1 < 1, then |x2km+2s+1| → 0, s ∈ {0,1, . . . , k−1}, asm→ ∞.
EJQTDE, 2013 No. 47, p. 7
(t) Ifαb+β=−(aβ+b)6= 0andv0+u−1=αb+β, then the sequencesx4km+2s+1 andx4km+2k+2s+1,s∈ {0,1, . . . , k−1}, are convergent.
(u) If αb+β = −(aβ+b) 6= 0 and uαb+β−1+v0 < 0, then |y2km+2s| → g−1(+∞), s∈ {0,1, . . . , k−1}, asm→ ∞.
(v) Ifαb+β=−(aβ+b)6= 0and u−1αb+β+v0 >0, theny2km+2s→0,s∈ {0,1, . . . , k− 1}, asm→ ∞.
(w) If αb+β =−(aβ+b)6= 0 and u−1 =−v0, then the sequences y4km+2s and y4km+2k+2s,s∈ {0,1, . . . , k−1}, are convergent.
(x) Ifαb+β=−(aβ+b)6= 0and u0+vαb+β−1+αb+β <0, then|y2km+2s+1| →g−1(+∞), s∈ {0,1, . . . , k−1}, asm→ ∞.
(y) If αb+β = −(aβ +b) 6= 0 and u0+vαb+β−1+αb+β > 0, then y2km+2s+1 → 0, s∈ {0,1, . . . , k−1}, asm→ ∞.
(z) If αb+β = −(aβ+b) 6= 0 and u0+v−1+αb+β = 0, then the sequences y4km+2s+1 andy4km+2k+2s+1,s∈ {0,1, . . . , k−1}, are convergent.
Proof. (a), (b)We have
m→∞lim
v−1+ (αb+β)(km+s) u0+ (aβ+b)(km+s) = lim
m→∞
v0+ (αb+β)(km+s)
u−1+ (aβ+b)(km+s+ 1) =αb+β aβ+b,
m→∞lim
u−1+ (aβ+b)(km+s) v0+ (αb+β)(km+s) = lim
m→∞
u0+ (aβ+b)(km+s)
v−1+ (αb+β)(km+s+ 1) = aβ+b αb+β. From these limits, formulae (21)–(24) and the continuity of functionsf andgthese two statements follow.
(c)–(n)By some calculations, and using the next known formulas
ln(1 +x) =x−x2/2 +O(x3) and (1 +x)−1= 1−x+O(x2), x→0 (31) (which we may assume that hold for all the terms in products (21)–(24)), we get
x2km+2s=f−1
f(x2s)
m
Y
j=1
1 + (αb+β)s+vkj(αb+β)−1
1 + u0kj(aβ+b)+(aβ+b)s
=f−1
f(x2s)
m
Y
j=1
1 + v−1−u0 kj(αb+β)+O
1 j2
=f−1
f(x2s) exp
m
X
j=1
v−1−u0 kj(αb+β)+O
1 j2
, (32) EJQTDE, 2013 No. 47, p. 8
f(x2km+2s+1) =f−1
f(x2s+1)
m
Y
j=1
1 + v0kj(αb+β)+(αb+β)s
1 +u−1+(aβ+b)(s+1) kj(aβ+b)
=f−1
f(x2s+1)
m
Y
j=1
1 + v0−u−1−(αb+β) kj(αb+β) +O
1 j2
=f−1
f(x2s+1) exp
m
X
j=1
v0−u−1−(αb+β) kj(αb+β) +O
1 j2
, (33) y2km+2s=g−1
g(y2s)
m
Y
j=1
1 + u−1kj(aβ+b)+(aβ+b)s 1 +v0kj(αb+β)+(αb+β)s
=g−1
g(y2s)
m
Y
j=1
1 + u−1−v0
kj(αb+β)+O 1
j2
=g−1
g(y2s) exp
m
X
j=1
u−1−v0 kj(αb+β)+O
1 j2
, (34)
y2km+2s+1=g−1
g(y2s+1)
m
Y
j=1
1 +u0kj(aβ+b)+(aβ+b)s
1 + v−1+(αb+β)(s+1) kj(αb+β)
=g−1
g(y2s+1)
m
Y
j=1
1 + u0−v−1−(αb+β) kj(αb+β) +O
1 j2
=g−1
g(y2s+1) exp
m
X
j=1
u0−v−1−(αb+β) kj(αb+β) +O
1 j2
, (35)
for everys∈ {0,1,2, . . . , k−1}.
Using (32)–(35), the relations
∞
X
j=1
1
j = +∞ and
+∞
X
j=1
O
1 j2
<+∞, (36)
and the continuity of the functionsf and g, these results easily follow.
(o)–(z)By some calculations and (31) (which we may also assume that hold for all the terms in products (21)–(24)), we get
EJQTDE, 2013 No. 47, p. 9
x2km+2s=f−1
f(x2s)(−1)m
m
Y
j=1
1 + (αb+β)s+vkj(αb+β)−1
1 +(αb+β)s−ukj(aβ+b)0
=f−1
f(x2s)(−1)m
m
Y
j=1
1 + v−1+u0 kj(αb+β)+O
1 j2
= (−1)mf−1
f(x2s) exp
m
X
j=1
v−1+u0 kj(αb+β)+O
1 j2
, (37)
x2km+2s+1=f−1
f(x2s+1)(−1)m
m
Y
j=1
1 + v0kj(αb+β)+(αb+β)s
1 + (αb+β)(s+1)−u−1
kj(αb+β)
=f−1
f(x2s+1)(−1)m
m
Y
j=1
1 +v0+u−1−(αb+β) kj(αb+β) +O
1 j2
= (−1)mf−1
f(x2s+1) exp
m
X
j=1
v0+u−1−(αb+β) kj(αb+β) +O
1 j2
, (38) y2km+2s=g−1
g(y2s)(−1)m
m
Y
j=1
1 +(αb+β)s−ukj(αb+β)−1
1 + v0kj(αb+β)+(αb+β)s
=g−1
g(y2s)(−1)m
m
Y
j=1
1− u−1+v0 kj(αb+β)+O
1 j2
= (−1)mg−1
g(y2s) exp
−
m
X
j=1
u−1+v0
kj(αb+β)+O 1
j2
, (39)
y2km+2s+1=g−1
g(y2s+1)(−1)m
m
Y
j=1
1 + (αb+β)s−ukj(αb+β)0
1 +v−1+(αb+β)(s+1) kj(αb+β)
=g−1
g(y2s+1)(−1)m
m
Y
j=1
1−u0+v−1+αb+β kj(αb+β) +O
1 j2
= (−1)mg−1
g(y2s+1) exp
−
m
X
j=1
u0+v−1+αb+β kj(αb+β) +O
1 j2
, (40) for everys∈ {0,1,2, . . . , k−1}.
EJQTDE, 2013 No. 47, p. 10
Using (37)–(40), relations (36) and the continuity of the functionsf andg, the
results easily follow.
Theorem 2. Assume that aα 6= 1, f, g :R →R are continuous, odd, increasing functions satisfying the conditions in (3), and (xn, yn)n≥−2k is a well-defined so- lution of system (13) such that x−i 6= 06=y−i, i= 1, . . . ,2k. Then the following statements are true.
(a) If|aα|>1,|v−1(1−aα)−αb−β|<|u0(1−aα)−aβ−b|, thenx2km+2s→0, s∈ {0,1, . . . , k−1} asm→ ∞.
(b) If|aα|>1,|v−1(1−aα)−αb−β|>|u0(1−aα)−aβ−b|, then|x2km+2s| → f−1(+∞),s∈ {0,1, . . . , k−1} asm→ ∞.
(c) If|aα|>1,v−1(1−aα)−αb−β =u0(1−aα)−aβ−b6= 0, then the sequences x2km+2s,s∈ {0,1, . . . , k−1} are convergent.
(d) If |aα| > 1, v−1(1−aα)−αb−β = −(u0(1−aα)−aβ −b) 6= 0, then the sequencesx4km+2s andx4km+2k+2s,s∈ {0,1, . . . , k−1} are convergent.
(e) If|aα|<1 and|αb+β|<|aβ+b|, then x2km+2s→0,s∈ {0,1, . . . , k−1} as m→ ∞.
(f ) If|aα|<1and|αb+β|>|aβ+b|, then|x2km+2s| →f−1(+∞),s∈ {0,1, . . . , k−
1} asm→ ∞.
(g) If|aα|<1andαb+β=aβ+b, then the sequencesx2km+2s,s∈ {0,1, . . . , k−1}
are convergent.
(h) If|aα|<1andαb+β=−(aβ+b), then the sequencesx4km+2sandx4km+2k+2s, s∈ {0,1, . . . , k−1} are convergent.
(i) Ifaα=−1, then
x2km+2s=f−1
f(x2s)
m
Y
j=1
αb+β+ (−1)kj+s(2v−1−αb−β) aβ+b+ (−1)kj+s(2u0−aβ−b)
. (41) Proof. Let
psm:= αb+β+ (aα)km+s(v−1(1−aα)−αb−β)
aβ+b+ (aα)km+s(u0(1−aα)−aβ−b) , m∈N0, s∈ {0,1, . . . , k−1}.
(a)Note that in this case
m→∞lim |psm|=|v−1(1−aα)−αb−β|
|u0(1−aα)−aβ−b| <1,
which along with formula (25), and the continuity of functionf, easily implies the result.
(b)In this case
m→∞lim |psm|=|v−1(1−aα)−αb−β|
|u0(1−aα)−aβ−b| >1,
from which, (25) and the continuity of functionf, the result easily follows.
EJQTDE, 2013 No. 47, p. 11
(c)Using (31) we have that for sufficiently largem psm=
1 + (aα)km+s(v−1αb+β(1−aα)−αb−β))
1 + (aα)km+s(vaβ+b
−1(1−aα)−αb−β))
= 1 + αb+β−aβ−b
(aα)km+s(v−1(1−aα)−αb−β))+ 1
(aα)km
. (42)
Employing (42) in (25), then using (31), the condition|aα|>1, and the continuity of functionf, the statement easily follows.
(d)Using (31) we have that for sufficiently largem psm=−1 +(aα)km+s(vαb+β
−1(1−aα)−αb−β))
1−(aα)km+s(vaβ+b
−1(1−aα)−αb−β))
=−
1 + αb+β+aβ+b
(aα)km+s(v−1(1−aα)−αb−β)+ 1
(aα)km
. (43)
Using (43) in (25), then (31), the condition|aα|>1, and the continuity of function f, the statement easily follows.
(e)In this case
m→∞lim |psm|=|αb+β|
|aβ+b| <1,
which along with (25) and the continuity of functionf, the result follows.
(f )In this case
m→∞lim |psm|=|αb+β|
|aβ+b| >1,
which along with (25) and the continuity of functionf, the result follows.
(g)Using (31) we have that for sufficiently largem psm=
1 + (aα)km+s(v−1(1−aα)−αb−β)) αb+β
1 + (aα)km+s(u0(1−aα)−αb−β) αb+β
= 1 +(aα)km+s(v−1−u0)(1−aα)
αb+β + ((aα)km). (44)
Employing (44) in (25), then using (31), the condition|aα|<1 and the continuity of functionf, the statement follows.
(h)Using (31) we have that for sufficiently largem psm=−
1 +(aα)km+s(v−1(1−aα)−αb−β)) αb+β
1−(aα)km+s(u0(1−aα)+αb+β) αb+β
=−
1 + (aα)km+s(v−1+u0)(1−aα)
αb+β + ((aα)km)
. (45)
Employing (45) in (25), then using (31), the condition|aα|<1, the continuity and oddness of functionf, the statement follows.
(i)By using the conditionaα=−1 in (25), formula (41) directly follows.
EJQTDE, 2013 No. 47, p. 12
Theorem 3. Assume that aα 6= 1, f, g :R →R are continuous, odd, increasing functions satisfying the conditions in (3), and that (xn, yn)n≥−2k is a well-defined solution of system (13)such that x−i6= 06=y−i,i= 1, . . . ,2k. Then the following statements are true.
(a) If|aα|>1,|v0(1−aα)−αb−β|<|aα||u−1(1−aα)−aβ−b|, thenx2km+2s+1→ 0,s∈ {0,1, . . . , k−1}asm→ ∞.
(b) If|aα|>1,|v0(1−aα)−αb−β|>|aα||u−1(1−aα)−aβ−b|, then|x2km+2s+1| → f−1(+∞),s∈ {0,1, . . . , k−1} asm→ ∞.
(c) If|aα|>1,v0(1−aα)−αb−β=aα(u−1(1−aα)−aβ−b), then the sequences x2km+2s+1,s∈ {0,1, . . . , k−1} converge.
(d) If|aα|>1,v0(1−aα)−αb−β =−aα(u−1(1−aα)−aβ−b), then the sequences x4km+2s+1 andx4km+2k+2s+1,s∈ {0,1, . . . , k−1} converge.
(e) If|aα|<1 and|αb+β|<|aβ+b|, then x2km+2s+1 →0,s∈ {0,1, . . . , k−1}
asm→ ∞.
(f ) If |aα| < 1 and |αb+β| > |aβ +b|, then |x2km+2s+1| → f−1(+∞), s ∈ {0,1, . . . , k−1} asm→ ∞.
(g) If|aα|<1andαb+β=aβ+b, then the sequencesx2km+2s+1,s∈ {0,1, . . . , k−
1} are convergent.
(h) If |aα| < 1 and αb+β = −(aβ +b), then the sequences x4km+2s+1 and x4km+2k+2s+1,s∈ {0,1, . . . , k−1} are convergent.
(i) Ifaα=−1, then x2km+2s+1=f−1
f(x2s+1)
m
Y
j=1
αb+β+ (−1)kj+s(2v0−αb−β) aβ+b+ (−1)kj+s+1(2u−1−aβ−b)
. (46) Proof. Let
rsm:= αb+β+ (aα)km+s(v0(1−aα)−αb−β)
aβ+b+ (aα)km+s+1(u−1(1−aα)−aβ−b), m∈N0, s∈ {0,1, . . . , k−1}.
(a)Note that in this case
m→∞lim |rsm|= |v0(1−aα)−αb−β|
|u−1(1−aα)−aβ−b||aα| <1,
which along with formula (26) and the continuity of functionf, easily implies the result.
(b)In this case
m→∞lim |rsm|= |v0(1−aα)−αb−β|
|u−1(1−aα)−aβ−b||aα| >1,
from which along with (26) and the continuity of functionf, the result follows.
(c)Using (31) we have that for sufficiently largem rms =
1 +(aα)km+s(vαb+β
0(1−aα)−αb−β)
1 + (aα)km+s+1(uaβ+b−1(1−aα)−aβ−b)
= 1 + αb+β−aβ−b
(aα)km+s(v0(1−aα)−αb−β))+ 1
(aα)km
. (47)
EJQTDE, 2013 No. 47, p. 13
Employing (47) in (26), then using (31), the condition|aα|>1 and the continuity of functionf, the statement easily follows.
(d)Using (31) we have that for sufficiently largem rms =−1 +(aα)km+s(vαb+β
0(1−aα)−αb−β)
1−(aα)km+s(vaβ+b
0(1−aα)−αb−β)
=−
1 + αb+β+aβ+b
(aα)km+s(v0(1−aα)−αb−β))+ 1
(aα)km
. (48)
Employing (48) in (26), then using (31), the condition|aα|>1 and the continuity of functionf, the statement easily follows.
(e)In this case
m→∞lim |rsm|= |αb+β|
|aβ+b| <1,
from which along with (26) and the continuity of functionf, the result follows.
(f )In this case
m→∞lim |rsm|= |αb+β|
|aβ+b| >1,
from which along with (26) and the continuity of functionf, the result follows.
(g)Using (31) we have that for sufficiently largem rms = 1 + (aα)km+s(v0(1−aα)−αb−β)
αb+β
1 + (aα)km+s+1(u−1(1−aα)−αb−β) αb+β
= 1 +(aα)km+s(v0−aαu−1−αb−β)(1−aα)
αb+β + ((aα)km). (49)
Employing (49) in (26), then using (31), the condition|aα|<1 and the continuity of functionf, the statement follows.
(h)Using (31) we have that for sufficiently largem rsm=− 1 + (aα)km+s(v0(1−aα)−αb−β)
αb+β
1−(aα)km+s+1(u−1(1−aα)+αb+β) αb+β
=−
1 + (aα)km+s(v0+αau−1−αb−β)(1−aα)
αb+β + ((aα)km)
. (50) Employing (50) in (26), then using (31), the condition|aα|<1, the continuity and oddness of functionf, the statement follows.
(i)By using the conditionaα=−1 in (26) formula (46) easily follows.
The proofs of the next two theorems use formulas (27) and (28), and are similar to those ones of Theorems 2 and 3, so they are omitted.
Theorem 4. Assume that aα 6= 1, f, g :R →R are continuous, odd, increasing functions satisfying the conditions in (3), and that (xn, yn)n≥−2k is a well-defined solution of system (13)such that x−i6= 06=y−i,i= 1, . . . ,2k. Then the following statements are true.
EJQTDE, 2013 No. 47, p. 14
(a) If|aα|>1,|v0(1−aα)−αb−β|>|u−1(1−aα)−aβ−b|, theny2km+2s→0, s∈ {0,1, . . . , k−1} asm→ ∞.
(b) If|aα|>1,|v0(1−aα)−αb−β|<|u−1(1−aα)−aβ−b|, then|y2km+2s| → g−1(+∞),s∈ {0,1, . . . , k−1} asm→ ∞.
(c) If |aα|>1, v0(1−aα)−αb−β =u−1(1−aα)−aβ−b, then the sequences y2km+2s,s∈ {0,1, . . . , k−1} converge.
(d) If|aα|>1,v0(1−aα)−αb−β=−(u−1(1−aα)−aβ−b), then the sequences y4km+2s andy4km+2k+2s,s∈ {0,1, . . . , k−1} converge.
(e) If|aα|<1 and|αb+β|>|aβ+b|, theny2km+2s→0,s∈ {0,1, . . . , k−1} as m→ ∞.
(f ) If|aα|<1and|αb+β|<|aβ+b|, then|y2km+2s| →g−1(+∞),s∈ {0,1, . . . , k−
1} asm→ ∞.
(g) If|aα|<1andαb+β=aβ+b, then the sequencesy2km+2s,s∈ {0,1, . . . , k−1}
are convergent.
(h) If|aα|<1andαb+β=−(aβ+b), then the sequencesy4km+2sandy4km+2k+2s, s∈ {0,1, . . . , k−1} are convergent.
(i) Ifaα=−1, then y2km+2s=g−1
g(y2s)
m
Y
j=1
aβ+b+ (−1)kj+s(2u−1−aβ−b) αb+β+ (−1)kj+s(2v0−αb−β)
m
. Theorem 5. Assume that aα 6= 1, f, g :R →R are continuous, odd, increasing functions satisfying the conditions in (3), and that (xn, yn)n≥−2k is a well-defined solution of system (13)such that x−i6= 06=y−i,i= 1, . . . ,2k. Then the following statements are true.
(a) If|aα|>1,|aα||v−1(1−aα)−αb−β|>|u0(1−aα)−aβ−b|, theny2km+2s+1→ 0,s∈ {0,1, . . . , k−1} asm→ ∞.
(b) If|aα|>1,|aα||v−1(1−aα)−αb−β|<|u0(1−aα)−aβ−b|, then|y2km+2s+1| → g−1(+∞),s∈ {0,1, . . . , k−1} asm→ ∞.
(c) If |aα| >1, aα(v−1(1−aα)−αb−β) = u0(1−aα)−aβ−b 6= 0, then the sequencesy2km+2s+1,s∈ {0,1, . . . , k−1} are convergent.
(d) If|aα|>1,aα(v−1(1−aα)−αb−β) =−(u0(1−aα)−aβ−b)6= 0, then the sequencesy4km+2s+1 andy4km+2k+2s+1,s∈ {0,1, . . . , k−1} are convergent.
(e) If |aα|<1 and |αb+β|>|aβ+b|, then y2km+2s+1→0, s∈ {0,1, . . . , k−1}
asm→ ∞.
(f ) If |aα| < 1 and |αb+β| < |aβ +b|, then |y2km+2s+1| → g−1(+∞), s ∈ {0,1, . . . , k−1} asm→ ∞.
(g) If|aα|<1andαb+β =aβ+b, then the sequencesy2km+2s+1,s∈ {0,1, . . . , k−
1} are convergent.
(h) If |aα| < 1 and αb +β = −(aβ +b), then the sequences y4km+2s+1 and y4km+2k+2s+1,s∈ {0,1, . . . , k−1} are convergent.
(i) Ifaα=−1, then y2km+2s+1=g−1
g(y2s+1)
m
Y
j=1
aβ+b+ (−1)kj+s(2u0−aβ−b) αb+β+ (−1)kj+s+1(2v−1−αb−β)
m
. EJQTDE, 2013 No. 47, p. 15
Theorems 2–5 and Lemma 1 yield the next corollary.
Corollary 1. Assume that |aα|<1, f, g:R→Rare continuous, odd, increasing functions satisfying the conditions in (3), and (xn, yn)n≥−2k is a well-defined so- lution of system (13) such that x−i 6= 06=y−i, i= 1, . . . ,2k. Then the following statements are true.
(a) Ifαb+β=aβ+b, then the solution(xn, yn)n≥−2kconverges to a, not necessarily prime,2k-periodic solution of system (13).
(b) If αb+β = −(aβ +b), then the solution (xn, yn)n≥−2k converges to a, not necessarily prime,4k-periodic solution of system (13).
Acknowledgements
This work was funded by the Deanship of Scientific Research (DSR), King Ab- dulaziz University, under grant No. (11-130/1433 HiCi). The authors, therefore, acknowledge technical and financial support of KAU.
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EJQTDE, 2013 No. 47, p. 17
