Báo cáo hóa học: " Research Article Global Attractivity Results for Mixed-Monotone Mappings in Partially Ordered Complete Metric Spaces" potx

Nieto We prove fixed point theorems for mixed-monotone mappings in partially ordered complete metric spaces which satisfy a weaker contraction condition than the classical Banach contrac

Volume 2009, Article ID 762478, 17 pages

doi:10.1155/2009/762478

Research Article

Global Attractivity Results for Mixed-Monotone Mappings in Partially Ordered Complete

Metric Spaces

D ˇz Burgi´c, 1 S Kalabuˇsi ´c, 2 and M R S Kulenovi ´c 3

1 Department of Mathematics, University of Tuzla, 75000 Tuzla, Bosnia and Herzegovina

2 Department of Mathematics, University of Sarajevo, 71000 Sarajevo, Bosnia and Herzegovina

3 Department of Mathematics, University of Rhode Island, Kingston, R I 02881-0816, USA

Correspondence should be addressed to M R S Kulenovi´c,mkulenovic@mail.uri.edu

Received 28 October 2008; Revised 17 January 2009; Accepted 9 February 2009

Recommended by Juan J Nieto

We prove fixed point theorems for mixed-monotone mappings in partially ordered complete metric spaces which satisfy a weaker contraction condition than the classical Banach contraction condition for all points that are related by given ordering We also give a global attractivity result for all solutions of the difference equation z n1  Fz n , z n−1 , n  2, 3, , where F satisfies

mixed-monotone conditions with respect to the given ordering

Copyrightq 2009 Dˇz Burgi´c et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction and Preliminaries

The following results were obtained first in 1 and were extended to the case of higher-order difference equations and systems in 2 6 For the sake of completeness and the readers convenience, we are including short proofs

Theorem 1.1 Let a, b be a compact interval of real numbers, and assume that

f : a, b × a, b −→ a, b 1.1

is a continuous function satisfying the following properties:

afx, y is nondecreasing in x ∈ a, b for each y ∈ a, b, and fx, y is nonincreasing in

y ∈ a, b for each x ∈ a, b;

b If m, M ∈ a, b × a, b is a solution of the system

fm, M  m, fM, m  M, 1.2

then m  M.

Then

x n1  fx n , x n−1



, n  0, 1, 1.3

has a unique equilibrium x ∈ a, b and every solution of 1.3 converges to x.

Proof Set

and for i  1, 2, set

M i  fM i−1 , m i−1



, m i  fm i−1 , M i−1



Now observe that for each i ≥ 0,

m0≤ m1≤ · · · ≤ m i ≤ · · · ≤ M i ≤ · · · ≤ M1≤ M0,

m i ≤ x k ≤ M i , for k ≥ 2i  1.

1.6

Set

m  lim

i → ∞ m i , M  lim

Then

M ≥ lim sup

i → ∞

x i≥ lim inf

and by the continuity of f,

m  fm, M, M  fM, m. 1.9 Therefore in view ofb,

from which the result follows

Theorem 1.2 Let a, b be an interval of real numbers and assume that

f : a, b × a, b −→ a, b 1.11

is a continuous function satisfying the following properties:

a fx, y is nonincreasing in x ∈ a, b for each y ∈ a, b, and fx, y is nondecreasing in

y ∈ a, b for each x ∈ a, b;

b the difference equation 1.3 has no solutions of minimal period two in a, b Then 1.3

has a unique equilibrium x ∈ a, b and every solution of 1.3 converges to x.

Proof Set

and for i  1, 2, set

M i  fm i−1 , M i−1



, m i  fM i−1 , m i−1



Now observe that for each i ≥ 0,

m0≤ m1≤ · · · ≤ m i ≤ · · · ≤ M i ≤ · · · ≤ M1≤ M0,

m i ≤ x k ≤ M i , for k ≥ 2i  1.

1.14

Set

m  lim

i → ∞ m i , M  lim

Then clearly1.8 holds and by the continuity of f,

m  fM, m, M  fm, M. 1.16

In view ofb,

from which the result follows

These results have been very useful in proving attractivity results for equilibrium or periodic solutions of 1.3 as well as for higher-order difference equations and systems of difference equations; see 2,7 12 Theorems1.1and1.2have attracted considerable attention

of the leading specialists in difference equations and discrete dynamical systems and have been generalized and extended to the case of maps inRn, see3, and maps in Banach space

with the cone see4 6 In this paper, we will extend Theorems 1.1and 1.2to the case of monotone mappings in partially ordered complete metric spaces

On the other hand, there has been recent interest in establishing fixed point theorems

in partially ordered complete metric spaces with a contractivity condition which holds for all points that are related by partial ordering; see 13–20 These fixed point results have been applied mainly to the existence of solutions of boundary value problems for differential equations and one of them, namely20, has been applied to the problem of solving matrix equations See also21, where the application to the boundary value problems for integro-differential equations is given and 22 for application to some classes of nonexpansive mappings and 23 for the application of the Leray-Schauder theory to the problems of

an impulsive boundary value problem under the condition of non-well-ordered upper and lower solutions None of these results is global result, but they are rather existence results In this paper, we combine the existence results with the results of the type of Theorems1.1and 1.2to obtain global attractivity results

2 Main Results: Mixed Monotone Case I

Let X be a partially ordered set and let d be a metric on X such that X, d is a complete metric space Consider X × X We will use the following partial ordering.

Forx, y, u, v ∈ X × X, we have

This partial ordering is well known as “south-east ordering” in competitive systems in the plane; see5,6,12,24,25

Let d1be a metric on X × X defined as follows:

d1x, y, u, v  dx, u  dy, v. 2.2 Clearly

d1x, y, u, v  d1y, x, v, u. 2.3

We prove the following theorem

Theorem 2.1 Let F : X × X → X be a map such that Fx, y is nonincreasing in x for all y ∈ X,

and nondecreasing in y for all x ∈ X Suppose that the following conditions hold.

i There exists k ∈ 0, 1 with

dFx, y, Fu, v ≤ k

2d1x, y, u, v ∀x, y  u, v. 2.4

ii There exists x0, y0∈ X such that the following condition holds:

x0≤ Fy0, x0



, y0≥ Fx0, y0



iii If {x n } ∈ X is a nondecreasing convergent sequence such that lim n → ∞ x n  x, then

x n ≤ x, for all n ∈ N and if {y n } ∈ Y is a nonincreasing convergent sequence such

that lim n → ∞ y n  y, then y n ≥ y, for all n ∈ N; if x n ≤ y n for every n, then

limn → ∞ x n≤ limn → ∞ y n

Then we have the following.

a For every initial point x0, y0 ∈ X × X such that condition 2.5 holds, F n x0, y0 →

x, F n y0, x0 → y, n → ∞, where x, y satisfy

x  Fy, x, y  Fx, y. 2.6

If x0≤ y0in condition2.5, then x ≤ y If in addition x  y, then {x n }, {y n } converge to

the equilibrium of the equation

x n1  Fy n , x n



, y n1  Fx n , y n



, n  1, 2, 2.7

b In particular, every solution {z n } of

z n1  Fz n , z n−1



, n  2, 3, 2.8

such that x0≤ z0, z1≤ y0converges to the equilibrium of 2.8.

c The following estimates hold:

d

F n

y0, x0



, x

≤ 1 2

k n

1− k



d

F

x0, y0



, y0



 dF

y0, x0



, x0



d

F n

x0, y0



, y

≤ 1 2

k n

1− k



d

F

y0, x0



, x0



 dF

x0, y0



, y0



Proof Let x1 Fy0, x0 and y1 Fx0, y0 Since x0≤ Fy0, x0  x1and y0≥ Fx0, y0  y1,

for x2  Fy1, x1, y2 Fx1, y1, we have

F2

y0, x0

 : FF

x0, y0



, F

y0, x0



 Fy1, x1



 x2,

F2

x0, y0

 : FF

y0, x0



, F

x0, y0



 Fx1, y1



 y2.

2.11

Now, we have

x2 F2

y0, x0



 Fy1, x1



≥ Fy0, x0



 x1,

y2 F2

x0, y0



 Fx1, y1



≤ Fx0, y0



 y1.

2.12

For n  1, 2, , we let

x n1  F n1

y0, x0



 FF n

x0, y0



, F n

y0, x0



,

y n1  F n1

x0, y0



 FF n

y0, x0



, F n

x0, y0



.

2.13

By using the monotonicity of F, we obtain

x0≤ Fy0, x0



 x1≤ F2

y0, x0



 x2≤ · · · ≤ F n1

y0, x0



≤ · · · ,

y0≥ Fx0, y0



 y1 ≥ F2

x0, y0



 y2≥ · · · ≥ F n1

x0, y0



≥ · · · 2.14 that is

x0≤ x1≤ x2≤ · · ·

We claim that for all n ∈ N the following inequalities hold:

d

x n1 , x n



 dF n1

y0, x0



, F n

y0, x0



≤ k n

2 d1



x1, y1



,

x0, y0



d

y n1 , y n



 dF n1

x0, y0



, F n

x0, y0



≤ k n

2 d1



x1, y1



,

x0, y0



Indeed, for n  1, using x0≤ Fy0, x0, y0 ≥ Fx0, y0, and 2.3, we obtain

d

x2, x1



 dF

y1, x1



, F

y0, x0



≤ k

2d1



y1, x1



,

y0, x0



 k

2d1



x1, y1



,

x0, y0



,

d

y2, y1



 dFx1, y1



, F

x0, y0



≤ k

2d1



x1, y1



,

x0, y0



.

2.18 Assume that2.16 holds Using the inequalities

F n1

y0, x0



≥ F n

y0, x0



,

F n1

x0, y0



≤ F n

x0, y0



,

2.19

and the contraction condition2.4, we have

d

x n2 , x n1



 dF n2

y0, x0



, F n1

y0, x0



 dF

F n1

x0, y0



, F n1

y0, x0



, F

F n

x0, y0



, F n

y0, x0



≤ k 2



d

F n1

x0, y0



, F n

x0, y0



 dF n1

y0, x0



, F n

y0, x0



≤ k 2



k n

2



d

F

x0, y0



, y0



 dF

y0, x0



, x0



 dF

y0, x0



, x0



 dF

x0, y0



, y0



 k n1

2 d1



x1, y1



,

x0, y0



.

2.20 Similarly,

d

y n2 , y n1



 dF n2

x0, y0



, F n1

x0, y0



≤ k n1

2 d1



x1, y1



,

x0, y0



This implies that{x n }  {F n y0, x0} and {y n }  {F n x0, y0} are Cauchy sequences in X.

Indeed,

d

F n

y0, x0



, F np

y0, x0



≤ dF n

y0, x0



, F n1

y0, x0



 · · ·

 dF np−1

y0, x0



, F np

y0, x0



≤ k n

2 dF

x0, y0



, y0



 dF

y0, x0



, x0



  · · · 

k np−1 2



d

F

x0y0



, y0



 dF

y0, x0



, x0



 k n 2



1 k  k2 · · ·  k p−1

d

F

x0, y0



, y0



 dFy0, x0



, x0



 k n 2

1− k p

1− k



d

F

x0, y0



, y0



 dF

y0, x0



, x0



.

2.22

Since k ∈ 0, 1, we have

d

x n , x np



 dF n

y0, x0



, F np

y0, x0



≤ 21 − kk n d1



x1, y1



,

x0, y0



Using 2.23, we conclude that {x n }  {F n y0, x0} is a Cauchy sequence Similarly, we conclude that{y n }  {F n x0, y0} is a Cauchy sequence Since X is a complete metric space, then there exist x, y ∈ X such that

lim

n → ∞ x n  lim

n → ∞ F n

y0, x0



 x , lim

n → ∞ y n lim

m → ∞ F m

x0, y0



Using the continuity of F, which follows from contraction condition 2.4, the equations

x n1  Fy n , x n



, y n1  Fx n , y n



2.25

imply2.6

Assume that x0≤ y0 Then, in view of the monotonicity of F

x1 Fy0, x0



≤ Fx0, y0



 y1,

x2 Fy1, x1



≤ Fx1, y1



 y2,

x3 Fy2, x2



≤ Fx2, y2



 y3.

2.26

By using induction, we can show that x n ≤ y n for all n Assume that x0 ≤ z0, z1 ≤ y0 Then,

in view of the monotonicity of F, we have

x1 Fy0, x0



≤ Fz1, z0



 z2≤ Fx0, y0



 y1,

x1 Fy0, x0



≤ Fz2, z1



 z3≤ Fx0, y0



 y1.

2.27

Continuing in a similar way we can prove that x i ≤ z k ≤ y i for all k ≥ 2i1 By using condition

iii we conclude that whenever limn → ∞ z kexists we must have

x ≤ lim

which in the case when x  y implies lim k → ∞ z k  x.

By letting p → ∞ in 2.23, we obtain the estimate 2.9

Remark 2.2 Propertyiii is usually called closedness of the partial ordering, see 6, and is

an important ingredient of the definition of an ordered L-space; see 17,19

Theorem 2.3 Assume that along with conditions (i) and (ii) of Theorem 2.1 , the following condition

is satisfied:

iv every pair of elements has either a lower or an upper bound.

Then, the fixed point x, y is unique and x  y.

Proof First, we prove that the fixed point x, y is unique Condition iv is equivalent to the

following For everyx, y, x∗, y∗ ∈ X × X, there exists z1, z2 ∈ X × X that is comparable

tox, y, x∗, y∗ See 16

Letx, y and x∗, y∗ be two fixed points of the map F.

We consider two cases.

Case 1 If x, y is comparable to x∗, y∗, then for all n  0, 1, 2, F n y, x, F n x, y is

comparable toF n y∗, x∗, F n x∗, y∗  x∗, y∗ We have to prove that

d1



x, y,x∗, y∗

Indeed, using2.2, we obtain

d1



x, y,x∗, y∗

 dx, x∗

 dy, y∗

 dF n y, x, F n

y∗, x∗

 dF n x, y, F n

x∗, y∗

We estimate dF n y, x, F n y∗, x∗, and dF n x, y, F n x∗, y∗.

First, by using contraction condition2.4, we have

d

Fy, x, F

y∗, x∗

≤ k 2



d

y, y∗

 dx, x∗

 k

2d1



x, y,x∗, y∗

,

d

Fx, y, F

x∗, y∗

≤ k 2



d

x, x∗

 dy, y∗

 k

2d1



x, y,x∗, y∗

.

2.31

Now, by using2.31 and 2.30, we have

d1



x, y,x∗, y∗

≤ kd1



x, y,x∗, y∗

< d1



x, y,x∗, y∗

which implies that

d1



x, y,x∗, y∗

Case 2 If x, y is not comparable to x∗, y∗, then there exists an upper bound or a

lower bound z1, z2 of x, y and x∗, y∗ Then, F n z2, z1, F n z1, z2 is comparable to

F n y, x, F n x, y and F n y∗, x∗, F n x∗, y∗.

Therefore, we have

d1



x, y,x∗, y∗

 d1



F n y, x, F n x, y,

F n

y∗, x∗

, F n

x∗, y∗

≤ d1



F n y, x, F n x, y,

F n

z2, z1



, F n

z1, z2



 d1



F n

z2, z1



, F n

z1, z2



,

F n

y∗, x∗

, F n

x∗, y∗

 dF n y, x, F n

z2, z1



 dF n

z2, z1



, F n

y∗, x∗

 dF n z1, z2



, F n

x∗, y∗

 dF n

z2, z1, F n

y∗, x∗

.

2.34

Now, we obtain

d1



x, y,x∗, y∗

 dF n y, x, F n

z2, z1



 dF n

z2, z1



, F n

y∗, x∗

 dF n

z1, z2



, F n

x∗, y∗

 dF n

z2, z1



, F n

y∗, x∗

.

2.35

We now estimate the right-hand side of2.35

First, by using

d

Fy, x, F

z2, z1



≤ k 2



d

y, z2



 dx, z1



we have

d

F2y, x, F2

z2, z1



 dFFx, y, Fy, x, F

F

z1, z2



, F

z2, z1



≤ k

2dFx, y, F

z1, z2



 dFy, x, F

z2, z1





≤ k 2



k

2



d

x, z1



 dy, z2



k 2



d

y, z2



 dx, z1



 k2 2



d

x, z1



 dy, z2



.

2.37 Similarly,

d

F2x, y, F2

z1, z2



 dFFy, x, Fx, y, F

F

z2, z1



, F

z1, z2



≤ k 2



d

Fy, x, F

z2, z1



 dFx, y

, F

z1, z2



≤ k 2



k

2



d

y, z2



 dx, z1



k 2



d

y, z2



 dx, z1



 k2 2



d

x, z1



 dy, z2



.

2.38 So,

d

F2y, x, F2

z2, z1



≤ k2 2



d

x, z1



 dy, z2



,

d

F2x, y, F2

z1, z2



≤ k2 2



d

x, z1



 dy, z2



.

2.39

Using induction, we obtain

d

F n y, x, F n

z2, z1



≤ k n 2



d

x, z1



 dy, z2



,

d

F n x, y, F n

z1, z2



≤ k n 2



d

x, z1



 dy, z2



,

d

F n

z2, z1



, F n

y∗, x∗

≤ k n 2



d

z1, x∗

, d

z2, y∗

,

d

F n

z1, z2



, F n

x∗, y∗

≤ k n 2



d

z1, x∗

, d

z2, y∗

.

2.40

Using2.40, relation 2.35 becomes

d1



x, y,x∗, y∗

≤ k n 2



d

x, z1



 dy, z2



k n 2



d

x, z1



 dy, z2



k n 2



d

z1, x∗

 dz2, y∗

 k n 2



d

z1, x∗

 dz2, y∗

 k n

d

x, z1



 dy, z2



 dz1, x∗

 dz2, y∗

−→ 0, n −→ ∞.

2.41 So,

d1



x, y,x∗, y∗

Finally, we prove that x  y We will consider two cases.

Case A If x is comparable to y, then Fy, x  x is comparable to Fx, y  y Now, we obtain

dx, y  dFy, x, Fx, y ≤ k

2dx, y  dy, x  kdx, y, 2.43

since k ∈ 0, 1, this implies

dx, y  0 ⇐⇒ x  y. 2.44

Case B If x is not comparable to y, then there exists an upper bound or alower bound of x

and y, that is, there exists z ∈ X such that x ≤ z, y ≤ z Then by using monotonicity character

of F, we have

Fx, y ≤ Fx, z, Fy, x ≤ Fy, z, Fx, y ≥ Fz, y, Fy, x ≥ Fz, x. 2.45 Now,

F2x, y  FFy, x, Fx, y ≤ FFz, x, Fx, z  F2x, z, 2.46 that is

Furthermore,

F2x, y  FFy, x, Fx, y ≥ FFy, z, Fz, y  F2y, z, 2.48 that is

Similarly,

F2y, x  FFx, y, Fy, x ≤ FFz, y, Fy, z  F2y, z, 2.50 that is

and

F2y, x  FFx, y, Fy, x ≥ FFx, z, Fz, x  F2z, x. 2.52

By using induction, we have

F n1 x, y ≤ F n1 x, z,

F n1 x, y ≥ F n1 y, z,

F n1 y, x ≤ F n1 y, z,

F n1 y, x ≥ F n1 z, x.

2.53