# Global Attractivity Results on Complete Ordered Metric Spaces for Third-Order Difference Equations

Authors: Mujahid Abbas; Maher Berzig
Journal: International Journal of Analysis
 (2013)

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## Abstract

We establish fixed-point theorems for mixed monotone mappings in the setting of ordered metric spaces which satisfy a contractive condition for all points that are related by a given ordering. We also give a global attractivity result for all solutions of the difference equationun+1=F(un,un-1,un-2),n=2,3,…, where F satisfies certain monotonicity conditions with respect to the given ordering. As an application of our obtained results, we present some iterative algorithms to solve a class of matrix equations. A numerical example is also presented to test the validity of the algorithms.


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## Body

## 1. Introduction

The following global attractivity result from [1] (see also [2]) is very useful in establishing convergence results in many situations.Theorem 1 (see [1]).
Let[a,b] be a closed and bounded interval of real numbers and let F∈C([a,b]3,[a,b]) satisfy the following conditions.(i)
The functionF(z1,z2,z3) is monotonic in each of its arguments.(ii)
For eachm,M∈[a,b] and for each i∈{1,2,3}, one defines
(1)Mi(m,M)={M,ifFisincreasinginzi,m,ifFisdecreasinginzi,mi(m,M)=Mi(M,m)
and assume that if(m,M) is a solution of the system
(2)M=F(M1(m,M),M2(m,M),M3(m,M)),m=F(m1(m,M),m2(m,M),m3(m,M)),
thenM=m.
Then there exists exactly one equilibriumx¯ of the equation
(3)xn+1=F(xn,xn-1,xn-2),n=0,1,…
and every solution of the above equation converges to x¯.The above result in Theorem1 attracted considerable attention of the leading specialists in difference equations and discrete dynamic systems and has been generalized and extended to the case of maps in ℝn, see [3], and maps in Banach space with cone, see [4–6].Moreover, 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, e.g., [7–19]). These fixed-point results have been applied mainly to the existence of solutions of boundary value problems for differential equations, namely [18], has been applied for solving a class of matrix equations.In [20], Burgić et al. obtained the following global attractivity result for mixed monotone mappings in partially ordered complete metric spaces (see also Gnana Bhaskar and Lakshmikantham [10]).Theorem 2 (see Burgić et al. [20]).
Let(X,⪯) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:X×X→X be a map such that F(x,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 existsk∈[0,1) with
(4)d(F(x,y),F(u,v))≤k2[d(x,u)+d(y,v)]∀x⪯u,y≽v.(ii)
There existsx0,y0∈X such that the following condition holds:
(5)x0⪯F(x0,y0),y0≽F(y0,x0).(iii)
If{xn}⊂X is a nondecreasing convergent sequence such that limn→∞xn=x, then xn⪯x, for all n∈ℕ and if {yn}⊂X is a nonincreasing convergent sequence such that limn→∞yn=y, then yn≽y, for all n∈ℕ; if xn⪯yn for every n, then limn→∞xn⪯limn→∞yn.
Then one has the following.(a)
For every initial pointx0,y0∈X such that condition (ii) holds, Fn(x0,y0)→x, Fn(y0,x0)→y, n→∞, where x,y∈X satisfy
(6)x=F(x,y),y=F(y,x).(b)
Ifx0⪯y0 in condition (ii), then x=y. If in addition x=y, then {xn}, {yn} converge to the equilibrium of the equation
(7)xn+1=F(xn,yn),yn+1=F(yn,xn),n=1,2,….(c)
In particular, every solution{zn} of
(8)zn+1=F(zn,zn-1),n=2,3,…
such thatx0⪯z0, z1⪯y0 converges to the equilibrium of (7).In this paper, motivated by the results and ideas in a recent work of Berinde and Borcut [9], we extend Theorem 2 to mappings F:X×X×X→X. Such extension allows us to study the third-order difference equation
(9)un+1=F(un,un-1,un-2),n=2,3,….
The presented theorems also extend and generalize the work in [9]. We use our obtained results to build some iterative algorithms to solve a class of matrix equations. A numerical example is also presented to test the validity of the algorithms.Now we introduce the following concepts.Definition 3.
LetX be a nonempty set and F:X×X×X→X a given mapping. One says that (x,y,z)∈X×X×X is a fixed-point of the third order if
(10)F(x,y,x)=x,F(y,x,y)=y,F(z,y,z)=z.Definition 4.
Let(X,⪯) be a partially ordered set and F:X×X×X→X a given mapping. one says that F has the mixed monotone property if F(x,y,z) is monotonously increasing in x and z and is monotonously decreasing in y; that is, for any x,y,z∈X,(i)
x1,x2∈X,x1⪯x2⇒F(x1,y,z)⪯F(x2,y,z),(ii)
y1,y2∈X,y1⪯y2⇒F(x,y1,z)≽F(x,y2,z), and(iii)
z1,z2∈X,z1⪯z2⇒F(x,y,z1)⪯F(x,y,z2).Through this paper we will use the following notations.Let(X,⪯) be a partially ordered set. (i)
Forx,y,z∈X×X×X the notation x∈[y,z] means that x≽y and x⪯z.(ii)
We endowX×X×X with the partial order that we denote also by ⪯, defined by
(11)(x,y,z),(u,v,w)∈X×X×X,(x,y,z)⪯(u,v,w)⟺x⪯u,y≽v,z⪯w.(iii)
LetF:X×X×X→X be a given mapping. For all x,y,z∈X, we denote
(12)F2(x,y,z)=F(F(x,y,x),F(y,x,y),F(z,y,z)),Fp+1(x,y,z)=F(Fp(x,y,x),Fp(y,x,y),Fp(z,y,z)),∀p≥2.


## 2. Main Result

Our first result is the following.Theorem 5.
Let(X,⪯) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:X×X×X→X be a mapping having the mixed monotone property on X. Suppose that the following conditions hold.(i)
There existsλ∈[0,1) with
(13)d(F(x,y,z),F(u,v,w))≤λmax{d(x,u),d(y,v),d(z,w),d(F(x,y,x),x),d(F(x,y,x),u),d(F(y,x,y),y),d(F(y,x,y),v),d(F(z,y,z),z),d(F(z,y,z),w)}for all(x,y,z)⪯(u,v,w).(ii)
There existx0,y0,z0∈X such that
(14)x0⪯F(x0,y0,x0),y0≽F(y0,x0,y0),z0⪯F(z0,y0,z0).(iii)
If{xn}⊂X is a nondecreasing convergent sequence such that limn→∞xn=x, then xn⪯x for all n∈ℕ, if {yn}⊂X is a nonincreasing convergent sequence such that limn→∞yn=y, then yn≽y for every n∈ℕ, and if xn⪯yn for every n, then limn→∞xn⪯limn→∞yn.
Then one has the following.(a)
For every initial point(x0,y0,z0)∈X×X×X such that condition (14) holds:
(15)xn=Fn(x0,y0,x0)⟶x,yn=Fn(y0,x0,y0)⟶y,zn=Fn(z0,y0,z0)⟶zasn⟶∞,wherex,y,andz satisfy
(16)x=F(x,y,x),y=F(y,x,y),z=F(z,y,z).Ifx0⪯y0 and z0⪯y0 in condition (14), then x=y=z and {xn}, {yn}, and {zn} converge to the equilibrium of the equation
(17)xn+1=F(xn,yn,xn),yn+1=F(yn,xn,yn),zn+1=F(zn,yn,zn),n=0,1,2….(b)
In particular, every solution{un} of
(18)un+1=F(un,un-1,un-2),n=2,3,…,such thatu0,u1,u2∈[x0,y0] (or [z0,y0]) converges to the equilibrium of (18).(c)
The following estimates hold:(19)d(x,Fn(x0,y0,x0))≤λn1-λmax{d(F(x0,y0,x0),x0),d(F(y0,x0,y0),y0),d(z0,F(z0,y0,z0))},d(y,Fn(y0,x0,y0))≤λn1-λmax{d(F(x0,y0,x0),x0),d(F(y0,x0,y0),y0),d(z0,F(z0,y0,z0))},d(z,Fn(z0,y0,z0))≤λn1-λmax{d(F(x0,y0,x0),x0),d(F(y0,x0,y0),y0),d(F(z0,y0,z0),z0)}.Proof.
Letx0,y0,z0∈X such that condition (14) is satisfied. Denote x1=F(x0,y0,x0), y1=F(y0,x0,y0), and z1=F(z0,y0,z0). Since x0⪯F(x0,y0,x0)=x1, y0≽F(y0,x0,y0)=y1, and z0⪯F(z0,y0,z0)=y1, from the mixed monotone property of F, we get that
(20)x0⪯x1=F(x0,y0,x0)⪯x2=F(x1,y1,x1),y0≽y1=F(y0,x0,y0)≽y2=F(y1,x1,y1),z0⪯z1=F(z0,y0,z0)⪯z2=F(z1,y1,z1).
Consider the sequences {xn}, {yn}, and {zn} defined by (17). By induction and using the mixed monotone property of F, we obtain easily that
(21)x0⪯x1⪯⋯⪯xn⪯⋯;y0≽y1≽⋯≽yn≽⋯;z0⪯z1⪯⋯⪯zn⪯⋯.
For the sake of clarity, for all n≥1, denote
(22)Dnx=d(xn-1,xn),Dny=d(yn-1,yn),Dnz=d(zn-1,zn).
We claim that, for all n≥1, we have
(23)Dn+1x≤λnmax{D1x,D1y,D1z},Dn+1y≤λnmax{D1x,D1y,D1z},Dn+1z≤λnmax{D1x,D1y,D1z}.
By (13) and (21), we obtain
(24)D2x=d(x1,x2)=d(F(x0,y0,x0),F(x1,y1,x1))≤λmax{d(x0,x1),d(y0,y1),d(x0,x1),d(F(x0,y0,x0),x0),d(F(x0,y0,x0),x1),d(F(y0,x0,y0),y0),d(F(y0,x0,y0),y1),d(F(x0,y0,x0),x0),d(F(x0,y0,x0),x1)}=λmax{d(x0,x1),d(y0,y1)},D2y=d(y1,y2)=d(F(y0,x0,y0),F(y1,x1,y1))≤λmax{d(y0,y1),d(x0,x1),d(y0,y1),d(F(y0,x0,y0),y0),d(F(y0,x0,y0),y1),d(F(x0,y0,x0),x0),d(F(x0,y0,x0),x1),d(F(y0,x0,y0),y0),d(F(y0,x0,y0),y1)}=λmax{d(y0,y1),d(x0,x1)},D2z=d(z1,z2)=d(F(z0,y0,z0),F(z1,y1,z1))≤λmax{d(z0,z1),d(y0,y1),d(z0,z1),d(F(z0,y0,z0),z0),d(F(z0,y0,z0),z1),d(F(y0,z0,y0),y0),d(F(y0,z0,y0),y1),d(F(z0,y0,z0),z0),d(F(z0,y0,z0),z1)}=λmax{d(z0,z1),d(y0,y1)}.
Thus we get that
(25)D2x≤λmax{D1x,D1y,D1z},D2y≤λmax{D1x,D1y,D1z},D2z≤λmax{D1x,D1y,D1z}.
Then our claim holds for n=1. Suppose now that (23) holds for some fixed n≥1. Similarly, by (13) and (21), we obtain
(26)Dn+2x=d(xn+1,xn+2)=d(F(xn,yn,xn),F(xn+1,yn+1,xn+1))≤λmax{Dn+1x,Dn+1y}≤λn+1max{D1x,D1y,D1y}.
Similarly, one can show that
(27)Dn+2y≤λn+1max{D1x,D1y,D1z},Dn+2z≤λn+1max{D1x,D1y,D1z}.
Then by the induction principle, (23) holds for all n≥1.
Now we will prove that{xn}, {yn}, and {zn} are Cauchy sequences in the metric space (X,d). Using (23) and the triangular inequality, for m>n, we have
(28)d(xm,xn)≤∑i=nm-1‍d(xi,xi+1)=∑i=nm-1‍Di+1x≤∑i=nm-1‍λimax{D1x,D1y,D1z}≤(∑i=nm-1‍λi)max{D1x,D1y,D1z}≤λn1-λmax{D1x,D1y,D1z}⟶0asn⟶∞.
This implies that {xn} is a Cauchy sequence. Similarly, one can prove that {yn} and {zn} are Cauchy sequences.
Since(X,d) is complete, there exist x,y,z∈X such that
(29)xn⟶x,yn⟶y,zn⟶zasn⟶∞.
From condition (iii) and (21), we get that
(30)xn⪯x,yn≽y,zn⪯z,∀n.
We claim that x=F(x,y,x). Indeed, we have
(31)d(x,F(x,y,x))≤d(x,xn+1)+d(xn+1,F(x,y,x))=d(x,xn+1)+d(F(xn,yn,xn),F(x,y,x))≤d(x,xn+1)+λmax{d(xn,x),d(yn,y),d(xn,x),d(F(xn,yn,xn),xn),d(F(xn,yn,xn),x),d(F(yn,xn,yn),yn),d(F(yn,xn,yn),y),d(F(xn,yn,xn),xn),d(F(xn,yn,xn),x)}=d(x,xn+1)+λmax{d(xn,x),d(yn,y)}⟶0asn⟶∞by(29).
Then, our claim holds; that is,x=F(x,y,x). Similarly, one can show that y=F(y,x,y) and z=F(z,y,z). Thus we proved (16).
On the other hand, from (28), for m>n with n being fixed, we have
(32)d(xm,xn)=d(xm,Fn(x0,y0,x0))≤λn1-λmax{D1x,D1y,D1z}.
Letting m→∞ in the above inequality, we obtain
(33)d(x,Fn(x0,y0,x0))≤λn1-λmax{d(x0,x1),d(y0,y1),d(z0,z1)};
that is,
(34)d(x,Fn(x0,y0,x0))≤λn1-λmax{d(x0,F(x0,y0,x0)),d(y0,F(y0,x0,y0)),d(z0,F(z0,y0,z0))}.
Similarly, one can show that
(35)d(y,Fn(y0,x0,y0))≤λn1-λmax{d(x0,F(x0,y0,x0)),d(y0,F(y0,x0,y0)),d(z0,F(z0,y0,z0))},d(z,Fn(z0,y0,z0))≤λn1-λmax{d(x0,F(x0,y0,x0)),d(y0,F(y0,x0,y0)),d(z0,F(z0,y0,z0))}.
Thus we proved (19).
Now, ifx0⪯y0 and z0⪯y0, we claim that, for all n∈ℕ,xn⪯yn and zn⪯yn. Indeed, by the mixed monotone property of F,
(36)x1=F(x0,y0,x0)⪯F(y0,x0,y0)=y1,z1=F(z0,y0,z0)⪯F(y0,x0,y0)=y1.
Assume that xn⪯yn and zn⪯yn for some n. Then,
(37)xn+1=F(xn,yn,xn)⪯F(yn,xn,yn)=yn+1,
and similarly for yn and zn. Thus we proved that
(38)xn⪯yn,zn⪯yn,∀n∈ℕ.
Next, from (38), we have
(39)d(x,y)≤d(x,xn+1)+d(xn+1,yn+1)+d(yn+1,y)≤d(x,xn+1)+d(F(xn,yn,xn),F(yn,xn,yn))+d(yn+1,y)≤d(x,xn+1)+λd(xn,zn)+d(yn+1,y)⋮≤d(x,xn+1)+λnd(x0,y0)+d(yn+1,y)⟶0asn⟶∞,
which implies that x=y. Similarly, we obtain that d(y,z)=0, that is, y=z. Then x=y=z.
Now, assume thatx0⪯y0. Then, in view of the monotonicity of F, we have
(40)x1=F(x0,y0,x0)⪯x2=F(x1,y1,x1)⪯F(y1,x1,y1)=y2⪯F(y0,x0,y0)=y1.
Continuing this process, one can show that
(41)xn⪯xn+1⪯yn+1⪯yn,∀n.
If we assume that u0,u1,u2∈[x0,y0], in view of the monotonicity of F, we have
(42)x1=F(x0,y0,x0)⪯F(u2,u1,u0)=u3⪯F(y0,x0,y0)=y1,x1=F(x0,y0,x0)⪯F(u3,u2,u1)=u4⪯F(y0,x0,y0)=y1,x1=F(x0,y0,x0)⪯F(u4,u3,u2)=u5⪯F(y0,x0,y0)=y1.
Continuing in a similar way, we can prove that
(43)xk⪯u3k,u3k+1,u3k+2⪯yk,∀k≥1.
Letting k→∞ and using (iii) and (29), we get that un→x as n→∞, where x is the equilibrium of (17). Similarly, if z0⪯y0 and u0,u1,u2∈[z0,y0], we obtain also the same result.Remark 6.
If we replace the condition, if{xn}⊂X is a nondecreasing convergent sequence such that limn→∞xn=x, then xn⪯x for all n∈ℕ and if {yn}⊂X is a nonincreasing convergent sequence such that limn→∞yn=y, then yn≽y for every n∈ℕ, by the continuity of F, we can check easily that the result of Theorem 5 holds also in this case.Theorem 7.
In addition to the hypotheses of Theorem5, suppose that for every (x,y,z), (x′,y′,z′)∈X×X×X, there exists (u,v,w) such that (x,y,z)⪯(u,v,w) and (x′,y′,z′)⪯(u,v,w). Then one obtains the uniqueness of the fixed point of the third order.Proof.
From (a) of Theorem5, we know that F admits a fixed point of the third order (x,y,z)∈X×X×X; that is,
(44)x=F(x,y,x),y=F(y,x,y),z=F(z,y,z),
where
(45)xn=Fn(x0,y0,x0)⟶x,yn=Fn(y0,x0,y0)⟶y,zn=Fn(z0,y0,z0)⟶zasn⟶∞.
Suppose that (x*,y*,z*)∈X×X×X is another fixed point of the third order of F; that is,
(46)x*=F(x*,y*,x*),y*=F(y*,x*,y*),z*=F(z*,y*,z*).
We will prove that
(47)η((x,y,z),(x*,y*,z*))=0,
where
(48)η((x,y,z),(x*,y*,z*))=max{d(x,x*),d(y,y*),d(z,z*)}.
From the hypothesis of Theorem 7, there exists (u,v,w) such that
(49)(x,y,z)⪯(u,v,w),(x*,y*,z*)⪯(u,v,w).
Since F is a mixed monotone operator, we have
(50)(Fn(x,y,x),Fn(y,x,y),Fn(z,y,z))=(x,y,z)⪯(Fn(u,v,w),Fn(v,u,v),Fn(w,v,w)),(Fn(x*,y*,x*),Fn(y*,x*,y*),Fn(z*,y*,z*))=(x*,y*,z*)⪯(Fn(u,v,w),Fn(v,u,v),Fn(w,v,w)).
We have
(51)η((x,y,z),(x*,y*,z*))=η((Fn(x,y,x),Fn(y,x,y),Fn(z,y,z)),(Fn(x*,y*,x*),Fn(y*,x*,y*),Fn(z*,y*,z*)))≤η((Fn(x,y,x),Fn(y,x,y),Fn(z,y,z)),(Fn(u,v,u),Fn(v,u,v),Fn(w,v,w)))+η((Fn(u,v,u),Fn(v,u,v),Fn(w,v,w)),(Fn(x*,y*,x*),Fn(y*,x*,y*),Fn(z*,y*,z*)))≤λnmax{d(x,u),d(y,v),d(z,w)}+λnmax{d(u,x*),d(v,y*),d(w,z*)}⟶0asn⟶∞,
which implies that
(52)η((x,y,z),(x*,y*,z*))=0;
that is,
(53)x=x*,y=y*,z=z*.
Example8. Let X=[0,1] be an ordered set with the natural ordering of real numbers and d a usual metric on X. Let F:X×X×X→X be defined by
(54)F(x,y,z)=1+x-y+2z8∀x,y,z∈X.
It is easy to check that F has the mixed monotone property on X. For x,y,z,u,v,w∈X with (x,y,z)⪯(u,v,w), we have
(55)d(F(x,y,z),F(u,v,w))=18|x-u-(y-v)+2(z-w)|≤18[|x-u|+|y-v|+2|z-w|]≤34max{|x-u|,|y-v|,|z-w|}=34max{d(x,u),d(y,v),d(z,w)}≤34max{d(x,u),d(y,v),d(z,w),d(F(x,y,x),x),d(F(x,y,x),u),d(F(y,x,y),y),d(F(y,x,y),v),d(F(z,y,z),z),d(F(z,y,z),w)}.
Thus (13) is satisfied for λ=3/4∈[0,1). Thus all the conditions of Theorems 5 and 7 are satisfied. Moreover, there exists a unique (x,y,z)=(1/6,1/6,1/6) in X×X×X such that
(56)F(x,y,x)=x,F(y,x,y)=y,F(z,y,z)=z.Corollary 9.
Let(X,⪯) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:X×X×X→X be a mapping having the mixed monotone property on X. Suppose that the following conditions hold.  (1)
There existsai's≥0 for i=1,2,…,9 and ∑i=19ai<1 with
(57)d(F(x,y,z),F(u,v,w))≤a1d(x,u)+a2d(y,v)+a3d(z,w)+a4d(F(x,y,x),x)+a5d(F(x,y,x),u)+a6d(F(y,x,y),y)+a7d(F(y,x,y),v)+a8d(F(z,y,z),z)+a9d(F(z,y,z),w)for all(x,y,z)⪯(u,v,w).(2)
There existx0,y0,z0∈X such that
(58)x0⪯F(x0,y0,x0),y0≽F(y0,x0,y0),z0⪯F(z0,y0,z0).(3)
If{xn}⊂X is a nondecreasing convergent sequence such that limn→∞xn=x, then xn⪯x for all n∈ℕ, if {yn}⊂X is a nonincreasing convergent sequence such that limn→∞yn=y, then yn≽y for every n∈ℕ, and if xn⪯yn for every n∈ℕ, then limn→∞xn⪯limn→∞yn.
Then one has the following.(a)
For every initial point(x0,y0,z0)∈X×X×X such that condition (14) holds,
(59)xn=Fn(x0,y0,x0)⟶x,yn=Fn(y0,x0,y0)⟶y,zn=Fn(z0,y0,z0)⟶zasn⟶∞,wherex,y,andz satisfy
(60)x=F(x,y,x),y=F(y,x,y),z=F(z,y,z).Ifx0⪯y0 and z0⪯y0 in condition (14), then x=y=z and {xn}, {yn}, and {zn} converge to the equilibrium of the equation
(61)xn+1=F(xn,yn,xn),yn+1=F(yn,xn,yn),zn+1=F(zn,yn,zn),n=0,1,2….(b)
In particular, every solution{un} of
(62)un+1=F(un,un-1,un-2),n=2,3,…such thatu0,u1,u2∈[x0,y0] (or [z0,y0]) converges to the equilibrium of (18).Proof.
Note that (14) implies that
(63)d(F(x,y,z),F(u,v,w))≤hmax{d(x,u),d(y,v),d(z,w),d(F(x,y,x),x),d(F(x,y,x),u),d(F(y,x,y),y),d(F(y,x,y),v),d(F(z,y,z),z),d(F(z,y,z),w)}
for all (x,y,z)⪯(u,v,w), where h=∑i=19ai<1 and the result follows from Theorem 5.Corollary 10.
In addition to the hypotheses of Corollary9, suppose that for every (x,y,z), (x′,y′,z′)∈X×X×X, there exists (u,v,w) such that (x,y,z)⪯(u,v,w) and (x′,y′,z′)⪯(u,v,w). Then one obtains the uniqueness of the fixed point of the third order.Corollary 11.
Let(X,⪯) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Let F:X×X×X→X be a mapping having the mixed monotone property on X. Suppose that the following conditions hold. (1)
There existsλ∈[0,1) with
(64)d(F(x,y,z),F(u,v,w))≤λmax{d(x,u),d(y,v),d(z,w)}∀(x,y,z)⪯(u,v,w).(2)
There existx0,y0,z0∈X such that
(65)x0⪯F(x0,y0,x0),y0≽F(y0,x0,y0),z0⪯F(z0,y0,z0).(3)
If{xn}⊂X is a nondecreasing convergent sequence such that limn→∞xn=x, then xn⪯x for all n∈ℕ; if {yn}⊂X is a nonincreasing convergent sequence such that limn→∞yn=y, then yn≽y for every n∈ℕ, and if xn⪯yn for every n, then limn→∞xn⪯limn→∞yn.
Then one has the following.(a)
For every initial point(x0,y0,z0)∈X×X×X such that condition (14) holds,
(66)xn=Fn(x0,y0,x0)⟶x,yn=Fn(y0,x0,y0)⟶y,zn=Fn(z0,y0,z0)⟶zasn⟶∞,wherex,y,andz satisfy
(67)x=F(x,y,x),y=F(y,x,y),z=F(z,y,z).Ifx0⪯y0 and z0⪯y0 in condition (14), then x=y=z and {xn}, {yn}, and {zn} converge to the equilibrium of the equation
(68)xn+1=F(xn,yn,xn),yn+1=F(yn,xn,yn),zn+1=F(zn,yn,zn),n=0,1,2….(b)
In particular, every solution{un} of
(69)un+1=F(un,un-1,un-2),n=2,3,…such thatu0,u1,u2∈[x0,y0] (or [z0,y0]) converges to the equilibrium of (18).(c)
The following estimates hold:(70)d(x,Fn(x0,y0,x0))≤λn1-λmax{d(F(x0,y0,x0),x0),d(F(y0,x0,y0),y0),d(z0,F(z0,y0,z0))},d(y,Fn(y0,x0,y0))≤λn1-λmax{d(F(x0,y0,x0),x0),d(F(y0,x0,y0),y0),d(z0,F(z0,y0,z0))},d(z,Fn(z0,y0,z0))≤λn1-λmax{d(F(x0,y0,x0),x0),d(F(y0,x0,y0),y0),d(F(z0,y0,z0),z0)}.Corollary 12.
In addition to the hypotheses of Corollary11, suppose that for every (x,y,z), (x′,y′,z′)∈X×X×X, there exists (u,v,w) such that (x,y,z)⪯(u,v,w) and (x′,y′,z′)⪯(u,v,w). Then one obtains the uniqueness of the fixed point of the third order.


## 3. Application: Solving a Class of Third-Order Difference Matrix Equations

In this section, we apply our main results to the study of a class of third-order difference matrix equations. At first, we start by fixing some notations and recalling some preliminaries.We will use the symbolH(N) for the set of all N×N Hermitian matrices. We denote by P(N) the set of all N×N Hermitian positive definite matrices. Instead of X∈P(N) we will also write X≻0. Furthermore, X≽0 means that X is positive semidefinite. As a different notation for X-Y≽0,X-Z≽0 and X⪯Z⪯Y we will use, respectively, X≽Y,X≽Z, and Z∈[X,Y]. The symbol ∥·∥ denotes the spectral norm, that is, ∥A∥=λmax1/2(A*A), the largest eigenvalue of A*A. We denote by ∥·∥1 the Ky Fan norm defined by ∥A∥1=∑j=1N‍sj(A), where sj(A),j=1,…,N are the singular values of A. For a given Q∈P(N), we define the modified norm ∥·∥1,Q given by ∥A∥1,Q=∥Q1/2AQ1/2∥1. The set H(N) endowed with this norm is a complete metric space for any positive definite matrix Q. For any N×N matrix A, we denote by tr(A) the trace of the matrix A.The following lemmas will be useful later.Lemma 13 (see [18]).
LetA≽0 and B≽0 be N×N matrices, then
(71)0≤tr(AB)≤∥A∥tr(B).Lemma 14 (See [21]).
LetA∈H(N) satisfy -I≺A≺I, then ∥A∥<1.Finally, we recall the well-known Schauder fixed-point theorem.Theorem 15 (The Schauder fixed-point theorem).
LetS be a nonempty, compact, and convex subset of a normed vector space. Every continuous function f:S→S mapping S into itself has a fixed point.Now, we consider the class of third-order difference matrix equations:(72)Un+1=Q+∑i=1m‍AiUnAi-∑i=1m‍BiUn-1Bi+∑i=1m‍CiUn-2Ci,n=2,3,…
for given U0,U1,U2≻0, where Q≻0 and Ai, Bi, and Ci are N×N Hermitian matrices. This type of difference equations often arises from many areas such as ladder networks [22, 23], dynamic programming [24, 25], and control theory [26, 27].


### 3.1. A Convergence Result

Theorem 16.
Suppose that(73)∑i=1m‍AiQAi≺Q4,∑i=1m‍BiQBi≺Q2,∑i=1m‍CiQCi≺Q4.
Then, one has the following.(i)
Equation (72) has one and only one equilibrium point X^>0.(ii)
X^∈[F(0,Q,0),F(2Q,0,2Q)], where
(74)F(0,2Q,0)=Q-2∑i=1m‍BiQBi,F(2Q,0,2Q)=Q+2∑i=1m‍AiQAi+2∑i=1m‍CiQCi.(iii)
The sequences(Xk) and (Yk) defined by X0=0=Z0, Y0=2Q, and
(75)Xk+1=Q+∑i=1m‍AiXkAi-∑i=1m‍BiYkBi+∑i=1m‍AiXkAi,Yk+1=Q+∑i=1m‍AiYkAi-∑i=1m‍BiXkBi+∑i=1m‍AiYkAi,Zk+1=Q+∑i=1m‍AiZkAi-∑i=1m‍BiYkBi+∑i=1m‍AiZkAi,k=0,1,2,…converge toX^, and the error estimation is given by
(76)max{∥Xk+1-X^∥1,Q,∥Yk+1-X^∥1,Q,∥Zk+1-Z^∥1,Q}≤λk(1-k)max{∥X1-X0∥1,Q,∥Y1-Y0∥1,Q,∥Z1-Z0∥1,Q},for allk=0,1,…, where λ is a certain constant in [0,1).(iv)
For everyU0,U1,U2∈[0,2Q], every solution {Un} of (72) converges to X^.Proof.
In order to make the proof easy, we divide it into several steps.
Step1. We claim that there exists a unique (X,Y,Z)∈H(N)×H(N)×H(N) solution to the system
(77)X=Q+∑i=1m‍AiXAi-∑i=1m‍BiYBi+∑i=1m‍CiXCi,Y=Q+∑i=1m‍AiYAi-∑i=1m‍BiXBi+∑i=1m‍CiYCi,Z=Q+∑i=1m‍AiZAi-∑i=1m‍BiYBi+∑i=1m‍CiZCi.
Consider the mapping F:H(N)×H(N)×H(N)→H(N) defined by
(78)F(X,Y,Z)=Q+∑i=1m‍AiXAi-∑i=1m‍BiYBi+∑i=1m‍CiZCi,
for all X,Y,Z∈H(N). It is clear that F is a mapping having the mixed monotone property with respect to the partial order ⪯. Let X,Y,Z,U,V,W∈H(N) such that X≽U, Y⪯V, and Z≽W. Using Lemma 13, we have
(79)∥F(X,Y,Z)-F(U,V,W)∥1,Q=tr(Q1/2(F(X,Y,Z)-F(U,V,W))Q1/2)=tr(Q1/2(∑i=1m‍AiXAi-∑i=1m‍BiYBi+∑i=1m‍CiZCi-∑i=1m‍AiUAi+∑i=1m‍BiVBi+∑i=1m‍CiWCi)Q1/2)=∑i=1m‍tr(Q1/2(Ai(X-U)Ai+Bi(V-Y)Bi+Ci(Z-W)Ci)Q1/2)=∑i=1m‍tr(Q1/2(Ai(X-U)Ai)Q1/2)+tr(Q1/2(Bi(V-Y)Bi)Q1/2)+tr(Q1/2(Ci(Z-W)Ci)Q1/2)=∑i=1m‍tr(AiQAi(X-U))+tr(BiQBi(V-Y))+tr(CiQCi(Z-W))=∑i=1m‍tr(AiQAiQ-1/2Q1/2(X-U)Q1/2Q-1/2)+tr(BiQBiQ-1/2Q1/2(V-Y)Q1/2Q-1/2)+tr(CiQCiQ-1/2Q1/2(Z-W)Q1/2Q-1/2)=∑i=1m‍tr(Q-1/2AiQAiQ-1/2Q1/2(X-U)Q1/2)+tr(Q-1/2BiQBiQ-1/2Q1/2(V-Y)Q1/2)+tr(Q-1/2CiQCiQ-1/2Q1/2(Z-W)Q1/2)=tr(∑i=1m‍Q-1/2AiQAiQ-1/2Q1/2(X-U)Q1/2)+tr(∑i=1m‍Q-1/2BiQBiQ-1/2Q1/2(V-Y)Q1/2)+tr(∑i=1m‍Q-1/2CiQCiQ-1/2Q1/2(Z-W)Q1/2)≤∥∑i=1m‍Q-1/2AiQAiQ-1/2∥∥X-U∥1,Q+∥∑i=1m‍Q-1/2BiQBiQ-1/2∥∥V-Y∥1,Q+∥∑i=1m‍Q-1/2CiQCiQ-1/2∥∥Z-W∥1,Q≤δmax{∥X-U∥1,Q,∥V-Y∥1,Q,∥X-U∥1,Q},
where
(80)δ=max{∥∑i=1m‍Q-1/2AiQAiQ-1/2∥,∥∑i=1m‍Q-1/2BiQBiQ-1/2∥,∥∑i=1m‍Q-1/2CiQCiQ-1/2∥}.
From (73) and Lemma 14, we have δ<1. Thus, the contractive condition of Theorem 5 is satisfied for all X,Y,Z,U,V,W∈H(N) with X≽U, Y⪯V, and Z≽W. Moreover, from (73), we have X0=Z0=0≺F(0,2Q,0) and F(2Q,0,2Q)≺2Q=Y0.
Now, all the hypotheses of Theorem5 are satisfied. Consequently, there exists (X,Y,Z)∈H(N)×H(N)×H(N) solution to (77). Since for every X,Y,Z∈H(N) there is a greatest lower bound and a least upper bound with respect to the partial order ⪯, we deduce from Theorem 7 the uniqueness of the solution to (77). Then, our claim holds.
Step2. We claim that X=Y=Z=X^.
SinceX0=Z0=0≺Y0=2Q (note that Q is a positive definite matrix), applying Theorem 5, we obtain the equality X=Y=Z. This proves our claim.
From Steps 1 and 2, we know that (72) has a unique equilibrium point X^∈H(N). Now, we need to prove that X^∈P(N). This is the goal of the next step.
Step3 (proof of (i)–(iii)). Define the mapping G:[F(0,2Q,0),F(2Q,0,2Q)]→H(N) by
(81)GX=Q+∑i=1m‍AiXAi-∑i=1m‍BiXBi+∑i=1m‍CiXCi,
for all X∈[F(0,2Q,0),F(2Q,0,2Q)].
From (73), one can show that G([F(0,2Q,0),F(2Q,0,2Q)])⊆[F(0,2Q,0),F(2Q,0,2Q)]. Now, applying the Schauder fixed-point theorem (see Theorem 15), we deduce the existence of a fixed point of G in [F(0,2Q,0),F(2Q,0,2Q)]. But a fixed point of G is an equilibrium point of (72), and from Step 2, we know that (72) has a unique equilibrium point in H(N). Consequently, 0≺F(0,2Q,0)⪯X^⪯F(2Q,0,2Q). This proves (i) and (ii). For the proof of (iii), we have only to apply the inequalities (19).
Step4. Let U0=0, U1=Q, and U2=2Q. Then, we have U0,U1,U2∈[0,2Q], thus by (b) Theorem 5, every solution {Un} to (72) converges to X^.


### 3.2. Numerical Example

Consider a third-order difference matrix equation(82)Un+1=Q+AUnA-BUn-1B+CUn-2C,
where Q, A, B, C, U0, U1, and U2 are given by
(83)Q=(1.50.30.30.31.50.30.30.31.5),A=(0.30.050.050.050.30.050.050.050.3),B=(0.10.20.10.20.10.20.10.20.1),C=(0.10.10.10.10.10.10.10.10.1),U0=0,U1=Q,U2=2Q,We are interested to approximate the unique positive equilibrium to (82).We use Algorithm (iii) of Theorem16 with X0=0=Z0 and Y0=2Q.After35 iterations, we get the unique solution X^ given by (84)X^=X35=Y35=Z35=(1.6156434875710.318746203393550.335643487570950.318746203393551.53742213234620.318746203393550.335643487570950.318746203393551.615643487571),U35=(1.61564348758170.318746203405830.335643487581710.318746203405831.53742213236040.318746203405830.335643487581710.318746203405831.6156434875817).The residual errors are(85)R(X)35=∥X35-(Q+A1X35A1+A2X35A2-B1X35B1-B2X35B2)∥=3.792×10-16,R(Y)35=∥Y35-(Q+A1Y35A1+A2Y35A2-B1Y35B1-B2Y35B2)∥=1.796×10-16,R(Z)35=∥Z35-(Q+A1Z35A1+A2Z35A2-B1Z35B1-B2Z35B2)∥=2.351×10-16,R(U)35=∥U35-(Q+A1U35A1+A2U35A2-B1U35B1-B2U35B2)∥=3.352×10-11.The convergence history is given by Figure1, where (a) corresponds to ∥Xk-F(Xk,Xk,Xk)∥, (b) corresponds to ∥Yk-F(Yk,Yk,Yk)∥, (c) corresponds to ∥Zk-F(Zk,Zk,Zk)∥, and (d) corresponds to ∥Uk-F(Uk,Uk,Uk)∥.Convergence history for (82).
(a)
(b)
(c)
(d)


## 3.1. A Convergence Result

Theorem 16.
Suppose that(73)∑i=1m‍AiQAi≺Q4,∑i=1m‍BiQBi≺Q2,∑i=1m‍CiQCi≺Q4.
Then, one has the following.(i)
Equation (72) has one and only one equilibrium point X^>0.(ii)
X^∈[F(0,Q,0),F(2Q,0,2Q)], where
(74)F(0,2Q,0)=Q-2∑i=1m‍BiQBi,F(2Q,0,2Q)=Q+2∑i=1m‍AiQAi+2∑i=1m‍CiQCi.(iii)
The sequences(Xk) and (Yk) defined by X0=0=Z0, Y0=2Q, and
(75)Xk+1=Q+∑i=1m‍AiXkAi-∑i=1m‍BiYkBi+∑i=1m‍AiXkAi,Yk+1=Q+∑i=1m‍AiYkAi-∑i=1m‍BiXkBi+∑i=1m‍AiYkAi,Zk+1=Q+∑i=1m‍AiZkAi-∑i=1m‍BiYkBi+∑i=1m‍AiZkAi,k=0,1,2,…converge toX^, and the error estimation is given by
(76)max{∥Xk+1-X^∥1,Q,∥Yk+1-X^∥1,Q,∥Zk+1-Z^∥1,Q}≤λk(1-k)max{∥X1-X0∥1,Q,∥Y1-Y0∥1,Q,∥Z1-Z0∥1,Q},for allk=0,1,…, where λ is a certain constant in [0,1).(iv)
For everyU0,U1,U2∈[0,2Q], every solution {Un} of (72) converges to X^.Proof.
In order to make the proof easy, we divide it into several steps.
Step1. We claim that there exists a unique (X,Y,Z)∈H(N)×H(N)×H(N) solution to the system
(77)X=Q+∑i=1m‍AiXAi-∑i=1m‍BiYBi+∑i=1m‍CiXCi,Y=Q+∑i=1m‍AiYAi-∑i=1m‍BiXBi+∑i=1m‍CiYCi,Z=Q+∑i=1m‍AiZAi-∑i=1m‍BiYBi+∑i=1m‍CiZCi.
Consider the mapping F:H(N)×H(N)×H(N)→H(N) defined by
(78)F(X,Y,Z)=Q+∑i=1m‍AiXAi-∑i=1m‍BiYBi+∑i=1m‍CiZCi,
for all X,Y,Z∈H(N). It is clear that F is a mapping having the mixed monotone property with respect to the partial order ⪯. Let X,Y,Z,U,V,W∈H(N) such that X≽U, Y⪯V, and Z≽W. Using Lemma 13, we have
(79)∥F(X,Y,Z)-F(U,V,W)∥1,Q=tr(Q1/2(F(X,Y,Z)-F(U,V,W))Q1/2)=tr(Q1/2(∑i=1m‍AiXAi-∑i=1m‍BiYBi+∑i=1m‍CiZCi-∑i=1m‍AiUAi+∑i=1m‍BiVBi+∑i=1m‍CiWCi)Q1/2)=∑i=1m‍tr(Q1/2(Ai(X-U)Ai+Bi(V-Y)Bi+Ci(Z-W)Ci)Q1/2)=∑i=1m‍tr(Q1/2(Ai(X-U)Ai)Q1/2)+tr(Q1/2(Bi(V-Y)Bi)Q1/2)+tr(Q1/2(Ci(Z-W)Ci)Q1/2)=∑i=1m‍tr(AiQAi(X-U))+tr(BiQBi(V-Y))+tr(CiQCi(Z-W))=∑i=1m‍tr(AiQAiQ-1/2Q1/2(X-U)Q1/2Q-1/2)+tr(BiQBiQ-1/2Q1/2(V-Y)Q1/2Q-1/2)+tr(CiQCiQ-1/2Q1/2(Z-W)Q1/2Q-1/2)=∑i=1m‍tr(Q-1/2AiQAiQ-1/2Q1/2(X-U)Q1/2)+tr(Q-1/2BiQBiQ-1/2Q1/2(V-Y)Q1/2)+tr(Q-1/2CiQCiQ-1/2Q1/2(Z-W)Q1/2)=tr(∑i=1m‍Q-1/2AiQAiQ-1/2Q1/2(X-U)Q1/2)+tr(∑i=1m‍Q-1/2BiQBiQ-1/2Q1/2(V-Y)Q1/2)+tr(∑i=1m‍Q-1/2CiQCiQ-1/2Q1/2(Z-W)Q1/2)≤∥∑i=1m‍Q-1/2AiQAiQ-1/2∥∥X-U∥1,Q+∥∑i=1m‍Q-1/2BiQBiQ-1/2∥∥V-Y∥1,Q+∥∑i=1m‍Q-1/2CiQCiQ-1/2∥∥Z-W∥1,Q≤δmax{∥X-U∥1,Q,∥V-Y∥1,Q,∥X-U∥1,Q},
where
(80)δ=max{∥∑i=1m‍Q-1/2AiQAiQ-1/2∥,∥∑i=1m‍Q-1/2BiQBiQ-1/2∥,∥∑i=1m‍Q-1/2CiQCiQ-1/2∥}.
From (73) and Lemma 14, we have δ<1. Thus, the contractive condition of Theorem 5 is satisfied for all X,Y,Z,U,V,W∈H(N) with X≽U, Y⪯V, and Z≽W. Moreover, from (73), we have X0=Z0=0≺F(0,2Q,0) and F(2Q,0,2Q)≺2Q=Y0.
Now, all the hypotheses of Theorem5 are satisfied. Consequently, there exists (X,Y,Z)∈H(N)×H(N)×H(N) solution to (77). Since for every X,Y,Z∈H(N) there is a greatest lower bound and a least upper bound with respect to the partial order ⪯, we deduce from Theorem 7 the uniqueness of the solution to (77). Then, our claim holds.
Step2. We claim that X=Y=Z=X^.
SinceX0=Z0=0≺Y0=2Q (note that Q is a positive definite matrix), applying Theorem 5, we obtain the equality X=Y=Z. This proves our claim.
From Steps 1 and 2, we know that (72) has a unique equilibrium point X^∈H(N). Now, we need to prove that X^∈P(N). This is the goal of the next step.
Step3 (proof of (i)–(iii)). Define the mapping G:[F(0,2Q,0),F(2Q,0,2Q)]→H(N) by
(81)GX=Q+∑i=1m‍AiXAi-∑i=1m‍BiXBi+∑i=1m‍CiXCi,
for all X∈[F(0,2Q,0),F(2Q,0,2Q)].
From (73), one can show that G([F(0,2Q,0),F(2Q,0,2Q)])⊆[F(0,2Q,0),F(2Q,0,2Q)]. Now, applying the Schauder fixed-point theorem (see Theorem 15), we deduce the existence of a fixed point of G in [F(0,2Q,0),F(2Q,0,2Q)]. But a fixed point of G is an equilibrium point of (72), and from Step 2, we know that (72) has a unique equilibrium point in H(N). Consequently, 0≺F(0,2Q,0)⪯X^⪯F(2Q,0,2Q). This proves (i) and (ii). For the proof of (iii), we have only to apply the inequalities (19).
Step4. Let U0=0, U1=Q, and U2=2Q. Then, we have U0,U1,U2∈[0,2Q], thus by (b) Theorem 5, every solution {Un} to (72) converges to X^.


## 3.2. Numerical Example

Consider a third-order difference matrix equation(82)Un+1=Q+AUnA-BUn-1B+CUn-2C,
where Q, A, B, C, U0, U1, and U2 are given by
(83)Q=(1.50.30.30.31.50.30.30.31.5),A=(0.30.050.050.050.30.050.050.050.3),B=(0.10.20.10.20.10.20.10.20.1),C=(0.10.10.10.10.10.10.10.10.1),U0=0,U1=Q,U2=2Q,We are interested to approximate the unique positive equilibrium to (82).We use Algorithm (iii) of Theorem16 with X0=0=Z0 and Y0=2Q.After35 iterations, we get the unique solution X^ given by (84)X^=X35=Y35=Z35=(1.6156434875710.318746203393550.335643487570950.318746203393551.53742213234620.318746203393550.335643487570950.318746203393551.615643487571),U35=(1.61564348758170.318746203405830.335643487581710.318746203405831.53742213236040.318746203405830.335643487581710.318746203405831.6156434875817).The residual errors are(85)R(X)35=∥X35-(Q+A1X35A1+A2X35A2-B1X35B1-B2X35B2)∥=3.792×10-16,R(Y)35=∥Y35-(Q+A1Y35A1+A2Y35A2-B1Y35B1-B2Y35B2)∥=1.796×10-16,R(Z)35=∥Z35-(Q+A1Z35A1+A2Z35A2-B1Z35B1-B2Z35B2)∥=2.351×10-16,R(U)35=∥U35-(Q+A1U35A1+A2U35A2-B1U35B1-B2U35B2)∥=3.352×10-11.The convergence history is given by Figure1, where (a) corresponds to ∥Xk-F(Xk,Xk,Xk)∥, (b) corresponds to ∥Yk-F(Yk,Yk,Yk)∥, (c) corresponds to ∥Zk-F(Zk,Zk,Zk)∥, and (d) corresponds to ∥Uk-F(Uk,Uk,Uk)∥.Convergence history for (82).
(a)
(b)
(c)
(d)



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*Source: 486357-2013-03-27.xml*