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Research article

Validation of a rapid test to dose SO2 in vinegar

  • Received: 09 June 2022 Revised: 15 September 2022 Accepted: 13 November 2022 Published: 02 December 2022
  • Sulfur dioxide is generally used in wine and vinegar production. It is employed to decrease the bacteria' growth, improve the wines' aroma (since it supports the extraction of polyphenols during maceration), protect the wines from chemical oxidation and the musts from chemical and enzymatic oxidation (blocking free radicals and oxidase enzymes such as tyrosinase and laccase). The composition and storage conditions (i.e., pH, temperature, and alcohol levels) affect oenological results. In various countries, competent authorities have imposed legal limits since it can have toxic effects on humans. It is crucial to dose SO2 levels to allow vinegar production and compliance with legal limits. The iodometric titration named "Ripper test" is the legal method used to dose it in vinegar. In this work, an automatized colorimetric test was validated using the international guidelines ISO/IEC (2017) to allow its use instead of the Ripper test. The test reliability was verified on white, red, and balsamic vinegar with low or high SO2 content. The automatized test showed linearity, precision, and reproducibility similar to the Ripper test, but the accuracy parameter was not respected for the vinegar with a low concentration of SO2. Therefore, the automatized colorimetric test can be helpful to dose SO2 in vinegar with high concentrations of SO2.

    Citation: Irene Dini, Antonello Senatore, Daniele Coppola, Andrea Mancusi. Validation of a rapid test to dose SO2 in vinegar[J]. AIMS Agriculture and Food, 2023, 8(1): 1-24. doi: 10.3934/agrfood.2023001

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  • Sulfur dioxide is generally used in wine and vinegar production. It is employed to decrease the bacteria' growth, improve the wines' aroma (since it supports the extraction of polyphenols during maceration), protect the wines from chemical oxidation and the musts from chemical and enzymatic oxidation (blocking free radicals and oxidase enzymes such as tyrosinase and laccase). The composition and storage conditions (i.e., pH, temperature, and alcohol levels) affect oenological results. In various countries, competent authorities have imposed legal limits since it can have toxic effects on humans. It is crucial to dose SO2 levels to allow vinegar production and compliance with legal limits. The iodometric titration named "Ripper test" is the legal method used to dose it in vinegar. In this work, an automatized colorimetric test was validated using the international guidelines ISO/IEC (2017) to allow its use instead of the Ripper test. The test reliability was verified on white, red, and balsamic vinegar with low or high SO2 content. The automatized test showed linearity, precision, and reproducibility similar to the Ripper test, but the accuracy parameter was not respected for the vinegar with a low concentration of SO2. Therefore, the automatized colorimetric test can be helpful to dose SO2 in vinegar with high concentrations of SO2.



    In this paper, we study the behavior of the solutions of the following three-dimensional system of difference equations of second order

    xn+1=f(yn,yn1),yn+1=g(zn,zn1),zn+1=h(xn,xn1) (1)

    where nN0, the initial values xi, yi and zi,i=0,1, are positive real numbers, the functions f,g,h:(0,+)2(0,+) are continuous and homogeneous of degree zero. We establish results on local and global stability of the unique positive equilibrium point. To do this we prove some general convergence theorems, that can be applied to generalize a lot of existing systems and to study new ones. Some results on existence of periodic and oscillatory solutions are also proved.

    Now, we explain our motivation for doing this work. Clearly if we take zi=xi,i=0,1, and hg, then the system (1), will be

    xn+1=f(yn,yn1),yn+1=g(xn,xn1). (2)

    Noting also that if we choose zi=yi=xi,i=0,1, and hgf, then System (1) will be

    xn+1=f(xn,xn1). (3)

    In [23], the behavior of the solutions of System (2) has been investigated. System (2) is a generalization of Equation (3), studied in [17]. The present System (1) is the three-dimensional generalization of System (2).

    In the literature there are many studies on difference equations defined by homogeneous functions, see for instance [1,2,5,7,11,16]. Noting that also a lot of studies are devoted to various models of difference equations and systems, not necessary defined by homogeneous functions, see for example [4,9,10,12,14,18,19,20,21,22,24,25,26,27,28,29].

    Before we state our results, we recall the following definitions and results. For more details we refer to the following references [3,6,8,13].

    Let F:(0,+)6(0,+)6 be a continuous function and consider the system of difference equations

    Yn+1=F(Yn),nN0, (4)

    where the initial value Y0(0,+)6. A point ¯Y(0,+)6 is an equilibrium point of (4), if it is a solution of ¯Y=F(¯Y).

    Definition 1.1. Let ¯Y be an equilibrium point of System (4), and let . any convenient vector norm.

    1. We say that the equilibrium point ¯Y is stable (or locally stable) if for every ϵ>0 there exists δ>0 such that for every initial condition Y0: Y0¯Y<δ implies Yn¯Y<ϵ. Otherwise, the equilibrium point ¯Y is unstable.

    2. We say that the equilibrium point ¯Y is asymptotically stable (or locally asymptotically stable) if it is stable and there exists γ>0 such that Y0¯Y<γ implies

    limnYn=¯Y.

    3. We say that the equilibrium point ¯Y is a global attractor if for every Y0,

    limnYn=¯Y.

    4. We say that the equilibrium point ¯Y is globally (asymptotically) stable if it is stable and a global attractor.

    Assume that F is C1 on (0,+)6. To System (4), we associate a linear system, about the equilibrium point ¯Y, given by

    Zn+1=FJ(¯Y)Zn,nN0,Zn=Yn¯Y,

    where FJ is the Jacobian matrix of the function F evaluated at the equilibrium point ¯Y.

    To study the stability of the equilibrium point ¯Y, we need the following theorem.

    Theorem 1.2. Let ¯Y be an equilibrium point of System (4). Then, the following statements are true:

    (i) If all the eigenvalues of the Jacobian matrix FJ lie in the open unit disk |λ|<1, then the equilibrium ¯Y is asymptotically stable.

    (ii) If at least one eigenvalue of FJ has absolute value greater than one, then the equilibrium ¯Y is unstable.

    Definition 1.3. A solution (xn,yn,zn)n1 of System (1) is said to be periodic of period pN if

    xn+p=xn,yn+p=yn,zn+p=zn,n1. (5)

    The solution (xn,yn,zn)n1 is said to be periodic with prime period pN, if it is periodic with period p and p is the least positive integer for which (1.3) holds.

    Definition 1.4. Let (xn,yn,zn)n1 be a solution of System (1). We say that the sequence (xn)n1 (resp. (yn)n1, (zn)n1) oscillates about ¯x (resp. ¯y, ¯z) with a semi-cycle of length one if: (xn¯x)(xn+1¯x)<0,n1 (resp. (yn¯y)(yn+1¯y)<0,n1, (zn¯z)(zn+1¯z)<0,n1).

    Remark 1. For every term xn0 of the sequence (xn)n1, the notation "+ " means xn0¯x>0 and the notation "" means xn0¯x<0. The same notations will be used for the terms of the sequences (yn)n1 and (zn)n1.

    Definition 1.5. A function Φ:(0,+)2(0,+) is said to be homogeneous of degree mR if we have

    Φ(λu,λv)=λmΦ(u,v)

    for all (u,v)(0,+)2 and for all λ>0.

    Theorem 1.6. Let Φ:(0,+)2(0,+) be a C1 function on (0,+)2.

    1. Then, Φ is homogeneous of degree m if and only if

    uΦu(u,v)+vΦv(u,v)=mΦ(u,v),(u,v)(0,+)2.

    (This statement is usually called Euler's Theorem).

    2. If Φ is homogeneous of degree m on (0,+)2, then Φu and Φv are homogenous of degree m1 on (0,+)2.

    A point (¯x,¯y,¯z)(0,+)3 is an equilibrium point of System (1) if it is a solution of the following system

    ¯x=f(¯y,¯y),¯y=g(¯z,¯z),¯z=g(¯x,¯x).

    Using the fact that f, g and h are homogeneous of degree zero, we get that

    (¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1))

    is the unique equilibrium point of System (1).

    Let F:(0,+)6(0,+)6 be the function defined by

    F(W)=(f1(W),f2(W),g1(W),g2(W),h1(W),h2(W)),W=(u,v,w,t,r,s)

    with

    f1(W)=f(w,t),f2(W)=u,g1(W)=g(r,s),g2(W)=w,h1(W)=h(u,v),g2(W)=r.

    Then, System (1) can be written as follows

    Wn+1=F(Wn),Wn=(xn,xn1,yn,yn1,zn,zn1)t,nN0.

    So, (¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1)) is an equilibrium point of system (1) if and only if

    ¯W=(¯x,¯x,¯y,¯y,¯z,¯z)=(f(1,1),f(1,1),g(1,1),g(1,1),h(1,1),h(1,1))

    is an equilibrium point of Wn+1=F(Wn).

    Assume that the functions f, g and h are C1 on (0,+)2. To System (1), we associate about the equilibrium point ¯W the following linear system

    Xn+1=JFXn,nN0

    where JF is the Jacobian matrix associated to the function F evaluated at

    ¯W=(f(1,1),f(1,1),g(1,1),g(1,1),h(1,1),h(1,1)).

    We have

    JF=(00fw(¯y,¯y)ft(¯y,¯y)001000000000gr(¯z,¯z)gs(¯z,¯z)001000hu(¯x,¯x)hv(¯x,¯x)0000000010)

    As f, g and h are homogeneous of degree 0, then using Part 1. of Theorem 1.6, we get

    ¯yfw(¯y,¯y)+¯yft(¯y,¯y)=0

    which implies

    ft(¯y,¯y)=fw(¯y,¯y).

    Similarly we get

    gs(¯z,¯z)=gr(¯z,¯z),hv(¯x,¯x)=hu(¯x,¯x).

    It follows that JF takes the form:

    JF=(00fw(¯y,¯y)fw(¯y,¯y)001000000000gr(¯z,¯z)gr(¯z,¯z)001000hu(¯x,¯x)hu(¯x,¯x)0000000010)

    The characteristic polynomial of the matrix JF is given by

    P(λ)=λ6hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y)λ3+3hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y)λ23hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y)λ+hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y).

    Now assume that

    |hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y)|<18

    and consider the two functions

    Φ(λ)=λ6,
    Ψ(λ)=hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y)λ3+3hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y)λ23hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y)λ+hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y).

    We have

    |Ψ(λ)|8|hu(¯x,¯x)gr(¯z,¯z)fw(¯y,¯y)|<1=|Φ(λ)|,λC:|λ|=1.

    So, by Rouché's Theorem it follows that all roots of P(λ) lie inside the unit disk.

    Hence, by Theorem 1.2, we deduce from the above consideration that the equilibrium point (¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1)) is locally asymptotically stable.

    Using Part 2. of Theorem 1.6 and the fact that the function f, g, h are homogeneous of degree zero, we get that fw, gr and hu are homogeneous of degree 1. So, it follows that

    fw(¯y,¯y)=fw(1,1)¯y,gr(¯z,¯z)=gr(1,1)¯z,hu(¯x,¯x)=hu(1,1)¯x.

    In summary, we have proved the following result.

    Theorem 2.1. Assume that f(u,v), g(u,v) and h(u,v) are C1 on (0,+)2. The equilibrium point

    (¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1))

    of System (1) is locally asymptotically stable if

    |fu(1,1)gu(1,1)hu(1,1)|<f(1,1)g(1,1)h(1,1)8.

    Now, we will prove some general convergence results. The obtained results allow us to deal with the global attractivity of the equilibrium point (¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1)) and so the global stability.

    Theorem 2.2. Consider System (1). Assume that the following statements are true:

    1. H1: There exist a,b,α,β,λ,γ(0,+) such that

    af(u,v)b,αg(u,v)β,λh(u,v)γ,(u,v)(0,+)2.

    2. H2: f(u,v),g(u,v),h(u,v) are increasing in u for all v and decreasing in v for all u.

    3. H3: If (m1,M1,m2,M2,m3,M3)[a,b]2×[α,β]2×[λ,γ]2 is a solution of the system

    m1=f(m2,M2),M1=f(M2,m2),m2=g(m3,M3),M2=g(M3,m3),m3=h(m1,M1),M3=h(M1,m1)

    then

    m1=M1,m2=M2,m3=M3.

    Then every solution of System (1) converges to the unique equilibrium point

    (¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1)).

    Proof. Let

    m01:=a,M01:=b,m02:=α,M02:=β,m03:=λ,M03:=γ

    and for each i=0,1,...,

    mi+11:=f(mi2,Mi2),Mi+11:=f(Mi2,mi2),
    mi+12:=g(mi3,Mi3),Mi+12:=g(Mi3,mi3),
    mi+13:=h(mi1,Mi1),Mi+13:=h(Mi1,mi1).

    We have

    af(α,β)f(β,α)b,
    αg(λ,γ)g(γ,λ)β,
    λh(a,b)h(b,a)γ,

    and so,

    m01=af(m02,M02)f(M02,m02)b=M01,
    m02=αg(m03,M03)g(M03,m03)β=M02,

    and

    m03=λh(m01,M01)g(M01,m01)γ=M03.

    Hence,

    m01m11M11M01,
    m02m12M12M02,

    and

    m03m13M13M03.

    Now, we have

    m11=f(m02,M02)f(m12,M12)=m21f(M12,m12)=M21f(M02,m02)=M11,
    m12=g(m03,M03)g(m13,M13)=m22g(M13,m13)=M22g(M03,m03)=M12,
    m13=h(m01,M01)h(m11,M11)=m23h(M11,m11)=M23h(M01,m01)=M13,

    and it follows that

    m01m11m21M21M11M01,
    m02m12m22M22M12M02,

    and

    m03m13m23M23M13M03.

    By induction, we get for i=0,1,..., that

    a=m01m11...mi11mi1Mi1Mi11...M11M01=b,
    α=m02m12...mi12mi2Mi2Mi12...M12M02=β,

    and

    λ=m03m13...mi13mi3Mi3Mi13...M13M03=γ.

    It follows that the sequences (mi1)iN0, (mi2)iN0, (mi3)iN0 (resp. (Mi1)iN0, (Mi2)iN0, (Mi3)iN0) are increasing (resp. decreasing) and also bounded, so convergent. Let

    m1=limi+mi1,M1=limi+Mi1,
    m2=limi+mi2,M2=limi+Mi2.
    m3=limi+mi3,M3=limi+Mi3.

    Then

    am1M1b,αm2M2β,λm3M3γ.

    By taking limits in the following equalities

    mi+11=f(mi2,Mi2),Mi+11=f(Mi2,mi2),
    mi+12=g(mi3,Mi3),Mi+12=g(Mi3,mi3),
    mi+13=h(mi1,Mi1),Mi+13=h(Mi1,mi1),

    and using the continuity of f, g and h we obtain

    m1=f(m2,M2),M1=f(M2,m2),m2=g(m3,M3),M2=g(M3,m3),m3=h(m1,M1),M3=h(M1,m1)

    so it follows from H3 that

    m1=M1,m2=M2,m3=M3.

    From H1, for n=1,2,, we get

    m01=axnb=M01,m02=αynβ=M02,m03=λznγ=M03.

    For n=2,3,..., we have

    m11=f(m02,M02)xn+1=f(yn,yn1)f(M02,m02)=M11,
    m12=g(m03,M03)yn+1=g(zn,zn1)g(M03,m03)=M12,

    and

    m13=h(m01,M01)zn+1=h(xn,xn1)h(M01,m01)=M13,

    that is

    m11xnM11,m12ynM12,m13znM13,n=3,4,.

    Now, for n=4,5,..., we have

    m21=f(m12,M12)xn+1=f(yn,yn1)f(M12,m12)=M21,

    and

    m22=g(m13,M13)yn+1=g(zn,zn1)g(M13,m13)=M22,
    m23=h(m11,M11)zn+1=h(xn,xn1)h(M11,m11)=M23,

    that is

    m21xnM21,m22ynM22,m23znM23,n=5,6,.

    Similarly, for n=6,7,..., we have

    m31=f(m22,M22)xn+1=f(yn,yn1)f(M22,m22)=M31,
    m32=g(m23,M23)yn+1=g(zn,zn1)g(M23,m23)=M32,

    and

    m33=h(m21,M21)zn+1=h(xn,xn1)h(M21,m21)=M33,

    that is

    m31xnM31,m32ynM32,m33znM33,n=7,8,.

    It follows by induction that for i=0,1,... we get

    mi1xnMi1,mi2ynMi2,mi3znMi3,n2i+1.

    Using the fact that i+ implies n+ and m1=M1,m2=M2,m3=M3, we obtain that

    limn+xn=M1,limn+yn=M2,limn+zn=M3.

    From (1) and using the fact that f, g and h are continuous and homogeneous of degree zero, we get

    M1=f(M2,M2)=f(1,1),M2=g(M3,M3)=g(1,1),M3=h(M1,M1)=h(1,1).

    Theorem 2.3. Consider System (1). Assume that the following statements are true:

    1. H1: There exist a,b,α,β,λ,γ(0,+) such that

    af(u,v)b,αg(u,v)β,λh(u,v)γ,(u,v)(0,+)2.

    2. H2: f(u,v),g(u,v) are increasing in u for all v and decreasing in v for all u and h(u,v) is decreasing in u for all v and increasing in v for all u.

    3. H3: If (m1,M1,m2,M2,m3,M3)[a,b]2×[α,β]2×[λ,γ]2 is a solution of the system

    m1=f(m2,M2),M1=f(M2,m2),m2=g(m3,M3),M2=g(M3,m3),m3=h(M1,m1),M3=h(m1,M1)

    then

    m1=M1,m2=M2,m3=M3.

    Then every solution of System (1) converges to the unique equilibrium point

    (¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1)).

    Proof. Let

    m01:=a,M01:=b,m02:=α,M02:=β,m03:=λ,M03:=γ

    and for each i=0,1,...,

    mi+11:=f(mi2,Mi2),Mi+11:=f(Mi2,mi2),
    mi+12:=g(mi3,Mi3),Mi+12:=g(Mi3,mi3),
    mi+13:=h(Mi1,mi1),Mi+13:=h(mi1,Mi1).

    We have

    af(α,β)f(β,α)b,
    αg(λ,γ)g(γ,λ)β,
    λh(b,a)h(a,b)γ,

    and so,

    m01=af(m02,M02)f(M02,m02)b=M01,
    m02=αg(m03,M03)g(M03,m03)β=M02,

    and

    m03=λh(M01,m01)h(m01,M01)γ=M03.

    Hence,

    m01m11M11M01,
    m02m12M12M02,

    and

    m03m13M13M03.

    Now, we have

    m11=f(m02,M02)f(m12,M12)=m21f(M12,m12)=M21f(M02,m02)=M11,
    m12=g(m03,M03)g(m13,M13)=m22g(M13,m13)=M22g(M03,m03)=M12,
    m13=h(M01,m01)h(M11,m11)=m23h(m11,M11)=M23h(m01,M01)=M13,

    and it follows that

    m01m11m21M21M11M01,
    m02m12m22M22M12M02,

    and

    m03m13m23M23M13M03.

    By induction, we get for i=0,1,..., that

    a=m01m11...mi11mi1Mi1Mi11...M11M01=b,
    α=m02m12...mi12mi2Mi2Mi12...M12M02=β,

    and

    λ=m03m13...mi13mi3Mi3Mi13...M13M03=γ.

    It follows that the sequences (mi1)iN0, (mi2)iN0, (mi3)iN0 (resp. (Mi1)iN0, (Mi2)iN0, (Mi3)iN0) are increasing (resp. decreasing) and also bounded, so convergent. Let

    m1=limi+mi1,M1=limi+Mi1,
    m2=limi+mi2,M2=limi+Mi2.
    m3=limi+mi3,M3=limi+Mi3.

    Then

    am1M1b,αm2M2β,λm3M3γ.

    By taking limits in the following equalities

    mi+11=f(mi2,Mi2),Mi+11=f(Mi2,mi2),
    mi+12=g(mi3,Mi3),Mi+12=g(Mi3,mi3),
    mi+13=h(Mi1,mi1),Mi+13=h(mi1,Mi1),

    and using the continuity of f, g and h we obtain

    m1=f(m2,M2),m2=f(M2,m2),m2=g(m3,M3),M2=g(M3,m3),m3=h(M1,m1),M3=h(m1,M1)

    so it follows from H3 that

    m1=M1,m2=M2,m3=M3.

    From H1, for n=1,2,, we get

    m01=axnb=M01,m02=αynβ=M02,m03=λznγ=M03.

    For n=2,3,..., we have

    m11=f(m02,M02)xn+1=f(yn,yn1)f(M02,m02)=M11,
    m12=g(m03,M03)yn+1=g(zn,zn1)g(M03,m03)=M12,

    and

    m13=h(M01,m01)zn+1=h(xn,xn1)h(m01,M01)=M13,

    that is

    m11xnM11,m12ynM12,m13znM13,n=3,4,.

    Now, for n=4,5,..., we have

    m21=f(m12,M12)xn+1=f(yn,yn1)f(M12,m12)=M21,

    and

    m22=g(m13,M13)yn+1=g(zn,zn1)g(M13,m13)=M22,
    m23=h(M11,m11)zn+1=h(xn,xn1)h(m11,M11)=M23,

    that is

    m21xnM21,m22ynM22,m23znM23,n=5,6,.

    Similarly, for n=6,7,..., we have

    m31=f(m22,M22)xn+1=f(yn,yn1)f(M22,m22)=M31,
    m32=g(m23,M23)yn+1=g(zn,zn1)g(M23,m23)=M32,

    and

    m33=h(M21,m21)zn+1=h(xn,xn1)h(m21,M21)=M33,

    that is

    m31xnM31,m32ynM32,m33znM33,n=7,8,.

    It follows by induction that for i=0,1,... we get

    mi1xnMi1,mi2ynMi2,mi3znMi3,n2i+1.

    Using the fact that i+ implies n+ and m1=M1,m2=M2,m3=M3, we obtain that

    limn+xn=M1,limn+yn=M2,limn+zn=M3.

    From (1) and using the fact that f, g and h are continuous and homogeneous of degree zero, we get

    M1=f(M2,M2)=f(1,1),M2=g(M3,M3)=g(1,1),M3=h(M1,M1)=h(1,1).

    Theorem 2.4. Consider System (1). Assume that the following statements are true:

    1. H1: There exist a,b,α,β,λ,γ(0,+) such that

    af(u,v)b,αg(u,v)β,λh(u,v)γ,(u,v)(0,+)2.

    2. H2: f(u,v) is increasing in u for all v and decreasing in v for all u and g(u,v), h(u,v) are decreasing in u for all v and increasing in v for all u.

    3. H3: If (m1,M1,m2,M2,m3,M3)[a,b]2×[α,β]2×[λ,γ]2 is a solution of the system

    m1=f(m2,M2),M1=f(M2,m2),m2=g(M3,m3),M2=g(m3,M3),m3=h(M1,m1),M3=h(m1,M1)

    then

    m1=M1,m2=M2,m3=M3.

    Then every solution of System (1) converges to the unique equilibrium point

    (¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1)).

    Proof. Let

    m01:=a,M01:=b,m02:=α,M02:=β,m03:=λ,M03:=γ

    and for each i=0,1,...,

    mi+11:=f(mi2,Mi2),Mi+11:=f(Mi2,mi2),
    mi+12:=g(Mi3,mi3),Mi+12:=g(mi3,Mi3),
    mi+13:=h(Mi1,mi1),Mi+13:=h(mi1,Mi1).

    We have

    af(α,β)f(β,α)b,
    αg(γ,λ)g(λ,γ)β,
    λh(b,a)h(a,b)γ,

    and so,

    m01=af(m02,M02)f(M02,m02)b=M01,
    m02=αg(M03,m03)g(m03,M03)β=M02,

    and

    m03=λh(M01,m01)h(m01,M01)γ=M03.

    Hence,

    m01m11M11M01,
    m02m12M12M02,

    and

    m03m13M13M03.

    Now, we have

    m11=f(m02,M02)f(m12,M12)=m21f(M12,m12)=M21f(M02,m02)=M11,
    m12=g(M03,m03)g(M13,m13)=m22g(m13,M13)=M22g(m03,M03)=M12,
    m13=h(M01,m01)h(M11,m11)=m23h(m11,M11)=M23h(m01,M01)=M13,

    and it follows that

    m01m11m21M21M11M01,
    m02m12m22M22M12M02,

    and

    m03m13m23M23M13M03.

    By induction, we get for i=0,1,..., that

    a=m01m11...mi11mi1Mi1Mi11...M11M01=b,
    α=m02m12...mi12mi2Mi2Mi12...M12M02=β,

    and

    λ=m03m13...mi13mi3Mi3Mi13...M13M03=γ.

    It follows that the sequences (mi1)iN0, (mi2)iN0, (mi3)iN0 (resp. (Mi1)iN0, (Mi2)iN0, (Mi3)iN0) are increasing (resp. decreasing) and also bounded, so convergent. Let

    m1=limi+mi1,M1=limi+Mi1,
    m2=limi+mi2,M2=limi+Mi2.
    m3=limi+mi3,M3=limi+Mi3.

    Then

    am1M1b,αm2M2β,λm3M3γ.

    By taking limits in the following equalities

    mi+11=f(mi2,Mi2),Mi+11=f(Mi2,mi2),
    mi+12=g(Mi3,mi3),Mi+12=g(mi3,Mi3),
    mi+13=h(Mi1,mi1),Mi+13=h(mi1,Mi1),

    and using the continuity of f, g and h we obtain

    m1=f(m2,M2),M1=f(M2,m2),m2=g(M3,m3),M2=g(m3,M3),m3=h(M1,m1),M3=h(m1,M1)

    so it follows from H3 that

    m1=M1,m2=M2,m3=M3.

    From H1, for n=1,2,, we get

    m01=axnb=M01,m02=αynβ=M02,m03=λznγ=M03.

    For n=2,3,..., we have

    m11=f(m02,M02)xn+1=f(yn,yn1)f(M02,m02)=M11,
    m12=g(M03,m03)yn+1=g(zn,zn1)g(m03,M03)=M13,

    and

    m13=h(M01,m01)zn+1=h(xn,xn1)h(m01,M01)=M13,

    that is

    m11xnM11,m12ynM12,m13znM13,n=3,4,.

    Now, for n=4,5,..., we have

    m21=f(m12,M12)xn+1=f(yn,yn1)f(M12,m12)=M21,

    and

    m22=g(M13,m13)yn+1=g(zn,zn1)g(m13,M13)=M22,
    m23=h(M11,m11)zn+1=h(xn,xn1)h(m11,M11)=M23,

    that is

    m21xnM21,m22ynM22,m23znM23,n=5,6,.

    Similarly, for n=6,7,..., we have

    m31=f(m22,M22)xn+1=f(yn,yn1)f(M22,m22)=M31,
    m32=g(M23,m23)yn+1=g(zn,zn1)g(m23,M23)=M32,

    and

    m33=h(M21,m21)zn+1=h(xn,xn1)h(m21,M21)=M33,

    that is

    m31xnM31,m32ynM32,m33znM33,n=7,8,.

    It follows by induction that for i=0,1,... we get

    mi1xnMi1,mi2ynMi2,mi3znMi3,n2i+1.

    Using the fact that i+ implies n+ and m1=M1,m2=M2,m3=M3, we obtain that

    limn+xn=M1,limn+yn=M2,limn+zn=M3.

    From (1) and using the fact that f, g and h are continuous and homogeneous of degree zero, we get

    M1=f(M2,M2)=f(1,1),M2=g(M3,M3)=g(1,1),M3=h(M1,M1)=h(1,1).

    Theorem 2.5. Consider System (1). Assume that the following statements are true:

    1. H1: There exist a,b,α,β,λ,γ(0,+) such that

    af(u,v)b,αg(u,v)β,λh(u,v)γ,(u,v)(0,+)2.

    2. H2: f(u,v), h(u,v) are increasing in u for all v and decreasing in v for all u and g(u,v) is decreasing in u for all v and increasing in v for all u.

    3. H3: If (m1,M1,m2,M2,m3,M3)[a,b]2×[α,β]2×[λ,γ]2 is a solution of the system

    m1=f(m2,M2),M1=f(M2,m2),m2=g(M3,m3),M2=g(m3,M3),m3=h(m1,M1),M3=h(M1,m1)

    then

    m1=M1,m2=M2,m3=M3.

    Then every solution of System (1) converges to the unique equilibrium point

    (¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1)).

    Proof. Let

    m01:=a,M01:=b,m02:=α,M02:=β,m03:=λ,M03:=γ

    and for each i=0,1,...,

    mi+11:=f(mi2,Mi2),Mi+11:=f(Mi2,mi2),
    mi+12:=g(Mi3,mi3),Mi+12:=g(mi3,Mi3),
    mi+13:=h(mi1,Mi1),Mi+13:=h(Mi1,mi1).

    We have

    af(α,β)f(β,α)b,
    αg(γ,λ)g(λ,γ)β,
    λh(a,b)h(b,a)γ,

    and so,

    m01=af(m02,M02)f(M02,m02)b=M01,
    m02=αg(M03,m03)g(m03,M03)β=M02,

    and

    m03=λh(m01,M01)h(M01,m01)γ=M03.

    Hence,

    m01m11M11M01,
    m02m12M12M02,

    and

    m03m13M13M03.

    Now, we have

    m11=f(m02,M02)f(m12,M12)=m21f(M12,m12)=M21f(M02,m02)=M11,
    m12=g(M03,m03)g(M13,m13)=m22g(m13,M13)=M22g(m03,M03)=M12,
    m13=h(m01,M01)h(m11,M11)=m23h(M11,m11)=M23h(M01,m01)=M13,

    and it follows that

    m01m11m21M21M11M01,
    m02m12m22M22M12M02,

    and

    m03m13m23M23M13M03.

    By induction, we get for i=0,1,..., that

    a=m01m11...mi11mi1Mi1Mi11...M11M01=b,
    α=m02m12...mi12mi2Mi2Mi12...M12M02=β,

    and

    λ=m03m13...mi13mi3Mi3Mi13...M13M03=γ.

    It follows that the sequences (mi1)iN0, (mi2)iN0, (mi3)iN0 (resp. (Mi1)iN0, (Mi2)iN0, (Mi3)iN0) are increasing (resp. decreasing) and also bounded, so convergent. Let

    m1=limi+mi1,M1=limi+Mi1,
    m2=limi+mi2,M2=limi+Mi2.
    m3=limi+mi3,M3=limi+Mi3.

    Then

    am1M1b,αm2M2β,λm3M3γ.

    By taking limits in the following equalities

    mi+11=f(mi2,Mi2),Mi+11=f(Mi2,mi2),
    mi+12=g(Mi3,mi3),Mi+12=g(mi3,Mi3),
    \begin{equation*} m_3^{i+1} = h(m_1^{i}, M_1^{i}), \, M_3^{i+1} = h(M_1^{i}, m_1^{i}), \end{equation*}

    and using the continuity of f , g and h we obtain

    \begin{eqnarray*} m_{1} & = &f(m_{2}, M_{2}), \, M_{1} = f(M_{2}, m_{2}), \, \\ m_{2} & = &g(M_{3}, m_{3}), \, M_{2} = g(m_{3}, M_{3}), \, \\ m_{3} & = &h(m_{1}, M_{1}), \, M_{3} = h(M_{1}, m_{1}) \end{eqnarray*}

    so it follows from H_3 that

    \begin{equation*} m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. \end{equation*}

    From H_1 , for n = 1, 2, \cdots , we get

    \begin{equation*} m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, . \end{equation*}

    For n = 2, 3, ..., we have

    \begin{equation*} m_1^{1} = f(m_2^{0}, M_{2}^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{0}, m_{2}^{0}) = M_{1}^{1}, \end{equation*}
    \begin{equation*} m_2^{1} = g(M_3^{0}, m_{3}^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_3^{0}, M_{3}^{0}) = M_{3}^{1}, \end{equation*}

    and

    \begin{equation*} m_3^{1} = h(m_1^{0}, M_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{0}, m_{1}^{0}) = M_{3}^{1}, \end{equation*}

    that is

    \begin{equation*} m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots. \end{equation*}

    Now, for n = 4, 5, ..., we have

    \begin{equation*} m_1^{2} = f(m_2^{1}, M_{2}^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{1}, m_{2}^{1}) = M_{1}^{2}, \end{equation*}

    and

    \begin{equation*} m_2^{2} = g(M_3^{1}, m_{3}^{1})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_3^{1}, M_{3}^{1}) = M_{2}^{2}, \end{equation*}
    \begin{equation*} m_3^{2} = h(m_1^{1}, M_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{1}, m_{1}^{1}) = M_{3}^{2}, \end{equation*}

    that is

    \begin{equation*} m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots. \end{equation*}

    Similarly, for n = 6, 7, ..., we have

    \begin{equation*} m_1^{3} = f(m_2^{2}, M_{2}^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{2}, m_{2}^{2}) = M_{1}^{3}, \end{equation*}
    \begin{equation*} m_2^{3} = g(M_3^{2}, m_{3}^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_3^{2}, M_{3}^{2}) = M_{2}^{3}, \end{equation*}

    and

    \begin{equation*} m_3^{3} = h(m_1^{2}, M_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{2}, m_{1}^{2}) = M_{3}^{3}, \end{equation*}

    that is

    \begin{equation*} m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots. \end{equation*}

    It follows by induction that for i = 0, 1, ... we get

    \begin{equation*} m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1. \end{equation*}

    Using the fact that i\rightarrow+\infty implies n\rightarrow+\infty and m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3 , we obtain that

    \begin{equation*} \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3. \end{equation*}

    From (1) and using the fact that f , g and h are continuous and homogeneous of degree zero, we get

    \begin{equation*} M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1). \end{equation*}

    Theorem 2.6. Consider System (1). Assume that the following statements are true:

    1. H_1 : There exist a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right) such that

    \begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}

    2. H_2 : f(u, v), \, g(u, v), \, h(u, v) are decreasing in u for all v and increasing in v for all u .

    3. H_{3} : If (m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2} is a solution of the system

    \begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(M_{3}, m_{3}), \, M_{2} = g(m_{3}, M_{3}), \, \\ m_{3} & = &h(M_{1}, m_{1}), \, M_{3} = h(m_{1}, M_{1}) \end{eqnarray*}

    then

    \begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}

    Then every solution of System (1) converges to the unique equilibrium point

    \begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}

    Proof. Let

    \begin{equation*} m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma \end{equation*}

    and for each i = 0, 1, ...,

    \begin{equation*} m_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \, M_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}), \end{equation*}
    \begin{equation*} m_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}), \, M_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \end{equation*}
    \begin{equation*} m_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}), \, M_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}). \end{equation*}

    We have

    \begin{equation*} a\leq f(\beta, \alpha)\leq f(\alpha, \beta)\leq b, \end{equation*}
    \begin{equation*} \alpha\leq g(\gamma, \lambda)\leq g(\lambda, \gamma)\leq \beta, \end{equation*}
    \begin{equation*} \lambda\leq h(b, a)\leq h(a, b)\leq \gamma \end{equation*}

    and so,

    \begin{equation*} m_{1}^{0} = a\leq f(M_{2}^{0}, m_2^{0})\leq f(m_{2}^{0}, M_2^{0})\leq b = M_1^{0}, \end{equation*}
    \begin{equation*} m_{2}^{0} = \alpha\leq g(M_{3}^{0}, m_3^{0})\leq g(m_{3}^{0}, M_3^{0})\leq \beta = M_2^{0}, \end{equation*}
    \begin{equation*} m_{3}^{0} = \lambda\leq h(M_{1}^{0}, m_1^{0})\leq h(m_{1}^{0}, M_1^{0})\leq \gamma = M_3^{0}. \end{equation*}

    Hence,

    \begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}
    \begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}

    and

    \begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}

    Now, we have

    \begin{equation*} m_{1}^{1} = f(M_{2}^{0}, m_2^{0})\leq f(M_{2}^{1}, m_2^{1}) = m_1^{2}\leq f(m_{2}^{1}, M_2^{1}) = M_1^{2}\leq f(m_{2}^{0}, M_2^{0}) = M_1^{1}, \end{equation*}
    \begin{equation*} m_{2}^{1} = g(M_{3}^{0}, m_3^{0})\leq g(M_{3}^{1}, m_3^{1}) = m_2^{2}\leq g(m_{3}^{1}, M_3^{1}) = M_2^{2}\leq g(m_{3}^{0}, M_3^{0}) = M_2^{1}, \end{equation*}
    \begin{equation*} m_{3}^{1} = h(M_{1}^{0}, m_1^{0})\leq h(M_{1}^{1}, m_1^{1}) = m_3^{2}\leq h(m_{1}^{1}, M_1^{1}) = M_3^{2}\leq h(m_{1}^{0}, M_1^{0}) = M_3^{1} \end{equation*}

    and it follows that

    \begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}
    \begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}
    \begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}

    By induction, we get for i = 0, 1, ..., that

    \begin{equation*} a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b, \end{equation*}
    \begin{equation*} \alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta, \end{equation*}

    and

    \begin{equation*} \lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma. \end{equation*}

    It follows that the sequences (m_1^{i})_{i\in\mathbb{N}_{0}} , (m_2^{i})_{i\in\mathbb{N}_{0}} (m_3^{i})_{i\in\mathbb{N}_{0}} (resp. (M_1^{i})_{i\in\mathbb{N}_{0}} , (M_2^{i})_{i\in\mathbb{N}_{0}} , (M_3^{i})_{i\in\mathbb{N}_{0}} ) are increasing (resp. decreasing) and also bounded, so convergent. Let

    \begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}
    \begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}
    \begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}

    Then

    \begin{equation*} a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma. \end{equation*}

    By taking limits in the following equalities

    \begin{equation*} m_1^{i+1} = f(M_2^{i}, m_2^{i}), \, M_1^{i+1} = f(m_2^{i}, M_2^{i}), \end{equation*}
    \begin{equation*} m_2^{i+1} = g(M_3^{i}, m_3^{i}), \, M_2^{i+1} = g(m_3^{i}, M_3^{i}), \end{equation*}
    \begin{equation*} m_3^{i+1} = h(M_1^{i}, m_1^{i}), \, M_3^{i+1} = h(m_1^{i}, M_1^{i}) \end{equation*}

    and using the continuity of f , g , and h we obtain

    \begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(M_{3}, m_{3}), \, M_{2} = g(m_{3}, M_{3}), \, \\ m_{3} & = &h(M_{1}, m_{1}), \, M_{3} = h(m_{1}, M_{1}) \end{eqnarray*}

    so it follows from H_3 that

    \begin{equation*} m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. \end{equation*}

    From H_1 , for n = 1, 2, \cdots , we get

    \begin{equation*} m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, . \end{equation*}

    For n = 2, 3, ..., we have

    \begin{equation*} m_1^{1} = f(M_{2}^{0}, m_2^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{0}, M_2^{0}) = M_{1}^{1}, \end{equation*}
    \begin{equation*} m_2^{1} = g(M_{3}^{0}, m_3^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{0}, M_3^{0}) = M_{2}^{1}, \end{equation*}
    \begin{equation*} m_3^{1} = h(M_{1}^{0}, m_1^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_{1}^{0}, M_1^{0}) = M_{3}^{1}, \end{equation*}

    that is

    \begin{equation*} m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots. \end{equation*}

    Now, for n = 4, 5, ..., we have

    \begin{equation*} m_1^{2} = f(M_{2}^{1}, m_2^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{1}, M_2^{1}) = M_{1}^{2}, \end{equation*}
    \begin{equation*} m_2^{2} = g(M_{3}^{1}, m_1^{3})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{1}, M_1^{3}) = M_{2}^{2}, \end{equation*}
    \begin{equation*} m_3^{2} = h(M_{1}^{1}, m_1^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_{1}^{1}, M_1^{1}) = M_{3}^{2}, \end{equation*}

    that is

    \begin{equation*} m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots. \end{equation*}

    Similarly, for n = 6, 7, ..., we have

    \begin{equation*} m_1^{3} = f(M_{2}^{2}, m_2^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{2}, M_2^{2}) = M_{1}^{3}, \end{equation*}
    \begin{equation*} m_2^{3} = g(M_{3}^{2}, m_3^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{2}, M_3^{2}) = M_{2}^{3}, \end{equation*}
    \begin{equation*} m_3^{3} = h(M_{1}^{2}, m_1^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_{1}^{2}, M_1^{2}) = M_{3}^{3}, \end{equation*}

    that is

    \begin{equation*} m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots. \end{equation*}

    It follows by induction that for i = 0, 1, ... we get

    \begin{equation*} m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1. \end{equation*}

    Using the fact that i\rightarrow+\infty implies n\rightarrow+\infty and m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3 , we obtain that

    \begin{equation*} \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3. \end{equation*}

    From (1) and using the fact that f , g and h are continuous and homogeneous of degree zero, we get

    \begin{equation*} M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1). \end{equation*}

    Theorem 2.7. Consider System (1). Assume that the following statements are true:

    1. H_1 : There exist a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right) such that

    \begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}

    2. H_2 : f(u, v), \, g(u, v) are decreasing in u for all v and increasing in v for all u , however h(u, v) is increasing in u for all v and decreasing in v for all u .

    3. H_{3} : If (m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2} is a solution of the system

    \begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(M_{3}, m_{3}), \, M_{2} = g(m_{3}, M_{3}), \, \\ m_{3} & = &h(m_{1}, M_{1}), \, M_{3} = h(M_{1}, m_{1}) \end{eqnarray*}

    then

    \begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}

    Then every solution of System (1) converges to the unique equilibrium point

    \begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}

    Proof. Let

    m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma

    and for each i = 0, 1, ...,

    m_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \, M_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}),
    m_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}), \, M_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}),
    m_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}), \, M_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}).

    We have

    a\leq f(\beta, \alpha)\leq f(\alpha, \beta)\leq b,
    \alpha\leq g(\gamma, \lambda)\leq g(\lambda, \gamma)\leq \beta,
    \lambda\leq h(a, b)\leq h(b, a)\leq \gamma,

    and so,

    m_{1}^{0} = a\leq f(M_{2}^{0}, m_2^{0})\leq f(m_{2}^{0}, M_2^{0})\leq b = M_1^{0},
    m_{2}^{0} = \alpha\leq g(M_{3}^{0}, m_3^{0})\leq g(m_{3}^{0}, M_3^{0})\leq \beta = M_2^{0},

    and

    m_{3}^{0} = \lambda\leq h(m_1^{0}, M_{1}^{0})\leq h(M_1^{0}, m_{1}^{0})\leq \gamma = M_3^{0}.

    Hence,

    m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0},
    m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0},

    and

    m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}.

    Now, we have

    m_{1}^{1} = f(M_{2}^{0}, m_2^{0})\leq f(M_{2}^{1}, m_2^{1}) = m_1^{2}\leq f(m_{2}^{1}, M_2^{1}) = M_1^{2}\leq f(m_{2}^{0}, M_2^{0}) = M_1^{1},
    m_{2}^{1} = g(M_{3}^{0}, m_3^{0})\leq g(M_{3}^{1}, m_3^{1}) = m_2^{2}\leq g(m_{3}^{1}, M_3^{1}) = M_2^{2}\leq g(m_{3}^{0}, M_3^{0}) = M_2^{1},
    m_{3}^{1} = h(m_1^{0}, M_{1}^{0})\leq h(m_1^{1}, M_{1}^{1}) = m_3^{2}\leq h(M_1^{1}, m_{1}^{1}) = M_3^{2}\leq h(M_1^{0}, m_{1}^{0}) = M_3^{1},

    and it follows that

    m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0},
    m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0},

    and

    m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}.

    By induction, we get for i = 0, 1, ..., that

    a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b,
    \alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta,

    and

    \lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma.

    It follows that the sequences (m_1^{i})_{i\in\mathbb{N}_{0}} , (m_2^{i})_{i\in\mathbb{N}_{0}} (m_3^{i})_{i\in\mathbb{N}_{0}} (resp. (M_1^{i})_{i\in\mathbb{N}_{0}} , (M_2^{i})_{i\in\mathbb{N}_{0}} , (M_3^{i})_{i\in\mathbb{N}_{0}} ) are increasing (resp. decreasing) and also bounded, so convergent. Let

    \begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}
    \begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}
    \begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}

    Then

    a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma.

    By taking limits in the following equalities

    m_1^{i+1} = f(M_2^{i}, m_2^{i}), \, M_1^{i+1} = f(m_2^{i}, M_2^{i}),
    m_2^{i+1} = g(M_3^{i}, m_3^{i}), \, M_2^{i+1} = g(m_3^{i}, M_3^{i}),
    m_3^{i+1} = h(m_1^{i}, M_1^{i}), \, M_3^{i+1} = h(M_1^{i}, m_1^{i}),

    and using the continuity of f , g , and h we obtain

    \begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(M_{3}, m_{3}), \, M_{2} = g(m_{3}, M_{3}), \, \\ m_{3} & = &h(m_{1}, M_{1}), \, M_{3} = h(M_{1}, m_{1}) \end{eqnarray*}

    so it follows from H_3 that

    m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3.

    From H_1 , for , n = 1, 2, \ldots , we get

    m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}.

    For n = 2, 3, ..., we have

    m_1^{1} = f(M_{2}^{0}, m_2^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{0}, M_2^{0}) = M_{1}^{1},
    m_2^{1} = g(M_{3}^{0}, m_3^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{0}, M_3^{0}) = M_{2}^{1},

    and

    m_3^{1} = h(m_1^{0}, M_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{0}, m_{1}^{0}) = M_{3}^{1},

    that is

    m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots.

    Now, for n = 4, 5, ..., we have

    m_1^{2} = f(M_{2}^{1}, m_2^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{1}, M_2^{1}) = M_{1}^{2},
    m_2^{2} = g(M_{3}^{1}, m_1^{3})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{1}, M_1^{3}) = M_{2}^{2},

    and

    m_3^{2} = h(m_1^{1}, M_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{1}, m_{1}^{1}) = M_{3}^{2},

    that is

    m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots.

    Similarly, for n = 6, 7, ..., we have

    m_1^{3} = f(M_{2}^{2}, m_2^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{2}, M_2^{2}) = M_{1}^{3},
    m_2^{3} = g(M_{3}^{2}, m_3^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{2}, M_3^{2}) = M_{2}^{3},

    and

    m_3^{3} = h(m_1^{2}, M_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{2}, m_{1}^{2}) = M_{3}^{3},

    that is

    m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots.

    It follows by induction that for i = 0, 1, ... we get

    m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1.

    Using the fact that i\rightarrow+\infty implies n\rightarrow+\infty and m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3 , we obtain that

    \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3.

    From (1) and using the fact that f , g and h are continuous and homogeneous of degree zero, we get

    M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1).

    Theorem 2.8. Consider System (1). Assume that the following statements are true:

    1. H_1 : There exist a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right) such that

    \begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}

    2. H_2 : f(u, v) is decreasing in u for all v and increasing in v for all u , however g(u, v), \, h(u, v) are increasing in u for all v and decreasing in v for all u .

    3. H_{3} : If (m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2} is a solution of the system

    \begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(m_{3}, M_{3}), \, M_{2} = g(M_{3}, m_{3}), \, \\ m_{3} & = &h(m_{1}, M_{1}), \, M_{3} = h(M_{1}, m_{1}) \end{eqnarray*}

    then

    \begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}

    Then every solution of System (1) converges to the unique equilibrium point

    \begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}

    Proof. Let

    m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma

    and for each i = 0, 1, ...,

    m_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \, M_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}),
    m_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \, M_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}),
    m_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}), \, M_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}).

    We have

    a\leq f(\beta, \alpha)\leq f(\alpha, \beta)\leq b,
    \alpha\leq g(\lambda, \gamma)\leq g(\gamma, \lambda)\leq \beta,
    \lambda\leq h(a, b)\leq h(b, a)\leq \gamma,

    and so,

    m_{1}^{0} = a\leq f(M_{2}^{0}, m_2^{0})\leq f(m_{2}^{0}, M_2^{0})\leq b = M_1^{0},
    m_{2}^{0} = \alpha\leq g(m_3^{0}, M_{3}^{0})\leq g(M_3^{0}, m_{3}^{0})\leq \beta = M_2^{0},

    and

    m_{3}^{0} = \lambda\leq h(m_1^{0}, M_{1}^{0})\leq g(M_1^{0}, m_{1}^{0})\leq \gamma = M_3^{0}.

    Hence,

    m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0},
    m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0},

    and

    m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}.

    Now, we have

    m_{1}^{1} = f(M_{2}^{0}, m_2^{0})\leq f(M_{2}^{1}, m_2^{1}) = m_1^{2}\leq f(m_{2}^{1}, M_2^{1}) = M_1^{2}\leq f(m_{2}^{0}, M_2^{0}) = M_1^{1},
    m_{2}^{1} = g(m_3^{0}, M_{3}^{0})\leq g(m_3^{1}, M_{3}^{1}) = m_2^{2}\leq g(M_3^{1}, m_{3}^{1}) = M_2^{2}\leq g(M_3^{0}, m_{3}^{0}) = M_2^{1},
    m_{3}^{1} = h(m_1^{0}, M_{1}^{0})\leq h(m_1^{1}, M_{1}^{1}) = m_3^{2}\leq h(M_1^{1}, m_{1}^{1}) = M_3^{2}\leq h(M_1^{0}, m_{1}^{0}) = M_3^{1},

    and it follows that

    m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0},
    m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0},

    and

    m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}.

    By induction, we get for i = 0, 1, ..., that

    a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b,
    \alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta,

    and

    \lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma.

    It follows that the sequences (m_1^{i})_{i\in\mathbb{N}_{0}} , (m_2^{i})_{i\in\mathbb{N}_{0}} , (m_3^{i})_{i\in\mathbb{N}_{0}} (resp. (M_1^{i})_{i\in\mathbb{N}_{0}} , (M_2^{i})_{i\in\mathbb{N}_{0}} , (M_3^{i})_{i\in\mathbb{N}_{0}} ) are increasing (resp. decreasing) and also bounded, so convergent. Let

    \begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}
    \begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}
    \begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}

    Then

    a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma.

    By taking limits in the following equalities

    m_1^{i+1} = f(M_2^{i}, m_2^{i}), \, M_1^{i+1} = f(m_2^{i}, M_2^{i}),
    m_2^{i+1} = g(m_3^{i}, M_3^{i}), \, M_2^{i+1} = g(M_3^{i}, m_3^{i}),
    m_3^{i+1} = h(m_1^{i}, M_1^{i}), \, M_3^{i+1} = h(M_1^{i}, m_1^{i}),

    and using the continuity of f , g and h we obtain

    \begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(m_{3}, M_{3}), \, M_{2} = g(M_{3}, m_{3}), \, \\ m_{3} & = &h(m_{1}, M_{1}), \, M_{3} = h(M_{1}, m_{1}) \end{eqnarray*}

    so it follows from H_3 that

    m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3.

    From H_1 , for n = 1, 2, \ldots , we get

    m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, .

    For n = 2, 3, ..., we have

    m_1^{1} = f(M_{2}^{0}, m_2^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{0}, M_2^{0}) = M_{1}^{1},
    m_2^{1} = g(m_3^{0}, M_{3}^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{0}, m_{3}^{0}) = M_{3}^{1},

    and

    m_3^{1} = h(m_1^{0}, M_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{0}, m_{1}^{0}) = M_{3}^{1},

    that is

    m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots.

    Now, for n = 4, 5, ..., we have

    m_1^{2} = f(M_{2}^{1}, m_2^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{1}, M_2^{1}) = M_{1}^{2},
    m_2^{2} = g(m_3^{1}, M_{3}^{1})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{1}, m_{3}^{1}) = M_{2}^{2},

    and

    m_3^{2} = h(m_1^{1}, M_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{1}, m_{1}^{1}) = M_{3}^{2},

    that is

    m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots.

    Similarly, for n = 6, 7, ..., we have

    m_1^{3} = f(M_{2}^{2}, m_2^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{2}, M_2^{2}) = M_{1}^{3},
    m_2^{3} = g(m_3^{2}, M_{3}^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{2}, m_{3}^{2}) = M_{2}^{3},

    and

    m_3^{3} = h(m_1^{2}, M_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{2}, m_{1}^{2}) = M_{3}^{3},

    that is

    m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots.

    It follows by induction that for i = 0, 1, ... we get

    m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1.

    Using the fact that i\rightarrow+\infty implies n\rightarrow+\infty and m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3 , we obtain that

    \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3.

    From (1) and using the fact that f , g and h are continuous and homogeneous of degree zero, we get

    M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1).

    Theorem 2.9. Consider System (1). Assume that the following statements are true:

    1. H_1 : There exist a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right) such that

    \begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}

    2. H_2 : f(u, v), \, h(u, v) are decreasing in u for all v and increasing in v for all u , however g(u, v) is increasing in u for all v and decreasing in v for all u .

    3. H_{3} : If (m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2} is a solution of the system

    \begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(m_{3}, M_{3}), \, M_{2} = g(M_{3}, m_{3}), \, \\ m_{3} & = &h(M_{1}, m_{1}), \, M_{3} = h(m_{1}, M_{1}) \end{eqnarray*}

    then

    \begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}

    Then every solution of System (1) converges to the unique equilibrium point

    \begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}

    Proof. Let

    m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma

    and for each i = 0, 1, ...,

    m_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \, M_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}),
    m_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \, M_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}),
    m_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}), \, M_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}).

    We have

    a\leq f(\beta, \alpha)\leq f(\alpha, \beta)\leq b,
    \alpha\leq g(\lambda, \gamma)\leq g(\gamma, \lambda)\leq \beta,
    \lambda\leq h(b, a)\leq h(a, b)\leq \gamma,

    and so,

    m_{1}^{0} = a\leq f(M_{2}^{0}, m_2^{0})\leq f(m_{2}^{0}, M_2^{0})\leq b = M_1^{0},
    m_{2}^{0} = \alpha\leq g(m_3^{0}, M_{3}^{0})\leq g(M_3^{0}, m_{3}^{0})\leq \beta = M_2^{0},

    and

    m_{3}^{0} = \lambda\leq h(M_1^{0}, m_{1}^{0})\leq h(m_1^{0}, M_{1}^{0})\leq \gamma = M_3^{0}.

    Hence,

    m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0},
    m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0},

    and

    m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}.

    Now, we have

    m_{1}^{1} = f(M_{2}^{0}, m_2^{0})\leq f(M_{2}^{1}, m_2^{1}) = m_1^{2}\leq f(m_{2}^{1}, M_2^{1}) = M_1^{2}\leq f(m_{2}^{0}, M_2^{0}) = M_1^{1},
    m_{2}^{1} = g(m_3^{0}, M_{3}^{0})\leq g(m_3^{1}, M_{3}^{1}) = m_2^{2}\leq g(M_3^{1}, m_{3}^{1}) = M_2^{2}\leq g(M_3^{0}, m_{3}^{0}) = M_2^{1},
    m_{3}^{1} = h(M_1^{0}, m_{1}^{0})\leq h(M_1^{1}, m_{1}^{1}) = m_3^{2}\leq h(m_1^{1}, M_{1}^{1}) = M_3^{2}\leq h(m_1^{0}, M_{1}^{0}) = M_3^{1},

    and it follows that

    m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0},
    m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0},

    and

    m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}.

    By induction, we get for i = 0, 1, ..., that

    a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b,
    \alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta,

    and

    \lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma.

    It follows that the sequences (m_1^{i})_{i\in\mathbb{N}_{0}} , (m_2^{i})_{i\in\mathbb{N}_{0}} , (m_3^{i})_{i\in\mathbb{N}_{0}} (resp. (M_1^{i})_{i\in\mathbb{N}_{0}} , (M_2^{i})_{i\in\mathbb{N}_{0}} , (M_3^{i})_{i\in\mathbb{N}_{0}} ) are increasing (resp. decreasing) and also bounded, so convergent. Let

    \begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}
    \begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}
    \begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}

    Then

    a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma.

    By taking limits in the following equalities

    m_1^{i+1} = f(M_2^{i}, m_2^{i}), \, M_1^{i+1} = f(m_2^{i}, M_2^{i}),
    m_2^{i+1} = g(m_3^{i}, M_3^{i}), \, M_2^{i+1} = g(M_3^{i}, m_3^{i}),
    m_3^{i+1} = h(M_1^{i}, m_1^{i}), \, M_3^{i+1} = h(m_1^{i}, M_1^{i}),

    and using the continuity of f , g and h we obtain

    \begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(m_{3}, M_{3}), \, M_{2} = g(M_{3}, m_{3}), \, \\ m_{3} & = &h(M_{1}, m_{1}), \, M_{3} = h(m_{1}, M_{1}) \end{eqnarray*}

    so it follows from H_3 that

    m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3.

    From H_1 , for n = 1, 2, \ldots , we get

    m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, .

    For n = 2, 3, ..., we have

    m_1^{1} = f(M_{2}^{0}, m_2^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{0}, M_2^{0}) = M_{1}^{1},
    m_2^{1} = g(m_3^{0}, M_{3}^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{0}, m_{3}^{0}) = M_{3}^{1},

    and

    m_3^{1} = h(M_1^{0}, m_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{0}, M_{1}^{0}) = M_{3}^{1},

    that is

    m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots.

    Now, for n = 4, 5, ..., we have

    m_1^{2} = f(M_{2}^{1}, m_2^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{1}, M_2^{1}) = M_{1}^{2},
    m_2^{2} = g(m_3^{1}, M_{3}^{1})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{1}, m_{3}^{1}) = M_{2}^{2},

    and

    m_3^{2} = h(M_1^{1}, m_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{1}, M_{1}^{1}) = M_{3}^{2},

    that is

    m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots.

    Similarly, for n = 6, 7, ..., we have

    m_1^{3} = f(M_{2}^{2}, m_2^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{2}, M_2^{2}) = M_{1}^{3},
    m_2^{3} = g(m_3^{2}, M_{3}^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{2}, m_{3}^{2}) = M_{2}^{3},

    and

    m_3^{3} = h(M_1^{2}, m_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{2}, M_{1}^{2}) = M_{3}^{3},

    that is

    m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots.

    It follows by induction that for i = 0, 1, ... we get

    m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1.

    Using the fact that i\rightarrow+\infty implies n\rightarrow+\infty and m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3 , we obtain that

    \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3.

    From (1) and using the fact that f , g and h are continuous and homogeneous of degree zero, we get

    M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1).

    The following theorem is devoted to global stability of the equilibrium point (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)) .

    Theorem 2.10. Under the hypotheses of Theorem 2.1 and one of Theorems 2.2– 2.9, the equilibrium point (\overline{x}, \overline{y}, \overline{z} ) = (f(1, 1), g(1, 1), h(1, 1)) is globally stable.

    Now as an application of the previous results, we give an example.

    Example 1. Consider the following system of difference equations

    \begin{equation} x_{n+1} = f(y_n, y_{n-1}), \, y_{n+1} = g(z_n, z_{n-1}), \, z_{n+1} = h(x_n, x_{n-1}), \, n \in\mathbb{N}_{0}, \end{equation} (6)

    where x_{-1} , x_{0} , y_{-1} , y_{0} , z_{-1} , z_{0} \in\left(0, +\infty\right) and

    \begin{equation*} f(u, v) = \frac{a_1u+b_1v}{\alpha_1u+\beta_1v}, \, g(u, v) = \frac{a_2u+b_2v}{ \alpha_2u+\beta_2v}, \, h(u, v) = \frac{a_3u+b_3v}{\alpha_3u+\beta_3v}. \end{equation*}

    Assume that r_i: = a_i\beta_i-\alpha_ib_i, \, i = 1, 2, 3 are positive. For all (u, v)\in\left(0, +\infty\right) , we have

    \begin{equation*} \frac{\partial f}{\partial u}(u, v) = \frac{r_1v}{(\alpha_1u+\beta_1v)^{2}}, \, \frac{\partial f}{\partial v}(u, v) = -\frac{r_1u}{(\alpha_1u+\beta_1v)^{2}} , \, a: = \frac{b_1}{\beta_1}\leq f(u, v)\leq b: = \frac{a_1}{\alpha_1} \end{equation*}
    \begin{equation*} \frac{\partial g}{\partial u}(u, v) = \frac{r_2v}{(\alpha_2u+\beta_2v)^{2}}, \, \frac{\partial g}{\partial v}(u, v) = -\frac{r_2u}{(\alpha_2u+\beta_2v)^{2}} , \, \alpha: = \frac{b_2}{\beta_2}\leq g(u, v)\leq \beta: = \frac{a_2}{\alpha_2} \end{equation*}
    \begin{equation*} \frac{\partial h}{\partial u}(u, v) = \frac{r_3v}{(\alpha_3u+\beta_3v)^{2}}, \, \frac{\partial h}{\partial v}(u, v) = -\frac{r_3u}{(\alpha_3u+\beta_3v)^{2}} , \, \lambda: = \frac{b_3}{\beta_3}\leq h(u, v)\leq \gamma: = \frac{a_3}{\alpha_3} \end{equation*}

    It follows from Theorem 2.1 that the unique equilibrium point

    \begin{equation*} \left(f(1, 1), g(1, 1), h(1, 1)\right) = \left(\frac{a_1+b_1}{\alpha_1+\beta_1}, \frac{a_2+b_2}{\alpha_2+\beta_2}, \frac{a_3+b_3}{\alpha_3+\beta_3}\right) \end{equation*}

    of System (6) will be locally stable if

    \begin{equation*} \frac{\partial f}{\partial u}(1, 1)\frac{\partial g}{\partial u}(1, 1)\frac{ \partial h}{\partial u}(1, 1) < \frac{f(1, 1)g(1, 1)h(1, 1)}{8} \end{equation*}

    which is equivalent to

    \begin{equation*} \frac{r_1r_2r_3}{(\alpha_1+\beta_1)^{2}(\alpha_2+\beta_2)^{2}(\alpha_3+ \beta_3)^{2}} < \frac{(a_1+b_1)(a_2+b_2)(a_3+b_3)} {8(\alpha_1+\beta_1)( \alpha_2+\beta_2)(\alpha_3+\beta_3)}. \end{equation*}

    Also, we have conditions H_1 and H_2 of Theorem 2.2 are satisfied. So, to prove the global stability of the equilibrium point (f(1, 1), g(1, 1), h(1, 1)) it suffices to check condition H_3 of Theorem 2.2.

    For this purpose, let (m_1, M_1, m_2, M_2, m_3, M_3)\in[a, b]\times[\alpha, \beta] \times[\gamma, \lambda] such that

    \begin{equation} m_1 = f(m_2, M_2) = \frac{a_1m_2+b_1M_2}{\alpha_1m_2+\beta_1M_2}, \, M_1 = f(M_2, m_2) = \frac{a_1M_2+b_1m_2}{\alpha_1M_2+\beta_1m_2}, \end{equation} (7)
    \begin{equation} m_2 = g(m_3, M_3) = \frac{a_2m_3+b_2M_3}{\alpha_2m_3+\beta_2M_3}, \, M_2 = g(M_3, m_3) = \frac{a_2M_3+b_2m_3}{\alpha_2M_3+\beta_2m_3}, \end{equation} (8)
    \begin{equation} m_3 = h(m_1, M_1) = \frac{a_3m_1+b_3M_1}{\alpha_3m_1+\beta_3M_1}, \, M_3 = h(M_1, m_1) = \frac{a_3M_1+b_3m_1}{\alpha_3M_1+\beta_3m_1}. \end{equation} (9)

    From (7)-(9), we get

    \begin{equation} m_1-M_1 = \frac{r_1(m_2-M_2)(m_2+M_2)}{(\alpha_1m_2+\beta_1M_2)(\alpha_1M_2+ \beta_1m_2)}, \end{equation} (10)
    \begin{equation} m_2-M_2 = \frac{r_2(m_3-M_3)(m_3+M_3)}{(\alpha_2m_3+\beta_2M_3)(\alpha_2M_3+ \beta_2m_3)}, \end{equation} (11)
    \begin{equation} m_3-M_3 = \frac{r_3(m_1-M_1)(m_1+M_1)}{(\alpha_3m_1+\beta_3M_1)(\alpha_3M_1+ \beta_3m_1)}. \end{equation} (12)

    Now, from (10)-(12), we get

    \begin{equation} (m_1-M_1)(m_2-M_2)(m_3-M_3) = (m_1-M_1)(m_2-M_2)(m_3-M_3) \Theta(m_1, M_1, m_2, M_2, m_3, M_3), \end{equation} (13)

    where \Theta(m_1, M_1, m_2, M_2, m_3, M_3) is equal to

    \begin{equation*} \frac{r_1r_2r_3(m_1+M_1)(m_2+M_2)(m_3+M_3)} {(\alpha_3m_1+\beta_3M_1)(\alpha_3M_1+\beta_3m_1)(\alpha_1m_2+\beta_1M_2)(\alpha_1M_2+\beta_1m_2)(\alpha_2m_3+\beta_2M_3)(\alpha_2M_3+\beta_2m_3)}. \end{equation*}

    It is not hard to see that

    \begin{equation*} 8\prod\limits_{i = 1}^{3}\frac{\alpha_i^{2}b_ir_i}{a_i^{2}\beta_i(\alpha_i+ \beta_i)^{2}}\leq\Theta(m_1, M_1, m_2, M_2, m_3, M_3) \leq8\prod\limits_{i = 1}^{3}\frac{ a_i\beta_i^{2}r_i}{\alpha_ib_i^{2}(\alpha_i+\beta_i)^{2}}. \end{equation*}

    Using the fact

    \begin{equation*} r_i\leq a_i\beta_i, \, (\alpha_i+\beta_i)^2\geq 2\alpha_i\beta_i, \, \alpha_ib_i < a_i\beta_i, \, i = 1, 2, 3, \end{equation*}

    we get

    \begin{equation*} 8\prod\limits_{i = 1}^{3}\frac{\alpha_i^{2}b_ir_i}{a_i^{2}\beta_i(\alpha_i+ \beta_i)^{2}}\leq\prod\limits_{i = 1}^{3}\frac{\alpha_ib_i}{a_i\beta_i} < 1. \end{equation*}

    If we choose the parameters a_i, \, b_i, \, \alpha_i, \, \beta_i, \, i = 1, 2, 3 such that

    \begin{equation*} \prod\limits_{i = 1}^{3}\frac{a_i\beta_i^{2}r_i}{\alpha_ib_i^{2}(\alpha_i+\beta_i)^{2} } < \frac{1}{8}, \end{equation*}

    then we get that

    \begin{equation*} \Theta(m_1, M_1, m_2, M_2, m_3, M_3)\neq 1 \end{equation*}

    and so it follows from (13) that

    \begin{equation*} (m_1-M_1)(m_2-M_2)(m_3-M_3) = 0. \end{equation*}

    Using this and (10)-(12), we obtain m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3 . That is the condition H_3 is satisfied.

    In summary we have the following result.

    Theorem 2.11. Assume that that parameters a_i, \, b_i, \, \alpha_i, \, \beta_i, \, i = 1, 2, 3 are such that

    C_1 : a_i\beta_i>\alpha_ib_i, \, i = 1, 2, 3 .

    C_2 :

    \begin{equation*} \prod\limits_{i = 1}^{3}\frac{a_i\beta_i-\alpha_ib_i}{(\alpha_i+\beta_i)^{2}} < \frac{1 }{8}\prod\limits_{i = 1}^{3}\frac{a_i+b_i}{\alpha_i+\beta_i}. \end{equation*}

    C_3 :

    \begin{equation*} \prod\limits_{i = 1}^{3}\frac{a_i\beta_i^{2}r_i}{\alpha_ib_i^{2}(\alpha_i+\beta_i)^{2} } < \frac{1}{8}. \end{equation*}

    Then the equilibrium point \left(f(1, 1), g(1, 1), h(1, 1)\right) = \left(\frac{ a_1+b_1}{\alpha_1+\beta_1}, \frac{a_2+b_2}{\alpha_2+\beta_2}, \frac{a_3+b_3}{ \alpha_3+\beta_3}\right) is globally stable.

    We visualize the solutions of System (6) in Figures 1-3. In Figure 1 and Figure 2, we give the solution and corresponding global attractor of System (6) for x_{-1} = 2.13, x_{0} = 3.1, y_{-1} = 4.03, y_{0} = 2.21, z_{-1} = 2.76, z_{0} = 3.12 and a_1 = 3, \alpha_1 = 3.2, b_{1} = 4, \beta_{1} = 5, a_2 = 1.2, \alpha_2 = 2, b_{2} = 3.2, \beta_{2} = 6, a_3 = 2.4, \alpha_3 = 2.3, b_{3} = 1.3, \beta_{3} = 1.3 , respectively. Note that the conditions C_1, C_2, C_3 is satisfied for these values.

    Figure 1.   .
    Figure 2.   .
    Figure 3.   .

    However Figure 3 shows the unstable solution corresponding to the values x_{-1} = 2.13, x_{0} = 3.1, y_{-1} = 4.03, y_{0} = 2.21, z_{-1} = 2.76, z_{0} = 3.12 and a_1 = 0.3, \alpha_1 = 4, b_{1} = 3, \beta_{1} = 1.1, a_2 = 1.2, \alpha_2 = 4, b_{2} = 3.7, \beta_{2} = 1, a_3 = 2.7, \alpha_3 = 0.2, b_{3} = 1.3, \beta_{3} = 3 , which do not satisfy the conditions C_1, C_2, C_3 .

    Here, we are interested in existence of periodic solutions for System (1). In the following result we will established a necessary and sufficient condition for which there exist prime period two solutions of System (1).

    Theorem 3.1. Assume that \alpha , \beta and \gamma are positive real numbers such that (\alpha-1)(\beta-1)(\gamma-1)\neq0 . Then, System (1) has a prime period two solution in the form of

    \begin{equation*} ..., (\alpha p, \beta q, \gamma r), \, (p, q, r), (\alpha p, \beta q, \gamma r), \, (p, q, r), ... \end{equation*}

    if and only if

    \begin{equation*} f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1). \end{equation*}

    Proof. Let \alpha , \beta , \gamma be positive real numbers such that (\alpha-1)(\beta-1)(\gamma-1)\neq0 and assume that

    \begin{equation*} ..., (\alpha p, \beta q, \gamma r), \, (p, q, r), (\alpha p, \beta q, \gamma r), \, (p, q, r), ... \end{equation*}

    is a solution for System (1). Then, we have

    \begin{equation} \alpha p = f(q, \beta q) = f(1, \beta) \end{equation} (14)
    \begin{equation} p = f(\beta q, q) = f(\beta, 1) \end{equation} (15)
    \begin{equation} \beta q = g(r, \gamma r) = g(1, \gamma) \end{equation} (16)
    \begin{equation} q = g(\gamma r, r) = g(\gamma, 1) \end{equation} (17)
    \begin{equation} \gamma r = h(p, \alpha p) = h(1, \alpha) \end{equation} (18)
    \begin{equation} r = h(\alpha p, p) = h(\alpha, 1). \end{equation} (19)

    From (14)-(19), it follows that

    \begin{equation*} f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1). \end{equation*}

    Now, assume that

    \begin{equation*} f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1). \end{equation*}

    and let

    \begin{equation*} x_{0} = f(\beta, 1), \, x_{-1} = f(1, \beta), \, y_{0} = g(\gamma, 1), \, y_{-1} = g(1, \gamma), \, z_{0} = h(\alpha, 1), \, z_{-1} = h(1, \alpha). \end{equation*}

    We have

    \begin{equation*} x_{1} = f(y_0, y_{-1}) = f(g(\gamma, 1), g(1, \gamma)) = f(g(\gamma, 1), \beta g(\gamma, 1)) = f(1, \beta) = x_{-1}, \end{equation*}
    \begin{equation*} y_{1} = g(z_0, z_{-1}) = g(h(\alpha, 1), h(1, \alpha)) = g(h(\alpha, 1), \gamma h(\alpha, 1)) = g(1, \gamma) = y_{-1}, \end{equation*}
    \begin{equation*} z_{1} = h(x_0, x_{-1}) = h(f(\beta, 1), f(1, \beta)) = h(f(\beta, 1), \alpha f(\beta, 1)) = h(1, \alpha) = z_{-1}, \end{equation*}
    \begin{equation*} x_{2} = f(y_1, y_{0}) = f(g(1, \gamma), g(\gamma, 1)) = f(\beta g(\gamma, 1), g(\gamma, 1)) = f(\beta, 1) = x_{0}, \end{equation*}
    \begin{equation*} y_{2} = g(z_1, z_{0}) = g(h(1, \alpha), h(\alpha, 1)) = g(\gamma h(\alpha, 1), h(\alpha, 1)) = g(1, \gamma) = y_{0}, \end{equation*}
    \begin{equation*} z_{1} = h(x_1, x_{0}) = h(f(1, \beta), f(\beta, 1)) = h(\alpha f(\beta, 1), f(\beta, 1)) = h(1, \alpha) = z_{0}. \end{equation*}

    By induction we get

    \begin{equation*} x_{2n-1} = x_{-1}, \, x_{2n} = x_0, \, y_{2n-1} = y_{-1}, \, y_{2n} = y_0, \, z_{2n-1} = z_{-1}, \, z_{2n} = z_0, \, n\in\mathbb{N}_{0}. \end{equation*}

    In the following, we apply our result in finding prime period two solutions of two special Systems.

    Consider the three dimensional system of difference equations

    \begin{equation} \left\{ \begin{array}{c} x_{n+1} = a_{1}+b_{1}\frac{y_{n}}{y_{n-1}}+ c_{1}\frac{y_{n-1}}{y_{n}}, \\ y_{n+1} = a_{2}+b_{2}\frac{z_{n}}{z_{n-1}}+ c_{2}\frac{z_{n-1}}{z_{n}}, \\ z_{n+1} = a_{3}+b_{3}\frac{x_{n}}{x_{n-1}}+ c_{3}\frac{x_{n-1}}{x_{n}} \end{array} \right. \end{equation} (20)

    where n\in \mathbb{N}_{0} , the initial values x_{-i}, \, y_{-i}, \, z_{-i}, \, i = 0, 1 and the a_{i}, b_i, \, c_i, \, i = 1, 2, 3 are positive real numbers.

    System (20) can be seen as a generalization of the system

    \begin{equation*} x_{n+1} = a_{1}+b_{1}\frac{y_{n}}{y_{n-1}}+ c_{1}\frac{y_{n-1}}{y_{n}} , \, y_{n+1} = a_{2}+b_{2}\frac{x_{n}}{x_{n-1}}+ c_{2}\frac{x_{n-1}}{x_{n}}, \end{equation*}

    studied in [23]. This last one is also a generalization of the equation

    \begin{equation*} x_{n+1} = a_{1}+b_{1}\frac{x_{n}}{x_{n-1}}+ c_{1}\frac{x_{n-1}}{x_{n}} \end{equation*}

    studied in [7] and [15].

    Corollary 1. Assume that (\alpha-1)(\beta-1)(\gamma-1)\neq0 , then System (20) has prime period two solution of the form

    \begin{equation*} ..., ( \alpha f(\beta, 1), \beta g(\gamma, 1), \gamma h(\alpha, 1)), \, ( f(\beta, 1), g(\gamma, 1), h(\alpha, 1)), ... \end{equation*}

    if and only if

    \begin{equation} \left\{ \begin{array}{c} (b_{1}\alpha-c_{1})\beta^2+a_{1}\beta(\alpha-1)+c_{1}\alpha-b_{1} = 0, \\ (b_{2} \beta-c_{2})\gamma^2+a_{2}\gamma(\beta-1)+c_{2}\beta-b_{2} = 0, \\ (b_{3}\gamma-c_{3})\alpha^2+a_{3}\alpha(\gamma-1)+c_{3}\gamma-b_{3} = 0. \end{array} \right . \end{equation} (21)

    Proof. System (20) can be written as

    x_{n+1} = f(y_n, y_{n-1}), \, y_{n+1} = g(z_{n}, z_{n-1}), \, z_{n+1} = h(x_{n}, x_{n-1}),

    where

    f(u, v) = a_{1}+b_{1}\frac{u}{v}+ c_{1}\frac{v}{u}, \, g(u, v) = a_{2}+b_{2}\frac{u}{v}+ c_{2}\frac{v}{u}, \, h(u, v) = a_{3}+b_{3}\frac{u}{v}+ c_{3}\frac{v}{u}.

    So, from Theorem 3.1,

    \begin{equation*} ..., ( \alpha f(\beta, 1), \beta g(\gamma, 1), \gamma h(\alpha, 1)), \, ( f(\beta, 1), g(\gamma, 1), h(\alpha, 1)), ... \end{equation*}

    will be a period prime two solution of System (20) if and only if

    f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1).

    Clearly this condition is equivalent to (21).

    Example 2. If we choose \alpha = 2, \, \beta = 3, \, \gamma = \frac{1}{2} then, condition (21) will be

    \begin{equation*} 3a_{1}+17b_{1}-7c_{1} = 0, \, 4a_{2}-b_{2}+11c_{2} = 0, \, -2a_{3}+2b_{3}-7c_{3} = 0. \end{equation*}

    The last condition is satisfied for the choice

    \begin{equation*} a_1 = \frac{4}{3}, \, b_1 = 1, \, c_1 = 3, \, a_2 = \frac{1}{4}, \, b_2 = 2, \, c_2 = \frac{1}{ 11}, \, a_3 = \frac{1}{2}, \, b_3 = 4, \, c_3 = 1 \end{equation*}

    of the parameters. The corresponding prime period two solution will be

    \begin{equation*} x_{2n-1} = x_{-1} = \alpha f(\beta, 1) = \frac{32}{3}, \end{equation*}
    \begin{equation*} y_{2n-1} = y_{-1} = \beta g(\gamma, 1) = \frac{189}{44}, \end{equation*}
    \begin{equation*} z_{2n-1} = z_{-1} = \gamma h(\alpha, 1) = \frac{9}{2}, \end{equation*}

    and

    \begin{equation*} x_{2n} = x_{0} = f(\beta, 1) = \frac{16}{3}, \end{equation*}
    \begin{equation*} y_{2n} = y_{0} = g(\gamma, 1) = \frac{63}{44}, \end{equation*}
    \begin{equation*} z_{2n} = z_{0} = h(\alpha, 1) = 9, \end{equation*}

    that is

    \begin{equation*} \left\{\left(\frac{32}{3}, \frac{189}{44}, \frac{9}{2} \right), \, \left(\frac{ 16}{3}, \frac{63}{44}, 9 \right), \, \left(\frac{32}{3}, \frac{189}{44}, \frac{9 }{2} \right), \, \left(\frac{16}{3}, \frac{63}{44}, 9 \right), \cdots\right\}. \end{equation*}

    Now, consider the following system of difference equations

    \begin{equation} \left\{ \begin{array}{c} x_{n+1} = a_{1}+b_{1}\frac{y_{n}}{y_{n-1}}+ c_{1}\left(\frac{y_{n-1}}{y_{n}} \right)^{2}, \\ y_{n+1} = a_{2}+b_{2}\frac{z_{n-1}}{z_{n}}+ c_{2}\left(\frac{ z_{n-1}}{z_{n}}\right)^{2}, \\ z_{n+1} = a_{3}+b_{3}\frac{x_{n-1}}{x_{n}}+ 3_{2}\left(\frac{x_{n-1}}{x_{n}}\right)^{2}, \end{array} \, n\in \mathbb{N}_{0} \right. \end{equation} (22)

    where the initial values x_{-i}, \, y_{-i}, \, z_{-i}, \, i = 0, 1 and a_{i}, b_i, \, c_i, \, i = 1, 2, 3 are positive real numbers.

    Corollary 2. Assume that (\alpha-1)(\beta-1)(\gamma-1)\neq0 , then System (22) has prime period two solution of the form

    \begin{equation*} ..., ( \alpha f(\beta, 1), \beta g(\gamma, 1), \gamma h(\alpha, 1)), \, (f(\beta, 1), g(\gamma, 1), h(\alpha, 1)), ... \end{equation*}

    if and only if

    \begin{equation} \left\{ \begin{array}{c} a_1\beta^2(\alpha-1)+b_1\beta(\alpha\beta^2-1)+c_1(\alpha-\beta^4) = 0, \\ a_2\gamma^2(\beta-1)+b_2\gamma(\beta\gamma^2-1)+c_2(\beta-\gamma^4) = 0, \\ a_3\alpha^2(\gamma-1)+b_3\alpha(\gamma\alpha^2-1)+c_3(\gamma-\alpha^4) = 0. \end{array} \right. \end{equation} (23)

    Proof. System (22) can be written as

    x_{n+1} = f(y_n, y_{n-1}), \, y_{n+1} = g(z_{n}, z_{n-1}), \, z_{n+1} = h(x_{n}, x_{n-1}),

    where

    \begin{equation*} \left\{ \begin{array}{c} f(u, v) = a_{1}+b_{1}\frac{u}{v}+ c_{1}\left(\frac{v}{u}\right)^{2}, \\ g(u, v) = a_{2}+b_{2}\frac{v}{u}+ c_{2}\left(\frac{v}{u}\right)^{2}, \\ h(u, v) = a_{3}+b_{3}\frac{v}{u}+ c_{3}\left(\frac{v}{u}\right)^{2}. \end{array} \right. \end{equation*}

    So, from Theorem 3.1,

    ..., ( \alpha f(\beta, 1), \beta g(\gamma, 1), \gamma h(\alpha, 1)), \, ( f(\beta, 1), g(\gamma, 1), h(\alpha, 1)), ...

    will be a period prime two solution of System (20) if and only if

    f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1).

    Clearly this condition is equivalent to

    \begin{equation*} \left\{ \begin{array}{c} a_1\beta^2(\alpha-1)+b_1\beta(\alpha\beta^2-1)+c_1(\alpha-\beta^4) = 0, \\ a_2\gamma^2(\beta-1)+b_2\gamma(\beta\gamma^2-1)+c_2(\beta-\gamma^4) = 0, \\ a_3\alpha^2(\gamma-1)+b_3\alpha(\gamma\alpha^2-1)+c_3(\gamma-\alpha^4) = 0. \end{array} \right. \end{equation*}

    Example 3. For \alpha = 3, \, \beta = 2, \, \gamma = \frac{1}{3} then, condition (23) will be

    \begin{equation*} 8a_1+22b_1-13c_1 = 0, \, 9a_2-21b_2+161c_2 = 0, \, 18a_3-18b_3+242c_3 = 0 . \end{equation*}

    The last condition is satisfied for the choice

    \begin{equation*} a_1 = 1, \, b_1 = 1, \, c_1 = \frac{30}{13}, \, a_2 = \frac{19}{9}, \, b_2 = 2, \, c_2 = \frac{1}{ 7}, \, a_3 = \frac{1}{6}, \, b_3 = \frac{7}{9}, \, c_3 = \frac{1}{22} \end{equation*}

    of the parameters. The corresponding prime period two solution will be

    \begin{equation*} x_{2n-1} = x_{-1} = 3 f(2, 1) = \frac{297}{26}, \end{equation*}
    \begin{equation*} y_{2n-1} = y_{-1} = 2 g(\frac{1}{3}, 1) = \frac{512}{63}, \end{equation*}
    \begin{equation*} z_{2n-1} = z_{-1} = \frac{1}{3} h(3, 1) = \frac{248}{297}, \end{equation*}

    and

    \begin{equation*} x_{2n} = x_{0} = f(2, 1) = \frac{93}{26}, \end{equation*}
    \begin{equation*} y_{2n} = y_{0} = g(\frac{1}{3}, 1) = \frac{256}{63}, \end{equation*}
    \begin{equation*} z_{2n} = z_{0} = h(3, 1) = \frac{744}{297}, \end{equation*}

    that is

    \begin{equation*} \left\{\left(\frac{297}{26}, \frac{512}{63}, \frac{248}{297} \right), \, \left( \frac{93}{26}, \frac{256}{63}, \frac{744}{297} \right), \, \left(\frac{297}{26}, \frac{512}{63}, \frac{248}{297} \right), \, \cdots\right\}. \end{equation*}

    Here, we are interested in the oscillation of the solutions of System (1) about the equilibrium point (\overline{x}, \overline{y}, \overline{z }) = (f(1, 1), g(1, 1), h(1, 1)) .

    Theorem 4.1. Let \left(x_n, y_n, z_n\right)_{n\geq-1} be a solution of System (1) and assume that f(x, y) , g(x, y) h(x, y) are decreasing in x for all y and are increasing in y for all x .

    1. If

    \begin{equation*} x_{0} < \overline{x}, \, x_{-1} > \overline{x}, \, y_{0} < \overline{y}, \, y_{-1} > \overline{y}, \, z_{0} < \overline{z}, \, z_{-1} > \overline{z}, \end{equation*}

    then we get

    \begin{equation*} x_{2n} < \overline{x}, \, x_{2n-1} > \overline{x}, \, y_{2n} < \overline{y}, \, y_{2n-1} > \overline{y}, \, z_{2n} < \overline{z}, \, z_{2n-1} > \overline{z}, \, n\in\mathbb{N} _0. \end{equation*}

    That is for the sequences (x_n)_{n\geq-1} , (y_n)_{n\geq-1} and (z_n)_{n\geq-1} we have semi-cycles of length one of the form

    \begin{equation*} +-+-+-\cdots. \end{equation*}

    2. If

    \begin{equation*} x_{0} > \overline{x}, \, x_{-1} < \overline{x}, \, y_{0} > \overline{y}, \, y_{-1} < \overline{y}, \, z_{0} > \overline{z}, \, z_{-1} < \overline{z}, \end{equation*}

    then we get

    \begin{equation*} x_{2n} > \overline{x}, \, x_{2n-1} < \overline{x}, \, y_{2n} > \overline{y}, \, y_{2n-1} < \overline{y}, \, z_{2n} > \overline{z}, \, z_{2n-1} < \overline{z}, \, n\in\mathbb{N} _0. \end{equation*}

    That is for the sequences (x_n)_{n\geq-1} , (y_n)_{n\geq-1} and (z_n)_{n\geq-1} we have semi-cycles of length one of the form

    \begin{equation*} -+-+-+\cdots. \end{equation*}

    Proof. 1. Assume that

    \begin{equation*} x_{0} < \overline{x}, \, x_{-1} > \overline{x}, \, y_{0} < \overline{y}, \, y_{-1} > \overline{y}, \, z_{0} < \overline{z}, \, z_{-1} > \overline{z}. \end{equation*}

    We have

    \begin{equation*} x_{1} = f(y_0, y_{-1}) > f(\overline{y}, y_{-1}) > f(\overline{y}, \overline{y}) = f(1, 1) = \overline{x}, \end{equation*}
    \begin{equation*} y_{1} = g(z_0, z_{-1}) > g(\overline{z}, z_{-1}) > g(\overline{z}, \overline{z}) = g(1, 1) = \overline{y}, \end{equation*}
    \begin{equation*} z_{1} = h(x_0, x_{-1}) > h(\overline{x}, x_{-1}) > h(\overline{x}, \overline{x}) = h(1, 1) = \overline{z}, \end{equation*}
    \begin{equation*} x_{2} = f(y_1, y_{0}) < f(\overline{y}, y_{0}) < f(\overline{y}, \overline{y}) = f(1, 1) = \overline{x}, \end{equation*}
    \begin{equation*} y_{2} = g(z_1, z_{0}) < g(\overline{z}, z_{0}) < g(\overline{z}, \overline{z}) = g(1, 1) = \overline{y}, \end{equation*}
    \begin{equation*} z_{2} = h(x_1, x_{0}) < h(\overline{x}, x_{0}) < h(\overline{x}, \overline{x}) = h(1, 1) = \overline{z}. \end{equation*}

    By induction, we get

    \begin{equation*} x_{2n} < \overline{x}, \, x_{2n-1} > \overline{x}, \, y_{2n} < \overline{y}, \, y_{2n-1} > \overline{y}, \, z_{2n} < \overline{z}, \, z_{2n-1} > \overline{z}, \, n\in\mathbb{N}_0. \end{equation*}

    2. Assume that

    \begin{equation*} x_{0} > \overline{x}, \, x_{-1} < \overline{x}, \, y_{0} > \overline{y}, \, y_{-1} < \overline{y}, \, z_{0} > \overline{z}, \, z_{-1} < \overline{z}. \end{equation*}

    We have

    \begin{equation*} x_{1} = f(y_0, y_{-1}) < f(\overline{y}, y_{-1}) < f(\overline{y}, \overline{y}) = f(1, 1) = \overline{x}, \end{equation*}
    \begin{equation*} y_{1} = g(z_0, z_{-1}) < g(\overline{z}, z_{-1}) < g(\overline{z}, \overline{z}) = g(1, 1) = \overline{y}, \end{equation*}
    \begin{equation*} z_{1} = h(x_0, x_{-1}) < h(\overline{x}, x_{-1}) < h(\overline{x}, \overline{x}) = h(1, 1) = \overline{z}, \end{equation*}
    \begin{equation*} x_{2} = f(y_1, y_{0}) > f(\overline{y}, y_{0}) > f(\overline{y}, \overline{y}) = f(1, 1) = \overline{x}, \end{equation*}
    \begin{equation*} y_{2} = g(z_1, z_{0}) > g(\overline{z}, z_{0}) > g(\overline{z}, \overline{z}) = g(1, 1) = \overline{y}, \end{equation*}
    \begin{equation*} x_{2} = h(x_1, x_{0}) > h(\overline{x}, x_{0}) > h(\overline{x}, \overline{x}) = h(1, 1) = \overline{y}. \end{equation*}

    By induction, we get

    \begin{equation*} x_{2n} > \overline{x}, \, x_{2n-1} < \overline{x}, \, y_{2n} > \overline{y}, \, y_{2n-1} < \overline{y}, \, z_{2n} > \overline{z}, \, z_{2n-1} < \overline{z}, \, n\in\mathbb{N}_0. \end{equation*}

    So, the proof is completed. Now, we will apply the results of this section on the following particular system.

    Example 4. Consider the system of difference equations

    \begin{equation} x_{n+1} = a_{1}+b_{1}\left(\frac{y_{n-1}}{y_{n}}\right)^{p}, \, y_{n+1} = a_{2}+ b_{2}\left(\frac{z_{n-1}}{z_{n}}\right)^{q}, \, z_{n+1} = a_{3}+ b_{3}\left( \frac{x_{n-1}}{x_{n}}\right)^{r}, \, n\in \mathbb{N}_{0}, \end{equation} (24)

    where p, \, q, \, r\in \mathbb{N} , the initial values x_{-i}, \, y_{-i}, \, z_{-i}, \, i = 0, 1 and the parameters a_i, \, b_i, \, i = 1, 2, 3 are positive real numbers.

    Let f , g and h be the functions defined by

    \begin{equation*} f(u, v) = a_{1}+b_{1}\left(\frac{v}{u}\right)^{p}, \, g(u, v) = a_{2}+b_{2}\left( \frac{v}{u}\right)^{q}, \, h(u, v) = a_{3}+b_{3}\left(\frac{v}{u} \right)^{r}, \, u, v\in (0, +\infty). \end{equation*}

    It is not hard to see that

    \begin{equation*} \frac{\partial f}{\partial u}(u, v) < 0, \, \frac{\partial f}{\partial v} (u, v) > 0, \, \frac{\partial g}{\partial u}(u, v) < 0, \, \frac{\partial g}{\partial v }(u, v) > 0\, \frac{\partial h}{\partial u}(u, v) < 0, \, \frac{\partial h}{\partial v }(u, v) > 0. \end{equation*}

    System (24) has the unique equilibrium point (\overline{x}, \overline{y }, \overline{z}) = (a_1+b_1, a_2+b_2, a_3+b_3) .

    Corollary 3. Let \left(x_n, y_n, z_n\right)_{n\geq-1} be a solution of System (24). The following statements holds true:

    1. Let

    \begin{equation*} x_{0} < \overline{x}, \, x_{-1} > \overline{x}, \, y_{0} < \overline{y}, \, y_{-1} > \overline{y}, \, z_{0} < \overline{z}, \, z_{-1} > \overline{z}. \end{equation*}

    Then the sequences (x_n)_n (resp. (y_n)_n , (z_n)_n ) oscillates about \overline{x} (resp. about \overline{y} , \overline{z} ) with semi-cycle of length one and every semi-cycle is in the form

    \begin{equation*} +-+-+-\cdots. \end{equation*}

    2. Let

    \begin{equation*} x_{0} > \overline{x}, \, x_{-1} < \overline{x}, \, y_{0} > \overline{y}, \, y_{-1} < \overline{y}, \, z_{0} > \overline{z}, \, z_{-1} < \overline{z}. \end{equation*}

    Then the sequences (x_n)_n (resp. (y_n)_n , (z_n)_n ) oscillates about \overline{x} (resp. about \overline{y} , \overline{z} ) with semi-cycle of length one and every semi-cycle is in the form

    \begin{equation*} -+-+-+\cdots. \end{equation*}

    Proof. 1. Let

    x_{0} < \overline{x}, \, x_{-1} > \overline{x}, \, y_{0} < \overline{y}, \, y_{-1} > \overline{y}, \, z_{0} < \overline{z}, \, z_{-1} > \overline{z}.

    We have

    \frac{y_{-1}}{y_{0}} > \frac{\overline{y}}{\overline{y}} = 1,

    which implies that

    x_1 = a_1+b_1\left(\frac{y_{-1}}{y_{0}}\right)^{p} > a_1+b_1 = \overline{x}.

    Using the fact that

    \frac{z_{-1}}{z_{0}} > \frac{\overline{z}}{\overline{z}} = 1,

    we get

    y_1 = a_2+b_2\left(\frac{z_{-1}}{z_{0}}\right)^{q} > a_2+b_2 = \overline{y}.

    Also as,

    \frac{x_{-1}}{x_{0}} > \frac{\overline{x}}{\overline{x}} = 1,

    we get

    z_1 = a_3+b_3\left(\frac{x_{-1}}{x_{0}}\right)^{r} > a_3+b_3 = \overline{z}.

    Now, as

    \frac{y_{0}}{y_{1}} < \frac{\overline{y}}{\overline{y}} = 1,

    we obtain

    x_2 = a_1+b_1\left(\frac{y_{0}}{y_{1}}\right)^{p} < a_1+b_1 = \overline{x}.

    Similarly,

    \frac{z_{0}}{z_{1}} < \frac{\overline{z}}{\overline{z}} = 1\Rightarrow y_2 = a_2+b_2\left(\frac{z_{0}}{z_{1}}\right)^{q} < a_2+b_2 = \overline{y},

    and

    \frac{x_{0}}{x_{1}} < \frac{\overline{x}}{\overline{x}} = 1\Rightarrow z_2 = a_3+b_3\left(\frac{x_{0}}{x_{1}}\right)^{r} < a_3+b_3 = \overline{z},

    and by induction we get that

    x_{2n}-\overline{x} < 0, \, y_{2n}-\overline{y} < 0, \, z_{2n}-\overline{z} < 0,
    x_{2n-1}-\overline{x} > 0, \, y_{2n-1}-\overline{y} > 0, \, z_{2n-1}-\overline{z} > 0,

    for n\in \mathbb{N}_{0} . That is, the sequences (x_n)_n (resp. (y_n)_n , (z_n)_n ) oscillates about \overline{x} (resp. about \overline{y} , \overline{z} ) with semi-cycle of length one and every semi-cycle is in the form

    +-+-+-\cdots

    and this is the statement of Part 1. of Theorem 4.1.

    2. Let

    x_{0} > \overline{x}, \, x_{-1} < \overline{x}, \, y_{0} > \overline{y}, \, y_{-1} < \overline{y}, \, z_{0} > \overline{z}, \, z_{-1} < \overline{z}.

    We have

    \frac{y_{-1}}{y_{0}} < \frac{\overline{y}}{\overline{y}} = 1,

    which implies that

    x_1 = a_1+b_1\left(\frac{y_{-1}}{y_{0}}\right)^{p} < a_1+b_1 = \overline{x}.

    Using the fact that

    \frac{z_{-1}}{z_{0}} < \frac{\overline{z}}{\overline{z}} = 1,

    we get

    y_1 = a_2+b_2\left(\frac{z_{-1}}{z_{0}}\right)^{q} < a_2+b_2 = \overline{y}.

    Also as,

    \frac{x_{-1}}{x_{0}} < \frac{\overline{x}}{\overline{x}} = 1,

    we get

    z_1 = a_3+b_3\left(\frac{x_{-1}}{x_{0}}\right)^{r} < a_3+b_3 = \overline{z}.

    Now, as

    \frac{y_{0}}{y_{1}} > \frac{\overline{y}}{\overline{y}} = 1,

    we obtain

    x_2 = a_1+b_1\left(\frac{y_{0}}{y_{1}}\right)^{p} > a_1+b_1 = \overline{x}.

    Similarly,

    \frac{z_{0}}{z_{1}} > \frac{\overline{z}}{\overline{z}} = 1\Rightarrow y_2 = a_2+b_2\left(\frac{z_{0}}{z_{1}}\right)^{q} > a_2+b_2 = \overline{y},

    and

    \frac{x_{0}}{x_{1}} > \frac{\overline{x}}{\overline{x}} = 1\Rightarrow z_2 = a_3+b_3\left(\frac{x_{0}}{x_{1}}\right)^{r} > a_3+b_3 = \overline{z}.

    Thus, by induction we get that

    x_{2n}-\overline{x} > 0, \, y_{2n}-\overline{y} > 0, \, z_{2n}-\overline{z} > 0,
    x_{2n-1}-\overline{x} < 0, \, y_{2n-1}-\overline{y} < 0, \, z_{2n-1}-\overline{z} < 0

    for n\in \mathbb{N}_{0} . That is the sequences (x_n)_n (resp. (y_n)_n , (z_n)_n ) oscillates about \overline{x} (resp. about \overline{y} , \overline{z} ) with semi-cycle of length one and every semi-cycle is in the form

    -+-+-+\cdots

    and this is the statement of Part 2. of Theorem 4.1.

    In this study, the global stability of the unique positive equilibrium point of a three-dimensional general system of difference equations defined by positive and homogeneous functions of degree zero was studied. For this, general convergence theorems were given considering all possible monotonicity cases in arguments of functions f , g and h . In addition, the periodic nature and oscillation of the general system considered was also discussed and successful results were obtained. It is noteworthy that the results obtained on our general three-dimensional system have high applicability.

    The authors thanks the two referees for their comments and suggestions. The work of N. Touafek and Y. Akrour was supported by DGRSDT (MESRS-DZ).



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