
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,yn−1),yn+1=g(zn,zn−1),zn+1=h(xn,xn−1) | (1) |
where
Now, we explain our motivation for doing this work. Clearly if we take
xn+1=f(yn,yn−1),yn+1=g(xn,xn−1). | (2) |
Noting also that if we choose
xn+1=f(xn,xn−1). | (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
Yn+1=F(Yn),n∈N0, | (4) |
where the initial value
Definition 1.1. Let
limn→∞Yn=¯Y. |
limn→∞Yn=¯Y. |
Assume that
Zn+1=FJ(¯Y)Zn,n∈N0,Zn=Yn−¯Y, |
where
To study the stability of the equilibrium point
Theorem 1.2. Let
(i) If all the eigenvalues of the Jacobian matrix
(ii) If at least one eigenvalue of
Definition 1.3. A solution
xn+p=xn,yn+p=yn,zn+p=zn,n≥−1. | (5) |
The solution
Definition 1.4. Let
Remark 1. For every term
Definition 1.5. A function
Φ(λu,λv)=λmΦ(u,v) |
for all
Theorem 1.6. Let
1. Then,
u∂Φ∂u(u,v)+v∂Φ∂v(u,v)=mΦ(u,v),(u,v)∈(0,+∞)2. |
(This statement is usually called Euler's Theorem).
2. If
A point
¯x=f(¯y,¯y),¯y=g(¯z,¯z),¯z=g(¯x,¯x). |
Using the fact that
(¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1)) |
is the unique equilibrium point of System (1).
Let
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,xn−1,yn,yn−1,zn,zn−1)t,n∈N0. |
So,
¯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
Assume that the functions
Xn+1=JFXn,n∈N0 |
where
¯W=(f(1,1),f(1,1),g(1,1),g(1,1),h(1,1),h(1,1)). |
We have
JF=(00∂f∂w(¯y,¯y)∂f∂t(¯y,¯y)001000000000∂g∂r(¯z,¯z)∂g∂s(¯z,¯z)001000∂h∂u(¯x,¯x)∂h∂v(¯x,¯x)0000000010) |
As
¯y∂f∂w(¯y,¯y)+¯y∂f∂t(¯y,¯y)=0 |
which implies
∂f∂t(¯y,¯y)=−∂f∂w(¯y,¯y). |
Similarly we get
∂g∂s(¯z,¯z)=−∂g∂r(¯z,¯z),∂h∂v(¯x,¯x)=−∂h∂u(¯x,¯x). |
It follows that
JF=(00∂f∂w(¯y,¯y)−∂f∂w(¯y,¯y)001000000000∂g∂r(¯z,¯z)−∂g∂r(¯z,¯z)001000∂h∂u(¯x,¯x)−∂h∂u(¯x,¯x)0000000010) |
The characteristic polynomial of the matrix
P(λ)=λ6−∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y)λ3+3∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y)λ2−3∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y)λ+∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y). |
Now assume that
|∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y)|<18 |
and consider the two functions
Φ(λ)=λ6, |
Ψ(λ)=−∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y)λ3+3∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y)λ2−3∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y)λ+∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y). |
We have
|Ψ(λ)|≤8|∂h∂u(¯x,¯x)∂g∂r(¯z,¯z)∂f∂w(¯y,¯y)|<1=|Φ(λ)|,∀λ∈C:|λ|=1. |
So, by Rouché's Theorem it follows that all roots of
Hence, by Theorem 1.2, we deduce from the above consideration that the equilibrium point
Using Part 2. of Theorem 1.6 and the fact that the function
∂f∂w(¯y,¯y)=∂f∂w(1,1)¯y,∂g∂r(¯z,¯z)=∂g∂r(1,1)¯z,∂h∂u(¯x,¯x)=∂h∂u(1,1)¯x. |
In summary, we have proved the following result.
Theorem 2.1. Assume that
(¯x,¯y,¯z)=(f(1,1),g(1,1),h(1,1)) |
of System (1) is locally asymptotically stable if
|∂f∂u(1,1)∂g∂u(1,1)∂h∂u(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
Theorem 2.2. Consider System (1). Assume that the following statements are true:
1.
a≤f(u,v)≤b,α≤g(u,v)≤β,λ≤h(u,v)≤γ,∀(u,v)∈(0,+∞)2. |
2.
3.
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
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
a≤f(α,β)≤f(β,α)≤b, |
α≤g(λ,γ)≤g(γ,λ)≤β, |
λ≤h(a,b)≤h(b,a)≤γ, |
and so,
m01=a≤f(m02,M02)≤f(M02,m02)≤b=M01, |
m02=α≤g(m03,M03)≤g(M03,m03)≤β=M02, |
and
m03=λ≤h(m01,M01)≤g(M01,m01)≤γ=M03. |
Hence,
m01≤m11≤M11≤M01, |
m02≤m12≤M12≤M02, |
and
m03≤m13≤M13≤M03. |
Now, we have
m11=f(m02,M02)≤f(m12,M12)=m21≤f(M12,m12)=M21≤f(M02,m02)=M11, |
m12=g(m03,M03)≤g(m13,M13)=m22≤g(M13,m13)=M22≤g(M03,m03)=M12, |
m13=h(m01,M01)≤h(m11,M11)=m23≤h(M11,m11)=M23≤h(M01,m01)=M13, |
and it follows that
m01≤m11≤m21≤M21≤M11≤M01, |
m02≤m12≤m22≤M22≤M12≤M02, |
and
m03≤m13≤m23≤M23≤M13≤M03. |
By induction, we get for
a=m01≤m11≤...≤mi−11≤mi1≤Mi1≤Mi−11≤...≤M11≤M01=b, |
α=m02≤m12≤...≤mi−12≤mi2≤Mi2≤Mi−12≤...≤M12≤M02=β, |
and
λ=m03≤m13≤...≤mi−13≤mi3≤Mi3≤Mi−13≤...≤M13≤M03=γ. |
It follows that the sequences
m1=limi→+∞mi1,M1=limi→+∞Mi1, |
m2=limi→+∞mi2,M2=limi→+∞Mi2. |
m3=limi→+∞mi3,M3=limi→+∞Mi3. |
Then
a≤m1≤M1≤b,α≤m2≤M2≤β,λ≤m3≤M3≤γ. |
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
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
m1=M1,m2=M2,m3=M3. |
From
m01=a≤xn≤b=M01,m02=α≤yn≤β=M02,m03=λ≤zn≤γ=M03. |
For
m11=f(m02,M02)≤xn+1=f(yn,yn−1)≤f(M02,m02)=M11, |
m12=g(m03,M03)≤yn+1=g(zn,zn−1)≤g(M03,m03)=M12, |
and
m13=h(m01,M01)≤zn+1=h(xn,xn−1)≤h(M01,m01)=M13, |
that is
m11≤xn≤M11,m12≤yn≤M12,m13≤zn≤M13,n=3,4,⋯. |
Now, for
m21=f(m12,M12)≤xn+1=f(yn,yn−1)≤f(M12,m12)=M21, |
and
m22=g(m13,M13)≤yn+1=g(zn,zn−1)≤g(M13,m13)=M22, |
m23=h(m11,M11)≤zn+1=h(xn,xn−1)≤h(M11,m11)=M23, |
that is
m21≤xn≤M21,m22≤yn≤M22,m23≤zn≤M23,n=5,6,⋯. |
Similarly, for
m31=f(m22,M22)≤xn+1=f(yn,yn−1)≤f(M22,m22)=M31, |
m32=g(m23,M23)≤yn+1=g(zn,zn−1)≤g(M23,m23)=M32, |
and
m33=h(m21,M21)≤zn+1=h(xn,xn−1)≤h(M21,m21)=M33, |
that is
m31≤xn≤M31,m32≤yn≤M32,m33≤zn≤M33,n=7,8,⋯. |
It follows by induction that for
mi1≤xn≤Mi1,mi2≤yn≤Mi2,mi3≤zn≤Mi3,n≥2i+1. |
Using the fact that
limn→+∞xn=M1,limn→+∞yn=M2,limn→+∞zn=M3. |
From (1) and using the fact that
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.
a≤f(u,v)≤b,α≤g(u,v)≤β,λ≤h(u,v)≤γ,∀(u,v)∈(0,+∞)2. |
2.
3.
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
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
a≤f(α,β)≤f(β,α)≤b, |
α≤g(λ,γ)≤g(γ,λ)≤β, |
λ≤h(b,a)≤h(a,b)≤γ, |
and so,
m01=a≤f(m02,M02)≤f(M02,m02)≤b=M01, |
m02=α≤g(m03,M03)≤g(M03,m03)≤β=M02, |
and
m03=λ≤h(M01,m01)≤h(m01,M01)≤γ=M03. |
Hence,
m01≤m11≤M11≤M01, |
m02≤m12≤M12≤M02, |
and
m03≤m13≤M13≤M03. |
Now, we have
m11=f(m02,M02)≤f(m12,M12)=m21≤f(M12,m12)=M21≤f(M02,m02)=M11, |
m12=g(m03,M03)≤g(m13,M13)=m22≤g(M13,m13)=M22≤g(M03,m03)=M12, |
m13=h(M01,m01)≤h(M11,m11)=m23≤h(m11,M11)=M23≤h(m01,M01)=M13, |
and it follows that
m01≤m11≤m21≤M21≤M11≤M01, |
m02≤m12≤m22≤M22≤M12≤M02, |
and
m03≤m13≤m23≤M23≤M13≤M03. |
By induction, we get for
a=m01≤m11≤...≤mi−11≤mi1≤Mi1≤Mi−11≤...≤M11≤M01=b, |
α=m02≤m12≤...≤mi−12≤mi2≤Mi2≤Mi−12≤...≤M12≤M02=β, |
and
λ=m03≤m13≤...≤mi−13≤mi3≤Mi3≤Mi−13≤...≤M13≤M03=γ. |
It follows that the sequences
m1=limi→+∞mi1,M1=limi→+∞Mi1, |
m2=limi→+∞mi2,M2=limi→+∞Mi2. |
m3=limi→+∞mi3,M3=limi→+∞Mi3. |
Then
a≤m1≤M1≤b,α≤m2≤M2≤β,λ≤m3≤M3≤γ. |
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
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
m1=M1,m2=M2,m3=M3. |
From
m01=a≤xn≤b=M01,m02=α≤yn≤β=M02,m03=λ≤zn≤γ=M03. |
For
m11=f(m02,M02)≤xn+1=f(yn,yn−1)≤f(M02,m02)=M11, |
m12=g(m03,M03)≤yn+1=g(zn,zn−1)≤g(M03,m03)=M12, |
and
m13=h(M01,m01)≤zn+1=h(xn,xn−1)≤h(m01,M01)=M13, |
that is
m11≤xn≤M11,m12≤yn≤M12,m13≤zn≤M13,n=3,4,⋯. |
Now, for
m21=f(m12,M12)≤xn+1=f(yn,yn−1)≤f(M12,m12)=M21, |
and
m22=g(m13,M13)≤yn+1=g(zn,zn−1)≤g(M13,m13)=M22, |
m23=h(M11,m11)≤zn+1=h(xn,xn−1)≤h(m11,M11)=M23, |
that is
m21≤xn≤M21,m22≤yn≤M22,m23≤zn≤M23,n=5,6,⋯. |
Similarly, for
m31=f(m22,M22)≤xn+1=f(yn,yn−1)≤f(M22,m22)=M31, |
m32=g(m23,M23)≤yn+1=g(zn,zn−1)≤g(M23,m23)=M32, |
and
m33=h(M21,m21)≤zn+1=h(xn,xn−1)≤h(m21,M21)=M33, |
that is
m31≤xn≤M31,m32≤yn≤M32,m33≤zn≤M33,n=7,8,⋯. |
It follows by induction that for
mi1≤xn≤Mi1,mi2≤yn≤Mi2,mi3≤zn≤Mi3,n≥2i+1. |
Using the fact that
limn→+∞xn=M1,limn→+∞yn=M2,limn→+∞zn=M3. |
From (1) and using the fact that
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.
a≤f(u,v)≤b,α≤g(u,v)≤β,λ≤h(u,v)≤γ,∀(u,v)∈(0,+∞)2. |
2.
3.
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
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
a≤f(α,β)≤f(β,α)≤b, |
α≤g(γ,λ)≤g(λ,γ)≤β, |
λ≤h(b,a)≤h(a,b)≤γ, |
and so,
m01=a≤f(m02,M02)≤f(M02,m02)≤b=M01, |
m02=α≤g(M03,m03)≤g(m03,M03)≤β=M02, |
and
m03=λ≤h(M01,m01)≤h(m01,M01)≤γ=M03. |
Hence,
m01≤m11≤M11≤M01, |
m02≤m12≤M12≤M02, |
and
m03≤m13≤M13≤M03. |
Now, we have
m11=f(m02,M02)≤f(m12,M12)=m21≤f(M12,m12)=M21≤f(M02,m02)=M11, |
m12=g(M03,m03)≤g(M13,m13)=m22≤g(m13,M13)=M22≤g(m03,M03)=M12, |
m13=h(M01,m01)≤h(M11,m11)=m23≤h(m11,M11)=M23≤h(m01,M01)=M13, |
and it follows that
m01≤m11≤m21≤M21≤M11≤M01, |
m02≤m12≤m22≤M22≤M12≤M02, |
and
m03≤m13≤m23≤M23≤M13≤M03. |
By induction, we get for
a=m01≤m11≤...≤mi−11≤mi1≤Mi1≤Mi−11≤...≤M11≤M01=b, |
α=m02≤m12≤...≤mi−12≤mi2≤Mi2≤Mi−12≤...≤M12≤M02=β, |
and
λ=m03≤m13≤...≤mi−13≤mi3≤Mi3≤Mi−13≤...≤M13≤M03=γ. |
It follows that the sequences
m1=limi→+∞mi1,M1=limi→+∞Mi1, |
m2=limi→+∞mi2,M2=limi→+∞Mi2. |
m3=limi→+∞mi3,M3=limi→+∞Mi3. |
Then
a≤m1≤M1≤b,α≤m2≤M2≤β,λ≤m3≤M3≤γ. |
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
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
m1=M1,m2=M2,m3=M3. |
From
m01=a≤xn≤b=M01,m02=α≤yn≤β=M02,m03=λ≤zn≤γ=M03. |
For
m11=f(m02,M02)≤xn+1=f(yn,yn−1)≤f(M02,m02)=M11, |
m12=g(M03,m03)≤yn+1=g(zn,zn−1)≤g(m03,M03)=M13, |
and
m13=h(M01,m01)≤zn+1=h(xn,xn−1)≤h(m01,M01)=M13, |
that is
m11≤xn≤M11,m12≤yn≤M12,m13≤zn≤M13,n=3,4,⋯. |
Now, for
m21=f(m12,M12)≤xn+1=f(yn,yn−1)≤f(M12,m12)=M21, |
and
m22=g(M13,m13)≤yn+1=g(zn,zn−1)≤g(m13,M13)=M22, |
m23=h(M11,m11)≤zn+1=h(xn,xn−1)≤h(m11,M11)=M23, |
that is
m21≤xn≤M21,m22≤yn≤M22,m23≤zn≤M23,n=5,6,⋯. |
Similarly, for
m31=f(m22,M22)≤xn+1=f(yn,yn−1)≤f(M22,m22)=M31, |
m32=g(M23,m23)≤yn+1=g(zn,zn−1)≤g(m23,M23)=M32, |
and
m33=h(M21,m21)≤zn+1=h(xn,xn−1)≤h(m21,M21)=M33, |
that is
m31≤xn≤M31,m32≤yn≤M32,m33≤zn≤M33,n=7,8,⋯. |
It follows by induction that for
mi1≤xn≤Mi1,mi2≤yn≤Mi2,mi3≤zn≤Mi3,n≥2i+1. |
Using the fact that
limn→+∞xn=M1,limn→+∞yn=M2,limn→+∞zn=M3. |
From (1) and using the fact that
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.
a≤f(u,v)≤b,α≤g(u,v)≤β,λ≤h(u,v)≤γ,∀(u,v)∈(0,+∞)2. |
2.
3.
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
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
a≤f(α,β)≤f(β,α)≤b, |
α≤g(γ,λ)≤g(λ,γ)≤β, |
λ≤h(a,b)≤h(b,a)≤γ, |
and so,
m01=a≤f(m02,M02)≤f(M02,m02)≤b=M01, |
m02=α≤g(M03,m03)≤g(m03,M03)≤β=M02, |
and
m03=λ≤h(m01,M01)≤h(M01,m01)≤γ=M03. |
Hence,
m01≤m11≤M11≤M01, |
m02≤m12≤M12≤M02, |
and
m03≤m13≤M13≤M03. |
Now, we have
m11=f(m02,M02)≤f(m12,M12)=m21≤f(M12,m12)=M21≤f(M02,m02)=M11, |
m12=g(M03,m03)≤g(M13,m13)=m22≤g(m13,M13)=M22≤g(m03,M03)=M12, |
m13=h(m01,M01)≤h(m11,M11)=m23≤h(M11,m11)=M23≤h(M01,m01)=M13, |
and it follows that
m01≤m11≤m21≤M21≤M11≤M01, |
m02≤m12≤m22≤M22≤M12≤M02, |
and
m03≤m13≤m23≤M23≤M13≤M03. |
By induction, we get for
a=m01≤m11≤...≤mi−11≤mi1≤Mi1≤Mi−11≤...≤M11≤M01=b, |
α=m02≤m12≤...≤mi−12≤mi2≤Mi2≤Mi−12≤...≤M12≤M02=β, |
and
λ=m03≤m13≤...≤mi−13≤mi3≤Mi3≤Mi−13≤...≤M13≤M03=γ. |
It follows that the sequences
m1=limi→+∞mi1,M1=limi→+∞Mi1, |
m2=limi→+∞mi2,M2=limi→+∞Mi2. |
m3=limi→+∞mi3,M3=limi→+∞Mi3. |
Then
a≤m1≤M1≤b,α≤m2≤M2≤β,λ≤m3≤M3≤γ. |
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
\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
\begin{equation*} m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. \end{equation*} |
From
\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
\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
\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
\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
\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
\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
\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.
\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.
3.
\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
\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
\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
\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
\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
\begin{equation*} m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. \end{equation*} |
From
\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
\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
\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
\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
\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
\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
\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.
\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.
3.
\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
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
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
\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
\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
m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. |
From
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
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
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
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
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
\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
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.
\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.
3.
\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
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
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
\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
\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
m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. |
From
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
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
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
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
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
\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
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.
\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.
3.
\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
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
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
\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
\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
m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. |
From
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
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
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
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
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
\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
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
Theorem 2.10. Under the hypotheses of Theorem 2.1 and one of Theorems 2.2– 2.9, the equilibrium point
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
\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
\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
For this purpose, let
\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
\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
\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
In summary we have the following result.
Theorem 2.11. Assume that that parameters
●
●
\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*} |
●
\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
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
However Figure 3 shows the unstable solution corresponding to the values
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
\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
\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
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*} |
Corollary 1. Assume that
\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
\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
Corollary 2. Assume that
\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
\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
Theorem 4.1. Let
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
\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
\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
Let
\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
Corollary 3. Let
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
\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
\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
+-+-+-\cdots |
and this is the statement of Part 1. of Theorem 4.1.
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
-+-+-+\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
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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