# On a System of Difference Equations of an Economic Model

Authors: M. F. Elettreby; H. El-Metwally
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We study some qualitative properties of the solutions of a system of difference equations, which describes an economic model. The study of the local stability of the equilibrium points is carried out. We give some important results of the invariant and the boundedness of the solutions to the considered system. The global convergence of the solutions is presented and investigated.


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## Body

## 1. Introduction

The increasing study of realistic mathematical models is a reflection of their use in helping to understand the dynamic processes involved in areas such as population dynamics, biology, epidemiology, ecology, and economy. More realistic models should include some of the past states of these systems; that is, ideally, a real system should be modeled by difference equations with time delays. Most of these models are described by nonlinear delay difference equations; see, for example, [1–4]. The subject of the qualitative study of the nonlinear delay population models is very extensive, and the current research work tends to center around the relevant global dynamics of the considered systems of difference equations such as oscillation, boundedness of solutions, persistence, global stability of positive steady sates, permanence, and global existence of periodic solutions. See [5–18] and the references therein.In this paper, we intend to cover some of these global aspects of the qualitative behavior of a system of a discrete model in the economy area, where we deal with the studying of some qualitative properties of solutions of the following system of difference equations:(1)xn+1=(1-α)xn+βxn(1-xn)e-(xn+yn),yn+1=(1-α)yn+βyn(1-yn)e-(xn+yn),yn+1=(1-α)ynβyn(1-yn)n=0,1,…,
where α and β∈ (0, ∞) with the initial conditions x0 and y0∈ (0, ∞). We study the boundedness and the invariant of the solutions of system (1) and also investigate global convergence for the solutions of system (1).System (1) is an important type of economic models which describes a discrete-time map generated by bounded rationally duopoly game with exponential demand function. See [19].The following theorem was presented in [6], and it will be useful in the investigation of the global stability of system (1).Theorem A.
Consider the following system of difference equations:(2)xn+1=f(xn,yn),yn+1=g(xn,yn),yn+1n=0,1,….
Suppose that(i)
f(x;y) is nondecreasing in xand is nonincreasing in y, and g(x;y) is nonincreasing in x and is nondecreasing in y,(ii)
ifm1=f(m1,M2); M1=f(M1,m2); m2=g(M1,m2) and M2=f(m1,M2) imply m1=M1 and m2=M2.
Then, system (2) has a unique positive equilibrium point (x,y),  and every solution of system (2) converges to (x,y).
The equilibrium points of system (1) are the solutions of the following system:
(3)x¯=(1-α)x¯+βx¯(1-x¯)e-(x¯+y¯),y¯=(1-α)y¯+βy¯(1-y¯)e-(x¯+y¯)
or(4)x¯=0,α=β(1-x¯)e-(x¯+y¯),y¯=0,α=β(1-y¯)e-(x¯+y¯).
El-Metwally and Elsadany [19] have shown that system (1) has the equilibrium points (0,0),(x,0),(0,x), and (u,u) where x and u satisfy α=β(1-x)e-x and α=β(1-u)e-2u, respectively, and proved that (i)
the equilibrium point(0,0) of system (1) is locally asymptotically stable if β<α<β+2 and unstable if β>α;(ii)
the equilibrium points(x,0) and (0,x) of system (1) are unstable;(iii)
the equilibrium point(u,u) of system (1) is stable if β<(4α2/(α-2+4-4α+9α2))e(1/2α)(2+3α-4-4α+9α2) and unstable if β>(4α2/(α-2+4-4α+9α2))e(1/2α)(2+3α-4-4α+9α2).


## 2. Boundedness and Invariant

In this section, we concern ourselves with the boundedness character of solutions of system (1). Under appropriate conditions, we give some bounded results related to system (1).Theorem 1.
Assume that{(xn,yn)}n=0∞ is a solution of system (1). Then,(i)
x0=y0 implies that xn=yn for all n=0,1,…,(ii)
1-α+β[1-(x0+y0)]e-(x0+y0)=0 implies that xn=yn for all n=0,1,…,(iii)
ifα+β/e2<1, then xn>yn(xn<yn)for all n=0,1,… provided that x0>y0(x0<y0).Proof.
It follows from (1) that
(5)xn+1-yn+1=(xn-yn)[1-α+β(1-(xn+yn))e-(xn+yn)].
So Cases (i) and (ii) are immediately proved. Now set
(6)f(x)=1-α+β(1-x)e-x.
Then,
(7)f′(x)=β(x-2)e-x.
Therefore, f(2) is the absolute minimum of f(x). That is,
(8)f(x)≥f(2)=1-α-βe2>0.
Note that (5) implies
(9)xn+1-yn+1=(xn-yn)f(xn+yn),
and, hence, (xn+1-yn+1) has the same sign of (xn-yn) for all n>0. The proof is so complete.Remark 2.
Theorem1 reduces system (1) into the following single difference equation:
(10)xn+1=xn[1-α+β(1-xn)]e-2xn.Theorem 3.
Assume thatα+β/e2<1. Then, every solution {(xn,yn)}n=0∞ of system (1) with x0>0 and y0>0 satisfies that xn>0 and yn>0 for all n>0.Proof.
LetH(x,y) be a continuous function defined by
(11)H(x,y)=1-α+β(1-x)e-(x+y).
Then, system (1) can be rewritten in the form
(12)xn+1=xnH(xn,yn),yn+1=ynH(yn,xn).
Now assume that {(xn,yn)}n=0∞ is a solution of system (1) with positive initial values. Then, it suffices to show that H(x,y) is positive for all x>0,y>0. Observe that
(13)∂H(x,y)∂x=β(x-2)e-(x+y),∂H(x,y)∂y=-β(1-x)e-(x+y).
Therefore, H has no positive critical points. Let a and b be arbitrary positive numbers and consider the domain
(14)D={(x,y):0≤x≤a,0≤y≤b}.
Then,
(15)H(0,y)=1-α+βe-y,0≤y≤b,H(x,0)=1-α+β(1-x)e-x,0≤x≤a,H(x,b)=1-α+β(1-x)e-(x+b),0≤x≤a,H(a,y)=1-α+β(1-a)e-(a+y),0≤y≤b.
Using elementary differential calculus, we obtain that the absolute minimum of each one of the above functions is 1-α-β/e2. Therefore, H(x,y)≥1-α-β/e2>0for all(x,y)∈D. Since a and b are arbitrary positive numbers, we can conclude that H(x,y)>0for all(x,y)∈(0,∞)2.Theorem 4.
Assume that{(xn,yn)}n=0∞ is a solution of system (1) with (xn0,yn0)∈(0,1]2 for some n0≥0. Assume also that one of the following statements is true. (i)
β≤e(1-α).(ii)
e(1-α)<β≤e.(iii)
(β-1)2≤α.Then,(xn,yn)∈(0,1]2 for all n≥n0.Proof.
Letn0≥0 be such that xn0∈(0,1]. It follows from system (1) that
(16)xn0+1≤(1-α)xn0+β(1-xn0)xn0e-xn0,(17)xn0+1≤(1-α)xn0+β(1-xn0)e-1xn0+1=(1-α-βe)xn0+βe,(18)xn0+1≤(1-α)xn0+β(1-xn0)xn0.
Set w(x)=(1-α)x+β(1-x)xe-x for x≤1. Then, it follows from (16) that xn0+1≤w(xn0). Also, we obtain that
(19)w′(x)=(1-α)+β(x2-3x+1)e-x,w′′(x)=β(-x2+5x-4)e-x=-β(x-1)(x-4)e-xw′′(x)≤0∀x∈(0,1].
Then, w′(x)≥w′(1)=1-α-β/e. If (i) holds, then w′(1)≥0, and hence w(x) is increasing on (0,1]. Therefore, xn0+1≤w(1)<1. If (ii) holds, then (17) yields xn0+1≤β/e<1.
Now suppose that (iii) holds. In this case, it follows from (18) that xn0+1≤p(xn0), where p(x)=(1-α)x+βx(1-x)for all x∈(0,1]. It is not difficult to see that p(x*) is the absolute maximum of p(x) on (0,1] where x*=(1-α+β)/2β. According to (iii) and since p(x*)=(1-α+β)2/4β≤1,xn0+1≤p(x*)≤1. That is in all cases we obtain that whenever xn0≤1 gives xn0+1≤1. So it is easy to prove by induction that xn∈(0,1] for all n≥1. The proof of yn is similar and so will be omitted. This completes the proof.Theorem 5.
For every solution{(xn,yn)}n=0∞ of system (1), the following statements hold. (i)
xn≤xn0(1-α)n-n0+(β/eα)(1-(1-α)n-n0), n≥n0≥0.(ii)
yn≤yn0(1-α)n-n0+(β/eα)(1-(1-α)n-n0), n≥n0≥0.Proof.
We obtain, forn0≥0, from (1) that
(20)xn+1≤(1-α)xn+βxne-xnxn+1≤(1-α)xn+βe,∀n≥n0.
Then, it follows by Theorems 3 and 4 that Case (i) is true. The proof of Case (ii) is similar and so will be omitted.The following corollaries are coming immediately from Theorem5.Corollary 6.
Assume that{(xn,yn)}n=0∞ is a positive solution of system (1) with (xn0,yn0)∈(0,β/αe]2 for n0≥0. Then, (xn,yn)∈(0,β/αe]2 for all n≥n0.Corollary 7.
Every positive solution{(xn,yn)}n=0∞ of system (1) is bounded. Moreover,
(21)limsupn→∞xn≤βαe,limsupn→∞yn≤βαe.
In Theorem 4, we gave conditions under which every positive solution {(xn,yn)}n=0∞ of system (1) to be in (0,1]2 provided that (xn0,yn0)∈(0,1]2. In the next result we show that every positive solution {(xn,yn)}n=0∞ of system (1) eventually lies in (0,1]2.Theorem 8.
Assume that{(xn,yn)}n=0∞ is a positive solution of system (1), and assume that one of the following conditions is true.(i)
β<αe.(ii)
2+ν2+2eν-4ν-νeν>0,1-α+βe-ν[1-ν(2e-ν+1)+ν2e-ν] and (1-α)ν+βνe-ν-βν2e-2ν<1, where ν=β/αe.Then, there existsn0≥0 such that (xn,yn)∈(0,1]2 for all n≥n0.Proof.
The proof of the theorem, when (i) holds, is followed by Corollary7. Now consider that (ii) is true. Then, it follows from Corollary 7 that for every constant ε>0, there exists n0≥0 such that xn≤β/αe+ε=γ,n≥n0. Set δ=e-γ. Since δ→e-ν when ε→0 and the inequalities in (ii) hold, depending on the continuity in ν of the left hand side of each inequality in (ii), one can choose ε so small such that
(22)2+γ2+2-γδ-4γ≥0,(23)1-α+βδ[1-(2δ+1)γ+δγ2]≥0,(24)(1-α)γ+βγδ-βγ2δ2≤1.
Now, we obtain from (1) that
(25)xn+1=(1-α)xn+βxne-(xn+yn)-βxn2e-(xn+yn)xn+1≤(1-α)xn+βxne-xn-βδxn2e-xnxn+1=K(xn),n≥n0,
where K(x)=(1-α)x+βe-x(x-δx2),x≤γ, and then
(26)K′(x)=1-α+βe-x[δx2-(2δ+1)x+1],K′′(x)=-βe-x[δx2-(4δ+1)x+2(δ+1)].
On the other hand, the equation
(27)δx2-(4δ+1)x+2(δ+1)=0,
has the positive roots
(28)x1=4δ+1+8δ2+12δ,x2=4δ+1-8δ2+12δ.
Observe that x2=2+(1/2δ)-2+1/4δ2≥γ if and only if (2+(1/2δ)-γ)2≥2+1/4δ2 which holds by (22). Therefore, x1≥x2≥γ. Consequently, K′′(x)<0 for all x≤γ which yields by (23) that K′(x)>K′(γ)≥0. Using the increasing property of K(x) on (0,γ) and inequality (24), we see that K(x)≤K(γ)≤1. Since xn≤γ, it follows that
(29)xn+1≤K(xn)≤K(γ)≤1∀n≥n0.
This completes the proof.Theorem 9.
Assume that{(xn,yn)}n=0∞ is a positive solution of system (1). If either
(30)(1-α+β)2<4βe-2ν
or
(31)βeν4+β(1-α)<1,
where ν=β/αe, then there exists n0≥0 such that (xn,yn)∈(0,1]2 for all n≥n0.Proof.
Assume thatγ,δ, and the function K(xn) are defined as in the previous proof. Then,
(32)K(xn)=(1-α+β)xn-βxn2δ2=K¯(xn),
where K¯(x)=(1-α+β)x-βx2δ2,x≤γ. Thus,
(33)K′¯(x)=1-α+β-2βxδ2.
Hence, K¯(x) attains its maximum value at x=(1-α+β)/2βδ2; that is,
(34)K¯(x)≤K¯(1-α+β2βδ2)=(1-α+β)24βδ2.
Also,
(35)K(xn)=-βδe-xn(xn-12δ)2+βe-xn4δ+β(1-α)K(xn)<β4δ+β(1-α),n≥n0.
Similar to the proof of Theorem 8, we can choose ε so small such that our assumptions imply that
(36)(1-α+β)22βδ2≤1,β4δ+β(1-α)≤1.
Therefore, we have either
(37)xn+1≤K¯(xn)≤(1-α+β)22βδ2≤1,n≥n0
or
(38)xn+1≤K(xn)≤β4δ+β(1-α)≤1,n≥n0,
which is our desired conclusion for xn. Similarly, one can accomplish the same conclusion for yn. The proof is so complete.


## 3. Global Stability Analysis

In this section, we are interested in establishing conditions under which the equilibrium points of system (1) are to be the attractors of the solutions of system (1).In the following theorem, we investigate the global attractivity of the equilibrium point(0,0) of system (1).Theorem 10.
Assume thatα≥β. Then, (0,0) is a global attractor of all positive solutions of system (1).Proof.
Let{(xn,yn)}n=0∞ be a solution of system (1). It follows from system (1) that
(39)xn+1=(1-α)xn+βxn(1-xn)e-(xn+yn)xn+1<(1-α+β)xn<xn,yn+1=(1-α)yn+βyn(1-yn)e-(xn+yn)yn+1<(1-α+β)yn<yn.
Then, there exist x≥0 and y≥0 such that limn→∞xn=x and limn→∞yn=y. Since the only possible values of (x,y) in the present case are (0,0),limn→∞xn=0 and limn→∞yn=0. This completes the proof.In the following theorems, we investigate the global attractivity of the positive equilibrium point(x¯;x¯) of system (1) where x¯ is given by α=β(1-x¯)e-2x¯.Theorem 11.
Assume thatα+βe-2<1. Then the unique positive equilibrium point (x¯;x¯) of system (1) is a global attractor of all positive solutions of system (1).Proof.
Let{(xn,yn)}n=0∞ be a solution of system (1), and let xn≤x¯ (the case xn≥x¯ is similar, and it will be left to the reader).
Now there are two cases to consider.
Case
1. Assume that x0≥y0. Then, it follows by Theorem 1 that xn<yn for all n≥1. Since xn≤x¯, then h(xn)≤0, where h(xn)=α-β(1-xn)e-2xn. Thus, α≤β(1-xn)e-2xn. Therefore, we obtain from system (1) that
(40)xn+1=(1-α)xn+βxn(1-xn)e-(xn+yn)xn+1≥(1-α)xn+αxne2xne-(xn+yn)xn+1=(1-α)xn+αxnexn-ynxn+1≥(1-α)xn+αxn=xn.
Then, the sequence {xn}n=0∞ is increasing, and since it was shown that it is bounded above, then it converges to the only positive equilibrium point x¯, and it follows by the comparison test of convergence for sequence that {yn}n=0∞ is also convergent to the only positive equilibrium point y¯=x¯:. Thus, {(xn,yn)}n=0∞ converges to (x¯;x¯).
Case
2. Assume that x0<y0. Then, it follows from system (1) and Theorem 1 that
(41)yn+1=(1-α)yn+βyn(1-yn)e-(xn+yn)yn+1≥(1-α)yn+αyne2yne-(xn+yn)yn+1=(1-α)yn+αyneyn-xnyn+1≥(1-α)yn+αyn=yn.
The rest of the proof is similar to Case 1, and it will be left to the reader.Theorem 12.
Assume thatβ(αe-β)≥α2e3. Then, the unique positive equilibrium point (x¯;x¯) of system (1) is a global attractor of all positive solutions of system (1).Proof.
Let{(xn,yn)}n=0∞ be a solution of system (1). It follows from system (1) that
(42)xn+1=(1-α)xn+βxn(1-xn)e-(xn+yn)xn+1≥(1-α)xn+βxn(1-xn)e-2.
Thus, we see from Corollary 7 that
(43)xn+1≥[1-α+β(1-βαe)e-2]xn≥xn.
Then, the sequence {xn}n=0∞ is increasing, and since it is bounded, then it converges to the only positive equilibrium point x¯. Similarly, it is easy to show that the sequence {yn}n=0∞ is also convergent to the unique positive equilibrium point y¯=x¯:. Therefore {(xn,yn)}n=0∞ converges to (x¯,x¯), and then the proof is so complete.Theorem 13.
Assume that one of the following conditions hold.(I)
5β≤4e2(1-α).(II)
α+β<1. Then the unique positive equilibrium point (x¯,x¯) of system (1) is a global attractor of all positive solutions of system (1).Proof.
Rewrite system (1) as follows:
(44)xn+1=F(xn,yn)xn+1=(1-α)xn+β(1-xn)xne-(xn+yn),yn+1=G(xn,yn)yn+1=(1-α)yn+β(1-yn)yne-(xn+yn),n=0,1,…,
where F(x,y)=(1-α)x+β(1-x)xe-(x+y) and G(x,y)=(1-α)y+β(1-y)ye-(x+y) are continuous functions. Now, consider the system
(45)m1=F(m1,M2),M1=F(M1,m2),m2=G(M1,m2),M2=F(m1,M2).
Then,
(46)m1=(1-α)m1+βm1(1-m1)e-(m1+M2),M1=(1-α)M1+βM1(1-M1)e-(M1+m2),m2=(1-α)m2+βm2(1-m2)e-(m2+M1),M2=(1-α)M2+βM2(1-M2)e-(M2+m1).
Thus, either m1=M2=m2=M2 or
(47)α=β(1-m1)e-(m1+M2),α=β(1-M1)e-(M1+m2),α=β(1-m2)e-(m2+M1),α=β(1-M2)e-(M2+m1).
Then, m1=M2, m2=M2 and
(48)(1-m1)e-2m1=(1-M1)e-2M1=(1-m2)e-2m2=(1-M2)e-2M2.
Now, since (1-m1)e-2m1=(1-M1)e-2M1, then e2(M1-m1)=(1-M1)/(1-m1); that is,
(49)2(M1-m1)=log(1-M1)-log(1-m1).
We claim that M1=m1; otherwise for the sake of contradiction assume that M1>m1 (the case where M1<m1 is similar and it will be left to the reader). Then, log(1-M1)-log(1-m1)<0, which contradicts that the left hand side of (49) is positive.
Now it easy to see that(50)∂F(x,y)∂x=1-α+β(x2-3x+1)e-(x+y),∂F(x,y)∂y=-xβ(1-x)e-(x+y),∂G(x,y)∂x=-βy(1-y)e-(x+y),∂G(x,y)∂y=1-α+β(y2-3y+1)e-(x+y).
Thus,
(51)∂F(x,y)∂x=1-α+β(x2-3x+1)e-(x+y)∂F(x,y)∂x≥1-α+β(x2-3x+1)e-2∂F(x,y)∂x=βe-2x2-3βe-2x+βe-2+1-α.
Now, there are two cases to consider.
Case
1. Suppose that 5β≤4e2(1-α). Therefore, the function w(x)=βe-2x2-3βe-2x+βe-2+1-α has no real roots. Thus, ∂F(x,y)/∂x≥0. Similarly, it is easy to prove that ∂G(x,y)/∂x≥0. Then, it follows by Theorem A that the equilibrium point (x¯,y¯)=(x¯,x¯) of system (1) is a global attractor of all positive solutions of system (1).
Case
2. Suppose that α+2β<1. Since 0≤x≤1,3≥3-x≥x(3-x)=3x-x2, or 2≥3x-x2-1, and since α+2β<1, then 1-α>2β>2βe-2≥2βe-(x+y)≥β(3x-x2-1)e-(x+y). Thus, ∂F(x,y)/∂x≥0. Similarly, it is easy to prove that ∂G(x,y)/∂x≥0. Then, it follows again by Theorem A that the equilibrium point (x¯,y¯)=(x¯,x¯) of system (1) is a global attractor of all positive solutions of system (1). Thus, the proof is now completed.



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