Complete Asymptotic Analysis of a Nonlinear Recurrence Relation with Threshold Control

doi:10.1155/2010/143849

Research Article

Complete Asymptotic Analysis of a Nonlinear

Recurrence Relation with Threshold Control

Qi Ge,

_{Chengmin Hou,}

_{and Sui Sun Cheng}

2_{Department of Mathematics, Tsing Hua University, Taiwan 30043, Taiwan}

Correspondence should be addressed to Qi Ge,[email protected] Received 23 November 2009; Accepted 10 January 2010

Academic Editor: Toka Diagana

Copyrightq2010 Qi Ge et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We consider a three-term nonlinear recurrence relation involving a nonlinear filtering function with a positive thresholdλ. We work out a complete asymptotic analysis for all solutions of this equation when the threshold varies from 0to∞. It is found that all solutions either tends to 0, a limit 1-cycle, or a limit 2-cycle, depending on whether the parameterλis smaller than, equal to, or greater than a critical value. It is hoped that techniques in this paper may be useful in explaining natural bifurcation phenomena and in the investigation of neural networks in which each neural unit is inherently governed by our nonlinear relation.

1. Introduction

LetN{0,1,2, . . .}.In1, Zhu and Huang discussed the periodic solutions of the following difference equation:

xnaxn−1 1−afλxn−k, n∈N, 1.1

wherea∈0,1, kis a positive integer, andf :R → Ris a nonlinear signal filtering function of the form

fλx ⎧ ⎨ ⎩

1, x∈0, λ,

0, x∈−∞,0∪λ,∞, 1.2

In this paper, we consider the following delay difference equation:

xnaxn−2bfλxn−1, n∈N, 1.3

where a ∈ 0,1and b > 0. Besides the obvious and complementary differences between

1.1and our equation, a good reason for studying 1.3 is that the study of its behavior is preparatory to better understanding of more general neuralnetwork models. Another one is that there are only limited materials on basic asymptotic behavior of discrete time dynamical systems with piecewise smooth nonlinearities! Besides 1, see 2–6. In particular, in2, Chen considers the equation

xnxn−1gxn−k−1, n∈N, 1.4

wherekis a nonnegative integer andg:R → Ris a McCulloch-Pitts type function

gξ

⎧ ⎨ ⎩

−1, ξ∈σ,∞,

1, ξ∈−∞, σ, 1.5

in whichσ∈Ris a constant which acts as a threshold. In3, convergence and periodicity of solutions of a discrete time network model of two neurons with Heaviside type nonlinearity are considered, while “polymodal” discrete systems in4are discussed in general settings. Therefore, a complete asymptotic analysis of our equation is essential to further development of polymodal discrete time dynamical systems.

We need to be more precise about the statements to be made later. To this end, we first note that givenx₋2, x−1 ∈R,we may compute from1.3the numbersx0, x1, x2, . . .in a unique

manner. The corresponding sequence{xn}∞n−2is called the solution of1.3determined by

the initial vectorx−2, x−1.For better description of latter results, we consider initial vectors

in different regions in the plane. In particular, we set

Ω x, y∈R2_{|x >0 ory >0, 1.6}

which is the complement of nonpositive orthant−∞,02 and contains the positive orthant

0,∞2_{.Note thatΩis the union of the disjoint sets}

U 0,∞2_{\0, λ}2_{, 1.7}

V 0, λ2_{∪−∞,0×0,∞∪0,∞×−∞,0. 1.8}

Recall also that a positive integerηis a period of the sequence{wn}∞nαifwηnwnfor

alln≥αand thatτis the least or prime period of{wn}∞nαifτ is the least among all periods

of{wn}∞nα.The sequence{wn}n∞αis said to beτ-periodic ifτ is the least period of{wn}∞nα.

w0_{, w}1_{, . . . , w}ω−1_{,whereωis a positive integer, such that}

lim

n→ ∞wωniw

i_{, i0,1, . . . , ω−1. 1.9}

In case {w0_{, w}1_{, . . . , w}ω−1_{, w}0_{, w}1_{, . . . , w}ω−1_{, . . .} is an ω-periodic sequence, we say}

thatw is an asymptoticallyω-periodic sequence tending to the limitω-cycleThis term is introduced since the underlying concept is similar to that of the limit cycle in the theory of ordinary differential equations. w0_{, w}1_{, . . . , w}ω−1_{.In particular, an asymptotically}

1-periodic sequence is a convergent sequence and conversely.

Note that1.3is equivalent to the following two-dimensionalautonomousdynamical system:

un1

vn1

vn

aunbfλvn

, n∈N, 1.10

by means of the identification xn un2 forn ∈ {−2,−1,0, . . .}. Therefore our subsequent

results can be interpreted in terms of the dynamics of plane vector sequences defined by

1.10. For the sake of simplicity, such interpretations will be left in the concluding section of this paper.

To obtain complete asymptotic behavior of1.3, we need to derive results for solutions of1.3determined by vectors in the entire plane. The following easy result can help us to concentrate on solutions determined by vectors inΩ.

Theorem 1.1. A solution{xn}∞n−2 of 1.3withx−2, x−1in the nonpositive orthant−∞,02 is nonpositive and tends to0.

Proof. Letx−2, x−1≤0.Then by1.3,

x0ax−2bfλx−1 ax−2 ≤0,

x1ax−1bfλx0 ax−1≤0,

x2ax0bfλx1 a2x−2≤0,

x3ax1bfλx2 a2x−1≤0,

1.11

and by induction, for anyk∈N,we have

x2kak1x−2≤0,

x2k1ak1x−1≤0.

1.12

Sincea∈0,1,we have

lim

n→ ∞xn0. 1.13

Note that if we try to solve for an equilibrium solution{x}of1.3, then

x b

1−afλx, 1.14

which has exactly two solutions x 0, b/1−a whenλ ≥ b/1−aand has the unique solutionx 0 whenλ ∈ 0, b/1−a.However, sincefλ is a discontinuous function, the

standard theories that employ continuous arguments cannot be applied to our equilibrium solutionsx0 orb/1−ato yield a set of complete asymptotic criteria. Fortunately, we may resort to elementary arguments as to be seen below.

To this end, we first note that our equation is autonomoustime invariant, and hence if{xn}∞n−2is a solution of1.3, then for anyk ∈ N,the sequence{yn}∞n−2,defined byyn

xnkforn−2,−1,0, . . . ,is also a solution. For the sake of convenience, we need to let

A0 λ−_ab,

aAj1bAj, j∈N.

1.15

Then

Aj λ−b

1a· · ·aj aj1

λ1−a−b aj1_1−a

b

1−a, j ∈N, 1.16

Aj1−Aj λ1−_aja₂−b, j∈N, 1.17

λaA0ba2A1abb· · ·aj1Ajajbaj−1b· · ·abb, j∈N. 1.18

We also let

B0−

b a, aBj1bBj, j∈N.

1.19

Then

Bj −b

1a· · ·aj aj1

−baj1_b

aj1_1−a, j∈N, 1.20

Bj1−Bj−

b

aj2, j ∈N, 1.21

aB0ba2B1abb· · ·aj1Bjajb· · ·abb, j∈N, 1.22

lim

2. The Case

λ > b/

1

−

a

Supposeλ > b/1−a.Then

lim

j→ ∞Aj jlim→ ∞

_λ

1−a−b aj1_1−a

b 1−a

∞. 2.1

We first show the following.

Lemma 2.1. Letλ > b/1−a.If{xn}∞n−2 is a solution of 1.3withx−2, x−1 ∈ Ω,then there exists an integerm∈ {−2,−1,0, . . .}such that0< xm, xm1≤λ.

Proof. From our assumption, we haveaλb < λ.Let {xn}∞n−2 be a solution of 1.3with

x−2, x−1∈Ω.Then there are eight cases.

Case 1. If 0< x−2, x−1≤λ,our assertion is true by takingm−2.

Case 2. Suppose x₋2, x−1 ∈ 0, λ×λ,∞.Thenλ−b/a > λ.Furthermore, in view of

1.17and2.1,

0, λ×λ,∞ 0, λ×λ,λ−b a

∪∞

k1

Ak−1, Ak

. 2.2

Ifx−1∈λ,λ−b/a,then by1.3,

x0ax−2bfλx−1 ax−2∈0, λ,

0< x1ax−1bfλx0 ax−1b≤λ−bbλ.

2.3

This means that our assertion is true by takingm0.Next, ifx₋1 ∈A0, A1 λ−b/a,λ−

b−ab/a2_{,then by1.3and1.18,}

x0 ax−2bfλx−1 ax−2∈0, λ,

x1 ax−1bfλx0 ax−1b > λ,

x2 ax0bfλx1 a2x−2 ∈0, λ,

0< x3ax1bfλx2 a2x−1abb

≤a2_A

1abbλ.

Thus our assertion holds by takingm2.Ifx−1∈Ap, Ap1,wherepis an arbitrary positive

integer, then by1.3,

x0ax−2bfλx−1 ax−2∈0, λ,

x1ax−1bfλx0 ax−1b > aApbAp−1> λ,

x2ax0bfλx1 a2x−2∈0, λ.

2.5

By induction,

x2pap1x−2∈0, λ,

x2p1ap1x−1apb· · ·abb > ap1Apapb· · ·abb

ap_A

p−1ap−1b· · ·abbaA0bλ,

x2p1ax2pbfλx2p1

ap2_x

−2∈0, λ,

x2p11ax2p1bfλ

x2p1

ap2_x

−1ap1b· · ·abb

≤ap2_A

p1ap1b· · ·abbλ.

2.6

Thus our assertion holds by takingm2p1.

Case 3. Supposex₋2, x−1∈λ,∞×λ,∞.We assert that there is a nonnegative integerμ

such thatxn > λforn−2,−1, . . . , μ−1 andxμ ∈0, λ.Otherwise we havexn ∈λ,∞for

n∈N.It follows that

x0 ax−2bfλx−1 ax−2> λ,

x1ax−1bfλx0 ax−1> λ,

x2ax0bfλx1 a2x−2> λ,

x3ax1bfλx2 a2x−1> λ.

2.7

By induction, for anyk∈N,we have

x2kak1x−2 > λ,

x2k1ak1x−1> λ,

2.8

which implies

lim

k→ ∞x2k0klim→ ∞x2k1. 2.9

Now that there exists an integerμ∈Nsuch thatx−2, x−1, . . . , xμ−1 ∈λ,∞andxμ ∈

0, λ,it then follows

0< xμ1axμ−1bfλxμaxμ−1b. 2.10

Ifxμ1≤λ,then our assertion holds by takingmμ.Ifxμ1> λ,thenxμ−1>λ−b/a.

Thus

0< xμ2axμbfλxμ1

axμ≤aλ < λ,

0< xμ3axμ1bfλxμ2

axμ1ba2xμ−1abb.

2.11

If 0 < xμ3 ≤ λ,then our assertion holds by takingm μ2.Ifxμ3 > λ,we havexμ−1 >

λ−b−ab/a2_.Hence

0< xμ4axμ2bfλxμ3

axμ2a2xμ< λ,

0< xμ5axμ3bfλxμ4

axμ3ba3xμ−1a2babb.

2.12

Repeating the procedure, we have

0< xμ2kakxμ< λ,

0< xμ2k1 ak1xμ−1akb· · ·abb.

2.13

If 0< xμ2k1≤λ,then our assertion holds by takingmμ 2k1.Otherwise,

xμ−1>

λ−b−ab− · · · −ak_b

ak1 2.14

for allk∈N.But this is contrary to2.1. Thus we conclude that 0< xμ2k1≤λfor somek.

Our assertion then holds by takingmμ2k.

Case 4. Supposex₋2, x−1∈λ,∞×0, λ.As in Case2,

λ,∞×0, λ λ,λ−b a

∪∞

k1

Ak−1, Ak

×0, λ. 2.15

Ifx−2∈λ,λ−b/a,then by1.3,

Thus our assertion holds takingm−1.Ifx−2∈A0, A1,then by1.3,

Thus our assertion holds by takingm2p1.

Thus our assertion holds by takingm1.Ifx−2 ∈Bp1, Bp,wherepis an arbitrary positive

arbitrary nonnegative integer, then by1.3, we have

Ifx−1∈−b/a,0,then by1.3,

0< x0ax−2bfλx−1 ax−2 < λ,

0< x1ax−1bfλx0 ax−1b≤b < λ.

2.28

Thus our assertion holds by takingm 0.Ifx−1 ∈ B1, B0 −b−ab/a2,−b/a,then by

1.3,

0< x0ax−2bfλx−1 ax−2 < λ,

−b a

−b−ab

a b < x1ax−1bfλx0 ax−1b≤0.

2.29

That is,x0, x1∈0, λ×−b/a,0.Thus our assertion holds by takingm2.

Ifx−1∈Bp1, Bp.wherepis an arbitrary positive integer, then by1.3, we have

0< x0ax−2bfλx−1 ax−2< λ ,

Bp< x1ax−1bfλx0 ax−1b≤Bp−1.

2.30

That is,x0, x1∈0, λ×Bp, Bp−1.Thus our assertion follows from induction. Case 8. Supposex−2, x−1∈λ,∞×−∞,0.Then

λ,∞×−∞,0

∞

k0 _λ

ak,

λ ak1

×−∞,0. 2.31

Ifx−2∈λ, λ/a,then by1.3,

0< x0ax−2bfλx−1 ax−2≤λ. 2.32

That is,x₋1, x0∈−∞,0×0, λ.We may now apply the assertion in Case5to conclude our

proof. Ifx−2∈λ/ap1, λ/ap2,wherepis an arbitrary nonnegative integer, then by1.3, we

have

λ

ap < x0 ax−2bfλx−1 ax−2≤

λ ap1 ,

x1ax−1bfλx0 ax−1≤0.

2.33

That is, x0, x1 ∈ λ/ap, λ/ap1 × −∞,0. We may thus complete our proof by

induction.

Proof. In view ofLemma 2.1, we may assume without loss of generality that 0< x−2, x−1≤λ.

From our assumption, we haveaλb < λ.Furthermore, by1.3,

0< x0ax−2bfλx−1 ax−2b≤aλb < λ,

0< x1ax−1bfλx0 ax−1b≤aλb < λ,

0< x2ax0bfλx1 a2x−2abb≤a2λabbaaλb b < aλb < λ,

0< x3ax1bfλx2 a2x−1abb≤a2λabbaaλb b < aλb < λ,

0< x4ax2bfλx3 a3x−2a2babb≤a3λa2babb

a2_{aλb abb < a}2_{λabb < λ,}

0< x5ax3bfλx4 a3x−1a2babb≤a3λa2babb

a2_{aλb abb < a}2_{λabb < λ.}

2.34

By induction, for anyk∈N,we have

0< x2kak1x−2akbak−1b· · ·abb

≤ak1_λak_bak−1_{b· · ·abba}k_{aλb a}k−1_{b· · ·abb}

< ak_λak−1_{b· · ·abba}k−1_{aλb a}k−2_{b· · ·abb}

< ak−1_λak−2_{b· · ·abb <· · ·< a}2_{λabb < λ,}

2.35

and similarly

0< x2k1ak1x−1akbak−1b· · ·abb < λ. 2.36

Thusx2k, x2k1∈0, λfor anyk∈Nand

lim

k→ ∞x2kklim→ ∞

ak1_x

−2b×1−

ak1

1−a

b

1−a,

lim

k→ ∞x2k1 klim→ ∞

ak1_x

−1b×1−

ak1

1−a

b

1−a.

2.37

The proof is complete.

3. The Case

λ

∈

0

, b/

1

−

a

Lemma 3.1. Let0 < λ < b/1−a.Ifx {xn}∞n−2 is a solution of 1.3with x−2, x−1 ∈ Ω, there exists an integerm ∈ {−2,−1,0, . . .}such that0 < xm ≤ λand xm1 > λ(or xm > λand

0< xm1≤λ).

Proof. From our assumption, we have aλ b > λ. Let {xn}∞n−2 be the solution of 1.3

determined byx₋2, x−1∈Ω.Then there are eight cases to show that there exists an integer

m∈ {−2,−1,0, . . .}such that 0< xm≤λandxm1> λ.

Case 1. Supposex−2, x−1∈0, λ×λ,∞.Then our assertion is true by takingm−2. Case 2. Supposex−2, x−1∈λ,∞×0, λ.By1.3

x0ax−2bfλx−1 ax−2b > aλb > λ. 3.1

This means that our assertion is true by takingm−1.

Case 3. Supposex−2, x−1∈0, λ×0, λ.Ifxn∈0, λfor anyn∈N,then by1.3,

x0ax−2bfλx−1 ax−2b,

x1ax−1bfλx0 ax−1b,

x2ax0bfλx1 a2x−2abb,

x3ax1bfλx2 a2x−1abb.

3.2

By induction, for anyk∈N,we have

x2kak1x−2akb· · ·abbak1x−2b× 1−a k1

1−a ,

x2k1ak1x−1akb· · ·abbak1x−1b×

1−ak1

1−a .

3.3

Hence

lim

k→ ∞x2k

b

1−aklim→ ∞x2k1. 3.4

But this is contrary to our assumption that 0 < λ < b/1−a.Hence there exists an integer μ ∈ {−1,0,1, . . .}such that x−2, x−1, . . . , xμ ∈ 0, λand xμ1 ∈ λ,∞.Thus our assertion

holds by takingmμ.

Case 4. Supposex₋2, x−1∈λ,∞×λ,∞.As in Case3ofLemma 2.1, we may show that

ifxn∈λ,∞for alln∈N,then it follows that

lim

But this is contrary to the fact thatxn∈λ,∞forn∈N.Hence there exists an integerμ∈N

such thatx₋2, x−1, . . . , xμ−1∈λ,∞andxμ∈0, λ,it then follows

xμ1axμ−1bfλxμaxμ−1b > aλb > λ. 3.6

This means that our assertion is true by takingmμ.

Case 5. Supposex₋2, x−1∈−∞,0×0, λ. Then by1.21and1.23,

−∞,0×0, λ

⎧ ⎨ ⎩

⎛ ⎝∞

j1

Bj, Bj−1 ⎞_⎠

∪

−_ab,0

⎫_⎬

⎭×0, λ. 3.7

Ifx−2∈−b/a,0,then by1.3,

x0ax−2bfλx−1 ax−2b >−bb0. 3.8

Whenλ≥b,we have

0< x0ax−2b≤b≤λ. 3.9

That is,x−1, x0∈0, λ×0, λ.We may thus apply the conclusion of Case3to deduce our

assertion.

Supposeλ < b.If−b/a < x−2≤λ−b/a <0,then we have

0< x0ax−2b≤λ. 3.10

We may apply the conclusion of Case3to deduce our assertion. Ifλ−b/a < x−2 ≤ 0,we

have

x0ax−2b > λ. 3.11

Thus our assertion holds by takingm−1.Ifx−2 ∈B1, B0 −b−ab/a2,−b/a,then by

1.3,

B0aB1b < x0ax−2bfλx−1 ax−2b≤0,

0< x1ax−1bfλx0 ax−1< λ.

That is,x0, x1∈−b/a,0×0, λ.In view of the above discussions, our assertion is true. If

x₋2∈Bp1, Bp,wherepis an arbitrary positive integer, then by1.3, we have

BpaBp1b < x0ax−2bfλx−1 ax−2b≤aBpbBp−1,

0< x1ax−1bfλx0 ax−1< λ.

3.13

That is,x0, x1∈Bp, Bp−1×0, λ.Therefore we may conclude our assertion by induction. Case 6. Suppose

x₋2, x−1∈−∞,0×λ,∞ −∞,0× ∞

k0 _λ

ak,

λ ak1

. 3.14

As in Case6ofLemma 2.1, ifx−1∈λ, λ/a,then by1.3, we havex0, x1∈−∞,0×0, λ.

We may thus apply the conclusion of Case5to deduce our assertion. Ifx₋1∈λ/ap1, λ/ap2,

where p is an arbitrary nonnegative integer, then by 1.3, we have x0, x1 ∈ −∞,0 ×

λ/ap_{, λ/a}p1_{.We may thus use induction to conclude our assertion.} Case 7. Supposex₋2, x−1∈0, λ×−∞,0.By1.3, we have

0< x0ax−2bfλx−1 ax−2< λ. 3.15

That is,x₋1, x0∈−∞,0×0, λ.We may thus apply the conclusion of Case5to deduce our

assertion.

Case 8. Supposex−2, x−1∈λ,∞×−∞,0.Then

λ,∞×−∞,0

∞

k0 _λ

ak,

λ ak1

×−∞,0. 3.16

As in Case8ofLemma 2.1, ifx−2∈λ, λ/a,then by1.3, we havex−1, x0∈−∞,0×0, λ.

We may now apply the assertion in Case5to conclude our proof. Ifx₋2 ∈ λ/ap1, λ/ap2,

wherepis an arbitrary nonnegative integer, then by1.3, we havex0, x1∈λ/ap, λ/ap1×

−∞,0.We may thus complete our proof by induction.

Proof. In view ofLemma 3.1, we may assume without loss of generality that 0< x−2≤λand

x₋1> λ.Then by1.3,

0< x0ax−2bfλx−1 ax−2< λ,

x1ax−1bfλx0 ax−1b > aλb > λ,

0< x2ax0bfλx1 a2x−2< λ,

x3ax1bfλx2 a2x−1abb > a2λabb

aaλb b > aλb > λ,

0< x4ax2bfλx3 a3x−2< λ,

x5ax3bfλx4 a3x−1a2babb > a3λa2babb

a2_{aλb abb > a}2_{λabb > λ.}

3.17

By induction, for anyk∈N,we have

0< x2kak1x−2< λ,

x2k1ak1x−1akb· · ·abb > ak1λakb· · ·abb

ak_{aλb a}k−1_{b· · ·abb > a}k_λak−1_{b· · ·abb}

>· · ·> a2_{λabb > λ.}

3.18

Thusx2k∈0, λandx2k1∈λ,∞for anyk∈N.Then

lim

k→ ∞x2kklim→ ∞a k1_x

−20,

lim

k→ ∞x2k1klim→ ∞

ak1_x −1b×

1−ak1

1−a

b

1−a.

3.19

4. The Case

λ

b/

1

−

a

Supposeλb/1−a.Thenλaλb > b.We need to consider solutions with initial vectors inUorV defined by1.7and1.8, respectively.

Lemma 4.1. Letλ b/1−a.If{xn}∞n−2 is a solution of 1.3withx−2, x−1 ∈ V, then there exists an integerm∈Nsuch that0< xm, xm1≤λ.

The proof is the same as the discussions in Cases5 through Case 8 in the proof of

Lemma 2.1, and hence is skipped.

Proof. In view ofLemma 4.1, we may assume without loss of generality that 0< x−2, x−1≤λ.

By1.3,

0< x0ax−2bfλx−1 ax−2b≤aλbλ,

0< x1ax−1bfλx0 ax−1b≤aλbλ,

0< x2ax0bfλx1 a2x−2abb≤a2λabbaaλb baλbλ,

0< x3ax1bfλx2 a2x−1abb≤a2λabbaaλb baλbλ,

0< x4ax2bfλx3 a3x−2a2babb≤a3λa2babb

a2_{aλb abba}2_λabbλ,

0< x5ax3bfλx4 a3x−1a2babb≤a3λa2babb

a2_{aλb abba}2_λabbλ.

4.1

By induction, for anyk∈N,we have

0< x2kak1x−2akbak−1b· · ·abb

≤ak1_λak_bak−1_{b· · ·abba}k_{aλb a}k−1_{b· · ·abb}

ak_λak−1_{b· · ·abba}k−1_{aλb a}k−2_{b· · ·abb}

ak−1_λak−2_{b· · ·abb· · ·a}2_λabbλ,

4.2

and similarly

0< x2k1ak1x−1akbak−1b· · ·abb≤λ. 4.3

Thusx2k, x2k1∈0, λfor anyk∈N.Thus2.37hold so that

lim

n→ ∞xn

b

1−a. 4.4

The proof is complete.

Proof. We first discuss the case, wherex−2, x−1∈0, λ×λ,∞.By1.3,

0< x0ax−2bfλx−1 ax−2< λ,

x1ax−1bfλx0 ax−1b > aλbλ,

0< x2ax0bfλx1 a2x−2< λ,

x3ax1bfλx2 a2x−1abb > a2λabb

aaλb baλbλ,

0< x4ax2bfλx3 a3x−2< λ,

x5ax3bfλx4 a3x−1a2babb > a3λa2babb

a2_{aλb abba}2_λabbλ.

4.5

By induction, for anyk∈N,we have

0< x2kak1x−2< λ,

x2k1ak1x−1akb· · ·abb > ak1λakb· · ·abbλ.

4.6

Thusx2k∈0, λandx2k1∈λ,∞for anyk∈N.Then

lim

k→ ∞x2k0,

lim

k→ ∞x2k1 klim→ ∞

ak1_x

−1b×1−a k1

1−a

b

1−a.

4.7

Ifx−2, x−1∈λ,∞×0, λ,then by1.3,

x0ax−2bfλx−1 ax−2b > aλbλ. 4.8

That is,x₋1, x0∈0, λ×λ,∞.We may thus apply the previous conclusion to deduce our

assertion.

Ifx−2, x−1∈λ,∞×λ,∞,then similar to the discussions of Case3ofLemma 2.1,

there exists an integerμ ∈ N such thatx−2, x−1, . . . , xμ−1 ∈ λ,∞andxμ ∈ 0, λ.That is,

xμ−1, xμ ∈ λ,∞×0, λ.In view of the previous case, our assertion holds. The proof is

complete.

5. Concluding Remarks

The results in the previous sections can be stated in terms of the two-dimensional dynamical system1.10. Indeed, a solution of1.10is a vector sequence of the form{un, vn†}∞n0that

Let us say that a solution{un, vn†}∞n0of1.10eventually falls into a plane regionΨ

ifun, vn† ∈Ψfor all largen; that it is eventually falls into two disjoint plane regionsΨ1and

Ψ2alternately if there is somem∈Nsuch thatum2i, vm2i†∈Ψ1andum2i1, vm2i1†∈Ψ2

for alli∈N; and that it approaches a limit 2-cycleα1, β1†,α2, β2†if there is somem∈N

such thatum2i, vm2i† → α1, β1†andum2i1, vm2i1† → α2, β2†asi → ∞.Then we

may restate the previous theorems as follows.

iThe vectors0,0†,0, b/1−a†,b/1−a, b/1−a†,andb/1−a,0†form the corners of a square in the plane.

iiA solution{un, vn†}∞n0of1.10withu0, v0†in the nonpositive orthant−∞,02

is nonpositive andtends to0,0†.

iiiSupposeλ > b/1−a, then a solution{un, vn†}∞n0of1.10withu0, v0† inΩ

willeventually falls into0, λ2andtend tob/1−a, b/1−a†.

ivSuppose 0 < λ < b/1−a, then a solution {un, vn†}∞n0 of1.10withu0, v0†

inΩwilleventually falls into0, λ×λ,∞andλ,∞×0, λalternately and approach the limit 2-cycle0, b/1−a†,b/1−a,0†.

vSuppose λ b/1−a, then a solution{un, vn†}∞n0 of1.10withu0, v0† inV

willeventually falls into0, λ2tend tob/1−a, b/1−a†.

viSupposeλ b/1−a, Then a solution{un, vn†}∞n0 of1.10withu0, v0†inU

willeventually falls into0, λ×λ,∞andλ,∞×0, λalternatelyapproach the limit 2-cycle0, b/1−a†,b/1−a,0†.

Since we have obtained a complete set of asymptotic criteria, we may deduce

bifurcationresults such as the following.

If 0< λ < b/1−a,then all solutions{un, vn†}∞n0originated from the positive orthant

approach the limit 2-cycle0, b/1−a†,b/1−a,0†; ifλ > b/1−a,then all solutions originated from the positive orthant tend to b/1−a, b/1−a†; if λ b/1 −a,then all solutions originated from the positive orthant tend tob/1−a, b/1−a†ifu0, v0† ∈

0, λ2_{and approach the limit cycle0, b/1−a}†_,b/1−a,0†_otherwise.

Roughly the above statements show that when the threshold parameterλis a relatively small positive parameter, all solutions from the positive orthant tend to a limit 2-cycle; when it reaches the critical valueb/1−a,some of these solutionsthose from0, b/1−a2switch away and tend to a limit 1-cycle, and whenλdrifts beyond the critical value, all solutions tend to the limit 1-cycle. Such an observation seems to appear in many natural processes and hence our model may be used to explain such phenomena. It is also expected that when a group of neural units interact with each other in a network where each unit is governed by evolutionary laws of the form 1.3, complex but manageable analytical results can be obtained. These will be left to other studies in the future.

Acknowledgment

References

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