# Bifurcation in a Discrete-Time Piecewise Constant Dynamical System

Authors: Chenmin Hou; Sui Sun Cheng
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

The study of recurrent neural networks with piecewise constant transition or control functions has attracted much attention recently because they can be used to simulate many physical phenomena. A recurrent and discontinuous two-state dynamical system involving a nonnegative bifurcation parameter is studied. By elementary but novel arguments, we are able to give a complete analysis on its asymptotic behavior when the parameter varies from 0 to∞. It is hoped that our analysis will provide motivation for further results on large-scale recurrent McCulloch-Pitts-type neural networks and piecewise continuous discrete-time dynamical systems.


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## Body

## 1. Introduction

It is generally accepted that the McCulloch-Pitts model of a neural network can be used as the components of computer-like systems. That is, where neural networks are commonly used to learn something, a McCulloch-Pitts neuron is constructed to do a particular job. Although the same job can be done by means of traditional Boolean components, it is interesting to see how it all works using components which are closer to “biological” components. One particular important component can be described by means of the step (activation) function defined by(1)fσ(u)={1ifu≤σ-1ifu>σ
with a nonnegative (threshold) real parameter σ. Roughly, a neuron may receive an inhibitory (indicated by a -1 sign) value if the input signal has strength exceeding the biological threshold σ and otherwise it remains intact with an excitory (indicated by a +1 sign) value.Therefore, if we letxn be the state value of a neural unit during the time period n, then the following recurrence relation:
(2)xn=F(fσ(xn-1),xn-2)
may be used to describe a one-neuron McCulloch-Pitts system where the state value is updated from the two most recent state values. There are now many studies that are concerned with such neural networks.For a neural network system that contains two or more neural units, things are much more complicated, and a variety of models can be designed. In usual practice, we can build a mathematical model and use it to generate simulation results, and from these results, we may understand the general properties of our models. What is now more difficult but important is to make mathematical conclusions that may provide a full understanding of all the properties observed.In this paper, we consider a two-neuron dynamical neural network system:(3)xn=axn-2+(1-a)fσ(yn-1),yn=ayn-2+(1-a)fσ(xn-1)
for n∈N={0,1,2,…}, where a∈(0,1), and fσ:R→R is the function in. Note that if we make the change of variables
(4)xn=1-abun,yn=1-abvn,
where b is an arbitrary positive number, then (3) becomes the “more general” system:
(5)un=aun-2+bfbσ/(1-a)(vn),vn=avn-2+bfbσ/(1-a)(un),
and hence all properties of (5) can be inferred from those of (3).In this paper, we are concerned with the asymptotic behavior of (3) when the nonnegative parameter σ is treated as a bifurcation variable and allowed to vary from 0 to ∞ (cf. [1, 2]). What is interesting is that a complete asymptotic and bifurcation analysis can be obtained. With such success, it is expected that more general discontinuous recurrent McCulloch-Pitts-type neural networks [3] can be dealt with to some extent in a similar manner.To simplify matters, note that if we letz=(x,y) and
(6)Fσ(z)=(fσ(y),fσ(x)),
then the above system (3) can be written as
(7)zn=azn-2+a′Fσ(zn-1),n∈N,
where we write zn=(xn,yn) and a′=1-a for the sake of convenience. The above vector equation is a three-term recurrence relation. Hence for given z-2 and z-1 in the plane, a unique sequence {zk}k=-2∞ can be calculated from it. Such a sequence is called a solution of (7) determined by z-2 and z-1.Depending on the locations ofz-2 and z-1, it is clear that the behavior of the corresponding solution may differ. For this reason, it is convenient to distinguish various parts of R2. Before doing so, it is convenient to adopt several simplifying conventions. First, we set
(8)Rσ-=(-∞,σ],Rσ+=(σ,∞).
Next, if I and J are real intervals, their cross-product I×J will be denoted by IJ instead, and we will assume that this product receives the priority attention in a mathematical expression. By means of these convections,
(9)Rσ+Rσ+=(σ,∞)2,Rσ+Rσ-=(σ,∞)×(-∞,σ],Rσ-Rσ-=(-∞,σ]2,Rσ-Rσ+=(-∞,σ]×(σ,∞).
Clearly, {Rσ+Rσ+,Rσ+Rσ-,Rσ-Rσ+,Rσ-Rσ-} is a partition of R2. Other subsets of the plane will be introduced in the subsequent sections.For solutions originated fromz-2 and z-1 in the above subsets, we will show that they are all “asymptotically periodic.” More precisely, we say that a positive integer η is a period of a scalar or vector sequence v={vn}n=α∞ if vη+n=vn for all n≥α and that τ is the least or prime period of {vn}n=α∞ if τ is the least among all periods of {vn}n=α∞. The sequence {vn}n=α∞ is said to be τ-periodic if τ is the least period of {vn}n=α∞. The sequence {vn}n=α∞ is said to be asymptotically periodic if there exist real numbers v(0),v(1),…,v(ω-1), where ω is a positive integer such that
(10)limn→∞vωn+i=v(i),i=0,1,…,ω-1.
In case {v(0),v(1),…,v(ω-1),v(0),v(1),…,v(ω-1),…} is also an ω-periodic sequence, we say that v is an asymptotically ω-periodic sequence tending to the limit ω-cycle (this term is introduced since the underlying concept is similar to that of the limit cycle in the theory of ordinary differential equations) 〈v(0),v(1),…,v(ω-1)〉. For the sake of convenience, for such an asymptotically ω-periodic sequence v, we write
(11)vn⟶〈v(0),v(1),…,v(ω-1)〉.
Note that in case v is asymptotically 1-periodic, then it is a sequence convergent to v(0) for some v(0). The converse is also true. In such a case, we simply write vn→v(0) instead of vn→〈v(0)〉.Having these terminologies at hand, our main issue is to show that each solution of (3) originated from R2 is either asymptotically 1- or 2-periodic. Note, however, that since fσ is a discontinuous function, the standard theories that employ continuous arguments cannot be applied to yield asymptotic criteria. Fortunately, we may resort to elementary arguments as to be seen below.Before doing so, let us make a few remarks. First note that our system (7) is autonomous (time invariant) and also symmetric in the sense that under two sets of “symmetric initial conditions,” the behaviors of the corresponding solutions are also “symmetric.” This statement can be made more precise in mathematical terms. However, a simple example is sufficient to illustrate this: if {zn}n=-2∞ is a solution of (3) with (z-2,z-1)∈Rσ+Rσ+×Rσ-Rσ-, then as will be seen below, z2n→(1,-1) and z2n+1→(-1,1). If we now replace the condition (z-2,z-1)∈Rσ+Rσ+×Rσ-Rσ- with the symmetric initial condition (z-2,z-1)∈Rσ-Rσ-×Rσ+Rσ+, then we will end up with the conclusion that z2n→(-1,1) and z2n+1→(1,-1). Such two conclusions will be referred to as dual results, and the principle of proof for either one can be applied to that of the other.In the sequel, we will first distinguish three different cases (i)0≤σ<1, (ii) σ=1, and (iii) σ>1 and then consider different z-2, z-1 in Rσ+Rσ+,Rσ+Rσ-,Rσ-Rσ+, or Rσ-Rσ- and the (asymptotic) behaviors of the corresponding solutions determined by them.We will need the following simple but useful results. First, let{xk}k=-2∞ be real scalar (or vector) sequences that satisfy
(12)x2k=ax2k-2+d,n∈N,(13)x2k+1=ax2k-1+d,k∈N,
where a∈(0,1), and d is a real number (resp., a real vector).(i)
If{xn}n=-2∞ is a sequence which satisfies (12), then
(14)x2k=ak+1x-2+(1-ak+1)1-ad,k∈N.(ii)
If{xn}n=-2∞ is a sequence which satisfies (13), then
(15)x2k+1=ak+1x-1+(1-ak+1)1-ad,k∈N.Second, the functionFσ in (6) satisfies
(16)Fσ(Rσ+Rσ+)=-k→,Fσ(Rσ+Rσ-)=i→,Fσ(Rσ-Rσ+)=-i→,Fσ(Rσ-Rσ-)=k→,
where i→=(1,-1) and k→=(1,1).Third, we need to consider various ordering arrangements for three nonnegative integersk, p, and l or four nonnegative integers k, p, l, and m. First, the ordering arrangements of three integers k, p, and l can be classified into 6 cases:
(17)(A1)k=p≤l(A2)l=k<p(A3)p=l<k(A4)k<min{p,l}(A5)p<min{k,l}(A6)l<min{k,p}.Indeed, leta,b,c∈R. Then, since either a<b, a=b or a>b, we see that
(18)a<b⇒c∈(-∞,a),c=a,c∈(a,b),c=borc∈(b,∞),a=b⇒c∈(-∞,a),c=aorc∈(a,∞),a>b⇒c∈(-∞,b),c=b,c∈(b,a),c=aorc∈(a,∞).
These are equivalent to
(19)a=b≤c,c=a<b,b=c<a,a<min{b,c},b<min{a,c},c<min{a,b},
by comparing the two sets of statements (18) and (19).By similar reasoning, there are 12 ordering arrangements for four nonnegative integersk, p, l, and m:
(20)(B1)k=p≤min{l,m}(B2)k<min{p,l,m}(B3)p<min{k,l,m}(B4)p=l<min{k,m}(B5)p=l=m<k(B6)l<min{p,k,m}(B7)l=m<min{k,p}(B8)l=m=k<p(B9)m<min{l,k,p}(B10)m=k<min{l,p}(B11)p=m<min{l,k}(B12)k=l<min{p,m}.In the sequel, ifα,β∈R and I,J are real intervals, then we adopt the convention that
(21)αI={αx∣x∈I},α+I={α+x∣x∈I},αIJ={αx∣x∈I}×{αy∣y∈J},(α,β)+IJ={α+x∣x∈I}×{β+y∣y∈J}.
On the other hand, since a∈(0,1),σ>0, we have
(22)σ<σ+(1-a)a<σ+(1-a2)a2<⋯<σ+(1-ak)ak⟶+∞.
If we let
(23)J(k)=(σ+(1-ak)ak,σ+(1-ak+1)ak+1],k∈N,J(-1)=aJ(0)-a′=(aσ-(1-a),σ],
then
(24)aJ(k)-a′=J(k-1),k≥0,J(-1)⊂Rσ-,Rσ+=⋃k=0∞‍J(k),(25)Rσ+Rσ+=⋃k,s=0∞‍J(k)J(s)=⋃k=0∞‍J(k)Rσ+=⋃s=0∞‍Rσ+J(s).


## 2. The Caseσ=1

In this section, we assume thatσ=1.Let{zn}n=-2∞ be a solution of (3). Let us consider first the case where (zn,zn+1)∈R1-R1-×R1-R1-. Then by (7) and (16), we see that
(26)zn+2=azn+a′k→,aR1-R1-+a′F1(R1-R1-)=aR1-R1-+a′k→=a((-∞,1]×(-∞,1])+(1-a)(1,1)=(-∞,1]×(-∞,1]=R1-R1-.
Hence, we may see further that zn+2∈R1-R1-. Similarly, since
(27)aJ(k)R1-+a′F1(R1-J(s))=aJ(k)R1-+a′(-i→)=a(J(k)×(-∞,1])+(1-a)(-1,1)=J(k-1)×(-∞,σ]=J(k-1)R1-,
we see that if (zn,zn+1)∈J(k)R1-×R1-J(s), then zn+2∈J(k-1)R1-. By similar considerations, we may build a “directed graph” represented by Table 1.Table 1
R
1
-
R
1
-
J
(
s
)
R
1
-
R
1
-
J
(
s
)
R
1
+
J
(
s
)
J
(
s
)
R
1
+
J
(
s
)
J
(
l
)
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
R
1
-
J
(
k
)
R
1
-
R
1
+
R
1
-
R
1
+
R
1
-
J
(
k
-
1
)
R
1
-
J
(
k
-
1
)
R
1
-
R
1
-
J
(
k
)
R
1
-
R
1
+
R
1
-
J
(
k
-
1
)
R
1
-
R
1
+
R
1
-
J
(
k
-
1
)
R
1
+
J
(
k
)
R
1
+
R
1
+
R
1
+
J
(
k
-
1
)
J
(
k
)
R
1
+
R
1
+
R
1
+
J
(
k
-
1
)
R
1
+
J
(
k
)
J
(
p
)
R
1
+
R
1
+
J
(
k
-
1
)
J
(
p
-
1
)In this table, we record the fact thataR1-R1-+a′F1(R1-R1-)⊆R1-R1- as the (R1-R1-,R1-R1-) entry and so forth.Next, we let{zn}n=-2∞ be a solution of (3) again. If (z-2,z-1)∈R1-R1-×R1-R1-, then from Table 1 we see that z0∈R1-R1-. Now that (z-1,z0)∈R1-R1-×R1-R1-, we may use Table 1 to infer that z1∈R1-R1- again. By induction, we see that zn∈R1-R1- for all n≥-2. By (7), we then see that
(28)zn=azn-2+a′k→,n∈N,
and hence by (14) and (15), we see that zn→k→ as n→∞. If (z-2,z-1)∈R1-R1+×R1-R1-, then from the (R1-R1+,R1-R1-) entry of Table 1 we see that z0∈R1-R1+. Now that (z-1,z0)∈R1-R1-×R1-R1+, by Table 1 again, we see that
(29)z1=az-1+a′(-i→)∈R1-R1-.
By induction, we see that z2k∈R1-R1+ and z2k+1∈R1-R1- for k≥-1. By (7), we may see further that
(30)z2n=az2n-2+a′k→,z2n+1=az2n-1-a′i→,
for n∈N, and hence by (14) and (15), z2n→k→ and z2n+1→-i→ as n→∞. By considering z-2 and z-1 in different parts of the plane, we may apply the same principle to obtain Table 2.Table 2
R
1
-
R
1
-
R
1
+
R
1
-
R
1
-
R
1
+
R
1
+
R
1
+
R
1
-
R
1
-
〈
k
→
〉
〈
i
→
,
k
→
〉
〈
-
i
→
,
k
→
〉
〈
-
k
→
,
k
→
〉
R
1
+
R
1
-
〈
k
→
,
i
→
〉
〈
i
→
〉
See  Table3
See  Table4
R
1
-
R
1
+
〈
k
→
,
-
i
→
〉
See  Table3
〈
-
i
→
〉
See  Table4
R
1
+
R
1
+
〈
k
→
,
-
k
→
〉
See  Table5
See  Table5
See  Table6In this table, there are seven indeterminate cases. Let us go through two cases. First, let(z-2,z-1)∈R1+R1-×R1-R1+. Then in view of (25), (z-2,z-1)∈J(k)R1-×R1-J(s) for some k,s∈N. There are then two subcases: (i) 0≤k≤s or (ii) 0≤s<k. In the former case, by Table 1,
(31)z0=az-2+a′F1(z-1)∈J(k-1)R1-,
and (hence)
(32)z1=az-1+a′F1(z0)∈R1-J(s-1),
and by induction,
(33)z2k∈J(-1)R1-⊆R1-R1-,z2k+1∈R1-R1+.
In the latter case, we may similarly show that (z2s,z2s+1)∈R1+R1-×R1-R1-. By considering (z-2,z-1) in different parts of the plane, we may then construct the following three self-explanatory Tables 3, 4, and 5.Table 3
(
z
-
2
,
z
-
1
)
∈
J
(
k
)
R
-
×
R
1
-
J
(
s
)
(
R
1
-
J
(
k
)
×
J
(
s
)
R
1
-
)
0
≤
k
≤
s
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
1
-
R
1
-
×
R
1
-
R
1
+
(
R
1
-
R
1
-
×
R
1
+
R
1
-
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
R
1
+
R
1
-
×
R
1
-
R
1
-
(
R
1
-
R
1
+
×
R
1
-
R
1
-
)Table 4
(
z
-
2
,
z
-
1
)
∈
J
(
k
)
R
-
×
R
1
+
J
(
s
)
(
R
1
-
J
(
k
)
×
J
(
s
)
R
1
+
)
0
≤
k
≤
s
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
1
-
R
1
-
×
R
1
+
R
1
+
(
R
1
-
R
1
-
×
R
1
+
R
1
+
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
R
1
+
R
1
-
×
R
1
+
R
1
-
(
R
1
-
R
1
+
×
R
1
-
R
1
+
)Table 5
(
z
-
2
,
z
-
1
)
∈
R
1
+
J
(
k
)
×
J
(
s
)
R
1
-
(
J
(
k
)
R
1
+
×
R
1
-
J
(
s
)
)
0
≤
k
≤
s
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
1
+
R
1
-
×
R
1
+
R
1
-
(
R
1
-
R
1
+
×
R
1
-
R
1
+
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
R
1
+
R
1
+
×
R
1
-
R
1
-
(
R
1
+
R
1
+
×
R
1
-
R
1
-
)As a consequence, if(z-2,z-1)∈J(k)R1-×R1-J(s) and 0≤k≤s, then the solution of (3) originated from z-2 and z-1 will satisfy (z2k,z2k+1)∈R1-R1-×R1-R1+. Then by Table 2, zn→〈-i→,k→〉.Next, let(z-2,z-1)∈R1+R1+×R1+R1+. Then by (25), (z-2,z-1)∈J(k)J(p)×J(l)J(m) for some k,p,l,m∈N.As mentioned before, there are twelve different ordering arrangements B1–B12 for these integers. In case B1 holds, we may make use of Table 1 repeatedly to show that (z2k,z2k+1)∈R1-R1-×R1+R1+ (which is recorded in self-explanatory Table 6), and then by Table 2, we finally see that zn→〈-k→,k→〉.Table 6
(
z
-
2
,
z
-
1
)
∈
J
(
k
)
J
(
p
)
×
J
(
l
)
J
(
m
)
B1⇒(z2k,z2k+1)∈R1-R1-×R1+R1+
B2⇒(z2k,z2k+1)∈R1-J(p-k-1)×J(l-k-1)R1+
B3⇒(z2p,z2p+1)∈J(k-p-1)R1-×R1+J(m-p-1)
B4⇒(z2p,z2p+1)∈J(k-p-1)R1-×R1+J(m-p-1)
B5⇒(z2p,z2p+1)∈R1+R1-×R1+R1-
B6⇒(z2l,z2l+1)∈J(k-l-1)R1+×R1-J(m-l-1)
B7⇒(z2l,z2l+1)∈R1+R1+×R1-R1-
B8⇒(z2l,z2l+1)∈R1-R1+×R1-R1+
B9⇒(z2m,z2m+1)∈R1+J(p-m-1)×J(l-m-1)R1-
B10⇒(z2k,z2k+1)∈R1-J(p-k-1)×J(l-k-1)R1+
B11⇒(z2p,z2p+1)∈R1+R1-×R1+R1-
B12⇒(z2l,z2l+1)∈R1-R1+×R1-R1+In conclusion, we have shown the following result.Theorem 1.
Suppose thatσ=1. Then, a solution of (3) must either be asymptotically 1-periodic tending to the limit cycles
(34)〈i→〉,〈-i→〉,or〈k→〉,
or asymptotically 2-periodic tending to the limit cycles
(35)〈i→,k→〉,〈-i→,k→〉,or〈-k→,k→〉.We remark that, as can be seen from the above exhaustive arguments, the “region of attraction” of each limit cycle can be given, here ifS is the set of all solutions of (3) that tend to the limit cycle G, then {(z-2,z-1)∈R2×R2|{zk}k=-2∞∈S} is called the region of attraction of G.Example 2.
Consider the region of attraction of the limit cycle〈i→,k→〉. By Table 2, we first see that solutions with initial values in R1-R1-×R1+R1- or R1+R1-×R1-R1- tend to the limit cycle 〈i→,k→〉 (and hence 〈k→,i→〉); and solutions with initial values in the seven indeterminate cases may or may not tend to the limit cycle 〈i→,k→〉.
Therefore, we next consider Table3. If (z-2,z-1)∈J(k)R1-×R1-J(s) where 0≤s<k, then (z2s,z2s+1)∈R1+R1-×R1-R1- and hence zn→〈k→,i→〉. If (z-2,z-1)∈R1-J(k)×J(s)R1- where 0≤k≤s, then (z2k,z2k+1)∈R1-R1-×R1+R1- and hence zn→〈i→,k→〉. Furthermore, no other cases in Tables 3, 4, and 5 can lead to (zm,zm+1)∈R1+R1-×R1+R1-,R1-R1+×R1-R1+, or R1+R1+×R1-R1-.
Next, we consider Table6. By direct examination, there is no case which leads to (xm,xm+1)∈R1+R1-×R1-R1- or R1-R1-×R1+R1-. There is no case, except possibly B10, that leads to (xm,xm+1)∈J(k)R1-×R1-J(s) with 0≤s<k, or (z-2,z-1)∈R1-J(k)×J(s)R1- with 0≤k≤s. However, if B10 holds and p-k-1≤l-k-1, then (z-2,z-1)∈J(k)J(p)×J(l)J(m) implies (z2k,z2k+1)∈R1-J(p-k-1)×J(l-k-1)R1+, which in turn implies (by Table 3) that (z2p,z2p+1)∈R1-R1-×R1+R1-. Hence, zn→〈i→,k→〉.
In summary, the region of attraction of〈i→,k→〉 is the union of the following sets:
(36)R1-R1-×R1+R1-,R1+R1-×R1-R1-,R1-J(k)×J(s)R1-(0≤k≤s),J(k)R1+×R1-J(s)(0≤s<k),J(k)J(P)×J(l)J(m)(m=k<min{l,p},p-k-1≤l-k-1).


## 3. The Caseσ>1

In this section, we assume thatσ>1.Lemma 3.
Suppose thatσ>1. Let {zn}n=-2∞ be a solution of (3) originated from z-2,z-1∈Rσ+Rσ+. Then, there exists m∈{-1,0,…} such that zm∈Rσ+Rσ+ and zm+1∈R2∖Rσ+Rσ+.Proof.
Let{zn}n=-2∞ be a solution of (3) originated from z-2,z-1∈Rσ+Rσ+. Assume to the contrary that zn∈Rσ+Rσ+ for all n∈N. Then,
(37)zn=azn-2+a′(-k→),n∈N.
By (14) and (15), we see that zn→-k→ as n→∞. But σ>1 implies -k→ is in the interior of Rσ-Rσ- which is disjointed from Rσ+Rσ+. A contradiction is, thus, obtained.In view of Lemma3, we only need to investigate the asymptotic behaviors of solutions {zn}n=-2∞ of (3) such that z-2 and z-1 are not simultaneously in Rσ+Rσ+. To this end, we first note that
(38)σ<σ±(1-a)a<σ±(1-a2)a2<⋯<σ±(1-ak)ak⟶+∞.
Hence, if we let
(39)I(k)=(σ-(1-ak)ak,σ-(1-ak+1)ak+1],k∈N,I(-1)=aI(0)+a′=(aσ+(1-a),σ],
then
(40)aI(k)+a′=I(k-1),k∈N,Rσ+=⋃k=0∞‍J(k)=⋃k=0∞‍I(k),Rσ+Rσ-=⋃k=0∞‍I(k)Rσ-=⋃k=0∞‍J(k)Rσ-,Rσ-Rσ+=⋃k=0∞‍Rσ-I(k)=⋃k=0∞‍Rσ-J(k),Rσ+Rσ+=⋃k,s=0∞‍I(k)J(s)=⋃k,s=0∞‍J(k)I(s)=⋃k,s=0∞‍I(k)I(s)=⋃k,s=0∞‍J(k)J(s),I(-1),J(-1)⊂Rσ-.
As in the previous section, we may build a table (consisting of two separate parts Tables 7 and 8) which plays the same role as Table 1.Table 7
R
σ
-
R
σ
-
I
(
s
)
R
σ
-
J
(
s
)
R
σ
-
R
σ
-
I
(
s
)
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
I
(
k
)
R
σ
-
I
(
k
-
1
)
R
σ
-
I
(
k
-
1
)
R
σ
-
I
(
k
-
1
)
R
σ
-
J
(
k
)
R
σ
-
J
(
k
-
1
)
R
σ
-
R
σ
-
I
(
k
)
R
σ
-
I
(
k
-
1
)
R
σ
-
I
(
k
-
1
)
R
σ
-
J
(
k
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
I
(
k
)
J
(
p
)
I
(
k
-
1
)
J
(
p
-
1
)
I
(
k
-
1
)
J
(
p
-
1
)
J
(
k
)
I
(
p
)
J
(
k
-
1
)
I
(
p
-
1
)
I
(
k
)
I
(
p
)
I
(
k
-
1
)
I
(
p
-
1
)
J
(
k
)
J
(
p
)Table 8
R
σ
-
J
(
s
)
I
(
s
)
J
(
l
)
J
(
s
)
I
(
l
)
I
(
s
)
I
(
l
)
J
(
s
)
J
(
l
)
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
I
(
k
)
R
σ
-
J
(
k
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
R
σ
-
R
σ
-
I
(
k
)
R
σ
-
I
(
k
-
1
)
R
σ
-
J
(
k
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
I
(
k
)
J
(
p
)
J
(
k
)
I
(
p
)
J
(
k
-
1
)
I
(
p
-
1
)
I
(
k
)
I
(
p
)
J
(
k
)
J
(
p
)
J
(
k
-
1
)
J
(
p
-
1
)
J
(
k
-
1
)
J
(
p
-
1
)
J
(
k
-
1
)
J
(
p
-
1
)
J
(
k
-
1
)
J
(
p
-
1
)Then, we may construct Tables9–14 which play the same role as  Tables 3–6.Table 9
(
z
-
2
,
z
-
1
)
∈
R
σ
-
R
σ
-
×
I
(
k
)
R
σ
-
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
-
R
σ
-
×
R
σ
-
R
σ
-
(
z
-
2
,
z
-
1
)
∈
R
σ
-
R
σ
-
×
R
σ
-
I
(
k
)
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
-
R
σ
-
×
R
σ
-
R
σ
-
(
z
-
2
,
z
-
1
)
∈
I
(
k
)
R
σ
-
×
R
σ
-
R
σ
-
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
-
R
σ
-
×
R
σ
-
R
σ
-
(
z
-
2
,
z
-
1
)
∈
R
σ
-
I
(
k
)
×
R
σ
-
R
σ
-
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
-
R
σ
-
×
R
σ
-
R
σ
-Table 10
(
z
-
2
,
z
-
1
)
∈
I
(
k
)
R
σ
-
×
I
(
s
)
R
σ
-
(
R
σ
-
I
(
s
)
×
R
σ
-
I
(
k
)
)
0
≤
k
≤
s
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
-
R
σ
-
×
I
(
s
-
k
-
1
)
R
σ
-
(
R
σ
-
I
(
s
-
k
-
1
)
×
R
σ
-
R
σ
-
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
I
(
k
-
s
-
1
)
R
σ
-
×
R
σ
-
R
σ
-
(
R
σ
-
R
σ
-
×
R
σ
-
I
(
k
-
s
-
1
)
)Table 11
(
z
-
2
,
z
-
1
)
∈
J
(
k
)
R
σ
-
×
R
σ
-
J
(
s
)
(
R
σ
-
J
(
k
)
×
J
(
s
)
R
σ
-
)
0
≤
k
≤
s
⇒
(
z
2
k
-
1
,
z
2
k
)
∈
R
σ
-
J
(
s
-
k
)
×
R
σ
-
R
σ
-
(
J
(
s
-
k
)
R
σ
-
×
R
σ
-
R
σ
-
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
J
(
k
-
s
-
1
)
R
σ
-
×
R
σ
-
R
σ
-
(
R
σ
-
J
(
k
-
s
-
1
)
×
R
σ
-
R
σ
-
)Table 12
(
z
-
2
,
z
-
1
)
∈
R
σ
-
R
σ
-
×
I
(
s
)
I
(
k
)
(
I
(
s
)
I
(
k
)
×
R
σ
-
R
σ
-
)
0
≤
k
≤
s
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
-
R
σ
-
×
I
(
s
-
k
-
1
)
R
σ
-
(
I
(
s
-
k
-
1
)
R
σ
-
×
R
σ
-
R
σ
-
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
R
σ
-
R
σ
-
×
R
σ
-
I
(
k
-
s
-
1
)
(
R
σ
-
I
(
k
-
s
-
1
)
×
R
σ
-
R
σ
-
)Table 13
(
z
-
2
,
z
-
1
)
∈
R
σ
-
J
(
k
)
×
J
(
p
)
I
(
l
)
(
J
(
k
)
R
σ
-
×
I
(
l
)
J
(
p
)
)
A1⇒(z2k-1,z2k)∈J(p-k)I(l-k)×Rσ-Rσ-
(
I
(
l
-
k
)
J
(
p
-
k
)
×
R
σ
-
R
σ
-
)
A2⇒(z2k-1,z2k)∈J(p-k)I(l-k)×Rσ-Rσ-
(
I
(
l
-
k
)
J
(
p
-
k
)
×
R
σ
-
R
σ
-
)
A3⇒(z2l,z2l+1)∈Rσ-J(k-l-1)×Rσ-Rσ-
(
J
(
k
-
l
-
1
)
R
σ
-
×
R
σ
-
R
σ
-
)
A4⇒(z2k-1,z2k)∈J(p-k)I(l-k)×Rσ-Rσ-
(
I
(
l
-
k
)
J
(
p
-
k
)
×
R
σ
-
R
σ
-
)
A5⇒(z2p,z2p+1)∈Rσ-J(k-p-1)×Rσ-I(l-p-1)
(
J
(
k
-
p
-
1
)
R
σ
-
×
I
(
l
-
p
-
1
)
R
σ
-
)
A6⇒(z2l,z2l+1)∈Rσ-J(k-l-1)×J(p-l-1)Rσ-
(
J
(
k
-
l
-
1
)
R
σ
-
×
R
σ
-
J
(
p
-
l
-
1
)
)Table 14
(
z
-
2
,
z
-
1
)
∈
I
(
s
)
J
(
p
)
×
J
(
k
)
R
σ
-
(
J
(
s
)
I
(
p
)
×
R
σ
-
J
(
k
)
)
0
=
s
<
p
⇒
(
z
-
1
,
z
0
)
∈
J
(
k
)
R
σ
-
×
R
σ
-
J
(
p
-
1
)
(
R
σ
-
J
(
k
)
×
R
σ
-
I
(
p
-
1
)
)
0
<
s
<
p
⇒
(
z
-
1
,
z
0
)
∈
J
(
k
)
R
σ
-
×
I
(
s
-
1
)
J
(
p
-
1
)
(
R
σ
-
J
(
k
)
×
J
(
s
-
1
)
I
(
p
-
1
)
)
0
=
s
=
p
⇒
(
z
-
1
,
z
0
)
∈
J
(
k
)
R
σ
-
×
R
σ
-
R
σ
-
(
R
σ
-
J
(
k
)
×
R
σ
-
R
σ
-
)
0
<
s
=
p
⇒
(
z
-
1
,
z
0
)
∈
J
(
k
)
R
σ
-
×
I
(
s
-
1
)
J
(
p
-
1
)
(
R
σ
-
J
(
k
)
×
J
(
s
-
1
)
I
(
p
-
1
)
)
0
=
p
<
s
⇒
(
z
-
1
,
z
0
)
∈
J
(
k
)
R
σ
-
×
I
(
s
-
1
)
R
σ
-
(
R
σ
-
J
(
k
)
×
J
(
s
-
1
)
R
σ
-
)
0
<
p
<
s
⇒
(
z
-
1
,
z
0
)
∈
J
(
k
)
R
σ
-
×
I
(
s
-
1
)
J
(
p
-
1
)
(
R
σ
-
J
(
k
)
×
J
(
s
-
1
)
I
(
p
-
1
)
)Then by Lemma3 and these tables, we may show the following result.Lemma 4.
Suppose thatσ>1. Let {zn}n=-2∞ be a solution of (3). Then, there exists m∈{-2,-1,…} such that zm,zm+1∈Rσ-Rσ-.Proof.
In view of Lemma3, we may assume that z-2∈R2 and z-1∈R2∖Rσ+Rσ+, or, z-2∈R2∖Rσ+Rσ+ and z-1∈Rσ+Rσ+.
Consider first the case wherez-2∈Rσ+Rσ- and z-1∈Rσ-Rσ+. Then, z-2∈J(k)Rσ- and z-1∈Rσ-J(s) for some k,s∈N. By Table 11, if 0≤k≤s, then (z2k-1,z2k)∈Rσ-J(s-k)×Rσ-Rσ-. Since J(s-k)⊂(σ,+∞), there exists some t∈N such that (z2k-1,z2k)∈Rσ-I(t)×Rσ-Rσ-. By the last entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-. If 0≤s<k, then (z2s,z2s+1)∈J(k-s-1)Rσ-×Rσ-Rσ-. Since J(k-s-1)⊂(σ,+∞), there exists some t∈N such that (z2s,z2s+1)∈I(t)Rσ-×Rσ-Rσ-. By the last entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-.
Consider next the case wherez-2∈Rσ-Rσ- and z-1∈Rσ+Rσ+. Then, z-1∈I(s)I(k) for some k,s∈N. By Table 12 if 0≤k≤s, then (z2k,z2k+1)∈Rσ-Rσ-×I(s-k-1)Rσ-. By the last entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-. If 0≤s<k, then (z2s,z2s+1)∈Rσ-Rσ-×Rσ-I(k-s-1). By the last entry in Table 9, we see that there is some m such that (zm,zm+1)∈Rσ-Rσ-.
Consider next the case wherez-2∈Rσ+Rσ- and z-1∈Rσ+Rσ+. Then, z-2∈J(k)Rσ- and z-1∈I(l)J(p) for some k,l,p∈N. By Table 13, if A1 holds, that is, 0≤k=p≤l, then (z2k-1,z2k)∈I(l-k)J(p-k)×Rσ-Rσ-. Then, there exists t∈N such that (z2k-1,z2k)∈I(l-k)I(t)×Rσ-Rσ-. By Table 12 again, if 0≤t≤l-k, then (z2t+2k+1,z2t+2k+2)∈I(l-k-t-1)Rσ-×Rσ-Rσ-. By the last entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-; if 0≤l-k<t, then (z2l+1,z2l+2)∈Rσ-I(t-l+k-1)×Rσ-Rσ-. By the last entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-. If A5 holds, that is, 0≤p<min{k,l}, then (z2p,z2p+1)∈J(k-p-1)Rσ-×I(l-p-1)Rσ-. Then, there exists t∈N such that (z2p,z2p+1)∈I(t)Rσ-×I(l-p-1)Rσ-. By Table 10 again, if 0≤t≤l-p-1, then (z2t+2p+2,z2t+2p+3)∈Rσ-Rσ-×I(l-p-1-t-1)Rσ-. By the last entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-; if 0≤l-p-1<t, then (z2l,z2l+1)∈Rσ-I(t-l+p)×Rσ-Rσ-. By the last entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-. The other cases are similarly proved.
Consider next the case wherez-2∈Rσ+Rσ+ and z-1∈Rσ+Rσ-. Then, z-2∈I(l)J(p) and z-1∈J(k)Rσ- for some k,l,p∈N. Next, we need to consider six cases: (i) 0=l<p, (ii) 0<l<p, (iii) 0=l=p, (iv) 0<l=p, (v) 0=p<l, and (vi) 0<p<l. Suppose that 0<l<p holds, by Table 14, then (z-1,z0)∈J(k)Rσ-×I(l-1)J(p-1). Furthermore, we need to consider six cases: A1–A6. Suppose that A3 holds, that is, p-1=l-1<k, then by Table 13, (z2l-1,z2l)∈J(k-l)Rσ-×Rσ-Rσ-. Then there exists t∈N such that (z2l-1,z2l)∈I(t)Rσ-×Rσ-Rσ-. By the third entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-. Suppose that A6 holds, that is, l-1<min{k,p-1}, then by Table 13, (z2l-1,z2l)∈J(k-l)Rσ-×Rσ-J(p-l-1). If k-l≤p-l-1, then by Table 11, (z2k,z2k+1)∈Rσ-J(p-k-l)×Rσ-Rσ-. Then, there exists t∈N such that (z2k,z2k+1)∈Rσ-I(t)×Rσ-Rσ-. By the last entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-; if p-l-1<k-l, then by Table 11, (z2p-1,z2p)∈J(k-p)Rσ-×Rσ-Rσ-. Then there exists some t∈N such that (z2p-1,z2p)∈I(t)Rσ-×Rσ-Rσ-. By the last entry in Table 9, we see that there is some m such that zm,zm+1∈Rσ-Rσ-. The other cases are similarly proved.
The proof is complete.Theorem 5.
Suppose thatσ>1. Then, every solution of (3) tends to k→.
Indeed, let{zn}n=-2∞ be a solution of (3). In view of Lemma 4, one may assume that z-2,z-1∈Rσ-Rσ-. Then by the (Rσ-Rσ-,Rσ-Rσ-) entry in Table 7, one may sees that z0∈Rσ-Rσ- and by induction, zn∈Rσ-Rσ- for all n≥-2. By (14) and (15), then sees that zn→k→ as required.


## 4. The Case0≤σ<1

Suppose that0≤σ<1. Then,
(41)σ>σ-(1-a)a>σ-(1-a2)a2>⋯>σ-(1-ak)ak⟶-∞.
If we let
(42)K(k)=(σ-(1-ak+1)ak+1,σ-(1-ak)ak],k∈N,K(-1)=aK(0)+a′=(σ,aσ+(1-a)],
then
(4)aK(k)+a′=K(k-1),k∈N,Rσ+=⋃k=0∞‍J(k),Rσ-=⋃k=0∞‍K(k),Rσ-Rσ-=⋃k=0∞‍K(k)Rσ-=⋃k=0∞‍Rσ-K(k)=⋃k,s=0∞‍K(k)K(s),Rσ+Rσ+=⋃k=0∞‍J(k)Rσ+=⋃k=0∞‍Rσ+J(k)=⋃k,s=0∞‍J(k)J(s),Rσ+Rσ-=⋃k=0∞‍J(k)Rσ-=⋃k=0∞‍Rσ+K(k)=⋃k,s=0∞‍J(k)K(s),Rσ-Rσ+=⋃k=0∞‍K(k)Rσ+=⋃k=0∞‍Rσ-J(k)=⋃k,s=0∞‍K(k)J(s).
Let {zn}n=-2∞ be a solution of (3). Let us consider first the case where zn,zn+1∈Rσ+Rσ-. Then by (7) and (16), we see that
(44)zn+2=azn+a′i→.
Note that (in view of our assumption that 0≤σ<1)
(45)aRσ+Rσ-+a′Fσ(Rσ+Rσ-)=aRσ+Rσ-+a′i→=a((σ,∞)×(-∞,σ])+(1-a)(1,-1)=(aσ+(1-a),∞)×(-∞,aσ-(1-a)]⊆Rσ+Rσ-.
Hence, we may see further that zn+2∈Rσ+Rσ-. Similarly, since
(46)aRσ+Rσ++a′Fσ(Rσ-Rσ-)=aRσ+Rσ++a′k→=a((σ,∞)×(σ,∞))+(1-a)(1,1)=(aσ+(1-a),∞)×(aσ+(1-a),∞)⊆Rσ+Rσ+,aRσ-Rσ-+a′Fσ(Rσ+Rσ+)=aRσ-Rσ--a′k→=a((-∞,σ]×(-∞,σ])-(1-a)(1,1)=(-∞,aσ-(1-a)]×(-∞,aσ-(1-a)]⊆Rσ-Rσ-,
we see that if (zn,zn+1)∈Rσ+Rσ+×Rσ-Rσ-, then zn+2∈Rσ+Rσ+ and zn+3∈Rσ-Rσ-. By similar considerations, we may build a table (consisting of two separate parts Tables 15 and 16) which plays the same role as Table 1.Table 15
K
(
s
)
R
σ
-
R
σ
-
K
(
s
)
K
(
s
)
K
(
l
)
J
(
s
)
R
σ
-
R
σ
+
K
(
l
)
J
(
s
)
K
(
l
)
K
(
k
)
R
σ
-
K
(
k
-
1
)
R
σ
-
K
(
k
-
1
)
R
σ
-
K
(
k
-
1
)
R
σ
-
R
σ
-
K
(
k
)
K
(
k
)
K
(
p
)
K
(
k
-
1
)
K
(
p
-
1
)
K
(
k
-
1
)
K
(
p
-
1
)
K
(
k
-
1
)
K
(
p
-
1
)
J
(
k
)
R
σ
-
R
σ
+
R
σ
-
R
σ
+
R
σ
-
R
σ
+
R
σ
-
J
(
k
)
K
(
s
)
R
σ
+
R
σ
-
R
σ
+
R
σ
-
R
σ
+
R
σ
-
R
σ
+
K
(
k
)
R
σ
+
K
(
k
-
1
)
R
σ
+
K
(
k
-
1
)
R
σ
+
K
(
k
-
1
)
R
σ
+
R
σ
-
R
σ
+
R
σ
-
R
σ
+
R
σ
-
R
σ
-
K
(
k
)
K
(
k
)
R
σ
+
K
(
k
-
1
)
R
σ
+
K
(
k
-
1
)
R
σ
+
K
(
k
-
1
)
R
σ
+
K
(
k
)
J
(
p
)
K
(
k
-
1
)
J
(
p
-
1
)
K
(
k
-
1
)
J
(
p
-
1
)
K
(
k
-
1
)
J
(
p
-
1
)
R
σ
+
J
(
k
)
R
σ
+
R
σ
+
R
σ
+
R
σ
+
R
σ
+
R
σ
+
R
σ
+
J
(
k
-
1
)
R
σ
+
J
(
k
-
1
)
R
σ
+
J
(
k
-
1
)
J
(
k
)
R
σ
+
R
σ
+
R
σ
+
R
σ
+
R
σ
+
R
σ
+
R
σ
+
J
(
k
)
J
(
p
)
R
σ
+
R
σ
+
R
σ
+
R
σ
+
R
σ
+
R
σ
+Table 16
R
σ
-
J
(
s
)
K
(
s
)
R
σ
+
K
(
s
)
J
(
l
)
R
σ
+
J
(
s
)
J
(
s
)
R
σ
+
J
(
s
)
J
(
l
)
K
(
k
)
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
K
(
k
)
R
σ
-
K
(
k
-
1
)
R
σ
-
K
(
k
-
1
)
R
σ
-
K
(
k
-
1
)
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
K
(
k
)
K
(
p
)
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
R
σ
-
J
(
k
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
)
K
(
s
)
J
(
k
-
1
)
K
(
s
-
1
)
J
(
k
-
1
)
K
(
s
-
1
)
J
(
k
-
1
)
K
(
s
-
1
)
R
σ
+
K
(
k
)
R
σ
-
J
(
k
)
R
σ
-
R
σ
+
R
σ
-
R
σ
+
R
σ
-
R
σ
+
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
R
σ
-
J
(
k
-
1
)
K
(
k
)
R
σ
+
R
σ
-
R
σ
+
R
σ
-
R
σ
+
R
σ
-
R
σ
+
K
(
k
)
J
(
p
)
R
σ
-
R
σ
+
R
σ
-
R
σ
+
R
σ
-
R
σ
+
R
σ
+
J
(
k
)
J
(
k
)
R
σ
+
J
(
k
-
1
)
R
σ
+
J
(
k
-
1
)
R
σ
+
J
(
k
-
1
)
R
σ
+
J
(
k
)
J
(
p
)
J
(
k
-
1
)
J
(
p
-
1
)
J
(
k
-
1
)
J
(
p
-
1
)
J
(
k
-
1
)
J
(
p
-
1
)Next, we let{zn}n=-2∞ be a solution of (3) again. If (z-2,z-1)∈Rσ+Rσ+×Rσ-Rσ-, then by the (Rσ+J(k),Rσ-I(s)) entry of Table 15, we see that z0∈Rσ+Rσ+. Now that (z-1,z0)∈Rσ-Rσ-×Rσ+Rσ+, we may use the (K(k)Rσ-,J(s)Rσ+) entry of Table 16 to infer z1∈Rσ-Rσ- again. By induction, we see that z2n∈Rσ+Rσ+,z2n+1∈Rσ-Rσ- for all n≥-2. By (7), we then see that
(47)z2n=azn-2+a′k→,n∈N,z2n+1=az2n-1-a′k→,n∈N,
and hence by (14) and (15), we see that z2n→k→,z2n+1→-k→ as n→∞. By considering z-2 and z-1 in different parts of the plane, we may apply the same principle to obtain Table 17.Table 17
R
σ
-
R
σ
-
R
σ
+
R
σ
-
R
σ
-
R
σ
+
R
σ
+
R
σ
+
R
σ
-
R
σ
-
See  Table23
See  Table18
See  Table18
〈
-
k
→
,
k
→
〉
R
σ
+
R
σ
-
See  Table20
〈
i
→
〉
See  Table22
See  Table21
R
σ
-
R
σ
+
See  Table20
See  Table22
〈
-
i
→
〉
See  Table21
R
σ
+
R
σ
+
〈
k
→
,
-
k
→
〉
See  Table19
See  Table19
See  Table23In this table, there are twelve indeterminate cases. Let us go through one case. Let(z-2,z-1)∈Rσ-Rσ-×Rσ+Rσ-. Then, (z-2,z-1)∈K(k)Rσ-×Rσ+K(s) for some k,s∈N. There are then two subcases: (i) 0≤k≤s or (ii) 0≤s<k. In the former case, by Table 15,
(48)z0=az-2+a′Fσ(z-1)∈K(k-1)Rσ-,
and (hence)
(49)z1=az-1+a′Fσ(z0)∈Rσ+K(s-1),
and so forth, and finally,
(50)z2k∈K(-1)Rσ-⊆Rσ+Rσ-,z2k+1∈Rσ+Rσ-.
In the latter case, we may similarly show that (z2s,z2s+1)∈Rσ-Rσ-×Rσ+Rσ+. By considering (z-2,z-1) in different parts, we may then construct four self-explanatory Tables 18, 19, 20, and 21.Table 18
(
z
-
2
,
z
-
1
)
∈
K
(
k
)
R
σ
-
×
R
σ
+
K
(
s
)
(
R
σ
-
K
(
k
)
×
K
(
s
)
R
σ
+
)
0
≤
k
≤
s
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
+
R
σ
-
×
R
σ
+
R
σ
-
(
R
σ
-
R
σ
+
×
R
σ
-
R
σ
+
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
R
σ
-
R
σ
-
×
R
σ
+
R
σ
+
(
R
σ
-
R
σ
-
×
R
σ
+
R
σ
+
)Table 19
(
z
-
2
,
z
-
1
)
∈
J
(
k
)
R
σ
+
×
R
σ
-
J
(
s
)
(
R
σ
+
J
(
k
)
×
J
(
s
)
R
σ
-
)
0
≤
k
≤
s
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
-
R
σ
+
×
R
σ
-
R
σ
+
(
R
σ
+
R
σ
-
×
R
σ
+
R
σ
-
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
R
σ
+
R
σ
+
×
R
σ
-
R
σ
-
(
R
σ
+
R
σ
+
×
R
σ
-
R
σ
-
)Table 20
(
z
-
2
,
z
-
1
)
∈
R
σ
+
K
(
k
)
×
K
(
s
)
R
σ
-
(
K
(
k
)
R
σ
+
×
R
σ
-
K
(
s
)
)
0
≤
k
≤
s
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
+
R
σ
+
×
R
σ
-
R
σ
-
(
R
σ
+
R
σ
+
×
R
σ
-
R
σ
-
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
R
σ
+
R
σ
-
×
R
σ
+
R
σ
-
(
R
σ
-
R
σ
+
×
R
σ
-
R
σ
+
)Table 21
(
z
-
2
,
z
-
1
)
∈
R
σ
-
J
(
k
)
×
J
(
s
)
R
σ
+
(
J
(
k
)
R
σ
-
×
R
σ
+
J
(
s
)
)
0
≤
k
≤
s
⇒
(
z
2
k
,
z
2
k
+
1
)
∈
R
σ
-
R
σ
-
×
R
σ
+
R
σ
+
(
R
σ
-
R
σ
-
×
R
σ
+
R
σ
+
)
0
≤
s
<
k
⇒
(
z
2
s
,
z
2
s
+
1
)
∈
R
σ
-
R
σ
+
×
R
σ
-
R
σ
+
(
R
σ
+
R
σ
-
×
R
σ
+
R
σ
-
)As a consequence, if(z-2,z-1)∈K(k)R-×R+K(s) and 0≤k≤s, then the solution of (3) originated from z-2 and z-1 will satisfy (z2k,z2k+1)∈Rσ+Rσ-×Rσ+Rσ-. Then by Table 17, zn→〈i→〉.Next, let(z-2,z-1)∈Rσ-Rσ+×Rσ+Rσ-. Then, (z-2,z-1)∈K(k)J(p)×J(l)K(m) for some k,p,l,m∈N.As mentioned before, there are twelve different ordering arrangements B1–B12 for the nonnegative integers k,p,l,andm. In case B2 holds, we may make use of Table 22 repeatedly to show that (z2k,z2k+1)∈Rσ+J(p-k-1)×J(l-k-1)Rσ- (which is recorded in self-explanatory Tables 22 and 23). If 0≤p-k-1≤l-k-1, then by Table 19, (z2p,z2p+1)∈Rσ+Rσ-×Rσ+Rσ-. By Table 17 again, zn→〈i→〉.Table 22
(
z
-
2
,
z
-
1
)
∈
K
(
k
)
J
(
p
)
×
J
(
l
)
K
(
m
)
(
J
(
k
)
K
(
p
)
×
K
(
l
)
J
(
m
)
)
B1⇒(z2k,z2k+1)∈Rσ+Rσ-×Rσ+Rσ-
(
R
σ
-
R
σ
+
×
R
σ
-
R
σ
+
)
B2⇒(z2k,z2k+1)∈Rσ+J(p-k-1)×J(l-k-1)Rσ-
(
R
σ
-
K
(
p
-
k
-
1
)
×
K
(
l
-
k
-
1
)
R
σ
+
)
B3⇒(z2p,z2p+1)∈K(k-p-1)Rσ-×Rσ+K(m-p-1)
(
J
(
k
-
p
-
1
)
R
σ
+
×
R
σ
-
J
(
m
-
p
-
1
)
)
B4⇒(z2p,z2p+1)∈K(k-p-1)Rσ-×Rσ+K(m-p-1)
(
J
(
k
-
p
-
1
)
R
σ
+
×
R
σ
-
J
(
m
-
p
-
1
)
)
B5⇒(z2p,z2p+1)∈Rσ-Rσ-×Rσ+Rσ+
(
R
σ
+
R
σ
+
×
R
σ
-
R
σ
-
)
B6⇒(z2l,z2l+1)∈K(k-l-1)Rσ+×Rσ-K(m-l-1)
(
J
(
k
-
l
-
1
)
R
σ
-
×
R
σ
+
J
(
m
-
l
-
1
)
)
B7⇒(z2l,z2l+1)∈Rσ-Rσ+×Rσ-Rσ+
(
R
σ
+
R
σ
-
×
R
σ
+
R
σ
-
)
B8⇒(z2l,z2l+1)∈Rσ+Rσ+×Rσ-Rσ-
(
R
σ
-
R
σ
-
×
R
σ
+
R
σ
+
)
B9⇒(z2m,z2m+1)∈Rσ-J(p-m-1)×J(l-m-1)Rσ+
(
R
σ
+
K
(
p
-
m
-
1
)
×
K
(
l
-
m
-
1
)
R
σ
-
)
B10⇒(z2m,z2m+1)∈Rσ+J(p-m-1)×J(l-m-1)Rσ-
(
R
σ
-
K
(
p
-
m
-
1
)
×
K
(
l
-
m
-
1
)
R
σ
+
)
B11⇒(z2p,z2p+1)∈Rσ-Rσ-×Rσ+Rσ+
(
R
σ
+
R
σ
+
×
R
σ
-
R
σ
-
)
B12⇒(z2l,z2l+1)∈Rσ+Rσ+×Rσ-Rσ-
(
R
σ
-
R
σ
-
×
R
σ
+
R
σ
+
)Table 23
(
z
-
2
,
z
-
1
)
∈
J
(
k
)
J
(
p
)
×
J
(
l
)
J
(
m
)
(
K
(
k
)
K
(
p
)
×
K
(
l
)
K
(
m
)
)
B1⇒(z2k,z2k+1)∈Rσ-Rσ-×Rσ+Rσ+
(
R
σ
+
R
σ
+
×
R
σ
-
R
σ
-
)
B2⇒(z2k,z2k+1)∈Rσ-J(p-k-1)×J(l-k-1)Rσ+
(
R
σ
+
K
(
p
-
k
-
1
)
×
K
(
l
-
k
-
1
)
R
σ
-
)
B3⇒(z2p,z2p+1)∈J(k-p-1)Rσ-×Rσ+J(m-p-1)
(
K
(
k
-
p
-
1
)
R
σ
+
×
R
σ
-
K
(
m
-
p
-
1
)
)
B4⇒(z2p,z2p+1)∈J(k-p-1)Rσ-×Rσ+J(m-p-1)
(
K
(
k
-
p
-
1
)
R
σ
+
×
R
σ
-
K
(
m
-
p
-
1
)
)
B5⇒(z2p,z2p+1)∈Rσ+Rσ-×Rσ+Rσ-
(
R
σ
-
R
σ
+
×
R
σ
-
R
σ
+
)
B6⇒(z2l,z2l+1)∈J(k-l-1)Rσ+×Rσ-J(m-l-1)
(
K
(
k
-
l
-
1
)
R
σ
-
×
R
σ
+
K
(
m
-
l
-
1
)
)
B7⇒(z2l,z2l+1)∈Rσ+Rσ+×Rσ-Rσ-
(
R
σ
-
R
σ
-
×
R
σ
+
R
σ
+
)
B8⇒(z2l,z2l+1)∈Rσ-Rσ+×Rσ-Rσ+
(
R
σ
+
R
σ
-
×
R
σ
+
R
σ
-
)
B9⇒(z2m,z2m+1)∈Rσ+J(p-m-1)×J(l-m-1)Rσ-
(
R
σ
-
K
(
p
-
m
-
1
)
×
K
(
l
-
m
-
1
)
R
σ
+
)
B10⇒(z2m,z2m+1)∈Rσ-J(p-m-1)×J(l-m-1)Rσ+
(
R
σ
+
K
(
p
-
m
-
1
)
×
K
(
l
-
m
-
1
)
R
σ
-
)
B11⇒(z2p,z2p+1)∈Rσ+Rσ-×Rσ+Rσ-
(
R
σ
-
R
σ
+
×
R
σ
-
R
σ
+
)
B12⇒(z2l,z2l+1)∈Rσ-Rσ+×Rσ-Rσ+
(
R
σ
+
R
σ
-
×
R
σ
+
R
σ
-
)In conclusion, we have shown the following result.Theorem 6.
Suppose that0≤σ<1. Then, a solution of (3) must be either asymptotically 1-periodic tending to the limit cycles 〈i→〉 or 〈-i→〉 or asymptotically 2-periodic tending to the limit cycle 〈k→,-k→〉.We remark that, as can be seen from the above exhaustive arguments and Example2, the region of attraction of each limit cycle can be given precisely (by implementing a simple computer program, if necessary).


## 5. Conclusions

In neural network terminologies, we have discussed a simple two-neuron recurrent McCulloch-Pitts-type neural network with a nonnegative thresholdσ. For σ>1, all solutions tend to the limiting state (1,1); for σ=1, all solutions must either be tending to the limiting states (-1,1),(1,-1) or (1,1), or asymptotically 2-periodic tending to the limit cycles 〈(1,-1),(1,1)〉,〈(-1,1),(1,1)〉, or 〈(-1,-1),(1,1)〉; and for 0≤σ<1, all solutions must either be tending to the steady states (-1,1), or (1,-1) or asymptotically 2-periodic tending to the limit cycle 〈(1,1),(-1,-1)〉. In all cases, the corresponding region of attraction of each limit cycle can be given.Our proofs show that more general multiple-neuron recurrent McCulloch-Pitts-type neural networks possess similar behaviors. However, the derivations may involve more delicate graph theoretic arguments and are better left for other studies.



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*Source: 492014-2013-04-21.xml*