Discrete Dynamics in Nature and Society

Discrete Dynamics in Nature and Society / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 745697 | 17 pages | https://doi.org/10.1155/2012/745697

Complete Asymptotic Analysis of a Two-Nation Arms Race Model with Piecewise Constant Nonlinearities

Academic Editor: Guang Zhang
Received30 Aug 2011
Accepted25 Oct 2011
Published05 Jan 2012

Abstract

A discrete time two-nation arms race model involving a piecewise constant nonlinear control function is formulated and studied. By elementary but novel arguments, we are able to give a complete analysis of its asymptotic behavior when the threshold parameter in the control function varies from 0+ to . We show that all solutions originated from positive initial values tend to limit one or two cycles. An implication is that when devastating weapons are involved, “terror equilibrium” can be achieved and escalated race avoided. It is hoped that our analysis will provide motivation for further studying of discrete-time equations with piecewise smooth nonlinearities.

1. Introduction

In [1, pages 87–90], a simple dynamical model of a two-nation arms race based on Richardson’s ideas in [2] is explained, and several interesting conclusions are drawn which can be used to explain stable and escalated arms races. Roughly, let 𝐍={0,1,2,}, and let 𝐴𝑛 and 𝐵𝑛 be the amount spent on armaments by two respective countries 𝐴 and 𝐵 in year 𝑛𝐍. Assuming, 𝐴 has a fixed amount of distrust of the other country, causing it to retain arms, then𝐴𝑛=1𝑟𝐴𝐴𝑛1+𝑠𝐴𝐵𝑛1+𝑢,(1.1) where the constant 𝑠𝐴 measures country 𝐴’s distrust of country 𝐵 in that it reacts to the way 𝐵 arms itself, 𝑟𝐴(0,1) is a measure of 𝐴’s own economy, and 𝑢 is the basic annual expenditure (e.g., maintenance expense). If we now assume a similar situation for country 𝐵, then we also have𝐵𝑛=1𝑟𝐵𝐵𝑛1+𝑠𝐵𝐴𝑛1+𝑣.(1.2) Under the assumption that 𝑟𝐴=𝑟𝐵=𝑟 and 𝑠𝐴=𝑠𝐵=𝑠, it is shown that if the initial total expenditure 𝐴1+𝐵1 is large and that the distrust factor is also so large that 𝑠>𝑟, then no two countries can sustain exponentially increasing expenditures on arms, and the alternative is war or negotiation. While this model is an oversimplification one, it could help to understand plausible reasons behind World War I (see [2, 3] in which various aspects of arms race modeling are discussed).

The above model cannot explain some of the observations we can make nowadays. Therefore, we need to build various models and analyze their asymptotic behaviors. In this paper, we will build one such model based on the idea that although the distrust factor is the same as in the previous model, the expenditure by the other country in year 𝑛1 in (1.1) is replaced by 𝑠𝐴𝑓𝜆(𝐵𝑛1) where𝑓𝜆𝐵𝑛1=1if𝐵𝑛1],(0,𝜆0if𝐵𝑛1](,0(𝜆,),(1.3) and the term 𝐴𝑛1 in (1.2) is replaced by a similar one. The “discontinuous” function 𝑓𝜆 has a clear physical meaning. Indeed, the positive parameter 𝜆 may be treated as a cutoff threshold indicator so that when the competitor is already spending an unreasonable amount of money (such as stocking of hundreds of nuclear missiles that can annihilate our mother earth) or is not spending any, there is no need to add the budget anymore. With this function at hand, we may rewrite our new model as follows:𝑥𝑛=𝑎𝑥𝑛𝛼+𝑏𝑓𝜆𝑦𝑛1𝑦+𝑐,𝑛=𝑟𝑦𝑛𝛽+𝑠𝑓𝜏𝑥𝑛1+𝑡,(1.4) where we have introduced two “delays” 𝛼 and 𝛽 in order to reflect the fact that the expenditure in a previous accounting period may not be recorded precisely, and hence historical expenditure records may be more reliable for use in making future decisions.

Although (1.4) may seem to be a simple model, there are still too many parameters involved. We therefore make further (reasonable) assumptions as follows:𝑐,𝑡=0,𝛼=𝛽=2,𝜏=𝜆>0,𝑟=𝑎(0,1),𝑏=𝑠=(1𝑎).(1.5) The assumption that 𝑐,𝑡=0 means that the fixed expenditures are relatively low in both countries, while we assume that 𝛼=𝛽=2 so as to use the best up-to-date and “reliable” accounting records 𝑥𝑛2 and 𝑦𝑛2. By choosing 𝑎(0,1) and 𝑏=1𝑎, country 𝐴 is making a decision based on a convex combination of the expenditures 𝑥𝑛2 and the blanket-ceiling sum 𝑓𝜆(𝑦𝑛1). If country 𝐵 takes on a similar decision policy, then we end up with𝑦𝑛=𝑎𝑦𝑛2+1𝑎𝑓𝜆𝑥𝑛1,(1.6) where 𝑎 may or may not differ from 𝑎. The case 𝑎𝑎 is only more technically difficult, and therefore, in this paper, we will assume the case 𝑎=𝑎 (which is already nontrivial as we will see) so that both countries play “symmetric” roles in the interactions.

By adopting these assumptions, we then settle on the following dynamical system:𝑥𝑛=𝑎𝑥𝑛2+(1𝑎)𝑓𝜆𝑦𝑛1,𝑦𝑛=𝑎𝑦𝑛2+(1𝑎)𝑓𝜆𝑥𝑛1,(1.7) for 𝑛𝐍, where in this model, 𝑎(0,1),𝜆>0. Note that if we let 𝑧=(𝑥,𝑦) and𝐹𝜆𝑓(𝑧)=𝜆(𝑦),𝑓𝜆(𝑥),(1.8) then the above system (1.7) can be written as𝑧𝑛=𝑎𝑧𝑛2+𝑎𝐹𝜆𝑧𝑛1,𝑛𝐍,(1.9) where we write 𝑧𝑛=(𝑥𝑛,𝑦𝑛) and 𝑎=1𝑎 for the sake of convenience.

The above vector equation is a three-term recurrence relation. Hence, for given 𝑧2 and 𝑧1 in the plane, a unique sequence {𝑧𝑘}𝑘=2 can be calculated from it. Such a sequence is called a solution of (1.9) determined by 𝑧2 and 𝑧1. Among different 𝑧2 and 𝑧1, those lying in the positive quadrant are of special interests since expenditures are always positive. Therefore, our subsequent interests are basically the asymptotic behaviors of all such solutions determined by 𝑧2 and 𝑧1 with positive components.

We remark that system (1.9) can be regarded as a discrete dynamical system with piecewise smooth nonlinearities. Such systems have not been explored extensively (see, e.g., the discussions on “polymodal” discrete systems in [4], and there are only several recent studies on scalar equations with piecewise smooth nonlinearities [59])! Therefore, a complete asymptotic analysis of our equation is essential in the further development of discontinuous (in particular, 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 given any 𝑧2,𝑧1 in the quadrant (0,)2, the solution {𝑧𝑛}𝑛2 determined by it also lies in the same quadrant (in the sense that 𝑧𝑛(0,)2 for 𝑛𝐍). Depending on the locations of 𝑧2 and 𝑧1, it is clear that the behavior of the corresponding solution may differ. For this reason, it is convenient to distinguish various parts of the first quadrant in the following manner:]𝐴=(0,𝜆2]],𝐵=(𝜆,)×(0,𝜆,𝐶=(𝜆,)×(𝜆,),𝐷=(0,𝜆×(𝜆,),(1.10) where 𝜆 is a fixed positive number, then𝔓={𝐴,𝐵,𝐶,𝐷}(1.11) is a partition of the quadrant (0,)2.

Note that these subsets depend on 𝜆, but this dependence is not emphasized in the sequel for the sake of convenience.

For any solution {𝑧𝑛}𝑛=2 originated from 𝑧2 and 𝑧1 in the above subsets, our main conclusion in this paper is that {𝑧2𝑛} will tend to some vector 𝑢 and {𝑧2𝑛+1} will tend to another vector 𝑣 (which may or may not be equal to 𝑢). This implication is important since it says that escalated arms race cannot happen and World War III should not happen if our model is correct!

For the sake of convenience, we record this fact by means of𝑧𝑛𝑢,𝑣.(1.12) In case 𝑢=𝑣,{𝑧𝑛} is convergent to 𝑢, and hence we may also write𝑧𝑛𝑢or𝑧𝑛𝑢.(1.13)

To arrive at our main conclusion, we note, however, that since 𝑓𝜆 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 (1.9) 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: suppose that 𝜆=1. If {𝑧𝑛}𝑛=2 is a solution of (1.9) with (𝑧2,𝑧1)𝐴×𝐵, then as will be seen below, 𝑧2𝑛(1,0) and 𝑧2𝑛+1(1,1). If we now replace the condition (𝑧2,𝑧1)𝐴×𝐵 with the symmetric initial condition (𝑧2,𝑧1)𝐴×𝐷, then we will end up with the conclusion that 𝑧2𝑛(0,1) and 𝑧2𝑛+1(1,1). Such two conclusions will be referred to as dual results. We will see some tables which contain some obvious dual results later.

Next, by (1.8), we may easily see that𝐹𝜆(𝐴)=𝐤,𝐹𝜆(𝐵)=𝐢,𝐹𝜆(𝐶)=𝟎,𝐹𝜆(𝐷)=𝐣,(1.14) where𝟎=(0,0),𝐢=(1,0),𝐣=(0,1),𝐤=(1,1).(1.15) Therefore, in case {𝑧𝑘} is a solution of (1.9) such that, say, 𝑧𝑘𝐴 for all large 𝑘, then (1.9) is reduced to𝑧𝑛=𝑎𝑧𝑛2+𝑎𝐤,(1.16) for all large 𝑛. Hence, linear systems and their related properties will also appear in later discussions. More precisely, two groups of bounding quantities 𝛼𝑗 and 𝛽𝑗 are needed:𝛼𝑗=1+𝜆1𝑎𝑗,𝛽𝑗=𝜆𝑎𝑗,𝑗𝐍.(1.17) Note that they satisfy 𝛼0=𝛽0=𝜆 and the recurrence relations 𝛼𝑗+1=1𝑎𝑎𝑗+𝑎1𝑎,𝛽𝑗+1=1𝑎𝛽𝑗,𝑗𝐍.(1.18) We also need the following two properties of linear systems. Let {𝑥𝑘}𝑘=2 be real scalar (or vector) sequences that satisfy𝑥2𝑘=𝑎𝑥2𝑘2+𝑑,𝑘𝐍,(1.19)𝑥2𝑘+1=𝑎𝑥2𝑘1+𝑑,𝑘𝐍,(1.20) where 𝑎(0,1), and 𝑑 is a real number (resp., a real vector).(i)If {𝑥𝑛}𝑛=2 is a sequence which satisfies (1.19), then 𝑥2𝑘=𝑎𝑘+1𝑥2+1𝑎𝑘+11𝑎𝑑,𝑘𝐍.(1.21)(ii)If {𝑥𝑛}𝑛=2 is a sequence which satisfies (1.20), then𝑥2𝑘+1=𝑎𝑘+1𝑥1+1𝑎𝑘+11𝑎𝑑,𝑘𝐍.(1.22)

Finally, we need to consider various ordering arrangements for three or four nonnegative integers 𝑘,𝑝,𝑙,and𝑚. First, the ordering arrangements of three integers 𝑘,𝑝,and𝑙 can be classified into 6 cases: (1) 𝑘=𝑝𝑙,(2)𝑙=𝑘<𝑝,(3)𝑝=𝑙<𝑘,(4)𝑘<min{𝑝,𝑙},(5)𝑝<min{𝑘,𝑙},and(6)𝑙<min{𝑘,𝑝}. In fact, let 𝑎,𝑏𝑅. Then either 𝑎<𝑏,𝑎=𝑏, or 𝑎>𝑏. Let 𝑎,𝑏,𝑐𝑅, then𝑎<𝑏𝑐(,𝑎),𝑐=𝑎,𝑐(𝑎,𝑏),𝑐=𝑏or𝑐(𝑏,),𝑎=𝑏𝑐(,𝑎),𝑐=𝑎or𝑐(𝑎,),𝑎>𝑏𝑐(,𝑏),𝑐=𝑏,𝑐(𝑏,𝑎),𝑐=𝑎or𝑐(𝑎,).(1.23) These are equivalent to𝑎=𝑏𝑐,𝑐=𝑎<𝑏,𝑏=𝑐<𝑎,𝑎<min{𝑏,𝑐},𝑏<min{𝑎,𝑐},𝑐<min{𝑎,𝑏},(1.24) by comparing the two sets of statements.

By similar reasoning, there are 12 ordering arrangements for four integers 𝑘,𝑝,𝑙,and𝑚: (1)𝑘=𝑝min{𝑙,𝑚},(2)𝑘<min{𝑝,𝑙,𝑚},(3)𝑝<min{𝑘,𝑙,𝑚},(4)𝑝=𝑙<min{𝑘,𝑚},(5)𝑝=𝑙=𝑚<𝑘,(6)𝑙<min{𝑝,𝑘,𝑚},(7)𝑙=𝑚<min{𝑘,𝑝},(8)𝑙=𝑚=𝑘<𝑝,(9)𝑚<min{𝑙,𝑘,𝑝},(10)𝑚=𝑘<min{𝑙,𝑝},(11)𝑝=𝑚<min{𝑙,𝑘},and(12)𝑘=𝑙<min{𝑝,𝑚}.

Our following plan is quite simple. We will treat our 𝜆 as a bifurcation parameter and distinguish four different cases (i) 𝜆>1, (ii) 𝜆=1, (iii) 0<𝜆<1𝑎, and (iv) 1𝑎𝜆<1, and consider different 𝑧2,𝑧1 in 𝐴,𝐵,𝐶, or 𝐷 and discuss the precise asymptotic behaviors of the corresponding solutions determined by them.

2. The Case 𝜆>1

This case is relatively simple.

Theorem 2.1. Suppose, 𝜆>1. Let {𝑧𝑘}𝑘=2 be any solution of (1.9) with (𝑧2,𝑧1)(0,)2. Then 𝑧𝑛𝐤.

Proof. By (1.9), we may see that 𝑥𝑛𝑎𝑥𝑛2+𝑎 and 𝑦𝑛𝑎𝑦𝑛2+𝑎 for 𝑛𝐍, then lim𝑛𝑥𝑛1<𝜆 and lim𝑛𝑦𝑛1<𝜆. Thus, there exists an integer 𝑚 such that (𝑧𝑘,𝑧𝑘+1)(0,𝜆]2 for all 𝑘𝑚. Therefore, 𝑧𝑘+2=𝑎𝑧𝑘+𝑎𝐤 for all 𝑘𝑚. In view of (1.21) and (1.22), 𝑧𝑛𝐤. The proof is complete.

3. The Case 𝜆=1

In this section, we assume that 𝜆=1. If {𝑧𝑘}𝑘=2 is a solution of (1.9) and if 𝑧𝑘𝐴 and 𝑧𝑘+1(0,)2, then in view of (1.14), 𝐹𝜆(𝑧𝑘+1){𝐢,𝐣,𝐤,𝟎}, and hence𝑧𝑘+2=𝑎𝑧𝑘+𝑎𝐹𝜆𝑧𝑘+1𝐴.(3.1) By similar reasoning, we may consider all other possible cases and collect our findings in tables. To simplify the description of these tables, we first note that {𝛽𝑗}𝑗=0={1/𝑎𝑗}𝑗=0 is a strictly increasing and divergent sequence. Therefore, if we let𝐼(𝑖)=𝛽𝑖,𝛽𝑖+1,𝐵(𝑖)=𝐼(𝑖)]×(0,1,𝐷(𝑖)]=(0,1×𝐼(𝑖),𝐶(𝑖,𝑗)=𝐼(𝑖)×𝐼(𝑗),(3.2) then(1,)=𝑖=0𝐼(𝑖),𝐵=𝑖=0𝐵(𝑖),𝐷=𝑖=0𝐷(𝑖),𝐶=𝑖,𝑗𝑁𝐶(𝑖,𝑗).(3.3)

First of all, the fact that 𝑧𝑘𝐴 and 𝑧𝑘+1𝐴 implies, 𝑧𝑘+2𝐴 is recorded as the (𝐴,𝐴) entry in Table 1. In this table, we may also find other entries which are self-explanatory.


𝐴 𝐵 ( 𝑠 ) 𝐶 ( 𝑠 , 𝑡 ) 𝐷 ( 𝑠 )

𝐴 𝐴 𝐴 𝐴 𝐴
𝐵 ( 𝑖 ) 𝐵 𝐵 𝑎 𝐵 ( 𝑖 ) 𝑎 𝐵 ( 𝑖 ) + 𝑎 𝐣
𝐶 ( 𝑖 , 𝑗 ) 𝐶 𝑎 𝐶 ( 𝑖 , 𝑗 ) + 𝑎 𝐢 𝑎 𝐶 ( 𝑖 , 𝑗 ) 𝑎 𝐶 ( 𝑖 , 𝑗 ) + 𝑎 𝐣
𝐷 ( 𝑖 ) 𝐷 𝑎 𝐷 ( 𝑖 ) + 𝑎 𝐢 𝑎 𝐷 ( 𝑖 ) 𝐷

By similar considerations, we may also obtain Tables 1 and 2.


Initial conditionConditionConclusion

( 𝑧 2 , 𝑧 1 ) 𝐵 ( 𝑘 ) × 𝐶 ( 𝑠 , 𝑝 ) 0 𝑘 𝑝 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐴 × 𝐶
( 𝑧 2 , 𝑧 1 ) 𝐵 ( 𝑘 ) × 𝐶 ( 𝑠 , 𝑝 ) 0 𝑝 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 × 𝐵
( 𝑧 2 , 𝑧 1 ) 𝐵 ( 𝑘 ) × 𝐷 ( 𝑝 ) 0 𝑘 𝑝 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐴 × 𝐷
( 𝑧 2 , 𝑧 1 ) 𝐵 ( 𝑘 ) × 𝐷 ( 𝑝 ) 0 𝑝 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 × 𝐴
( 𝑧 2 , 𝑧 1 ) 𝐶 ( 𝑠 , 𝑘 ) × 𝐵 ( 𝑝 ) 0 𝑘 𝑝 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐵 × 𝐵
( 𝑧 2 , 𝑧 1 ) 𝐶 ( 𝑠 , 𝑘 ) × 𝐵 ( 𝑝 ) 0 𝑝 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐶 × 𝐴
( 𝑧 2 , 𝑧 1 ) 𝐷 ( 𝑘 ) × 𝐶 ( 𝑝 , 𝑠 ) 0 𝑘 𝑝 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐴 × 𝐶
( 𝑧 2 , 𝑧 1 ) 𝐷 ( 𝑘 ) × 𝐶 ( 𝑝 , 𝑠 ) 0 𝑝 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐷 × 𝐷
( 𝑧 2 , 𝑧 1 ) 𝐷 ( 𝑘 ) × 𝐵 ( 𝑝 ) 0 𝑘 𝑝 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐴 × 𝐵
( 𝑧 2 , 𝑧 1 ) 𝐷 ( 𝑘 ) × 𝐵 ( 𝑝 ) 0 𝑝 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐷 × 𝐴
( 𝑧 2 , 𝑧 1 ) 𝐶 ( 𝑘 , 𝑠 ) × 𝐷 ( 𝑝 ) 0 𝑘 𝑝 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐷 × 𝐷
( 𝑧 2 , 𝑧 1 ) 𝐶 ( 𝑘 , 𝑠 ) × 𝐷 ( 𝑝 ) 0 𝑝 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐶 × 𝐴

For instance, let us show the third data row in Table 2. Suppose that 𝑧2𝐵(𝑘) and 𝑧1𝐷(𝑝), where 0𝑘𝑝. By Table 1, if 0=𝑘𝑝, then𝑧0=𝑎𝑧2+𝑎𝐹𝜆𝑧1=𝑎𝑧2+𝑎𝐣𝑎𝐵(0)+𝑎𝑧𝐣𝐴,1=𝑎𝑧1+𝑎𝐹𝜆𝑧0=𝑎𝑧1+𝑎𝐤𝑎𝐷(𝑝)+𝑎𝐤𝐷.(3.4) If 0<𝑘𝑝, then𝑧0=𝑎𝑧2+𝑎𝐹𝜆𝑧1=𝑎𝑧2+𝑎𝐣𝑎𝐵(𝑘)+𝑎𝐣𝐵(𝑘1),𝑧1=𝑎𝑧1+𝑎𝐹𝜆𝑧0=𝑎𝑧1+𝑎𝐢𝑎𝐷(𝑝)+𝑎𝐢𝐷(𝑝1),𝑧2=𝑎𝑧0+𝑎𝐹𝜆𝑧1=𝑎𝑧0+𝑎𝐣𝑎𝐵(𝑘1)+𝑎𝐣𝐵(𝑘2),𝑧3=𝑎𝑧1+𝑎𝐹𝜆𝑧2=𝑎𝑧1+𝑎𝐢𝑎𝐷(𝑝1)+𝑎𝐢𝐷(𝑝2),(3.5) and by induction,𝑧2𝑘=𝑎𝑧2𝑘2+𝑎𝐹𝜆𝑧2𝑘1=𝑎𝑧2𝑘2+𝑎𝐣𝑎𝐵(0)+𝑎𝑧𝐣𝐴,2𝑘+1=𝑎𝑧2𝑘1+𝑎𝐹𝜆𝑧2𝑘=𝑎𝑧2𝑘1+𝑎𝐤𝑎𝐷(𝑝𝑘)+𝑎𝐤𝐷(𝑝𝑘1),(3.6) that is, 𝑧2𝑘𝐴 and 𝑧2𝑘+1𝐷.

As another example, let us show the second data row in Table 3. Suppose that (𝑧2,𝑧1)𝐶(𝑘,𝑝)×𝐶(𝑙,𝑚), where 0𝑘<min{𝑙,𝑚,𝑝}, then by Table 1, if 0=𝑘<min{𝑙,𝑚,𝑝},𝑧0=𝑎𝑧2+𝑎𝐹𝜆𝑧1=𝑎𝑧2𝑎𝐶(0,𝑝)𝐷(𝑝1),𝑧1=𝑎𝑧1+𝑎𝐹𝜆𝑧0=𝑎𝑧1+𝑎𝐣𝑎𝐶(𝑙,𝑚)+𝑎𝐣𝐼(𝑙1)×(1,).(3.7) If 0<𝑘<min{𝑙,𝑚,𝑝}, then𝑧0=𝑎𝑧2+𝑎𝐹𝜆𝑧1=𝑎𝑧2𝑎𝐶(𝑘,𝑝)𝐶(𝑘1,𝑝1),𝑧1=𝑎𝑧1+𝑎𝐹𝜆𝑧0=𝑎𝑧1𝑎𝐶(𝑙,𝑚)𝐶(𝑙1,𝑚1),𝑧2=𝑎𝑧0+𝑎𝐹𝜆𝑧1=𝑎𝑧0𝑎𝐶(𝑘1,𝑝1)𝐶(𝑘2,𝑝2),𝑧3=𝑎𝑧1+𝑎𝐹𝜆𝑧2=𝑎𝑧1𝑎𝐶(𝑙1,𝑚1)𝐶(𝑙2,𝑚2),(3.8) and by induction,𝑧2𝑘=𝑎𝑧2𝑘2+𝑎𝐹𝜆𝑧2𝑘1=𝑎𝑧2𝑘2𝑎𝐶(0,𝑝𝑘)𝐷(𝑝𝑘1),𝑧2𝑘+1=𝑎𝑧2𝑘1+𝑎𝐹𝜆𝑧2𝑘=𝑎𝑧2𝑘1+𝑎𝐣𝑎𝐶(𝑙𝑘,𝑚𝑘)+𝑎𝐣𝐼(𝑙𝑘1)×(1,),(3.9) that is, 𝑧2𝑘𝐷(𝑝𝑘1) and 𝑧2𝑘+1𝐶(𝑙𝑘1,𝑡) for some 𝑡𝑁.


ConditionConclusion

0 𝑘 = 𝑝 m i n { 𝑙 , 𝑚 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐴 × 𝐶
0 𝑘 < m i n { 𝑙 , 𝑚 , 𝑝 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐷 ( 𝑝 𝑘 1 ) × 𝐶 ( 𝑙 𝑘 1 , 𝑡 ) for some 𝑡 𝐍
0 𝑝 < m i n { 𝑘 , 𝑙 , 𝑚 } ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 ( 𝑘 𝑝 1 ) × 𝐶 ( 𝑡 , 𝑚 𝑝 1 ) for some 𝑡 𝐍
0 𝑝 = 𝑙 < m i n { 𝑚 , 𝑘 } ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 ( 𝑘 𝑝 1 ) × 𝐶 ( 𝑡 , 𝑚 𝑝 1 ) for some 𝑡 𝐍
0 𝑝 = 𝑙 = 𝑚 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 × 𝐵
0 𝑙 < m i n { 𝑚 , 𝑘 , 𝑝 } ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) 𝐶 ( 𝑘 𝑙 1 , 𝑝 𝑙 1 ) × 𝐷 ( 𝑚 𝑙 1 )
0 𝑙 = 𝑚 < m i n { 𝑝 , 𝑘 } ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) 𝐶 × 𝐴
0 𝑙 = 𝑘 = 𝑚 < 𝑝 ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) 𝐷 × 𝐷
0 𝑚 < m i n { 𝑙 , 𝑘 , 𝑝 } ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐶 ( 𝑘 𝑚 1 , 𝑝 𝑚 1 ) × 𝐵 ( 𝑙 𝑚 1 )
0 𝑘 = 𝑚 < m i n { 𝑙 , 𝑝 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐷 ( 𝑝 𝑘 1 ) × 𝐶 ( 𝑙 𝑘 1 , 𝑡 ) for some 𝑡 𝐍
0 𝑝 = 𝑚 < m i n { 𝑙 , 𝑘 } ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 × 𝐵
0 𝑘 = 𝑙 < m i n { 𝑝 , 𝑚 } ( 𝑧 2 𝑘 , 𝑧 2 𝑙 + 1 ) 𝐷 × 𝐷

By means of the information obtained so far, let {𝑧𝑛}𝑛=2 be a solution of (1.9), we will be able to show the following result.

Theorem 3.1. Suppose that 𝜆=1. Let {𝑧𝑛}𝑛=2 be a solution of (1.9) originated from (0,)2. Then 𝑧𝑛𝐢,𝐣,𝐤,𝟎,𝐤,𝐤,𝟎,𝐢,𝐤,𝐤,𝐢,𝐣,𝐤or𝐤,𝐣.(3.10)

To this end, let us consider first the case where (𝑧2,𝑧1)𝐴×𝐴, then by the (𝐴,𝐴) entry in Table 1, 𝑧0𝐴. By induction, we may then see that 𝑧𝑘𝐴 for all 𝑘2. Hence, by (1.9), we see that𝑧𝑛=𝑎𝑧𝑛2+(1𝑎)𝐤,𝑛𝑁,(3.11) from which we easily obtain𝑧2𝑘=𝑎𝑘+1𝑧2+1𝑎𝑘+1𝑧𝐤,𝑛𝑁,2𝑘+1=𝑎𝑘+1𝑧1+1𝑎𝑘+1𝐤,𝑛𝑁.(3.12) By (1.21) and (1.22), we see that 𝑧2𝑘,𝑧2𝑘+1𝐤 so that 𝑧𝑛𝐤. We record this conclusion in the (𝐴,𝐴) entry of Table 4.


𝐴 𝐵 ( 𝑠 ) 𝐶 ( 𝑠 , 𝑡 ) 𝐷 ( 𝑠 )

𝐴 𝐤 𝐢 , 𝐤 𝟎 , 𝐤 𝐣 , 𝐤
𝐵 ( 𝑖 ) 𝐤 , 𝐢 𝐢 𝟎 , 𝐤 or 𝐢 𝐣 , 𝐤 or 𝐤 , 𝐢
𝐶 ( 𝑖 , 𝑗 ) 𝐤 , 𝟎 𝐢 or 𝐤 , 𝟎 𝟎 , 𝐤 , 𝐤 , 𝟎 , 𝐢 or 𝐣 𝐣 or 𝐤 , 𝟎
𝐷 ( 𝑖 ) 𝐤 , 𝐣 𝐢 , 𝐤 or 𝐤 , 𝐢 𝟎 , 𝐤 or 𝐢 𝐣

Consider another case where 𝑧2𝐴 and 𝑧1𝐵(𝑗). Then by the (𝐴,𝐵(𝑗)) entry of Table 1, we see that 𝑧0𝐴. Since 𝑧1𝐵(𝑗) and 𝑧0𝐴, then by Table 1 again,𝑧1𝑎𝐵(𝑗)+𝑎𝐤𝐵(𝑖),(3.13) for some 𝑖𝑁. By induction, we may then see that 𝑧2𝑘𝐴 and 𝑧2𝑘+1𝐵 for 𝑘1. Hence, by (1.9), we see that𝑧2𝑛=𝑎𝑧2𝑛2+𝑎𝑧𝐢,2𝑛+1=𝑎𝑧2𝑛1+𝑎𝐤,(3.14) for 𝑛𝑁. We may then easily see that 𝑧2𝑛𝐢 and 𝑧2𝑛+1𝐤.

By similar arguments, we may then derive the (𝐴,𝐶(𝑠,𝑡)),(𝐴,𝐷(𝑠)),(𝐵(𝑖),𝐴),(𝐵(𝑖),𝐵(𝑠)),(𝐶(𝑖,𝑗),𝐴),(𝐷(𝑖),𝐴) and (𝐷(𝑖),𝐷(𝑠)) entries in Table 4.

To see why the other entries are correct, we consider a typical case where 𝑧2𝐵(𝑖) and 𝑧1𝐷(𝑠) for some 𝑖,𝑠𝑁. Suppose that 0𝑖𝑠. By Table 2, 𝑧2𝑖𝐴 and 𝑧2𝑖+1𝐷. Hence by the (𝐴,𝐷) entry in Table 4, we see that 𝑧2𝑘𝐣 and 𝑧2𝑘+1𝐤. While if 0𝑠<𝑖, then 𝑧2𝑠𝐵 and 𝑧2𝑠+1𝐴. Hence, by the (𝐵,𝐴) entry in Table 4, we see that 𝑧2𝑘𝐤 and 𝑧2𝑘+1𝐢.

4. The Case 𝜆(0,1𝑎)

Suppose that 𝜆(0,1𝑎). Then 𝑎𝜆+1𝑎>𝜆, and {𝛼𝑗}𝑗=0 is a strictly decreasing sequence which diverges to . Hence, there exists 𝑀𝐍 such that 𝛼0,,𝛼𝑀>0 and 𝛼𝑀+10. Let 𝐷𝑗=𝛼𝑗 for 𝑗=0,,𝑀, and let 𝐷𝑀+1=0, then(]=0,𝜆𝑀𝑗=0𝐷𝑗+1,𝐷𝑗.(4.1)

We denote𝐴(𝑝,𝜆)=𝐷𝑝+1,𝐷𝑝]×(0,𝜆,𝐴(𝜆,𝑝)]×𝐷=(0,𝜆𝑝+1,𝐷𝑝,𝐴(𝑝,𝑞)=𝐷𝑝+1,𝐷𝑝×𝐷𝑞+1,𝐷𝑞,𝐵(𝜆,𝑝)=𝐷(𝜆,)×𝑝+1,𝐷𝑝,𝐵(𝑝,𝜆)=𝐼(𝑝)×](0,𝜆,𝐵(𝑝,𝑞)=𝐼(𝑝)×𝐷𝑞+1,𝐷𝑞,𝐶(𝜆,𝑝)=(𝜆,)×𝐼(𝑝),𝐶(𝑝,𝜆)=𝐼(𝑝)×(𝜆,),𝐶(𝑝,𝑞)=𝐼(𝑝)×𝐼(𝑞),𝐷(𝜆,𝑝)]=(0,𝜆×𝐼(𝑝),𝐷(𝑝,𝜆)=𝐷𝑝+1,𝐷𝑝×(𝜆,),𝐷(𝑝,𝑞)=𝐷𝑝+1,𝐷𝑝×𝐼(𝑞),(4.2) then 𝐴=𝑀𝑝=0𝐴(𝑝,𝜆)=𝑀𝑝=0𝐴(𝜆,𝑝)=𝑀𝑀𝑝=0𝑞=0𝐴(𝑝,𝑞),𝐵=𝑝=0𝐵(𝑝,𝜆)=𝑀𝑝=0𝐵(𝜆,𝑝)=𝑀𝑝=0𝑞=0𝐵(𝑝,𝑞),𝐶=𝑝=0𝐶(𝑝,𝜆)=𝑝=0𝐶(𝜆,𝑝)=𝑝=0𝑞=0𝐶(𝑝,𝑞),𝐷=𝑀𝑝=0𝐷(𝑝,𝜆)=𝑝=0𝐷(𝜆,𝑝)=𝑀𝑝=0𝑞=0𝐷(𝑝,𝑞).(4.3)

In Table 5, we record the fact that 𝑧𝛼𝐵 and 𝑧𝛼+1𝐵 which implies 𝑧𝛼+2𝐵 as the (𝐵,𝐵) entry, and so forth.


𝐴 𝐵 𝐶 𝐷

𝐴 𝐶 𝐵 𝐴 𝐷
𝐵 𝐶 𝐵 𝑎 𝐵 𝑎 𝐵 + 𝑎 𝐣
𝐶 𝐶 𝑎 𝐶 + 𝑎 𝐢 𝑎 𝐶 𝑎 𝐶 + 𝑎 𝐣
𝐷 𝐶 𝑎 𝐷 + 𝑎 𝐢 𝑎 𝐷 𝐷

Tables 6 and 7 are similar to Tables 2 and 3.


Initial conditionConditionConclusion

( 𝑧 2 , 𝑧 1 ) 𝐶 ( 𝜆 , 𝑝 ) × 𝐵 ( 𝑚 , 𝜆 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 × 𝐵
( 𝑧 2 , 𝑧 1 ) 𝐶 ( 𝜆 , 𝑝 ) × 𝐵 ( 𝑚 , 𝜆 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐶 × 𝐴
( 𝑧 2 , 𝑧 1 ) 𝐵 ( 𝑝 , 𝜆 ) × 𝐶 ( 𝜆 , 𝑚 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐴 × 𝐶
( 𝑧 2 , 𝑧 1 ) 𝐵 ( 𝑝 , 𝜆 ) × 𝐶 ( 𝜆 , 𝑚 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐵 × 𝐵
( 𝑧 2 , 𝑧 1 ) 𝐷 ( 𝜆 , 𝑘 ) × 𝐵 ( 𝑝 , 𝜆 ) 0 𝑘 𝑝 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐵 × 𝐵
( 𝑧 2 , 𝑧 1 ) 𝐷 ( 𝜆 , 𝑘 ) × 𝐵 ( 𝑝 , 𝜆 ) 0 𝑝 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐶 × 𝐴
( 𝑧 2 , 𝑧 1 ) 𝐶 ( 𝑝 , 𝜆 ) × 𝐷 ( 𝜆 , 𝑚 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐷 × 𝐷
( 𝑧 2 , 𝑧 1 ) 𝐶 ( 𝑝 , 𝜆 ) × 𝐷 ( 𝜆 , 𝑚 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐶 × 𝐴
( 𝑧 2 , 𝑧 1 ) 𝐷 ( 𝜆 , 𝑝 ) × 𝐶 ( 𝑚 , 𝜆 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐴 × 𝐶
( 𝑧 2 , 𝑧 1 ) 𝐷 ( 𝜆 , 𝑝 ) × 𝐶 ( 𝑚 , 𝜆 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐷 × 𝐷
( 𝑧 2 , 𝑧 1 ) 𝐵 ( 𝑘 , 𝜆 ) × 𝐷 ( 𝜆 , 𝑝 ) 0 𝑘 𝑝 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐷 × 𝐷
( 𝑧 2 , 𝑧 1 ) 𝐵 ( 𝑘 , 𝜆 ) × 𝐷 ( 𝜆 , 𝑝 ) 0 𝑝 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐶 × 𝐴


ConditionConclusion

0 𝑘 = 𝑝 m i n { 𝑙 , 𝑚 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐴 × 𝐶
0 𝑘 < m i n { 𝑙 , 𝑚 , 𝑝 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐷 ( 𝜆 , 𝑝 𝑘 1 ) × 𝐶 ( 𝑙 𝑘 1 , 𝜆 )
0 𝑝 < m i n { 𝑘 , 𝑙 , 𝑚 } ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 ( 𝑘 𝑝 1 , 𝜆 ) × 𝐶 ( 𝜆 , 𝑚 𝑝 1 )
0 𝑝 = 𝑙 < m i n { 𝑚 , 𝑘 } ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 ( 𝑘 𝑝 1 , 𝜆 ) × 𝐶 ( 𝜆 , 𝑚 𝑝 1 )
0 𝑝 = 𝑙 = 𝑚 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 × 𝐵
0 𝑙 < m i n { 𝑚 , 𝑘 , 𝑝 } ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) 𝐶 ( 𝑘 𝑙 1 , 𝜆 ) × 𝐷 ( 𝜆 , 𝑚 𝑙 1 )
0 𝑙 = 𝑚 < m i n { 𝑝 , 𝑘 } ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) 𝐶 × 𝐴
0 𝑙 = 𝑘 = 𝑚 < 𝑝 ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) 𝐷 × 𝐷
0 𝑚 < m i n { 𝑙 , 𝑘 , 𝑝 } ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐶 ( 𝜆 , 𝑝 𝑚 1 ) × 𝐵 ( 𝑙 𝑚 1 , 𝜆 )
0 𝑘 = 𝑚 < m i n { 𝑙 , 𝑝 } ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐷 ( 𝜆 , 𝑝 𝑘 1 ) × 𝐶 ( 𝑙 𝑘 1 , 𝜆 )
0 𝑝 = 𝑚 < m i n { 𝑙 , 𝑘 } ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 × 𝐵
0 𝑘 = 𝑙 < m i n { 𝑝 , 𝑚 } ( 𝑧 2 𝑘 , 𝑧 2 𝑙 + 1 ) 𝐷 × 𝐷

For example, let us show the first data row in Table 6. Suppose that 𝑧2𝐶(𝜆,𝑝) and 𝑧1𝐵(𝑚,𝜆) where 0𝑝𝑚. Then by Table 5, if 0=𝑝𝑚,𝑧0=𝑎𝑧2+𝑎𝐹𝜆𝑧1=𝑎𝑧2+𝑎𝐢𝑎𝐶(𝜆,0)+𝑎𝑧𝐢𝐵,1=𝑎𝑧1+𝑎𝐹𝜆𝑧0=𝑎𝑧1+𝑎𝐢𝑎𝐵(𝑚,𝜆)+𝑎𝐢𝐵.(4.4) If 0<𝑝𝑚, then𝑧0=𝑎𝑧2+𝑎𝐹𝜆𝑧1=𝑎𝑧2+𝑎𝐢𝑎𝐶(𝜆,𝑝)+𝑎𝐢𝐶(𝜆,𝑝1),𝑧1=𝑎𝑧1+𝑎𝐹𝜆𝑧0=𝑎𝑧1𝑎𝐵(𝑚,𝜆)𝐵(𝑚1,𝜆),𝑧2=𝑎𝑧0+𝑎𝐹𝜆𝑧1=𝑎𝑧0+𝑎𝐢𝑎𝐶(𝜆,𝑝1)+𝑎𝐢𝐶(𝜆,𝑝2),𝑧3=𝑎𝑧1+𝑎𝐹𝜆𝑧2=𝑎𝑧1𝑎𝐵(𝑚1,𝜆)𝐵(𝑚2,𝜆),(4.5) and by induction,𝑧2𝑝=𝑎𝑧2𝑝2+𝑎𝐹𝜆𝑧2𝑝1=𝑎𝑧2𝑝2+𝑎𝐢𝑎𝐶(𝜆,0)+𝑎𝑧𝐢𝐵,2𝑝+1=𝑎𝑧2𝑝1+𝑎𝐹𝜆𝑧2𝑝=𝑎𝑧2𝑝1+𝑎𝐢𝑎𝐵(𝑚𝑝,𝜆)+𝑎𝐢𝐵,(4.6) that is, 𝑧2𝑝𝐵 and 𝑧2𝑝+1𝐵.

As another example, let us show the second data row in Table 7. Suppose that (𝑧2,𝑧1)𝐶(𝑘,𝑝)×𝐶(𝑙,𝑚), where 0𝑘<min{𝑙,𝑚,𝑝}. Then by Table 5, if 0=𝑘<min{𝑙,𝑚,𝑝},𝑧0=𝑎𝑧2+𝑎𝐹𝜆𝑧1=𝑎𝑧2𝑎𝐶(0,𝑝)𝐷(𝜆,𝑝1),𝑧1=𝑎𝑧1+𝑎𝐹𝜆𝑧0=𝑎𝑧1+𝑎𝐣𝑎𝐶(𝑙,𝑚)+𝑎𝐣𝐼(𝑙1)×(𝜆,).(4.7) If 0<𝑘<min{𝑙,𝑚,𝑝}, then𝑧0=𝑎𝑧2+𝑎𝐹𝜆𝑧1=𝑎𝑧2𝑎𝐶(𝑘,𝑝)𝐶(𝑘1,𝑝1),𝑧1=𝑎𝑧1+𝑎𝐹𝜆𝑧0=𝑎𝑧1𝑎𝐶(𝑙,𝑚)𝐶(𝑙1,𝑚1),𝑧2=𝑎𝑧0+𝑎𝐹𝜆𝑧1=𝑎𝑧0𝑎𝐶(𝑘1,𝑝1)𝐶(𝑘2,𝑝2),𝑧3=𝑎𝑧1+𝑎𝐹𝜆𝑧2=𝑎𝑧1𝑎𝐶(𝑙1,𝑚1)𝐶(𝑙2,𝑚2),(4.8) and by induction,𝑧2𝑘=𝑎𝑧2𝑘2+𝑎𝐹𝜆𝑧2𝑘1=𝑎𝑧2𝑘2𝑎𝐶(0,𝑝𝑘)𝐷(𝜆,𝑝𝑘1),𝑧2𝑘+1=𝑎𝑧2𝑘1+𝑎𝐹𝜆𝑧2𝑘=𝑎𝑧2𝑘1+𝑎𝐣𝑎𝐶(𝑙𝑘,𝑝𝑘)+𝑎𝐣𝐼(𝑙𝑘1)×(𝜆,),(4.9) that is, 𝑧2𝑘𝐷(𝜆,𝑝𝑘1) and 𝑧2𝑘+1𝐶(𝑙𝑘1,𝜆).

Theorem 4.1. Suppose that 𝜆(0,1𝑎). Let {𝑧𝑛}𝑛=2 be a solution of (1.9) originated from (0,)2. Then 𝑧𝑛𝐢,𝐣,𝟎,𝐤or𝐤,𝟎.(4.10)

As in the proof of Theorem 3.1, we may construct Table 8.


𝐴 𝐵 𝐶 𝐷

𝐴 𝐤 , 𝟎 𝐢 𝟎 , 𝐤 𝐣
𝐵 𝐤 , 𝟎 𝐢 𝟎 , 𝐤 o r 𝐢 𝐣 or 𝐤 , 𝟎
𝐶 𝐤 , 𝟎 𝐢 or 𝐤 , 𝟎 𝟎 , 𝐤 , 𝐤 , 𝟎 , 𝐢 o r 𝐣 𝐣 or 𝐤 , 𝟎
𝐷 𝐤 , 𝟎 𝐢 or 𝐤 , 𝟎 𝟎 , 𝐤 or 𝐣 𝐣

For example, the (𝐵,𝐵) entry states that if (𝑧2,𝑧1)𝐵×𝐵, then the solution {𝑧𝑛} of (1.9) originated from it will tend to 𝐢. Indeed, by Table 5, 𝑧0𝐵, and then by induction, 𝑧𝑘𝐵 for all 𝑘2. Hence, by (1.9), we see that𝑧𝑛=𝑎𝑧𝑛2+(1𝑎)𝐢,𝑛𝑁,(4.11) from which we easily obtain𝑧2𝑘=𝑎𝑘+1𝑧2+1𝑎𝑘+1𝑧𝐢,𝑛𝑁,2𝑘+1=𝑎𝑘+1𝑧1+1𝑎𝑘+1𝐢,𝑛𝑁.(4.12) Hence 𝑧2𝑘,𝑧2𝑘+1𝐢 so that 𝑧𝑛𝐢. By similar reasoning, we may show the validity of the (𝐴,𝐴),(𝐴,𝐵),(𝐴,𝐶),(𝐴,𝐷), (𝐵,𝐴),(𝐶,𝐴),(𝐷,𝐴), and (𝐷,𝐷) entries.

Next, suppose that (𝑧2,𝑧1)𝐵(𝑝,𝜆)×𝐶(𝜆,𝑚). Then the solution {𝑧𝑛} of (1.9) originated from it will tend to 𝟎,𝐤 or 𝐢. Indeed, by Table 6, if 0𝑝𝑚, then (𝑧2𝑝,𝑧2𝑝+1)𝐴×𝐶 and by induction, (𝑧2𝑛+2𝑝,𝑧2𝑛+2𝑝+1)𝐴×𝐶 for all 𝑛𝑁. Hence, by (1.9), we have𝑧2𝑛+2𝑝=𝑎𝑛𝑧2𝑝𝑧,𝑛𝑁,2𝑛+2𝑝+1=𝑎𝑛𝑧2𝑝+1+(1𝑎𝑛)𝐤,𝑛𝑁.(4.13) Hence, 𝑧2𝑛𝟎,𝑧2𝑛+1𝐤. If 0𝑚<𝑝, then (𝑧2𝑚,𝑧2𝑚+1)𝐵×𝐵. By previous argument, we see 𝑧𝑛𝐢. By similar reasoning, we may show the correctness of the other entries. The proof is complete.

5. The Case 𝜆[1𝑎,1)

Suppose that 𝜆[1𝑎,1), then 𝑎𝜆+1𝑎>𝜆. Therefore, we may continue to use the notations described in the previous case 𝜆(0,1𝑎) and proceed as in the previous two sections and derive Tables 9, 10, 11, 12, 13 and 14.


AssumptionConclusion

( 𝑧 2 , 𝑧 1 ) 𝐵 × 𝐵 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐵 × 𝐵
( 𝑧 2 , 𝑧 1 ) 𝐴 × 𝐶 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐴 × 𝐶
( 𝑧 2 , 𝑧 1 ) 𝐷 × 𝐷 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐷 × 𝐷
( 𝑧 2 , 𝑧 1 ) 𝐶 × 𝐴 ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐶 × 𝐴


Initial conditionConditionConclusion

𝐴 ( 𝑝 , 𝜆 ) × 𝐵 ( 𝜆 , 𝑚 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 × 𝐵
𝐴 ( 𝑝 , 𝜆 ) × 𝐵 ( 𝜆 , 𝑚 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐴 × 𝐶
𝐶 ( 𝜆 , 𝑝 ) × 𝐵 ( 𝑚 , 𝜆 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐵 × 𝐵
𝐶 ( 𝜆 , 𝑝 ) × 𝐵 ( 𝑚 , 𝜆 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐶 × 𝐴
𝐷 ( 𝑝 , 𝜆 ) × 𝐴 ( 𝜆 , 𝑚 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐶 × 𝐴
𝐷 ( 𝑝 , 𝜆 ) × 𝐴 ( 𝜆 , 𝑚 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐷 × 𝐷
𝐵 ( 𝑝 , 𝜆 ) × 𝐶 ( 𝜆 , 𝑚 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐴 × 𝐶
𝐵 ( 𝑝 , 𝜆 ) × 𝐶 ( 𝜆 , 𝑚 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐵 × 𝐵
𝐴 ( 𝜆 , 𝑝 ) × 𝐷 ( 𝑚 , 𝜆 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐷 × 𝐷
𝐴 ( 𝜆 , 𝑝 ) × 𝐷 ( 𝑚 , 𝜆 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐴 × 𝐶
𝐶 ( 𝑝 , 𝜆 ) × 𝐷 ( 𝜆 , 𝑚 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐷 × 𝐷
𝐶 ( 𝑝 , 𝜆 ) × 𝐷 ( 𝜆 , 𝑚 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐶 × 𝐴
𝐵 ( 𝜆 , 𝑝 ) × 𝐴 ( 𝑚 , 𝜆 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐶 × 𝐴
𝐵 ( 𝜆 , 𝑝 ) × 𝐴 ( 𝑚 , 𝜆 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐵 × 𝐵
𝐷 ( 𝜆 , 𝑝 ) × 𝐶 ( 𝑚 , 𝜆 ) 0 𝑝 𝑚 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐴 × 𝐶
𝐷 ( 𝜆 , 𝑝 ) × 𝐶 ( 𝑚 , 𝜆 ) 0 𝑚 < 𝑝 ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐷 × 𝐷


ConditionConclusion

0 𝑘 = 𝑝 m i n { 𝑙 , 𝑚 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐶 × 𝐴
0 𝑝 = 𝑙 = 𝑚 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐷 × 𝐷
0 𝑙 = 𝑚 < m i n { 𝑘 , 𝑝 } ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) 𝐴 × 𝐶
0 𝑙 = 𝑚 = 𝑘 < 𝑝 ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) 𝐵 × 𝐵
0 𝑝 = 𝑚 < m i n { 𝑘 , 𝑙 } ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐷 × 𝐷
0 𝑘 = 𝑙 < m i n { 𝑝 , 𝑚 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐵 × 𝐵
0 𝑘 < m i n { 𝑝 , 𝑙 , 𝑚 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐵 ( 𝜆 , 𝑝 𝑘 1 ) × 𝐴 ( 𝑙 𝑘 1 , 𝜆 )
0 𝑝 < m i n { 𝑘 , 𝑙 , 𝑚 } ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐷 ( 𝑘 𝑝 1 , 𝜆 ) × 𝐴 ( 𝜆 , 𝑚 𝑝 1 )
0 𝑝 = 𝑙 < m i n { 𝑘 , 𝑚 } ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) 𝐷 ( 𝑘 𝑝 1 , 𝜆 ) × 𝐴 ( 𝜆 , 𝑚 𝑝 1 )
0 𝑙 < m i n { 𝑝 , 𝑘 , 𝑚 } ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) 𝐴 ( 𝑘 𝑙 1 , 𝜆 ) × 𝐵 ( 𝜆 , 𝑚 𝑙 1 )
0 𝑚 < m i n { 𝑝 , 𝑘 , 𝑙 } ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐴 ( 𝜆 , 𝑝 𝑚 1 ) × 𝐷 ( 𝑙 𝑚 1 , 𝜆 )
0 𝑘 = 𝑚 < m i n { 𝑝 , 𝑙 } ( 𝑧 2 𝑚 , 𝑧 2 𝑚 + 1 ) 𝐵 ( 𝜆 , 𝑝 𝑘 1 ) × 𝐴 ( 𝑙 𝑘 1 , 𝜆 )


ConditionConclusion

0 𝑘 = 𝑝 m i n { 𝑙 , 𝑚 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) 𝐴 × 𝐶