Research Article | Open Access

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

Chengmin Hou, Sui Sun Cheng, "Complete Asymptotic Analysis of a Two-Nation Arms Race Model with Piecewise Constant Nonlinearities", Discrete Dynamics in Nature and Society, vol. 2012, Article ID 745697, 17 pages, 2012. 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 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  is explained, and several interesting conclusions are drawn which can be used to explain stable and escalated arms races. Roughly, let , and let and be the amount spent on armaments by two respective countries and in year . Assuming, has a fixed amount of distrust of the other country, causing it to retain arms, then where the constant measures country ’s distrust of country in that it reacts to the way arms itself, is a measure of ’s own economy, and is the basic annual expenditure (e.g., maintenance expense). If we now assume a similar situation for country , then we also have Under the assumption that and , it is shown that if the initial total expenditure is large and that the distrust factor is also so large that , 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 in (1.1) is replaced by where and the term in (1.2) is replaced by a similar one. The “discontinuous” function has a clear physical meaning. Indeed, the positive parameter 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: where we have introduced two “delays” and 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: The assumption that means that the fixed expenditures are relatively low in both countries, while we assume that so as to use the best up-to-date and “reliable” accounting records and . By choosing and , country is making a decision based on a convex combination of the expenditures and the blanket-ceiling sum . If country takes on a similar decision policy, then we end up with where may or may not differ from . The case is only more technically difficult, and therefore, in this paper, we will assume the case (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: for , where in this model, . Note that if we let and then the above system (1.7) can be written as where we write and for the sake of convenience.

The above vector equation is a three-term recurrence relation. Hence, for given and in the plane, a unique sequence can be calculated from it. Such a sequence is called a solution of (1.9) determined by and . Among different and , 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 and 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 , and there are only several recent studies on scalar equations with piecewise smooth nonlinearities )! 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 in the quadrant , the solution determined by it also lies in the same quadrant (in the sense that for ). Depending on the locations of and , 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: where is a fixed positive number, then is a partition of the quadrant .

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

For any solution originated from and in the above subsets, our main conclusion in this paper is that will tend to some vector and will tend to another vector (which may or may not be equal to ). 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 of In case , is convergent to , and hence we may also write

To arrive at our main conclusion, we note, however, that since 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 . If is a solution of (1.9) with , then as will be seen below, and . If we now replace the condition with the symmetric initial condition , then we will end up with the conclusion that and . 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 that where Therefore, in case is a solution of (1.9) such that, say, for all large , then (1.9) is reduced to for all large . Hence, linear systems and their related properties will also appear in later discussions. More precisely, two groups of bounding quantities and are needed: Note that they satisfy and the recurrence relations We also need the following two properties of linear systems. Let be real scalar (or vector) sequences that satisfy where , and is a real number (resp., a real vector).(i)If is a sequence which satisfies (1.19), then (ii)If is a sequence which satisfies (1.20), then

Finally, we need to consider various ordering arrangements for three or four nonnegative integers . First, the ordering arrangements of three integers can be classified into 6 cases: (1) . In fact, let . Then either , or . Let , then These are equivalent to by comparing the two sets of statements.

By similar reasoning, there are 12 ordering arrangements for four integers : .

Our following plan is quite simple. We will treat our as a bifurcation parameter and distinguish four different cases (i) , (ii) , (iii) , and (iv) , and consider different in , or 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, . Let be any solution of (1.9) with . Then .

Proof. By (1.9), we may see that and for , then and . Thus, there exists an integer such that for all . Therefore, for all . In view of (1.21) and (1.22), . The proof is complete.

#### 3. The Case 𝜆=1

In this section, we assume that . If is a solution of (1.9) and if and , then in view of (1.14), , and hence 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 is a strictly increasing and divergent sequence. Therefore, if we let then

First of all, the fact that and implies, is recorded as the 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 condition Condition Conclusion ( 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 and , where . By Table 1, if , then If , then and by induction, that is, and .

As another example, let us show the second data row in Table 3. Suppose that , where , then by Table 1, if , If , then and by induction, that is, and for some .

 Condition Conclusion 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 be a solution of (1.9), we will be able to show the following result.

Theorem 3.1. Suppose that . Let be a solution of (1.9) originated from . Then

To this end, let us consider first the case where , then by the entry in Table 1, . By induction, we may then see that for all . Hence, by (1.9), we see that from which we easily obtain By (1.21) and (1.22), we see that so that . We record this conclusion in the 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 and . Then by the entry of Table 1, we see that . Since and , then by Table 1 again, for some . By induction, we may then see that and for . Hence, by (1.9), we see that for . We may then easily see that and .

By similar arguments, we may then derive the and entries in Table 4.

To see why the other entries are correct, we consider a typical case where and for some . Suppose that . By Table 2, and . Hence by the entry in Table 4, we see that and . While if , then and . Hence, by the entry in Table 4, we see that and .

#### 4. The Case 𝜆∈(0,1−𝑎)

Suppose that . Then , and is a strictly decreasing sequence which diverges to . Hence, there exists such that and . Let for , and let , then

We denote then

In Table 5, we record the fact that and which implies as the 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 condition Condition Conclusion ( 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
 Condition Conclusion 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 and where . Then by Table 5, if , If , then and by induction, that is, and .

As another example, let us show the second data row in Table 7. Suppose that , where . Then by Table 5, if , If , then and by induction, that is, and .

Theorem 4.1. Suppose that . Let be a solution of (1.9) originated from . Then

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 entry states that if , then the solution of (1.9) originated from it will tend to . Indeed, by Table 5, , and then by induction, for all . Hence, by (1.9), we see that from which we easily obtain Hence so that . By similar reasoning, we may show the validity of the , , and entries.

Next, suppose that . Then the solution of (1.9) originated from it will tend to or . Indeed, by Table 6, if , then and by induction, for all . Hence, by (1.9), we have Hence, . If , then . By previous argument, we see . By similar reasoning, we may show the correctness of the other entries. The proof is complete.

#### 5. The Case 𝜆∈[1−𝑎,1)

Suppose that , then . Therefore, we may continue to use the notations described in the previous case and proceed as in the previous two sections and derive Tables 9, 10, 11, 12, 13 and 14.

 Assumption Conclusion ( 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 condition Condition Conclusion 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
 Condition Conclusion 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 )
 Condition Conclusion 0 ≤ 𝑘 = 𝑝 ≤ m i n { 𝑙 , 𝑚 } ( 𝑧 2 𝑘 , 𝑧 2 𝑘 + 1 ) ∈ 𝐴 × 𝐶 0 ≤ 𝑝 = 𝑙 = 𝑚 < 𝑘 ( 𝑧 2 𝑝 , 𝑧 2 𝑝 + 1 ) ∈ 𝐵 × 𝐵 0 ≤ 𝑙 = 𝑚 < m i n { 𝑘 , 𝑝 } ( 𝑧 2 𝑙 , 𝑧 2 𝑙 + 1 ) ∈ 𝐶 × 𝐴 0 ≤ 𝑙 = 𝑚 = 𝑘 < 𝑝 ( 𝑧 2 𝑙 , 𝑧 2