Convergence and Divergence of the Solutions of a Neutral Difference Equation

Abstract

We investigate the asymptotic behavior of the solutions of a neutral type difference equation of the form Δ[𝑥(𝑛)+𝑐𝑥(𝜏(𝑛))]+𝑝(𝑛)𝑥(𝜎(𝑛))=0, where 𝜏(𝑛) is a general retarded argument, 𝜎(𝑛) is a general deviated argument (retarded or advanced), 𝑐∈ℝ, (𝑝(𝑛))_𝑛≥0 is a sequence of positive real numbers such that 𝑝(𝑛)≥𝑝, 𝑝∈ℝ₊, and Δ denotes the forward difference operator Δ𝑥(𝑛)=𝑥(𝑛+1)−𝑥(𝑛). Also, we examine the asymptotic behavior of the solutions in case they are continuous and differentiable with respect to 𝑐.

1. Introduction

Neutral type differential equations are differential equations in which the highest-order derivative of the unknown function appears in the equation both with and without delays (or delays advanced). See Driver [1], Bellman and Cooke [2], and Hale [3] for questions of existence, uniqueness, and continuous dependence.

It is to be noted that, in general, the theory of neutral differential equations presents extra complications, and basic results which are true for delay differential equations may not be true for neutral equations. For example, Snow [4] has shown that, even though the characteristic roots of a neutral differential equation may all have negative real parts, it is still possible for some solutions to be unbounded.

The discrete counterparts of neutral differential equations are called neutral difference equations, and it is a well-known fact that there is a similarity between the qualitative theories of neutral differential equations and neutral difference equations.

Besides its theoretical interest, strong interest in the study of the asymptotic and oscillatory behavior of solutions of neutral type equations (difference or differential) is motivated by the fact that they arise in many areas of applied mathematics, such as circuit theory [5, 6], bifurcation analysis [7], population dynamics [8, 9], stability theory [10], and dynamical behavior of delayed network systems [11]. See, also, Driver [12], Hale [3], Brayton and Willoughby [13], and the references cited therein. This is the reason that during the last few decades these equations are in the main interest of the literature.

In the present paper, we are interested in the first-order neutral type difference equation of the formΔ[𝑥(𝑛)+𝑐𝑥(𝜏(𝑛))]+𝑝(𝑛)𝑥(𝜎(𝑛))=0,𝑛≥0 (E) where (𝑝(𝑛))_𝑛≥0 is a sequence of positive real numbers such that 𝑝(𝑛)≥𝑝, 𝑝∈ℝ₊, 𝑐∈ℝ, (𝜏(𝑛))_𝑛≥0 is an increasing sequence of integers which satisfies𝜏(𝑛)≤𝑛−1∀𝑛≥0,lim_𝑛→∞𝜏(𝑛)=+∞, (1.1) and (𝜎(𝑛))_𝑛≥0 is an increasing sequence of integers such that 𝜎(𝑛)≤𝑛−1∀𝑛≥0,lim_𝑛→∞𝜎(𝑛)=+∞, (1.2) or 𝜎(𝑛)≥𝑛+1∀𝑛≥0. (1.3)

Define𝑘₁=−min_𝑛≥0𝜏(𝑛),𝑘₂=−min_𝑛≥0𝜎(𝑛),𝑘=max𝑘₁,𝑘₂. (1.4) (Clearly, 𝑘 is a positive integer.)

By a solution of the neutral type difference equation (E), we mean a sequence of real numbers (𝑥(𝑛))_{𝑛≥−𝑘} which satisfies (E) for all 𝑛≥0. It is clear that, for each choice of real numbers 𝑐_−𝑘,𝑐_−𝑘+1,…,𝑐₋₁,𝑐₀, there exists a unique solution (𝑥(𝑛))_{𝑛≥−𝑘} of (E) which satisfies the initial conditions 𝑥(−𝑘)=𝑐_−𝑘,𝑥(−𝑘+1)=𝑐_−𝑘+1,…,𝑥(−1)=𝑐₋₁, 𝑥(0)=𝑐₀.

A solution (𝑥(𝑛))_{𝑛≥−𝑘} of the neutral type difference equation (E) is called oscillatory if for every positive integer 𝑛 there exist 𝑛₁,𝑛₂≥𝑛 such that 𝑥(𝑛₁)𝑥(𝑛₂)≤0. In other words, a solution (𝑥(𝑛))_{𝑛≥−𝑘} is oscillatory if it is neither eventually positive nor eventually negative. Otherwise, the solution is said to be nonoscillatory.

In the special case where 𝜏(𝑛)=𝑛−𝑎 and 𝜎(𝑛)=𝑛−𝑏, 𝑎,𝑏∈ℕ, (E) takes the formΔ[𝑥(𝑛)+𝑐𝑥(𝑛−𝑎)]+𝑝(𝑛)𝑥(𝑛−𝑏)=0,𝑛≥0. (E1)

Equation (E) represents a discrete analogue of the neutral type differential equations 𝑑[𝑥(𝑡)(𝜏(𝑡(𝑡)𝑥(𝜎(𝑡𝑑𝑡+𝑐𝑥))]+𝑝))=0,𝑡≥0, (E2) which, in the special case where 𝜏(𝑡)=𝑡−𝜏₀ and 𝜎(𝑡)=𝑡−𝜎₀, 𝜏₀,𝜎₀∈ℝ₊, takes the form (see e.g., [14, 15]) 𝑑𝑥(𝑡)𝑑𝑡+𝑐𝑥𝑡−𝜏₀+𝑝(𝑡)𝑥𝑡−𝜎₀=0,𝑡≥0. (E3)

The search for the asymptotic behavior and, especially, for oscillation criteria and stability of neutral type (difference or differential) equations has received a great attention in the last few years. Hence, a large number of related papers have been published. See [4, 9, 10, 14–52] and the references cited therein. Most of these papers are concerning the special case of the delay difference equations (E₁) and (E₃) where the algebraic characteristic equation gives useful information about oscillation and stability.

The purpose of this paper is to investigate the convergence and divergence of the solutions of (E) in the case of a general delay argument 𝜏(𝑛) and of a general deviated (retarded or advanced) argument 𝜎(𝑛).

2. Some Preliminaries

Assume that (𝑥(𝑛))_{𝑛≥−𝑘} is a nonoscillatory solution of (E). Then it is either eventually positive or eventually negative. As (−𝑥(𝑛))_{𝑛≥−𝑘} is also a solution of (E), we may (and do) restrict ourselves only to the case where 𝑥(𝑛)>0 for all large 𝑛. Let 𝑛₁≥−𝑘 be an integer such that 𝑥(𝑛)>0 for all 𝑛≥𝑛₁. Then, there exists 𝑛₀≥𝑛₁ such that𝑥(𝜏(𝑛)),𝑥(𝜎(𝑛))>0forevery𝑛≥𝑛₀. (2.1) Set𝑧(𝑛)=𝑥(𝑛)+𝑐𝑥(𝜏(𝑛)). (2.2) Then, in view of (E) and taking into account the fact that 𝑝(𝑛)≥𝑝>0, we haveΔ𝑧(𝑛)+𝑝(𝑛)𝑥(𝜎(𝑛))=0 (2.3) orΔ𝑧(𝑛)=−𝑝(𝑛)𝑥(𝜎(𝑛))≤−𝑝𝑥(𝜎(𝑛))<0∀𝑛≥𝑛₀, (2.4) which means that the sequence (𝑧(𝑛))_𝑛≥𝑛0 is strictly decreasing, regardless of the value of the real constant 𝑐.

Throughout this paper, we are going to use the following notation:𝜏∘𝜏=𝜏²,𝜏∘𝜏∘𝜏=𝜏³,andsoon. (2.5) Let the domain of 𝜏 be the set 𝐷(𝜏)=ℕ_𝑛∗={𝑛_∗,𝑛_∗+1,𝑛_∗+2,…}, where 𝑛_∗ is the smallest natural number that 𝜏 is defined with. Then for every 𝑛>𝑛_∗ it is clear that there exists a natural number 𝑚(𝑛) such that𝑥𝜏^𝑚(𝑛)(𝑛)=𝑥𝜏𝑛_∗,lim_𝑛→∞𝑚(𝑛)=+∞ (2.6) since (𝑚(𝑛)) is increasing and unbounded function of 𝑛.

The following lemma provides some tools which are useful for the main results.

Lemma 2.1. Assume that (𝑥(𝑛))_{𝑛≥−𝑘} is a positive solution of (E). Then, one has the following.(i)If 𝑐≠0 and ^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=𝑆₀<+∞, (2.7) then lim_𝑛→∞𝑧(𝑛)=𝑐𝐿=𝑐lim_𝑛→∞𝑥(𝜏(𝜎(𝑛))). (2.8)(ii)If ^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=+∞, (2.9) then 𝑧(𝑛)<0eventually. (2.10)(iii)If 𝑐≥−1, then ^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=𝑆₀<+∞. (2.11)(iv)If 𝑐<−1, then ^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=+∞. (2.12)

Proof. Summing up (2.3) from 𝑛₀ to 𝑛, we obtain 𝑧(𝑛+1)−𝑧𝑛₀+^𝑛_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=0 (2.13) or 𝑧(𝑛+1)=𝑧𝑛₀−^𝑛_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖)). (2.14) For the above relation, there are only two possible cases:^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=𝑆₀<+∞ (2.15a) or ^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=+∞. (2.15b)Assume that (2.15a) holds. Since 𝑝(𝑛)≥𝑝>0, we have +∞>𝑆₀=^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))≥𝑝^∞_𝑖=𝑛0𝑥(𝜎(𝑖)). (2.16) The last inequality guarantees that ^∞_𝑖=𝑛0𝑥(𝜎(𝑖))<+∞ (2.17) and, consequently, lim_𝑛→∞𝑥(𝜎(𝑛))=0. (2.18) Also, (2.15a) guarantess that lim_𝑛→∞𝑧(𝑛) exists as a real number. Now, assume that lim_𝑛→∞𝑧(𝑛)=ℓ∈ℝ,𝑐≠0. (2.19) Since (𝑧(𝜎(𝑛))) is a subsequence of (𝑧(𝑛)), it is obvious that lim_𝑛→∞𝑧(𝜎(𝑛))=ℓ (2.20) or lim_𝑛→∞[𝑥(𝜎(𝑛))+𝑐𝑥(𝜏(𝜎(𝑛)))]=ℓ, (2.21) and, in view of (2.18), we obtain lim_𝑛→∞𝑥(𝜏(𝜎(𝑛=ℓ)))𝑐∶=𝐿. (2.22) Hence, lim_𝑛→∞𝑧(𝑛)=ℓ=𝑐𝐿=𝑐lim_𝑛→∞𝑥(𝜏(𝜎(𝑛))). (2.23) The proof of part (i) of the lemma is complete.
Assume that (2.15b) holds. Then, by taking limits on both sides of (2.14) we obtain lim_𝑛→∞𝑧(𝑛)=−∞, (2.24) which, in view of the fact that the sequence (𝑧(𝑛)) is strictly decreasing, means that 𝑧(𝑛)<0eventually. (2.25) The proof of part (ii) of the lemma is complete.
Assume that −1≤𝑐<0, and suppose, for the sake of contradiction, that ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=+∞. Then, in view of part (ii), we have 𝑧(𝑛)<0 eventually. Thus, 𝑥(𝑛)<−𝑐𝑥(𝜏(𝑛)) (2.26) or 𝑥(𝑛)<−𝑐−𝑐𝑥𝜏²(𝑛). (2.27) Repeating the above procedure we obtain 𝑥(𝑛)<(−𝑐)^𝑚(𝑛)𝑥𝜏^𝑚(𝑛)(𝑛)=(−𝑐)^𝑚(𝑛)𝑥𝜏𝑛_∗. (2.28) If −1<𝑐<0, clearly, (−𝑐)^𝑚(𝑛)→0 since 𝑚(𝑛)→∞ as 𝑛→∞. Since 𝑥(𝑛)>0 for all large 𝑛, (2.28) guarantees that lim_𝑛→∞𝑥(𝑛)=0. (2.29) This implies that lim_𝑛→∞𝑧(𝑛)=0, (2.30) and consequently ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))<∞, which contradicts our assumption.
If 𝑐=−1, by (2.28) we obtain 𝑥(𝑛)<𝑥𝜏𝑛_∗, (2.31) which means that (𝑥(𝑛)) is bounded and therefore (𝑧(𝑛)) is bounded. Hence, ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))<∞, which contradicts our assumption.
Assume that 𝑐≥0 and that ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=∞. In view of part (ii), (2.10) holds, that is, 𝑧(𝑛)<0 eventually. This contradicts 𝑧(𝑛)=𝑥(𝑛)+𝑐𝑥(𝜏(𝑛))>0. Therefore ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=𝑆₀<+∞. The proof of part (iii) of the lemma is complete.
In the remainder of this proof, it will be assumed that 𝑐<−1.
If (2.15a) holds, that is, ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=𝑆₀<+∞, then, in view of part (i), (2.8) is satisfied, that is, lim_𝑛→∞𝑧(𝑛)=𝑐𝐿,where𝐿=lim_𝑛→∞𝑥(𝜏(𝜎(𝑛))). (2.32) If 𝐿>0, then lim_𝑛→∞𝑧(𝑛)=lim_𝑛→∞(𝑥(𝑛)+𝑐𝑥(𝜏(𝑛)))=𝑐𝐿<0. (2.33) Since (𝑧(𝑛)) is strictly decreasing, we have 𝑧(𝑛)𝑥(𝑛)(𝜏(𝑛>𝑐𝐿,>−𝑐𝑥))+𝑐𝐿, (2.34)(2.35) or 𝑥(𝑛)>−𝑐−𝑐𝑥𝜏²(𝑛)+𝑐𝐿+𝑐𝐿. (2.36) Repeating this procedure 𝑚(𝑛_ℓ) times we have 𝑥(𝑛)>−𝑐−𝑐𝑥𝜏²(𝑛=())+𝑐𝐿+𝑐𝐿−𝑐²𝑥𝜏²(𝑛)−𝑐²()𝐿+𝑐𝐿>⋯>−𝑐^{𝑚(𝑛ℓ)}𝑥𝜏^{𝑚(𝑛ℓ)}(𝑛)+𝑐𝐿−𝑐²𝐿+⋯±𝑐^{𝑚(𝑛ℓ)}𝐿=(−𝑐)^{𝑚(𝑛ℓ)}𝑥𝜏𝑛_𝜆()+𝑐𝐿−𝑐^{𝑚(𝑛ℓ)}−1=()−1−𝑐−𝑐^{𝑚(𝑛ℓ)}𝑥𝜏𝑛_𝜆−𝑐𝐿)1+𝑐(−𝑐^{𝑚(𝑛ℓ)}=()−1−𝑐^{𝑚(𝑛ℓ)}𝑥𝜏𝑛_𝜆−𝑐𝐿+1+𝑐𝑐𝐿.1+𝑐 (2.37) Now, if for this index 𝜆 we have 𝑥(𝜏(𝑛_𝜆))−(𝑐𝐿/(1+𝑐))>0, then, as 𝑛→∞, 𝑥(𝑛)→∞, which contradicts (2.18). Therefore, there exists an index 𝑠 such that 𝑥𝜏𝑛_𝑠=𝑥𝜏^{𝑚(𝑛𝜇)}(𝑛) (2.38) with 𝑥𝜏𝑛_𝑠−𝑐𝐿1+𝑐<0. (2.39) Then since lim_𝑛→∞𝑧(𝑛)=𝑐𝐿, for every 𝜀>0 there exists 𝑛₂(𝜀) such that 𝑧(𝑛)<𝑐𝐿+𝜀forevery𝑛≥max𝜏𝑛_𝑠,𝑛₂=𝑛₃, (2.40) where 𝜏(𝑛_𝑠) is satisfying the previous inequality.
Hence, 𝑥(𝑛)<−𝑐𝑥(𝜏(𝑛))+𝑐𝐿+𝜀forevery𝑛≥𝑛₃, (2.41) or 𝑥(𝑛)<−𝑐−𝑐𝑥𝜏²(𝑛)+𝑐𝐿+𝜀+𝑐𝐿+𝜀forevery𝑛≥𝑛₃. (2.42) Repeating this procedure 𝑚(𝑛_𝜇) times we have 𝑥(𝑛)<−𝑐−𝑐𝑥𝜏²(𝑛=())+𝑐𝐿+𝜀+𝑐𝐿+𝜀−𝑐²𝑥𝜏²(𝑛)−𝑐²()𝐿−𝑐𝜀+𝑐𝐿+𝜀<⋯<−𝑐^{𝑚(𝑛𝜇)}𝑥𝜏𝑛_𝑠−𝑐𝐿)1+𝑐(−𝑐^{𝑚(𝑛𝜇)}+()−1−𝑐^{𝑚(𝑛𝜇)}−1𝜀=()−1−𝑐−𝑐^{𝑚(𝑛𝜇)}𝑥𝜏𝑛_𝑠−𝑐𝐿−𝜀1+𝑐+1+𝑐𝑐𝐿+𝜀1+𝑐1+𝑐 (2.43) or 𝑥(𝑛)<(−𝑐)^{𝑚(𝑛𝜇)}𝑥𝜏𝑛_𝑠−𝑐𝐿−𝜀1+𝑐+1+𝑐𝑐𝐿+𝜀1+𝑐.1+𝑐 (2.44) Then, for sufficiently large 𝑛 the above inequality gives 𝑥(𝑛)<0, (2.45) which condradicts 𝑥(𝑛)>0.
If 𝐿=0, then lim_𝑛→∞𝑧(𝑛)=0 (2.46) or lim_𝑛→∞(𝑥(𝑛)+𝑐𝑥(𝜏(𝑛)))=0, (2.47) and since (𝑧(𝑛)) is stricly decreasing, it is obvious that 𝑧(𝑛)>0 eventually or 𝑥(𝑛)>−𝑐𝑥(𝜏(𝑛)). (2.48) In view of (2.6), the last inequality becomes 𝑥(𝑛)>(−𝑐)^𝑚(𝑛)𝑥𝜏𝑛_∗, (2.49) and consequently lim_𝑛→∞𝑥(𝑛)≥lim_𝑛→∞(−𝑐)^𝑚(𝑛)𝑥𝜏𝑛_∗=+∞, (2.50) which contradicts (2.18). Hence, it is clear that ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=+∞. The proof of part (iv) of the lemma is complete.
The proof of the lemma is complete.

3. Main Results

The asymptotic behavior of the solutions of the neutral type difference equation (E) is described by the following theorem.

Theorem 3.1. For (E) one has the following.(I)Every nonoscillatory solution is unbounded if 𝑐<−1.(II)Every solution oscillates if 𝑐=−1.(III)Every nonoscillatory solution tends to zero if −1<𝑐<1.(IV)Every nonoscillatory solution is bounded if 𝑐≥1.Furthermore, if any solution of (E) is continuous with respect to 𝑐, one has the following.(V)Every solution is zero if 𝑐≤−1.(VI)Every solution tends to zero if −1<𝑐≤1.(VII)If, additionally, any solution of (E) has continuous derivatives of any order and convergent Taylor series for every 𝑐∈ℝ, then the solution is zero.

Proof. Assume that (𝑥(𝑛))_{𝑛≥−𝑘} is a nonoscillatory solution of (E). Then it is either eventually positive or eventually negative. As (−𝑥(𝑛))_{𝑛≥−𝑘} is also a solution of (E), we may (and do) restrict ourselves only to the case where 𝑥(𝑛)>0 for all large 𝑛. Let 𝑛₁≥−𝑘 be an integer such that 𝑥(𝑛)>0 for all 𝑛≥𝑛₁. Then, there exists 𝑛₀≥𝑛₁ such that 𝑥(𝜏(𝑛)),𝑥(𝜎(𝑛))>0∀𝑛≥𝑛₀, (3.1) which, in view of the previous section, means that the sequence (𝑧(𝑛))_𝑛≥𝑛0 is strictly decreasing, regardless of the value of the real constant 𝑐.
Assume that 𝑐<−1. Then, in view of part (iv) of Lemma 2.1, we have ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=+∞, and, consequently, in view of part (ii) of Lemma 2.1, 𝑧(𝑛)<0 eventually. Therefore, lim_𝑛→∞𝑧(𝑛)=−∞ (3.2) or lim_𝑛→∞(𝑥(𝑛)+𝑐𝑥(𝜏(𝑛)))=−∞. (3.3) Since 𝑐<−1, the last relation guarantees that lim_𝑛→∞𝑥(𝜏(𝑛))=+∞, (3.4) which means that (𝑥(𝑛)) is unbounded. The proof of part (I) of the theorem is complete.
Assume that 𝑐=−1. Then, in view of part (iii) of Lemma 2.1, we have ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))<∞, and, therefore, in view of part (i) of Lemma 2.1, we obtain lim_𝑛→∞[𝑥(𝑛)−𝑥(𝜏(𝑛))]=−𝐿. (3.5) Assume that 𝐿>0. Then there exists a natural number 𝑛_𝜆 such that 𝑧(𝑛)<0 for every 𝑛≥𝑛_𝜆, and therefore 𝑥(𝑛)<𝑥(𝜏(𝑛))<⋯<𝑥𝜏^{𝑚(𝑛ℓ)}𝑛_𝜆, (3.6) which means that (𝑥(𝑛)) is bounded. Since (𝑥(𝑛)) is bounded, let 𝑀=limsup𝑥(𝑛),where𝑀≥𝐿. (3.7) Then there exists a subsequence (𝑥(𝜃(𝑛))) of (𝑥(𝑛)) such that lim_𝑛→∞𝑥(𝜃(𝑛))=𝑀. (3.8) Thus, lim_𝑛→∞𝑧(𝜃(𝑛))=−𝐿,lim_𝑛→∞[𝑥(𝜃(𝑛))−𝑥(𝜏(𝜃(𝑛)))]=−𝐿, (3.9)(3.10) or lim_𝑛→∞𝑥(𝜏(𝜃(𝑛)))=𝑀+𝐿>𝑀, (3.11) which contradicts (3.7). Therefore, 𝐿>0 is not valid. Hence, 𝐿=0 and consequently lim_𝑛→∞𝑧(𝑛)=0. Furthermore, taking into account the fact that the sequence (𝑧(𝑛)) is strictly decreasing we conclude that 𝑧(𝑛)>0 or equivalently 𝑥(𝑛)>𝑥(𝜏(𝑛))>⋯>𝑥𝜏𝑛_∗. (3.12) Since (𝑥(𝑛)) has a lower bound greater than zero, it cannot have any subsequence that tends to zero. Thus, lim_𝑛→∞𝑥(𝜎(𝑛))=0 is not valid, and therefore ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=∞ which contradicts our previous conclusions. Hence, if 𝑐=−1, (𝑥(𝑛)) oscillates. The proof of part (II) of the theorem is complete.
Assume that −1<𝑐<0.
Assume that 𝐿>0. Then there exists a natural number 𝑛_𝜆 such that 𝑧(𝑛)<0 for every 𝑛≥𝑛_𝜆, and therefore 𝑥(𝑛)<(−𝑐)𝑥(𝜏(𝑛))<⋯<(−𝑐)^{𝑚(𝑛ℓ)}𝑥𝜏^{𝑚(𝑛ℓ)}𝑛_𝜆, (3.13) which means that (𝑥(𝑛)) tends to zero as 𝑛→∞. Therefore, (𝑧(𝑛)) tends to zero as 𝑛→∞, that is, 𝐿=0, which contradicts 𝐿>0. Hence 𝐿=0. Taking into account the fact that the sequence (𝑧(𝑛)) is strictly decreasing, it is obvious that 𝑧(𝑛)>0. Hence, for every 𝜖>0 there exists a natural number 𝑛₄ such that for every 𝑛≥𝑛₄𝑥(𝑛)+𝑐𝑥(𝜏(𝑛))<𝜖 (3.14) or 𝑥(𝑛)<−𝑐𝑥(𝜏(𝑛))+𝜖<−𝑐−𝑐𝑥𝜏²(𝑛)+𝜖+𝜖=𝑐²𝑥𝜏²(𝑛)−𝑐𝜖+𝜖. (3.15) For sufficiently large 𝑛, after 𝑚-steps we obtain 𝑥(𝑛)<𝑐^𝑚+1𝑥𝜏^𝑚+1(𝑛)+𝜖−𝑐𝜖+⋯+(−𝑐)^𝑚𝜖. (3.16) As 𝑛→∞, clearly 𝑚→∞, and therefore lim_𝑛→∞𝑥(𝑛)≤lim_𝑚→∞𝜖−𝑐𝜖+⋯+(−𝑐)^𝑚𝜖=𝜖.1+𝑐 (3.17) Since 𝜖 is an arbitrary real positive number and, taking into account the fact that (𝑥(𝑛))>0, it is clear that lim_𝑛→∞𝑥(𝑛)=0. (3.18) Let 𝑐=0. In view of part (iii) of Lemma 2.1, we have ^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))<+∞. (3.19) This guarantees that (𝑧(𝑛)) is bounded, and therefore, since 𝑧(𝑛)=𝑥(𝑛), (𝑥(𝑛)) is bounded. Also, since (𝑧(𝑛)) is stricly decreasing, lim_𝑛→∞𝑧(𝑛) exists, that is, lim_𝑛→∞𝑥(𝑛) exists. In view of (2.18), we conclude that lim_𝑛→∞𝑥(𝑛)=0. (3.20) Assume that 0<𝑐<1, then lim_𝑛→∞𝑧(𝑛)=𝑐𝐿,wherelim_𝑛→∞𝑥(𝜏(𝜎(𝑛)))=𝐿 (3.21) or lim_𝑛→∞[𝑥(𝑛)+𝑐𝑥(𝜏(𝑛))]=𝑐𝐿. (3.22) Therefore, lim_𝑛→∞𝑧(𝜏(𝜎(𝑛)))=𝑐𝐿,lim_𝑛→∞𝑥(𝜏(𝜎(𝑛)))+𝑐𝑥𝜏²(𝜎(𝑛))=𝑐𝐿, (3.23)(3.24) or lim_𝑛→∞𝑥𝜏²(𝜎(𝑛𝐿))=𝐿−𝑐≤0, (3.25) which means that 𝐿=0. Hence, lim_𝑛→∞𝑧(𝑛)=0, (3.26) and since 𝑧(𝑛)≥𝑥(𝑛)>0, (3.27) we have lim_𝑛→∞𝑥(𝑛)=0. (3.28) The proof of part (III) of the theorem is complete.
Assume that 𝑐≥1. In view of Part (iii) of Lemma 2.1, it is clear that ∑^∞_𝑖=𝑛0𝑝(𝑖)𝑥(𝜎(𝑖))=𝑆₀<+∞ which combined with (2.14) implies that (𝑧(𝑛)) is bounded and, therefore (𝑥(𝑛)) is bounded. The proof of part (IV) of the theorem is complete.
In the remainder of this proof, it will be assumed that 𝑥(𝑛) is a continuous function with respect to 𝑐, and, therefore, instead of 𝑥(𝑛) we will write 𝑥(𝑐,𝑛).
Let 𝑐=−1. Then, in view of part (II), (𝑥(−1,𝑛)) oscillates. On the other hand, since 𝑥(𝑐,𝑛) is continuous, we have lim_{𝑐→−1−}𝑥(𝑐,𝑛)=𝑥(−1,𝑛). (3.29) But 𝑥(𝑐,𝑛)>0 for all large 𝑛, and therefore its limit is always nonnegative. Thus, 𝑥(−1,𝑛)>0 for all large 𝑛 which contradicts that (𝑥(−1,𝑛)) oscillates. Therefore, 𝑥(−1,𝑛)=0 eventually.
Let 𝑐<−1. In view of part (I) we have (𝑥(𝑐,𝑛)) is unbounded, and therefore 𝑥(𝑐,𝜏(𝑛))→∞. Let 𝑀>0, then there exists an index 𝑛₅ such that, for every 𝑛≥𝑛₅, 𝑥(𝑐,𝜏(𝑛))>𝑀. Since the function 𝑥(𝑐,𝑛) is continuous, 𝑥(𝑐,𝜏(𝑛)) is continuous, and therefore lim_{𝑐→−1−}𝑥(c,𝜏(𝑛))=𝑥(−1,𝜏(𝑛))=0forevery𝑛≥𝑛₅. (3.30) Hence, there exists ℎ>0 so that, if 𝑐>−1−ℎ then 𝑥(𝑐,𝜏(𝑛))<𝑀 for every 𝑛≥𝑛₅, which contradicts 𝑥(𝑐,𝜏(𝑛))>𝑀. This implies that, there exists an interval (𝑎,−1) such that 𝑥(𝑐,𝑛)=0forevery𝑐∈(𝑎,−1)eventually. (3.31) Let 𝐴=inf{𝑐∣𝑥(𝑐,𝑛)=0}. (3.32) Then, with the same argument as above, we conclude that 𝑥(𝐴,𝑛)=0eventually. (3.33) So, by a similar procedure we obtain an interval (𝐵,𝐴] such that 𝑥(𝑐,𝑛)=0forevery𝑐∈(𝐵,𝐴]eventually. (3.34) This contradicts our assumption for 𝐴. Thus, if 𝑐<−1, we have 𝑥(𝑐,𝑛)=0 eventually. The proof of part (V) of the theorem is complete.
Assume that −1<𝑐≤1. Taking into account part (III), it is enough to show part (VI) only for 𝑐=1. In view of parts (iii) and (i) of Lemma 2.1, we have lim_𝑛→∞𝑥(𝑐,𝜏(𝜎(𝑛)))=𝐿. (3.35) Assume, for the sake of contradiction, that 𝐿>0. Taking into account the fact that lim_𝑛→∞𝑥(𝑐,𝑛)=0 when 0<𝑐<1, there exists 𝑛₆∈ℕ such that, for every 𝑛≥𝑛₆, we have 𝑥((𝜎(𝑛<𝐿𝑐,𝜏)))2, (3.36) and, since 𝑥(𝑐,𝜏(𝜎(𝑛))) is continuous, lim_𝑐→1−𝑥(𝑐,𝜏(𝜎(𝑛)))=𝑥(1,𝜏(𝜎(𝑛))). (3.37) Hence, 𝑥((𝜎(𝑛≤𝐿1,𝜏)))2,∀𝑛≥𝑛₆ (3.38) which contradicts lim_𝑛→∞𝑥(𝑐,𝜏(𝜎(𝑛)))=𝐿. Therefore, 𝐿>0 is impossible. Thus, 𝐿=0, and, in view of (2.8), we conclude that lim_𝑛→∞𝑧(1,𝑛)=0, (3.39) which means that lim_𝑛→∞𝑥(1,𝑛)=0. (3.40) The proof of part (VI) of the theorem is complete.
Finally, since 𝑥(𝑐,𝑛) has convergent Taylor series, we have 𝑥(𝑐,𝑛)=^∞_𝑚=0𝑥^(𝑚)(𝑎,𝑛)𝑐𝑚!^𝑚forevery𝑎∈ℝ. (3.41) Choose 𝑎<−1. Then 𝑥^(𝑚)(𝑎,𝑛)=0 and, therefore, 𝑥(𝑐,𝑛)=0. The proof of part (VII) of the theorem is complete.
The proof of the theorem is complete.

By way of illustration and for purely pedagogical purposes, the asymptotic behavior of nonoscillatory solutions of (E) is presented in Figure 1.

Figure 1 

Corollary 3.2. For (E₁) one has the following.(i)Every nonoscillatory solution tends to ±∞ if 𝑐<−1.(ii)Every solution oscillates if 𝑐=−1.(iii)Every nonoscillatory solution tends to zero if 𝑐>−1.Furthermore, if any solution of (E₁) is continuous with respect to 𝑐, one has the following.(iv)Every solution is zero if 𝑐≤−1.(v)If, additionally, any solution of (E₁) has continuous derivatives of any order and convergent Taylor series for every 𝑐∈ℝ, then the solution is zero.

Proof. In part (I) of Theorem 3.1 we have proved that lim_𝑛→∞𝑥(𝜏(𝑛))=+∞ (3.42) and consequently lim_𝑛→∞𝑥(𝑛−𝑎)=+∞, (3.43) which means that lim_𝑛→∞𝑥(𝑛)=+∞. (3.44) The proof of part (i) of the corollary is complete.
Part (ii) is direct from part (II) of Theorem 3.1.
As we have proved in parts (III) and (IV) of Theorem 3.1, if 𝑐>−1, then lim_𝑛→∞𝑥(𝜎(𝑛))=0 (3.45) and consequently lim_𝑛→∞𝑥(𝑛−𝑏)=0, (3.46) which means that lim_𝑛→∞𝑥(𝑛)=0. (3.47) The proof of part (iii) of the corollary is complete.
Parts (iv) and (v) are direct from parts (V) and (VII) of Theorem 3.1.
The proof of the corollary is complete.