Discrete Dynamics in Nature and Society
Volume 2008 (2008), Article ID 867623, 18 pages
doi:10.1155/2008/867623
Research Article

Asymptotic Behavior of Solutions for Nonlinear Volterra Discrete Equations

E. Messina,2 Y. Muroya,1 E. Russo,2 and A. Vecchio3

1Department of Mathematics, Waseda University, 3-4-1 Ohkubo Shinjuku-ku, Tokyo 169-8555, Japan
2Dipartimento di Matematica e Applicazioni, Università degli Studi di Napoli “Federico II”, Complesso Monte S. Angelo, Via Cintia, 80126 Napoli, Italy
3Istituto per le Applicazioni del Calcolo “Mauro Picone”, Sede di Napoli, Via P. Castellino 111, 80131 Napoli, Italy

Received 19 June 2007; Revised 14 January 2008; Accepted 5 May 2008

Academic Editor: A. Matsumoto

Copyright © 2008 E. Messina et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We consider nonlinear difference equations of unbounded order of the form 𝑥 𝑖 = 𝑏 𝑖 − ∑ 𝑖 𝑗 = 0 𝑎 𝑖 , 𝑗 𝑓 𝑖 − 𝑗 ( 𝑥 𝑗 ) , 𝑖 = 0 , 1 , 2 , … , where 𝑓 𝑗 ( 𝑥 ) ( 𝑗 = 0 , … , 𝑖 ) are suitable functions. We establish sufficient conditions for the boundedness and the convergence of 𝑥 𝑖 as 𝑖 → + ∞ . Some of these conditions are interesting mainly for studying stability of numerical methods for Volterra integral equations.

1. Introduction

We consider the following nonlinear Volterra discrete equation of nonconvolution type: 𝑥 𝑖 = 𝑏 𝑖 − 𝑖 ∑ 𝑗 = 0 𝑎 𝑖 , 𝑗 𝑓 𝑖 − 𝑗 ( 𝑥 𝑗 𝑥 ) , 𝑖 = 0 , 1 , 2 , … , 𝑖 , 𝑏 𝑖 , 𝑎 𝑖 , 𝑗 ∈ 𝐼 𝑅 , 𝑏 𝑖 ≠ 0 , ∀ 𝑖 = 0 , 1 , 2 , … . ( 1 . 1 ) The existence problem for solution of Volterra discrete equations arises in the nonlinear implicit case. For linear implicit equations and nonlinear explicit equations, the problem is easily solved. Recently, some local and global existence theorems for Volterra discrete equations in the general case are given in [1, 2].

From now on, we assume that there exists a strictly increasing function 𝑓 ( 𝑥 ) such that 0 ≤ 𝑓 𝑖 ( 𝑥 ) ≤ 𝑓 ( 𝑥 ) f o r 𝑥 ≥ 0 , 0 ≥ 𝑓 𝑖 ( 𝑥 ) ≥ 𝑓 ( 𝑥 ) f o r 𝑥 ≤ 0 , 𝑖 ≥ 0 . ( 1 . 2 ) Note that (1.2) implies that 𝑓 ( 0 ) = 0 , 𝑓 𝑖 ( 0 ) = 0 , 𝑖 ≥ 0 . ( 1 . 3 ) The above difference equation can be considered as the discrete counterpart of the Volterra integral equation whose importance in the applications is well known (see, e.g., [3, 4]), and arises also in the application of numerical methods to Volterra integral and integrodifferential equations. The theory of the qualitative behavior of this type of nonlinear difference equation is very important, in particular for the study of numerical stability of such methods (see, e.g., [5–11] and the references therein).

In this paper, we study some sufficient conditions for the boundedness of the solutions (if they exist) of (1.1), subject to (1.2), and their asymptotic behavior as 𝑖 → + ∞ . In particular, in Section 2 we investigate the asymptotic behavior when | 𝑓 𝑖 ( 𝑥 ) | is upper bounded by a linear function. The case of nonnegative coefficients is investigated in Section 3 and, with additional monotonicity assumptions, in Section 4.

2. Case of | 𝑓 𝑖 ( 𝑥 ) | ≤ | 𝑥 | , 𝑖 ≥ 0

Assume that, in (1.1) with (1.2), the following additional hypotheses hold: i n f 𝑖 ≥ 0 𝑎 𝑖 , 𝑖 | | | 𝑓 > − 1 , 𝑖 | | | ≤ | | 𝑥 | | ( 𝑥 ) , f o r a n y 𝑥 ∈ ( − ∞ , + ∞ ) , 𝑖 ≥ 0 . ( 2 . 1 ) Observe that the second part of (2.1) is true if in (1.2) 𝑓 ( 𝑥 ) = 𝑥 . The following lemma can be easily proved.

Lemma 2.1. If 𝑎 𝑖 𝑖 | | 𝑥 | | ≤ | | | ≥ 0 , 𝑡 ℎ 𝑒 𝑛 𝑥 + 𝑎 𝑖 𝑖 𝑓 0 | | | , 𝑎 ( 𝑥 ) 𝑖 𝑖 ≤ 0 , 𝑡 ℎ 𝑒 𝑛 1 + 𝑎 𝑖 𝑖 | | 𝑥 | | ≤ | | | 𝑥 + 𝑎 𝑖 𝑖 𝑓 0 | | | , ( 𝑥 ) ( 2 . 2 ) for all 𝑥 ∈ ℝ .

Here and in the sequel we assume a sum with a negative superscript to be zero. By using (2.1) and Lemma 2.1, from (1.1), we have that | | | 𝑥 𝑖 | | | ≤ | | | 𝑏 𝑖 | | | + 𝑖 − 1 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | | | | 𝑥 𝑗 | | | , 𝑖 ≥ 0 , ( 2 . 3 ) and we set 𝑏 𝑖 = 𝑏 𝑖 1 + m i n 0 , 𝑎 𝑖 , 𝑖 , 𝑎 𝑖 , 𝑗 = 𝑎 𝑖 , 𝑗 1 + m i n 0 , 𝑎 𝑖 , 𝑖 , 0 ≤ 𝑗 ≤ 𝑖 − 1 . ( 2 . 4 ) This inequality will be useful in order to find sufficient conditions for the boundedness of 𝑥 𝑖 and for its convergence to zero as 𝑖 tends to infinity.

Theorem 2.2. Consider (1.1) with (1.2) and (2.1), if there exists a positive constant 𝐴 such that s u p 0 ≤ 𝑗 ≤ 𝑖 − 1 | | | 𝑎 𝑖 , 𝑗 | | | ≤ 𝐴 < + ∞ , ∀ 𝑖 ≥ 1 , 𝐵 = s u p 𝑖 ≥ 0 | | | 𝑏 𝑖 | | | < + ∞ , 𝐴 0 = s u p 𝑖 ≥ 𝑖 0 𝑖 − 1 ∑ 𝑗 = 𝑖 0 | | | 𝑎 𝑖 , 𝑗 | | | < 1 ( 2 . 5 ) for some positive integer 𝑖 0 , then 𝑥 𝑖 is bounded and | | | 𝑥 𝑖 | | | ≤ ( 1 + 𝐴 ) 𝑖 0 𝐵 1 − 𝐴 0 < + ∞ , 𝑖 ≥ 𝑖 0 . ( 2 . 6 ) Moreover, if l i m 𝑖 → ∞ 𝑏 𝑖 = 0 , l i m 𝑖 → ∞ | | | 𝑎 𝑖 , 𝑗 | | | = 0 ∀ 𝑗 ≥ 0 , ( 2 . 7 ) then l i m 𝑖 → ∞ 𝑥 𝑖 = 0 .

Proof. Let us consider (2.3), by using (2.5), we have that | | | 𝑥 0 | | | ≤ | | | 𝑏 0 | | | | | | 𝑥 ≤ 𝐵 , 1 | | | ≤ | | | 𝑏 1 | | | + | | | 𝑎 1 , 0 | | | | | | 𝑥 0 | | | | | | 𝑥 ≤ 𝐵 + 𝐴 𝐵 = ( 1 + 𝐴 ) 𝐵 , 2 | | | ≤ | | | 𝑏 2 | | | + | | | 𝑎 2 , 0 | | | | | | 𝑥 0 | | | + | | | 𝑎 2 , 1 | | | | | | 𝑥 1 | | | ≤ 𝐵 + 𝐴 𝐵 + 𝐴 ( 1 + 𝐴 ) 𝐵 = ( 1 + 𝐴 ) 2 | | | 𝑥 𝐵 , … , 𝑖 | | | ≤ | | | 𝑏 𝑖 | | | + 𝑖 − 1 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | | | | 𝑥 𝑗 | | | ≤ 𝐵 + 𝑖 − 1 𝑗 = 0 𝐴 ( 1 + 𝐴 ) 𝑗 𝐵 = 𝐵 + ( 1 + 𝐴 ) 𝑖 − 1 𝐵 = ( 1 + 𝐴 ) 𝑖 𝐵 , 𝑖 ≥ 0 . ( 2 . 8 ) In particular, assume that the third part of (2.5) holds, then | | 𝑥 𝑖 | | | | | 𝑏 ≤ { 𝑖 | | | + 𝑖 0 − 1 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | | | | 𝑥 𝑗 | | | } + 𝑖 − 1 𝑗 = 𝑖 0 | | | 𝑎 𝑖 , 𝑗 | | | | | | 𝑥 𝑗 | | | , ∀ 𝑖 ≥ 𝑖 0 , ( 2 . 9 ) and thus, | | | 𝑥 𝑖 | | | | | | 𝑏 ≤ { 𝑖 | | | + 𝑖 0 − 1 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | | | | 𝑥 𝑗 | | | } + 𝐴 0 m a x 𝑖 0 ≤ 𝑗 ≤ 𝑖 | | | 𝑥 𝑗 | | | . ( 2 . 1 0 ) Hence, the following inequalities hold for each 𝑘 ≤ 𝑖 : | | | 𝑥 𝑘 | | | | | | 𝑏 ≤ { 𝑖 | | | + 𝑖 0 − 1 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | | | | 𝑥 𝑗 | | | } + 𝐴 0 m a x 𝑖 0 ≤ 𝑗 ≤ 𝑘 | | | 𝑥 𝑗 | | | | | | 𝑏 ≤ { 𝑖 | | | + 𝑖 0 − 1 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | | | | 𝑥 𝑗 | | | } + 𝐴 0 m a x 𝑖 0 ≤ 𝑗 ≤ 𝑖 | | | 𝑥 𝑗 | | | . ( 2 . 1 1 ) For this reason, m a x 𝑖 0 ≤ 𝑗 ≤ 𝑖 | | | 𝑥 𝑗 | | | | | | 𝑏 ≤ { 𝑖 | | | + 𝑖 0 − 1 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | | | | 𝑥 𝑗 | | | } + 𝐴 0 m a x 𝑖 0 ≤ 𝑗 ≤ 𝑖 | | | 𝑥 𝑗 | | | , ( 2 . 1 2 ) from which we obtain that m a x 𝑖 0 ≤ 𝑗 ≤ 𝑖 | | | 𝑥 𝑗 | | | ≤ ∑ { 𝐵 + 𝑖 0 − 1 𝑗 = 0 𝐴 ( 1 + 𝐴 ) 𝑗 𝐵 } 1 − 𝐴 0 = ( 1 + 𝐴 ) 𝑖 0 𝐵 1 − 𝐴 0 < + ∞ , 𝑖 ≥ 𝑖 0 . ( 2 . 1 3 ) Thus, | 𝑥 𝑖 | is bounded and satisfies (2.6).
Assume that 𝑥 = l i m s u p 𝑖 → ∞ | 𝑥 𝑖 | > 0 and put 𝑟 = s u p 𝑖 ≥ 𝑖 0 ∑ 𝑖 − 1 𝑗 = 𝑖 0 | 𝑎 𝑖 , 𝑗 | , 𝛾 = 1 + m i n ( 0 , i n f 𝑖 ≥ 𝑖 0 𝑎 𝑖 , 𝑖 ) and 𝑀 = s u p 𝑖 ≥ 0 | 𝑥 𝑖 | . Then, since | 𝑥 𝑖 | is bounded and the third of (2.5) holds, we have that 𝑀 < + ∞ and 𝑟 < 𝛾 . Let's take any 𝜖 > 0 and consider a continuous function 𝐹 ( 𝑥 ) = ( 𝑟 + 𝑥 ) ( 𝑥 + 𝑥 ) + 2 𝑥 on [ 0 , 𝜖 ] . Then, by 𝐹 ( 0 ) = 𝑟 𝑥 < 𝛾 𝑥 , there exists a constant 0 < 𝜖 0 < 𝜖 such that 𝐹 𝜖 0 = 𝑟 + 𝜖 0 𝑥 + 𝜖 0 + 2 𝜖 0 < 𝛾 𝑥 . ( 2 . 1 4 ) For the above 𝜖 0 > 0 , there exists a positive integer 𝑖 1 ≥ 𝑖 0 such that | 𝑥 𝑖 | < 𝑥 + 𝜖 0 and ∑ 𝑖 − 1 𝑗 = 𝑖 0 | 𝑎 𝑖 , 𝑗 | < 𝑟 + 𝜖 0 , for any 𝑖 ≥ 𝑖 1 . By assumption (2.7), we have that for 𝜖 1 = 𝜖 0 / ( 1 + 𝑀 ) > 0 , there exists a positive integer 𝑖 2 ≥ 𝑖 1 such that | | | 𝑏 𝑖 | | | < 𝜖 1 , 𝑖 2 − 1 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | < 𝜖 1 f o r a n y 𝑖 ≥ 𝑖 2 . ( 2 . 1 5 ) Then, for 𝑖 ≥ 𝑖 2 and 𝑐 𝑖 = 𝑏 𝑖 − ∑ 𝑖 2 − 1 𝑗 = 0 𝑎 𝑖 , 𝑗 𝑓 𝑖 − 𝑗 ( 𝑥 𝑗 ) , we have that | | | 𝑐 𝑖 | | | ≤ | | | 𝑏 𝑖 | | | + ( 𝑖 2 − 1 ∑ 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | ) 𝑀 < 𝜖 1 + 𝜖 1 𝑀 = 𝜖 0 , | | | 𝑥 𝑖 | | | ≤ 𝑖 − 1 ∑ 𝑗 = 𝑖 2 | | | 𝑎 𝑖 , 𝑗 | | | | | | 𝑓 𝑖 − 𝑗 𝑥 𝑗 | | | + | | | 𝑐 𝑖 | | | < 𝑟 + 𝜖 0 𝑥 + 𝜖 0 + 𝜖 0 < 𝑥 − 𝜖 0 , 𝑖 ≥ 𝑖 2 , ( 2 . 1 6 ) which is a contradiction with the l i m s u p definition. Hence, 𝑥 = 0 and we obtain l i m 𝑖 → ∞ 𝑥 𝑖 = 0 .

Note that the third part of (2.5) is equivalent to 𝑟 = s u p 𝑖 ≥ 𝑖 0 ∑ 𝑖 − 1 𝑗 = 𝑖 0 | 𝑎 𝑖 , 𝑗 | < 𝛾 = 1 + m i n ( 0 , i n f 𝑖 ≥ 𝑖 0 𝑎 𝑖 , 𝑖 ) .

The theorem above gives some conditions on the coefficients 𝑎 𝑖 𝑗 of (1.1) for the boundedness of 𝑥 𝑖 which supplement the results in [12, Theorem 2.1]. Moreover, it worths while to compare our result with the ones in [9, Theorem 3.1] and [5, Theorem 4.1]. In order to do that, we assume 𝑎 𝑖 𝑖 = 0 , 𝑏 𝑖 = 0 and then 𝑥 0 is given. In this case, following the line of the proof of Theorem 2.2, we can still show that 𝑥 𝑖 vanishes as 𝑖 → ∞ provided that (2.5) and the second part of (2.7) hold. Observe that this represents an additional result with respect to [9, Theorem 3.1] and [5, Theorem 4.1] which, involving the sum of the coefficients 𝑎 𝑖 , 𝑗 on the second index, enlarges the set of conditions for 𝑥 𝑖 to be bounded and convergent to zero. As an example, for equation 𝑥 𝑖 = 𝑖 − 1 𝑗 = 0 1 ( 𝑖 + 1 ) 2 𝑖 − 𝑗 𝑥 𝑗 , ( 2 . 1 7 ) (3.2) in [9] or the sufficient condition in [5] is not satisfied, however (2.5) is fulfilled. Moreover, it is easy to see that, in the convolution case 𝑎 𝑖 , 𝑗 = 𝑎 𝑖 − 𝑗 , the third of (2.5) coincides with the known one [5, 10] + ∞ 𝑙 = 0 | | | 𝑎 𝑙 | | | < 1 , ( 2 . 1 8 ) and the second part of (2.7) is implied by (2.5).

Theorem 2.2 turns out to be quite useful in the linear case when (1.1) represents the linearized equation for the global error of a numerical method applied to a Volterra integral equation. In this case, 𝑏 𝑖 represents the local truncation error of the method at the step 𝑖 . Thus, if 𝑏 𝑖 is bounded for all 𝑖 and if (2.5) holds, then the error 𝑥 𝑖 is bounded and the bound is given in (2.6).

The following theorem provides some sufficient conditions on the coefficients of (1.1) for the summability of { 𝑥 𝑖 } + ∞ 𝑖 = 0 , which turn out to be less restrictive of those stated by [13, Theorem 2.8].

Theorem 2.3. For (1.1) with (1.2), assume (2.1). If 𝐵 = ∞ 𝑖 = 0 | | | 𝑏 𝑖 | | | < + ∞ , 𝐴 = s u p ∞ 𝑗 ≥ 0 𝑖 = 𝑗 + 1 | | | 𝑎 𝑖 , 𝑗 | | | < 1 , ( 2 . 1 9 ) then ∑ ∞ 𝑖 = 0 | 𝑥 𝑖 | ≤ 𝐵 / ( 1 − 𝐴 ) < + ∞ , and consequently, l i m 𝑖 → ∞ 𝑥 𝑖 = 0 .

Proof. By (2.3), 𝑖 𝑘 = 0 | | | 𝑥 𝑘 | | | ≤ 𝑖 𝑘 = 0 | | | 𝑏 𝑘 | | | + 𝑖 𝑘 = 0 𝑘 − 1 𝑗 = 0 | | | 𝑎 𝑘 , 𝑗 | | | | | | 𝑥 𝑗 | | | ≤ ( 𝑖 𝑘 = 0 | | | 𝑏 𝑘 | | | ) + 𝑖 − 1 𝑗 = 0 ( 𝑖 𝑘 = 𝑗 + 1 | | | 𝑎 𝑘 , 𝑗 | | | ) | | | 𝑥 𝑗 | | | ≤ ( 𝑖 𝑘 = 0 | | | 𝑏 𝑘 | | | ) + ( s u p ∞ 𝑗 ≥ 0 𝑘 = 𝑗 + 1 | | | 𝑎 𝑘 , 𝑗 | | | ) ( 𝑖 − 1 𝑗 = 0 | | | 𝑥 𝑗 | | | ) . ( 2 . 2 0 ) Therefore, by (2.19), we have that 𝑖 𝑘 = 0 | | | 𝑥 𝑘 | | | ≤ ∑ 𝑖 𝑘 = 0 | | | 𝑏 𝑘 | | | { 1 − ( s u p 𝑗 ≥ 0 ∑ ∞ 𝑘 = 𝑗 + 1 | | | 𝑎 𝑘 , 𝑗 | | | ≤ ) } 𝐵 1 − 𝐴 < + ∞ , ( 2 . 2 1 ) and then, l i m 𝑖 → ∞ 𝑥 𝑖 = 0 .

In the case (1.1) is linear, 𝑥 𝑖 = 𝑏 𝑖 − 𝑖 𝑗 = 0 𝑎 𝑖 , 𝑗 𝑥 𝑗 , 𝑖 ≥ 0 , ( 2 . 2 2 ) the following theorem is easily proved.

Theorem 2.4. For the linear equation (2.22), assume i n f 𝑖 ≥ 0 𝑎 𝑖 , 𝑖 > − 1 , and for 𝛾 = 1 + m i n 0 , i n f 𝑖 ≥ 0 𝑎 𝑖 , 𝑖 , 𝑐 𝑖 , 𝑗 = 𝑎 𝑖 − 1 , 𝑗 − 𝑎 𝑖 , 𝑗 𝛾 , 𝑖 − 2 ≥ 𝑗 ≥ 0 , 𝑐 𝑖 , 𝑖 − 1 = 1 + 𝑎 𝑖 − 1 , 𝑖 − 1 − 𝑎 𝑖 , 𝑖 − 1 𝛾 𝑑 , 𝑖 ≥ 1 , 0 = 𝑏 0 𝛾 , 𝑑 𝑖 = 𝑏 𝑖 − 𝑏 𝑖 − 1 𝛾 , 𝑖 ≥ 1 . ( 2 . 2 3 )
(i)Suppose that s u p 0 ≤ 𝑗 ≤ 𝑖 − 1 | | | 𝑐 𝑖 , 𝑗 | | | = 𝐶 < + ∞ , 𝐷 = s u p 𝑖 ≥ 0 | | | 𝑑 𝑖 | | | < + ∞ . ( 2 . 2 4 ) Then, | 𝑥 𝑖 | ≤ ( 1 + 𝐶 ) 𝑖 𝐷 , 𝑖 ≥ 0 . In particular, if there exists a positive integer 𝑖 0 such that 𝐶 0 = s u p 𝑖 ≥ 𝑖 0 𝑖 − 1 𝑗 = 𝑖 0 | | | 𝑐 𝑖 , 𝑗 | | | < 1 , ( 2 . 2 5 ) then 𝑥 𝑖 is bounded and | | 𝑥 𝑖 | | ≤ ( 1 + 𝐶 ) 𝑖 0 𝐷 1 − 𝐶 0 < + ∞ , 𝑖 ≥ 𝑖 0 . ( 2 . 2 6 ) Moreover, if l i m 𝑖 → ∞ 𝑏 𝑖 − 𝑏 𝑖 − 1 = 0 , l i m 𝑖 → ∞ ( 𝑖 0 − 1 𝑗 = 0 | | | 𝑎 𝑖 , 𝑗 | | | ) = 0 , ( 2 . 2 7 ) then l i m 𝑖 → ∞ 𝑥 𝑖 = 0 .(ii)If 𝐶 = s u p ∞ 𝑗 ≥ 0 𝑖 = 𝑗 + 1 | | | 𝑐 𝑖 , 𝑗 | | | < 1 , ∑ 𝐷 = ∞ 𝑖 = 0 | | | 𝑏 𝑖 + 1 − 𝑏 𝑖 | | | + | | | 𝑏 0 | | | 𝛾 < + ∞ , ( 2 . 2 8 ) then ∑ ∞ 𝑖 = 0 | 𝑥 𝑖 | ≤ 𝐷 / ( 1 − 𝐶 ) < + ∞ , and consequently, l i m 𝑖 → ∞ 𝑥 𝑖 = 0 .

Proof. By (2.22), we obtain that 1 + 𝑎 𝑖 , 𝑖 𝑥 𝑖 = 𝑏 𝑖 − 𝑏 𝑖 − 1 + 1 + 𝑎 𝑖 − 1 , 𝑖 − 1 − 𝑎 𝑖 , 𝑖 − 1 𝑥 𝑖 − 1 + 𝑖 − 2 𝑗 = 0 𝑎 𝑖 − 1 , 𝑗 − 𝑎 𝑖 , 𝑗 𝑥 𝑗 , 𝑖 ≥ 1 . ( 2 . 2 9 ) Then, we have that | | | 𝑥 𝑖 | | | ≤ | | | 𝑑 𝑖 | | | + 𝑖 − 1 𝑗 = 0 | | | 𝑐 𝑖 , 𝑗 | | | | | | 𝑥 𝑗 | | | , 𝑖 ≥ 1 . ( 2 . 3 0 ) Thus, analogously to the proofs of Theorems 2.2 and 2.3, we obtain the conclusion of this theorem.

3. Nonnegative Coefficients

In this section, we focus on the solutions of (1.1) with (1.2) and 𝑎 𝑖 , 𝑗 ≥ 0 , 𝑖 ≥ 𝑗 ≥ 0 . ( 3 . 1 ) Such discrete equations are useful, above all, in the investigations on the behavior of the solution of some numerical methods when used to solve nonlinear heat flow in a material with memory (see [14] and the bibliography therein). Let us start with the following lemma, which describes some aspects of the solution of (1.1)-(1.2) with (3.1) when 𝑥 𝑖 has a sign eventually constant for all 𝑖 ≥ 0 . The utility of this lemma is not in itself, but as an instrument to prove some of the next theorems (see Theorems 3.4, 4.1 and 4.3).

Lemma 3.1. Let { 𝑥 𝑖 } ∞ 𝑖 = 0 be the solution of (1.1) and assume that
(i) | 𝑏 𝑖 | ≤ 𝐵 and 𝑎 𝑖 , 𝑗 ≤ 𝐴 𝑗 for each 𝑖 , 𝑗 ≥ 0 ;(ii)there exists 𝑖 0 > 0 such that 𝑥 𝑖 ≥ 0 (resp., 𝑥 𝑖 ≤ 0 ) for any 𝑖 ≥ 𝑖 0 ,
then | | 𝑐 𝑖 | | ≤ 𝐶 , 0 ≤ 𝑥 𝑖 ≤ 𝑐 𝑖 ( 𝑟 𝑒 𝑠 𝑝 . , 0 ≥ 𝑥 𝑖 ≥ 𝑐 𝑖 ) ∀ 𝑖 ≥ 𝑖 0 , ( 3 . 2 ) where 𝑐 𝑖 = 𝑏 𝑖 − ∑ 𝑖 0 − 1 𝑗 = 0 𝑎 𝑖 , 𝑗 𝑓 𝑖 − 𝑗 ( 𝑥 𝑗 ) and 𝐶 is a positive constant.
Moreover, assume that one of the following conditions holds:
(iii1) l i m 𝑖 → ∞ 𝑐 𝑖 = 0 ,(iii2) 𝑎 = l i m i n f 𝑗 → ∞ ( l i m i n f 𝑖 → ∞ 𝑎 𝑖 , 𝑗 ) > 0 and there exists a strictly increasing function 𝑓 ( 𝑥 ) on ( − ∞ , + ∞ ) such that 𝑓 ( 0 ) = 0 and i n f 𝑗 ≥ 0 𝑓 𝑗 ( 𝑥 ) ≥ 𝑓 ( 𝑥 ) , 𝑥 ∈ [ 0 , + ∞ ) (resp., i n f 𝑗 ≥ 0 𝑓 𝑗 ( 𝑥 ) ≤ 𝑓 ( 𝑥 ) , 𝑥 ∈ ( − ∞ , 0 ] ), then l i m 𝑖 → ∞ 𝑥 𝑖 = 0 .
Furthermore, if, in addition to ( 𝑖 𝑖 𝑖 2 ) , there exists a positive constant 𝛿 such that 𝑓 ( 𝑥 ) ≥ 𝛿 𝑥 , 𝑥 ∈ [ 0 , + ∞ ) (resp., 𝑓 ( 𝑥 ) ≤ 𝛿 𝑥 , 𝑥 ∈ ( − ∞ , 0 ] ), then ∑ ∞ 𝑖 = 𝑖 0 | 𝑥 𝑖 | ≤ ( s u p 𝑖 ≥ 𝑖 0 | 𝑐 𝑖 | ) / ( 𝛿 𝑎 ) < + ∞ .

Proof. Since 𝑎 𝑖 , 𝑗 ≤ 𝐴 𝑗 , | 𝑏 𝑖 | ≤ 𝐵 and 𝑓 ( 𝑥 ) is a continuous function, then | 𝑐 𝑖 | is bounded. Assume that there exists a nonnegative integer 𝑖 0 such that 𝑥 𝑖 ≥ 0 for any 𝑖 ≥ 𝑖 0 (the analysis of the case 𝑥 𝑖 ≤ 0 for all 𝑖 ≥ 𝑖 0 is analogous). Then, by the fact that, for the main hypothesis (1.2), 𝑓 𝑖 − 𝑗 ( 𝑥 ) > 0 whenever 𝑥 > 0 , we have 0 ≤ 𝑥 𝑖 = { 𝑏 𝑖 − 𝑖 0 − 1 𝑗 = 0 𝑎 𝑖 , 𝑗 𝑓 𝑖 − 𝑗 𝑥 𝑗 } − 𝑖 𝑗 = 𝑖 0 𝑎 𝑖 , 𝑗 𝑓 𝑖 − 𝑗 𝑥 𝑗 ≤ 𝑐 𝑖 , 𝑖 ≥ 𝑖 0 . ( 3 . 3 ) Hence, the first part of the lemma is proved. Consider now the two cases ( i i i 1 ) and ( i i i 2 ) separately.
Case (iii1): l i m 𝑖 → ∞ 𝑐 𝑖 = 0 of course implies l i m 𝑖 → ∞ 𝑥 𝑖 = 0 .
Case (iii2): put l i m s u p 𝑖 → ∞ 𝑥 𝑖 = 𝑥 . Assume that 𝑥 > 0 , and let { 𝑥 𝑖 𝑘 } ∞ 𝑘 = 0 be a subsequence of { 𝑥 𝑖 } ∞ 𝑖 = 𝑖 0 such that l i m 𝑘 → ∞ 𝑥 𝑖 𝑘 = 𝑥 . Then, one can prove that l i m s u p 𝑘 → ∞ ∑ 𝑖 𝑘 𝑗 = 𝑖 0 𝑓 ( 𝑥 𝑗 ) = + ∞ . By (1.1) and assumptions, we have 𝑥 𝑖 𝑘 + 𝑖 𝑘 𝑗 = 𝑖 0 𝑎 𝑖 𝑘 , 𝑗 𝑓 𝑥 𝑗 ≤ 𝑐 𝑖 𝑘 . ( 3 . 4 ) Therefore, + ∞ > l i m s u p 𝑘 → ∞ 𝑐 𝑖 𝑘 ≥ l i m s u p 𝑘 → ∞ { 𝑥 𝑖 𝑘 + 𝑖 𝑘 𝑗 = 𝑖 0 𝑎 𝑖 𝑘 , 𝑗 𝑓 𝑥 𝑗 } ≥ 𝑥 + l i m i n f 𝑗 → ∞ l i m i n f 𝑖 → ∞ 𝑎 𝑖 , 𝑗 l i m s u p 𝑖 𝑘 → ∞ 𝑘 𝑗 = 𝑖 0 𝑓 𝑥 𝑗 ( 3 . 5 ) which is a contradiction because l i m i n f 𝑗 → ∞ ( l i m i n f 𝑖 → ∞ 𝑎 𝑖 , 𝑗 ) > 0 and l i m s u p 𝑘 → ∞ ∑ 𝑖 𝑘 𝑗 = 𝑖 0 𝑓 ( 𝑥 𝑗 ) = + ∞ . Hence, we have 𝑥 = l i m 𝑖 → ∞ 𝑥 𝑖 = 0 .
In addition, suppose that there exists a positive constant 𝛿 such that 𝑓 ( 𝑥 ) ≥ 𝛿 𝑥 , 𝑥 ∈ [ 0 , + ∞ ) . Then, we have that 𝑥 𝑖 + 𝛿 𝑖 𝑗 = 𝑖 0 𝑎 𝑖 , 𝑗 𝑥 𝑗 ≤ 𝑐 𝑖 , 𝑖 ≥ 𝑖 0 . ( 3 . 6 ) Thus, from 𝑎 = l i m i n f 𝑗 → ∞ ( l i m i n f 𝑖 → ∞ 𝑎 𝑖 , 𝑗 ) > 0 , we conclude that ∑ 0 ≤ ∞ 𝑖 = 𝑖 0 𝑥 𝑖 ≤ ( s u p 𝑖 ≥ 𝑖 0 𝑐 𝑖 ) / ( 𝛿 𝑎 ) < + ∞ .
The proof is completely analogous when there exists a nonnegative integer 𝑖 0 such that 𝑥 𝑖 ≤ 0 for any 𝑖 ≥ 𝑖 0 .

Remark 3.2. Observe that in the linear case (2.22), the last conditions of Lemma 3.1 are satisfied whenever 𝛿 = 1 and 𝑓 ( 𝑥 ) = 𝑥 .

Hereafter, we investigate on the boundedness of the solution of (1.1)-(1.2) when 𝑓 ( 𝑥 ) ≠ 𝑥 , l i m 𝑥 → − ∞ 𝑓 ( 𝑥 ) > − ∞ . ( 3 . 7 )

Lemma 3.3. Let { 𝑥 𝑖 } ∞ 𝑖 = 0 be the solution of (1.1) with (1.2) and (3.7), and assume that s u p 𝑖 ≥ 0 | | | 𝑏 𝑖 | | | < + ∞ , 𝜆 = s u p 𝑖 𝑖 ≥ 0 𝑗 = 0 𝑎 𝑖 , 𝑗 < + ∞ , ( 3 . 8 ) then | 𝑥 𝑖 | is bounded.

Proof. Let 𝑏 be the bound for | 𝑏 𝑖 | , 𝑖 ≥ 0 and l i m 𝑥 → − ∞ 𝑓 ( 𝑥 ) = − 𝛽 > − ∞ . Let us write ∑ 𝑖 𝑗 = 0 𝑎 𝑖 , 𝑗 𝑓 𝑖 − 𝑗 ( 𝑥 𝑗 ) as the sum of the following two contributions: 𝑥 𝑖 = 𝑏 𝑖 − 𝑖 𝑗 = 0 𝑎 𝑖 , 𝑗 𝑓 𝑖 − 𝑗 𝑥 𝑗 = 𝑏 𝑖 − 𝑖 𝑝 𝑙 = 1 𝑎 𝑖 , 𝑗 𝑙 𝑓 𝑖 − 𝑗 𝑙 𝑥 𝑗 𝑙 − 𝑖 𝑛 𝑙 = 1 𝑎 𝑖 , 𝑘 𝑙 𝑓 𝑖 − 𝑘 𝑙 𝑥 𝑘 𝑙 , ( 3 . 9 ) where 𝑓 𝑖 − 𝑗 𝑙 ( 𝑥 𝑗 𝑙 ) ≥ 0 , 𝑙 = 1 , … , 𝑖 𝑝 , 𝑓 𝑖 − 𝑘 𝑙 ( 𝑥 𝑘 𝑙 ) < 0 , 𝑙 = 1 , … , 𝑖 𝑛 , and 𝑖 𝑝 + 𝑖 𝑛 = 𝑖 + 1 . Therefore, since (1.2), (3.1), (3.7), and (3.8) hold, we have that 𝑥 𝑖 ≤ 𝑏 − 𝑖 𝑛 𝑙 = 1 𝑎 𝑖 , 𝑘 𝑙 𝑓 𝑖 − 𝑘 𝑙 𝑥 𝑘 𝑙 ≤ 𝑏 − 𝑖 𝑛 𝑙 = 1 𝑎 𝑖 , 𝑘 𝑙 𝑓 𝑥 𝑘 𝑙 ≤ 𝑥 𝑏 + 𝜆 𝛽 ; 𝑖 ≥ − 𝑏 − 𝑖 𝑝 𝑙 = 1 𝑎 𝑖 , 𝑗 𝑙 𝑓 𝑖 − 𝑗 𝑙 𝑥 𝑗 𝑙 ≥ − 𝑏 − 𝑓 𝑏 + 𝛽 𝜆 𝑖 𝑝 𝑙 = 1 𝑎 𝑖 , 𝑗 𝑙 ≥ − 𝑏 − 𝜆 𝑓 . 𝑏 + 𝛽 𝜆 ( 3 . 1 0 ) Thus, 𝑥 𝑖 is bounded and the proof is complete.

As an example we consider the equation 𝑥 𝑖 = 1 − 𝑖 𝑗 = 0 1 ( 𝑖 + 1 ) 2 𝑖 − 𝑗 𝑒 𝑥 𝑗 − 1 1 + 𝑥 2 𝑗 , ( 3 . 1 1 ) in this case 𝑓 ( 𝑥 ) = 𝑒 𝑥 − 1 , 𝑏 = s u p 𝑖 ≥ 0 | 𝑏 𝑖 | = 1 , 𝜆 = s u p 𝑖 ≥ 0 ∑ 𝑖 𝑗 = 0 1 / ( ( 𝑖 + 1 ) 2 𝑖 − 𝑗 ) = 1 , and 𝛽 = l i m 𝑥 → − ∞ ( 𝑒 𝑥 − 1 ) = − 1 . Hence, − 1 𝑒 ≤ 𝑥 𝑖 ≤ 0 . ( 3 . 1 2 ) Another example is given by the explicit equation 𝑥 𝑖 = ( − 1 ) 𝑖 − 1 1 0 𝑖 − 1 𝑗 = 0 1 ( 𝑖 + 1 ) 2 𝑖 − 𝑗 𝑒 𝑥 𝑗 − 1 1 + 𝑥 2 𝑗 . ( 3 . 1 3 ) Here 𝑏 = 1 and 𝜆 = 1 / 4 0 , hence 1 − 1 − 𝑒 4 0 − 1 / 4 0 − 1 ≤ 𝑥 𝑖 ≤ 3 9 4 0 . ( 3 . 1 4 ) From Figure 1 it is clear that the bounds established by Lemma 3.3 (represented by dotted lines) may be quite sharp. We are able to prove the following result.

867623.fig.001
Figure 1: Plot of (3.13) and its bounds given by (3.14).

Theorem 3.4. Assume that 𝑓 ( 𝑥 ) ≠ 𝑥 is continuous on ( − ∞ , + ∞ ) , 𝑟 = l i m s u p 𝑖 → ∞ 𝑖 − 1 𝑗 = 0 𝑎 𝑖 , 𝑗 < + ∞ , − − 𝑟 𝑓 𝑟 𝑓 ( 𝐿 ) > 𝐿 , 𝑓 𝑜 𝑟 𝑎 𝑛 𝑦 𝐿 < 0 , ( 3 . 1 5 ) l i m 𝑖 → ∞ 𝑏 𝑖 = 0 , l i m 𝑖 → ∞ 𝑎 𝑖 , 𝑗 = 0 , ( 3 . 1 6 ) 𝑗 ≥ 0 for all l i m 𝑖 → ∞ 𝑥 𝑖 = 0 . Then 𝑥 = l i m i n f 𝑖 → ∞ 𝑥 𝑖 .

Proof. Let 𝑥 < 0 and assume | 𝑥 𝑖 | . Since we are in the hypotheses of Lemma 3.3, 𝑀 =