# Asymptotic Behavior of Solutions for Nonlinear Volterra Discrete Equations

Authors: E. Messina; Y. Muroya; E. Russo; A. Vecchio
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

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


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## Body

## 1. Introduction

We consider the following nonlinear Volterra discrete
equation of nonconvolution type:(1.1)xi=bi−∑j=0iai,jfi−j(xj),i=0,1,2,…,xi,bi,ai,j∈IR,bi≠0,∀i=0,1,2,…. 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 functionf(x) such that (1.2)0≤fi(x)≤f(x)forx≥0,0≥fi(x)≥f(x)forx≤0,i≥0. Note that (1.2) implies
that (1.3)f(0)=0,fi(0)=0,i≥0. 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 |fi(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|fi(x)|≤|x|,i≥0

Assume that, in (1.1) with (1.2), the
following additional hypotheses hold: (2.1)inf⁡i≥0ai,i>−1,|fi(x)|≤|x|,foranyx∈(−∞,+∞),i≥0. 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(2.2)aii≥0,then|x|≤|x+aiif0(x)|,aii≤0,then(1+aii)|x|≤|x+aiif0(x)|, 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 (2.3)|xi|≤|bi|+∑j=0i−1|ai,j||xj|,i≥0, and we
set (2.4)bi=bi1+min⁡(0,ai,i),ai,j=ai,j1+min⁡(0,ai,i),0≤j≤i−1. This inequality will be useful
in order to find sufficient conditions for the boundedness of xi 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 (2.5)sup⁡0≤j≤i−1|ai,j|≤A<+∞,∀i≥1,B=sup⁡i≥0|bi|<+∞,A¯0=sup⁡i≥i0∑j=i0i−1|ai,j|<1 for some positive integer i0,
then xi is bounded and (2.6)|xi|≤(1+A)i0B1−A¯0<+∞,i≥i0. Moreover, if (2.7)lim⁡i→∞bi=0,lim⁡i→∞|ai,j|=0∀j≥0, then limi→∞xi=0.Proof.
Let us consider (2.3), by using (2.5), we have
that (2.8)|x0|≤|b0|≤B,|x1|≤|b1|+|a1,0||x0|≤B+AB=(1+A)B,|x2|≤|b2|+|a2,0||x0|+|a2,1||x1|≤B+AB+A(1+A)B=(1+A)2B,…,|xi|≤|bi|+∑j=0i−1|ai,j||xj|≤B+∑j=0i−1A{(1+A)jB}=B+{(1+A)i−1}B=(1+A)iB,i≥0. In particular, assume that the
third part of (2.5) holds, then (2.9)|xi|≤{|bi|+∑j=0i0−1|ai,j||xj|}+∑j=i0i−1|ai,j||xj|,∀i≥i0, and thus, (2.10)|xi|≤{|bi|+∑j=0i0−1|ai,j||xj|}+A¯0max⁡i0≤j≤i|xj|. Hence, the following
inequalities hold for each k≤i: (2.11)|xk|≤{|bi|+∑j=0i0−1|ai,j||xj|}+A¯0max⁡i0≤j≤k|xj|≤{|bi|+∑j=0i0−1|ai,j||xj|}+A¯0max⁡i0≤j≤i|xj|. For this reason, (2.12)max⁡i0≤j≤i|xj|≤{|bi|+∑j=0i0−1|ai,j||xj|}+A¯0max⁡i0≤j≤i|xj|, from which we obtain
that (2.13)max⁡i0≤j≤i|xj|≤{B+∑j=0i0−1A(1+A)jB}1−A¯0=(1+A)i0B1−A¯0<+∞,i≥i0. Thus, |xi| is bounded and satisfies (2.6).
Assume thatx¯=limsupi→∞|xi|>0 and put r¯=supi≥i0∑j=i0i−1|ai,j|,γ=1+min(0,infi≥i0ai,i) and M=supi≥0|xi|.
Then, since |xi| 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)+2x on [0,ϵ].
Then, by F(0)=r¯x¯<γx¯,
there exists a constant 0<ϵ0<ϵ such that (2.14)F(ϵ0)=(r¯+ϵ0)(x¯+ϵ0)+2ϵ0<γx¯. For the above ϵ0>0,
there exists a positive integer i1≥i0 such that |xi|<x¯+ϵ0 and ∑j=i0i−1|ai,j|<r¯+ϵ0,
for any i≥i1.
By assumption (2.7), we have that for ϵ1=ϵ0/(1+M)>0,
there exists a positive integer i2≥i1 such that (2.15)|bi|<ϵ1,∑j=0i2−1|ai,j|<ϵ1foranyi≥i2. Then, for i≥i2 and ci=bi−∑j=0i2−1ai,jfi−j(xj),
we have that (2.16)|ci|≤|bi|+(∑j=0i2−1|ai,j|)M<ϵ1+ϵ1M=ϵ0,|xi|≤∑j=i2i−1|ai,j||fi−j(xj)|+|ci|<(r¯+ϵ0)(x¯+ϵ0)+ϵ0<x¯−ϵ0,i≥i2, which is a contradiction with
the limsup definition. Hence, x¯=0 and we obtain limi→∞xi=0.Note that the third part
of (2.5) is equivalent to r¯=supi≥i0∑j=i0i−1|ai,j|<γ=1+min(0,infi≥i0ai,i).The theorem above gives some conditions on the
coefficientsaij of (1.1) for the boundedness of xi 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 aii=0, bi=0 and then x0 is given. In this case, following the line of
the proof of Theorem 2.2, we can still show that xi 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 ai,j on the second index, enlarges the set of conditions
for xi to be bounded and convergent to zero. As an
example, for equation (2.17)xi=∑j=0i−11(i+1)2i−jxj, (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 ai,j=ai−j,
the third of (2.5) coincides with the known one [5, 10] (2.18)∑l=0+∞|al|<1, and the second
part of (2.7) is implied by (2.5).Theorem2.2 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, bi represents the local truncation error of the
method at the step i.
Thus, if bi is bounded for all i and if (2.5) holds, then the error xi 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 {xi}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 (2.19)B¯=∑i=0∞|bi|<+∞,A¯=sup⁡j≥0∑i=j+1∞|ai,j|<1, then ∑i=0∞|xi|≤B¯/(1−A¯)<+∞,
and consequently, limi→∞xi=0.Proof.
By (2.3), (2.20)∑k=0i|xk|≤∑k=0i|bk|+∑k=0i∑j=0k−1|ak,j||xj|≤(∑k=0i|bk|)+∑j=0i−1(∑k=j+1i|ak,j|)|xj|≤(∑k=0i|bk|)+(sup⁡j≥0∑k=j+1∞|ak,j|)(∑j=0i−1|xj|). Therefore, by (2.19), we have
that (2.21)∑k=0i|xk|≤∑k=0i|bk|{1−(supj≥0⁡∑k=j+1∞|ak,j|)}≤B¯1−A¯<+∞, and then, limi→∞xi=0.In the case (1.1) is linear, (2.22)xi=bi−∑j=0iai,jxj,i≥0, the following theorem is easily
proved.Theorem 2.4.
For the linear equation (2.22),
assume infi≥0ai,i>−1,
and for (2.23)γ=1+min⁡(0,inf⁡i≥0ai,i),ci,j=ai−1,j−ai,jγ,i−2≥j≥0,ci,i−1=1+ai−1,i−1−ai,i−1γ,i≥1,d0=b0γ,di=bi−bi−1γ,i≥1.
(i)
Suppose that(2.24)sup⁡0≤j≤i−1|ci,j|=C<+∞,D=sup⁡i≥0|di|<+∞.Then, |xi|≤(1+C)iD,i≥0.
In particular, if there exists a positive integeri0such that(2.25)C¯0=sup⁡i≥i0∑j=i0i−1|ci,j|<1,thenxiis bounded and(2.26)|xi|≤(1+C)i0D1−C¯0<+∞,i≥i0.Moreover, if(2.27)lim⁡i→∞(bi−bi−1)=0,lim⁡i→∞(∑j=0i0−1|ai,j|)=0,thenlimi→∞xi=0.(ii)
If(2.28)C¯=sup⁡j≥0∑i=j+1∞|ci,j|<1,D¯=∑i=0∞|bi+1−bi|+|b0|γ<+∞,then∑i=0∞|xi|≤D¯/(1−C¯)<+∞,
and consequently, limi→∞xi=0.Proof.
By (2.22), we obtain that (2.29)(1+ai,i)xi=(bi−bi−1)+(1+ai−1,i−1−ai,i−1)xi−1+∑j=0i−2(ai−1,j−ai,j)xj,i≥1. Then, we have
that (2.30)|xi|≤|di|+∑j=0i−1|ci,j||xj|,i≥1. 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 (3.1)ai,j≥0,i≥j≥0. 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 xi 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{xi}i=0∞ be the solution of (1.1) and assume that
(i)
|bi|≤Bandai,j≤Ajfor eachi,j≥0;(ii)
there existsi0>0such thatxi≥0 (resp., xi≤0) for anyi≥i0,
then(3.2)|ci|≤C,0≤xi≤ci(resp.,0≥xi≥ci)∀i≥i0, where ci=bi−∑j=0i0−1ai,jfi−j(xj) and C is a positive constant.
Moreover,
assume that one of the following conditions holds:
(iii1)
limi→∞ci=0,(iii2)
a=liminfj→∞(liminfi→∞ai,j)>0and there exists a strictly increasing
functionf¯(x) on (−∞,+∞)such thatf¯(0)=0andinfj≥0fj(x)≥f¯(x),x∈[0,+∞) (resp., infj≥0fj(x)≤f¯(x),x∈(−∞,0]), then limi→∞xi=0.
Furthermore, if, in addition to(iii2),
there exists a positive constant δ such that f¯(x)≥δx,x∈[0,+∞) (resp., f¯(x)≤δx,x∈(−∞,0]), then ∑i=i0∞|xi|≤(supi≥i0|ci|)/(δa)<+∞.Proof.
Sinceai,j≤Aj,|bi|≤B and f(x) is a continuous function, then |ci| is bounded. Assume that there exists a
nonnegative integer i0 such that xi≥0 for any i≥i0 (the analysis of the case xi≤0 for all i≥i0 is analogous). Then, by the fact that, for the
main hypothesis (1.2), fi−j(x)>0 whenever x>0, we have (3.3)0≤xi={bi−∑j=0i0−1ai,jfi−j(xj)}−∑j=i0iai,jfi−j(xj)≤ci,i≥i0. Hence, the first part of the
lemma is proved. Consider now the two cases (iii1) and (iii2) separately.
Case (iii1): limi→∞ci=0 of course implies limi→∞xi=0.
Case (iii2): put limsupi→∞xi=x¯.
Assume that x¯>0,
and let {xik}k=0∞ be a subsequence of {xi}i=i0∞ such that limk→∞xik=x¯.
Then, one can prove that limsupk→∞∑j=i0ikf¯(xj)=+∞.
By (1.1) and assumptions, we have (3.4)xik+∑j=i0ikaik,jf¯(xj)≤cik. Therefore, (3.5)+∞>limsup⁡k→∞cik≥limsup⁡k→∞{xik+∑j=i0ikaik,jf¯(xj)}≥x¯+{liminf⁡j→∞(liminf⁡i→∞ai,j)}limsup⁡k→∞∑j=i0ikf¯(xj) which is a contradiction because liminfj→∞(liminfi→∞ai,j)>0 and limsupk→∞∑j=i0ikf¯(xj)=+∞.
Hence, we have x¯=limi→∞xi=0.
In addition, suppose that there exists a positive
constantδ such that f¯(x)≥δx,x∈[0,+∞).
Then, we have that (3.6)xi+δ∑j=i0iai,jxj≤ci,i≥i0. Thus, from a=liminfj→∞(liminfi→∞ai,j)>0,
we conclude that 0≤∑i=i0∞xi≤(supi≥i0ci)/(δa)<+∞.
The proof is completely analogous when there exists a
nonnegative integeri0 such that xi≤0 for any i≥i0.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 (3.7)f(x)≠x,lim⁡x→−∞f(x)>−∞.Lemma 3.3.
Let{xi}i=0∞ be the solution of (1.1) with (1.2) and (3.7),
and assume that (3.8)sup⁡i≥0|bi|<+∞,λ=sup⁡i≥0∑j=0iai,j<+∞, then |xi| is bounded.Proof.
Letb¯ be the bound for |bi|,i≥0 and limx→−∞f(x)=−β>−∞.
Let us write ∑j=0iai,jfi−j(xj) as the sum of the following two
contributions: (3.9)xi=bi−∑j=0iai,jfi−j(xj)=bi−∑l=1ipai,jlfi−jl(xjl)−∑l=1inai,klfi−kl(xkl), where fi−jl(xjl)≥0,l=1,…,ip,fi−kl(xkl)<0,l=1,…,in, and ip+in=i+1.
Therefore, since (1.2), (3.1), (3.7), and (3.8) hold, we have
that (3.10)xi≤b¯−∑l=1inai,klfi−kl(xkl)≤b¯−∑l=1inai,klf(xkl)≤b¯+λβ;xi≥−b¯−∑l=1ipai,jlfi−jl(xjl)≥−b¯−f(b¯+βλ)∑l=1ipai,jl≥−b¯−λf(b¯+βλ). Thus, xi is bounded and the proof is complete.As an example we consider the equation(3.11)xi=1−∑j=0i1(i+1)2i−jexj−11+xj2, in this case f(x)=ex−1,b¯=supi≥0|bi|=1,λ=supi≥0∑j=0i1/((i+1)2i−j)=1, and β=limx→−∞(ex−1)=−1.
Hence, (3.12)−1e≤xi≤0. Another example is given by the
explicit equation (3.13)xi=(−1)i−110∑j=0i−11(i+1)2i−jexj−11+xj2. Here b¯=1 and λ=1/40,
hence (3.14)−1−140(e−1/40−1)≤xi≤3940. 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.Figure 1
Plot of (3.13) and its
bounds given by (3.14).Theorem 3.4.
Assume thatf(x)≠x is continuous on (−∞,+∞), (3.15)r¯=limsup⁡i→∞∑j=0i−1ai,j<+∞,−r¯f(−r¯f(L))>L,foranyL<0,(3.16)lim⁡i→∞bi=0,lim⁡i→∞ai,j=0, for all j≥0.
Then limi→∞xi=0.Proof.
Letx¯=liminfi→∞xi and assume x¯<0.
Since we are in the hypotheses of Lemma 3.3, |xi| is bounded and then M=supi≥0|f(xi)|<+∞.
For any fixed ϵ>0,
consider a continuous function F(x)=−(r¯+x)f(−(r¯+x)f(x¯−x)+x)−2x on [0,ϵ].
Then, by F(0)=−r¯f(−r¯f(x¯))>x¯,
there exists a constant 0<ϵ0<ϵ such that (3.17)F(ϵ0)=−(r¯+ϵ0)f(−(r¯+ϵ0)f(x¯−ϵ0)+ϵ0)−2ϵ0>x¯. For the above ϵ0>0,
there exists a positive integer i0 such that xi>x¯−ϵ0 and ∑j=0i−1ai,j<r¯+ϵ0,
for any i≥i0.
By Assumption (3.16), we have that for ϵ1=ϵ0/(1+M)>0,
there exists a positive integer i1≥i0 such that (3.18)|bi|<ϵ1,∑j=0i1−1ai,j<ϵ1foranyi≥i1. Then, for i≥i1 and ci=bi−∑j=0i1−1ai,jfi−j(xj),
we have that (3.19)|ci|≤|bi|+(∑j=0i1−1ai,j)M<ϵ1+ϵ1M=ϵ0. Let us rewrite (1.1) in the
following form: (3.20)xi=ci−∑l=1ipai,jlfi−jl(xjl)−∑l=1inai,klfi−kl(xkl), where fi−jl(xjl)≥0,
for l=1,…,ip and fi−kl(xkl)<0,
for l=1,…,in and ip+in=i−i1+1.
Thus, (3.21)xi≤ci−∑l=0inai,klf(xkl)<ϵ0−(r¯+ϵ0)f(x¯−ϵ0),∀i≥i1 and, since f(x) is an increasing function, we have that, for
all jl≥i1, (3.22)f(xjl)<f(ϵ0−(r¯+ϵ0)f(x¯−ϵ0)). Since we are in the hypothesis
that the coefficients ai,j are nonnegative, it follows
that (3.23)−∑l=0ipai,jlf(xjl)>−∑l=0ipai,jlf(ϵ0−(r¯+ϵ0)f(x¯−ϵ0))>−(r¯+ϵ0)f(ϵ0−(r¯+ϵ0)f(x¯−ϵ)). In conclusion,
from (3.24)−∑l=0ipai,jlf(xjl)>−(r¯+ϵ0)f(ϵ0−(r¯+ϵ0)f(x¯−ϵ0))=F(ϵ0)+2ϵ0 and by using (3.20), (3.19), and
(3.17), the following inequality holds: (3.25)xi≥ci−∑l=0ipai,jlf(xjl)>−ϵ0−(r¯+ϵ0)f(ϵ0−(r¯+ϵ0)f(x¯−ϵ))=F(ϵ0)+ϵ0>x¯+ϵ0. This result contradicts the liminf definition. Hence, x¯≥0,
so xi are eventually nonnegative. Since it is easy
to see that we are in the hypotheses of Lemma 3.1 (case (iii1)), then limi→∞xi=0.Remark 3.5.
Once again, in the convolution
case, the first part of (3.15) implies the second
one of (3.16).For the special casef(x)=(eαx−1)/α,α>0,
we establish the following sufficient condition from Theorem 3.4.Theorem 3.6.
Suppose thatf(x)=(eαx−1)/α,α>0 and assume that (3.26)(3.16),thefirstpartof(3.15)withr¯≤1 hold, then the solution xi of (1.1) tends to zero as i tends to infinity.Proof.
Putφ(x)=−r¯f(x),−∞<x<+∞. Since φ'(x)=−r¯eαx<0 for x∈(−∞,+∞),
hence φ(x) is a strictly monotone decreasing function in (−∞,+∞).
Now, we will prove that φ(φ(x))>x,
for −∞<x<0.
Letg(x)=φ(φ(x))−x for −∞<x<0.
Then we have that (3.27)g(x)=r¯{1−exp⁡(−r¯(eαx−1))}α−x,g'(x)=r¯{r¯eαxexp⁡(r¯−r¯eαx)}−1. By recalling that r¯<1 and x<0,
we have 0≤r¯eαx<1.
Since the function ye−y is increasing for 0≤y≤1,
there results (r¯eαx)e−r¯eαx≤e−1.
Thus, g'(x)≤r¯exp(r¯−1)−1≤r¯−1≤0.
Hence, we have that φ(φ(x))>x,
for −∞<x<0.
Thus, for f(x)=(eαx−1)/α with α>0,
the second of (3.15) is true and, by Theorem 3.4, we have limi→∞xi=0.Remark 3.7.
From the proof of Theorem3.6 it is
clear that the second part of (3.15) is satisfied by f(x)=(eαx−1)/α(α>0). By playing with α,
this allows us to consider a wide variety of functions fi which satisfy (1.2). For instance, in the
cases α=1/2 and α=1 the stained areas in Figure 2 represent the
admissible regions for the functions fi,i=0,1,…,
respectively (the solid lines show, as an example, the graphs of fi(x)=x/(1+x2) and fi(x)=(ex−1)/(1+x2)).Plots of some admissible regions
forfi according to Theorem 3.6.
(a)(b)


## 4. Monotonic Nonnegative Coefficients

In this
section, for (1.1), first, we consider the case that (4.1)0≤ai,j≤ai−1,j,0≤j≤i−1,|fi−1(x)|≥|fi(x)|,i≥1. We provide the following theorem
which generalizes [15, Theorem 2.1] to the nonlinear case.Theorem 4.1.
In addition to condition (4.1), suppose that (4.2)bi≥bi−1≥0(resp.,bi≤bi−1≤0),i≥1. Then, any solution xi of (1.1) satisfies 0≤xi≤bi (resp., 0≥xi≥bi), i≥0.
Moreover, if |bi|≤B,
for all i≥0,a=liminfj→∞(limi→∞ai,j)>0 and there exists a strictly increasing
function f¯(x) on (−∞,+∞) such that f¯(0)=0 and infj≥0fj(x)≥f¯(x),x∈[0,+∞) (resp., infj≥0fj(x)≤f¯(x),x∈(−∞,0]), then limi→∞xi=0.
In addition, if there exists a positive constantδ such that f¯(x)≥δx,x∈[0,+∞) (resp., f¯(x)≤δx,x∈(−∞,0]), then ∑i=0∞|xi|≤(supi≥0|bi|)/(δa)<+∞.Proof.
We prove the theorem in the casebi≥bi−1≥0,i≥1,
the proof for bi≤bi−1≤0,i≥1 is perfectly symmetric. Then by (1.1), we have
that (4.3)x0+a0,0f0(x0)=b0≥0, hence, for the properties of fi(x) described in (1.2), it has to be x0≥0.
Proceeding by induction, suppose thatxj≥0,0≤j≤i−1,i≥1.
By (1.1), (4.4)xi+ai,if0(xi)=bi−∑j=0i−1ai,jfi−j(xj),0=xi−1−bi−1+∑j=0i−1ai−1,jfi−j−1(xj), and hence, by adding the two
relations and taking into account that, for the second
part of (4.1), fi−j−1(x)≥fi−j(x),
we have that (4.5)xi+ai,if0(xi)≥(bi−bi−1)+xi−1+∑j=0i−2(ai−1,j−ai,j)fi−j(xj)≥bi−bi−1≥0,i≥1. So we have that xi≥0,i≥0 and, from (1.1), xi≤bi,i≥0.
Thus, we are in the hypotheses of Lemma3.1 part (iii2) and, hence, we get the thesis.Observe that when (1.1) is linear, the last condition
of Theorem 4.1 is satisfied by choosing δ=1 and f¯(x)=x.
In this case, the hypotheses of Theorem 4.1 include, as particular cases, those
of [15, Theorem 2.1]. In particular we note that, as Theorem 4.1 prove the
summability of the solution xi,
it is interesting when applied to the equation satisfied by the fundamental
matrix of a Volterra difference equation (see, e.g., [15, equation (1.4)]). Namely, in
[15] it is underlined
that such a result can be employed in the study of the stability of some
numerical methods.A simple application of Theorem4.1 in the linear case
is given by the following example.Example 4.2.
Let us consider the difference
equation(4.6)xi=b−∑j=0icxj,i≥0, whose solution is given
by (4.7)xi=b(1+c)i+1,i≥0. Then, for a>0 and b≥0,
all the conditions in Theorem 4.1 are satisfied with δ=1,
which implies limi→∞xi=0 and ∑i=0+∞xi≤(b/a)<+∞. Observe that in this case the bound coincides with the exact value of the sum of the series.Next, we provide the following theorem whose proof is
a direct extension of the proof of Crisci et al. [6, Theorem 2.1], which gives a
priori bound for the solution xi of (1.1) depending on the forcing terms bi.Theorem 4.3.
In addition to the conditions (4.1), assume that (4.8)sup⁡i≥1∑j=0i−1|bj+1−bj|<+∞. Then, any solution xi of (1.1) is bounded and satisfies (4.9)|xi|≤∑j=0i−1|bj+1−bj|+|b0|,i≥0. Moreover, suppose that a=liminfj→∞(limi→∞ai,j)>0 and there exists a strictly increasing function f¯(x) on (−∞,+∞) such that f¯(0)=0 and infj≥0fj(x)≥f¯(x),x∈[0,+∞) (resp., infj≥0fj(x)≤f¯(x),x∈(−∞,0]), then limi→∞xi=0.
In addition, if there exists a positive constantδ such that f¯(x)≥δx,x∈[0,+∞) (resp., f¯(x)≤δx,x∈(−∞,0]), then ∑i=0∞|xi|≤(2/δa)(∑j=0∞|bj+1−bj|+|b0|)<+∞.Proof.
Consider the two possible subcases: (a)x0≥0,
and (b) x0<0.
(a) Assumex0≥0.
Ifxj≥0,0≤j≤i,
then by (1.1), we get bi≥0 and 0≤xi≤bi.
Hence, (4.9) holds if {xj}j=0i is oscillatory about 0.
Let m0=−1 and denote by l1 the time moment of the first passage of the
solution xi through the zero, that is, (4.10)xj≥0form0+1=0≤j≤l1,xl1+1<0. The time moment of the following
passage through the zero of the solution after l1 is denoted by m1,
that is, (4.11)xj≤0forl1+1≤j≤m1,xm1+1>0. In a similar way, we introduce
the indexes lp,mp,p≥1 as follows: (4.12)xj≥0formp−1+1≤j≤lp,xlp+1<0,xj≤0forlp+1≤j≤mp,xmp+1>0.
(1) Consider thatmk+1≤i≤lk+1,
hence xi≥0.
Then, from (1.1), we have that (4.13)xi=bi−∑j=mk+1iai,jfi−j(xj)−∑j=0mkai,jfi−j(xj),0=xmk−bmk+∑j=lk+1mkamk,jfmk−j(xj)+∑j=0lkamk,jfmk−j(xj),0=−xlk+blk−∑j=mk−1+1lkalk,jflk−j(xj)−∑j=0mk−1alk,jflk−j(xj),0=xmk−1−bmk−1+∑j=lk−1+1mk−1amk−1,jfmk−1−j(xj)+∑j=0lk−1amk−1,jfmk−1−j(xj),⋯0=xm1−bm1+∑j=l1+1m1am1,jfm1−j(xj)+∑j=0l1am1,jfm1−j(xj),0=−xl1+bl1−∑j=0l1al1,jfl1−j(xj), where every summation of the
type ∑j=mr−1+1lr involves only positive xj,
while ∑j=lr+1mr the negative ones. Now observe that, for lk+1≤j≤mk,
by using (4.1) and the fact that mk<i,
we have |fmk−j(xj)|≥|fi−j(xj)|,
furthermore xj≤0, because of (4.12), hence both fi−j(xj) and fmk−j(xj) are less than or equal to zero, thus fmk−j(xj)≤fi−j(xj).
By using these considerations it is easy to see that the following inequality
holds: (4.14)∑j=lk+1mk[amk,jfmk−j(xj)−ai,jfi−j(xj)]≤∑j=lk+1mk(amk,j−ai,j)fi−j(xj). With analogues considerations we
get (4.15)−∑j=mk−1+1lk[alk,jflk−j(xj)−amk,jfmk−j(xj)+ai,jfi−j(xj)]≤−∑j=mk−1+1lk(alk,j−amk,j+ai,j)fi−j(xj),⋯−∑j=0l1[al1,jfl1−j(xj)−am1,jfm1−j(xj)+⋯+ai,jfi−j(xj)]≤−∑j=0l1(al1,j−am1,j+⋯+ai,j)fi−j(xj). By adding each side of (4.13) and
taking into account (4.14), (4.15), it comes out that (4.16)0≤xi≤{bi−bmk+blk−bmk−1+⋯−bm1+bl1}+{xmk−xlk+xmk−1−⋯+xm1−xl1}−∑j=mk+1iai,jfi−j(xj)+∑j=lk+1mk(amk,j−ai,j)fi−j(xj)−∑j=mk−1+1lk(alk,j−amk,j+ai,j)fi−j(xj)+∑j=lk−1+1mk−1(amk−1,j−alk,j+amk,j−ai,j)fi−j(xj)⋯+∑j=l1+1m1(am1,j−al2,j+am2,j−al3,j+⋯+amk,j−ai,j)fi−j(xj)−∑j=0l1(al1,j−am1,j+al2,j−am2,j+⋯−amk,j+ai,j)fi−j(xj). By using the monotonicity of ai,j stated by (4.1) and the main hypothesis (1.2),
taking into account (4.12), we have that (4.17)0≤xi≤bi−bmk+blk−bmk−1+⋯−bm1+bl1≤∑j=0i−1|bj−bj−1|+|b0|.
(2) Consider thatlk+1≤i≤mk,
hence xi≤0.
Proceeding as above, we have (4.18)0≥xi≥−∑j=0i−1|bj−bj−1|+|b0|. Hence, from (1) and (2), we
obtain (4.9). Part (b) of the proof is essentially mirror-like of part (a) and
leads once again to (4.9). Thus, any solution xi of (1.1) is bounded and satisfies (4.9).
Moreover, suppose thata=liminfj→∞(limi→∞ai,j)>0 and there exists a strictly increasing
function f¯(x) on (−∞,+∞) such that f¯(0)=0 and infj≥0fj(x)≥f¯(x),x∈[0,+∞) (resp., infj≥0fj(x)≤f¯(x),x∈(−∞,0]).
If there exists a nonnegative integeri0 such that xi≥0 (resp., xi≤0) for any i≥i0,
since |bi|≤∑j=0i−1|bj−bj−1|+|b0| and 0≤ai,j≤a0,j, for all j,
we are in the hypotheses of Lemma 3.1 part (iii2) and we obtain limi→∞xi=0.
On the contrary, if such an index does not exist, let x0≥0 and consider the extract subsequence {xip}p=0∞ of all the positive values in {xi}i=0∞.
Assume that x¯=limsupp→+∞xip>0,
then (4.19)limsup⁡p→+∞∑j=0pf¯(xij)=+∞. Taking into account (4.12), there
exists an index k≥0 such that mk+1≤ip≤lk+1,
then xip plays the role of xi in (4.13) and, analogously to (4.16), we
have (4.20)xip+∑j=mk+1ipaip,jfip−j(xj)+∑j=mk−1+1lk(alk,j−amk,j+aip,j)fip−j(xj)+⋯+∑j=0l1(al1,j−am1,j+al2,j−am2,j+⋯−amk,j+aip,j)fip−j(xj)≤{bip−bmk+blk−bmk−1+⋯−bm1+bl1}+{xmk−xlk+xmk−1−⋯+xm1−xl1}+∑j=lk+1mk(amk,j−aip,j)fip−j(xj)+∑j=lk−1+1mk−1(amk−1,j−alk,j+amk,j−aip,j)fip−j(xj)⋯+∑j=l1+1m1(am1,j−al2,j+am2,j−al3,j+⋯+amk,j−aip,j)fip−j(xj)≤bip−bmk+blk−bmk−1+⋯−bm1+bl1≤∑j=0ip−1|bj+1−bj|+|b0|<+∞. Hence, since f¯(x)≤fj(x),
for all x≥0,
we have that (4.21)xip+∑j=mk+1ipaip,jf¯(xj)+∑j=mk−1+1lk(alk,j−amk,j+aip,j)f¯(xj)+⋯+∑j=0l1(al1,j−am1,j+al2,j−am2,j+⋯−amk,j+aip,j)f¯(xj)≤∑j=0ip−1|bj+1−bj|+|b0|<+∞, and so, since only positive
quantities are involved, we get (4.22)a(∑j=mk+1ipf¯(xj)+∑j=mk−1+1lkf¯(xj)+⋯+∑j=0l1f¯(xj))≤∑j=0ip−1|bj+1−bj|+|b0|<+∞.Passing to the limsup as p→+∞,
we have that(4.23)alimsup⁡p→+∞(∑j=mk+1ipf¯(xj)+∑j=mk−1+1lkf¯(xj)+⋯+∑j=0l1f¯(xj))<+∞. Taking into account that all the xj involved in the summations above form the
extract {xip}p=0+∞ of the positive values in {xi}i=0+∞,
we get limsupp→+∞∑j=0pf¯(xij)<+∞,
which is a contradiction with (4.19), so x¯=0.
An analogous proof on the extract subsequence of all negative values of {xi}i=0+∞ leads to liminfn→+∞xin=0.
The same happens when x0<0.
Hence, in conclusion, we obtain that limi→∞xi=0.
In addition, suppose that there exists a positive constantδ such that f¯(x)≥δx,x∈[0,+∞).
Then, by (4.22) and the fact that a is strictly positive, we conclude that 0≤∑i=0∞max(0,xi)≤1/δa(∑j=0∞|bj+1−bj|+|b0|).
Similarly, we obtain that 0≥∑i=0∞min(0,xi)≥−1/δa(∑j=0∞|bj+1−bj|+|b0|).
Hence, ∑i=0∞|xi|≤2/δa(∑j=0∞|bj+1−bj|+|b0|)<+∞.



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*Source: 867623-2008-05-22.xml*