# Boundedness and Asymptotic Stability for the Solution of Homogeneous Volterra Discrete Equations

Authors: E. Messina; A. Vecchio
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

We consider homogeneous linear Volterra Discrete Equations and we study the asymptotic behaviour of their solutions under hypothesis on the sign of the coefficients and of the first- and second-order differences. The results are then used to analyse the numerical stability of some classes of Volterra integrodifferential equations.


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## Body

## 1. Introduction

Linear Volterra Discrete Equations (VDEs) are usually represented according to two types of formulae (see, e.g., [1] and references therein, [2, XIII-10], [3, Chap. 7]):(1)xn+1-xn=gn+1+an+1xn+1+∑j=0n+1bn+1,jxj,n≥0,x0given,(2)xn=fn+∑j=0ncn,jxj,n≥1,x0given.Even if each of the equations above can be easily transformed into the other, we read in the literature (see, e.g., [4]) that (1) is the discrete analogue of a Volterra Integrodifferential Equation (VIDE), whereas (2) is seen as the discrete version of a second kind Volterra Integral Equation (VIE). This is due to the fact that the simple position(3)cn+1,n+1=an+1,cn+1,n=1+bn+1,n,cn+1,j=bn+1,j,j=0,…,n-2,which transforms (2) into (1) is not meaningful when we are dealing with numerical analysis of Volterra equations.To be more specific, a simple numerical method for the VIDE,y′(t)=g(t)+A(t)y(t)+∫0tK(t,s)y(s)ds, has the form(4)yn+1-yn=hgtn+1+hAtn+1yn+1+h2∑j=0n+1wn+1,jKtn+1,tjyj,where h is the stepsize, tn=nh,wn,j are given weights, and yn≈y(tn).Using (3), namely, cn+1,n+1=hA(tn+1), cn+1,n=h2wn+1,nK(tn+1,tn)-1, and cn+1,j=h2wn+1,jK(tn+1,tj), (4) turns into the form of (2), the analysis of which would be complicated by the fact that the coefficients do not have the same dependence on h. For such a reason, in this paper we focus on the following homogeneous VDE:(5)xn+1-xn=an+1xn+1+∑j=0n+1bn+1,jxj,n≥0,where x0 is given and bn,j=0 for j>n, and we study its asymptotic properties exploiting its particular form.For the sake of completeness, there is also another type of VDE widely used in literature (see, e.g., [5, 6])(6)xn+1=Anxn+∑j=0nBn,jxj,n≥0,x0given.This is an explicit equation which can be recasted in the form (5) with bn+1,n+1=an+1=0, by imposing An=1.Asymptotic analysis of difference equations of the form (5) or its explicit version often appeared in literature in the last decades. Some of them deal with the convolution case (bn,j=bn-j); see, for instance, [6] and the references therein and [7–12]. Most of the known results for the nonconvolution case are based on the hypothesis of double summability of the coefficients (∑n=0+∞∑j=0nbn,j<+∞); see [1, 4, 13–19]. Another interesting approach, resembling the study of continuous VIDE (see, e.g., [20, 21]), basically requires that the coefficient an+1 of (5), assumed to be negative, in some sense “prevails” on the summation of the remaining coefficients bnj. Here we would like to add another piece to the framework regarding the analysis of VDE behaviour, by considering hypotheses based on the sign of the coefficients and of their first and second differences.Since (5) is homogeneous, it has always the trivial solution. Therefore, all the results that follow are valid automatically and no assumptions are necessary when x0=0. From now on we assume that the given datum x0 is different from zero and we want to analyse the behaviour of the corresponding solution with respect to the trivial one. In Section 2 we report our main results on the asymptotic behaviour of the nontrivial solution to (5) which are then used, in Section 3, to prove the boundedness of the solution and the convergence to zero in some cases of interest.In the whole paper it is assumed the empty sum convention∑j=mMvj=0, if M<m.


## 2. Main Results

Letvn,j be a double-indexed sequence and define Δ1vn,j=vn+1,j-vn,j, Δ2vn,j=vn,j+1-vn,j, and Δ12vn,j=Δ1(Δ2vn,j). Our main result gives sufficient conditions for (5) to have a solution xn vanishing at infinity.Theorem 1.
Consider (5) and assume that(i)
∃n¯>0 such that 2an+bn,n≤0,n≥n¯,(ii)
bn,0≤0,n≥2,Δ1bn,0≥0,n≥n¯,(iii)
Δ2bn,j≤0,n≥2,j≤n-2,(iv)
Δ12bn,j≥0,n≥n¯≥1,j≤n-1.
Then, for anyx0∈R, there exists x∗ such that |xn|≤x∗,∀n>0. If, in addition,(v)
2an+bn,n≤-a∗<0,n>n¯≥1,
or(vbis)
Δ12bn,j≥b∗>0,n≥n¯≥1,j≤n-1,
then, for anyx0∈R, limn→+∞xn=0.Proof.
Setαn=an+1xn+1+∑j=0n+1bn+1,jxj, then(7)xn+12-xn2=2xn+1αn-αn2,and hence(8)2xn+1αn=2xn+12an+1+bn+1,n+1+2xn+1∑j=0nbn+1,jxj.The second addendum in the right-hand side of (8) can be written as(9)2xn+1∑j=0n+1bn+1,jxj=2xn+1∑j=0nbn+1,jΔXj,where(10)Xn+1=∑j=0nxj,n≥0,X0=0.Applying the summation by parts rule, we have(11)2xn+1∑j=0nbn+1,jxj=2xn+1bn+1,n+1Xn+1-∑j=0nXj+1Δ2bn+1,j.By adding and subtracting Xn+1∑j=0nΔ2bn+1,j in the right-hand side and by setting(12)Vn,j=Xn+1-Xj+1=∑i=j+1nxj,j≤n-1,Vn,j=0forj≥n,we get(13)2xn+1∑j=0nbn+1,jxj=2xn+1bn+1,n+1Xn+1+∑j=0nVn,jΔ2bn+1,j-Xn+1∑j=0nΔ2bn+1,j=2xn+1∑j=0nVn,jΔ2bn+1,j+Xn+1bn+1,0.Now, taking into account the fact that(14)2xn+1∑j=mnxj=∑j=mn+1xj2-∑j=mnxj2-xn+12,0≤m≤n,we have(15)2xn+1∑j=0nbn+1,jxj=∑j=0nΔ2bn+1,jΔ1Vn,j2-xn+12+bn+1,0ΔXn+12-xn+12=∑j=0nΔ2bn+1,jΔ1Vn,j2-xn+12bn+1,n+1+bn+1,0ΔXn+12.By (8) and (15), (7) becomes(16)xn+12-xn2=xn+122an+1+bn+1,n+1+∑j=0nΔ2bn+1,jΔ1Vn,j2+bn+1,0ΔXn+12-αn2.Summing up over n, for all N>n¯≥0, we have(17)xN+12-x02=∑n=0Nxn+122an+1+bn+1,n+1+∑n=0N∑j=0nΔ2bn+1,jΔ1Vn,j2+∑n=0Nbn+1,0ΔXn+12-∑n=0Nαn2.Now, let us consider the double summation at the right-hand side. By inverting the summation order, applying the summation by part rule and recalling that Vj,j=0, it becomes(18)∑n=0N∑j=0nΔ2bn+1,jΔ1Vn,j2=∑j=0NΔ2bN+2,jVN+1,j2-∑n=0N∑j=0nΔ12bn+1,jVn+1,j2.Taking account of this and applying the summation by part rule also to the third addendum in (17), we get(19)xN+12-x02=∑n=0Nxn+122an+1+bn+1,n+1+∑j=0NΔ2bN+2,jVN+1,j2-∑n=0N∑j=0nΔ12bn+1,jVn+1,j2+bN+2,0XN+22-b1,0X22-∑n=0NΔ1bn+1,0Xn+22-∑n=0Nαn2.In view of the first group of hypotheses (i)–(iv), this implies(20)xN+12≤x02+∑n=0n¯-2xn+122an+1+bn+1,n+1-∑n=0n¯-2∑j=0nΔ12bn+1,jVn+1,j2-b1,0x0+x12-∑n=0n¯-1Δ1bn+1,0Xn+22.As the whole right-hand side does not depend on N, (20) assures the boundedness of |xn| and the first part of the theorem is proved.
In order to prove the second part of our result, let us proceed by contradiction. Assume that(21)limN→+∞∑n=n¯Nxn2=+∞.From (19) and (20) we have xN+12≤C+∑n=n¯N+1(2an+bn,n)xn2 and in view of (v) xN+12≤C-a∗∑n=n¯N+1xn2, which leads to an absurd because of (21). So the series in (21) converges and limn→+∞xn2=0.
Now consider hypothesis (vbis) and once again proceed by contradiction. Assume that the series ∑n=0+∞xn does not converge, therefore ∃ϵ>0 and two increasing sequences of integers {ni} and {li} with ni>li,l0>n¯ such that ∑j=li+1ni+1xj≥ϵ, and hence(22)Vni+1,li2≥ϵ2,∀i≥0.Again, using (19) and (20), we write(23)xN+12≤C-∑n=n¯N∑j=0nΔ12bn+1,jVn+1,j2.Because of (iv) we can write(24)∑n=n¯N∑j=0nΔ12bn+1,jVn+1,j2≥∑i=0mNΔ12bni+1,liVni+1,li2,with m(N) such that nm(N)<N<nm(N)+1. This together with (22) and (vbis) leads to xN+12≤C-ϵ2b∗m(N). Since m(N)→+∞, as N increases, this is absurd. Hence, the series ∑n=0+∞xn converges and the desired result follows.It is well known that one of the most used tools in the stability analysis of VDEs is the Lyapunov approach [22–28]. As already mentioned in the introduction, among the results that can be obtained by Lyapunov techniques, the most popular are based on the hypothesis that the coefficients are summable (e.g., the result in [28, Th. 2] applied to (5) requires, among other hypotheses, that ∑i=0n-1∑j=n+1+∞bj,i/(1-aj-bj,j)<+∞). Our attempt to construct new functionals for the form (5) leads inevitably to this type of hypothesis. Very few are the cases where no summability requirements are made. One of these can be found in [26, Th. 2.2], where a Lyapunov functional is constructed which allows the stability analysis of an explicit equation, provided that some conditions on the sign of the coefficients and of their Δ’s are satisfied. In order to compare the technique developed in this paper with the Lyapunov one, we refer precisely to this theorem and consider (5) with bn,n=0 and an=a<0. In this situation the hypotheses of Theorem 1 proved above guarantee that the solution vanishes also in few cases not covered by Theorem 2.2 in [26] (just to mention one example, the coefficient of  xn,  which should be negative in [26], is allowed to assume whatever sign here).Remark 2.
It is easy to see that if hypothesis (iii) in Theorem1 holds also for j=n-1, then (vbis) assures bnn≤-b∗, for all n>n¯. Therefore, if we assume an≤0, n≥n¯, hypothesis (vbis) becomes sufficient for (v). So (vbis) does no more represent an alternative with respect to (v) and can be dropped out. In this case Theorem 1 can be stated as follows.Corollary 3.
Consider (5) and assume that an≤0, for n≥n¯,Δ2bn,n-1≤0, and that (ii)–(v) hold. Then, for any x0∈R, limn→+∞xn=0.Furthermore, we point out that checking assumption (vbis) of Theorem 1 may be difficult; hence the following result can be useful.Corollary 4.
Consider (5) and assume that (i)–(iv) hold and(25)Δ12bn,n-1≥b∗,n≥n¯,with b∗>0. Then, for any x0∈R, limn→+∞xn=0.Proof.
From (iv) and the definition ofVn,j in (12),(26)∑n=n¯N∑j=0nΔ12bn+1,jVn+1,j2≥∑n=n¯NΔ12bn+1,nVn+1,n2=∑n=n¯NΔ12bn+1,nxn+12.The desired result is readily obtained by using (23) and (25).We want to underline that Theorem1 is strongly inspired by [29] where the asymptotic behaviour of a nonlinear VIDE is studied and that “in some sense” our result can be viewed as its discrete analogue. This will be illustrated in the following section.Remark 5.
Observe that, whenΔ12bn,j is of convolution type, hypothesis (iv) in Theorem 1 becomes Δ2bn≤0, so that the advantage of using hypothesis (iv), which allowed Δ12bn,j to have a constant sign only definitely with respect to n, is completely lost. This drawback can be overcome if we know that the sign of xn is definitely constant, as it is shown in the following theorem.Theorem 6.
Consider (5) and assume that(a)
∃n¯>0 such that xn≥0(<0), for n≥n¯,(b)
∃n∗>n¯ such that βn=2an+2bn,n-bn,n¯≤0, for n≥n∗≥n¯,(c)
bn,0≤0,n≥0,Δ1bn,0≥0,n≥n∗≥n¯,(d)
Δ2bn,j≤0,n≥1,j<n,(e)
∃p≥0 such that Δ12bn,j≥0,n≥n¯+p,0≤j≤n¯-2,(f)
βn≤-a∗<0,n≥n∗≥n¯,
orfbis
Δ12bn,j≥b∗>0,n≥n¯+p,0≤j≤n¯-2.
Then, for anyx0∈R,limn→+∞xn=0.Proof.
First of all observe that (a) assures2xn+1Vn,j≥0,n≥j≥n¯-1. From here and (13) we derive(27)2xn+1∑j=0nbn+1,jxj≤∑j=0n¯-2Δ2bn+1,j2xn+1Vn,j+2xn+1Xn+1bn+1,0.Now, proceeding as in the proof of Theorem 1, we arrive to(28)xN+12-x02≤∑n=0Nβn+1xn+12+∑j=0n¯-2Δ2bN+2,jVN+1,j2-∑n=0N∑j=0n¯-2Δ12bn+1,jVn+1,j2+bN+2,0XN+22-b1,0X12-∑n=0NΔ1bn+1,0Xn+22-∑n=0Nαn2,which, taking into account (b), (c), and (e), assures(29)xN+12≤C+∑n=n∗-1Nβn+1xn+12,or(30)xN+12≤C-∑n=n¯+p-1N∑j=0n¯-2Δ12bn+1,jVn+1,j2,which corresponds to (14) and (23) of Theorem 1, respectively. The desired result follows as in the proof of Theorem 1.As a consequence of this result, the following can be easily proved.Corollary 7.
Under assumptions (b)–(f) of Theorem6 a sequence xn, obtained by (5) with x0∈R, cannot diverge and if it is convergent then its limit is zero.Remark 8.
If, in Theorem6, n¯=1, then hypothesis (e) can be removed, and the theorem assumes a simplified form.


## 3. Examples of Applications

Consider the following theoretical examples of application of Corollaries3 and 4.It can be easily seen that (5), with the choices for an and bn,j(31)an=0,bnj=-e-n-j-1n+12;(32)an=12nn+1+1n+12,bnj=-e-n-j-1n+12;(33)an=-1,bnj=-e-2n-jjn-12;(34)an=-1,bnj=-e-2n-jjn-12-1,satisfies the assumptions of the corollaries in the previous section.To be more specific (31) fulfills both corollaries with n¯=0. Equation (33) satisfies only Corollary 3 with n¯=3, whereas (32) satisfies Corollary 4 but not 2.1, because bnn+an→0. Equation (34) is only a slight modification of (33) and, like (33), it fulfills all the hypotheses of Corollary 3; furthermore it can be easily seen that ∑j=0nbnj is an unbounded sequence. So (5) with coefficients as in (34) is an example of VDE with vanishing solution and nonsummable coefficients.As a counterexample, consider (5) with coefficients given by(35)an=1+12n+11+2n-1n+12,bnj=-e-n-j-1n+12.Here condition (i) for the coefficients an and bn,n in Theorem 1 is violated and the boundedness of the solution of (5) is not guaranteed any more. In fact, this is clear in Figure 1, which shows the actual behaviour of xn.Figure 1
Problem (35): unbounded xn.Theorem6 can be applied to the following example:(36)an=-20,x0=-10,bn,j=-12n+1e-n+1n+12,j=0,-n+1-je-n-j+1n+12,1≤j≤n.First of all we need to show that (a) holds with n¯=2. Starting from the initial condition given in (36), by simple computation, we have x1=-0.003<0,0<x2<1. Our aim is to prove that 0<xn≤1, for n≥2. Let us proceed by induction on n. Assume 0≤xj≤1,j=2,…,n-1 and verify that the same is true for xn given by(37)1+20-bn,nxn=bn,0x0+bn,1x1+∑j=2n-2bn,jxj+1+bn,n-1xn-1.From the definition of bn,j in (36), it easily follows that -4<1+∑j=0n-1bn,j<0. Then taking into account the induction hypothesis and that bn,n-1≤-1, we obtain(38)-4<∑j=0n-2bn,jxj+1+bn,n-1xn-1<0.As x0=-10 and x1<0, it turns out that the right-hand side of (37) is positive, then xn>0. On the other hand (38) implies xn<(bn,0x0+bn,1x1)/21,n≥2, with bn,0≤-1/2,bn,1≤-1, which assures 0<xn<1. We conclude that hypothesis (a) of Theorem 6 is satisfied. Since (d) is true and (c), (f) are obvious with n∗=0, it remains to prove (e). In our case (e) corresponds to ∃p≥0 such that Δ12bn,j≥0, for n≥2+p and j=0. Observe that Δ12bn,0 can be written as Δ12bn,0=bn+1,1-bn,1-2(bn+1,0-bn,0)+Δ1bn,0. So Δ12bn,0≥0,n≥2, since (c) holds and bn+1,1-bn,1-2(bn+1,0-bn,0)=(e-1)e-(n+1)(n(e-1)-2). In conclusion, all the hypotheses of Theorem 6 are fulfilled and xn→0, as can be seen in Figure 2.Figure 2
Problem (36): vanishing xn.Remark 9.
We want explicitly to mention that Theorem1 cannot be applied to (36) because Δ12bn,n-1=e2-4e+3/e2<0,∀n≥2, and hypothesis (iv) is not satisfied.Finally, we observe that, if in (36) we choose bn,0 according to the remaining coefficients for j≠0, that is, bn,0=-((n+1)e-n+1/(n+1)2), then Theorem 6 is still valid with n¯=1. So we are in the case of Remark 8 and the assumption (e) of Theorem 6 can be ignored.A more practical application of our results is the study of the longtime behaviour of the numerical solution to VIDEs. Let us consider the homogeneous problem(39)y′t=Atyt+∫0tKt,sysds,t≥0,y0=y0,and a simple method of family (4), the Backward Euler method (see [30], [12, (3.8)])(40)yn+1-yn=hAtn+1yn+1+h2∑j=1n+1Ktn+1,tjyj,n≥0,where h>0 is the stepsize. With the help of the results of the previous section we can prove the following.Theorem 10.
Consider (40) and assume that(i)
∃t¯ such that A(t)≤0,t>t¯;(ii)
K(t,0)≤0,t≥0,∂K(t,0)/∂t≥0,t>t¯;(iii)
∂K(t,s)/∂s≤0,t>0;(iv)
∂2K(t,s)/∂t∂s≥0,t>t¯.
Then the solutionyn of (40) is bounded ∀y0∈R. If in addition(v)
A(t)≤-A∗<0,t>t¯,
or(vbis)
∂2K(t,s)/∂t∂s≥K∗>0,t>t¯,s<t¯,
thenlimn→+∞yn=0.Proof.
Note that (5) coincides with (40) whenever an=hA(tn),n≥1, and bnj=h2K(tn,tj),n≥1,j≤n. Now, assumptions (i)–(iv) immediately assure (i)–(iv) of Theorem 1 for any fixed h and n¯(h) such that n¯(h)h=t¯. Moreover, (v) implies (v) of Theorem 1 with a∗=hA∗, so that Corollary 3 holds. In order to exploit (vbis) note that it is equivalent to ∂2K(t,s)-K∗ts/∂t∂s>0,t>t¯,s<t¯, which in turn implies that, for any fixed h, the function Γ(t,s,h)=K(t+h,s)-K(t,s)-K∗ts is increasing with respect to s, so that (vbis) of Theorem 1 is fulfilled with b∗=h4K∗ and n¯ given above. Once again all the hypotheses of Theorem 1 are satisfied and the desired result follows.Remark 11.
Theorem10 would be particularly interesting whenever it is known that, under the same hypotheses, also the analytical solution y(t) of (39) goes to zero as t tends to infinity.As we mentioned in the previous section, the analogue of Theorem1 in the continuous case can be obtained following the line of the proof of Theorem 1 in [29]. The following is a reformulation of such a theorem suited to our case.Theorem 12.
Assume that hypotheses (i)–(iv) of Theorem10 are valid, then the solution y(t) of (39) is bounded, for any choice of y0∈R. Furthermore, if hypothesis (v) or (vbis) of Theorem 10 holds and(41)yisuniformlycontinuous,then limt→+∞y(t)=0.Remark 13.
From Theorems1 and 10 it is clear that the asymptotic behaviour of the solution to (39) is preserved both in a generic discretization of the kind (5), where bnj represents the samples K(tn,tj), and in the numerical solution obtained by the Backward Euler method (40).
Moreover, it is worth noting that if in (39)(42)∫0tKt,sds<+∞,then assumption (41) is automatically verified; nevertheless in the discrete case, the summability of the coefficients, which is the analogous of (42), is not required as showed in example (34).In the literature we sometimes encounter VIDEs with the following structure (see, e.g., [31]):(43)y′t=Atyt+∫0tft-s∫styτdτds,t≥0,y0=y0.This kind of equation can be easily recast in the form (39), with K(t,s)=∫t-stf(τ)dτ, and the method (40) for it will read(44)yn+1-yn=hAtn+1yn+1+h2∑i=1n+1∑j=n+1-in+1ftjyj,n≥0.Theorem 10 immediately becomes as follows.Corollary 14.
Consider (44) and assume that(a1)
∃t¯>0:A(t)≤0,t>t¯,(a2)
f(t)≤0,t≥0,(a3)
f′(t)≥0,t≥0,(a4)
A(t)≤A∗<0,t≥t¯,
or(a4bis)
f′(t)≥K∗>0,t>0.
Then, for anyy0∈R,limn→+∞yn=0.A comparison to Theorem12 reveals that, contrarily to (iv), hypothesis (a3) is required to hold in the whole integration range. This is due to the fact that (see Remark 5 in case of (43)) ∂2K(t-s)/∂t∂s is of convolution type. The application of Theorem 6 leads to the following result.Theorem 15.
Consider (44) and assume that(HA)
∃n¯≥0:yn≥0(<0),n>n¯;(HB)
∃t~>0:A(t)≤0,t>t~;(HC)
f(t)≤0,t>0;(HD)
f′(t)≥0,t>t~.
Then, for anyy0∈R, limn→+∞yn=0.Proof.
We want to prove that all the hypotheses of Theorem6 are fulfilled with an=hA(tn) and bnj=h2∑i=n-jnfti. Let h>0 be fixed and let n~:=n~(h) be such that n~(h)h=t~. Hypothesis (a) of Theorem 6 is obviously true because of (HA). In order to prove (b), observe that (HB) together with (HC) assure that, for n≥n∗=max⁡{n~,n¯},(45)βn=2hAtn+2h2∑j=0nftj-h2∑j=n-n¯nftj<h2ft~≤0.As bn,0=h2f(tn), (HC) implies the first condition in (c), and since Δ1bn,0=h2(f(tn+1)-f(tn)), (HD) immediately assures the second condition of (c). Furthermore, since Δ2bn,j=h2f(tn-j-1), (HC) also implies (d) and, taking into account the fact that Δ1,2bn+1,j=h2(f(tn+1-j)-f(tn-j)), we have that (HD) implies (e) with p=max⁡{0,n~-2}. Finally, whenever f is not identically zero, we can always assume, with no loss of generality, that t~ is such that f(t~)<0, so that (45) assures (f) with A∗=-h2f(t~), and the proof is complete.As already mentioned above, in [31] equation (3.32) represents the velocity of the centre of mass of a system of N particles in collective motion under alignment and chemotaxis effect. Here A(t)=0 and(46)ft=-c1e-c2te-c3/tt2,with c1,c2, and c3 depending on the number of particles, their dimension, and the dynamic of the motion. For significative values of c1,c2, and c3 hypotheses (a1), (a2), (a3), and (a4bis) of Theorem 15 are satisfied. Hence, we expect that in the numerical simulation of (43) obtained by using the Backward Euler method (44), any convergent numerical solution vanishes at infinity. Figure 3 shows exactly this behaviour when integrating (43)–(46) with c1=105,c2=10-2, and c3=10-4.Figure 3
Problem (43)–(46): vanishing yn.



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