¹¹institutetext: Ricardo Baccas ²²institutetext: Department of Mathematics, University of the West Indies, Mona, Kingston 7, Jamaica.
²²email: ricardo.baccas02@uwimona.edu.jm ³³institutetext: Cónall Kelly ⁴⁴institutetext: School of Mathematical Sciences, University College Cork, Ireland.
⁴⁴email: conall.kelly@ucc.ie ⁵⁵institutetext: Alexandra Rodkina ⁶⁶institutetext: Department of Mathematics, University of the West Indies, Mona, Kingston 7, Jamaica.
⁶⁶email: alexandra.rodkina@uwimona.edu.jm

On cubic difference equations with variable coefficients and fading stochastic perturbations

Ricardo Baccas Cónall Kelly and Alexandra Rodkina

Abstract

We consider the stochastically perturbed cubic difference equation with variable coefficients

x_{n+1}=x_{n}(1-h_{n}x_{n}^{2})+\rho_{n+1}\xi_{n+1},\quad n\in\mathbb{N},\quad x_{0}\in\mathbb{R}.

Here $(\xi_{n})_{n\in\mathbb{N}}$ is a sequence of independent random variables, and $(\rho_{n})_{n\in\mathbb{N}}$ and $(h_{n})_{n\in\mathbb{N}}$ are sequences of nonnegative real numbers. We can stop the sequence $(h_{n})_{n\in\mathbb{N}}$ after some random time $\mathcal{N}$ so it becomes a constant sequence, where the common value is an $\mathcal{F}_{\mathcal{N}}$ -measurable random variable. We derive conditions on the sequences $(h_{n})_{n\in\mathbb{N}}$ , $(\rho_{n})_{n\in\mathbb{N}}$ and $(\xi_{n})_{n\in\mathbb{N}}$ , which guarantee that $\lim_{n\to\infty}x_{n}$ exists almost surely (a.s.), and that the limit is equal to zero a.s. for any initial value $x_{0}\in\mathbb{R}$ .

1 Introduction

In this paper we analyse the global almost sure (a.s.) asymptotic behaviour of solutions of a cubic difference equation with variable coefficients and subject to stochastic perturbations

x_{n+1}=x_{n}(1-h_{n}x_{n}^{2})+\rho_{n+1}\xi_{n+1},\quad n\in\mathbb{N},\quad x_{0}\in\mathbb{R}.

(1)

Here $(\xi_{n})_{n\in\mathbb{N}}$ is a sequence of independent identically distributed random variables, $(\rho_{n})_{n\in\mathbb{N}}$ is a sequence of nonnegative reals, and $(h_{n})_{n\in\mathbb{N}}$ is a sequence of nonnegative reals.

When $(\xi_{n})_{n\in\mathbb{N}}$ is an independent sequence of standard Normal random variables, (1) can be interpreted as the Euler-Maruyama discretisation of the Itô-type stochastic differential equation

dX_{t}=-bX_{t}^{3}dt+g(t)dW_{t},\quad n\in\mathbb{N},\quad X_{0}\in\mathbb{R},

(2)

where $(W_{t})_{t\geq 0}$ is a standard Wiener process, $b>0$ is some constant, $g:[0,\infty)\to[0,\infty)$ is a continuous function. It was shown in CW that when $\lim_{t\to\infty}\rho^{2}(t)\ln t=0$ solutions of stochastic differential equation (2) are globally a.s. asymptotically stable, i.e. $\lim_{t\to\infty}X_{t}=0$ , almost surely, for any initial value $X_{0}\in\mathbb{R}$ .

There is an extensive literature on the global a.s. asymptotic behaviour of solutions of nonlinear stochastic difference equations, and the most relevant publications for our purposes are: [1, 2, 3, 4, 5, 7, 14, 15]. However, if the timestep sequence in Eq. (1) is constant, so that $h_{n}\equiv h$ , the global dynamics of (2) are not preserved and convergence of solutions to zero will only occur on a restricted subset of initial values. An early attempt to address local dynamics in an equation with bounded noise can be found in F ; general results for equations with fading, state independent noise may be found in ABR . In AKMR a complete description is given of these local dynamics (see also ABR and AMcR ). It was proved that the set of initial values can be partitioned into a “stability” region, within which solutions converge asymptotically to zero, an “instability” region, within which solutions rapidly grow without bound, and a region of unknown dynamics that is in some sense small. In the first two cases, the dynamic holds with probability at least $1-\gamma$ for $\gamma\in(0,1)$ .

In the same article, it was shown that for any initial value $x_{0}\in\mathbb{R}$ , the behaviour of solution of the difference equation can be made consistent with the corresponding solution of the differential equation, with probability $1-\gamma$ , by choosing the stepsize parameter $h$ sufficiently small. This observation motivates the approach taken in this article, wherein the stepsize parameter is allowed to decrease over a random interval in order to capture trajectories within the basin of attraction of the point at zero long enough to ensure asymptotic convergence.

Several recent publications are devoted to the use of adaptive timestepping in a explicit Euler-Maruyama discretization of nonlinear equations: for example AKR2010 ; KL2017 ; LM ; KRR . In KL2017 (see also Giles2017 ) it was shown that suitably designed adaptive timestepping strategies could be used to ensure strong convergence of order $1/2$ for a class of equations with non-globally Lipschitz drift, and globally Lipschitz diffusion. These strategies work by controlling the extent of the nonlinear drift response in discrete time and required that the timesteps depend on solution values. In KRR an extension of that idea allows an explicit Euler-Maruyama discretisation to reproduce dynamical properties of a class of nonlinear stochastic differential equations with a unique equilibrium solution and non-negative, non-globally Lipschitz drift and diffusion coefficients. The a.s. asymptotic stability and instability of the equilibrium at zero is closely reproduced, and positivity of solutions is preserved with arbitrarily high probability.

An element that these articles have in common is that the variable time-step depends upon the value of the solution. By contrast, in the present paper the sequence $(h_{n})_{n\in\mathbb{N}}$ does not, and will be the same for any given initial value $x_{0}\in\mathbb{R}$ . However since the values of $h_{n}$ can become arbitrarily small, it is not necessarily the case that $x_{n}$ converges to zero: in fact if the stepsize sequence is summable we will show that the limit is necessarily nonzero a.s. So we freeze the sequence $(h_{n})_{n\in\mathbb{N}}$ at an appropriate random moment $\mathcal{N}$ , i.e. all step-sizes after $\mathcal{N}$ are the same: $h_{n}=h_{\mathcal{N}}$ for $n\geq\mathcal{N}$ . The time at which this occurs depends on the initial value $x_{0}$ , and is chosen to ensure that $x_{n}$ converges to zero a.s., as required.

The structure of the article is as follows. Some necessary technical results are stated in Section 2. In Section 3 we construct a timestep sequence $(h_{n})_{n\in\mathbb{N}}$ that ensures solutions of the unperturbed cubic difference equation converge to a finite limit, and show that the summability of $(h_{n})_{n\in\mathbb{N}}$ determines whether or not that limit is zero. In Section 4 we examine the convergence of solutions under the influence of a deterministic perturbation, and in Section 5 we consider two kinds of stochastic perturbation; one with bounded noise, and one with Gaussian noise. Illustrative numerical examples are provided in Section 6.

2 Mathematical preliminaries

Everywhere in this paper, let $(\Omega,{\mathcal{F}},{\mathbb{P}})$ be a complete probability space. A detailed discussion of probabilistic concepts and notation may be found, for example, in Shiryaev Shiryaev96 . We will use the following elementary inequality: for each $a,b>0$ and $\alpha\in(0,1)$

(a+b)^{\alpha}\leq a^{\alpha}+b^{\alpha}.

(3)

The following lemmas also present additional useful technical results:

Lemma 1

Let $f:[0,\infty)\to[0,\infty)$ be a decreasing continuous function, then

\int_{0}^{n+1}f(x)dx>\sum_{i=1}^{n}f(i)>\int_{1}^{n+1}f(x)dx>\sum_{i=2}^{n+1}f(i).

Lemma 2

(i)

$\ln(1-x)<-x$ for $-\infty<x<0$ ;
(ii)

For $0<x<\frac{1}{2}$ the following estimate holds

$\ln(1-x)>-2x.$ (4)

Lemma 3

Let $q_{n}\in[0,1)$ for all $n\in\mathbb{N}$ . Then $\prod_{n=1}^{\infty}(1-q_{n})$ converges to non zero limit if and only if $\sum_{n=1}^{\infty}q_{n}$ converges.

We adopt the convention $\displaystyle\prod_{n=i}^{j}1=1$ if $i>j$ from here forwards. The next result can be found in (Shiryaev96, , Ch. 4.4, Ex. 1).

Lemma 4

Let $(\xi_{n})_{n\in\mathbb{N}}$ be a sequence of independent $\mathcal{N}(0,1)$ distributed random variables. Then

\mathbb{P}\left\{\limsup_{n\to\infty}\frac{\xi_{n}}{\sqrt{2\ln n}}=1\right\}=1.

(5)

We will use the following notation throughout the article:

Definition 1

Denote, for $k\in\mathbb{N}$ ,

\begin{split}&e_{[k]}^{a}=\exp\{\exp\{\dots\{a\underbrace{\}\dots\}}_{k\,\,times}\quad\mbox{for each}\quad a\in\mathbb{R},\quad e_{[0]}^{a}=1;\\ &\ln_{k}b=\ln[\ln[\dots[\ln b\underbrace{]\dots]]}_{k\,\,times}\quad\mbox{for each}\quad b\geq e_{[k]}^{1},\quad\ln_{0}b=b.\end{split}

(6)

Corollary 1

For all $n,k\in\mathbb{N}$ ,

\sum_{i=j}^{n}\frac{1}{(i+1)\ln(i+1)\dots\ln_{k}\left(i+e_{[k]}^{1}\right)}\\ >\int_{j}^{n+1}\frac{dy}{(y+e_{[k]}^{1})\ln(y+e_{[k]}^{1})\dots\ln_{k}\left(y+e_{[k]}^{1}\right)}\\ =\ln_{k+1}(n+1+e_{[k]}^{1})-\ln_{k+1}(j+e_{[k]}^{1}),

(7)

and

\sum_{i=1}^{n+1}\frac{1}{\left(i+e_{[k]}^{1}\right)\ln\left(i+e_{[k]}^{1}\right)\dots\ln_{k}\left(i+e_{[k]}^{1}\right)}\\ <\int_{0}^{n+2}\frac{dy}{(y+e_{[k]}^{1})\ln(y+e_{[k]}^{1})\dots\ln_{k}\left(y+e_{[k]}^{1}\right)}\\ =\ln_{k+1}(n+2+e_{[k]}^{1})-\ln_{k+1}(e_{[k]}^{1})=\ln_{k+1}(n+2+e_{[k]}^{1}).

(8)

Proof

Applying Lemma 1 to the decreasing, continuous function

f(x)=\frac{1}{(x+1)\ln(x+1)\dots\ln_{k}\left(x+e_{[k]}^{1}\right)}

yields the result.

3 The unperturbed deterministic cubic equation

Consider

x_{n+1}=x_{n}(1-h_{n}x_{n}^{2}),\quad x_{0}\in\mathbb{R},\quad n\in\mathbb{N}.

(9)

Everywhere in this paper we assume that $(h_{n})_{n\in\mathbb{N}}$ is a non-increasing sequence of positive numbers. We derive an estimate on each $|x_{n}|$ and present a time-step sequence $(h_{n})_{n\in\mathbb{N}}$ which provides convergence of the solution for any initial value $x_{0}\in\mathbb{R}$ .

3.1 Preliminary lemmata on solutions of Eq. (9)

Lemma 5

Let $x_{n}$ be a solution to equation (9). Assume that

there exists

N\in\mathbb{N}

such that

h_{N}x_{N}^{2}<2

(10)

Then,

(a)

the sequence $(|x_{n}|)_{n\in\mathbb{N}}$ is non-increasing and $h_{n}x_{n}^{2}<2$ for each $n\geq N$ ;
(b)

the sequence $(|x_{n}|)_{n\in\mathbb{N}}$ converges to a finite limit.

Proof

(a) Since $h_{N}x_{N}^{2}<2$ implies that $1-h_{N}x_{N}^{2}\in(-1,1)$ we have

|x_{N+1}|=|x_{N}||1-h_{N}x^{2}_{N}|<|x_{N}|.

(11)

Since $(h_{n})_{n\in\mathbb{N}}$ is a non-increasing sequence, we have $h_{N}\geq h_{N+1}$ and

h_{N+1}x^{2}_{N+1}<h_{N}x^{2}_{N}<2.

The remainder of the proof of (a) follows by induction. To prove (b) we note that the sequence $(|x_{n}|)_{n\in\mathbb{N}}$ is non-increasing and bounded below by $0$ , and therefore it converges to a finite limit.

Lemma 6

Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to equation (9). Assume that there exist $N\in\mathbb{N}$ such that

2>h_{N}x_{N}^{2}>1.

(12)

Then there exists $N_{1}>N$ such that $h_{N_{1}}x_{N_{1}}^{2}\leq 1$ .

Proof

By Lemma 5, the sequence $(|x_{n}|)_{n\in\mathbb{N}}$ is non-increasing. Furthermore, Lemma 5 part (b) implies that, for some $L\in\mathbb{R}$ ,

\lim_{n\to\infty}x_{n}^{2}=L^{2}.

(13)

Proceed by contradiction and assume that $h_{n}x^{2}_{n}>1$ for all $n\geq N$ . If either $L=0$ or $\lim_{n\to\infty}h_{n}=0$ , it follows that $\lim_{n\to\infty}h_{n}x^{2}_{n}=0$ . So $L\neq 0$ and $\lim_{n\to\infty}h_{n}=K\neq 0$ . Since $h_{n}x_{n}^{2}$ is not increasing and by (12) we have

1\leq L^{2}K<h_{n}x^{2}_{n}<2.

So it is only possible that either

(i)

$2>L^{2}K>1$ or
(ii)

$L^{2}K=1$ .

For case (i), $1-L^{2}K\in(-1,0)$ . Since $\lim_{n\to\infty}h_{n}x^{2}_{n}=L^{2}K$ , there exists $\delta\in(0,1)$ and $N_{1}\in\mathbb{N}$ such that $|1-h_{n}x^{2}|<\delta$ , for all $n\geq N_{1}$ , implying

|x_{n+1}|<\delta|x_{n}|,\quad n\geq N_{1}.

(14)

Passing to the limit of both sides of (14) as $n\to\infty$ , we get $L<\delta L$ . Since $\delta\in(0,1)$ , case (i) leads to a contradiction.

For case (ii), we have

\lim_{n\to\infty}|x_{n+1}|=\lim_{n\to\infty}|x_{n}|\lim_{n\to\infty}|1-h_{n}x_{n}^{2}|=0,

which implies that $\lim_{n\to\infty}|x_{n}|=0$ . Hence, case (ii) also leads to a contradiction. This completes the proof.

Lemma 7

Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to (9) with arbitrary initial condition $x_{0}\neq 0$ . If

\text{ there exists $N\in\mathbb{N}$ such that $h_{N}x_{N}^{2}<1$},

(15)

then

(a)

terms of the sequence $(x_{n})_{n\geq N}$ do not change sign;
(b)

the sequence $(x_{n})_{n\in\mathbb{N}}$ converges to a finite limit.

Proof

(a) Since (15) implies (10), we conclude that the sequence $(|x_{n}|)_{n\in\mathbb{N}}$ is non-increasing and therefore convergent, $1-h_{n}x_{n}^{2}\in(0,1)$ for all $n\geq N$ and then $x_{n}x_{n+1}>0$ for all $n\geq N$ . So the sign of $x_{n}$ stops changing for $n\geq N$ , which implies that the sequence $(x_{n})_{n\in\mathbb{N}}$ converges to a finite limit.

Remark 1

From Lemma 6 we conclude that condition (10) implies (15). So without loss of generality we refer to (15) instead of (10) for the remainder of the article.

Remark 2

In the case where $h_{N}x_{N}^{2}=1$ for some $N\in\mathbb{N}$ , we have $x_{n}=0$ for all $n>N$ , ensuring that $\lim_{n\to\infty}x_{n}=0$ . In the case when $h_{N}x_{N}^{2}=2$ for some $N\in\mathbb{N}$ , we have

x_{N+1}=x_{N}(1-h_{N}x^{2}_{N})=-x_{N},

which implies that $x_{N+k}=(-1)^{k}x_{N}$ . In this case $\lim_{n\to\infty}|x_{n}|=|x_{N}|$ but $\lim_{n\to\infty}x_{n}$ does not exist.

3.2 Timestep summability and the limit of solutions

In this section we show that if (15) holds, then solutions converge to a nonzero limit if the stepsize sequence is summable. If not, solutions converge asymptotically to zero.

Lemma 8

Let $(x_{n})_{n\in\mathbb{N}}$ be a solution of (9) with initial condition $x_{0}\neq 0$ . Suppose that (15) holds and that $\sum_{j=1}^{\infty}h_{j}=S<\infty$ . Then, $\lim_{n\to\infty}x_{n}=L\neq 0$ .

Proof

Since (15) holds for some $N\in\mathbb{N}$ , by Lemmata 5 and 7 we have, for all $k\in\mathbb{N}$ ,

x^{2}_{N+k}<x_{N}^{2},\quad 1-h_{N+i}x_{N+i}^{2}>0.

(16)

Then, for all $k\in\mathbb{N}$ ,

\begin{split}x_{N+k}=&x_{N+k-1}(1-h_{N+k-1}x_{N+k-1}^{2})\\ &=x_{N+k-2}(1-h_{N+k-2}x_{N+k-2}^{2})(1-h_{N+k-1}x_{N+k-1}^{2})\\ &=x_{N}\prod_{i=0}^{k-1}\left(1-h_{N+i}x_{N+i}^{2}\right).\end{split}

This implies

x_{N+k}=x_{N}e^{\sum_{i=0}^{k-1}\ln\left(1-h_{N+i}x_{N+i}^{2}\right)}.

(17)

By Lemma 5, part (a),

\sum_{i=0}^{k-1}h_{N+i}x_{N+i}^{2}<x_{N}^{2}\sum_{i=0}^{k-1}h_{N+i}<x_{N}^{2}S.

By Lemma 7, part (b), for some $L\in\mathbb{R}$ we have $\lim_{n\to\infty}x_{n}=L$ . Also, $\lim_{j\to\infty}h_{j}=0$ , since $\sum_{j=1}^{\infty}h_{j}<\infty$ . So there exists $N_{1}\in\mathbb{N}$ such that $h_{n}x^{2}_{n}<\frac{1}{2}$ for all $n\geq N_{1}$ . Without loss of generality we may therefore suppose that $N_{1}=N$ . Part (ii) of Lemma 2 applies, and so for all $i\in\mathbb{N}$ ,

\ln\left(1-h_{N+i}x_{N+i}^{2}\right)>-2h_{N+i}x_{N+i}^{2}.

(18)

Let $x_{N}>0$ . By applying (18) to (17), and by (16), we have

x_{N+k}>x_{N}e^{-2\sum_{i=0}^{k-1}h_{N+i}x_{N+i}^{2}}\geq x_{N}e^{-2x_{N}^{2}\sum_{i=0}^{k-1}h_{N+i}}>x_{N}e^{-2x_{N}^{2}S}>0.

Passing to the limit for $k\to\infty$ in above inequality we get

L=\lim_{n\to\infty}x_{n}>x_{N}e^{-2x_{N}^{2}S}>0.

Similarly, for $x_{N}<0$ , we have

x_{N+k}<x_{N}e^{-2\sum_{i=0}^{k-1}h_{N+i}x_{N+i}^{2}}\leq x_{N}e^{-2x_{N}^{2}\sum_{i=0}^{k-1}h_{N+i}}<x_{N}e^{-2x_{N}^{2}S}<0.

In both cases $\lim_{n\to\infty}x_{n}\neq 0$ , proving the statement of the Lemma.

Lemma 9

Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to (9) with the initial value $x_{0}\neq 0$ . Suppose that (15) holds and that $\sum_{j=1}^{\infty}h_{j}=\infty$ . Then $\lim_{n\to\infty}x_{n}=0$ .

Proof

First, (15) implies (16). So, by Lemma 2 part (i), for each $k\in\mathbb{N}$ ,

\ln(1-h_{N+k}x_{N+k}^{2})<-h_{N+k}x_{N+k}^{2}.

(19)

Proceed by contradiction, and suppose that $\lim_{n\to\infty}x_{n}^{2}=L^{2}$ for some $L>0$ . Since the sequence $(|x_{n}|)_{n\in\mathbb{N}}$ is non-increasing, we have $x_{N}^{2}>x_{N+i}^{2}\geq L^{2}$ . Applying (17) and (19) we obtain

\begin{split}|x_{N+k}|&=|x_{N}|e^{\sum_{i=0}^{k-1}\ln\left(1-h_{N+i}x_{N+i}^{2}\right)}<|x_{N}|e^{\sum_{i=0}^{k-1}\left(-h_{N+i}x_{N+i}^{2}\right)}\\ &<|x_{N}|e^{-L^{2}\sum_{i=0}^{k-1}h_{N+i}}.\end{split}

(20)

Passing to the limit in (20) as $k\to\infty$ , we arrive at

L^{2}<|x_{N}|e^{-L^{2}\sum_{j=1}^{\infty}h_{j}}=0,

yielding the desired contradiction.

3.3 Estimation of $|x_{n}|$

In this section we establish a useful estimate for each $|x_{n}|$ when there exists $\bar{N}\in\mathbb{N}$ such that

\frac{1}{h_{n}x_{n}^{2}}\in(0,1),\text{ for all }n\leq\bar{N}.

(21)

Lemma 10

If (21) holds for some $\bar{N}\in\mathbb{N}$ , then for all $n\leq\bar{N}$

|x_{n}|<|x_{0}|^{{3}^{n}}\prod_{i=0}^{n-1}h_{n-1-i}^{{3}^{i}},\quad n\in\mathbb{N}.

(22)

Proof

For $n=0$ we have,

x_{1}=x_{0}(1-h_{0}x_{0}^{2})=-h_{0}x_{0}^{3}\left(1-\frac{1}{h_{0}x_{0}^{2}}\right),

which, by (21), implies that

|x_{1}|=\left|h_{0}x_{0}^{3}\left(1-\frac{1}{h_{0}x_{0}^{2}}\right)\right|=h_{0}|x_{0}|^{3}\left|1-\frac{1}{h_{0}x_{0}^{2}}\right|<h_{0}|x_{0}|^{3}.

So (22) holds for $n=1$ . Assume that (22) holds for some $k<\bar{N}$ . By (21), $|x_{k+1}|<h_{k}|x_{k}|^{3},$ which implies that

|x_{k+1}|<h_{k}|x_{k}|^{3}<h_{k}|x_{0}|^{3^{k+1}}\prod_{i=0}^{k-1}h_{k-1-i}^{{3}^{i+1}}=|x_{0}|^{3^{k+1}}\prod_{i=-1}^{k-1}h_{k-1-i}^{{3}^{i+1}},

which demonstrates that (22) holds for $k+1$ , and concludes the proof for all $n\leq\bar{N}$ by induction.

Lemma 11

Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to (9) with arbitrary $x_{0}\in\mathbb{R}$ and with $(h_{n})_{n\in\mathbb{N}}$ satisfying the following condition

\sum_{j=0}^{\infty}3^{-j}\ln h^{-1}_{j}=\infty.

(23)

Then there exists $\bar{N}=\bar{N}(x_{0})$ such that (15) holds.

Proof

Suppose that (15) fails to hold for any $\bar{N}$ . Then $1/{h_{n}x_{n}^{2}}\in(0,1)$ , for all $n\in\mathbb{N}$ . For an arbitrary $\bar{N}$ , we can apply Lemma 10, making the change of variables

j=\bar{N}-1-i,\quad i=\bar{N}-1-j,\quad i=0,\dots,\bar{N}-1,\quad j=\bar{N}-1,\dots,0,

to get

|x_{\bar{N}}|<|x_{0}|^{{3}^{\bar{N}}}\prod_{j=0}^{\bar{N}-1}h_{j}^{{3}^{\bar{N}-1-j}}.

(24)

Set

F(\bar{N}):=h_{\bar{N}}\left||x_{0}|^{{3}^{\bar{N}}}\prod_{j=0}^{\bar{N}-1}h_{j}^{{3}^{\bar{N}-1-j}}\right|^{2}.

(25)

Squaring both sides of (24) and multiplying throughout by $h_{\bar{N}}$ , we obtain $h_{\bar{N}}|x_{\bar{N}}|^{2}<F(\bar{N})$ . Then

\begin{split}\ln\left[F(\bar{N})\right]&=\ln h_{\bar{N}}+2\cdot 3^{\bar{N}}\ln|x_{0}|+\sum_{j=0}^{\bar{N}-1}\ln h_{j}^{2\cdot 3^{\bar{N}-1-j}}\\ &=\ln h_{\bar{N}}+2\cdot 3^{\bar{N}}\ln|x_{0}|+\frac{2}{3}\cdot\sum_{j=0}^{\bar{N}-1}3^{\bar{N}-j}\ln h_{j}.\end{split}

(26)

Without loss of generality we can assume that $\ln h_{\bar{N}}<0$ , so $\ln h_{\bar{N}}<\frac{2}{3}\ln h_{\bar{N}}$ and, continuing from (26),

\begin{split}\ln\left[F(\bar{N})\right]&\leq 2\cdot 3^{\bar{N}}\ln|x_{0}|+\frac{2}{3}\cdot 3^{\bar{N}}\sum_{j=0}^{\bar{N}}3^{-j}\ln h_{j}\\ &=\frac{2}{3}\cdot 3^{\bar{N}}\left[\ln|x_{0}|^{3}+\sum_{j=0}^{\bar{N}}3^{-j}\ln h_{j}\right].\end{split}

(27)

The expression in the square brackets is negative for any $x_{0}\in\mathbb{R}$ with $\bar{N}$ sufficiently large if condition (23) holds. In this case for each $x_{0}\in\mathbb{R}$ we can find $\bar{N}=\bar{N}(x_{0})$ s.t.

\sum_{j=0}^{\bar{N}}3^{-j}\ln h^{-1}_{j}>\ln|x_{0}|^{3}.

Then $F(\bar{N})<1$ which means that $|x_{\bar{N}}|<1$ as well as $h_{\bar{N}}x_{\bar{N}}^{2}<1$ . So condition (15) holds for $\bar{N}=\bar{N}(x_{0})$ . Obtained contradiction proves the result.

Lemmata 9 and 11 imply the following corollary.

Corollary 2

Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to (9) with arbitrary $x_{0}\in\mathbb{R}$ and with $(h_{n})_{n\in\mathbb{N}}$ satisfying condition (23). Then $\lim_{n\to\infty}x_{n}=0$ .

Lemma 12

Condition (23) holds if

(i)

$h_{n}\leq e^{-3^{n}}$ ;
(ii)

$h_{n}\leq e^{-\frac{3^{n}}{n}}$ ;
(iii)

$h_{n}\leq e^{-\frac{3^{n}}{n\ln n}}$ ;
(iv)

$h_{n}\leq e^{-\frac{3^{n}}{n\ln n\ln_{2}n\dots\ln_{k}n}}$ .

Proof

Case (i): we have $3^{-j}\ln h^{-1}_{j}\geq 1$ . Case (ii): we have $3^{-j}\ln h^{-1}_{j}\geq\frac{1}{j}$ . Cases (iii) and (iv): we have $3^{-j}\ln h^{-1}_{j}\geq\frac{1}{j\ln j},\dots$ etc. Note that the series

\sum^{\infty}3^{-j}\ln h^{-1}_{j}=\infty,

for $h_{j}$ defined by each of (i)-(iv). The lower limit of summation should be chosen according to $h_{j}$ in order to avoid zero denominators.

Remark 3

Applying Lemma 1 we conclude that for $h_{j}$ defined by each of (i)-(iv), the corresponding $\bar{N}(x_{0})$ can be estimated as

(i)

$\bar{N}(x_{0})>\ln|x_{0}|^{3}$ ;
(ii)

$\ln\bar{N}(x_{0})>\ln|x_{0}|^{3}$ , so $\bar{N}(x_{0})>|x_{0}|^{3}$ ;
(iii)

$\ln[\ln[\bar{N}(x_{0})]]>\ln|x_{0}|^{3}$ , so $\bar{N}(x_{0})>e^{|x_{0}|^{3}}$ ;
(iv)

$\ln_{k-1}[\bar{N}(x_{0})]>\ln|x_{0}|^{3}$ , so $\bar{N}(x_{0})>e^{e^{\dots{}^{|x_{0}|^{3}}}}$ .

4 The perturbed deterministic cubic difference equation

Consider the perturbed difference equation

x_{n+1}=x_{n}(1-h_{n}x_{n}^{2})+u_{n+1},\quad x_{0}\in\mathbb{R}.

(28)

where $(u_{n})_{n\in\mathbb{R}}$ is a real-valued sequence. We begin by providing an estimate for solutions of (28) under condition (21).

Lemma 13

Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to equation (28) and let condition (21) hold. Then, for $n\leq\bar{N}$ ,

\begin{split}|x_{n+1}|^{\frac{1}{3^{n+1}}}&<|x_{0}|\prod_{i=0}^{n}h_{i}^{\frac{1}{3^{i+1}}}+\sum_{i=1}^{n+1}\prod_{j=i}^{n}h_{j}^{\frac{1}{3^{j+1}}}|u_{i}|^{\frac{1}{3^{i}}}\\ &=\prod_{i=0}^{n}h_{i}^{\frac{1}{3^{i+1}}}\left[|x_{0}|+\sum_{i=1}^{n+1}\prod_{j=0}^{i-1}h_{j}^{-\frac{1}{3^{j+1}}}|u_{i}|^{\frac{1}{3^{i}}}\right].\end{split}

(29)

Proof

By condition (21), for each $n\leq\bar{N}$ we have

\begin{split}|x_{n+1}|&\leq|x_{n}(1-h_{n}x_{n}^{2})|+|u_{n+1}|\\ &\leq h_{n}|x_{n}|^{3}\left|1-\frac{1}{h_{n}x_{n}^{2}}\right|+|u_{n+1}|\\ &\leq h_{n}|x_{n}|^{3}+|u_{n+1}|.\end{split}

(30)

Applying the inequality (3) with $\alpha_{1}=\frac{1}{3}$ , to (30) with $n=0$ , we get

|x_{1}|^{\frac{1}{3}}\leq h_{0}^{\frac{1}{3}}|x_{0}|+|u_{1}|^{\frac{1}{3}}.

(31)

Applying the inequality (3) with $\alpha_{2}=\frac{1}{3^{2}}$ , to (30) with $n=1$ , and substituting (31), we get

|x_{2}|^{\frac{1}{3^{2}}}\leq h_{1}^{\frac{1}{3^{2}}}|x_{1}|^{\frac{1}{3}}+|u_{2}|^{\frac{1}{3^{2}}}\leq h_{1}^{\frac{1}{3^{2}}}h_{0}^{\frac{1}{3}}|x_{0}|+h_{1}^{\frac{1}{3^{2}}}|u_{1}|^{\frac{1}{3}}+|u_{2}|^{\frac{1}{3^{2}}}.

(32)

Continue this process inductively, and applying the inequality (3) with $\alpha_{n}=\frac{1}{3^{n+1}}$ we get

\begin{split}|x_{n+1}|^{\frac{1}{3^{n+1}}}&\leq h_{n}^{\frac{1}{3^{n+1}}}h_{n-1}^{\frac{1}{3^{n}}}\dots h_{1}^{\frac{1}{3^{2}}}h_{0}^{\frac{1}{3}}|x_{0}|+h_{n}^{\frac{1}{3^{n+1}}}h_{n-1}^{\frac{1}{3^{n}}}\dots h_{2}^{\frac{1}{3^{2}}}h_{1}^{\frac{1}{3}}|u_{1}|^{\frac{1}{3}}\\ &+h_{n}^{\frac{1}{3^{n+1}}}h_{n-1}^{\frac{1}{3^{n}}}\dots h_{3}^{\frac{1}{3^{4}}}h_{2}^{\frac{1}{3^{3}}}|u_{2}|^{\frac{1}{3^{2}}}+\dots+h_{n}^{\frac{1}{3^{n+1}}}|u_{n}|^{\frac{1}{3^{n}}}+|u_{n+1}|^{\frac{1}{3^{n+1}}},\end{split}

which completes the proof.

4.1 Boundedness of $(|x_{n}|)_{n\in\mathbb{N}}$ for particular $(h_{n})_{n\in\mathbb{N}}$ and $(u_{n})_{n\in\mathbb{N}}$

In this section we consider two special cases of $h_{n}$ and $u_{n}$ each of which guarantees the boundedness of the sequence $(|x_{n}|)_{n\in\mathbb{N}}$ . Both forms of $h_{n}$ were introduced in Lemma 12: the first corresponds to (ii)- (iv), the second corresponds to (i). Estimates of $|u_{n}|$ are chosen relative to the estimates for $|h_{n}|$ .

Case 1

Let $e_{[k]}^{1}$ and $\ln_{k}(\cdot)$ be defined as in (6). Assume that, there exists $k\in\mathbb{N}$ and $\beta\in(0,1)$ such that

\begin{split}&h_{n}\leq\exp\left\{{-\frac{3^{n+1}}{\left(n+e_{[k]}^{1}\right)\ln\left(n+e_{[k]}^{1}\right)\dots\ln_{k}\left(n+e_{[k]}^{1}\right)}}\right\},\\ &(h_{n})_{n\in\mathbb{N}}\quad\mbox{is a decreasing sequence},\end{split}

(33)

and

|u_{n}|:\leq\left(\frac{\beta}{\left(n+e_{[k]}^{1}\right)\ln\left(n+e_{[k]}^{1}\right)\dots\ln_{k}\left(n+e_{[k]}^{1}\right)}\right)^{3^{n}}.

(34)

Lemma 14

Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to equation (28) and let $(h_{n})_{n\in\mathbb{N}}$ and $(u_{n})_{n\in\mathbb{N}}$ satisfy (33) and (34), respectively. Then

(i)

there exists $N_{1}$ such that $|x_{N_{1}+1}|<1$ , and (15) holds;
(ii)

$|x_{N_{1}+i}|$ is uniformly bounded for all $i\in\mathbb{N}$ .

Proof

Suppose to the contrary that (21) holds for all $n$ . Then, by Lemma 13, estimate (29) holds for all $n\in\mathbb{N}$ .

Substituting the values of $h_{n}$ from (33) and $u_{n}$ from (34) into (29) we get

\begin{split}&|x_{n+1}|^{\frac{1}{3^{n+1}}}\leq\exp\left\{-\sum_{i=0}^{n}\frac{1}{\left(i+e_{[k]}^{1}\right)\ln\left(i+e_{[k]}^{1}\right)\dots\ln_{k}\left(i+e_{[k]}^{1}\right)}\right\}|x_{0}|\\ &+\sum_{i=1}^{n+1}\exp\left\{-\sum_{j=i}^{n}\frac{1}{(j+e_{[k]}^{1})\ln(j+e_{[k]}^{1})\dots\ln_{k}\left(n+2+e_{[k]}^{1}\right)}\right\}|u_{j}|^{\frac{1}{3^{i}}}.\end{split}

Now we apply the inequalities from (7) and (8) and get

\exp\left\{-\sum_{i=j}^{n}\frac{1}{\left(i+e_{[k]}^{1}\right)\ln\left(i+e_{[k]}^{1}\right)\dots\ln_{k}\left(i+e_{[k]}^{1}\right)}\right\}\leq\frac{\ln_{k}(j+e_{[k]}^{1})}{\ln_{k}(n+1+e_{[k]}^{1})},

and

\exp\left\{-\sum_{i=0}^{n}\frac{1}{\left(i+e_{[k]}^{1}\right)\ln\left(i+e_{[k]}^{1}\right)\dots\ln_{k}\left(i+e_{[k]}^{1}\right)}\right\}\leq\frac{1}{\ln_{k}(n+1+e_{[k]}^{1})}.

Applying all the above we arrive at

\begin{split}|x_{n+1}|^{\frac{1}{3^{n+1}}}&\leq\frac{|x_{0}|}{\ln_{k}(n+1+e_{[k]}^{1})}+\frac{\sum_{j=1}^{n+1}\ln_{k}(j+e_{[k]}^{1})|u_{j}|^{\frac{1}{3^{j}}}}{\ln_{k}(n+2+e_{[k]}^{1})}\\ &\leq\frac{|x_{0}|\ln_{k}}{\ln_{k}(n+2+e_{[k]}^{1})}+\frac{\sum_{j=1}^{n+1}\frac{\beta}{(j+e_{[k]}^{1})\ln(j+e_{[k]}^{1})\dots\ln_{k-1}(j+e_{[k]}^{1})}}{\ln_{k}(n+2+e_{[k]}^{1})}\\ &=\frac{|x_{0}|}{\ln_{k}(n+1+e_{[k]}^{1})}+\beta\left(\ln_{k}(n+2+e_{[k]}^{1})\right)^{-1}\ln_{k}(n+2+e_{[k]}^{1})\\ &=\frac{|x_{0}|}{\ln_{k}(n+1+e_{[k]}^{1})}+\beta.\end{split}

(35)

So for each $\beta\in(0,1)$ we can find $N_{1}$ such that, for $n\geq N_{1}$ ,

\frac{|x_{0}|}{\ln_{k}(n+1+e_{[k]}^{1})}+\beta<1,

which implies $|x_{N_{1}+1}|<1$ . Assume now that $N_{2}>2$ is such that, for $n\geq N_{2}$ , we have

\left(n+e_{[k]}^{1}\right)\ln\left(n+e_{[k]}^{1}\right)\dots\ln_{k}\left(n+e_{[k]}^{1}\right)\leq 3^{\frac{n}{2}}.

Then, for $n\geq N_{2}$ ,

h_{n}\leq e^{-\frac{3^{n+1}}{\left(n+e_{[k]}^{1}\right)\ln\left(n+e_{[k]}^{1}\right)\dots\ln_{k}\left(n+e_{[k]}^{1}\right)}}<e^{-3^{\frac{n}{2}+1}}\leq e^{-3^{2}}=e^{-9}.

(36)

Without loss of generality we can assume that $N_{1}\geq N_{2}$ . We have

0<1-h_{N_{1}+1}x_{N_{1}+1}^{2}<1,\quad|x_{N_{1}+2}|<|x_{N_{1}+1}|+|u_{N_{1}+2}|.

Also

x_{N_{1}+2}^{2}<2x_{N_{1}+1}^{2}+2u_{N_{1}+2}^{2},\quad\mbox{and}\quad|u_{n}|<1,\quad\forall n\in\mathbb{N},

\begin{split}&h_{N_{1}+2}x_{N_{1}+2}^{2}<2h_{N_{1}+2}\left[x_{N_{1}+1}^{2}+u_{N_{1}+2}^{2}\right]=2e^{-9}\left[x_{N_{1}+1}^{2}+1\right]\\ &=4e^{-9}\approx 0.00049<1.\end{split}

(37)

Based on that we get

|x_{N_{1}+3}|<|x_{N_{1}+2}|+|u_{N_{1}+3}|<|x_{N_{1}+1}|+|u_{N_{1}+2}|+|u_{N_{1}+3}|.

Applying induction, assume that, for some $k\in\mathbb{N}$ ,

|x_{N_{1}+2+k}|\leq|x_{N_{1}+1}|+\sum_{i=1}^{k}|u_{N_{1}+2+i}|\quad\mbox{and}\quad x^{2}_{N_{1}+2+k}h_{N_{1}+2+k}<1,

(38)

and prove that relations in (38) hold for $k+1$ . In order to do so we first get the estimate of $\sum_{i=1}^{k}|u_{N_{1}+2+i}|$ . For all $n\in\mathbb{N}$ , we have

|u_{n}|\leq\left(\frac{\beta}{\left(n+e_{[k]}^{1}\right)\ln\left(n+e_{[k]}^{1}\right)\dots\ln_{k}\left(n+e_{[k]}^{1}\right)}\right)^{3^{n}}<\left(\frac{\beta}{n+e_{[k]}^{1}}\right)^{3^{n}}.

Then, for $n\geq N_{1}+2\geq 4$ ,

|u_{n}|\leq\left(\frac{\beta}{n}\right)^{3^{n}}<\left(\frac{\beta}{n}\right)^{n}\leq\left(\frac{\beta}{4}\right)^{n}

and

\sum_{i=1}^{k}|u_{N_{1}+2+i}|<\sum_{i=1}^{\infty}|u_{N_{1}+2+i}|\leq\sum_{n=4}^{\infty}\left(\frac{\beta}{4}\right)^{n}=\frac{\left(\frac{\beta}{4}\right)^{4}}{1-\frac{\beta}{4}}<\frac{4\left(\frac{1}{4}\right)^{4}}{4-\beta}<\frac{1}{4^{3}\times 3}<1.

(39)

Now,

|x_{N_{1}+2+k+1}|\leq|x_{N_{1}+2+k}|+|u_{N_{1}+2+k+1}|\leq|x_{N_{1}+1}|+\sum_{i=1}^{k+1}|u_{N_{1}+2+i}|,

proving the first part of (38) for each $k\in\mathbb{N}$ , and

\begin{split}h_{N_{1}+2+k+1}x^{2}_{N_{1}+2+k+1}&\leq 2h_{N_{1}+2+k+1}|x_{N_{1}+1}|^{2}+2h_{N_{1}+2+k+1}\left(\sum_{i=1}^{k+1}|u_{N_{1}+2+i}|\right)^{2}\\ &\leq 2e^{-9}\left[1+1\right]\leq 4e^{-9}<1,\end{split}

proving the second part of (38) for each $k\in\mathbb{N}$ . This completes the proof of Part (i).

From (38) and (39) we have

\displaystyle|x_{N_{1}+2+k}|<|x_{N_{1}+1}|+1,

for each $k\in\mathbb{N}$ , which completes the proof of Part (ii).

Case 2

Assume that, for some $\beta\in(0,1)$ ,

h_{n}\leq e^{-3^{n+1}},\quad|u_{n}|\leq\left[\frac{\beta(e-1)}{e}\right]^{3^{n}}.

(40)

Lemma 15

The statement of Lemma 14 holds if, instead of conditions (33)–(34), we assume that condition (40) holds.

Proof

The proof is analogous to the proof of Lemma 14. Instead of (35) we obtain

\begin{split}|x_{n+1}|^{\frac{1}{3^{n+1}}}&\leq e^{-(n+1)}|x_{0}|+\frac{\beta(e-1)}{e}[e^{-n}+e^{-n+1}+\dots+e^{-1}+1]\\ &=e^{-(n+1)}|x_{0}|+\frac{\beta(e-1)}{e}\frac{1-e^{-n-1}}{1-e^{-1}}\leq e^{-(n+1)}|x_{0}|+\beta.\end{split}

(41)

Taking $N_{1}\geq\ln|x_{0}|-\ln[1-\beta]$ we get $|x_{n+1}|<1$ for $n\geq N_{1}$ . Instead of (LABEL:est:nx2) we have

\begin{split}&h_{N_{1}+2}x_{N_{1}+2}^{2}<2h_{N_{1}+2}\left[x_{N_{1}+1}^{2}+u_{N_{1}+2}^{2}\right]=2e^{-3^{N_{1}+3}}\left[x_{N_{1}+1}^{2}+\left[\frac{\beta(e-1)}{e}\right]^{2\cdot 3^{N_{1}+2}}\right]\\ &\leq 2e^{-3^{N_{1}+3}}\left[1+1\right]\leq 4e^{-3^{N_{1}+3}}<4e^{-3^{4}}<1,\end{split}

and instead of (39) we have

\begin{split}&\sum_{i=1}^{k}|u_{N_{1}+2+i}|=\sum_{i=1}^{k}\left[\frac{\beta(e-1)}{e}\right]^{3^{N_{1}+2+i}}\leq\sum_{j=4}^{k}\left[\frac{\beta(e-1)}{e}\right]^{3^{j}}\\ &<\sum_{j=4}^{k}\left[\frac{\beta(e-1)}{e}\right]^{j}=\left[\frac{\beta(e-1)}{e}\right]^{3^{4}}\frac{1}{1-\frac{\beta(e-1)}{e}}\\ &<\left[\frac{\beta(e-1)}{e}\right]^{3^{4}}e<1.\end{split}

The last inequality holds true since, in particular,

\left[\frac{(e-1)}{e}\right]^{3^{4}}\approx(0.6321)^{8}1<0.3678\approx e^{-1}

The rest of the proof is similar to the proof of Lemma 14.

4.2 Convergence of $(x_{n})_{n\in\mathbb{N}}$ to a finite limit.

Lemma 16

Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to equation (28) and let $(h_{n})_{n\in\mathbb{N}}$ and $(u_{n})_{n\in\mathbb{N}}$ satisfy either conditions (33)-(34) or condition (40). Then the sequence $(x_{k})_{k\in\mathbb{N}}$ converges to a finite limit as $k\to\infty$ .

Proof

It is sufficient to consider only the terms $\{x_{N_{1}+2+k}\}_{k\in\mathbb{N}}$ . Since the sequence $\{x_{N_{1}+2+k}\}_{k\in\mathbb{N}}$ is bounded, it has a convergent subsequence $\{x_{N_{1}+2+k_{l}}\}_{l\in\mathbb{N}}$ ,

\lim_{l\to\infty}x_{N_{1}+2+k_{l}}=L.

We now show that

\lim_{m\to\infty}x_{N_{1}+2+m}=L

follows. For each $m\in\mathbb{N}$ denote $l_{m}\in\mathbb{N}$

l_{m}=\sup\{l:N_{2}+2+k_{l}\leq m\}.

Then

N_{2}+2+k_{l_{m}}\leq m\leq N_{2}+2+k_{l_{m}+1}

and

|x_{N_{1}+2+m}|\leq|x_{N_{1}+2+m-1}|+|u_{N_{1}+2+m}|\leq|x_{N_{1}+2+k_{l_{m}}}|+\sum_{i=k_{l_{m}}}^{m}|u_{N_{1}+2+i}|,

(42)

|x_{N_{1}+2+k_{l_{m}+1}}|\leq|x_{N_{1}+2+m}|+\sum^{k_{l_{m}+1}}_{i=m}|u_{N_{1}+2+i}|

(43)

Passing to the limit in (42) and (43) we obtain, respectively,

\limsup_{m\to\infty}x_{N_{1}+2+m}\leq L,\quad\mbox{and}\quad L\leq\liminf_{m\to\infty}x_{N_{1}+2+m}.

This implies that $\lim_{m\to\infty}x_{N_{1}+2+m}$ exists and equal to $L$ .

When condition (40) holds it is possible that solutions of (28) converge to a nonzero limit. Example 1 below demonstrates that $\lim_{n\to\infty}x_{n}$ can be either zero or nonzero.

Example 1

We show that the limit of solutions of (28) can be positive, zero, or negative. For all three cases below, choose $h_{n}=e^{-3^{n+1}}$ .

(i)

Zero limit ( $L=0$ ). Set

$u_{1}=-e^{-3}\approx-0.0498,\quad u_{n}=0\quad\text{for all}\quad n\geq 2.$

Then (40) is satisfied for $\beta\in(1/(e-1),1)$ . The continuous function

$f(x)=x-e^{-3}x^{3}.$

takes its maximum $f_{m}=\frac{2}{3\sqrt{3e^{-3}}}\approx 1.724>0.0498\approx-u_{1}$ at the point $x_{m}=\frac{1}{\sqrt{3e^{-3}}}\approx 2.586$ , and $f(0)=0$ . So the equation

$x-e^{-3}x^{3}=e^{-1},$

has a solution $x_{0}$ on the interval $\left(0,\frac{1}{\sqrt{3e^{-3}}}\right)\approx(0,2.586)$ . Consider now the equation (28) with this specific initial value. We get $x_{1}=0$ and since all $u_{n}=0$ for $n\geq 2$ , we have $x_{n}=0$ for $n\geq 2$ . Therefore $\lim_{n\to\infty}x_{n}=0$ .
(ii)

Positive limit ( $L>0$ ). Set

$u_{1}=e^{-3}\approx 0.0498,\quad u_{n}>0,\quad\text{for all}\quad n\geq 2,$

so that (40) is satisfied. Suppose also that $x_{0}>0$ is chosen as in case (i). Then,

$x_{1}=2u_{1}+\underbrace{x_{0}(1-h_{0}x_{0}^{2})-u_{1}}_{=0}=2u_{1}=2e^{-3}>0.$

Moreover, note that $h_{1}x_{1}^{2}=2e^{-12}<1/2$ . We can also write

$x_{n+1}\geq x_{n}(1-h_{n}x_{n}^{2})\geq x_{1}\prod_{i=1}^{n}(1-h_{i}x_{i}^{2}).$

The same approach as in Lemma 8 with $N=1$ gives that $\lim_{n\to\infty}x_{n}>0$ .
(iii)

Negative limit ( $L<0$ ). Set

$u_{1}=-2e^{-3}\approx-0.0996,\quad u_{n}<0,\quad\text{for all}\quad n\geq 2,$

so that (40) is satisfied, and choose $x_{0}>0$ as in Cases (i) and (ii). Then

$x_{1}=\underbrace{x_{0}(1-h_{0}x_{0}^{2})+\frac{u_{1}}{2}}_{=0}+\frac{u_{1}}{2}<0.$

Again, we see that $h_{1}x_{1}^{2}=e^{-18}<1/2$ , and we can write for all $n\geq 1$

$x_{n+1}\leq x_{n}(1-h_{n}x_{n}^{2})\leq x_{1}\prod_{i=1}^{n}(1-h_{i}x_{i}^{2}).$

The same approach as in Lemma 8 with $N=1$ gives that $\lim_{n\to\infty}x_{n}<0$ .

4.3 Modified process with a stopped timestep sequence $(h_{n})_{n\in\mathbb{N}}$

Based on Example 1 and Lemma 8 we cannot expect that, in general, the finite limit $L$ will be zero. In order to obtain a sequence that converges to zero we modify the timestep sequence $(h_{n})_{n\in\mathbb{N}}$ further by stopping it (preventing terms from varying further) after $N_{3}$ steps:

\hat{h}_{n}=\left\{\begin{array}[]{cc}h_{n},&n<N_{3},\\ h_{N},&n\geq N_{3},\end{array}\right.

(44)

where $N_{3}$ is such that

|x_{N_{3}}|\leq 1.

(45)

Note that under the conditions of Lemmas 14 and 15 we would have $N_{3}=N_{1}$ . Note that $N_{3}$ is not necessarily the first moment where (45) holds, and that (45) implies $x_{N_{3}}^{2}h_{N_{3}}<1$ , but the converse does not necessarily hold.

Consider

x_{n+1}=x_{n}(1-\hat{h}_{n}x_{n}^{2})+u_{n+1},\quad x_{0}\in\mathbb{R}.

(46)

Lemma 17

Let $(h_{n})_{n\in\mathbb{N}}$ and $(u_{n})_{n\in\mathbb{N}}$ satisfy either conditions (33)-(34) or condition (40). Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to equation (46) with $(\hat{h}_{n})_{n\in\mathbb{N}}$ defined by (44). Then $\lim_{n\to\infty}x_{n}=0$ for any initial value $x_{0}\in\mathbb{R}$ .

Proof

Choose $N_{1}$ defined as in Lemmata 14 or 15 and set $N_{3}=N_{1}$ . To prove that

x_{n}^{2}\hat{h}_{n}<1,\quad\mbox{for all}\quad n>N_{3},

we follow the approach taken in the proofs of Lemma 14, Part (i), and Lemma 15, Part (i).

Let assume first that conditions (33)–(34) hold, so we use $N_{1}$ from Lemma 14. We have $N_{3}=N_{1}>2$ , $|x_{N_{3}}|<1$ , $\hat{h}_{N_{3}+1}<e^{-3^{\frac{N_{3}}{2}+1}}$ ,

|x_{N_{3}+1}|\leq|x_{N_{3}}|+|u_{N_{3}+1}|

and

\begin{split}&\hat{h}_{N_{3}+1}x_{N_{3}+1}^{2}<2\hat{h}_{N_{3}+1}\left[x_{N_{3}}^{2}+u_{N_{3}+1}^{2}\right]=2e^{-3^{\frac{N_{3}}{2}+1}}\left[x_{N_{3}}^{2}+\left(\frac{\beta}{4}\right)^{N_{3}}\right]\\ &\leq 2e^{-3^{2}}\left[1+1\right]=4e^{-3^{2}}<1.\end{split}

This gives us

|x_{N+2}|\leq|x_{N+1}|+|u_{N+2}|,

which, as above, leads to

	$\displaystyle\hat{h}_{N+2}x_{N+2}^{2}$	$\displaystyle<$	$\displaystyle 2\hat{h}_{N_{3}}\left[x_{N_{3}}^{2}+u_{N_{3}+1}^{2}\right]\leq 2e^{-3^{\frac{N_{3}+1}{2}+1}}\left[1+\left(\frac{\beta}{4}\right)^{N_{3}+1}\right]$
		$\displaystyle<$	$\displaystyle 4e^{-3^{2}}<1.$

Now we complete the proof by induction and arrive at

|x_{N+k}|\leq|x_{N}|+\sum_{i=1}^{k}|u_{N+i}|,

(47)

which implies the boundedness of the sequence $(x_{n})_{n\in\mathbb{N}}$ . Note that Lemma 16 also holds when, instead of $(h_{n})_{n\in\mathbb{N}}$ we have a stopped sequence $(\hat{h}_{n})_{n\in\mathbb{N}}$ , since its proof uses only (47) and convergence of the series $\sum_{i=1}^{\infty}u_{i}$ . So we conclude that $\lim_{n\to\infty}x_{n}=L$ . Passing to the limit in equation (46) we obtain the equality

L=L(1-\hat{h}_{N}L),

which holds only for $L=0$ .

If condition (40) hold, we use $N_{1}$ from Lemma 15. The proof of this case is similar to that of the first, except that $\hat{h}_{n}\leq 3^{N_{3}+1}$ .

Remark 4

Convergence of the solutions of equation (46) with stopped time-step sequence $(\hat{h}_{n})_{n\in\mathbb{N}}$ may be slow, either if $h_{N_{3}}$ is very small, or if $N_{3}$ is large. Alternative strategies for stopping the sequence $(\hat{h}_{n})_{n\in\mathbb{N}}$ are as follows:

(i)

Define

$N_{4}=\inf\{n\in\mathbb{N}:x_{n}^{2}h_{n}<1\},$ (48)

and assume that $x_{N_{4}}\neq 0$ . Define

$\quad\hat{h}_{n}=\left\{\begin{array}[]{cc}h_{n},&n<N_{4},\\ \frac{1}{x^{2}_{N_{4}}},&n\geq N_{4}.\end{array}\right.$

Then, $|x_{N_{4}+1}|=|u_{N_{4}+1}|<1,$ and the conditions of Lemma 17 hold. If $x_{N_{4}}=0$ , we also have $|x_{N_{4}+1}|=|u_{N_{4}+1}|<1.$

(ii)

Assume that $|u_{n+1}|\leq h_{n}$ for all $n\in\mathbb{N}$ . Define again $N_{4}$ by (48). If $|x_{N_{4}}|\leq 1$ the conditions of Lemma 17 hold. If $|x_{N_{4}}|>1$ , we have

|x^{3}_{N_{4}}|>1\geq\frac{|u_{N_{4}+1}|}{h_{N_{4}}},\,\,\mbox{or}\,\,|x^{3}_{N_{4}}|h_{N_{4}}\geq|u_{N_{4}+1}|.

Then,

|x_{N_{4}+1}|\leq|x_{N_{4}}|(1-x^{2}_{N_{4}}h_{N_{4}})+|u_{N_{4}+1}|\\ =|x_{N_{4}}|-|x^{3}_{N_{4}}|h_{N_{4}}+|u_{N_{4}+1}|\leq|x_{N_{4}}|.

x^{2}_{N_{4}+1}\hat{h}_{N_{4}+1}\leq x^{2}_{N_{4}}h_{N_{4}}\leq 1.

By induction it can be shown that $x^{2}_{N_{4}+k}\hat{h}_{N_{4}+k}\leq 1$ for all $k\in\mathbb{N}$ . Now, applying the same reasoning as before we can prove that $(|x_{N_{4}+k}|)_{k\in\mathbb{N}}$ is uniformly bounded and converges to zero.

Lemma 18

Let $(h_{n})_{n\in\mathbb{N}}$ and $(u_{n})_{n\in\mathbb{N}}$ satisfy either conditions (33)-(34) or condition (40) with $\beta<\frac{1}{e-1}$ , in all cases with equality instead of inequality in the conditions placed upon each $h_{n}$ . Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to equation (46) with initial value $x_{0}\in\mathbb{R}$ and $(\hat{h}_{n})_{n\in\mathbb{N}}$ defined by (44) and (48). Then $\lim_{n\to\infty}x_{n}=0$ .

Proof

By Lemmas 14, 15 and Remark (4), Part (ii), it is sufficient to show that $|u_{n+1}|\leq h_{n}$ . Denote

Q(n):=\ln\beta-\sum_{i=1}^{k+1}\ln_{i}\left(n+e^{1}_{[k]}\right)+\left(\prod_{i=0}^{k}\ln_{i}\left(n+e^{1}_{[k]}\right)\right)^{-1}

Note that, for $n\geq 1$ ,

\begin{split}Q(n)&<\ln\beta-\ln\left(n+e^{1}_{[k]}\right)+\frac{1}{\left(n+e^{1}_{[k]}\right)}\\ &\leq\ln\beta-\ln 2+\frac{1}{2}\\ &\approx\ln\beta-0.1931<0.\end{split}

When conditions (33)–(34) hold we have, for $n\geq 1$ ,

\begin{split}\frac{|u_{n+1}|}{h_{n}}\leq\exp\left\{3^{n+1}Q(n)\right\}\leq 1.\end{split}

When condition (40) holds with $\beta(e-1)\leq 1$ , we have, for $n\geq 1$ ,

\frac{|u_{n+1}|}{h_{n}}\leq(\beta(e-1))^{3^{n+1}}\leq 1.

5 The stochastically perturbed cubic difference equation

In this section we consider a stochastic difference equation

x_{n+1}=x_{n}(1-h_{n}x_{n}^{2})+\rho_{n+1}\xi_{n+1},\quad n\in\mathbb{N},\quad x_{0}\in\mathbb{R},

(49)

where $(\xi_{n})_{n\in\mathbb{N}}$ is a sequence of independent identically distributed random variables. We discuss only two cases: $|\xi_{n}|\leq 1$ and $\xi_{n}\sim\mathcal{N}(0,1)$ . Denoting

u_{n}:=\rho_{n}\xi_{n}

we can apply the results of Section 4 pathwise to solutions of (49) for almost all $\omega\in\Omega$ .

We also consider a stochastically perturbed equation with stopped timestep sequence $(\hat{h}_{n})_{n\in\mathbb{N}}$

x_{n+1}=x_{n}(1-\hat{h}_{n}x_{n}^{2})+\rho_{n+1}\xi_{n+1},\quad n\in\mathbb{N},\quad x_{0}\in\mathbb{R},

(50)

where $\hat{h}_{n}$ is defined by (44) with $N_{3}$ selected as equal to $N_{1}$ from Lemmas 14, 15 or as equal to $N_{4}$ from Remark 4. Note that since solutions of (49) are stochastic processes, $N_{1}$ and $N_{4}$ are a.s. finite $\mathbb{N}$ -valued random variables, which we therefore denote by $\mathcal{N}_{1}$ and $\mathcal{N}_{4}$ , respectively.

5.1 Case 1: bounded noise ( $|\xi_{n}|\leq 1$ )

In this case, for all $\omega\in\Omega$ and all $n\in\mathbb{N}$ , we have

|u_{n}|=|\rho_{n}\xi_{n}|\leq|\rho_{n}|.

for all $\omega\in\Omega$ . So we may apply the results of Section 4 to each trajectory, arriving at

Theorem 5.1

Let $(h_{n})_{n\in\mathbb{N}}$ and $(\rho_{n})_{n\in\mathbb{N}}$ satisfy either conditions (33)-(34) or condition (40) ( $\rho_{n}$ satisfying the constraint for $u_{n}$ ). Let $(\xi_{n})_{n\in\mathbb{N}}$ be a sequence of random variables s.t. $|\xi_{n}|\leq 1$ for all $n\in\mathbb{N}$ . Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to (49), $(\hat{h}_{n})_{n\in\mathbb{N}}$ defined as in (44), and $(\hat{x}_{n})_{n\in\mathbb{N}}$ a solution to (50). Then, a.s.,

(i)

$\lim_{n\to\infty}x_{n}=L$ , where $L$ is an a.s. finite random variable;
(ii)

$\lim_{n\to\infty}\hat{x}_{n}=0$ .

5.2 Case 2: unbounded noise ( $\xi_{n}\sim\mathcal{N}(0,1)$ ).

Theorem 5.2

Let $(h_{n})_{n\in\mathbb{N}}$ and $(\rho_{n})_{n\in\mathbb{N}}$ satisfy either conditions (33)-(34) or condition (40) ( $\rho_{n}$ satisfying the constraint for $u_{n}$ ). Let $(\xi_{n})_{n\in\mathbb{N}}$ be a sequence of mutually independent $\mathcal{N}(0,1)$ random variables. Let $(x_{n})_{n\in\mathbb{N}}$ be a solution to (49), $(\bar{h}_{n})_{n\in\mathbb{N}}$ as defined in (44), and $(\hat{x}_{n})_{n\in\mathbb{N}}$ a solution to (50). Then, a.s.,

(i)

$\lim_{n\to\infty}x_{n}=L$ , where $L$ is an a.s. finite random variable;
(ii)

$\lim_{n\to\infty}\hat{x}_{n}=0$ .

Proof

If (40) holds for $\beta\in(0,1)$ , then for some $\beta_{1}\in(\beta,1)$ we have

\left[\frac{\beta(e-1)}{e}\right]^{3^{n}}=\left[\frac{\beta_{1}(e-1)}{e}\right]^{3^{n}}\times\left[\frac{\beta}{\beta_{1}}\right]^{3^{n}},

and, for each $\varsigma>0$ ,

\lim_{n\to 0}\left[\frac{\beta}{\beta_{1}}\right]^{3^{n}}\ln^{\frac{1}{2}+\zeta}n=0,

Applying Lemma 4 we conclude that there exists $\mathcal{N}$ such that for all $n\geq\mathcal{N}$ ,

\left|\frac{1}{(\ln n)^{1/2+\varsigma}}\xi_{n}\right|<1.

Then, for all $n\geq\mathcal{N}$ ,

|u_{n+1}|=\left|\left[\frac{\beta_{1}(e-1)}{e}\right]^{3^{n}}\times\left[\frac{\beta}{\beta_{1}}\right]^{3^{n}}\xi_{n}\right|\leq\left[\frac{\beta_{1}(e-1)}{e}\right]^{3^{n}}.

If (34) hold holds for $\beta\in(0,1)$ , then for some $\beta_{1}\in(\beta,1)$ we use the estimate

|u_{n+1}|\leq\left(\frac{\beta_{1}}{\left(n+e_{[k]}^{1}\right)\ln\left(n+e_{[k]}^{1}\right)\dots\ln_{k}\left(n+e_{[k]}^{1}\right)}\right)^{3^{n}}\left[\frac{\beta}{\beta_{1}}\right]^{3^{n}}|\xi_{n+1}|,

and apply the same reasoning as above.

Define for a.a. $\omega\in\Omega$

y_{m}:=x_{m+\mathcal{N}(\omega)},\quad u_{m+1}:=\rho_{m+\mathcal{N}(\omega)}\xi_{m+\mathcal{N}(\omega)}(\omega),\quad{\rm h}_{m}:=h_{m+\mathcal{N}(\omega)},

and consider the deterministic stochastic equation

y_{m+1}=y_{m}(1-{\rm h}_{m}y_{m}^{2})+u_{m+1},\quad m\in\mathbb{N},\quad y_{0}=x_{\mathcal{N}(\omega)}.

(51)

Equation (51) satisfies the conditions of either Lemma 14 or Lemma 15. So there exists $N_{1}$ (which depends on $\omega$ ) such that $h_{N_{1}}x_{N_{1}}^{2}<1$ . The remainder of the proof follows by the same argument as that in Section 4.

6 Illustrative numerical examples

In this section we illustrate the asymptotic behaviour of solutions of the unperturbed equation (9) with summable and non-summable timestep sequences, as described in Lemmas 8 & 9, and the stochastically perturbed equation (49) with unbounded Gaussian noise as described in Theorem 5.2.

Figure 1, parts (a) and (b) provide three solutions of the unperturbed deterministic equation (9) corresponding to the initial values $x_{0}=1.1,0.5,-1.1$ , with timestep sequence $h_{n}=1/n^{10}$ , so that $\sum_{i=1}^{\infty}h_{i}<\infty$ . We observe that all three solutions appear to converge to different finite limits, as predicted by Lemma 8.

Parts (c) and (d) provide three solutions of (9) with the same initial values and with timestep sequence $h_{n}=1/n^{0.1}$ , so that $\sum_{i=1}^{\infty}h_{i}=\infty$ . Note that we have selected values of $x_{0}$ that are sufficiently small for (15) to hold with this choice of $h_{n}$ , hence avoiding the possibility of blow-up. All three solutions appear to converge to a zero limit, as predicted by Lemma 9.

Figure 2, part (a) and (b) provide three solution trajectories of the stochastic equation (49) each corresponding to initial value given by $x_{0}=2.5,0.5,-2.5$ with timestep sequence $h_{n}=e^{-\frac{3^{n+1}}{n+e}}$ , satisfying (33) for $k=1$ , $(\xi_{n})_{n\in\mathbb{N}}$ a sequence of i.i.d. $N(0,1)$ random variables, and

\rho_{n}=\left(\frac{\beta}{n+e}\right)^{3^{n}},

(52)

with $\beta=0.5$ satisfying (34) with $k=1$ . We observe that all three solutions approach different nonzero limits, as predicted by Theorem 5.2.

Parts (c) and (d) repeat the computation, but with the timestep sequence stopped so that its values become fixed when $h_{n}x_{n}^{2}<1$ is satisfied for the first time. Solutions demonstrate behaviour consistent with asymptotic convergence to zero, also as predicted by Theorem 5.2.

Note that $\beta\in(0,1)$ in Condition (34), but that in Figure 2 the effect of the stochastic perturbation decays too rapidly for differences between trajectories to be visible. Therefore in each part of Figure 3 we choose larger values of $\beta$ and generate fifteen trajectories of (49) with $(\xi_{n})_{n\in\mathbb{N}}$ a sequence of i.i.d. $\mathcal{N}(0,1)$ random variables, timestep sequence $h_{n}=e^{-\frac{3^{n+1}}{n+e}}$ stopped when $h_{n}x_{n}^{2}<1$ is satisfied for the first time, $x_{0}=2.5$ , and each $\rho_{n}$ chosen to satisfy (52). Parts (a) and (b) show that, when $\beta=3/2$ , trajectories appear to converge to zero. However, Parts (c) and (d) show that, when $\beta=3$ and $\beta=5$ respectively, trajectories may converge to a random limit that is not necessarily zero a.s.

Refer to caption — Figure 1: Solutions of (9) with summable (Parts (a) and (b)) and non-summable (Parts (c) and (d)) timestep sequences. Short term and long term dynamics are given in the first and second columns, respectively.

7 Acknowledgment

The third author is grateful to the organisers of the 23rd International Conference on Difference Equations and Applications, Timisoara, Romania, who supported her participation. Discussions at the conference were quite beneficial for this research.

References

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On cubic difference equations with variable coefficients and fading stochastic perturbations

Abstract

1 Introduction

2 Mathematical preliminaries

Lemma 1

Lemma 2

Lemma 3

Lemma 4

Definition 1

Corollary 1

Proof

3 The unperturbed deterministic cubic equation

3.1 Preliminary lemmata on solutions of Eq. (9)

Lemma 5

Proof

Lemma 6

Proof

Lemma 7

Proof

Remark 1

Remark 2

3.2 Timestep summability and the limit of solutions

Lemma 8

Proof

Lemma 9

Proof

3.3 Estimation of |xn||x_{n}|

Lemma 10

Proof

Lemma 11

Proof

Corollary 2

Lemma 12

Proof

Remark 3

4 The perturbed deterministic cubic difference equation

Lemma 13

Proof

4.1 Boundedness of (|xn|)n∈ℕ(|x_{n}|)_{n\in\mathbb{N}} for particular (hn)n∈ℕ(h_{n})_{n\in\mathbb{N}} and (un)n∈ℕ(u_{n})_{n\in\mathbb{N}}

Case 1

Lemma 14

Proof

Case 2

Lemma 15

Proof

4.2 Convergence of (xn)n∈ℕ(x_{n})_{n\in\mathbb{N}} to a finite limit.

Lemma 16

Proof

Example 1

4.3 Modified process with a stopped timestep sequence (hn)n∈ℕ(h_{n})_{n\in\mathbb{N}}

Lemma 17

Proof

Remark 4

Lemma 18

Proof

5 The stochastically perturbed cubic difference equation

5.1 Case 1: bounded noise (|ξn|≤1|\xi_{n}|\leq 1)

Theorem 5.1

5.2 Case 2: unbounded noise (ξn∼𝒩​(0,1)\xi_{n}\sim\mathcal{N}(0,1)).

Theorem 5.2

Proof

6 Illustrative numerical examples

7 Acknowledgment

References

3.3 Estimation of $|x_{n}|$

4.1 Boundedness of $(|x_{n}|)_{n\in\mathbb{N}}$ for particular $(h_{n})_{n\in\mathbb{N}}$ and $(u_{n})_{n\in\mathbb{N}}$

4.2 Convergence of $(x_{n})_{n\in\mathbb{N}}$ to a finite limit.

4.3 Modified process with a stopped timestep sequence $(h_{n})_{n\in\mathbb{N}}$

5.1 Case 1: bounded noise ( $|\xi_{n}|\leq 1$ )

5.2 Case 2: unbounded noise ( $\xi_{n}\sim\mathcal{N}(0,1)$ ).