- Research
- Open Access

# On the convergence of ${x}_{n}=f({x}_{n-2},{x}_{n-1})$ when $f(x,y)<x$

- Gabor Nyerges
^{1}Email author

**2014**:8

https://doi.org/10.1186/1687-1847-2014-8

© Nyerges; licensee Springer. 2014

**Received:**23 September 2013**Accepted:**11 December 2013**Published:**7 January 2014

## Abstract

Given the second-order difference equation ${x}_{n}=f({x}_{n-2},{x}_{n-1})$, if $f\in C[{(0,\mathrm{\infty})}^{2},(0,\mathrm{\infty})]$ and $f(x,y)<x$, then $({x}_{2n},{x}_{2n+1})$ tends either to $(L,0)$ or to $(0,L)$ for some $L\ge 0$. In this paper we show that if $f(x,y)$ decreases in *y*, then for any ${x}_{0}$ there is an ${x}_{1}$ such that ${x}_{n}$ monotonically decreases to 0. We also prove that if $x-y\ge f(x,y)-f(y,f(x,y))$, then for any $L\ge 0$ and ${x}_{0}>L$ there is an ${x}_{1}$ such that $({x}_{2n},{x}_{2n+1})\to (L,0)$ and similarly, for any ${x}_{0}$ there is an ${x}_{1}$ such that $({x}_{2n},{x}_{2n+1})\to (0,L)$. The class of functions satisfying the latter condition includes any function of the form $f(x,y)=\frac{x}{1+h(x,y)y}$, where *h* is symmetric and increases in *y*.

**MSC:**39A11, 39A23.

## Keywords

- second-order difference equation
- convergence to zero
- convergence to prime period two sequences
- approximating initial conditions for convergence

## 1 Introduction

- (A)
$f(x,y)<x$.

We note that by ‘decreasing’ and ‘increasing’ we mean non-increasing and non-decreasing, respectively.

- (a)
$f\in {C[[0,\mathrm{\infty})}^{2},[0,\mathrm{\infty})]$ and its restriction $\overline{f}$ to ${(0,\mathrm{\infty})}^{2}$ is in ℱ,

- (b)
$f(x,0)=x$,

- (c)
$\overline{f}$ strictly increases in

*x*and strictly decreases in*y*.

*f*be of this type. Then, since $\overline{f}\in \mathcal{F}$, every positive solution tends either to $(L,0,L,0,\dots )$ or to $(0,L,0,L,\dots )$ for some $L\ge 0$, the only periodic solutions of equation (1), with $(0,0,0,0,\dots )$ being the unique equilibrium solution (see,

*e.g.*[1]). The question of the existence of positive solutions converging to the equilibrium solution in the special case $f(x,y)=\frac{x}{1+\alpha y}$ ($x,y\ge 0$) was raised by Kulenovic and Ladas in [2]. Kent gave an affirmative answer in [1] by showing that, in general, if

*f*satisfies (a), (b), (c) and the condition that

- (d)
*f*is differentiable and there is a differentiable function*g*such that ${x}_{n-2}=g({x}_{n-1},{x}_{n})$, with some further properties detailed in [1],

then for any ${x}_{0}>0$ there is an ${x}_{1}>0$ such that ${x}_{n}$ monotonically decreases to 0. We note that Janssen and Tjaden [3] had previously proved this for $f(x,y)=\frac{x}{1+y}$ and ${x}_{0}=1$. As we shall see, Kent’s conclusion follows from much weaker assumptions, namely that (a) holds and $\overline{f}(x,y)$ decreases in *y*.

To be more precise, we show (*cf.* Theorem 2.1) that if $f\in {\mathcal{F}}_{1}$, then for any ${x}_{0}$ there is an ${x}_{1}$ such that ${x}_{n}$ monotonically decreases to 0. Also, for any ${x}_{1}$, if ${x}_{1}\le f(x,{x}_{1})$ for some *x*, then there is an ${x}_{0}$ such that ${x}_{n}$ monotonically decreases to 0. Theorem 2.1 also applies to $f(x,y)=x\frac{\alpha y+\beta x}{Ay+\beta x}$, investigated by Chan *et al.* in [4]. In their paper, they prove, *inter alia*, that if $0<\alpha <A$ and $0<\beta $, then there are positive initial values for which ${x}_{n}$ in equation (1) monotonically decreases to 0. Since the restriction of *f* to ${(0,\mathrm{\infty})}^{2}$ is in ${\mathcal{F}}_{1}$, their result also follows from Theorem 2.1 in a stronger form.

For functions in ${\mathcal{F}}_{2}$ we show (*cf.* Theorem 2.2) that if $L\ge 0$, then for any ${x}_{0}>L$ there is an ${x}_{1}$ such that $({x}_{2n},{x}_{2n+1})\to (L,0)$ and for any ${x}_{1}$, if ${x}_{1}+L\le f(x,{x}_{1})$ for some *x*, then there is an ${x}_{0}$ such that $({x}_{2n},{x}_{2n+1})\to (L,0)$. Similar results hold for convergence to $(0,L)$.

The class ${\mathcal{F}}_{2}$ includes several types of functions of interest. For instance, as we shall see later on, ${\mathcal{F}}_{2}\cap {\mathcal{F}}_{1}$ contains any function of the form $f(x,y)=\frac{x}{1+h(x,y)y}$, where $h\in C[{(0,\mathrm{\infty})}^{2},(0,\mathrm{\infty})]$ is symmetric, *i.e.* $h(x,y)=h(y,x)$, and increases in *y*. In particular, ${\mathcal{F}}_{2}\cap {\mathcal{F}}_{1}$ includes $f(x,y)=\frac{x}{1+\alpha y}$ for any $\alpha >0$. Actually, functions of this form, defined on ${[0,\mathrm{\infty})}^{2}$, belong to another subset of the set of functions satisfying (a), (b) and (c) above, for which a result similar to, but stronger than Theorem 2.2 was proved in [5]. Given any function in this set, for any $L\ge 0$, the set of positive initial values $({x}_{0},{x}_{1})$ for which $({x}_{2n},{x}_{2n+1})\to (L,0)$ is a surjective, strictly increasing (and hence continuous) function from $(L,\mathrm{\infty})$ to $(0,\mathrm{\infty})$.

## 2 Main results

*n*, and by (A)

In particular, for $n\ge 1$, ${r}_{n}$ satisfies (A), hence ${r}_{n}\in \mathcal{F}$, and ${s}_{n}$ satisfies the condition

(A^{∗}) $f(x,y)<y$.

We shall write *R* and *S* for the pointwise limits of ${r}_{n}$ and ${s}_{n}$, respectively. *R* and *S* map ${(0,\mathrm{\infty})}^{2}$ to $[0,\mathrm{\infty})$ and, since ${r}_{n}(x,y)>R(x,y)$, ${s}_{n}(x,y)>S(x,y)$, *R* and *S* satisfy (A) and (A^{∗}), respectively. However, as we shall see in Example 3.4, *R* and *S* are not necessarily continuous. If $L=R(x,y)$ and $M=S(x,y)$ were both positive, then, by the recursive relationship between ${r}_{n}$ and ${s}_{n}$ and the continuity of *f*, $L=f(L,M)$, contradicting (A). Therefore, either $R(x,y)=0$ or $S(x,y)=0$, *i.e.* $R(x,y)\cdot S(x,y)=0$ for any *x* and *y*. We also observe that *R* and *S* have the same range, since $R(x,y)=S(y,f(x,y))$ and $S(x,y)=R(y,f(x,y))$. In particular, 0 is in the range of both functions.

*R*and

*S*, the sets of initial conditions under which $({x}_{2n},{x}_{2n+1})$ in equation (1) tends to $(L,0)$ or to $(0,L)$. For any $L\ge 0$, let us put ${R}_{L}$ and ${S}_{L}$, respectively, for these sets, so that

We note that for $L>0$, since $R(x,y)\cdot S\cdot S(x,y)=0$, ${R}_{L}$ and ${S}_{L}$ are simply the *L*-level sets of *R* and *S*, respectively. Also, ${R}_{0}$ is the intersection of the 0-level sets of *R* and *S* and ${R}_{0}={S}_{0}$.

Regarding ${R}_{L}$ and ${S}_{L}$ as relations, the theorems below characterise their domain and their range, Theorem 2.1 for $L=0$ and any $f\in {\mathcal{F}}_{1}$ and Theorem 2.2 for any $L\ge 0$ and any $f\in {\mathcal{F}}_{2}$.

**Theorem 2.1**

*Let*$f\in {\mathcal{F}}_{1}$.

*Then*

- (i)
*for any**a**there is a**b**in the interval*$[f(a,a),a]$*for which*$R(a,b)=S(a,b)=0$, - (ii)
*for any**b*,*if there is a*${b}^{\prime}$*such that*$b\le f({b}^{\prime},b)$,*then there is an**a**in the interval*$[b,{b}^{\prime}]$*for which*$R(a,b)=S(a,b)=0$.

**Theorem 2.2**

*Let*$f\in {\mathcal{F}}_{2}$

*and*$L\ge 0$.

*Then*

- (i)
*for any*$a>L$*there is a**b**in the interval*$[f(a-L,a),a-L]$*for which*$R(a,b)=L$, $S(a,b)=0$, - (ii)
*for any**b*,*if there is a*${b}^{\prime}$*such that*$b+L\le f({b}^{\prime},b)$,*then there is an**a**in the interval*$[b+L,{b}^{\prime}]$*for which*$R(a,b)=L$, $S(a,b)=0$, - (iii)
*for any**a**there is a**b**in the interval*$[f(a+L,a),a+L]$*for which*$R(a,b)=0$, $S(a,b)=L$, - (iv)
*for any*$b>L$,*if there is a*${b}^{\prime}$*such that*$b-L\le f({b}^{\prime},b)$,*then there is an**a**in the interval*$[b-L,{b}^{\prime}]$*for which*$R(a,b)=0$, $S(a,b)=L$.

**Remark 2.1** If ${lim}_{x\to \mathrm{\infty}}f(x,b)=\mathrm{\infty}$, then of course there is a ${b}^{\prime}$ satisfying the condition in (ii) of Theorem 2.1 and in (ii), (iv) of Theorem 2.2. The condition is not necessary, as shown later on by Example 3.1.

*n*

*i.e.*${r}_{n}(a,b)-{s}_{n}(a,b)-L\ge 0$ and ${s}_{n}(a,b)-{r}_{n+1}(a,b)+L\ge 0$. This is sufficient, since either $R(a,b)=0$ or $S(a,b)=0$. Similarly, a sufficient condition for $R(a,b)=0$, $S(a,b)=L$ is that for all

*n*

*i.e.* ${r}_{n}(a,b)-{s}_{n}(a,b)+L\ge 0$ and ${s}_{n}(a,b)-{r}_{n+1}(a,b)-L\ge 0$. In particular, if $L=0$, then ${r}_{n}(a,b)\ge {s}_{n}(a,b)\ge {r}_{n+1}(a,b)$ and so $R(a,b)=S(a,b)=0$. We also observe that if $f\in {\mathcal{F}}_{2}$, then the first condition is also necessary for $R(a,b)=L$, $S(a,b)=0$ and the second one is also necessary for $R(a,b)=0$, $S(a,b)=L$. This is because if $f\in {\mathcal{F}}_{2}$, then for any *x*, *y* and *n*, ${r}_{n}(x,y)-{s}_{n}(x,y)\ge {r}_{n+1}(x,y)-{s}_{n+1}(x,y)$ and ${s}_{n}(x,y)-{r}_{n+1}(x,y)\ge {s}_{n+1}(x,y)-{r}_{n+2}(x,y)$.

According to the discussion above, in order to prove that given *a* there is a *b* such that $R(a,b)=L$ and $S(a,b)=0$, it is sufficient to show that there is a *b* for which the functions ${p}_{n}(y)={r}_{n}(a,y)-{s}_{n}(a,y)-L$ and ${q}_{n}(y)={s}_{n}(a,y)-{r}_{n+1}(a,y)+L$ are both non-negative for all *n*. Similarly, given *a* and ${p}_{n}(y)={r}_{n}(a,y)-{s}_{n}(a,y)+L$, ${q}_{n}(y)={s}_{n}(a,y)-{r}_{n+1}(a,y)-L$, if both functions are non-negative for all *n*, then $R(a,b)=0$ and $S(a,b)=L$.

Lemma 2.1 below provides sufficient conditions ensuring that, in general, the functions in two sequences of continuous real functions are all non-negative for some argument. Lemma 2.2 uses Lemma 2.1 to show that if $L\ge 0$ and $f\in \mathcal{F}$ has the property that $f(x,y)-f(y,f(x,y))\le \pm L$ whenever $x-y=\pm L$, then statements (i), (ii), (iii) and (iv) in Theorem 2.2 are true for *L* and *f*. Since any function in ${\mathcal{F}}_{1}$ has this property for $L=0$, Theorem 2.1 follows from Lemma 2.1, and since any function in ${\mathcal{F}}_{2}$ has this property for any $L\ge 0$, so does Theorem 2.2.

**Lemma 2.1**

*Let*${p}_{n},{q}_{n}\in C[[a,b],(-\mathrm{\infty},\mathrm{\infty})]$, $n=0,1,\dots $ ,

*and put*

- (i)
*If*- (a)
*for all**n**and**x*,*either*${p}_{n}(x)>0$*or*${q}_{n}(x)>0$, - (b)
*for all**n**and**x*,*if*${p}_{n}(x)=0$*then*${p}_{n+1}(x)\le 0$*and if*${q}_{n}(x)=0$*then*${q}_{n+1}(x)\le 0$, - (c)
${p}_{0}(b)\le 0$

*and*${q}_{0}(a)\le 0$,

*then*${S}_{ab}\ne \mathrm{\varnothing}$. - (a)
- (ii)
*If*(a)*holds and*(b), (c)*are replaced by*

(b′) *for all* *n* *and* *x*, ${p}_{n}(x)\ge {p}_{n+1}(x)$ *and* ${q}_{n}(x)\ge {q}_{n+1}(x)$,

(c′) ${p}_{N}(b)\le 0$ *and* ${q}_{M}(a)\le 0$ *for some* $N,M\ge 0$,

*then* ${S}_{ab}\ne \mathrm{\varnothing}$.

*Proof* (i) Let $p,q\in C[[a,b],(-\mathrm{\infty},\mathrm{\infty})]$ be any two functions such that for any *x* either $p(x)>0$ or $q(x)>0$ and $p(b)\le 0$, $q(a)\le 0$. Then $q(a)\le 0$ implies that $p(a)>0$ and so, since $p(b)\le 0$, *p* has a smallest root ${b}^{\prime}$ in $(a,b]$. Therefore, $p>0$ on $[a,{b}^{\prime})$ and $p({b}^{\prime})=0$. But then $q({b}^{\prime})>0$ and so, since $q(a)\le 0$, *q* has a greatest root ${a}^{\prime}$ in $[a,{b}^{\prime})$. Hence, $q>0$ on $({a}^{\prime},{b}^{\prime}]$ and $q({a}^{\prime})=0$. Thus, there is an interval $[{a}^{\prime},{b}^{\prime}]\subseteq [a,b]$ such that $p>0$ on $[{a}^{\prime},{b}^{\prime})$ and $q>0$ on $({a}^{\prime},{b}^{\prime}]$ and $p({b}^{\prime})=q({a}^{\prime})=0$.

Let us now assume that the hypothesis of (i) holds. Then, by the above, there is an interval $[{a}_{0},{b}_{0}]\subseteq [a,b]$ such that ${p}_{0}>0$ on $[{a}_{0},{b}_{0})$ and ${q}_{0}>0$ on $({a}_{0},{b}_{0}]$ and ${p}_{0}({b}_{0})={q}_{0}({a}_{0})=0$. Now suppose that there is an interval $[{a}_{n},{b}_{n}]$ such that ${p}_{n}>0$ on $[{a}_{n},{b}_{n})$, ${q}_{n}>0$ on $({a}_{n},{b}_{n}]$ and ${p}_{n}({b}_{n})={q}_{n}({a}_{n})=0$. Then, by (b), ${p}_{n+1}({b}_{n})\le {p}_{n}({b}_{n})=0$ and ${q}_{n+1}({a}_{n})\le {q}_{n}({a}_{n})=0$. Therefore, by the above, there is an interval $[{a}_{n+1},{b}_{n+1}]\subseteq [{a}_{n},{b}_{n}]$ such that ${p}_{n+1}>0$ on $[{a}_{n+1},{b}_{n+1})$, ${q}_{{}_{n+1}}>0$ on $({a}_{n+1},{b}_{n+1}]$ and ${p}_{n+1}({b}_{n+1})={q}_{n+1}({a}_{n+1})=0$. Hence, by induction, there is a descending sequence $[{a}_{0},{b}_{0}]\supseteq [{a}_{1},{b}_{1}]\supseteq [{a}_{2},{b}_{2}]\supseteq \cdots $ of intervals such that ${p}_{n}>0$ on $[{a}_{n},{b}_{n})$, ${q}_{n}>0$ on $({a}_{n},{b}_{n}]$ and ${p}_{n}({b}_{n})={q}_{n}({a}_{n})=0$.

*I*is a non-empty closed interval and $I\subseteq {S}_{ab}$. We note that if $int(I)\ne \mathrm{\varnothing}$, then ${p}_{n}$ and ${q}_{n}$ are both positive on $int(I)$.

- (ii)If, say, $M\le N$, then ${p}_{N}(a)\le 0$ by (b′). Therefore, by (i)$\{x:{p}_{n}(x)\ge 0\text{and}{q}_{n}(x)\ge 0\text{for}n\ge N\}\ne \mathrm{\varnothing}$

and by (b′), $\{x:{p}_{n}(x)\ge 0\text{and}{q}_{n}(x)\ge 0\text{for}n\ge N\}={S}_{ab}$. □

**Remark 2.2** (Approximating an element of ${S}_{ab}$)

Suppose that ${p}_{n}$, ${q}_{n}$ satisfy the conditions of Lemma 2.1(ii) on an interval $[a,b]$, so that ${S}_{ab}\ne \mathrm{\varnothing}$. Choose a $c\in (a,b)$, say $c=(a+b)/2$. Suppose that either ${p}_{N}(c)\le 0$ or ${q}_{N}(c)\le 0$ for some $N\ge 0$. Then, by Lemma 2.1(ii), ${p}_{n}$ and ${q}_{n}$ satisfy the conditions of Lemma 2.1(ii) either on $[a,c]$ if ${p}_{N}(c)\le 0$, or on $[c,b]$ if ${q}_{N}(c)\le 0$. Therefore, if ${p}_{N}(c)\le 0$, then ${S}_{ac}\ne \mathrm{\varnothing}$ and if ${q}_{N}(c)\le 0$, then ${S}_{cb}\ne \mathrm{\varnothing}$. Since ${S}_{ac},{S}_{cb}\subseteq {S}_{ab}$, in either case we have a better approximation of an element of ${S}_{ab}$. The method will fail if ${p}_{n}(c)>0$ and ${q}_{n}(c)>0$ for all $n\ge 0$ (in which case $c\in {S}_{ab}$).

**Lemma 2.2**

*Let*$f\in \mathcal{F}$.

*Suppose that for an*$L\ge 0$

*and any*

*x*

*and*

*y*,

*Then statements* (i), (ii), (iii) *and* (iv) *in Theorem * 2.2 *are true for* *f* *and* *L*.

*Proof*First we observe that

- (i)
Let $a>L$. Put ${p}_{n}(y)={r}_{n}(a,y)-{s}_{n}(a,y)-L$ and ${q}_{n}(y)={s}_{n}(a,y)-{r}_{n+1}(a,y)+L$. Then ${p}_{0}(a-L)=a-(a-L)-L=0$. By (2), $f(a-L,a)-f(a,f(a-L,a))\le -L$ and so ${q}_{0}(f(a-L,a))=f(a-L,a)-f(a,f(a-L,a))+L\le 0$. Together with (3) and (4) this means that ${p}_{n}$ and ${q}_{n}$ satisfy the hypothesis of Lemma 2.1(i) on the interval $[f(a-L,a),a-L]$. Therefore, there is a

*b*in the interval $[f(a-L,a),a-L]$ such that ${p}_{n}(b)\ge 0$ and ${q}_{n}(b)\ge 0$ for all $n\ge 0$ and so $R(a,b)=L$, $S(a,b)=0$. - (ii)
Given

*b*, suppose that there is a ${b}^{\prime}$ such that $b+L\le f({b}^{\prime},b)$. Put ${p}_{n}(x)={s}_{n}(x,b)-{r}_{n+1}(x,b)+L$ and ${q}_{n}(x)={r}_{n}(x,b)-{s}_{n}(x,b)-L$. Then ${p}_{0}({b}^{\prime})=b-f({b}^{\prime},b)+L\le 0$ and ${q}_{0}(b+L)=(b+L)-b-L=0$. Together with (3) and (4) this means that ${p}_{n}$ and ${q}_{n}$ satisfy the hypothesis of Lemma 2.1(i) on the interval $[b+L,{b}^{\prime}]$. Therefore, there is an*a*in the interval $[b+L,{b}^{\prime}]$ such that ${p}_{n}(b)\ge 0$ and ${q}_{n}(b)\ge 0$ for all $n\ge 0$ and so $R(a,b)=L$, $S(a,b)=0$. - (iii)
Given

*a*, put ${p}_{n}(y)={r}_{n}(a,y)-{s}_{n}(a,y)+L$ and ${q}_{n}(y)={s}_{n}(a,y)-{r}_{n+1}(a,y)-L$. Then ${p}_{0}(a+L)=a-(a+L)+L=0$. By (2), $f(a+L,a)-f(a,f(a+L,a))\le L$ and so ${q}_{0}(f(a+L,a))=f(a+L,a)-f(a,f(a+L,a))-L\le 0$. Together with (3) and (4) this means that ${p}_{n}$ and ${q}_{n}$ satisfy the hypothesis of Lemma 2.1(i) on the interval $[f(a+L),a+L]$. Therefore, there is a*b*in the interval $[f(a+L,a),a+L]$ such that ${p}_{n}(b)\ge 0$ and ${q}_{n}(b)\ge 0$ for all $n\ge 0$ and so $R(a,b)=0$, $S(a,b)=L$. - (iv)
Let $b>L$ and suppose that there is a ${b}^{\prime}$ such that $b-L\le f({b}^{\prime},b)$. Put ${p}_{n}(x)={s}_{n}(x,b)-{r}_{n+1}(x,b)-L$ and ${q}_{n}(x)={r}_{n}(x,b)-{s}_{n}(x,b)+L$. Then ${p}_{0}({b}^{\prime})=b-f({b}^{\prime},b)-L\le 0$ and ${q}_{0}(b-L)=(b-L)-b+L=0$. Together with (3) and (4) this means that ${p}_{n}$ and ${q}_{n}$ satisfy the hypothesis of Lemma 2.1(i) on the interval $[b-L,{b}^{\prime}]$. Therefore there is an

*a*in the interval $[b-L,{b}^{\prime}]$ such that ${p}_{n}(b)\ge 0$ and ${q}_{n}(b)\ge 0$ for all $n\ge 0$ and so $R(a,b)=0$, $S(a,b)=L$. □

We can now prove Theorem 2.1 and Theorem 2.2.

*Proof of Theorem 2.1 and Theorem 2.2* If $f\in {\mathcal{F}}_{1}$, then $f(x,x)-f(x,f(x,x))\le 0$, since $f(x,x)<x$. Thus *f* satisfies the hypothesis of Lemma 2.2 for $L=0$ and so Theorem 2.1 holds.

If $f\in {\mathcal{F}}_{2}$ then $x-y\ge f(x,y)-f(y,f(x,y))$ and so *f* satisfies the hypothesis of Lemma 2.2 for any $L\ge 0$ and Theorem 2.2 follows. □

**Remark 2.3** (Approximating initial conditions for convergence)

Let $f\in {\mathcal{F}}_{2}$ and $L\ge 0$. Given any $a>L$, define ${p}_{n}$ and ${q}_{n}$ as in (i) of the proof of Lemma 2.2 above. Then, using the bisection method described in Remark 2.2, we can (usually) approximate a *b* such that $R(a,b)=L$ and $S(a,b)=0$ to any degree of accuracy. Similarly, if we define ${p}_{n}$ and ${q}_{n}$ as in (iii) of the proof of Lemma 2.2, then, given any *a*, we can (usually) approximate a *b* such that $R(a,b)=0$ and $S(a,b)=L$.

## 3 Applications

In this section, we give a few examples of applying the results of Section 2. We also show that *R* (and hence *S*) is not necessarily continuous for functions in ℱ. We note that, in general, the elements of ℱ can be written as $f(x,y)=\frac{x}{1+g(x,y)}$, with $g\in C[{(0,\mathrm{\infty})}^{2},(0,\mathrm{\infty})]$ and *vice versa*.

We shall use the following proposition in the examples.

**Proposition 3.1**

- (i)
*Let*${\mathcal{F}}_{21}$*be the set of functions of the form*$f(x,y)=\frac{x}{1+h(x,y)y},\phantom{\rule{1em}{0ex}}x,y>0,$*where*$h\in C[{(0,\mathrm{\infty})}^{2},(0,\mathrm{\infty})]$*is symmetric and increases in**y*(*and hence also in**x*).*Then*${\mathcal{F}}_{21}\subseteq {\mathcal{F}}_{2}\cap {\mathcal{F}}_{1}$. - (ii)
*Let*${\mathcal{F}}_{22}$*be the set of functions of the form*$f(x,y)=\frac{x}{1+g(y)},\phantom{\rule{1em}{0ex}}x,y>0,$*where*$g:(0,\mathrm{\infty})\to (0,\mathrm{\infty})$*is differentiable*, $g(y)\to 0$*as*$y\to 0$, $y\le g(y)$*and*${g}^{\prime}(y)\le {(1+g(y))}^{2}$.*Then*${\mathcal{F}}_{22}\subseteq {\mathcal{F}}_{2}$.

*Proof*(i) First of all, functions in ${\mathcal{F}}_{21}$ (strictly) decrease in

*y*and so they are in ${\mathcal{F}}_{1}$. Next we note that if $f\in \mathcal{F}$ is written as $f(x,y)=\frac{x}{1+g(x,y)}$, then $x-f(x,y)=f(x,y)g(x,y)$. Therefore, if $f\in {\mathcal{F}}_{21}$, then

- (ii)
Let us put $e(x,y)=(x-y)-(f(x,y)-f(y,f(x,y)))$. Then, for any

*y*, $e(x,y)\to 0$ as $x\to 0$, and so in order to show that $e(x,y)\ge 0$,*i.e.*that ${\mathcal{F}}_{22}\subseteq {\mathcal{F}}_{2}$, it will be sufficient to demonstrate that for any*y*, $e(x,y)$ increases in*x*.

If ${g}^{\prime}(y)\le {(1+g(y))}^{2}$ and $y\le g(y)$, then $1+y\frac{{g}^{\prime}(f(x,y))}{{(1+g(f(x,y)))}^{2}}\le 1+y\le 1+g(y)$ and so ${e}_{x}(x,y)\ge 0$, *i.e.* $e(x,y)$ increases in *x* for any *y*, as claimed. □

Functions in ${\mathcal{F}}_{21}$ do not have to be monotonic in *x*. For instance, $f(x,y)=\frac{x}{1+{x}^{2}{y}^{3}}$ is not monotonic in *x* (for any *y*). Also, functions in ${\mathcal{F}}_{22}$ strictly increase in *x*, but they are not necessarily monotonic in *y* (*cf.* Example 3.2). Finally, we note that $f(x,y)=\frac{x}{1+y}$ is in ${\mathcal{F}}_{21}\cap {\mathcal{F}}_{22}$.

**Example 3.1** Let $f(x,y)=\frac{x}{1+x{y}^{2}}$, $x,y>0$. We note that if we extend *f* to ${[0,\mathrm{\infty})}^{2}$, then it satisfies (a), (b) and (c) of Section 1.

Since *f* is in ${\mathcal{F}}_{21}$, Theorem 2.2 holds for *f* by Proposition 3.1(i). For instance, if $L=3$ and $a=4$, then there is a *b* in the interval $[f(4-3,4),4-3]=[\frac{1}{17},1]$ such that $R(4,b)=3$. Using the bisection method described in Remark 2.3, we can get a better approximation for such a *b*. After seven iterations we find that the interval $[0.264706,0.272059]$ contains a *b* such that $R(4,b)=3$.

Similarly, after seven iterations of the bisection method we find that there is a *b* in the interval $[1.01538,1.04615]$ such that $R(4,b)=S(4,b)=0$. Given such a *b*, $f(x,b)<\frac{1}{{b}^{2}}<1<b$ for any *x*, showing that the condition in (ii) and (iv) of Lemma 2.2 - and hence in (ii) of Theorem 2.1 and (ii) and (iv) of Theorem 2.2 - is not necessary.

*b*and

*L*such that $L\ge \frac{1}{{b}^{2}}$, there is no

*a*satisfying $R(a,b)=L$. In fact, there is a $B>1$ such that if $b\ge B$ then $R(x,b)=0$ and $S(x,b)>0$ for any

*x*. In order to show this, let us put $h(u,v)=v-f(u,v)$ and let $z=\{(u,v):h(u,v)=0\}$ be the level set of

*h*for 0. As $h(u,v)$ (strictly) decreases in

*u*and strictly increases in

*v*,

*z*is a strictly increasing function. Since $h(u,v)$ strictly increases in

*v*,

*i.e.*

*z*maps $(0,\mathrm{\infty})$ to $(0,1)$. Also, for any $v\in (0,1)$, $h(\frac{v}{1-{v}^{3}},v)=0$ and so

*z*is surjective. Thus

*z*is a strictly increasing surjective function from $(0,\mathrm{\infty})$ to $(0,1)$, hence it is also continuous. Therefore, since ${lim}_{u\to 0}z(u)=0$ and ${lim}_{u\to \mathrm{\infty}}z(u)=1$, $z(u)=\frac{1}{{u}^{2}}$ has a unique solution $B>1$, and then

Since $f(x,b)<\frac{1}{{b}^{2}}$ for any *x* and, by (6), $\frac{1}{{b}^{2}}\le z(b)$ for any $b\ge B$, we see that if $b\ge B$, then $f(x,b)<z(b)$ for any *x*. Therefore, by (5), if $b\ge B$ then ${r}_{1}(x,b)=f(x,b)<f(b,f(x,b))={s}_{1}(x,b)$ for any *x*. Since $f\in {\mathcal{F}}_{2}$, this means that if $b\ge B$ then $R(x,b)=0$ and $S(x,b)>0$ for any *x*.

**Example 3.2** Let $f(x,y)=\frac{x}{1+y+\frac{1}{2}y{sin}^{2}(y)}$, $x,y>0$. Put $g(y)=y+\frac{1}{2}y{sin}^{2}(y)$. Then $g(y)\to 0$ as $y\to 0$, $y\le g(y)$ and ${g}^{\prime}(y)=1+\frac{1}{2}{sin}^{2}(y)+\frac{1}{2}ysin(2y)\le 1+y\le 1+g(y)\le {(1+g(y))}^{2}$. Therefore, $f\in {\mathcal{F}}_{22}$ and so, by Proposition 3.1(ii), Theorem 2.2 holds for *f*. We observe that $g(y)$ is not monotonic in *y* (for any *x*), so neither is $f(x,y)$.

**Example 3.3**Let $f(x,y)=\frac{x}{1+y/x}$, $x,y>0$. Since

*f* is in ${\mathcal{F}}_{2}$ and so Theorem 2.2 is true for *f*. We also note that for any *b*, ${lim}_{x\to \mathrm{\infty}}f(x,b)=\mathrm{\infty}$ and so the consequent of Theorem 2.2(ii) holds for any $L\ge 0$ and *b* and that of Theorem 2.2(iv) for any $L\ge 0$ and $b>L$.

Finally, we show that *R* (and hence *S*) is not necessarily continuous for functions in ℱ. Our counterexample uses Proposition 3.2 below, which is a kind of comparison test for functions in ℱ. In what follows, we use superscripts (*e.g.* ${r}_{n}^{f}$) to distinguish between ${r}_{n}$, ${s}_{n}$, *R* and *S* for different functions.

**Proposition 3.2**

*Let*$f,g\in \mathcal{F}$, $C>0$

*and assume that*

- (i)
$g(x,y)$

*increases in**x**and decreases in**y*, - (ii)
*if*$C<x$*then*$g(x,y)\le f(x,y)$*and if*$x\le C$*then*$g(x,y)\ge f(x,y)$.

*Let* $C\le L<a$, *where* *L* *is a value of* ${R}^{g}$. *Then there is a* ${b}^{\prime}\le C$ *such that* ${R}^{f}(a,b)\ge L$ *for all* $b\le {b}^{\prime}$.

*Proof* Let ${x}^{\prime}\le x$, ${y}^{\prime}\ge y$. Then, by (i), $g({x}^{\prime},{y}^{\prime})\le g(x,y)$ and so, by (ii), if $C<x$ then $g({x}^{\prime},{y}^{\prime})\le g(x,y)\le f(x,y)$. Similarly, if ${x}^{\prime}\ge x$, ${y}^{\prime}\le y$ and $x\le C$ then $g({x}^{\prime},{y}^{\prime})\ge g(x,y)\ge f(x,y)$.

*L*is a value of ${R}^{g}$, then ${R}^{g}({a}^{\prime},{b}^{\prime})=L$ for some ${a}^{\prime}$, ${b}^{\prime}$ such that $C\le L<{a}^{\prime}\le a$, $C\ge {b}^{\prime}$. Then, by the above, for any

*b*such that ${b}^{\prime}\ge b$

and also, $C\le L\le L<g({a}^{\prime},{b}^{\prime})$, $C\ge g({b}^{\prime},g({a}^{\prime},{b}^{\prime}))$.

Hence, by induction, for all $n\ge 0$, ${r}_{n}^{g}({a}^{\prime},{b}^{\prime})\le {r}_{n}^{f}(a,b)$ (and ${s}_{n}^{g}({a}^{\prime},{b}^{\prime})\ge {s}_{n}^{f}(a,b)$). Therefore, ${R}^{f}(a,b)\ge {R}^{g}(a,b)=L$, as claimed. □

**Example 3.4**Let $f\in \mathcal{F}$ be defined as

and put $g(x,y)=\frac{x}{1+y}$. We note that *f* is actually in ${\mathcal{F}}_{1}$, in fact it strictly decreases in *y* and strictly increases in *x*.

By Theorem 2.2, the range of ${R}^{g}$ is $[0,\mathrm{\infty})$. Therefore, by Proposition 3.2, ${R}^{f}$ has values ≥1, since $f(x,y)\le g(x,y)$ for $x\le 1$ and $f(x,y)=g(x,y)$ for $x>1$. If ${R}^{f}(x,y)<1$, then eventually ${r}_{n}^{f}(x,y)<1$ and hence ${R}^{f}(x,y)=\frac{{R}^{f}(x,y)}{1+(1-{R}^{f}(x,y))+{S}^{f}(x,y)}$. Therefore, ${R}^{f}(x,y)=0$, *i.e.* ${R}^{f}$ has no positive values <1. Since 0 is a value of ${R}^{f}$, ${R}^{f}$ is not continuous. Since ${R}^{f}$ and ${S}^{f}$ have the same range, ${S}^{f}$ is not continuous either. We also note that if $x,y\le 1$, then ${R}^{f}(x,y)={S}^{f}(x,y)=0$, since then ${R}^{f}(x,y)<1$ and ${S}^{f}(x,y)<1$.

## Declarations

### Acknowledgements

The author would like to thank the referees for their helpful comments and suggestions.

## Authors’ Affiliations

## References

- Kent CM: Convergence of solutions in a nonhyperbolic case.
*Nonlinear Anal.*2001, 47: 4651–4665. 10.1016/S0362-546X(01)00578-8MathSciNetView ArticleMATHGoogle Scholar - Kulenovic MRS, Ladas G:
*Dynamics of Second Order Rational Difference Equations: With Open Problems and Conjectures*. Chapman & Hall/CRC, Boca Raton; 2001.View ArticleMATHGoogle Scholar - Janssen AJEM, Tjaden DLA: Solution to problem 86–2.
*Math. Intell.*1987, 9: 40–43. 10.1007/BF03023954View ArticleGoogle Scholar - Chan DM, Kent CM, Ortiz-Robinson NL: Convergence results on a second-order rational difference equation with quadratic terms.
*Adv. Differ. Equ.*2009., 2009: Article ID 985161Google Scholar - Kalikow S, Knopf PM, Huang YS, Nyerges G:Convergence properties in the nonhyperbolic case ${x}_{n+1}={x}_{n-1}/(1+f({x}_{n}))$.
*J. Math. Anal. Appl.*2007, 326: 456–467. 10.1016/j.jmaa.2006.03.018MathSciNetView ArticleMATHGoogle Scholar

## Copyright

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.