A Proof of Conjecture 1 in Kulenovi´c, Ladas and Overdeep (2003)

1. Introduction

The qualitative behaviour of rational difference equations has been studied extensively because such equations arise in discrete-time modelling and often exhibit rich global dynamics. A systematic treatment of many second-order rational recurrences can be found in the monograph [2]. Further background on nonlinear difference equations, higher-order rational recurrences, and standard qualitative methods may be found in [3,4]. In a short note of open problems and conjectures, Kulenović, Ladas and Overdeep [1] proposed, among other questions, the following conjecture.

 Remark 1

(Conjecture 1 in [1]). Show that every positive solution of

$$y_{n+1}={\displaystyle \frac{y_n{(1+y_n)}^2}{y_n(1+y_n)+(1+y_{n-1})}},n=0,1,\dots ,$$

(1)

with initial conditions $y_{-1},y_0>0$ has a finite limit.

A special feature of (1) is that every positive number is an equilibrium. Therefore, a direct passage to the limit in (1) gives only the identity

$$L={\displaystyle \frac{L{(1+L)}^2}{L(1+L)+(1+L)}}=L,$$

and does not by itself determine which equilibrium is selected by the initial data. This makes it important to keep track of monotonicity and of quantities that enclose the limiting value along the orbit.

We note that Conjecture 1 asks for global convergence without requiring a closed-form expression for the limit. The argument below is based on a short difference identity that shows the sign of successive increments $y_{n+1}-y_n$ cannot change with n. This immediately forces monotonicity and, together with a simple estimate on the increments in the increasing case, yields boundedness and convergence. After proving convergence, we introduce an auxiliary sequence that provides nested two-sided enclosures of the limit and, more importantly, an exact invariant-series formula whose value is precisely the selected limiting equilibrium. This gives a rigorous determination of the limit from the initial orbit and supplies a certified stopping criterion for numerical computation.

The purpose of this note is to confirm the conjecture. The proof is elementary: we first show that the sign of $y_{n+1}-y_n$ equals the sign of $y_n-y_{n-1}$ for all $n\ge 0$, which forces monotonicity. The decreasing case then converges immediately by monotone convergence. In the increasing case we prove that the increments decay at least geometrically, which provides an explicit upper bound and yields convergence to a finite limit.

2. Main Result

Throughout we assume $y_{-1},y_0>0$ and define ${\left\{y_n\right\}}_{n=-1}^{\infty }$ by (1). Note that positivity is forward invariant: if $y_{n-1},y_n>0$, then the right-hand side of (1) is positive, so $y_{n+1}>0$ for all $n\ge 0$. Set

$$D_n:=y_n(1+y_n)+(1+y_{n-1})>0.$$

 Lemma 1

(Difference identity). For every $n\ge 0$,

$$y_{n+1}-y_n={\displaystyle \frac{y_n(y_n-y_{n-1})}{D_n}}.$$

(2)

Consequently,

$$sgn(y_{n+1}-y_n)=sgn(y_n-y_{n-1}).$$

Proof. 

Starting from (1),

$$y_{n+1}-y_n=y_n\left({\displaystyle \frac{{(1+y_n)}^2}{D_n}}-1\right)=y_n{\displaystyle \frac{{(1+y_n)}^2-D_n}{D_n}}.$$

Since

$${(1+y_n)}^2-D_n=(1+2y_n+y_n^2)-(y_n+y_n^2+1+y_{n-1})=y_n-y_{n-1},$$

we obtain (2). The sign statement follows because $y_n>0$ and $D_n>0$.    □

 Lemma 2

(Monotonicity). Every positive solution of (1) is monotone. More precisely: if $y_0>y_{-1}$, then ${\left\{y_n\right\}}_{n\ge 0}$ is strictly increasing; if $y_0<y_{-1}$, then it is strictly decreasing; and if $y_0=y_{-1}$, then $y_n\equiv y_0$.

Proof. 

If $y_0=y_{-1}$, then (2) gives $y_1-y_0=0$, and by induction $y_{n+1}-y_n=0$ for all n, so the solution is constant.

Assume $y_0\ne y_{-1}$. Then $y_1-y_0$ has the same sign as $y_0-y_{-1}$ by Lemma 2.1. Applying Lemma 2.1 iteratively shows that the sign of $y_{n+1}-y_n$ is constant in n, hence the sequence is strictly monotone.    □

 Lemma 3

(An initial-data-dependent upper bound in the increasing case). Assume $y_0>y_{-1}$ so that ${\left\{y_n\right\}}_{n\ge 0}$ is increasing. Let $d_n:=y_n-y_{n-1}$ for $n\ge 0$ (so $d_0=y_0-y_{-1}>0$). Then

$$0<d_n\le {\displaystyle \frac{d_0}{{(1+y_0)}^n}},n=0,1,2,\dots ,$$

(3)

and, in particular,

$$y_n\le y_0+{\displaystyle \frac{y_0-y_{-1}}{y_0}}=y_0+1-{\displaystyle \frac{y_{-1}}{y_0}}\mathrm{for}\mathrm{all}n\ge 0.$$

(4)

Proof. 

From (2),

$$d_{n+1}=y_{n+1}-y_n={\displaystyle \frac{y_n{d}_n}{D_n}}<{\displaystyle \frac{y_n{d}_n}{y_n(1+y_n)}}={\displaystyle \frac{d_n}{1+y_n}},$$

where we used $D_n=y_n(1+y_n)+(1+y_{n-1})>y_n(1+y_n)$. Since $\left\{y_n\right\}$ is increasing, $y_n\ge y_0$ for all $n\ge 0$, hence

$$d_{n+1}\le {\displaystyle \frac{d_n}{1+y_0}}.$$

Iterating yields (3). Summing (3) over $k\ge 1$ gives

$$y_n=y_0+{\displaystyle \sum _{k=1}^n}{d}_k\le y_0+{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{d_0}{{(1+y_0)}^k}}=y_0+{\displaystyle \frac{d_0}{y_0}},$$

which is (4).    □

 Remark 2.

The estimate (4) is not uniform over the whole set of positive initial data. It is an explicit orbitwise bound depending on the chosen initial values $y_{-1}$ and $y_0$. This is the only meaning in which boundedness is used below: each individual positive solution is bounded.

We now state and prove the main theorem, which confirms Conjecture 1 in [1].

 Theorem 1

(Global convergence). Every positive solution of (1) converges to a finite limit.

Proof. 

By Lemma 2.2, the sequence ${\left\{y_n\right\}}_{n\ge 0}$ is monotone.

If it is decreasing, then $y_n\ge 0$ for all n and therefore $\left\{y_n\right\}$ is bounded below; hence it converges to a finite limit by the monotone convergence theorem.

If it is increasing, then Lemma 2.3 provides the explicit upper bound (4), so $\left\{y_n\right\}$ is increasing and bounded above, and therefore converges to a finite limit.    □

 Remark 3.

Theorems 3.2 and 3.4 below provide two complementary descriptions of the selected limiting equilibrium. The first gives nested two-sided bounds and an effective stopping criterion. The second gives an exact invariant-series expression for L from the initial orbit. This addresses the limit-selection issue caused by the continuum of equilibria of (1).

3. A Two-Sided Characterisation and an Invariant Formula for the Limit

The proof of Theorem 2.4 guarantees the existence of a finite limit

$$L:=\underset{n\to \infty }{lim}{y}_n\in [0,\infty ).$$

Because (1) has a continuum of equilibria, the value of L cannot be recovered by solving an equilibrium equation alone. The following auxiliary sequence gives an orbit-based replacement. It first produces nested intervals whose unique common point is precisely L, and then yields an invariant-series formula whose value is exactly the selected equilibrium.

 Lemma 4

(An auxiliary sequence). For $n\ge 0$ define

$$t_n:={\displaystyle \frac{y_n(1+y_n)}{1+y_{n-1}}}.$$

(5)

Then, for every $n\ge 0$,

$$t_n-y_n={\displaystyle \frac{y_n(y_n-y_{n-1})}{1+y_{n-1}}},$$

(6)

and, for every $n\ge 0$,

$$t_{n+1}-t_n={\displaystyle \frac{y_n^3{(1+y_n)}^2(y_{n-1}-y_n)}{(1+y_{n-1}){(1+y_{n-1}+y_n+y_n^2)}^2}}.$$

(7)

Consequently,

$$sgn(t_{n+1}-t_n)=sgn(y_{n-1}-y_n)=-sgn(y_n-y_{n-1}).$$

Proof. 

Identity (6) follows immediately from (5):

$$t_n-y_n=y_n\left({\displaystyle \frac{1+y_n}{1+y_{n-1}}}-1\right)={\displaystyle \frac{y_n(y_n-y_{n-1})}{1+y_{n-1}}}.$$

To prove (7), write (1) as

$$y_{n+1}={\displaystyle \frac{y_n{(1+y_n)}^2}{1+y_{n-1}+y_n+y_n^2}}.$$

Then

$$t_{n+1}={\displaystyle \frac{y_{n+1}(1+y_{n+1})}{1+y_n}}={\displaystyle \frac{y_{n+1}}{1+y_n}}(1+y_{n+1}).$$

Substituting the above expression for $y_{n+1}$ and simplifying yields (7). The sign statement follows because all factors in the denominator of (7) are positive.    □

 Theorem 2 (Two-sided bounds for the limit).

Let $\left\{y_n\right\}$ be any positive solution of (1) and let $t_n$ be defined by (5). Then:

• If $y_0>y_{-1}$, then ${\left\{y_n\right\}}_{n\ge 0}$ is strictly increasing, ${\left\{t_n\right\}}_{n\ge 0}$ is strictly decreasing, and

  $$y_n<L<t_n\mathrm{for}\mathrm{all}n\ge 0.$$
  Moreover,

  $$L=\underset{n\ge 0}{sup}{y}_n=\underset{n\ge 0}{inf}{t}_n=\underset{n\to \infty }{lim}{t}_n.$$
• If $y_0<y_{-1}$, then ${\left\{y_n\right\}}_{n\ge 0}$ is strictly decreasing, ${\left\{t_n\right\}}_{n\ge 0}$ is strictly increasing, and

  $$t_n<L<y_n\mathrm{for}\mathrm{all}n\ge 0,$$
  with

  $$L=\underset{n\ge 0}{inf}{y}_n=\underset{n\ge 0}{sup}{t}_n=\underset{n\to \infty }{lim}{t}_n.$$

Proof. 

Assume $y_0>y_{-1}$. By Lemma 2.2, $\left\{y_n\right\}$ is strictly increasing and converges to L by Theorem 2.4. Identity (6) implies $t_n>y_n$ for all n. By (7), $\left\{t_n\right\}$ is strictly decreasing. Since $t_n>y_n\ge y_0$, it is bounded below and hence has a finite limit $\tau$. Passing to the limit in (5) using $y_{n-1}\to L$ and $y_n\to L$ gives

$$\tau =\underset{n\to \infty }{lim}{\displaystyle \frac{y_n(1+y_n)}{1+y_{n-1}}}={\displaystyle \frac{L(1+L)}{1+L}}=L.$$

Thus $t_n\searrow L$ and $y_n\nearrow L$, proving the claimed inequalities and the extremal characterisation of L.

The case $y_0<y_{-1}$ is analogous: $\left\{y_n\right\}$ decreases to L, (6) yields $t_n<y_n$, and (7) implies that $\left\{t_n\right\}$ increases. Taking limits in (5) again gives ${lim}_{n\to \infty }{t}_n=L$.    □

 Theorem 3

(Nested-interval determination of the limit). For $n\ge 0$, set

$$J_n:=\left[min\{y_n,t_n\},max\{y_n,t_n\}\right].$$

Then the intervals $J_n$ are nested, their lengths are

$$|J_n|=|t_n-y_n|,$$

  and 

$${\displaystyle \bigcap _{n=0}^{\infty }}{J}_n=\left\{L\right\}.$$

Equivalently, L is the unique real number lying in every interval $J_n$.

Proof. 

If $y_0>y_{-1}$, then $J_n=[y_n,t_n]$. By Theorem 3.2, $y_n$ increases to L and $t_n$ decreases to L, so $J_{n+1}\subset J_n$ and the intersection is $\left\{L\right\}$. If $y_0<y_{-1}$, then $J_n=[t_n,y_n]$, with $t_n\nearrow L$ and $y_n\searrow L$, and the same conclusion follows. If $y_0=y_{-1}$, then the solution is constant and the assertion is immediate. The formula for $|J_n|$ follows from the definition of $J_n$.    □

 Theorem 4

(An invariant-series formula for the selected equilibrium). Let $\left\{y_n\right\}$ be a positive solution of (1), and let

$$t_n:={\displaystyle \frac{y_n(1+y_n)}{1+y_{n-1}}},n\ge 0.$$

Set

$$g_n:=t_n-y_n,{\rho }_n:={\displaystyle \frac{t_n}{{(1+t_n)}^2}}.$$

Then

$$y_{n+1}={\displaystyle \frac{t_n(1+y_n)}{1+t_n}},t_{n+1}={\displaystyle \frac{t_n(1+y_{n+1})}{1+t_n}},$$

(8)

and consequently

$$y_{n+1}-y_n={\displaystyle \frac{g_n}{1+t_n}},g_{n+1}={\rho }_n{g}_n.$$

(9)

Since $t_n>0$,

$$0<{\rho }_n={\displaystyle \frac{t_n}{{(1+t_n)}^2}}\le {\displaystyle \frac{1}{4}}.$$

For $n\ge 0$, define

$$I_n:=y_n+g_n{\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{1+t_{n+j}}}{\displaystyle \prod _{m=0}^{j-1}}{\rho }_{n+m},$$

(10)

where the empty product is understood to be 1. Then the series is absolutely convergent,

$$I_{n+1}=I_n\mathrm{for}\mathrm{all}n\ge 0,$$

and

$$I_n=\underset{k\to \infty }{lim}{y}_k=L.$$

In particular, the limiting equilibrium selected by the initial values $(y_{-1},y_0)$ is given by

$$L=y_0+\left({\displaystyle \frac{y_0(1+y_0)}{1+y_{-1}}}-y_0\right){\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{1+t_j}}{\displaystyle \prod _{m=0}^{j-1}}{\displaystyle \frac{t_m}{{(1+t_m)}^2}}.$$

(11)

Proof. 

By the definition of $t_n$,

$$y_n(1+y_n)=t_n(1+y_{n-1}).$$

Substituting this identity into (1) gives

$$y_{n+1}={\displaystyle \frac{t_n(1+y_n)}{1+t_n}}.$$

Furthermore,

$$t_{n+1}={\displaystyle \frac{y_{n+1}(1+y_{n+1})}{1+y_n}}={\displaystyle \frac{t_n(1+y_{n+1})}{1+t_n}},$$

because $y_{n+1}/(1+y_n)=t_n/(1+t_n)$. Thus (8) holds. It follows that

$$y_{n+1}-y_n={\displaystyle \frac{t_n(1+y_n)}{1+t_n}}-y_n={\displaystyle \frac{t_n-y_n}{1+t_n}}={\displaystyle \frac{g_n}{1+t_n}}.$$

Also,

$$g_{n+1}=t_{n+1}-y_{n+1}={\displaystyle \frac{t_n}{1+t_n}}(y_{n+1}-y_n)={\displaystyle \frac{t_n}{{(1+t_n)}^2}}{g}_n={\rho }_n{g}_n.$$

Since $t_n>0$, the elementary inequality

$${\displaystyle \frac{t_n}{{(1+t_n)}^2}}\le {\displaystyle \frac{1}{4}}$$

shows that the series in (10) is absolutely convergent; indeed, each product is bounded above by $4^{-j}$.

Let

$$S_n:={\displaystyle \sum _{j=0}^{\infty }}{\displaystyle \frac{1}{1+t_{n+j}}}{\displaystyle \prod _{m=0}^{j-1}}{\rho }_{n+m}.$$

Then

$$S_n={\displaystyle \frac{1}{1+t_n}}+{\rho }_n{S}_{n+1}.$$

Using (9), we obtain

$$I_n=y_n+g_n{S}_n=y_n+{\displaystyle \frac{g_n}{1+t_n}}+g_n{\rho }_n{S}_{n+1}=y_{n+1}+g_{n+1}{S}_{n+1}=I_{n+1}.$$

Thus $I_n$ is invariant along the orbit.

Finally, (9) and $0<{\rho }_n\le 1/4$ give

$$|g_n|\le |g_0|4^{-n}.$$

Moreover,

$$0<S_n\le {\displaystyle \sum _{j=0}^{\infty }}{4}^{-j}={\displaystyle \frac{4}{3}}.$$

Hence $I_n-y_n=g_n{S}_n\to 0$. Since $I_n$ is independent of n and $y_n\to L$ by Theorem 2.4, it follows that $I_n=L$ for every $n\ge 0$. Taking $n=0$ in (10) gives (11).    □

 Remark 4

(A practical stopping criterion). Theorem 3.3 implies that, for any $n\ge 0$,

$$L\in \left[min\{y_n,t_n\},max\{y_n,t_n\}\right],max\{y_n,t_n\}-min\{y_n,t_n\}=|t_n-y_n|.$$

Thus one may approximate L together with an a posteriori error bound by iterating until $|t_n-y_n|$ is below a prescribed tolerance.

 Remark 5

(On the nature of the invariant). The invariant in Theorem 3.4 is an exact convergent-series invariant. It plays the same limit-selection role as the elementary invariant $2y_n+y_{n-1}$ for the linear averaging equation $y_{n+1}=(y_n+y_{n-1})/2$: after passing to the limit in the invariant identity, one obtains the selected equilibrium. The formula above is not a finite algebraic first integral; finding an even simpler finite expression would be a further refinement beyond the convergence assertion of Conjecture 1.

4. Numerical Experiments

We briefly illustrate the theoretical conclusions with numerical simulations.

4.1. Trajectories and Convergence of Increments

Figure 1 shows representative trajectories for several initial conditions, including a “large-data” example $(y_{-1},y_0)=(100,1000)$. In each case the solution is monotone and converges quickly. Figure 2(a) plots the successive increments $|y_n-y_{n-1}|$ on a semilogarithmic scale for the increasing case $(y_{-1},y_0)=(1,2)$, illustrating the geometric decay predicted by Lemma 2.3.

4.2. Two-Sided Bounds and Initial-Value Sweep

Figure 2(b) plots $y_n$ together with the auxiliary sequence $t_n$ for both an increasing example $(y_{-1},y_0)=(1,2)$ and a decreasing example $(y_{-1},y_0)=(2,1)$. In agreement with Theorem 3.2, the two sequences are monotone in opposite directions and rapidly squeeze the common limit. Figure 2(c) shows the gap $|t_n-y_n|$, which is the width of the rigorous enclosure interval for L.

Although equation (1) has no external parameter, one may view the initial value $y_0$ as a varying input when $y_{-1}$ is fixed. For each $y_0$ in a logarithmic grid on $[{10}^{-3},{10}^3]$ (400 points) and fixed $y_{-1}=1$, we computed an approximation of $L(y_{-1},y_0)={lim}_{n\to \infty }{y}_n$ by iterating until the enclosure width $|t_n-y_n|<{10}^{-12}$ (see Remark 3.5) was satisfied, and then taking the midpoint of the resulting enclosure interval. Figure 2(d) shows the resulting initial-value sweep $y_0\mapsto L(1,y_0)$. No qualitative changes are observed over six orders of magnitude, consistent with the global convergence established in Theorem 2.4. All simulations and plots were generated in Python (NumPy/Matplotlib); the script numerics.py included with the source reproduces the figures.

5. Conclusion

We confirmed Conjecture 1 of Kulenović, Ladas and Overdeep [1] by showing that every positive solution of (1) is monotone and bounded, and hence convergent. The boundedness in the increasing case is supplied by an explicit initial-data-dependent estimate. Moreover, the auxiliary sequence $t_n$ provides nested two-sided enclosures whose unique common point is the limit, and Theorem 3.4 supplies an exact invariant-series formula for the selected equilibrium. Thus the limiting value is determined by

$$\left\{L\right\}={\displaystyle \bigcap _{n=0}^{\infty }}\left[min\{y_n,t_n\},max\{y_n,t_n\}\right]$$

and also by the invariant identity

$$L=I_0.$$

This completes the convergence result and gives an explicit orbit-based determination of the limiting value selected by the initial data.

6. Consequences for Dynamical Systems and Chaos Theory

Equation (1) defines the continuous planar map

$$F(x,y)=\left(y,{\displaystyle \frac{y{(1+y)}^2}{y(1+y)+(1+x)}}\right)$$

on the positive quadrant. This is the standard state-space representation of a second-order difference equation [5]. Theorem 2.4 and Lemma 2.2 imply that every positive forward orbit

$$F^n(y_{-1},y_0)=(y_{n-1},y_n)$$

converges to a point $(L,L)$ on the equilibrium diagonal

$$\mathcal{E}=\left\{\right(s,s):s\ge 0\}$$

in the closure of the positive quadrant.

 Corollary 1.

For the positive dynamics generated by (1):

(i)

the omega-limit set of every orbit is the singleton $\left\{\right(L,L\left)\right\}$;

(ii)

there is no positive periodic orbit of prime period greater than one;

(iii)

there is no Li–Yorke scrambled pair, and therefore no Li–Yorke chaos in the positive phase space in the sense introduced in [6];

(iV)

the map cannot be Devaney chaotic on any compact invariant subset of the positive quadrant containing more than one point, in the standard sense of [7,8].

Proof. 

The first assertion follows directly from $(y_{n-1},y_n)\to (L,L)$. A periodic orbit that converges must be constant, which proves the second assertion. For two positive initial states whose limits are L and $\tilde{L}$, respectively, the distance between their nth states converges to $\sqrt{2}|L-\tilde{L}|$. Hence the lower and upper limiting distances coincide, so no pair can satisfy the defining separation condition of a Li–Yorke scrambled pair. Finally, the periodic points of F are precisely the fixed points on the equilibrium diagonal. If periodic points were dense in a compact invariant set K, then, because the equilibrium diagonal is closed, one would have K contained in that diagonal. The restriction ${F|}_K$ would then be the identity map, which is not topologically transitive when K contains more than one point. Thus the defining conditions for Devaney chaos cannot hold on such a set.    □

Thus the proved conjecture has a direct dynamical interpretation: although (1) is nonlinear and rational and possesses a continuum of non-hyperbolic equilibria, its positive dynamics are globally ordered and asymptotically non-chaotic. These conclusions concern positive initial data only; they do not address sign-changing extensions, parameter perturbations, or nonautonomous variants of the recurrence.