Dynamic Systems and Applications 15 (2006) 89-96

GLOBAL ATTRACTIVITY IN A NONLINEAR DIFFERENCE EQUATION AND ITS APPLICATIONS

JOHN R. GRAEF^{1} AND CHUANXI QIAN^{2}

1Department of Mathematics, University of Tennessee at Chattanooga Chattanooga, TN 37403 USA. E-mail: John-Graef@utc.edu

2Department of Mathematics and Statistics, Mississippi State University Mississippi State, MS 39762 USA. E-mail: qian@math.msstate.edu

Dedicated to John Baxley on the occasion of his retirement from Wake Forest University.

ABSTRACT. The authors consider the nonlinear difference equation

x_{n+1} = x_{n}f(x_{n−k}),

where f ∈ C[[0, ∞), (0, ∞)], f(x̄) = 1 for some x̄ > 0, and k is a nonnegative integer. They obtain a sufficient condition for the positive equilibrium x̄ to be a global attractor of all positive solutions of the equation. By applying this result, they obtain a global stability criterion for the following difference equation derived from mathematical biology

x_{n+1} = αx_{n} + \frac{βx_{n}}{1 + x_{n−k}^{r}} ,

where 0 ≤ α < 1 < α + β and r ∈ (0, ∞).

AMS (MOS) Subject Classification. 39A10, 34A11.

1. INTRODUCTION

Consider the difference equation

x_{n+1} = αx_{n} + \frac{βx_{n}}{1 + x_{n−k}^{r}} , (1.1)

where

0 ≤ α < 1 < α + β, r ∈ (0, ∞), and k ∈ {0, 1, ⋯ }. (1.2)

When k = 0, (1.1) reduces to

x_{n+1} = αx_{n} + \frac{βx_{n}}{1 + x_{n}^{r}} (1.3)

Received November 4, 2004               1056-2176 $15.00 \begin{matrix}c \\ ◯ \end{matrix}Dynamic Publishers, Inc.

which was proposed by Milton and Belar [5] as a model for the bobwhite quail population of northern Wisconsin. It was shown in [5] that if

r < \frac{2β}{(1 − α)(α + β − 1)}, (1.4)

then the positive equilibrium x̄ = (\frac{α+β−1}{1−α})^{1/r} is locally asymptotically stable. The global stability of (1.3) and (1.1) was discussed in [1], and some sufficient conditions for the global stability of x̄ were established. However, these conditions are complicated and hard to verify in some cases. Our main aim in this paper is to establish some new criteria, which are easy to check, for the global stability of (1.3) and (1.1).

Observe that Eq.(1.1) can be written in the form

x_{n+1} = x_{n}f(x_{n−k}), (1.5)

where

(1.6) f ∈ C[[0, ∞), (0, ∞)] and f(x̄) = 1 for some x̄ > 0.

Global attractivity of Eq.(1.5) was studied in [2] and the following result was obtained.

Theorem A. Assume that k is a positive integer and (1.6) holds. Suppose also that f satisfies the following properties:

(i) f(u) is nonincreasing in u; (ii) x̄ is the only positive solution of the equation f(x) = 1 and

[xf(x) − x̄][x − x̄] > 0 for x ≠ x̄; (1.7)

(iii) The only solution of the equation

x̄[f(x̄(f(x))^{k})]^{k} = x, (1.8)

in the interval 0 ≤ x ≤ x̄(f(0))^{k}, is x = x̄.

Then x̄ is a global attractor of all positive solutions of Eq.(1.5).

This result can be applied to some difference equations derived from mathematical biology. However, it is not easy to apply this theorem to Eq.(1.1) to obtain an explicit condition for the positive equilibrium x̄ to be a global attractor of all positive solutions since it is hard to find suitable conditions such that the hypothesis (iii) is satisfied. In this paper, we will establish a global attractivity result for Eq.(1.5) which is quite different from Theorem A. Our result can be applied to Eq.(1.1) and (1.3) to obtain some explicit conditions for the global stability of these equations.

GLOBAL ATTRACTIVITY 91

2. GLOBAL ATTRACTIVITY OF EQ.(1.5) In this section, we establish a criteria for the global attractivity of Eq.(1.5) by

using the following result obtained in [4]. Lemma 2.1. ([4; Theorem 1.1]) Consider the difference equation y_{n+1} − y_{n} + p_{n}g(y_{n−k}) = 0, (2.1) where {p_{n}} is a nonnegative sequence, g : R → R is a continuous function with xg(x) > 0 if x ≠ 0, and k is a nonnegative integer. Assume that |g(y)| < |y| for y ≠ 0, (2.2)

∑_{i=n−k}^{n} p_{i} ≤ \frac{3}{2} + \frac{1}{2(k + 1)} for n ≥ k, (2.3)

and

∑_{i=0}^{∞} p_{i} = ∞. (2.4)

Then every solution of Eq.(2.1) tends to zero as n → ∞.

Our attractivity result is the following. Theorem 2.2. Consider Eq.(1.5) and assume that (1.6) holds and that k is a nonnegative integer. Suppose also that (f(x) − 1)(x − x̄) < 0 for x ≠ x̄, (2.5) and h(x) = x(f(x))^{\frac{2(k+1)^{2}}{3k+4}} , x > 0, (2.6) is a increasing function. Then every positive solution {x_{n}} of Eq.(1.5) tends to x̄ as n → ∞. Proof. Let x_{n} = x̄e^{y_{n}}. Then Eq.(1.5) becomes y_{n+1} − y_{n} − ln f(x̄e^{y_{n−k}}) = 0. (2.7) To show that x_{n} → x̄ as n → ∞, it suffices to show that y_{n} → 0 as n → ∞. To this end, rewrite this equation in the form y_{n+1} − y_{n} + p(−q ln f(x̄e^{y_{n−k}})) = 0, (2.8) where p = \frac{1}{k+1}(\frac{3}{2} + \frac{1}{2(k+1)}) and q = 1/p. Let

g(y) = −q ln f(x̄e^{y}), −∞ < y < ∞.

By noting (2.5), we see that yg(y) > 0 for y ≠ 0. Next, we show that y − g(y) is a increasing function. Let z = x̄e^{y} and observe that

y − g(y) = ln(\frac{1}{x̄}z) + q ln f(z) = ln \frac{1}{x̄} + ln(z(f(z))^{q}), z > 0.

Since z(f(z))^{q} is increasing by (2.6), ln(z(f(z))^{q}) is increasing. Hence, y − g(y) is a increasing function, and since g(0) = 0, we see that

|g(y)| < |y| for y ≠ 0.

Let p_{n} ≡ p. Then

∑_{i=n−k}^{n} p_{i} = \frac{3}{2} + \frac{1}{2(k + 1)}

and (2.4) holds automatically. Therefore, the hypotheses of Lemma 2.1 are satisfied, so y_{n} → 0 as n → ∞. It then follows that x_{n} → x̄ as n → ∞, and this completes the proof of the theorem.

Remark 2.3. From the proof of this theorem, we see that by using the same argument, a similar result may be obtained for the following more general difference equation

x_{n+1} = x_{n}[f(x_{n−k})]^{p_{n}}, (2.9)

where {p_{n}} is a positive sequence.

3. GLOBAL STABILITY OF EQ.(1.1)

In this section, we apply Theorem 2.2 and linearized stability theory to Eq.(1.1) to establish a sufficient condition for its positive equilibrium to be globally asymptotically stable.

Theorem 3.1. Consider Eq.(1.1) and assume that (1.2) holds. Suppose also that

r < \frac{2α + β + 2\sqrt{α^{2} + αβ}}{β} · \frac{3k + 4}{2(k + 1)^{2}} . (3.1)

Then the positive equilibrium x̄ = (\frac{α+β−1}{1−α})^{1/r} of Eq.(1.1) is globally asymptotically stable.

Proof. First, we show that x̄ is a global attractor of all positive solutions of Eq.(1.1). Let

f(x) = α + \frac{β}{1 + x^{r}} , x > 0.

Clearly, f satisfies (2.5). We claim that the function h(x) = x[f(x)]^{\frac{2(k+1)^{2}}{3k+4}} is increasing. To this end, it suffices to show that h′(x) > 0. For the sake of simplicity, let q = \frac{2(k+1)^{2}}{3k+4} . Observe that

h′(x) = (α + \frac{β}{1 + x^{r}} )^{q−1} \frac{α(1 + x^{r})^{2} + β(1 + x^{r}) − qβrx^{r}}{(1 + x^{r})^{2}} , x > 0.

Hence, we only need to show that

α(1 + x^{r})^{2} + β(1 + x^{r}) − qβrx^{r} > 0, x > 0. (3.2)

By noting that

α(1 + x^{r})^{2} + β(1 + x^{r}) − qβrx^{r} = α(x^{r})^{2} + (2α + β − qβr)x^{r} + α + β,

we see that if r ≤ \frac{2α+β}{qβ}, then (3.2) holds. Next, we show that (3.2) also holds if r > \frac{2α+β}{qβ} and (3.1) is satisfied. To this end, let y = x^{r} and

s(y) = αy^{2} + (2α + β − qβr)y + α + β, y > 0.

Hence, it suffices to show that s(y) > 0 for y > 0. Since

s(y) = (\sqrt{α}y + \frac{2α + β − qβr}{2\sqrt{α}})^{2} + \frac{4α^{2} + 4αβ − (2α + β − qβr)^{2}}{4α} ,

we only need to show that

4α^{2} + 4αβ − (2α + β − qβr)^{2} > 0. (3.3)

Observe that

4α^{2} + 4αβ − (2α + β − qβr)^{2} = −β(βq^{2}r^{2} − (4αq + 2βq)r + β).

Clearly, if

\frac{4αq + 2βq − \sqrt{(4αq + 2βq)^{2} − 4β^{2}q^{2}}}{2βq^{2}} < r < \frac{4αq + 2βq + \sqrt{(4αq + 2βq)^{2} − 4β^{2}q^{2}}}{2βq^{2}} ,

that is, if

\frac{2α + β − 2\sqrt{α^{2} + αβ}}{βq} < r < \frac{2α + β + 2\sqrt{α^{2} + αβ}}{βq} ,

then (3.3) holds. Since r > \frac{2α+β}{βq}, q = \frac{2(k+1)^{2}}{3k+4} , and (3.1) holds, we see that (3.3) holds, and so it follows that (3.2) holds. Combining this with the fact that we previously showed (3.2) holds if r ≤ \frac{2α+β}{βq}, we conclude that (3.2) holds under condition (3.1). Hence, h′(x) > 0, and so h(x) is increasing. Thus, by Theorem 2.2, x̄ is a global attractor of all positive solutions of Eq.(1.1).

Next, we show that x̄ is also stable. To this end, let x_{n} = x̄e^{y_{n}}. Then Eq.(1.1) becomes

y_{n+1} − y_{n} − ln(α + \frac{β}{1 + x̄^{r}e^{ry_{n−k}}}) = 0. (3.4)

The linearized equation of (3.4) is

z_{n+1} − z_{n} + \frac{(1 − α)(α + β − 1)r}{β}z_{n−k} = 0. (3.5)

It is well-known (for example, see [3; Theorem 1.3.6]) that the zero solution of the linear equation (3.5) is asymptotically stable if and only if

r < \frac{2β}{(1 − α)(α + β − 1)} cos \frac{kπ}{2k + 1}. (3.6)

Hence, by linearized stability theory, if (3.6) holds, then the trivial solution of Eq.(3.4) is locally asymptotically stable. Now, we show that (3.1) implies (3.6), that is,

\frac{2α + β + 2\sqrt{α^{2} + αβ}}{β} · \frac{3k + 4}{2(k + 1)^{2}} ≤ \frac{2β}{(1 − α)(α + β − 1)} cos \frac{kπ}{2k + 1}. (3.7)

In fact, from the proof of Theorem 2 in [6], we have that

\frac{3k + 4}{2(k + 1)^{2}} ≤ 2 cos \frac{kπ}{2k + 1}

for any k ≥ 0. Hence, to show that (3.7) holds, it suffices to show that

\frac{2α + β + 2\sqrt{α^{2} + αβ}}{β} ≤ \frac{β}{(1 − α)(α + β − 1)}. (3.8)

To this end, let β = cα and observe that

\frac{2α + β + 2\sqrt{α^{2} + αβ}}{β} = \frac{2 + c + 2\sqrt{1 + c}}{c}

and

\frac{β}{(1 − α)(α + β − 1)} = \frac{cα}{(1 − α)((1 + c)α − 1)}.

We see that (3.8) is equivalent to

\frac{2 + c + 2\sqrt{1 + c}}{c} ≤ \frac{cα}{(1 − α)((1 + c)α − 1)} (3.9)

that is,

\frac{(1 − α)((1 + c)α − 1)}{α} ≤ \frac{c^{2}}{2 + c + \sqrt{1 + c}} (3.10)

Since

\frac{(1 − α)((1 + c)α − 1)}{α} = 2 + c − ((1 + c)α + 1/α)

and

min_{α>0} \{(1 + c)α + \frac{1}{α}\} = 2\sqrt{1 + c},

it follows that

\frac{(1 − α)((1 + c)α − 1)}{β} ≤ 2 + c − 2\sqrt{1 + c}.

Hence,

(2 + c + \sqrt{1 + c})\frac{(1 − α)((1 + c)α − 1)}{α} ≤ (2 + c + 2\sqrt{1 + c})(2 + c − 2\sqrt{1 + c}) = c^{2}.

That is, (3.9) holds and so (3.8) holds. Therefore, the zero solution of (3.4) is stable and it follows that the positive equilibrium x̄ of Eq.(1.1) is stable. This completes the proof of the theorem.

The following result is a direct sequence of the above theorem.

Corollary 3.2. Consider Eq.(1.3) and assume that (1.2) holds with k = 0. Suppose also that

r < 2\frac{2α + β + 2\sqrt{α^{2} + αβ}}{β} . (3.11)

Then the positive equilibrium x̄ of Eq.(1.3) is globally asymptotically stable.

Remark 3.3. (i) For Eq.(1.1) with k > 0, Graef, Qian, and Spikes [1; Theorem 2(a)] showed that if

r ≤ 1 and (α + β)^{k+1} ≤ 2, (3.12)

then the positive equilibrium x̄ of Eq.(1.1) is globally asymptotically stable. For the case r > 1, the authors give a much more complicated condition for global asymptotic stability (see [1; Theorem 2(b)]); an explicit upper bound for r can not be found from their result. Condition (3.1) obtained here is different from the conditions obtained in [1] and is much easier to check when (3.12) is not satisfied.

(ii) For Eq.(1.3), it was proved in [1; Theorem 1(a)] that if

r < \frac{2(α + 1) + β + 2\sqrt{(α + 1)(α + β + 1)}}{β}, (3.13)

then the positive equilibrium x̄ is globally asymptotically stable. For the case where (3.13) is not satisfied, much more complicated conditions were obtained (see [1; Theorem 1 parts (b) and (c)]). Once again, an explicit upper bound for r could not be obtained. Condition (3.11) obtained here is different from those established in [1]; when β = \frac{1}{α} − α, (3.11) is the sharpest since (3.11) and (1.4) are same in this case.

We conclude with two examples to illustrate our results.

Example 3.4. Consider the difference equation

x_{n+1} = \frac{3}{4}x_{n} + \frac{x_{n}}{1 + x_{n−2}^{r}} . (3.14)

where r is a positive constant. x̄ = 3^{1/r} is the unique positive equilibrium of this equation. Observe that

\frac{2α + β + 2\sqrt{α^{2} + αβ}}{β} · \frac{3k + 4}{2(k + 1)^{2}} = \frac{5(5 + \sqrt{21})}{18}.

By Theorem 3.1, we see that x̄ is globally asymptotically stable when r < ^{\frac{5(5+\sqrt{21})}{18}}. However, we can not apply the results in [1] to obtain a range on r for the global stability of this equation.

Example 3.5. The difference equation

x_{n+1} = \frac{2}{3}x_{n} + \frac{4x_{n}}{1 + x^{4}} (3.15)

has the unique positive equilibrium x̄ = 11^{1/4}. Observe that

2\frac{2α + β + 2\sqrt{α^{2} + αβ}}{β} = \frac{8 + \sqrt{28}}{3} > 4 = r.

Hence, by Corollary 3.2, x̄ is globally asymptotically stable. The results in [1] do not apply to this equation.

ACKNOWLEDGMENTS

The research of J. R. Graef was supported in part by the University of Tennessee at Chattanooga Center of Excellence for Computer Applications.

REFERENCES

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[4] H. Masunaga, T. Hara and S. Sakata, Global attractivity for a nonlinear difference equation with delay, Comput. Math. Appl. 41 (2001), 543–551.

[5] G. G. Milton and J. Belair, Chaos, noise and extinction in models of population growth, Theoret. Population Biol. 37 (1990), 273–290.

[6] C. Qian, Global stability in a nonautonomous genotype selection model, Quart. Appl. Math. 61 (2003), 265–277.