Applied Mathematics E-Notes, 1(2001), 56-64 \begin{matrix}c \\ ◯ \end{matrix} Available free at mirror sites of http://math2.math.nthu.edu.tw/∼amen/

Global Attractivity for a Differential-Difference Population Model ^{∗}

Yuji Liu^{†}^{‡}

Received 28 December 2000

Abstract

In this paper, sufficient conditions are found for the positive equilibrium of the differential-difference equation (1) to be globally attractive.

Consider the delay differential equation

N′(t) = r(t)N(t) (\frac{1 − N(t − τ)}{1 + λ(t)N(t − τ)})^{α} , t ≥ 0,  (1)

where r(t) ∈ C([0, ∞), (0, ∞)), λ(t) ∈ C([0, ∞), (0, ∞)), τ > 0 and α is a ratio of two odd integers so that α ≥ 1. If α = 1, λ(t) ≡ λ then (1) becomes the so called “food-limited” model

N′(t) = r(t)N(t)\frac{1 − N(t − τ)}{1 + λN(t − τ)}, t ≥ 0,  (2)

which has been studied by many authors, see [1-6]. Equation (1) was proposed in [1] but has not been studied. In this paper, we shall prove two theorems related to attractivity of the positive equilibrium 1. Before stating our results, let us first note that equation (1) under the initial condition

N(t) = ϕ(t), t ∈ [−τ, 0], ϕ ∈ C([−τ, 0], [0, ∞)), ϕ(0) > 0.  (3)

has a unique positive solution N(t) on [−τ, ∞).

THEOREM 1. Suppose 0 < λ(t) ≤ 1 for t ≥ 0,

∫_{0}^{+∞} \frac{r(t)}{[1 + λ(t)]^{α}} dt = ∞,  (4)

and

lim sup_{t→+∞} ∫_{t−τ}^{t} \frac{r(s)}{(λ(s))^{α}} ds ≤ 3.  (5)

∗Mathematics Subject Classifications: 34K20, 92D25.

†Department of Mathematics, Yueyang Normal College, Yueyang, Hunan 414000, P. R. China

‡Supported by the NNSF of P. R. China

Then every solution of (1) and (3) tends to 1 as t tends to +∞.

THEOREM 2. Suppose λ(t) ≥ 1 for t ≥ 0, (4) holds and

lim sup_{t→+∞} ∫_{t−τ}^{t} r(s)ds ≤ 3,  (6)

then every solution of (1) and (3) tends to 1 as t tends to +∞.

To prove our results, we need the following lemmas.

LEMMA 1. For any v ∈ [0, 1), ln(2e^{−v(1−v/2)} − 1) ≥ −2v, and for any u ∈ [0, ∞), ln(2e^{u(1+u/2)} − 1) ≥ 2u.

Indeed, let f(v) = 2e^{−v(1−v/2)} − e^{−2v} and g(v) = (1 − v)e^{v(1+v/2)}. It is easy to see that g(0) = 1, g′(v) = −v^{2}e^{v(1+v/2)} ≤ 0, and for some ξ ∈ (0, v),

f′(v) = 2e^{−2v}[1 − g(v)] = −2e^{−2v}g′(ξ)v ≥ 0.

It follows that f(v) ≥ f(0) = 1 for v ∈ [0, 1). The other assertion is similarly proved. LEMMA 2. Assume that v ∈ (0, 1). Then for any x ∈ [0, ∞),

ln \frac{1 + [2e^{−v(1−v/2)} − 1]e^{−vx}}{1 + e^{−vx}} ≤ −v (1 − \frac{v}{2} ) + \frac{v^{2}}{2} x.  (7)

PROOF. Set a = 2e^{−v(1−v/2)} − 1 and f(x) = ln ((1 + ae^{−vx}) / (1 + e^{−vx})). Observe that f(0) = −v (1 − v/2) , f′(0) = v[e^{v(1−v/2)} − 1]/2 and

f′′(x) = [\frac{a}{(a + e^{vx})^{2}} − \frac{1}{(1 + e^{vx})^{2}} ] v^{2}e^{vx}.

Since a ≤ 1, it follows that f′′(x) ≤ 0 for x ≥ 0. By the mean value theorem and the fact that e^{x(1−x/2)} ≤ 1 + x for x ≥ 0, we have

f(x) ≤ f(0) + f′(0)x = −v (1 − \frac{v}{2} ) + \frac{vx}{2}[e^{v(1−v/2)} − 1]
≤ −v (1 − \frac{v}{2} ) + \frac{v^{2}}{2} x.

LEMMA 3. The system of inequalities

ln \frac{1 + u}{1 − u} ≤ 2v, − ln \frac{1 − v}{1 + v} ≤ 2u,  (8)

has a unique solution (u, v) = (0, 0) in the region {(u, v) : −1 < v ≤ 0 ≤ u < 1}. PROOF. Set g(x) = exp (2(1 − x)/(1 + x)) , f(x) = x − g(g(x)), and

h(x) = (1 + x)^{2}[1 + g(x)]^{2} − 16g(x)g(g(x)).

Observe that h(1) = 0,

f′(x) = 1 − g′(x)g′(g(x)) = 1 − \frac{16g(x)g(g(x))}{(1 + x)^{2}[1 + g(x)]^{2}} ,

and for x > 1

h′(x) = 2[1 + g(x)][(1 + x)(1 + g(x)) − 4g(x)]

+\frac{64}{(1 + x)^{2}} g(x)g(g(x))\frac{[1 − g(x)]^{2}}{[1 + g(x)]^{2}} > 0.

It follows that h(x) > h(1) = 0 for x > 1, and so f′(x) > 0 for x > 1. This shows that f(x) > f(1) = 0 for x > 1. On the other hand, from (8), we have g(μ) ≤ λ ≤ 1 ≤ μ ≤ g(λ), where λ = (1 − v)/(1 + v) and μ = (1 + u)/(1 − u). If u > 0, then μ > 1, and so μ ≤ g(λ) ≤ g(g(μ)) < μ. This contradiction implies that u = v = 0. The proof is complete.

LEMMA 4. Suppose (4) holds. Then every solution of (1) and (3) that does not oscillate about 1 tends to 1 as t → ∞.

The proof is similar to the arguments presented in [2] and is thus omitted.

LEMMA 5. Suppose 0 < λ(t) ≤ 1 for t ≥ 0 and (5) holds. Let N(t) = N(t; 0, ϕ) be a solution of (1) and (3) which is oscillatory about 1. Then N(t) is bounded above and is strictly bounded below by 0.

PROOF. Let t_{0} > 0 be large enough so that

∫_{t−τ}^{t} \frac{r(s)}{(λ(t))^{α}} ds ≤ 4,

for all t ≥ t_{0}. Let t^{∗} be a local maximum point of N(t) for t ≥ t_{0} + τ. Then N′(t^{∗}) = 0 and by (1), N(t^{∗} − τ) = 1. Integrating (1) from t^{∗} − τ to t^{∗},

N(t^{∗}) = exp (∫_{t^{∗}−τ}^{t^{∗}} r(s)N(s) [\frac{1 − N(s − τ)}{λ(s)N(s − τ)}]^{α} ds)

≤ exp (∫_{t^{∗}−τ}^{t^{∗}} r(s)ds) ≤ e^{4}.

Consequently, lim sup_{t→∞} N(t) ≤ e^{4}. Next, let t_{∗} be a local minimum point of N(t) for t ≥ t_{0} + 3τ. Then N′(t_{∗}) = 0 and N(t_{∗} − τ) = 1. Proceeding as before and using the fact that

\frac{1 − N(t − τ)}{1 + λ(t)N(t − τ)} ≥ \frac{1 − e^{4}}{1 + λ(t)e^{4}} ≥ \frac{1 − e^{4}}{λ(t)(1 + e^{4})}

for t ≥ t_{0} + τ, we have

N(t_{∗}) ≥ exp (∫_{t_{∗}−τ}^{t^{∗}} \frac{r(s)}{λ^{α}(s)} [\frac{1 − e^{4}}{1 + e^{4}} ]^{α} ds) ≥ exp (4 (\frac{1 − e^{4}}{1 + e^{4}} )^{α}) .

Hence

lim inf_{t→∞} N(t) ≥ exp (4 (\frac{1 − e^{4}}{1 + e^{4}} )^{α}) > 0.

The proof is complete.

LEMMA 6. Assume that λ(t) ≥ 1 for t ≥ 1 and (6) holds. Let N(t) = N(t, 0, ϕ) be a solution of (1) and (3) which is oscillatory about 1. Then N(t) is bounded above and is strictly bounded below by 0.

The proof is similar to the proof of Lemma 5 and is thus omitted.

We now turn to the proof of Theorem 1. Let u = lim sup_{t→∞} N(t) and v = lim inf_{t→∞} N(t). Then by Lemma 5, 0 < v ≤ 1 and u ≥ 1. It suffices to show that u = v = 1. For any ε ∈ (0, v), choose t_{0} = t_{0}(ε) such that

v_{1} ≡ v − ε < N(t − τ) < u + ε ≡ u_{1}, t ≥ t_{0},  (9)

and

∫_{t−τ}^{t} \frac{r(s)}{λ^{α}(t)}ds ≤ 3 + ε, t ≥ t_{0} − τ.  (10)

Note that (1 − x)/(1 + λx) ≤ (1 − x)/(λ(1 + x)) for x ≤ 1 and (1 − x)/(1 + λx) ≥ (1 − x)/(λ(1 + x)) for x ≥ 1. Thus

N′(t) ≤ r(t)N(t) (\frac{1 − v_{1}}{1 + λ(t)v_{1}} )^{α} ≤ r(t)N(t) (\frac{1 − v_{1}}{λ(t)(1 + v_{1})})^{α} , t ≥ t_{0}, (11)

and

N′(t) ≥ r(t)N(t) (\frac{1 − u_{1}}{1 + λ(t)u_{1}} )^{α} ≥ r(t)N(t) (\frac{1 − u_{1}}{λ(t)(1 + u_{1})})^{α} , t ≥ t_{0}.  (12)

Consequently,

N′(t) ≤ \frac{r(t)}{λ^{α}(t)}N(t)\frac{1 − v_{1}}{1 + v_{1}} , t ≥ t_{0},  (13)

and

N′(t) ≥ \frac{r(t)}{λ^{α}(t)}N(t)\frac{1 − u_{1}}{1 + u_{1}} , t ≥ t_{0}.  (14)

Set R(t) = r(t)/λ^{α}(t). Let {p_{n}} be an increasing sequence such that p_{n} ≥ t_{0} + τ, lim_{n→∞} p_{n} = +∞, N′(p_{n}) = 0 and lim_{n→∞} N(p_{n}) = u. By (1), N(p_{n} − τ) = 1. For p_{n} − τ ≤ t ≤ p_{n}, by integrating (13) from t − τ to p_{n} − τ, we get

N(t − τ) ≥ exp (−\frac{1 − v_{1}}{1 + v_{1}} ∫_{t−τ}^{p_{n}−τ} R(s)ds) , p_{n} − τ ≤ t ≤ p_{n}.

Substituting this into (1), if N(t − τ) ≤ 1, we have

N′(t) ≤ R(t)N(t) (\frac{1 − N(t − τ)}{1 + N(t − τ)})^{α} ≤ R(t)N(t)\frac{1 − N(t − τ)}{1 + N(t − τ)}

≤ R(t)N(t)\frac{1 − exp (−\frac{1−v_{1}}{1+v_{1}} ∫_{t−τ}^{p_{n}−τ} R(s)ds)}{1 + exp (−\frac{1−v_{1}}{1+v_{1}} ∫_{t−τ}^{p_{n}−τ} R(s)ds)}.

If N(t − τ) > 1, by (1), N′(t) < 0, thus

N′(t) ≤ R(t)N(t)\frac{1 − exp (−\frac{1−v_{1}}{1+v_{1}} ∫_{t−τ}^{p_{n}−τ} R(s)ds)}{1 + exp (−\frac{1−v_{1}}{1+v_{1}} ∫_{t−τ}^{p_{n}−τ} R(s)ds)}.

If t ∈ (p_{n} − τ, p_{n}), we have

N′(t) ≤ min \{R(t)N(t)\frac{1 − v_{1}}{1 + v_{1}} , R(t)N(t)\frac{1 − exp (−\frac{1−v_{1}}{1+v_{1}} ∫_{t−τ}^{p_{n}−τ} R(s)ds)}{1 + exp (−\frac{1−v_{1}}{1+v_{1}} ∫_{t−τ}^{p_{n}−τ} R(s)ds)}\} . (15)

Since 0 < x = (1 − v_{1})/(1 + v_{1}) < 1, it follows from Lemma 1 that ln 2e^{−x(1−x/2)−1} ≥ −2x, and so 0 < −\frac{1}{x} ln (2e^{−x(1−x/2)} − 1) ≤ 2. There are two possibilities. First of all, consider the case

∫_{p_{n}−τ}^{p_{n}} R(s)ds ≤ −\frac{1}{v_{n}} ln (2e^{−v_{0}(1−v_{0}/2)} − 1) ≡ A ≤ 3 + ε,

where v_{0} = (1 − v_{1})/(1 + v_{1}). Then

ln N(p_{n})
≤ ∫_{p_{n}−τ}^{p_{n}} \frac{R(t) [1 − exp (−v_{0} ∫_{t−τ}^{p_{n}−τ} R(s)ds)]}{1 + exp (−v_{0} ∫_{t−τ}^{p_{n}−τ} R(s)ds)}dt

= ∫_{p_{n}−τ}^{p_{n}} \frac{R(t) \{1 − exp [−v_{0} (∫_{t−τ}^{t} r(s)ds − ∫_{p_{n}−τ}^{t} R(s)ds)]\}}{1 + exp [−v_{0} (∫_{t−τ}^{t} R(s)ds − ∫_{p_{n}−τ}^{t} R(s)ds)]}dt

≤ ∫_{p_{n}−τ}^{p_{n}} \frac{R(t) \{1 − exp [−v_{0} (3 + ε − ∫_{p_{n}−τ}^{t} R(s)ds)]\}}{1 + exp [−v_{0} (3 + ε − ∫_{p_{n}−τ}^{t} R(s)ds)]}dt

= ∫_{p_{n}−τ}^{p_{n}} R(s)ds − \frac{2}{v_{0}} ln \frac{1 + exp [−v_{0} (3 + ε − ∫_{p_{n}−τ}^{t} R(s)ds)]}{1 + e^{−(3+ε)v_{0}}}.

The function f(x) = x − (2 ln [1 + e^{−v_{1}(3+ε−x)}]) /v_{1} is increasing in [0, 3 + ε]. Thus, by Lemmas 1 and 2, we have

ln N(p_{n}) ≤ A − \frac{2}{v_{0}} ln \frac{1 + e^{−v_{0}(3+ε−A)}}{1 + e^{−(3+ε)v_{0}}}
= A + \frac{2}{v_{0}} ln \frac{1 + [2e^{−v_{0}(1−v_{0}/2)} − 1] e^{−v_{0}(3+ε−A)}}{1 + e^{−v_{0}(3+ε−A)}}
≤ A + \frac{2}{v_{0}} [−v_{0} (1 − \frac{v_{0}}{2} ) + \frac{v_{0}^{2}}{2} (3 + ε − A)]
= −2 + (4 + ε)v_{0} − \frac{1 − v_{0}}{v_{0}} ln (2e^{−v_{0}(1−v_{0}/2)} − 1)
≤ (2 + ε)v_{1}.

Next, suppose

A < ∫_{p_{n}−τ}^{p_{n}} R(s) ≤ 3 + ε.

Choose ξ_{n} ∈ (p_{n} − τ, p_{n}) such that

∫_{ξ_{n}}^{p_{n}} R(s)ds ≡ A.

Then by (15) and Lemma 1,

ln N(p_{n})
≤ v_{0} ∫_{p_{n}−τ}^{ξ_{n}} R(s)ds + ∫_{ξ_{n}}^{p_{n}} \frac{R(t) [1 − exp (−v_{0} ∫_{t−τ}^{p_{n}−τ} R(s)ds)]}{1 + exp (−v_{0} ∫_{t−τ}^{p_{n}−τ} R(s)ds)}dt

≤ v_{0} ∫_{p_{n}−τ}^{ξ_{n}} R(s)ds + ∫_{ξ_{n}}^{p_{n}} \frac{R(t) \{1 − exp [−v_{0} (3 + ε − ∫_{p_{n}−τ}^{t} R(s)ds)]\}}{1 + exp [−v_{0} (3 + ε − ∫_{p_{n}−τ}^{t} R(s)ds)]}dt

= v_{0} ∫_{p_{n}−τ}^{ξ_{n}} R(s)ds + ∫_{ξ_{n}}^{p_{n}} R(s)ds − \frac{2}{v_{0}} ln \frac{1 + exp [−v_{0} (3 + ε − ∫_{p_{n}−τ}^{p_{n}} R(s)ds)]}{1 + exp [−v_{0} (3 + ε − ∫_{p_{n}−τ}^{ξ_{n}} R(s)ds)]}

= v_{0} ∫_{p_{n}−τ}^{p_{n}} R(s)ds + (1 − v_{0})A − \frac{2}{v_{0}} ln \frac{1 + exp [−v_{0} (3 + ε − ∫_{p_{n}−τ}^{p_{n}} R(s)ds)]}{1 + exp [−v_{0} (3 + ε − ∫_{p_{n}−τ}^{ξ_{n}} R(s)ds)]}

≤ (3 + ε)v_{0} + (1 − v_{0})A − \frac{2}{v_{0}} ln \frac{2}{1 + e^{−Av_{0}}}
= −2 + (4 + ε)v_{0} − \frac{1 − v_{0}}{v_{0}} ln (2e^{−v_{0}(1−v_{0}/2)} − 1)
≤ (2 + ε)v_{0},

where we have used the fact that the function

g(x) = −\frac{2}{v_{1}} ln \frac{1 + exp [−v_{1} (3 + ε − x)]}{1 + exp [−v_{1} (3 + ε + A − x)]} + v_{1}x

is increasing on [0, 3 + ε]. In either cases, we have proved that ln N(p_{n}) ≤ (2 + ε)v_{1} for n = 1, 2, ... . Letting n → ∞ and ε → 0, we have

ln u ≤ 2\frac{1 − v}{1 + v} .  (16)

Next, let {q_{n}} be an increasing sequence such that q_{n} ≥ t_{0} + τ, lim_{n→∞} q_{n} = +∞, N′(q_{n}) = 0 and lim_{n→∞} N(q_{n}) = −v. By (1), N(q_{n} − τ) = 1. For q_{n} − τ ≤ t ≤ q_{n}, integrating (14) form t − τ to q_{n} − τ, we have

N(t − τ) ≤ exp (−\frac{1 − u_{1}}{1 + u_{1}} ∫_{t−τ}^{p_{n}−τ} R(s)ds) , q_{n} − τ ≤ t ≤ q_{n}.

62 Global Attractivity Substituting this into (1), if N(t − τ) ≥ 1, we have

N′(t) = r(t)N(t) (\frac{1 − N(t − τ)}{1 + λ(t)N(t − τ)})^{α}
≥ R(t)N(t)\frac{1 − N(t − τ)}{1 + λ(t)N(t − τ)}
≥ R(t)N(t)\frac{1 − exp (−u_{0} ∫_{t−τ}^{q_{n}−τ} R(s)ds)}{1 + exp (−u_{0} ∫_{t−τ}^{q_{n}−τ} R(s)ds)}

for q_{n} − τ ≤ t ≤ q_{n}. If N(t − τ) < 1, then by (1), N′(t) > 0, thus

N′(t) ≥ R(t)N(t)\frac{1 − exp (−u_{0} ∫_{t−τ}^{q_{n}−τ} R(s)ds)}{1 + exp (−u_{0} ∫_{t−τ}^{q_{n}−τ} R(s)ds)},

where u_{0} = (1 − u_{1})/(1 + u_{1}). Thus

−N′(t) ≤ min \{−R(t)N(t)u_{0}, −R(t)N(t)\frac{1 − exp (−u_{0} ∫_{t−τ}^{q_{n}−τ} R(s)ds)}{1 + exp (−u_{0} ∫_{t−τ}^{q_{n}−τ} R(s)ds)}\}  (17)

for q_{n} − τ ≤ t ≤ q_{n}. Noting that 0 < −u_{0} < 1, one can easily see that

0 < −\frac{1}{u_{0}} ln (2e^{−u_{0}(1−u_{0}/2)} − 1) < 3.

There are two cases to consider. In the first case,

B ≡ ∫_{q_{n}−τ}^{q_{n}} R(s)ds ≤ (3 + ε) + \frac{1}{u_{0}} ln (2e^{−u_{0}(1−u_{0}/2)} − 1) .

By (17) and Lemma 1,

− ln N(q_{n}) ≤ −u_{0} ∫_{q_{n}−τ}^{q_{n}} R(s)ds ≤ −(3 + ε)u_{0} − ln (2e^{−u_{0}(1−u_{0}/2)} − 1)
≤ −(1 + ε)u_{0}.

In the second case,

B < ∫_{q_{n}−τ}^{q_{n}} R(s)ds ≤ 3 + ε.

We choose η_{n} ∈ (q_{n} − τ, q_{n}) such that B ≡ ∫_{q_{n}−τ}^{η_{n}} R(s)ds. Then by (7) and Lemma 1,

− ln N(q_{n})

≤ −u_{0} ∫_{q_{n}−τ}^{η_{n}} R(s)ds + ∫_{η_{n}}^{q_{n}} \frac{R(t) [exp (−u_{0} ∫_{t−τ}^{q_{n}−τ} R(s)ds) − 1]}{1 + exp (−u_{0} ∫_{t−τ}^{q_{n}−τ} R(s)ds)}dt

≤ −u_{0} ∫_{q_{n}−τ}^{η_{n}} R(s)ds + ∫_{η_{n}}^{q_{n}} R(t)\frac{exp [−u_{0} (3 + ε − ∫_{q_{n}−τ}^{t} R(s)ds)] − 1}{1 + exp [−u_{0} (3 + ε − ∫_{q_{n}−τ}^{t} R(s)ds)]}dt

= −u_{0} ∫_{q_{n}−τ}^{η_{n}} R(s)ds − ∫_{η_{n}}^{q_{n}} R(s)ds − \frac{2}{u_{0}} ln \frac{1 + exp [−u_{0} (3 + ε − ∫_{q_{n}−τ}^{q_{n}} R(s)ds)]}{1 + exp [−u_{0} (3 + ε − ∫_{q_{n}−τ}^{η_{n}} R(s)ds)]}

= (1 − u_{0})B − ∫_{η_{n}}^{q_{n}} R(s)ds + 2 (1 − \frac{u_{0}}{2} )

+\frac{2}{u_{0}} ln \frac{1 + exp [−u_{0} (3 + ε − ∫_{q_{n}−τ}^{q_{n}} R(s)ds)]}{2}

≤ 2 − (4 + ε)u_{0} + \frac{1 − u_{0}}{u_{0}} ln (2e^{−u_{0}(1−u_{0}/2)−1})
≤ −(2 + ε)u_{0},

where we have used the fact that

h(x) = −x − \frac{2}{u_{0}} ln \frac{1 + exp (−u_{0}(3 + ε − x))}{2}

is increasing on [0, 3 + ε]. In either cases, we have proved that − ln N(q_{n}) ≤ −(2 + ε)u_{0} for n = 1, 2, ... . Letting n → ∞ and ε → 0, we have

− ln v ≤ −2\frac{1 − u}{1 + u}.  (18)

Let y = −(1−u)/(1+u) and x = (1−v)/(1+v), then in view of (16), (18) and Lemma 3, we get x = y = 0. This then shows that u = v = 1. The proof is complete.

By means of methods similar to those in the proof of Theorem 1, and by noting that if λ ≥ 1, then (1 − x)/(1 + λx) ≤ (1 − x)/(1 + x) for x ≤ 1, and (1 − x)/(1 + λx) ≥ (1 − x)/(1 + x) for x ≥ 1, we may prove Theorem 2. The details are omitted.

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[3] Y. Kuang, Delay Differential Equations with Applications in Population Dynamics, Academic Press, Boston, 1993.

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