Asymptotic behaviour of positive solutions of the model which describes cell differentiation

Svetlin Georgiev Georgiev

University of Veliko Tarnovo, Department of Mathematical analysis and applications, 5000 Veliko Tarnovo, Bulgaria e-mail : sgg2000bg@yahoo.com

1. Introduction

In this paper we will study the asymptotic behaviour of positive solutions to the system

(1)  | \begin{matrix}x′_{1}(t) = \frac{A(t)}{1+x_{2}^{n}(t)} − x_{1}(t) \\ x′_{2}(t) = \frac{B(t)}{1+x_{1}^{n}(t)} − x_{2}(t), \end{matrix}

where A and B belong to C_{+} and C_{+} is the set of continuous functions g : R ⟶ R, which are bounded above and below by positive constants. n is fixed natural number. The system (1) describes cell differentiation, more precisely - its passes from one regime of work to other without loss of genetic information. The variables x_{1} and x_{2} make sense of concentration of specific metabolits. The parameters A and B reflect degree of development of base metabolism. The parameter n reflects the highest row of the repression’s reactions. For more details on the interpretation of (1) one may see [1]. With C_{∘} we denote the space of continuous and bounded functions g : R ⟶ R. For g ∈ C_{∘} we define

g_{L}(∞) = lim inf_{t⟶∞}g(t), g_{M}(∞) = lim sup_{t⟶∞}g(t),
g_{L} = inf{g(t) : t ∈ R}, g_{M} = sup{g(t) : t ∈ R}.

2. Preliminary results

Here and further next lemmas will pay important role.

Lemma 1.[2]Let g : (α, ∞) ⟶ R be a bounded and differentiable function. Then there exists a sequence {t_{n}}_{n=1}^{∞} such that t_{n}⟶_{n⟶∞}∞, g′(t_{n}) ⟶_{n⟶∞} 0, g(t_{n}) ⟶_{n⟶∞} g_{M}(∞) (resp. g(t_{n}) ⟶_{n⟶∞} g_{L}(∞)).

Lemma 2.[2]Let g ∈ C_{∘} be a differentiable function. Then there exists a sequence {t_{n}}_{n=1}^{∞} such that g′(t_{n}) ⟶_{n⟶∞} 0, g(t_{n}) ⟶_{n⟶∞} g_{M} (resp. g(t_{n}) ⟶_{n⟶∞} g_{L}).

Proposition 1.Let (x_{1}, x_{2}) be a positive solution of (1) and A(t), B(t) ∈ C_{+}. Then

\frac{A_{L}(∞)}{1 + B_{M}^{n}(∞)} ≤ x_{1L}(∞) ≤ x_{1M}(∞) ≤ A_{M}(∞),

\frac{B_{L}(∞)}{1 + A_{M}^{n}(∞)} ≤ x_{2L}(∞) ≤ x_{2M}(∞) ≤ B_{M}(∞).

Proof. From lemma 1 there exists a sequence {t_{m}}_{m=1}^{∞} ⊂ R for which t_{m}⟶_{m⟶∞}∞, x′_{1}(t_{m}) ⟶_{m⟶∞} 0, x_{1}(t_{m}) ⟶_{m⟶∞} x_{1M}(∞). Then from

x′_{1}(t_{m}) = \frac{A(t_{m})}{1+x_{2}^{n}(t_{m})} − x_{1}(t_{m}),

as m ⟶ ∞, we get

0 = lim_{m⟶∞} \frac{A(t_{m})}{1 + x_{2}^{n}(t_{m})} − x_{1M}(∞) ≤ A_{M}(∞) − x_{1M}(∞),

i. e.

x_{1M}(∞) ≤ A_{M}(∞).

Let now {t_{m}}_{m=1}^{∞} be a sequence of R such that t_{m}⟶_{m⟶∞}∞, x′_{2}(t_{m}) ⟶_{m⟶∞} 0, x_{2}(t_{m}) ⟶_{m⟶∞} x_{2L}(∞). From

x′_{2}(t_{m}) = \frac{B(t_{m})}{1 + x_{1}^{n}(t_{m})} − x_{2}(t_{m}),

as m ⟶ ∞, we find that

0 = lim_{m⟶∞} \frac{B(t_{m})}{1 + x_{1}^{n}(t_{m})} − x_{2L}(∞) ≥ \frac{B_{L}(∞)}{1 + A_{M}^{n}(∞)} − x_{2L}(∞)

or

x_{2L}(∞) ≥ \frac{B_{L}(∞)}{1 + A_{M}^{n}(∞)}.

Let {t_{m}}_{m=1}^{∞} ⊂ R is susch that t_{m}⟶_{m⟶∞}∞, x′_{2}(t_{m}) ⟶_{m⟶∞} 0, x_{2}(t_{m}) ⟶_{m⟶∞} x_{2M}(∞). From

x′_{2}(t_{m}) = \frac{B(t_{m})}{1 + x_{1}^{n}(t_{m})} − x_{2}(t_{m}),

as m ⟶ ∞, we get

0 = lim_{m⟶∞} \frac{B(t_{m})}{1 + x_{1}^{n}(t_{m})} − x_{2M}(∞) ≤ B_{M}(∞) − x_{2M}(∞).

Consequently

x_{2M}(∞) ≤ B_{M}(∞).

Let {t_{m}}_{m=1}^{∞} be a sequence of R such that t_{m}⟶_{m⟶∞}∞, x′_{1}(t_{m}) ⟶_{m⟶∞} 0, x_{1}(t_{m}) ⟶_{m⟶∞} x_{1L}(∞). From equality

x′_{1}(t_{m}) = \frac{A(t_{m})}{1 + x_{2}^{n}(t_{m})} − x_{1}(t_{m}),

as m ⟶ ∞, we get

0 = lim_{m⟶∞} \frac{A(t_{m})}{1 + x_{2}^{n}(t_{m})} − x_{1L}(∞) ≥ \frac{A_{L}(∞)}{1 + B_{M}^{n}(∞)} − x_{1L}(∞)

or

x_{1L}(∞) ≥ \frac{A_{L}(∞)}{1 + B_{M}^{n}(∞)}.

This completes the proof.

Remark. Proposition 1 shows that (1) is permanent, i. e. there exist positive constants α and β such that

0 < α ≤ lim inf_{t⟶∞}x_{i}(t) ≤ lim sup_{t⟶∞} x_{i}(t) ≤ β < ∞, i = 1, 2,

where (x_{1}(t), x_{2}(t)) is a positive solution of (1). In [3] was proved that permanence implies existence of positive periodic solutions of (1), when A(t) and B(t) are continuous positive periodic functions.

Let X_{1} be a positive solution of the equation

x′(t) = A(t) − x(t),

and X_{2} be a positive solution of the equation

x′(t) = B(t) − x(t).
Proposition 2.Let X_{1}, X_{2} be as above and A(t), B(t) ∈ C_{+}. Then
A_{L}(∞) ≤ X_{1L}(∞) ≤ X_{1M}(∞) ≤ A_{M}(∞),
B_{L}(∞) ≤ X_{2L}(∞) ≤ X_{2M}(∞) ≤ B_{M}(∞).

Proof. From lemma 1 there exists a sequence {t_{m}}_{m=1}^{∞} of R for which t_{m}⟶_{m⟶∞}∞, X′_{1}(t_{m}) ⟶_{m⟶∞} 0, X_{1}(t_{m}) ⟶_{m⟶∞} X_{1L}(∞). Then from

X′_{1}(t_{m}) = A(t_{m}) − X_{1}(t_{m}),

as m ⟶ ∞, we get

0 = lim_{m⟶∞} A(t_{m}) − X_{1L}(∞) ≥ A_{L}(∞) − X_{1L}(∞),

i. e.

X_{1L}(∞) ≥ A_{L}(∞).

Let {t_{m}}_{m=1}^{∞} be a sequence of R such that t_{m}⟶_{m⟶∞}∞, X′_{1}(t_{m}) ⟶_{m⟶∞} 0, X_{1}(t_{m}) ⟶_{m⟶∞} X_{1M}(∞). From

X′_{1}(t_{m}) = A(t_{m}) − X_{1}(t_{m}),

as m ⟶ ∞, we find that

0 = lim_{m⟶∞} A(t_{m}) − X_{1M}(∞) ≤ A_{M}(∞) − X_{1M}(∞)

or

X_{1M}(∞) ≤ A_{M}(∞).

In the same way we may prove other pair of inequalities.

3.Asymptotic behaviour of positive solutions

The results which are formulated and proved below are connected to (1) and to

(1_{∗})  | \begin{matrix}x′_{1}(t) = \frac{A_{∗}(t)}{1+x_{2}^{n}(t)} − x_{1}(t) \\ x′_{2}(t) = \frac{B_{∗}(t)}{1+x_{1}^{n}(t)} − x_{2}(t), \end{matrix}

where A_{∗}, B_{∗} ∈ C_{+} and A(t) − A_{∗}(t) ⟶_{t⟶∞} 0, B(t) − B_{∗}(t) ⟶_{t⟶∞} 0.

We notice that every solution to (1)(resp. (1_{∗})) with positive initial data x(t_{∘}) = (x_{1}(t_{∘}), x_{2}(t_{∘})) > 0 (x_{∗}(t_{∘}) = (x_{1∗}(t_{∘}), x_{2∗}(t_{∘})) > 0) is defined and positive in [t_{∘}, ∞).

Theorem 1. Let A, B, A_{∗}, B_{∗} ∈ C_{+} and

A(t) − A_{∗}(t) ⟶_{t⟶∞} 0, B(t) − B_{∗}(t) ⟶_{t⟶∞} 0.

Let also

\frac{n^{2}A_{M}^{n}(∞)B_{M}^{n}(∞)(1 + A_{M}^{n}(∞))^{2n}(1 + B_{M}^{n}(∞))^{2n}}{[A_{L}^{n}(∞) + (1 + B_{M}^{n}(∞))^{n}]^{2}[B_{L}^{n}(∞) + (1 + A_{M}^{n}(∞))^{n}]^{2}} < 1.

If (x_{1}(t), x_{2}(t)) and (x_{1∗}(t), x_{2∗}(t)) are positive solutions respectively of (1) and (1_{∗}), then (x_{1}(t) − x_{1∗}(t), x_{2}(t) − x_{2∗}(t)) ⟶_{t⟶∞} (0, 0).

Proof. Let h_{1}(t) = x_{1}(t) − x_{1∗}(t), h_{2}(t) = x_{2}(t) − x_{2∗}(t). We have

h′_{1}(t) = x′_{1}(t) − x′_{1∗}(t) =

= \frac{A(t)}{1 + x_{2}^{n}(t)} − x_{1}(t) − \frac{A_{∗}(t)}{1 + x_{2∗}^{n}(t)} + x_{1∗}(t) =

= −A(t)\frac{(x_{2}(t) − x_{2∗}(t))(x_{2}^{n−1}(t) + x_{2}^{n−2}(t)x_{2∗}(t) + ⋯ + x_{2∗}^{n−1}(t))}{(1 + x_{2}^{n}(t))(1 + x_{2∗}^{n}(t))}−

−h_{1}(t) + \frac{A(t) − A_{∗}(t)}{(1 + x_{2∗}^{n}(t))}.

Let

α(t) = A(t)\frac{x_{2}^{n−1}(t) + x_{2}^{n−2}(t)x_{2∗}(t) + ⋯ + x_{2∗}^{n−1}(t)}{(1 + x_{2}^{n}(t))(1 + x_{2∗}^{n}(t))}, β(t) = \frac{A(t) − A_{∗}(t)}{(1 + x_{2∗}^{n}(t))}.

We notice that β(t) ⟶_{t⟶∞} 0. For h_{1}(t) we get the equation

h′_{1}(t) = −h_{1}(t) − α(t)h_{2}(t) + β(t).

On the other hand

h′_{2}(t) = x′_{2}(t) − x′_{2∗}(t) =

= \frac{B(t)}{1 + x_{1}^{n}(t)} − x_{2}(t) − \frac{B_{∗}(t)}{1 + x_{1∗}^{n}(t)} + x_{2∗}(t) =

= −B(t)\frac{(x_{1}(t) − x_{1∗}(t))(x_{1}^{n−1}(t) + x_{1}^{n−2}(t)x_{1∗}(t) + ⋯ + x_{1∗}^{n−1}(t))}{(1 + x_{1}^{n}(t))(1 + x_{1∗}^{n}(t))}−

−h_{2}(t) + \frac{B(t) − B_{∗}(t)}{(1 + x_{1∗}^{n}(t))}.

Let

γ(t) = B(t)\frac{x_{1}^{n−1}(t) + x_{1}^{n−2}(t)x_{1∗}(t) + ⋯ + x_{1∗}^{n−1}(t)}{(1 + x_{1}^{n}(t))(1 + x_{1∗}^{n}(t))}, δ(t) = \frac{B(t) − B_{∗}(t)}{(1 + x_{1∗}^{n}(t))},

δ(t) ⟶_{t⟶∞} 0. Then

h′_{2}(t) = −γ(t)h_{1}(t) − h_{2}(t) + δ(t).

For h_{1}(t) and h_{2}(t) we find the system

| \begin{matrix}h′_{1}(t) = −h_{1}(t) − α(t)h_{2}(t) + β(t) \\ h′_{2}(t) = −γ(t)h_{1}(t) − h_{2}(t) + δ(t). \end{matrix}

Let h(t) = (h_{1}(t), h_{2}(t)) and |h|(t) = |h(t)|. We assume that |h_{1}|_{M}(∞) > 0. From lemma 1 there exists a sequence {t_{m}}_{m=1}^{∞} of R such that t_{m} ⟶_{m⟶∞} ∞, h′_{1}(t_{m}) ⟶_{m⟶∞} 0, |h_{1}|(t_{m}) ⟶_{m⟶∞} |h_{1}|_{M}(∞). From

|h′_{1}(t_{m})| = | − h_{1}(t_{m}) − α(t_{m})h_{2}(t_{m}) + β(t_{m})|,

as m ⟶ ∞, we have

0 ≥ |h_{1}|_{M}(∞) − α_{M}(∞)|h_{2}|_{M}(∞),

i. e.

(2)  |h_{1}|_{M}(∞) ≤ α_{M}(∞)|h_{2}|_{M}(∞).

Since |h_{1}|_{M}(∞) > 0 then |h_{2}|_{M}(∞) > 0. Let now {t_{m}}_{m=1}^{∞} ⊂ R is such that t_{m} ⟶_{m⟶∞} ∞, h′_{2}(t_{m}) ⟶_{m⟶∞} 0, |h_{2}|(t_{m}) ⟶_{m⟶∞} |h_{2}|_{M}(∞). As m ⟶ ∞, from

|h′_{2}(t_{m})| = | − γ(t_{m})h_{1}(t_{m}) − h_{2}(t_{m}) + δ(t_{m})|,

we get

0 ≥ |h_{2}|_{M}(∞) − γ_{M}(∞)|h_{1}|_{M}(∞)

or

|h_{2}|_{M}(∞) ≤ γ_{M}(∞)|h_{1}|_{M}(∞).

From last inequality and (2) we find that

|h_{1}|_{M}(∞)|h_{2}|_{M}(∞) ≤ α_{M}(∞)γ_{M}(∞)|h_{1}|_{M}(∞)|h_{2}|_{M}(∞),

from where

1 ≤ α_{M}(∞)γ_{M}(∞).

Since

α_{M}(∞) = (A(t)\frac{x_{2}^{n−1}(t) + x_{2}^{n−2}(t)x_{2∗}(t) + ⋯ + x_{2∗}^{n−1}(t)}{(1 + x_{2}^{n}(t))(1 + x_{2∗}^{n}(t))})_{M} (∞) ≤

≤ A_{M}(∞).\frac{n.B_{M}^{n−1}(∞)}{(1 + \frac{B_{L}^{n}(∞)}{(1+A_{M}^{n} (∞))^{n}} )^{2}} = \frac{n.A_{M}(∞).B_{M}^{n−1}(∞)(1 + A_{M}^{n}(∞))^{2n}}{[(1 + A_{M}^{n}(∞))^{n} + B_{L}^{n}(∞)]^{2}} ,

γ_{M}(∞) = (B(t)\frac{x_{1}^{n−1}(t) + x_{1}^{n−2}(t)x_{1∗}(t) + ⋯ + x_{1∗}^{n−1}(t)}{(1 + x_{1}^{n}(t))(1 + x_{1∗}^{n}(t))})_{M} (∞) ≤

≤ B_{M}(∞).\frac{n.A_{M}^{n−1}(∞)}{(1 + \frac{A_{L}^{n}(∞)}{(1+B_{M}^{n} (∞))^{n}} )^{2}} = \frac{n.B_{M}(∞).A_{M}^{n−1}(∞)(1 + B_{M}^{n}(∞))^{2n}}{[(1 + B_{M}^{n}(∞))^{n} + A_{L}^{n}(∞)]^{2}} .

Therefore we get the contradiction

1 ≤ \frac{n^{2}.A_{M}^{n}(∞).B_{M}^{n}(∞)(1 + A_{M}^{n}(∞))^{2n}(1 + B_{M}^{n}(∞))^{2n}}{[(1 + B_{M}^{n}(∞))^{n} + A_{L}^{n}(∞)]^{2}.[(1 + A_{M}^{n}(∞))^{n} + B_{L}^{n}(∞)]^{2}} .

The proof is complete.

Let

r_{1} = \frac{A_{M}^{2}(∞)}{A_{L}^{2}(∞)}, r_{2} = \frac{B_{M}^{2}(∞)}{B_{L}^{2}(∞)},

p_{1} = \frac{1}{r_{1}} \frac{1}{1 + B_{M}^{n}(∞)r_{2}^{n}} , p_{2} = \frac{1}{r_{2}} \frac{1}{1 + A_{M}^{n}(∞)r_{1}^{n}} .

Theorem 2.Let A, B, A_{∗}, B_{∗} ∈ C_{+}, A(t)−A_{∗}(t) ⟶_{t⟶∞} 0, B(t)−B_{∗}(t) ⟶_{t⟶∞} 0. If (x_{1}(t), x_{2}(t)) and (x_{1∗}(t), x_{2∗}(t)) are positive solutions respectively to (1) and (1_{∗}) and

\frac{n^{2}r_{1}^{n}r_{2}^{n}A_{M}^{n}(∞)B_{M}^{n}(∞)}{(1 + A_{L}^{n}(∞)p_{1}^{n})^{2}(1 + B_{L}^{n}(∞)p_{2}^{n})^{2}} < 1,

then (x_{1}(t) − x_{1∗}(t), x_{2}(t) − x_{2∗}(t)) ⟶_{t⟶∞} (0, 0).

Proof. Let x = \frac{x_{1}}{X_{1}} , y = \frac{x_{2}}{X_{2}} , where X_{1} and X_{2} as in proposition 2. Then

x′(t) = \frac{1}{X_{1}(t)}.x′_{1}(t) − \frac{x_{1}(t)}{X_{1}^{2}(t)}.X′_{1}(t) =

= \frac{1}{X_{1}(t)}. [\frac{A(t)}{1 + x_{2}^{n}(t)} − x_{1}(t)] − \frac{x_{1}(t)}{X_{1}^{2}(t)}[A(t) − X_{1}(t)] =

= \frac{1}{X_{1}(t)}.\frac{A(t)}{1 + X_{2}^{n}(t)y^{n}(t)} − \frac{A(t)}{X_{1}(t)}.\frac{x_{1}(t)}{X_{1}(t)} = −\frac{A(t)}{X_{1}(t)}.x(t) + \frac{A(t)}{X_{1}(t)[1 + X_{2}^{n}(t)y^{n}(t)]},

i. e.

x′(t) = −\frac{A(t)}{X_{1}(t)}.x(t) + \frac{A(t)}{X_{1}(t)[1 + X_{2}^{n}(t)y^{n}(t)]}.

y′(t) = \frac{1}{X_{2}(t)}.x′_{2}(t) − \frac{x_{2}(t)}{X_{2}^{2}(t)}.X′_{2}(t) =

= \frac{1}{X_{2}(t)}. [\frac{B(t)}{1 + x_{1}^{n}(t)} − x_{2}(t)] − \frac{x_{2}(t)}{X_{2}^{2}(t)}[B(t) − X_{2}(t)] =

= \frac{1}{X_{2}(t)}.\frac{B(t)}{1 + X_{1}^{n}(t)x^{n}(t)} − \frac{B(t)}{X_{2}(t)}.\frac{x_{2}(t)}{X_{2}(t)} = −\frac{B(t)}{X_{2}(t)}.y(t) + \frac{B(t)}{X_{2}(t)[1 + X_{1}^{n}(t)x^{n}(t)]},

i. e.

y′(t) = −\frac{B(t)}{X_{2}(t)}.y(t) + \frac{B(t)}{X_{2}(t)[1 + X_{1}^{n}(t)x^{n}(t)]}.

Consequently for x(t) and y(t) we get the system

(3)  | \begin{matrix}x′(t) = −\frac{A(t)}{X_{1}(t)}.x(t) + \frac{A(t)}{X_{1}(t)[1+X_{2}^{n}(t)y^{n}(t)]} \\ y′(t) = −\frac{B(t)}{X_{2}(t)}.y(t) + \frac{B(t)}{X_{2}(t)[1+X_{1}^{n}(t)x^{n}(t)]}. \end{matrix}

Let us consider the system

(3_{∗})  | \begin{matrix}x′(t) = −\frac{A_{∗}(t)}{X_{1∗}(t)}.x(t) + \frac{A_{∗}(t)}{X_{1∗}(t)[1+X_{2∗}^{n}(t)y^{n}(t)]} \\ y′(t) = −\frac{B_{∗}(t)}{X_{2∗}(t)}.y(t) + \frac{B_{∗}(t)}{X_{2∗}(t)[1+X_{1∗}^{n}(t)x^{n}(t)]}, \end{matrix}

where X_{1∗} and X_{2∗} be a positive solutions respectively to x′(t) = A_{∗}(t) − x(t),

x′(t) = B_{∗}(t) − x(t).

Let (x(t), y(t)) be a positive solution to (3) and let (x_{∗}(t), y_{∗}(t)) be a positive solution to (3_{∗}). At first, we will prove that

(x(t) − x_{∗}(t), y(t) − y_{∗}(t)) ⟶_{t⟶∞} (0, 0).

Let h(t) = x(t) − x_{∗}(t), k(t) = y(t) − y_{∗}(t). We have

h′(t) = x′(t) − x′_{∗}(t) =

= −\frac{A(t)}{X_{1}(t)}.x(t) + \frac{A(t)}{X_{1}(t)[1 + X_{2}^{n}(t)y^{n}(t)]} + \frac{A_{∗}(t)}{X_{1∗}(t)}.x_{∗}(t) − \frac{A_{∗}(t)}{X_{1∗}(t)[1 + X_{2∗}^{n}(t)y_{∗}^{n}(t)]} =

= −\frac{A(t)}{X_{1}(t)}.x(t) + \frac{A(t)}{X_{1}(t)}.x_{∗}(t) − \frac{A(t)}{X_{1}(t)}.x_{∗}(t) + \frac{A(t)}{X_{1∗}(t)}.x_{∗}(t) − \frac{A(t)}{X_{1∗}(t)}.x_{∗}(t)+

+\frac{A_{∗}(t)}{X_{1∗}(t)}.x_{∗}(t) + \frac{A_{∗}(t)}{X_{1∗}(t)[1 + X_{2∗}^{n}(t)y^{n}(t)]} − \frac{A_{∗}(t)}{X_{1∗}(t)[1 + X_{2∗}^{n}(t)y_{∗}^{n}(t)]}

−\frac{A_{∗}(t)}{X_{1∗}(t)[1 + X_{2∗}^{n}(t)y^{n}(t)]} + \frac{A(t)}{X_{1}(t)[1 + X_{2}^{n}(t)y^{n}(t)]} =

= −\frac{A(t)}{X_{1}(t)}[x(t) − x_{∗}(t)] − A(t)x_{∗}(t) [\frac{1}{X_{1}(t)} − \frac{1}{X_{1∗}(t)}] − \frac{x_{∗}(t)}{X_{1∗}(t)}[A(t) − A_{∗}(t)]+

EJQTDE, 2000 No. 6, p. 8

+\frac{A(t)}{X_{1}(t)[1 + X_{2}^{n}(t)y^{n}(t)]} − \frac{A_{∗}(t)}{X_{1∗}(t)[1 + X_{2∗}^{n}(t)y^{n}(t)]}−

−\frac{A_{∗}(t)X_{2∗}^{n}(t)[y(t) − y_{∗}(t)][y^{n−1}(t) + y^{n−2}(t)y_{∗}(t) + ⋯ + y_{∗}^{n−1}(t)]}{X_{1∗}(t)[1 + X_{2∗}^{n}(t)y_{∗}^{n}(t)][1 + X_{2∗}^{n}(t)y^{n}(t)]}.

Let

α(t) = \frac{A(t)}{X_{1}(t)},

β(t) = \frac{A_{∗}(t)X_{2∗}^{n}(t)[y^{n−1}(t) + y^{n−2}(t)y_{∗}(t) + ⋯ + y_{∗}^{n−1}(t)]}{X_{1∗}(t)[1 + X_{2∗}^{n}(t)y_{∗}^{n}(t)][1 + X_{2∗}^{n}(t)y^{n}(t)]},

γ(t) = A(t).x_{∗}(t) [\frac{1}{X_{1}(t)} − \frac{1}{X_{1∗}(t)}] + \frac{x_{∗}(t)}{X_{1∗}(t)}[A(t) − A_{∗}(t)]−

−\frac{A(t)}{X_{1}(t)[1 + X_{2}^{n}(t)y^{n}(t)]} + \frac{A_{∗}(t)}{X_{1∗}(t)[1 + X_{2∗}^{n}(t)y^{n}(t)]}.

Since X_{1}(t) − X_{1∗}(t) ⟶_{t⟶∞} 0, A(t) − A_{∗}(t) ⟶_{t⟶∞} 0, then γ(t) ⟶_{t⟶∞} 0. For h(t) we get the equation

h′(t) = −α(t)h(t) − β(t)k(t) − γ(t).

In the same way we find the equation

k′(t) = −δ(t)h(t) − σ(t)k(t) − ρ(t),

where

σ(t) = \frac{B(t)}{X_{2∗}(t)},

δ(t) = \frac{B_{∗}(t)X_{1∗}^{n}(t)[x^{n−1}(t) + x^{n−2}(t)x_{∗}(t) + ⋯ + x_{∗}^{n−1}(t)]}{X_{2∗}(t)[1 + X_{1∗}^{n}(t)x_{∗}^{n}(t)][1 + X_{1∗}^{n}(t)x^{n}(t)]},

ρ(t) = B(t).y_{∗}(t) [\frac{1}{X_{2}(t)} − \frac{1}{X_{2∗}(t)}] + \frac{y_{∗}(t)}{X_{2∗}(t)}[B(t) − B_{∗}(t)]−

−\frac{B(t)}{X_{2}(t)[1 + X_{1}^{n}(t)x^{n}(t)]} + \frac{B_{∗}(t)}{X_{2∗}(t)[1 + X_{1∗}^{n}(t)x^{n}(t)]}.

Since X_{2}(t) − X_{2∗}(t) ⟶_{t⟶∞} 0, B(t) − B_{∗}(t) ⟶_{t⟶∞} 0, then ρ(t) ⟶_{t⟶∞} 0. Therefore for h(t) and k(t) we get the system

| \begin{matrix}h′(t) & = & −α(t)h(t) − β(t)k(t) − γ(t) \\ k′(t) & = & −δ(t)h(t) − σ(t)k(t) − ρ(t). \end{matrix}

We assume that |h|_{M}(∞) > 0. From lemma 1 there exists a sequence {t_{m}}_{m=1}^{∞} of R such that t_{m} ⟶_{m⟶∞} ∞, |h′(t_{m})| ⟶_{m⟶∞} 0, |h|(t_{m}) ⟶_{m⟶∞} |h|_{M}(∞). From

|h′(t_{m})| = | − α(t_{m})h(t_{m}) − β(t_{m})k(t_{m}) − γ(t_{m})|,

as m ⟶ ∞, we find that

0 ≥ |h|_{M}(∞) − (\frac{β}{α})_{M} (∞)|k|_{M}(∞),

i. e.

(4)  |h|_{M}(∞) ≤ (\frac{β}{α})_{M} (∞)|k|_{M}(∞).

Consequently |k|_{M}(∞) > 0. Let {t_{m}}_{m=1}^{∞} be a sequence of R for which t_{m} ⟶_{m⟶∞} ∞, |k′(t_{m})| ⟶_{m⟶∞} 0, |k|(t_{m}) ⟶_{m⟶∞} |k|_{M}(∞). As m ⟶ ∞ in

|k′(t_{m})| = | − δ(t_{m})h(t_{m}) − σ(t_{m})k(t_{m}) − ρ(t_{m})|,

we get

0 ≥ |k|_{M}(∞) − (\frac{δ}{σ} )_{M} (∞)|h|_{M}(∞),

i. e.

(5)  |k|_{M}(∞) ≤ (\frac{δ}{σ} )_{M} (∞)|h|_{M}(∞).

From here and from (4) we find that

(6)  1 ≤ (\frac{β}{α})_{M} (∞) (\frac{δ}{σ} )_{M} (∞).

Lemma 3.Let (x, y) be a positive solution of (3). The following inequalities hold

p_{1} ≤ x_{L}(∞) ≤ x_{M}(∞) ≤ r_{1},
p_{2} ≤ y_{L}(∞) ≤ y_{M}(∞) ≤ r_{2}.

Proof of Lemma 3. Let {t_{m}}_{m=1}^{∞} be a sequence of R for which t_{m} ⟶_{m⟶∞} ∞, x′(t_{m}) ⟶_{m⟶∞} 0, x(t_{m}) ⟶_{m⟶∞} x_{M}(∞). From the equality

x′(t_{m}) = −\frac{A(t_{m})}{X_{1}(t_{m})}x(t_{m}) + \frac{A(t_{m})}{X_{1}(t_{m})[1 + X_{2}^{n}(t_{m})y^{n}(t_{m})]},

as m ⟶ ∞, we get

0 ≤ −\frac{A_{L}(∞)}{X_{1M}(∞)}x_{M}(∞) + \frac{A_{M}(∞)}{X_{1L}(∞)},

i. e.

x_{M}(∞) ≤ \frac{A_{M}^{2}(∞)}{A_{L}^{2}(∞)}

or x_{M}(∞) ≤ r_{1}. In the same way we get y_{M}(∞) ≤ r_{2}. Let {t_{m}}_{m=1}^{∞} be a sequence of R for which t_{m} ⟶_{m⟶∞} ∞, x′(t_{m}) ⟶_{m⟶∞} 0, x(t_{m}) ⟶_{m⟶∞} x_{L}(∞). From

x′(t_{m}) = −\frac{A(t_{m})}{X_{1}(t_{m})}x(t_{m}) + \frac{A(t_{m})}{X_{1}(t_{m})[1 + X_{2}^{n}(t_{m})y^{n}(t_{m})]},

as m ⟶ ∞, we find that

0 ≥ −\frac{A_{M}(∞)}{X_{1L}(∞)}x_{L}(∞) + \frac{A_{L}(∞)}{X_{1M}(∞)}.\frac{1}{1 + X_{2M}^{n}(∞)y_{M}^{n}(∞)},

i. e.

x_{L}(∞) ≥ \frac{A_{L}(∞)}{A_{M}(∞)}.\frac{X_{1L}(∞)}{X_{1M}(∞)} \frac{1}{1 + X_{2M}^{n}(∞)y_{M}^{n}(∞)} ≥

≥ (\frac{A_{L}(∞)}{A_{M}(∞)})^{2} .\frac{1}{1 + B_{M}^{n}(∞)r_{2}^{n}} = p_{1},

or x_{L}(∞) ≥ p_{1}. In the same way we get y_{L}(∞) ≥ p_{2}.

Now we will continue with the proof of Theorem 2. We have

(\frac{β}{α})_{M} (∞) ≤ \frac{β_{M}(∞)}{α_{L}(∞)}.

Using the lemma 3 we get

(\frac{β}{α})_{M} (∞) ≤ (\frac{A_{M}(∞)}{A_{L}(∞)})^{2} .\frac{nr_{2}^{n−1}(∞)B_{M}^{n}(∞)}{(1 + B_{L}^{n}(∞)p_{2}^{n})^{2}}.

On the other hand

δ_{M}(∞) ≤ \frac{B_{M}(∞)nr_{1}^{n−1}A_{M}^{n}(∞)}{B_{L}(∞)[1 + A_{L}^{n}(∞)p_{1}^{n}]^{2}} ,

σ_{L}(∞) ≥ \frac{B_{L}(∞)}{X_{2M}(∞)} ≥ \frac{B_{L}(∞)}{B_{M}(∞)}.

Consequently

(\frac{δ}{σ} )_{M} (∞) ≤ (\frac{B_{M}(∞)}{B_{L}(∞)})^{2} .\frac{nr_{1}^{n−1}A_{M}^{n}(∞)}{[1 + A_{L}^{n}(∞)p_{1}^{n}]^{2}} .

From (6) we get the contradiction

1 ≤ \frac{n^{2}r_{1}^{n}r_{2}^{n}A_{M}^{n}(∞)B_{M}^{n}(∞)}{[1 + A_{L}^{n}(∞)p_{1}^{n}]^{2}[1 + B_{L}^{n}(∞)p_{2}^{n}]^{2}}

with the condition of Theorem 2. Consequently |h|_{M}(∞) = 0. From here x(t) − x_{∗}(t) ⟶_{t⟶∞} 0. We assume that |k|_{M}(∞) > 0. From (5) we find that |h|_{M}(∞) > 0, which is contradiction. Consequently |k|_{M}(∞) = 0. From here y(t) − y_{∗}(t) ⟶_{t⟶∞} 0. Since

x(t) − x_{∗}(t) = \frac{x_{1}(t)}{X_{1}(t)} − \frac{x_{1∗}(t)}{X_{1∗}(t)} ⟶_{t⟶∞} 0

and X_{1}(t) − X_{1∗}(t) ⟶_{t⟶∞} 0, then x_{1}(t) − x_{1∗}(t) ⟶_{t⟶∞} 0. From

y(t) − y_{∗}(t) = \frac{x_{2}(t)}{X_{2}(t)} − \frac{x_{2∗}(t)}{X_{2∗}(t)} ⟶_{t⟶∞} 0

and X_{2}(t) − X_{2∗}(t) ⟶_{t⟶∞} 0, therefores x_{2}(t) − x_{2∗}(t) ⟶_{t⟶∞} 0.

Theorem 3. Let A, B ∈ C_{+}, and

\frac{n^{2}A_{M}^{n}B_{M}^{n}(1 + A_{M}^{n})^{2n}(1 + B_{M}^{n})^{2n}}{[A_{L}^{n} + (1 + B_{M}^{n}))^{n}]^{2}[B_{L}^{n} + (1 + A_{M}^{n})^{n}]^{2}} < 1.

Then the system (1) has exactly one positive solution (x_{10}(t), x_{20}(t)) such that x_{10}(t) ∈ C_{+}, x_{20}(t) ∈ C_{+}. In particular

(x_{1}(t) − x_{10}(t), x_{2}(t) − x_{20}(t)) ⟶_{t⟶∞} (0, 0),

for every positive solution (x_{1}(t), x_{2}(t)) of (1).

Proof. In C_{+} we define the metric

∥g − f∥ = sup{|g(t) − f(t)| : t ∈ R} for f, g ∈ C_{+}.

Let

X_{x_{1}} = \{x(t) ∈ C_{+} : \frac{A_{L}}{1 + B_{M}^{n}} ≤ x(t) ≤ A_{M}\},

X_{x_{2}} = \{x(t) ∈ C_{+} : \frac{B_{L}}{1 + A_{M}^{n}} ≤ x(t) ≤ B_{M}\}.

Let also x_{2}(t) ∈ X_{x_{2}} be fixed and x_{1x_{2}}(t) is the unique solution to the problem

u′(t) = \frac{A(t)}{1 + x_{2}^{n}(t)} − u(t), u ∈ C_{+}.

If x_{1x_{2}}(t) attains its maximum when t = t_{1}, then x′_{1x_{2}}(t_{1}) = 0. From

0 = x′_{1x_{2}}(t_{1}) = \frac{A(t_{1})}{1 + x_{2}^{n}(t_{1})} − x_{1x_{2}}(t_{1}),

we get

x_{1x_{2}}(t) ≤ A_{M} ∀t ∈ R.

If x_{1x_{2}}(t) attains its minimum in t = t_{1}, then x′_{1x_{2}}(t_{1}) = 0 and

0 = x′_{1x_{2}}(t_{1}) = \frac{A(t_{1})}{1 + x_{2}^{n}(t_{1})} − x_{1x_{2}}(t_{1}) ≥ \frac{A_{L}}{1 + B_{M}^{n}} − x_{1x_{2}}(t_{1}).

Consequently

x_{1x_{2}}(t) ≥ \frac{A_{L}}{1 + B_{M}^{n}} , ∀t ∈ R,

i. e. x_{1x_{2}}(t) ∈ X_{x_{1}}, ∀t ∈ R. From here, the map

P : X_{x_{2}} ⟶ X_{x_{1}}, P x_{2} = x_{1x_{2}},

is well defined. Analogously, the map

Q : X_{x_{1}} ⟶ X_{x_{2}}, Qx_{1}(t) = x_{2x_{1}}(t),

where x_{2x_{1}}(t) is the unique positive bound solution of

u′(t) = \frac{B(t)}{1 + x_{1}^{n}(t)} − u(t), u ∈ C_{+},

is well defined. Let v_{0}(t), v_{1}(t) ∈ X_{x_{2}}. Then

[P (v_{0}) − P (v_{1})]′ = P′(v_{0}) − P′(v_{1}) =

= \frac{A(t)}{1 + v_{0}^{n}(t)} − P (v_{0}) − \frac{A(t)}{1 + v_{1}^{n}(t)} + P (v_{1}) =

= \frac{A(t)[v_{1}^{n}(t) − v_{0}^{n}(t)]}{[1 + v_{0}^{n}(t)][1 + v_{1}^{n}(t)]} − [P (v_{0}) − P (v_{1})] =

= −\frac{A(t)[v_{1}^{n−1}(t) + v_{1}^{n−2}(t)v_{0}(t) + ⋯ + v_{0}^{n−1}(t)]}{[1 + v_{0}^{n}(t)][1 + v_{1}^{n}(t)]}(v_{0}(t) − v_{1}(t)) − [P (v_{0}) − P (v_{1})].

From lemma 2 there exists a sequence {t_{m}}_{m=1}^{∞} of R for which [P (v_{0}) − P (v_{1})]′(t_{m}) ⟶_{m⟶∞} 0, |P (v_{0}) − P (v_{1})|(t_{m}) ⟶_{m⟶∞} ∥P (v_{0}) − P (v_{1})∥. From

|[P (v_{0}) − P (v_{1})]′(t_{m})| =

= | − \frac{A(t_{m})[v_{1}^{n−1}(t_{m}) + v_{1}^{n−2}(t_{m})v_{0}(t_{m}) + ⋯ + v_{0}^{n−1}(t_{m})]}{[1 + v_{0}^{n}(t_{m})][1 + v_{1}^{n}(t_{m})]}(v_{0}(t_{m}) − v_{1}(t_{m}))−

−[P (v_{0}) − P (v_{1})](t_{m})|,

as m ⟶ ∞, we get

∥P (v_{0}) − P (v_{1})∥ ≤ \frac{n.A_{M}.B_{M}^{n−1}(1 + A_{M}^{n})^{2n}}{[B_{L}^{n} + (1 + A_{M}^{n})^{n}]^{2}} ∥v_{0} − v_{1}∥.

In the same way, we find that for v_{0}(t), v_{1}(t) ∈ X_{x_{1}}

∥Q(v_{0}) − Q(v_{1})∥ ≤ \frac{n.B_{M}.A_{M}^{n−1}(1 + B_{M}^{n})^{2n}}{[A_{L}^{n} + (1 + B_{M}^{n})^{n}]^{2}} ∥v_{0} − v_{1}∥.

Let R = Q.P : X_{x_{2}} ⟶ X_{x_{2}}. The map R is well defined and we have for it(v_{0}, v_{1} ∈ X_{x_{2}})

∥R(v_{0}) − R(v_{1})∥ = ∥Q.P (v_{0}) − Q.P (v_{1})∥ ≤

≤ \frac{n.B_{M}.A_{M}^{n−1}(1 + B_{M}^{n})^{2n}}{[A_{L}^{n} + (1 + B_{M}^{n})^{n}]^{2}} ∥P (v_{0}) − P (v_{1})∥ ≤

≤ \frac{n^{2}.A_{M}^{n}.B_{M}^{n}(1 + A_{M}^{n})^{2n}.(1 + B_{M}^{n})^{2n}}{[A_{L}^{n} + (1 + B_{M}^{n})^{n}]^{2}.[B_{L}^{n} + (1 + A_{M}^{n})^{n}]^{2}} ∥v_{0} − v_{1}∥ < ∥v_{0} − v_{1}∥.

Consequently R is a contraction map. Let x_{20} ∈ X_{x_{2}} be a unique fixed point of R. Then (P x_{20}, x_{20}) is a solution to (1) in C_{+} ⨂ C_{+}. If (x_{1}, x_{2}) is another solution to (1) in C_{+} ⨂ C_{+}, then x_{2} ∈ X_{x_{2}} and x_{1} ∈ X_{x_{1}}. But P (x_{2}) = x_{1}, Q(x_{1}) = x_{2} and R(x_{2}) = x_{2}. From here x_{2} = x_{20} and x_{1} = x_{10}. Therefore (1) has exactly one solution in C_{+} ⨂ C_{+}. The rest of the proof follows from theorem 1 since its conditions are hold.

Theorem 4. Let A, B, A_{∗}, B_{∗} ∈ C_{+},

A(t) − A_{∗}(t) ⟶_{t⟶∞} 0, B(t) − B_{∗}(t) ⟶_{t⟶∞} 0

and

\frac{nA_{M}^{n−1}(∞)B_{M}(∞)(1 + A_{M}^{n}(∞))(1 + B_{M}^{n}(∞))^{2n}}{B_{L}(∞)((1 + B_{M}^{n}(∞))^{n} + A_{L}^{n}(∞))^{2}} < \frac{A_{L}(∞)}{A_{M}^{2}(∞)(1 + B_{M}^{n}(∞))},

\frac{nB_{M}^{n−1}(∞)A_{M}(∞)(1 + B_{M}^{n}(∞))(1 + A_{M}^{n}(∞))^{2n}}{A_{L}(∞)((1 + A_{M}^{n}(∞))^{n} + B_{L}^{n}(∞))^{2}} < \frac{B_{L}(∞)}{B_{M}^{2}(∞)(1 + A_{M}^{n}(∞))}.

Then, if (x_{1}(t), x_{2}(t)) and (x_{1∗}(t), x_{2∗}(t)) are positive solutions respectively to (1) and (1_{∗}), then (x_{1}(t) − x_{1∗}(t), x_{2}(t) − x_{2∗}(t)) ⟶_{t⟶∞} (0, 0).

Proof. Let (x_{1}(t), x_{2}(t)), (x_{1∗}(t), x_{2∗}(t)) be positive solutions respectively to (1) and (1_{∗}), which are defined in (T, ∞)(T > 0). For t > T we define

r(t) = |ln\frac{x_{1∗}(t)}{x_{1}(t)}| + |ln\frac{x_{2∗}(t)}{x_{2}(t)}| .

There exists X ⊂ (T, ∞) such that r(t) is differentiable in X and for t ∈ X we have

r′(t) = sg ln \frac{x_{1∗}(t)}{x_{1}(t)} (ln \frac{x_{1∗}(t)}{x_{1}(t)})′ + sg ln \frac{x_{2∗}(t)}{x_{2}(t)} (ln \frac{x_{2∗}(t)}{x_{2}(t)})′ =

= sg ln \frac{x_{1∗}(t)}{x_{1}(t)}.\frac{x_{1}(t)}{x_{1∗}(t)} \frac{x′_{1∗}(t)x_{1}(t) − x_{1∗}(t)x′_{1}(t)}{x_{1}^{2}(t)}+

+sg ln \frac{x_{2∗}(t)}{x_{2}(t)}.\frac{x_{2}(t)}{x_{2∗}(t)} \frac{x′_{2∗}(t)x_{2}(t) − x_{2∗}(t)x′_{2}(t)}{x_{2}^{2}(t)} =

= sg(x_{1∗}(t) − x_{1}(t)) [\frac{x′_{1∗}(t)}{x_{1∗}(t)} − \frac{x′_{1}(t)}{x_{1}(t)}] + sg(x_{2∗}(t) − x_{2}(t)) [\frac{x′_{2∗}(t)}{x_{2∗}(t)} − \frac{x′_{2}(t)}{x_{2}(t)}] =

= sg(x_{1∗}(t) − x_{1}(t))[\frac{A_{∗}(t)}{x_{1∗}(t)[1 + x_{2∗}^{n}(t)]} − 1 − \frac{A(t)}{x_{1}(t)[1 + x_{2}^{n}(t)]} + 1]+

+sg(x_{2∗}(t) − x_{2}(t))[\frac{B_{∗}(t)}{x_{2∗}(t)[1 + x_{1∗}^{n}(t)]} − 1 − \frac{B(t)}{x_{2}(t)[1 + x_{1}^{n}(t)]} + 1] =

= sg(x_{1∗}(t) − x_{1}(t))[\frac{A_{∗}(t)}{x_{1∗}(t)[1 + x_{2∗}^{n}(t)]} − \frac{A_{∗}(t)}{x_{1∗}(t)[1 + x_{2}^{n}(t)]} + \frac{A_{∗}(t)}{x_{1∗}(t)[1 + x_{2}^{n}(t)]}−

−\frac{A_{∗}(t)}{x_{1}(t)[1 + x_{2}^{n}(t)]} + \frac{A_{∗}(t)}{x_{1}(t)[1 + x_{2}^{n}(t)]} − \frac{A(t)}{x_{1}(t)[1 + x_{2}^{n}(t)]}]+

+sg(x_{2∗}(t) − x_{2}(t))[\frac{B_{∗}(t)}{x_{2∗}(t)[1 + x_{1∗}^{n}(t)]} − \frac{B_{∗}(t)}{x_{2∗}(t)[1 + x_{1}^{n}(t)]} + \frac{B_{∗}(t)}{x_{2∗}(t)[1 + x_{1}^{n}(t)]}−

−\frac{B_{∗}(t)}{x_{2}(t)[1 + x_{1}^{n}(t)]} + \frac{B_{∗}(t)}{x_{2}(t)[1 + x_{1}^{n}(t)]} − \frac{B(t)}{x_{2}(t)[1 + x_{1}^{n}(t)]}] =

= sg(x_{1∗}(t) − x_{1}(t))[\frac{A_{∗}(t)}{x_{1∗}(t)}.\frac{x_{2}^{n}(t) − x_{2∗}^{n}(t)}{(1 + x_{2∗}^{n}(t))(1 + x_{2}^{n}(t)}+

+\frac{A_{∗}(t)(x_{1}(t) − x_{1∗}(t))}{x_{1}(t)x_{1∗}(t))(1 + x_{2}^{n}(t))} + \frac{A_{∗}(t) − A(t)}{x_{1}(t)(1 + x_{2}^{n}(t))}]+

+sg(x_{2∗}(t) − x_{2}(t))[\frac{B_{∗}(t)}{x_{2∗}(t)}.\frac{x_{1}^{n}(t) − x_{1∗}^{n}(t)}{(1 + x_{1∗}^{n}(t))(1 + x_{1}^{n}(t)}+

+\frac{B_{∗}(t)(x_{2}(t) − x_{2∗}(t))}{x_{2}(t)x_{2∗}(t))(1 + x_{1}^{n}(t))} + \frac{B_{∗}(t) − B(t)}{x_{2}(t)(1 + x_{1}^{n}(t))}] ≤

≤ −\frac{A_{L}(∞)}{A_{M}^{2}(∞)(1 + B_{M}^{n}(∞))}|x_{1}(t) − x_{1∗}(t)| − \frac{B_{L}(∞)}{B_{M}^{2}(∞)(1 + A_{M}^{n}(∞))}|x_{2}(t) − x_{2∗}(t)|+

+ \frac{B_{M}(∞)}{\frac{B_{L}(∞)}{1+A_{M}^{n} (∞)}} .\frac{n.A_{M}^{n−1}(∞)}{(1 + \frac{A_{L}^{n}(∞)}{(1+B_{M}^{n} (∞))^{n}} )^{2}} |x_{1}(t) − x_{1∗}(t)|+

+ \frac{A_{M}(∞)}{\frac{A_{L}(∞)}{1+B_{M}^{n} (∞)}} .\frac{n.B_{M}^{n−1}(∞)}{(1 + \frac{B_{L}^{n}(∞)}{(1+A_{M}^{n} (∞))^{n}} )^{2}} |x_{2}(t) − x_{2∗}(t)| + ϵ(t),

where ϵ(t) ⟶_{t⟶∞} 0. From the conditions of theorem 4 there exists a positive constant α for which

r′(t) ≤ −α(|x_{1}(t) − x_{1∗}(t)| + |x_{2}(t) − x_{2∗}(t)|) + ϵ(t), t ∈ X.

From the mean value theorem there exists a positive constant m > 0 such that

|lnx_{1}(t) − lnx_{1∗}(t)| ≤ m^{−1}|x_{1}(t) − x_{1∗}(t)|, |lnx_{2}(t) − lnx_{2∗}(t)| ≤ m^{−1}|x_{2}(t) − x_{2∗}(t)|

and we get

r′(t) ≤ −αmr(t) + ϵ(t), t ∈ X.

Therefore r(t) ⟶_{t⟶∞} 0, from where

x_{1}(t) − x_{1∗}(t) ⟶_{t⟶∞} 0, x_{2}(t) − x_{2∗}(t) ⟶_{t⟶∞} 0.

Remark. Theorems 1, 2, 3, 4 show that the system (1) is globally stable, i. e. any two positive solutions attract each other as t ⟶ ∞. We notice that the condition of theorem 3 is stronger than the condition in [3].

4. Examples

Systems

(7)  | \begin{matrix}x′(t) = \frac{2+sin^{2} \sqrt{t+1}}{5(1+y^{3}(t))} − x(t) \\ y′(t) = \frac{2+cos^{2} \sqrt{t+1}}{5(1+x^{3}(t))} − y(t), \end{matrix}

(7_{∗})  | \begin{matrix}x′(t) = \frac{2+sin^{2} \sqrt{t}}{5(1+y^{3}(t))} − x(t) \\ y′(t) = \frac{2+cos^{2} \sqrt{t}}{5(1+x^{3}(t))} − y(t), \end{matrix}

satisfy all conditions of theorems 1, 2, 3. Systems

| \begin{matrix}x′_{1}(t) = \frac{\frac{1}{3} + \frac{1}{t^{2}+1}}{1+x_{2}^{2}(t)} − x_{1}(t) \\ x′_{2}(t) = \frac{\frac{1}{3} + \frac{1}{t^{2}+1}}{1+x_{1}^{2}(t)} − x_{2}(t) \end{matrix}

and

| \begin{matrix}x′_{1}(t) = \frac{\frac{1}{3}}{1+x_{2}^{2}(t)} − x_{1}(t) \\ x′_{2}(t) = \frac{\frac{1}{3}}{1+x_{1}^{2}(t)} − x_{2}(t) \end{matrix}

satisfy conditions of theorem 4.

References

[1] Romanovskii, U., N. Stepanova, D. Chernavskii. Mathematical biopysic, Nauka, Moscow, 1984.

[2] Tineo, A. Asymptotic behaviour of positive solutions of the nonautonomous Lotka-Volterra competetion equations, Differential and Integral Equations, Vol. 6, 1993, 449-457.

[3] Georgiev, S., D. Tsvetkov. A model of cell differentiation, Russe University, 57-65, 1999.