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Permanence and extinction in competitive Lotka–Volterra systems with delays
Nonlinear Analysis: Real World Applications 12 (2011) 2130–2141
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Permanence and extinction in competitive Lotka–Volterra systems with delays Zhanyuan Hou ∗ Faculty of Computing, London Metropolitan University, North Campus, 166-220 Holloway Road, London N7 8DB, UK
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Article history: Received 3 May 2010 Accepted 29 December 2010
In this paper, a class of competitive Lotka–Volterra systems are considered that have distributed delays and constant coefficients on interaction terms and have time dependent growth rate vectors with an asymptotic average. Under the assumption that all proper subsystems are permanent, it is shown that the asymptotic behaviour of the system is determined by the relationship between an equilibrium and a nullcline plane of the corresponding autonomous system: if the equilibrium is below the plane then the system is permanent; if the equilibrium is above the plane then this species will go extinct in an exponential rate while the other species will survive. Similar asymptotic behaviour is also retained under an alternative assumption. © 2011 Elsevier Ltd. All rights reserved.
Keywords: Permanence Strong persistence Extinction Lotka–Volterra Competitive systems Distributed delays
1. Introduction Lotka–Volterra type systems of differential equations have been used in modeling population dynamics of a community of species since 1920s. For this reason, the study of such systems has been comprehensive and is still attracting a large number of researchers. The simplest Lotka–Volterra system has the form x′i = xi (bi − Ai x),
i ∈ IN ,
(1)
where, for any positive integer m, Im = {1, 2, . . . , m} and for any positive integer N and each i ∈ IN , Ai = (ai1 , . . . , aiN ) is the ith row of a matrix A = (aij ) and the bi and aij are real constants. We assume that
∀i, j ∈ IN ,
aij ≥ 0,
aii > 0,
bi > 0
(2)
so that (1) reflects the competitive nature. Since xi (t ) denotes the population size of the ith species at time t, as usual, we restrict the study of such systems to the nonnegative cone RN+ = {x ∈ RN : ∀i ∈ IN , xi ≥ 0}. In particular, solutions of (1) are mostly considered in int RN+ = {x ∈ RN+ : ∀i ∈ IN , xi > 0}, the interior of RN+ . Ahmad and Lazer [1] investigated the system x′i (t ) = xi (t )[ri (t ) − Ai x(t )],
i ∈ IN ,
(3)
where the ri (t ) are continuous from R0 = (−∞, ∞), (c , ∞) or [c , ∞) for some c ∈ R, to R satisfying
∀t ∈ R0 , ∀i ∈ IN , 0 < riL ≤ ri (t ) ≤ riM < ∞, ∀i ∈ IN , lim m(ri , t0 , t ) = bi > 0 (uniformly for t0 ∈ R0 ). t →∞
∗
Tel.: +44 20 73203124. E-mail address: [email protected].
1468-1218/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2010.12.027
(4) (5)
Z. Hou / Nonlinear Analysis: Real World Applications 12 (2011) 2130–2141
Here the notation m(g , t0 , t ) is defined by m(g , t0 , t ) = t −1t 0 xi = 0 if and only if i ∈ I }. Under the condition that
∀ i ∈ IN − 1 ,
−
riL >
aij
j∈IN \{i}
riM ajj
t t0
2131
g (s)ds. For any proper subset I ⊂ IN , let CI0 = {x ∈ RN+ :
,
(6)
it is shown (see [2] or [3, Corollary 4.4]) that the (N − 1)-dimensional subsystem of (3) with xN ≡ 0 has a solution x¯ (t ) ∈ C{0N } with x¯ i (t ) ≥ δ > 0 for all i ∈ IN −1 and t ∈ R0 . Note that (5) and (6) imply
∀ i ∈ IN − 1 ,
−
bi >
aij
j∈IN \{i}
bj ajj
,
(7)
which implies the existence of a unique equilibrium x∗ of (1) in C{0N } . Then the relative position of x∗ and the Nth nullcline plane γN = {x ∈ RN+ : AN x = bN } completely determines the global asymptotic behaviour of the solutions of (3) in int RN+ . Under assumptions (4)–(6), Ahmad and Lazer [1] proved the following conclusions for (3): (a) AN x∗ > bN (x∗ is above γN ) H⇒ x¯ is globally attractive; (b) AN x∗ = bN (x∗ is on γN ) H⇒ lim inft →∞ xN (t ) = 0; (c) AN x∗ < bN (x∗ is below γN ) H⇒ (3) is strongly persistent. When r (t ) in (3) is periodic, Tineo [4] relaxed assumptions (4) and (6)–(7) and proved the global attractiveness of a periodic solution (x¯ ∈ C{0N } if AN x∗ ≥ bN or x˜ ∈ int RN+ if AN x∗ < bN with an extra condition). If we drop the global attractiveness from these results and observe only the behaviour of the Nth component xN (t ) of any solution of (3) in int RN+ , then (a) AN x∗ > bN H⇒ limt →∞ xN (t ) = 0; (b) AN x∗ = bN H⇒ lim inft →∞ xN (t ) = 0; (c) AN x∗ < bN H⇒ lim inft →∞ xN (t ) > 0. For (3) with a general r (t ) satisfying (4), (5) and (7), Ahmad and Lazer [5] actually achieved the above (and more). Clearly, AN x∗ > bN is a threshold condition for the Nth species to die out. Other conditions for extinction can be found in [6–12] and the references quoted there. Ecologically, one concern about the community is whether all species can coexist. This is mathematically gauged by the concepts of persistence and permanence. A system is called strongly persistent if every solution in int RN+ satisfies lim inft →∞ xi (t ) > 0 for all i ∈ IN . A system is called permanent if there are r > 0 and M > r such that every solution of the system in int RN+ satisfies r ≤ xi (t ) ≤ M for all i ∈ IN and all sufficiently large t. A system is called strongly permanent if all small perturbations of the system are permanent. Definition 1. A vector b ∈ int RN+ and a matrix A = (aij ) ∈ RN+×N are said to satisfy the (I − J )-conditions if for each proper subset I ⊂ IN with J = IN \ I, if there exists an x∗J ∈ CJ0 such that
∀i ∈ I ,
Ai x∗J = bi
(8)
∀j ∈ J ,
Aj x∗J < bj .
(9)
then
Ahmad and Lazer [6] proved that all subsystems of (3) are strongly permanent if and only if b and A satisfy the (I − J )conditions. Further, under the (I − J )-conditions for the subsystem of (3) with xN = 0, there is an x∗ ∈ C{0N } such that Ai x∗ = bi for all i ∈ IN −1 . It is also shown [6, Theorem 4.1] that AN x∗ > bN implies limt →∞ xN (t ) = 0 provided xN (t0 ) is small enough. In a recent paper [13], the author investigated the following system with distributed delays,
xi (t ) = xi (t ) ri (t ) − ′
N − j =1
∫
0
xj (t + θ )dξij (θ ) ,
aij
i ∈ IN ,
(10)
−τ
where the ri (t ) and aij are the same as in (3), τ > 0, and the ξij are nondecreasing functions from any open interval containing [−τ , 0] to R+ satisfying
∀i, j ∈ IN ,
ξij (0+ ) − ξij (−τ − ) = 1.
(11)
The theorem below can be viewed as an extension of a result of [6] from (3) to (10). Theorem 1 ([13, Theorems 4,5]). If b and A satisfy the (I − J )-conditions, then all subsystems of (10) and its small perturbations are permanent. Conversely, if all subsystems of (10) and its small perturbations on A are permanent, then b and A satisfy the (I − J )-conditions.
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In this paper, we are concerned with the global asymptotic behaviour of solutions of (10) when the (I − J )-conditions are slightly violated or the (I − J )-conditions for an (N − 1)-dimensional subsystem of (10) are satisfied. We shall prove that, under certain assumptions, a threshold condition AN x∗ > bN will drive the Nth species to go extinct in an exponential rate while the other species will survive, and that the reversed inequality can make the system permanent or strongly persistent. 2. Main results and examples We consider system (10) with assumptions (2), (4), (5) and (11). Theorem 2. Assume that b and A satisfy the (I − J )-conditions except I ̸= IN −1 . Then every k-dimensional subsystem of (10) with 1 ≤ k < N is permanent and there is a unique x∗{N } ∈ C{0N } such that
∀ i ∈ IN − 1 ,
Ai x∗{N } = bi .
(12)
Moreover, the following conclusions hold. (a) If AN x∗{N } = bN then every solution of (10) in int RN+ satisfies
∀ i ∈ IN − 1 ,
lim inf xi (t ) > 0
(13)
t →∞
and lim inft →∞ xN (t ) = 0. (b) If δ0 = AN x∗{N } − bN > 0, then every solution of (10) in int RN+ satisfies (13) and
∀δ ∈ (0, δ0 ),
xN (t ) = o(e−δ t ) (t → ∞).
(14)
(c) System (10) is permanent if and only if AN x∗{N } < bN . To give a geometric interpretation of the conditions, let
πi = {x ∈ RN+ : xi = 0},
i ∈ IN ,
γi = {x ∈ RN+ : Ai x = bi },
(15)
i ∈ IN .
(16)
We note that γi defined by (16) is the ith nullcline plane of (1) in R+ and πi defined by (15) is the ith coordinate plane. For any (N − 1)-dimensional plane γ , if the origin 0 ̸∈ γ , then γ divides RN+ into three mutually exclusive connected sets γ − , γ and γ + with 0 ∈ γ − . A point x ∈ RN+ is said to be below (on or above) γ if x ∈ γ − (x ∈ γ or x ∈ γ + ). A set S is said to be below (on or above) γ if every point in S is so. N
Remark 1. Under assumption (2), it is shown in [13, Remarks 1,2] that the following conditions are equivalent: (a) b and A satisfy the (I − J )-conditions. (b) For each proper subset I ⊂ IN with J = IN \ I, (1) has a unique equilibrium x∗J ∈ CJ0 such that
∀j ∈ J ,
Aj x∗J < bj .
(c) For each i ∈ IN every equilibrium of (1) on πi is below γi . Remark 2. The condition of Theorem 2 is equivalent to each of the following: (a) For each proper subset I ⊂ IN (I ̸= IN −1 ) with J = IN \ I, (1) has a unique equilibrium x∗J ∈ CJ0 such that Aj x∗J < bj holds for all j ∈ J. (b) For each i ∈ IN , every equilibrium of (1) on πi (except the equilibrium x∗{N } ∈ πN ) is below γi . Obviously, under this condition, b and A satisfy the (I − J )-conditions if and only if x∗{N } is below γN . Remark 3. Let us compare the condition of Theorem 2 with condition (7). Let Yi = bi /aii for all i ∈ IN and let [0, Y ] be the set of x ∈ RN+ satisfying 0 ≤ xi ≤ Yi for all i ∈ IN . Then it is clear that all equilibria of (1) fall into the cell [0, Y ]. The geometric interpretation of (7) is that the set πi ∩ [0, Y ] is below γi for all i ∈ IN −1 . Thus, (7) implies that for each i ∈ IN −1 every equilibrium of (1) on πi is below γi . But (7) does not imply any relationship between γN and the equilibria on πN . Hence, neither of (7) and the condition of Theorem 2 completely implies the other. If b and A satisfy the (I − J )-conditions, then by [6, Lemma 2.7] or [9, Lemma 5], (1) in int RN+ has a unique equilibrium x = A−1 b. Thus, if the (I − J )-conditions for the subsystem of (1) with xN = 0 are satisfied then (1) in C{0N } has a unique equilibrium x∗{N } . Consider the following (N − 1)-dimensional auxiliary system ∗
′
[
xi = xi bi − daiN −
N −1 −
]
aij xj ,
i ∈ IN − 1 ,
(17)
j =1
where the bi and aij are the same as in (1) and d ∈ [0, bN /aNN ] is a constant. By saying that (17) satisfies the (I − J )-conditions, we mean bi − daiN > 0 for all i ∈ IN −1 and the obvious replacements of N by N − 1 and bi by bi − daiN in Definition 1.
Z. Hou / Nonlinear Analysis: Real World Applications 12 (2011) 2130–2141
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Theorem 3. Assume that (17) satisfies the (I − J )-conditions for both d = 0 and d = bN /aNN . Then the (N − 1)-dimensional subsystem of (10) with xN ≡ 0 is permanent and the following conclusions hold. (a) If AN x∗{N } = bN then lim inft →∞ xN (t ) = 0 and (13) holds for every solution of (10) in int RN+ . (b) If δ0 = AN x∗{N } − bN > 0, then every solution of (10) in int RN+ satisfies (13) and (14). (c) System (10) is strongly persistent if and only if AN x∗{N } < bN . Remark 4. Note that condition (7) can be rearranged as
∀ i ∈ IN − 1 ,
−
bi − daiN >
aij
j∈IN −1 \{i}
bj ajj
,
d=
bN aNN
.
(18)
This implies that for each i ∈ IN −1 every equilibrium of (17) with d = bN /aNN on the ith coordinate plane is below the ith nullcline plane. Thus, (7) implies that (17) with d = bN /aNN satisfies the (I − J )-conditions. Also, (7) implies
∀ i ∈ IN − 1 ,
bi >
−
aij
j∈IN −1 \{i}
bj ajj
,
(19)
which implies that (17) with d = 0 also satisfies the (I − J )-conditions. Therefore, (7) implies the condition of Theorem 3. On the other hand, however, Example 1 given below shows that the condition of Theorem 3 does not imply (7). This shows that Theorem 3 even in the case of τ = 0 covers some of the results given in [5] as a special case. Remark 5. The two examples given below show that the condition of Theorem 2 is independent of the condition of Theorem 3. Remark 6. Just as commented in [13], the asymptotic behaviour of the solutions of (10) conveyed by Theorems 2 and 3 is not affected by the delays since the conditions of the theorems do not involve the delays at all. Remark 7. Although Theorems 2 and 3 are stated in terms of N ∈ IN and x∗{N } only, this is purely for the convenience of formulating the statements and presenting the proofs. In practice, if the concerned component is xi for some i ∈ IN rather than xN , then we can always swap the position of xi with xN before applying the theorems to the system. In the following examples, we present the coordinates of any point x ∈ RN+ in a row form although it is viewed as a column vector or an N × 1 matrix whenever it appears with matrix operations. Example 1. Consider systems (1) and (10) with
2 1 A= 1 1
1 2 1 1
1 1 3 2
2 2 , 1 4α
1 1 b = , 1.5
where α is a positive constant. For d ∈
α
0,
1 4
, the corresponding system (17) has the following nontrivial equilibria on
∂ R3+ : p1 = (0.5 − d, 0, 0), p2 = (0, 0.5 − d, 0), p3 = 0, 0, 0.5 − 3d , p4 = 1−32d , 1−32d , 0 , p5 = (0.3 − d, 0, 0.4), p6 = (0, 0.3 − d, 0.4). On π1 we have p2 , p3 , p6 , which are all below γ1 = {x ∈ R3+ : 2x1 + x2 + x3 = 1 − 2d}. On π2 we have p1 , p3 , p5 , which are all below γ2 = {x ∈ R3+ : x1 + 2x2 + x3 = 1 − 2d}. On π3 we have p1 , p2 , p4 , which are below γ3 = {x ∈ R3+ : x1 + x2 + 3x3 = 1.5 − d}. Therefore, (17) satisfies the (I − J )-conditions for all d ∈ 0, 14 so the condition 3 3 5 of Theorem 3 is met. The equilibrium x∗{4} for (1) is 14 , 14 , 14 , 0 . As A4 x∗{4} − b4 = 87 − α , by Theorem 3, (10) is strongly 8 8 persistent if α > 7 , (13) holds if 0 < α ≤ 7 , lim inft →∞ x4 (t ) = 0 if α = 87 and (14) holds if 0 < α < 87 . Note that a12 b2 a22
+
a13 b3 a33
+
a14 b4 a44
=
1 2
+
1.5 3
+
2α 4α
=
3 2
> 1 = b1 .
Thus, condition (7) is not satisfied. Note also that (1) on π4 has an equilibrium p = A4 p = 2/3 ≥ α = b4
if 0 < α ≤ 2/3.
Thus, from Remark 2, the condition of Theorem 2 is not fulfilled if α ∈ 0,
Example 2. Consider systems (1) and (10) with
3 2 A= 2
α
2 3 2
α
2 2 3
α
1 1 , 2 4
1
1 b = , 1 1
2 3
.
1 3
, 13 , 0, 0 but
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where α is a constant in 1,
. On ∂ R4+ system (1) has the following nontrivial equilibria:
5 2
1 1 1 , 0, 0, 0 , p2 = 0, , 0, 0 , p3 = 0, 0, , 0 , p4 = 0, 0, 0, , 3 3 3 4 1 1 1 3 1 1 1 3−α p5 = , , 0, 0 , p6 = 0, , , 0 , p7 = , 0, , 0 , p8 = , 0, 0, , 5 5 5 5 5 5 12 − α 12 − α 3 3 3−α 1 3−α 3 5 − 2α p9 = 0, , 0, , p10 = 0, 0, , , p11 = , , 0, , 12 − α 12 − α 6 − α 2(6 − α) 20 − 2α 20 − 2α 20 − 2α 5−α α 5 − 2α 5−α α 5 − 2α 1 1 1 p12 = , 0, , , p13 = 0, , , , x∗{4} = , , ,0 . 20 − 3α 20 − 3α 20 − 3α 20 − 3α 20 − 3α 20 − 3α 7 7 7 p1 =
1
On π1 we have p2 , p3 , p4 , p6 , p9 , p10 and p13 and they are all below γ1 defined by 3x1 + 2x2 + 2x3 + x4 = 1 in R4+ . On π2 we have p1 , p3 , p4 , p7 , p8 , p10 and p12 , which are below γ2 defined by 2x1 +3x2 +2x3 +x4 = 1. On π3 we have p1 , p2 , p4 , p5 , p8 , p9 and p11 , which are all below γ3 defined by 2x1 + 2x2 + 3x2 + 2x4 = 1. On π4 we have p1 , p2 , p3 , p5 , p6 , p7 and x∗{4} which, except x∗{4} , are below γ4 given by α x1 +α x2 +α x3 + 4x4 = 1. Thus, by Remark 2, the condition of Theorem 2 is satisfied. Since A4 x∗{4} − b4 = 37α − 1, by Theorem 2 system (10) is permanent if 1 < α < if α = p=
7 and (14) holds 3 3 3 0 but 20 20
,
,
if
7 3
7 , 3
(13) holds if
< α < 25 . Note that the corresponding system (17) with d =
7 3 1 4
≤ α < 52 , lim inft →∞ x4 (t ) = 0 on π3 in R3+ has an equilibrium
(a31 , a32 , a33 )p = 0.6 > 0.5 = b3 − da34 . Thus, (17) with d =
1 4
does not satisfy the (I − J )-conditions so the condition of Theorem 3 is not met.
3. Proof of Theorem 3 Lemma 1 ([13, Lemmas 1,2]). There exist σ > 0, ρ > 0 and T = T (ϕ) > 0 (for each ϕ ∈ C ([−τ , 0], RN+ ) with ϕ(0) ̸= 0) such that the solution of (10) with xt0 = ϕ for all t0 ∈ R0 satisfies
∀t ≥ t0 + T ,
x1 (t ) + · · · + xN (t ) ≥ σ
and,
∀i ∈ IN , xi (t , t0 , ϕ) < ρ.
Lemma 2. If x(t ) is a solution of (10) in int RN+ on [t0 , ∞), then
∀i ∈ IN , ∀t ≥ T ≥ t0 ,
lim sup m(xi , T , t ) ≤
(t −T )→∞
bi aii
.
(20)
Proof. Let x(t ) ∈ int RN+ be a solution of (10) for t ≥ t0 and let i ∈ IN be fixed. Case 1. Suppose lim inft →∞ xi (t ) > 0. Then, by Lemma 1, ln[xi (t )/xi (T )] is bounded for all t ≥ T ≥ t0 so 1
lim
(t −T )→∞
t −T
ln
xi (t ) xi ( T )
= 0.
(21)
By Lemma 1 and (11), for all k, j ∈ IN , 1 t −T
∫ t∫ T
0
∫ 0 ∫
1
xj (s + θ )dξkj (θ )ds =
t −T
−τ
−τ
t +θ
xj (s)ds dξkj (θ )
T +θ
= m(xj , T , t ) + o(1) ((t − T ) → ∞).
(22)
Dividing (10) by xi (t ), integrating the equation from T to t and dividing through by t − T , we obtain
∫ t ∫ 0 N − aij ln = m(ri , T , t ) − xj (s + θ )dξij (θ ) ds t −T xi (T ) t −T T −τ j=1 ∫ ∫ 1
xi ( t )
≤ m(ri , T , t ) −
t
aii
t −T
0
xi (s + θ )dξii (θ ) ds.
T
(23)
−τ
Then combination of (21)–(23) gives aii m(xi , T , t ) ≤ m(ri , T , t ) + o(1)
((t − T ) → ∞).
From this and (5) the inequality in (20) follows.
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Case 2. If lim inft →∞ xi (t ) = 0 and lim(t −T )→∞ m(xi , T , t ) = 0 then the inequality in (20) is obvious. Case 3. Suppose lim inft →∞ xi (t ) = 0 but lim sup xi (t ) ≥ lim sup m(xi , T , t ) = σ0 > 0. (t −T )→∞
t →∞
Then there is a sequence {[sn , tn ]} with tn − sn → ∞ as n → ∞ such that m(xi , sn , tn ) > σ0 /2 for all n ≥ 1 and limn→∞ m(xi , sn , tn ) = σ0 . If there is a subsequence {nℓ } ⊂ {n} such that xi (snℓ ) ≤ xi (tnℓ ) holds for all ℓ ≥ 1, then replacing t by tnℓ and T by snℓ in (22)–(24) and letting ℓ → ∞, we obtain the inequality in (20). Otherwise, xi (sn ) > xi (tn ) holds for some integer K ≥ 1 and all n ≥ K . (i) Suppose there is a subsequence {nℓ } ⊂ {n} such that snℓ = t0 for all ℓ ≥ 1. Then lim supt →∞ m(xi , t0 , t ) = σ0 so lim supt →∞ m(xi , T , t ) = σ0 for any fixed T ≥ t0 . Since lim inft →∞ xi (t ) = 0, we can choose a T1 ≥ t0 and {tn′ } ⊂ [T1 , ∞) with tn′ ↑ ∞ as n → ∞ such that xi (T1 ) < σ0 /2, m(xi , T1 , tn′ ) > σ0 /2 for all n ≥ 1, and limn→∞ m(xi , T1 , tn′ ) = σ0 . If there is ′ a subsequence {mℓ } ⊂ {n} such that xi (tm ) ≥ xi (T1 ) for all ℓ ≥ 1, then (20) follows from (22)–(24) after the replacements ℓ ′ of T by T1 and t by tmℓ . Otherwise, there is an integer K1 > 0 such that xi (tn′ ) < xi (T1 ) holds for all n ≥ K1 . Then, for each fixed n ≥ K1 , there is a vn ∈ (T1 , tn′ ) satisfying xi (t ) < xi (vn ) = xi (T1 ) for all t ∈ (vn , tn′ ]. Thus,
(tn′ − T1 )m(xi , T1 , tn′ ) < (vn − T1 )m(xi , T1 , vn ) + (tn′ − vn )xi (T1 ). If {vn } is bounded then
vn − T1 tn′ − vn → 0 and → 1 (n → ∞) tn′ − T1 tn′ − T1 so the above inequality implies σ0 = limn→∞ m(xi , T1 , tn′ ) ≤ xi (T1 ) < σ0 /2, a contradiction to σ0 > 0. This contradiction shows that {vn } is unbounded. Without loss of generality, we may assume vn ↑ ∞ as n → ∞. Then, as m(xi , T1 , vn ) > m(xi , T1 , tn′ ) +
tn′ − vn
vn − T1
[m(xi , T1 , tn′ ) − xi (T1 )]
> m(xi , T1 , tn′ ), we have limn→∞ m(xi , T1 , vn ) = σ0 . Hence, (20) follows from (22)–(24) with the replacements of T by T1 and t by vn . (ii) If (i) is not true, then there is an integer K2 > K such that
∀n ≥ K2 , sn > t0 ,
xi (sn ) > xi (tn ).
For each n ≥ K2 , by continuity, xi (t + sn ) > xi (t + tn ) holds, so m(xi , t + sn , t + tn ) > m(xi , sn , tn ) when t < 0 and |t | is sufficiently small. Then there are s′n ∈ [t0 , sn ) and tn′ = s′n + tn − sn satisfying m(xi , s′n , tn′ ) = max m(xi , t , t + tn − sn ) > m(xi , sn , tn ). t ∈[t0 ,sn ]
Clearly, we have either xi (s′n ) ≤ xi (tn′ ) or s′n = t0 . If there is a subsequence {nℓ } ⊂ {n} such that s′nℓ = t0 for all ℓ ≥ 1, then (20) follows from (i). Otherwise, replacing t by tn′ and T by s′n in (22)–(24) and letting n → ∞, we obtain the inequality in (20). Let A˜ be the (N − 1) × (N − 1) submatrix of A obtained by deleting the last row and the last column of A and let b˜ be the (N − 1)-dimensional vector obtained from b by deleting its last component. For any proper subset J ⊂ IN −1 and any (N − 1)-dimensional (row or column) vector v , let vJ be the vector obtained from v by deleting its jth component for all j ∈ J and let A˜ J denote the submatrix of A˜ obtained by deleting the jth row and the jth column of A˜ for all j ∈ J. Let A˜ i be the ith row of A˜ and set CN A = (a1N , . . . , a(N −1)N )T . Lemma 3. Assume that system (17) satisfies the (I − J )-conditions for both d = 0 and d = bN /aNN . Then (17) satisfies the (I − J )-conditions for every d ∈ [0, bN /aNN ]. Proof. For each proper subset J ⊂ IN −1 , we show that the linear algebraic system A˜ J xJ = bJ − d(CN A)J
(25)
for each fixed d ∈ [0, bN /aNN ] has a unique solution xJ (d) with positive components satisfying
∀j ∈ J ,
(A˜ j )J xJ (d) < bj − dajN .
(26)
Since (17) with d = 0 satisfies the (I − J )-conditions, by Remark 2 there is a unique x¯ J with positive components satisfying A˜ J x¯ J = bJ ,
∀j ∈ J ,
(A˜ j )J x¯ J < bj .
(27)
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Z. Hou / Nonlinear Analysis: Real World Applications 12 (2011) 2130–2141
By the (I − J )-conditions for (17) with d = bN /aNN and Remark 2, there is a unique x˜ J with positive components satisfying A˜ J x˜ J = bJ − (bN /aNN )(CN A)J ,
∀j ∈ J ,
(28)
(A˜ j )J x˜ J < bj − (bN /aNN )ajN .
(29)
1 Then the uniqueness of x¯ J and x˜ J implies the existence of A˜ − and the unique solution xJ (d) of (25). Thus, J
∀d ∈ [0, bN /aNN ],
1 xJ (d) = A˜ − J (bJ − d(CN A)J ).
(30)
Since both xJ (0) = x¯ J and xJ (bN /aNN ) = x˜ J have positive components only and xJ (d) is linear in d, all components of xJ (d) are positive for all d ∈ [0, bN /aNN ]. Now consider the linear function Fj (d) = bj − dajN − (A˜ j )J xJ (d),
d ∈ [0, bN /aNN ], j ∈ J .
(31)
Since Fj (0) > 0 by (27) and Fj (bN /aNN ) > 0 by (29), we have Fj (d) > 0 for all d ∈ [0, bN /aNN ] and j ∈ J. Thus, for each d ∈ [0, bN /aNN ], the unique solution xJ (d) of (25) given by (30) satisfies (26). Therefore, (17) for any fixed d ∈ [0, bN /aNN ] satisfies the (I − J )-conditions. Lemma 4. Assume that (17) satisfies the (I − J )-conditions for both d = 0 and d = bN /aNN . Then (1) in C{0N } has a unique equilibrium x∗{N } and the following conclusions hold. (a) AN x∗{N } < bN H⇒ Ax∗ = b has a unique solution x∗ ∈ int RN+ ; (b) AN x∗{N } = bN H⇒ Ax∗ = b has a unique solution x∗ = x∗{N } in RN+ ; (c) AN x∗{N } > bN H⇒ Ax ̸= b for any x ∈ RN+ .
˜ = b˜ Proof. Since the (I − J )-conditions hold for (17) with both d = 0 and d = bN /aNN , by [6, Lemma 2.7] or [9, Lemma 5], Ax ˜ = b˜ −(bN /aNN )(CN A) has a unique solution y˜ = A˜ −1 (b˜ −(bN /aNN )(CN A)) ∈ has a unique solution y¯ = A˜ −1 b˜ ∈ int RN+−1 and Ax int RN+−1 . Define x¯ and x˜ in RN+ by x¯ N = 0, Then x¯ =
x∗{N }
x˜ N = bN /aNN ,
∀i ∈ IN −1 ,
Ai x∗{N } = bi ,
Ai x˜ = bi .
x¯ i = y¯ i ,
x˜ i = y˜ i .
(32)
and
∀ i ∈ IN − 1 ,
(33)
This shows that the line segment x∗{N } x˜ connecting x∗{N } to x˜ lies in ∩i∈IN −1 γi . Since the existence of A˜ −1 implies that ∩i∈IN −1 γi is a line segment containing x∗{N } x˜ , if Ax = b has a solution in RN+ then 0 ≤ xN ≤ bN /aNN = x˜ N so x must be on x∗{N } x˜ and satisfy AN x = bN . (a) If AN x∗{N } < bN , since AN x˜ ≥ aNN x˜ N = bN , there is a unique x∗ ∈ x∗{N } x˜ \ {x∗{N } } ⊂ int RN+ such that AN x∗ = bN so that Ax = b. (b) If AN x∗{N } = bN then there is an i0 ∈ IN −1 such that aNi0 > 0 so AN x˜ ≥ aNi0 x˜ i0 + aNN x˜ N > bN . Thus, x = x∗{N } is the unique point on x∗{N } x˜ meeting AN x = bN and Ax = b. ∗
(c) If AN x∗{N } > bN , then there is an i0 ∈ IN −1 such that aNi0 > 0 so AN x˜ ≥ aNi0 x˜ i0 + aNN x˜ N > bN . Thus, AN x > bN holds for
all x ∈ x∗{N } x˜ . Therefore, Ax = b has no solution in RN+ .
Lemma 5. Assume that (17) satisfies the (I − J )-conditions for both d = 0 and d = bN /aNN . Then every solution of (10) in int RN+ satisfies (13). Proof. Suppose (10) has a solution x(t ) ∈ int RN+ on [t0 , ∞) satisfying, for some i0 ∈ IN −1 , lim inft →∞ xi0 (t ) = 0. Then, by Lemma 1, there is a subset J ⊂ IN −1 fulfilling lim inf max{xj (t ) : j ∈ J } = 0,
(34)
∀i ∈ I ,
(35)
t →∞
lim inf max{xj (t ) : j ∈ J ∪ {i}} > 0, t →∞
where I = IN −1 \ J if J ̸= IN −1 or I = {N } otherwise. By Lemma 1, there is a ρ0 > 0 such that xi (t ) ≤ ρ0 for all i ∈ IN and t ≥ t0 . Thus, (35) implies the existence of δ1 ∈ (0, ρ0 ) such that
∀i ∈ I , ∀t ≥ t0 ,
δ1 ≤ max{xj (t ) : j ∈ J ∪ {i}} ≤ ρ0 .
(36)
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Put a0 = max{aij : i, j ∈ IN },
(37)
ρ1 = min{δ1 , xi (t0 ) : i ∈ IN }.
(38)
As ρ1 > 0, by (34) there is a sequence {tn } with tn ↑ ∞ as n → ∞ satisfying
∀n ≥ 1 ,
max{xj (tn ) : j ∈ J } <
ρ1 2n
e−nN ρ0 a . 0
(39)
By (38), (39) and continuity, there are sn ∈ (t0 , tn ) and jn ∈ J such that
∀n ≥ 1 ,
max{xj (sn ) : j ∈ J } = xjn (sn ) =
∀n ≥ 1, ∀t ∈ (sn , tn ],
ρ1 2n
max{xj (t ) : j ∈ J } <
,
(40)
ρ1 2n
.
(41)
Then (36), (40) and (41) imply that
∀i ∈ I , ∀t ∈ [sn , tn ],
δ1 ≤ xi (t ) ≤ ρ0 .
(42)
From (10), (11), (37) and (40), we have
∫
tn
xjn (tn ) = xjn (sn ) exp
[
rjn (t ) −
N −
sn
ρ1
>
2n
∫
xj (t + θ )dξjn j (θ ) dt
aj n j −τ
j=1
0 e−N ρ0 a (tn −sn ) ,
]
0
∀n ≥ 1.
Combining this with (39), we obtain
∀n ≥ 1 ,
tn − sn > n.
(43)
Now from (22) and the way of obtaining (23) from (10), we derive, as n → ∞,
∀i ∈ I , 1 tn − sn
1
ln
tn − sn ln
xjn (tn )
xi (tn ) xi (sn )
= m(ri , sn , tn ) −
aij m(xj , sn , tn ) + o(1),
(44)
j =1
= m(rjn , sn , tn ) −
xjn (sn )
N −
N −
ajn j m(xj , sn , tn ) + o(1).
(45)
j =1
Since from (40) and (41) follows
∀j ∈ J ,
lim m(xi , sn , tn ) = 0
(46)
n→∞
and (42) and (43) imply
∀i ∈ I ,
lim
1
n→∞ tn
− sn
ln
xi (tn ) xi (sn )
= 0,
(47)
by (40) and (41) again, we can rewrite (44) and (45) as
∀i ∈ I ,
−
aij m(xj , sn , tn ) = m(ri , sn , tn ) + o(1) (n → ∞),
(48)
j∈I ∪{N }
−
ajn j m(xj , sn , tn ) > m(rjn , sn , tn ) + o(1)
(n → ∞).
(49)
j∈I ∪{N }
Note that for all n ≥ 1, jn ∈ J , m(xj , sn , tn ) ∈ [δ1 , ρ0 ] for all j ∈ I and, if I ̸= {N }, lim supn→∞ m(xN , sn , tn ) ∈ [0, bN /aNN ] by Lemma 2. Then there is a subsequence {nℓ } ⊂ {n} such that jnℓ = j0 ∈ J for all ℓ ≥ 1 and
∀j ∈ I ∪ {N },
lim m(xj , snℓ , tnℓ ) = x∗j ,
ℓ→∞
where x∗j ∈ [δ1 , ρ0 ] for all j ∈ I and, if I ̸= {N }, x∗N ∈ [0, bN /aNN ]. With the replacement of n by nℓ in (48) and (49) and by assumption (5), a limit process as ℓ → ∞ results in
∀i ∈ I ,
− j∈I ∪{N }
aij x∗j = bi ,
− j∈I ∪{N }
aj0 j x∗j ≥ bj0 .
(50)
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When I = {N }, (50) becomes aj0 N x∗N ≥ bj0 .
aNN x∗N = bN ,
(51)
From this follows aj0 N bN /aNN ≥ bj0 . This is impossible since, by the (I − J )-conditions for (17) with d = bN /aNN , we require bi − aiN bN /aNN > 0 for all i ∈ IN −1 . When I ̸= {N }, (50) becomes
−
∀i ∈ I ,
−
aij x∗j = bi − aiN x∗N ,
j∈I
aj0 j x∗j ≥ bj0 − aj0 N x∗N .
(52)
j∈I
From Lemma 3 we know that (17) with d = x∗N also satisfies the (I − J )-conditions. Thus, for the above I ⊂ IN −1 and J = IN −1 \ I we should have
−
∀i ∈ I ,
aij x∗j = bi − aiN x∗N ,
∀k ∈ J ,
j∈I
−
akj x∗j < bk − akN x∗N .
(53)
j∈I
Clearly, (52) and (53) are in conflict. This shows the truth of the conclusion of the lemma.
Proof of Theorem 3. The permanence of the subsystem
xi (t ) = xi (t ) ri (t ) − ′
N −1 −
∫
xj (t + θ )dξij (θ ) ,
aij
j =1
0
i ∈ IN − 1
−τ
follows from the (I − J )-conditions for (17) with d = 0 and Theorem 1. (a) By Lemma 5, every solution of (10) in int RN+ satisfies (13). Suppose (10) in int RN+ has a solution satisfying lim inft →∞ xN (t ) > 0. Then, by Lemma 1, this solution on [t0 , ∞) satisfies
∀t ≥ t0 , ∀i ∈ IN ,
0 < δ1 ≤ xi (t ) ≤ ρ0 < ∞.
(54)
By the method of obtaining (21)–(24), we have N −
∀ i ∈ IN ,
aij m(xj , t0 , t ) = m(ri , t0 , t ) + o(1) (t → ∞).
(55)
j =1
From (5), (54) and (55) we obtain an x∗ ∈ int RN+ such that Ax∗ = b. But this contradicts Lemma 4(b). Therefore, we must have lim inft →∞ xN (t ) = 0 for every solution of (10) in int RN+ . (b) By Lemmas 4 and 5(c), the proof of part (a) is still valid to show that every solution of (10) in int RN+ satisfies (13) and lim inft →∞ xN (t ) = 0. Suppose lim supt →∞ xN (t ) > 0 for a particular solution. Then we can choose a T ≥ t0 and an increasing sequence {tn } ⊂ [T , ∞) such that
∀n ≥ 1,
xN (tn ) > xN (T ),
∀ i ∈ IN − 1 ,
(56)
lim m(xi , T , tn ) = x∗i > 0,
(57)
n→∞
lim m(xN , T , tn ) = x∗N ≥ 0.
(58)
n→∞
Then, by (56) and Lemmas 1 and 5, ln[xi (tn )/xi (T )] is bounded for all i ∈ IN and n ≥ 1. By the method of obtaining (55), we have
∀i ∈ IN , ∀n ≥ 1,
N −
aij m(xj , T , tn ) = m(ri , T , tn ) + o(1) (n → ∞).
(59)
j =1
This leads to Ax∗ = b as n → ∞, a contradiction to Lemma 4(c). Hence, every solution of (10) in int RN+ satisfies limt →∞ xN (t ) = 0. Now suppose (14) does not hold for a solution of (10) in int RN+ . Then, for some σ1 ∈ (0, δ0 ), lim supt →∞ xN (t )eσ1 t > 0 so for any fixed δ ∈ (σ1 , δ0 ), lim supt →∞ xN (t )eδ t = ∞. Let y(t ) = xN (t )eδ t . From the Nth component equation of (10) we obtain, as t → ∞, 1 t − t0
ln
y(t ) y(t0 )
= δ + m(rN , t0 , t ) −
N −
aNj m(xj , t0 , t ) + o(1).
(60)
j =1
Take a sequence {tn } with tn ↑ ∞ as n → ∞ such that y(tn ) > y(t0 ) so that ln[y(tn )/y(t0 )] > 0 for all n ≥ 1. By taking subsequences if necessary, we may assume that (57) holds. Note that limt →∞ m(xN , t0 , t ) = 0 by limt →∞ xN (t ) = 0. Then
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(60) with the replacement of t by tn gives N −1 −
aNj m(xj , t0 , tn ) < δ + m(rN , t0 , tn ) + o(1)
(n → ∞).
(61)
j=1
From (10) we also have
∀ i ∈ IN − 1 ,
N −1 −
aij m(xj , t0 , tn ) = m(ri , t0 , tn ) + o(1) (n → ∞).
(62)
j=1
Then a limit process as n → ∞ in (61) and (62) gives
∀ i ∈ IN − 1 ,
A i x∗ = bi ,
AN x∗ ≤ δ + bN ,
(63)
(I − J )-conditions for (17) with d = 0 imply that is the unique solution of Ai x = bi for x∗ = x∗{N } in (63). Thus, since δ0 = AN x∗{N } − bN > 0, the inequality in (63) gives δ0 ≤ δ , a contradiction to δ ∈ (σ1 , δ0 ). Therefore, every solution of (10) in int RN+ satisfies (14). (c) Suppose AN x∗{N } < bN and (10) has a solution x(t ) ∈ int RN+ satisfying lim inft →∞ xN (t ) = 0. Since Lemmas 1 and 5 ensure the existence of ρ0 > σ0 > 0 such that where
all i ∈
x C{0N } . Since the IN −1 in C{0N } , we have ∗
x∗{N }
∈
∀i ∈ IN −1 , ∀t ≥ t0 ,
0 < σ0 ≤ xi (t ) ≤ ρ0 ,
(64)
using the same technique as that used in the proof of Lemma 5, we can choose a sequence {[sn , tn ]} with tn − sn → ∞ as n → ∞ such that
∀n ≥ 1, ∀t ∈ (sn , tn ], ∀ i ∈ IN − 1 ,
xN (sn ) > xN (t ),
lim xN (sn ) = 0,
(65)
n→∞
lim m(xi , sn , tn ) = x∗i ∈ [σ0 , ρ0 ].
(66)
n→∞
Since (65) implies ln[xN (tn )/xN (sn )] < 0 and limn→∞ m(xN , sn , tn ) = 0, from (10), (65) and (66), a limit process as n → ∞ produces
∀ i ∈ IN − 1 , ∗
A i x∗ = bi ,
A N x∗ ≥ bN ,
(67)
C{0N } .
∗
x∗{N }
AN x∗{N }
where x ∈ By the same reasoning as that used in part (b), we have x = so ≥ bN , a contradiction to the assumption. Therefore, (10) is strongly persistent. Conversely, if (10) is strongly persistent then from parts (a) and (b) we must have AN x∗{N } < bN . 4. Proof of Theorem 2 Lemma 6. Assume that, for some i0 ∈ IN , the (N − 1)-dimensional subsystem
[
x′i = xi bi −
−
]
aij xj ,
i ∈ IN \ { i 0 }
j∈IN \{i0 }
of (1) satisfies the (I − J )-conditions. If every equilibrium of (1) on πi0 is below γi0 , then every solution of (10) in int RN+ satisfies lim inft →∞ xi0 (t ) > 0. Outline of the Proof. Since the proof of this lemma is similar to that of Lemma 5 and is almost verbatim from that of [13, Lemma 7], we only give an outline. Without loss of generality, suppose i0 = N and (10) has a solution x(t ) ∈ int RN+ on [t0 , ∞) satisfying lim inft →∞ xN (t ) = 0. Then a proper J ⊂ IN with N ∈ J and I = IN \ J can be chosen to satisfy (34)–(47). Then we can rewrite (44) and (45) as
∀i ∈ I ,
−
aij m(xj , sn , tn ) = m(ri , sn , tn ) + o(1) (n → ∞),
(68)
j∈I
−
ajn j m(xj , sn , tn ) > m(rjn , sn , tn ) + o(1)
(n → ∞)
(69)
j∈I
instead of (48) and (49). Then choosing a subsequence {nℓ } of {n} if necessary and letting ℓ → ∞ we obtain
∀i ∈ I ,
− j∈I
aij x∗j = bi ,
−
aj0 j x∗j ≥ bj0
(70)
j∈I
instead of (50). Since I ⊂ IN −1 and j0 ∈ J, if j0 < N then (70) contradicts the (I − J )-conditions for the subsystem of (10) with xN = 0. If j0 = N then (70) contradicts the assumption that every equilibrium of (1) on πN is below γN . Therefore, the conclusion holds.
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Fig. 1. Geometric demonstration in the proof of Lemma 7.
Lemma 7. Assume that for each proper subset J ⊂ IN , the linear system Ai x = bi ,
i ∈ I = IN \ I ∗
has a unique solution xJ in
∀j ∈ J ,
CJ0 .
(71) ∗
Assume also that xJ satisfies
Aj x∗J < bj .
(72)
if J ̸= {N }. Then (a) AN x∗{N } < bN H⇒ Ax∗ = b has a unique solution x∗ ∈ int RN+ ; (b) AN x∗{N } = bN H⇒ Ax∗ = b has a unique solution x∗ = x∗{N } in RN+ ; (c) AN x∗{N } > bN H⇒ Ax ̸= b for any x ∈ RN+ . Proof. (a) In this case, for each i ∈ IN , every equilibrium on πi is below γi so b and A satisfy the (I − J )-conditions. By [6, Lemma 2.7] or [9, Lemma 5], Ax = b has a unique solution x∗ ∈ int RN+ . (b) Clearly, x∗{N } is a solution of Ax = b in RN+ . Suppose there is an x∗ ∈ RN+ such that Ax∗ = b but x∗ ̸= x∗{N } . Then Ax = b
has infinitely many solutions in RN+ so rank(A, b) = rank(A) < N. For J = {1}, since (71) has a unique solution x∗{1} in C{01} , the submatrix of A obtained by deleting the first row and the first column of A has an inverse so rank(A, b) = rank(A) = rank((AT2 , . . . , ATN )T ) = N − 1. This shows the equivalence of Ax = b and (71) with J = {1}. Thus, x∗{1} is also a solution of Ax = b. In particular, A1 x∗{1} = b1 , which contradicts (72). Therefore, x∗{N } is the unique solution of Ax = b in RN+ .
(c) Suppose Ax = b has a solution x∗ ∈ RN+ . Obviously, x∗ ̸= 0 as bi > 0 for all i ∈ IN . If x∗ ̸∈ int RN+ then there is a proper subset J ⊂ IN such that x∗ ∈ CJ0 . Since Ax∗ = b implies that x∗ is also a solution of (71) and (71) in CJ0 has a unique solution x∗J , we must have x∗ = x∗J so Ax∗J = b. This contradicts (72) if J ̸= {N }. If J = {N } then AN x∗{N } = bN , which contradicts the assumption AN x∗{N } > bN . Hence, we must have x∗ ∈ int RN+ . Now consider the line ℓ1 in RN+ containing x∗ and x∗{N } . We claim the existence of an ε1 > 1 such that x¯ = ε1 x∗ + (1 − ε1 )x∗{N } = x∗{N } + ε1 (x∗ − x∗{N } ) ∈ ∂ RN+
(73)
so ℓ1 = x∗{N } x¯ . Indeed, since x∗{N } x∗ ⊂ ℓ1 and x∗ ∈ int RN+ , if x = ε x∗ + (1 − ε)x∗{N } ∈ ℓ1 for an ε > 1, then by assumption, aNN xN ≤ AN x = ε AN x∗ + (1 − ε)AN x∗{N } < ε bN + (1 − ε)bN = bN so xN = ε x∗N < bN /aNN and ε < bN /(aNN x∗N ). This shows the existence of ε1 ∈ (1, bN /(aNN x∗N )) satisfying (73). The assumption of the lemma implies that ∩j∈IN −1 γj is a line segment in RN+ containing x∗ and x∗{N } so
γi ∩ (∩j∈IN \{i,N } γj ) = ∩j∈IN −1 γj = ℓ1 = x∗{N } x¯ . As x¯ N = ε1 x∗N > 0, there is an i ∈ IN −1 such that x¯ i = 0 and x¯ j ≥ 0 for j ∈ IN \ {i}. The assumption also implies that ∩j∈IN \{i,N } γj is a two-dimensional plane in RN+ (N ≥ 3) and πi ∩ (∩j∈IN \{i,N } γj ) is a line segment containing x∗{i,N } x¯ . Since {x∗{i} } = πi ∩ (∩j∈IN \{i} γj ) ⊂ πi ∩ (∩j∈IN \{i,N } γj ), x∗{i} , x∗{i,N } and x¯ are on the same line. By (72), Ai x∗{i,N } < bi , Ai x∗{i} < bi but Ai x¯ = bi . Thus, x∗{i} and x∗{i,N } lie on the same side of x¯ . Since the Nth component of x∗{i} is positive whereas x∗{i,N } ∈ πN , x∗{i} must lie between x∗{i,N } and x¯ (see Fig. 1 below). By (73) and the assumption, AN x¯ = AN (ε1 x∗ + (1 − ε1 )x∗{N } ) = ε1 bN + (1 − ε1 )AN x∗{N } < bN so x¯ is below γN . As AN x∗{i} = bN so x∗{i} is on γN , x∗{i,N } must be above γN , i.e. AN x∗{i,N } > bN . This contradicts (72). Therefore, Ax = b has no solution in RN+ .
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Proof of Theorem 2. By the assumption, every (N − 1)-dimensional subsystem of (1) satisfies the (I − J )-conditions. By Theorem 1, every k-dimensional (k < N) subsystem of (10) is permanent. The existence of a unique x∗{N } ∈ C{0N } satisfying (12) follows from [6, Lemma 2.7] or [9, Lemma 5]. (a) Applying Lemma 6 to (10) for every i0 ∈ IN −1 , we obtain (13). If (10) has a solution in int RN+ that satisfies lim inft →∞ xN (t ) > 0, then the same reasoning as that used in the proof of Theorem 3(a) leads to the existence of x∗ ∈ int RN+ satisfying Ax∗ = b, a contradiction to Lemma 7(b). Thus, every solution of (10) in int RN+ satisfies lim inft →∞ xN (t ) = 0. (b) The conclusion follows from Lemmas 6 and 7(c) and the proof of Theorem 3(b). (c) If AN x∗{N } < bN then, for each i ∈ IN , every equilibrium of (1) on πi is below γi , so b and A satisfy the (I − J )-conditions. By Theorem 1, (10) is permanent. Conversely, if (10) is permanent then by parts (a) and (b) we must have AN x∗{N } < bN . References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
S. Ahmad, A.C. Lazer, Necessary and sufficient average growth in a nonautonomous Lotka–Volterra system, Nonlinear Analysis 34 (1998) 191–228. S. Ahmad, A.C. Lazer, On the nonautonomous N-competing species problem, Applicable Analysis 57 (1995) 309–323. A. Tineo, An iterative scheme for the N-competing species problem, Journal of Differential Equations 116 (1995) 1–15. A. Tineo, Necessary and sufficient conditions for extinction of one species, Advanced Nonlinear Studies 5 (2005) 57–71. S. Ahmad, A.C. Lazer, Average growth and extinction in a competitive Lotka–Volterra system, Nonlinear Analysis 62 (2005) 545–557. S. Ahmad, A.C. Lazer, Average growth and total permanence in a competitive Lotka–Volterra system, Annali di Matematica Pura ed Applicata 185 (2006) S47–S67. S. Ahmad, I.M. Stamova, Partial persistence and extinction in N-dimensional competitive systems, Nonlinear Analysis 60 (2005) 821–836. Z. Hou, Global attractor in competitive Lotka–Volterra systems with retardation, Nonlinear Differential Equations and Applications 9 (2002) 397–417. Z. Hou, Global attractor in competitive Lotka–Volterra systems, Mathematische Nachrichten 282 (7) (2009) 995–1008. Z. Hou, Vanishing components in autonomous competitive Lotka–Volterra systems, Journal of Mathematical Analysis and Applications 359 (2009) 302–310. J. Zhao, J. Jiang, Average conditions for permanence and extinction in nonautonomous Lotka–Volterra system, Journal of Mathematical Analysis and Applications 299 (2004) 663–675. J. Zhao, J. Jiang, A.C. Lazer, The permanence and global attractivity in a nonautonomous Lotka–Volterra system, Nonlinear Analysis: Real World Applications 5 (2004) 265–276. Z. Hou, On permanence of all subsystems of competitive Lotka–Volterra systems with delays, Nonlinear Analysis: Real World Applications 11 (2010) 4285–4301.
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Nonlinear Analysis: Real World Applications journal homepage: www.elsevier.com/locate/nonrwa
Permanence and extinction in competitive Lotka–Volterra systems with delays Zhanyuan Hou ∗ Faculty of Computing, London Metropolitan University, North Campus, 166-220 Holloway Road, London N7 8DB, UK
article
abstract
info
Article history: Received 3 May 2010 Accepted 29 December 2010
In this paper, a class of competitive Lotka–Volterra systems are considered that have distributed delays and constant coefficients on interaction terms and have time dependent growth rate vectors with an asymptotic average. Under the assumption that all proper subsystems are permanent, it is shown that the asymptotic behaviour of the system is determined by the relationship between an equilibrium and a nullcline plane of the corresponding autonomous system: if the equilibrium is below the plane then the system is permanent; if the equilibrium is above the plane then this species will go extinct in an exponential rate while the other species will survive. Similar asymptotic behaviour is also retained under an alternative assumption. © 2011 Elsevier Ltd. All rights reserved.
Keywords: Permanence Strong persistence Extinction Lotka–Volterra Competitive systems Distributed delays
1. Introduction Lotka–Volterra type systems of differential equations have been used in modeling population dynamics of a community of species since 1920s. For this reason, the study of such systems has been comprehensive and is still attracting a large number of researchers. The simplest Lotka–Volterra system has the form x′i = xi (bi − Ai x),
i ∈ IN ,
(1)
where, for any positive integer m, Im = {1, 2, . . . , m} and for any positive integer N and each i ∈ IN , Ai = (ai1 , . . . , aiN ) is the ith row of a matrix A = (aij ) and the bi and aij are real constants. We assume that
∀i, j ∈ IN ,
aij ≥ 0,
aii > 0,
bi > 0
(2)
so that (1) reflects the competitive nature. Since xi (t ) denotes the population size of the ith species at time t, as usual, we restrict the study of such systems to the nonnegative cone RN+ = {x ∈ RN : ∀i ∈ IN , xi ≥ 0}. In particular, solutions of (1) are mostly considered in int RN+ = {x ∈ RN+ : ∀i ∈ IN , xi > 0}, the interior of RN+ . Ahmad and Lazer [1] investigated the system x′i (t ) = xi (t )[ri (t ) − Ai x(t )],
i ∈ IN ,
(3)
where the ri (t ) are continuous from R0 = (−∞, ∞), (c , ∞) or [c , ∞) for some c ∈ R, to R satisfying
∀t ∈ R0 , ∀i ∈ IN , 0 < riL ≤ ri (t ) ≤ riM < ∞, ∀i ∈ IN , lim m(ri , t0 , t ) = bi > 0 (uniformly for t0 ∈ R0 ). t →∞
∗
Tel.: +44 20 73203124. E-mail address: [email protected].
1468-1218/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2010.12.027
(4) (5)
Z. Hou / Nonlinear Analysis: Real World Applications 12 (2011) 2130–2141
Here the notation m(g , t0 , t ) is defined by m(g , t0 , t ) = t −1t 0 xi = 0 if and only if i ∈ I }. Under the condition that
∀ i ∈ IN − 1 ,
−
riL >
aij
j∈IN \{i}
riM ajj
t t0
2131
g (s)ds. For any proper subset I ⊂ IN , let CI0 = {x ∈ RN+ :
,
(6)
it is shown (see [2] or [3, Corollary 4.4]) that the (N − 1)-dimensional subsystem of (3) with xN ≡ 0 has a solution x¯ (t ) ∈ C{0N } with x¯ i (t ) ≥ δ > 0 for all i ∈ IN −1 and t ∈ R0 . Note that (5) and (6) imply
∀ i ∈ IN − 1 ,
−
bi >
aij
j∈IN \{i}
bj ajj
,
(7)
which implies the existence of a unique equilibrium x∗ of (1) in C{0N } . Then the relative position of x∗ and the Nth nullcline plane γN = {x ∈ RN+ : AN x = bN } completely determines the global asymptotic behaviour of the solutions of (3) in int RN+ . Under assumptions (4)–(6), Ahmad and Lazer [1] proved the following conclusions for (3): (a) AN x∗ > bN (x∗ is above γN ) H⇒ x¯ is globally attractive; (b) AN x∗ = bN (x∗ is on γN ) H⇒ lim inft →∞ xN (t ) = 0; (c) AN x∗ < bN (x∗ is below γN ) H⇒ (3) is strongly persistent. When r (t ) in (3) is periodic, Tineo [4] relaxed assumptions (4) and (6)–(7) and proved the global attractiveness of a periodic solution (x¯ ∈ C{0N } if AN x∗ ≥ bN or x˜ ∈ int RN+ if AN x∗ < bN with an extra condition). If we drop the global attractiveness from these results and observe only the behaviour of the Nth component xN (t ) of any solution of (3) in int RN+ , then (a) AN x∗ > bN H⇒ limt →∞ xN (t ) = 0; (b) AN x∗ = bN H⇒ lim inft →∞ xN (t ) = 0; (c) AN x∗ < bN H⇒ lim inft →∞ xN (t ) > 0. For (3) with a general r (t ) satisfying (4), (5) and (7), Ahmad and Lazer [5] actually achieved the above (and more). Clearly, AN x∗ > bN is a threshold condition for the Nth species to die out. Other conditions for extinction can be found in [6–12] and the references quoted there. Ecologically, one concern about the community is whether all species can coexist. This is mathematically gauged by the concepts of persistence and permanence. A system is called strongly persistent if every solution in int RN+ satisfies lim inft →∞ xi (t ) > 0 for all i ∈ IN . A system is called permanent if there are r > 0 and M > r such that every solution of the system in int RN+ satisfies r ≤ xi (t ) ≤ M for all i ∈ IN and all sufficiently large t. A system is called strongly permanent if all small perturbations of the system are permanent. Definition 1. A vector b ∈ int RN+ and a matrix A = (aij ) ∈ RN+×N are said to satisfy the (I − J )-conditions if for each proper subset I ⊂ IN with J = IN \ I, if there exists an x∗J ∈ CJ0 such that
∀i ∈ I ,
Ai x∗J = bi
(8)
∀j ∈ J ,
Aj x∗J < bj .
(9)
then
Ahmad and Lazer [6] proved that all subsystems of (3) are strongly permanent if and only if b and A satisfy the (I − J )conditions. Further, under the (I − J )-conditions for the subsystem of (3) with xN = 0, there is an x∗ ∈ C{0N } such that Ai x∗ = bi for all i ∈ IN −1 . It is also shown [6, Theorem 4.1] that AN x∗ > bN implies limt →∞ xN (t ) = 0 provided xN (t0 ) is small enough. In a recent paper [13], the author investigated the following system with distributed delays,
xi (t ) = xi (t ) ri (t ) − ′
N − j =1
∫
0
xj (t + θ )dξij (θ ) ,
aij
i ∈ IN ,
(10)
−τ
where the ri (t ) and aij are the same as in (3), τ > 0, and the ξij are nondecreasing functions from any open interval containing [−τ , 0] to R+ satisfying
∀i, j ∈ IN ,
ξij (0+ ) − ξij (−τ − ) = 1.
(11)
The theorem below can be viewed as an extension of a result of [6] from (3) to (10). Theorem 1 ([13, Theorems 4,5]). If b and A satisfy the (I − J )-conditions, then all subsystems of (10) and its small perturbations are permanent. Conversely, if all subsystems of (10) and its small perturbations on A are permanent, then b and A satisfy the (I − J )-conditions.
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In this paper, we are concerned with the global asymptotic behaviour of solutions of (10) when the (I − J )-conditions are slightly violated or the (I − J )-conditions for an (N − 1)-dimensional subsystem of (10) are satisfied. We shall prove that, under certain assumptions, a threshold condition AN x∗ > bN will drive the Nth species to go extinct in an exponential rate while the other species will survive, and that the reversed inequality can make the system permanent or strongly persistent. 2. Main results and examples We consider system (10) with assumptions (2), (4), (5) and (11). Theorem 2. Assume that b and A satisfy the (I − J )-conditions except I ̸= IN −1 . Then every k-dimensional subsystem of (10) with 1 ≤ k < N is permanent and there is a unique x∗{N } ∈ C{0N } such that
∀ i ∈ IN − 1 ,
Ai x∗{N } = bi .
(12)
Moreover, the following conclusions hold. (a) If AN x∗{N } = bN then every solution of (10) in int RN+ satisfies
∀ i ∈ IN − 1 ,
lim inf xi (t ) > 0
(13)
t →∞
and lim inft →∞ xN (t ) = 0. (b) If δ0 = AN x∗{N } − bN > 0, then every solution of (10) in int RN+ satisfies (13) and
∀δ ∈ (0, δ0 ),
xN (t ) = o(e−δ t ) (t → ∞).
(14)
(c) System (10) is permanent if and only if AN x∗{N } < bN . To give a geometric interpretation of the conditions, let
πi = {x ∈ RN+ : xi = 0},
i ∈ IN ,
γi = {x ∈ RN+ : Ai x = bi },
(15)
i ∈ IN .
(16)
We note that γi defined by (16) is the ith nullcline plane of (1) in R+ and πi defined by (15) is the ith coordinate plane. For any (N − 1)-dimensional plane γ , if the origin 0 ̸∈ γ , then γ divides RN+ into three mutually exclusive connected sets γ − , γ and γ + with 0 ∈ γ − . A point x ∈ RN+ is said to be below (on or above) γ if x ∈ γ − (x ∈ γ or x ∈ γ + ). A set S is said to be below (on or above) γ if every point in S is so. N
Remark 1. Under assumption (2), it is shown in [13, Remarks 1,2] that the following conditions are equivalent: (a) b and A satisfy the (I − J )-conditions. (b) For each proper subset I ⊂ IN with J = IN \ I, (1) has a unique equilibrium x∗J ∈ CJ0 such that
∀j ∈ J ,
Aj x∗J < bj .
(c) For each i ∈ IN every equilibrium of (1) on πi is below γi . Remark 2. The condition of Theorem 2 is equivalent to each of the following: (a) For each proper subset I ⊂ IN (I ̸= IN −1 ) with J = IN \ I, (1) has a unique equilibrium x∗J ∈ CJ0 such that Aj x∗J < bj holds for all j ∈ J. (b) For each i ∈ IN , every equilibrium of (1) on πi (except the equilibrium x∗{N } ∈ πN ) is below γi . Obviously, under this condition, b and A satisfy the (I − J )-conditions if and only if x∗{N } is below γN . Remark 3. Let us compare the condition of Theorem 2 with condition (7). Let Yi = bi /aii for all i ∈ IN and let [0, Y ] be the set of x ∈ RN+ satisfying 0 ≤ xi ≤ Yi for all i ∈ IN . Then it is clear that all equilibria of (1) fall into the cell [0, Y ]. The geometric interpretation of (7) is that the set πi ∩ [0, Y ] is below γi for all i ∈ IN −1 . Thus, (7) implies that for each i ∈ IN −1 every equilibrium of (1) on πi is below γi . But (7) does not imply any relationship between γN and the equilibria on πN . Hence, neither of (7) and the condition of Theorem 2 completely implies the other. If b and A satisfy the (I − J )-conditions, then by [6, Lemma 2.7] or [9, Lemma 5], (1) in int RN+ has a unique equilibrium x = A−1 b. Thus, if the (I − J )-conditions for the subsystem of (1) with xN = 0 are satisfied then (1) in C{0N } has a unique equilibrium x∗{N } . Consider the following (N − 1)-dimensional auxiliary system ∗
′
[
xi = xi bi − daiN −
N −1 −
]
aij xj ,
i ∈ IN − 1 ,
(17)
j =1
where the bi and aij are the same as in (1) and d ∈ [0, bN /aNN ] is a constant. By saying that (17) satisfies the (I − J )-conditions, we mean bi − daiN > 0 for all i ∈ IN −1 and the obvious replacements of N by N − 1 and bi by bi − daiN in Definition 1.
Z. Hou / Nonlinear Analysis: Real World Applications 12 (2011) 2130–2141
2133
Theorem 3. Assume that (17) satisfies the (I − J )-conditions for both d = 0 and d = bN /aNN . Then the (N − 1)-dimensional subsystem of (10) with xN ≡ 0 is permanent and the following conclusions hold. (a) If AN x∗{N } = bN then lim inft →∞ xN (t ) = 0 and (13) holds for every solution of (10) in int RN+ . (b) If δ0 = AN x∗{N } − bN > 0, then every solution of (10) in int RN+ satisfies (13) and (14). (c) System (10) is strongly persistent if and only if AN x∗{N } < bN . Remark 4. Note that condition (7) can be rearranged as
∀ i ∈ IN − 1 ,
−
bi − daiN >
aij
j∈IN −1 \{i}
bj ajj
,
d=
bN aNN
.
(18)
This implies that for each i ∈ IN −1 every equilibrium of (17) with d = bN /aNN on the ith coordinate plane is below the ith nullcline plane. Thus, (7) implies that (17) with d = bN /aNN satisfies the (I − J )-conditions. Also, (7) implies
∀ i ∈ IN − 1 ,
bi >
−
aij
j∈IN −1 \{i}
bj ajj
,
(19)
which implies that (17) with d = 0 also satisfies the (I − J )-conditions. Therefore, (7) implies the condition of Theorem 3. On the other hand, however, Example 1 given below shows that the condition of Theorem 3 does not imply (7). This shows that Theorem 3 even in the case of τ = 0 covers some of the results given in [5] as a special case. Remark 5. The two examples given below show that the condition of Theorem 2 is independent of the condition of Theorem 3. Remark 6. Just as commented in [13], the asymptotic behaviour of the solutions of (10) conveyed by Theorems 2 and 3 is not affected by the delays since the conditions of the theorems do not involve the delays at all. Remark 7. Although Theorems 2 and 3 are stated in terms of N ∈ IN and x∗{N } only, this is purely for the convenience of formulating the statements and presenting the proofs. In practice, if the concerned component is xi for some i ∈ IN rather than xN , then we can always swap the position of xi with xN before applying the theorems to the system. In the following examples, we present the coordinates of any point x ∈ RN+ in a row form although it is viewed as a column vector or an N × 1 matrix whenever it appears with matrix operations. Example 1. Consider systems (1) and (10) with
2 1 A= 1 1
1 2 1 1
1 1 3 2
2 2 , 1 4α
1 1 b = , 1.5
where α is a positive constant. For d ∈
α
0,
1 4
, the corresponding system (17) has the following nontrivial equilibria on
∂ R3+ : p1 = (0.5 − d, 0, 0), p2 = (0, 0.5 − d, 0), p3 = 0, 0, 0.5 − 3d , p4 = 1−32d , 1−32d , 0 , p5 = (0.3 − d, 0, 0.4), p6 = (0, 0.3 − d, 0.4). On π1 we have p2 , p3 , p6 , which are all below γ1 = {x ∈ R3+ : 2x1 + x2 + x3 = 1 − 2d}. On π2 we have p1 , p3 , p5 , which are all below γ2 = {x ∈ R3+ : x1 + 2x2 + x3 = 1 − 2d}. On π3 we have p1 , p2 , p4 , which are below γ3 = {x ∈ R3+ : x1 + x2 + 3x3 = 1.5 − d}. Therefore, (17) satisfies the (I − J )-conditions for all d ∈ 0, 14 so the condition 3 3 5 of Theorem 3 is met. The equilibrium x∗{4} for (1) is 14 , 14 , 14 , 0 . As A4 x∗{4} − b4 = 87 − α , by Theorem 3, (10) is strongly 8 8 persistent if α > 7 , (13) holds if 0 < α ≤ 7 , lim inft →∞ x4 (t ) = 0 if α = 87 and (14) holds if 0 < α < 87 . Note that a12 b2 a22
+
a13 b3 a33
+
a14 b4 a44
=
1 2
+
1.5 3
+
2α 4α
=
3 2
> 1 = b1 .
Thus, condition (7) is not satisfied. Note also that (1) on π4 has an equilibrium p = A4 p = 2/3 ≥ α = b4
if 0 < α ≤ 2/3.
Thus, from Remark 2, the condition of Theorem 2 is not fulfilled if α ∈ 0,
Example 2. Consider systems (1) and (10) with
3 2 A= 2
α
2 3 2
α
2 2 3
α
1 1 , 2 4
1
1 b = , 1 1
2 3
.
1 3
, 13 , 0, 0 but
2134
Z. Hou / Nonlinear Analysis: Real World Applications 12 (2011) 2130–2141
where α is a constant in 1,
. On ∂ R4+ system (1) has the following nontrivial equilibria:
5 2
1 1 1 , 0, 0, 0 , p2 = 0, , 0, 0 , p3 = 0, 0, , 0 , p4 = 0, 0, 0, , 3 3 3 4 1 1 1 3 1 1 1 3−α p5 = , , 0, 0 , p6 = 0, , , 0 , p7 = , 0, , 0 , p8 = , 0, 0, , 5 5 5 5 5 5 12 − α 12 − α 3 3 3−α 1 3−α 3 5 − 2α p9 = 0, , 0, , p10 = 0, 0, , , p11 = , , 0, , 12 − α 12 − α 6 − α 2(6 − α) 20 − 2α 20 − 2α 20 − 2α 5−α α 5 − 2α 5−α α 5 − 2α 1 1 1 p12 = , 0, , , p13 = 0, , , , x∗{4} = , , ,0 . 20 − 3α 20 − 3α 20 − 3α 20 − 3α 20 − 3α 20 − 3α 7 7 7 p1 =
1
On π1 we have p2 , p3 , p4 , p6 , p9 , p10 and p13 and they are all below γ1 defined by 3x1 + 2x2 + 2x3 + x4 = 1 in R4+ . On π2 we have p1 , p3 , p4 , p7 , p8 , p10 and p12 , which are below γ2 defined by 2x1 +3x2 +2x3 +x4 = 1. On π3 we have p1 , p2 , p4 , p5 , p8 , p9 and p11 , which are all below γ3 defined by 2x1 + 2x2 + 3x2 + 2x4 = 1. On π4 we have p1 , p2 , p3 , p5 , p6 , p7 and x∗{4} which, except x∗{4} , are below γ4 given by α x1 +α x2 +α x3 + 4x4 = 1. Thus, by Remark 2, the condition of Theorem 2 is satisfied. Since A4 x∗{4} − b4 = 37α − 1, by Theorem 2 system (10) is permanent if 1 < α < if α = p=
7 and (14) holds 3 3 3 0 but 20 20
,
,
if
7 3
7 , 3
(13) holds if
< α < 25 . Note that the corresponding system (17) with d =
7 3 1 4
≤ α < 52 , lim inft →∞ x4 (t ) = 0 on π3 in R3+ has an equilibrium
(a31 , a32 , a33 )p = 0.6 > 0.5 = b3 − da34 . Thus, (17) with d =
1 4
does not satisfy the (I − J )-conditions so the condition of Theorem 3 is not met.
3. Proof of Theorem 3 Lemma 1 ([13, Lemmas 1,2]). There exist σ > 0, ρ > 0 and T = T (ϕ) > 0 (for each ϕ ∈ C ([−τ , 0], RN+ ) with ϕ(0) ̸= 0) such that the solution of (10) with xt0 = ϕ for all t0 ∈ R0 satisfies
∀t ≥ t0 + T ,
x1 (t ) + · · · + xN (t ) ≥ σ
and,
∀i ∈ IN , xi (t , t0 , ϕ) < ρ.
Lemma 2. If x(t ) is a solution of (10) in int RN+ on [t0 , ∞), then
∀i ∈ IN , ∀t ≥ T ≥ t0 ,
lim sup m(xi , T , t ) ≤
(t −T )→∞
bi aii
.
(20)
Proof. Let x(t ) ∈ int RN+ be a solution of (10) for t ≥ t0 and let i ∈ IN be fixed. Case 1. Suppose lim inft →∞ xi (t ) > 0. Then, by Lemma 1, ln[xi (t )/xi (T )] is bounded for all t ≥ T ≥ t0 so 1
lim
(t −T )→∞
t −T
ln
xi (t ) xi ( T )
= 0.
(21)
By Lemma 1 and (11), for all k, j ∈ IN , 1 t −T
∫ t∫ T
0
∫ 0 ∫
1
xj (s + θ )dξkj (θ )ds =
t −T
−τ
−τ
t +θ
xj (s)ds dξkj (θ )
T +θ
= m(xj , T , t ) + o(1) ((t − T ) → ∞).
(22)
Dividing (10) by xi (t ), integrating the equation from T to t and dividing through by t − T , we obtain
∫ t ∫ 0 N − aij ln = m(ri , T , t ) − xj (s + θ )dξij (θ ) ds t −T xi (T ) t −T T −τ j=1 ∫ ∫ 1
xi ( t )
≤ m(ri , T , t ) −
t
aii
t −T
0
xi (s + θ )dξii (θ ) ds.
T
(23)
−τ
Then combination of (21)–(23) gives aii m(xi , T , t ) ≤ m(ri , T , t ) + o(1)
((t − T ) → ∞).
From this and (5) the inequality in (20) follows.
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2135
Case 2. If lim inft →∞ xi (t ) = 0 and lim(t −T )→∞ m(xi , T , t ) = 0 then the inequality in (20) is obvious. Case 3. Suppose lim inft →∞ xi (t ) = 0 but lim sup xi (t ) ≥ lim sup m(xi , T , t ) = σ0 > 0. (t −T )→∞
t →∞
Then there is a sequence {[sn , tn ]} with tn − sn → ∞ as n → ∞ such that m(xi , sn , tn ) > σ0 /2 for all n ≥ 1 and limn→∞ m(xi , sn , tn ) = σ0 . If there is a subsequence {nℓ } ⊂ {n} such that xi (snℓ ) ≤ xi (tnℓ ) holds for all ℓ ≥ 1, then replacing t by tnℓ and T by snℓ in (22)–(24) and letting ℓ → ∞, we obtain the inequality in (20). Otherwise, xi (sn ) > xi (tn ) holds for some integer K ≥ 1 and all n ≥ K . (i) Suppose there is a subsequence {nℓ } ⊂ {n} such that snℓ = t0 for all ℓ ≥ 1. Then lim supt →∞ m(xi , t0 , t ) = σ0 so lim supt →∞ m(xi , T , t ) = σ0 for any fixed T ≥ t0 . Since lim inft →∞ xi (t ) = 0, we can choose a T1 ≥ t0 and {tn′ } ⊂ [T1 , ∞) with tn′ ↑ ∞ as n → ∞ such that xi (T1 ) < σ0 /2, m(xi , T1 , tn′ ) > σ0 /2 for all n ≥ 1, and limn→∞ m(xi , T1 , tn′ ) = σ0 . If there is ′ a subsequence {mℓ } ⊂ {n} such that xi (tm ) ≥ xi (T1 ) for all ℓ ≥ 1, then (20) follows from (22)–(24) after the replacements ℓ ′ of T by T1 and t by tmℓ . Otherwise, there is an integer K1 > 0 such that xi (tn′ ) < xi (T1 ) holds for all n ≥ K1 . Then, for each fixed n ≥ K1 , there is a vn ∈ (T1 , tn′ ) satisfying xi (t ) < xi (vn ) = xi (T1 ) for all t ∈ (vn , tn′ ]. Thus,
(tn′ − T1 )m(xi , T1 , tn′ ) < (vn − T1 )m(xi , T1 , vn ) + (tn′ − vn )xi (T1 ). If {vn } is bounded then
vn − T1 tn′ − vn → 0 and → 1 (n → ∞) tn′ − T1 tn′ − T1 so the above inequality implies σ0 = limn→∞ m(xi , T1 , tn′ ) ≤ xi (T1 ) < σ0 /2, a contradiction to σ0 > 0. This contradiction shows that {vn } is unbounded. Without loss of generality, we may assume vn ↑ ∞ as n → ∞. Then, as m(xi , T1 , vn ) > m(xi , T1 , tn′ ) +
tn′ − vn
vn − T1
[m(xi , T1 , tn′ ) − xi (T1 )]
> m(xi , T1 , tn′ ), we have limn→∞ m(xi , T1 , vn ) = σ0 . Hence, (20) follows from (22)–(24) with the replacements of T by T1 and t by vn . (ii) If (i) is not true, then there is an integer K2 > K such that
∀n ≥ K2 , sn > t0 ,
xi (sn ) > xi (tn ).
For each n ≥ K2 , by continuity, xi (t + sn ) > xi (t + tn ) holds, so m(xi , t + sn , t + tn ) > m(xi , sn , tn ) when t < 0 and |t | is sufficiently small. Then there are s′n ∈ [t0 , sn ) and tn′ = s′n + tn − sn satisfying m(xi , s′n , tn′ ) = max m(xi , t , t + tn − sn ) > m(xi , sn , tn ). t ∈[t0 ,sn ]
Clearly, we have either xi (s′n ) ≤ xi (tn′ ) or s′n = t0 . If there is a subsequence {nℓ } ⊂ {n} such that s′nℓ = t0 for all ℓ ≥ 1, then (20) follows from (i). Otherwise, replacing t by tn′ and T by s′n in (22)–(24) and letting n → ∞, we obtain the inequality in (20). Let A˜ be the (N − 1) × (N − 1) submatrix of A obtained by deleting the last row and the last column of A and let b˜ be the (N − 1)-dimensional vector obtained from b by deleting its last component. For any proper subset J ⊂ IN −1 and any (N − 1)-dimensional (row or column) vector v , let vJ be the vector obtained from v by deleting its jth component for all j ∈ J and let A˜ J denote the submatrix of A˜ obtained by deleting the jth row and the jth column of A˜ for all j ∈ J. Let A˜ i be the ith row of A˜ and set CN A = (a1N , . . . , a(N −1)N )T . Lemma 3. Assume that system (17) satisfies the (I − J )-conditions for both d = 0 and d = bN /aNN . Then (17) satisfies the (I − J )-conditions for every d ∈ [0, bN /aNN ]. Proof. For each proper subset J ⊂ IN −1 , we show that the linear algebraic system A˜ J xJ = bJ − d(CN A)J
(25)
for each fixed d ∈ [0, bN /aNN ] has a unique solution xJ (d) with positive components satisfying
∀j ∈ J ,
(A˜ j )J xJ (d) < bj − dajN .
(26)
Since (17) with d = 0 satisfies the (I − J )-conditions, by Remark 2 there is a unique x¯ J with positive components satisfying A˜ J x¯ J = bJ ,
∀j ∈ J ,
(A˜ j )J x¯ J < bj .
(27)
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Z. Hou / Nonlinear Analysis: Real World Applications 12 (2011) 2130–2141
By the (I − J )-conditions for (17) with d = bN /aNN and Remark 2, there is a unique x˜ J with positive components satisfying A˜ J x˜ J = bJ − (bN /aNN )(CN A)J ,
∀j ∈ J ,
(28)
(A˜ j )J x˜ J < bj − (bN /aNN )ajN .
(29)
1 Then the uniqueness of x¯ J and x˜ J implies the existence of A˜ − and the unique solution xJ (d) of (25). Thus, J
∀d ∈ [0, bN /aNN ],
1 xJ (d) = A˜ − J (bJ − d(CN A)J ).
(30)
Since both xJ (0) = x¯ J and xJ (bN /aNN ) = x˜ J have positive components only and xJ (d) is linear in d, all components of xJ (d) are positive for all d ∈ [0, bN /aNN ]. Now consider the linear function Fj (d) = bj − dajN − (A˜ j )J xJ (d),
d ∈ [0, bN /aNN ], j ∈ J .
(31)
Since Fj (0) > 0 by (27) and Fj (bN /aNN ) > 0 by (29), we have Fj (d) > 0 for all d ∈ [0, bN /aNN ] and j ∈ J. Thus, for each d ∈ [0, bN /aNN ], the unique solution xJ (d) of (25) given by (30) satisfies (26). Therefore, (17) for any fixed d ∈ [0, bN /aNN ] satisfies the (I − J )-conditions. Lemma 4. Assume that (17) satisfies the (I − J )-conditions for both d = 0 and d = bN /aNN . Then (1) in C{0N } has a unique equilibrium x∗{N } and the following conclusions hold. (a) AN x∗{N } < bN H⇒ Ax∗ = b has a unique solution x∗ ∈ int RN+ ; (b) AN x∗{N } = bN H⇒ Ax∗ = b has a unique solution x∗ = x∗{N } in RN+ ; (c) AN x∗{N } > bN H⇒ Ax ̸= b for any x ∈ RN+ .
˜ = b˜ Proof. Since the (I − J )-conditions hold for (17) with both d = 0 and d = bN /aNN , by [6, Lemma 2.7] or [9, Lemma 5], Ax ˜ = b˜ −(bN /aNN )(CN A) has a unique solution y˜ = A˜ −1 (b˜ −(bN /aNN )(CN A)) ∈ has a unique solution y¯ = A˜ −1 b˜ ∈ int RN+−1 and Ax int RN+−1 . Define x¯ and x˜ in RN+ by x¯ N = 0, Then x¯ =
x∗{N }
x˜ N = bN /aNN ,
∀i ∈ IN −1 ,
Ai x∗{N } = bi ,
Ai x˜ = bi .
x¯ i = y¯ i ,
x˜ i = y˜ i .
(32)
and
∀ i ∈ IN − 1 ,
(33)
This shows that the line segment x∗{N } x˜ connecting x∗{N } to x˜ lies in ∩i∈IN −1 γi . Since the existence of A˜ −1 implies that ∩i∈IN −1 γi is a line segment containing x∗{N } x˜ , if Ax = b has a solution in RN+ then 0 ≤ xN ≤ bN /aNN = x˜ N so x must be on x∗{N } x˜ and satisfy AN x = bN . (a) If AN x∗{N } < bN , since AN x˜ ≥ aNN x˜ N = bN , there is a unique x∗ ∈ x∗{N } x˜ \ {x∗{N } } ⊂ int RN+ such that AN x∗ = bN so that Ax = b. (b) If AN x∗{N } = bN then there is an i0 ∈ IN −1 such that aNi0 > 0 so AN x˜ ≥ aNi0 x˜ i0 + aNN x˜ N > bN . Thus, x = x∗{N } is the unique point on x∗{N } x˜ meeting AN x = bN and Ax = b. ∗
(c) If AN x∗{N } > bN , then there is an i0 ∈ IN −1 such that aNi0 > 0 so AN x˜ ≥ aNi0 x˜ i0 + aNN x˜ N > bN . Thus, AN x > bN holds for
all x ∈ x∗{N } x˜ . Therefore, Ax = b has no solution in RN+ .
Lemma 5. Assume that (17) satisfies the (I − J )-conditions for both d = 0 and d = bN /aNN . Then every solution of (10) in int RN+ satisfies (13). Proof. Suppose (10) has a solution x(t ) ∈ int RN+ on [t0 , ∞) satisfying, for some i0 ∈ IN −1 , lim inft →∞ xi0 (t ) = 0. Then, by Lemma 1, there is a subset J ⊂ IN −1 fulfilling lim inf max{xj (t ) : j ∈ J } = 0,
(34)
∀i ∈ I ,
(35)
t →∞
lim inf max{xj (t ) : j ∈ J ∪ {i}} > 0, t →∞
where I = IN −1 \ J if J ̸= IN −1 or I = {N } otherwise. By Lemma 1, there is a ρ0 > 0 such that xi (t ) ≤ ρ0 for all i ∈ IN and t ≥ t0 . Thus, (35) implies the existence of δ1 ∈ (0, ρ0 ) such that
∀i ∈ I , ∀t ≥ t0 ,
δ1 ≤ max{xj (t ) : j ∈ J ∪ {i}} ≤ ρ0 .
(36)
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Put a0 = max{aij : i, j ∈ IN },
(37)
ρ1 = min{δ1 , xi (t0 ) : i ∈ IN }.
(38)
As ρ1 > 0, by (34) there is a sequence {tn } with tn ↑ ∞ as n → ∞ satisfying
∀n ≥ 1 ,
max{xj (tn ) : j ∈ J } <
ρ1 2n
e−nN ρ0 a . 0
(39)
By (38), (39) and continuity, there are sn ∈ (t0 , tn ) and jn ∈ J such that
∀n ≥ 1 ,
max{xj (sn ) : j ∈ J } = xjn (sn ) =
∀n ≥ 1, ∀t ∈ (sn , tn ],
ρ1 2n
max{xj (t ) : j ∈ J } <
,
(40)
ρ1 2n
.
(41)
Then (36), (40) and (41) imply that
∀i ∈ I , ∀t ∈ [sn , tn ],
δ1 ≤ xi (t ) ≤ ρ0 .
(42)
From (10), (11), (37) and (40), we have
∫
tn
xjn (tn ) = xjn (sn ) exp
[
rjn (t ) −
N −
sn
ρ1
>
2n
∫
xj (t + θ )dξjn j (θ ) dt
aj n j −τ
j=1
0 e−N ρ0 a (tn −sn ) ,
]
0
∀n ≥ 1.
Combining this with (39), we obtain
∀n ≥ 1 ,
tn − sn > n.
(43)
Now from (22) and the way of obtaining (23) from (10), we derive, as n → ∞,
∀i ∈ I , 1 tn − sn
1
ln
tn − sn ln
xjn (tn )
xi (tn ) xi (sn )
= m(ri , sn , tn ) −
aij m(xj , sn , tn ) + o(1),
(44)
j =1
= m(rjn , sn , tn ) −
xjn (sn )
N −
N −
ajn j m(xj , sn , tn ) + o(1).
(45)
j =1
Since from (40) and (41) follows
∀j ∈ J ,
lim m(xi , sn , tn ) = 0
(46)
n→∞
and (42) and (43) imply
∀i ∈ I ,
lim
1
n→∞ tn
− sn
ln
xi (tn ) xi (sn )
= 0,
(47)
by (40) and (41) again, we can rewrite (44) and (45) as
∀i ∈ I ,
−
aij m(xj , sn , tn ) = m(ri , sn , tn ) + o(1) (n → ∞),
(48)
j∈I ∪{N }
−
ajn j m(xj , sn , tn ) > m(rjn , sn , tn ) + o(1)
(n → ∞).
(49)
j∈I ∪{N }
Note that for all n ≥ 1, jn ∈ J , m(xj , sn , tn ) ∈ [δ1 , ρ0 ] for all j ∈ I and, if I ̸= {N }, lim supn→∞ m(xN , sn , tn ) ∈ [0, bN /aNN ] by Lemma 2. Then there is a subsequence {nℓ } ⊂ {n} such that jnℓ = j0 ∈ J for all ℓ ≥ 1 and
∀j ∈ I ∪ {N },
lim m(xj , snℓ , tnℓ ) = x∗j ,
ℓ→∞
where x∗j ∈ [δ1 , ρ0 ] for all j ∈ I and, if I ̸= {N }, x∗N ∈ [0, bN /aNN ]. With the replacement of n by nℓ in (48) and (49) and by assumption (5), a limit process as ℓ → ∞ results in
∀i ∈ I ,
− j∈I ∪{N }
aij x∗j = bi ,
− j∈I ∪{N }
aj0 j x∗j ≥ bj0 .
(50)
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When I = {N }, (50) becomes aj0 N x∗N ≥ bj0 .
aNN x∗N = bN ,
(51)
From this follows aj0 N bN /aNN ≥ bj0 . This is impossible since, by the (I − J )-conditions for (17) with d = bN /aNN , we require bi − aiN bN /aNN > 0 for all i ∈ IN −1 . When I ̸= {N }, (50) becomes
−
∀i ∈ I ,
−
aij x∗j = bi − aiN x∗N ,
j∈I
aj0 j x∗j ≥ bj0 − aj0 N x∗N .
(52)
j∈I
From Lemma 3 we know that (17) with d = x∗N also satisfies the (I − J )-conditions. Thus, for the above I ⊂ IN −1 and J = IN −1 \ I we should have
−
∀i ∈ I ,
aij x∗j = bi − aiN x∗N ,
∀k ∈ J ,
j∈I
−
akj x∗j < bk − akN x∗N .
(53)
j∈I
Clearly, (52) and (53) are in conflict. This shows the truth of the conclusion of the lemma.
Proof of Theorem 3. The permanence of the subsystem
xi (t ) = xi (t ) ri (t ) − ′
N −1 −
∫
xj (t + θ )dξij (θ ) ,
aij
j =1
0
i ∈ IN − 1
−τ
follows from the (I − J )-conditions for (17) with d = 0 and Theorem 1. (a) By Lemma 5, every solution of (10) in int RN+ satisfies (13). Suppose (10) in int RN+ has a solution satisfying lim inft →∞ xN (t ) > 0. Then, by Lemma 1, this solution on [t0 , ∞) satisfies
∀t ≥ t0 , ∀i ∈ IN ,
0 < δ1 ≤ xi (t ) ≤ ρ0 < ∞.
(54)
By the method of obtaining (21)–(24), we have N −
∀ i ∈ IN ,
aij m(xj , t0 , t ) = m(ri , t0 , t ) + o(1) (t → ∞).
(55)
j =1
From (5), (54) and (55) we obtain an x∗ ∈ int RN+ such that Ax∗ = b. But this contradicts Lemma 4(b). Therefore, we must have lim inft →∞ xN (t ) = 0 for every solution of (10) in int RN+ . (b) By Lemmas 4 and 5(c), the proof of part (a) is still valid to show that every solution of (10) in int RN+ satisfies (13) and lim inft →∞ xN (t ) = 0. Suppose lim supt →∞ xN (t ) > 0 for a particular solution. Then we can choose a T ≥ t0 and an increasing sequence {tn } ⊂ [T , ∞) such that
∀n ≥ 1,
xN (tn ) > xN (T ),
∀ i ∈ IN − 1 ,
(56)
lim m(xi , T , tn ) = x∗i > 0,
(57)
n→∞
lim m(xN , T , tn ) = x∗N ≥ 0.
(58)
n→∞
Then, by (56) and Lemmas 1 and 5, ln[xi (tn )/xi (T )] is bounded for all i ∈ IN and n ≥ 1. By the method of obtaining (55), we have
∀i ∈ IN , ∀n ≥ 1,
N −
aij m(xj , T , tn ) = m(ri , T , tn ) + o(1) (n → ∞).
(59)
j =1
This leads to Ax∗ = b as n → ∞, a contradiction to Lemma 4(c). Hence, every solution of (10) in int RN+ satisfies limt →∞ xN (t ) = 0. Now suppose (14) does not hold for a solution of (10) in int RN+ . Then, for some σ1 ∈ (0, δ0 ), lim supt →∞ xN (t )eσ1 t > 0 so for any fixed δ ∈ (σ1 , δ0 ), lim supt →∞ xN (t )eδ t = ∞. Let y(t ) = xN (t )eδ t . From the Nth component equation of (10) we obtain, as t → ∞, 1 t − t0
ln
y(t ) y(t0 )
= δ + m(rN , t0 , t ) −
N −
aNj m(xj , t0 , t ) + o(1).
(60)
j =1
Take a sequence {tn } with tn ↑ ∞ as n → ∞ such that y(tn ) > y(t0 ) so that ln[y(tn )/y(t0 )] > 0 for all n ≥ 1. By taking subsequences if necessary, we may assume that (57) holds. Note that limt →∞ m(xN , t0 , t ) = 0 by limt →∞ xN (t ) = 0. Then
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(60) with the replacement of t by tn gives N −1 −
aNj m(xj , t0 , tn ) < δ + m(rN , t0 , tn ) + o(1)
(n → ∞).
(61)
j=1
From (10) we also have
∀ i ∈ IN − 1 ,
N −1 −
aij m(xj , t0 , tn ) = m(ri , t0 , tn ) + o(1) (n → ∞).
(62)
j=1
Then a limit process as n → ∞ in (61) and (62) gives
∀ i ∈ IN − 1 ,
A i x∗ = bi ,
AN x∗ ≤ δ + bN ,
(63)
(I − J )-conditions for (17) with d = 0 imply that is the unique solution of Ai x = bi for x∗ = x∗{N } in (63). Thus, since δ0 = AN x∗{N } − bN > 0, the inequality in (63) gives δ0 ≤ δ , a contradiction to δ ∈ (σ1 , δ0 ). Therefore, every solution of (10) in int RN+ satisfies (14). (c) Suppose AN x∗{N } < bN and (10) has a solution x(t ) ∈ int RN+ satisfying lim inft →∞ xN (t ) = 0. Since Lemmas 1 and 5 ensure the existence of ρ0 > σ0 > 0 such that where
all i ∈
x C{0N } . Since the IN −1 in C{0N } , we have ∗
x∗{N }
∈
∀i ∈ IN −1 , ∀t ≥ t0 ,
0 < σ0 ≤ xi (t ) ≤ ρ0 ,
(64)
using the same technique as that used in the proof of Lemma 5, we can choose a sequence {[sn , tn ]} with tn − sn → ∞ as n → ∞ such that
∀n ≥ 1, ∀t ∈ (sn , tn ], ∀ i ∈ IN − 1 ,
xN (sn ) > xN (t ),
lim xN (sn ) = 0,
(65)
n→∞
lim m(xi , sn , tn ) = x∗i ∈ [σ0 , ρ0 ].
(66)
n→∞
Since (65) implies ln[xN (tn )/xN (sn )] < 0 and limn→∞ m(xN , sn , tn ) = 0, from (10), (65) and (66), a limit process as n → ∞ produces
∀ i ∈ IN − 1 , ∗
A i x∗ = bi ,
A N x∗ ≥ bN ,
(67)
C{0N } .
∗
x∗{N }
AN x∗{N }
where x ∈ By the same reasoning as that used in part (b), we have x = so ≥ bN , a contradiction to the assumption. Therefore, (10) is strongly persistent. Conversely, if (10) is strongly persistent then from parts (a) and (b) we must have AN x∗{N } < bN . 4. Proof of Theorem 2 Lemma 6. Assume that, for some i0 ∈ IN , the (N − 1)-dimensional subsystem
[
x′i = xi bi −
−
]
aij xj ,
i ∈ IN \ { i 0 }
j∈IN \{i0 }
of (1) satisfies the (I − J )-conditions. If every equilibrium of (1) on πi0 is below γi0 , then every solution of (10) in int RN+ satisfies lim inft →∞ xi0 (t ) > 0. Outline of the Proof. Since the proof of this lemma is similar to that of Lemma 5 and is almost verbatim from that of [13, Lemma 7], we only give an outline. Without loss of generality, suppose i0 = N and (10) has a solution x(t ) ∈ int RN+ on [t0 , ∞) satisfying lim inft →∞ xN (t ) = 0. Then a proper J ⊂ IN with N ∈ J and I = IN \ J can be chosen to satisfy (34)–(47). Then we can rewrite (44) and (45) as
∀i ∈ I ,
−
aij m(xj , sn , tn ) = m(ri , sn , tn ) + o(1) (n → ∞),
(68)
j∈I
−
ajn j m(xj , sn , tn ) > m(rjn , sn , tn ) + o(1)
(n → ∞)
(69)
j∈I
instead of (48) and (49). Then choosing a subsequence {nℓ } of {n} if necessary and letting ℓ → ∞ we obtain
∀i ∈ I ,
− j∈I
aij x∗j = bi ,
−
aj0 j x∗j ≥ bj0
(70)
j∈I
instead of (50). Since I ⊂ IN −1 and j0 ∈ J, if j0 < N then (70) contradicts the (I − J )-conditions for the subsystem of (10) with xN = 0. If j0 = N then (70) contradicts the assumption that every equilibrium of (1) on πN is below γN . Therefore, the conclusion holds.
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Fig. 1. Geometric demonstration in the proof of Lemma 7.
Lemma 7. Assume that for each proper subset J ⊂ IN , the linear system Ai x = bi ,
i ∈ I = IN \ I ∗
has a unique solution xJ in
∀j ∈ J ,
CJ0 .
(71) ∗
Assume also that xJ satisfies
Aj x∗J < bj .
(72)
if J ̸= {N }. Then (a) AN x∗{N } < bN H⇒ Ax∗ = b has a unique solution x∗ ∈ int RN+ ; (b) AN x∗{N } = bN H⇒ Ax∗ = b has a unique solution x∗ = x∗{N } in RN+ ; (c) AN x∗{N } > bN H⇒ Ax ̸= b for any x ∈ RN+ . Proof. (a) In this case, for each i ∈ IN , every equilibrium on πi is below γi so b and A satisfy the (I − J )-conditions. By [6, Lemma 2.7] or [9, Lemma 5], Ax = b has a unique solution x∗ ∈ int RN+ . (b) Clearly, x∗{N } is a solution of Ax = b in RN+ . Suppose there is an x∗ ∈ RN+ such that Ax∗ = b but x∗ ̸= x∗{N } . Then Ax = b
has infinitely many solutions in RN+ so rank(A, b) = rank(A) < N. For J = {1}, since (71) has a unique solution x∗{1} in C{01} , the submatrix of A obtained by deleting the first row and the first column of A has an inverse so rank(A, b) = rank(A) = rank((AT2 , . . . , ATN )T ) = N − 1. This shows the equivalence of Ax = b and (71) with J = {1}. Thus, x∗{1} is also a solution of Ax = b. In particular, A1 x∗{1} = b1 , which contradicts (72). Therefore, x∗{N } is the unique solution of Ax = b in RN+ .
(c) Suppose Ax = b has a solution x∗ ∈ RN+ . Obviously, x∗ ̸= 0 as bi > 0 for all i ∈ IN . If x∗ ̸∈ int RN+ then there is a proper subset J ⊂ IN such that x∗ ∈ CJ0 . Since Ax∗ = b implies that x∗ is also a solution of (71) and (71) in CJ0 has a unique solution x∗J , we must have x∗ = x∗J so Ax∗J = b. This contradicts (72) if J ̸= {N }. If J = {N } then AN x∗{N } = bN , which contradicts the assumption AN x∗{N } > bN . Hence, we must have x∗ ∈ int RN+ . Now consider the line ℓ1 in RN+ containing x∗ and x∗{N } . We claim the existence of an ε1 > 1 such that x¯ = ε1 x∗ + (1 − ε1 )x∗{N } = x∗{N } + ε1 (x∗ − x∗{N } ) ∈ ∂ RN+
(73)
so ℓ1 = x∗{N } x¯ . Indeed, since x∗{N } x∗ ⊂ ℓ1 and x∗ ∈ int RN+ , if x = ε x∗ + (1 − ε)x∗{N } ∈ ℓ1 for an ε > 1, then by assumption, aNN xN ≤ AN x = ε AN x∗ + (1 − ε)AN x∗{N } < ε bN + (1 − ε)bN = bN so xN = ε x∗N < bN /aNN and ε < bN /(aNN x∗N ). This shows the existence of ε1 ∈ (1, bN /(aNN x∗N )) satisfying (73). The assumption of the lemma implies that ∩j∈IN −1 γj is a line segment in RN+ containing x∗ and x∗{N } so
γi ∩ (∩j∈IN \{i,N } γj ) = ∩j∈IN −1 γj = ℓ1 = x∗{N } x¯ . As x¯ N = ε1 x∗N > 0, there is an i ∈ IN −1 such that x¯ i = 0 and x¯ j ≥ 0 for j ∈ IN \ {i}. The assumption also implies that ∩j∈IN \{i,N } γj is a two-dimensional plane in RN+ (N ≥ 3) and πi ∩ (∩j∈IN \{i,N } γj ) is a line segment containing x∗{i,N } x¯ . Since {x∗{i} } = πi ∩ (∩j∈IN \{i} γj ) ⊂ πi ∩ (∩j∈IN \{i,N } γj ), x∗{i} , x∗{i,N } and x¯ are on the same line. By (72), Ai x∗{i,N } < bi , Ai x∗{i} < bi but Ai x¯ = bi . Thus, x∗{i} and x∗{i,N } lie on the same side of x¯ . Since the Nth component of x∗{i} is positive whereas x∗{i,N } ∈ πN , x∗{i} must lie between x∗{i,N } and x¯ (see Fig. 1 below). By (73) and the assumption, AN x¯ = AN (ε1 x∗ + (1 − ε1 )x∗{N } ) = ε1 bN + (1 − ε1 )AN x∗{N } < bN so x¯ is below γN . As AN x∗{i} = bN so x∗{i} is on γN , x∗{i,N } must be above γN , i.e. AN x∗{i,N } > bN . This contradicts (72). Therefore, Ax = b has no solution in RN+ .
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Proof of Theorem 2. By the assumption, every (N − 1)-dimensional subsystem of (1) satisfies the (I − J )-conditions. By Theorem 1, every k-dimensional (k < N) subsystem of (10) is permanent. The existence of a unique x∗{N } ∈ C{0N } satisfying (12) follows from [6, Lemma 2.7] or [9, Lemma 5]. (a) Applying Lemma 6 to (10) for every i0 ∈ IN −1 , we obtain (13). If (10) has a solution in int RN+ that satisfies lim inft →∞ xN (t ) > 0, then the same reasoning as that used in the proof of Theorem 3(a) leads to the existence of x∗ ∈ int RN+ satisfying Ax∗ = b, a contradiction to Lemma 7(b). Thus, every solution of (10) in int RN+ satisfies lim inft →∞ xN (t ) = 0. (b) The conclusion follows from Lemmas 6 and 7(c) and the proof of Theorem 3(b). (c) If AN x∗{N } < bN then, for each i ∈ IN , every equilibrium of (1) on πi is below γi , so b and A satisfy the (I − J )-conditions. By Theorem 1, (10) is permanent. Conversely, if (10) is permanent then by parts (a) and (b) we must have AN x∗{N } < bN . References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
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