Abstract
In this paper, we propose the stochastic Lotka–Volterra model with delay disturbed by G-Brownian motion Under a natural assumption on noise, we study existence and uniqueness of the global positive solution for the system and its asymptotic pathwise moment behavior and prove that the solution does not explode to infinity in a finite time.
1. Introduction
Since the Lotka–Volterra model (LVM in short) was provided by Lotka [1] and Volterra [2], there were extensive works concerned with the dynamics of this system and global stability and the stochastic Lotka–Volterra population model, and in here, we only mention [3, 4] (for deterministic situation) and [5–8] (for stochastic situation). The well-known two-dimensional delay Lotka–Volterra ecological population model driven by Brownian motion is
Bahar and Mao in [9] proved that the solution of (1) is almost surely nonnegative and finite. Wu and Xu in [10] investigated stochastic LVM with infinite delay. Global asymptotic stability for a stochastic delay LVM was obtained in [11].
Peng first established the stochastic analysis theory under the G-expectation framework in references [12–14]. Peng’s G-expectation space is an essential extension for probability measure space. Since then, many important theoretical results in this field are obtained, for example, SLL for sublinear expectations are obtained in [15], capacity theory results are discussed in [16] and [17–19], and other related technologies in [20–22]. Inspired by these results, we investigate a stochastic delay Lotka–Volterra model disturbed by G-Brownian motion:with , where is a n-dimensional vector, xi(t) is the population size of species i at time t(t ≥ 0), , bi is the species i’s growth rate, is a n × n community matrix, aij(I ≠ j) is the interspecific interaction effect, and aii is the intraspecific interaction effect. We assume that the interaction effect in this system was disturbed by a G-Brownian motion with and , where is a constant matrix, representing the total interference intensity matrix for the system; B(t) has a variance-uncertainty but not mean-uncertainty; 〈B〉(t) has a mean-uncertainty property. Therefore, (〈B〉, B) is used to characterise the disturbed growth rate, disturbed interspecific, or intraspecific interactions and interference intensity at the same time. We think the model (2) considers the stochastic interference from both mean-uncertainty and variance-uncertainty, but the traditional stochastic model cannot describe this property. Indeed, we prove the solution of (2) is quasi-surely nonnegative and finite. Some asymptotic pathwise moment estimations for the solutions of this system are presented.
2. Stochastic Delay Lotka–Volterra Model Driven by G-Brownian Motion
Definitions about sublinear expectations, G-Brownian motions, and quadratic variation process 〈B〉(t) and notations, as well as more details can also be found in [12–14]. For a matrix A, we denote and ∥A∥ = sup{|Ax|: |x| = 1}. denotes the family of continuous functions from [−τ, 0] to . We assume the matrix σ satisfies the following assumption:
The assumption (A) was first assumed by Mao et al. in [5], and it is also necessary in our framework.
Theorem 1. If the matrix σ in system (2) satisfies assumption (A), then ∀A ∈ Rn×n, b ∈ Rn and {x(s): s ∈ [−τ, 0]}, then there exists a unique solution x of equation (2). Furthermore, for all s ≥ −τ quasi-surely, namely, .
Proof. Because the coefficients of equation (2) are locally Lipschitz continuous, there exist a unique local solution x(s) on s ∈ [−τ, τe), where τe is called explosion time. To see it is also global, we must show τ∞ = ∞ q.s. Suppose k0(k0 > 0) is large enough s.t. x(t)(t ∈ [−τ, 0]) satisfies 1/k0 < min |x(t)|, max |x(t)| < k0. For any k(k ≥ k0), set τk = inf{s ∈ [0, τe): xi(s)∉(1/k, k), 1 < i ≤ n}, where inf ∅ = ∞. Noting that τk is increasing when k ⟶ ∞, let , then τ∞ ≤ τe q.s. If we can prove τ∞ = ∞ q.s., then τe = ∞ q.s. and q.s., t ≥ 0. If τ∞ ≠ ∞ q.s., then ∃ a constant T > 0 s.t. V(ω: τ∞(ω) ≤ T) ≥ ɛ for any ɛ > 0, namely, ∃ an integer k1(k1 ≥ k0) s.t. V(Ak) ≔ V(ω: τk(ω) ≤ T) ≥ ɛ for all k ≥ k1. Let be . Set k ≥ k0 and T > 0. Using the G-Itô lemma for , t ∈ [0, τk ∧ T], we getand noting thatand, as well as by the assumption (A). Thus,Denotesince we note that there is K s.t. f(x, a, b, σ) is bounded, namely, f(x, a, b, σ) < K, thenthen . From the definition of τk, we know ∀ω ∈ Ak, ∃ some i s.t. xi(τk, ω) ∉ (1/k, k), namely, xi(τk) ≤ 1/k, or xi(τk) ≥ k < ∞. Noting that the function U(xi) is decreasing when 0 < xi ≤ 1 and is increasing when xi > 1, hence U(x(τk)) ≥ U(1/k, …, 1/k) and U(x(τk)) ≥ U(k, …, k), namely, . Therefore, we haveSetting k ⟶ ∞, we have the contradiction therefore, we have τ∞ = ∞ q.s., namely, τe = ∞ q.s., so .
3. Asymptotic Behaviors of the Solution
Theorem 2. Under the assumption (A), if , for any β ∈ (0, 1) and δ ∈ (0, 1), ∃C0 = C(δ) > 0 s.t. the solution x(t) of equation (2) is as follows:
Proof. LetUsing the G-Itô lemma for U(x) and notingwe haveSinceand from equation (13), we getthenWe setthen F1(x) is bounded in , say K1, from (16),namely,where U0 = U(x(0)), and noting that satisfies , by (19), thentherefore,In addition, we noteso∀δ > 0, let , thenHence,
Theorem 3. Suppose the (A) is true, and there exists K(K > 0) is independent of {x(s): s ∈ [−τ, 0]}, then
Proof. Write (7) as = − |x|2, wherethen ≤ K − |x|2, where . Taking expectation from 0 to τk ∧ T on both sides of equation (6), we haveLetting k ⟶ ∞ yieldsTherefore, setting T ⟶ ∞,
4. Asymptotic Moment Estimations
Theorem 4. If condition (A) is true, then ∀ {x(s): s ∈ [−τ, 0]}, x(t) in (2) satisfieswhere .
Proof. Let then by G-Itô’s lemma of Reference [13], we havewhere C0 = log(Ṽ(x(0))). Noting that, from Lemma 3.1 in reference [19], for any integer k ≥ 1, we havesoapplying Lemma 2 in [15], we know for all but finitely many k,quasi-surely true, i.e., ∃Ωi ⊂ Ω ( = 1) s.t. ∀ω ∈ Ωi and ki = ki(ω) s.t.k ≥ ki(ω). From equation (32) and inequality (37),t ∈ [0, ki(ω)], k ≥ ki(ω), in other words,where . Taking G-expectation for (39), and then , from (39), we getwhere . Set max{ki(ω), i ∈ [1, n]} = k0(ω), then , t ∈ [k − 1, k], k ≥ k0(), it gets from (40):Letting ɛ tend to zero and noting that yieldThe proof is complete.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest to this work.
Acknowledgments
This research was supported by the NSF-China (nos. 11761028 and 11501009), NSF-Anhui Province (1508085 JGD10), HTIT and RTFS-Yunnan Province, and Honghe University (nos. 2015HB061, 2014HB0204, and 2018JS480).