Journal Pre-proof Stationary distribution and extinction of a stochastic ratio-dependent predator–prey system with stage structure for the predator Xin Zhao, Zhijun Zeng
PII: DOI: Reference:
S0378-4371(19)31854-0 https://doi.org/10.1016/j.physa.2019.123310 PHYSA 123310
To appear in:
Physica A
Received date : 16 May 2019 Revised date : 15 September 2019 Please cite this article as: X. Zhao and Z. Zeng, Stationary distribution and extinction of a stochastic ratio-dependent predator–prey system with stage structure for the predator, Physica A (2019), doi: https://doi.org/10.1016/j.physa.2019.123310. This is a PDF file of an article that has undergone enhancements after acceptance, such as the addition of a cover page and metadata, and formatting for readability, but it is not yet the definitive version of record. This version will undergo additional copyediting, typesetting and review before it is published in its final form, but we are providing this version to give early visibility of the article. Please note that, during the production process, errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.
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Journal Pre-proof
Stationary distribution and extinction of a stochastic ratio-dependent predator-prey system with stage structure for the predator Xin Zhao, Zhijun Zeng∗
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Northeast Normal Univ, Sch Math Stat, 5268 Renmin St, Changchun 130024, Jilin, Peoples R China.
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Abstract
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In this paper, we deal with a stochastic predator-prey model with stage structure for predator population and ratio-dependent functional response. The proposed mathematical model consists of a system of three stochastic differential equations to stimulate the interactions between prey population, immature predator and mature predator population. We first establish sufficient conditions for the existence and uniqueness of the positive solutions by constructing an appropriate Lyapunov function. Then we extend the existence of stationary distribution under certain parametric restrictions. We also obtain the sufficient conditions for extinction of the predator populations. Finally, numerical simulations have been carried out to validate our analytical findings. Keywords: Stationary distribution; Extinction; Stochastic predator-prey system; Stage structure 2000 MSC: 34D20, 39A50, 92B99
1. Introduction
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The dynamic relationship between predators and their preys has been favored by both biologists and mathematicians due to its worldwide existence and significance in population dynamics since the well-known LotkaVolterra model was brought forward [1, 2, 3, 4]. To describe a predator-prey relationship, functional response has been came up to specify the rate of prey consumption by an average predator, which has been a crucial concept in modeling predator-prey interactions [5, 6, 7]. During the past decades, the investigation of predatorprey models with different functional responses have received great attention. Traditionally, the functional response can be classified as prey-dependent (Michaelis-Menten or Holling type II [8, 9, 10, 11] ) and predatordependent for a long time. The difference lies in the functional response is determined by prey density solely or by both predator and prey populations. It has been universally acknowledged that predator-prey systems with prey-dependent functional response cannot explain the laboratory observations and have been raised doubts by several ecologists [12, 13]. In order to address these issues, Arditi and Ginzburg [14] shifted their attention to seek alternative forms for functional responses and suggested that a ratio-dependent functional response is considered as a reasonable option to model predation. The ratio-dependence is a particular type of predator dependence in which the per capita predator growth rate is a function of the ratio of prey to predator abundance [15]. This kind of interaction has been found to be typical in cases when predators have to search, share, and/or compete for food [16]. In recent years, predator-prey models with ratio-dependent functional response have revealed a wide variety of rich dynamical behaviors and are supported by numerous laboratory experiments which made the plenty of results based on ratio-dependent theory biologically more reliable. Ruan [17] studied the versal unfolding of a predator-prey system with ratio-dependent functional response and obtained all possible ∗
Corresponding author. fax: 043185098237. Email address:
[email protected] (Zhijun Zeng)
Preprint submitted to Elsevier
September 15, 2019
Journal Pre-proof
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phase portraits for its perturbations. Shi [18] explored the global asymptotic stability of a diffusive predatorprey model with ratio-dependent functional response and showed that the prey and predator will be spatially homogeneously distributed as time converges to infinities. But as we all know, there are many factors that affect dynamical properties of predator-prey systems and influence the outcomes of population evolution such as the functional response, stage structure, refuge, harvesting, etc., especially the joint effect of these factors. Earlier models and approaches primarily absorbed in the simplified situations, assuming that the predators consume the preys of all ages omitting the predator eating preferences within the prey populations or ignoring the fact that only the mature predators are equipped with the ability of predation. The species represented by mammalian population whose generalized life structure is divided into two sub populations such as immature and mature. As a consequence, it is practical to introduce stage structure into the predator-prey models. The predator-prey models with stage structure for prey or predator or even for both prey and predators have been considered by several researchers [19, 20, 21, 22] since the pioneering paper of Aiello and Freedman [23]. Here we provide a ratio-dependent predator-prey system with Holling type II functional response and stage structure for the predator. Many researchers began to study based on the following system such as Deng [24]. ! dx αy2 (t) , = x(t) r − ax(t) − dt my2 (t) + x(t) dy1 βx(t)y2 (t) (1) = − by1 (t) − dy1 (t), dt my (t) + x(t) 2 dy 2 = dy1 (t) − cy2 (t), dt where x(t) represents the population density of the prey at time t, y1 (t) and y2 (t) represent the population density of the immature and mature predator at time t, respectively. All the parameters involved in system (1) are assumed to be positive from the viewpoint of ecology, r is the intrinsic growth rate of prey, a is the intra-specific competition rate of the prey, α represents the capturing rate and αβ is the conversion rate of the mature predator, m denotes the half saturation constant, b and c are the death rate of the immature and mature predator, respectively, d is interpreted as the rate of immature predator becoming mature predator. As a matter of fact, in the real world, most natural phenomena can not be explained by deterministic laws but are always affected by the environmental noise which is an inevitable attribute of dynamics of any ecosystems ( see [25]-[30]). So the deterministic models which do not incorporate the effect of environmental noise have plenty of limitations in mathematical modelling of ecological systems and encounter difficulties in fitting data ideally and predicting the future dynamics of the system precisely. It has been shown in large number of literatures that many researchers introduced stochastic environmental variation described by Brownian motion into parameters in the deterministic model to establish stochastic predator-prey models. May [31] pointed out that the birth rates, carrying capacity, competition coefficients or other parameters involved in the model exhibit random fluctuation to a greater or lesser extent due to the environmental fluctuation. Mao [32] revealed that even a tiny environmental noise can suppress a potential population explosion. Liu [33] developed a stochastic predator-prey model with stage structure for predator and Holling type II functional response in which established sufficient conditions for the existence and uniqueness of an ergodic stationary distribution and obtained sufficient conditions for extinction of the predator populations. Deng [24] focused on the Hopf bifurcation analysis for a ratio-dependent predator-prey system with two delays and stage structure for the predator and established sufficient conditions for the local stability and the existence of Hopf bifurcation with respect to both delays. Motivated by the referred works, our objective of this paper is to investigate and analyze a stochastic stage structure predator-prey model with the functional response of ratio-dependent type. There are certain methods to introduce the stochastic fluctuations into population models. Here we follow the approach adopted 2
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in [34]-[38]. We assume the intrinsic growth rate of prey and the death rate of predators are not constants but are subject to environmental noise. Suppose that r, −b and −c are stochastically perturbed with .
r → r + σ1 w1 ,
.
− b → −b + σ2 w2 ,
.
− c → −c + σ3 w3 ,
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where σ21 , σ22 and σ23 stand for the intensities of the white noise, w1 and w2 denote the independent standard Brownian motions. And then corresponding to the deterministic system (1), the stochastic version can be rewritten as the following form ! αy (t) 2 dt + σ1 x(t)dw1 (t), dx(t) = x(t) r − ax(t) − my2 (t) + x(t) ! βx(t)y2 (t) (2) − by1 (t) − dy1 (t) dt + σ2 y1 (t)dw2 (t), dy1 (t) = my2 (t) + x(t) dy2 (t) = (dy1 (t) − cy2 (t))dt + σ3 y2 (t)dw3 (t). We give some basic theory in stochastic differential equations which is introduced in [25]. Let x(t) be a d-dimensional Itˆo’s process on t ≥ 0 with the stochastic differential
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dx(t) = f (x(t), t)dt + g(x(t), t)dw(t),
where f ∈ L1 (R+ ; Rd ) and g ∈ L2 (R+ ; Rd×m ). Let V ∈ C 2,1 (Rd × R+ ; R), where C 2,1 (Rn × R+ ; R) represents for the family of all real-valued nonnegative functions V(x, t) on Rd ×R+ which are continuously twice differentiable in x and once in t. We define a diffusion operator LV : Rd × R+ → R of the Itˆo’s process associated with the C 2,1 -function V by 1 LV(x(t), t) = Vt (x(t), t) + V x (x(t), t) f (x(t), t) + trace(gT (x(t), t)V xx (x(t), t)g(x(t), t)), 2
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then the Itˆo’s formula can be written as
dV(x(t), t) = LV(x(t), t)dt + V x (x(t), t)g(x(t), t)dw(t) a.s.
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Throughout this paper, unless otherwise specified, let (Ω, F , {Ft }t≥0 , P) be a complete probability space with a filtration {Ft }t≥0 satisfying the usual conditions (i.e., it is right continuous and increasing while F0 contains all P-null sets). Let w1 (t), w2 (t) and w3 (t) denote the independent standard Brownian motions defined on this probability space. We denote by Rd+ = {x = (x1 , · · · , xd ) ∈ Rd : xi > 0, 1 ≤ i ≤ d}. We also set inf ∅ = ∞. If A is a vector or matrix, its transpose is denoted by AT . This paper is structured in the following way. In Section 2, we demonstrate system (2) has a unique global positive solution. In Section 3, we prove system (2) has a stationary distribution under certain parametric restrictions. In Section 4, we obtain sufficient conditions for extinction of the predator populations in two cases. In Section 5, the results of numerical simulations are given to support the above theoretical results. 2. Global positive solution
In this section, we first verify system (2) has a unique global positive solution in order to investigate the dynamical behavior of system (2). Due to the coefficients of system (2) satisfy the local Lipschitz condition instead of linear growth condition, therefore, there exists a unique maximal local solution to system (2). In other words, the solution may exist from R3+ space at a finite time. We will show that the solution of system (2) is positive and global. This forms the following theorem. 3
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Theorem 1. There is a unique global positive solution X = (x(t), y1 (t), y2(t)) of system (2) on t ≥ 0 and the solution will remain in R3+ for any given initial value (x(0), y1 (0), y2 (0)) ∈ R3+ with probability one.
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Proof. Since the coefficients of system (2) satisfy the local Lipschitz condition, then for any initial value (x(0), y1 (0), y2 (0)) ∈ R3+ , system (2) has a unique local solution (x(t), y1 (t), y2 (t)) ∈ R3+ on [0, τe ) a.s., where τe is the explosion time. We aim to prove this solution is global, i.e. τe = ∞ a.s. Let n0 ≥ 0 be sufficiently large such that x(0), y1 (0) and y2 (0) are lying within the interval [ n10 , n0 ]. For each integer n ≥ n0 , we define the stopping time ) ( 1 τn = inf t ∈ [0, τe ) : min{x(t), y1 (t), y2 (t)} ≤ or max{x(t), y1 (t), y2 (t)} ≥ n . n
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It’s apparently that τn is increasing as n → ∞. Set τ∞ = limn→∞ τn , whence τ∞ ≤ τe a.s. Hence, we only need to prove that τ∞ = ∞ a.s. If it is false, then there exists two constants T > 0 and δ ∈ (0, 1) such that P {τ∞ ≤ T } > δ. Therefore, there exists an integer n1 ≥ n0 such that
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P {τn ≤ T } ≥ δ, ∀n ≥ n1 . ˜ R3+ → R+ of the form We define a C 2 -function V:
! α α x(t) ˜ + (y1 (t) − 1 − ln y1 (t)) + (y2 (t) − 1 − ln y2 (t)) . V(x(t), y1 (t), y2 (t)) = x(t) − α − α ln α β β
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By using Itˆo’s formula, we can get ! ! ! α α 2 1 α 2 1 α α α ˜ dV = 1 − dx(t) + σ1 dt + 1− dy1 (t) + σ2 dt + 1− dy2 (t) + σ23 dt x(t) 2 β y1 (t) 2β β y2 (t) 2β α α ˜ = LV(x(t), y1 (t), y2(t))dt + (x(t) − α)σ1 dw1 (t) + (y1 (t) − 1)σ2 dw2 (t) + (y2 (t) − 1)σ3 dw3 (t), β β where
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! ! ! αy2 (t) 1 αx(t)y2 (t) α α ˜ LV = (x(t) − α) r − ax(t) − + 1− − by1 (t) − dy1 (t) my2 (t) + x(t) y1 (t) my2 (t) + x(t) β β ! ! 2 2 2 ασ1 ασ2 ασ3 α 1 α + 1− dy1 (t) − cy2 (t) + + + y2 (t) β β 2 2β 2β α αx(t)y2 (t) α α2 y2 (t) − by1 (t) − + (b + c + d) = rx(t) − ax2 (t) − αr + aαx + my2 (t) + x(t) β (my2 (t) + x(t))y1 (t) β 2 2 2 αdy1 (t) ασ1 ασ2 ασ3 α + + + − cy2 (t) − β βy2 (t) 2 2β 2β 2 2 2 ασ1 ασ22 ασ23 (aα + r) α α ≤ + + (b + c + d) + + + 4a m β 2 2β 2β := N, where N is a positive constant, it follows that ˜ dV(x(t), y1 (t), y2 (t)) ≤ Ndt + (x(t) − α)σ1 dw1 (t) + 4
α α (y1 (t) − 1)σ2 dw2 (t) + (y2 (t) − 1)σ3 dw3 (t). β β
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By integrating it from 0 to τn ∧ T and taking expectation, we obtain ˜ ˜ ˜ EV(X(τ n ∧ T )) ≤ V(X(0)) + NE(τn ∧ T ) ≤ V(X(0)) + NT.
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Let Ωn = {ω ∈ Ω : τn = τn (ω) ≤ T }, so we get P(Ωn ) ≥ δ. Consequently, for all ω ∈ Ωn , x(τn , ω), y1 (τn , ω) or y2 (τn , ω) equals either n or 1n . As a consequence, it infers that n ˜ ˜ V(X(0)) + NT ≥ E(IΩn V(X(τ ))) ≥ δ min n − α − α ln , n − 1 − ln n , n α or ) ( 1 1 ˜ ˜ − α + α ln(nα), − 1 + ln n , V(X(0)) + NT ≥ E(IΩn V(X(τ n ))) ≥ δ min n n where IΩn denotes the indicator function of Ωn . Let n → ∞, then we see that ˜ ∞ > V(X(0)) + NT = ∞,
which leads to a contradiction. So we have τ∞ = ∞. This completes the proof.
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3. The existence of stationary distribution
It was known that random perturbation may destroy the stability of the equilibria existing in the deterministic systems and lead to a stochastic weak stability named stationary distribution. Recently, the study of stationary distribution has attracted many researchers (see [40]-[48] and the references therein) since it is applied extensively in each realm. In this section, we shall consider there exists a unique ergodic stationary distribution of system (2) under certain conditions. Next, we introduce a lemma in order to state and prove the existence and uniqueness of a stationary distribution from Has’minskii [49].
Theorem 2. If
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Lemma 1. The Markov process X(t) has a unique ergodic stationary distribution ν(·) if there exists a bounded open set Ω of Rd with regular boundary U such that the following properties hold: A1. The diffusion matrix A(x) is strictly positive definite for all x ∈ Ω. A2. There exists a nonnegative C 2 − function V such that LV is negative on Rd \ Ω. r− βd
c+
σ23 2
σ21 α > 2 m
! σ22 > b+d+ , 2
(3)
(4)
then system (2) has a unique ergodic stationary distribution for any initial value (x(0), y1 (0), y2 (0)) ∈ R3+ . Proof. Direct computation shows that the diffusion matrix of system (2) is given by 2 2 σ1 x 0 0 σ22 y21 0 . A = 0 0 0 σ23 y22
It is clearly that matrix A is positive definite for any compact subset of R3+ . We have verified that condition A1 in Lemma 1 is satisfied. In the next moment, we define a C 2 − function V(x, y1 , y2 ) such that LV ≤ −1 on 5
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R3+ \ Ω, where Ω is an open bounded set. This guarantees condition A2 in Lemma 1 is satisfied. Then we can show system (2) has a unique ergodic stationary distribution. Firstly, we define
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V1 (x(t)) = x−θ (t), V2 (y1 (t), y2 (t)) = − ln y1 (t) − k ln y2 (t), !γ+1 α α(b + d) 1 x(t) + y1 (t) + y2 (t) , V3 (x(t), y1 (t), y2 (t)) = γ+1 β 2βd V4 (y1 (t)) = − ln y1 (t), V(x(t), y1 (t), y2 (t)) = V1 + MV2 + V3 + V4 ,
where θ > 0 is a sufficiently small number, M > 0 is a positive number, k =
βd
c+
γ > 0 which satisfies the following condition
σ2 3 2
!2
is also a positive number,
) b+d c . γ3 < min , 4σmax 2σmax We define λ :=
βd σ23 2
σ2 − b+d+ 2 2
and (
(5)
!
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c+
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(
γ
!γ ) a γ+2 γ3γ α α(b + d) αγ+1 (b + d) γ+1 αγ+1 c(b + d)γ+1 γ+1 γ+1 − x + σmax x + rx x + y1 + y2 − y1 − γ+3 γ+1 γ+1 y2 , B = sup 2 2 β 2βd 8βγ+1 2 β d (x,y1 ,y2 )∈R3+ n o where σmax := max σ21 , σ22, σ23 . Applying the generalized Itˆo’s formula, we arrive that θ(θ + 1) −θ−2 x (t)(dx(t))2 2 ! θ(θ + 1) −θ−2 αy2 (t) −θ−1 dt − θx−θ−1 (t)σ1 x(t)dw1 (t) + x (t)σ21 x2 (t)dt. = −θx (t)x(t) r − ax(t) − my2 (t) + x(t) 2
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dV1 (x(t)) = −θx−θ−1 (t)dx(t) +
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Hence, it yields that
−θ
LV1 (x(t)) = −θx (t)
= −θx−θ (t)
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≤ −θx−θ (t)
= −θx−θ (t)
! αy2 (t) θ(θ + 1) 2 −θ r − ax(t) − + σ1 x (t) my2 (t) + x(t) 2 ! θ+1 2 αθx−θ (t)y2 (t) r− σ1 + aθx1−θ (t) + 2 my2 (t) + x(t) ! α θ+1 2 σ1 + aθx1−θ (t) + θx−θ (t) r− 2 m ! θ+1 2 α + aθx1−θ (t). σ1 − r− 2 m
Similarly, it can be derived that
1 k k 1 dy1 (t) + 2 (dy1 (t))2 − dy2 (t) + 2 (dy2 (t))2 dV2 (y1 (t), y2 (t)) = − y1 (t) y2 (t) 2y1 (t) 2y2 (t) " # ! βx(t)y2 (t) y1 (t) 1 k 2 2 2 2 = − + b + d + 2 σ2 y1 (t) − kd − ck + 2 σ3 y2 (t) dt (my2 (t) + x(t))y1 (t) y2 (t) 2y1 (t) 2y2 (t) − σ2 dw2 (t) − cσ3 dw3 (t). 6
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Thus, it follows that
=
=
≤
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=
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≤
Pr e-
LV2 (y1 (t), y2 (t)) =
! σ23 σ22 y1 βx(t)y2 (t) − +b+d+ − kd + k c + (my2 (t) + x(t))y1 (t) 2 y2 2 s ! σ23 σ22 βkdx(t) +b+d+ +k c+ −2 my2 (t) + x(t) 2 2 s ! ! p p σ23 σ22 βkdx(t) +k c+ + 2 βkd − 2 −2 βkd + b + d + 2 2 my2 (t) + x(t) s ! ! p p σ23 σ22 x(t) +k c+ + 2 βkd 1 − −2 βkd + b + d + 2 2 my2 (t) + x(t) ! ! x(t) 1 − my2 (t)+x(t) p p σ23 σ22 −2 βkd + b + d + +k c+ + 2 βkd q 2 2 x(t) 1 + my2 (t)+x(t) ! ! p p σ23 σ22 my2 (t) +k c+ + 2 βkd , −2 βkd + b + d + 2 2 my2 (t) + x(t)
√ where in the second inequality, we have received the result with the help of basic inequality A + B ≥ 2 AB. If and only if A = B the equality holds. We are going to plug k = βdσ2 !2 into the above inequality, we get c+
3 2
c+
σ23 2
! σ22 2βd my2 (t) + . + b+d+ 2 σ 2 c + 23 my2 (t) + x(t)
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LV2(y1 (t), y2 (t)) ≤ −
βd
By means of Itˆo’s formula, we obtain that
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!γ ! α α(b + d) α α(b + d) dV3 (x(t), y1 (t), y2 (t)) = x(t) + y1 (t) + y2 (t) dx(t) + dy1 (t) + dy2 (t) β 2βd β 2βd !γ−1 !2 α α(b + d) α α(b + d) γ y2 (t) dx(t) + dy1 (t) + dy2 (t) . + x(t) + y1 (t) + 2 β 2βd β 2βd
7
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Then we can derive that
≤
=
of
Jo
=
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≤
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≤
Pr e-
LV3(x(t), y1 (t), y2 (t)) =
!γ ! α α(b + d) α(b + d) cα(b + d) 2 x(t) + y1 (t) + y2 (t) rx(t) − ax (t) − y1 (t) − y2 (t) β 2βd 2β 2βd !γ−1 ! α α(b + d) α2 2 2 α2 (b + d)2 2 2 γ 2 2 y2 (t) σ1 x (t) + 2 σ2 y1 (t) + σ3 y2 (t) + x(t) + y1 (t) + 2 β 2βd β 4β2d 2 !γ ! α(b + d) α(b + d) cα(b + d) α 2 y2 (t) rx(t) − ax (t) − y1 (t) − y2 (t) x(t) + y1 (t) + β 2βd 2β 2βd !γ+1 γ α α(b + d) + σmax x(t) + y1 (t) + y2 (t) 2 β 2βd !γ α(b + d) αγ+1 (b + d) γ+1 α y2 (t) − axγ+2 (t) − y1 (t) rx(t) x(t) + y1 (t) + β 2βd 2βγ+1 !γ+1 cαγ+1 (b + d)γ+1 γ+1 γ α α(b + d) − γ+1 γ+1 γ+1 y2 (t) + σmax x(t) + y1 (t) + y2 (t) 2 β d 2 β 2βd αγ+1 (b + d) γ+1 cαγ+1 (b + d)γ+1 γ+1 a y (t) − y (t) − xγ+2 (t) − 1 2 4βγ+1 2γ+2 βγ+1 d γ+1 2 a cαγ+1 (b + d)γ+1 γ+1 αγ+1 (b + d) γ+1 − xγ+2 (t) − y (t) − y (t) 1 2 4βγ+1 2γ+2 βγ+1 d γ+1 2 !γ !γ+1 α α(b + d) γ α α(b + d) +rx(t) x(t) + y1 (t) + y2 (t) + σmax x(t) + y1 (t) + y2 (t) β 2βd 2 β 2βd a αγ+1 (b + d) γ+1 cαγ+1 (b + d)γ+1 γ+1 − xγ+2 (t) − y (t) − y (t) 1 2 4βγ+1 2γ+2 βγ+1 d γ+1 2 a αγ+1 (b + d) γ+1 cαγ+1 (b + d)γ+1 γ+1 − xγ+2 (t) − y (t) − y (t) 1 2 4βγ+1 2γ+2 βγ+1 d γ+1 2 !γ+1 !γ+1 γ+1 α α(b + d) γ3γ σmax x (t) + y1 (t) + y2 (t) + 2 β 2βd !γ α α(b + d) +rx(t) x(t) + y1 (t) + y2 (t) β 2βd αγ+1 (b + d) γ+1 cαγ+1 (b + d)γ+1 γ+1 a γ+2 γ3γ a y (t) − y (t) − x (t) + σmax xγ+1 (t) − xγ+2 (t) − 1 2 4βγ+1 2γ+2 βγ+1 d γ+1 2 2 2 !γ α α(b + d) αγ+1 (b + d) γ+1 cαγ+1 (b + d)γ+1 γ+1 +rx(t) x(t) + y1 (t) + y2 (t) − y (t) − y (t) 1 β 2βd 8βγ+1 2γ+3 βγ+1 d γ+1 2 ! ! αγ+1 (b + d) γ3γ αγ+1 σmax γ+1 cαγ+1 (b + d)γ+1 γ3γ αγ+1 (b + d)γ+1 σmax γ+1 − y1 − y2 − − 8βγ+1 2βγ+1 2γ+3 βγ+1 d γ+1 2γ+2 βγ+1 d γ+1 a αγ+1 (b + d) γ+1 cαγ+1 (b + d)γ+1 γ+1 a γ+2 γ3γ − xγ+2 (t) − y (t) − y (t) − x (t) + σmax xγ+1 (t) 1 2 4βγ+1 2γ+2 βγ+1 d γ+1 2 2 2 !γ α(b + d) αγ+1 (b + d) γ+1 cαγ+1 (b + d)γ+1 γ+1 α y2 (t) − y (t) − y (t) +rx(t) x(t) + y1 (t) + 1 β 2βd 8βγ+1 2γ+3 βγ+1 d γ+1 2 a αγ+1 (b + d) γ+1 cαγ+1 (b + d)γ+1 γ+1 − xγ+2 (t) − y (t) − y (t) + B, 1 2 4βγ+1 2γ+2 βγ+1 d γ+1 2
≤
≤
8
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where in the last equality, we can obtain αγ+1 (b + d) γ3γ αγ+1 σmax − >0 8βγ+1 2βγ+1 and
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cαγ+1 (b + d)γ+1 γ3γ αγ+1 (b + d)γ+1 σmax − > 0, 2γ+3 βγ+1 d γ+1 2γ+2 βγ+1 d γ+1
which follows from condition (5). According to Itˆo’s formula, it holds that
σ2 βx(t)y2 (t) + b + d + 2. (my2 (t) + x(t))y1 (t) 2
p ro
LV4 (y1 (t)) = −
Therefore, the following inequality can be estimated
Pr e-
2 ! 2Mβd my2 (t) σ θ+1 2 α βd 2 + + aθx1−θ (t) − M σ1 − LV ≤ −θx−θ (t) r − − b + d + σ23 σ2 2 m 2 c+ 2 c + 23 my2 (t) + x(t) !
σ22 a γ+2 αγ+1 (b + d) γ+1 cαγ+1 (b + d)γ+1 γ+1 βx(t)y2 (t) − x (t) − y1 (t) − γ+2 γ+1 γ+1 y2 (t) + B − +b+d+ 2 4βγ+1 2 β d (my2 (t) + x(t))y1 (t) 2 ! γ+1 2Mβd my2 (t) θ+1 2 α a α (b + d) γ+1 + aθx1−θ (t) + ≤ −θx−θ (t) r − σ1 − − xγ+2 (t) − y1 (t) 2 σ 2 m 4βγ+1 c + 3 my2 (t) + x(t) 2 2
σ22 βx(t)y2 (t) cα (b + d) γ+1 y (t) − − Mλ + B + b + d + . 2γ+2 βγ+1 d γ+1 2 (my2 (t) + x(t))y1 (t) 2 γ+1
−
γ+1
Jo
urn
al
We aim to prove that LV ≤ −1 on R3+ \ Ω, once this is established, the Theorem 2 follows from Lemma 1, where Ω is an open bounded set defined as ) ( 1 2 1 1 4 3 Ω = (x, y1 , y2 ) ∈ R+ : ε < x < , ε < y1 < 4 , ε < y2 < 2 . ε ε ε S S S S S Next, we divide Ωc into six domains: Ωc = Ωc1 Ωc2 Ωc3 Ωc4 Ωc5 Ωc6 to show LV ≤ −1 actually holds, in which n o 1. Ωc1 = (x, y1 , y2 ) ∈ R3+ : 0 < x ≤ ε , o n 2. Ωc2 = (x, y1 , y2 ) ∈ R3+ : x > ε, 0 < y2 ≤ ε2 , o n 3. Ωc3 = (x, y1 , y2 ) ∈ R3+ : x > ε, 0 < y1 ≤ ε4 , y2 ≥ ε2 , n o 4. Ωc4 = (x, y1 , y2 ) ∈ R3+ : x > 1ε , o n 5. Ωc5 = (x, y1 , y2 ) ∈ R3+ : y1 > ε14 , n o 6. Ωc6 = (x, y1 , y2 ) ∈ R3+ : y2 > ε12 , and 0 < ε < 1 is a constant satisfying the following conditions ! 2 σ θ θ+1 2 α 2Mβd a γ+2 2 1−θ +C1 ≤ −1, C1 = sup − θ r− σ1 − x (t) + aθx (t) + − . (6) − Mλ + B + b + d + 2 σ ε 2 m 2 2 x∈Ωc1 c + 23 9
Journal Pre-proof
2Mβdmε c+
σ23 2
+ C2 − Mλ ≤ −1,
C2 = sup
x,y2 ∈Ωc2
(
) σ22 a γ+2 1−θ . − x (t) + aθx (t) + B + b + d + 2 2
(7)
a cαγ+1 (b + d)γ+1 ε2γ+2 1 − εγ+2 − − + C3 ≤ −1, 4 2γ+2 βγ+1 d γ+1 mε3 + ε2 2 σ 2Mβd a γ+2 2 1−θ x (t) + aθx (t) + . − Mλ + B + b + d + C3 = sup − σ2 2 4 x,y1 ,y2 ∈Ωc3 c+ 3 a 1 4 ε
+ C3 ≤ −1.
p ro
−
of
2
!γ+2
(8)
Pr e-
!γ+1 cαγ+1 (b + d)γ+1 1 + C1 ≤ −1. − γ+2 γ+1 γ+1 2 β d ε2 !γ+1 αγ+1 (b + d) 1 − + C1 ≤ −1. 4βγ+1 ε4 Now we have to start splitting in the following situation. I. If (x, y1 , y2 ) ∈ Ωc1 ,
(9)
(10)
(11)
! σ22 2Mβd my2 (t) θ+1 2 α a γ+2 + aθx1−θ (t) + LV ≤ −θx (t) r − σ1 − − x (t) − Mλ + B + b + d + σ2 2 m 2 c + 23 my2 (t) + x(t) 2 ! θ+1 2 α −θ + C1 σ1 − ≤ −θx (t) r − 2 m ! θ θ+1 2 α ≤− θ r− + C1 σ1 − ε 2 m ≤ −1,
urn
which follows from (6). II. If (x, y1 , y2 ) ∈ Ωc2 ,
al
−θ
LV ≤ aθx1−θ (t) +
c+
σ23 2
σ2 a my2 (t) − xγ+2 (t) − Mλ + B + b + d + 2 my2 (t) + x(t) 2 2
2Mβd mε2 + C2 − Mλ σ2 c+ 3 ε
Jo
≤
2Mβd
2
≤ −1,
which follows from (7).
10
Journal Pre-proof
III. If (x, y1 , y2 ) ∈ Ωc3 , LV ≤ aθx
(t) +
2Mβd c+
σ23 2
a γ+2 cαγ+1 (b + d)γ+1 γ+1 βx(t)y2 (t) my2 (t) − x (t) − γ+2 γ+1 γ+1 y2 (t) − my2 (t) + x(t) 2 2 β d (my2 (t) + x(t))y1 (t)
σ22 − Mλ + B + b + d + 2 a γ+2 cαγ+1 (b + d)γ+1 2γ+2 1 ≤ − ε − γ+2 γ+1 γ+1 ε − + C3 3 4 2 β d mε + ε2 ≤ −1,
LV ≤ aθx
a 1 ≤− 4 ε ≤ −1,
(t) + !γ+2
2Mβd c+
σ23 2
σ22 a γ+2 a γ+2 my2 (t) − x (t) − x (t) − Mλ + B + b + d + my2 (t) + x(t) 4 4 2
+ C3
Pr e-
1−θ
p ro
which follows from (8). IV. If (x, y1 , y2 ) ∈ Ωc4 ,
which follows from (9). V. If (x, y1 , y2 ) ∈ Ωc5 , 2Mβd σ23 2 γ+1
c+
cαγ+1 (b + d) 2γ+2 βγ+1 d γ+1 ≤ −1,
≤−
σ22 a cαγ+1 (b + d)γ+1 my2 (t) − xγ+2 (t) − γ+2 γ+1 γ+1 yγ+1 (t) − Mλ + B + b + d + 2 my2 (t) + x(t) 2 2 β d 2 !γ+1 1 + C1 ε2
al
LV ≤ aθx1−θ (t) +
of
1−θ
urn
which follows from (10). VI. If (x, y1 , y2 ) ∈ Ωc6 ,
LV ≤ aθx1−θ (t) +
2Mβd
c+
σ23 2
Jo
αγ+1 (b + d) 1 ≤− 4βγ+1 ε4 ≤ −1,
σ22 a αγ+1 (b + d) γ+1 my2 (t) − xγ+2 (t) − y − Mλ + B + b + d + 1 my2 (t) + x(t) 2 4βγ+1 2
!γ+1
+ C1
which follows from (11). Based on the analysis above-mentioned, we come to the conclusion that LV ≤ −1 on R3+ \ Ω. According to Lemma 1, we know system (2) admits a stationary distribution. This completes the proof. 4. Extinction of the predator populations Theorem 3. Let (x(t), y1 (t), y2 (t)) be the solution of system (2) with any initial value (x(0), y1 (0), y2 (0)) ∈ R3+ . If q √ √ σ2 −2 −1 r > 21 and min{b + d, c} R − 1 I{ √R≤1} + max{b + d, c} R − 1 I{ √R>1} + cσr 1 Rr − (2(σ−2 < 0, 2 + σ3 )) 2 11
Journal Pre-proof
where R =
βd , (b+d)c
then the predator populations will die out, that is to say, lim y1 (t) = 0,
lim y2 (t) = 0.
t→∞
t→∞
Moreover, the distribution of x(t) converges weakly a.s. to the measure which has the density −2+ 2r2 − 2a2 x
χ(x) = Gσ−2 1 x "
σ
1
e
σ
1
, x ∈ (0, ∞),
of
2 2r2 −1 #−1 R∞ σ σ 2r 1 1 where G = σ−2 − 1 is a constant satisfying χ(x)dx = 1. This statement highlights the Γ 2 1 2a σ 0 1
lim
t→∞
R+
f (x)ϑt (x)dx =
p ro
relationship of the probability density ϑt (x) of the prey population at time t and any sufficiently smooth function f (x) is Z Z f (x)χ(x)dx.
R+
Pr e-
Proof. Since we have obtained the solution of system (2) is positive for any initial value (x(0), y1 (0), y2 (0)) ∈ R3+ , so we get dx(t) ≤ x(t)(r − ax(t))dt + σ1 x(t)dw1 (t). We consider the following 1-dimensional stochastic differential equation with the initial value X(0) = x(0) > 0 dX(t) = X(t)(r − aX(t))dt + σ1 X(t)dw1 (t),
al
then according to the comparison theorem of stochastic differential equation one can get x(t) ≤ X(t) for any t ≥ 0 a.s. Let √ R(ω1 , ω2 ) = (ω1 , ω2 )D, √ β where (ω1 , ω2 ) = R, b+d and ! β 0 b+d D= d . 0 c
urn
We define a C 2 − function V¯ : R2+ → R+ as
¯ 1 (t), y2 (t)) = py1 (t) + qy2 (t), V(y
where p =
ω1 , b+d
q=
ω2 . c
By using Itˆo’s formula, we can derive that
Jo
1 ¯ 1 ¯ ¯ 2 d(ln V(t)) = dV(t) − (dV(t)) ¯ ¯ V(t) 2V 2(t) ! p p βx(t)y2 (t) = − by1 (t) − dy1 (t) dt + σ2 y1 (t)dw2 (t) ¯ ¯ my2 (t) + x(t) V(t) V(t) q 1 q (dy1 (t) − cy2 (t))dt + σ3 y2 (t)dw3 (t) − (p2 σ22 y21 (t) + q2 σ23 y22 (t))dt + ¯ ¯ ¯ V(t) V(t) 2V 2 (t) p q ¯ = LV(t)dt + σ2 y1 (t)dw2 (t) + σ3 y2 (t)dw3 (t), ¯ ¯ V(t) V(t) where ! p 1 βx(t)y2 (t) q ¯ LV(t) = (dy1 (t) − cy2 (t)) − (p2 σ22 y21 (t) + q2 σ23 y22 (t)). − by1 (t) − dy1 (t) + ¯ ¯ my2 (t) + x(t) V(t) V(t) 2V¯ 2(t) 12
Journal Pre-proof
In addition, one can get 1 1 V (t) = pσ2 y1 (t) + qσ3y2 (t) σ2 σ3 ¯2
!2
≤ (p
2
σ22 y21 (t)
+q
2
σ23 y22 (t))
! 1 1 + , σ22 σ23
and
Pr e-
p ro
of
( ! ) 1 βx(t)y2 (t) p − by1 (t) − dy1 (t) + q(dy1 (t) − cy2 (t)) ¯ my2 (t) + x(t) V(t) ( ! ) ! β ar βx(t) βr py2 (t) 1 p − y2 (t) − (b + d)y1 (t) + q(dy1 (t) − cy2 (t)) = + ¯ ¯ my2 (t) + x(t) my2 (t) + ar may2 (t) + r V(t) V(t) ! ) ( my2 (t) x(t) − ar pβy2 (t) 1 βr ω2 ω1 = + y2 (t) − (b + d)y1 (t) + (dy1 (t) − cy2 (t)) ¯ ¯ (my2 (t) + x(t))(my2 (t) + ar ) V(t) b + d may2 (t) + r c V(t) r 1 pβy2 (t) X(t) − a ω2 ω1 + ≤ (βy2 (t) − (b + d)y1 (t)) + (dy1 (t) − cy2 (t)) r ¯ ¯ c V(t) V(t) b + d a pβa r 1 X(t) − + ≤ (ω1 , ω2) D(y1 (t), y2 (t))T − (y1 (t), y2 (t))T ¯ qr a V(t) r 1 √ pβa X(t) − + = R − 1 (ω1 y1 (t) + ω2 y2 (t)) ¯ qr a V(t) √ r 1 pβa X(t) − + R − 1 [(b + d)py1 (t) + cqy2 (t)] = qr a py1 (t) + qy2 (t) √ √ r pβa X(t) − + min{b + d, c} R − 1 I{ √R≤1} + max{b + d, c} R − 1 I{ √R>1} . ≤ qr a
Hence, we have
al
√ √ pβa X(t) − ¯ ≤ min{b + d, c}( R − 1)I{ √R≤1} + max{b + d, c}( R − 1)I{ √R>1} + LV(t) qr
r −2 −1 − (2(σ−2 2 + σ3 )) . a
urn
It follows that " √ √ pβa √ √ ¯ X(t) − d ln V(t) ≤ min{b + d, c}( R − 1)I{ R≤1} + max{b + d, c}( R − 1)I{ R>1} + qr p q + σ2 y1 (t)dw2 (t) + σ3 y2 (t)dw3 (t). ¯ ¯ V(t) V(t)
# r −2 −2 −1 − (2(σ2 + σ3 )) dt a (12)
By integrating it from 0 to t and dividing by t on both sides of (12), we get
Jo
Z √ √ ¯ − ln V(0) ¯ ln V(t) pβa t √ √ X(s) − ≤ min{b + d, c}( R − 1)I{ R≤1} + max{b + d, c}( R − 1)I{ R>1} + t qrt 0 Z Z 1 t pσ2 y1 (s) 1 t qσ3 y2 (s) −2 −2 −1 − (2(σ2 + σ3 )) + dw2 (s) + dw3 (s) ¯ ¯ t 0 V(s) t 0 V(s) Z √ √ pβa t √ √ X(s) − ≤ min{b + d, c}( R − 1)I{ R≤1} + max{b + d, c}( R − 1)I{ R>1} + qrt 0 ¯ ¯ M(t) N(t) −2 −1 − (2(σ−2 + σ )) + + , 2 3 t t 13
r ds a r ds a (13)
Journal Pre-proof
¯ where M(t) := follows
Rt
pσ2 y1 (s) dw2 (s) ¯ V(s) 0
¯ ¯ h M(t), M(t)i t =
σ22
¯ := and N(t)
Z
t 0
py1 (s) ¯ V(s)
Rt 0
!2
qσ3 y2 (s) dw3 (s) ¯ V(s)
ds ≤
are local martingales with quadratic variations as
¯ ¯ t= hN(t), N(t)i
σ22 t,
σ23
Applying the strong law of large numbers for local martingales, we obtain
t→∞
0
Z t X(s) − 0
ds ≤ σ23 t.
¯ N(t) = 0 a.s. t→∞ t lim
χ(x)dx < ∞, we have 1 lim t→∞ t
!2
Z ∞ r x − r χ(x)dx ds = a a 0 # 12 "Z ∞ r 2 χ(x)dx ≤ x− a 0 !1 rσ21 2 = , 2a2
p ro
R∞
0
qy2 (s) ¯ V(s)
Pr e-
Since X(t) is ergodic and
t
of
¯ M(t) = 0, t
lim
Z
where a detailed calculation of this integral has been accomplished by Liu [33]. Taking the superior limit on both sides of (13), we have
al
1 2 !2 √ √ ¯ rσ pβa ln V(t) 1 −2 −1 − (2(σ−2 ≤ min{b + d, c}( R − 1)I{ √R≤1} + max{b + d, c}( R − 1)I{ √R>1} + lim sup 2 + σ3 )) 2 t qr 2a t→+∞ r √ √ cσ1 Rr −2 −1 √ √ ≤ min{b + d, c}( R − 1)I{ R≤1} + max{b + d, c}( R − 1)I{ R>1} + − (2(σ−2 2 + σ3 )) r 2 < 0,
which is equivalent to
lim sup It implies that
urn
t→+∞
ln y1 (t) < 0, t
lim y1 (t) = 0,
t→∞
lim sup t→+∞
ln y2 (t) < 0. t
lim y2 (t) = 0 a.s.
t→∞
This completes the proof.
Jo
Theorem 4. Let (x(t), y1 (t), y2 (t)) be the solution of system (2) with any initial value (x(0), y1 (0), y2 (0)) ∈ R3+ . If σ2 r < 21 and β < min{b, c}, then the prey and predator populations will die out, that is to say, lim x(t) = 0,
t→∞
lim y1 (t) = 0,
t→∞
lim y2 (t) = 0.
t→∞
Proof. By using Itˆo’s formula to the first equation of system (2), we can derive that ! σ21 αy2 (t) dt + σ1 dw1 (t). − d(ln x(t)) = r − ax(t) − my2 (t) + x(t) 2
14
(14)
Journal Pre-proof
By integrating it from 0 to t and dividing by t on both sides of (14), we get Z Z σ21 a t y2 (s) α t σ1 w1 (t) ln x(t) − ln x(0) x(s)ds − =r− − ds + t 2 t 0 t 0 my2 (s) + x(s) t 2 σ σ1 w1 (t) . ≤r− 1 + 2 t
(15)
lim
t→∞
w1 (t) = 0, a.s. t
lim sup t→+∞
p ro
Taking the superior limit on both sides of (15), we have
of
Applying the strong law of large numbers for local martingales, we obtain
σ2 ln x(t) ≤ r − 1 < 0 a.s., t 2
which implies that
lim x(t) = 0.
t→∞
Pr e-
Making use of Itˆo’s formula to ln(y1 (t) + y2 (t)), one can get βx(t)y2 (t) my (t)+x(t) − by1 (t) − cy1 (t) σ2 y2 (t) + σ2 y2 (t) 3 3 dt + σ2 y1 (t) dw2 (t) + σ3 y2 (t) dw3 (t) d ln(y1 (t) + y2 (t)) = 2 − 2 2 2 y1 (t) + y2 (t) 2(y1 (t) + y2 (t)) y1 (t) + y2 (t) y1 (t) + y2 (t) ! βx(t)y2 (t) σ3 y2 (t) σ2 y1 (t) ≤ − min{b, c} dt + dw2 (t) + dw3 (t). (my2 (t) + x(t))(y1 (t) + y2 (t)) y1 (t) + y2 (t) y1 (t) + y2 (t)
urn
al
According to lim x(t) = 0, we know there exists T > 0 such that x(t) < ǫ when t > T . Hence, a direct calculation t→∞ shows that ! βǫ σ3 y2 (t) σ2 y1 (t) d ln(y1 (t) + y2 (t)) ≤ − min{b, c} dt + dw2 (t) + dw3 (t) my2 (t) + ǫ y1 (t) + y2 (t) y1 (t) + y2 (t) (16) σ3 y2 (t) σ2 y1 (t) dw2 (t) + dw3 (t). ≤ (β − min{b, c}) dt + y1 (t) + y2 (t) y1 (t) + y2 (t) Rt R t σ y (s) 2 y1 (s) 3 2 Noting that M(t) := 0 y1σ(s)+y dw (s) and N(t) := dw3 (s) are local martingales with quadratic varia2 (s) 0 y1 (s)+y2 (s) 2 tions as follows !2 !2 Z t Z t y2 (s) y1 (s) 2 2 2 ds ≤ σ2 t, hN(t), N(t)it = σ3 ds ≤ σ23 t. hM(t), M(t)it = σ2 y1 (s) + y2 (s) y1 (s) + y2 (s) 0 0
Jo
Applying the strong law of large numbers for local martingales, we obtain lim
t→∞
M(t) = 0, t
N(t) = 0 a.s. t→∞ t lim
By integrating it from 0 to t and dividing by t on both sides of (16), we get Z Z 1 t σ2 y1 (s) 1 t σ3 y2 (s) ln(y1 (t) + y2 (t)) − ln(y1 (0) + y2 (0)) ≤ β − min{b, c} + dw2 (s) + dw3 (s) t t 0 y1 (s) + y2 (s) t 0 y1 (s) + y2 (s) M(t) N(t) + . ≤ β − min{b, c} + t t (17) 15
Journal Pre-proof
Taking the superior limit on both sides of (17), we have lim sup t→+∞
ln(y1 (t) + y2 (t)) ≤ β − min{b, c} < 0 a.s., t
which implies that lim y1 (t) = 0,
t→∞
lim y2 (t) = 0 a.s.
t→∞
of
This completes the proof. 5. Numerical test
Pr e-
p ro
In this section, we will carry out some numerical simulations to illustrate our theoretical results. We shall corroborate all the analytical findings with the help of numerical simulations accomplished with MATLAB. The initial value of system (2) is given by (0.2, 0.2, 0.2). We use Milstein’s higher-order method mentioned in Higham [50] to obtain the following discretization transformation of system (2) ! i √ αy σ21 i 2 2 i i i+1 i i ∆t + σ1 x ∆tε1,i + x (ε1,i − 1)∆t, x = x + x r − ax − i 2 my2 + xi ! √ i+1 βxi yi2 σ22 i 2 i i i i (18) y1 (ε2,i − 1)∆t, ∆tε + y = y + − by − dy ∆t + σ y 2,i 2 1 1 1 1 1 i i 2 my + x 2 √ σ23 i 2 i i i i yi+1 y (ε − 1)∆t, = y + (dy − cy )∆t + σ y ∆tε + 3 2 3,i 2 2 1 2 2 2 3,i
where the time increment ∆t is positive and εk,i are the Gaussian random variables which follow the distribution N(0, 1), k = 1, 2, 3.
find r = 0.6 > 0.5 =
σ21 2
+
α m
and
al
Example 1. We numerically simulate the model for the following set of parameter values: σ21 = 0.04, σ22 = 0.34, σ23 = 0.08. Other values of the system parameters can be seen from Table 1. By simple computations, we βd
σ2 c+ 23
= 4 > 1 = b+d+
σ22 . 2
It is obvious that this group of parameters satisfy
urn
condition (3) and (4). From Theorem 2, there exists an ergodic distribution ν(·) of system (2) which is clearly shown in Figure 1. Table 1: List of parameters
Description Intrinsic growth rate of the prey Intra-specific competition rate of the prey Death rate of immature predator Death rate of mature predator The rate of immature predator becoming mature predator Capture rate of the predator Conversion rate of the mature predator Half saturation constant
Jo
Parameters r a b c d α β m
16
Values 0.6 0.2 0.03 0.04 0.8 0.6 0.4 1.25
Journal Pre-proof
4
8
5 4
6 x(t)
x 10
3 4 2 2 0
1 0
1000 2000 3000 4000 5000 Time t
0
0
1
2
3
4
4
4
10
2
5
x 10
0
1000
2000 3000 Time t
4000
5000
0
0
0.5
1
1.5
2
2.5
p ro
0
of
15
y1(t)
6
y2(t)
5
20
4
15
3
10
2
5
1
0
1000
2000 3000 Time t
4000
5000
0
0
100
200
300
Pr e-
0
x 10
Figure 1: The left column shows the paths of x(t), y1 (t) and y2 (t) of system (2) with initial value (x(0), y1(0), y2 (0)) = (0.2, 0.2, 0.2) under the noise intensities σ21 = 0.04, σ22 = 0.34 and σ23 = 0.08. The blue lines represent the solution to system (2) and the red lines represent the solution to the corresponding undisturbed system (1). The right column displays the histogram of the probability density functions of x, y1 , y2 populations.
urn
al
Example 2. We numerically simulate the model for the following set of parameter values: σ21 = 0.04, σ22 = 15, σ23 = 15. Other values of the system parameters can be seen from Table 2. By simple computations, we find q √ σ21 βd cσ1 Rr √ r = 0.5 > 0.02 = 2 , R = (b+d)c = 16 > 1, max{b + d, c} = 1 and max{b + d, c}( R − 1)I{ R>1} + r − 2 −2 −2 −1 (2(σ2 + σ3 )) = 3.008 − 3.75 < 0. Obviously this group of parameters obey Theorem 3, y1 (t) and y2 (t) will become extinct as t → ∞ with probability 1. This is consistent with the results in Theorem 3. It also explains the big noise can cause the extinction of species which is clearly shown in Figure 2. Table 2: List of parameters
Description Intrinsic growth rate of the prey Intra-specific competition rate of the prey Death rate of immature predator Death rate of mature predator The rate of immature predator becoming mature predator Capture rate of the predator Conversion rate of the mature predator Half saturation constant
Jo
Parameters r a b c d α β m
17
Values 0.5 0.2 0.6 0.01 0.4 0.6 0.4 1.25
Journal Pre-proof
8
6
6 y1(t)
x(t)
4 4
2 2 0
0
1000
2000 3000 Time t
4000
5000
0
1000
2000 3000 Time t
4000
5000
0
0
1000 2000 3000 4000 5000 Time t
of
20
y2(t)
15 10
Pr e-
0
p ro
5
Figure 2: The paths of x(t), y1 (t) and y2 (t) of system (2) with initial value (x(0), y1(0), y2 (0)) = (0.2, 0.2, 0.2) under the noise intensities σ21 = 0.04, σ22 = 15 and σ23 = 15. The blue lines represent the solution to system (2) and the red lines represent the solution to the corresponding undisturbed system (1).
Example 3. We numerically simulate the model for the following set of parameter values: σ21 = 2, σ22 = 15, σ23 = 15. Other values of the system parameters can be seen from Table 3. By simple computations, we σ2
urn
al
find r = 0.5 < 1 = 21 and β = 0.4 < 0.5 = min{b, c}. It is obvious that this group of parameters satisfy the conditions of Theorem 4. It indicates that the big noise can cause both species of system (2) to die out ultimately and it may happen when a serious disease or severe weather occurs in the real life. This is clearly shown in Figure 3. Table 3: List of parameters
Description Intrinsic growth rate of the prey Intra-specific competition rate of the prey Death rate of immature predator Death rate of mature predator The rate of immature predator becoming mature predator Capture rate of the predator Conversion rate of the mature predator Half saturation constant
Jo
Parameters r a b c d α β m
18
Values 0.5 0.2 0.6 0.5 0.7 0.6 0.4 1.25
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8
6
6 y1(t)
x(t)
4 4
2 2 0
0
1000 2000 3000 4000 5000 Time t
0
0
1000 2000 3000 4000 5000 Time t
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y2(t)
15 10
0
1000 2000 3000 4000 5000 Time t
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0
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5
Figure 3: The paths of x(t), y1 (t) and y2 (t) of system (2) with initial value (x(0), y1(0), y2 (0)) = (0.2, 0.2, 0.2) under the noise intensities σ21 = 2, σ22 = 15 and σ23 = 15. The blue lines represent the solution to system (2) and the red lines represent the solution to the corresponding undisturbed system (1).
6. Acknowledgment
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References
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This work was supported by the Fundamental Research Funds for the Central Universities (Grant No. 2412017FZ004).
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A stochastic ratio-dependent predator-prey system with stage structure for the predator is studied.
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We establish sufficient conditions for the existence of a unique
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ergodic stationary distribution.
We obtain sufficient conditions for extinction which is discussed
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in two cases.