# Dynamics of an Impulsive Stochastic Nonautonomous Chemostat Model with Two Different Growth Rates in a Polluted Environment

Authors: Yajie Li; Xinzhu Meng
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

This paper proposes a novel impulsive stochastic nonautonomous chemostat model with the saturated and bilinear growth rates in a polluted environment. Using the theory of impulsive differential equations and Lyapunov functions method, we first investigate the dynamics of the stochastic system and establish the sufficient conditions for the extinction and the permanence of the microorganisms. Then we demonstrate that the stochastic periodic system has at least one nontrivial positive periodic solution. The results show that both impulsive toxicant input and stochastic noise have great effects on the survival and extinction of the microorganisms. Furthermore, a series of numerical simulations are presented to illustrate the performance of the theoretical results.


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## Body

## 1. Introduction

The chemostat is a continuous culture device that keeps the flow of the nutrient solution constant and makes the microorganism reproduce under the condition of its maximum growth rate. In a chemostat, the density of microorganisms is controlled by the concentration of the growth-limiting nutrient, and the growth rate is controlled by the washout rate which can be adjusted arbitrarily. The chemostat model is mainly used for laboratory theoretical research. So far, many scholars have obtained significant results for chemostat models [1–11]. While the populations in nature suffer from instantaneous discontinuous interference (for example, toxic input, seasonal harvest, and spraying pesticides), this interference phenomenon can be described as an impulse mathematically [12–16]. Furthermore, the theory and method of impulsive differential equations are widely used in many domains of biological science [17–20]. In reality, many populations are inevitably affected by stochastic disturbances [21–28], and many scholars have explored many stochastic dynamic systems and obtained some new results [29–33]. Recently, impulsive stochastic dynamics models attract the research interests of scholars [34–37]. Therefore, we also consider the influence of stochastic disturbances and impulsive toxic input on the chemostat.In previous studies, there are almost works on research of chemostat models with one of some different growth rates. In fact, microorganisms take different forms to absorb nutrients at different times, so the growth rates of microorganisms are different. For example, some take saturated growth rate and some adopt bilinear growth rate. Consequently, a model can utilize the transfer function asT(S,x)=∑i=1npiTi(S,x), where pi(i=1,2,…,n) represent the probability of occurrence of Ti(S,x)(i=1,2,…,n) and ∑i=1npi=1 [38]. In this paper, we consider two types of growth rates, bilinear growth rate and saturated growth rate. Hence we assume T(S,x)=pβ1Sx+(1-p)β2Sx/a+S, where p(0<p<1) denotes the probability of occurrence of bilinear growth rate and 1-p is correspondingly the probability of occurrence of saturated growth rate. In addition, we also consider the impact of time on the system. Then the system becomes nonautonomous: (1)dStdt=DtS0t-St-ptβ1tStxt-1-ptβ2tStxtat+St,dxtdt=ptβ1tStxt+1-ptβ2tStxtat+St-Dtxt,where S(t) represents the concentration of the unconsumed nutrient at time t and x(t) represents the biomass of the population of microorganism at time t. D(t) is the washout rate at time t, S0(t) is the concentration of the growth-limiting nutrient at time t, T(S,x) represents the nutrient intake rate of each individual in microbial population, and it also describes the conversion rate from nutrient to microorganism. a(t) is the semisaturated rate at time t.As we know, environmental pollution is one of the most important social and ecological problems at present. It affects the quality of life of human beings, the persistence of species, and the ecological vicissitude of habitat. In order to reasonably apply and control toxic substances, we must evaluate the degree of toxicant’s damage to the population. In recent years, many scholars have already studied the effects of toxicant on various ecosystems [39–44]. In this paper, we use impulsive differential equations to describe the effects of toxicant on microbial population. Besides toxicant, we also consider the effects of environmental noises. In this paper, we assume that fluctuations in the environment will manifest themselves mainly as fluctuations in two types of growth response rate, so that βi→βi+σiB˙i(i=1,2). Then, a nonautonomous impulsive chemostat model with stochastic perturbation in a polluted environment takes the following form:(2)dSt=DtS0t-St-ptβ1tStxt-1-ptβ2tStxtat+Stdt-ptσ1tStxtdB1t-1-ptσ2tStxtat+StdB2t,dxt=ptβ1tSt+1-ptβ2tStat+St-rtC0t-Dtxtdt+ptσ1tStxtdB1t+1-ptσ2tStxtat+StdB2t,dC0t=kCet-gC0t-mC0tdt,dCet=-hCetdt,t≠nτ,n∈Z+,ΔSt=0,Δxt=0,ΔC0t=0,ΔCet=u,t=nτ,n∈Z+,where C0(t) and Ce(t) denote the concentrations of the toxicant in the organism and in the environment at time t, respectively. r(t)>0 is the rate of decrease of the intrinsic growth rate at time t because of toxicant, k represents environmental toxicant uptake rate per unit mass organism, g and m are organismal net ingestion and depuration rates of toxicant, respectively, h denotes the loss rate of toxicant from the environment itself by volatilization, and u is the amount of pulsed input concentration of the toxicant at each τ. Moreover, all the above parameters are positive and S(t), x(t), C0(t), and Ce(t) are both positive functions according to its biomathematics meaning, and D(t), S0(t), p(t), β1(t), β2(t), a(t), r(t), σ1(t), and σ2(t) are both periodic functions with T=nτ.The paper is organized as follows. Preliminaries are provided in Section2. Existence and uniqueness of the global positive solution are demonstrated in Section 3. In Section 4, we explore the sufficient conditions for the extinction and the permanence of system (2). In Section 5, we show that the stochastic system has at least one nontrivial positive periodic solution by constructing a suitable Lyapunov function and a rectangular set. Finally, some numerical simulations are developed to illustrate the performance of the theoretical results.


## 2. Preliminaries

In this section, we give some notations and lemmas which will be used for the following reasoning process. Throughout this paper, we assume thatS(t), x(t), and C0(t) are continuous at t=nτ and Ce(t) is left continuous at t=nτ and Ce(nτ+)=limt→nτ+Ce(t), and let (Ω,F,Ft≥0,P) be a complete probability space with a filtration Ft≥0 satisfying the usual conditions (i.e., it is increasing and right continuous while F0 contains all P-null sets). For an integrable function f(t) on [0,∞), define fu=supt∈[0,∞)f(t), fl=inft∈[0,∞)f(t), and ftt=1/t∫0tf(s)ds; here f(t) is a bounded function on [0,∞). Due to the properties of continuous positive periodic functions, it is obvious that the coefficients of the system (2) satisfy max⁡{Du,S0u,pu,β1u,β2u,au,ru,σ1u,σ2u}<∞ and min⁡{Dl,S0l,pl,β1l,β2l,al,rl,σ1l,σ2l}≥0.Lemma 1.
For any positive solution(S(t),x(t),C0(t),Ce(t)) of system (2) with initial value (S(0),x(0),C0(0),Ce(0+))∈R+4, we have limsupt→+∞S(t)≤S0u, limsupt→+∞x(t)≤S0u.Proof.
From the first two equations of system (2), we have (3)dSt+xtdt=DtS0t-rtC0txt-Dtst+xt≤DtS0t-Dtst+xt≤DtS0u-Dtst+xt,which implies that limsupt→+∞S(t)≤S0u, limsupt→+∞x(t)≤S0u. This completes the proof of Lemma 1.Lemma 2 (see [3]).
(i) The microorganismx(t) is said to be extinctive if limt→+∞⁡x(t)=0.
(ii) The speciesx(t) is said to be permanent in the mean if there exists a positive constant λ such that limt→+∞xt≥λ.Lemma 3 (see [45]).
Assume that
(i) system has a unique global solution;
(ii) there is a functionV(t,x)∈C2 which is τ-periodic in t and satisfies the following conditions:(4)infx>R⁡Vt,x→∞asR→∞,and(5)LV≤-1outsidesomecompactset,where the operator L is given by(6)LV=Vtt,x+Vxt,xft,x+12tracegTt,xVxxt,xgt,x.Then the system has a τ-periodic solution.Now we consider the following subsystem of system (2):(7)dC0t=kCet-gC0t-mC0tdt,dCet=-hCetdt,t≠nτ,n∈Z+,ΔC0t=0,ΔCet=u,t=nτ,n∈Z+.Lemma 4 (see [3, 20]).
System (7) has a unique positive τ-periodic solution (C0∗(t),Ce∗(t))T and for each solution (C0(t),Ce(t))T of (7), C0(t)→C0∗(t), and Ce(t)→Ce∗(t) as t→+∞. Moreover, C0(t)>C0∗(t), Ce(t)>Ce∗(t) for all t≥0 if C0(0)>C0∗(0), and Ce(0)>Ce∗(0), where (8)C0∗t=C0∗0e-g+mt-nτ+kue-g+mt-nτ-e-ht-nτh-g-m1-e-hτ,Ce∗t=ue-ht-nτ1-e-hτ,C0∗0=kue-g+mτ-e-hτh-g-m1-e-g+mτ1-e-hτ,Ce∗0=u1-e-hτ,for t∈(nτ,(n+1)τ] and n∈Z+.From Lemma4, we have (9)limt→+∞⁡C0tt=kuhg+mτ≜C0¯.


## 3. Existence and Uniqueness of the Global Positive Solution

This section discusses the existence and uniqueness of the global positive solution as an important prerequisite for further research. We take into account the following subsystem of system (2): (10)dSt=DtS0t-St-ptβ1tStxt-1-ptβ2tStxtat+Stdt-ptσ1tStxtdB1t-1-ptσ2tStxtat+StdB2t,dxt=ptβ1tStxt+1-ptβ2tStxtat+St-rtC0txt-Dtxtdt+ptσ1tStxtdB1t+1-ptσ2tStxtat+StdB2t,where the bounded function C0(t) is defined by system (7).Theorem 5.
For any initial value(S(0),x(0))∈R+2, there is a unique positive solution (S(t),x(t)) of system (2) on t≥0, and the solution will remain in R+2 with probability one.Proof.
Owing to the coefficients of the system (10) obey local Lipschitz conditions; there exists a unique solution on [0,τe), where τe denotes the explosion time. Now, let us show that (S(t),x(t)) is global, i.e., τe=∞.
Letn0≥1 be sufficiently large such that S(t) and x(t) all lie within the interval [1/n0,n0], and define the stopping time (11)τn=inf⁡t∈0,τe:min⁡St,xt≤1normax⁡St,xt≥n,for each integer n>n0. We set inf⁡∅=∞(∅ denotes the empty set), and τn is distinctly increasing as n→∞. Let τ∞=limn→∞τn, and then τ∞≤τe a.s. If τ∞=∞ is true, then τe=∞ a.s. and (S(t),x(t))∈R+2 a.s. for t≥0. If this hypothesis is not true, there exist a constant T>0 and an ε∈(0,1) such that (12)Pτ∞≤T>ε.Hence, there is an integer n1≥n0 such that (13)Pτk≤T≥εfor  alln≥n1.Define a C2 function V:R+2→R+, by (14)Vt=S-γ-γln⁡Sγ+x-γ-γln⁡xγ,where γ=S0u. The nonnegativity of this function can be seen from (15)e-1-ln⁡e≥0for  anye>0.Applying the Itô’s formula to V leads to (16)LV=1-γStDtS0t-St-ptβ1tStxt-1-ptβ2tStxtat+St+1-γxtptβ1tSt+1-ptβ2tStat+St-rtC0t-Dtxt-γ2p2tσ12tx2t-γ21-pt2σ22tx2tat+St2+γ2p2tσ12tS2t+γ21-pt2σ22tS2tat+St2≤DtS0t+Dtγ+ptβ1txtγ+1-ptβ2txtγat+St+rtC0t+Dtγ+γ2p2tσ12tS2t+γ21-pt2σ22tS2tat+St2≤3Duγ+puβ1uγ2+1-plβ2uγ2al+ruC0∗0γ+γ32p2uσ12u+γ321-pl2σ22ua2l≕K.Here K is a positive constant which is independent of S(t) and x(t). Therefore, (17)dV≤Kdt-St-γptσ1txtdB1t-St-γ1-ptσ2txtat+StdB2t+xt-γptσ1tStdB1t+xt-γ1-ptσ2tStat+StdB2t.Then, in the same way with [34], we get (18)∞>VS0,x0+KT=∞,n→∞,which is a contradiction, and then we have τ∞=∞. This completes the proof.


## 4. Extinction and Persistence in Mean

### 4.1. Extinction

In this section, we establish sufficient conditions for the extinction of the microorganism, which implies that microculture failed.Theorem 6.
Let(S(t),x(t),C0(t),Ce(t)) be the solution of system (2) with any initial value (S(0),x(0),C0(0),Ce(0+))∈R+4. If one of the conditions holds(i)
σ1(t)>β1l/puS0u,σ2(t)>alβ2l/(1-pl)S0u, and R11=β12uσ22l+β22uσ12l/2σ12lσ22l(Dl+rlC0¯)<1,(ii)
σ1(t)>β1l/puS0u,σ2(t)<alβ2u/(1-pu)S0u, and R12=β12ual+2(1-pl)β2uS0uσ12l/2σ12lal(Dl+rlC0¯)<1,(iii)
σ1(t)<β1u/plS0u,σ2(t)>alβ2l/(1-pl)S0u, and R21=2puβ1uS0uσ22l+β22u/2σ22l(Dl+rlC0¯)<1,(iv)
σ1(t)<β1u/plS0u,σ2(t)<alβ2u/(1-pu)S0u, and R22=[puβ1ual+(1-pl)β2u]S0u/al(Dl+rlC0¯)<1,
then the microorganism will die out almost surely, i.e.,limt→∞x(t)=0, moreover, limt→∞S(t)=S0u, limt→∞C0(t)=C0∗(t), and limt→∞Ce(t)=Ce∗(t).Proof.
LetV(t)=ln⁡x. Applying Itô’s formula to system (2), we have (19)dVt=ptβ1tSt+1-ptβ2tStat+St-rtC0t-Dt-12p2tσ12tS2t-121-pt2σ22tS2tat+St2+ptσ1tStdB1t+1-ptσ2tStat+StdB2t=β12t2σ12t-p2tσ12t2St-β1tptσ12t2+β22t2σ22t-1-pt2σ22t2Stat+St-β2t1-ptσ22t2-rtC0t-Dt+ptσ1tStdB1t+1-ptσ2tStat+StdB2t.Integrating, respectively, from 0 to t and dividing both sides of (19) by t, one has (20)ln⁡xtt=1t∫0tβ12t2σ12t-p2tσ12t2St-β1tptσ12t2+β22t2σ22t-1-pt2σ22t2Stat+St-β2t1-ptσ22t2-rtC0t-Dtdt+M1tt+M2tt+ln⁡x0t,where M1(t)=∫0tp(θ)σ1(θ)S(θ)dB1(θ), M2(t)=∫0t(1-p(θ))σ2(θ)S(θ)/a(θ)+S(θ)dB2(θ), and by the strong law of large numbers [46] we have limt→∞M1(t)/t=0 and limt→∞M2(t)/t=0.
Case 1. When σ1(t)>β1l/puS0u,σ2(t)>alβ2l/(1-pl)S0u and R11<1, applying (20) we have (21)ln⁡xtt≤1t∫0tβ12t2σ12t+β22t2σ22t-Dtdt-rtt∫0tC0sds+M1tt+M2tt+ln⁡x0t≤β12u2σ12l+β22u2σ22l-Dl-rlC0tt+M1tt+M2tt+ln⁡x0t=Dl+rlC0ttβ12uσ22l+β22uσ12l2σ12lσ22lDl+rlC0tt-1+M1tt+M2tt+ln⁡x0t.Taking the limit superior on both sides of (21), we have (22)limsupt→+∞⁡ln⁡xtt≤Dl+rlC0¯R11-1<0.Thus limt→∞x(t)=0.
Case 2. When σ1(t)>β1l/puS0u,σ2(t)<alβ2u/(1-pu)S0u and R12<1, applying (20) we have(23)ln⁡xtt≤1t∫0tβ12t2σ12t+1-ptβ2tStat+St-Dtdt-rtt∫0tC0sds+M1tt+M2tt+ln⁡x0t≤β12u2σ12l+1-plβ2uS0ual-Dl-rlC0tt+M1tt+M2tt+ln⁡x0t=Dl+rlC0ttβ12ual+21-plβ2uS0uσ12l2σ12lalDl+rlC0tt-1+M1tt+M2tt+ln⁡x0t.Taking the limit superior on both sides of (23), we have (24)limsupt→+∞⁡ln⁡xtt≤Dl+rlC0¯R12-1<0.Thus limt→∞x(t)=0.
Case 3. When σ1(t)<β1u/plS0u,σ2(t)>alβ2l/(1-pl)S0u, with R21<1, applying (20) we have (25)ln⁡xtt≤1t∫0tptβ1tSt+β22t2σ22t-Dtdt-rtt∫0tC0sds+M1tt+M2tt+ln⁡x0t≤puβ1uS0u+β22u2σ22l-Dl-rlC0tt+M1tt+M2tt+ln⁡x0t=Dl+rlC0tt2puβ1uS0uσ22l+β22u2σ22lDl+rlC0tt-1+M1tt+M2tt+ln⁡x0t.Taking the limit superior on both sides of (25), we have (26)limsupt→+∞⁡ln⁡xtt≤Dl+rlC0¯R21-1<0.Thus limt→∞x(t)=0.
Case 4. When σ1(t)<β1u/plS0u,σ2(t)<alβ2u/(1-pu)S0u, with R22<1, applying (20) we have (27)ln⁡xtt≤1t∫0tptβ1tSt+1-ptβ2tStat+St-Dtdt-rtt∫0tC0sds+M1tt+M2tt+ln⁡x0t≤puβ1uS0u+1-plβ2uS0ual-Dl-rlC0tt+M1tt+M2tt+ln⁡x0t=Dl+rlC0ttpuβ1ual+1-plβ2uS0ualDl+rlC0tt-1+M1tt+M2tt+ln⁡x0t.Taking the limit superior on both sides of (27), we have (28)limsupt→+∞⁡ln⁡xtt≤Dl+rlC0¯R22-1<0.This implies limt→+∞x(t)=0, a.s.
Since the limit system of (2) is(29)dSt=DtS0t-Stdt,dC0t=kCet-gC0t-mC0tdt,dCet=-hCetdt,t≠nτ,n∈Z+,ΔSt=0,ΔC0t=0,ΔCet=u,t=nτ,n∈Z+,by Lemma 4, it is clear that limt→∞S(t)=S0u,limt→∞C0(t)=C0∗(t), limt→∞Ce(t)=Ce∗(t). This completes the proof of Theorem 6.


### 4.2. Persistence in Mean

Theorem 7.
For any initial value(S(0),x(0),C0(0),Ce(0+))∈R+4, system (2) is permanent in the mean if (30)R1∗=DlS0lplβ1l+1-puβ2l/au+S0uDuruC0¯+Du+1/2p2uσ12uS02u+1/21-pl2σ22uS02u/a2l>1,and the solution (S(t),x(t),C0(t),Ce(t)) of system (2) satisfies (31)liminft→+∞⁡xtt≥ruC0¯+Du+1/2p2uσ12uS02u+1/21-pl2σ22uS02u/a2lruC0∗0+Du1+plβ1lau+S0u+1-puβ2l/Duau+S0uR1∗-1.Proof.
From system (2),(32)δt≜St-S0t+xt-x0t=1t∫0tDsS0s-Ss-rC0s+Dsxsds≥1t∫0tDlS0l-DuSs-ruC0∗0+Duxsds≥DlS0l-DuStt-ruC0∗0+DuxttObviously, limt→+∞δ(t)=0, and we obtain (33)Stt≥DlS0lDu-ruC0∗0+DuDuxtt-δtDu.Applying Itô’s formula gives(34)dln⁡xt+xt=1+xtptβ1tSt+1-ptβ2tStat+St-rtC0t-Dt-12p2tσ12tS2t-121-pt2σ22tS2tat+St2dt+1+xtptσ1tStdB1t+1+xt1-ptσ2tStat+StdB2t≥1+xtplβ1lSt+1-puβ2lStau+S0u-ruC0t-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2ldt+1+xtptσ1tStdB1t+1+xt1-ptσ2tStat+StdB2t≥-ruC0t-Du-ruC0∗0+Duxt+plβ1lSt+1-puβ2lStau+S0u-12p2uσ12uS02u-121-pl2σ22uS02ua2ldt+1+xtptσ1tStdB1t+1+xt1-ptσ2tStat+StdB2t.Then(35)ln⁡xt-ln⁡x0t+xt-x0t≥-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l-ruC0∗0+Duxtt+plβ1l+1-puβ2lau+S0uStt+M3tt+M4tt≥-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lplβ1lDu+1-puDlS0lβ2lDuau+S0u-ruC0∗0+Du1+plβ1lau+S0u+1-puβ2lDuau+S0uxtt-δtDuplβ1l+1-puβ2lau+S0u+M3tt+M4tt,where limt→+∞M3(t)/t=0 and limt→+∞M4(t)/t=0. The inequality (35) can be rewritten as (36)xtt≥1Δ-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lplβ1lDu+1-puDlS0lβ2lDuau+S0u-δtDuplβ1l+1-puβ2lau+S0u+M3tt+M4tt-ln⁡xt-ln⁡x0t-xt-x0t≥1Δ-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lplβ1lDu+1-puDlS0lβ2lDuau+S0u-δtDuplβ1l+1-puβ2lau+S0u+M3tt+M4tt+ln⁡x0t-xt-x0t0<xt<1;1Δ-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lplβ1lDu+1-puDlS0lβ2lDuau+S0u-δtDuplβ1l+1-puβ2lau+S0u+M3tt+M4tt-ln⁡xt-ln⁡x0t-xt-x0t1≤xt<S0u,where Δ=(ruC0∗(0)+Du)1+plβ1l(au+S0u)+(1-pu)β2l/Du(au+S0u). According to Lemma 1, we see that x(t)≤S0u; thus one has limt→+∞x(t)/t=0, limt→+∞ln⁡x(t)/t=0 as x(t)≥1, and limt→+∞δ(t)=0.
Taking the inferior limit of both sides of (36), one can yield that (37)liminft→+∞⁡xt≥1Δ-ruC0¯-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lDuplβ1l+1-puβ2lau+S0u=ruC0¯+Du+1/2p2uσ12uS02u+1/21-pl2σ22uS02u/a2lΔR1∗-1>0.By (ii) of Lemma 2, this completes the proof of Theorem 7.


## 4.1. Extinction

In this section, we establish sufficient conditions for the extinction of the microorganism, which implies that microculture failed.Theorem 6.
Let(S(t),x(t),C0(t),Ce(t)) be the solution of system (2) with any initial value (S(0),x(0),C0(0),Ce(0+))∈R+4. If one of the conditions holds(i)
σ1(t)>β1l/puS0u,σ2(t)>alβ2l/(1-pl)S0u, and R11=β12uσ22l+β22uσ12l/2σ12lσ22l(Dl+rlC0¯)<1,(ii)
σ1(t)>β1l/puS0u,σ2(t)<alβ2u/(1-pu)S0u, and R12=β12ual+2(1-pl)β2uS0uσ12l/2σ12lal(Dl+rlC0¯)<1,(iii)
σ1(t)<β1u/plS0u,σ2(t)>alβ2l/(1-pl)S0u, and R21=2puβ1uS0uσ22l+β22u/2σ22l(Dl+rlC0¯)<1,(iv)
σ1(t)<β1u/plS0u,σ2(t)<alβ2u/(1-pu)S0u, and R22=[puβ1ual+(1-pl)β2u]S0u/al(Dl+rlC0¯)<1,
then the microorganism will die out almost surely, i.e.,limt→∞x(t)=0, moreover, limt→∞S(t)=S0u, limt→∞C0(t)=C0∗(t), and limt→∞Ce(t)=Ce∗(t).Proof.
LetV(t)=ln⁡x. Applying Itô’s formula to system (2), we have (19)dVt=ptβ1tSt+1-ptβ2tStat+St-rtC0t-Dt-12p2tσ12tS2t-121-pt2σ22tS2tat+St2+ptσ1tStdB1t+1-ptσ2tStat+StdB2t=β12t2σ12t-p2tσ12t2St-β1tptσ12t2+β22t2σ22t-1-pt2σ22t2Stat+St-β2t1-ptσ22t2-rtC0t-Dt+ptσ1tStdB1t+1-ptσ2tStat+StdB2t.Integrating, respectively, from 0 to t and dividing both sides of (19) by t, one has (20)ln⁡xtt=1t∫0tβ12t2σ12t-p2tσ12t2St-β1tptσ12t2+β22t2σ22t-1-pt2σ22t2Stat+St-β2t1-ptσ22t2-rtC0t-Dtdt+M1tt+M2tt+ln⁡x0t,where M1(t)=∫0tp(θ)σ1(θ)S(θ)dB1(θ), M2(t)=∫0t(1-p(θ))σ2(θ)S(θ)/a(θ)+S(θ)dB2(θ), and by the strong law of large numbers [46] we have limt→∞M1(t)/t=0 and limt→∞M2(t)/t=0.
Case 1. When σ1(t)>β1l/puS0u,σ2(t)>alβ2l/(1-pl)S0u and R11<1, applying (20) we have (21)ln⁡xtt≤1t∫0tβ12t2σ12t+β22t2σ22t-Dtdt-rtt∫0tC0sds+M1tt+M2tt+ln⁡x0t≤β12u2σ12l+β22u2σ22l-Dl-rlC0tt+M1tt+M2tt+ln⁡x0t=Dl+rlC0ttβ12uσ22l+β22uσ12l2σ12lσ22lDl+rlC0tt-1+M1tt+M2tt+ln⁡x0t.Taking the limit superior on both sides of (21), we have (22)limsupt→+∞⁡ln⁡xtt≤Dl+rlC0¯R11-1<0.Thus limt→∞x(t)=0.
Case 2. When σ1(t)>β1l/puS0u,σ2(t)<alβ2u/(1-pu)S0u and R12<1, applying (20) we have(23)ln⁡xtt≤1t∫0tβ12t2σ12t+1-ptβ2tStat+St-Dtdt-rtt∫0tC0sds+M1tt+M2tt+ln⁡x0t≤β12u2σ12l+1-plβ2uS0ual-Dl-rlC0tt+M1tt+M2tt+ln⁡x0t=Dl+rlC0ttβ12ual+21-plβ2uS0uσ12l2σ12lalDl+rlC0tt-1+M1tt+M2tt+ln⁡x0t.Taking the limit superior on both sides of (23), we have (24)limsupt→+∞⁡ln⁡xtt≤Dl+rlC0¯R12-1<0.Thus limt→∞x(t)=0.
Case 3. When σ1(t)<β1u/plS0u,σ2(t)>alβ2l/(1-pl)S0u, with R21<1, applying (20) we have (25)ln⁡xtt≤1t∫0tptβ1tSt+β22t2σ22t-Dtdt-rtt∫0tC0sds+M1tt+M2tt+ln⁡x0t≤puβ1uS0u+β22u2σ22l-Dl-rlC0tt+M1tt+M2tt+ln⁡x0t=Dl+rlC0tt2puβ1uS0uσ22l+β22u2σ22lDl+rlC0tt-1+M1tt+M2tt+ln⁡x0t.Taking the limit superior on both sides of (25), we have (26)limsupt→+∞⁡ln⁡xtt≤Dl+rlC0¯R21-1<0.Thus limt→∞x(t)=0.
Case 4. When σ1(t)<β1u/plS0u,σ2(t)<alβ2u/(1-pu)S0u, with R22<1, applying (20) we have (27)ln⁡xtt≤1t∫0tptβ1tSt+1-ptβ2tStat+St-Dtdt-rtt∫0tC0sds+M1tt+M2tt+ln⁡x0t≤puβ1uS0u+1-plβ2uS0ual-Dl-rlC0tt+M1tt+M2tt+ln⁡x0t=Dl+rlC0ttpuβ1ual+1-plβ2uS0ualDl+rlC0tt-1+M1tt+M2tt+ln⁡x0t.Taking the limit superior on both sides of (27), we have (28)limsupt→+∞⁡ln⁡xtt≤Dl+rlC0¯R22-1<0.This implies limt→+∞x(t)=0, a.s.
Since the limit system of (2) is(29)dSt=DtS0t-Stdt,dC0t=kCet-gC0t-mC0tdt,dCet=-hCetdt,t≠nτ,n∈Z+,ΔSt=0,ΔC0t=0,ΔCet=u,t=nτ,n∈Z+,by Lemma 4, it is clear that limt→∞S(t)=S0u,limt→∞C0(t)=C0∗(t), limt→∞Ce(t)=Ce∗(t). This completes the proof of Theorem 6.


## 4.2. Persistence in Mean

Theorem 7.
For any initial value(S(0),x(0),C0(0),Ce(0+))∈R+4, system (2) is permanent in the mean if (30)R1∗=DlS0lplβ1l+1-puβ2l/au+S0uDuruC0¯+Du+1/2p2uσ12uS02u+1/21-pl2σ22uS02u/a2l>1,and the solution (S(t),x(t),C0(t),Ce(t)) of system (2) satisfies (31)liminft→+∞⁡xtt≥ruC0¯+Du+1/2p2uσ12uS02u+1/21-pl2σ22uS02u/a2lruC0∗0+Du1+plβ1lau+S0u+1-puβ2l/Duau+S0uR1∗-1.Proof.
From system (2),(32)δt≜St-S0t+xt-x0t=1t∫0tDsS0s-Ss-rC0s+Dsxsds≥1t∫0tDlS0l-DuSs-ruC0∗0+Duxsds≥DlS0l-DuStt-ruC0∗0+DuxttObviously, limt→+∞δ(t)=0, and we obtain (33)Stt≥DlS0lDu-ruC0∗0+DuDuxtt-δtDu.Applying Itô’s formula gives(34)dln⁡xt+xt=1+xtptβ1tSt+1-ptβ2tStat+St-rtC0t-Dt-12p2tσ12tS2t-121-pt2σ22tS2tat+St2dt+1+xtptσ1tStdB1t+1+xt1-ptσ2tStat+StdB2t≥1+xtplβ1lSt+1-puβ2lStau+S0u-ruC0t-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2ldt+1+xtptσ1tStdB1t+1+xt1-ptσ2tStat+StdB2t≥-ruC0t-Du-ruC0∗0+Duxt+plβ1lSt+1-puβ2lStau+S0u-12p2uσ12uS02u-121-pl2σ22uS02ua2ldt+1+xtptσ1tStdB1t+1+xt1-ptσ2tStat+StdB2t.Then(35)ln⁡xt-ln⁡x0t+xt-x0t≥-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l-ruC0∗0+Duxtt+plβ1l+1-puβ2lau+S0uStt+M3tt+M4tt≥-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lplβ1lDu+1-puDlS0lβ2lDuau+S0u-ruC0∗0+Du1+plβ1lau+S0u+1-puβ2lDuau+S0uxtt-δtDuplβ1l+1-puβ2lau+S0u+M3tt+M4tt,where limt→+∞M3(t)/t=0 and limt→+∞M4(t)/t=0. The inequality (35) can be rewritten as (36)xtt≥1Δ-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lplβ1lDu+1-puDlS0lβ2lDuau+S0u-δtDuplβ1l+1-puβ2lau+S0u+M3tt+M4tt-ln⁡xt-ln⁡x0t-xt-x0t≥1Δ-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lplβ1lDu+1-puDlS0lβ2lDuau+S0u-δtDuplβ1l+1-puβ2lau+S0u+M3tt+M4tt+ln⁡x0t-xt-x0t0<xt<1;1Δ-ruC0tt-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lplβ1lDu+1-puDlS0lβ2lDuau+S0u-δtDuplβ1l+1-puβ2lau+S0u+M3tt+M4tt-ln⁡xt-ln⁡x0t-xt-x0t1≤xt<S0u,where Δ=(ruC0∗(0)+Du)1+plβ1l(au+S0u)+(1-pu)β2l/Du(au+S0u). According to Lemma 1, we see that x(t)≤S0u; thus one has limt→+∞x(t)/t=0, limt→+∞ln⁡x(t)/t=0 as x(t)≥1, and limt→+∞δ(t)=0.
Taking the inferior limit of both sides of (36), one can yield that (37)liminft→+∞⁡xt≥1Δ-ruC0¯-Du-12p2uσ12uS02u-121-pl2σ22uS02ua2l+DlS0lDuplβ1l+1-puβ2lau+S0u=ruC0¯+Du+1/2p2uσ12uS02u+1/21-pl2σ22uS02u/a2lΔR1∗-1>0.By (ii) of Lemma 2, this completes the proof of Theorem 7.


## 5. The Existence of Nontrivial Positive Periodic Solution

In this section, based on the theory of Has’minskii [45], we testify the existence of the nontrivial positive periodic solution.Define(38)R0s≔ptβ1tS0tTrtC0t+Dt+1/2p2uσ12uS02u+1/21-pl2σ22uS02u/a2lT.Theorem 8.
Assuming thatR0s>1, then for any initial value (S(0),x(0))∈R+2, system (10) has at least one positive τ-periodic solution.Proof.
Define a nonnegativeC2-function V~: R+2→R+ as follows: (39)V~=M-ln⁡St-αln⁡xt+ωt+1θ+1St+xtθ+1-ln⁡St≕MV1+ωt+V2+V3,where V1=-ln⁡St-aln⁡x(t), V2=1/θ+1(S(t)+x(t))θ+1, V3=-ln⁡S(t), α is positive constant which is defined in the following discussion, θ>1 is a positive constant satisfying (40)Dl-θS02u4p2uσ12u+1-pl2σ22ua2l>0,namely, (41)1<θ<4DlS02up2uσ12u+1-pl2σ22u/a2l,and M>0 obeys (42)-Mλ~+E≤-2,where(43)λ~=DTR0s-1>0and(44)B=supS,x∈R+2⁡DuS0u-rlS0lC0lS+xθ-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lS+xθ+1<∞,C=Du+S0upuβ1u+1-plβ2ual+S02u2p2uσ12u+1-pl2σ22ua2l,E=supS,x∈R+2⁡-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+xθ+1+B+C.In addition, V~(S,x) must have a minimum point (S0,x0)∈R+2 as V~(S,x) is a continuous function; then we can define a nonnegative C2-function V: R+2→R+ as follows: (45)VS,x=V~S,x-V~S0,x0.Applying Itô’s formula, we can get(46)LV1=-DtS0tSt+Dt+ptβ1txt+1-ptβ2txtat+St+12p2tσ12tx2t+121-pt2σ22tx2tat+St2-αptβ1tSt-α1-ptβ2tStat+St+αrtC0t+αDt+α2p2tσ12tS2t+α21-pt2σ22tS2tat+St2≤-2αDtptβ1tS0t+αrtC0t+Dt+12p2uσ12uS02u+121-pl2σ22uS02ua2l+Dt+puβ1u+1-plβ2ualx+12p2uσ12u+1-pl2σ22ua2lx2≕R0t+puβ1u+1-plβ2ualx+12p2uσ12u+1-pl2σ22ua2lx2.Define a positive τ-periodic function ω(t) satisfying(47)ω′t=R0T-R0t,and then we obtain (48)LV1+ωt≤R0T+puβ1u+1-plβ2ualx+12p2uσ12u+1-pl2σ22ua2lx2.We choose (49)α=Dpβ1S0TrC0+D+1/2p2uσ12uS02u+1/21-pl2σ22uS02u/a2lT,then (50)R0T=Dpβ1S0TrC0+D+1/2p2uσ12uS02u+1/21-pl2σ22uS02u/a2lT+DT≕-DTR0s-1.Similarly,(51)LV2=St+xtθDtS0t-DtSt-rtC0txt-Dtxt+θSt+xtθ-1p2tσ12tStxt+1-pt2σ22tStxtat+St2Stxt≤S+xθDuS0u-rlC0lS0l-DlS+x+θ4S+xθ+1p2uσ12uS02u+1-pl2σ22uS02ua2l=DuS0u-rlC0lS0lS+xθ-Dl-θS02u4p2uσ12u+1-pl2σ22ua2lS+xθ+1≤B-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lS+xθ+1≤B-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+xθ+1,and(52)LV3=-DtS0tSt+Dt+ptβ1txt+1-ptβ2txtat+St+12p2tσ12tx2t+121-pt2σ22tx2tat+St2≤-DlS0lS+Du+puβ1uS0u+1-plβ2uS0ual+12p2uσ12uS02u+121-pl2σ22uS02ua2l≕-DlS0lS+C.Therefore, we get (53)LV≤-MDTR0s-1+Mpuβ1u+1-plβ2ualx+M2p2uσ12u+1-pl2σ22ua2lx2-DlS0lS-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+xθ+1+B+C.Define a bounded closed set (54)Uϵ=S,x∈R+2:ϵ≤S≤1ϵ,ϵ≤x≤1ϵ,and here we can choose ϵ∈(0,1) sufficiently small which satisfies (55)-DlS0lϵ+A≤-1,(56)-Mλ~+Mpuβ1u+1-plβ2ualϵ+M2p2uσ12u+1-pl2σ22ua2lϵ2≤-1,(57)-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2l1ϵθ+1+F≤-1,(58)-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2l1ϵθ+1+G≤-1,where A, F, and G are both positive constants, which are defined in the following reasoning process. And for a sufficiently small ϵ, it is easy to see that if (42) holds, then (56) holds.
DivideR+2∖Uϵ into four domains: (59)U1=S,x∈R+2:0<S<ϵ,U2=S,x∈R+2:0<x<ϵ,U3=S,x∈R+2:S>1ϵ,U4=S,x∈R+2:x>1ϵ,noting that Uϵc=U1∪U2∪U3∪U4. Next we prove that LV≤-1 on Uϵc.
Case 1. If (S,x)∈U1, then(60)LV≤-DlS0lS+Mpuβ1u+1-plβ2ualx+M2p2uσ12u+1-pl2σ22ua2lx2-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+xθ+1+B+C≤-DlS0lS+A≤-DlS0lϵ+A≤-1,where(61)A=supS,x∈R+2⁡Mpuβ1u+1-plβ2ualx+M2p2uσ12u+1-pl2σ22ua2lx2-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+xθ+1+B+C.According to (55), we get (62)LV≤-1,S,x∈U1.Case 2. If (S,x)∈U2, then (63)LV≤-MDTR0s-1+Mpuβ1u+1-plβ2ualx+M2p2uσ12u+1-pl2σ22ua2lx2-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+xθ+1+B+C≤-Mλ~+Mpuβ1u+1-plβ2ualϵ+M2p2uσ12u+1-pl2σ22ua2lϵ2+E≤-1.By means of (56), we can conclude that (64)LV≤-1,S,x∈U2.Case 3. If (S,x)∈U3, then(65)LV≤-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+Mpuβ1u+1-plβ2ualx+M2p2uσ12u+1-pl2σ22ua2lx2-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lxθ+1+B+C≤-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+F≤-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2l1ϵθ+1+F≤-1,where (66)F=supS,x∈R+2⁡-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+Mpuβ1u+1-plβ2ualx+M2p2uσ12u+1-pl2σ22ua2lx2-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lxθ+1+B+C.According to (57), one can get that (67)LV≤-1,S,x∈U3.Case 4. If (S,x)∈U4, then (68)LV≤-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2lxθ+1-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2lxθ+1+Mpuβ1u+1-plβ2ualx+M2p2uσ12u+1-pl2σ22ua2lx2-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+B+C≤-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2lxθ+1+G≤-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2l1ϵθ+1+G≤-1,where (69)G=supS,x∈R+2⁡-14Dl-θS02u4p2uσ12u+1-pl2σ22ua2lxθ+1+Mpuβ1u+1-plβ2ualx+M2p2uσ12u+1-pl2σ22ua2lx2-12Dl-θS02u4p2uσ12u+1-pl2σ22ua2lSθ+1+B+C.In view of (58), we have (70)LV≤-1,S,x∈U4.Thus, we have LV≤-1 on Uϵc from the above four cases. By Lemma 3, we can get that system (2) admits a positive τ-periodic solution. This completes the proof of Theorem 8.


## 6. Conclusions and Simulations

This paper investigates the dynamics of an impulsive stochastic chemostat model with two different growth rates in a polluted environment. The advantage of this paper is to take two different variable growth rates of microorganisms. We propose the novel model with the transfer function asT(S,x)=∑i=1npiTi(S,x), where pi(i=1,2,…,n) represent the probability of occurrence of Ti(S,x)(i=1,2,…,n) and ∑i=1npi=1. Hence we assume T(S,x)=pβ1Sx+(1-p)β2Sx/a+S, and it shows that the microorganism has two types of growth rates which are bilinear and saturated growth rate. Furthermore, if p=0, this means the growth rate of the microorganism has only one growth rate, which is saturated growth rate. On the contrary, if the growth rate of the microorganism is only bilinear growth rate, we can assume p=1. Thus our hypothesis is more extensive. And in this paper, by using the theory of impulsive differential equations and Lyapunov functions method, we establish the conditions for the extinction and the permanence of the microorganisms. Then we demonstrate that the stochastic system has at least one positive τ-periodic solution. The main results suggest that(1) If one of the four conditions of Theorem6 is satisfied, then limt→+∞x(t)=0a.s. This implies that the microorganism goes to extinction.(2) IfR1∗>1, system (2) is permanent in the mean.(3) IfR0s>1, system (10) exists a nontrivial positive periodic solution.Some numerical simulations are introduced to illustrate our main results [47].In Figure1, we set the parameter values in system (2) as u=0.6,k=0.02,g=0.01,m=0.01,h=0.02,τ=1,(S(0),x(0),C0(0),Ce(0))=(2,2,1,1), and (a)σ1l>0.202,σ2l>0.128 and R11=0.908<1; (b)σ1l>0.285,σ2u<0.187 and R12=0.982<1; (c)σ1u<1.104,σ2l>0.114 and R21=0.999<1; (d)σ1u<1.104,σ2u<0.187 and R22=0.994<1. It meets the conditions of Theorem 6 and the microorganisms in figure are all going extinct, so this is consistent with the conclusion in Theorem 6. In order to verify the condition of Theorems 7 and 8, we choose the parameter values as follows: u=0.4,k=0.1,g=0.8,m=0.5,h=1,τ=1,(S(0),x(0),C0(0),Ce(0))=(1.5,1.5,0.5,0.5),1<θ<1.617 in Figure 2. It obtains R1∗=1.060>1 and R0s=1.545>1, and the microorganism in Figure 2 is persistence in mean, which is in line with the conclusion in Theorems 7 and 8, respectively.Figure 1
The solution of system (2) for extinction of the microorganism.
(a) (b) (c) (d)Figure 2
Numerical simulations of persistence in mean of system (2).
(a) (b)The results show that both impulsive toxicant input and stochastic noise have great effects on the survival and extinction of the microorganisms. High concentration of toxicant or large intensity of noises can lead to extinction of microorganisms. Appropriate toxin input and noise intensity will make the microorganisms show periodic changes.In this paper, we considered the effects of white noises. However, there are some other random perturbations in the environment, for example, regime-switching and Markovian switching. In fact, stochastic regime-switching or Markovian switching population models were studied by several authors recently [27, 43, 48, 49]. Therefore, it is interesting to consider regime-switching and Markovian switching models, and we will leave these cases as our future work.



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