# Dynamical Analysis of Two-Microorganism and Single Nutrient Stochastic Chemostat Model with Monod-Haldane Response Function

Authors: Mengnan Chi; Wencai Zhao
Journal: Complexity
 (2019)

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## Abstract

In this paper, we formulate and investigate a two-microorganism and single nutrient chemostat model with Monod-Haldane response function and random perturbation. First, for the corresponding deterministic system, we introduce the conditions of the stability of the equilibrium points. Then, using Lyapunov function and Itô’s formula, we investigate the existence and uniqueness of the global positive solution of the stochastic chemostat model. Furthermore, we explore and obtain the criterions of the extinction and the permanence for the stochastic model. Finally, numerical simulations are carried out to illustrate our main results.


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## Body

## 1. Introduction

The chemostat is a classic bioreactor for continuous microbial culture. It has been widely used in microbiology and bioengineering [1–3]. The device is mainly composed of three parts: nutrient bottle, incubator, and collector, generally by using pump or an overflow device to keep the chemostat volume constant. Chemostat model has attracted great interests of many scholars since Monod [4]. Monod [5], Novick et al. [6], and Herbert et al. [7] considered the basic theory of microbial interaction in the chemostat. The simple chemostat model described by ordinary differential equations (ODEs) takes the following form: (1)dStdt=S0-StD-mStxta+St,dxtdt=mSta+St-Dxt,where S(t) and x(t) stand for the concentrations of the nutrient and the microorganism at time t, respectively. S0 and D are positive constants, which represent the input concentration of the nutrient and the common washout rate respectively. The function mS(t)/a+St denotes the Monod growth functional response, m is called the maximal growth rate, and a is the half-saturation (or Michaelis-Menten) constant. Based on the above model, many scholars have studied the chemostat model with different functional response in which a single microbe feeds on a single nutrient [8–11].Taylor and Williams [12] considered the coexistence of two different microbe populations feeding on single nutrient in the chemostat and proposed the following model: (2)dStdt=DS0-St-m1Stxta1+St-m2Styta2+St,dxtdt=m1Stxta1+St-Dxt,dytdt=m2Styta2+St-Dyt,where x(t) and y(t) denote the concentrations of two different microorganisms at time t. And then many scholars have investigated the competition of microorganisms in chemostat [13–16].Taking into account more complex inhibitory effects of high substrate concentrations on microbial growth, for example, nitrite and ammonia can lead to the inhibition of Nitrobacter and Nitrosomonas, respectively. Andrews [17] proposed the following nonmonotonic response function called Monod-Haldane growth rate (inhibitory rate):(3)μSt=mSta+St+KS2t,where m>0 is the maximal growth rate and a>0 is the Michaelis-Menten constant, assuming that the term KS2(t) is an inhibitor and K is a half-saturation parameter. Based on [17], Bush and Cook [18] improved the inhibition function to a general functional response which retains the particular features of the Andrews function. Recently, Wang et al. [19] proposed and investigated the stochastic dynamic behaviors of a stochastic chemostat model with Monod-Haldane response function in which a single microbe feeds on a single nutrient.Considering the nutritional substitution, Chi and Zhao [20] established a single microorganism and multinutrient chemostat model with impulsive toxicant input in a polluted environment as follows: (4)dS1t=QS10-S1t-μ1S1txtδ1a1+xtdt-σ1S1txtδ1a1+xtdB1t,dS2t=QS20-S2t-μ2S2txtδ2a2+xtdt-σ2S2txtδ2a2+xtdB2t,dxt=μ1S1txta1+xt+μ2S2txta2+xt-Qxt-rP0txtdt+σ1S1txta1+xtdB1t+σ2S2txta2+xtdB2t,dP0t=kPet-gP0t-mP0tdt,dPet=-hPetdt,t≠nT,n∈Z+,ΔS1t=0,ΔS2t=0,Δxt=0,ΔP0t=0,ΔPet=u,t=nT,n∈Z+.In the above model, a single species (x(t)) feeds on two substitutable resources (S1(t) and S2(t)) in a polluted environment. For the system without stochastic effect, by using the theory of impulsive differential equations, the authors proved that the model system has a globally asymptotically stable ‘microorganism-extinction’ periodic solution for R1<1 and the system is permanent for R2>1. And for the system with stochastic effect, by using the theory of stochastic differential equations, the authors obtained the conditions for the persistence and extinction of microorganisms. Their results showed that stochastic disturbance and toxicant can affect the survival of microorganism.While, in reality, there are generally various microorganisms coexisting in lakes and oceans, which might depend on single nutrient in the region. Different from the model in Chi and Zhao [20], in present paper, we consider two different microbes to compete for a nutrient, by introducing Monod-Haldane functional response; we get the model as follows:(5)dStdt=DS0-St-m1Stxta1+St+K1S2t-m2Styta2+St+K2S2t,dxtdt=m1Stxta1+St+K1S2t-Dxt-c1x2t,dytdt=m2Styta2+St+K2S2t-Dyt-c2y2t,where c1 and c2 denote the intraspecies competition rates. The meanings of other parameters are the same as those described above.It is well known that nature is often affected by random factors [21–25], unavoidably the process of microbial culture is affected by the interference of random factors [26–28], such as the degradation of microbial strains, the existence of an inducer or inhibitor on the growth, cultivation temperature changes, and the species and concentrations of inorganic salts changes. To understand the phenomenon of stochastic perturbations deeply [29–31], many scholars have studied the effect of the noise on the dynamical behavior of the stochastic chemostat models [32–37].A parameter of the system is often subject to random disturbance [38–42]; thus, in this paper, we assume that the maximal growth rates mi(i=1,2) are perturbed by environment noise on the basis of the approaches used in [20, 26, 43]. In this case, mi(i=1,2) change to random variables m~i(i=1,2), and m~i=mi+σiB˙i(t)(i=1,2), where Bi(t)(i=1,2) are independent standard Brownian motions with intensity σi≥0(i=1,2). In summary, we replace mi(i=1,2) in deterministic model (5) with mi+σiB˙i(t)(i=1,2) to get the following stochastic differential equations model: (6)dSt=DS0-St-m1Stxta1+St+K1S2t-m2Styta2+St+K2S2tdt-σ1Stxta1+St+K1S2tdB1t-σ2Styta2+St+K2S2tdB2t,dxt=m1Stxta1+St+K1S2t-Dxt-c1x2tdt+σ1Stxta1+St+K1S2tdB1t,dyt=m2Styta2+St+K2S2t-Dyt-c2y2tdt+σ2Styta2+St+K2S2tdB2t.The remaining part of this paper is organized as follows. In Section2, we give some notations, definitions, and lemmas which will be used in the following section. In Section 3, we explore the sufficient conditions of system (6) for the extinction and persistence of the two microorganisms. Finally, some conclusions and numerical simulations are given in Section 4.


## 2. Preliminary Knowledge

In this paper, we denoteR+3=S,x,y∈R3:S>0,x>0,y>0. For an integrable function V on [0,+∞), define Vt=1/t∫0tV(θ)dθ.Definition 1.
(i) The microorganismx(t) and y(t) are said to be extinctive if limt→+∞x(t)=0 and limt→+∞y(t)=0.
(ii) The microorganismx(t) and y(t) are said to be permanent in mean if there exist two positive constants λ1 and λ2 such that liminft→+∞〈x(t)〉≥λ1 and liminft→+∞〈y(t)〉≥λ2.Lemma 2.
For any positive solution(S(t),x(t),y(t)) of systems (5) or (6) with initial value (S(0),x(0),y(0))∈R+3, we have(7)max⁡limsupt→+∞⁡St,limsupt→+∞⁡xt,limsupt→+∞⁡yt≤S0,a.s.Proof.
From system (5) or (6), we have(8)dSt+xt+ytdt≤DS0-DSt+xt+yt.This implies that(9)limt→+∞⁡St+xt+yt≤S0.Then obviously we obtain(10)limsupt→+∞⁡St≤S0,limsupt→+∞⁡xt≤S0,limsupt→+∞⁡yt≤S0,a.s.This completes the proof of Lemma 2.Lemma 3.
(a) System (5) has at most three equilibrium points on the boundary of the positive quadrant. The trivial equilibrium point E0(S0,0,0) always exists. It is a locally asymptotically stable ‘microorganism-extinction’ equilibrium point when m1S0/(a1+S0+K1S02)-D<0 and m2S0/a2+S0+K2S02-D<0.
(b) Ifm1S0/a1+S0+K1S02-D>0, then there exists a equilibrium point E1(θ0,θ1,0) with θ1>0, and θ0 and θ1 satisfy the equation c1θ12+Dθ1-D(S0-θ0)=0. It is a locally asymptotically stable equilibrium point when m2θ0/a2+θ0+K2θ02-D<0 and (θ1-θ0)(a1-K1θ02)-θ02>0.
(c) Ifm2S0/a2+S0+K2S02-D>0, then there exists a equilibrium point E2(α0,0,α2) with α2>0, and α0 and α2 satisfy the equation c2α22+Dα2-D(S0-α0)=0. It is a locally asymptotically stable equilibrium point when m1α0/a1+α0+K1α02-D<0 and (α2-α0)(a2-K2α02)-α02>0.Proof.
(I) Firstly, we prove the existence of boundary equilibrium point. In system (5), let (11)f1≔DS0-St-m1Stxta1+St+K1S2t-m2Styta2+St+K2S2t=0,f2≔m1Stxta1+St+K1S2t-Dxt-c1x2t=0,f3≔m2Styta2+St+K2S2t-Dyt-c2y2t=0.By the above formula, we get the following equilibria: (12)E0:S0,0,0.Let y(t)=0,x(t)≠0 and S(t)≠0; we get the following equation through system (5):(13)DS0-St-m1Stxta1+St+K1S2t=0,m1Stxta1+St+K1S2t-Dxt-c1x2t=0.Let m1S(t)/a1+St+K1S2t=Z(t); then, by simplifying (13) we get(14)m1Sta1+St+K1S2t=Zt,-1c1Z2t+Dc1Zt+DS0-St=0.For system (14), if m1S0/a1+S0+K1S02-D>0, we can see that the two functional images have unique intersection point, which proves that system (13) has a positive solution (θ0,θ1) and satisfies the equation c1θ12+Dθ1-D(S0-θ0)=0. So it proves the existence of E1(θ0,θ1,0).
Similarly, we can prove the existence ofE2(α0,0,α2).
(II) Secondly, we prove the stability of the boundary equilibrium point. The Jacobian matrix associating to the equilibrium of system (5) is (15)J=γ11γ12γ13γ21γ22γ23γ31γ32γ33,where (16)γ11=-D-a1m1x-K1m1xS2a1+S+K1S22-a2m2y-K2m2yS2a2+S+K2S22,γ12=-m1Sa1+S+K1S2,γ13=-m2Sa2+S+K2S2,γ21=a1m1x-K1m1xS2a1+S+K1S22,γ22=m1Sa1+S+K1S2-D-2c1x,γ23=0,γ31=a2m2y-K2m2yS2a2+S+K2S22,γ32=0,γ33=m2Sa2+S+K2S2-D-2c2y.
(i) The stability of the ‘microorganism-extinction’ equilibrium point(S0,0,0) of system (5) is determined by the Jacobian (17)J0=-D-m1S0a1+S0+K1S02-m2S0a2+S0+K2S020m1S0a1+S0+K1S02-D000m2S0a2+S0+K2S02-D,which has the following eigenvalues(18)λ1=-D<0,λ2=m1S0a1+S0+K1S02-D,λ3=m2S0a2+S0+K2S02-D.According to stability theory [44, 45], (S0,0,0) is stable if λ2<0 and λ3<0; i.e., m1S0/a1+S0+K1S02-D<0 and m2S0/a2+S0+K2S02-D<0.
(ii) The functional matrix at(θ0,θ1,0) is equal to (19)J1=-D-a1m1θ1-K1m1θ1θ02a1+θ0+K1θ022-m1θ0a1+θ0+K1θ02-m2θ0a2+θ0+K2θ02a1m1θ1-K1m1θ1θ02a1+θ0+K1θ022m1θ0a1+θ0+K1θ02-D-2c1θ1000m2θ0a2+θ0+K2θ02-D,which has the following eigenvalue:(20)λ1=m2θ0a2+θ0+K2θ02-D.The other eigenvalues satisfy the following equations:(21)λ2+γλ+ν=0,where(22)γ=2D+2c1θ1+a1m1θ1-K1m1θ1θ02a1+θ0+K1θ022-m1θ0a1+θ0+K1θ02,ν=D2+2c1θ1D+2c1θ1a1m1θ1-K1m1θ1θ02a1+θ0+K1θ022+Da1m1θ1-K1m1θ1θ02a1+θ0+K1θ022-m1θ0a1+θ0+K1θ02.If (θ1-θ0)(a1-K1θ02)-θ02>0, then γ>0 and ν>0. Hence, it follows that any root of (21) has negative real part owing to the Routh-Hurwitz criterion. According to stability theory [44], E1(θ0,θ1,0) is stable if m2θ0/a2+θ0+K2θ02-D<0 and (θ1-θ0)(a1-K1θ02)-θ02>0.
Similarly, we can prove the stability ofE2(α0,0,α2).
This finishes the proof of Lemma3.Remark 4.
We also can obtain the conditionsm1S0/a1+S0+K1S02-D>0 and m2S0/a2+S0+K2S02-D>0 for the existence of equilibrium points E1(θ0,θ1,0) and E2(α0,0,α2) from the Proposition 1 and Proposition 2 in [34].Lemma 5.
For any initial value(S0,x0,y0)∈R+3, system (6) has a unique solution (S(t),x(t),y(t)) on t≥0, and the solution remains in R+3 with probability one; namely, (S(t),x(t),y(t))∈R+3 for all t≥0 almost surely.Proof.
Since the coefficients of system (6) are locally Lipschitz continuous, for any given initial value (S0,x0,y0)∈R+3, there is a unique local solution (S(t),x(t),y(t)) on t∈[0,τe), where τe is the explosion time. If we can prove τe=∞ a.s., then the solution will be global.
First we setinf⁡∅=∞ and ∅ is an empty set. Select ψ0>0 such that S0>ψ0,x0>ψ0,y0>ψ0. For any positive ψ(ψ≤ψ0), define the stopping time as follows:(23)τψ=inf⁡t∈0,τe:St≤ψorxt≤ψoryt≤ψ.Clearly τψ is increasing as ψ→0. Assuming τ0=limψ→0τψ, obviously τ0≤τe. We declare that τ0=∞ holds a.s.; thus, τe=∞  a.s. and (S(t),x(t),y(t))∈R+3 for all t≥0. So we only need to prove τ0=∞ a.s.
If this assertion is false, then there is a pair of constantsT>0 and η∈(0,1) such that Pτ0≤T>η. Hence, there is a positive constant ψ1≤ψ0 such that Pτψ≤T>η for any positive ψ≤ψ1.
From system (6), the total biomass N(t)=S(t)+x(t)+y(t) satisfies the equation dN(t)≤D(S0-N(t))dt then leads to that for all t<τe,(24)Nt≤max⁡S0+x0+y0,S0≔h1.
Define aC2-function V: R+3→R+ by(25)VSt,xt,yt=-ln⁡Sth1-ln⁡xth1-ln⁡yth1.Obviously, V is positive defined. Applying Itô formula, we obtain(26)dV=LVdt+σ1xt-Sta1+St+K1S2tdB1t+σ2yt-Sta2+St+K2S2tdB2t,where (27)LV=-DS0St+3D+m1xt-Sta1+St+K1S2t+m2yt-Sta2+St+K2S2t+σ12x2t+S2t2a1+St+K1S2t2+σ22y2t+S2t2a2+St+K2S2t2+c1xt+c2yt≤3D+m1h1a1+m2h1a2+σ12h12a12+σ22h12a22+c1h1+c2h1≔h2.So we get(28)dVt≤h2dt+σ1xt-Sta1+St+K1S2tdB1t+σ2yt-Sta2+St+K2S2tdB2t.Integrating both sides from 0 to τψ∧T, and then taking expectations, we have(29)EVSτψ∧T,xτψ∧T,yτψ∧T≤VS0,x0,y0+h2T.Denote Ωψ=τψ≤T and P(Ωψ)>η for any positive ψ≤ψ1. Note that for every ϱ∈Ωψ, at least, one of S(τψ,ϱ),x(τψ,ϱ) or y(τψ,ϱ) equals to ψ, and then(30)VSτψ,ϱ,xτψ,ϱ,yτψ,ϱ≥-ln⁡ψh1.So we obtain (31)VS0,x0,y0+h2T≥EIΩψϱVSτψ,ϱ,xτψ,ϱ,yτψ,ϱ=PΩψVSτψ,ϱ,xτψ,ϱ,yτψ,ϱ>-ηln⁡ψh1,where IΩψ(ϱ) is the indicator function of Ωψ. Letting ψ→0 leads to the contradiction(32)∞>VS0,x0,y0+h2T=∞.So τ0=∞ almost surely. The proof is complete.


## 3. Extinction and Persistence of System (6)

### 3.1. Extinction

In this section, we will try to give conditions that lead to the extinction of microorganism. Let(33)Ri=mi1+2aiKiD-σi22D11+2aiKi2,Ri∗=miS0ai+S0+KiS02D-σi2S022Dai+S0+KiS022,i=1,2.Then we get the following theorem.Theorem 6.
Let(S(t),x(t),y(t)) be a solution of system (6) with initial value (S(0),x(0),y(0))∈R+3. Then if one of the following holdsA11:S0≥a1/K1,1/1+2a1K1≥m1/σ12 and σ12>m12/2D, orA21:S0<a1/K1,S0/a1+S0+K1S02≥m1/σ12 and σ12>m12/2D, orA31:S0≥a1/K1,1/1+2a1K1<m1/σ12 and R1<1, orA41:S0<a1/K1,S0/a1+S0+K1S02<m1/σ12 and R1∗<1,
the microorganismx(t) goes to extinction almost surely, i.e., limt→+∞x(t)=0, a.s.Proof.
Applying Itô’s formula to system (6) yields(34)dln⁡xt=m1Sta1+St+K1S2t-D-c1xt-σ12S2t2a1+St+K1S2t2dt+σ1Sta1+St+K1S2tdB1t.Integrating from 0 to t and dividing by t on both sides of (34) yields (35)ln⁡xtt=1t∫0tm1Sta1+St+K1S2t-D-c1xt-σ12S2t2a1+St+K1S2t2dt+Mtt+ln⁡x0t≤1t∫0t-σ12S2t2a1+St+K1S2t2+m1Sta1+St+K1S2t-Ddt+Mtt+ln⁡x0t,
where the functionM(t)=∫0t(σ1S(τ)/a1+Sτ+K1S2τ)dB1(τ). By the strong law of large number and Lemma 2, we get(36)limt→+∞⁡Mtt=0,a.s.Then four cases should be discussed.
A 11 : since S0≥a1/K1,1/1+2a1K1≥m1/σ12 and σ12>m12/2D, then we can easily see from (35) that(37)ln⁡xtt≤-D-m122σ12+Mtt+ln⁡x0t.Taking the superior limit on both sides of (37) yields(38)limt→+∞⁡sup⁡ln⁡xtt≤-D-m122σ12<0,a.s.,which implies limt→+∞x(t)=0, a.s.
Similarly, we can prove that microorganism also tends to extinction under conditionA21, and we omit it here.
A 31 : S 0 ≥ a 1 / K 1 , 1 / 1 + 2 a 1 K 1 < m 1 / σ 1 2 and R1<1. From (35), we have(39)ln⁡xtt≤-σ122+8a1K1+8a1K1+m11+2a1K1-D+Mtt+ln⁡x0t.Taking the superior limit on both sides of (39) leads to(40)limt→+∞⁡sup⁡ln⁡xtt≤DR1-1<0,a.s.,which implies limt→+∞x(t)=0, a.s.
A 41 : S 0 < a 1 / K 1 , S 0 / a 1 + S 0 + K 1 S 02 < m 1 / σ 1 2 and R1∗<1. In this case, we can see from (35) that(41)ln⁡xtt≤-σ12S022a1+S0+K1S022+m1S0a1+S0+K1S02-D+Mtt+ln⁡x0t.Taking the superior limit on both sides of (41) yields(42)limt→+∞⁡sup⁡ln⁡xtt≤DR1∗-1<0,a.s.,which implies limt→+∞x(t)=0, a.s.
This completes the proof of Theorem6.The same discussion can be used iny(t); we have the following.Theorem 7.
Let(S(t),x(t),y(t)) be a solution of system (6) with initial value (S(0),x(0),y(0))∈R+3. Then if one of the following holdsA12:S0≥a2/K2,1/(1+2a2K2)≥m2/σ22 and σ22>m22/2D, orA22:S0<a2/K2,S0/a2+S0+K2S02≥m2/σ22 and σ22>m22/2D, orA32:S0≥a2/K2,1/1+2a2K2<m2/σ22 and R2<1, orA42:S0<a2/K2,S0/a2+S0+K2S02<m2/σ22 and R2∗<1,
the microorganismy(t) goes to extinction almost surely, i.e., limt→+∞y(t)=0, a.s.


### 3.2. Permanence in Mean

For system (6), let(43)li=mic1+c2S02+D2ai+S0+KiS02,i=1,2,and (44)R¯i=DmiS0li-Dσi2S022aili,i=1,2.Then we get the following theorem.Theorem 8.
Let(S(t),x(t),y(t)) be a solution of system (6) with initial value (S(0),x(0),y(0))∈R+3, then we have the following.
(i) IfR¯1>1, y(t) satisfies one of the conditions A12,A22,A32, and A42 of Theorem 7, then the microorganism y(t) goes extinct and the microorganism x(t) is permanent in mean; moreover, x(t) satisfies (45)liminft→+∞⁡xt≥l1R¯1-1Dc1a1+S0+K1S02+m1.
(ii) IfR¯2>1, x(t) satisfies one of the conditions A11,A21,A31, and A41 of Theorem 6, then the microorganism x(t) goes extinct and the microorganism y(t) is permanent in mean; moreover, y(t) satisfies (46)liminft→+∞⁡yt≥l2R¯2-1Dc2a2+S0+K2S02+m2.(iii) If R¯1>1 and R¯2>1, then two microorganisms x(t) and y(t) are permanent in mean; moreover, x(t) and y(t) satisfy (47)liminft→+∞⁡xt+yt≥1DΔmax∑i=12liR¯i-1,where (48)Δmax=max⁡m1+m2+c1a1+S0+K1S02,m1+m2+c2a2+S0+K2S02.Proof.
Case (i): by Theorem7, since y(t) satisfies one of the conditions A12,A22,A32, and A42 of Theorem 7, then limt→+∞y(t)=0. Since R¯1>1, for ε small enough, such that 0<y(t)<ε for all t large enough and(49)Dm1S0-εl1-Dσ12S022a1l1>1.Integrating from 0 to t and dividing by t on both sides of system (6) yields (50)ϵt≜St-S0t+xt-x0t+yt-y0t=DS0-DSt-Dxt-Dyt-c1t∫0tx2tdt-c2t∫0ty2tdt≥DS0-DSt-Dxt-Dε-c1+c2S02,then one can get (51)St≥S0-xt-ε-c1+c2DS02-ϵtD.Applying Itô’s formula gives(52)da1+S0+K1S02ln⁡xt=a1+S0+K1S02m1Sta1+St+K1S2t-Da1+S0+K1S02-c1a1+S0+K1S02xt-a1+S0+K1S02σ12S2t2a1+St+K1S2t2dt+a1+S0+K1S02σ1Sta1+St+K1S2tdB1t≥m1St-Da1+S0+K1S02-c1a1+S0+K1S02xt-σ12S022a1dt+a1+S0+K1S02σ1Sta1+St+K1S2tdB1t.Integrating from 0 to t and dividing by t on both sides of (52) yields(53)a1+S0+K1S02ln⁡xt-ln⁡x0t≥m1St-Da1+S0+K1S02-c1a1+S0+K1S02xt-σ12S022a1+M1tt≥m1S0-xt-ε-c1+c2DS02-ϵtD-Da1+S0+K1S02-c1a1+S0+K1S02xt-σ12S022a1+M1tt=m1S0-ε-m1c1+c2DS02-m1ϵtD-Da1+S0+K1S02-c1a1+S0+K1S02+m1xt-σ12S022a1+M1tt,where M1(t)=∫0ta1+S0+K1S02σ1Sτ/a1+Sτ+K1S2τdB1(τ). The inequality (53) can be rewritten as (54)a1+S0+K1S02ln⁡xtt≥m1S0-ε-m1c1+c2DS02-m1ϵtD-Da1+S0+K1S02-c1a1+S0+K1S02+m1xt-σ12S022a1+M1tt+a1+S0+K1S02ln⁡x0t.
By the strong law of large numbers and Lemma2, we have limt→+∞M1t/t=0 and limt→+∞ϵ(t)=0. Taking the inferior limit of both sides of (54) and using Lemma 2.5 in [46], we can get (55)liminft→+∞⁡xt≥l1Dc1a1+S0+K1S02+m1Dm1S0-εl1-Dσ12S022a1l1-1>0,a.s.Let ε→0, then we have(56)liminft→+∞⁡xt≥l1R¯1-1Dc1a1+S0+K1S02+m1,a.s.The same discussion can be used in Case (ii), and we omit it here.
Case (iii): notice that(57)St≥S0-xt-yt-c1+c2DS02-ϵtD.Define(58)Vt=ρ1ln⁡xt+ρ2ln⁡yt,where ρ1=a1+S0+K1S02 and ρ2=a2+S0+K2S02. Then we have (59)D+Vt=a1+S0+K1S02m1a1+St+K1S2t+a2+S0+K2S02m2a2+St+K2S2tSt-∑i=12Dai+S0+KiS02-c1a1+S0+K1S02xt-c2a2+S0+K2S02yt-∑i=12ai+S0+KiS02σi2S2t2ai+St+KiS2t2dt+∑i=12ai+S0+KiS02σiStai+St+KiS2tdBit≥m1+m2St-∑i=12Dai+S0+KiS02-c1a1+S0+K1S02xt-∑i=12σi2S022ai-c2a2+S0+K2S02ytdt+∑i=12ai+S0+KiS02σiStai+St+KiS2tdBit.Integrating from 0 to t and dividing by t on both sides of (59) yields (60)Vtt-V0t≥m1+m2St-∑i=12Dai+S0+KiS02-c1a1+S0+K1S02xt-c2a2+S0+K2S02yt-∑i=12σi2S022ai+∑i=12Mitt≥m1+m2S0-m1+m2c1+c2DS02-∑i=12Dai+S0+KiS02-m1+m2ϵtD-m1+m2+c1a1+S0+K1S02xt-m1+m2+c2a2+S0+K2S02yt-∑i=12σi2S022ai+∑i=12Mitt≥m1+m2S0-m1+m2c1+c2DS02-∑i=12Dai+S0+KiS02-m1+m2ϵtD-Δmaxxt+yt-∑i=12σi2S022ai+∑i=12Mitt,where Mi(t)=∫0tai+S0+KiS02σiSτ/ai+Sτ+KiS2τdBi(τ). The inequality (60) can be rewritten as (61)xt+yt≥1Δmaxl1DDm1S0l1-Dσ12S022a1l1-1-m1+m2ϵtD+∑i=12Mitt+l2DDm2S0l2-Dσ22S022a2l2-1-Vt-V0t.
By the strong law of large numbers and Lemma2, we have limt→+∞Mit/t=0(i=1,2) and limt→+∞ϵ(t)=0. Taking the inferior limit of both sides of (61) yields(62)liminft→+∞⁡xt+yt≥1DΔmax∑i=12liR¯i-1>0,a.s.This proof is completed.


## 3.1. Extinction

In this section, we will try to give conditions that lead to the extinction of microorganism. Let(33)Ri=mi1+2aiKiD-σi22D11+2aiKi2,Ri∗=miS0ai+S0+KiS02D-σi2S022Dai+S0+KiS022,i=1,2.Then we get the following theorem.Theorem 6.
Let(S(t),x(t),y(t)) be a solution of system (6) with initial value (S(0),x(0),y(0))∈R+3. Then if one of the following holdsA11:S0≥a1/K1,1/1+2a1K1≥m1/σ12 and σ12>m12/2D, orA21:S0<a1/K1,S0/a1+S0+K1S02≥m1/σ12 and σ12>m12/2D, orA31:S0≥a1/K1,1/1+2a1K1<m1/σ12 and R1<1, orA41:S0<a1/K1,S0/a1+S0+K1S02<m1/σ12 and R1∗<1,
the microorganismx(t) goes to extinction almost surely, i.e., limt→+∞x(t)=0, a.s.Proof.
Applying Itô’s formula to system (6) yields(34)dln⁡xt=m1Sta1+St+K1S2t-D-c1xt-σ12S2t2a1+St+K1S2t2dt+σ1Sta1+St+K1S2tdB1t.Integrating from 0 to t and dividing by t on both sides of (34) yields (35)ln⁡xtt=1t∫0tm1Sta1+St+K1S2t-D-c1xt-σ12S2t2a1+St+K1S2t2dt+Mtt+ln⁡x0t≤1t∫0t-σ12S2t2a1+St+K1S2t2+m1Sta1+St+K1S2t-Ddt+Mtt+ln⁡x0t,
where the functionM(t)=∫0t(σ1S(τ)/a1+Sτ+K1S2τ)dB1(τ). By the strong law of large number and Lemma 2, we get(36)limt→+∞⁡Mtt=0,a.s.Then four cases should be discussed.
A 11 : since S0≥a1/K1,1/1+2a1K1≥m1/σ12 and σ12>m12/2D, then we can easily see from (35) that(37)ln⁡xtt≤-D-m122σ12+Mtt+ln⁡x0t.Taking the superior limit on both sides of (37) yields(38)limt→+∞⁡sup⁡ln⁡xtt≤-D-m122σ12<0,a.s.,which implies limt→+∞x(t)=0, a.s.
Similarly, we can prove that microorganism also tends to extinction under conditionA21, and we omit it here.
A 31 : S 0 ≥ a 1 / K 1 , 1 / 1 + 2 a 1 K 1 < m 1 / σ 1 2 and R1<1. From (35), we have(39)ln⁡xtt≤-σ122+8a1K1+8a1K1+m11+2a1K1-D+Mtt+ln⁡x0t.Taking the superior limit on both sides of (39) leads to(40)limt→+∞⁡sup⁡ln⁡xtt≤DR1-1<0,a.s.,which implies limt→+∞x(t)=0, a.s.
A 41 : S 0 < a 1 / K 1 , S 0 / a 1 + S 0 + K 1 S 02 < m 1 / σ 1 2 and R1∗<1. In this case, we can see from (35) that(41)ln⁡xtt≤-σ12S022a1+S0+K1S022+m1S0a1+S0+K1S02-D+Mtt+ln⁡x0t.Taking the superior limit on both sides of (41) yields(42)limt→+∞⁡sup⁡ln⁡xtt≤DR1∗-1<0,a.s.,which implies limt→+∞x(t)=0, a.s.
This completes the proof of Theorem6.The same discussion can be used iny(t); we have the following.Theorem 7.
Let(S(t),x(t),y(t)) be a solution of system (6) with initial value (S(0),x(0),y(0))∈R+3. Then if one of the following holdsA12:S0≥a2/K2,1/(1+2a2K2)≥m2/σ22 and σ22>m22/2D, orA22:S0<a2/K2,S0/a2+S0+K2S02≥m2/σ22 and σ22>m22/2D, orA32:S0≥a2/K2,1/1+2a2K2<m2/σ22 and R2<1, orA42:S0<a2/K2,S0/a2+S0+K2S02<m2/σ22 and R2∗<1,
the microorganismy(t) goes to extinction almost surely, i.e., limt→+∞y(t)=0, a.s.


## 3.2. Permanence in Mean

For system (6), let(43)li=mic1+c2S02+D2ai+S0+KiS02,i=1,2,and (44)R¯i=DmiS0li-Dσi2S022aili,i=1,2.Then we get the following theorem.Theorem 8.
Let(S(t),x(t),y(t)) be a solution of system (6) with initial value (S(0),x(0),y(0))∈R+3, then we have the following.
(i) IfR¯1>1, y(t) satisfies one of the conditions A12,A22,A32, and A42 of Theorem 7, then the microorganism y(t) goes extinct and the microorganism x(t) is permanent in mean; moreover, x(t) satisfies (45)liminft→+∞⁡xt≥l1R¯1-1Dc1a1+S0+K1S02+m1.
(ii) IfR¯2>1, x(t) satisfies one of the conditions A11,A21,A31, and A41 of Theorem 6, then the microorganism x(t) goes extinct and the microorganism y(t) is permanent in mean; moreover, y(t) satisfies (46)liminft→+∞⁡yt≥l2R¯2-1Dc2a2+S0+K2S02+m2.(iii) If R¯1>1 and R¯2>1, then two microorganisms x(t) and y(t) are permanent in mean; moreover, x(t) and y(t) satisfy (47)liminft→+∞⁡xt+yt≥1DΔmax∑i=12liR¯i-1,where (48)Δmax=max⁡m1+m2+c1a1+S0+K1S02,m1+m2+c2a2+S0+K2S02.Proof.
Case (i): by Theorem7, since y(t) satisfies one of the conditions A12,A22,A32, and A42 of Theorem 7, then limt→+∞y(t)=0. Since R¯1>1, for ε small enough, such that 0<y(t)<ε for all t large enough and(49)Dm1S0-εl1-Dσ12S022a1l1>1.Integrating from 0 to t and dividing by t on both sides of system (6) yields (50)ϵt≜St-S0t+xt-x0t+yt-y0t=DS0-DSt-Dxt-Dyt-c1t∫0tx2tdt-c2t∫0ty2tdt≥DS0-DSt-Dxt-Dε-c1+c2S02,then one can get (51)St≥S0-xt-ε-c1+c2DS02-ϵtD.Applying Itô’s formula gives(52)da1+S0+K1S02ln⁡xt=a1+S0+K1S02m1Sta1+St+K1S2t-Da1+S0+K1S02-c1a1+S0+K1S02xt-a1+S0+K1S02σ12S2t2a1+St+K1S2t2dt+a1+S0+K1S02σ1Sta1+St+K1S2tdB1t≥m1St-Da1+S0+K1S02-c1a1+S0+K1S02xt-σ12S022a1dt+a1+S0+K1S02σ1Sta1+St+K1S2tdB1t.Integrating from 0 to t and dividing by t on both sides of (52) yields(53)a1+S0+K1S02ln⁡xt-ln⁡x0t≥m1St-Da1+S0+K1S02-c1a1+S0+K1S02xt-σ12S022a1+M1tt≥m1S0-xt-ε-c1+c2DS02-ϵtD-Da1+S0+K1S02-c1a1+S0+K1S02xt-σ12S022a1+M1tt=m1S0-ε-m1c1+c2DS02-m1ϵtD-Da1+S0+K1S02-c1a1+S0+K1S02+m1xt-σ12S022a1+M1tt,where M1(t)=∫0ta1+S0+K1S02σ1Sτ/a1+Sτ+K1S2τdB1(τ). The inequality (53) can be rewritten as (54)a1+S0+K1S02ln⁡xtt≥m1S0-ε-m1c1+c2DS02-m1ϵtD-Da1+S0+K1S02-c1a1+S0+K1S02+m1xt-σ12S022a1+M1tt+a1+S0+K1S02ln⁡x0t.
By the strong law of large numbers and Lemma2, we have limt→+∞M1t/t=0 and limt→+∞ϵ(t)=0. Taking the inferior limit of both sides of (54) and using Lemma 2.5 in [46], we can get (55)liminft→+∞⁡xt≥l1Dc1a1+S0+K1S02+m1Dm1S0-εl1-Dσ12S022a1l1-1>0,a.s.Let ε→0, then we have(56)liminft→+∞⁡xt≥l1R¯1-1Dc1a1+S0+K1S02+m1,a.s.The same discussion can be used in Case (ii), and we omit it here.
Case (iii): notice that(57)St≥S0-xt-yt-c1+c2DS02-ϵtD.Define(58)Vt=ρ1ln⁡xt+ρ2ln⁡yt,where ρ1=a1+S0+K1S02 and ρ2=a2+S0+K2S02. Then we have (59)D+Vt=a1+S0+K1S02m1a1+St+K1S2t+a2+S0+K2S02m2a2+St+K2S2tSt-∑i=12Dai+S0+KiS02-c1a1+S0+K1S02xt-c2a2+S0+K2S02yt-∑i=12ai+S0+KiS02σi2S2t2ai+St+KiS2t2dt+∑i=12ai+S0+KiS02σiStai+St+KiS2tdBit≥m1+m2St-∑i=12Dai+S0+KiS02-c1a1+S0+K1S02xt-∑i=12σi2S022ai-c2a2+S0+K2S02ytdt+∑i=12ai+S0+KiS02σiStai+St+KiS2tdBit.Integrating from 0 to t and dividing by t on both sides of (59) yields (60)Vtt-V0t≥m1+m2St-∑i=12Dai+S0+KiS02-c1a1+S0+K1S02xt-c2a2+S0+K2S02yt-∑i=12σi2S022ai+∑i=12Mitt≥m1+m2S0-m1+m2c1+c2DS02-∑i=12Dai+S0+KiS02-m1+m2ϵtD-m1+m2+c1a1+S0+K1S02xt-m1+m2+c2a2+S0+K2S02yt-∑i=12σi2S022ai+∑i=12Mitt≥m1+m2S0-m1+m2c1+c2DS02-∑i=12Dai+S0+KiS02-m1+m2ϵtD-Δmaxxt+yt-∑i=12σi2S022ai+∑i=12Mitt,where Mi(t)=∫0tai+S0+KiS02σiSτ/ai+Sτ+KiS2τdBi(τ). The inequality (60) can be rewritten as (61)xt+yt≥1Δmaxl1DDm1S0l1-Dσ12S022a1l1-1-m1+m2ϵtD+∑i=12Mitt+l2DDm2S0l2-Dσ22S022a2l2-1-Vt-V0t.
By the strong law of large numbers and Lemma2, we have limt→+∞Mit/t=0(i=1,2) and limt→+∞ϵ(t)=0. Taking the inferior limit of both sides of (61) yields(62)liminft→+∞⁡xt+yt≥1DΔmax∑i=12liR¯i-1>0,a.s.This proof is completed.


## 4. Conclusions and Simulations

In this paper, we propose and analyse the dynamics of two-microorganism and single nutrient stochastic chemostat models with Monod-Haldane response function. First, we discuss the existence and locally asymptotical stability of boundary equilibria of the system neglecting stochastic effect. Then, we investigate the dynamics of the system under stochastic effect and obtain the conditions which determine the persistence and extinction of the microorganisms with stochastic effect. Our results show that large stochastic noise can lead to microbial extinction (see Theorems6 and 7), and small stochastic noise is beneficial to the survival of microorganisms (see Theorem 8).In order to verify the theoretical results obtained in this paper, we give some numerical simulation. We choose the parameters in model (5) and model (6) as follows:(63)S0=2.1,D=1,m1=2.2,m2=2.2,a1=0.1,a2=0.1,K1=0.1,K2=0.2,c1=0.05,c2=0.08,σ1=1.7,σ2=1.9.Note that(64)2.1=S0>a1K1=1,0.8333=11+2a1K1>m1σ12=0.7612,2.8900=σ12>m122D=2.4200,2.1=S0>a2K2=0.7071,0.7795=11+2a2K2>m2σ22=0.6094,3.6100=σ22>m222D=2.4200.Theorems6 and 7 show that the microorganism die out under a large white noise disturbance intensity (see Figure 1(a) with σ1=1.7,σ2=1.9). Figure 1 shows that the persistent microorganism of a deterministic system (see Figure 1(b)) can become extinct due to the white noise stochastic disturbance. Therefore, the large white noise stochastic disturbance intensity is detrimental to the survival of microorganisms.Figure 1
Computer simulation of the pathsS(t),x(t),y(t) for the chemostat model, where (a) is the stochastic system for σ1=1.7,σ2=1.9. (b) is the corresponding deterministic system of (a) for σ1=0,σ2=0.
(a) (b)Theorem8 indicates that the microorganism can become extinct or persistent under a small white noise disturbance intensity. We keep the system parameters same as those in Figure 1, and let σ1 and σ2 take different parameter values. When σ1 is larger and σ2 is smaller (σ1=1.6,σ2=0.1), here 0.8594=m1/σ12>1/1+2a1K1=0.8333,R1=0.9444<1,R¯2=1.0129>1. Thus, the microorganism x(t) tends to die out, and the microorganism y(t) is persistent (see Figure 2(a)). Conversely, when σ1 is smaller and σ2 is larger (σ1=0.15,σ2=1.65), here 0.8081=m2/σ22>1/1+2a2K2=0.7795,R2=0.8878<1,R¯1=1.0568>1, Figure 2(b) shows that the microorganism y(t) tends to die out, and the microorganism x(t) is persistent. Furthermore, let σ1 and σ2 take small parameter values (σ1=0.05,σ2=0.05); here R¯1=1.1698>1,R¯2=1.0510>1, then both x(t) and y(t) are persistent (see Figure 2(c)); that is, the two microorganisms can coexist. Although σ1 and σ2 still take small parameter values (σ1=0.4,σ2=0.7), when 7.8750=m1/σ12>1/1+2a1K1=0.8333,R1=0.9944<1,3.0000=m2/σ22>1/1+2a2K2=0.7795,R2=0.9970<1, both x(t) and y(t) can become extinct (see Figure 2(d)). The above simulations support our results in the article.Figure 2
Numerical simulation of the pathsS(t),x(t),y(t) for the chemostat stochastic system (6), where (a) σ1=1.6,σ2=0.1.R1=0.9444<1,R¯2=1.0129>1. (b) σ1=0.15,σ2=1.65.R2=0.8878<1,R¯1=1.0568>1. (c) σ1=0.05,σ2=0.05.R¯1=1.1698>1,R¯2=1.0510>1. (d) m1=1.26,m2=1.47.σ1=0.4,σ2=0.7.R1=0.9944<1,R2=0.9970<1.
(a) (b) (c) (d)



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