# Threshold Dynamics of a Stochastic Chemostat Model with Two Nutrients and One Microorganism

Authors: Jian Zhang; Anqi Miao; Tongqian Zhang
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

A new stochastic chemostat model with two substitutable nutrients and one microorganism is proposed and investigated. Firstly, for the corresponding deterministic model, the threshold for extinction and permanence of the microorganism is obtained by analyzing the stability of the equilibria. Then, for the stochastic model, the threshold of the stochastic chemostat for extinction and permanence of the microorganism is explored. Difference of the threshold of the deterministic model and the stochastic model shows that a large stochastic disturbance can affect the persistence of the microorganism and is harmful to the cultivation of the microorganism. To illustrate this phenomenon, we give some computer simulations with different intensity of stochastic noise disturbance.


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## Body

## 1. Introduction

Chemostat is commonly used to describe the dynamics of a microbial population in a continuous bioreactor in which microorganisms grow on a substrate and has attracted great interest of many scholars [1–8], since it was first introduced by Monod [9]. A single simple species chemostat model with Michaelis-Menten-Monod functional response was proposed by [9] as follows:(1)dStdt=DS0-St-rδStXtK+St,dXtdt=rStXtK+St-DXt,where St is the concentration of the nutrient, Xt is the concentration of the organism, D is the dilution (or washout) rate, r is the maximal growth rate, K is the Michaelis-Menten (or half-saturation) constant with units of concentration, and δ is a “yield” constant reflecting the conversion of nutrient to organism.However, experimental results have indicated that the microorganisms depend on a variety of nutrition substances such as carbon, nitrogen, energy, growth factors, inorganic salts, and water. Then the model of microorganisms species growth in the chemostat on two nutrients is considered by [10–14]. A model of single-species growth in the chemostat on two substitutable resources with Michaelis-Menten-Monod functional response was proposed by [14] as follows:(2)dS1tdt=DS10-S1t-r1S1tXtK1+S1t,dS2tdt=DS20-S2t-r2S2tXtK2+S2t,dXtdt=r1S1tXtK1+S1t+r2S2tXtK2+S2t-DXt.However, it is now well known that stochastic noise is widely present in biological systems and so on [15–33] and microorganisms are inevitably influenced by some random factors in the process of cultivation. To better understand the dynamic behavior of the chemostat, a host of scholars proposed a slice of stochastic chemostat models and studied the effect of the random noise on the dynamic behavior of the stochastic models. As an example, Imhof and Walcher [34] proposed a stochastic chemostat model for a single microorganism species consuming a single nutrient. They found that random effects may lead to extinction in scenarios where the deterministic model predicts persistence. Recently, Xu and Yuan [35] established a stochastic chemostat model in which the maximal growth rate is influenced by the white noise in environment as follows:(3)dSt=DS0-St-mStXta+Stdt-αStXtdBta+St,dXt=mStXta+St-DXtdt+αStXtdBta+St.They got an analogue break-even concentration involving the white noise which can determine the exclusion and persistence of the microorganism. And more stochastic chemostat models can be found in [36–39].Motivated by the papers mentioned above, in this paper, we further consider a model of single-species growth in the chemostat on two supplementary resources with Michaelis-Menten-Monod functional response and environmental noise. We assume that the maximal growth raterii=1,2 is perturbed by white noises so that (4)ri⟶ri+σiB˙it,where Bit is a standard Brownian motion with intensity σi>0. Then the resultant model takes the following form:(5)dS1t=DS10-S1t-r1S1tXtK1+S1tdt-σ1S1tXtdB1tK1+S1t,dS2t=DS20-S2t-r2S2tXtK2+S2tdt-σ2S2tXtdB2tK2+S2t,dXt=r1S1tXtK1+S1t+r2S2tXtK2+S2t-DXtdt+σ1S1tXtdB1tK1+S1t+σ2S2tXtdB2tK2+S2t.Our main objective in the rest of this paper is to investigate the threshold dynamics of stochastic chemostat model (5) and explore the conditions under which microorganisms will die out or exist.


## 2. Preliminaries

In this section, we will give some notations, definitions, and lemmas which will be used for analyzing our main results. To this end, throughout this paper, we letΩ,F,Ft≥0,P be a complete probability space with a filtration Ftt≥0 satisfying the usual conditions: it is increasing and right continuous while F0 contains all P-null sets; we use Bt to represent a scalar Brownian motion defined on the complete probability space Ω; also let R+2=xi>0,i=1,2. If for an integrable function f on 0,+∞, define (6)ft=1t∫0tfθdθ.Then we have the following.Definition 1.
For system (5),(i)
the microorganismXt is said to be extinctive if limt→+∞⁡Xt=0,(ii)
the microorganismXt is said to be permanent in mean if there exists a positive constant λ such that lim⁡inft→+∞⁡Xt≥λ.Then, one can show the following lemmas.Lemma 2.
The solutionS1t,S2t,Xt of model (2) or (5) with the initial condition S10,S20,X0∈R+3 is ultimately bounded; that is, (7)lim⁡supt→∞⁡S1t≤S10,lim⁡supt→∞⁡S2t≤S20,lim⁡supt→∞⁡Mt≤S10+S20,where Mt=S1t+S2t+Xt.Proof.
LettingMt=S1t+S2t+Xt, from system (2) or system (5), we have (8)dMtdt=DS10+S20-DMt.This implies that (9)limt→+∞⁡Mt=S10+S20.Thus, we have (10)lim⁡supt→∞⁡S1t≤S10,lim⁡supt→∞⁡S2t≤S20,lim⁡supt→∞⁡Mt=S10+S20.This completes the proof of Lemma 2.By Lemma2 and the strong law of large numbers for martingales [40], we can obtain the following lemma.Lemma 3.
LettingS1t,S2t,Xt be a solution of system (5) with initial value S10,S20,X0∈R+3, then (11)limt→+∞⁡1t∫0tσ1S1τK1+S1τdBτ=0,limt→+∞⁡1t∫0tσ2S2τK2+S2τdBτ=0.


## 3. Dynamics of Deterministic System (2)

In this section, we will focus on the deterministic system (2). It is easy to see that the equilibria point of (2) satisfy(12)DS10-S1t-r1S1tXtK1+S1t=0,DS20-S2t-r2S2tXtK2+S2t=0,r1S1tXtK1+S1t+r2S2tXtK2+S2t-DXt=0,and, obviously, model (2) has a microorganism extinction equilibrium E0S10,S20,0. Let E∗S1∗,S2∗,X∗ be the coexistence equilibrium of model (2), which satisfies(13)fS2≕aS23+bS22+cS2+d=0,where (14)a=r2-Dr1+r2-D,b=r1S20D+DS10r2-2r1K2D-r22S10+r1K2r2-S20r22-r22K1+2D2K2-S20D2-S10r2r1-2r2DK2+2S20r2D+DK1r2-r1S20r2,c=-2S20D2K2+r2K1DK2-K22r1D+S10r2DK2-K2S20r1r2+2DS20K2r1+D2K22-S10r2r1K2+2S20r2DK2,d=K22S20Dr1-D.Then we have that (15)f0=K22S20Dr1-D,fS20=DS10r2S202-r22S10S202-r22K1S202-S10r2r1S202+DK1r2S202+r2K1DK2S20+S10r2DK2S20-S10r2r1K2S20=r2S20DS10+K1S20+K2-S10r1K2+S20-r2S20S10+K1.Denote(16)R=1D1δ1+1δ2,where δ1=K1+S10/r1S10, δ2=K2+S20/r2S20.Obviously, IfR>1, we have fS20<0. If r1>D, we have f0>0. Thus, equation has one positive root S2 at least, and S2∈0,S20.From the second equation of (12), one gets(17)X=DS20-S2K2+S2r2S2.Substituting (17) into the first equation of (12), we have(18)r2S2S12+r1S20-S2K2+S2-r2S10-K1S2S1-r2S2S10K1=0.Let (19)gS1=r2S2S12+r1S20-S2K2+S2-r2S10-K1S2S1-r2S2S10K1.It is easy to see that (20)g0=-r2S2S10K1<0,gS10=r1S10S20-S2K2+S2>0.Thus, (18) has one positive root S1 at least, and S1∈0,S10.From the third equation of (12), we have 0<X<S10+S20. Then we have the following theorem.Theorem 4.
Ifr1>D and R>1, then system (2) has unique positive equilibrium E∗.Regarding the stability of these equilibria, we have the following theorem.Theorem 5.
Then for system (2), one has the following. (i)
IfR<1, microorganism extinction equilibrium E0 is locally stable; if R>1 it is unstable.(ii)
Ifr1>D and R>1, the coexistence equilibrium E∗ is locally stable.Proof.
Linearizing the system at the equilibriumES1∘,S2∘,X∘ gives the Jacobian(21)J=-D-a10-a40-D-a2-a3a1a2a3+a4-D,where (22)a1=K1r1X∘K1+S1∘2,a2=K2r2X∘K2+S2∘2,a3=r2S2∘K2+S2∘,a4=r1S1∘K1+S1∘.The characteristic equation gives(23)λ+Dλ2+Aλ+B=0,where (24)A=2D+a1+a2-a3-a4,B=D2+a1D+a2D+a1a2-a3D-a4D-a1a3-a2a4.Obviously, we have, at E0, (25)a1=0,a2=0,a3=r2S20K2+S20,a4=r1S10K1+S10.Then we have (26)A=2D-a3-a4,B=DD-a3+a4,and thus if R<1, all the eigenvalues of (23) have negative real part; then, by the stability theory, E0 is stable.
And, atE∗, we have (27)A=2D+a1+a2-a3-a4=D+a1+a2>0,B=D2+a1D+a2D+a1a2-a3D-a4D-a1a3-a2a4=a1a2+a1D-a3+a2D-a4>0;here a3+a4=D is used. Then all the eigenvalues of (23) have negative real part; thus, by the stability theory, the diseases equilibrium is stable as long as it exists.


## 4. Dynamics of Stochastic System (5)

### 4.1. Extinction

In this section, we explore the conditions leading to the extinction of the two infectious diseases. Denote(28)R∗=R-1Dσ122S10K1+S102-σ222S20K2+S202,where R is introduced in (16). Then we have the following.Theorem 6.
For system (5), if one of the following holds, (i)
σ12>δ1, σ22>δ2, and r12/2σ12+r22/2σ22<D,(ii)
σ12>δ1, σ22<δ2, and -D+r12/2σ12+r2S20/K2+S20-σ22/2(S20/K2+S20)2<0,(iii)
σ12<δ1, σ22>δ2, and -D+r22/2σ22+r1S10/K1+S10-σ12/2(S10/K1+S10)2<0,(iv)
σ12<δ1, σ22<δ2, and R∗<1,then the microorganism Xt of system (5) goes to extinction almost surely. Moreover, (29)limt→+∞⁡S1t=S10,limt→+∞⁡S2t=S20,almost surely.Proof.
LetS1t,S2t,Xt be a solution of system (5) with initial value S10,S20,X0∈R+3. Applying Itô’s formula to system (5) results in(30)dln⁡Xt=r1ϕ1t+r2ϕ2t-D-σ122ϕ12t-σ222ϕ22tdt+σ1S1tdBtK1+S1t+σ2S2tdBtK2+S2t,where ϕ1t=S1tXt/K1+S1t, ϕ2t=S2tXt/K2+S2t.
Integrating both sides of (30) from 0 to t gives(31)ln⁡Xt=∫0tr1ϕ1τ+r2ϕ2τ-σ122ϕ12τ-σ222ϕ22τdτ-Dt+M1t+M2t+ln⁡X0,where (32)M1t=∫0tσ1S1tdBτK1+S1t,M2t=∫0tσ2S2tdBτK2+S2t,known as the local continuous martingale, and M0=0. Obviously, we need to estimate the maximum value of r1ϕ1t+r2ϕ2t-σ12/2ϕ12t-σ22/2ϕ22t.
Let us consider quadratic function(33)gz=az-σ22z2,z∈0,bK+b,b>0.It is easy to verify that when σ2>δ=K+b/ab, gz reaches its maximum value gmax=a2/2σ2 at z=a/σ2; and when σ2<δ, gz achieve its maximum value gmax=δ3=ab/K+b-σ2/2(b/K+b)2 at z=b/K+b. Then, in (31), we have four cases to be discussed, depending on whether σ12>δ1 or σ22>δ2, which are as follows: Case 1: σ12>δ1, σ22>δ2; Case 2: σ12>δ1, σ22<δ2; Case 3: σ12<δ1, σ22>δ2; and Case 4: σ12<δ1, σ22<δ2.
For Case 1, sinceσ12>δ1, σ22>δ2, then r1ϕ1t+r2ϕ2t-σ12/2ϕ12t-σ22/2ϕ22t achieve the maximum value r12/2σ12+r22/2σ22. Then we can easily see from (31) that(34)ln⁡Xt≤r122σ12+r222σ22t-Dt+M1t+M2t+ln⁡X0.Dividing both sides of (34) by t>0, we have(35)ln⁡Xtt≤-D-r122σ12-r222σ22+Mtt+ln⁡X0tand, by Lemma 3, we have (36)limt→+∞⁡Mtt=0.Then, taking the limit superior on both sides of (35) leads to (37)lim⁡supt→+∞⁡ln⁡Xtt≤-D-r122σ12-r222σ22<0,which implies limt→+∞⁡Xt=0, and here r12/2σ12+r22/2σ22<D is used.
Case
2. σ12>δ1=K+S10/r1S10, σ22<δ2=K+S20/r2S20.
In this case, we can easily see from (31) that(38)ln⁡Xt≤r122σ12+r2S20K2+S20-σ222S20K2+S202t-Dt+M1t+M2t+ln⁡X0.Dividing both sides of (38) by t>0, we have(39)ln⁡Xtt≤-D-r122σ12-r2S20K2+S20+σ222S20K2+S202+Mtt+ln⁡X0tand, by Lemma 3, we have (40)limt→+∞⁡Mtt=0.Then, taking the limit superior on both sides of (38) leads to (41)lim⁡supt→+∞⁡ln⁡Xtt≤-D-r122σ12-r2S20K2+S20+σ222S20K2+S202<0which implies limt→+∞⁡Xt=0.
The same discussion can be used in Case 3; here we omit it.
Next, we consider Case 4:σ12<δ1=K+S10/r1S10, σ22<δ2=K+S20/r2S20. From (31), we have(42)ln⁡Xt≤r1S10K1+S10-σ122S10K1+S102+r2S20K2+S20-σ222S20K2+S202-Dt+M1t+M2t+ln⁡X0.Dividing both sides of (42) by t>0, we have(43)ln⁡Xtt≤-D-r1S10K1+S10+σ122S10K1+S102-r2S20K2+S20+σ222S20K2+S202+Mtt+ln⁡X0tand, by Lemma 3, we have (44)limt→+∞⁡Mtt=0.Then, taking the limit superior on both sides of (43) leads to (45)lim⁡supt→+∞⁡ln⁡Xtt≤-D-r1S10K1+S10+σ122S10K1+S102-r2S20K2+S20+σ222S20K2+S202<0which implies limt→+∞⁡Xt=0.
Next, we prove the last conclusion. Given0<ε≪1, since limt→+∞⁡Xt=0, we have 0<Xt<ε for t large enough. By the first equation of system (5), we have (46)dS1t≥DS10-S1t-r1S1tεK1+S1tdt-σ1S1tεdBtK1+S1t,dS1tdt≥DS10-D+r1εK1+σ1εBtK1S1t.Then when ε→0 we have(47)lim⁡inft→+∞⁡S1t≥S10.On the other hand from the proof of Lemma 2, we have (48)limt→+∞⁡S1t≤S10+ε.Let ε→0. Then one has(49)lim⁡supt→+∞⁡S1t≤S10.From (47) and (49), we have (50)limt→+∞⁡S1t=S10almost surely.
By employing the method similar above, it then follows that(51)limt→+∞⁡S2t=S20almost surely. This completes the proof of Theorem 6.


### 4.2. Permanence in Mean

Theorem 7.
IfR>1, then the microorganism Xt is permanent in mean; moreover, Xt satisfies (52)lim⁡inft→+∞⁡Xt≥DΔR∗-1,where Δ=min⁡K1+S10D/r1,K2+S20D/r2.Proof.
Integrating from0 to t and dividing by t on both sides of system (5) yield(53)Θt≜S1t-S10t+S2t-S20t+Xt-X0t=DS10+S20-DS1t+S2t-DXt.Then one can get(54)DXt=DS10+S20-DS1t+S2t-Θt.Applying Itô’s formula gives(55)dln⁡Xt=r1ϕ1t+r2ϕ2t-D-σ122ϕ12t-σ222ϕ22tdt+σ1S1tdBtK1+S1t+σ2S2tdBtK2+S2t≥r1ϕ1t+r2ϕ2t-D-σ122S10K1+S102-σ222S20K2+S202dt+σ1S1tdBtK1+S1t+σ2S2tdBtK2+S2t.Integrating from 0 to t and dividing by t on both sides of (55) yields (56)ln⁡Xt-ln⁡X0t≥r11t∫0tϕ1θdθ+r21t∫0tϕ2θdθ-D+σ122S10K1+S102+σ222S20K2+S202+M1tt+M2tt,where M1t=∫0tσ1S1τ/K1+S1τdBτ, M2t=∫0tσ2S2τ/K2+S2τdBτ. Noticing that (57)∫0tϕ1θdθ=∫0tS1θK1+S1θdθ≥∫0tS1θK1+S10dθ,∫0tϕ2θdθ=∫0tS2θK2+S2θdθ≥∫0tS2θK2+S20dθ,then we have(58)ln⁡Xt-ln⁡X0t≥-D+σ122S10K1+S102+σ222S20K2+S202+r1K1+S10S1t+r2K2+S20S2t+M1tt+M2tt=-D+σ122S10K1+S102+σ222S20K2+S202+r2K2+S20S10+S20+r1K1+S10-r2K2+S20S1t-r2K2+S20Xt-r2K2+S20ΘtD+M1tt+M2tt.If r1/K1+S10≤r2/K2+S20, we can get (59)ln⁡Xt-ln⁡X0t≥-D+σ122S10K1+S102+σ222S20K2+S202+r2K2+S20S10+S20+r1K1+S10-r2K2+S20S10-r2K2+S20Xt-r2K2+S20ΘtD+M1tt+M2tt≥-D+σ122S10K1+S102+σ222S20K2+S202+r2S20K2+S20+r1S10K1+S10-r2K2+S20Xt-r2K2+S20ΘtD+M1tt+M2tt.By inequality (59), we have(60)Xt≥K2+S20r2r1S10K1+S10+r2S20K2+S20-σ122S10K1+S102-σ222S20K2+S202-D-K2+S20r2ln⁡Xt-ln⁡X0t+r2K2+S20ΘtD+M1tt+M2tt.By Lemma 3, we get that limt→+∞⁡Mt/t=0. According to Lemma 2, one sees that lim⁡supt→+∞⁡S1t≤S10, lim⁡supt→+∞⁡S2t≤S20, and lim⁡supt→+∞⁡Xt≤C0+S20, and then one has limt→+∞⁡ln⁡X(t)/t=0 and limt→+∞⁡Θt=0. Thus taking the inferior limit of both sides of (60) yields(61)lim⁡inft→+∞⁡Xt≥K2+S20Dr2R∗-1.And if r1/K1+S10≥r2/K2+S20, we can get (62)ln⁡Xt-ln⁡X0t≥-D+σ122S10K1+S102+σ222S20K2+S202+r1K1+S10S10+S20-S2t-Xt-ΘtD+r2K2+S20S2t+M1tt+M2tt=-D+σ122S10K1+S102+σ222S20K2+S202+r1S10K1+S10+r2S20K2+S20-r1K1+S10Xt-r2K2+S20ΘtD+M1tt+M2tt,where M1t=∫0tσ1S1τ/K1+S1τdBτ, M2t=∫0tσ2S2τ/K2+S2τdBτ.
Inequality (62) can be rewritten as(63)Xt≥K2+S20r2r1S10K1+S10+r2S20K2+S20-σ122S10K1+S102-σ222S20K2+S202-D-K2+S20r2ln⁡Xt-ln⁡X0t+r1K1+S10ΘtD+M1tt+M2tt.Taking the inferior limit of both sides of (63) yields(64)lim⁡inft→+∞⁡Xt≥K1+S10Dr1R∗-1.Let Δ=min⁡K1+S10D/r1,K2+S20D/r2, and we get from (61) and (64)(65)lim⁡inft→+∞⁡Xt≥DΔR∗-1.This completes the proof of Theorem 7.Remark 8.
Theorems6 and 7 show that the condition for the microorganism to go to extinction or permanence depends on the intensity of the noise disturbances completely. And small noise disturbances will be beneficial to the cultivation of the microorganism; conversely, large white noise disturbance is harmful to the cultivation of the microorganism.


## 4.1. Extinction

In this section, we explore the conditions leading to the extinction of the two infectious diseases. Denote(28)R∗=R-1Dσ122S10K1+S102-σ222S20K2+S202,where R is introduced in (16). Then we have the following.Theorem 6.
For system (5), if one of the following holds, (i)
σ12>δ1, σ22>δ2, and r12/2σ12+r22/2σ22<D,(ii)
σ12>δ1, σ22<δ2, and -D+r12/2σ12+r2S20/K2+S20-σ22/2(S20/K2+S20)2<0,(iii)
σ12<δ1, σ22>δ2, and -D+r22/2σ22+r1S10/K1+S10-σ12/2(S10/K1+S10)2<0,(iv)
σ12<δ1, σ22<δ2, and R∗<1,then the microorganism Xt of system (5) goes to extinction almost surely. Moreover, (29)limt→+∞⁡S1t=S10,limt→+∞⁡S2t=S20,almost surely.Proof.
LetS1t,S2t,Xt be a solution of system (5) with initial value S10,S20,X0∈R+3. Applying Itô’s formula to system (5) results in(30)dln⁡Xt=r1ϕ1t+r2ϕ2t-D-σ122ϕ12t-σ222ϕ22tdt+σ1S1tdBtK1+S1t+σ2S2tdBtK2+S2t,where ϕ1t=S1tXt/K1+S1t, ϕ2t=S2tXt/K2+S2t.
Integrating both sides of (30) from 0 to t gives(31)ln⁡Xt=∫0tr1ϕ1τ+r2ϕ2τ-σ122ϕ12τ-σ222ϕ22τdτ-Dt+M1t+M2t+ln⁡X0,where (32)M1t=∫0tσ1S1tdBτK1+S1t,M2t=∫0tσ2S2tdBτK2+S2t,known as the local continuous martingale, and M0=0. Obviously, we need to estimate the maximum value of r1ϕ1t+r2ϕ2t-σ12/2ϕ12t-σ22/2ϕ22t.
Let us consider quadratic function(33)gz=az-σ22z2,z∈0,bK+b,b>0.It is easy to verify that when σ2>δ=K+b/ab, gz reaches its maximum value gmax=a2/2σ2 at z=a/σ2; and when σ2<δ, gz achieve its maximum value gmax=δ3=ab/K+b-σ2/2(b/K+b)2 at z=b/K+b. Then, in (31), we have four cases to be discussed, depending on whether σ12>δ1 or σ22>δ2, which are as follows: Case 1: σ12>δ1, σ22>δ2; Case 2: σ12>δ1, σ22<δ2; Case 3: σ12<δ1, σ22>δ2; and Case 4: σ12<δ1, σ22<δ2.
For Case 1, sinceσ12>δ1, σ22>δ2, then r1ϕ1t+r2ϕ2t-σ12/2ϕ12t-σ22/2ϕ22t achieve the maximum value r12/2σ12+r22/2σ22. Then we can easily see from (31) that(34)ln⁡Xt≤r122σ12+r222σ22t-Dt+M1t+M2t+ln⁡X0.Dividing both sides of (34) by t>0, we have(35)ln⁡Xtt≤-D-r122σ12-r222σ22+Mtt+ln⁡X0tand, by Lemma 3, we have (36)limt→+∞⁡Mtt=0.Then, taking the limit superior on both sides of (35) leads to (37)lim⁡supt→+∞⁡ln⁡Xtt≤-D-r122σ12-r222σ22<0,which implies limt→+∞⁡Xt=0, and here r12/2σ12+r22/2σ22<D is used.
Case
2. σ12>δ1=K+S10/r1S10, σ22<δ2=K+S20/r2S20.
In this case, we can easily see from (31) that(38)ln⁡Xt≤r122σ12+r2S20K2+S20-σ222S20K2+S202t-Dt+M1t+M2t+ln⁡X0.Dividing both sides of (38) by t>0, we have(39)ln⁡Xtt≤-D-r122σ12-r2S20K2+S20+σ222S20K2+S202+Mtt+ln⁡X0tand, by Lemma 3, we have (40)limt→+∞⁡Mtt=0.Then, taking the limit superior on both sides of (38) leads to (41)lim⁡supt→+∞⁡ln⁡Xtt≤-D-r122σ12-r2S20K2+S20+σ222S20K2+S202<0which implies limt→+∞⁡Xt=0.
The same discussion can be used in Case 3; here we omit it.
Next, we consider Case 4:σ12<δ1=K+S10/r1S10, σ22<δ2=K+S20/r2S20. From (31), we have(42)ln⁡Xt≤r1S10K1+S10-σ122S10K1+S102+r2S20K2+S20-σ222S20K2+S202-Dt+M1t+M2t+ln⁡X0.Dividing both sides of (42) by t>0, we have(43)ln⁡Xtt≤-D-r1S10K1+S10+σ122S10K1+S102-r2S20K2+S20+σ222S20K2+S202+Mtt+ln⁡X0tand, by Lemma 3, we have (44)limt→+∞⁡Mtt=0.Then, taking the limit superior on both sides of (43) leads to (45)lim⁡supt→+∞⁡ln⁡Xtt≤-D-r1S10K1+S10+σ122S10K1+S102-r2S20K2+S20+σ222S20K2+S202<0which implies limt→+∞⁡Xt=0.
Next, we prove the last conclusion. Given0<ε≪1, since limt→+∞⁡Xt=0, we have 0<Xt<ε for t large enough. By the first equation of system (5), we have (46)dS1t≥DS10-S1t-r1S1tεK1+S1tdt-σ1S1tεdBtK1+S1t,dS1tdt≥DS10-D+r1εK1+σ1εBtK1S1t.Then when ε→0 we have(47)lim⁡inft→+∞⁡S1t≥S10.On the other hand from the proof of Lemma 2, we have (48)limt→+∞⁡S1t≤S10+ε.Let ε→0. Then one has(49)lim⁡supt→+∞⁡S1t≤S10.From (47) and (49), we have (50)limt→+∞⁡S1t=S10almost surely.
By employing the method similar above, it then follows that(51)limt→+∞⁡S2t=S20almost surely. This completes the proof of Theorem 6.


## 4.2. Permanence in Mean

Theorem 7.
IfR>1, then the microorganism Xt is permanent in mean; moreover, Xt satisfies (52)lim⁡inft→+∞⁡Xt≥DΔR∗-1,where Δ=min⁡K1+S10D/r1,K2+S20D/r2.Proof.
Integrating from0 to t and dividing by t on both sides of system (5) yield(53)Θt≜S1t-S10t+S2t-S20t+Xt-X0t=DS10+S20-DS1t+S2t-DXt.Then one can get(54)DXt=DS10+S20-DS1t+S2t-Θt.Applying Itô’s formula gives(55)dln⁡Xt=r1ϕ1t+r2ϕ2t-D-σ122ϕ12t-σ222ϕ22tdt+σ1S1tdBtK1+S1t+σ2S2tdBtK2+S2t≥r1ϕ1t+r2ϕ2t-D-σ122S10K1+S102-σ222S20K2+S202dt+σ1S1tdBtK1+S1t+σ2S2tdBtK2+S2t.Integrating from 0 to t and dividing by t on both sides of (55) yields (56)ln⁡Xt-ln⁡X0t≥r11t∫0tϕ1θdθ+r21t∫0tϕ2θdθ-D+σ122S10K1+S102+σ222S20K2+S202+M1tt+M2tt,where M1t=∫0tσ1S1τ/K1+S1τdBτ, M2t=∫0tσ2S2τ/K2+S2τdBτ. Noticing that (57)∫0tϕ1θdθ=∫0tS1θK1+S1θdθ≥∫0tS1θK1+S10dθ,∫0tϕ2θdθ=∫0tS2θK2+S2θdθ≥∫0tS2θK2+S20dθ,then we have(58)ln⁡Xt-ln⁡X0t≥-D+σ122S10K1+S102+σ222S20K2+S202+r1K1+S10S1t+r2K2+S20S2t+M1tt+M2tt=-D+σ122S10K1+S102+σ222S20K2+S202+r2K2+S20S10+S20+r1K1+S10-r2K2+S20S1t-r2K2+S20Xt-r2K2+S20ΘtD+M1tt+M2tt.If r1/K1+S10≤r2/K2+S20, we can get (59)ln⁡Xt-ln⁡X0t≥-D+σ122S10K1+S102+σ222S20K2+S202+r2K2+S20S10+S20+r1K1+S10-r2K2+S20S10-r2K2+S20Xt-r2K2+S20ΘtD+M1tt+M2tt≥-D+σ122S10K1+S102+σ222S20K2+S202+r2S20K2+S20+r1S10K1+S10-r2K2+S20Xt-r2K2+S20ΘtD+M1tt+M2tt.By inequality (59), we have(60)Xt≥K2+S20r2r1S10K1+S10+r2S20K2+S20-σ122S10K1+S102-σ222S20K2+S202-D-K2+S20r2ln⁡Xt-ln⁡X0t+r2K2+S20ΘtD+M1tt+M2tt.By Lemma 3, we get that limt→+∞⁡Mt/t=0. According to Lemma 2, one sees that lim⁡supt→+∞⁡S1t≤S10, lim⁡supt→+∞⁡S2t≤S20, and lim⁡supt→+∞⁡Xt≤C0+S20, and then one has limt→+∞⁡ln⁡X(t)/t=0 and limt→+∞⁡Θt=0. Thus taking the inferior limit of both sides of (60) yields(61)lim⁡inft→+∞⁡Xt≥K2+S20Dr2R∗-1.And if r1/K1+S10≥r2/K2+S20, we can get (62)ln⁡Xt-ln⁡X0t≥-D+σ122S10K1+S102+σ222S20K2+S202+r1K1+S10S10+S20-S2t-Xt-ΘtD+r2K2+S20S2t+M1tt+M2tt=-D+σ122S10K1+S102+σ222S20K2+S202+r1S10K1+S10+r2S20K2+S20-r1K1+S10Xt-r2K2+S20ΘtD+M1tt+M2tt,where M1t=∫0tσ1S1τ/K1+S1τdBτ, M2t=∫0tσ2S2τ/K2+S2τdBτ.
Inequality (62) can be rewritten as(63)Xt≥K2+S20r2r1S10K1+S10+r2S20K2+S20-σ122S10K1+S102-σ222S20K2+S202-D-K2+S20r2ln⁡Xt-ln⁡X0t+r1K1+S10ΘtD+M1tt+M2tt.Taking the inferior limit of both sides of (63) yields(64)lim⁡inft→+∞⁡Xt≥K1+S10Dr1R∗-1.Let Δ=min⁡K1+S10D/r1,K2+S20D/r2, and we get from (61) and (64)(65)lim⁡inft→+∞⁡Xt≥DΔR∗-1.This completes the proof of Theorem 7.Remark 8.
Theorems6 and 7 show that the condition for the microorganism to go to extinction or permanence depends on the intensity of the noise disturbances completely. And small noise disturbances will be beneficial to the cultivation of the microorganism; conversely, large white noise disturbance is harmful to the cultivation of the microorganism.


## 5. Conclusion and Numerical Simulation

This paper proposes and investigates a new stochastic chemostat model with two substitutable nutrients and one microorganism. Then main objective in this paper is to investigate the threshold dynamics of stochastic chemostat model (5) and explore the conditions which can determine the extinction and permanence of the microorganism using two substitutable nutrients. Firstly, for the corresponding deterministic model, the threshold for extinction or existence of the microorganism is obtained by analyzing the stability of the equilibria. Then the threshold of the stochastic chemostat for the extinction and the permanence in mean of the microorganism is explored. The results show that there exists a significant difference between the threshold of the deterministic system and the stochastic system, which makes the persistent microorganism of a deterministic system become extinct due to large stochastic disturbance. That is, large stochastic disturbance is harmful to the cultivation of the microorganism. It is worth mentioning that this paper is a promotion of the work of Xu and Yuan [35].Next, using the Euler Maruyama (EM) method [40], we give some numerical simulation to illustrate the extinction and persistence of the microorganism in stochastic system and corresponding deterministic system for comparison.Firstly, we begin from a deterministic system; the basic parameters are set asD=3.8, S10=1, S20=2, r1=2.5, r2=4, K1=1, and K2=1. Direct calculation shows that δ1=0.8, δ2=0.375, and R=1.0307>1. Then according to Theorems 4 and 5, the deterministic system has a unique stable positive equilibrium E∗0.9295785859,1.848231322,0.2221900926, which is locally stable and the deterministic system is permanent (see Figure 1).Figure 1
Time series for the pathsS1t, S2t, Xt for the deterministic system with D=3.8, S10=1, S20=2, r1=2.5, r2=4, K1=1, K2=1, R=1.0307>1.Next, we consider the influence of stochastic disturbance on the above deterministic system. According to Theorem6, different parameters are chosen to give insights into the reasonability of the results stated in Theorem 6.We choose different value of parametersσ1 and σ2 and discuss below five different cases.Case
1. Choose σ1=1.6, σ2=1.8, by direct calculation; we have r12/2σ12+r22/2σ22=3.6898<D=3.8. Then, by Theorem 6, the microorganism eventually tends to be extinct (see Figure 2(a)).Figure 2
Time series for the pathsS1t, S2t, Xt for the stochastic system.
(a)
σ
1
=
1.6, σ2=1.8
(b)
σ
1
=
0.7, σ2=1.3
(c)
σ
1
=
0.7, σ2=1.3
(d)
σ
1
=
0.7, σ2=0.6, R′=0.9935<1Case
2. Choose σ1=1, σ2=0.5, by direct calculation; we have r12/2σ12+r2S20/K2+S20-σ22/2S20/K2+S202=0.5139<D=3.8. Then, by Theorem 6, the microorganism eventually tends to be extinct (see Figure 2(b)).Case
3. Choose σ1=0.7, σ2=1.3, by direct calculation; we have r22/2σ22+r1S10/K1+S10-σ12/2S10/K1+S102=3.5450<D=3.8. Then, by Theorem 6, the microorganism eventually tends to be extinct (see Figure 2(c)).Case
4. Choose σ1=0.7, σ2=0.6, by direct calculation; we have R′=0.9935<1. Then, by Theorem 6, the microorganism eventually tends to be extinct (see Figure 2(d)).Case
5. Choose σ1=0.2, σ2=0.3, by direct calculation; we have R′=1.0241>1. Then, by Theorem 7, the microorganism is persistent (see Figure 3).Figure 3
Time series for the pathsS1t, S2t, Xt for the stochastic system with σ1=0.2, σ2=0.3, R′=1.0241>1.



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*Source: 5217027-2017-09-25.xml*