# Stochastic Characteristics and Optimal Control for a Stochastic Chemostat Model with Variable Yield

Authors: Rong Yan; Shulin Sun
Journal: Complexity
 (2020)

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## Abstract

In this paper, a stochastic chemostat model with variable yield and Contois growth function is investigated. The yield coefficient depends on the limiting nutrient, and the environmental noises are given by independent standard Brownian motions. First, the existence and uniqueness of global positive solution are proved. Second, by using stochastic Lyapunov function, Itô’s formula, and some important inequalities, stochastic characteristics for the stochastic model are studied, including the extinction of micro-organism, the strong persistence in the mean of micro-organism and, the existence of a unique stationary distribution of the stochastic model. Third, the necessary condition of an optimal stochastic control for the stochastic model is established by Hamiltonian function. In addition, some numerical simulations are carried out to illustrate the theoretical results and the influence of the variable yield on the microorganism.


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## Body

## 1. Introduction

The chemostat occupies a central place in mathematical ecology. It has been widely used in modeling natural ecosystems such as lakes and establishing waste-water treatment mathematical model. Many works related to the chemostat model have been published in the journal of mathematical, biological, and chemical engineering (see [1–5], etc.).A large number of scholars are attracted to yield coefficient reflecting the conversion of nutrient to microorganism. Most of the models assume that the yield coefficient is a constant (see [1, 2]). Smith and Waltman [1] have described a chemostat model with constant yield as follows:(1)S′t=S0−StD−pS,xxtγ,x′t=xtpS,x−D,where all the parameters are positive constants. St and xt stand for the concentrations of the nutrient and the microorganism at time t in culture vessel, respectively. S0 is the input nutrient concentration. D is the common dilution rate. γ is a “yield” constant. pS,x represents the growth function. Sun and Yin [2] and Wang et al. [3] have studied the chemostat model with Monod growth function. Bayen et al. [4] and Sun and Chen [5] have investigated the chemostat model with Contois growth function. Based on model (1), choosing Contois growth function pS,x=mSt/kxt+St, the model takes the following form:(2)S′t=S0−StD−mStkxt+Stxtγ,x′t=xtmStkxt+St−D,where m>0 represents the maximal growth rate and k>0 stands for the growth coefficient of the Contois function.However, experimental data indicate that a constant yield may fail to explain the observed oscillatory behavior in the vessel (see [6]). This leads to the formulation of the variable yield model, for example [7–9]. The chemostat model with variable yield takes the following form:(3)S′t=S0−StD−mStkxt+StxtA+BSt,x′t=xtmStkxt+St−D,where γS=A+BS is the variable yield which is considered in [9]. Both A and B are positive constants.Actually, the above models are all deterministic models, but almost all ecosystems are inevitably perturbed by various types of environmental noises. Taking into account the realistic and biological significance, many scholars have investigated stochastic chemostat models (see [3, 10–14]) and stochastic population models (see [15–19]). In the chemostat model, the micro-organism may be affected by distribution of nutrient, temperature, humidity, etc, which can be described by continuous white noise. There are different possible approaches to include random effects in the model, both from a biological and mathematical perspective. Some authors have superimposed white noise processes on the dilution rate [20]; others have considered the linear white noise [11, 21]. Our basic approach is analogous to that of Beddington and May [21]. By this method, the white noise are directly proportional to St and xt, influenced on S′t and x′t. A stochastic chemostat model with variable yield and Contois growth function will be investigated, and the model (3) will be rewritten as the form:(4)dSt=S0−StD−mStkxt+StxtA+BStdt+σ1StdB1t,dxt=xtmStkxt+St−Ddt+σ2xtdB2t,where σi>0i=1,2 are intensities of the white noise. Biti=1,2 are independent standard Brownian motions defined on a complete probability space Ω,ℱ,ℱtt≥0,ℙ with the filtration ℱtt≥0 satisfying the usual conditions. The model (4) turns into the corresponding deterministic model (3) if the noise intensities σi=0i=1,2.The initial conditions of (4) are given as(5)S0=S0,x0=x0,where S0,x0 are positive random variables, and denote ℝ+2=x1,x2xi>0,i=1,2.Because the solutions of the deterministic model (3) are no longer the solutions of the stochastic model (4), the dynamic behavior of the deterministic model (3) is distinct from that of the stochastic model (4). For further details on the stochastic model, please see [10–14], they have investigated the stochastic characteristics of these stochastic models. However, the nonlinear term of the model (4) is different from that of their models. Therefore, it is valuable to investigate the stochastic characteristics of the stochastic chemostat model with variable yield and Contois growth function. On the other hand, the optimal control plays an important role in practical application [4, 22]. Many scholars have paid more attention to the optimal stochastic control problem, which covers all aspects of physics, biology, economics, etc. Ding et al. [23] have solved the distributed H∞ state estimation problem for a class of discrete time-varying nonlinear system with both stochastic parameters and stochastic nonlinearities. Guo et al. [24] have studied the near-optimal control of a stochastic SIRS epidemic model that includes a nonmonotone incidence rate. Framstad et al. [25] have proved a sufficient maximum principle for the optimal control of jump diffusions and showed its connections to dynamic programming and given applications to financial optimization problems in a market described by such processes. Thus, it is necessary to study the optimal stochastic control of a stochastic chemostat model. Throughout the whole paper, Lyapunov function, Itô formula, and other basic methods can be referred to these monographs [26–32].This paper is organized as follows. In Section2, the existence of the unique global positive solution for the model (4) is studied. In Section 3, the stochastic characteristics of the stochastic model (4) are investigated, including the extinction at an exponential rate of the microorganism, the strong persistence in the mean of the microorganism, and the stationary distribution of the model (4). In Section 4, the necessary condition of an optimal stochastic control for the stochastic model (4) is investigated, and the near-optimal stochastic control is mentioned. In Section 5, the numerical simulations conclude the paper, and the influence of the variable yield on the microorganism is explained by taking the different parameters.


## 2. Preliminaries

First of all, the notations are described for the whole paper as follows:(i)
Ω: a set of the elementary events(ii)
ℱ: a family of the subsets of Ω(iii)
ℱtt≥0: a family of increasing sub−σ−algebras of ℱ(iv)
ℙω: the probability of events ω(v)
EX: the expectation of X(vi)
∅: the empty set(vii)
IA: the indicator function of a set A, i.e., IAx=1 if x∈A or otherwise 0(viii)
a.s.: almost surely(ix)
En: the n−dimensional Euclidean space(x)
ℝ: the set of all real numbers(xi)
ℝ+: the set of all nonnegative real numbers, i.e., ℝ+=0,∞(xii)
ℝ+d: x∈ℝd:xi>0,1≤i≤d, i.e. the positive cone(xiii)
U: an open domain(xiv)
U¯: a close domainIn view of biological significance and dynamical behavior, the first concerned thing is whether the solution is unique, global, and positive. Hence, to further study the stochastic chemostat model with variable yield and Contois growth function (4), the first problem to be solved is the existence of the unique global positive solution, namely, there is no explosion in a finite time under the initial value (5). If the coefficients of the equations are generally required to satisfy the linear growth condition and the local Lipschitz condition (see [28, 30]), the stochastic differential equations for any given initial value have a unique global solution. However, the stochastic model (4) may allow the solutions to explode at a finite time because the coefficients of the stochastic model (4) do not satisfy the linear growth condition. So, we need to search for the positive solutions. The following theorem assures that the solution of the model (4) with the initial value (5) is unique, global, and positive. First, a lemma and a remark are given.Lemma 1.
(see [19]). The following inequality holds:(6)u≤2u+1−lnu−4−2ln2,∀u>0.Remark 1.
Based on Lemma1, since the inequality u≤2u+1−lnu−4−2ln2, we can obtain u≤2u+1−lnu holds for all u>0.
Next, the existence of the unique global positive solution for the model (4) is proved.Theorem 1.
For any initial valueS0,x0∈ℝ+2, there exists a unique positive solution St,xt to the model (4) for t≥0, and the solution will remain in ℝ+2 with probability one (i.e., St,xt∈ℝ+2 for all t≥0 a.s.).Proof.
Since the coefficients of the stochastic differential equations (4) satisfy the local Lipschitz condition, the model (4) has a unique positive local solution St,xt on t∈0,τe for any initial value S0,x0∈ℝ+2, where τe is the explosion time. In order to show the solution is global, we just need to prove that τe=∞ a.s.. Select k0≥1 sufficiently large such that S0 and x0 all lie within the interval 1/k0,k0. For ∀k≥k0, where k is a integer, define the stopping time(7)τk=inft∈0,τeSt∉1k,korxt∉1k,k.
Throughout this paper, we setinf∅=∞. τk is increasing when k⟶∞. Let τ∞=limk⟶+∞τk, clearly, τ∞≤τe a.s. τe=∞ and St,xt∈ℝ+2 a.s. for all t≥0 if τ∞=∞ can be verified. Next, for τ∞=∞, the proof process is as follows:
Define aC2−function V:ℝ+2⟶ℝ+ by(8)VS,x=S+1−lnS+x+1−lnx,for ∀T>0, by applying Itô’s formula on t∈0,τk∧T, we have(9)dVS,x=1−1SdS+1−1xdx+121S2dS2+121x2dx2=1−1SS0−SD−S−1mxkx+SA+BS+x−1mSkx+S−D+12σ12+12σ22dt+σ1S−1dB1t+σ2x−1dB2t=LVdt+σ1S−1dB1t+σ2x−1dB2t,by using Lemma 1(10)LV=S0D−SD−S0SD+D−mSxkx+SA+BS+mxkx+SA+BS+mSxkx+S−xD−mSkx+S+D+12σ12+12σ22≤S0D+2D+12σ12+12σ22+mxkx+SA+BS+mSxkx+S≤S0D+2D+12σ12+12σ22+mkA+mx≤S0D+2D+12σ12+12σ22+mkA+2mx+1−lnx+2S+1−lnS≜N1+N2VS,x,so,(11)dVS,x≤N1+N2VS,xdt+σ1S−1dB1t+σ2x−1dB2t,where(12)N1=S0D+2D+12σ12+12σ22+mkA,N2=2m∨2.
Integrating equation (11) both sides from 0 to τk∧T and taking expectation,(13)EVSτk∧T,xτk∧T≤VS0,x0+N1T+N2E∫0τk∧TVSt,xtdt.
By using Gronwall’s inequality, we have(14)EVSτk∧T,xτk∧T≤VS0,x0+N1TeN2T.
For∀ω∈τk≤T, it exists that Sτk,ω or xτk,ω equals either k or 1/k,(15)VSτk,ω,xτk,ω≥k+1−lnk∧1k+1+lnk≜Gk.
Thus,(16)VS0,x0+N1TeN2T≥EIτk≤TωVSτk,ω,xτk,ω≥ℙτk≤TGk,where Iτk≤T is the indicator function of τk≤T.
Lettingk⟶∞ gives limk⟶∞ℙτk≤T=0, and hence, ℙτ∞≤T=0. Since T>0 is arbitrary, we must have(17)ℙτ∞<∞=0,so ℙτ∞=∞=1 as required.
The proof is thus complete.


## 3. Stochastic Characteristics

In this section, the stochastic characteristics of the stochastic model (4) are discussed. As is known to all, the equilibriums of the deterministic model (3) are no longer the equilibriums of the stochastic model (4), then what happens in the stochastic model (4)? Next, the sufficient conditions of the extinction and persistence of microorganism and the stationary distribution of the model (4) are investigated.In order to study the model (4), we introduce the following definitions and lemmas, and denote x=1/t∫0txsds.Lemma 2.
(see [14]). For any initial value (5), the solutions St and xt of the model (4) have the properties that(18)limsupt⟶∞St<∞,limsupt⟶∞xt<∞,a.s..That is, there are two positive constantsH1 and H2 such that(19)St≤H1,xt≤H2,for allt≥0,a.s..Lemma 3.
(see [28]). Let M=Mtt≥0 be a real-valued continuous local martingale vanishing at t=0. Then(20)limsupt⟶∞M,Mtt<∞,a.s.⇒limt⟶∞Mtt=0,a.s..Next, the extinction at an exponential rate of the microorganism is studied.Theorem 2.
If the condition holds,(21)σ22>2m−D,then for any initial value (5), the solution St,xt of the model (4) satisfies(22)limsupt⟶∞lnxtt≤m−D−12σ22<0,a.s.,which means(23)limt⟶∞xt=0,a.s..That is, the microorganism will be extinct at an exponential rate with probability one.Proof.
Define a function(24)V=lnx,using Itô formula(25)dV=mSkx+S−D−12σ22dt+σ2dB2t≤m−D−12σ22dt+σ2dB2t,a.s..
Integrating both sides from 0 tot, we can obtain(26)Vtt−V0t≤m−D−12σ22+σ2tB2t,then(27)lnxtt≤m−D−12σ22+φ1t,a.s.,where(28)φ1t=σ2tB2t+lnx0t,limt⟶∞φ1t=0,a.s..
According to equation (27), we have(29)limsupt⟶∞lnxtt≤m−D−12σ22<0,a.s.,which implies that(30)limt⟶∞xt=0,a.s..
Thus, the proof is completed.Remark 2.
Theorem2 shows that the condition σ22>2m−D makes the microorganism extinct, but it has nothing to do with m<D or m>D, which means that large noises can lead to the extinction of the microorganism, although the microorganism is persistent in the corresponding deterministic model (3) (see Figures 1 and 2)
A definition and a lemma are given before discussing the strong persistence in the mean of the microorganism.Figure 1
The solutions of deterministic model (3) and stochastic model (4). (a) Sample paths of St with A=2,B=0. (b) Sample paths of xt with A=2,B=0. (c) Sample paths of St with A=2,B=3. (d) Sample paths of xt with A=2,B=3.
(a)
(b)
(c)
(d)Figure 2
The solutions of deterministic model (3) and stochastic model (4). (a) Sample paths of St with A=2,B=0. (b) Sample paths of xt with A=2,B=0. (c) Sample paths of St with A=2,B=3. (d) Sample paths of xt with A=2,B=3.
(a)
(b)
(c)
(d)Definition 1.
(see [17]). The population xt is said to be strong persistence in the mean if(31)x∗>0,where x∗=liminft⟶∞x.Lemma 4.
(see [18]). Let f∈C0,∞×Ω,0,∞. If there exist positive constants λ0, λ such that(32)lnft≥λt−λ0∫0tfsds+Ft,a.s.,for all t≥0, where F∈C0,∞×Ω,ℝ and limt⟶∞Ft/t=0,a.s., then(33)f∗≥λλ0,a.s..Theorem 3.
For any initial value (5), if the condition holds:(34)12σ22<ζ≜minmkH2+1−D,mS0kH2+1−D,then the solution St,xt of the model (4) satisfies(35)xt∗≥ADkH2+1mm+ADζ−12σ22>0.That is, the microorganism is almost surely strongly persistent in the mean.Proof.
On the one hand, based on the model (4), we have(36)dS+x=S0D−SD−mSxkx+SA+BS+mSxkx+S−xDdt+σ1SdB1t+σ2xdB2t≥S0D−SD−mxA−xDdt+σ1SdB1t+σ2xdB2t=S0D−SD−m+ADAxdt+σ1SdB1t+σ2xdB2t,so,(37)St+xtt−S0+x0t≥S0D−DSt−m+ADAxt+σ1tM1t+σ2tM2t,where(38)M1t=∫0tSsdB1s,M2t=∫0txsdB2s.
Thus,(39)St≥S0−m+ADADxt−φ2t,where(40)φ2t=1DSt+xtt−S0+x0t−σ1tM1t−σ2tM2t.
SinceM1t and M2t are local continuous martingales with M10=0 and M20=0, from Theorem 5.14 in Mao’s monograph [28], Lemmas 2 and 3, we have(41)limsupt⟶∞M1,M1tt=∫0tS2sdst≤H12<∞,limsupt⟶∞M2,M2tt=∫0tx2sdst≤H22<∞,and(42)limt⟶∞φ2t=0.
On the other hand, we will discuss the long time behavior of the microorganismx in two cases, we define(43)V=lnx.Case 1.
IfSt<1, we get(44)dV=mSkx+S−D−12σ22dt+σ2dB2t≥mSkH2+1−D−12σ22dt+σ2dB2t,so,(45)Vtt−V0t≥mkH2+1St−D−12σ22+σ2tB2t.
Substituting (39) into (45), we obtain(46)Vtt−V0t≥mkH2+1S0−m+ADADxt−φ2t−D−12σ22+σ2tB2t.
WhenS0<1, we have(47)lnxt≥mS0kH2+1−D−12σ22t−mm+ADADkH2+1xtt−mtφ2tkH2+1+σ2B2t+lnx0≜λ1t−λ0∫0txsds+F1t,where(48)λ1=mS0kH2+1−D−12σ22,λ0=mm+ADADkH2+1,F1t=−mtφ2tkH2+1+σ2B2t+lnx0,take note of the condition of Theorem 3, we have λ1>0. Thus, from (42), Lemmas 3 and 4, we can obtain that(49)xt∗≥ADkH2+1mm+ADmS0kH2+1−D−12σ22>0.
WhenS0≥1, we have(50)lnxt≥mkH2+1−D−12σ22t−mm+ADADkH2+1xtt−mtφ2tkH2+1+σ2B2t+lnx0≜λ2t−λ0∫0txsds+F1t,where λ2=m/kH2+1−D−1/2σ22, similar to the above, we can obtain that(51)xt∗≥ADkH2+1mm+ADmkH2+1−D−12σ22>0.Case 2.
IfSt≥1, we get(52)dV=mSkx+S−D−12σ22dt+σ2dB2t≥mkH2+1−D−12σ22dt+σ2dB2t,so,(53)Vtt−V0t≥mkH2+1−D−12σ22+σ2tB2t.
Obviously,(54)Vtt−V0t≥mkH2+1−D−12σ22−mm+ADADkH2+1xtt+σ2tB2t.
WhenS0<1, we have(55)lnxt≥mS0kH2+1−D−12σ22t−mm+ADADkH2+1xtt+σ2B2t+lnx0≜λ1t−λ0∫0txsds+F2t,where F2t=σ2B2t+lnx0, similarly, we have(56)xt∗≥ADkH2+1mm+ADmS0kH2+1−D−12σ22>0.
WhenS0≥1, we have(57)lnxt≥mkH2+1−D−12σ22t−mm+ADADkH2+1xtt+σ2B2t+lnx0≜λ2t−λ0∫0txsds+F2t,
Similar to the above, we have(58)xt∗≥ADkH2+1mm+ADmkH2+1−D−12σ22>0.
In conclusion, no matter whatSt chooses, we have(59)xt∗>0.
This completes the proof of Theorem3.Remark 3.
It follows from Theorem3 that the microorganism will be strongly persistent in the mean if the condition 1/2σ22<ζ≜minm/kH2+1−D,mS0/kH2+1−D is satisfied. That is, when the intensity of the white noise is small enough, the microorganism will survive (see Figures 3 and 4).
Stationary distribution is one of the most significant dynamical characteristics of the stochastic model; that is, the stochastic model has a stationary distribution which represents the persistence of microorganism in the future. Therefore, in the rest of this section, we will make a positive decision by the existence of the stationary distribution. Before proving the main result, several known results are given for the stochastic differential equations.
AssumeXt is a regular time-homogeneous Markov Process in n−dimensional Euclidean space En. The stochastic differential equation takes the following form:(60)dXt=hXdt+∑r=1ngrXdBrt.
The diffusion matrix of the processXt is defined as follows:(61)Ax=αijx,αijx=∑r=1ngrixgrjx.
Define the differential operatorL associated with (60) by(62)L=∂∂t+∑i=1nhix∂∂xi+12∑i,j=1nαijx∂2∂xi∂xj.Figure 3
The solutions of deterministic model (3) and stochastic model (4). (a) Sample paths of St with A=2,B=0. (b) Sample paths of xt with A=2,B=0. (c) Sample paths of St with A=2,B=3. (d) Sample paths of xt with A=2,B=3.
(a)
(b)
(c)
(d)Figure 4
The solutions of deterministic model (3) and stochastic model (4). (a) Sample paths of St with A=2,B=0. (b) Sample paths of xt with A=2,B=0. (c) Sample paths of St with A=2,B=3. (d) Sample paths of xt with A=2,B=3.
(a)
(b)
(c)
(d)Lemma 5.
(see [30]). Assume that there exists a bounded open domain U⊂En with regular boundary Γ, having the following properties:(1)
In the domainU and some neighborhood thereof, the smallest eigenvalue of the diffusion matrix A is bounded away from zero.(2)
Ifx∈En\U, the mean time τ at which a path issuing from x reaches the set U is finite, and supx∈KExτ<∞ for every compact subset K⊂En.
Then the Markov processXt has a stationary distribution μ⋅, and it is unique.Remark 4.
Since we investigate the stochastic model (4) in the space ℝ+2, to validate condition (1) of Lemma 5, we need to prove that for any bounded open domain U⊂ℝ+2, there is a positive number M0 such that ∑i,j=12αijxξiξj≥M0ξ2, x∈U¯,ξ∈ℝ+2 (see [31, 32]). To validate condition (2) of Lemma 5, we need to show that there exists a neighborhood U and a non-negative C2-function V such that for any S,x∈ℝ+2∖U, LV is negative.
The following theorem shows that the model (4) has a unique stationary distribution for any given initial value (5).Theorem 4.
Letη=c2S0D−c1mS0/kA−c1σ12S0/2−D−σ22/2 where c2=mkH2/DkH2+S02. If there exists a constant c1>mkH2S0/DkH2+S03 such that η is positive, then for any given initial value (5), the solution of the model (4) admits a unique stationary distribution μ⋅.Proof.
According to the model (4), we get(63)dSx=S0−SD−mSxkx+SA+BSmSxkx+S−xDdt+σ1S00σ2xdB1tdB2t.
So, the diffusion matrix is(64)AS,x=σ12S200σ22x2.
LetU be any bounded open domain in ℝ+2, then there exists a positive constant(65)M0=minσ12S2,σ22x2,S,x∈U¯,such that(66)∑i,j=12αijS,xξiξj=σ12S2ξ12+σ22x2ξ22≥M0ξ2,for all S,x∈U¯,ξ∈ℝ+2, which implies that condition (1) of Lemma 5 is satisfied.
Next, we need to construct a non-negativeC2-function VS,x and a open set U∈ℝ+2, such that(67)supS,x∈ℝ+2\ULVS,x<0.
Then, we selectθ∈0,1 to be a sufficiently small constant such that(68)A.1D−θ2σ12∨σ22>0,choosing a large enough positive constant N such that(69)A.2ψ1u−Nη≤−2,where ψ1u=supS,x∈ℝ+2\Uψ1S,x, and ψ1S,x is given in (A.3).
Define a nonnegativeC2-Lyapunov function(70)VS,x=HS,x−HS∗,x∗,where HS,x=NV1+V2+V3,V1=c1S−S0−S0lnS/S0−lnx−c2S+x,V2=1/θ+1S+xθ+1,V3=−lnS, and HS,x is a C2-function with a unique minimum value point S∗,x∗.
Using Itô formula forV1,V2,V3, we estimate LV1 firstly.(71)dS−S0−S0lnSS0=−S−S02DS−mxS−S0kx+SA+BS+12σ12S0dt+σ1S−S0dB1t,dlnx=mSkx+S−D−12σ22dt+σ2dB2t,dS+x=S0−SD−mSxkx+SA+BS+mSxkx+S−xDdt+σ1SdB1t+σ2xdB2t.
So,(72)dV1=c1−S−S02DS−mxS−S0kx+SA+BS+12σ12S0dt−mSkx+S−D−12σ22dt−c2S0−SD−mSxkx+SA+BS+mSxkx+S−xDdt+c1σ1S−S0dB1t−σ2dB2t−c2σ1SdB1t+σ2xdB2t,thus,(73)LV1=−c1DS−S02S−c1mxS−S0kx+SA+BS+c12σ12S0−mSkx+S+D+12σ22−c2DS0+c2DS+c2mAx+c2Dx−c2DS0−S+c2mSxkx+SA+BS−c2mSxkx+S+c2Dx≤−c1DS−S02S+c1mS0kA+c12σ12S0−mSkH2+S+D+12σ22=−c1DS−S02S−mSkH2+S+c2DS+c2mA+c2Dx−c2DS0−c1mS0kA−c12σ12S0−D−12σ22≜lS+c2mA+c2Dx−η,where(74)lS=−c1DS−S02S−mSkH2+S+c2DS,η=c2DS0−c1mS0kA−c12σ12S0−D−12σ22.
Therefore, according to the previous functionlS, we have(75)l′S=−c1DS2−S02S2−mkH2kH2+S2+c2D,l′′S=−2c1DS02S3+2mkH2kH2+S3.
Obviously,(76)l′SS=S0=0,forc2=mkH2DkH2+S02,l′′SS=S0=−2c1DkH2+S03+2mkH2S0S0kH2+S03<0,forc1>mkH2S0DkH2+S03.
So,(77)lS≤lS0=c2DS0−mS0kH2+S0=−mS02kH2+S02<0,thus,(78)LV1≤−η+c2mA+c2Dx.
Now, we estimateLV2 and LV3,(79)dV2=S+xθS0−SD−mSxkx+SA+BS+mSxkx+S−xDdt+θ2S+xθ−1σ12S2+σ22x2dt+S+xθσ1SdB1t+σ2xdB2t=LV2dt+S+xθσ1SdB1t+σ2xdB2t.
So,(80)LV2=S+xθS0−SD−mSxkx+SA+BS+mSxkx+S−xD+θ2S+xθ−1σ12S2+σ22x2≤S+xθS0D−S+xθ+1D+mH2S+xθ+θ2S+xθ−1σ12∨σ22S+x2=S0D+mH2S+xθ−D−θ2σ12∨σ22S+xθ+1,where the positive constant H2 is mentioned in Lemma 2.(81)dV3=−1SS0D−SD−mSxkx+SA+BSdt+12σ12dt−σ1dB1t=LV3dt−σ1dB1t.
So,(82)LV3=−S0DS+D+mxkx+SA+BS+12σ12≤−S0DS+D+mkA+12σ12,thus, from VS,x=NV1+V2+V3−HS∗,x∗, we have(83)LV≤N−η+c2mA+c2Dx+S0D+mH2S+xθ−D−θ2σ12∨σ22S+xθ+1−S0DS+D+mkA+12σ12≜ψ1S,x+ψ2x,where(84)A.3ψ1S,x=S0D+mH2S+xθ−D−θ2σ12∨σ22S+xθ+1−S0DS+D+mkA+12σ12,ψ2x=N−η+c2mA+c2Dx.
Through observation, we find that(85)ψ1S,x+ψ2x=−∞,asS⟶+∞,x∈ℝ+,ψ1S,x+ψ2x=−∞,asx⟶+∞,S∈ℝ+,ψ1S,x+ψ2x=−∞,asS⟶0+,x∈ℝ+,ψ1S,x+ψ2x≤−2,asx⟶0+,S∈ℝ+.
LetU=ε,1/ε×ε,1/ε, where ε is a small enough constant. It follows that(86)LV<0,S,x∈ℝ+2\U.
This implies that the model (4) admits a unique stationary distribution μ⋅.
Thus, the proof is completed.Remark 5.
By Theorem4, we known that the model (4) has a unique stationary distribution if the condition holds, which means that the microorganism is persistent in the future.


## 4. Stochastic Maximum Principle

In this section, the existence of the optimal stochastic control for the model (4) is proved. Our aim is to find an optimal stochastic control under given the initial conditions such that the production of the microorganism is maximized at a given time T. We select the dilution rate D as stochastic control variable Dt. Several preliminaries are given.For convenience, the model (4) can be rewritten as(87)dSt=f1St,xt,Dtdt+σ1StdB1t,dxt=f2St,xt,Dtdt+σ2xtdB2t,where(88)f1St,xt,Dt=S0−StDt−mStkxt+StxtA+BSt,f2St,xt,Dt=xtmStkxt+St−Dt.The stochastic control variableDt is a non-negative continuous bounded function. Let the constant V denote the volume of the culture vessel, and let Ft be the volumetric flow rate, then the dilution rate Dt=Ft/V. For convenience, we define V=1, so Dt=Ft, the production function is expressed as(89)Mt=∫0tℳtdt,where ℳt=xtFt=xtDt is a continuous function denoting the production of unit time of the microorganism. Since we will focus on the maximum microorganism production problem in the interval 0,T, the objective functional is given as(90)JD⋅=EMT=E∫0TxtDtdt.Denote a bounded nonempty closed setA⊂ℝ, and the set of admissible controls is defined as follows:(91)Aad≜D:0,T×Ω⟶ADisℱtt≥0−adapted.For anyD∈Aad, equation (87) is a stochastic differential equation with random coefficients, which has a unique strong solution St,xt called an admissible state trajectory, and St,xt,Dt is called an admissible triple.The objective functional (90) can be rewritten as(92)JD⋅≜E∫0TLSt,xt,Dtdt,E∫0TLSt,xt,Dtdt<∞,where LSt,xt,Dt=xtDt, the optimal control condition is to select a nonanticipative decision that maximize the objective functional, that is to seek D^∈Aad such that(93)JD^=supD∈AadJD,where D^ is called an optimal stochastic control. S^,x^ is called the corresponding solution of the model (87) and S^,x^,D^ is called an optimal triple.Define a Hamiltonian functionH:ℝ+×ℝ+×A×ℝ×ℝ×ℝ×ℝ⟶ℝ by(94)HSt,xt,Dt,p1t,p2t,q1t,q2t=f1St,xt,Dtp1t+f2St,xt,Dtp2t+σ1Stq1t+σ2xtq2t+LSt,xt,Dt.The adjoint equation corresponding to the admissible tripleSt,xt,Dt on the unknown ℱtt≥0−adapted processes p1t,p2t,q1t,q2t is the backward stochastic differential equation with the natural boundary conditions:(95)dp1t=−g1St,xt,Dt,p1t,p2t,q1t,q2tdt+q1tdB1t,dp2t=−g2St,xt,Dt,p1t,p2t,q1t,q2tdt+q2tdB2t,p1T=0,p2T=0,where(96)g1S,x,D,p1,p2,q1,q2=−D+mxAkx−BS2kx+S2A+BS2p1+mkx2kx+S2p2+σ1q1,g2S,x,D,p1,p2,q1,q2=−mS2kx+S2A+BSp1+mS2kx+S2−Dp2+σ2q2+D.Next, the necessary condition of the optimal stochastic control for the model (4) is established.Theorem 5.
Suppose that there exists an adapted solutionp^1,p^2,q^1,q^2 of the corresponding adjoint equation (95) such that for all D∈Aad, we have(97)E∫0TS^−S2q^12+x^−x2q^22dt<∞,and(98)E∫0Tσ12S2p^12+σ22x2p^22dt<∞.Furthermore, if the following conditions hold,(i)
The HamiltonianHS,x,D,p^1,p^2,q^1,q^2 with respect to S,x,D is concave for all t∈0,T(ii)
HS^,x^,D^,p^1,p^2,q^1,q^2=maxD∈AHS^,x^,D,p^1,p^2,q^1,q^2,for allt∈0,Tthen, S^,x^,D^ is an optimal triple and D^ is an optimal stochastic control.Proof.
Proof. For any control tripleS,x,D∈Aad, we consider(99)JD^−JD=E∫0TLS^,x^,D^dt−E∫0TLS,x,Ddt=E∫0TLS^,x^,D^−LS,x,Ddt≜I.
By the Hamiltonian functionH and the assumption of p^1,p^2,q^1,q^2, we have(100)LS,x,D=HS,x,D,p^1,p^2,q^1,q^2−f1S,x,Dp^1−f2S,x,Dp^2−σ1Sq^1−σ2xq^2,therefore(101)I=E∫0THS^,x^,D^,p^1,p^2,q^1,q^2−HS,x,D,p^1,p^2,q^1,q^2dt−E∫0Tf1S^,x^,D^p^1−fS,x,Dp^1dt−E∫0Tf2S^,x^,D^p^2−f2S,x,Dp^2dt−E∫0Tσ1S^q^1−σ1Sq^1dt−E∫0Tσ2x^q^2−σ2xq^2dt.
Since the concavity ofH in S, x and D, we get(102)I≥E∫0TS^−SHSS^,x^,D^,p^1,p^2,q^1,q^2+x^−xHxS^,x^,D^,p^1,p^2,q^1,q^2dt+E∫0TD^−DHDS^,x^,D^,p^1,p^2,q^1,q^2dt−E∫0Tf1S^,x^,D^p^1−f1S,x,Dp^1dt−E∫0Tf2S^,x^,D^p^2−f2S,x,Dp^2dt−E∫0Tσ1S^q^1−σ1Sq^1dt−E∫0Tσ2x^q^2−σ2xq^2dt≜I1+I2,where(103)I1=E∫0TS^−SHSS^,x^,D^,p^1,p^2,q^1,q^2+x^−xHxS^,x^,D^,p^1,p^2,q^1,q^2dt−E∫0Tf1S^,x^,D^p^1−f1S,x,Dp^1dt−E∫0Tf2S^,x^,D^p^2−f2S,x,Dp^2dt−E∫0Tσ1S^q^1−σ1Sq^1dt−E∫0Tσ2x^q^2−σ2xq^2dt,I2=E∫0TD^−DHDS^,x^,D^,p^1,p^2,q^1,q^2dt.
Conditions (97) and (98) ensure that the stochastic integrals have zero mean. Next, we will estimate I1 by Itô product formula and the natural boundary conditions of (95).(104)I1=E∫0TS^−SHSS^,x^,D^,p^1,p^2,q^1,q^2−f1S^,x^,D^−f1S,x,Dp^1−S^−Sσ1q^1dt+E∫0Tx^−xHxS^,x^,D^,p^1,p^2,q^1,q^2−f2S^,x^,D^−f2S,x,Dp^2−x^−xσ2q^2dt=−E∫0TS^−Sdp^1+∫0Tp^1dS^−S+∫0Tdp^1dS^−S−E∫0Tx^−xdp^2+∫0Tp^2dx^−x+∫0Tdp^2dx^−x=−ES^T−STp^1T−Ex^T−xTp^2T=0.
Applying the condition (ii) of theorem,(105)ED^t−DtHDS^,x^,D^,p^1,p^2,q^1,q^2=D^t−DtEHDS^,x^,D,p^1,p^2,q^1,q^2D=D^≥0,then(106)I2=E∫0TD^−DHDS^,x^,D^,p^1,p^2,q^1,q^2dt≥0.
Hence, we haveI=I1+I2≥0; that is,(107)JD^≥JD.
ConsideringS,x,D is arbitrary, then S^,x^,D^ is an optimal triple and D^ is an optimal stochastic control. This completes the proof of Theorem 5.
Since the control variableD appears linearly in the Hamiltonian function HS,x,D,p1,p2,q1,q2. This indicates that the problem may have an optimal stochastic singular control. However, it is usually difficult to find the exact optimal stochastic singular control. To this end, we will introduce a small perturbation to turn the singular control into a nonsingular control and call it a near-optimal stochastic control of this problem.
Let the objective functional is(108)J∗D⋅=E∫0TxtDt−12εDt2dt.where 1/2εDt2 represents a small perturbation, and the positive constant ε is small enough to ensure xtDt−1/2εDt2>0. Thus, the Hamiltonian function can be rewritten as(109)H∗St,xt,Dt,p1t,p2t,q1t,q2t=f1St,xt,Dtp1t+f2St,xt,Dtp2t+σ1Stq1t+σ2xtq2t+xtDt−12εDt2.
The necessary conditions for the near-optimal stochastic control areHD∗=0 and HDD∗<0 at the point S^,x^,D^,p1^,p2^,q1^,q2^. Hence, we get the following result.Corollary 1.
The model (4) has a near-optimal stochastic control:(110)D^=1εS0−S^p1^−x^p2^+x^,and the corresponding optimal triple S^,x^,D^.


## 5. Numerical Simulations

Numerical simulations are presented for supporting our theoretical results of the model (4) by Milstein method [33]. The discretized equations are listed as follows:(111)Si+1=Si+S0−SiD−mSikxi+SixiA+BSiΔt+Siσ1ξiΔt+σ122ξi2−1Δt,xi+1=xi+ximSikxi+Si−DΔt+xiσ2ζiΔt+σ222ζi2−1Δt,where ξi,ζi,i=1,2,… are independent ℕ0,1−distributed Gaussian random variables.Next, we present numerical simulations to demonstrate the results.Example 1.
For the model (3) and (4), choose two sets of parameters:S0=2,m=0.5,k=1,D=1,A=2,B=0,σ1=0.1,σ2=1.2, Figures 1(a) and 1(b)andS0=2,m=0.5,k=1,D=1,A=2,B=3,σ1=0.1,σ2=1.2, Figures 1(c) and 1(d)and the initial point isS0,x0=0.4,0.5. Obviously,(112)m<D,which means that when the microorganism is extinct without reference to the intensity of noise (see Figure 1).Example 2.
For the model (3) and (4), choose two sets of parameters:S0=2,m=1.45,k=1,D=1,A=2,B=0,σ1=0.1,σ2=1.2, Figures 2(a) and 2(b)andS0=2,m=1.45,k=1,D=1,A=2,B=3,σ1=0.1,σ2=1.2, Figures 2(c) and 2(d)and the initial point isS0,x0=0.4,0.5. Since(113)σ22=1.44,2m−D=0.9,obviously,(114)σ22>2m−D,which means that when the condition of Theorem 2 satisfies, then the microorganism is extinct, though the microorganism xt in deterministic model (3) is persistent (see Figure 2).Example 3.
For the model (3) and (4), choose two sets of parameters:S0=1.1,m=1.4,k=1,D=0.1,H2=10,A=2,B=0,σ1=0.15,σ2=0.15, Figures 3(a) and 3(b)andS0=1.1,m=1.4,k=1,D=0.1,H2=10,A=2,B=3,σ1=0.15,σ2=0.15, Figures 3(c) and 3(d)and the initial point isS0,x0=0.4,0.5. Since(115)12σ22≈0.0113,mkH2+1−D≈0.0273,mS0kH2+1−D=0.0400,obviously,(116)12σ22<mkH2+1−D<mS0kH2+1−D,which means that when the condition of Theorem 3 satisfies, then the microorganism is almost surely strongly persistent in the mean (see Figure 3).Example 4.
For the model (3) and (4), choose two sets of parameters:S0=0.9,m=1.4,k=1,D=0.1,H2=10,A=2,B=0,σ1=0.15,σ2=0.15, Figures 4(a) and 4(b)andS0=0.9,m=1.4,k=1,D=0.1,H2=10,A=2,B=3,σ1=0.15,σ2=0.15, Figures 4(c) and 4(d)and the initial point isS0,x0=0.4,0.5. Since(117)12σ22≈0.0113,mS0kH2+1−D≈0.0145,mkH2+1−D≈0.0273,obviously,(118)12σ22<mS0kH2+1−D<mkH2+1−D,which means that when the condition of Theorem 3 satisfies, then the microorganism is almost surely strongly persistent in the mean (see Figure 4).
Through the above numerical simulations, we know that the parametersB rarely affect the extinction of the microorganism xt and have a significant impact on the persistence of the microorganism xt, that is, the concentration of the microorganism xt increases with the parameter B (see Figures 3(b), 3(d), 4(b), and 4(d)). In conclusion, the intensities of white noise are disadvantageous to the growth of the microorganism, and the variable yield γS=A+BS is advantageous to the growth of the microorganism.



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