# Stationary Distribution and Extinction of a Stochastic Brucellosis Model with Standard Incidence

Authors: Dilnaray Iskandar; Xamxinur Abdurahman; Ahmadjan Muhammadhaji
Journal: Computational and Mathematical Methods in Medicine
 (2022)

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## Abstract

In this paper, we proposed a stochastic SVEI brucellosis model with stage structure by introducing the effect of environmental white noise on transmission dynamics of brucellosis. By Has’minskii theory and constructing suitable Lyapunov functions, we established sufficient conditions on the existence of ergodic stationary distribution for the considered model. Moreover, we also established sufficient condition for extinction of the disease. Finally, two examples with numerical simulations are given to illustrate the main results of this paper.


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## Body

## 1. Introduction

Brucellosis, which is recognized as a major public health problem, is a serious and economically devastating zoonosis which can infect animals, such as sheep, cattle, pig, and dogs. The disease is caused by bacteria of the genusBrucella, of which there are six species: B. abortus, B. melitensis, B. suis, B. ovis, B. canis, and B. neotomae [1]. Brucella can survive for long periods in dust, dung, water, slurry, aborted fetuses, soil, meat, and dairy products. In animals, brucellosis can be infected by contact with the infected animals (direct way of infection) and by contact of polluted environment (indirect way of transmission), and the disease mainly affects reproduction and fertility and reduces survival of newborns [2]. Brucellosis also can infect human being; the main transmission sources of human brucellosis include exposure to a contaminated environment by infected animals, direct contact with infected animals, and the ingestion of fresh milk or dairy products prepared from unpasteurized milk and unheated meat and animal liver [3]; there is no recorded cases of the infection between humans. Most of the human brucellosis cases are infected by Brucella melitensis (which is infected in sheep and goats), accounting for 84.5% of the total cases [4]. In humans, disease-related mortality is negligible, but the illness can last for several years [5]. Therefore, the key to solve the problem of this public health problem is the elimination of animal brucellosis.It is worth noting that, mathematical models are widely used not only to study the transmission dynamics of brucellosis, but also to study the epidemiological characteristics of brucellosis [1–8]. Recently, the authors in [6] presented a sheep brucellosis model with immigration and proportional birth, considering both direct and indirect transmission. In [7], the authors proposed a multigroup SEIRV dynamical model with bidirectional mixed cross infection between cattle and sheep and investigated the influence of cross infection of mixed feeding on the brucellosis transmission. In [8], the authors proposed the following deterministic brucellosis transmission model:
(1)dS1dt=A+bS2+V1+τS2+V−μ1+d+η1S1,dS2dt=η1S1−βS2I−θS2−μ2S2+δV,dVdt=θS2−εβVI−μ2+δV,dEdt=βεVI+βS2I−μ2+η2E,dIdt=η2E−μ2+cI,and studied the dynamical behavior of the model, where sheep population is classified into five compartments: the susceptible young sheep S1t, the susceptible adult (or sexually mature) sheep S2t, the vaccinated sheep Vt, the exposed sheep Et, and the infectious sheep It. A and b are the input number of young sheep and the natural birth rate of sheep, and τ is the extent of the birth being delayed. μ1 and μ2 are the young sheep natural mortality rate and the elimination rate of adult sheep. η1 and η2 are the transfer rate from young sheep to adult sheep and exposed sheep to infected sheep. d is the output number of young sheep, δ is the vaccination rate, β is sheep-to-sheep transmission rate, ε is the ineffective vaccination rate, and c is the elimination rate caused by brucellosis. All the parameters are assumed to be positive.However, the epidemics in the real world are often disturbed by some uncertain factors, such as environmental white noises. Therefore, it is difficult to describe these epidemic dynamics by using determined differential equation [9]. Thus, the deterministic models has some limitations in mathematical modeling of epidemics, and it is quite difficult to fitting data perfectly and to predicting the future dynamics of the epidemic system. In the past years, there has been a lot of researchers who are interested in the stochastic dynamical models [3, 9–28]. In particular, the stochastic epidemic models have been extensively studied [3, 10–25]. For example, in [3], the authors proposed and studied a periodic stochastic brucellosis model and obtained some conditions on the existence of nontrivial positive periodic solution of the model. In [11], the authors studied a stochastic SIRS epidemic model with standard incidence rate and partial immunity and obtained sufficient conditions on the extinction and existence of a stationary probability measure for the disease of the system. In [12], the authors studied a kind of stochastic SEIR epidemic model with standard incidence and obtained sufficient conditions for the existence of stationary distribution and the extinction of the disease in the system. In [24], the authors discussed a stochastic SIRS epidemic model with logistic growth and nonlinear incidence and obtained sufficient conditions on the ergodic stationary distribution and extinction of the considered model.On the other hand, there are different approaches used in the literature to introduce random perturbations into population models, both from a mathematical and biological perspective [3, 9–28]. In this paper, in light of the above analysis and reasons, we consider the stochastic perturbations for deterministic system (1) and we employ the approach used in Mao et al. [26] and assume that the parameters involved in the model always fluctuate around some average value due to continuous fluctuations in the environment. This approach is reasonable and well justified biologically [27, 28]. By this approach, we study a stochastic S1S2VEI brucellosis model with standard incidence, and we assume that the environmental white noise affects the natural mortality rate, the elimination rate, transfer rate, and transmission rate. In order to obtain the stochastic S1S2VEI brucellosis model, we let Xt=S1t,S2t,Vt,Et,ItT, and then it is appropriate to model X=S1,S2,V,E,IT as a Markov process; thus, from [15] and model (1), we can get the following properties when 0≤Δt≪1, the conditional mean
(2)ES1t+Δt−S1t∣X=x≈A+bS2+V1+τS2+V−μ1+d+η1S1Δt,ES2t+Δt−S2t∣X=x≈η1S1−βS2I−θS2−μ2S2+δVΔt,EVt+Δt−Vt∣X=x≈θS2−εβVI−μ2+δVΔt,EEt+Δt−Et∣X=x≈βεVI+βS2I−μ2+η2EΔt,EIt+Δt−It∣X=x≈η2E−μ2+cIΔt,and the conditional covariance
(3)VarS1t+Δt−S1t∣X=x≈σ12S12Δt,VarS2t+Δt−S2t∣X=x≈σ22S22Δt,VarVt+Δt−Vt∣X=x≈σ32V2Δt,VarEt+Δt−Et∣X=x≈σ42E2Δt,VarIt+Δt−It∣X=x≈σ52I2Δt.Then, we derive the following stochastic form of system (1)
(4)dS1t=A+bS2+V1+τS2+V−μ1+d+η1S1dt+σ1S1dB1t,dS2t=η1S1−βS2I−θS2−μ2S2+δVdt+σ2S2dB2t,dVt=θS2−εβVI−μ2+δVdt+σ3VdB3t,dEt=βεVI+βS2I−μ2+η2Edt+σ4EdB4t,dIt=η2E−μ2+cI+σ5IdB5t,whereB1t,B2t,B3t,B4t, and B5t are the standard one-dimensional independent Brownian motions and σi2>0i=1,2,3,4,5 are the intensity of the white noises.The main purpose of the paper is to obtain the conditions for the existence of ergodic stationary distribution and extinction of the disease for model (4).This paper is organized as follows. In Section2, we present some preliminaries which will be used in the following analysis. In Section 3, we show that there is a unique global positive solution of system (4). In Section 4, we prove the existence of ergodic stationary distribution for system (4) under certain conditions. In Section 5, we establish sufficient conditions for the disease extinction.


## 2. Preliminaries

Throughout this paper, letΩ,F,ℙ be a complete probability space with a filtration Ftt≥0 satisfying the usual conditions (i.e., it is increasing and right continuous, while F0 contains all ℙ -null sets); Biti=1,2,3,4 are defined on this complete probability space, and also let ℝ+d=x∈ℝd:xi>0,1≤i≤d.In general, consider thed-dimensional stochastic differential equation
(5)dxt=fxt,tdt+gxt,tdBtfort0,∞,with initial value x0=x0∈ℝd. Bt denotes an n-dimensional standard Brownian motion defined on the complete probability space Ω,F,Ftt≥0,ℙ. C2,1ℝd×t0,∞;ℝ+ denotes the family of all nonnegative functions Vx,t defined on ℝd×t0,∞ such that they are continuously twice differentiable in x and once in t. The differential operator L of equation (5) is defined by [16].
(6)Ł=∂∂t+∑i=1dfix,t∂∂xi+12∑i,j=1dgTx,tgx,tij∂2∂xi∂xj.IfL acts on a function V∈C2,1ℝd×t0,∞;ℝ+, then
(7)LVx,t=Vtx,t+Vxx,tfx,t+12tracegTx,tVxxx,tgx,t,where Vt=∂V/∂t,Vx=∂V/∂x1,⋯,∂V/∂xd,Vxx=∂2V/∂xi∂xjd×d. By Itô’s formula, if xt∈ℝd, then
(8)dVxt,t=LVxt,tdt+Vxxt,tgxt,tdBt.Next, we present a result about the existence of stationary distribution (see Has’minskii [17]).LetXt be a homogeneous Markov process in Ed (Ed denotes d-dimensional Euclidean space) and be described by the following stochastic differential equation:
(9)dXt=bXdt+∑r=1kgrXdBrt.The diffusion matrix is defined as follows:(10)Ax=aijx,aijx=∑r=1kgrixgrjx.Lemma 1.
The Markov processXt has a unique ergodic stationary distribution Π· if there exists a bounded domain D⊂Ed with regular boundary Γ and
A1: there is a positive number M such that ∑i,j=1daijxξiξj≥Mξ2,x∈D,ξ∈ℝd.
A2: there exists a nonnegative C2-function V such that LV is negative for any Ed\D. Then
(11)ℙxlimT⟶∞1T∫0Tfxtdt=∫EdfxΠdx=1,for all x∈Ed, where f· is a function integrable with respect to the measure π.


## 3. Main Results

### 3.1. Existence and Uniqueness of the Positive Solution

In studying the dynamical behavior of an epidemic model, the first importance is whether the solution is global and positive. Hence, in the following theorem, we will study the existence and uniqueness of the global positive solution, which is a prerequisite for researching the long-term behavior of model (4).Theorem 2.
For any initial valueX0=S10,S20,V0,E0,I0∈ℝ+5, there is a unique solution Xt=S1t,S2t,Vt,Et,It of system (4) on t≥0, and the solution will remain in ℝ+5 with probability one.Proof.
Since the coefficients of system (4) satisfy the local Lipschitz condition, then for any initial value S10,S20,V0,E0,I0∈ℝ+5, there is a unique local solution S1t,S2t,Vt,Et,It on 0,τe, where τe is the explosion time [16]. To show this solution is global, we only need to prove that τe=∞ a.s. To this end, let n0>0 be sufficiently large such that every component of X0 lying within the interval 1/n0,n0. For each integer n>n0, define the stopping time as follows:
(12)τn=inft∈0,τe:minS1t,S2t,Vt,Et,It≤1normaxS1t,S2t,Vt,Et,It≥n.Throughout this paper, we setinf∅=∞ (as usual ∅ denotes the empty set). It is easy to see that τn is increasing as n⟶∞. Let τ∞=limn⟶∞τn, then τ∞≤τe a.s. In what follows, we need to verify τ∞=∞ a.s. If this assertion is violated, there is a constant T>0 and an ε∈0,1 such that ℙτ∞≤T>ε. As a result, there exists an integer n1≥n0 such that
(13)ℙτn≤T≥ε,n≥n1.Define aC2-function V: ℝ+5⟶ℝ+ by
(14)VS1,S2,V,E,I=S1−1−lnS1+S2−1−lnS2+V−1−lnV+E−1−lnE+I−1−lnI.Using Ito^’s formula, we have
(15)dVS1S2,V,E,I=LVS1S2,V,E,Idt+σ1S1−1dB1t+σ2S2−1dB2t+σ3V−1dB3t+σ4E−1dB4t+σ5I−1dB5t,where
(16)LV=1−1S1A+bS2+V1+τS2+V−μ1+d+η1S1+1−1S2η1S1−βS2I−θS2−μ2S2+δV+1−1VθS2−εβVI−μ2+δV+1−1EβεVI+βS2I−μ2+η2E+1−1Iη2E−μ2+cI+σ12+σ22+σ32+σ42+σ522.By applying the following invariant set of model (1) which is obtained in [8]
(17)Ω=S1,S2,V,E,I∈R+5:S1+S2+V+E+I≤Aτ+bμτ,and from the following inequalities
(18)bS2+V1+τS2+V≤bτ,Iβε+1≤βε+1Aτ+bμτ,and also cancel the items less than zero, so we have
(19)LV=A+bS2+V1+τS2+V+μ1+d+η1+βI+θ+4μ2+εβI+δ+η2+c−μ1+dS1−AS1−bS2+VS11+τS2+V−μ2S2+V+E+I−η1S1S2−δVS2−θS2V−βεVIE−βS2IE−cI−η2EI+σ12+σ22+σ32+σ42+σ522≤A+bτ+βε+1Aτ+bμτ+μ1+d+η1+θ+4μ2+δ+η2+c+σ12+σ22+σ32+σ42+σ522=ζ.Sinceζ is positive constant which is independent of S1,S2,V,E,I, and t, we can get
(20)dVS1S2,V,E,I≤ζdt+σ1S1−1dB1t+σ2S2−1dB2t+σ3V−1dB3t+σ4E−1dB4t+σ5I−1dB5t.Integrating both sides (20) from 0 to T∧τε and taking expectations, then we can obtain
(21)EVS1τε∧T,S2τε∧T,Vτε∧T,Eτε∧T,Iτε∧T≤VS10,S20,V0,E0,I0+ζT<∞.SetΩε=τε≤t for n≥n1 by (13), PΩn≥ε. Notice that for every ω∈Ωε, there is at least one of S1τε,ω,S2τε,ω,Vτε,ω,Eτε,ω, and Iτε,ω that equal either n or 1/n. Hence, S1τε,ω,S2τε,ω,Vτε,ω,Eτε,ω, and Iτε,ω are no less than
(22)n−1−lognor1n−1−logn.Consequently,(23)VS1τε,ω,S2τε,ω,Vτε,ω,Eτε,ω,Iτε,ω≥n−1−logn∧1n−1−logn,where a∧b donates the minimum of a and b. In view of (21) and (23) we have
(24)VS10,S20,V0,E0,I0+ζT≥E1ΩωVS1τε∧T,S2τε∧T,Vτε∧T,Eτε∧T,Iτε∧T≥δ~n−1−logn∧1n−1+logn,where 1Ωω is the indicator function of Ωn. Let n⟶∞ leads to the contradiction
(25)∞>VS10,S20,V0,E0,I0+ζT=∞.Therefore, we must haveτ∞=∞ a.s.


### 3.2. Stationary Distribution and Ergodicity

The difference between model (1) and the stochastic model is that the stochastic model does not have the endemic equilibrium. Hence, we cannot study the persistence of the disease by studying the stability of the endemic equilibrium and turn to check out the existence and uniqueness of the stationary distribution for the system (4) which implies the persistence of the disease in some sense. In this section, based on the theory of Has’minskii [17], we verify that there is an ergodic stationary distribution, which reveals the persistence of the disease.Define a parameter(26)R0s=η1η2βεθμ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/2.Theorem 3.
Assume thatR0s>1, then system (4) has a unique stationary distribution Π· and it has the ergodic property.Proof.
In view of Theorem2, we have obtained that for any initial value S10,S20,V0,E0,I0∈ℝ+5, there is a unique global solution S1,S2,V,E,I∈ℝ+5..
The diffusion matrix of system (4) is given by
(27)A=σ12S1200000σ22S2200000σ32V200000σ42E200000σ52I2.ChooseM=minS1,S2,V,E,I∈D¯σ⊂R+5σ12S12,σ12S22,σ32V2,σ42E2,σ52I2; one can get that
(28)∑i,j=15aijS1,S2,V,E,Iξiξj=σ12S12ξ12+σ12S22ξ22+σ32V2ξ32+σ42E2ξ42+σ52I2ξ52≥Mξ2,S1,S2,V,E,I∈D¯σ,ξ=ξ1,ξ2,ξ3,ξ4,ξ5∈ℝ+5.Then the conditionA1 in Lemma 1 is satisfied.Construct aC2-function Q:ℝ+5⟶ℝ in the following from
(29)QS1,S2,V,E,I=MS1+S2+V+E+I−c1lnS1−c2lnS2−c3lnE−c4lnI−lnV+1χ+1S1+S2+V+E+Iχ+1−lnS1−lnS2−lnE−lnV+S1+S2+V+E+I=MV1+V2+V3+V4+V5+V6,where χ is a constant satisfying 0<χ<2μ/σ12∨σ22∨σ32∨σ42∨σ52,
(30)c1=Aμ1+d+η1+σ12/2,c2=Aθ+μ2+σ22/2,c3=Aμ2+η2+σ42/2,c4=Aμ2+c+σ52/2,and M>0 satisfies the following condition
(31)−Mλ+C≤−2,where
(32)λ=5Aη1η2βεθμ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/21/5−1=5AR0s1/5−1>0,(33)C=supS1,S2,V,E,Iεℝ+5−12μ−12σσ12∨σ22∨σ32∨σ42∨σ52·S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422.It is easy to check that(34)liminfk⟶∞,S1,S2,V,E,I∈ℝ+5\UkQS1,S2,V,E,I=∞,where Uk=1/k,k×1/k,k×1/k,k×1/k,k×1/k,k. Furthermore, QS1,S2,V,E,I is a continuous function. Hence, QS1,S2,V,E,I must have a minimum point S¯10,S¯20,V¯0,E¯0,I¯0 in the interior of ℝ+5. Then we define a nonnegative C2-function V:ℝ+5⟶ℝ+ as follows:
(35)VS1,S2,V,E,I=QS1,S2,V,E,I−QS¯10,S¯20,V¯0,E¯0,I¯0.Making use of Ito^’s formula, we have
(36)LV1=−μ2S2+V+E+I+μ1+dS1+c1AS1+c1bS2+VS11+τS2+V+c2η1S1S2+c2δVS2+c3βεVIE+c3βS2IE+c4η2EI+θS2V−cI+A+bS2+V1+τS2+V+εβI+c2βI+μ2+δ+σ322+c1μ1+d+η1+σ122+c2θ+μ2+σ222+c3μ2+η2+σ422+c4μ2+c+σ522.Using the inequalitya+b+c+d+e≥5abcde5,a,b,c,d,e>0 leads to
(37)LV1≤−c1AS1+c2η1S1S2+c3βεVIE+c4η2EI+θS2V+A+bτ+εβI+c2βI+μ2+δ+σ322+c1μ1+d+η1+σ122+c2θ+μ2+σ222+c3μ2+η2+σ422+c4μ2+c+σ522≤−5c1c2c3c4Aη1βεη2θ1/5+A+bτ+εβI+c2βI+μ2+δ+σ322+c1μ1+d+η1+σ122+c2θ+μ2+σ222+c3μ2+η2+σ422+c4μ2+c+σ522=−5A5η1βεη2θμ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/21/5+5A+μ2+δ+bτ+βIc2+ε+σ322=−5Aη1βεη2θμ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/21/5−1+μ2+δ+bτ+βIc2+ε+σ322=−λ+μ2+δ+bτ+βIc2+ε+σ322,where λ is defined in (32).Similarly(38)LV2=S1+S2+V+E+IχA+bS2+V1+τS2+V−μ1+dS1−μ2S2+V+E+I−cI+12χS1+S2+V+E+Iχ−1×σ12S12+σ22S22+σ32V2+σ42E+σ52I2≤S1+S2+V+E+IχA+bτ−μS1+S2+V+E+I+12χS1+S2+V+E+Iχ+1σ12∨σ22∨σ32∨σ42∨σ52=A+bτS1+S2+V+E+Iχ−μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1+S2+V+E+Iχ+1.Then(39)LV2≤B−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1+S2+V+E+Iχ+1≤B−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1,where
(40)B=supS1,S2,V,E,I∈ℝ+5A+bτS1+S2+V+E+Iχ−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1+S2+V+E+Iχ+1<∞.We can also get(41)LV3=−AS1−bS2+VS11+τS2+V+μ1+d+η1+σ122,(42)LV4=−η1S1S2+βI+θ+μ2−δVS2+σ222,(43)LV5=−βεVIE−βS2IE+μ2+η2+σ422,(44)LV6=−θS2V+εβI+μ2+δ+σ322,(45)LV7=A+S2+V1+τS2+V−μ1+dS1−μ2S2+V+E+I−cI≤A+bτ−μ1+dS1−μ2S2+V+E+I.Hence, by (37)-(45), we obtain
(46)LV≤−Mλ+MβIc2+ε+1+εM+B−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1−AS1−bS2+VS11+τS2+V−η1S1S2−δVS2−βεVIE−βS2IE−θS2V−μ1+dS1−μ2S2+V+E+I+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422.Thus, we can construct a compact subsetD such that the condition A2 in Lemma 1 holds. Define the bounded closed set
(47)D=ε1≤S1≤1ε1,ε2≤S2≤1ε2,ε3≤I≤1ε3,ε4≤E≤1ε4,ε5≤V≤1ε5,where εi>0i=1,2,3,4,5 are sufficiently small constants satisfying the following conditions:
(48)−1ε1+F≤−1,(49)−Mλ−ε5ε2+C≤−1,(50)−Mλ+Mβε31+εM+c2+ε+C≤−1,(51)−βε2ε3ε4+F≤−1,(52)−θS2ε5+F≤−1,(53)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε1χ+1+G≤−1,(54)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε2χ+1+H≤−1,(55)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε4χ+1+J≤−1,(56)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε3χ+1+K≤−1,(57)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε5χ+1+L≤−1,where F,G,H,J,K, and L are positive constants which can be seen from (60), (68), (70), (72), (74), and (76), respectively. Note that for sufficiently small εi,i=1,2,3,4,5. For convenience, we divide ℝ+5\D into ten domains
(58)D1=S1,S2,V,E,I∈ℝ+5:0<S1<ε1,D2=S1,S2,V,E,I∈ℝ+5:0<S2<ε2,V≥ε5,D3=S1,S2,V,E,I∈ℝ+5:0<I<ε3,D4=S1,S2,V,E,I∈ℝ+5:S2≥ε2,I≥ε3,0<E<ε4,D5=S1,S2,V,E,I∈ℝ+5:0<V<ε5S2≥ε2,D6=S1,S2,V,E,I∈ℝ+5:S1>1ε1,D7=S1,S2,V,E,I∈ℝ+5:S2>1ε2,D8=S1,S2,V,E,I∈ℝ+5:I>1ε3,D9=S1,S2,V,E,I∈ℝ+5:E>1ε4,D10=S1,S2,V,E,I∈ℝ+5:V>1ε5.Next, we will show thatLVS1,S2,V,E,I≤−1 on ℝ+5\D, which is equivalent to proving it on the above ten domains.Case 1.
IfS1,S2,V,E,I∈D1, one can get that
(59)LV≤−AS1+MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422≤−AS1+F≤−Aε1+F≤−1,
where(60)F=supS1,S2,V,E,I∈ℝ+5MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422.According to (48), we have
(61)LV≤−1,foranyS1,S2,V,E,I∈D1.Case 2.
IfS1,S2,V,E,I∈D2, we have
(62)LV≤−Mλ−δVS2−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422≤−Mλ−δVS2+C≤−Mλ−ε5ε2+C,where C is defined in (33).
In view of (49), we can obtain that for sufficiently small εii=2,5, LV≤−1 for any S1,S2,V,E,I∈D2.Case 3.
IfS1,S2,V,E,I∈D3, one can see that
(63)LV≤−Mλ+MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422.We obtain that(64)LV≤−1foranyS1,S2,V,E,I∈D3.Case 4.
IfS1,S2,V,E,I∈D4, one can see that
(65)LV≤−βS2IE+MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422≤−βS2IE+F≤−βε2ε3ε4+F.In view of (51), we can obtain that for sufficiently small εii=3,4,LV≤−1 for any S1,S2,V,E,I∈D4.Case 5.
IfS1,S2,V,E,I∈D5, one can see that
(66)LV≤−θS2V+MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422≤−θS2V+F≤−θε2ε5+F.We can obtain that for sufficiently smallεii=2,5, LV≤−1 for any S1,S2,V,E,I∈D5.Case 6.
IfS1,S2,V,E,I∈D6, one can see that
(67)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+G≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε1χ+1+G,where
(68)G=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.By (53), we conclude that LV≤−1onD6.Case 7.
IfS1,S2,V,E,I∈D7, one can see that
(69)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S2χ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S2χ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S2χ+1+H≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε2χ+1+H,where
(70)H=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S2χ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.Together with (54), we can deduce that LV≤−1onD7.Case 8.
IfS1,S2,V,E,I∈D8, one can see that
(71)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Eχ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×Eχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Eχ+1+J≤−14μ−12θσ12∨σ22∨σ32∨σ42∨σ521ε4χ+1+J,where
(72)J=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Eχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Vχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.which together with (55) implies that LV≤−1onD8.Case 9.
IfS1,S2,V,E,I∈D9, we obtain
(73)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Iχ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×Iχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Iχ+1+K≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε3χ+1+K,where
(74)K=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Iχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Vχ+1+Eχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.By (56), we can conclude that LV≤−1 on D9.Case 10.
IfS1,S2,V,E,I∈D10, it follows that
(75)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Vχ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×Vχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Vχ+1+L≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε5χ+1+L,where
(76)L=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Vχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.Combining with (57) yields LV≤−1 on D10.Obviously,A2 in Lemma 2.1 is satisfied. According to Lemma 2.1, we can obtain that system (4) is ergodic and has a unique stationary distribution.Remark 4.
Theorem3 reveals that system (4) has a unique ergodic stationary distribution π· if R0s=η1η2βεθ/μ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/2>1. Note that the expression of R0s coincide with the threshold R0 of the deterministic system (1) if there is no stochastic perturbation. This shows that we generalize the result of the deterministic system.


### 3.3. Extinction of the Disease

As it is well known, one of the main concern of epidemiology is how we regulate the disease dynamics in order to eradicate the disease in the long term. Moreover, in [10], Allen et al. proposed and studied several types of stochastic epidemic models and pointed out that the stochastic models should suit the question of disease extinction better. Hence, in this section, we shall establish some sufficient conditions for extinction of the disease in stochastic model (4).Theorem 5.
LetS1t,S2t,Vt,Et,It be the solution of system (4) with any initial value S10,S20,V0,E0,I0∈ℝ+5. If βε+1Aτ+bμ2+η2/μτη2<σ52/2+μ2+c∧σ42/2, then the disease It will extinct exponentially with probability one, i.e., moreover
(77)limt⟶∞sup1tlogE+μ2+η2η2I≤βη2ε+1Aτ+bμ2+η2μτ−η2μ2+η22σ522+μ2+c∧σ422.Proof.
Applying Ito^’s formula to logE+μ2+η2/η2I, we have
(78)dlogE+μ2+η2η2I=βεV+S2IE+μ2+η2/η2I−μ2+η2μ2+cIE+μ2+η2/η2Iη2−σ42E2+σ52μ2+η2/η2I22E+μ2+η2/η2I2dt+σ4EE+μ2+η2/η2IdB4t+σ5μ2+η2Iη2E+μ2+η2/η2IdB5t=βεV+S2IE+μ2+η2/η2Idt−1E+μ2+η2/η2I2μ2+η2η2μ2+cIE+σ522+μ2+cμ2+η2η22I2+σ42E22dt+σ4EE+μ2+η2/η2IdB4t+σ5μ2+η2Iη2E+μ2+η2/η2IdB5t≤βεV+S2η2μ2+η2dt−1E+μ2+η2/η2I2σ522+μ2+cμ2+η2η22I2+σ42E22dt+σ4EE+μ2+η2/η2IdB4t+σ5μ2+η2Iη2E+μ2+η2/η2IdB5t≤βη2ε+1Aτ+bμ2+η2Aτdt−η2μ2+η22σ522+μ2+c∧σ422dt+σ4EE+μ2+η2/η2IdB4t+σ5μ2+η2Iη2E+μ2+η2/η2IdB5t.Integrating the above inequality from 0 tot, and the fact that limt⟶∞Bit/t=0,i=4,5 [16], yields
(79)limt⟶∞sup1tlogE+μ2+η2η2I≤βη2ε+1Aτ+bμ2+η2μτ−η2μ2+η22σ522+μ2+c∧σ422=x0.For anyε¯>0, and almost ω∈Ω,∃T=Tω such that
(80)It≤η2μ2+η2expx0+ε¯t,∀t≥T.Remark 6.
Theorem5 suggests that the disease will become extinct if βε+1Aτ+bμ2+η2/μτη2<σ52/2+μ2+c∧σ42/2.


## 3.1. Existence and Uniqueness of the Positive Solution

In studying the dynamical behavior of an epidemic model, the first importance is whether the solution is global and positive. Hence, in the following theorem, we will study the existence and uniqueness of the global positive solution, which is a prerequisite for researching the long-term behavior of model (4).Theorem 2.
For any initial valueX0=S10,S20,V0,E0,I0∈ℝ+5, there is a unique solution Xt=S1t,S2t,Vt,Et,It of system (4) on t≥0, and the solution will remain in ℝ+5 with probability one.Proof.
Since the coefficients of system (4) satisfy the local Lipschitz condition, then for any initial value S10,S20,V0,E0,I0∈ℝ+5, there is a unique local solution S1t,S2t,Vt,Et,It on 0,τe, where τe is the explosion time [16]. To show this solution is global, we only need to prove that τe=∞ a.s. To this end, let n0>0 be sufficiently large such that every component of X0 lying within the interval 1/n0,n0. For each integer n>n0, define the stopping time as follows:
(12)τn=inft∈0,τe:minS1t,S2t,Vt,Et,It≤1normaxS1t,S2t,Vt,Et,It≥n.Throughout this paper, we setinf∅=∞ (as usual ∅ denotes the empty set). It is easy to see that τn is increasing as n⟶∞. Let τ∞=limn⟶∞τn, then τ∞≤τe a.s. In what follows, we need to verify τ∞=∞ a.s. If this assertion is violated, there is a constant T>0 and an ε∈0,1 such that ℙτ∞≤T>ε. As a result, there exists an integer n1≥n0 such that
(13)ℙτn≤T≥ε,n≥n1.Define aC2-function V: ℝ+5⟶ℝ+ by
(14)VS1,S2,V,E,I=S1−1−lnS1+S2−1−lnS2+V−1−lnV+E−1−lnE+I−1−lnI.Using Ito^’s formula, we have
(15)dVS1S2,V,E,I=LVS1S2,V,E,Idt+σ1S1−1dB1t+σ2S2−1dB2t+σ3V−1dB3t+σ4E−1dB4t+σ5I−1dB5t,where
(16)LV=1−1S1A+bS2+V1+τS2+V−μ1+d+η1S1+1−1S2η1S1−βS2I−θS2−μ2S2+δV+1−1VθS2−εβVI−μ2+δV+1−1EβεVI+βS2I−μ2+η2E+1−1Iη2E−μ2+cI+σ12+σ22+σ32+σ42+σ522.By applying the following invariant set of model (1) which is obtained in [8]
(17)Ω=S1,S2,V,E,I∈R+5:S1+S2+V+E+I≤Aτ+bμτ,and from the following inequalities
(18)bS2+V1+τS2+V≤bτ,Iβε+1≤βε+1Aτ+bμτ,and also cancel the items less than zero, so we have
(19)LV=A+bS2+V1+τS2+V+μ1+d+η1+βI+θ+4μ2+εβI+δ+η2+c−μ1+dS1−AS1−bS2+VS11+τS2+V−μ2S2+V+E+I−η1S1S2−δVS2−θS2V−βεVIE−βS2IE−cI−η2EI+σ12+σ22+σ32+σ42+σ522≤A+bτ+βε+1Aτ+bμτ+μ1+d+η1+θ+4μ2+δ+η2+c+σ12+σ22+σ32+σ42+σ522=ζ.Sinceζ is positive constant which is independent of S1,S2,V,E,I, and t, we can get
(20)dVS1S2,V,E,I≤ζdt+σ1S1−1dB1t+σ2S2−1dB2t+σ3V−1dB3t+σ4E−1dB4t+σ5I−1dB5t.Integrating both sides (20) from 0 to T∧τε and taking expectations, then we can obtain
(21)EVS1τε∧T,S2τε∧T,Vτε∧T,Eτε∧T,Iτε∧T≤VS10,S20,V0,E0,I0+ζT<∞.SetΩε=τε≤t for n≥n1 by (13), PΩn≥ε. Notice that for every ω∈Ωε, there is at least one of S1τε,ω,S2τε,ω,Vτε,ω,Eτε,ω, and Iτε,ω that equal either n or 1/n. Hence, S1τε,ω,S2τε,ω,Vτε,ω,Eτε,ω, and Iτε,ω are no less than
(22)n−1−lognor1n−1−logn.Consequently,(23)VS1τε,ω,S2τε,ω,Vτε,ω,Eτε,ω,Iτε,ω≥n−1−logn∧1n−1−logn,where a∧b donates the minimum of a and b. In view of (21) and (23) we have
(24)VS10,S20,V0,E0,I0+ζT≥E1ΩωVS1τε∧T,S2τε∧T,Vτε∧T,Eτε∧T,Iτε∧T≥δ~n−1−logn∧1n−1+logn,where 1Ωω is the indicator function of Ωn. Let n⟶∞ leads to the contradiction
(25)∞>VS10,S20,V0,E0,I0+ζT=∞.Therefore, we must haveτ∞=∞ a.s.


## 3.2. Stationary Distribution and Ergodicity

The difference between model (1) and the stochastic model is that the stochastic model does not have the endemic equilibrium. Hence, we cannot study the persistence of the disease by studying the stability of the endemic equilibrium and turn to check out the existence and uniqueness of the stationary distribution for the system (4) which implies the persistence of the disease in some sense. In this section, based on the theory of Has’minskii [17], we verify that there is an ergodic stationary distribution, which reveals the persistence of the disease.Define a parameter(26)R0s=η1η2βεθμ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/2.Theorem 3.
Assume thatR0s>1, then system (4) has a unique stationary distribution Π· and it has the ergodic property.Proof.
In view of Theorem2, we have obtained that for any initial value S10,S20,V0,E0,I0∈ℝ+5, there is a unique global solution S1,S2,V,E,I∈ℝ+5..
The diffusion matrix of system (4) is given by
(27)A=σ12S1200000σ22S2200000σ32V200000σ42E200000σ52I2.ChooseM=minS1,S2,V,E,I∈D¯σ⊂R+5σ12S12,σ12S22,σ32V2,σ42E2,σ52I2; one can get that
(28)∑i,j=15aijS1,S2,V,E,Iξiξj=σ12S12ξ12+σ12S22ξ22+σ32V2ξ32+σ42E2ξ42+σ52I2ξ52≥Mξ2,S1,S2,V,E,I∈D¯σ,ξ=ξ1,ξ2,ξ3,ξ4,ξ5∈ℝ+5.Then the conditionA1 in Lemma 1 is satisfied.Construct aC2-function Q:ℝ+5⟶ℝ in the following from
(29)QS1,S2,V,E,I=MS1+S2+V+E+I−c1lnS1−c2lnS2−c3lnE−c4lnI−lnV+1χ+1S1+S2+V+E+Iχ+1−lnS1−lnS2−lnE−lnV+S1+S2+V+E+I=MV1+V2+V3+V4+V5+V6,where χ is a constant satisfying 0<χ<2μ/σ12∨σ22∨σ32∨σ42∨σ52,
(30)c1=Aμ1+d+η1+σ12/2,c2=Aθ+μ2+σ22/2,c3=Aμ2+η2+σ42/2,c4=Aμ2+c+σ52/2,and M>0 satisfies the following condition
(31)−Mλ+C≤−2,where
(32)λ=5Aη1η2βεθμ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/21/5−1=5AR0s1/5−1>0,(33)C=supS1,S2,V,E,Iεℝ+5−12μ−12σσ12∨σ22∨σ32∨σ42∨σ52·S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422.It is easy to check that(34)liminfk⟶∞,S1,S2,V,E,I∈ℝ+5\UkQS1,S2,V,E,I=∞,where Uk=1/k,k×1/k,k×1/k,k×1/k,k×1/k,k. Furthermore, QS1,S2,V,E,I is a continuous function. Hence, QS1,S2,V,E,I must have a minimum point S¯10,S¯20,V¯0,E¯0,I¯0 in the interior of ℝ+5. Then we define a nonnegative C2-function V:ℝ+5⟶ℝ+ as follows:
(35)VS1,S2,V,E,I=QS1,S2,V,E,I−QS¯10,S¯20,V¯0,E¯0,I¯0.Making use of Ito^’s formula, we have
(36)LV1=−μ2S2+V+E+I+μ1+dS1+c1AS1+c1bS2+VS11+τS2+V+c2η1S1S2+c2δVS2+c3βεVIE+c3βS2IE+c4η2EI+θS2V−cI+A+bS2+V1+τS2+V+εβI+c2βI+μ2+δ+σ322+c1μ1+d+η1+σ122+c2θ+μ2+σ222+c3μ2+η2+σ422+c4μ2+c+σ522.Using the inequalitya+b+c+d+e≥5abcde5,a,b,c,d,e>0 leads to
(37)LV1≤−c1AS1+c2η1S1S2+c3βεVIE+c4η2EI+θS2V+A+bτ+εβI+c2βI+μ2+δ+σ322+c1μ1+d+η1+σ122+c2θ+μ2+σ222+c3μ2+η2+σ422+c4μ2+c+σ522≤−5c1c2c3c4Aη1βεη2θ1/5+A+bτ+εβI+c2βI+μ2+δ+σ322+c1μ1+d+η1+σ122+c2θ+μ2+σ222+c3μ2+η2+σ422+c4μ2+c+σ522=−5A5η1βεη2θμ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/21/5+5A+μ2+δ+bτ+βIc2+ε+σ322=−5Aη1βεη2θμ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/21/5−1+μ2+δ+bτ+βIc2+ε+σ322=−λ+μ2+δ+bτ+βIc2+ε+σ322,where λ is defined in (32).Similarly(38)LV2=S1+S2+V+E+IχA+bS2+V1+τS2+V−μ1+dS1−μ2S2+V+E+I−cI+12χS1+S2+V+E+Iχ−1×σ12S12+σ22S22+σ32V2+σ42E+σ52I2≤S1+S2+V+E+IχA+bτ−μS1+S2+V+E+I+12χS1+S2+V+E+Iχ+1σ12∨σ22∨σ32∨σ42∨σ52=A+bτS1+S2+V+E+Iχ−μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1+S2+V+E+Iχ+1.Then(39)LV2≤B−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1+S2+V+E+Iχ+1≤B−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1,where
(40)B=supS1,S2,V,E,I∈ℝ+5A+bτS1+S2+V+E+Iχ−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1+S2+V+E+Iχ+1<∞.We can also get(41)LV3=−AS1−bS2+VS11+τS2+V+μ1+d+η1+σ122,(42)LV4=−η1S1S2+βI+θ+μ2−δVS2+σ222,(43)LV5=−βεVIE−βS2IE+μ2+η2+σ422,(44)LV6=−θS2V+εβI+μ2+δ+σ322,(45)LV7=A+S2+V1+τS2+V−μ1+dS1−μ2S2+V+E+I−cI≤A+bτ−μ1+dS1−μ2S2+V+E+I.Hence, by (37)-(45), we obtain
(46)LV≤−Mλ+MβIc2+ε+1+εM+B−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1−AS1−bS2+VS11+τS2+V−η1S1S2−δVS2−βεVIE−βS2IE−θS2V−μ1+dS1−μ2S2+V+E+I+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422.Thus, we can construct a compact subsetD such that the condition A2 in Lemma 1 holds. Define the bounded closed set
(47)D=ε1≤S1≤1ε1,ε2≤S2≤1ε2,ε3≤I≤1ε3,ε4≤E≤1ε4,ε5≤V≤1ε5,where εi>0i=1,2,3,4,5 are sufficiently small constants satisfying the following conditions:
(48)−1ε1+F≤−1,(49)−Mλ−ε5ε2+C≤−1,(50)−Mλ+Mβε31+εM+c2+ε+C≤−1,(51)−βε2ε3ε4+F≤−1,(52)−θS2ε5+F≤−1,(53)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε1χ+1+G≤−1,(54)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε2χ+1+H≤−1,(55)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε4χ+1+J≤−1,(56)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε3χ+1+K≤−1,(57)−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε5χ+1+L≤−1,where F,G,H,J,K, and L are positive constants which can be seen from (60), (68), (70), (72), (74), and (76), respectively. Note that for sufficiently small εi,i=1,2,3,4,5. For convenience, we divide ℝ+5\D into ten domains
(58)D1=S1,S2,V,E,I∈ℝ+5:0<S1<ε1,D2=S1,S2,V,E,I∈ℝ+5:0<S2<ε2,V≥ε5,D3=S1,S2,V,E,I∈ℝ+5:0<I<ε3,D4=S1,S2,V,E,I∈ℝ+5:S2≥ε2,I≥ε3,0<E<ε4,D5=S1,S2,V,E,I∈ℝ+5:0<V<ε5S2≥ε2,D6=S1,S2,V,E,I∈ℝ+5:S1>1ε1,D7=S1,S2,V,E,I∈ℝ+5:S2>1ε2,D8=S1,S2,V,E,I∈ℝ+5:I>1ε3,D9=S1,S2,V,E,I∈ℝ+5:E>1ε4,D10=S1,S2,V,E,I∈ℝ+5:V>1ε5.Next, we will show thatLVS1,S2,V,E,I≤−1 on ℝ+5\D, which is equivalent to proving it on the above ten domains.Case 1.
IfS1,S2,V,E,I∈D1, one can get that
(59)LV≤−AS1+MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422≤−AS1+F≤−Aε1+F≤−1,
where(60)F=supS1,S2,V,E,I∈ℝ+5MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422.According to (48), we have
(61)LV≤−1,foranyS1,S2,V,E,I∈D1.Case 2.
IfS1,S2,V,E,I∈D2, we have
(62)LV≤−Mλ−δVS2−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422≤−Mλ−δVS2+C≤−Mλ−ε5ε2+C,where C is defined in (33).
In view of (49), we can obtain that for sufficiently small εii=2,5, LV≤−1 for any S1,S2,V,E,I∈D2.Case 3.
IfS1,S2,V,E,I∈D3, one can see that
(63)LV≤−Mλ+MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422.We obtain that(64)LV≤−1foranyS1,S2,V,E,I∈D3.Case 4.
IfS1,S2,V,E,I∈D4, one can see that
(65)LV≤−βS2IE+MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422≤−βS2IE+F≤−βε2ε3ε4+F.In view of (51), we can obtain that for sufficiently small εii=3,4,LV≤−1 for any S1,S2,V,E,I∈D4.Case 5.
IfS1,S2,V,E,I∈D5, one can see that
(66)LV≤−θS2V+MβIc2+ε+1+εM−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+σ12+σ22+σ32+σ422≤−θS2V+F≤−θε2ε5+F.We can obtain that for sufficiently smallεii=2,5, LV≤−1 for any S1,S2,V,E,I∈D5.Case 6.
IfS1,S2,V,E,I∈D6, one can see that
(67)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+G≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε1χ+1+G,where
(68)G=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S2χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.By (53), we conclude that LV≤−1onD6.Case 7.
IfS1,S2,V,E,I∈D7, one can see that
(69)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S2χ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S2χ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S2χ+1+H≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε2χ+1+H,where
(70)H=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52S2χ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+Vχ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.Together with (54), we can deduce that LV≤−1onD7.Case 8.
IfS1,S2,V,E,I∈D8, one can see that
(71)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Eχ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×Eχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Eχ+1+J≤−14μ−12θσ12∨σ22∨σ32∨σ42∨σ521ε4χ+1+J,where
(72)J=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Eχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Vχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.which together with (55) implies that LV≤−1onD8.Case 9.
IfS1,S2,V,E,I∈D9, we obtain
(73)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Iχ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×Iχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Vχ+1+Eχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Iχ+1+K≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε3χ+1+K,where
(74)K=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Iχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Vχ+1+Eχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.By (56), we can conclude that LV≤−1 on D9.Case 10.
IfS1,S2,V,E,I∈D10, it follows that
(75)LV≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Vχ+1−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52×Vχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52S1χ+1+S2χ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Vχ+1+L≤−14μ−12χσ12∨σ22∨σ32∨σ42∨σ521ε5χ+1+L,where
(76)L=supS1,S2,V,E,I∈ℝ+5−14μ−12χσ12∨σ22∨σ32∨σ42∨σ52Vχ+1−12μ−12χσ12∨σ22∨σ32∨σ42∨σ52×S1χ+1+S2χ+1+Eχ+1+Iχ+1+μ1+d+η1+θ+3μ2+η2+δ+A+bτ+Mσ322+μ2+δ+bτ+MβIc2+ε+1+εM+σ12+σ22+σ32+σ422.Combining with (57) yields LV≤−1 on D10.Obviously,A2 in Lemma 2.1 is satisfied. According to Lemma 2.1, we can obtain that system (4) is ergodic and has a unique stationary distribution.Remark 4.
Theorem3 reveals that system (4) has a unique ergodic stationary distribution π· if R0s=η1η2βεθ/μ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/2>1. Note that the expression of R0s coincide with the threshold R0 of the deterministic system (1) if there is no stochastic perturbation. This shows that we generalize the result of the deterministic system.


## 3.3. Extinction of the Disease

As it is well known, one of the main concern of epidemiology is how we regulate the disease dynamics in order to eradicate the disease in the long term. Moreover, in [10], Allen et al. proposed and studied several types of stochastic epidemic models and pointed out that the stochastic models should suit the question of disease extinction better. Hence, in this section, we shall establish some sufficient conditions for extinction of the disease in stochastic model (4).Theorem 5.
LetS1t,S2t,Vt,Et,It be the solution of system (4) with any initial value S10,S20,V0,E0,I0∈ℝ+5. If βε+1Aτ+bμ2+η2/μτη2<σ52/2+μ2+c∧σ42/2, then the disease It will extinct exponentially with probability one, i.e., moreover
(77)limt⟶∞sup1tlogE+μ2+η2η2I≤βη2ε+1Aτ+bμ2+η2μτ−η2μ2+η22σ522+μ2+c∧σ422.Proof.
Applying Ito^’s formula to logE+μ2+η2/η2I, we have
(78)dlogE+μ2+η2η2I=βεV+S2IE+μ2+η2/η2I−μ2+η2μ2+cIE+μ2+η2/η2Iη2−σ42E2+σ52μ2+η2/η2I22E+μ2+η2/η2I2dt+σ4EE+μ2+η2/η2IdB4t+σ5μ2+η2Iη2E+μ2+η2/η2IdB5t=βεV+S2IE+μ2+η2/η2Idt−1E+μ2+η2/η2I2μ2+η2η2μ2+cIE+σ522+μ2+cμ2+η2η22I2+σ42E22dt+σ4EE+μ2+η2/η2IdB4t+σ5μ2+η2Iη2E+μ2+η2/η2IdB5t≤βεV+S2η2μ2+η2dt−1E+μ2+η2/η2I2σ522+μ2+cμ2+η2η22I2+σ42E22dt+σ4EE+μ2+η2/η2IdB4t+σ5μ2+η2Iη2E+μ2+η2/η2IdB5t≤βη2ε+1Aτ+bμ2+η2Aτdt−η2μ2+η22σ522+μ2+c∧σ422dt+σ4EE+μ2+η2/η2IdB4t+σ5μ2+η2Iη2E+μ2+η2/η2IdB5t.Integrating the above inequality from 0 tot, and the fact that limt⟶∞Bit/t=0,i=4,5 [16], yields
(79)limt⟶∞sup1tlogE+μ2+η2η2I≤βη2ε+1Aτ+bμ2+η2μτ−η2μ2+η22σ522+μ2+c∧σ422=x0.For anyε¯>0, and almost ω∈Ω,∃T=Tω such that
(80)It≤η2μ2+η2expx0+ε¯t,∀t≥T.Remark 6.
Theorem5 suggests that the disease will become extinct if βε+1Aτ+bμ2+η2/μτη2<σ52/2+μ2+c∧σ42/2.


## 4. Numerical Examples

In this section, we will give two numerical examples to illustrate the main theoretical results obtained in this paper. The numerical simulation method can be found in [9, 22, 23]. The following is a corresponding discrete equations of system (4):
(81)S1k+1=S1k+A+bS2k+Vk1+τS2k+Vk−μ1+d+η1S1kΔt+S1kσ1ξkΔt+12σ12ξk2−1Δt,S2k+1=S2k+η1S1k−βS2kIk−θS2k−μ2S2k+δVkΔt+S2kσ2ξkΔt+12σ22ξk2−1Δt,Vk+1=Vk+θS2k−εβVkIk−μ2+δVkΔt+Vkσ3ξkΔt+12σ32ξk2−1Δt,Ek+1=Ek+βεVkIk+βS2kIk−μ2+η2EkΔt+Ekσ4ξkΔt+12σ42ξk2−1Δt,Ik+1=Ik+η2Ek−μ2+cIkΔt+Ikσ5ξkΔt+12σ52ξk2−1Δt,where ξkk=1,2,⋯ are the Gaussian random variables which follow standard normal distribution N0,1, and σi,1≤i≤5, are intensities of white noises.Example 1.
We take parameters asA=45, β=0.05, μ1=0.05, μ2=0.06, ε=3.2, τ=0.01, b=0.3, η1=0.25, η2=0.5, and θ=1.1, δ=0.01, c=0.02, σ1=0.2, σ2=0.05, σ3=0.31, σ4=0.03, and σ5=0.02. It is clear that conditions of Theorem 3 are satisfied; by calculating, we have the basic reproduction number R0s=η1η2βεθ/μ1+d+η1+σ12/2θ+μ2+σ22/2μ2+η2+σ42/2μ2+c+σ52/2=1.17080128The histogram and the smoothing curves of the probability density functions ofS1t,S2t,Vt,Et,It are given in Figure 1.Example 2.
We take parameters asA=1000, β=0.0001, μ1=0.1, μ2=0.25ε=0.18, τ=0.002, b=1.5, η1=1.06, η2=3.4, and θ=0.1, δ=0.4, c=0.05, σ1=0.2, σ2=0.5, σ3=0.31, σ4=1.65, and σ5=1.5. It is clear that conditions of Theorem 5 are satisfied.Figure 1
Dynamic behaviors of the system.
(a)(b)(c)(d)(e)The curves on the persistence ofS1t,S2t,Vt,Et and extinction of It for stochastic model (4) are given in Figure 2, where the initial value is S10,S20,V0,E0,I0=1,1.5,1,1,1.Remark 7.
In this paper, we consider the stochastic perturbations for deterministic model (1) and derived model (4). Thus, model (4) can be specialized as models (1). Hence, model (4) can be seen as a general model compared to model (1), and the theoretical results obtained in this article can be seen as the extensions and supplements of the model and the theoretical results obtained in [8].Figure 2
Dynamic behaviors of the system.
(a)(b)(c)(d)(e)


## 5. Conclusion

In this paper, firstly, we have considered the stochastic perturbations for deterministic system (1) and established corresponding stochastic system (4). Secondly, under the condition R0s>1 and applying the theory of stochastic differential equations, Has’minskii theory, Ito’s formula, and Lyapunov function method, we obtained some sufficient conditions on the existence of ergodic stationary distribution of model (4). We also established sufficient conditions on the extinction of the disease. Finally, two examples are presented to validate the main results of this paper. The results obtained in this paper suggest that stochastic perturbations have remarkable effects on the disease in model (4). Especially, from the numerical simulations, we can see that, under the stochastic perturbations, the disease in the stochastic system will become extinct more quickly than the corresponding deterministic one.



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