# Dynamics of a Stochastic SIRS Epidemic Model with Regime Switching and Specific Functional Response

Authors: Amine EL Koufi; Abdelkrim Bennar; Noura Yousfi
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

The purpose of this work is to investigate the dynamic behaviors of the SIRS epidemic model with nonlinear incident rate under regime switching. We establish the existence of a unique positive solution of our system.  Furthermore, we obtain the conditions for the extinction of diseases, and we show the existence of the stationary distribution for our stochastic SIRS model under regime switching. Numerical simulations are employed to illustrate our theoretical analysis.


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## Body

## 1. Introduction

Several of mathematicians have developed various epidemic models to prevent and control the spread of transmissible diseases in the community.The classical SIR model presented by Kermack and McKendrick [1] has played an important role in mathematical epidemiology. The SIR model are used to study the disease spread between three groups of population to know the susceptible S, the infective I, and the recovered R.In this work, we introduce a switched stochastic SIRS epidemic model with specific functional response. Then, we consider the following deterministic SIRS epidemic model with specific functional response:(1)dSdt=Λ−μS−βSI1+α1S+α2I+α3SI+γR,dIdt=βSI1+α1S+α2I+α3SI−μ+λ+δI,dRdt=λI−μ+γR,where St denotes the number of susceptible individuals, It denotes the number of infective individuals, and Rt represents the number of removed individuals. Λ is the recruitment rate of the population, μ is the natural death rate of the population, γ is the rate at which recovered individuals loss immunity and return to the susceptible class, λ denotes the natural recovery rate of the infectious individuals, and δ denotes the disease inducing death rate. The infection transmission process in (1) is modeled by the specific functional response βSI/1+α1S+α2I+α3SI, where β is the transmission coefficient between compartments S and I, and α1, α2, α3≥0 are the saturation factors measuring the psychological or inhibitory effect. In addition, this functional response generalizes many common types existing in the literature such as the Crowley–Martin functional response introduced in [2] and used in [3] when α3=α1α2 and the Beddington–DeAngelis functional response proposed in [4] and used in [5] when α3=0.Environmental fluctuations have been indicated to play an important role in the propagation of disease [6, 7]. In effect, disease infestation is highly stochastic, and stochastic noise can raise the probability of disease extinction in the early phase of epidemics. By running an ODE system, we can get only a certain sample solution, whereas by running an SDE system, we can obtain the stochastic distribution of disease dynamics [8]. Lately, dynamic modeling of infectious diseases based on stochastic differential equations (SDE) has received considerable attention from experts and academics [9–11]. The SIRS epidemic model with white noise is expressed by(2)dS=Λ−μS−βSI1+α1S+α2I+α3SI+γRdt+σ1SdB1t,dI=βSI1+α1S+α2I+α3SI−μ+λ+δIdt+σ2IdB2t,dR=λI−μ+γRdt+σ3RdB3t,where B1t,B2t, andB3t are independent Brownian motions and σ1,σ2, andσ3 are their intensities. Besides environment white noise, in this paper, we will also consider another noise, namely, telegraph noise ([12–15]). The latter can be described as a switching between two or more regimes of environment, which differ in terms of factors such as nutrition, climatic characteristics, or sociocultural factors. The switching among different environments is memoryless and the waiting time for the next switch is exponentially distributed. The regime switching can hence be modeled by a finite-state Markov chain ςt,t≥0 taking values in a finite-state space M=1,2,…,N. The stochastic system (1) with regime switching can be described by the following model:(3)dS=Λςt−μςtS−βςtSI1+α1ςtS+α2ςtI+α3ςtSI+γςtRdt+σ1ςtSdB1t,dI=βςtSI1+α1ςtS+α2ςtI+α3ςtSI−μςt+λςt+δςtIdt+σ2ςtIdB2t,dR=λςtI−μςt+γςtRdt+σ3ςtRdB3t.Throughout this paper, we letΩ,ℱ,ℱtt≥0,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). Let ςt,t≥0 be a right-continuous Markov chain on the probability space Ω,ℱ,ℱtt≥0,ℙ taking values in a finite-state space M=1,2,…,N with the generator Φ=ϕuv1≤u,v≤N given, for δ>0, by(4)Pςt+δ=vςt=u=ϕuvδ+oδ,ifu≠v,1+ϕuuδ+oδ,ifu=v.Here,ϕuv is the transition rate from u to v and ϕuv≥0 if u≠v, while(5)ϕuu=−∑u≠vϕuv.Suppose that the Markov chainςt is independent of the Brownian motion B⋅ and it is irreducible. Under this condition, the Markov chain has a unique stationary (probability) distribution π=π1,…,πN, which can be determined by solving the linear equation πΦ=0, subject to ∑i=1Nπi=1, and πi>0, ∀i∈M. Thereafter, for any vector h=h1,…,hNT, let h^=i∈Mminhi and hˇ=i∈Mmaxhi.The rest of the paper is organized as follows. In Section2, we show that there exists a unique global positive solution of system (3). In Section 3, we give sufficient conditions for the extinction of the disease. In Section 4, sufficient conditions for the existence of the ergodic stationary distribution are established for model (3). Finally, numerical simulations are carried out to support the theoretical results.


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

In this section, we will prove that model (3) has a unique global positive solution. We also denote(6)ℝ+3=x1,x2,x2xi>0,i=1,2,3.Thus, we established the following theorem.Theorem 1.
For any given initial value,X0=S0,I0,R0,ς0∈ℝ+3×M, there is a unique positive solution Xt=St,It,Rt,ςt∈ℝ+3×M of model (3) on t≥0 and the solution will remain in ℝ+3×M with probability 1, namely, St,It,Rt,ςt∈ℝ+3×M for all t≥0 almost surely.Proof.
Since the coefficients of system (3) are locally Lipshitz continuous, for any initial value X0∈ℝ+3, there exists a unique local solution Xt on t∈0,τe, where τe is the explosion time. We need to show that this solution is global almost surely that is, τe=∞a.s. Let m0 be sufficiently large such that every component of X0 lies within the interval 1/m0,m0. For each integer m≥m0, define a sequence of stopping times by(7)τm=inft∈0,τe:St∉1m,morIt∉1m,morRt∉1m,m.
We setinf∅=∞, where ∅ denotes the empty set. Obviously, τm is increasing as m⟶∞. Set τ∞=limm⟶∞τm with τ∞≤τea.s. Now, we need to show τ∞=∞a.s. If this statement is violated, then there exist T>0 and ε∈0,1 such that(8)ℙτ∞≤T>ε.
Hence, there is an integerm1≥m0 such that(9)ℙτ∞≤T≥εfor allm≥m1.
Define aC2−function V1:ℝ+3⟶ℝ+ as(10)V1S,I,R=S−1−lnS+I−1−lnI+R−1−lnR.
By Itô’s formula, we have(11)dV1=ℒV1dt+S−1dB1t+I−1dB2t+R−1dB3t,where(12)ℒV1=1−1SΛk−μkS−βkSI1+α1kS+α2kI+α3kSI+γkR+1−1IβkSI1+α1kS+α2kI+α3kSI−μk+λk+δkI+1−1RλkI−μk+γkR+σ12k2+σ22k2+σ32k2,which implies that(13)ℒV1=Λk−μkS−ΛkS+μk+βkI1+α1kS+α2kI+α3kSI−γkRS+σ12k2−μk+δkI−βkS1+α1kS+α2kI+α3kSI+μk+λk+δk+σ22k2−μkR−λkIR+μk+γk+σ32k2≤Λk+3μk+λk+δk+γk+σ12k+σ22k+σ32k2βkI1+α1kS+α2kI+α3kSI≤Λˇ+3μˇ+λˇ+δˇ+γˇ+σˇ12+σˇ22+σˇ322+βˇα^2≔K.
Hence,(14)dV1S,I,R≤Kdt+S−1dB1t+I−1dB2t+R−1dB3t.
Integrating both sides of the above inequality from 0 toτm∧T, and taking the expectations, we get(15)EV1Sτm∧T,Iτm∧T,Rτm∧T≤V1S0,I0,R0+KT.
SetΩm=τm≤T for m≥m1 and by (4), we have ℙτ∞≤T≥ε for each m≥m1. For every ω∈Ωm, we have(16)V1Sτm∧T,Iτm∧T,Rτm∧T≥minm−1−lnm,1m−1−1ln1m≔Lm.
Then, we obtain(17)V1S0,I0,R0+KT≥E1ΩmV1Sτm∧T,Iτm∧T,Rτm∧T≥εLm,where 1Ωm is the indicator function of Ωm. Letting m⟶+∞ leads to the contradiction ∞=V1S0,I0,R0+KT<∞. So, we must therefore have τ∞=∞a.s. This completes the proof.


## 3. Extinction

Our goal in this section is to study the extinction and give the extinction threshold of system (3). Then, the following theorem gives a sufficient condition for extinction of the disease.Theorem 2.
Ifℛ0s=∑k=1Nπkβk/α^1∑k=1Nπkμk+λk+δk+σ2k/2<1, then the disease I tends to zero exponentially with probability one, i.e.,(18)lim supt⟶∞lnItt≤∑k=1Nπkμk+λk+δk+σ2k2ℛ0s−1.Proof.
Applying Itô’s formula, we can get(19)d lnIt=βςtS1+α1ςtS+α2ςtI+α3ςtSI−μςt+λςt+δςt+σ2ςt2+σ2ςtdB2t.
Integrating (19) from 0 to t and then dividing by t into both sides leads to(20)lnIt−lnI0t=1t∫0tβςsS1+α1ςsS+α2ςsIs+α3ςtSsIsds−1t∫0tμςs+λςs+δςs+σ2ςs2ds+1t∫0tσ2ςsdB2s≤1t∫0tβςsα^1−μςs+λςs+δςs+σ2ςs2ds+1t∫0tσ2ςsdB2s=1t∫0tβςsα^1−μςs+λςs+δςs+σ2ςs2ds+Mtt,where Mt is a local martingale defined by Mt=∫0tσ2ςsdB2s, whose quadratic variation is M,Mt=∫0tσ22ςsds≤σ22ˇt. Making use of the strong law of large numbers for martingales ([16]) yields(21)limt⟶∞Mtt=0,a.s.
Taking the superior limit on both sides of (20) and applying the ergodicity of Markov chain ςt, we get(22)lim supt⟶∞lnItt≤∑k=1Nπkβkα^1−μk+λk+δk+σ2k2=∑k=1Nπkμk+λk+δk+σ2k2×∑k=1Nπkβkα^1∑k=1Nπkμk+λk+δk+σ2k/2−1=∑k=1Nπkμk+λk+δk+σ2k2ℛ0s−1<0,which implies that limt⟶∞It=0. This completes the proof of theorem.


## 4. Existence of Ergodic Stationary Distribution

In this section, we shall discuss sufficient conditions for the existence of an ergodic stationary distribution to model (3). The following lemma gives a criterion for positive recurrence in terms of lyapunov function [17].LetXt,ςt is the diffusion Markov process and satisfy the following equation(23)dXt=fXt,ςtdt+σXt,ςtdBt,X0=X0,ς0=ς,where f:ℝn×M⟶ℝn, σ:ℝn×M⟶ℝn×d satisfying σx,kσTx,k=dijx,k. For each k∈M, and for any twice continuously differentiable function V.,k, the operator ℒ can be defined by(24)ℒVx,k=∑i=1nfix,k∂Vx,k∂xi+12∑i,j=1ndijx,k∂2Vx,k∂xi∂xj+∑l=1nϕklVx,l.Lemma 1.
If the following conditions are satisfied:(1)
ϕij>0 for any i≠j.(2)
For eachk∈M, Dx,k=dijx,k is symmetric and satisfies(25)λζ2≤Dx,kζ,ζ≤λ−1ζ2for allζ∈ℝn,with some constantλ∈0,1 for all x∈ℝn.(3)
There exists a nonempty open setD with compact closure, satisfying that, for each k∈M, there is a nonnegative function V.,k : Dc⟶ℝ such that V.,k is twice continuously differential and that for some α>0,(26)LVx,k≤−αfor allx,k∈Dc×M.Then,Xt,ςt of system (23) is positive recurrent and ergodic. That is to say, there exists a unique stationary distribution μ.,. such that for any Borel measurable function f:ℝn×M⟶ℝ satisfying(27)∑k=1N∫ℝnfx,kμdx,k<∞,we have(28)ℙlimt⟶∞1t∫0tfXs,ςsds=∑k=1N∫ℝnfx,kμdx,k=1.Theorem 3.
Assume thatℛ0s>1, then for any initial value S0,I0,R0,ς0∈ℝ+3×M, the solution St,It,Rt,ςt of system (3) admits a unique ergodic stationary distribution.Proof.
In order to prove Theorem3, it is sufficient to prove conditions (1), (2) and (3) in Lemma 1. Assumption ϕij>0 for any i≠j, in Section 1 implies that condition (1) in Lemma 1 is satisfied. To verify condition (2), consider the bounded open subset(29)D=1η,η×1η,η×1η,η⊂ℝ+3,where η is a sufficiently large number. Then, D¯⊂ℝ+3. We have DS,I,R,k=QS,I,R,kQTS,I,R,k, where QS,I,R,k=diagSσ1k,Iσ2k,Rσ3k, k∈M. Then, DS,I,R,k is positive semi-definite, and since RS,I,R,k is a nonsingular matrix, we deduce that DS,I,R,k is positive definite. Hence,(30)λmaxDS,I,R,k≥λminDS,I,R,k>0,in addition, we have for all ζ∈D,(31)λminDS,I,R,kζ2≤ζTDS,I,R,kζ≤λmaxDS,I,R,kζ2.
It is easy to see thatλminDS,I,R,k and λmaxDS,I,R,k are two continuous functions of S,I, and R. Therefore, λ^=minD¯×MλminDS,I,R,k>0 and λ^=maxD¯×MλmaxDS,I,R,k>0, which implies that(32)λζ2≤ζTDS,I,R,kζ≤λ−1ζ2,where λ=minλ^,λˇ−1. Then, condition (2) in Lemma 1 is verified.
Now, we verify condition3. Define a C2−function V˜:ℝ+3×M⟶ℝ by(33)V˜S,I,R,k=−MlnI+ωk−lnS−lnR+S+I+R.
In addition,V˜S,I,R,k is a continuous function on with respect to S,I,R. So there is a unique minimum value point V˜S¯,I¯,R¯,k¯ of V˜S,I,R,k in the interior of ℝ+3×M; then, we define a nonnegative C2-function ℋ: ℝ+3×M⟶ℝ+ as follows:(34)ℋS,I,R,k=V˜S,I,R,k−V˜S¯,I¯,R¯,k¯=−MlnI+ωk−lnS−lnR+S+I+R−V˜S¯,I¯,R¯,k¯≔−MV2−V3−V4+V5,where S,I,R,k∈1/n,n×1/n,n×1/n,n×M and n>1 is a sufficiently large number, V2=−MlnI+ωk, V3=−lnS,V4=−lnR,V5=S+I+R−V˜S¯,I¯,R¯,k¯, and M>0 satisfies the following condition:(35)−M∑k=1Nπkμk+λk+δk+σ2k2ℛ0s−1+Mβˇα^1+2μˇ+γˇ+Λˇ+σˇ12+σˇ32≤−2.
Applying Itô’s formula toV2 leads to(36)ℒV2=−βkS1+α1kS+α2kI+α3kSI+μk+λk+δk+σ2k2−∑l=1Nϕklωl=−βkα^1+μk+λk+δk+σ2k2−∑l=1Nϕklωl+βk1α^1−S1+α1kS+α2kI+α3kSI≤−βkα^1+μk+λk+δk+σ2k2−∑l=1Nϕklωl+βkα^1≔−R0k−∑l=1Nϕklωl+βkα^1,where R0k=βk/αˇ1−μk−λk−δk−σ2k/2. Since the generator matrix Φ is irreducible, then for R0=R01,R02,…,R0NT, there exists ω=ω1,ω2,…,ωNT solution of the Poisson system ([18]):(37)Φω=−R0+∑l=1NπlR0lF⟶,where F⟶ denotes the column vector with all its entries equal to 1. Then,(38)ℒV2≤−∑k=1NπkR0k+βkαˇ1=−∑k=1Nπkμk+λk+δk+σ2k2ℛ0s−1+βˇα^1.
Next, we calculateℒV3(39)ℒV3=−ΛkS+μk+βkI1+α1kS+α2kI+α3kSI−γkRS+σ1k2≤−Λ^S+μˇ+βˇI1+α^1S+α^2I+α^3SI−γ^RS+σˇ12,and ℒV4(40)ℒV4=−λkIR+μk+γk+σk322≤−λ^IR+μˇ+γˇ+σ32ˇ2.
We have(41)ℒV5=Λk−μkS+I+R−δkI≤Λk−μkS+I+R≤Λˇ−μ^S+I+R.
The differential operatorℒ acting on the function ℋ yields(42)ℒℋ≤−M∑k=1Nπkμk+λk+δk+σ2k2ℛ0s−1+Mβˇα^1−Λ^S+βˇI1+α^1S+α^2I+α^3SI−γ^RS−μ^S+I+R−λ^IR+Λˇ+2μˇ+γˇ+σˇ322+σˇ122.
Defining the following compact set(43)Dε=S,I,R∈ℝ+3:ε≤S≤1ε,ε2≤I≤1ε2,ε3≤R≤1ε3,where ε is a sufficiently small number. In the set ℝ+3/Dε, we can choose ε sufficiently small such that the following conditions hold:(44)−Λ^ε+C1≤−1,ε≤α^1βˇ,−λ^ε+C≤−1,−μ^ε+C≤−1,−μ^ε2+C≤−1,−μ^ε3+C≤−1,C=β^α^2+Mβˇα^1+Λˇ+2μˇ+γˇ+σˇ322+σˇ122.
Next, we can divideℝ+3/Dε into the following six domains:(45)D1=S,I,R∈ℝ+3:S≤ε,D2=S,I,R∈ℝ+3:ε≤S,≤I≤ε2,D3=S,I,R∈ℝ+3:ε2≤I,R≤ε3,D4=S,I,R∈ℝ+3:S>1ε,D5=S,I,R∈ℝ+3:I>1ε2,D6=S,I,R∈ℝ+3:R>1ε3.
Clearly,Dεc=D1∪D2∪D3∪D4∪D5∪D6. Next, we will prove that(46)ℒℋS,I,R,k≤−1onℝ+3/Dε×M.Case 1.
IfS,I,R,k∈D1×M, we have that(47)ℒℋ≤−Λ^S+βˇI1+α^1S+α^2I+α^3SI+Mβˇα^1+Λˇ+2μˇ+γˇ+σˇ322+σˇ122≤−Λ^S+C≤−Λ^ε+C,which together with 9 implies that(48)ℒℋ≤−1for allS,I,R,k∈D1×M.Case 2.
IfS,I,R,k∈D2×M, we have(49)ℒℋ≤−M∑k=1Nπkμk+λk+δk+σ2k2ℛ0s−1+Mβˇα^1+βˇI1+α^1S+α^2I+α^3SI+Λˇ+2μˇ+γˇ+σˇ322+σˇ122≤−M∑k=1Nπkμk+λk+δk+σ2k2ℛ0s−1+Mβˇα^1+2μˇ+γˇ+Λˇ+σˇ12+σˇ32+βˇεα^1≤−2+βˇεα^1.
Combining with10, it can be achieved that(50)ℒℋ≤−1for allS,I,R,k∈D2×M.Case 3.
IfS,I,R,k∈D3×M, we get(51)ℒℋ≤−λ^IR+βˇI1+α^1S+α^2I+α^3SI+Mβˇα^1+Λˇ+2μˇ+γˇ+σˇ322+σˇ122≤−λ^IR+C≤−λ^ε+C≤−1,which follows from 11.Case 4.
IfS,I,R,k∈D4×M, it follows that(52)ℒℋ≤−μ^S+βˇI1+α^1S+α^2I+α^3SI+Mβˇα^1+Λˇ+2μˇ+γˇ+σˇ322+σˇ122≤−μ^S+C≤−μ^ε+C,which together with (44) implies that(53)ℒℋ≤−1for allS,I,R,k∈D4×M.Case 5.
IfS,I,R,k∈D5×M, we obtain(54)ℒℋ≤−μ^I+βˇI1+α^1S+α^2I+α^3SI+Mβˇα^1+Λˇ+2μˇ+γˇ+σˇ322+σˇ122≤−μ^I+C≤−μ^ε2+C≤−1,which follows from (44). Hence,(55)ℒℋ≤−1for allS,I,R,k∈D5×M.Case 6.
IfS,I,R,k∈D6×M, we have(56)ℒℋ≤−μ^R+βˇI1+α^1S+α^2I+α^3SI+Mβˇα^1+Λˇ+2μˇ+γˇ+σˇ322+σˇ122≤−μ^R+C≤−μ^ε3+C.
In view of (44), we arrive at(57)ℒℋ≤−1for allS,I,R,k∈D6×M.
Therefore, we have proof that(58)ℒℋ≤−1for allS,I,R,k∈ℝ+3/Dε×M.
Thus, condition 3 in Lemma1 has been satisfied, and system (3) has a unique stationary distribution and ergodicity holds. This completes the proof.Remark 1.
Assume the conditionℛ0s<1 holds. Disease I goes to extinction exponentially with probability one, Theorem 2, and if ℛ0s>1 there is a unique ergodic stationary distribution μ(⋅,⋅) of system (3), which implies that disease I persists Theorem 3. Then, the number ℛ0s can be considered as a threshold to identifying the stochastic extinction and persistence of system (3).


## 5. Simulations

Numerical solutions of stochastic differential equations are very important in the study of real examples of epidemic. In this section, we present some numerical results to illustrate the theoretical one. For numerical simulations of the SDEMS model (3), we use the Euler–Maruyama (EM) method ([19]).Letςt be a right-continuous Markov chain taking values on the state space ℳ=1,2,3 with the generator(59)Γ=−5141−1013−4.Obviously, the Markov chainςt has a unique stationary distribution π=π1,π2,π3=0.1666,0.6666,0.1666. Given a step size Δ=0.0001, the Markov chain can be simulated by computing the one-step transition probability matrix P=eΔΓ ([20]), and the transition probability matrix is given by(60)P=0.99950.00010.00040.00010.999900.00010.00030.9996.Figure1 shows a result of one simulation run of the Markov chain ςt.Figure 1
Computer simulation of a single path ofςt with initial value ς0=2.Example 1.
To illustrate Theorem2, we choose the parameter values in system (3) as follows:(61)Λ1=1,μ1=0.2,β1=0.1,γ1=0.25,λ1=0.1,δ1=0.05,α11,α21,α31=0.1,0.1,0.1,σ11,σ21,σ31=0.5,0.7,0.8,Λ2=1.1,μ2=0.21,β2=0.12,γ2=0.26,λ2=0.09,δ2=0.056,α12,α22,α32=0.1,0.1,0.1,σ12,σ22,σ32=0.7,0.8,0.6,Λ3=1.3,μ3=0.19,β3=0.13,γ3=0.24,λ3=0.11,δ3=0.058,α13,α23,α33=0.1,0.1,0.1,σ13,σ23,σ33=0.8,0.6,0.7.
Simple computations result,(62)ℛ0s=∑i=1Nπiβiα^1∑i=1Nπiμi+λi+δi+σ2i/2=0.9505<1,as a consequence result of Theorem 2. Disease I dies out exponentially with probability one. Figure 2 confirms this.Figure 2
The solution (S(t), I(t), and R(t)) of stochastic model (3) with (S(0), I(0), and R(0)=(0.7, 0.2, and 0).
(a)
(b)
(c)Example 2.
For this example, we have(63)Λ1=1,μ1=0.2,β1=0.1,γ1=0.25,λ1=0.1,δ1=0.05,α11,α21,α31=0.14,0.1,0.1,σ11,σ21,σ31=0.12,0.1,0.12,Λ2=1.1,μ2=0.21,β2=0.12,γ2=0.26,λ2=0.09,δ2=0.056,α12,α22,α32=0.14,0.1,0.1,σ12,σ22,σ32=0.15,0.18,0.19,Λ3=1.3,μ3=0.19,β3=0.13,γ3=0.24,λ3=0.11,δ3=0.058,α13,α23,α33=0.14,0.1,0.1,σ13,σ23,σ33=0.14,0.17,0.13.
By calculating, we find(64)ℛ0s=∑i=1Nπiβiα^1∑i=1Nπiμi+λi+δi+σ2i/2=1.9624>1.
Then according to Theorem3, the solution (S(t), I(t), and R(t)) of system (3) with any initial value (S(0), I(0), and R(0)) = (0.7, 0.6, and 0) has a unique stationary distribution, and it has the ergodic property, that is, the epidemic disease is permanent. Figures 3–5 confirm this.Figure 3
The solution S(t) of stochastic model (3) and its histogram.
(a)
(b)Figure 4
The solution I(t) of stochastic model (3) and its histogram.
(a)
(b)Figure 5
The solution R(t) of stochastic model (3) and its histogram.
(a)
(b)


## 6. Conclusion

This article discusses the dynamic behavior of a SIRS epidemic model with a regime switching and nonlinear incidence rate. We obtain sufficient conditions for the extinction of system (3) if ℛ0s<1. We prove the stochastic system (3) under regime switching has a unique stationary distribution which is ergodic and positive recurrent by using the Lyapunov function method. In future works, it is interesting to study the effect of Lévy noise and a color noise (telegraph noise) in the stochastic SIRS epidemic model (2). We will investigate this case in our future works.



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*Source: 5898456-2020-04-27.xml*