
In this paper, a stochastic SIB(Susceptible-Infected-Vibrios) cholera model with saturation recovery rate and Ornstein-Uhlenbeck process is investigated. It is proved that there is a unique global solution for any initial value of the model. Furthermore, the sufficient criterion of the stationary distribution of the model is obtained by constructing a suitable Lyapunov function, and the expression of probability density function is calculated by the same condition. The correctness of the theoretical results is verified by numerical simulation, and the specific expression of the marginal probability density function is obtained.
Citation: Buyu Wen, Bing Liu, Qianqian Cui. Analysis of a stochastic SIB cholera model with saturation recovery rate and Ornstein-Uhlenbeck process[J]. Mathematical Biosciences and Engineering, 2023, 20(7): 11644-11655. doi: 10.3934/mbe.2023517
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In this paper, a stochastic SIB(Susceptible-Infected-Vibrios) cholera model with saturation recovery rate and Ornstein-Uhlenbeck process is investigated. It is proved that there is a unique global solution for any initial value of the model. Furthermore, the sufficient criterion of the stationary distribution of the model is obtained by constructing a suitable Lyapunov function, and the expression of probability density function is calculated by the same condition. The correctness of the theoretical results is verified by numerical simulation, and the specific expression of the marginal probability density function is obtained.
Cholera is an acute intestinal infectious disease caused by Vibrio cholerae, which is mainly transmitted through unclean water and food (see [1]). According to news reports, more than 20 countries and regions (southwestern Cameroon, Afghanistan, the Republic of Mozambique, northern Iraq, Haiti, Malawi, etc.) have experienced or are experiencing cholera epidemics in the past two years. Therefore, many scholars have studied the cholera transmission model, such as differential dynamics system [2,3,4,5], age-structured transmission model [6], reaction-convection-diffusion equations [7], generalized fractional model [8].
In [9], the scholars investigated the following SIB cholera model:
S′(t)=A−βSBK+B−μ1S+γI+cIb+I,I′(t)=βSBK+B−(μ1+γ+α)I−cIb+I,B′(t)=ηI−μ2B. | (1.1) |
where S(t) and I(t) denote the numbers of susceptible individuals and infected individuals at time t, respectively. B(t) denotes the concentration of vibrios in contaminated water at time t. Besides, The parameter A is the recruitment rate, and parameter μ1 is the natural human death rate. The parameters β is the transmission coefficients of environment-to-human pathways. The parameter K is the concentration of vibrios in contaminated water that yields 50% chance of catching cholera. The parameter γ is the recovery rate of infected individuals. c is the maximum recovery per unit of time, and b is the infected size at 50% saturation. α is the disease-induced human death rate. The parameter η is the contribution rate of each infected individual to the concentration of vibrios, and μ2 is the net death rate of vibrios. All parameters are usually also assumed to be nonnegative. For system (1.1), the basic reproduction number is defined by
R0=S0bβηKμ2(rb+bμ1+bα+c), |
where S0=Aμ1. In [9], the scholars obtained that if R0<1, model has only a disease-free equilibrium E0=(S0,0,0) that is locally asymptotically stable; if R0>1, system (1.1) has a unique endemic equilibrium E+=(S+,I+,B+) that is locally asymptotically stable.
In the real environment, the spread of cholera is disturbed by random factors, so many scholars have proposed a kind of stochastic differential equation cholera spread model with random perturbation, such as [10,11], stochastic cholera model between communities linked by migration [12], stochastic model with Lˊevy process[13], stochastic model under regime switching [14]. At present, the O-U process is popular in various random interference channels (see [15,16,17,18]). Therefore, in order to reveal the impact of environmental noise on the transmission rate, we assume that it is a random variable and satisfies the following form [19,20,21,22]:
dγ=λ1(ˉγ−γ(t))dt+σ1dB1(t),dc=λ2(ˉc−c(t))dt+σ2dB2(t). |
where ˉγ,ˉc are measure the long-time mean levels of the infection rates γ,c; λi(i=1,2) are the speeds of reversion. Bi(t) are independent standard Brownian motion parameters defined on a complete probability space (Ω,F,P), and parameter σi>0(i=1,2) represents the intensity of Bi(t). All parameters are usually also assumed to be nonnegative. In order to discuss the need for positivity of the stochastic model, variable max{γ(t),0},max{c(t),0} is used instead of variable γ(t),c(t) in [19]. Then, we investigated the following stochastic SIB model with O-U process:
S′(t)=A−βSBK+B−μ1S+max{γ(t),0}I+max{c(t),0}Ib+I,I′(t)=βSBK+B−(μ1+max{γ(t),0}+α)I−max{c(t),0}Ib+I,B′(t)=ηI−μ2B,dγ=λ1(ˉγ−γ(t))dt+σ1dB1(t),dc=λ2(ˉc−c(t))dt+σ2dB2(t). | (1.2) |
Throughout this paper, we define R3+={(x1,x2,x3):xi>0,i=1,2,3}. For an integrable function f(t) defined on [0,∞), define ⟨f(t)⟩=1t∫t0f(s)ds. For the numbers a and b, we define a∨b=max{a,b},a∧b=min{a,b}.
In the next section, we verify the existence and uniqueness of a global solution of the model (1.2) with any initial value. In Section 3, the criterion on the ergodicity and existence of unique stationary distribution for any solution of model (1.2) is stated and proved. In Section 4, We obtain the probability density function of model (1.2) around the positive equilibrium point E∗. In Section 5, the numerical examples are carried out to illustrate the main theoretical results.
In this section, we will discuss the existence and uniqueness of a global solution for model (1.2).
Theorem 2.1. For any initial value (S(0),I(0),B(0),γ(0),c(0))∈R3+×R2, model (1.2) has a unique global solution (S(t),I(t),B(t),γ(t),c(t)). That is, solution (S(t),I(t),B(t),γ(t),c(t)) is defined for all t≥0 and remains in R3+×R2with probability one.
Proof. Since the coefficients of model (1.2) satisfy the local Lipschitz conditions, for any initial value (S(0),I(0),B(0),γ(0),c(0))∈R3+×R2, there exists a unique local solution (S(t),I(t),B(t),γ(t),c(t)) on t∈[0,τe), where τe denotes the explosion time. To show this solution is global, we only need to prove that τe=∞ a.s.. To this end, let k0≥1 be sufficiently large such that S(0),I(0),B(0), eγ(0) and ec(0)all lie within the interval [1k0,k0]. For each integer k≥k0, define the stopping time
τk=inf{t∈[0,τe):min{S(t),I(t),B(t),eγ(t),ec(t)}≤1kormax{S(t),I(t),B(t),eγ(t),ec(t)}≥k}, |
where throughout this paper, we set inf∅=∞ (as usual ∅ represents the empty set). Clearly, τk is increasing as k→∞. Let τ∞=limk→∞τk, whence τ∞≤τe a.s. If τ∞=∞ a.s. is not false, then τe=∞ a.s. and (S(t),I(t),B(t),γ(t),c(t))∈R3+×R2 a.s. for all t>0. That is to say, if we want to finish the proof, we only need to show τ∞=∞ a.s. If this assertion is not true, then there is a pair of constants T>0 and ε∈(0,1) such that
p{τ∞≤T}>ε. |
Consequently, there exists an integer k1≥k0 such that
p{τk≤T}≥εforallk≥k1. | (2.1) |
Define a Lyapunov function
V=S−1−lnS+I−1−lnI+B−1−lnB+12(γ2+c2). |
By calculating, we have
LV=A−μ1(S+I)−αI+ηI−μ2B−1B[ηI−μ2B]−1I[βSBK+B−(μ1+max{γ(t),0}+α)I−max{c(t),0}Ib+I]−1S[A−βSBK+B−μ1S+max{γ(t),0}I+max{c(t),0}Ib+I]+12(σ21+σ22)+λ1γ(ˉγ−γ)+λ2c(ˉc−c)≤A+ηI+μ2+μ1+|γ|+α+|c|1b+12(σ21+σ22)+βBK+B+μ1+λ1γ(ˉγ−γ)+λ2c(ˉc−c). | (2.2) |
By model (1.2), we have
d(S+I)dt=A−μ1(S+I)−αI≤A−μ1(S+I). | (2.3) |
This implies that
S+I≤{S(0)+I(0),ifS(0)+I(0)≥Aμ1,Aμ,ifS(0)+I(0)<Aμ1,≤˜N, |
where ˜N=max{Aμ1,S(0)+I(0)}.
B′(t)=ηI−μ2B≤η˜N−μ2B, |
which implies that
B≤{B(0),ifB(0)≥Aημ1μ2,Aξμμ2,ifB(0)<Aημ1μ2,≤ˉN, |
where ˉN=max{Aημ1μ2,B(0)}. Then,
LV≤A+η˜N+μ2+2μ1+α+βˉNK+ˉN+supγ∈R{|γ|+λ1γ(ˉγ−γ)}+12(σ21+σ22)+supc∈R{|c|1b+λ2c(ˉc−c)}:=˜K. | (2.4) |
Therefore, integrating both sides of (2.4) from 0 to τk∧T=min{τk,T} for any k≥k1 and then taking the expectations result in
EV(S(τk∧T),I(τk∧T),B(τk∧T),γ(τk∧T),c(τk∧T))≤V(S(0),I(0),B(0),γ(0),c(0))+KE(τk∧T). |
Thus,
EV(S(τk∧T),I(τk∧T),B(τk∧T),γ(τk∧T),c(τk∧T))≤V(S(0),I(0),B(0),γ(0),c(0))+KT. | (2.5) |
Set Ωk={τk≤T} for k≥k1, and in view of (2.1), we get P(Ωk)≥ε. Notice that for every ω∈Ωk, there exists S(τk,ω),I(τk,ω),B(τk,ω),γ(τk,ω) or c(τk,ω), which equals either k or 1k. Therefore, V(S(τk,ω),I(τk,ω),B(τk,ω),γ(τk,ω),c(τk,ω)) is no less than either
(k−1−lnk)∧ln2k2or(1k−1+lnk)∧ln2k2. |
Thereby, we can obtain
V(S(τk,ω),I(τk,ω),B(τk,ω),γ(τk,ω),c(τk,ω))≥[k−1−lnk]∧[1k−1+lnk]∧ln2k2. |
By (2.5), if follows that
V(S(0),I(0),B(0),γ(0),c(0))+KT≥E[IΩk(ω)V(S(τk,ω),I(τk,ω),B(τk,ω),γ(τk,ω),c(τk,ω))]≥ε[k−1−lnk]∧[1k−1+lnk]∧ln2k2, |
where IΩk denotes the indicator function of Ωk. Letting k→∞, then one can get that
∞>V(S(0),I(0),B(0),γ(0),c(0))+KT=∞ |
which is a contradiction, and then we derive τ∞=∞. The proof is complete. Define the set Γ as follows:
Γ={(S,I,B,γ,c)∈R3+×R2:S+I≤Aμ1,B≤Aημ1μ2}. |
Corollary 2.2. For any initial value x(0)=(S(0),I(0),B(0),γ(0),c(0))∈R3+×R2 theglobal solution x(t)=(S(t),I(t),B(t),γ(t),c(t)) of model (1.2) ultimately enters into region Γ with probability one as t→∞, and when x(0)∈Γ, then x(t)∈Γ with probability one for all t≥0.
In this section, we study the existence of the stationary distribution of model (1.2). Define
Rs0=AbβηKμ1μ2[b(μ1+ˉγ+α)+ˉc+12√π(bσ1√λ1+σ2√λ2)]. |
Theorem 3.1. Assume that Rs0>1, and then model (1.2) has at least one stationary distribution and ergodic property.
Proof. Define the Lyapunov function as follows:
V(S,I,B,γ,c)=MV1(S,I,B)+V2(S,I,B,γ,c), |
where
V1=−lnI−c1lnS−c2lnB+c3B+c1ˉβ1Kμ2B,V2=−lnS−lnB−ln(Aμ1−S−I)+c2+γ22. |
By calculating, we have
LV1=−1I[βSBK+B−(μ1+max{γ(t),0}+α)I−max{c(t),0}Ib+I]−c11S[A−βSBK+B−μ1S+max{γ(t),0}I+max{c(t),0}Ib+I]+c1βKμ2[ηI−μ2B]−c21B[ηI−μ2B]+c3[ηI−μ2B]≤−βSBI(K+B)−c2ηIB−c1AS−c3μ2(B+K)+c1μ1+c2μ2+c3μ2K+c3ηI+(μ1+max{γ(t),0}+α)+max{c(t),0}b+I+c1βBK+B+c1βKμ2[ηI−μ2B]≤−44√βc2ηc1Ac3μ2+c1μ1+c2μ2+c3μ2K+(μ1+ˉγ+α)+ˉcb+(y1(t)∨0)+y2(t)∨0b+c1βKμ2ηI+c3ηI |
where y1(t)=γ(t)−ˉγ,y2(t)=c(t)−ˉc and
c1=AβηKμ21μ2,c2=AβηKμ1μ22,c3=AβηK2μ1μ22. |
We have the following stochastic differential equation:
dyi=−λiyi(t)dt+σidBi(t),i=1,2. |
yi(t) has the ergodic property and density as follows:
˜πi(x)=√λi√πσie−λix2σ2i,x∈R,i=1,2. |
∫∞−∞(x∨0)˜πi(x)dx=∫∞0x√λi√πσ1e−λix2σ2idx=σi2√πλi. | (3.1) |
LV1≤−AβηKμ1μ2+(μ1+ˉγ+α+ˉcb)+12√π(σ1√λ1+σ2b√λ2)+c1βηKμ2I+c3ηI+((y1(t)∨0)−∫∞0x˜π1(x)dx)+1b((y2(t)∨0)−∫∞0x˜π2(x)dx)=−[μ1+ˉγ+α+ˉcb+12√π(σ1√λ1+σ2b√λ2)](Rs0−1)+c1βηKμ2I+c3ηI+((y1(t)∨0)−∫∞0x˜π1(x)dx)+1b((y2(t)∨0)−∫∞0x˜π2(x)dx), |
and
LV2=−1B[ηI−μ2B]−1S[A−βSBK+B−μ1S+max{γ(t),0}I+max{c(t),0}Ib+I]+1Aμ1−S−I[A−(μ1+α)I−μ1S]+λ1γ(ˉγ−γ)+λ2c(ˉc−c)+12(σ21+σ22)≤−AS−ηIB−αIAμ1−S−I+μ2+2μ1+βAηKμ2μ1+Aη+12(σ21+σ22)+λ1γ(ˉγ−γ)+λ2c(ˉc−c). |
Then,
LV≤M{−[μ1+ˉγ+α+ˉcb+12√π(σ1√λ1+σ2b√λ2)](Rs0−1)+c1βηKμ2I+c3ηI+((y1(t)∨0)−∫∞0x˜π1(x)dx)+1b((y2(t)∨0)−∫∞0x˜π2(x)dx)}−AS−ηIB−αIAμ1−S−I+μ2+2μ1+AηβKμ2μ1+Aη+12(σ21+σ22)+λ1γ(ˉγ−γ)+λ2c(ˉc−c):=F(S,I,B,γ,c)+M((y1(t)∨0)−∫∞0x˜π1(x)dx)+Mb((y2(t)∨0)−∫∞0x˜π2(x)dx), | (3.2) |
where
F(S,I,B,γ,c)=−M[μ1+ˉγ+α+ˉcb+12√π(σ1√λ1+σ2b√λ2)](Rs0−1)+μ2+2μ1+(c1βKμ2I+c3)MηI−αIAμ1−S−I+AηβKμ2μ1+Aη+σ21+σ222−AS−ηIB−12λ1γ2−12λ2c2+sup(γ,c)T∈R2{λ1γ(ˉγ−12γ)+λ2c(ˉc−12c)}. |
Define the bounded closed set
U={(S,I,B,γ,c)∈Γ|S+I≤Aμ1−ε2,ε≤S,ε≤I,ε≤B,|γ|≤1ε,|c|≤1ε}, |
where ε are small enough positive constants, which will be determined later.
For convenience, we divide R3+×R2∖U into six domains.
U1={(S,I,B,γ,c)∈R3+×R2,0<S<ε},U2={(S,I,B,γ,c)∈R3+×R2,0<I<ε},U3={(S,I,B,γ,c)∈R3+×R2,0<B<ε2,I>ε},U4={(S,I,B,γ,c)∈R3+×R2,S+I>Aμ1−ε2,I>ε},U5={(S,I,B,γ,c)∈R3+×R2,|γ|≥1ε},U6={(S,I,B,γ,c)∈R3+×R2,|c|≥1ε}. |
We will prove that F(S,I,B,γ,c)≤−1 on R3+×R2∖U, which is equivalent to show it on the above seven domains.
Case 1. If (S,I,B,γ,c)∈U1, we can obtain
F(S,I,B,γ,c)≤−AS+G1≤−Aε+G1, |
where
G1=−M[μ1+ˉγ+α+ˉcb+12√π(σ1√λ1+σ2b√λ2)](Rs0−1)+(c1βKμ2I+c3)MηI+2μ1+μ2+AηβKμ2μ1+Aη+12(σ21+σ22)+sup(γ,c)T∈R2{λ1γ(ˉγ−12γ)+λ2c(ˉc−12c)}. |
We choose a constant ε>0 small enough such that −Aε+G1≤−1, and then it follows that
F(S,I,B,γ,c)≤−1for all(S,I,B,γ,c)∈U1. | (3.3) |
Case 2. If (S,I,B,γ,c)∈U2, we can obtain
F(S,I,B,γ,c)≤(c1βKμ2+c3)MηI+G2≤(c1βKμ2+c3)Mηε+G2, |
where
G2=−M[μ1+ˉγ+α+ˉcb+12√π(σ1√λ1+σ2b√λ2)](Rs0−1)+12(σ21+σ22)+2μ1+μ2+AηβKμ2μ1+Aη+sup(γ,c)T∈R2{λ1γ(ˉγ−12γ)+λ2c(ˉc−12c)}. |
Choose constants M>0 large enough and ε>0 small enough such that (c1βKμ2+c3)Mηε+G2≤−1, and then it follows that
F(S,I,B,γ,c)≤−1for all(S,I,B,γ,c)∈U2. | (3.4) |
Case 3. If (S,I,B,γ,c)∈U3, we can obtain
F(S,I,B,γ,c)≤−ηIB+G1≤−ηεε2+G1. |
Choose a constant ε>0 small enough such that −ηε+G1≤−1, and then it follow that
F(S,I,B,γ,c)≤−1for all(S,I,B,γ,c)∈U3. | (3.5) |
Case 4. If (S,I,B,γ,c)∈U4, we can obtain
F(S,I,B,γ,c)≤−αIAμ1−S−I+G1≤−αεε2+G1. |
Choose a constant ε>0 small enough such that −αε+G1≤−1, and then we have
F(S,I,B,γ,c)≤−1for all(S,I,B,γ,c)∈U4. | (3.6) |
Case 5. If (S,I,B,γ,c)∈U5, we can obtain
F(S,I,B,γ,c)≤−12γ2+G1≤−12ε2+G1. |
Choose a constant ε>0 small enough such that −12ε2+G1≤−1, and then we get
F(S,I,B,γ,c)≤−1for all(S,I,B,γ,c)∈U5. | (3.7) |
Case 6. If (S,I,B,γ,c)∈U6, we can obtain
F(S,I,B,γ,c)≤−12c2+G1≤−12ε2+G1. |
Choose a constant ε>0 small enough such that −12ε2+G1≤−1, and then we get
F(S,I,B,γ,c)≤−1for all(S,I,B,γ,c)∈U6. | (3.8) |
Finally, from (3.3)-(3.7) we obtain
F(S,I,B,γ,c)≤−1for all(S,I,B,γ,c)∈R3+×R2∖U. | (3.9) |
Define
ˉV(S,I,B,γ,c)=V(S,I,B,γ,c)−V(S0,I0,B0,γ0,c0), |
where (S0,I0,B0,γ0,c0) is the point in the interior R3+×R2, such that V(S,I,B,γ,c) will be minimized. By (2.2), we have
LˉV≤F(S,I,B,γ,c)+M((y1(t)∨0)−∫∞0x˜π1(x)dx)+Mb((y2(t)∨0)−∫∞0x˜π2(x)dx). | (3.10) |
Taking the mathematical expectation and Itˆo's integral of (3.10), for any initial value (S(0),I(0),B(0),γ(0),c(0)), one has
0≤EˉV(S(t),I(t),B(t),γ(t),c(t))t≤EˉV(S(0),I(0),B(0),γ(0),c(0))t+1t∫t0E(F(S(s),I(s),B(s),γ(s),c(s)))ds+M[E(1t∫t0(y1∨0)ds−∫∞0x˜π1(x)dx)+1bE(1t∫t0(y2∨0)ds−∫∞0x˜π2(x)dx)]. | (3.11) |
By the ergodicity of yi(i=1,2) and the strong law of large numbers, one has
limt→∞E[1t∫t0(yi∨0)ds−∫∞0x˜πi(x)dx]=E[∫∞0x˜πi(x)dx]−∫∞0x˜πi(x)dx=0a.s. | (3.12) |
By (3.12), we have the inferior limit of (3.11),
0≤lim inft→∞EˉV(S(0),I(0),B(0),γ(0),c(0))t+lim inft→∞1t∫t0E(F(S(s),I(s),B(s),γ(s),c(s)))ds=lim inft→∞1t∫t0E(F(S(s),I(s),B(s),γ(s),c(s)))ds. | (3.13) |
From (3.9), we have
lim inft→∞1t∫t0E(F(S(s),I(s),B(s),γ(s),c(s)))ds=lim inft→∞1t∫t0E(F(S(s),I(s),B(s),γ(s),c(s)))1(S(s),I(s),B(s),γ(s),c(s))∈Uds+lim inft→∞1t∫t0E(F(S(s),I(s),B(s),γ(s),c(s)))1(S(s),I(s),B(s),γ(s),c(s))∈(R3+×R2∖U)ds≤ˇFlim inft→∞1t∫t01(S(s),I(s),B(s),γ(s),c(s))∈Uds+lim inft→∞1t∫t01(S(s),I(s),B(s),γ(s),c(s))∈(R3+×R2∖U)ds≤−1+(ˇF+1)lim inft→+∞1t∫t01(S(s),I(s),B(s),γ(s),c(s))∈Uds. | (3.14) |
By (3.13) and (3.14), we have
lim inft→∞1t∫t01(S(s),I(s),B(s),γ(s),c(s))∈Uds≥1ˇF+1>0a.s. | (3.15) |
In view of the definition of event probability and Fatou's lemma (see[23]), the result of (3.15) is
lim inft→∞1t∫t0P(s,(S(s),I(s),B(s),γ(s),c(s)),U)ds≥1ˇF+1>0a.s. | (3.16) |
where P(t,(S,I,B,γ,c),U) is the transition probability of (S(t),I(t),B(t),γ(t),c(t)) belonging to set U. This shows that the solution (S(t),I(t),B(t),γ(t),c(t)) of model (1.2) has the Feller and ergodic property. This completes the proof.
The positive equilibrium P∗=(S∗,I∗,B∗,γ∗,c∗) of system (1.2) satisfied the following equations:
{0=A−βS∗B∗K+B∗−μ1S∗+max{γ∗,0}I+max{c∗,0}I∗b+I∗,0=βS∗B∗K+B∗−(μ1+max{γ∗,0}+α)I∗−max{c∗,0}Ib+I∗,0=ηI∗−μ2B∗,0=λ1(ˉγ−γ∗),0=λ2(ˉc−c∗). | (4.1) |
When R0>1, we have
I∗=I+,S∗=S+,B∗=B+,γ∗=ˉγ,c∗=ˉc. |
Let (x1,x2,x3,y1,y2)=(S−S∗,I−I∗,B−B∗,γ−γ∗,c−c∗). The linearization model of system (1.2) is:
{dx1(t)=[−a11x1+a12x2−a13x3+a14y1+a15y2]dt,dx2(t)=[a21x1−a22x2+a13x3−a14y1−a15y2]dt,dx3(t)=[ηx2−μ2x3]dt,dy1(t)=−λ1y1dt+σ1dB1(t),dy2(t)=−λ2y2dt+σ2dB2(t), | (4.2) |
where
a11=βB∗K+B∗+μ1,a12=γ∗+c∗b(b+I∗)2,a13=S∗Kβ(K+B∗)2,a14=I∗,a15=I∗b+I∗,a21=βB∗K+B∗,a22=(μ1+γ∗+α)+c∗b(b+I∗)2. |
Define
D=(−a11a12−a13a14a15a21−a22a13−a14−a150η−μ200000−λ100000−λ2),Q=(000000000000000000σ100000σ2). |
Let X(t)=(x1(t),x2(t),x3(t),y1(t),y2(t))T,P(t)=(0,0,0,0,B1(t),B2(t))T, and then the linear system (4.2) can be written as follows:
dX(t)=DX(t)dt+QdP(t). |
The characteristic equation of
ϕ(λA)=(λ+λ1)(λ+λ2)(λ3−a1λ2−a2λ−a3)=0, |
where
a1=a11+a22+μ2>0,a2=(a11+a22)μ2−a13η+a11a22−a21a12>0,a3=μ2(a11a22−a21a12)+(a21−a11)a13η>0, | (4.3) |
and
a2a1−a3=(a11+a22)[(a11+a22)μ2−a13η+a11a22−a21a12]+μ2[(a11+a22)μ2−a13η]+(a11−a21)a13η>0. | (4.4) |
Lemma 4.1. For the five-dimensional algebraic equation P20+D0Σ0+Σ0DT0=0, where P0=diag(1,0,0,0,0), Σ0 is a symmetrical matrix,
(1)
D0=(−a1−a2−a3−a4−a51000001000000ˉa00000˜a). |
If D0 is a Hurwitz matrix, that is, ai>0(i=1,2,3),a2a1−a3>0, then Σ0 is positive definite and has the following expression:
Σ0=(a22(a1a2−a3)0−12(a1a2−a3)00012(a1a2−a3)000−12(a1a2−a3)0a12a3(a1a2−a3)000000000000). |
(2)
D0=(−a1−a2−a3−a4−a51000001000001000000a). |
If D0 is a Hurwitz matrix, that is, ai>0(i=1,2,3,4),a2a1−a3>0 and, a1a2a3−a23−a21a4>0, then Σ0 is positive semi-definite and has the following expression:
Σ0=(a2a3−a1a42(a1a2a3−a21a4−a23)0−a32(a1a2a3−a21a4−a23)000a32(a1a2a3−a21a4−a23)0−a12(a1a2a3−a21a4−a23)0−a32(a1a2a3−a21a4−a23)0a12(a1a2a3−a21a4−a23)000−a12(a1a2a3−a21a4−a23)0a1a2−a32a4(a1a2a3−a21a4−a23)000000). |
Theorem 4.2. If Rs0>1, the stationary solution (S(t),I(t),B(t),γ(t),c(t)) of model (1.2) around (S∗,I∗,B∗,γ∗,c∗)T has a unique normal density function, which takes the form
Φ(S,I,B,γ,c)=(2π)−52|Σ|−12e−12(S−S∗,I−I∗,B−B∗,γ−γ∗,c−c∗)Σ−1(S−S∗,I−I∗,B−B∗,γ−γ∗,c−c∗)T, |
with the positive definite matrix Σ=Σ1+Σ2.
(1) μ1≠μ2
Σ1=(a14η(a21−a11+μ2)σ1)2(H1H2H3H4H5)−1Σ11[(H1H2H3H4H5)−1]T. |
Σ2=(a15η(a21−a11+μ2)σ2)2(H8H2H3H4H9)−1Σ21[(H8H2H3H4H9)−1]T. |
(2) μ1=μ2
Σ1=(a14ασ1)2(H1H2H3H6H7)−1Σ12[(H1H2H3H6H7)−1]T. |
Σ2=(a15ασ2)2(H8H2H3H6H10)−1Σ22[(H8H2H3H6H10)−1]T. |
Hi(i=1,2,3,4,5,6,7,8,9,10), Σij(j,i=1,2) are given in the following proof.
Proof. Let the symmetric matrices Σ=(rij)5×5, where rji=rij. Σ is determined by the following algebraic equation:
Q2+DΣ+ΣDT=0. | (4.5) |
Furthermore, let Σ=Σ1+Σ2, Q1=diag(0,0,0,0,σ1,0),Q2=diag(0,0,0,0,0,σ2), and then equation (4.5) can be decomposed into the following two equations: Q2i+DΣi+ΣiDT=0(i=1,2). We will calculate the matrices Σi(i=1,2) in the following two steps.
Step (1) Q21+DΣ1+Σ1DT=0. Let D1=H1DH−11, where
H1=(0001001000100000010000001),D1=(−λ10000−a14−a22a21a13−a15a14a12−a11−a13a150η0−μ200000−λ2). |
Let D2=H2D1H−12, where
H2=(1000001000011000001000001),D2=(−λ10000−a14−a22−a21a21a13−a150a12−a22−(a21−a11)a21−a11000η0−μ200000−λ2). |
where a12−a22−(a21−a11)=−α. Let D3=H3D2H−13, where
H3=(10000010000010000ηα1000001),D3=(−λ10000−a14−a22−a21a21−ηαa13a13−a150−αa21−a110000ηα(a21−a11+μ2)−μ200000−λ2). |
Case 1. Let a21−a11+μ2≠0, that is, μ2≠μ1. Let D4=H4D3H−14, where
H4=(100000−η(a21−a11+μ2)ηα[(a21−a11)2−μ22]μ22000ηα(a21−a11+μ2)−μ200001000001). |
D4=(−λ10000a14η(a21−a11+μ2)−a1−a2−a3a15η(a21−a11+μ2)01000001000000−λ2). |
Let D5=H5D4H−15, where
H5=(a14η(a21−a11+μ2)−a1−a2−a3a15η(a21−a11+μ2)01000001000001000001), |
where a1,a2,a3 are given in (4.3).
D5=(−(a1+λ1)−(a1λ1+a2)−(a2λ1+a3)−a3λ1−a15η(a21−a11+μ2)(a1+λ2)1000a15η(a21−a11+μ2)01000001000000−λ2). |
Then, by Lemma 4.1(2), for the matrix equation
Q20+D5Σ11+Σ11DT5=0, |
with Q0=diag(1,0,0,0,0), we can obtain the positive definite matrix as follows:
Σ11=((a1a2−a3)λ21+a22λ1+a2a32H110−a2λ1+a32H11000a2λ1+a32H110−a1+λ12H110−a2λ1+a32H110a1+λ12H11000−a1+λ12H110a1a2−a3+a1λ1(a1+λ1)2a3λ1H11000000), |
where H11=(a1a2−a3)(a3+λ1a2+λ21a1+λ31). The element a15η(a21−a11+μ2) in the second row and fifth column of matrix D5 does not affect the calculation results using the method of Lemma 4.1(2).
Then, we have that Q21+AΣ1+Σ1AT=0 is equivalent to
(H1H2H3H4H5)Q21(H1H2H3H4H5)T+D5(H1H2H3H4H5)Σ1(H1H2H3H4H5)T+(H1H2H3H4H5)Σ1(H1H2H3H4H5)TDT5=0, |
which is further equivalent to
Q20+(−a14η(a21−a11+μ2)σ1)−2(D5(H1H2H3H4H5)Σ1(H1H2H3H4H5)T+(H1H2H3H4H5)Σ1(H1H2H3H4H5)TDT5)=0. |
Therefore, we finally have that
Σ1=(a14η(a21−a11+μ2)σ1)2(H1H2H3H4H5)−1Σ11[(H1H2H3H4H5)−1]T. |
Thus, Σ1 is calculated and is positive semi-definite.
Case 2. μ2=μ1. Let D6=H6D3H−16, where
H6=(100000−αa21−a1100001000001000001),D6=(−λ10000a14α−a11−a22−D∗−a13αa15α01000000−μ200000−λ2), |
where D∗=(a22+a21)(a11−a21)+a21α−a13η. Let D7=H7D6H−17, where
H7=(a14α−a11−a22−D∗−a13αa15α01000001000001000001), |
D7=(−(a11+a22+λ1)−((a11+a22)λ1+D∗)−D∗λ1D71D72100−a13αa15α01000000−μ200000−λ2), |
where D71=−a13α(a11+a22+μ2),D72=−a15α(a11+a22−λ2). Then, by Lemma 4.1(1), for the matrix equation
Q20+D7Σ12+Σ12DT7=0, |
we can obtain the positive define matrix as follows:
Σ12=((a11+a22)λ1+D∗2H120−12H1200012H12000−12H120a11+a22+λ12D∗λ1H12000000000000). |
H12=(a11+a22+λ1)a11+a22λ1+a11+a22D∗. The elements −a13α,a15α in the second row, fourth column and fifth column of matrix D5 do not affect the calculation results using the method of Lemma 4.1(1). Then, we have that Q21+DΣ1+Σ1DT=0 is equivalent to
(H1H2H3H6H7)Q21(H1H2H3H6H7)T+D7(H1H2H3H6H7)Σ1(H1H2H3H6H7)T+(H1H2H3H6H7)Σ1(H1H2H3H6H7)TDT7=0, |
which is further equivalent to
Q20+(a14ασ1)−2(D7(H1H2H3H6H7)Σ1(H1H2H3H6H7)T+(H1H2H3H6H7)Σ1(H1H2H3H6H7)TDT7)=0. |
Therefore, we finally have that
Σ1=(a14ασ1)2(H1H2H3H6H7)−1Σ12[(H1H2H3H6H7)−1]T. |
Thus, Σ1 is calculated and is positive semi-definite.
Step (2) Q22+DΣ2+Σ2DT=0. Let D8=H8DH−18, D9=H2D8H−12, D10=H3D9H−13, where H2,H3 are given in step (1), and
H8=(0000101000100000010000010),D8=(−λ20000−a15−a22a21a13−a14a15a12−a11−a13a140η0−μ200000−λ1), |
D9=(−λ20000−a15−a22−a21a21a13−a140−αa21−a11000η0−μ200000−λ1), |
D10=(−λ20000−a15−a22−a21a21−ηαa13a13−a140−αa21−a110000ηα(a21−a11+μ2)−μ200000−λ1). |
Case 1. Let μ2≠μ1. Let D11=H4D10H−14, where H4 is given in step (1), and
D11=(−λ20000a15η(a21−a11+μ2)−a1−a2−a3a14η(a21−a11+μ2)01000001000000−λ1), |
where a1,a2,a3 are given in (4.3). Let D12=H9D11H−19, where
H9=(a15η(a21−a11+μ2)−a1−a2−a3a14η(a21−a11+μ2)01000001000001000001), |
D12=(−(a1+λ2)−(a1λ2+a2)−(a2λ2+a3)−a3λ2a14η(a21−a11+μ2)(a1+λ1)1000a14η(a21−a11+μ2)01000001000000−λ1). |
Then, by Lemma 4.1(2), for the matrix equation:
Q20+D12Σ21+Σ21DT12=0, |
we can obtain the positive define matrix as follows:
Σ21=((a1a2−a3)λ22+a22λ2+a2a32H210−a2λ2+a32H21000a2λ2+a32H210−a1+λ22H210−a2λ2+a32H210a1+λ22H21000−a1+λ22H210a1a2−a3+a1λ2(a1+λ2)2a3λ2H21000000), |
where H21=(a1a2−a3)(a3+λ2a2+λ22a1+λ32).
Then, we have that Q22+DΣ2+Σ2DT=0 is equivalent to
(H8H2H3H4H9)Q22(H8H2H3H4H9)T+D12(H8H2H3H4H9)Σ2(H8H2H3H4H9)T+(H8H2H3H4H9)Σ2(H8H2H3H4H9)TDT12=0, |
which is further equivalent to
Q20+(a15η(a21−a11+μ2)σ2)−2(D12(H8H2H3H4H9)Σ2(H8H2H3H4H9)T+(H8H2H3H4H9)Σ2(H8H2H3H4H9)TDT12)=0. |
Therefore, we finally have that
Σ2=(a15η(a21−a11+μ2)σ2)2(H8H2H3H4H9)−1Σ21[(H8H2H3H4H9)−1]T. |
Thus, Σ2 is calculated and is positive semi-definite.
Case 2. μ1=μ2. Let D13=H6D10H−16, where H6 is given in step (1), and
D13=(−λ20000a15α−a11−a22−D∗−a13αa14α01000000−μ200000−λ1). |
Let D14=H10D13H−110, where
H10=(a15α−a11−a22−D∗−a13αa14α01000001000001000001). |
D14=(−(a11+a22+λ2)−((a11+a22)λ2+D∗)−D∗λ2D71D73100−a13(γ+α)−a14(γ+α)01000000−μ200000−λ1), |
where D73=−a14α(a11+a22−λ1). Then, by Lemma 4.1(1), for the matrix equation
Q20+D14Σ22+Σ22DT14=0, |
we can obtain the positive define matrix as follows:
Σ22=((a11+a22)λ2+D∗2H220−12H2200012H22000−12H220a11+a22+λ22D∗λ2H22000000000000). |
where H22=(a11+a22+λ2)(a11+a22)λ2+(a11+a22)D∗. Then, we have that Q22+DΣ2+Σ2DT=0 is equivalent to
(H8H2H3H6H10)Q22(H8H2H3H6H10)T+D14(H8H2H3H6H10)Σ2(H8H2H3H6H10)T+(H8H2H3H6H10)Σ2(H8H2H3H6H10)TDT14=0, |
which is further equivalent to
Q20+(a15ασ2)−2(D14(H8H2H3H6H10)Σ2(H8H2H3H6H10)T+(H8H2H3H6H10)Σ2(H8H2H3H6H10)TDT14)=0. |
Therefore, we finally have that
Σ2=(a15ασ2)2(H8H2H3H6H10)−1Σ22[(H8H2H3H6H10)−1]T. |
Thus, Σ2 is calculated and is positive semi-definite. Then, Σ=Σ1+Σ2 is positive define. Therefore, the expression of a normal density function around the quasi-endemic equilibrium of model (1.2) is obtained by
Φ(S,I,B,γ,c)=(2π)−52|Σ|−12e−12(S−S∗,I−I∗,B−B∗,γ−γ∗,c−c∗)Σ−1(S−S∗,I−I∗,B−B∗,γ−γ∗,c−c∗)T. |
This completes the proof.
In model (1.2), we take the parameters A=2, β=0.5, K=0.4, μ1=0.2, μ2=0.5, α=0.2, η=0.2, θ1=θ2=0.2, ˉc=0.06, ˉγ=0.1, b=1 and σ1=σ2=0.02. The numerical simulation of solution (S(t),I(t),B(t)) is done with initial value (S(0),I(0),B(0))=(3,1.5,1.5). We obtain RS0=8.5436>1. The positive equilibrium of model (1.2) is E∗=(S∗,I∗,B∗,γ∗,c∗)≈(4.08,2.96,1.18,0.1,0.06). (See Fig. 1) The conditions in Theorem 3.1 and Theorem 4.2 are satisfied. Therefore, there exists the stationary distribution of model (1.2) (See Fig. 2), and there is a unique normal density function near equilibrium E∗. Then, the positive definite matrix Σ is calculated as
Σ=(0.7302−0.4415−0.15440.16360.0413−0.44150.29710.0880−0.1090−0.0275−0.15440.08800.0352−0.0312−0.00790.1636−0.1090−0.03120.050000.0413−0.0275−0.007900.0500), |
and density function
Φ(S,I,B,γ,c)=1.0100×10−6e−12(S−4.08,I−2.96,B−1.18,γ−0.1,c−0.06)×Σ−1(S−4.08,I−2.96,B−1.18,γ−0.1,c−0.06)T. |
Then, S, I and B have the following marginal probability densities:
∂Φ∂S=0.1863e−49.4615(S−1.6875)2,∂Φ∂I=0.2920e−28.5588(I−1.3961)2,∂Φ∂B=0.8483e−408.5348(B−1.3961)2. |
The numerical simulation of the marginal probability densities for S, I and B is given in Figure 3.
In this paper, we investigated a stochastic SIBS cholera model with saturation recovery rate and Ornstein-Uhlenbeck process. First, we proved that there is a unique global solution for any initial value of model. Secondly, the sufficient criterion of the stationary distribution of the model was obtained by constructing a suitable Lyapunov function, and the expression of the probability density function was calculated by the same condition. Finally, the correctness of the theoretical results is verified by numerical simulation, and the specific expression of the marginal probability density function is obtained.
This paper analyzes the interference of random disturbance in the process of recovering the infected person to the susceptible person during the transmission of cholera, but in fact, other parameters will also be more or less affected by random disturbance. In addition, people in many countries and regions have been vaccinated against cholera. In this process, cholera transmission is also affected by random factors. Therefore, there is still much work worthy of further study.
This research is supported by the National Natural Science Foundation of China (Grant No. 12201274, 12171004, 12001305), the Natural Science Foundation of Liaoning Province of China (Grant No. 2022-BS-287), the Basic Scientific Research Projects of Department of Education of Liaoning Province of China (Grant No. LJKQZ20222424) and the Scientific Research Foundation of Higher Education Institutions of Ningxia (Grant No. NGY2020005).
The authors declare there is no conflict of interest.
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