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The spatial dynamics of a zebrafish model with cross-diffusions

  • Received: 01 April 2016 Published: 01 August 2017
  • MSC : 34D05, 34D20

  • This paper investigates the spatial dynamics of a zebrafish model with cross-diffusions. Sufficient conditions for Hopf bifurcation and Turing bifurcation are obtained by analyzing the associated characteristic equation. In addition, we deduce amplitude equations based on multiple-scale analysis, and further by analyzing amplitude equations five categories of Turing patterns are gained. Finally, numerical simulation results are presented to validate the theoretical analysis. Furthermore, some examples demonstrate that cross-diffusions have an effect on the selection of patterns, which explains the diversity of zebrafish pattern very well.

    Citation: Hongyong Zhao, Qianjin Zhang, Linhe Zhu. The spatial dynamics of a zebrafish model with cross-diffusions[J]. Mathematical Biosciences and Engineering, 2017, 14(4): 1035-1054. doi: 10.3934/mbe.2017054

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  • This paper investigates the spatial dynamics of a zebrafish model with cross-diffusions. Sufficient conditions for Hopf bifurcation and Turing bifurcation are obtained by analyzing the associated characteristic equation. In addition, we deduce amplitude equations based on multiple-scale analysis, and further by analyzing amplitude equations five categories of Turing patterns are gained. Finally, numerical simulation results are presented to validate the theoretical analysis. Furthermore, some examples demonstrate that cross-diffusions have an effect on the selection of patterns, which explains the diversity of zebrafish pattern very well.


    Anthrax is a contagious disease caused by bacteria produced by gram-positive endospores and can be spread between animals and humans [1]. The disease can infect all warm-blooded animals, especially herbivores such as cattle, sheep, goats, and horses [2,3,4,5]. When herbivores ingest a sufficient number of spores from soil and plants, they are usually infected with anthrax. Once infected, the animal will die within 24–48 hours [6,7]. Humans are infected with anthrax primarily through contact with diseased animals. Once the bacteria enters the human body, it can cause symptoms such as fever, skin lesions and breathing difficulties [7]. Anthrax spores have a strong ability to survive, which can result in contaminated soil and water becoming long-lasting sources of infection [9]. Anthrax occurs worldwide in 2004, a massive outbreak of anthrax killed up to 500 animals in Zimbabwe, and 109 animal anthrax deaths were reported in North Dakota in 2005. Between August 2009 and October 2010, there was an outbreak of 273 human cutaneous anthrax cases and 140 animal anthrax deaths in Bangladesh [10,11,12,13,14]. In a limited resource environment, anthrax has caused great harm to the development of animal husbandry and human health. As a result, anthrax has attracted widespread attention, and some progress has been made in disease control and vaccine research. In order to efficiently control the spread of anthrax, it is imperative to investigate the dynamics of the disease.

    In the past few decades, various mathematical models have been intensively studied on anthrax dynamics. In previous studies, Hahn and Furniss analyzed how anthrax spreads and determined the threshold of the disease [15]. In relation to animal migration, Friedman and Yakubu addressed the basic reproduction number of the model. They examined the sufficient conditions for the presence of an endemic equilibrium point in the diffusion model [16]. Mushayabasa et al. investigated the global asymptotic stability of the endemic equilibrium point by analyzing a model that considers the vectorial impacts of anthrax transmission [17]. In fact, the spread of the virus is linked to climate, season, temperature and humidity. Due to the complexity of the environment system, the parameters of the system inevitably change randomly due to perturbations, thus making it random. Incorporating randomness into mathematical models of biological and biochemical processes is essential for gaining a deeper comprehension of the mechanisms that regulate biological systems [18]. From [19,20], Cai and Wang examined how changes in the environment affect the spread of diseases by analyzing the random fluctuations in a susceptible-infected-susceptible (SIS) model that includes media coverage. However, few studies have constructed stochastic processes into anthrax models, which facilitates the study in this paper.

    Soil renovation, heavy rains, and flash floods can lead to a rise in spore levels in the surroundings, heightening the likelihood of infection [21]. It is crucial to take into account environmental disturbances when studying anthrax. Based on the existing research [7], we present a stochastic model of anthrax to explore the impact of stochastic disturbances on threshold dynamics. The stability of the new model and the probability density function in the case of disease persistence are worth studying. It is well known that the endemic equilibrium point and the basic reproduction number can reflect the persistence of the disease for identified systems. However, environmental perturbations are unpredictable, so there is no positive equilibrium in the stochastic system [22]. In this way, a stochastic anthrax model is investigated for the stationary distribution of disease persistence. In practice, the statistical nature of the model can lead to prevention and control the disease easier [22,23]. However, the resolution of the high-dimensional Fokker-Planck equation presents a challenging task, and only a limited number of research endeavors have been dedicated to addressing the computation of the probability density function. To close this gap, in this paper we employ the four-dimensional Fokker-Planck equation to investigate the probability density function for the stochastic anthrax model. Our article has the following contributions:

    ● On the basis of existing research [7], we propose a stochastic anthrax model. Sufficient conditions for the existence of an ergodic stationary distribution of globally positive solutions are established.

    ● The probability density function expression has been explored by few studies due to the difficulty of solving the high-dimensional Fokker-Planck equation. Therefore, this study investigates the precise representation of the probability density function in the vicinity of the quasi-equilibrium point of the model when a persistent disease is present.

    The rest of our paper is arranged as follows. Section 2 introduces the mathematical models and gives the relevant notation and lemma. Section 3 proves the unique ergodic stationary distribution existence when Rs>1. In Section 4, when Rs>1, the probability density function is obtained by solving the Fokker-Planck equation in four dimensions. Section 5 provides the numerical simulations for our results of the analysis. The main results are discussed in Section 6.

    Understanding the transmission mechanisms of anthrax epidemics requires the use of mathematical models as an important tool. As it is known that there are different forms of anthrax models, here we consider the following anthrax model [7]:

    {dS(t)dt=rN(t)(1N(t)K)ηiS(t)I(t)N(t)ηaA(t)S(t)ηcC(t)S(t)μS(t)+τI(t),dI(t)dt=ηiS(t)I(t)N(t)+ηaA(t)S(t)+ηcC(t)S(t)(γ+μ+τ)I(t),dA(t)dt=αA(t)+βC(t),dC(t)dt=(γ+μ)I(t)δ(S(t)+I(t))C(t)κC(t), (2.1)

    where S(t) and I(t) are the number of susceptible and infected animals at time t, respectively. N(t)=S(t)+I(t) denotes the total live animal population at time t, A denotes grams of spores in the environment, and C denotes the number of infected carcasses. The birth rate of animals is r>0, K is the animal carrying capacity, and rN(1NK) is born into the susceptible class per unit time. ηa is the transmission rate of environment to susceptible animals, ηc is the transmission rate of infected carcasses to susceptible animals, and ηi is the transmission rate of infected animals to susceptible animals, so ηa>0,ηc>0,ηi>0. μ is the death rate, γ is the disease death rate, α is the spore decay rate, β is the spore growth rate, δ the is carcass consumption rate, κ is the carcass decay rate, and τ denotes the recovery rate of infected animals.

    Assume that ηiγ+μ+τ<1, otherwise the anthrax increase is unbounded. On account of the total live animal population as N(t)=S(t)+I(t), adding the equation S(t) and I(t) results in

    dN(t)dt=rN(t)(1N(t)K)μN(t)γI(t)rN(t)(1N(t)K)μN(t),

    and if r>μ, this implies that lim suptN(t)K(1μr)=S0.

    Assuming 0<N(0)S0, the positive invariant is

    Γ={(S,I,A,C)R4+|0S+IS0,0AβS0(γ+μ)α(δS0+κ),0C(γ+μ)S0δS0+κ}.

    The disease-free equilibrium point E0 and the endemic equilibrium point E are obtained by solving the equation. E0=(S0,0,0,0),E=(S,I,A,C), where

    S=ND,I=NS,A=β(γ+μ)Iα(δN+κ),C=(γ+μ)IδN+κ,

    N=S0Kγr(11D) with D=ηiγ+μ+τ+γ+μγ+μ+τηcNδN+κ+γ+μγ+μ+τβηaNα(δN+κ), and the basic reproduction number of Model (2.1) is

    R0:=ηiγ+μ+τ+γ+μγ+μ+τηcS0δS0+κ+γ+μγ+μ+τβηaS0α(δS0+κ).

    According to Saad-Roy et al. [7], the main results about the stability of the equilibrium point are as follows:

    (i) If R0<1, the disease-free eqilibrium E0=(S0,0,0,0) is globally asymptotically stable; thus a small outbreak of anthrax is eradicated. If R0>1, the disease-free equilibrium is unstable; thus, anthrax persists.

    (ii) If R0>1, there has a unique endemic equilibrium with E=(S,I,A,C), which is globally asymptotically stable in Γ.

    In real life, the dynamic behaviors of anthrax are disturbed by random factors in nature. In this paper, we consider random noise on the basis of the Model (2.1) and adopt the method used by Imhof et al. [24]. Our hypothesis posits a direct correlation between stochastic fluctuations and S(t),I(t),A(t),C(t). Thus, the stochastic model for anthrax transmission is to be described in the following form:

    {dS(t)=[rN(t)(1N(t)K)ηiS(t)I(t)N(t)ηaA(t)S(t)ηcC(t)S(t)μS(t)+τI(t)]dt+σ1S(t)dB1(t),dI(t)=[ηiS(t)I(t)N(t)+ηaA(t)S(t)+ηcC(t)S(t)(γ+μ+τ)I(t)]dt+σ2I(t)dB2(t),dA(t)=[αA(t)+βC(t)]dt+σ3A(t)dB3(t),dC(t)=[(γ+μ)I(t)δ(S(t)+I(t))C(t)κC(t)]dt+σ4C(t)dB4(t), (2.2)

    where σi (i=1,2,3,4) represents the intensities of the white noises and the definition of Bi(t) (i=1,2,3,4) is based on complete probability space and is independent of standard Brownian motions. The transmission diagram illustrating the dynamics of anthrax transmission is shown in Figure 1. After giving the basic information of the anthrax model, the relevant lemma is given in the following section.

    Throughout this paper, we use X(t)=(S(t),I(t),A(t),C(t)) as the solution of Model (2.2) with the initial value X(0)=(S(0),I(0),A(0),C(0)). Let (Ω,F,{Ft}t0,P) be the probability space with the filtration {Ft}t0 satisfying the usual conditions that are right continuous, and F0 contains all P-null sets. Let R4+ be an-dimensional Euclidean space and Un=[1nn]×[1nn]×[1nn]×[1nn].

    Our focus in this section is on the existence and uniqueness of a globally positive solution. In order to achieve this goal, Lemma 2.1 is introduced to demonstrate the existence of a positive solution to the Model (2.2). The proof of the existence theorem for positive solution can be obtained in terms of Mao and Bahar [25,26], so the proof is omitted here.

    Lemma 2.1. For any initial value X(0)=(S(0),I(0),A(0),C(0))R4+, Model (2.2) has a unique global positive solution X(t)=(S(t),I(t),A(t),C(t)),t0, and the solution will remain in R4+ with probability one.

    Based on the existence of the positive solution, the next step we will concentrate on is the asymptotic estimate of the Model (2.2).

    Lemma 2.2. If for all ϵ>1, the following inequalities hold

    ϵ(ηi+μ+τ)>τ+ϵ(ϵ1)σ212+1,ϵ2(γ+μ+τ)>(γ+μ+τ)+ϵ2(ϵ1)σ222+1,ϵα>(ϵ1)β+ϵ(ϵ1)σ232+1,ϵκ>(ϵ1)(γ+μ)+β+ϵ(ϵ1)σ242+1, (2.3)

    there exists a positive constant ˆT(ϵ) such that the solution X(t)=(S(t),I(t),A(t),C(t)) with initial value X(0)R4+ has the following property lim suptE|X(t)|ˆT(ϵ). For this expression, the solution X(t) is stochastically ultimately bounded.

    The proof of Lemma 2.2 is shown in the Appendix.

    Let X(t) be a regular time-homogeneous Markov process in Rl described by the following stochastic differential eqaution: dX(t)=b(X)dt+kr=1gr(X)dBr(t). The diffusion matrix is defined as follows: B(X)=((bi,j(X))),bi,j(X)=kr=1girgjr(X), where X(t) is nonsingular [27], then we have the following lemma.

    Lemma 2.3. (Khasminskii [27]). The Markov process X(t) has a unique ergodic stationary distribution v() if there exists a bounded open domain URl with regular boundary Γ and if it has the following properties:

    (i) There exists a positive ˉE such that ni,j=1bij(X)ξiξjˉE|ξ|2, XU, ˉERl;

    (ii) There exists a nonnegative C2-function V such that LV is negative for any XRlU.

    Thus the Markov process X(t) has a stationary distribution v(). Let f(x) be a function integrable with respect to the measure π, for all xRl. We can get

    P{limT1TT0f(X(t))dt=Rlf(x)π(dx)}=1.

    According to the Routh-Hurwitz stability principle and Zhou et al. [28], the following algebraic equation solving theory is given.

    Lemma 2.4 (Zhou et al. [28]). For the four-dimensional real matrix D=diag(σ1,σ2,σ3,σ4), Z=(aij)4×4, the characteristic polynomial of Z is ωZ(ζ)=ζ4+s1ζ3+s2ζ2+s3ζ+s4. Assuming that Λ=(νij)4×4 is a symmetric matrix, we have the following algebraic equation

    D2+ZΛ+ΛZT=0.

    If Z has all negative real part eigenvaues, that is,

    s1>0,s3>0,s1s2s3>0,

    then Λ is a positive definite matrix.

    Using Lemmas 2.5–2.7, we will develop solutions for the four dimensions as described by Tan et al. [23] and Zhou et al. [29].

    Lemma 2.5 (Zhou et al. [29]). For the algebraic equation G20+B1X1+X1BT1=0 and G0=diag(1,0,0,0),

    B1=[a1a2a3a4100001000010],X1=[r110r1300r220r24r130r3300r240r44], (2.4)

    where r22=a32(a1(a2a3a1a4)a23),r13=r22,r33=a1a3r22,r11=a2a3a1a4a3r22,r24=r33,r44=a1a2a3a3a4r22. If a1>0, a3>0,a4>0 and a1a2a3>0,a1(a2a3a1a4)a23>0, then the matrix X1 is postive definite.

    Lemma 2.6 (Tan et al. [23]). For the algebraic equation G20+B2X2+X2BT2=0 and G0=diag(1,0,0,0),

    B2=[b1b2b3b410000100000b44],X2=[˜r110˜r1300˜r2200˜r130˜r3300000],

    where ˜r22=12(b1b2b3),˜r13=˜r22,˜r11=b2˜r22,˜r33=b1b3˜r22. If b1>0,b3>0 and b1b2b3>0, then the matrix X2 is semi-postive definite.

    Lemma 2.7 (Tan et al. [23]). For the algebraic equation G20+B2X2+X2BT2=0 and G0=diag(1,0,0,0),

    B3=[c1c2c3c4100000c33c3400c43c44],X3=[˜r110000˜r220000000000],

    where ˜r11=12c1,˜r22=12c1c2. If c1>0 and c2>0, then the matrix X3 is semi-postive definite.

    For a stochastic model, stability is the basis for studying other properties, so the stationary distribution is an important research content. Thus, the ergodic stationary distribution of the Model (2.2) is studied in the following section.

    Due to the complex structure of nonlinear systems, it is generally difficult to solve their state equations. Therefore, the Lyapunov method has been widely applied [30,31,32,33,34,35]. In this section, within the framework of the stochastic Lyapunov method, we will provide sufficient conditions for the ergodicity of the global positive solution and the existence of a stationary distribution.

    Define

    Rs:=ηiγ+μ+τ+σ222+γ+(μ+σ212)γ+μ+τ+σ222ηcS0δS0+(κ+σ242)+γ+(μ+σ212)γ+μ+τ+σ222βηaS0(α+σ232)[δS0+(κ+σ242)],

    where Rs is a stochastic threshold for Model (2.2). Next, its ergodicity is discussed using threshold.

    Theorem 3.1. If Rs>1, for any initial value (S(0),I(0),A(0),C(0))R4+, then Model (2.2) has a unique stationary distribution v(·) and it follows the ergodic property.

    Proof. The diffusion matrix of Model (2.2) is given by

    B(X)=(bi,j(X))4×4=[σ21S20000σ22I20000σ23A20000σ24C2], (3.1)

    where X=(S,I,A,C). Choose ˉE=min(SI,A,C)ˉUR4+{σ21S2,σ22I2,σ23A2,σ24C2}, then we have ˉE>0, where ˉU=[1nn]×[1nn]×[1nn]×[1nn]. For any (S,I,A,C)ˉU and (ξ1,ξ2,ξ3,ξ4)R4+, from (3.1), we have

    4i,j=1bij(X)ξiξj=σ21S2ξ21+σ22I2ξ22+σ23A2ξ23,+σ24C2ξ24ˉE|ξ|2,

    where |ξ|=(ξ21+ξ22+ξ23+ξ24)12, then the condition (i) in Lemma 2.3 holds.

    Construct the following function:

    W1=S+I+A+Cm1lnSm2lnIm2m3lnAm2m4lnC,

    and the positive constants mi (i=1,2,3,4) will be determined at a later stage. By Itˆo's fomular, it can be obtained that

    LW1=m1S[rN(1NK)ηaASηcCSηiSINμS+τI]m2I[ηaAS+ηcCS+ηiSIN(γ+μ+τ)I]m2m3A(αA+βC)m2m4C[(γ+μ)Iδ(S+I)CκC]+m1σ21S22S2+m2σ22I22I2+m2m3σ23A22A2+m2m4σ24C22C2+rN(1NK)μSαA+βCκCδ(S+I)C
    rN(1NK)[m1SrN(1NK)+m2ηiSN+δNC][m2m4(γ+μ)IC+m2ηcCSI][m1τIS+m2ηaASI+m2m3βCA]+m1ηcC+m1μ+m2(γ+μ+τ)+m2m3α+m2m4δ(S+I)+m2m4κ+m1σ212+m2σ222+m2m3σ232+m2m4σ242+ηim1INμSαA+βAκCrN(1NK)33m1m2rN(1NK)ηiδC33m1m22m3τηaβC2m22m4(γ+μ)ηcS+m1(ηaA+ηcC+μ+σ212)+m2(γ+μ+τ+σ222)+m2m3(α+σ232)+m2m4[δ(S+I)+κ+σ242]μSαA+βCκC+ηim1IN.

    Let

    m4[δ(S+I)+κ+σ24]2=(γ+μ)ηcS,m3(α+σ332)2=m1τηaβC,m2[(γ+μ+τ+σ222)+rN(1NK)τηaβC(ηaA+ηcC+μ+σ212)(α+σ232)+(γ+μ)ηcSδ(S+I)+κ+σ242]=m1(ηaA+ηcC+μ+σ212)=rN(1NK),

    then we can get

    m1=rN(1NK)ηaA+ηcC+μ+σ212,m2=rN(1NK)[(γ+μ+τ+σ212)+rN(1NK)τηaβC(ηaA+ηcC+μ+σ212)(α+σ232)+(γ+μ)ηcSδ(S+I)+κ+σ242],m3=rN(1NK)τηaβC(ηaA+ηcC+μ+σ212)(α+σ232)2,m4=(γ+μ)ηcS[δ(S+I)+κ+σ242]2.

    Thus,

    LW13rN(1NK)(3Rt1)+ηim1IN+βC.

    Define the function

    W(S,I,A,C)=M(W1+W2)+W3+W4+W5+W6,

    where W2=S+I,W3=lnS,W4=lnI,W5=lnA, and W6=1p+1(S+I)p+1. Select a suitable constant M>0 satisfying the following condiction

    3MrN(1NK)(3Rs1)+F12,

    where

    F1=sup(S,I,A,C)R4+{μ4Sp+1μ+γ4Ip+1+MrN(1NK)+(Mm1+1)ηi+p2(σ21Sp+1+σ21Sp+1)μ2Sp+1μ+γ2Ip+1+2μ+τ+γ+Mσ21S22+Mσ22I22+σ212+σ222+σ232}<.

    It is easy to check that

    lim infn(S,I,A,C)R4+UnW(S,I,A,C)=+,

    where Un=[1nn]×[1nn]×[1nn]×[1nn]. Furthermore, W(S,I,A,C) is a continuous function. Hence, (S,I,A,C) must have a minimum point (ˉS0,ˉI0,ˉA0,ˉC0) in the interior of R4+, then we define a nonnegative function ˉW:R4+R as follows

    ˉW(S,I,A,C)=W(S,I,A,C)W(ˉS0,ˉI0,ˉA0,ˉC0).

    By utilizing Itˆo's formula, we can derive the following result

    LW2=rN(1NK)μS(γ+μ)I+σ21S22+σ22I22rN(1NK)+σ21S22+σ22I22,LW3=1S[rN(1NK)ηaASηcCSηiSINμS+τI]+σ212rN(1NK)S+ηiIN+ηaA+ηcC+μ+σ212,LW4=1I[ηiIN+ηaA+ηcC(γ+μ+τ)I]+σ222ηaAIηcCI+(γ+μ+τ)+σ222,LW5=1A(αA+βC)+σ232AβCA+σ232,

    and

    LW6=(S+I)p[rN(1NK)μS(γ+μ)I]+p2(S+I)p1(σ21S2+σ22I2)rN(1NK)(S+I)pμ(S+I)p+1γIp+1+p2(σ21Sp+1+σ22Ip+1).

    Choose 0<p<min{1,μσ21μ+γσ22}, then we can see that

    LW6μ2Sp+1μ+γ2Ip+1+rN(1NK)(S+I)P+p2(σ21Sp+1+σ22Ip+1)μ2Sp+1μ+γ2Ip+1,

    hence,

    LˉW3MrN(1NK)(3Rt1)+m1ηiIMN+MrN(1NK)+MβC+ηaA+ηcC+A+ηiINrN(1NK)SηaAIηcCIμ2Sp+1μ+γ2Ip+1+rN(1NK)(S+I)P+p2(σ21Sp+1+σ22Ip+1)μ2Sp+1μ+γ2Ip+1+2μ+γ+τ+Mσ21S22+Mσ22I22+σ212+σ222+σ232,3MrN(1NK)(3Rt1)+MrN(1NK)+MβC+(Mm1+1)ηirN(1NK)SηaAIηcCI+ηaA+ηcC+Aμ2Sp+1μ+γ2Ip+1+rN(1NK)(S+I)P+p2(σ21Sp+1+σ22Ip+1)μ2Sp+1μ+γ2Ip+1+2μ+γ+τ+Mσ21S22+Mσ22I22+σ212+σ222+σ232.

    Next, we form a condensed subset D in order to satisfy the condition (ii) as outlined in Lemma 2.3. Define the bounded closed set

    D:={ξS1ξ,ξ2I1ξ2,ξA1ξ,ξC1ξ},

    where ξ is a suffciently small constant. Choose ξ satisfying the following conditions

    3MrN(1NK)(3Rt1)+βMξ+F11,rN(1NK)ξ+F21,ηaξ+F21,ηcξ+F21,μ4ξp+1+F21,μ+γ4ξ2(p+1)+F21,

    where

    F2=sup(S,I,A,C)R4+{μ4Sp+1μ+γ4Ip+1+MrN(1NK)+MβC+(Mm1+1)ηi+p2(σ21Sp+1+σ21Sp+1)μ2Sp+1μ+γ2Ip+1+2μ+τ+γ+Mσ21S22+Mσ22I22+σ212+σ222+σ232}<.

    Divide R4+D into six domains:

    D1={(S,I,A,C)R4+,0<C<ξ},D2={(S,I,A,C)R4+,0<S<ξ},D3={(S,I,A,C)R4+,0<I<ξ2,Aξ},D4={(S,I,A,C)R4+,0<I<ξ2,Cξ},D5={(S,I,A,C)R4+,0<S<1ξ},D6={(S,I,A,C)R4+,0<I<1ξ2}.

    We aim to demonstrate that the inequality LˉW(S,I,A,C)1 holds true on the domain R4+D. This is equivalent to establishing its validity on the six specified domains.

    Case 1: If (S,I,A,C)D1, one can see that

    LˉW3MrN(1NK)(3Rt1)+βMC+F13MrN(1NK)(3Rt1)+βMξ+F11.

    Case 2: If (S,I,A,C)D2, one can see that

    LˉWrN(1NK)S+F2rN(1NK)ξ+F21.

    Case 3: If (S,I,A,C)D3, one can see that

    LˉWηaAI+F2ηaξ+F21.

    Case 4: If (S,I,A,C)D4, one can see that

    LˉWηcCI+F2ηcξ+F21.

    Case 5: If (S,I,A,C)D5, one can see that

    LˉWμ4Sp+1+F2μ4ξp+1+F21.

    Case 6: If (S,I,A,C)D6, one can see that

    LˉWμ+γ4Ip+1+F2μ+γ4ξ2(p+1)+F21.

    Hence, we obtain LˉW1 for all (S,I,A,C)R4+D. According to Lemma 2.3, it can be deduced that the Model (2.2) possesses a unique stationary distribution denoted as v(·). This serves to conclude the proof.

    Remark 1. As we know, a deterministic system disease persistence is measured according to the basic reproduction number R0 and the endemic equilibrium point E. However, in a stochastic model, there is no positive equilibrium due to unpredictable noise perturbations. Therefore, the persistence of the disease in a stochastic system is reflected by the presence of an ergodicity stationary distribution. According to Theorem 3.1, Model (2.2) has a unique stationary distribution when Rs>1. Comparing the expressions for R0 and Rs, we find that RsR0 and the presence of disease in both the model and the stochastic model can be assessed using a consistent measure of Rs>1.

    Remark 2. Model (2.2) characterizes the impact of random noise on the spread of anthrax, which is an extension of Model (2.1). When random noise is absent, i.e., σi=0 (i=1,2,3,4), Model (2.2) degenerates into Model (2.1).

    This section aims to derive the precise formulation of the density function for Model (2.2) at a quasi-equilibrium point. It is important to emphasize that the probability density function serves as a representation of the dynamic attributes of a stochastic system with statistical relevance. Our main goal is to ascertain the probability density function for the purpose of examining the dynamics of the model from a statistical perspective.

    Before proving the probability density function, two equivalent differential equation forms for Model (2.2) are given. Let x1=lnS,x2=lnI,x3=lnA,x4=lnC. By Itˆo's formula, an equivalent model can be obtained:

    {dx1=[rN(1NK)ex1ηiex2ex1+ex2ηaex3ηcex4+τex2ex1μσ212]dt+σ1dB1(t),dx2=[ηiex1ex1+ex2+ηaex1ex3ex2+ηcex1ex4ex2(γ+μ+τ)σ222]dt+σ2dB2(t),dx3=[βex4ex3ασ232]dt+σ3dB3(t),dx4=[(γ+μ)ex2ex4δ(ex1+ex2)κσ242]dt+σ4dB4(t). (4.1)

    Let b1=μ+σ212,b2=γ+μ+τ+σ222,b3=α+σ232,b4=κ+σ242. If

    ˉRs:=ηiδC[δ(S+I)+b4]b3b3(ηaA+ηcC+b1)[(γ+μ)ηcS+b2δ(S+I)+b2b4]+τηaβrN(1NK)C[δ(S+I)+b4]>1

    holds, Model (2.2) admits a quasi-steady ˉE1=(ˉS1,ˉI1,ˉA1,ˉC1)=(ex1,ex2,ex3,ex4), where ˉI1 is the positive root of the equtaion

    F(I)=[b21b2b3+b1b2b3δ(τb2)+ηaβ(γ+μ)(b2τ)(τb2)+ηcb3(γ+μ)(b2τ)(τb2)]ˉI21+[b21b2b3b4+b1b2b3δrN(1NK)+b1b3(b2ηi)+b3δ(b2ηi)(τb2)b1ηaβ(γ+μ)+2ηaβ(γ+μ)rN(1NK)(b2τ)+2b3ηc(γ+μ)rN(1NK)(b2τ)+b1b3ηc(γ+μ)(b2τ)]ˉI1{b3(ηib2)δrN(1NK)+ηaβ(γ+μ)[rN(1NK)]2+b1ηa(γ+μ)rN(1NK)+b3ηc(γ+μ)[rN(1NK)]2+b1b3ηc(γ+μ)rN(1NK)},

    and

    ˉS1=rN(1NK)+ˉI1(b2τ)b1,ˉC1=b1(γ+μ)ˉI1b1b4+b1ˉI1+δ[rN(1NK)+ˉI1(b2τ)],ˉA1=b1(γ+μ)βˉI1b1b3b4+b1b3ˉI1+b3δ[rN(1NK)+ˉI1(b2τ)].

    In addition, assume if Rs>1, then ˉRs>1, which means that Model (2.2) admits a quasi-steady ˉE1 under the condition Rs>1. Let

    yi=xixi(i=1,2,3,4), (4.2)

    where x1=lnˉS1,x2=lnˉI1,x3=lnˉA1,x4=lnˉC1, then the linearized system of (4.1) is as follows:

    {dy1=(a11y1a12y2a13y3a14y4)dt+σ1dB1(t),dy2=(a21y1a21y2+a23y3+a24y4)dt+σ2dB2(t),dy3=(a33y3+a33y4)dt+σ3dB3(t),dy4=(a41y1a42y2a44y4)dt+σ4dB4(t), (4.3)

    where

    a11=μ+σ212+ηiˉI21ˉN21+ηaˉA1+ηcˉC1,a12=ηiˉS1ˉI1ˉN21τˉI1ˉS1,a13=ηaˉA1,a14=ηcˉC1,a21=ηiˉS1ˉI1ˉN21+ηaˉS1ˉA1+ηcˉS1ˉC1ˉI1,a23=ηaˉS1ˉA1ˉI1,a24=ηcˉS1ˉC1ˉI1,a33=βˉC1ˉA1,a41=δˉS1,a42=δˉI1(γ+μ)ˉI1ˉC1,a44=(γ+μ)ˉI1ˉC1.

    By employing the above lemmas, we will demonstrate the probability density function of the Model (2.2) near the quasi-steady equilibrium ˉE1.

    Let Y=(y1,y2,y3,y4)T,D=diag(σ1,σ2,σ3,σ4), and B(t)=(B1(t),B2(t),B3(t),B4(t))T. We can rewrite Model (4.3) as the following form:

    dY=ZYdt+DdB(t),

    where

    Z=[a11a12a13a14a21a21a23a2400a33a33a41a420a44],

    then the linearized Model (4.3) can be simiplified to dY=ZYdt+DdB(t). Based on the pertinent theory proposed by Zhou et al., it is postulated that there is a probability density function denoted as Φ(y1,y2,y3,y4) in the vicinity of the quasi-endemic equilibrium ˉE1.

    Theorem 4.1. [Zhou et al. [22]] Assuming that Rs>1 for any initial value (S(0),I(0),A(0),C(0)R4+, the solution (S(t),I(t),A(t),C(t)) of the Model (2.2) follows the unique log-normal probability density function Φ(S,I,A,C) around the quasi-endemic equilibrium ˉE1, which is described by

    Φ(S,I,A,C)=(2π)2|Λ|12e12(lnSˉS1+lnIˉI1+lnAˉA1+lnCˉC1)Λ1(lnSˉS1+lnIˉI1+lnAˉA1+lnCˉC1)T,

    where Λ is a positive definite matrix.

    From Gardiner [36], the density function U(Y)=U(y1,y2,y3,y4) of the quasi-stationary distribution of Model (4.3) at the origin Y=(0,0,0,0) can be described by the four-dimensional Fokker-Planck equation as follows:

    4i=1σ2i22Uy2i+y1[(a11y1+a12y2a13y3a14y4)U]+y2[(a21y1a21y2+a23y3+a24y4)U]+y3[(a33y3+a33y4)U]+y4[(a41y1+a42y2a44y4)U]=0,

    which can be approximated by a Gaussian distribution

    U(Y)=ˉcexp{12(YY)ˉM(YY)T},

    ˉM is a real symmetric matrix, which is described by ˉMD2ˉM+ZTˉM+ˉMZ=0. If ˉM is positive definite, let ˉM1=Λ, then

    D2+ZΛ+ΛZT=0. (4.4)

    According to Tian et al. [37], (4.4) can be rewriten as the sum of the four equations as follows:

    D2i+ZΛi+ΛiZT=0(i=1,2,3,4),

    where D1=(σ1,0,0,0),D2=(0,σ2,0,0),D3=(0,0,σ3,0),D4=(0,0,0,σ4),Λ=Λ1+Λ2+Λ3+Λ4, D2=D21+D22+D23+D24. Consider the corresponding characteristic equation of Z as follows:

    ωZ(ζ)=ζ4+s1ζ3+s2ζ2+s3ζ+s4, (4.5)

    where

    s1=a11+a21+a33+a44,s2=a11(a21+a33+a44)+a21(a12+a33+a44)+a33a44+a24a42a14a41,s3=a11a21(a33+a44)+a33a44(a11+a21)+a12a21(a33+a44)a14a21(a41+a42)+a24a41(a11a12)a33a41(a13+a14)+a23a33a42+a24a33a41,s4=a21a33a44(a11+a12)+a33a42(a11a23+a11a24a33a41a14a21a13a21)a12a33a41(a23+a24)a21a33a41(a13+a14).

    In s2, the following formula can be calculated

    a21a44+a24a42a14a41=ηi(γ+μ)ˉS1ˉI21ˉC1+ηa(γ+μ)ˉS1ˉA1ˉC1>0,a12a21=ηiˉI21ˉN21(τηiˉS21ˉN21)+(ηaˉA1+ηcˉC1)(τηiˉS21ˉN21)>0.

    In s3, the following formula can be calculated

    a21a33a44+a23a33a42a14a33a41a13a33a41=ηi(γ+μ)βˉS1ˉI21ˉN21ˉA1+ηcβˉS1ˉC1ˉA1(γ+μδˉC1)>0,a12a21a44=ηi(γ+μ)ˉI31ˉN21ˉC1(τηiˉS21ˉN21)+ηa(γ+μ)ˉI1ˉA1ˉC1(τηiˉS21ˉN21)+ηc(γ+μ)ˉI1(τηiˉS21ˉN21)>0,a12a21a33=ηiβˉI21ˉC1ˉN21ˉA1(τηiˉS21ˉN21)+ηaβˉC1(τηiˉS21ˉN21)+ηcβˉC21ˉA1(τηiˉS21ˉN21)>0,a14a21a42=ηiηcˉS1ˉI21ˉN21(γ+μδˉC1)+ηaηcˉS1ˉA1(γ+μδˉC1)+η2cˉS1ˉC1(γ+μδˉC1)>0.

    In s4, the following formula can be calculated

    a14a21a33a42=ηiηcβˉS1ˉI21ˉC1ˉN21ˉA1(γ+μδˉC1)+ηaηcδˉS1ˉC1(γ+μδˉC1)+η2cβˉS1ˉC21ˉA1(γ+μδˉC1)>0,a2a21a33a44=ηiβ(γ+μ)ˉI31ˉN21ˉA1(τηiˉS21ˉN21)+ηaβ(γ+μ)ˉI1(τηiˉS21ˉN21)+ηcβ(γ+μ)ˉI1ˉC1ˉA1(τηiˉS21ˉN21)>0,

    where τηiˉS21ˉN21>0 and

    (γ+μδˉC1)=b1b4(γ+μ)+[b1ˉI1(γ+μ)δb1ˉI1(γ+μ)]+δ(γ+μ)[rN(1NK)+ˉI1(τb2)]b1b4+b1ˉI1+δ[rN(1NK)+ˉI1(τb2)]>0,

    while s1>0, s2>0, and s3>0. We can calculate s1s2s3>0 and s1(s2s3s1s4)s23>0. As per the Routh-Hurwitz stability criterion [38], it can be deduced that the matrix Z possesses eigenvalues with negative real parts. According to the theory of matrix similar transformations, it is evident that ωZ(ζ) is a similarity invariant, indicating that s1,s2,s3, and s4 are also similarity invariants.

    Next, the corresponding proof for the positive definiteness of Λ is divided into four steps.

    Proof. Step 1. Consider the algebraic equation

    D21+ZΛ1+Λ1ZT=0, (4.6)

    Let Z1=U1ZU11, where

    U1=[1000010000100a41a2101],

    and

    Z1=[a11a14a41a12a21a21a13a14a21a221+a24a41a21a23a240w1a33a330w2a23a41a21a24a41a21a44a21],

    where w1=a33a41a21,w2=a41a42+a41a44a21a24a241a221.

    Case 1: When w10 and w20, there exists the elimination matrix U11 such that Z11=U11Z1U111, where

    U11=[10000100001000w2w11],

    then there is

    Z11=[a11a14a41a12a21a21a13a14a21a221+a24a41a21a23+a24w2w1a240w1a33+a33w2w1a3300w3a24a41a21a44a21a33w2w1],

    where w3=w2w1a33w1a33w2w1+w2w1a24a41a21a44a21+a23a41a21.

    Case 1-1: If w30, in order to find the standard transformation matrix Q111 such that T111=Q111Z11Q1111, where

    Q111=[m1m2m3m40w1w3a24a41a21a44a21a33a33w3(a24a41a21a44a21a33w2w1)200w3a24a41a21a44a21a33w2w10001],

    with

    m1=a21w1w3,m2=(a24a41+a24a41a21a44a221a33a21)w1w3,m3=w3[a23w1+a24w2+a33w3+(a33(w2w1)w2)(a24a41a21a44a21a33w2w1)+(a33(w2w1)w2)2+(a24a41a21a44a21a33w2w1)2],m4=[a33w3+(a24a41a21a44a21a33w2w1)2](a24a41a21a44a21a33w2w1)+a24w1w3+a33w3(a24a41a21a44a21a33a21).

    Thus, we can get

    T111=[s1s2s3s4100001000010],

    Meanwhile, (4.6) has the following form:

    (Q111U11U1)D21(Q111U11U1)T+[(Q111U11U1)Z(Q111U11U1)1](Q111U11U1)Λ1(Q111U11U1)T+(Q111U11U1)Λ1(Q111U11U1)T[(Q111U11U1)Z(Q111U11U1)1]T=0.

    That is, D20+T111Θ1+Θ1TT111=0, where Θ1=1ρ2111(Q111U11U1)Λ1(Q111U11U1)T, ρ111=a21w1w3σ1.

    According to Lemma 2.5, we can obtain the semipositive definite matrix

    Θ1=[s2s3s1s42[s1(s2s3s1s4)s23]0s32[s1(s2s3s1s4)s23]00s32[s1(s2s3s1s4)s23]0s12[s1(s2s3s1s4)s23]s32[s1(s2s3s1s4)s23]0s12[s1(s2s3s1s4)s23]00s12[s1(s2s3s1s4)s23]0s1s2s32[s1(s2s3s1s4)s23]],

    Therefore, it can be calculated that

    Λ1=ρ2111(Q111U11U1)1Θ1[(Q111U11U1)1]T (4.7)

    is positive definite.

    Case 1-2: If w3=0, we can find the standard transformation matrix Q112 such that T112=Q112Z11Q1112, where

    Q112=[m1m2m3m40w1a33(w2w1)w1a3300100001],

    with

    m1=a21w1,m2=a33w2a33w1w1a221+a24a41a21,m3=a23w1+a24w2+(a33w2a33w1w1)2,m4=a24w1+a33(a33w2a33w1w1)+a33(a24a41a21a44a21a33w2w1).

    We have

    T112=[b1b2b3b410000100000a24a41a21a44a21a33w2w1],

    where bi(i=1,2,3,4) are marks, and we only care about their traits. According to the uniqueness of ωZ(ζ), it can be obtained that

    b1=a11+a33+a44a21a42a41>0,b2=a11a33+a11a44+a33a44+a221a42a41+a221a242a241a11a21a42a41a21a33a42a41a21a42a44a41>0,b3=a11a33a44+a12a21(a33+a44)+a24a41(a11a12)a33a41(a13+a14)+a23a33a42+a24a33a41+a21a33a42(a21a11)a41+a21a42a44(a21a11)a41+a11a221a242a241+a221a33a242a241+a221a242a44a241>0.

    The following results can be obtained that b1b2b3>0, then (4.6) can be transformed into the following form:

    (Q112U11U1)D21(Q112U11U1)T+[(Q112U11U1)Z(Q112U11U1)1](Q112U11U1)Λ1(Q112U11U1)T+(Q112U11U1)Λ1(Q112U11U1)T[(Q112U11U1)Z(Q112U11U1)1]T=0.

    That is, D20+T112Θ2+Θ2TT112=0, where Θ2=1ρ2112(Q112U11U1)Λ1(Q112U11U1)T, ρ112=a21w1σ1.

    According to the Lemma 2.6, we can obtain the semipositive definite matrix

    Θ2=[b22(b1b2b3)012(b1b2b3)0012(b1b2b3)0012(b1b2b3)0b12b3(b1b2b3)00000],

    Therefore, by calculation,

    Λ1=ρ2111(Q111U11U1)1Θ2[(Q111U11U1)1]T (4.8)

    is semi-postive definite.

    Case 2: When w10 and w2=0, the objective is to determine the standard transformation matrix Q12 that satisfies the equation T12=Q12Z1Q112, where

    Q12=[a23a33a241a21m2m3m40a23a33a241a221(a33+a24a41a21a44a21)a23a41a21[a33+(a24a41a21a44a21)2]a23a41a2100a23a41a21a24a41a21a44a210001],

    with

    m2=(a221+a24a41a21+a33a24a41a21a44a21)a23a33a241a221,m3=a23a41a21[a233a33a24a41a21a44a21+(a24a41a21a44a21)2],m4=a23a24a33a241a221+(a33+a24a41a21a44a21)a23a33a41a21.

    We can obtain that T12=T111, and (4.6) can be transformed into

    D20+T12Θ1+Θ1TT12=0.

    where Θ1=1ρ212(Q12U1)Λ1(Q12U1)T, ρ12=a23a33a241a21σ1. Consequently,

    Λ1=ρ212(Q12U1)1Θ1[(Q12U1)1]T (4.9)

    is a positive definite matrix.

    Case 3: When w1=0 and w20, there exists the transformation matrix U13 such that Z13=U13Z1U113, where

    U13=[1000010000010010],

    then we have

    Z13=[a11a14a41a12a21a21a14a13a21a221+a24a41a21a24a230w2a24a41a21a44a21a23a41a2100a33a33],

    we are able to find the standard transformation matrix Q13 such that T13=Q13Z13Q113, where

    Q13=[m1m2m3m40a33(a221+a24a41a21)a33(a33+a24a41a21a44a21)a233+a23a33a41a2100a33a330001],

    with

    m1=a21a33w2,m2=a33(a24a41a21a44a21a221+a24a41a21a33)a24a41a21a44a21,m3=a33[a24w2+a23a33a41a21a33(a24a41a21a44a21)+a233+(a24a41a21a44a21)2],m4=a23a33w2+a23a33a41a21(a24a41a21a44a21a33)a33(a23a33a41a21+a233).

    We can obtain that T13=T111, and (4.6) can be transformed into

    D20+T13Θ1+Θ1TT13=0,

    where Θ1=1ρ213(Q13U13U1)Λ1(Q13U13U1)T, ρ13=a21w2σ1. Consequently,

    Λ1=ρ213(Q13U13U1)1Θ1[(Q13U13U1)1]T (4.10)

    is a positive definite matrix.

    Case 4: When w1=0 and w2=0, there exists the transformation matrix U14 such that T14=Q14Z1Q114, where

    Q14=[m1m2m3m4a21a221+a24a41a21a23a2400100001],

    with

    m1=a21(a11a21+a221+a24a41a21),m2=a14a41a12a21+(a221+a24a41a21)2,m3=a13a21a23a33a23a221a21,m4=a23a33a14a21a24a21a24a44.

    We then have

    T14=[c1c2c3c4100000a33a3300a23a41a21a24a41a21a44a21],

    c1=a11+a21+a24a41a21>0, c2=(a11+a12)a21+(a41+a42)a24a14a41+a11a24a41a21+a23a33a41a21+a224a241a221>0. Thus, (4.6) can be transformed into

    D20+T14Θ3+Θ3TT14=0,

    where Θ3=1ρ214(Q14U1)Λ1(Q14U1)T, ρ14=a21σ1. With Lemma 2.7, we can obtain the semi-postive definite matrix

    Θ3=[12c1000012c1c20000000000].

    Therefore, by calculation,

    Λ1=ρ214(Q14U1)1Θ3[(Q14U1)1]T (4.11)

    is semi-postive definite.

    Step 2. For the following algebraic equation

    D22+ZΛ2+Λ2ZT=0, (4.12)

    consider the corresponding order matrix J2 and elimination matrix U2:

    J2=[0100001000011000],U2=[100001000a44a33100a14a3301].

    Let Z2=(U2J2)Z(U2J2)1, then we can obtain that

    Z2=[a24a21a33+a21a44a14a23a33a12a23a33a140a330r1a42a440r2a12a13a14],

    where r1=a41+a14a44+a42a44a33, r2=a11+a13a14+a12a44+a214a33.

    Case 1: When r10,r20, there exists the elimination matrix U21 such that Z21=U21Z2U121, where

    U21=[10000100001000r2r11],

    then we have

    Z21=[a24a21+a21a44a14a23a33a12+a23r2r1a23a33a14a33r2r1a330r1a42a44r2r1a4400r3a13a14+a44r2r1],

    where r3=(a42+a44r2r1)r2r1a12(a13+a14)r2r1.

    Case 1-1: If r30, the standard transformation matrix Q211 can be determined in order to obtain the transformation T211 as T211=Q211Z21Q1211, where

    Q211=[n1n2n3n40r1r3(a13+a14+a42)r3(a44r2r1a13a14)2a44r300r3a44r2r1a13a140001],

    with

    n1=a33r1r3,n2=(a13a42)r1r3,n3=[a33r2a44r3+(a42+a44r2r1)(a13+a14a44r2r1)+(a42+a44r2r1)2+(a44r2r1a13a14)2]r3,n4=a33r1r3+a44(a13+a14+a42)r3+[a44r3+(a44r2r1a13a14)2](a44r2r1a13a14).

    As a result, we have T211=T111, and (4.12) has the following form:

    (Q211U21U2J2)D22(Q211U21U2J2)T+[(Q211U21U2J2)Z(Q211U21U2J2)1](Q211U21U2J2)Λ2(Q211U21U2J2)T+(Q211U21U2J2)Λ2(Q211U21U2J2)T[(Q211U21U2J2)Z(Q211U21U2J2)1]T=0.

    That is, D20+T211˜Θ1+˜Θ1T211=0, where ˜Θ1=1ρ2211(Q211U21U2J2)Λ2(Q211U21U2J2)T, ρ211=a33r1r3σ2. According to Lemma 2.5, we can obtain the semipositive definite matrix

    ˜Θ1=[s2s3s1s42[s1(s2s3s1s4)s23]0s32[s1(s2s3s1s4)s23]00s32[s1(s2s3s1s4)s23]0s12[s1(s2s3s1s4)s23]s32[s1(s2s3s1s4)s23]0s12[s1(s2s3s1s4)s23]00s12[s1(s2s3s1s4)s23]0s1s2s32[s1(s2s3s1s4)s23]].

    Therefore, it can be obtained by calculation that

    Λ2=ρ2211(Q211U21U2J2)1˜Θ1[(Q211U21U2J2)1]T. (4.13)

    Case 1-2: If r3=0, we can find the standard transformation matrix Q212 such that T212=Q212Z21Q1212, where

    Q212=[a33r1(a14a42a44r2r1)r1(a42+a44r2r1)2a33r2(a13+a14+a42)a44a33r10r1(a42+a44r2r1)a4400100001],

    then we have

    T212=[˜b1˜b2˜b3˜b410000100000a44r2r1a13a14],

    where

    ˜b1=1a14a44+a42a44a33a41(a214a44a12a244+a11a33a44+a13a242+a13a33a41+a214a44)+1a14a44+a42a44a33a41(a11a14a44+a11a42a44a11a33a41+a14a21a44a14a33a41)+1a14a44+a42a44a33a41(a14a33a44+a33a42a44a233a41+a14a244+a42a244)+1a14a44+a42a44a33a41(+a14a42a44+a21a42a44a21a33a41a33a41a44)>0,˜b2=[a11(a21+a33+a44)+a21(a12+a33+a44)+a33a44+a24a42a14a41]+1(a14a44+a42a44a33a41)2(a214a44a12a244+a11a33a44)(a13a242+a13a33a41+a214a44)+1(a14a44+a42a44a33a41)2(a13a14a44+a214a44+a12a244a11a33a44)(a14a42a44a14a33a41)+1(a14a44+a42a44a33a41)2(a13a14a44+a214a44+a12a244a11a33a44)+1(a14a44+a42a44a33a41)2(a13a14a44+a214a44+a12a244a11a33a44)>0,˜b3=a14a44+a42a44a33a41a13a14a44+a214a44+a12a244a13a33a44[a33a42(a11a23+a11a24a33a41a14a21a13a21)]+a14a44+a42a44a33a41a13a14a44+a214a44+a12a244a13a33a44(+a21a33a44(a11+a12)+a21a33a41(a13+a14))+a14a44+a42a44a33a41a13a14a44+a214a44+a12a244a13a33a44[a12a33a41(a23+a24)]>0.

    We can obtain ˜b1˜b2˜b3>0, then (4.12) can be transformed into the following form:

    (Q212U21U2J2)D22(Q212U21U2J2)T+[(Q212U12U2J2)Z(Q212U21U2J2)1](Q212U21U2J2)Λ2(Q212U21U2J2)T+(Q212U21U2J2)Λ2(Q212U21U2J2)T[(Q212U21U2J2)Z(Q212U21U2J2)1]T=0.

    That is, D20+T212˜Θ2+˜Θ2TT212=0, where ˜Θ2=1ρ2212(Q212U21U2J2)Λ2(Q212U21U2J2)T, ρ212=a33r1σ2. According to Lemma 2.6, we can obtain the semipositive definite matrix

    ˜Θ2=[˜b22(˜b1˜b2˜b3)012(˜b1˜b2˜b3)0012(˜b1˜b2˜b3)0012(˜b1˜b2˜b3)0˜b12˜b3(˜b1˜b2˜b3)00000].

    Therefore, it can be obtained by calculation that

    Λ2=ρ2212(Q212U21U2J2)1˜Θ2[(Q212U21U2J2)1]T (4.14)

    is semi-postive definite.

    Case 2: When r10 and r2=0, there exists the transformation matrix Q22 such that T22=Q22Z2Q122, where

    Q22=[a12a33r1a12(a13+a42)r1n3n40a12r1a12(a13+a14+a42)(a13+a14)2+a12a4400a12a13a140001],

    with

    n3=a12[a12a44+a42(a13+a14)+a242+(a13+a14)2],n4=a12a33r1a12a44(a13+a14+a42)[a12a44+(a13+a14)2](a13+a14).

    It can be obtained that T22=T111, and (4.12) can be transformed into

    D20+T22˜Θ1+˜Θ1TT22=0,

    where ˜Θ1=1ρ222(Q22U2J2)Λ2(Q22U2J2)T, ρ22=a21a33r1σ2. Consequently,

    Λ2=ρ222(Q22U2J2)1˜Θ1[(Q22U2J2)1]T (4.15)

    is a positive definite matrix.

    Case 3: When r1=0 and r20, there exists the transformation matrix U23 such that Z23=U23Z2U123

    U23=[1000010000010010],

    then we have

    Z23=[a24a21+a21a44a14a23a33a23a21a33a14a3300r2a13a14a1200a44a42],

    We can find the standard transformation matrix Q23 such that T23=Q23Z23Q123, where

    Q23=[a33a44r2a44(a13+a42)r2m3m40a44r2a44(a13+a14+a42)a242+a12a4400a44a420001],

    with

    m3=a44[a33r2+a12a44+a42(a13+a14)+a224+(a13+a14)2]m4=a12a33r1a12a44(a13+a14+a42)[a12a44+(a13+a14)2](a13+a14).

    It can be obtained by calculation that T23=T111, and (4.12) can be transformed into

    D20+T23˜Θ1+˜Θ1TT23=0,

    where ˜Θ1=1ρ223(Q23U23U2J2)Λ2(Q23U23U2J2)T, ρ23=a33a44r2σ2. Consequently,

    Λ2=ρ223(Q23U23U2J2)1˜Θ1[(Q23U23U2J2)1]T (4.16)

    is a positive definite matrix.

    Case 4: When r1=0 and r2=0, there exists the transformation matrix Q24 such that T24=Q24Z2Q124, where

    Q24=[a33(a14+a24)a21a33+a21a44a14a23+a214a21a33+a12a33(a13+a23)a33a33a140a3300100001],

    then we have

    T24=[˜c1˜c2˜c3˜c4100000a42a4400a12a13a14],

    where c1=a11+a21+a33+a44a13a14a42>0, c2=s4a13a42+a14a42a12a44>0. Thus, (4.12) can be transformed into

    D20+T24˜Θ3+˜Θ3TT24=0,

    where ˜Θ3=1ρ224(Q24U2J2)Λ2(Q24U2J2)T, ρ24=a33σ2. With Lemma 2.7, we can obtain the positive semidefinite matrix

    ˜Θ3=[12˜c1000012˜c1˜c20000000000].

    Therefore, it can be obtained by calculation that

    Λ2=ρ224(Q24U2J2)1˜Θ3[(Q24U2J2)1]T (4.17)

    is semi-postive definite.

    Step 3. For the following algebraic equation

    D23+ZΛ3+Λ3ZT=0, (4.18)

    consider the corresponding order matrix J3 and elimination matrix U3 :

    J3=[0010000110000100],U3=[10000100001000a23a131].

    Let Z3=(U3J3)Z(U3J3)1. It can be obtained by calculation that

    Z3=[a330a330a13a12a23a11a13a13a14a120a23a42a13a41a13a44a420k1k2a12a23+a13a21a13],

    where k1=a21+a21a23a11a23a13+a12a223a213, k2=a24a14a23a13.

    Case 1: When k1=0 and k20, there exists the standard transformation matrix Q31 such that T31=Q31Z3Q131, where

    Q31=[q1q2q3q40(a41+a23a42a13)k2(a44+a12a23+a13a21a13)k2(a12a23+a13a21a13)2a42k200k2a12a23+a13a21a130001],

    with

    q1=(a41+a23a42a13)a13k2,q2=(a11a44+a13a21a13)(a41+a23a42a13)k2,q3=[a14a41a14a23a42a13a42k2+a12a23a44+a13a21a44a13+a244+(a12a23+a13a21a13)2]k2,q4=(a41a23a42a13)a12k2+(a44+a12a23+a13a21a13)a42k2+[a42k2(a12a23+a13a21a13)2]a12a23+a13a21a13.

    We can obtain that T31=T111, and (4.18) can be transformed into

    D20+T31ˉΘ1+ˉΘ1TT31=0,

    where ˉΘ1=1ρ231(Q31U3J3)Λ3(Q31U3J3)T, ρ31=(a13a41a23a42)k2σ3, and ˉΘ1=Θ1. Consequently,

    Λ3=ρ231(Q31U3J3)1ˉΘ1[(Q31U3J3)1]T (4.19)

    is positive definite matrix.

    Case 2: When k_1\neq0 , k_2 = 0 , there exists the transformation matrix U_{32} such that Z_{32} = U_{32}Z_3U_{32}^{-1} , where

    \begin{equation*} U_{32} = \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&-\frac{a_{13}k_1}{a_{23}a_{42}-a_{13}a_{41}}&1 \end{bmatrix}, \end{equation*}

    then we have

    \begin{equation*} Z_{32} = \begin{bmatrix} -a_{33}&0&a_{33}&0\\ -a_{13}&-a_{11}+\frac{a_{12}a_{23}}{a_{13}}&-a_{14}+\frac{a_{12}a_{13}k_1}{a_{23}a_{42}-a_{13}a_{41}}&a_{12}\\ 0&-a_{41}+\frac{a_{23}a_{42}}{a_{13}}&-a_{44}+\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}}&a_{42}\\ 0&0&k_3&\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}}-\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}} \end{bmatrix}, \end{equation*}

    where k_3 = \frac{(a_{13}a_{44}+a_{12}a_{23}+a_{13}a_{21})k_1}{a_{23}a_{42}-a_{13}a_{41}}-\frac{a_{13}^2a_{42}k_1^2}{(a_{23}a_{42}-a_{13}a_{41})^2} .

    Case 2-1: If k_3\neq0 , we can find the standard transformation matrix Q_{321} such that T_{321} = Q_{321}Z_{32}T_{321}^{-1} , where

    \begin{equation*} Q_{321} = \begin{bmatrix} q_1&q_2&q_3&q_4\\ 0&(-a_{41}+\frac{a_{23}a_{24}}{a_{13}})k_3&(-a_{44}+\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}})k_3&(\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}}-\frac{a_{13}a_{42}k_1}{a_{23}a_{24}-a_{13}a_{41}})^2+a_{42}k_3\\ 0&0&k_3&\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}}-\frac{a_{13}a_{42}k_1}{a_{23}a_{24}-a_{13}a_{41}}\\ 0&0&0&1 \end{bmatrix}, \end{equation*}

    with

    \begin{equation*} \begin{aligned} q_1 = &(a_{13}a_{41}-a_{23}a_{42})k_3, \; q_2 = (-a_{11}-a_{44}+\frac{a_{12}a_{23}}{a_{13}}+\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}})a_{42}k_3, \\ q_3 = &k_3[(-a_{41}+\frac{a_{23}a_{24}}{a_{13}})(-a_{14}+\frac{a_{12}a_{13}k_1}{a_{23}a_{42}-a_{13}a_{41}})+a_{42}k_2+(-a_{44}+\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}})\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}}\\ &-(-a_{44}+\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}})\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}}+(-a_{44}+\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}})^2\\ &+(\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}}-\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}})^2], \\ q_4 = &[a_{42}k_3+(\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}}-\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}})^2](\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}}-\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}})\\ &+(-a_{41}+\frac{a_{23}a_{24}}{a_{13}})a_{12}k_3 +(-a_{44}+\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}})a_{42}k_3. \end{aligned} \end{equation*}

    It can be obtained by calculation that T_{321} = T_{111} , and ( 4.18 ) can be transformed into

    \begin{equation*} D_0^2+T_{321}\bar{\Theta}_1+\bar{\Theta}_1T_{321}^T = 0, \end{equation*}

    where \bar{\Theta}_1 = \frac{1}{\rho_{321}^2}(Q_{321}U_{32}U_3J_3)\Lambda_3(Q_{321}U_{32}U_3J_3)^T , \rho_{321} = (a_{13}a_{41}-a_{23}a_{42})k_3\sigma_3 . Consequently,

    \begin{equation} \Lambda_3 = \rho_{321}^2(Q_{321}U_{32}U_3J_3)^{-1}\bar{\Theta}_1[(Q_{321}U_{32}U_3J_3)^{-1}]^T \end{equation} (4.20)

    is positive definite matrix.

    Case 2-2: If k_3 = 0 , we can find the standard transformation matrix Q_{322} such that T_{322} = Q_{322}Z_{32}Q_{322}^{-1} , where

    \begin{equation*} Q_{322} = \begin{bmatrix} q_1&q_2&q_3&q_4\\ 0&-a_{41}+\frac{a_{23}a_{24}}{a_{13}}&-a_{44}+\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}}&a_{42}\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}, \end{equation*}

    with

    \begin{equation*} \begin{aligned} q_1 = &a_{13}a_{41}-a_{23}a_{42}, \; q_2 = (-a_{11}-a_{44}+\frac{a_{12}a_{23}}{a_{13}}+\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}})(-a_{41}+\frac{a_{23}a_{24}}{a_{13}}), \\ q_3 = &(-a_{44}+\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}})^2+(-a_{14}+\frac{a_{12}a_{13}k_1}{a_{23}a_{42}-a_{13}a_{41}})(-a_{41}+\frac{a_{23}a_{24}}{a_{13}}), \\ q_4 = &(-a_{44}+\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}})a_{42}+(-a_{41}+\frac{a_{23}a_{24}}{a_{13}})a_{12}. \end{aligned} \end{equation*}

    We can find

    \begin{equation*} T_{322} = \begin{bmatrix} -\bar{b}_1&-\bar{b}_2&-\bar{b}_3&-\bar{b}_4\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&0&\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}}-\frac{a_{13}a_{42}k_1}{a_{23}a_{42}-a_{13}a_{41}} \end{bmatrix}, \end{equation*}

    which means that

    \begin{equation*} \begin{aligned} \bar{b}_1 = &a_{13}^2a_{21}a_{42}+a_{12}a_{23}^2a_{42}+a_{13}a_{21}a_{23}a_{42}+a_{13}a_{21}a_{23}a_{42}+a_{13}a_{23}a_{33}a_{42}\\&+a_{13}a_{23}a_{41}a_{44}-a_{13}^2a_{41}(a_{11}+a_{21}+a_{33}+a_{44}) > 0, \\ \bar{b}_2 = & (a_{13}^2a_{21}a_{42}+a_{12}a_{23}^2a_{42}+a_{13}a_{21}a_{23}a_{42}+a_{13}a_{23}a_{33}a_{42})(a_{13}^2a_{21}a_{42}+a_{12}a_{23}^2a_{42}+a_{13}a_{21}a_{23}a_{42})\\ &+(a_{13}a_{21}a_{23}a_{42}+a_{13}a_{23}a_{41}a_{44})(a_{13}^2a_{21}a_{42}+a_{12}a_{23}^2a_{42}+a_{13}a_{21}a_{23}a_{42}-a_{11}a_{13}a_{23}a_{42})\\ &-a_{13}^2a_{41}(a_{11}+a_{21}+a_{33}+a_{44})(a_{13}^2a_{21}a_{42}+a_{12}a_{23}^2a_{42}+a_{13}a_{21}a_{23}a_{42}-a_{11}a_{13}a_{23}a_{42})\\ &+a_{11}(a_{21}+a_{33}+a_{44})+a_{21}(a_{12}+a_{33}+a_{44})+a_{33}a_{44}+a_{24}a_{42}-a_{14}a_{41} > 0, \\ \bar{b}_3 = &\frac{a_{13}(a_{23}a_{42}-a_{13}a_{41})}{a_{13}^2a_{21}a_{42}+a_{12}a_{23}^2a_{42}+a_{13}a_{21}a_{23}a_{42}-a_{11}a_{13}a_{23}a_{42}}[a_{33}a_{42}(a_{11}a_{23} +a_{11}a_{24}a_{33}a_{41}-a_{14}a_{21}-a_{13}a_{21})]\\ &+\frac{a_{13}(a_{23}a_{42}-a_{13}a_{41})}{a_{13}^2a_{21}a_{42}+a_{12}a_{23}^2a_{42}+a_{13}a_{21}a_{23}a_{42}-a_{11}a_{13}a_{23}a_{42}}[a_{21}a_{33}a_{44}(a_{11}+a_{12})-a_{21}a_{33}a_{41}(a_{13}+a_{14})]\\ &+\frac{a_{13}(a_{23}a_{42}-a_{13}a_{41})}{a_{13}^2a_{21}a_{42}+a_{12}a_{23}^2a_{42}+a_{13}a_{21}a_{23}a_{42}-a_{11}a_{13}a_{23}a_{42}}[-a_{12}a_{33}a_{41}(a_{23}+a_{24})] > 0, \\ \end{aligned} \end{equation*}

    we can find that \bar{b}_1\bar{b}_2-\bar{b}_3 > 0 , and ( 4.18 ) can be transformed into the following form :

    \begin{equation*} \begin{aligned} &(Q_{322}U_{32}U_3J_3)D_3^2(Q_{322}U_{32}U_3J_3)^T\\ &+[(Q_{322}U_{32}U_3J_3)Z(Q_{322}U_{32}U_3J_3)^{-1}](Q_{322}U_{32}U_3J_3)\Lambda_3(Q_{322}U_{32}U_3J_3)^T\\ &+(Q_{322}U_{32}U_3J_3)\Lambda_3(Q_{322}U_{32}U_3J_3)^T[(Q_{322}U_{32}U_3J_3)Z(Q_{322}U_{32}U_3J_3)^{-1}]^T = 0. \end{aligned} \end{equation*}

    That is, D_0^2+T_{322}\bar{\Theta}_2+\bar{\Theta}_2T_{322}^T = 0 , where \bar{\Theta}_2 = \frac{1}{\rho_{322}^2}(Q_{322}U_{32}U_3J_3)\Lambda_3(Q_{322}U_{32}U_3J_3)^T , \rho_{322} = (a_{13}a_{41}-a_{23}a_{42})\sigma_3 . According to Lemma 2.6, we are able to obtain the semipositive definite matrix

    \begin{equation*} \bar{\Theta}_2 = \begin{bmatrix} \frac{\bar{b}_2}{2(\bar{b}_1\bar{b}_2-\bar{b}_3)}&0&-\frac{1}{2(\bar{b}_1\bar{b}_2-\bar{b}_3)}&0\\ 0& \frac{1}{2(\bar{b}_1\bar{b}_2-\bar{b}_3)}& 0&0\\ -\frac{1}{2(\bar{b}_1\bar{b}_2-\bar{b}_3)} & 0&\frac{\bar{b}_1}{2\bar{b}_3(\bar{b}_1\bar{b}_2-\bar{b}_3)}&0\\ 0 & 0&0&0 \end{bmatrix}. \end{equation*}

    Therefore, it can be obtained by calculation that

    \begin{equation} \Lambda_3 = \rho_{322}^2(Q_{322}U_{32}U_3J_3)^{-1}\bar{\Theta}_2[(Q_{322}U_{32}U_3J_3)^{-1}]^T, \end{equation} (4.21)

    is semi-postive definite.

    Case 3: When k_1 = 0 and k_2 = 0 , there exists the transformation matrix Q_{33} such that T_{33} = Q_{33}Z_3Q_{33}^{-1} , where

    \begin{equation*} Q_{33} = \begin{bmatrix} a_{13}(a_{11}+a_{33}-\frac{a_{12}a_{23}}{a_{13}})&(-a_{11}-\frac{a_{12}a_{23}}{a_{13}})^2&n_3^*&n_4^*\\ -a_{13}&-a_{11}+\frac{a_{12}a_{23}}{a_{13}}&-a_{14}&-a_{12}\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}, \end{equation*}

    with n_3^* = -a_{13}a_{33}+a_{14}(a_{11}+a_{44}-\frac{a_{12}a_{23}}{a_{13}}) and n_4^* = a_{12}(a_{11}+\frac{a_{13}a_{21}}{a_{13}})+a_{14}a_{42} , then we can find

    \begin{equation*} T_{33} = \begin{bmatrix} -\bar{c}_1&-\bar{c}_2&-\bar{c}_3&-\bar{c}_4\\ 1&0&0&0\\ 0&0&-a_{44}&-a_{42}\\ 0&0&0&-\frac{a_{12}a_{23}+a_{13}a_{21}}{a_{13}} \end{bmatrix}, \end{equation*}

    where

    \begin{equation*} \begin{aligned} \bar{c}_1 = &a_{11}+a_{33}-\frac{a_{12}a_{23}}{a_{13}} > 0, \\ \bar{c}_2 = &\frac{a_{13}}{a_{12}a_{23}+a_{13}a_{21}+a_{13}a_{44}}[a_{21}a_{33}a_{44}(a_{11}+a_{12})+a_{33}a_{42}(a_{11}a_{23}+a_{11}a_{24}a_{33}a_{41}-a_{14}a_{21}-a_{13}a_{21})]\\ &-\frac{a_{13}}{a_{12}a_{23}+a_{13}a_{21}+a_{13}a_{44}}[a_{12}a_{33}a_{41}(a_{23}+a_{24})+a_{21}a_{33}a_{41}(a_{13}+a_{14})] > 0. \end{aligned} \end{equation*}

    Thus, ( 4.18 ) can be transformed into

    \begin{equation*} D_0^2+T_{33}\bar{\Theta}_3+\bar{\Theta}_3T_{33}^T = 0, \end{equation*}

    where \bar{\Theta}_3 = \frac{1}{\rho_{33}^2}(Q_{33}U_3J_3)\Lambda_3(Q_{33}U_3J_3)^T , \rho_{33} = -a_{13}\sigma_3 . With Lemma 2.7, we can obtain the positive semi-definite matrix

    \begin{equation*} \bar{\Theta}_3 = \begin{bmatrix} \frac{1}{2\bar{c}_1}&0&0&0\\ 0&\frac{1}{2\bar{c}_1\bar{c}_2}&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}. \end{equation*}

    Therefore, it can be obtained by calculation that

    \begin{equation} \Lambda_3 = \rho_{33}^2(Q_{33}U_3J_3)^{-1}\bar{\Theta}_3[(Q_{33}U_3J_3)^{-1}]^T \end{equation} (4.22)

    is semi-postive definite.

    Step 4. For the following algebraic equation

    \begin{equation} D_4^2+Z\Lambda_4+\Lambda_4Z^T = 0, \end{equation} (4.23)

    consider the corresponding order matrix J_4 and elimination matrix U_4 :

    \begin{equation*} J_4 = \begin{bmatrix} 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0 \end{bmatrix}, \quad U_4 = \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&-\frac{a_{21}}{a_{12}}&1&0\\ 0&0&0&1 \end{bmatrix}. \end{equation*}

    Let Z_4 = (U_4J_4)Z(U_4J_4)^{-1} . It can be obtained that

    \begin{equation*} Z_4 = \begin{bmatrix} 0&-\frac{a_{21}a_{44}}{a_{12}}&-a_{44}&-a_{41}\\ -a_{12}&-a_{13}-\frac{a_{14}a_{21}}{a_{12}}&-a_{41}&-a_{11}\\ 0&\frac{a_{13}a_{21}}{a_{12}}+\frac{a_{14}a_{21}^2}{a_{12}^2}&a_{24}+\frac{a_{21}a_{41}}{a_{12}}&a_{21}+\frac{a_{11}a_{21}}{a_{12}}\\ 0&p_1&p_2&0 \end{bmatrix}, \end{equation*}

    where p_1 = -a_{33}+\frac{a_{21}a_{33}}{a_{12}} , p_2 = a_{33} .

    Case 1: When p_1 = 0 and p_2\neq0 , it is possible to determine the standard transformation matrix Q_{41} in order to satisfy the equation T_{41} = Q_{41}Z_4Q_{41}^{-1} , where

    \begin{equation*} Q_{41} = \begin{bmatrix} l_1&l_2&l_3&l_4\\ 0&(\frac{a_{13}a_{21}}{a_{12}}+\frac{a_{14}a_{21}^2}{a_{12}^2})a_{33}&a_{24}a_{33}+\frac{a_{21}a_{33}a_{41}}{a_{12}}&a_{21}a_{33}+\frac{a_{11}a_{21}a_{33}}{a_{12}}\\ 0&0&a_{33}&0\\ 0&0&0&1 \end{bmatrix}, \end{equation*}

    with

    \begin{equation*} \begin{aligned} l_1 = &(-a_{13}a_{21}-\frac{a_{14}a_{21}^2}{a_{12}})a_{33}, \; l_2 = (-a_{13}+a_{24}+\frac{a_{21}a_{41}-a_{14}a_{21}}{a_{12}})(\frac{a_{13}a_{21}}{a_{12}}+\frac{a_{14}a_{21}^2}{a_{12}^2})a_{33}, \\ l_3 = &a_{33}[-a_{41}(\frac{a_{13}a_{21}}{a_{12}}+\frac{a_{14}a_{21}^2}{a_{12}^2})+a_{21}a_{33}+\frac{a_{11}a_{21}a_{33}}{a_{12}}+(a_{24}+\frac{a_{21}a_{41}}{a_{12}})^2], \\ l_4 = &-a_{11}a_{33}(\frac{a_{13}a_{21}}{a_{12}}+\frac{a_{14}a_{21}^2}{a_{12}^2})+(a_{24}+\frac{a_{21}a_{41}}{a_12})(a_{21}+\frac{a_{11}a_{21}}{a_{12}})a_{33}, \end{aligned} \end{equation*}

    then it can be obtained by calculation that T_{41} = T_{111} , and ( 4.23 ) can be transformed into

    \begin{equation*} D_0^2+T_{41}\hat{\Theta}_1+\hat{\Theta}_1T_{41}^T = 0, \end{equation*}

    where \hat{\Theta}_1 = \frac{1}{\rho_{41}^2}(Q_{41}U_4J_4)\Lambda_4(Q_{41}U_4J_4)^T , \rho_{41} = (a_{13}a_{21}-\frac{a_{14}a_{21}^2}{a_{12}})a_{33}\sigma_4 , and \hat{\Theta}_1 = \Theta_1 . Consequently,

    \begin{equation} \Lambda_4 = \rho_{41}^2(Q_{41}U_4J_4)^{-1}\hat{\Theta}_1[(Q_{41}U_4J_4)^{-1}]^T \end{equation} (4.24)

    is positive definite matrix.

    Case 2: When p_1\neq0 and p_2 = 0 , there exists the transformation matrix U_{42} , then Z_{42} = U_{42}Z_4U_{42}^{-1} , where

    \begin{equation*} U_{42} = \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&\frac{a_{12}(a_{12}a_{33}-a_{21}a_{33})}{a_{12}a_{13}a_{21}+a_{14}a_{21}^2}&1 \end{bmatrix}, \end{equation*}

    then we can find

    \begin{equation*} Z_{42} = \begin{bmatrix} 0&-\frac{a_{21}a_{44}}{a_{12}}&-a_{44}+\frac{a_{12}a_{33}a_{41}(a_{12}-a_{21})}{a_{12}a_{13}a_{21}+a_{14}a_{21}^2}&-a_{41}\\ -a_{12}&-a_{13}-\frac{a_{14}a_{21}}{a_{12}}&-a_{41}+\frac{a_{11}a_{12}a_{33}(a_{12}-a_{21})}{a_{12}a_{13}a_{21}+a_{14}a_{21}^2}&-a_{11}\\ 0&\frac{a_{12}a_{13}a_{21}+a_{14}a_{21}^2}{a_{12}^2}&\frac{a_{12}a_{24}+a_{21}a_{41}}{a_{12}}-\frac{(a_{12}+a_{12})(a_{12}a_{33}-a_{21}a_{33})}{a_{12}a_{13}+a_{14}a_{21}}&\frac{a_{21}(a_{11}+a_{12})}{a_{12}}\\ 0&0&p_3&\frac{(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}a_{13}+a_{14}a_{21}} \end{bmatrix}, \end{equation*}

    where p_3 = a_{33}+\frac{a_{33}(a_{12}-a_{21})(a_{12}a_{24}+a_{21}a_{41})}{a_{12}a_{13}+a_{14}a_{21}}-\frac{a_{12}a_{33}^2(a_{11}+a_{12})(a_{12}-a_{21})^2}{a_{21}(a_{12}a_{13}+a_{14}a_{21})^2} .

    Case 2-1: If p_3\neq0 , we can find the standard transformation matrix Q_{421} such that T_{421} = Q_{421}Z_{42}Q_{421}^{-1} , where

    \begin{equation*} Q_{421} = \begin{bmatrix} l_1&l_2&l_3&l_4\\ 0&\frac{p_3a_{21}(a_{12}a_{13}+a_{14}a_{21})}{a_{12}^2}&l_5&\frac{(a_{11}+a_{12})^2(a_{12}-a_{21})^2}{(a_{12}a_{13}+a_{14}a_{21})^2}+\frac{p_3a_{21}(a_{11}+a_{12})}{a_{12}}\\ 0&0&p_3&\frac{(a_{11}+a_{12})(a_{12}-a_{21})}{(a_{12}a_{13}+a_{14}a_{21})}\\ 0&0&0&1 \end{bmatrix}, \end{equation*}

    with

    \begin{equation*} \begin{aligned} l_1 = &\frac{-a_{21}a_{33}(a_{12}-a_{21})(a_{12}a_{24}+a_{24}a_{41})}{a_{12}}, \\ l_2 = &[\frac{a_{21}(-a_{12}a_{13}-a_{14}a_{21}+a_{21}a_{24}+a_{21}a_{41})(a_{12}a_{13}+a_{14}a_{21})}{a_{12}^3}+\frac{a_{21}(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}^2}]p_3\\ &-[\frac{a_{21}a_{33}(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}^2}]p_3, \\ l_3 = &[\frac{-a_{41}a_{21}(a_{12}a_{13}+a_{14}a_{21})}{a_{12}^2}+\frac{a_{11}a_{33}(a_{12}-a_{21})}{a_{12}}-\frac{a_{33}(a_{11}+a_{12})^2(a_{12}-a_{21})^2}{(a_{12}a_{13}+a_{14}a_{21})^2}]p_3\\ &+[\frac{(a_{11}+a_{12})(a_{12}-a_{21})(a_{12}a_{24}+a_{21}a_{41})}{a_{12}(a_{12}a_{13}+a_{14}a_{21})}+(\frac{a_{12}a_{24}+a_{21}a_{41}}{a_{12}}-\frac{(a_{11}+a_{12})(a_{12}a_{33}-a_{21}a_{33})}{a_{12}a_{13}+a_{14}a_{21}})^2\\ &+\frac{p_3a_{21}(a_{11}+a_{12})}{a_{12}}+\frac{(a_{11}+a_{12})^2(a_{12}-a_{21})^2}{(a_{12}a_{13}+a_{14}a_{21})^2}]p_3, \\ l_4 = &(\frac{a_{12}a_{24}+a_{21}a_{41}}{a_{12}}-\frac{(a_{11}+a_{12})(a_{12}a_{33}-a_{21}a_{33})}{a_{12}a_{13}+a_{14}a_{21}}+\frac{(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}a_{13}+a_{14}a_{21}})\frac{p_3a_{21}(a_{11}+a_{12})}{a_{12}}\\ &-\frac{a_{11}a_{21}a_{33}(a_{12}-a_{21})(a_{12}a_{24}+a_{21}a_{41})}{a_{12}^2}+\frac{p_3a_{21}(a_{11}+a_{12})}{a_{12}}\frac{(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}a_{13}+a_{14}a_{21}}\\ &+\frac{(a_{11}+a_{12})^2(a_{12}-a_{21})^2}{(a_{12}a_{13}+a_{14}a_{21})^2}\frac{(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}a_{13}+a_{14}a_{21}}, \\ l_5 = &[\frac{a_{12}a_{24}+a_{21}a_{41}}{a_{12}}-\frac{(a_{11}+a_{12})(a_{12}a_{33}-a_{21}a_{33})}{a_{12}a_{13}+a_{14}a_{21}}+\frac{(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}a_{13}+a_{14}a_{21}}], \end{aligned} \end{equation*}

    then we can obtain that T_{421} = T_{111} , and ( 4.23 ) can be transformed into

    \begin{equation*} D_0^2+T_{421}\hat{\Theta}_1+\hat{\Theta}_1T_{421}^T = 0, \end{equation*}

    where \hat{\Theta}_1 = \frac{1}{\rho_{421}^2}(Q_{421}U_{42}U_4J_4)\Lambda_4(Q_{421}U_{42}U_4J_4)^T , \rho_{421} = \frac{-a_{21}(a_{12}a_{13}+a_{14}a_{21})}{a_{12}}\sigma_4 . Consequently,

    \begin{equation} \Lambda_4 = \rho_{421}^2(Q_{421}U_{42}U_4J_4)^{-1}\hat{\Theta}_1[(Q_{421}U_{42}U_4J_4)^{-1}]^T \end{equation} (4.25)

    is positive definite matrix.

    Case 2-2: If p_3 = 0 , we are able to find the standard transformation matrix Q_{422} such that T_{422} = Q_{422}Z_{42}Q_{422}^{-1} , where

    \begin{equation*} Q_{422} = \begin{bmatrix} l_1&l_2&l_3&l_4\\ 0&\frac{a_{12}a_{13}a_{21}+a_{14}a_{21}^2}{a_{12}^2}&\frac{a_{12}a_{24}+a_{21}a_{41}}{a_{12}}-\frac{a_{33}(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}a_{13}+a_{14}a_{21}}&\frac{a_{21}(a_{11}+a_{12})}{a_{12}}\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}, \end{equation*}

    with

    \begin{equation*} \begin{aligned} l_1 = &-\frac{a_{21}(a_{12}a_{13}+a_{14}a_{21})}{a_{12}}, \\ l_2 = &[-a_{13}-\frac{a_{12}a_{24}+a_{21}a_{41}-a_{14}a_{21}}{a_{12}}-\frac{(a_{11}+a_{12})(a_{12}a_{13}-a_{21}a_{33})}{a_{12}a_{13}+a_{14}a_{21}}]\frac{a_{21}(a_{12}a_{13}+a_{14}a_{21})}{a_{12}^2}, \\ l_3 = &[\frac{a_{12}a_{24}+a_{21}a_{41}}{a_{12}}-\frac{a_{33}(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}a_{13}+a_{14}a_{21}}]^2-\frac{a_{41}(a_{12}a_{33}+a_{14}a_{21})}{a_{12}^2}+\frac{a_{11}a_{33}(a_{12}-a_{21})}{a_{12}}, \\ l_4 = &\frac{a_{21}(a_{11}+a_{12})}{a_{12}}[\frac{a_{12}a_{24}+a_{21}a_{41}}{a_{12}}-\frac{(a_{11}+a_{12})(a_{12}a_{33}-a_{21}a_{33})}{a_{12}a_{13}+a_{14}a_{21}}+\frac{(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}a_{13}+a_{14}a_{21}}]\\ &-\frac{a_{11}a_{21}(a_{12}a_{13}+a_{14}a_{21})}{a_{12}^2}. \end{aligned} \end{equation*}

    We then have

    \begin{equation*} T_{422} = \begin{bmatrix} -\hat{b}_1&-\hat{b}_2&\hat{b}_3&\hat{b}_4\\ 1&0&0&\hat{b}_5\\ 0&1&0&\hat{b}_6\\ 0&0&0&\hat{b}_7 \end{bmatrix}, \end{equation*}

    which means that

    \begin{equation*} \begin{aligned} \hat{b}_1 = &\frac{1}{a_{12}a_{13}+a_{14}a_{21}}[(a_{12}a_{13}+a_{14}a_{21})(a_{11}+a_{21}+a_{33}+a_{44})+(a_{11}+a_{12})(a_{12}-a_{21})] > 0, \\ \hat{b}_2 = &\frac{(a_{11}+a_{12})(a_{12}-a_{21})}{(a_{12}a_{13}+a_{14}a_{21})^2}[(a_{12}a_{13}+a_{14}a_{21})(a_{11}+a_{21}+a_{33}+a_{44})+(a_{11}+a_{12})(a_{12}-a_{21})]\\ &+a_{11}(a_{21}+a_{33}+a_{44})+a_{21}(a_{12}+a_{33}+a_{44})+a_{33}a_{44}+a_{24}a_{42}-a_{14}a_{41} > 0, \\ \hat{b}_3 = &\frac{(a_{11}+a_{12})(a_{12}-a_{21})}{a_{12}a_{13}+a_{14}a_{21}}[a_{11}(a_{21}+a_{33}+a_{44})+a_{21}(a_{12}+a_{33}+a_{44})+a_{33}a_{44}+a_{24}a_{42}-a_{14}a_{41}]\\ &+\frac{(a_{11}+a_{12})^2(a_{12}-a_{21})^2}{(a_{12}a_{13}+a_{14}a_{21})^3}[(a_{12}a_{13}+a_{14}a_{21})(a_{11}+a_{21}+a_{33}+a_{44})+(a_{11}+a_{12})(a_{12}-a_{21})]\\ &+a_{11}a_{21}(a_{33}+a_{44})+a_{33}a_{44}(a_{11}+a_{21})+a_{12}a_{21}(a_{33}+a_{44})-a_{14}a_{21}(a_{41}+a_{42})\\ &+a_{24}a_{41}(a_{11}-a_{12})-a_{33}a_{41}(a_{13}+a_{14}) +a_{23}a_{33}a_{42}+a_{24}a_{33}a_{41} > 0, \\ \end{aligned} \end{equation*}

    \hat{b}_1\hat{b}_2-\hat{b}_3 > 0 , and ( 4.23 ) can be transformed into the following form :

    \begin{equation*} \begin{aligned} &(Q_{422}U_{42}U_4J_4)D_4^2(Q_{422}U_{42}U_4J_4)^T\\ &+[(Q_{422}U_{42}U_4J_4)Z(Q_{422}U_{42}U_4J_4)^{-1}](Q_{422}U_{42}U_4J_4)\Lambda_4(Q_{422}U_{42}U_4J_4)^T\\ &+(Q_{422}U_{42}U_4J_4)\Lambda_4(Q_{422}U_{42}U_4J_4)^T[(Q_{422}U_{42}U_4J_4)Z(Q_{422}U_{42}U_4J_4)^{-1}]^T = 0. \end{aligned} \end{equation*}

    That is, D_0^2+T_{422}\hat{\Theta}_2+\hat{\Theta}_2T_{422}^T = 0 , where \hat{\Theta}_2 = \frac{1}{\rho_{422}^2}(Q_{422}U_{42}U_4J_4)\Lambda_4(Q_{422}U_{42}U_4J_4)^T , \rho_{422} = \frac{-a_{21}(a_{12}a_{13}+a_{14}a_{21})}{a_{21}}\sigma_4 . According to Lemma 2.6, we can find the semipositive definite matrix

    \begin{equation*} \hat{\Theta}_2 = \begin{bmatrix} \frac{\hat{b}_2}{2(\hat{b}_1\hat{b}_2-\hat{b}_3)}&0&-\frac{1}{2(\hat{b}_1\hat{b}_2-\hat{b}_3)}&0\\ 0& \frac{1}{2(\hat{b}_1\hat{b}_2-\hat{b}_3)}& 0&0\\ -\frac{1}{2(\hat{b}_1\hat{b}_2-\hat{b}_3)} & 0&\frac{\hat{b}_1}{2\hat{b}_3(\hat{b}_1\hat{b}_2-\hat{b}_3)}&0\\ 0 & 0&0&0 \end{bmatrix}. \end{equation*}

    Therefore, we can obtain that

    \begin{equation} \Lambda_4 = \rho_{422}^2(Q_{422}U_{42}U_4J_4)^{-1}\hat{\Theta}_2[(Q_{422}U_{42}U_4J_4)^{-1}]^T \end{equation} (4.26)

    is semi-postive definite.

    Case 3: When p_1 = 0 and p_2 = 0 , there exists the transformation matrix Q_{42} such that T_{43} = Q_{42}Z_{4}Q_{42} , where

    \begin{equation*} Q_{42} = \begin{bmatrix} l_1&l_2&l_3&l_4\\ -a_{12}&-a_{13}-\frac{a_{14}a_{21}}{a_{12}}&-a_{41}&-a_{11}\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix}, \end{equation*}

    with

    \begin{equation*} \begin{aligned} &l_1 = a_{12}a_{13}+a_{14}a_{21}, \; l_2 = a_{21}a_{44}+(a_{13}+\frac{a_{14}a_{21}}{a_{12}})^2, \; l_3 = a_{12}a_{44}-a_{41}(-a_{13}-\frac{a_{14}a_{21}}{a_{12}}+a_{24}+\frac{a_{21}a_{41}}{a_{12}}), \\ &l_4 = a_{12}a_{41}+a_{11}a_{13}+\frac{a_{11}a_{14}a_{21}}{a_{12}}-a_{21}a_{41}-\frac{a_{11}a_{21}a_{41}}{a_{12}}. \end{aligned} \end{equation*}

    We then have

    \begin{equation*} T_{43} = \begin{bmatrix} -\hat{c}_1&-\hat{c}_2&-\hat{c}_3&-\hat{c}_4\\ 1&0&0&0\\ 0&0&a_{24}+\frac{a_{21}a_{41}}{a_{21}}&a_{21}+\frac{a_{11}a_{21}}{a_{12}}\\ 0&0&0&0 \end{bmatrix}, \end{equation*}

    where

    \begin{equation*} \begin{aligned} \hat{c}_1 = &a_{11}+a_{21}+a_{33}+a_{44}+a_{24}+\frac{a_{21}a_{41}}{a_{12}} > 0, \\ \hat{c}_2 = &a_{21}a_{33}a_{44}(a_{11}+a_{12})+a_{33}a_{42}(a_{11}a_{23}+a_{11}a_{24}a_{33}a_{41}-a_{14}a_{21}-a_{13}a_{21})-a_{12}a_{33}a_{41}(a_{23}+a_{24}) \end{aligned} \end{equation*}
    \begin{equation*} \begin{aligned} &+(a_{11}+a_{21}+a_{33}+a_{44}+a_{24}+\frac{a_{21}a_{41}}{a_{12}})(a_{24}+\frac{a_{21}a_{41}}{a_{12}})-a_{21}a_{33}a_{41}(a_{13}+a_{14}) > 0. \end{aligned} \end{equation*}

    Thus, ( 4.23 ) can be transformed into

    \begin{equation*} D_0^2+T_{43}\hat{\Theta}_3+\hat{\Theta}_3T_{43}^T = 0, \end{equation*}

    where \hat{\Theta}_3 = \frac{1}{\rho_{43}^2}(Q_{43}U_4J_4)\Lambda_4(Q_{43}U_4J_4)^T , \rho_{43} = -a_{12}\sigma_4 . With Lemma 2.7, we can obtain the positive semidefinite matrix \hat{\Theta}_3

    \begin{equation*} \hat{\Theta}_3 = \begin{bmatrix} \frac{1}{2\hat{c}_1}&0&0&0\\ 0&\frac{1}{2\hat{c}_1\hat{c}_2}&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}. \end{equation*}

    Therefore, it can be obtained by calculation that

    \begin{equation} \Lambda_4 = \rho_{43}^2(Q_{43}U_4J_4)^{-1}\hat{\Theta}_3[(Q_{43}U_4J_4)^{-1}]^T \end{equation} (4.27)

    is semi-postive definite.

    In summary, \Lambda_3 , \Lambda_4 are positive definite and \Lambda_1, \Lambda_1 are at least semi-positive definite. Hence, \Lambda = \Lambda_1+\Lambda_2+\Lambda_3+\Lambda_4 is a positive definite matrix, and there is a unique and precise density function surrounding the quasi-endemic equilibrium \bar{E} when R^s > 1 . Considering the transformation of (y_1, y_2, y_3, y_4) = (\ln{\frac{S}{\bar{S}_1}}, \ln{\frac{I}{\bar{I}_1}}, \ln{\frac{A}{\bar{A}_1}}, \ln{\frac{C}{\bar{C}_1}}) and (S, I, A, C) , the unique ergodic stationary distribution v(\cdot) of Model (2.2) approximately follows a unique log-normal probability density function

    \begin{equation*} \begin{aligned} \Phi(S, I, A, C) = &(2\pi)^{-2}\left| \Lambda\right| ^{-{\frac{1}{2}}}e^{-\frac{1}{2}(\ln{\frac{S}{\bar{S}_1}}+\ln{\frac{I}{\bar{I}_1}}+\ln{\frac{A}{\bar{A}_1}}+\ln{\frac{C}{\bar{C}_1}})\Lambda^{-1}(\ln{\frac{S}{\bar{S}_1}}+\ln{\frac{I}{\bar{I}_1}}+\ln{\frac{A}{\bar{A}_1}}+\ln{\frac{C}{\bar{C}_1}})^T}. \end{aligned} \end{equation*}

    This completes the proof of Theorem 4.1.

    Remark 3. By combining Theorems 3.1 and 4.1, it is possible to find evidence that the stationary distribution of Model (2.2) around the equilibrium \bar {E} _1 is compatible with the probability density function \Phi(S, I, A, C) . Hence, a reasonable stochastic criterion for disease persistence can be achieved when R^s > 1 . From this we can see that accurately representing the density function can be an efficient method for preventing and managing numerous epidemics.

    In this section, we apply theoretical findings to analyze the dynamics of anthrax and investigate the impact of environmental disturbance on the transmission of the disease. Using by Milstein's method [], the discretization equation of Model (2.2) is given by

    \begin{equation} \left\{ \begin{aligned} S_{k+1}& = S_k+\left[rN_k(1-\frac{N_k}{K})-\eta_aA_kS_k-\eta_cC_kS_k-\eta_i\frac{S_kI_k}{N_k}-\mu S_k+\tau I_k\right] \Delta t\\ &+\sigma_1S_k\sqrt{\Delta t}\chi_k+\frac{\sigma_1^2}{2}S_k^2(\chi_k^2-1)\Delta t, \\ I_{k+1}& = I_k+\left[\eta_aA_kS_k+\eta_cC_kS_k+\eta_i\frac{S_kI_k}{N_k}-(\gamma+\mu+\tau)I_k\right]\Delta t+\sigma_2I_k\sqrt{\Delta t}\pi_k+\frac{\sigma_2^2}{2}I_k^2(\pi_k^2-1)\Delta t, \\ A_{k+1}& = A_k+(-\alpha A_k+\beta C_k )\Delta t+\sigma_3A_k\sqrt{\Delta t}\omega_k+\frac{\sigma_3^2}{2}A_k^2(\omega_k^2-1)\Delta t, \\ C_{k+1}& = C_k+\left[(\gamma+\mu)I_k-\delta(S_k+I_k)C_k-\kappa C_k\right]\Delta t+\sigma_4C_k\sqrt{\Delta t}\zeta_k+\frac{\sigma_4^2}{2}C_k^2(\zeta_k^2-1)\Delta t, \end{aligned} \right. \end{equation} (5.1)

    where the time increment is \Delta t > 0 , and \chi_k, \pi_k, \omega_k, \zeta_k\:(k = 1, ..., n) are the independent Gaussian random variables following the distribution N(0, 1) . As stated by Saad-Roy et al. [7] and Mushayabasa et al. [17], Table 1 displays the relevant biological parameters and the initial value of Model (2.2). Next, our objective is to perform empirical examples to concentrate on the following three aspects:

    Table 1.  List of biological parameters and initial of Model (2.2).
    Parameters Description Value Units Ref.
    K Animal carrying capacity 100 animals [7]
    r Birth rate 1/300 Day ^{-1} [7]
    \mu Death rate 1/600 Day ^{-1} [7]
    \kappa Carcass decay rate 1/10 Day ^{-1} [7,40]
    \alpha Spore decay rate 1/10 Day ^{-1} [7,17]
    \gamma Disease death rate 1/10 Day ^{-1} [7,14]
    \eta_i Transmission rate of infected animals 1/100 Day ^{-1} . [7]
    \eta_a Transmission rate of environment to susceptible animals 1/2 Day ^{-1} gm spore ^{-1} . [7]
    \eta_c Transmission rate of infected carcasses to susceptible animals 1/10 Day ^{-1} carcass ^{-1} [7]
    \tau Recovery rate of infected animals 1/10 Day ^{-1} [7,41]
    \beta Spore growth rate 1/500 Spores carcass ^{-1} day ^{-1} [7]
    \delta Carcass consumption rate 1/20 Day ^{-1} animal ^{-1} [7,41]

     | Show Table
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    1. If R^s > 1 , there is a unique ergodicity property stationary distribution.

    2. Stochastic fluctuations have an impact on disease persistence in the Model (2.2).

    3. Focus on the probability density function of Model (2.2).

    The main parameters are given in Table 1. Let the stochastic perturbations be (\sigma_1, \sigma_2, \sigma_3, \sigma_4) = (0.005, 0.005, 0.005, 0.005) in Model (2.2) then we can calculate that

    \begin{equation*} \begin{aligned} \mathit{R_0} : = & \frac{\eta_i}{\gamma+\mu+\tau}+\frac{\gamma+\mu}{\gamma+\mu+\tau}\frac{\eta_cS_0}{\delta S_0+\kappa}+\frac{\gamma+\mu}{\gamma+\mu+\tau}\frac{\beta\eta_aS_0}{\alpha(\delta S_0+\kappa)} = 4.3560 > 1, \\ \mathit{R^s} : = &\frac{\eta_i}{\gamma+\mu+\tau+\frac{\sigma_2^2}{2}}+\frac{\gamma+(\mu+\frac{\sigma_1^2}{2})}{\gamma+\mu+\tau+\frac{\sigma_2^2}{2}}\frac{\eta_cS_0}{\delta S_0+(\kappa+\frac{\sigma_4^2}{2})}+\frac{\gamma+(\mu+\frac{\sigma_1^2}{2})}{\gamma+\mu+\tau+\frac{\sigma_2^2}{2}}\frac{\beta\eta_aS_0}{(\alpha+\frac{\sigma_3^2}{2})[\delta S_0+(\kappa+\frac{\sigma_4^2}{2})]}\\ = &1.1545 > 1. \end{aligned} \end{equation*}

    The disease has a tendency to persist in both deterministic and stochastic models according to Theorem 3.1. It can be inferred that Model (2.2) has a stationary distribution v(\cdot) . In Figure 2, we give the time series diagram of the model solution and its corresponding histogram. In the time series diagram in Figure 2, the curve fluctuates up and down, but the overall trend is stable. The corresponding interval of frequency can be seen in the histogram on the right. It is shown that the Model (2.2) has a unique ergodic stationary distribution, which is consistent with the result of Theorem 3.1. This indicates that the disease will endure. In real life, anthrax is not easy to be completely eradicated, but in a relatively stable state. Next, the dynamical behavior is observed when the parameters of the stochastic and deterministic models are identical. It is shown that the stochastic path fluctuates around the deterministic path, which indicates that the stochastic Model (2.2) has an ergodic stationary distribution. Figure 3 confirms this.

    Figure 1.  The flow chart of anthrax transmission.
    Figure 2.  The left column shows the paths of S(t) , I(t) , A(t) and C(t) for Model (2.2) with initial values (S(0), I(0), A(0), C(0)) = (10, 0.1, 0.001, 0.01) under noise intensities \sigma_1 = 0.005, \sigma_2 = 0.005, \sigma_3 = 0.005, \sigma_4 = 0.005 , respectively. The right column shows the histograms of S(t) , I(t) , A(t) and C(t) .
    Figure 3.  The column shows the paths of S, I, A, C , respectively, with \sigma_i = 0.005\; (i = 1, 2, 3, 4) . The red line represents the result of deterministic Model (2.1), and the blue line represents the result of stochastic Model (2.2), repectively.

    Mathematical limitations make it tough to establish sufficient conditions for disease extinction in the stochastic anthrax model [42,43]. Therefore, for a comprehensive discussion, we present numerical simulation of disease elimination with high noise. For example, to prevent the spread of anthrax, people can increase the culling rate or vaccination rate, which will effectively eliminate the spread of anthrax.

    We use numerical simulations to prove that excessive environmental noise can lead to the extinction of anthrax. Let \sigma = \sigma_1 = \sigma_2 = \sigma_3 = \sigma_4 , \sigma be taken as 0.002, 0.005, and 0.01, respectively. The other parameters and initial value are identical to those in Table 1 and the simple calculations show that \mathit{R^s} > 1 . It can be concluded that the disease persists; if we let \sigma = 0.02, 0.05, 0.1 , studies have shown that large enough environment noise will eradicate the disease, and the numerical simulation clearly shows our conclusion in Figure 4.

    Figure 4.  The paths of I(t) for stochastic Model (2.2) with different noise intensities (\sigma_1, \sigma_2, \sigma_3, \sigma_4) . The red line represents the result of deterministic Model (2.1), and the blue line represents the result of stochastic Model (2.2), repectively.

    In this part, the numerical representation of the probability density function of Model (2.2) will be our focus. From Theorem 3.1, we can calculate that if (\sigma_1, \sigma_2, \sigma_3, \sigma_4) = (0.005, 0.005, 0.005, 0.005) , the Model (2.2) has a unique stationary distribution. Moreover, through meticulous computation, it is determined that the variables' values in Table 2 are all positive. By the definition of \bar{E}_1 , we obtain that (\bar{S}_1, \bar{I}_1, \bar{A}_1, \bar{C}_1) = (24.8194, 0.4094, 0.00044, 0.02852) . According to Theorem 4.1, we know that the distribution has a probability density function around the quasi-steady state \bar{E}_1 , and the stationary distribution v(\cdot) obeys a log-normal density function \Phi(S, I, A, C) . Its function expression is:

    \begin{equation*} \Lambda = 10^{-3} \times \begin{bmatrix} 0.0511&-0.0271&0.0134&0.0283\\ -0.0271&1.4113& 0.4421&0.1271\\ 0.0134&0.4421&0.2216&0.0932\\ 0.0283&0.1271&0.0932&0.1014\\ \end{bmatrix}, \end{equation*}
    Table 2.  Value of variables.
    w_1 w_2 w_3 r_1 r_2 r_3
    -0.8739 1.7655 -0.8033 -17.2838 -0.0217 -0.0003
    k_1 k_2 k_3 p_1 p_2 p_3
    11.1646 -0.0001 -0.0352 -16.5267 0.1312 -10.7746

     | Show Table
    DownLoad: CSV

    and the corresponding marginal density functions are :

    \begin{equation*} \begin{aligned} F(S)& = \frac{\partial \Phi}{\partial S} = 0.08\exp[-9.79(\ln(S-24.8194))^2], \quad F(I) = \frac{\partial \Phi}{\partial I} = 2.22\exp[-0.35(\ln(I-0.4094))^2], \\ F(A)& = \frac{\partial \Phi}{\partial A} = 0.35\exp[-2.26(\ln(A-0.00044))^2], \quad F(C) = \frac{\partial \Phi}{\partial C} = 0.16\exp[-4.93(\ln(C-0.2852))^2]. \end{aligned} \end{equation*}

    By constructing frequency histograms and fitting curves for the variables S, I, A, and C , the marginal density functions F(S) , F(I) , F(A) , and F(C) were plotted. The numerical findings depicted in Figure 5 validate the assertion that the fitting curves closely align with the corresponding theoretical marginal density functions.

    Figure 5.  The marginal density functions and frequency fitting curves of S, I, A , and C in Model (2.2), respectively. All of the parameter values are the same as in the above Figure.

    To explore the possibility of an ergodic stationary distribution and a representation of the probability density function, we propose the stochastic anthrax model in this study. We will discuss our primary contributions in the two areas.

    By considering the effect of external environmental perturbations, we focus on the stochastic anthrax model. First, we calculate the basic reproduction number of the deterministic model \mathit{R}_0 , if \mathit{R_0} < 1 , which means that the anthrax will be extinct in the animal population, while \mathit{R_0} > 1 implies that anthrax will persistence. For stochastic Model (2.2), the uniform boundedness of the model is presented. Second, we demonstrate that the stochastic anthrax model has a unique ergodic stationary distribution at the threshold R_s > 1 in Lemma 2.3 and Theorem 3.1 by building appropriate Lyapunov functions and applying ergodic theory. From the similar threshold and basic reproduction number, \mathit{R}^s > 1 can be seen as a unified criterion to ensure the persistence of disease in deterministic and stochastic models. Comparing the expressions for R_0 and R^s , we find that R^s\leqslant R_0 , so R^s > 1 can be viewed as a uniform measure to determine the persistence of the disease in both the model and the stochastic model. In addition, by setting up noise perturbations of different intensity, we found that the higher the noise level, the faster the disease extinction. That is, external disturbances play an important role in disease development, such as environmental elimination, culling of infected animals, and vaccination.

    Understanding the statistical characteristics is crucial for managing infectious diseases, but it is challenging to ascertain the statistical properties of disease persistence within an ergodic stationary distribution. To study the dynamic behavior of disease, we construct the endemic equilibrium point, through the equivalence between the Model (4.1) and the corresponding linear Model (4.3) and deduced the log-normal accurate expression of \Phi(S, I, A, C) four-dimensional density function. Furthermore, the covariance matrix \Lambda is determined through the use of an algebraic equation D^2+Z\Lambda+\Lambda Z^T = 0 . The precision of \Lambda can be verified by applying the stability theory to the zero solution of the linear equation, as discussed in the previous research [39]. Furthermore, our method and theory can demonstrate that \Lambda is positive definite, even in cases where D is a semi-definite matrix.

    Finally, there are several important themes that deserve further study. To begin, the stochastic model established in this paper only considers the effect of white noise. However, L \acute{e} vy noise can describe the model more accurately when there is a sudden change in the environment. On the other hand, time delay is an important factor affecting the stability of ecological dynamic systems, as time delay may alter the stability of the system and lead to system fluctuations [43,44]. Therefore, considering L \acute{e} vy noise and delay effects has theoretical value and practical significance. We leave these for future work.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The research was supported by the Ningxia Natural Science Foundation (No. 2022AAC03265), and Natural Science Foundation of China (No. 12161068).

    The authors declare there is no conflicts of interest.

    Proof. Denote the Lyapunov function : V(S, I, A, C) = S^\epsilon+\epsilon I^\epsilon+A^\epsilon+C^\epsilon , for \epsilon > 1 , applying It \hat{o} 's formula

    \begin{equation*} \mbox{D}V(S, I, A, C) = \mbox{L}V(S, I, A, C)\mbox{d}t+\epsilon\sigma_1 S^\epsilon \mbox{d}B_1(t)+\epsilon \sigma_2I^\epsilon \mbox{d}B_2(t)+\epsilon \sigma_3 A^\epsilon \mbox{d}B_3(t)+\epsilon\sigma_4 C^\epsilon \mbox{d}B_4(t), \end{equation*}

    where

    \begin{equation*} \begin{aligned} \mbox{L}&V(S, I, A, C)\\ & = \epsilon S^{\epsilon-1}[rN(1-\frac{N}{K})-\eta_aAS-\eta_cCS-\eta_i\frac{SI}{N}-\mu S+\tau I]\\ &+\epsilon^2I^{\epsilon-1}[\eta_aAS+\eta_cCS+\eta_i\frac{SI}{N}-(\gamma+\mu )I-\tau I]+\epsilon A^{\epsilon-1}[-\alpha A+\beta C]\\ &+\epsilon C^{\epsilon-1}[(\gamma+\mu)I-\delta(S+I)C-\kappa C]+\frac{\epsilon(\epsilon-1)}{2}\sigma_1^2S^\epsilon+\frac{\epsilon^2(\epsilon-1)}{2}\sigma_2^2I^\epsilon\\ &+\frac{\epsilon(\epsilon-1)}{2}\sigma_3^2A^\epsilon+\frac{\epsilon(\epsilon-1)}{2}\sigma_4^2C^\epsilon\\ &\leqslant-S^\epsilon[\epsilon\eta_i+\mu\epsilon+\tau(\epsilon-1)-\frac{\epsilon(\epsilon-1)}{2}\sigma_1^2S^\epsilon]\\ &-I^\epsilon[-(\gamma+\mu+\tau)(1-\epsilon^2)-\frac{\epsilon^2(\epsilon-1)}{2}\sigma_2^2I^\epsilon]-A^\epsilon[\alpha\epsilon-\beta(\epsilon-1)-\frac{\epsilon(\epsilon-1)}{2}\sigma_3^2A^\epsilon]\\ &-C^\epsilon[-(\epsilon-1)(\gamma+\mu)-\beta+\kappa\epsilon-\frac{\epsilon(\epsilon-1)}{2}\sigma_4^2C^\epsilon]+\epsilon rN(1-\frac{N}{K})S^{\epsilon-1}. \end{aligned} \end{equation*}

    Let

    \begin{equation*} \begin{aligned} F&(S, I, A, C)\\ & = -S^\epsilon[\epsilon\eta_i+\mu\epsilon+\tau(\epsilon-1)-1-\frac{\epsilon(\epsilon-1)}{2}\sigma_1^2S^\epsilon] -I^\epsilon[-(\gamma+\mu+\tau)(1-\epsilon^2)-1-\frac{\epsilon^2(\epsilon-1)}{2}\sigma_2^2I^\epsilon]\\ &-A^\epsilon[\alpha-\beta(\epsilon-1)-1-\frac{\epsilon(\epsilon-1)}{2}\sigma_3^2A^\epsilon] -C^\epsilon[-(\epsilon-1)(\gamma+\mu)-\beta+\kappa\epsilon-1-\frac{\epsilon(\epsilon-1)}{2}\sigma_4^2C^\epsilon]\\ &+\epsilon rN(1-\frac{N}{K})S^{\epsilon-1}. \end{aligned} \end{equation*}

    It follows from ( 2.3 ) that there exists the positive constant \hat{T}_2 such that F(S, I, A, C)\leqslant\hat{T}_2 . Hence, we can obtain that

    \begin{equation*} \mbox{L}V(S, I, A, C)\leqslant F(S, I, A, C)-V(S, I, A, C)\leqslant\hat{T}_2-V(S, I, A, C), \end{equation*}

    which yields that \mbox{d}V(S, I, A, C)\leqslant[\hat{T}_2-V(S, I, A, C)]\mbox{d}t +\epsilon\sigma_1 S^\epsilon \mbox{d}B_1(t)+\epsilon \sigma_2I^v \mbox{d}B_2(t)+\epsilon\sigma_3 A^\epsilon \mbox{d}B_3(t)+\epsilon \sigma_4 C^\epsilon \mbox{d}B_4(t) . Integrating both sides of the equation \rm{d[e^tV(S, I, A, C)]} and taking expectations leads to the following

    \begin{equation*} \mathbb{E}[e^tV(S, I, A, C)]\leqslant V(S(0), I(0), A(0), C(0))+\hat{T}_2e^t. \end{equation*}

    Thus, \limsup_{t\to \infty }\mathbb{E}V(S, I, A, C)\leqslant\hat{T}_2 , as a result, \limsup_{t\to \infty }\mathbb{E}|X(t)|^\epsilon\leqslant\frac{\hat{T}^2}{4^{(1-\frac{\epsilon}{2})\wedge0}}: = \hat{T}(\epsilon). In addition, when \epsilon = 2 , we have \limsup_{t\to \infty }\mathbb{E}|X(t)|^2\leqslant\hat{T}_2 , which is the second moment of the solution of Model ( 2.2 ). Now, for any \varepsilon > 0 , let \Psi = \sqrt{\frac{\hat{T}_2}{\varepsilon}} . By Chebyshev's inequality, \mathbb{P}\{|X(t)| > \Psi\}\leqslant\frac{\mathbb{E}|X(t)|^2}{\Psi^2} , which yields that \limsup_{t\to \infty }\mathbb{P}\{|X(t)| > \Psi\}\leqslant\frac{\hat{T}_2}{\Psi^2} = \varepsilon . Therefore, we have \limsup_{t\to \infty }\mathbb{P}\{|X(t)|\leqslant\Psi\}\geqslant 1-\varepsilon . In other words, the solution of Model ( 2.2 ) is stochastically ultimately bounded.

    Through the use of invertible linear transformations, we will derive the corresponding standardized transformation matrices of standard B_1 . The matrix of B_1 : For the algebraic equation D_1^2+Z\Lambda_1+\Lambda_1 Z^T = 0 , where D = \mbox{diag}(\sigma_1, 0, 0, 0) and

    \begin{equation*} Z = \begin{pmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ 0&a_{32}&a_{33}&a_{34}\\ 0&0&a_{43}&a_{44} \end{pmatrix}. \end{equation*}

    We assume that a_{21}\neq0 , a_{32}\neq0 and a_{43}\neq0 . Define X = (x_1, x_2, x_3, x_4) , which follows \mbox{d}X = ZX\mbox{d}t . Considering the following vector Y = (y_1, y_2, y_3, y_4)^T ,

    \begin{equation*} \begin{aligned} &y_4 = x_4, y_3 = y_4' = a_{43}x_3+a_{44}x_4, \\ &y_2 = y_3' = a_{32}a_{43}x_2+(a_{33}+a_{44})a_{43}x_3+(a^2_{44}+a_{34}a_{43})x_4, \\ &y_1 = y_2' = a_{21}a_{32}a_{43}x_1+[(a_{22}+a_{33}+a_{44})a_{32}a_{43}]x_2+[a_{43}(a_{23}a_{32}+a_{34}a_{43}+a_{33}a_{44})+a^2_{33}+a_{44}^2]x_3\\ &+[a_{24}a_{32}a_{43}+(a_{33}+a_{44})a_{34}a_{43}+(a_{34}a_{43}+a_{44}^2)a_{44}]x_4\\ &: = n_1x_1+n_2x_2+n_3x_3+n_4x_4, \end{aligned} \end{equation*}

    then the corresponding strandardized transformation matrix is given by

    \begin{equation*} N = \begin{pmatrix} n_1&n_2&n_3&n_4\\ 0&a_{32}a_{43}&(a_{33}+a_{44})a_{43}&a_{44}^2+a_{34}a_{43}\\ 0&0&a_{43}&a_{44}\\ 0&0&0&1 \end{pmatrix}. \end{equation*}

    We derive that Y = NX , which means \mbox{d}Y = N\mbox{d}X = NZX\mbox{d}t = (NAN^{-1})Y\mbox{d}t , then we have

    \begin{equation*} \mbox{d}Y = \mbox{d} \begin{pmatrix} y_1\\ y_2\\ y_3\\ y_4 \end{pmatrix} = \begin{pmatrix} -a_1&-a_2&-a_3&-a_4\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0 \end{pmatrix} \begin{pmatrix} y_1\\ y_2\\ y_3\\ y_4 \end{pmatrix}\mbox{d}t. \end{equation*}

    Therefore, we can obtain that the corresponding standard B_1 matrix is NZN^{-1}: = Z_1 , which refers to Lemma 2.5. Let \rho = a_{21}a_{32}a_{43}\sigma and \Theta = \rho^{-2}N\Lambda N^T .

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