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Biological occupational allergy: Protein microarray for the study of laboratory animal allergy (LAA)

  • Background: Laboratory Animal Allergy (LAA) has been considered a risk for the workers since 1989 by the NIOSH. About one third of the Laboratory Animal Workers (LAWs) can manifest symptoms to LAA as asthma, rhinitis, conjunctivitis and cutaneous reactions. The prevalence of LAA-induced clinical symptoms has been estimated with a great variability (4–44%) also due to the different methodologies applied. Objective: Evaluate the prevalence of IgE positivity to mouse and rat allergens in LAWs and assess which factors are predisposing to sensitization among subjects exposed to laboratory animals in the workplace. Methods: One hundred LAWs were invited to fill out a questionnaire regarding current allergic symptoms, atopic history, home environment, previous and current occupational history. IgE reactivity versus specific allergens was evaluated with ImmunoCAP ISAC. Results: Out of one hundred LAWs, 18% had a serum susceptibility to mouse and/or rat allergens and 42% reported to have occupational allergy symptoms. Combining the results acquired by ImmunoCAP ISAC and questionnaire, 17% of LAWs have been defined as LAWs-LAA positive since they present a positive IgE response and allergy symptoms, 1% LAWs-LAA sensitized, 25% LAWs-LAA symptomatic and 57% LAWs-LAA negative. Presence of previous allergy symptoms in work and life environment were significantly related to LAWs-LAA positive/sensitized. Conclusions: The study aimed to define the immunological profile of LAWs using the proteomic array as an innovative approach in the study of environmental and occupational exposure to allergens. We suggested a definition of LAWs-LAA considering serum IgE response and presence of allergy symptoms. The proposed approach has the advantage to provide a standard methodology for evaluating the specific IgE responsiveness to animal allergens in specific workplace also considering the immunological profile of workers referred to exposure in life and occupational environment.

    Citation: Maria C D’Ovidio, Annarita Wirz, Danila Zennaro, Stefania Massari, Paola Melis, Vittoria M Peri, Chiara Rafaiani, Maria C Riviello, Adriano Mari. Biological occupational allergy: Protein microarray for the study of laboratory animal allergy (LAA)[J]. AIMS Public Health, 2018, 5(4): 352-365. doi: 10.3934/publichealth.2018.4.352

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  • Background: Laboratory Animal Allergy (LAA) has been considered a risk for the workers since 1989 by the NIOSH. About one third of the Laboratory Animal Workers (LAWs) can manifest symptoms to LAA as asthma, rhinitis, conjunctivitis and cutaneous reactions. The prevalence of LAA-induced clinical symptoms has been estimated with a great variability (4–44%) also due to the different methodologies applied. Objective: Evaluate the prevalence of IgE positivity to mouse and rat allergens in LAWs and assess which factors are predisposing to sensitization among subjects exposed to laboratory animals in the workplace. Methods: One hundred LAWs were invited to fill out a questionnaire regarding current allergic symptoms, atopic history, home environment, previous and current occupational history. IgE reactivity versus specific allergens was evaluated with ImmunoCAP ISAC. Results: Out of one hundred LAWs, 18% had a serum susceptibility to mouse and/or rat allergens and 42% reported to have occupational allergy symptoms. Combining the results acquired by ImmunoCAP ISAC and questionnaire, 17% of LAWs have been defined as LAWs-LAA positive since they present a positive IgE response and allergy symptoms, 1% LAWs-LAA sensitized, 25% LAWs-LAA symptomatic and 57% LAWs-LAA negative. Presence of previous allergy symptoms in work and life environment were significantly related to LAWs-LAA positive/sensitized. Conclusions: The study aimed to define the immunological profile of LAWs using the proteomic array as an innovative approach in the study of environmental and occupational exposure to allergens. We suggested a definition of LAWs-LAA considering serum IgE response and presence of allergy symptoms. The proposed approach has the advantage to provide a standard methodology for evaluating the specific IgE responsiveness to animal allergens in specific workplace also considering the immunological profile of workers referred to exposure in life and occupational environment.


    Avian influenza is an animal infectious disease caused by the transmission of influenza A viruses. Influenza A viruses are divided into subtypes according to two proteins on the surface of the virus: Hemagglutinin (HA) and neuraminidase (NA) [1]. Most avian influenza viruses infect only certain species and do not infect humans. However, a few avian influenza viruses have crossed the species barrier to infect humans and even kill them, such as H5N1, H7N1, H7N2, H7N3, H7N7, H9N2 and H7N9. Among them, H5N1 is a highly pathogenic avian influenza virus, which was first detected in human in Hong Kong in 1997. After that, humans infection with avian influenza have occurred from time to time. As of December 2019, the global cumulative number of cases of human infection with H5N1 avian influenza arrives 861, with 455 deaths. Unlike H5N1, H7N9 is classified as a low pathogenicity avian influenza virus [2]. In March 2013, there was the first case of human infection with the H7N9 avian influenza virus in Shanghai, China. In the following weeks, this virus spread to several provinces and municipalities in mainland China. As of May 2017, H7N9 has resulted in 1263 human cases in China, of whom 459 died, with a mortality rate of nearly 37%. The frequent outbreak of avian influenza in the world not only brings a serious threat to human health, but also causes psychological panic and huge social impact, and brings a huge blow to the national economy. Therefore, it has been important to understand the dynamical behavior of avian influenza and to predict what may occur. Mathematical modeling has been a useful tool to describe the dynamical behavior of avian influenza and to obtain a better understanding of transmission mechanisms. Recently, many avian influenza models have been built from different perspectives (see [2,3,4,5,6,7,8,9,10,11,12] and references therein).

    As we all know, there exist time delays during the spread of avian influenza, which can be used to describe not only the infection period of avian influenza virus in poultry (human) population, but also the incubation period of avian influenza in poultry (human) population and the immune period of recovered human to avian influenza. Therefore, the time delays should be considered such that the avian influenza models are more realistic. Generally speaking, delayed differential equations exhibit more complex dynamical behavior than differential equations without delay because time delay can make a stable equilibrium position to be unstable [13,14,15,16]. Consequently, it is of great interest to describe the transmission mechanism of avian influenza by introducing time delay into the models. For example, Liu et al. [7] and Kang et al. [12] established avian influenza models with different time delays in the poultry and human populations by considering the incubation periods of avian influenza virus and the survival probabilities of infected poultry and humans. By considering the existence of intracellular delay between initial infection of a cell and the release of new virus particles, Samanta [17] established a non-autonomous ordinary differential equation with distributed delay to characterize the spread of avian influenza between poultry and humans. These surveys imply that the research of time delay on avian influenza is a meaningful issue and is still open for study.

    On the other hand, many existing literatures only focus on the deterministic avian influenza models that do not consider the impact of environmental noise. However, in the real world, the spread of avian influenza is often affected by the variations of environmental factors, such as humidity, temperature and so on [18,19]. Due to the fluctuations in the environment, an actual avian influenza system would not remain in a stable state, which would interfere with this stable state by acting directly on the density or indirectly affecting the parameter values. Therefore, it is of great significance to reveal the impact of environmental noise on avian influenza model by using stochastic model, so as to obtain more real benefits and accurately predict the future dynamics of avian influenza. To better understand the transmission dynamics of avian influenza, some authors have introduced stochastic perturbations into the deterministic models [20,21,22]. Zhang et al. [20] constructed a stochastic avian-human influenza model with logistic growth for avian population, and discussed the dynamical behavior of this model. Further, Zhang et al. [21] investigated a stochastic avian-human influenza epidemic model with psychological effect in human population and saturation effect within avian population. On the basis of the deterministic model established by Iwami et al. [3], Zhang et al. [22] established the corresponding stochastic model by introducing density disturbance. All the papers mentioned above only focused on the extinction and persistence of stochastic avian influenza models. However, to the best of our knowledge, there is no results related to the asymptotic behavior of stochastic avian influenza model around the equilibria of the corresponding deterministic model.

    Motivated by the above discussions, in this paper, we investigate the asymptotic behavior of a stochastic delayed avian influenza model with saturated incidence rate. This work differs from existing results [7,12,17,20,21,22] in that (a) time delays and white noise are taken into account to describe the latency period of avian influenza virus in both poultry and human population and the environmental fluctuations; (b) asymptotic behavior of a stochastic delayed avian influenza model is studied. Overview of the rest of the article is as follows: In section 3, we show that there exists a unique global positive solution of system (2.3) with the given initial value (2.4). In section 4, we prove that the solution of system (2.3) is going around $ E^{0} $ under certain conditions. Further, we derive that the solution of system (2.3) is going around $ E^{*} $ under certain conditions in section 5. In section 6, some numerical examples are introduced to illustrate the effectiveness of theoretic results. Finally, some conclusions are given in section 7.

    Although the avian influenza virus spreads between wild birds and poultry, and between poultry and humans, we will only consider the transmission dynamics of avian influenza between poultry and humans because poultry is the main source of infection. Moreover, we assume that the virus is not spread between humans and mutate. We denote the total population of poultry and humans at time $ t $ by $ N_{a}(t) $ and $ N_{h}(t) $, respectively. When the susceptible poultry contact with the infected poultry closely, there is usually no quick way to detect whether they are infected or the detection cost is too high, which makes it impossible to distinguish whether the close contacts of poultry are infected with the avian influenza virus. Therefore, the poultry population is divided into three sub-populations depending on the state of the disease: susceptible poultry $ S_{a}(t) $, exposed poultry $ E_{a}(t) $ and infected poultry $ I_{a}(t) $. The total poultry population at time $ t $ is denoted by $ N_{a}(t) = S_{a}(t) + E_{a}(t) + I_{a}(t) $. The human population is divided into three sub-populations, which are susceptible human $ S_{h}(t) $, infected human with avian influenza $ I_{a}(t) $ and recovered human from avian influenza $ R_{h}(t) $. The total population of human at time $ t $ is given by $ N_{h}(t) = S_{h}(t) + I_{h}(t) + R_{h}(t) $.

    The reason why we do not consider the exposed class for human population is that the close contacts of human beings are usually isolated and tested to determine whether they are infected with the avian influenza virus. The poultry in $ E_a $ either shows symptoms after incubation period and move to $ I_a $, or always stays in $ E_a $ until natural death. The number of susceptible poultry (human) is increased by new recruitment, but decreases by natural death and infection (moving to class $ I_{a} $ ($ I_{h} $)). The number of infected poultry (human) is increased by the infection of susceptible poultry (human) and reduced through natural and disease-related death. In addition, the number of infected humans is also reduced by recovery from the disease (moving to class $ R_{h} $). Based on the above discussions, we obtain the schematic diagram of our model (see Figure 1).

    Figure 1.  Schematic diagram of the model (2.1).

    The corresponding avian influenza model can be represented by the following equations:

    $ \left\{ dSa(t)dt=ΛaμaSa(t)βaSa(t)Ia(t)1+α1Ia(t),dEa(t)dt=βaeμaτaSa(tτa)Ia(tτa)1+α1Ia(tτa)(μa+γa)Ea(t),dIa(t)dt=γaEa(t)(μa+δa)Ia(t),dSh(t)dt=ΛhμhSh(t)βhSh(t)Ia(t)1+α2Ia(t),dIh(t)dt=βheμhτhSh(tτh)Ia(tτh)1+α2Ia(tτh)(μh+δh+θh)Ih(t),dRh(t)dt=θhIh(t)μhRh(t). \right. $ (2.1)

    All parameters in model (2.1) are assumed non-negative and described in Table 1.

    Table 1.  Parameters description in the model (2.1).
    Parameter Description
    $ \Lambda_a $ new recruitment of the poultry populations
    $ \Lambda_{h} $ new recruitment of the human population
    $ \beta_{a} $ the transmission rate from infective poultry to susceptible poultry
    $ \beta_{h} $ the transmission rate from infective poultry to susceptible human
    $ \mu_{a} $ the natural death rate of poultry populations
    $ \mu_{h} $ the natural death rate of human populations
    $ \delta_{a} $ the disease-related death rate of poultry populations
    $ \delta_{h} $ the disease-related death rate of humans populations
    $ \gamma_{a} $ the transfer rate of exposed poultry to infected poultry
    $ \theta_{h} $ the recovery rate of the infective human
    $ \alpha_{i}(i=1, 2) $ parameters that measure the inhibitory effect

     | Show Table
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    Because the removed human populations $ R_{h}(t) $ has no effect on the dynamics of the first five equations, system (2.1) can be decoupled to the following system:

    $ \left\{ dSa(t)dt=ΛaμaSa(t)βaSa(t)Ia(t)1+α1Ia(t),dEa(t)dt=βaeμaτaSa(tτa)Ia(tτa)1+α1Ia(tτa)(μa+γa)Ea(t),dIa(t)dt=γaEa(t)(μa+δa)Ia(t),dSh(t)dt=ΛhμhSh(t)βhSh(t)Ia(t)1+α2Ia(t),dIh(t)dt=βheμhτhSh(tτh)Ia(tτh)1+α2Ia(tτh)(μh+δh+θh)Ih(t). \right. $ (2.2)

    A realistic avian influenza system would not remain in this stable state due to environmental fluctuations. In this paper, we will reveal how the environmental white noise affects the spread of avian influenza through investigating the dynamics of a stochastic delayed avian influenza model with saturated incidence rate. Taking the same approach as the literatures [23,24], we assume that the environmental white noise is directly proportional to the variables $ S_a(t) $, $ E_a(t) $, $ I_a(t) $, $ S_h(t) $ and $ I_h(t) $, respectively. Then, corresponding to system (2.2), the stochastic avian influenza model with time delay is of the following form

    $ \left\{ dSa(t)=(ΛaμaSa(t)βaSa(t)Ia(t)1+α1Ia(t))dt+σ1Sa(t)dB1(t),dEa(t)=(βaeμaτaSa(tτa)Ia(tτa)1+α1Ia(tτa)(μa+γa)Ea(t))dt+σ2Ea(t)dB2(t),dIa(t)=(γaEa(t)(μa+δa)Ia(t))dt+σ3Ia(t)dB3(t),dSh(t)=(ΛhμhSh(t)βhSh(t)Ia(t)1+α2Ia(t))dt+σ4Sh(t)dB4(t),dIh(t)=(βheμhτhSh(tτh)Ia(tτh)1+α2Ia(tτh)(μh+δh+θh)Ih(t))dt+σ5Ih(t)dB5(t), \right. $ (2.3)

    in which $ B_{i}(t)\; (i = 1, 2, \cdots, 5) $ are mutually independent standard Brownian motions defined on a complete probability space $ (\Omega, \mathcal{F}, \bf{P}) $ with a filtration$ \{\mathcal{F}_{t}\}_{t \geq 0} $ satisfying the usual conditions (i.e., it is increasing and right continuous while $ \mathcal{F}_{0} $ contains all $ \bf{P} $-null sets), $ \sigma_{i}\; (i = 1, 2, \cdots, 5) $ denote the intensities of the white noises. The initial value of system (2.3) are

    $ \left\{Sa(θ)=φ1(θ),Ea(θ)=φ2(θ),Ia(θ)=φ3(θ),Sh(θ)=φ4(θ),Ih(θ)=φ5(θ),φi(θ)C([τ,0],R5+),i=1,2,3,4,5,τ=max{τa,τh},\right. $ (2.4)

    where $ \mathcal{C} $ is the Banach space $ \mathcal{C}([-\tau, 0]; \mathbb{R}_+^5) $ of continuous functions mapping the interval $ [-\tau, 0] $ into $ \mathbb{R}_+^5 $, and $ \mathbb{R}_+^5 = \{x = (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}): x_{i} > 0, i = 1, 2, 3, 4, 5\} $. By a biological meaning, we assume that $ \varphi_i(0) > 0\; (i = 1, 2, 3, 4, 5) $.

    In this section, we prove that the solution of system (2.3) is global and positive for any initial value (2.4).

    Theorem 1. For any initial value (2.4), system (2.3) has a unique positive solution $ (S_a(t), E_a(t), I_a(t), S_h(t), I_h(t)) $ on $ t\geq 0 $ and the solution will remain in $ \mathbb{R}_+^5 $ with probability one, in other words, $ (S_a(t), E_a(t), I_a(t), S_h(t), I_h(t))\in \mathbb{R}_+^5 $ for all $ t\geq 0 $ almost surely.

    Proof. Since the coefficients of system (2.3) satisfy the local Lipschitz conditions, then for any initial value (2.4), there exists a unique local solution $ (S_a(t), E_a(t), I_a(t), S_h(t), I_h(t)) $ on $ t\in [-\tau, \tau_e) $, where $ \tau_e $ is the explosive time. To show this solution is global, we only need to show that $ \tau_e = \infty $ a.s. To this end, let $ k_0 \geq 1 $ be sufficiently large such that $ (S_a(\theta), E_a(\theta), I_a(\theta), S_h(\theta), I_h(\theta))\; (\theta \in [-\tau, 0]) $ all lie within the interval $ [\frac{1}{k_0}, k_0] $. For each integer $ k\geq k_0 $, define the stopping time as

    $ τk=inf{t[0,τe):Sa(t)(1k,k) or Ea(t)(1k,k) or Ia(t)(1k,k) or Sh(t)(1k,k) or Ih(t)(1k,k)}. $

    We set $ \inf \varnothing = \infty $. Obviously, $ \tau_k $ increasing when $ k\rightarrow \infty $. Let $ \tau_\infty = \lim_{k\rightarrow \infty} \tau_k $, where $ \tau_\infty \leq \tau_e $ a.s. If we can verify $ \tau_\infty = \infty $ a.s., then $ \tau_e = \infty $ and $ (S_a(t), E_a(t), I_a(t), S_h(t), I_h(t))\in \mathbb{R}_+^5 $ a.s. for all $ t\geq 0 $. That is to say, to complete the proof we only need to show that $ \tau_\infty = \infty $ a.s. If this assertion is not true, then there is a pair of constants $ T > 0 $ and $ \varepsilon \in (0, 1) $ such that

    $ P{τT}>ε. $

    There exists an integer $ k_1 \geq k_0 $ such that

    $ P{τkT}ε for all kk1. $ (3.1)

    Define a $ \mathcal{C}^2 $-function $ V $: $ \mathbb{R}_+^5 \rightarrow \mathbb{R}_+ $ by

    $ V(Sa,Ea,Ia,Sh,Ih)=eμaτa(SaaalnSaa)+(Ea1lnEa)+(Ia1lnIa)+βaeμaτattτaSa(s)Ia(s)1+α1Ia(s)ds+eμhτh(ShbblnShb)+(Ih1lnIh)+βheμhτhttτhSh(s)Ia(s)1+α2Ia(s)ds, $

    in which $ a $ and $ b $ are positive constants to be determined later. The nonnegativity of this function can be derived from $ x- 1- \ln x \geq 0 $ for any $ x > 0 $. Applying the Itô's formula to $ V $, we get

    $ dV=eμaτa(1aSa)dSa+eμaτaa2S2a(dSa)2+(11Ea)dEa+a2E2a(dEa)2+(11Ia)dIa+a2I2a(dIa)2+βaeμaτaSaIa1+α1IaβaeμaτaSa(tτa)Ia(tτa)1+α1Ia(tτa)+eμhτh(1bSh)dSh+eμhτhb2S2h(dSh)2+(11Ih)dIh+12I2h(dIh)2+βheμhτhShIh1+α2IaβheμhτhSh(tτh)Ia(tτh)1+α2Ia(tτh)=LVdt+eμaτaσ1(Saa)dB1(t)+σ2(Ea1)dB2(t)+σ3(Ia1)dB3(t)+eμhτhσ4(Shb)dB4(t)+σ5(Ih1)dB5(t), $ (3.2)

    where

    $ LV=eμaτa(1aSa)(ΛaμaSa)(11Ea)(μa+γa)Ea+(11Ia)(γaEa(μa+δa)Ia)+eμhτh(1bSh)(ΛhμhSh)(11Ih)(μh+δh+θh)Ih+eμaτaaσ212+σ222+σ232+eμhτhbσ242+σ252eμaτaΛa+aμaeμaτa+aσ212eμaτa+2μa+δa+γa+12σ22+12σ23+eμhτhΛh+bμheμhτh+μh+δh+θh+bσ242eμhτh+12σ25+(aβaeμaτa+bβheμhτh(μa+δa))Ia. $

    Choose $ a = \frac{\mu_a e^{\mu_a \tau_a}}{\beta_a} $ and $ b = \frac{\delta_a e^{\mu_h \tau_h}}{\beta_h} $ or $ a = \frac{\delta_a e^{\mu_a \tau_a}}{\beta_a} $ and $ b = \frac{\mu_a e^{\mu_h \tau_h}}{\beta_h} $ such that

    $ a \beta_a e^{-\mu_a \tau_a} + b \beta_h e^{-\mu_h \tau_h}- (\mu_a + \delta_a) = 0. $

    Then, we can get

    $ LV(Sa,Ea,Ia,Sh,Ih)eμaτaΛa+aμaeμaτa+eμhτhΛh+bμheμhτh+2μa+γa+δa+μh+δh+θh+aσ212eμaτa+bσ242eμhτh+12(σ22+σ23+σ25)=:K, $

    where $ K $ is a positive constant. It thus follows from (3.2) that

    $ dV(Sa,Ea,Ia,Sh,Ih)Kdt+eμaτaσ1(Saa)dB1(t)+σ2(Ea1)dB2(t)+σ3(Ia1)dB3(t)+eμhτhσ4(Shb)dB4(t)+σ5(Ih1)dB5(t). $ (3.3)

    Integrating both sides of (3.3) from $ 0 $ to $ \tau_k\wedge T = \min\{\tau_k, T\} $ and then taking the expectation results in

    $ EV(Sa(τkT),Ea(τkT),Ia(τkT),Sh(τkT),Ih(τkT))V(Sa(0),Ea(0),Ia(0),Sh(0),Ih(0))+KE(τkT)V(Sa(0),Ea(0),Ia(0),Sh(0),Ih(0))+KT. $ (3.4)

    Set $ \Omega_k = \{\tau_k \leq T\} $ for $ k\geq k_1 $, and according to (3.1), we have $ P(\Omega_k) \geq \varepsilon $. For every $ \omega \in \Omega_k $, there exists $ S_a(\tau_k, \omega) $ or $ E_a(\tau_k, \omega) $ or $ I_a(\tau_k, \omega) $ or $ S_h(\tau_k, \omega) $ or $ I_h(\tau_k, \omega) $ equals either $ k $ or $ \frac{1}{k} $. Therefore, $ V(S_a(\tau_k, \omega), E_a(\tau_k, \omega), I_a(\tau_k, \omega), S_h(\tau_k, \omega), I_h(\tau_k, \omega)) $ is no less either $ k-1-\ln k $ or $ \frac{1}{k}- 1- \ln \frac{1}{k} $ or $ k-a-a\ln \frac{k}{a} $ or $ \frac{1}{k}- a+ a\ln ak $ or $ k-b-b\ln \frac{k}{b} $ or $ \frac{1}{k}- b+ b\ln bk $.

    Therefore, we have

    $ V(Sa(τk,ω),Ea(τk,ω),Ia(τk,ω),Sh(τk,ω),Ih(τk,ω))(k1lnk)(1k1+lnk)(kaalnka)(1ka+alnak)(kbblnkb)(1kb+blnbk). $

    It follows from (3.4) that

    $ V(Sa(0),Ea(0),Ia(0),Sh(0),Ih(0))+KTE[1ΩkV(Sa(τk,ω),Ea(τk,ω),Ia(τk,ω),Sh(τk,ω),Ih(τk,ω))]ε[(k1lnk)(1k1+lnk)(kaalnka)(1ka+alnak)(kbblnkb)(1kb+blnbk)], $

    where $ 1_{\Omega_k} $ denotes the indicator function of $ \Omega_k $. Letting $ k\rightarrow \infty $, then

    $ >V(Sa(0),Ea(0),Ia(0),Sh(0),Ih(0))+KT=, $

    which leads to the contradiction. This completes the proof.

    In this section, we will investigate the solution of system (2.3) around disease-free equilibrium $ E^0 $ under certain conditions. It is worthwhile to mention that, if $ \mathscr{R}_{0} = \frac{\beta_a \gamma_a\Lambda_a e^{-\mu_a \tau_a}}{\mu_a(\mu_a +\delta_a)(\mu_a +\gamma_a)} < 1 $, the deterministic system (2.2) is globally asymptotically stable around the unique disease-free equilibrium $ E^{0} = (S_{a}^{0}, 0, 0, S_{h}^{0}, 0) = (\frac{\Lambda_a}{\mu_a}, 0, 0, \frac{\Lambda_h}{\mu_h}, 0) $, but $ E^0 $ is not the equilibrium of the stochastic system (2.3). Thus, the result concerning the solution of stochastic system (2.3) around $ E^0 $ is presented by the following theorem.

    Theorem 2. Let $ (S_a(t), E_a(t), I_a(t), S_h(t), I_h(t)) $ be the solution of system (2.3) with the initial value (2.4). If $ \mathscr{R}_{0} < 1 $ and the following conditions hold

    $ \sigma_1 ^2 \lt \mu_a, \sigma_2 ^2 \lt \mu_a+\gamma_a, \sigma_3 ^2 \lt \mu_a+\delta_a, \sigma_4 ^2 \lt \mu_h, \sigma_5 ^2 \lt \mu_h+\delta_h+\theta_h, $

    then,

    $ lim supt1tEt0(SaΛaμa)2dsσ21Λ2aμ2a(μaσ21),lim supt1tEt0(E2a+I2a)dsP1M1,lim supt1tEt0(ShΛhμh)2dsΛ2hμ2h(μhσ24)(σ24+βhα2),lim supt1tEt0I2hdsP2, $

    where

    $ M1=min{μa+γaσ224,(μa+γaσ22)(μa+δaσ23)(μa+δa)4γ2a},P1=e2μaτaσ21Λ2aμ2a[1μaσ21(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)+1],P2=2e2μhτhΛhμ2h(μh+δh+θhσ25)[α2σ24+βhα2(μhσ24)(2μ2h+2μhδh+2μhθh+(δh+θh)22(μh+δh+θh)+σ24)+σ24]. $

    Proof. Since $ (S_a^0, 0, 0, S_h^0, 0) $ is the disease-free equilibrium of system (2.2), then

    $ Λa=μaS0a,Λh=μhS0h. $

    According to system (2.3), we can obtain that

    $ dSa(t)=[μa(SaΛaμa)βaSaIa1+α1Ia]dt+σ1SadB1(t)=[μa(SaΛaμa)βa(SaΛaμa)Ia1+α1IaβaΛaμaIa1+α1Ia]dt+σ1SadB1(t), $ (4.1)

    and

    $ d[Ea(t+τa)+μa+γaγaIa(t+τa)]=dEa(t+τa)+μa+γaγadIa(t+τa)[βaeμaτa(SaΛaμa)Ia1+α1Ia(μa+γa)(μa+δa)γaIa(t+τa)+βaeμaτaΛaμaIa]dt+σ2Ea(t+τa)dB2(t)+σ3(μa+γa)γaIa(t+τa)dB3(t)[βaeμaτa(SaΛaμa)Ia1+α1Ia+(μa+γa)(μa+δa)γa(Ia(t)Ia(t+τa))]dt+σ2Ea(t+τa)dB2(t)+σ3(μa+γa)γaIa(t+τa)dB3(t). $ (4.2)

    Let $ V_1 = \frac{1}{2} \left(S_a -\frac{\Lambda_a}{\mu_a}\right)^2 $, then applying the Itô's formula to $ V_1 $, together with (4.1), we have

    $ dV1=[(SaΛaμa)(μa(SaΛaμa)βa(SaΛaμa)Ia1+α1IaβaΛaμaIa1+α1Ia)+12σ21S2a]dt+σ1Sa(SaΛaμa)dB1(t)=[μa(SaΛaμa)2βa(SaΛaμa)2Ia1+α1IaβaΛaμa(SaΛaμa)Ia1+α1Ia+12σ21S2a]dt+σ1Sa(SaΛaμa)dB1(t)=:LV1dt+σ1Sa(SaΛaμa)dB1(t), $

    where

    $ LV1μa(SaΛaμa)2βaΛaμa(SaΛaμa)Ia1+α1Ia+σ21(SaΛaμa)2+σ21Λ2aμ2a=(μaσ21)(SaΛaμa)2βaΛaμa(SaΛaμa)Ia1+α1Ia+σ21Λ2aμ2a. $ (4.3)

    Similarly, let $ V_2 = E_a(t+\tau_a)+\frac{\mu_a+\gamma_a}{\gamma_a}I_a(t+\tau_a) + \frac{(\mu_a+\gamma_a)(\mu_a+\delta_a)}{\gamma_a} \int_t^{t+\tau_a} I_a(s)ds $, it follows from (4.2) that

    $ dV2βaeμaτa(SaΛaμa)Ia1+α1Ia+σ2Ea(t+τa)dB2(t)+σ3(μa+γa)γaIa(t+τa)dB3(t). $

    Define $ \bar{V} = e^{-\mu_a \tau_a} V_1 + \frac{\Lambda_a}{\mu_a}V_2 $, then

    $ dˉV[eμaτa(μaσ21)(SaΛaμa)2+eμaτaσ21Λ2aμ2a]dt+σ1Sa(SaΛaμa)dB1(t)+σ2Ea(t+τa)dB2(t)+σ3(μa+γa)γaIa(t+τa)dB3(t). $ (4.4)

    Integrating both sides of (4.4) from $ 0 $ to $ t $ and taking expectation, we get

    $ EˉV(t)EˉV(0)eμaτa(μaσ21)Et0(SaΛaμa)2ds+eμaτaσ21Λ2aμ2at. $

    Therefore, we can obtain

    $ lim supt1tEt0(SaΛaμa)2dsσ21Λ2aμ2a(μaσ21). $

    Similarly, we define

    $ V3=12[eμaτa(SaΛaμa)+Ea(t+τa)]2, $

    then,

    $ LV3=e2μaτaμa(SaΛaμa)2eμaτa(2μa+γa)(SaΛaμa)Ea(t+τa)(μa+γa)E2a(t+τa)+12e2μaτaσ21S2a+12σ22E2a(t+τa)e2μaτaμa(SaΛaμa)2+μa+γa2E2a(t+τa)+(2μa+γa)2e2μaτa2(μa+γa)(SaΛaμa)2(μa+γa)E2a(t+τa)+e2μaτaσ21(SaΛaμa)2+e2μaτaσ21Λ2aμ2a+12σ22E2a(t+τa)=e2μaτa(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)(SaΛaμa)212(μa+γaσ22)E2a(t+τa)+e2μaτaσ21Λ2aμ2a. $

    Let $ V_4 = V_3 + \frac{1}{2} (\mu_a + \gamma_a -\sigma_2 ^2)\int_t^{t+\tau_a} E_a^2(s)ds $, we get

    $ LV4e2μaτa(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)(SaΛaμa)212(μa+γaσ22)E2a+e2μaτaσ21Λ2aμ2a. $

    Let $ V_5 = \frac{1}{2} I_a^2 $, the derivative of $ V_5 $ can be calculated as

    $ LV5=γaEaIa(μa+δa)I2a+12σ23I2aμa+δa2I2a+γ2a2(μa+δa)E2a(μa+δa)I2a+12σ23I2a=γ2a2(μa+δa)E2a12(μa+δaσ23)I2a. $

    The Young's inequality is used above. Let

    $ ˜V=V4+eμaτaμaσ21(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)ˉV+(μa+γaσ22)(μa+δa)2γ2aV5, $

    which implies that

    $ L˜V12(μa+γaσ22)E2a+e2μaτaσ21Λ2aμ2a+e2μaτaσ21Λ2aμ2a(μaσ21)(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)+14(μa+γaσ22)E2a(μa+γaσ22)(μa+δaσ23)(μa+δa)4γ2aI2a=14(μa+γaσ22)E2a(μa+γaσ22)(μa+δaσ23)(μa+δa)4γ2aI2a+e2μaτaσ21Λ2aμ2a[1μaσ21(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)+1]. $ (4.5)

    Integrating both sides of (4.5) from $ 0 $ to $ t $ and then taking expectation yields

    $ E˜V(t)E˜V(0)14(μa+γaσ22)Et0E2a(s)ds(μa+γaσ22)(μa+δaσ23)(μa+δa)4γ2aEt0I2a(s)ds+e2μaτaσ21Λ2aμ2a[1μaσ21(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)+1]t. $

    Consequently, we can obtain

    $ lim supt1tEt0(E2a(s)+I2a(s))dsP1M1, $

    where $ M_1 $ and $ P_1 $ are defined in Theorem 2. Further, according to system (2.3), we have

    $ dSh(t)=[μh(ShΛhμh)βhShIa1+α2Ia]dt+σ4ShdB4(t)=[μh(ShΛhμh)(ShΛhμh)βhIa1+α2IaβhΛhIaμh(1+α2Ia)]dt+σ4ShdB4(t), $ (4.6)

    and

    $ dIh(t+τh)=[βheμhτhShIa1+α2Ia(μh+δh+θh)Ih(t+τh)]dt+σ5Ih(t+τh)dB5(t)[βheμhτhIa1+α2Ia(ShΛhμh)+βhΛheμhτhα2μh(μh+δh+θh)Ih(t+τh)]dt+σ5Ih(t+τh)dB5(t). $ (4.7)

    Let $ V_6 = \frac{1}{2} \left(S_h-\frac{\Lambda_h}{\mu_h}\right)^2 $. Noting (4.6), we have

    $ LV6=μh(ShΛhμh)2βh(ShΛhμh)2Ia1+α2IaβhΛhμh(ShΛhμh)Ia1+α2Ia+12σ24S2hμh(ShΛhμh)2βhΛhμh(ShΛhμh)Ia1+α2Ia+σ24(ShΛhμh)2+σ24Λ2hμ2h=(μhσ24)(ShΛhμh)2βhΛhμh(ShΛhμh)Ia1+α2Ia+σ24Λ2hμ2h. $

    Let $ V_7 = e^{-\mu_h \tau_h} V_6 + \frac{\Lambda_h}{\mu_h} I_h(t+\tau_h) $, it follows from (4.7) that

    $ LV7eμhτh(μhσ24)(ShΛhμh)2+eμhτhσ24Λ2hμ2h+βhΛ2heμhτhα2μ2hΛhμh(μh+δh+θh)Ih(t+τh)eμhτh(μhσ24)(ShΛhμh)2+eμhτhΛ2hμ2h(σ24+βhα2). $ (4.8)

    Integrating both sides of (4.8) from $ 0 $ to $ t $ and then taking the expectation yields

    $ EV7(t)EV7(0)eμhτh(μhσ24)Et0(ShΛhμh)2ds+eμhτhΛ2hμ2h(σ24+βhα2)t, $

    therefore, we can get

    $ lim supt1tEt0(ShΛhμh)2dsΛ2hμ2h(μhσ24)(σ24+βhα2). $

    Let $ V_8 = \frac{1}{2} \left[ e^{-\mu_h \tau_h} \left(S_h-\frac{\Lambda_h}{\mu_h}\right) + I_h(t+\tau_h)\right]^2 $, then

    $ LV8=(eμhτh(ShΛhμh)+Ih(t+τh))[eμhτh(ΛhμhSh)(μh+δh+θh)Ih(t+τh)]+12e2μhτhσ24S2h+12σ25I2h(t+τh)e2μhτhμh(ShΛhμh)2+(2μh+δh+θh)2e2μhτh2(μh+δh+θh)(ShΛhμh)2+μh+δh+θh2I2h(t+τh)(μh+δh+θh)I2h(t+τh)+e2μhτhσ24(ShΛhμh)2+e2μhτhσ24Λ2hμ2h+12σ25I2h(t+τh)=e2μhτh(2μ2h+2μhδh+2μhθh+(δh+θh)22(μh+δh+θh)+σ24)(ShΛhμh)212(μh+δh+θhσ25)I2h(t+τh)+e2μhτhσ24Λ2hμ2h. $

    Defining

    $ V9=V8+eμhτhμhσ24(2μ2h+2μhδh+2μhθh+(δh+θh)22(μh+δh+θh)+σ24)V7+12(μh+δh+θhσ25)t+τhtI2h(s)ds, $

    we get

    $ LV912(μh+δh+θhσ25)I2h+e2μhτhΛ2hμ2h[1μhσ24(σ24+βhα2)(2μ2h+2μhδh+2μhθh+(δh+θh)22(μh+δh+θh)+σ24)+σ24]. $ (4.9)

    Integrating both sides of (4.9) from $ 0 $ to $ t $ and taking expectation, we obtain

    $ EV9(t)EV9(0)12(μh+δh+θhσ25)Et0I2h(s)ds+e2μhτhΛ2hμ2h[1μhσ24(σ24+βhα2)(2μ2h+2μhδh+2μhθh+(δh+θh)22(μh+δh+θh)+σ24)+σ24]t. $

    Consequently, we can obtain

    $ lim suptEt0I2h(s)dsP2, $

    where $ P_{2} $ is defined in Theorem 2. This completes the proof.

    If $ \mathscr{R}_{0} > 1 $, there exists an endemic equilibrium $ E^* = (S_a^*, E_a^*, I_a^*, S_h^*, I_h^*) $ of system (2.2), but it is not the equilibrium of system (2.3), where $ S_a^* = \frac{\Lambda_a(1+\alpha_1 I_a^*)}{\mu_a(1+\alpha_1 I_a^*)+ \beta_aI_a^*} $, $ E_a^* = \frac{\beta_a \Lambda_a e^{-\mu_a \tau_a}I_a^*}{(\mu_a+\gamma_a)[\mu_a(1+\alpha_1 I_a^*)+ \beta_aI_a^*]} $, $ I_a^* = \frac{\mu_a (\mathscr{R}_{0}-1)}{\alpha_1 \mu_a + \beta_a} $, $ S_h^* = \frac{\Lambda_h(1+\alpha_2I_a^*)}{\mu_h(1+\alpha_2I_a^*)+ \beta_hI_a^*} $, $ E_a^* = \frac{\beta_h e^{-\mu_h \tau_h}S_h^*I_a^*}{(\mu_h+\delta_h+\theta_h)(1+\alpha_2I_a^*)} $. In this section, we show that the solution of system (2.3) is going around $ E^* $ under certain conditions.

    Theorem 3. Let $ (S_a(t), E_a(t), I_a(t), S_h(t), I_h(t)) $ be the solution of system (2.3) with initial value (2.4). If $ \mathscr{R}_{0} > 1 $ and the following conditions hold

    (i) $ \sigma_1^2 < \mu_a, \sigma_2 ^2 < \frac{1}{2} (\mu_a+\gamma_a), \sigma_3 ^2 < \frac{1}{2} (\mu_a+\delta_a), \sigma_4^2 < \mu_h, \sigma_5 ^2 < \mu_h+\delta_h+\theta_h $;

    (ii) $ \max(\sqrt{P_{3}}, \sqrt{P_{4}}, \sqrt{P_{5}}, \sqrt{P_{6}}) < d(E^{*}, E^{0}) $,

    then

    $ lim supt1tEt0(SaSa)2dsP3,lim suptEt0[(Ea(s)Ea)2+(Ia(s)Ia)2]dsL1L2=:P4,lim suptEt0(ShSh)2dsP5,lim suptEt0(IhIh)2dsP6, $

    where

    $ d(E,E0)=(SaΛaμa)2+(Ea)2+(Ia)2+(ShΛhμh)2+(Ih)2P3=1μaσ21[σ21(Sa)2+σ21SaL32μa+(eμaτaSa+L3μaeμaτa)(12σ22Ea+μa+γa2γaσ23Ia)],P4=L1L2,P5=σ24(Sh)2μhσ24,P6=σ24L24(μhσ24)(μh+δh+θhσ25)2+2σ25(Ih)2μh+δh+θhσ25,L1=eμaτaμaσ21(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)[σ21(Sa)2+σ21SaL32μa+(eμaτaSa+L3μaeμaτa)(12σ22Ea+12μa+γaγaσ23Ia)]+e2μaτaσ21(Sa)2+σ22(Ea)2+σ23(μa+δa)(μa+γa2σ22)2γ2a(Ia)2,L2=min{14(μa+γa2σ22),(μa+δa)(μa+γa2σ22)(μa+δa2σ23)4γ2a},L3=βaSaIa1+α1Ia,L4=βhShIa1+α2Ia. $

    Proof. Since $ (S_a^*, E_a^*, I_a^*, S_h^*, I_h^*) $ is the interior equilibrium of system (2.2), then

    $ Λa=μaSa+βaSaIa1+α1Ia,(μa+γa)Ea=βaeμaτaSaIa1+α1Ia,IaEa=γaμa+δa,Λh=μhSh+βhShIa1+α2Ia,(μh+δh+θh)Ih=βheμhτhShIa1+α2Ia. $ (5.1)

    Define the Lyapunov function $ W_1 $ as $ W_1 = S_a-S_a^*-S_a^* \ln \frac{S_a}{S_a^*} $, from which we have

    $ dW1=(ΛaμaSaβaSaIa1+α1IaΛaSaSa+μaSa+βaSaIa1+α1Ia+12Saσ21)dt+σ1(SaSa)dB1(t)=[(μaSa+βaSaIa1+α1Ia)(2SaSaSaSa)+βaSaIa1+α1Ia(SaIa(1+α1Ia)SaIa(1+α1Ia)+SaSa+Ia(1+α1Ia)Ia(1+α1Ia)1)+12Saσ21]dt+σ1(SaSa)dB1(t)=LW1dt+σ1(SaSa)dB1(t), $

    where

    $ LW1=(μa+βaIa1+α1Ia)(SaSa)2Saβa(SaSa)(Ia1+α1IaIa1+α1Ia)+12Saσ21. $ (5.2)

    Similarly, we can define $ W_2 $ as

    $ W2=Ea(t+τa)EaEalnEa(t+τa)Ea+μa+γaγa(Ia(t+τa)IaIalnIa(t+τa)Ia). $

    By using the Itô's formula, the derivative of $ W_2 $ is calculated as follows

    $ LW2=(1EaEa(t+τa))(βaeμaτaSaIa1+α1Ia(μa+γa)Ea(t+τa))+μa+γaγa(1IaIa(t+τa))(γaEa(t+τa)(μa+δa)Ia(t+τa))+12σ22Ea+μa+γa2γaσ23Ia=βaeμaτaIa1+α1Ia(SaSa)(1+α1IaIaIa1+α1Ia1)+βaeμaτaSaIa1+α1Ia(SaSa1+α1IaSaIaSaIa1+α1IaEaEa(t+τa)+1+α1IaIaIa1+α1IaIa(t+τa)IaEa(t+τa)EaIaIa(t+τa))+12σ22Ea+12μa+γaγaσ23Ia. $ (5.3)

    Since $ x - 1 - \ln x \geq 0 $ for $ x > 0 $, the following estimate can be obtained

    $ 1+α1IaSaIaSaIa1+α1IaEaEa(t+τa)1+ln(1+α1IaSaIaSaIa1+α1IaEaEa(t+τa))=1+lnSaSalnIa(t+τa)Ia+lnIa(1+α1Ia)Ia(1+α1Ia)lnEa(t+τa)EaIaIa(t+τa). $ (5.4)

    Substituting (5.4) into (5.3), we can get

    $ LW2βaeμaτaIa1+α1Ia(SaSa)(1+α1IaIaIa1+α1Ia1)+βaeμaτaSaIa1+α1Ia(SaSa1lnSaSa+lnIa(t+τa)IalnIa(1+α1Ia)Ia(1+α1Ia)+lnEa(t+τa)EaIaIa(t+τa)+1+α1IaIaIa1+α1IaIa(t+τa)IaEa(t+τa)EaIaIa(t+τa))+12σ22Ea+12μa+γaγaσ23Ia=βaeμaτaIa1+α1Ia(SaSa)(1+α1IaIaIa1+α1Ia1)+βaeμaτaSaIa1+α1Ia[(SaSalnSaSa)(Ia(t+τa)IalnIa(t+τa)Ia)+(Ia(1+α1Ia)Ia(1+α1Ia)lnIa(1+α1Ia)Ia(1+α1Ia))(Ea(t+τa)EaIaIa(t+τa)lnEa(t+τa)EaIaIa(t+τa))1]+12σ22Ea+12μa+γaγaσ23Ia. $ (5.5)

    Choose $ W_3 = W_2 + \frac{\beta_a e^{-\mu_a \tau_a} S_a^*I_a^*}{1+\alpha_1 I_a^*} \int_t^{t+\tau_a} \left(\frac{I_a(s)}{I_a^*} - \ln\frac{I_a(s)}{I_a^*} - 1\right)ds $. Therefore, $ LW_3 $ can be obtained as follows by using (5.5):

    $ LW3βaeμaτaIa1+α1Ia(SaSa)(Ia(1+α1Ia)Ia(1+α1Ia)1)+βaeμaτaSaIa1+α1Ia[(SaSalnSaSa)(Ia(t+τa)IalnIa(t+τa)Ia)+(Ia(1+α1Ia)Ia(1+α1Ia)lnIa(1+α1Ia)Ia(1+α1Ia))(Ea(t+τa)EaIaIa(t+τa)lnEa(t+τa)EaIaIa(t+τa))1]+12σ22Ea+12μa+γaγaσ23Ia+βaeμaτaSaIa1+α1Ia(Ia(t+τa)IalnIa(t+τa)Ia1)βaeμaτaSaIa1+α1Ia(IaIalnIaIa1)βaeμaτaIa1+α1Ia(SaSa)(Ia(1+α1Ia)Ia(1+α1Ia)1)+βaeμaτaSaIa1+α1Ia[SaSa+SaSa1IaIa+Ia(1+α1Ia)Ia(1+α1Ia)+lnIa(1+α1Ia)Ia(1+α1Ia)IaIa1]+12σ22Ea+12μa+γaγaσ23Ia. $ (5.6)

    Noting that $ x - 1 - \ln x \geq 0 $ holds for $ x > 0 $, we also have

    $ IaIa+Ia(1+α1Ia)Ia(1+α1Ia)+lnIa(1+α1Ia)Ia(1+α1Ia)IaIaIaIa+Ia(1+α1Ia)Ia(1+α1Ia)+Ia(1+α1Ia)Ia(1+α1Ia)IaIa1Ia(1+α1Ia)Ia(1+α1Ia)IaIa(Ia(1+α1Ia)Ia(1+α1Ia)IaIa1)(Ia(1+α1Ia)Ia(1+α1Ia)1)=(1+α1Ia)(1+α1Ia)Ia(11+α1Ia11+α1Ia)(Ia1+α1IaIa1+α1Ia)<0, $ (5.7)

    substituting (5.7) into (5.6) and using $ \frac{S_a}{S_a^*} + \frac{S_a^*}{S_a} -2 = \frac{(S_a -S_a^*)^2}{S_aS_a^*} $, we know that

    $ LW3βaeμaτaIa1+α1Ia(SaSa)(Ia(1+α1Ia)Ia(1+α1Ia)1)+βaeμaτaIa1+α1Ia(SaSa)2Sa+12σ22Ea+12μa+γaγaσ23Ia. $ (5.8)

    Let $ W_4 = W_1 + \frac{1+\alpha_1 I_a^*}{\beta_a e^{-\mu_a \tau_a} I_a^*} \left(\mu_a + \frac{\beta_a I_a^*}{1+\alpha_1 I_a^*} \right)W_3 $. Applying the Itô's formula, together with (5.2) and (5.8), derives that

    $ LW4=LW1+1+α1IaβaeμaτaIa(μa+βaIa1+α1Ia)LW3(μa+βaIa1+α1Ia)(SaSa)2Saβa(SaSa)(Ia1+α1IaIa1+α1Ia)+12σ21Sa+1+α1IaβaeμaτaIa(μa+βaIa1+α1Ia)[βaeμaτaIa1+α1Ia(SaSa)(Ia(1+α1Ia)Ia(1+α1Ia)1)+βaeμaτaIa1+α1Ia(SaSa)2Sa+12σ22Ea+12μa+γaγaσ23Ia]=(μa+βaIa1+α1Ia)(SaSa)(Ia(1+α1Ia)Ia(1+α1Ia)1)βa(SaSa)(Ia1+α1IaIa1+α1Ia)+12σ21Sa+1+α1IaβaeμaτaIa(μa+βaIa1+α1Ia)(12σ22Ea+12μa+γaγaσ23Ia)=(SaSa)(Ia1+α1IaIa1+α1Ia)[(μa+βaIa1+α1Ia)1+α1IaIaβa]+12σ21Sa+1+α1IaβaeμaτaIa(μa+βaIa1+α1Ia)(12σ22Ea+12μa+γaγaσ23Ia)=μa(1+α1Ia)Ia(SaSa)(Ia1+α1IaIa1+α1Ia)+12σ21Sa+1+α1IaβaeμaτaIa(μa+βaIa1+α1Ia)(12σ22Ea+12μa+γaγaσ23Ia). $ (5.9)

    Choose Lyapunov function $ W_5 $ as $ W_5 = \frac{(S_a-S_a^*)^2}{2} $, then its derivative is

    $ LW5=(SaSa)[ΛaμaSaβaSaIa1+α1Ia]+12σ21S2a=(SaSa)[μaSaμaSa+βaSaIa1+α1IaβaSaIa1+α1Ia]+12σ21S2a=μa(SaSa)2βaSa(SaSa)(Ia1+α1IaIa1+α1Ia)βa(SaSa)2Ia1+α1Ia+12σ21S2aμa(SaSa)2βaSa(SaSa)(Ia1+α1IaIa1+α1Ia)+σ21(SaSa)2+σ21(Sa)2=(μaσ21)(SaSa)2βaSa(SaSa)(Ia1+α1IaIa1+α1Ia)+σ21(Sa)2. $

    Let $ \bar{W} = W_5 + \frac{\beta_a S_a^* I_a^*}{\mu_a(1+\alpha_1 I_a^*)}W_4 $, one can derive that

    $ LˉW(μaσ21)(SaSa)2βaSa(SaSa)(Ia1+α1IaIa1+α1Ia)+σ21(Sa)2+βaSaIaμa(1+α1Ia)[μa(1+α1Ia)Ia(SaSa)(Ia1+α1IaIa1+α1Ia)+12σ21Sa+1+α1IaβaeμaτaIa(μa+βaIa1+α1Ia)(12σ22Ea+12μa+γaγaσ23Ia)]=(μaσ21)(SaSa)2+σ21(Sa)2+βaSaIa2μa(1+α1Ia)σ21Sa+(eμaτaSa+βaSaIaμaeμaτa(1+α1Ia))(12σ22Ea+12μa+γaγaσ23Ia). $ (5.10)

    Integrating both sides of (5.10) from $ 0 $ to $ t $ and then taking expectation yields

    $ EˉW(t)EˉW(0)(μaσ21)Et0(Sa(s)Sa)2ds+[σ21(Sa)2+βaSaIa2μa(1+α1Ia)σ21Sa+(eμaτaSa+βaSaIaμaeμaτa(1+α1Ia))(12σ22Ea+12μa+γaγaσ23Ia)]t. $

    Then, we can get

    $ lim supt1tEt0(Sa(s)Sa)2dsP3, $

    where $ P_3 $ is defined in Theorem 3. Defining $ W_6 = \frac{1}{2} \left[ e^{-\mu_a \tau_a}(S_a-S_a^*) + E_a(t+\tau_a) -E_a^* \right]^2 $, the use of Itô's formula yields that

    $ LW6=μae2μaτa(SaSa)2(μa+γa)(Ea(t+τa)Ea)2(2μa+γa)eμaτa(SaSa)(Ea(t+τa)Ea)+12e2μaτaσ21S2a+12σ22E2a(t+τa)μae2μaτa(SaSa)2(μa+γa)(Ea(t+τa)Ea)2+μa+γa2(Ea(t+τa)Ea)2+(2μa+γa)2e2μaτa2(μa+γa)(SaSa)2+e2μaτaσ21(SaSa)2+e2μaτaσ21(Sa)2+σ22(Ea(t+τa)Ea)2+σ22(Ea)2=e2μaτa(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)(SaSa)2(μa+γa2σ22)(Ea(t+τa)Ea)2+e2μaτaσ21(Sa)2+σ22(Ea)2. $

    Let $ W_7 = W_6 + \left(\frac{\mu_a+\gamma_a}{2} -\sigma_2 ^2 \right) \int_t^{t+\tau_a}(E_a(s) -E_a^*)^2ds $ and $ W_8 = \frac{1}{2} (I_a -I_a^*)^2 $. We have

    $ LW7e2μaτa(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)(SaSa)2(μa+γa2σ22)(EaEa)2+e2μaτaσ21(Sa)2+σ22(Ea)2, $ (5.11)

    and

    $ LW8=(IaIa)(γaEa(μa+δa)Ia)+12σ23I2a=γa(EaEa)(IaIa)(μa+δa)(IaIa)2+12σ23I2aμa+δa2(IaIa)2+γ2a2(μa+δa)(EaEa)2(μa+δa)(IaIa)2+σ23(IaIa)2+σ23(Ia)2=γ2a2(μa+δa)(EaEa)2(μa+δa2σ23)(IaIa)2+σ23(Ia)2. $ (5.12)

    Let $ \tilde{W} = W_7+ \frac{e^{-\mu_a \tau_a}}{\mu_a - \sigma_1 ^2} \left(\frac{2\mu_a^2 +2\mu_a\gamma_a +\gamma_a^2}{2(\mu_a +\gamma_a)} +\sigma_1 ^2 \right)\bar{W} + \frac{ (\mu_a+\delta_a) (\mu_a+\gamma_a-2\sigma_2 ^2)}{2\gamma_a^2} W_8 $. Making use of (5.10), (5.11) and (5.12) yields that

    $ L˜W=LW7+eμaτaμaσ21(2μ2a+2μaγa+γ2a2(μa+γa)+σ21)LˉW+(μa+δa)(μa+γa2σ22)2γ2aLW814(μa+γa2σ22)(EaEa)2(μa+δa)(μa+γa2σ22)(μa+δa2σ23)4γ2a(IaIa)2+L1. $ (5.13)

    Integrating both sides of (5.13) from $ 0 $ to $ t $ and then taking expectation yields

    $ E˜W(t)E˜W(0)14(μa+γa2σ22)Et0(Ea(s)Ea)2ds(μa+δa)(μa+γa2σ22)(μa+δa2σ23)4γ2aEt0(Ia(s)Ia)2ds+L1t. $

    Therefore, we can obtain

    $ lim suptEt0[(Ea(s)Ea)2+(Ia(s)Ia)2]dsL1L2=:P4, $

    where $ L_{1}, L_{2} $ have been defined in Theorem 3. Taking $ U_1 = \frac{1}{2}(S_h - S_h^*)^{2} $, we have

    $ LU1=(ShSh)(ΛhμhShβhShIa1+α2Ia)+12σ24Sh=(ShSh)[μhShμhSh+βhShIa1+α2IaβhShIa1+α2Ia]+12σ24Sh=(μh+βhIa1+α2Ia)(ShSh)2+σ24(ShSh)2+σ24(Sh)2(μhσ24)(ShSh)2+σ24(Sh)2. $ (5.14)

    Integrating both sides of (5.14) from $ 0 $ to $ t $ and then taking expectation, we get

    $ EU1(t)EU1(0)(μhσ24)Et0(ShSh)2ds+σ24(Sh)2t. $

    Therefore, we can obtain

    $ lim suptEt0(ShSh)2dsσ24(Sh)2μhσ24. $

    Let $ U_2 = \frac{1}{2}[I_h(t+\tau_{h}) - I_h^*]^{2} $, we have

    $ LU2=(Ih(t+τh)Ih)[βhShIa1+α2Ia(μh+δh+θh)Ih(t+τh)]+12σ25I2h(t+τh)=βhIa1+α2Ia(Ih(t+τh)Ih)(ShSh)(μh+δh+θh)(Ih(t+τh)Ih)2+12σ25I2h(t+τh)β2h(Ia)22(1+α2Ia)2(μh+δh+θhσ25)(ShSh)2μh+δh+θhσ252(Ih(t+τh)Ih)2(μh+δh+θh)(Ih(t+τh)Ih)2+σ25(Ih(t+τh)Ih)2+σ25(Ih)2=β2h(Ia)2(ShSh)22(1+α2Ia)2(μh+δh+θhσ25)μh+δh+θhσ252(Ih(t+τh)Ih)2+σ25(Ih)2. $

    Let $ \overline{U} = \frac{\beta_h^{2}(I_a^*)^{2}}{2(\mu_h-\sigma_4 ^2)(1+\alpha_2I_a^*)^{2} (\mu_h+\delta_{h} +\theta_{h} -\sigma_5 ^2)}U_{1}+U_{2} $, then

    $ L¯U=μh+δh+θhσ252(Ih(t+τh)Ih)2+β2h(Ia)2σ24(Sh)22(μhσ24)(1+α2Ia)2(μh+δh+θhσ25)+σ25(Ih)2. $

    Let $ U_{3} = \frac{\mu_h+\delta_{h}+\theta_{h}-\sigma_5 ^2}{2}\int_t^{t+\tau_{h}} (I_h(s) - I_h^*)^{2} ds $, we obtain

    $ LU3=μh+δh+θhσ252[(Ih(t+τh)Ih)2+(IhIh)2]. $

    Let $ \widetilde{U} = \overline{U}+U_{3} $, then,

    $ L˜U=μh+δh+θhσ252(Ih(t+τh)Ih)2+β2h(Ia)2σ24(Sh)22(μhσ24)(1+α2Ia)2(μh+δh+θhσ25)+σ25(Ih)2+μh+δh+θhσ252[(Ih(t+τh)Ih)2+(IhIh)2]=μh+δh+θhσ252(IhIh)2+β2h(Ia)2σ24(Sh)22(μhσ24)(1+α2Ia)2(μh+δh+θhσ25)+σ25(Ih)2. $ (5.15)

    Integrating both sides of (5.15) from $ 0 $ to $ t $ and then taking expectation, we have

    $ E˜U(t)E˜U(0)μh+δh+θhσ252Et0(IhIh)2ds+β2h(Ia)2σ24(Sh)22(μhσ24)(1+α2Ia)2(μh+δh+θhσ25)t+σ25(Ih)2t. $

    Therefore, we can obtain

    $ lim suptEt0(IhIh)2dsP6, $

    where $ P_{6} $ has been defined in Theorem 3. The proof is completed.

    This section is devoted to illustrating the theoretical results by numerical examples. The parameters of system (2.3) are selected as in Table 2, $ \alpha_1 $ and $ \alpha_2 $ are varying parameters that is taken value from $ 0.001 $ to $ 0.1 $, and $ \sigma_1 = 0.01, \, \sigma_2 = \sigma_3 = \sigma_5 = 0.04, \, \sigma_4 = 0.008 $. The initial conditions of system (2.3) are $ S_a(\theta) = 3,000,000, E_a(\theta) = 1,000, I_a(\theta) = 10, S_h(\theta) = 1,000, I_h(\theta) = 5, \, \theta\in[-\tau, 0] $. The Milstein method [25] is used to obtain the discrete form of system (2.3) as follows:

    $ \left\{ Sa(k+1)=Sa(k)+(ΛaμaSa(k)βaSa(k)Ia(k)1+α1Ia(k))Δt+σ1Sa(k)Δtξ1(k)+12σ21Sa(k)(ξ21(k)1)Δt,Ea(k+1)=Ea(k)+(βaeμaτaSa(kτaΔt)Ia(kτaΔt)1+α1Ia(kτaΔt)(μa+γa)Ea(k))Δt+σ2Ea(k)Δtξ2(k)+12σ22Ea(k)(ξ22(k)1)Δt,Ia(k+1)=Ia(k)+(γaEa(k)(μa+δa)Ia(k))Δt+σ3Ia(k)Δtξ3(k)+12σ23Ia(k)(ξ23(k)1)Δt,Sh(k+1)=Sh(k)+(ΛhμhSh(k)βhSh(k)Ia(k)1+α2Ia(k))Δt+σ4Sh(k)Δtξ4(k)+12σ24Sh(k)(ξ24(k)1)Δt,Ih(k+1)=Ih(k)+(βheμhτhSh(kτhΔt)Ia(kτhΔt)1+α2Ia(kτhΔt)(μh+δh+θh)Ih(k))Δt+σ5Ih(k)Δtξ5(k)+12σ25Ih(k)(ξ25(k)1)Δt, \right. $ (6.1)
    Table 2.  Parameter values used in numerical simulations for model (2.3).
    Parameter Value Source of data
    $ \Lambda_a $ $ 30000 $ Assumed
    $ \Lambda_{h} $ $ \mu_h\times1000 $ Assumed
    $ \beta_{a} $ $ (0.5 $---$ 12.5)\times10^{-6} \text{day}^{-1} $ [10]
    $ \beta_{h} $ $ 3\times10^{-4} $ [10]
    $ \mu_{a} $ $ 1/100 \text{day}^{-1} $ [10]
    $ \mu_{h} $ $ 200/(70\times365)\text{day}^{-1} $ Assumed
    $ \delta_{a} $ $ 5 \text{day}^{-1} $ [10]
    $ \delta_{h} $ $ 0.03 \text{day}^{-1} $ [10,11]
    $ \gamma_{a} $ $ 0.3 \text{day}^{-1} $ [11]
    $ \theta_{h} $ $ 0.16 \text{day}^{-1} $ [11]
    $ \tau_a $ $ 7 $ day Assumed
    $ \tau_h $ $ 14 $ day Assumed

     | Show Table
    DownLoad: CSV

    where $ \xi_i(k) \sim N(0, 1) (i = 1, \cdots, 5; k = 1, 2, \cdots) $ are independent Gaussian random variables. Initially, we study the effect of $ \mathscr{R}_0 $, which, by Theorems 4.1 and 5.1, can govern the asymptotic behavior.

    Example 1 Effect of basic reproduction number $ \mathscr{R}_0 $.

    Choose different $ \beta_a $ such that $ \mathscr{R}_0 $ take different values, which are shown in Table 3. Since $ \sigma_1^2 = 10^{-4} < \mu_a = 10^{-2}, \sigma_2^2 = 0.0016 < \frac{1}{2} (\mu_a+\gamma_a) = 0.155, \sigma_3^2 = 0.0016 < \frac{1}{2} (\mu_a+\delta_a) = 5.01, \sigma_4^2 = 0.000064 < \mu_h = 0.0078, \sigma_5^2 = 0.0016 < \mu_h+\delta_h+\theta_h = 0.1978 $, the condition (ⅰ) of Theorem 3 is satisfied. From Table 3, we see that for each $ \mathscr{R}_0 $, the inequality $ P_{m} < d_E $ holds, which means the condition (ⅱ) of Theorem 3 is also satisfied. Thus, all the conclusions of Theorem 3 hold. It follows from Table 3 that the change of $ \mathscr{R}_0 $ can result in different values of $ E^* $, which also illustrate the value of $ E^* $ is related to $ \mathscr{R}_0 $. By the discrete form of system (2.3), the numerical results under different $ \mathscr{R}_0 $ are presented by Figures 2 and 3 when $ \mathscr{R}_0 > 1 $, which show that the solution of system (2.3) goes around the endemic equilibrium $ E^* $. The effectiveness of Theorem 3 is also indicated by these two figures. In addition, we can see from Figures 2 and 3 and Table 3 that the number of infected poultry and humans will reduce with the decrease of $ \mathscr{R}_0 $. On the other hand, in order to explore if the results of Theorem 3 hold, we enhance the intensity of perturbation as $ \sigma = (\sigma_1, \cdots, \sigma_5) = (0.02, 0.08, 0.08, 0.016, 0.08) $ (Case Ⅰ: condition (ⅰ) of Theorem 3 is satisfied but condition (ⅱ) is not satisfied), $ \sigma = (0.06, 0.24, 0.24, 0.048, 0.24) $ (Case Ⅱ: Both conditions (ⅰ) and (ⅱ) are not satisfied) and $ \sigma = (0.10, 0.40, 0.40, 0.080, 0.40) $ (Case Ⅲ: Both conditions (ⅰ) and (ⅱ) are not satisfied). The simulation results are presented in Figure 4, which are obtained by computing the average of 800 simulations. The equilibrium of corresponding deterministic model is $ E^* = (2.3931\times10^{6}, 1.8254\times10^{4}, 0.7812\times10^{3}, 227.3975, 11.5855) $. From Figure 4, we see that the curves will move away from the equilibrium point $ E^* $ with the increasing of intensity of perturbation, which violate the conclusions of Theorem 3.

    Table 3.  Value of $ E^{*} (S_{a}^{*}, E_{a}^{*}, I_{a}^{*}, S_{h}^{*}, I_{h}^{*}) $ under different $ \mathscr{R}_0 $.
    $ \mathscr{R}_0 $ $ S_a^*(\times10^{6}) $ $ E_a^*(\times10^{4}) $ $ I_a^*(\times10^{3}) $ $ S_h^* $ $ I_h^* $ $ P_{m} \, (\times 10^5)^{\sharp} $ $ d_E \, (\times 10^5)^{\sharp} $
    4.8269 0.7977 6.6239 2.8348 212.6795 11.8062 2.5226 22.033
    3.2823 1.1336 5.6137 2.4024 213.7049 11.7908 2.7378 18.673
    1.9308 1.7948 3.6249 1.5513 217.3720 11.7359 3.2294 12.057
    1.3515 2.3931 1.8254 0.7812 227.3975 11.5855 3.7266 6.072
    $^{\sharp}\, P_{m}=\max(\sqrt{P_{3}}, \sqrt{P_{4}}, \sqrt{P_{5}}, \sqrt{P_{6}})$, $d_E = d(E^*,E^0)$.

     | Show Table
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    Figure 2.  The behavior of avian population under different $ \mathscr{R}_0 > 1 $.
    Figure 3.  The behavior of human population under different $ \mathscr{R}_0 > 1 $.
    Figure 4.  The trajectories of $ E_a, I_a $ and $ I_h $ for large perturbation when $ \mathscr{R}_0 = 1.3515 > 1 $.

    According to the values of $ \sigma_1, \cdots, \sigma_5 $ and the parameters values in Table 2, we easily verify the conditions of Theorem 2 are satisfied. Therefore, from Theorem 2 we know that the solution of system (2.3) will go around the disease-free equilibrium $ E^0 $ when $ \mathscr{R}_0 < 1 $. The numerical simulation results of $ \mathscr{R}_0 $ are presented in Figures 5 and 6. These figures show $ E_a, I_a $ and $ I_h $ all go to zero when $ \mathscr{R}_0 < 1 $, which illustrate the effectiveness of the theoretical results in Theorem 2. Meanwhile, Figures 5 and 6 also show that the rate of $ E_a, I_a $ and $ I_h $ converges to zero is increasing with the decrease of $ \mathscr{R}_0 $. The conditions of Theorem 2 are only a sufficient ones, so we want to know whether the conclusions of Theorem 2 hold when the intensity of perturbation increase such that these conditions are not satisfied. Thus, we choose $ \sigma = (\sigma_1, \cdots, \sigma_5) = (0.02, 0.08, 0.08, 0.016, 0.08) $ (Case I), $ \sigma = (0.06, 0.24, 0.24, 0.048, 0.24) $ (Case II) and $ \sigma = (0.12, 0.48, 0.48, 0.096, 0.48) $ (Case III), and the simulation results are presented in Figure 7. Figure 7 shows $ E_a, I_a $ and $ I_h $ converge to zero for each cases, so the results of Theorem 2 also hold.

    Figure 5.  The behavior of infected avian population under different $ \mathscr{R}_0 < 1 $.
    Figure 6.  The behavior of infected human population under different $ \mathscr{R}_0 < 1 $.
    Figure 7.  The trajectories of $ E_a, I_a $ and $ I_h $ for large perturbation when $ \mathscr{R}_0 = 0.7723 < 1 $.

    Example 2 Effect of time delays $ \tau_a $ and $ \tau_h $.

    In order to study the effect of time delays, we consider the average peak values of $ E_a $, $ I_a $ and $ I_h $, and the time of reaching average peak values by $ 300 $ simulation runs. The simulation results are shown in Figures 8 and 9. It follows from Figure 8 that the increase of time delay $ \tau_a $ or $ \tau_h $ can reduce the peak value of both infected poultry and human population. Meanwhile, from Figure 9, we know that the large time delay also lead to the delay of reaching peak value. Thus, we may conclude that time delays have significate influence for the spread of avian influenza. According to the practical meaning of $ \tau_a $ and $ \tau_h $, related department can adopt some measures to increase the spread delay to suppress the outbreak of influenza, such as isolation. In addition, the adopting of those control measures will win time for taking drug control.

    Figure 8.  The average peak values of $ E_a $, $ I_a $ and $ I_h $ under different $ \tau_a $ and $ \tau_h $ by $ 300 $ simulation runs.
    Figure 9.  The time of reaching average peak values of $ E_a $, $ I_a $ and $ I_h $ by $ 300 $ simulation runs.

    Example 3 Effect of saturation constants $ \alpha_1 $ and $ \alpha_2 $.

    According to the analysis of Introduction, the saturation constants $ \alpha_1 $ and $ \alpha_2 $ are important parameters for avian influenza. We thus explore the effects of $ \alpha_1 $ and $ \alpha_2 $ in this example. In order to explore the effect of $ \alpha_1 $ under fixed $ \alpha_2 $, we run 1000 simulations and take their average values. The results are shown in Figure 10. It follows from Figure 10 that $ \alpha_1 $ can influence the rate of convergence to the equilibria of the poultry population, while it can not significantly influence the rate of convergence to the equilibria of the human population. In addition, we study the influence of $ \alpha_2 $ under fixed $ \alpha_1 $. The simulation results are presented in Figure 11, which implies that $ \alpha_2 $ can not change the rate of convergence to the equilibria of the poultry population. Figure 11 also means that $ \alpha_2 $ can not increase the rate of convergence to the equilibria of the human population, but it can evidently reduce the peak value of $ I_h(t) $. In summary, $ \alpha_1 $ and $ \alpha_2 $ have evidently influence to the spreading of avian influenza among both avian and human population.

    Figure 10.  The effect of $ \alpha_1 $ and $ \alpha_2 $ by $ 1000 $ simulation runs (for fixed $ \alpha_2 $).
    Figure 11.  The effect of $ \alpha_1 $ and $ \alpha_2 $ by $ 1000 $ simulation runs (for fixed $ \alpha_1 $).

    In this paper, we establish a stochastic delayed avian influenza model with saturated incidence rate. To begin with, we investigate the existence and uniqueness of the global positive solution to the system (2.3) with any positive initial value (2.4). Since there is no equilibrium point in the system (2.3) at this time, thus, the asymptotic behaviors of the disease-free equilibrium and the endemic equilibrium are given by constructing some suitable Lyapunov functions and applying the Young's inequality and Hölder's inequality. Theorem 2 shows that if $ \mathscr{R}_0 < 1 $, then the solution of system (2.3) is going around $ E^0 $ while from Theorem 3, we obtain that if $ \mathscr{R}_0 > 1 $, then the solution of system (2.3) is going around $ E^* $. Finally, some numerical examples are given to illustrate the accuracy of the theoretical results.

    There are some interesting issues deserve further investigations. On the one hand, we can formulate some more realistic but complex avian influenza models, such as considering the effects of Lévy jumps or impulsive perturbations on system (2.3). On the other hand, the coefficients in our model studied in this paper are all constants. If the coefficients are with Markov switching, how will the properties change? We leave these investigations as our future work.

    The research was supported by the National Natural Science Foundation of China (11661064), Ningxia Natural Science Foundation Project (2019AAC03069) and the Funds for Improving the International Education Capacity of Ningxia University (030900001921).

    The authors declare that they have no conflict of interest.

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