Dynamic Behavior of a Stochastic Avian Influenza Model with Two Strains of Zoonotic Virus

Abstract

In this paper, a stochastic avian influenza model with two different pathogenic human–avian viruses is studied. The model analyzes the spread of the avian influenza virus from poultry populations to human populations in a random environment. The dynamic behavior of the stochastic avian influenza model is analyzed. Firstly, the existence and uniqueness of a global positive solution are obtained. Secondly, under the condition of high pathogenic virus extinction, the persistence in the mean and extinction of the infected avian population with a low pathogenic virus is analyzed. Thirdly, the sufficient conditions for the existence and uniqueness of the ergodic stationary distribution in the stochastic avian influenza model are derived. We find the threshold of the stochastic model to determine whether the disease spreads when the white noise is small. The analysis results show that random white noise is effective for disease control. Finally, the theoretical results are verified by numerical simulation, and the numerical simulation analysis is carried out for the cases that cannot be theoretically deduced.

1. Introduction

Avian influenza is an infectious disease caused by a subtype of the influenza A virus (also known as avian influenza virus). It is classified as a Class A infectious disease by the International Veterinary Epidemic Bureau, and is also known as true chicken fever or European chicken fever. Avian influenza can infect a variety of birds, including chickens, ducks, turkeys, and geese. The virus can be transmitted among commercially farmed, wild, and pet birds, but in rare cases, it can cross species barriers to infect people. Infected birds do not necessarily get sick, but those that appear ’healthy’ still pose a threat to the humans they come into contact with. Since the discovery of human infection with avian influenza in Hong Kong in 1997, the disease has attracted the attention of the World Health Organization. Since then, the disease has broken out sporadically in Asia, but since December 2003, avian influenza in East Asian countries, mainly in Vietnam, South Korea, and Thailand, has seen serious outbreaks and caused the death of many patients in Vietnam. Now, many countries in Eastern Europe also have cases [1,2,3]. Avian influenza is divided into three categories: highly pathogenic, low pathogenic, and non-pathogenic avian influenza, according to the pathogenicity of its pathogens. Non-pathogenic avian influenza does not cause obvious symptoms and only causes infected birds to produce virus antibodies. Low pathogenic avian influenza can cause mild respiratory symptoms in poultry, reduced food intake, decreased egg production, and sporadic deaths. Highly pathogenic avian influenza is the most serious, with high morbidity and mortality. Among them, the common highly pathogenic avian influenza viruses, including H5N1 and H7N9, have caused a wide range of epidemics worldwide, leading to the deaths of many people [4,5].

In March 2013, human infection with the H7N9 avian influenza virus was first reported in China [6]. As of 30 June 2017, the H7N9 avian influenza virus has caused a total of five waves of epidemics, including 1523 laboratory-confirmed cases and 600 deaths. The number of cases infected with the H7N9 avian influenza virus has exceeded the number of infections caused by any other type of avian influenza, including H5N1. Although the H7N9 virus prevalent in the first four waves of the epidemic has a high mortality rate for humans, it is less pathogenic or non-pathogenic to poultry. A total of 730 laboratory-confirmed cases were reported in mainland China during the fifth wave of the epidemic, which began on 1 October 2016 [7]. It is by far the most widely distributed and most infected wave of human infection with H7N9 avian influenza. In the fifth wave of the epidemic, polybasic amino acid insertions appeared in the hemagglutinin protein hydrolysis site of some H7N9 viruses. This molecular marker suggested that these H7N9 viruses were mutated from low pathogenic or non-pathogenic avian influenza viruses to highly pathogenic avian influenza viruses. The spread of multiple strains in poultry has led to the risk of humans being infected with different strains at the same time. Moreover, the increase in the number and geographical distribution of the fifth wave of avian influenza A(H7N9) infections emphasizes that the transmission and control of human infections of both zoonotic influences are well worth studying. Because of its great threat to public health, the corresponding genetic analysis, epidemiology, and disease control have been studied extensively. The infectious disease model of multi-virus infection is also common. For example, in [8], based on the COVID-19 Big Data Hub, the authors established a COVID-19 dynamic model with multiple virus strains and distributed delays, used the COBYLA algorithm to identify the parameters, and simulated with the help of Julia high-performance computing to predict the studied model. This paper introduces a good practice in Big Data related to epidemiological analysis based on time-delay dynamic system modeling. Guo et al. [9] investigated the global dynamics of a human–poultry H7N9 avian epidemic model. Tuncer et al. [10] introduced the transmission of two different pathogenic avian influenza viruses in wild and domestic bird populations, and their results emphasized that these two viruses can coexist in two populations. Kuddus et al. [11] studied a two-strain disease model with amplification and analyzed the epidemic patterns of drug-sensitive and drug-resistant strains.

Since 2013, the H7N9 avian influenza virus, which had previously only infected poultry, has begun to infect humans. Since then, new avian influenza virus transmission has occurred every winter, which has not only hit the development of the poultry industry but also seriously endangered public health and safety. In view of the spread of the epidemic range and the surge in the number of infected people between 2016 and 2017, taking into account the fact that humans are infected with the original low pathogenic and newly emerging highly pathogenic A(H7N9) virus, Chen et al. [12] established an avian–human epidemic model with two strain viruses, as follows:

$$\left\{\begin{array}{c}{\dot{S}}_a\left(t\right)=(1-a)A-{\displaystyle \frac{{\beta }_1{S}_a\left(t\right)E_a\left(t\right)}{1+{\alpha }_1{E}_a\left(t\right)}}-{\displaystyle \frac{{\beta }_2{S}_a\left(t\right)I_a\left(t\right)}{1+{\alpha }_2{I}_a\left(t\right)}}-d{S}_{{}_a}\left(t\right),\hfill \\ {\dot{E}}_a\left(t\right)=aA+{\displaystyle \frac{{\beta }_1{S}_a\left(t\right)E_a\left(t\right)}{1+{\alpha }_1{E}_a\left(t\right)}}-d{E}_a\left(t\right),\hfill \\ {\dot{I}}_a\left(t\right)={\displaystyle \frac{{\beta }_2{S}_a\left(t\right)I_a\left(t\right)}{1+{\alpha }_2{I}_a\left(t\right)}}-d{I}_a\left(t\right)-\xi I_a\left(t\right),\hfill \\ {\dot{S}}_h\left(t\right)=B-{\displaystyle \frac{{\eta }_1{E}_a\left(t\right)S_h\left(t\right)}{1+{\upsilon }_1{E}_a^2\left(t\right)}}-{\displaystyle \frac{{\eta }_2{I}_a\left(t\right)S_h\left(t\right)}{1+{\upsilon }_2{I}_a^2\left(t\right)}}-\rho S_h\left(t\right),\hfill \\ {\dot{I}}_h\left(t\right)={\displaystyle \frac{{\eta }_1{E}_a\left(t\right)S_h\left(t\right)}{1+{\upsilon }_1{E}_a^2\left(t\right)}}+{\displaystyle \frac{{\eta }_2{I}_a\left(t\right)S_h\left(t\right)}{1+{\upsilon }_2{I}_a^2\left(t\right)}}-\delta I_h\left(t\right)-\gamma I_h\left(t\right)-\rho I_h\left(t\right),\hfill \\ {\dot{R}}_h\left(t\right)=\gamma I_h\left(t\right)-\rho R_h\left(t\right),\hfill \end{array}\right.$$

(1)

here, the number of susceptible poultry populations is denoted by $S_a\left(t\right)$, the number of infected poultry with low pathogenic A(H7N9) virus is denoted by $E_a\left(t\right)$ because of very mild or no disease symptoms, and the number of infected poultry with high pathogenic A(H7N9) virus is denoted by $I_a\left(t\right)$ at time t. Similarly, $S_h\left(t\right)$, $I_h\left(t\right)$, and $R_h\left(t\right)$, respectively, represent the numbers of susceptible, infected, and recovered human populations at time t. All of the parameters here are positive. A and B represent the constant input rates of poultry and human populations, respectively. Poultry infected with highly pathogenic A(H7N9) virus are easy to distinguish, while poultry infected with low pathogenic A(H7N9) virus show disease-free characteristics. Therefore, we believe that susceptible poultry and poultry infected with low pathogenic A(H7N9) virus are included in the recruitment of poultry. $0\le a\le 1$ represents the proportion of poultry infected with low pathogenic A(H7N9) virus in recruitment. d is the output rate in the poultry industry, including natural deaths and sales. $\rho$ is the natural mortality of human beings. $\xi$ is the mortality of $I_a\left(t\right)$ caused by highly pathogenic A(H7N9) virus infection. $\delta$ is the mortality rate caused by human infection with avian influenza virus. $\gamma$ represents the recovery rate of human cases. $\frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}$ and $\frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}$ are the saturated incidence functions in poultry $({\beta }_1<{\beta }_2)$. The corresponding infection rate functions from poultry to humans are $\frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}$ and $\frac{{\eta }_2{I}_a{S}_h}{1+{\upsilon }_2{I}_a^2}.$ In [12], the local asymptotic stability of the equilibrium point of system (1) is analyzed by using the Routh–Hurwitz criterion. Furthermore, the global stability analysis is performed by constructing a suitable Lyapunov function.

In fact, infectious diseases are bound to be affected by various environmental noises in the process of their transmission. Therefore, the important factor of environmental noise must be considered when establishing the corresponding mathematical model of infectious diseases. Compared with the deterministic infectious disease model, the discussion of the stochastic model is more in line with the law of disease transmission [13,14,15,16,17,18]. Inspired by [19,20,21,22,23], we assume that the random perturbation under the influence of white noise is proportional to the variables $S_a,E_a,I_a,S_h,I_h$, and $R_h$ in the model (1) to obtain a stochastic model

$$\left\{\begin{array}{c}\mathrm{d}{S}_a=\left((1-a)A-{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-{\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{S}_{{}_a}\right)\mathrm{d}t+{\sigma }_1{S}_a\mathrm{d}{B}_1\left(t\right),\hfill \\ \mathrm{d}{E}_a=\left(aA+{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-d{E}_a\right)\mathrm{d}t+{\sigma }_2{E}_a\mathrm{d}{B}_2\left(t\right),\hfill \\ \mathrm{d}{I}_a=\left({\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{I}_a-\xi I_a\right)\mathrm{d}t+{\sigma }_3{I}_a\mathrm{d}{B}_3\left(t\right),\hfill \\ \mathrm{d}{S}_h=\left(B-{\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}-{\displaystyle \frac{{\eta }_2{I}_a{S}_h}{1+{\upsilon }_2{I}_a^2}}-\rho S_h\right)\mathrm{d}t+{\sigma }_4{S}_h\mathrm{d}{B}_4\left(t\right),\hfill \\ \mathrm{d}{I}_h=\left({\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}+{\displaystyle \frac{{\eta }_2{I}_a{S}_h}{1+{\upsilon }_2{I}_a^2}}-\delta I_h-\gamma I_h-\rho I_h\right)\mathrm{d}t+{\sigma }_5{I}_h\mathrm{d}{B}_5\left(t\right),\hfill \\ \mathrm{d}{R}_h=(\gamma I_h-\rho R_h)\mathrm{d}t+{\sigma }_6{R}_h\mathrm{d}{B}_6\left(t\right).\hfill \end{array}\right.$$

(2)

where ${\sigma }_i^2$ is the intensity of the white noise, $B_i\left(t\right)$ is an independent standard Brownian motion defined on a complete probability space $(\Omega ,\mathcal{F},{\left\{{\mathcal{F}}_t\right\}}_{t\ge 0},P)$ with the filtration ${\left\{{\mathcal{F}}_t\right\}}_{t\ge 0}$ satisfying the usual conditions, and here $i=1,2,3,4,5,6$. When the noise intensity ${\sigma }_i=0$, model (2) becomes the deterministic model (1). Moreover, we assign $R_{+}^6=\{(x_1,x_2,x_3,x_4,x_5,x_6)\in R^6|x_i>0,i=1,2,3,4,5,6\}$.

The content of this article is arranged as follows. In Section 2, the existence and uniqueness of the global positive solution for model (2) are proved. In Section 3, by analyzing the random bird subsystem, we obtain the thresholds for extinction and persistence in the mean of poultry infected with low pathogenic A(H7N9) virus. In Section 4, we analyze the dynamic behavior of the complete stochastic avian influenza model (2). In Section 5, we prove that there is an ergodic stationary distribution in the stochastic avian–human system (2), which means that two avian influenza viruses will coexist. In Section 6, the theoretical results are verified by numerical simulation and the influence of white noise in random environments on disease transmission is analyzed. Some conclusions we have summarized are put in Section 7. The situation that cannot be theoretically deduced in the conclusion is analyzed by numerical simulation in Section 8.

2. Existence and Uniqueness of the Positive Solution

In this section, the existence and uniqueness of the global positive solution of model (2) will be proven.

Theorem 1.

For any initial value $\left(S_a\left(0\right),E_a\left(0\right),I_a\left(0\right),S_h\left(0\right),I_h\left(0\right),R_h\left(0\right)\right)\in R_{+}^6$, there exists a unique positive solution $\left(S_a\left(t\right),E_a\left(t\right),I_a\left(t\right),S_h\left(t\right),I_h\left(t\right),R_h\left(t\right)\right)$ to model (2) for $t\ge 0$, and the solution will remain in $R_{+}^6$ with a probability of one.

Proof. 

Define a $C^2$-function V: $R_{+}^6\to R_{+}$ by

$$\begin{array}{cc}\hfill V(S_a,E_a,I_a,S_h,I_h,R_h)=& \left(S_a-b-bln\frac{S_a}{b}\right)+(E_a-1-ln{E}_a)+(I_a-1-ln{I}_a)\hfill \\ \hfill & +\left(S_h-1-ln{S}_h\right)+(I_h-1-ln{I}_h)+(R_h-1-ln{R}_h)\hfill \end{array}$$

By using Itô formula [24], we have

$$\begin{array}{cc}\hfill LV=& \left(1-{\displaystyle \frac{b}{S_a}}\right)\left((1-a)A-{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-{\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{S}_{{}_a}\right)+\left(1-{\displaystyle \frac{1}{E_a}}\right)\left(aA+{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-d{E}_a\right)\hfill \\ \hfill & +\left(1-{\displaystyle \frac{1}{I_a}}\right)\left({\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{I}_a-\xi I_a\right)+\left(1-{\displaystyle \frac{1}{S_h}}\right)\left(B-{\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}-{\displaystyle \frac{{\eta }_2{I}_a{S}_h}{1+{\upsilon }_2{I}_a^2}}-\rho S_h\right)\hfill \\ \hfill & +\left(1-{\displaystyle \frac{1}{I_h}}\right)\left({\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}+{\displaystyle \frac{{\eta }_2{I}_a{S}_h}{1+{\upsilon }_2{I}_a^2}}-\delta I_h-\gamma I_h-\rho I_h\right)+\left(1-{\displaystyle \frac{1}{R_h}}\right)(\gamma I_h-\rho R_h)+{\displaystyle \frac{b}{2}}{\sigma }_1^2\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}{\sigma }_2^2+{\displaystyle \frac{1}{2}}{\sigma }_3^2+{\displaystyle \frac{1}{2}}{\sigma }_4^2+{\displaystyle \frac{1}{2}}{\sigma }_5^2+{\displaystyle \frac{1}{2}}{\sigma }_6^2\hfill \\ \hfill \le & A-d{E}_a-(d+\xi )I_a+B+b{\beta }_1{E}_a+b{\beta }_2{I}_a+bd+2d+\xi +\rho +\frac{{\eta }_1}{2\sqrt{v_1}}+\frac{{\eta }_2}{2\sqrt{v_2}}\hfill \\ \hfill & +\delta +\gamma +2\rho +{\displaystyle \frac{b}{2}}{\sigma }_1^2+{\displaystyle \frac{1}{2}}{\sigma }_2^2+{\displaystyle \frac{1}{2}}{\sigma }_3^2+{\displaystyle \frac{1}{2}}{\sigma }_4^2+{\displaystyle \frac{1}{2}}{\sigma }_5^2+{\displaystyle \frac{1}{2}}{\sigma }_6^2\hfill \\ \hfill =& A+B+(b+2)d+\xi +3\rho +\delta +\gamma +\frac{{\eta }_1}{2\sqrt{v_1}}+\frac{{\eta }_2}{2\sqrt{v_2}}+{\displaystyle \frac{b}{2}}{\sigma }_1^2+{\displaystyle \frac{1}{2}}{\sigma }_2^2+{\displaystyle \frac{1}{2}}{\sigma }_3^2+{\displaystyle \frac{1}{2}}{\sigma }_4^2+{\displaystyle \frac{1}{2}}{\sigma }_5^2\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}{\sigma }_6^2+\left(b{\beta }_1-d\right)E_a+\left[b{\beta }_2-(d+\xi )\right]I_a\hfill \end{array}$$

taking $b={\displaystyle \frac{d}{{\beta }_2}}>0$, by ${\beta }_1<{\beta }_2$, we have

$$\begin{array}{cc}\hfill LV\le & A+B+(b+2)d+\xi +3\rho +\delta +\gamma +\frac{{\eta }_1}{2\sqrt{v_1}}+\frac{{\eta }_2}{2\sqrt{v_2}}+{\displaystyle \frac{b}{2}}{\sigma }_1^2+{\displaystyle \frac{1}{2}}{\sigma }_2^2+{\displaystyle \frac{1}{2}}{\sigma }_3^2\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}{\sigma }_4^2+{\displaystyle \frac{1}{2}}{\sigma }_5^2+{\displaystyle \frac{1}{2}}{\sigma }_6^2:=K.\hfill \end{array}$$

the rest of the proof is similar to Theorem 1 in [20] and we omit it here. □

3. Analysis of the Stochastic Avian-Only Subsystem

In this section, we will discuss the dynamics of stochastic poultry subsystems independent of human systems:

$$\left\{\begin{array}{c}\mathrm{d}{S}_a=\left((1-a)A-{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-{\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{S}_{{}_a}\right)\mathrm{d}t+{\sigma }_1{S}_a\mathrm{d}{B}_1\left(t\right),\hfill \\ \mathrm{d}{E}_a=\left(aA+{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-d{E}_a\right)\mathrm{d}t+{\sigma }_2{E}_a\mathrm{d}{B}_2\left(t\right),\hfill \\ \mathrm{d}{I}_a=\left({\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{I}_a-\xi I_a\right)\mathrm{d}t+{\sigma }_3{I}_a\mathrm{d}{B}_3\left(t\right).\hfill \end{array}\right.$$

(3)

The focus of our study is to analyze the extinction and persistence of $E_a\left(t\right)$. First, we introduce some notations, as follows: if $h\left(t\right)$ is an integral function on $[0,\infty )$, define ${\langle h\rangle }_t=\frac{1}{t}{\int }_0^{t}h\left(s\right)ds$. If $h\left(t\right)$ is a bounded function on $[0,\infty )$, define $h^l={inf}_{t\in [0,\infty )}h\left(t\right)$, $h^u={sup}_{t\in [0,\infty )}h\left(t\right)$. Now, we introduce the following lemma, which will be used later.

Lemma 1

([17]). Let $\left(S_a\left(t\right),E_a\left(t\right),I_a\left(t\right),S_h\left(t\right),I_h\left(t\right),R_h\left(t\right)\right)$ be the solution of system (2) with any positive initial value $\left(S_a\left(0\right),E_a\left(0\right),I_a\left(0\right),S_h\left(0\right),I_h\left(0\right),R_h\left(0\right)\right)\in R_{+}^6$. Then

$$\underset{t\to \infty }{lim}{\displaystyle \frac{S_a\left(t\right)}{t}}=0,\underset{t\to \infty }{lim}{\displaystyle \frac{E_a\left(t\right)}{t}}=0,\underset{t\to \infty }{lim}{\displaystyle \frac{I_a\left(t\right)}{t}}=0,\underset{t\to \infty }{lim}{\displaystyle \frac{S_h\left(t\right)}{t}}=0,\underset{t\to \infty }{lim}{\displaystyle \frac{I_h\left(t\right)}{t}}=0,\underset{t\to \infty }{lim}{\displaystyle \frac{R_h\left(t\right)}{t}}=0,a.s.$$

$$\underset{t\to \infty }{lim}{\displaystyle \frac{ln{S}_a\left(t\right)}{t}}=0,\underset{t\to \infty }{lim}{\displaystyle \frac{ln{E}_a\left(t\right)}{t}}=0,\underset{t\to \infty }{lim}{\displaystyle \frac{ln{I}_a\left(t\right)}{t}}=0,\underset{t\to \infty }{lim}{\displaystyle \frac{ln{S}_h\left(t\right)}{t}}=0,$$

$$\underset{t\to \infty }{lim}{\displaystyle \frac{ln{I}_h\left(t\right)}{t}}=0,\underset{t\to \infty }{lim}{\displaystyle \frac{ln{R}_h\left(t\right)}{t}}=0a.s.$$

And,

$${\displaystyle \underset{t\to \infty }{lim}{\displaystyle \frac{1}{t}}{\int }_0^t{S}_a\left(r\right)\mathrm{d}{B}_1\left(r\right)=0,\underset{t\to \infty }{lim}{\displaystyle \frac{1}{t}}{\int }_0^t{E}_a\left(r\right)\mathrm{d}{B}_2\left(r\right)=0,\underset{t\to \infty }{lim}{\displaystyle \frac{1}{t}}{\int }_0^t{I}_a\left(r\right)\mathrm{d}{B}_3\left(r\right)=0,}$$

$${\displaystyle \underset{t\to \infty }{lim}{\displaystyle \frac{1}{t}}{\int }_0^t{S}_h\left(r\right)\mathrm{d}{B}_4\left(r\right)=0,\underset{t\to \infty }{lim}{\displaystyle \frac{1}{t}}{\int }_0^t{I}_h\left(r\right)\mathrm{d}{B}_5\left(r\right)=0,\underset{t\to \infty }{lim}{\displaystyle \frac{1}{t}}{\int }_0^t{R}_h\left(r\right)\mathrm{d}{B}_6\left(r\right)=0.a.s.}$$

Lemma 2

([20]). Suppose that $Z\left(t\right)\in C(\Omega \times [0,\infty ),R_{+})$.

(I) If there are two positive constants T and ${\delta }_0$ such that

$$lnZ\left(t\right)\le \delta t-{\delta }_0{\int }_0^{t}Z\left(s\right)\mathrm{d}s+\sum _{i=1}^n{\alpha }_{{}_i}B\left(t\right)a.s.$$

for all $t>T$, where ${\alpha }_{{}_i},\delta$ are constants, then

$$\left\{\begin{array}{c}\underset{t\to \infty }{lim\; sup}{\langle Z\rangle }_t\le \frac{\delta }{{\delta }_0}a.s.,if\delta \ge 0;\hfill \\ \underset{t\to \infty }{lim}Z\left(t\right)=0a.s.,if\delta <0.\hfill \end{array}\right.$$

(II) If there exist three positive constants $T,\delta ,{\delta }_0$ such that

$$lnZ\left(t\right)\ge \delta t-{\delta }_0{\int }_0^{t}Z\left(s\right)\mathrm{d}s+\sum _{i=1}^n{\alpha }_{{}_i}B\left(t\right)a.s.$$

for all $t>T$, then $\underset{t\to \infty }{lim\; inf}{\langle Z\rangle }_t\ge \frac{\delta }{{\delta }_0}a.s.$

Next, we analyze the persistence and extinction of the disease in model (3). First, we define the random basic reproduction numbers as follows:

$$R_0={\displaystyle \frac{{\beta }_{2}A}{d(d+\xi +\frac{1}{2}{\sigma }_3^2)}},R_1={\displaystyle \frac{{\beta }_{1}A}{d(d+\frac{1}{2}{\sigma }_2^2)}},R_2={\displaystyle \frac{{\beta }_{1}A+{\alpha }_{1}adA}{d(d+\frac{1}{2}{\sigma }_2^2)}}.$$

Here, $R_0$ represents the threshold of whether the highly pathogenic virus is transmitted in the poultry system; $R_1$ represents the threshold value of whether the low pathogenic virus is transmitted in the poultry system under condition $a=0$; and $R_2$ represents the threshold value of whether the low pathogenic virus is transmitted in poultry under condition $a\ne 0$.

Theorem 2.

Let $\left(S_a\left(t\right),E_a\left(t\right),I_a\left(t\right)\right)$ be the solution of system (3) with any positive initial value $\left(S_a\left(0\right),E_a\left(0\right),I_a\left(0\right)\right)\in R_{+}^3$.

(i) When $a=0$, if $R_0<1$ and $R_1<1$, then

$$\underset{t\to \infty }{lim}{E}_a\left(t\right)=0,\underset{t\to \infty }{lim}{I}_a\left(t\right)=0,\underset{t\to \infty }{lim}{\langle S_a\rangle }_t=\frac{A}{d}.a.s.$$

the disease of system (3) will be extinct at an exponential rate with probability one. In other words, the bird flu virus will not spread in the poultry world.

(ii) When $a\ne 0$, if $R_0<1$ and $R_2>1$, then

$$\underset{t\to \infty }{lim}{I}_a\left(t\right)=0,\underset{t\to \infty }{lim\; inf}{\langle E_a\rangle }_t\ge \frac{(d+\frac{1}{2}{\sigma }_2^2)(R_2-1)}{{\beta }_1+d{\alpha }_1}>0.a.s.$$

this means that poultry infected with the highly pathogenic A(H7N9) virus will go extinct, while poultry infected with low pathogenic A(H7N9) virus will persist in the mean. That is to say, avian influenza will spread in the poultry world.

Proof. 

From stochastic model (3),

$$\mathrm{d}(S_a+E_a+I_a)=[A-d(S_a+E_a+I_a)-\xi I_a]\mathrm{d}t+{\sigma }_1{S}_a\mathrm{d}{B}_1\left(t\right)+{\sigma }_2{E}_a\mathrm{d}{B}_2\left(t\right)+{\sigma }_3{I}_a\mathrm{d}{B}_3\left(t\right).$$

(4)

Calculating the integral from 0 to t on both sides of Equation (4) and dividing by t,

$$\begin{array}{cc}\hfill {\displaystyle {\displaystyle \frac{1}{t}}{\int }_0^t\mathrm{d}\left(S_a\left(r\right)+E_a\left(r\right)+I_a\left(r\right)\right)\le }& A-{\displaystyle \frac{d}{t}}{\int }_0^t\left(S_a\left(r\right)+E_a\left(r\right)+I_a\left(r\right)\right)\mathrm{d}r+{\displaystyle \frac{{\sigma }_1}{t}}{\int }_0^t{S}_a\left(r\right)\mathrm{d}{B}_1\left(r\right)\hfill \\ \hfill & {\displaystyle +{\displaystyle \frac{{\sigma }_2}{t}}{\int }_0^t{E}_a\left(r\right)\mathrm{d}{B}_2\left(r\right)+{\displaystyle \frac{{\sigma }_3}{t}}{\int }_0^t{I}_a\left(r\right)\mathrm{d}{B}_3\left(r\right).}\hfill \end{array}$$

So,

$$\begin{array}{cc}\hfill {\displaystyle \frac{S_a\left(t\right)+E_a\left(t\right)+I_a\left(t\right)}{t}}\le & {\displaystyle A+{\displaystyle \frac{S_a\left(0\right)+E_a\left(0\right)+I_a\left(0\right)}{t}}-{\displaystyle \frac{d}{t}}{\int }_0^t\left(S_a\left(r\right)+E_a\left(r\right)+I_a\left(r\right)\right)\mathrm{d}r}\hfill \\ \hfill & {\displaystyle +{\displaystyle \frac{{\sigma }_1}{t}}{\int }_0^t{S}_a\left(r\right)\mathrm{d}{B}_1\left(r\right)+{\displaystyle \frac{{\sigma }_2}{t}}{\int }_0^t{E}_a\left(r\right)\mathrm{d}{B}_2\left(r\right)+{\displaystyle \frac{{\sigma }_3}{t}}{\int }_0^t{I}_a\left(r\right)\mathrm{d}{B}_3\left(r\right).}\hfill \end{array}$$

From Lemma 1,

$$\begin{array}{c}{\displaystyle \underset{t\to \infty }{lim\; sup}{\displaystyle \frac{1}{t}}{\int }_0^t(S_a\left(r\right)+E_a\left(r\right)+I_a\left(r\right))\mathrm{d}r\le \frac{A}{d}}\hfill \end{array}.$$

(5)

Applying Itô formula to the third equation of (3), we obtain

$$\begin{array}{cc}\hfill \mathrm{d}ln{I}_a=& \left[{\displaystyle \frac{{\beta }_2{S}_a}{1+{\alpha }_2{I}_a}}-\left(d+\xi +{\displaystyle \frac{1}{2}}{\sigma }_3^2\right)\right]\mathrm{d}t+{\sigma }_3\mathrm{d}{B}_3\left(t\right)\hfill \\ \hfill \le & \left[{\beta }_2{S}_a-\left(d+\xi +{\displaystyle \frac{1}{2}}{\sigma }_3^2\right)\right]\mathrm{d}t+{\sigma }_3\mathrm{d}{B}_3\left(t\right).\hfill \end{array}$$

(6)

Integrating Equation (6) on both sides from 0 to t and dividing by t, we can obtain

$${\displaystyle \frac{ln{I}_a\left(t\right)}{t}}\le {\beta }_2{\langle S_a\rangle }_t-\left(d+\xi +{\displaystyle \frac{1}{2}}{\sigma }_3^2\right)+{\displaystyle \frac{ln{I}_a\left(0\right)}{t}}+\frac{{\sigma }_3{B}_3\left(t\right)}{t},$$

By the strong law of large numbers [25], combining with (5), we have

$$\begin{array}{c}\underset{t\to \infty }{lim\; sup}{\displaystyle \frac{ln{I}_a\left(t\right)}{t}}\le {\displaystyle \frac{{\beta }_{2}A}{d}}-\left(d+\xi +{\displaystyle \frac{1}{2}}{\sigma }_3^2\right)=\left(d+\xi +{\displaystyle \frac{1}{2}}{\sigma }_3^2\right)(R_0-1),a.s.\hfill \end{array}$$

Therefore, if $R_0<1$, then

$$\underset{t\to \infty }{lim}{I}_a\left(t\right)=0,a.s.$$

(7)

(i) When $a=0$, applying Itô formula to the second equation of (3), we obtain

$$\begin{array}{cc}\hfill \mathrm{d}ln{E}_a=& \left[{\displaystyle \frac{{\beta }_1{S}_a}{1+{\alpha }_1{E}_a}}-\left(d+{\displaystyle \frac{1}{2}}{\sigma }_2^2\right)\right]\mathrm{d}t+{\sigma }_2\mathrm{d}{B}_2\left(t\right)\hfill \\ \hfill \le & \left[{\beta }_1{S}_a-\left(d+{\displaystyle \frac{1}{2}}{\sigma }_2^2\right)\right]\mathrm{d}t+{\sigma }_2\mathrm{d}{B}_2\left(t\right).\hfill \end{array}$$

Similar to the discussion above, we obtain

$$\begin{array}{c}\underset{t\to \infty }{lim\; sup}{\displaystyle \frac{ln{E}_a\left(t\right)}{t}}\le {\displaystyle \frac{{\beta }_{1}A}{d}}-\left(d+{\displaystyle \frac{1}{2}}{\sigma }_2^2\right)=\left(d+{\displaystyle \frac{1}{2}}{\sigma }_2^2\right)(R_1-1),a.s.\hfill \end{array}$$

If $R_1<1$, then

$$\underset{t\to \infty }{lim}{E}_a\left(t\right)=0,a.s.$$

(8)

Hence, combining (7) and (8), as $t\to \infty$, then

$$\mathrm{d}{S}_a=(A-d{S}_{{}_a})\mathrm{d}t+{\sigma }_1{S}_a\mathrm{d}{B}_1\left(t\right),$$

(9)

Integrating Equation (9) on both sides from 0 to t and dividing by t, we can obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{S_a\left(t\right)-S_a\left(0\right)}{t}}=A-d{\langle S_a\rangle }_t+{\displaystyle \frac{{\sigma }_1}{t}}{\int }_0^t{S}_a\left(r\right)\mathrm{d}{B}_1\left(r\right),\end{array}$$

From Lemma 1,

$$\underset{t\to \infty }{lim}{\langle S_a\rangle }_t=\frac{A}{d}.a.s.$$

Next we consider (ii). Assume $a\ne 0$, when $R_0<1$, as $t\to \infty$,

$$\mathrm{d}(S_a+E_a)=[A-d(S_a+E_a)]\mathrm{d}t+{\sigma }_1{S}_a\mathrm{d}{B}_1\left(t\right)+{\sigma }_2{E}_a\mathrm{d}{B}_2\left(t\right).$$

(10)

Calculating the integral from 0 to t on both sides of Equation (10) and dividing by t,

$$\begin{array}{c}\hfill {\displaystyle \frac{S_a\left(t\right)-S_a\left(0\right)}{t}}+{\displaystyle \frac{E_a\left(t\right)-E_a\left(0\right)}{t}}=A-d{\langle S_a\rangle }_t-d{\langle E_a\rangle }_t+{\displaystyle \frac{{\sigma }_1}{t}}{\int }_0^t{S}_a\left(r\right)\mathrm{d}{B}_1\left(r\right)+{\displaystyle \frac{{\sigma }_2}{t}}{\int }_0^t{E}_a\left(r\right)\mathrm{d}{B}_2\left(r\right),\end{array}$$

This implies that

$${\langle S_a\rangle }_t={\displaystyle \frac{A}{d}}-{\langle E_a\rangle }_t+\phi \left(t\right),$$

(11)

where $\phi \left(t\right)=\frac{1}{d}\left[\frac{{\sigma }_1}{t}{\int }_0^t{S}_a\left(r\right)\mathrm{d}{B}_1\left(r\right)+\frac{{\sigma }_2}{t}{\int }_0^t{E}_a\left(r\right)\mathrm{d}{B}_2\left(r\right)-\frac{S_a\left(t\right)-S_a\left(0\right)}{t}-\frac{E_a\left(t\right)-E_a\left(0\right)}{t}\right].$ According to Lemma 1, we can get $\underset{t\to \infty }{lim}\phi \left(t\right)=0.$

Let $V=ln{E}_a+{\alpha }_1{E}_a$, applying Itô formula, we have

$$\begin{array}{cc}\hfill \mathrm{d}V=& \left({\displaystyle \frac{1}{E_a}}+{\alpha }_1\right)\mathrm{d}{E}_a-{\displaystyle \frac{1}{2E_a^2}}{\left(\mathrm{d}{E}_a\right)}^2\hfill \\ \hfill =& \left(\frac{1+{\alpha }_1{E}_a}{E_a}\right)\left[\left(aA+{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-d{E}_a\right)\mathrm{d}t+{\sigma }_2{E}_a\mathrm{d}{B}_2\left(t\right)\right]-\frac{1}{2}{\sigma }_2^2\mathrm{d}t\hfill \\ \hfill \ge & \left({\alpha }_{1}aA+{\beta }_1{S}_a-d-d{\alpha }_1{E}_a-\frac{1}{2}{\sigma }_2^2\right)\mathrm{d}t+{\alpha }_1{\sigma }_2{E}_a\mathrm{d}{B}_2\left(t\right)+{\sigma }_2\mathrm{d}{B}_2\left(t\right)\hfill \end{array}$$

(12)

Integrating Equation (12) on both sides from 0 to t and dividing by t, we can obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{ln{E}_a\left(t\right)-ln{E}_a\left(0\right)}{t}}+{\alpha }_1{\displaystyle \frac{E_a\left(t\right)-E_a\left(0\right)}{t}}\hfill \\ \hfill \ge & {\displaystyle \left({\alpha }_{1}aA-d-\frac{1}{2}{\sigma }_2^2\right)+{\beta }_1{\langle S_a\rangle }_t-d{\alpha }_1{\langle E_a\rangle }_t+{\displaystyle \frac{{\alpha }_1{\sigma }_2}{t}}{\int }_0^t{E}_a\left(r\right)\mathrm{d}{B}_2\left(r\right)+\frac{{\sigma }_2{B}_2\left(t\right)}{t},}\hfill \end{array}$$

(13)

Substituting (11) into (13), we can obtain

$$\begin{array}{c}{\displaystyle \frac{ln{E}_a\left(t\right)-ln{E}_a\left(0\right)}{t}}+{\alpha }_1{\displaystyle \frac{E_a\left(t\right)-E_a\left(0\right)}{t}}\hfill \\ \ge {\displaystyle \left({\alpha }_{1}aA+{\displaystyle \frac{{\beta }_{1}A}{d}}-d-\frac{1}{2}{\sigma }_2^2\right)-({\beta }_1+d{\alpha }_1){\langle E_a\rangle }_t+{\displaystyle \frac{{\alpha }_1{\sigma }_2}{t}}{\int }_0^t{E}_a\left(r\right)\mathrm{d}{B}_2\left(r\right)+\frac{{\sigma }_2{B}_2\left(t\right)}{t}+{\beta }_1\phi \left(t\right).}\hfill \end{array}$$

By Lemma 2,

$$\underset{t\to \infty }{lim\; inf}{\langle E_a\rangle }_t\ge {\displaystyle \frac{(d+\frac{1}{2}{\sigma }_2^2)(R_2-1)}{{\beta }_1+d{\alpha }_1}}.a.s.$$

That is to say, the infected poultry with low pathogenic A(H7N9) virus will be persistent in the mean when $R_2>1$. The proof is complete. □

4. Analysis of the Full Stochastic System

The dynamic behavior of stochastic system (2) will be analyzed in this part. Since the variable $R_h\left(t\right)$ is independent of the first five equations of system (2), in order to simplify the discussion, we omit the last equation. We only need to analyze the dynamical behavior of the following equivalent system:

$$\left\{\begin{array}{c}\mathrm{d}{S}_a=\left((1-a)A-{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-{\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{S}_{{}_a}\right)\mathrm{d}t+{\sigma }_1{S}_a\mathrm{d}{B}_1\left(t\right),\hfill \\ \mathrm{d}{E}_a=\left(aA+{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-d{E}_a\right)\mathrm{d}t+{\sigma }_2{E}_a\mathrm{d}{B}_2\left(t\right),\hfill \\ \mathrm{d}{I}_a=\left({\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{I}_a-\xi I_a\right)\mathrm{d}t+{\sigma }_3{I}_a\mathrm{d}{B}_3\left(t\right),\hfill \\ \mathrm{d}{S}_h=\left(B-{\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}-{\displaystyle \frac{{\eta }_2{I}_a{S}_h}{1+{\upsilon }_2{I}_a^2}}-\rho S_h\right)\mathrm{d}t+{\sigma }_4{S}_h\mathrm{d}{B}_4\left(t\right),\hfill \\ \mathrm{d}{I}_h=\left({\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}+{\displaystyle \frac{{\eta }_2{I}_a{S}_h}{1+{\upsilon }_2{I}_a^2}}-\delta I_h-\gamma I_h-\rho I_h\right)\mathrm{d}t+{\sigma }_5{I}_h\mathrm{d}{B}_5\left(t\right).\hfill \end{array}\right.$$

(14)

In this section, we define the random basic reproduction number

$$R_3={\displaystyle \frac{{\eta }_{1}B}{\rho (1+{\upsilon }_1{H}^2)(\delta +\gamma +\rho +\frac{1}{2}{\sigma }_5^2)}},$$

here, $R_3$ represents the threshold of whether the avian influenza virus is transmitted to the human system.

Theorem 3.

Let $\left(S_a\left(t\right),E_a\left(t\right),I_a\left(t\right),S_h\left(t\right),I_h\left(t\right)\right)$ be the solution of system (14) with any positive initial value $\left(S_a\left(0\right),E_a\left(0\right),I_a\left(0\right),S_h\left(0\right),I_h\left(0\right)\right)\in R_{+}^5$.

(i) When $a=0$, if $R_0<1$ and $R_1<1$, then

$$\begin{array}{c}\hfill \underset{t\to \infty }{lim}{E}_a\left(t\right)=0,\underset{t\to \infty }{lim}{I}_a\left(t\right)=0,\underset{t\to \infty }{lim}{I}_h\left(t\right)=0,\underset{t\to \infty }{lim}{\langle S_a\rangle }_t=\frac{A}{d},\underset{t\to \infty }{lim}{\langle S_h\rangle }_t=\frac{B}{\rho }.a.s.\end{array}$$

This is to tell us that the avian influenza virus will not spread in the human world.

(ii) When $a\ne 0$, if $R_0<1$, $R_2>1$, $R_3>1$, then

$$\underset{t\to \infty }{lim}{I}_a\left(t\right)=0,\underset{t\to \infty }{lim\; inf}{\langle E_a\rangle }_t\ge \frac{(d+\frac{1}{2}{\sigma }_2^2)(R_2-1)}{{\beta }_1+d{\alpha }_1}>0,a.s.$$

$$\underset{t\to \infty }{lim\; inf}{\langle I_h\rangle }_t\ge {\displaystyle \frac{\rho (1+{\upsilon }_1{H}^2)(\delta +\gamma +\rho +\frac{1}{2}{\sigma }_5^2)}{\left[{\eta }_1+\rho (1+{\upsilon }_1{H}^2)\right](\delta +\gamma +\rho )}}(R_3-1)>0.a.s.$$

This conclusion tells us that the infected poultry with high pathogenic A(H7N9) virus goes to extinction, but the infected poultry with low pathogenic A(H7N9) virus will be persistent in the mean. That is to say, low the pathogenic A(H7N9) virus will spread in the human world.

Proof. 

(i) According to Theorem 2, $\underset{t\to \infty }{lim}{E}_a\left(t\right)=0,\underset{t\to \infty }{lim}{I}_a\left(t\right)=0,\underset{t\to \infty }{lim}{\langle S_a\rangle }_t=\frac{A}{d}$ a.s. when $R_0<1,R_1<1$. Letting $t\to \infty$, the transmission dynamic of the avian influenza in human populations is given by the following system

$$\left\{\begin{array}{c}\mathrm{d}{S}_h=\left(B-\rho S_h\right)\mathrm{d}t+{\sigma }_4{S}_h\mathrm{d}{B}_4\left(t\right),\hfill \\ \mathrm{d}{I}_h=\left(-\delta I_h-\gamma I_h-\rho I_h\right)\mathrm{d}t+{\sigma }_5{I}_h\mathrm{d}{B}_5\left(t\right).\hfill \end{array}\right.$$

(15)

Integrating the second equation of system (15) from 0 to t and dividing by t, we have

$${\displaystyle {\displaystyle \frac{I_h\left(t\right)-I_h\left(0\right)}{t}}=-(\delta +\gamma +\rho ){\langle I_h\rangle }_t+{\displaystyle \frac{{\sigma }_5}{t}}{\int }_0^t{I}_h\left(r\right)\mathrm{d}{B}_5\left(r\right),}$$

Therefore,

$${\displaystyle {\langle I_h\rangle }_t={\displaystyle \frac{1}{\delta +\gamma +\rho }}\left[{\displaystyle \frac{{\sigma }_5}{t}}{\int }_0^t{I}_h\left(r\right)\mathrm{d}{B}_5\left(r\right)-{\displaystyle \frac{I_h\left(t\right)-I_h\left(0\right)}{t}}\right].}$$

By Lemma 1, $\underset{t\to \infty }{lim}{\langle I_h\rangle }_t=0.a.s.$ Thus, we conclude that $\underset{t\to \infty }{lim}{I}_h\left(t\right)=0.a.s.$ In fact, if $\underset{t\to \infty }{lim}{I}_h\left(t\right)=ca.s.(c>0)$, then there exists $T>0$ such that $I_h\left(t\right)\ge {\displaystyle \frac{c}{2}}$ for almost all $t>T$. Therefore, we can obtain

$$\begin{array}{cc}\hfill {\displaystyle {\displaystyle \frac{1}{t}}{\int }_0^t{I}_h\left(r\right)\mathrm{d}r=}& {\displaystyle {\displaystyle \frac{1}{t}}\left[{\int }_0^T{I}_h\left(r\right)\mathrm{d}r+{\int }_T^t{I}_h\left(r\right)\mathrm{d}r\right]}\hfill \\ \hfill \ge & {\displaystyle \frac{1}{t}}·{\displaystyle \frac{c}{2}}(t-T)\hfill \\ \hfill =& {\displaystyle \frac{c}{2}}-{\displaystyle \frac{cT}{2t}}.\hfill \end{array}$$

Let $t\to \infty$, then

$${\displaystyle \underset{t\to \infty }{lim}{\displaystyle \frac{1}{t}}{\int }_0^t{I}_h\left(r\right)\mathrm{d}r\ge \underset{t\to \infty }{lim}\left({\displaystyle \frac{c}{2}}-{\displaystyle \frac{cT}{2t}}\right)=\frac{c}{2}>0.}$$

which contradicts $\underset{t\to \infty }{lim}{\langle I_h\rangle }_t=0a.s.$ Therefore, by using the non-negativity of $I_h\left(t\right)$, we can prove

$$\underset{t\to \infty }{lim}{I}_h\left(t\right)=0.a.s.$$

Applying Itô formula to the first equation of (15),

$${\displaystyle {\displaystyle \frac{S_h\left(t\right)-S_h\left(0\right)}{t}}=B-\rho {\langle S_h\rangle }_t+{\displaystyle \frac{{\sigma }_4}{t}}{\int }_0^t{S}_h\left(r\right)\mathrm{d}{B}_4\left(r\right),}$$

Therefore,

$$\underset{t\to \infty }{lim}{\langle S_h\rangle }_t=\frac{B}{\rho }.a.s.$$

(ii) Form Theorem 2, when $a\ne 0$, $R_0<1$ and $R_2>1$ then $\underset{t\to \infty }{lim}{I}_a\left(t\right)=0,\underset{t\to \infty }{lim\; inf}{\langle E_a\rangle }_t\ge \frac{(d+\frac{1}{2}{\sigma }_2^2)(R_2-1)}{{\beta }_1+d{\alpha }_1}>0.a.s.$ By the Lemma 2 of [26], there exists a positive constant H to satisfy:

$$1\le E_a\left(t\right)\le H,\mathrm{for}\mathrm{all}\mathrm{t}\ge 0,\mathrm{a}.\mathrm{s}.$$

(16)

Letting $t\to \infty$, we have

$$\left\{\begin{array}{c}\mathrm{d}{S}_h=\left(B-{\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}-\rho S_h\right)\mathrm{d}t+{\sigma }_4{S}_h\mathrm{d}{B}_4\left(t\right),\hfill \\ \mathrm{d}{I}_h=\left({\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}-\delta I_h-\gamma I_h-\rho I_h\right)\mathrm{d}t+{\sigma }_5{I}_h\mathrm{d}{B}_5\left(t\right).\hfill \end{array}\right.$$

(17)

Form stochastic model (17)

$$\mathrm{d}(S_h+I_h)=[B-\rho S_h-(\delta +\gamma +\rho )I_h]\mathrm{d}t+{\sigma }_4{S}_h\mathrm{d}{B}_4\left(t\right)+{\sigma }_5{I}_h\mathrm{d}{B}_5\left(t\right).$$

(18)

Calculating the integral from 0 to t on both sides of Equation (18) and dividing by t,

$$\begin{array}{cc}\hfill {\displaystyle \frac{S_h\left(t\right)-S_h\left(0\right)}{t}}+{\displaystyle \frac{I_h\left(t\right)-I_h\left(0\right)}{t}}=& B-\rho {\langle S_h\rangle }_t-(\delta +\gamma +\rho ){\langle I_h\rangle }_t+{\displaystyle \frac{{\sigma }_4}{t}}{\int }_0^t{S}_h\left(r\right)\mathrm{d}{B}_4\left(r\right)\hfill \\ \hfill & +{\displaystyle \frac{{\sigma }_5}{t}}{\int }_0^t{I}_h\left(r\right)\mathrm{d}{B}_5\left(r\right),\hfill \end{array}$$

This implies that

$${\langle S_h\rangle }_t={\displaystyle \frac{B}{\rho }}-{\displaystyle \frac{\delta +\gamma +\rho }{\rho }}{\langle I_h\rangle }_t+\psi \left(t\right),$$

(19)

where $\psi \left(t\right)=\frac{1}{\rho }\left[\frac{{\sigma }_4}{t}{\int }_0^t{S}_h\left(r\right)\mathrm{d}{B}_4\left(r\right)+\frac{{\sigma }_5}{t}{\int }_0^t{I}_h\left(r\right)\mathrm{d}{B}_5\left(r\right)-\frac{S_h\left(t\right)-S_h\left(0\right)}{t}-\frac{I_h\left(t\right)-I_h\left(0\right)}{t}\right].$ According to Lemma 1, we can get $\underset{t\to \infty }{lim}\psi \left(t\right)=0.$

Let $V=ln{I}_h+I_h$, applying Itô formula, form (16) and we have

$$\begin{array}{cc}\hfill \mathrm{d}V=& {\displaystyle \frac{1+I_h}{I_h}}\mathrm{d}{I}_h-{\displaystyle \frac{1}{2I_h^2}}{\left(\mathrm{d}{I}_h\right)}^2\hfill \\ \hfill =& {\displaystyle \frac{1+I_h}{I_h}}\left[\left({\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}-(\delta +\gamma +\rho )I_h\right)\mathrm{d}t+{\sigma }_5{I}_h\mathrm{d}{B}_5\left(t\right)\right]-\frac{1}{2}{\sigma }_5^2\mathrm{d}t\hfill \\ \hfill \ge & \left[{\displaystyle \frac{{\eta }_1}{1+{\upsilon }_1{H}^2}}{S}_h-(\delta +\gamma +\rho )I_h-\left(\delta +\gamma +\rho +\frac{1}{2}{\sigma }_5^2\right)\right]\mathrm{d}t+{\sigma }_5{I}_h\mathrm{d}{B}_5\left(t\right)+{\sigma }_5\mathrm{d}{B}_5\left(t\right)\hfill \end{array}$$

(20)

Integrating Equation (20) on both sides from 0 to t and dividing by t, we can obtain

$$\begin{array}{cc}\hfill & {\displaystyle \frac{ln{I}_h\left(t\right)-ln{I}_h\left(0\right)}{t}}+{\displaystyle \frac{I_h\left(t\right)-I_h\left(0\right)}{t}}\hfill \\ \hfill \ge & {\displaystyle {\displaystyle \frac{{\eta }_1}{1+{\upsilon }_1{H}^2}}{\langle S_h\rangle }_t-(\delta +\gamma +\rho ){\langle I_h\rangle }_t-\left(\delta +\gamma +\rho +\frac{1}{2}{\sigma }_5^2\right)+\frac{{\sigma }_5}{t}{\int }_0^t{I}_h\left(r\right)\mathrm{d}{B}_5\left(r\right)+\frac{{\sigma }_5{B}_5\left(t\right)}{t},}\hfill \end{array}$$

(21)

Substituting (19) into (21), we can obtain

$$\begin{array}{c}{\displaystyle \frac{ln{I}_h\left(t\right)-ln{I}_h\left(0\right)}{t}}+{\displaystyle \frac{I_h\left(t\right)-I_h\left(0\right)}{t}}\hfill \\ \ge {\displaystyle \frac{{\eta }_{1}B}{\rho (1+{\upsilon }_1{H}^2)}}-\left[\frac{({\eta }_1+\rho (1+{\upsilon }_1{H}^2))(\delta +\gamma +\rho )}{\rho (1+{\upsilon }_1{H}^2)}\right]{\langle I_h\rangle }_t-\left(\delta +\gamma +\rho +\frac{1}{2}{\sigma }_5^2\right)\hfill \\ {\displaystyle +\frac{{\sigma }_5}{t}{\int }_0^t{I}_h\left(r\right)\mathrm{d}{B}_5\left(r\right)+\frac{{\sigma }_5{B}_5\left(t\right)}{t}+{\displaystyle \frac{{\eta }_1}{1+{\upsilon }_1{H}^2}}\psi \left(t\right).}\hfill \end{array}$$

By Lemma 2,

$$\underset{t\to \infty }{lim\; inf}{\langle I_h\rangle }_t\ge {\displaystyle \frac{\rho (1+{\upsilon }_1{H}^2)(\delta +\gamma +\rho +\frac{1}{2}{\sigma }_5^2)}{\left[{\eta }_1+\rho (1+{\upsilon }_1{H}^2)\right](\delta +\gamma +\rho )}}(R_3-1).a.s.$$

That is to say, the low pathogenic A(H7N9) virus will spread in the human world when $R_3>1$. The proof is complete. □

5. Existence of Ergodic Stationary Distribution

In this section, we investigate the conditions for the existence of a unique ergodic stationary distribution.

Lemma 3

([18]). Assume that there exists a bounded domain $U\subset R^d$ with a regular boundary Γ, and

(1) there is a positive number M satisfied that ${\Sigma }_{i,j=1}^d{a}_{ij}\left(x\right){\xi }_i{\xi }_j\ge M{\left|\xi \right|}^2,x\in \overline{U},\xi \in R^d.$

(2) there exists a non-negative $C^2$-function V such that $LV$ is negative for any $R^d\setminus U$.

Then, the Markov process $X\left(t\right)$ has an ergodic stationary distribution $\mu (·)$, and it is unique.

Theorem 4.

If $R_0>1$, for any initial value $\left(S_a\left(0\right),E_a\left(0\right),I_a\left(0\right),S_h\left(0\right),I_h\left(0\right)\right)\in R_{+}^5$, the solution of model (14) admits a unique stationary distribution, which is ergodic.

Proof. 

First, the diffusion matrix of (14) is as follows:

$$\begin{array}{c}A\left(X\right)=\left(\begin{array}{ccccc}{\sigma }_1^2{S}_a^2& 0& 0& 0& 0\\ 0& {\sigma }_2^2{E}_a^2& 0& 0& 0\\ 0& 0& {\sigma }_3^2{I}_a^2& 0& 0& \\ 0& 0& 0& {\sigma }_4^2{S}_h^2& 0\\ 0& 0& 0& 0& {\sigma }_5^2{I}_h^2\end{array}\right).\end{array}$$

Let U be any bounded open domain in $R_{+}^5$, $\exists M>0\left(M=min\{{\sigma }_1^2{S}_a^2,{\sigma }_2^2{E}_a^2,{\sigma }_3^2{I}_a^2,{\sigma }_4^2{S}_h^2,{\sigma }_5^2{I}_h^2,(S_a,E_a,I_a,S_h,I_h)\in \overline{U}\}\right)$ that satisfies

$$\sum _{i,j=1}^5{a}_{ij}\left(X\right){\xi }_i{\xi }_j={\sigma }_1^2{S}_a^2{\xi }_1^2+{\sigma }_2^2{E}_a^2{\xi }_2^2+{\sigma }_3^2{I}_a^2{\xi }_3^2+{\sigma }_4^2{S}_h^2{\xi }_4^2+{\sigma }_5^2{I}_h^2{\xi }_5^2\ge M{\left|\xi \right|}^2,$$

for all $(S_a,E_a,I_a,S_h,I_h)\in \overline{U},\xi =({\xi }_1,{\xi }_2,{\xi }_3,{\xi }_4,{\xi }_5)\in R_{+}^5$. So, Condition (1) in Lemma 3 is satisfied.

Then, we are going to verify Condition (2). Consider a non-negative $C^2$-function $V(S_a,E_a,I_a,S_h,I_h):R_{+}^5\to R_{+}$ with

$$V=H(S_a,E_a,I_a,S_h,I_h)-H(S_{a0},E_{a0},I_{a0},S_{h0},I_{h0}),$$

where $(S_a,E_a,I_a,S_h,I_h)\in U$, $H(S_{a0},E_{a0},I_{a0},S_{h0},I_{h0})$ is a unique minimum value point of the $C^2$-function $H(S_a,E_a,I_a,S_h,I_h)$.

Denote

$$H(S_a,E_a,I_a,S_h,I_h)=M{V}_1+V_2+V_3,$$

there, $V_1=-ln{I}_a-{\alpha }_2{I}_a-{\displaystyle \frac{{\beta }_2}{d}}(S_a+E_a),$$V_2=-ln{S}_a-ln{E}_a-ln{S}_h-ln{I}_h,V_3=(1/(\theta +1)){(S_a+E_a+I_a+I_h+R_h)}^{\theta +1},$ $0<\theta <1$ satisfy:

$$l:=(d\wedge \rho )-\frac{\theta }{2}({\sigma }_1^2\vee {\sigma }_2^2\vee {\sigma }_3^2\vee {\sigma }_4^2\vee {\sigma }_5^2)>0,$$

and a sufficiently large positive constant M satisfies:

$$M_0+f_1^u+f_2^u+K_0-M\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)\le -2,$$

where positive constants $M_0$, $K_0$ and functions $f_1^u,f_2^u$ will be determined later. Using Itô formula, one has

$$\begin{array}{cc}\hfill L{V}_1=& -{\displaystyle \frac{1+{\alpha }_2{I}_a}{I_a}}\left({\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{I}_a-\xi I_a\right)-{\displaystyle \frac{{\beta }_2}{d}}\left(A-{\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{S}_a-d{E}_a\right)+\frac{1}{2}{\sigma }_3^2\hfill \\ \hfill \le & d+\xi +\frac{1}{2}{\sigma }_3^2-{\displaystyle \frac{{\beta }_{2}A}{d}}+\frac{{\beta }_2^2}{d{\alpha }_2}{S}_a+{\beta }_2{E}_a+{\alpha }_2(d+\xi )I_a\hfill \\ \hfill =& -\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)+\frac{{\beta }_2^2}{d{\alpha }_2}{S}_a+{\beta }_2{E}_a+{\alpha }_2(d+\xi )I_a,\hfill \end{array}$$

Similarly, we can get

$$\begin{array}{cc}\hfill L{V}_2=& -\frac{1}{S_a}\left[(1-a)A-{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-{\displaystyle \frac{{\beta }_2{S}_a{I}_a}{1+{\alpha }_2{I}_a}}-d{S}_{{}_a}\right]-\frac{1}{E_a}\left(aA+{\displaystyle \frac{{\beta }_1{S}_a{E}_a}{1+{\alpha }_1{E}_a}}-d{E}_a\right)\hfill \\ \hfill & -\frac{1}{S_h}\left(B-{\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}-{\displaystyle \frac{{\eta }_2{I}_a{S}_h}{1+{\upsilon }_2{I}_a^2}}-\rho S_h\right)-\frac{1}{I_h}\left({\displaystyle \frac{{\eta }_1{E}_a{S}_h}{1+{\upsilon }_1{E}_a^2}}+{\displaystyle \frac{{\eta }_2{I}_a{S}_h}{1+{\upsilon }_2{I}_a^2}}-\delta I_h-\gamma I_h-\rho I_h\right)\hfill \\ \hfill & +\frac{1}{2}{\sigma }_1^2+\frac{1}{2}{\sigma }_2^2+\frac{1}{2}{\sigma }_4^2+\frac{1}{2}{\sigma }_5^2\hfill \\ \hfill \le & -\frac{(1-a)A}{S_a}-\frac{aA}{E_a}-\frac{{\beta }_1{S}_a}{1+{\alpha }_1{E}_a}-\frac{B}{S_h}-\frac{{\eta }_2{I}_a}{(1+{\upsilon }_2{I}_a^2)}\frac{S_h}{I_h}+{\eta }_1{E}_a+{\eta }_2{I}_a+2d+2\rho +\delta +\gamma \hfill \\ \hfill & +\frac{{\beta }_1}{{\alpha }_1}+\frac{{\beta }_2}{{\alpha }_2}+\frac{1}{2}{\sigma }_1^2+\frac{1}{2}{\sigma }_2^2+\frac{1}{2}{\sigma }_4^2+\frac{1}{2}{\sigma }_5^2,\hfill \end{array}$$

$$\begin{array}{cc}\hfill L{V}_3=& {(S_a+E_a+I_a+S_h+I_h)}^{\theta }\left[A+B-d(S_a+E_a+I_a)-\xi I_a-\rho (S_h+I_h)-(\delta +\gamma )I_h\right]\hfill \\ \hfill & +{\displaystyle \frac{\theta }{2}}{(S_a+E_a+I_a+S_h+I_h)}^{\theta -1}({\sigma }_1^2{S}_a^2+{\sigma }_2^2{E}_a^2+{\sigma }_3^2{I}_a^2+{\sigma }_4^2{S}_h^2+{\sigma }_5^2{I}_h^2)\hfill \\ \hfill \le & (A+B){(S_a+E_a+I_a+S_h+I_h)}^{\theta }-\left[(d\wedge \rho )-\frac{\theta }{2}({\sigma }_1^2\vee {\sigma }_2^2\vee {\sigma }_3^2\vee {\sigma }_4^2\vee {\sigma }_5^2)\right]\hfill \\ \hfill & ·{(S_a+E_a+I_a+S_h+I_h)}^{\theta +1}\hfill \\ \hfill \le & Q-\frac{l}{2}{(S_a+E_a+I_a+S_h+I_h)}^{\theta +1}\hfill \\ \hfill \le & Q-\frac{l}{2}(S_a^{\theta +1}+E_a^{\theta +1}+I_a^{\theta +1}+S_h^{\theta +1}+I_h^{\theta +1})\hfill \end{array}$$

there $Q={sup}_{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5}\left\{(A+B){(S_a+E_a+I_a+S_h+I_h)}^{\theta }-{\displaystyle \frac{l}{2}}{(S_a+E_a+I_a+S_h+I_h)}^{\theta +1}\right\}.$

Therefore,

$$\begin{array}{cc}\hfill LV=& ML{V}_1+L{V}_2+L{V}_3\hfill \\ \hfill \le & M\left[-\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)+\frac{{\beta }_2^2}{d{\alpha }_2}{S}_a+{\beta }_2{E}_a+{\alpha }_2(d+\xi )I_a\right]-\frac{(1-a)A}{S_a}-\frac{aA}{E_a}\hfill \\ \hfill & -\frac{{\beta }_1{S}_a}{1+{\alpha }_1{E}_a}-\frac{B}{S_h}-\frac{{\eta }_2{I}_a}{(1+{\upsilon }_2{I}_a^2)}\frac{S_h}{I_h}+{\eta }_1{E}_a+{\eta }_2{I}_a+2d+2\rho +\delta +\gamma +\frac{{\beta }_1}{{\alpha }_1}+\frac{{\beta }_2}{{\alpha }_2}\hfill \\ & +\frac{1}{2}{\sigma }_1^2+\frac{1}{2}{\sigma }_2^2+\frac{1}{2}{\sigma }_4^2+\frac{1}{2}{\sigma }_5^2+Q-\frac{l}{2}(S_a^{\theta +1}+E_a^{\theta +1}+I_a^{\theta +1}+S_h^{\theta +1}+I_h^{\theta +1})\hfill \\ \hfill =& M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)-\frac{(1-a)A}{S_a}-\frac{aA}{E_a}-\frac{{\beta }_1{S}_a}{1+{\alpha }_1{E}_a}-\frac{B}{S_h}-\frac{{\eta }_2{I}_a}{(1+{\upsilon }_2{I}_a^2)}\frac{S_h}{I_h}\hfill \\ \hfill & -\frac{l}{2}{S}_h^{\theta +1}-\frac{l}{2}{I}_h^{\theta +1}.\hfill \end{array}$$

where

$M_0=Q+2d+2\rho +\delta +\gamma +{\displaystyle \frac{{\beta }_1}{{\alpha }_1}}+{\displaystyle \frac{{\beta }_2}{{\alpha }_2}}+{\displaystyle \frac{1}{2}}{\sigma }_1^2+{\displaystyle \frac{1}{2}}{\sigma }_2^2+{\displaystyle \frac{1}{2}}{\sigma }_4^2+{\displaystyle \frac{1}{2}}{\sigma }_5^2,$

$f_1\left(S_a\right)={\displaystyle \frac{M{\beta }_2^2}{d{\alpha }_2}}{S}_a-{\displaystyle \frac{l}{2}}{S}_a^{\theta +1},$

$f_2\left(E_a\right)=M{\beta }_2{E}_a+{\eta }_1{E}_a-{\displaystyle \frac{l}{2}}{E}_a^{\theta +1},$

$f_3\left(I_a\right)=M\left[-\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)+{\alpha }_2(d+\xi )I_a\right]+{\eta }_2{I}_a-{\displaystyle \frac{l}{2}}{I}_a^{\theta +1}$.

Next, we construct a bounded closed set as follows:

$$U=\left\{ϵ\le S_a\le 1/ϵ,{ϵ}^2\le E_a\le 1/{ϵ}^2,ϵ\le I_a\le 1/ϵ,ϵ\le S_h\le 1/ϵ,{ϵ}^3\le I_h\le 1/{ϵ}^3\right\}$$

here, $ϵ$ is a sufficiently small positive constant and satisfies the following inequalities:

$$M_0+f_1^u+f_2^u+f_3^u-\frac{(1-a)A}{ϵ}\le -1,$$

(22)

$$M_0+f_1^u+f_2^u+f_3^u-\frac{aA}{{ϵ}^2}\le -1,$$

(23)

$$M_0+f_1^u+f_2^u+K_0-M\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)-\frac{{\beta }_1ϵ}{(1+{\alpha }_1{ϵ}^2)}\le -1,$$

(24)

$$M_0+f_1^u+f_2^u-M\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)+\left[M{\alpha }_2(d+\xi )+{\eta }_2\right]ϵ\le -1,$$

(25)

$$M_0+f_1^u+f_2^u+f_3^u-\frac{B}{ϵ}\le -1,$$

(26)

$$M_0+f_1^u+f_2^u+f_3^u-{\displaystyle \frac{{\eta }_2}{ϵ(1+{\upsilon }_2)}}\le -1,$$

(27)

$$M_0+B_1-\frac{l}{4{ϵ}^{\theta +1}}\le -1,$$

(28)

$$M_0+B_2-\frac{l}{4{ϵ}^{2\theta +2}}\le -1,$$

(29)

$$M_0+B_3-\frac{l}{4{ϵ}^{\theta +1}}\le -1,$$

(30)

$$M_0+f_1^u+f_2^u+f_3^u-\frac{l}{2{ϵ}^{\theta +1}}\le -1,$$

(31)

$$M_0+f_1^u+f_2^u+f_3^u-\frac{l}{2{ϵ}^{3\theta +3}}\le -1,$$

(32)

where constants $K_0,B_1,B_2$ and $B_3$ will be determined later. Then,

$$R_{+}^5\setminus U=U_1^c\cup U_2^c\cup U_3^c\cup U_4^c\cup U_5^c\cup U_6^c\cup U_7^c\cup U_8^c\cup U_9^c\cup U_{10}^c,$$

and

$$\begin{array}{cc}\hfill & U_1^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|0<S_a<ϵ\right\},U_2^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|0<E_a<{ϵ}^2,S_a\ge ϵ\right\},\hfill \\ \hfill & U_3^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|0<I_a<ϵ\right\},U_4^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|0<S_h<ϵ\right\},\hfill \\ \hfill & U_5^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|0<I_h<{ϵ}^3,I_a\ge ϵ,S_h\ge ϵ\right\},\hfill \\ \hfill & U_6^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|S_a>\frac{1}{ϵ}\right\},U_7^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|E_a>\frac{1}{{ϵ}^2}\right\},\hfill \\ \hfill & U_8^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|I_a>\frac{1}{ϵ}\right\},U_9^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|S_h>\frac{1}{ϵ}\right\},\hfill \\ \hfill & U_{10}^c=\left\{(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5|I_h>\frac{1}{{ϵ}^3}\right\}.\hfill \end{array}$$

Now, we verify the negativity of $LV(S_a,E_a,I_a,S_h,I_h)$ for any $(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5\setminus U$.

Case 1. If $(S_a,E_a,I_a,S_h,I_h)\in U_1^c$ form (22), we have

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)-\frac{(1-a)A}{S_a}\hfill \\ \hfill \le & M_0+f_1^u+f_2^u+f_3^u-\frac{(1-a)A}{ϵ}\le -1.\hfill \end{array}& \hfill \end{array}$$

Case 2. If $(S_a,E_a,I_a,S_h,I_h)\in U_2^c$,

(a) When $a\ne 0$, inequality (23) implies that

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)-\frac{aA}{E_a}\hfill \\ \hfill \le & M_0+f_1^u+f_2^u+f_3^u-\frac{aA}{{ϵ}^2}\le -1.\hfill \end{array}& \hfill \end{array}$$

(b) When $a=0$, inequality (24) implies that

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)-\frac{{\beta }_1{S}_a}{1+{\alpha }_1{E}_a}\hfill \\ \hfill \le & M_0+f_1^u+f_2^u+K_0-M\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)-\frac{{\beta }_1ϵ}{1+{\alpha }_1{ϵ}^2}\le -1.\hfill \end{array}& \hfill \end{array}$$

where $K_0={sup}_{I_a\in (0,\infty )}\left\{M{\alpha }_2(d+\xi )I_a+{\eta }_2{I}_a-{\displaystyle \frac{l}{2}}{I}_a^{\theta +1}\right\}.$

Case 3. If $(S_a,E_a,I_a,S_h,I_h)\in U_3^c$, we obtain form (25)

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)\hfill \\ \hfill \le & M_0+f_1^u+f_2^u-M\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)+\left[M{\alpha }_2(d+\xi )+{\eta }_2\right]ϵ\le -1.\hfill \end{array}& \hfill \end{array}$$

Case 4. If $(S_a,E_a,I_a,S_h,I_h)\in U_4^c$ form (26), it follow that

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)-\frac{B}{S_h}\hfill \\ \hfill \le & M_0+f_1^u+f_2^u+f_3^u-\frac{B}{ϵ}\le -1.\hfill \end{array}& \hfill \end{array}$$

Case 5. If $(S_a,E_a,I_a,S_h,I_h)\in U_5^c$, then, by using (27), we have

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)-{\displaystyle \frac{{\eta }_2{I}_a}{(1+{\upsilon }_2{I}_a^2)}}{\displaystyle \frac{S_h}{I_h}}\hfill \\ \hfill \le & M_0+f_1^u+f_2^u+f_3^u-{\displaystyle \frac{{\eta }_2}{ϵ(1+{\upsilon }_2{ϵ}^2)}}\hfill \\ \hfill \le & M_0+f_1^u+f_2^u+f_3^u-{\displaystyle \frac{{\eta }_2}{ϵ(1+{\upsilon }_2)}}\le -1.\hfill \end{array}& \hfill \end{array}$$

Case 6. If $(S_a,E_a,I_a,S_h,I_h)\in U_6^c$, then (28) implies that

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)\hfill \\ \hfill \le & M_0+f_2^u+f_3^u+{\displaystyle \frac{M{\beta }_2^2}{d{\alpha }_2}}{S}_a-{\displaystyle \frac{l}{2}}{S}_a^{\theta +1}\hfill \\ \hfill \le & M_0+B_1-\frac{l}{4{ϵ}^{\theta +1}}\le -1.\hfill \end{array}& \hfill \end{array}$$

where $B_1={sup}_{S_a\in (0,\infty )}\left\{f_2^u+f_3^u+{\displaystyle \frac{M{\beta }_2^2}{d{\alpha }_2}}{S}_a-{\displaystyle \frac{l}{4}}{S}_a^{\theta +1}\right\}$.

Case 7. If $(S_a,E_a,I_a,S_h,I_h)\in U_7^c$ form (29), it follows that

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)\hfill \\ \hfill \le & M_0+f_1^u+f_3^u+(M{\beta }_2+{\eta }_1)E_a-{\displaystyle \frac{l}{2}}{E}_a^{\theta +1}\hfill \\ \hfill \le & M_0+B_2-\frac{l}{4{ϵ}^{2\theta +2}}\le -1.\hfill \end{array}& \hfill \end{array}$$

where $B_2={sup}_{E_a\in (0,\infty )}\left\{f_1^u+f_3^u+(M{\beta }_2+\eta )E_a-{\displaystyle \frac{l}{4}}{E}_a^{\theta +1}\right\}$.

Case 8. If $(S_a,E_a,I_a,S_h,I_h)\in U_8^c$, then (30) tells us that

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)\hfill \\ \hfill \le & M_0+f_1^u+f_2^u-M\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)+\left[M{\alpha }_2(d+\xi )+{\eta }_2\right]I_a-{\displaystyle \frac{l}{2}}{I}_a^{\theta +1}\hfill \\ \hfill \le & M_0+B_3-\frac{l}{4{ϵ}^{\theta +1}}\le -1.\hfill \end{array}& \hfill \end{array}$$

where $B_3={sup}_{I_a\in (0,\infty )}\left\{f_1^u+f_2^u-M\left(d+\xi +\frac{1}{2}{\sigma }_3^2\right)(R_0-1)+\left[M{\alpha }_2(d+\xi )+{\eta }_2\right]I_a-{\displaystyle \frac{l}{4}}{I}_a^{\theta +1}\right\}$.

Case 9. If $(S_a,E_a,I_a,S_h,I_h)\in U_9^c$, (31) implies that

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)-\frac{l}{2}{S}_h^{\theta +1}\hfill \\ \hfill \le & M_0+f_1^u+f_2^u+f_3^u-\frac{l}{2{ϵ}^{\theta +1}}\le -1.\hfill \end{array}& \hfill \end{array}$$

Case 10. If $(S_a,E_a,I_a,S_h,I_h)\in U_{10}^c$, inequality (32) implies that

$$\begin{array}{cc}\hfill \begin{array}{cc}\hfill LV\le & M_0+f_1\left(S_a\right)+f_2\left(E_a\right)+f_3\left(I_a\right)-\frac{l}{2}{I}_h^{\theta +1}\hfill \\ \hfill \le & M_0+f_1^u+f_2^u+f_3^u-\frac{l}{2{ϵ}^{3\theta +3}}\le -1.\hfill \end{array}& \hfill \end{array}$$

From the above analysis, it follows that

$$LV(S_a,E_a,I_a,S_h,I_h)\le -1,(S_a,E_a,I_a,S_h,I_h)\in R_{+}^5\setminus U.$$

Therefore, Condition (2) in Lemma 3 also satisfies. Hence, model (14) has a unique stationary distribution and it is ergodic. The proof is complete. □

6. Numerical Simulations

Using the Milstein method [27], we obtain the numerical analysis which supports the theoretical results in model (14). The discretized equations are listed as follows:

$$\left\{\begin{array}{c}{S}_a^{k+1}=S_a^k+\left((1-a)A-{\displaystyle \frac{{\beta }_1{S}_a^k{E}_a^k}{1+{\alpha }_1{E}_a^k}}-{\displaystyle \frac{{\beta }_2{S}_a^k{I}_a^k}{1+{\alpha }_2{I}_a^k}}-d{S}_a^k\right)\Delta t+S_a^k\left({\sigma }_1{\xi }_k\sqrt{\Delta t}+{\displaystyle \frac{{\sigma }_1^2}{2}}({\xi }_k^2-1)\Delta t\right),\hfill \\ E_a^{k+1}=E_a^k+\left(aA+{\displaystyle \frac{{\beta }_1{S}_a^k{E}_a^k}{1+{\alpha }_1{E}_a^k}}-d{E}_a^k\right)\Delta t+E_a^k\left({\sigma }_2{\eta }_k\sqrt{\Delta t}+{\displaystyle \frac{{\sigma }_2^2}{2}}({\eta }_k^2-1)\Delta t\right),\hfill \\ I_a^{k+1}=I_a^k+\left({\displaystyle \frac{{\beta }_2{S}_a^k{I}_a^k}{1+{\alpha }_2{I}_a^k}}-d{I}_a^k-\xi I_a^k\right)\Delta t+I_a^k\left({\sigma }_3{\zeta }_k\sqrt{\Delta t}+{\displaystyle \frac{{\sigma }_3^2}{2}}({\zeta }_k^2-1)\Delta t\right),\hfill \\ S_h^{k+1}=S_h^k+\left(B-{\displaystyle \frac{{\eta }_1{E}_a^k{S}_h^k}{1+{\upsilon }_1{\left(E_a^k\right)}^2}}-{\displaystyle \frac{{\eta }_2{I}_a^k{S}_h^k}{1+{\upsilon }_2{\left(I_a^k\right)}^2}}-\rho S_h^k\right)\Delta t+S_h^k\left({\sigma }_4{\varrho }_k\sqrt{\Delta t}+{\displaystyle \frac{{\sigma }_4^2}{2}}({\varrho }_k^2-1)\Delta t\right),\hfill \\ I_h^{k+1}=I_h^k+\left({\displaystyle \frac{{\eta }_1{E}_a^k{S}_h^k}{1+{\upsilon }_1{\left(E_a^k\right)}^2}}+{\displaystyle \frac{{\eta }_2{I}_a^k{S}_h^k}{1+{\upsilon }_2{\left(I_a^k\right)}^2}}-\delta I_h^k-\gamma I_h^k-\rho I_h^k\right)\Delta t+I_h^k\left({\sigma }_5{\varsigma }_k\sqrt{\Delta t}+{\displaystyle \frac{{\sigma }_5^2}{2}}({\varsigma }_k^2-1)\Delta t\right).\hfill \end{array}\right.$$

where time increment $\Delta t>0$ and ${\xi }_k,{\eta }_k,{\zeta }_k,{\varrho }_k,{\varsigma }_k$ are independent $N(0,1)-$distributed Gaussian random variables.

The values of the parameters are chosen as [12]:

$$A=15,B=7,{\alpha }_1={\alpha }_2=0.0001,{\eta }_1={\eta }_2=0.0001,{\upsilon }_1={\upsilon }_2=0.0005,d=0.1033,\xi =0.6,$$

$$\rho =1/(75\times 365),\delta =0.08,\gamma =0.04,a=0.007,{\beta }_1=0.005,{\beta }_2=0.034,H=500.$$

Using the above parameters, according to the conclusion of [11], it can be concluded that both viruses of system (1) will spread to the human world. Now, we use the initial value $(S_a\left(0\right),E_a\left(0\right),I_a\left(0\right),S_h\left(0\right),I_h\left(0\right))=(100,40,20,50,1)$ to simulate stochastic model (2).

Example 1.

Let $a=0$, the environmental white noise intensities be ${\sigma }_1=0.1,{\sigma }_2=1.2,{\sigma }_3=3,{\sigma }_4=0.1$, and ${\sigma }_5=0.6$. Then, $R_0=\frac{{\beta }_{2}A}{d(d+\xi +\frac{1}{2}{\sigma }_3^2)}=0.9488<1$ and $R_1=\frac{{\beta }_{1}A}{d(d+\frac{1}{2}{\sigma }_2^2)}=0.8819<1$, the conditions (i) of Theorem 3 are satisfied. Theorem 3 tells us that diseases of stochastic system (2) become extinct, see Figure 1b. From Figure 1a, we can see that compared with the deterministic system (1), lager random white noise will cause the extinction of the diseases.

Example 2.

Let $a=0.007\ne 0$ and the environmental white noise intensities be ${\sigma }_1=0.1,{\sigma }_2=0.1,{\sigma }_3=3,{\sigma }_4=0.1$, and ${\sigma }_5=0.1$. Then, $R_0=0.9488<1$, $R_2=\frac{{\beta }_{1}A+{\alpha }_{1}adA}{d(d+\frac{1}{2}{\sigma }_2^2)}=6.7041>1$ and $R_3=\frac{{\eta }_{1}B}{\rho (1+{\upsilon }_1{H}^2)(\delta +\gamma +\rho +\frac{1}{2}{\sigma }_5^2)}=1.2163>1$, the conditions (ii) of Theorem 3 are satisfied. Theorem 3 tells us that the infected poultry with high pathogenic A(H7N9) virus go to extinction, but the low pathogenic virus has spread to the human world, see Figure 2b. The simulation of the corresponding system (1) is shown in Figure 2a.

Example 3.

Let the environmental white noise intensities be ${\sigma }_1=0.1,{\sigma }_2=0.1,{\sigma }_3=0.1,{\sigma }_4=0.1$, and ${\sigma }_5=0.1$. Then, $R_0=6.9703>1$ and Theorem 4 tells us that there is a unique stationary distribution in the stochastic avian–human influenza system (2), which means that small-intensity white noise can lead to the persistence of avian influenza. Figure 3 confirms this.

7. Conclusions

Avian influenza is an acute infectious disease in humans and poultry caused by the influenza A virus. The Avian influenza virus can be divided into high pathogenicity, low pathogenicity, and no pathogenicity according to pathogenicity. This paper analyzes the dynamic behavior of a random avian influenza model with two different zoonotic viruses. By applying Itô formula and constructing a Lyapunov function, we get the following conclusions (

Table 1. The analysis results of model (2).

Case	Threshold	Conclusion
$a=0$	$R_0<1$, $R_1<1$	Disease extinction
$a\ne 0$	$R_0<1$, $R_2>1$ and $R_3>1$	Diseases spread to the human world
$0\le a\le 1$	$R_0>1$	Admits a unique stationary distribution

):

On the basis of theoretical research, it is found by numerical simulation that under certain conditions, smaller noise can lead to the appearance of stationary distribution. When the noise intensity is large, it can lead to the extinction of the disease. The highlights of this paper are as follows: first, from the theoretical point of view, the development trend of the avian influenza virus under random white noise interference is analyzed, and the application of the stochastic differential equation theory in infectious disease model is promoted; second, compared with the deterministic model (1), the study of model (2) is more consistent with the transmission state of the disease during the disease outbreak, and the research results of the random model has a certain theoretical guiding significance for the disease control.

Of course, our work should be further expanded. According to Reference [18], the first possible extension is that the parameters of the model are independent of each other, and different stochastic processes can be used to describe their dynamics. Obtaining or estimating parameters from the cases of infectious diseases that have occurred, as well as conducting numerical simulation studies through existing models so as to discuss the differences in the results obtained, is another direction for our next efforts.

8. Conjectures from Numerical Simulations

In Table 1 of the conclusion, we summarize the sufficient conditions for the extinction or persistence of the avian influenza virus in stochastic system (2). For the case that cannot be theoretically derived in Table 1, we will explore from the perspective of numerical simulation. In the following examples, except for a, the values of the other parameters are consistent with the values in the numerical simulation of the sixth part.

Case 1. When $a=0$, the case of $R_0<1$ and $R_1>1$.

Example 4.

Let $a=0$, the environmental white noise intensities be ${\sigma }_1=0.1,{\sigma }_2=0.9,{\sigma }_3=3,{\sigma }_4=0.1$, and ${\sigma }_5=0.6$, then $R_0=0.9488<1$ and $R_1=1.4284>1$. From the simulation results in Figure 4a, it can be seen that the poultry infected with highly pathogenic viruses are extinct. Although the low pathogenic virus has spread to the human world, it will not eventually break out in the population due to its recruitment ratio a = 0 in poultry and its own low pathogenicity.

Case 2. When $a\ne 0$, the case of $R_0<1$ and $R_2<1$.

Example 5.

Let $a=0.007\ne 0$, the environmental white noise intensities be ${\sigma }_1=0.1,{\sigma }_2=1.2,{\sigma }_3=3,{\sigma }_4=0.1$, and ${\sigma }_5=0.1$, then $R_0=0.9488<1$, $R_2=0.8819<1$. From the simulation results in Figure 4b, it can be seen that the disease in stochastic system (2) is eventually extinct.

Case 3. When $a\ne 0$, the case of $R_0<1,$$R_2>1$ and $R_3<1$.

Example 6.

Let $a=0.007\ne 0$, the environmental white noise intensities be ${\sigma }_1=0.1,{\sigma }_2=0.1,{\sigma }_3=3,{\sigma }_4=0.1$ and ${\sigma }_5=1$, then $R_0=0.9488<1$, $R_2=6.7041>1$, and $R_3=0.2453<1$. From the simulation results in Figure 4c, it can be seen that the poultry infected with highly pathogenic viruses are extinct. Although the low pathogenic virus spread to the human world, compared with Figure 2b, the disease eventually became extinct in the human world under the interference of large random white noise.