Dynamics of a kind of stochastic SIRS models with two different nonlinear incidences

Abstract

A new kind of stochastic SIRS models with two different nonlinear incidences are extended. The obtained results can be expressed in two dimensions. In mathematics, the threshold values $R_1^s$ and $R_2^s$ which ensure permanent or extinct disease are presented, respectively. More concretely, when $R_i^s>1(i=1,2)$ , the two diseases are persistence in mean. When $R_i^s<1$ or $R_i^s>1(i=1,2)$ , the two diseases will either be extinct or be permanent, respectively. What’s more interesting is the numerical results which show that the two diseases go to extinction at a large time, when $R_1^s=1$ and $R_2^s=1$ . Furthermore, the sufficient conditions for the diseases that are extinct and permanent are given. In addition, the stochastic boundedness of the model is investigated. Epidemiologically, we can conclude that large noise fluctuations can effectively prevent the outbreak of disease, which can supply us beneficial methods to control the spreading of two diseases. Finally, a few numerical simulations are presented to demonstrate and verify the findings obtained.

Introduction

According to the World Health Organization’s published data in 2017, of the 56.4 million deaths worldwide in 2015, more than 10% were due to infectious diseases. ¹ Therefore, analysis and control of the spread of the epidemic disease have become an important topic. Thanks to the groundbreaking research of Kermack and McKendrick, ² mathematical modeling is an advisable method for us to thoroughly investigate the spreading of epidemic disease. In the early work, many well-known classic model which investigated disease control was deterministic.^3,4 In 1927, the SIRS epidemic model was proposed and studied. From then on, the SIRS model has been pursued by many researchers. As we all know that the spreading of disease is unpredictable due to the impact of environmental variations, proper and realistic stochastic differential equation models are proposed for studying the spreading of disease under various environmental conditions.^5–8 An increasing number of researchers investigate and improve the stochastic SIRS models, such as one susceptible individual with one infectious individual,^9–15 two susceptible individuals with one infectious individual, ¹⁶ and multiple susceptible individuals with one infectious individual. ¹⁷ In the previous papers, the susceptible individuals be infected with only a disease, whether multifarious or onefold species. However, that susceptible individuals can be infected by multiple diseases at the same time in the real world. Recently, a stochastic SIRS model with two infectious individuals was proposed, ¹⁸ the model is given by

$$\{\begin{array}{c}\mathrm{d}S(t)=(\mathrm{\Lambda }-\mu S(t)-\frac{{\beta }_{1}S(t)I_1(t)}{{\alpha }_1+I_1(t)}-\frac{{\beta }_{2}S(t)I_2(t)}{{\alpha }_2+S(t)}+\delta R(t))\mathrm{d}t\hfill \\ -\frac{{\sigma }_{1}S(t)I_1(t)}{{\alpha }_1+I_1(t)}\mathrm{d}{B}_1(t)-\frac{{\sigma }_{2}S(t)I_2(t)}{{\alpha }_2+S(t)}\mathrm{d}{B}_2(t)\hfill \\ \mathrm{d}{I}_1(t)=(\frac{{\beta }_{1}S(t)I_1(t)}{{\alpha }_1+I_1(t)}-(\mu +d_1+{\gamma }_1)I_1(t))\mathrm{d}t+\frac{{\sigma }_{1}S(t)I_1(t)}{{\alpha }_1+I_1(t)}\mathrm{d}{B}_1(t)\hfill \\ \mathrm{d}{I}_2(t)=(\frac{{\beta }_{2}S(t)I_2(t)}{{\alpha }_2+S(t)}-(\mu +d_2+{\gamma }_2)I_2(t))\mathrm{d}t+\frac{{\sigma }_{2}S(t)I_2(t)}{{\alpha }_2+S(t)}\mathrm{d}{B}_2(t)\hfill \\ \mathrm{d}R(t)=({\gamma }_1{I}_1(t)+{\gamma }_2{I}_2(t)-(\mu +\delta )R(t))\mathrm{d}t\hfill \end{array}$$

(1)

where $S(t)$ represents the number of individuals susceptible; $I_1(t)$ and $I_2(t)$ represent the number of individuals infected by diseases $A_1$ and $A_2$ ; $R(t)$ denotes the number of individuals recovered; ${\gamma }_1$ and ${\gamma }_2$ are recovery rates of two infective individuals; $\mathrm{\Lambda }$ is the influx rate for individuals; $\mu$ denotes the natural death rate for the population; $\delta$ denotes the lost immunity rate for removed population which turns back to the susceptible population; ${\beta }_1$ and ${\beta }_2$ are the contact coefficient of the two infective individuals; and $d_1$ and $d_2$ are the disease-inducing death rate. The functions $(({\beta }_{1}S(t)I_1(t))/{\alpha }_1+I_1(t))$ and $(({\beta }_{2}S(t)I_2(t))/({\alpha }_2+S(t)))$ denote incidence rates for $I_1(t)$ and $I_2(t)$ , respectively; and $B_1(t)$ and $B_2(t)$ are the Brownian motion with intensities ${\sigma }_1$ and ${\sigma }_2$ .

There arise following problems: how does the noise fluctuations impact the spreading of diseases for the model which one susceptible individual was infected by n group of disease species? Whether the noise fluctuations can effectively prevent the outbreak of diseases?

In this article, for convenience, we research a new stochastic SIRS model: one susceptible individual with two infectious individuals. Moreover, the data could determine the type of incidence for real world. ¹⁹ For the purpose of more realistic and the “psychological” effects,^20,21 we investigate the general nonlinear incidences of the forms ${\beta }_{1}f(I_1(t))S(t)$ and ${\beta }_{2}g(S(t))I_2(t)$ . The function $f(I_1(t))$ increased with the decrease in $I_1(t)$ , while it decreases with the increase in $I_1(t)$ . The function $g(S(t))$ increases monotonically with respect to $S(t)$ , due to the contact that could increase for a sufficient large number of susceptible individuals. Inspired by the discussion above, we investigate a new stochastic SIRS model with two nonlinear incidences. The model is given by

$$\{\begin{array}{c}\mathrm{d}S(t)=[\mathrm{\Lambda }-\mu S(t)-{\beta }_{1}f(I_1(t))S(t)-{\beta }_{2}g(S(t))I_2(t)+\delta R(t)]\mathrm{d}t\hfill \\ -{\sigma }_{1}f(I_1(t))S(t)\mathrm{d}{B}_1(t)-{\sigma }_{2}g(S(t))I_2(t)\mathrm{d}{B}_2(t)\hfill \\ \mathrm{d}{I}_1(t)=[{\beta }_{1}f(I_1(t))S(t)-(\mu +d_1+{\gamma }_1)I_1(t)]\mathrm{d}t+{\sigma }_{1}f(I_1(t))S(t)\mathrm{d}{B}_1(t)\hfill \\ \mathrm{d}{I}_2(t)=[{\beta }_{2}g(S(t))I_2(t)-(\mu +d_2+{\gamma }_2)I_2(t)]\mathrm{d}t+{\sigma }_{2}g(S(t))I_2(t)\mathrm{d}{B}_2(t)\hfill \\ \mathrm{d}R(t)=[{\gamma }_1{I}_1(t)+{\gamma }_2{I}_2(t)-(\mu +\delta )R(t)]\mathrm{d}t\hfill \end{array}$$

(2)

whose state space is ${\mathbb{R}}_{+}^4=\{(S,I_1,I_2,R):S>0,I_1\ge 0,I_2\ge 0,R\ge 0\}$ . Functions ${\beta }_{1}f(I_1(t))S(t)$ and ${\beta }_{2}g(S(t))I_2(t)$ represent two different types of incidence functions. Other parameters are taken as in model (1). The two functions $f(I_1(t))$ and $g(S(t))$ satisfy the following assumptions:

• $(\mathbf{H}1)$ $f(I_1(t))$ is nonnegative bounded and continuously twice differentiable and the derivative of f is bounded, $(f(I_1(t))/I_1(t))$ decreases monotonically with respect to $I_1(t)$ , $f(0)=0$ , and $\underset{I_1(t)\to 0}{lim}(f(I_1(t))/I_1(t))=f\prime (0)>0$ .

• $(\mathbf{H}2)$ $g(S(t))$ is differentiable continuously and increases monotonically with respect to $S(t)$ , for any $S(t)>0$ , $g(0)=0$ , $H\stackrel{\Delta }{=}\underset{0<S(t)<\frac{\mathrm{\Lambda }}{\mu }}{sup}(g(S(t))/S(t))<+\infty$ .

The article is organized as follows. In section “Preliminaries,” we present several preliminaries and notions. The existence or uniqueness and stochastic boundedness of the global solution for model (2) are discussed in section “Existence or uniqueness and stochastic boundedness of global positive solution”. In section “Extinction of diseases”, sufficient conditions for the two diseases that are extinct are given are established. In section “Persistence in mean”, we obtain the sufficient conditions for the two diseases that are persistence for model (2). In section “Almost sure exponential stability,” a sufficient condition for the almost sure exponentially stable disease-free equilibrium is established. In section “Numerical simulations and examples,” some simulations are given to demonstrate the obtained results. Finally, a brief conclusion is present in section “Conclusion.”

Preliminaries

In the section, we give several preliminaries as follows. Throughout the article, consider the process $X(t)$ defined in the state space $(\Omega ,\mathcal{F},P)$ . Let $X(t)=(S(t),I_1(t),I_2(t),R(t))$ denotes the solution for model (2). $X(0)=(S(0),I_1(0),I_2(0),R(0))$ denotes initial value for stochastic model (2). $\mathrm{\Lambda }$ , $\mu$ , ${\beta }_1$ , ${\beta }_2$ , ${\sigma }_1$ , ${\sigma }_2$ , $\delta$ , ${\gamma }_1$ , ${\gamma }_1$ , $d_1$ , and $d_2$ are positive constants. $|X(t)|=\sqrt{S^2(t)+I_1^2(t)+I_2^2(t)+R^2(t)}$ represents the norm of $X(t)$ .

Definition 1

Meng et al. ²² define $〈h(t)〉=\frac{1}{t}{\int }_0^{t}h(s)\mathrm{d}s$ , where $h(t)$ be an integrable function on $[0,+\infty )$ .

1. $I_i(t)$ are named extinct, if $\underset{t\to +\infty }{lim}{I}_i(t)=0,(i=1,2)$ .

2. $I_i(t)$ are named permanent in mean, if $\underset{t\to +\infty }{liminf}〈I_i(t)〉\ge {\lambda }_i,(i=1,2)$ , for some constants ${\lambda }_i>0$ .

Definition 2

For any initial value $(S(0),I_1(0),I_2(0),R(0))\in {\mathbb{R}}_{+}^4$ .^23,24 Assume for any $0<\epsilon <1$ , there has a constant $\rho =\rho (\epsilon )>0$ , such that the inequality in following holds

$$\underset{t\to \infty }{limsup}P\{|X(t)|>\rho \}<\epsilon$$

Then solution $X(t)=(S(t),I_1(t),I_2(t),R(t))$ for the stochastic model (2) is named as stochastically ultimately bounded.

Definition 3

For any initial value $(S(0),I_1(0),I_2(0),R(0))\in {\mathbb{R}}_{+}^4$ .^24,25 Assume for any $0<\epsilon <1$ , there has two constants $\rho =\rho (\epsilon )>0$ and $\varpi =\varpi (\epsilon )>0$ , such that the following inequalities hold

$$\begin{array}{c}\multicolumn{1}{c}{\underset{t\to \infty }{lim\inf }P(|X(t)|\le \rho )\ge 1-\epsilon \mathrm{and}\underset{t\to \infty }{liminf}P(|X(t)|\ge \varpi )}\\ \multicolumn{1}{c}{\ge 1-\epsilon }\end{array}$$

Then solution $X(t)=(S(t),I_1(t),I_2(t),R(t))$ for the stochastic model (2) is named stochastically permanent.

Lemma 1

Set $(S(t),I_1(t),I_2(t),R(t))$ be a solution for model (2) with any given initial value $(S(0),I_1(0),I_2(0),R(0))\in {\mathbb{R}}_{+}^4$ , one obtains

$$\underset{t\to \infty }{limsup}(S(t)+I_1(t)+I_2(t)+R(t))\le \frac{\mathrm{\Lambda }}{\mu }$$

Proof

From model (2), one can obtain

$$\begin{array}{l}\mathrm{d}(S(t)+I_1(t)+I_2(t)+R(t))\\ \le \mathrm{\Lambda }-\mu \left(S(t)+I_1(t)+I_2(t)+R(t)\right)\mathrm{d}t\end{array}$$

By integration, we obtain

$$\begin{array}{c}(S(t)+I_1(t)+I_2(t)+R(t))\hfill \\ \le ((S(0)+I_1(0)+I_2(0)+R(0))-\frac{\mathrm{\Lambda }}{\mu })e^{-\mu t}+\frac{\mathrm{\Lambda }}{\mu }\hfill \end{array}$$

Hence, one has

$$\underset{t\to \infty }{limsup}(S(t)+I_1(t)+I_2(t)+R(t))\le \frac{\mathrm{\Lambda }}{\mu }.$$

□

Remark 1

By Lemma 1, one could conclude that for solution $(S(t),I_1(t),I_2(t),R(t))$ for model (2) with any given initial value $(S(0),I_1(0),I_2(0),R(0))\in {\mathbb{R}}_{+}^4$ , there is $T_1>0$ such that

$$S(t)+I_1(t)+I_2(t)+R(t)\le \frac{\mathrm{\Lambda }}{\mu },t\ge T_1$$

Therefore, one has

$$I_1(t)\le \frac{\mathrm{\Lambda }}{\mu },I_2(t)\le \frac{\mathrm{\Lambda }}{\mu },R(t)\le \frac{\mathrm{\Lambda }}{\mu },\mathrm{and}S(t)\le \frac{\mathrm{\Lambda }}{\mu }$$

for $t\ge T_1$ .

Existence or uniqueness and stochastic boundedness of global positive solution

Theorem 1

For any given initial value $(S(0),I_1(0),I_2(0),R(0))\in {\mathbb{R}}_{+}^4$ , model (2) has a unique positive solution $(S(t),I_1(t),I_2(t),R(t))$ and it stays in ${\mathbb{R}}_{+}^4$ with probability one for all $t\ge 0$ .

Proof

Because coefficients for stochastic model (2) satisfy locally Lipschitz continuous, model (2) has a unique local solution $(S(t),I_1(t),I_2(t),R(t))$ for $0\le t<{\tau }_e$ , where ${\tau }_e$ be a explosion time.^26,27 For the globality of the solution, one demonstrates that ${\tau }_e=\infty$ a.s.

Choose $n_0>0$ is large sufficiently such that $S(0)>0$ , $I_1(0)>0$ , $I_2(0)>0$ , and $R(0)>0$ in a interval $[(1/n_0),n_0]$ . For any $n>n_0$ , defining a stop-time

$${\tau }_n=inf\{t\in [0,{\tau }_e):S(t)\notin (\frac{1}{n},n)\mathrm{or}{I}_1(t)\notin (\frac{1}{n},n)\mathrm{or}{I}_2(t)\notin (\frac{1}{n},n)\mathrm{or}R(t)\notin (\frac{1}{n},n)\}$$

Let $infØ=\infty$ (Ø denotes a empty set). Apparently, ${\tau }_n$ increases as $n\to \infty$ . Set ${\tau }_{\infty }=\underset{n\to \infty }{lim}{\tau }_n$ , hence ${\tau }_{\infty }\le {\tau }_e$ . If one can demonstrate that ${\tau }_{\infty }=\infty$ , it can lead to ${\tau }_e=\infty$ . Assuming that ${\tau }_{\infty }\ne \infty$ , then there are two positive constants T and $0<\eta <1$ such that $P\{{\tau }_{\infty }\le T\}\ge \eta .$ Therefore, for each integer $n_1\ge n_0$ , one can obtain

$$P({\tau }_n\le T)\ge \eta ,n\ge n_1$$

(3)

Denoting

$$N(t)=S(t)+I_1(t)+I_2(t)+R(t)$$

(4)

From model (2), one can obtain

$$\frac{\mathrm{d}N(t)}{\mathrm{d}t}=\mathrm{\Lambda }-d_1{I}_1(t)-d_2{I}_2(t)-\mu N(t)$$

One has $(\mathrm{d}N(t)/\mathrm{d}t)\le \mathrm{\Lambda }-\mu N(t)$ , then for all $t<{\tau }_n$ , one can obtain

$$N(t)\le e^{-\mu t}(N(0)-\frac{\mathrm{\Lambda }}{\mu })+\frac{\mathrm{\Lambda }}{\mu }\le \max \{\frac{\mathrm{\Lambda }}{\mu },N(0)\}:=k_1$$

Defining a Liapunov function

$$\begin{array}{c}{V}_1(S(t),I_1(t),I_2(t),R(t))=(S(t)-1-\log S(t))\hfill \\ +(I_1(t)-1-\log I_1(t))+(I_2(t)-1-\log I_2(t))\hfill \\ +(R(t)-1-\log R(t))\hfill \end{array}$$

Using the Itô’s formula of $V_1$ , we obtain

$$\begin{array}{c}\mathrm{d}{V}_1(S(t),I_1(t),I_2(t),R(t))=L{V}_1\mathrm{d}t+{\sigma }_{1}f(I_1(t))\hfill \\ \frac{(I_1(t)-S(t))}{I_1(t)}\mathrm{d}{B}_1(t)+{\sigma }_{2}g(S(t))\frac{(I_2(t)-S(t))}{S(t)}\mathrm{d}{B}_2(t)\hfill \end{array}$$

where

$$\begin{array}{c}\multicolumn{1}{c}{L{V}_1=\mathrm{\Lambda }+4\mu +d_1+d_2+{\gamma }_1+{\gamma }_2+\delta }\\ +{\beta }_{1}f(I_1(t))+{\beta }_2{I}_2(t)\frac{g(S(t))}{S(t)}\hfill \\ +{\sigma }_1^2\frac{f^2(I_1(t))}{2}+{\sigma }_2^2{I}_2^2(t)\frac{g^2(S(t))}{2S^2(t)}\hfill \\ +{\sigma }_1^2{S}^2(t)\frac{f^2(I_1(t))}{2I_1^2(t)}+{\sigma }_2^2\frac{g^2(S(t))}{2}\hfill \\ -\frac{\mathrm{\Lambda }}{S(t)}-\frac{\delta R(t)}{S(t)}-\frac{{\gamma }_1{I}_1(t)+{\gamma }_2{I}_2(t)}{R(t)}\hfill \\ -{\beta }_{1}S(t)\frac{f(I_1(t))}{I_1(t)}-{\beta }_{2}g(S(t))\hfill \\ -d_1{I}_1(t)-d_2{I}_2(t)-\mu (S(t)+I_1(t)+I_2(t)+R(t)).\hfill \end{array}$$

From assumptions (H1) and (H2), one has $f(I_1(t))\le f\prime (0)I_1(t)$ and $g(S(t))\le g(k_1)$ . Thus

$$\begin{array}{c}\multicolumn{1}{c}{L{V}_1\le \mathrm{\Lambda }+4\mu +d_1+d_2+{\gamma }_1+{\gamma }_2+\delta }\\ +{\beta }_{1}f\prime (0)k_1+{\beta }_{2}H{k}_1+{\sigma }_1^2(f\prime (0)k_1{)}^2\hfill \\ +\frac{{\sigma }_2^2}{2}(H{k}_1{)}^2+\frac{{\sigma }_2^2}{2}{g}^2(k_1):=k_2\hfill \end{array}$$

Then, one can obtain

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}{V}_1(t)\le k_2\mathrm{d}t+{\sigma }_{1}f(I_1(t))\frac{(I_1(t)-S(t))}{I_1(t)}\mathrm{d}{B}_1(t)}\\ +{\sigma }_{2}g(S(t))\frac{(I_2(t)-S(t))}{S(t)}\mathrm{d}{B}_2(t)\hfill \end{array}$$

This implies that

$$\begin{array}{c}\underset{0}{\overset{{\tau }_n\wedge T}{\int }}\mathrm{d}{V}_1(S(s),I_1(s),I_2(s),R(s))\le \underset{0}{\overset{{\tau }_n\wedge T}{\int }}{k}_2\mathrm{d}t\hfill \\ +\underset{0}{\overset{{\tau }_n\wedge T}{\int }}{\sigma }_{1}f(I_1(s))\frac{(I_1(s)-S(s))}{I_1(s)}\mathrm{d}{B}_1(s)\hfill \\ +\underset{0}{\overset{{\tau }_n\wedge T}{\int }}{\sigma }_{2}g(S(s))\frac{(I_2(s)-S(s))}{S(s)}\mathrm{d}{B}_2(s)\hfill \end{array}$$

where ${\tau }_n\wedge T=min\{{\tau }_n,T\}$ . Taking the expectations of the above inequality, we obtain

$$\begin{array}{c}\mathbb{E}\underset{0}{\overset{{\tau }_n\wedge T}{\int }}{\sigma }_{1}f(I_1(s))\frac{(I_1(s)-S(s))}{I_1(s)}\mathrm{d}{B}_1(s)=0\hfill \\ \mathbb{E}\underset{0}{\overset{{\tau }_n\wedge T}{\int }}{\sigma }_{2}g(S(s))\frac{(I_2(s)-S(s))}{S(s)}\mathrm{d}{B}_2(s)=0\hfill \end{array}$$

Then, one has

$$\begin{array}{c}\mathbb{E}{V}_1(S({\tau }_n\wedge T),I_1({\tau }_n\wedge T),I_2({\tau }_n\wedge T),R({\tau }_n\wedge T))\hfill \\ \le V_1(S(0),I_1(0),I_2(0),R(0))+k_{2}T.\hfill \end{array}$$

Let ${\Omega }_n=\{{\tau }_n\le T\}$ for $n\ge n_1$ . From equation (3), one can obtain $P({\Omega }_n)\ge \eta$ . By the definition of the stopping time, for any $\omega \in {\Omega }_n$ , there is $S({\tau }_n,\omega )$ or $I_1({\tau }_n,\omega )$ or $I_2({\tau }_n,\omega )$ or $R({\tau }_n,\omega )$ equaling (1/n) or n, as a result

$$\begin{array}{c}{V}_1(S({\tau }_n,\omega ),I_1({\tau }_n,\omega ),I_2({\tau }_n,\omega ),R({\tau }_n,\omega ))\ge \hfill \\ ((n-1-\log n)\wedge (\frac{1}{n}-1-\log \frac{1}{n}))\hfill \end{array}$$

Hence, one can obtain

$$\begin{array}{c}{V}_1(S(0),I_1(0),I_2(0),R(0))+k_{2}T\hfill \\ \ge \mathbb{E}[I_{{\Omega }_n}{V}_1(S({\tau }_n,\omega ),I_1({\tau }_n,\omega ),I_2({\tau }_n,\omega ),R({\tau }_n,\omega ))]\hfill \\ =P({\Omega }_n)V_1(S({\tau }_n,\omega ),I_1({\tau }_n,\omega ),I_2({\tau }_n,\omega ),R({\tau }_n,\omega ))\hfill \\ \ge \eta ((n-1-\log n)\wedge (\frac{1}{n}-1-\log \frac{1}{n}))\hfill \end{array}$$

where $I_{{\Omega }_n}$ be a indicator function for ${\Omega }_n$ . Letting $n\to \infty$ of the above inequality, one has

$$\infty >V_1(S(0),I_1(0),I_2(0),R(0))+k_{2}T=\infty ,\mathrm{a}.\mathrm{s}.$$

As a consequence, one can have ${\tau }_{\infty }=\infty$ . □

Theorem 2

For any $(S(0),I_1(0),I_2(0),R(0))\in {\mathbb{R}}_{+}^4$ , the solution $X(t)=(S(t),I_1(t),I_2(t),R(t))$ for the stochastic model (2) is stochastically ultimately bounded and permanent.

Proof

Define a function as follows

$$V_2(t)=\frac{1}{N(t)}+N(t)$$

where $N(t)$ is given as in equation (4). From assumptions (H1) and (H2), one can obtain $f(I_1(t))\le f\prime (0)k_1$ and $g(S(t))\le g(k_1)$ . By the Itô’s formula, we have

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}(e^{\mu t}{V}_2(t))=[\mu e^{\mu t}(N(t)+\frac{1}{N(t)})+e^{\mu t}(1-\frac{1}{N^2(t)})(\mathrm{\Lambda }-\mu N(t)-d_1{I}_1(t)-d_2{I}_2(t))}\\ +2e^{\mu t}\frac{{\sigma }_1^2{f}^2(I_1(t))S^2(t)+{\sigma }_2^2{g}^2(S(t))I_2^2(t)}{N^3(t)}]\mathrm{d}t\hfill \\ \le e^{\mu t}[\mathrm{\Lambda }-\frac{\mathrm{\Lambda }}{N^2(t)}+\frac{d_1{I}_1(t)+d_2{I}_2(t)}{N^2(t)}+\frac{2\mu }{N(t)}\hfill \\ +2\frac{{\sigma }_1^2{f}^2(I_1(t))S^2(t)+{\sigma }_2^2{g}^2(S(t))I_2^2(t)}{N^3(t)}]\mathrm{d}t\hfill \\ \le e^{\mu t}[\mathrm{\Lambda }-\frac{\mathrm{\Lambda }}{N^2(t)}+\frac{2\mu +d_1+d_2+2{\sigma }_1^2{f\prime }^2(0)k_1^2+2{\sigma }_2^2{g}^2(k_1)}{N(t)}]\mathrm{d}t\le e^{\mu t}K\mathrm{d}t\hfill \end{array}$$

where $K=((4{\mathrm{\Lambda }}^2+(2\mu +d_1+d_2+2{\sigma }_1^{2}f{\prime }^2(0)k_1^2+2{\sigma }_2^2{g}^2(k_1){)}^2)/4\mathrm{\Lambda })$ .

By integrating the above inequality, we obtain

$$e^{\mu t}{V}_2(t)\le K\underset{0}{\overset{t}{\int }}{e}^{\mu s}\mathrm{d}s+V_2(0)=\frac{K}{\mu }(e^{\mu t}-1)+V_2(0)$$

Hence, one has

$$\begin{array}{c}\multicolumn{1}{c}{\mathbb{E}[e^{\mu t}{V}_2(t)]\le \mathbb{E}[\frac{K}{\mu }(e^{\mu t}-1)+V_2(0)]}\\ =\frac{K}{\mu }(e^{\mu t}-1)+V_2(0)\hfill \end{array}$$

This implies that

$$\mathbb{E}{V}_2(t)=\frac{K}{\mu }(1-e^{-\mu t})+e^{-\mu t}{V}_2(0)\le \frac{K}{\mu }+V_2(0):=G$$

Let a constant $\rho$ large enough such that $(G/\rho )<1$ . Applying Chebyshev’s inequality, one has

$$P(N(t)+\frac{1}{N(t)}>\rho )\le \frac{1}{\rho }\mathbb{E}(N(t)+\frac{1}{N(t)})\le \frac{G}{\rho }\stackrel{\Delta }{=}\epsilon$$

Then, one gets

$$1-\epsilon \le P(N(t)+\frac{1}{N(t)}\le \rho )\le P(\frac{1}{\rho }\le N(t)\le \rho )$$

Applying the inequality $a^2+b^2\ge 2\mathrm{ab}$ , we obtain

$$\begin{array}{c}\multicolumn{1}{c}{N^2(t)=S^2(t)+I_1^2(t)+I_2^2(t)+R^2(t)+2[S(t)I_1(t)+S(t)I_2(t)}\\ +S(t)R(t)+I_1(t)R(t)+I_2(t)R(t)+I_1(t)I_2(t)]\hfill \\ \le 4S^2(t)+4I_1^2(t)+4I_2^2(t)+4R^2(t)\le 4N^2(t)\hfill \end{array}$$

This implies that

$$N^2(t)\le 4|X(t){|}^2\le 4N^2(t)$$

where $|X(t)|$ is given in the “Preliminaries” section.

Then, one can obtain

$$P(\frac{1}{2\rho }\le \frac{N(t)}{2}\le |X(t)|\le N(t)\le \rho )\ge 1-\epsilon$$

By Definitions 2 and 3, one can obtain the conclusion. □

Extinction of diseases

Theorem 3

For any given initial value $(S(0),I_1(0),I_2(0),R(0))\in {\mathbb{R}}_{+}^4$ . Let $R_1^s=(({\beta }_1\mathrm{\Lambda }f\prime (0)/\mu (\mu +d_1+{\gamma }_1))-({\sigma }_1^2{\mathrm{\Lambda }}^{2}f{\prime }^2(0)/2{\mu }^2(\mu +d_1+{\gamma }_1)))$ and $R_2^s=(({\beta }_{2}g(\mathrm{\Lambda }/\mu )/\mu +d_2+{\gamma }_2)-({\sigma }_2^2{g}^2(\mathrm{\Lambda }/\mu ))/2(\mu +d_2+{\gamma }_2)))$ . Suppose conditions that

• ${\sigma }_i>({\beta }_i/\sqrt{2(\mu +d_i+{\gamma }_i)})$ , $(i=1,2)$ ;

• ${\sigma }_1\le \sqrt{({\beta }_1\mu /f\prime (0)\mathrm{\Lambda })}$ , ${\sigma }_2\le \sqrt{({\beta }_2/(g(\mathrm{\Lambda }/\mu )))}$ , $R_1^s<1$ and $R_2^s<1$ .

If condition (a) or (b) is satisfied, then the two diseases $I_1(t)$ and $I_2(t)$ of the stochastic model (2) both are extinct. Furthermore

$$\begin{array}{c}{lim}_{t\to \infty }{I}_1(t)=0,\underset{t\to \infty }{lim}{I}_2(t)=0,\underset{t\to \infty }{lim}R(t)=0,\hfill \\ \mathrm{and}\underset{t\to \infty }{lim}S(t)=\frac{\mathrm{\Lambda }}{\mu }\hfill \end{array}$$

Proof

• Case (a).

Using the Itô’s formula to the stochastic model (2), one can obtain

$$\begin{array}{c}\mathrm{d}\ln I_1(t)=\hfill \\ (\frac{{\beta }_{1}S(t)f(I_1(t))}{I_1(t)}-(\mu +d_1+{\gamma }_1)-\frac{{\sigma }_1^2{S}^2(t)f^2(I_1(t))}{2I_1^2(t)})\mathrm{d}t\hfill \\ +\frac{{\sigma }_{1}S(t)f(I_1(t))}{I_1(t)}\mathrm{d}{B}_1(t)\hfill \end{array}$$

Integrating the above equality, one can obtain

$$\ln I_1(t)$$

(5)

$$\begin{array}{c}=\underset{0}{\overset{t}{\int }}(\frac{{\beta }_{1}S(s)f(I_1(s))}{I_1(s)}-(\mu +d_1+{\gamma }_1)-\frac{{\sigma }_1^2{S}^2(s)f^2(I_1(s))}{2I_1^2(s)})\mathrm{d}s\hfill \\ +\underset{0}{\overset{t}{\int }}\frac{{\sigma }_{1}S(s)f(I_1(s))}{I_1(s)}\mathrm{d}{B}_1(s)+\ln I_1(0)\hfill \\ =-\frac{{\sigma }_1^2}{2}\underset{0}{\overset{t}{\int }}{(\frac{S(s)f(I_1(s))}{I_1(s)}-\frac{{\beta }_1}{{\sigma }_1^2})}^2\mathrm{d}s-(\mu +d_1+{\gamma }_1)t+\frac{{\beta }_1^2}{2{\sigma }_1^2}t\hfill \\ +\underset{0}{\overset{t}{\int }}\frac{{\sigma }_{1}S(s)f(I_1(s))}{I_1(s)}\mathrm{d}{B}_1(s)+\ln I_1(0)\hfill \\ \le -(\mu +d_1+{\gamma }_1)t+\frac{{\beta }_1^2}{2{\sigma }_1^2}t\hfill \\ +\underset{0}{\overset{t}{\int }}\frac{{\sigma }_{1}S(s)f(I_1(s))}{I_1(s)}\mathrm{d}{B}_1(s)+\ln I_1(0)\hfill \end{array}$$

(6)

Let $M_1(t)={\int }_0^t(({\sigma }_{1}S(s)f(I_1(s)))/(I_1(s)))\mathrm{d}{B}_1(s)$ , since

$$\begin{array}{c}\frac{{〈M_1,M_1〉}_t}{t}=\frac{1}{t}\underset{0}{\overset{t}{\int }}\frac{{\sigma }_1^2{S}^2(s)f^2(I_1(s))}{I_1^2(s)}\mathrm{d}s\hfill \\ \le \frac{1}{t}\underset{0}{\overset{t}{\int }}{({\sigma }_1{k}_{1}f\prime (0))}^2\mathrm{d}s={({\sigma }_1{k}_{1}f\prime (0))}^2<+\infty \hfill \end{array}$$

Applying the strong law of large numbers for martingales,^28,29 we obtain $\underset{t\to \infty }{limsup}(M_1(t)/t)=0$ .

Dividing t and $t\to \infty$ as in equation (5), then

$$\begin{array}{c}\underset{t\to \infty }{lim\sup }\frac{\ln I_1(t)}{t}\le -(\mu +d_1+{\gamma }_1)\hfill \\ +\frac{{\beta }_1^2}{2{\sigma }_1^2}<0,\mathrm{when}{\sigma }_1>\frac{{\beta }_1}{\sqrt{2(\mu +d_1+{\gamma }_1)}}\hfill \end{array}$$

Similarly, from the stochastic model (2), thus

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}\ln I_2(t)=({\beta }_{2}g(S(t))-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2(S(t))}{2})\mathrm{d}t}\\ +{\sigma }_{2}g(S(t))\mathrm{d}{B}_2(t)\hfill \end{array}$$

Integrating the above equality, one can obtain

$$\ln I_2(t)$$

(7)

$$\begin{array}{c}=\underset{0}{\overset{t}{\int }}({\beta }_{2}g(S(s))-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2(S(s))}{2})\mathrm{d}s\hfill \\ +\underset{0}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)+\ln I_2(0)\hfill \\ =-\frac{{\sigma }_2^2}{2}\underset{0}{\overset{t}{\int }}{(g(S(s))-\frac{{\beta }_2}{{\sigma }_2^2})}^2\mathrm{d}s-(\mu +d_2+{\gamma }_2)t+\frac{{\beta }_2^2}{2{\sigma }_2^2}t\hfill \\ +\underset{0}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)+\ln I_2(0)\hfill \\ \le -(\mu +d_2+{\gamma }_2)t+\frac{{\beta }_2^2}{2{\sigma }_2^2}t\hfill \\ +\underset{0}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)+\ln I_2(0)\hfill \end{array}$$

(8)

Let $M_2(t)={\int }_0^t{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)$ , since

$$\begin{array}{c}\multicolumn{1}{c}{\frac{{〈M_2,M_2〉}_t}{t}=\frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_2^2{g}^2(S(s))\mathrm{d}s}\\ \multicolumn{1}{c}{\le \frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_2^2{g}^2(k_1)\mathrm{d}s={\sigma }_2^2{g}^2(k_1)<+\infty }\end{array}$$

Analogously, we can obtain $\underset{t\to \infty }{limsup}((M_2(t))/t)=0$ . Dividing by t and taking the limit of equation (7), one has

$$\begin{array}{c}\underset{t\to \infty }{limsup}\frac{\ln I_2(t)}{t}\le -(\mu +d_2+{\gamma }_2)+\frac{{\beta }_2^2}{2{\sigma }_2^2}<0,\hfill \\ \mathrm{when}{\sigma }_2>\frac{{\beta }_2}{\sqrt{2(\mu +d_2+{\gamma }_2)}}\hfill \end{array}$$

• Case (b).

First, one concentrates on $I_1(t)$ . In the view of equation (5), dividing by t, one gets

$$\begin{array}{c}\multicolumn{1}{c}{\frac{\ln I_1(t)}{t}=-\frac{{\sigma }_1^2}{2}\frac{1}{t}\underset{0}{\overset{t}{\int }}{(\frac{S(s)f(I_1(s))}{I_1(s)}-\frac{{\beta }_1}{{\sigma }_1^2})}^2\mathrm{d}s}\\ -(\mu +d_1+{\gamma }_1)+\frac{{\beta }_1^2}{2{\sigma }_1^2}\hfill \\ +\frac{1}{t}\underset{0}{\overset{t}{\int }}\frac{{\sigma }_{1}S(s)f(I_1(s))}{I_1(s)}\mathrm{d}{B}_1(s)+\frac{\ln I_1(0)}{t}\hfill \end{array}$$

(9)

Defining a function as follows

$$D_1(u)=-\frac{{\sigma }_1^2}{2}{(u-\frac{{\beta }_1}{{\sigma }_1^2})}^2$$

It is easy to see that $D_1(u)$ is increasing monotonously for $0<u\le ({\beta }_1/{\sigma }_1^2)$ , and $D_1(u)$ is decreasing monotonously for $u>({\beta }_1/{\sigma }_1^2)$ . When ${\sigma }_1\le \sqrt{({\beta }_1\mu /f\prime (0)\mathrm{\Lambda })}$ , it follows that $((f\prime (0)\mathrm{\Lambda })/\mu )\le ({\beta }_1/{\sigma }_1^2)$ . From Lemma 1 and assumption (H1), one can get $((S(t)f(I_1(t)))/I_1(t))\in (0,((f\prime (0)\mathrm{\Lambda })/\mu )]$ for all $t\ge T_1$ . Then $D_1((S(t)f(I_1(t)))/I_1(t))\le D_1((f\prime (0)\mathrm{\Lambda })/\mu )$ for all $t\ge T_1$ .

Hence, when $t\ge T_1$ , from equation (9), one has

$$\begin{array}{c}\multicolumn{1}{c}{\frac{\ln I_1(t)}{t}\le -\frac{{\sigma }_1^2}{2}\frac{1}{t}\underset{0}{\overset{T_1}{\int }}{(\frac{S(s)f(I_1(s))}{I_1(s)}-\frac{{\beta }_1}{{\sigma }_1^2})}^2\mathrm{d}s-\frac{{\sigma }_1^2}{2}\frac{1}{t}\underset{T_1}{\overset{t}{\int }}{(\frac{\mathrm{\Lambda }f\prime (0)}{\mu }-\frac{{\beta }_1}{{\sigma }_1^2})}^2\mathrm{d}s}\\ -(\mu +d_1+{\gamma }_1)+\frac{{\beta }_1^2}{2{\sigma }_1^2}+\frac{1}{t}\underset{0}{\overset{t}{\int }}\frac{{\sigma }_{1}S(s)f(I_1(s))}{I_1(s)}\mathrm{d}{B}_1(s)+\frac{\ln I_1(0)}{t}\hfill \\ =-\frac{{\sigma }_1^2}{2}\frac{1}{t}{(\frac{\mathrm{\Lambda }f\prime (0)}{\mu }-\frac{{\beta }_1}{{\sigma }_1^2})}^2(t-T_1)-(\mu +d_1+{\gamma }_1)+\frac{{\beta }_1^2}{2{\sigma }_1^2}\hfill \\ -\frac{{\sigma }_1^2}{2}\frac{1}{t}\underset{0}{\overset{T_1}{\int }}{(\frac{S(s)f(I_1(s))}{I_1(s)}-\frac{{\beta }_1}{{\sigma }_1^2})}^2\mathrm{d}s+\frac{M_1(t)}{t}+\frac{\ln I_1(0)}{t}\hfill \end{array}$$

Then, one can obtain that

$$\begin{array}{c}\underset{t\to \infty }{lim\sup }\frac{\ln I_1(t)}{t}\le (\frac{\mathrm{\Lambda }{\beta }_{1}f\prime (0)}{\mu }-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^2{f\prime }^2(0)}{2{\mu }^2}-(\mu +d_1+{\gamma }_1))\hfill \\ =(\mu +d_1+{\gamma }_1)(\frac{\mathrm{\Lambda }{\beta }_{1}f\prime (0)}{\mu (\mu +d_1+{\gamma }_1)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^2{f\prime }^2(0)}{2{\mu }^2(\mu +d_1+{\gamma }_1)}-1)\hfill \\ =(\mu +d_1+{\gamma }_1)(R_1^s-1)<0,\mathrm{when}{R}_1^s<1\hfill \end{array}$$

It implies that $I_1(t)$ goes to extinction exponentially. Therefore

$$\underset{t\to \infty }{lim}{I}_1(t)=0$$

(10)

Finally, one concentrates on $I_2(t)$ . In the view of equation (7), dividing by t, then

$$\begin{array}{c}\multicolumn{1}{c}{\frac{\ln I_2(t)}{t}=-\frac{{\sigma }_2^2}{2}\frac{1}{t}\underset{0}{\overset{t}{\int }}{(g(S(s))-\frac{{\beta }_2}{{\sigma }_2^2})}^2\mathrm{d}s}\\ -(\mu +d_2+{\gamma }_2)+\frac{{\beta }_2^2}{2{\sigma }_2^2}\hfill \\ +\frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)+\frac{\ln I_2(0)}{t}\hfill \end{array}$$

(11)

Define a function as follows

$$D_2(u)=-\frac{{\sigma }_2^2}{2}{(u-\frac{{\beta }_2}{{\sigma }_2^2})}^2$$

Similarly, $D_2(u)$ is increasing monotonously for $0<u\le ({\beta }_2/{\sigma }_2^2)$ and $D_2(u)$ is decreasing monotonously for $u>({\beta }_2/{\sigma }_2^2)$ . When ${\sigma }_2\le \sqrt{({\beta }_2/(g(\mathrm{\Lambda }/\mu )))}$ , it leads to $g(\mathrm{\Lambda }/\mu )\le ({\beta }_2/{\sigma }_2^2)$ . From Lemma 1 and assumption (H2), one can have $g(S(t))\in (0,g(\mathrm{\Lambda }/\mu )]$ for all $t\ge T_1$ , then $D_2(g(S(t))\le D_2(g(\mathrm{\Lambda }/\mu ))$ for all $t\ge T_1$ .

When $t\ge T_1$ , from equation (11), it follows that

$$\begin{array}{c}\multicolumn{1}{c}{\frac{\ln I_2(t)}{t}\le -\frac{{\sigma }_2^2}{2}\frac{1}{t}\underset{0}{\overset{T_1}{\int }}{(g(S(s))-\frac{{\beta }_2}{{\sigma }_2^2})}^2\mathrm{d}s}\\ -\frac{{\sigma }_2^2}{2}\frac{1}{t}\underset{T_1}{\overset{t}{\int }}{(g(\frac{\mathrm{\Lambda }}{\mu })-\frac{{\beta }_2}{{\sigma }_2^2})}^2\mathrm{d}s\hfill \\ -(\mu +d_2+{\gamma }_2)+\frac{{\beta }_2^2}{2{\sigma }_2^2}\hfill \\ +\frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)+\frac{\ln I_2(0)}{t}\hfill \\ \multicolumn{1}{c}{=-\frac{{\sigma }_2^2}{2}\frac{1}{t}{(g(\frac{\mathrm{\Lambda }}{\mu })-\frac{{\beta }_2}{{\sigma }_2^2})}^2(t-T_1)-(\mu +d_2+{\gamma }_2)+\frac{{\beta }_2^2}{2{\sigma }_2^2}}\\ -\frac{{\sigma }_2^2}{2}\frac{1}{t}\underset{0}{\overset{T_1}{\int }}{(g(S(s))-\frac{{\beta }_2}{{\sigma }_2^2})}^2\mathrm{d}s+\frac{M_2(t)}{t}+\frac{\ln I_2(0)}{t}\hfill \end{array}$$

It leads to the following

$$\begin{array}{c}\multicolumn{1}{c}{\underset{t\to \infty }{lim\sup }\frac{\ln I_2(t)}{t}\le ({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2}-(\mu +d_2+{\gamma }_2))}\\ \multicolumn{1}{c}{=(\mu +d_2+{\gamma }_2)(\frac{{\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{\mu +d_2+{\gamma }_2}-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2(\mu +d_2+{\gamma }_2)}-1)}\\ \multicolumn{1}{c}{=(\mu +d_2+{\gamma }_2)(R_2^s-1)<0,\mathrm{when}{R}_2^s<1}\end{array}$$

This implies that $I_2(t)$ goes to extinction exponentially. Therefore

$$\underset{t\to \infty }{lim}{I}_2(t)=0$$

(12)

From the stochastic model (2), one can get

$$R(t)=e^{-(\mu +\delta )t}(R(0)+\underset{0}{\overset{t}{\int }}({\gamma }_1{I}_1(s)+{\gamma }_2{I}_2(s))e^{(\mu +\delta )s}\mathrm{d}s)$$

Taking the limit of the above equality, in the view of equations (10) and (12), it leads to

$$\underset{t\to \infty }{lim}R(t)=0$$

Directly from model (2), one can get

$$\mathrm{d}N(t)=[\mathrm{\Lambda }-\mu N(t)-(d_1{I}_1(t)+d_2{I}_2(t))]\mathrm{d}t$$

where $N(t)$ is given in equation (4).

Integrating this equation, we obtain

$$N(t)=(\underset{0}{\overset{t}{\int }}{e}^{\mu s}[\mathrm{\Lambda }-(d_1{I}_1(s)+d_2{I}_2(s))]\mathrm{d}s+N(0))e^{-\mu t}$$

Taking the limit of the above equality, from equations (10) and (12), then

$$\underset{t\to \infty }{lim}N(t)=\frac{\mathrm{\Lambda }}{\mu }-\underset{t\to \infty }{lim}\frac{d_1{I}_1(t)+d_2{I}_2(t)}{\mu }=\frac{\mathrm{\Lambda }}{\mu }$$

Hence, one has $\underset{t\to \infty }{lim}S(t)=(\mathrm{\Lambda }/\mu )$ . □

Remark 2

By Theorem 3, when ${\sigma }_i>({\beta }_i/\sqrt{2(\mu +d_i+{\gamma }_i)})$ (i = 1, 2), the diseases $I_1(t)$ and $I_2(t)$ for model (2) are extinct a.s. Accordingly, one can conclude that stochastic fluctuations can prevent the outbreak of diseases effectively.

Remark 3

When functions $f(I_1(t))=((I_1(t))/({\alpha }_1+I_1(t)))$ and $g(S(t))=((S(t))/({\alpha }_2+S(t)))$ , the stochastic model (2) degenerates into model (1) which has been investigated in Chang et al. ¹⁸ Compute that $R_1^s=(({\beta }_1\mathrm{\Lambda })/(\mu {\alpha }_1(\mu +d_1+>{\gamma }_1)))-(({\sigma }_1^2{\mathrm{\Lambda }}^2)/(2{\mu }^2{\alpha }_1^2(\mu +d_1+{\gamma }_1)))$ and $R_2^s=(({\beta }_2\mathrm{\Lambda })/((\mu {\alpha }_2+\mathrm{\Lambda })(\mu +d_2+{\gamma }_2)))-(({\sigma }_2^2{\mathrm{\Lambda }}^2)/(2(\mu {\alpha }_2+\mathrm{\Lambda }{)}^2(\mu +d_2+{\gamma }_2)))$ . Assume that

• ${\sigma }_i>({\beta }_i/(\sqrt{2(\mu +d_i+{\gamma }_i)}))$ , (i = 1, 2);

• $R_i^s<1$ and ${\sigma }_1\le \sqrt{(\mu {\beta }_1{\alpha }_1/\mathrm{\Lambda })}$ , ${\sigma }_2\le \sqrt{(({\beta }_2(\mu {\alpha }_2+\mathrm{\Lambda }))/\mathrm{\Lambda })}$ .

If condition (a) or (b) holds, then the two diseases of model (1) go to extinction.

This result has been obtained in Chang et al. ¹⁸ Obviously, Theorem 3 is an extension of Theorem 3.2 of Chang et al. ¹⁸

Persistence in mean

Theorem 4

For the stochastic model (2) with initial value $(S(0),I_1(0),I_2(0),R(0))\in {\mathbb{R}}_{+}^4$ , there are following properties

• (a) When $R_2^s<1$ , ${\sigma }_2\le \sqrt{({\beta }_2/(g(\mathrm{\Lambda }/\mu )))}$ and $R_1^s>1$ , the disease $I_1(t)$ to be persistent in mean while disease $I_2(t)$ dies out. Furthermore

$$\underset{t\to \infty }{lim\inf }〈I_1(t)〉\ge (f\prime (0)W_1{)}^{-1}(\mu +d_1+{\gamma }_1)(R_1^s-1),a.s.$$

• (b) When $R_2^s>1,R_1^s<1$ and ${\sigma }_1\le \sqrt{({\beta }_1\mu /f\prime (0)\mathrm{\Lambda })}$ , the disease $I_1(t)$ goes to extinction and the disease $I_2(t)$ is persistent in mean. Furthermore

$$\underset{t\to \infty }{lim\inf }〈I_2(t)〉\ge W_2^{-1}(\mu +d_2+{\gamma }_2)(R_2^s-1),a.s.$$

• (c) When $R_1^s>1$ and $R_2^s>1$ , the two diseases $I_1(t)$ and $I_2(t)$ are persistent in mean a.s. Furthermore, one gets

$$\begin{array}{c}{lim\inf }_{t\to \infty }(〈I_1(t)〉+〈I_2(t)〉)\ge {W_3}^{-1}\hfill \\ [{(f\prime (0))}^{-1}(\mu +d_1+{\gamma }_1)(R_1^s-1)+(\mu +d_2+{\gamma }_2)(R_2^s-1)],a.s.\hfill \end{array}$$

where

$$\begin{array}{c}\multicolumn{1}{c}{R_1^s=(\frac{{\beta }_1\mathrm{\Lambda }f\prime (0)}{\mu (\mu +d_1+{\gamma }_1)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^2{f\prime }^2(0)}{2{\mu }^2(\mu +d_1+{\gamma }_1)})}\\ \multicolumn{1}{c}{R_2^s=(\frac{{\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{\mu +d_2+{\gamma }_2}-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2(\mu +d_2+{\gamma }_2)})}\\ \multicolumn{1}{c}{W_1={\beta }_1(\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta })+(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_1\le \frac{\mathrm{\Lambda }}{\mu }}\frac{f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1)}{f^2({\xi }_1)}}\\ \multicolumn{1}{c}{W_2={\beta }_2\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_2\le \frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2)(\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta })}\\ \multicolumn{1}{c}{W_3=max\{({\beta }_1+{\beta }_2\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_2\le \frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2))(\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta })}\\ +(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_1\le \frac{\mathrm{\Lambda }}{\mu }}\frac{f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1)}{f^2({\xi }_1),}({\beta }_1+{\beta }_2\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_2\le \frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2))(\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta })\}\hfill \end{array}$$

Proof

By integrating of the model (2), we obtain

$$\begin{array}{c}\frac{1}{t}(S(t)-S(0))=\mathrm{\Lambda }-{\beta }_1〈S(t)f(I_1(t))〉-{\beta }_2〈I_2(t)g(S(t))〉-\mu 〈S(t)〉\hfill \\ +\delta 〈R(t)〉-\frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_{1}S(s)f(I_1(s))\mathrm{d}{B}_1(s)\hfill \\ -\frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_2{I}_2(s)g(S(s))\mathrm{d}{B}_2(s)\hfill \\ \frac{1}{t}(I_1(t)-I_1(0))={\beta }_1〈S(t)f(I_1(t))〉-(\mu +d_1+{\gamma }_1)〈I_1(t)〉\hfill \\ +\frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_{1}S(s)f(I_1(s))\mathrm{d}{B}_1(s)\hfill \\ \frac{1}{t}(I_1(t)-I_1(0))={\beta }_1〈S(t)f(I_1(t))〉-(\mu +d_1+{\gamma }_1)〈I_1(t)〉\hfill \\ +\frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_{1}S(s)f(I_1(s))\mathrm{d}{B}_1(s)\hfill \\ \frac{1}{t}(I_2(t)-I_2(0))={\beta }_2〈I_2(t)g(S(t))〉-(\mu +d_2+{\gamma }_2)〈I_2(t)〉\hfill \\ +\frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_2{I}_2(s)g(S(s))\mathrm{d}{B}_2(s)\hfill \\ \frac{1}{t}(R(t)-R(0))={\gamma }_1〈I_1(t)〉+{\gamma }_2〈I_2(t)〉-(\mu +\delta )〈R(t)〉\hfill \end{array}$$

It leads to

$$\begin{array}{c}\frac{S(t)-S(0)}{t}+\frac{I_1(t)-I_1(0)}{t}+\frac{I_2(t)-I_2(0)}{t}\hfill \\ +\frac{\delta }{\mu +\delta }\frac{R(t)-R(0)}{t}\hfill \\ =\mathrm{\Lambda }-(\mu +d_1+\frac{{\gamma }_1\mu }{\mu +\delta })〈I_1(t)〉\hfill \\ -(\mu +d_2+\frac{{\gamma }_2\mu }{\mu +\delta })〈I_2(t)〉-\mu 〈S(t)〉\hfill \end{array}$$

Then

$$\begin{array}{c}\multicolumn{1}{c}{\frac{\mathrm{\Lambda }}{\mu }-〈S(t)〉=(\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta })〈I_1(t)〉}\\ +(\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta })〈I_2(t)〉+\Theta (t)\hfill \end{array}$$

(13)

where

$$\Theta (t)=(\frac{1}{t}(S(t)-S(0))+\frac{1}{t}(I_1(t)-I_1(0))+\frac{1}{t}(I_2(t)-I_2(0))+\frac{\delta (R(t)-R(0))}{t(\mu +\delta )}){\mu }^{-1}$$

From Theorem 3, one can get that if ${\sigma }_1\le \sqrt{({\beta }_1\mu /f\prime (0)\mathrm{\Lambda })}$ and $R_1^s<1$ , then $\underset{t\to \infty }{lim}{I}_1(t)=0$ . Thus, for any ${\epsilon }_1>0$ , there exists a constant $T_2>0$ , such that $0<I_1(t)\le {\epsilon }_1$ , for $t\ge T_2$ . Then, for $t\ge T_2$

$$\begin{array}{c}\multicolumn{1}{c}{\frac{\mathrm{\Lambda }}{\mu }-〈S(t)〉\le (\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta }){\epsilon }_1}\\ +(\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta })〈I_2(t)〉+\Theta (t)\hfill \end{array}$$

(14)

Analogously, if ${\sigma }_2\le \sqrt{({\beta }_2/(g(\mathrm{\Lambda }/\mu )))}$ and $R_2^s<1$ , then $\underset{t\to \infty }{lim}{I}_2(t)=0$ . Hence, for any ${\epsilon }_2>0$ , there exists a constant $T_3>0$ , such that $0<I_2(t)\le {\epsilon }_2$ , for $t\ge T_3$ . It leads to the following, for $t\ge T_3$

$$\begin{array}{c}\multicolumn{1}{c}{\frac{\mathrm{\Lambda }}{\mu }-〈S(t)〉\le (\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta })〈I_1(t)〉}\\ +(\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta }){\epsilon }_2+\Theta (t)\hfill \end{array}$$

(15)

Choose $T_4=\max \{T_1,T_2,T_3\}$ .

• Case (a).

Define a Liapunov function

$$U_1(t)=\underset{I_1(0)}{\overset{I_1(t)}{\int }}\frac{1}{f(I_1(s))}\mathrm{d}{I}_1(s)$$

(16)

Applying Itô’s formula, one has

$$\begin{array}{c}\mathrm{d}{U}_1(t)=\hfill \\ ({\beta }_{1}S(t)-(\mu +d_1+{\gamma }_1)\frac{I_1(t)}{f(I_1(t))}-\frac{{\sigma }_1^2{S}^2(t)f\prime (I_1(t))}{2})\mathrm{d}t\hfill \\ +{\sigma }_{1}S(t)\mathrm{d}{B}_1(t)\hfill \end{array}$$

From assumption (H1), one can see that $(f(I_1(t))/I_1(t))$ is decreasing monotonically, it follows that $f\prime (I_1(t))<(f(I_1(t))/I_1(t))<f\prime (0)$ . When $t\ge T_4$ , then

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}{U}_1(t)\ge ({\beta }_{1}S(t)-(\mu +d_1+{\gamma }_1)\frac{I_1(t)}{f(I_1(t))}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})\mathrm{d}t}\\ +{\sigma }_{1}S(t)\mathrm{d}{B}_1(t)\hfill \end{array}$$

By integrating, we obtain

$$\begin{array}{c}\frac{1}{t}(U_1(t)-U_1(T_4))\hfill \\ \ge \frac{1}{t}\underset{T_4}{\overset{t}{\int }}({\beta }_{1}S(s)-(\mu +d_1+{\gamma }_1)\frac{I_1(s)}{f(I_1(s))}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})\mathrm{d}s\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)\hfill \\ =\frac{1}{t}({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})(t-T_4)\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}({\beta }_1(S(s)-\frac{\mathrm{\Lambda }}{\mu })-(\mu +d_1+{\gamma }_1)(\frac{I_1(s)}{f(I_1(s))}-\frac{1}{f\prime (0)}))\mathrm{d}s\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)\hfill \end{array}$$

(17)

Define a function $F(I_1(t))=(I_1(t)/f(I_1(t))),I_1(t)>0$ , from assumption (H1), one gets $F(0)=\underset{I_1(t)\to 0}{lim}F(I_1(t))=(1/f\prime (0))$ . By mean value theorem to $F(I_1(t))-F(0)$ of equation (17), for $t\ge T_4$ , one can obtain

$$\begin{array}{c}\frac{1}{t}(U_1(t)-U_1(T_4))\hfill \\ \ge \frac{1}{t}({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})(t-T_4)\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}({\beta }_1(S(s)-\frac{\mathrm{\Lambda }}{\mu })-(\mu +d_1+{\gamma }_1)F\prime ({\xi }_1)I_1(s))\mathrm{d}s\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)\hfill \\ \ge \frac{1}{t}({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})(t-T_4)\hfill \\ -(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_1\le \frac{\mathrm{\Lambda }}{\mu }}F\prime ({\xi }_1)(〈I_1(t)〉-\frac{1}{t}\underset{0}{\overset{T_4}{\int }}{I}_1(s)\mathrm{d}s)\hfill \\ -{\beta }_1(\frac{\mathrm{\Lambda }}{\mu }-〈S(t)〉-\frac{T_4\mathrm{\Lambda }}{t\mu }+\frac{1}{t}\underset{0}{\overset{T_4}{\int }}S(s)\mathrm{d}s)+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)\hfill \end{array}$$

(18)

Where $F\prime ({\xi }_1)=((f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1))/(f^2({\xi }_1)))$ , since

$$\underset{I_1(t)\to 0}{lim}\frac{f(I_1(t))-I_1(t)f\prime (I_1(t))}{f^2(I_1(t))}=-\frac{f″(0)}{2{f\prime }^2(0)}<\infty$$

Then

$$0\le \mathrm{ma}{\mathrm{x}}_{0\le {\xi }_1\le \frac{\mathrm{\Lambda }}{\mu }}\frac{f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1)}{f^2({\xi }_1)}<\infty$$

Therefore, $0\le \mathrm{ma}{\mathrm{x}}_{0\le {\xi }_1\le \frac{\mathrm{\Lambda }}{\mu }}F\prime ({\xi }_1)<\infty$ .

Combining equations (15) and (18), for $t\ge T_4$ , we obtain

$$\begin{array}{c}\frac{1}{t}(U_1(t)-U_1(T_4))\hfill \\ \ge \frac{1}{t}({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})(t-T_4)\hfill \\ -(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_1\le \frac{\mathrm{\Lambda }}{\mu }}F\prime ({\xi }_1)(〈I_1(t)〉-\frac{1}{t}\underset{0}{\overset{T_4}{\int }}{I}_1(s)\mathrm{d}s)\hfill \\ -{\beta }_1((\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta }){\epsilon }_2+\Theta (t)-\frac{T_4\mathrm{\Lambda }}{t\mu }+\frac{1}{t}\underset{0}{\overset{T_4}{\int }}S(s)\mathrm{d}s)\hfill \\ -{\beta }_1(\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta })〈I_1(t)〉+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)\hfill \end{array}$$

(19)

When $t\ge T_4$ , from equation (16), one gets

$$\frac{U_1(t)}{t}=\frac{1}{t}\underset{I_1(0)}{\overset{I_1(t)}{\int }}\frac{1}{f(I_1(s))}\mathrm{d}{I}_1(s)\le \frac{1}{t}\underset{I_1(0)}{\overset{\frac{\mathrm{\Lambda }}{\mu }}{\int }}\frac{1}{f(I_1(s))}\mathrm{d}{I}_1(s)$$

In the view of equation (16)

$$〈I_1(t)〉\ge W_1^{-1}(C_1+{\mathrm{\Psi }}_1(t))$$

(20)

where

$$\begin{array}{c}{C}_1=\frac{1}{t}({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})(t-T_4)\hfill \\ {\mathrm{\Psi }}_1(t)=-{\beta }_1((\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta }){\epsilon }_2+\Theta (t)-\frac{T_4\mathrm{\Lambda }}{t\mu }+\frac{1}{t}\underset{0}{\overset{T_4}{\int }}S(s)\mathrm{d}s)\hfill \\ +(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_1\le \frac{\mathrm{\Lambda }}{\mu }}\frac{f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1)}{f^2({\xi }_1)}\frac{1}{t}\underset{0}{\overset{T_4}{\int }}{I}_1(s)\mathrm{d}s\hfill \\ -\frac{1}{t}\underset{I_1(0)}{\overset{\frac{\mathrm{\Lambda }}{\mu }}{\int }}\frac{1}{f(I_1(s))}\mathrm{d}{I}_1(s)+\frac{1}{t}\underset{I_1(0)}{\overset{I_1(T_4)}{\int }}\frac{1}{f(I_1(s))}\mathrm{d}{I}_1(s)+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)\hfill \\ W_1={\beta }_1(\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta })\hfill \\ +(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_1\le \frac{\mathrm{\Lambda }}{\mu }}\frac{f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1)}{f^2({\xi }_1)}\hfill \end{array}$$

Let $M_3(t)={\int }_{T_4}^t{\sigma }_{2}S(s)\mathrm{d}{B}_2(s)$ , since

$$\begin{array}{c}\frac{{〈M_3,M_3〉}_t}{t}=\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_2^2{S}^2(s)\mathrm{d}s\hfill \\ \le \frac{1}{t}\underset{0}{\overset{t}{\int }}{\sigma }_2^2{k}_1^2\mathrm{d}s={\sigma }_2^2{k}_1^2<+\infty \hfill \end{array}$$

Thus, one gets

$$\underset{t\to \infty }{lim}\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)=0$$

Since $\underset{t\to \infty }{lim}\Theta (t)=0$ and the arbitrariness of ${\epsilon }_2$ , letting ${\epsilon }_2\to 0$ . Then, $\underset{t\to \infty }{lim}{\mathrm{\Psi }}_1(t)=0$ .

Taking the limit of equation (20) and letting $t\to \infty$ , one finally has

$$\begin{array}{c}\underset{t\to \infty }{lim\inf }〈I_1(t)〉\ge W_1^{-1}\hfill \\ ({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})\hfill \\ =(f\prime (0)W_1{)}^{-1}(\mu +d_1+{\gamma }_1)(R_1^s-1)>0,\mathrm{when}{R}_1^s>1.\hfill \end{array}$$

• Case (b).

Define a Liapunov function

$$U_2(t)=\ln I_2(t).$$

Applying Itô’s formula, one gets

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}{U}_2(t)=({\beta }_{2}g(S(t))-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2(S(t))}{2})\mathrm{d}t}\\ +{\sigma }_{2}g(S(t))\mathrm{d}{B}_2(t)\hfill \end{array}$$

When $t\ge T_4$ , then

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}{U}_2(t)\ge ({\beta }_{2}g(S(t))-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})\mathrm{d}t}\\ +{\sigma }_{2}g(S(t))\mathrm{d}{B}_2(t)\hfill \end{array}$$

By integrating, we obtain

$$\begin{array}{c}\frac{1}{t}(U_2(t)-U_2(T_4))\hfill \\ \ge \frac{1}{t}\underset{T_4}{\overset{t}{\int }}({\beta }_{2}g(S(s))-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})\mathrm{d}s\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \\ =\frac{1}{t}({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})(t-T_4)\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\beta }_2(g(S(s))-g\left(\frac{\mathrm{\Lambda }}{\mu }\right))\mathrm{d}s+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \end{array}$$

By mean value theorem $g(S(s))-g(\mathrm{\Lambda }/\mu )$ of the above inequality, one can obtain

$$\begin{array}{c}\frac{1}{t}(U_2(t)-U_2(T_4))\hfill \\ \ge \frac{1}{t}({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})(t-T_4)\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\beta }_{2}g\prime ({\xi }_2)(S(s)-\frac{\mathrm{\Lambda }}{\mu })\mathrm{d}s+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \\ \ge \frac{1}{t}({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})(t-T_4)+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \\ -{\beta }_2\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_2\le \frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2)(\frac{\mathrm{\Lambda }}{\mu }-〈S(t)〉-\frac{T_4\mathrm{\Lambda }}{t\mu }+\frac{1}{t}\underset{0}{\overset{T_4}{\int }}S(s)\mathrm{d}s)\hfill \end{array}$$

(21)

Combining equations (14) and (21), for $t\ge T_4$ , we obtain

$$\begin{array}{c}\frac{1}{t}(U_2(t)-U_2(T_4))\hfill \\ \ge \frac{1}{t}({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})(t-T_4)\hfill \\ -{\beta }_2\mathrm{ma}{\mathrm{x}}_{0<{\xi }_2<\frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2)((\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta }){\epsilon }_1+\Theta (t)-\frac{T_4\mathrm{\Lambda }}{t\mu }+\frac{1}{t}\underset{0}{\overset{T_4}{\int }}S(s)\mathrm{d}s)\hfill \\ -{\beta }_2\mathrm{ma}{\mathrm{x}}_{0<{\xi }_2<\frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2)(\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta })〈I_2(t)〉+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \end{array}$$

It leads to

$$〈I_2(t)〉\ge W_2^{-1}(C_2+{\mathrm{\Psi }}_2(t)),$$

(22)

where

$$\begin{array}{c}{C}_2=\frac{1}{t}({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})(t-T_4)\hfill \\ W_2={\beta }_2\mathrm{ma}{\mathrm{x}}_{0<{\xi }_2<\frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2)(\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta })\hfill \\ {\mathrm{\Psi }}_2(t)=-{\beta }_2\mathrm{ma}{\mathrm{x}}_{0<{\xi }_2<\frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2)\hfill \\ ((\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta }){\epsilon }_1+\Theta (t)-\frac{T_4\mathrm{\Lambda }}{t\mu }+\frac{1}{t}\underset{0}{\overset{T_4}{\int }}S(s)\mathrm{d}s)\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)-\frac{\ln I_2(t)-\ln I_2(T_4)}{t}\hfill \end{array}$$

Similarly, one also gets

$$\underset{t\to \infty }{lim}\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)=0$$

From Lemma 1, one can obtain $I_2(t)\le (\mathrm{\Lambda }/\mu )$ , for $t\ge T_4$ , then $\underset{t\to \infty }{lim}((\ln I_2(t)-\ln I_2(T_4))/t)=0$ , for $I_2(t)>0$ . Since the arbitrariness of ${\epsilon }_1$ , letting ${\epsilon }_1\to 0$ . Hence, one has $\underset{t\to \infty }{lim}{\mathrm{\Psi }}_2(t)=0$ .

Taking the limit of (22) and letting $t\to \infty$ , one finally obtains

$$\begin{array}{c}{lim\inf }_{t\to \infty }〈I_2(t)〉\hfill \\ \ge W_2^{-1}({\beta }_{2}g(\mathrm{\Lambda }/\mu )-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2(\mathrm{\Lambda }/\mu )}{2})\hfill \\ =W_2^{-1}(\mu +d_2+{\gamma }_2)(R_2^s-1)>0,\mathrm{when}{R}_2^s>1\hfill \end{array}$$

• Case (c).

Define a Liapunov function

$$U_3(t)=\underset{I_1(0)}{\overset{I_1(t)}{\int }}\frac{1}{f(I_1(s))}\mathrm{d}{I}_1(s)+\ln I_2(t)$$

Applying Itô’s formula, one has

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}{U}_3(t)=({\beta }_{1}S(t)-(\mu +d_1+{\gamma }_1)\frac{I_1(t)}{f(I_1(t))}-\frac{{\sigma }_1^2{S}^2(t)f\prime (I_1(t))}{2})\mathrm{d}t}\\ +({\beta }_{2}g(S(t))-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2(S(t))}{2})\mathrm{d}t\hfill \\ +{\sigma }_{1}S(t)\mathrm{d}{B}_1(t)+{\sigma }_{2}g(S(t))\mathrm{d}{B}_2(t).\hfill \end{array}$$

When $t\ge T_4$ , then

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}{U}_3(t)\ge ({\beta }_{1}S(t)-(\mu +d_1+{\gamma }_1)\frac{I_1(t)}{f(I_1(t))}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})\mathrm{d}t}\\ +({\beta }_{2}g(S(t))-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})\mathrm{d}t\hfill \\ +{\sigma }_{1}S(t)\mathrm{d}{B}_1(t)+{\sigma }_{2}g(S(t))\mathrm{d}{B}_2(t)\hfill \end{array}$$

By integrating, we obtain

$$\begin{array}{c}\frac{1}{t}(U_3(t)-U_3(T_4))\hfill \\ \ge \frac{1}{t}\underset{T_4}{\overset{t}{\int }}({\beta }_{1}S(s)-(\mu +d_1+{\gamma }_1)\frac{I_1(s)}{f(I_1(s))}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})\mathrm{d}s\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}({\beta }_{2}g(S(s))-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})\mathrm{d}s\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \\ =\frac{1}{t}({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})(t-T_4)\hfill \\ +\frac{1}{t}{\int }_{T_4}^t({\beta }_1(S(s)-\frac{\mathrm{\Lambda }}{\mu })-(\mu +d_1+{\gamma }_1)(\frac{I_1(s)}{f(I_1(s))}-\frac{1}{f\prime (0)}))\mathrm{d}s\hfill \\ +\frac{1}{t}({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})(t-T_4)\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \\ \frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\beta }_2(g(S(s))-g\left(\frac{\mathrm{\Lambda }}{\mu }\right))\mathrm{d}s\hfill \end{array}$$

Similarly, applying Lagrange’s mean value theorem of the above inequality, in the view of equation (13), one can obtain

$$\begin{array}{c}\frac{1}{t}(U_3(t)-U_3(T_4))\hfill \\ \ge \frac{1}{t}({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})(t-T_4)\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \\ +\frac{1}{t}({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})(t-T_4)\hfill \\ -[({\beta }_1+{\beta }_2\mathrm{ma}{\mathrm{x}}_{0<{\xi }_2<\frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2))(\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta })\hfill \\ +(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0<{\xi }_1<\frac{\mathrm{\Lambda }}{\mu }}F\prime ({\xi }_1)]\times 〈I_1(t)〉\hfill \end{array}$$

$$\begin{array}{c}-({\beta }_1+{\beta }_2\mathrm{ma}{\mathrm{x}}_{0<{\xi }_2<\frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2))(\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta })\times 〈I_2(t)〉\hfill \\ -({\beta }_1+{\beta }_2\mathrm{ma}{\mathrm{x}}_{0<{\xi }_2<\frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2))(\Theta (t)-\frac{T_4\mathrm{\Lambda }}{t\mu }+\frac{1}{t}\underset{0}{\overset{T_4}{\int }}S(s)\mathrm{d}s)\hfill \\ +(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0<{\xi }_1<\frac{\mathrm{\Lambda }}{\mu }}\frac{f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1)}{f^2({\xi }_1)}\frac{1}{t}\underset{0}{\overset{T_4}{\int }}{I}_1(s)\mathrm{d}s\hfill \\ \ge \frac{1}{t}({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})(t-T_4)\hfill \\ +\frac{1}{t}({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})(t-T_4)\hfill \\ +\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \\ -({\beta }_1+{\beta }_2\mathrm{ma}{\mathrm{x}}_{0<{\xi }_2<\frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2))(\Theta (t)-\frac{T_4\mathrm{\Lambda }}{t\mu }+\frac{1}{t}\underset{0}{\overset{T_4}{\int }}S(s)\mathrm{d}s)\hfill \\ +(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0<{\xi }_1<\frac{\mathrm{\Lambda }}{\mu }}\frac{f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1)}{f^2({\xi }_1)}\frac{1}{t}\underset{0}{\overset{T_4}{\int }}{I}_1(s)\mathrm{d}s\hfill \\ -W_3[〈I_1(t)〉+〈I_2(t)〉]\hfill \end{array}$$

Accordingly

$$〈I_1(t)〉+〈I_2(t)〉\ge W_3^{-1}(C_3+{\mathrm{\Psi }}_3(t)),$$

(23)

where

$$\begin{array}{c}\multicolumn{1}{c}{C_3=\frac{1}{t}({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})(t-T_4)}\\ +\frac{1}{t}({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})(t-T_4)\hfill \\ \multicolumn{1}{c}{{\mathrm{\Psi }}_3(t)=-({\beta }_1+{\beta }_2\mathrm{ma}{\mathrm{x}}_{0<{\xi }_2<\frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2))}\\ (\Theta (t)-\frac{T_4\mathrm{\Lambda }}{t\mu }+\frac{1}{t}\underset{0}{\overset{T_4}{\int }}S(s)\mathrm{d}s)+\frac{1}{t}\underset{T_4}{\overset{t}{\int }}{\sigma }_{1}S(s)\mathrm{d}{B}_1(s)+\frac{1}{t}\underset{T_1}{\overset{t}{\int }}{\sigma }_{2}g(S(s))\mathrm{d}{B}_2(s)\hfill \\ +(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0<{\xi }_1<\frac{\mathrm{\Lambda }}{\mu }}\frac{f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1)}{f^2({\xi }_1)}\frac{1}{t}\underset{0}{\overset{T_4}{\int }}{I}_1(s)\mathrm{d}s\hfill \\ -\frac{1}{t}\underset{I_1(0)}{\overset{\frac{\mathrm{\Lambda }}{\mu }}{\int }}\frac{1}{f(I_1(s))}\mathrm{d}{I}_1(s)+\frac{1}{t}\underset{I_1(0)}{\overset{I_1(T_4)}{\int }}\frac{1}{f(I_1(s))}\mathrm{d}{I}_1(s)-\frac{\ln I_2(t)-\ln I_2(T_4)}{t}\hfill \\ \multicolumn{1}{c}{W_3=max\{({\beta }_1+{\beta }_2\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_2\le \frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2))(\frac{\mu +d_1}{\mu }+\frac{{\gamma }_1}{\mu +\delta })}\\ +(\mu +d_1+{\gamma }_1)\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_1\le \frac{\mathrm{\Lambda }}{\mu }}\frac{f({\xi }_1)-{\xi }_{1}f\prime ({\xi }_1)}{f^2({\xi }_1),}({\beta }_1+{\beta }_2\mathrm{ma}{\mathrm{x}}_{0\le {\xi }_2\le \frac{\mathrm{\Lambda }}{\mu }}g\prime ({\xi }_2))(\frac{\mu +d_2}{\mu }+\frac{{\gamma }_2}{\mu +\delta })\}.\hfill \end{array}$$

In the view of Cases (a) and (b), one obtains $\underset{t\to \infty }{lim}{\mathrm{\Psi }}_3(t)=0$ . Taking the limit of equation (23) and letting $t\to \infty$ , one finally obtains

$$\begin{array}{c}\underset{t\to \infty }{lim\inf }(〈I_1(t)〉+〈I_2(t)〉)\ge W_3^{-1}\hfill \\ [({\beta }_1\frac{\mathrm{\Lambda }}{\mu }-(\mu +d_1+{\gamma }_1)\frac{1}{f\prime (0)}-\frac{{\sigma }_1^2{\mathrm{\Lambda }}^{2}f\prime (0)}{2{\mu }^2})\hfill \\ +({\beta }_{2}g\left(\frac{\mathrm{\Lambda }}{\mu }\right)-(\mu +d_2+{\gamma }_2)-\frac{{\sigma }_2^2{g}^2\left(\frac{\mathrm{\Lambda }}{\mu }\right)}{2})]\hfill \\ =W_3^{-1}[{(f\prime (0))}^{-1}(\mu +d_1+{\gamma }_1)(R_1^s-1)\hfill \\ +(\mu +d_2+{\gamma }_2)(R_2^s-1)]>0\hfill \end{array}$$

when $R_1^s>1$ and $R_2^s>1$ . □

Almost sure exponential stability

It is easy to see that model (2) has a disease-free equilibrium $E_0=(\mathrm{\Lambda }/\mu ,0,0,0)$ . In this section, a few discussion about the almost sure exponentially stable for the disease-free equilibrium $E_0=(\mathrm{\Lambda }/\mu ,0,0,0)$ of model (2) are carried out. Finally, a sufficient condition for the almost sure exponentially stable disease-free equilibrium $E_0$ is established.

Theorem 5

If ${\sigma }_1^2>({\beta }_1^2/\mu )$ and ${\sigma }_2^2>({\beta }_2^2/\mu )$ hold, then the disease-free equilibrium $E_0=(\mathrm{\Lambda }/\mu ,0,0,0)$ of model (2) is almost sure exponentially stable in $\mathrm{\Gamma }$ . Where

$$\begin{array}{c}\mathrm{\Gamma }=\hfill \\ \{(S,I_1,I_2,R):S\ge 0,I_1\ge 0,I_2\ge 0,R\ge 0,0<S+I_1+I_2+R\le \frac{\mathrm{\Lambda }}{\mu }\}\hfill \end{array}$$

is a positive invariant set of model (2).

Proof

The proof is similar to Zhou et al. ³⁰ One defines the function

$$V=\ln [(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R]$$

Applying Itô’s formula, one has

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}V=\frac{1}{(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R}[-\mathrm{\Lambda }+\mu S+2{\beta }_{1}f(I_1)S+2{\beta }_{2}g(S)I_2-(\mu +2\delta )R-(\mu +d_1)I_1-(\mu +d_2)I_2]\mathrm{d}t}\\ -\frac{2{\sigma }_1^2{f}^2(I_1)S^2+2{\sigma }_2^2{g}^2(S)I_2^2}{{[(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R]}^2}\mathrm{d}t+\frac{2{\sigma }_{1}f(I_1)S}{(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R}\mathrm{d}{B}_1(t)+\frac{2{\sigma }_{2}g(S)I_2}{(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R}\mathrm{d}{B}_2(t)\hfill \\ =\frac{1}{(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R}[-\frac{\mathrm{\Lambda }}{2}+\frac{\mu S}{2}+2{\beta }_{1}f(I_1)S-\frac{(\mu +2\delta )R}{2}-\frac{(\mu +d_1)I_1}{2}-\frac{(\mu +d_2)I_2}{2}]\mathrm{d}t\hfill \\ -\frac{2{\sigma }_1^2{f}^2(I_1)S^2}{{[(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R]}^2}\mathrm{d}t+\frac{2{\sigma }_{1}f(I_1)S}{(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R}\mathrm{d}{B}_1(t)\hfill \\ +\frac{1}{(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R}[-\frac{\mathrm{\Lambda }}{2}+\frac{\mu S}{2}+2{\beta }_{2}g(S)I_2-\frac{(\mu +2\delta )R}{2}-\frac{(\mu +d_1)I_1}{2}-\frac{(\mu +d_2)I_2}{2}]\mathrm{d}t\hfill \\ -\frac{2{\sigma }_2^2{g}^2(S)I_2^2}{{[(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R]}^2}\mathrm{d}t+\frac{2{\sigma }_{2}g(S)I_2}{(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R}\mathrm{d}{B}_2(t)\hfill \end{array}$$

Let $Y_1=\frac{f(I_1)S}{(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R}$ , $Y_2=\frac{g(S)I_2}{(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R}$ .

Then, one obtains

$$\begin{array}{c}\multicolumn{1}{c}{\mathrm{d}V=[-2{\sigma }_1^2{Y}_1+2{\beta }_1{Y}_1-\frac{\mu (\frac{\mathrm{\Lambda }}{\mu }-S)+(\mu +2\delta )R+(\mu +d_1)I_1+(\mu +d_2)I_2}{2[(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R]}]\mathrm{d}t}\\ +[-2{\sigma }_2^2{Y}_2+2{\beta }_2{Y}_2-\frac{\mu (\frac{\mathrm{\Lambda }}{\mu }-S)+(\mu +2\delta )R+(\mu +d_1)I_1+(\mu +d_2)I_2}{2[(\frac{\mathrm{\Lambda }}{\mu }-S)+I_1+I_2+R]}]\mathrm{d}t\hfill \\ +2{\sigma }_1{Y}_1\mathrm{d}{B}_1(t)+2{\sigma }_2{Y}_2\mathrm{d}{B}_2(t)\hfill \\ \le [-2{\sigma }_1^2{Y}_1+2{\beta }_1{Y}_1-\frac{\mu }{2}]\mathrm{d}t+[-2{\sigma }_2^2{Y}_2+2{\beta }_2{Y}_2-\frac{\mu }{2}]\mathrm{d}t\hfill \\ +2{\sigma }_1{Y}_1\mathrm{d}{B}_1(t)+2{\sigma }_2{Y}_2\mathrm{d}{B}_2(t)\hfill \\ \le [\frac{{\beta }_1^2-\mu {\sigma }_1^2}{2{\sigma }_1^2}+\frac{{\beta }_2^2-\mu {\sigma }_2^2}{2{\sigma }_2^2}]\mathrm{d}t+2{\sigma }_1{Y}_1\mathrm{d}{B}_1(t)+2{\sigma }_2{Y}_2\mathrm{d}{B}_2(t)\hfill \end{array}$$

Integrating both sides of the above inequation from 0 to t, one has

$$\begin{array}{c}\ln [(\frac{\mathrm{\Lambda }}{\mu }-S(t))+I_1(t)+I_2(t)+R(t)]\hfill \\ \le \ln [(\frac{\mathrm{\Lambda }}{\mu }-S(0))+I_1(0)+I_2(0)+R(0)]\hfill \\ +[\frac{{\beta }_1^2-\mu {\sigma }_1^2}{2{\sigma }_1^2}+\frac{{\beta }_2^2-\mu {\sigma }_2^2}{2{\sigma }_2^2}]t\hfill \\ +\underset{0}{\overset{t}{\int }}2{\sigma }_1{Y}_1(s)\mathrm{d}{B}_1(s)+\underset{0}{\overset{t}{\int }}2{\sigma }_2{Y}_2(s)\mathrm{d}{B}_2(s)\hfill \end{array}$$

By the strong law of large number for local martingales, one gets

$$\underset{t\to \infty }{lim}\frac{1}{t}\underset{0}{\overset{t}{\int }}2{\sigma }_1{Y}_1(s)\mathrm{d}{B}_1(s)=0,\underset{t\to \infty }{lim}\frac{1}{t}\underset{0}{\overset{t}{\int }}2{\sigma }_2{Y}_2(s)\mathrm{d}{B}_2(s)=0$$

Under the condition that ${\sigma }_1^2>({\beta }_1^2/\mu )$ and ${\sigma }_2^2>({\beta }_2^2/\mu )$ , one finally has

$$\begin{array}{c}\underset{t\to \infty }{lim\sup }\frac{1}{t}\ln [(\frac{\mathrm{\Lambda }}{\mu }-S(t))+I_1(t)+I_2(t)+R(t)]\hfill \\ \le \frac{{\beta }_1^2-\mu {\sigma }_1^2}{2{\sigma }_1^2}+\frac{{\beta }_2^2-\mu {\sigma }_2^2}{2{\sigma }_2^2}<0\hfill \end{array}$$

Numerical simulations and examples

Now, numerical simulations can be demonstrated by the findings obtained. One considers functions $f(I_1(t))=(I_1(t)/e^{I_1(t)})$ , and $g(S(t))=(S(t)/(1+S(t)))$ . Motivated by Cui et al. ³¹ Then, the corresponding deterministic model (2) is

$$\{\begin{array}{c}\frac{\mathrm{d}S(t)}{\mathrm{d}t}=\mathrm{\Lambda }-\mu S(t)-\frac{{\beta }_1{I}_1(t)S(t)}{e^{I_1(t)}}-\frac{{\beta }_2{I}_2(t)S(t)}{1+S(t)}+\delta R(t)\hfill \\ \frac{\mathrm{d}{I}_1(t)}{\mathrm{d}t}=\frac{{\beta }_1{I}_1(t)S(t)}{e^{I_1(t)}}-(\mu +d_1+{\gamma }_1)I_1(t)\hfill \\ \frac{\mathrm{d}{I}_2(t)}{\mathrm{d}t}=\frac{{\beta }_2{I}_2(t)S(t)}{1+S(t)}-(\mu +d_2+{\gamma }_2)I_2(t)\hfill \\ \frac{\mathrm{d}R(t)}{\mathrm{d}t}={\gamma }_1{I}_1(t)+{\gamma }_2{I}_2(t)-(\mu +\delta )R(t).\hfill \end{array}$$

(24)

The stochastic model for model (24) is given as follows

$$\{\begin{array}{c}\mathrm{d}S(t)=(\mathrm{\Lambda }-\mu S(t)-\frac{{\beta }_1{I}_1(t)S(t)}{e^{I_1(t)}}-\frac{{\beta }_2{I}_2(t)S(t)}{1+S(t)}+\delta R(t))\mathrm{d}t\hfill \\ -\frac{{\sigma }_1{I}_1(t)S(t)}{e^{I_1(t)}}\mathrm{d}{B}_1(t)-\frac{{\sigma }_2{I}_2(t)S(t)}{1+S(t)}\mathrm{d}{B}_2(t)\hfill \\ \mathrm{d}{I}_1(t)=(\frac{{\beta }_1{I}_1(t)S(t)}{e^{I_1(t)}}-(\mu +d_1+{\gamma }_1)I_1(t))\mathrm{d}t+\frac{{\sigma }_1{I}_1(t)S(t)}{e^{I_1(t)}}\mathrm{d}{B}_1(t)\hfill \\ \mathrm{d}{I}_2(t)=(\frac{{\beta }_2{I}_2(t)S(t)}{1+S(t)}-(\mu +d_2+{\gamma }_2)I_2(t))\mathrm{d}t+\frac{{\sigma }_2{I}_2(t)S(t)}{1+S(t)}\mathrm{d}{B}_2(t)\hfill \\ \mathrm{d}R(t)=({\gamma }_1{I}_1(t)+{\gamma }_2{I}_2(t)-(\mu +\delta )R(t))\mathrm{d}t.\hfill \end{array}$$

(25)

It is easy to verify that $f(I_1(t))$ and $g(S(t))$ satisfy two assumptions (H1) and (H2). For models (24) and (25), consider the initial value $(S(0),I_1(0),I_2(0),R(0))=(2,2,2,2)\in {\mathbb{R}}_{+}^4$ . The parameters for models (24) and (25) are given by

$$\begin{array}{c}\mathrm{\Lambda }=1,\mu =0.2,{\beta }_1=0.13,{\beta }_2=0.5,\delta =0.3,\hfill \\ {\gamma }_1=0.1,{\gamma }_2=0.15,d_1=0.05,d_2=0.05\hfill \end{array}$$

(26)

Compute that ${\sigma }_1=0.17>({\beta }_1/\sqrt{2(\mu +d_1+{\gamma }_1)})=0.1554,{\sigma }_2=0.6>({\beta }_2/\sqrt{2(\mu +d_2+{\gamma }_2)})=0.559,(\mathrm{\Lambda }/\mu )=5$ . For the deterministic model (24), the two diseases are persistent (see Figure 1(a)). However, when the intensity for the noise ${\sigma }_1>({\beta }_1/\sqrt{2(\mu +d_1+{\gamma }_1)})$ and ${\sigma }_2>({\beta }_2/\sqrt{2(\mu +d_2+{\gamma }_2)})$ , the two diseases are extinct of the stochastic model (25) (see Figure 1(b), ${\sigma }_1=0.17$ , ${\sigma }_2=0.6$ ). This means that the persistent deterministic model may become extinct due to the noise fluctuations. Therefore, a large stochastic fluctuations can suppress the outbreak of diseases. This confirms Case (a) of Theorem 3.

Figure 1.

Some paths for $S(t),I_1(t),I_2(t)$ and $R(t)$ with $\mathrm{\Lambda }=1,{\beta }_1=0.13,\delta =0.3,{\gamma }_1=0.1,{\beta }_2=0.5,{\gamma }_2=0.15,\mu =0.2,d_1=0.05$ and $d_2=0.05$ . (a) The deterministic model (24); (b) the stochastic model (25) with ${\sigma }_1=0.17$ and ${\sigma }_2=0.6$ ; and (c) the stochastic model (25) with ${\sigma }_1=0.16$ and ${\sigma }_2=0.5$ .

Considering the stochastic model (25) with the intensity of environmental noise ${\sigma }_1$ and ${\sigma }_2$ being large. Let ${\sigma }_1=0.16<\sqrt{({\beta }_1\mu /(f\prime (0)\mathrm{\Lambda }))}=0.1612$ , ${\sigma }_2=0.5<\sqrt{{\beta }_2/g(5)}=0.7746$ , and other parameters be as in equation (26). Compute that $R_1^s=0.9429<1$ , $R_2^s=0.8247<1$ . Figure 1(c) shows that $I_1,I_2,R$ tend to zero, S tends to 5, separately. This means the diseases $I_1(t)$ and $I_1(t)$ die out of the stochastic model (25). This confirms Case (b) of Theorem 3.

Considering the stochastic model (25) with the environmental noise ${\sigma }_1$ being small, while ${\sigma }_2$ being large. The parameters are taken as in equation (26). Let ${\sigma }_1=0.1$ and ${\sigma }_2=0.3<\sqrt{{\beta }_2/g(5)}=0.7746$ . Compute that $R_1^s=1.500>1$ and $R_2^s=0.9635<1$ . Figure 2(a) shows that the disease $I_2(t)$ goes to extinction, the disease $I_1(t)$ is persistent. This confirms Case (a) of Theorem 4.

Figure 2.

Some paths for $S(t),I_1(t),I_2(t)$ , and $R(t)$ of the stochastic model (25) with $\mathrm{\Lambda }=1,\mu =0.2,{\beta }_1=0.13,{\beta }_2=0.5,\delta =0.3,{\gamma }_1=0.1,{\gamma }_2=0.15,d_1=0.05$ , and $d_2=0.05$ . (a) ${\sigma }_1=0.1$ and ${\sigma }_2=0.3$ ; (b) ${\sigma }_1=0.16$ and ${\sigma }_2=0.1$ ; and (c) ${\sigma }_1=0.05$ and ${\sigma }_2=0.06$ .

Consider the stochastic model (25) with the environmental noise ${\sigma }_1$ being large, while ${\sigma }_2$ being small. Let ${\sigma }_1=0.16<\sqrt{({\beta }_1\mu /f\prime (0)\mathrm{\Lambda })}=0.1612$ , ${\sigma }_2=0.1$ . Compute that $R_1^s=0.9429<1$ and $R_2^s=1.0330>1$ . Similarly, the disease $I_1(t)$ goes to extinction, the disease $I_2(t)$ is persistent of the model (25) (see Figure 2(b)). This confirms Case (b) of Theorem 4.

Consider the stochastic model (25) with the intensity of environmental noise ${\sigma }_1$ and ${\sigma }_2$ being both small. Let ${\sigma }_1=0.05$ and ${\sigma }_2=0.06$ . Compute that $R_1^s=1.7679>1$ and $R_2^s=1.0385>1$ . Obviously, the two diseases are persistent of the model (25) (see Figure 2(c)). What’s more, we keep the intensity of noise unchange and present the probability density functions to show the persistent of the two diseases. With the help of MATLAB software, we run the numerical simulation 50,000 times and collect the values of $S(t),I_1(t),I_2(t)$ , and $R(t)$ . The probability density functions in Figure 3, where the smoothed curves are the probability density functions of $S(t),I_1(t),I_2(t)$ and $R(t)$ , respectively. From Figure 3, we can see that $S(t),I_1(t),I_2(t)$ and $R(t)$ are persistent around $4.10,0.35,0.20$ , and 0.15, respectively. This confirms Case (c) of Theorem 4.

Figure 3.

The probability density function of $S(t)$ , $I_1(t)$ , $I_2(t)$ , and $R(t)$ for stochastic model (25) with $\mathrm{\Lambda }=1,\mu =0.2,{\beta }_1=0.13,{\beta }_2=0.5,\delta =0.3,{\gamma }_1=0.1,{\gamma }_2=0.15,d_1=0.05,d_2=0.05,{\sigma }_1=0.05$ and ${\sigma }_2=0.06$ . (a) for $S(t)$ ; (b) for $I_1(t)$ ; (c) for $I_2(t)$ ; and (d) for $R(t)$ .

And there naturally comes an interesting and significant question: how do the two diseases spreading with $R_1^s=1$ and $R_2^s=1$ ? It is arduous to prove theoretically. Therefore, we give some numerical results about this question. Consider model (25) with $(S(0),I_1(0),I_2(0),R(0))=(2,2,2,2)\in {\mathbb{R}}_{+}^4$ . Let ${\sigma }_1=0.15492,{\sigma }_2=0.21910$ and other parameters be as in equation (26). Compute that $R_1^s=1$ and $R_2^s=1$ . Figure 4(a) shows that the two diseases go to extinction.

Figure 4.

Some paths for $S(t),I_1(t),I_2(t)$ and $R(t)$ of the stochastic model (25) with $\mathrm{\Lambda }=1,\mu =0.2,\delta =0.3,{\gamma }_1=0.1,{\gamma }_2=0.15,d_1=0.05$ , and $d_2=0.05$ . (a) ${\beta }_1=0.13,{\beta }_2=0.5,{\sigma }_1=0.15492$ , and ${\sigma }_2=0.21910$ ; (b) ${\beta }_1=0.2,{\sigma }_1=0.228033,{\beta }_2=0.8$ , and ${\sigma }_2=0.876330$ ; and (c) ${\sigma }_1=0.3,{\sigma }_2=1.2$ and the parameters are given in equation (26).

We consider the other sets of parameters as follows

$$\begin{array}{c}\mathrm{\Lambda }=1,{\beta }_1=0.2,{\beta }_2=0.8,\delta =0.3,{\gamma }_1=0.1,\hfill \\ {\gamma }_2=0.15,d_1=0.05,d_2=0.05,\mu =0.2\hfill \end{array}$$

$${\sigma }_1=0.228033,{\sigma }_2=0.876330$$

In this case, $R_1^s=1$ and $R_2^s=1$ . Figure 4(b) also shows that $I_1,I_2,R$ tend to zero, S tends to 5, separately. This means that the two diseases are extinct.

Furthermore, we consider the stochastic model (25) with the parameters (26) and the initial value $(S(0),I_1(0),I_2(0),R(0))=(2,2,2,2)\in \mathrm{\Gamma }$ . And we choose ${\sigma }_1=0.3,{\sigma }_2=1.2$ . Compute that ${\sigma }_1^2=0.09>({\beta }_1^2/\mu )=0.0845,{\sigma }_2^2=1.4400>({\beta }_2^2/\mu )=1.2500$ Then the sufficient condition in Theorem 5 is satisfied. According to Theorem 5, the disease-free equilibrium $E_0=(5,0,0,0)$ is almost sure exponentially stable in $\mathrm{\Gamma }$ (see Figure 4(c)). From Figure 4(c), we can see that S tends to 5 and $I_1,I_2,R$ tend to 0, separately. And they do not change with time. It implies that the disease-free equilibrium $E_0$ is almost sure exponentially stable in $\mathrm{\Gamma }$ . This confirms Theorem 5.

Conclusion

In the article, a new kind of stochastic SIRS models with two different nonlinear incidences is extended. The model is introduced with multiplicative noise to the valid contact coefficients ${\beta }_1$ and ${\beta }_2$ . First, we present the model that has a global positive solution. Next, sufficient conditions of the diseases that die out for model (2) are discussed. Furthermore, two thresholds $R_1^s$ and $R_2^s$ for the model be established. Dynamic behaviors are governed by the thresholds. More concretely, when $R_1^s>1$ and $R_2^s>1$ , the two infectious diseases $I_1(t)$ and $I_2(t)$ are persistent in mean; when $R_1^s>1$ , $R_2^s<1$ , the infectious diseases $I_1(t)$ is permanent and the infectious diseases $I_2(t)$ are extinct; when $R_1^s<1$ , $R_2^s>1$ , the infectious diseases $I_1(t)$ is extinct and the infectious diseases $I_2(t)$ are permanent. In addition, an interesting result is presented. That is, when $R_1^s=1$ and $R_2^s=1$ , then diseases $I_1(t)$ and $I_2(t)$ for model (2) die out at a large time. Finally, the results in this article are an extension of the corresponding results established in Chang et al. ¹⁸ It is worthy to mention that, when the intensity of environmental noise ${\sigma }_i>({\beta }_i/\sqrt{2(\mu +d_i+{\gamma }_i)})$ , the two infectious diseases are extinct. This implies that noise can effectively prevent the outbreak of diseases, which can provide us some beneficial strategies to control the spreading of diseases. For the dynamical analysis in this article, we mainly adopt the definition of extinct, permanent in mean, and the strong law of large number for local martingales to establish two thresholds of the model. The thresholds can be used to control the dynamic behaviors of the model. Especially, the stochastically boundedness and the almost sure exponentially stable disease-free equilibrium $E_0$ of the model are investigated. These are the differences present in the article. ¹⁸