A stochastic mutualism model with saturation effect and impulsive toxicant input in a polluted environment

Abstract

In this paper, we use a mean-reverting Ornstein–Uhlenbeck process to model the stochastic perturbations in the environment, and then a stochastic mutualism model with saturation effect and pulse toxicant input in a polluted environment is proposed. A set of sufficient conditions including exponential extinction, persistence in the mean, permanence in time average and stochastic permanence are derived. Numerical simulations are worked out to support the analysis results. Also, we look at the effects of the volatile intensity, reversion rate, impulsive period and impulsive toxicant input amount on the survival of two species.

Appendices

Appendix A: The proof of Lemma 2.4

Consider the first two equations of model (1.6):

$$\begin{aligned} \left\{ \begin{array}{ll} du(t) =u(t)\left[ R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)\right. \\ \ \ \ \ \ \ \ \ \ \ -\left. a_1u(t)+\frac{b_1v(t)}{1+v(t)}\right] dt+\xi _1(t)u(t)dW_1(t),\\ dv(t)=v(t)\left[ R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)\right. \\ \ \ \ \ \ \ \ \ \ \ -\left. a_2v(t)+\frac{b_2u(t)}{1+u(t)}\right] dt+\xi _2(t)v(t)dW_2(t). \end{array}\right. \end{aligned}$$

(A.1)

By the theory of stochastic differential equation, (A.1) has a unique local solution (u(t),  v(t)) on $[0,~t_e)$, where $t_e$ denotes the explosion time. If we can show that $t_e=+\infty$ a.s., then $(u(t),~v(t))\in R_+^2$ a.s. for all $t\ge 0$. To achieve this objective, let $k_0>0$ be sufficiently large, and u(0) and v(0) lying in the interval $(1/{k_0},~k_0)$. Then define a stopping time as follows

$$\begin{aligned} t_k:=\inf \big \{t\in [0,t_e):u(t)\notin (\frac{1}{k_0},k_0),~v(t)\notin (\frac{1}{k_0},k_0)\big \}, \end{aligned}$$

where we set $\inf \emptyset =+\infty$ ($\emptyset$ denotes the empty set). Let $t_{+\infty }=\lim _{k\rightarrow +\infty }t_k$, obviously, $t_{+\infty }\le t_e$. If one can show that $t_{+\infty }=+\infty$ a.s., then $t_e=+\infty$ a.s. and $(u(t),~v(t))\in R_+^2$ a.s. for all $t\ge 0$. To complete the proof, it is sufficient to show that $t_{+\infty }=+\infty$ a.s.

Consider a $C^2$-function $V:R_+^2\rightarrow R_+$ with the form $V(u,v)=u-1-\ln u+v-1-\ln v$. An application of Itô’s formula can show that

$$\begin{aligned} \begin{array}{ll} dV(u,v)&{}=(1-\frac{1}{u})du+\frac{1}{2u^2}(du)^2+(1-\frac{1}{v})dv+\frac{1}{2v^2}(dv)^2\\ &{}=LV(u,v)dt+(1-\frac{1}{u})\xi _1(t)u(t)dW_1(t)\\ &{}\quad +\,(1-\frac{1}{v})\xi _2(t)v(t)dW_2(t), \end{array} \end{aligned}$$

(A.2)

where

$$\begin{aligned} \begin{array}{lll} \displaystyle LV &{}=(1-\frac{1}{u})u\left[ R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)-a_1u(t)+\frac{b_1v(t)}{1+v(t)}\right] +\frac{\xi _1^2(t)}{2}\\ &{}\quad +\,(1-\frac{1}{v})v\left[ R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)-a_2v(t)+\frac{b_2u(t)}{1+u(t)}\right] +\frac{\xi _2^2(t)}{2}\\ &{}\le -a_1u^2-\left[ R_{1e}-\frac{\xi _1^2(t)}{2}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)\right] \\ &{}\quad +\,[R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)+a_1+b_1]u\\ &{}\quad -\,a_2v^2-\left[ R_{2e}-\frac{\xi _2^2(t)}{2}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)\right] \\ &{}\quad +\,\left[ R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)+a_2+b_2\right] v\\ &{}\le -\left[ R_{1e}-\frac{\xi _1^2(t)}{2}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)\right] \\ &{}\quad -\,\left[ R_{2e}-\frac{\xi _2^2(t)}{2}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)\right] +Q_1\\ &{}=-\left[ R_{1e}-\frac{\beta _1^2}{4\alpha _1}(1-e^{-2\alpha _1t})+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)\right] \\ &{}\quad -\,\left[ R_{2e}-\frac{\beta _2^2}{4\alpha _2}(1-e^{-2\alpha _2t})+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)\right] +Q_1, \end{array} \end{aligned}$$

(A.3)

and

$$\begin{aligned} \begin{array}{lll} Q_1 &{}=\sup _{(u,v)\in R_+^2}\{-a_1u^2+[R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)+a_1+b_1]u\\ &{}\quad -\,a_2v^2+[R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)+a_2+b_2]v\}. \end{array} \end{aligned}$$

It is not difficult to prove that LV is bounded. Substituting (A.3) into (A.2) yields that

$$\begin{aligned} \begin{array}{ll} dV(u,v) &{}\le \left\{ -\left[ R_{1e}-\frac{\beta _1^2}{4\alpha _1}(1-e^{-2\alpha _1t})+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)\right] \right. \\ &{}\quad {-}\,\left. \left[ R_{2e}{-}\frac{\beta _2^2}{4\alpha _2}(1-e^{-2\alpha _2t})+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)\right] {+}Q_1\right\} dt\\ &{}\quad {+}\,\left( 1-\frac{1}{u}\right) \xi _1(t)udW_1(t)+\left( 1-\frac{1}{v}\right) \xi _2(t)vdW_2(t). \end{array} \end{aligned}$$

The rest of the proof can be obtained by almost the same method in [45], and the details are omitted. This completes the proof.

Appendix B: The proof of Theorem 3.1

From (2.1) and Lemma 2.2, for any $\varepsilon >0$, there exists a constant $T>0$ such that for all $t\ge T$,

$$\begin{aligned} {\bar{q}}_i-\frac{\varepsilon }{2}\le \langle q_i(t)\rangle \le {\bar{q}}_i+\frac{\varepsilon }{2},\ {\tilde{C}}_0(t)-\frac{\varepsilon }{2r_i}\le C_0(t)\le {\tilde{C}}_0(t)+\frac{\varepsilon }{2r_i},\ i=1,2. \end{aligned}$$

(B.1)

Applying Itô’s formula to (1.6) gives

$$\begin{aligned} d\ln u(t)= & {} \left[ R_{1e}-\frac{1}{2}\xi _1^2(t)+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)-a_1u(t)\right. \\&+\left. \,\frac{b_1v(t)}{1+v(t)}\right] dt+\xi _1(t)dW_1 \end{aligned}$$

and

$$\begin{aligned} d\ln v(t)= & {} \left[ R_{2e}-\frac{1}{2}\xi _2^2(t)+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)-a_2v(t)\right. \\&+\left. \,\frac{b_2u(t)}{1+u(t)}\right] dt+\xi _2(t)dW_2. \end{aligned}$$

Integrating both sides from 0 to t, one has

$$\begin{aligned} \begin{array}{ll} \ln \frac{u(t)}{u(0)} &{}=\int _0^t[R_{1e}-\frac{1}{2}\xi _1^2(s)]ds+\int _0^t(R_1(0)-R_{1e})e^{-\alpha _1s}ds\\ &{}\quad -\,r_1\int _0^tC_0(s)ds-a_1\int _0^tu(s)ds+b_1\int _0^t\frac{v(s)}{1+v(s)}ds+\int _0^t\xi _1(s)dW_1(s) \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{ll} \ln \frac{v(t)}{v(0)} &{}=\int _0^t[R_{2e}-\frac{1}{2}\xi _2^2(s)]ds+\int _0^t(R_2(0)-R_{2e})e^{-\alpha _2s}ds\\ &{}\quad -\,r_2\int _0^tC_0(s)ds-a_2\int _0^tv(s)ds+b_2\int _0^t\frac{u(s)}{1+u(s)}ds+\int _0^t\xi _2(s)dW_2(s). \end{array} \end{aligned}$$

Dividing t to both sides of above two equations yields that

$$\begin{aligned} \frac{1}{t}\ln \frac{u(t)}{u(0)}= & {} \big \langle q_1(t)\big \rangle +f_1(t)-r_1\big \langle C_0(t)\big \rangle -a_1\big \langle u(t)\big \rangle \nonumber \\&+\, b_1\big \langle \frac{v(t)}{1+v(t)}\big \rangle +\frac{1}{t}\int _0^t\xi _1(s)dW_1(s) \end{aligned}$$

(B.2)

and

$$\begin{aligned} \frac{1}{t}\ln \frac{v(t)}{v(0)}= & {} \big \langle q_2(t)\big \rangle +f_2(t)-r_2\big \langle C_0(t)\big \rangle -a_2\big \langle v(t)\big \rangle \nonumber \\&+\, b_2\big \langle \frac{u(t)}{1+u(t)}\big \rangle +\frac{1}{t}\int _0^t\xi _2(s)dW_2(s), \end{aligned}$$

(B.3)

where

$$\begin{aligned} f_i(t)=(R_i(0)-R_{ie})(1-e^{-\alpha _it})/\alpha _it,\ i=1,2. \end{aligned}$$

Obviously, $\lim _{t\rightarrow +\infty }f_i(t)=0$.

• (i) Substituting (B.1) into (B.2), we can deduce that

  $$\begin{aligned} \begin{array}{lll} \frac{1}{t}\ln \frac{u(t)}{u(0)} &{}\le {\bar{q}}_1+\frac{\varepsilon }{2}+f_1(t)-r_1\langle {\tilde{C}}_0(t)-\frac{\varepsilon }{2r_1}\rangle +b_1+\frac{1}{t}\int _0^t\xi _1(s)dW_1(s)\\ &{}\le {\bar{q}}_1+\varepsilon +f_1(t)-r_1\langle {\tilde{C}}_0(t)\rangle +b_1+\frac{1}{t}\int _0^t\xi _1(s)dW_1(s). \end{array} \end{aligned}$$

  (B.4)

  Note that $\lim _{t\rightarrow +\infty }f_i(t)=0$. Using the strong law of large number of local martingale (see [46]) yields $\lim _{t\rightarrow +\infty }t^{-1}\int _0^t\xi _1(s)dW_1(s)=0$. It follows from (B.4) that $\limsup _{t\rightarrow +\infty }t^{-1}\ln u(t) \le {\bar{q}}_1-r_1\delta +b_1$. If ${\bar{q}}_1-r_1\delta +b_1<0$, then u(t) will be extinct exponentially. Similarly, when ${\bar{q}}_2-r_2\delta +b_2<0$, specie v(t) is also extinct exponentially.

• (ii) If ${\bar{q}}_1-r_1\delta +b_1<0$, then it follows from (1) in Definition 2.1 that $\lim _{t\rightarrow +\infty }u(t)=0$ a.s. Thus, for any $\varepsilon >0$, there is a constant T such that for all $t\ge T$,

  $$\begin{aligned} -\frac{\varepsilon }{2}\le b_2\langle \frac{u}{1+u}\rangle \le \frac{\varepsilon }{2},\ \ -\frac{\varepsilon }{2}\le t^{-1}\ln v(0)\le \frac{\varepsilon }{2}. \end{aligned}$$

  (B.5)

  Combining (B.1), (B.3) and (B.5) yields

  $$\begin{aligned} \begin{array}{lll} \ln v(t) &{}\le \frac{\varepsilon }{2}t+({\bar{q}}_2+\frac{\varepsilon }{2})t+tf_2(t)-r_2\langle {\tilde{C}}_0(t)-\frac{\varepsilon }{2r_2}\rangle t\\ &{}\quad -\,a_2\int _0^tv(s)ds+\frac{\varepsilon }{2}t+\int _0^t\xi _2(s)dW_2(s)\\ &{}\le tf_2(t)+({\bar{q}}_2-r_2\langle {\tilde{C}}_0(t)\rangle +2\varepsilon )t-a_2\int _0^tv(s)ds\\ &{}\quad +\,\int _0^t\xi _2(s)dW_2(s) \end{array} \end{aligned}$$

  (B.6)

  and

  $$\begin{aligned} \ln v(t)\ge & {} tf_2(t)+({\bar{q}}_2-r_2\langle {\tilde{C}}_0(t)\rangle -2\varepsilon )t -a_2\int _0^tv(s)ds\nonumber \\&+\,\int _0^t\xi _2(s)dW_2(s). \end{aligned}$$

  (B.7)

  Noticing that the function $tf_i(t)=(R_i(0)-R_{ie})(1-e^{-\alpha _it})/\alpha _i$ is bounded as $t\rightarrow +\infty$. Making use of (1) in Lemma 2.3 to (B.6), we can deduce that $\langle v(t)\rangle ^*\le ({\bar{q}}_2-r_2\delta +2\varepsilon )/a_2$. Similarly, we have from (2) in Lemma 2.3 and (B.7) that $\langle v(t)\rangle _*\ge ({\bar{q}}_2-r_2\delta -2\varepsilon )/a_2$. Since $\varepsilon$ is arbitrary, we obtain a desired assertion that $\lim _{t\rightarrow +\infty }\langle v(t)\rangle =({\bar{q}}_2-r_2\delta )/a_2$.

• (iii) The proof of (iii) coincides with that of (ii), and hence is omitted.

• (iv) Substituting (B.1) into (B.2), one has for $t\ge T$,

  $$\begin{aligned} \ln u(t)\le & {} \ln u(0)+tf_1(t)+\big ({\bar{q}}_1-r_1\langle {\tilde{C}}_0(t)\rangle +b_1+\varepsilon \big )t\\&\quad -\,a_1\int _0^tu(s)ds+\int _0^t\xi _1(s)dW_1(s) \end{aligned}$$

  and

  $$\begin{aligned} \ln u(t)\ge & {} \ln u(0)+tf_1(t)+({\bar{q}}_1-r_1\langle {\tilde{C}}_0(t)\rangle -\varepsilon )t\\&\quad -\,a_1\int _0^tu(s)ds+\int _0^t\xi _1(s)dW_1(s). \end{aligned}$$

  Applying (1) and (2) in Lemma 2.3 to the above two inequalities, we can obtain that

  $$\begin{aligned} ({\bar{q}}_1-r_1\delta )/{a_1}\le \langle u(t)\rangle _*\le \langle u(t)\rangle ^*\le ({\bar{q}}_1-r_1\delta +b_1)/{a_1}. \end{aligned}$$

  Similarly, we can prove

  $$\begin{aligned} ({\bar{q}}_2-r_2\delta )/{a_2}\le \langle v(t)\rangle _*\le \langle v(t)\rangle ^*\le ({\bar{q}}_2-r_2\delta +b_2)/{a_2}, \end{aligned}$$

  which completes the claim.

Appendix C: The proof of Theorem 3.2

Let’s first prove $\liminf _{t\rightarrow +\infty }{\mathcal {P}}\{\sqrt{u^2(t)+v^2(t)}\ge \varphi \}\ge 1-\varepsilon$. Define $V_1(u,v)=1/(u+v)$. Applying Itô’s formula leads to

$$\begin{aligned} \begin{array}{ll} dV_1 &{}=-V_1^2(du+dv)+V_1^3(du+dv)^2\\ &{}=-V_1^2\big \{u(t)\left[ R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)-a_1u(t)+\frac{b_1v(t)}{1+v(t)}\right] \\ &{}\quad +\,\xi _1(t)u(t)dW_1(t)+v(t)\left[ R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)-a_2v(t)\right. \\ &{}\quad +\left. \,\frac{b_2u(t)}{1+u(t)}\right] dt+\xi _2(t)v(t)dW_2(t)\big \}+V_1^3[\xi _1^2(t)u^2(t)+\xi _2^2(t)v^2(t)]dt. \end{array} \end{aligned}$$

By the assumption of Theorem 3.2, we can choose a constant $\mu >0$ such that

$$\begin{aligned} \min _{i=1,2}\{R_{ie}-r_i{\widetilde{C}}_0^*\}-\frac{1}{2}\max _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} >\frac{\mu }{2}\max _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} . \end{aligned}$$

Define $V_2=(1+V_1)^{\mu }$. By Itô’s formula we get that

$$\begin{aligned} \begin{array}{ll} dV_2 &{}=\mu (1+V_1)^{\mu -1}dV_1+\frac{1}{2}\mu (\mu -1)(1+V_1)^{\mu -2}(dV_1)^2\\ &{}=\mu (1+V_1)^{\mu -2}[(1+V_1)dV_1+\frac{1}{2}(\mu -1)(dV_1)^2]\\ &{}=\mu (1+V_1)^{\mu -2}\left\{ -(1+V_1)V_1^2\left[ u\left( R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)-a_1u(t)\right. \right. \right. \\ &{}\quad +\left. \left. \,\frac{b_1v(t)}{1+v(t)}\right) +v\left( R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)-a_2v(t)+\frac{b_2u(t)}{1+u(t)}\right) \right] \\ &{}\quad +\left. \,(1+V_1)V_1^3[\xi _1^2(t)u^2(t)+\xi _2^2(t)v^2(t)]+\frac{1}{2}(\mu -1)V_1^4[\xi _1^2(t)u^2(t)+\xi _2^2(t)v^2(t)]\right\} dt\\ &{}\quad -\,\mu (1+V_1)^{\mu -1}V_1^2\big (\xi _1(t)u(t)dW_1+\xi _2(t)v(t)dW_2\big )\\ &{}=\mu (1+V_1)^{\mu -2}J_1(u,v)dt-\mu (1+V_1)^{\mu -1}V_1^2\big (\xi _1(t)u(t)dW_1+\xi _2(t)v(t)dW_2\big ), \end{array} \end{aligned}$$

where

$$\begin{aligned} \begin{array}{ll} \displaystyle J_1(u,v) &{}\displaystyle =-(1+V_1)V_1^2\left[ u\left( R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1C_0(t)-a_1u(t)\right. \right. \\ &{}\displaystyle \ \ \ +\left. \frac{b_1v(t)}{1+v(t)}\right) +\left. v\left( R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2C_0(t)-a_2v(t)+\frac{b_2u(t)}{1+u(t)}\right) \right] \\ &{}\displaystyle \ \ \ +V_1^3[\xi _1^2(t)u^2(t)+\xi _2^2(t)v^2(t)]+\frac{\mu +1}{2}V_1^4[\xi _1^2(t)u^2(t)+\xi _2^2(t)v^2(t)]\\ &{}\displaystyle \le -(1+V_1)V_1^2[u\big (R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t}-r_1{\widetilde{C}}_0^*(t)-\varepsilon \big )\\ &{}\displaystyle \ \ \ +v\big (R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t}-r_2{\widetilde{C}}_0^*(t)-\varepsilon \big )]\\ &{}\displaystyle \ \ \ -(1+V_1)V_1^2\left( -a_1u^2(t)+u(t)\frac{b_1v(t)}{1+v(t)}-a_2v^2(t)+v(t)\frac{b_2u(t)}{1+u(t)}\right) \\ &{}\displaystyle \ \ \ +V_1\max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} +\frac{\mu +1}{2}V_1^2\max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} \\ &{}\displaystyle \le -(1+V_1)V_1^2\left[ \min _{i=1,2}\{R_{ie}+(R_i(0)-R_{ie})e^{-\alpha _it}-r_i{\widetilde{C}}_0^*-\varepsilon \}(u(t)+v(t))\right] \\ &{}\displaystyle \ \ \ +(1+V_1)V_1^2\left( a_1u^2(t)-u\frac{b_1v(t)}{1+v(t)}+a_2v^2(t)-v(t)\frac{b_2u(t)}{1+u(t)}\right) \\ &{}\displaystyle \ \ \ +V_1\max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} +\frac{\mu +1}{2}V_1^2\max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} \\ &{}\displaystyle \le \max \{a_1,a_2\}-V_1^2\left[ \min _{i=1,2}\left\{ R_{ie}-r_i{\widetilde{C}}_0^*-\varepsilon \right\} -\frac{\mu +1}{2}\max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} \right. \\ &{}\displaystyle \ \ \ +\left. \min _{i=1,2}\left\{ (R_i(0)-R_{ie})e^{-\alpha _it}\right\} \right] +V_1\left( \max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} +\max \{a_1,a_2\}\right) . \end{array} \end{aligned}$$

Assign $\kappa >0$ sufficiently small such that

$$\begin{aligned} \min _{i=1,2}\{R_{ie}-r_i{\widetilde{C}}_0^*-\varepsilon \}-\frac{1+\mu }{2}\max _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\}>\frac{\kappa }{\mu }>0. \end{aligned}$$

(C.1)

We continue to define $V_3=e^{\kappa t}V_2$. The Itô’s formula then gives that

$$\begin{aligned} \begin{array}{ll} \displaystyle dV_3 &{}\displaystyle =\kappa e^{\kappa t}V_2dt+e^{\kappa t}dV_2\\ &{}\displaystyle \le \kappa e^{\kappa t}(1+V_1)^{\mu }dt+e^{\kappa t}\mu (1+V_1)^{\mu -2}\left\{ \max \{a_1,a_2\} +V_1\left( \max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} \right. \right. \\ &{}\displaystyle \ \ \ +\left. \max \{a_1,a_2\}\right) -V_1^2\left[ \min _{i=1,2}\{R_{ie}-r_i{\widetilde{C}}_0^*-\varepsilon \} -\frac{\mu +1}{2}\max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} \right. \\ &{}\displaystyle \ \ \ +\left. \left. \min _{i=1,2}\{(R_i(0)-R_{ie})e^{-\alpha _it}\}\right] \right\} dt-e^{\kappa t}\mu (1+V_1)^{\mu -1}V_1^2\big (\xi _1(t)u(t)dW_1+\xi _2(t)v(t)dW_2\big )\\ &{}\displaystyle \le e^{\kappa t}(1+V_1)^{\mu -2}\left\{ \kappa (1+V_1)^2-\mu V_1^2\left[ \min _{i=1,2}\{R_{ie}-r_i{\widetilde{C}}_0^*-\varepsilon \}\right. \right. \\ &{}\displaystyle \ \ \ -\left. \frac{\mu +1}{2}\max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} +\min _{i=1,2}\{(R_i(0)-R_{ie})e^{-\alpha _it}\}\right] \\ &{}\displaystyle \ \ \ +\left. \mu V_1\left( \max \limits _{i=1,2}\{\frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\}+\max \{a_1,a_2\}\right) +\mu \max \{a_1,a_2\}\right\} dt\\ &{}\displaystyle \ \ \ -\mu e^{\kappa t}(1+V_1)^{\mu -2}V_1^2\big (\xi _1(t)u(t)dW_1+\xi _2(t)v(t)dW_2\big )\\ &{}\displaystyle =e^{\kappa t}J_2(u,v)dt-\mu e^{\kappa t}(1+V_1)^{\mu -2}V_1^2\big (\xi _1(t)u(t)dW_1+\xi _2(t)v(t)dW_2\big ), \end{array} \end{aligned}$$

where

$$\begin{aligned} \begin{array}{ll} \displaystyle J_2(u,v) &{}\displaystyle =\mu (1+V_1)^{\mu -2}\left\{ \frac{\kappa }{\mu }(1+V_1)^2-V_1^2\left[ \min _{i=1,2}\{R_{ie}-r_i{\widetilde{C}}_0^*-\varepsilon \}\right. \right. \\ &{}\displaystyle \ \ \ -\left. \frac{\mu +1}{2}\max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} +\min _{i=1,2}\left\{ (R_i(0)-R_{ie})e^{-\alpha _it}\right\} \right] \\ &{}\displaystyle \ \ \ +\left. V_1\left( \max \limits _{i=1,2}\left\{ \frac{\beta _i^2}{2\alpha _i}(1-e^{-2\alpha _it})\right\} +\max \{a_1,a_2\}\right) +\max \{a_1,a_2\}\right\} . \end{array} \end{aligned}$$

It is easy to verify that $J_2(u,v)$ is upper bounded as $t\rightarrow +\infty$, denoted as $J:=\sup _{(u,v)\in R_+^2}J_2(u,v)<+\infty$. Thus,

$$\begin{aligned} dV_3\le e^{\kappa t}Jdt-\mu e^{\kappa t}(1+V_1)^{\mu -2}V_1^2(\xi _1udW_1+\xi _2vdW_2). \end{aligned}$$

(C.2)

Integrating both sides of (C.2) and then taking expectation, one has

$$\begin{aligned} {\mathbb {E}}[V_3(u(t),v(t))]={\mathbb {E}}[e^{\kappa t}(1+V_1)^{\mu }]\le (1+V_1(u(0),v(0)))^{\mu }+\frac{J}{\kappa }e^{\kappa t}. \end{aligned}$$

Dividing both sides of $e^{\kappa t}$ gives that

$$\begin{aligned} \limsup _{t\rightarrow +\infty }{\mathbb {E}}[V_1^{\mu }]\le \limsup _{t\rightarrow +\infty }{\mathbb {E}}[(1+V_1)^{\mu }]\le \frac{J}{\kappa }. \end{aligned}$$

Therefore

$$\begin{aligned} (u+v)^{\mu }\le 2^{\mu }(\max {\{u,v\}})^{\mu }=2^{\mu }(\max {\{u^2,v^2\}})^{\frac{\mu }{2}}\le 2^{\mu }\Big (\sqrt{u^2(t)+v^2(t)}\Big )^{\mu }, \end{aligned}$$

from which we conclude that

$$\begin{aligned} \limsup _{t\rightarrow +\infty }{\mathbb {E}}[\frac{1}{2^{\mu }\big (\sqrt{u^2(t)+v^2(t)}\big )^{\mu }}]\le \limsup _{t\rightarrow +\infty }{\mathbb {E}}[V_1^{\mu }]\le \frac{J}{\kappa }. \end{aligned}$$

Then

$$\begin{aligned} \limsup \limits _{t\rightarrow +\infty }{\mathbb {E}}[\frac{1}{\sqrt{u^2(t)+v^2(t)}^{\mu }}]\le 2^{\mu }\frac{J}{\kappa }:=K. \end{aligned}$$

For any given $\varepsilon >0$, let $\varphi =\varepsilon ^{\frac{1}{\mu }}/K^{\frac{1}{\mu }}$, by Chebyshev’s inequality (see [46]), we have

$$\begin{aligned} {\mathcal {P}}\big (\sqrt{u^2(t)+v^2(t)}<\varphi \big ){=}{\mathcal {P}}\{\sqrt{u^2(t){+}v^2(t)}^{{-}\mu }>\varphi ^{-\mu }\} \le {\mathbb {E}}[\sqrt{u^2(t)+v^2(t)}^{-\mu }]/\varphi ^{-\mu }, \end{aligned}$$

and then $\limsup _{t\rightarrow +\infty }{\mathcal {P}}\{\sqrt{u^2(t)+v^2(t)}<\varphi \}\le \varphi ^{\mu }K=\varepsilon$. Thus, a desirable result,

$\liminf _{t\rightarrow +\infty }{\mathcal {P}}\{\sqrt{u^2(t)+v^2(t)}\ge \varphi \}\ge 1-\varepsilon$, is obtained. Next, we will verify that $\liminf _{t\rightarrow +\infty }{\mathcal {P}}\{\sqrt{u^2(t)+v^2(t)}\le \phi \}\ge 1-\varepsilon$. Define $V_4(u,v)=u^{\theta }+v^{\theta }$, $\theta >1$. Using Itô’s formula we have

$$\begin{aligned} \begin{array}{lll} dV_4 &{}=\theta u^{\theta -1}du+\frac{1}{2}\theta (\theta -1)u^{\theta -2}(du)^2 +\theta v^{\theta -1}dv+\frac{1}{2}\theta (\theta -1)v^{\theta -2}(dv)^2\\ &{}=\theta u^\theta \left[ R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t} -r_1C_0(t)-a_1u(t)+\frac{b_1v(t)}{1+v(t)}+\frac{1}{2}(\theta -1)\xi _1^2(t)\right] dt\\ &{}\quad +\,\theta v^\theta \left[ R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t} -r_2C_0(t)-a_2v(t)+\frac{b_2u(t)}{1+v(t)}+\frac{1}{2}(\theta -1)\xi _2^2(t)\right] dt\\ &{}\quad +\,\theta u^\theta \xi _1(t)dW_1(t)+\theta v^\theta \xi _2(t)dW_2(t)\\ &{}\le \theta u^\theta \big [R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t} -a_1u+b_1+\frac{1}{2}(\theta -1)\xi _1^2(t)\big ]dt\\ &{}\quad +\,\theta v^\theta \big [R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t} -a_2v+b_2+\frac{1}{2}(\theta -1)\xi _2^2(t)\big ]dt\\ &{}\quad +\,\theta u^\theta \xi _1(t)dW_1(t)+\theta v^\theta \xi _2(t)dW_2(t). \end{array} \end{aligned}$$

We finally define $V_5(u,v)=e^tV_4$. Applying Itô’s formula, one can show that

$$\begin{aligned} \begin{array}{lll} dV_5 &{}=e^tV_4dt+e^tdV_4\\ &{}\le e^t(u^\theta +v^\theta )dt+e^t\big \{\theta u^\theta \left[ R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t}-a_1u+b_1\right. \\ &{}\quad +\left. \,\frac{1}{2}(\theta -1)\xi _1^2(t)\right] dt+\theta v^\theta \left[ R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t}-a_2v+b_2\right. \\ &{}\quad +\left. \,\frac{1}{2}(\theta -1)\xi _2^2(t)\right] dt+\theta u^\theta \xi _1(t)dW_1(t)+\theta v^\theta \xi _2(t)dW_2(t)\big \}\\ &{}\le e^t\left\{ \theta u^\theta \left[ R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t} -a_1u+b_1+\frac{1}{2}(\theta -1)\xi _1^2(t)+\frac{1}{\theta }\right] \right. \\ &{}\quad +\,\theta v^\theta \left[ R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t} \left. -a_2v+b_2+\frac{1}{2}(\theta -1)\xi _2^2(t)+\frac{1}{\theta }\right] dt\right\} \\ &{}\quad +\,\theta e^t(u^\theta \xi _1(t)dW_1(t)+v^\theta \xi _2(t)dW_2(t))\\ &{}=e^tJ_3(u,v)dt+\theta e^t(u^\theta \xi _1(t)dW_1(t)+v^\theta \xi _2(t)dW_2(t)), \end{array} \end{aligned}$$

(C.3)

where

$$\begin{aligned} \begin{array}{lll} J_3(u,v) &{}=-a_1\theta u^{\theta +1}+u^\theta \theta \left[ R_{1e}+(R_1(0)-R_{1e})e^{-\alpha _1t} +b_1+\frac{1}{2}(\theta -1)\xi _1^2(t)+\frac{1}{\theta }\right] \\ &{}\quad -\,a_2\theta v^{\theta +1}+v^\theta \theta \left[ R_{2e}+(R_2(0)-R_{2e})e^{-\alpha _2t} +b_2+\frac{1}{2}(\theta -1)\xi _2^2(t)+\frac{1}{\theta }\right] . \end{array} \end{aligned}$$

It is easy to know that $J_3(u,v)$ is upper bounded, denoted as $K_1:=\sup _{(u,v)\in R_+^2}J_3<+\infty$.

Integrating both sides of (C.3) shows that

$$\begin{aligned} e^tV_4(u(t),v(t))-V_4(u(0),v(0))\le & {} \int _0^te^sK_1ds+\theta \int _0^te^su^\theta (s)\xi _1(s)dW_1\\&\quad +\theta \int _0^te^sv^\theta (s)\xi _2(s)dW_2. \end{aligned}$$

Taking expectation, one has

$$\begin{aligned} {\mathbb {E}}[e^tV_4]-V_4(u(0),v(0))\le K_1{\mathbb {E}}[e^t]. \end{aligned}$$

Namely, $\limsup _{t\rightarrow +\infty }{\mathbb {E}}[u^\theta +v^\theta ]\le K_1$. Assign $\phi =(2^{\theta -1}K_1)^\frac{1}{\theta }/\varepsilon ^{\frac{1}{\theta }}$, we derive, by Chebyshev’s inequality (see [46]) again, that

$$\begin{aligned} {\mathcal {P}}\{\sqrt{u^2(t)+v^2(t)}^\theta >\phi ^\theta \}\le \phi ^{-\theta }{\mathbb {E}} [\sqrt{u^2(t)+v^2(t)}^\theta ]\le \phi ^{-\theta }{\mathbb {E}}[(u+v)^\theta ]\le 2^{\theta -1}\phi ^{-\theta }{\mathbb {E}}[u^\theta +v^\theta ]. \end{aligned}$$

Then we know that $\limsup _{t\rightarrow +\infty }{\mathcal {P}}\{\sqrt{u^2(t)+v^2(t)}>\phi \}\le \varepsilon$. So the required assertion follows as $\liminf _{t\rightarrow +\infty }{\mathcal {P}}\{\sqrt{u^2(t)+v^2(t)}\le \phi \}\ge 1-\varepsilon$.